Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve the equation $x^3-6x-6=0$
Evaluate the roots of
$$x^3-6x-6=0$$
I solved it using Cardano's method, but I'm looking for other elementary approaches through substitutions and properties of polynomials.
Thanks.
| $\text{Let } x = z + \frac{2}{z}\\\begin{align*}\left(z+\frac{2}{z}\right)^3 - 6\left(z+\frac{2}{z}\right) - 6 &= 0\\z^6 - 6z^3 + 8 &= 0\\z^3 &= 3\pm1\, (\text{using the quadratic formula})\\z^3 &= 4, 2,\\z &= \sqrt[3]{2},\sqrt[3]{4}\end{align*}$
Now just substitute back one value for $z$ to get one root:
$x = \sqrt[3]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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What is the radius of convergence of: What is the radius of convergence of the series
$$1+z+{z}^2/2^2+z^3/3!+z^4/2^4+z^5/5!+...$$
I used ratio test but then it gives two values of $R$
the values of $R$ are 0 and $\sqrt2$,
| $$1 + z + \frac{z^2}{2^2} + \frac{z^3}{3!} + \frac{z^4}{2^4} + \cdots
= \left(1 + \frac{z^2}{2^2} + \frac{z^4}{2^4} + \cdots\right) + \left(z + \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots \right)$$
We can rearrange terms like this as long as the series is absolutely convergent. So let's investigate that. First note t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the sum of series $\sum_{n=0}^∞ \frac{2^n + 3^n}{6^n}$ I am being asked to find the sum of the following convergent series :
$$\sum_{n=0}^∞ \frac{2^n}{6^n} + \frac{3^n}{6^n}$$
Attempting to generalize from partial sums yields nothing of interest:
$s_1 = \frac{5}{6}$
$s_2 = \frac{5}{6} + \frac{13}{36} = \frac{43... | It is just the sum of two geometric series in disguise.
$$
\begin{aligned}
\underset{i=0}{\overset{\infty}{\sum}}\frac{2^{n}+3^{n}}{6^{n}}&=\underset{i=0}{\overset{\infty}{\sum}}\frac{2^{n}}{6^{n}}+\frac{3^{n}}{6^{n}}\\
&=\underset{i=0}{\overset{\infty}{\sum}}(1/3)^{n}+\underset{i=0}{\overset{\infty}{\sum}}(1/2)^{n}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Write $π = (3, 2, 5)(2, 5, 4)$ in “table” notation? Isn't this impossible...? Because this permutation goes from 3 --> 2 ---> 5 ---> 3 according to the first cycle, but goes from 2 --> 5 ---> 4 ---> 2 according to the second cycle. So 5 can't go to 3 and 4.
| Assuming you do multiplication left to right, then
$$(3\quad2\quad5)(2\quad5\quad4) =
\begin{pmatrix}1&2&3&4&5\\
1&5&2&4&3
\end{pmatrix}\begin{pmatrix}1&2&3&4&5\\
1&5&3&2&4
\end{pmatrix} = \begin{pmatrix}1&2&3&4&5\\
1&4&5&2&3\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
f(x) is a function such that $\lim_{x\to0} f(x)/x=1$ $f(x)$ is a function such that $$\lim_{x\to0} \frac{f(x)}{x}=1$$ if
$$\lim_{x \to 0} \frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}=1$$
Find $a$ and $b$
Can I assume $f(x)$ to be $\sin(x)$ since $\sin$ satisfies the given condition?
| Since $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=1$, we have
$$\lim_{x\to0}\frac{x}{f(x)}
=\lim_{x\to0}\frac{1}{\frac{f(x)}{x}}
=\frac{1}{\displaystyle\lim_{x\to0}\frac{f(x)}{x}}
=\frac{1}{1}=1.$$
Now, by using L'Hopital's rule, we have
\begin{align}
1&=\lim_{x \to 0} \frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}\\
&=\lim_{x \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} = \tan^{-1}\frac{n}{n+2}$ Prove that
$$ \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + ... + \tan^{-1}\frac{1}{n^2+n+1} = \tan^{-1}\frac{n}{n+2}$$
I have been trying to solve it step by like $ \tan^{-1}\frac{1}{3} + \tan^{-1}\f... | Hint: Observe last term of $L.H.S.$ carefully and note that $\tan^{-1}a-\tan^{-1} b=\tan^{-1}\frac{a-b}{1+ab}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of $\frac{n^4}{\binom{4n}{4}}$ as $n \rightarrow \infty$ $\frac{n^4}{\binom{4n}{4}}$
$= \frac{n^4 4! (4n-4)!}{(4n)!}$
$= \frac{24n^4}{(4n-1)(4n-2)(4n-3)}$
$\rightarrow \infty$ as $n \rightarrow \infty$
However, the answer key says that
$\frac{n^4}{\binom{4n}{4}}$
$= \frac{6n^3}{(4n-1)(4n-2)(4n-3)}$ this ... | Hint:
$$
\begin{align}
\frac{n^4}{\binom{4n}{4}}
&=\frac{n^4}{\frac{4n(4n-1)(4n-2)(4n-3)}{4!}}\\
&=\frac{n^4}{\frac{4^4n^4\left(1-\frac1{4n}\right)\left(1-\frac2{4n}\right)\left(1-\frac3{4n}\right)}{4!}}\\
&=\frac{4!}{4^4}\frac1{\left(1-\frac1{4n}\right)\left(1-\frac2{4n}\right)\left(1-\frac3{4n}\right)}
\end{align}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $A$ is normal and upper triangular then it is diagonal
Let $A$ be a normal matrix in Mat$_{n\times n}(\mathbb C)$, if $A$ is upper triangular then it is diagonal
(Normal means $AA^*=A^*A$, where $A^*$ is the conjugate transpose of $A$)
If I consider the diagonal of $AA^*$, let denote $(a_{ij})=A$ and $(â_{ij})_{i... | Let's prove this by induction. Suppose $A$ is an $n\times n$ upper-triangular, normal matrix.
If $n=1$, this is trivial.
If $n=2$, we have
$$ A = \begin{pmatrix}
a & b\\
0 & c
\end{pmatrix}$$
So
\begin{align*}
0 = AA^* - A^*A &= \begin{pmatrix}
a & b\\
0 & c
\end{pmatrix}\begin{pmatrix}
\overline{a} & 0\\
\overline... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$ If we have
$$\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$$
Then, what will be the set of $x$ for which this equation is true?
I tried to solve it by putting $x = \sin a$ or $\cos a$ but got no result.
I am totally stuck on how to do it.
| Note that if
$$\arcsin(2x\sqrt{1-x^2})=2\arcsin(x) \tag 1$$
then taking the sine of both sides of $(1)$ yields
$$\begin{align}
2x\sqrt{1-x^2}&=\sin(2\arcsin(x))\\\\
&=2\sin(\arcsin(x))\cos(\arcsin(x))\\\\
&=2x\sqrt{1-x^2}
\end{align}$$
However, taking the cosine of the left-hand side of $(1)$ yields
$$\cos(\arcsin(2x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Let $α$ and $β$ be the roots of equation $px^2+qx+r=0,p≠0$ Let $α$ and $β$ be the roots of equation $px^2+qx+r=0,p≠0$, If $p,q,r$ are in A.P and $\dfrac{1}{α}+\dfrac{1}{β}=4$, then the value of $|α−β|$ is $:$
*
*$\dfrac{\sqrt{61}}{9} $
*$\dfrac{2\sqrt{17}}{9}$
*$\dfrac{\sqrt{34}}{9}$
*$\dfrac{2\sqrt{13}}{9}$
... | Let $\alpha$ and $\beta$ be the roots of the equation $ax^2+bx+c=0$. By the usual formula, the difference of the roots is $\frac{\sqrt{b^2-4ac}}{a}$. Hence it's enough for us to find the correct $a,b,c$ for which the roots are $\alpha$ and $\beta$.
But here's the trick: So we know that $\frac{1}{\alpha} + \frac{1}{\bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding all solutions for $3^c=2^a+2^b+1$ Given the equation:
$3^c=2^a+2^b+1$
Find all solutions for $a,b,c$ - given that they are positive integers and $b>a$.
Any ideas?
| It's still relatively simple to handle such exponential equations using local arguments. If $a \geq 5$, then we have $3^c \equiv 1 \pmod{32}$ and hence $8 \mid c$. It follows that $3^c \equiv 1 \pmod{41}$ and so $2^{b-a} \equiv -1 \pmod{41}$. We therefore have $b-a \equiv 10 \pmod{20}$ which implies that $2^{b-a} \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Evaluate the following limit: Find
$$\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left[\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right]$$
MY TRY:
$$
\begin{align}
\lim_{n \to \infty} &\frac{1}{\sqrt{n}} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}}
+ \frac{1}{\sqrt{4}+\sqrt{6}}
+ \... | $$\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=\frac{\sqrt{2k+2}-\sqrt{2k}}{2}$$
The sum telescopes to $\displaystyle \frac{1}{\sqrt n} \frac {\sqrt{2n+2} - \sqrt 2 }{2}$ which converges to $1/\sqrt{2}$
If you miss this, you can still use integrals to get bounds on $\sum_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Only valid for Pythagoraean triples $\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$? $$\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$$
Where (a,b,c) are the Pythagoraean Triples and are satisfy by the Pythagoras theorem $a^2+b^2=c^2$
An example of Pythagoraean t... | Just compute your left side:
$$x=\sqrt{2}+\frac{b}{x}$$
$$x(x-\sqrt{2})=b$$
$$x^2-\sqrt2 x-b=0$$
$$x=\frac{\sqrt2+ \sqrt{2+4b}}{2}$$
So now you have
$$\sqrt2+\sqrt{2+4b}=2\sqrt{a+c}$$
$$2+2+4b+2\sqrt{4+8b}=4(a+c)$$
$$a+c-b-1=\sqrt{1+2b}$$
$$c=1+b-a+\sqrt{1+2b}$$
For that to be a whole number, $b=(x^2-1)/2$ where $x$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find all real numbers $a,b$ such that $|a|+|b|\geq\frac{2}{\sqrt{3}}$ and $|a\sin x+b\sin{2x}|\leq 1$ for all real $x$. Find all real numbers $a,b$ such that $|a|+|b|\geqslant\frac{2}{\sqrt{3}}$ and $|a\sin(x)+b\sin(2x)|\leqslant 1$ for all real $x$.
