Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\lim_{n\to \infty} F(k)=\frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{3})}$ Find F(5) and F(6) Find the value of F(5) & F(6).It is given that
$$F(k)= \lim_{n\to \infty} \frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{... | note that $$\sum_{i=1}^ni^2=\frac{1}{6}n(n+1)(2n+1)$$
$$\sum_{i=1}^n i^3=\frac{1}{4}n^2(n+1)^2$$
$$\sum_{i=1}^ni^5=\frac{1}{12}n^2(n+1)^2(2n^2+2n-1)$$
and we get $$\lim_{n\to \infty}\frac{\frac{1}{12}n^2(n+1)^2(2n^2+2n-1)}{\frac{1}{6}n(n+1)(2n+1)\frac{1}{4}n^2(n+1)^2}$$
the searched limit is $0$
note for the next:
$$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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closed form solution for $ \displaystyle \sum_{x=1}^{k-1} {\frac{1}{x(k-x)}} $ Is there any closed form solution for this summation?
$$ \sum_{x=1}^{k-1} {\frac{1}{x(k-x)}} $$
$k$ is a finite integer constant.
| We have that $\frac{1}{x(k-x)} = \frac{1}{k} \left( \frac{1}{x} + \frac{1}{k-x} \right)$. Therefore,
$$\sum_{x=1}^{k-1} \frac{1}{x(k-x)} = \frac{1}{k} \left( \sum_{x=1}^{k-1} \frac{1}{x} + \sum_{x=1}^{k-1} \frac{1}{k-x} \right).$$
By reversing the order of summation in the second sum, we get that this is equal to $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\sqrt{x}$ grows faster than $\ln{x}$. So I have the limit $$\lim_{x\rightarrow \infty}\left(\frac{1}{2-\frac{3\ln{x}}{\sqrt{x}}}\right)=\frac{1}2,$$ I now want to motivate why $(3\ln{x}/\sqrt{x})\rightarrow0$ as $x\rightarrow\infty.$ I cam up with two possibilites:
*
*Algebraically it follows that ... | For any $a > c > 0, x > 1$,
we have
$\begin{array}\\
\ln(x)
&=\int_1^x \frac{dt}{t}\\
&\lt\int_1^x \frac{dt}{t^{1-c}}\\
&=\int_1^x t^{c-1}dt\\
&=\dfrac{t^c}{c}\big|_1^x\\
&=\dfrac{x^c-1}{c}\\
&<\dfrac{x^c}{c}\\
\text{so}\\
\dfrac{\ln(x)}{x^a}
&<\dfrac1{x^a}\dfrac{x^c}{c}\\
&=\dfrac{x^{c-a}}{c}\\
&\to 0
\qquad\text{as }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding common vectors in $R^5$ of both spans
Determine which vectors in $R^5$ belong to both $sp${$(1, 0, 3, 5, 1),
(0, 2, 0, 1, 5)$} and $sp${$(1, 0, 2, 5, 1), (1, 3, 1, 1, 0), (2, 7,
0, 0, 1)$}.
My approach:
I've interpreted this question as a particular vector $v$ $ϵ$ $R^5$, such that
$v$ $ϵ$ $sp${$(1, 0, 3, 5... | Hint: You're having a system of equations with $5$ variables. Bring it back to matrix $\mathbf{Ax}=0$ and solve it. The answer will be all vectors $\alpha \cdot \mathbf{x}$ where $\alpha \in \mathbf{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}$
Let $c\ne0$ be a constant. Consider the limit of $$f(n)=e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}$$ as $n \to \infty$.
I think it is zero because for large $n$,
$$e^{n^{3/4}} ((1- \frac{c}{n^{1/4}})^{n^{1/4}})^{\frac{n^{3/4}}{c}} \approx e^{n^{3/... | If we set $ h=\frac{1}{n^{1/4}} $ then
\begin{split}\lim_{n\to \infty}f(n)&=&\lim_{n\to \infty}e^{n^{3/4}} ((1- c/n^{1/4})^{n^{1/4}})^{n^{3/4}/c}\\
&=& \lim_{n\to \infty}e^{n^{3/4}} (1- c/n^{1/4})^{n/c}
\\&=&\lim_{h\to 0}e^{ \frac{1}{ h^3}}\left( 1-ch\right)^{1/ch^4}
\\&=&\lim_{h\to 0}\exp\left( \frac{1}{ h^3}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2483346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Concrete Mathematics: Clarification on Euclidean algorithm Let $m$ and $n$ be two positive integers, then:
$$m'm + n'n = gcd(m, n)$$
We can rewrite this as:
$$\overline{r}r + \overline{m}m = gcd(r, m)$$
with $r = n - \lfloor n/m \rfloor m$. Inserting the value of $r$ in the above equation, we have:
$$\overline{r}\left(... | In the previous page it is mentioned that, we can calculate $m'$ and $n'$ recursively until $m \neq 0$. I missed the crucial point as I over read it.
Here's what is written
Euclid's algorithm also gives us more.: We can extend it so that it will compute integers $m'$ and $n'$ satisfying
$$m'm + n'n = gcd(m, n).$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\int_{0}^{\infty}e^{-x}(x^2+3x+3)\cos(x){\mathrm dx\over (1+x)^3}=1?$ How to show that $(1)$ is
$$\int_{0}^{\infty}e^{-x}(x^2+3x+3)\cos(x){\mathrm dx\over (1+x)^3}=1?\tag1$$
$$x^2+3x+3=\left(x+{3\over 2}\right)^2+{3\over 4}$$
$$\int_{0}^{\infty}e^{-x}\cos(x){\mathrm dx\over (1+x)^3}+\int_{0}^{\infty}... | Start with partial fractions:
$$ \frac{x^2+3x+3}{(1+x)^3} = \frac{1}{1+x} + \frac{1}{(1+x)^2} + \frac{1}{(1+x)^3} $$
Now let
$$ J_k = \int_0^\infty e^{-(1+i)x}\frac{dx}{(1+x)^k} $$
Integrating by parts,
$$ J_k = \frac{1-i}{2} - \frac{((1-i)k}{2} J_{k+1} $$
Using this for $k=1$ and $k=2$, $J_1 + J_2 + J_3$ simplifies t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Non unique factorization in $\mathbb{Z}_5$
Prove $3X^3+4X^2+3=(X+2)^2(3X+2)=(X+2)(X+4)(3X+1)$ in $\mathbb{Z}_5$.
I see that $2$ is a root (since $-24+16+3=-5$) , then $3X^3+4X^2+3=(X+2)(3X^2)$, and also $-4$ (since -$64\cdot3+16\cdot4+3=-125$) and although $x=-1/3$ is not in $\mathbb{Z}_5$, because it also works as a... | Factorization is only unique up to associates; e.g. over the reals, we can factor $4x^2 - 1$ in many different ways
$$ 4x^2 - 1 = (2x-1)(2x+1) = 4\left(x - \frac{1}{2}\right)\left(x + \frac{1}{2}\right) = (4x-2)\left(x + \frac{1}{2} \right)$$
In $k[x]$, we most commonly normalize by factoring out an overall unit so as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to determine which of these numbers is greater: $\sqrt{2}^{\sqrt{3}}$ or $\sqrt{3}^{\sqrt{2}}$ without approximating anything? Is it possible to determine which of these numbers is greater: $\sqrt{2}^{\sqrt{3}}$ or $\sqrt{3}^{\sqrt{2}}$?
I tried to express both in terms of powers of 3, but in the first n... | We compare
$$\sqrt{2}^\sqrt{3} \text{ vs. } \sqrt{3}^\sqrt{2}$$
or (taking natural logarithms)
$$(\sqrt{3}/2) \ln 2 \text{ vs. } (\sqrt{2}/2) \ln 3$$ or (multiplying by $2$) $$\sqrt{3} \ln 2 \text{ vs. } \sqrt{2} \ln 3$$ or (rearranging) $$\sqrt{3/2} \text{ vs. } (\ln3)/(\ln2) = \log_2 3.$$
Now $2^{\sqrt{3/2}} < 2^{3/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Factorising $(4 + 3i)z^2 + 26iz + (-4+3i)$? Quadratic formula attempt included. I'm trying to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$.
I tried to use the quadratic formula to factorise $(4 + 3i)z^2 + 26iz + (-4+3i)$, where $b = 26i$, $a = (4 + 3i)$, $c = (-4+3i)$. This gives us the roots $z = \dfrac{-4i - 12}{25}$ and... | Another way, let $a = 4+3i$, then the equation is $az^2 + (a \overline a+1)iz - \overline{a}=0$.
$$\implies (az)^2 + (a \overline ai+i)(az) - a\overline{a}=0$$
Knowing Vieta, it is easy to observe the roots are $az \in \{ -a\overline ai, -i\}$, which implies $z \in \{-\overline ai, -\dfrac{i}a \} = \{-3-4i, -\frac3{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $(a+ \dfrac {1}{a})^3 +(b+ \dfrac {1}{b})^3 \ge \dfrac {125}{4}$
If $a,b$ are positive real numbers, such that $a+b=1$, then prove that $$\bigg(a+ \dfrac {1}{a}\bigg)^3 +\bigg(b+ \dfrac {1}{b}\bigg)^3 \ge \dfrac {125}{4}$$
I just learnt to prove I... | Let $a=\cos^2 t,b=\sin^2 t$.
Note
$$ \cos^6t+\sin^6t=(\cos^2t+\sin^2t)(\cos^4t-\cos^2t\sin^2t+\sin^4t)=1-3\cos^2t\sin^2t. $$
Then
\begin{eqnarray}
&&(a+\frac1a)^3+(b+\frac1b)^3\\
&=&(\cos^2t+\sec^2t)^3+(\sin^2t+\csc^2t)^3\\
&=&\cos^6t+3\cos^4t\sec^2t+3\cos^2t\sec^4t\\
&&+\sin^6t+3\sin^4t\csc^2t+3\sin^2t\csc^4t+\csc^6t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2493834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that if $p|x^p+y^p$ then $p^2|x^p+y^p$ I can show that $5|x^5+y^5$, by considering $(x+y)^5$ and using binomial expansion. But I am not sure how to show that $25|x^5+y^5$.
