Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is the series $\sum_{n=1}^\infty\frac{1}{n}\left(\sum_{k=1}^n\frac{1}{k}\left(\frac{1}{2}\right)^{n-k}\right)$ convergent? $$
\sum_{n = 1}^\infty\dfrac{1}{n}\left(\sum_{k = 1}^n\dfrac{1}{k}\left(\dfrac{1}{2}\right)^{n - k}\right)
$$
Does the series converge?
I calculate it using Matlab, and it seems that the sum conver... | Let
$$S_n = \sum_{k=1}^n \frac{1}{k} \left( \frac{1}{2} \right)^{n-k}$$
Let's show that $(nS_n)$ is eventually decreasing. By direct computation, for $n > 2$
\begin{align*} (n-1)S_{n-1} – nS_n &= \left( \sum_{k=1}^{n-1} \frac{n-1}{k} \left( \frac{1}{2} \right)^{n-1-k}\right)- \left(\sum_{k=1}^n \frac{n}{k} \left( \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Find function $ f $ such that $f(\frac{x-3}{x+1})+f(\frac{x+3}{1-x})=x$ I am looking for functions $ f:\Bbb R \to \Bbb R $ satisfying
$$f\Big(\frac{x-3}{x+1}\Big)+f\Big(\frac{x+3}{1-x}\Big)=x$$
I used the substitution $ x=\cos(2t) $ for $ x\in (0,2\pi) $, to use the identities
$$1+x=2\cos^2(t) \text{ and } 1-x=2\sin^2(... | Set $u= \frac{x-3}{x+1}$ the equation becomes $f(u)+f(\frac{u+3}{1-u})=\frac{u+3}{1-u}$
Set $v=\frac{x+3}{1-x}$ the equation becomes $f(\frac{v-3}{v+1})+f(v)=\frac{v-3}{v+1}$
Now replace all dummy variables with $y$ and solve for $f(x)$,
*
*$$A+B=y$$
2.$$B+C=\frac{y+3}{1-y}$$
*$$A+C=\frac{y-3}{y+1}$$
Can you finish... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How does $ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$ reduce to $ 1 - \frac{1}{\sqrt{n+2}}\;$? $$ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$$
Reduces to:
$$ 1 - \frac{1}{\sqrt{n+2}} $$
I have no clue how. What is the exact trick here and how can I prac... | HINT
Here is a different approach for the sake of curiosity (as suggested by Paul in the comments).
Precisely, we shall multiply the numerator and the denominator by the conjugate of the denominator:
\begin{align*}
\frac{1}{(n+2)\sqrt{n+1} + (n+1)\sqrt{n+2}} & = \frac{(n+2)\sqrt{n+1} - (n+1)\sqrt{n+2}}{(n + 2)^{2}(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Let $a$ and $b$ be positive integers such that $an + 1$ s a cube if and only if $ bn + 1$ is a cube. Prove that $a = b.$ Let $a$ and $b$ be positive integers such that $an + 1$ is a cube if and only if $bn + 1$ is a cube. Prove that $a = b.$
By choosing $p_n^3 \equiv 1 \mod b$ , we find that there are infinitely many n... | As noted in the answer above by @richrow, the crux is showing that, if $a \not =b$, there is an integer $M$ such that $ab^2M^3+3abM^2+3aM+1$ is not a perfect cube. Note also that we may also assume that $a<b$. We give an alternative proof of this, that does not use the Kronecker Theorem.
Case 1: $ab^2$ is not a perfect... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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How to prove this inequality with the simplest means? $$ x^2+5y^2+6z^2 \geq 4\sqrt{5xyz^2},\ \text{ if }\ xy>0 $$
I was trying to prove it. The right hand side should be nonnegative, so I can square it on both sides.
But once I have done it I get to a point, where I do not see how to show, that the statement is greater... | Looks like the AM.GM inequality will by useful. We might as well assume $x,y>0$. (As $xy>0$ then either $x,y > 0$ or $x,y < 0$. If $x,y < 0$ then we can replace $x,y$ with $|x|, |y|$ and the result will be the same.... Or... I'll just use $|x|,|y|$).
By AM.GM we have $x^2 + 5y^2 \ge 2\sqrt{x^2\cdot 5y}=2|x||y|\sqrt 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tiling of a deficient $7\times7$ chessboard with L trominoes
Prove that a $7\times7$ chessboard with one square removed can always be tiled by $L$ trominoes.
I'm looking for a reasonably simple proof. I was able to prove some specific cases, For instance, when the central square is deleted, the chessboard can be part... | Consider this (i am going draw some formal pictures after I have taken a nap.)
$$\begin{array}{|c|c|c|c|c|c|}
\hline
X&X&\circ&\circ&\triangle&\triangle&\circ\\
\hline
X&X&\circ&\square&\triangle&\circ&\circ\\
\hline
\circ&\circ&\square&\square&\blacksquare &\blacksquare &\square\\
\hline
\circ&\triangle&\circ&\circ&\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding integer solutions to $(x-y)^2+(y-z)^2+(z-x)^2=2022$ This was from round A of the Awesomemath Summer Program application. The deadline was yesterday and we can discuss the problems now.
Find all integer triples $(x,y,z)$ which satisfy
$$(x-y)^2+(y-z)^2+(z-x)^2=2022$$
I tried a couple of things. First I noticed... | We have:
$$(x-y)^2+(y-z)^2+(z-x)^2=2(x^2+y^2+z^2)-2(xy+yz+zx)$$
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
Summing these relations we get:
$$(x-y)^2+(y-z)^2+(z-x)^2+(x+y+z)^2=3(x^2+y^2+z^2)$$
$(x-y)^2+(y-z)^2+(z-x)^2=3\times 2\times 337$
$\Rightarrow 3\times 2\times 337+(x+y+z)^2=3(x^2+y^2+z^2)\space\space\space\space\space... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove that $\sum \frac{a^3}{a^2+b^2}\le \frac12 \sum \frac{b^2}{a}$
Let $a,b,c>0$. Prove that
$$ \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\le \frac12 \left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).\tag{1}$$
A idea is to cancel the denominators, but in this case Muirhead don't work because th... | A funny solution (which seems correct): note that for $t>0$
$$
\frac{1}{1+t^2} \leq \frac{1}{2}t^2 + \frac{3}{2}(1-t).
$$
To prove this note first that $p(t) = (t^2 - 3t + 3)(t^2 + 1)$ has no real roots. Next, $p''(t) = 12t^2 -18t + 8 > 0$, implying that $p$ is strictly convex. Finally $p'(t) = 4t^3 - 9t^2 + 8t -3$, an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
solving $\sin(5x)+\sin(x)=0$ using the sum formulas so symbolab says its just
$$2 \sin(3x) \cos(2x)$$
But after applying the angle sum formula I get
$$\sin(5x)+\sin(x)=\sin(3x+2x)+\sin(x)=\sin(3x)\cos(2x)+\sin(2x)\cos(3x)+\sin(x)$$
How does this reduce more?
| Using the compound angle formula, $$
\sin A+\sin B=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2},
$$
we have
$$
\begin{array}{l}
\sin 5 x+\sin x=0 \\
2 \sin 3 x \cos 2 x=0 \\
\sin 3 x=0 \text { or } \cos 2 x=0 \\
\displaystyle 3 x=n \pi \text { or } 2 x=n \pi+\frac{\pi}{2} \\
\displaystyle x=\frac{n \pi}{3} \text { or } \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Six people (half are female, half are male) for seven chairs. Problem:
Suppose there are $7$ chairs in a row. There are $6$ people that are going to randomly
sit in the chairs. There are $3$ females and $3$ males. What is the probability that
the first and last chairs have females sitting in them?
Answer:
Let $p$ be th... | An alternative approach is that you have the outer seats both occupied with probability $\frac57$ and the conditional probability that they both have a woman is $$\frac{\binom32}{\binom62}=\frac15.$$
So the probability is $\frac57\cdot \frac15=\frac17.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What is wrong with this projection onto two basis vectors? Given
\begin{align}
b_1 &= \begin{bmatrix} 2 \\ -1 \end{bmatrix} \\
b_2 &= \begin{bmatrix} 1 \\ 3 \end{bmatrix} \\
v &= \begin{bmatrix} 3 \\ 2 \end{bmatrix}
\end{align}
it is clear that
\begin{align}
v &= b_1 + b_2 \\
&= \begin{bmatrix} b_1 & b_2 \end{bmatrix} ... | Your formula will only work when $b_1$ and $b_2$ are orthogonal. Here they are not, so you get an incorrect result. In general the projection matrix is $A(A^\top A)^{-1} A^\top$ where the columns of $A$ are $b_1$ and $b_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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To find the number of ways to put $14$ identical balls into $4$ bins with the condition that no bin can hold more than $7$ balls. To find the number of ways to put $14$ identical balls into $4$ bins with the condition that no bin can hold more than $7$ balls.
I have tried the following:
The total no of ways to distrib... | What you have is correct. Here’s a slightly alternative approach. You want to count the nonnegative integer solutions to $x_1+x_2+x_3+x_4=14$ such that $x_i\le 7$ for all $i$. By inclusion-exclusion and stars-and-bars, we have
$$\binom{14+4-1}{4-1}-\binom{4}{1}\sum_{k=8}^{14}\binom{14-k+3-1}{3-1}$$ solutions. By the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve for theta , $0<\theta<90^\circ$
Solve $$\cos60=\frac{\cos(2θ)+\frac32\sin(2θ))}{\sqrt{4-\sin^2 (2θ)}},$$ where $0<θ<90^\circ $
Question: Is the following solution correct?