We could write the inequality as
$$
\left|\frac{a}{\sqrt{a^2+b^2}}\si... | If $a$ and $b$ are of the same sign then $|a|+|b|=|a+b|\geqslant\frac{2}{\sqrt3}$. It's easy to guess a value of $x$ for which inequality $|a\sin x+b\sin 2x|\leqslant 1$ becomes equality - for example, $x=\frac{\pi}{3}$. Therefore this point is a local extremum for $f(x)=a\sin x+b\sin 2x$, so $$f'(\pi/3)=a\cos\pi/3+2b\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, ... | It's actually possible (though somewhat tedious) to calculate $\exp(tA)$ using the series definition. The definition states that,
$$\exp(tA) = \sum_{n=0}^\infty \frac{t^n}{n!}A^n.$$
Now, since $\exp(tA)$ is a matrix, it is enough to find the values of $\exp(tA)e_i$ for $i=1,2,3$, where the $e_i$ are the standard basis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 6
} |
What is the minimum value of a radical sum? How would you find the minimum value of $\sqrt {x^2+16} + \sqrt {x^2-12x+37}$ through algebraic manipulation? Graphing will clearly show that the minimum value is 4.8, but how do you get the answer manually?
In general, how would you find the minimum value(if it exists) of th... | Another way.
By C-S $$\sqrt{x^2+16}+\sqrt{x^2-12x+37}=$$
$$=\frac{1}{\sqrt{2.44}}\sqrt{(1.2^2+1)^2(x^2+4^2)}+\frac{1}{\sqrt{2.44}}\sqrt{(1.2^2+1^2)((6-x)^2+1^2)}\geq$$
$$=\frac{1}{\sqrt{2.44}}\left(1.2x+4+1.2(6-x)+1\right)=\frac{12.2}{\sqrt{2.44}}=\sqrt{61}.$$
The equality occurs for $(x,4)||(6-x,1)$ or for $x=4.8,$ wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Estimate of Fresnel-like integral Does it correct that if $x\in\mathbb{R}_+$ then
$$\left | \int_{x^2}^{(x+1)^2}\frac{\sin t}{\sqrt{t}}dt \right |\le \frac{2}{x}?$$
| Assume $x>0$.
One may integrate by parts:
$$
\begin{align}
\left| \int_{x^2}^{(x+1)^2}\frac{\sin t}{\sqrt{t}}dt\right|&=\left|\left[-\frac{1}{\sqrt{t}}\cos t\right]_{x^2}^{(x+1)^2}-\frac{1}{2}\int_{x^2}^{(x+1)^2}\frac{1}{t\sqrt{t}}\cos t\:dt\right|
\\\\&= \left|\frac{\cos (x^2)}x-\frac{\cos ((x+1)^2)}{x+1}-\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculate $ \lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx}$ My attempt:
\begin{align*}
\lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx} &= (\frac{\ln1}{0}) \text{ (we apply L'Hopital's rule)} \\
&= \lim_{x \to0}\frac{\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{x^n+x^{n-1}+\dots+1}}{n} \\
&= \lim_{x \to0}\frac{nx^{n-1}+(n-1)x... | Hint: $$S=1+x+x^2+\cdots+x^n$$ expand with $x$ to get $$xS=x+x^2+\cdots+x^{n+1}$$
Now, subtract both equations and solve for $S$:
$$S-xS=1-x^{n+1}$$
$$S=\frac{1-x^{n+1}}{1-x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Intuitive explanation of $(1-x)^{-a-1}=\sum_{j=0}^{\infty}{{a+j} \choose j}x^j$ Could anyone please explain me the reasoning behind this formula?
$(1-x)^{-a-1}=\sum_{j=0}^{\infty}{{a+j} \choose j}x^j$
Thanks so much!
| From the sum of geometric series, we know that $1/(1-x)=1+x^2+x^3+\cdots$, for $|x|<1$.
Thus, one can perceive $(1-x)^{-a-1}=1/(1-x)^{a+1}$ as the $a+1$th power of $1/(1-x)$, given $a$ is an integer,
$$
\underbrace{\frac{1}{1-x}\cdot \frac{1}{1-x} \cdots \frac{1}{1-x}}_\text{a+1} \\
=\underbrace{(1+x^2+x^3+\cdots)\time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Deadly integral $\int_0^1\frac{x^{2}+x+1}{x^{4}+x^{3}+x^{2}+x+1}dx$. How to solve this question
$$\int\limits_0^1\frac{x^{2}+x+1}{x^{4}+x^{3}+x^{2}+x+1}dx$$
. Please help me in solving this short way
my approach is in the answer
Is it correct and can it be solved in a shorter way ?
| Multiply top and bottom by $(-(x-1)$, we have $\displaystyle \int_0^1 \frac{1-x^3}{1-x^5} \, dx$
Notice that the integrand can be written as a sum of a convergent geometric series with common ratio $x^5$
$\begin{eqnarray} \displaystyle \int_0^1 \frac{1-x^3}{1-x^5} \, dx &=& \displaystyle \int_0^1 \displaystyle \sum_{r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Equation with limit $\lim\limits_{n\to \infty}\sqrt{1+\sqrt{x+\sqrt{x^2+...+\sqrt{x^n}}}}=2$
I had never faced with problems like this. Give me, please, a little hint.
| Hint.
$$f(x) = \sqrt{x+\sqrt{x^2+\sqrt{x^3+\sqrt{x^4+\ldots}}}} $$
is a continuous and increasing function on $\mathbb{R}^+$, as well as:
$$ f_k(x) = \sqrt{x^k+\sqrt{x^{k+1}+\sqrt{x^{k+2}+\ldots}}}$$
It is enough to check that $f(4)=3$, or:
$$ f_1(4) = 2\cdot 4^0+1, $$
or:
$$ f_2(4) = f_1(4)^2-4 = (2\cdot 4^0+1)^2-2^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to calculate $\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$? Evaluate this limit
$$\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$$
What's the method?
The answer is $1000$.
| Since $\sin(n+1)x=2\sin nx\cos x-\sin(n-1)x$, we have
$$\begin{align}\sin3x&=2\sin2x\cos x-\sin x=2(2\sin x\cos x)\cos x-\sin x\\
&=4\sin x(1-\sin^2 x)-\sin x=3\sin x-4\sin^3x\end{align}$$
Then $\sin x=\sin(3(x/3))=3\sin(x/3)-4\sin^3(x/3)$. Then
$$\begin{align}\frac{x^n-\sin^nx}{x^2\sin^nx}&=\frac{(3(x/3))^n-(3(\sin(x/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 8,
"answer_id": 0
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Find the sum of the infinite series $\sum n(n+1)/n!$ How do find the sum of the series till infinity?
$$ \frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+\cdots$$
I know that it gets reduced to $$\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}$$
But I don't know how to proceed further.
| $$xe^x = \sum\frac{x^{n+1}}{n!}\\
\frac d{dx} x e^x = e^x + x e^x = \sum \frac{(n+1)x^n}{n!}\\
\frac {d^2}{dx^2} x e^x =2 e^x + x e^x =\sum \frac{n(n+1)x^{n-1}}{n!}$$
and as $x$ approaches $1$
$$3 e=\sum \frac{n(n+1)}{n!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Prove Why $B^2 = A$ exists?
Define
$$A =
\begin{pmatrix}
8 & −4 & 3/2 & 2 & −11/4 & −4 & −4 & 1 \\
2 & 2 & 1 & 0 & 1 & 0 & 0 & 0 \\
−9 & 8 & 1/2 & −4 & 31/4 & 8 & 8 & −2 \\
4 & −6 & 2 & 5 & −7 & −6 & −6 & 0 \\
−2 & 0 & −1 & 0 & 1/2 & 0 & 0 & 0 \\
−1 & 0 & −1/2 & 0 & −3/4 & 3 & 1 & 0 \\
1 & 0 & 1/2 & 0 & 3/4 & −1 & 1... | I borrow (and slightly modify) an explanation that can be found in the excellent thread
https://mathoverflow.net/questions/14106/finding-the-square-root-of-a-non-diagonalizable-positive-matrix
$f$ being any function, sufficient derivable, one can write, for example for a $4 \times 4$ Jordan block:
$$f\left( \left[
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Minimize $a^5+b^5+c^5+d^5+e^5 = p^4+q^4+r^4+s^4 = x^3+y^3+z^3 = m^2 + n^2$ with distinct positive integers
Find the minimum value of the following:
$$a^5+b^5+c^5+d^5+e^5 = p^4+q^4+r^4+s^4 = x^3+y^3+z^3 = m^2 + n^2$$
where all numbers are different/distinct positive integers.
I know the answer (see below), but wan... | Thanks Will Jagy for sharing the code.
I had used following VBA code in excel
Sub Power5()
For u = 1 To 10
For v = u + 1 To 10
For w = v + 1 To 10
For x = w + 1 To 10
For y = x + 1 To 10
mysum = u ^ 5 + v ^ 5 + w ^ 5 + x ^ 5 + y ^ 5
For p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Right Triangle and Circle Theorem Let $ABC$ be a triangle such that $\angle BAC$ is a right angle. Suppose $D$ is a point lying on $BC$ such that $BD=1$, $DC =3$ and $\angle ADB=60^{\circ}$, find the length of $AC$.
