Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Total Derivative at a point Let $f(x,y)=x^{3}+y^{3}$ .
How do I find the total derivative at (0,0) by using the definition?
I am confused of what to take as the Error function.
| $$ (x+h)^3 +(y+k)^3 -x^3 -y^3 -3x^2 h -3y^2k = 3xh^2 +3yk^2 +h^3 +k^3 =\sqrt{h^2 +k^2} \left[(x+y)\sqrt{h^2 +k^2} +\frac{h^3 +k^3}{\sqrt{h^2 +k^2 }}\right] $$
setting
$r(h,k) =(x+h)^3 +(y+k)^3 -x^3 -y^3 -3x^2 h -3y^2k$
$\phi (h,k )=(x+y)\sqrt{h^2 +k^2} +\frac{h^3 +k^3}{\sqrt{h^2 +k^2 }}$
we have $$\frac{|f((x,y) +(h,k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the set of values of x for which $\frac{x+1}{2x-3}<\frac{1}{x-3}$ Here's what I've done:
$\frac{x+1}{2x-3}<\frac{1}{x-3}$
$x+1<\frac{2x-3}{x-3}$
$(x+1)(x-3)<2x-3$
$x^2-2x-3<2x-3$
$x^2-4x<0$
$x(x-4)<0$
$0<x<4$
However this clearly fails because when $x$ is $2$, for example, the inequality fails.
Where have I gone w... | Taking in account the comment of @SteamyRoot and resolve it again, you will have the answer $x\in I=]0,\frac{3}{2}[~\cup~]3,4[$.
Clearly the example you have taken ($x=2$) doesn't belong to I.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Induction Proof of Taylor Series Formula I'm attempting to prove a formula for the taylor series of function from a differential equation.
The equation is
$$f(0)=1$$ $$f'(x) = 2xf(x)$$
I have found empirically that
$$f(x) = \sum_{k=0}^{\infty}\frac{x^{2k}}{k!}$$
I need to prove that this general formula works via ind... | Note: This proof relies on a formula for the $n^{th}$ derivative of $f(x)$ that can be found HERE
The proof is rather extensive, and would roughly quadruple the length of this answer.
Proof of The Taylor Expansion of $f(x)$
$$\mbox{Prove true that } \ 1 + x^2 + \frac{x^4}{4} + \frac{x^6}{6} + \cdots + \frac{x^{2k}}{k!... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Edwards Differential Calculus for Beginners, 1896. Chapter 4, Question 82 This question asks for an evaluation of an infinite series assuming only knowledge of basic differential calculus. I couldn't figure it out, but user 'Dr. MV' gave me a hint which was sufficient for me to find the answer. I hope this makes the qu... | HINT:
$$\frac{2^kx^{2^k-1}-2^{k+1}x^{2^{k+1}-1}}{1-x^{2^k}+x^{2{k+1}}} =-\frac{d}{dx}\log\left(1-x^{2^k}+x^{2^{k+1}}\right)\tag 1$$
Then, write the argument of the logarithm in $(1)$ as
$$1-x^{2^k}+x^{2^{k+1}}=\frac{1+x^{2^{k+1}}+x^{2^{k+2}}}{1+x^{2^k}+x^{2^{k+1}}}$$
Then, telescope and differentiate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $ \int \frac{1+x\cos(x)}{x(1-x^2(e^{2\sin(x)}))}dx $ $$ \int \frac{1+x\cos x}{x(1-x^2e^{2\sin x})}dx $$
Attempt:
I substituted $(1-xe^{2\sin(x)})$ by $u$ and tried from there by differentiating it. But I get stuck midway.
| Substitute $1-x^2e^{2\sin x} = u$
This gives $$-[2xe^{2\sin x} + 2x^2 \cos x e^{2\sin x}] dx = du \\
-2xe^{2 \sin x} (1+x\cos x) dx = du$$
Putting the value of $dx$ in your integral, we get
$$\int \frac{du}{-2x^2e^{2\sin x}(u)}$$
We know that $$\begin{align} 1-x^2 e^{2\sin x} &= u \\
x^2 e^{2\sin x} &= 1-u \\
-2x^2 e^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Dividing an integer into a fixed number of integers What is the formula for dividing an integer into a fixed number of integers where the greatest distance between consecutive integers is 1.
Dividing 10 into 4 integers we can get:
[3, 3, 2, 2]
Dividing 7 into 4 integers we can get:
[2, 2, 2, 1]
Dividing 10 into 8 int... | Allowing fractions, you would split exactly as
$$\left[\frac{10}4,\frac{10}4,\frac{10}4,\frac{10}4\right].$$
Adding those fractions from left to right, you get
$$\left[\frac{10}4,\frac{20}4,\frac{30}4,\frac{40}4\right].$$
Then taking the integer parts,
$$[2,5,7,10].$$
And finally the pairwise differences
$$[2,3,2,3].$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1791795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\sum\limits_{cyc}\left(\frac{a^4}{a^3+b^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c^{\frac34}}{2^{\frac34}}$ When $a,b,c > 0$, prove
$$\left(\frac{a^4}{a^3+b^3}\right)^{\frac34}+\left(\frac{b^4}{b^3+c^3}\right)^ {\frac34}+\left(\frac{c^4}{c^3+a^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^... | Define:
$$
x = \frac{a^{3/4}}{2^{3/4}} = \left(\frac{a}{2}\right)^{3/4} \quad ; \quad
y = \frac{b^{3/4}}{2^{3/4}} = \left(\frac{b}{2}\right)^{3/4} \quad ; \quad
z = \frac{c^{3/4}}{2^{3/4}} = \left(\frac{c}{2}\right)^{3/4}
$$
Then:
$$
a = 2\,x^{4/3} \quad ; \quad b = 2\,y^{4/3} \quad ; \quad c = 2\,z^{4/3}
$$
And:
$$
\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 1
} |
$12^7+8^8$ divided by $13$ I Need to find what the remainder is when $12^7+8^8$ is divided by $13$
I have a solution, but don't know if it is right.
$12=-1\mod13$
$12^7=-1\mod13$
$8=8\mod13$
$8^2=6\mod13$
$8^4=10\mod13$
$8^8=9\mod13$
Then I did $-1\mod13+9\mod13=8\mod13$, so the remainder is $8$.
If someone could tell... | By Fermat's Little Theorem we have $2^{12}=1\bmod13$. So $8^4=1\bmod13$ and hence $8^8=1\bmod13$. As you said, $12^7=(-1)^7=-1\bmod13$, so $12^7+8^8=0\bmod13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding non-negative integer solutions to $3^x + 5^y = a^2$ $3^x+5^y=a^2$ ($x, y, a$ are non-negative integers)
Find all pairs $(x, y)$ which satisfy the equation.
I have found the trivial solution $x=1, y=0$, and I have tried with congruences, but it didn't really get me anywhere.
| By observing modulo 4 one could easily prove that $x$ is an odd number. If $x=1$, by observing modulo 3 it is easy to see that $y$ is an even number. If $x>1$, then $x\ge 3$. So, $27 \mid a^2-5^y$. Looking at modulo 27, we have $y$ is an even number (check every possibility for $0<y<18$ odd and $0<a<14$). Let $y=2t$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $x+y^2+z^3 \geqslant x^2y+y^2z+z^2x$ for $xy+yz+zx=1$
Let $x,y,z \geqslant 0$ and $xy+yz+zx=1$. Prove that
$$x+y^2+z^3 \geqslant x^2y+y^2z+z^2x.$$
What I try:
$$x+y^2+z^3 \geqslant x^2y+y^2z+z^2x$$
$$\Leftrightarrow x(xy+yz+zx)+y^2+z^3- x^2y-y^2z-z^2x \geqslant 0$$
$$\Leftrightarrow \left(x^2z+\frac{z^3}{4}-z^2... | As you wrote, from the identity:
$$x(xy+yz+zx)+y^2+z^3-x^2y-y^2z-z^2x=z(\frac{z}{2}-x)^2+(y^2+\frac{3}{4}z^3-y^2z)+xyz$$
It suffices to check $y^2+\frac{3}{4}z^3\ge y^2z$ given $yz\le 1$. We take $3$ cases:
If $z\le 1$, then $y^2\ge y^2z$ already.
If $z\ge 1$ and $y\le \frac{1}{4}$, then $\frac{3}{4}z^3\ge \frac{3}{4}z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Using L'hopital's rule to prove differentiability
Question: Define
$$
f(x) = \left\{\begin{aligned}
& \frac{2x\cos(x)}{x+\sin(x)} &&: x \ne 0 \\
&1 &&: x = 0
\end{aligned}
\right.$$
Show $f'(0)$ exist and find the value of it.
My Attempt:
After applying L'Hôpital's rule once I get $$\lim \limits_{h \to 0} \frac{2\... | You can simply do it as follows
\begin{align}
f'(0) &= \lim_{x \to 0}\frac{f(x) - f(0)}{x}\notag\\
&= \lim_{x \to 0}\frac{2x\cos x - x - \sin x}{x^{2} + x\sin x}\notag\\
&= \lim_{x \to 0}\dfrac{2x\cos x - x - \sin x}{x^{2}\left(1 + \dfrac{\sin x}{x}\right)}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\dfrac{2x\cos x - 2x + x -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve the following solvable equation? The following equation is solvable by radicals. How to get all the roots in terms of radicals?
$ x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 = 0$
The transcendental solution is given by
$ \cos(k\pi/13)+i\sin(k\pi/13)$ for $ k = 1, 2,\dots.,12$. Only a solutio... | Let $\zeta = e^{2\pi i / 13}$.
