Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Integer as sum of 6 positive squares Can every integer be written as sum of exactly $6$ positive squares?
I am also curious to know when an integer can be written as sum of exactly 8 squares.
I know the problem is related to Waring-Hilbert Theorem, but I couldnn't find anything specific.
If it is not possible for e... | Here's a way to show that every sufficiently large integer is a sum of eight positive squares.
Legendre's three-square theorem guarantees (among much else) that every number congruent to $3$ mod $8$ is the sum of three squares, and it's easy to see that all three squares must be odd, hence nonzero. Any sufficiently la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\lim_{x\to1}=\frac{x^2+x-2}{1-\sqrt{x}}$? let $f(x)=\dfrac{x^2+x-2}{1-\sqrt{x}}$
How do I solve this limit?
$$\lim_{x\to1}f(x)$$
I can replace the function with its content
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}$$
Then rationalizing the denominator
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}\cdot\dfrac{1... | Alternatively you can alwasy factor the hard way:
$\frac {x^2 + x - 2}{-\sqrt x + 1}=$
$\frac {x^{\frac 32}(\sqrt{x} - 1) + x^{\frac 32} + x - 2}{-\sqrt x + 1}=$
$-x^{\frac 32} + \frac {x(\sqrt x-1) + 2x - 2}{-\sqrt x+ 1}=$
$-x^{\frac 32} -x + \frac {2\sqrt x(\sqrt x -1) +2\sqrt x - 2}{-\sqrt x +1}=$
$-x^{\frac 32} - x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Factoring $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$ I am trying to factor the expressions $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$.
I am rather stuck though. Is there a general method for going about this? I always find myself having to g... | Note that
$$(x+y+z)^3xyz-(xz+xy+yz)^3=(x^2-yz)(y^2-xz)(z^2-xy)$$
and
$$(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3=3 (b-a) (c-a) (a-x) (b-c) (b-x) (c-x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find all functions such that $f(1+xf(y))=yf(x+y)$ where $x,y \in R^+$
Find all functions run over positive real numbers such that
$f(1+xf(y))=yf(x+y)$ where $x,y\in R^+$
MY ANSWER:
Putting $x=y=0$,we get,
$f(1)=0$
Putting $x=0$ we get,
$f(1)=yf(y)$
or,$yf(y)=0$
or,$f(y)=0$ (since $y\ne 0$., $y \in \mathbb R^+$)
... | For reference, here is CanVQ's solution taken from here.
Replacing $x$ by $\frac{x}{f(y)}$ in $(1),$ we have
$$f(1+x)=y\cdot f\left(\frac{x}{f(y)}+y\right),\quad \forall x ,\,y \in \mathbb R^+. \quad (2)$$
Next, we replace $y$ by $\frac{f(1+x)}{y}$ in $(2)$ to get
$$y=f\left(\frac{x}{f\left(\frac{f(1+x)}{y}\rig... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{x^2y}{1+x^4+y^2}$ has no global minimum I am not sure how to approach this problem. The usual methods do not work to find a minimum.
I can see that, but how to show that there must not exist a minimum?
| $$f(x,y) = \frac{x^2y}{1+x^4+y^2}$$
$$\frac{\partial f}{\partial x} = \frac{2xy(1+x^4+y^2) - 4x^5y}{(1+x^4+y^2)^2}$$
$$\frac{\partial f}{\partial y} = \frac{x^2(1+x^4+y^2) - 2x^2y^2}{(1+x^4+y^2)^2}$$
$$\frac{\partial f}{\partial x} = 0 \implies 1+x^4+y^2 = 2x^4 \lor x=0 \lor y=0$$
$$\frac{\partial f}{\partial y} = 0 \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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Show that $\sin (x/2)+2\sin (x/4)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$
Given that $$f (x)=\sin (x/2)+2\sin (x/4)$$
Show that $f (x)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$.
My try
$f (x)=\sin (x/2)+2\sin (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}... | By AM-GM we obtain:
$$\sin\frac{x}{2}+2\sin\frac{x}{4}=2\sin\frac{x}{4}\left(1+\cos\frac{x}{4}\right)=\frac{2}{\sqrt3}\sqrt{\left(3-3\cos\frac{x}{4}\right)\left(1+\cos\frac{x}{4}\right)^3}\leq$$
$$\leq\frac{2}{\sqrt3}\sqrt{\left(\frac{3-3\cos\frac{x}{4}+3\left(1+\cos\frac{x}{4}\right)}{4}\right)^4}=\frac{3\sqrt3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\cos(3x) = \cos(2x)$ I'm struggling with solving given trigonometric equation
$$\cos(3x) = \cos(2x)$$
Let's take a look at the trigonometric identities we can use:
$$\cos(2x) = 2\cos^2-1$$
and
$$\cos(3x) = 4\cos^3(x) -3\cos(x)$$
Plugging into the equation and we have that
$$4\cos^3(x) -3\cos(x) = 2\cos^2(x)-1... | The equality $\cos(3x)=\cos(2x)$ is obviously true when $x=0$ and thus when $t=1.$ Therefore the polynomial
$$
4t^3-2t^2-3t +1
$$
has $t=1$ as one of its zeros. Consequently it can be factored:
$$
4t^3-2t^2-3t +1 = (t-1)(\cdots\cdots\cdots)
$$
The other zeros are those of a quadratic polynomial, written here as $(\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Finding limit using Euler number $$\lim_{x\rightarrow \infty }\left ( 1+\frac{3}{x+2} \right )^{3x-6}$$
I've tried to factor and simplfy the expression. I got:
$${\left ( 1+\frac{3}{x+2} \right )^{\frac{1}{x+2}}}^{3({x^2-4})}$$
I set $x$ to $1/t$ I get:
$${\left ( 1+\frac{3}{\frac{1}{t}+2} \right )^{\frac{1}{\frac{1}{t... | $\lim_\limits{x\to\infty}\left(1+\frac {3}{x+2}\right)^{3x-6}$
$y = x+2$
$\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{3y-12}\\
\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{3y}\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{-12}$
Let't attack these separately
$\lim_\limits{y\to\infty}\left(1+\frac ... | {
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"url": "https://math.stackexchange.com/questions/2826327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the number of ways in which I can select 17 balls from an urn containing 30 balls? I have an urn that contains 30 balls with 10 coloured white, 10 coloured black and the remaining coloured red. Each ball is numbered, from 1 to 10: that is I have red balls numbered 1 to 10, black balls numbered 1 to 10 and white... | There are 21 composition of 17:
$17 = 4+4+9 , 17 = 4+5+8, .... , 17=9+4+4$
However, some of have the same structure are equivalent and we may take 3 or 6 of them at a time:
$ 3 \binom{10}{4} \binom{10}{4} \binom{10}{9} +
3 \binom{10}{5} \binom{10}{5} \binom{10}{7} +
3 \binom{10}{6} \binom{10}{6} \binom{10}{5} ... | {
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"url": "https://math.stackexchange.com/questions/2826877",
"timestamp": "2023-03-29T00:00:00",
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Showing that $4b^2+4b = a^2+a$ has no non-zero integer solutions? The problem is
Show that $4b^2+4b = a^2+a$ has no integer solutions where none of $a, b$ are zero.
I have a solution but I think there must be some better ways:
My Solution:
$$4b^2+4b = a^2+a$$
$$(2b+a)(2b-a)+4b-a= 0$$
Now letting $x = 2b + a$ and $y =... | Left side looks almost like $(2b+1)^2$, indeed adding $1$ we can write $(2b+1)^2=a^2+a+1$. It is known that right side can be a perfect square only in few cases - see Integral value of $n$ that makes $n^2+n+1$ a perfect square.
Specifically, when $a>0$, the right side lies between two consecutive squares: $a^2<a^2+a+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Why doesn't the quadratic equation contain $2|a|$ in the denominator? When deriving the quadratic equation as shown in the Wikipedia article about the quadratic equation (current revision) the main proof contains the step:
$$
\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}
$$
the square root is taken fro... | If you put
$x
=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$
into
$ax^2+bx+c$,
since
$x^2
=\dfrac{b^2\mp2b\sqrt{b^2-4ac}+(b^2-4ac)}{4a^2}
=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a^2}
$
you get
$\begin{array}\\
ax^2+bx+c
&a\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a^2}
+b\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}+c\\
&=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 3
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For the differential equation given by $(x^2-y^2)dx+3xydy=0$ the general solution is $(x^2+2y^2)^3 =cx^2$
For the differential equation given by $$(x^2-y^2)dx+3xy \ dy=0$$ the general soultion is $$(x^2+2y^2)^3 =cx^2$$ Check this solution by differentiating and rewriting to get the original function.
I solved for $\f... | The hint says you should solve for $c$ before differentiating, so let's do that
$$ \frac{(x^2+2y^2)^3}{x^2} = c $$
Performing implicit differentiation on this form will eliminate $c$
$$ \frac{3x^2(x^2+2y^2)^2\left(2x + 4y\frac{dy}{dx}\right)-2x(x^2+2y^2)^3}{x^4} = 0 $$
Now you can solve for $\frac{dy}{dx}$ and finish t... | {
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"timestamp": "2023-03-29T00:00:00",
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Reciprocal solutions of a differential equation
I need to show that if $a$ is a constant and $b(x)$ is a function, then
$$y''+\frac{b'(x)}{b(x)}y'-\frac{a^2}{[b(x)]^2}y=0$$
has a pair of linearly independent solutions which are reciprocal and then find them.
