Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$ How to prove that
$$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})}
{512}$$
This problem was proposed by @... | Using W.A.as well as my previous calculations on harmonics sums.
I find $$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad -\frac{\pi^2G}{4}-2G+\frac{7}{4}\zeta(3)-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi}{2}-\frac{\pi^2}{4}+\frac{\pi^3}{16}+\frac{\pi^4}{64}+\ln2-2-3\beta(4)+\frac{\Psi^{(3)}(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Find all real values of the parameter a for which the equation $x^4+2ax^3+x^2+2ax+1=0$ has Find all real values of the parameter a for which the equation $x^4+2ax^3+x^2+2ax+1=0$ has
1) exactly two distinct negative roots
2) at least two distinct negative roots
I tried to factorize it but didn't get any breakthrough.
| Hint:
This is a reciprocal equation, so set $y=x+\dfrac1x$. Dividing the equation by $x^2$, the equation can be rewritten as
$$x^2+2ax+1+\frac{2a}x+\frac1{x^2}=x^2+\frac1{x^2}+2a\Bigl(x+\frac1x\Bigl)+1=y^2+2ay-1=0.$$
Now, as the reduced discriminant is $\Delta'=a^2+1>0$, this equation in $x$ has two roots with opposite... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that the inequality $\frac{n^3}{3} < 3n-3$ applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold I have been given the following task:
Prove that the following inequality applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold:
$$\frac{n^3}{3} < 3n-3$$
M... | Consider the opposite inequality:$$3n-3\leqslant\frac{n^3}3.$$As you proved, it holds for $n=3$. Now, assume that it holds for some $n\in\mathbb N\setminus\{1,2\}$. Then\begin{align}3(n+1)-3&=3n\\&=3n-3+3\\&\leqslant\frac{n^3}3+3\\&=\frac{n^3+9}3\\&\leqslant\frac{n^3+3n^2+3n+1}3\text{ (because $n>2$)}\\&=\frac{(n+1)^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.
I struggle to even start this question.
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ ... | Let $f(x)=\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}$.
As $g(x)=x^3$ is increasing, the inverse function $h(x)=x^{1/3}$ is also increasing, so $f(x)$, the sum of 3 increasing functions, is also increasing. By the way, $f(x)$ is also continuous.
Inspection shows that $x=-3$ is a solution. As the function $f(x)$ is both... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Probability of euclidean distance between two random points inside a unit circle/sphere greater than 1 Problem: Say there are two points inside the circle; A and B, and they are both randomly drawn according to a uniform distribution where the boundary is the circumference of the unit circle/the surface of the unit sph... | This is the solution for the 2D problem
Fix $A$ to be a distance $r$ from the center of the circle (i.e. fix $|A| = r$). Note that the probability density function of $|A|$ is
$$f_{|A|}(r) = \frac{2\pi r}{\pi} = 2r \qquad r \in [0,1]$$
The $2\pi r$ is the "area" of the circle representing all possible $A$ for which $|A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Scalar potential in origo Find a scalar potential φ for the vector field F such that $\varphi(0,0)=5$ when
$$\mathbf{F}(x,y) = {3\cdot x\cdot \left(2\cdot y+1\right)}\mathbf{i}+{3\cdot x^2}\mathbf{j}$$
First we need to check that vector field is conservative so
$$\frac{\delta F_1}{\delta y}=\frac{\delta F_2}{\delta x}$... | We have that
$$f_x ={3\cdot x\cdot \left(2\cdot y+1\right)} \implies f= 3\cdot x^2 \cdot y+\frac{3\cdot x^2}{2}+g(y)$$
$$f_y=3\cdot x^2+g'(y)=3\cdot x^2\implies g(y)=k$$
and therefore
$$f(x,y)= 3\cdot x^2 \cdot y+\frac{3\cdot x^2}{2}+k$$
but in any case $k=5$ for the given condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for all even positive integers $n = x^2 + 3y^2$ $(x, y \in \mathbb Z)$, $\dfrac{n}{4} = x'^2 + 3y'^2$ $(x, y \in \mathbb Z)$.
Prove that for all even positive integers $n$ that can be written in the form of $x^2 + 3y^2$ $(x, y \in \mathbb Z)$, $\dfrac{n}{4}$ can be also written in the form of $x'^2 + 3y'^2$... | The basic method you have used is correct but complications have been introduced by using $n$ for two different variables (see line 2 of your answer). A simple answer is as follows:-
If $n$ is even then, as in your proof, we can suppose $x$ and $y$ are both odd. Then $4$ is a factor of either $x+y$ or $x-y$ and, by ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$?
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.
My attempt
Obviously, we want to reach a stat... | Write
$$
\frac{3n-2}{3n+1} = 1-\frac{3}{3n+1}
$$
Recall that
$$
\left(1-\frac{3}{3n+1}\right)^{3n+1} \to e^{-3}
$$
Then
$$
\begin{align}
\left(1-\frac{3}{3n+1}\right)^{3n+1}
= &\left(1-\frac{3}{3n+1}\right)^{3n} \left(1-\frac{3}{3n+1}\right)
\\ \implies&
\left(1-\frac{3}{3n+1}\right)^{3n} \to e^{-3}
\\ \implies&
\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine all $P(x) \in \mathbb R[x]$ such that $P(x^2) + x\big(mP(x) + nP(-x)\big) = \big(P(x)\big)^2 + (m - n)x^2$, $\forall x \in \mathbb R$.
Determine all polynomials $P(x) \in \mathbb R[x]$ knowing that $$ P(x^2) + x\big(mP(x) + nP(-x)\big) = \big(P(x)\big)^2 + (m - n)x^2, \forall x \in \mathbb R\,.$$
Replacing ... | We conjecture that for odd degree $s\geq 3$ and above, the only polynomials that work are
$$P(x)=x^s+\frac{m-n}{2}x\text{ for }(m - n) (-2 + m - n)=0$$
while for even degree $s>3$ the only polynomials that work are
$$P(x)=x^s+\frac{m+n}{2}x\text{ for }(m - 3 n) (-2 + m + n)=0$$
In order to prove this, let
$$P(x)=\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$ $$\lim_{x\to 0} \frac{x\sin(3x)}{1-\cos(6x)}$$
I tried the following but it doesn't seem to work...
$$= \lim_{x\to 0} \frac{x}{2} \cdot \frac{\sin(3x)}{3x}\cdot\frac{6x}{1-\cos(6x)}$$
$$= 0$$
But the result of this limit is $\frac{1}{6}$.
Am I missing something or did I make... | More generally:
All you need is
$\dfrac{\sin(x)}{x} \to 1$
and
$1-\cos(x)
=2\sin^2(x/2)
$.
$\begin{array}\\
\dfrac{x\sin(ax)}{1-\cos(bx)}
&=ax^2\dfrac{\sin(ax)}{ax}\dfrac{1}{2\sin^2(bx/2)}\\
&=\dfrac{ax^2}{2(b/2)^2}\dfrac{\sin(ax)}{ax}\dfrac{(b/2)^2}{\sin^2(bx/2)}\\
&=\dfrac{a}{b^2/2}\dfrac{\sin(ax)}{ax}\dfrac{(bx/2)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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How can I solve the following inequality? I have the inequality:
$lg((x^3-x-1)^2) < 2 lg(x^3+x-1)$
And I'm not sure how should I go about solving it. I wrote it like this:
$2lg(x^3-x-1) < 2lg(x^3+x-1)$
$lg(x^3-x-1) < lg(x^3 + x - 1)$ (*)
Here I have the conditions:
$x^3-x-1 > 0$
$x^3+x-1 > 0$
At first this stumpe... | Better not get rid of the squares -- they're your friend. So we have $$\log{(x^3-x-1)^2}< \log{(x^3+x-1)^2}.$$ Since the logarithm is monotonic, this implies $$(x^3-x-1)^2< (x^3+x-1)^2,$$ or that $$(x^3-x-1)^2- (x^3+x-1)^2<0.$$ This is easily factored to give $$(x^3-x-1+x^3+x-1)(x^3-x-1-x^3-x+1)<0,$$ or $$(2x^3-2)(-2x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
In the range $0\leq x \lt 2\pi$ the equation has how many solutions $\sin^8 {x}+\cos^6 {x}=1$ In the range $0\leq x \lt 2\pi$ the equation has how many solutions
$$\sin^8 {x}+\cos^6 {x}=1$$
What i did
$\cos^6 {x}=1-\sin^8 {x}=(1-\sin^4 {x})(1+\sin^4 {x})=(1-\sin^2 {x})(1+\sin^2 {x})(1+\sin^4 {x})$
$\cos^4 {x}=(1+\sin^2... | We don't need any factorization, indeed since for $x\neq k\frac \pi 2$ we have that $0<|\sin x|<1$ and $0<|\cos x|<1$ then
$$\sin^8 {x}+\cos^6 {x}<\sin^2 {x}+\cos^2 {x}=1$$
we need only to check for the solutions $x= k\frac \pi 2$ which indeed are all the solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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For triangle sides $a,b,c\in\Bbb N$, find semiperimeter $s$ if $\frac1{36}(s-a)^2+\frac19(s-b)^2+\frac14(s-c)^2=\frac1{49}s^2$ and $HCF(a,b,c)=1$
Let $a,b,c$ be the lengths of sides opposite to vertex $A$, $B$, and $C$ respectively in triangle $ABC$ . If
$$\frac{(s-a)^2}{36}+\frac{(s-b)^2}{9}+\frac{(s-c)^2}{4}=\frac... | We have the case of the equality in the inequality of Cauchy-Schwarz:
$$
s^2
=\left(
\frac{(s-a)^2}{6^2}+\frac{(s-b)^2}{3^2}+\frac{(s-c)^2}{2^2}
\right)
\left(6^2+3^2+2^2\right)
\ge
\Big(\ (s-a)+(s-b)+(s-c)\ \Big)^2
=s^2\ .
$$
So the two vectors $\Big(\ \frac 16(s-a),\ \frac 13(s-b),\ \frac 12(s-c)\ \Big)$ and $(6,3,2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(1+a^2)(1+b^2)\geq 4ab$ for all $a,b>0$ without limits I'm facing a problem where I need to prove that $(1+a^2)(1+b^2)\geq 4ab$ for all $a,b>0$ without any use of limits.