$(\mathbb{Z}/13\mathbb{Z})^{\times}$ is cyclic of order $12$. You get something cubic over $\mathbb{Q}$ by adding $\zeta^k$ where $k$ runs through the subgroup of index $3$ or its cosets:
$$A := \zeta^1 + \zeta^5 + \zeta^8 + \zeta^{12}, \; B := \zeta^2 + \zeta^3 + \zeta^{10} + \zeta^{11},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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build absolute value equations know solution We have absolute value equations with unknown coefficients:
$$|x + a| = b$$
and we know the solutions:
$$x = 11 \text{ and } x = 5$$
We need to find $a$ and $b$. From
$$11 + a = b \\
5 + a = -b$$
we get $a = -8$ and $b = 3$.
But we can try another way:
$$11 + a = -b \\
5 + a... | Plug $x=11$ into the original equation and square both sides. Do the same with $x=5$. We obtain
$$(11+a)^2=b^2=(5+a)^2.\tag1$$
Expanding (1) and solving for $a$ we get
$$121+22a+a^2=25+10a+a^2\ \Longleftrightarrow\ a=-8$$
Plugging this value of $a$ back into the original equation, this gives
$$b=\vert 11+a\vert=\vert 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a formula for a kth element in a sequence I've setup a recurrence relation as part of a numerical analysis problem, and found that $$x_{n+1} = \frac{x_n+1}{2}$$
The notes then say that for $x_0=0$, it is easy to show that $$x_k = 1 - 2^{-k}$$
Plugging in a few numbers shows this is clearly correct, and equating... | Here's one possible process for discovering the general formula: try applying the recurrence a few times:
$$
\begin{aligned}
x_1&=\frac{x_0+1}{2}=\frac{x_0}{2}+\frac{1}{2}\\
x_2&=\frac{x_1+1}{2}=\frac{x_1}{2}+\frac{1}{2}=\frac{x_0}{4}+\frac{1}{4}+\frac{1}{2}=\frac{x_0}{4}+\frac{3}{4}\\
x_3&=\frac{x_2+1}{2}=\frac{x_2}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluating the inverse trigonometric limit $\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}$
$$
\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}
$$
I was doing some questions on limits, I saw one in which there is $\arc... | That $\sqrt{1-x^2}$ shouts "Trigonometric functions!!!". So:
Put $x=\sin\theta$, which means that $\sqrt{1-x^2}=\cos\theta$. Then $$\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}\equiv \lim_{\theta \to \frac{\pi}{4}} \frac{\arccos \left(\sin 2\theta \right)}{\sin \thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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A question about $ax = b$ I am studying inequality and come across the following statement. I don't understand it and want to believe the book must have made mistakes. I am going to copy what the book says here exactly.
A linear inequality with one variable is in the form: $ax>b$. When $a>0$, the solution is $x>b/a$, ... | Your book is correct.
When $a>0$, multiply both sides of the inequality by $\frac{1}{a}$. Since this is a positive number, the direction of the inequality doesn't change:
\begin{align*}
ax&>b
\\\frac{1}{a}\cdot ax &> \frac{1}{a} \cdot b\\
x&>\frac{b}{a}
\end{align*}
When $a<0$, again multiply both sides by $\frac{1}{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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proof that $\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$ diverges, by comparsion I need to prove that
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$$
diverges by comparsion.
The way I did was to use
$$\frac{n}{n^2+2n+1}>\frac{n}{n^2+2n^2+n^2} = \frac{n}{4n^2} = \frac{1}{4n}$$
which diverges. Can I do that? Because $$2n+1<2n^2+n... | $$\sum_{n = 1}^{\infty} \frac{n}{(n+1)^2} \geq \sum_{n = 1}^{\infty} \frac{n}{(n+n)^2} = \sum_{n = 1}^{\infty} \frac{n}{4n^2} = \frac{1}{4}\sum_{n = 1}^{\infty} \frac{1}{n} = \infty. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How Do The Integration of The Given Expression Is $27\pi$ Basically, this is a double integration question from the book Calculus 2 by Howard Anton. Half of the question is resolved and now I'm at this stage where I can't prove the left hand side to the right hand side.
$$
\int_{-3}^3 (6\sqrt {9-x^2} - 2x \sqrt{9-x^2})... | Observe that $\sqrt{9 - x^2}$ is an even function and $x \sqrt{9 - x^2}$ is an odd function. Therefore, the given integral is equal to
\begin{equation*}
2 \times \int_0^3 6 \sqrt{9 - x^2} \,\mathrm dx = 12 \int_0^3 \sqrt{9 - x^2} \,\mathrm dx.
\end{equation*}
Now, substituting $x = 3 \sin t$, the integral becomes
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$
$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove
$$\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$$
A natural though is that from the condition $x^2+y^2+z^2+xyz=4$, I tried a trig substitutions $x=2\cdot \cos A$, $y=2\cdot \cos B$ and $z=2\cdot \cos C$, where $A,... | The condition $\;x^2+y^2+z^2+xyz=4\;$ can be rewritten as:
$$
z^2+xy\cdot z + (x^2+y^2-4) = 0 \quad \Longrightarrow \quad
z = - \frac{xy}{2} \pm \sqrt{\left(\frac{xy}{2}\right)^2-(x^2+y^2-4)}
$$
From $z \ge 0$ it follows that:
$$
z = - \frac{xy}{2} + \sqrt{\left(\frac{xy}{2}\right)^2-(x^2+y^2-4)} \quad \mbox{and} \quad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
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Choosing Functions for the Squeeze Theorem
Evaluate $$\lim_{n\to \infty}\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$$
$$$$
I came across the the question on this site itself but had a few doubts on the given solution. As I do not yet have 50 reputation points, I cannot comment over ther... | Limits don't necessarily preserve strict inequalities. For example, $1-\frac{1}{n}<1+\frac{1}{n}$, yet they have the same limit as $n$ goes to $\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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If $a,b>0$ and $a+b=1\;,$ Then minumum value of $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ is If $a,b>0$ and $a+b=1\;,$ Then minumum value of $\displaystyle \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2$ is
$\bf{My\; Try::}$ Let $a=\sin^2 \theta$ and $b=\cos^2 \theta\;,$ Then We have to minimize
$$\displaystyl... | Here is an alternative solution: take $f(x) = \left(x + \frac{1}{x}\right)^2$ then $f''(x) = 2 + \frac{6}{x^4} > 0$ so $f$ is a convex function and therefore satisfy
$$f(a) + f(b) \geq 2f\left(\frac{a+b}{2}\right) = 2f\left(\frac{1}{2}\right) = \frac{25}{2}$$
This argument can also be used to solve other similar types... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Given $\tan 3x=4$, find the value of $\tan^2 x+\tan ^2(120+x)+\tan^2 (60+x)$ Given $\tan 3x=4$, find the value of $S=\tan^2 x+\tan ^2(120+x)+\tan^2 (60+x)$
I expanded each of $\tan (120+x)$ and $\tan (60+x)$ getting as
$$S=\tan^2 x+\left(\frac{\tan 120+\tan x }{1-\tan 120 \tan x}\right)^2+\left(\frac{\tan 60+\tan x }{1... | Hint:
Observe the following :
$1)$ $3\tan(3x)=\tan(x)+\tan({x+60^{\circ}})+\tan({x+120^{\circ}})$
$2)$ $\tan{60^{\circ}=\tan[(x+60^{\circ})-60^{\circ}}]$ and similar results
$3)$ $\tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$
$4)$ $\tan(60^{\circ})=\sqrt{3}$ and $\tan(120^{\circ})=-\sqrt{3}$
From $(1)$, we get ... | {
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"url": "https://math.stackexchange.com/questions/1806826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $a_1,a_{100}, a_i=a_{i-1}a_{i+1}$, what's $a_1+a_2$? I've been given the following puzzle
Let $a_1, a_{100}$ be given real numbers. Let $a_i=a_{i-1}a_{i+1}$ for $2\leq i \leq 99$. Further suppose that the product of the first $50$ is $27$, and the product of all the $100$ numbers is also $27$.
Find $a_1+a_2$.
