Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to express $2x^3-x^2-3x+2$ as a linear combination of Legendre polynomials I have used the formula
\begin{align}p_0(x)&=1\\
p_1(x)&=x\\
p_2(x)&=\frac12(3x^2−1)\\
p_3(x)&=\frac12(5x^3−3x)
\end{align}
$$2x^3-x^2-3x+2=Ap_3(x)+Bp_2(x)+Cp_1(x)+Dp_0(x)$$
EDIT-
\begin{align}&=\frac A2(5x^3−3x) + \frac B2(3x^2−1) + Cx +D \... | In matrix form,
$$\begin{bmatrix} p_0\\ p_1\\ p_2\\ p_3\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ -\frac{1}{2} & 0 & \frac{3}{2} & 0\\ 0 & -\frac{3}{2} & 0 & \frac{5}{2}\end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\\ x^3\end{bmatrix}$$
We want to find a weight vector $\mathrm{w}$ such that
$$2 x^3 - x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Continuity of $\frac{x^3y^2}{x^4+y^4}$ at $(0,0)$? Suppose a function $f$ is defined as follows:
$$f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}$$
Is this function continuous at $(0,0)$? How is this shown? I've tried considering limits for different ... | Since $x^4-2x^2y^2+y^4= (x^2-y^2)^2 \ge 0$, we have $2x^2y^2 \le x^4+y^4$.
Therefore, $\dfrac{x^2y^2}{x^4+y^4} \le \dfrac{1}{2}$ for all $(x,y) \neq (0,0)$.
Also, $\dfrac{x^2y^2}{x^4+y^4} \ge 0$ for all $(x,y) \neq (0,0)$.
From the above inequalities, we have that $\left|\dfrac{x^2y^2}{x^4+y^4}\right| \le \dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to solve this 1st order ODE $$ \frac{dy}{dx} = \frac{2x^2 + 3y^2 - 7}{3x^2 + 4y^2 + 8}$$
This does not satisfy the exactness and I can't find any integrating factor to transform it. I can't make it homogeneous too.
Thanks in advance for the replies.
edit: The question was written incorrectly by the instructor, so i... | Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=164:
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial fraction decomposition of $\pi\cdot \tan(\pi z)$
Evaluate the partial fraction decomposition of $\pi \tan(\pi z)$
$$2\pi \tan(\pi z)=\cot\left(\frac{\pi}{2}-\pi z\right)-\cot\left(\frac{\pi}{2}+\pi z\right)$$ $$=\frac{2}{1-2z}+\sum_{k=1}^\infty \frac{1-2z}{(\frac{1}{2}-z)^2-k^2}-\frac{2}{1+2z}-\sum_{k=1}^\in... | You may exploit the fact that $\tan(z)=\frac{d}{dz}\log(\cos z)$ and:
$$ \cos(z)=\prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2\pi^2}\right)\tag{1}$$
(Weierstrass product) so:
$$ \tan(z) = \sum_{n\geq 0}\frac{-8z}{4z^2-(2n+1)^2\pi^2}\tag{2}$$
and:
$$\begin{eqnarray*}\pi\tan(\pi z) = \sum_{n\geq 0}\frac{-2z}{z^2-\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$ How to prove this inequality?
If $a^{2}+b^{2}+c^{2}\leq 3$ and $a,b,c\in \Bbb R^+$, then
$$\left( a+b+c\right) \left( a+b+c-abc\right)\geq 2\left( a^{2}b+b^{2}c+c^{2}a\right) $$
I tried AM>GM but I couldn't get result
| Another way.
We need to prove that
$$(a+b+c)^2\geq2\sum_{cyc}a^2b+abc(a+b+c)$$ and since $$\frac{3}{a^2+b^2+c^2}\geq1$$ and $$1\geq\frac{a^2+b^2+c^2}{3},$$ it's enough to prove that:
$$(a+b+c)^2\sqrt{\frac{a^2+b^2+c^2}{3}}\geq2\sum_{cyc}a^2b+\sqrt{\frac{3}{a^2+b^2+c^2}}abc(a+b+c)$$ or
$$(a+b+c)^2(a^2+b^2+c^2)\geq2\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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How can I solve this hard system of equations? Solve the system below
\begin{align}
&\sqrt {3x} \left( 1+\frac {1}{x+y} \right) =2\\
&\sqrt {7y} \left( 1-\frac{1}{x+y} \right) =4\sqrt{2}
\end{align}
Frankly I am disappointed, because I spent around 2 hours in solving this equation, but, finally I didn't do it so, I... | Square both equations. Then let $u=\frac{1}{x+y}$. Note then $x+y=\frac{1}{u}$
$$3x(1+\frac{1}{x+y})^2=4$$
$$7y(1-\frac{1}{x+y})^2=32$$
So.
$$x(1+u)^2=\frac{4}{3}$$
$$y(1-u)^2=\frac{32}{7}$$
And for $u \neq 1$
$$x=\frac{4}{3}(1+u)^{-2}$$
$$y=\frac{32}{7}(1-u)^{-2}$$
Adding both equations we have:
$$\frac{1}{u}=\frac{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2 \;=\; 4+4\sqrt{3} + 3 \;=\; 7 + 4\sqrt{3}.$$
However, the only way I can thi... | $\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$
$a+\sqrt{b}=x+y+2\sqrt{xy}$
$a+\sqrt{b}=(x+y)+\sqrt{4xy}$
Then:
$a=x+y$ and $b=4xy$
$a^2=x^2+y^2+2xy$
$a^2-b=(x-y)^2$
If $x>y$ then:
$\sqrt{a^2-b}=x-y$
Let's take $c=\sqrt{a^2-b}$ then $c=x-y$
$c=x-y$ and $a=x+y$ then:
$x=\frac{a+c}{2}$ and $y=\frac{a-c}{2}$
Then:
$\sqrt{a+\sqrt{b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Upper bound for $\gcd(a,b)$ if $\frac{a+1}{b}+\frac{b+1}{a}\in\Bbb{N}$
Suppose that $a,b$ are two positive integers so that $\frac{a+1}{b}+\frac{b+1}{a}$ is also a positive integer.Find the best upper bound for $\gcd(a,b)$.
My work:
$\frac{a+1}{b}+\frac{b+1}{a}=\frac{a(a+1)+b(b+1)}{ab} \in \Bbb{N} \implies ab|a^2+b... | Fortunately with the help of one of my friends I got to the right answer as follows:
Assuming $d=\gcd(a,b),a=a'd,b=b'd$ we have: $$\frac{a+1}{b}+\frac{b+1}{a}=\frac{a'^2d+b'^2d+a'+b'}{a'b'd}$$
Now it's required that $d|a'+b'$ so that the expression is an integer , then:
$$d^2|a'd+b'd\implies d^2|a+b \implies d^2\leqs... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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What is the Fourier Sine Transform of $\frac{1}{x(a^2+x^2)}$? (Please check my solution) My textbook says that the answer is $\frac{\pi}{2a^2}(1-e^{-as})$ while my answer is $\frac{1}{a^2}\sqrt{\frac{\pi}{2}}(1-\frac{e^{-as}}{2}+\frac{e^{as}}{2})$. Here is my solution:
\begin{align}
\hat{f_s} & = \sqrt{\frac{2}{\pi}} \... | When doing the change of variable $u = sx$, it is implicitly assumed that $s \neq 0$ (otherwise, the integral evaluates to $0$). This means that the middle expression of $y$ and the subsequent chain of differentiations (hence, also, the obtained expression for $y''$) are only valid for $s \neq 0$. To obtain a formula f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution of given equation for $x$ Solve the given equation for $x$
$$\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3}=125$$
I solved the question by taking ${x^2-2x+3}=t$, and squaring twice and finally solving ${x^2-2x+3}=t$ but it required very hectic calculations. I wonder if someone can suggest better approach.
| As $8-3=x^2-2x+8-(x^2-2x+3)$
$=(\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3})(\sqrt{x^2-2x+8}-\sqrt{x^2-2x+3})$
$$\sqrt{x^2-2x+8}+\sqrt{x^2-2x+3}=125\iff(\sqrt{x^2-2x+8}-\sqrt{x^2-2x+3})=\dfrac{8-3}{125}$$
Adding we get $2\sqrt{x^2-2x+8}=125+\dfrac1{25}=?$
Now square both sides.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Compute $|z|$ , $z = \frac{(2+i)^7(1-2i)^3}{(1+2i)^8}$ Compute $|z|$ , $z = \frac{(2+i)^7(1-2i)^3}{(1+2i)^8}$,
if $z = a+ib$ then,
I tried to do that with $|z| = (a+ib)(a-ib)$
then i multipled it $z$ with $z^-$ and then I got stuck. answer is $|z| = 5$
| Hint. One may write
$$
|z|^2=z\cdot \bar z=\frac{|2+i|^{14}|1-2i|^6}{|1+2i|^{16}}=\frac{(2^2+1^2)^7(1^2+2^2)^3}{(1^2+2^2)^8}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways
Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways.
I took extremely long to solve this
I got
$50= 7^2 + 1^2 ... | If $p$ is a prime and $p \equiv 1 \bmod 4$, then $p=a^2+b^2$, for some $a,b \in \mathbb N$.
In this case, $2p^2$ has two representations: $p^2+p^2$ and $(a-b)^2+(a+b)^2$.
The second representation ones from $2p^2=|(1+i)(a+bi)^2|^2$.
