Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Why is this nested sum formula true I've been trying to get this sum:
$\sum_{i}^{n} \sum_{j=0}^{n-i}j$ into a closed formula but couldn't really understand how to "unpack" that nested sum.
It occured to me that the answer is:
$$\sum_{i=1}^n \left(\sum_{j=0}^{n-i} j\right) = \frac16 n (n^2-1)$$
But I can't figure out why.
| From sum of first n natural numbers, we get
$$\sum_{i = 1}^{n}\sum_{j = 0}^{n - i}j = \sum_{i = 1}^{n}\frac{(n - i)(n - i + 1)}{2}$$
$$=\frac{1}{2}\sum_{i = 1}^{n}n^2 - (2n + 1)i +i^2+n$$
$$=\frac{n^3}{2} + \frac{n^2}{2} - (2n + 1)\frac{1}{2}\sum_{i = 1}^{n}i + \frac{1}{2}\sum_{i = 1}^{n}i^2$$
Using sum of first n natural numbers and sum of first n squares, we get
$$=\frac{n^3}{2} + \frac{n^2}{2} - \frac{n(n+1)(2n + 1)}{4} + \frac{n(n+1)(2n+1)}{12}$$
$$=\frac{n^3}{2} + \frac{n^2}{2} - \frac{n(n+1)(2n + 1)}{6}$$
$$=\frac{n^3}{2} + \frac{n^2}{2} - \frac{2n^3 + 3n^2 + n}{6}$$
$$= \frac{3n^3 + 3n^2 - 2n^3 - 3n^2 - n}{6}$$
$$= \frac{n^3 - n}{6}$$
$$= \frac{n(n^2 - 1)}{6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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area of figure in sector of intersecting circles I need to find an area of blue part of figure APBC. I draw line segments between B and C, between C and A, and got equilateral triangle. I'm stuck here. Please help. Thanks.
|AB| = a, P is midpoint of segment AB
| As achille hui says in the coments, $$(1-r)^2 = |AF|^2 = \frac{1}{2^2} + |PF|^2 = \frac{1}{2^2} + |DF|^2 - \frac{1}{4^2} = \frac{1}{2^2} + \left(r + \frac14\right)^2 - \frac{1}{4^2}.$$ Solve for $r$ and you get $r=\dfrac{3}{10}.$
Since area of intersection between the two larger circles is $\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}$ (see here), shaded area is
\begin{align}
&\dfrac{1}{2}\left(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\right)-\pi\left(\dfrac{1}{4}\right)^2-\pi\left(\dfrac{3}{10}\right)^2\\
\\
& = \dfrac{217 \pi }{1200}-\dfrac{\sqrt{3}}{4}\\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple Trigonometric Problem Here is a Trig problem, and I am missing some understanding of some basic algebraic rule:
$if \sin\theta= \frac{m^2 +2mn}{m^2+2mn+2n^2}$
then prove that $\tan\theta = \frac {m^2 +2mn}{2mn+2n^2},$ In the picture below can anyone explain me how did they achieve the step from 1 to 2 highlighted in red.
| \begin{align*}
\sqrt{1-\left(\frac{m^2+2mn}{m^2+2mn+2n^2}\right)^2}
&= \sqrt{\frac{(m^2+2mn+2n^2)^2 - (m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\frac{4(m^2+2mn)2n^2 + 4n^4}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\frac{4n^2(m^2+2mn+n^2)}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\frac{4n^2(m+n)^2}{(m^2+2mn+2n^2)^2}}\\
&= \sqrt{\left(\frac{2n(m+n)}{m^2+2mn+2n^2}\right)^2}\\
&= \frac{2n(m+n)}{m^2+2mn+2n^2}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I prove $\pi ^2=\sum_{n=0}^{\infty }\frac{1}{(2n+1+\frac{a}{3})^2}+\frac{1}{(2n+1-\frac{a}{3})^2}$ Proving this formula
$$
\pi^{2}
=\sum_{n\ =\ 0}^{\infty}\left[\,{1 \over \left(\,2n + 1 + a/3\,\right)^{2}}
+{1 \over \left(\, 2n + 1 - a/3\,\right)^{2}}\,\right]
$$
if $a$ an even integer number so that
$$
a \geq 4\quad\mbox{and}\quad{\rm gcd}\left(\,a,3\,\right) = 1
$$
| Start with the well known? expansion of $\cot z$ and differentiate,
$$\cot z = \sum_{n=-\infty}^\infty \frac{1}{z - n\pi}
\implies \frac{1}{\sin(z)^2} = \sum_{n=-\infty}^\infty \frac{1}{(z - n\pi)^2}
\implies \frac{\pi^2}{\sin(\pi z)^2} = \sum_{n=-\infty}^\infty \frac{1}{(z - n)^2}
$$
Substitute $z$ by $-\frac{1+\alpha}{2}$, we get
$$\frac{\pi^2}{4\cos(\frac{\pi\alpha}{2})^2} = \sum_{n=-\infty}^\infty\frac{1}{(2n+1+\alpha)^2}\tag{*1}$$
In the RHS of above sum, if we index those negative $n$ as $-(m+1)$, we have
$$\frac{1}{(2n+1 + \alpha)^2} = \frac{1}{(2m+1 - \alpha)^2}$$
This means one can rewrite $(*1)$ as
$$\frac{\pi^2}{4\cos(\frac{\pi\alpha}{2})^2} = \sum_{n=0}^\infty \left(\frac{1}{(2n+1 + \alpha)^2} + \frac{1}{(2n+1 - \alpha)^2}\right)\tag{*2}$$
When $\alpha = \frac{a}{3}$ where $a$ is an even integer with $\gcd(a,3) = 1$,
$\cos(\frac{\pi\alpha}{2}) = \pm \frac12$.
RHS$(*2)$ reduces to $\frac{\pi^2}{4\left(\frac12\right)^2} = \pi^2$ as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $\sin x \sin 2x \sin 4x$ converted to a sum? Converting production to sum. I tried to find the formulas but i could find the formulas only for two identities not for three that is why I couldn't solve. Anybody help?
| $$ \begin{align*} \sin x \sin 2x \sin 4x &= \frac{1}{2}(\cos x - \cos 3x) \sin 4x \\&= \frac{1}{2}\cos x \sin 4x - \frac{1}{2}\cos 3x \sin 4x \\&= \frac{1}{4}(\sin 5x + \sin 3x) - \frac{1}{4}(\sin 7x + \sin x) \\&= \frac{1}{4} (\sin 5x + \sin 3x - \sin 7x - \sin x) \end{align*}$$
| {
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which answer is correct and why? $\arcsin \left(\sin\frac{11\pi}{4}\right)$ Find the value of $\arcsin\left(\sin\dfrac{11\pi}4\right)$
I'm confused if the answer should be $\dfrac{\pi}4$ or $\dfrac{3\pi}4$.
my calculator says it's $\dfrac{\pi}4$ however I don't understand why that is correct? Is that because of the range?
| The range of the arcsine function is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Thus,
$$\arcsin\left[\sin\left(\frac{11\pi}{4}\right)\right] \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$
Since
\begin{align*}
\sin\left(\frac{11\pi}{4}\right) & = \sin\left(2\pi + \frac{3\pi}{4}\right)\\
& = \sin\left(\frac{3\pi}{4}\right)\\
& = \sin\left(\frac{\pi}{4}\right)
\end{align*}
we may conclude that
$$\arcsin\left[\sin\left(\frac{11\pi}{4}\right)\right] = \arcsin\left[\sin\left(\frac{\pi}{4}\right)\right] = \frac{\pi}{4}$$
| {
"language": "en",
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Prove that if a rectangle's sides are all odd, then it's diagonal is irrational? In trying to write an alternate and simple proof that at least one leg of a right triangle is a multiple of 4 using Dickson's method of generating triples, I came across quite an interesting observation that ifall the sides of a rectangle is odd, then it's diagonal is irrational. For example, consider a rectangle of length 5 units and breadth 3 units. It's diagonal,by the Pythagorean theorem is 34^0.5, which is irrational. This is the same for lots of rectangles. Is there a general proof for this, or can this whole thing be disproved?
| If the sides of the rectangle are integers, then the square of the length of the diagonal is also an integer since
$d^2 = l^2 + w^2$
and the integers are closed under multiplication and addition.
The only integers that have rational square square roots are perfect squares. Since
$d = \sqrt{l^2 + w^2}$
the length of the diagonal is irrational unless $d^2 = l^2 + w^2$ is a perfect square.
If an integer is a perfect square, then it is either the square of an even number or an odd number.
If the diagonal is an even number, then $d = 2k$ for some $k \in \mathbb{N}$, so $d^2 = (2k)^2 = 4k^2$.
If the diagonal is an odd number, then $d = 2k - 1$ for some $k \in \mathbb{N}$, so $d^2 = (2k - 1)^2 = 4k^2 - 4k + 1 = 4(k^2 - k) + 1$.
Hence, the square of any integer will have remainder $0$ or $1$ when divided by $4$.
If both the length and width are odd numbers, then $l^2 + w^2$ has remainder $2$ when divided by $4$. To see this, suppose that $l = 2m - 1$ and $w = 2n - 1$, where $m$ and $n$ are positive integers. Then
\begin{align*}
d^2 & = l^2 + w^2\\
& = (2m - 1)^2 + (2n - 1)^2\\
& = 4m^2 - 4m + 1 + 4n^2 - 4n + 1\\
& = 4(m^2 - m + n^2 - n) + 2
\end{align*}
Since $d^2 = l^2 + w^2$ has remainder $2$ when divided by $4$, it is not a perfect square. Thus, $d = \sqrt{l^2 + w^2}$ is irrational.
What I have not shown is why the only integers that have rational square roots are perfect squares. It hinges on unique factorization. You can show that $n$ is a perfect square if and only if each prime in its prime factorization appears an even number of times. For example, $$144 = 2^4 \cdot 3^2 = (2^2 \cdot 3)^2 = 12^2$$ while $12 = 2^2 \cdot 3$ is not a perfect square. If $n$ is not a perfect square, then some prime in its factorization must appear an odd number of times. Suppose $n$ is not a perfect square and $$\sqrt{n} = \frac{r}{s}$$ where $r, s$ are integers with $s \neq 0$. Then
\begin{align*}
n & = \frac{r^2}{s^2}\\
ns^2 & = r^2
\end{align*}
If $p$ is a prime that appears in the prime factorization of $n$ an odd number of times, then its appears in the prime factorization of $ns^2$ an odd number of times, while it appears in the prime factorization of $r^2$ an even number of times, a contradiction of unique factorization.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I prove the maximum of this function I have the function
$$y = x - \sqrt{x^2 - 1}$$
which must have a maximum of $1$ at $x = 1$, as after that you're taking $x$ and subtracting something slightly smaller than $x$, tending to $0$ as $x$ tends to infinity, however its derivative of
$$1 - \frac{x}{\sqrt{x^2 - 1}}$$
is undefined at $x = \pm 1$, as is its second derivative.
How can I prove this function is bounded above by 1, and that the absolute value of y doesn't exceed 1 at some point 0 < x < 1?
| To get maximum of this function, clearly, $x \ge 0$. So we have:
$$x-\sqrt{x^2-1} = x-\sqrt{x^2-1} \times \frac{x+\sqrt{x^2-1} }{x+\sqrt{x^2-1} } = \frac{1 }{x+\sqrt{x^2-1} } \ge \frac{1}{1+\sqrt{1^2-0}}= 1,$$
Note that $x+\sqrt{x^2-1}$ is an increasing function at $[1, +\infty]$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Problem involving cube roots of unity Given that $$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2\omega^2\;\;\;\;\;(1)$$ $$\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2\omega\;\;\;\;\;(2)$$
Find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=?\;\;\;\;\;\;(3)$$
I tried adding the given two equations, and simplified them. I'll show the working for one term here. $$\frac{1}{a+\omega}+\frac{1}{a+\omega^2}=\frac{2a-1}{a^2-a+1}=\frac{(2a-1)(a+1)}{(a^3+1)}=\frac{2a^2+a-1}{a^3+1}$$
$$\frac{1}{a+1}=\frac{a^2-a+1}{a^3+1}=\frac{1}{2}\left[\frac{(2a^2+a-1)-3(a-1)}{a^3+1}\right]=\frac12\left[\frac{2a^2+a-1}{a^3+1}\right]-\frac32\left[\frac{a-1}{a^3+1}\right]$$
Now, I cannot eliminate that extra term which I get at the end of the above expression. Is there hope beyond this? Or is there a better alternative?
| Clearly, the two of the three roots of $$\frac1{a+x}+\frac1{b+x}+\frac1{c+x}=\frac2x$$ are $\omega,\omega^2$
On simplification, $$2x^3+2(a+b+c)x^2+2(ab+bc+ca)x+2abc=3x^3+2x^2(a+b+c)+x(ab+bc+ca)$$
$$\iff x^3+x^2(0)-(ab+bc+ca)x-2abc=0$$
If $a$ is the third root, using Veita's formula $$a+\omega^2+\omega=-\dfrac01$$
Hope the rest should be easy to deal with.
| {
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Evaluating $\displaystyle\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x} \frac{t^3\ln(1-t)}{t^4 + 4}\,dt$
Evaluate the following limit: $$\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x}
\frac{t^3\ln(1-t)}{t^4 + 4}\,dt$$
Any advice on how to tackle this problem ?
| Using Fundamental Theorem of Calculus
$$\frac{d}{dx}\int_0^{x}
\frac{t^3\ln(1-t)}{t^4 + 4}\,dt=
\frac{x^3\ln(1-x)}{x^4 + 4}$$
$$\begin{align}
\lim_{x\space\to\space0} \dfrac{1}{x^5}\int_0^{x}\dfrac{t^3\ln(1-t)}{t^4 + 4}\,dt&=\lim_{x\space\to\space0} \frac{1}{5x^4}\cdot
\frac{x^3\ln(1-x)}{x^4 + 4}\tag{1}\\
&=\lim_{x\space\to\space0}
\frac{\ln(1-x)}{(5x)(x^4 + 4)}\\
&=\lim_{x\space\to\space0}
-\frac{\ln(1-x)}{-x}\cdot\frac{1}{5(x^4 + 4)}\\
&=-\frac{1}{20}\\
\end{align}$$
$$\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x}
\frac{t^3\ln(1-t)}{t^4 + 4}\,dt=-\frac{1}{20}$$
Explanation : $(1)$ Use L'Hopital's Rule
| {
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Find $E(X)$ and $Var(X)$ In a box there are $30$ balls, $20$ are black and $10$ are red.
Let $X$ be the number of red in a selection of two balls drawn without replacement then $$X=I_1 + I_2$$ where $I_1 = 1$ if red is drawn first, else 0 the same thing applies to $I_2$ in the second draw. Find $$E(I_1),\,\, E(I_2),\,\, E(I^2 _1),\,\, E(I^2 _2),\,\, E(I_1 I_2), \,\,E(X)$$ and $Var(X)$.
| The variables $I_1, I_2$ are indicator variables and therefore you can use that $$E[I_i]=P(I_i=1)$$
for $i=1,2$. In more detail, you have that
$$E[I_1]=1\cdot P(I_1=1)+0\cdot P(I_1=0)=1\cdot\frac{10}{30}=\frac{1}{3}$$
and by conditiong on $I_1$ also that
$$\begin{align*}E[I_2]&=E[E[I_2|I_1]]=E[I_2|I_1=1]P(I_1=1)+E[I_2|I_1=0]P(I_1=0)\\\\&=\left(1\cdot\frac{9}{29}+0\cdot\frac{20}{29}\right)\frac{10}{30}+\left(1\cdot\frac{10}{29}+0\cdot\frac{19}{29}\right)\frac{20}{30}=\frac{9(10)+10(20)}{29(30)}=\frac{1}{3}\end{align*}$$
Now, since $I_1=I_1^2$ and $I_2=I_2^2$ (due to $1^2=1$ and $0^2=0$) you have that
$$E[I_1^2]=1^2\cdot P(I_1=1)+0^2\cdot P(I_1=0)=E[I_1]=\frac{1}{3}$$ and similarly $E[I_2^2]=\frac{1}{3}$. Finally
$$E[I_1I_2]=1\cdot1\cdot P(I_1=1,I_2=1)+0=1\cdot P(I_2=1|I_1=1)P(I_1=1)=1\cdot\frac{9}{29}\cdot\frac{1}{3}=\frac{3}{29}$$
and therefore, using the linearity of expectation and the formula for the variance of the sum of random variables you obtain that $$E[X]=E[I_1+I_2]=E[I_1]+E[I_2]=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$$ and $$\begin{align*}\operatorname{Var}(X)&=\operatorname{Var}(I_2)+\operatorname{Var}(I_2)+2\operatorname{Cov}(I_1,I_2)=\\&=E[I_1^2]-E[I_1]^2+E[I_2^2]-E[I_2]^2+2(E[I_1I_2]-E[I_1]E[I_2])=\\&=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3}-\frac{1}{3^2}+2\left(\frac{3}{29}-\frac{1}{3}\frac{1}{3}\right)=\frac{4}{9}+2\left(\frac{-2}{261}\right)=\frac{112}{261}\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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elementary properties of cyclotomic polynomials How can one rewrite $1+x^2+x^4+x^8+\cdots x^{2^n}$ as a product of cyclotomic polynomials? more general how can we express $1+x^p+\cdots+x^{p^n}$, where $p$ is a prime, in term of product of cyclotomic polynomials?
| Let $p(x)=1+x^2+x^4+x^6...+x^{2n}$. Assume that $n>1$, otherwise when $n=1$, $p(x)=x^2+1$, which is the only irreducible polynomial in this case. The case that $n$ is even, $p(x)=(1+x+x^2+x^3+x^n)(1-x+x^2-x^3+x^n)$. When $n$ is odd, the best possible factorization is $p(x)=(x^2+1)(1+x^4+x^8+x^{4k})$ where $2k+1=n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$ How to find the sum of the following series:
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$
Any hints.
| I am applying Robert Israel's suggestion -
I have learned that this is generally a good thing to do.
The general term in
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot$
seems to be
$\begin{array}\\
a_n
&=(-1)^n\dfrac{\prod_{k=1}^n (3k-2)}{5^n n!}\\
&=\dfrac{\prod_{k=1}^n (2-3k)}{5^n n!}\\
&=\dfrac{3^n\prod_{k=1}^n (\frac23-k)}{5^n n!}\\
&=\dfrac{3^n\prod_{k=0}^{n-1} (\frac23-(k+1))}{5^n n!}\\
&=\dfrac{3^n\prod_{k=0}^{n-1} (-\frac13-k)}{5^n n!}\\
&=\left(\dfrac{3}{5}\right)^n\dfrac{\prod_{k=0}^{n-1} (-\frac13-k)}{ n!}\\
\end{array}
$
The expansion of
$(1+t)^r$
is
$\begin{array}\\
(1+t)^r
&=\sum_{n=0}^{\infty} t^n \dfrac{\prod_{k=0}^{n-1} (r-k)}{n!}\\
\end{array}
$
If $t=\frac35$
and
$r=-\frac13$,
this gives
$(1+\frac35)^{-1/3}
=\sum_{n=0}^{\infty} (\frac35)^n \dfrac{\prod_{k=0}^{n-1} (-\frac13-k)}{n!}
=1-\sum_{n=1}^{\infty} a_n
$.
Therefore
$\sum_{n=1}^{\infty} a_n
=1-(1+\frac35)^{-1/3}
=1-(\frac85)^{-1/3}
=1-(\frac58)^{1/3}
=1-\frac{5^{1/3}}{2}
$.
Cowabunga!!!!!