Solution attempt:
$$\sqrt{4-\sin^2 (2θ)}=2\cos(2θ)+3\sin(2θ)$$
$$4-\sin^2 (2θ)=(2\cos(2θ)+3\sin(2θ))^2$$
$$4-\sin^2 (2θ)=4\cos^2 (2θ)... | Noting that $\mathbf{0^\circ<2\theta<180^\circ}$ and continuing from
$$\sin(2θ)=0 \text{ or } \sin(2θ)+2 \cos(2θ)=0$$
$2\theta=\pi\;$ or $\;\big[\tan(2θ){=}-2\;$ or $\;\cos2\theta=0\big]$
$\displaystyle\theta=\frac\pi2\;$ or $\;\big[2\theta=180^\circ{-}\arctan2\;$ or $\displaystyle\;\frac\pi2\big]$
$\theta\displ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$(n^1+n)/2$ sequence This is going to be hard to explain so I'll just give an example
Let's say we have a standard arithmetic sequence that goes up by 1 each time
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
The last number (in this case, 10) is n
Scenario z:
Pick 2 numbers in the sequence and multiply them. The product should be equ... | $\left(n^2 + n + 2\right) \times \left(n^2 - n + 1\right) =
n^4 + 2n^2 - n + 2.$
Therefore
$$\left[\frac{n^4 + 3n^2 + 2}{2} - \frac{n^2 + n}{2}\right] \div \frac{n^2 + n + 2}{2} = (n^2 - n + 1).$$
$$S = \sum_{i=1}^{n^2 + 1} i = \frac{(n^2 + 1)(n^2 + 2)}{2} = \frac{n^4 + 3n^2 + 2}{2}.$$
So, for a particular value of $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find all intgers $a,b,c$ such that $a\cdot 4^n+b\cdot 6^n+c\cdot 9^n$ is a perfect square Find all intgers $a,b,c$ such that
$$a\cdot 4^n+b\cdot 6^n+c\cdot 9^n$$ is a perfect square for all sufficiently large $n$
This problem is creat by Vesselin Dimitrov,From the (Problems from the books chapter 17).
My approach:it is... | We’re given that $x_n = \sqrt{a4^n + b6^n + c9^n}$ is an integer for sufficiently large $n$. We have
\begin{multline*}
(6x_n + 5x_{n+1} + x_{n+2})(6x_n + 5x_{n+1} - x_{n+2}) \\ (-6x_n + 5x_{n+1} + x_{n+2})(-6x_n + 5x_{n+1} - x_{n+2}) = 900(b^2 - 4ac)36^n,
\end{multline*}
and since $x_n = \sqrt c⋅3^n\left[1 ± O{\left(\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $\exp(x)-1 > x^2 ,\forall x>0$ I'm trying to prove that
$\exp(x)-1 > x^2 ,\forall x>0$
What I've tried so far:
for $0<x<1$ I know that the equation is valid because:
$\exp(x)-1= x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$
so considering only the first term ($x$) I compared $x^2$ with it by subtracting both
$x... | $x+\frac{x^2}{2} + \frac{x^3}{6} > x^2$ reduces to
$$\frac{1}{6}x(x^2-3x+6) > 0,$$
which is true for all $x>0$ since the discriminant of the quadratic is negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4387167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to solve simultaneous equations of the form $\frac{a}{x}+\frac{b}{y}=1$ really fast? Context:
*
*Taking the major and minor axes of an ellipse as the $x$ and $y$ axes respectively find the equation of the ellipse passing through the points $(1,\sqrt{6})$ and $(3,0)$.
*Taking the major and minor axes of an ellips... | Use the substitution $x = \dfrac{1}{p} , y = \dfrac{1}{q} $, then your equations become:
$ 4 x + 16 y = 1 $
$ 25 x + 2 y = 1 $
And these can solved very easily, using for example, elimination
$(1) - 8 \times (2) $:
$- 196 x = - 7 \Rightarrow x = \dfrac{7}{196} \Rightarrow p = \dfrac{196}{7} = 28 $
And then, from equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati) $y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati)
(a) Find the solutions.
(b) $y(x_0)=y_0$, prove two cases:
$$0<y_0<x_0 \implies \text{solution's domain is} [x_0,\infty) $$
$$0<x_0<y_0 \implies \text{solution's domain is} [x_0,x_0+\alpha) , \alpha \in \mathbb {R}.$$
I will b... | $$z'=\frac{z}{x}+\frac{2z^2}{x^2}$$
$$\dfrac {z'}{z^2}=\frac{1}{zx}+\frac{2}{x^2}$$
$$-\left(\dfrac {1}{z}\right)'=\frac{1}{zx}+\frac{2}{x^2}$$
Looks like there is a little sign mistake. But all what you did looks really good.
$$u'=\color{red}{-\frac{u}{x}}-\frac{2}{x^2}$$
$$(ux)'=-\frac{2}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4389227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Minimize $xy$ over $x^2+y^2+z^2=7$ and $xy+xz+yz=4$.
Let $x,$ $y,$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 7$ and $xy + xz + yz = 4.$ Find the smallest possible value of $xy.$
I used Cauchy to get $$(x^2+y^2+z^2)(1^2+1^2+1^2)\geq (x+y+z)^2$$ and $$(x^2+y^2+z^2)(y^2+z^2+x^2)\geq(xy+xz+yz)^2,$$ but this d... | I would start with noticing that
\begin{align*}
(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + xz + yz) = 15
\end{align*}
Therefore $x + y + z = \pm\sqrt{15} =: k$. Better saying, $z = k - x - y$.
Then we arrive at the relation
\begin{align*}
xy + xz + yz & = xy + z(x + y)\\\\
& = xy + (k - (x + y))(x + y)\\\\
& = xy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4394067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What am I doing wrong with this Euler substitution? I want to evaluate the integral $\displaystyle\int\frac{1}{x\sqrt{x^2+2}}dx$ using the substitution $\sqrt{x^2+2}=-x+t$.
So I get from the substitution $x=\frac{t^2-2}{2t}$, $dx=\frac{t^2+2}{2t^2}dt$ and $\sqrt{x^2+2}=\frac{t^2+2}{2t}$. Substituting all this in I get:... | Both the answers are correct as $
{\left(\frac{x+\sqrt{x^2+2}-\sqrt{2}}{x+\sqrt{x^2+2}+\sqrt{2}}\right)}^2={\left(\frac{x^2+2+x\sqrt{x^2+2}-\sqrt{2x^2+4}-\sqrt{2}x}{x^2+2+x\sqrt{x^2+2}+\sqrt{2x^2+4}+\sqrt{2}x}\right)}={\left(\frac{\sqrt{x^2+2}-\sqrt{2}}{\sqrt{x^2+2}+\sqrt{2}}\right)}$
It can be proved by using compendo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How do I determine the value of $d$ from the tangent graph of the form $y = a\times \tan (bx + c) + d$?
I understand that $d$ is the vertical shift, and I have already worked out the values of $a$, $b$ and $c$ to be the following (please tell me if I am wrong):
$\pmb{a}\quad$ Given point is $(-1, -7)$ so I substituted... | You are correct in that $b = \frac{\pi}{4}$, but your reasoning isn't so clear. The function's period is $4$, and since $f(x) = \tan (\alpha x)$ has period $\pi / \alpha$, $b$ is $\frac{\pi}{4}$ by simple arithmetic.
So far we have $f(x) = a\tan(\frac{\pi}{4}x + c) + d$.
The function $g(x) = \tan(\frac{\pi}{4}x)$ has a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4397994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding max{f(x)} without derivative Consider the function $f(x)=\frac{\sqrt{x^3+x}}{x^2+x+1}$
the question is about: to find max{f} without using derivative.
I can find max with derivation and it is not hard to find. it is $f'(x)=0 \to x=1 $ so $max\{f\}=\frac{\sqrt 2}{3}$
but I am looking for an idea to do as the que... | Using AM-GM, we have
\begin{align*}
\frac{\sqrt{x^3 + x}}{x^2 + x + 1} &= \frac{\sqrt{2}\sqrt x \sqrt{\frac{x^2 + 1}{2}}}{x^2 + x + 1}\\
&\le \frac{\sqrt{2}\cdot\frac12\left(x + \frac{x^2 + 1}{2}\right)}{x^2 + x + 1}\\
&= \frac{\sqrt 2 (x^2 + 2x + 1)}{4(x^2 + x + 1)}\\
&= \frac{\sqrt{2}}{3} - \frac{\sqrt 2 (x - 1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to evaluate $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{12n+1} $? First of all, I am going to convert this series into a definite integral.
For $|x|<1,$ we have
$\displaystyle \quad \frac{1}{1+x^{12}}=\displaystyle \sum_{n=0}^{\infty}\left(-x^{12}\right)^{n}=\displaystyle \sum_{n=0}^{\infty}(-1)^{n} x^{12 n} \tag*{} $
Int... | Similar to the original integral, using the same technique yields
$\displaystyle \quad \frac{1}{1+x^{n}}=\displaystyle \sum_{k=0}^{\infty}\left(-x^{n}\right)^{k}=\displaystyle \sum_{k=0}^{\infty}(-1)^{k} x^{nk} ,\tag*{}
$
we have
Integrating both sides from $x=0$ to $1$, we can relate the target series and the definite... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Volume enclosed by the surface $\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}.$
Compute the volume enclosed by the surface $$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)^2=\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}.$$
My attempt:
I ... | $$\cos\left(\pi-\frac12\cos^{-1}(-\rho^2)\right) = $$
If $0 \le \rho^2 \le -\cos(2\theta)$, then $\frac{(4n-3)\pi}4 \le \theta \le \frac{(4n-1)\pi}4 \text{ for } n\in\mathbb Z$.
The polar angle must generally be between $0$ and $\pi$, so we take $\theta\in\left[\frac\pi4,\frac{3\pi}4\right]$. Then the volume is
$$\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Taylor series of $\sqrt{1-x}$ This is problem 6.7.6 in Abbott's Understanding Analysis 2nd ed. The section is called The Weierstrass approximation theorem.
a) Let $c_{n} = \frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots 2n}$ for $n\geq 1$. Show $c_{n} < \frac{2}{\sqrt{2n+1}}$.
b) Use a) to show that $\sum_{n... | Using the hint given by Gary in the comments of the question and the MSE linked in the question, I will answer the whole exercise.
a) First, let's start with a lemma.
For all $k \in \mathbb{N}$, $\frac{k}{k+1} < \frac{k+1}{k+2}$ holds.