So, the smallest solution is with $p=5$, for which $a=2$ and $b=1$ and $(a+bi)^2=3-4i$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Find all pairs such that $x^2 y + x + y$ is divisible by $xy^2 + y + 7$
Find all pairs $(x, y)$ of positive integers such that $x^2 y + x + y$ is divisible by $xy^2 + y + 7$. If there are too many to write, write a generic form.
I was thinking of rewriting the divisibility in a simpler way such as $xy^2 + y + 7\mid x... | We have $x=7n^2,y=7n$ as a class of solutions because then $n(xy^2+7+y)=x^2y+y+x$.
There is one other solution $x=11,y=1$, giving $xy^2+7+y=19,x^2y+y+x=7\cdot19$.
We have $y(x^2y+x+y)-x(xy^2+y+7)=y^2-7x$, so $xy^2+y+7$ must divide $y^2-7x$.
One possibility is $y^2=7x$. That implies 7 divides $y$ and hence also $x$. So ... | {
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"timestamp": "2023-03-29T00:00:00",
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prove that $\sum_{n=1}^\infty \frac{2^n+n^2+n}{2^{n+1}n(n+1)}$ is convergent and find the limit when $n \to \infty$ Does the following sumatory converges? If yes find the limit when $n \to \infty$ $$\sum_{n=1}^\infty \frac{2^n+n^2+n}{2^{n+1}n(n+1)}$$
ideas?
I have tried by the comparison test.
| Observe that we can break up the sum to:
$$ \sum_{n=1}^{\infty} \frac{2^n}{2^{n+1} n(n+1)} + \frac{n^2+n}{2^{n+1}n(n+1)} $$
Which then becomes two sums:
$$ \sum_{n=1}^{\infty} \frac{1}{2n(n+1)} + \sum_{n=1}^{\infty}\frac{1}{2^{n+1}} $$
Now observe the left sum decays quadratically (any function f(n) which has the prope... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\f... | You looked at each of the three terms separately. You have $|x-1|<\frac{1}{10}$. You should have $|x+1|<\frac{21}{10}$ and you should have $\frac{1}{|x+3|}<\frac{10}{39}$. The critical point for your method is that you have to take the worst case each time.
The three inequalities then multiply together to give $\left|... | {
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"timestamp": "2023-03-29T00:00:00",
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Value of $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
If $a+b+c=0,$ Then value of $\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\bf{My\; Try::}$ Given $$\displa... | We have $$A = \left(\frac{a}{b-c} + \frac{b}{c-a} + \frac{c}{a-b}\right)\frac{bc(b-c) + ca(c-a) + ab(a-b)}{abc} = \frac{-3abc - a^3 -b^3 -c^3+ a^2(c-b) + b^2(a-c) +c^2(b-a)}{abc}.$$
Note that $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a)$ or $a^3+b^3+c^3 = 3abc$.
Then $$A = \frac{(a-b)(c-b)(a-c)}{abc}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the value of $h$ if $x^2 + y^2 = h$
Consider equation $x^2 + y^2 = h$ that touches the line $y=3x+2$ at some point $P$. Find the value of $h$
I know that $x^2 + y^2 = h$ is a circle with radius $\sqrt{h}$.
Also, since $y = 3x + 2 $ is a tangent, we know that the slope of the radius perpendicular to the tangent i... | There are couple of approaches already mentioned in other answers, but I will try to summarize them and add (probably unwarranted) generality. Also, I will describe approach using calculus.
If you are not interested in this generalities, please skip to the last paragraph.
We are given a line $y = ax + b$ and a circle... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 7
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Difference between $\cos(x)$= positive and $\cos(x)$= negative I'm confusing myself here so need some clarification. If I was to work out the solutions to $\cos(x) = 1/\sqrt2$ , I know the solutions would be $\pi/4, 2\pi - \pi/4, 2\pi + \pi/4...$ etc.
What is the difference when working out the solution to $\cos(x) = -... | If $\cos x > 0$, then the terminal side of angle $x$ lies in the first quadrant, the fourth quadrant, or the positive $x$-axis. If $\cos x < 0$, then the terminal side of angle $x$ lies in the second quadrant, the third quadrant, or the negative $x$-axis.
One solution of the equation $\cos x = -\dfrac{1}{\sqrt{2}}$ is... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$ I've been trying to prove
$$\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}$$
I've tried perturbation and inversion but still nothing. I've even tried expanding the sum to try and find some pattern that could relate this to the harmonic seri... | Suppose we seek to verify that
$$\sum_{k=1}^n {n\choose k} (-1)^{k+1} \frac{1}{k} = H_n.$$
Now observe that
$$\frac{1}{k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} \log\frac{1}{1-z} \; dz$$
where we take $\epsilon < 1$ so that the logarithmic term is analytic.
We get for the sum
$$\frac{1}{2\pi i}
\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Induction Sum of all odd Numbers Show that
$\sum_{k=1}^{n}(2k-1)=n^2$
Beginning: n=1
$\sum_{k=1}^{1}(2k-1)=(2*1-1)=1=1^2$
Let $\sum_{k=1}^{n}(2k-1)=n^2$ be true, then for n=p+1
$\sum_{k=1}^{p+1}(2k-1)=(p+1)^2$ has to be true too.
$\sum_{k=1}^{p+1}(2k-1)=1+3+...+(2(p+1)-3)+(2(p+1)-1)$
$=1+3+...+(2p-1)+(2p+1)$
$=1+3+...+... | Induction step:
Assume $\sum_{k=1}^n$ $(2k - 1)$ = $n^2$. Then $\sum_{k=1}^{n+1}$ $(2k - 1)$ = $\sum_{k=1}^n$ $(2k - 1)$ + [$(2(n + 1) - 1)$] = $n^2 + [2(n + 1) - 1]$ = $n^2 + 2n + 1$ = $(n + 1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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On the proof that $\sum\limits_{k=0}^{n-1}\frac {a^k}{(1+a^k x) (1+ a^{k+1}x)}=\frac 1 {1-a} \left( \frac 1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$ Question:-
Find the sum to $n$ terms of the following series
$$\frac{1}{(1+x) (1+ax)} + \frac{a}{(1+ax) (1+a^2 x)} + \frac{a^2}{(1+a^2 x) (1+a^3 x)} + \cdots$$
My solution:-... | Your result is the same as the answer from the book. Proof below :
$$f(a,x)=\left( \frac 1 {1-a} \right) \left( \frac 1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$$
$$f(a,x)=\left( \frac 1 {(1-a)x} \right) \left( \frac x {1+x} -\frac {a^n x}{ 1+a^n x }\right)$$
$\frac x {1+x}=1-\frac 1 {1+x}$
$\frac {a^n x}{ 1+a^n x } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Eliminate $\theta$
Eliminate $\theta$ in
$$\sin \theta + \mbox{cosec} \, \theta = m$$
$$\sec \theta - \cos \theta = n$$
My approach-
I multiplied the first equation by $\sin \theta$ and the second equation by $\cos \theta$ but it doesn't give me the desired answer..
| We have $\dfrac{1-\cos^2\theta}{\cos\theta}=n\implies\sin^2\theta=n\cos\theta\ \ \ \ (1)$
$\iff\cos^2\theta+n\cos\theta-1=0\ \ \ \ (2)$
and multiply $\sin\theta$ with the both sides of $\sin \theta +\csc\theta = m$, $$\sin^2\theta=m\sin\theta-1\ \ \ \ (3)$$
$(1),(3)\implies n\cos\theta+1=m\sin\theta$
Squaring ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
} |
Find a polynomial with integer coefficients which has a global minimum equal to (a)$- \sqrt{2}$, (b)$\sqrt{2}$ Find a polynomial with integer coefficients which has a global minimum equal to (a)$- \sqrt{2}$, (b)$\sqrt{2}$.
It it a high-school math contest problem. The answer is given:
$$(a) ~~~~~~~P(x)=N(2x^2-1)^2-2x^3... | For part (a), it can be easily deduced that $f$ does not have degree 2. Further, $f$ cannot have degree 3, as an odd degree polynomial has no global minimum. So assume
$$f(x) = sx^4 + bx^3 + cx^2 + dx + e$$
with $f(a) = -\sqrt{2}, f'(a) = 0$. As one last piece of guesswork, let $a = k\sqrt{2}$. Then we have
$$4sk^4 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove: $\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive
$$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$
$$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$
$$\frac{x(\sqrt... | Alternatively, apply AM-GM inequality with $a = \sqrt{x}, b = \sqrt{y}$:
$\dfrac{a^2}{b} + b \geq 2\sqrt{\dfrac{a^2}{b}\cdot b} = 2a, \dfrac{b^2}{a} + a \geq 2\sqrt{\dfrac{b^2}{a}\cdot a} = 2b$. Adding these $2$ inequalities to get the answer.
Generalization: The same technique could be used to prove: If $x, y, z > 0 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Is difference of two consecutive sums of consecutive integers (of the same length) always square? I am an amateur who has been pondering the following question. If there is a name for this or more information about anyone who has postulated this before, I would be interested about reading up on it. Thanks.