I have tried 2 different methods that I'm not really sure about:
1) Taking out factors $a$ and $b$ so I get: $$ab(a+\frac{1}{a})(b+\frac... | Writing
$$1+a^2+b^2+a^2b^2\geq 4ab$$ and this is
$$a^2+b^2-2ab+1+a^2b^2-2ab\geq 0$$ and this is $$(a-b)^2+(ab-1)^2\geq 0$$
Or $$ab+\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}\geq 2+2=4$$
since $$x+\frac{1}{x}\geq 2$$ if $$x\geq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2... | Use $$2(x^2+y^2)^2-(x^2+2xy-y^2)^2=(x^2-2xy-y^2)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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how the following formula or any other explicit formula for computing Bernoulli numbers can be derived? how the following formula
$$B_n=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}$$
can be derived?
I know how to use the formula but still have not seen any proof for the given formula.
|
The recurrence relation
\begin{align*}
B_n&=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}\qquad n\geq 0\\
\end{align*}
specifies a variation of the Bernoulli numbers with $B_1=+\frac{1}{2}$ instead of the more common value $B_1=-\frac{1}{2}$.
We therefore start with the generating function
\begin{align*}
\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3414892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Proving an inequality using AM-GM
Let $ a,b,c > 0$ such that $ abc = 1$. Prove that
$$\frac {ab}{a^2 + b^2 + \sqrt {c}} + \frac {bc}{b^2 + c^2 + \sqrt {a}} + \frac {ca}{c^2 + a^2 + \sqrt {b}}\le 1.$$
$$\frac{ab}{a^2+b^2+\sqrt{c}}=\frac{1}{\frac{a}{b}+\frac{b}{c}+\sqrt{c^3}}$$
(AM-GM): $$\frac{a}{b}+\frac{b}{a}+\sq... | Here is an easier way to prove the inequality. Just prove:
$$\frac {ab}{a^2 + b^2 + \sqrt {c}} \leq \frac {a+b}{2a + 2b + 2{c}} $$ and you are done! It is equivalent to $$a^2b+ab^2+2\leq a^3+b^3+ a\sqrt{c}+b\sqrt{c}$$ which is easy to see since $2\leq a\sqrt{c}+b\sqrt{c}$ and $ a^2b+ab^2\leq a^3+b^3$ are true by Am-Gm... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Limits in multivariable Calculus $$\lim_{(x,y,z)\to(0,0,0)} \dfrac{(x+y+z)^2}{x^2 + y^2 + z^2}$$
According to me the answer should be $3$ but it says limit does not exist. Pls let me know where am I wrong.
The expression simplifies to $$1 + \dfrac{2\cdot(xy+yz+zx)}{x^2+y^2+z^2}$$
And for $x=y=z$ (say $x = y = z = a$) ... | You have determined the limit for the special case $x=y=z=t \to 0$ but for $x+y+z=0$, that is for example $x=t$, $y=-t-t^2$, $z=t^2$ we obtain
$$\frac{(x+y+z)^2}{x^2+y^2+z^2}=0$$
thereofore the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove $\lim_{x \to 2} \frac{ x(x^2-1) }{x+3}=6/5$. Prove by $\epsilon$-$\delta$ definition that $\displaystyle\lim_{x \to 2} \frac{ x(x^2-1) }{x+3}=\dfrac{6}{5}$.
I know this is simple but I am stuck, If we assume $\delta <1$ and $|x-2|< \delta$ we can get rid of the factor $x-2$ in the following expression $$\dfrac{... | From here
$$\left|\frac{ x(x^2-1) }{x+3}-6/5\right|=\left|\frac{(x-2)(5x^2+10x+9)}{5x+15}\right|$$
the trick is consider wlog $|x-2|<1 \iff 1<x<3$ and therefore
$$\left|\frac{(x-2)(5x^2+10x+9)}{5x+15}\right|=|x-2|\left|\frac{5x^2+10x+9}{5x+15}\right|\le \frac{84}{20}|x-2|\le5|x-2|$$
therefore it suffices to take $\delt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3419082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding maximum with two constraints Let $a$,$b$,$c$ be positive real numbers satisfying $$a+b+c=1$$ $$a^2+b^2+c^2=\frac{3}{8}$$ Find the maximum value of $$a^3+b^3+c^3$$
Using the well known $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, the problem changes to finding maximum of $abc$ but in this case AM-GM, cauchy... | Let $a+b+c=3u,$ $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $u=\frac{1}{3}$, $$2(ab+ac+bc)=(a+b+c)^2-(a^2+b^2+c^2),$$ which gives
$$v^2=\frac{5}{48}.$$
Now, $$(a-b)^2(a-c)^2(b-c)^2\geq0$$ gives
$$3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0,$$ which gives
$$3uv^2-2u^3-2\sqrt{(u^2-v^2)^3}\leq w^3\leq3uv^2-2u^3+2\sqrt{(u^2-v^2)^3},$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Intersection of the blowing up and the exceptional curve for a concrete example Let $f(x,y)=x^2y+xy^2 - x^4-y^4=0$ be an affine curve.
Consider its blowing up at the origin, namely $x^2y+xy^2 = x^4+y^4$ and $xu=ty$ in $\mathbb{A}^2 \times \mathbb{P}^1$.
If $t \neq 0$ then we can set $y=xu$ and substituting this to th... | As Lazzaro Campoetti says, you've made a mistake in your calculations. After substituting $y = xz$ and simplifying, I find the equation
$$
0 = x^3(z(1+z) - x(1+z^4)) \, .
$$
The factor $x^3$ corresponds to the exceptional locus. However, it is not a point, but a line: since $y = xz$, then $x = 0$ implies $y = 0$ and le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding the value of $c$ to make $-\dfrac{1}{2}x^2+x+c$ a perfect square trinomial
I should find the value of $c$ to make: $$-\dfrac{1}{2}x^2+x+c$$ a
perfect square trinomial.
I really messed up: $-\dfrac{1}{2}x^2+x+c=x-\dfrac{1}{2}x^2+c$, but it seems like $-\dfrac{1}{2}x^2$ isn't $a^2\boldsymbol{-2ab}+b^2$. Can ... | Extract the $-1/2$ from the equation:
$$ -\frac{1}{2}x^2 + x + c = -\frac{1}{2}( x^2 - 2x - 2c ) $$
The expression $ x^2 - 2x - 2c $ is similar to the expression $ x^2-2x + 1 = (x-1)^2 $ which gives that
$$ -2c = 1 \rightarrow c = -\frac{1}{2} $$
and the final expression is:
$$ -\frac{1}{2}(x-1)^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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determining a prime factorization What is the prime factorizatíon of
$$N:= \sum_{n=1}^{10^8 -1} n^3 $$
writing the sum out, didn't help me..
therefore..any help is really appreciated!
| In general, $\sum_{n=1}^k n^3 =\frac {k^2(k+1)^2}4$
In your case $k=10^8 -1$
The sum is $\frac {(10^8-1)^2(10^8)^2}4$
$10^8-1=99999999=3^2 \times 11 \times 73 \times 101 \times 137$
$\frac {(10^8)^2}4=\frac {10^{16}}4=\frac{2^{16} \times 5^{16}}{2^2}=2^{14} \times 5^{16}$
So $\frac {(10^8-1)^2(10^8)^2}4=2^{14} \times 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Simple continued fraction of $\sqrt{d}$ with period of shortest length $3$ This is the problem:
Does there exist positive integer $d$ ( which is not a perfect square ) such that the length of the least period in the simple continued fraction of $\sqrt{d}$ is $3$?
Consider the following theorem
Theorem : If the pos... | Yes there are infinitely many. And it is not difficult to find them.
We seek continued fractions of the form
$\sqrt{N}=[a,\overline{b,c,2a}]$
First off, add $a$ to get a "pure" periodic expression. We shall call the quadratic surd $x$:
$x=a+\sqrt{N}=[\overline{2a,b,c}]$
We may then render
$x=2a+\dfrac{1}{b+\dfrac{1}{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Solving for terms of the sequence given by $ x^{2(2^k-1)}(x-1)=\ln2$ or gain some insight I'm interested in properties of the sequence given by (int k) :
$$ x^{2(2^k-1)}(x-1)= \ln2$$
$k = 1$: $x^2(x-1) = \ln2$ or $x^3-x^2 = \ln2$ :$x \approx 1.3737...$
$k=2$: $x^6(x - 1) = \ln2$ or $x^7 - x^6 = \ln2$ : $x\app... | One can make approximations of the solution building, around $x=1$, the $[1,n]$ Padé approximant of
$$x^{2(2^k-1)}(x-1)- \log(2)$$
To fit in the page, I limited to $n=4$ and $a=\log(2)$. The approximation is
$$x=1+\frac{4 a^4 \left(2^k-2\right) \left(2^k-1\right) \left(2^{k+1}-3\right)+12 a^3
\left(-9\ 2^k+4^{k+1}+5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What's the easiest way to find if this series is monotonic? This is my sequence:
$$a_n=\frac{3^n-4^n}{2^n-n+1}$$
So I am supposed to check how it is monotonic. Thats what i tried to run using wolfram, but honestly it gets complicated easily
$$\frac{3^{n+1}-4^{n+1}}{2^{n+1}-n}-\frac{3^n-4^n}{2^n-n+1}$$
I honestly don't ... | Let consider
$$a_n=\frac{3^n-4^n}{2^n-n+1}=\frac{3^n-4^n}{4^n}\cdot \frac{4^n}{2^n-n+1}=b_n \cdot c_n$$
and
$$b_n=\frac{3^n-4^n}{4^n}=\left(\frac34\right)^n-1$$
is negative strictly decreasing while
$$c_n=\frac{4^n}{2^n-n+1}$$
is positive strictly increasing indeed
$$\frac{4^{n+1}}{2^{n+1}-n}-\frac{4^n}{2^n-n+1}=4^n\l... | {
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"url": "https://math.stackexchange.com/questions/3432980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$ Let $a$,$b$,$c$ and $d$ be non-zero, pairwise different real numbers such that $ \frac{a}{b} +\frac{b}{c} +\frac{c}{d} + \frac{d}{a}=4$ and $ac=bd$ .
Show that $\frac{a}{c} + \frac{b}{d} +\frac{c}{a} + \frac{d}{b}\le-12$ and that $-12$ is the max... | The hint.
