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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:
$x = \sqrt[3]{2} + 2\sqrt[3]{4}$
The rest of the roots can now be gotten through long division, or by using the imaginary cube roots of $2$ or $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
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 that
$$ \left(1 + \frac{z^2}{2^2} + \frac{z^4}{2^4} + \cdots\right) = \sum_{n=0}^{+\infty} \left(\frac{z}{2}\right)^{2n}$$
and
$$ \left(z + \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots \right) = \sum_{n=0}^{+\infty} \frac{z^{2n+1}}{(2n+1)!}$$
Let $R_1$ be the radius of convergence of the first series and $R_2$ be the radius of convergence of the second series.
Apply the ratio test to the first series to get $R_1 = 2$. Note that if $|z| = 2$ then the first series will diverge. So the disk of convergence for the first series is all $z \in \mathbb{C}$ such that $|z| < 2$. Apply the ratio test to the second series to get $R_2 = +\infty$. So the second series converges on all of $\mathbb{C}$.
We need both of these series to be absolutely convergent in order for the original series to converge. Therefore the original series will converge on the intersection of the domains of convergence for the other two series. Since the second series converges on all of $\mathbb{C}$ and the first series converges on $\{z \in \mathbb{C} : |z| < 2\}$, then the original series converges on $\{z \in \mathbb{C} : |z| < 2\}$. Therefore the radius of convergence of the original series is 2.
EDIT - Verifying $R_1$:
$a_n = \dfrac{z^{2n}}{2^{2n}} = \dfrac{z^{2n}}{4^n}$. Therefore $a_{n+1} = \dfrac{z^{2n+2}}{4^{n+1}} = \dfrac{z^{2n}z^2}{4^n4}.$
$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{z^{2n}z^2}{4^n4} \cdot \frac{4^n}{z^{2n}}\right| = \frac{|z|^2}{4}$$
Take the limit as $n \to +\infty$, still get $|z|^2 / 4$. Solve $|z|^2 / 4 < 1$ to get $|z| < 2$.
| {
"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
} |
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:
$s_1 = \frac{5}{6}$
$s_2 = \frac{5}{6} + \frac{13}{36} = \frac{43}{36}$
$s_3 = \frac{43}{36} + \frac{35}{216} = \frac{293}{216}$
$s_4 = \frac{293}{216} + \frac{97}{1296} = \frac{1855}{1296} $
I do not see a pattern here...
How must I proceed to find the sum of this series?
| It is just the sum of two geometric series in disguise.
$$
\begin{aligned}
\underset{i=0}{\overset{\infty}{\sum}}\frac{2^{n}+3^{n}}{6^{n}}&=\underset{i=0}{\overset{\infty}{\sum}}\frac{2^{n}}{6^{n}}+\frac{3^{n}}{6^{n}}\\
&=\underset{i=0}{\overset{\infty}{\sum}}(1/3)^{n}+\underset{i=0}{\overset{\infty}{\sum}}(1/2)^{n}\\
&=\frac{1}{1-\frac{1}{3}}+\frac{1}{1-\frac{1}{2}}\\
&=\frac{3}{2}+2\\
&=\frac{7}{2}.
\end{aligned}
$$
| {
"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
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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 0}\left[\frac{x^3}{f(x)^3}\cdot\frac{x(1+a\cos(x))-b\sin(x)}{x^3}\right]\\
&=\left(\lim_{x \to 0}\frac{x}{f(x)}\right)^3\cdot
\lim_{x\to0}\frac{x(1+a\cos(x))-b\sin(x)}{x^3}\\
&\stackrel{\rm H}{=}
1^3\cdot\lim_{x\to0}\frac{1+a\cos(x)-ax\sin(x)-b\cos(x)}{3x^2}\tag{1}\\
&\stackrel{\rm H}{=}
\lim_{x\to0}\frac{-a\sin(x)-a\sin(x)-ax\cos(x)+b\sin(x)}{6x}\\
&\stackrel{\rm H}{=}
\lim_{x\to0}\frac{-2a\cos(x)-a\cos(x)+ax\sin(x)+b\cos(x)}{6}\\
&=\frac{-3a+b}{6}\tag{2}.
\end{align}
By $(1)$, we have to force the limit of the numerator to be zero, so $1+a-b=0$. Also, by $(2)$ we naturally get $-3a+b=6$. Hence we conclude that
$a=-\frac{5}{2}$ and $b=-\frac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"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}\frac{1}{7}=\tan^{-1}\frac{1}{2}$ and so on but cannot observe any pattern. Could someone suggest something?
| 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",
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} |
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 is the part I don't understand
$\rightarrow \frac{6}{32}$
How did the numerator simplify to $6n^3$?
| 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 $(â_{ij})_{i,j}=AA^*$ then, since $AA^*=A^*A$
$â_{ii}=\sum\limits_{k=1}^na_{ik}\overline{a}_{ik}=\sum\limits_{k=1}^n\overline{a_{ki}}{a}_{ki}$
$\implies\sum\limits_{k=1}^n|a_{ik}|^2=\sum\limits_{k=1}^n|a_{ki}|^2$.
If I take $i=n$ then it follows that $a_{in}=0, \forall 1\le i\le n-1$ and continuing in this manner the upper diagonal entries are zero, Is this correct ?
Can I show it in another way, because in a previous exercise I had to show that ''If A is normal and nilpotent then $A=0$'' so using this can I decompose $A$ into diagonal and nilpotent matrix, then show that the nilpotent part is zero ?
| 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{b} & \overline{c}
\end{pmatrix}- \begin{pmatrix}
\overline{a} & 0\\
\overline{b} & \overline{c}
\end{pmatrix}\begin{pmatrix}
a & b\\
0 & c
\end{pmatrix} \newline
&= \begin{pmatrix}
|a|^2 + |b|^2 & b\overline{c}\\
\overline{b}c & |c|^2
\end{pmatrix} - \begin{pmatrix}
|a|^2 & \overline{a}b \\
a\overline{b} & |b|^2+|c|^2
\end{pmatrix} \newline
&= \begin{pmatrix}
|b|^2 & b\overline{c}-\overline{a}b\\
\overline{b}c -a\overline{b} & |b|^2
\end{pmatrix}_.
\end{align*}
Therefore $|b|^2 = 0$. Thus $b = 0$ and $A$ is diagonal.
Now assume the result holds for $n-1$.
Then $$A = \begin{pmatrix}
a & B\\
0 & C
\end{pmatrix}$$
where $B$ is an $1\times (n-1)$ matrix and $C$ is an $(n-1)\times(n-1)$ upper-triangular matrix.
Since $A$ is normal,
\begin{align*}
0 = A^*A-AA^* &= \begin{pmatrix}
\overline{a} & 0\\
B^* & C^*
\end{pmatrix}\begin{pmatrix}
a & B\\
0 & C
\end{pmatrix} - \begin{pmatrix}
a & B\\
0 & C
\end{pmatrix}\begin{pmatrix}
\overline{a} & 0\\
B^* & C^*
\end{pmatrix} \newline
&= \begin{pmatrix}
|a|^2 & \overline{a}B\\
aB^* & B^*B + C^*C
\end{pmatrix} - \begin{pmatrix}
|a|^2 + BB^* & BC^*\\
CB^* & CC^*
\end{pmatrix} \newline
&= \begin{pmatrix}
-|B|^2 & \overline{a}B-BC^*\\
aB^*-CB^* & B^*B + C^*C-CC^*
\end{pmatrix}_.
\end{align*}
Thus $|B| = 0$ and so $B=0$. Hence $C^*C-CC^* = 0$, i.e. $C$ is normal. By the hypothesis of induction, $C$ is diagonal. Finally,
$$ A = \begin{pmatrix}
a & 0\\
0 & C
\end{pmatrix}_.$$
So $A$ is diagonal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
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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
$$\cos(\arcsin(2x\sqrt{1-x^2}))=|1-2x^2|$$
while taking the cosine of the right-hand side of $(1)$ yields
$$\cos(2\arcsin(x))=1-2x^2$$
Therefore, the equality
$$\arcsin(2x\sqrt{1-x^2})=2\arcsin(x)$$
is valid for $|x|\le 1/\sqrt 2$.
| {
"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} $
*$\dfrac{2\sqrt{17}}{9}$
*$\dfrac{\sqrt{34}}{9}$
*$\dfrac{2\sqrt{13}}{9}$
$\text{Somewhere it explained as:}$
$$\frac{1}{\alpha}+\frac{1}{\beta}=4$$
$$2q = p + r$$
$$\Rightarrow \space\space-2(\alpha+\beta)=1+\alpha\beta$$
$$\Rightarrow-2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\frac{1}{\alpha\beta}+1\cdots(1)$$
$$\Rightarrow\frac{1}{\alpha\beta}=-9$$
$\text{Equation having roots $α$ and $β$ is:}$ $$9x^2\space +\space 4x\space -1=0$$
$$\alpha,\space \beta=\frac{-4\underline{+}\sqrt{16+36}}{2\times 9}$$
$$\left|\alpha-\beta\right|=\frac{2\sqrt{13}}{9}$$
$\text{As I know:} $$$x^2-(\alpha+\beta)+(\alpha\times\beta)=0\cdots(2)$$
$\text{My question is:}$
Can you explain in formal/alternative way, please?
| 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 $\frac{1}{\alpha} + \frac{1}{\beta} = 4$. That means that if we look at the equation $2q = p + r$, we get that $\frac{1}{\alpha \beta} = -9$. So if $ex^2+fx+g=0$is the equation satisfied by $\frac{1}{\alpha}$ and $\frac{1}{\beta}$, then $e=1$, $f$ is the negative of the sum which is $-4$, and $g$ is the product, which is $-9$. Putting together, this gives us $x^2-4x-9=0$ is the equation satisfied by $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. To get the equation satisfied by $\alpha$ and $\beta$, all we do is invert the $x$ by making it $\frac{1}{x}$, and then that gives us $9x^2+4x-1=0$. Finally, using the formula we know above of the difference of the roots, we get the answer as $\frac{2 \sqrt{13}}{9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $2^{b-a} \equiv -1 \pmod{25}$ and $3^c \equiv 1 \pmod{25}$, whereby $5 \mid c$ and $2^{b-a} \equiv -1 \pmod{11}$. This last congruence implies that $b-a \equiv 5 \pmod{10}$, contradicting the parity of $b-a$.
The cases where $a=0, 2$ and $4$ are (more) easily treated (leading to the solutions with $(a,b)=(0,0), (2,2)$ and $(4,6)$).
| {
"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
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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:
$$
\begin{align}
\lim_{n \to \infty} &\frac{1}{\sqrt{n}} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}}
+ \frac{1}{\sqrt{4}+\sqrt{6}}
+ \cdots + \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr] \\
&= \lim_{n \to \infty} \frac{\sqrt{n}}{n} \biggl[
\frac{1}{\sqrt{2}+\sqrt{4}}
+ \frac{1}{\sqrt{4}+\sqrt{6}} + \cdots
+ \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr]
\end{align}
$$
Now using Cauchy first thm of limits
$$
a_n = \frac{\sqrt{n}}{\sqrt{2n}+\sqrt{2n+1}}
$$
The answer should be $\frac{1}{2\sqrt{2}}$.
But the answer is $1/\sqrt{2}$.
| $$\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=\frac{\sqrt{2k+2}-\sqrt{2k}}{2}$$
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",
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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 triple (3,4,5)
It is true that this continued fraction is only valid for Pythagoraean Triple only?
I try other numbers and it seem that the only numbers that are valid are the Pythagoraean Triples. Can anyone verify this? Or show some examples where is also work for other numbers.
| 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 odd.
Note that it's valid for number combinations that are not pythagorean triples. That's because you only ever restrict $a+c$, and there's no rule to prescribe how to split it into $a$ and $c$. The above condition for $b$ is enough to make this valid. Every pythagorean triple that has $b$ of the above form, is a candidate. Use the standard $n$,$m$ parameterization of pythagorean triples to get all solutions (within pythagorean triples, there are other solutions).
$$a=n^2-m^2$$
$$b=2nm$$
$$c=n^2+m^2$$
we have a condition for $b=(x^2-1)/2$:
$$4nm=(x-1)(x+1)$$
and
$$a+c=2n^2=1+b+x=(1+x)^2/2$$
$$4n^2=(1+x)^2$$
From this:
$$n=(x+1)/2$$
$$m=(x-1)/2$$
for odd $x$, or rather...
$$n-m=1$$
So... for every $m\in\mathbb{N}$, triples
$$(a,b,c)=(2m+1,2(m+1)m,2m^2+2m+1)$$
satisfy your relation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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
$$
\left|\frac{a}{\sqrt{a^2+b^2}}\sin (x)+\frac{b}{\sqrt{a^2+b^2}}\sin(2x)\right|\leqslant \frac{1}{\sqrt{a^2+b^2}}
$$
and let
$$
\cos (y)=\frac{a}{\sqrt{a^2+b^2}}
$$
and
$$
\sin (y) = \frac{b}{\sqrt{a^2+b^2}}.
$$
The inequality is equivalent to
$$
\left|\cos (y)\cos\left(\frac{\pi}{4}-x\right)-\sin (y)\sin(2x)\right|\leqslant \frac{1}{\sqrt{a^2+b^2}},
$$
but the terms are different, so we cannot use the cosine addition formula.
| 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 $$f'(\pi/3)=a\cos\pi/3+2b\cos2\pi/3 = a/2 - b = 0,$$
i.e. $a=2b$. Using now relation $|a+b|=\frac{2}{\sqrt3}$ you can find two pairs $(a,b)$, and both of them are suitable.
The case of $a$ and $b$ of different sign can be considered in a similar way (here $|a|+|b|=|a-b|$), and leads to another two pairs.
| {
"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, $A$ is not diagonalizable.
How does one find the exponential of a non-diagonalizable matrix?
My attempt:
Write
$\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix} = M + N$,
with $M = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$ and $N = \begin{pmatrix}
0 & 1 & 2 \\
0 & 0 & -4 \\
0 & 0 & 0
\end{pmatrix}$.
We have $N^3 = 0$, and therefore $\forall x > 3$, $N^x = 0$. Thus:
$$\begin{aligned}
e^{At}
&= e^{(M+N)t} = e^{Mt} e^{Nt} \\
&= \begin{pmatrix}
e^t & 0 & 0 \\
0 & e^t & 0 \\
0 & 0 & e^t
\end{pmatrix} \left(I + \begin{pmatrix}
0 & t & 2t \\
0 & 0 & -4t \\
0 & 0 & 0
\end{pmatrix}+\begin{pmatrix}
0 & 0 & -2t^2 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}\right) \\
&= e^t \begin{pmatrix}
1 & t & 2t \\
0 & 1 & -4t \\
0 & 0 & 1
\end{pmatrix} \\
&= \begin{pmatrix}
e^t & te^t & 2t(1-t)e^t \\
0 & e^t & -4te^t \\
0 & 0 & e^t
\end{pmatrix}.
\end{aligned}$$
Is that the right answer?
| 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 vectors.
