Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove $\Sigma_{cyc}(\frac{a}{b-c}-3)^4\ge193$ The inequality is expected original question of this MSE question. The exact statement is "If $a$, $b$ and $c$ are positive real numbers and none of them are equal pairwise, prove the following inequality."
$$\Sigma_{cyc}\left(\frac{a}{b-c}-3\right)^4\ge193$$
Full expanding... | It remains to make two steps only.
*
*For $$\frac{a}{b-c}=\frac{b}{c-a}$$ or
$$c=\frac{a^2+b^2}{a+b}$$ it's enough to prove that
$$2\left(\frac{a}{b-\frac{a^2+b^2}{a+b}}-3\right)^2+\left(\frac{a^2+b^2}{a^2-b^2}-3\right)^4\geq193.$$
Now, let $a=tb$.
Thus, we need to prove that
$$2\left(\frac{t}{b-\frac{t^2+1}{t+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Help on solutions of the congruence $f(x)=x^3+4x+8 \equiv 0 \pmod {15}$ I'm doing a little exercise, solve the congruence $f(x)=x^3+4x+8 \equiv 0 \pmod {15}$.
I know that $15=3 \times 5$ and they are relatively prime, so I can split the congruence into:
a) $f(x) \equiv 0 \pmod {3}$
b) $f(x) \equiv 0 \pmod {5}$
I procee... | As other answrers note, there are no solutions because the equation fails $\bmod 5$.
Here an alternate method for solving cubic equations $\bmod 5$ is explored. Suppose the equation has the form
$ax^2+bx^2+cx+d\equiv 0\bmod 5$ Eq. 1
If $d\equiv 0$ then $x\equiv 0$ is a root and the quadratic equation $ax^2+bx+c=0$ ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3078065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How do I solve for $a$ and $b$ in $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \d... | Hint: you have to expand $\log (1+y)$ up to the term in $y^{2}$ to answer this question. $\log (1+y)=y-\frac {y^{2}} 2+0(y^{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluating $\lim_{x\to0^+}\frac{2x(\sin x)^2+\frac{2x^7+x^8}{3x^2+x^4}-\arctan(2x^3)}{\ln(\frac{1+x^2}{1-x^2})-2x^2+xe^{-{1\over x}}}$ To start with, the term $xe^{-{1\over x}}$ can be ignored.Then splitting the term $\ln(\frac{1+x^2}{1-x^2})=\ln(1+x^2)-\ln(1-x^2)$ and expanding both of them up to order 3 we have in th... | The Taylor series approach suggested by @Matteo and @WillJagy is the way to go. Here is that approach made a bit more explicit:
*
*Divide the numerator and denominator by x^6.
*Construct the resulting first order Taylor series for both numerator and denominator.
*Observe what the limits must be.
$$\lim_{x\to 0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that for all integers $n$ if $3 \mid n^2$, then $3 \mid n$ Prove that for all integers $n$ if $3$ | $n^2$, then $3$ | $n$.
I figured using contrapostive was the best method by using the definition "an integer $k$ is not divisible by 3 if and only if there exists an integer $k$ such that $n=3k+1$ or $n=3k+2$. Also... | Presumably you have had the division theorem.
For $n$ and integer there exist integer $k, r$ so that $n = 3k + r$ where $0 \le r < 3$. So $r = 0, 1$ or $2$.
Can you accept that?
If $r=0$ then $3|n$.
If $r = 1$ then $n^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$ and $3\not \mid n^2$ and that's a contradiction.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I ... | If $a=-b$ so $c=-d$ and we are done.
Let $a+b\neq0.$
Thus, $$(a+b)^3-3ab(a+b)=(c+d)^3-3cd(c+d)$$ or
$$ab(a+b)=cd(c+d)$$ or $$ab=cd$$ and the rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
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Find the numerical value of this expression If $x$ is a complex number such that $x^2+x+1=0$, then the numerical value of $(x+\frac{1}{x})^2+(x^2+\frac{1}{x^2})^2+(x^3+\frac{1}{x^3})^2+\ldots+(x^{27}+\frac{1}{x^{27}})^{2}$ is equal to?
A) 52 . B) 56 . C) 54. D)58 . E)None of these
Where is this question from? I... | Multiply both sides by $x-1\ne 0$:
$$x^2+x+1=0 \Rightarrow (x-1)(x^2+x+1)=0 \Rightarrow x^3-1=0 \Rightarrow x^3=1\Rightarrow \\
x^{54}=1;x^{-54}=1 \quad (1)$$
Expand:
$$(x+\frac{1}{x})^2+(x^2+\frac{1}{x^2})^2+(x^3+\frac{1}{x^3})^2+\ldots+(x^{27}+\frac{1}{x^{27}})^{2}=\\
[x^2+x^4+\cdots+x^{54}]+[x^{-2}+x^{-4}+\cdots+x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\int\limits_0^\infty {x^4 \over (x^4-x^2+1)^4}\ dx$ I want to calculate
$$\int\limits_0^\infty \frac{x^4}{(x^4-x^2+1)^4}dx$$
I have searched with keywords "\frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results
| Here is another way to get to the same point as what @Sangchul Lee gives.
Let
$$I = \int_0^\infty \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx.$$
Then
$$I = \int_0^1 \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx + \int_1^\infty \frac{x^4}{(x^4 - x^2 + 1)^4} \, dx.$$
Enforcing a substitution of $x \mapsto 1/x$ in the right most integral l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is it possible to reach the initial arrangement?
We have a stack of $n$ books piled on each other, and labeled by $1, 2, ..., n$. In each round we make $n$ moves in the following manner: In the $i$-th move of each turn, we turn over the $i$ books at the top, as a single book. After each round we start a new round simi... | For clear reference, here is a complete cycle of moves. Negative number represents book face down.
1 2 3 4
-1 2 3 4
-2 1 3 4
-3 -1 2 4
-4 -2 1 3
4 -2 1 3
2 -4 1 3
-1 4 -2 3
-3 2 -4 1
3 2 -4 1
-2 -3 -4 1
4 3 2 1
-1 -2 -3 -4
1 -2 -3 -4
2 -1 -3 -4
3 1 -2 -4
4 2 -1 -3
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3083542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Integral of $\int{\frac{1}{\sqrt{x(1+x^2)}}dx}$ I was trying to solve the following question:
Evaluate: $$\int{\frac{1}{\sqrt{x(1+x^2)}}dx}$$
This is an unsolved question in my sample papers book and so I believe it should have an elementary primitive.
But, I don't know how to start this. I want a hint to get started... | We can also express the solution in terms of the Beta Function and the Incomplete Beta Function as I cover here:
\begin{align}
I&= \int \frac{1}{\sqrt{x\left(1 + x^2\right)}}\:dx = \int_0^x \frac{t^{-\frac{1}{2}}}{\left(t^2 + 1 \right)^{\frac{1}{2}}} \:dt\\
&=\frac{1}{2} \left[ B\left(\frac{1}{2} - \frac{-\frac{1}{2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $n$ such that polynomial is divisible Find $n \in N$, such that:
$$(x^2+x+1)^2 | x^n+(x+1)^n+1 = P(x)$$
If I let $Q(x) = x^2 + x + 1$, I have that $x^3 \equiv 1$ mod$Q(x)$, so I can work out the cases for remainder of $n$ when divided by $3$ ..
But what to do next, because I need $Q^2(x)$ dividing $P(x)$ ?
I suppo... | Since $(x^2+x+1)^2$ divides $P(x):=x^n+(x+1)^n+1$, we get that $Q(x):=x^2+x+1$ divides its derivative $P'(x)=n(x^{n-1}+(x+1)^{n-1})$.
Let us see when $n(x^{n-1}+(x+1)^{n-1})$ is divisible by $Q$ by working modulo $Q(x)$. Since $x^2+x+1=0$ we get $x^2=-(x+1)$ and
$$x^3=x^2x=-(x+1)x=-x^2-x=(x+1)-x=1,$$
hence
$$x^{3k}=1, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to solve $\lim \left(\frac{n^3+n+4}{n^3+2n^2}\right)^{n^2}$ I can't seem to find a way to solve:
$$\lim \left(\dfrac{n^3+n+4}{n^3+2n^2}\right)^{n^2}$$
I've tried applying an exponential and logaritmic to take the $n^2$ out of the exponent, I've tried dividing the expression, but I don't get anywhere that brings lig... | $$
\begin{align*}
L &= \lim_{n\to\infty}\left(\frac{n^3+n+4}{n^3+2n^2}\right)^{n^2} \\
&= \lim_{n\to\infty}\left(\frac{n^3 +2n^2 - 2n^2+n+4}{n^3+2n^2}\right)^{n^2}\tag1 \\
&= \lim_{n\to\infty}\left(1 + \frac{- 2n^2+n+4}{n^3+2n^2}\right)^{n^2} \tag2 \\
&= \lim_{n\to\infty}\left(1 + \frac{- 2n^2+n+4}{n^3+2n^2}\right)^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$ $$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$$
We know that : $\arctan{x} - \arctan{y} = \arctan{\frac{x-y}{1+xy}}$ for every $ xy > 1 $
I need to find two numbers which satisfy: $ab = n^2+2n+3 $ and $ a-b =2$ in order to telescope.