Prove that $$\frac{a}{b}+\frac{c}{d}=\frac{d}{c}+\frac{c}{d}\leq-2$$ or
$$\frac{b}{c}+\frac{d}{a}=\frac{a}{d}+\frac{d}{a}\leq-2$$ and use
$$\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=\left(\frac{a}{b}+\frac{c}{d}\right)\left(\frac{b}{c}+\frac{d}{a}\right).$$
A full solution.
Let $\frac{a}{b}+\frac{c}{d... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Sum of the given series: $(1^2 - 1 + 1)(1!) + (2^2 - 2 + 1)(2!) + \cdots + (n^2 - n + 1)(n!) $
Find the sum of the following series:
$$(1^2 - 1 + 1)(1!) + (2^2 - 2 + 1)(2!) + \cdots + (n^2 - n + 1)(n!) $$
I tried simplifying the $n^{th}$ term to use the method of telescoping to see if most of the terms get cancell... | We have
$$\sum_{k=1}^n (k^2-k+1)k!=\sum_{k=1}^n [(k+1)^2-3k]k!=\sum_{k=1}^n (k+1)(k+1)!-\sum_{k=1}^n kk!-2\sum_{k=1}^n kk!=$$
$$=(n+1)(n+1)!-1-2\sum_{k=1}^n kk!$$
and
$$\sum_{k=1}^n kk!=\sum_{k=1}^n (k+1-1)k!=\sum_{k=1}^n (k+1)!-\sum_{k=1}^n k!=(n+1)!-1$$
therefore
$$\sum_{k=1}^n (k^2-k+1)k!=(n+1)(n+1)!-1-2((n+1)!-1)=(... | {
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"url": "https://math.stackexchange.com/questions/3436047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this process map every fraction to the golden ratio? Start with any positive fraction $\frac{a}{b}$.
First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$
Then add the (new) numerator to the denominator:
$$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$
So $\frac{2}{5} \rightarrow ... | Your numerators and denominators follow the same recursive relationship that defines the Fibonacci sequence. I.e. each time you make a new number (either a new numerator or a new denominator), the new number is equal to the sum of the two most recent previously made numbers.
Any sequence that follows this recursive rel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 5,
"answer_id": 3
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Closed form of the Fibonacci sequence: solving using the characteristic root method Here is the official theorem I'll use:
Since the Fibonacci sequence is defined as $F_n=F_{n-1}+F_{n-2}$, we solve the equation $x^2-x-1=0$ to find that $r_1 = \frac{1+\sqrt 5}{2}$ and $r_2 = \frac{1-\sqrt 5}{2}$
So we have $F_n = c_1\l... | Let's see...
$$f_n = \left\{
\begin{array}{ll}
0 & \text{ for } n = 0 \\
1 & \text{ for } n = 1 \\
f_{n-1} + f_{n-2} & \text{ for } n>1
\end{array}
\right.$$
Now, the recursion can be written as $$f_n - f_{n-1} - f_{n-2} = 0,$$ so characteristic equation is $$x^2-x-1=0.$$
Now, the roots of the equation are $$X_{1,2} ... | {
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"url": "https://math.stackexchange.com/questions/3441296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$. I have to find the limit:
$$\lim\limits_{n \to \infty} \sqrt[3]{n^3+2n^2+1}-\sqrt[3]{n^3-1}$$
I tried multiplying with the conjugate of the formula:
$$(a-b)(a^2+ab+b^2)=a^3-b^3$$
So I got:
$$\lim\limits_{n \to \infty} \dfrac{n^3+2n^2+1-n^3+1}{\sqrt... | The denominator is in the form
$$\sqrt[3]{n^6+...}+ \sqrt[3]{n^6+...}+ \sqrt[3]{n^6+...}\sim 3n^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove derangement formula by induction I found this in my math book. I have solved a). Exercise b) is to prove the derangement sum by induction.
A derangement of $n$ elements is a permutation where none of the elements keep its original placement. Let $a_n$ be the number of possible derangements of n elements.
a) Show... | It is actually straightforward. It is clear that the formula holds for 1 and 2. Assume the formula holds for $1 \leq k \leq n $ and show that it holds for $n+1$
\begin{align}
a_{n+1} &= n \cdot (a_n + a_{n-1})\\
&= n \cdot ( n! \cdot [ 1 - \frac1{1!} + \cdots + (-1)^{n} \cdot \frac1{n!}] + (n-1)! \cdot [ 1 - \frac1{1!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the $n^{th}$ derivative of $y=\dfrac {x^n}{x-1}$. Find the $n^{th}$ derivative of $y=\dfrac {x^n}{x-1}$.
My Attempt:
$$y=\dfrac {x^n}{x-1}$$
$$y=x^n\cdot(x-1)^{-1}$$
Differentiating both sides,
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdot n\cdot x^{(n-1)}$$
$$y_{1}=x^n\cdot(-1)\cdot(x-1)^{-2}+(x-1)^{-1}\cdo... | $$\frac{x^n}{x-1}=\frac{x^n-1}{x-1}+\frac1{x-1}=P(x)+\frac1{x-1}$$ where $P$ is a polynomial of degree $n-1$. The $n^{th}$ derivative of this polynomial vanishes and we are left with the $n^{th}$ derivative of $\dfrac1{x-1}$,
$$\frac{(-1)(-2)\cdots(-n)}{(x-1)^{n+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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General expression of polynomial roots sequence. Let $$P_n(X) = X^n-X^{n-1}-\cdots-1 = X^n - \sum_{k=0}^{n-1} X^k$$
The question is : Find the general expression of the sequence $(u_n)_{n \in \mathbb{N}}$ where $u_k$ is the greater root of $P_{k+1}(X)$.
We aldready know that $$u_0 = 1$$, $$u_1 = \frac{1+\sqrt{5}}{2}$$ ... | By Descartes rule of signs, $P_n(x)$ has one sign change, so it has one root on the positive half of the real line. $P_n(2) = 1$ and $P_n \left( 2-\frac{1}{n} \right) = \frac{n - \left( 2-\frac{1}{n} \right)^n}{n-1}$. Since $n-1>0$ for $n > 1$, the sign of this latter expression is controlled by the numerator. If we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $x,y$ $\begin{cases}\sin^2x+\sin^2y=\frac{3}{4}\\x+y=\frac{5\pi}{12}\end{cases}$
Designate $x,y\in\left(0,\frac{\pi}{2}\right)$ which fulfills
$\begin{cases}\sin^2x+\sin^2y=\frac{3}{4}\\x+y=\frac{5\pi}{12}\end{cases}$
My proof:
$\cos2x=1-2\sin^2x\Rightarrow \sin^2x=\frac{1}{2}-\frac{1}{2}\cos2x$
$\sin^2x+\s... | From $\sin^2x+\sin^2y=\frac{3}{4}$, we get
$$\cos 2x +\cos 2y = \frac 12$$
or
$$\cos(x+y)\cos(x-y) = \frac14 \implies
\cos \frac {5\pi}{12}\cos(x-y) =\frac12\sin\frac\pi{6}$$
which leads to,
$$\cos(x-y) = \frac{\sin\frac\pi6}{2\cos\frac{5\pi}{12}}=\frac{2\sin\frac\pi{12}\cos\frac\pi{12}}{2\sin\frac\pi{12}}=\cos\frac\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove:$\frac{1}{(x-1)^2+(y-1)^2}+\frac{1}{(x+1)^2+(y-1)^2}+\frac{1}{(x-1)^2+(y+1)^2}+\frac{1}{(x+1)^2+(y+1)^2}\geq2 $ ,if $-1
Prove:
$$\frac{1}{(x-1)^2+(y-1)^2}+\frac{1}{(x+1)^2+(y-1)^2}+\frac{1}{(x-1)^2+(y+1)^2}\\+\frac{1}{(x+1)^2+(y+1)^2}\geq2 $$
if $-1< x,y< 1$.
I tried Cauchy-Schwarz inequality and I proved Le... | We need to prove that:
$$\left(\frac{1}{(x-1)^2+(y-1)^2}+\frac{1}{(x+1)^2+(y+1)^2}\right)+$$
$$+\left(\frac{1}{(x-1)^2+(y+1)^2}+\frac{1}{(x+1)^2+(y-1)^2}\right)\geq2$$ or
$$\left(\frac{1}{x^2+y^2+2-2(x+y)}+\frac{1}{x^2+y^2+2+2(x+y)}\right)+$$
$$+\left(\frac{1}{x^2+y^2+2-2(x-y)}+\frac{1}{x^2+y^2+2+2(x-y)}\right)\geq2$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given the sequence $(a_n)_{n \ge 1}$ with $a_1=2$ and $a_{n+1} = \frac{n^2-1}{a_n} + 2$ for $n \ge 1$, find the following limits. I have the sequence $(a_n)_{n \ge 1}$, such that:
$$a_1 = 2, \hspace{1.5cm} a_{n+1} = \dfrac{n^2-1}{a_n}+2 \hspace{.25cm}, \forall n \ge 1$$
And I have to find $2$ limits:
$$\lim\limits_{n \... | An easy proof by induction that $n\le a_n\le n+1$ with inductive step$$k\le a_k\le k+1\implies k+1=\frac{k^2-1}{k+1}+2\le a_{k+1}\le\frac{k^2-1}{k}+2<k+2$$implies the first limit is $1$ and the second is $\frac14$, since $\sum_{k=1}^nk^3=\frac14n^4+o(n^4)$. In particular, $1\le\frac{a_n}{n}\le1+\frac1n$ gets the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factorize $xy(xy + 1) + (xy + 3) - 2(x + y + 1/2) - (x + y - 1)^2$ Factorize $xy(xy + 1) + (xy + 3) - 2(x + y + \frac{1}{2}) - (x + y - 1)^2$
When I substitute $x = 1$, $x = -1$, $y = 1$, $y = -1$, It gives a result of $0$. So the result will be $(x+1)(x-1)(y+1)(y-1)$ after checking.
However, it seems that the question... | $$xy(xy+1)+(xy+3)-2(x+y+1/2)-(x+y-1)^2$$
$$xy(xy+1)+(xy+1)+2-2(x+y+1/2)-(x+y-1)^2$$
$$(xy+1)^2+2-2(x+y+1/2)-(x+y-1)^2$$
$$(xy+1)^2+2-2x-2y-1-(x+y-1)^2$$
$$(xy+1)^2-2(x+y-1)-1-(x+y-1)^2$$
$$(xy+1)^2-1-(x+y-1)(x+y+1)$$
$$(xy+1)^2-1-((x+y)^2-1)$$
$$(xy+1)^2-(x+y)^2$$
$$\Big[xy+1-(x+y)\Big]\Big[xy+1+(x+y)\Big]$$
$$\Big[y(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$ Inspired by the recent question if the series $\sum_{k=1}^{\infty} \frac{\sqrt{k}-\left\lfloor \sqrt{k}\right\rfloor}{k}$ diverges (which is the case) I became interested in the alternating series which is convergent ... | Update
We may use the convergence acceleration of alternating series developed by Cohen, Villegas, and Zagier.