If $x$ is th... | Consider the $n$th triangular number $T(n)$, which is the sum of $n$ natural numbers $1$ to $n$, and which has the closed form $T(n)=\tfrac{1}{2}n(n+1)=\binom{n+1}{2}$ where $\binom{n+1}{2}$ is a binomial coefficient (see https://en.wikipedia.org/wiki/Triangular_number). Then we can look at the differences of triangul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 9,
"answer_id": 5
} |
Find all $a, b, c, d$ Find all $a, b, c, d$ satisfying
$$\frac{x^4+ax^3+bx^2-8x+4}{(x^2+cx+d)^2} = 1$$
My answers are:
$$\begin{cases}
d=2\\
d=-2
\end{cases}
\quad \begin{cases}
a=4\\
a=-4
\end{cases} \quad
\begin{cases}
c=2\\
c=-2
\end{cases} \quad
\begin{cases}
b=8\\
b=0
\end{cases}$$
Is it right?
| You can write it as $x^4+ax^3+bx^2-8x+4=(x^2+cx+d)^2$, which you want to be true for all $x$. This requires that the coefficients of each power match. You should expand the square on the right and match the coefficients. This will give you each of $a,b,c,d$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the 11th term of $\frac{7x^2+2x+6}{(x+2)(x^2+1)}$ Taylor expansion
Given $f(x)=\frac{7x^2+2x+6}{(x+2)(x^2+1)}$, find $f^{(11)}(0).$
I understood that we first need to use partial fractions to simplify the function.
$$\frac{7x^2+2x+6}{(x+2)(x^2+1)}=\frac{A}{(x+2)}+\frac{Cx+B}{(x^2+1)}=\frac{6}{(x+2)}+\frac{x}{... | We have
$$ f(x)=\frac{7x^2+2x+6}{(x+2)(x^2+1)}=\frac{6}{x+2}+\frac{x}{1+x^2}\tag{1}$$
through the residue theorem, then:
$$ f(x) = 3\sum_{n\geq 0}\frac{(-1)^n}{2^n}x^{n}+\sum_{n\geq 0}(-1)^n x^{2n+1} \tag{2}$$
by exploiting geometric series. It follows that:
$$ [x^{11}]\,f(x) = -\frac{3}{2^{11}}-1 \tag{3} $$
so:
$$ f^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
chain rule for multi-variables The monthly demand for the Instant Pie Maker is given by
$$D(x,y)= \frac1{125}xe^{xy/1000} \text{ units}$$
where $x$ dollars are spent on infomercials and $y$ dollars are spent on in-person demonstrations. If $t$ months from now $$x=20+t^{2/3}$$ dollars are spent on infomercials and $$y=... | If I understood your problem correctly, then you have to first plug $x = 20 + t^{\frac{2}{3}}$ and $y = t\ln(1 + t)$ into $D(x,y)$ and then calculate it's derivative with respect to $t$ at the point $t = 8$. Here is a possible solution:
First we define
\begin{equation}
\begin{split}
\tilde D(t) := D(20 + t^{\frac{2}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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where is my mistake with this equation ot hyperbolic problem? Let parabola $\Gamma_{1}:$$y^2=4x$,and hyperbolic curve $\Gamma_{2}$: $x^2-y^2=1$.it is well known this two is symmetric.so the two point $A$ and $B$ about $x$ axial symmetry,or mean $x_{A}=x_{B}$.see figure,
but we use
$$\begin{cases}
y^2=4x\\
x^2-y^2=1... | The confusion arises from ignoring an implicit condition of the first equation, $y^2 = 4x$. Since $y^2 \geq 0$, we have the condition $x \geq 0$.
Thus of the two solutions for $x$ you have, the positive $x = 2 + \sqrt{5}$ is the only solution.
We see that there are two values of $y$ that satisfy the equations with the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove that $\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$
Prove that
$$\cos\frac {2\pi}{7}+ \cos\frac {4\pi}{7}+ \cos\frac {8\pi}{7}=-\frac{1}{2}$$
My attempt
\begin{align}
\text{LHS}&=\cos\frac{2\pi}7+\cos\frac{4\pi}7+\cos\frac{8\pi}7\\
&=-2\cos\frac{4\pi}7\cos\frac\pi7+2\cos^2\frac... | $cos(2\pi/7)$+$cos(4\pi/7)$+$cos(8\pi/7)$
= $cos(2\pi/7)$+$cos(4\pi/7)$+$cos(6\pi/7)$ (angles add to give $2\pi$, thus one is $2\pi$ minus the other)
At this point, we'll make an observation
$cos(2\pi/7)$$sin(\pi/7)$ = $\frac{sin(3\pi/7) - sin(\pi/7)}{2}$ ..... (A)
$cos(4\pi/7)$$sin(\pi/7)$ = $\frac{sin(5\pi/7) - sin(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
how do I integrate a modulus function. is it possible? I am solving a question in which after calculating, my instantaneous velocity is $f(t)=\frac{ab\sin(bt)}{\sqrt{2-2\cos(bt)}}$.
So I need to find distance in time $T$. For finding distance I need to take its modulus and integrate
$|f(t)|$dt.
So how do I find integ... | \begin{align*}
\left| \frac{ab\sin bt}{\sqrt{2-2\cos bt}} \right| &=
\left|
\frac{2ab\sin \frac{bt}{2} \cos \frac{bt}{2}}{\sqrt{4\sin^{2} \frac{bt}{2}}}
\right| \\
&= \left| ab\cos \frac{bt}{2} \right| \\
s &= \int_{0}^{t} \left| ab\cos \frac{bt}{2} \right| dt \\
&= 2a E \left( \frac{bt}{2}, 1 \right)
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find $\tan x $ from $(a+1)\cos x + (a-1)\sin x=2a+1$? How do I find $\tan x$ from this equation?
$$(a+1)\cos x + (a-1)\sin x=2a+1$$
Thanks for any help!!
| Expressing the trigonometric functions $\sin(x)$ and $\cos(x)$ as a function of $\tan(\frac{x}{2})$, we get:
$\tan(\frac{x}{2})=+\frac{a-1+\sqrt{1-4a-2a^2}}{3a+2}$,
$\tan(\frac{x}{2})=-\frac{-a+1+\sqrt{1-4a-2a^2}}{3a+2}$.
Applying the tangent duplication formula:
$\tan(x)=\tan(\frac{x}{2}+\frac{x}{2})=\frac {\tan (\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Applying the Cartesian Coordinate Find the surface area of the portion of $2x+y+z=8$ in the first octant.
I know that $f(x,y)=8-y-2x$, $x=0$ to $4$ and $y= 0$ to $8-2x$, but I'm having trouble solving the problem in Cartesian form. The answer is supposedly $16\sqrt6$
| This is an analytic proof:
$$\int_0^4\int_0^{8-2x} \sqrt{(-1)^2+(-2)^2+1} \ dydx=\sqrt{6}\int_0^4 8-2x dx=16\sqrt6$$
This is a geometric proof: the surface is a triangle, whose three vertices lie on the three axes. The three vertices have the following coordinates: $(4,0,0);(0,8,0);(0,0,8)$ which yields the following ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to partial fraction $\frac{1}{(x+1)^2}$ I need to integrate $\frac{1}{(x^2+2x+1)}$, so I need to use partial fraction as the polynomial can be factored as $\frac{1}{(x+1)^2}$. This is what I've tried:
$$\frac{A}{(x+1)} + \frac{B}{(x+1)^2}$$
$$A\cdot(x+1)^2 + B\cdot(x+1)$$
$$Ax^2+2Ax+A+Bx+B$$
$$Ax^2+(2A+B)x+(A+B)$$
... | Just lookoing at the line $\frac{A}{(x+1)} + \frac{B}{(x+1)^2} = \frac{1}{(x+1)^2}$ we clearly see that $A = 0, B = 1$. As for your mistake, you multiplied by $(x+1)$ one time too many when removing the denominators.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Integrate $\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$ Integrate using integrating by parts :
$$\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
My attempt :
$$I=\int \frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}dx$$
$$I=\int \frac{e^x}{(1+x^2)^2}\cdot(1-x)^2dx$$
$$I=\frac{e^x}{(1+x^2)^2}\cdot\frac{(1-x)^3}{-3}-\int \frac{(1-x)^... | Hint
$$\frac{(1-x)^2 \cdot e^x}{(1+x^2)^2}=\frac{e^x}{1+x^2}-\frac{2xe^x}{(1+x^2)^2}$$
Indeed
$$(1-x)^2=(1+x^2)-2x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, and $k=1$, does it follow that $\frac{\sigma(n^2)}{n^2} \geq 2 - \frac{5}{3q}$? Let $\sigma=\sigma_{1}$ be the classical sum-of-divisors function.
If $N = q^k n^2$ is an odd perfect number with Euler prime $q$ (i.e., $q$ satisfies $q \equiv k \equiv 1 \pmo... | If $N$ is perfect then $2N = 2q^k n^2 = \sigma(q^k n^2) = \sigma(q^k)\sigma(n^2) = \frac{q^{k+1}-1}{q-1} \sigma(n^2)$, which gives
\begin{equation}
\frac{\sigma(n^2)}{n^2} = \frac{2q^k(q-1)}{q^{k+1}-1} = 2-2\frac{q^k-1}{q^{k+1}-1}
\end{equation}
This being $\ge 2-\frac{5}{3q}$ works out to being equivalent to $q^{k+1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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If $n$ is a positive integer, then $(-2^n)^{-2} + (2^{-n})^2 = 2^{-2n+1}$ I'm not sure why $$(-2^n)^{-2} + (2^{-n})^2=2^{-2n+1}$$
I have been going over this equation for a while now, noticing, and have successfully got quite far in the equation, finding that
$$ (-2^n)^{-2} + (2^{-n})^2 \implies \frac{1}{(-2^n)^2} + \... | $(-2^n)^{-2} + (2^{-n})^2=$
$\frac 1{(-2^n)^2} + (\frac{1}{2^n})^2 = $
$\frac 1{(2^n)^2} + \frac 1 {(2^n)^2} = $
$\frac 1{2^{2n}} + \frac 1{2^{2n}} = $
$\frac 2{2^{2n}} = $
$\frac 1{2^{2n-1}} = $
$2^{-(2n-1)} = $
$2^{-2n+1}$
Or if you are more comfortable with the rules of exponents:
$(-a)^{b_{\text{even}}} = a^b$ so
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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If $5\sqrt [ x ]{ 125 } =\sqrt [ x ]{ { 5 }^{ -1 } } $, then $x$ equals $-4$ For the equation
$$5\sqrt [ x ]{ 125 } =\sqrt [ x ]{ { 5 }^{ -1 } } $$
$x$ is equal to $-4$, but I'm not sure why.