More generally, if p is a prime and $p>2$, how do I prove that if $p|x^p+y^p$ then $p^2|x^p+y^p$?
| Assuming $x,y\in \mathbb{Z}$, if we have $5\mid x^5 + y^5$ then by binomial expansion it follows that $5\mid (x+y)^5$. But by prime factorization, it is obvious that $5^5\mid (x+y)^5$ and hence $25\mid (x+y)^5$. To see that $25\mid x^5 + y^5$, it is enough to show that $5x^4y + 10x^3y^2 + 10x^2y^3 +5xy^4$ is divisible ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2494388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find roots of $x^4 -6x^3 + 12x^2 - 12 x + 4 = 0$ the original equation is:
$$(x^2 + 2)^2 -6x(x^2+2) + 8x^2=0.$$
cannot see how to go solving this. I tried following way to factorise:
$$(x^2+2)(x^2-6x+2) + 8x^2 = 0.$$
But this has no help to solve.
Thank you people, but I need the thinking process, not the answer.
| One way to find the factorizations others have given is to complete the square. If you recognize $y^2-6y+9=(y-3)^2$ you can do $(x^2+2)^2-6(x^2+2)+8x^2=((x^2+2)-3x)^2-x^2=(x^2+2-4x)(x^2+2-2x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2494962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Finding the Inverse of a function with natural logs The function is
$$f(x)= \frac{7 e^{x} +2}{5 e^{x} - 6}$$
and I am suppose to find the inverse, so I switched the $x$'s and $y$'s.
I know I am suppose to multiple the denominator out and I end up with $5 \, e^{y} \, x - 6 x = 7 \, e^{y} + 2$. From there I do not know... | Consider:
$$f(x) = \frac{7 \, e^{x} + 2}{5 \, e^{x} - 6}$$
for which
\begin{align}
y &= \frac{7 \, e^{x} + 2}{5 \, e^{x} - 6} \\
5 y \, e^{x} - 6 y &= 7 \, e^{x} + 2 \\
(5 y - 7) \, e^{x} &= 6 y + 2 \\
e^{x} &= \frac{6 y + 2}{5y - 7} \\
x &= \ln\left( \frac{6 y + 2}{5y - 7} \right)
\end{align}
It can now be stated that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that if $n \geq 2$ is even, then: $\frac {2n}{3} \leq \sum_{k=1}^{n} \frac{p(k)}{k} \leq \frac {2(n+1)}{3}$ Show that if $n \geq 2$ is even, then: $$\frac {2n}{3} \leq \sum_{k=1}^{n} \frac{p(k)}{k} \leq \frac {2(n+1)}{3}$$ where $p(k)$ is the greatest odd integer that divides $k$.
I think I'm almost done but cann... | As you found, we have
$$\frac{p(k)}{k} = 2^{-v_2(k)}.$$
Thus we can compute the sum if we know how many $k \leqslant n$ have $v_2(k) = m$ for every $m$. That's the number of multiples of $2^m$ not exceeding $n$ that are not multiples of $2^{m+1}$. The number of multiples of $r$ not exceeding $x$ is $\bigl\lfloor \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2498494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to find the Laplace Transform of $\frac{1-\cos(t)}{ t}$? What I tried is to find the transform of $f(t) = \frac{1-\cos(t)}{t}$ with $$\int_{s}^{\infty }\frac{1}{u} du + \int_{s}^{\infty }\frac{u}{u^2+1}du$$
$\int_{s}^{\infty }\frac{1}{u} du = \ln(\infty)-\ln(s)$. Is it a valid integral if I get an infinite value as... | Define the Laplace transform,
$$f(s) = \int_{0}^{\infty} e^{- s t} f(t) \, dt,$$
with the short notation $f(t) \doteqdot f(s)$.
The long method:
\begin{align}
\frac{1 - \cos(a t)}{t} &\doteqdot \int_{0}^{\infty} e^{- s t} \, \frac{1 - \cos(a t)}{t} \, dt \\
&\doteqdot \int_{s}^{\infty} \int_{0}^{\infty} e^{- u t} \, (1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solving Laplace's equation for PDE problem: Contradiction Between Fourier Analysis and Instructor's Solution? Confusion with How to Proceed. I am told to find a general solution for the following PDE (partial differential equation):
$\dfrac{\partial^2{u}}{\partial{x}^2} + \dfrac{\partial^2{u}}{\partial{y}^2} = 0, u(x,... | Before the "having problem zone" $u(x,y)$ is
$$u(x,y) = \sum_{n=1}^{\infty} B_{n} \, \sinh\left(\frac{n \pi \, x}{2}\right) \, \sin\left(\frac{n \pi \, y}{2}\right).$$
Now let
$$C_{n} = B_{n} \, \sinh\left(\frac{3 n \pi }{2}\right)$$
to obtain
$$u(3,y) = f(y) = \sum_{n=1}^{\infty} C_{n} \sin\left(\frac{n \pi \, y}{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove That $ \sqrt{\frac xy + \frac yz}+ \sqrt{\frac yz + \frac zx}+ \sqrt{\frac zx + \frac xy}\ge 3 $ for x, y, z $\gt 0$, prove that: $$ \sqrt{\frac xy + \frac yz}+ \sqrt{\frac yz + \frac zx}+ \sqrt{\frac zx + \frac xy}\ge 3 $$
| By AM-GM inequality we have
$$
\begin{align*}
\frac{x}{y}+\frac{y}{z} &\ge 2\sqrt{\frac{x}{z}}\\
\frac{y}{z}+\frac{z}{x} &\ge 2\sqrt{\frac{y}{x}}\\
\frac{z}{x}+\frac{x}{y} &\ge 2\sqrt{\frac{x}{z}}.
\end{align*}
$$
Then again by AM-GM inequality we have
$$
\begin{align*}
\frac{1}{3}\left(\sqrt{\frac{x}{y}+\frac{y}{z}}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $2\int_{0}^{\pi\over 2}\ln^2(\tan^2{x})\mathrm dx$ How to show that $(1)$
$$2\int_{0}^{\pi\over 2}\ln^2(\tan^2{x})\mathrm dx=\pi^3?\tag1$$
$u=\tan^2{x}$ then $\mathrm dx={\mathrm du\over \sqrt{u}(u-1)}$
$$\int_{0}^{\infty}\ln^2{u}{\mathrm du\over \sqrt{u}(u-1)}\tag2$$
$$-\sum_{n=0}^{\infty}\int_{0}^{\infty}u... | Setting $I = \int^{\pi/2}_0 \ln^2 (\tan^2 x) \, dx$, after letting $u = \tan x$ the integral becomes
$$I = 4 \int^\infty_0 \frac{\ln^2 u}{1 + u^2} \, du.$$
Writing this integral as
$$I = 4 \int^1_0 \frac{\ln^2 u}{1 + u^2} \, du + 4 \int^\infty_1 \frac{\ln^2 u}{1 + u^2} \, du,$$
if, in the second of these integrals we s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Conditional Expected Value of a Joint Probability Density Function I was wondering if anyone could help me with the following question specifically:
The continuous random variables $X_1$ and $X_2$ have the following joint probability density function: $$f(x_1,x_2)=\frac2{27}$$ over $0< x_1< 3$ and $0< x_2< 9-3x_1.$
Fin... | From the definition of conditional expectation,
\begin{equation}
\mathbb{E}(X_{1}|X_{2} = x_{2}) = \int_{-\infty}^{\infty}x_{1}f_{X_{1}|X_{2}}(x_{1}|x_{2})\, dx_{1}.
\end{equation}
Furthermore,
\begin{equation}
f_{X_{1}|X_{2}}(x_{1}|x_{2}) = \dfrac{f_{X_{1}, X_{2}}(x_{1}, x_{2})}{f_{X_{2}}(x_{2})},
\end{equation}
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Express the rectangles area as a function of $x$.
Translation: 3214 C A rectangle is drawn inside a semicircle as the figure shows.
FIG (See picture)
a) Express the rectangles area as a function of $v$.
b) Express the rectangles area as a function of $x$.
c) Determine the largest value of the rectangles area.
My... | Let $y$ be the side adjacent to the angle $v$. Then, the area of the rectangle would be equal to $x\cdot 2y$.
a)
$$
\sin{v}=\frac{x}{12}\implies x=12\cdot \sin{v}\\
\cos{v}=\frac{y}{12}\implies y=12\cdot \cos{v}\
$$
Therefore, the area of the rectangle expressed as a function of $v$ is $A(v)=2\cdot 12^2\cdot \sin{v} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Sum of the multiplication of two geometric progressions I want to calculate the convolution of two geometric progressions by the convolution definition:
$$2^n*3^n=\sum_{k=0}^{n}f_1[k]f_2[n-k]=\sum_{k=0}^{n}2^k3^{n-k}=3^n\sum_{k=0}^n 2^k 3^{-k}=3^n\sum_{k=0}^n \bigg(\frac{2}{3}\bigg)^k=3^n\frac{1-\big(\frac{2}{3}\big)^n... | Use
\begin{eqnarray*}
\sum_{k=0}^{n} r^k = \frac{1-r^{n+1}}{1-r}
\end{eqnarray*}
with $r=\frac{2}{3}$. So we have
\begin{eqnarray*}
2^n*3^n=\sum_{k=0}^{n}f_1[k]f_2[n-k]=\sum_{k=0}^{n}2^k3^{n-k}=3^n\sum_{k=0}^n 2^k 3^{-k} =3^n \frac{1-(\frac{2}{3})^{n+1}}{1-\frac{2}{3}} =3^{n+1}-2^{n+1}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Let $f(x) =|x|+|x+1|+...+|x+99|$ then find: $f'\left(\frac{-9}{2}\right)=?$
Let $f(x) =|x|+|x+1|+...+|x+99|$ then find :
$$f'\left(\frac{-9}{2}\right)=?$$
My Try :
$$\lim_{x\to \frac{-9}{2}}\dfrac{f(x)-f\left(\frac{-9}{2}\right)}{\left(x+\frac{9}{2}\right)}
=
\lim_{x\to \frac{-9}{2}}\dfrac{(90x+95(52)-10)-f\left(\fr... | we have $f(x)=\left| x\right| +\left| x+1\right| +\left| x+2\right| +\left| x+3\right| +\left| x+4\right|+\ldots +\left| x+99\right|$
In the interval $\left[-4.6,-4.4\right]$ the function is
$f(x)=-x-x-1-x-2-x-3-x-4+(x+5)+(x+6)+(x+7)+(x+8)+(x+9) +\ldots+x+99$
the first $10$ terms containing $-x$ and $x$ simplify to $25... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $x$ is divisible by $4$, then $x=a^2-b^2$
If $x\in Z$ and divisible by $4$, then there are existing $a,b \in Z$ , so: $x=a^2-b^2$
I had been trying to figure out which way this should be solved.