Let
$$s = \ln 2 + \sum_{n=1}^\infty (-1)^n n
\sum_{i=1}^{n} \frac{1}{(n^2 + 2i-1)(n^2+2i)}$$
and
$$s_n = \ln 2 + \sum_{k=1}^n \frac{c_{n,k}}{d_n}\sum_{i=1}^k \frac{k}{(k^2 + 2i-1)(k^2+2i)}$$
where
\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Show function is continuous at $(0,0)$ Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function with
$$\space f(x, y) = \begin{equation}
\begin{cases}
\dfrac{y^2\log(1+x^2y^2)}{\sqrt{x^4+y^4}}, & (x, y) \neq (0, 0)\\
0, & (x, y) = 0
\end{cases}\end{equation}$$
Show that f is continious.
I've already shown that f is ... | in the numerator $0\le y^2\log(1+x^2y^2)\le x^2y^4$
In the denominator $\sqrt{x^4+y^4} \ge \max (x^2,y^2)\ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Limit of this sequence $\lim_{n\to \infty}\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}$ I am trying to calculate the limit of this sequence :
$$\lim_{n\to \infty}\frac{ \sqrt{2n+2} - \sqrt{2n-2}}{\sqrt{3n+1} - \sqrt{3n}}$$
I tried two methods and the two methods leaded me to infinity or 4/0.
Anything woul... | Call your limit $L$. Since $\sqrt{a+b}-\sqrt{a}=\frac{b}{\sqrt{a+b}+\sqrt{a}}$, $$L=\lim_{n\to\infty}\frac{\sqrt{2n+2}-\sqrt{2n-2}}{\sqrt{3n+1}-\sqrt{3n}}=\lim_{n\to\infty}\frac{4(\sqrt{3n+1}+\sqrt{3n})}{\sqrt{2n+2}+\sqrt{2n-2}}.$$Since $\sqrt{an+b}\sim\sqrt{an}$ for large $n>0$ and $a>0$,$$L=\lim_{n\to\infty}\frac{8\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Calculation of the limit $\lim_{x\to 0} (\cos x)^{1/x^2}$ without De l'Hospital/Landau's symbols/asymptotic comparison I have calculate this limit
$$\lim_{x\to 0}\ (\cos x)^{1/x^2}$$
with these steps. I have considered that:
$$(\cos x)^{1/x^2}=(\cos x -1+1)^{1/x^2}=\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{... | So you are assuming the fact $(1+u)^{1/u}\rightarrow e$ as $u\rightarrow 0$. With such, we also have $\dfrac{1}{u}\cdot\log(1+u)\rightarrow 1$, then
\begin{align*}
\dfrac{1}{x^{2}}\cdot\log(\cos x)&=\dfrac{1}{\cos x-1}\cdot\log(1+(\cos x-1))\cdot\dfrac{\cos x-1}{x^{2}}\\
&=\dfrac{1}{\cos x-1}\cdot\log(1+(\cos x-1))\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Proving that $8^x+4^x\geq 5^x+6^x$ for $x\geq 0$. I want to prove that $$8^x+4^x\geq 6^x+5^x$$ for all $x\geq 0$. How can I do this?
My attempt:
I try by AM-GM: $$8^x+4^x\geq 2\sqrt{8^x4^x}=2(\sqrt{32})^x.$$
However, $\sqrt{32}\approx 5.5$ so I am not sure if $$2(\sqrt{32})^x\geq 5^x+6^x$$ is true.
Also, I try to compu... | The best and the easiest way to prove it is using Induction .
Base Case :
For $x=1$ , $ 8 + 4 \gt 6+5$
Induction Step :
Let us assume that $8^k +4^k \ge5^k + 6^k$ . This implies that $$8^k \ge (5^k + 6^k) - 4^k$$
Multiplying both the sides by $8$ , we get
$$\color{#f14}{8^{k+1} \ge 8\cdot5^k +8\cdot6^k -8\cdot4^k} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
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Closed form of sum $\sum\limits_{k=1}^{\infty } \frac{(-1)^{k+1}}{\left\lfloor \sqrt{k}\right\rfloor}$ In two previous problems (Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$ and Closed expression for sum $\sum_{k = 1}^{\infty} \frac{\left\lfloor \sqrt{k} \right ... | OK. Here's my generalization
which ran into some resistance
when posted as a separate problem.
Let $f(x)$ be such that
$f(1) = 1,
f'(x) > 0, f''(x) < 0,
f(x) \to \infty,
n \in \mathbb{N}
\implies f^{(-1)}(n)\in \mathbb{N}
$.
($f^{(-1)}(n)$
is the inverse function of $f$)
What can we say about
$$S=\sum_{k=1}^{\infty} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solve without L'Hopital's rule: $\lim_{x\to0}\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}$ Solve without L'Hopital's rule:
$$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}$$
My work:
$\displaystyle\lim_{x\to0}{\frac{\sqrt{\cosh{(3x^2)}}\cdot e^{4x^3}-1}{x^2\tan(2x)}}=\... | \begin{align*}
\dfrac{\sqrt{\cosh(3x^{2})}e^{4x^{3}}-1}{x^{2}\tan(2x)}&=\dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}\dfrac{2x}{\tan(2x)}\dfrac{1}{\sqrt{\cosh(3x^{2})}e^{4x^{3}}+1},
\end{align*}
so the issue is to compute
\begin{align*}
\lim_{x\rightarrow 0}\dfrac{\cosh(3x^{2})e^{8x^{3}}-1}{x^{2}(2x)}.
\end{align*}
Note... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How to find matrix $B$?
let $A$ be a $2 \times 2$ matrix and $I$ be the identity matrix . Assume that the null spaces of $A-4I$ and $A-I$ respectively spanned by $\begin{bmatrix} 3 \\ 2\end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1\end{bmatrix}$ respectively.Find a matrix $B $ such that $B^2= A$
My attempt :
Acc... | You know that $A.\left[\begin{smallmatrix}3\\2\end{smallmatrix}\right]=4\left[\begin{smallmatrix}3\\2\end{smallmatrix}\right]$ and that $A.\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]=\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]$. So, if $P=\left[\begin{smallmatrix}3&1\\2&1\end{smallmatrix}\right]$, t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Function that requires differentiation Differentiate:
$$\ln\left(\dfrac{x^2\sqrt{2x^2+3}}{\left(x^4+x^2\right)^3}\right)$$
I have tried to figure it out here:
The steps are too long so I tidy up as an image
After the steps from the image, these are the final steps of simplifying:
$$=\dfrac{\frac{2x^5\left(x^2+1\right)^... | To simplify we can use that
$$f(x)=\ln\left(\dfrac{x^2\sqrt{2x^2+3}}{\left(x^4+x^2\right)^3}\right)=\ln x^2+\ln\sqrt{2x^2+3}-\ln \left(x^4+x^2\right)^3=$$
$$=2\ln x+\frac12\ln(2x^2+3)-3\ln(x^4+x^2) $$
and therefore
$$f'(x)=2\cdot \frac1x+\frac12\cdot\frac{4x}{2x^2+3}-3\cdot\frac{4x^3+2x}{x^4+x^2}=\frac 2x+\frac{2x}{2x^... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^... | Option:
$1\le (n^2+n)^{1/n} \le (2n^2)^{1/n} =$
$2^{1/n}(n)^{1/n}(n)^{1/n}.$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
number of such real values of $a.$
Let $a$ be a real number in the interval $[0,314]$ such that
$$\displaystyle \int^{3\pi+a}_{\pi+a}|x-a-\pi|\sin \frac{x}{2}dx=-16.$$
Determine the number of such real values of $a.$
What I tried:
Put $x-a-\pi=t$. Then,
$$\displaystyle \int^{2\pi}_{0}|t|\sin \bigg(\frac{t+a+\pi}{2}\b... | You already have
$$\int^{2\pi}_{0}t\cos\bigg(\frac{t+a}{2}\bigg)dt=-16$$
From here, we have
$$2t\sin\frac{t+a}{2}\bigg|^{2\pi}_{0}-2\int_0^{2\pi}\sin\frac{t+a}{2}dt=-16$$
$$\iff -4\pi\sin\frac a2-2\int_0^{2\pi}\bigg(\sin\frac t2\cos\frac a2+\cos\frac t2\sin\frac a2\bigg)dt=-16$$
$$\iff -4\pi\sin\frac a2-2\cos\frac a2\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of $f(x) = \sum_{k=0}^\infty a_kx^k$ with $a_0 := a_1 := 1$ and $a_{n+2} := a_n+a_{n+1}$ A recursion is defined by $a_0 := a_1 := 1$ and $a_{n+2} := a_n+a_{n+1}$ for $n \geq 2$.
How can one prove that the power series
$$f(x) = \sum_{k=0}^\infty a_kx^k$$ converges absolutely for $|x| < \frac{1}{2}$ and
$$f(x) = -... | Note that $|a_n| \le 2^k$, hence ${1 \over R} = \limsup_n \sqrt[n]{|a_n|} \le 2$.
To obtain the functional form we have $a_{n+2} = a_{n+1} + a_n$, so
$\sum_{n=0}^\infty a_{n+2} x^{n+2} = \sum_{n=0}^\infty a_{n+2} x^{n+2} + \sum_{n=0}^\infty a_{n+2} x^{n+2} $ from which we get
$f(x)-1-x = x(f(x)-1) + f(x) x^2$, and from... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Without use of derivatives, prove that $e^{-x}\left(1+x+\frac{x^2}{2}\right)$ is strictly decreasing
Prove, without use of derivatives, that function $x\mapsto e^{-x}\left(1+x+\frac{x^2}{2}\right)$ is strictly decreasing.
Attempt. Functions $x\mapsto x^ne^{-x}$ for $n=0,1,2$ are not strictly decreasing (in order to... | I will assume the properties $e^x > 1 +x + \frac{x^2}{2}$ and $e^{x+y} = e^x e^y$ are given.