Proof:
We can write $k^{2} +2k < k^{2} + 2k + 1$. This implies $k(k+2) < (k+1)^{2}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Idempotent relations in a ring Let $(A,+,.)$ be a ring such that, if $x \in A$ with $6x = 0$, then $x=0$. Let $a,b,c \in A$ such that $a-b$ , $b-c$ , $c-a$ are idempotent. Prove that $a=b=c$.
Unfortunately, I haven't made any big progress on this one. I noticed that $(a+b+c)^2 = 3(a^2+b^2+c^2)$ and I tried finding an e... | Note that
$\begin{align*}
(a-c)^2=(a-c)&=((a-b)+(b-c))^2\\
&=(a-b)^2+(b-c)^2+(a-b)(b-c)+(b-c)(a-b)\\
&=a-b+b-c+ab-ac-b^2+bc+ba-b^2-ca+cb
\end{align*}$
It follows that
$\begin{align*}
0=&ab-ac-b^2+bc+ba-b^2-ca+cb\\
=&\color{blue}{(ab+ba)-b^2} +\color{red}{(bc+cb)-b^2}-(ac+ca)\\
=&\color{blue}{a^2-a+b}+\color{red}{c^2-c+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4407967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the equation of the Parabola The parabola $y = x^2 + bx + c$ has the following properties:
*
*The point on the parabola closest to $(12,3)$ is the intersection with the $y$ axis of the parabola.
*The parabola passes through $(-5,0).$
How can I find $(b, c)$?
Here is my attempt:
The point $(0, c)$ is the int... |
The parabola $y = x^2 + bx + c$
*
*The point on the parabola closest to $(12,3)$ is the intersection of the $y$ axis and the parabola
and we have the equation $-5b +25+c =0$.
*
*Let $P$ be the parabola's $y$-intercept $(0,c)$ and $Q$ be $(12,3).$
Then the parabola's tangent at $P$ is perpendicular to $PQ,$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Given $z^2-1\mid x^2z^2-1$, prove $\frac{x^2z^2-1}{z^2-1}$ is never prime, for $x$, $z$ integers such that $x>z>1$.
Given $z^2-1\mid x^2z^2-1$, prove that $\frac{x^2z^2-1}{z^2-1}$ can never be prime, assuming $x$, $z$ are integers such that $x>z>1$.
So far I have tried taking mod a lot of different numbers, but I can... | Answered for my own benefit, as I already had seen the above two answers first******
Note that $x^2z^2-1 = (xz-1)(xz+1)$. Then, as $\frac{x^2z^2-1}{z^2-1} = \frac{(xz-1)(xz+1)}{z^2-1}$ is an integer, it follows that $z^2-1$ can be written as follows: $z^2-1 = ab$ for some positive integers $a$ and $b$, that satisfy the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
On the log-cosine integral $\int\limits_0^{\pi/2}\ln(\cos(x))\,dx$ I know that
$$\mathcal I = \int\limits_0^{\pi/2}\ln(\cos(x))\,dx = -\frac\pi2 \ln(2)$$
and I am aware of a few different clever ways to demonstrate this result (e.g. MSE 4065767 and MSE 1992462). I would like to know if there's any way to wrap up the me... | With $x\to (1-x)/(1+x)$,
$$\int_0^1 \frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{1+x^2} dx=\int_0^1 \frac{\ln\left(\frac{2x}{1+x^2}\right)}{1+x^2}dx$$
$$=\int_0^1\frac{\ln(2)}{1+x^2}dx+\int_0^1\frac{\ln(x)}{1+x^2}dx-\int_0^1\frac{\ln(1+x^2)}{1+x^2}dx$$
$$=\frac{\pi}{4}\ln(2)-G+\sum_{n=0}^\infty\frac{(-1)^nH_n}{2n+1}.\tag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Integral $\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$ I tried to integrate
$$\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$$
by multipling by $\sqrt{2x^2 + 2x + 1}$ in the numerator and the denominator to break it into $5$ fractions
The answer is possible but it's too long
Another better solution
| Notice that
$$
\left(3x^2 + 5x +1\right)\sqrt{2x^2 + 2x+1} = \frac{1}{4\sqrt{2}}\left(3(2x+1)^2 + 4(2x+1) -3\right)\sqrt{(2x+1)^2+1}
$$
under substitution $2x +1 = \tan(\alpha)$ simplifies your integral to
\begin{align*}
&\int\left(3x^2 + 5x +1\right)\sqrt{2x^2 + 2x+1} \, \mathrm{d}x \\
&=\frac{1}{8\sqrt{2}} \int \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Calculate $\gcd(a^2b^2, a^2 + ab + b^2)$ Given $\gcd(a, b) = 1$, calculate $d =\gcd(a^2b^2, a^2 + ab + b^2)$ in terms of $a$ and $b$.
I have tried some manipulations of the terms arriving to some expressions such as that $d$ divides $a^4 + b^4$ or that $d$ divide s $(a+b)^4$ but those haven't given me much help. I have... | It is easier in such problems to consider first the case when $d=p$ is a prime number (instead of an arbitary postive integer).
Indeed, suppose that $p\mid\gcd(a^2b^2,a^2+ab+b^2)$. Then, $p\mid a^2b^2$ and $p\mid a^2+ab+b^2$. Can you now find out for which prime $p$ it is possible given that $\gcd(a,b)=1$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Range of $\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^2}$ on $0 \le a,$ $b,$ $c,$ $d \le 1.$ Let $0 \le a,$ $b,$ $c,$ $d \le 1.$ Find the possible values of the expression
$$\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^... |
The max must be $4$ (while $a=b=c=d=1$) and min be $a=c$ and $b=d$).
Define
$$L(a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2,\lambda_1,\lambda_2,\lambda_3,\lambda_4)=\sqrt{a_1^2+b_2^2}+\sqrt{b_1^2+c_2^2}+\sqrt{c_1^2+d_2^2}+\sqrt{d_1^2+a_2^2}+\lambda_1(a_1+a_2-1)+\lambda_2(b_1+b_2-1)+\lambda_3(c_1+c_2-1)+\lambda_4(d_1+d_2-1).$$
Le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4420384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
calculate one determinant to show 【National College Entrance Exam, old China】
$ \omega =\frac{1 \pm \sqrt{3} i } {2}$
Please calculate and show:
$ \begin{vmatrix} 1 &\omega& \omega^2 & 1\\ \omega& \omega^2 & 1 &1\\ \omega^2& 1& 1 & \omega \\ 1& 1& \omega& \omega^2\\ \end{vmatri... | You typed $\omega=\frac{1\pm\sqrt3\,i}{2}$ but I think you mean
$$\omega=\frac{-1+\sqrt3\,i}{2}\ .$$
In that case, subtracting a suitable multiple of row 1 from every other row (and not bothering to calculate irrelevant entries) gives
$$\eqalign{\det\pmatrix{1&\omega&\omega^2&1\cr \omega&\omega^2&1&1\cr \omega^2&1&1&\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can we prove that $\int_0^{2\pi} e^{r \cos \theta} \cos (r \sin \theta+n \theta) d \theta=0$ without complex analysis? In textbooks on complex analysis, a typical example using Cauchy Integral Theorem is shown that
$$
\int_{0}^{2 \pi} e^{r \cos \theta} \cos (r \sin \theta+\theta) d \theta=0
$$
Later, I generalized it i... | With
$$I_n(r)=\int_{0}^{2\pi}e^{r\cos\theta}\cos(n\theta+r\sin\theta)d\theta
$$
it is straightforward to establish $I_{n+1}(r) =I_n’(r)$, along with
$$I_1(r)= \int_{0}^{2\pi}e^{r\cos\theta}\cos(\theta+r\sin\theta)d\theta=\frac1r e^{r\cos\theta}\cos(r\sin\theta)\bigg|_{0}^{2\pi} =0$$
Then
\begin{align}
I_{n+1}(r) = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4422037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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If $|ax^2+bx+c|\leq 2\ \ \forall x\in[-1,1]$ then find the maximum value of $|cx^2+2bx+4a|\ \ \forall x\in [-2,2]$.
If
$$\left|ax^2+bx+c\right|\leq 2\quad \forall x\in[-1,1]$$
then find the maximum value of
$$\left|cx^2+2bx+4a\right|\quad \forall x\in [-2,2].$$
My Attempt
Let $f(x)=ax^2+bx+c$, then
$$|cx^2+2bx+4a|=x^... | We have
\begin{align*}
|a - b + c| &\le 2, \tag{1}\\
|a + b + c| &\le 2, \tag{2}\\
|c| &\le 2. \tag{3}
\end{align*}
Using (1) and (2), we have
$$|a + c| + |b| \le 2. \tag{4}$$
(Note: If $(a + c)b \ge 0$, then $|a + c| + |b| = |a + c + b| \le 2$. If $(a + c)b < 0$, then $|a + c| + |b| = |a + c - b| \le 2$.)
Using (1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Find $k\in\mathbb{R}$ given $w = k+i$ and $z=-4+5ki$ and $\arg(w+z)$ I am working on problem 2B-11 from the book Core Pure Mathematics, Edexcel 1/AS. The question is:
The complex numbers $w$ and $z$ are given by $w = k + i$ and $z = -4 + 5ki$ where $k$ is a real constant. Given at $\arg(w + z) = \frac{2\pi}{3}$, find t... |
$k\in\Bbb R,\,w:=k+i,\,z:=-4+i5k$ and $\arg(w+z)=\frac{2\pi}{3}$
Of course $w+z=(k-4)+i(5k+1)$. Since $0\lt\frac{2\pi}{3}\le\pi$ the (principal) argument tells us it is in the second quadrant (the top left), where $\arg(x+iy)=\pi-\arctan\frac{y}{-x}$ ("$-x$" because $x$ is negative and $-\arctan$ since we begin at $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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If $f(x)=\frac {x^ 2 -2x +4}{ x^ 2 +2x+4}$ for $x \in \mathbb{R}$, prove that the range of $f(x)$ is $[1/3, 3]$ One method to solve it would by putting $y = f(x)$ then multiplying the denominator with $y$ hence making a quadratic equation in x then we can just use the inequalities for $x$ being real to prove it.
