Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find limit without using l'Hospital rule $\lim\limits_{x\rightarrow0}\frac{\tan x-\sin x}{x^3}$ How to find the following limit without using l'Hospital rule
$$\lim_{x\rightarrow0}\frac{\tan x-\sin x}{x^3}$$
Using l'Hospital I got $1\over2$. Thanks for your help.
| $$
\begin{aligned}
\lim _{x\to 0}\left(\frac{\tan \left(x\right)-\sin \left(x\right)}{x^3}\right)
& = \lim _{x\to 0}\left(\frac{\left(x+\frac{x^3}{3}+o\left(x^3\right)\right)-\left(x-\frac{x^3}{3!}+o\left(x^3\right)\right)}{x^3}\right)
\\& = \lim _{x\to 0}\left(\frac{x+\frac{x^3}{3}-x+\frac{x^3}{3!}+o\left(x^3\right)}{x^3}\right)
\\& = \color{red}{\frac{1}{2}}
\end{aligned}
$$
Solved with Taylor expansion
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Probability in cards that $4$ people each get queen and king? A $52$ card deck is shuffled and then dealt out to $4$ people $($each person gets $13$ cards$)$. What is the chance that each person gets an a Queen & a King?
My attempt: I know there's $\binom{52}{13}$ different hands that can be shuffled out. The chances of getting a Queen is $4/52$ and the chances of getting a King is $4/52$. This makes the chance of having both $16/(52*52)$ = $16/2704$. Since there are $4$ people, does that mean the overall chance of each person having a Queen and King in their $13$ card deck $(\frac{16}{2704})^4$?
| There are $\binom{52}{13}\binom{52-13}{13}\binom{52-13-13}{13}$ ways of dealing 4 hands of 13 cards each.
Given that each hand has to have one king and queen, there are $\binom{52-8}{13-2}\binom{52-13-2-8}{13-2}\binom{52-13-2-13-2-8}{13-2}$ for the remaining cards, and $(4\cdot 3\cdot 2)\cdot (4\cdot 3\cdot 2)$ ways for the king and queen to go, So probability is $\frac{24\cdot 24\cdot \binom{52-8}{13-2}\binom{52-13-2-8}{13-2}\binom{52-13-2-13-2-8}{13-2}}{\binom{52}{13}\binom{52-13}{13}\binom{52-13-13}{13}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the sum of the power series below? For $$\sum_{n=1}^{\infty}\frac{(n+2)}{n(n+1)}x^n$$
What is the sum of it?
| Since
\begin{align}
\frac{n+2}{n(n+1)} = \frac{2}{n} - \frac{1}{n+1}
\end{align}
then the series is as follows.
\begin{align}
\sum_{n=1}^{\infty} \frac{(n+2) \ x^{n} }{ n(n+1)} &= 2 \sum_{n=1}^{\infty} \frac{x^{n}}{n} - \sum_{n=2}^{\infty} \frac{x^{n-1}}{n} \\
&= - 2 \ln(1-x) - \frac{1}{x} \left( - \ln(1-x) - x \right) \\
&= \left( \frac{1}{x} - 2 \right) \ln(1-x) + 1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819514",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Trying to evaluate integral$\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx$ I am trying to integrating
$$
\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx\equiv I
$$
but cannot get the answer, I would like a proof not a numerical answer. My attempt at proof:
$$
y=\sqrt{1-e^{-x}}\\
y(0)=0, \ y(\infty)=1\\
y^2=1-e^{-x}\\
2ydy=e^{-x}dx\\
e^{-x}=1-y^2\\
x=\ln\frac{1}{1-y^2} \rightarrow\\
I=2\int_0^1 y^2\ln\frac{1}{1-y^2} dy=\\
-2\int_0^1y^2\ln(1-y^2)\,dy=\\2\int_0^1y^2\sum_{k=1}^\infty \frac{y^{2k}}{k}=\\
2\sum_{k=1}^\infty \frac{1}{k}\int_0^1 y^{2(k+1)}dy=\\
2\sum_{k=1}^\infty \frac{1}{k(3+2k)}=\\
2\sum_{k=1}^\infty \left(\frac{1}{3k}-\frac{2}{3(2k+3)} \right)
$$
but this diverges because $\sum_k\frac{1}{k}\to\infty$? Mistakes I made...
Please help if can on doing the sum or integral. Thank you, Grazie
| By partial integration we have
$$
\int x \sqrt{1-e^{-x}}e^{-x}dx = \frac{2}{3} x (1-e^{-x})^{3/2} -\frac{2}{3}\int dx (1-e^{-x})^{3/2}
$$
now by letting $y=e^{-x}$, $-\ln y =x$, $dx = -dy / y$ we have
$$
\int dx (1-e^{-x})^{3/2} = \int \frac{-dy}{y}(1-y)^{3/2}
$$
now let $\sqrt{1-y} = \tau$, $y= 1-\tau^2$, $dy = -2\tau d \tau$ so
$$
\int \frac{+2\tau d \tau}{1-\tau^2}\tau^3 = 2\int d\tau \frac{\tau^4-\tau^2+\tau^2-1+1}{1-\tau^2} = -2 \int \tau^2 d\tau - 2\int d\tau +2\int \frac{d\tau}{1-\tau^2}=\\
-\frac{2\tau^3}{3}-2\tau + \ln \frac{1+\tau}{1-\tau}.
$$
We have to put everything back together in orderly fashion:
$$
\frac{2}{3}x(1-e^{-x})^{3/2}-\frac{2}{3}\left[-\frac{2}{3}(1-e^{-x})^{3/2}-2\sqrt{1-e^{-x}}+\ln \frac{1+\sqrt{1-e^{-x}}}{1-\sqrt{1-e^{-x}}}\right].
$$
Now, as $x\to 0$, the primitive approaches $0$. As $x\to \infty$ we have to handle an indeterminate form:
$$
\lim_{x\to \infty}\left[\ln\frac{1+\sqrt{1-e^{-x}}}{1-\sqrt{1-e^{-x}}}-x(1-e^{-x})^{3/2}\right]=\\
=\lim_{y\to 0}\left[\ln\frac{1+\sqrt{1-y}}{1-\sqrt{1-y}}+\ln y \cdot (1-y)^{3/2}\right]=\\
=\lim_{y\to 0}\left[\ln\frac{1+1-y/2}{1-1+y/2}+\ln y \cdot (1-3y/2)\right] =\lim_{y\to 0}\ln\frac{(2-y/2)y}{y/2} = \ln 4.
$$
If we have done everything right our result is:
$$
\frac{2}{3}(-\ln 4+2+\frac{2}{3}) = \frac{2}{9}(8-3\ln 4).
$$
| {
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"url": "https://math.stackexchange.com/questions/820635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the limit of $ \lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} $ I need to evaluate the limit without using L'Hopital's rule.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$
| Here is another solution, not using Taylor series, but based on multiplying by conjugates.
I wanted to see if it would work.
Note $x^3-y^3=(x-y)(x^2+xy+y^2)$
so $$\sqrt{x+2}-\sqrt[3]{x+20}=\frac{\sqrt{(x+2)^3}-(x+20)}{x+2+ \sqrt{x+2}\sqrt[3]{x+20}+\sqrt[3]{(x+20)^2}}
$$and
$$\sqrt{(x+2)^3}-(x+20)=\frac{(x+2)^3-(x+20)^2}{\sqrt{(x+2)^3}+(x+20)}$$
and
$$(x+2)^3-(x+20)^2=(x-7)(x^2+12x+56)$$
For the denominator we have
$$\sqrt[4]{x+9}-2=\frac{\sqrt{x+9}-4}{\sqrt[4]{x+9}+2}
=\frac{x-7}{(\sqrt[4]{x+9}+2)(\sqrt{x+9}+4)}
$$
So all together,
$$\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
=\frac{(x^2+12x+56)(\sqrt[4]{x+9}+2)(\sqrt{x+9}+4) }{(x+2+ \sqrt{x+2}\sqrt[3]{x+20}+\sqrt[3]{(x+20)^2})(\sqrt{(x+2)^3}+(x+20))}$$
Now if you substitute $x=7$ you get $\frac{112}{27}$ like the other answers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/821855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
| $1 - x + \sqrt{2 + 2x + x^2} = \dfrac{(1-x)^2 - (2 + 2x + x^2)}{1 - x - \sqrt{2 +2x + x^2}} = \dfrac{-1 - 4x}{1 - x - \sqrt{2 + 2x + x^2}} = \dfrac{-\dfrac{1}{x} - 4}{\dfrac{1}{x} - 1 - \sqrt{\dfrac{2}{x^2} + \dfrac{2}{x} + 1}} \to \dfrac{-4}{-1-1} = 2$ when $x \to \infty$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Sum Involving Bernoulli Numbers : $\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}=\frac{2n-1}{2n+1}$ How can we prove that
$$\sum_{r=1}^n \binom{2n}{2r-1}\frac{B_{2r}}{r}=\frac{2n-1}{2n+1}$$
where $B_{2r}$ are the Bernoulli numbers?
$$\begin{array}{c|c|c|} n & \frac{2n-1}{2n+1} & \sum_{r=1}^n \binom{2n}{2r-1} \frac{B_{2r}}{r} \\ \hline 1 &\frac{1}{3} & \frac{1}{3} \\ 2 &\frac{3}{5} & \frac{3}{5} \\ 3 &\frac{5}{7} &\frac{5}{7} \\ 4 & \frac{7}{9} & \frac{7}{9} \\ 5 & \frac{9}{11} &\frac{9}{11}\end{array}$$
The formula appears to be correct for a lot of values of $n$.
| This identity also has a proof using the technique of annihilated
coefficient extractors (ACE).
First observe that it is equivalent to
$$\sum_{r=1}^n \frac{2n+1}{2r} {2n\choose 2r-1} B_{2r}
= n - \frac{1}{2}.$$
The left simplifies to
$$\sum_{r=1}^n {2n+1\choose 2r} B_{2r}.$$
Introduce the following generating function $f(z)$ for this quantity,
which is
$$f(z) = \sum_{n\ge 1} \frac{z^{2n}}{(2n+1)!}
\sum_{r=1}^n {2n+1\choose 2r} B_{2r}.$$
By the generating function of the Bernoulli numbers we have that
$f(z)$ is
$$\sum_{n\ge 1} \frac{z^{2n}}{(2n+1)!}
\sum_{r=1}^n {2n+1\choose 2r} (2r)! [w^{2r}] \frac{w}{e^w-1}.$$
Switch summations to get
$$\sum_{r\ge 1}
\left( [w^{2r}] \frac{w}{e^w-1} \right)
\sum_{n\ge r} \frac{z^{2n}}{(2n+1-2r)!}$$
which is
$$\sum_{r\ge 1}
\left( [w^{2r}] \frac{w}{e^w-1} \right)
\sum_{n\ge 0} \frac{z^{2n+2r}}{(2n+1)!}.$$
This in turn simplifies to
$$\sum_{r\ge 1}
z^{2r} \left( [w^{2r}] \frac{w}{e^w-1} \right)
\sum_{n\ge 0} \frac{z^{2n}}{(2n+1)!}.$$
The first term is the promised annihilated coefficient extractor and
the second is $\sinh(z)/z$ so we get
$$f(z) = \left(-1 + \frac{1}{2} z + \frac{z}{e^z-1}\right)
\frac{\sinh(z)}{z}.$$
We extract coefficients from the three components. First,
$$(2n+1)! [z^{2n}] \left(-\frac{\sinh(z)}{z}\right)
= -(2n+1)! [z^{2n+1}] \sinh(z) = -1.$$
Second,
$$(2n+1)! [z^{2n}] \left(\frac{1}{2} z \frac{\sinh(z)}{z}\right)
= (2n+1)! [z^{2n}] \frac{1}{2} \sinh(z) = 0.$$
And third,
$$(2n+1)! [z^{2n}] \frac{\sinh(z)}{e^z-1}
= (2n+1)! [z^{2n}] \frac{1}{2}\frac{e^z-e^{-z}}{e^z-1}.$$
This last one needs some rewriting as in
$$\frac{e^z-e^{-z}}{e^z-1}
= 1 + \frac{-e^z + 1 + e^z - e^{-z}}{e^z-1}
\\= 1 + \frac{1 - e^{-z}}{e^z-1}
= 1 + e^{-z} \frac{e^z - 1}{e^z-1} = 1 + e^{-z}.$$
Therefore the third component is
$$(2n+1)! [z^{2n}] \frac{1}{2} (1 + e^{-z})
\\ = (2n+1)! \times \frac{1}{2} \times \frac{(-1)^{2n}}{(2n)!}
= (2n+1)\times \frac{1}{2} = n + \frac{1}{2}.$$
The sum of the three contributions is
$$n + \frac{1}{2} + (0) + (-1)
= n - \frac{1}{2}$$
precisely as was to be shown.
There is another annihilated coefficient extractor at this
MSE link I and yet another one at this MSE link II.
| {
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Given $2^7 \equiv 2 \mod n$ and $3^7 \equiv 3 \mod n$, prove for all $a$ in $\mathbb{Z}$: $a^7 \equiv a \mod n$ Let $n$ be a positive whole number. Given $2^7 \equiv 2 \mod n$ and $3^7 \equiv 3 \mod n$, prove for all $a$ in $\mathbb{Z}$: $a^7 \equiv a \mod n$, without using a computer.
$n$ must be greater than 3 and less than $2^6=64$, so using a computer it is easy to verify that $n$ must be 6,7,14,21 or 42.
| From the given congruences, $n$ is a common factor of
$$\eqalign{
2^7-2=2(2^3+1)(2^3-1)&=2\times3^2\times7\cr
\hbox{and}\quad 3^7-3=3(3^3+1)(3^3-1)&=2^3\times3\times7\times13\ ,\qquad\cr}$$
so $n\mid 2\times3\times7$. Since for any $a$ we have
$$\eqalign{
a^2\equiv a\pmod2\quad\Rightarrow\quad a^7\equiv(a^2)^3a\equiv a^4\pmod2\quad\Rightarrow\quad &a^7\equiv a\pmod2\cr
a^3\equiv a\pmod3\quad\Rightarrow\quad a^7\equiv(a^3)^2a\equiv a^3\pmod3
\quad\Rightarrow\quad &a^7\equiv a\pmod3\cr
&a^7\equiv a\pmod 7\ ,\cr}$$
it follows that $a^7\equiv a$ modulo any number which is a product of $2,3$ and $7$ at most once each. All the $n$ we are considering are of this form.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration by parts - $\int \ln (2x+1) \text{dx}$ Use integration by parts to find
$\int \ln (2x+1) \text{dx}$.
So far I have:
$$x\ln(2x+1)-\int\dfrac{2x}{2x+1}dx+c$$
Using integration by substitution to find the integral
$$u=2x+1\Rightarrow\text{du}=2\text{dx}$$
$$\int\dfrac{2x}{2x+1}\cdot\dfrac{1}{2}\text{du}=\int xu^{-1}$$
$$=\int \left(\dfrac{u}{2}-\dfrac{1}{2}\right)u^{-1}\text{du}=\int\left[\dfrac{1}{2}-\dfrac{1}{2}u^{-1}\right]\text{du}$$
$$=\dfrac{1}{2}x-\dfrac{1}{2} \ln \left|2x+1\right|$$
Looking at the answer in the back, this is wrong.
The answer is $x \ln(2x+1)-x+\dfrac{1}{2}\ln(2x+1)+c$.
What have I done wrong?
| You made a mistake when evaluating the last integral. One approach to evaluate the integral of $\tfrac{2x}{2x+1}$ is to write it as: $$\eqalign{\int\dfrac{2x}{2x+1}\mathrm dx&=\int\dfrac{-1+(2x+1)}{2x+1}\mathrm dx\\&=\int\left[\dfrac{-1}{2x+1}+\dfrac{2x+1}{2x+1}\right]\mathrm dx\\
&=\int\left[\dfrac{-1}{2x+1}+1\right]\mathrm dx\\
&=\int\dfrac{-1}{2x+1}\mathrm dx+\int1\,\mathrm dx\\
&=\int-\dfrac{1}{2x+1}\mathrm dx+x.\\
&=-\int\dfrac{1}{2x+1}\mathrm dx+x.\\
}$$
To evaluate the remaining integral use the substitution $u=2x+1$, then $\mathrm du=2\,\mathrm dx\ldots$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove the continuity and differentiability of parametric integration $$F(\alpha )=\int_{0}^{+ \infty } \frac{\cos x}{1+(x+\alpha )^{2} } dx$$
Prove the function F is continuous and differentiable on the interval $[0, +\infty )$
| $|F(\alpha) - F(\beta)| \leq \displaystyle \int_{0}^\infty \left|\dfrac{cosx}{1+(x+\alpha)^2} - \dfrac{cosx}{1+(x+\beta)^2}\right|dx \leq |\beta - \alpha|\cdot \displaystyle \int_{0}^\infty \dfrac{(x+\alpha)+(x+\beta)}{(1+(x+\alpha)^2)(1+(x+\beta)^2)}dx \leq |\alpha - \beta|\cdot \left(\displaystyle \int_{0}^\infty \dfrac{(x+\alpha)}{(1+(x+\alpha)^2)(1+(x+\beta)^2)}dx + \displaystyle \int_{0}^\infty \dfrac{(x+\beta)}{(1+(x+\alpha)^2)(1+(x+\beta)^2)}dx\right) \leq \dfrac{|\alpha - \beta|}{2}\cdot \left(\displaystyle \int_{0}^\infty \dfrac{1}{1+(x+\beta)^2}dx + \displaystyle \int_{0}^\infty \dfrac{1}{1+(x+\alpha)^2}dx\right) = \dfrac{|\alpha - \beta|}{2}\cdot \left(\displaystyle \int_{\beta}^\infty \dfrac{1}{1+x^2}dx + \displaystyle \int_{\alpha}^\infty \dfrac{1}{1+x^2}dx\right) = \dfrac{|\alpha - \beta|}{2}\cdot \left(\dfrac{\pi}{2} - tan^{-1}\beta + \dfrac{\pi}{2} - tan^{-1}\alpha\right) \leq \dfrac{|\alpha - \beta|}{2}\cdot \left(\dfrac{\pi}{2} + \dfrac{\pi}{2} + \dfrac{\pi}{2} + \dfrac{\pi}{2}\right) = \pi\cdot |\alpha - \beta|$. This shows $F$ is Lipschitz and therefore is uniformly continuous, hence continuous on the indicated interval. You can easily show it is differentiable there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving the exponential inequality: $x^y+y^x\gt1$ How can the following inequality be proven?
$$x^y+y^x\gt1$$
for $(x\gt0,y\lt1)$
| Use this Bernoulli inequality
$$(1+x)^a\le 1+ax,0<a<1,x>-1$$
It is clear we only prove $0<x,y<1$
then
$$x^y=\dfrac{1}{\left(\dfrac{1}{x}\right)^y}=\dfrac{1}{\left(1+\dfrac{1-x}{x}\right)^y}>\dfrac{1}{1+\dfrac{1-x}{x}\cdot y}=\dfrac{x}{x+y-xy}>\dfrac{x}{x+y}$$
Similarly
$$y^x>\dfrac{y}{x+y}$$
so
$$x^y+y^x>\dfrac{x}{x+y}+\dfrac{y}{x+y}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/827754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
} |
Solving $x^2 - x - 1 > 0$ I am having problems understanding how to solve:
$ x^2 - x - 1 > 0 $.