For $x+h > x > 0$, we have $\left(h+ \frac{h^2}{2}\right)\left(1+x+ \frac{x^2}{2}\right)> h + xh+\frac{h^2}{2}$, and, hence,
$$e^h > 1+ h + \frac{h^2}{2} > 1 + \frac{h + xh+\frac{h^2}{2}}{1+x+ \frac{x^2}{2}} = \frac{1 + (x+h... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\int_0^1 \frac{x^{p}}{x^{p+1}+(1-x)^{p+1}} dx=?$ $$\int_0^1 \frac{x^{p}}{x^{p+1}+(1-x)^{p+1}} dx=?$$
I tried to use
$$\int_0^1 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}} dx=\frac{1}{2}$$ and integration by parts. I do not know if there is any restriction on p.in original question p=2014,Question from Jalil Hajimir.
| Assume $p$ is a positive integer. Of course, you don't need it this strong.
First, observe that
\begin{align}
I(p) =& \int^1_0 \frac{x^p}{x^{p+1}+(1-x)^{p+1}}\ dx = \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{dx}{x}\\
=& \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{(1-x)}{x} dx+ \int^1_0 \frac{x^{p+1... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)... | If you know harmonic numbers
$$S_n=\sum_{i=i}^n \frac 1 {n^2+i}=H_{n^2+n}-H_{n^2}$$ Using the asympotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
p^2}+O\left(\frac{1}{p^3}\right)$$ apply it twice and continue with Taylor series to get
$$n S_n=1-\frac{1}{2 n}-\frac{1}{6 n^2}+O\left(\frac{1}{n^3}\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
evaluate $\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}$ $$
\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}
$$
I have tried the following:
$$
\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^... | Just write
*
*$\frac{n^3-n^2+2n+3}{n^3+2n^2-1} = 1 - \underbrace{\frac{3n^2-2n-4}{n^3+2n^2-1}}_{a_n :=}$ and note that
*$\lim_{n\to \infty}(1-a_n)^{\frac 1{a_n}} = \frac 1e$
Now, you need only the limit
*
*$\lim_{n\to\infty}a_n\frac{8n^3-3n^2+2n-2}{10n^2+3n-1} = \lim_{n\to\infty}\frac{3n^2-2n-4}{n^3+2n^2-1}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3468124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show $x^2+y^2=9z+3$ has no integer solutions
Show $x^2+y^2=9z+3$ has no integer solutions
So I know that $x^2+y^2=3(3z+1)$
And since $3\mid (9z+3)$ and $3\not\mid (3z+1)$ then $9z+3$ cannot be a square since it has a prime divisor which has a power less then $2$.
So does know this isn't a pythagorean triple $(x,y,\sq... | Since $x^2 + y^2 = 9z+3$ must be true, it must also be true that $$x^2+y^2 \equiv 3 \pmod 9$$
However, if we list out $x^2 \pmod 9$ for $0 \le x < 9$, we get $$0, 1, 4, 0, 7, 7, 0, 4, 1$$
There is no way of adding two numbers from $0, 1, 4, 7$ such that the sum is equal to $0 \pmod 9$.
Therefore, $x^2 + y^2 = 9z + 3$ h... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Is this a decreasing sequence? In this nice question , the OP asks, or wants to prove that the sequence: $$a_{n+1} = \sqrt[n]{a_{1}+\dots+a_{n}}, \\a_1=1$$ is increasing. After playing a bit with Mathematica Cloud I realised that this is not the case as you may see in the following diagram:
Actually, it seems that it... | For any integer $m \geqslant 3$,
\begin{align*}
\left(1 + \frac1m\right)^m & = 1 + \binom{m}1m^{-1} + \binom{m}2m^{-2} + \cdots + m^{-m+2} + m^{-m} \\
& \leqslant 1 + 1 + 1 + \cdots + \frac1m + \frac1m \quad (m + 1 \text{ terms}) \\
& < m.
\end{align*}
Therefore, $(m + 1)^m < m^{m+1}$. (I'm almost sure I once gave a sh... | {
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"url": "https://math.stackexchange.com/questions/3471506",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
value of $k$ in binomial sum
If $\displaystyle (1-x)^{\frac{1}{2}}=a_{0}+a_{1}x+a_{2}x^2+\cdots \cdots +\infty.$ and $a_{0}+a_{1}+a_{2}+\cdots+a_{10}=\frac{\binom{20}{10}}{k^{10}}.$ Then $k$ is
what i try
$(1-x)^{\frac{1}{2}}=1-\frac{1}{2}x+\frac{1}{2}\bigg(\frac{1}{2}-1\bigg)\frac{x^2}{2!}+\frac{1}{2}\bigg(\frac{1}... | We have
$$\sum_{n=0}^{1}(-1)^n\binom{\frac 12}{n}=\frac 12=\frac 24=\frac{\binom{2}{1}}{4^1},\quad \sum_{n=0}^{2}(-1)^n\binom{\frac 12}{n}=\frac 38=\frac{6}{16}=\frac{\binom{4}{2}}{4^2}$$
$$\sum_{n=0}^{3}(-1)^n\binom{\frac 12}{n}=\frac{5}{16}=\frac{20}{64}=\frac{\binom{6}{3}}{4^3},\quad \sum_{n=0}^{4}(-1)^n\binom{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472143",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the area enclosed by the locus of the vertex of the rectangle at which the normals meet. Let a and b be the lengths of the semimajor and semiminor axes of an ellipse respectively.
Draw a rectangle whose two sides are tangent to the ellipse and the other two are normal to the ellipse.
I want to find the area enc... | Let $A(u,v)$ be the outer vertex of the rectangle which should lies on the director circle, hence
$$u^2+v^2=a^2+b^2$$
Refer to another answer of mine here, the mid-point of the polar (the chord $BD$) is
$$\frac{1}{\dfrac{u^2}{a^2}+\dfrac{v^2}{b^2}}
\begin{pmatrix}
u \\ v
\end{pmatrix}$$
Since diagonals bisect each ot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\int_0^\infty \frac{\sin^n x}{x^m}dx$ could be expressed via $\pi$ or $\log$ I want to show some results first (they were computed by MMA)
$$
\int_0^\infty \frac{\sin^5 x}{x^3} dx =\frac{5}{32}{\color{Red}\pi} \quad
\int_0^\infty \frac{\sin^5 x}{x^5} dx =\frac{115}{384} {\color{Red}\pi} \\
\int_0^\infty \frac{\sin^5 ... | If $n=2p$ is even then note that
$$\sin^n(x)=\frac1{2^n}\binom np+\frac1{2^{n-1}}\sum_{k=0}^{p-1}\binom nk(-1)^{p-k}\cos((n-2k)x)$$
If $n=2p+1$ is odd then note that
$$\sin^n(x)=\frac1{2^{n-1}}\sum_{k=0}^p\binom nk(-1)^{p-k}\sin((n-2k)x)$$
By rewriting the $\sin^n(x)$, we get an integral of a sum of $\cos(jx)/x^m$ or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Bounded sequence of reals Suppose that $\{a_{n} \}$ is a sequence of real numbers that satisfy $a_{i} + a_{j} \geq a_{i+j}$.
Then prove $$a_{n} \leq \sum_{i=1}^{n} \frac{a_{i}}{i}$$.
I tried to use straight-up induction, but I am getting stuck. What I have is:
$a_{n+1} \leq a_{1} + a_{n} \leq a_{1} + \sum_{i=1}^{n} \... | From the inequality you easily obtain that for all integers $i_1, ..., i_k$, one has: $a_{i_1} + ... + a_{i_k} \geq a_{i_1 + ... + i_k}$. By taking them all equal, one gets: $a_{kl} \leq l \cdot a_k$, so $\frac{a_{kl}}{l} \leq a_k$.
We have:
$$
a_1 + a_{n-1} \geq a_n \\
... \\
a_{n-1} + a_{1} \geq a_n \\
$$
Summing all... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3479081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
For $a$, $b$, $c$ the sides of a triangle, show $ 7(a+b+c)^3-9(a+b+c)\left(a^2+b^2+c^2\right)-108abc\ge0$
If $a$, $b$, and $c$ are the three sidelengths of an arbitrary triangle, prove that the following inequality is true, with equality for equilateral triangles.