Edit: I am very sorry ... | $$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}} = \sum\limits_{n=1}^{\infty}\arctan{\frac{2(n+1)-2n}{n(n+1)(1+\frac{4}{n(n+1)})}} =
\sum\limits_{n=1}^{\infty}\arctan{\frac{\frac{2}{n}-\frac{2}{n+1}}{1+\frac{2}
{n}\cdot\frac{2}{n+1}}} =\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n}-\arctan{\frac{2}{n+1}}} =\arctan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to prove that $\csc^2x=\sum_{-\infty}^{\infty}\frac{1}{(x-n\pi)^2}$ I was reading the book The Princeton Companion to Mathematics
On page $293$, there is a statement
$$\csc^2x=\sum_{n=-\infty}^{\infty}\frac{1}{(x-n\pi)^2}\tag{1}$$
Here is one method from the book:
Observe first that
$$h(x)= \sum_{n=-\infty}^{\i... | METHODOLOGY $1$: FOURIER SERIES
We begin by writing the Fourier series,
$$\cos(xy)=a_0/2+\sum_{n=1}^\infty a_n\cos(nx) \tag1$$
for $x\in [-\pi/\pi]$. The Fourier coefficients are given by
$$\begin{align}
a_n&=\frac{2}{\pi}\int_0^\pi \cos(xy)\cos(nx)\,dx\\\\
&=\frac1\pi (-1)^n \sin(\pi y)\left(\frac{1}{y +n}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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The greatest value of $|z|$ if $\Big|z+\frac{1}{z}\Big|=3$ where $z\in\mathbb{C}$
$\bigg|z+\dfrac{1}{z}\bigg|=3$ then the greatest value of $|z|$ is ___________
My Attempt
$$
\bigg|z+\frac{1}{z}\bigg|=\bigg|\dfrac{z^2+1}{z}\bigg|=\frac{|z^2+1|}{|z|}=3\\
\bigg|z+\frac{1}{z}\bigg|=3\leq|z|+\frac{1}{|z|}\implies |z|^2-3... | The Joukowsky transform $z\to z+1/z$ transforms circles of radius $r>1$ into ellipses of semi-axes $r-1/r$ and $r+1/r$. The max value of $|z|$ corresponds to the radius of the circle such that the smallest semi-axis $r-1/r=3$. Solving this equation $r^2-3r-1=0$ for $r>0$ you get the result $r=(3+\sqrt{13})/2$ which agr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3091795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How do I prove this relationship between positive terms of a G.P.?
$a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:
$$\frac1{ab} + \frac1{cd} > 2 \left(\frac1{bd} + \frac1{ac} - \frac1{ad}\right)$$
This question was listed under the section on Arithmetic, Geometric, an... | So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$\frac1{a^2x} + \frac1{a^2x^5} > 2 (\frac1{a^2x^4} + \frac1{a^2x^2} - \frac1{a^2x^3})$$
or $$1 + \frac1{x^4} > 2 (\frac1{x^3} + \frac1{x} - \frac1{x^2})$$
or $$\boxed{x^4+1>2(x+x^3-x^2)}$$
or
$$x^4-2x^3+2x^2-2x+1>0\iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ Find the remainder of the division of $x^n+5$ with $x^3+10x^2+25x$ over $\mathbb{Q}$
What I tried to do is to write $x^n+5=p(x)(x^3+10x^2+25x)+Ax^2+Bx+C$, where $p(x)$ is a polynomial of degree $n-3$. If I set $x=0$ I obtain that $C=5$.
Now $x^3+10x^2+... | Hint $\ $ Subtracting $\, C = 5\,$ then cancelling $x$ yields
$\qquad f(x) := x^{\large n-1} = B + Ax + (x+5)^{\large 2} p(x) $
Thus $\, B + Ax = f(-5) + f'(-5)(x+5)\,$ by Taylor expansion at $\,x = -5\,$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I simplify this fraction problem? I have the problem $\frac{x^2}{x^2-4} - \frac{x+1}{x+2}$ which should simplify to $\frac{1}{x-2}$
I have simplified $x^2-4$, which becomes:
$\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}$
However, if I combine the fractions I get, $x^2-x-1$ for the numerator, which can't be factored... | Write $$\frac{x^2}{(x-2)(x+2)}-\frac{x+1}{x+2}=\frac{x^2}{(x-2)(x+2)}-\frac{(x+1)(x-2)}{(x+2)(x-2)}=…$$
Note that it must be $$x\ne 2,-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?
For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13
if x = 8, y = 325 which is divisible by 1... | Hint $\bmod 13\!:\,\ 0\equiv -3(4x^2\!+\!8x+5)\equiv x^2\! +\!2x\!-\!15\equiv (x\!+\!5)(x\!-\!3)\ $ so $\,x\equiv -5,3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3095147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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How to prove $\frac{(x+y)(y+z)(z+x)}{4xyz}≥\frac{x+z}{y+z}+\frac{y+z}{x+z}$ for $x,y,z>0$?
Prove that for $x,y,z$ positive numbers:
$$
\frac{(x+y)(y+z)(z+x)}{4xyz}≥\frac{x+z}{y+z}+\frac{y+z}{x+z}
$$
I tried to apply MA-MG inequality:
$x+y≥2\sqrt{xy}$ and the others and multiply them but it becomes
$\frac{(x+y)(y+z)... | We need to prove that
$$\frac{(x+y)(x+z)(y+z)}{4xyz}-2\geq\frac{x+z}{y+z}+\frac{y+z}{x+z}-2$$ or
$$\sum_{cyc}\frac{(x-y)^2}{xy}\geq\frac{4(x-y)^2}{(y+z)(x+z)},$$ which is true by C-S:
$$\sum_{cyc}\frac{(x-y)^2}{xy}\geq\frac{(x-y+x-z+z-y)^2}{xy+xz+yz}=$$
$$=\frac{4(x-y)^2}{xy+xz+yz}\geq\frac{4(x-y)^2}{z^2+xy+xz+yz}=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Calculate limit with squares $\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4$
$$\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4 $$
What I did was to multiply it and I got $\frac{1}{2}$ as the final result. Could someone confirm i... | The hint:
Since $$a-b=\frac{a^3-b^3}{a^2+ab+b^2},$$ we obtain:
$$\left(\sqrt[3]{n^3+2n-1}-\sqrt{n^3+2n-3}\right)^6(1+3n+2n^3)^4=$$
$$=\left(\tfrac{n^3+2n-1-(n^3+2n-3)}{\sqrt[3]{(n^3+2n-1)^2}+\sqrt[3]{(n^3+2n-1)(n^3+2n-3)}+\sqrt[3]{n^3+2n-3)^2}}\right)^6(1+3n+2n^3)^4=$$
$$=\frac{64(1+3n+2n^3)^4}{\left(\sqrt[3]{(n^3+2n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem... | If $x^2 - x -2$ is indeed a factor of $x^4 - 2x^3 + x -2$, that means I can write:
$$ x^4 - 2x^3 + x - 2 = P(x)(x^2 -x - 2)$$
In order to reproduce the $x^4$ term, that means that the leading term of $P(x)$ is $x^2$. Now
$$ x^2(x^2 -x -2) = x^4 -x^3 -2x^2 $$
which is not the same as what we started with. Let's try to s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
How to solve this recurrence relation using generating functions: $a_n = 10 a_{n-1}-25 a_{n-2} + 5^n\binom{n+2}2$? How can we solve the following recurrence relation using GF?
$a_n = 10 a_{n-1}-25 a_{n-2} + 5^n {n+2 \choose 2}$ , for each $n>2, a_0 = 1, a_1 = 15$
I think that most of it is pretty straightforward. What ... | Hint for finding the GF.
Note that $(n+1)(n+2)x^n=\frac{d^2}{dx^2}\left(x^{n+2}\right)$
and therefore
$$\sum_{n=2}^\infty {n+2 \choose 2}5^nx^n = \frac{1}{2\cdot 5^2}\frac{d^2}{dx^2}\left(
\sum_{n=2}^\infty (5x)^{n+2}
\right).$$
Then recall that $\sum_{n=0}^{\infty}z^n=\frac{1}{1-z}$.
Moreover in your attempt it shou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Minimum point of $x^2+y^2$ given that $x+y=10$ How do you I approach the following question:
Find the smallest possible value of $x^2 + y^2$ given that $x + y = 10$.
I can use my common sense and deduce that the minimum value is $5^2 + 5^2 = 50$. But how do you approach this mathematically?
Thanks in advance.
|
To minimize $f(x,y) = x^2+y^2$; given that $y=10-x$.
Then:
$$f(x,10-x)=x^2+(x-10)^2$$
$f$ is minimized for $f'=0$, so:
$$f'=2x+2(x-10)=0$$
Gives:
$$x=5\land y=5$$
Thus we have:
$$\min (x^2+y^2)=5^2+5^2=50$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3108814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 9,
"answer_id": 7
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Studying the convergence of the series $\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$ Studying the convergence of the series $$\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$$
I saw this problem and I tried to do it my own way but I don't know what I'm doing wrong because I'm getting a d... | I think what is meant is that the sequence $a_0 = \sqrt{2}, a_{n+1} = \sqrt{2 + a_n}$ converges. This can be shown by proving that for all $n \in \mathbb{N}$ we have $a_n \leq 2$ by induction. It is clear that the sequence is monotonically increasing. Since the sequence is monotonically increasing and has upper bound $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proving inequality involving radical I have tried many examples for the following inequality in Mathematica. It is likely true. I need some help proving it.