I've taken the right side of the equation ${ \left( \frac { 1 }{ 5 } \right) }^{ \frac { 1 }{ x } }$ and converted it to ... | $$5\cdot125^{\frac{1}{x}}=\left(\frac{1}{5}\right)^{\frac{1}{x}}\Longleftrightarrow5\cdot125^{\frac{1}{x}}=\frac{1^{\frac{1}{x}}}{5^{\frac{1}{x}}}\Longleftrightarrow5\cdot125^{\frac{1}{x}}=\frac{1}{5^{\frac{1}{x}}}\Longleftrightarrow$$
Use (real solution) $5^y=125\Longleftrightarrow y=\log_5(125)=3$:
$$5\cdot(5^3)^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Finding total elements in a series I have a confusion.
How many terms in the following series are needed to make a sun greater than 5/2?
1/2 + 1/3 + 1/4 + 1/5 + 1/6 + .........
Is there any shortcut technique to find this out?
| Observe that, for $k\geq 1$,
\begin{align*}
\sum_{n=2}^{2^k}\frac{1}{n}&=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\cdots+\frac{1}{2^k}\\
&>\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\cdots+\frac{1}{2^k} \\
&=\frac{1}{2}+\frac{1}{2}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\f... | $$\frac { a_{ n+1 } }{ a_{ n } } =\frac { \left( 1+\frac { 1 }{ n+1 } \right) }{ { \left( 1+\frac { 1 }{ n } \right) }^{ n } } ^{ n+1 }=\left( 1-\frac { 1 }{ (n+1)^{ 2 } } \right) ^{ n+1 }\frac { n+1 }{ n } >\left( 1-\frac { 1 }{ n+1 } \right) \frac { n+1 }{ n } =1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Solve the following using AM-GM inequality The least value of $a \in R$ for which $4ax^2 + \frac{1}{x} \ge 1 $for all $x \gt 0 $, is
Using AM-GM inequality
$$\frac{4ax^2 + \frac{1}{2x} + \frac{1}{2x}}{3} \ge \sqrt[3]{a}$$
$$4ax^2 + \frac{1}{x} \ge 3\sqrt[3]{a}$$
Now my question start from here .
Can I do that for le... |
Can I do that for least value of $a$, the value of $3\sqrt[3]{a} $ must greater than minimum value of $4ax^2 + \frac{1}{x}$
I find this strange because "minimum value of $4ax^2 + \frac{1}{x}$" is $3\sqrt[3]{a}$. So I guess that you meant that "the value of $3\sqrt[3]{a} $ must greater than $1$". Then, it is correc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solve $\sec (x) + \tan (x) = 4$ $$\sec{x}+\tan{x}=4$$
Find $x$ for $0<x<2\pi$.
Eventually I get $$\cos x=\frac{8}{17}$$
$$x=61.9^{\circ}$$
The answer I obtained is the only answer, another respective value of $x$ in $4$-th quadrant does not solve the equation, how does this happen? I have been facing the same problem ... | If $\sec x+ \tan x=4 \\
\implies \frac{1+\sin x}{\cos x}=4\\
\implies 1+\sin x=4\cos x$
That does not help! So let's square it both sides
$\implies 1+\sin^2x+2\sin x=16\cos^2x=16-16\sin^2x
\implies 17\sin^2x+2\sin x-15=0$
Assuming, $\sin x=y$, makes the above equation quadratic in y having 2 solutions. So,
$17y^2+17y-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 12,
"answer_id": 11
} |
Evaluating the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer
How to evaluate the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer?
For $k=1$, the series does not converge.
When $k=2$, I can prove that:
$$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{... | Again, this is not going to be a full answer. However since my approach is slightly different from what was presented here so far and since my result is similar to the result given by Tito Piezas III I will post my answer.
\begin{eqnarray}
&&S_k:=\sum\limits_{n=1}^\infty \frac{1}{n \binom{ k n}{n}}=\\
&&\sum\limits_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 2
} |
Iterated Integral with variable substitution I need to calculate the double integral of the function $f(x,y) = (x+y)^9(x-y)^9$:
$\int_0^{1/2} \int_x^{1-x} (x+y)^9(x-y)^9 dydx$
I have a solution but I definitely arrived at it after a sloppy attempt. I got -0.0025. To begin with, I know you need to separate the variab... | It's convenient to solve this integral with substitution. Since we want to separate the variables in the Integral
\begin{align*}
\int_0^{1/2} \int_x^{1-x} (x+y)^9(x-y)^9 dy\,dx\tag{1}
\end{align*}
it's reasonable to use the substitution
\begin{align*}
\left.
\begin{matrix}
u=x+y\\
v=x-y\\
\end{matrix}
\right\}
\quad\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.... | Let $\alpha = \sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}$. Then
$$ \alpha^2 = 12+2\sqrt{36-20} = 20 $$
hence $\color{red}{\alpha=2\sqrt{5}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Chemical reaction modeled by a differential equation I am badly stuck on the following question. Thus, I am asking for some help :)
Consider a chemical reaction in which compounds $A$ and $B$ combine to form a third compound $X$. The reaction can be written as
$$A + B \xrightarrow{k} X$$
If $2 \, \rm{g}$ of $A$ and $... | The given ODE can be rewritten in the form
$$\dot x = \frac{2 \kappa}{9} \left( x - \frac{3 a}{2} \right) \left( x - 3 b \right)$$
or, alternatively, in the form
$$\left( \frac{1}{x - \frac{3 a}{2}} - \frac{1}{x - 3 b} \right) \, \mathrm{d} x = \left(\frac{a - 2 b}{3}\right) \kappa \,\mathrm{d} t$$
Integrating,
$$\ln \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integrate the following : $\displaystyle\int \frac{x^n+1}{(x^n-1) \sqrt {x^4+1}} dx$ ; n is odd. My original Question :
$\displaystyle\int \frac{x^3+1}{(x^3-1) \sqrt {x^4+1}} dx $
What to substitute? How to do it?
I want to add that I have tried doing this using substitutions as :
$x = \frac{1}{t}$ and considered the ... | First, introduce variables $y, z$ such that $x + x^{-1} = y = \frac{1}{\sqrt{2}}(z+z^{-1})$, we have
$$\begin{align}
& \frac{x+1}{x-1}\frac{dx}{\sqrt{x^4+1}}
= \frac{x+1}{x-1}\frac{1}{\sqrt{x^2+x^{-2}}}\frac{dx}{x}
= \frac{x+1}{x-1}\frac{1}{\sqrt{(x+x^{-1})^2-2}}\frac{d(x+x^{-1})}{x-x^{-1}}\\
= & \frac{dy}{(y-2)\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A Problem with the Generating Function of Fibonacci. So basically I want to find the closed form of $G_n = \sum_{k = 1}^n \binom{n+k - 1}{2k-1}$.
After checking for $n = 1,2,3,4$ the values are $1, 3, 8, 21$ respectively. I claim that it is $F_{2n}$ where $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$. My idea was to... | Suppose we seek the closed form of
$$G_n = \sum_{k=1}^n {n+k-1\choose 2k-1}.$$
Observe that the binomial coefficient
$${n+k-1\choose 2k-1} = {n+k-1\choose n-k}$$
has the property that its integral representation
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} (1+z)^{n+k-1} \; dz$$
vanishes when $k\gt n$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Landau symbol and taylor series $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{1}{2}$$
I saw the calculation $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=$$...$$=\sqrt x (\sqrt{1+\sqrt{1/x+\sqrt{1/x^3}}}-1)= $$ $$\sqrt x(1+\frac{1}{2}\sqrt{1/x+\sqrt{1/x^3}}+ \color{red}{\mathca... | If you want to do it with an asymptotic expansion, you have to be cautious with the order of expansion. The scale of comparison will be the powers of $\sqrt x$, and we'll expand $\sqrt{1+\sqrt{\dfrac1x+\sqrt{\dfrac1{x^3}}}}$ at order $2\;$ (i.e. up to a term in $1/x$).
*
*$\sqrt{\dfrac1x+\sqrt{\dfrac1{x^3}}}=\dfrac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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What is the value of the following determinant: given that $q^2- pr<0, p>0$ then evaluate \begin{vmatrix}
p &q &px+qy\\
q &r &qx+ry\\
px+qy &qx+ry &0
\end{vmatrix}
I've tried this:
\begin{vmatrix}
p &q &px+qy\\
q &r &qx+ry\\
px+qy &qx+ry &0
\end{vmatrix}
then, i modified the rows as $R_3=R_3-(xR_1+yR_2)$ to get \begin{... | First, $pr-q^2$ is positive. What about $px^2+2qxy+ry^2$? This is the quadratic form $\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}p&q\\q&r\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$. Whether this is a positive quantity comes down to whether the eigenvalues of the matrix are positive (and whether $x,y$ are nonze... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}... | $$I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\cdots$$
$$2I=1+1+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\cdots$$
$$2I-I=1+\left(1-\frac 12 \right)+\left(\frac 34 -\frac 24 \right)+\left(\frac 48 -\frac 38 \right)+\left(\frac {5}{16} -\frac {4}{16} \right)+\cdots$$
$$I=1+\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 1
} |
Prove that $\frac{1}{\sqrt{nx+1}}+\frac{1}{\sqrt{nx+2}}+\ldots+\frac{1}{\sqrt{nx+n}}=\sqrt n$ has a unique positive solution Prove that for each $n\in \mathbb Z$, $n\ge 1$ there is exactly one $x>0$ satisfying
$$\frac{1}{\sqrt{nx+1}}+\frac{1}{\sqrt{nx+2}}+\ldots+\frac{1}{\sqrt{nx+n}}=\sqrt n$$
Please give me a hint. I... | Note that $n=1$ gives $x=0$.
