Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\int_0^\pi e^{\cos x}\cos(x-\sin x) dx$ I’m not quite sure how to approach this problem using complex analysis. Mainly I’m confused int the contour to use after changing the integral by changing
$\cos(x-\sin x)$
to become
$Re(e^{i(x-\sin x})$, so that the integral became
$$Re\int_0^\pi e^{\cos x}e^{i(x-\sin x)} dx$... | \begin{align}\int_0^\pi \text{e}^{\cos x}\cos(x-\sin x) dx&=\int_0^\pi\text{e}^{\cos x}\cos x\cos(\sin x)dx+\int_0^\pi\text{e}^{\cos x}\sin x\sin(\sin x)dx\\
&=\Big[\sin(\sin x)\text{e}^{\cos x}\Big]_0^\pi+2\int_0^\pi\text{e}^{\cos x}\sin x\sin(\sin x)dx\\
&=2\int_0^\pi\text{e}^{\cos x}\sin x\sin(\sin x)dx\\
&=2\int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\lim\limits_{x\to 2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}$ without l'Hospital's rule
How to evaluate the following limit?
$$
\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}
$$
I factored the denominator into $(x-2)(x^2+2x+4)$, but I couldn't go on from there.
| For a start, $$ x^9 - 1 = (x-1) (x^2+x+1)(x^6 + x^3 + 1)$$
This leads us to
$$ a^9 - b^9 = (a-b)(a^2 + ab + b^2)(a^6 + a^3 b^3 + b^6) $$
Now take $a,b$ to be the two cube roots in your fraction
After you cancel the $(a-b)$ factors, you can simply set $a,b$ equal, so the denominator becomes $3 b^2 3 b^6 = 9 b^8$ with $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
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Eigenvalues of a Hermitian tridiagonal matrix. I want to know how to calculate the eigenvalues of the following Hermitian tridiagonal $(N+1)\times(N+1)$ matrix,
$$
A=\begin{pmatrix}
N+1&i\sqrt{N}\\
-i\sqrt{N}&N+1&i\sqrt{2(N-1)}\\
&-i\sqrt{2(N-1)}&\ddots&i\sqrt{3(N-2)}\\
&&-i\sqrt{3(N-2)}&\ddots&\ddots\\
&&&\ddots&N+1&i... | (For convenience, all matrices below are zero-indexed.)
Let $B=A-(N+1)I$. Then $B$ is a tridiagonal complex skew-symmetric matrix. When $0\le k<N$, we have
\begin{align}
b_{k,k+1}&=i\sqrt{(k+1)(N-k)},\\
b_{k+1,k}&=-b_{k,k+1}.
\end{align}
(Note that the above formula is different from the one in the OP because our matri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Solve : $2\sin^3 (x) = \cos (x)$
Solve : $2\sin^3 (x) = \cos (x)$
How to solve the above equation?
I tried, but failed to succeed.
Graph of the following equation
But how to solve the equation without graph?
| Here's how I solved it by hand:
Notice that if we square both sides, we get $$4 \sin^6 x = \cos^2 x$$
Using the Pythagorean identity,
$$4 \sin^6 x + \sin^2 x - 1 = 0$$
Let $y = \sin^2 x$. Our equation becomes $$4y^3 + y - 1 = 0$$ Normally, we would have to use the cubic equation formula, but according to the Rational R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Computing $\sum \frac{1}{n^2(n^2+a^2)}$ using the residue theorem I have to find a closed form for $\sum_{n=1}^{\infty}\frac{1}{n^2(n^2+a^2)}$ (with $a\in \mathbb{R}$).
My idea was to use the residue theorem on $f(z)=\frac{\pi \cot(\pi z)}{z^2(z^2+a^2)}$, thus obtaining:
\begin{equation}
\text{Res}_{0}(f)+\text{Res}_{... | We first decompose the series in two parts:
$\sum_{n=1}^{\infty} \dfrac{1}{n^{2}(n^{2}+a^{2})} = \dfrac{1}{a^{2}} \sum_{n=1}^{\infty} \dfrac{1}{n^{2}} - \dfrac{1}{a^{2}} \sum_{n=1}^{\infty} \dfrac{1}{n^{2} + a^{2}}$.
For the first series we use $\sum_{n=1}^{\infty} \dfrac{1}{n^{2}} = \dfrac{\pi^{2}}{6}$.
For the secon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find $a,b,c \in \mathbb{R^+}$ such that $a^2+b^2+c^2+abc=4$ and $a+b+c=3$
Find $a,b,c \in \mathbb{R^+}$ such that $$a^2+b^2+c^2+abc=4\;\;\;{\rm and}\;\;\;a+b+c=3$$
Seeing $a^2+b^2+c^2+abc=4$,I substituted $a=2\cos{A}$,$b=2\cos{B},c=\cos{C}$ with $A+B+C=180°$.
Therefore,$cosA+cosB+cosC=\frac 32$ have thousands of so... | It is $$\cos(C)+\cos(B)+\cos(A)\le \frac{3}{2}$$ and this Comes from
$$\cos(A)+\cos(B)+\cos(C)=1+\frac{r}{R}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve $2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0$?
How to solve $$2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0?$$
I just have no idea and I'have some knowledge about the polynomial equations. Here, just nothing.
| The only real root is
$$
\alpha=-\frac{1}{\sqrt{2}}=- 0.707106781187\cdots
$$
The polynomial divided by $x-\alpha$ then has degree $4$ and has roots $\alpha\pm i\beta$, $\alpha\pm i \gamma$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find the ratio of the area $EDPQ$ to the area of ABC In triangle $ABC$,points $E$ and $D$ are on side $AC$ and point $F$ is on side $BC$ such that AE=ED=DC and $BF:FC$ =2:3. $AF$ intersects $BD$ and $BE$ at points $P$ and $Q$, respectively. Find the ratio of the area $EDPQ$ to the area of $ABC$
taken from the 2017 IMC ... | Let $k\in DC$ such that $FK||BD$.
Thus, $DK:KC=2:3.$
Let $DC=5x$.
Thus, $DK=2x$ and $AD=10x$, which gives
$$\frac{AP}{PF}=\frac{AD}{DK}=\frac{10x}{2x}=5.$$
Similarly, let $M\in EC$ such that $FM||BE.$
Thus, $EM:MC=2:3,$ which gives $EM=4x$ and
$$\frac{AQ}{QF}=\frac{AE}{EM}=\frac{5x}{4x}=\frac{5}{4}.$$
From here we obta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is $\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}$? I encountered the following hard problem in a math olympiad book:
Evaluate $$\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot2018\cdot2019}.$$
I tried to evaluate $\sum^{k}_{n=1}\frac{n(n+1)(n+2)(n+3)}{2016\cdot2017\cdot20... | I will use the following Lemma (that you can prove by induction):
Lemma. For all $N\in\Bbb N:$ $$\sum_{n=1}^N n(n+1)(n+2)(n+3)=\frac15 N(N+1)(N+2)(N+3)(N+4).$$
In our case, we have $$\sum_{n=1}^{2016} n(n+1)(n+2)(n+3) = 2016\cdot2017\cdot2018\cdot2019\cdot2020\cdot\frac15.$$
So $$\sum^{2016}_{n=1}\frac{n(n+1)(n+2)(n+3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rearranging the formula Transpose this formula to make $y$ the subject.
$$x=\sqrt{x^2y^2+1-y}$$
My try:
$$x^2=x^2y^2+1-y$$
$$x^2-x^2y^2=1-y$$
$$x^2(1-y^2)=1-y$$
Here I got 2 $y$ terms, but I am not sure what to do next.
| If you can't see the factorization, treat the thing as a quadratic in $y$:
$$
x^2y^2-y+(1-x^2)=0
$$
The discriminant is
$$
1-4x^2(1-x^2)=4x^4-4x^2+1=(2x^2-1)^2
$$
Apply the quadratic formula:
$$
y=\frac{1\pm(2x^2-1)}{2x^2}
$$
and you find either $y=1$ or
$$
y=\frac{1-2x^2+1}{2x^2}=\frac{1-x^2}{x^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Regularization of $\sum_{n=2}^\infty (-1)^n \log n$ I accidentally stumbled on the following regularization of this divergent series:
$$\sum_{n=2}^\infty (-1)^n \log n "=" \frac{1}{2} \log \frac{\pi}{2}$$
I'm not familiar enough with regularization, so I wanted to ask if this result agrees with any other known regular... | We can "calculate" the sum using the Wallis product formula.
Denoting our sum by
$$f = \sum_{n=1}^\infty (-1)^n \log(n)$$
we have
$$f= \log(2) - \log(3) + \log(4) \mp ...$$
and
$$e^f = \frac{2}{3} \frac{4}{5} \frac{6}{7} ...$$
Taking the square gives
$$e^{2f} = \frac{2}{3} \frac{2}{3} \frac{4}{5} \frac{4}{5} \frac{6}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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If $f(x) =5x^2 - 2kx + 1 < 0$ has exactly one integral solution, find the sum of all positive integral values of $k$. Question : If $f(x) =5x^2 - 2kx + 1 < 0$ has exactly one integral solution, find the sum of all positive integral values of $k$.
My Attempt
Corresponding Equation of Inequality, $f(x) =5x^2 - 2kx + 1 =... | Using your notations:
$$\alpha=\frac{k-\sqrt{k^2-5}}{5}; \beta=\frac{k+\sqrt{k^2-5}}{5}.$$
The integral solution of $5x^2 - 2kx + 1 < 0$ must be in $(\alpha,\beta)$.