First I found out that if $x=20$, then $a^2 = 6^2$ and $b^2 = 4^2$ (of course $20=36-16$ ).
now I'm not sure... | As $a+b\pm(a-b)$ are even,
$a+b, a-b$ have the same parity
Case$\#1:$ If $a+b$ is even, so will be $a-b\implies(a+b)(a-b)$ must be divisible by $4$
Let $(a+b)(a-b)=4m\iff\dfrac{a+b}2\cdot\dfrac{a-b}2=m$
If $m=pq,$ $\dfrac{a+b}2=p,\dfrac{a-b}2=q$
$\implies a=p+q, b=p-q$
Trivially choose $p$ or $q=1$
Case$\#2:$ If $a+b$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sum\limits_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)}=3\zeta(3)$, where, for every positive $n$, $H_n=\sum\limits_{k=1}^n\frac1k$ The problem I was considering about is the evaluation of the following series:
\begin{align*}
\sum_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)}
\end{align*}
The attempt I could make was to... | using the identity $$ \sum_{n=1}^{\infty}x^n \left(H_n^2-H_n^{(2)}\right)=\frac{\ln^2(1-x)}{1-x}$$
divide both sides by $x$ then integrate w.r.t $x$ from $x=0$ to $y$ , we get
$$ \sum_{n=1}^{\infty}\frac{y^n}{n}\left(H_n^2-H_n^{(2)}\right)=\int_0^y\frac{\ln^2(1-x)}{1-x}\ dx-\frac13 {\ln^3(1-y)}$$
integrate both sides ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How to solve $e^{2x - 1} = -x^2 + 3x - 0.25$ I know the answer is $x=0.5$, but I only know it because my calculator showed me the intersection point of the two graphs. What I want to know is how to get to this result algebraically. I've tried to use the Lambert-W function, but I ended up with $$W((1-2x)/(-x^2+3x-0.25))... | For
$$e^{2x-1} + x^2 = 3 x - \frac{1}{4}$$
it can be seen that
\begin{align}
e^{2x-1} + x^2 &= 3 x - \frac{1}{4} \\
e^{2x -1} + x^{2} - x + \frac{1}{4} &= 2 x = 2 \left(x - \frac{1}{2}\right) + 1 \\
e^{2 x -1} + \left(x - \frac{1}{2}\right)^{2} - 2 \left( x - \frac{1}{2} \right) + 1 &= 2 \\
e^{2 x - 1} + \left( \left(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving $(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3$ for all $n\in\mathbb N$ and all $a\ge -1.$ I was asked to prove the the following without induction. Could someone please verify whether my proof is right? Thank you in advance.
For any real number $a\ge -1$ and every natural number n, the statem... | proof-verification:
Proof. From the binomial theorem we can see that,
$$
(1+a)^n\ge 1+na+\frac{n(n-1)}{2}a^{2}+\frac{n(n-1)(n-2)}{6}a^3,
$$
becomes ("is equivalent to"),
$$
\sum^{n}_{k=0}\binom{n}{k}a^{k}\ge\sum^{n}_{k=0}\binom{n}{k}a^{k}-\sum^{n}_{k=4}\binom{n}{k}a^{k}.
$$
Also, if $a\lt b$ (I would use o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to show this function is a metric: $d(x,y) = \frac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}$? Consider $( \mathbb{R}, d )$ where $$d(x,y) = \frac{2|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}.$$
The problem arises in showing triangle inequality. I can deal with the numerator only with the help of triangle inequality for the mod in ... | Let $x\geq y\geq z$.
Hence, $$\frac{|x-z|}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{|x-y|}{\sqrt{1+x^2}+\sqrt{1+y^2}}$$ it's
$$\frac{x-z}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+y^2}}$$ or
$$\frac{y-z}{\sqrt{1+x^2}+\sqrt{1+z^2}}\geq\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+y^2}}-\frac{x-y}{\sqrt{1+x^2}+\sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2510186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove $ \frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}< \frac{1}{40} $ Prove:
$$
\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}< \frac{1}{40}
$$
i have tried to write $1/40$ as $(1/40^{1/2007})^{2007}$ and prove $(1... | Let
$$
A=\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2007}{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}
$$
$$
B=\frac{2 \cdot 4 \cdot 6 \cdot \dots \cdot 2008}{1\cdot 3 \cdot 5 \cdot 7 \cdot \dots \cdot 2009}
$$
Then $$A<B$$
Then $$A^2<AB=\frac1{2009}$$
Then $$A<\frac1{\sqrt{2009}}<\frac1{\sqrt{1600}}=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Prove $ ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} )^{1/4} + ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} )^{1/4}\ge 68^{1/4}$
Let $0<\theta<\pi/2$. Prove that $$\left ( \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta} \right )^{1/4}+\left ( \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta} \r... | By Minkowski (see here: https://en.wikipedia.org/wiki/Minkowski_inequality) we obtain
$$\sqrt[4]{ \frac{\sin ^{2}\theta}{2}+\frac{2}{\cos ^{2}\theta}}+\sqrt[4]{ \frac{\cos ^{2}\theta}{2}+\frac{2}{\sin ^{2}\theta}}\geq\sqrt[4]{\frac{1}{2}(\sqrt{\sin\theta}+\sqrt{\cos\theta})^4+2\left(\frac{1}{\sqrt{\sin\theta}}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2513277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Show that the lines joining the origin to the point of intersection of the line Show that the lines joining the origin to the point of intersection of the line $x+y=1$ with the curve $4x^2+4y^2+4x-2y-5=0$ are at right angles to each other.
How do I approach this? Please help.
| Solving of the system gives
$$4x^2+4(1-x)^2+4x-2(1-x)-5=0$$ or
$$8x^2-2x-3=0,$$ which gives the following points: $\left(\frac{3}{4},\frac{1}{4}\right)$ and $\left(-\frac{1}{2},\frac{3}{2}\right)$.
Now, $$m_1=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$$ and
$$m_2=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3$$ and since $m_1m_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Real root of $f(x) = 1+2x+3x^2 +4x^3$ Consider the polynomial $f(x) = 1+2x+3x^2 +4x^3$. Let $s$ be the sum of all distinct real roots of $f(x)$ and let $t = |s|$.
The real number $s$ lies in the interval
A) $\ \ (-\frac{1}{4},0)$
B) $ \ \ (−11,−\frac{3}{4}) $
C) $\ \ (−\frac{3}{4},−\frac{1}{2})$
D) $\ \ (0,\frac{1}{4}... | f(x) is a polynomial function and so it is continuous and differentiable through out. We have by intermediate value theorem, if f is continous in $[a,b]$ and if $f(a)<0, f(b)>0$ then there exists $c \in (a,b)$ such that f(c) = 0
in above options, consider (C):
$f(-1/2) = positve$
$f(-3/4) = negative$
So there should ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prime factorization of integers Find the prime factorization of the following integers:
$e) 2^{30} -1$
Click here for the solutions
I used the formulas:
$a^2-1^2=(a-1)(a+1)$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$a^3-b^3=(a-b)(a^2+ab+b^2)$
And I became:
$2^{30} -1$
$=(2^{15}-1)(2^{15}+1)$
$=(2^5-1)(2^{10}+2^5+1)(2^5+1)(2^{10}-2... | One good method to obtain some factors of numbers of the form $a^n-1$ is to
look at the factorization of $x^{n}-1$ into cyclotomic polynomials:
$$
x^n - 1 = \prod_{d\mid n}\Phi_d(x)
$$
$
x^6-1
= \Phi_{1} \Phi_{2} \Phi_{3} \Phi_{6}
=(x - 1) (x + 1) (x^2 + x + 1) (x^2 - x + 1)
$.
Therefore, $10^6-1=9 \cdot 11 \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
The limit of a sum: $\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)$
Evaluate the following limit:
$$
\lim_{n\to \infty} \sum_{k=1}^{n} \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
$$
I haven't ever taken the limit of the sum... Where do I start?
Do I start taking the sum?
| With Riemann sums:
We have
$$
\sum_{k=1}^n \left(\frac{k}{n^2}-\frac{k^2}{n^3}\right)
= \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}-\left(\frac{k}{n}\right)^2\right)
= \frac{1}{n}\sum_{k=1}^n \frac{k}{n}\left(1-\frac{k}{n}\right)
$$
which is a Riemann sum for $f\colon[0,1]\to\mathbb{R}$ defined by $f(x)=x(1-x)$. Therefor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Ordinary Generating Function for the number of solution :$ x_1 + x_2 + \cdot\cdot\cdot + x_k = n$ Let $a_n$ be the number of solutions of the equation:
$$x_1 + x_2 + \cdot\cdot\cdot + x_k = n$$
where $x_i$ is the positive odd integer.
Hereto I would like to find the ordinary generating function for the sequence $a_n$ ... | Let $X$ be the set of positive odd numbers.
For any formal power series $f(t) = \sum_{j=0}^\infty f_k t^k$, let $[t^k]f(t)$ be the coefficient $f_k$.
When one expand following product into a series of $t$.
$$\left(\sum_{x_1\in X} t^{x_1}\right)\cdots\left(\sum_{x_k\in X} t^{x_k}\right)
= \sum_{x_1,\ldots,x_k \in X} t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Coefficient of $x^{n-2}$ in $(x-1)(x-2)(x-3)\dotsm(x-n)$ Question
Find the coefficient of $x^{n-2}$ in the expression $$(x-1)(x-2)(x-3)\dots(x-n)~~.$$
My approach
The coefficient of $x^n$ is $1$. The coefficient of $x^{n-1}$ is $- \frac{n(n+1)}{2}$
But I cannot proceed from here.