For an... | HINT
I would recommend you to notice that
\begin{align*}
\frac{x^{2} - 2x + 4}{x^{2} + 2x + 4} & = \frac{(x^{2} + 2x + 4) - 4x}{x^{2} + 2x + 4}\\\\
& = 1 - \frac{4x}{x^{2} + 2x + 4}
\end{align*}
Consequently, the problem reduces to study the critical points of the last expression.
In order to do so, let us take its der... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4424161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How do I find the parabola parameter equation? Show that $x(t)=\cos^4t, y(t)=\sin^4t$ is a parametrization of the parabola $(x-y-1)^2=4y$.
I think that to solve this problem, we need to know how to find the parametric equation of the parabola. So I searched through books and internet search, but I haven't found out yet... | Given the conic of Cartesian equation:
$$
(x-y-1)^2 = 4\,y
\quad \quad \Rightarrow \quad \quad
x^2+y^2-2\,x\,y-2\,x-2\,y+1=0
$$
we can rewrite this polynomial equation in the following two ways:
$$
\small
\begin{bmatrix}
x & y & 1 \\
\end{bmatrix}
\underbrace{\begin{bmatrix}
1 & -1 & -1 \\
-1 & 1 & -1 \\
-1 & -1 & 1 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit/asymptotic of a modification of the exponent's Taylor series Consider the following function on $\left[0,\infty\right)\times\left[0,1\right]$
$$f\left(x,y\right)=\sum_{n=0}^{\infty}\dfrac{x^{n}}{n!}y^{n^{2}}$$
I am interested in the limit/asymptotic behavior of the following expression for $x\rightarrow\infty$
$$... | Step 1. Let us parametrize $x = e^{s}$ and $y = e^{-\varepsilon}$ for $s \in \mathbb{R}$ and $\varepsilon > 0$. Fix the values of $k$ and $\varepsilon$, and then write
$$ f(xy^k, y) = \sum_{n=0}^{\infty} e^{-h(n)}, \qquad \text{where} \quad h(n) = \varepsilon n^2 + k\varepsilon n - sn + \log n! $$
Regarding $h(n)$ as a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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minimizing the distance between values
Find all values of $a$ that minimize the expression $|a- x_1| + |a-x_2|+\cdots + |a-x_n|$, where $x_1\leq x_2\leq \cdots \leq x_n$.
I know the minimum is achieved when $a=x_{\lfloor n/2\rfloor + 1}$ if $n$ is odd and when $a\in [x_{\lfloor n/2\rfloor}, x_{\lfloor n/2\rfloor + 1... | Note that
$$ |x - c| = \begin{cases}
c - x & \text{for $x \leq c$} \\
x - c & \text{for $x \geq c$}
\end{cases} $$
So, adopting the conventions that $x_0 = -\infty$ and $x_{n+1} = \infty$, then for any $x$ satisfying $ x_k \leq x \leq x_{k+1} $,
$$ f(x) = \sum_{i=1}^{n} |x - x_i| = \sum_{i=1}^{k} (x - x_i) + \sum_{i=k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding closed form expression for $n^{th}$ term of sequence with generating function $F(x)$? I have been asked the above question regarding the generating function
$$F(x) = \frac{x^2(1-x)}{(1-x)^3}$$
I have no idea what procedure this type of question follows. The solution gives that $F(x)$ can be written as
$$\frac{x... |
Denoting with $[x^n]$ the coefficient of $x^n$ of a series we obtain
\begin{align*}
\color{blue}{[x^n]F(x)}&=[x^n]\frac{x^2(1-x)}{(1-x)^3}\\
&=[x^n]\left(\frac{x^2}{(1-x)^3}-\frac{x^3}{(1-x)^3}\right)\tag{1}\\
&\,\,\color{blue}{=[x^{n-2}]\frac{1}{(1-x)^3}-[x^{n-3}]\frac{1}{(1-x)^3}}\tag{2}
\end{align*}
In the last lin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How can I find this limit that is supposed to converge to e? I made a formula to get an approximation $x-1$ based on $x$; $f(x) = \frac{(-2 + x)^{3/2 - x} x^{-1/2 + x}}{e^2}$. I got this formula by looking at $\frac{n!}{n(n-2)!}$ which equals $n-1$, then substituting the factorials with Stirling approximants. After sim... | $$ f(x)=\frac{(x-2)^{\frac{3}{2}-x} x^{x-\frac{1}{2}}}{e^2}\implies \log[f(x)]=\left(\frac{3}{2}-x\right)\log(x-2)-\left(x-\frac{1}{2}\right)\log(x)-2$$ Expanding as series for large values of $x$
$$\log[f(x)]=\log (x)-\frac{1}{x}-\frac{1}{3 x^2}+O\left(\frac{1}{x^4}\right)$$
$$f(x)=e^{\log[f(x)]}=x-1+\frac{1}{6 x}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Explanation for a solution: Howard Anton, Elementary Linear Algebra I am currently working with the 1st edition of Howard Anton's "Elementary Linear Algebra". I tried the following problem:
Excercise Set 1.2 (p. 17), Problem 12:
For which values of $a$ will the following system have no solutions? Exactly one solution? ... | Alternative approach:
Compute the determinant of the matrix.
If the determinant is non-zero, then there will automatically be exactly $1$ solution.
If the determinant is $0$, then there will either be $0$ solutions or an infinite number of solutions, depending on whether there is an inconsistency (explained at the end ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Is there a mistake in solving this limit? I want to solve this:
\begin{equation}
L=\lim_{x\rightarrow 0} \frac{\sum\limits_{m=1}^{M}a_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}} } {\sum\limits_{m=1}^{M} c_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}}}
\end{equation}
wh... | If $x^2 \to 0$ then you certainly can't say that $\exp(\frac{b}{x^2}) \to 1$. Quite the contrary, this term goes to infinity very fast, the larger $b$ the faster.
You started with neglecting a smaller order term, that's all and well to get a rough idea, but that's certainly not rigorous, all the more inside an exponent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Diagonal element of traceless hermitian matrix? In physics, we are familiar with a set of traceless hermitian matrices named Pauli matrices:
$$
{\displaystyle {\begin{aligned}\sigma _{1}=\sigma _{\mathrm {x} }&={\begin{pmatrix}0&1\\1&0\end{pmatrix}}\\\sigma _{2}=\sigma _{\mathrm {y} }&={\begin{pmatrix}0&-i\\i&0\end{pma... | Yes. Let $H$ be any $n\times n$ traceless Hermitian matrix and $UDU^\ast$ be its unitary diagonalisation. Let $Q$ be a real orthogonal matrix whose last column is $\frac{1}{\sqrt{n}}(1,1,\ldots,1)^T$. The last diagonal element of $H':=Q^TU^\ast HUQ=Q^TDQ$ is then the mean of all diagonal elements of $H$, which is zero.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral Basis of $O_k$ Let $K=Q(\sqrt 6,\sqrt{11})$. Write $α ∈ O_K$ and its conjugates in terms of a $Q$-basis.
And show that an integral basis of $O_K$ is given by
${1,\sqrt 6,\sqrt {11},\frac{\sqrt 6+\sqrt{66}}2 }$, from first principles.
I'm really not sure how to do this, if anyone could help I'd really appreciat... | You can consider the traces of $\alpha = a+b\sqrt{6}+c\sqrt{11}+d\sqrt{66}$ over the 3 quadratic subfields, noting that this gives an algebraic integer (in that subfield).The three subfields are $L_1 = \mathbb{Q}(\sqrt{6})$, $L_2=\mathbb{Q}(\sqrt{11})$ and $L_3=\mathbb{Q}(\sqrt{66})$ having integers $\mathbb{Z}[6]$, $\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
To prove the inequality of positive rational numbers Show that: $$ \left(\frac{a+b}{a+b+c}\right)^{c} \left(\frac{b+c}{a+b+c}\right)^{a} \left(\frac{a+c}{a+b+c}\right)^{b}< \left(\frac{2}{3}\right)^{a+b+c} ,a\ne b\ne c$$
PS: I am supposed to use weighted AM > GM > HM for this problem.
I tried by expressing $$\frac{a+b}... | Edit: Fixed algebra error.
Consider the weighted AM-GM with weights: $\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}$. We attain:
$$(\frac{a(b+c) + b(a+c) + c(a+b)}{a+b+c})^{a+b+c} > (a+b)^c (b+c)^a (a+c)^b.$$
And thus:
$$(\frac{a(b+c) + b(a+c) + c(a+b)}{(a+b+c)^2})^{a+b+c} > (\frac{a+b}{a+b+c})^c (\frac{b+c}{a+b+c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to factorize $2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c$? If $a,b,c$ are in AP $\implies \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are in AP $\implies
\frac{ab+bc+ca}{bc}, \frac{ab+bc+ca}{ca}, \frac{ab+bc+ca}{ab}$ are in AP $\implies \frac{bc+ca}{bc},\frac{ab+bc}{ca},\frac{bc+ca}{ab}$ are in AP $\implies \frac{bc}{bc+ca}, \f... | It's tricky!
\begin{array}{l}
2ab^2 + 2b^2 c - ac^2 - bc^2 - a^2 b - a^2 c + 2abc- 2abc = \\
\\
= (2ab^2 + 2b^2 c + 2abc) - \left( {a^2 b + a^2 c + abc} \right) - \left( {ac^2 + bc^2 + abc} \right) = \\
\\
= 2b(ab + bc + ac) - a(ab + bc + ac) - c(ab + bc + ac) = \\
\\
= - (ab + bc + ac)\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Algebra of o-symbols In theorem $7.8$, Tom Apostol in his Calculus Vol. $1$ gave and proved the following basic rules for algebra of o-symbols
*
*$o(g(x)) \pm o(g(x)) = o(g(x)) $
*$o(cg(x)) = o(g(x)) = o(g(x)) $, if $c \ne 0$
*$f(x) \cdot o(g(x)) = o(f(x)g(x)) $
*$o(o(g(x))) = o(g(x)) $
*$\frac{1}{1 + g(x)} = 1 -... | The best way to deal with the $"o"$ notation is to write it as
$$F(x)=o(g(x)) \; (x\to 0)$$
means that
$$F(x)=g(x)\epsilon(x)$$
with
$$\lim_{x\to 0}\epsilon(x)=0$$
So, your last equality gives
$$o(x^2)+o(x^3)=$$
$$x^2\epsilon_1(x)+x^3\epsilon_2(x)=$$
$$x^2(\epsilon_1(x)+x\epsilon_2(x))=$$
$$x^2\epsilon(x)=o(x^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4442024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $abc=1$ ,and $n$ is a natural number,prove $ \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2} $ If $a, b, c$ are distinct positive real numbers such that $abc=1$,and $n$ is a natural number,prove
$$
\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \g... | Hint :
Using Fuchs's inequality wich is an extension of Karamata's inequality one can show that for ($f(x)=x^n$),$a\geq b \geq c>0$ such that $abc=1$:
$$\left(\frac{a^{n}}{\left(a+c\right)\left(a+c\right)}-1\right)a\cdot\frac{1}{\left(a-b\right)\left(a-c\right)}\geq \left(\frac{b^{n}}{\left(b+c\right)\left(b+a\right)}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How does this expression simplification make sense? My simplification does not equal what is given. I attached an image of the expression I'm working with
I have no idea how $A(x)$ is simplified to $\displaystyle 5x - \left(\frac{1+\pi}{2}\right) x^2$.