Consider only $k>0$ by the requirement and note that:
$$\\
0<\alpha<1 \iff 0<k-\sqrt{k^2-5}<5 \iff k-5<\sqrt{k^2-5} \Rightarrow k>0
$$
Hence:
$$\beta>1 \R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\sum_{r=1}^n \frac{\tan(x/2^r)}{2^{r-1}\cos(x/2^{r-1})}$
I was asked to find the sum of
$$\sum_{r=1}^n \dfrac{\tan \dfrac{x}{2^r}}{2^{r-1}\cos\dfrac{x}{2^{r-1}}}$$
I proceeded by breaking $\tan x$ into $\sin x$ and $\cos x$ and writing numerator as follows
$$\sin\left(\frac{x}{2^{r-1}}-\frac{x}{2^r} \r... | $$\sum^{n}_{k=1}\frac{\tan(x/2^k)}{2^{k-1} \cos(x/2^{k-1})} = \sum^{n}_{k=1}\frac{\sin^2(x/2^k)}{2^{k-1}\sin (x/2^k)\cos(x/2^k)\cos(x/2^{k-1})}$$
$$=2\sum^{n}_{k=1}\frac{1-\cos(x/2^{k-1})}{2^{k-1}\sin(x/2^{k-1})\cos(x/2^{k-1})}$$
$$= 2\sum^{n}_{k=1}\bigg[\frac{1}{2^{k-2}\sin(x/2^{k-2})}-\frac{1}{2^{k-1}\sin(x/2^{k-1})}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
ambiguity in solving algebraic equations. Given $x+\frac{1}{x}=2 \tag{1}$
$x^3+\frac{1}{x^4}=1 \tag{2}$
Find the value of $$x^4+\frac{1}{x^3}$$
If equation $\bf{2}$ is multiplied by $\bf x$ we get $x^4+\frac{1}{x^3}=x$ now, we just need to find the value of $x$ which from $\bf{1}$ is equal to $1$.
But the answer to t... | If $x=3$ and $x$ is solution to $(1)$, we have
$$
x + \frac{1}{x} = 2 \iff 3 + \frac 1 3 = 2
$$
which is obviously wrong. There must be some mistake in the solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Arc length of an implicit function Find the arc length of $$y=\frac{x^4}{8} + \frac{1}{4y^2}$$ for the region between $~x=1~$ to $~x=2~$.
I have tried implicitly deriving it to find $~\frac{dy}{dx}~$, but since $~\frac{dy}{dx}~$ is dependent on two variables, you cant integrate through the normal definition of $$\int... | Why not to look at $x$ as a function of $y$ and then integrate along $y$ ?
For the range of interest, we have
$$x=\frac{\left(8 y^3-2\right)^{1/4}}{\sqrt{y}}$$
$$x'=\frac{2 y^3+1}{y^{3/2} \left(8 y^3-2\right)^{3/4}}$$ and we face the problem of
$$L=\int_a^b \sqrt{1+\frac{\left(2 y^3+1\right)^2}{y^3 \left(8 y^3-2\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$. $$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\} \tag{$\star$}$$
My proof of $(\star)$ is as follows:
\begin{align}
10^{2n+1}+1
&= 10\cdot10^{2n}+1 \\
&= (11-1)\cdot10^{2n}+1 \\
&= 11\cdot10^{2n}-10^{2n}+1 \\
&= 11\cdot10^{2n}-\left(10^{2n}-1\... | Not as easy as the other answers, but this is a classic induction question.
For $n = 0$, we have $10^{2n + 1} + 1 = 11$, which $11$ divides, confirming the base case.
Suppose $11$ divides $10^{2n + 1} + 1$ for some $n$. Then some $k$ exists such that $10^{2n + 1} + 1 = 11k$. Then,
$$10^{2(n + 1) + 1} + 1 = 100 \cdot 10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Conjecture: ${_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right)=\prod_{n=2}^\infty \frac{(2n-2+x)(2n-1+x)}{2n (2n-3+2x)}$
Is the following conjecture for $0<x<1$ true and how do we prove it?
$${_2 F_1} \left(\frac{1}{2}- \frac{x}{2},1- \frac{x}{2};2;1 \right)=\prod_{n=2}^\infty \frac{(2n-2+x)(2n-1+x)}... | The conjecture is indeed true for all $x\in\Bbb C$ when $\Re\, x>-1/2$.
If $\Re\, x>-1/2$ then DLMF $15.4.20$ tells us that
$$
F\left(\tfrac{1-x}{2},\tfrac{2-x}{2};2;1\right)=\frac{\Gamma(2)\Gamma(x+\frac{1}{2})}{\Gamma(\frac{x}{2}+\frac{3}{2})\Gamma(\frac{x}{2}+1)}.
$$
Now notice that the sum of the arguments in the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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What is the ratio the areas of △$PST$ and quadrilateral $STRQ$? I tried to answer this question:
What is the ratio of the areas of △$PST$ and quadrilateral $STRQ$ if $∠1\cong ∠2$
Since $∠1\cong ∠2$, △$PST\sim$ △$PQR$. And since the two triangles are similar, the ratio of their areas is $\frac{1}{25}$. The area of the ... | $\triangle PST$ and $\triangle PQR$ are similar due to AA (two angles congruent). I will assume you can prove that.
Given that the area of a triangle is $A=\frac{1}{2}ab\sin{C}$, we can calculate the area for both triangles, then find the ratio. Let the side $\overline {PT}=y$. Then we need to calculate $\overline {PR}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
I saw instantly that $$h(x - \frac{1}{x})= x^2 - \frac{1}{x^2} = \left( x -\frac{1}{x} \right)\left... | instead of writing x+1/x as a+2/x,
$ \sqrt{(x-1/x)^2+4} = \sqrt{a^2+4} = |x+1/x| = x+1/x$ for x>0 and $-(x+1/x)$ for x<0$
$h(a)=a \sqrt{a^2+4}$ for x>0 and $h(a)= -a \sqrt{a^2+4}$ for x<0, where a=x-1/x.
Try simplifying this into a function in terms of x, considering cases for x>0 and x<0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Prove that if $x$ is odd, then $x^2$ is odd
Prove that if $x$ is odd, then $x^2$ is odd
Suppose $x$ is odd. Dividing $x^2$ by 2, we get:
$$\frac{x^2}{2} = x \cdot \frac{x}{2}$$
$\frac{x}{2}$ can be rewritten as $\frac{x}{2} = a + 0.5$ where $a \in \mathbb Z$. Now, $x\cdot\frac{x}{2}$ can be rewritten as:
$$x\cdot\fra... | Let $x = 2k + 1 \in \mathbb{Z}$ be odd. So, $$ x^{2} = (2k + 1)^2 = 4k^{2} + 4k + 1 = 2(2k^{2} + 2k) + 1$$ Let $t = 2k^{2} + 2k \in \mathbb{Z}$, thus: $ x^{2} = 2t + 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Where is the mistake in the equality of roots? I have this statement:
If $f(x) = \frac{\sqrt[3]{x}}{\sqrt{2}}, g(x) =
> \frac{\sqrt[6]{8x^2}}{2}$, with $x < 0$ is $f(x) = g(x)$ ?
My attempt was:
$(1)$ $f(x) = \frac{\sqrt[3]{x}}{\sqrt{2}} = \frac{\sqrt{2}\sqrt[3]{x}}{2}$
$(2)$ $g(x) = \frac{\sqrt[6]{8x^2}}{2}$ usin... | with $x<0$, $\sqrt[3]x<0$, but $\sqrt[6]{x^2}>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the following equation: $\sin x \cos x = \frac{1}{2}$ I am required to solve the following equation:
$$\sin x \cos x = \frac{1}{2}$$
My attempt:
Rewriting $\cos x$
$$\sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}$$
Squaring both sides
$$\bigl(\sin x \sqrt{1 - \sin^2 x}\bigr)^2 = \bigl(\frac{1}{2}\bigr)^2$$
$$\sin^2 x... | Your solution is incorrect, as others pointed out. The reason is because you've made two "irreversible" operations, and did not took care of the branches they create.
The first one is:
$$
\sin x \cos x = \frac{1}{2} \Rightarrow \sin x \sqrt{1 - \sin^2 x} = \frac{1}{2}
$$
This is is simply wrong, because if your are wor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
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Floor function simplification identities I can't seem to find any identity(if any)for division/multiplication involving floor functions: for example
$$\lfloor{\frac{n-1}{2}}\rfloor\cdot 2$$
I know does not simplify down to $$n-1$$.
| You could consider separately what happens when $n$ is an even number and when $n$ is an odd number.
If $n=2k$, then
\begin{align}
2\left\lfloor \dfrac{n-1}{2} \right\rfloor
&= 2\left\lfloor \dfrac{2k-1}{2} \right\rfloor\\
&= 2\left\lfloor k - \dfrac 12 \right\rfloor\\
&= 2(k-1) \\
&= 2\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Formula for the first base-$n$ digit of $n^{d+1}-\sum_{i=1}^{n}i^d$ Let $k,n,d \in \mathbb{N}$ and $n>1$.
Converting $k$ in a base $n$,
$$k = (n_{ l} ~\cdots~ n_2~ n_1)_{n} \quad\text{where}\quad n_{l} \ne 0$$
Show that, if
$$k = n^{d+1} - \sum_{i=1}^n i^d$$ then
$$
n_l=n - \left\lceil\frac{\sum_{i=1}^n i^d}{n^d}\... | Your conjecture doesn't always hold based on the stated conditions. For example, $n = 2, d = 3$ results in
$$\sum_{i=1}^{n}i^d = \sum_{i=1}^{2}i^3 = 1 + 2^3 = 9 \tag{1}\label{eq1A}$$
$$k = n^{d+1} - \sum_{i=1}^{n}i^d = 2^4 - 9 = 7 = (111)_2 \tag{2}\label{eq2B}$$
Thus, $l = 3$ and $n_l = n_3 = 1$. However,
$$n - \left\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $a_1,a_2,a_3....$ be an arithmetic progression. Question is
Let $a_1,a_2,a_3....$ be an arithmetic progression. If $\frac{a_1+a_2+a_3+...+a_p}{a_1+a_2+a_3+...+a_q}=\frac{p^2}{q^2},p\not=q$, then find $\frac{a_6}{a_{21}}$
Answer: $\frac{11}{41}$
All options: $\frac{7}{2}$, $\frac{2}{7}$, $\frac{11}{41}$, $\frac{41... | We have that
$$\frac{a_1+a_2}{a_1}=\frac{2^2}{1^2}=4$$
$$a_1+a_2=4a_1$$
$$a_2=3a_1$$
So the common difference is
$$d=a_2-a_1=2a_1$$
Hence our final answer is
$$\frac{a_6}{a_{21}}=\frac{a_1+(6-1)(2a_1)}{a_1+(21-1)(2a_1)}=\frac{11}{41}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Area boundes by two circles
Find the area bounded by $r \le 3\sin(t)$ and $r \le -3 \sqrt3 \cos(t)$.