I would have thought I could just substitute in $y=u(x)+\df... | You got so close. Suppose that $y_1=u$ is a solution, then $y_2=1/u$ must also be a solution. Substituting this, we obtain
$$ u'' -2\frac{(u')^2}{u} + \frac{b'}{b}u' + \frac{a^2}{b^2}u = 0 $$
But we already know that
$$ u'' + \frac{b'}{b}u' = \frac{a^2}{b^2}u $$
From the given assumption. Therefore
$$ 2\frac{a^2}{b^2}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the length of AB in Triangle ABC In $\Delta ABC , m \angle A = 2 m \angle C$ , side $BC$ is 2 cm longer than side $AB$ . $AC = 5 $What is $AB$ ?
Well I thought you can use trigonometry or Complete Pythagoras theorem , but I don't really know how to apply it
| Let $|AB|=c$, $|BC|=a=c+2$,
$|AC|=b=5$,
$\angle BCA=\gamma$,
$\angle CAB=\alpha=2\gamma$
By the sine rule we have
\begin{align}
\frac{\sin\alpha}{a}
&=
\frac{\sin\beta}{b}
=\frac{\sin\gamma}{c}
,\\
\frac{\sin2\gamma}{c+2}
&=
\frac{\sin(\pi-3\gamma)}{5}
=\frac{\sin\gamma}{c}
.
\end{align}
By the rules based on
comp... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Doubt with an inequality involving integrals I have to prove this inequality about integrals. I did it but I'm not sure if my arguments were correct. Take a look:
Prove:
$$0\leq\int_{-1}^1\frac{1-x^2}{x^4+1}dx\leq\int_{-1}^1\frac{2+x^4}{x^4+1}dx$$
My try is this:
Let $f(x)=\frac{1-x^2}{x^4+1}$ with $f:[-1,1]\to\mat... | $$\int\limits_{-1}^1\frac{2+x^4}{x^4+1}dx-\int\limits_{-1}^1\frac{1-x^2}{x^4+1}dx=\int\limits_{-1}^1\frac{x^4+x^2+1}{x^4+1}dx>0$$
$$\int\frac{1-x^2}{x^4+1}dx=-\int\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx=-\int\frac{1}{\left(x+\frac{1}{x}\right)^2-2}d\left(x+\frac{1}{x}\right)=$$
$$=\frac{1}{2\sqrt2}\int\left(\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding probability of ball being white using Bayes theorem? From an urn containing 3 white and 5 black balls, 4 balls are transferred into an empty urn. From this urn 2 balls are taken and they both happen to be white. They are then replaced. What is the probability that the third ball taken from the same urn will be ... | Let $X$ denote the number of white balls initially transferred into the urn from which you are drawing balls. Then, before you have drawn any balls, $X$ has a hypergeometric distribution: $$\Pr[X = x] = \frac{\binom{3}{x}\binom{5}{4-x}}{\binom{8}{4}}, \quad x \in \{0, 1, 2, 3\}.$$ After you draw two balls from this ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In how many ways can a committee of $5$ members be formed from $4$ women and $6$ men such that at least $1$ woman is a member of the committee.
In how many ways can a committee of $5$ members be formed from $4$ women and $6$ men such that at least $1$ woman is a member of the committee.
I know that the correct answer... | The number of groups with exactly $k$ of the four women and $5 - k$ of the six men is
$$\binom{4}{k}\binom{6}{5 - k}$$
so the number of groups with exactly $k$ women is
\begin{align*}
\sum_{k = 1}{4} \binom{4}{k}\binom{6}{5 - k} & = \binom{4}{1}\binom{6}{4} + \binom{4}{2}\binom{6}{3} + \binom{4}{3}\binom{6}{2} + \bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2837701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $5^{2^n} - 1$ is divisible by $2^{n+1}$ for all $n ≥ 1$ I made the following using induction:
If $n=1$ then the proposition is true: $5^{2^1} - 1=24$ is divisible by $2^{1+1} = 4$
Now I suppose that for a natural number $k$, $5^{2^k} - 1$ is divisible by $2^{k+1}$ is true. And I want to prove (using this... | First assume that $5^{2^k} - 1$ is divisible by $2^{k + 1}$ so $2^{k + 1}\,|\,5^{2^k} - 1$. Now consider
$$5^{2^{k + 1}} - 1 = 5^{2\cdot 2^k} - 1 = (5^{2^k})^2 - 1 = (5^{2^k} - 1)(5^{2^k} + 1)$$
Now we need to show that $2^{(k + 1) + 1} = 2^{k + 2}$ divides $(5^{2^k} - 1)(5^{2^k} + 1)$. Our inductive assumption allows ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine Minimum Value.
Find the minimum value of
$$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$$
for $x>0$.
When $x=1$, $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=6$$
I tried to plot some points on a graph and I observed that the minimum value is $6$.
Any hints would be sufficient.
Thanks... | A nice trick:
As $x>0$, $$f(x)=\frac{\left(x+\cfrac1x\right)^6-\left(x^6+\cfrac1{x^6}\right)-2}{\left(x+\cfrac1x\right)^3+\left(x^3+\cfrac1{x^3}\right)}=\frac{6x^4+15x^2+18+\cfrac{15}{x^2}+\cfrac6{x^4}}{2x^3+3x+\cfrac3x+\cfrac2{x^3}}$$ giving $$f(x)=\frac{6x^8+15x^6+18x^4+15x^2+6}{2x^7+3x^5+3x^3+2x}=\frac{6x^8+9x^6+9x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Correct way of finding $\delta $ for $\lim_{x \to a} \sqrt{x} = \sqrt{a}$ Prove: $\lim_{x \to a} \sqrt{x} = \sqrt{a}$ using $\epsilon$-$\delta$.
My solution:
We have that $0 < |x-a| < \delta $.
Also, $ |\sqrt{x} - \sqrt{a}| < \epsilon $
$ \therefore -\epsilon < \sqrt{x} - \sqrt{a} < \epsilon $
$ \therefore \sqrt{a} -\... | The OP's logic can be salvaged. One problem is they assume on their third $\therefore$ that $u \lt v$ implies $u^2 \lt v^2$ without knowing for sure that $u \ge 0$.
Here is a 'logic patch':
If $a = 0$, simple algebra shows that $\delta = \varepsilon^2$ works.
To show continuity when $a \gt 0$, we can take a few easy st... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum $e$ where $a,b,c,d,e$ are reals such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ I have a question about this 1978 USAMO problem:
Given that $a,b,c,d,e$ are real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the maximum value that $e$ can attain.
I had the following solution:
Let $a+... | By the RMS-AM inequality:
$$
\dfrac{a^2+b^2+c^2+d^2}{4} \ge \left(\dfrac{a+b+c+d}{4}\right)^2 = \dfrac{(8-e)^2}{16} \tag{1}
$$
Therefore:
$$
e^2 = 16-(a^2+b^2+c^2+d^2) \le 16 - \dfrac{(8-e)^2}{4} \;\;\iff\;\; e(5e-16) \le 0 \tag{2}
$$
Equality is in fact attained for $\,e = \dfrac{16}{5}\,$ and $\,a=b=c=d\,$.
Where am... | {
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"source": "stackexchange",
"question_score": "3",
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If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ are $7$ and $1/7$ , then the value of $c$ is?
I cross multiplied the equation and tried to find it's discriminant, but I don't think it gets me anywhere. A little hint would be appre... | Solution
First, we should constrain the value of $c$ such that $x^2+3x+c \neq 0$ for all $x \in \mathbb{R}.$ Otherwise, there necessarily exists at least one infinite discontinuity for $y=f(x)$, and if so, there exists no minimum or maximum value for $y$. For this purpose, let $\Delta=9-4c<0$, i.e. $c>\dfrac{9}{4},$... | {
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Error Correcting Polynomial Code I apologize for the screenshot, it's quite a long passage. I've linked the textbook below, page number 428-429, example 16.4.3 http://faculty.uml.edu/klevasseur/ads2
I constructed the power table for $GF(8)$, but honestly I'm extremely confused on the two things being asked of me:
Sup... | $\newcommand{\b}{\beta}$We bear in mind $\b^3 = \b + 1$.
Given $1001$, we need to reduce $1 \cdot \b^3 + 0 \cdot \b^4 + 0 \cdot \b^5 + 1 \cdot \b^6$.
We compute $\b^6+\b^3 = \b^4+\b^3+\b^3 = \b^4 = \b^2 + \b$, so we transmit $0111001$.
We received $1010101$.
We check if $\b^0+\b^2+\b^4+\b^6 = 0$:
$\b^0+\b^2+\b^4+\b^6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Triples $(x, y, z)$ that satisfy a set of equations
Suppose that $a$ is a fixed (but unknown) real number, with $a^2 \neq 1$. Determine all triples $(x, y, z)$ of real numbers that satisfy the system of equations:
$x + y + z = a$
$xy + yz + xz = -1$
$xyz = -a$
I've tried making substitutions but don't seem to be ab... | A more systematic way: Use the third equation to eliminate $z$ in the second equation,
$$xy-a/x -a/y = -1$$
$$z = -a/xy \Rightarrow x+y-\frac{a}{xy} = a$$
Now spot $-a/x - a/y = -a\frac{x+y}{xy}$.