Considering
$$\int_{0}^{1} \frac{dt}{\sqrt{\frac{9}{16}+t}} = 1$$
Since $\frac{1}{\sqrt{\frac{9}{16}+t}}$ is monotonic decreasing,
the integral is sandwiched by the upper and lower Riemann sums: $U_{n}<U_{n+1}<1<L_{n+1}<L_{n}$
$$\sum_{k=1}^{n} \frac{1}{n\sqrt{\frac{9}{16}+\frac{k}{n}}} < 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$a_1 + a_2 + \dots a_n = 1$ find min of $a_1^2 +\frac{a_2^2}{2} + \dots + \frac{a_n^2}n.$ Given $n$ numbers $a_1$, $\cdots$ and such that $a_1$, $a_2$, $\cdots$, $a_n > 0$ and their sum is $1$, I want to find the minimum value of
$$a_1^2 + \frac{a_2^2}{2} + \cdots + \frac{a_n^2}{n}.$$
I have tried using weighted AM-GM... | By the Cauchy-Schwarz inequality we have
$$ \left(\sum_{i=1}^{n}i\right)\left(\sum_{i=1}^{n}\frac{a_i^2}{i}\right) \geq \left(\sum_{i=1}^{n}\sqrt{i}\cdot\frac{a_i}{\sqrt{i}}\right)^2 = 1 \tag{1}$$
hence with our hypothesis we have:
$$ \sum_{i=1}^{n}\frac{a_i^2}{i}\geq \frac{2}{n(n+1)}\tag{2} $$
and equality is achieved... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Prove the inequalitiy
If $a, b, c$ are positive reals such that:
$\frac a {b + c + 1} + \frac b {a + c + 1} + \frac c {b + a + 1} \le
1$ then:
$\frac 1 {b + c + 1} + \frac 1 {a + c + 1} + \frac 1 {b + a + 1} \ge
1$
I tried using inequality of arithmetic and harmonic means, to no avail.
Any help is appreciated.
| We'll write the condition in the following form:
$$a+b+c+1\geq a^3+b^3+c^3+abc+\sum\limits_{cyc}(a^2-ab)$$
and since $\sum\limits_{cyc}(a^2-ab)\geq0$ we obtain
$$a+b+c+1\geq a^3+b^3+c^3+abc$$
In another hand, we need to prove that
$$2(a+b+c+1)\geq(a+b)(a+c)(b+c)$$
Thus, it remains to prove that
$$2(a^3+b^3+c^3+abc)\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_\frac12 ^2 \frac1x\tan(x-\frac1x)dx $ $$\int\limits_\frac{1}{2} ^2 \frac{1}{x}\tan\left(x-\frac{1}{x}\right)\mathrm{d}x$$
I have tried substitution and by parts and it seems failed at all. Can anyone give me some hints?
| Let $$ I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\cdot \tan \left(x-\frac{1}{x}\right)\,dx$$
Now Let $\displaystyle\left(x-\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)\,dx = dt\Rightarrow \left(x+\frac{1}{x}\right)\,dx = x\,dt$
and Changing Limits
Now Using $$ \left(x+\frac{1}{x}\right)^2 -\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Elementary matrix operations I am taking an introductory/first year course in Linear Algebra and I am at my wits' end with the following problem.
I am asked to find both the nullspace and the general solution of the following systems $A\tilde{x}=\tilde{b}$.
The first is
$$
A=
\begin{pmatrix}
1 & 1 & 0 & 2\\
2 & 1 & 1 &... | This is correct! You can check if you found solutions of the system simply by replacing $x$ in $Ax=b$ by what you found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to break $\frac{1}{z^2}$ into real and imaginary parts? $$ \frac{1}{(x+iy)^2}=\frac{1}{x^2+i2xy-y^2}=\frac{x^2}{(x^2+y^2)}-\frac{2ixy}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
So I thought I could just say:
$$ Re(\frac{1}{z^2})=\frac{x^2}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$
and
$$ Im(\frac{1}{z^2})=-\frac{2ixy}{(x^2+y^2)}... |
Notice, when $z\in\mathbb{C}$:
$$z=\Re[z]+\Im[z]i$$
So, we get (in steps):
*
*$$z^2=\left(\Re[z]+\Im[z]i\right)^2=\Re^2[z]-\Im^2[z]+2\Re[z]\Im[z]i$$
*$$\overline{z^2}=\overline{\Re^2[z]-\Im^2[z]+2\Re[z]\Im[z]i}=\Re^2[z]-\Im^2[z]-2\Re[z]\Im[z]i$$
*$$z^2\cdot\overline{z^2}=|z|^4=\left(\sqrt{\Re^2[z]+\Im^2[z]}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How does $\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$? How does $$\pi - \arctan(\frac{8}{x}) - \arctan(\frac{13}{20-x}) = \arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})$$
| Use the fact that for $t\not=0$,
$\displaystyle \arctan(t)+\arctan(1/t)=\mbox{sgn}(t)\cdot\frac{\pi}{2}$.
See here for the details: $\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$
Then for $x\not=0$ and $x\not=20$,
$$\arctan(\frac{8}{x})+\arctan(\frac{13}{20-x})+\arctan(\frac{x}{8}) + \arctan(\frac{20-x}{13})\\
=\mbox{sg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Complex number fraction Find the real and imaginary number:
$$\frac{1}{z}=\frac{2}{2+j3}+\frac{1}{3-j2}$$
How do I invert $\frac{1}{z}$ to $z$ so that I can start solving it?
| The denominators of the right hand side are closely related.
We obtain
\begin{align*}
\frac{1}{z}&=\frac{2}{2+j3}+\frac{1}{3-j2}=\frac{2}{2+j3}+\frac{j}{2+j3}
\end{align*}
It follows
\begin{align*}
z=\frac{2+j3}{2+j}=\frac{1}{5}(2+j3)(2-j)=\frac{1}{5}(7+j4)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Continuity and differentiability of $f(x, y) = \frac{2x^3 + 3y^3}{x^2 + y^2}$ I'm learning multivariable calculus, specifically multivariable continuity and differentiability, and need help with the following exercise:
Let
$$f(x,y) = \begin{cases} \frac{2x^3 + 3y^3}{x^2 + y^2}, & (x,y) \neq (0, 0) \\ 0, & (x,y) = (0, ... | Your proof about continuity is correct.
In order to check differentiability fistly find partial derivatives so that
$$ { f }_{ x }^{ \prime }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( x,0 \right) -f\left( 0,0 \right) }{ x } } =\lim _{ x\rightarrow 0 }{ \frac { \frac { 2x^{ 3 } }{ x^{ 2 } } }{ x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find the fourier series of a piecewise continuous function
Consider the function
$f(x) =
\begin{cases}
0 & & x \in [-\pi, 0) \\
1 & & x \in [0, \frac{\pi}{2})\\
0 & & x \in [\frac{\pi}{2}, \pi)\\
\end{cases}$
Calculate its fourier series
I am unsure if this is correct but i have:
$a_0=\frac{1}{2\pi} \int_{-\pi}... | The setup is alright but the evaluations are not correct. For example
$$
\dfrac{1}{\pi} \int_0^{\pi/2} \sin(nx) \; dx = \dfrac{1 - \cos(n \pi/2)}{n \pi} \neq - \dfrac{1}{n \pi}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proving that a given line is tangent to a hyperbola The question is: a line $x \cos\theta + y\sin\theta = p$ is given such that $a^2\cos^2\theta - b^2\sin^2\theta =p^2$.
I have to prove that it touches a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. I do not see how to proceed. I thought I could proceed by elimin... | The tangent at $(x_1,y_1)$ to the hyperbola is
\begin{align*}
\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1
\end{align*}
If this is the given line, we have
\begin{align*}
\frac{x_1/a^2}{\cos \theta} = \frac{-y_1/b^2}{\sin \theta} = \frac{1}{p}
\end{align*}
Hence $x_1 = \frac{a^2\cos\theta}{p}, y_1 = -\frac{b^2\sin\theta}{p}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Loxodromics (curves with fixed angle with meridians) of a revolution surface I am trying to find the curves with a fixed angle $\phi$ with meridians for the revolution surface given by the revolution of the graph of $z=\frac{1}{x}$ around the $z$ axis. The surface is easily parametrized as
$$(x,\theta)\mapsto(x \cos \t... | Well, actually you can also use the vector $(0,1)$, since the tangent vectors to the coordinate lines $x$ and $\theta$ are orthogonal at each point. Then your curve measures a constant angle $\alpha_0$ with the tangent $x-$vector if and only if it also measures a constant angle $\pi/2 - \alpha_0$ with the tangent $\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Constant when integrating I was integrating $(t+2)^2$ by using the substitution method and by expanding the function. By expanding it then integrating I got the same answer as the book ($\frac{1}{3}(t^3+6t^2+12t)+C$), but when I substituted I got $\frac{1}{3}\cdot(t+2)^3 + C = \frac{1}{3} \cdot (t^3+6t^2+12t+8) + C$.