I... | Such a sequence is periodic with period length 6, one cycle
$$ A, B, \frac{B}{A}, \frac{1}{A}, \frac{1}{B}, \frac{A}{B} $$
Note that the product of all six is $1.$ Therefore the product of the first 96 is one, and 27 is $\frac{B^2}{A}.$ The product of the first 48 is one, and 27 is also $AB.$ That is
$$ 27 A = B^2,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Prove $12\int_{0}^{\infty}{x\over (e^{x^2}+1)(e^{x^2}+3)}dx=\ln(2)$ $$I=12\int_{0}^{\infty}{x\over (e^{x^2}+1)(e^{x^2}+3)}dx=\ln(2)$$
$u=e^{x^2}\rightarrow du=2xe^{x^2}dx$
$x\rightarrow \infty,\, u=\infty$
$x\rightarrow 0,\, u=1$
$$6\int_{1}^{\infty}{2x\over (u+1)(u+3)}\cdot{du\over 2xe^{x^2}}=6\int_{1}^{\infty}{1\over... | Following Claude Leibovici's comment, assuming $\text{Re}(a),\text{Re}(b)>-1$,
$$\begin{eqnarray*}J(a,b)=\int_{0}^{+\infty}\frac{2x\,dx}{(e^{x^2}+a)(e^{x^2}+b)}&=&\int_{0}^{+\infty}\frac{du}{(e^{u}+a)(e^{u}+b)}\\&=&\frac{1}{b-a}\int_{0}^{+\infty}\left(\frac{1}{e^u+a}-\frac{1}{e^u+b}\right)\,du\\&=&\frac{1}{b-a}\int_{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int \frac{dx}{(1+x^2)^2}$? I need to integrate this to finish an old STEP problem I'm doing, but I'm stuck here, at the very end:
$$\int_0^\infty \frac{dx}{(1+x^2)^2}$$
The result should be $\pi\over 4$ . I don't know how to approach this. *Somehow, this question doesn't seem to've been posted here e... | $$
\begin{aligned}
\int \frac{d x}{\left(1+x^2\right)^2} &=-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{1+x^2}\right) \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \int\left(-\frac{1}{x^2}\right) \frac{1}{1+x^2} d x \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \int\left(\frac{1}{1+x^2}-\frac{1}{x^2}\right) d ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 7
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Find the element in $\mathbb{Z}/143\mathbb{Z}$ whose image is $(\overline{10},\overline{11})$ under the Chinese remainder theorem Find the element in $\mathbb{Z}/143\mathbb{Z}$ whose image is $(\overline{10},\overline{11})$ in $\mathbb{Z}/11\mathbb{Z} \times \mathbb{Z}/13\mathbb{Z}$ under the Chinese remainder theorem
... | By Bezout's identity we know that we can find $ x, y$ such that $ 11x + 13y = 1 $. Using the Euclidean algorithm, we have:
$$ 13 = 11\cdot 1 + 2 $$
$$ 11 = 2 \cdot 5 + 1 $$
$$ 2 = 2 \cdot 1 + 0 $$
and now we substitute back:
$$ 1 = 11 - 2 \cdot 5 = 11 - (13 - 11) \cdot 5 = 11 \cdot 6 - 13 \cdot 5 $$
This helps us becau... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$ Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$
According to Fermat's theorem:
$$1^7+7^7+13^7+19^7+23^7\equiv{1+7+13+19+23}\pmod{7}\equiv{63}\pmod{7}\equiv{0}\pmod{7}$$
Now we need to show: $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{9}$ , but how??
| Just as Fermat's theorem tells you that $x^6\equiv1\pmod{7}$, Euler's theorem (which generalizes Fermat's) tells you that $x^6\equiv1\pmod{9}$ whenever $\gcd(x,9)=1$. It follows that
$$1^7+7^7+13^7+19^7+23^7\equiv1+7+13+19+23\equiv0\pmod{9}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Finding the second derivative of$f(x)=x^2\sqrt{4-x}$
Find the second derivative of the function following:
$$f(x)=x^2\sqrt{4-x}$$
Here I go...
$$f(x)= x^2(4-x)^{1\over 2}$$
\begin{align*}
f'(x) &= 2x(4-x)^{1\over 2}+{1\over 2}x^2(4-x)^{-{1\over 2}}(-1)\\
&= 2x(4-x)^{1\over 2} - {1 \over 2}x^2(4-x)^{-{1\over 2}}\\
&... | Your first derivative is 100% correct. The second derivative indeed has an error, as you must do the product rule within a product rule if you don't simplify your first derivative as in Ángel Mario Gallegos answer.
$$f''(x)=(\frac{1}{2}x(4-x)^{\frac{-1}{2}})(-5)+(16-5x)(\frac{1}{2}x[\frac{-1}{2}(4-x)^{\frac{-3}{2}}(-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $\sin \theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$, then $\sin 2\theta$ is a root of $x^2 -44x +36=0$ My own bonafide attempt.
$$ 0<\theta<\pi/2$$
and $$\sin\theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$$
then show that $\sin 2\theta$ is a root of the equation $$x^2 -44x +3... | With an obvious short notation and setting $x=2cs$, we reduce to the common denominator
$$s+c+\frac sc+\frac cs+\frac 1s+\frac 1c=\frac{sx+cx+2+2c+2s}{x}=\frac{(x+2)(c+s)+2}x=7.$$
As we know that $x\ne0$ and with $(c+s)^2=x+1$, we rewrite
$$(x+2)^2(x+1)=(7x-2)^2,$$
$$x^3-44x^2+36x=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Equation of a tangent on a circle given the gradient and equation of the circle My maths teacher told me this problem was impossible without knowledge of implicit differentiation: is she right?
You are given the equation of the circle $\left(x+2\right)^2+\left(y-2\right)^2=16$ , what are the equations of the lines with... | $$ (x+2)^2 + (y-2)^2 = 16 $$
$$ \Rightarrow x^2 + y^2+4x-4y+4+4 = 16 $$
$$ \Rightarrow x^2 + y^2+4x-4y-8= 0 $$
$$ \text{centre} =(-2,2), \text{ radius}=4 $$
for tangent the line must satisfies the condition, $r=d$
we have line $y=2x+c$ here gradient$=2$ need to find $c=$?
$$ \pm4 =\frac {2(-2)-2+c}{2^2+1} $$
$$ \pm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Prove $\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}}$ for $x,y,z \geq 0$ On the left we have arithmetic mean of pairwise quadratic means, which obeys:
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{ \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2... | You're almost done! $\sqrt{2(x^2+y^2)} \ge x+y$, and so $2\sqrt{x^2+y^2}\sqrt{x^2+z^2} \ge x^2+xy+xz+yz$. Summing gives the inequality you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove this $|a_{n}|\le 2$ Let sequence such $$|a_{n}-\dfrac{a_{n+1}}{2}|\le 1$$
if $|a_{n}|\le \dfrac{3^n}{2^n}$, show that
$$|a_{n}|\le 2$$
since
$$2a_{n}-2\le a_{n+1}\le 2a_{n}+2$$
so we
$$a_{n+1}-2\ge 2(a_{n}-2)\ge\cdots \ge 2^n(a_{1}-2)$$
| Suppose $a_N>2$. Put $a_N=2+k$ where $k>0$. Then we have $1+k\le\frac{1}{2} a_{N+1}$ and so $a_{N+1}\ge2+2k$. By a trivial induction $a_{N+m}\ge2+2^mk$. Now we claim that for sufficiently large $m$ we have $a_{N+m}>\frac{3^{N+m}}{2^{N+m}}$.
For $\left(\frac{4}{3}\right)^m>\frac{1}{k}\left(\frac{3}{2}\right)^N$ for suff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$ Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$
My Attempt:
Let the equation of the sphere be $x^2+y^2+z^2+2ux+2vy+2wz+d=... | Hints:
(1) The given sphere is $\;(x+1)^2+(y-3)^2+z^2=9\;$
(2) The line through the above sphere's center and the given tangency point is $\;(-1,3,0)+t(1,2,-2)\;,\;\;t\in\Bbb R\;$
(3) In a similar way as in plane geometry, two tangent spheres' centers are joined by a line passing through the tangency point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Is $(x^2+y^2+z^2) \ \sin \frac{1}{\sqrt{x^2+y^2+z^2}}$ Differentiable in $(0,0,0)$? $$f(x,y,z)=\begin{cases} (x^2+y^2+z^2) \ \sin \frac{1}{\sqrt{x^2+y^2+z^2}} \qquad (x,y,z) \ne (0,0,0) \\ \\ 0 \qquad (x,y,z)=(0,0,0) \end{cases} $$
At first, I study the continuity in the origin.
I apply the concept of sequential cont... | You can't study the continuity of $f$ at the origin by choosing one sequence $\mathbf{p_n} = (x_n,y_n,z_n)$ that approaches $\mathbf{0}$ in a very specific way ($x_n = y_n = z_n = \frac{1}{n}$) and checking that $f(\mathbf{p_n}) \rightarrow f(\mathbf{0})$. You need to check that this works for all sequences converging ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Calculating a limit with series Good evening to everyone. I have a limit that gave me a lot of trouble and I couldn't find a way to solve it. I tried solving it with series but I couldn't arrive at a result.
$$
\lim _{x\to 0+}\left(\frac{\left(e^{-\frac{1}{x^2}}\cos \left(\log _e\left(x\right)\right)+\cos \left(\arctan... | tl;dr: This is going to be such a hoot.
*
*Let us start with the denominator:
$$\begin{align}
\ln\left(1+x^2\right)-\sin \left(x^2\right)
&=
x^2-\frac{x^4}{2} + o(x^4) - (x^2+o(x^4)) = -\frac{x^4}{2} + o(x^4)
\end{align}
$$
*Now, the meat — the numerator, piece by piece:
*
*No Taylor for the first, it does not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Using AM-GM inequality to prove prove that $$x^4 + y^4 + z^4 \geq xyz(x+y+z)$$
This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.
| AM–GM is invoked in two steps as follows:
$\begin{array}{rcl}x^4+y^4+z^4 &=& \dfrac{x^4+y^4}2+\dfrac{y^4+z^4}2+\dfrac{z^4+x^4}2 \\\\ &\ge& x^2y^2+y^2z^2+z^2x^2 \\\\ &=& \dfrac{x^2y^2+y^2z^2}2+\dfrac{y^2z^2+z^2x^2}2+\dfrac{z^2x^2+x^2y^2}2 \\\\ &\ge& xy^2z+yz^2x+zx^2y \\\\ &=& xyz(x+y+z)\end{array}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that $4$ divides $3^{2m+1} - 3$
Prove that $4$ divides $3^{2m+1} - 3$.
By plugging in numbers I can see this is true, but I can't figure out a way to prove this, I was thinking maybe proving first that it is divisible by $2$, and concluding its divisible by $4$.
| There are generally several ways to approach these types of divisibility problems. I am showing two of them.