For $e_1$, we have
$$\exp(tA)e_1 = \sum_{n=0}^\infty \frac{t^n}{n!}A^ne_1 = \sum_{n=0}^\infty \frac{t^n}{n!} e_1 = \exp(t)e_1$$
so the first column of $\exp(tA)$ is
$$\left[ \exp(t)\quad 0\quad 0\right]^T$$
For $e_2$, we have that $Ae_2 = e_1 + e_2$, so
$$A^n e_2 = A^{n-1}e_1 + a^{n-1}e_2 = e_1 + A^{n-1}e_n = \cdots = ne_1 + e_2$$
and consequently,
$$\exp(tA)e_2 = \sum_{n=0}^\infty \frac{t^n}{n!}A^ne_2 = \sum_{n=0}^\infty \frac{t^n}{n!} ne_1 + \sum_{n=0}^\infty \frac{t^n}{n!} ne_2 = t\sum_{n=1}^\infty \frac{t^n}{(n-1)!} e_1 + \exp(t)e_2 = t\exp(t)e_1 + \exp(t)e_2$$
and the second row of the matrix is
$$\left[t\exp(t)\quad \exp(t)\quad 0\right]^T$$
Finally,
$$A e_3 = 2e_1 + -4e_2 + e_3$$
so
$$A^n e_3 = 2A^{n-1}e_1 - 4A^{n-1}e_2 + A^{n-1}e_3 = 2e_1 - 4((n-1)e_1 + e_2) + A^{n-1} e_3=\cdots$$ $$ \cdots = 2n e_1 - 4\left(\frac{n(n-1)}{2}e_1 + ne_2\right) + e_3 = -(2n^2+4n) e_1 - 4ne_2 + e_3.$$
Thus,
$$\exp(tA)e_3 = \sum_{n=0}^\infty \frac{t^n}{n!}A^ne_1 = \sum_{n=0}^\infty \frac{t^n}{n!}\left(-(2n^2-4n) e_1 - 4ne_2 + e_3\right)$$
$$ = -2\sum_{n=0}^\infty \frac{t^n}{n!}n(n-2)e_1 - 4\sum_{n=0}^\infty \frac{t^n}{n!}ne_2 + \sum_{n=0}^\infty \frac{t^n}{n!} e_3$$
$$= -2t\sum_{n=1}^\infty \frac{t^n}{(n-1)!}(n-2)e_1 -4 t\exp(t)e_2 + \exp(t)e_3$$
$$= -2t\sum_{n=0}^\infty \frac{t^n}{n!}(n-1)e_1 -4 t\exp(t)e_2 + \exp(t)e_3 $$
$$ = -2t\sum_{n=0}^\infty \frac{t^n}{n!}ne_1 + 2t\sum_{n=0}^\infty \frac{t^n}{n!}e_3 -4 t\exp(t)e_2 + \exp(t)e_3$$
$$= -2t \sum_{n=1}^\infty \frac{t^n}{(n-1)!} e_1 + 2t\exp(t)e_1 - 4t\exp(t)e_2 + \exp(t)e_3$$
$$= -2t^2 \sum_{n=0}^\infty \frac{t^n}{n!} e_1 + 2t\exp(t)e_1 - 4t\exp(t)e_2 + \exp(t)e_3$$
$$= -2t(t-1)\exp(t)e_1 -4t\exp(t)e_2 + \exp(t)e_3$$
so the final column of $\exp(tA)$ is
$$\left[-2t(t-1) \exp(t) \quad -4t\exp(t) \quad \exp(t)\right]^T$$
Thus,
$$\exp(tA) = \exp(t)\begin{bmatrix}
1 & t & -2t(t-1)\\
0 & 1 & -4t\\
0 & 0 & 1\\
\end{bmatrix}$$
| {
"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 the expression $\sqrt[n_1] {P_1(x)} + \sqrt[n_2] {P_2(x)} + \sqrt[n_3] {P_3(x)} + \cdots + \sqrt[n_i] {P_i(x)}$, where $P_1(x), P_2(x), P_3(x), \ldots, P_i(x)$ are polynomials, and $n_1, n_2, n_3, \ldots,n_i$ are whole numbers? On the other hand, how would you prove that no minimum value for such a sum exists?
| 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 $x=4.8,$ which says that we got a minimal value.
| {
"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:
$$
\begin{align}
\left| \int_{x^2}^{(x+1)^2}\frac{\sin t}{\sqrt{t}}dt\right|&=\left|\left[-\frac{1}{\sqrt{t}}\cos t\right]_{x^2}^{(x+1)^2}-\frac{1}{2}\int_{x^2}^{(x+1)^2}\frac{1}{t\sqrt{t}}\cos t\:dt\right|
\\\\&= \left|\frac{\cos (x^2)}x-\frac{\cos ((x+1)^2)}{x+1}-\frac{1}{2}\int_{x^2}^{(x+1)^2}\frac{1}{t\sqrt{t}}\cos t\:dt\right|
\\\\&\leq\left|\frac{\cos (x^2)}x-\frac{\cos ((x+1)^2)}{x+1}\right|+\frac{1}{2}\int_{x^2}^{(x+1)^2}\left|\frac{1}{t\sqrt{t}}\cos t\right|\:dt
\\\\&\leq\left|\frac{(x+1)\cos (x^2)-x\cos ((x+1)^2)}{x(x+1)}\right|+\frac{1}{2}\int_{x^2}^{(x+1)^2}\frac{1}{t\sqrt{t}}\:dt
\\\\&\leq\frac{2x+1}{x(x+1)}+\frac{1}{x(x+1)}
\\\\&\leq\frac{2}{x}.
\end{align}
$$
| {
"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
} |
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^{n-2}+\dots+2x+1}{n(1+x+\dots+x^n)} \\
&= \frac{1}{n}.
\end{align*}
Are my steps correct? Thanks.
| 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,
$$
\underbrace{\frac{1}{1-x}\cdot \frac{1}{1-x} \cdots \frac{1}{1-x}}_\text{a+1} \\
=\underbrace{(1+x^2+x^3+\cdots)\times(1+x^2+x^3+\cdots)\times \cdots \times (1+x^2+x^3+\cdots)}_\text{a+1} \tag{*}
$$
From $(*)$, we know that the coefficient of $x^j$ is the number of solutions for the integer equation $m_1+m_2+\cdots+m_{a+1}=j$, with $m_k=0,1,\ldots$, which is ${a+j \choose j}$.
Therefore, we can deduce that $(1-x)^{-a-1}=\sum_{j=0}^{\infty}{{a+j} \choose j}x^j$.
| {
"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 $x^5$
$\begin{eqnarray} \displaystyle \int_0^1 \frac{1-x^3}{1-x^5} \, dx &=& \displaystyle \int_0^1 \displaystyle \sum_{r=0}^\infty (1-x^3) x^{5r} \, dx \\ \displaystyle &=& \sum_{r=0}^\infty \int_0^1 (x^{5r} - x^{3+5r} ) \, dx \\ \displaystyle &=& \sum_{r=0}^\infty \left( \frac1{5r+1} - \frac1{5r+4}\right ) \\ \displaystyle &=& \sum_{r=0}^\infty \frac3{25r^2+25r+4} \qquad (\star) \\ \end{eqnarray}$
We consider the logarithmic differentiation of the Weiestrass Product:
$[\displaystyle \cos(\pi x) = \prod_{n=0}^\infty \left( 1 - \frac{4x^2}{(2n+1)^2} \right)$
to get
$\displaystyle \sum_{n=0}^\infty \frac1{(2n+1)^2 - (2x)^2} = \frac{\pi}{8x} \tan(\pi x ) \qquad (\star \star)$
Notice that $ \frac3{25r^2+25r+4} = \frac{12}{25} \cdot \frac1{(2r+1)^2 - \left (2\cdot \frac3{10} \right)^2}$ Thus $x=\frac3{10}$ for $(\star \star)$
Hence, the integral in question equals to $\dfrac{12}{25} \cdot \dfrac{\pi}{8\cdot 3/10} \tan\left(\pi \cdot \dfrac3{10}\right) = \dfrac\pi5 \tan\left(\dfrac3{10}\pi\right)$
Now, we just need to evaluate $\tan\left(\frac3{10}\pi\right)$ .Apply the identity $\tan(x) = \sqrt{\frac{1-\cos(2x)}{1+\cos(2x)}}$ for $x = \frac3{10}\pi$
This means we need to find what is the value of $\cos\left(\frac35\pi \right)$ Let $y$ denote this value. Because $\cos\left(\frac{1}5\pi\right) = -\cos\left( 4\times \frac15\pi \right)$. Apply the double angle formula twice yields $\cos\left( \frac15\pi\right) = \frac{1+\sqrt5}4$. Apply the triple angle formula to get $y = \frac{1-\sqrt5}4$
Substitution yields the answer of $\displaystyle \pi \sqrt{\dfrac{5+2\sqrt5}{125}} \approx 0.8648 \ \square$
| {
"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
} |
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:
$$ f_2(4) = f_1(4)^2-4 = (2\cdot 4^0+1)^2-2^2=4+1 $$
or:
$$ f_3(4)=f_2(4)^2-4^2 = (4+1)^2-4^2 = 2\cdot 4+1 $$
or:
$$ f_4(4)=f_3(4)^2-4^3 = (2\cdot 4+1)^2 - 4^3 = 4^2+1 $$
or:
$$ f_5(4)=f_4(4)^2-4^4 = (4^2+1)^2-(4^2)^2 = 2\cdot 4^2+1 $$
$\ldots$ A pattern emerged. Notice that
$$ f_{2n+1}(4) = 2\cdot 4^n+1 \qquad f_{2n}(4) = 4^n+1 \tag{1}$$
are equivalent, since
$$ f_{2n+2}(x)=f_{2n+1}(x)^2-x^{2n+1}, \qquad f_{2n+1}(x)=f_{2n}(x)^2-x^{2n}. $$
So it is enough to prove that
$$ \lim_{n\to +\infty}\frac{f_{2n}(4)}{4^n+1}=\lim_{n\to +\infty}\frac{f_{2n+1}(4)}{2\cdot 4^n+1}=1 \tag{2}$$
hold by squeezing.
| {
"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
$$\begin{align}\frac{x^n-\sin^nx}{x^2\sin^nx}&=\frac{(3(x/3))^n-(3(\sin(x/3)-(4/3)\sin^3(x/3)))^n}{x^2\sin^nx}\\
&=\frac{(3(x/3))^n-3^n\sin^n(x/3)+n\cdot3^n(4/3)\sin^{n+2}(x/3)+O(\sin^{n+4}(x/3))}{x^2\sin^nx}\\
&=\frac{(3(x/3))^n-3^n\sin^n(x/3)}{x^2\sin^nx}+\frac{n\cdot3^n(4/3)\sin^{n+2}(x/3)}{x^2\sin^nx}+\frac{O(\sin^{n+4}(x/3)}{x^2\sin^nx}\\
&=\frac{(3(x/3))^n-3^n\sin^n(x/3)}{(3(x/3))^2(3^n\sin^n(x/3))+O(\sin^{n+2}(x/3)))}\\
&+\frac{n\cdot3^n(4/3)\sin^{n+2}(x/3)}{(3(x/3))^2(3^n\sin^n(x/3))+O(\sin^{n+2}(x/3)))}+\frac{O(\sin^{n+4}(x/3)}{x^2\sin^nx}\end{align}$$
So
$$\begin{align}\lim_{x\rightarrow0}\frac{x^n-\sin^nx}{x^2\sin^nx}&=\lim_{x\rightarrow0}\frac{3^n}{3^{n+2}}\frac{(x/3)^n-\sin^n(x/3)}{(x/3)^2\sin^n(x/3)}\frac{1}{\left(1+\frac{O(\sin^{n+2}(x/3))}{3^n\sin^n(x/3)}\right)}\\&+\lim_{x\rightarrow0}\frac{4n\cdot3^{n-1}}{3^{n+2}}\frac{\sin^{n+2}(x/3)}{(x/3)^2\sin^n(x/3)}\frac{1}{\left(1+\frac{O(\sin^{n+2}(x/3))}{3^n\sin^n(x/3)}\right)}+\lim_{x\rightarrow0}\frac{O(\sin^{n+4}(x/3)}{x^2\sin^nx}\\
&=\frac19\lim_{x\rightarrow0}\frac{x^n-\sin^nx}{x^2\sin^nx}\frac1{1+0}+\frac{4n}{27}\frac1{1+0}+0=\frac98\frac{4n}{27}=\frac n6=\frac{6000}6=1000\end{align}$$
Sort of like the Taylor series approach but using trigonometric identities instead. That $O(\sin^{n+4}(x/3))$ term represents further terms of the binomial expansion which have at least the factor $\sin^{n+4}(x/3)$, so their limits when divided merely by $\sin^{n+2}(x/3)$ are all zero. We are, of course, using
$$\lim_{x\rightarrow0}\frac{\sin x}x=\lim_{x\rightarrow0}\frac{\sin(x/3)}{(x/3)}=1$$
| {
"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
} |
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
$$A =
\begin{pmatrix}
8 & −4 & 3/2 & 2 & −11/4 & −4 & −4 & 1 \\
2 & 2 & 1 & 0 & 1 & 0 & 0 & 0 \\
−9 & 8 & 1/2 & −4 & 31/4 & 8 & 8 & −2 \\
4 & −6 & 2 & 5 & −7 & −6 & −6 & 0 \\
−2 & 0 & −1 & 0 & 1/2 & 0 & 0 & 0 \\
−1 & 0 & −1/2 & 0 & −3/4 & 3 & 1 & 0 \\
1 & 0 & 1/2 & 0 & 3/4 & −1 & 1 & 0 \\
2 & 0 & 1 & 0 & 0 & 0 & 0 & 5 \\
\end{pmatrix} \in M_8(\mathbb R)$$
Justify why there is a matrix $B$ such that $B^2 = A$.
I think it has something to do with the Jordan normal form so I found it.
\begin{pmatrix}
2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 2 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 2 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 5 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 5 \\
\end{pmatrix}
(I know the ones are under the diagonal, but I think it doesn't make a difference.)
I have no idea what should I do next, a help will be appreciated!
Thanks!
| 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:
$$f\left( \left[
\begin{array}
[c]{cccc}%
\lambda & 1 & & \\
& \lambda & 1 & \\
& & \lambda & 1\\
& & & \lambda
\end{array}
\right] \right) =\left[
\begin{array}
[c]{cccc}%
f\left( \lambda\right) & f^{\prime}\left( \lambda\right) & \frac{1}
{2!}f^{\prime\prime}\left( \lambda\right) & \frac{1}{3!}f^{\prime
\prime\prime}\left( \lambda\right) \\
& f\left( \lambda\right) & f^{\prime}\left( \lambda\right) & \frac
{1}{2!}f^{\prime\prime}\left( \lambda\right) \\
& & f\left( \lambda\right) & f^{\prime}\left( \lambda\right) \\
& & & f\left( \lambda\right)
\end{array}
\right]$$
(it is based on a certain Taylor expansion of $I+N$ where $N$ is a nilpotent matrix).
It suffices now to take $f(x)=\sqrt{x}$ to get for the square root of a Jordan block of similar form as above but $3 \times 3$:
$$\left[
\begin{array}
[c]{ccc}%
\sqrt{\lambda} & \dfrac{1}{2\sqrt{\lambda}} & -\dfrac{1}{8 (\lambda)^{3/2}} \\
& \sqrt{\lambda} & \dfrac{1}{2\sqrt{\lambda}} \\
& & \sqrt{\lambda}
\end{array}
\right]$$
| {
"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 want to confirm the same.
Is there any way to prove following conjecture?
Conjecture. There is always unique way to write down $\sum_{i=1}^{n} a_{i}^n$ for any arbitrary value of $n$ such that it gives same value for all values of $n$.
Answer is given below, spoiler alert:
$$1^5+2^5+4^5+7^5+9^5 = 3^4+6^4+10^4+16^4 = 17^3+20^3+40^3 = 88^2 + 263^2 = 76913$$
| 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 = 1 To 100
For q = p + 1 To 100
For r = q + 1 To 100
For s = r + 1 To 100
If p ^ 4 + q ^ 4 + r ^ 4 + s ^ 4 = mysum Then
For i = 1 To 100
For j = i + 1 To 100
For k = j + 1 To 100
If i ^ 3 + j ^ 3 + k ^ 3 = mysum Then
For a = 1 To 1000
For b = a + 1 To 1000
If a ^ 2 + b ^ 2 = mysum Then
n = n + 1: Cells(n, 1) = u & "| " & v & "| " & w & "| " & x & "| " & y & "<|> " & p & "| " & q & "| " & r & "| " & s & "<|> " & i & "| " & j & "| " & k & "<|> " & a & "| " & b
GoTo Mynext
End If
Next
Next
End If
Next
Next
Next
End If
Next
Next
Next
Next
Next
Next
Next
Mynext:
Next
Next
End Sub
| {
"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 $\triangle ABC$. Since $\angle BAC=90^{\circ}$, $BC$ must be a diameter. Let $M$ be the midpoint point $BC$, then it's the centre of the circumcircle. But then I couldn't see anything substantial, please helps,
| 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:
$$\frac{\sin{\angle DAM}}{|DM|} = \frac{\sin{\angle ADM}}{|AM|} \implies \sin{\angle DAM} = \frac12 \sin{120^{\circ}} = \frac{\sqrt{3}}{4} $$
We may therefore find $\angle AMD = 180^{\circ} - \angle DAM - \angle ADM$ and therefore $\angle AMC = 180^{\circ} - \angle AMD = \angle DAM + \angle ADM$. We may now use the law of cosines to find $|AC|$:
$$\begin{align}|AC|^2 &= |AM|^2+|MC|^2 - 2 |AC| |AM| \cos{\angle AMC}\\&= 2^3 [1- \cos{(120^{\circ}+\angle DAM)}]\\&= 16 \sin^2{\left (60^{\circ}+\frac12 \angle DAM \right )}\\ &= 4 \left (\sin^2{\left (\frac12 \angle DAM \right )}+3 \sin^2{\left (\frac12 \angle DAM \right )} + \sqrt{3} \sin{\angle DAM}\right ) \\ &= 4 \left (\frac{4-\sqrt{13}}{8} + \frac{12+3 \sqrt{13}}{8} + \frac34 \right ) \\ &= 11+\sqrt{13} \end{align}$$
Thus
$$|AC| = \sqrt{11+\sqrt{13}} $$
ADDENDUM
This is even easier when one sees that $\angle ABC = 1/2 \angle AMC$. Also, the previous result I had was correct but not simplified enough.
| {
"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.
$$
\begin{align}
\lim \limits_{x \rightarrow 0} \frac{(a+x)\sec(a+x) - a \sec(a)}{x}
& = \lim \limits_{x \rightarrow 0} \frac{a\sec(a+x) - a \sec(a)}{x} + \lim \limits_{x \rightarrow 0} \frac{x\sec(a+x)}{x} \\
& = A + B
\end{align}
$$
$$
\begin{align}
A & = \lim \limits_{x \rightarrow 0} \frac{a\sec(a+x) - a \sec(a)}{x} \\
& = a\lim \limits_{x \rightarrow 0} \frac{\cos(a) - \cos(a+x)}{\cos(a)\cos(a+x)x} \\
& = a\lim \limits_{x \rightarrow 0} 2\frac{\sin(a + x/2) sin(x/2)}{\cos(a)\cos(a+x)x} \\
& = a\lim \limits_{x \rightarrow 0} \; \frac{\sin(a + x/2)}{\cos(a)\cos(a+x)} \frac {\sin(x/2)}{x/2} \\
& = a\tan(a)sec(a)
\end{align}
$$
$$
B = \sec(a)
$$
| {
"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
} |
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 $$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 + (a - b)^4 + (a - c)^4 + (a - d)^4 + (b - c)^4 + (b - d)^4 + (c - d)^4 = 6(a^2 + b^2 + c^2 + d^2)^2.$$
But this just comes out of the blue. Are there any other solutions?