For $x_1, x_2, y_1, y_2 \geq 0$,
\begin{align*}
6(x_1^5y_1 + x_1y_1^5) + 4(x_2^3y_2 + x_2y_2^3) \leq 6(x_1^6 + x_2^4)^{\frac{5}{6}}(y_1^6 + y_2^4)^{\frac{1}{6}}... | Note by Hölder's inequality
$$(x_1^6+x_2^4)^{5/6} (y_1^6+y_2^4)^{1/6} \geqslant x_1^5y_1+x_2^{10/3}y_2^{2/3}$$
and
$$(x_1^6+x_2^4)^{1/6} (y_1^6+y_2^4)^{5/6} \geqslant x_1y_1^5+x_2^{2/3}y_2^{10/3}$$
Using this, it is left to show
$$3(x_2^{10/3}y_2^{2/3}+x_2^{2/3}y_2^{10/3}) \geqslant 2(x_2^3y_2 + x_2y_2^3)$$
In fact as ... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finite double sum $\sum_{k=0}^N\sum_{l=0}^M\left\lfloor\frac{k+l}{c}\right\rfloor$; any advanced summation technique?
Let $M,N,c$ be positive integer. It was astonishing when trying to solve $\sum_{k=0}^N\sum_{l=0}^M\left\lfloor\frac{k+l}{c}\right\rfloor$ to obtain this rather complex looking result
\begin{align*}
... | Let $$S_n=\sum_{k=0}^{n-1}\left\lfloor \frac kc\right\rfloor$$
and
$$ T_n=\sum_{k=0}^{n-1}S_k.$$
Then we are looking for
$$ \begin{align}\sum_{k=0}^N\sum_{l=0}^M\left\lfloor\frac{k+l}{c}\right\rfloor
&=\sum_{k=0}^N(S_{M+k+1}-S_k)\\
&=T_{N+M+2}-T_{M+1}-T_{N+1}\end{align}$$
Note that for $n=qc+r$ with $0\le r<c$, we hav... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving that $\lim_{v\rightarrow \infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} +\cdots + \frac{v^2}{v^3+v}\right]= 1 $ I wonder if my solution that $\lim_{v\to\infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + \cdots + \frac{v^2}{v^3+v}\right]= 1 $ is correct.
$$\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} ... | I think that its good idea to use the sandwich theorem.
Let $a_v $ the sequence that you posted. Then
$$a_v \leq \frac{v^2}{v^3 + v} + \cdots + \frac{v^2}{v^3 +v} = b_v $$
And
$$b_v = v\left(\frac{v^2}{v^3+v}\right) \to 1$$
On the other hand you must use
$$c_v = \frac{v^2}{v^3 +1} + \cdots + \frac{v^2}{v^3 +1} \leq a_v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114947",
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Prove that $a^2+b^2+c^2\geqslant\frac{1}{3}$ given that $a\gt0, b\gt0, c\gt0$ and $a+b+c=1$, using existing AM GM inequality Using the AM and GM inequality, given that
$a\gt0, b\gt0, c\gt0$ and $a+b+c=1$ prove that
$$a^2+b^2+c^2\geqslant\frac{1}{3}$$
| HINT: You can use your idea of squaring $a+b+c$, but also note that $\color{blue}{ab+bc+ca \le a^2 + b^2 + c^2}$, which you can prove with the help of AM-GM. (Hint for proving this: the AM-GM inequality tells us what about $a^2 + b^2, b^2+c^2$ and $c^2+a^2$?)
One more hint (based on a suggestion from user qsmy): let $x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate this question based on series and limits.
For $a \in \mathbb R,a≠-1$
$$\lim_{n\to\infty}\frac{1^a+2^a+\cdots +n^a}{(n+1)^{a-1}[(na+1)+(na+2)+(na+3)+\cdots+(na+n)]}=\frac{1}{60}$$
Then find the values of $a $.
I tried to solve this problem using approximation but I got the the value of $a$ as $\dfrac{-3}{2}... | For $a>-1$ and $a\neq -\frac 12$ we have
\begin{align}
\frac{1^a+2^a+\cdots +n^a}{(n+1)^{a-1}[(na+1)+(na+2)+(na+3)+\cdots+(na+n)]}
&\sim\frac {n^{a+1}/(a+1)}{n^{a-1}(a+1/2)n^2}\\
\to\frac 2{(a+1)(2a+1)}
\end{align}
from which $a=7$.
For $a<-1$ the numerator converges to positive value, hence the limit is $-\infty $.
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Combining simultaneous linear recurrence relations I was studying the sequence A249665, which I will call $a_n$, and came up with a second sequence $b_n$ for which I could prove that the following recurrence holds.
\begin{align}
a_n&=a_{n-1}+a_{n-4}+a_{n-5}+b_{n-2}+b_{n-3}
\\b_n&=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+b_{n-2}... | I think I figured it out. At least for this specific case. I think the idea applies pretty generally, but can become much more complicated. This case is just a bit of algebraic manipulation.
Note that from the first recurrence formula we get $b_n+b_{n-1}=a_{n+2}-a_{n+1}-a_{n-2}-a_{n-3}$. We find
\begin{align}
a_n
&=a_{... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Is my integral in fully reduced form?
I have to integrate this:
$$\int_0^1 \frac{x-4}{x^2-5x+6}\,dx$$
Now
$$\int_0^1 \frac{x-4}{(x-3)(x-2)}\, dx$$
and by using partial fractions we get
$$\frac{x-4}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$$
$$x-4 = A(x-2) + B(x-3)$$
$$= Ax - 2A + Bx - 3B$$
$$x-4 = (A+B)x - 2A - ... | It is fine, except that I don't see where that $3$ at the last line comes from. Anyway,\begin{align}-\bigl(\log(2)-\log(3)\bigr)+2\bigl(\log(1)-\log(2)\bigr)&=\log(3)-3\log(2)\\&=\log(3)-\log(2^3)\\&=\log\left(\frac38\right)\\&=-\log\left(\frac83\right).\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?
The number of pairs of $(a,b)$ of positive real numbers satisfying
$a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2
Solution:We have $a,b>0$,
According to the given situation,$0<a^4+b^... | Since the inequalities are strict, options (ii) and (iii) are ruled out by general notions of continuity: As soon as you have a single solution $(a,b)$ in the first quadrant, you have an entire open ball of solutions of some (possibly very small) radius. So it suffices to exhibit a single solution. The values $a=b=0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Finding powers of 2 and 3 in modular arithmetic
Find all the powers of $2$ and $3$ modulo $17$.
How would you solve this question and explain the steps please!
| Recall that:
$$2^m\equiv k \pmod{17} \implies 2^{m+1}\equiv 2k \pmod {17}$$
We can then list, remembering that if our $k$ is greater than $17$, we subtract $17$ from it.
$$1\equiv 1 \pmod {17}$$
$$2\equiv 2 \pmod {17}$$
$$4\equiv 4 \pmod {17}$$
$$8\equiv 8 \pmod {17}$$
$$16\equiv 16 \pmod {17}$$
$$32\equiv 15 \pmod {17... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can you help me solve this algebra problem? Hi I need to solve this problem and I don’t know how. I’d appreciate your help.
If $x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2}$ and $x\neq y\neq z$, then $$x - \frac{ayz}{x^2} = y - \frac{azx}{y^2} = z - \frac{axy}{z^2} = x + y + z - a$$
I think I need... | Denote $k=xyz$ and $b$ the common value of $x-\frac{ayz}{x^2}=y-\frac{azx}{y^2}=z-\frac{axy}{z^2}$. We can see that the equation
$$
t-\frac{ak}{t^3}=b\tag{*}
$$ is satisfied by $t=x,y,z$. Hence $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a polynomial equation of degree 4
$$
t^4-b t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Stuck on a Geometry Problem
$ABCD$ is a square, $E$ is a midpoint of side $BC$, points $F$ and $G$ are on the diagonal $AC$ so that $|AF|=3\ \text{cm}$, $|GC|=4\ \text{cm}$ and $\angle{FEG}=45 ^{\circ}$. Determine the length of the segment $FG$.
How can I approach this problem, preferably without trigonometry?
|
Let $H$ be the midpoint of $AC$ and $\angle EIC= 90^{\circ}$. We can observe that $$FH+3=HG+4,\quad FH+HG=x.$$ So we obtain $HG=\frac{x-1}2$. Since two corresponding angles are congruent; $\angle FEG =\angle EHG=45^{\circ}$ and $\angle EGF=\angle HGE$, we have that $\triangle FEG$ and $\triangle EHG$ are similar to ea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 12,
"answer_id": 4
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integrate sin(x)cos(x) using trig identity. Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$
\sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B)
$$
$$
\begin{split}
\int \sin(x)\cos(x) dx
&= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin... | Because
$$\frac{1}{2} \sin^2(x) = \frac{1-\cos(2x)}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to deduce the following complex number problem I am stuck with the following problem that says:
Using the result $$x^n-1=(x^2-1)\prod_{k=1}^{(n-2)/2}[x^2-2x\cos \frac{2k\pi}{n}+1],$$ if $n$ be an even positive integer, deduce that $$\sin \frac{\pi}{32}\sin \frac{2\pi}{32}\sin \frac{3\pi}{32}.........\sin \frac{15... | Fixing $n=32$, we can rewrite the equation like this:
$$\frac{x^{32}-1}{x^2-1} = \prod_{k=1}^{15}\left(x^2+1-2x\cos\frac{k\pi}{16}\right)$$
for all $x\neq\pm1$.
We will take the limit as $x\to 1$ of both sides of the equality.
First, we have (using L'Hôpital's rule)
$$\lim\limits_{x\to 1} \frac{x^{32}-1}{x^2-1} = \lim\... | {
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"url": "https://math.stackexchange.com/questions/3129039",
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"source": "stackexchange",
"question_score": "2",
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Find $\lim\limits_{t \to \infty} \int_{0}^{t} \frac{\mathrm dx}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}$
Find $$\lim_{t \to \infty} \int_{0}^{t} \frac{1}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}\mathrm dx$$ where $a,b,c$ are strict positive and dinstinct real numbers.