I was told that there is a neat way to solve the question by considering the circumcircle of $\triangle ... | The circumcircle provides a good approach. $|DM|=1$, $|AM|=2$, and $|MC|=2$. Because $\angle ADB = 60^{\circ}$, $\angle ADM = 120^{\circ}$. We may then find $\angle DAM$ using the law of sines:
$$\frac{\sin{\angle DAM}}{|DM|} = \frac{\sin{\angle ADM}}{|AM|} \implies \sin{\angle DAM} = \frac12 \sin{120^{\circ}} = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How can I find the limit of given trigonometric function? I am messed up on solving this question. What should I do first in order to get the answer ?
This is the trigonometric function
$$ \lim \limits_{x \rightarrow 0} \frac{(a+x)\sec(a+x) - a \sec(a)}{x} $$
| Changing into cosines greatly eases the manipulation of terms.
$$
\begin{align}
\lim \limits_{x \rightarrow 0} \frac{(a+x)\sec(a+x) - a \sec(a)}{x}
& = \lim \limits_{x \rightarrow 0} \frac{a\sec(a+x) - a \sec(a)}{x} + \lim \limits_{x \rightarrow 0} \frac{x\sec(a+x)}{x} \\
& = A + B
\end{align}
$$
$$
\begin{align}
A &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Show that $(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6$ for $a^2 + b^2 + c^2 + d^2 = 1$. For $a, b, c, d \in \Bbb R$ such that $a^2 + b^2 + c^2 + d^2 = 1$, show that $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6.$$
The answer uses the mysterious identity $... | I give two solutions.
Solution 1:
After homogenization, it suffices to prove that
$$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6(a^2+b^2+c^2+d^2)^2.$$
The Buffalo Way works. WLOG, assume that $d \le c \le b \le a$.
Let $c = d + s, b = d + s + t, a = d + s + t + r$ for $s, t, r\ge 0$. We h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
sum of all Distinct solution of the equation $ \sqrt{3}\sec x+\csc x+2(\tan x-\cot x) = 0\;,$
The sum of all Distinct solution of the equation $\displaystyle \sqrt{3}\sec x+\csc x+2(\tan x-\cot x) = 0\;,$
Where $x\in (-\pi,\pi)$ and $\displaystyle x\neq 0,\neq \frac{\pi}{2}.$
$\bf{My\; Try::}$ We can write equation a... |
Where i have done wrong
In the following part :
We can write equation as $$\frac{\sqrt{3}\sin x+\cos x}{\sin x\cos x}-2\frac{\cos 2x}{\sin x\cos x} = 0$$
So we get $$2\frac{\cos \left(x+\frac{\pi}{3}\right)}{\sin x\cos x}-2\frac{\cos 2x}{\sin x\cos x}=0$$
This is wrong. It should be the following (a sign mistake) :... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Check convergence of $\sum^{\infty}_{n=1} \frac{2^{n} +n ^{2}}{3^{n} +n ^{3}}$ Zoomed version: $$\sum^{\infty}_{n=1} \frac{2^{n} +n ^{2}}{3^{n} +n ^{3}}$$
So, I've seen similar example at Convergence or divergence of $\sum \frac{3^n + n^2}{2^n + n^3}$
And I liked that answer :
$$3^n+n^2\sim_03^n,\quad2^n+n^3\sim_\inf... | The notation $$3^n+n^2\sim_{+\infty}3^n$$ means $$\lim_{n \to +\infty} \frac{3^n+n^2}{3^n}=1.$$ You can see this is true since $$\lim_{n \to +\infty} \frac{3^n+n^2}{3^n} = 1 + \lim_{n\to +\infty} \frac{n^2}{3^n}.$$ The second limit is well-known to be equal to $0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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A cube and a sphere have equal volume. What is the ratio of their surface areas? The answer is supposed to be $$ \sqrt[3]{6} : \sqrt[3]{\pi} $$
Since $$ \ a^3 = \frac{4}{3} \pi r^3 $$
I have expressed it as: $$ \ a = \sqrt[3]{ \frac{4}{3} \pi r^3} $$
and,
$$ \ 6 \left( \sqrt[3]{ \frac{4}{3} \pi r^3 } \right) ^2 : 4 ... | Another way to approach this is to write the surface areas of each solid in terms of its volume:
cube -- $$ S_c \ = \ 6 \ a^2 \ \ , \ \ V_c \ = \ a^3 \ \ \Rightarrow \ \ a \ = \ V_c^{1/3} \ \ \Rightarrow \ \ S_c \ = \ 6 \ (V^{1/3})^2 \ = \ 6 \ V_c^{2/3} \ \ ; $$
sphere -- $$ S_s \ = \ 4 \ \pi \ r^2 \ \ , \ \ V_s \ = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$ Is there an easy to solve the problem? The way I did it is to find the value of $a$ from the second expression and then use it to find the value of the first expression. I believe there must be an simple and elegant approach to tackle the prob... | Since
$$
\frac{1}{3}=\frac{12}{36}=\frac{12a^2}{(a^2+a+1)^2}=\frac{12a^2}{a^4+a^2+1+2a(a^2+a+1)}=\frac{12a^2}{(a^4+a^2+1)+12a^2},
$$
we have
$$
\frac{1}{2}=\frac{1}{3-1}=\frac{12a^2}{[(a^4+a^2+1)+12a^2]-12a^2}=\frac{12a^2}{a^4+a^2+1}.
$$
Hence
$$
\frac{a^2}{a^4+a^2+1}=\frac{1}{24}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Is a diagonalization of a matrix unique? I was solving problems of diagonalization of matrices and I wanted to know if a diagonalization of a matrix is always unique? but there's nothing about it in the books nor the net.
I was trying to look for counter examples but I found none.
Any hint would be much appreciated
Th... | Actually it is not unique. Even without changing the order of eigenvalues. For example, the matrix $$A=\begin{pmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0&0&1\end{pmatrix}=\begin{pmatrix}1&0& -1 \\ 1 & 0 & -1 \\ 0&1&0\end{pmatrix}\begin{pmatrix}1&0& 0 \\ 0 & 1 & 0 \\ 0&0&-1\end{pmatrix}\begin{pmatrix}1&1& 0 \\ 0 & 0 & 1 \\ -1&1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Approximation of $\sin\left(\frac{x}{n}\right)$ I wrote in my analysis notes the following: $\sin\left(\dfrac{x}{n}\right) = -\dfrac{x}{n} + \omicron\left(\left| \dfrac{x}{n} \right|\right)$.
I'm guessing it comes from Taylor's formula but I don't understand how they got that result.
Also, it said: $\sin\left(\dfrac{x... | The taylor series for $\sin y$ around $y=0$ is
$$
\sin y=y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}+\cdots.
$$
Plug in $y=x/n$ to get
$$
\sin\frac{x}{n}=\frac{x}{n}\underbrace{-\frac{x^{3}}{n^{3}3!}+\frac{x^{5}}{n^{5}5!}+\cdots.}_{o(|\frac{x}{n}|)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve the following equation for $x$,$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$ I am not great at transposition and wolfram alpha confused me so I would like to see the steps in solving for x.
$$\left(\frac{\frac{1}{2}\cdot(n-x^2)}{x}\right)^2 =\frac{1}{2}\cdot(n-x^2)$$
Wolfram alpha g... | Both of the solutions are possible since if you let $x = \pm\sqrt{n}$, then $x^2 = n$
Now, substiute in the equation you got
$$\left(\frac{\frac{1}{2}\cdot(n-n)}{\pm\sqrt{n}}\right)^2 =\frac{1}{2}\cdot(n-n)$$
$$\left(\frac{\frac{1}{2}\cdot(0)}{\pm\sqrt{n}}\right)^2 =\frac{1}{2}\cdot(0)$$
Which simply is $0=0$
That i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How do you solve $x^2 - 4 \equiv 0 \mod 21$ There is an example in my textbook of how you solve:
$$ x^2 -4\equiv 0 \mod 21 \Leftrightarrow x^2-4\equiv 0 \mod 3 \times 7$$
and then 2 congruences can be formed out of this equation if:
$$x^2-4\equiv0 \mod 3 \\ x^2-4 \equiv 0 \mod 7$$
and from these 2 congruences result 2... | Your ‘4 solutions’ are not solutions modulo $21$, but pairs of solutions $\bmod3$ on one hand, $\bmod 7$ on the other hand.
From these pairs of solutions, you recover solutions modulo $3\times 7$ with the Chinese remainder theorem.
Start from the Bézout's relation $\;5\cdot 3-2\cdot 7=1$. Then the solution correspondi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove $(x+y+z) \cdot \left( \frac1x +\frac1y +\frac1z\right) \geqslant 9 + \frac{4(x-y)^2}{xy+yz+zx}$ $x,y,z >0$, prove
$$(x+y+z) \cdot \left( \frac1x +\frac1y +\frac1z\right) \geqslant 9 + \frac{4(x-y)^2}{xy+yz+zx}$$
The term $\frac{4(x-y)^2}{xy+yz+zx}$ made this inequality tougher. It remains me of this inequality. I... | We may recall that Lagrange's identity gives a proof of the Cauchy-Schwarz inequality:
$$ (a^2+b^2+c^2)(d^2+e^2+f^2) = (ad+be+cf)^2 + (ae-bd)^2+(af-cd)^2+(bf-ce)^2 $$
and set $(a,b,c)=(\sqrt{x},\sqrt{y},\sqrt{z})$, $(d,e,f)=\left(\frac{1}{\sqrt{x}},\frac{1}{\sqrt y},\frac{1}{\sqrt z}\right)$ to get:
$$ (x+y+z)\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Partial Fractions with Quadratic Factor I understand that if we have a quadratic factor such as in
$\frac{8}{(x^2 + 1)(2x-3)} $
and we want to decompose, we should have a linear factor above $ x^2 +1$. Is the reason behind this primarily for calculus purposes, since if we have the numerator in linear form it will b... | Hint. Observe that, we have
$$
\frac{a}{x^2+1}+\frac{c}{2x-3}=\frac{c x^2+2ax+c-3 a}{\left(1+x^2\right)(2 x-3) }
$$ and the latter fraction can not be equal to
$$
\frac{8}{\left(1+x^2\right)(2 x-3) }
$$ for all appropriate $x$. Do you see why?