Any help would be much appreciated.
| Since $x^2-x-1=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}$ we must have $|x-\frac{1}{2}|>\frac{\sqrt{5}}{2}$ so $x<\frac{1-\sqrt{5}}{2}$ or $x>\frac{1+\sqrt{5}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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How do you solve two equal absolute value expressions? I'm having trouble understanding how the following is solved.
$$|x+1| = |x-2|$$
| So we know what absolute value means. It means that we want the positive version of $x+1$ and $x-2$.
So we don't know what the signs of $x-2$ and $x+2$ are.
Thus, we can make four different equalities, of which $0$, $1$, $2$, $3$, or $4$ may be true. There may be multiple values of $x$ or there might not be any value of $x$ that satisfies.
First, $x+1 = x-2$ simply implies that $x+3 = x$ which is not possible.
Secondly, $-(x+1) = x-2$ implies that $2x=1$, so $x = 1/2$.
Thirdly, $(x+1)=-(x-2)$. This case is the same as the second case because we can multiply both sides by $-1$.
Fourthly, $-(x+1)=-(x-2)$ implies that $x+1=x-2$, which is the same as the first case.
Thus, the only possible answer is in cases 2 and 3 for which $x = 1/2$. To check, note that $1+1/2 = 3/2$ and $1/2-2= -3/2$. $3/2$ and $-3/2$ have the same absolute value.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the inhomogeneous solution $x_{n+2} = x_{n+1} + 20x_n + n^2 + 5^n \text{ with } x_0 = 0 \text{ and } x_1 = 0$
How would I find the inhomogeneous solution for this since the homogenous solution is 0 given initial conditions?
| Define $y_n = x_{n+1} - 5x_n$
Then $y_{n+2} = y_{n+1} + 20 y_n + ((n+1)^2-5n^2) + 5^{n+1} - 5^{n+1} = y_{n+1} + 20 y_n + (-4n^2+2n+1)$.
There, that "erased" the $5^n$ term in the recurrence. The $n^2.1^n$ term spilled a bit, but no biggie, we can do the same to those, by putting $z_n = y_{n+1} - y_n$, then $u_n = z_{n+1} - z_n$ and finally $v_n = u_{n+1} - u_n$. Each time the term of the form $P(n).1^n$ loses a degree, until it disappears completely.
Hence $v_{n+2} = v_{n+1} + 20v_n$. Now if you unfold back everything, you obtain a big recurrence relation on $x_{n+1}$, whose characteristic polynomial is $(X^2-X-20)(X-1)^3(X-5) = (X+4)(X-1)^3(X-5)^2$.
From there you should know that $x_n$ should be of the form $a.4^n + (bn+c).5^n + (dn^2+en+f)$
| {
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Proving Definite Integral Strong Inequation I got the following question:
Prove:
$$\int^{1}_0 \frac{x^9}{\sqrt{1+x}} < \frac{1}{10}$$
I proved the weak inequation like so:
$x \geq 0 \Rightarrow \sqrt{1+x} \geq 1$ (monotonicity of square root)
$\Rightarrow \frac{x^9}{\sqrt{1+x}} \leq x^9$
$\Rightarrow \int^{1}_0 \frac{x^9}{\sqrt{1+x}} \leq \int^{1}_0 x^9 = \frac{x^{10}}{10}|_0^1 = \frac{1}{10}$ (integral monotonicity)
However, I am stumped as how to prove a strong inequation here. There doesn't seem to be a number less than $\frac{1}{10}$ that I can limit this function to, so that technique doesn't work, and finding the actual integral is not an option.
| If you like something stronger, consider that $\frac{1}{\sqrt{1+x}}$ is a convex function over $[0,1]$, hence:
$$\frac{1}{\sqrt{1+x}}\leq 1-\left(1-\frac{1}{\sqrt{2}}\right)x\tag{1}$$
and:
$$\int_{0}^{1}\frac{x^9}{\sqrt{1+x}}\,dx\leq \int_{0}^{1}x^9\,dx-\left(1-\frac{1}{\sqrt{2}}\right)\int_{0}^{1}x^{10}\,dx = \frac{1}{10}-\frac{1}{11}\left(1-\frac{1}{\sqrt{2}}\right),$$
so:
$$\int_{0}^{1}\frac{x^9}{\sqrt{1+x}}\,dx\leq\frac{1}{11\sqrt{2}}+\frac{1}{110}<\frac{1}{13}.\tag{2}$$
| {
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} |
Prove that limit $\lim_{n\to\infty}\sqrt{4+\frac{1}{n^2}}+\sqrt{4+\frac{2}{n^2}}+\cdots+ \sqrt{4+\frac{n}{n^2}}-2n=\frac{1}{8}$
Let $$a_{n}=\sqrt{4+\dfrac{1}{n^2}}+\sqrt{4+\dfrac{2}{n^2}}+\cdots+
\sqrt{4+\dfrac{n}{n^2}}-2n,$$
show that
$$\lim_{n\to\infty}a_{n}=\dfrac{1}{8}$$
My attempt:
Since
$$\sqrt{4+\dfrac{i}{n^2}}-2=\dfrac{\dfrac{i}{n^2}}{\sqrt{4+\dfrac{i}{n^2}}+2}=\dfrac{1}{\sqrt{4n^2+i}+2}\dfrac{i}{n}$$
then I can't work it. Thank you.
| Let $0 < \alpha < \frac{1}{4}$ then for $n$ large enough and $k \in \{1,\ldots,n\}$ $$2 + \frac{\alpha \, k}{n^2} < \sqrt{4+\frac{k}{n^2}} < 2 + \frac{k}{4 n^2}.$$ The limit now follows from $$1+2+\ldots+n = \frac{n(n+1)}{2}$$ and taking $\alpha \uparrow \frac{1}{4}$.
| {
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Find all positive integer solutions $(x,y,z)$ that satisfy $5^x \cdot 7^y +4= 3^z$? This is another contest math-problem.
The only problem that I cannot find the way to tackle this problem.
Can anybody try to provide the solution to solve this problem?
Thanks
| Given; $5^x \cdot 7^y +4= 3^z$
we get, $(-1)^x +1 \equiv 0 \pmod3$ hence $x$ must be odd, again, $(-1)^y \equiv (-1)^z \pmod 4$ hence $y$ and $z$ have the same parity. Again $-1 \equiv 3^z \pmod 5$ or $1 \equiv 3^{2z} \pmod 5$ hence by Euler-Fermat theorem $2|z$. Again since $y$ and $z$ are even we have, $5^x \equiv 4 \pmod 8$, since $x$ is odd, we have two cases; either $x=4k+1$ or $x=4k+3$, note that $5^{4k} \equiv 1 \pmod 8$, the two cases each yield $5^x \equiv 5 \pmod 8$ a contradiction, hence $x$ cannot be greater than $0$, i.e $x=0$. The given equation then becomes $7^{2m} +4= 3^{2n}$ or $(7^m)^2 +2^2= (3^n)^2$, a pythagora's equation with solution, $2=2uv$, $7^m=u^2-v^2$ and $3^n=u^2+v^2$ from which it is obvious that $m$ and $n$ do not exist, and hence, $y(=2m)$ and $z(=2n)$ do not exist
| {
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From any arbitrary point $P$ on $y =\cos x$ tangents $PA$ and $PB$ are drawn to a circle which passes through From any arbitrary point $P$ on $y =\cos x$ tangents $PA$ and $PB$ are drawn to a circle which passes through the points $(1,0)$ and $(3,0)$ and touches the circle $x^2+y^2-2x-8=0$ and have its centre in first quadrant. Find locus of circumcentre of $\triangle PAB$
My approach :
The given circle is having centre at $(1,0)$ and radius $3$. Please suggest how to proceed in this problem. thanks.
| Given Circle
The fixed circle described in your problem is actually
$$(x-2)^2+(y-\frac{\sqrt{5}}{2})^2=\frac{9}{4}$$
This is because its tangent circle is $(x-1)^2+y^2=9$ is centered in $(1,0)$ which lies on the given circle, which means that its diameter is equal to the radius of the tangent circle, so it is $\frac{3}{2}$.
The $x$ coordinate $a$ of the center is 2 because $(1,0)$ and $(3,0)$ lie on the circle.
The $y$ coordinate $b$ of the center can be found by plugging $(1,0)$ into the equation $(x-2)^2+(y-b)^2=\frac{9}{4}$.
Circumcenter
The circumcenter of $\triangle PAB$ is the midpoint of $PO$ where $O=(a,b)$ is the center of the given circle. This is because $\angle PAO=\frac{\pi}{2}$.
Locus
For the point $P=(x,y=cos(x))$, the circumcenter is $C=(\frac{x+a}{2},\frac{y+b}{2})$ so its locus is a modified cosine line:
$$ Y = \frac{\cos(2X-a)+b}{2}$$
where $a=2$ and $b=\frac{\sqrt{5}}{2}$.
| {
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Linear inequalities with one unknown $$3x+\frac{2}{4} \leq x+\frac{7}{2}$$ What is the solution to the inequality?
I started multiplying both sides by 4
which gave me $3x+2\leq14$.
Then I subtracted two from both sides obtaining
$$3x\leq12$$
which then I divided by 3 to obtain the solution
$$x\leq4\ .$$
This solution does not match for my possible answers on my practice guide.
| Just subtract $x+\frac{2}{4}$ on both sides...
$$3x+\frac{2}{4} \leq x+\frac{7}{2} \iff 2x \leq \frac{7}{2}-\frac{2}{4} = \frac{14}{4}-\frac{2}{4} = \frac{12}{4} = 3$$
and finally divide both sides by $2$ to find $x \leq \frac{3}{2}$.
Note:
I started multiplying both sides by $4$ which gave me $3x+2\leq 14$.
You should have got $12x+2 \leq 4x+14 \iff 8x \leq 12$...
| {
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Easier Proof of $\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$ I am curious to see whether anybody can give me a proof that takes less steps.
Here is how I did it:
$$\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$$
LHS $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + (\cos^2\theta - \sin^2\theta)\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + \sin\theta(\cos^2\theta - \sin^2\theta + 1)\\
&= \sin2\theta\cos\theta + \sin\theta(2\cos^2\theta)\\
&= \sin2\theta\cos\theta + 2\sin\theta\cos^2\theta\\
&= \sin2\theta\cos\theta + \cos\theta(\sin\theta\cos\theta + \sin\theta\cos\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= 2\sin2\theta\cos\theta.}$$
| Use the trigonometric identity about $\sin A + \sin B= 2\sin \frac{A+B}{2} \cos\frac{A-B}{2}$.
| {
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Expressing sums of powers I am trying to solve this question but I'm not really understanding how to continue, I would greatly appreciate some kind of tip.
The Question
The formula for $1^2 + \cdots + n^2$ may be derived as follows. We begin with the formula $$(k+1)^3 - k^3 = 3k^2 + 3k + 1$$
Writing this formula for $k=1,\ldots,n$ and adding we obtain an expression.
Thus we can find $\sum_{k=1}^{n}k^2$ if we already know $\sum_{k=1}^n k$.
Use this method to find $1^3 + \cdots + n^3$.
I am confused about this method. How did they get the formula to begin with, and what formula am I supposed to use for $1^3 + \cdots + n^3$?
EDIT: I noted that they did already hint at $k^2$ being expressed from $k^1$, but we can't square $k^1$ into $k^3$
| The formula is the Binomial Theorem, or just multiplication: $(k+1)^3=k^3+3k^2+3k+1$. This can be rewritten as
$$(k+1)^3-k^3=3k^2+3k+1.$$
The corresponding formula for fourth powers is $(k+1)^4=k^4+4k^3+6k^2+4k+1$. It yields the identity
$$(k+1)^4-k^4=4k^3+6k^2+4k+1.$$
That can be used to give a telescoping argument for the sum of the first $n$ cubes, much like the telescoping argument for the sum of the first $n$ squares. Sum both sides from $k=1$ to $k=n$. On the right, there is almost total cancellation, and we get
$$(n+1)^4-1=4\sum_1^n k^3+6\sum_1^n k^2+4\sum_1^n k +\sum_1^n 1.\tag{1}$$
Now a fair bit of messy algebra gets us $\sum_1^n k^3$, since we have formulas for every other sum in (1).
In general, the Binomial Theorem is the assertion that if $m$ is a positive integer, then
$$\small (x+y)^m=\binom{m}{0}x^m +\binom{m}{1}x^{m-1}y+\binom{m}{2}x^{m-2}y^2+\cdots +\binom{m}{m-1}xy^{m-1}+\binom{m}{m}y^m.$$
Remark: It will turn out that
$$1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2.$$
Once we have guessed this formula, it can be proved more easily by induction.
| {
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Solvability of a variant Fermat equation If
$p$
is an odd prime,
are there any pairwise coprime integers
$a, b, c > 0$
such that
$$(a-b+c)^{p} + (a+b-c)^{p} = (a+b+c)^{p}$$
[a proof without Fermat's last theorem]
| Fermat's Last Theorem states that there are no non-trivial (i.e. $x,y,z\neq 0$) positive integer solutions to $x^n+y^n=z^n$ for $n\geq 3$. The first odd prime is 3, so there are no solutions to your equation.
The only possibilities are for at least one of $a-b+c$, $a+b-c$, or $a+b+c$ to be non-positive. If one is zero, all must be zero, but $a+b+c>0$ since $a,b,c>0$.
Thus, we need only check the case when some of them are negative. Since $a,b,c>0$, it follows that $a+b+c>0$. Thus, with $x=a-b+c$, $y=a+b-c$ and $z=a+b+c$, we know that $z^p>0$. Thus, it cannot be that both $x,y<0$ as $x^p,y^p<0$ and $x^p+y^p<0<z^p$. Thus, only one of $x$ or $y$ may be negative. Suppose without $x<0$. Then $y^p=(-x)^p+z^p$ contradicts Fermat's Last Theorem. Similarly, if $y<0$, then $x^p=(-y)^p+z^p$, again giving us a contradiction.
For this reason, there can be no solutions.
| {
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Extrema of $f(x)=\frac{\sin (5x)} 5 - \frac{2\sin(3x)} {3} + \sin (x)$. (a) I need help in finding maxima and minima of the following funcion:
$$f(x)=\frac{\sin (5x)} 5 - \frac{2\sin(3x)} {3} + \sin (x)$$
therefore I need to find the roots of $f'(x)=\cos(5x)-2\cos(3x)+\cos(x)$.
(b) I need to find the minima of $$f(x)=4x+\frac{9\pi^2} x + \sin x.$$
How can I find the roots of $$f'(x)=4-\frac{9\pi^2 }{x^2} + \cos x?$$
Thank you.
| Hints:
a) $\cos(5x) \equiv 16\cos^5(x)-20\cos^3(x)+5\cos(x)$
$\cos(3x) \equiv 4\cos^3(x)-3\cos(x)$
So, we've got:
Now, let $c \equiv \cos(x)$.
We've got:
$$[16c^5-10c^3+5c]-2[4c^3-3c]+c=0 \iff \boxed{2c\underbrace{[8c^4-9c^2+6=0]}_{\textrm{quadratic in} \ c^2}} \tag{A}$$
Solve $(A)$!
b) Solving this analytically is very hard, so use the bisection method (noting that $4-\frac{9\pi^2}{x^2}=\cos(x)=0$ is continuous on $\mathbb{R \setminus\{0\}}$).
The exact answer to b) is $\boxed{\pm \frac{3\pi}{2}}$.
Now, your job is to evaluate the second derivative of these extrema, in order to determine their nature (maxima or minima).
| {
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Real part of Complex Function I've this function
$$f(k,\theta) = \frac{1}{k}\frac{1}{\cot\delta_0(k) -i }$$
and i know that $k\cot\delta_0(k) = -\frac{1}{a} + \frac{1}{2}r_ek^2 + \cdots$
it is an expansion.
How can i get that the $\Re{f(k,\theta)} = -a + a^2(a-\frac{1}{2}r_e)k^2$ ?
| You probably already solved this by yourself, but to close off this question with an answer, I followed through on my comment.
$$
\Re f(k,\theta) = \frac{1}{k}\frac{\cot \delta_0(k)}{\cot^2 \delta_0(k) +1}
$$
multiplying top and bottom by $k^2$ we obtain
$$
\Re f(k,\theta) =\frac{k\cot \delta_0(k)}{k^2\cot^2 \delta_0(k) +k^2}
$$
using the expansion for $k\cot \delta_0(k) = -\frac{1}{a} + \frac{1}{2}r_ek^2$ the denominator becomes
$$
k^2\cot^2 \delta_0(k) +k^2 = \frac{1}{a^2} - \frac{1}{a}r_ek^2 + k^2 = \frac{1}{a^2} +\left(1 -\frac{r_e}{a}\right)k^2
$$
therefore
$$
\Re f(k,\theta) = \frac{-\frac{1}{a} + \frac{1}{2}r_ek^2}{\frac{1}{a^2} +\left(1 -\frac{r_e}{a}\right)k^2} = \frac{-a +\frac{a^2}{2}r_ek^2}{1 +a^2\left(1 -\frac{r_e}{a}\right)k^2}
$$
using
$$
\frac{1}{1+x} = 1 - x +...
$$
we obtain
$$
\Re f(k,\theta) = \left(-a +\frac{a^2}{2}r_ek^2\right)\left(1 - a^2\left(1 -\frac{r_e}{a}\right)k^2 + ..O(k^4)\right)\\
= -a +a^3\left(1 -\frac{r_e}{a}\right)k^2 + \frac{a^2}{2}r_ek^2\\
=-a + a^3k^2 -a^2r_ek^2 + \frac{a^2}{2}r_ek^2 = -a +a^3k^2 -\frac{a^2}{2}r_ek^2
$$
This yields the result as required
$$
\Re f(k,\theta) = -a +a^2\left(a - \frac{r_e}{2}\right)k^2.
$$
| {
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Given a sequence $ (a_{n})_{n \in \mathbb{N}} $, find $ \sum_{n = 1}^{\infty} \frac{1}{a_{n} + 2} $. I would appreciate it if somebody could help me with the following problem:
If the sequence $ (a_{n})_{n \in \mathbb{N}} $ satisfies
$$
a_{1} = 1
\quad \text{and} \quad
\forall n \in \mathbb{N}, n \ge 2: \quad
a_{n} = \frac{(a_{1} + 2) (a_{2} + 2) \cdots (a_{n-1} + 2)}{2^{n}},
$$
find $ \displaystyle \sum_{n = 1}^{\infty} \frac{1}{a_{n} + 2} $.