Well, I've found the intersection point, which is $t=\frac{2 \pi}{3}$ . Then, I set up two integrals
$$
A=\frac{1}{2} \int_{\pi/2}^{2\pi/3} (3\sin(t))^2 dt
$$
and
$$
A=\frac{1}{2} \int_{2\pi/3}^{\pi}(-3 \sqrt3 \cos(t... |
$$
\text{Area}_1=\frac12\,\frac\pi3\left(\frac{3\sqrt3}2\right)^2=\frac{9\pi}8
$$
$$
\text{Area}_2=\frac12\,\frac{2\pi}3\left(\frac32\right)^2=\frac{3\pi}4
$$
$$
\text{Area}_3=\frac32\,\frac{3\sqrt3}2=\frac{9\sqrt3}4
$$
$$
\begin{align}
\text{Area}
&=\text{Area}_1+\text{Area}_2-\text{Area}_3\\[3pt]
&=\frac{9\pi}8+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $(\sqrt{3} + 1)b/2 < a < 1 + (\sqrt{3} + 1)b/2$? I'm starting with the assumption that $a, b > 0$, and $$\frac{a - 1}{a + b - 1} \cdot \frac{a}{a + b} = \frac{1}{3}$$
I am trying to show that $$(\sqrt{3} + 1)b/2 < a < 1 + (\sqrt{3} + 1)b/2.$$
Since $$\frac{a - 1}{a + b - 1} < \frac{a}{a + b}$$ I can w... | The condition gives:
$$2a^2-2a(b+1)-b^2+b=0,$$ which gives
$$a=\frac{b+1+\sqrt{3b^2+1}}{2}$$ or
$$a=\frac{b+1-\sqrt{3b^2+1}}{2}.$$
In the first case we need to prove that
$$\frac{(1+\sqrt3)b}{2}<\frac{b+1+\sqrt{3b^2+1}}{2}<1+\frac{(1+\sqrt3)b}{2}$$ or
$$\sqrt3b<1+\sqrt{1+3b^2}<2+\sqrt3b,$$ which is obvious.
In the seco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Taylor series of $(1+3x) \cdot \ln(1+x)$ I have to find Taylor series of $(1+3x) \cdot \ln(1+x)$. I know Taylor series of $(1+3x) \cdot \ln(1+x)$ but I do not know hot to simplify.
Any help?
| We have that
$$\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
therefore
\begin{align}(1+3x)\ln(1+x)&=\ln(1+x)+3x\ln(1+x) \\&= \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}~+~\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(3x)x^n}{n}\\&=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(3x+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Find all natural number $n$ for which $3^9+3^{12}+3^{15}+3^n$ is a perfect cube.
Find all natural number $n$ for which $3^9+3^{12}+3^{15}+3^n$ is a perfect cube.
What I have tried.
$3^9+3^{12}+3^{15}+3^n$
$=3^9(757+3^{n-9})$
Let $757+3^{n-9}=a^3$
Taking modulo $3$:
$a^3\equiv 1 \pmod 3$
$\implies a\equiv 1 \pmod 3$
A... | Check manually if there exist any solution with $n \le 9$.
Now assume $n>9$, and look at the equation $3^m+757=a^3$ modulo 7 (with $m=n-9$):
$3^m+1 \equiv 3^m+757 \equiv a^3 \equiv \{0,-1,1\} \pmod 7$
$3^m \equiv \{0,-1,-2\} \pmod 7$
$n \equiv \{3,5\} \pmod 6$
If $m=6k+3$,
$ 757 = a^3-(3^{2k+1})^3 = (a-b)(a^2+ab+b^2)$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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} |
simplifying $\frac{15^4+105^2-225\cdot210}{130^2-50^2}$ $x=\frac{15^4+105^2-225\cdot210}{130^2-50^2}$
Since $225 = 15^2$ and $210 = 2\cdot105$ the numerator is $(15^2-105)^2$ which is $120^2 = 14400$.
The denominator is $(130-50)(130+50)$ which is $14400$. The answer is $1$. The solutions for the exam have proven to be... | $\dfrac{15^4+105^2-225\cdot210}{130^2-50^2}=\dfrac{(15^2-105)^2}{(130+50)(130-50)}=\dfrac{120^2}{180\times80}=\dfrac{12^2}{18\times8}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to get the following summation of the series $\sum\limits_{n=0}^{\infty}\frac{1}{n!(n^4+n^2+1)}$ I am trying to find the sum $$\sum\limits_{n=0}^{\infty}\frac{1}{n!(n^4+n^2+1)}$$ I had factorized the sum as $$\frac{1}{2n(n!)}\left(\frac{1}{n(n-1)+1}-\frac{1}{n(n+1)+1}\right)$$ From this step, how to proceed?
| Given your decomposition of the main term, summation by parts ensures that the series equals
$$1+\frac{1}{2}\sum_{n\geq 1}\left(\frac{1}{n\cdot n!}-\frac{1}{(n+1)\cdot(n+1)!}\right)\left(1-\frac{1}{n^2+n+1}\right)$$
or
$$ 1+\frac{1}{2}\sum_{n\geq 1}\frac{n^2+n+1}{n(n+1)(n+1)!}\cdot\frac{n(n+1)}{n^2+n+1}=1+\frac{1}{2}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the recursive formula for $a_{n}={b^n}$ for $n>2$ How can I solve it,
and can anybody help me to find recursive formula for $a_{n}=3n^3$ for $n\geq 0$.
| Set $b_n=a_{n+1}-a_n=3(n+1)^3-3n^3=9n^2+9n+3$
Set $c_n=b_{n+1}-b_n=9(n+1)^2+9(n+1)+3-(9n^2+9n+3)=18n+18$
So $c_{n+1}=c_n+18$ and $c_0=18$
Then $b_{n+2}-b_{n+1}=b_{n+1}-b_n+18$
then $a_{n+3}-a_{n+2}-(a_{n+2}-a_{n+1})=a_{n+2}-a_{n+1}-(a_{n+1}-a_n)+18$
So $a_{n+3}=3a_{n+2}-3a_{n+1}+a_n+18$, with $a_0=0, a_1=3, a_2=24$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equations of the sides an equilateral triangle with centroid at the origin and one side is $x+y=1$
An equilateral triangle has its centroid at the origin and one side is $x+y=1$. Find the equations of the other sides.
My Attempt
$$
OD=\frac{1}{\sqrt{2}}\implies OC=\sqrt{2}\implies C=(-1,-1)\\
m_{AB}=m_1=-1\implies\t... | The coordinates of the vertices of the triangle are found to be $$A=(\frac {1+\sqrt 3}{2}, \frac {1-\sqrt 3}{2})$$
$$B=(\frac {1-\sqrt 3}{2}, \frac {1+\sqrt 3}{2})$$
$$C= (-1,-1)$$
The equations of lines passing through these points are
$$ CA: y+1=\frac {\sqrt 3 -1}{\sqrt 3+1}(x+1)$$
$$ CB: y+1=\frac {\sqrt 3 +1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Geometry - Find angle ∠C of a triangle given angle bisector and two segment ratios. In the triangle ABC, one has $∠A = 70^o$.
The point $D$ is chosen on the segment
$AC$ such that the angle bisector $AE$ intersects the segment $BD$ at the point $H$, with the following ratios
$$\dfrac{AH}{HE}=\dfrac 31, \>\>\> \dfrac{B... |
Examine the two ratios as follows,
$$\frac{5}{3}= \frac{BH}{HD} = \frac{\triangle ABE} {\triangle AED} $$
$$=\frac{\frac{BE}{BC}\triangle ABC}{\frac{EC}{BC}\frac{AD}{AC}\triangle ABC}= \frac{BE}{EC}\left(\frac{CD}{AD}+1\right)\tag{1}$$
Similarly,
$$\frac{3}{1} = \frac{AD}{CD}\left(\frac{EC}{BE}+1\right)\tag{2}$$
Combi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $
I need help with this limit:
$$ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $$
I've tried using the $$ \tan(a-b) = \frac{\tan a-\tan b}{1+\tan a\tan b} $$ with $(x+a)$ as $a$ and $(x)$ as $b$ but having difficulty getting beyond that.
| $$\lim_{a \to 0} \frac{\tan(x+a)- \tan x}{a}= \lim_{a \to 0} \frac{\frac{\sin(x+a)}{\cos(x+a)}- \frac{\sin x}{\cos x}}{a}\\
=\lim_{a \to 0} \frac{\cos x \sin(x+a)-\sin x \cos (x+a)}{a(\cos x \cos (x+a))}\\
= \lim_{a \to 0} \frac{\cos x (\sin x \cos a + \cos x \sin a) - \sin x (\cos x \cos a - \sin x \sin a)}{a\cos x \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3344722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The polynomial $x^3 + 2ax^2 + (2a^2 + b)x + c$ has three (real and not necessairily distinct) roots and $b$ is one of them. Prove that $(ac)^2 \le 3$.
The polynomial $$\large x^3 + 2ax^2 + (2a^2 + b)x + c$$ has three (real and not necessairily distinct) roots and $b$ is one of them. Prove that $(ac)^2 \le 3$.