Whereas,
$$
\frac{8}{\left(1+x^2\right)(2 x-3) }=\frac{ax+b}{x^2+1}+\frac{c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The geometric construction of the $90^\circ, 87^\circ, 3^\circ$ triangle The construction of the $90^\circ, 45^\circ, 45^\circ$ and the $90^\circ, 60^\circ, 30^\circ$ triangles is well known.
How can be constructed a triangle with angles $90^\circ, 87^\circ, 3^\circ$ without using regular polygons? By "construction" I... |
Is it possible to use the exterior angle theorem and the triangles $90^\circ, 72^\circ, 18^\circ$ and $90^\circ, 75^\circ, 15^\circ$ to construct geometrically the angle $18^\circ - 15^\circ= 3^\circ$? How this construction can be done?
Certainly if you can construct $75^\circ$ and $72^\circ$, then you can construct ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I'm stuck in a logarithm question: $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ If $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ so $x + 2y= ?$
I've tried this far, and I'm stuck
$$\begin{align}4^{y+3x}&= 64 \\
4^{y+3x} &= 4^3 \\
y+3x &= 3 \end{align}$$
$$\begin{align}\log_x (x+12)- 3 \log_x 4 &= -1 \\
\log_x ... | Solve the Equation System
\begin{align}
I&& 4^{y+3x}&&=64\\
II&&\log_x(x+12)-3\log_x(4)&&=-1
\end{align}
Step 1: Solving for $x$
The key trick is apply the power-transform $x^h$ to both sides of equation $II$ which will yield a quadratic equation with elementary solutions. Here are the algebrarics:
*
*Multiply the eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Find f'(x) at the given value of x Find $f^\prime (x)$ at the given value of $x$:
$f(x)=3\sqrt{x}$
Find $f^\prime (5)$
For this one in my attempt to find the derivative I ended up with "$9/0$" which would lead me to believe that the value does not exist at $f^\prime (5)$. I was wondering if this was correct and if not ... | Since you're using the formal definition of the derivative here, given $f(x)=3\sqrt{x}$, we know that
$$
\begin{align*}
f'(x)&=\lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\
&= \lim_{h\to0} \frac{3\sqrt{x+h}-3\sqrt{x}}{h}\\
&= 3 \cdot \lim_{h\to0} \frac{\sqrt{x+h}-\sqrt{x}}{h}\\
\end{align*}
$$
We can then multiply the numerato... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why is this way of doing $\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$ I did it this way :
$$\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$$
now as $x \to \frac{\pi}{2}$, $\tan(2*x) \to \tan(\pi)$ and $x - \frac{\pi}{2} \to 0$
As $\tan(\pi) = 0$
$$\therefore \tan(2*x) \to 0$$
$... | $$\dfrac{\tan2x}{x-\dfrac\pi2}=\dfrac{\tan2\left(x-\dfrac\pi2\right)}{x-\dfrac\pi2}=2\cdot\dfrac{\tan(2x-\pi)}{2x-\pi}$$
Actually we have $$\lim_{(h)\to0}\dfrac{\tan(h)}{(h)}=\lim_{h\to}\dfrac{\sin h}h\cdot\dfrac1{\lim_{h\to0}\cos h}=1$$
Observe that all the $h$ in parenthesis must be same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to express $(-1+ \sqrt{3}i)^5$ in rectangle form? The question is "Express $(-1+ \sqrt{3}i)^5$ in rectangle form."
I've never heard of rectangle form before and it's apparently gonna be on the final exam. I couldn't find anything on google to help.
| \begin{align*}
(-1+\sqrt{3}\,i)^{5} &=
2^{5} \left( -\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{5} \\
&= 2^{5} (e^{2\pi i/3})^{5} \\
&= 2^{5} e^{10\pi i/3} \\
&= 32\, e^{4\pi i/3} \\
&= 32\left( -\frac{1}{2}-\frac{\sqrt{3}}{2} i\right) \\
&= -16-16\sqrt{3} \, i
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
General Two-State Markov Chain: $P(X_{n}=1)=\frac{b}{a+b}+(1-a-b)^n \big(P(X_0=1)-\frac{b}{a+b}\big)$ Consider a general chain with the state space $S=\{1,2\}$ and write the transition probability as
$$\begin{pmatrix}
1-a&a\\
b&1-b\end{pmatrix}$$
Use the Markov property to show that
$$P(X_{n}=1)=\dfrac{b}{a+b}+(1-a-b)^... | From
$$P(X_{n+1}=1)=(1-a-b)P(X_n=1)+b,$$
You can conclude that
\begin{align}
P(X_{n+1}=1)&=(1-a-b)\left[(1-a-b)P(X_{n-1}=1)+b\right]+b\\
&=(1-a-b)^2P(X_{n-1}=1)+(1-a-b)b+b
\end{align}
Repeating the procedure, we have
\begin{align}
P(X_{n}=1)&=(1-a-b)^nP(X_0=1)+b \sum_{i=0}^{n-1}(1-a-b)^i\\
&=(1-a-b)^nP(X_0=1)+b \left[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Range of $xyz\;,$ If $x+y+z=4$ and $x^2+y^2+z^2=6$
If $x,y,z\in \mathbb{R}$ and $x+y+z=4$ and $x^2+y^2+z^2=6\;,$ Then range of $xyz$
$\bf{My\; Try::}$Using $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
So we get $$16=6+2(xy+yz+zx)\Rightarrow xy+yz+zx = -5$$ and given $x+y+z=4$
Now let $xyz=c\;,$ Now leyt $t=x,y,z$ be the roo... | I thought it might be worth showing the Lagrange-multiplier method applied to this problem, largely for illustrating how the system of "Lagrange equations" can be handled, and for showing the interesting character of the solution.
With the constraints $ \ x^2 \ + \ y^2 \ + \ z^2 \ = \ 6 \ $ and $ \ x \ + \ y \ + \ z \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Showing $\int_{0}^{1}{1-x^2\over x^2}\ln\left({(1+x^2)^2\over 1-x^2}\right)dx=2$ Showing
$$I=\int_{0}^{1}{1-x^2\over x^2}\ln\left({(1+x^2)^2\over 1-x^2}\right)dx=2\tag1$$
$$I=\int_{0}^{\infty}{1-x^2\over x^2}\ln\left({1+x^2\over 1-x^2}\right)+\int_{0}^{\infty}{1-x^2\over x^2}\ln(1+x^2)dx\tag2$$
Let $$J=\int_{0}^{\in... | We have $$I=\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(\frac{\left(1+x^{2}\right)^{2}}{1-x^{2}}\right)dx$$ $$=2\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(1+x^{2}\right)dx-\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(1-x^{2}\right)dx
$$ Let us analyze the first integral. We have $$\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Finding solution of the irrational equation Given equation $\sqrt{x + 3 - 2\sqrt{x + 2}} + \sqrt{x + 27 - 10\sqrt{x + 2}} = 4$, find its solution(s).
At first, finding the domain of the function. Noting that $\sqrt{x + 2} \geq 0 \implies x \geq -2$. Then, solving inequalities for $x + 3 - 2\sqrt{x + 2} \geq 0$ and $x +... |
Solving the equation via substitution $t = \sqrt{x + 2}$ we get:
$$\sqrt{x + 3 - 2t} + \sqrt{x + 27 - 10t} = 4$$
$$x + 3 - 2t = 16 - 8\sqrt{x + 27 - 10t} + x + 27 - 10t$$
Here, you have to have
$$\sqrt{x+3-2t}=4-\sqrt{x+27-10t}\color{red}{\ge 0}$$
$$t^2 - 10t + 25 = x + 27 - 10t$$
Here, you have to have
$$\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find $3^{333} + 7^{777}\pmod{ 50}$ As title say, I need to find remainder of these to numbers. I know that here is plenty of similar questions, but non of these gives me right explanation. I always get stuck at some point (mostly right at the beginning) and don't have idea how to start.
Thanks in advance.
| You can seperate and do it $\pmod{2}$ and $\pmod{25}$ and use chinese remainder:
They are both odd so their sum is even and thus $\equiv 0 \pmod{2}$.
The euler function of 25 gives 20 and thus $3^{333}\equiv 3^{13}\pmod{25}$ and $7^{777}\equiv 7^{17} \pmod{25}$.
Now, $7^2=49\equiv -1 \pmod{25}$. Thus $7^{17}=7^{16}\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Computing $\underset{x^2+y^2+(z-2)^2\le 1}{\int\int\int}{1\over x^2+y^2+z^2}dxdydz$ in Spherical Coordinates Compute: $\underset{x^2+y^2+(z-2)^2\le 1}{\int\int\int}{1\over x^2+y^2+z^2}dxdydz$.
Hint given: show that
*
*$\cos \theta> {r^2+3\over 4r}$
*$1<r<3$
What I already did: I shift the unit sphere two coordin... | I think the aim of the hint is to guide you through finding the limits of integration. To me, there is no need for shifting the coordinate system. You can proceed as follows
$$\begin{align}{}
I
&=\underset{x^2+y^2+(z-2)^2\le 1}{\int\int\int}{1\over x^2+y^2+z^2}dxdydz \\
&= \int_{\phi=0}^{2 \pi} \int_{r=1}^{3} \int_{\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate $\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\cdots$ $$\frac{0!}{4!}+\frac{1!}{5!}+\frac{2!}{6!}+\frac{3!}{7!}+\frac{4!}{8!}+\frac{5!}{9!}+\frac{6!}{10!}+\cdots$$
This goes up to infinity. Trying finite cases may help.