I would appreciate any help.
| For $~x^{n-1}~$ the coefficient would be sum of the roots that is $~1+2+3+4+5+\cdots+n~$ for $~x^{n-2}~$ coefficient would be sum of roots taken two at a time that is $~1.2+2.3+3.4+4.5+\cdots+(n-1)n~$ so let it be $~S~$ them $~S~$ can be find using steps below
$$(1+2+3+4+5+\cdots+n)^2=(1^2+2^2+3^3+\cdots+n^2)+2(1.2+2.3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2527894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
Help with proving the surjectivity of a function satisfying $ f ( x + y ) \big( f ( x ) + f ( y ) \big) = f ( x y ) $ I have the following functional equation for injective $ f : \mathbb R ^ * \to \mathbb R ^ * $
$$ x + y \ne 0 \implies f ( x + y ) \big( f ( x ) + f ( y ) \big) = f ( x y ) $$
Found $ f ( 2 ) = \frac 1 ... | You can show that the only functions satisfying (I will assume that $ x , y \ne 0 $ everywhere I use them)
$$ x + y \ne 0 \Longrightarrow f ( x + y ) \big( f ( x ) + f ( y ) \big) = f ( x y ) \tag 0 \label 0 $$
where $ f : \mathbb R ^ * \to \mathbb R ^ * $, are $ f ( x ) = \frac 1 x $ and $ f ( x ) = \frac 1 2 $. It's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of odd terms of a binomial expansion: $\sum\limits_{k \text{ odd}} {n\choose k} a^k b^{n-k}$ Is it possible to find a closed form expression for the sum
$$\sum_{k \text{ odd}} {n\choose k} a^k b^{n-k}$$
in terms of $a$ and $b$ ?
| I found a rather interesting alternate method to solve this problem (from a problem in communication theory). Please share your thoughts.
Let
$
A =
\left[ {\begin{array}{cc}
a & b \\
b & a \\
\end{array} } \right]
$
Now, $$[A^n]_{1,1} = \sum_{k \text{ even}} {n \choose k} a^k b^{n-k}$$
and $$[A^n]_{1,2} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Integration by parts of $x^5\ln(x)$ I have to integrate the function: $x^5\ln(x)$
My Attempt
$$\int(x^5\ln(x))dx$$
$u=\ln(x)$ and $du=\frac{1}{x}dx$
$dv=x^5dx$ and
$v=\frac{x^6}{6}$
Using this I can then integrate the function using the method
$$uv-\int v du$$
Which then substituting I get:
$$\frac{x^6\ln(x)}{6}-\int\f... | Setting $u = x^{5}$, $v' = \ln x$ and noting that $$\int \ln x = x \ln x - x + C$$ yields
\begin{align}
I &= \int x^{5} \ln x dx \\
&= x^{6} \ln x - x^{6} - 5 \int (x^{5} \ln x - x^{5}) dx \\
&= x^{6} \ln x - x^{6} - 5 I + \frac{5 x^{6}}{6} + C \\
&= x^{6} \ln x - \frac{x^{6}}{6} - 5 I + C \\
\implies 6I &= x^{6} \ln x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Calculate the length of the portion of the hypocycloid $x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$ Calculate the length of the portion of the hypocycloid $x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$ in the first quadrant from the point $(\frac{1}{8},\frac{3\sqrt{3}}{8})$ to the point (1.0)
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$
$y^{\frac{... | Formula for arc length is
$$L=\int_{\frac18}^1\sqrt{1+\left(y'\right)^2}dx$$
It is easy to find the derivative:
$$y'=-\frac{\sqrt{1-x^{\frac23}}}{x^\frac13}$$
The length is now:
$$\begin{align}L&=\int_{\frac18}^1\sqrt{1+\left(-\frac{\sqrt{1-x^{\frac23}}}{x^\frac13}\right)^2}dx\\&=\int_{\frac18}^1x^{-\frac13}dx\\&=\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find limit $\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$
Find the following limit without using L'Hopital's rule:
$$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$$
My attempt:
$$x-1=u\implies x=u+1$$
So we have:
$$\lim_{u\to 0} \dfrac{4(u+1)^2\sqrt{u+4}-17(u+1)+9}{(u)^2}$$
What now?
| $$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}=\lim_{x\to 1} \dfrac{16x^4(x+3)-(17x-9)^2}{(x-1)^2(4x^2\sqrt{x+3}+17x-9)}=$$
$$=\lim_{x\to 1} \dfrac{16x^3+80x^2+144x-81}{4x^2\sqrt{x+3}+17x-9}=\frac{159}{16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\lim_{n\to\infty} 2^n \sqrt{2-x_n} = \pi$ Given the sequence $x_1 = 0$, $x_{n+1} = \sqrt{2+x_n}$, proove:
$$\lim_{n\to\infty} 2^n \sqrt{2-x_n} = \pi$$
I have used the relations
$$2\cos({\frac x 2}) = \sqrt{2 + 2\cos(x)}$$
and
$$2\sin({\frac x 2}) = \sqrt{2 - 2\cos(x)}$$
Observe:
$x_1 = 0 = 2\cos(\alpha) \iff \al... | $\cos(x) = \cos(2\cdot \frac{x}{2})= \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = -1+ 2\cos^2(\frac{x}{2})$ and hence $\cos(\frac{x}{2})= \sqrt{\frac{1}{2}+\frac{1}{2}\cos(x)} = \frac{1}{2}\sqrt{2+2\cos(x)}$. That is the inductive step for the calculation.
Geometrically, you can check that $2^n \sqrt{2-x_n}$ is half the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2534794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Lie group representation, exponential, and $\theta$-periodicty
SU(2)
I know we can view the group element in the SU(2) Lie group as
$$ g = \exp\left(\theta\sum_{k=1}^{3} i t_k \frac{\sigma_k}{2}\right) $$
where $(t_1,t_2,t_3)$ forms a unit vector [effectively pointing in some direction on a unit 2-sphere $S^2$], an... | At least for SU(3), the exponential of arbitrary Lie-algebra elements in the fundamental representation is a bit more complicated, and, as a consequence of the Cayley-Hamilton theory, it also has a term bilinear in the Lie algebra elements, unlike the simple SU(2) expression you may be hankering for. See Curtright & Za... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2538747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Compute $\int_0^x \sqrt{s(2-s)}\, ds $ Could anyone shed some light on how to compute the following two integrals?
$$ \int_0^x \sqrt{s(2-s)}\, ds \ \ \textrm{ for $0<x<1$}, $$
and
$$\int_x^0 \sqrt{s(s-2)}\, ds \ \ \textrm{ for $-1<x<0$}. $$
| Solution without Finding Antiderivative: Observe
\begin{align}
\int^x_0\sqrt{2s-s^2}\ ds = \int^x_0\sqrt{1-(1-s)^2}\ ds = \int^{1}_{1-x} \sqrt{1-s^2}\ ds
\end{align}
which is just the area underneath the unit semi-circle from $1-x$ to $1$. But that area is given by sector of the disk minus a right triangle, i.e. we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Factorization of polynomial $2x^4 - 2x^2 + 4$ I'm struggling with the polynomial factorization of the polynomial function $2x^4 - 2x^2 + 4$ in fields $\mathbb{R}, \mathbb{Q}, \mathbb{Z}, \mathbb{Z}_5, \mathbb{Z}_7$.
So far I have determined that it cannot be factored into something*function of degree 1 since it has no ... | You can get disregard, for a while, the common factor $2$ and look for the factorization of $x^4-x^2+2$.
First let's look at $\mathbb{Q}$ and $\mathbb{R}$ that are related. Since the polynomial $y^2-y+2$ has no real roots, we can complete the square in a different way than usual:
$$
x^4-x^2+2=x^4+2\sqrt{2}x^2+2-(1+2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I calculate the limit $\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$ without L'Hospital's rule? I have a problem with calculation of the limit:
$$\lim\limits_{x \to 7} \frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}$$
Is there a way to calculate it? How can I do it?
| With Taylor's formula at order $1$:
Set $x=7+h$. You obtain
\begin{align}
\frac{\sqrt{x+2} - \sqrt[3]{x+20}}{\sqrt[4]{x+9} - 2}&=\frac{\sqrt{9+h} - \sqrt[3]{27+h}}{\sqrt[4]{16+h} - 2}=\frac{3\sqrt{1+\frac h9} - 3\sqrt[3]{1+\frac h{27}}}{2\sqrt[4]{1+\frac h{16}} - 2}\\
&=\frac{\bigl(3+\frac h6+o(h)\bigr)-\bigl(3+\frac h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 1
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Error in $\int\limits_0^{\infty}dx\,\frac {\log^2 x}{a^2+x^2}$ This is my work for solving the improper integral$$I=\int\limits_0^{\infty}dx\,\frac {\log^2x}{x^2+a^2}$$I feel like I did everything write, but when I substitute values into $a$, it doesn’t match up with Wolfram Alpha.
First substitute $x=\frac {a^2}u$ so... | It appears that you wrote
$$
\log\left(\frac{a^2}u\right)^2=2\log(a)^2-\log(u)^2
$$
when it should be
$$
4\log(a)^2-4\log(a)\log(u)+\log(u)^2
$$
Using the Dirichlet beta function as evaluated in this answer
$$
\begin{align}
&\int_0^\infty\frac{\log(x)^2}{a^2+x^2}\,\mathrm{d}x\tag1\\
&=\frac1a\int_0^\infty\frac{\log(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
prove that $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ convergies using Cauchy criterion Prove using Cauchy criterion that the sum of the series $$\sum_{n=1}^\infty \frac{(-1)^n}{n}$$
converges.
The cauchy criterion says that $\sum_{i=0}^\infty a_{i}$ converges if and only if for every $\epsilon>0$ exists $N$ such that $\fora... | Hint:
$$\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \frac{1}{n+4} \pm \ldots \\= \frac{1}{n+1} - \left(\frac{1}{n+2} - \frac{1}{n+3}\right) - \left(\frac{1}{n+4} - \frac{1}{n+5}\right) - \ldots \\\leqslant \frac{1}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $x$ such that $2^x+3^x-4^x+6^x-9^x=1$ The question:
Find values of $x$ such that $2^x+3^x-4^x+6^x-9^x=1$, $\forall x \in \mathbb R$.