A... | No worries,
$$\left(\frac{1}{3}(t^3+6t^2+12t)+C\right)'=t^2+4t+4$$
and
$$\left(\frac{1}{3}(t+2)^3 + C \right)'=(t+2)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $A$ and $B$ are two matrices such that $AB=B$ and $BA=A$ then how to show that $A^2+B^2$ equals $A+B$?
If $A$ and $B$ are two matrices such that $AB=B$ and $BA=A$ then $A^2+B^2$ equals ?
(a) $2AB$
(b) $2BA$
(c) $A+B$
(d) $AB$
I tried
$(A+B)^2=A^2+B^2+AB+BA$
or,$A^2+B^2=(A+B)^2-AB-BA$
$=(A+B)^2-A-B$
$
=(A+B)^2-(A... | First, here is how you can figure out the answer by process of elimination. We can see that if $A$ and $B$ are both identity, then the condition is satisfied, and $A^2 + B^2 = 2I$ in that case. This rules out option (d). Next, note that $A^2 + B^2$ is symmetric in $A$ and $B$, so if (a) were correct, then by symmetry (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
$(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ I need to show that $(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ and $a,b,c > 0$ using means of univariate Analysis.
It is intuitively clear that $(a+2)^3+(b+2)^3+(c+2)^3$ is at its minimum (when $a+b+c=3$) if $a,b,c$ have "equal weights", i.e. $a=b=c=1$.
To show t... | Alternately, here's a hint
$$(x+2)^3-27 -27(x-1) =(x-1)^2(x+8)\geqslant 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find $ \int\frac{\mathrm{d}x}{(1-x^2)^{3/2}}$ $$
\int\frac{1}{(1-x^2)^{3/2}}\mathrm{d}x=?
$$
I am aware that you can do it with $x=\sin u$ but how do you do it with $x=\tanh u$? I kept ending with with $x^2+1$ in my denominator instead of the correct $x^2-1$.
| If you set $x=\tanh u$, $\;\mathrm d\mkern1mu x=\dfrac1{\cosh^2u}\,\mathrm d\mkern1mu u$, you get
\begin{align*}\int\frac{1}{(1-x^2)^{3/2}}\mathrm{d}x&=\int\cosh^3u\dfrac1{\cosh^2u}\,\mathrm d\mkern1mu u=\int\cosh u\,\mathrm d\mkern1mu u\\
&=\sinh u =\tanh u\cosh u=\frac{\tanh u}{\sqrt{1-\tanh^2 u}}\\
&=\frac x{\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to simplify this expression of square roots? If one simplifies the infinite nested radical
$$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}$$
Does one get $$\sqrt x\frac{1+\sqrt5}2$$
Any help would be appreciated!
| $$\sqrt{x+\sqrt{x^2+\sqrt{x^4+\sqrt{x^8+\dots}}}}=\sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
$$\sqrt{x+x\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}=\sqrt{x}\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
now let
$$y=\sqrt{1+1\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}}$$
so
$$y^2-1=y$$
$$y=\frac{1\pm\sqrt{5}}{2}$$
select the $$y=\frac{1+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Study extreme values of functions of several variables. Well, I have to solve the following problem:
$\textit{Study the extreme values of}$ $f(x,y)=x^2+y^2+\alpha x^3y^3$ $\textit{depending on}$ $\alpha$, $where $ $\alpha\in \mathbb{R}$.
It's easy to find that the extreme values of the function must verify that
$$\left... | For the case $\alpha<0$, let $t=\sqrt[4]\frac{-2}{3\alpha}$, so the critical points are given by $(0,0), (t,t), \text{ and } (-t,-t)$.
As remarked above, there is a relative minimum at $(0,0)$,
and there are saddle points at $(t,t)$ and $(-t,-t)$ since
$\;\;\;f_{xx}=2+6\alpha xy^3=2+6\alpha\big(\frac{-2}{3\alpha}\big)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
By really long division i got :-
$$Q=ax^{15} + (a+b)x^{14} + \dots +(610a + 377b) $$
$$R = x(987a+610b)+1+610a+377b$$
since remainder is $0 $,
$$987a+610b = 0$$
$$1+610a+377b =... | Alternatively, observe that $x^2-x-1$, $x^3-2x^2+1$, and $2x^4-3x^3-1$ are the only polynomials degree less than $5$ divisible by $x^2-x-1$ which are of the form $A\,x^{n+1}+B\,x^{n}+C$ with $A,B,C\in\mathbb{Z}$, $A>0$, and $\gcd(A,B,C)=1$. We suppose that $a_n\,x^{n+1}-b_n\,x^n+(-1)^n$ is divisible by $x^2-x-1$ for e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Complex Numbers - Given $(a+b) +i(a-b) = (1+i)^2 + i(2+i)$ obtain the values of $a$ and $b$. This would be a very easy complex number question to someone I understand of most of it its just one of those questions I should know but I've stared at it so much I'm stuck! could someone please explain it to me! here it is:
Q... | Just do it.
$(1+i)^2 + i(2+i) =$
$(1 + 2i + i^2)+(2i + i^2)=$
$(1 + 2i -1) + (2i - 1)=$
$-1 + 4i$
So $-1 + 4i = (a+b) + (a+b)i$
So $a + b = -1$ and $a+b = 4$.
Well, no wonder you are stuck! You were given an impossible equation.
====
From your commments it sounds like what you we actually given was
$(a+b) + (a-b)i = (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove this inequality $\sum\limits_\text{cyc}\sqrt{1-xy}\ge 2$ Let $x,y,z\ge 0$, and $x+y+z=2$, show that
$$\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\ge 2.$$
Mt try: $$\Longleftrightarrow 3-(xy+yz+zx)+2\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge 4$$
or
$$\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge\dfrac{1}{2}(1+xy+yz+zx)$$
| Lets assume $z=\min\{x,y,z\}$ then note $\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\geq\sqrt{1-\frac{(2-z)^2}{4}}+2\sqrt[4]{(1-yz)(1-xz)}=\sqrt{z-\frac{z^2}{4}}+2\sqrt[4]{1-z(2-z)+z^2xy}\geq\sqrt{z-\frac{z^2}{4}}+2\sqrt[4]{(1-z)^2+z^4}\geq 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate the integral $\int \sqrt{1+\sin(x)} dx$ To find:
$$\int \sqrt{1+\sin(x)} dx$$
What I tried:
I put $\tan(\frac{x}{2}) = t$, using which I got it to:
$$I = 2\int \dfrac{1+t}{(1+t^2)^{\frac{3}{2}}}dt$$
Now I am badly stuck. There seems no way to approach this one. Please give a hint. Also, can we initiall... | \since, $$1=\sin^2\left(\frac{x}{2}\right)+\cos^2\left(\frac{x}{2}\right)$$
$$\sin (x)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$
Therefore, we can put the above values of $1$ and $\sin (x)$ in the question.
Now, we have. $$\int \sqrt{\sin^2\left(\frac{x}{2}\right)+ \cos^2\left(\frac{x}{2}\right)+2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Sum of the series $1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$ Find the sum of $n$ terms of following series:
$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$
I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way... | An alternative to finding the formula for $\sum k^4$ is to instead write the $k$th term as $$(k+2)(k+1)^2k=(k+3)(k+2)(k+1) k-2(k+2)(k+1)k.$$
That is, one trades sums of integer powers for sums of consecutive products. To see the advantage, note for example that
$$1\cdot 2\cdot 3 +2\cdot 3\cdot 4+3\cdot 4\cdot 5+4\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Faster way to get a rational integral $\int_2^3 \frac{x^{100}+1}{x^3+1} \, dx$ So my problem is calculating this integral:
$$
\int_{2}^{3} \frac{x^{100}+1}{x^3+1} \, dx
$$
I know it can be done by polynomial division but that is really tedious I would have to divide $\approx 30$ times.
And i know that Mathematica would... | Well...maybe the division isn't that bad. Do you know about geometric series? What's
$$
1 + r + r^2 + \ldots + r^{32}?
$$
It's
$$
\frac{r^{33} - 1}{r-1}
$$
If you apply this to $r = -x^3$, you get
$$
\frac{-x^{99} - 1}{-(x^3)-1}
= \frac{x^{99} + 1}{x^3 + 1}
$$
I know that's not what you wanted, but if you do a little... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
if the sum of the digits of $a+b$, $b+c$ and $a+c$ is $<5$, then is the sum of the digits of $a+b+c$ less or equal than 50? This is a problem from a math olympiad:
Let $S(x)$ denote the sum of the digits of a natural number $x$ in its decimal representation. Do there exist three natural numbers $a$, $b$ and $c$ with $... | The answer to the question in the title is no -- it is possible that $S(a+b+c) = 51$. In particular, your proof is incorrect.
Take
\begin{align*}
a &= 545\;545\;545\;5 \\
b &= 554\;554\;554\;5 \\
c &= 455\;455\;455\;5.
\end{align*}
(Spaces to aid readability, so you can tell what's going on here.) Then we have
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $(A-B)^2=O_2$ then $\det(A^2 - B^2)=(\det(A) - \det(B))^2$
Let $A,B \in M_2(\mathbb{R})$ be two matrices such that $(A-B)^2=O_2$.