My Attempt:It seems that it is going to be $\frac{1}{18}$. My calc... | Just a tip; For natural number $n$,
$$\sum_{k=1}^n\frac{1}{k(k+1)\cdots(k+m)}=\frac{1}{m}\left(\frac{1}{1\cdot 2\cdot \cdots\cdot m} -\frac{1}{(n+1)\cdot (n+2)\cdot \cdots\cdot (n+m)}\right) $$
Now, the given infinite sum is equal to
$$\sum_{k=1}^n \frac{(k-1)!}{(k+3)!}=\sum_{k=1}^n \frac{1}{k(k+1)(k+2)(k+3)}$$
by plu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
Prove that $ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $ Prove:
$$ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $$
Hypothesis:
$$ F(x) = 1+2q+3q^2+...+xq^{x-1} = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} $$
Proof:
$$ P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x... | Hint: Consider the finite geometric series given by
\begin{equation}
Q(q,p)=\sum_{k=1}^p q^k=\frac{1-q^{p+1}}{1-q}
\end{equation}
Consider $\frac{d}{dq}G(q,n)$ and see where it takes you...
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Why does an argument similiar to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$ show that $2+4+8+...=-2$ See how to prove $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$
$x=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
$2x=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
Then:
$x=1$
Now I use the same argument to prove $2+4+8+...=-2$
$x... | $\def\bR{\mathbb{R}}\def\bQ{\mathbb{Q}}$Well, $\sum_{k\geq1} 2^k$ converges to $-2$, but not in $\bR$, it does so in $\bQ_2$, the $2$-adic completion of $\bQ$. (In $\bR$, it obviously diverges.)
Let me prove that $\sum_{k\geq0} 2^k\to-1$, which is the same (just multiply by $2$). We have, for the partial sums, $A_n:=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Suppose that $a$ and $b$ satisfy $a^2b|a^3+b^3$. Prove that $a=b$.
Suppose that $a$ and $b \in \mathbb{Z}^+$ satisfy $a^2b|a^3+b^3$. Prove that $a=b$.
I have reduced the above formulation to these two cases. Assuming $b = a + k$. Proving that any of the below two implies that $a=b$ will be enough.
$$a^2b|(a+b)^3 - 3a... | Let $a=da_1$, $b=db_1$, where $d=\gcd(a,b)$, $a_1,b_1\in\mathbb Z^+$. Then $\gcd(a_1,b_1)=1$, $\gcd\left(a_1^2,b\right)=1$ and $$a^2b\mid a^3+b^3$$
$$\iff d^3a_1^2b_1\mid d^3a_1^3+d^3b_1^3$$
$$\iff a_1^2b_1\mid a_1^3+b_1^3$$
$$\iff \begin{cases}a_1^2\mid a_1^3+b_1^3\\b_1\mid a_1^3+b_1^3\end{cases}\iff \begin{cases}a_1^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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prove $\lim_{n\to\infty}{\frac{a^n}{n!}}=0$ I have the proof but i don't understand one part. The proof (for $a>1$) goes as follows:
there exists $k \in N$ such that$a<k, \frac{a}{k}<1$ and since $\frac{a}{k+i}<\frac{a}{k}, i \in N$
$$\frac{a^n}{n!}=\frac{a}{1} \cdot \frac{a}{2} \cdot...\cdot\frac{a}{k}\cdot\frac{a}{k+... | We consider $\frac{a}{1}.\frac{a}{2}\ldots \frac{a}{k}\frac{a}{k+1}\ldots \frac{a}{n-2}\frac{a}{n-1}\frac{a}{n}$
Now since $\frac{a}{n}<\frac{a}{n-1}$ and $\frac{a}{n-1}<\frac{a}{n-2}$ and so on till we reach $\frac{a}{n-(n-k)}<\frac{a}{n-(n-k-1)}$ i.e. $\frac{a}{k}<\frac{a}{k-1}$. When we repeatedly use this inequalit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Finding all $z\in \mathbb{C}$ such that the series $\sum\limits_{n=1}^{\infty} \frac{1}{1+z^n}$ converges I am trying to find out all $z\in \mathbb{C}$ such that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{1+z^n}$ converges.
I notice that for $\left|z\right|\leq 1$, we have $\left|1+z^n\right|\leq 1+\left|z... | Suppose $|z|>1$. Suppose $z=r(\cos\theta + i \sin \theta)$ with $r>1$. Then $\displaystyle |1+z^n|=\sqrt{1+2r^n\cos n\theta+r^{2n}}>(r^n-1)$, so $\displaystyle \frac{1}{|1+z^n|}<\frac{1}{r^n-1}$. We will try to prove that $\displaystyle \sum \frac{1}{r^n-1}$ is convergent which will give us our desired result by the co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0$ without $\varepsilon - \delta$. Unlike Multivariable Delta Epsilon Proof $\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}$ --- looking for a hint I would like to avoid the $\varepsilon - \delta$ criterium.
Prove $$\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0 ... | You have:
$$(x^2-y^2)^2 \ge 0 \Rightarrow x^4+y^4 \ge 2x^2y^2$$
So also:
$$x^2y^2 \le x^4+y^4$$
Which means:
$$\left| \frac{x^2y^3}{x^4+y^4} \right| \le \left| \frac{x^2y^3}{x^2y^2} \right| = |y|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Differential equation $\left(x^2+xy\right)y'=x\sqrt{x^2-y^2}+xy+y^2$ I am not sure which type of differential equation this falls into:
$$\left(x^2+xy\right)y'=x\sqrt{x^2-y^2}+xy+y^2$$ any hints? P.S. I first tried reornazing it so I have $y'$ alone, and hoping that I would get a homogeneous equation, but no such luck.... | Divide both side to the $x^{ 2 }$ $$\left( x^{ 2 }+xy \right) y'=x\sqrt { x^{ 2 }-y^{ 2 } } +xy+y^{ 2 }\\ \left( 1+\frac { y }{ x } \right) { y }^{ \prime }=\sqrt { 1-\frac { { y }^{ 2 } }{ { x }^{ 2 } } } +\frac { y }{ x } +\frac { { y }^{ 2 } }{ { x }^{ 2 } } \\ y=zx\\ { y }^{ \prime }={ z }^{ \prime }x+z\\ \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve for $X$ in matrix equation How can I solve for $X$ in this matrix equation?
$$\begin{bmatrix}-3&-8\\-9&5\end{bmatrix} X + \begin{bmatrix}4&-7\\3&-2\end{bmatrix} = \begin{bmatrix}5&8\\-1&-1\end{bmatrix} X$$
First I tried $AX + B = CX$ but then I don't know how to solve for $X$ because no matter what I do I end up ... | Did you consider, since these are 2 by 2 matrices, just writing X as $\begin{bmatrix}a & b \\ c & d \end{bmatrix}$ and then doing the indicated operations?
$\begin{bmatrix}-3 & -8 \\ -9 & 5\end{bmatrix}$$\begin{bmatrix}a & b \\ c & d \end{bmatrix}+ \begin{bmatrix}4 & -7 \\ 3 & -2 \end{bmatrix}$$= \begin{bmatrix}5 & 8 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Minimum value of $\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$ Recently I was solving one question, in which I was solving for the smallest value of this expression
$$f(\theta)=\cos^2\theta-6\sin\theta \cos\theta+3\sin^2\theta+2$$
My first attempt:
$$\begin{align}
f(\theta) &=3+2\sin^2\theta-6\sin\theta \cos\... | $f(\theta) = 4 - \sqrt{10}$ is correct.
so what is the error here:
$f(\theta) = 3(\sin \theta-\cos\theta)^2 + 2\sin^2\theta$
That part is a true statement but then you say.
Hence the minimum value of $f(θ)=2\sin^2θ,$ when $θ=π/4$ hence minimum value of $f(θ)=1.$
It is a logical jump that was a step too far.
If you had ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the congruence $6x+15y \equiv 9 \pmod {18}$ Solve the congruence $6x+15y \equiv 9\pmod {18}$
Approach:
$(6,18)=6$, so $$15y \equiv 9\pmod 6$$
$$15y \equiv 3\pmod 6$$
So the equation will have $(15,6)$ solutions. Now we divide by 3
$$5y \equiv 1\pmod 2$$.
Solving the Diophantine equation we get $y \equiv1\pmod 2 ... | I hope my answer is okay.