By Mathematical Induction: Putting $m=0$ we have $3^{2m+1}-3=0$, which is obviously divisible by $4$. Now let $4\mid 3^{2m+1}-3$ for some $m\in N_0$. Let $3^{2m+1}-3=4k$, for some $k\in N_0$. Now $3^{2(m+1)+1}-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 1
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Proving $\cos A \cdot \cos 2 A \cdot \cos 4 A \cdots \cos 2^{n-1} A = \frac{\sin 2^n A}{2^n \sin A}$ Just a bit of background on the question:
When proving:
$$\cos\frac{\pi}{15}\cdot \cos\frac{2\pi}{15} \cdot \cos\frac{3\pi}{15}\cdot \cos\frac{4\pi}{15} \cdot \cos\frac{5\pi}{15} \cdot \cos\frac{6\pi}{15}\cdot \cos\frac... | By the sine duplication formula $\sin(2x)=2\sin(x)\cos(x)$ it follows that:
$$ \cos(2^n A) = \frac{\sin(2^{n+1} A)}{2\sin(2^n A)}\tag{1} $$
hence:
$$ \prod_{n=0}^{N-1}\cos(2^n A) = \frac{\sin(2^N A)}{2^N \sin(A)}\tag{2}$$
comes from a telescopic product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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System of diophantine equations $x^2+3y=u^2$, $y^2+3x=v^2$
Solve the following system of Diophantine equations(the unknowns are positive integers):
$$
\left\{
\begin{array}{c}
x^2+3y=u^2 \\
y^2+3x=v^2
\end{array}
\right.
$$
I worked as follows:
subtract the two equations to get: $4x^2-4y^2-12(x-y)=9y^2-9x^2\ \... | We can without (much) loss of generality assume that $y\le x$. Note that $x^2+3y$ is a perfect square greater than $x^2$.
Thus we have $x^2+3y\ge (x+1)^2$. But $x^2+3y\lt (x+2)^2$. It follows that $3y=2x+1$.
Since $y^2+3x$ is a perfect square, so is $9y^2+27x$, that is, $4x^2+31x+1$. This has to be a square, so $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Re-write a quadratic equation in another form? $x^2 + \sqrt{2}x = \frac{1}{2}$
I need to find the real solutions for this equation and write it in this form:
$$\frac{-\sqrt{A} \pm B}{C}$$
So when I work the problem out with the quadratic equation I get: $x = 0.118121$
I had no idea how to even put that in the form de... | Form of any quadratic Equation is $ax^2+bx+c=0$
Here for you $ a=1 , b= \sqrt{2}$ and $c = -\frac{1}{2}$, $$x^2 + \sqrt{2}x + \left(-\frac{1}{2} \right)=0$$
Solution is $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-\sqrt{2}\pm \sqrt{(\sqrt{2})^2-4\cdot 1 \cdot \left(-\frac{1}{2}\right)}}{2 \cdot 1}$$
$$x=\frac{-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Deriving the Airy functions from first principles I have just started reading about the Airy functions and am stuck on a particular step of their derivation. But first here is some background information to give this question some meaning, more information can be found from a previous question of mine:
The general sol... | At first note, that $i=\sqrt{-1}$ is the imaginary unit in (9) and we see a plain multiplication with $i$.
We start with the representation (5) and use (7) to obtain
\begin{align*}
y&=x^{\frac{1}{2}}Z_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\\
&=x^{\frac{1}{2}}\left[AJ_{\frac{1}{3}}\left(\frac{2}{3}ix^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$ Prove that
$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$
$$\text{if and only if}$$
$$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after ge... | Note that we have
$$\begin{align}
a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}&=\sqrt{x}\left(a\sqrt{1+\frac{1}{x}}+b\sqrt{1+\frac{2}{x}}+c\sqrt{1+\frac{3}{x}}\right)\\\\
&=a\sqrt{x}\left(1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)\right)\\\\
&+b\sqrt{x}\left(1+\frac{1}{x}+O\left(\frac{1}{x^2}\right)\right)\\\\
&+c\sqrt{x}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
problem proving: $(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$ I'm trying to prove this, and it is really frustrating, because it seems a really easy problem to prove, however, I'm having a little problem with exponents:
$$(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$$
Hyp... | Induction on $n$.
For brevity let $\prod_{j=0}^n(1+q^{2^j})=P(q,n)$.
For the case $n=0$ we have $(1-q)P(q,0)=(1-q)(1+q)=1-q^2=1-q^{2^{0+1}}.$
If $(1-q)P(q,n)= 1-q^{2^{n+1}}$ then $$(1-q)P(q,n+1)=(1-q)P(q,n)(1+q^{2^{n+1}})=$$ $$=(1-q^{2^{n+1}})(1+q^{2^{n+1}})=1-(q^{2^{n+1}})^2=1-q^{2^{n+2}}.$$
For example $$(1-q)(1+q)(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solution of $(n+1)^{1/3}-n^{1/3}=\frac{1}{12}$ Solve the given equation for $n$
$(n+1)^{1/3}-n^{1/3}=\frac{1}{12}$
How to approach this particular question? Sorry cannot show any work because the only approach I can see is take cube on both sides but that is complicating the equation.
| Let $x=\sqrt[3]{n+1}$ and $y=\sqrt[3]{n}$. Note that $(x-y)^3=x^3-y^3-3xy(x-y)=1-3xy(x-y)$. So
$$\frac{1}{12^3}=1-\frac{xy}{4}.$$
Solve for $xy$. We get $xy=a$, where $a$ is a mildly messy number.
We also have $x-y=b$, where $b=\frac{1}{12}$.
So $(x+y)^2=b^2+4a$, and now we know that $x+y=\pm\sqrt{b^2+4a}$.
We know $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Which number is greater, $11^{11}$ or $9^{12}$?
Which number is greater than $11^{11}$ or $9^{12}$?
My work so far:
$11^{11}=285311670611>9^{12}=282429536481$.
But to verify the validity of equality should be in the range of easily verifiable calculations.
| Just for the heck of it, here's another approach using the fact that
$$9^3=6\times11^2+3\qquad\text{ and }\qquad 6^4<11^3-3\times11.$$
It is by no means the slickest way;
\begin{eqnarray*}
9^{12}&=& (6\times11^2+3)^4=3^4\times(2\times11^2+1)^4\\
&=& 3^4\times(2^4\times11^8+4\times2^3\times11^6+6\times2^2\times11^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
| Using the rule $a\sqrt b=\sqrt{a^2b}$ for $a,b>0$ one checks the signs of numerator and denominator to be both positive for the first fraction and both negative for the second fraction. This is therefore a comparison of two positive numbers (so no verdict yet), and by flipping the signs of numerator and denominator on ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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How to prove this inequality $x\sin^2{A}+y\sin^2{B}\ge xy\sin^2{C}$ In $\Delta ABC$,if $x,y>0$ and $x+y=1$.show that
$$x\sin^2{A}+y\sin^2{B}\ge xy\sin^2{C}$$
I have looked at the simpler methods,? Here is one solution
$$\dfrac{\sin^2{A}}{y}+\dfrac{\sin^2{B}}{x}\ge\dfrac{(\sin{A}+\sin{B})^2}{x+y}=\dfrac{\sin^2{(A+B... | It's $\frac{4S^2x}{b^2c^2}+\frac{4S^2y}{a^2c^2}\geq\frac{4S^2xy}{a^2b^2}$ or
$(x+y)(a^2x+b^2y)\geq xyc^2$ or $a^2x^2+b^2y^2+(a^2+b^2-c^2)xy\geq0$.
$\Delta=(a^2+b^2-c^2)^2-4a^2b^2=-16S^2<0$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Compute $\int \sqrt{1+4x^2} \, dx$ with Euler substitution In this post:
Computing $\int \sqrt{1+4x^2} \, dx$
someone mentioned Euler substitution to compute the following integral:
$$\int \sqrt{1+4x^2} \, dx$$
I tried to follow this advice and got very nice result, namely I substituted $\sqrt{1+4x^2}=t-2x$ which after... | If you set $\sqrt{1+4x^2}=t-2x$, you have
$$
1+4x^2=t^2-4tx+4x^2
$$
so $4tx=t^2-1$ and therefore
$$
x=\frac{t^2-1}{4t}=\frac{t}{4}-\frac{1}{4t}
$$
Thus
$$
dx=\left(\frac{1}{4}+\frac{1}{4t^2}\right)\,dt=\frac{t^2+1}{4t^2}\,dt
$$
and
$$
\sqrt{1+4x^2}=t-\frac{t}{2}+\frac{1}{2t}=\frac{t^2+1}{2t}
$$
so the integral becomes
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solutions of $\sin^2\theta = \frac{x^2+y^2}{2xy} $ If $x$ and $y$ are real, then the equation
$$\sin^2\theta = \frac{x^2+y^2}{2xy}$$
has a solution:
*
*for all $x$ and $y$
*for no $x$ and $y$
*only when $x \neq y \neq 0$
*only when $x = y \neq 0$
| HINT:
$$(x/y)^2-2(x/y)\sin^2\theta+1=0$$
As $x/y$ is real,the discriminant is $$(2\sin^2\theta)^2-4\ge0\iff\sin^4\theta\ge1\iff\sin^2\theta\ge1\iff\cos^2\theta\le0$$
this is only possible if $\cos^2\theta=0\iff\sin^2\theta=?$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $∇(r^2/(2z(1+a/z^2)))$ in cylindrical coordinates How to calculate $$∇\bigg(\frac{(ρ^2)}{2z(1+\frac{a}{z^2})}\bigg)$$ where the function is in cylindrical coordinates $$ρ^2=x^2+z^2$$
$$∇\bigg(\frac{x^2+z^2}{2z(1+\frac{a}{z^2})}\bigg)$$
Is the answer in Cartesian coordinates a vector
$$[x/(z(1+a/z^2))),... | You need the gradient operator for cylindrical coordinates, e.g. see here.
$$
\nabla f =
\frac{\partial f}{\partial \rho} e_\rho +
\frac{1}{\rho}\frac{\partial f}{\partial \varphi} e_\varphi +
\frac{\partial f}{\partial z} e_z
$$
and apply it to
$$
f(\rho, \phi, z) = \frac{\rho^2}{2z \left( 1 + \frac{a}{z^2} \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the Derivatives of $g(x) = \sqrt{3-2x^2}$ and $h(x) = \ln {(x^2 – x)}$ I am asked to find the derivatives of $g(x) =\sqrt{3-2x^2}$ and $h(x) = \ln{(x^2 – x)} $
For:
$g(x)h(x)$
and
$\dfrac{h(x)}{g(x)}$
and
$h^3 (x)$
First off I am not sure if my derivatives are correct. Here is what I have..