| Since $a_1=1$ and
$$
a_{n+1}=\prod_{k=1}^n\left(1+\frac{a_k}2\right)\tag{1}
$$
we have that
$$
a_n\ge1\tag{2}
$$
and
$$
a_{n+1}=a_n\left(1+\frac{a_n}2\right)\tag{3}
$$
Subtracting $a_n$ from both sides of $(3)$ gives
$$
a_{n+1}-a_n=\frac{a_n^2}2\tag{4}
$$
and since $a_1=1$, $(1)$ and $(2)$ show that
$$
a_n\ge\left(\frac32\right)^{n-1}\tag{5}
$$
Using $(3)$ and then $(4)$ yields
$$
\begin{align}
\frac1{a_n+2}
&=\frac12\frac{a_n}{a_{n+1}}\\
&=\frac{a_{n+1}-a_n}{a_{n+1}a_n}\\
&=\frac1{a_n}-\frac1{a_{n+1}}\tag{6}
\end{align}
$$
Summing $(6)$ yields
$$
\begin{align}
\sum_{k=1}^n\frac1{a_k+2}
&=\sum_{k=1}^n\left(\frac1{a_k}-\frac1{a_{k+1}}\right)\\
&=\frac1{a_1}-\frac1{a_{n+1}}\tag{7}
\end{align}
$$
Taking the limit of $(7)$ and applying $(5)$ gives
$$
\begin{align}
\sum_{k=1}^\infty\frac1{a_k+2}
&=\frac1{a_1}\\
&=1\tag{8}
\end{align}
$$
Original Sequence
It has been pointed out that the recursion was and is again different than when I wrote my answer. We have
$$
a_1=1, a_2=\frac34, a_3=\frac{33}{32}
$$
$(3)$ and $(4)$ still hold, but since $(2)$ only holds for $n\ge3$, $(5)$ should read
$$
a_n\ge\left(\frac32\right)^{n-3}
$$
and $(6)$ only holds for $n\ge2$. Furthermore, $(7)$ should read
$$
\begin{align}
\sum_{k=2}^n\frac1{a_k+2}
&=\sum_{k=2}^n\left(\frac1{a_k}-\frac1{a_{k+1}}\right)\\
&=\frac1{a_2}-\frac1{a_{n+1}}
\end{align}
$$
and $(8)$ should read
$$
\begin{align}
\frac1{a_1+2}+\sum_{k=2}^\infty\frac1{a_k+2}
&=\frac1{a_1+2}+\frac1{a_2}\\
&=\frac53
\end{align}
$$
As it is in JimmyK4542's answer.
| {
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Contour integral $\int^\pi_{-\pi}(a-\cos\theta)^b\exp(c\cos\theta)d\theta$ assuming $a>1$, $b>0$, $c>0$ Under the condition $a>1$, $b>0$, $c>0$, is there any good function to express the following integral?
$$
\int^\pi_{-\pi}\left(a-\cos\theta\right)^b\exp\left(c\cos\theta\right)d\theta
$$
I guess some hypergeometric function can do this.
| $\int_{-\pi}^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$
$=\int_{-\pi}^0(a-\cos\theta)^be^{c\cos\theta}~d\theta+\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$
$=\int_\pi^0(a-\cos(-\theta))^be^{c\cos(-\theta)}~d(-\theta)+\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$
$=\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta+\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$
$=2\int_0^\pi(a-\cos\theta)^be^{c\cos\theta}~d\theta$
$=2\int_1^{-1}(a-x)^be^{cx}~d(\cos^{-1}x)$
$=2\int_{-1}^1\dfrac{(a-x)^be^{cx}}{\sqrt{1-x^2}}dx$
$=2\int_0^2\dfrac{(a-(x-1))^be^{c(x-1)}}{\sqrt{1-(x-1)^2}}d(x-1)$
$=2e^{-c}\int_0^2x^{-\frac{1}{2}}(2-x)^{-\frac{1}{2}}(a+1-x)^be^{cx}~dx$
$=2e^{-c}\int_0^1(2x)^{-\frac{1}{2}}(2-2x)^{-\frac{1}{2}}(a+1-2x)^be^{2cx}~d(2x)$
$=2(a+1)^be^{-c}\int_0^1x^{-\frac{1}{2}}(1-x)^{-\frac{1}{2}}\left(1-\dfrac{2x}{a+1}\right)^be^{2cx}~dx$
$=2\pi(a+1)^be^{-c}~\Phi_1\left(\dfrac{1}{2},-b,1;\dfrac{2}{a+1},2c\right)$ (according to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions)
| {
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Trigonometric Triangle Equality $A, B, C$ are the angles of a triangle then $tan^2(A/2)+tan^2(B/2)+tan^2(C/2)$ is
always greater than what integral value.
| Since $A,B,C$ are angles of a triangle, we have $0<\dfrac{A}{2},\dfrac{B}{2},\dfrac{C}{2}<\dfrac{\pi}{2}$. For this range of values, $\tan^2 x$ is a convex function. Hence, from Jensen's inequlaity,
$$\tan^2\left(\frac{1}{3}\frac{A}{2}+\frac{1}{3}\frac{B}{2}+\frac{1}{3}\frac{C}{2}\right) \leq \frac{1}{3}\tan^2\frac{A}{2}+\frac{1}{3}\tan^2\frac{B}{2}+\frac{1}{3}\tan^2\frac{C}{2}$$
$$\tan^2\left(\frac{A+B+C}{6}\right) \leq \frac{1}{3}\left(\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2}\right)$$
Since $A+B+C=\pi$, hence
$$\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2} \geq 1$$
$\blacksquare$
| {
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Limit of a matrix multiplication How can I calculate this limit:
$\displaystyle\lim_{n\to\infty}\begin{bmatrix}0.9 & 0.2\\0.1 & 0.8\end{bmatrix}^n$
What is the tool that i need to aply? eigenvalues and eigenvectors? diagonalization? canonical form?
(This came in a contest and was the only problem i cannont have an idea for solve it).
| Because that is a stochastic matrix you just need to fine the fixed probability vector or steady state vector.
$$(t1,t2)\cdot\left(
\begin{array}{cc}
0.9 & 0.1 \\
0.2 & 0.8 \\
\end{array}
\right)=(t1,t2)$$
Notice I have transposed A for convenience and this yields the simultaneous set of equations
$$.9 \cdot t1+.2 \cdot t2=\text{t1}$$
$$.1 \cdot t1+.8 \cdot t2=\text{t2}$$
$$t1 + t2 = 1$$
this is easily solved to get
$t1=\frac{2}{3}$
$t2=\frac{1}{3}$
Because this is a regular Markov chain all the rows of $A^{\infty} $ are equal to $(t1,t2)$, so
$$A^{\infty} = \left(
\begin{array}{cc}
\frac{2}{3} & \frac{1}{3} \\
\frac{2}{3} & \frac{1}{3} \\
\end{array}
\right)$$
Transpose A back:
$$A^{\infty} = \left(
\begin{array}{cc}
\frac{2}{3} & \frac{2}{3} \\
\frac{1}{3} & \frac{1}{3} \\
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/843994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Diagonalization of a Toeplitz matrix Let $0<\lambda\leq1$ so that the $n \times n$ matrix
$$\Sigma = \begin{pmatrix}
1&1-\lambda& \cdots &1-\lambda\\
1-\lambda&\ddots&\ddots& \vdots\\
\vdots &\ddots&\ddots&1-\lambda\\
1-\lambda&\cdots&1-\lambda&1\\
\end{pmatrix}$$
is positive definite. I believe we can orthogonally diagonalize $\Sigma$ as
$$\Sigma = VDV^T$$
where
$$ V = \begin{pmatrix}
\frac{-1}{\sqrt{2 \cdot 1}} & \frac{-1}{\sqrt{3 \cdot 2}}&\cdots&\cdots&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\
0&\cdots&\cdots&0&\sqrt{\frac{n-1}{n}}&\frac{1}{\sqrt{n}}\\
0&\cdots&0&\sqrt{\frac{n-2}{n-1}}&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\
\vdots&\cdots&\cdots&\frac{-1}{\sqrt{(n-1)(n-2)}}&\vdots&\vdots\\
0&\cdots&\cdots&\vdots&\vdots&\vdots\\
\sqrt{\frac{1}{2}}&\frac{-1}{\sqrt{3 \cdot 2}}&\cdots&\frac{-1}{\sqrt{(n-1)(n-2)}}&\frac{-1}{\sqrt{n(n-1)}}&\frac{1}{\sqrt{n}}\\
\end{pmatrix}$$
$$D = \begin{pmatrix}
\lambda&0& \cdots &0\\
0&\ddots&\ddots& \vdots\\
\vdots &\ddots&\lambda&0\\
0&\cdots&0&n-(n-1)\lambda\\
\end{pmatrix}$$
I am having some trouble showing this result, can someone offer a suggestion for the proof?
| We can rewrite $\Sigma$ as $(1 - \lambda)M + \lambda I$, where $M$ is given by
$$
M = \pmatrix{
1&\cdots & 1\\
\vdots &\ddots&\vdots\\
1&\cdots &1
}
$$
It follows that $V^T\Sigma V$ will be diagonal iff $V^T M V$ will be diagonal. And so, it suffices to show that
$M = VDV^T$ with
$$
D = \pmatrix{
0&&&\\\
&\ddots&&\\
&&0&\\
&&&n
}
$$
Which is to say that $MV = VD$.
In fact, all we need to say in order to show this is the case is that $M$ is symmetric with rank $1$. From there, it suffices to show that $v$, the last column of $V$, is length one and satisfies $Mv = n\cdot v$, and that all other columns are length one, mutually orthogonal, and orthogonal to $v$.
| {
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} |
Apostol (6.22.10): Finding $\int \frac{\arcsin x}{x^2} dx$ I'm trying to solve another integral from Apostol (Chapter 6, Section 6.22, Question 10) which says to show the following:
$$
\int \frac {\arcsin x}{x^2}dx = \log|{\frac {1-\sqrt{1-x^2}}{x}}| - \frac {\arcsin x}{x} +C
$$
I have tried the following:
With $\int \frac {\arcsin x}{x^2}dx$ I started using integration by parts, choosing $u = \arcsin x $ and $dv = 1/x$ which lead to
$$
\int \frac {\arcsin x}{x^2}dx = -\frac{\arcsin x}{x} + \int \frac{dx}{x\sqrt{1-x^2}}
$$
$$
= -\frac{\arcsin x}{x} + \int \frac{x.dx}{x^2.sqrt{1-x^2}}
$$
With the Integral on the RHS, I set $u^2 = 1 - x^2$ which implied $x^2 = 1- u^2$ and $xdx = -u. du$. This meant
$$
\int \frac{xdx}{x^2\sqrt{1-x^2}} = \int \frac{-u du}{u (1-u^2)} = \int \frac{-du}{ 1-u^2} = -\int \frac{du}{1-u^2}
$$
For the integral $\int \frac{du}{1-u^2}$, I set $u = \sin \theta$ which meant $du = \cos\theta. d\theta$ which implied
$$
\int \frac{du}{1-u^2} = \int \frac{\cos\theta}{1 -\sin^2\theta}d\theta = \int\sec\theta\,d\theta = \log|\sec \theta + \tan \theta| = \log|\frac{1+\sin \theta}{\cos \theta}|
$$
Now, since $u = \sin \theta$ in the last substitution and $\sin^2\theta + \cos^2\theta = 1$ it follows that $\cos\theta = \sqrt{1-u^2}$ which means
$$
\log|\frac{1+\sin \theta}{\cos \theta}| = \log |\frac{1+u}{\sqrt{1-u^2}}|
$$
$$
= \log|\frac{1+u}{\sqrt{1-u}\sqrt{1+u}}|
$$
$$
= \log|\frac{\sqrt{1+u}}{\sqrt{1-u}}| = \frac{1}{2}\log|\frac{1+u}{1-u}|
$$
That is
$$
\int \frac{du}{1-u^2} =\frac{1}{2}\log|\frac{1+u}{1-u}|
$$
Now, in the first substitution $u^2 = 1 - x^2$ implies $u = \sqrt{1-x^2}$ which means
$$
\frac{1}{2}\log|\frac{1+u}{1-u}| = \frac{1}{2}\log|\frac{1+ \sqrt{1-x^2}}{1 - \sqrt{1-x^2}}|
$$
$$
= \frac{1}{2}\log|\frac{(1+\sqrt{1-x^2})(1+\sqrt{1-x^2})}{(1 - \sqrt{1-x^2})(1+\sqrt{1-x^2})}|
$$
$$
= \frac{1}{2}\log|\frac{2 - x^2 +2\sqrt{1-x^2}}{x^2}|
$$
$$
= \log|\frac{1+\sqrt{1-x^2}}{x}|
$$
That is
$$
\int \frac{xdx}{x^2\sqrt{1-x^2}} = - \log|\frac{1+\sqrt{1-x^2}}{x}|
$$
with respect to the original integral means
$$
\int \frac {\arcsin x}{x^2}dx = -\frac{\arcsin x}{x} - \log|\frac{1+\sqrt{1-x^2}}{x}| + C
$$
which is not the answer as stated in Apostol's book. I've probably made some error in the calculations which I just haven't been able to find, so if someone can point out the error in my calculations that would be appreciated
| Note that $(1+\sqrt{1-x^2})(1-\sqrt{1-x^2}) = 1 - (1-x^2) = x^2$, so
$$\log\left|\dfrac{1 + \sqrt{1-x^2}}{x}\right| = \log \left|\dfrac{x}{1-\sqrt{1-x^2}}\right| = - \log \left|\dfrac{1-\sqrt{1-x^2}}{x}\right|$$
You and Apostol have the same answer after all.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Distribution of a binomial variable squared If I know $X$ is a binomial random variable, how can I find the distribution of $X$ squared (I know that $P(Y=y=x^2) = p(X=x)$ but does this distribution have a standard name)? In particular, how can I find its expected value?
Thanks!
EDIT:
Thank you all! (-:
| $\begin{align}
\text{E}(X^2) & = \sum\limits_{k = 0}^n k^2P[X = k]\\
& = \sum\limits_{k = 0}^n k^2\,^nC_kp^kq^{n - k}\\
& = \sum\limits_{k = 1}^n k^2\, \dfrac{n!}{k! (n - k)!} p^kq^{n - k}\\
& = \sum\limits_{k = 1}^n k \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k}\\
& = \sum\limits_{k = 1}^n (k - 1 + 1) \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k}\\
& = \sum\limits_{k = 1}^n (k - 1) \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k} + \sum\limits_{k = 1}^n \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k}\\
& = \sum\limits_{k = 2}^n \dfrac{n!}{(k - 2)! (n - k)!} p^kq^{n - k} + \sum\limits_{k = 1}^n \dfrac{n!}{(k - 1)! (n - k)!} p^kq^{n - k}\\
& = n(n-1)p^2\sum\limits_{k = 2}^n \dfrac{(n-2)!}{(k - 2)! (n - k)!} p^{k-2}q^{(n-2) - (k-2)} +\\& \qquad np\sum\limits_{k = 1}^n \dfrac{(n-1)!}{(k - 1)! (n - k)!} p^{k-1}q^{(n-1) - (k-1)}\\
& = n(n-1)p^2\sum\limits_{k=2}^n \, ^{n-2}C_{k-2}p^{k-2}q^{(n-2) - (k-2)} + np\sum\limits_{k = 1}^n \, ^{n-1}C_{k-1} p^{k-1}q^{(n-1) - (k-1)}\\
& = n(n - 1)p^2(p + q)^{n - 2} + np(p + q)^{n - 1}\\
& = n^2p^2 - np^2 + np\\
& = \boxed{n^2p^2 + npq}
\end{align}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
} |
The convergence of $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$ I would like to know if this sequence converges $\sqrt {2+\sqrt {2+\sqrt {2+\ldots}}}$.
I know this sequence is increasing monotone, but I couldn't prove it's bounded.
Thanks
| As written it's not clear this is a sequence at all but I'm assuming your sequence is
$\sqrt{2}$, $\sqrt{2 + \sqrt{2}}$, $\sqrt{2 + \sqrt{2+ \sqrt{2}}}$, $\sqrt{2 + \sqrt{2+ \sqrt{2+\sqrt{2}}}}$, $\sqrt {2+\sqrt {2+\sqrt {2+ \sqrt{2 +\ldots}}}}$.
Then it should be obvious that each entry in the sequence is positive and the sequence increases each time.
Given this the final entry in the infinite sequence would be $x = \sqrt{2 + x}$ which we can easily solve.
$$\begin{align}
x &= \sqrt{2+x} \\
x^2 &= 2 + x\\
x^2 - x -2 &= 0\\
x &= \dfrac{1 \pm \sqrt{1+8}}{2} \\
x &= \dfrac{4}{2} = 2
\end{align}$$
The sequence has a lower bound of $\sqrt{2}$ and an upper bound of $2$ increasing monotonically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/849274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Rewriting this complex square root for some reason I can to figure out how to rewrite this square root.
I have:
$\sqrt{2+i}$
And I need to rewrite it into:
$\frac{\sqrt{2(\sqrt{5} + 2)} + \sqrt{-2(\sqrt{5} - 2)}}{2}$
Can anybody show me I to do this? I've been trying for an hour now..
| One can use trigonometry for sure but I thing the fastest way to solve this is just to solve the quadratic.
$$(x+yi)^2=2+i$$ so
$$x^2-y^2=2$$
$$2xy=1$$
and $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=4+1=5$$
So $$x^2+y^2=\sqrt{5}$$
(positive since $x$ and $y$ real)
and
$$x^2=\frac{\sqrt{5}+2}{2}$$
$$y^2=\frac{\sqrt{5}-2}{2}$$
so
$$x=\pm\sqrt{\frac{\sqrt{5}+2}{2}}$$
$$y=\pm\sqrt{\frac{\sqrt{5}-2}{2}}$$
Both $x$ and $y$ have same sign since $2xy=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/849747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $ I have a question. I tried so to solve it, but there is a problem.
that is i don't have any idea to findout how can i work with degrees 4,5,7 ...
this is the problem :
let $ x , y $ and $ z $ three real numbers such $ x+y+z = 0 $.
prove : $ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $
Please think and write your solutions! ; )
| If $x=y=z=0$, then it is trivial.
WLOG, $x\ne 0$ (at least one of variables $\ne 0$).
Denote
$y=ax$,
$z=-x(1+a)$.
It is enough to prove, that
$$
\left(\frac{1^7+a^7-(1+a)^7}{7}\right)^2 = \left( \frac{1^5+a^5-(1+a)^5}{5} \right)^2 \cdot \left(\frac{1^4+a^4+(1+a)^4}{2}\right);
$$
$$
\left(\frac{7a+21a^2+35a^3+35a^4+21a^5+7a^6}{7}\right)^2 = \left( \frac{5a+10a^2+10a^3+5a^4}{5} \right)^2 \cdot \left(\frac{2+4a^2+6a^3+4a^4+2a^5}{2}\right);
$$
$$
(1+3a+5a^2+5a^3+3a^4+a^5)^2 = (1+2a+2a^2+a^3)^2 \cdot (1+2a+3a^2+2a^3+a^4);
$$
$$
(1+3a+5a^2+5a^3+3a^4+a^5) = (1+2a+2a^2+a^3) \cdot (1+a+a^2).
$$
Last identity is obvious.
| {
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How to prove $\frac{1}{4}(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a})\ge \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}$ Let $a,b,c,d>0$, show that
$$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}$$
I know this is interesting inequality,and Mathlinks can't solution,Thank you for you help
also can see this:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=595079
today,when I see this secrets in inequalitys Volume 1,(Pham kim Hung) page 205
let $a,b,c,d>0$ prove that
$$\sum_{cyc}\dfrac{a^2}{b}\ge 2\sqrt{2}\sqrt[4]{a^4+b^4+c^4+d^4}$$
I can't prove it,can you help me? Thank you
| The above answer by johannesvalks "is not yet completed", as johannesvalks himself points out. Here is a new answer without calculus.
Rewrite the question as follows: Show that
$$
f = (a b c d)^4\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^4 - 64 (a b c d)^4\left({{a^4+b^4+c^4+d^4}}\right) \ge 0
$$
Due to cyclicity, we can demand that $d = \min\{a,b,c,d\}$. Due to homogeneity, we can demand that $d=1$. Hence we need to show
$$
f = \left(a^3c+ab^3 + abc^3 + bc\right)^4 - 64 (a b c)^4\left({{a^4+b^4+c^4+1}}\right) \ge 0
$$
Now this has become an inhomogeneous polynomial of degree 20.