So the first equation can be written as $$a(x+y) = (xy)^{2}+(xy)$$
Getting rid of $x+y$, the second equation becomes
$$(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Evaluate the Determinants A? Evaluate the Determinants
$$A=\left(\begin{matrix}
1 & 1 & 0 & 0 & 0\\
-1 & 1 & 1 & 0 & 0\\
0 & -1& 1 & 1 & 0\\
0& 0 & -1 & 1 & 1\\
0 & 0 & 0 & -1 & 1\\
\end{matrix}\right)$$
My attempst : I was thinking abouts the
Schur complement https://en.wikipedia.org/wiki/Schur_co... | After adding the first line to the second one, your matrix becomes$$\begin{pmatrix}1 & 1 & 0 & 0 & 0\\ 0 & 2 & 1 & 0 & 0\\ 0 & -1& 1 & 1 & 0\\ 0& 0 & -1 & 1 & 1\\ 0 & 0 & 0 & -1 & 1\end{pmatrix}$$and therefore$$\det A=\begin{vmatrix}2 & 1 & 0 & 0\\ -1& 1 & 1 & 0\\ 0 & -1 & 1 & 1\\ 0 & 0 & -1 & 1\end{vmatrix}=2\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Difficulty finding Lagrange multiplier because of $\leq$
Let $f: \mathbb R^3 \to \mathbb R$ be defined by $$f(x,y,z)=x-y+z$$
and $$E:=\{(x,y,z)\in \mathbb R^{3} \mid x^2+2y^2+2z^2\leq1\}$$ Find the extrema of $f$ on $E$.
Path: I have already proven that $E$ is compact and then defined $g_\lambda:=f-\lambda h$ wher... | You can handle the inequality condition introducing a slack variable $\epsilon$ such that
$$
E\to x^2+2y^2+z^2+\epsilon^2 = 1
$$
and now the lagrangian formulation gives
$$
L(x,y,z,\lambda,\epsilon) = f(x,y,z) + \lambda(x^2+2y^2+z^2+\epsilon^2 -1)
$$
and the stationary conditions give
$$
\nabla L = \left\{
\begin{array... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What can we say about the series $\sum_{n=1}^{\infty}n\cdot (1-\cos(\frac{a}{n}))$ Series $\sum_{n=1}^{\infty}n\cdot (1-\cos(\frac{a}{n}))$
$$\sum_{n=1}^{\infty}n\cdot(1-\cos(\frac{a}{n}))=\sum_{n=1}^{\infty}n\cdot[1- \{1-\frac{a^2}{2!\cdot n^2} + O(\frac{1}{n^4})\}] = \sum_{n=1}^{\infty} \frac{a^2}{2n} + O(\frac{1}{n^... | Another way, using
$\frac{2x}{\pi}
\le \sin(x)
\le x$
for
$0 \le x \le \pi/2$:
$\begin{array}\\
\sum_{n=1}^{m}n(1-\cos(\frac{a}{n}))
&=\sum_{n=1}^{m}n(2\sin^2(\frac{a}{2n}))\\
&=2\sum_{n=1}^{m}n\sin^2(\frac{a}{2n})\\
&\ge 2\sum_{n=1}^{m}n(\frac{2a}{2\pi n})^2
\qquad\text{for }\frac{a}{2n} \le \frac{\pi}{2}
\text{ or }n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Series Convergence of Harmonic Means
Let $\{x_n\}$ be a sequence of real numbers such that $0< x_1 <x_2$. If $$x_n= \frac{2}{\frac{1}{x_{n-1}}+\frac{1}{x_{n-2}}}, $$then show that
$$\lim_{n\to\infty}x_n=\frac{3x_1x_2}{2x_1+x_2}.$$
| Solution
Since $$x_n = \frac{2}{\dfrac{1}{x_{n-1}} + \dfrac{1}{x_{n-2}}},$$
then $$ \frac{2}{x_{n}}=\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}.$$
Thus, $$2y_n=y_{n-1}+y_{n-2},$$ where $y_n=\dfrac{1}{x_n}$.
From $(1)$, we may obtain $$y_n-y_{n-1} = -\frac{1}{2}\left(y_{n-1}-y_{n-2}\right),$$ for $n=3,4,\cdots.$
Hence，$$y_n-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area of a quadrilateral inside a square In the image, the segments inside the square go from a vertex to the middle point of the opposite side. If the length of the sides of the square is $1$, the area of $ABCD$ is?
Any hints?
I tried some things matching the bigger triangles but i can't get the area of $ABCD$.
| If we move the figure to a $xy$ axis, we can solve this problem by finding the line equations and intersection points. Consider this image:
Since we know the point coordinates in the image, we can easily find the line equation of lines 1, 2 and 3 (we can substitute the point values in the slope-intercept form equation... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got a... | Note that $$u=\frac{b+1}{\sqrt{1-b^2}+b-2}$$ so $$\frac{du}{db}=\frac{1\big(\sqrt{1-b^2}+b-2\big)-(b+1)\left(-\frac{2b}{\sqrt{1-b^2}}+1\right)}{\big(\sqrt{1-b^2}+b-2\big)^2}$$ and setting to zero gives $$-3\sqrt{1-b^2}+b+1=0\implies 1-b^2=\frac{b^2+2b+1}9\implies 5b^2+b-4=0$$ and we see that $b=4/5,-1$ are roots.
Check... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
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Minimizing in 3 variables
Find the least possible value of the fraction $\dfrac{a^2+b^2+c^2}{ab+bc}$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3\sqrt[3]{-2ac/b(a+c)}$
And I cant somehow move on.
| Write
$$\frac{2a^2+b^2+b^2+2c^2}{2}=\frac{2a^2+b^2}{2}+\frac{b^2+2c^2}{2}\geq \sqrt{2}ab+\sqrt{2}bc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Trigonometry and quadratics : Possible mismatch? There’s this problem I came across, gives me an invalid answer by using general quadratic formula. Wonder why?
$2\sin^2{x} -5\cos{x} -4 =0 $
Here’s what I did:
$2\sin^2{x} -5\cos{x} -4 =0 $
$2(1-\cos^2{x}) - 5 \cos{x} - 4 = 0$
$2 \cos^2{x} + 5 \cos{x} + 2 =0$
This is a q... | Yes from here (you are right) we have:
$$2\sin^2x -5\cos x -4 =0 \implies 2\cos^2 x+5\cos x +2=0$$
that is by quadratic formula (here was your mistake):
$$\cos x = \frac{-5\pm \sqrt{25-16}}{4}= \frac{-5\pm 3}{4}\implies \cos x=-\frac12$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$
Calculate:$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$$
Even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.
My own solution
$$\lim_{n\to\i... | Two alternative ideas:
(1) Write
$$
\sqrt{5n^2+4} - \sqrt{5n^2+n} =
\sqrt 5n(\sqrt{1 + 4/5n^2} - \sqrt{1 + 1/5n})
$$
and use Taylor ($\sqrt{1 + x} = 1 + x/2 - x^2/8 + \cdots$).
(2) Let be $f(x) = \sqrt x$. By the Mean Value Theorem, for some $c_n\in(5n^2 + 4,5n^2 + n)$
$$
\sqrt{5n^2 + 4} - \sqrt{5n^2 + n} =
f'(c_n)((5n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to prove that $\pi=\pi$? I am trying to prove that indeed $\pi=\pi$. More precisely, that:
$$6\sum_{n=0}^\infty \frac{(2n)!}{4^n (n!)^2 (2n+1) 2^{2n+1}}=\pi$$
Where the definition of $\pi:$
$$\pi=4\sum_{n=1}^\infty\frac{(-1)^{(n+1)}}{(2n-1)}$$
Using the epsilon delta definition, we should prove:
$$\forall \epsilon_... | Consider the expression
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{2n+1}}{2n+1}.
\end{eqnarray*}
Now use
\begin{eqnarray*}
\frac{x^{2n+1}}{2n+1}= \int_0^x x^{2n} dx
\end{eqnarray*}
and
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n}{n} x^{2n} = \frac{1}{\sqrt{1-4x^2}}.
\end{eqnarray*}
Invert the or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Help summing the telescoping series $\sum_{n=2}^{\infty}\frac{1}{n^3-n}$. I know a priori that the series $$\sum_{n=2}^{\infty}\frac{1}{n^3-n}$$ converges. However, I am tasked with summing the series by treating it as a telescoping series.
By partial fraction decomposition, the series can be written as: $$\sum_{n=2}^... | Alt. hint: it is not necessary to do a full partial fractions decomposition, it's enough to telescope:
$$
\frac{1}{n^3-n} = \frac{1}{2}\frac{(n+1) - (n-1)}{(n-1)n(n+1)} = \frac{1}{2}\left(\frac{1}{(n-1)n} - \frac{1}{n(n+1)}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $... | Use the Maclaurin series
$$\cos x = 1-1/2 \; x^2+1/24 \; x^4 +O(x^6)$$
$$\cos 3x = 1-9/2 \; x^2+27/8 \; x^4 +O(x^6)$$
$$\sin x^2 = x^2-1/6\; x^6+O(x^8)$$
$$\sin 3x^2 = 3x^2-9/2 \; x^6+O(x^8)$$
then quotient is
$$\frac{-1/2 + 9/2}{3-1} + O(x^2)$$
and therefore the limit is 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 4
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Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2 \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{... | We wish to find the values of $c$ such that the following simultaneous equations have real solutions to $a$ and $b$:
$$\begin{cases}a+b&=&2-c\\a^2+b^2&=&12-c^2\end{cases}$$
Labelling the equations as $(1)$ and $(2)$ respectively, $(1)^2 - (2)$ gives $2ab
= -8-4c+2c^2$, so $(a-b)^2 = a^2+b^2 - 2ab = (12-c^2) - (-8-4c+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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Application of Chebyschev inequality
I want to prove the inequality above. On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble.
My Attempt:
Since Chebyschevs inequalty is
We can square the left hand side of our inequality with the... | As you've noted here the use of the Chebyshev's Inequality doesn't help much. However we can make use of the AM-GM inequality.