Edit (for those who say this is lacking context): This comes from the beginning of a book named Algebraic Inequalities (not in English). After the preface to the first chapter, where the author introduced that many inequalities can be proved by using the non-negativity of $x^2$, this is the first example (without developing any other methods yet). Then the author notes: the reader is supposed to memorize identities like that.
| 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 have
\begin{align}
\mathrm{RHS} - \mathrm{LHS} &=
3 r^4+4 r^3 s+8 r^3 t+6 r^2 s^2+12 r^2 s t+12 r^2 t^2+4 r s^3\\
&\quad +12 r s^2 t +12 r s t^2+8 r t^3+3 s^4+8 s^3 t+12 s^2 t^2+8 s t^3+4 t^4.
\end{align}
We are done.
$\phantom{2}$
Solution 2:
We can use SOS (Sum of Squares) method without the identity, with the help of computer.
After homogenization, it suffices to prove that
$$(a + b)^4 + (a + c)^4 + (a + d)^4 + (b + c)^4 + (b + d)^4 + (c + d)^4 \le 6(a^2+b^2+c^2+d^2)^2.$$
We have
\begin{align}
\mathrm{RHS} - \mathrm{LHS} &=
(-a b+a c+b d-c d)^2+\frac{1}{3} (a b+a c-2 a d-2 b c+b d+c d)^2 \\
&\quad +\frac{1}{2} (-a^2+2 a d+b^2-2 b c+c^2-d^2)^2\\
&\quad +(a b-a c-b^2+b d+c^2-c d)^2+(a^2-a b-a c+b d+c d-d^2)^2\\
&\quad +\frac{1}{6} (-3 a^2+2 a b+2 a c+2 a d-3 b^2+2 b c+2 b d-3 c^2+2 c d-3 d^2)^2.
\end{align}
We are done.
| {
"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 as $$\frac{\sqrt{3}\sin x+\cos x}{\sin x\cos x}-2\frac{\cos 2x}{\sin x\cos x} = 0$$
So we get $$2\frac{\cos \left(x+\frac{\pi}{3}\right)}{\sin x\cos x}-2\frac{\cos 2x}{\sin x\cos x}=0$$
So we get $$\cos (2x) = \cos\left(x+\frac{\pi}{3}\right)$$
So we Get $\displaystyle 2x=2n\pi\pm (x+\frac{\pi}{3})\;,n\in \mathbb{Z}$
So we get $\displaystyle x= 2n\pi+\frac{\pi}{3}$ or $\displaystyle x = \frac{2n\pi-\frac{\pi}{3}}{3}\;,$ Where $n\in \mathbb{Z}$
So we get $$x=\frac{\pi}{3} = \frac{3\pi}{9}\;\;,-\frac{5\pi}{9}\;\;,-\frac{7\pi}{9}$$
So Sum of distinct Roots is $\displaystyle = -\pi$
But it is wrong, Where i have done wrong, Help me
Thanks
|
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) :
$$2\frac{\cos \left(x\color{red}{-}\frac{\pi}{3}\right)}{\sin x\cos x}-2\frac{\cos 2x}{\sin x\cos x}=0$$
I think you can continue from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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 :
$$3^n+n^2\sim_03^n,\quad2^n+n^3\sim_\infty2^n,\enspace\text{hence}\quad \frac{3^n+n^2}{2^n+n^3}\sim_\infty\Bigr(\frac32\Bigl)^n,$$
which doesn't even tend to $0$.
Could you explain why does $3^n+n^2\sim_03^n$ and $2^n+n^3\sim_\infty2^n$ when we have only $n \to \infty$ ?
| 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",
"url": "https://math.stackexchange.com/questions/1796952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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,
$$ \ 6 \left( \sqrt[3]{ \frac{4}{3} \pi r^3 } \right) ^2 : 4 \pi r^2 $$
But I am not really sure how to arrive at the desired result. I have tried to simplify it, but apparently I am missing some step in the process and come to a result that is far from the correct one.
Could you please help? Thank you.
| 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 -- $$ S_s \ = \ 4 \ \pi \ r^2 \ \ , \ \ V_s \ = \ \frac{4}{3} \ \pi \ r^3 \ \ \Rightarrow \ \ r \ = \ \left(\frac{3 \ V_s}{4 \ \pi} \right)^{1/3} $$ $$ \Rightarrow \ \ S_s \ = \ 4 \ \pi \ \left[\left(\frac{3 \ V_s}{4 \ \pi} \right)^{1/3} \right]^2 \ = \ (4 \ \pi)^{1/3} \cdot 3^{2/3} \cdot V_s^{2/3} \ = \ \ (36 \ \pi)^{1/3} \ V_s^{2/3} \ \ . $$
The ratio of the surface areas is then
$$ \frac{S_c}{S_s} \ \ = \ \ \frac{6 \ V_c^{2/3}}{(36 \ \pi)^{1/3} \ V_s^{2/3}} \ \ ; $$
upon equating the volumes, we have
$$ \frac{S_c}{S_s} \ \ = \ \ \frac{6 \ V ^{2/3}}{(36 \ \pi)^{1/3} \ V ^{2/3}} \ \ = \ \ \frac{6 }{6^{2/3} \ \pi^{1/3} } \ \ = \ \ \left(\frac{6 }{ \pi } \right)^{1/3} \ \ = \ \ \frac{6^{1/3} }{ \pi^{1/3} } \ \ \text{or} \ \ \frac{\sqrt[3]{6} }{ \sqrt[3]{ \pi} } \ \ . $$
| {
"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 problem. Any help is appreciated.
Find the value of $$\frac{a^2}{a^4+a^2+1}$$ if $$\frac{a}{a^2+a+1}=\frac{1}{6}$$
| 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
Thanks!
| Actually it is not unique. Even without changing the order of eigenvalues. For example, the matrix $$A=\begin{pmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0&0&1\end{pmatrix}=\begin{pmatrix}1&0& -1 \\ 1 & 0 & -1 \\ 0&1&0\end{pmatrix}\begin{pmatrix}1&0& 0 \\ 0 & 1 & 0 \\ 0&0&-1\end{pmatrix}\begin{pmatrix}1&1& 0 \\ 0 & 0 & 1 \\ -1&1&0\end{pmatrix}=\begin{pmatrix}\frac{1}{\sqrt{2}}&0& -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0&1&0\end{pmatrix}\begin{pmatrix}1&0& 0 \\ 0 & 1 & 0 \\ 0&0&-1\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}& 0 \\ 0 & 0 & 1 \\ -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\end{pmatrix}.$$
| {
"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: $\sin\left(\dfrac{x}{n}\right)^2 = \dfrac{x^2}{n^2} + \omicron\left(\left| \dfrac{x^2}{n^2} \right|\right)$
Do we simply assume that since $ \omicron\left(\left| \dfrac{x}{n} \right|\right) \rightarrow 0 \implies$ $(\sin\left(\dfrac{x}{n}\right) = -\dfrac{x}{n} + \omicron\left(\left| \dfrac{x}{n} \right|\right))^2 = \sin\left(\dfrac{x}{n}\right)^2 = \dfrac{x^2}{n^2} + \omicron\left(\left| \dfrac{x^2}{n^2} \right|\right)$
| 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
} |
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 gave me two incompatible answers
$$x=\pm\sqrt{n}, \ \ \ \ \ \ \ \ \sqrt{n}\not=0$$
$$x=\pm\frac{\sqrt{n}}{\sqrt{3}}, \ \ \ \ \ \ \ \ \sqrt{n}\not=0$$
The first answer is wrong the second is correct.
Can anyone explain why the first solution is wrong and how to derive the second?
| 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 is why the first solution is correct.
Now let $x = \pm\frac{\sqrt{n}}{\sqrt{3}}$
$$\left(\frac{\frac{1}{2}\cdot(n-\frac{n}{3})}{\pm\frac{\sqrt{n}}{\sqrt{3}}}\right)^2 =\frac{1}{2}\cdot(n-\frac{n}{3})$$
$$\frac{\frac{1}{4}\cdot{(n-\frac{n}{3})}^2}{\frac{n}{3}}=\frac{1}{2}\cdot(n-\frac{n}{3})$$
$$\frac{3(n-\frac{n}{3})^2}{4n} = \frac{1}{2}n - \frac{n}{6}$$
$$\frac{3(n^2-\frac{2}{3}n^2+\frac{n^2}{9})}{4n} = \frac{1}{2}n - \frac{n}{6}$$
$Multiply\ both\ sides\ by\ 4n$
$$3n^2-2n^2+\frac{n^2}{3} = 2n^2 - \frac{2}{3}n^2$$
$$\frac{4}{3}n^2=\frac{4}{3}n^2$$
which implies that solution is also correct
| {
"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 more congruences, for each:
$$x - 2 \equiv 0 \mod 3 \Rightarrow x_1 = 2\\ x + 2 \equiv 0 \mod 3 \Rightarrow x_2 = 1\\ x - 2 \equiv 0 \mod 7 \Rightarrow x_3 = 2\\ x + 2\equiv 0 \mod 7 \Rightarrow x_4 = 5$$
and then 4 systems of linear congruences are formed:
$$\begin{cases} x \equiv 2 \mod 3 \\ x \equiv 2 \mod 7\end{cases}$$
$$\begin{cases} x \equiv 2 \mod 3 \\ x \equiv 5 \mod 7\end{cases}$$
$$\begin{cases} x \equiv 1 \mod 3 \\ x \equiv 2 \mod 7\end{cases}$$
$$\begin{cases} x \equiv 1 \mod 3 \\ x \equiv 5 \mod 7\end{cases}$$
What is the purpose of these systems? I already have the 4 solutions ($x_1, x_2, x_3, x_4$) of the congruence. Why do I need to form these systems?
| 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 corresponding to the pair $(\color{red}1\bmod3,\color{red}5\bmod7)$, for instance, will be
$$x\equiv\color{red}5\cdot5\cdot 3-\color{red}1\cdot2\cdot 7=61\equiv19\mod 21.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 think one can prove it by expanding everything, but is there any elegant solution for this problem ?
| 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:
$$ (x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)= 9 + \sum_{cyc}\left(\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}\right)^2 \tag{1} $$
so it is enough to prove that:
$$ \sum_{cyc}\frac{(x-y)^2}{xy}\geq 4\frac{(x-y)^2}{xy+xz+yz}\tag{2} $$
but that is a consequence of the CS inequality in the form:
$$ \frac{A^2}{a}+\frac{B^2}{b}+\frac{C^2}{c}\geq \frac{(A+B+C)^2}{a+b+c}\tag{3} $$
also known as Titu's lemma:
$$\frac{(x-z)^2}{xz}+\frac{(z-y)^2}{zy}+\frac{(y-x)^2}{xy}\geq \frac{\left(|x-z|+|z-y|+|y-x|\right)^2}{xy+xz+yz}. \tag{4}$$
| {
"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
} |
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 be easier to take the anti derivative? Theoretically, we could solve as if there is a constant in the numerator, though then we would have integration problems down the road?
| 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,
$$
\frac{8}{\left(1+x^2\right)(2 x-3) }=\frac{ax+b}{x^2+1}+\frac{c}{2x-3}=\frac{(2a+c) x^2+(2b-3a)x+c-3 b}{\left(1+x^2\right)(2 x-3) }
$$ admits a valid solution:
$$
a=-\frac{16}{13}, \quad b=-\frac{24}{13},\quad c=\frac{32}{13}.
$$
Remark. Another way to see what happens, is to decompose in $\mathbb{C}(x)$:
$$
\frac{8}{\left(1+x^2\right)(2 x-3) }=\frac{A}{x-i}+\frac{B}{x+i}+\frac{C}{2x-3}.
$$
| {
"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 mean the determination of the proportion of its sides and the required internal angles by the common operations in plane geometry, like in the other triangles.
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?
|
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 $75^\circ - 72^\circ = 3^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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
$$\begin{align}4^{y+3x}&= 64 \\
4^{y+3x} &= 4^3 \\
y+3x &= 3 \end{align}$$
$$\begin{align}\log_x (x+12)- 3 \log_x 4 &= -1 \\
\log_x (x+12)- \log_x 4^3 &= -1 \\
\log_x(x+12)- \log_x 64 &= -1 \end{align}$$
then I substituted $4^{y+3x} = 64$
$\log_x (x+12) - \log_x 4^{y+3x} = -1$
I don't know what should I do next. any ideas?
| 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 equation with $-1$: $3\log_x 4-\log_x(x+12)=1$
*Take both sides of the equation to the power of $x$, i.e. $x^{LHS}=x^{RHS}$ and solve for $x$ using the properties of the logarithm function:
\begin{align}
&&x^{3\log_x(4)-\log_x(x+12)}&=x^1\\
\leftrightarrow&&\frac{x^{3\log_x(4)}}{x^{\log_x(x+12)}}&=x
\end{align}
Observe that applying a logarithmic term with (its) inverse (i.e. power function) we obtain the identity function, that is $h^{\log_h (s)}=s$. Thus, the left handside of the equation also satisfies $\frac{x^{3\log_x(4)}}{x^{\log_x(x+12)}}=\frac{4^3}{x+12}$. Thus, we finally have
$$
\frac{4^3}{x+12x}=x
$$
*Pushing all terms to the right-hand side, we get the quadratic equation
$$
x^2+12x-64=0.
$$
The roots of a quadratic equation can be found using the $p/q$ formula, i.e.
$$
x^*=-6\pm\sqrt{100}=-6\pm 10
$$
Notice that we have two solutions, for this equation $x^*\in\{4,-16\}$. But negative values of $x$ can be excluded as the expressions above, specifically $\log_x$ are not defined for $x<0$. Hence, the solution must be
$$
x^*=4
$$
Step 2: Solving for $y$
Substitute $x^*$ into equation $I$:
\begin{align}
&&4^{y+3*x^*}&=64\\
&&4^{y+12}&=64
\end{align}
Take the $log_4$ to on both sides of the equation,
\begin{align}
y+12=3
\end{align}
This gives
$$
y^*=-9.
$$
| {
"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 how I would approach this question differently.
$\lim_{h\to0} \frac{3\sqrt{x+h}-3\sqrt{x}}{h} \cdot \frac{3\sqrt{x+h}-3\sqrt{x}}{3\sqrt{x+h}-3\sqrt{x}}$
$\lim_{h\to0} \frac{9(x+h)-9x}{h(3\sqrt{x+h}{-3\sqrt{x}})}$
| 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 numerator and the denominator by the conjugate:
$$
f'(x)= 3 \cdot \lim_{h\to0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\
$$
Try simplifying from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $\tan(\pi) = 0$
$$\therefore \tan(2*x) \to 0$$
$\therefore$ we can write
$$\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$$
As
$$\lim_{x \to 0} \frac{\tan(x)}{x} = 1$$
Which is incorrect answer.
My second attempt :
$$\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$$
$$\lim_{x \to \frac{\pi}{2}} \frac{\sin(2*x)}{\cos(2*x)(x - \frac{\pi}{2})}$$
$$\lim_{x \to \frac{\pi}{2}} \frac{\sin(2*x)}{(x - \frac{\pi}{2})} * \lim_{x \to \frac{\pi}{2}} (\frac{1}{\cos(2*x)})$$
$$-\lim_{x \to \frac{\pi}{2}} \frac{\sin(2*x)}{(x - \frac{\pi}{2})}$$
$$\lim_{x \to \frac{\pi}{2}} \frac{\sin(2*x)}{\color{red}{(\frac{\pi}{2} - x)}}$$
$$\lim_{x \to \frac{\pi}{2}} \frac{2 * \sin(2*x)}{\pi - 2 *x}$$
$$2 * \lim_{x \to \frac{\pi}{2}} \frac{\sin(\pi - 2*x)}{\pi - 2 *x} $$
Because, $$\sin(x) = \sin(\pi - x)$$
$$\therefore 2 * \lim_{x \to \frac{\pi}{2}} \frac{\sin(\pi - 2*x)}{\pi - 2 *x} = 2 $$
Which is correct and makes sense.