Generalizations:
If $a, b, c, d$ are non-zero integers
such that
$gcd(a, b)
=gcd(c, d)
=1
$,
$\begin{array}\\
(1+\frac{a}{b})^{c/d}
&=\sum_{n=0}^{\infty} \left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (\frac{c}{d}-k)}{n!}\\
&=\sum_{n=0}^{\infty} \left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (c-kd)}{d^n n!}\\
&=\sum_{n=0}^{\infty} \dfrac{a^n\prod_{k=0}^{n-1} (c-kd)}{(bd)^n n!}\\
\end{array}
$
Putting
$-c/d$ for $c/d$,
$\begin{array}\\
(1+\frac{a}{b})^{-c/d}
&=\sum_{n=0}^{\infty} \left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (-\frac{c}{d}-k)}{n!}\\
&=\sum_{n=0}^{\infty} (-1)^n\left(\frac{a}{b}\right)^n \dfrac{\prod_{k=0}^{n-1} (c+kd)}{d^n n!}\\
&=\sum_{n=0}^{\infty} (-1)^n\dfrac{a^n\prod_{k=0}^{n-1} (c+kd)}{(bd)^n n!}\\
\end{array}
$
| {
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Two squares and angle can someone answer this please?
I have two equal squares (picture). We know the lengths of three lines inside them. One of them is upside down. Find angle alpha. The picture is not perfect.
Thanks
| Following is one boring way of solving the problem using coordinate geometry.
Even though I suspect there are more geometry intuitive and clever solutions,
a solution is a solution and here we go...
Choose a coordinate system such that the lower left corner of the upper square is the origin.
Let $\ell$ be the side and $(x,y)$ be the intersection of the three 3 rays of/in the upper square.
We have
$$
\begin{eqnarray}
x^2 + y^2 &= 2^2 = 4 &\tag{*1a}\\
(x-\ell)^2 + y^2 &= 3^2= 9 &\tag{*1b}\\
x^2 + (y-\ell)^2 &= 1^2 = 1 &\tag{*1c}\\
\end{eqnarray}
$$
"Subtract" $(*1a)$ from $(*1b)$ and $(*1c)$, we get
$$
\begin{cases}
\ell^2 - 2\ell x &= 9-4 = 5\\
\ell^2 - 2\ell y &= 1-4 -3
\end{cases}
\quad\implies\quad
\begin{cases}
x &= \displaystyle\;\frac{\ell^2 - 5}{2\ell}\\
y &= \displaystyle\;\frac{\ell^2 + 3}{2\ell}
\end{cases}
$$
Substitute this back into $(*1a)$, we get
$$
(\ell^2 - 5)^2 + (\ell^2 + 3)^2 = 4(2\ell)^2
\iff 2 \ell^4 - 20 \ell^2 + 34 = 0
\iff \ell^2 = 5 \pm 2\sqrt{2}
$$
Since $x > 0$, we find $\ell^2 = 5 + 2\sqrt{2}$.
Apply cosine rule to the triangle with side $2,3$ and $\ell$, we find
$$\begin{align}
&\cos\alpha = \frac{2^2 + 3^2 - \ell^2}{2(2)(3)} = \frac{13 - (5 + 2\sqrt{2})}{12} = \frac{4 - \sqrt{2}}{6}\\
\implies & \alpha = \cos^{-1}\left(\frac{4 - \sqrt{2}}{6}\right) \approx 64.47122063449069^\circ
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integration giving different answers (trig substitution) Integrating $\sin^3x\cos^5x$, i get 2 different answers, using techniques that should both be valid.
| HINT:
Establish that $$\frac{\cos^8x-\sin^8x}8-\frac{\cos^6x-\sin^6x}6-\frac{\sin^4x}4$$ is a constant
If $u_n=\cos^nx-\sin^nx,$
$$(\cos^nx-\sin^nx)(\cos^2x+\sin^2x)=\cdots$$
$$\implies u_n=u_{n+2}+\cos^2x\sin^2x\cdot u_{n-2}$$
$$\implies u_{n+2}=u_n-\cos^2x\sin^2x\cdot u_{n-2}$$
Now, $u_0=0,u_2=\cos^2x-\sin^2x$
$$n=2\implies u_4=u_2-\cos^2x\sin^2x\cdot u_0=u_2$$
$$n=4\implies u_6=u_4-\cos^2x\sin^2x\cdot u_2=u_2(1-\cos^2x\sin^2x)$$
$$n=6\implies u_8=u_6-\cos^2x\sin^2x\cdot u_4=u_2(1-\cos^2x\sin^2x)-u_2=-u_2(\cos^2x\sin^2x)$$
If $f(x)=g(x)+C, f'(x)=g'(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Maximum and minimum of $f(x,y)=xy$ when $x^2 + y^2 + xy =1$ It is asked to find the maximum and minimum points of the function
$$f(x,y)=xy$$
when $x^2 + y^2 + xy=1$
I've tried Lagrange and obtained
$$\lambda = \frac{y}{2x+y}=\frac{x}{2y+x}$$
but what should I do with this? Any other suggestion?
Thanks!
| An elementary approach: using $2xy\leq x^2+y^2$, you can write
$$
xy=\frac{1}{3}(2xy+xy)\le\frac{1}{3}(x^2+y^2+xy)=\frac{1}{3}1=\frac{1}{3}
$$
with equality iff $x=y=\frac{1}{\sqrt{3}}$. For minimum, note that
$$
xy+1=xy+(x^2+y^2+xy)=(x+y)^2\geq 0\implies xy\geq-1
$$
where equality realizes when $(x,y)=(1,-1)$ or $(x,y)=(-1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove $\lim_{x \to 0}$ $\sqrt{3x^2 + 4} = 2$ using the definition We have $\lim_{x \to 0}$ $\sqrt{3x^2 + 4} = 2$
Proof: $\vert \sqrt{3x^2+4} - 2 \vert = \cdots$ I got $\left \vert \frac{3x^2}{\sqrt{3x^2+4}+2} \right \vert = \frac{\vert x \vert \vert 3x \vert}{\sqrt{3x^2+4}+2}$
Then what I do next?
$\vert 3x \vert$ $\lt 6$ ?
| You found
$$\vert \sqrt{3x^2+4} - 2 \vert = \left| \frac{3x^2}{\sqrt{3x^2+4}+2} \right|. $$
Hence,
$$\vert \sqrt{3x^2+4} - 2 \vert = \left| \frac{3x^2}{\sqrt{3x^2+4}+2} \right| \leqslant \frac{3}{4}|x|^2$$
and $\vert \sqrt{3x^2+4} - 2 \vert < \epsilon$ if $|x| < \delta = 2\sqrt{\epsilon/3}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Limit, having trouble applying L'Hospital rule I have been asked to solve for this limit:
$$\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}$$
I try to find the derivative of the numerator and denominator:
$$\frac{6\sin(3x)\cos(3x)}{2\sin(2x)}$$ but that still gives me a $\frac{0}{0}$. How do I solve this
|
Using L'Hospital Again
$$\begin{align}\lim_{x \to 0} \frac{6\sin(3x)\cos(3x)}{2\sin(2x)}
&=\lim_{x \to 0} \frac{18 \cos ^2(3 x)-18 \sin ^2(3 x)}{4\cos(2x)}\\
&=\lim_{x \to 0} \frac{18 \cos ^2(3 x)}{4\cos(2x)}-\frac{18 \sin ^2(3 x)}{4\cos(2x)}\\
&=\frac{18}{4}+0\\
&=\frac{9}{2}\\
\end{align}$$
You can do it without L'Hospital
$$\begin{align}
\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}
&=\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos^2(x)+\sin^2 x}\tag{1}\\
&=\lim_{x \to 0} \frac{\sin^2(3x)}{\sin^2(x)+\sin^2x}\tag{2}\\
&=\lim_{x \to 0} \frac{\sin^2(3x)}{2\sin^2(x)}\tag{3}\\
&=\lim_{x \to 0} \frac{3^2x^2\sin^2(3x)}{2\cdot 3^2x^2\sin^2(x)}\tag{4}\\
&=\lim_{x \to 0} \frac{3^2}2\cdot\frac{\sin^2(3x)}{(3x)^2}\cdot\frac{x^2}{\sin^2(x)}\tag{5}\\
&=\lim_{x \to 0} \frac{3^2}2\left(\frac{\sin(3x)}{3x}\right)^2\left(\frac{x}{\sin(x)}\right)^2\tag{6}\\
&=\lim_{x \to 0} \frac{3^2}2\cdot1^2\cdot1^2\tag{7}\\
\end{align}$$
$$\large\lim_{x \to 0} \frac{\sin^2(3x)}{1-\cos(2x)}=\frac{9}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Closed form of $\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$ Is it possible to get a closed form of the following integral
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$$
| For real $a,b\neq 0$ Mathematica gives
$$
\Phi(a,b) = \frac{1}{a\sqrt{b^2-a^2}}\cdot\arctan\left(\frac{\sqrt{b^2-a^2}}{a\sqrt{1+b^2}}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Triple Integral to Find Volume Question: Use a triple integral to find the volume of the solid enclosed by the parabaloids $y=x^2+z^2$ and $y=8-x^2-z^2$.
My attempt: The best I can figure, this object looks kind of like a football oriented along the $y$-axis from $y=0$ to $y=8$ and is symmetric about the $y$-axis and the plane $y=4$.
It seems best to integrate first with respect to $y$, and $x^2+z^2 \le y \le 8-(x^2+z^2)$.
The widest part of the football is at $y=4$; substitute that into both of the equations above to find that the projection onto the $xz$-plane is $x^2+z^2=4$, or a circle of radius 2, so my bounds for $z$ are $-\sqrt{4-x^2} \le z \le \sqrt{4-x^2}$ and my bounds for $x$ are $-2 \le x \le 2$.
But I believe I can make this easier by integrating from $0\le z \le \sqrt{4-x^2}$ and multiplying by 2 and integrating from $0 \le x \le 2$ and multiplying by another 2. (I actually think I can integrate $y$ from $4 \le y \le 8-(x^2+z^2)$ and multiply by another 2, but that doesn't seem to simplify anything.)
Since I'm finding the volume, the function I integrate is one. I come up with this
$$4\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-(x^2+z^2)} 1\ dy\ dz\ dx.$$
(Is this right so far?)
If nothing's wrong yet, I still can't finish this integral
$$4\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-\left(x^2+z^2\right)} 1\ dy\ dz\ dx \\
4\int_0^2 \int_0^{\sqrt{4-x^2}} y \Big|_{x^2+z^2}^{8-\left(x^2+z^2\right)} dz\ dx \\
4\int_0^2 \int_0^{\sqrt{4-x^2}} \left(8-\left(x^2+z^2\right)\right)-\left(x^2+z^2\right) dz\ dx \\
4\int_0^2 \int_0^{\sqrt{4-x^2}} \left(8-2x^2-2z^2\right)\ dz\ dx \\
4\int_0^2 \left[\left(8-2x^2\right)z-\frac 23 z^3\right]_0^{\sqrt{4-x^2}} \ dx \\
4\int_0^2 \left[ \left(8-2x^2\right)\sqrt{4-x^2} - \frac 23 \sqrt{4-x^2}^3 \right]\ dx \\
-\frac{16}3\int_0^2 \left[ \left(x^2 -4\right)\sqrt{4-x^2} \right]\ dx \\
\vdots \\ ??$$
| The volume will be represented by the triple integral:
$\int \int \int _{\Omega} dV = \Gamma$ where $ \Omega$ is the solid enclosed by the two paraboloids. The paraboloids intersect in $y = 4$ So:
$\Gamma = \int \int_{\Omega'} \int_{x^{2}+z^{2}}^{8-(x^{2}+z^{2})}dV$. Where $\Omega'= $$\{(x,z) \lvert x^{2} + z^{2} \leq 4 \}$.
So:
$\Gamma = \int \int_{\Omega´}8-2(x^{2}+z^{2})dA$
Using the substitution:
$x = r cos(\theta)$ and $z = r sin(\theta)$, $0 \leq r \leq 2$ and $0\leq \theta \leq 2\pi$.
$\Gamma = \int_{0}^{2\pi} \int_{0}^{2}r(8-2r^{2})drd\theta$
So $\Gamma = 2\pi(4(2^2) - (2)^3) = 16\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$$ with the help of the identity $$\sum_{i=1}^ni = \frac{n(n+1)}{2}$$ My proof of the induction step goes as follows (supposing equality holds for all $k \in \{1,2,\dots n \})$: $$\sum_{i=1}^{n+1} i^3 = \sum_{i=1}^{n} i^3+(n+1)^3 \\ = \left( \sum_{i=1}^ni\right)^2+(n+1)^3 \\ = \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3 \\ = \frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4} \\ = \frac{n^4+2n^3+n^2}{4}+\frac{4n^3+12n^2+12n+4}{4} \\ = \frac{n^4+6n^3+13n^2+12n+4}{4} \\ = \frac{(n^2+3n+2)^2}{4} \\ = \frac{[(n+1)(n+2)]^2}{4} \\ = \left(\frac{(n+1)(n+2)}{2}\right)^2 \\ = \left(\sum_{i=1}^{n+1}i\right)^2$$ I was a little disappointed in my proof because the algebra got really hairy. It took me a long while to see that I could twice unfoil the polynomial $n^4+6n^3+13n^2+12n+4$ and all in all the solution seems pretty inelegant. Does anyone have a smoother way to prove the induction step or bypass the algebra? I feel like their must be some concise way to get the same result.
| Induction $n-1\to n$ makes it more compact.
$$\left(\sum_1^{n-1}i+n\right)^2=\left(\sum_1^{n-1}i\right)^2+2n\sum_1^{n-1}i+n^2=\left(\sum_1^{n-1}i\right)^3+n^2(n-1)+n^2=\left(\sum_1^{n-1}i\right)^3+n^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
When is a particular sum $\Theta(n)$? Define
$$S_n = \prod_{x=1}^{\lceil\frac{n}{\ln{n} }\rceil} \left(\frac{1}{\sqrt{n}} + \frac{2x}{n}\left(z_n-\frac{1}{\sqrt{n}} \right)\right) .$$
I am trying to work out necessary and sufficient conditions for a simply stated, positive,
non-increasing function $z_n$ with $1/\sqrt{n} \leq z_n \leq 1/2$ so that
$$- \ln{S_n} = \Theta(n).$$
Clearly $z_n = 1/\sqrt{n}$ works and I know that $z_n = 1/2$ does not work for example.
Here is my attempt so far $$\prod_{x=1}^{k} (A + Bx) = \frac{B^k \Gamma(k+1+A/B)}{\Gamma(1+A/B)}.$$
In our case $A = 1/\sqrt{n}$ and $B= (2\sqrt{n} z_n - n)/n^{3/2}$ and
$$A/B = \frac{n}{2\sqrt{n} z_n - n}$$
We know from Asymptotics of $\prod_{x=1}^{\lceil\frac{n}{\log_2{n} }\rceil} \left(\frac{1}{\sqrt{n}} + x\left(\frac{1}{n}-\frac{2}{n^\frac{3}{2}} \right)\right) $ that
\begin{align}
\ln S_n&=k\ln B+\ln \Gamma(1+k+A/B)-\ln \Gamma(1+A/B)\approx\\
&\approx k\ln B+ \left(k+\frac AB\right)\left[\ln\left(k+\frac AB\right)-1\right]+\frac12\ln\left(k+\frac AB\right)\\&\qquad -\frac AB\left[\ln\frac AB -1\right]-\frac12\ln\frac AB\\
&\approx \color{red}{k\ln B}+\left(\color{red}{k}+\frac AB\right)\left[\color{red}{\ln k} +\frac{A}{Bk}-\frac12\left(\frac{A}{Bk}\right)^2-\color{red}{1}\right]+\frac12\ln k\\
&\qquad -\frac AB\left[\ln\frac AB -1\right]-\frac12\ln\frac AB.
\end{align}
It feels like with the right log approximation this might be close but I am not sure where to take it from here.
| Note that
$$
\prod_{x=1}^{\lceil n/\ln n \rceil} \max\bigg\{ \frac{1}{\sqrt{n}}, \frac{2x}{n}\bigg(z_n-\frac{1}{\sqrt{n}} \bigg)\bigg\} \le S_n \le \prod_{x=1}^{\lceil n/\ln n \rceil} 2\max\bigg\{ \frac{1}{\sqrt{n}}, \frac{2x}{n}\bigg(z_n-\frac{1}{\sqrt{n}} \bigg)\bigg\},
$$
and so
\begin{align*}
-\ln S_n &= \sum_{x=1}^{\lceil n/\ln n \rceil} \min\bigg\{ \tfrac12\ln n, \ln \frac{n^{3/2}}{2x} - \ln (z_n\sqrt n-1)\bigg\} + O\bigg( \frac{n\ln 2}{\ln n} \bigg) \\
&= \sum_{1\le x\le \min\{n/2(z_n\sqrt n-1),\lceil n/\ln n \rceil\}} \tfrac12\ln n \\
&\qquad{}+ \sum_{n/2(z_n\sqrt n-1)<x\le\lceil n/\ln n \rceil} \bigg( \ln \frac{n^{3/2}}{2x} - \ln (z_n\sqrt n-1) \bigg) + O\bigg( \frac n{\ln n} \bigg).
\end{align*}
If $(z_n\sqrt n-1)/\ln n$ is bounded away from $0$, then the first term is already $\Omega(n)$. If $(z_n\sqrt n-1)/\ln n \to 0$, then the first term is $o(n)$ and so
$$
-\ln S_n = \sum_{n/2(z_n\sqrt n-1)<x\le\lceil n/\ln n \rceil} \bigg( \ln \frac{n^{3/2}}{2x} - \ln (z_n\sqrt n-1) \bigg) + o(n).
$$
Ran out of time - see if someone can take it from here....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$ How do I evaluate
$$\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$$
I've started doing this problem by taking $u=\tan x$.
| There are already 7 nice solutions. I want to generalize the result further, by rationalization and substitutions, of the given integral to $$
I(m ,n):=\int \frac{\sec ^m x}{(\sec x+\tan x)^{n}} d x= \int \frac{\sec ^{m} x(\sec x-\tan x)^{n}}{\left(\sec ^{2} x-\tan ^{2} x\right)^{n}} d x=\int \frac{(1-\sin x)^{n}}{\cos ^{m+n} x} d x
$$
Letting $y=\frac{\pi}{2}-x$ transforms the integral into
$$
I(m, n)=-\int \frac{\left(2\sin ^{2} \frac{y}{2}\right)^{n}}{2 ^{m+n} \sin ^{m+n} \frac{y}{2} \cos ^{m+n} \frac{y}{2}} d x$$
Letting $z=2y$ transforms the integral into
$$I(m,n) =-\frac{1}{2^{m-1}} \int \tan ^{n-m}z \sec ^{2m}z d z$$
Letting $t=\tan z$ transforms the integral into
$$
\begin{aligned}
I(m, n)&=-\frac{1}{2^{m-1}} \int^{n} t^{n-m}\left(1+t^{2}\right)^{m-1} d t \\
&=-\frac{1}{2^{m-1}} \sum_{\substack{k=0}}^{m-1} \int t^{n-m+2 k} d t\\ &=-\frac{1}{2^{m-1}} \sum_{k=0}^{m-1} \frac{t^{n-m+2 k+1}}{n-m+2 k+1}+C \\ &=-\frac{1}{2^{m-1}} \sum_{k=0}^{m-1} \frac{(\sec x-\tan x)^{n-m+2 k+1}}{n-m+2 k+1}+C
\end{aligned}
$$
$$\boxed{I(m, n)=-\frac{1}{2^{m-1}} \sum_{k=0}^{m-1} \frac{(\sec x-\tan x)^{n-m+2 k+1}}{n-m+2 k+1}+C}$$
as $t=\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}=\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}=\sec x-\tan x.$
Back to our original integral, putting $m=2$ and $n=\frac 92$ yields $$\begin{aligned}
\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx&=-\frac{1}{7} (\sec x-\tan x) ^{\frac{7}{2}}-\frac{1}{11} (\sec x-\tan x) ^{\frac{11}{2}}+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 7
} |
If $a,b,c$ are positive real numbers, then $\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq \frac{3}{2}$
If $a$, $b$, and $c$ are positive real numbers such that $abc=1$, then prove that
$$\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq \frac{3}{2}$$
Progress
I think the relevant concept would be the application of AM/GM inequality.
| Writing $a=x/y$, $b=y/z$, and $c=z/x$, we simplify the LHS:
$$
\frac{xz}{xy+yz}+\frac{xy}{xz+yz}+\frac{yz}{xy+xz}=\frac{r}{s+t}+\frac{s}{r+t}+\frac{t}{s+r}
$$
where $r=xz$, $s=xy$, and $t=yz$. It remains to invoke Nesbitt's inequality.