$g'(x) = \dfrac{2x}... | Your $h'$ is correct, but your $g'$ should be $\dfrac 12 \cdot \dfrac{-4x}{\sqrt{3-2x^2}}=\dfrac {-2x}{\sqrt{3-2x^2}}$
So from the product rule we have $(h \cdot g)'=h' \cdot g + h \cdot g'=\dfrac {2x-1}{x^2-1} \cdot \sqrt{3-2x^2} + \ln (x^2-x) \cdot \dfrac {-2x}{\sqrt{3-2x^2}}$. You can't really do much to simplify it... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $2^x+4^x=2$ This is the equation, but the result is different from wolframalpha:
$$2^x+4^x=2$$
$$2^x+2^{2x}=2^1$$
$$x+2x=1$$
$$x=\frac{1}{3}$$
WolframAlpha: $x=0$
Where is the error?
| $$2^x+2^{2x}=2$$
Now put $2^x=t$
$$t+t^{2}=2$$
$$t^{2}+t-2=0$$
$$(t-1)(t+2)=0$$
Thus $t=1$ or $t=-2$
$2^x=1$ or $2^x=-2$
Since $2^x>0 $ for all real $x$ , $2^x=1=2^0$
Therefore $x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Inequality with square root $x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}}$ Good morning to everyone! The inequality is the following:$$ x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} $$. I don't know how to solve it. Here's what I tried: $$x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} \rightarrow 2x^2-11x+9+\sqrt... | Hint:
The domain of the inequation is defined by
$$x^2-10x+9\ge 0\iff (x-5)^2\ge 16\iff x\ge 9\quad\text{or}\quad x\le 1.$$
$u\ge \sqrt u\iff u\ge 1$ or $u=0$. Here
$$u=0 \iff\sqrt{x^2-10x+9}=-x\iff -10x+9=0\quad\text{and}\quad x\le 0,$$
which has no solution. Hence the inequation comes down to
$$\sqrt{x^2-10x+9}\ge 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the chain rule for cos and sin functions I am having issues with derivatives containing chain rules. I know there is multiple threads already but after reading a few, I still find myself confused. I also checked the actual answer following a step by step website without success.
Derivate: $$h(x)=\sin ( x^6 - cos^... | $f'(g(x))\cdot g'(x)$ means, in your case, $\sin'(x^6-\cos^3{x^2})\cdot (x^6-\cos^3(x^2))'$ and not what you have written. This would give the result to be $\cos(x^6-\cos^3(x^2))\cdot (6x^5-(3\cos^2(x^2)(-\sin (x^2))(2x))$ which equals the real answer given, i.e. $\cos(x^6-\cos^3(x^2))\cdot(6x^5+6x\cos^2(x^2)\sin(x^2))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Proving that $\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+\cdots+\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}$ by induction I've proved the base case where $n=1$ and made the assumption that $n=k$ is true, but I'm stuck on the $n=k+1$ part. I just cannot seem to get the algebra to work in my favor.
Here is the original:
... | Let $S(n)$ be the statement: $\dfrac{1}{4(5)}+\dfrac{1}{5(6)}+\dfrac{1}{6(7)}+\cdots+\dfrac{1}{(n+3)(n+4)}=\dfrac{n}{4\hspace{1 mm}(n+4)}$; $n\in\mathbb{N}$
Basis step: $S(1)$:
LHS: $\dfrac{1}{\big((1)+3\big)\big((1)+4\big)}=\dfrac{1}{4\times{5}}$
$\hspace{47.5 mm}=\dfrac{1}{20}$
RHS: $\dfrac{n}{4\hspace{1 mm}(n+4)}=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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The value of $x^2+y^2+z^2+w^2$ Let$x,y,z,w$ satisfy
$$\frac{x^2}{2^2 - 1^2} +\frac{y^2}{2^2 - 3^2} +\frac{z^2}{2^2 - 5^2} +\frac{w^2}{2^2 - 7^2} =1$$
$$\frac{x^2}{4^2 - 1^2} +\frac{y^2}{4^2 - 3^2} +\frac{z^2}{4^2 - 5^2} +\frac{w^2}{4^2 - 7^2} =1$$
$$\frac{x^2}{6^2 - 1^2} +\frac{y^2}{6^2 - 3^2} +\frac{z^2}{6^2 - 5^2} +\... | Set up a matrix equation
$$\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34}\\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{bmatrix} = \begin{bmatrix} 1\\1\\1\\1\end{bmatrix}$$
and then solve. You ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Fibonacci summation Can anyone help me to prove the following relation.
$$\sum_{k=1}^{\infty} \frac{F_{2k}H^{(2)}_{k-1}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt{5}}$$
I was studying recently about Fibonacci and Lucas numbers.
And I came through the above relationship. I tried applying golden ratio but nothing wor... | It is well known that $$\sum_{k=1}^{\infty}\frac{x^{2k}}{k^{2}\dbinom{2k}{k}}H_{k-1}^{\left(2\right)}=\frac{2\arcsin^{4}\left(\frac{x}{2}\right)}{3},\,\left|x\right|\leq2$$ (see here or here for a proof), then from the Binet formula we get $$\sum_{k=1}^{\infty}\frac{F_{2n}}{k^{2}\dbinom{2k}{k}}H_{k-1}^{\left(2\right)}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Factoring $x^4-11x^2y^2+y^4$ I am brushing up on my precalculus and was wondering how to factor the expression
$$
x^4-11x^2y^2+y^4
$$
Thanks for any help!
| For emphasis, if you have a quadratic or quadratic form, $a x^2 + bx + c$ or $a x^2 + bxy + c y^2,$it factors over the integers if and only if the discriminant from the Quadratic Formula, $b^2 - 4 a c,$ is an integer squared.
Things are rather different for quartics with only EVEN exponents, either
$f^2 x^4 + g x^2 +... | {
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"url": "https://math.stackexchange.com/questions/1847983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$
Find the number of solutions of the equation
$$\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,$$
... | Let
$$
f(x) = \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor
+ \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor.
$$
For integers $n$ we have $f(n)=63n$, but for a small number to the left of $n$ all terms $\lfloor x \rfloor, \lfloor 2x \rfloor,\dots$ decrease by one. So at every integer $n$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem
What is the remainder when $10^{35}$ is divided by $... | You can use the fact that the remainder of $ab$ is the remaineder of $a$ multiplied by the remaineder of $b$. So forgeting factors of $7$, we have $3^{35}=27*3^{32}=6*(9)^{16}=6*2^{16}=6*(16)^4=6*2^4=6*2=5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
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Polynomial ,divides and Induction Proof?
$\text{The polynomial } x-y \;\text{divides the polynomial}\; x^2-y^2 \text{ and } x^3-y^3 \text{because}\; x^2-y^2 = (x+y)(x-y) \text{ and } x^3-y^3=(x-y)(x^2+xy+y^2.) \; \text{for every natural number n } \quad x-y \;\text{ divides }\; x^n-y^n \text{ prove by induction... | Let say ${ x }^{ n }-{ y }^{ n }$ is dividing by $x-y$ then we have $${ x }^{ n }-{ y }^{ n }=\left( x-y \right) P\left( x \right) $$,where $P(x)$ is some polynomial.Now we should prove that ${ x }^{ n+1 }-{ y }^{ n+1 }$ is also dividing by $x-y$
$${ x }^{ n+1 }-{ y }^{ n+1 }={ x }^{ n+1 }-{ x }^{ n }y+{ x }^{ n }y-{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$1+2+3+45+6+78+9=144$ what are other combinations Note that $$1+2+3+45+6+78+9 = 144$$ In how many other ways is it possible to make a total of $144$ using only $1, 2, 3, 4, 5, 6, 7, 8,$ and $9$ in that order and addition signs?
Sorry I am only in high school so dont over complicate the explanation. Thank you
| Inspired by the idea: "1+2+..9=45 therefore we need 99 more to get 144. The two digits 'ab'=10a+b so changing a+b to 'ab' adds 9a to the total. Therefore an extra eleven 9s are required. This means that the the possibilites are (3,8) (6,4,1) etc."
We need find all the combinations that add up to 11. But in these combin... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $a + b + c$, given that $(a+1)^{1/2} - a + (b+2)^{1/2} \cdot 2 - b + (c+3)^{1/2} \cdot 3 - c = \frac{19}{2}.$
Let $a,b,c$ be real number such that $$(a+1)^{1/2} - a + (b+2)^{1/2} \cdot 2 - b + (c+3)^{1/2} \cdot 3 - c = \frac{19}{2}.$$
Find $a + b + c$.
The answer is: $-\frac{5}{2}$. Please give me some clues o... | The function $(a+1)^{1/2}-a$ has derivative $\frac{1}{2(a+1)^{1/2}}-1$, so reaches a max of $5/4$ at $a=-3/4$.