Squaring both sides and expanding them we get that the inequality is equivalent to: $(a+b+c)^2 \ge 3(ab+bc+ca)$, which is equivalent to $a^2 + b^2 + c^2 \ge ab + bc + ca$.
The last equality is ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is the domain of the function $f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$?
What is the domain of the function
$f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$?
I started using the fact $-\frac{\pi}{2}\le f(x) \le \frac{\pi}{2}\implies-1 \le\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\le1$.Now,... | Hint:
Let $3^{x-2}=a\implies a>0$ for real $x$
$$-1\le\dfrac{8a}{1-9a^2}\le1$$
$$\dfrac{8a}{1-9a^2}\le1\iff0\le\dfrac{9a^2+8a-1}{9a^2-1}=\dfrac{(9a-1)(a+1)}{(3a-1)(3a+1)}$$
As $a.0,$ we need $\dfrac{9a-1}{3a-1}\ge0$
$\implies$ either $9a-1=0\iff a=?$ or $ a>$max$\left(\dfrac19,\dfrac13\right)$ or $a<$min$\left(\dfrac19... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get i... | Hint: The solutions of your polynomial are given by
$$x_{1,2}=-\frac{3}{2}\pm\frac{\sqrt{13}}{2}$$
plugging $x_1$ into the other equation we get
$$b\sqrt{13}+3c\sqrt{13}-3\sqrt{13}+2a+3b+11c+11=0$$
the rest is for you!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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determine whether the following series converge or diverge? determine whether the following series converge or diverge ?
$a)$ $\sum_{n=1}^{\infty} (\frac {n}{n+1})^{n(n+1)}$
$b)$ $\sum_{n=1}^{\infty} (\frac {n^2 +1 }{n^2 +n +1})^{n^2}$
My attempts : for $a)$ $(\frac {n}{n+1})^{n(n+1)}= (1- \frac{1}{n+1})^{n(n+1)}= ... | By root test we have that
$$\sqrt[n]{a_n}=\left(\frac {n}{n+1}\right)^{n+1}=\left(1-\frac {1}{n+1}\right)^{n+1}\to \frac1e$$
$$\sqrt[n]{b_n}=\left(\frac {n^2 +1 }{n^2 +n +1}\right)^{n}=\left[\left(1-\frac {n}{n^2+n+1}\right)^{\frac{n^2+n+1}{n}}\right]^{\frac{n^2}{n^2+n+1}} \to \frac1e$$
therefore $\sum a_n$ and $\sum b... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that: $xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$ Let $x$,$y$ and $z$ are positive and $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq 3$$
Prove that: $$xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$$
The things I have done so far
$$3\geq \sum \limits_{cyc}\frac{1}{x}\geq \frac{9}{\sum \limits_{cyc}x}\Rightarrow \sum \li... | We need to prove that
$ xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y}\geq x+y+z $
$\Leftrightarrow \frac{xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y}}{xyz}\geq \frac{x+y+z}{xyz}$
$\Leftrightarrow \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\geq \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}$
$\Leftrightarrow \frac{1}{x^{2}}+\frac{1}{y^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Evaluate $I_n=\int_0^{\pi/2} \frac{1}{\left( a\cos^2x+b\sin^2x\right)^n} \, dx$ I would like to evaluate (using elementary methods if possible) : (for $a>0,\ b>0$)
$$
I_n=\int_0^{\pi/2} \frac{1}{( a\cos^2x+b\sin^2x)^n} \, dx,\quad \ n=1,2,3,\ldots
$$
I thought about using $u=\tan(x)$ or $u=\frac{\pi}{2}-x$ but did n... | This is not an answer but it is too long for a comment.
For the antiderivative, Wolfram Alpha (and other CAS) return, for $$J_n(x)=(n-1)(a-b) I_n(x)$$s the messy expression
$$2^{n-1} \csc (2 x) \sqrt{\frac{(a-b) \sin ^2(x)}{a}} \sqrt{\frac{(b-a) \cos ^2(x)}{b}}$$ $$
((a-b) \cos (2 x)+a+b)^{1-n}$$ $$
F_1\left(1-n;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the quotient and the remainder of $(n^6-7)/(n^2+1)$
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.
So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
| We have $n^6-7 = (n^2+1)q(n)+an+b$, with $a,b \in \mathbb Z$, because $n^2+1$ is monic.
Plugging $n=\pm i$, we get
$$
-8 = \pm ai+b
$$
which implies $a=0$ and $b=-8$. Therefore,
$$
q(n) = \frac{n^6-7+8}{n^2+1} = \frac{n^6+1}{n^2+1} = n^4 - n^2+ 1
$$
Thus,
$$
n^6-7 = (n^2+1)(n^4 - n^2+ 1)-8
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2... | Hint:
$$\dfrac2{x^4+ax^2+1}=\dfrac{1-1/x^2}{x^2+a+1/x^2}+\dfrac{1+1/x^2}{x^2+a+1/x^2}$$
$$x^2+1/x^2=(x-1/x)^2+?=(x+1/x)^2-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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How to prove that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4$? Using the Cardano formula, one can show that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$ is a real root of the depressed cubic
$f(x)=x^3-6x-40$. Actually, one can show by the calculating the determinant that this is the only real root. On the ... | Can we express $\sqrt[3]{20+14\sqrt2}$ as $a+b\sqrt2$, with $a,b\in\mathbb Z$? In other words, are thre integers $a$ and $b$ such that $(a+b\sqrt2)^3=20+14\sqrt2$? Note that\begin{align}(a+b\sqrt2)^3=20+14\sqrt2&\iff a^3+3\sqrt2a^2b+6ab^2+2\sqrt2b^3=20+14\sqrt2\\&\iff a^3+6ab^2+(3a^2b+2b^3)\sqrt2=20+14\sqrt2.\end{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the roots of $3x^3-4x-8$ It is given that $\alpha$, $\beta$ and $\gamma$ are the roots of the polynomial $3x^3-4x-8$.
I have been asked to calculate the value of $\alpha^2 + \beta^2 + \gamma^2$.
However I am unsure how to find these roots, seeing as though I haven't been given a root to start with.
I began by ide... | $ x(3x^2-4)=8$
Squaring we get $x^2(3x^2-4)^2=8^2$
Let $x^2=y\implies y(3y-4)^2=64\iff9y^3-24y^2+16y-64=0$
whose roots are $a^2,b^2,c^2$
$\implies a^2+b^2+c^2=\dfrac{24}9=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Differential equation: $(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $
$(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $
Attempt:
After expanding, everything is neat except: $3x^2y dy - 3y^2 x dx$
I can't convert it to exact differential.
Also, there's weird symmetry in the equation wrt the coefficients of $x^2$ and... | Given $(2x^2+3y^2-7)x\ dx=(3x^2+2y^2-8)y\ dy$
Let us take $$X=x^2$$$$Y=y^2$$and we get$$\dfrac{dY}{dX}=\dfrac{2X+3Y-7}{3X+2Y-8}$$Again let us consider $$X=p+a$$$$Y=q+b$$$$\implies\dfrac{dq}{dp}=\dfrac{2p+2a+3q+3b-7}{3p+3a+2q+2b-8}=\dfrac{2p+3q}{3p+2q}$$From $2a+3b-7=0$ and $3a+2b-8=0$ we get $\implies a=2,b=1$
$$\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Solve $ \binom{a}{2} + \binom{b}{2} = \binom{c}{2} $ with $a,b,c \in \mathbb{Z}$ I am trying to solve the Diophantine equation:
$$ \binom{a}{2} + \binom{b}{2} = \binom{c}{2} $$
Here's what it looks like if you expand, it's variant of the Pythagorean triples:
$$ a \times (a-1) + b \times (b-1) = c \times (c-1) $$
I was ... | After some algebra you can rewrite the equation as
$$
a(a-1)=(c-b)(c+b-1)
$$
Note that $c-b$ and $c+b-1$ have opposite parity. Conversely, if $p$ and $q$ are integers with opposite parity, then we can find integers $b,c$ such that $p=c-b$, $q=c+b-1$: namely, $b=\frac{q+1-p}{2}$ and $c=\frac{q+1+p}{2}$.
That is to say, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Partial fraction of $\frac{2s+12}{ (s^2 + 5s + 6)(s+1)} $ then inverse transform it
Find the inverse Laplace transform of
$$\mathcal{L}^{-1}\left(\frac{2s+12}{ (s^2 + 5s + 6)(s+1)}\right)$$
I recognise I need to use partial fractions to solve it and that is where I got stuck. Here’s my working,
After factoring the... | There is a shortcut way here, in this simple case. Let to mutiply the relation to $\color{red}{s+1}$, then
$$\frac{2s+12}{ (s+2)(s+3)(s+1)} \color{red}{(s+1)}= \frac{A\color{red}{(s+1)}}{s+2} + \frac{B\color{red}{(s+1)}}{s+3} + \frac{C\color{red}{(s+1)}}{s+1}$$
or
$$\frac{2s+12}{ (s+2)(s+3)} = \frac{A\color{red}{(s+1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Roots of unity and large expression
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}.$$
I have tried combining the first and third terms & first and last ter... | $\frac{w}{1+w^2 }+\frac{w^3 }{1+w} +\frac{w^2 }{1 +w^4}+\frac{w^4}{1+w^3 }=\frac{w+w^2 +w^3 +w^5}{(1+w)(1+w^2) }+\frac{w^2 +w^5+w^4+w^8}{(1+w^4)(1+w^3) }=\frac{w+w^2 +w^3 +w^5}{w+w^2 +w^3 +1}+\frac{w^2 +1+w^4+w^3 }{1+w^3 +w^4+w^2} =2$
Because $w^5 =1\Rightarrow w^7=w^2$ and $w ^8 = w^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there a smarter way to differentiate the function $f(x) = \sin^{-1} \frac{2x}{1+x^2}$?