We now distinguish cases for $a,b,c$. As $1 = d = \min\{a,b,c,d\}$, we write $a = 1+A$, $b = 1+B$, $c = 1+C$, with $A,B,C \ge 0$. Indeed, we are not going to use $A,B,C$. When doing so, and fully expanding (by computer) $f$, we obtain terms with negative sign, which (without further processing) makes the result inconclusive.
Remark: "Standard" Buffalo Way doesn't succeed here, since we have to construct all 6 cases of the ordering of $A,B,C$. In the two cases where we let $a=1+x$; $c = 1+x+y$; $b=1+x+y+z$, and $c=1+x$; $a = 1+x+y$; $b=1+x+y+z$, with $x,y,z \ge 0$, we obtain after fully expanding (by computer) $f$, terms with negative sign.
Instead, we consider the following 4 cases:
Case 1: $A<B+C$; $B<A+C$; $C<A+B$. Visually, these three conditions are met if $A,B,C$ are "similar" in value. In this case, we can write $A = x+y$, $B= y+z$, $C= x+z$, with $x,y,z \ge 0$. This is possible, since $x = \frac{A+C-B}{2}$, and cyclically for $y,z$. Inserting $a = 1 + x+y$ etc. into $f$ and fully expanding (by computer) gives the expression $f_1$ in the appendix below, which clearly is nonnegative since all terms have positive sign.
Further cases: Note that for $A,B,C \ge 0$, only one of the conditions $A<B+C$; $B<A+C$; $C<A+B$ can be violated. Proof: suppose the first two conditions indeed are reverted to $A \ge B +C$ and $B\ge A+C$; then we add these two conditions and have $C \le 0$ which is a contradiction. The same argument holds cyclically. Hence we only need to consider one reverted inequality in $A,B,C$ (3 more cases). This covers all possible cases.
Case 2: Visually, this condition is met if $a$ is "largest" in value. $A \ge B +C$; $B<A+C$; $C<A+B$. Writing $B=x$; $C = y$; $A = x+y+z$, with $x,y,z \ge 0$, satisfies these conditions. Inserting $a = 1 + x+y+z$, $b = 1 + x$, $c = 1 + y$ into $f$ and fully expanding (by computer) gives the expression $f_2$ in the appendix below, which clearly is nonnegative since all terms have positive sign.
Case 3: As case 2, with $b$ the largest value. Inserting $b = 1 + x+y+z$ etc. into $f$ and fully expanding (by computer) gives the expression $f_3$ in the appendix below, which clearly is nonnegative since all terms have positive sign.
Case 4: As case 2, with $c$ the largest value. Inserting $c = 1 + x+y+z$ etc. into $f$ and fully expanding (by computer) gives the expression $f_4$ in the appendix below, which clearly is nonnegative since all terms have positive sign.
This proves the claim. $\qquad \Box$
Appendix:
The four cases are given as a diary file from MATLAB (The "*"s have been removed). The diary comes as files, as the number of letters exceeds the limit allowed to be put in directly as text. MATLAB diary allows to re-perform the calculations, if wished. Obviously, the results are VERY lengthy, however the relevant fact to notice is that there are no terms with negative signs in the expanded versions of $f_1, f_2, f_3, f_4$, which can safely be verified by actually searching the files for minus signs (and not finding any in the expanded versions).
| {
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Max. and Min. value of $|z|$ in $\left|z+\frac{2}{z}\right| = 2\,$ If $z$ is a complex no. such that $\displaystyle \left|z+\frac{2}{z}\right| = 2\,$ Then find max. and min. value of $\left|z\right|$.
$\bf{My\; Try:}$ Given $\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow \left|z+\frac{2}{z}\right|^2 = 2^2=4$.
So $\displaystyle \left(z+\frac{2}{z}\right)\cdot \left(\bar{z}+\frac{2}{\bar{z}}\right) = 4\Rightarrow \left|z\right|^2+2\left(\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right)+\frac{1}{|z|^2} = 4$.
Now how can I find the max. and min. values of $|z|$?
Help me please.
Thanks
| $$|z|-\frac{2}{|z|}\le \left|z+\frac{2}{z}\right|=2 \implies\\ |z|^2-2|z|-2\le0\implies\\ 1-\sqrt{3}<0\le|z|\le1+\sqrt{3}$$
Maximum value is $1+\sqrt{3}$ provided we show that equality holds.
Recall that equality holds when $z$, $-\frac{2}{z}$ have same argument. Hence $\arg{z}=-\frac{\pi}{2}$.
Similarly using $\frac{2}{|z|}-|z|\le \left|z+\frac{2}{z}\right|=2$ we can show that the minimum is $\frac{2}{1+\sqrt{3}}=\sqrt{3}-1$
| {
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$2^{1/4} \times 4^{1/8} \times 8^{1/16} \times 16^{1/32} \times \ldots\to2$ $2^{1/4} \times 4^{1/8} \times 8^{1/16} \times 16^{1/32} \times \ldots\to2$
How can I explain this to a school student who doesn't know what a limit is?
| $$2^{1/4} \times 4^{1/8} \times 8^{1/16} \times 16^{1/32} \times \ldots=\prod^{\infty}_{n=1}(2^n)^{1/2^{(n+1)}}=x$$
Log by base 2 both sides and we have:
$$1/4+1/4+3/16+1/8+5/64..=\sum^{\infty}_{n=1} n/2^{(n+1)}$$
This sum is $1$. To show it we need to find
$$S=\lim \sum^{\infty}_{n=1} n/2^{(n+1)}=1$$
Let's multiply it by $2$ and substitute $S_0=2S$
$$1/2+2/4+3/8+4/16...=S_0=\lim \sum^{\infty}_{n=1} n/2^{n}=2$$
Now let's take
$$f(x)=\frac{1}{1-x} =1+x+x^2+x^3... $$
$$f'(x)=\frac{1}{(1-x)^2}=1+2*x+3*x^2+4*x^3...$$
Multiply it by $x$
$$g(x)=f'(x)*x=\frac{x}{(1-x)^2}=1*x+2*x^2+3*x^3+4*x^4...$$
And finally the magic, watching $g(1/2)$
$$S_0=g(1/2)=2=1/2+2/4+3/8+4/16...$$
Going back from $S_0$ to $S$ and from $S$ to logarithms solving $\log_2 x = S=1 $and get final answer $$x =\prod^{\infty}_{n=1}(2^n)^{1/2^{(n+1)}}=2$$
No ideas how to show it without Taylor series and derivative.
| {
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Volume of sphere inside a cylinder I have homework question:
Calculate the volume of $\{(x,y,z) \mid x^2+y^2+z^2\le (2a)^2\} \cap \{(x,y,z) \mid (x-a)^2 + y^2 \le a^2\}$
in two ways:
1) Polar coordinates
2) Spherical coordinates
How can I solve it?
Edit: What I tried:
$$2\int_0^{2\pi}\int_0^a r\sqrt{4a^2-x^2-y^2}\,dr\,d\theta = 2\int_0^{2\pi}\int_0^a r\sqrt{4a^2-(a+r\cos\theta)^2-r\sin^2\theta}\,dr\,d\theta \\
= 2\int_0^{2\pi}\int_0^a r\sqrt{3a^2-2ar\cdot\cos\theta - r^2}\,dr\,d\theta = \: ?$$
Edit: I need answer with one of the two ways.
| For cylindrical coordinates:
$$z=\pm \sqrt{4a^2-(x^2+y^2)}$$
Using polar coordinates
$$x=r\cos\phi \ , \ y=r\sin \phi $$
we end up with
$$z=\pm \sqrt{4a^2-r^2}$$
now, just integrate
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{2a\cos \phi}\int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}}rdzdrd\phi=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^{2a\cos \phi}r\sqrt{4a^2-r^2}dzdrd\phi$$
For spherical coordinates:
$$ x=r\cos\phi \sin \theta \ , \ y = r\sin \phi \sin \theta \ , \ z = r\cos \theta \ , \ J=r^2\sin \theta $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/857989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that the following series is less than $4 \pi^2 / 3$. Show: For any $k = 0,1,2,...$,
$$ \sum_{i=0}^{i=k} \frac{(k+1)^2}{(i+1)^2 (k-i+1)^2} \leq \frac{4 \pi^2}{3}. $$
| Note that
$$
\frac{1}{i+1}+\frac{1}{k-i+1}=\frac{k+2}{(i+1)(k-i+1)}
$$
So, (because $(a+b)^2\leq 2a^2+2b^2$):
$$
\frac{(k+2)^2}{(i+1)^2(k-i+1)^2}\leq\frac{2}{(i+1)^2}+\frac{2}{(k-i+1)^2}
$$
Adding we get
$$
\sum_{i=0}^{k}\frac{(k+2)^2}{(i+1)^2(k-i+1)^2}
\leq 4\sum_{i=0}^{k}\frac{1}{(1+i)^2}\leq 4\zeta(2)=\frac{2\pi^2}{3}
$$
So
$$
\sum_{i=0}^{k}\frac{(k+1)^2}{(i+1)^2(k-i+1)^2}
\leq \frac{(k+1)^2}{(k+2)^2}\frac{2\pi^2}{3}<\frac{2\pi^2}{3}< \frac{4\pi^2}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the Value of Integral Find the Value of $$\begin{align}I=\int_{0}^{1}\frac{\ln(x)\,dx}{1-x^2}\end{align}$$
I have tried like this: We have $$\begin{align}2I=\int_{0}^{1}\frac{\ln(x^2)\,dx}{1-x^2}=\int_{0}^{1}\frac{\ln(1-(1-x^2))\,dx}{1-x^2}\end{align}$$ So
$$2I=\begin{align}\int_{0}^{1}\frac{-(1-x^2)-\frac{(1-x^2)^2}{2}-\frac{(1-x^2)^3}{3}-\cdots }{1-x^2}\end{align}=$$
I need help from here..
| Let
$$ I(\alpha)=\int_{0}^{1}\frac{x^{\alpha}dx}{1-x^2}. $$
Then $I'(0)=I$. Now
\begin{eqnarray*}
I(\alpha)=\lim_{a\to 1^-}\int_{0}^{a}\sum_{n=0}^\infty x^{\alpha+2n}dx=\lim_{a\to 1^-}\sum_{n=0}^\infty \frac{1}{\alpha+2n+1}a^{\alpha+2n+1}
\end{eqnarray*}
and hence
\begin{eqnarray*}
I'(\alpha)&=&\lim_{a\to 1^-}\sum_{n=0}^\infty \left(\frac{-1}{(\alpha+2n+1)^2}a^{\alpha+2n+1}+\frac{1}{\alpha+2n+1}a^{\alpha+2n+1}\ln a\right)\\
&=&-\sum_{n=0}^\infty\frac{1}{(\alpha+2n+1)^2}
\end{eqnarray*}
Thus
$$ I=I'(0)=-\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=-\frac{\pi^2}{8}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/860091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find the limit of $\lim\limits_{x \to+\infty} \frac{4x-3}{2x+5}$ and $\lim \limits_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1}$ Hello guys help me with these:
1.) $\lim \limits_{x \to +\infty} frac{4x-3}{2x+5}$
$=\lim \frac{4-\frac{3}{x}}{2+\frac{5}{x}}$
$= \lim \frac{4-3\frac{1}{x}}{2+5\frac{1}{x}}$
=$\frac{\lim 4- \lim 3 \lim \frac{1}{x}}{\lim 2+ \lim 5 \lim\frac {1}{x}}$ // lim of $\frac{1}{x}$=0
=$\frac{4 - 3 * 0}{2 + 5 * 0}$
=$\frac{4 - 0}{2 + 0}$
=$\frac{4}{2}
=2$
lim of the function as $x$ approaches positive infinity is $2$.
Is it correct?
If then what should I do if the given is:
$\lim \limits_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1}$
| $$\lim_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1} =\lim_{x \to -\infty} \frac{2x^2(1-\frac{1}{2x}+\frac{5}{2x^2})}{4x^3(1-\frac{1}{4x^3})} =\lim_{x \to -\infty} \frac{(1-\frac{1}{2x}+\frac{5}{2x^2})}{2x(1-\frac{1}{4x^3})}$$
It is clear that
$$\lim_{x \to -\infty} \frac{1}{4x^3} =\lim_{x \to -\infty} \frac{1}{2x} = \lim_{x \to -\infty} \frac{5}{2x^2}= 0,$$
and so
$$\lim_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1} = \lim_{x \to -\infty} \frac{1}{2x} = 0.$$
Note: Using the same process it easy to observe that
$$\lim_{x \to -\infty} \frac{2x^3-x+5}{4x^3-1} = \lim_{x \to -\infty} \frac{2x^3(1-\frac{1}{2x^2}+\frac{5}{2x^3})}{4x^3(1-\frac{1}{4x^3})} = \lim_{x \to -\infty} \frac{(1-\frac{1}{2x^2}+\frac{5}{2x^3})}{2(1-\frac{1}{4x^3})} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Show that $\frac{1}{2}n^2-3n=\Theta{(n^2)}$ Show that $$\frac{1}{2}n^2-3n=\Theta{(n^2)}$$
$$$$
$\displaystyle{\frac{1}{2}n^2-3n=\Theta{(n^2)}: \\ \exists c_1, c_2 >0 , \ \ \exists n_0 \geq 1 \text{ such that } \forall n \geq n_0 \\ 0<c_1 n^2 \leq \frac{1}{2}n^2-3n \leq c_2 n^2}$
$$$$
That means that I have to find the values of $n_0, c_1, c_2$ such that
$$0<c_1 n^2 \leq \frac{1}{2}n^2-3n \leq c_2 n^2$$
$$\Rightarrow 0<c_1 \leq \frac{1}{2}-\frac{3}{n} \leq c_2$$
*
*$\displaystyle{0<c_1 \leq \frac{1}{2}-\frac{3}{n}} :$
It must be $\displaystyle{\frac{1}{2}-\frac{3}{n}>0 \Rightarrow \frac{3}{n} < \frac{1}{2} \Rightarrow n>6 \Rightarrow n \geq 7}$
So, $\displaystyle{n_1=7}$.
$\displaystyle{n \geq 7 \Rightarrow \frac{1}{7} \geq \frac{1}{n} \Rightarrow \frac{3}{7} \geq \frac{3}{n} \Rightarrow -\frac{3}{n} \geq -\frac{3}{7} \Rightarrow \frac{1}{2}-\frac{3}{n} \geq \frac{1}{2}-\frac{3}{7} \Rightarrow \frac{1}{2}-\frac{3}{n} \geq \frac{1}{14}}$
So $\displaystyle{c_1=\frac{1}{14}}$.
*
*$\displaystyle{\frac{1}{2}-\frac{3}{n} \leq c_2} :$
How can I find a $\displaystyle{n_2}$ such that $\displaystyle{\forall n \geq n_2:\frac{1}{2}-\frac{3}{n} \leq c_2}$ ??
And how can I find this $\displaystyle{c_2}$ ??
| $$n\geqslant12\implies\frac14n^2\leqslant\frac12n^2-3n\leqslant\frac12n^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find expression for : $ S_n =\sum_{i=1}^{n} \frac{i}{i^4+i^2+1} $ I want to find a formula for the sum of this series using its general term.
How to do it?
Series
$$
S_n = \underbrace{1/3 + 2/21 + 3/91 + 4/273 + \cdots}_{n \text{ terms}}
$$
General Term
$$
S_n = \sum_{i=1}^{n} \frac{i}{i^4+i^2+1}
$$
| Using partial fractions, we get $\dfrac{i}{i^4+i^2+1} = \dfrac{\tfrac{1}{2}}{i^2-i+1} - \dfrac{\tfrac{1}{2}}{i^2+i+1}$.
Thus, $S_n = \displaystyle\sum_{i = 1}^{n}\dfrac{i}{i^4+i^2+1} = \dfrac{1}{2}\sum_{i = 1}^{n}\dfrac{1}{i^2-i+1} - \dfrac{1}{i^2+i+1}$.
Since $(i+1)^2-(i+1)+1 = i^2+i+1$, this sum telescopes to $S_n = \dfrac{1}{2}\left(1 - \dfrac{1}{n^2+n+1}\right)$.
Taking the limit as $n \to \infty$ gives $S = \dfrac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/865442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solving a trigonometric limit $\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$ First off, please excuse my n00bishness I have only just begun learning about algebraic manipulation of limits so this is probably a really dumb or obvious question.
I'm trying to solve the following limit:
$$
\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}
$$
This limit is $0/0$ if evaluated directly, so I tried multiplying by the conjugate of the denominator:
$$
\begin{align}
& = \lim_{x\to\pi/6}\frac{(1 - 2\sin{x})(2\sqrt{3}\cos{x} + 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{x} + 3)}\\
& = \lim_{x\to\pi/6}\frac{(2\sin{x} - 1)(2\sqrt{3}\cos{x} - 3)}{(2\sqrt{3}\cos{x} - 3)(2\sqrt{3}\cos{3} + 3)} \\
& = \lim_{x\to\pi/6}\frac{2\sin{x} - 1}{2\sqrt{3}\cos{x} + 3}\\
& = \frac{2(1/2) - 1}{2\sqrt{3}\frac{\sqrt{3}}{2} + 3}\\
& = \frac{0}{6}\\
& = 0
\end{align}
$$
But according to WolframAlpha this is incorrect, and the limit should be 1. What have I done wrong?
Also, as I have only just begun I am unfamiliar with L'Hopital's rule.
| $$\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$$
$$=-\frac2{2\sqrt3}\lim_{x\to\pi/6}\frac{\sin x-\sin\frac\pi6}{\cos x-\cos\frac\pi6}$$
$$=-\frac2{2\sqrt3}\lim_{x\to\pi/6}\frac{\sin x-\sin\frac\pi6}{x-\frac\pi6}\cdot\frac1{\lim_{x\to\pi/6}\dfrac{\cos x-\cos\frac\pi6}{x-\frac\pi6}}$$
$$=-\frac2{2\sqrt3}\frac{\dfrac{d(\sin x)}{dx}_{(\text{ at } x=\frac\pi6)}}{\dfrac{d(\cos x)}{dx}_{(\text{ at } x=\frac\pi6)}}$$
$$=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/865509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Pythagorean triple means Just out of curiosity, do there exist two positive integers whose arithmetic mean ($A $), geometric mean ($G $) and harmonic mean ($H $) constitute a Pythagorean triple?
That is, $A $, $G $ and $H $ are positive integers, and $H^2 + G^2 = A^2$.
| Let $a$ and $b$ be positive integers with $a \ge b$; then you're asking whether we can ever have
$$\left(\frac{2}{\frac 1 a + \frac 1 b}\right)^2 + \Big(\sqrt{ab}\Big)^2 = \left(\frac{a + b}{2}\right)^2$$
or upon some simplification,
$$\frac{4a^2 b^2}{(a+b)^2} + ab = \frac{(a + b)^2}{4}$$
Rearranging,
$$\frac{4a^2 b^2}{(a + b)^2} = \frac{(a - b)^2}{4}$$
Taking square roots,
$$\frac{2ab}{a + b} = \frac{a - b}{2}$$
$$4ab = a^2 - b^2$$
We now show there are no positive integer solutions to this. Suppose there was; since this is homogeneous, we can cancel any common factors in $a$ and $b$, so they're relatively prime. Clearly $a$ and $b$ must have the same parity, so they're both odd.