Prove $\det(A^2 - B^2)=(\det(A) - \det(B))^2$.
OBS. $O_2$ is the zero matrix
Let $D=A-B$ then $D^2=O_2$ therefore, using Cayley Hamilton theorem, we get $tr(D)=\det(D)=0$. It follows $... | Since $D=A-B$ is nilpotent, it is similar to the matrix $\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$. So without loss of generality assume $D = \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$. If $B = \left(\begin{matrix} a & b \\ c & d \end{matrix}\right)$, then $A = B+D = \left(\begin{matrix} a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve an inequality with absolute values on both sides? I have the following inequality:
$$|x+3| \geq |x-1| $$
Following this answer I get:
$$
|x+3|=\left\{ \begin{align}
x+3 & \text{ , if }x\geq -3 \\
-x-3 & \text{ , if }x <-3
\end{align}
\right\}
$$
$$
|x-1|=\left\{ \begin{align}
x-1 & \text{ , if }x\ge... | You didn't do anything wrong. You just misinterpreted what you were trying to do and what the information tells you:
You have 3 ranges to consider and in each range you have a set of inequalities to interpret.
If $x \le -3$ (which it might or might not be).
We have $-x - 3 \ge -x + 1$ so $-3 \ge 1$. This is impossibl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Does this constitute a valid proof that $\frac{x^2}{1+x^4} \leq \frac{1}{2}$?
Prove that $$\frac{x^2}{1+x^4} \leq \frac{1}{2}.$$
First off, we observe that the expression on the LHS is positive for all $x \in \Bbb R,$ and equality is achieved iff $x \in \{-1, 1 \}$. That being said, we start by manipulating the expre... | Yes it's Ok, you could also just multiply both sides by $2(1+x^4)$ having to prove that
$$
2x^2\le x^4+1
$$ that is
$$
0\le(x^2-1)^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Largest divisor of $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$
Let $P(n) = (n + 1)(n + 3)(n + 5)(n + 7)(n + 9)$. What is the largest integer that is a divisor of $P(n)$ for all positive even integers $n$?
I can see that obviously it is divisible by $3$ and $5$ and hence answer is $15$ but I can't prove it.
| We have that $\gcd(P(1),P(2))=15$, hence a number $N$ with the property that $N\mid P(n)$ for every $n$ has to be a divisor of $15$. We may notice that $3\mid P(n)$ for every $n$ since exactly one number among $n+3,n+1,n+5$ is a multiple of three. In a similar way, exactly one number among $n+5,n+1,n+7,n+3,n+9$ is a mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
If $a$ and $b$ are consecutive integers, prove that $a^2 + b^2 + a^2b^2$ is a perfect square. Problem is as stated in the title. Source is Larson's 'Problem Solving through Problems'. I've tried all kinds of factorizations with this trying to get it to the form $$k^2l^2$$ but nothing's clicking. I tried Bézout but the ... | Let $p=\frac{a+b}{2}$, that is if $a<b$, $p=(2a+1)/2$.
\begin{align}(p-\frac{1}{2})^2+(p+\frac12)^2+(p-\frac12)^2(p+\frac12)^2&=(2p^2+\frac12)+(p^2-\frac14)^2\\
&=p^4+\frac32 p^2+\frac{9}{16}\\
&=(p^2+\frac{3}{4})^2
\end{align}
since $p^2=\frac{4a^2+4a+1}{4}$,
$$p^2+\frac{3}{4}=a^2+a+1.$$
Hence, $$a^2+b^2+a^2b^2=(a^2+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1909974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
if $AB=AC=EF,AE=BE,\angle BAC=120^{\circ}$ then find $\angle ABE$ If $$AB=AC=EF, AE=BE, \angle BAC=120^{\circ}, \angle ADB=\angle BEF=90^{\circ}$$
find $\angle ABE$
Following is one method:
Let $\angle ABE=a, \angle AFE=\angle DBE=30^{\circ}-a, \angle EAF=60^{\circ}-a$. Use sine thereom, we have
$$\dfrac{\sin{a}}{\sin... |
Let $\angle EBA = \alpha$.
Let $A$ and $X$ be symmetric with respect to $BE$. Then $BX=AB$. Also note that $BE=AE=EX$, so $A,B,X$ lie on the circle with center $E$ and radius $AE$.
Let $Y$ be on $AF$ and $AY=AB$. Since $\angle BAY = 60^\circ$, $\triangle BAY$ is equilateral. Thus $BY=AB$. Therefore $A,X,Y$ lie on the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Partial Fraction Expansion of $12\frac{x^3+4}{(x^2-1)(x^2+3x+2)}$ Find the vector $(A,B,C,D)$ if $A$, $B$, $C$, and $D$ are the coefficients of the partial fractions expansion of
$$12\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2}.$$
After plugging in a few values fo... | You should not begin with solving systems of linear equations. Rewrite the partial fractions decomposition so as to remove all denominators. Multiplying both sides with $(x-1)(x+2)(x+1)^2$, you get
$$12(x^3+4)=A(x+2)(x+1)^2+B(x-1)(x+1)^2 +C(x-1)(x+2)(x+1)+D(x-1)(x+2).$$
Now set
*
*$x=1$: it yields at once $12\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$ I'm trying to prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$, which is part of a larger proof. We can write:
$n^3+2n^2+n=2(1+2+...+n)(n+1)$
$n^3+2n^2+n=(1+2+...+n)(2n+2)$
$n^3+2n^2+n=(2n+4n+6n+...+2n^2)+(2+4+6+...+n)$
I have no idea how to proceed, though.
| The sum $P(n):=1+2+\cdots n$ must be a quadratic polynomial in $n$, because $P(n)-P(n-1)=n$ is a linear polynomial.
Then as both members are cubic polynomials, it suffices to check equality for four distinct values of $n$.
$$\begin{align}n=0&\to 1^3=2()\cdot1+1^2&=1,\\
n=1&\to 2^3=2(1)\cdot2+2^2&=8,\\
n=2&\to 3^3=2(1+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that
$$\frac{(a+b+c)^3\sum\limits_{cyc}(a^2+3ab)}{\prod\limits_{cyc}(a+b)}+3(ab+ac+bc)\geq\frac{11}{2}(a+b+c)^2$$
which is fifth degree, which says that we need to prove a linear inequality of $w^3$,
which says that it's enough to prove the last inequality... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
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How to find the locus of this equation? I am a beginner in math, and am stuck by this problem. The problem is,
Find the locus of the point of intersection of the lines, $x\cos\alpha$ + $y\sin\alpha$ = a and $x\sin\alpha$ - $y\cos\alpha$ = b , where "alpha" is a variable.
What they did, is that they eliminated alpha ... | You goal is to find the set of all $\{x,y\}$ that meet the following criteria:
$x \cos \alpha + y \sin \alpha = a\\
x \sin \alpha - y \cos \alpha = b$
These are perpendicular lines that intersect in a point. But as we allow $\alpha$ to move, that point of intersection moves.
If we can eliminate $\alpha$ then we would ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to show that the roots of $-x^3+3x+\left(2-\frac{4}{n}\right)=0$ are real (and how to find them) I'm trying to find the three distinct and real roots of
$$-x^3+3x+\left(2-\frac{4}{n}\right)=0,$$
where $n>0$ (we could say $n\geq 2$ if that helps), but I'm not able to get very far:
Using the notation of the Wikipedi... | For the calculation of the roots of the depressed cubic
$$
y^{\,3} + p\,y + q = 0
$$
where $p$ and $q$ are real or complex,
I personally adopt a method indicated in this work by A. Cauli, by which putting
$$
u = \sqrt[{3\,}]{{ - \frac{q}
{2} + \sqrt {\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}}} }}\quad v = - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Number of roots of $x^4 + x^3 \sin(x) + x^2 \cos(x) = 0$ How many roots do the following polynomial have?
$$x^4 + x^3 \sin(x) + x^2 \cos(x)$$
Obviously $0$ is a root. Dividing by $x^2$, I am left with
$$x^2 + x\sin(x) + \cos(x)$$
How do I know this polynomial has no root?
| Look for extremum of $f(x)=x^2 + x\sin(x) + \cos(x)$
$$
\frac{d}{dx}(x^2 + x\sin(x) + \cos(x))=2x+x\cos(x)=x(2+\cos(x))=0
$$
Then the only extremum we have is at $x=0$
We know that $x(2+\cos(x))<0$ for $x<0$ and $x(2+\cos(x))>0$ for $x>0$, also $f(0)=1$, which means that $f(x)\ge1$ for all $x$, hence it have no roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why $4a^3+27b^2<0 \iff x^3+ax+b=0$ has exactly three solutions? According to Exc. 8 Sec. 4.3 of the book Advanced Calculus by Fitzpatrick,
For numbers $a$ and $b$, prove that the following equation has exactly three solutions if and only $4a^3+27b^2<0$: $$ x^3+ax+b=0, \ \ \ \ \ x \ \text{in} \ \mathbb R.$$
Long tim... | A third degree-polynomial $p(x)=x^3+ax+b$ has three real roots iff it has two stationary points $x_1,x_2$ and $p(x_1)\,p(x_2)<0$. Such stationary points are located at the roots of $p'(x)=3x^2+a$, hence at $\pm\sqrt{-\frac{a}{3}}$ ($a<0$ is a necessary condition, otherwise $p(x)$ is an increasing function and has only ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Proof for formula for sum of sequence $3 + 9 + 27 + 81...+ 3^n$ Trying to formulate a proof for that sequence as practice. After reading this question's answer and lecture on this, I decided to try and practice with this sequence.