Given $f(x) = \sin^{-1} \frac{2x}{1+x^2}$,
Prove that $$f'(x) = \begin{cases}\phantom{-}\frac{2}{1+x^2},\,|x|<1 \\\\ -\frac{2}{1+x^2},\,|x|>1 \end{cases}$$
Obviously the standard approach would be to use the chain rule and simpl... | We have
$$ \frac{t}{1+\sqrt{1-t^2}} \Bigg \rvert_{t = \frac{2x}{1+x^2}} = \frac{2x}{1+x^2 + \lvert 1-x^2\rvert} = \begin{cases} x &\!\!\!, |x|<1 \\ \frac{1}{x} &\!\!\!, |x|>1 \end{cases} \, .$$
Therefore we can use
$$ \arcsin(t) = 2 \arctan\left(\frac{t}{1+\sqrt{1-t^2}}\right) \, ,$$
which follows from the half-angle f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Prove that $\frac{1}{2}+ \sum_{k=1}^n \cos (k\theta) = \sin((n+ \frac{1}{2})\theta)/2\sin\frac{\theta}{2}$ Suppose $\sin \frac{\theta}{2} \neq 0$ . Prove that
$$\frac{1}{2}+ \sum_{k=1}^n \cos (k\theta) = \frac{\sin[(n+ \frac{1}{2})\theta]}{2\sin\frac{\theta}{2}}$$
The question also give the hint,
$$z=\cos\theta + i\sin... | \begin{equation}
\cos k \theta
=
\frac{e^{jk\theta} + e^{-jk\theta}}{2}
\end{equation}
Let's use $\sum\limits_{k=0}^{N-1 }r^k= \frac{1-r^N}{1-r} $
\begin{equation}
\sum \cos k \theta
=
\frac{1}{2}
\sum\limits_{k=1}^{N}
e^{jk\theta} +
\frac{1}{2}
\sum\limits_{k=1}^{N}
e^{-jk\theta}
=
\frac{1}{2}
\big(
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$
Solve
$$8\sin x=\dfrac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$$
My approach is as follow
$8 \sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}$
On squaring we get
$64 \sin^2 x+\frac{1}{\sin^2 x}-16=\frac{3}{\cos^2 x}$
$(64\sin^4 x-16\sin^2 x+1)(1-\sin^2 x)=3 \sin^2 ... | Nice job so far.
Notice that every exponent is even, so you can further substitute $t^2=y$, to obtain a cubic equation in $y$, which you can solve by the Cardano formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
The remainder when $x^2$ $+ ax +4$ is divided by $x-4$ is four times the remainder when the same expression is divided by $x-1$. Find the value of $a$ According to my textbook, the answer to this is $-2$.
Isn't this true for any value of a though?
I started by doing the following:
$4a+20=4x$
$a+5 = x$
I'm not sure... | (This is not an answer, but it is too long for a comment.)
Let $f(x) = x^2 + ax + 4$.
I can't make any sense of what you claim to get on dividing $f(x)$ by $x-4$, "$(x-4)(x+4+a)(4a+20)$" or on dividing $f(x)$ by $x-1$, "$(x-1)(x+1+a)(a+5)$". You also claim the first is the second multiplied by four, which is evidently... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x... | As
$$
a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=\sqrt{a}-\sqrt{b}
$$
then with $a\ne b$ it gives $\sqrt{a}+\sqrt{b}=1$. Now you have to maximize
$$
(\sqrt{a})^2+(\sqrt{b})^2
$$
subject to $\sqrt{a}+\sqrt{b}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Show that $3y''+4xy'-8y=0$ has an integral which is a polynomial in x. Deduce the general solution.
Show that $3y''+4xy'-8y=0$ has an integral which is a polynomial in x. Deduce the general solution.
In this problem, i tried in direct method. But i could not do. So, i did in following way
Since order is 2, i am assum... | If $y(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ is a solution with $a_n\neq 0$, then note that $3\,y''(x)$ is of degree $n-2$, but $4x\,y'(x)$ and $-8\,y(x)$ are of the same degree $n$. As $3\,y''(x)+4x\,y'(x)-8\,y(x)=0$, the leading terms of $4x\,y'(x)$ and $-8\,y(x)$ must cancel. This proves that
$$4n\,a_n\,x^n-8\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate the integral $\int \frac{a\tan^2{x}+b}{a^2\tan^2{x}+b^2} dx$? How to calculate the integral $\displaystyle\int \dfrac{a\tan^2{x}+b}{a^2\tan^2{x}+b^2} dx$?
It seems like a $\arctan$ of what else... but I can not work it out.
I have previously asked how to calculate the definite integral between $0$ and ... | Note that:
$$\begin{align}
\left(a+b\right)\left(a\tan^2{x}+b\right) & =a^2\tan^2{x}+b^2+ab\left(1+\tan^2{x}
\right) \\[5pt] & = \color{blue}{a^2\tan^2{x}+b^2}+ab\sec^2x\end{align}$$
So we then have:
$$\begin{align}\int \frac{a\tan^2{x}+b}{\color{blue}{a^2\tan^2{x}+b^2}} \,\mbox{d}x
& = \frac{1}{a+b}\int\frac{\color{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Solve the system $x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$
Find all real numbers $x,\ y,\ z$ that satisfy $$x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$$
First natural move would be rewriting the system as:
$$x^{2}+y^{2}+2xy=4z-1$$
$$x^{2}+z^{2}+2xz=4y-1$$
$$z^{2}+y^{2}+2zy=4x-1$$
Thus,
$$2x^{2}+2... | Note that these equations are defined only for $x,y,z\ge 1/4$.
Consider the first two equations
$$x+y=\sqrt{4z-1}$$
$$x+z=\sqrt{4y-1}$$
Subtracting these two equations gives us
$$y-z=\sqrt{4z-1}-\sqrt{4y-1}$$
If $y\gt z$, then the LHS is positive but the RHS is negative, so this cannot be. Similarly, if $y\lt z$, then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Find out when $(x + y)^5 = x^5 + y^5$ (Spivak's Calculus Book) I know that I can see that this match with $y = 0$, $x = 0$ and $y = -x$.
But, the author of the book says:
"Hint: From the assumption $(x + y)^5 = x^5 + y^5$ you should be able to derive the equation $x^3 + 2x^2y + 2xy^2 + y^3 = 0$ if $xy \neq 0$. This imp... |
Hint: From the assumption $(x + y)⁵ = x⁵ + y⁵$ you should be able to derive the equation $x³ + 2x²y + 2xy² + y³ = 0$ if $xy \neq 0$.
Using the binomial expansion:
$$\require{cancel}
\begin{align}
(x+y)^5=x^5+y^5 \;\;&\iff\;\; \cancel{x^5} + 5x^4y +10x^3y^2 + 10x^2y^3 + 5xy^4 + \bcancel{y^5} = \cancel{x^5}+\bcancel{y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\c... | If $f(x)\ge0$ and $g(x)\ge0$, for all $x$, then
$$
f(x)+g(x)=0
\qquad\text{if and only if}\qquad
f(x)=0\text{ and }g(x)=0
$$
Take $f(x)=(x^2-1)^2$. This equals zero only at $-1$ and $1$.
If $g(x)=2x^2\cos^2\bigl(\frac{\pi x}{2}\bigr)$, is $g(1)=0$ or $g(-1)=0$? No other values of $x$ can make $f(x)+g(x)=0$.
A different... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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Stirling type formula for Sum on $\ln(n)^2$ Is there a similar formula like the Stirling one on the sum over $\ln(n)$ (take logarithms on its factorial representation):
$$\sum_{n=1}^N \ln(n) = N \ln(N)-N+\ln(N)/2+\ln(2\pi)/2+\mathcal{O}(\ln(N)/N)$$
but on the sum over its squares?
$$\sum_{n=1}^N (\ln(n))^2$$
I alread... | We can use the Euler-Maclaurin formula to obtain the asymptotic expansion
$$ f_N (a) \equiv \sum \limits_{n=1}^N n^a \sim \zeta(-a) + \sum \limits_{k=0}^\infty \frac{B_k^*}{a+1} {a+1 \choose k} N^{a+1-k} $$
for $N \in \mathbb{N}$ and $a \in \mathbb{R} \setminus \{-1\}$ . Here $(B_k^*)_{k\in \mathbb{N}_0}$ are the Berno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \c... | Recall that $w\in \mathbb{C}$ is purely imaginary $\iff w=-\bar w$, then
$$\frac{z-1}{z+1}=-\overline{\left(\frac{z-1}{z+1}\right)} =-\frac{\bar z-1}{\bar z+1} \iff (z-1)(\bar z+1)=-(\bar z-1)(z+1)$$
$$z\bar z+z-\bar z-1=-z\bar z+z-\bar z+1$$
$$2z\bar z=2 \iff|z|^2=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if $q=3$?