A similar calculation shows that $2(b+2)^{1/2}-b$ reaches a max of $3$ at $b=-1$, and the function $3(c+3)^{1/2}$ reaches a max of $21/4$ at $c=-3/4$.
Add up. The sum is $19/2$. What a coincidence! Thus $a+b+... | {
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"url": "https://math.stackexchange.com/questions/1857129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int \frac{\sqrt{64x^2-256}}{x}\,dx$ QUESTION
Evaluate $$\int \frac{\sqrt{64x^2-256}}{x}\,dx$$
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
MY ATTEMPT
*
*Typed
$\newcommand{\dd}{\... | $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Show that $\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}n+\mathcal{O}\left(\frac{1}{n^{3/2}}\right)$
How can i prove that $$\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}{n} +\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\tag{$*$}$$
using the following method :
note that :
$(1+x)^{\alpha}=1+\alpha x+\mathcal{O}(x^... | I would initially ignore the
$(-1)^n$
and do
$\begin{array}\\
\frac{1}{n+(-1)^n\sqrt{n+1}}-\frac{1}{n}
&=\frac{n-(n+(-1)^n\sqrt{n+1})}{n(n+(-1)^n\sqrt{n+1})}\\
&=\frac{-(-1)^n\sqrt{n+1}}{n(n+(-1)^n\sqrt{n+1})}\\
&=\frac1{n^{3/2}}\frac{(-1)^{n+1}\sqrt{1+1/n}}{1+(-1)^n\sqrt{1/n+1/n^2})}\\
\text{so}\\
\big|\frac{1}{n+(-1)... | {
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"url": "https://math.stackexchange.com/questions/1858316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\tfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\tfrac{1}{n^{3/2}} \right)$
I would like to show that :
$$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$
by starting from the left side and... | Simply using series expansion will do:
\begin{align*}
\sqrt{n+1}-\sqrt{n} &= \sqrt{n} \left( \sqrt{1+\frac{1}{n}}-1 \right) \\
&=\sqrt{n} \left( 1+\frac{1}{2n}-\frac{1}{8n^2}+\ldots-1 \right) \\
&= \frac{1}{2\sqrt{n}}-\frac{1}{8n^{3/2}}+\ldots
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/1859125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find all even numbers that can be represented as a difference of squares in only two ways I am currently working on this proof. I am looking to find (with proof) all even numbers that can be represented as a difference of squares in only two ways.
My thoughts thus far.
I examined the first 40 natural numbers and found ... |
My claim is that all numbers $x$ for which $x=8k+16$ will have exactly two representations.
Counterexample:
\begin{align}
120 &= 31^2-29^2 \\
&= 17^2 - 13^2 \\
&= 13^2-7^2 \\
&= 11^2-1^2
\end{align}
Note that while I just picked a number divisible by $8$ with plenty of factors, the claim actually falls at $48$:
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 5
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Linear combination issue I have 4 vectors:
$u_1 = \begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix} $, $\; u_2 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{pmatrix} $, $\; u_3 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{p... | Note that $(4,-2,1)$ by itself cannot be the right answer since you need four coefficients to give a linear combination of your four vectors.
Since your reduced matrix has a non-leading (non-pivot) column, the system will have infinitely many solutions. You could take the third coefficient to be zero, giving the solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiate and simplify. $m(x) = \frac{x}{\sqrt{4x-3}}$ My work so far is:
\begin{align}
m'(x) &= \frac{(1)(\sqrt{4x-3})-(x)(1/2)(4x-3)^{-1/2}(4)}{(\sqrt{4x-3})^2} \\
&= \frac{\sqrt{4x-3} - 2x(4x-3)^{1/2}}{4x-3}
\end{align}
and now I'm stuck on how to simplify further
| In the right hand side of your second line, you wrote
$(4x-3)^{1/2}$
Note the exponent here is incorrect: it should still be $(4x-3)^{-1/2}$. Let's write it that way and try to proceed.
$$m'(x)=\frac{\sqrt{4x-3}-2x(4x-3)^{-1/2}}{4x-3}=\frac{\sqrt{4x-3}-\frac{2x}{\sqrt{4x-3}}}{4x-3}$$
Now, in the numerator, we can si... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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In AB + BC + AC = N, how can I find all possibilities for A, B and C in less than n³ computational time? The problem is the one on the title. Given a N, find all possibilies for A, B and C that make this true: $AB+BC+AC = N$when $A \ge B \ge C$.
This code in C do the job:
int ans = 0;
for(int a=1; a<=N; a++){
... | $$AB+AC+BC=N\tag{1}$$
is equivalent to:
$$(2A+B+C)^2-4A^2-(C-B)^2 = 4N\tag{2}$$
hence we may simply look for the integers $q\geq\sqrt{4N}$ that ensure that $q^2-4N$ is the sum of two squares, i.e. is not divided with odd multiplicity by a prime $p\equiv 3\pmod{4}$. For instance, let $N=10$. If we take $q=7$ we have $q^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem 1, Ch. 6 in Piskunov's, Differential and Integral calculus
Find the curvature of the curve at indicated points
$b^2x^2+a^2y^2=a^2b^2$ at $(0,b)$ and $(a,0)$
My attempt
$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$
Differentiating the implicit equ... | Use chain rule for double differentiation.
I think you should multiply by dy/dx in the expression for y''.
For example,let y=1/x
So y"=2/x^3
But the way you have done would give y"=-2y=-2/x
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Maximum Likelihood Estimate of a a discrete r.d - I spent more than 4 hours on this questions, help!! Suppose $X$ is a discrete r.d with the following p.d.f:
$$
\begin{array}{c|lc}
X & \text{0} & \text{1} & \text{2} & \text{3} \\
\hline
p(x) & 2\theta/2 & \theta/3 & 2(1-\theta)/3 & (1-\theta)/3
\end{array}
$$
where $0\... | Your $p(x)$ is wrong: its sum is $4/3$, not $1$. And who would write $2\theta/2$ instead of $\theta$? I suspect that $2\theta/2$ should be $2\theta/3$ and $2(1-\theta)/2$ should be $2(1-\theta)/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help solving this recurrence relation I wanted to resolve the determinant of the next (nxn) matrix via recurrence relations:
$$
\begin{vmatrix}
a & 1 & 0 & 0 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0\\
1 & a & 1 & 0 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0 \\
0 & 1 & a & 1 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & a & 1 & 0 &.... ... | You plug your initial conditions in, so you get
$$D_1=a = C_1*(\frac{a}{2} + \frac{\sqrt{a^2-4}}{2}) + C_2*(\frac{a}{2} - \frac{\sqrt{a^2-4}}{2})\\D_2=a^2-1 = C_1*(\frac{a}{2} + \frac{\sqrt{a^2-4}}{2})^2 + C_2*(\frac{a}{2} - \frac{\sqrt{a^2-4}}{2})^2$$
These are two equations in the two unknowns $C_1,C_2$. The usual s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\prod\limits_{i=1}^{n}(x-4i+2)(x-4i+1)>\prod\limits_{i=1}^{n}(x-4i+3)(x-4i)$ for all $x\in\mathbb{R}$? I would like to prove that for $n\in\mathbb{N}$ we have $f_n(x):=\prod\limits_{r=1}^{n}(x-4r+2)(x-4r+1)>\prod\limits_{r=1}^{n}(x-4r+3)(x-4r)=:g_n(x)$ for all $x\in\mathbb{R}$ (actually it would suffice f... | Let $a_r(x)=(x-(4r-2))(x-(4r-3)), b_r(x)=(x-(4r-1))(x-4r),
c_r(x)=\frac{a_r(x)}{b_r(x)}$,
and $A_n(x)=\prod_{r=1}^{n}a_r(x),B_n(x)=\prod_{r=1}^{n}b_r(x)$.
We wish to show that $A_n(x)>B_n(x)$ for any $n$ and $x$.
Note that when $n=1$, the result follows from $A_1-B_1=2$. Suppose now that
$n\geq 2$. For $k\in[|1,n|]$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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I was trying to find out the intervals where $\sin ^{-1}x > \cos ^{-1}x$ I was trying to find out the intervals where $\sin ^{-1}x > \cos ^{-1}x$
The easiest way was to just look at the graph and I found out that the region is $x \in ({1\over \sqrt{2}} , 1]$
But I tried to prove the statement algebraically also but cou... | About your method you have to note that, unfortunately, $\arccos x=\arcsin\sqrt{1-x^2}$ is a false identity in general; it's true only for $0\le x\le 1$.
However, for $-1\le x<0$ we have $-\pi/2\le\arcsin x<0$ and $\pi/2<\arccos x\le\pi$, so we can exclude this interval, where the inequality certainly doesn't hold.
So ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Fourier transform property(uniformly converges) proof Suppose that f is a 2π-periodic function that satisfies the estimate
\begin{equation}
|f(x)-f(y)|\leqslant M|x-y|^\alpha
\end{equation}
for an 0< $\alpha$ <1
Show that $S_n(x)$ converges uniformly to f (x) for all real x.
\begin{equation}
S_n(x)=\int_0^{2\pi} f(y)\f... | $$
f(x)-S_n(x) = \frac{1}{2\pi}\int_{x-\pi}^{x+\pi}\left[\frac{f(x)-f(y)}{\sin{\frac{1}{2}(x-y)}}\right]\sin((n+1/2)(x-y))dy
$$
Because $|f(x)-f(y)| \le M|x-y|^{\alpha}$, the expression enclosed by square brackets is absolutely integrable. So $S_n(x)\rightarrow f(x)$ for all $x$ by the Riemann-Lebesgue Lemma. The u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why are the limits different? Consider $\lim_{ x \to -\infty} \sqrt{x^2-x+1}+x$
Rationalising, one will get, $\lim_{x \to -\infty} \frac{1-x}{\sqrt{x^2-x+1}-x}$, which after taking x common and cancelling out gives $-\infty$.