Add $2b^2$ to both sides, leading to
$$4ab + 2b^2 = a^2 + b^2$$
$$2ab + 2b^2 = (a - b)^2$$
$$2b (a + b) = (a - b)^2$$
Here we have the desired contradiction. We have that $b | (a - b)^2$, and upon expanding, this leads to $b | a^2$. As $b$ and $a$ are relatively prime, this forces $b = 1$, so that $4a = a^2 - 1$. It's easy to verify that this has no solutions in the integers, and we're done.
Actually, it's simpler to just add $b^2$ to both sides, leading to $$a^2 = b(4ab + b)$$
hence $b | a^2$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/865832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$12\frac{\sin 45^\circ}{\sin 60^\circ}$ Need help breaking this down. Otherwise known as $12\dfrac{\left(\frac{1}{\sqrt2}\right)}{\left(\frac{\sqrt3}{2}\right)}$
How do you simplify this multi level fractional radical expression into $4\sqrt{6}$.
| $$12\dfrac{\left(\frac{1}{\sqrt2}\right)}{\left(\frac{\sqrt3}{2}\right)}=12\cdot\frac{1}{\sqrt2}\div\frac{\sqrt3}{2}=12\cdot\frac{2}{\sqrt{6}}=\frac{24}{\sqrt{6}}=\frac{24}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=\frac{24\sqrt{6}}{6}=4\sqrt{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Integration practice of $\int \frac{\sqrt{25-y^2}}{y}dy$ I need to solve $\int \frac{\sqrt{25-y^2}}{y}dy$.
I originally thought IBP, but that led to a very large and confusing algebra problem.
Then I started to look at the $\sqrt{25-y^2}$ and started to consider doing a $u$-substituion, and letting $u=\sin\theta$ and $du=\cos\theta d\theta$. Once I started playing a bit, the intuition was lost.
Any and all help is appreciated.
| Trigonometric substitutions are not necessary. Let $u^2 = 25 - y^2$, so that $2u \, du = -2y \, dy$. Then $$\begin{align*} \int \frac{\sqrt{25-y^2}}{y} \, dy &= - \int \frac{-y \sqrt{25-y^2}}{y^2} \, dy \\ &= - \int \frac{u \sqrt{u^2}}{25-u^2} \, du \\ &= \int \frac{u^2 - 25 + 25}{u^2 - 25} \, du \\ &= u + 25 \int \frac{du}{u^2 - 25} \\ &= u + \frac{5}{2} \int \left(\frac{1}{u-5} - \frac{1}{u+5} \,\right) du\end{align*}$$ and the rest is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/867085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Let $x,y,z$ be positive integers such that $\frac{1}{x}-\frac{1}{y}=\frac{1}{z}$. Let $h=\gcd(x,y,z)$, Prove that $hxyz,h(y-x)$ are perfect squares Let $x,y,z$ be positive integers such that $\frac{1}{x}-\frac{1}{y}=\frac{1}{z}$. Let $h=\gcd(x,y,z)$, Prove that $hxyz,h(y-x)$ are both perfect squares.
My attempt:
Let $x=ha,y=xb,z=xc$, then $a,b,c$ are positive integers such that $\gcd(a,b,c)=1$.
Now, suppose $\gcd(a,b)=g$, so, $a=ga',b=gb'$, where $\gcd(a',b')=\gcd(a',a'-b')=\gcd(b',a'-b')=1$. So, we have $c(b'-a')=ga'b'$, so $g\mid c$, and $g=1$.
Now, I do not know what to do. Please help. Thank you.
| Let us consider
$$
\frac{1}{x} - \frac{1}{y} = \frac{1}{z},
$$
and
$$
h = \textrm{gcd}(x,y,z).
$$
Say that
$$
\begin{eqnarray}
x &=& h x',\\
y &=& h y',\\
z &=& h z',\\
\end{eqnarray}
$$
then we obtain
$$
\frac{1}{x'} - \frac{1}{y'} = \frac{1}{z'},
$$
where
$$
\textrm{gcd}(x',y',z') = 1.
$$
We can write
$$
\Big(y'-x'\Big)z' = x'y'.
$$
Let us look to odd and even:
$$
\begin{array}{ccccl}
x' & y' & z'\\
\textrm{even} & \textrm{even} & \textrm{even} & \Rightarrow &
\textrm{gcd}(x',y',z') > 1.\\
\textrm{odd} & \textrm{odd} & \textrm{odd} & \Rightarrow&
\textrm{contradiction as $xy$ is odd but $(x-y)z$ is even}.\\
\textrm{odd} & \textrm{even} & \textrm{even}\\
\textrm{even} & \textrm{odd} & \textrm{even}
\end{array}
$$
So we write
$$
\begin{eqnarray}
y' &=& x' + 2p + 1,\\
z' &=& 2q,
\end{eqnarray}
$$
whene
$$
x'^2 + \Big(2p+1\Big)x' = \Big(2p+1\Big)2q,
$$
so
$$
x' = - \frac{2p+1}{2} + \frac{2p+1}{2} \sqrt{1 + \frac{8q}{2p+1}},
$$
meaning that
$$
q = \frac{1}{2} r \Big( r + 1 \Big) \Big( 2p + 1 \Big),
$$
so that
$$
\begin{eqnarray}
x' &=& r \Big( 2p + 1 \Big),\\
y' &=& \Big( r + 1 \Big) \Big( 2p + 1 \Big),\\
z' &=& r \Big( r + 1 \Big) \Big( 2p + 1 \Big).\\
\end{eqnarray}
$$
However $\textrm{gcd}(x',y',z') = 1$, thus $2p+1=1$, whence
$$
\begin{eqnarray}
x &=& h r,\\
y &=& h \Big( r + 1),\\
z &=& h r \Big( r + 1 \Big).\\
\end{eqnarray}
$$
It is clear that
$$
\begin{eqnarray}
hxyz &=& \left[ h^2 r \Big( r + 1 \Big) \right]^2,\\
h(y-x) &=& \Big[ h r \Big]^2,\\
\end{eqnarray}
$$
thus both $hxyz$ and $h(y-x)$ are perfect squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/867392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Another integral for $\pi$ Here is a new integral for $\pi$.
$$\int_{0}^{1}\sqrt{\frac{\left\{1/x\right\}}{1-\left\{1/x\right\}}}\, \frac{\mathrm{d}x}{1-x} = \pi $$
where $\left\{x\right\}$ denotes the fractional part of $x$.
Do you have any proof?
| With the substitution $1/x=t$, the integral is:
$$I=\int_1^{\infty} \sqrt{\frac{\{t\}}{1-\{t\}}}\frac{dt}{t(t-1)}=\sum_{n=1}^{\infty}\int_n^{n+1}\sqrt{\frac{t-n}{n+1-t}}\frac{dt}{t(t-1)}$$
Next, use the substitution $\dfrac{t-n}{n+1-t}=x^2 \Rightarrow t=n+1-\dfrac{1}{1+x^2} \Rightarrow dt=\dfrac{2x}{(1+x^2)^2}\,dx$ i.e
$$I=2\sum_{n=1}^{\infty} \int_0^{\infty} \frac{x^2}{(x^2(n+1)+n)(nx^2+n-1)}\,dx$$
$$\Rightarrow I=2\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\int_0^{\infty}\frac{x^2}{\left(x^2+\frac{n}{n+1}\right)\left(x^2+\frac{n-1}{n}\right)}\,dx$$
The above integral can be evaluated by decomposing into partial fractions i.e
$$\begin{aligned}
I &=\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\frac{\pi}{\left(\sqrt{\frac{n}{n+1}}+\sqrt{\frac{n-1}{n}}\right)} \\
&=\pi\sum_{n=1}^{\infty} \frac{1}{\sqrt{n(n+1)}(n+\sqrt{n^2-1})}\\
&=\pi\sum_{n=1}^{\infty} \frac{n-\sqrt{n^2-1}}{\sqrt{n(n+1)}} \\
&=\pi \sum_{n=1}^{\infty} \left(\sqrt{\frac{n}{n+1}}-\sqrt{\frac{n-1}{n}}\right)
\end{aligned}$$
The final sum telescopes and its value is $1$. Hence,
$$\boxed{I=\pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/870680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
} |
QR transformation with Householder transformation It's a task i do to understand minimizing the error including the QR transformation with the help of Householder transformation. I think i really do something wrong but i dont get it running i hope you can help me.
Given the task to minimizing the error of
$$ A = \begin{pmatrix}
1 &1\\
1 & 0\\
1 & 1 \\
1 & 4
\end{pmatrix}
b=\left[2,\frac{1}{2},\frac{3}{2},5\right]^T
$$
So at first getting $A^TA = \begin{pmatrix} 4 & 6 \\ 6 &18 \end{pmatrix}$
after this calculate the householder vector $v_1=a_1+\alpha*e_1$.
$\alpha = sign(a_{1,1})*\lVert a_{1}\rVert = \sqrt{52} $
so
$$ v_1 = \begin{pmatrix} 4 \\ 6 \end{pmatrix} + \sqrt{52}*\begin{pmatrix}1 \\ 0 \end{pmatrix}$$
BUT the problem is if i calculate Q ouf of this:
$$Q = I- \frac{2*v*v^T}{v^T*v} $$
I would be done. It would be $A^TA = QR$ with $ R= QA^TA$. (Sure i am resulting in some matix like $R=\begin{pmatrix} x &y\\0&z\end{pmatrix}$
But if i calculate that with Matlab i get incorrect results ($R=A^Tb$). Even the calculation of $v$ seems to be incorrect since it should ne just simple in fact that no calculater is allowed. If i simply calculate $A^TAx=A^Tb$ it's a simple calculation resulting in $x_1=\frac{7}{12} x_2=\frac{10}{9}$ which i dont get with the QR in any way.
| Based on the comments, this is probably what you're supposed to do.
To transform $A$ to the upper triangular form requires an application of two Householder transformations. Take
$$
Q_1=I-2\frac{v_1v_1^T}{v_1^Tv_1}=
\frac{1}{6}
\begin{bmatrix}
-3 & -3 & -3 & -3 \\
-3 & 5 & -1 & -1 \\
-3 & -1 & 5 & -1 \\
-3 & -1 & -1 & 5 \\
\end{bmatrix}
\quad\text{with}\quad v_1=[3,1,1,1]^T.
$$
This gives
$$
Q_1^TA=\begin{bmatrix}
-2 & -3 \\
0 & -4/3 \\
0 & -1/3 \\
0 & 8/3
\end{bmatrix}.
$$
Next, take
$$
Q_2=I-2\frac{v_2v_2^T}{v_2^Tv_2}
=\frac{1}{117}\begin{bmatrix}
117 & 0 & 0 & 0 \\
0 & -52 & -13 & 104 \\
0 & -13 & 116 & 8 \\
0 & 104 & 8 & 53
\end{bmatrix}
\quad\text{with}\quad v_2=[0,-13/3,-1/3,8/3]^T
$$
to get
$$
R:=Q^TA:=Q_2^TQ_1^TA=
\begin{bmatrix}
-2 & -3 \\
0 & 3 \\
0 & 0 \\
0 & 0
\end{bmatrix}.
$$
Apply $Q^T$ to $b$ to get
$$
Q^Tb=
\begin{bmatrix}
-9/2 \\
10/3 \\
-11/39 \\
-19/78
\end{bmatrix}.
$$
Hence, the least squares problem $Ax=b$ is equivalent to the least squares problem
$$
\begin{bmatrix}
-2 & -3 \\
0 & 3 \\
0 & 0 \\
0 & 0
\end{bmatrix}x
=
\begin{bmatrix}
-9/2 \\
10/3 \\
-11/39 \\
-19/78
\end{bmatrix}.
$$
The residual 2-norm is minimal, if the first two components of the residual vector are zero, that is,
the least squares solution is given by the solution of
$$
\begin{bmatrix}
-2 & -3 \\
0 & 3
\end{bmatrix}x
=
\begin{bmatrix}
-9/2 \\
10/3
\end{bmatrix}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$ The following result discussed by Ramanujan is very famous: $$\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}\tag {1}$$ and can be easily proved by cubing both sides and using $x = \sqrt[3]{2}$ for simplified typing.
Ramanujan established many such denesting of radicals such as $$\sqrt{\sqrt[5]{\frac{1}{5}} + \sqrt[5]{\frac{4}{5}}} = \sqrt[5]{1 + \sqrt[5]{2} + \sqrt[5]{8}} = \sqrt[5]{\frac{16}{125}} + \sqrt[5]{\frac{8}{125}} + \sqrt[5]{\frac{2}{125}} - \sqrt[5]{\frac{1}{125}}\tag {2}$$$$\sqrt[3]{\sqrt[5]{\frac{32}{5}} - \sqrt[5]{\frac{27}{5}}} = \sqrt[5]{\frac{1}{25}} + \sqrt[5]{\frac{3}{25}} - \sqrt[5]{\frac{9}{25}}\tag {3}$$$$\sqrt[4]{\frac{3 + 2\sqrt[4]{5}}{3 - 2\sqrt[4]{5}}} = \frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}\tag{4}$$$$\sqrt[\color{red}6]{7\sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}\tag{5}$$$$\sqrt[6]{4\sqrt[3]{\frac{2}{3}} - 5\sqrt[3]{\frac{1}{3}}} = \sqrt[3]{\frac{4}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{1}{9}}\tag{6}$$
$$\sqrt[8]{1\pm\sqrt{1-\left(\frac{-1+\sqrt{5}}{2}\right)^{24}}} = \frac{-1+\sqrt{5}}{2}\,\frac{\sqrt[4]{5}\pm 1}{\sqrt{2}}\tag{7}$$
with the last one found in Ramanujan's Notebooks, Vol 5, p. 300. Most of these radical expressions are units (a unit is an algebraic integer $\alpha$ such that $\alpha\beta = 1$ where $\beta$ is another algebraic integer).
For me the only way to establish these identities is to raise each side of the equation to an appropriate power using brute force algebra and then check the equality. However for higher powers (for example equation $(2)$ above) this seems very difficult.
Is there any underlying structure in these powers of units which gives rise to such identities or these are mere strange cases which were noticed by Ramanujan who used to play with all sorts of numbers as a sort of hobby? I believe (though not certain) that perhaps Ramanujan did have some idea of such structure which leads to some really nice relationships between units and their powers. I wonder if there is any sound theory of such relationships which can be exploited to give many such identities between nested and denested radicals.
| A general identity: $$\sum_{k=0}^{3n-1}\frac{(-2)^{k/3}}{9^{1/3}} = \frac{1-(-2)^n}{3}\sqrt[3]{\sqrt[3]2-1}$$ so Ramanujan's case was $n=1$, yielding the famous $$\sqrt[3]{\sqrt[3]2-1}=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}$$ and possibly sheds light on how he found other like $(2)$, $(3)$, $(5)$ and $(6)$. There are other denestations like $$(1)=\sqrt[8]{4\sqrt[3]{\frac 23}-5\sqrt[3]{\frac 13}}=\sqrt[15]{19-5\sqrt[3]2-8\sqrt[3]4}$$ and very similarly, $$\sqrt[3]{\frac 19}+2\sqrt[3]{\frac 29}-2\sqrt[3]{\frac 49}=\sqrt{4\sqrt[3]{\frac 23}-5\sqrt[3]{\frac 13}}\tag{$\star$}$$ and that $$\sqrt[4]{(\star)}=\sqrt[5]{\sqrt[3]9-\sqrt[3]{\frac 23}-\sqrt[3]{\frac 43}}=\sqrt[6]{1-2\sqrt[3]2+\sqrt[3]4}$$
The fact that we predominantly have the numerators in geometric progression, for some reason (throughout my experimentation), seems to make radicals combine and cancel out very elegantly for abnormally high powers.
Regarding $(2)$, by taking out the numerators, I noticed they can be factored when arranged as an alternating sum. Id est, $$\sqrt[5]1-\sqrt[5]2+\sqrt[5]8-\sqrt[5]{16}=(\sqrt[5]2+1)(\sqrt[5]2-1)(1-\sqrt[5]2+\sqrt[5]4)$$ Not sure if this adds anything, though.
| {
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Find the $n$th term of $1, 2, 5, 10, 13, 26, 29, ...$ How would you find the $n$th term of a sequence like this?
$1, 2, 5, 10, 13, 26, 29, ...$
I see the sequence has a group of three terms it repeats: Double first term to get second term, add three to get third term, repeat.
What about: $2, 3, 6, 7, 14, 15, 30,... $?
Again the sequence has a group of three terms it repeats: Add one to first term to get second term, then double second term to get third term.
How do you compute the $n$th term of such sequences directly, without iterating through all preceding terms?
| Denote the $n$-th term by $x_{n}$ and have a look at the terms with
odd index $n$. Based on what you remarked yourself:
$x_{1}=1$
$x_{3}=2x_{1}+3=2+3$
$x_{5}=2x_{3}+3=2\left(2+3\right)+3=2^{2}+2.3+3$
$x_{7}=2x_{5}+3=2\left(2^{2}+2.3+3\right)+3=2^{3}+2^{2}.3+2.3+3$
et cetera.
This starts to look like
$x_{2n+1}=2^{n}+\left(2^{n-1}+2^{n-2}+\cdots+1\right)3=2^{n}+\left(2^{n}-1\right)3=2^{n+2}-3$
This can be proved by induction and its easy now to find that $x_{2n}=x_{2n+1}-3=2^{n+2}-6$.
On a sortlike way you come to $y_{2n+1}=2^{n+2}-2$ and $y_{2n}=2^{n+1}-1$ where $y_{n}$
denotes the $n$-th term of the second sequence.
| {
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Show that $\frac{x^4 +7x^3+5}{4x+1}$ is big-theta($x^3$) I'm having trouble grasping how to set these types of problems. There are a lot of related questions but it's difficult to abstract a general procedure on finding constants that give the given function bounding constraints to make it big-theta(general function).
so $\frac{x^4 +7x^3+5}{4x+1}$ is $ \Theta (x^3) $
to show this, we need to find constants such that.
$$ |c_1|(x^3) \leq \frac{x^4 +7x^3+5}{4x+1} \leq |c_2|(x^3)$$
In addition, there also has to be a $k$ such that for all values $x >k $ the argument holds.
start with one inequality
$$ |c_1|(x^3) \leq \frac{x^4 +7x^3+5}{4x+1}$$
$$ = |c_1| \leq \frac{x^4 +7x^3+5}{4x^4+x^3}$$
$$ = |c_1| \leq \frac{x^4}{x^3(4x+1)} + \frac{7x^3}{x^3(4x+1)} + \frac{5}{x^3(4x+1)}$$
so basically for $x > 0$, $$ |c_1| \leq \frac{1}{4} + 0 + 0$$
I'm assuming after I take the limit as x goes to infinity, i could choose any $c_1$ less than or equal to $\frac{1}{4}$? The other way would then have the same procedure? What would I set $k$ to?
| Yes, you could take $c_1=\frac{1}{4}$, then you have the following:
$$ \frac{1}{4}x^3 \leq \frac{x^4 +7x^3+5}{4x+1} \Rightarrow \frac{4x^4+x^3}{4} \leq x^4+7x^3+5 \Rightarrow x^4+\frac{1}{4}x^3 \leq x^4+7x^3+5 \\ \Rightarrow \frac{27}{4}x^3 \geq -5 \Rightarrow x^3 \geq -\frac{20}{27} \Rightarrow x \geq - \sqrt[3]{\frac{20}{27}}$$
Since $x>0$ and $x \geq - \sqrt[3]{\frac{20}{27}}$, it must be $x>0$, therefore $k_1=1$.