Question: If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if
$q=3.$
I am certain the answer to the question is yes, and I believe I have worked out a solution - although I am not certain it ... | You have $q=4m+3$, so $m=\frac{q-3}{4}$
\begin{align}
{q+1 \choose 2}={4m+4 \choose 2}&=(2m+2)(4m+3)=2(m+1)(4m+3)\\
&=2q(m+1)
\end{align}
When $m=0$ we have ${3+1 \choose 2}=6=2\cdot3$ as required. Now you always get $2$ and $q$ as two distinct factors along with $m+1$. Now $m+1$ is never equal to $q$, and equals $2$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Find a formula for $\left(\begin{smallmatrix} -4 & -15 \\ 2 & 7 \end{smallmatrix}\right)^n$ We're going to consider the matrix $\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$
(a) Let $\mathbf{P} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$. Find the $2 \times 2$ matrix $\mathbf{D}$ such that
$\mathbf{P}^{-1} \mat... | Let consider
$$\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} x & y \\ z & w \end{pmatrix}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}=\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\\$$
$$\iff
D=\begin{pmatrix} x & y \\ z & w \end{pmatrix}=\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} -4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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$ABCD$ a tetrahedron. $P$ and $Q$ be points on the sides $AB$ and $CD$ respectively. What is the locus of the midpoint of the line segment $PQ$? Let $ABCD$ a tetrahedron. $P$ and $Q$ be points on the sides $AB$ and $CD$ respectively. What is the locus of the midpoint of the line segment $PQ$?
The locus is the set of po... | Denote points of tetrahedron with $A(0, 0, 0)$, $B(a, 0, 0)$, $C(\frac{a}2,\frac{a\sqrt 3}{2}, 0)$, $D(\frac{a}2,\frac{a\sqrt3}{6},\frac{a\sqrt6}3)$
Arbitrary points $P\in AB$ and $Q\in CD$ have the following coordinates:
$$P(pa,0,0), Q(\frac{a}2, \frac{a\sqrt3}{6}(3-2q),\frac{aq\sqrt6}3)$$
...with:
$$p,q\in[0,1]$$
Coo... | {
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"url": "https://math.stackexchange.com/questions/2901367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Prove that $3^{16} -33$ and $3^{15} +5$ is divisible by 4 by means of binomial theorem This is a question that I found in a textbook:
Given that $p=q+1$, $p$ and $q$ are integers, then show that
$p^{2n} - 2nq-1$ is divisible by $q^2$ given that $n$ is a positive integer.
By taking a suitable value of $n$, $p$ and $q$,... | Use the Euclid lemma in last implication:
$$\begin{eqnarray}
3(3^{15}-11)=4[120+1120+...]&\implies &3\mid 4[120+1120+...] \\
&\implies & 3\mid 120+1120+...
\end{eqnarray}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Show that $\iint_S (x^2+y^2) dA = 9 \pi /4$ In exam it was asked to show that
$$\iint_S x^2+y^2 dA = 9 \pi /4$$ for $$S = {\{(x, y, z) | x>0, y>0,3>z>0, z^2 = 3(x^2 + y^2)}\}$$
I have tried many times but I don't get the $9 \pi /4$.
$$\begin{align}
\iint_S\sqrt{1+f_x^2+f_y^2}\,dA &=\int_0^{\sqrt3}\int_0^{2\pi} r^2\... | Your problem arises from using the fact that $f_x = 6x$ and $f_y = 6y$. Indeed this isn't the case, as $z^2 = 3(x^2+y^2)$, not $z$. Thus you would have:
$$z=f(x,y) = \sqrt{3(x^2+y^2)}$$
$$f_x = \frac{\sqrt{3}x}{\sqrt{x^2+y^2}} \quad \quad f_y = \frac{\sqrt{3}y}{\sqrt{x^2+y^2}}$$ Then we have:
$$\iint_S (x^2+y^2)\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
prove that $\lim _{x\to 1}f(x)=3$ if $f(x)=\frac{x^3-1}{x-1}$ Prove that $\lim _{x\to 1}f(x)=3$ where $f:(0,\infty)\to \mathbb{R}$ is given by $f(x)=\frac{x^3-1}{x-1}$.
I proved by definition of the limit
$$|f(x)-3|=\left|\frac{x^3-1}{x-1}-3\right|=|x^2+x-2|=|x-1||x+2|\leq |x-1|(|x|+2)$$
how to processed from this
| $x^3-1=(x-1)(x^2+x+1)$.
Hence $\lim_{x\to1}\frac{x^3-1}{x-1}=\lim_{x\to1}x^2+x+1=1^2+1+1=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x ... | By geometric series we have
$$\frac x {x^2+k}=\frac1x\frac 1 {1+k/x^2}=$$$$=\frac1x\left(1-\frac{k}{x^{2}}+\left(-\frac{k}{x^{2}}\right)^2+\ldots\right)=\frac1x-\frac{k}{x^{3}}+\frac{k^2}{x^{5}}+\ldots$$
and therefore
$$\sum_{k=1}^x \frac x {x^2+k}=\sum_{k=1}^x \left(\frac1x-\frac{k}{x^{3}}+\frac{k^2}{x^{5}}+\ldots\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
How to prove that for all natural $n$, $133|11^{n+2}+12^{2n+1}$
How to prove that for every natural $n$, the number
$$11^{n+2} +12^{2n+1}$$
is divisible by $133$?
I tried Induction method, so assuming that
$$
n=k \implies
A = (11^{k+2} +12^{2k+1} ) \pmod{133} \equiv 0
$$
Then trying to prove that
$$
n=k+1 \impli... | $\displaystyle 11^{k + 3} + 12^{2k + 3} = 11 \times 11^{k + 2} +
\underbrace{\qquad 144\qquad}_{\displaystyle =\ 133 + 11} \times 12^{2k + 1} =
11\left(\color{red}{11^{k + 2} + 12^{2k + 1}}\right) + \color{red}{133} \times 12^{2k + 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
The number of natural number $n\leq 50$ such that
$\displaystyle \sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
Try: Let $\displaystyle x=\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}$
So $\displaystyle x... | By the rational root theorem, a solution of $$x^3-x-n=0$$ must be a divisor of $-n.$ Since $n\leq50$ such a solution can't be very large. By definition of $x$ we have $x\geq0$. If $x=4,$ then $x^3 =64$ and there's already no hope. So $x$ must be one of $0, 1, 2, 3.$
$$\begin{align}
x&=0\implies n=0\\
x&=1\implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Solve $a^2 - 2b^2 - 3 c^2 + 6 d^2 =1 $ over integers $a,b,c,d \in \mathbb{Z}$ Are we able to completely solve this variant of Pell equation?
$$ x_1^2 - 2x_2^2 - 3x_3^2 + 6x_4^2 = 1 $$
This has an interpretation as is related to the fundamental unit equation of $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}[x,y]/(x^2 - 2,... | $$x^2-2y^2-3z^2+6q^2=1$$
Use what any decision $a^2-2b^2=1$ and $c^2-2d^2-3k^2+6t^2=1$
$$x=ac\pm{2bd}$$
$$y=ad\pm{bc}$$
$$z=ak\pm{2bt}$$
$$q=at\pm{bk}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Calculate $A^5 - 27A^3 + 65A^2$, where $A$ is the matrix defined below. If $A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$, find $A^5 - 27A^3 + 65A^2$
$$A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$$
Let $\lambda$ be its eigenvalue, then
$$(A-\lambda I) = \begin{bmatrix} 0-\lambd... | To add to the above answer, using $A^3 = 5A^2 - 6A + 5$ to simplify further $A^2(5A^2 -33A + 70) = A(5(5A^2 - 6A + 5) - 33A^2 + 70A) = A(-8A^2 + 40A + 25) = -8(5A^2 -33A + 70) + 40A^2 + 25A = 289A - 560$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Solve: $\log_3(5(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})+2^{64})=?$
$$\log_3(5(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})+2^{64})=?$$
At a first glance, it seems that I need to do this:
$$\log_3((2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})+2^{64})=?$$
But afterwards, I... | Since
\begin{align*}
&\phantom{==}(3-2)(3+2)(3^2+ 2^2)(3^4+2^4)\cdots(3^{32}+2^{32})\\
&= (3^2-2^2)(3^2 + 2^2)(3^4+2^4)\cdots(3^{32}+2^{32})\\
&= (3^4-2^4)(3^4+2^4)(3^8+2^8)\cdots(3^{32}+2^{32})\\
&= \cdots\\
&= 3^{64}-2^{64},
\end{align*}
the result is $\log_3(3^{64}) = 64$. The trick is $(a-b)(a+b)= a^2 - b^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
knowing: $ \tan x=2-\sqrt{3}$ , obtain: $ \cos 2x$
Knowing: $$ \tan x=2-\sqrt{3} $$
Obtain: $$\cos2x$$
I tried converting $\tan x$ into it's sinus and cosine form and trying to square both sides to try to get the form of:
$$\cos^2x-\sin^2x$$
But I can't really get to this form without having extra expressions of sine... | Draw a right triangle with the opposite side being $2 - \sqrt3$, and the adjacent side being $1$ so that $\tan x = 2 - \sqrt{3}$. Then the hypotenuse is $\sqrt{1^2 + (2 - \sqrt{3})^2} = \sqrt{1 + 4 - 4\sqrt3 + 3} = \sqrt{8 - 4 \sqrt{3}}$, and so $\cos^2 x = \frac{1}{8 - 4 \sqrt{3}}$ as $\cos x$ is adjacent/hypotenuse.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How to prove $b \ll a (a^{2}b^2)^{2.1/k} $ In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is
written on page $9$ that
It follows from (13) and Lemma 2 with $\eta=0.1 $ we have that,
$$b \ll a (a^{2}b^2)^{2.1/k} $$
The image of page 9 :
How do we prove $b \ll a (a^{2}b^2)^{2.... | We know,
$ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b} \cdots (13)$
Here, $xz= ((a+1)(ab^2+1))^\frac{1}{k}, y^2= (ab+1)^\frac{2}{k}.$
So, $ \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab+1)^2} \right|\leq \frac{1}{b} \cdots (a)$
Since, $(ab+1)>ab\implies(ab+1)^\frac{2}{k}>(ab)^\frac{2}{k}$
$\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lfloor n / \lfloor n / \lfloor \sqrt n \rfloor \rfloor \rfloor$ for positive integers $n$. I'm considering:
$$
\left\lfloor {n \over \lfloor n / \lfloor \sqrt n \rfloor\rfloor} \right\rfloor \:\:\:\: \forall n \in \mathbf N^+
$$
which seems to be $\lfloor \sqrt n \rfloor$.