The linked wikipedia article contains a standard Cauchy-Schwarz proof:
$$
[(s+t)+(r+t)+(s+r)]\left[\frac{1}{s+t}+\frac{1}{r+t}+\frac{1}{s+r}\right]\geq9
$$
which after the simplification of the LHS implies
$$
\frac{r}{s+t}+\frac{s}{r+t}+\frac{t}{s+r}\geq\frac{3}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area of the region that is enclosed by the cardioid $r=2+2\sin(\theta)$ We just learned polar integration, so I know that's how we're supposed to do it. I have a problem though: I'm getting a negative answer.
What I did:
Using the graph, which is:
I figured out that every $2\pi$ it repeated, so I did
$$\int_0^{2\pi} {\frac{{(2+2\sin \theta)}^2}{2}d\theta}$$
Evaluating it I got $-6\pi$, which is negative and definitly not right. Can I have help please? Thanks!
| Your integral is correct, but you must have made a mistake in integration.
\begin{array} \\
\frac{1}{2} \int_0^{2\pi} (2 + 2\sin \theta)^2 \, \mathrm{d}\theta &= \frac{1}{2} \int_0^{2\pi} (4 + 8 \sin \theta + 4 \sin^2 \theta) \, \mathrm{d}\theta \\
&= \frac{1}{2} \int_0^{2\pi} \left(4 + 8 \sin \theta + 4 \left(\frac{1 - \cos (2 \theta)}{2} \right)\right) \, \mathrm{d}\theta \\
&= \frac{1}{2} \int_0^{2\pi} (6 + 8 \sin \theta - 2 \cos (2 \theta)) \, \mathrm{d}\theta \\
&= \left. \frac{1}{2} (6 \theta - 8 \cos \theta - \sin(2\theta)) \right|_{\theta=0}^{2\pi} \\
&= \frac{1}{2} ((12 \pi - 8) - (-8)) \\
&= 6 \pi
\end{array}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the first two non vanishing maclaurin terms Find the first two nonvanishing terms in the Maclaurin series of $\sin(x + x^3)$.
Suggestion: use the Maclaurin series of $\sin(y)$ and write $y = x + x^3$
Using
this result, find
$\lim\limits_{x\to 0}\frac{\sin(x + x^3)−x}{x^3}$
$\sin(y)= y-\frac{y^3}{3!}+\frac{y^5}{5!}+\frac{y^7}{7!}$
$y= x+x^3$
$x+x^3-\frac{(x+x^3)^3}{3!}+\frac{(x+x^3)^5}{5!}-\frac{(x+x^3)^7}{7!}$
Now what do I do?
| The limit in question is more easily solved by a little bit of algebra. Note that $$\frac{\sin(x+x^3)-x}{x^3}=\frac{\sin (x+x^3)-(x+x^3)}{(x+x^3)^3}\cdot (1+x^2)^3+1$$ Putting $t=x+x^3$ we can see that $t\to 0$ and hence the first factor $(\sin t - t) /t^3$ in first term tends to $-1/6$ and the desired limit is thus $(-1/6)+1=5/6$.
There is no need to find the Maclaurin series for $\sin(x+x^3)$, just the well known series for $\sin x $ is sufficient to tackle the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Find an integral with fractions How to find the integral $$\int_0^\infty \frac{e^{-x^2}}{(x^2+1/2)^2}dx?$$
I find it is difficult to do if I integrate by parts...What's the trick?
| $$\int\dfrac{e^{-x^2}}{\left(x^2+\dfrac{1}{2}\right)^2}dx=-\int\dfrac{e^{-x^2}}{2x}d\left(\dfrac{1}{x^2+\dfrac{1}{2}}\right)=-\dfrac{e^{-x^2}}{2x\left(x^2+\dfrac{1}{2}\right)}-\int\dfrac{e^{-x^2}}{x^2}$$
and
$$-\int\dfrac{e^{-x^2}}{x^2}=\int e^{x^2}d\dfrac{1}{x}=\dfrac{e^{-x^2}}{x}+2\int e^{-x^2}dx$$
so
$$I=\int_{0}^{+\infty}\dfrac{e^{-x^2}}{\left(x^2+\dfrac{1}{2}\right)^2}dx
=\dfrac{xe^{-x^2}}{x^2+\dfrac{1}{2}}|_{0}^{+\infty}+2\int_{0}^{+\infty}e^{-x^2}dx=\sqrt{\pi}$$
solution2:
since
$$e^{-x^2}=\int_{x}^{+\infty}2te^{-t^2}dt$$
so
$$I=\int_{0}^{+\infty}\left[\int_{x}^{+\infty}\dfrac{2te^{-t^2}}{
\left(x^2+\dfrac{1}{2}\right)^2}dt\right]dx=\int_{0}^{+\infty}2te^{-t^2}\left(\sqrt{2}\arctan{\sqrt{2}t}+\dfrac{t}{t^2+1/2}\right)dt$$
so
$$I=\int_{0}^{+\infty}\dfrac{2t^2}{t^2+1/2}e^{-t^2}-\int_{0}^{+\infty}\sqrt{2}\arctan{\sqrt{2}t}de^{-t^2}=2\int_{0}^{+\infty}e^{-t^2}dt=\sqrt{\pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving $1+2^n+3^n+4^n$ is divisible by $10$ How can I prove
$$1+2^n+3^n+4^n$$ is divisible by $10$ if $$n\neq 0,4,8,12,16.....$$
| HINT : You can consider each right-most digit.
$$1:1,1,1,1,1,\cdots$$
$$2^n:2,4,8,6,2,\cdots$$
$$3^n:3,9,7,1,3,\cdots$$
$$4^n:4,6,4,6,4,\cdots$$
So, we have
$$1+2+3+4=1\color{red}{0},1+4+9+6=2\color{red}{0},1+8+7+4=2\color{red}{0},1+6+1+6=14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
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To prove $(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 \ge 9$ I used the following way but got wrong answer
$$A.M. \ge G.M.$$
$$ \frac{\sin \theta + \csc \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$
Squaring both sides,
\begin{equation*}
(\sin\theta + \csc\theta )^2 \ge 4 \tag{1}
\end{equation*}
Similarly
\begin{equation*}
(\cos\theta + \sec\theta )^2 \ge 4 \tag{2}
\end{equation*}
Adding equation (1) and (2)
\begin{equation*}
(\sin\theta + \csc\theta )^2+(\cos\theta + \sec\theta )^2 \ge 8
\end{equation*}
What is wrong?
| $\begin{align}(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 & =5+\sec^2 x+\csc^2 x\\&=5+1+\tan^2 x+1+\cot^2x\\&=7+\tan^2x+\cot^2x\\&\geq9\end{align}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Improper integral: $\int_1^{+\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)}$ I've tried many ways, but it seems that it didn't work:
$$
\int_1^{+\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)}
= \int_0^1\frac{x^{n-1}}{(x+1)(2x+1)\cdots(nx+1)}\mathrm dx
= \cdots
$$
Any help would be appreciated!
| We have the partial fraction decomposition
$$\frac{1}{x(x+1)(x+2)\cdots(x+n)}=\frac{a_0}{x}+\frac{a_1}{x+1}+\frac{a_2}{x+2}+\ldots+\frac{a_n}{x+n}$$
Now as $x\to 0$, the part with $a_0$ dominates, and we have
$$a_0=\lim_{x\to0}\frac{x}{x(x+1)(x+2)\cdots(x+n)}=\frac1{n!}$$
Similarily, as $x\to -1$, the part with $a_1$ dominates, and we have
$$a_1=\lim_{x\to-1}\frac{x+1}{x(x+1)(x+2)\cdots(x+n)}=-\frac1{(n-1)!}$$
and if you were to continue this idea, you'd find that
$$a_k=\lim_{x\to -k}\frac{x+k}{x(x+1)(x+2)\cdots(x+n)}= \frac{(-1)^k}{k!(n-k)!}=\frac{(-1)^k {n\choose k}}{n!}$$
so that
$$\frac{1}{x(x+1)(x+2)\cdots(x+n)}=\frac1{n!}\sum_{k=0}^n\frac{(-1)^k {n\choose k}}{(x+k)} $$
and so
$$\int_1^{\infty}\frac{\mathrm dx}{x(x+1)(x+2)\cdots(x+n)}
=\frac{1}{n!}\int_1^{\infty}\sum_{k=0}^n\frac{(-1)^k {n\choose k}}{(x+k)}\\
=\frac1{n!}\left(\sum_{k=0}^n(-1)^k {n\choose k}\ln(x+k)\right)\left. \right|_1^\infty$$
For $x=\infty$, consider the binomial expansion of $(1-1)^k$. It implies that the sum of the positive binomial terms is equal to the sum of the negative binomial terms, and so if we combine all the logs into one, we get the logarithm of a polynomial divided by a polynomial of the same degree, so the limit to infinity is hence $\ln(1) = 0$. So then the integral is thus equal to the negative of the antiderivative above at $x=1$, so
$$\frac{1}{x(x+1)(x+2)\cdots(x+n)}=\frac1{n!}\sum_{k=0}^n(-1)^{k+1} {n\choose k}\ln(k+1)\\=\frac1{n!}\sum_{k=2}^{n+1}(-1)^{k} {n\choose k-1}\ln(k)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Easy way to calculate the determinant of a big matrix? Given this matrix:
\begin{matrix}
2 & 3 & 0 & 9 & 0 & 1 & 0 & 1 & 1 & 2 & 1 \\
1 & 1 & 0 & 3 & 0 & 0 & 0 & 9 & 2 & 3 & 1 \\
1 & 4 & 0 & 2 & 8 & 5 & 0 & 3 & 6 & 1 & 9 \\
0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0 \\
2 & 2 & 4 & 1 & 1 & 2 & 1 & 6 & 9 & 0 & 7 \\
0 & 0 & 0 & 6 & 0 & 7 & 0 & 1 & 0 & 0 & 0 \\
2 & 5 & 0 & 7 & 0 & 4 & 6 & 8 & 5 & 1 & 3 \\
0 & 0 & 0 & 1 & 0 & 4 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 8 & 0 & 2 & 0 & 0 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 2 & 1 & 1 \\
2 & 6 & 0 & 1 & 0 & 30 & 0 & 2 & 3 & 2 & 1 \\
\end{matrix}
Is there an efficient way of calculating its determinant using the matrix minors? it just seems that there's lots of zeroes inside so it has to ease up the operation and I'm missing something.
Thanks.
| It is possible to notice that the third column has only one non-zero value, $4$. Hence we cancel out the third column and the fifth row. Now the (original) fourth row has only one non-zero value, $5$, hence we cancel out the fourth row and the sixth column and continue this way. After six steps, i.e. after using the elements in positions $(5,3),(4,6),(3,5),(7,7),(9,4),(8,8)$ as "pivots", we can notice that no non-zero element on the sixth row remains, so the determinant is zero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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graphing $\frac{x^3-x+1}{x^2}$ I want to graph:
$$f(x) = \frac{x^3-x+1}{x^2}$$
so I took the first derivative:
$$f'(x) = \frac{x^3+x-2}{x^3}$$
but this function is hard to find the signals. In other words, it's hard to find where the function is increasing or decreasing. Also, the second derivative is even worse. I'm assuming it shouldn't be all that hard to find the signals, because it's as exercise on a book.
Any ideas? Do I really have to find the roots of the numerator?
| Study of sign of $f':$
Write the numerator as
$$x^3+x-2=(x-1)(x^2+x+2).$$ Now, $(x-1)(x^2+x+2)=0\iff x=1.$ So,
$$\begin{array}{ccc} x & (-\infty,1) & (1,\infty) \\ \mathrm{sign}(x^3+x-2) & - & +\end{array}$$
It is clear that $x^3=0\iff x=0.$ Now
$$\begin{array}{ccc} x & (-\infty,0) & (0,\infty) \\ \mathrm{sign}(x^3) & - & +\end{array}$$
Thus
$$\begin{array}{cccc} x & (-\infty,0)& (0,1) & (1,\infty) \\ \mathrm{sign}(f') & + & - & +\end{array}$$
So, $f$ is increasing in $(-\infty,0),$ decreasing in $(0,1)$ and increasing in $(1,\infty).$ Since it changes from decreasing to increasing at $x=1$ it has a local minimum at $x=1.$ (Note that $0$ doesn't belong to the domain.)
Note that the study of $$f''(x)=\frac{2(3-x)}{x^4}$$ is simpler. Repeating the same process:
$$2(3-x)=0\iff x=3$$ and
$$\begin{array}{ccc} x & (-\infty,3) & (3,\infty) \\ \mathrm{sign}(2(3-x)) & + & -\end{array}$$
$$x^4=0\iff x=0$$ and
$$\begin{array}{ccc} x & (-\infty,0) & (0,\infty) \\ \mathrm{sign}(x^4) & + & -\end{array}$$ Finally
$$\begin{array}{cccc} x & (-\infty,0) & (0,3) & (3,\infty) \\ \mathrm{sign}(f'') & + & + & -\end{array}$$ Thus $f$ is convex on $(-\infty,0)$ and $(0,3)$ and concave on $(3,\infty).$ Since it changes curvature at $x=3$ $f$ has an inflection point at $x=3.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find Minimum value of $P=\frac{1}{1+2x}+\frac{1}{1+2y}+\frac{3-2xy}{5-x^2-y^2}$ Given: $x,y\in (-\sqrt2;\sqrt2)$ and $x^4+y^4+4=\dfrac{6}{xy}$
Find Minimum value Of $$P=\frac{1}{1+2x}+\frac{1}{1+2y}+\frac{3-2xy}{5-x^2-y^2}$$
Could someone help me ?
| You could try this approach, although I think there is a simpler way.
Firstly, apply the change of variables
$$
v=x^4+y^4, \ w = \frac{1}{xy}
$$
in order to simplify the domain $D$ to
$$
D = \{\ v(w) = 6w-4\ : w \in (-\infty,-1/2) \cup (1/ 2,\infty)\ \}.
$$
The minimal value is now given by invoking the chain rule
\begin{align}
0 =& \frac{dP}{dw}\\
=& \frac{dP}{dv} \cdot \frac{dv}{dw} \\
=& \Big(\frac{\partial P}{\partial x} \cdot \frac{dx}{dv} + \frac{\partial P}{\partial y} \cdot \frac{dy}{dv}\Big) \cdot \frac{dv}{dw} \\
=& \Big( \frac{\partial P}{\partial x} \cdot \frac{dv}{dx}^{-1} + \frac{\partial P}{\partial y} \cdot \frac{dv}{dy}^{-1}\Big) \cdot \frac{dv}{dw},
\end{align}
taking into account that the solution lies in the domain $D$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How can I understand solving the equation? $$\begin{align}
&\left[(\sqrt[4]{p}-\sqrt[4]{q})^{-2} + (\sqrt[4]{p}+\sqrt[4]{q})^{-2}\right] : \frac{\sqrt{p} + \sqrt{q}}{p-q} \\
&= \left(\frac{1}{(\sqrt[4]{p}-\sqrt[4]{q})^{2}}+\frac{1}{(\sqrt[4]{p}+\sqrt[4]{q})^{2}}\right) \times \frac{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqrt{q})}{\sqrt{p} + \sqrt{q}} \\
&= \frac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}} \times \frac{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqrt{q})}{(\sqrt{p} + \sqrt{q})}\end{align}$$
How can I get this expression? $$\frac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}}$$
Only this solving I can't understand.
| If it helps, try expressing it in another way, recall that we can also express $\sqrt[n]{{a}^m}$ as $a^{m/n}$, so we have
$$\begin{align}\left(\frac{1}{(\sqrt[4]{p}-\sqrt[4]{q})^{2}}+\frac{1}{(\sqrt[4]{p}+\sqrt[4]{q})^{2}}\right)&=\left(\frac{1}{(p^{1/4}-q^{1/4})^2}+\frac{1}{(p^{1/4}+q^{1/4})^2}\right)\\
\end{align}
$$
Then, use the properties of quotients $\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}$ and exponents $(a^m)^n=a^{mn}$ to see that
$$\begin{align}\left(\frac{1}{(p^{1/4}-q^{1/4})^2}+\frac{1}{(p^{1/4}+q^{1/4})^2}\right)&=\frac{(p^{1/4}+q^{1/4})^2+(p^{1/4}-q^{1/4})^2}{(p^{1/4}+q^{1/4})^2*(p^{1/4}-q^{1/4})^2}\\
&=\frac{(p^{1/4}+q^{1/4})^2+(p^{1/4}-q^{1/4})^2}{[(p^{1/4}+q^{1/4})*(p^{1/4}-q^{1/4})]^2}\\
&=\frac{(p^{1/4}+q^{1/4})^2+(p^{1/4}-q^{1/4})^2}{(p^{1/2}-q^{1/2})^2}\\
&=\frac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}}
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factorial identity $\left(\tfrac{1}{2}\right)!$ to get Waallis I asked the wrong question here, my fault :(
How does one see, using $n! = \prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^n \frac{k}{k+n}$, that
$$\left(\frac{1}{2}\right)! = \prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^{\tfrac{1}{2}}\left( \frac{k}{k+\tfrac{1}{2}}\right) = \tfrac{1}{2}\sqrt{2 \cdot \left(\frac{2 \cdot 2}{1 \cdot 3}\frac{4 \cdot 4}{3 \cdot 5}\dotsm \right)}? $$
In other words, derive
$$ \prod_{k=1}^{\infty}\left(\frac{(2k)^2}{(2k+1)(2k-1)}\right)$$
while manipulating
$$\prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^{\tfrac{1}{2}}\left( \frac{k}{k+\tfrac{1}{2}}\right)$$
Note this is a way to 'derive' the Wallis product
$$ \tfrac{1}{2}\sqrt{2 \cdot \left(\frac{2 \cdot 2}{1 \cdot 3}\frac{4 \cdot 4}{3 \cdot 5}\dotsm \right)} = \frac{\sqrt{2 \tfrac{\pi}{2}}}{2} = \frac{\sqrt{ \pi}}{2}$$
It should follow, "after a bit of manipulation", but it just doesn't! I must be missing something!
Edit: I now think it is not possible directly!
| Looks like you need to show that
$$\prod_{k=1}^{\infty}\left(\frac{(2k)^2}{(2k+1)(2k-1)}\right) = \frac{\pi}{2}.$$
This Wikipedia article uses Euler's infinite product for the sine function:
$$\frac{\sin x}{x} = \prod_{k=1}^{\infty}\left(1 - \frac{x^2}{k^2 \pi^2}\right).$$
The result follows quickly by applying $x = \pi/2.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Evaluating $\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$ Evaluating $$\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$$ using $ux=\sqrt{x^2-1}$
UPDATE 'official' solution
$$u^2x^2=x^2-1$$
$$x^2=\frac{-1}{u^2-1}$$
$$x^2+1=\frac{u^2-2}{u^2-1}$$
$$2xdx=\frac{-2u}{(u^2-1)^2}$$
$$\int{\frac{x}{x\sqrt{x^2-1}(x^2+1)}dx}$$
$$\int{\frac{1}{\frac{-1}{u^2-1}u\frac{u^2-2}{u^2-1}}\left(\frac{-u}{(u^2-1)^2}\right)du} $$
$$\int{\frac{1}{(u^2-2)}du} $$
$$\frac{\log \left(\sqrt{2}-x\right)-\log \left(x+\sqrt{2}\right)}{2 \sqrt{2}}$$
$$\frac{\log \left(\sqrt{2}-\sqrt{x^2-1}/x\right)-\log \left(\sqrt{x^2-1}/x+\sqrt{2}\right)}{2 \sqrt{2}}$$
However at mathematica I get $$\frac{\log \left(-3 x^2-2 \sqrt{2} \sqrt{x^2-1} x+1\right)-\log \left(-3 x^2+2 \sqrt{2} \sqrt{x^2-1} x+1\right)}{4 \sqrt{2}}$$
Where is the error?
| $$u=\frac{\sqrt{x^2-1}}x$$
$$\frac{du}{dx}=-\frac{\sqrt{x^2-1}}{x^2}+\frac{2x}{x\cdot2\sqrt{x^2-1}}$$
$$=\frac{-(x^2-1)+x^2}{x^2\sqrt{x^2-1}}$$
$$\implies\int\frac{dx}{(1+x^2)\sqrt{x^2-1}}=\int\frac{x^2}{1+x^2} \frac{dx}{x^2\sqrt{x^2-1}}$$
Now $u^2=\dfrac{x^2-1}{x^2}\implies\dfrac1{x^2}=1-u^2$
$\implies\dfrac{x^2}{1+x^2}=\dfrac1{1+1/x^2}=\dfrac1{1+1-u^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integral evaluation (step-by-step) I'm trying to evaluate the integral by exponent. Could you help me with following steps?