Notice the numbers $4$, $6$ and $9$ can be expressed as powers of $2$ and/or $3$. Hence let $a = 2^x$ and $b=3^x$.
\begin{align}
1 & = 2^x+3^x-4^x+6^x-9^x \\
& = 2^x + 3^x - (2^2)^... | A sum of squares equals zero if and only if each of the squares equals zero. So you get $a=b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Find locus of point vertex of right angle triangle whose hypotenuse is chord of circle. Let $AB$ be a variable chord of length 5 to the circle $$x^2+y^2=\frac{25}{2}$$. A triangle $ABC$ is constructed such that $BC=4$ and $CA=3$. If the locus of C is $x^2+y^2=a$ find all possible values of $a$
I tried using the paramet... | Consider the particular position of $A\left(-\frac52,\frac52\right);\;B\left(\frac52,\frac52\right)$
The line $AC_2$ has equation $y-\frac{5}{2}=\frac{4}{3} \left(x+\frac{5}{2}\right)$
And the circle with center $A$ and radius $3$ has equation $\left(x+\frac{5}{2}\right)^2+\left(y-\frac{5}{2}\right)^2=9$
The intersecti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{x \to \infty} x(\ln(1+x) - \ln(x))$ without l'Hopital I solved the limit
$$\lim_{x \to \infty} x(\ln(1+x) - \ln(x))$$
by writing it as $\lim_{x \to \infty} \frac{\ln(\frac{1+x}{x})}{\frac{1}{x}}$ and applying l'Hopital rule but is it possible to solve it without using l'Hopital rule?
| $$\log(x+1)-\log(x)=\log\left(\frac{x+1}{x}\right)=\log\left(1+\frac{1}{x}\right)$$
We know that $\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}+O(t^4)$ (for $t \in (-1,1)$); substituting $t=\frac{1}{x}$ we get: $\log\left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2} \frac{1}{x^2}+\frac{1}{3} \frac{1}{x^3}+O\left(\frac{1}{x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in R$ will be real if $a(a+7b+49c)+c(a-b+c)<0$
Prove that the roots of $ax^2+bx+c=0$, $a\neq 0$ and $a,b,c\in \mathbb{R}$ will be real if $$a(a+7b+49c)+c(a-b+c)<0$$
My Attempt:
Given
\begin{align}
a(a+7b+49c)+c(a-b+c) &< 0 \\
49a \left( \dfrac {a}{49} + ... | a) Suppose that $a$ and $c$ have opposite sign. Then $b^2-4ac\geq 0$, and we are done.
b) Suppose that $f(1/7)$ and $f(-1)$ have opposite sign. Then by the intermediate value theorem, $ax^2+bx+c=0$ have a real root, hence two real roots, and we are done.
c) Suppose now that $a,c$ have the same sign, say $e_1$, and that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the limit of a series of fractions starting with $\frac{1}{2}, \, \frac{1/2}{3/4}, \, \frac{\frac{1}{2}/\frac{3}{4}}{\frac{5}{6}/\frac{7}{8}}$ Problem
Let $a_{0}(n) = \frac{2n-1}{2n}$ and $a_{k+1}(n) = \frac{a_{k}(n)}{a_{k}(n+2^k)}$ for $k \geq 0.$
The first several terms in the series $a_k(1)$ for $k \geq 0$ are:... | Only a remark, not a complete answer.
For $q\in \mathbb{N}$, put $s_2(q)=$ the sum of the digits of the base two expansion of $q$, ie $s_2(3)=2$, $s_2(4)=1$, etc. The following formula can be proven by induction :
$$a_m(n)=\prod_{0\leq q<2^m}\left(1-\frac{1}{2n+2q}\right)^{(-1)^{s_2(q)}}$$
For $m=0$, we have only $q=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2555815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Number of ways to pick 12 elements from 3 different kinds of different no of elements Question:
A florist has 5 aspidistras, 6 buttercups, and 7 chrysanthemums. How many different kinds of bouquets of a dozen flowers (it is not required to use all types of flowers) can she make from these?
My Solution:
The solution I c... | Let $a$ be the number of aspidistras, $b$ be the number of buttercups, and $c$ be the number of chrysanthemums. The number of ways of forming a bouquet of a dozen flowers is the number of solutions of the equation
$$a + b + c = 12 \tag{1}$$
in the nonnegative integers subject to the constraints $a \leq 5$, $b \leq 6$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2560842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Set of values of $x$
If the inequality
$$(1-a^2)x^2+(2a-3)x+1<0$$
is true for all values of $a$ then the set of values of $x$ is?
I took two cases (the parabola opens upwards and the parabola opens downwards). For the first case the value of $x$ is all real except the interval containing roots. For the second cas... | $$\left(1-a^2\right)x^2+(2 a-3) x+1<0$$
Expand and collect wrt $a$
$$-a^2 x^2+2 a x+\left(x^2-3 x+1\right)<0$$
Multiply both sides by $-1$ and change the $<$ into $>$
$$a^2x^2-2ax-\left(x^2-3 x+1\right)>0$$
This is true for any $a$ if and only if the discriminant $\Delta$ of the polynomial
$P(a)=a^2x^2-2ax-\left(x^2-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find : $\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$ with l'Hôpital's rules I want to calculate :
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$$
with l'Hôpital's rules.
I get
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) = \lim\l... | You can also use Taylor series:
$$\sin(x)=x-\frac{x^3}{6}+O(x^4)$$
So your function:
$$\frac{1}{\sin(x)}-\frac{1}{x}=\frac{x-\sin(x)}{x\sin(x)}=\frac{\frac{x^3}{6}+O(x^4)}{x^2-\frac{x^4}{6}+O(x^5)}=\frac{\frac{x}{6}+O(x^2)}{1-\frac{x^2}{6}+O(x^3)} \to \frac{0}{1}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
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Convergence of $\sum_{n=0}^{+ \infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$
Does $$\sum_{n=0}^{+\infty} \frac{n^3 - 5n^2 + \pi}{2n^5 - \sqrt{3}}$$
converge?
My attempt:
$$\forall n \in \mathbb{N} \setminus\{0\}:\frac{n^3-5n^2 + \pi }{2n^5 -\sqrt{3}} = \frac{1}{n^2}\frac{1-5/n + \pi/n^2}{2- \sqrt{3}/n^5}$$
And bec... | A rational expression $\dfrac{P(n)}{Q(n)}$ is asymptotic to $\dfrac{p_p}{q_q}n^{p-q}$ where $p,q$ are the respective degrees and $p_p,q_q$ the leading coefficients. (You can check this by factoring out the powers $n^p$ and $n^q$ from the fraction.)
And as you know, the summation of $n^{p-q}$ converges for $p-q<-1$. The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Can someone explain this proof to me for $2n < 2^n - 1? \,\, n >= 3$ Prove $2n < 2^n - 1;n \geq 3$
Prove for $n = 3$
$2\cdot 3 < 2^6 - 1$
$6 < 7$
Prove for $n \mapsto n + 1$
$$2(n + 1) = 2n + 2$$
$$< 2^n - 1) + 2$$
$$= 2^n + 1$$
$$< 2^n + 2^n - 1$$
$$ = 2^{n + 1} - 1$$
Can someone explain this proof to me? I get... | The inductive step: assuming that $2n < 2^n - 1$, show that $2(n+1) < 2^{n+1} - 1$.
*
*$2(n+1) = 2n + 2\qquad$distribution
*$2n + 2 < (2^n -1) + 2\qquad$ by the inductive hypothesis that $2n < 2^n -1$. (Just add two to both sides of inductive hypothesis to see that this is true.)
*$(2^n -1) + 2 = 2^n -1 + 2 = 2^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
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Convergence and divergence of an infinite series The series is
$$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$
I just stuck over the nth term finding and once I get nth term than I can do diff... | Hint:
\begin{align}
\frac{1\cdot 3 \cdot 5 \cdot 7 \cdot 9}{2\cdot 4 \cdot 6 \cdot8\cdot10}.\frac{x^6}{12} &=\frac{(1\cdot 3 \cdot 5 \cdot 7 \cdot 9)(2\cdot 4 \cdot 6 \cdot8\cdot10)}{(2\cdot 4 \cdot 6 \cdot8\cdot10)^2}.\frac{x^6}{12} \\
&=\frac{10!}{2^{10}(1\cdot 2 \cdot 3 \cdot4\cdot 5)^2} \frac{x^6}{12}\\
&=\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Is the curve $ z = e^{i\theta}\left(\frac{7}{8} + \frac{1}{4} e^{6i\theta}\right) $ algebraic? Is this spirograph curve algebraic? I an only write it in polar coordinates:
$$ z = e^{i\theta}\left(\frac{7}{8} + \frac{1}{4} e^{6i\theta}\right) $$
and here is a picture. It is a six-sided rose-shaped curve, a hypotrochoi... | Given
$$ x=\frac{7}{8}\cos\theta+\frac{1}{4}\cos(7\theta),\qquad y=\frac{7}{8}\sin\theta+\frac{1}{4}\sin(7\theta) $$
we have $x^2+y^2=\frac{1}{64}\left(53+28\cos(6\theta)\right)$ and the given hypotrochoid is an algebraic curve since we may eliminate the $t$ variable in
$$ \left\{\begin{array}{ccc} x& =& \frac{7}{8} T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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Integral involving $\phi$
$$\int_{0}^{\pi/2}\arctan\left({2\over \cos^2{x}}\right)\mathrm dx=\pi\arctan\left({1\over \sqrt{\phi}}\right)\tag1$$
$\phi$ is the golden ratio
$2\sec^2{x}=2\tan^2{x}+2$
$u=\sec^2{x}$ then $\mathrm du=2{\tan{x}\over \cos^2{x}}\mathrm dx$
$$\int{\cos^2{x}\over \tan{x}}\arctan\left({2\over u... | $$\begin{eqnarray*}\int_{0}^{\pi/2}\arctan\left(\frac{2}{\cos^2\theta}\right)\,d\theta&=&\int_{0}^{\pi/2}\arctan\left(\frac{2}{\sin^2\theta}\right)\,d\theta\\&=&\int_{0}^{1}\frac{\arctan\frac{2}{u^2}}{\sqrt{1-u^2}}\,du\\&=&\int_{0}^{1}\frac{\frac{\pi}{2}-\arctan\frac{u^2}{2}}{\sqrt{1-u^2}}\,du\\&=&\frac{\pi^2}{4}-\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides I have been asked to prove:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
Which I can easily do by converting the LHS to index form, then squaring it and simplifying it down to get 2, which is equal to the RHS squared, hence proved.
H... | You can square both sides in a proof if you note the extraneous solutions are added.
Example $\sqrt{2 + \sqrt{3}} -\sqrt{2 - \sqrt{3}} = k$.
First $2 > \sqrt 3$ so $\sqrt{2 - \sqrt{3}} > 0$ so $k > 0$. !!!TAKE NOTE OF THAT!!!
$(\sqrt{2 + \sqrt{3}} -\sqrt{2 - \sqrt{3}})^2 = k^2$
$2 + \sqrt3 + 2-\sqrt 3 - 2(\sqrt{2 + \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 12,
"answer_id": 10
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Solving a congruence using primitive roots Suppose we know that $3$ is a primitive root of $17$.