Is there any way to prove or dispro... | Let $n$ be a real number such that $n\geq 1$, and $m:=\left\lfloor\sqrt{n}\right\rfloor$. Then, we have
$$m^2\leq n< (m+1)^2=m^2+2m+1\leq m^2+3m\,.$$
Thus,
$$m\leq \frac{n}{m}< m+3\,.$$
Set $k:=\left\lfloor\dfrac{n}{m}\right\rfloor$, so that $k\in\{m,m+1,m+2\}$. If $k=m$, then $m^2\leq n<m(m+1)$. That is,
$$m\le\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Finding a closed form for $\sum_{n = 1}^{\infty} \frac{1}{(1 + x^2)^{n}}$ So I got the sum
$$\sum_{n = 1}^{\infty} \frac{1}{(1 + x^2)^{n}}$$
and I want a closed form.
Can I just do
$$\frac{1}{1 - r} = \frac{1}{1 - \frac{1}{1 + x^2}} = \frac{1}{x^2} + 1 $$
for a closed form?
| Another way of solving it is by bringing it to the standard form, then you can use the formula for infinite geometric series considering $x \ne 0$
take $n=m+1$, so you have:
$$ \sum_{m = 0}^{\infty} \frac{1}{(1 + x^2)^{m+1}}= \frac{1}{(1 + x^2)} \sum_{m = 0}^{\infty} \frac{1}{(1 + x^2)^{m}}=\frac{1}{(1 + x^2)}\cdot \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$ If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$.
Is there a way to prove it without using the Binomial Theorem? Is it possible to use the Bernoulli's Inequality to prove it? If yes, please show me.
I tried to prove... | As an alternative we can proceed by induction for the stronger
$$(1+x)^{n}>1+nx+\frac12n(n-1)x^2>\frac12n(n-1)x^2$$
that is for the induction step
$$(1+x)^{n+1}=(1+x)(1+x)^{n}\stackrel{Ind.Hyp.}>1+nx+\frac12n(n-1)x^2+x+nx^2+\frac12n(n-1)x^3>$$$$\stackrel{?}>1+(n+1)x+\frac12(n+1)nx^2$$
therefore we need to prove that
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Value of 'a' for which $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b $ is injective Let $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b \forall x \in R$.
Find the least value of a for which $f(x)$ is injective function.
My approach , for $f(x)$ to be injective either f(x) should be increasing or decreasing function.'
$Y=f'(x)=... | Your idea of computing $f'$ is fine, but your computation is wrong. In fact$$f'(x)=x^2+x+a=\left(x+\frac12\right)^2+a-\frac14.$$Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Ask a about hard integral of $\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$ I want to evaluate the integral:
$$I(a,b)=\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$$
Attempt:$$\frac{\partial ^2I}{\partial a\partial b}=4ab\int_{0}^{\infty}\frac{\log x}{(a^2+x^... | One way is to replace $\ln x$ with $x^p$, then the integrand becomes a product of two linear Meijer G-functions after the change of variables $t = 1/x^2$. We obtain
$$I(p) = \int_0^\infty x^p
\ln \left(1 + \frac {a^2} {x^2} \right)
\ln \left(1 + \frac {b^2} {x^2} \right) dx = \\
\frac 1 2 \int_0^\infty t^{(-3-p)/2}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Polar form of $z^2+1$, $z\in \mathbb C$ Let $z\in \mathbb C$. Can we write $z^2+1 = re^{i\theta}$, for some $r\in \mathbb R$ and $\theta \in (-\pi, \pi]$?
We have $z^2+1 = r^2e^{i2\theta}+e^{i0}$ or we can write $z^2+1 = x^2-y^2+1 + i2xy$. We can use formulas to calculate $r$ and $\theta$ of $z^2+1$, but the process se... | First of all, notice that $z^{2} + 1 = (z-i)(z+i)$. Thus we have:
\begin{align*}
|z^{2} + 1| = |z-i||z+i|
\end{align*}
If we agree that $z = \rho e^{i\theta}$, we get
\begin{align*}
z - i = \rho\cos(\theta) + i(\rho\sin(\theta) - 1) \Rightarrow |z - i|^{2} & = \rho^{2}\cos^{2}(\theta) + (\rho\sin(\theta) - 1)^{2}\\
& =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Tricks to find a closed form $x_n$ of recurrence relation $x_{n+1} = \frac{1}{1+x_n}$ where $x_1 = a > 0$, $n \in \mathbb N$ With the help of you guys I've been able to learn how to solve various recurrence equations, but today I came across one I couldn't handle:
Find a closed form of the sequence $\{x_n\}$ of recurr... | When stuck, there's always the option of computing the first few terms of the sequence, to see if a pattern appears.
Computing the first $6$ terms, we get
\begin{align*}
x_1&=a\\[4pt]
x_2&=\frac{1}{1+a}\\[4pt]
x_3&=\frac{1+a}{2+a}\\[4pt]
x_4&=\frac{2+a}{3+2a}\\[4pt]
x_5&=\frac{3+2a}{5+3a}\\[4pt]
x_6&=\frac{5+3a}{8+5a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Choosing an interval of the CDF to find each quartile I have a random variable $X$ which has the following CDF:
$$F(y) = \left\{\begin{array}{ll}
0 & : y \lt 0\\
\frac{y}{30} & : 0 \le y \lt 20\\
\frac{2}{3} + \frac{y-20}{60} & : 20 \le y \lt 40\\
1 & : y \ge 40
\end{array}
\right.$$
To find the median of $X$, I know I... | If your random variable $X$ has the cdf
$$F(y) = \left\{\begin{array}{ll}
0 & : y \lt 0\\
\frac{y}{30} & : 0 \le y \lt 20\\
\frac{2}{3} + \frac{y-20}{60} & : 20 \le y \lt 40\\
1 & : y \ge 40
\end{array}
\right.$$
you're supposed to find the inverse so you get $F(y) = \frac{y}{30} $ for $y$. then the inverse is $30y$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Classifying singular points of $\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}$
I am trying to classify the singular points of the function $$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}.$$
My attempt:
$$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}=\frac{\sin(z^2)}{z^2\left(z-\frac{\pi}{4}\right)}.$$
Hence the singular points ... | It is correct.
Also you can notice that
$$\lim_{z\to 0} \frac{\sin(z^2)}{z^2\left(z-\frac{\pi}{4}\right)}= \lim_{z\to 0} \frac{\sin(z^2)}{z^2} \cdot \lim_{z\to 0}\frac1{z-\frac\pi4} = 1 \cdot \frac1{-\frac\pi4} = -\frac4\pi$$
so $z = 0$ is a removable singularity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove by induction that, for all positive integers n, the following inequality holds: (1-2-1)(1-2-2)...(1-2-n) ≥ 1/4 + 2-(n+1)
My solution so far:
After verifying the base case for n = 1, I began the inductive step:
(1-2-1)(1-2-2)...(1-2-(k+1)) ≥ 1/4 + 2-(k+2)
Subtracting 1/4 + 2-(k+1) on the RHS as it's an inequality:... | I'd avoid starting the induction step from what you need to prove; rather
$$
\left(1-\frac{1}{2}\right)\left(1-\frac{1}{2^2}\right)\dots
\left(1-\frac{1}{2^k}\right)\left(1-\frac{1}{2^{k+1}}\right)\ge
\left(\frac{1}{4}+\frac{1}{2^{k+1}}\right)\left(1-\frac{1}{2^{k+1}}\right)
$$
by the induction hypothesis.
The right ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding value of $\int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx$ without contour Integration
Finding value of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx\;\;, a>b>0$ without contour Integration
Try: Let $\displaystyle I =\int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx=2\int^{\pi}_{0}\frac{\sin^2 x}{a-b\cos... | $$I=\int_0^{2\pi}\frac{\sin^2(x)}{a-b\cos(x)}dx=\int_0^\pi\frac{\sin^2(x)}{a-b\cos(x)}dx+\int_\pi^{2\pi}\frac{\sin^2(x)}{a-b\cos(x)}dx$$
now using substitution:
$$t=\tan\frac{x}{2}$$
$$dx=\frac{2}{1+t^2}dt$$
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
because of the discontinuities of the function, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Evaluating $\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)}$ I am trying to evaluate this integral
$$\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)}$$
with $\mathcal{P}$ the principal value and $a,b>0$.
I already know the answer to be $$ - \frac{a\pi}{2(1 + a^2b^2)}$$
after fiddling with Mathem... | Thanks to both of you for the help!
So indeed, this contour is a good one to evaluate this integral.
The residues at $b$ and $-b$ just cancel each other.
The residue at $i/a$ gives $$ - \frac{a}{2i\left( a^2 b^2 +1 \right)}$$
Thus we have
$$ 2\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)} = 2\pi i \left(- ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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In how many ways can ten indistinguishable cookies be distributed to four friends with restrictions?
You have ten indistinguishable cookies that you’ve baked for four of your friends Jianbei, Sione, Sina and Julia. You want to give away all ten of your cookies to your friends, and for each friend to get at least one c... | Your strategy is sound, but your answer is indeed incorrect.