Integral: $$\int \frac{1}{4+\sin(x)} dx$$
$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$
$$\int \frac{1}{4+sin(x)} dx = \int \frac{1}{4+\frac{e^{ix}-e^{-ix}}{2i}} dx = \int \frac{2i}{8i+e^{ix}-e^{-ix}} dx$$
Is it possible? Or any ways exist?
Thanks in advance.
| Let be $t = \tan(\frac{x}{2})$ so that $$\sin x=\frac{2t}{1+t^2},\quad\cos x=\frac{1-t^2}{1+t^2},\quad\operatorname{d}\!x=\frac{2 \operatorname{d}\!t}{1 + t^2}$$
and the integral becomes
$$
\int \frac{\operatorname{d}\!t}{2t^2+t+2}=\frac{1}{2}\int \frac{\operatorname{d}\!t}{\left(t+\frac{1}{4}\right)^2+\frac{15}{16}}=8\int \frac{\operatorname{d}\!t}{\left(\frac{4t+1}{\sqrt{15}}\right)^2+1}
$$
Then putting $u=\frac{4t+1}{\sqrt{15}},\,\operatorname{d}\!u=\frac{4}{\sqrt{15}}\operatorname{d}\!t$ we obtain
$$
\frac{2}{\sqrt{15}}\int \frac{\operatorname{d}\!t}{u^2+1}=\frac{2}{\sqrt{15}}\arctan u+\text{constant}.
$$
Finally we have $u=\frac{4t+1}{\sqrt{15}}=\frac{4\tan(\frac{x}{2})+1}{\sqrt{15}}$ and the integral is
$$
\int\frac{\operatorname{d}\!x}{4+\sin x}=\frac{2}{\sqrt{15}}\arctan\left(\frac{4\tan(\frac{x}{2})+1}{\sqrt{15}}\right)+\text{constant}
$$
| {
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"url": "https://math.stackexchange.com/questions/1077806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A problem from KVS 2014 I was doing the the following problem-
Prove that $$\frac { \sqrt { a+b+c } +\sqrt { a } }{ b+c } +\frac { \sqrt { a+b+c } +\sqrt { b } }{ c+a } +\frac { \sqrt { a+b+c } +\sqrt { c } }{ a+b } \ge \frac { 9+3\sqrt { 3 } }{ 2\sqrt { a+b+c } }. $$
I normalized this to $a+b+c=1$ and simplified to get-
$$\frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9+3\sqrt { 3 } }{ 2 }.$$
By Titu's lemma,
$$\Rightarrow \quad \frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9 }{ 3-(\sqrt { a } +\sqrt { b } +\sqrt { c } ) } .$$
However the RHS is maximized when $\sqrt { a } +\sqrt { b } +\sqrt { c } $ is maximized which is at $\sqrt { 3 } $,
$$\Rightarrow \quad \frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9 }{ 3-(\sqrt { a } +\sqrt { b } +\sqrt { c } ) } \le \frac { 9 }{ 3-\sqrt { 3 } } =\frac { 9+3\sqrt { 3 } }{ 2 } .$$
What went wrong?
You can view the rest of the problems here.
| we consider $f(x)=\frac{1}{1-\sqrt{x}}$ for $0<x<1$ and the tangent line at the point $x=\frac{1}{3}$ we get $y=\frac{3}{4}\sqrt{3}(2+\sqrt{3})x+\frac{3}{4}$. We have for $0<x<1$ $$f(x)\geq \frac{3}{4}\sqrt{3}(2+\sqrt{3})x+\frac{3}{4}$$. Thus we get $$\frac{1}{1-\sqrt{a}}+\frac{1}{1-\sqrt{b}}+\frac{1}{1-\sqrt{c}}\geq \frac{3}{4}\sqrt{3}(2+\sqrt{3})(a+b+c)+\frac{9}{4}=\frac{1}{2}(9+3\sqrt{3})$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:
*
*$\sqrt{4x-1} = \sqrt{x+2}-3$
*$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
*$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
*$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x+2})^2-3\sqrt{x+2}-3\sqrt{x+2}+9$
*$\sqrt{(4x-1)^2} = \sqrt{(x+2)\times(x+2)}-6\sqrt{x+2}+9$
*$\underline{4x-1} = \sqrt{(x+2)^2}-6\sqrt{x+2}+9$
*$4x-1 = \underline{x+2}-6\sqrt{x+2}+9$
How did the underlined expressions in steps 6 and 7 not include $\pm$? I thought that $\sqrt{x^2} = \pm x$.
| You have $x^2 = y^2 \implies x = \pm y$. However, if $x = y$, then $x^2 = y^2$.
Also note that $\sqrt{x^2} = |x|$, since we adopt the convention that $\sqrt{x}$ is the nonnegative square root of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
| the trouble with your approach stems from $0 \times \infty$ being indeterminate.
here is an explanation.
(a) what you are doing is approximating $\sqrt{x^2+1} = x + \cdots$
(b) $\sqrt{x^2 + 1} - x = 0 + \cdots$
(c) $x(\sqrt{x^2+1} - x) = x(0 + \cdots) = 0?$
you think the answer is 0. but, perhaps $x \times \cdots \neq 0?$
now, i will go back and elaborate on that $\cdots$ by using binomial thorem
$$(BIG + small)^{1/2} = BIG^{1/2} + {1 \over 2}BIG^{-1/2}small + \cdots$$
$(x^2+1)^{1/2} = x + {1 \over 2x} + \cdots$
now, if you go thru the steps (a)-(c) again, you see that $x(\sqrt{x^2+1} - x) = x({1 \over 2x}+ \cdots) = 1/2$
the reason is that $\cdots = const . 1/x^2 + \cdots$ and $x \times \cdots = 0$ as $x \to \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Why do we have $u_n=\frac{1}{\sqrt{n^2-1}}-\frac{1}{\sqrt{n^2+1}}=O(\frac{1}{n^3})$? Why do we have
*
*$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}=O\left(\dfrac{1}{n^3}\right)$
*$u_n=e-\left(1+\frac{1}{n}\right)^n\sim \dfrac{e}{2n}$
any help would be appreciated
| $$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}=$$
$$\dfrac{\sqrt{n^2+1}-\sqrt{n^2-1}}{\sqrt{n^2-1}\sqrt{n^2+1}}=$$
$$\dfrac{(\sqrt{n^2+1}-\sqrt{n^2-1})(\sqrt{n^2+1}+\sqrt{n^2-1})}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$$
$$\dfrac{n^2+1-n^2+1}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$$
$$\dfrac{2}{\sqrt{n^2-1}\sqrt{n^2+1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=$$
$$\dfrac{2}{\sqrt{n^4-1}(\sqrt{n^2+1}+\sqrt{n^2-1})}=O\left(\frac{1}{n^3}\right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$? How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?
I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
| $\frac{1}{\sqrt{2}}=2^{-\frac{1}{2}}$
$\frac{\sqrt{2}}{2}=\frac{2^{\frac{1}{2}}}{2^1}=2^{\frac{1}{2}-1}=2^{-\frac{1}{2}}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$ $a,b,c \in \mathbb{R}$ and $a+b+c=0$.
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$
I think that $2^{a}.2^{b}.2^{c}=1$, but i don't know what to do next
| We have
$$
8^a+1+1\geq3\sqrt[3]{8^a}=3\times 2^a,
$$
$$
8^b+1+1\geq3\sqrt[3]{8^b}=3\times 2^b,
$$
$$
8^c+1+1\geq3\sqrt[3]{8^c}=3\times 2^c,
$$
$$
2^a+2^b+2^c\geq 3\sqrt[3]{2^{a+b+c}}=3.
$$
It follows that
\begin{eqnarray}
8^a+8^b+8^c&\geq&3(2^a+2^b+2^c)-6\\
&=&(2^a+2^b+2^c)+2(2^a+2^b+2^c-3)\\
&\geq&2^a+2^b+2^c.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Removing the root squares from this expression? I would like to understand how to remove the root squares from this expression:
$$x = \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}$$
How to do it?
| $\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}=\frac {\sqrt{2}+ \sqrt{3}- \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}=\frac{\sqrt{2}+ \sqrt{3} - \sqrt5}{2\sqrt{6}}$ I think you can do the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Asymptotics of $\int_{0}^{+\infty}\!\!\frac{dx}{\sinh^2(\epsilon \sqrt{x^2+1}) } $ for $\epsilon$ near $0$ How to find an asymptotic expansion, for $\epsilon$ near $0$, of the following integral
$$
I(\epsilon):=\int_{0}^{+\infty}\frac 1{\sinh^2 (\epsilon \sqrt{x^2+1}) } {\rm d}x.
$$
As $\epsilon \rightarrow 0$, I have easily obtained
$$
I(\epsilon) \sim \frac{\pi}{2\:\epsilon^2}.
$$
Some clear steps leading to an extended expansion would be appreciated.
| My calculation shows that
$$I(\epsilon)
:= \int_{0}^{\infty} \frac{dx}{\sinh^{2}(\epsilon\sqrt{x^{2}+1})}
= \frac{\pi}{2\epsilon^{2}} - \frac{1}{\epsilon} + \pi \epsilon \sum_{n=1}^{\infty} \frac{1}{(\pi^{2}n^{2} + \epsilon^{2})^{3/2}} \tag{1}. $$
In particular, if we expand the infinite sum on the RHS, we get
$$ I(\epsilon) = \frac{\pi}{2\epsilon^{2}} - \frac{1}{\epsilon} + \sum_{n=0}^{\infty} \binom{-3/2}{n} \frac{\zeta(2n+3)}{\pi^{2n+2}} \epsilon^{2n+1}. $$
Indeed, from the following expansion
$$ \frac{1}{\sinh^{2}z}
= \sum_{n=-\infty}^{\infty} \frac{1}{(z-i\pi n)^{2}}
= \frac{1}{z^{2}} + 2 \sum_{n=1}^{\infty} \frac{z^{2} - \pi^{2}n^{2}}{(z^{2} + \pi^{2} n^{2})^{2}}, $$
it follows from term-wise integration that
\begin{align*}
\int_{0}^{R} \frac{dx}{\sinh^{2}(\epsilon\sqrt{x^{2}+1})}
= \frac{\arctan R}{\epsilon^{2}} + \sum_{n=1}^{\infty}
&\Bigg( \frac{2\epsilon \arctan \left( R\epsilon \big/ \sqrt{\pi^{2}n^{2} + \epsilon^{2}} \right)}{(\pi^{2}n^{2} + \epsilon^{2})^{3/2}} \\
&\quad + \frac{2}{R}\frac{1}{\pi^{2}n^{2}+\epsilon^{2}} \\
&\quad - \frac{2(R^{2}+1)}{R(\pi^{2}n^{2} + (R^{2}+1)\epsilon^{2})} \Bigg). \tag{2}
\end{align*}
Here, term-wise integration is possible from Fubini's theorem together with the following estimate:
$$ \int_{0}^{R} \left| \frac{\epsilon^{2}(x^{2}+1) - \pi^{2}n^{2}}{(\epsilon^{2}(x^{2}+1) + \pi^{2} n^{2})^{2}} \right| = \frac{\arctan \left( R\epsilon \big/ \sqrt{\pi^{2}n^{2} + \epsilon^{2}} \right)}{\epsilon\sqrt{\pi^{2}n^{2}+\epsilon^{2}}}
\lesssim_{\epsilon, R} \frac{1}{n^{2}}. $$
(Notice that the arctan term also contributes to order $n^{-1}$. It means that this argument fails if we consider $R = \infty$. This is why we consider proper integral first.)
Finally, taking $R \to \infty$ to (2) yields (1). When doing this, the only non-trivial calculation is to check that
$$ \lim_{R\to\infty} \sum_{n=1}^{\infty} \frac{2(R^{2}+1)}{R(\pi^{2}n^{2} + (R^{2}+1)\epsilon^{2})} = \frac{1}{\epsilon}. $$
But this follows from the squeezing lemma combined with the following inequality
$$ C(1, R) \leq \sum_{n=1}^{\infty} \frac{2(R^{2}+1)}{R(\pi^{2}n^{2} + (R^{2}+1)\epsilon^{2})} \leq C(0, R), $$
where
\begin{align*}
C(a, R)
&:= \int_{a}^{\infty} \frac{2(R^{2}+1)}{R(\pi^{2}x^{2} + (R^{2}+1)\epsilon^{2})} \, dx \\
&= \frac{\sqrt{R^{2}+1}}{R\epsilon} \frac{\arctan\left( \epsilon\sqrt{R^{2}+1} \big/ a\pi \right)}{\pi/2}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate: $\int e^{x^4}(x+x^3+2x^5)e^{x^2} dx$ Evaluate:
$$\int e^{x^4}(x+x^3+2x^5)e^{x^2} dx$$
I know the answer of this integral but got stuck at how to solve this. It seems to be the form like $ \int e^x(f(x)+f''(x))dx = e^x f(x)+C$
| $$\begin{aligned}
\int e^{x^4+x^2}\left ( x+x^3+2x^5 \right )\,dx &=\frac{1}{2}\int e^{x^4+x^2}\left ( 2x+2x^3+4x^5 \right )\,dx \\
&= \frac{1}{2}\int \left ( 2xe^{x^4+x^2}+x^2e^{x^4+x^2}\left ( 4x^3+2x \right ) \right )\,dx\\
&= \frac{1}{2}\int \left [ \left ( x^2 \right )'e^{x^2+x^4}+x^2 \left ( e^{x^4+x^2} \right )' \right ]\,dx\\
&= \frac{1}{2}\int \left ( x^2e^{x^2+x^4} \right )'\,dx\\
&= \frac{1}{2}x^2e^{x^2+x^4}+c, \;\; c \in \mathbb{R}
\end{aligned}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Summation of the telescoping series $\sum_{n=1}^N \frac{x}{(1+(n-1)x)(1+nx)}$ $(i)$ Verify that $$\frac{1}{1+(n-1)x} - \frac{1}{1+nx} = \frac{x}{(1+(n-1)x)(1+nx)}$$
$(ii)$ Hence show that for $x \ne 0$, $$\sum_{n=1}^N \frac{x}{(1+(n-1)x)(1+nx)}=\frac{N}{1+Nx}$$
Deduce that the infinite series $\frac{1}{1.\frac{3}{2}} +\frac{1}{2.\frac{3}{2}} +\frac{1}{2.\frac{5}{2}}.....$ is convergent and state the sum to infinity.
I need some help with the last part. I tried using the identity in the first part by letting $x=1$ and then varying $n$ but that did not produce the series that they have given. I also tried the same thing after setting $x=2$ but again could not get the series that they have given.
| From the LHS (The series), you can see that for:
At n=1, we have $\frac{x}{1+x}$
At n=2, we have $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}=\frac{2}{1+2x}$ (By simplifying)
At n=3, we have $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}=\frac{3}{1+3x}$ (Again by simplifying)
At n=4, we have $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}+\frac{x}{(1+3x)(1+4x)}=\frac{4}{1+4x}$ (You see the partern?)
Therefore, By mathematical induction, we have......
At n=N, $\frac{x}{1+x}+\frac{x}{(1+x)(1+2x)}+\frac{x}{(1+2x)(1+3x)}+\frac{x}{(1+3x)(1+4x)}+.....+\frac{x}{(1+(N-1)x)(1+Nx)}=\frac{N}{1+Nx}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Inequality $\frac 1{\sqrt{1+xy}}+\frac 1{\sqrt{1+yz}}+\frac 1{\sqrt{1+zx}}\ge \frac 9{\sqrt {10}}$ Let $x,y,z$ be non-negative real numbers such that $x+y+z=1$ , then is the following true? $$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}\ge \dfrac 9{\sqrt {10}}$$
| By the inequality between harmonic and arithmetic means we have that
$$
\frac{1}{\sqrt{1+xy}}+\frac{1}{\sqrt{1+yz}}+\frac{1}{\sqrt{1+xz}}\ge \frac{9}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}}.
$$
Now if we prove that
$$
\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}\le\sqrt{10}
$$
we are done. but by Cauchy-Schwarz we have
$$
\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}\le\sqrt{3}(3+xy+yz+zx)^{\frac{1}{2}},
$$
so we just need to prove that
$$
xy+yz+zx\le\frac{1}{3},
$$
but it is the same to prove that
$$
x^2+y^2+z^2\ge\frac{1}{3}.
$$
The last inequality can be easily proved using Cauchy-Schwarz.
| {
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"url": "https://math.stackexchange.com/questions/1085590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Solve $x^4+3x^3+6x+4=0$... easier way? So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$
Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but none of these work.
Because there were no rational linear factors, I had to assume that the quartic separated into two quadratic equations yielding either imaginary or irrational "pairs" of roots. My initial attempt was to "solve for the coefficients of these factors".
I assumed that $x^4+3x^3+6x+4=0$ factored into something that looked like this
$$\left(x^2+ax+b\right)\left(x^2+cx+d\right)=0$$
because the coefficient of the first term is one. Expanding this out I got
$$x^4+ax^3+cx^3+bx^2+acx^2+dx^2+adx+bcx+bd=0$$
$$x^4+\left(a+c\right)x^3+\left(b+ac+d\right)x^2+\left(ad+bc\right)x+bd=0$$
Equating the coefficients of both equations
$$a+c = 3$$
$$b+ac+d = 0$$
$$ad+bc = 6$$
$$bd = 4$$
I found these relationships between the various coefficients. Solving this system using the two middle equations:
$$\begin{cases} b+a\left(3-a\right)+\frac4b=0 \\ a\frac4b+b\left(3-a\right)=6 \end{cases}$$
From the first equation: $$a = \frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}$$
Substituting this into the second equation:
$$\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\cdot\frac4b+b\cdot\left(3-\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\right)=6$$
$$3\left(b-2\right)^2 = \left(b^2-4\right)\cdot\pm\sqrt{9+4b+\frac{16}b}$$
$$0 = \left(b-2\right)^2\cdot\left(\left(b+2\right)^2\left(9+4b+\frac{16}b\right)-9\left(b-2\right)^2\right)$$
So $b = 2$ because everything after $\left(b-2\right)^2$ did not really matter in this case. From there it was easy to get that $d = 2$, $a = -1$ and $c = 4$. This meant that
$$x^4+3x^3+6x+4=0 \to \left(x^2-x+2\right)\left(x^2+4x+2\right)=0$$
$$x = \frac12\pm\frac{\sqrt7}{2}i,\space x = -2\pm\sqrt2$$
These answers worked! I was pretty happy at the end that I had solved the equation which had taken a lot of work, but my question was if there was a better way to solve this?
| In general any quartic equation can be solved. If the polynomial happens to be the product of two irreducible polynomials over $\mathbb Z$, the following method allows to find the coefficients easily. Reducing the coefficients mod $2$, one has $$f(x):=x^4+3x^3+6x+4=x^4+x^3=x^2(x^2+x).$$ Lifting back to ${\mathbb Z}$-coefficients, one may assume that $$f(x)=(x^2+2ax\pm 2)(x^2+bx\pm 2),$$ where $a,b\in {\mathbb Z}$ and $b$ is odd. By comparison of coefficients, one has $$2a+b=3,2ab\pm 4=0,\pm 4a\pm 2b=6,$$ which shows immediately that one needs to take the ‘plus’ sign from $\pm$. From the first two equations, by eliminating $a$, one has $b=4$ or $b=-1$, so $b=-1$ since it is odd. It follows that $a=2,b=-1$ and $$f(x)=(x^2+4x+2)(x^2-x+2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 4,
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Simple limit problem without L'Hospital's rule $$\lim_{x \to 1}\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1}$$
We are not supposed to use any derivatives yet, but I can't find any formula that helps here. It's a $\frac{0}{0}$ indeterminate form, and all I think of doing is
$$\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} = \frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} \cdot \frac{\sqrt{x^4 + 1} + \sqrt{2}}{\sqrt{x^4 + 1} + \sqrt{2}} = \frac{x^4-1}{(\sqrt[3]{x}-1)\cdot(\sqrt{x^4+1}+\sqrt{2})}$$
but I don't see if this leads anywhere.
| Let $\sqrt[3]x-1=y\implies x=(1+y)^3,x^4=[(1+y)^3]^4=(1+y)^{12}$
$$\lim_{x \to 1}\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1}=\lim_{y\to0}\frac{\sqrt{(1+y)^{12}+1}-\sqrt2}y$$
$$=\lim_{y\to0}\frac{(1+y)^{12}+1-2}{(\sqrt{(1+y)^{12}+1}+\sqrt2)y}$$
$$=\lim_{y\to0}\frac{1+1+12y+O(y^2)-2}y\cdot\frac1{\lim_{y\to0}(\sqrt{(1+y)^{12}+1}+\sqrt2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Solve logarithmic equation for $x$ to find the inverse of $f(x)= \ln(x+\sqrt{x^2+1})$
Let $f(x)= \ln(x+\sqrt{x^2+1})$. Find $f^{-1}(x)$.