$6x+15y \equiv 9(mod 18)$
$=> 2x+5y \equiv 3(mod 6)$
$=> 2x-y \equiv 3(mod 6)$
i.e $2x-y-3 \equiv 0(mod 2)$ and $\equiv 0(mod 3)$
this means $y+1 \equiv 0(mod 2)$ and $x+y \equiv 0(mod 3)$
Write $y=2p+1$ and so $x-p +1 \equiv 0(mod 3)$
Then use parametric equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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An inequality of three positive variables Suppose that $x,y,z>0$ and $xyz=1$. Why does the inequality $$\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{x+z+1}\leq 1$$ holds? I couldn't see anything useful as I tried Jensen's inequality and calculus methods.
| Hint
If $a,b$ and $c$ be positive then
$$\frac{1}{{{a}^{3}}+{{b}^{3}}+abc}+\frac{1}{{{b}^{3}}+{{c}^{3}}+abc}+\frac{1}{{{c}^{3}}+{{a}^{3}}+abc}\le \frac{1}{abc}$$
set
$$a=\sqrt[3]{x}\quad,\quad b=\sqrt[3]{y}\quad,\quad c=\sqrt[3]{z}$$
Edit
$$I=\frac{1}{{{a}^{3}}+{{b}^{3}}+abc}\le \frac{1}{a^2b+b^2a+abc}$$
$$J=\frac{1}{{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$
\sin 2x = \sin x; \\ 0 \le x < 2 \pi $$
My method:
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$ \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\le... | Other people have already pointed out we can rewrite like this:
$$2\sin(t)\cos(t) = \sin(t)$$
Now what most other people have not done, is this to substitute $\sin(t)$ and $\cos(t)$ for $y$ and $x$ and rewrite the problem to an algebraic variety:
$$\cases{2yx - y = 0\\
x^2+y^2=0}$$
This particular one is really silly e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$ How to find ?$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$$
I tried using the substitution $x^2=z$.But that did not help much.
| Let $$I = \int\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx = \int\frac{x^2-1}{x^3\cdot x^2\sqrt{2-2x^{-2}+x^{-4}}}dx$$
So $$I=\int\frac{x^{-3}-x^{-5}}{\sqrt{2-2x^{-2}+x^{-4}}}dx$$
Now Put $2-2x^{-2}+x^{-4} = t^2\;,$ Then $4(x^{-3}-x^{-5})dx = 2tdt$
So $$I = \frac{1}{2}\int\frac{t}{t}dt = \frac{1}{2}t+\mathcal{C}=\frac{1}{2}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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What is $e^{A}$ where A is an anti-diagonal matrix I am trying to get a closed form for the matrix produced by the following operation: $$e^A$$ where $A$ is an anti diagonal matrix, say, of size $2\times 2$:
$$A=\begin{pmatrix}
0 &b \\
c &0
\end{pmatrix}$$
Using Mathematica MatrixExp I got
$$ \left(
\begin{array}{cc}... | $$A=\begin{pmatrix}
0 &b \\
c &0
\end{pmatrix}$$
So
$$A^2 = bc I$$
Then
$$e^A = \sum_{k=0}^\infty\frac{A^k}{k!} = \sum_{k=0}^\infty\frac{A^{2k}}{(2k)!} + \frac{A^{2k + 1}}{(2k+1)!} \\
= \sum_{k=0}^\infty\frac{(bc)^k}{(2k)!} I + \frac{(bc)^k }{(2k+1)!} A $$
And you can then sum this component-wise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Abuse of notation for infimum and supremum I would like to take the infimum and supremum of two sets $(\frac{1}{2} e^{8m+4} - 1, e^{8m+4} - 1)$ and $(\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})$, but writing
$\sup((\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})) < \inf((\frac{1}{2}
e^{8m+4} - 1, e^{8m+4} - 1)) \log(\inf((... | This isn't a terrible abuse of notation, but it might take some thought for your reader to figure out what you mean. So I think a better way to do this would be to just give names to the sets themselves, rather than elements of them. If you define $K=(\frac{3}{2} e^{8m+4} - 1, 2e^{8m+4} - 1)$, $R=(\frac{1}{2} e^{8m+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving that $1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } $ by induction
Prove that
$$1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 }. $$
I can get to $1/3(k+1)(k+2) + (k+1)(k+2)$ but then finishing off an... | When $n=1$, it is trival.
Suppose it is true for $n=k$, ie. $\sum \limits_i^k i(i+1)=\frac{k(k+1)(k+2)}{3}$.
Consider $n=k+1$.
$$\begin{align}
\sum \limits_i^{k+1} i(i+1)=&(k+1)(k+2)+\sum \limits_i^{k} i(i+1) \\
=&(k+1)(k+2)+\frac{k(k+1)(k+2)}{3} \\
=&\frac{(k^2+4k+3)(k+2)}{3} \\
=&\frac{(k+1)(k+1+1)(k+1+2)}{3}.
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Solving what Mathematica could not Right, so as the final step of my project draws near and after having made a bad layout sort of question, I am posting a new one very clear and unambiguous. I need to find this specific definite integral which Mathematica could not solve:
$$ \int_{x=0}^\pi \frac{\cos \frac{x}{2}}{\sq... | Let $\mathcal{I}$ be the integral at hand and $B = \frac{1+A}{\sqrt{A}}\omega$. Introduce variables $y, t, \theta, z$ such that
$$y = \sin\frac{x}{2},\quad t = \tanh\theta = \frac{\cos\frac{x}{2}}{\sqrt{1+A\sin^2\frac{x}{2}}}\quad\text{ and }\quad z = e^\theta$$
Notice
$$t = \sqrt{\frac{1-y^2}{1+Ay^2}} \implies
y = \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
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Proof of the square root inequality $2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}$ I stumbled on the following inequality: For all $n\geq 1,$
$$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}.$$
However I cannot find the proof of this anywhere.
Any ideas how to proceed?
Edit: I posted a... | $$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}\iff\sqrt{n^2+n}-n\lt\frac 12\lt n-\sqrt{n^2-1}$$
Put $X=n+\frac 12$ and $Y=n-\frac 12$ so you get the evidences
$$\sqrt{X^2-\frac 14}\lt X\iff X^2-\frac 14\lt X^2$$ and $$\sqrt{Y^2-\frac 14}\lt Y\iff Y^2-\frac 14\lt Y^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
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Show that $u_1^3+u_2^3+\cdots+u_n^3$ is a multiple of $u_1+u_2+\cdots+u_n$
*
*Let $k$ be a positive integer.
*Define $u_0 = 0\,,\ u_1 = 1\ $ and $\ u_n = k\,u_{n-1}\ -\ u_{n-2}\,,\
n \geq 2$.
*Show that for each integer $n$, the number
$u_{1}^{3} + u_{2}^{3} + \cdots + u_{n}^{3}\ $ is a multiple of
$\ ... | It appears that $$y_n = \dfrac{u_1^3 + \ldots u_n^3}{u_1 + \ldots + u_n}$$
satisfies the recurrence relation
$$ y_n = (k^2+k-1)(y_{n-1} - k y_{n-2} + k y_{n-3} - y_{n-4}) + y_{n-5} \ \text{for}\ n \ge 6$$
Given that $y_1, \ldots, y_5$ are integers, this would imply that all $y_n$ are integers.
EDIT: Writing $\cos(\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Is the polynomial $f(x) = x^4 + tx^3 + (t^2 + 1)x^2 + (t^3 + t)x + (t^4 + t^2)$ irreducible over $k(t)$? Let $k$ be an algebraically closed field of characteristic 2 and let $k(t)$ be rational function field of one variable. Consider the polynomial $f(x) = x^4 + tx^3 + (t^2 + 1)x^2 + (t^3 + t)x + (t^4 + t^2) \in k(t)[x... | First, by Gauss' lemma, $f(x)$ is irreducible over $k(t)$ iff it's irreducible over $k[t]$, so we may do all our calculating in the ring $k[t]$.
The polynomial has no linear factors $(x-a)$, as if it did then $a|t^4+t^2=t^2(t+1)^2$ by the rational root theorem, and each of the 9 possibilities may be plugged in to the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trouble finding the inverse of $f(x) = x + \frac{1}{x}$ . Let $ f: \Bbb R - \{0\} \rightarrow \Bbb R \;\text{ given by } f(x) = x + \frac{1}{x} . \text{Find} $
$f(f^{-1}(\Bbb R))$ , $\Bbb R = \text{real numbers}$.
For this problem I know one needs to find the inverse in order to solve.
$ y = x+ \frac{1}{x} \Rightarr... | You might not need to compute an inverse in order to solve this, but it may help to think about it. I would say
$$\begin{align*}
f(x) &= x + \frac{1}{x}\\
&= \frac{x^2}{x} + \frac{1}{x}\\
&= \frac{x^2 + 1}{x}\\
\\
y &\equiv \frac{x^2 + 1}{x}\\
xy &= x^2 + 1\\
0 &= x^2 - xy + 1\\
\end{align*}$$
Then solve for $x$ as a f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the value of $I=\lim_{n \to \infty} \int_0^1 {{1 + nx^2}\over{(1 + x^2)^n}} \log(2 + \cos(x/n))\,dx.$? Find the integral $I$.....it looks like a good problem which I was not able to solve ....please help...
$$I=\lim_{n \to \infty} \int_0^1 {{1 + nx^2}\over{(1 + x^2)^n}} \log(2 + \cos(x/n))\,dx.$$
| Note that after enforcing the substitution $x\to x/\sqrt{n}$, we have
$$\begin{align}
\left|\int_0^1 \frac{1+nx^2}{(1+x^2)^n}\log\left(2+\cos\left(\frac{x}{n}\right)\right)\,dx\right|&=\frac{1}{\sqrt{n}}\int_0^\sqrt{n} \frac{1+x^2}{(1+x^2/n)^n}\log\left(2+\cos\left(\frac{x}{n^{3/2}}\right)\right)\,dx\\\\
&\le \frac{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Factoring polynomials with a 2nd degree coefficient greater than $1$ I'm learning how to factor polynomials, but I'm having a hard time understanding the approach when the 2nd degree coefficient is greater than $1$.