Every time I work out the equation I get $$A(x) = 5x - x^2 - (\pi ... | So, step-by-step
\begin{align*}
A(x)&=xy+\pi\left(\frac{x}{2}\right)^{2},\\&=x\left(5-x-\frac{\pi}{2}x\right)+\pi\left(\frac{x}{2}\right)^{2},\\&=5x-x^{2}-\frac{\pi}{2}x^{2}+\frac{\pi}{2^{2}}x^{2},\\&=5x+\left(\frac{\pi}{2^{2}}-\frac{\pi}{2}-1\right)x^{2},\\&=5x+\left(\frac{\pi}{2^2}-\frac{2\pi}{2^{2}}-\frac{2^{2}\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Showing the value the expression takes using another approach Let $1, \omega, \omega^{2}$ be the cube roots of unity. Then the product
$$
\left(1-\omega+\omega^{2}\right)\left(1-\omega^{2}+\omega^{2^{2}}\right)\left(1-\omega^{2^{2}}+\omega^{2^{3}}\right) \cdots\left(1-\omega^{2^{9}}+\omega^{2^{10}}\right)
$$ is equal t... | Note that, $\omega^2 + \omega +1=0$. Also note that,
\begin{equation*}
2^n \equiv \begin{cases}
1 (\text{ mod } 3), \hspace{1cm} \text{if $n$ is even}\\
2 (\text{ mod } 3), \hspace{1cm} \text{if $n$ is odd}
\end{cases}
\end{equation*}
Establishing above statement is easy. Using this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4446948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $ k $ if both roots of this equation $ x^2-2kx+k^2+k-5=0 $ are less than 5 If both roots of this equation $ x^2-2kx+k^2+k-5=0 $ are less than $ 5$, then find the limit of $ k$.
Method-1
Let's say,
\begin{align} f(x)=x^2-2kx+k^2+k-5 \end{align}
\begin{align} f(0)=0 →\alpha,\beta \end{align}
$ a=1 \gt 0 ,$ So grap... |
Find $ k $ if both roots of this equation $ x^2-2kx+k^2+k-5=0 $ are less than 5
Alternative approach:
Use the quadratic equation directly, calculate the value of the two roots, in terms of $k$, and then apply the constraint that both roots are less than $5$.
The roots are
$~\displaystyle \frac{1}{2}\left[2k \pm \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving $\tan ^{-1}(\frac{1-x}{1+x})=\frac{1}{2}\tan ^{-1}(x)$ I was solving the following equation,$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
But I missed a solution (don't know where's the mistake in my work).
Here's my work:
$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\ta... |
$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
Putting $x = \tan(\theta)$
Here, it is useful to impose the restriction $$\theta\in\left(-\frac\pi2,-\frac\pi4\right)\cup\left(-\frac\pi4,\frac\pi2\right);\tag1$$ since $\tan\theta$ for which the given equation is defined is surjective on ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Regarding the solution of finding the remainder of $g(x^{12})$ divided by $g(x)$
Let $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1.$ What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$?
I'm reading the solution for this and I don't understand how can someone come up with it.
It states
W... |
$(x - 1)g(x) = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1) = x^6 - 1$
That product is "telescoping":
$$\begin{array}{ccccccc}
(x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1)
&= &x^6 &+ x^5 &+ x^4 &+ x^3 &+ x^2 &+ x \\
& & &- x^5 &- x^4 &- x^3 &- x^2 &- x &-1 \\
&=& x^6&&&&&&-1
\end{array}$$
More generally: $$\sum_{k=0}^n x^k =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Simpler proof that $y^3[d^2y/dx^2]$ is a constant if $y^2=ax^2+bx+c$? here's my question
If $y^2=ax^2+bx+c$ then prove that $y^3[d^2y/dx^2]$ is a constant .
I have solved this using the conventional method, taking square root, differentiating w.r.t to x and using chain and quotient rule
But can't think of some alter... | Using differentials, it is easy to see that
$$
2y dy = (2ax+ b) dx \tag{1}
$$
from which we can write
$$
u = \frac{dy}{dx} = \frac{2ax+b}{2y}
$$
Using again differentials and product rule,
\begin{eqnarray}
du
&=& \frac{2a}{2y} dx - \frac{2ax+b}{2y^2} dy
= \left[ \frac{a}{y} - \frac{(2ax+b)^2}{4y^3} \right] dx
\end{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
What did I do wrong in solving $\int\sec^{-1} x\,{dx}$? I used integration by parts:
let u=$\sec^{-1}\,x$, dv=dx,
then du=$\frac{1}{|x|\sqrt{x^2-1}}$, v=x.
I = $x\sec^{-1}\,x\;-\;\int\frac{x}{|x|\sqrt{x^2-1}}dx\\$
Integration of $\int\frac{x}{|x|\sqrt{x^2-1}}dx$:
Let x=$\sec\theta$, then dx = $\sec\theta\tan\theta\,d\t... |
When $\theta\in(0, \frac{\pi}2),$
Then I = $x\sec^{-1}\,x\;-\;\ln|x+\sqrt{x^2-1}|+c$.
Small correction: $\theta\in[0, \frac{\pi}2).$ This corresponds to $x\ge1.$
When $\theta\in(\frac{\pi}2, \pi),$
Then I = $x\sec^{-1}\,x\;\boldsymbol{+}\;\ln|x+\sqrt{x^2-1}|+c$
Small correction: $\theta\in(\frac{\pi}2, \pi].$ This ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Let $a$,$b$ be the roots of $x^2-6cx-7d=0$, and let $c$, $d$ be the roots of $x^2-6ax-7b=0$. Find $b+d$.
Let $a$,$b$ be the roots of the equation $x^{2}-6 c x-7 d=0$ and let $c$,$d$ be the roots of the equation $x^{2}-6ax-7 b=0$. $a,b,c,d$ are distinct. Find value of $b+d$.
Using the relation between roots and coeffi... | I like to work with a bit of generality whenever possible, so take the equations to be
$$x^2+pcx+qd=0 \qquad x^2+pax+qb=0$$
By Vieta's formula for the linear coefficients, we have a solvable linear system in $a$ and $c$:
$$
\left.\begin{align}
a+b=-pc \\ c+d=-pa
\end{align}\right\}\quad\to\quad a=\frac{b-dp}{p^2-1}\qqu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluate: $\prod_{n=1}^{n=9}(\sin(20n-10^\circ))$ The given answer is $2^{-8}$
My attempt
$$=\sin(10^\circ) \sin(30^\circ) \sin(50^\circ)\cdots\sin(170^\circ)$$
after this I multiply and divide to fill in the even multiples of $10^\circ$
$\dfrac{\sin(10)\sin(20)\sin(30) \sin(40)\cdots \sin(160) \sin(170)}{\sin(20)\sin(... | $$A=\sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) \sin(90^\circ) \sin(110^\circ) \sin(130^\circ) \sin(150^\circ) \sin(170^\circ)$$
$$\sin(180^\circ-n)=\sin n \Rightarrow A=\sin^2(10^\circ)\sin^2(30^\circ)\sin^2(50^\circ)\sin^2(70^\circ)\sin(90^\circ)$$
$$\sin(30^\circ)=\frac12,\sin(90^\circ)=1\Rightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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The integral $\int_{0}^{1} \frac{ \log (1-x)}{1+x^2}dx$ Recently a very interesting result
$\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$
has been proved in a more than elegant way. See Show that $\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathr... | Split the integrand\begin{align}
\int_{0}^1 \frac{\ln(1-x)}{1+x^2} dx
=& \int_{0}^1 \underset{= \frac\pi8 \ln2} {\frac{\ln \sqrt2}{1+x^2} }dx
+ \int_{0}^1 \underset{=-G}{\frac{\ln \frac{1-x}{1+x}}{1+x^2}}\overset{x\to\frac{1-x}{1+x}}{dx}
+ \int_{0}^1 \underset{K=- K=0}{\frac{\ln \frac{1+x} {\sqrt2 }}{1+x^2} }\overset... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Number of ways to form $24$ using $7$,$3$ and $2$
$\text{What is the number of ways to form }24\text{ using }7,2,\text{ and }3,\text{ zero or more times?}$
We can write $24$ as $7a+3b+2c$ . Since $\lfloor\frac{24}{7}\rfloor = 3,$ $\lfloor\frac{24}{3}\rfloor = 8$ and $\lfloor\frac{24}{2}\rfloor = 12$ ,We can say $0 \l... | Note that once you've found $a$ and $b$, you can calculate $c$ directly:
$$c = \frac{24-7a-3b}{2}$$
This limits your search space to the 36 possible combinations of $(a, b)$ with $0 \le a \le 3$ and $0 \le b \le 8$.