$$ 7\left(a+b+c\right)^3-9\left(a+b+c\right)\left(a^2+... | By AM-GM
$$\frac{a_1+\cdots+a_n}{n}\geq\sqrt[n]{a_1\cdots a_n}$$
Since $a,~ b,~ c$ are positive real numbers
$$\frac{a^3+b^3+c^3}{3}\geq\sqrt[3]{a^3b^3c^3}=abc$$
$$a^3+b^3+c^3\geq3abc$$
$$2(a^3+b^3+c^3)\geq6abc\tag{1}$$
Now we want to prove that
$$2a^2(b + c) + 2b^2(c + a) + 2c^2(a + b) ≥ a^3 + b^3 + c^3 + 9abc\tag{2}$... | {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Trying to find $\cot^{-1}x-\cot^{-1}y$ for $\forall$ $x,y$ Suppose we want to calculate $\cot^{-1}x-\cot^{-1}y$ for $\forall~x,y$
$$\cot^{-1}x-\cot^{-1}y=\theta\tag{1}$$
Let's find range of $\theta$, assuming $x$ and $y$ to be independent variables
$$\theta\in(-\pi,\pi)$$
Taking $\cot$ on both sides of equation $1$
$$... | $$0<\cot^{-1}x+\cot^{-1}y<2\pi$$
So, $\cot^{-1}x+\cot^{-1}y$ will be $\cot^{-1}\dfrac{xy-1}{x+y}$ if $\cot^{-1}x+\cot^{-1}y<\pi$
$\iff\tan^{-1}x+\tan^{-1}y>0\iff\tan^{-1}x>-\tan^{-1}y=\tan^{-1}(-y)$
As $\tan^{-1}x$ is strictly increasing in $\left(-\dfrac\pi2,\dfrac\pi2\right),$ we need $x>-y\iff x+y>0$
So, $\cot^{-1}... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$... | $$
(x^2-1)(x^4+x^2+1) = x^6 - 1
$$
$x^6-1=0$ if and only if $x$ is one of the $6$th roots of $1.$
$(x^2-1)(x^4+x^2+1)=0$ if and only if either $x^2-1=0,$ in which case $x=\pm 1,$ or $x^4+x^2+1=0,$ which fails to hold if $x=\pm 1,$ since $(\pm 1)^4+(\pm 1)^2 + 1 = 3\ne0.$ Thus the roots of $x^4+x^2+1=0$ are the $6$th ro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
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Limit of sum $x^3+x^5+x^7+x^9+...)$ I am asked to give the limit of:
$$ x^3+x^5+x^7+x^9+... \quad x\in(-1,1)$$
So I do the following:
The sum of the first $n$ terms will be equal to:
$$x^3+x^5+x^7+...+x^{3+2(n-1)}$$
I factor out $x^3$, I get:
$$x^3(1+x^2+x^4+..+x^{2(n-1)})$$
I also factor out $x^2$, I get:
$$x^3 x^2(1/... | Note that
$$1+x^2+x^4+\dots =\sum_{n=0}^\infty x^{2n}=\sum_{n=0}^\infty (x^2)^n$$
is convergent for $x\in (-1,1)$. Thus, call
$$S(x)=1+x^2+x^4+\dots$$
Then we have
$$S(x)=1+x^2\left(1+x^2+x^4+\dots\right)=1+x^2S(x)$$
Solving for $S(x)$ gives us
$$S(x)=\frac{1}{1-x^2}$$
(you could also get $S(x)$ by noting that it is a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Solving: $\int\cos^4(x)\sin^4(x)\ dx$ I've been trying to solve this problem - $\int\cos^4(x)\sin^4(x)\ dx$, but I don't seem to be succeeding. This is what I've done: $$\int\left(\frac{1+\cos(2x)}{2}\right)^2\left(\frac{1-\cos(2x)}{2}\right)^2\ dx\\=\frac1{16}\int\left(1+2\cos(2x)+\cos^2(2x)\right)\left(1-2\cos(2x)+\c... | Hint: An alternative approach is to use the identity
$$\sin(2x) = 2\sin(x)\cos(x)$$
Then we get that
\begin{align*}
\int \cos^4(x)\sin^4(x)dx &= \int \big(\cos(x)\sin(x)\big)^4 dx\\
&= \int \left(\frac{\sin(2x)}{2}\right)^4 dx \\
&= \frac{1}{16} \int \sin^2(2x) \sin^2(2x)dx \\
&= \frac{1}{16} \int \left(\frac{1-\cos(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Reducing $\frac{\left(\left(\sqrt{\frac{a + b}{a - b}}\right)^2+1\right)\cdot c}{2c\sqrt{\frac{a + b}{a - b}}}$ to $\frac{1}{\sqrt{1 - b^2/a^2}} $ I want to rearrange
$$ \frac{\left(\left(\sqrt{\frac{a + b}{a - b}}\right)^2+1\right) \cdot c}{2c \cdot \sqrt{\frac{a + b}{a - b}}} $$
into $$ \frac{1}{\sqrt{1 - b^2/a^2}} ... | HINTS: After cancelling the $c$ as you mention, we have
\begin{align}
\frac{\left(\sqrt{\frac{a + b}{a - b}}\right)^2+1}{2 \sqrt{\frac{a + b}{a - b}}}
= \frac{\left(\sqrt{\frac{a + b}{a - b}}\right)^2}{2 \sqrt{\frac{a + b}{a - b}}} + \frac{1}{{2 \sqrt{\frac{a + b}{a - b}}}}
\end{align}
Now you can easily cancel stuff ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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$\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$
I should solve the following system: $$\begin{cases} x^4+x^2y^2+y^4=21 \\
x^2+xy+y^2=3 \end{cases}$$
by reducing the system to a system of second degree.
What can I look for in such situations? What is the way to solve this kind of systems? The only th... | I found $$9=(x^2+xy+y^2)^2=$$ $$=(x^4+x^2y^2+y^4)+2(x^2y^2+x^3y+xy^3)=$$ $$=21+2xy(xy+x^2+y^2)=21+2xy(3)=$$ $$=21+6xy$$ giving $xy=-2.$
Then $(x+y)^2=(x^2+xy+y^2)+xy=3+xy=3-2=1.$
Knowing $(S,P)=(\pm 1,-2),$ we can find all possible $\{x,y\}$ as the solutions of $0=z^2-Sz+P=0=z^2\pm z-2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Sum of the reciprocal $P$-smooth numbers for $P = \{ p, q \}$ ?? For distinct primes $p$ and $q$ (where $p < q$), the following sum
$$\sum_{n=0}^{\infty} \sum_{k=0}^n \frac{1}{p^n q^k}$$
yields the expression $\frac{f}{g}$ such that
$$f = p^2 q$$
and
$$g \equiv 1 \ (\text{mod} \ p).$$
The problem I'm having is finding ... | Wikipedia's Geometric series article gives the formulas for the sums of a finite & infinite geometric series of
$$\sum_{k = 0}^{n-1}ar^k = a\left(\frac{1-r^n}{1-r}\right) \tag{1}\label{eq1A}$$
$$\sum_{k = 0}^{\infty}ar^k = \frac{a}{1-r}, \text{ for } \left|r\right| \lt 1 \tag{2}\label{eq2A}$$
Using \eqref{eq1A} and \eq... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $2018^{2018}$ mod $20$? The answer is $4$.
I saw someone answered in $5$ seconds. How did they do that?
| Since $2018 \equiv -2 \pmod{20}$ we have $$2018^{2018} \equiv (-2)^{2018} \equiv 4^{1009}\pmod{20}$$
and then
$$4^1 \equiv 4 \pmod{20} \\ 4^2 = 16 \equiv -4 \pmod{20} \\ 4^3 \equiv -4\cdot 4 \equiv 4 \pmod{20} \\ \vdots$$
Since $1009$ is odd and the powers of $4$ alternate $4, -4, 4, -4, \ldots$, the answer is $4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Solving logarithm leaving in terms of $p$ and $q$ I would like to check the steps if Part a) is done correctly.
For Part b), how do I continue from below? I seem to stuck for $\log_{10}(5)$…
Here is the problem:
Given that $p = \log_{10} 2$ and $q = \log_{10} 7$, express the following in terms of $p$ and $q$.
a) $\log... | Part a) is perfect (nice job)
For part b) the $\log_{10} 5$ is throwing a monkey wrench and you can't express $5$ in terms of $2$ and $7$ using only multiplication and division and exponents.
BUt you have two other values at your disposal: $\log_{10} 10 = 1$ and $\log_{10} 1 = 0$.
Can you express $5$ in terms of $2, 7... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove no two numbers besides 1 in the sequences are identical Question:
Given that {$a_n$} and {$b_n$} are two sequences of integers defined by
$$a_1=1,a_2=10,a_{n+1}=2a_n+3a_{n-1}$$
$$b_1=1,b_2=8,b_{n+1}=3b_n+4b_{n-1}$$
for $n=2,3,4,...$
Prove that, besides the number '1' , no two numbers in the sequences are id... | Using the method of characteristic equation and roots, given in Linear difference equation,
gives for the first sequence the characteristic equation of
$$\begin{equation}\begin{aligned}
\lambda^{2} & = 2\lambda + 3 \\
\lambda^{2} - 2\lambda - 3 & = 0 \\
(\lambda - 3)(\lambda + 1) & = 0
\end{aligned}\end{equation}\tag{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Set of vectors defined by an equation Let $W$ be a set of vectors given by an equation $2x_1 + x_2 - 2x_3 + 3x_4 = 5$. I need to write this into a affine subspace form like $W = p + span(s)$ and I have absolutely no idea how should I proceed with this kind of definition.
| Note that
\begin{align}
\textsf{W} &= \left\{
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \in \mathbb{R}^4 :\, 2x_1 + x_2 - 2x_3 + 3x_4 = 5
\right\} \\
&= \left\{
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \in \mathbb{R}^4 :\, x_2 = 5 - 2x_1 - 2x_3 - 3x_4
\right\} \\
&= \left\{
\begin{pmatrix}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Problem in deducing an identity related to Logarithm to the base e . I have a problem in deriving this inequality about the logarithm to base $e$ .
Can someone please give hints.
If $x>1$ then prove that $$\log x + \log\frac{x}2+ \log \frac {x} {2^2} + \log \frac{x} {2^3} +\dots < \log^2 (x) .$$
Can someone ple... | Because
$$\begin{align}
&\frac{\ln^2x}{\ln x+\ln\frac x2+\ln\frac x{2^2}+\cdots}\\
= &\frac{\ln^2x}{\ln\left(\displaystyle\prod_{i=0}^n\frac x{2^i}\right)}\\
= &\frac{\ln^2x}{\ln\left(\frac{x^n}{2^\frac{n(n-1)}2}\right)}\\
= &\frac{\ln^2x}{n\ln x-\frac{n^2-n}2\ln 2}\\
= &\frac{1}{\frac n{\ln x}-\frac{n^2-n}2\cdot\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^\pi\frac{\sin\left(n+\frac{1}{2}\right)x}{\sin \frac{x}{2}}dx$
Evaluate $$
\int_0^\pi\frac{\sin\Big(n+\frac{1}{2}\Big)x}{\sin \frac{x}{2}}dx
$$
$$
\int_0^\pi\frac{\sin\Big(n+\frac{1}{2}\Big)x}{\sin \frac{x}{2}}dx=\int_0^\pi\frac{\sin\Big(nx+\frac{x}{2}\Big)}{\sin \frac{x}{2}}dx=\int_0^\pi\frac{\sin n... | Note that
$$2\sin\frac x2\cos x = \sin\frac32x -\sin\frac12x$$
$$2\sin\frac x2\cos 2x = \sin\frac52x -\sin\frac32x$$
$$…$$
$$2\sin\frac x2\cos nx =\sin(n+\frac12)x -\sin(n-\frac12)x $$
Sum up both sides,
$$2\sin\frac x2 (\cos x + \cos 2x + … +\cos nx )= \sin(n+\frac12)x - \sin\frac x2 $$
Therefore,
$$
\int_0^\pi\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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if $mx+3|x+4|-2=0$ has no solutions, solve for $m$
If $mx+3|x+4|-2=0$ has no solutions, which of the following value could be $m$?
(A)5
(B)$-\frac{1}{2}$
(C)-3
(D)-6
(E)$\frac{10}{3}$
my attempt:
$$mx-2=-3|x+4| \\ m^2x^2-4mx+4=9x^2+72x+144 \\ (9-m^2)x^2+(4m+72)x+140=0$$
because the equation has no solutions, therefor... | Consider the fact that $|x+4|$ has 2 different possible values. For $x\geq-4$, it becomes $x+4$. For $x<-4$, it becomes $-x-4$. Divide the problem into 2 those two different cases.