From the other inequality you will find $k_2$, such that $\frac{x^4 +7x^3+5}{4x+1} \leq |c_2|(x^3), \forall x \geq k_2$.
Then $k=\max \{k_1, k_2 \}$.
| {
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Proof of an inequality
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c} + \frac{\sqrt{a+b+c}+\sqrt{b}}{c+a} + \frac{\sqrt{a+b+c}+\sqrt{c}}{a+b} \geq \frac{9+3\sqrt{3}}{2\sqrt{a+b+c}}$$
| We need to prove that
$$\sum_{cyc}\frac{a+b+c+\sqrt{a(a+b+c)}}{b+c}\geq\frac{9+3\sqrt3}{2}$$ or
$$\sum_{cyc}\frac{a}{b+c}+\sum_{cyc}\frac{\sqrt{a(a+b+c)}}{b+c}\geq\frac{3+3\sqrt3}{2}.$$
But by C-S $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{\sqrt{a}}{b+c}\geq\frac{3\sqrt{3}}{2\sqrt{a+b+c}}.$$
Now, by Holder
$$\left(\sum_{cyc}\frac{\sqrt{a}}{b+c}\right)^2\sum_{cyc}a^2(b+c)^2\geq(a+b+c)^3.$$
Id est, it remains to prove that
$$4(a+b+c)^4\geq27\sum_{cyc}a^2(b+c)^2,$$
which is true by AM-GM:
$$27\sum_{cyc}a^2(b+c)^2=108\sum_{cyc}a\left(a\left(\frac{b+c}{2}\right)^2\right)\leq$$
$$\leq108\sum_{cyc}a\left(\frac{a+\frac{b+c}{2}+\frac{b+c}{2}}{3}\right)^3=4\sum_{cyc}a(a+b+c)^3=4(a+b+c)^4.$$
Done!
| {
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Calculate $\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$ I'm an eight-grader and I need help to answer this math problem.
Problem:
Calculate $$\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$$
This one is very hard for me. It seems unsolvable. How to calculate the series without using Wolfram Alpha? Please help me. Grazie!
| This sum can be represented in the form $$S=\sum_{k=0}^\infty (2k+1)x^{2k+1}$$$$=>S=x\sum_{k=0}^\infty (2k+1)x^{2k}$$ where $x=\frac 15$. Now we look at the following geometric progression $S(x)=\sum_{k=0}^\infty x^{2k+1}$ where $|x|\lt 1$.hence $S(x)=\frac a{1-r}$ where a=x and $r=x^2$. Therefore $S(x)=\frac x{1-x^2}$. Therefore $$\sum_{k=0}^\infty x^{2k+1}=\frac x{1-x^2}$$$$=>\sum_{k=0}^\infty \frac d{dx}(x^{2k+1})=\frac d{dx} \left(\frac x{1-x^2}\right)$$$$=>\sum_{k=0}^\infty (2k+1)x^{2k}=\left[\frac {1+x^2}{(1-x^2)^2}\right]$$$$=>\sum_{k=0}^\infty (2k+1)x^{2k+1}=x\left[\frac {1+x^2}{(1-x^2)^2}\right]$$
Now in the question $x=\frac 15$. Therefore $$S=\left(\frac 15\right)\left[\frac {1+\left(\frac 15\right)^2}{(1-\left(\frac 15\right)^2)^2}\right]=\left(\frac 15\right)\left(\frac {5^2+1}{(5-\frac 15)^2}\right)=\left(\frac 15\right)\left(\frac {25*26}{576}\right)=\frac {65}{288}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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prove $\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$
Show that if $a$, $b$, $c$ and $d$ are non-negatives and $ab+bc+cd+da=1$ then:
$$\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$$
yet again it should be solved with Cauchy inequality.
thing i have done so far:
$(\sum\limits_{cyc} \frac {a^3} {b+c+d})\times(\sum\limits_{cyc} a(b+c+d)) \geq (\sum\limits_{cyc} a^2)^2$
so my problem is simplified to proving this:
$\frac {(\sum\limits_{cyc} a^2)^2} {(\sum\limits_{cyc} a(b+c+d))} \geq \frac {1} {3}$
$3 \times (\sum\limits_{cyc} a^2)^2 \geq \sum\limits_{cyc} a(b+c+d)$
$3 \times (a^2+b^2+c^2+d^2)^2 \geq 2(ab+ac+ad+bc+bd+cd)$
someone said to me if i play around with AM-GM it could be solved and i'm almost there
my idea is this right now:
prove $(a^2+b^2+c^2+d^2) \geq ab+bc+cd+da=1$ proved(with help of Jineon Baek hint)
prove $3(a^2+b^2+c^2+d^2) \geq 2(ab+ac+ad+bc+bd+cd)$ proved(with help of Jineon Baek hint)
| This solution does not use Cauchy inequality, but I think it is interesting by its own.
Multiply the right-hand side by $ab+bc+cd+da$. Now the inequality is homogeneous so we can
assume $a+b+c+d=1$ instead of $ab+bc+cd+da = 1$ by multiplying a common constant to all variables. Now we have to show this.
$$\sum_{cyc} \frac{a^3}{1-a} \geq \frac{ab+bc+cd+da}{3}$$
By Cauchy or rearrangement inequality or whatever, $\sum_{cyc} a^2 \geq \sum_{cyc} ab$. So we only need to prove this.
$$\sum_{cyc} \left( \frac{a^3}{1-a} - \frac{a^2}{3} \right) \geq 0$$
The function $f(x) = \frac{x^3}{1-x} - \frac{x^2}{3}$ is convex on $[0, 1]$. We just assumed $a+b+c+d=1$, so Jensen's inequality proves this.
@Macavity noticed that the function is not convex on whole interval, so this solution is not correct :( Since the interval where the function is convex is quite large, maybe we can divide the case by whether all variables are large enough or not. But that solution would be a mass.
@Macavity just gave a full solution for this. Instead of showing that $f$ is convex, it is enough to show that the tangent line of $f(x)$ at $x=1/4$ is less or equal to $f$. It seems to be a decent technique for proving competition-style inequalities.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Antiderivative of $\frac{1}{1+\sin {x} +\cos {x}}$ How do we arrive at the following integral
$$\displaystyle\int\dfrac{dx}{1+\sin {x}+\cos {x}}=\log {\left(\sin {\frac{x}{2}}+\cos {\frac{x}{2}}\right)}-\log {\left(\cos {\frac{x}{2}}\right)}+C\ ?$$
| Hint :
Rewrite the integrand as
\begin{align}
\frac{1}{1+\color{blue}{\sin x}+\color{red}{\cos x}}&=\frac{1}{\sin^2\frac x2+\cos^2\frac x2+\color{blue}{2\sin\frac x2\cos\frac x2}+\color{red}{\cos^2\frac x2-\sin^2\frac x2}}\\
&=\frac{1}{2\cos\frac x2\left(\sin\frac x2+\cos\frac x2\right)}\\
&=\frac{\sin\frac x2}{2\cos\frac x2}+\frac{\cos\frac x2-\sin\frac x2}{2\left(\sin\frac x2+\cos\frac x2\right)}.
\end{align}
Then use
$$
\int\frac{f'(x)}{f(x)}\ dx=\ln\left|f(x)\right|+C.
$$
I think this way will obtain a shorter way than a standard Weierstrass substitution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$ $$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$$
Prove that L.H.S.$=$R.H.S.
This type of questions always creates problem when in right hand side some trigonometry function is given then it is bit easy to think how to proceed further.
Can some one help me not only to solve the problem but also how to tackle this type of other problem (when R.H.S. is $1$)?
| HINT
Simplify as follows : substitute $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$. By this, we will get first term of LHS is
$\frac{\sin \theta \ \cos \theta }{1 + \sin \theta - \cos \theta }$. Similar if we substitute \second term of LHS, we will get LHS =
$\frac{\sin \theta \ \cos \theta }{1 + \sin \theta - \cos \theta }$ + $\frac{\sin \theta \ \cos \theta }{1 + \cos \theta - \sin \theta }$
$= \sin \theta \ \cos\theta [\frac {1}{1 + \sin \theta - \cos \theta} + \frac {1}{1 + \cos \theta - \sin \theta } ]$
$=\sin \theta \ \cos\theta [\frac {1}{1 + \sin \theta - \cos \theta} + \frac {1}{1 - (\sin \theta - \cos \theta) } ]$
Now simplifying $[\frac {1}{1 + \sin \theta - \cos \theta} + \frac {1}{1 - (\sin \theta - \cos \theta) } ]$ we get $\frac {1}{\cos \theta \sin \theta}$. But it is with product of $\sin \theta \cos \theta$. Hence the answer is $1$, Which is LHS. Hence proved.
And we should tackle these problems always by substituting in $\sin \theta$ and $\cos \theta$ when LHS is a number.
| {
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"url": "https://math.stackexchange.com/questions/877936",
"timestamp": "2023-03-29T00:00:00",
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Calculating a cubic spline goes wrong I am trying to solve a old exam and really stuck at the cubic splines.
We have the function $f(x) = \cos^2(\frac{x}{2})$ and the points $x_0 = \frac{\pi}{2}$, $x_1=0$ and $x_2 = \frac{\pi}{2}$.
With some simple math we get the points:
$(-\frac{\pi}{2},\frac{1}{2}),(0,1),(\frac{\pi}{2},\frac{1}{2})$.
The splines have the form:
$$ p_i(x) = a_i + b_i(x-x_i) + c_i(x-x_i)^2 + d_i(x-x_i)^3$$
So we get the $a_i$ simply by $a_i = y_i$.
some formulae that are needed
$$
\begin{align}
h_i &= x_{i+1} - x_i\\
b_i &= \frac{1}{h_i}(a_{i+1}-a_i) - \frac{h_i}{3}(c_{i+1}+2c_i)\\
d_i &= \frac{1}{3h_i}(c_{i+1}-ci)\\
c_1 &= 0, \ \ c_n=0
\end{align}$$
So next step for the calculation is to get the $c_i$ out of $Ac=r$
with
$$
A=\left[
2(h_1+h_2) \dots
\right]
$$
Since we just got 3 points it is an $1\times 1$ matrix (else it would be bigger). So I calculated the matrix $A=(2\pi)$ since $h_1 = h_2 = \frac{\pi}{2}$ with
$$
r = \frac{3}{h_2}(a_3-a_2)-\frac{3}{h1}(a_2-a_1)
$$
Out of this we get that $c_2 = -\frac{2}{3}$ if I am not wrong. And now we can calculate the left over coefficients.
$$
\begin{align}
b_1 &= \frac{1}{\pi} + \frac{9\pi}{8} \\
b_2 &= -\frac{1}{\pi} - \frac{18\pi}{8} \\
d_1 &= -\frac{6}{12\pi}\\
d_2 &= \frac{6}{12\pi}
\end{align}
$$
BUT the splines are not correct. For example for the first interval I get
$$
p_1(x) = \frac{1}{2} + (\frac{1}{\pi} + \frac{9\pi}{8})(x-\frac{\pi}{2}) - \frac{6}{12\pi}(x-\frac{\pi}{2})^3\\
p_2(x) = \frac{1}{2} + (\frac{1}{\pi} + \frac{9\pi}{8})(x-0) - \frac{6}{12\pi}(x-0)^3
$$
So what have i done wrong? Most of the stuff is just using formulae but I don't get the right solution out of it.
In the end thanks for reading this long post!
| I hope that the mistake I pointed out in a comment will help to straighten out the computation. Meanwhile, let me present a different approach, which may be more illuminating that "put these numbers into these formulas".
The given data points are symmetric about the $y$-axis. So will be the spline. In particular, the spline will have derivative zero at $x=0$. By symmetry, we only need to find it on the interval $[-\pi/2,0]$.
First, find the piecewise linear function through given points: on the interval $[-\pi/2,0]$ it is $l(x)=1+x/\pi$. On the whole interval the piecewise linear interpolant looks like this.
Second, look for the cubic correction term $c(x)$, which must be zero at the nodes of interpolation, have zero second derivative at $-\pi/2$, and have derivative $-1/\pi$ at $x=0$ (to cancel out the slope of linear term, which creates a corner on the plot above). It is convenient to use Taylor form at $-\pi/2$, in the following fashion:
$$c(x) = a\left(\frac{2}{\pi}x + 1\right)^3 + b\left(\frac{2}{\pi}x + 1\right)$$
This choice simplifies the algebra. Indeed, $c(0)=0$ implies $b=-a$ and $c'(0)=-1/\pi$ yields
$$ 3\frac{2}{\pi}a - \frac{2}{\pi}a = -\frac{1}{\pi}$$
hence $a=-1/4$.
Final result: for $-\pi/2\le x\le 0$ the spline is
$$
p(x) = 1+\frac{1}{\pi}x - \frac14 \left(\frac{2}{\pi}x + 1\right)^3 + \frac14\left(\frac{2}{\pi}x + 1\right)
$$
On the other interval it's symmetric: $p(x)=p(-x)$.
Here is the plot of piecewise linear interpolant (blue), cubic correction term (red), and their sum (black), which is the desired spline.
Derivation from basic principles is generally more enlightening than mimicking a spline-generating computer routine.
| {
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How to show $n(n+1)(2n+1) \equiv 0 \pmod 6$? I've been asked to show that:
$n(n+1)(2n+1) \equiv 0 \pmod 6$
I found in a previous question that:
$n(n+1)$ was divisible by $2$ and resulted in an even number e.g
$n(n+1) \equiv 0 \pmod 2$
so I figured I needed to find:
$(2n+1) \equiv 0 \pmod 3$ in order to complete
$n(n+1)(2n+1) \equiv 0 \pmod 6$
but I am unsure on how to find $(2n+1) \equiv 0 \pmod 3$
Is this the right way to find the mod 6 and if so could you tell me how I could find
$(2n+1) \equiv 0 \pmod 3?$
| The numbers $2n$, $2n+1$, and $2n+2$ are three consecutive numbers, so one of them is divisible by $3$. If $2n+1$ is not, then one of $2n$ or $2n+2$ is. But if $2n$ is divisible by $3$, so is $n$. And if $2n+2$ is divisible by $3$, so is $n+1$. Thus one of $n$, $2n+1$, and $n+1$ is divisible by $3$.
| {
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"source": "stackexchange",
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"answer_count": 5,
"answer_id": 2
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Show that there is an angle $\theta$ such that $a=\cos\theta$ and $b=\sin\theta$ My problem is from Israel Gelfand's Trigonometry textbook.
Page 50. Exercise 3: Suppose that $\alpha$ is some angle. If $a=4\cos^3\alpha-3\cos\alpha$ and $b=3\sin\alpha-4\sin^3\alpha$, show that there is an angle $\theta$ such that $a=\cos\theta$ and $b=\sin\theta$
The attempt at a solution:
In order to show that, I understand that I have to show that $a^2+b^2=1$, now i have expanded $(4\cos^3\alpha-3\cos\alpha)^2+(3\sin\alpha-4\sin^3\alpha)^2$, but all I get is messy expression. I would appreciate some hints, thank you in advance.
| To solve it using the approach the OP wants to take, it helps to use the abbreviations $c$ and $s$ for $\cos\alpha$ and $\sin\alpha$. With these abbreviations, the desired identity becomes
$$16(c^6+s^6)-24(c^4+s^4)+9(c^2+s^2)=1$$
The key thing we know is that $c^2+s^2=1$. Squaring this gives
$$1=(c^2+s^2)^2=c^4+2c^2s^2+s^4\implies c^4+s^4=1-2c^2s^2$$
Cubing it gives
$$1=(c^2+s^2)^3=c^6+3c^4s^2+3c^2s^4+s^6\implies c^6+s^6=1-3c^2s^2(c^2+s^2)=1-3c^2s^2$$
And now we're ready to plug these in:
$$\begin{align}
16(c^6+s^6)-24(c^4+s^4)+9(c^2+s^2)
&=16(1-3c^2s^2)-24(1-2c^2s^2)+9\\
&=16-24+9-48c^2s^2+48c^2s^2\\
&=1
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/880924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove $\int_{\mathbb{R^{+}}} \frac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\frac{\pi}{32} (3\pi-4)^2$ How do you arrive at the result $$I=\displaystyle\int_{\mathbb{R^{+}}} \dfrac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\dfrac{\pi}{32} (3\pi-4)^2\ ?$$
Wolfram Alpha agrees numerically.
I tried replacing $\sin p$ by $\dfrac{e^{ip}-e^{-ip}}{2i}$ and similarly for $\cos p$, but in vain.
| Here I present another way to solve/generalize the problem. It is much easier. Define
$$ I_q(p)=\frac{1}{2}\int_0^\infty\frac{\sin^3(pt)\cos(qt)}{t^3}dt,p\ge0,q\ge0. $$
Clearly $I_q(0)=I_q'(0)=I_q''(0)=0$ and
\begin{eqnarray}
I_q''(p)&=&\frac{3}{8}\int_0^\infty\frac{(3\sin(3pt)-\sin(pt))\cos(qt)}{t}dt\\
&=&\frac{3}{16}\int_0^\infty\frac{3\sin((3p+q)t)+3\sin((3p-q)t)-\sin((p+q)t)-\sin((p-q)t)}{t}dt\\
&=&\left\{\begin{array}{l}
0, \text{ if }p<\frac{q}{3},\\
\frac{9\pi}{32}, \text{ if }p=\frac q3,\\
\frac{9\pi}{16}, \text{ if }\frac{q}{3}<p<q,\\
\frac{15\pi}{32}, \text{ if }p=q,\\
\frac{3\pi}{8}, \text{ if }p>q.
\end{array}\right.
\end{eqnarray}
Here we used
$$ \int_0^\infty\frac{\sin(\alpha x)}{x}dx=\text{sgn}(\alpha)\frac{\pi}{2}. $$
Thus
$$ I_q'(p)=\left\{\begin{array}{l}
0, \text{ if }p\le\frac{q}{3},\\
\frac{9\pi}{16}(p-\frac{q}{3}), \text{ if }\frac{q}{3}<p\le q,\\
\frac{3p\pi}{8}, \text{ if }p>q,
\end{array}\right. $$
and hence
$$ I_q(p)=\left\{\begin{array}{l}
0, \text{ if }p\le\frac{q}{3},\\
\frac{\pi}{32}(3p-q)^2, \text{ if }\frac{q}{3}<p\le q,\\
\frac{(3p^2-q^2)\pi}{16}, \text{ if }p>q.