I know it should be something with arctangent but I don't know how to ge... | Using partial fraction decomposition, we have that
$$\begin{align*}
\frac1{(x^2+a^2)(x^2+b^2)(x^2+c^2)}&=\frac1{c^2-a^2}\left(\frac1{(x^2+a^2)(x^2+b^2)}-\frac1{(x^2+b^2)(x^2+c^2)}\right)\\&=\frac1{c^2-a^2}\left(\frac{1}{b^2-a^2}\left(\frac1{x^2+a^2}-\frac1{x^2+b^2}\right)\right)\\&-\frac1{c^2-a^2}\left(\frac{1}{c^2-b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Calculate the sum of fractionals
Let $n \gt 1$ an integer. Calculate the sum: $$\sum_{1 \le p \lt q \le
n} \frac 1 {pq} $$ where $p, q$ are co-prime such that $p + q > n$.
Calculating the sum for several small $n$ value I found out that the sum is always $\frac 1 2$.
Now, I'm trying to prove the sum is $\frac 1 2$ ... | If $p+q=n+1$ and $(p,n+1)=1$, then it follows from Euclidean Algorithm that $(p,q)=1$. Now, we are to prove:
$$\frac{1}{n+1} \cdot \sum_{1 \leqslant p<n+1}{\frac{1}{p}}=\sum_{1 \leqslant p<n+1}{\frac{1}{pq}}=\frac{1}{2}\sum{\frac{1}{p(n+1-p)}}$$
This can be seen easily by writing: $$\frac{1}{p(n+1-p)}=\frac{1}{n+1}\big... | {
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Find the number of ordered triplets satisfying $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$ Find the number of ordered triplets $(x,y,z)$ of real numbers satisfying $$5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$$
and
$$xy+yz+zx=1$$
My ... | From the $xy+yz+zx=1$ we get $x = \frac{1-yz}{y+z}$. Substitute that into $5 (x+1/x)-12 (y+1/y)$. Taking the numerator of this and the numerator of $12 (y + 1/y) - 13 (z + 1/z)$, you can eliminate $y$ and get $z (z^4-1) = 0$. Since $z=0$ is not allowed, $z = \pm 1$. From that you can find the $y$ and $x$ values.... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Confusion in this algebraic limit approaching infinity
Question : Evaluate $\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}}$.
My working:
$$\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}}$$
Dividing and transforming each fra... | Two different answers with different methods in solving limits.
here you know the two important things are there one is $x\rightarrow\infty$ and and also the terms are variying as 1,2,3.....
you mean by individualy taking them as constant even when they are going to $\infty$
for example this below
similar misleading qu... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Help finding the derivative of $f(x) = \cos{\left(\sqrt{e^{x^5} \sin{x}}\right)}$ I am trying to find the derivative of $f(x) = \cos(\sqrt{(e^{x^5} \sin(x)})$.
I keep getting the wrong answer, and I'm not sure what I'm doing wrong.
$$\frac{d}{dx} e^{x^5} = e^{x^5} \cdot 5x^4$$
$$\frac{d}{dx} \sin(x) = \cos(x)$$
$$\fra... | It may be a bit overkill but you can actually integrate your result and show it is equivalent to what you started with
$$ -\frac{1}{2}\int \left(\sin(\sqrt{e^{x^5} \sin(x)} \right) \frac{5x^4e^{x^5} \sin(x) + \cos(x) e^{x^5}}{\sqrt{e^{x^5} \sin(x)}} dx,$$
let
$$u(x) = \sqrt{e^{x^5} \sin(x)}, \qquad du = \frac{5x^4e... | {
"language": "en",
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"source": "stackexchange",
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How many numbers of the form $m^2 + \sqrt{2} n^2$ are between $1 \times 10^6$ and $2 \times 10^6$? I have a purely computational question today. How many numbers of the form $m^2 + \sqrt{2} n^2$ with $m,n \in \mathbb{Z}$ are between $1 \times 10^6$ and $2 \times 10^6$ ?
$$ \# \big\{ (m,n): 1 \times 10^6 < m^2 + \sqrt... | COMMENT.- Without calculator the problem is
tedious but not difficult.
The required points are enclosed between the ellipses of equations $m^2 + \sqrt{2} n^2=10^6$ and $m^2 + \sqrt{2} n^2=2\cdot10^6$. Define the following sets in the first quadrant:
$$A=\{(0,y)\text { with } y\gt0 \}\\B=\{(x,0)\text { with } x\gt0 \}... | {
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Find the maximum integer $n$ such that $e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 2\sqrt{n}-1$
If $$e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 2\sqrt{n}-1 \quad (n \in \mathbb{N}^+),$$
find the maximum of $n$.
I can prove that the inequality holds when $n=4$.
My proof:
When $n=4$:
$$e^{x-1}\ln x+\frac{e^x+2x-1}{x^2} > 3\quad(n ... | We will prove that $\max n=4$.
Let $$f(x)=1+e^{x-1}\ln x+\frac{e^x+2x-1}{x^2}\tag1.$$ We wish to determine $\max n$ such that $$\min f(x)=2\sqrt{\max n}\implies \max n=\frac14[\min f(x)]^2\tag2.$$ Then $$f'(x)=\frac{e^x\ln x}e+\frac{e^x}{ex}+\frac{(e^x+2)x^2-2x(e^x+2x-1)}{x^4}=0\tag3$$ for critical points, which can b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Multi variable calculus, equation tangent to plane question So i think i may have it right but now sure...please check and help me to see if i got it right thanks!
Question:
$f(x,y) = 1 + x^2 + y^2$, find vector $v$ tangent to plane of graph at $f(1,1,3)$.
Answer:
$f(x,y) = 1 + x^2 + y^2$ and $g(x,y,z) = 1 + x^... | No, your gradient is not correct (and, in your statement, $f(1,1,3)$ is misleading since $f$ is a function of TWO variables).
Let $f(x,y) = 1 + x^2 + y^2$ then
the gradient of
$$g(x,y,z):=f(x,y)-z=1 + x^2 + y^2 – z$$
at $(1,1,f(1,1))=(1,1,3)$, is
$$\left(\frac{\partial g}{\partial x},\frac{\partial g}{\partial y},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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minimum value of $2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$ Let $\alpha,\beta$ be real numbers ; find the minimum value of
$2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$
What I tried :
$\bigg|4\cos \beta+(2\cos \alpha+3\sin \alpha)\sin \beta\bigg|^2 \leq 4^2+(2\cos \alpha+3\sin \alpha)^2$... |
Property to note : $a\cos x + b\sin x = \pm\sqrt{a^2 +b^2}$, So,
$$2\cos\alpha + 3\sin\alpha = \pm\sqrt{13}$$
Taking the minimum value of the expression,
$$-\sqrt{13}\sin\beta +4\cos\beta = \pm\sqrt{29}$$
Therefore, the minimum value of the expression is $-\sqrt{29}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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A simpler method to prove $\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1?$ This is the way I think about it:
$
1 = \log_aa = \log_bb = \log_cc \\~\\
\textbf{Using the ‘change of base rule':} \\
\log_{a}{b} = \frac{\log_{b}{b}}{\log_{b}{a}}, \hspace{0.3cm} \log_{b}{c} = \frac{\log_{c}{c}}{\log_{c}{b}}, \hspace{0... | You can generalize this result with logarithm rule:
$\log_{b} a = \dfrac{\log_{e}a}{\log_{e}b} = \dfrac{\ln a}{\ln b}$. That is, the identity is valid for all $a_i$ positives reals numbers and differents of $1$:
$$
\log_{a_2}a_1 \cdot \log_{a_3}a_2 \cdot \log_{a_4}a_3 \cdots \log_{a_n}a_{n-1}\cdot \log_{a_1}a_n = 1.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How is $\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $ equivalent to $\frac{ y dx + xdy - zdz}{0}=\frac{ xdx - ydy -zdz}{0}$? In the book of PDE by Kumar, it is given that
However, I couldn't figure out how is
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $$
is equivalent ... | Some notes about solving PDE
$$z(x+y)z_x+z(x-y)z_y=x^2+y^2\qquad (1)$$
*
*see
https://www.math24.net/first-integrals-page-2/
Example 4
*In our case two independent first integrals is:
$$y^2+2xy-y^2=C_1, \quad z^2-2xy=C_2$$
*General solution of PDE $(1)$ is
$$z^2=2xy+f(y^2+2xy-x^2)$$
*Let $u=z^2$. Then from $(1)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to derive this sequence: $1^3+5^3+3^3=153,16^3+50^3+33^3=165033,166^3+500^3+333^3=166500333,\cdots$? I found it is interesting but I don't know how the R.H.S is coming from the L.H.S,
i.e, how to derive this sequence?
The sequence is as follows:
$$1^3+5^3+3^3=153$$
$$16^3+50^3+33^3=165033$$
$$166^3+500^3+333^3=1665... | First some notations, let express the repunits $r_n=11\cdots 1=\dfrac{(10^n-1)}9$ in term of $k=10^n$
$\begin{cases}
a_n&=16\cdots 6&=10^{n-1}+6\ r_{n-1}&=\frac{k-4}6\\
b_n&=50\cdots 0&=5\times 10^{n-1}&=\frac{k}2\\
c_n&=33\cdots 3&=3\,r_n&=\frac{k-1}3\end{cases}$
The sum of cubes
$f(n)=a_n^3+b_n^3+c_n^3=\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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What is wrong with this proof of $3\arcsin x$? We know that
\begin{align}
2\arcsin x&= \arcsin \left(2x\sqrt{1-x^2}\right) \tag{1}\\
\arcsin x + \arcsin y &= \arcsin \left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right] \tag{2}\\
3\arcsin x &= \arcsin x + 2\arcsin x \tag{3}
\end{align}
Thus $x=x, y=2x\sqrt{1-x^2}$
using ($1$)... | A geometric point of view might be illuminating.