Now i did this question one more way
$$\lim_{x \to \frac{\pi}{2}} \frac{\tan(2*x)}{x - \frac{\pi}{2}}$$
$$\lim_{x \to \frac{\pi}{2}} \frac{2 * \tan(2*x)}{2x - \pi}$$
$$2 * \lim_{x \to \frac{\pi}{2}} -\frac{\tan(\pi - 2*x)}{2*x - \pi}$$
$$2 * \lim_{x \to \frac{\pi}{2}} \frac{\tan(\pi - 2*x)}{\pi-2*x}$$
Now we observe that
now as $x \to \frac{\pi}{2}$, $\tan(\pi -2*x) \to 0$ and $\pi - 2*x \to 0$
Thus we can write
$$2 * \lim_{x \to \frac{\pi}{2}} \frac{\tan(\pi - 2*x)}{\pi-2*x} = 2 * \lim_{x \to 0} \frac{\tan(x)}{x} = 2$$
Which is also correct, Can anyone please tell me what is incorrect with 1 approach ?
| $$\dfrac{\tan2x}{x-\dfrac\pi2}=\dfrac{\tan2\left(x-\dfrac\pi2\right)}{x-\dfrac\pi2}=2\cdot\dfrac{\tan(2x-\pi)}{2x-\pi}$$
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
$$P(X_{n}=1)=\dfrac{b}{a+b}+(1-a-b)^n \left(P(X_0=1)-\dfrac{b}{a+b}\right)$$.
So, I started off with
\begin{align*}
P(X_{n+1}=1)&=P(X_{n+1}=1|X_n=1)P(X_n=1)+P(X_{n+1}=1|X_n=2)P(X_n=2)\\
&=(1-a)P(X_n=1)+bP(X_n=2)\\
&=(1-a)P(X_n=1)+b(1-P(X_n=1))\\
&=(1-a-b)P(X_n=1)+b
\end{align*}
Then, I manipulated the form to show that
\begin{align*}
P(X_{n+1}=1)-\dfrac{b}{a+b}&=(1-a-b)P(X_n=1)-\dfrac{(1-a-b)b}{a+b}\\
P(X_{n+1}=1)&=(1-a-b)P(X_n=1)+b
\end{align*}
Any attempt I made to show that the original statement is true from the result that I got led me nowhere.
| 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[\frac{1-(1-a-b)^n}{1-(1-a-b)}\right]\\
&=(1-a-b)^nP(X_0=1)+b \left[\frac{1-(1-a-b)^n}{a+b}\right]\\
&=\frac{b}{a+b}+(1-a-b)^n\left[P(X_0=1)-\frac{b}{a+b}\right]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 roots of cubic equation, Then
$$\displaystyle (t-x)(t-y)(t-z)=0\Rightarrow t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$$
So we get $\displaystyle t^3-4t^2-5t-c=0$
Now let $f(t)=t^3-4t^2-5t-c\;,$ Then $f'(t)=3t^2-8t-5$
and $f''(t)=6t-8.$ Now for max. and Min.$f'(t)=0\Rightarrow 3t^2-8t-5=0$
So we get $\displaystyle t=\frac{8\pm \sqrt{64+60}}{2\cdot 3}=$
Now How can I solve it after that, Help required, Thanks
| 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 \ = \ 4 \ $ on the function $ \ f (x, \ y, \ z ) \ = \ xyz \ $ , the equations using two "multipliers" are
$$ yz \ = \ \lambda \cdot 2x \ + \ \mu \cdot 1 \ \ , \ \ xz \ = \ \lambda \cdot 2y \ + \ \mu \cdot 1 \ \ , \ \ xy \ = \ \lambda \cdot 2z \ + \ \mu \cdot 1 \ \ , $$
permitting us to write
$$ \mu \ = \ yz \ - \ 2 \lambda x \ = \ xz \ - \ 2 \lambda y \ = \ xy \ - \ 2 \lambda z \ \ . $$
Re-arranging the first implied equation produces
$$ \ (y \ - \ x) \ z \ + \ (y \ - \ x ) \ 2 \lambda \ = \ 0 \ \ \Rightarrow \ \ (y \ - \ x) \ ( z \ + \ 2 \lambda) \ = \ 0 \ \ ; $$
the other equated pairs of terms give us similar relations.
One solution then is to use $ \ x \ = \ y \ \ , \ \ x \ = \ z \ \ , $ and $ \ y \ = \ z \ $ in turn to obtain from the constraint equations
$$ x \ + \ x \ + \ z \ = \ 4 \ \ \Rightarrow \ \ x^2 \ + \ x^2 \ + \ ( 4 \ - \ 2x)^2 \ = \ 6x^2 \ - \ 16x \ + \ 16 \ = \ 6 $$
$$ 3x^2 \ - \ 8x \ + \ 5 \ = \ 0 \ \ \Rightarrow \ \ x \ = \ \frac{4 \ \pm \ 1}{3} \ = \ 1 \ \ , \ \ \frac{5}{3} \ \ , $$
as already described in the comments for the posted question, which the corrected cubic equation of OP would yield. So we find three ordered triples, $ ( 1, \ 1, \ 2) \ , \ ( 1, \ 2, \ 1) \ , $ and $ \ ( 2, \ 1, \ 1) \ $ which give the same value of $ \ 2 \ $ for $ \ xyz \ $ and another three,
$ ( \frac{5}{3}, \ \frac{5}{3}, \ \frac{2}{3}) \ , \ ( \frac{5}{3}, \ \frac{2}{3}, \ \frac{5}{3}) \ , $ and $ \ ( \frac{2}{3}, \ \frac{5}{3}, \ \frac{5}{3}) \ $ , for all of which $ \ xyz \ = \ \frac{50}{27} \ $ .
The alternative of using $ \ x \ \ne \ y \ , \ z \ = \ - 2 \lambda \ $ (and analogously for the other combinations of the three variables) produces the equation $ \ yz \ - \ 2 \cdot \left( -\frac{z}{2} \right) \cdot x \ = \ xz \ - \ 2 \cdot \left( -\frac{z}{2} \right) \cdot y \ $ $ \Rightarrow \ \ yz \ + \ xz \ = \ xz \ + \ yz \ $ , so no further information is gained. We conclude that we have found the extremal values for the functions already and that our function has the (constrained) range $$ \ \frac{50}{27} \ \le \ f (x, \ y, \ z ) \ \le \ 2 \ \ . $$
The graph below shows the geometrical interpretation of a sphere intersected by a tilted plane, so that we are seeking extremal values of the function on a circle. Since the function has symmetry about the line $ \ x \ = \ y \ = \ z \ $ , it may be expected that the maxima and minima number three each and are arranged symmetrically around the "constraint circle", the center of which lies at $ \ ( \frac{4}{3}, \ \frac{4}{3}, \ \frac{4}{3}) \ $ , which is connected to the appearance of $ \ \frac{4}{3} \ $ in the solution calculation using the quadratic formula in the original post.
The yellow "vertical" lines emerge at the positions of the maximal values for the function, while the red lines mark the locations of the minimal values.
| {
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"url": "https://math.stackexchange.com/questions/1816807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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 $$J=\int_{0}^{\infty}{1-x^2\over x^2}\ln\left({1+x^2\over 1-x^2}\right)dx\tag3$$
Apply integration by part to (3)
$u=\ln\left({1+x^2\over 1-x^2}\right)\rightarrow du={4x\over (1-x^2)^2}dx$
$dv=1-x^{-2}\rightarrow v=x-x^{-1}$
$$J=\left.-{x+1\over x}\ln\left({1+x^2\over 1-x^2}\right)\right|_{0}^{1}+\int_{0}^{1}{4\over (1+x)(1-x)^2}dx\tag4$$
$$J=\left.-{x+1\over x}\ln\left({1+x^2\over 1-x^2}\right)\right|_{0}^{1}+\int_{0}^{1}\left({1\over 1+x}+{1\over 1-x}+{2\over (1-x)^2}\right)dx\tag5$$
$$J=\left.-{x+1\over x}\ln\left({1+x^2\over 1-x^2}\right)\right|_{0}^{1}+\left.\ln\left({1+x\over 1-x}\right)-{2\over 1-x}\right|_{0}^{1}\tag6$$
I am going to stop here. Evaluate these limits is not valid because of ${1\over 0}$. I have try substitution it looks more messier and complicated than this. I need some help, thank.
| 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 $$\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(1+x^{2}\right)dx=\int_{0}^{1}\frac{\log\left(1+x^{2}\right)}{x^{2}}dx-\int_{0}^{1}\log\left(1+x^{2}\right)dx
$$ and integrating by parts we get $$\int_{0}^{1}\frac{\log\left(1+x^{2}\right)}{x^{2}}dx=-\log\left(2\right)+2\int_{0}^{1}\frac{1}{1+x^{2}}=-\log\left(2\right)+\frac{\pi}{2}.
$$ For the other integral we can integrate by parts again $$\int_{0}^{1}\log\left(1+x^{2}\right)dx=\log\left(2\right)-2\int_{0}^{1}\frac{x^{2}}{x^{2}+1}dx
$$ $$=\log\left(2\right)+\frac{\pi}{2}-2
$$ hence $$\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(1+x^{2}\right)dx=2-2\log\left(2\right).
$$ Fro the second integral we can do essentially the same argument $$\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(1-x^{2}\right)dx=2-4\log\left(2\right)
$$ hence $$\int_{0}^{1}\frac{1-x^{2}}{x^{2}}\log\left(\frac{\left(1+x^{2}\right)^{2}}{1-x^{2}}\right)=2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $x + 27 - 10\sqrt{x + 2} \geq 0$ we get that their domain $D = \mathbb{R}$, because quadratics which arise from those inequalities have exactly one root - $S_1 = \{-1\}$ and $S_2 = \{23\}$. Because all expressions are greater than $0$, I am omitting absolute values.
Solving the equation via substitution $t = \sqrt{x + 2}$ we get:
$$\sqrt{x + 3 - 2t} + \sqrt{x + 27 - 10t} = 4$$
$$x + 3 - 2t = 16 - 8\sqrt{x + 27 - 10t} + x + 27 - 10t$$
$$t^2 - 10t + 25 = x + 27 - 10t$$
$$t^2 = x + 2$$
$$x + 2 = x + 2 \implies S = [-2, \infty).$$
But according to my textbook (which I use as a preparation for entrance exams) the solution must be $S = [-1, 23]$, where -1 and 23 are, obviously not coincidently, solutions to the quadratics, which we obtain while solving expressions under square roots. Where do I have a mistake, then?
|
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
$$\sqrt{x+27-10t}=5-t\color{red}{\ge 0}$$
By the way, the equation can be written as
$$\sqrt{(\sqrt{x+2}-1)^2}+\sqrt{(\sqrt{x+2}-5)^2}=4,$$
i.e.
$$|\sqrt{x+2}-1|+|\sqrt{x+2}-5|=4$$
which should be easy to deal with.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 $7^{17}=7^{16}\cdot 7 \equiv 7\pmod{25}$.
$3^3=27\equiv 2 \pmod{25}$. Thus $3^{13}=3^{12}\cdot 3\equiv 2^4\cdot 3 \equiv -2 \pmod{25}$.
Thus their sum is $5\pmod {25}$.
In conclusuion, $3^{333}+7^{777} \equiv 30 \pmod{50}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 coordinates upwards. With $z=r\cos \theta$, assisted by a general drawn line from the sphere (suppose from $a=(x,y,z)$) to the origin, we get that $x^2+y^2$ equals $a$'s projection on the $z$-axis, which by simple computations equals $(r\sin\theta)^2$.
Then: $$x^2+y^2+z^2-4z+4=r^2-4r\cos\theta+4\le 1\Rightarrow \cos \theta \ge {r^2+3\over 4r}$$ (How is the strong inequality achieved? I tried showing it but the RHS are variables.) Obviously $1<r<3$. The change of variables is injective and surjective using the inequality in the hint, and the remaining term is $r^2\sin \theta$, but I have no how $\theta$ acts in a single round. I find myself starting from $\theta=0$ (r=1), and again $\theta =0 $ for $r=3$. Do you have any idea how one can proceed from here? While I do understand how to bound the variables in the unit sphere(or, obviously any sphere centered at the origin), here it all seems to collapse.
Final answer is told to be: $\pi(2-{3\over 2}\log 3)$, for those who are interested or need it.
| 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
$$\begin{align}{}
I
&=\underset{x^2+y^2+(z-2)^2\le 1}{\int\int\int}{1\over x^2+y^2+z^2}dxdydz \\
&= \int_{\phi=0}^{2 \pi} \int_{r=1}^{3} \int_{\theta=0}^{ \cos^{-1} (\frac{r^2+3}{4r})} \frac{1}{r^2} r^2 \sin \theta \, d\theta \, dr \, d \phi \\
&= \int_{\phi=0}^{2 \pi} \int_{r=1}^{3} \int_{\theta=0}^{\cos^{-1} (\frac{r^2+3}{4r})} \sin \theta \, d\theta \, dr \, d \phi \\
&= \int_{\phi=0}^{2 \pi} \int_{r=1}^{3} -\cos \theta|_{0}^{\cos^{-1} (\frac{r^2+3}{4r})} \, dr \, d \phi \\
&= \int_{\phi=0}^{2 \pi} \int_{r=1}^{3} (1-\frac{r^2+3}{4r}) \, dr \, d \phi \\
&=(\int_{0}^{2 \pi} d\phi)(\int_{r=1}^{3} (1-\frac{1}{4}r-\frac{3}{4}\frac{1}{r}) \, dr) \\
&= (2 \pi) (1-\frac{3}{4}\log3) \\
&=\pi (2-\frac{3}{2}\log3)
\end{align} \\
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 calculations show that its going near $\frac{1}{18}$.
| 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 plugging in $m=3$ to the tip, we would get
$$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)(k+3)}=\frac{1}{3} \left( \frac{1}{1\cdot2\cdot3} -\frac{1}{(n+1)(n+2)(n+3)}\right)$$
Take limit, $n \to \infty$.
$$\sum_{k=1}^\infty \frac{1}{k(k+1)(k+2)(k+3)}=\frac{1}{3} \left( \frac{1}{1\cdot2\cdot3} -\lim_{n \to \infty}\frac{1}{(n+1)(n+2)(n+3)}\right)$$
$$=\frac{1}{18}$$
How to decompose the fraction;
$$\frac{1}{k(k+1)\cdots(k+m)}=\frac{1}{m}\left(\frac{1}{k\cdot (k+1)\cdot \cdots\cdot (k+m-1)} -\frac{1}{(k+1)\cdot (k+2)\cdot \cdots\cdot (k+m)}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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:
$$ P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2} $$
$$ P2 | \frac{1-(x+1)q^x+xq^{x+1}+[(x+1)(1-q)^2]q^x}{(1-q)^2} = \frac{1-(x+2)q^{x+1}+xq^{x+2}}{(1-q)^2} $$
$$ P3| \frac{x\color{red}{q^{x+1}}+[-(x+1)]\color{red}{q^x}+1+[(x+1)(1-q)^2]\color{red}{q^x}}{(1-q)^2} = \frac{x\color{red}{q^{x+2}}-(x+2)\color{red}{q^{x+1}}+1}{(1-q)^2} | $$
Here I just reorganize both sides of the equation, so LHS is explicity an expression with a degree of x+1, while the degree of RHS is x+2. Both LHS' $\color{red}{q^x}$ are added next.
$$P4| \frac{xq^{x+1}+[-(x+1)+(x+1)(<1^2q^0+\binom{2}{1}1q-1^0q^2>)]q^x+1}{(1-q)^2}=\frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $$
$$P5 | \frac{xq^{x+1}+[2xq-xq^2+2q-q^2]q^x+1}{(1-q)^2} = \frac{xq^{x+2}-(x+2)q^{x+1}+1}{(1-q)^2} $$
I get stuck at this point. I don't know if i'm approaching the problem the right way. So, any help would be appreciated.
Thanks in advance.
| 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...
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $2+4+8+...=-2$
$x=2+4+8+...$
$2x=4+8+16+...$
Then:
$x=-2$
| $\def\bR{\mathbb{R}}\def\bQ{\mathbb{Q}}$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, $A_n:=\sum_{k=0}^{n-1} 2^k= 2^n-1$ and then $A_n-(-1)=2^n$ and $2^n\to0$ in the $2$-adic sense.
So what you discovered is that series that are divergent in $\bR$ need not be divergent in other completions of $\bQ$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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.
$$a^2b|(a+b)^3 - 3ab^2$$
$$a^2b|2a^3+3a(a+k)+k^3$$
I can't proceed from here. How should I proceed from here?