How can that help us solving $7^x \equiv 6 \pmod {17}$?
| Use Discrete Logarithm wrt primitive root $3\pmod{17},$
$x$ind$_37\equiv$ind$_3(2\cdot3)\pmod{\phi(17)}\equiv$ind$_32+1$
Now $3^3\equiv10\pmod{17},3^5\equiv9\cdot10\equiv5\implies7\equiv5^{-1}\equiv3^{-5}\equiv3^{16-5}$
$2\equiv(-1)3\cdot5\pmod{17}\equiv3^{8+1+5}$ as $3$ is a primitive root, $3^{(17-1)/2}\equiv-1\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
| We have $\dfrac {1+i \sqrt 7}{2} = \sqrt 2 \left(\cos \theta + i \sin \theta \right)$ where $ \cos \theta = \dfrac{1}{2 \sqrt 2}$
We need to calculate $2 \Re \left(\dfrac {1+i \sqrt 7}{2} \right)^4 = 2 \times (\sqrt 2)^4 \times \cos 4\theta = 8 \cos 4 \theta$
We have $\cos 2 \theta = 2 \cos^2 \theta -1 = -\dfrac{3}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
Argue that $f$ is not uniformly continuous Let $f(x)=x^3$. We want to show that $f$ is not uniformly continuous.
Can someone please explain from $|f(x_n)-f(y_n)=\big|(n+\frac{1}{n})^3 - n^3)\big| = 3n+\displaystyle\frac{3}{n}$? How does $\big|(n+\frac{1}{n})^3-n^3\big| $ yield $3n+\displaystyle\frac{3}{n}$? ${}{}{}{}$... | You're missing the $\frac{1}{n^3}$ term. Indeed, $$\left(n+\frac{1}{n}\right)^3-n^3=n^3+3n^2\frac{1}{n}+3n\frac{1}{n^2}+\frac{1}{n^3}-n^3=3n+3\frac{1}{n}+\frac{1}{n^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$ Find the limits :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)$$
My Try :
$$\lim_{x\to 0} \left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\lim_{x\to 0}\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}$$
Now what do I do ?
| $$\left(\frac{1}{1-\cos x}-\frac{2}{x^2}\right)=\frac{x^2-2(1-\cos x)}{x^2(1-\cos x)}=\frac{x^2(1+\cos x)-2\sin^2 x}{x^2\sin^2 x}=$$
from here by Taylor series:
$\begin{cases}\cos x=1-\frac{x^2}{2}+o(x^2)\\\sin x=x-\frac{x^3}{6}+o(x^3)\implies \sin^2 x=x^2-\frac{x^4}{3}+o(x^4)\end{cases}$
thus:
$$=\frac{x^2(1+1-\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a base and dimension to a polynomial space under conditions I need to find a base and dimension for the polynomial space $V$ while the factor of $x$ is 58: $$V=\left \{ \right.p(x)\in P_{4}[x]: p(1)=p''(1)=0\left. \right \}$$
What I did is to create a general polynome: $p(x)=e+dx+cx^2+bx^3+ax^4$
Then I calcu... | Solution:
Let's write a general polynomial: $p(x)=ax^4+bx^3+cx^2+dx+e$
Then: $p''(x)=12ax^2+6bx+2c$
Under the condition $p''(1)=p(1)=0$: $$a+b+c+d+e=0$$ $$12a+6b+c=0$$
Let's express $a$ and $b$ using $c,d=58,e$: $$a=2/3c+58+e$$ $$b=-5/3c-116-2e$$
$$\Rightarrow p(x)=(2/3c+58+e)x^4+(-5/3c-116-2e)x^3+cx^2+58x+e$$
$$\Right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find the range of the function : $y=(3\sin 2x-4\cos 2x)^2-5$ find the range of the function :
$$y=(3\sin 2x-4\cos 2x)^2-5$$
My try :
$$y=9\sin^22x+16\cos^22x-24\sin 2x\cos 2x-5\\y=9+7\cos^22x-12\sin4x-5$$
now what do I do؟
| Hints:
Every real number $t$ can be expressed as $2x$ for some $x$, so the range of the function
$$f(x) = (3\sin 2x-4\cos 2x)^2-5$$
is the same as the range of
$$g(t)=(3\sin t-4\cos t)^2-5$$
But the expression
$$3\sin t-4\cos t$$
can be expressed as
$$5\left({\small{\frac{3}{5}}}\sin t-{\small{\frac{4}{5}\cos t}}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. I've updated my proof to be complete now, edited for proof-verification!
We know that for the partial sums with even an uneven terms, the following holds:
$S_{2N}=\sum_{n=1}^{2N} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\fra... | As you computed, it's the alternating harmonic series in a different order. In fact the first N terms are just the first N terms of the alternating harmonic series permuted (edit: with an extra term which decays like 1/N in case N is odd), so the sequence of partial sums is the same.
| {
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"source": "stackexchange",
"question_score": "4",
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How to I approach this problem? Trigonometric or Algebraic computations? The radicals as denominator is not all that encouraging too and I don't see any promising method to evaluate it.
Solve for $n$ in the equation below:
$$\frac{\sin\left(\frac{90^\circ}{2^n}\right)}{\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots\sq... | Use the half-angle formula:
$$\cos{x}=2\cos^2\frac{x}{2}-1$$
So
$$2+2\cos{x}=4\cos^2\frac{x}{2}$$
Then, for $0\leq x\leq\frac{\pi}{2}$,
$$2\cos\frac{x}{2}=\sqrt{2+2\cos x}$$
So
$$2\cos\frac{x}{2^{k+1}}=\sqrt{2+2\cos \frac{x}{2^k}}=\sqrt{2+\sqrt{2+2\cos \frac{x}{2^{k-1}}}}=\ldots$$
We have the denominator
$${\underbrace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ I am trying to find the derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ by only using the formula $\arctan\left(u\left(x\right)\right)'=\frac{u'\left(x\right)}{u\left(x\right)^2+1}$. I don't honestly understand this fo... | We have
\begin{eqnarray*}
\frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2}
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{d}{dx} \frac{1+x}{1-x} = \frac{ (1-x)--(1+x)}{(1-x)^2}= \frac{2 }{(1-x)^2}.
\end{eqnarray*}
Now we need to do a $3-$ fold c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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2nd solution of $\cos x \cos 2x\cos 3x= \frac 1 4 $
$\cos x \cos 2x\cos 3x= \dfrac 1 4 $
Attempt explained:
$(2\cos x \cos 3x)\cos 2x = \frac1 2 $
$(\cos 4x +\cos 2x )\cos 2x = \frac 1 2 \\\cos ^2y + \cos y (2\cos^2y- 1)= \frac1 2 \\ $
(Let, y = 2x)
$\implies 4\cos^3 y+2\cos^2y- 2\cos y-1=0$
I solved this equation us... | Remember that a cubic equation can have up to three real solutions. You did find that $\cos y = -\dfrac{1}{2}$ is one of the solutions, but you still need to find the other two.
Since $\cos y = -\dfrac{1}{2}$ is a root of $4\cos^3 y+2\cos^2y- 2\cos y-1=0$, we can factor the equation as:
\begin{align*}
4\cos^3 y+2\cos^2... | {
"language": "en",
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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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$\sin^2 x \cos^2 x + \sin x \cos x -1 = 0$
$\sin^2 x \cos^2 x + \sin x \cos x -1 = 0$
Attempt:
$\sin^2 2x + 2\sin 2x -4 = 0 \\ \implies \sin 2x= \sqrt 5 - 1$
Now, using the formula $\sin 2 x = \dfrac {2\tan x}{1+\tan^2x}$, I couldn't express the answer in terms of $\tan()$
The answer given is in terms of tan. How do... | Hint:
Observe the roots of $t^2+2t-4=0$ are $-1\pm\sqrt 5$ (no denominator), and their absolute values are greater than $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sqrt{s(s-a)(s-b)(s-c)}=A$ In 50 AD, the Heron of Alexandria came up with the well-known formula, that, given the three side lengths of a triangle (or even two and an angle, thanks to trigonometry) you can get the area of said triangle by using this formula:
$$
\text{if } s=\frac{a+b+c}{2},\\
\text{then} A=... | Consider a triangle $ABC$.
$[ABC]=\frac{ab}{2}\sin C$
$=\frac{ab}{2}\sqrt{1-\cos^2 C}$
$=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$
$=\sqrt{\frac{a^2b^2}{4}\left[1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\right]}$
$=\sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$
$=\sqrt{\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine the Eigenvalues and Eigen vectors of the operator $ \ A \ $ Consider the vector space of real polynomials of degree not greater than $ \ n \ $ given by $ \large \mathbb{R}_n[x]=\{\sum_{i=0}^{n} a_i x^i \ | a_i \in \mathbb{R}, \ \forall i \} \ $ .
Define an operator $ \ A \in End (\mathbb{R}_n[x] ) \ $ by $ \... | I like what you have so far.
The eigenvalues of an upper-triangle matrix are the values of the main diagonal
The first few eigenvectors
$\pmatrix{1&1&1&2\\&1&2&3\\&&1&3\\&&&1}$
Well that looks like Pascals triangle.
or $(x+1)^n$ appears to describe the set of eigenvectors.
$A((x+1)^n) = (2x+2)^n = 2^n(x+1)^n$ and that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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product of terms taken $3$ at a time in polynomial expression
Finding product of terms taken $3$ at a time in $\displaystyle \prod^{100}_{r=1}(x+r)$
Try:
$$\displaystyle \prod^{100}_{r=1}(x+r)=x^{100}+(1+2+3+\cdots +100)x^{99}+(1\cdot 2+1\cdot 3+\cdots+100\cdot 99)x^{98}+(1\cdot 2\cdot 3+2\cdot 3 \cdot 4+\cdot\cdot... | The coefficient of $x^{97}$ is
$$
\begin{align}
\sum_{i=1}^{100}\sum_{j=1}^{i-1}\sum_{k=1}^{j-1}ijk
&=\sum_{i=1}^{100}\sum_{j=1}^{i-1}\sum_{k=1}^{j-1}ij\binom{k}{1}\\
&=\sum_{i=1}^{100}\sum_{j=1}^{i-1}i\,((j-2)+2)\binom{j}{2}\\
&=\sum_{i=1}^{100}\sum_{j=1}^{i-1}i\left[3\binom{j}{3}+2\binom{j}{2}\right]\\
&=\sum_{i=1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Having trouble of finding this integral in the standard integral list I am facing difficult problem to deal with this integral
$$\int_{0}^{\pi/2}\ln(9-4\cos^2\theta)\,\mathrm{d}\theta \tag*{(1)}$$
note $(3)^2-(2\cos{\theta})^2=(3-2\cos{\theta})(3+2\cos{\theta}) \tag*{(2)}$
also note that, $$\log(AB)=\log A+\log B \tag... | Here a slightly different approach will be considered. We will make use of the generating function for the central binomial coefficients of
\begin{align*}
\sum_{n=0}^{\infty}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}, \qquad |x|<\frac{1}{4}, \tag1
\end{align*}
a result that follows directly from the binomial series.