Suppose $0\leq x \leq \frac{1}{\sqrt 2}$. Consider the figure below, where we start from a right-angled triangle $\triangle ABC$ with sides $\overline{AB} = \sqrt{1-x^2}$ and $\overline{BC} = x$, and hypotenuse $\overline{AC}=1$. The choice of $x$ we have made guarantees... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\lim_{n \to \infty} \int_0^n \frac{dx}{1+n^2\cos^2x}$ Find $$\lim_{n \to \infty} \int_0^n \frac{dx}{1+n^2\cos^2x}$$
I tried:
*
*mean value theorem.
*variable change with $ \tan x = t $ but I need to avoid the points which are not in the domain of $\tan$ and it's complicated.
| Using the substitution $u=\tan{(x)}\Rightarrow du=\sec^2{(x)}dx$ turns the integral into
$$\int_0^\infty \frac{du}{u^2+n^2+1}+\bigg\lfloor\frac{n-\frac\pi2}{\pi}\bigg\rfloor\int_{-\infty}^\infty \frac{du}{u^2+n^2+1}+\int_{-\infty}^{\tan{(n)}}\frac{du}{u^2+n^2+1}$$
$$=\bigg[\frac1{\sqrt{n^2+1}}\arctan{\bigg(\frac{u}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Summing up the series $\sum_{n=0}^\infty C_{2n}^n r^{2n} $ Here $|r|<1/2$, so that the series converge.
I can do it by using contour inregration.
$$ S =\sum_n r^{2n} \frac{1}{2\pi i } \oint_C \frac{1}{z}(z+ \frac{1}{z})^{2n} dz
\\
= \frac{1}{2\pi i } \oint_C \frac{1}{z} r^{2n} (z+ \frac{1}{z})^{2n} dz \\
= \fra... |
We can use the binomial series expansion. In order to do so we recall the binomial identity
\begin{align*}
\binom{2n}{n}=(-4)^n\binom{-\frac{1}{2}}{n}\tag{1}
\end{align*}
and obtain
\begin{align*}
\color{blue}{\sum_{n=0}^\infty \binom{2n}{n}r^{2n}}&=\sum_{n=0}^\infty (-4)^n\binom{-\frac{1}{2}}{n}r^{2n}\\
&=\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find a set of points in the given complex plane Here's the Question: Find a set of points in the complex plane that satisfies:
$$|z-i|+|z+i| = 1$$
Now from triangle inequality I found:
$$|z+z+i-i|=|2z|\geq1 $$
Which refers that there's no soluton and the set should be empty.
But if I substitute z=x+iy and simplify th... | There is no such point because for all $z \in \mathbb{C}$ you have
$$|z-i|+|z+i| \geq |z-i - (z+i)| =|2i| = 2 > 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Automorphism of a matrices ring
Let $R$ be the ring of $3 \times 3$ matrices with coefficients in $\Bbb Z_5$.
For every $g \in GL_3(\Bbb Z_5)$ prove that the function $$f\colon R \rightarrow R$$ defined as $$x \mapsto g^{-1}xg$$ is an automorphism of $R$.
If I choose the matrix $$g = \begin{pmatrix}
0 & 0 & 1 \\
0... | This is much more general: suppose $R$ is a ring and $g$ is invertible in $R$. Then the map $f\colon R\to R$, $f(x)=gxg^{-1}$ is an automorphism of $R$.
Indeed, $f(x+y)=g(x+y)g^{-1}=gxg^{-1}+gyg^{-1}=f(x)+f(y)$ and
$$
f(xy)=gxyg^{-1}=gxg^{-1}gyg^{-1}=f(x)f(y)
$$
Obviously, $f(1)=1$.
Since the map $x\mapsto g^{-1}xg$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Recursion Question using Generating Functions Here is my question:
Consider the recurrence,
$$a_{n+1}=2a_n+(-1)^n$$
with initial condition,
$$a_0=0$$
Find and prove a formula for $a_n$.
I don't really know how to prove this formula
I tried going with a generating function method, but that kind of led nowhere.
| The sequence generated by the recurrence relation is;
$$0, 1, 1, 3, 5, 11, 21, \dots$$
Write the recurrence relation as;
$$a_n-2a_{n-1}=(-1)(-1)^n$$
Get the generating function, $GF$ in the standard way;
$$GF=0+x+x^2+3x^3+5x^4+11x^5+21x^6+\dots$$
$$-2xGF=0+0x-2x^2-2x^3-6x^4-10x^5-22x^6+\dots$$
$$(1-2x)GF=0+x-x^2+x^3-x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Sum of all values that satisfy $\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$.
What is the sum of all values of $x$ that satisfy the equation
$\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$?
I start off by cross multiplying.
$$x^2=x^{x-3+\frac{4}{x}}$$
Taking the square root of both sides gives me:
$$x=\pm x^{\frac{1}{2}x... | I start off by cross multiplying.
$$x^2=x^{x-3+\frac{4}{x}}$$
Taking the square root of both sides gives me:
$$x=\pm x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$
I start with the positive side first:
$$1=\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}$$$$x^2-5x+4=0$$$$x=\boxed{4}, \boxed{1}$$
Now, I start with the negative side:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Compute $I = \int_{0}^{\infty} \frac{8}{\gamma^{4}} x^{3} e^{-\frac{2}{\gamma^{2}}x^{2}} e^{jxt} dx$ I want to evaluate the following integral:
$$I = \int_{0}^{\infty} \frac{8}{\gamma^{4}} x^{3} e^{-\frac{2}{\gamma^{2}}x^{2}} e^{jxt} dx $$
Where $\gamma \in \mathbb{R}$ and $j = \sqrt{-1}$.
The first thing I do is let... | Well, in general we have:
$$\mathcal{I}_\text{n}:=\int_0^{\infty}x^\text{n}\cdot\exp\left(\alpha \cdot x^{\text{n}-1}\right)\cdot\exp\left(\beta\cdot x\right)\space\text{d}x\tag1$$
Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:
$$\mathcal{I}_\text{n}=\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Need to compute $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$. Is my solution correct? $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$$
Since $$\tan^{-1}x=\int \frac{1}{1+x^{2}} dx=\int (1-x^{2}+x^{4}+...)dx=x-\frac{x
^3}{3}+\frac{x^5}{5}+...$$
so$$\int \tan^{-1}x dx=\int (x-\frac{x^3}{3}+\frac{x^5}{5}... | You series equals
$$ \frac{1}{3}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)3^n}-\sum_{n\geq 0}\frac{(-1)^n}{(2n+2)3^{n+1}} $$
or
$$ \frac{1}{\sqrt{3}}\arctan\frac{1}{\sqrt{3}}+\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^n}{n 3^n}=\frac{\pi}{6\sqrt{3}}-\frac{1}{2}\log\left(1+\frac{1}{3}\right). $$
No integrals, just partial fraction de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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induction proof: number of compositions of a positive number $n$ is $2^{n-1}$ So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
... | Let's return to the original problem. A composition of a positive integer $n$ is a way of writing $n$ as the sum of a sequence of strictly positive integers.
For $n = 5$, there are indeed $2^{5 - 1} = 2^4 = 16$ compositions since each composition corresponds to placing or omitting an addition sign in the four spaces b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Equation of plane containing a point and a line
Find the equation of the plane containing the point $A(0,1,-1)$ and the line
$(d) : \begin{cases} 2x - y + z + 1 = 0 \\ x + y + 1 = 0 \end{cases}$
Where should I start? I was thinking about writing the normal vectors for the line and make their cross product or someth... | A plane in $\mathbb{R}^{3}$ satisfies an equation of the type
\begin{equation}
ax+by+cz+d=0. \tag{1}
\end{equation}
You can reduce this problem to that of finding a plane defined by 3 points. So just find 2 points on the line $ d $, using the given equations. For $x=0$ and $x=1$ you get, respectively, the points $B\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find minimum perimeter of the triangle circumscribing semicircle The following diagram shows triangle circumscribing a semi circle of unit radius. Find minimum perimeter of triangle
My try:
Letting $$AP=AQ=x$$
By power of a point we have:
$$BP^2=OB^2-1$$ where $O$ is center of the circle.
Also $$CQ^2=OC^2-1$$
Let $$OB=... |
Let us calculate the perimeter from the figure.
$OB = \dfrac{1}{\sin\alpha}, PB=\dfrac{1}{\tan\alpha},AP= \dfrac{1}{\tan(\pi/2 -\dfrac{\alpha+\theta}{2})},AQ= \dfrac{1}{\tan(\pi/2 -\dfrac{\alpha+\theta}{2})},QC=\dfrac{1}{\tan\theta}, OC =\dfrac{1}{\sin\theta} $
So total perimeter is $ p = \dfrac{1}{\sin\alpha}+\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3166836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving a sequence has limit using an epsilon - N argument I have two separate sequences that I using this approach on.