For $x\geq-4$, the equation becomes $mx+3(x+4)-2=mx+3x+10=(m+3)x+10=0$. We obtain that $x=-\frac{10}{m+3}$. In order to make there exist no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Stirling Numbers of the Second Kind Proof
Prove that
\begin{align*}
\sum_{n=1}^\infty S(n,n-2)x^n=\dfrac{x^3(1+2x)}{(1-x)^5}
\end{align*}
My guess is that I have to take the LHS and simply it, as well as take the RHS and simplify it, but not sure how to exactly do that.
Any help, tips, or a fully worked out... |
Answer $1$ : Using the recurrence relation.
The Stirling numbers of the second kind satisfy the recurrence relation
\begin{eqnarray*}
{n \brace k}= k {n-1 \brace k}+ {n-1 \brace k-1}.
\end{eqnarray*}
With $k=n$ we have ${n \brace n}= {n-1 \brace n-1}=1$ and
\begin{eqnarray*}
S_{0}(x)=\sum_{n=1}^{\infty} {n \brace n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Need Help with Mistake in Generating Function for Nonlinear Recurrence relation I'm having a bit of trouble finding the generating function for the following recurrence relation:
$$
w_n -1 = \sum _{k=1}^{n-1} w_k w_{n-k}, \quad n \geq 2, \; w_0 = 0, \; w_1 = 1.
$$
I set out to find a generating function $F$ such that
$... | Multiplying both sides of the recurrence by $x^n$ and summing for $n=2$ to infinity:
$$
\sum_{n=2}^\infty w_nx^n = \sum_{n=2}^\infty \sum_{k=1}^{n-1} w_kw_{n-k}x^n + \sum_{n=2}^\infty x^n.
$$
Because $w_0=0$, $w_1=1$, and $w_n$ is a sum of products of $w_1,\ldots,w_{n-1}$ plus one, $w_n$ is nonnegative for all $n$. So ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine the conditions for $n$ and $\theta\neq0+2k\pi$ such that $(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta})^n \in\mathbb{R}$ Let $z=1+\cos\theta+i\sin\theta=|z|(\cos\alpha+i\sin\alpha)$ and $z'=1-\cos\theta+i\sin\theta=|z'|(\cos\alpha'+i\sin\alpha')$.
My line of reasoning is to convert the numerator... | Easier is to simplify the denominator using conjugates. \begin{align*}
&\frac{1 + \cos \theta + \mathrm{i} \sin \theta}{1 - \cos \theta + \mathrm{i} \sin \theta} \cdot \frac{1 - \cos \theta - \mathrm{i} \sin \theta}{1 - \cos \theta - \mathrm{i} \sin \theta} \\
&\quad{}= \frac{2 \sin \theta}{2 - 2 \cos \theta} (-\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this a correct proof of the divergence of a series? I wanted to show if
$$ \sum_{n=2}^\infty \frac{\sqrt{n}}{3^n\ln(n)} $$
converges or diverges. By the ratio test, I could solve for $a_n=\frac{\sqrt{n}}{3^n\ln(n)}$ the limit
$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}$$
and see what the result tells me.
$Proof:$
$i)$ L... | It's correct. It can also be done without switching to functional limits and using L'Hôpital's rule:
$$\frac{a_{n+1}}{a_n}=\frac 13\sqrt{1+\frac 1n}\frac{\ln n}{\ln(n+1)}\to \frac 13\cdot 1\cdot 1=\frac 13<1 $$
To handle the fraction with the logarithms
$$\frac{\ln(n+1)}{\ln n}=\frac{\ln n+\ln\left(1+\frac 1n\right)}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Logarithmic inequality $5\times \frac{3^{x-2}}{3^{x}-2^{x}}\geq1+(\frac{2}{3})^x$ Can somebody help me with this problem?
Solve inequation
$$5\times \frac{3^{x-2}}{3^{x}-2^{x}}\geq1+(\frac{2}{3})^x$$
I am trying to multiply both sides of inequation with
$$\frac{3^{x}}{3^{x}+2^{x}}$$
and I get
$$\frac{5\times3^{2x}}{9\... | You can rewrite it as follows:
$$5 \geq
(1 + \left(\frac{2}{3}\right)^x) \frac{3^x - 2^x}{3^{x-2}}=
(1 + \left(\frac{2}{3}\right)^x) ( 9 - \frac{1}{9} \left(\frac{2}{3}\right)^x) $$
Now if we set $a = \left(\frac{2}{3}\right)^x$, we're reducing the problem to this binomial inequality:
$$(1 + a)(9 - \frac{1}{9}a) \leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521235",
"timestamp": "2023-03-29T00:00:00",
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What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$ Is there closed form for
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.
My approach,
In this paper page $95$ Eq $(5)$ we have
$$\sum... | From this paper page $101$ we have
$$\sum_{n=1}^\infty\overline{H}_n\frac{x^{n+1}}{(n+1)^2}=\operatorname{Li}_3\left(\frac{2x}{1+x}\right)-\operatorname{Li}_3\left(\frac{x}{1+x}\right)-\operatorname{Li}_3\left(\frac{1+x}{2}\right)-\operatorname{Li}_3(x)$$
$$+\ln(1+x)\left[\operatorname{Li}_2(x)+\operatorname{Li}_2\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For △ABC, prove $\frac a{h_a} + \frac b{h_b} + \frac c{h_c} \ge 2 (\tan\frac{\alpha}2+ \tan\frac{\beta}2 + \tan\frac{\gamma}2)$
Given $\triangle ABC$, (using the main parameters and notation), prove that $$ \frac{a}{h_a} + \frac{b}{h_b} + \frac{c}{h_c} \ge 2 \cdot \left(\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \ta... | Using the double-angle and half-angle trigonometric identities, we have that $$2 \cdot \sum_{cyc}\tan\frac{\beta}{2} = 2 \cdot \sum_{cyc}\frac{\sin\beta}{\cos^2\dfrac{\beta}{2}} = \frac{1}{R} \cdot \sum_{cyc}\frac{b}{\cos\beta + 1} = \frac{2abc}{R} \cdot \sum_{cyc}\frac{1}{(c + a)^2 - b^2}$$
$$ = \frac{8 \cdot [ABC]}{a... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ when $a+b+c=0$
Let $a,b,c$ such that $$a + b + c =0$$ and
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$$ is defined.
Find the value of $P$.
This is a very queer problem.
| Put $c=-a-b$ in
$$2a^2+bc=2a^2+b(-a-b)=a^2-ab+a^2-b^2=(a-b)(2a+b)=(a-b)(b-c)$$
Similarly $2b^2+ca=(b-c)(c-a)$ and
$2c^2+ab=(c-a)(c-b)$
So expression reduce into
$$\frac{a^2(c-a)+b^2(a-b)+c^2(b-c)}{(a-b)(b-c)(c-a)}=1$$
Factorisation of Numerator
$a^2(c-a)+b^2(a-b)+c^2(b-c)$ is zero when $a=b$ or $b=c$ or $c=a$
$$a^2(b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How can I calculate this (rather tricky) limit? $$\lim\limits_{n \to \infty} \cos^{n^2} \left (\frac{2x}{n} \right)$$
Any hints and/or help is greatly appreciated.
| Consider $$a_n=\cos^{n^2} \left (\frac{2x}{n} \right)\implies \log(a_n)=n^2\log\left( \cos\left (\frac{2x}{n} \right)\right)$$ Now use the series expansion
$$\cos\left (\frac{2x}{n} \right)=1-\frac{2 x^2}{n^2}+\frac{2 x^4}{3 n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log\left( \cos\left (\frac{2x}{n} \right)\right)=-\frac{2... | {
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"url": "https://math.stackexchange.com/questions/3527791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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integration of. $\int \frac{x}{x^3-3x+2}$ I am trying to integrate :
$\Large \int \frac{x}{x^3-3x+2}dx$
I decomposed the fraction and got :
$\Large \frac{x}{x^3-3x+2} = \frac {x}{(x-1)^2(x+2)}$
Then I tried to get two different fractions:
$ \Large \frac {x}{(x-1)^2(x+2)} = \frac {Ax}{(x-1)^2} + \frac {B}{(x+2)}$
W... | When the denominator of a fraction has an irreducible factor $\bigl(p(x)\bigr)^m$ with order of multiplicity $m>1$, its contribution to the decomposition into partial fractions is not
$$\frac{A(x)}{\bigl(p(x)\bigr)^m} \qquad (\deg A(x)<\deg p(x)),$$
but
$$\frac{A_1(x)}{p(x)\vphantom{\Big)}}+\frac{A_2(x)}{\bigl(p(x)\big... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimize $\frac{2}{1-a}+\frac{75}{10-b}$
Let $a,b>0$ and satisfy $a^2+\dfrac{b^2}{45}=1$. Find the minimum
value of $\dfrac{2}{1-a}+\dfrac{75}{10-b}.$
WA gives the result that $\min\left(\dfrac{2}{1-a}+\dfrac{75}{10-b}\right)=21$ with $a=\dfrac{2}{3},b=5$.
Consider making a transformation like this
$$\dfrac{2}{1-a... | There is another way using parameterisation. Let $a = \cos t, b = \sqrt{45} \sin t$, then we have:
$$f(t) = \frac{2}{1-\cos t} + \frac{75}{10-\sqrt{45} \sin t}$$
$$f'(t) = -\frac{2 \sin t}{(1-\cos t)^2} + \frac{225 \sqrt5 \cos t}{(10-3 \sqrt{5} \sin t)^2}$$
The condition that $a,b > 0$ translates to $0 < t < \frac{\pi}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the remainder when $x^{100}$ is divided by $x^8 - x^6 + x^4 - x^2 + 1.$ I need help in the problem:
Find the remainder when $x^{100}$ is divided by $x^8 - x^6 + x^4 - x^2 + 1.$
I have tried factoring $x^{100}-1$ and adding 1 to that, but that hasn't helped. Could someone please help with this?
| Hint
Replace $x^2$ with to find the divisor $$=\dfrac{y^5+1}{y+1}$$
$y^{50}=(y^5+1-1)^{10}=1+$ terms divisible by $y^5+1$
| {
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Proof by induction:$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$ In the very beginning I'm going to refer to similar posts with provided answers:
Induction Inequality Proof with Product Operator $\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3k+1}}$ (answered by Öz... | I concur with everyone else that it's basically right. Everyone has their own style, but the following is how I would probably write up the "meat and potatoes" of the proof:
Let's start with some preliminary observations. Note that
\begin{align}
\frac{4n+3}{4n+5}\cdot\sqrt{\frac{3}{4n+3}}
&=\frac{4n+3}{4n+5}\cdot\frac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $a^2$ cannot be congruent to 2 or 3 mod 5 for any integer a) Show that $a^2$ cannot be congruent to $2$ or $3 \bmod 5$ for any integer.