Here is what I got so far: $y= \ln(x+\sqrt{x^2+1})$, rewrite as $x= \ln(y+\sqrt{y^2+1})$,
then $$e^x= y+\sqrt{y^2+1}$$ $$e^x-y= \sqrt{y^2+1}$$ $$ y^2+ e^{2x}-2(e^x)y= 1$$
So if $e^x= a$, then $a^2-2ay-1= 0$
| Let $y=\ln(x+\sqrt{x^2+1})$, then:
$$e^{y}=x+\sqrt{x^2+1}$$
$$e^{y}-x=\sqrt{x^2+1}$$
$$(e^{y}-x)^2=(\sqrt{x^2+1})^2$$
$$e^{2y}-2xe^{y}+x^2=x^2+1$$
$$e^{2y}-2xe^{y}=1$$
$$e^{2y}-1=2xe^{y}$$
$$x=\frac{e^{y}-e^{-y}}{2}$$
So $\displaystyle f^{-1}(y)=\frac{e^{y}-e^{-y}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solving $\tan x-\tan(2x)=2\sqrt{3}$ $$\tan x-\tan(2x)=2\sqrt{3}$$
TRY #1
$$\begin{align*}
\tan x-\tan(2x)=2\sqrt{3}&\implies\tan x=2\sqrt{3}+\tan{2x}\\
&\implies \tan^2x=\tan^2(2 x)+4 \sqrt{3} \tan(2 x)+12\\
&\implies\tan^2x=(\frac{2\tan x}{1-\tan^2 x})^2+4\sqrt{3}\frac{2\tan x}{1-\tan^2x}+12
\end{align*}$$
but this will give me an equation with $\tan^4$ which needs quartic formula, too difficult!!
TRY #2
$$\begin{align*}
\tan x-\tan(2x)=2\sqrt{3} &\implies \frac{\sin x}{\cos x}-\frac{\sin 2x}{\cos 2x}=2\sqrt3 \\
&\implies\frac{\sin x\cos 2x-\sin 2x\cos x}{\cos x\cos 2x}=2\sqrt{3}\\
&\implies\frac{-\sin x}{\cos x\cos 2x}=2\sqrt{3}\\
&\implies\frac{-\sin x-2\sqrt{3}\cos x\cos 2x}{1}=0
\end{align*}
$$ then i can't!!
can anyone help me?
| How about
$$\tan x-\frac{2\tan x}{1-\tan^2 x}=2\sqrt 3$$
Removing the fraction will give you a cubic equation. At least that is easier than the quartic!
| {
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How can I show that $\lim_{n\to\infty} 2^n \left( \frac{n}{n+1} \right ) ^{n^2} = 0$? How to calculate limit of the following expression:
$$2^n \left( \frac{n}{n+1} \right ) ^{n^2} $$
I know that limit of this sequence is equal to zero, but how to show that?
| $(\frac{n}{n+1})^{n^{2}} = ( 1 -\frac{1}{n+1})^{n^{2}}$. For sufficiently large $n$,
$(1 - \frac{1}{n+1})^{n+1}$ is very close to $e^{-1}$, so is less than $\frac{2}{5}$ as $e >2.5.$ Then $( 1 -\frac{1}{n+1})^{n^{2}} < ( 1 -\frac{1}{n+1})^{n^{2}-1} < (\frac{2}{5})^{n-1}$. So $2^{n}(\frac{n}{n+1})^{n^{2}} < \frac{2^{2n-1}}{5^{n}} < 2 \times (\frac{4}{5})^{n-1}$, and the rightmost expression tends to $0$ as $n \to \infty.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Permutation in discrete math Is the permutation
$$\begin{pmatrix} 1& 2 &3 &4 &5 &6&7 \\ 7 & 4 & 2 & 1 & 3 & 6 & 5 \end{pmatrix}$$
even or odd?
The product of disjoint cycles is $$\begin{pmatrix}1& 7&5&3&2&4\end{pmatrix}\begin{pmatrix}6\end{pmatrix}$$
and the transposition are $$\begin{pmatrix}1 & 7 \end{pmatrix}\begin{pmatrix}1 & 5 \end{pmatrix}\begin{pmatrix}1 & 3\end{pmatrix}\begin{pmatrix}1 & 2 \end{pmatrix}\begin{pmatrix}1 & 4 \end{pmatrix}\begin{pmatrix}6 \end{pmatrix}$$ is it correct?
and the answer is even??
I feel confused about this..
| This is as you said $$(1,7,5,3,2,4),$$
and then it is equal to
$$(1,7)(7,5)(5,3)(3,2)(2,4),$$
and hence it is odd.
| {
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Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}.$$
I know that $$\sum_{n=0}^\infty\frac{1}{n!}=e$$
so $$\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1$$
But, what can I do for $$\sum_{n=1}^\infty\frac{1}{(n+1)!}$$ ?
Am I allowed to start a sum for $n=-1$ ? How can I bring to a something similar to $$\sum_{n=0}^\infty\frac{1}{n!}$$?
| $$
\sum\limits_{n = 1}^{ + \infty } {\frac{1}{{\left( {n - 1} \right)!\left( {n + 1} \right)}}} = \sum\limits_{n = 1}^{ + \infty } {\frac{n}{{\left( {n + 1} \right)!}}} = \sum\limits_{n = 0}^{ + \infty } {\frac{n}{{\left( {n + 1} \right)!}}} = 1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?
| $x=-2$ is a solution of the equation $$x^3+4x^2+x-6$$ thus you can divide $$x^3+4x^2+x-6$$ by $$x+2$$
| {
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Working on sequence, possibly recursive I am working on this problem which asks to find if the sequence converges or not and if so the value it converges to. I am not sure how to deal with this type of question, but I feel like it may be a recursive relation. It is $a_n=\dfrac{1^2}{n^3}+\dfrac{2^2}{n^3}+ \cdots +\dfrac{n^2}{n^3}$
What I have tried was calculating the first couple terms, $a_1=1$, $a_2=0.625$, $a_3=0.5185$ etc.I also tried writing it as $\dfrac{1^2+2^2+..+n^2}{n^3}$, but I am just really lost on where to go with this.
Thanks all
| First, note that $1^2+2^2+\cdots+n^2 = \frac{n(n+1)(2n+1)}{6}$. Then:
\begin{align}
\frac{1^2}{n^3} + \cdots + \frac{n^2}{n^3} &= \frac{\frac{n(n+1)(2n+1)}{6}}{n^3}\\
&\underset{n\to\infty}{=} \frac{2n^3}{6n^3}\\
&= \frac{1}{3}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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irreducibility in $\mathbb Q[X]$ Are these polynomials irreducible in $\mathbb Q[X]$:
1) $x^4+3x^3+x^2-2x+1$
What I did: I reduced it modulo $3$, then saw that it has no roots, so then I checked all $9$ monic polynomials of $\mathbb F_3$ and got $3$ of them, then checked their values under multiplication (as I calculated them allready to determin if they are reducible) with each other to conclude that they have different values then 1) for atleast one $a\in\mathbb F_3$. This way seems very inefficient, is it possible to do this more clever?
2) $2x^4+200x^3+2000x^2+20000x+20$
Here I divided by $2$ and then I could use Schönemann's criterion.
3) $x^2y+xy^2-x-y+1$ in $\mathbb Q[X,Y]$
$x y^2+(x+1)(x-1)y+(-x+1)$, so after Schönemann's criterion it's irreducible.
| (3) To "see" it better I would change $x=x+y$ and $y=x-y$ to get
$$P(x,y)=2x^3-2yx^2-2(y^2-2)x+2y^3+1$$
So, a non-trivial factorization would have to be of the form $2(x+p(y))(x^2+q(y)x+r(y))$.
From this $p(y)r(y)=2y^3+1$. But $2y^3+1$ is irreducible over $\mathbb{Q}$. So, either $p(y)$ or $r(y)$ is a constant.
If $p(y)$ where a constant $c$, then $P(c,y)=0$ but this can't be since $P(c,y)$ has degree $3$ in $y$.
Therefore we would need $p(y)=-a(2y^3+1)$ and $r(y)=-a^{-1}$ for some non-zero $a$. But when we try $$0\equiv P(a(2y^3+1),y)=16a^3y^9+24a^3y^6-8 a^2 y^7-4ay^5-8a^2y^4+(12a^3-4a+2)y^3-2ay^2-2a^2y-2a+1+2 a^3$$
We see that $a$ must be zero.
(1) It doesn't have rational roots since the only candidates $\pm1$ do not work. Therefore a factorization would have to be of the form
$$\begin{align}x^4+3x^3+x^2−2x+1&=(x^2+ax+b)(x^2+px+q)\\&=x^4+(a+p)x^3+(ap+b+q)x^2+(aq+bp)x+bq\end{align}$$
Therefore either $b=q=1$ or $b=q=-1$.
Now we can use the equations (coming from equating coefficients above) $$\begin{align}a+p&=3\\ap&=1-(b+q)=-1\text{ or }3\end{align}$$
to find what should be the values of $a,p$.
They should be roots of the polynomial $$x^2-3x+(-1\text{ or }3)$$ but none of these has rational roots.
| {
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Evaluating the limit $\lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$ Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$
I have a question about the following solution:
We may write it in the form:
$$ \frac{1}{n} \left[ \frac{1}{1+(\frac{1}{n})^2} + \ldots + \frac{1}{1+(\frac{n}{n})^2} \right] $$
Somehow I need to figure out that the limit is actually the Riemann sum of $\frac{1}{1 + x^2}$ on $[0,1]$ for $\pi = 0 < \frac{1}{n} < \ldots < \frac{n}{n}$.
Can you explain to me to reach this conclusion?
| The number you seek is
\begin{equation}
L
= \lim_{n\to\infty}\sum_{k=1}^n\frac{n}{n^2+k^2}
= \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^2}\cdot\frac{1}{n}\tag{1}
\end{equation}
Now, letting
\begin{align*}
\Delta x &= \frac{1}{n} & x_k &= \frac{k}{n}
\end{align*}
allows us to rewrite (1) as
$$
L = \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{1+x_k^2}\Delta x
$$
If you're familiar with the standard construction of the Riemann integral, you will recognize this as
$$
L=\int_0^1\frac{1}{1+x^2}\,dx
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the remainder when polynomial $f(x)$ is divided by $(x+1)(x-3)$ when $f(-1) = -4$ and $f(3) = 2$? A polynomial $f(x)$ gives remainder $2$ when divided by $(x-3)$ and gives a remainder $-4$ when divided by $(x+1)$. What is the remainder when $f(x)$ is divided by $(x^2 - 2x - 3)$?
I have shortened the question by:
1.Showing that $f(-1) = -4$ and $f(3) = 2$ by remainder theorem respectively.
2.Figured out that $(x^2 - 2x - 3)= (x-1)(x-3)$.
| You are on the right track.
Hint: For some polynomial $s(x),$ and for some $a,b \in \mathbb{R},$
$f(x) = (x^2 - 2x - 3) \cdot s(x)+(ax+b) = (x+1)(x-3) \cdot s(x)+(ax+b).$
Can you see why?
How then can you use this and everything else you know about $f$?
| {
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Implicit Differentiation: $(x/y)+(y/x) =1$ Hi can anyone please tell me where I goes wrong with this question:
Find $ \frac{dy}{dx} $ for the curves defines by this equation:
\begin{align}
\frac{x}{y} + \frac{y}{x} = 1
\end{align}
Here is what I did:
\begin{align}
&\frac{y-xy'}{y^2} + \frac{y'x - y}{x^2} =0 \\
&\frac{yx^2 - x^3 y'+y'y^2 x - y^3}{x^2 y^2} = 0 \\
&yx^2 + (-x^3 +y^2 x)y' -y^3 =0 \\
&\therefore y'= \frac{y^3 -yx^2}{-x^3 +y^2x}
\end{align}
The answer say it should be: $ y' = \frac{y}{x} $ but I had no clue how to proceed from there.
Please help, Thanks.
| if you going to do implicit differentiation you might as well multiply by $xy$ to get $x^2 + y^2 = xy$ before differencing. now differencing gives you $2xdx + 2ydy = xdy + ydx.$ this can also be written as
$$\dfrac{dy}{dx} = \dfrac{y - 2x}{2y-x} = \dfrac{y(y-2x)}{y(2y-x)}=
\dfrac{y(y-2x)}{2y^2 - xy} = \dfrac{y(y-2x)}{2xy - 2x^2 - xy} = \dfrac{y(y-2x)}{x(y-2x)} = \dfrac{y}{x} \tag 1$$
in implicit differentiation you don't have a unique answer. you always have to carry the constraint $\dfrac{x}{y} + \dfrac{y}{x} = 1$ along or an equivalent one like $x^2 + y^2 = xy$ along with the solution. in the equality (1) any one of them could be an answer.
| {
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$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$ Understand the representation $$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$$=$$\frac{1}{2} \left(-\frac{2}{n+1}+\frac{1}{n+2}+\frac{1}{n}\right)$$
$$s_n=\frac{1}{2} \left(\left(\frac{1}{n+2}+\frac{1}{n}-\frac{2}{n+1}\right)+\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}\right)+\ldots +\left(1 \frac{1}{1}-\frac{2}{2}+\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{3}+\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{2}{7}+\frac{1}{8}\right)\right)$$
Im sorry for not being able to get the term in right order, but as you can see the first two terms in the parenthesis should be the last for a accurate representation.
$$\left(
\begin{array}{cc}
1 & 1-1+\frac{1}{3} \\
2 & \frac{1}{2}-\frac{2}{3}+\frac{1}{4} \\
3 & \frac{1}{3}-\frac{1}{2}+\frac{1}{5} \\
4 & \frac{1}{4}-\frac{2}{5}+\frac{1}{6} \\
5 & \frac{1}{5}-\frac{1}{3}+\frac{1}{7} \\
6 & \frac{1}{6}-\frac{2}{7}+\frac{1}{8} \\
\end{array}
\right)$$
I think that my pure understanding of the representation is what makes me confused. From the table I see that a pattern in the denominators emerge, when n=1, then (1-2+3), when n=2 then, 2-3+4. However given the part of sn where $$\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}$$ I fail to see how the first part of expression right above is not undefined when n=1.
I am convinced that it is my lack of knowledge of what the representation actually means. And I am also struggling to see what cancels to give$$\frac{1}{2} \left(\frac{1}{n+2}-\frac{1}{n+1}+\frac{1}{2}\right)$$
However given that this is true I am able to understand that the answer is (1/4). But as I said it is the representation I do not get. Could someone help me as this would be very useful for my next chapter in my textbook!
| Your diagram is perfect. All these cancel.. It's a telescoping sum
| {
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Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n>0\ \forall\ n\ge1$.
Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.
I can't show that $a_n > 0\ \forall n\ge1$. I tried using induction but it wouldn't work.
Attempt:
$$
\begin{align}
2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}&=(2\sqrt{n}-(\sqrt{n-1}+\sqrt{n+1}))\cdot {2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\\&={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\end{align}$$
(The calculations are true for sure. No check is desired.)
Denote
$$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\text{ and }b_n={1\over n^{2}}.$$
Then
$$\begin{align*}
\lim_{n\to \infty}{a_n\over b_n} & =\lim_{n\to \infty}n^{2}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=\lim_{n\to \infty}n^{2}\lim_{n\to \infty}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=0
\end{align*}$$
By the comparison test for series convergence, since $\lim_{n\to \infty}{a_n\over b_n}=0$, then if $b_n$ converges, which it does, so does $a_n$.
| We have, $$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1} =\frac{1}{\sqrt{n} + \sqrt{n-1}} - \frac{1}{\sqrt{n+1} + \sqrt{n}} \ge 0$$
and $\displaystyle \sum a_n$ telescoping!
| {
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Find sum of the roots of quadratic polynomials
The zeroes of a quadratic polynomial $x^2+ax+b$ are $c$ and $d$ and the zeroes of a quadratic polynomial $x^2+cx+d$ are $a$ and $b$. Find the value of $a+b+c+d$.
The thing doesn't make sense how to use the two equations to get the sum?
Should I use the Vieta's formulas to find it?
| Hint You know that
$$x^2 + ax + b = (x-c)(x-d) = x^2 - (c+d)x + cd\\
x^2 + cx + d = (x-a)(x-b) = x^2 - (a+b)x + ab$$
Now compare the coefficients to get four equations.
| {
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"timestamp": "2023-03-29T00:00:00",
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indefinite integration the result is -1 instead of $1 \over 2$ Below is from a book,
When 0 $\le$ x < 1, F(x) = $\int_0^x$ t dt = $x^2 \over 2 $;
When 1 $\le$ x < 2, F(x) = $\int_0^1$ t dt + $\int_1^x$(2-t) dt
= -$x^2 \over 2$ + $2x$ $\color{blue}{-1}$;
However, I think the last part of above in blue, the -1 should be $1 \over 2 $, because
$\int_0^1$ t dt = F(1) - F(0) = $1^2 \over 2$ - 0 = $1 \over 2$,
$\int_1^x$(2-t) dt = $\int_1^x2$ dt - $\int_1^x$ t dt = $2x$ - $x^2 \over 2$
Please help me on this.
| The problem is in your last integral:
$$
\int_1^x (2-t)dt = 2t - \frac{t^2}{2} \bigg|_1^x = \left( 2x-\frac{x^2}{2} \right) -\left( 2-\frac{1}{2} \right) = -\frac{x^2}{2} + 2x -\frac{3}{2}.
$$
Then, we add the term $\int_0^1 t dt = \frac{1}{2}$, which yields that
$$
F(x) = \int_0^1 t dt + \int_1^x (2-t)dt = \frac{1}{2} + \left( -\frac{x^2}{2} + 2x -\frac{3}{2} \right) = -\frac{x^2}{2} + 2x -1.
$$
| {
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Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a}$ Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a},a \geq 0.$
One way to do it is using the formula
$$
a^3+2 - a^2-2 \sqrt{a}=(\sqrt{a}-1)^2(1+(a+1)(\sqrt{a}+1)^2) \geq 0.
$$
But I hope there is a better way.
| You can do it with the AM-GM inequality:
$$
\frac{a^3+a^3+1}{3}\ge \left(a^3\cdot a^3\cdot 1\right)^{\frac{1}{3}}=a^2
$$
$$
2\cdot\frac{a^3+1+1+1+1+1}{6}\ge 2\cdot\left(a^3\cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \right)^{\frac{1}{6}}=2\sqrt{a}
$$
Adding these inequalities yields the desired result.
| {
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$\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = ?$ $S_1=\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = ?$
Attempt: $S_1=\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = \dfrac {2n+2-1}{n^2(n+1)^2}$
$S_1=2 \sum_{n=1}^\infty [\dfrac {n+1}{n^2(n+1)^2} ] - \sum_{n=1}^\infty \dfrac {1}{n^2(n+1)^2}$
$=2 \sum_{n=1}^\infty [\dfrac {1}{n^2(n+1) } ] - \sum_{n=1}^\infty \dfrac {1}{n^2(n+1)^2}$
I am stuck on how to move ahead. Please guide me.