For example, when I begin to factor $12k^4 + 22k^3 - 70k^2$, I first break it down to $2k^2(6k^2 + 11k -... | To split the linear term of $6k^2 + 11k - 35$, we must find two numbers with product $6 \cdot -35 = -210$ and sum $11$. They are $21$ and $-10$. Hence,
\begin{align*}
6x^2 + 11k - 35 & = \color{blue}{6}x^2 \color{green}{+ 21}x \color{green}{- 10}x \color{blue}{- 35} && \text{split the linear term}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Evaluation of $\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$ Evaluate the following limit:
$$L=\lim_{x\to 0} \frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}$$
Using $\ln(1+x)=x-x^2/2+x^3/3-\cdots$
I got $(1+x)^{1/x}=e^{1-x/2+x^2/3-\cdots}$
Could some tell me how to proceed further?
| You can proceed in the following manner
\begin{align}
L &= \lim_{x \to 0}\dfrac{(1 + x)^{1/x} - e + \dfrac{ex}{2}}{x^{2}}\notag\\
&= e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1 + \dfrac{x}{2}}{x^{2}}\notag\\
&= e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - \exp\left(\log\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to find $ \tan \left(\frac{x}{2}\right) $ knowing that $\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $ Good evening to everyone. I don't know how to find $ \tan \left(\frac{x}{2}\right) $ knowing that $$\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $$ and x$\in (0,\frac{\pi}{3})$ Here's what I've tried... | Since:
$$ \cos(x)=\frac{1-\tan^2\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)},\qquad\sin(x) =\frac{2\tan\left(\frac{x}{2}\right)}{1+\tan^2\left(\frac{x}{2}\right)}\tag{1}$$
we get:
$$ \left(1-\tan\left(\frac{x}{2}\right)\right)^2 = \frac{7}{5}\left(1+\tan^2\left(\frac{x}{2}\right)\right)\tag{2}$$
and $\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Asymptotic solution of the equation $\gamma_{i+2} + 4\gamma_{i+1} + \gamma_{i} = \frac{Kh^2}{12}$ I'm struggling with the following equation, I'm interested in an asymptotic solution:
$$\gamma_{i+2} + 4\gamma_{i+1} + \gamma_{i} = \frac{Kh^2}{12}$$
Where $K$ is known constant, when $h \rightarrow 0$ I guess I can assume... | Assuming $h \neq h(i)$, i.e. $h$ is not a function of $i$, you can start by solving a difference equation: write out an expression for $ \gamma_{i+1}$ and subtract it from the one you have. Thus you will get rid of the constant term and work with $b_{i+2} = \gamma_{i+2} - \gamma_{i+1}$. From there you can solve it usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$
*
*$\sin x + \sin y = 1$
*$\cos x + \cos y = 0$
Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.
I got the question from chapter 26 of a comic called Yamada-kun.
How can I solve this equa... | There are so many different ways to solve it the real question is which way.
What reaches out to grab me is:
$\cos x + \cos y = 0$
$\cos x = - \cos y$ which means either $y = \pi - x$ (within a period of $2\pi$) or $y = x + \pi$ (within a period of $2\pi$).
If $y = x + \pi$ then $\sin y = - \sin x$ and $\sin y + \sin x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 0
} |
Compute $1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \cdots + n \cdot \frac {1}{2^n} + \cdots $ I have tried to compute the first few terms to try to find a pattern but I got
$$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$
but I still don't see any obvious pattern... | $$I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\cdots$$
$$2I=1+1+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\cdots$$
$$2I-I=1+\left(1-\frac 12 \right)+\left(\frac 34 -\frac 24 \right)+\left(\frac 48 -\frac 38 \right)+\left(\frac {5}{16} -\frac {4}{16} \right)+\cdots$$
$$I=1+\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Evaluation of $\sum^{\infty} _{n=1} \arctan\left(\frac{4n}{n^4-2n^2+2}\right)$
Evaluate $\displaystyle\sum^{\infty} _{n=1} \arctan\left(\frac{4n}{n^4-2n^2+2}\right).$
I know we know to convert it in the of $\arctan\left(\frac{a-b}{1+ab}\right)$ but I am not able to do so here.
Could someone give me some hint?
| Hint. One may write, for $n\ge2$,
$$
\frac{4n}{n^4-2n^2+2}=\frac{4n}{(n^2-1)^2+1}=\frac{\frac{4n}{(n^2-1)^2}}{1+\frac1{(n^2-1)^2}}=\frac{\frac1{(n-1)^2}-\frac1{(n+1)^2}}{1+\frac1{(n-1)^2(n+1)^2}}
$$ giving here
$$
\arctan\frac{4n}{n^4-2n^2+2}=\arctan\frac1{(n-1)^2}-\arctan\frac1{(n+1)^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving $28^x \equiv 2 \pmod{43}$
How do we solve $28^x \equiv 2 \pmod{43}$?
I know there are not generally efficient methods for computing the discrete logarithm which are defined for an invertible $a$ modulo $q$ by $$a \equiv t^x \pmod{q}, \quad 0 \leq x \leq q-1,$$ but I was wondering if there was a way to compute... | We try to make $\,2\equiv 45\equiv \color{#c00}5\cdot \color{#0a0}3^2\,$ using factors of $\,3\,$ and $\,5\,$ from small powers of $\,28.\,$
$\qquad\qquad 28^3\equiv 22\equiv -\color{#0a0}3\cdot 7$
$\qquad\qquad \color{#c00}{28^5}\equiv \color{#c00}5$
$\qquad\qquad 28^7\equiv 7$
So we have $\,\color{#c00}5\,$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Partial fractions and using values not in domain I'm studying partial fraction decomposition of rational expression. In this video the guy decompose this rational expression:
$$ \frac{3x-8}{x^2-4x-5}$$
this becomes:
$$\frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1} $$
$$[(x-5)(x+1)]\times \frac{3x-8}{(x-5)(x+1)... | When you get to the stage $3x-8 = A(x+1) + B(x-5)$, you've forgotten about the partial fractions and all you're trying to do is determine a polynomial identity. i.e: what $A$ and $B$ will make $A(x+1) + B(x-5)$ identical to $3x-8$ for all values of $x$, this includes $x=-1$ and $5$.
Once you've got those values, then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Prove using induction the following equation is true. If $$(1-x^2)\frac{dy}{dx} - xy - 1 = 0$$
Using induction prove the following for any positive integer n$$(1-x^2)\frac{d^{n+2}y}{dx^{n+2}} - (2n+3)x\frac{d^{n+1}y}{dx^{n+1}} - (n+1)^2\frac{d^ny}{dx^n} = 0$$
I know Leibtniz can be used to solve it easier but I need th... | You have
$(1-x^2)y' - xy - 1 = 0
$
and you want to show that
$(1-x^2)y^{(n+2)} - (2n+3)xy^{(n+1)} - (n+1)^2y^{(n)} = 0
$.
Rewrite these as
$(1-x^2)y'
= xy + 1
$
and
$(1-x^2)y^{(n+2)}
= (2n+3)xy^{(n+1)} + (n+1)^2y^{(n)}
$.
Differentiating the first one,
$(1-x^2)y''-2xy'
= xy'+y
$
or
$(1-x^2)y''
= 3xy'+y
$,
which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$ where $a,b \gt 0$ Evaluate
$$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$$ where $a,b \gt 0$
I tried using $y=e^x$, but I still can't solve it.
I get $\displaystyle\int_0^\infty \frac y{ay^3+b} \, dy.$
Is there any different method to solve it?
| $$
ay^3 + b = a\left( y^3 + \frac b a \right)
$$
$$
y^3 + \frac b a = y^3 +c^3 = (y+c)(y^2 - yc + c^2) \quad\text{where } c = \sqrt[3]\frac b a.
$$
So use partial fractions to get
$$
\frac \bullet {y+c} + \frac {((\bullet\, y) + \bullet)} {y^2 - yc + c^2}.
$$
To integrate the second term, you have
$$
\frac {ey+f}{y^2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Find the total number of different integers that the function takes
Find the total number of different integers that the function $$f(x) = [x]+[2x]+\left[\dfrac{5x}{3}\right]+[3x]+[4x]$$ takes for $0 \leq x \leq 100$. (Note: $[x]$ denotes the greatest integer not exceeding $x$.)
Is my solution below correct?
Attempt:... | Your approach is correct, though your logic is written in a way that seems somewhat ambiguous. The most notable thing is that the condition for $f$ to experience a jump at $x$ is that $x$ is a (integer) multiple of one of $1,\,\frac{1}2,\,\frac{1}3,\,\frac{1}4,\,\frac{3}5$. If you need more formality, this a convenien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
coefficient for interpolation of data $(\frac{\pi}{2},1), (0,-1)$ with $P(x)=c_1p_1(x)+c_2p_2(x)$? Equation: $P_1(x)=cos^2x$, $p_2(x)=sin^2y$
Goal is: interpolation of data $(\frac{\pi}{2},1), (0,-1)$ with
$P(x)=c_1p_1(x)+c_2p_2(x)$
Question: find $c_1, c_2$.
Answer: $c_1=-1, c_2=1$
My question is via the answer of ... | Since $P(x)=c_1p_1(x)+c_2p_2(x)$ so
$\begin{pmatrix}p_1(x) & p_2(x)\\ p_1(y) & p_2(y) \end{pmatrix} \begin{pmatrix}c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix}P(x) \\ P(y) \end{pmatrix}$
then
$\begin{pmatrix}p_1(\frac{\pi}{2}) & p_2(\frac{\pi}{2})\\ p_1(0) & p_2(0) \end{pmatrix} \begin{pmatrix}c_1 \\ c_2 \end{pmatrix} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve the limit $\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{n+1}{n}\right)\right]$? Hi I got an examination at the school which was so arduous that I'm stumped.