I'm unc... | That $b$ is a root implies $-c = b^3+2ab^2+b^2+2a^2b$. Then $x^3+2ax^2+(2a^2+b)x+c = (x-b)(x^2+(2a+b)x+(2a^2+b+2ab+b^2)$ is valid. Since both roots of the quadratic are real, the discriminant must be non-negative: $(2a+b)^2-4(2a^2+b+2ab+b^2) \ge 0$, which is equivalent to $a \in [\frac{-b-\sqrt{-2b^2-4b}}{2},\frac{-b+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Evaluating $ \lim_{x\to \infty} x \left({{\left(\frac{x}{x+1}\right)}^{x}-\frac{1}{e}}\right)$
Evaluate the following:
$$
\lim_{x\to \infty}
x \left({{\left(\frac{x}{x+1}\right)}^{x}-\frac{1}{e}}\right)$$
I tried to first solve the interior portion as $1^\infty$ indeterminate form but ended up getting a different... | Let $$y={\left(\frac{x}{x+1}\right)}^{x}\implies \log(y)=x \log\left(\frac{x}{x+1}\right)=-x \log\left(1+\frac 1 x\right)$$
Now, using Taylor expansion
$$\log\left(1+\frac 1 x\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
x^3}+O\left(\frac{1}{x^4}\right)$$
$$\log(y)=-1+\frac{1}{2 x}-\frac{1}{3 x^2}+O\left(\frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the value of an infinite sum from a definite integral: It can be computed that $$\int_0^1 \frac{x^2+1}{x^4+x^2+1}dx = \frac{\pi}{2\sqrt{3}}.$$
The integral can be done using $$\frac{2(x^2+1)}{x^4+x^2+1} = \frac{1}{x^2-x+1}+\frac{1}{x^2+x+1}.$$
But I got stuck in the following part:
We need to show that $$1-\fr... | $$\frac{x^2+1}{x^4+x^2+1}
=\frac{(1-x^2)(x^2+1)}{(1-x^2)(x^4+x^2+1)}
=\frac{1-x^4}{1-x^6}=1-x^4+x^6-x^{10}+x^{12}-x^{16}+\cdots.$$
Now integrate this termwise.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Adding $\left(\frac{b}{c}\right)^{1/2}+\left(\frac{c}{b}\right)^{3/2}$ I am trying to refresh my algebra skills by reading the book Elements Of Algebra by Leonhard Euler. I came across a problem:
$$\text{Add}\;\left(\frac{b}{c}\right)^{1/2}\;\text{to}\;\left(\frac{c}{b}\right)^{3/2}. \qquad\qquad \text{Ans.}\;\frac{b^... | \begin{align*}
\left(\frac{b}{c}\right)^{1/2} + \left(\frac{c}{b}\right)^{3/2} = \frac{\sqrt{b}}{\sqrt{c}} + \frac{\sqrt{c^{3}}}{\sqrt{b^{3}}} = \frac{\sqrt{b}\times\sqrt{b^{3}} + \sqrt{c}\times\sqrt{c^{3}}}{\sqrt{cb^{3}}} = \frac{b^{2}+c^{2}}{b\sqrt{bc}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proper way to solve $\lim\limits_{x\rightarrow\infty}\left[\frac{6}{x}\right]\frac{x}{3}$ I made a problem for my friends, but not everyone agreed with my answer.
$$\lim_{x\rightarrow\infty}\left[\frac{6}{x}\right]\frac{x}{3}+\lim_{x\rightarrow\infty}\frac{6}{x}\left[\frac{x}{3}\right]+\lim_{x\rightarrow0}\left[\frac{6... | Note that when $x=6,\left \lfloor \frac{6}{x} \right \rfloor\frac{x}{3} = \left \lfloor \frac{6}{6} \right \rfloor\frac{6}{3} =2$.
While when $x>6$, say $x=8, \left \lfloor \frac{6}{x} \right \rfloor\frac{x}{3} = \left \lfloor \frac{6}{8} \right \rfloor\frac{8}{3} =0\cdot \frac{8}{3}=0$.
Similarly for any value of $x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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logarithm inequality: 2log5(x) - logx(125) I was given the following problem:
$2\log_{5}x$ - $\log_{x}125$ $< 1$
I tried to use the exponent law and the change of base law and I got this.
$\frac{\log_{}x^2}{\log_{}5}$ - $\frac{3log_{}5}{\log_{}x}$ $< 1$
Which I then changed back to:
$\frac{2log_{}x}{\log_{}5}$ - $\frac... | $2\log_{5}x - \log_{x}125 < 1$
$2\log_{5}x - 3\log_{x}5 < 1$
$2\log_{5}x - 3\frac{1}{\log_{5}x} < 1$At this point you need to be careful as $\log_5 x$ can be negative (which is where your mistake is).
If $\log_5 x > 0$ we have $2(\log_{5}x)^2 - \log_5 x -3 <0 \rightarrow 0 < \log_5 x < 1.5 \rightarrow 1 < x < 5\sqrt{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For $x\geq 2$,prove that$(x+1)\cos\left(\frac{\pi}{x+1}\right)-x\cos\left(\frac{\pi}{x}\right)>1$ For $x\geq 2$,prove that$$(x+1)\cos\left(\frac{\pi}{x+1}\right)-x\cos\left(\frac{\pi}{x}\right)>1$$
My Attempt:
Let $f(x)=x\cos\left(\frac{\pi}{x}\right)$
$$f'(x)=\cos\left(\frac{\pi}{x}\right)+\frac{\pi}{x}\sin\left(\frac... | Another possible way of proving your inequality is rearranging and then applying the MVT as follows:
\begin{eqnarray*}
(x+1)\cos\left(\frac{\pi}{x+1}\right)-x\cos\left(\frac{\pi}{x}\right) & > & 1\\
& \Leftrightarrow & \\
x\left( \cos \frac{\pi}{x+1} - \cos \frac{\pi}{x} \right) & > & 1 - \cos \frac{\pi}{x+1} \\
& \sta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Simplifying $\frac{a^x}{a^x+\sqrt{a}} + \frac{a^{1-x}}{a^{1-x}+\sqrt{a}}$ While solving a problem related to functions, I came across the following algebraic expression:
$$\frac{a^x}{a^x+\sqrt{a}} + \frac{a^{1-x}}{a^{1-x}+\sqrt{a}}$$
Although I tried, I couldn't find an easy way to simplify it.
Can anyone simplify th... | $\ {a ^ x \over a^x + \sqrt a} + {a ^ {1-x} \over
a ^ {1-x} + \sqrt a}$
Well, often it's a good idea to put over common denominator:
$\frac {a^x(a^{1-x} + \sqrt a) + a^{1-x}(a^x+\sqrt a)}{(a^x + \sqrt a)(a^{1-x} + \sqrt a)} =$
$\frac {(a^{x + 1-x} + a^x\sqrt a) + (a^{1-x+x} + a^{1-x}\sqrt a)}{a^{x+1-z} + a^x\sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find the value of $a$ and $b$, so the linear system stays consistant I was wondering if someone could take a look at my solution and tell me if i'm wrong. I feel like my solution is incorrect.
Given
$ \left[
\begin{array}{ccc|c}
1&1&2&2a-b+1\\
2&1&1&3a-b+1\\
1&2&5&3a-2b+6
\end{array}
\right] $ find the values of ... | Multiplying the first equation by-2 and adding to the second we get
$$-x_2-3x_3=-a+b-1$$
and the first by -1 and adding to the third
$$x_2+3x_3=a-b+5$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does each positive integer can be represent in difference of power sum? Let $a,b,m,n$ are positive integers with, if $m= 1$ then $n\ne 1$ and vice versa.
Let define
$S_m(a)=1^m+2^m+...+a^m$
and $S_n(b)=1^n+2^n+...+b^n$
Question
Show that every $k\in \mathbb{N}$ can be represented as
$$ k = S_m(a)-S_n(b)$$
Example
... | Probably, most positive integers $k$ cannot be represented in this way, though proving such a result is, I believe, rather out of range of current technology. Perhaps $k=163$ is the first non-representable $k$, though I didn't try very hard to verify this (actually proving this for even one fixed $k$ is likely extremel... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $~\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32~ $, find $~ a+b=~?$ If $$~\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32~ ,$$ find $~ a+b=~?$
I get $a-16b = 24$,
But how to get $a+b$?
a,b is positive integer
| $\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32$
Define $k(x) \equiv
\sqrt{64x^2 + ax + 7} - 8x$
$k(x)= (\sqrt{64x^2 + ax + 7} - 8x) \frac{\sqrt{64x^2 + ax + 7} + 8x}{\sqrt{64x^2 + ax + 7} + 8x}$
$$=\frac{64x^2 + ax + 7 - 64x^2}{\sqrt{64x^2 + ax + 7} + 8x}$$
$$=\frac{a + \frac7x}{\sqrt{64 + \frac ax + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of $f(x, y)$ at $(0, 0)$ I need to show
$$\lim_{(x, y) \rightarrow (0,0)} \frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} = 0
$$
Not really sure how to go about this without using the epsilon delta definition, which I would prefer not to. Any sort of help is appreciated.
Edit: I do have the inequality:
$$\frac{xy(x^2-y^2)}{... | Let $y=tx$. Then
$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}=\frac{x^2tx^2(1-t^2)}{x^3(1+t^2)^{3/2}}=x\frac{t(1-t^2)}{(1+t^2)^{3/2}}.$$
As the fraction remains bounded for all values of $t$, the limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
how to solve $\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}$? $\displaystyle\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+1}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+n}}\right)^{n}$
left=$\displaystyle\lim_{n\to\infty}{\l... | Following your solution from here
$$...= \lim_{n\to\infty}{n\ln{n \cdot2 \left(\sqrt{1+\frac{1}{n}}-1\right)}}=...$$
we have that
$$\sqrt{1+\frac{1}{n}}-1=\frac1{2n}-\frac1{8n^2}+O\left(\frac1{n^3}\right)$$
and therefore
$$\lim_{n\to\infty}{n\ln{n \cdot2 \left(\sqrt{1+\frac{1}{n}}-1\right)}}=\lim_{n\to\infty}{n\ln{\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Proof of the identity $F_{(n+1)k}=\frac{F_{2k}}{F_k}F_{nk}+(-1)^{k+1}F_{(n-1)k}\quad\forall k,n\in\mathbb{N}$ I'm reading the solution to an exercise I have been given in which they use the identity
$$F_{(n+1)k}=\frac{F_{2k}}{F_k}F_{nk}+(-1)^{k+1}F_{(n-1)k}\quad\forall k,n\in\mathbb{N}$$
where $F_n$ denotes the $n$th F... | $$ \varphi = \frac{1+\sqrt5}{2}$$
$$ F_n = \frac{\varphi^n +(-1)^{n+1}\cdot \varphi^{-n}}{\sqrt5} $$
Example: $$ F_{13} = \frac{\varphi^{13} + \varphi^{-13}}{\sqrt5} = 233 $$
Then: $$ \frac{F_{2k}}{F_k}=\frac{\frac{\varphi^{2k} +(-1)^{2k+1}\cdot \varphi^{-2k}}{\sqrt5}}{\frac{\varphi^k +(-1)^{k+1}\cdot \varphi^{-k}}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3370773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor=2018$, find $x$ If $x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor=2018$, find $x$.