$a_n=\frac{n^3-2n^2+3}{2n^3+7n}$ & $a_n=\frac{n^3}{2^n}$
Proof 1
$a_n=\frac{n^3-2n^2+3}{2n^3+7n}\to \frac{1}{2}$
$$\text{Let } \epsilon \gt 0 \,\, \exists N \;\;\forall n\geq N \; \left\lvert {a_n-\f... | On the first proof:
It would be simplest to show that $\left|\dfrac{4n^2+1}{4n^3+14n}\right|<\dfrac{1}{n}$ since, clearly $4n^3+n<4n^3+14n$.
On the second proof:
For $n\ge16$ we have that $n^4\le2^n$ from which it follows that $\dfrac{n^3}{2^n}\le\dfrac{1}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2$ Suppose $a>0, b>0, c>0$.
Prove that:
$$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$
Hence or otherwise prove:
$$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$
| Hint: Substitute $$b+c=x,a+c=y,a+b=z$$ so $$a=\frac{-x+y+z}{2}$$
$$b=\frac{x-y+z}{2}$$
$$c=\frac{x+y-z}{2}$$
And we get
$$\frac{-x+y+z}{2x}+\frac{x-y+z}{2y}+\frac{x+y-z}{2z}\geq \frac{3}{2}$$
Can you finish?
And we get $$\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}\geq 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove that $a+b+2\sqrt{ab+c^2}$ cannot be a prime number Prove that the number
$$a+b+2\sqrt{ab+c^2}$$
cannot be a prime number for any positive integer numbers $a,b,c$.
My attempt:
Suppose that $p=a+b+2\sqrt{ab+c^2}$ is a prime. WLOG assume that $a \geq b$. From the equality we have
$$\left(a+b\right)^2+p^2-2p\left(a... | Your work has almost finished the problem. $p|(a-b-2c)$ is the key property.
If $a-b-2c>0$, then $a-b-2c\ge p$. Hence $a\ge b+2c+p$, a contradiction since $a<p$.
If $a-b-2c<0$, then $a-b-2c\le -p$. Hence $a+p\le b+2c$, so $2a+b+2\sqrt{ab+c^2}\le b+2c$, so $a+\sqrt{ab+c^2}\le c$, again a contradiction.
Finally, if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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show this inequality to $\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $ Let $a,b$ and $c$ be positive real numbers. Prove that
$$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$
This problem is from Iran 3rd round-2017-Algebra final exam-P3,Now I can't find... | Using binomial inequality for
$$|c-a|<2a+2b+c,$$
one can get
$$\dfrac1{(3a+2b)^3}=\dfrac1{(2a+2b+c)^3}\left(1-\dfrac{c-a}{2a+2b+c}\right)^{-3}
\ge \dfrac1{(2a+2b+c)^3}\left(1+3\dfrac{c-a}{2a+2b+c}\right),$$
$$\dfrac{a^3b}{(3a+2b)^3}-\dfrac{a^2bc}{(2a+2b+c)^3} \ge \dfrac{a^2b}{(2a+2b+c)^3}\left(a-c-3a\dfrac{c-a}{2a+2b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3172778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Interpreting images representing geometric series I understand the formula for infinite geometric series as
$$S = \frac{a_{1}}{1-r}$$ if $0<r<1$
However I'm having trouble applying it to these images
It seems to me that in the first image, the first square represents 1/4 of the entire square
For the second and third ... | 1) The area of the first shaded square is a fourth part of the original square: $\frac{1}{4}$. The area of the second shaded square is a fourth of a fourth of the original area: $\frac{1}{4}\cdot\frac{1}{4}$. The area of the third square would be a fourth of that: $\frac{1}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}$. Do you s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation... | If you prefer using parametric form or your equations are already in parametric form, this is how you can proceed:
We know that $(0,18)$ is a solution to $3x+2y=36$ and $(0,16)$ is a solution to $5x + 4y = 64$. Therefore the equations in parametric form become:
$${ \begin{pmatrix} 0 \\ 18 \\ \end{pmatrix} } + t_1 { \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 13
} |
Show that a determinant equals the product of another determinant and a polynomial function without calculating Show without calculating the determinant, that
$$
\det\left(\begin{bmatrix}
a_{1}+b_{1}x & a_{1}-b_{1}x & c_{1}\\
a_{2}+b_{2}x & a_{2}-b_{2}x & c_{2}\\
a_{3}+b_{3}x & a_{3}-b_{3}x& c_{3}\\
... | It is indeed multilinearity you want to use. Write
$$
a =
\begin{bmatrix}
a_{1}\\
a_{2}\\
a_{3}
\end{bmatrix},
\quad
b =
\begin{bmatrix}
b_{1}\\
b_{2}\\
b_{3}
\end{bmatrix},
\quad
c =
\begin{bmatrix}
c_{1}\\
c_{2}\\
c_{3}
\end{bmatrix},
$$
then
\begin{align*}
\det(\begin{bmatrix}a + b x & a - b x & c\end{bmatrix})
&=
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Use generating functions to solve the recurrence relation Use generating functions to solve $a_n = 3a_{n-1} - 2a_{n-2} + 2^n + (n+1)3^n$.
What I have so far, not sure if I forgot to do something or am missing out on something obvious:
Define $$G(x) = \sum_{n=0}^{\infty} a_nx^n$$
Then, $G(x) = a_0 + a_1x + a_2x^2 + \su... | After making these replacements, you will have $G(x)$ on both sides, and a couple of other summations you can (hopefully?) find closed forms for. Then you can solve for $G(x)$.
$$
\sum_{n=1}^\infty a_nx^n=G(x)-a_0\hspace{1.3cm}
$$
$$
\sum_{n=2}^\infty a_nx^n=G(x)-a_0-a_1x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}$
Question: How can we evaluate $$\sum_{n=1}^\infty\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n},$$where $H_n=\frac11+\frac12+\cdots+\frac1n$?
Quick Results
This series converges because $$\frac{(H_n)^2}{n}\frac{\binom{2n}n}{4^n}=O\left(\frac{\ln^2n}{n^{... | From this paper, Eq $(13)$ page 4, we have
$$\sum_{n=1}^\infty H_n\binom{2n}n x^n=\frac{2}{\sqrt{1-4x}}\ln\left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right).$$
Replace $x$ by $\frac{x}{4}$, multiply both sides by $-\frac{\ln(1-x)}{x}$ then integrate using $-\int_0^1 x^{n-1} \ln(1-x)dx=\frac{H_n}{n}$, we have
$$\sum_{n=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3184675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
$f$ dividing by $x + 1$ have remainder 4, when dividing with $x^2 + 1$ have remainder 2x+3. Find remainder dividing polynomial with($x+1$)($x^2+1$) Problem: The polynomial $f$ dividing by ($x + 1$) gives the remainder 4, and when dividing with ($x^2 + 1$) gives the remainder (2x+3). Determine the remainder when dividin... | Upon applying the law: $\,ab\bmod ac = a(b\bmod c) = $ Mod Distributive Law we obtain
$(f\!-\!(2x\!+\!3))\bmod{(x^2\!+\!1)(x\!+\!1)}\,=\, (x^2\!+\!1){\Huge[}\dfrac{\overbrace{f\!-\!(2x\!+\!3)}^{\large 4\,-\,(2(\color{#c00}{\bf -1})+3)}}{\underbrace{x^2\!+\!1}_{\large (\color{#c00}{\bf -1})^2+1\ \ \ \ }}\underbrace{\bm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Multiplying two generating functions I am trying to complete exercise 10 from here. It says to find $a_7$ of the sequence with generating function $\frac{2}{(1−x)^2} \cdot \frac{x}{1−x−x^2}$. I wrote down the first $7$ numbers of both sequences and got $2, 4, 6, 8, 10, 12, 14$ and $1, 1, 2, 3, 5, 8, 13, 21$. I then tri... | The two sequences are,
$$2, 4, 6, 8, 10, 12, 14, 16, \dots$$
and,
$$0, 1, 1, 2, 3, 5, 8, 13, 21, \dots$$
Note that with generating functions the initial term is $a_0$, "term zero".
So, $a_7$ is actually the $8^{th}$ term.
The convolution of these two sequences, also called the Cauchy product, is found by multiplying th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of real roots of $p_n(x)=1+2x+3x^2+....+(n+1)x^n$ if $n$ is an odd integer
If $n$ be an odd integer. Then find the number of real roots of the polynomial equation $p_n(x)=1+2x+3x^2+....+(n+1)x^n$
$$
p_n(x)=1+2x+3x^2+....+(n+1)x^n\\
x.p_n(x)=x+2x^2+....nx^n+(n+1)x^{n+1}\\
p(x)[1-x]=1+x+x^2+....+x^n-(n+1)x^{n+1}... | $$
p(x)=1+2x+3x^2+\ldots+(n+1)x^n;\\
xp(x)=x+2x^2+3x^3+\ldots+(n+1)x^{n+1};\\
(1-x)p(x)=1+x+x^2+\ldots+x^n-(n+1)x^{n+1}=\frac{1-x^{n+1}}{1-x}-(n+1)x^{n+1}\\
=\frac{1-x^{n+1}-(1-x)(n+1)x^{n+1}}{1-x}=0.\\
\Rightarrow1-x^{n+1}-(1-x)(n+1)x^{n+1}=0
$$
Note that $x=1$ is clearly not a solution of the original equation. As a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Definite Integral of $\int_0^1\frac{dx}{\sqrt {x(1-x)}}$ We have to calculate value of the following integral :
$$\int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \qquad \qquad (2)$$
What i've done for (2) :
\begin{align}
& = \int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \\
& = \int_0^1\cfrac{dx}{\sqrt {x-x^2}} \\
& = \int_0^1\cfrac{dx}{\s... | Letting $x=\sin^2 \theta$ yields
$$
\begin{aligned}
\int_0^1 \frac{d x}{\sqrt{x(1-x)}} &=\int_0^{\frac{\pi}{2}} \frac{2 \sin \theta \cos \theta d \theta}{\sin \theta \cos \theta} =[2 \theta]_0^{\frac{\pi}{2}} =\pi
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solution to $\sqrt{\sqrt{x + 5} + 5} = x$ There are natural numbers $a$, $b$, and $c$ such that the solution to the equation
\begin{equation*}
\sqrt{\sqrt{x + 5} + 5} = x
\end{equation*}
is $\displaystyle{\frac{a + \sqrt{b}}{c}}$. Evaluate $a + b + c$.