Case 1: Sione receives two cookies.
Let $x_i$, $1 \leq i \leq 4$, be the number of cookies distributed to the $i$th friend. Let $x_4$ denote the number of cookies given to Sione. Since a total of ten cookies are distributed and Sione receives two,
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with proof by induction and divisibility I have a question about induction (I'm a little fuzzy on it).
*
*Prove $3n+1<n^2$ for all integers $n\geq 4$.
*Assume $k^2+k$ is an even integer for any $k\in\mathbb{Z}$. Prove $n^3-n$ is divisible by 6 for all $n\in\mathbb{N}$.
For 1, I have: Base Case: Let $n=4$. E... | 1)
$3(k+1)+1 = 3k + 4$ and $3k+1 < k^2$ so $3k + 4 < k^2 +3$.
And $(k+1)^2 = k^2 + 2k + 1$. ANd $k\ge 4$ so $2k + 1 \ge 9$.
So $k^2 + 3 < k^2 + 9 \le k^2 + 2k +1 = (k+1)^2$
2)
$n^3 - n = 0$ if $n = 1$. And $0$ is divisible by $6$ (because $0 = 6*0$) so, no, you are not stuck.
But if you want to do $n =2$ for backup... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2945926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Distance between the Orthocenter and the Lemoine Point (Symmedian Point) I need help to solve this problem. I have no idea how to solve.
Is there previously solved copy of this problem?
Thank you very much.
Let $H$ be the orthocenter of a triangle $ABC$ with circumradius $R$. Show that the distance $HK$ between $H$... | You can solve this question by analytic geometry, using two lemmas:
Lemma 1: In any triangle ABC these relations hold true:
$$a^2+{AH}^2=b^2+{BH}^2=c^2+{CH}^2=4R^2$$
Lemma 2: If $A=(x_1,y_1)$, $B=(x_2,y_2)$, $C=(x_3,y_3)$ are the coordinates of vertices of triangle ABC, then the coordinates of its Lemoine point are $$K... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2947510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proving $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers. I would like to verify whether my proof is correct. The answer sheet used a much more intuitive and logical approach but I think mine is correct also.
To prove: $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers
Proof: We proceed by induction and s... | Rewrite $13\cdot 6^n+8\cdot 13^n$ in base $7$:
$$16\cdot 6^n+11\cdot 16^n$$
Note that $6\cdot 6=51$ in base $7$.
The first term ends with $1$ if $n$ is even and ends with $6$ if $n$ is odd. Conversely, the second term ends with $1$ if $n$ is even and ends with $6$ if $n$ is odd. Therefore, the sum ends with $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find $\sum_{n=1}^\infty\frac{2^{f(n)}+2^{-f(n)}}{2^n}$, where $f(n)=\left[\sqrt n +\frac 12\right]$ denotes greatest integer function
Question: Let $f(n)=\left[\sqrt n +\dfrac 12\right]$, where $[\cdot]$ denotes greatest integer function, $\forall n\in \Bbb N$. Then,
$$\sum_{n=1}^\infty\frac{2^{f(n)}+2^{-f(n)}}{2^... | Note that $f(n) = \left\lfloor \sqrt{n}+\tfrac{1}{2}\right\rfloor = k$ iff $k^2-k+\tfrac{1}{4} \le n < k^2+k+\tfrac{1}{4}$, i.e. $k^2-k+1 \le n \le k^2+k$.
Therefore, we have:
\begin{align*}
\sum_{n = 1}^{\infty}\dfrac{2^{f(n)}+2^{-f(n)}}{2^n} &= \sum_{k = 1}^{\infty}\sum_{f(n) = k}\dfrac{2^{f(n)}+2^{-f(n)}}{2^n}
\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Finding the $2n+1$ th derivative of $\frac{y^{2n+1}xy}{1-x^2y^2}$ with respect to $x$. $f(x,y) = \frac{y^{2n+1}xy}{1-x^2y^2}$.
I made the following table:
\begin{align}
& 2n+1 = 1 \implies f^{(1)} = \frac{1!y^2(1+x^2y^2)}{(1-x^2y^2)^2}\\
& 2n+1 = 3 \implies f^{(3)} = \frac{3!y^6(x^4y^4+6x^2y^2+1)}{(1-x^2y^2)^4} \\
& 2n... | You may write $z = xy\Rightarrow z'(x) = y$.
$$f(x) = \frac{y^{2n+1}xy}{1-x^2y^2} = y^{2n+1}\color{blue}{\frac{z}{1-z^2}} $$
$$\Rightarrow f(x) = g(z(x)) \Rightarrow f'(x) = g'(z)\cdot z'(x) = g'(z)\cdot y \Rightarrow \boxed{f^{(k)}(x) = g^{(k)}(z)\cdot y^k}$$
$$\color{blue}{\frac{z}{1-z^2} = -\frac{1}{2}\left( \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Dual Numbers and Automatic Differentiation I have the following equation. I would like to solve this at the point x = a, but using dual numbers.
$$f\left(x\right) = \dfrac{1}{x} + \sin\left(\dfrac{1}{x}\right)$$
Now, the derivative of this function is below, and that is the function we'd like to reach in our answer.
... | Break your problem into steps. Let's say $f(x) = g(x) + h(x)$ then $f' = g' + h'$ (omitting $x$). In general even with dual variables you should use the chain rule.
First solve the derivative of $1/x$ at $x=a$.
$$\begin{align} \frac{1}{a + \epsilon} = \frac{1}{a} \frac{1}{1+ \epsilon/a} = \frac{1}{a} \big(1 - \frac{\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding range of expression $f(x,y)=x^2+y^2$
Finding range of $f(x,y)=x^2+y^2$ subjected to
the condition $2x^2+6xy+5y^2=1$ without Calculus
Try: Let $k=x^2+y^2,$ Then $\displaystyle k=\frac{x^2+y^2}{2x^2+6xy+5y^2}$
Now put $\displaystyle \frac{y}{x}=t(x\neq 0)$ and $t\in \mathbb{R}$
$$k=\frac{1+t^2}{2+6t+5t^2}\Right... | Yes, of course. We can make it.
For example, we need to prove that
$$x^2+y^2\geq\frac{7-3\sqrt5}{2}(2x^2+6xy+5y^2)$$ or
$$(2\sqrt5-4)x^2-2(7-3\sqrt5)xy+(5\sqrt5-11)y^2\geq0$$ or
$$2x^2-2(7-3\sqrt5)(2+\sqrt5)xy+(5\sqrt5-1)(2+\sqrt5)y^2\geq0$$ or
$$2x^2-2(-1+\sqrt5)xy+(3-\sqrt5)y^2\geq0$$ or
$$4x^2-4(-1+\sqrt5)xy+(6-2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve $9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}$ $$9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}.$$
The equation states solve for $x$.
What I first did was put like bases together.
$$3^{2x}+3^{2x-1}= 2^{x+\frac{7}{2}}+ 2^{x+\frac{1}{2}}.$$
Then I factored $3^{2x}$ and $2^x$
$$3^{2x}(1+\frac{1}{3})=2^x(2^{\fr... | If you let $x=u+{1\over2}$, you can get rid of the pesky square roots: the expression simplifies to
$$9^{u+1/2}-2^{u+1}=2^{u+4}-3^{2u}$$
or
$$3\cdot9^u-2\cdot2^u=16\cdot2^u-9^u$$
This simplifies first to $4\cdot9^u=18\cdot2^u$ and then to $9^{u-1}=2^{u-1}$, which clearly implies $u=1$, i.e., $x=3/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Purchasing Donuts
At closing time at a Dunkin Donuts store they still have $10$ vanilla,
$20$ custard, $24$ cinnamon, and $30$ chocolate donuts available.
Donuts of the same kind are regarded as identical.
(a) How many ways are there to purchase $8$ donuts?
(b) How many ways are there to purchase $15$ donuts wi... |
(c) How many ways are there to purchase $15$ donuts?
The question can be formulated as:
$$x_1+x_2+x_3+x_4=15,\\
0\le x_1\le 10,\\
0\le x_2\le 20,\\
0\le x_3\le 24,\\
0\le x_4\le 30,$$
which is equivalent to:
$$x_1+x_2+x_3+x_4=15,\\
0\le x_1\le 10,\\
0\le x_2\le 15,\\
0\le x_3\le 15,\\
0\le x_4\le 15.$$
which is the d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $f(x)=x^n(1-x)\lt \frac{1}{ne}$ for all $n\in \mathbb{N},x\in(0,1)$
Prove $f(x)=x^n(1-x)\lt \frac{1}{ne}$ for all $n\in \mathbb{N},x\in(0,1)$.
My try: $f'(x)=nx^{n-1}-(n+1)x^n=x^{n-1}(n-(n+1)x)=0\Rightarrow x=\frac{n}{n+1} $, hence $\max_{x\in(0,1)} f(x)=f(\frac{n}{n+1})=(1-\frac{1}{n+1})^n\cdot\frac{1}{n+1}$. ... | \begin{equation}
\max_{x\in(0,1)} f(x)=f(\frac{n}{n+1})=(1-\frac{n}{n+1})^n\cdot\frac{n}{n+1}
=
\frac{n}{(n+1)^{n+1}}
\tag{1}
\end{equation}
Now
\begin{equation}
\frac{n}{(n+1)^{n+1}} - \frac{1}{ne}
=
\frac{1}{n}( \frac{n^2}{(n+1)^{n+1}} - \frac{1}{e})
=
\frac{1}{n}( \frac{en^2 - (n+1)^{n+1}}{e(n+1)^{n+1}})
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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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.