Now, replace $x$ by $-x$, so the limit becomes, $\lim_{x \to \infty} \frac{1+x}{\sqrt{x^2+x+1... | \begin{align*}
\lim_{x \to -\infty} \left(\sqrt{x^2 - x + 1} + x\right) & = \lim_{x \to -\infty} \left(\sqrt{x^2 - x + 1} + x\right) \cdot \frac{\sqrt{x^2 - x + 1} - x}{\sqrt{x^2 - x + 1} - x}\\
& = \lim_{x \to -\infty} \frac{x^2 - x + 1 - x^2}{\sqrt{x^2 - x + 1} - x}\\
& = \lim_{x \to -\infty} \frac{-x + 1}{\sqrt{x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solution of Inequality $ \frac{1}{x-6}\le 3$
Solve the inequality: $\displaystyle \frac{1}{x-6}\le 3$
solution:
\begin{align*}\frac{1}{x-6}& \le 3 \\ x-6& \le \frac{1}{3}
\\x& \le 6+\frac{1}{3}\\
x&\le19/3\end{align*}
but, for values of $x\le 6$ also, inequality holds true as left hand side provides some negative val... | Find the critical values: The values where the domain is undefined and when the function $f(x)= \frac{1}{x-6} -3$ is equal to $0$.
Here, the critical values are $6, \frac{19}{3} $
Use these values to divide the $x-axis$ into intervals: $(-\infty, 6), (6, \frac{19}{3}), (\frac{19}{3}, \infty) $
Test each intervals by p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Considering the complex number $z = m+i$ for which values of $m$ do we have $ \left|\overline{z}+\frac{2}{z}\right| \ge 1 $ Good evening to everyone. I have the following problem that I tried to solve but my mathematical instinct tells me that I didn't solve it right:
Considering the complex number $z = m+i$ for which ... | you have mistakes ,it should be $$\left| \overline { z } +\frac { 2 }{ z } \right| \ge 1\rightarrow \left| m-i+\frac { 2 }{ m+i } \right| \ge 1\rightarrow \left| \frac { m^{ 2 }+3 }{ m+i } \right| \ge 1\rightarrow \frac { \left| { m }^{ 2 }+3 \right| }{ \sqrt { { m }^{ 2 }+1 } } \ge 1\\ { m }^{ 2 }+3\ge \sqrt { { ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Why does this $6\times 6$ matrix has a null determinant? about this matrix:
$$
\left(
\begin{array}{ c c c c c c}
1& 0& 0& \mathit{1}& 0& 0\\
0.5& 0.5& 0& 0.5& \mathit{0.5}& 0 \\
0.5& 0& 0.5& 0.5& 0& \mathit{0.5} \\
\mathit{0.25}& 0.5& 0.25& 0& 1& 0 \\
0 & \mathit{0.5}& 0.5& 0& 0.5& 0.5 \\ ... | Looking at the first row, I notice that there are exactly two columns with an $1$ there, column 1 and column 4. Therefore my first thought was to take the difference of those:
$$\begin{pmatrix} 1\\0.5\\0.5\\0.25\\0\\0 \end{pmatrix}
- \begin{pmatrix} 1\\0.5\\0.5\\0\\0\\0 \end{pmatrix}
= \begin{pmatrix} 0\\0\\0\\0.25\\0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$ I am working on the following exercise. Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$.
I have a couple of ideas for going about this exercise.
$1)$ By moving $1$ to the other side of the equation we obtain:
$y^4-1=x^4-x^3+x^2-x \rightar... | Hint: note that for $x$ big enough the following holds:
$$(x-1)^4 < x^4-x^3+x^2-x+1 < x^4.$$
Similarly, for $x$ small enough we have
$$(x-1)^4 > x^4-x^3+x^2-x+1 > x^4.$$
Therefore you are left with a finite number of possible values of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to determine the convergence behavior of the sequence $x_{n}=\sqrt{n^{2}+11n+21}-\sqrt{n^{2}+6}$? How to determine the convergence behavior of the sequence $x_{n}=\sqrt{n^{2}+11n+21}-\sqrt{n^{2}+6}$ and its limit?
My effort: I don't know where to begin.
| Hint: multiply and divide $x_n$ by $\sqrt{n^2+11n+21}+\sqrt{n^2+6}$
$\sqrt{n^2+11n+21}-\sqrt{n^2+6}=
(\sqrt{n^2+11n+21}-\sqrt{n^2+6})({\sqrt{n^2+11n+21}+\sqrt{n^2+6}\over {\sqrt{n^2+11n+21}+\sqrt{n^2+6}}}$
$={{11n+15}\over{\sqrt{n^2+11n+21}+\sqrt{n^2+6}}}$. So the limit is $11/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proving $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit
Prove $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit.
My question is, I've picked $\delta\lt1$, and I've found that $\delta \lt \min(1,\sqrt{\epsilon})$. Was picking $1$ pro... | Here is a more "gimmicky" kind of solution (Battani's is definitely more elegant and natural): We find a positive constant $C$ such that $\left|\frac{x-1}{x+1}\right|<C\Rightarrow |x-1|\left|\frac{x-1}{x+1}\right|<C|x-1|$, and we can make $C|x-1|<\epsilon$ by taking $|x-1|<\frac{\epsilon}{C}=\delta$. We restrict $x$ to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate the following integral: $\int_0^1\frac{\log(1+x)}{1+x^2}dx$ I would like to evaluate:
$$\displaystyle\int_0^1 \frac{\log(1+x)}{1+x^2}\, dx$$
I have tried to evaluate this using integration by parts but failed. How can I evaluate it?
| The previously presented solution is perhaps the simplest. Here is my favorite.
We define a function $$f(t) = \int^1_0 \frac{\ln (xt+1)}{x^2+1} \text{ d}x.$$ The goal is to evaluate $f(1)$. We notice that $f(0)=0$. Differentiating $f$ with respect to $t$ (and differentiating under the integral; allowable by Leibniz's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Evaluation of given limit when $f(x)=\sum^{n}_{k=1} \frac{1}{\sin 2^kx}$ and $g(x)=f(x)+\frac{1}{\tan 2^nx}$
Question Statement:-
If $\displaystyle f(x)=\sum^{n}_{k=1} \frac{1}{\sin 2^kx}$ and $g(x)=f(x)+\dfrac{1}{\tan 2^nx}$, then find the value of
$$\lim_{x\to 0} \bigg( (\cos x)^{g(x)}+\bigg(\frac{1}{\cos x} \bigg)^... | The following holds:
$$g(x)=\sum^{n}_{k=1} \frac{1}{\sin 2^kx}+\dfrac{1}{\tan 2^nx},$$
$$g(x)=\sum^{n-1}_{k=1} \frac{1}{\sin 2^kx}+(\frac{1}{\sin 2^nx}+\dfrac{1}{\tan 2^nx}),$$
$$g(x)=\sum^{n-1}_{k=1} \frac{1}{\sin 2^kx}+\dfrac{1}{\tan 2^{n-1}x},$$
$$g(x)=\sum^{n-2}_{k=1} \frac{1}{\sin 2^kx}+(\dfrac{1}{\sin{2^{n-1}x}}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Find all rational solutions to $x^2+y^2=2$
Find all rational solutions to $x^2+y^2=2$
I rewrote the equation to $x= \sqrt{2-y^2}$ and thought that $x$ is rational if and only if $2-y^2$ is a square. So the only solution to the first problem is $x=1$ or $-1$ and $y=1$ or $-1$.
Is there another approach? Thank you.
Ed... | Given any rational $$ u^2 + v^2 = 1,$$ we find
$$ (u-v)^2 + (u+v)^2 = 2 $$
Meanwhile, given any integer Pythagorean triple $$ a^2 + b^2 = c^2, $$ we get rational
$$ \left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 = 1 $$
From the comments above, I guess I should add that all primitive solutions in intege... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove inequality $\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$
Let $x,y,z>0$ and $x+y+z=1$. Prove that
$$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$$
My work so far:
I use Titu's Lemma:
$$\frac{x^3}y+\frac{y^3}z+\... | $$S=\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x=\frac{x^4+y^4}{xy}+\frac{y^4+z^4}{yz}+\frac{z^4+x^4}{zx},$$
$$S=\frac{(x^2+y^2)^2-2x^2y^2}{xy}+\frac{(y^2+z^2)^2-2y^2z^2}{yz}+\frac{(z^2+x^2)^2-2z^2x^2}{zx}.$$
Since $(a^2+b^2)^2=(a^2+b^2)(a^2+b^2)\geq 2ab(a^2+b^2)$ by AM-GM, we have
$$S\geq 4(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Computing $\lim_{n \to \infty} \left(\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\cdots+\frac{n}{(n+n)^2}\right)$
Calculate $$\lim_{n \to \infty} \left(\dfrac{n}{(n+1)^2}+\dfrac{n}{(n+2)^2}+\cdots+\dfrac{n}{(n+n)^2}\right).$$
I tried turning this into a Riemann sum, but didn't see how since we get $\dfrac{1}{n} \cdot \dfrac{... | Just another way to do it, probably just for your curiosity.
$$\sum_{i=1}^n \frac 1{(n+i)^2}=\psi ^{(1)}(n+1)-\psi ^{(1)}(2 n+1)$$ where appears the first derivative of the digamma function.