Thank you for your help.
| $$\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}(n+1)^{2}}=\sum_{n=1}^{\infty}\left[\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right]$$
$$=\bigl[\frac{\pi^{2}}{6}-\bigl(\frac{\pi^{2}}{6}-1\bigr)\bigr]=1.$$
using the fact that $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$.
Otherwise:
$$\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}(n+1)^{2}}=\sum_{n=1}^{\infty}\left[\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right]=\sum_{n=1}^{\infty}\frac{1}{n^{2}}-\sum_{n=1}^{\infty}\frac{1}{(n+1)^{2}}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n^{2}}-\left[\sum_{n=1}^{\infty}\frac{1}{n^{2}}-1\right]=1.$$
As, the series $\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ is convergent( by p-test) so it cancel out.
| {
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Finding $\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$ How would I go about solving this following limit?
$$\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$$
My attempts:
Direct substitution yields the limit to be undefined, also ruling out the possibility of using L'Hospital's Rule.
I don't see any clever substitutions that can be made with this limit.
Would squeeze theorem help here? Maybe using the trig. identities:
$$-1 \le \cos x \le 1$$
and
$$-1 \le \cos x \le 1$$
EDIT
I attempted to break the limit down term by term.
So, for the first one:
$$y = \lim_{x\to 0} (1 + \tan x)^{1/x}$$
Taking the natural log of both sides:
$$\ln y = \lim_{x\to 0} \frac{\ln(1+\tan x)}{x}$$
Direct sub. yields $0/0$. Using L'Hospital's rule:
$$\ln y = \lim_{x\to 0} \frac{\frac{\sec^2{x}}{1+\tan x}}{1} = \frac{\sec^2{x}}{1+\tan x} = 1$$
Thus, $\ln y = 1$, so $y= e$
EDIT #2
Thanks to a random comment, it actually does help me:
$$\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$$
$$\lim_{x\to 0} \frac{e-e}{0} = \frac{0}{0}$$
Thus, we can use L'Hospitals here:
$$\lim_{x\to 0} \frac{(\tan x+1)^{1/x} \left(\frac{\sec^2 x}{x(\tan(x)+1)}-\frac{\ln(\tan(x)+1))}{x^2}\right)}{1} = (\tan x+1)^{1/x} \left(\frac{\sec^2 x}{x(\tan(x)+1)}-\frac{\ln(\tan(x)+1))}{x^2}\right)$$
I haven't made any further progress, sadly.
Any help would be appreciated.
| i will use the maclaurin series $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3}+\cdots, \ \tan x = x + \frac{x^3}{3} + \cdots.$
now we can expand
$\begin{align}
\ln(1 + \tan x) &= \tan x - \frac{\tan^2 x}{2} +\frac{\tan^3 x}{3} + \cdots \\
&= x + \frac{x^3}{3}+\cdots -\frac{x^2}{2}+\cdots + \frac{x^3}{3} + \cdots\\
&= x - \frac{x^2}{2} + \cdots
\end{align}$
therefore $\frac{1}{x} \ln(1 + \tan x) = 1 - \frac{x}{2} + \cdots$ exponentiating
the last result gives $$(1 + \tan x)^{1/x} = ee^{- x/2 + \cdots} = e\{1- x/2 + \cdots \}$$
finally, $$\lim_{x \to 0}\frac{(1 + \tan x)^{1/x} - e}{x} = -\frac{e}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 3
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Find the range of $ x-\sqrt{4-x^2}$ $Y=x-\sqrt{4-x^2}$. How to find these types of functions' range?
I just know that the answer is $R=\{y\in\mathbb{R}\mid-2\sqrt{2}\leq y\leq 2\}$, but I have no idea how to find it step by step.
| We have $y=x-\sqrt{4-x^2}$. Function domain
\begin{equation*}
4-x^2\ge 0,~~|x|\le 2
\end{equation*}
Now find the value of function at the ends of the segment
\begin{equation*}
y(-2)=-2, ~~y(2)=2
\end{equation*}
Find the extrema
\begin{equation*}
y^{\prime} = 1-\frac{-2x}{2\sqrt{4-x^2}}=0
\end{equation*}
\begin{equation*}
\frac{-x}{\sqrt{4-x^2}}=1
\end{equation*}
Because $\sqrt{...}>0$ and $1>0$, so $x$ must be negative. Now solve the last equation, we get:
\begin{equation*}
\frac{x^2}{4-x^2}=1~\to~x^2=4-x^2~\to~x=\pm\sqrt{2}~\to~x=-\sqrt{2}
\end{equation*}
We get that the minimal value of function is
\begin{equation*}
y(-\sqrt{2})=-2\sqrt{2}
\end{equation*}
Remember that $x=-\sqrt{2}$ is the only one extrema and $y(-2)<y(-\sqrt{2})$, we see that $y$ decrease at $x\in[-2;-\sqrt{2}]$ and increase at $x\in[-\sqrt{2};2]$. After comparison of values of function at the ends of the segment we get, that the maximal value of function is $y(2)=2$, so
\begin{equation*}
-2\sqrt{2}\leq y\leq 2
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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CHKMO 2015 and cubic equations Let $a,b,c$ be distinct real numbers. If the equations $E_1: ax^3+bx+c=0, E_2: bx^3+cx+a=0$ and $E_3: cx^3+ax+b=0$ have a common root, prove that at least one of these equations has three real roots(not necessarily distinct).
Suppose $r$ is a common root, then adding the three equations implies $r$ is also a root of $(a+b+c)x^3+(a+b+c)x+(a+b+c)=0$. If $a+b+c\not=0,$ then $r$ is a root of $x^3+x+1=0$, which has one real root and two complex roots. If $r$ is complex, then we can easily derive a contradiction by using viete theorem, hence $r$ is real. Further more observation shows that if $p_i$ is a complex root of $E_i$, then $Re (p_i)=Re (p_j)$. But then I can't really move on, please helps.
| If $abc = 0$, WLOG $a = 0$, and so the common root must b$ x = - \frac{c}{b}$. However, it is easy to check that $ 0 = b \frac{-c^3}{b^2} + c \frac{-c}{b} = \frac{ -c^3-c^2b } { b^2} = - \frac{ c^2 (c+b) } {b^2}$, hence we must have $ c = -b$, which gives us $x=1$. Then, $ 0 = bx^3 + cx + a = bx^3 - bx = bx(x-1)(x+1)$ would have 3 real roots and we are done. Henceforth, $abc \neq 0$.
If $ a+b+c\neq 0$, then $r^3+r+1 = 0$ and $ r\neq 0$. WLOG, let $c = \min(a,b,c)$, and form the first equation we get $ (a-c)r^3 + (b-c)r = 0 \Rightarrow (a-c)r^2 + (b-c) = 0 $. However, since $ a-c > 0, b -c > 0$, we must have $ (a-c)r^+ (b-c) > 0 $ which is a contradiction. hence we must have $a+b+c=0$.
Thus, $r= 1$ is a root. Observe that $(ax^3 + bx+c) / (x-1) = ax^2 + ax + a + b $. For this to have 2 real roots, it is necessary and sufficient that $ \frac{ -b} {a}\geq \frac{3}{4}$.
WLOG, $ab>0$. WLOG, $a, b,-c > 0 $. We want to show that we have either $ \frac{-c}{b} \geq \frac{3}{4} $ or $ \frac{a}{-c} \geq \frac{3}{4}$ or $\frac{-b}{a} \geq \frac{3}{4}$. Suppose not, then we have $ 4 (-c) < 3b, \ldots$. However, summing these up we get $ a + b + c < 0$, which is a contradiction.
Hence, one of these inequalities must hold and the corresponding equation has 3 real roots.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generating function for Pell numbers Problem: The Pell numbers $p_n$ are defined by the recurrence relation \begin{align*} p_{n+1} = 2p_n + p_{n-1} \end{align*} for $n \geq 1$. The initial conditions are $p_0 = 0$ and $p_1 = 1$.
a) Determine the generating function \begin{align*} P(x) = \sum_{n=0}^{\infty} p_n x_n \end{align*} for the Pell numbers. What is the radius of convergence?
b) Determine (on the basis of the found generating function) an explicit formula for $p_n$.
Solution: Together with the initial conditions we have \begin{align*} P(x) &= 0 + x + \sum_{n=2}^{\infty} p_n x_n \\ &= x + \sum_{n=1}^{\infty} p_{n+1} x^{n+1} \\ &= x + \sum_{n=1}^{\infty} (2p_n + p_{n-1}) x^{n+1} \\ &= x + \sum_{n=1}^{\infty} 2p_n x^{n+1} + \sum_{n=1}^{\infty} p_{n-1} x^{n+1} \\ &= x + 2x \sum_{n=1}^{\infty} p_n x^n + x \sum_{n=1}^{\infty} p_{n-1} x^n. \end{align*}
Now we look at each series separately to see what we've got. The first series on the left expands as $(x + p_2 x^2 + p_3 x^3 + ...)$. This is nothing but the original $P(x)$ (because we can ignore the constant term $0$ right?). So for the first series we've got $2x P(x)$.
The second series on the right expands as $(0 + x^2 + p_2 x^3 + ...)$. We can factorize $x$ out such that we get $P(x)$ again. So everything together we have:
\begin{align*} P(x) = x + 2xP(x) + x^2 P(x), \end{align*} which gives us \begin{align*} P(x) (1-2x-x^2) = x, \end{align*} or \begin{align*} P(x) = \frac{x}{(1-2x-x^2)}. \end{align*}
But then I don't know how to determine the radius of convergence, and how to do b).
Any help would be appreciated.
| The terms in your sum are $p_nx^n$. If you solve the recurrence relation, you can find that $p_n$ grows as $r^n$ for some $r$, the larger root of the characteristic equation. You need the terms to decease faster than $\frac 1n$, so need $|x| \lt \frac 1r$
| {
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"source": "stackexchange",
"question_score": "4",
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The equation $x^4+y^4=z^2$ has no integer solution The equation $$x^4+y^4=z^2$$ has no integer solution for $(x, y, z), x \cdot y \neq 0 , z >0$.
We suppose that there is a solution $(x, y, z)$.
We consider the set $$M=\{z \in \mathbb{N} | \exists x, y \in \mathbb{Z}: x^4+y^4=z^2, x \cdot y \neq 0 \} \subseteq \mathbb{N}$$
Without loss of generality, we suppose that $(x, y, z)=1$.
One of $x$ and $y$ must be even and the other one odd.
We suppose that $x=2k+1, k \in \mathbb{Z}$ and $y=2l, l \in \mathbb{Z}$,
$\Rightarrow z=2m+1>0, m \in \mathbb{Z}$.
$$x^4+y^4=z^2 \Rightarrow y^4=z^2-x^4=(z-x^2)(z+x^2)$$
It stands that $gcd(z-x^2, z+x^2)=2$.
To show that $gcd(z-x^2, z+x^2)=2$, is the following the only way??
Let $(z-x^2, z+x^2)=d>1$. Then it has a prime divisor, let $p$.
$p \mid d , d \mid z-x^2 \Rightarrow p \mid z-x^2$
$p \mid d , d \mid z+x^2 \Rightarrow p \mid z+x^2$
So $p \mid 2x^2 \Rightarrow p \mid 2 \text{ OR } p \mid x$
and $p \mid 2z \Rightarrow p \mid 2 \text{ OR } p \mid z$
When $p \mid x \text{ AND } p \mid z$, we have that $p \mid y^4 \Rightarrow p \mid y$, so $p \mid (x, y, z)=1$, a contradiction.
So, it should be $p \mid 2 \Rightarrow p=2$.
Is this correct so far?? How can we continue to show that $d=2$ ??
$$$$
We have the following two cases
$$(1): \begin{cases}
z-x^2=2a^4\\
z+x^2=8b^4
\end{cases} \text{ with } (a, b)=1, a \equiv 1 \pmod 2 , a>0$$
$$(2): \begin{cases}
z-x^2=8b^4\\
z+x^2=2a^4
\end{cases} \text{ with } (a, b)=1, a \equiv 1 \pmod 2 , a>0$$
Why do we have these two cases??
| Since the discussion in the comments is getting complicated, let's do this directly.
Let $d=\gcd(z-x^2, z+x^2)$. Clearly $2|d$ since $2|z-x^2$ and $2|z+x^2$, seeing as how each of $z,x^2$ are odd, so we may write $d=2k$.
Now, for any divisor $f|(z-x^2)$ and $f|(z+x^2)$ then $f$ divides sums and differences of these things, so for our gcd we have
$$d|(z-x^2)+(z+x^2)=2z\qquad d|(z+x^2)-(z-x^2)=2x^2$$
Then as $d=2k$ we see that $k|x^2$ and $k|z$. Because $\gcd(x^2,z)=1$, there are integers $u,v$ such that $ux^2+vz=1$, hence $k|(ux^2+vz)\implies k|1$, i.e. $k=\pm 1$. We conclude $d=2$.
| {
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"url": "https://math.stackexchange.com/questions/1123796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Generating Functions and Polynomial Expansions Give a formula similar to:
$\frac{1-x^{m+1}}{1-x} = 1 + x + x^2 + ... + x^m$
For the following
(a) $1 + x^4 + x^8 + ... + x^{24}$
(b) $x^{20} + x^{40} + ... + x^{180}$
Workings
a. $1 + x^4 + x^8 + ... + x^{24}$
$g(x) = \frac{1-x^{25}}{1-x^4}$
b. $x^{20} + x^{40} + ... + x^{180}$
$g(x) = x^{20}(1 + x^{20} + ... + x^{160})$
$g(x) = x^{20}\frac{1-x^{161}}{1-x^{20}}$
I'm wondering if I am correct. Any help wil be appreciated.
| You almost got it! But are just a little bit off. One way to check each of these is to introduce $y = x^{4}$ for a) and $z = x^{20}$ for b). This means $$1 + x^4 + x^8 + ... + x^{24} = 1+y+y^2+\ldots +y^6 \\ = \frac{1-y^{7}}{1-y} \\ = \frac{1-x^{28}}{1-x^4}$$ and $$x^{20} + x^{40} + ... + x^{180} = z+z^2+\ldots +z^8 + z^9 \\ =(1+z+z^2+\ldots +z^8+ z^9)-1 \\ = \frac{1-z^{10}}{1-z}-1 \\ = \frac{1-z^{10}}{1-z}-\frac{1-z}{1-z} \\ = \frac{z(1-z^9)}{1-z} \\ = \frac{x^{20}(1-x^{180})}{1-x^{20}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with $\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1})$ $\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1}) = ?$
I don't know how to solve the indetermination there... is it possible to rearrange the expression in brackets in order to use L'Hospital or Taylor Series?
| You may write
$$
x^2 - \sqrt{x^4 - x^2 + 1}=\frac{(x^2)^2 - (x^4 - x^2 + 1)}{x^2 + \sqrt{x^4 - x^2 + 1}}=\frac{1 - \frac{1}{x^2}}{1 + \sqrt{1 - \frac{1}{x^2} + \frac{1}{x^4}}} \to \frac12
$$ as $x \to +\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Parametrization of solutions of diophantine equation $x^2 + y^2 = z^2 + w^2$ I need integer solutions of $x^2 + y^2 = z^2 + w^2$ parametrized. Can it be done? Thanks.
| [Partial solution.]
Look for rational solutions to $$x_1^2+y_1^2-z_1^2=1\tag{1}$$ first.
We know that $p_0=(-1,0,0)$ is a solution. Let $(a,b,c)$ be any set of integers. Then solve $p_0+t(a,b,c)=(x_1,y_1,z_1)$. $(-1+at)^2+(bt)^2-(ct)^2 =1$, or $1-2at+a^2t^2+b^2t^2-c^2t^2=1$ or $2at = a^2t^2+b^2t^2-c^2t^2$. Now, $t=0$ corresponds to $p_0$. Assuming $t\neq 0$, this means:
$$t=\frac{2a}{a^2+b^2-c^2}$$
This means that $$(x_1,y_1,z_1)=\left(\frac{a^2-b^2+c^2}{a^2+b^2-c^2},\frac{2ab}{a^2+b^2-c^2},\frac{2ac}{a^2+b^2-c^2}\right)$$
That gives a complete solution set to $(1)$.
That gives parametric solutions for the original equation:
$$(x,y,z,w)=(a^2-b^2+c^2,2ab,2ac,a^2+b^2-c^2)$$
Note that if $a^2+b^2+c^2$ is even, then all terms are even, so we can divide by $2$, so this doesn't give reduces solutions even when $\gcd(a,b,c)=1$.
Example, $(x,y,z,w)=(8,1,4,7)$. Then $(x_1,y_1,z_1)=\left(\frac{8}{7},\frac{1}{7},\frac{4}{7}\right)$. So $(-1,0,0)+\frac{1}{7}(15,1,4)$ so $(a,b,c)=(15,1,4)$. That yields $$(a^2-b^2+c^2,2ab,2ac,a^2+b^2-c^2)=(240,30,120,210)=30(8,1,4,7).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Legendre polynomial to show identity, can't spot mistake Using Legendre polynomial generating function
\begin{equation}
\sum_{n=0}^\infty P_n (x) t^n = \frac{1}{\sqrt{(1-2xt+t^2)}}
\end{equation}
Or $$ P_n(x)=\frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2-1)^n] $$
Show$$ P_{2n}(0)=\frac{(-1)^n (2n)!}{(4)^n (n!)^2} $$
And $$ P_{2n+1}(0)=0$$
I expressed $$(x^2 -1)^{2n}= \sum_{k=0}^{2n} {2n \choose k}x^{4n-2k}(-1)^k$$
And using second formula given, the only term that remains after differentiating 2n times and substituting x=0 is where $$4n-2k=2n$$ so $$2n=2k$$ k=n i.e $$(-1)^n \frac{(2n)!}{(n!)^2} $$ but multiplying this by$$ \frac{1}{2^{2n} (2n)!} $$ doesn't give desired solution. Where have I gone wrong?
| This one can also be done using complex variables.
Starting from the generating function
$$\sum_{n\ge 0} P_n(x) t^n
= \frac{1}{\sqrt{1-2xt+t^2}}$$
Call $P_n(0) = Q_n$ so that
$$\sum_{n\ge 0} Q_n t^n
= \frac{1}{\sqrt{1+t^2}}.$$
Now using Lagrange Inversion we get
$$Q_{2n} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2n+1}}
\frac{1}{\sqrt{1+z^2}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2n+2}}
\frac{1}{\sqrt{1+z^2}} \; z \; dz.$$
Put $1+z^2 = w^2$ so that $z\; dz = w\; dw$ to obtain
$$2\times \frac{1}{2\pi i}
\int_{|w-1|=\epsilon} \frac{1}{(w^2-1)^{n+1}}
\frac{1}{w} \; w\; dw.$$
The scalar two in front is because when $z$ makes one turn around the
origin, $w$ makes two, but the contour only counts one turn.