Thanks.
| 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 $$a^2b\mid a^3+b^3$$
$$\iff d^3a_1^2b_1\mid d^3a_1^3+d^3b_1^3$$
$$\iff a_1^2b_1\mid a_1^3+b_1^3$$
$$\iff \begin{cases}a_1^2\mid a_1^3+b_1^3\\b_1\mid a_1^3+b_1^3\end{cases}\iff \begin{cases}a_1^2\mid b_1^3\\ b_1\mid a_1^3\end{cases}$$
Since $\gcd\left(a_1^2,b_1^3\right)=\gcd\left(b_1,a_1^3\right)=1$ and $a_1,b_1\in\mathbb Z^+$, we get $a_1=b_1=1$. Therefore $a=b=\gcd(a,b)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\frac{a}{k+i}<\frac{a}{k}, i \in N$
$$\frac{a^n}{n!}=\frac{a}{1} \cdot \frac{a}{2} \cdot...\cdot\frac{a}{k}\cdot\frac{a}{k+1}\cdot...\cdot\frac{a}{n}<\frac{a^k}{k!}(\frac{a}{k})^{n-k}=\frac{a^k}{k!}\cdot \frac{a^na^{-k}}{k^nk^{-k}}$$
$$0<\frac{a^n}{n!}<\frac{k^k}{k!}(\frac{a}{k})^n$$
then $$0 \leq \lim_{n\to\infty}{\frac{a^n}{n!}}\le\frac{k^k}{k!}\lim_{n\to\infty}{(\frac{a}{k})^n}$$
Now, the part I don't understand is in the expansion of $\frac{a^n}{n!}$. How does one come with $<\frac{a^k}{k!}(\frac{a}{k})^{n-k}$ part, to be more precise, how does one come up with $(\frac{a}{k})^{n-k}$ part of the inequality?
| 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 inequality, we finally will get $\left(\frac{a}{k}\right)^{n-k}$ because there are $n-k$ such terms.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\left|1+z^n\right|\leq 1+\left|z\right|^n\leq 1 + 1=2$ and hence $\limsup_{n\to \infty}\frac{1}{1+z^n}\ge 1/2$ which means that $\frac{1}{1+z^n}$ does not go to zero and so the series does not converge for $|z|\leq 1$.
$\left|\frac{1}{1+z^n}\right|=\sqrt{1/\left(1+2\Re(z^n)+\left|z\right|^{2n}\right)}$ and I suspect $1/\left(1+2\Re(z^n)+\left|z\right|^{2n}\right)^{1/2n}$ goes to $1/\left|z\right|$ but I am unable to prove that.
Edit:
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 comparison test. Note that $\displaystyle \frac{r^n-1}{r^{n+1}-1}=\frac{r^{n-1}+\dots+1}{r^n+\dots+1}=1- (r-1)\frac{r^n}{r^{n+1}-1}$ which goes to $1/r<1$ and hence $\displaystyle \sum \frac{1}{r^n-1}$ is convergent by the ratio test.
| 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 comparison test. Note that $\displaystyle \frac{r^n-1}{r^{n+1}-1}=\frac{r^{n-1}+\dots+1}{r^n+\dots+1}=1- (r-1)\frac{r^n}{r^{n+1}-1}$ which goes to $1/r<1$ and hence $\displaystyle \sum \frac{1}{r^n-1}$ is convergent by the ratio test.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $$\lim_{(x,y)\to (0,0)} \frac{x^2 y^3}{x^4 + y^4} =0 \,.$$
Approaching this limit from $y=0$, $x=0$, $y=x$, $y=x^2$ etcetera all yields 0 as value, so my proposal is that this limit is indeed 0.
I have been able to solve most similar limits so far by finding some convergent upper bound for the absolute limit, but with this one the difference between the numerator and the denominator is so small I can't find anything to fit inbetween. For example, $(x,y)\to(0,0)$,
$$ \left| \frac{x^2 y}{x^2 + y^2} \right| \le \left| \frac{(x^2 + y^2)y}{x^2 + y^2} \right| \to 0 \,. $$
Also, Continuity of $\frac{x^3y^2}{x^4+y^4}$ at $(0,0)$? and Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. contain some helpful hints.
| 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",
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"timestamp": "2023-03-29T00:00:00",
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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 $x^{ 2 }$ $$\left( x^{ 2 }+xy \right) y'=x\sqrt { x^{ 2 }-y^{ 2 } } +xy+y^{ 2 }\\ \left( 1+\frac { y }{ x } \right) { y }^{ \prime }=\sqrt { 1-\frac { { y }^{ 2 } }{ { x }^{ 2 } } } +\frac { y }{ x } +\frac { { y }^{ 2 } }{ { x }^{ 2 } } \\ y=zx\\ { y }^{ \prime }={ z }^{ \prime }x+z\\ \left( 1+z \right) \left( { z }^{ \prime }x+z \right) =\sqrt { 1-{ z }^{ 2 } } +z+{ z }^{ 2 }\\ { z }^{ \prime }x+z+z{ z }^{ \prime }x+{ z }^{ 2 }=\sqrt { 1-{ z }^{ 2 } } +z+{ z }^{ 2 }\\ { z }^{ \prime }x\left( 1+z \right) =\sqrt { 1-{ z }^{ 2 } } \\ \int { \frac { \left( 1+z \right) dz }{ \sqrt { 1-{ z }^{ 2 } } } } =\int { \frac { dx }{ x } } $$
substitute now :$z=\sin { \theta } $
$$\int { \frac { 1+\sin { \theta } }{ \cos { \theta } } \cos { \theta } d\theta } =\ln { Cx } \\ \theta -\cos { \theta } =\ln { Cx } \\ \arcsin { z-\sqrt { 1-{ z }^{ 2 } } } =\ln { Cx } $$
so the final aswer is :
$$\arcsin { \frac { y }{ x } } -\frac { \sqrt { { x }^{ 2 }-{ y }^{ 2 } } }{ x } =\ln { Cx } $$
| {
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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 getting rid of the $X$.
| 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 \\ -1 & -1\end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix}$.
$\begin{bmatrix}-3a- 8c+ 4 & -3b- 8d- 7 \\ -9a+ 5b+ 3 & -9b+ 5d- 2\end{bmatrix}= \begin{bmatrix}5a+ 8c & 5b+ 8d \\ -a- c & -b- d\end{bmatrix}$.
So we have the system of four equations
$$-3a- 8c+ 4= 5a+ 8c\\-9a+ 5b+ 3= -a- c\\-3b- 8d- 7= 5b+ 8d\\-9b+ 5d- 2= -b- d$$
Which is easy to solve for $a, b, c,$ and $d$.
| {
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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:
$$\begin{align}
f(\theta) &=3+2\sin^2\theta-6\sin\theta \cos\theta \\
&=3(1-2\sin\theta \cos\theta)+2\sin^2\theta \\
&=3(\sin\theta-\cos\theta)^2 + 2\sin^2\theta
\end{align}$$
Hence the minimum value of $f(\theta)=2\sin^2\theta$ when $\theta=\pi/4$
hence minimum value of $f(\theta)=1$.
But then again I tried to do question differently by making substitutions in order to change the whole $f(\theta)$ in the form of $\cos x+\sin x$
Then $f(\theta)$ came out to be
$$f(\theta)= 4-(\cos(2\theta)+3\sin(2\theta))$$
The minimum value of this expression is surely $$(f(\theta))_{min}=4-\sqrt{10}$$
Can anybody explain me algebraically why my first method gave the wrong result?
| $f(\theta) = 4 - \sqrt{10}$ 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 had,
$f(\theta) = 3(\sin \theta-\cos\theta)^2 + k$
it would be okay to zero out the term in parentheses. It must be greater than or equal to zero, so set it to zero.
But in your actual expression
$f(\theta) = 3(\sin \theta-\cos\theta)^2 + 2\sin^2\theta$
Minimizing either term doesn't minimize the sum.
Furthermore, when you say the minimum of $f(θ)=2\sin^2θ,$ occurs when $θ=\pi/4$ that is just wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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 $, so $y=1+2m$
$$6x \equiv 9-15y\pmod {18}$$
$$6x \equiv 9-15(1+2m)\pmod {18}$$
$$6x \equiv -6-30m\pmod {18}$$
Divide by 6
$$x \equiv -1-5m\pmod 3$$
The right solution is $x-m \equiv 2\pmod 3$. I have $x+5m \equiv 2\pmod 3$
$$x-m+6m \equiv 2\pmod 3$$ $$x-m \equiv 2\pmod 3$$
Wolfram alpha says $y=1+2t$ and $x=t+3d+2$
| 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",
"answer_count": 3,
"answer_id": 1
} |
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
$$I=\frac{1}{{{a}^{3}}+{{b}^{3}}+abc}\le \frac{1}{a^2b+b^2a+abc}$$
$$J=\frac{1}{{{b}^{3}}+{{c}^{3}}+abc}\le \frac{1}{b^2c+c^2b+abc}$$
$$K=\frac{1}{{{c}^{3}}+{{a}^{3}}+abc}\le \frac{1}{c^2a+a^2c+abc}$$
we have
$$I+J+K\le\frac{1}{a^2b+b^2a+abc}+\frac{1}{b^2c+c^2b+abc}+\frac{1}{c^2a+a^2c+abc}$$
$$I+J+K\le\frac{1}{a+b+c}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)$$
$$I+J+K\le\frac{1}{abc}$$
| {
"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:
$$ 2\sin\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) = 0 $$
$$ \sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right) = 0 $$
Here one of the factors has to be $0$,
$$ \sin x = 0 \ \Rightarrow \ x = 0 \ or \ x = \pi $$
$$ \sin\left(\frac{x}{2}'\right) = 0 \ \Rightarrow \ x = 0 ;\ x \text{ can't be } \pi \text{ because of its range} $$
$$ \cos x = 0 \ \Rightarrow \ x = \frac{\pi}{2} \text{ or } \ x = \frac{3\pi}{2} $$
$$ \cos\left(\frac{3x}{2}\right) = 0 => x = \frac{\pi}{3} \text{ or } x = \pi $$
So the solutions are : $$ 0, \pi, \frac{\pi}{3} $$
I have seen other methods to solve this, so please don't post them. I'm really interested what's wrong with this one.
| 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 easy to factor ( compared to systems of polynomial equations in general ):
$$\cases{y(2x-1) = 0\\x^2+y^2=0}$$
So either $2x-1 = 0$ or $y=0$
We recognize these equations as lines in the plane. The solutions are where they intercept the unit circle. Now it will be an easy exercise to substitute back and find all solutions.
| {
"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
} |
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 $$I = \frac{1}{2}\int\frac{t}{t}dt = \frac{1}{2}t+\mathcal{C}=\frac{1}{2}\sqrt{2-2x^{-2}+x^{-4}}+\mathcal{C}$$
So $$I = \frac{\sqrt{2x^4-2x^2+1}}{2x^2}+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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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
$$ \left(
\begin{array}{cc}
\frac{1}{2} e^{-\sqrt{b} \sqrt{c}}+\frac{1}{2} e^{\sqrt{b} \sqrt{c}} & \frac{\sqrt{b} e^{\sqrt{b} \sqrt{c}}}{2 \sqrt{c}}-\frac{\sqrt{b} e^{-\sqrt{b} \sqrt{c}}}{2 \sqrt{c}} \\
\frac{\sqrt{c} e^{\sqrt{b} \sqrt{c}}}{2 \sqrt{b}}-\frac{\sqrt{c} e^{-\sqrt{b} \sqrt{c}}}{2 \sqrt{b}} & \frac{1}{2} e^{-\sqrt{b} \sqrt{c}}+\frac{1}{2} e^{\sqrt{b} \sqrt{c}} \\
\end{array}
\right)
$$
But here one can find a formula for the computation on a general $2\times 2$ matrix, at the bottom of the page. Using that formula I got a different result.
What is the correct answer?
| $$A=\begin{pmatrix}
0 &b \\
c &0
\end{pmatrix}$$
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",
"answer_count": 3,
"answer_id": 1
} |
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
$\sup((\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})) < \inf((\frac{1}{2}
e^{8m+4} - 1, e^{8m+4} - 1)) \log(\inf((\frac{1}{2}
e^{8m+4} - 1, e^{8m+4} - 1))) < p_r$ and
$\inf((\frac{1}{2} e^{8m+4},
\frac{3}{2}e^{8m+4}))\log(\inf((\frac{1}{2} e^{8m+4},
\frac{3}{2}e^{8m+4}))) > e^{8m+4}$ for all $m > \frac{1}{4}$.
looks absolutely terrible. What I would like to write is this:
If we let $k \in (\frac{3}{2} e^{8m+4} - 1, 2e^{8m+4} - 1)$ and $r \in (\frac{1}{2} e^{8m+4} - 1, e^{8m+4} - 1)$ then we have $c = k - r + 1 \in (\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})$. We observe that
$\sup(c) < \inf(r) \log(\inf(r)) < p_r$ and $\inf(c)\log(\inf(c)) > e^{8m+4}$ for all $m > \frac{1}{4}$.
These statements aren't equivalent given that $r$ and $c$ are elements of the respective sets and not the sets themselves. However, I fear that both the first approach and assigning variables to the given sets would make this statement significantly longer than it needs to be.
Would this be considered a meaningful abuse of notation, or should I perhaps rewrite this another way?
Thank you.
| 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 $K=(\frac{3}{2} e^{8m+4} - 1, 2e^{8m+4} - 1)$, $R=(\frac{1}{2} e^{8m+4} - 1, e^{8m+4} - 1)$, and $C=(\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})$, then you can write with no abuse of notation:
If we let $k \in K$ and $r \in R$ then we have $k - r + 1 \in C$. We observe that
$\sup C < (\inf R) \log(\inf R) < p_r$ and $(\inf C) \log(\inf C) > e^{8m+4}$ for all $m > \frac{1}{4}$.
Combined with the definitions of the sets, this is only slightly longer than your proposed version, and it is significantly clearer what it means.
Alternatively, it sounds from context like you might not really care about the sups and infs themselves but, and care only about the relationships between the specific numbers $k$, $r$, and $c$ chosen. So you could avoid talking about sups and infs at all, and instead write something like:
If we let $k \in (\frac{3}{2} e^{8m+4} - 1, 2e^{8m+4} - 1)$ and $r \in (\frac{1}{2} e^{8m+4} - 1, e^{8m+4} - 1)$ then we have $c = k - r + 1 \in (\frac{1}{2} e^{8m+4}, \frac{3}{2}e^{8m+4})$. We observe that
$c < r \log(r) < p_r$ and $c\log(c) > e^{8m+4}$ for all $m > \frac{1}{4}$. The first inequality folows from $c<\frac{3}{2}e^{8m+4}$ and $r>\frac{1}{2} e^{8m+4} - 1$, and the second inequality follows from $c>\frac{1}{2} e^{8m+4}$.
| {
"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
} |
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 and rearranging the problem to $1/3(k+2)(k+3)$ is where I always get stuck.
| 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 $n=k+1$.
$$\begin{align}
\sum \limits_i^{k+1} i(i+1)=&(k+1)(k+2)+\sum \limits_i^{k} i(i+1) \\
=&(k+1)(k+2)+\frac{k(k+1)(k+2)}{3} \\
=&\frac{(k^2+4k+3)(k+2)}{3} \\
=&\frac{(k+1)(k+1+1)(k+1+2)}{3}.
\end{align}
$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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:
$$ \int_{x=0}^\pi \frac{\cos \frac{x}{2}}{\sqrt{1 + A \sin^2 \frac{x}{2}}} \sin \left( \frac{1+A}{\sqrt{A}} \omega \tanh^{-1} \frac{\cos \frac{x}{2}}{\sqrt{1 + A \sin^2 \frac{x}{2}}} \right) \, dx.$$
where $ 0<A<1 $ and $ \omega > 0 $ are parameters of the problem.
I tried to use a substitution of the argument of the hyperbolic arctan but that seemed to make it worse. I was posting here in the hopes of receiving help, if someone could tell me if it is at all possible to solve it analytically via a trick of sorts or a clever substitution, or maybe it is an elliptic integral in disguise. I thank all helpers.