$a^2≡2,3 \pmod5$
$a≡0,±1,±2\pmod5 )⟹a^2≡0,1,4$.
But $2,3$ are not congruent to $0,1,4\pmod5$.
I am not sure if I did it right, please check it for me
(b) Show that if $5\... | You did a) just find. As $(-2,-1,0,1,2)$ is a complete residue class $\mod 5$ it must be that $a\equiv 0,\pm 1, \pm 2\pmod 5$ and therefore $a^2\equiv 0, 1,4\equiv -1$ none are congruent to $2$ or to $3\equiv -1$.
As for $b$ we have $x^2,y^2, z^2 \equiv 0, 1,-1\mod 5$ so we mus have some how that $x^2 + y^2 + z^2 \equ... | {
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"url": "https://math.stackexchange.com/questions/3539582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the inequality stated below:
If $a, b, c, d > 0$ then prove:
$$\frac{a}{b+c+d} + \frac{b+c}{a+d} + \frac{d}{b+a} > 1$$
I tried to consider the difference and lead to a common denominator, but in the numerator, there remains ${1}$ term with a minus which is less than ${0}$
$a^3 - a^2 c + b (b + c)^2 + c d^2 + d... | Notice that if $m,n,k>0$ then $${m\over n} >{m\over n+k}$$ so
$$\frac{a}{b+c+d} + \frac{b+c}{a+d} + \frac{d}{b+a} >\frac{a}{\color{red}{a}+b+c+d} + \frac{b+c}{\color{red}{b+c}+a+d} + \frac{d}{\color{red}{c+d}+b+a}$$ $$= \frac{a+b+c+d}{a+b+c+d}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
We have to make this function defined, by change the cosine into sine or tan. So,
$$\lim_{t \to 0} \left(\frac{a}{t^2} - \fr... | Just use Taylor expansion and simplify first $\cos^2 3t = \frac{1+\cos 6t}{2}$.
Hence,
$$\frac{at\cos^2 3t - \sin 6t}{t^3\cos^2 3t}= \frac{at(1+\cos 6t) - 2\sin 6t}{t^3(1+\cos 6t)}$$
$$= \frac{at + at(1- 18t^2+o(t^3)) - 2(6t-36t^3+o(t^4))}{t^3(1+\cos 6t)}$$
$$= \frac{1}{(1+\cos 6t)}\left(\frac{2a-12}{t^2} -18a+72 + o... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $xy''-y'-x^3y=0$ I want to solve $xy''-y'-x^3y=0$. My solution:
$y = z\cdot \exp(kx^2)$
$y' = z'\cdot \exp(kx^2) + z\cdot 2kx\exp(kx^2)$
$y'' = z''\cdot \exp(kx^2) + z'\cdot 4kx\exp(kx^2) + z\cdot (2k+4k^2x^2)\exp(kx^2)$
Plug in:
$z''\exp(kx^2)+z'(4kx-\frac{1}{x})\exp(kx^2)+z(2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2)\e... | Dividing by $x^2$, the equation is
$$\left(\frac{y'}x\right)'-xy=0$$
or
$$\frac{y'}x\left(\frac{y'}x\right)'-yy'=0.$$
After integration,
$$\left(\frac{y'}x\right)^2-y^2=c,$$
which is separable.
$$\frac{y'}{\sqrt{y^2+c}}=\pm x$$
gives
$$\text{arsinh}\frac y{\sqrt{|c|}}=c'\pm\frac{x^2}2$$ or
$$\text{arcosh}\frac y{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the equation $\sqrt[3]{15-x^3+3x^2-3x}=2\sqrt{x^2-4x+2}+3-x$. Solve the equation $\sqrt[3]{15-x^3+3x^2-3x}=2\sqrt{x^2-4x+2}+3-x$.
I have tried to solve for x by Casio and try to make the equation to $u.v=0$ but the solution is not in $\mathbb{Q}$. Any help is appreciated. Thanks
| For the square root to be defined we need:
$$x^2-4x+2\geq 0$$
Therefore, we have:
$$2\sqrt{x^2-4x+2} = x-3+\sqrt[3]{15-x^3+3x^2-3x}=$$
$$=\frac{(x-3)^3+15-x^3+2x^2-3x}{(x-3)^2-(x-3)\sqrt[3]{15-x^3+3x^2-3x}+\sqrt[3]{(15-x^3+3x^2-3x)^2}}$$
$$=\frac{-6(x^2-4x+2)}{(x-3)^2-(x-3)\sqrt[3]{15-x^3+3x^2-3x}+\sqrt[3]{(15-x^3+3x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3545501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$... | Assuming that h(x)= f(x)/g(x), I would not actually calculate h itself. Instead, use the "quotient rule": $\frac{h'(x)= f'(x)g(x)- f(x)g'(x)}{g^2(x)}$. That will be 0 if and only if the numerator is 0: $f'(x)g(x)- f(x)g'(x)= 0$.
$f(x)= x^2+ x^{-2}$ so $f'(x)= 2x+ 2x^{-3}$, $g(x)= x- x^{-1}$ so $g'(x)= 1- x^{-2}$.
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The area of interior quadrilateral formed by connecting trisect points and vertices of larger quadrilateral I drew a figure on GeoGebra to explore the area of the smaller quadrilateral formed by joining the vertices of the larger quadrilateral and the trisect points on the edges of the larger quadrilateral.
Here is wha... | Some example caculation for retangular case. Let $ABCD$ be a rectangular with $AB=6,AD=3$.
Let $TQ=x$,$XM=y$ and $\angle ZTX=\alpha$, then $XM=ZP=x$; $TQ=YN=y$; $DX=BZ=3x$;$AT=CY=3y$ and $XYZT$ is parallelogram rhombus with $XY=6x$ and $XT=6y$
We have $37a^2=(6a)^2+a^2=AB^2+AQ^2=BQ^2=100x^2$, then $x=\frac{\sqrt{37}}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trouble with trig substitution I have a few questions and a request for an explanation.
I worked this problem for a quite a while last night. I posted it here.
Here is the original question: $$\int\frac{-7 x^2}{\sqrt{4x-x^2}} dx$$
And here is the work that I did on it:
Help with trig sub integral
Sorry that the negati... | The mistake in your original attempt has been identified:
the substitution for $dx.$
In particular,
$$ \frac{d}{d\theta} 4 \sin^2 \theta = 8 \sin\theta\cos\theta \neq 2 \cos\theta.$$
Patching your original attempt by making the correct substitution,
\begin{align}
-7 \int\frac{x^2}{\sqrt{4x-x^2}} dx
&= -7 \int\frac{x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Increasing each element of a set, what is the sum of the medians of the two sets. Consider an ordered set of six consecutive integers in increasing order. To create a new set of six integers, the first, third, and fifth elements are each multiplied by two, and the second, fourth, and sixth elements are each increased b... | The original sequence of numbers is $a,a+1,a+2,a+3,a+4,a+5$, where $a$ is a positive integer. Let $M_1$ be the median of the old sequence. Note that $M_1=a+2.5$.
The new sequence of numbers is $2a,a+3,2a+4,a+5,2a+8,a+7$. Let $M_2$ be the median of the new sequence.
Note that $2a$ will eventually be bigger than $a+7$, i... | {
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"url": "https://math.stackexchange.com/questions/3548177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does the joint distribution of $X$ and $Y$ is the standard bivariate normal distribution using the change of variables We have $$ X = \sqrt{-2 \log(U)} \cos(2 \pi V)$$ and $$Y = \sqrt{-2 \log(U)} \sin(2 \pi V)$$
where $U$ and $V$ are independent uniform random variables over $[0,1]$.
I started solving it using the ... | \begin{align}
x & = \sqrt{-2\log u}\, \cos(2\pi v) \\[8pt]
y & = \sqrt{-2\log u}\, \sin(2\pi v)
\end{align}
\begin{align}
& \frac{\partial x}{\partial u} = \frac{-\cos(2\pi v)}{u\sqrt{-2\log u}} & & \frac{\partial y}{\partial u} = \frac{-\sin(2\pi v)}{u\sqrt{-2\log u}} \\[12pt]
& \frac{\partial x}{\partial v} = \sqrt{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solve the recurrence relation $a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$ Solve the recurrence
$a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$
with initial conditions $a_1 = 1, a_2 =5$ and $a_3 = 17$ How this was calculated? any hint or idea highly appreciated
| Let $a_n=b_n \, 3^n$ to make
$$b_n-\frac{5}{3} b_{n-1}+\frac{8}{9} b_{n-2}-\frac{4}{27} b_{n-3}=1$$ Now, make $b_n=c_n+k$ to make
$$c_n-\frac{5}{3} c_{n-1}+\frac{8}{9} c_{n-2}-\frac{4}{27} c_{n-3}=1-\frac{2 k}{27}$$ Chose $k=\frac{27}{2}$ to make the rhs equal to $0$.
Solve
$$c_n-\frac{5}{3} c_{n-1}+\frac{8}{9} c_{n-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $n^4 \mod 8$ is identically equal to either 0 or 1, $\forall \ n\in \mathbb{N}$. I feel pretty confident about the first half of my solution for this, however I don't like how I used the induction hypothesis on the case for even integers, it feels like it isn't doing anything useful since it is really easy t... | Your proof is unnecessarily complicated.
If $n$ is even, then we can say $n = 2k$
$(2k)^4 = 16k^4 = 8(2k^4)$ which is divisible by $8.$
$(2k)^4 \equiv 0 \pmod 8$
If $n$ is odd, then $n = 2k+1$
$(2k+1)^4 = 16k^4 + 4(2k)^3 + 6(2k)^2 + 4(2k)+ 1$
Each of the first $4$ terms is divisible by $8$
$(2k+1)^4 \equiv 1 \pmod 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
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