Continuing we have
$$\frac{2}{2\pi i}
\int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n+1}}
\frac{1}{(w+1)^{n+1}} \; dw
\\ = \frac{2}{2\pi i}
\int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n+1}}
\frac{1}{(2+w-1)^{n+1}} \; dw
\\ = \frac{1}{2^{n}} \frac{1}{2\pi i}
\int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n+1}}
\frac{1}{(1+(w-1)/2)^{n+1}} \; dw
\\ = \frac{1}{2^{n}} \frac{1}{2\pi i}
\int_{|w-1|=\epsilon} \frac{1}{(w-1)^{n+1}}
\sum_{q\ge 0} (-1)^q {q+n\choose n}
\frac{(w-1)^q}{2^q} \; dw
.$$
Extracting coefficients this gives
$$\frac{1}{2^{n}} (-1)^n {2n\choose n} \frac{1}{2^n}.$$
This is
$$\frac{(-1)^n}{4^n} \frac{(2n)!}{(n!)^2}$$
as claimed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing a sequence defined recursively is convergent Given the recursively defined sequence
$$ a_1 = 0, a_{n+1} = \frac 1{2+a^n} $$
Show it converges. I'm working with Cauchy sequences, and proved in a previous question that any sequence of real numbers $(a_n)$ satisfying:
$$|a_n - a_{n+1}| \leq \frac1{2^n} $$
is convergent (by showing $(a_n)$ was Cauchy), and was told that it may be helpful to use induction to prove that the above recursive sequence satisfies the inequality and is therefore convergent.
Work:
Base Case: When $n = 1 , a_1 = 0, a_2 = \frac12,$ so $$ \left|0-\frac12\right| \leq \frac1{2^{1}}$$ and the base case checks out
Inductive Step: Want to show $$|a_{n+1} - a_{n+2}| \leq \frac1{2^{n+1}} $$
So $$\left|\frac1{2+a_n} - \frac1{2+a_{n+1}}\right| \leq \frac1{2^{n+1}}$$
And I'm not sure where to proceed from here. Any help would be greatly appreciated, thanks.
| Continuing as Thomas Andrews suggested,
$a_{n+1}-a_{n+2}
=\frac1{2+a_n}-\frac1{2+a_{n+1}}
=\frac{(2+a_{n+1})-(2+a_n)}{(2+a_n)(2+a_{n+1})}
=\frac{a_{n+1}-a_n}{(2+a_n)(2+a_{n+1})}
$
so
$|a_{n+1}-a_{n+2}|
=\big|\frac{a_{n+1}-a_n}{(2+a_n)(2+a_{n+1})}\big|
<|\frac{a_{n+1}-a_n}{4}|
$.
From this,
$|a_{n+k}-a_{n+k+1}|
<|\frac{a_{n+1}-a_n}{4^k}|
$.
Putting
$n = 0$,
$|a_{k}-a_{k+1}|
<|\frac{a_{1}-a_0}{4^k}|
$
which is more than enough
to get convergence.
It is interesting
that this shows that
the convergence is at least
$\frac1{4^k}$,
not just
$\frac1{2^k}$.
To find the limit:
$|a_{n+1}-a_{n}|
=|a_n-\frac1{2+a_n}|
=|\frac{a_n(2+a_n)-1}{2+a_n}|
=|\frac{a^2_n+2a_n-1}{2+a_n}|
=|\frac{(a_n+1)^2-2}{2+a_n}|
$.
Since $a_n$ converges,
$(a_n+1)^2-2
\to 0
$
or
$a_n \to \sqrt{2}-1$.
To find the true rate of convergence,
since
$a_n \to \sqrt{2}-1$,
$|a_{n+1}-a_{n+2}|
=\big|\frac{a_{n+1}-a_n}{(2+a_n)(2+a_{n+1})}\big|
\approx \big|\frac{a_{n+1}-a_n}{(2+\sqrt{2}-1)(2+\sqrt{2}-1)}\big|
= \big|\frac{a_{n+1}-a_n}{(\sqrt{2}+1)^2)}\big|
= \big|\frac{a_{n+1}-a_n}{3+\sqrt{2}}\big|
$,
so the convergence
is like
$\frac1{(3+\sqrt{2})^k}
$.
| {
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Differentiation using the Chain Rule $$y=\frac { \cos(1+x) }{ 1+\cos(x) } $$
Steps I took:
$$y'=\frac { (-\sin(1+x))(1+\cos(x))-(\cos(1+x))(-\sin(x)) }{ (1+\cos(x))^2 } $$
$$y'=\frac { (\cos(1+x))(\sin(x)) }{ (1+\cos(x))^2 } -\frac { (\sin(1+x))(1+\cos(x)) }{ (1+\cos(x))(1+\cos(x)) } $$
$$y'=\frac { (\cos(1+x))(\sin(x)) }{ (1+\cos(x))^2 } -\frac { \sin(1+x) }{1+\cos(x)}$$
How would I further simplify my answer from here? I imagine that I would have to use the trigonometric addition formulas but as soon as I start breaking the problem down to do that, I end up with the incorrect answer. I was wondering if there was a more efficient way of getting to the most simplified solution.
| you can do this with without chain rule. let us see.
$\begin{align}
\dfrac{\cos (1+x)}{1+\cos x} &= \dfrac{\cos 1 \cos x - \sin 1 \sin x}{1+\cos x}\\
&= \dfrac{\cos 1 (1+\cos x) -\cos 1 - \sin 1 \sin x}{1+\cos x}\\
&= \cos 1 - \dfrac{\cos 1}{1 + \cos x} -\sin 1\tan(x/2)
\end{align}$
we can now take the derivative of
$\begin{align}
\dfrac{d}{dx}\left(\dfrac{\cos (1+x)}{1+\cos x}\right) &=
-\dfrac{\cos 1\sin x}{(1 + \cos x)^2} -\dfrac{\sin 1}{2\cos^2 x/2}\\
&=-\dfrac{\cos 1\sin x}{(1 + \cos x)^2} -\dfrac{\sin 1}{1+\cos x}\\
&=-\dfrac{\cos 1\sin x + \sin 1 + \sin 1\cos x}{(1 + \cos x)^2}\\
&=-\dfrac{\sin (1+x) + \sin 1 }{(1 + \cos x)^2}\\
\end{align}$
i used the calculator to check the derivative and it seems correct.
| {
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"url": "https://math.stackexchange.com/questions/1135844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the equation $4x=y^2+z^2+1$ has no integer solution
Show that the equation
$$4x=y^2+z^2+1$$
has no integer solution.
I divided throughout by $4$ to get
$$x=\frac{y^2}{4}+\frac{z^2}{4}+\frac{1}{4}$$
but not sure if that is correct
| $$4x=y^2+z^2+1$$4x is even ,so other side must be even ,so one of y or z is odd and ohter is even
suppose y is odd and z is even $$y=2k+1 \rightarrow y^2=4k^2+4k+1\\$$$$z=2q \rightarrow z^2=4q^2$$put them in furmula $$4x=(4k^2+4k+1)+(4q^2)+1\\4x=4(k^2+k+q^2)+2 $$now divide both side by 2 $$2x=2(k^2+k+q^2)+1$$ and it is impossible ,because left hand side is even ,but right hand side is odd
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this system without WolframAlpha How can I solve this system without using WolframAlpha or any other program?
$$\begin{equation}
\begin{cases}
2x_1+\lambda x_2+3x_3+4x_4=1\\x_1-x_2+9x_3+7x_4=3\\3x_1+5x_2+\lambda x_3+5x_4=1\\x_1+2x_2+x_4=0
\end{cases}
\end{equation}$$
| i will use $k$ for $\lambda.$ look at the augmented matrix
$\begin{align}\pmatrix{2&k&3&4&1&|a\\1&-1&9&7&3&|b\\3&5&k&5&1&|c\\1&2&0&1&1&|d}
&\to \pmatrix{1&2&0&1&1&|d\\1&-1&9&7&3&|b\\3&5&k&5&1&|c\\2&k&3&4&1&|a} \\
&\to \pmatrix{1&2&0&1&1&|d\\0&-3&9&6&-2&|b-d\\0&-1&k&3&-2&|c-3d\\0&k-4&3&2&-1&|a-2d} \\
&\to \pmatrix{1&2&0&1&1&|d\\0&1&-k&-3&2&|-c+3d\\0&0&9-3k&0&4&|b-d+3(-c+3d)\\
0&0&3-k(4-k)&2-3(4-k)&-1+2(4-k)&|a-2d+(4-k)(-c+3d)}
\end{align}$
now you have to look at different cases. for example if $k = 3.$
| {
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System of 4 tedious nonlinear equations: $ (a+k)(b+k)(c+k)(d+k) = $ constant for $1 \le k \le 4$ It is given that
$$(a+1)(b+1)(c+1)(d+1)=15$$$$(a+2)(b+2)(c+2)(d+2)=45$$$$(a+3)(b+3)(c+3)(d+3)=133$$$$(a+4)(b+4)(c+4)(d+4)=339$$ How do I find the value of $(a+5)(b+5)(c+5)(d+5)$. I could think only of opening each expression and then manipulating, and I also thought of integer solutions(none exist). How do I solve it then?
| According to Maple, there are no real solutions. There are complex solutions.
They have $a$ as a root of the irreducible quartic $x^4+4 x^2-3 x+7$.
But as for the value of $(a+5)(b+5)(c+5)(d+5)$: note that
$(a+x)(b+x)(c+x)(d+x)$ is a monic quartic polynomial in $x$. Find a
monic quartic with values $15,\;45,\;133,\;339$ at $1,2,3,4$. This part involves linear equations, or you might use a table of differences...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
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} |
Finding $\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
Find the limit: $\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
My attempt:
$\begin {align}\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}} &=
\lim _{n\to \infty}\frac{\sqrt{n+n^4\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\ &=
\lim _{n\to \infty}\frac{n^2\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\&=
\lim_{n\to \infty}\frac{\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{\sqrt{3+\frac{1}{n^2}}}\end{align}$
Which looks like is equal to $\infty$ because of the $n^3\cdot 1$ but it's wrong.
What am I doing wrong here?
| When you take a factor out of the square-root, it still gets square-rooted, but you have been squaring them. For example, $\sqrt{n^2+1}=\sqrt{n^2(1+1/n^2)}=\sqrt{n^2}\sqrt{1+1/n^2}=n\sqrt{1+1/n^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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For each of the following rules, either prove that it is true in every group $G$, or give a counterexample to show that it is false in some groups. Let $J$ be a group that consists of six matrices:
$I = \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}, A = \begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}, B = \begin{pmatrix} 0 & 1 \\-1 & -1 \end{pmatrix}, C = \begin{pmatrix} -1 & -1 \\0 & 1 \end{pmatrix}, D = \begin{pmatrix} -1 & -1 \\1 & 0 \end{pmatrix}, K = \begin{pmatrix} 1 & 0 \\-1 & -1 \end{pmatrix}$.
$ \text { If $x^2 = e$, then $x = e$.}$
$A^2 = I.$ So the rule above doesn't hold in group $J.$
$\text { If $x^2 = a^2$, then $x = a$.}$
$A^2 = I = C^2$, but $A \neq C.$ So the rule above doesn't hold in group $J.$
$ (ab)^2 = a^2b^2.$
$BA = K, K^2 = I.$
$A^2 = I, B^2 = D, DI = D.$
So the rule above doesn't hold in group $J.$
$\text { If $x^2 = x$, then $x = e$}$.
Let $x^2 = x$. Then, $xxx^{-1} = xx^{-1}.$ So, $x = e.$ The rule above holds for all groups $G$.
$ \text { For every $x$ in $G$, there is some $y$ in $G$ such that $x = y^2$}$.
Let $x = K$. Then, $K = y^2 = (\sqrt{K})^2 = K.$ So the rule above holds.
$ \text { For any two elements $x$ and $y$ in $G$, there is an element $z$ in $G$ such that $y = xz$}$.
Given: $xzx^{-1} = z.$ Let $x = A.$ Then $AzA \neq z.$ At least I wasnt able to show it. So, this rule fails.
Please, check my work.
| The second to last statement is false. Consider for example the integers with multiplication.
The last statement is true. Let $z=x^{-1}y.$
The other answers look good.
| {
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summation of a trigonometric series How to evaluate $\tan^2(1) + \tan^2(3) + \tan^2(5) + \tan^2(7) + \ldots + \tan^2(89)$?
Angles are given in degrees.
I tried converting $\tan(89)$ as $\cot(1)$ and then tried combining $\tan(1)$ and $\cot(1)$, but later got stuck.
| My answer is based on a rather beautiful answer given to a related question, that can be found here.
Expressing the summation in radians, we wish to evaluate
$$S=\sum_{m=0}^{45-1}\tan^2\left\{\frac{(2m+1)\pi}{2\cdot90}\right\}=\sum_{m=0}^{45-1}\tan^2\left\{\frac{(m+\frac{1}{2})\pi}{2\cdot45}\right\}$$
Using the well-known identity that
$$\left(\cos{\frac{k\pi}{2n}} + i\sin{\frac{k\pi}{2n}}\right)^{2n}=(-1)^{k}$$ and by setting $k=(m+\frac{1}{2})$ we have a purely imaginary expression
$$\left(\cos{\frac{(m+\frac{1}{2})\pi}{2n}} + i\sin{\frac{(m+\frac{1}{2})\pi}{2n}}\right)^{2n}=(-1)^{m}i$$
Calculating the binomial expansion, where $n=45$, we have
$$\sum_{t=0}^{2n}{2n \choose t}\left[\cos\frac{(m+\frac{1}{2})\pi}{2n}\right]^t\left[i\sin\frac{(m+\frac{1}{2})\pi}{2n}\right]^{2n-t}$$
Considering only the real terms, which should be $0$, we have
$$\sum_{r=0}^{n}{2n \choose 2r}\left[\cos\frac{(m+\frac{1}{2})\pi}{2n}\right]^{2r}(-1)^{n-r}\left[\sin\frac{(m+\frac{1}{2})\pi}{2n}\right]^{2n-2r}=0$$
dividing by $\left[\cos\frac{(m+\frac{1}{2})\pi}{2n}\right]^{2n}$ results in
$$\sum_{r=0}^{n}{2n \choose 2r}(-1)^{n-r}\left[\tan\frac{(m+\frac{1}{2})\pi}{2n}\right]^{2n-2r}=0$$
In a manner similar to the linked answer, we use Vieta's formula, as the summation we wish to evaluate is simply the sum of the roots of the polynomial. Thus, the sum
is ($n=45$)
$$S=\frac{{2n \choose 2}}{{2n \choose 0}}=\frac{(2n)(2n-1)}{2}=\frac{90\cdot89}{2}=4005$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I prove that $2/9$x$ and $y$ are real numbers.
Given that $1<x^2-xy+y^2<2$, how can I show that $\frac 29<x^4+y^4<8$ ?
Then can I use that to prove that for any natural number $n>3$
$$x^{2^n}+y^{2^n}>\frac 2{3^{2^n}} \text{?}$$
| $0\leq x^2+y^2<2+xy$. Squaring it, you get $x^4+y^4+2x^2y^2<4+4xy+x^2y^2\Rightarrow x^4+y^4<4+4xy-x^2y^2=8-(4+x^2y^2-4xy)=8-(2-xy)^2\leq 8$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Probability of drawing a yellow ball after 1 or 2 have been drawn. The question is as follows:
An opaque urn contains ten balls, five red, four yellow, and one green. Three balls are drawn from the urn in sequence but at random, without replacement.
What is the conditional probability of drawing a yellow ball given that a yellow ball has been drawn? (Note: “a yellow ball has been drawn” means at least one yellow ball has turned up, in the first draw, the second, or both.)
I have been trying to solve this problem using Bayes' Theorem the following way:
$
P(drawing\:yellow\:last\: |\: 1\: or\: 2\: drawn) = \frac{P(1\: or\: 2\: drawn\: | \:drawing \:yellow \:last)\cdot P(drawing \:yellow \:last)}{P(1 \:or \:2 \:drawn)}
$
Where $P(1 \:or \:2\: drawn\: | \:yellow\: drawn \:last)$ will be equal to $P(1\: or\: 2 \:drawn)$ because it is not a conditional probability (this may be where I'm going wrong).
$$
P(1 \:or \:2 \:drawn) = 2(\frac{4}{10}\cdot\frac{6}{9} ) + (\frac{4}{10}\cdot\frac{3}{9} ) = .666666
$$
$$
P(drawing \:yellow \:last) = 2\cdot \frac{4}{10}\cdot\frac{6}{9} \cdot \frac{3}{8} + \frac{4}{10}\cdot\frac{3}{9}\cdot \frac{2}{8} =.23333
$$
Resulting in .233333 which is not the correct answer of precisely 1/2 in the solutions given to me by my TA.
This problem is driving me insane! If you could steer me in the right direction, I'd be super grateful. Thanks!
EDIT: Here is the TA solution:
| Let $T$ be the event yellow on third, and let $A$ be the event at least one yellow in the first two trials. We want $\Pr(T|A)$. By the definition of conditional probability, we have
$$\Pr(T|A)=\frac{\Pr(A\cap T)}{\Pr(A)}.\tag{1}$$
We compute the two probabilities on the right-hand side of (1).
The event $A$ can happen in $3$ ways, which we can call YN, NY, and YY.
The sum of the probabilities of these is
$$\frac{4}{10}\cdot \frac{6}{9}+\frac{6}{10}\cdot\frac{4}{9}+\frac{4}{10}\cdot\frac{3}{9}.$$
The event $A\cap T$ can also happen in $3$ ways, which we can call YNY, NYY, and YYY. The probability is therefore
$$\frac{4}{10}\cdot \frac{6}{9}\cdot\frac{3}{8}+\frac{6}{10}\cdot\frac{4}{9}\cdot\frac{3}{8}+\frac{4}{10}\cdot\frac{3}{9}\cdot\frac{2}{8}.$$
Now we use (1) and in principle are finished, though you are probably expected to simplify. It does simplify, a lot.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$ How to prove this trigonometric identities ?
$$\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$$
Thank you.
| COMMENT.-In attention to the note by Community above I show the following:
$$LHS=\frac{\sin(\frac{15\pi}{18})}{\sin(\frac{11\pi}{18})\sin(\frac{4\pi}{18})}\\$$ Hence one has $$\sin(\frac{15\pi}{18})=4\cos(\frac{4\pi}{18})\sin(\frac{11\pi}{18})\sin(\frac{\pi}{18})$$ Note that $\dfrac{15\pi}{18}=150^{\circ}=180^{\circ}-30^{\circ}$ so one has $$\sin(\frac{15\pi}{18})=\frac 12$$ Besides $$2\cos(\frac{4\pi}{18})\sin(\frac{11\pi}{18})=\sin(\frac{15\pi}{18})+\sin(\frac{7\pi}{18})$$ so we have $$\frac12=2\left(\frac12+\sin(\frac{7\pi}{18})\right)\sin(\frac{\pi}{18})$$ and because of $\frac{7\pi}{18}=90^{\circ}-20^{\circ}$ and $\frac{\pi}{18}=10^{\circ}$ we can reduce all to elementary calculations.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim_{(x,y)\rightarrow (0,0)} \frac{\tan(x^2+y^2)}{\arctan(\frac{1}{x^2+y^2})} $ I'm having trouble understanding how the $\displaystyle\lim_{(x,y)\rightarrow (0,0)} \frac{\tan(x^2+y^2)}{\arctan(\frac{1}{x^2+y^2})}$. I used the product law to set it up as $\displaystyle\frac{\lim_{(x,y)\rightarrow (0,0)} \tan(x^2+y^2)}{\lim_{(x,y)\rightarrow (0,0)}\arctan(\frac{1}{x^2+y^2})}$ then found $\lim_{(x,y)\rightarrow (0,0)} \tan(x^2+y^2)=0$ but got caught up on $\lim_{(x,y)\rightarrow (0,0)}\arctan(\frac{1}{x^2+y^2})$. I looked up the solution and $\lim_{(x,y)\rightarrow (0,0)}\arctan(\frac{1}{x^2+y^2})= \pi/2$. Can someone explain how to solve the second limit?
| No need to transform into polar form.
Put, $x^2+y^2=z$. Then , $z\to 0$ as $(x,y)\to (0,0)$.
Now, $$\lim_{(x,y)\to (0,0)}\frac{\tan (x^2+y^2)}{\arctan \left(\frac{1}{x^2+y^2}\right)}$$
$$=\lim_{z\to 0}\frac{\tan z}{\arctan (1/z)}=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$
Prove that the following inequality
$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$$
holds for arbitrary real numbers $a$, $b$ and $c.$
Someone says, "It's very easy problem. It can also be proved by AM–GM inequality." But I can't reason out the answer to this question.
| This is best explained with a picture:
The length you have on the LHS is the length of the dotted red path in the unit square.