My working: if $x$ is positive then by estimation it must be in $(6,7)$ and for this interval I : $x\Big\lfloor x \big\lfloor x\lfloor x \r... | $$x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor=2018$$
Let $f(x) = x\Big\lfloor x \big\lfloor x\lfloor x \rfloor \big\rfloor \Big\rfloor$
Let $\xi = -\dfrac{6862735}{1037232} = -6 \dfrac{639343}{1037232}$
Then $f(\xi) - 2018 = -\dfrac{1}{1037232} = -\dfrac{1}{2^4 3^3 7^4}$
Let $x = \xi + \delta$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Nice olympiad inequality :$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$ I have this to solve :
Let $x,y,z>0$ such that $x+y+z=3$ then we have :
$$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$$
I try to use Jensen's inequality but the function $f(x)=\fra... | Using AM-GM, we have
$$4y^3+3=y^3+y^3+y^3+y^3+1+1+1\geq 7\left((y^3)^41^3\right)^{1/7}=7y^{12/7}.$$
So
$$\frac{xy^2}{4y^3+3}\leq\frac{xy^2}{7y^{12/7}}=\frac{xy^{2/7}}{7}.$$
Now, note by Holder's inequality that
$$\sum_{cyc}xy^{2/7}\leq (x+y+z)^{5/7}(xy+yz+zx)^{2/7}.$$
By Cauchy, $xy+yz+zx\leq \frac{(x+y+z)^2}{3}$. Thu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Partial derivatives problem, find $\frac{\partial z}{\partial x}$ Assume that $z$ and $w$ are differentiable functions of $x$ and $y$, satisfying the equations $xw^3+yz^2+z^3=-1$ and $zw^3-xz^3+y^2w=1.$ Then, at $\left(x,y,z,w\right)=\left(1,-1,-1,1\right)$, $\ \frac{\partial z}{\partial x}=$?
My solution
$xw^3+yz^2+z... | Bavo, your solution is good. The method is correct and simple.
If an even simpler method is wanted, Mr.Mathematician asks his friend Mr.Physicist. But the purists will not be fully happy :
$$\begin{cases}
xw^3+yz^2+z^3=-1\\
zw^3-xz^3+y^2w=1
\end{cases}\quad\to\quad
\begin{cases}
(1+dx)(1+dw)^3+(-1)(-1+dz)^2+(-1+dz)^3=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \cos^2x\sin^4x\mathrm{d}x$
Evaluate integral $$\int \cos^2x\sin^4x\mathrm{d}x.$$
Attempt. Setting $\tan x=t$, gives:
$$\int \cos^2x\sin^4x\mathrm{d}x =\int \frac{1}{1+t^2} \,\left(\frac{t^2}{1+t^2}\right)^2 \frac{\mathrm{d}t}{1+t^2}=\int \frac{t^4}{(1+t^2)^4} \mathrm{d}t,$$
which does not seem to be el... | Use that $$\cos^2(x)=1-\sin^2(x)$$ and your integral will be
$$\int\sin^4(x)-\sin^6(x)dx$$
For your work:
$$\int \sin(x)^4dx=-1/4\, \left( \left( \sin \left( x \right) \right) ^{3}+3/2\,\sin
\left( x \right) \right) \cos \left( x \right) +3/8\,x
+C$$
and
$$\int\sin(x)^6dx=-1/6\, \left( \left( \sin \left( x \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
Proving $\sum _{k=1}^{\infty } J_k(k z)^2=\frac{1}{2 \sqrt{1-z^2}}-\frac{1}{2}$ How to prove
$$\sum _{k=1}^{\infty } J_k(k z){}^2=\frac{1}{2 \sqrt{1-z^2}}-\frac{1}{2},\ |z|<1$$
The supplementary one $\sum _{k=1}^{\infty } J_k(k z)=\frac{z}{2 (1-z)}$ might be useful. Any help will be appreciated.
| We can use the integral representation for the product of Bessel functions DLMF
\begin{equation}
J_{\mu}\left(x\right)J_{\nu}\left(x\right)=\frac{2}{\pi}\int_{0}^{\pi/2}J_{\mu+\nu}\left(2x\cos\theta\right)\cos\left((\mu-\nu)\theta\right)\mathrm{d}\theta
\end{equation}
with $\mu=\nu=k$ and $x=kz$:
\begin{equation}
J_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Verifying the integral solution: $I= \int \frac{\sin x}{\sqrt{1+2\cos x}}dx$
Evaluate:
$$
I= \int \frac{\sin x}{\sqrt{1+2\cos x}}dx
$$
I got an answer which does not match the keys section of the book I'm solving and want to verify which one is correct.
Start by recalling:
$$
\cos 2x = \cos^2 x-\sin^2 x
$$
Therefor... | Take $u=1+2\cos{x}$
$\frac{-1}{2}du=\sin{x}dx$
So the integral becomes $$\frac{-1}{2}\int \frac{1}{\sqrt{u}}du=-\sqrt{u}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the principal root of $\sqrt{-i}$? Where is the mistake in this solution?
$$\sqrt{-i}=\sqrt {\cos \frac{3 \pi}2+i\sin \frac{3 \pi}2}=\cos \frac{3 \pi}4+i\sin \frac{3 \pi}4=-\frac{\sqrt 2}2+i \frac{\sqrt 2}2$$
WA gives me different result:
$$\frac{\sqrt 2}2-i \frac{\sqrt 2}2$$
Why the principal root must be ... | Let $\sqrt{-i} = a + bi$, so $-i = (a+bi)^2 = a^2 - b^2 + 2abi$.
Equating real values, $0 = a^2 - b^2$, and equating imaginery parts, $-1 = 2ab$.
Solving these simultaneously gives us either $a=\frac{1}{\sqrt 2}, b=\frac{-1}{\sqrt 2}$ or $a=\frac{-1}{\sqrt 2}, b=\frac{1}{\sqrt 2}$.
Thus, $\sqrt{-i} = \pm \frac{1}{\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Is $\sum_{k=1}^{n} \cos(k^ {1/3} )$ bounded by a constant M? I understand what $\sum_{k=1}^{n} \cos(k)$ is bounded by a constant, but I don't have any idea how to prove if $\sum_{k=1}^{n} \cos(k^ {1/3} )$ is bounded or not.
| Not bounded. The blocks where $\cos(^{1/3})>1/2$
have length going to $\infty$.
First, $\cos(z)>1/2$ if $-\pi/3 < z < \pi/3$.
For each $n \in \mathbb Z$ we have $\cos(z)>1/2$ if $2\pi n-\pi/3 < z < 2\pi n+\pi/3$.
$\cos(k^{1/3}) > 1/2$ if $2\pi n-\pi/3 < k^{1/3} < 2\pi n+\pi/3$
$\cos(k^{1/3}) > 1/2$ if $(2\pi n-\pi/3)... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find number of distinct terms in a multinomial expansion?
Find the number of distinct terms in the expansion of
$$\left(x+\frac{1}{x}+\frac{1}{x^2}+x^2\right)^{15}$$
(with respect to powers of $x$)
I saw that the formula for the number of distinct terms (or dissimilar) in a multinomial expansion $(x_1+x_2+x_3+... | Write it as:
$$\left(x+\frac{1}{x}+\frac{1}{x^2}+x^2\right)^{15}=\frac{(x^4+x^3+x+1)^{15}}{x^{30}}$$
So, it is equivalent to:
$$(x^4+x^3+x+1)^{15}=(x^3+1)^{15}(x+1)^{15}=\\
(x^{45}+15x^{42}+\cdots+15x^3+1)(x^{15}+15x^{14}+\cdots+15x+1)=\\
x^{60}+30x^{59}+\cdots15x+1.$$
Hence, there are $61$ distinct powers of $x$.
Note... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3380253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Volume of $\{(x,y,z)\mid 3x^2-4y+2y^2+2z-3=0, z>0\}$ Consider S=$\{(x,y,z)\mid| 3x^2-4y+2y^2+2z-3=0\}$
How can I determine the volume of $\{(x,y,z)\mid 3x^2-4y+2y^2+2z-3=0, z>0\}$? That is the part of $S$ over the x-y axis.
My wrong approach was to integrate $z^2\pi=(\frac{1}{2}(3-2y^2+4y-3x^2))^2\pi$ from $z=0$ to th... | The first thing I would do is "complete the square" (that's the second time today I have written that!). $2y^2- 4y= 2(y^2- 2y)= 2(y^2- 2y+ 1- 1)= 2(y^2- 2y+ 1)- 2= 2(y- 1)^2- 2$. So we can write the formula as $3x^2+ 2(y- 1)^2- 2+ 2z- 3= 0$, $z= 5/2- (3/2)x^2- 2(y- 1)^2$. Now let u= y- 1. We have $z= 5/2- (3/2)x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the coefficient of $x^9$ in the expansion of $(1+9x+27x^2+27x^3)^6$ Find the coefficient of $x^9$ in the expansion of $(1+9x+27x^2+27x^3)^6$.
Consider,
$$(1+9x+27x^2+27x^3)^6 = \sum \frac{6!}{a!\cdot b!\cdot c!\cdot d!}(1^{a}\cdot9x^{b} \cdot (27x^2)^{c} \cdot (27^3)^d)$$
Power of $x=x^{b+2c+d}$
So,
*
*$b +2c... | First note that $1+9x+27x^2+27x^3=(1+3x)^3$. So the problem is the same as calculating the coefficient of $x^9$ in $(1+3x)^{18}$. Which is much easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find a parametric representation for the line of intersection of the planes
Find a parametric representation for the line of intersection of the planes with equations $x+y+z = 3$ and $2x-y-z = 5$, and find the point where the line intersects the plane with equation $x-y = 2$.
First I get the two norm vectors of the g... | We are working with
\begin{cases}
\pi_1: & x+y+z=3\\
\pi_2: & 2x-y-z=5\\
\end{cases}
Since their normal vectors are non-parallel, we know that the planes are not parallel. As such, we know they intersect in a line at some point with some direction vector.