\end{array}\right. $$
For $p=\pi,q=4$, since $\frac{q}{3}<p<q$, we have
$$ I_4(\pi)=\frac{\pi}{32}(3\pi-4)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/881289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int x^2 \sqrt{x^2-1} dx$ How do I evaluate the following indefinite integral?
$$\int x^2 \sqrt{x^2-1} dx$$
Through integration of parts, I have obtained
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int (x^2-1)^{3/2} dx $$
I've attempted evaluating the second term through substitution, where
$$ x = \sec(u)$$
However, I am stuck with
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int \tan^4(u) \sec (u) du $$
What would be my next step?
| $$\int x^{2}\sqrt{x^{2}-1}\,dx\\
x=\cosh{y}\Rightarrow dx=\sinh{y}dy\\
\int x^{2}\sqrt{x^{2}-1}\,dx=\int \cosh^{2}{y}\sinh^{2}{y}\,dy=\frac{1}{4}\int\sinh^{2}{2y}\,dy\\
=\frac{1}{4}\int\frac{\cosh{4y}-1}{2}\,dy=\frac{\sinh{4y}}{2}-\frac{y}{8}+C\\
=\sinh{2y}\cosh{2y}-\frac{\cosh^{-1}x}{8}+C\\
=2(\cosh{y})(\cosh{2y})(\sinh{y})-\frac{\cosh^{-1}x}{8}+C\\
=2x(2x^{2}-1)\sqrt{x^{2}-1}-\frac{\cosh^{-1}x}{8}+C\\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Why does the imaginary number $i$ satisfy $i\times 0=0$? Why does the imaginary number $i$ satisfy $i\times 0=0$?
I mean, we don't really know what $i$ is. How could we be sure about that? I think there's a reason behind why mathematicians decided that.
| Ignore this answer if you've never heard of matrix multiplication. Better yet, learn matrix multiplication and then read this answer!
The answer to your question depends on what definition of complex numbers you're using.
For example, I like to think of the complex numbers as matrices of the form
$$
\begin{pmatrix}
a & -b \\ b & a
\end{pmatrix}
$$
where $a$ and $b$ are real numbers. This allows us to define two complex numbers
\begin{align*}
\mathbf 0
&=
\begin{pmatrix}
0 & 0\\ 0 & 0
\end{pmatrix} &
i
&=
\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}
\end{align*}
Hence we have the identity
$$
\mathbf 0\cdot i
=
\begin{pmatrix}
0 & 0\\ 0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}
=
\begin{pmatrix}
0\cdot 0+0\cdot 1 & 0\cdot(-1)+0\cdot 0\\
0\cdot 0+0\cdot 1 & 0\cdot (-1)+0\cdot 0
\end{pmatrix}
=
\begin{pmatrix}
0 & 0\\ 0 & 0
\end{pmatrix}
=
\mathbf0
$$
One of the nice things about defining the complex numbers this way is that we avoid the confusing equation $i=\sqrt{-1}$. We also don't have to resort to using silly words like "imaginary."
Note, however, that we do have
$$
i^2
=
\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}^2
=
\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}\begin{pmatrix}
0 & -1 \\ 1 & 0
\end{pmatrix}
=
\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}
$$
So, if we define the complex number
$$
\mathbf{1}=
\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}
$$
then we recover the formula
$$
i^2=-\mathbf{1}
$$
To convince yourself that our definition of the complex number $\mathbf1$ is not arbitrary, note that $\mathbf 1$ enjoys the property
$$
\mathbf 1
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
=
\begin{pmatrix}
1 & 0\\ 0 & 1
\end{pmatrix}
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
=
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
$$
This is remarkably similar to the celebrated equation $1\cdot a=a$. It's also worth noting that the complex number $\mathbf0$ satisfies
$$
\mathbf0+
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
=
\begin{pmatrix}
0&0\\0&0
\end{pmatrix}+
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
=
\begin{pmatrix}
0+a & 0-b \\
0+b & 0+a
\end{pmatrix}
=
\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
$$
which is similar to our usual equation $0+a=a$.
The point here is that complex-arithmetic "feels" like ordinary arithmetic but is indeed different. As @Hurkyl points out, algebraic gadgets that behave like this are called rings.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Polygamma function series: $\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2$ Applying the Copson's inequality, I found:
$$S=\displaystyle\sum_{k=1}^{\infty }\left(\Psi^{(1)}(k)\right)^2\lt\dfrac{2}{3}\pi^2$$ where
$\Psi^{(1)}(k)$ is the polygamma function.
Is it known any sharper bound for the sum $S$?
Thanks.
| Continuing from Olivier Oloa's answer,
$$ \begin{align} \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} &= \int_{0}^{1} \int_{0}^{1} \frac{\ln u \ln v}{(1-uv)(1-u)(1-v)} \ du \ dv \\ &= \int_{0}^{1} \frac{\ln v}{(1-v)^{2}} \int_{0}^{1} \left(\frac{\ln u}{1-u} - \frac{v \ln u}{1-vu} \right) \ du \ dv \\ &= \int_{0}^{1} \frac{\ln v}{(1-v)^{2}} \left( \int_{0}^{1} \frac{\ln u}{1-u} \ du - v \int_{0}^{1} \frac{\ln u}{1-vu} \ du\right) \ dv \end{align}$$
where $$ \int_{0}^{1} \frac{\ln u}{1-u} \ du = \int_{0}^{1} \frac{\ln (1-w)}{w} = -\text{Li}_{2}(1) = -\zeta(2) $$
and
$$ \int_{0}^{1} \frac{\ln u}{1-vu} \ du = - \frac{1}{v} \ln(1-vu) \ln u \Bigg|^{1}_{0} + \frac{1}{v} \int_{0}^{1} \frac{\ln (1-vu)}{u} \ du = - \frac{\text{Li}_{2}(v)}{v} .$$
Therefore,
$$ \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} = \int_{0}^{1} \frac{\ln v}{(1-v)^2} \Big(\text{Li}_{2}(v) - \zeta(2) \Big) \ dv .$$
Then integrating by parts,
$$ \begin{align} &= \big(\text{Li}_{2}(v) - \zeta(2) \big) \left(\ln (1-v) + \frac{v \ln v}{1-v} \right)\Bigg|^{1}_{0} + \int_{0}^{1} \frac{\ln^{2}(1-v)}{v} \ dv + \int_{0}^{1} \frac{\ln(1-v) \ln v}{1-v} \ dv \\ &= \int_{0}^{1} \frac{\ln^{2}(1-v)}{v} \ dv + \int_{0}^{1} \frac{ \ln(1-v) \ln v}{1-v} \ dv \end{align}$$
where $$ \begin{align} \int_{0}^{1} \frac{\ln^{2}(1-v)}{v} \ dt &= \ln^{2}(1-v)\ln v \Bigg|^{1}_{0} + 2 \int_{0}^{1} \frac{\ln(1-v) \ln v}{1-v} \ dv \\ &= 2 \int_{0}^{1} \frac{ \ln(1-v) \ln v}{1-v} \ dv . \end{align} $$
Therefore, $$ \begin{align} \sum_{k=1}^{\infty} \big(\psi^{(1)} (k) \big)^{2} &= 3 \int_{0}^{1} \frac{\ln(1-v) \ln v}{1-v} \ dv \\ &= -3 \int_{0}^{1} \ln v \sum_{n=1}^{\infty} H_{n}v^{n} \ dv \\ & = -3 \sum_{n=1}^{\infty} H_{n} \int_{0}^{1} v^{n} \ln v \ dv \\ &= 3 \sum_{n=1}^{\infty} \frac{H_{n}}{(n+1)^{2}} \\ &= 3 \left(\sum_{n=1}^{\infty} \frac{H_{n+1}}{(n+1)^{2}} - \sum_{n=1}^{\infty} \frac{1}{(n+1)^{3}} \right) \\ &= 3 \left( \sum_{n=1}^{\infty} \frac{H_{n}}{n^{2}} -1 - \zeta(3) + 1 \right) \\ &= 3 \big(2 \zeta(3) - \zeta(3) \big) \tag{1} \\ &= 3 \zeta(3) .\end{align} $$
$ $
$(1)$ Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 4,
"answer_id": 3
} |
Integrate a division of polynomials Hi I have the following integral:
$$\int \frac{2x}{x^2+6x+3}\, dx$$
I made some changes like:
$$\int \dfrac{2x+6-6}{x^2+6x+3}\, dx$$
then I have:
$$\int \dfrac{2x+6}{x^2+6x+3}\, dx -\int\dfrac{6}{x^2+6x+3}\, dx$$
and thus: $$\ln(x^2+6x+3)-\int\dfrac{6}{x^2+6x+3}\, dx$$
Ok, I have decomposed $$\frac{2x}{x^2+6x+3} $$ in: $$ \frac{3+\sqrt6}{\sqrt6(x+\sqrt 6+3)} + \frac{3-\sqrt6}{\sqrt6 (-x+\sqrt6-3)}$$
How can I integrate this expressions?
| $$\frac{2x}{x^2+6x+3}=\frac{A}{x-(-3-\sqrt{6})}+\frac{B}{x-(\sqrt{6}-3)}$$
Find $A$ and $B$ and then:
$$\int \frac{2x}{x^2+6x+3} dx=\int \frac{A}{x-(-3-\sqrt{6})} dx+\int \frac{B}{x-(\sqrt{6}-3)} dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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prove $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$
If $a,b,c$ are positive real numbers and $a+b+c=1$,Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$$
Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have done so far:
Using Cauchy inequalities I can write:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})(b+c+a) \geq (a+b+c)^2$$
$a+b+c=1$,So:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})\geq 1$$
| In this case you want to have quadratic terms in the RHS, so use Cauchy in the following form:
$$\left(\sum_{cyc} \frac{a^2}b \right) \left(\sum_{cyc} a^2b \right) \ge \left(a^2+b^2+c^2\right)^2 \tag{1}$$
Then it is enough to show that
$$a^2+b^2+c^2 \ge 3\sum_{cyc} a^2b \tag{2}$$
Homogenising,
$$\iff \left( a^2+b^2+c^2\right) (a+b+c) \ge 3\sum_{cyc} a^2b $$
$$\iff \sum_{cyc} a^3+ \sum_{cyc} ab^2 \ge 2\sum_{cyc} a^2b $$
which follows from AM-GM as $a^3+ab^2 \ge 2a^2b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int_0^{\sqrt{5}} \sqrt{4x^4 + 16 x^2} dx$ $$\int_0^{\sqrt{5}} \sqrt{4x^4 + 16 x^2} dx$$
The square root really got me confused here. I've tried using the standard trick with $x = x + \sqrt{x^2+4}$ but failed. That might still be the best approach.
What's the best way to solve this and how to do it?
| Hint
\begin{align}
\int^\sqrt{5}_0\sqrt{4x^4+16x^2} \ dx
&=2\int^{\sqrt{5}}_0 x\sqrt{x^2+4} \ dx \\
\end{align}
Now let $x=2\tan{u}$, then $dx=2\sec^2{u} \ du$.
\begin{align}
2\int^{\sqrt{5}}_0 x\sqrt{x^2+4} \ dx
&=8\int^{\arctan\left(\frac{\sqrt{5}}{2}\right)}_0 \tan{u}\sec^3{u} \ du\\
\end{align}
Use the substitution $y=\sec{u} \implies dy=\sec{u}\tan{u} \ du$. You should be able to take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int x\sqrt{(a^2 - x^2)}dx$ I need to find $$\int x\sqrt{(a^2 - x^2)}dx$$
I tried putting $x=a cos(t)$ but I ended up getting a very complicated expression, so any tips?
| $-\frac{1}{3}(a^2-x^2)^\frac{3}{2}$ is one of the origional function because $\frac{d}{dx}\bigg(-\frac{1}{3}(a^2-x^2)^\frac{3}{2}\bigg)=x\sqrt{a^2-x^2}.$
Why did I come up with it because $\int x\sqrt{a^2-x^2}dx=\int\sqrt{a^2-x^2}d(x^2)=-\frac{1}{3}(a^2-x^2)^\frac{3}{2}$+C.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How can I calculate $AB$ in a quadrilateral figure $(ABCD)$ $AD=3cm$, $DC=4cm$, $BC=5cm$. The diagonals of this quad figure must also be perpendicular. Find $AB$.
| Let $O$ be the intersecting point between the two diagonals $AC$ and $BD$.
Then by Pythagoras' Theorem we have the following identities:
$$\begin{cases}
AO^2 + DO^2 = AD^2 = 9 & (1)\\
DO^2 + CO^2 = CD^2 = 16 & (2)\\
CO^2 + BO^2 = BC^2 = 25 & (3)\\
BO^2 + AO^2 = AB^2 & (4)
\end{cases}$$
Note that $(1) + (3) - (2) = (4)$ thus $AB^2 = 9+25-16 = 18\Rightarrow AB = 3\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/886983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to find $\int_{0}^{\pi/2} \log ({1+\cos x}) dx$ using real-variable methods? How do you find the value of this integral, using real methods?
$$I=\displaystyle\int_{0}^{\pi/2} \log ({1+\cos x}) dx$$
The answer is $2C-\dfrac{\pi}{2}\log {2}$ where $C$ is Catalan's constant.
| $$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \ln (1+\cos x) d x &=\int_{0}^{\frac{\pi}{2}} \ln \left(2 \cos ^{2} \frac{x}{2}\right) d x \\
&=\int_{0}^{\frac{\pi}{2}} \ln 2 d x+2 \int_{0}^{\frac{\pi}{2}} \ln \left(\cos \frac{x}{2}\right) d x \\
&=\frac{\pi}{2} \ln 2+4 \int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x \\
&=\frac{\pi}{2} \ln 2+4\left(-\frac{\pi}{2} \ln 2+\frac{G}{2}\right)
\end{aligned}
$$ where the last integral comes from my post:
$\int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x=-\frac{\pi}{4} \ln 2+\frac{G}{2}$.
Hence we can conclude that
$$
\boxed{\int_{0}^{\frac{\pi}{2}} \ln (1+\cos x) d x= \frac{G}{2}-\frac{\pi}{2} \ln 2,}
$$
where G is the Catalan’s constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Find $ \int \sqrt{\frac{1}{\theta^2}+ \frac{1}{\theta^4}} d\theta$ Any Ideas ! is this integrable function
| Just for fun, an alternate approach: If $x = 1 + \frac{1}{\theta^2}$, then $\frac{d x}{d \theta} = - \frac{2}{\theta^3} = \frac{1}{\theta} \frac{-2}{\theta^2}$. So, the integral of interest is transformed to
$$
\int \frac{1}{\theta} \sqrt{ 1 + \frac{1}{\theta^2} } d \theta
= -2 \int (x-1) \sqrt{ x } d x
$$
Where $u = \sqrt{x}$, the integrand transforms to $\frac{(u^2 -1)u}{2u} du = \frac{1}{2} (u^2 - 1) \, du $. Hence,
$$
\int \frac{1}{\theta} \sqrt{ 1 + \frac{1}{\theta^2} } d \theta
= \int 1 - u^2 du = u - \frac{u^3}{3} + C
= \frac{1}{3} \sqrt{1 + \frac{1}{\theta^2}} \left( 2 - \frac{1}{\theta^2} \right) + C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
For a triangle $ABC$, $a^2+b^2+c^2=8R^2$ then it is a right triangle? $ABC$ is a triangle, $a^2+b^2+c^2=8R^2$ then how do we prove it is a right triangle?
| Let $S$ be the area of the triangle.
Since we have $4RS=abc$ and
$$\begin{align}16S^2&=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\\&=-(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)\end{align}$$
if $a^2+b^2+c^2=8R^2$, then we have
$$\begin{align} 0 &=\frac{a^2+b^2+c^2}{R^2}-8\\&=\frac{a^2+b^2+c^2}{a^2b^2c^2}(-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2)-8\\&=\frac{(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}{a^2b^2c^2}.\end{align}$$
$$\Rightarrow (-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)=0$$
So, the triangle has to be a right triangle.
P.S. Actually, $a^2+b^2+c^2=8R^2\iff\text{The triangle $ABC$ is a right triangle}$.
Also, we can see the followings :
$$a^2+b^2+c^2\gt 8R^2\iff \text{The triangle $ABC$ is an acute‐angled triangle}.$$
$$a^2+b^2+c^2\lt 8R^2\iff \text{The triangle $ABC$ is an obtuse triangle}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/889946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplifying a trigonometric identity Simplify $1 + \tan^2x$
My attempt:
$$\begin{align}1 + \tan^2x&\\
&= \frac{1}{1} + \frac{\sin^2x}{\cos^2x}\\
&= \frac{1(\cos^2x)}{1(\cos^2x)} +\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x + \sin^2x}{\cos^2x\cos^2x}\\
&= \frac{\sin^2x}{\cos^2x}\\
&= \tan^2x\end{align}$$
The correct answer, however..is $sec^2x$ Wherever I went wrong, please show.
| Firstly, I think that everyone (including yourself) disagrees with the assertion that in general
$$\frac{a}{a}+\frac{b}{a}=\frac{a+b}{a^2}$$
by the distributive law
$$\frac{a}{a}+\frac{b}{a}=\frac{1}{a}(a+b)=\frac{a+b}{a}$$
Secondly, I think that $\tan^2x+1=\sec^2x$ should be considered to be just as basic of an identity as $\sin^2x+\cos^2x=1$ is. My reasoning is as follows. Taking the Pythagorean relationship
$$opposite^2+adjacent^2=hypotenuse^2$$
and dividing both sides by $hypotenuse^2$ (yielding $\sin^2 x+\cos^2x=1$) is just as simple as dividing said relationship by $adjacent^2$, yielding ($\tan^2x+1=\sec^2x$). Because of that reasoning, I would either just state the identity
$$\tan^2x+1=\sec^2x$$
or perform a more elemental proof
$$\begin{array}{lll}
opposite^2+adjacent^2&=&hypotenuse^2\\
\frac{opposite^2+adjacent^2}{adjacent^2}&=&\frac{hypotenuse^2}{adjacent^2}\\
\frac{opposite^2}{adjacent^2}+\frac{adjacent^2}{adjacent^2}&=&\frac{hypotenuse^2}{adjacent^2}\\
\bigg(\frac{opposite}{adjacent}\bigg)^2+\bigg(\frac{adjacent}{adjacent}\bigg)^2&=&\bigg(\frac{hypotenuse}{adjacent}\bigg)^2\\
\tan^2x+1&=&\sec^2x&
\end{array}$$
unless instructed to do otherwise
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/892065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
any other method for evaluating $\int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx$? I tried below and its getting tedious :
$\begin{align}\\ \int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx &= \int\limits \frac{(2x-1)+ x^2-x+2 }{ \sqrt{x^2-x+1} } dx \\~\\ &= \int\limits \frac{(2x-1)dx}{ \sqrt{x^2-x+1} } + \int\limits \frac{ (x^2-x+1 )dx}{ \sqrt{x^2-x+1} } + \int\limits \frac{ dx }{ \sqrt{x^2-x+1} } \\~\\&\cdots\\~\\ \end{align}$
wolfram shows very much simplified answer :
http://www.wolframalpha.com/input/?i=%5Cint+%28x%5E2%2Bx%2B1%29%2F%28sqrt%28x%5E2-x%2B1%29%29
I'm wondering if there is any nice way to work this
| HINT:
I think using Trigonometric substitution of
$\displaystyle x^2-x+1=\frac{(2x-1)^2+(\sqrt3)^2}4$
with $2x-1=\sqrt3\tan\theta$ won't be a bad way to start with
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of all possible remainders when $2^n$, where n is a nonnegative integer, is divided by 1000
Let $R$ be the set of all possible remainders when a number of the
form $2^n$, $n$ a nonnegative integer, is divided by $1000$. Let $S$
be the sum of all elements in $R$. Find the remainder when $S$ is
divided by $1000$.