Let
$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove by induction that for every $n \ge 0$ the term $2^{2n+1} - 9n^2 + 3n -2$ is divisible by $54$ What I get is $4 \cdot 54k + 27(n(n-1)) - 2$.
I understand first two terms but $-2$ is giving me a headache. I suppose to be true every term should be divisible by $54$. Or did I make wrong conclusion?
P. S: Is there mat... | Let $f(n) = 2^{2n+1}-9n^2+3n-2$; then
\begin{align*}
f(n+1)-4f(n) &= 2^{2n+3}-9(n+1)^2+3(n+1)-2-4(2^{2n+1}-9n^2+3n-2) \\
&= 4\cdot 2^{2n+1} - 9n^2 - 18n - 9 + 3n+3-2-4(2^{2n+1}-9n^2+3n-2) \\
&= 27n^2-27n = 27n(n-1).
\end{align*}
Since either $n$ or $n-1$ is even, this expression is divisible by $54$. Sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality : $\sum_{cyc}\frac{\sqrt{2}a^2b}{2a+b} \leq \sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$ Let $a, b, c$ be positive real number such that $a+b+c = 3$.
Prove that
$$\displaystyle\sum_{cyc}\frac{\sqrt{2}a^2b}{2a+b} \leq \displaystyle\sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$$
My attempt :
By AM-GM,
$\displaystyle... | By C-S $$\sqrt2\sum_{cyc}\frac{2a^2b}{2a+b}\leq\frac{\sqrt2}{(2+1)^2}\sum_{cyc}a^2b\left(\frac{2^2}{2a}+\frac{1^2}{b}\right)=\frac{\sqrt2}{9}\sum_{cyc}(2ab+a^2)=\sqrt2.$$
Thus, it's enough to prove that
$$\sum_{cyc}\frac{\sqrt{a^2+b^2}}{1+2ab}\geq\sqrt2.$$
Now, by C-S again
$$\sum_{cyc}\frac{\sqrt{a^2+b^2}}{1+2ab}=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Double integration including e? I would like to calculate the following expression:
$\iint_D x^2e^\sqrt{x^2+y^2} dxdy$
for D = ${ a^2\le x^2+y^2 \le b^2} $ and $ 0 \le a \le b$
How would you do it, in a step-by-step, please?
| $$\iint_D x^2e^\sqrt{x^2+y^2} dxdy=\iint_D (rcos\phi)^2e^r\cdot r drd\phi= \iint_D r^3cos^2\phi e^r drd\phi=$$
$$=\int^{2\pi}_0d\phi \int^b_a r^3cos^2\phi e^r dr=\int^{2\pi}_0cos^2\phi d\phi \int^b_a r^3 e^r dr=$$
$$=\int^{2\pi}_0cos^2\phi d\phi ( r^3-3r^2+6r-6) e^r|^b_a =\int^{2\pi}_0(( b^3-3b^2+6b-6) e^b-(a^3-3a^2+6a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $1 - \frac{\binom{n^2}{1}}{\binom{n+1}{1}} + \frac{\binom{n^2}{2}}{\binom{n+2}{2}} - \frac{\binom{n^2}{3}}{\binom{n+3}{3}} + ..$ How to evaluate $1 - \frac{\binom{n^2}{1}}{\binom{n+1}{1}} + \frac{\binom{n^2}{2}}{\binom{n+2}{2} } - \frac{\binom{n^2}{3}}{\binom{n+3}{3}} + \frac{\binom{n^2}{4}}{\binom{n+4}... | Below is a quite brutal approach : whenever I see inverses of binomial coefficients, I try to use the following relation between the Gamma and Beta functions (I can provide a link if needed) :
$$\mbox{With } 0 < m \le n,\quad \quad \ \frac{1}{\binom{m+n}{m}} = \frac{mn}{m+n} \cdot \displaystyle{\int_0^1} t^{m-1}(1-t)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Find $P(XY+Z-XZ=2)$ when $Z$~$N(0,1)$, $Y$~$B(3, \frac{1}{2})$, $X$~$B(1, \frac{1}{3})$ Let $X,Y,Z$ be independent variables such that: \begin{gather*}
Z \sim N(0,1) \,\text{(normal distribution)},\\
Y \sim B(3, \frac{1}{2}) \,\text{(binomial distribution)},\\
X \sim B(1, \frac{1}{3}) \,\text{(binomial distribution)}.... | \begin{align*}
P(XY + Z - XZ = 2) &= P(XY + Z - XZ = 2 \mid X = 0) + P(XY + Z - XZ = 2 \mid X = 1)\\
&= P(Z = 2 \mid X = 0) + P(Y = 2 \mid X = 1)\\
&= P(Z = 2) P(X = 0) + P(Y = 2)P(X = 1)\\
&= 0 \cdot \frac{2}{3} + \frac{3}{8} \cdot \frac{1}{3} = \frac{1}{8}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588592",
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"source": "stackexchange",
"question_score": "1",
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The Jacobi-Madden equation $a^4+b^4+c^4+d^4 = (a+b+c+d)^4$ and disguised Pythagorean triples I. The Jacobi-Madden equation,
$$a^4+b^4+c^4+d^4 = (a+b+c+d)^4$$
is equivalent to a disguised Pythagorean triple,
$$(a^2+ab+b^2)^2+(c^2+cd+d^2)^2 = \big((a+b)^2+(a+b)(c+d)+(c+d)^2\big)^2$$
II. A special case of the Descartes' ... | Not sure if this is what you had in mind, but suppose we have two distinct triangular numbers whose product is a square. Say $[u(u+1)/2][v(v+1)/2] = t^2$, where $u > v > 0$. Define $a = 4t$, $b_1 = u - v$, $b_2 = u + v + 1$, $c = 2uv + u + v$. Then a routine calculation gives
$$a^2 + b_1^2 = c^2,\quad a^2 + b_2^2 = (c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$.
Is it even possible to be done without a computer?
| Let $x_{n+1} =f(x_n)$ that is $x_{n+1} =\frac{x_n}{3}+3 $
Letting $$u_n =x_n -\frac{3}{2}\implies u_{n+1}=\frac13 u_n \implies u_n = \frac{u_1}{3^{n-1}} \implies x_n = \frac{3}{2}+\frac{x_1 -\frac{3}{2}}{3^{n-1}}$$
Hence, we have $$ \color{blue}{ \underbrace{f(f(f(f(f(f(\cdots f(x)))))))}_{n~~times} = x_n = \frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 6
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How to evaluate $\tan^{-1}\left(\frac4x\right)-\tan^{-1}\left(\frac3x\right)$ This is what I did but please correct me if I am wrong.
Let $y=\tan^{-1}\Big(\dfrac{4}{x}\Big)-\tan^{-1}\Big(\dfrac{3}{x}\Big)$.
$\tan(y)\\
=\dfrac{\tan\Big(\tan^{-1}\Big(\dfrac{4}{x}\Big)\Big)-\tan\Big(\tan^{-1}\Big(\dfrac{3}{x}\Big)\Big)}{1... | Yes, your approach is correct but it must be completed. First of all, what you found is $\tan(y)$ in the end so with the correct formula $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}$$
the answer becomes $$y = \tan^{-1}\bigg(\frac{x}{x^2+12}\bigg)$$
And here is another approach by using geometry:
Here, notice that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Minimise $|z_az_b + z_cz_d|$ where $\{z\}$ are the roots of $x^4 + 14x^3 + 52x^2 + 56x + 16$ Let $f(x) = x^4 + 14x^3 + 52x^2 + 56x + 16.$ Let $z_1, z_2, z_3, z_4$ be the four roots of $f$. Find the smallest possible value of $|z_az_b + z_cz_d|$ where ${a, b, c, d} = {1, 2, 3, 4}.$
So I have tried Vieta, but it is try ... | I claim the minimum is $8$. I will use dxiv suggestion. Let $g(y) = y^4+7y^3+13y^2+7y+1$ Obviously all roots are real and negative since we have
\begin{array}{cccccc}
y & -5 & -2 & -1 & -0.5 & 0 \\
g(y) & + & - & + & - & + \\
\end{array}
Let $a,b,{1\over a},{1\over b}$ be all roots for $g$ where $a,b,$ are negativ... | {
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"source": "stackexchange",
"question_score": "15",
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How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac... | In order to prove that $f(x)=\frac{x}{1+x}$ is sublinear on $\mathbb{R}^+$ it is enough to notice that $f'(x)=\frac{1}{(1+x)^2}$ leads to
$$ \frac{f(a+b)-f(a)}{f(b)-f(0)} = \frac{\int_{a}^{a+b}\frac{dx}{(1+x)^2}}{\int_{0}^{b}\frac{dx}{(1+x)^2}}\leq 1 $$
since $f'(x)$ is decreasing. In other terms, the sublinearity is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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How to compute $\lim_{n\to \infty}\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n} $
Find
$$ \lim_{n\to \infty}\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n}.$$
I tried with ln composition but ineffectively. Any idea?
| Following the suggestion by Paramanand Singh let indicate
$$a_n=\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)^{1/n}=\sqrt[n] b_n \quad b_n=\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \right)$$
then
$$\frac{b_{n+1}}{b_n}=\frac{\prod_{k=1}^{2n+2} \left(1+ \frac{ k}{n+1} \right)}{\prod_{k=1}^{2n} \left(1+ \frac{ k}{n} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $f(1)+f(3)+f(5)+\dots+ f(999)$ where $f$ is a given function
Given $$f(x)= \frac {1}
{\sqrt[3] {x^2+2x+1} + \sqrt[3] {x^2-1} + \sqrt[3] {x^2-2x+1}}$$
and
$$E= f(1)+f(3)+f(5)+\dots+ f(999).$$ Then find the value of $E$.