** My question on the Melnikov integral can be found here I just used trig identities to soften it up
| 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
$$t = \sqrt{\frac{1-y^2}{1+Ay^2}} \implies
y = \sqrt{\frac{1-t^2}{1+At^2}}
\implies
\frac{dy}{\sqrt{1+Ay^2}} = - \frac{t\sqrt{A+1}dt}{\sqrt{1-t^2}(1+At^2)}
$$
and $dt = (1-t^2)d\theta$, we have
$$\begin{align}
\mathcal{I} &= 2\int_0^1 \sin(B\theta)\frac{dy}{\sqrt{1+Ay^2}}
= 2\sqrt{A+1}\int_0^1\sin(B\theta)\frac{t\sqrt{1-t^2}}{1+At^2}\frac{dt}{1-t^2}\\
&= 2\sqrt{A+1}\int_0^\infty \frac{\sin(B\theta)\sinh\theta}{\cosh^2\theta + A\sinh^2\theta} d\theta
= -i\sqrt{A+1}\int_{-\infty}^\infty \frac{e^{iB\theta}\sinh\theta}{\cosh^2\theta + A\sin^2\theta} d\theta\\
&= -i2\sqrt{A+1}\int_0^\infty \frac{z^{iB}(z^2-1)}{(z^2+1)^2 + A(z^2-1)^2} dz
\end{align}
$$
Let $\phi = \tan^{-1}\sqrt{A}$, this can be simplified as
$$
\mathcal{I} =
-\frac{2i}{\sqrt{A+1}}\int_0^\infty \frac{z^{iB}(z^2-1)}{z^4 + 2\left(\frac{1-A}{1+A}\right) z^2 + 1} dz
= -2i\cos\phi\int_0^\infty \frac{z^{iB}(z^2-1)}{(z^2+e^{2i\phi})(z^2+e^{-2i\phi})} dz$$
Consider following contour integral
$$\mathcal{J(\epsilon,R)} \stackrel{def}{=} \oint_{C(\epsilon,R)} \frac{(-z)^{iB}(z^2-1)}{(z^2+e^{2i\phi})(z^2+e^{-2i\phi})}
dz
\tag{*1}$$
where
*
*$\arg(-z) = 0$ on negative real axis.
*$(-z)^{iB}$ has a branch cut along positive real axis.
*$C(\epsilon,R)$ is the contour consists of
*
*$C_1$ : line segment from $\epsilon \to R$ above the positive real axis.
*$C_2$ : circular arc $Re^{iu}$ for $u$ from $0 \to 2\pi$.
*$C_3$ : line segment $R \to \epsilon$ below the positive real axis.
*$C_4$ : circular arc $\epsilon e^{iu}$ for $u$ from $2\pi \to 0$.
It is easy to see in $\mathcal{J}(\epsilon,R)$,
*
*the contribution from $C_1$ and $C_3$ adds up to
$$(e^{\pi B} - e^{-\pi B})\int_\epsilon^R \frac{z^{iB}(z^2-1)}{(z^2+e^{2i\phi})(z^2+e^{-2i\phi})} dz$$
*the contribution from $C_4$ vanishes as $\epsilon \to 0$.
*Since $B > 0$ is a real number, $|(-z)^{iB}|$ is bounded from above by $e^{\pi B}$ and the contribution from $C_2$ behaves as $O(R^{-1})$ as $R \to \infty$.
Combine these, we find
$$\mathcal{I} = -i\frac{\cos\phi}{\sinh(\pi B)}\lim_{\epsilon \to 0,R \to \infty} \mathcal{J}(\epsilon,R)$$
The integrand in $(*1)$ has 4 poles inside the contour: $\; e^{i(\frac{\pi}{2} \pm \phi)}\;$ and $\;e^{i(\frac{3\pi}{2} \pm \phi)}$.
*
*The residues at $e^{i(\frac{\pi}{2} \pm \phi)}$ are
$\displaystyle\;(e^{-(-\frac{\pi}{2} \pm \phi)B})\frac{-e^{\pm 2i\phi} - 1}{4i e^{\pm i\phi }(-e^{\pm 2i \phi} + \cos(2\phi))} = \mp\frac{e^{(\frac{\pi}{2}\mp \phi)B}}{4\sin\phi}$
*The residues at $e^{i(\frac{3\pi}{2} \pm \phi)}$ are
$\displaystyle\;(e^{-(\frac{\pi}{2} \pm \phi)B})\frac{-e^{\pm 2i\phi} - 1}{-4i e^{\pm i\phi }(-e^{\pm 2i \phi} + \cos(2\phi))} = \pm\frac{e^{(-\frac{\pi}{2}\mp \phi)B}}{4\sin\phi}$
This implies
$$\begin{align}
\mathcal{I} &= \left(-i \frac{\cos\phi}{\sinh\pi B}\right)\left(\frac{2\pi i}{4\sin\phi}\right)\left[
-e^{(\frac{\pi}{2}-\phi)B}
+e^{(\frac{\pi}{2}+\phi)B}
+e^{(-\frac{\pi}{2}-\phi)B}
-e^{(-\frac{\pi}{2}+\phi)B}
\right]\\
&= \left(\frac{\cos\phi}{\sinh\pi B}\right)\left(\frac{2\pi}{4\sin\phi}\right)
(e^{\frac{\pi}{2}B} - e^{-\frac{\pi}{2}B})
(e^{\phi B} - e^{-\phi B})
= \frac{\pi}{\tan\phi}\frac{\sinh(B\phi)}{\cosh\left(\frac{\pi}{2}B\right)}\\
&= \frac{\pi}{\sqrt{A}}\frac{\sinh(B\tan^{-1}\sqrt{A})}{\cosh\left(\frac{\pi}{2}B\right)}
\end{align}
$$
Treating $A, B$ as two independent parameters, in the limiting case $A \to 0$, we find
$$\lim_{A\to 0} \mathcal{I} = \frac{\pi B}{\cosh\left(\frac{\pi}{2}B\right)}$$
This matches what first pointed out by @nospoon in the comments.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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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 follow-up question about generalizations of this inequality here:
Square root inequality revisited
| $$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}\iff\sqrt{n^2+n}-n\lt\frac 12\lt n-\sqrt{n^2-1}$$
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
$\ u_{1} + u_{2} + \cdots + u_{n}$.
Computing a few terms I found
\begin{align*}u_0 &= 0\\u_1 &= 1\\u_2 &= k\\u_3 &= k^2-1\\u_4 &= k(k^2-1)-k = k^3-2k\\u_5 &= k(k^3-2k)-(k^2-1) = k^4-3k^2+1\\u_6 &= k(k^4-3k^2+1)-(k^3-2k) = k^5-4k^3+3k.\end{align*}
I am not sure how we can use this to solve the question, but I think it may help. Cubing these expressions seems very computational so there must be an easier way.
| 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 $\cos(\theta) = k/2$, we have
$$ u_n = \frac{\sin(n\theta)}{\sin(\theta)}$$
which can be verified by induction. Don't worry about $\theta$ being real only for $|k|\le 2$.
Using this we can obtain closed-form formulas for $u_1 + \ldots + u_n$ and $u_n^3 + \ldots + u_n$, and $y_n$ (it's rather tedious if working by hand, but elementary)
$$ y_n = \frac{-\cos((2n+1)\theta) + 2 \cos(\theta) - \cos((n+1)\theta) - \cos(n\theta) + 1}{\cos(\theta) - \cos(3\theta) - \cos(2\theta)+1} $$
and it can be verified directly that this satisfies the recurrence above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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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 $f(x) = x^4 + tx^3 + (t^2 + 1)x^2 + (t^3 + t)x + (t^4 + t^2) \in k(t)[x]$. Is $f(x)$ irreducible over $k(t)$?
| 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 polynomial to get a non-zero result.
This leaves the scenario that the polynomial factors as a product of quadratics:
$$f(x)=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+ bd$$
This gives us four equations: $$\begin{eqnarray} a+c&=& t \\ bd&=&t^4+t^2\\ ad+bc &=&t^3+t\\ ac+b+d&=&t^2+1\end{eqnarray}$$
By the mention of Gauss' lemma in the first sentence, we may list the 9 possible values for $b,d$ assuming they lie in $k[t]$ and then check each case.
By symmetry, we have only 5 cases to consider: $b=1,t,t+1,t^2,t(t+1)$ (these give $d=t^4+t^2,t^3+t,t^3+t^2,t^2+1,t(t+1)$ respectively).
We can eliminate most of these situations by considering the third equation, $ad+bc=t^3+t$.
Case 1 gives $a(t^4+t^2)+c=t^3+t$, which may be dismissed by degree considerations- if $a$ is of degree $n$ in $t$, then $c$ must be of degree $4+n$ in $t$, but also is at most degree $n+1$ in $t$ by the equation $a+c=t$.
Case 2 gives $a(t^3+t)+ct=t^3+t$, or $at^3+(a+c)t=at^3+t^2\neq t^3+t$, as $a+c=t$. So this case is impossible.
Case 3 gives $a(t^3+t^2)+c(t+1)=t^3+t$. On the other hand, the equation $ac+b+d=t^2+1$ will help here- plugging in, we have that $ac=t^3+t$, so $a=t^p(t+1)^q$ where $p\in\{0,1,2\}$ and $q\in\{0,1\}$. For each choice, it is clear that $a+c\neq t$, so this case is impossible.
Case 4 gives $a(t^2+1)+ct^2=t^3+t$, or $(a+c)t^2+a=t^3+a=t^3+t$ so $a=t$, $c=0$. Plugging this in to the remaining equation $ac+b+d=t^2+1$, we have $t\cdot0+t^2+t^2+1=1\neq t^2+1$. So this is impossible.
Case 5 may be immediately discarded, as it reduces to $(a+c)(t^2+t)=t(t^2+t)=t^3+t^2\neq t^3+t$.
Therefore the given polynomial is irreducible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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} \Rightarrow y^2 = \left( x+\frac{1}{x} \right)^{2}$
$y^2 = x^2+2 + \frac{1}{x^2} $
one should arrive at the conclusion. $$f(x)^{-1} = \frac{x \pm \sqrt{x^2-4} }{2}$$
Yet I am having trouble at arriving at that conclusion. Could anyone show me how to do it arithmetically.
| 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 function of $y$ using, for example, the quadratic formula.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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
$$\begin{align}
\left|\int_0^1 \frac{1+nx^2}{(1+x^2)^n}\log\left(2+\cos\left(\frac{x}{n}\right)\right)\,dx\right|&=\frac{1}{\sqrt{n}}\int_0^\sqrt{n} \frac{1+x^2}{(1+x^2/n)^n}\log\left(2+\cos\left(\frac{x}{n^{3/2}}\right)\right)\,dx\\\\
&\le \frac{\log(3)}{\sqrt{n}} \int_0^\infty \frac{1+x^2}{\left(1+\frac{x^2}{n}\right)^n}\,dx \tag 1\\\\
&\le \frac{\log(3)}{\sqrt{n}} \int_0^\infty \frac{1+x^2}{\left(1+\frac{x^2}{2}\right)^2}\,dx \tag 2\\\\
&=\frac{\log(3)}{\sqrt{n}}\,\left(\frac{3\sqrt{2}\pi}{4}\right)\tag 3
\end{align}$$
where we used the fact that $\left(1+\frac {x^2}n\right)^n\ge \left(1+\frac{x^2}{2}\right)^2$ for $n\ge2$ in going from $(1)$ to $(2)$.
The right-hand side of $(3)$ clearly approaches zero as $n\to \infty$. Therefore, the squeeze theorem guarantees that the integral of interest approaches zero also. And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"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 $2k^2(6k^2 + 11k - 35)$.
I would think that I'd want to find two numbers that sum up to $11$ and have a product of $-35$, but instead I'm told we need to multiply $-35$ by $6$ so that I now have to find two numbers that sum up to $11$ and have a product of $-210$.
Can anyone help me understand why $-35$ is multiplied by the coefficient $6$? Why isn't $11k$ also multiplied by $6$?
| 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}\\
& = 3x(2x + 7) - 5(2x + 7) && \text{factor by grouping}\\
& = (3x - 5)(2x + 7) && \text{extract the common factor}
\end{align*}
Notice that in the expression $\color{blue}{6}x^2 \color{green}{+ 21}x \color{green}{- 10}x \color{blue}{- 35}$, the product of the quadratic and constant coefficient is equal to the product of the two linear coefficients whose sum is the linear coefficient of $6x^2 + 11k - 35$, that is,
$$(\color{blue}{6})(\color{blue}{-35}) = (\color{green}{21})(\color{green}{-10}) = -210$$
Let's examine the general case. Suppose $ax^2 + bx + c$ factors with respect to the rationals as $(rx + s)(tx + u)$. Then
\begin{align*}
ax^2 + bx + c & = (rx + s)(tx + u)\\
& = rx(tx + u) + s(tx + u)\\
& = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{green}{su}\\
& = \color{blue}{rt}x^2 + (\color{green}{ru} + \color{green}{st})x + \color{blue}{su}
\end{align*}
Matching coefficients, we have $a = \color{blue}{rt}$, $b = \color{green}{ru + st}$, and $c = \color{blue}{su}$. Notice that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients whose sum is $b$, that is,
$$ac = (\color{blue}{rt})(\color{blue}{su}) = (\color{green}{ru})(\color{green}{st})$$
Thus, to split the linear term, we must find two numbers whose product is $ac$ and whose sum is $b$.
| {
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"url": "https://math.stackexchange.com/questions/1862611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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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\left(1 - \dfrac{x}{2}\right)\right)}{x^{2}}\notag\\
&= e\lim_{x \to 0}\exp\left(\log\left(1 - \dfrac{x}{2}\right)\right)\cdot\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)\right) - 1}{x^{2}}\notag\\
&= e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)\right) - 1}{\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)}\cdot\dfrac{\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)}{x^{2}}\notag\\
&= e\lim_{x \to 0}\dfrac{\dfrac{\log(1 + x)}{x} - 1 - \log\left(1 - \dfrac{x}{2}\right)}{x^{2}}\notag\\
&= e\lim_{x \to 0}\dfrac{\left(1 - \dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right) - 1 + \left(\dfrac{x}{2} + \dfrac{x^{2}}{8} + o(x^{2})\right)}{x^{2}}\text{ (using Taylor series)}\notag\\
&= e\left(\frac{1}{3} + \frac{1}{8}\right)\notag\\
&= \frac{11e}{24}\notag
\end{align}
The above approach uses the standard limit $$\lim_{t \to 0}\frac{\exp(t) - 1}{t} = 1$$ Taylor series is used only when necessary and this approach avoids any multiplication/division of infinite series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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:
$$\tan \left(\frac{x}{2}\right) = \frac{1-\cos \left(x\right)}{\frac{7}{5}- \cos \left(x\right)}$$ But I don't know what to do from here. Can someone explain to me how to solve this? Thanks for any answers.
| 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 $\tan\left(\frac{x}{2}\right)\color{red}{\in\left\{\frac{1}{2},\frac{1}{3}\right\}}$ can be found by solving a quadratic equation.
As an alternative, we may simply consider the (right) triangle with side lenghts $3,4,5$ and find its (unit) inradius.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 the equation is something like:
$$\gamma_{i+2} + 4\gamma_{i+1} + \gamma_{i} = K'$$
where $K' << 1$, much smaller than $1$.
Is there a way to solve such equation, I want to find the forced response, I've tried to use the z transform but somehow I end up with divergent solutions, because of the eigenvalues calculation.
I'm quite sure there's an easier way to study the equation.
If it is of any interest I could report the derivation of such equation.
Update:
Here is my new attempt
$$
\left\{
\begin{array}{l}
\gamma_{i+2} + 4\gamma{i+1} + \gamma_{i} = K' \\
\gamma_{i+3} + 4\gamma{i+2} + \gamma_{i+1} = K'
\end{array}
\right. \Rightarrow \Delta \gamma_{i+2} + 4 \Delta \gamma_{i+1} + \Delta \gamma_{i} = 0
$$
Using the characteristic equation I find:
$$
\begin{multline}
\Delta \gamma_{k} = c_0 \left(2 - \sqrt{3} \right)^k + c_1 \left(2 + \sqrt{3} \right)^k \Rightarrow \\
\sum_{j=0}^{k-1} \Delta \gamma_j = c_0 \sum_{j=0}^{k-1} \left(2 - \sqrt{3} \right)^j + c_1 \sum_{j=0}^{k-1} \left(2 + \sqrt{3} \right)^j \Rightarrow \\
\gamma_k - \gamma_0 = c_0 \left[ \frac{\left(2 - \sqrt{3} \right)^k - 1}{1- \sqrt{3}} \right] + c_1 \left[ \frac{\left(2 + \sqrt{3} \right)^k - 1}{1+ \sqrt{3}} \right] \Rightarrow \\
\gamma_k = c_0 \left[ \frac{\left(2 - \sqrt{3} \right)^k - 1}{1- \sqrt{3}} \right] + c_1 \left[ \frac{\left(2 + \sqrt{3} \right)^k - 1}{1+ \sqrt{3}} \right] + \gamma_0
\end{multline}
$$
Since $\gamma_k = \gamma_k(h)$ we have
$$
\gamma_k(h) = c_0(h) \left[ \frac{\left(2 - \sqrt{3} \right)^k - 1}{1- \sqrt{3}} \right] + c_1(h) \left[ \frac{\left(2 + \sqrt{3} \right)^k - 1}{1+ \sqrt{3}} \right] + \gamma_0(h)
$$
Is it correct so far?
| 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 using GFs or characteristic equation.