My try:
Base case $n = 1$
$S(n) = 3^n $
Induction step
$S(n+1) = 3^{n+1}$
Then
$Sn + n+1 ... | $$\begin{align}
\require{cancel}S & =3+9+27+\dots3^n \\
3S & =3(3+9+27+\dots3^n) \\
& = 9+27+81+\dots3^{n+1} \\
\hline S-3 & =(3+9+27+\dots3^n)-3 \\
& = 9+27+\dots3^n \\
\hline 3S-(S-3) & = \left(\color{blue}{9+27+\dots3^n}+3^{n+1}\right)-\left(\color{blue}{9+27+\dots3^n}\right) \\
2S+3 & = \left(\xcancel{9+27+\dots3^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\sum\limits_{k=0}^{n-1} \sin^t(\pi k/2n)$? How to evaluate $\displaystyle\sum_{k=0}^{n-1} \sin^t\left(\frac{\pi k}{2n}\right)$?
$t,n$ are constants $\in \Bbb{Z}$.
My try:
$$\begin{align}
\zeta:=e^{i\pi/2n} \implies & \sum_{k=0}^{n-1}\sin^t\left(\frac{\pi k}{2n}\right) \\
= &\frac{1}{2^t}\sum_{k=0}^{n-1... | This integral remind me of Matsubara sum. Let me use contour integral to provide another perspective of the case $t = 2s$.
Define $a_n = \sum_{k = 1}^{n-1} \sin ^{t} \left( \frac{\pi k }{n} \right) $. What OP is looking for is
\begin{equation}
\sum_{k=1}^{n-1} \sin^{t} \left( \frac{\pi k }{2n} \right) = \frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
How to solve $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
Solve the equation $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
I was thinking on it for a few minutes and came up with a few ideas (none of them worked).
My first idea: Use the quadratic formula. Is that allowed? If so, I got to this:
... | For a general approach, note that
$$
\begin{array}{l}
\sqrt z = \sqrt {x + i\,y} = \sqrt {\left| z \right|} \;e^{\,i\,\arg (z)/2} = \\
= \pm \left( {\sqrt {\frac{{\left| z \right| + {\mathop{\rm Re}\nolimits} (z)}}{2}} + \;i\,{\rm sign}^{\rm *} \left( {{\mathop{\rm Im}\nolimits} (z)} \right)\sqrt {\frac{{\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Cube root of a binomial The cube of a certain binomial is $8y^3-36y^2+54y-27$. Find the binomial.
I know that $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and that$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ but don't know how to go further...
| Simply look at the last two terms. $$(a-b)^3=a^3-3a^2b+3ab^2-b^3\tag{1}$$ and from the binomial $$8y^3-36y^2+54y-27\tag{2}$$
We see that $$\begin{cases}8y^3=a^3\\27=b^3\end{cases}\tag{3}$$
Solving, we see that $a=2y$ and $b=3$. So the binomial factors into $$(2y-3)^3\tag{4}$$
Expanding out $(4)$ to check, we get: $$8y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Minimum value of $ \frac{((x-a)^2 + p )^{\frac 12}}{v_1} + \frac{( (x-b)^2 + q )^{\frac12}} {v_2}$ I need to calculate the minimum value of the expression
$ y = \frac{((x-a)^2 + p )^{\frac 12}}{v_1} + \frac{( (x-b)^2 + q )^{\frac12}} {v_2}$
I'm calculating the first derivative , then calculting the corresponding value... | It's derivation of Snell's Law from Fermat's Principle
Assume $a<b$,
\begin{align}
T(x) &= \frac{\sqrt{(x-a)^2+p}}{v_1}+\frac{\sqrt{(x-b)^2+q}}{v_2} \\
T'(x) &=\frac{x-a}{v_1 \sqrt{(x-a)^2+p}}-
\frac{b-x}{v_2 \sqrt{(x-b)^2+q}} \\
0 &= \frac{\sin \theta_{1}}{v_{1}}-\frac{\sin \theta_{2}}{v_{2}} \\
s &= \sin \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Differentiation and proving Prove that if the curve $y=\frac{x^3}{3} + px + q$ is tangent to the straight line $y = x$, then $4(p-1)^3 + 9q^2 = 0$
I have differentiated the equation of the curve to get $x^2 + p$ and equated it to $1$.
But i have no idea how to proceed to get the subsequent equation.
| $$y=\frac{x^3}{3} + px + q$$
Therefore $$y'=x^2 + p $$
Thus the gradient of $y=x$ must be equal to $x^2+p$.
That is $$x^2+p=1$$.
Since $y=x$ at each tangent points $x=\sqrt{1-p}$ and $x=-\sqrt{1-p}$ ,
$$x=\frac{x^3}{3} + px + q$$
$$x^2=\frac{x^4}{3} + px^2 + qx$$
$$(1-p)=\frac{{(1-p)}^2}{3} + p(1-p) + qx$$
$$(1-p)-p(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I go from $33/64 - 2x + x^2 \longrightarrow \bigg(x - \frac{2 + \sqrt{4 +33/16}}{2}\bigg)\bigg(x - \frac{2 - \sqrt{4 +33/16}}{2}\bigg)$?
How do I go from $\frac{33}{64} - 2x + x^2 \longrightarrow\bigg(x - \frac{2 + \sqrt{4 +33/16}}{2}\bigg)\bigg(x - \frac{2 - \sqrt{4 +33/16}}{2}\bigg)$ and then to $\big(x - 1-\... | The quadratic eqation:
If $ax^2 +bx +c =0$ then
$x^2 +\frac {b }{a}x =-c/a$
$x^2 + 2\frac {b}{2a}x +\frac {b}{2a}^2 = -c/a +\frac {b}{2a}^2=\frac {b^2 - 4ac}{4a^2} $
$(x+b/2a)^2 = \frac {b^2 - 4ac}{4a^2} $
$x+ b/2a = \pm \frac {\sqrt { b^2 - 4ac }}{2a} $
$x=\frac {-b\pm \sqrt { b^2 - 4ac } }{2a}$
And therefore:
$ax^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Principal value of Carlson elliptic integral $R_C$ $$R_C(x,y)={1\over2}\int_0^\infty\frac{dt}{(t+y)\sqrt{t+x}},\quad(x\ge0)$$
$R_C$ has a singularity in the denominator when $y$ is negative. Wikipedia says in that case $R_C$ can be evaluated as follows
$$
PV\;R_C(x,-y)=\sqrt{\frac{x}{x+y}}R_C(x+y,y),\quad(y>0)
$$
Q: H... | One proves this identity by doing two things: 1) evaluating the integral for $y \gt 0$ and then 2) setting up the Cauchy PV integral in the complex plane.
1) The evaluation of the Carson integral is straightforward. Let's consider the case $x \gt y$:
$$\begin{align}R_C(x,y) &= \frac12 \int_0^{\infty} \frac{dt}{(t+y) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $p$ is a divisor of numbers $m^2 + 1$, $n^2 + 2$, if and only if it is a divisor of some number of the form $k^4 + 1$. Show that $p$ is a divisor of numbers $m^2 + 1$, $n^2 + 2$, if and only if it is a divisor of some number of the form $k^4 + 1$.
Since $p \mid m^2 + 1$, we have $m^2 \cong -1( \mod p)$ and h... | We have in $\Bbb F_p$
$$x^2=-1\iff (-1)^{\frac{p-1}{2}}=1\Rightarrow p=4s+1\\x^2=-2\iff (-2)^{\frac{p-1}{2}}=1\Rightarrow p=8t\pm 1$$
Hence we must have $p=8k+1$ (the cases $p=8t-1$ must be discarded)
On the other hand $$(-1)^{\frac{p-1}{2}}=((-1)^{\frac{p-1}{4}})^2$$
This means, since $\frac{8t+1-1}{4}$ is a natural i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $A+B+C=\pi$, prove: $\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$
If $A+B+C=\pi$, prove that:
$$\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$$
My Attempt:
Here, $A+B+C=\pi$
Now,
$$\begin{align}
LHS &=\cos(B+2C)+\... | Let $C-A=2x,B-C=2y,A-B=2z\implies2(x+y+z)=0$
$$F=\cos2x+\cos2y+\cos2z=2\cos(x+y)\cos(x-y)+2\cos^2z-1$$
Now as $\cos(x+y)=\cos(-z)=\cos z,$
$$F=2\cos z\cos(x-y)+2\cos z\cdot\cos(x+y)-1$$
$$=2\cos z\{\cos(x+y)+\cos(x-y)\}-1=2\cos z\{2\cos x\cos y\}-1=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the sum $\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \cdots$ Suppose $|x| < 1$. Can you give any ideas on how to find the following sum?
$$
\frac{x}{x+1} + \frac{2x^2}{x^2+1} + \frac{4x^4}{x^4+1} + \frac{8x^8}{x^8+1} + \cdots
$$
| Evaluating $$\sum_{q\ge 0} \frac{2^q x^{2^q}}{1+x^{2^q}}$$
we obtain
$$\sum_{q\ge 0} 2^q \sum_{k\ge 0} (-1)^k x^{(k+1)2^q}
= \sum_{n\ge 1} x^n \sum_{2^q|n} 2^q (-1)^{n/2^q-1}.$$
Now observe that
$$\sum_{2^q|n} 2^q (-1)^{n/2^q-1}
= \sum_{p=0}^{v_2(n)} 2^p (-1)^{n/2^p-1}$$
where $v_2(n)$ is the exponent of the highes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"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.