Now use some reflections in order to "straighten" that path. You will get:
$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \sqrt{(1+a)^2+(2-a)^2}$$
and now it is trivial that the RHS has a minimum for $a=\frac{1}{2}$, whose value is $\frac{3}{2}\sqrt{2}$.
Another option is to use the QM-AM (also known as Cauchy-Schwarz) inequality as orangeskid did. With the previous line, we can also improve the original inequality up to:
$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \sqrt{2}\left(\frac{3}{2}+\frac{\left|a-\frac{1}{2}\right|^2}{3+\left|a-\frac{1}{2}\right|}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let p be prime. Prove that: Let p be prime:
$p^2\choose p$ is congruent to p (mod $p^2$) and $2p\choose p$ is congruent to 2(mod$p^2)$
I know that when p is prime p|$p\choose k$ where $p\choose k$ can be defined as $p!/k!(p-k)!$ and have tried playing around from there but cannot seem to make it work. Any help would be greatly appreciated!
| Doing the case for $p=2$ can be done by hand, so I assume $p$ is odd.
$\binom{p^2}{p}=\frac{1\cdot 2 \cdot 3\dots p^2}{(1\cdot 2 \cdot 3 \cdot p)(1\cdot 2\cdot 3\dots \cdot p^2-p)}=\frac{(p^2-p+1)(p^2-p+2)\dots p^2}{1\cdot 2\cdot \dots p}$ If we cancel all of the factors of $p$ we get a complete residue class mod $p^2$ on the top and on the bottom we get a complete residue class, except instead of having the last $p-1$ terms we have their negatives. Luckily $p-1$ is even, so the product is the same as the complete residue class.
For $\binom{2p}{p}$ we just write $\frac{1\cdot2\cdot\dots 2p}{(1\cdot2\cdot3\dots\cdot p)(1\cdot2\cdot3\dots\cdot p)}$ cancelling $p$ with $p$ and $2p$ with $p$ we get $2\frac{1\cdot 2\cdot \dots (p-1) (p+1)\dots (2p-1)}{(1\cdot2\cdot3\dots\cdot p-1)(1\cdot2\cdot3\dots\cdot p-1)}$. Since we have to complete residue classes at both the top and bottom we get this is congruent to $2$.
| {
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Proof by induction that $\frac1{n+1} + \frac1{n+2} + \ldots + \frac1{2n} \geq \frac7{12}$ The question is prove that for every integer greater than or equal to 2
$$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} \geq \frac{7}{12}$$
So far I have
Base case let $p(2)$
\begin{align*}
\frac{1}{2}+ \frac{1}{2+2} + \frac{1}{4} & \geq \frac{7}{12}\\
\frac{5}{6} & \geq \frac{7}{12}
\end{align*}
therefore base case is true.
Let's assume it is a true statement for $p(n+1)$ for all $n$ greater than or equal to $2.$
\begin{align*}
\frac{1}{(n+1)+1} + \frac{1}{(n+1)+2} + \frac{1}{2(n+1)} \geq \frac{7}{12}\\
\frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{2n+2} \geq \frac{7}{12}
\end{align*}
then I wrote that if $n$ is greater than or equal to $2$ is a true statement then the inequality is true. This is a proof by induction and I'm not sure if I concluded properly
| Consider $P(n+1) - P(n)$. This difference has only three terms, all the middle terms cancel. And looking at those three terms you can easily show $P(n+1)-P(n)>0$. That proves that if $P(n)>7/12$ then $P(n+1)$ must also be.
| {
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"timestamp": "2023-03-29T00:00:00",
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System of Recurrence Relations Solve the following System of Recurrence Relation:
$$a_n = 2a_{n-1} - b_{n-1} + 2, a_0 = 0$$
$$b_n = -a_{n-1} + 2b_{n-1} - 1, b_0 = 1$$
Workings:
$b_n - 2b_{n-1} = -a_{n-1} - 1$
$a_n = 2a_{n-1} - b_{n-1} + 2$
$a_{n+1} = 2a_n - b_n + 2$
$-2a_n = -4a_{n-1} + 2b_{n-1} - 4$
$a_{n+1} + 2a_n = (2a_n - b_n + 2) + (4a_{n-1} - 2b_{n-1} + 4) $
$a_{n+1} + 2a_n = 2a_n + 4a_{n+1} - b_n -2b_{n-1} + 6$
$a_{n+1} + 2a_n = 2a_n + 4a_{n-1} - (-a_{n-1} - 1) + 6$
$a_{n+1} + 2a_n = 2a_n + 4a_{n-1} + a_{n-1} + 7$
$a_{n+1} = 5a_{n-1} + 7 (*)$
Now I'm not sure what to next. Can I shift $(*)$ down to $a-n$.
Any help will be appreciated.
| hint: add the two equations to get: $(a_n+b_n) = (a_{n-1} + b_{n-1}) + 1\to a_n+b_n = n+1$. Can you take it from here?
| {
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Find $f(2)$ if $f$ satisfies $2f(x)-3f(\frac1x)=x^2$ The following expression is given, and we are asked to find $f(2)$.
\begin{equation}
2f(x)-3f\left(\frac{1}{x}\right) =x^2
\end{equation}
Does a unique and well defined answer exist? Why? and what is it?
We have that $f(1)=f(-1)=-1$.
| Hint: Note that
$$
\begin{bmatrix}
2&-3\\
-3&2
\end{bmatrix}
\begin{bmatrix}
f(x)\\
f(1/x)
\end{bmatrix}
=
\begin{bmatrix}
x^2\\
1/x^2
\end{bmatrix}
$$
and
$$
\det\begin{bmatrix}
2&-3\\
-3&2
\end{bmatrix}=-5\ne0
$$
Completed Answer
Now that over a year has passed, I think it is appropriate to finish the answer.
$$
\begin{align}
\begin{bmatrix}
f(x)\\
f(1/x)
\end{bmatrix}
&=
\begin{bmatrix}
2&-3\\
-3&2
\end{bmatrix}^{-1}
\begin{bmatrix}
x^2\\
1/x^2
\end{bmatrix}\\
&=
-\frac15\begin{bmatrix}
2&3\\
3&2
\end{bmatrix}
\begin{bmatrix}
x^2\\
1/x^2
\end{bmatrix}\\
&=-\frac15\begin{bmatrix}
2x^2+3/x^2\\
3x^2+2/x^2
\end{bmatrix}\\
\end{align}
$$
Therefore,
$$
f(x)=-\frac{2x^4+3}{5x^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1157703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
the total differential equation $(y^2z - y^3 +x^2y)dx - (x^2z +x^3 -xy^2)dy +(x^2y - xy^2)dz=0$ In solving the total differential equation $$(y^2z - y^3 +x^2y)dx - (x^2z +x^3 -xy^2)dy +(x^2y - xy^2)dz=0$$
Substituting $x =uz$ and $y=vz$ and reached at a position where, i face the following
$$(v^2 - v^3 +u^2v)du - (u^2+u^3-uv^2)dv=0$$
I am stuck here please help!!
| Your equation $(v^2 - v^3 +u^2v)du - (u^2+u^3-uv^2)dv=0$ is OK. To continue, see below :
Instead of such a complicated calculus, a little bit of observation would have given the integrating factor $\frac{1}{u^2v^2}$ for $(v^2 - v^3 +u^2v)du - (u^2+u^3-uv^2)dv=0$
$$\left(\frac{1}{u^2}-\frac{v}{u^2}+\frac{1}{v}\right)du+\left(-\frac{1}{v^2}-\frac{u}{v^2}+\frac{1}{u}\right)dv=0$$
$$d\left(-\frac{1}{u}+\frac{v}{u}+\frac{u}{v}+\frac{1}{v}\right)=0$$
$$-\frac{1}{u}+\frac{v}{u}+\frac{u}{v}+\frac{1}{v}=constant$$
$$-\frac{z}{x}+\frac{y}{x}+\frac{x}{y}+\frac{z}{y}=constant$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help with conditional probability Assume a box contains 11 balls: 5 red, 4 blue, and 2 yellow. A ball is drawn and its color noted. If the ball is yellow, it is replaced; otherwise, it is not. A second ball is then drawn and its color is noted.
What is the probability that the first ball was yellow, given that the second was red?
I believe I can use Bayes' theorem to solve this problem. I calculated the probabilities of the different scenarios but I can't figure out this one. Any advice is appreciated.
| There are $5$ cases where $\color{red}{\text{the second ball is red}}$, out of which, in $3$ cases $\color{green}{\text{the first ball is yellow}}$:
*
*$\color{red }{P( RR)= \frac{5}{11}\cdot\frac{4}{10}}$
*$\color{black}{P( RB)= \frac{5}{11}\cdot\frac{4}{10}}$
*$\color{black}{P( RY)= \frac{5}{11}\cdot\frac{2}{10}}$
*$\color{red }{P( BR)= \frac{4}{11}\cdot\frac{5}{10}}$
*$\color{black}{P( BB)= \frac{4}{11}\cdot\frac{3}{10}}$
*$\color{black}{P( BY)= \frac{4}{11}\cdot\frac{2}{10}}$
*$\color{green}{P(YRR)=\frac{2}{11}\cdot\frac{5}{11}\cdot\frac{4}{10}}$
*$\color{black}{P(YRB)=\frac{2}{11}\cdot\frac{5}{11}\cdot\frac{4}{10}}$
*$\color{black}{P(YRY)=\frac{2}{11}\cdot\frac{5}{11}\cdot\frac{2}{10}}$
*$\color{green}{P(YBR)=\frac{2}{11}\cdot\frac{4}{11}\cdot\frac{5}{10}}$
*$\color{black}{P(YBB)=\frac{2}{11}\cdot\frac{4}{11}\cdot\frac{3}{10}}$
*$\color{black}{P(YBY)=\frac{2}{11}\cdot\frac{4}{11}\cdot\frac{2}{10}}$
*$\color{green}{P(YYR)=\frac{2}{11}\cdot\frac{2}{11}\cdot\frac{5}{10}}$
*$\color{black}{P(YYB)=\frac{2}{11}\cdot\frac{2}{11}\cdot\frac{4}{10}}$
*$\color{black}{P(YYY)=\frac{2}{11}\cdot\frac{2}{11}\cdot\frac{1}{10}}$
So the probability that the first ball was yellow given that the second ball was red is:
$$\frac{\color{green}{P(YRR)}+\color{green}{P(YBR)}+\color{green}{P(YYR)}}{\color{red}{P(RR)}+\color{red}{P(BR)}+\color{green}{P(YRR)}+\color{green}{P(YBR)}+\color{green}{P(YYR)}}=\frac{100}{540}\approx18.5\%$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the maximum and minimum of $x^2+2y^2$ if $x^2-xy+2y^2=1$.
Find the maximum and minimum of $x^2+2y^2$ if $x,y\in\mathbb R$ and $$x^2-xy+2y^2=1$$
My attempt:
Clearly, since $x^2-x(y)+(2y^2-1)=0$ and $2y^2-y(x)+(x^2-1)=0$, we have that
$$\Delta_1=y^2-8y^2+4=4-7y^2\ge 0$$
and
$$\Delta_2=x^2-8x^2+8=8-7x^2\ge 0$$
so that $y^2\le \frac{4}{7}$ and $x^2\le \frac{8}{7}$.
Thus $x^2+2y^2\le \frac{8}{7}+\frac{8}{7}=\frac{16}{7}$.
But clearly $x^2+2y^2\neq \frac{16}{7}$, since if $\begin{cases}x^2=\frac{8}{7}\\y^2=\frac{4}{7}\end{cases}$, then $x^2-xy+2y^2\neq 1$
so that equality can't be achieved and we only have that $x^2+2y^2<\frac{16}{7}$.
| You already have a great answer from Kim. Another way would be to use Cauchy-Schwarz inequality to get the same bounds on $xy$:
$$(1+xy)^2 = (x^2+2y^2)(2y^2+x^2) \ge 8x^2y^2$$
$$\implies -1-xy \le 2\sqrt2 xy \le 1+xy$$
$$\implies -\frac1{2\sqrt2+1} \le xy \le \frac1{2\sqrt2-1}$$
$$\implies 1-\frac1{2\sqrt2+1}\le 1+xy=x^2+2y^2 \le 1+\frac1{2\sqrt2-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $a^2+2b+c$, $b^2+2c+a$, and $c^2+2a+b$ are all perfect squares
Let $a,b,c$ (with $a,b,c>1$) be postive integers,and such that $\color{#0a0}{\text{$a^2+2b+c$}}$, $b^2+2c+a$, and $c^2+2a+b$ are all perfect squares.
Show that:
$$a+b+c=276$$
We note that
$$a^2+2b+c\ge (a+1)^2,b^2+2c+a\ge (b+1)^2,c^2+2a+b\ge (c+1)^2$$
| Without loss of generality let $a=\min(a,b,c)$. Then there are two cases: $a \leq b \leq c$ and $a \leq c \leq b$.
*
*Suppose $a \leq b \leq c$. Then $c^2 < c^2+2a+b \leq c^2 + 3c < (c+2)^2$, hence $c^2+2a+b=(c+1)^2$ yielding $2a+b=2c+1$. We also have $b^2 < b^2+2c+a = b^2+2a+b-1+a \leq b^2+4b - 1 < (b+2)^2$, hence $b^2+2c+a=(b+1)^2$ and $2c+a=2b+1$. However, since $c \geq b$ and $a \geq 1$ this would imply $a=1$, contradiction.
*Suppose $a \leq c \leq b$. Then $b^2 < b^2+2c+a \leq b^2 + 3b < (b+2)^2$, hence $b^2+2c+a=(b+1)^2$ and $2c+a=2b+1$. This implies $4a+2b = 4a+2c+a-1 \leq 7c - 1 < 8c+8$, so $2a+b < 4c+4$ and $c^2 < c^2+2a+b < (c+2)^2$. It follows that $2a+b=2c+1$. Combining this with $2c+a=2b+1$, we find that $(a,b,c)=(a,3a-2,\frac{5a-3}{2})$. Hence $a$ is odd. Writing $a=2t+1$ with $t \geq 1$ yields $(a,b,c)=(2t+1,6t+1,5t+1)$. Now $b^2+2c+a$ and $c^2+2a+b$ are squares by construction so we only need $a^2+2b+c=4t^2+21t+4$ to be a square. For $t \geq 3$ we have $(2t+4)^2 < 4t^2+21t+4<(2t+6)^2$, so it then follows that $4t^2+21t+4 = (2t+5)^2$ and $t=21$. Hence $t \in \{1,2,21\}$. Since $4t^2+21t+4$ is not a square for $t=1$ and $t=2$, we find $t=21$ and $(a,b,c)=(43,127,106)$ with $a+b+c=276$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1170097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\sum\limits_{n}\frac{1}{z+n}$ diverges
How to bound the sequence $\sum\limits_{n\ge0}\frac{1}{z+n}$ from below ?
Actually the problem is from complex analysis, and it is given that $z\in U=\mathbb C\setminus\mathbb R_{\le0}$, but I think it can be treated as in the real case.
So for $z>0, n\in \mathbb N$, I have to show the divergence of the sequence
Is it OK, if I write $\displaystyle\frac{1}{z+n}=\frac{1}{n}\cdot\frac{1}{1+\frac zn}\ge\frac{1}{n}\cdot\frac{1}{1+z},\quad\forall n$
then $\displaystyle \sum\limits_{n\ge0}\frac{1}{z+n}\ge\left(\frac{1}{1+z}\cdot \sum\limits_{n\ge1}\frac{1}{n}\right)+\frac1z\longrightarrow\infty$
$\bf{EDIT:}$ Original task is to show that $\sum_{n\ge0}f_n$ with $f_n=\frac{(-1)^n}{z+n}$ and $z\in\mathbb C\setminus\mathbb R_{\le0}=:U$ does not converge locally normally on $U$.
| You've only shown that $\sum_{n\ge 0}^\infty \frac{1}{z + n}$ diverges for $z > 0$. However, you missed the case when $z = a + bi$ with $b\neq 0$. Write
$$\frac{1}{z + n} = \frac{1}{(a + n) + bi} = \frac{a + n}{(a + n)^2 + b^2} - \frac{b}{(a + n)^2 + b^2}i.\tag{*} $$
The first term on the right-most side of $(*)$ dominates $\frac{c}{a + n}$ (with $c = \frac{a^2}{a^2 + 4b^2}$) for $n \ge \frac{|a|}{2}$. Indeed, if $n \ge \frac{|a|}{2}$, then $n + a \ge n - |a| \ge \frac{|a|}{2}$. So
$$\frac{a + n}{(a + n)^2 + b^2} = \frac{1}{a + n}\cdot \frac{1}{1 + \frac{b^2}{(a + n)^2}} \ge \frac{1}{a + n}\cdot \frac{1}{1 + \frac{4b^2}{a^2}} = \frac{c}{a + n}.$$
Since $\sum_{n\ge 1} \frac{c}{a + n}$ diverges, it follows that $\sum_{n\ge 0} \frac{a + n}{(a + n)^2 + b^2}$ diverges. The second term is dominated by $\frac{1}{b(a + n)^2}$, and $\sum_{n \ge 0} \frac{1}{b(a + n)^2}$ converges. So $\sum_{n\ge 1} \frac{b}{(a + n)^2 + b^2}$ converges. Therefore, $\sum_{n \ge 0} \frac{1}{z + n}$ diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the slope of the tangent line to the graph of $f^{-1}$ Given function $f$, find the slope of the line tangent to the graph $f^{-1}$ at the point on the graph $f^{-1}$. $f(x)=\sqrt{5x}$; $(4,\frac{16}{5})$?
Here is what I have thus far:
$f'(x)= \frac{\sqrt{5}}{2\sqrt{x}}$
can it also be $f'(x)= \frac{2.5}{\sqrt{5x}}$?
We then need to plug in for $x$ at $x=\frac{16}{5}$
$f'(x)= \frac{\sqrt{5}}{2\sqrt{\frac{16}{5}}}=\frac{5}{8}$
We then have to reciprocate $\frac{5}{8}$
Let $M$ be the slope of the tangent line $f^{-1}$
$M=\frac{8}{5}$
Are these steps and arithmetic correct?
| look at what the function $$y=f(x) = (5x)^{1/2} \text{ does at } x = 16/5, y = 4.$$ the derivative of $f$ is $$f'(x) = \frac12 (5x)^{-1/2}\times 5,\quad
f'(16/5) = \frac 52 \times \frac 14= \frac58 $$
so far we have at $$x = {16}/{5}, y=f(16/5) = 4, f'(16/5) = \frac58.$$ therefore
$$ y = 4, f^{-1}(4) = 16/5, \left(f^{-1}\right)'(4) = \frac85$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_\limits{x\to 0}\left({\tan x\over x}\right)^{1\over 1-\cos x}$. Find $\lim_\limits{x\to 0}\left({\tan x\over x}\right)^{1\over 1-\cos x}$. Is there a way to do it without differentiating so many times? That is exhausting and confusing and will probably cause errors. I would really appreciate your help with this.
| Using basic Taylor series as $x \to 0$,
$$
\frac{1}{1-\cos x} = \frac{1}{1-(1-\frac{x^2}{2} + o(x^2))} \sim \frac{2}{x^2}
$$
and
$$
\log\left(\frac{\tan x}{x}\right) = \log\left(1 + \frac{x^2}{3} + o(x^2)\right) \sim \frac{x^2}{3}.
$$
Therefore
$$
\frac{\log\left(\frac{\tan x}{x}\right)}{1-\cos x} \sim \frac{2}{x^2}\frac{x^2}{3} = \frac{2}{3}
$$
and the result follows by taking exponentials
$$
\lim_{x\to 0} \left(\frac{\tan x}{x}\right)^{\frac{1}{1-\cos x}} = \lim_{x\to 0} \exp\left[\frac{\log\left(\frac{\tan x}{x}\right)}{1-\cos x}\right] = e^{2/3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.