This problem is the toughest for me :
$$\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\righ... | We can proceed as follows
\begin{align}
L &= \lim_{n \to \infty}n^{2}\left\{\left(1 + \frac{1}{n(n + 2)}\right)^{n} - \frac{n + 1}{n}\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\left\{\left(\frac{(n + 1)^{2}}{n(n + 2)}\right)^{n} - \frac{n + 1}{n}\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\left\{\exp\left(n\log\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Convergence/Divergence of $\sum_{n=1}^{\infty}\frac{n+n^2+\cdots+n^n}{n^{n+2}}$
$$\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}$$
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \su... | We have: $n+n^2+\cdots + n^n = n(1+n+n^2+\cdots + n^{n-1})= \dfrac{n(n^n-1)}{n-1}\implies \displaystyle \sum_{n=1}^\infty \dfrac{n+n^2+\cdots n^n}{n^{n+2}} < \displaystyle \sum_{n=2}^\infty \dfrac{1}{n(n-1)}$, and this one converges .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Formula for $1^k+2^k+3^k...n^k$ for $n,k \in \mathbb{N}$ So I've been looking for a formula where I can input the parameter $k$ and it will give me a formula for $1^k+2^k+3^k...+ n^k$ with $n,k \in \mathbb{N}$. The result is always a polynomial with $k+1$ as highest power. I've taken the time to calculate the polynomes... | https://en.m.wikipedia.org/wiki/Faulhaber%27s_formula seems to deliver exactly what i needed. Original answer by MooS
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Decide if the series converges and prove it using comparison test: $\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}$
Decide if the series converges and prove it using comparison test:
$\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}$
$$\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}< \frac{k^{2}+k}{k^{4... | Following your idea I would write
$$\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}< \frac{3k^{2}+3k}{k^{4}+k^{3}} = \frac{3k}{k^{3}} = \frac{3}{k^{2}}.$$
As regards your inequalities, how do you justify that
$$\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}< \frac{k^{2}+k}{k^{4}+k^{3}}?$$
Notice that if $a$, $b$, $c$, $d$ are positive numbers
$$a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$
Calculate $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$
\begin{align}\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}&=3\sum_{k=2}^{\infty}\frac{1}{5^{k-1}}\\&=3\sum_{k=2}^{\infty}\frac{1}{5^{k}}\cdot\frac{1}{5^{-1}}\\&=3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5\\&= 15\lef... | You shoould obtain a positive number. Maybe you could follow this simpler approach:
$$\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}=\frac{3}{5}\sum_{k=2}^{\infty}\frac{1}{5^{k-2}}=\frac{3}{5}\sum_{j=0}^{\infty}\frac{1}{5^{j}}=\frac{3}{5}\cdot\frac{1}{1-1/5}=\frac{3}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to compute $\sum_{k=1}^n \frac{1}{\sin^2(\frac{k\pi}{2n+1})}$ efficiently?
Compute $\sum_{k=1}^n \frac{1}{\sin^2(\frac{k\pi}{2n+1})}$ in terms of $n$.
Here's my way of doing it: using the classic expansion of $\sin((2n+1)x)$ as a polynomial in $\sin(x)$ we have:
$$\sin((2n+1)x) = \sum_{k=0}^n \binom{2n+1}{2k+1}(-... | It is enough to recall Cauchy's proof of the Basel problem, relying on the identity:
$$ \binom{2n+1}{1}t^n-\binom{2n+1}{3}t^{n-1}+\ldots+(-1)^n\binom{2n+1}{2n+1}=\prod_{k=1}^{n}\left(t-\cot^2\frac{k\pi}{2n+1}\right)\tag{1} $$
By Vieta's formulas,
$$ \sum_{k=1}^{n}\cot^2\left(\frac{\pi k}{2n+1}\right) = \binom{2n+1}{3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $r^l, r^{-l-1}$ are solutions to $\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)=l(l+1)$ Here is a standard textbook question that is causing great difficulty (even though the question says it is easy):
I'm only interested in finding the solution to the second of the above $2$ differential equations by... | Did has already explained in the comments that it makes more sense to use a single power ansatz than a power series. If you really want to use a power series, you have to include negative powers if you want to get the $r^{-l-1}$ solution, since this can't be written as a series of non-negative powers at $r=0$.
In the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Classify up to similarity all the matrices A $\in Q^{3 \times 3}$ such that $A^3$ = I I know the minimal polynomial $p_A$ must divide $(x - 1)(x^2 + x + 1)$ since $A^3 - I = 0$.
If $p_A$ = (x - 1), then A = I.
If $p_A = (x - 1)(x^2 + x + 1)$, then $p_A$ is equal to the characteristic polynomial, since it is of degree... | Hint:
The minimal polynomial $p_A(x)$ is a divisor of $x^3-1$, so it is either $x-1$ or $x^2+x+1$ or $x^3-1$
Now the minimal polynomial and the characteristic polynomial have the same irreducible factors, so:
*
*if $p_A(x)=x-1$, $\chi_A(x)=(x-1)^3$ (the matrix is $I$);
*the case $p_A(x)=x^2+x+1$ cannot happen, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Better way to evaluate $\int \frac{dx}{\left (a +b\cos x \right)^2}$ $$\int \frac{dx}{\left (a +b\cos x \right)^2}$$
$$u=\frac{b +a \cos x}{a +b\cos x }$$
$$du=\frac{\sin x\left(b^2 -a^2\right)}{ \left (a +b.\cos x \right)^2}$$
$$\frac{du}{\sin x\left(b^2 -a^2\right)}=\frac{dx}{\left (a +b\cos x \right)^2}$$
$$\cos x=\... | I think the tangent half-angle substitution should work. Here's what I have:
\begin{align*}
\int \frac{\mathrm{d}x}{(a+b\cos{x})^2} &= \int \frac{2\,\mathrm{d}t}{(1+t^2)(a+b\frac{1-t^2}{1+t^2})^2}, \qquad t = \tan{(x/2)} \\
&= 2 \int \frac{(1+t^2)dt}{a^2(1+t^2)^2+2ab(1-t^2)(1+t^2)+b^2(1-t^2)^2}\\
&= 2\int \frac{(1+t^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find Maclaurin series of $\frac{x^2+3}{x^2-x-6}$
Find the Maclaurin series of $\frac{x^2+3}{x^2-x-6}$.
So far I have:
$$\frac{x^2+3}{x^2-x-6}=1+\frac{x+9}{(x-3)(x+2)}=1+\frac{x+2}{(x-3)(x+2)}+\frac{7}{(x-3)(x+2)}=1-\frac{1}{(3-x))}+\frac{7}{(x-3)(x+2)}$$
How should I continue?
| Hint:
Set
$$
\frac{7}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}
$$
And solve for $A$ and $B$.
This is called a "partial fraction decomposition". To solve for $A$ and $B$, generally you multiply both sides by $(x-3)(x+2)$ and plug in values of $x$: $x = 3$ and $x = -2$ will work well.
Finally, you need to know how to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Let $a,b,c$ be non-negative reals, and $k$ is the best possible constant Prove the inequality Let $a,b,c$ be nonnegative real numbers. Prove that
$$a^3+b^3+c^3-3abc\geq k|(a-b)(b-c)(c-a)|$$
where $k=\left(\frac{27}{4}\right)^{1/4}(1+\sqrt{3})$ and that $k$ is the best possible constant.
Not sure how to go about provin... | The problem amounts to minimizing:
$$ R(a,b,c)= \frac{a^3+b^3+c^3 - 3 a bc }{(b-a)(c-b)(c-a)} $$
on the set $0\leq a <b< c$.
The directional derivative:
$$ \frac{d}{dt}_{|t=0} R(a+t,b+t,c+t)= \frac{3}{2}\frac{(b-a)^2+(c-b)^2+(c-a)^2 }{(b-a)(c-b)(c-a)} $$ is non-negative so any minimum must occur for $a=0$. By homogenei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How to solve this determinant equation in a simpler way
Question Statement:-
Solve the following equation
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$$
My Solution:-
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=
\begin{vmatrix}
x+5 & 2 & 3 \\
x+5 & x & 1 ... | Notice that the first two columns are proportional, hence linearly dependent, if
$$\frac x 2 = \frac 4 x$$
which is the same as $x = \pm 2 \sqrt{2}$ after solving. Convince yourself that the third column can never be written as a linear combination of the first two by noticing that the first and second columns have equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Find all integral solutions of $ x^4 + y^4 + z^4 -w^4 = 1995 $ Find all integral solutions of $ x^4 + y^4 + z^4 -w^4 = 1995 $.
Attempt:
From FLT it can be concluded that either all of $ x , y , z$ and $w$ are multiples of 5 ( which is not possible since that would lead to $ x^4 + y^4 + z^4 -w^4 $ being a multiple of... | Just an addendum to wythagoras fine (+1) answer. $x^4\in\{0,1\}\pmod{16}$ can be proved by observing that if $x$ is even, $x^4$ is clearly a multiple of $16$, while
$$ (2k+1)^4-1 = 2k(2k+2)(4k^2+4k+2) = 16\binom{k+1}{2}(2k^2+2k+1) $$
gives $(2k+1)^4\equiv 1\pmod{16}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all prime numbers satisfying... Find all prime numbers p, q, r such that
$p(p+1)+q(q+1)=r(r+1)$.
I tried out with $6k\pm1$ form of prime numbers but got no useful result, other methods too proved futile.
| There is only one solution $p=q=2$ & $r=3$ , we will prove it for a positive integer $ n$
We deduce that , $p(p+1)=n(n+1)-q(q+1)=(n-q)(n+q+1)$ and we must have $ n>q$
Since $p $ is a prime we must have $ p|(n-q)$ or $p|(n+q+1)$ , Now if $p|(n-q)$ then $p\le (n-q)$ which implies $p(p+1)\le (n-q)(n-q+1)\implies (n-q)(n+q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all the solutions of this equation:$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $ in reals. Find all the solutions of this equation
$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $
My attempt:In the hint of the question it was written that show $x,y,z \in [-1,1]$ then add all equations and conclude $(x,y,z)=(-1,-1,-1),(0,0,0),(1,... | Brute force. If we use $y=4z^3-3z$ and $z=4x^3-3x$ in $x=4y^3-3y$, we obtain a huge polynomial of $27$th degree in $x$ which can be factorized (by some not-human algebraic manipulator) as
$$4x(x-1)(x+1)(8x^3+4x^2-4x-1)(8x^3-4x^2-4x+1)\\(64x^6-112x^4+56x^2-7)(64x^6-32x^5-80x^4+32x^3+24x^2-6x-1)\\(64x^6+32x^5-80x^4-32x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
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.