Suppose the direction vector can be written as some scalar mul... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Substitution integral with $t=x^3$
$$
\int \frac{3}{t(3+\sqrt[3]{t})} \mathrm{d} t . \text { Substitution } t=x^{3}
$$
$$
\int \frac{3}{x^3(3+\sqrt[3]{x^3})}3x^2
$$
$$
\int \frac{9}{x(3+{x})}
$$
$$
9\int \frac{1}{3x+{x^2}}
$$
Now i'm stuck
my calculation so far
| $$\int \frac{3}{t(3 + t^{1/3})} \, dt = \int \frac{1}{3 + t^{1/3}} \cdot 9t^{-1/3} \cdot \frac{1}{3} t^{-2/3} \, dt,$$ so with the substitution $$u = 3 + t^{1/3}, \quad du = \frac{1}{3} t^{-2/3} \, dt, \\ t^{-1/3} = \frac{1}{u-3},$$ we obtain $$\int \frac{3}{t(3+t^{1/3})} \, dt = \int \frac{1}{u} \cdot \frac{9}{u-3} \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
On the exponentiation of three-dimensional numbers. There exists a general formula to calculate $({a+bi})^{c+di}$. However, it there a formula to calculate exponentiation with three-dimensional numbers, such as $({a+bi+cj})^{d+ei+fj}$ ?
| Yes. But what exactly 3-dimensional algebra are you talking about? There are several. In the linked post are descriptions of 3 3-dimensional commutative number systems.
If we define
$1=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$
$j=\left(
\begin{array}{ccc}
0 & 1 & 0 \\
0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Difference quotient of differentiable functions $Q. \text { For a differentiable function } f : \mathbb { R } \rightarrow \mathbb { R } \text { define the differential quotient}$
\begin{array} { l }{ \left( D _ { x } f \right) ( h ) = \frac { f ( x + h ) - f ( x ) } { h } ; h > 0 } \\ { \text { consider numbers of the ... | Not sure if this is what your looking for, but your last implication is incorrect since
$\frac{f'(x)}{1+ \varepsilon} = f'(x)$ and $\varepsilon >0$ implies $$f'(x) = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sqrt{2} +\sqrt{n}$ is irrational Suppose $\sqrt{2} + \sqrt{n}$ is rational,
so it can be represented like so: $\sqrt{2} + \sqrt{n} = {x \over y}$
then $\sqrt{2} = {{x \over y} - \sqrt{n}}$
so $\sqrt{2} = {x - y\sqrt{n}\over y}$
but $\sqrt{2}$ is irrational, which leads to a contradiction.
Does this proof ... | The polynomial $$\left(x^2-\left(\sqrt{2}-\sqrt{n}\right)^2\right)
\left(x^2-\left(\sqrt{2}+\sqrt{n}\right)^2\right) = x^4-2(n+2)\,x^2+(n-2)^2$$
has $\sqrt2+\sqrt n$ as one of its roots. Any rational divisor would have to be an integer factor of $(n-2)^2$ by the rational roots theorem.
But $\sqrt 2+\sqrt n$ is not a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
reducing $\cos{x} + 2\sin{x}$ to $R\cos{x - a}$ finds strange phase-shift So, I'm analysing the function $\cos{x} + 2\sin{x} (*)$ and determining whether it is convertible to the single function $R\cos{x - a}$ for $a \in (0, \frac{\pi}{2}); R > 0$; so I begin by observing the maximum/minimum of $(*)$ and determining it... | Let's say you want to write, with $r\ge0$:
$$a\cos \theta+b\sin\theta=r\cos(\theta-\phi)=r\cos \phi\cos \theta+r\sin \phi\sin \theta$$
We are done if $a=r\cos\phi$ and $b=r\sin\phi$, hence $a^2+b^2=r^2$, which gives $r=\sqrt{a^2+b^2}$.
Then any $\phi$ such that $\cos\phi=\dfrac{a}{\sqrt{a^2+b^2}}$ and $\sin\phi=\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Can we simply further the expression? We have the set $\left \{z\in \mathbb{C}\mid z\neq 1, \ \left |\frac{z}{z-1}\right |\leq 1\right \}$
Let $z=x+yi$.
We have that \begin{align*}\frac{z}{z-1}&=\frac{x+yi}{(x-1)+yi}=\frac{\left (x+yi\right )\left ((x-1)-yi\right )}{\left ((x-1)+yi\right )\left ((x-1)-yi\right )} \... | We have
$$\left |\frac{z}{z-1}\right |=\frac{|z|}{|z-1|}=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2-2x+1}}\leq 1 \iff x\le \frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
General formula for the partial sum of the series $\sum_{i=1}^\infty \ln\frac{k(k+2)}{(k+1)^2}$
How would I find the formula for the partial sum $S_n $ of this series? $$\sum_{k=1}^\infty \ln\frac{k(k+2)}{(k+1)^2}$$
I know that $S_1=\ln(3/4) $, $S_2=\ln(2/3) $, $S_3 =\ln(5/8)$, and $S_4=\ln(3/5)$, but I don't see any... | Consider the partial sum,
$$S_{n} = \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) $$
which can be evaluated in two ways. The first being:
\begin{align}
S_{n} &= \sum_{k=1}^{n} \ln\left(\frac{k \, (k+2)}{(k+1)^{2}}\right) \\
&= \sum_{k=1}^{n} ( \ln(k) + \ln(k+2) - 2 \, \ln(k+1) ) \\
&= \sum_{k=2}^{n} \ln(k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$ƒ(x,y)=\sqrt{x^2+y^2}+\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-2y+1}+\sqrt{x^2+y^2-6x-8y+25}$ QUESTION: Let $ƒ(x,y)=\sqrt{x^2+y^2}+\sqrt{x^2+y^2-2x+1}+\sqrt{x^2+y^2-2y+1}+\sqrt{x^2+y^2-6x-8y+25}$
(A) Minimum value of $ƒ(x,y)= 5+\sqrt2$
(B) Minimum value of $ƒ(x,y)= 5-\sqrt2$
(C) Minimum value occurs of $ƒ(x,y)$ for $x=\frac... | Hint. Note that the centers of the four circles are:
$$O(0,0),A(0,1),B(1,0),C(3,4)$$
For the shortest distance, the point $(x,y)$ must lie on the intersection of the line passing through $(0,0)$ and $(3,4)$ and the line passing through $(0,1)$ and $(1,0)$. Hence:
$$|OC|+|AB|=5+\sqrt2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Three polynomials which have the same value for a variable
Let $P_1(x)= ax^2-bx-c$, $P_2(x)=bx^2-cx-a$, and $P_3(x)= cx^2-ax-b$ be three quadratic polynomials, where $a,b$, and $c$ are non-zero real numbers. Suppose that there exists a real number $k$ such that $$ P_1(k) = P_2(k)= P_3(k) $$ Prove that $a=b=c$.
Hello ... | You can't do that because we need to show that this is the unique solution.
As an alternative, since the conditions
*
*$(a-b)k^2-(b-c)k-(c-a)=0$
*$(b-c)k^2-(c-a)k-(a-b)=0 $
*$(c-a)k^2 - (a-b)k - (b-c)=0$
are equivalent to the system $Mx=0$
$$\begin{bmatrix}A&-B&-C\\B&-C&-A\\C&-A&-B\end{bmatrix}\begin{bmatrix}k^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Primitive instances where $c^3|(a^3+b^3)$ with $a,b,c\in\mathbb{N}$ It is known that by Fermat's Last Theorem there are no solutions to $a^3+b^3=c^3$ for $a,b,c\in\mathbb{N}$. I wondered about how multiplying the $c^3$ by a constant would change this fact.
Accordingly, I have been looking into instances where $c^3|(a^3... | It's actually very easy to identify an infinite number of solutions. Suppose that $m^3+1$ is divisible by $3^3=27$, whichnisctrue for $m=8$. Then
$(m+9)^3+1=(m^3+1)+(27m^2+243m+729)$
is also divisible by $27$. Having $1$ as one of the cubes guarantees relative primarily for all these solutions, but we can reasonably... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Expected hitting time of a certain state in a Markov chain
Let's consider the following Markov chain. If we start at state $2$, compute the probability to hit $4$ and the expected time until it happens.
The probability to hit $4$ in $n$ steps starting at $2$ is $Pr(X_n=4|X_0=2)=(\frac{1}{2})^n$ if $n$ is even, and $... | The general way for computing mean hitting times of any absorbing state is well known, rather easily proven and in this Wikipedia article.
To get the mean hitting time of a specific state (or subset of states), we simply condition the outcome on hitting that state. For this question, the modified chain looks like:
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is the closed form of this sum? What is the closed form of this sum?
$$
S = \sum_{k\ge1, r>s\ge 1}\frac{1}{k^2(r^2 + s^2)^2}
$$
Note: Though originally posted for Pythagorean triplets, their was an flaw in the question which changed the meaning of the question and was answered accordingly. I will post a separate q... | \begin{align*}
S &= \sum_{k \geq 1} \frac{1}{k^2} \sum_{r>s\geq 1} \frac{1}{(r^2 + s^2)^2} \\
&= \frac{\pi^2}{6} \sum_{r>s\geq 1} \frac{1}{(r^2 + s^2)^2}
\end{align*}
Note that $\frac{1}{(r^2 + s^2)^2}$ is unchanged by $\{r \mapsto -r\}$, $\{s \mapsto -s\}$, and $\{r \leftrightarrow s\}$. Applying these symmetri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 1,
"answer_id": 0
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Evaluate $\sum_{n=2}^{\infty}\frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$ Evaluate $\sum_{n=2}^{\infty}\frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$
I am not able to convert it into telescopic function.
| So the series
$$
\sum_{n=2}^{\infty}a_n\tag{1}
$$
has to be written in the form
$$
\sum_{n=2}^{\infty} (b_{n} - b_{n-1});\tag{2}
$$
therefore
$$
b_1 = \sum_{n=2}^{\infty} a_n, \\
b_2 = \sum_{n=3}^{\infty} a_n, \\
b_3 = \sum_{n=4}^{\infty} a_n, \\
\cdots \tag{3}
$$
If search $b_n$ in the form $$b_n = \dfrac{1}{2^n}\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Diophantine equations with cubic roots.
Find all positive, integral $x, y$ such that
$$x^3-y^3=xy+61$$
Little work:
$$(x-y)(x^2+xy+y^2)-xy=61$$
Also $x^3-xy-y^3$ seems to to be unable to be "completed by the square," or factored, so I'm stuck. Thanks!
| More generally,
suppose
$x^3-y^3=xy+m
$
where $x, y, m > 0$.
I will show that
$m \ge x^2+y^2$,
$x \ge y+1$,
$x \lt \sqrt{m}$
and
$y
\le \frac12(\sqrt{2m-1}-1)
$.