Since, for large values of $x$ (see the Wikipedia page)
$$\psi^{(0)}(x) = \log(x) -\frac{1}{2 x}-\frac{1}{12
x^2}+O\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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The image of a circle under $ z \mapsto \frac{z- \frac{1}{2}}{\frac{1}{2}z-1} $ Let $C_1 = \{ z: |z| = \frac{1}{7} \}$ be a circle inside the unit circle $C_0 = \{ z: |z|=1\}$. The fractional linear transformation:
$$ \phi: z \mapsto \frac{z- \frac{1}{2}}{\frac{1}{2}z-1} $$
is a map from the unit cirle to itself, $\p... | Since at this point I am solving my own question, let me note that:
$$ \phi: \frac{1}{7} \mapsto \frac{ \frac{1}{7}- \frac{1}{2}}{\frac{1}{2}\times\frac{1}{7}-1}, \hspace{0.25in}
-\frac{1}{7} \mapsto \frac{ -\frac{1}{7}- \frac{1}{2}}{-\frac{1}{2}\times\frac{1}{7}-1}, \hspace{0.25in}
\frac{i}{7} \mapsto \frac{ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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another inequality $\frac{1}{xy+z}+\frac{1}{yz+x}+\frac{1}{zx+y}\le\frac{1}{2}$
Let $x,y,z>0$ and such $xyz\ge 2+x+y+z$, show that
$$\dfrac{1}{xy+z}+\dfrac{1}{yz+x}+\dfrac{1}{zx+y}\le\dfrac{1}{2}\tag{1}$$
I have posted this problem. Unfortunately, it doesn't hold, but I guess $(1)$ may hold.
The theory basis of spe... | I agree with you. Your inequality is true!
Indeed, the condition gives $\sum\limits_{cyc}\frac{1}{x+1}\leq1$.
Let $x=a$, $y=b$ and $z=kc$, where $k>0$ and $\sum\limits_{cyc}\frac{1}{a+1}=1$. Hence,
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{kc+1}$$
which gives $k\geq1$.
Thus, $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$
Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$
I tried assuming it was equal to $\frac p q $.
I get $2^nq^2-q^2 = p^2 $
But I don't see where to go from there.
| To continue your idea, rather than to restart and do one of the other correct answers.
$2^nq^2 - q^2 = p^2$.
Assume $p = 1$
The $2^nq^2 - q^2 = 1$. Then $2^n - 1 = 1/q^2$ is an integer. So $q=1$. So $2^n - 1 = 1$. So $2^n = 2$ so $n = 1 \not > 1$.
Assume $p \ne 1$.
Let $k$ be a prime factor of $p$. Then either $k|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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$\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square
Prove that there exist infinitely many positive integers $n$ such that $\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square. Obviously, $1$ is the least integer having this property. Find the next two least integers with this property.
The given condition is equivalen... | This can be solved by using the quadratic formula on $2n^2 + 3n + (1-6p^2)=0$ to convert to the Pell's equation $r^2 - 48p^2 = 1$. Standard techniques can be used to generate infinitely many solutions, though we need to be careful to keep only those with $r \equiv 3\pmod{4}$.
For example, we can arrive at infinitely m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Solving the Integral $\int_0^{\infty} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) \, \mathrm{cos}(b \, x) \,\mathrm{d}x$ Is there any possibily to solve the following integral
$$\int_0^{\infty} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) \, \mathrm{cos}(b \, x) \,\mathrm{d}x$$
with $a>0$, ... | First, we simplify the parameters:
$$\int_0^{\infty} \, \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}} \, \frac{\mathrm{sin}(b \, x)}{b} \,\mathrm{d}x= \frac{a}{by} \int_0^\infty \frac{t \sin rt ~dt}{(1+t^2) \sqrt{c^2+t^2}}$$
$$c^2=1+\frac{a^2}{y^2}, \qquad r=b y$$
I assume $b$ is real for this calculation (which may no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What is the smallest value of $x$ such that the matrix $A$ is not invertible?
$$A = \begin{pmatrix}
-1 & -3 & x \\
-2 & -7 & -1 \\
x & -6 & 1
\end{pmatrix}$$
So far I tried taking the determinant by expanding down the last column but i still cant get the right answer.
x * det $ \begin{pmatrix}
-2 & -7 \\x & -6
\en... | You're on the right way continue by finding the roots
$x_1=-\frac{15}{14} - \frac{\sqrt{29}}{14}$
$x_2=-\frac{15}{14} + \frac{\sqrt{29}}{14}$
And taking the $\min\{x_1,x_2\} = -\frac{15}{14} - \frac{\sqrt{29}}{14}$
That's it that is your answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given $\tan a = \frac{1}{7}$ and $\sin b = \frac{1}{\sqrt{10}}$, show $a+2b = \frac{\pi}{4}$.
Given
$$\tan a = \frac{1}{7} \qquad\text{and}\qquad \sin b = \frac{1}{\sqrt{10}}$$
$$a,b \in (0,\frac{\pi}{2})$$
Show that $$a+2b=\frac{\pi}{4}$$
Does exist any faster method of proving that, other than expanding $\sin... | $\sin b = \dfrac{1}{\sqrt{10}}$, $b\in(0,\pi/2)$ $\;\Rightarrow\;$ $b\in(0,\pi/6)$, i.e. $2b\in (0,\pi/3)$, since $1/\sqrt{10}<1/2$, and
$$\sin 2b = 2\sin b \cos b = 2 \dfrac{1}{\sqrt{10}} \dfrac{3}{\sqrt{10}} = \dfrac{3}{5}.$$
Then
$$
\tan 2b = \dfrac{\sin 2b}{\cos 2b} = \dfrac{3/5}{4/5} = \dfrac{3}{4}.
$$
Now use for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$ Could it be possible to find the solution for the following series?
$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Thanks in advance!
| A slight variant that requires no explicit cancelling of $1-b$ (so you needn't handle the $b=1$ case with a continuity argument) rearranges the double summation, viz. $$\sum_{n=0}^\infty\left(a^n\sum_{k=0}^n b^k\right)=\sum_{k=0}^\infty b^k\sum_{n=k}^\infty a^n=\sum_{k=0}^\infty b^ka^k\frac{1}{1-a}=\frac{1}{\left( 1-a\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Box contains 2 different coins, one is chosen randomly and tossed n times, head came all n times. Probability that n+1 toss is a head too? A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). One coin is choose at random and tossed $n$ times.
If the first n coin tosses result in heads, What ... | First we find the probability that the coin is fair.
To do this we use Bayes theorem, we want $P(A|B)$, where $A$ is the probability the coin is fair and $B$ is the probability the first $n$ flips are heads.
By Bayes theorem:
$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$, the numerator is $\frac{1}{2^{n}}\times \frac{1}{2}$ and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$2^n > n^4$ proof by induction This is what I came up with so far:
Inductive step: assume $2^n > n^4$.
Need to prove $2^{n+1} > (n+1)^4$
$$
2^{n+1} = 2 \cdot 2^n > 2 \cdot n^4\\
(2 \cdot n^4)^{1/4} = (2)^{1/4} \cdot n > n+1 \implies 2n^4 > (n+1)^4 \implies 2^n > (n+1)^4
$$
Is there a better way to solve this problem?
| Here is a nice little non-induction proof that $n^4\lt2^n$ for $n\gt16$.
Note that
$$n^4=\left(\sqrt n\over2\right)^8\cdot2^8\cdot1^{n-16}$$
So if $n\gt16$, the arithmetic-geometric mean inequality tells us
$$\sqrt[n]{n^4}\lt{8\cdot\left(\sqrt n\over2\right)+8\cdot2+(n-16)\cdot1\over n}={4\over\sqrt n}+1\lt2$$
hence $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Matrix Inverse and Adjoint Method I have the following matrix:
$A = \begin{bmatrix}2 & 1 & 1\\ 1 & 2 & 1\\1 & 1 & 2 \end{bmatrix}$
If I use the Excel MINVERSE function it returns:
$A^{-1} = \begin {bmatrix}0.75 & -0.25 & -0.25\\ -0.25 & 0.75 & -0.25 \\ -0.25 & -0.25 & 0.75 \end{bmatrix}$
However if I calculate the mino... | Determinant:
$det(A) = (2\cdot2\cdot 2 + 1\cdot1\cdot1 + 1\cdot1\cdot 1) - (1\cdot2\cdot1 + 1\cdot1\cdot2 + 2\cdot1\cdot1) = (8+1+1)-(2+2+2) = 10-6 = 4$
Cofactor Matrix:
$C_{11} = C_{22} = C_{33} = (-1)^{1+1}\,(2\cdot2 - 1\cdot1) = 3$
$C_{12} = C_{21} = C_{32} = C_{23} = (-1)^{1+2}\,(2\cdot1 - 1\cdot1) = -1$
$C_{13} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $7$ Divides $\binom{2^n}{2}+1$ then $n =3k+2$ for some positive integer $k$ It is a straightforward question.
If $$ 7 \text{ }\Bigg | \binom{2^n}{2}+1$$ then $n=3k+2$ for some
positive integer $k$.
This is just curiosity no motivation just rummaging through some old question in a notebook. A simple counter exam... | $7|\binom{2^n}{2}+1\iff \binom{2^n}{2}\equiv -1 \bmod 7\iff 2^n(2^{n}-1)2^{-1}\equiv -1 \bmod 7\iff $
$2^n(2^n-1)\equiv -2\equiv 5\bmod 7$.
Now notice that $2^n\bmod 7$ repeats with period $3$, so $2^n(2^n-1)$ also repeats with period three.
In particular it goes: $2,5,0,2,5,0\dots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
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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.