I am not sure where I saw this problem. My guess is that it was fro... | Note that if $x = \sqrt{x+5}$ then $x = \sqrt{\sqrt{x+5}+5}$.
So, try solving $x = \sqrt{x+5}$. This is a quadratic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How does one prove such an equation? The problem occurred to me while I was trying to solve a problem in planimetry using analytic geometry.
for $b$ between $-\frac{1}2$ and $1$ :
$\sqrt{2+\sqrt{3-3b^2}+b} = \sqrt{2-2b}+ \sqrt{2-\sqrt{3-3b^2}+b}$
| Hint:
set $b=\cos t$:
\begin{align}
\sqrt{2+\cos t+\sqrt3\sqrt{1-\cos^2t}}
&\overset?=
\sqrt2\sqrt{\vphantom{\sqrt3}1-\cos t}+ \sqrt{2+\cos t-\sqrt3\sqrt{1-\cos^2t}}
,\\
\sqrt{2+\cos t+\sqrt3\sin t}
&\overset?=
\sqrt2\sqrt{\vphantom{\sqrt3}1-\cos t}+ \sqrt{2+\cos t-\sqrt3\sin t}
,\\
\sqrt{1+\tfrac12\cos t+\tfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Complex number cannot arive at $\frac{9}{2}-\frac{9}{2}i$ with problem $\frac{4+i}{i}+\frac{3-4i}{1-i}$ I am asked to evaluate: $\frac{4+i}{i}+\frac{3-4i}{1-i}$
The provided solution is: $\frac{9}{2}-\frac{9}{2}i$
I arrived at a divide by zero error which must be incorrect. My working:
$\frac{4+i}{i}$, complex conjugat... | You've missed a minus sign there:
$\frac{4+i}{i}=\frac{4i-1}{-1}=1-4i$
The second one is
$\frac{3-4i}{1-i}=\frac{(3-4i)(1+i)}{(1-i)(1+i)}=\frac{7-i}{2}$
Hence the sum is $1-4i+\frac{7}{2}-\frac{i}{2}=\frac{9}{2}-\frac{9i}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
to prove $x^2 + y^2+1\ge xy + y + x$ $$x^2 + y^2+1\ge xy + y + x$$
$x$ and $ y$ belong to all real numbers
my attempt
$(u-2)^2\ge0\Rightarrow \frac{u^2}{4}+1\ge u $
let $u=x+y\Rightarrow \frac{(x+y)^2}{4}+1\ge x+y$
$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)$
$but \frac{(x+y)^2}{4} \ge xy $ by AM-GM inequality
$... | Prove
$$f(x,y)=x^2+y^2+1-x-xy-y\geq 0.$$
$$f_x=2x-1-y$$
$$f_y=2y-1-x$$
$$f_x=0\implies y=2x-1$$
$$f_y=0\implies x=2y-1$$
$$y=2(2y-1)-1=4y-3\implies y=1$$
$$x=2(2x-1)-1=4x-3\implies x=1$$
There's a stationary point at $(1,1)$.
$$f_{xx}=2,f_{xy}=f_{yx}=-1,f_{yy}=2.$$
Since $$f_{xx}f_{yy}-f_{xy}^2=4-(-1)^2=3>0\text{ and }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ .
Given three non-negatve numbers $a, b, c$. Prove that:
$$1+ a^{2}+ b^{2}+ c^{2}+ 4abc\geqq a+ b+ c+ ab+ bc+ ca$$
Let $t= a+ b+ c$, we have to prove
$$\left(\!\frac{1}{t^{3}}- \frac{1}{t^{2}}+ \frac{1}{t}\!\right)\sum ... | Another way.
Since $$\prod_{cyc}(2a-1)^2=\prod_{cyc}((2a-1)(2b-1))\geq0$$ and our inequality is symmetric, we can assume that $$(2a-1)(2b-1)\geq0,$$ which gives
$$c(2a-1)(2b-1)\geq0$$ or
$$4abc\geq 2ac+2bc-c.$$
Thus, $$1+a^2+b^2+c^2+4abc-a-b-c-ab-ac-bc\ge$$
$$\geq1+c^2+a^2+b^2-ab+ac+bc-a-b-2c\geq$$
$$\geq1+c^2+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Jordan normal as transformation with respect to the basis of eigenvectors I have the following matrix
$$A = \begin{pmatrix}
2 & 0 & 1 & -3 \\
0 & 2 & 10 & 4 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 3 \\
\end{pmatrix}$$
and its Jordan normal form is
$$J = \begin{pmatrix}
2 & 0 & 0 & 0 \\
... | B= $\big\{ v_{1},v_{2}, v_{3}, v_{4} \big\} $
$P^{-1}_{B} [A \mathbf{v_{3}}] = [A \mathbf{v_{3}}]_{B} = \begin{bmatrix}
0 \\
1 \\
2 \\
0\\
\end{bmatrix}$
... from the change of basis formula: $x = P_{B}[x]_{B}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Minimum value of $\frac{2-\cos(x)}{\sin(x)}$ without differentiation I have to find the minimum value of the expression
$$\frac{2 - \cos x}{ \sin x}$$
Also $x$ lies between $0$ to $\pi$. One way is to find the minima using differentiation. But it is not taught in my grade so my teacher asked me to do it without differ... | You can use AM-GM as follows for $x \in (0,\pi)$:
$$\frac{2-\cos(x)}{\sin(x)}= \frac{2-(\cos^2(x/2) - \sin^2(x/2))}{2\sin(x/2)\cos(x/2)}$$
$$= \frac{\cos^2(x/2) +3\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}\stackrel{AM-GM}{\geq} \frac{\sqrt{\cos^2(x/2) \cdot 3\sin^2(x/2)}}{\sin(x/2)\cos(x/2)} = \sqrt{3}$$
Additional note after c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prime numbers of the form $p=m^2+n^2$ such that $p \mid m^3+n^3-4$
Find all prime numbers $p$, for which there are positive integers $m$ and $n$ such that $p=m^2+n^2$ and $p \mid m^3+n^3-4$.
I simplified this a little bit.
$$m^2+n^2 \mid m^3+n^3-4 =(m+n)(m^2+n^2-mn)-4 \\ \Longrightarrow m^2+n^2 \mid (m+n)mn+4 $$
The ... | There are no primes with this property other than $p=2$ and $p=5$; here is a proof.
Suppose that $m,n>1$ and $p=m^2+n^2$ is a prime dividing $m^3+n^3-4$. As you have observed, we have then
$$ mn(m+n) \equiv -4 \pmod p. $$
We apply two operations to this congruence: squaring and multiplying by $m+n$; keeping in mind... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can this determinant expression ever equal $0$? My issue is that when calculating the eigenvalues of a matrix, there are cases where the eigenvalue is $0$, and so I was trying to confirm this by calculating the determinant and checking when it is equal to $0$.
The setup is: $a^2 - 4bc \neq 0$, $bc \neq 0$, $(a,b,c) \i... | From $$\left(\frac{a+\sqrt{a^2-4bc}}2\right)^{n+1}=\left(\frac{a-\sqrt{a^2-4bc}}2\right)^{n+1} $$
we get, by dividing by the non-zero(!) rught hand side
$$ \left(\frac{a+\sqrt{a^2-4bc}}{a-\sqrt{a^2-4bc}}\right)^{n+1}=1,$$
or after making the denominator rational,
$$ \left(\frac{(a+\sqrt{a^2-4bc})^2}{4bc}\right)^{n+1}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3207065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove an inequality using mathematical induction? I have to prove the following:
$$ \sqrt{x_1} + \sqrt{x_2} +...+\sqrt{x_n} \ge \sqrt{x_1 + x_2 + ... + x_n}$$
For every $n \ge 2$ and $x_1, x_2, ..., x_n \in \Bbb N$
Here's my attempt:
Consider $P(n): \sqrt{x_1} + \sqrt{x_2} +...+\sqrt{x_n} \ge \sqrt{x_1 + x_2 +... | Without induction. For $any$ non-negative reals $x_1,...,x_n$ let $x_j=(y_j)^2$ for each $j,$ with each $y_j\ge 0.$ The inequality is then $$y_1+...+y_n\ge \sqrt {(y_1)^2+...+(y_n)^2}.$$ Since both sides are non-negative reals, this is equivalent to $$(y_1+...+y_n)^2\ge (y_1)^2+...+(y_n)^2.$$ Expand the LHS of this and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compute Moore-Penrose Pseudoinverse of a 3x3 matrix
Given a 3x3 matrix A= \begin{pmatrix}
1 & 1 & 1\\
1 & 1 &1 \\
1 & 1 & 1
\end{pmatrix} Find Psuedoinverse of the above matrix
For the above matrix I am getting it as singular.