EDIT: you need two boundary values: $a_0$ and $a_1$.
$$
a_{n+2} + 4 a_{n+1} + a_{n} = C \\
a_{n+1} + 4 a_n + a_{n-1} = C\\
b_{n+2} + 4 b_{n+1} + b_n = 0
$$
Use generating function to solve for $b_n$:
$$
\frac{G(z) - b_1 z -b_2 z^2}{z^2} + \frac{G(z) - b_1 z }{z} + G(z) = 0
$$
Solve this for $G(z)$ and extract expression for $b_n$. From that, since $b_n = a_n - a_{n-1}$, use a telescoping sum to get on LHS ana expression for $a_n$, and on RHS some sum that you will need to solve for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 equation?
| 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 $\sin y + \sin x = 0 \ne 1$ which is impossible.
If $y = x - \pi$ then $\sin y = \sin x$ and $\sin y + \sin x = 2 \sin x$. If this is so (and it's our only option) then $\sin x = 1/2$ which means $x = \{\pi/6, 5\pi/6\}$.
So $(x,y) = (\pi/6, 5\pi/6)$ or $(x,y)= (5\pi/6, \pi/6)$ (within periods of $2\pi$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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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(s). I also tried to look for a pattern in the question, but I cannot see any pattern (possibly because I'm overthinking it?) Please help me with this problem.
| $$I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\cdots$$
$$2I=1+1+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\cdots$$
$$2I-I=1+\left(1-\frac 12 \right)+\left(\frac 34 -\frac 24 \right)+\left(\frac 48 -\frac 38 \right)+\left(\frac {5}{16} -\frac {4}{16} \right)+\cdots$$
$$I=1+\frac 12+\frac 14+\frac 18+\cdots=2$$
| {
"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
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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
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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 it efficiently here.
| 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 we can make $\ {-}\color{#0a0}3\equiv 28^3/28^7\equiv \color{#0a0}{28^{-4}}$
Therefore $\ 2\equiv \color{#c00}5\cdot (\color{#0a0}{-3})^2 \equiv \color{#c00}{28^5}(\color{#0a0}{28^{-4}})^2\equiv 28^{-3}\equiv 28^{39},\ $ by $\,28^{42}\equiv 1\,$ by Fermat.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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:
$$\frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1} $$
$$[(x-5)(x+1)]\times \frac{3x-8}{(x-5)(x+1)} = \frac{A}{x-5} + \frac{B}{x+1}\times[(x-5)(x+1)]$$
$$3x-8 = A(x+1) + B(x-5)$$
then he substitutes $x$ with $-1$ this is when I feel something is wrong.
because $x=-1$ is not in the domain of the original function and all the statement below the original function hold true only if they comply within the domain of the original function isn't it? so at the time this guy substitute $x=-1$ the statement become false?
I'm newbie in math, please explain to me with easy understanding and step by step.
| 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 you divide both sides of the polynomial identity by $(x+1)(x-5)$ with the added restriction that $x\neq -1, 5$. So this gives you $$\frac{3x-8}{(x+1)(x-5)} = \frac{A(x-5) + B(x+1)}{(x+1)(x-5)} = \frac{A}{x-5} + \frac{B}{x+1}$$
which holds for all $x \neq -1, 5$. (which is what let us do the cancelling bit on the RHS)
Some explanation: the motivation for using $x=-1$ and $5$ is that is efficient in the sense that it makes one of the terms in the identity $0$, speeding up the process of finding the values of the coefficients. If you wanted to, you could use any other two values such as $x=3$ and $4$ to get two simultaneous equations in $A$ and $B$, allowing you to determine their values, just not as efficiently.
| {
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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 the proof to use induction.
| 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 is what you want to show
for $n=0$.
Differentiating that,
$(1-x^2)y'''-2xy''
= 3(xy''+y')+y'
$,
or
$(1-x^2)y'''
= 5xy''+4y'
$,
which is what you want to show
for $n=1$.
This establishes the basis
for the induction.
Suppose
$(1-x^2)y^{(n+2)}
= (2n+3)xy^{(n+1)} + (n+1)^2y^{(n)}
$.
Differentiating this,
$\begin{array}\\
(1-x^2)y^{(n+3)}-2xy^{(n+2)}
&= (2n+3)(xy^{(n+2)}+y^{(n+1)}) + (n+1)^2y^{(n+1)}\\
\text{or}\\
(1-x^2)y^{(n+3)}
&= (2n+5)xy^{(n+2)}+(2n+3)y^{(n+1)} + (n+1)^2y^{(n+1)}\\
&= (2n+5)xy^{(n+2)} + (n^2+2n+1+2n+3)y^{(n+1)}\\
&= (2n+5)xy^{(n+2)} + (n^2+4n+4)y^{(n+1)}\\
&= (2(n+1)+3)xy^{(n+2)} + (n+2)^2y^{(n+1)}\\
\end{array}
$
This is the induction step,
and shows that the result is true.
Actually,
looking at this,
I didn't need to show that
the result was true
for $n=1$.
| {
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} |
$ \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
$$
\frac {ey+f}{y^2 - yc + c^2},
$$
and letting $u=y^2-yc+c^2$ so $du = (2y-c)\,dy$. Multiplying both sides of the last equality by $e/2$, we get
$$
\frac e 2 \, du = \left( ey+\frac{ce}{2} \right) \, dy.
$$
So
$$
(ey+f)\,dy = \left( \underbrace{\left( ey + \frac{ce}2 \right)\,dy} + \left( f - \frac{ce}2 \right) \, dy \right).
$$
The substitution handles the part over the $\underbrace{\text{underbrace}}.$ Then you have
$$
\int \frac{dy}{y^2-yc+c^2}.
$$
Complete the square
$$
y^2 - yc + c^2 = \left( y^2 - yc + \frac{c^2} 4 \right) + \frac{3c^2} 4 = \left( y - \frac c 2 \right)^2 + \frac{3c^2} 4.
$$
So when you integrate, you get an arctangent.
| {
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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:
We see that the number of integer values of $f(x)$ where $x \in [0,3)$ is the same as the number of integer values of $f(x)$ where $x \in [3,6)$ and so on. The number of different integer values of $f(x)$ where $x \in [0,3)$ is equal to the number of places $f(x)$ changes value from where it was before plus one (its initial starting position at $x = 0$). Thus, the places where $f(x)$ changes its value can occur only when, expressed in simplest form, the numerator is a multiple of $3$ or when the denominator is a multiple of $3,4,$ or $5$. We then obtain the following $22$ fractions: $$0,1,2,\dfrac{1}{2},\dfrac{3}{2},\dfrac{5}{2},\dfrac{1}{3},\dfrac{2}{3},\dfrac{4}{3},\dfrac{5}{3},\dfrac{7}{3},\dfrac{8}{3},\dfrac{1}{4},\dfrac{3}{4},\dfrac{5}{4},\dfrac{7}{4},\dfrac{11}{4},\dfrac{3}{5},\dfrac{6}{5},\dfrac{9}{5},\dfrac{12}{5}.$$ Similarly, in the interval $[99,100)$ we see that are $8$ integer values of $f(x)$ generated here. Thus, there are $33 \times 22+8 = 734$ different integer values of $f(x)$ taken on this range.
| 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 convenient formulation, since you can prove that each summand of the form $[ax]$ is constant except where $x=\frac{c}a$ for some integer $c$. You can conclude that $f(y)$ is different from $f(x)$ exactly when there is some number of the form $\frac{c}a$ between them for $a$ being one of $1,\,2,\,3,\,4,\,\frac{5}3$ - i.e. the given coefficients.
Note that this has little to do with the divisibility of the numerator or denominator. Your current phrasing suggests that $\frac{1}8$ should be included in the list, as the denominator is a multiple of $4$, and that $\frac{3}7$ should be included in the list, as the numerator is a multiple of $3$. While your list is correct, the criteria you used for inclusion in it is different than what you stated.
The other small detail in your proof that seems suspicious is where you talk about "the number of places $f(x)$ changes value from where it was before plus one" - it seems to imply that all the jumps in $f(x)$ have a height of $1$, however, one can note that the jumps in the constituent functions overlap. For instance, $f(x)$ takes the value of $30$ when $x$ is in $[3-1/5,3)$, but jumps to $35$ at $3$, creating a jump with height $5$ due to every constituent function's jump lining up.
| {
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} |
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 this problem, how we can numerically
find $c_1$ and $c_2$?
| 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
$\begin{pmatrix}p_1(\frac{\pi}{2}) & p_2(\frac{\pi}{2})\\ p_1(0) & p_2(0) \end{pmatrix} \begin{pmatrix}c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix}P(\frac{\pi}{2}) \\ P(0) \end{pmatrix}$
then
$\begin{pmatrix}0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix}c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix}1 \\ -1 \end{pmatrix}$
By matrix multiplication:
$\begin{pmatrix}c_2 \\ c_1 \end{pmatrix} = \begin{pmatrix}1 \\ -1 \end{pmatrix}$
So $c_2 = 1$ and $c_1 = -1$
| {
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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 :
$$\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{n+1}{n}\right)\right]$$
I could not any approaching in order to solve problem . I need help!
| 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(\frac{(n + 1)^{2}}{n(n + 2)}\right)\right) - \exp\log\frac{n + 1}{n}\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\cdot\frac{n + 1}{n}\left\{\exp\left(n\log\left(\frac{(n + 1)^{2}}{n(n + 2)}\right) - \log\frac{n + 1}{n}\right) - 1\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\dfrac{\exp\left(n\log\left(\dfrac{(n + 1)^{2}}{n(n + 2)}\right) - \log\dfrac{n + 1}{n}\right) - 1}{n\log\left(\dfrac{(n + 1)^{2}}{n(n + 2)}\right) - \log\dfrac{n + 1}{n}}\cdot\left(n\log\left(\dfrac{(n + 1)^{2}}{n(n + 2)}\right) - \log\dfrac{n + 1}{n}\right)\notag\\
&= \lim_{n \to \infty}n^{2}\left(n\log\left(\dfrac{(n + 1)^{2}}{n(n + 2)}\right) - \log\dfrac{n + 1}{n}\right)\notag\\
&= \lim_{n \to \infty}n^{2}\left\{n\log\left(1 + \frac{1}{n(n + 2)}\right) - \log\left(1 + \frac{1}{n}\right)\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\left\{n\left(\frac{1}{n(n + 2)} - \frac{1}{2n^{2}(n + 2)^{2}} + o(1/n^{4})\right) - \left(\frac{1}{n} - \frac{1}{2n^{2}} + o(1/n^{2})\right)\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\left\{\left(\frac{1}{n + 2} - \frac{1}{2n(n + 2)^{2}} + o(1/n^{3})\right) - \left(\frac{1}{n} - \frac{1}{2n^{2}} + o(1/n^{2})\right)\right\}\notag\\
&= \lim_{n \to \infty}n^{2}\left(\frac{1}{n + 2} - \frac{1}{n}\right) + \frac{1}{2} + o(1)\notag\\
&= -2 + \frac{1}{2}\notag\\
&= -\frac{3}{2}\notag
\end{align}
We have used the standard limit $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ which shows that
\begin{align}
F(n) &= n\log\left(1 + \frac{1}{n(n + 2)}\right) - \log\frac{n + 1}{n}\notag\\
&= \frac{1}{n + 2}\cdot n(n + 2)\log\left(1 + \frac{1}{n(n + 2)}\right) - \log\frac{n + 1}{n}\notag\\
&\to 0\cdot 1 - 0 = 0\notag
\end{align}
Another standard limit $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$ is used to get rid of the exponential function and finally we use Taylor expansion $$\log(1 + x) = x - \frac{x^{2}}{2} + o(x^{2})$$ when $x \to 0$.
| {
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Convergence/Divergence of $\sum_{n=1}^{\infty}\frac{n+n^2+\cdots+n^n}{n^{n+2}}$
$$\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}$$
$$\sum_{n=1}^\infty \frac{1}{n^{n}}=\sum_{n=1}^{\infty}\frac{n^2}{n^{n+2}}=\sum_{n=1}^\infty \frac{n+n+\cdots+n}{n^{n+2}}\leq\sum_{n=1}^\infty \frac{n+n^2+\cdots+n^n}{n^{n+2}}\leq \sum_{n=1}^{\infty}\frac{n^n+n^n+\cdots+n^n}{n^{n+2}}=\sum_{n=1}^\infty \frac{n^{n+1}}{n^{n+2}}=\sum_{n=1}^\infty \frac{1}{n}$$
But is still does not help to conclude about converges/diverges
| 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 .
| {
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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 for $k=1$ to $k=10$ by hand and using the interpolation feature of Wolfram Alpha. Here are the results (I'll only show the coefficients for the sake of clarity. the coefficients are always from $n^{k+1}$ to $n^1$. the constant is always $0$. So $\frac{1}{2},-\frac{1}{2}$ becomes $\frac{1}{2}n^2-\frac{1}{2}n$):
*
*$k=1$ : $\frac{1}{2},-\frac{1}{2}$
*$k=2$ : $\frac{1}{3},\frac{1}{2},\frac{1}{6}$
*$k=3$ : $\frac{1}{4},\frac{1}{2},\frac{1}{4},0$
*$k=4$ : $\frac{1}{5},\frac{1}{2},\frac{1}{3},0,-\frac{1}{30}$
*$k=5$ : $\frac{1}{6},\frac{1}{2},\frac{5}{12},0,-\frac{1}{12},0$
*$k=6$ : $\frac{1}{7},\frac{1}{2},\frac{1}{2},0,-\frac{1}{6},0,\frac{1}{42}$
*$k=7$ : $\frac{1}{8},\frac{1}{2},\frac{7}{12},0,-\frac{7}{24},0,\frac{1}{12},0$
*$k=8$ : $\frac{1}{9},\frac{1}{2},\frac{2}{3},0,-\frac{7}{15},0,\frac{2}{9},0,-\frac{1}{30}$
*$k=9$ : $\frac{1}{10},\frac{1}{2},\frac{3}{4},0,-\frac{7}{10},0,\frac{1}{2},0,-\frac{3}{20},0$
*$k=10$ : $\frac{1}{11},\frac{1}{2},\frac{5}{9},0,1,0,1,0,-\frac{1}{2},0,\frac{5}{66}$
There are a few things i notice: Firstly, the coefficient of the highest power seems to be $\frac{1}{k+1}$. Secondly, the coefficient of the second highest power seems to be $\frac{1}{2}$ with the exception of $k=1$. Thirdly, all coefficients of the fourth, sixth, eight highest power and so on seem to be $0$.
What is the formula that will output the coefficients for any value of $k$?
| https://en.m.wikipedia.org/wiki/Faulhaber%27s_formula seems to deliver exactly what i needed. Original answer by MooS
| {
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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:
$\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}$
$$\sum_{k=1}^{\infty}\frac{3k^{2}+k+1}{k^{4}+k^{3}+4}< \frac{k^{2}+k}{k^{4}+k^{3}} < \frac{k^{2}}{k^{4}} \leq \frac{1}{k^{2}}$$
We know (from our readings) that $\sum_{k=1}^{\infty}\frac{1}{k^{2}}$ is a converging series.
Thus the complete series will converge.
Did I do everything correctly?
| 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
$$a<c\quad \mbox{and}\quad d<b\Rightarrow \frac{a}{b}<\frac{c}{d}.$$
| {
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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\left( \sum_{k=0}^{\infty}\left( \frac{1}{5} \right )^{k}-2 \right )\\&=15\left( \frac{1}{1-\frac{1}{5}}-2 \right )\\&=-\frac{45}{4}\end{align}
Did I do it correctly?
| You shoould obtain a positive number. Maybe you could follow this simpler approach:
$$\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}=\frac{3}{5}\sum_{k=2}^{\infty}\frac{1}{5^{k-2}}=\frac{3}{5}\sum_{j=0}^{\infty}\frac{1}{5^{j}}=\frac{3}{5}\cdot\frac{1}{1-1/5}=\frac{3}{4}.$$
| {
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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:
$$\sin((2n+1)x) = \sum_{k=0}^n \binom{2n+1}{2k+1}(-1)^k\cos^{2(n-k)}(x)\sin^{2k+1}(x)$$
As a result, the $\sin(\frac{k\pi}{2n+1})$ with $k\in\{-n,\ldots,n\}$ are roots of the polynomial $$\displaystyle P(X) = \sum_{k=0}^n \binom{2n+1}{2k+1}(-1)^k(1-X^2)^{n-k}X^{2k+1}$$
Note that $\deg P=2n+1$ and we have found $2n+1$ distinct roots.
Consequently, the $\displaystyle \frac{1}{\sin(\frac{k\pi}{2n+1})}$ with $k\in\{-n,\ldots,n\}\setminus\{0\}$ are roots of $Q$, the reciprocal polynomial of $P$ (Note that $\deg Q = 2n $).