Also, we need $c \ge 0$, which means that once we've set a value for $a$, we can constrain $7a + 3b \le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4466715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Classification of singularity and Laurent series We have the following complex function:
$$f(z) = \frac{1}{z^2+4}$$
We need to get the classification of singularities and also the main part (The one with negative coefficients) of Laurent series around the point $2i$.
What we have done is the following:
$$z^2 +4 = (z+2i... | This technique is a convenient shortcut to determine the residuum of a Laurentseries expansion at a simple pole $z_0$. Doing it the long way we obtain
\begin{align*}
\color{blue}{f(z)=\frac{1}{z^2+4}}&=\frac{1}{(z-2i)(z+2i)}\tag{1}\\
&=\frac{1}{z-2i}\,\frac{1}{4i+(z-2i)}\\
&=\frac{1}{z-2i}\,\frac{1}{4i\left(1+\frac{z-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \... | Given $a$ and $b$ be the roots of the equation $x^2-7x+2=0$. Sum of roots, $$a+b=7$$
Product of roots, $$ab=2.$$
Now, $(a-b)^2=(a+b)^2-4ab=49-8=41$.
$\Rightarrow a-b=\pm\sqrt{41}$.
Taking $a-b=\sqrt{41}$.
On solving equation $a+b=7$ and $a-b=\sqrt{41}$.
We obtain, $a=\frac{7+\sqrt{41}}{2}$ and $b=\frac{7-\sqrt{41}}{2}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 7
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Find the minimum value of a trigonometric function If the minimum value of $f\left(x\right)=\left(1+\frac{1}{\sin ^6\left(x\right)}\right)\left(1+\frac{1}{\cos ^6\left(x\right)}\right),\:x\:∈\:\left(0,\:\frac{\pi }{2}\right)$ is $m$, find $\sqrt m$.
How do I differentiate this function without making the problem unnece... | Without calculus, let $\,a = \sin^2(x)\,$, $\,b = \cos^2(x)\,$, then $\,f(x) = \left(1+\dfrac{1}{a^3}\right)\left(1+\dfrac{1}{b^3}\right)\,$, where $\,a,b \ge 0\,$ and $\,a+b=1\,$. By the generalized means inequalities:
*
*$\sqrt{ab} \le \dfrac{a+b}{2} = \dfrac{1}{2} \quad\implies\quad \dfrac{1}{ab} \ge 4 \quad\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Getting two different answers on differentiating $\cos^{-1}(\frac{3x+4\sqrt{1-x^2}}{5})$
Question given in my book asks to find $\frac{dy}{dx} $ from the following equation.$$y=\cos^{-1}\left(\frac{3x+4\sqrt{1-x^2}}{5}\right)$$
My Attempt:
Starting with substitutions,
*
*Putting $\frac35=\cos\alpha\implies \frac45... | $\cos^{-1}(\cos x)$ needn’t always be equal to $x$. See the graph below for $\cos^{-1}(\cos x)$ :
When $x < 0$ then $\cos^{-1}(\cos x) = -x$.
Now if $x < 0.6$ then $\beta >\alpha$ because $\cos^{-1} x$ is a decreasing function in $[-1, 1]$. So ideally there should be two cases, $x < 0.6$ and $x > 0.6$. Also note t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Find $\int \sin^4x dx$ In my textbook, I came upon the following problem:
Problem: Find $\int \sin^4x dx$
Since there was only given a solution sketch and I came to a different conclusion in one step, I ask if my following solution is correct.
Solution: In my solution, I use the following identities:
$$\sin^2x = \frac... | I like to rewrite things so as to reduce errors; as $\sin^4 x$ is even of small degree (as a polynomial in $\sin, \cos$), it can be written in terms of $\cos 2 x$ and $\cos 4x.$ After that the integral is immediate, finally we may wish to expand that back into a polynolial in sine and cosine.
First
$$ \cos 2x = 2 \co... | {
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"url": "https://math.stackexchange.com/questions/4476854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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When are matrices $AB$ and $BA$ similar? If $A$ or $B$ is invertible, it is easy to see that $AB$ and $BA$ are similar. I'm curious about when (maybe a strong sufficient or necessary condition) they become similar.
Here's what I have done:
There exists invertible matrices $P, Q$ such that $A=P\begin{pmatrix} I_r &\\ &0... | In general, two matrices $X$ and $Y$ are similar over an algebraically closed field $F$ if an only if $\operatorname{rank}\left((\lambda I-X)^k\right)=\operatorname{rank}\left((\lambda I-Y)^k\right)$ for all $\lambda\in F$ and $k\ge1$. However, in case $X=AB$ and $Y=BA$ for some square matrices $A$ and $B$, one only ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Show $2\mid (n^7 -n), \forall n \in\mathbb{N}$. Show $2\mid (n^7 -n), \forall n \in\mathbb{N}$.
$n^7 \equiv n \bmod 2\implies n^6 \equiv 1 \bmod 2$
Using Fermat's little theorem, it is easy to see that:
$n^1 \equiv 1\bmod 2\implies n^a \equiv 1^a\bmod 2, \forall a \in \mathbb{N}$
Algebraic approach:
$(n^3-1)(n^3+1)\equ... | $n^7-n = n(n^6-1) = n(n^3-1)(n^3+1)=n(n-1)(n^2+n+1)(n^3+1)$. Observe the product consists of $2$ consecutive integers $n,n-1$ hence divisible by $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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A specific binomial summation identity Let $m$ and $n$ be positive integers with $m < n$. Prove
\begin{equation}
\left(\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{n-k} \right)\left(\sum_{k=0}^m \binom{n}{k}\frac{(-1)^k}{k+1} \right) =\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(n-k)(k+1)}
\end{equation}
I'm not sure how to a... | We seek to prove for $n,m\ge 1$ with $m\lt n$ that
$$\sum_{k=0}^m {m\choose k} \frac{(-1)^k}{n-k}
\sum_{k=0}^m {n\choose k} \frac{(-1)^k}{k+1}
= \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{(n-k)(k+1)}.$$
The RHS is
$$\frac{1}{n+1} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{n-k}
+ \frac{1}{n+1} \sum_{k=0}^m {m\choose k} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Quick evaluation of definite integral $\int_0^\pi \frac{\sin 5x}{\sin x}dx$
Find$$\int_0^{\pi}\frac{\sin 5x}{\sin x}dx$$
I can solve it by involving polynomials in sine and cosine as shown in the links below, but it’s huge (doing double angle formulas twice; I noticed that using polynomials in cosine is better becaus... | We can generalized for positive integer k:
$\displaystyle \frac{\sin((2k-1)\,x)}{\sin(x)}
= 1 + 2\cos(2x) + 2\cos(4x) \;+\;...\;+\; 2\cos((2k-2)\,x)
$
Prove by induction:
Base case, $k=1$, is obviously true: $\displaystyle \frac{\sin x}{\sin x} = 1$
Assume formula is true for $n=k$, prove it is true for $n=k+1$
$\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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From 7 consonants and 5 vowels, how many words can be formed consisting of 4 consonants and 3 vowels if letters may be repeated? cannot be repeated?
From $7$ consonants and $5$ vowels, how many words can be formed consisting of $4$ consonants and $3$ vowels if:
*
*Any letter can be repeated.
*No letter can be repea... |
From $7$ consonants and $5$ vowels, how many words can be formed consisting of $4$ consonants and $3$ vowels if any letter can be repeated?
Your answer $7^4 \cdot 5^3$ is the number of ways the first four positions can be filled with consonants and the last three positions can be filled with vowels when letters may b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Expand the function $f(z) = \frac{1+2z^2}{z^2+z^4} $ into power series of $z$ in all areas of convergence. I'm studying for my upcoming complex analysis qualifying exam by working through problems in past exams. For this problem, I'd like to know (1) if my answer is correct and complete (i.e. whether I've made any erro... | Since we are looking for a power series we are looking for a representation
\begin{align*}
f(z)=\sum_{k=0}^\infty a_k\left(z-z_0\right)^k
\end{align*}
evaluated at center $z_0$.
We start with a partial fraction decomposition
\begin{align*}
\color{blue}{f(z)}&=\frac{1+2z^2}{z^2\left(1+z^2\right)}\\
&\,\,\color{blue}{=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding the range of $\frac {2x^2+x-3}{x^2+4x-5}$ I solved for the range as follows:
Setting $f(x)=\frac {2x^2+x-3}{x^2+4x-5} =y$, I rearranged it to get a quadratic in x.
$$(y-2)x^{2}+ (4y-1)x +(3-5y)=0$$
Next, using $\Delta \ge 0$, I got
$$(4y-1)^2-4(y-2)(3-5y) \ge 0$$
Which boiled down to
$$(6y-5)^2 \ge 0$$
And this... | Alternative approach, along the lines of R. J. Mathar in the comments.
For all $x\in\mathbb{R}\setminus\{1,-5\},$
$$ \frac{2x^2 + x -3}{x^2 + 4x - 5} \equiv \frac{2(x^2 + 4x - 5) - 7x + 7}{x^2 + 4x - 5} \equiv 2 - \frac{7x-7}{x^2 + 4x - 5} $$
$$ \equiv 2 - \frac{7(x-1)}{(x+5)(x-1)} \equiv 2 - \frac{7}{(x+5)}. $$
Def... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find the number of solutions to the equation $\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {2-x}}4\right)=\sqrt 2$ The following question appeared in a JEE Mock exam, held two days ago.
Question:
Find the number of solutions to the equation $\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {... | From the comments, the domain is $[0,2]$.
Let $f(x)=\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {2-x}}4\right)-\sqrt 2$
$f'(x)=\frac1{2\sqrt x}\frac\pi4\cos\left(\frac{\pi\sqrt x}4\right)+\frac1{2\sqrt{2-x}}\frac\pi4\sin\left(\frac\pi4\sqrt{2-x}\right)$
Angles of sine and cosine are varying from $0$ to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is $ \frac{x}{\sqrt{x^{2}+1}}=x^{4}-x$ considered impossible to solve? This is from Stewart's Transcendentals.