Also the rank of the above matrix is 1 so tried to calculate left and right inverse of A ... | Note that, if $A$ is the given matrix, then $A^2 = 3A$, and $A^3 = 9A$. Also, $A$ is symmetric. From this information, I claim that $B = \frac{1}{9}A$ is the pseudoinverse of $A$. Note that
\begin{align*}
ABA &= \frac{1}{9}A^3 = A \\
BAB &= \frac{1}{9^2}A^3 = \frac{1}{9}A = B \\
AB &= \frac{1}{3}A \\
BA &= \frac{1}{3}A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving limit without L'Hôpital's rule: $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$ How can I solve this limit without L'Hôpital's rule?
$$\begin{align}\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\lim\limits_{x \to 0} \frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x... | We can work asymptotically also:
We denote the limit by $L$
As $x \to 0$
$$L=\frac{(\sqrt{1+\tan x}-1) -(\sqrt{1+\sin x}-1)}{x^3} \sim \frac{1}{2} \,\,\left[ \frac{\tan x (1-\cos x)}{x^3} \right] \sim \frac{1}{4}\,\,\frac{\tan x}{x} \sim \frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How do I determine the limit of this sequence? I need a refresher.
I would assume that the answer would be $\infty$, but I am not quite sure.
The problem reads:
Determine the limit of this sequence? $c_n$ = $\sqrt {n^2+n} - \sqrt {n^2 - n}$ .
I would need a refresher on this. That's all thank you.
| Multiply $\sqrt{n^2+n} - \sqrt {n^2 - n}$ by its conjugate (it looks exactly the same way except that it has a plus sign instead of a minus sign between the square roots):
$$\require{cancel}
\frac{\sqrt {n^2+n} - \sqrt {n^2 - n}}{1}\cdot\frac{\sqrt {n^2+n} + \sqrt {n^2 - n}}{\sqrt {n^2+n} + \sqrt {n^2 - n}}=\\
\frac{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\fr... | Multiplying both sides by $x^2$ will result in
$$4x^2+x-1=0$$
Now, with $a=4, b=1, c=-1$ use the quadratic formula and let us know what you get.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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An algebra problem: Let $a+b+c=0$. Prove that $a^2+b^2+c^2 =6/5$ Let $a,b,c$ be nonzero real numbers such that $a+b+c=0$ and $a^3+b^3+c^3 = a^5 +b^5 +c^5$. Prove that $a^2+b^2+c^2 =6/5$
I tried to expand $(a+b+c)^5$ but I can't get term of $a^2+b^2+c^2$.
| Hint: With $$c=-a-b$$ we get in $$a^3+b^3+c^3-a^5-b^5-c^5=0$$ the equation
$$ab(a+b)(5a^2+5ab+5b^2-3)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$? I need to remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$, and find out what $f(7)$ equals. I am not sure what I've done wrong, but I'm getting 33, which the website I'm using tells me is wrong.
Please help me figure out what I... | x=7 is a root of $x^2-11x+28$. So write $x^2-11x+28=(x-7)p(x)$. Then $f(7)=p(7)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove that $\sum ^{n} _{k=1} \sin(kx) \le \frac{1}{|\sin(\frac{x}{2})|}$
Prove that $\sum ^{n} _{k=1} \sin(kx) \le \frac{1}{|\sin(\frac{x}{2})|}$
I am doing some task and I wanted to see solution which is in my book. However there is inequality $\sum ^{n} _{k=1} \sin(kx) \le \frac{1}{|\sin(\frac{x}{2})|}$ and I don't... | $\sum\limits_{k=1}^n\sin{kx}\sin{\frac{x}{2}}=\frac{1}{2}\sum\limits_{k=1}^n(\cos(kx-\frac{x}{2})-\cos(kx+\frac{x}{2}))=\frac{1}{2}(\cos(\frac{x}{2})-\cos(\frac{3x}{2})+\cos(\frac{3x}{2})-\cos(\frac{5x}{2})+...)=\frac{1}{2}(\cos{\frac{x}{2}}-\cos\frac{(2n+1)x}{2})$
So $|\sum\limits_{k=1}^n\sin{kx}|\leq \frac{\frac{1+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof that $ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $ for $p>5$ Proof that $$ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $$
for $p>5$ and $p$ is prime.
$\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}$
My try
Let show that
$$p^2 - 1 \equiv 0 \Mod{30} \vee p^2 - 1\equiv 18 \Mod{30}$$
Let check $$p^2 -1 = (p... | To extend your approach:
Let $p=5k+r$ with $r=1,2,-1,-2.$
As you note, when $r=1,-1$ you get $p^2\equiv 1\pmod{5}$ and hence $p^2\equiv 1\pmod{30},$ since you've already shown $p^2\equiv 1\pmod 6.$
In the other cases, you need to deduce additional properties about $k,$ because just $p=5k\pm 2$ doesn't let us deduce it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Convergence of powers of matrices in Jordan Canonical Form (Jordan Normal Form) I've actually been stuck on this for a bit while studying for an exam, so would appreciate any help.
The problem involves testing whether $\lim\limits_{m \to \infty}$ $A^m$ exists.
From lecture notes, I know that the limit above exists for... | For the first matrix, the Jordan blocks are:
$$J(A_1)=
\begin{pmatrix}
\color{red}{1/2} & \color{red}1 & \color{red}0 & 0 & 0 \\
\color{red}0 & \color{red}{1/2} & \color{red}1 & 0 & 0 \\
\color{red}0 & \color{red}0 & \color{red}{1/2} & 0 & 0 \\
0 & 0 & 0 & \color{blue}1 & 0 \\
0 & 0 & 0 & 0 & \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do I evaluate the following integral using residue theory? This is how I've approached solving the integral so far$$\int_{0}^{2\pi}\frac{d\theta}{4-2\sqrt{3}\cos(\theta)}=\frac{1}{2}\int_{0}^{2\pi}\frac{d\theta}{2-\sqrt{3}\cos(\theta)}$$
First, I substituted $$\cos(\theta)=\bigg(z+\frac{1}{z}\bigg)$$ into the given... | You need to make the substitution
$$\cos\theta=\frac12\left(z+\frac1z\right).$$
You then get the contour integral
$$\text{constant}\int_C\frac{dz}{z^2-(4/\sqrt3)z+1}$$
where $C$ is the unit circle. Now the function has one pole inside
and one outside $C$ and you'll get a non-zero answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left... | Another Solution
\begin{align*}
&\lim_{x \to 0+}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}\\
=&\lim_{x \to 0+}\frac{e^{(1+x)^{\frac{1}{x}}}-e^{\frac{e}{x}\ln(1+x)}}{x^2}\\
=&\lim_{x \to 0+}\left[e^{\frac{e}{x}\ln(1+x)}\cdot\frac{e^{(1+x)^{\frac{1}{x}}-\frac{e}{x}\ln(1+x)}-1}{x^2}\right]\\
=&e^e\cdot \lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
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The integral solution of $x^{2}-y^{3}=1 (x>1,y>1) $? I know it's a special case of catalan's conjecture.Wiki says its only solution is $(3,2)$.But I cannot work it out.Any help will be appreciated.
| You can re-write the expression as $y^3=x^2-1=(x+1)(x-1)$. Now we can do some analysis. The only common factor that $x+1$ and $x-1$ can have is $2$.
*
*Case 1- They have no common factor. In that case, both $x+1$ and $x-1$ are cubes. This is impossible, as the smallest difference between cubes is $2^3-1^3=7$.
*Cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Compute polynomial $p(x)$ if $x^5=1,\, x\neq 1$ [reducing mod $\textit{simpler}$ multiples] The following question was asked on a high school test, where the students were given a few minutes per question, at most:
Given that,
$$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$
and,
$$Q(x)=x^4+x^3+x^2+x+1$$
what is the re... | Let $a$ be zero of $x^4+x^3+x^2+x+1=0$. Obviously $a\ne 1$. Then $$a^4+a^3+a^2+a+1=0$$
so multiply this with $a-1$ we get $$a^5=1$$ (You can get this also from geometric series $$a^n+a^{n-1}+...+a^2+a+1 = {a^{n+1}-1\over a-1}$$ by putting $n=4$).
But then \begin{eqnarray} Q(a) &=& a^{100}\cdot a^4+a^{90}\cdot a^3+a^{80... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "63",
"answer_count": 6,
"answer_id": 0
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Pattern in Squared Numbers and their Digit Sum So this has been boggling my mind for some time now. On spring break I was really bored and started messing around with numbers when I noticed something. I was squaring each number (1-9) when I realized I could get the products down to a single digit using the digit sum.
1... | When you take the repeated digit sum you are finding the remainder on division by $9$ except you are getting $9$ instead of $0$. This is the classic divisibility test for $9$. When you take the powers, you are effectively doing that $\bmod 9$. That shows why you get stuck with $3,6,9.$ Once you square it you have t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I order these numbers without a calculator?
*
*Classify the following numbers as rational or irrational. Then place them in order on a number line:
$$\pi^2, -\pi^3, 10, 31/13, \sqrt{13}, 2018/2019, -17, 41000$$
I know $\pi$ is irrational so $\pi^2$ and $-\pi^3$ are irrational. 10, 31/30, 2018/2019, -17, 410... | For the second list we can see that $5 \gt 2^2$ as $5 \gt 4$ so if we take both sides to the power of $12$ we get $5^{12} \gt 2^{24}$. Then we can compare $10^8$ and $2^{24}$, we can split $10$ into $2*5$, so $10^8=2^8*5^8$. We claim $10^8 \gt 2^{24}$, which leads to $2^8*5^8 \gt 2^{24}$, then taking $2^8$ over, $5^8 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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.