Since we're interested in the sum of squares of the roots of $Q$, it is enough to look for the coefficients of $X^{2n}$ and $X^{2n-2}$ in $Q$, which are the coefficients of $X$ and $X^3$ in $P$.
A tedious computation shows that these coefficients are respectively $2n+1$ and $-\binom{2n+1}{3}-n(2n+1)$.
Noting that the roots of $Q$ are symmetric and using the standard expansion $\displaystyle \left(\sum_i x_i\right)^2 = \sum_i x_i^2 + 2\sum_{i<j} x_i$ yields $$\sum_{k=1}^n \frac{1}{\sin^2(\frac{k\pi}{2n+1})} = \frac 23 n(n+1) $$
The path I followed looks like it can be shortened. Is there a quicker way to find a suitable polynomial that annihilates all the $\displaystyle \frac{1}{\sin^2(\frac{k\pi}{2n+1})}$ ?
Is there a shorter way altogether ?
| 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,
$$ \sum_{k=1}^{n}\cot^2\left(\frac{\pi k}{2n+1}\right) = \binom{2n+1}{3} /\binom{2n+1}{1}=\frac{n(2n-1)}{3}\tag{2}$$
so:
$$ \sum_{k=1}^{n}\frac{1}{\sin^2\left(\frac{\pi k}{2n+1}\right)} = \frac{n(2n-1)}{3}+n=\color{red}{\frac{2}{3}n(n+1)}. \tag{3}$$
| {
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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 using a power series method:
$$\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)=l(l+1)\tag{1}$$
which has solutions $$R=\begin{cases} r^l \\ r^{-l-1} \end{cases}\tag{2}$$ (as printed in the back of the book)
Attempt #1:
Let $$R=\sum_{n=0}^\infty C_n\cdot r^n\tag{3}$$ where the $C_n$ are the expansion coefficients.
Then $$\frac{dR}{dr}=\sum_{n=1}^\infty C_n\cdot n\,r^{n-1}\tag{4}$$
and $$\frac{d^2R}{dr^2}=\sum_{n=2}^\infty C_n \cdot n(n-1)r^{n-2}\tag{5}$$
Re-writing $(1)$ via the product rule
$$2r\frac{dR}{dr}+r^2\frac{d^2R}{dr^2}=R\,l(l+1)\tag{6}$$
Now substituting $(3)$, $(4)$ and $(5)$ into $(6)$
$$2r\sum_{n=1}^\infty C_n \cdot n\,r^{n-1}+r^2\sum_{n=2}^\infty C_n \cdot n(n-1)\,r^{n-2}=l(l+1)\sum_{n=0}^\infty C_n \cdot r^{n}\tag{7}$$
I am unsure how to proceed from here to show that $$R=\begin{cases} r^l \\ r^{-l-1} \end{cases}\tag{2}$$
Could someone please give me some hints or advice on how I can proceed to reach $(2)$?
EDIT:
Proceeding with the advice given in the first comment by using the method of Frobenius:
Attempt #2:
Letting $$R=\sum_{n=0}^\infty C_n\cdot r^{s+n}\tag{A}$$ where the $C_n$ are the expansion coefficients.
Then $$\frac{dR}{dr}=\sum_{n=0}^\infty C_n\cdot (s+n)\,r^{s+n-1}\tag{B}$$
and $$\frac{d^2R}{dr^2}=\sum_{n=0}^\infty C_n \cdot (s+n)(s+n-1)\,r^{s+n-2}\tag{C}$$
Writing out the sums explicitly:
$$R=C_0\cdot r^s+C_1\cdot r^{s+1}+C_2\cdot r^{s+2}+C_3\cdot r^{s+3}+\cdots+C_n\cdot r^{s+n}$$
$$\frac{dR}{dr} = C_0\cdot s\,r^{s-1}+C_1\cdot(s+1)\,r^s + C_2\cdot(s+2)\,r^{s+1}+C_3\cdot(s+3)\,r^{s+2}\quad+\cdots + C_n\cdot(s+n)\, r^{s+n-1}$$
$$\frac{d^2R}{dr^2} = C_0\cdot s(s-1)\,r^{s-2}+C_1\cdot s(s+1)\,r^{s-1} + C_2\cdot(s+1)(s+2)\,r^s\quad +C_3\cdot(s+2)(s+3)\,r^{s+1}+\cdots + C_n\cdot(s+n)(s+n-1)r^{s+n-2}$$
Therefore
$$2r\frac{dR}{dr}=2C_0\cdot sr^s+2C_1\cdot (s+1)r^{s+1}+2C_2\cdot(s+2) r^{s+2}+2C_3\cdot(s+3)r^{s+3}\quad+\cdots+2C_n\cdot(s+n) r^{s+n}$$
$$r^2\frac{d^2R}{dr^2} = C_0\cdot s(s-1)\,r^{s}+C_1\cdot s(s+1)\,r^{s+1} + C_2\cdot(s+1)(s+2)\,r^{s+2}\quad + C_3\cdot(s+2)(s+3)\,r^{s+3}+\cdots + C_n\cdot(s+n)(s+n-1)\, r^{s+n}$$
$$l(l+1)R=l(l+1)C_0\cdot r^s+l(l+1)C_1\cdot r^{s+1}+l(l+1)C_2\cdot r^{s+2}+l(l+1)C_3\cdot r^{s+3}+\cdots+l(l+1)C_n\cdot r^{s+n}$$
Now rewriting equation $(6)$:
$$2r\frac{dR}{dr}+r^2\frac{d^2R}{dr^2}=R\,l(l+1)$$
in terms of summations gives
$$2r\sum_{n=0}^\infty C_n \cdot (s+n)\,r^{s+n-1}+r^2\sum_{n=0}^\infty C_n \cdot (n+s)(n+s-1)\,r^{s+n-2}=l(l+1)\sum_{n=0}^\infty C_n \cdot r^{s+n}$$
by bringing in the $r$'s into the sums for the first two summations on the LHS I find that
$$2\sum_{n=-1}^\infty C_{n+1} (s+n+1)r^{s+n}+\sum_{n=-2}^\infty C_{n+2}(s+n+2)(s+n+1)r^{s+n}=l(l+1)\sum_{n=0}^\infty C_n r^{s+n}$$
by shifting the index from $n\rightarrow n-1$ and $n\rightarrow n-2$ for the first and second sums respectively.
Now that the powers of $r$ are the same I can compare coefficients:
For $$n \ge -1\implies 2C_{n+1}\cdot(s+n+1)=l(l+1)\cdot C_n$$
For $$n \ge -2\implies 2C_{n+1}\cdot(s+n+1)+C_{n+2}\cdot(s+n+2)(s+n+1)=l(l+1)\cdot C_n$$
But how do I get the Indical equation from here to find the value of $s$?
Thank you.
| 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 equation you derived, pulling the extra powers of $r$ into the sums leads to all three sums having the same power $r^n$. Then equating the coefficients for all powers yields
$$
C_n(2n+n(n-1))=C_nl(l+1)\;,
$$
or
$$
C_n(n(n+1)-l(l+1))=0\;.
$$
It follows that for each $n$ either $C_n=0$ or $n(n+1)=l(l+1)$. The latter equation has the two solutions $n=l$ and $n=-l-1$, so those are the only two powers that can have non-zero coefficients, and the general solution is a linear combination of the two.
| {
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"url": "https://math.stackexchange.com/questions/1884830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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 three. Hence, $Q^3$ has an A-cyclic vector, and A is similar to the companion matrix \begin{bmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \\ \end{bmatrix}
What can I say about the case if $p_A =(x^2 + x + 1)$?
| 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, since $\chi_A(x)$ should be a power of $p_A(x)$ of degree $3$
*if $p_A(x)=x^3-1$, it is equal to $\chi_A(x)$, and the matrix is diagonalisable over $\mathbf C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Better way to evaluate $\int \frac{dx}{\left (a +b\cos x \right)^2}$ $$\int \frac{dx}{\left (a +b\cos x \right)^2}$$
$$u=\frac{b +a \cos x}{a +b\cos x }$$
$$du=\frac{\sin x\left(b^2 -a^2\right)}{ \left (a +b.\cos x \right)^2}$$
$$\frac{du}{\sin x\left(b^2 -a^2\right)}=\frac{dx}{\left (a +b\cos x \right)^2}$$
$$\cos x=\frac{au -b}{a - bu} $$
$$\sin x=\sqrt{1-\left(\frac{au -b}{a - bu}\right)^2}$$
It is becoming messy with this .I also tried it using half angle formula but didn't find it good.
| 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)\mathrm{d}t}{(a^2-2ab+b^2)t^4 + 2(a^2-b^2)t^2+a^2+2ab+b^2}\\
&=2\int \frac{(1+t^2)\mathrm{d}t}{((a-b)t^2 + (a+b))^2} \\
&=2\int \frac{(1+\frac{a+b}{a-b}\tan^2{\theta})\sqrt{\frac{a+b}{a-b}}\sec^2{\theta} \, \mathrm{d}\theta}{(a+b)^2\sec^4{\theta}} \qquad t=\sqrt{\frac{a+b}{a-b}}\tan{\theta} \\
&= 2\sqrt{\frac{a+b}{a-b}}\frac{1}{(a+b)^2}\int(\cos^2{\theta}+\frac{a+b}{a-b}\sin^2{\theta})\mathrm{d}\theta \\
&= \sqrt{\frac{a+b}{a-b}}\frac{1}{(a+b)^2}\left[\frac{2a}{a-b}\theta - \frac{2b}{a-b}\sin{\theta}\cos{\theta}\right]
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
"answer_count": 3,
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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 find the McLaurin series of $\frac{c}{ax + b}$. To do so, note that
$$
\frac{c}{ax + b} = (c/b) \cdot \frac{1}{1 - \left[-\frac{a}{b} x\right]}
$$
and plug in the McLaurin series for $\frac{1}{1 - y}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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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 proving this inequality, I have not dealt with a type of inequality that includes a best possible constant $k$.
Any help to solve and understand this would be greatly appreciated.
| 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 homogeneity we may assume $c=1$. So we are reduced to minimizing for $0<b<1$:
$$ R(0,b,1)=\frac{1+b^3}{(1-b)b} .$$
And this happens (set derivative=0) when
$$0=b^4-2b^3-2b+1=(b^2-(1+\sqrt{3})b+1)(b^2-(1-\sqrt{3})b+1)$$
Only the first factor has real roots from which:
$$ b=\frac{1+\sqrt{3}}{2}- \sqrt{\frac{\sqrt{3}}{2}} \ \ \left( \ \mbox{and}
\ \ \ \frac{1}{b}=\frac{1+\sqrt{3}}{2}+ \sqrt{\frac{\sqrt{3}}{2}} \right)$$
Inserting into the previous it yields (after reduction) the cited constant $k$. To see this note that:
$$ (1-b) \left[ (1-b)(1+b^2) \right] =(1-b)^2(1+b^2) = 1-2b+2b^2-2b^3+b^4 = 2 b^2$$
(use that $b$ is the root of the 4'th degree polynomial above).
Using also $1+b^3=(1+b)(1-b+b^2)=(1+b) \sqrt{3} b$ we get for the minimal value:
$$ \frac{1+b^3}{b(1-b)} = \frac{(1-b)(1+b^2) (1+b)\sqrt{3}}{2 b^2}=\frac{\sqrt{3}}{2}\left( \frac{1}{b}+b\right) \left( \frac{1}{b} - b\right)=\sqrt{3} \left( 1+ \sqrt{3}\right) \sqrt{\frac{\sqrt{3}}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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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:-
$$\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=
\begin{vmatrix}
x+5 & 2 & 3 \\
x+5 & x & 1 \\
x+7 & 2 & 5 \\
\end{vmatrix} \tag{$C_1\rightarrow C_1+C_2+C_3$}$$
$$=\begin{vmatrix}
0 & 2 & 3 \\
0 & x & 1 \\
2 & 2 & 5 \\
\end{vmatrix}+
(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}\tag{1}$$
On opening the first determinant in the last step above we get $2(2-3x)$.
On simplifying the secind determinant we get,
$$(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}=(x+5)\begin{vmatrix}
1 & 2 & 3 \\
0 & x-2 & -2 \\
0 & 0 & 2 \\
\end{vmatrix}
(R_2\rightarrow R_2-R_1)
(R_3\rightarrow R_3-R_1)$$
$=2(x+5)(x-2)$
Substituting the values obtained above in $(1)$, we get
$$=\begin{vmatrix}
0 & 2 & 3 \\
0 & x & 1 \\
2 & 2 & 5 \\
\end{vmatrix}+
(x+5)\begin{vmatrix}
1 & 2 & 3 \\
1 & x & 1 \\
1 & 2 & 5 \\
\end{vmatrix}=2(2-3x)+2(x+5)(x-2)=2(2-3x+x^2+3x-10)=2(x^2-8)$$
Now, as $\begin{vmatrix}
x & 2 & 3 \\
4 & x & 1 \\
x & 2 & 5 \\
\end{vmatrix}=0$, $\therefore 2(x^2-8)=0\implies x=\pm2\sqrt2$
As you can see there was lot of work in my solution so if anyone can provide me with some techniques to solve it faster, or a technique which includes less amount of pen and more thinking.
| 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 equal components in rows 1 and 3; you can work this into rigorous proof that the above equation gives all solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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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 $5^4$ which it is not since it's equal to 1995) or
$ w^4 $ and exactly one of $x^4, y^4, z^4 $ are of the form $5k+1$ .
So, there can be 3 cases with each having either $x^4, y^4 $ or $ z^4 $, along with $w^4$ , of the form $5k+1$ .
I am unable to find the solutions from there.
| 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",
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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 $p(p+1)\le (n-q)(n-q+1)\implies (n-q)(n+q+1)\le(n-q)(n-q+1)$ and therefore $(n+q+1)\le(n-q+1)$ which is impossible and thus $p|(n+q+1)$ .
For some positive integer $k$ we have $n+q+1=kp$ & thus putting it into original equation we yeild $ \require{cancel} \cancel{p}(p+1)=k\cancel{p}(n-q)$ which is $ p+1=k(n-q)$ , For $k=1$ we have $p+q+1=n$ & $n+q+1=p$ which yields $p-n=n-p $ on subtracting which is absurd and thus $k>1$ .
Note that $kp-1=n+q,p=k(n-q)-1$ & so we have the identity $2q=(n+q)-(n-q)=kp-1-(n-q)=k(k(n-q)-1)-(n-q)=(k+1)[(k-1)(n-q)-1]$ , With these and the condition that $k\ge2\implies k+1\ge 3$ .
$2q$ has divisors $1,2,q,2q$ only which implies $k+1=q$ or $k+1=2q$ only.
If $k+1=q$ , $(k-1)(n-q)=3$ hence $(q-2)(n-q)=3$ which akes either $q-2=1$ & $n-q=3$ which yields the solutions $(p,q,k)=(5,3,2)$ and thus $n=6$ .
Else if $q-2=3$ & $ n-q=1$ which makes $p=3,q=5,n=6$ .
Now if $k+1=2q$ then $(k-1)(n-k)=2$ and $2(q-1)(n-q)=2$ which leads to $q-1=1$ & $n-q=1$ and thus $p=2,q=2,n=3$
Thus we have the following solutions in positive integer $n$ & primes $p,q$ which are $(p,q,n)=(5,3,6),(3,5,6),(2,2,3)$ and it is clear that only one prime solution exist $p=q=2,r=3$ .
Hope it helps !
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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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 $(x,y,z)=(-1,-1,-1),(0,0,0),(1,1,1)$ are the solutions.I can show that $x,y,z \in [-1,1]$like below:
By symmetry consider $x \ge y \ge z$ then if $x>1$ we have:
$4x^3-4x>0 \Rightarrow 4x^3-3x>x \Rightarrow z>x$
which is wrong.Now consider $x<-1$ we have:
$x<-1 \Rightarrow y<-1 \Rightarrow 4y^3-4y<0 \Rightarrow 4y^3-3y<y \Rightarrow x<y$
By the same way we can get $x,y,z \in [-1,1]$.By adding the equations we can get:
$x+y+z=x^3+y^3+z^3$
But what should I do now?
| 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
$$4x(x-1)(x+1)(8x^3+4x^2-4x-1)(8x^3-4x^2-4x+1)\\(64x^6-112x^4+56x^2-7)(64x^6-32x^5-80x^4+32x^3+24x^2-6x-1)\\(64x^6+32x^5-80x^4-32x^3+24x^2+6x-1).$$
If we are interested in the integers solutions then we have only $x=0$, $x=1$ and $x=-1$ because the other factors do not have integer roots (check each constant term).
Now $x\in\{-1,0,1\}$ implies that $z=4x^3-3x=x$, $y=4z^3-3z=x$ and therefore the integer solutions of the system are: $(0,0,0)$, $(1,1,1)$ and $(-1,-1,-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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