My work :-
Let $\sqrt[3] {x+1}= a$, $\sqrt[3] {x-1}=b $
Then the equation reduces to
$... | Hint. Since $(a^3-b^3)=(a-b)(a^2+ab+b^2)$, it follows that
$$f(x)=\frac{1}{a^2+ab+b^2}=\frac{a-b}{a^3-b^3}=\frac{\sqrt[3] {x+1}-\sqrt[3] {x-1}}{2}.$$
Therefore the given sum is telescopic:
$$\sum_{k=0}^{n-1}f(2k+1)=\frac{1}{2}\left(\sum_{k=0}^{n-1}\sqrt[3] {2(k+1)}-\sum_{k=0}^{n-1}\sqrt[3] {2k}\right)=\frac{\sqrt[3] {2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Definite integral of a rational fraction Can I find the value of $$\int_3^{\infty}\frac{x-1}{(x^2-2x-3)^2}dx$$ by just factoring the fraction?
I tried to wrote:
$$\frac{x-1}{(x^2-2x-3)^2}=\frac{x-1}{(x^2-2x+1-4)^2}=\frac{x-1}{[(x-1)^2-2^2]^2}=\frac{x-1}{(x+1)^2\cdot(x-3)^2}$$ but didn't work out. Any ideas?
| With $u=x^2-2x-3$ we have $du = 2(x-1)\,dx$, so that
$$\int \frac{x-1}{(x^2-2x-3)^2}\,dx = \frac{1}{2}\int\frac{1}{u^2}\,du
= -\frac{1}{2u} + c= -\frac{1}{2(x^2-2x-3)}+c.$$
| {
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"url": "https://math.stackexchange.com/questions/2601228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a_{1}>2$and $a_{n+1}=a_{n}^{2}-2$ then Find $\sum_{n=1}^{\infty}$ $\frac{1}{a_{1}a_{2}......a_{n}}$ Question If $a_{1}>2$and $\left\{ a_{n}\right\} be$ a recurrsive
sequence defined by setting $a_{n+1}=a_{n}^{2}-2$ then Find
$\sum_{n=1}^{\infty}$$\frac{1}{a_{1}a_{2}......a_{n}}$
Book's Answer
I have mentioned my p... | You have proven that
$$a_n^2-4 = a_1^2 \ldots a_{n-1}^2(a_1^2-4)$$
Divide both sides by $a_1^2 \ldots a_{n-1}^2$ and then take square root on both sides,
We have
$$\sqrt{\frac{a_n^2-4}{a_1^2 \ldots a_{n-1}^2}} = \sqrt{a_1^2-4}$$
Now take limit $n \to \infty$,
\begin{align}\lim_{n \to \infty}\sqrt{\frac{a_n^2-4}{a_1^... | {
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"url": "https://math.stackexchange.com/questions/2604612",
"timestamp": "2023-03-29T00:00:00",
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A congruence with the Euler's totient function and number of divisors function Can you provide a proof or a counterexample to the claim given below ?
Inspired by the congruence $1.3$ in this paper I have formulated the following claim :
Let $n$ be a natural number , let $\tau(n)$ be number of divisors function and let... | We have
$$n = \sum_{d\mid n} \varphi(d)\,,$$
so
$$\varphi(n) + \tau(n) = n - \sum_{\substack{d\mid n \\ d < n}} \varphi(d) + \sum_{d \mid n} 1 = (n+1) - \sum_{\substack{d \mid n \\ 1 < d < n}}\bigl(\varphi(d) - 1\bigr)\,.$$
If $n$ is composite and has a prime factor $\geqslant 5$, it follows that $\varphi(n) + \tau(n) ... | {
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"url": "https://math.stackexchange.com/questions/2607585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What can be said about the series $\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]$ This is a sequel to this question.
I recently was browsing through Hansen's "A Table of Series and Products", and I miraculously found the sum that I was looking for:
$$
\sum_{n=1}^\infty K_{0}\left( n z \r... | The approximation you mentioned is just fine. In a very simple (or elementary) way we have
$$\sqrt{n^2+x^2}\geq n \Rightarrow 0\leq \frac{1}{n} - \frac{1}{\sqrt{n^2+x^2}}$$
so (from partials sums an taking limits)
$$F(x) \geq 0, \forall x \tag{1}$$
Next thing to observe is
$$\frac{1}{n} - \frac{1}{\sqrt{n^2+x^2}}=\fra... | {
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"url": "https://math.stackexchange.com/questions/2610207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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An algebraic inequality involving $\sum_{cyc} \frac1{(a+2b+3c)^2}$ I was reading through the proof of an inequality posted on a different website and the following was mentioned as being easily proven by AM-GM:
Let $a,\ b,\ c>0$, then $$\frac{1}{(a+2b+3c)^2} + \frac{1}{(b+2c+3a)^2} + \frac{1}{(c+2a+3b)^2} \le \frac{1}{... | Hint: Using AM-GM,
$$(a+c+2(b+c))^2=(a+c)^2+4(a+c)(b+c)+4(b+c)^2\geqslant 6(a+c)(b+c)+3(b+c)^2=3(b+c)(2a+b+3c)$$
Hence it is enough to show:
$$\sum_{cyc} \frac{ab+bc+ca}{(b+c)(2a+b+3c)} \leqslant \frac34$$
$$\iff \sum_{cyc}\frac{(b+c)^2+2c^2}{(b+c)(2a+b+3c)}\geqslant\frac32$$
which follows almost directly from CS ineq... | {
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"url": "https://math.stackexchange.com/questions/2612329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Calculate limit with L'Hopital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.
I have done the following:
It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.
So, we can use L'Hopital's rule:
\be... | L'Hospital's rule is not the alpha and omega of limits computation! When it works, Taylor's formula at order $1$ also works, and it is less dangerous.
This being said, in the present case, doing some asymptotic analysis gives you a fast answer:
Near $0$, $\;\sin x \sim x$ and $ \cos \frac1x$ is bounded, so
$$\frac{x^... | {
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"url": "https://math.stackexchange.com/questions/2614160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show that $Q=X^2+5X+7$ divides $P=(X+2)^m+(X+3)^{2m+3}$ for any $m\in\Bbb N$ Let $$P=(X+2)^m+(X+3)^{2m+3}$$ and $$Q=X^2+5X+7.$$ I need to show that $Q$ divides $P$ for any $m$ natural.
I said like this: let $a$ be a root of $X^2+5X+7=0$. Then $a^2+5a+7=0$.
Now, I know I need to show that $P(a)=0$, but I do not know i... | Note that
$$(X+3)^3=X^3+9X^2+27X+27=(X^2+5X+7)(X+4)-1 $$
and
$$(X+3)^2 = X^2+6X+9=(X^2+5X+7)+(X+2).$$
| {
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"url": "https://math.stackexchange.com/questions/2619185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Real ordered pair $(a,b)$ in equation
If $$a^2+5b^2+2b=6a+2ab-10$$
then all real ordered pair of $(a,b)$ is?
Try: I am trying to convert it into $$(..)^2+(..)^2=0$$ or $$(..)^2+(..)^2+(..)^2=0.$$
So $$(a-b)^2+(2a-0.75)^2+10-(0.75)^2-6a=0.$$
Could some help me to solve it? Thanks.
| $a^2+5b^2+2b-6a-2ab+10=0$
Let's suppose we had a $(a+kb+j)^2$ term and that were the only term with $a $. We'd have $(a+kb+j)^2=a^2+k^2b^2+j^2+2kab+2aj+2kjb $. So $k=-1;j=-3$.
So
$a^2+5b^2+2b-6a-2ab+10=0$
$a^2+b^2+9-2ab-6a+6b+4b^2-4b+1=0$
$(a-b-3)^2+(2b-1)^2=0$
So for real solutions
$2b-1=0$ and $b=\frac 12$
And $a-... | {
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"url": "https://math.stackexchange.com/questions/2621439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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u-substitution for a definite integral I want to solve the following integral:
$$\int_{-2}^2 \sqrt{4- x^2} dx$$
I am thinking of doing a u-substitution of the whole term inside the root. If I do this, I need to change the limits of the integral. Doing so, the new limits would be equal (0=0). Doesn't this mean that the ... | The integral can be found with the substitution $x=\sin \theta$.
If we let $u=4-x^2$. Then $du=-2xdx$.
Note that $x=\sqrt{4-u}$ if $x\ge 0$ and $x=-\sqrt{4-u}$ if $x<0$.
So, $\displaystyle dx=\frac{du}{-2\sqrt{4-u}}$ if $x\ge 0$ and $\displaystyle dx=\frac{du}{2\sqrt{4-u}}$ if $x\le 0$.
So we actually do not have $\dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluate the integral $ I=\frac{1}{2\pi i}\int_{\vert z \vert =R}(z-3)\sin \left(\frac{1}{z+2}\right)dz$, Evaluate the integral
$$ I=\frac{1}{2\pi i}\int_{\vert z \vert =R}(z-3)\sin\left(\frac{1}{z+2}\right)dz$$
where $R \geq 4$
My work: Here the zeros of $\sin\left(\frac{1}{z+2}\right)$ is $\frac{1}{n \pi}-2$, so $-2... | The integrand function has just ONE singularity at $z=-2$. For any non-zero integer $n$ then by letting $z=\frac{1}{n \pi}-2$, we have that
$$\sin\left(\frac{1}{z+2}\right)=\sin(n\pi)=0.$$
So, for $R>2$, the singularity $z=-2$ is inside the circle $|z|=R$, and, by the Residue Theorem,
$$\begin{align}\frac{1}{2\pi i}\in... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Trigonometry problem, Evaluate: $\frac {1}{\sin 18°}$ My problem is,
Evaluate: $$\frac {1}{\sin 18°}$$
I tried to do something myself.
It is obvious,
$$\cos 18°= \sin 72°$$
I accept $\left\{18°=x \right\}$ for convenience and here, $\sin (x)>0$
$$\cos (x)=\sin (4x)$$
$$\cos (x)=2× \sin(2x) \cos (2x)$$
$$\cos (x)=2× ... | Not an ugly solution at all, but one can do better. Here's an analytical solution based on complex numbers.
Let $\alpha=72^\circ=2\pi/5$, so we can consider $z=e^{\alpha}=\cos\alpha+i\sin\alpha$ that satisfies $z^5-1=0$. Since $z\ne1$, we can deduce
$$
z^4+z^3+z^2+z+1=0
$$
and also, dividing by $z^2$,
$$
z^2+\frac{1}{z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.