If we were to try to find the exact intersection points, we
would have to solve the equation $$ \frac{x}{\sqrt{x^{2}+1}}=x^{4}-x$$
This looks like a very difficult equation to solve exactly (in
fact, it's... | Of course it is possible to solve this equation, but answer cannot be written in closed form.
$$\frac{x}{\sqrt{x^{2}+1}}=x^{4}-x,x\in\mathbb{R}$$
$$\frac{x}{\sqrt{x^2+1}}\left((x^3-1)\sqrt{x^2+1}-1\right)=0$$
$$x=0\lor (x^3-1)\sqrt{x^2+1}=1$$
Let find real root other than $x=0$.
$$(x^3-1)\sqrt{x^2+1}=1,x\in\mathbb{R}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the probability of drawing balls until you get 3 matching colors? A bag has $50$ balls - $37$ red, $10$ white, and $3$ blue. I draw balls without replacement until I pull out $3$ of the same color. What is the probability of each color that it is the first to be drawn 3 times?
My attempt: I know I have at least... | You want the probability of, say, the third red being drawn before the third white, or third blue ; when there are 37 red, 10 white, 3 blue.
The probability that the third ball is red and the first two also red is $\left.{\binom{37}1\binom{36}{2}}\middle/{\binom{50}{1}\binom{49}{2}}\right.$
The probability that the fou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the u... | Let $J(a)=\int_0^\infty \frac{\tan^{-1}(ax)\ln x}{1+x^2}$
\begin{align}
J’(a)=& \int_0^\infty \frac{x\ln x}{(1+x^2)(1+a^2x^2)} \overset{x\to \frac1{ax}}{dx}\\
= &
-\frac1{2}\int_0^\infty \frac{x\ln a}{(1+x^2)(1+{a^2}x^2)} {dx}
= \frac{\ln^2 a}{2(1-a^2)}
\end{align}
which leads to
\begin{align}\int_0^\infty \frac{\arc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
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$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive 4 real roots.
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive four real roots.
$$(x-a)(x-b)(x-c)(x-d) \\ +(x-a)(x-b)(x-c)(x-e) \\ +(x-a)(x-b)(x-d)(x-e) \\ +(x-a)(x-c)(x-d)(x-e) \\ + (x-b)(x-c)(x-d)... | I found an answer from the comment.
\begin{align}
& \text{WLOG } a<b<c<d<e. \\
\ \\
& f(a)>0, f(b)<0, f(c)>0, f(d)<0, f(e)>0. \\
\ \\
& \therefore \text{ There exists 4 roots, one each in } [a, b], [b, c], [c, d], [d, e] \qquad \text{by the Intermediate Value Theorem}.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generating irreducible polynomials for binary numbers From a paper, there is a discussion about generating irreducible polynomials for a certain degree as can be seen below.
The reference is wolfram, but it is not clear how those polynomials are generated. For example, 37 and 41 are shown below.
5 4 3 2 1 0
37 = ... | For small degrees you can basically do this by hand with a variant of the sieve of Eratosthenes, starting with all polynomials and then successively eliminating those that are divisible by polynomials of lower degree. For polynomials of degree $6$ it suffices to eliminate polynomials which are divisible by irreducible ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Abel's binomial theorem Abel's generalization of the binomial theorem goes as follows
$$(a+x)^n=x^n+ \sum_{i=1}^n \binom{n}{i} a(a+iz)^{i-1}(x-iz)^{n-i}$$
The objective is to prove the generalization by induction. My first step following the textbook was to set $n=k$ and integrate both sides with respect to $x$, in ord... | Here we complete OPs proof by induction. Note, we will employ the induction hypothesis twice. Assuming $a\ne 0$ we can write OPs version of Abel's binomial theorem somewhat more convenient as:
\begin{align*}
(a+x)^k&=\sum_{q=0}^k\binom{k}{q}a(a+qz)^{q-1}(x-qz)^{k-q}\tag{1.1}
\end{align*}
The induction step we want to s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $a^{4/a} + b^{4/b} + c^{4/c} \ge 3$
Let $a, b, c > 0$ with $a + b + c = 3$. Prove that
$$a^{4/a} + b^{4/b} + c^{4/c} \ge 3.$$
This question was posted recently, closed and then deleted, due to missing of contexts etc.
By https://approach0.xyz/, the problem was proposed by Grotex@AoPS.
My strategy is to spl... | Before the actual proof, let's start with a discussion of $f(x) = x^{4/x}$ which is needed later.
We have
$ df(x) /dx = 4 x^{4/x - 2} (1 - \log(x))$ so $f(x)$ is rising for all $0 < x < e$ which is all we need to consider below. Further,
$$ d^2f(x) /dx^2 = 4 x^{4/x - 4} (-3 x + 4 \log^2(x) + 2 (x - 4) \log(x) + 4)$$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 0
} |
Finding the maximum value of $f(x)=\frac{|x|-2-x^2}{|x|+1},x\in\mathrm R$ Question:
Find the maximum value of $f(x)=\frac{|x|-2-x^2}{|x|+1},x\in\mathrm R$
My Attempt:
I took two cases. One where $x\ge0$, so, $|x|=x$ and another where $x\le0$, so, $|x|=-x$. Then I cross multiplied, made a quadratic in $x$ and wrote disc... | Perhaps surprisingly, $ \ f(x) \ = \ \large{\frac{|x| \ - \ 2 \ - \ x^2}{|x| \ + \ 1} } \ $ is an even function(!). If we consider $ \ f_{+}(x) \ = \ \large{\frac{ x \ - \ 2 \ - \ x^2}{ x \ + \ 1} } $ $ = \ (-x + 2) - \frac{4}{x + 1} \ \ , \ $ the corresponding curve has a vertical asymptote of $ \ x \ = \ -1 \ $ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Summation of reciprocal products When studying summation of reciprocal products I found some interesting patterns.
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)}=\frac{1}{1\cdot1!}-\frac{1}{1\cdot(N+1)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)}=\frac{1}{2\cdot2!}-\frac{1}{2\cdot(N+1)(N+2)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdo... | This identity is a matter of telescoping. Let $q$ be a positive integer.
We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^N}&\color{blue}{\frac{1}{k(k+1)\cdots(k+q)}}\\
&=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q)}\left((k+q)-k\right)\tag{1.1}\\
&=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q-1)}
-\frac{1}{q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Series expansion of $(1-cx)^{1/x}$ I am trying to understand the series expansion of $$(1-cx)^{1/x}$$ The wolframalpha seems to solve the problem by using taylor series for $ x\rightarrow 0$ and Puiseux series for $ x\rightarrow \infty$. Any ideas how can I calculate them ?
https://www.wolframalpha.com/input?i=%281-cx%... | As already said in comments
$$y=(1-c x)^{\frac{1}{x}}\quad \implies \quad \log(y)=\frac{1}{x}\log(1-cx)$$ For small values of $cx$
$$\log(y)=-\sum_{n=1}^\infty \frac {c^n} n\,x^{n-1}$$ Now, using $y=e^{\log(y)}$
$$\color{blue}{y=1 +c\,e^{-c}\sum_{n=1}^\infty(-1)^n \frac{ c^{n}}{\alpha_n}\,P_n(c)\,x^n}$$ The $\alpha_n$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How many methods to tackle the integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x ?$ $ \text{We are going to evaluate the integral}$
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \tag*{} \\$
by letting $ y=\frac{\pi}{4}-x. $ Then
$$\begin{aligned} \displaystyle ... | This is similar to your original solution but a bit quicker using sum-to-product formula:$$\begin{aligned} \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \displaystyle &=\sqrt2\int_{0}^{\frac{\pi}{4}} \frac{\sin \left(\frac{\pi}{4}+x\right)}{9+16 \sin 2x}d x \\
\displaystyle &=\sqrt2\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Upper triangular determinants Given that
$$A=
\begin{bmatrix}
1&211&311&327&337\\
0&2&427&438&491\\
0&0&3&547&551\\
0&0&0&2&672\\
0&0&0&0&3
\end{bmatrix}
$$
Find $|2A^{-1}+I_5|$. The solution is $\frac{100}{3}$ but I can't figure out how to do it. Any hints? My best guess is to use properties of the determinants of upp... | Hints:
*
*Eigenvalues of an upper triangular matrix are precisely the entries of the main diagonal.
*If $A$ is invertible then $\lambda$ is an eigenvalue of $A$ iff $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$
*Eigenvalues are polynomial invariant i.e $\lambda$ is an eigenvalue of $A$ implies $p(\lambda) $ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding all possible complex solutions to the system $xy = \sqrt 2(x + y)$, $yz = \sqrt 3(y + z)$, $zx = \sqrt 5(z + x)$
Find all possible complex solutions to:
$$\begin{cases}xy = \sqrt 2(x + y) \\
yz = \sqrt 3(y + z)\\
zx = \sqrt 5(z + x)\end{cases}$$
I have some ideas:
$$xy-\sqrt 2y=\sqrt 2x\implies y=\frac{\sqrt ... | The problem is not so bad.