If $m=61$ then
$x \le 7$
and
$y
\le \frac12(\sqrt{120}-1)
\approx 4.97
$
so
$y \le 4
$
and
$x \le \sqrt{61-y^2}
$.
We must have
$x > y$
so that
$x \ge y+1$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Please help with this limit of a sequence: $\lim\limits_{n\to\infty} \bigg(n\sqrt[3]{1+\frac1n} - n\sqrt[3]{1+\frac1{n^2}}\bigg)$ I have this limit of a sequence: $$\lim_{n\to\infty} \bigg(n\sqrt[3]{1+\frac1n} - n\sqrt[3]{1+\frac1{n^2}}\bigg).$$
Can I modify this expression to this expression by knowing that fractions ... | Let $a=\sqrt[3]{n^3+n^2},\,b=\sqrt[3]{n^3+n}$ so $a-b=\frac{a^3-b^3}{a^2+ab+b^2}=\frac{n^2-n}{a^2+ab+b^2}\approx\frac{n^2}{n^2+n^2+n^2}=\frac13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Showing $8(\arctan(\sqrt2-1)) = \pi$ My friend is saying $$8(\arctan(\sqrt2-1)) = \pi$$
I know $4\arctan(1) = \pi$
I thought of $4\arctan(\sqrt2-1) + 4\arctan(\sqrt2-1)$
But I got stuck somewhere
How to solve?
| After taking the $ \tan $ on both sides you get:
$$ \tan\frac{\pi}{8} = \sqrt{2}-1 $$
Note these two double angle formulas $ \sin2x = 2\sin x \cos x $ and $ \cos 2x = 2\cos^2x-1 $. Now we have:
$$ \tan\frac{\pi}{8} = \frac{\sin\frac{\pi}{8}}{\cos\frac{\pi}{8}} = \frac{\sin\frac{\pi}{8}}{\cos\frac{\pi}{8}} \cdot \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Counting the ways on grid if one can move from $(x,y)$ to $(x+a, x+b)$ for arbitrary $x,y,a,b\geq 0$. On 2 dimensional grid, consider the situation that one can move from $(p,q)$ to $(p+α,q+β)$ at once for arbitrary integer $p,q,α,β\geq 0 \land (α,β)\neq(0,0)$.
I want to count how many ways are there to move from (0,0)... | We consider non-negative integers $x,y$ and to get a first impression we start calculating the first few values of
\begin{align*}
\sum_{j\geq 0}\binom{x}{j}\binom{y}{j}2^{x+y-{j+1}}\tag{1}
\end{align*}
We write $j\geq 0$ and recall $\binom{p}{q}=0$ if $q>p$. The values of (1) are given in the picture below and we o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve the "four" variables problem Given x, y, z, and w are real numbers which satisfy these three equation:
$$x^{2} + 5z^{2} = 10$$
$$yz - xw = 5$$
$$xy + 5zw = \sqrt{105}$$
Find the value of $y^{2} + 5w^{2}$
Can anyone give a hint? I don't even know where to start
| Method$\#1:$
Solve the two simultaneous equations in $x,z$
$$z=\dfrac{aw+5y}{y^2+5w^2},x=\dfrac{ay-25w}{y^2+5w^2}$$
Replace the values of $z,w$ in $$10=x^2+5z^2$$
$$10=\dfrac{a^2y^2+625w^2-50ayw+5(a^2w^2+25y^2+10ayw)}{(y^2+5w^2)^2}$$
$$=\dfrac{(a^2+125)y^2+(625+5a^2)w^2}{(y^2+5w^2)^2}=\dfrac{a^2+125}{y^2+5w^2}$$
where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\int_ \frac\pi6^\frac\pi2 \frac {3dx}{2\sin2x+1}$ $$\int_ \frac\pi6^\frac\pi2 \frac {3dx}{2\sin2x+1}$$
I tried using this substitute:
$$(\tan x=u),(sin2x= \frac{2\tan x}{1+\tan^2x} =\frac{2u}{u^2+1}), (dx=\frac{1}
{u^2+1}du)$$
and after long answer I get:
$$\frac{6+3\sqrt{3}}{6+4\sqrt{3}}.\ln(\frac{\tan x+2
-\sqrt{... | Since
$$\lim_{x\to\frac{\pi}{2}}\frac{\tan x +2-\sqrt{3}}{\tan x +2+\sqrt{3}}=\lim_{x\to\frac{\pi}{2}}\frac{1+\frac{2-\sqrt{3}}{\tan x}}{1+\frac{2+\sqrt{3}}{\tan x}}=1.$$
Then you can get what you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos\sqrt{x}}}$ without using the l'Hospital rule The task is to evaluate
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}}.$$
I tried to use some trigonometric identities such as
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)... | Using $\operatorname{sinc}y:=\frac{\sin y}{y}$ so $\lim_{y\to0}\operatorname{sinc}y=1$, the last expression you obtained is$$\frac12\lim_{x\to0}\frac{x\operatorname{sinc}^2\frac{\sqrt{x^2}}{2}}{\operatorname{sinc}^2\frac{\sqrt{x}}{2}}=\frac12\lim_{x\to0}\frac{x\cdot1}{1}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
What are the roots of this equation? (Quadratic Equation) $\frac{x}{\sqrt{1-x^2}}=A \Rightarrow \frac{x^2}{1-x^2}=A^2$
$\Rightarrow x^2=A^2(1-x^2)=A^2-A^2x^2$
$\Rightarrow x^2+A^2x^2=A^2$
$\Rightarrow x^2(1+A^2)=A^2$
$\Rightarrow x^2(1+A^2)-A^2=0$
I have tried this;
$\Delta=b^2-4ac=-4(1+A^2)(-A^2)=4(1+A^2)(A^2)$
$x_{1,... | Your algebra is fine. Another way of writing your answer (in a more simplified form) would be
$$x=\pm\frac{A}{\sqrt{1+A^2}}$$
since $$\frac{\sqrt{t}}t=\frac1{\sqrt{t}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}$ using the definition of $e$
Evaluate $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}.$
I know that $e^x=\lim\limits_{n\to\infty}\left(1+\dfrac{x}{n}\r... | Your proof is nice, we have indeed
$$\left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}=\\=\left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-2n}\cdot\left(1+\frac{1}{n+1}\right)^{-1}=$$
$$=\left(\frac{(n+1)^2}{n(n+2)}\right)^{n^2}\cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Calculating the exponential of an upper triangular matrix
Find the matrix exponential $e^A$ for
$$ A = \begin{bmatrix}
2 & 1 & 1\\
0 & 2 & 1\\
0 & 0 & 2\\
\end{bmatrix}.$$
I think we should use the proberty
If $AB = BA$ then $e^{A+B} = e^A e^B$.
We can use that
$$\begin{bmatrix}
2 & 1 & 1\\
0 & 2 & 1\\
0 & 0 & 2\\
... | Edit: As pointed out by @XTChen there is a much easier way to do this but I detail evaluation of the requested exponential nonetheless.
Note that
$$\pmatrix{1&1&1\\0&1&1\\0&0&1}^n=\pmatrix{1&n&n(n+1)/2\\0&1&n\\0&0&1}$$
\begin{align}
\sum_{n=0}^\infty & \frac1{n!}\pmatrix{1&1&1\\0&1&1\\0&0&1}^n \\
&=\sum_{n=0}^\infty\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Find $p, q$ such that polynomial $P(x) = 6x^4 - 7x^3 + px^2 + 3x + 2$ is divisible by $x^2 - x + q$ I'm having difficulties with this problem:
Find $p, q$ such that polynomial $P(x) = 6x^4 - 7x^3 + px^2 + 3x + 2$ is divisible by $x^2 - x + q$.
I'm aware of the Bezout's Theorem, but I don't know how to use it in this pr... | We can write
$$6x^4 - 7x^3 + px^2 + 3x + 2=(ax^2+bx+c)(x^2-x+q)$$
Comparing $x^4$ coefficients we see that $a=6$. Now comparing $x^3$ coefficients we get $b=-1$:
$$6x^4 - 7x^3 + px^2 + 3x + 2=(6x^2-x+c)(x^2-x+q)$$
Looking at the $x^0$ and $x^1$ coefficients we have $cq=2$ and $q+c=-3$. By Viète's relations $c,q$ are th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3414682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Proof of equality of complex numbers with conditions
I have the next question:
Proof that $\operatorname{Re}\left \{ z_{1}\cdot \bar{z_{2}} \right \}=|z_{1}|\cdot |z_{2}|$ only if: $\arg z_{1} -\arg z_{2} =2\cdot \pi \cdot n$ ( $\;(n\in \mathbb{Z})$.
This is my solution:
$z_{1}=x+iy, z_{2}=a+ib$
$z_{1}\cdot \ba... | There is a hidden thing you should be careful of when you square both sides you assuming that $ax+by>0$ so the equality make sense . But this thing means $\cos \phi \cos \theta +\sin \phi \sin \theta >0$ hence $\cos(\theta -\phi)>0$ so the difference is either in first or fourth quarter but if the difference is multi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How can I find the eigenvalues of $\alpha$? The question is given below:
My trial is:
Let $\lambda $ be an eigenvalue of $\alpha,$ then
$$ \alpha (
\left[ {\begin{array}{cc}
a \\
b \\
c \\
d \\
\end{array} } \right]
) = \lambda
\left[ {\begin{array}{cc}
a \\
b \\
c \\
d \\
\end{arr... | Another way to do:
$$\alpha(a,b,c,d)=(b+c,c,0,0).$$
Assuming that $e_1,\dots, e_4$ is the canonical basis of $\mathbb{R}^4$. Then,
$$\alpha(e_1)=(0,0,0,0)$$
$$\alpha(e_2)=(1,0,0,0)$$
$$\alpha(e_3)=(1,1,0,0)$$
$$\alpha(e_4)=(0,0,0,0)$$
The matrix associated to $\alpha$ is:
$$[\alpha]=\begin{pmatrix}
0 & 1 & 1 & 0 \\
0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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A general formula to generate functions of power series I have been playing with Maclaurin series lately, I have been able to come across this:
$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5...$
$\dfrac{1}{(1+x)^2}=1-2x+3x^2-4x^3+5x^4-6x^5+7x^6...$
I found out by accident that:
$\dfrac{1-x}{(1+x)^3}=1-2^2x+3^2x^2-4^2x^3+5^2x^4+6^... | Given any analytic function $f ,f(x)=\sum_{n=0}^{\infty}f^n(0)x^n/n!$.You are guaranteed to get a series representation iff $R_n(x)=f(x)-\sum_{k=0}^{n}f^k(0)x^k/k! \to 0$ pointwise as $n\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Methods of evaluating $ \sum_{k=1}^\infty \frac{(m+k)!}{k!}\frac{1}{5^k}$? I am interested in ways of evaluating the following infinite seris:
$$
\sum_{k=1}^\infty \frac{(m+k)!}{k!}\frac{1}{5^k}.
$$
I already know the answer from Wolfram Alpha but I would like to see some methods of evaluating it as I haven't been able... | This is the non-calculus way. Let the sum be $S_m$. Then,
\begin{align}
& 5S_m-S_m =\sum_{k=0}^{\infty}\frac{(m+k+1)\cdots(k+2)}{5^k}-\sum_{k=1}^{\infty}\frac{(m+k)\cdots(k+1)}{5^k}\\
=&(m+1)!+m\sum_{k=1}^{\infty}\frac{(m+k)\cdots(k+2)}{5^k}\\
=&(m+1)!+5m\sum_{k=2}^{\infty}\frac{(m-1+k)\cdots(k+1)}{5^k}\\
=& (m+1)!+5m\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Compute $\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx$ How to prove
$$I=\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx=\frac16\left(\frac{\pi^3}{12}-\pi\operatorname{Li}_2\left(\frac13\right)\right)$$
This problem is proposed by Cornel which can be f... | Evaluating the blue integral:
First we write
$$\frac1{1+3\cos^2(x)}=\frac{1}{5+3\cos(2x)}$$
Using the same identity in the post body
$$\sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}, \ |p|<1\tag1$$
But lets manipulate the denominator to have it in the form of $\frac1{5+3\cos(x)}$:
$$\frac1{1-2p\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
How can we continue at the induction step? I want to show by induction that $$\sum_{i=1}^n i^2\geq 3n+2$$ for all $n\in \mathbb{N}$ with $n\geq 3$.
I have done the following:
Base Case: For $n=3$ we have that $\displaystyle{\sum_{i=1}^3i^2=1^2+2^2+3^2=1+2+9=12}$ and $\displaystyle{3n+2=3\cdot 3+2=9+2=11}$. It holds ... | We just need to show that
$$\sum_{i=1}^{k+1}i^2=\sum_{i=1}^{k}i^2+(k+1)^2\overset{(IH)}{\geq}3k+2+(k+1)^2\overset{(?)}\ge 3(k+1)+2$$
which is true indeed
$$3k+2+(k+1)^2\ge 3(k+1)+2$$
$$3k+2+(k+1)^2\ge 3k+5$$
$$(k+1)^2\ge 3$$
for $k\ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$ Evaluate :$$ \int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$
using residue theorem. The integral is even and it can be written as :
$$ \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$
The only pole with a positive imaginary part is $ \frac{i}{\sqrt2} $... | Note first that$$\frac{x}{1+2x^2}=\frac14\left(\frac{1}{x-i/\sqrt{2}}+\frac{1}{x+i/\sqrt{2}}\right).$$Hence$$\left(\frac{x}{1+2x^2}\right)^2=\frac{1}{16}\left(\frac{1}{(x-i/\sqrt{2})^2}+\frac{1}{(x+i/\sqrt{2})^2}+\frac{2}{x^2+1/2}\right)\\\frac{1}{16}\left(\frac{1}{(x-i/\sqrt{2})^2}+\frac{1}{(x+i/\sqrt{2})^2}+i\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $4x \equiv 7 \pmod{15}$ and $3x \equiv 5 \pmod{16}$ (different exercises) This is how I solved each:
$$4x \equiv 7\pmod {15} \\
4x - 7 \equiv 0\pmod{15} \\
4x-7 = 15k \Leftrightarrow 4x-15k= 7 \\
$$
$$15 = 4*3+3 \\
4=3*1+1\\
3=3*1+0$$
$$1 = 4-3*1 \\
3 = 15-4*3 \\
1 = 4 - (15-4*3)*1 \\
1 = 4-15+4*3 \\
1 = 4*4-15*1... | With such small coefficients ($4$ and $3$), here is a quick and easy way to solve the congruences:
Mod $15$: $4x\equiv7\equiv22\equiv37\equiv\color{red}{52}=4\times13\implies x\equiv 13$
Mod $16$: $3x\equiv5\equiv\color{red}{21}=3\times7\implies x\equiv 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find two elementary matrices $E_1$ and $E_2$ such that $E_2E_1A=B$. I have two matrices $A$ and $B$, and I need to find two matrices $E_1$ and $E_2$ that satisfy the question stated in the title.
\begin{align*}
A&=\begin{bmatrix}
1&2&-1\\
1&1&1\\
1&-1&0\\
\end{bmatrix} &
B&=\beg... | The aim is to find $E_1$ and $E_2$ such that $B = E_2E_1A$. The first thing is to interchange the first and the last row of the matrix $A$, so $$E_1 = \begin{bmatrix}0 & 0 & 1\\0 & 1 & 0\\ 1 & 0 & 0\end{bmatrix}, \quad A_1 := E_1A = \begin{bmatrix}1 & -1 & 0\\1 & 1 & 1\\ 1 & 2 & -1\end{bmatrix}.$$ Then, $B$ can be obta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve this: I was doing my homework and I stumbled over this particular exercise. I would've known how to solve it, if it had been the same thing under square root in both cases.
$$
\lim_{n\to \infty} \left(\sqrt{4n^2+3n+2}-\sqrt{4n^2+n-1}\right)
$$
| You can multiply both numerator and denominator by $\sqrt{4n^{2}+3n+2}+\sqrt{4n^{2}+n-1}$. Then you will find $$\sqrt{4n^{2}+3n+2}-\sqrt{4n^{2}+n-1}=\frac{4n^{2}+3n+2-4n^{2}-n+1}{\sqrt{4n^{2}+3n+2}+\sqrt{4n^{2}+n-1}} = \frac{2n(1+3/2n)}{2n\bigg{[}\sqrt{1+\frac{3}{4n}+\frac{1}{4n^{2}}}+\sqrt{1+\frac{1}{4n}-\frac{1}{4n^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to compute $\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}$?
How to compute $\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}$?
My process:
\begin{align}
\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}... | We have that by $y=x/a\to 1$
$$\frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}=a \frac{\sqrt{2 y-y^{4}}-\sqrt[3]{y}}{1-\sqrt[4]{y^3}}=$$
$$= -a \frac{y-1}{\sqrt[4]{y^3}-1}\left( \frac{\sqrt{2 y-y^{4}}-1}{y-1}- \frac{\sqrt[3]{y}-1}{y-1}\right)=f(y)$$
then use the definition of derivative to obtain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof that a sequence decreases past a certain point
I would like to try to prove that the sequence
$$
\frac{\sqrt{6}^{x}}{x!}
$$
for $x=0, 1, 2, \cdots$, is strictly decreasing when $x\geq 2$.
It appears to be the case after listing out a few terms. Here's my attempt at a proof:
To the contrary, suppose $\frac{... | The given sequence IS strictly decreasing for $x\geq 2$. If we follow your approach in a correct way, we find that the inequality
$$\frac{\sqrt{6}^{x}}{x!} > \frac{\sqrt{6}^{x+1}}{(x+1)!},$$ is equivalent to
$$\frac{(x+1)!}{x!} > \frac{\sqrt{6}^{x+1}}{\sqrt{6}^{x}}$$
which reduces to
$$x+1> \sqrt{6}$$
which is satisfi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$(p-1)(p+1)/24 \in \mathbb N$ for all primes $p \geq 5$ I want to show
\begin{align}
\frac{(p-1)(p+1)}{24} \in \mathbb N \quad \text{for all primes} \quad p \geq 5 \tag{1}.
\end{align}
I can show $(1)$, if the following statement is true.
Let $a,b,c,d \in \mathbb N$ and $a \geq b \cdot c \cdot d$.
\begin{align}
\te... | Primes greater than or equal to $5$ can't be divisible by $3$, so must have the form $3k+1$ or $3k+2$ , so one of $p-1$ or $p+1$ has a factor of $3$. Also, $p-1$ and $p+1$ are consecutive even numbers and one of them is divisible by $4$, so $2$ divides $(p-1)(p+1)$ with at least multiplicity $3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Find the 4 roots of the equation: $z^{4}+4=0$ Find the 4 roots of the equation: $z^{4}+4=0$
Note: I can't use the form: $e^{i\theta}$
My attempt:
Note we have: $z=(-4)^{\frac{1}{4}}$, consider $w=-4$, then $w^{1/4}=z$ this implies $w=z^4$.
Consider the polar form of $z$ and $w$:
$|w|=r=4$,
$Arg(w)=\pi=\theta$
Then,
$$... | Consider the factorization of $z^4+4$:
$$z^4+4=z^4\color{blue}{+4z^2}+4\color{blue}{-4z^2}=(z^2+2)^2-(2z)^2=(z^2+2z+2)(z^2-2z-2)$$
Then we get $z^2+2z+2=0$ or $z^2-2z+2=0$, then finish it by using quadratic formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the greatest $a \in \mathbb{Z}$ such that $x^2-ax-\ln x+e^{\sin x}-1>0$ holding for every $x>0$.
Find the greatest $a \in \mathbb{Z}$ such that $$x^2-ax-\ln x+e^{\sin
x}-1>0$$ holding for every $x>0$. (Reference data: $\sin 1 \approx 0.84, \ln2 \approx 0.693)$
Notice that the inequality is equivalent to
$$f(x)... | Answer: greatest $a=2$. Proof.
If $a=3$ and $x=e$ then $x^2-ax-\ln x+e^{\sin x}-1<0$.
Let $f(x)=x^2-2x-\ln x+e^{\sin x}-1$. We need to prove that $f(x)>0$ for all $x$.
Cases:
I. Let $x \ge e$. Let $g(x)=x^2-2x-\ln x-1$. Then $g'(x)=2x-2-\frac{1}{x}=\frac{2}{x} \left( x-\frac{\sqrt3+1}{2}\right)\left( x+\frac{\sqrt3-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.