From the first equation $y=\frac{\sqrt{2} x}{x-\sqrt{2}}$; from the third $z=\frac{\sqrt{5} x}{x-\sqrt{5}}$
Plug in the second
$$\frac{x \left(\left(\sqrt{6}-\sqrt{10}+\sqrt{15}\right) x-2
\sqrt{30}\right)}{\left(x-\sqrt{2}\right) \left(x-\sqrt{5}\right)}=0$$ So, if $x\neq 0$, then
$$x=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\vec a,\vec b,\vec c$ be three vectors such that $|\vec a|=1,|\vec b|=2,|\vec c|=4$ and then find the value of $|2\vec a+3\vec b+4\vec c|$ If $\vec a,\vec b,\vec c$ be three vectors such that
$\vert \vec a\vert =1,\vert \vec b\vert =2,\vert \vec c\vert=4$
and
$\vec a \cdot \vec b+\vec b \cdot \vec c+\vec c \cdot\ve... | Recall that $\|\vec x\|^2 = \vec x \cdot \vec x$. Then
$$\begin{align*}
\|2\vec a + 3\vec b + 4\vec c\|^2 &= (2\vec a + 3\vec b + 4\vec c) \cdot (2\vec a + 3\vec b + 4\vec c) \\
&= 4\|\vec a\|^2 + 9\|\vec b\|^2 + 16\|\vec c\|^2 + 2 \left(6\vec a\cdot\vec b + 8\vec a\cdot\vec c + 12\vec b\cdot \vec c\right)\\
&= (4\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Implicit Differentiation - $x^my^n = (x+y)^{m+n}$
Use implicit differentiation to find $\frac{\mbox{dy}}{\mbox{dx}}$ if $$x^my^n = (x+y)^{m+n}$$
Differentiation both sides with respect to $x$:
$$mx^{m-1}y^n + x^mny^{n-1}y' =(m+n)(x+y)^{m+n-1}(1 + y')$$
$$y' = \frac{(m+n)(x+y)^{m+n-1} - mx^{m-1}y^n}{x^mny^{n-1}-(m+n)(... | Simpler way: Take logarithms (base e unless specified). $$m\log x+n\log y=(m+n)\log (x+y)$$ Now differentiate: $$\frac mx +\frac ny\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)$$ Therefore $$\frac mx-\frac{m+n}{x+y}=\left(\frac{m+n}{x+y}-\frac ny\right)\frac{dy}{dx}$$$$\implies \frac{my-nx}{x(x+y)}= \frac{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
S... | Trick:
$$
\sqrt {3-\sqrt{5}}=\frac{\sqrt {6-2\sqrt{5}}}{\sqrt 2}=\frac{\sqrt {(\sqrt 5)^2-2\sqrt{5}+1}}{\sqrt 2}=\frac{\sqrt {(\sqrt 5-1)^2}}{\sqrt 2}
=\frac{\sqrt 5-1}{\sqrt 2}.
$$
Similarly we see $\sqrt {3+\sqrt{5}}
=\frac{\sqrt 5+1}{\sqrt 2}$. Thus the sum is $2\cdot \frac{\sqrt 5}{\sqrt 2}=\sqrt {10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 1
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Proving $30x + 3y^2 + \frac{2z^3}{9} + 36 (\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}) \ge 84$
If $x, y, z$ are positive real numbers, prove that $$30x + 3y^2 + \frac{2z^3}{9} + 36 \left(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}\right) \ge 84.$$
I genuinely have no clue on how to proceed. Is it proved using repeated... | Partial answer:
If we want to find the minimum, I think we can create a system from the implicit derivatives. We can multiply through by $xyz$, then subtract the right side to get:
$$f(x,y,z) = 30x^2yz+3xy^3z+\textstyle{\frac29}xyz^4+36x+36y+36z-84xyz \geqslant 0$$
And we want to show that this function has a minimum o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\frac{x^2+y^2+x+y-1}{xy-1}$ is an integer for positive integers $x$ and $y$, then its value is $7$. I saw this on quora
and haven't been able to solve it.
If $\dfrac{x^2+y^2+x+y-1}{xy-1}$
is an integer for positive integers
$x$ and $y$,
then its value is $7$.
If
$y=1$ this is
$\dfrac{x^2+x+1}{x-1}
= x+2+\dfrac{3}{x... | If $f(x,y)=\dfrac{x^2+y^2+x+y-1}{xy-1}$ when integer must be equal to $7$ then
$$g(x,y)=\frac{x^2+y^2+x+y-6xy+5}{xy-1}$$ must be equal to $1$ because $f(x,y)=6+g(x,y)$ so we can solve the equation
$$x^2+y^2+x+y-7xy+6=0$$ which admits several infinite sets of solutions $(x,y)$ like to them of Pell-Fermat equations (look... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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If the matrix A is given, then find the value of $A^{2022}$ If it is given that matrix $ A = \begin{bmatrix}
\dfrac{5}{2} & \dfrac{3}{2}\\
-\dfrac{3}{2} & -\dfrac{1}{2}
\end{bmatrix}$ then find the value of $A^{2022}$.
Here is my try on it : $$\\$$
If we look at the pattern while multliplying A with itself it ... | The matrix has determinant 1 and trace 2, so it is parabolic, so a conjugate of an upper triangular matrix (with diagonal elements equal to 1). Computing powers of parabolic matrices is fun and easy (though your solution is essentially that).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).... | We will use a few identities.
First is the Pythagorean Identity:
$$\sin^2\alpha +\cos^2\alpha = 1$$
Second is a simple factorization:
$$a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = (a-b)(a^4 + b^4 + ab(a^2+b^2+ab))$$
Third will help us write things of fourth degree in terms of second and first degree terms:
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
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Prove that: $\sum_{\text {sym}} a^3+3\sum_{\text{sym}} a^2b\ge 18$ Let $a,b,c$ be the sides of triangle, such that $abc=1$, then prove that:
$$\sum_{\text{sym}} a^3+3\sum_{\text{sym}} a^2b\ge 18.$$
Find at least one case where equality is possible.
My attempts:
I tried the well-known formula with triangle inequalitie... | $$3(a^2b+c^2b+b^2c+a^2c+c^2a+b^2a)=$$ $$=3abc((a/c+c/a)+(b/a+a/b)+(c/b+b/c))\ge$$ $$\ge 3(6)=18$$ because $abc=1$ and because if $x\in\{a/c,b/a,c/b\}$ then $x+1/x=(\sqrt x\,-1/\sqrt x)^2+2\ge 2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac{3}{b+c+d}+ \frac{3}{c+d+a}+\frac{3}{d+a+b}+\frac{3}{a+b+c} \ge \frac{16}{a+b+c+d}$ for $a,b,c,d>0$ Prove that
$$\frac{3}{b+c+d}+ \frac{3}{c+d+a}+\frac{3}{d+a+b}+\frac{3}{a+b+c}
\ge \frac{16}{a+b+c+d}$$
if $a,b,c,d>0$.
My attempt:
I've put this in the answers.
See also comments under this question.
| My attempt:
Let p=a+b+c+d then
$$
\frac{1}{b+c+d}+ \frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\equiv\\{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d}
$$
By the AM-GM inequality,
$$
{1\over p-a}+{1\over p-b}+{1\over p-c}+{1\over p-d} \geq4\sqrt[4]{1\over(p-a)(p-b)(p-c)(p-c)}.
$$
Also by AM-GM inequality,
$$
\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4529609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Analytic guess for integral $\int_0^\infty \frac{\ln^2(\sqrt{x^2+1}+x) }{\sqrt{x^2 + 1}} \ln \frac{\cosh w+\sqrt{x^2+1} }{1+\sqrt{x^2+1} } dx$ (I was directed here from mathoverflow.)
I am not able to do the following integral analytically, but after numerical evaluations was able to guess the result for all real $w$ v... | Substitute $x=\sinh t$ to reduce the integral
\begin{align}
I(w)=&\int_0^\infty \frac{\ln^2(\sqrt{x^2+1}+x) }{\sqrt{x^2 + 1}} \ln \frac{\cosh w+\sqrt{x^2+1} }{1+\sqrt{x^2+1} } dx\\
= &\int_0^\infty t^2 \ln \frac{\cosh w+\cosh t}{1+\cosh t}dt
\end{align}
and evaluate
\begin{align}
I’(w) =& \int_0^\infty \frac{t^2 \sinh ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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If $A^2=A,\,B^2=B,\,C^2=C$ and $A+B+C=I,$ prove $AB=BC=AC=\mathbb{O}$.
Let $A,\,B,\,C$ be square matrices such that $A^2=A,\,B^2=B,\,C^2=C$ and $A+B+C=I.$
Prove that $AB=BC=CA=\mathbb{O}.$
Attempt. We have $A(B+C)=A(I-A)=A-A^2=\mathbb{O}$ and similarly we get
$B(A+C)=C(A+B)=\mathbb{O}$, but this is where i get so far... | Roland's answer gives a great and simple linear algebra argument.
Here is a purely algebraic development:
\begin{align*}
(A+B+C)^2 & = A^2 + B^2 + C^2 + AB + AC + BC + BA + CA + CB \\
& = I + A(B+C) + BC + (B+C)A + CB \\
& = I + A(I - A) + (I-A)A + BC + CB \\
& = I + A - A^2 + A - A^2 + BC + CB \\
& = I + BC + CB \\
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
How to solve this logarithm problem If $x$ and $y$ are positive real numbers and $x^{\log _yx}=2$ And $y^{\log _xy}=16$ Find $\log _yx$ and find $x$.
The way I tried to solve it was to take $\log _x$ on the base $x$ one and $\log _y$ for the $y$ one. Then something really messy came up. I got that $\log _yx=\sqrt[3]{\d... | Note the following:
$$x^{\log_y x} = 2 \implies x^{\frac{\log_2 x}{\log_2 y}} = 2$$
$$\implies 2^{\frac{\log^2_2 x}{\log_2 y}}=2 \implies \frac{\log^2 x}{\log_2 y} = \log_2 2 = 1,$$ or in particular,
$$\frac{\log^2_2 x}{\log_2 y} = 1.$$
Similarly,
$$y^{\log_x y} = 16 \implies \frac{\log^2_2 y}{\log_2 x} = \log_2 16 = 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Reducing $ax^6-x^5+x^4+x^3-2x^2+1=0$ to a cubic equation using algebraic substitutions
Use algebraic substitutions and reduce the sextic equation to the cubic equation, where $a$ is a real number:
$$ax^6-x^5+x^4+x^3-2x^2+1=0$$
My attempts.
First, I tried to use the Rational root theorem, when $a$ is an integer $x=\pm... | Define $t,u=\frac{1\pm\sqrt{1-4a}}{2a}$ as the roots of $ax^2-x+1$. Then the sextic splits into two cubics over $\mathbb Q(t)$:
$$ax^6-x^5+x^4+x^3-2x^2+1=a(x^3-tx^2+t)(x^3-ux^2+u)$$
This can be verified by re-expanding. Here the substitution is $a=\frac{t-1}{t^2}$ and not in $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.