I am trying to understand the provided solution:
Consider the subset $R'$ of $R$ consisting of only those numbers which
are divisible by $8$, the highest power of $2$ dividing $1000$. Since
$\gcd(2,125) = 1$, by the Chinese Remainder Theorem the elements of
$R'$ cycle $\text{mod } 125$. Hence $R'$ stays the same $\text{mod }
> 1000$ when we multiply all elements by $2$. This means that if $S'$ is
the sum of the elements of $R'$, then $S' \equiv 2S' \pmod{1000}$, so
$S'$ is a multiple of $1000$.
Since $S = 1 + 2 + 4 + S'$, $S$ is equivalent to $\boxed{7} \text{ mod
> } 1000$.
I do not understand how they applied the Chinese remainder theorem to arrive at the fact that the elements cycle mod $125$. Further, how does this allow them to conclude that $R'$ remains the same mod 1000 when all elements are multiplied by 2?
Finally, how did they get that $S = 1 + 2 + 4 + S'$?
Also, an alternate solution here relies on the fact that $2^0, 2^1,\ldots, 2^{99}$ are distinct modulo 125. They prove this as follows:
Suppose for the sake of contradiction that they are not. Then, we must
have at least one of $2^{20}\equiv 1\pmod{125}$ or $2^{50}\equiv 1\pmod{125}$. However, writing $2^{10}\equiv 25 - 1\pmod{125}$, we can
easily verify that $2^{20}\equiv -49\pmod{125}$ and $2^{50}\equiv -1\pmod{125}$, giving us the needed contradiction.
Here, how did they arrive at the fact that at least one will be true: $2^{20}\equiv 1\pmod{125}$ or $2^{50}\equiv 1\pmod{125}$?
| Well for the second solution they used Euler's theorem
$$2^{\varphi(125)}\equiv2^{100}\equiv1\pmod{125}$$
So either $2^{50}\equiv-1\pmod{125}$ or $2^{20}\equiv 1\pmod{125}$
First part can be proved by euler's theorem since $2^{100k+n}\equiv2^{100k}\cdot2^{n}\equiv1\cdot2^{n}\pmod{125}$ it means it cycles $\pmod{125}$ note that maybe some divisor of $100$ raised to $2$ may also give residue $1$ divided by $\pmod{125}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$ Here is a challenging one maybe some would like a go at.
Show that:
$$\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx=\frac{-9\pi^{4}}{256}+\frac{\pi^{3}}{32}\ln2+\frac{\pi^{2}}{6}G-\frac{1}{1536}\left[\psi_{3}\left(\frac34\right)-\psi_{3}\left(\frac14\right)\right]$$
| Following the same idea of @Cody we have
$$I=\int_{0}^{1}\frac{x\ln^{3}x}{(1-x)(1+x^2)}\ dx+\int_{0}^{1}\frac{\ln^{2}x\ln(1+x)}{1+x^2}\ dx=K+J$$
\begin{align}
K&=\int_{0}^{1}\frac{x\ln^{3}x}{(1-x)(x^{2}+1)}\ dx\\
&=\frac12\int_0^1\frac{x\ln^3x}{1+x^2}\ dx-\frac12\int_0^1\frac{\ln^3x}{1+x^2}\ dx+\frac12\int_0^1\frac{\ln^3x}{1-x}\ dx\\
&=\frac12\left(-\frac{21}{64}\zeta(4)\right)-\frac12\left(-6\beta(4)\right)+\frac12\left(-6\zeta(4)\right)\\
&=-\frac{405}{128}\zeta(4)+3\beta(4)
\end{align}
The integral $J$ is evaluated here in two methods
$$J=\int_0^1\frac{\ln^2x\ln(1+x)}{1+x^2}\ dx=\frac{\pi^3}{32}\ln2+\zeta(2)G-2\beta(4)$$
Combining the results of $K$ and $J$ we have
$$I=-\frac{405}{128}\zeta(4)+\frac{\pi^3}{32}\ln2+\zeta(2)G+\beta(4)$$
Substituting $\beta(4)=\frac1{768}\left(\psi_3(1/4)-8\pi^4\right)$ along with $\zeta(4)=\frac{\pi^4}{90}$ and $\zeta(2)=\frac{\pi^2}{6}$ we get
$$I=\frac{\pi^2}{6}G+\frac{\pi^{3}}{32}\ln2-\frac{35}{768}\pi^4+\frac{1}{768}\psi_{3}(1/4)$$
| {
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"url": "https://math.stackexchange.com/questions/895660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
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} |
Solving $x^3+y^3=x^2y^2+1$ in non-negative integers I wanted to solve $x^3+y^3=x^2y^2+1$ in non-negative integers.
First I set $a=x+y$ and $b=xy$ to get $b^2+3ab+1=a^3$. View as a quadratic in $b$, the discriminant = $4a^3+9a^2-4$, which needs to be a perfect square.
Secondly, rearranging the quadratic in $b$ we get $4a^3+9a^2-4=(2b+3a)^2$.
So the discriminant is always a perfect square. Therefore we have (quadratic formula):
$b=\frac{-3a\pm (2b+3b)}{2}$ so $b\in \{b,-\frac{3a}{2}\}$.
Since we want $a,b\ge 0$, the only possibility is $a=b=0$ to give $x=y=0$.
This is the unique solution.
Note: I wasn't sure if it works, I never tried this way before.
Thanks!
| I think the following fact has been violated:-
In the quadratic equation, f(X) = 0, its discriminant ($\triangle$) should be free from X.
It is alright to have $\triangle = 4a^3 + 9a^2 – 4$, but rewriting it as $(2b + 3a)^2$ and use it to solve f(b) = 0 is not allowable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
System of equations in Lagrange multiplier problem Continuing from Confounding Lagrange multiplier problem:
I'm having trouble solving the system of equations below arisen from a Lagrange multiplier problem where we are to optimize $f(x,y,z) = 4x^2 + 3y^2 + 5z^2$ over $g(x,y,z) = xy + 2yz + 3xz = 6$.
$$ \begin{cases} 8x = \lambda (y + 3z) \\ 6y = \lambda (x + 2z) \\ 10 z = \lambda (2y + 3x) \\ xy + 2yz + 3xz = 6 \end{cases} $$
One suggestion I have got is to eliminate the terms $xy$, $yz$ and $xz$, however I have been unable to figure out how to do so. Help much appreciated!
| Adding the equations $8x=\lambda(y+3z)$, $6y=\lambda(x+2z)$, and $10z=\lambda(2y+3x)$ gives
$\lambda(4x+3y+5z)=8x+6y+10z$, so either $\lambda=2$ or $4x+3y+5z=0$.
$\textbf{1)}$ If $\lambda=2$, we have $8x=y+6z$ and $6y=2x+4z$, so $4x-y-3z=0$ and $x-3y+2z=0$.
Then $3(4x-y-3z)-(x-3y+2z)=0\implies 11x-11z=0\implies z=x$, and then
$x-3y+2z=0\implies y=x$.
Substituting into the constraint gives $6x^2=6$, so $x^2=1$ and $x=\pm1$.
Thus points where extrema can occur are $(1,1,1)$ and $(-1,-1,-1)$.
$\textbf{2)}$ If $4x+3y+5z=0$, then $z=-\frac{1}{5}(4x+3y)\implies xy-\frac{2}{5}y(4x+3y)-\frac{3}{5}x(4x+3y)=6$
$\implies
5xy-8xy-6y^2-12x^2-9xy=30\implies -12x^2-12xy-6y^2=30\implies$ $2x^2+2xy+y^2=-5\implies(y+x)^2+x^2=-5,$ so there is no solution in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Tangent to curve $x^3+y^3=a^3$ meets it again. Tangent to curve $x^3+y^3=a^3$ at $(x_1,y_1)$ meets it again at $(x_2,y_2)$.How to prove that $$\frac{x_2}{x_1}+\frac{y_2}{y_1}+1=0$$
Since $y'=-\frac{x_1^2}{y_1^2}$
$$\frac{y_2-y_1}{x_2-x_1}=-\frac{x_1^2}{y_1^2}=\frac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}$$
Solving we get:
$$\frac{x_2}{x_1}+\frac{y_2}{y_1}+\frac{x_2^2}{x_1^2}+\frac{y_2^2}{y_1^2}+2=0$$
Or
$$\left(\frac{x_2}{x_1}+\frac{y_2}{y_1}\right)+\left(\frac{x_2^2}{x_1^2}+\frac{y_2^2}{y_1^2}\right)^2=2\frac{x_2y_2-x_1y_1}{x_1y_1}$$
Which isn't what is nedded to prove. can someone guide me?
| $$\frac{y_2-y_1}{x_2-x_1} = \frac{y_2^3-y_1^3}{x_2^3-x_1} \times \frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2} = (-)\frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2}$$
And you have written $$\frac{y_2-y_1}{x_2-x_1} = \left(\frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2}\right) $$
Which is wrong.
So we have
$$\frac{x_1^2}{y_1^2} = \frac{x_2^2+x_1x_2+x_1^2}{y_2^2+y_1y_2+y_1^2} $$
This gives,
$$ x_1^2y_2^2+x_1^2y_1y_2+x_1^2y_1^2=y_1^2x_1x_2+x_1^2y_1^2+x_2^2y_1^2 $$
$$ x_1^2y_2^2+x_1^2y_1y_2=y_1^2x_1x_2+x_2^2y_1^2 $$
$$(x_1y_2+x_2y_1)(x_1y_2-x_2y_1)= x_1y_1(x_2y_1-y_2x_1)$$
$$ \frac{x_2}{x_1}+\frac{y_2}{y_1}+1 = 0 \Box $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Initial value $\left ( \frac{dy}{dt} \right )+3y=11$, $y(0)=1$ I have never done an initial value problem, and would like some help on how to start this please.
| $$\frac{dy}{dt} + 3y = 11\ \ ... (1)
$$
homogeneous form is
$$
\frac{dy}{dt} + 3y = 0\\
\int \frac{dy}{y} = -3\int dt\\
\log{|y|} = -3t + C\\
y = \exp(-3t + C)\\
y = C\exp(-3t)\\
$$
variation of Parameters
C => u(t)
$$
y = u(t)\exp(-3t)\ \ ...(2)\\
\frac{dy}{dt} = \frac{du}{dt}\exp(-3t) - 3u\exp(-3t)\ \ ...(3)
$$
(2), (3) => (1)
$$ \frac{du}{dt}\exp(-3t)-3u\exp(-3t) + 3u\exp(-3t) = 11\\
\frac{du}{dt}\exp(-3t) = 11\\
\int du = 11\int \exp(3t)dt\\
u = \frac{11}{3}\exp(3t) + C
$$
therefore,
$$
y = (\frac{11}{3}\exp(3t) + C)\exp(-3t)\\
y = C\exp(-3t) + \frac{11}{3}
$$
now, y(0) = 1
$$
y(0) = 1 = C + \frac{11}{3}\\
C = -\frac{8}{3}
$$
so,
$$
y = -\frac{8}{3}\exp(-3t) + \frac{11}{3}
$$
| {
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"url": "https://math.stackexchange.com/questions/898873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Where did the negative answer come from? The question is to evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$
$$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$$
$$x^2=2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$$
$$x^2=2+x$$
$$x^2-x-2=0$$
$$(x-2)(x+1)=0$$
$$x=2,-1$$
because $x$ is positive $x=2$ is the answer. but where did the $x=-1$ come from ?
| By squaring both sides, you turn an equation with one solution into an equation with two solutions; you can end up creating an extraneous solution. In this case, you come to find that $-1$ satisfies $x^2 = 2 + x$, but does not satisfy $x = \sqrt{2 + \sqrt{2 + \cdots}}$, and so is extraneous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898964",
"timestamp": "2023-03-29T00:00:00",
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A $n\cdot n$ square grid problem? I thought of this problem when I was playing a game called BINGO with my friend. The game basically is like this:
Suppose $2$ people are playing the game(can be played with any no of people though). Both make a $5\cdot 5$ square grid and and fill the numbers from $1$ to $25$ in the grid at random, i.e. you can fill any numbers anywhere inside the grid. So each has his own grid and doesn't show it to other.
The first player now calls any number between $1$ to $25$. The second player then can call any of the remaining $24$ numbers. The players call the numbers alternately and keep circling the numbers on their grid. The first player to get $5$ groups of $5$ numbers across the row, column or diagonal (overlapping of numbers allowed) wins the game. Bingo!
So the final grid may look something like this:
Now, this game can easily be extended to $6\cdot 6$ or $7\cdot 7$ (you may call this BAZINGA). In genreal $n\cdot n$ grid where you have to get $n$ group of $n$ numbers across row,column or diagonal.
Finally, the question is this:
What is the minimum number of numbers to be used to make the group of numbers as required by the game. For $5\cdot 5$ grid, at least 17 entries have to be used (as in the image above). How do I generalise the result for $n\cdot n$ grid?
I think there exists a recursive relation. Although this wouldn't help me in winning the game, your solution is appreciated.
| It depends on $n$ being odd or even so we'll deal with these separately:
Odd:
The minimum is as how you've drawn it for $n=5$. We have:
Cells down the first $\frac{n-1}{2}$ vertical lines: $\frac{n-1}{2} \times n$.
Extra cells across the first $\frac{n-1}{2}$ horizontal lines: $\frac{n-1}{2} \times \frac{n+1}{2}$.
Extra cells for the diagonal: $1$.
Summing:
\begin{eqnarray*}
\mbox{Total} &=& \frac{n-1}{2} \times n + \frac{n-1}{2} \times \frac{n+1}{2} + 1 \\
&& \\
&=& \frac{n^2 - n}{2} + \frac{n^2-1}{4} + 1 \\
&& \\
&=& \frac{1}{4} \left(3n^2 - 2n + 3\right)
\end{eqnarray*}
This gives $17$ when $n=5$, which agrees with your answer.
Even:
The minimum is with the first $\frac{n}{2}$ cols, the first $\frac{n}{2} - 1$ rows plus the diagonal (from top-right corner). We have:
Cells down the first $\frac{n}{2}$ vertical lines: $\frac{n}{2} \times n$.
Extra cells across the first $\frac{n}{2} - 1$ horizontal lines: $\frac{n}{2} - 1 \times \frac{n}{2}$.
Extra cells for the diagonal: $1$.
Summing:
\begin{eqnarray*}
\mbox{Total} &=& \frac{n}{2} \times n + \left(\frac{n}{2} - 1\right) \times \frac{n}{2} + 1 \\
&& \\
&=& \frac{1}{2}n^2 + \frac{1}{4}n^2 - \frac{n}{2} + 1 \\
&& \\
&=& \frac{1}{4} \left(3n^2 - 2n + 4\right)
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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When does $2^n+n \mid 8^n+n$? How to find all positive integers $n$ such that $2^n+n$ divides $8^n+n$ ?
| Since $$2^n\equiv -n\pmod{2^n+n}$$ we deduce $$8^n = (2^n)^3 \equiv (-n)^3 \pmod {2^n+n}$$
So $2^n+n\mid 8^n+n$ if and only if $2^n+n\mid n-n^3$.
For $n\geq 10$, $2^n>n^3$ so $2^n+n$ cannot divide $n^3-n=-(n-n^3)$.
Clearly, if $n=0,1$, $n^3-n=0$ so $2^n+n\mid n^3-n$.
So you really only need to check additionally $n=2,3,4,5,6,7,8,9$ by hand.
You get $n=0,1,2,4,6$. (Technically, we should exclude $n=0$ since the question asked for positive integers...)
| {
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Why does $(a+b)^2= a^2+b^2 + 2ab$? Why is the $2ab$ there? When I was doing research on finding the derivative I came across something strange.
If $f(x) = x^2$ you find the derivative by going
$$\frac{f(x+h)^2-f(x)^2}{h}
=\frac{x^2+2xh+h^2-x^2}{h}.$$
Why is there the $2xh$? Can someone explain the logic behind this? I'm assuming you plus $x$ and $h$ together then square it but why?
| \begin{align*}
(a+b)(a + b) &= a(a + b) + b(a + b)\quad &(\text{addition distributes over multiplication})\\
&= aa + ab + ba + bb\quad &(\text{multiplication distributes over addition})\\
&= a^2 + ab + ba + b^2\quad &(xx = x^2)\\
&= a^2 + ab + ab + b^2\quad &(\text{multiplication is commutative})\\
&= a^2 + 2ab + b^2
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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square of complex numbers I have this equation from here:
but it is not equal to:
$$(a + bi)^2 = a^2 + 2abi + (bi)^2.$$
could someone explain me what is the difference between this two calcultion?
| $|a+bi|^2 = a^2+b^2$, as you say. And $(a + bi)^2 = a^2 + 2abi + (bi)^2$, as you say.
You would expect that $|a+bi|^2 = |(a+bi)^2|$, and that's what happens:
$(a+bi)^2 = (a^2-b^2) + (2ab)i$
So |$(a+bi)^2| = \sqrt{(a^2-b^2)^2 + (2ab)^2} = \sqrt{a^4+b^4 + 2a^2b^2} = \sqrt{(a^2+b^2)^2} = a^2 + b^2$
| {
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How solve this equation $\sin x\cdot \sin20=2\sin(110-x) (\sin10)^2$ let
$0<x<90$, and such
$$\sin x\cdot \sin20=2\sin{(110-x)}(\sin10)^2$$
find the $x$
my idea: since
$$\sin x\cdot 2\sin10\cos10=2\sin(70+x)(\sin10)^2$$
so
$$\cot10=\dfrac{\sin(70+x)}{\sin x}$$
then How find it?
| $2\sin(x)\cdot\sin10\cdot\cos10=2[\cos20\cdot\cos x+\sin20\cdot\sin x]\cdot(\sin10)^2$
$\dfrac{\sin x\cdot\cos10}{\sin10}=\cos(20-x)$
$\dfrac{\cos10}{\sin10}=\dfrac{\cos(20-x)}{\sin x}$
$\Longrightarrow x=10$
| {
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"answer_id": 2
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Calculate $\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS$ where $a^2+b^2+c^2<1$.
Let $a^2+b^2+c^2 < 1$ and $S^2$ be unit sphere in $R^3$. Calculate $$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS$$
Let $(x,y,z)=(\cos\theta \cos\phi,\cos\theta \sin\phi, \sin\theta)$. By definition,
$$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS\\
=\int_{0}^{2\pi}\int_{0}^{\pi}\frac{1}{\sqrt{(\cos\theta \cos\phi-a)^2+(\cos\theta \sin\phi-b)^2+(\sin\theta-c)^2}}\sin\theta d\theta d\phi$$
It is too complicate.
This is calculus exam problem that I took yesterday. Is there any good idea?
| taking my suggested simplification a little further, since $y^2+z^2=1-x^2$ on the surface of the sphere, we obtain:
$$
I = 2\pi \int_{-1}^1 \frac{1}{\sqrt{1-2xr+r^2}}dx
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/907439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
The general solution for this differential equation? Find the general solution of this differential equation:
$$ \frac{dy}{dx} = \frac{3x^5 y^3}{4} $$
Here's what I've done so far:
$ dy=\frac{3x^5 y^3 dx}{4} $
$ 4dy = 3x^5 y^3 dx $
$ \frac{4dy}{y^3} = 3x^5 dx $
Integrating both sides:
$ \int \frac{4dy}{y^3} = \int 3x^5 dx $
$ \int \frac{4dy}{y^3} = \frac{3x^6}{6} + C $
$ 4\int \frac{dy}{y^3} = \frac{x^6}{2} + C $
To solve the left side, let $ u=y^3 $, therefore $ du=3y^2 dy $, so:
$ \frac{4}{3} \int \frac{3y^2 dy}{y^5} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{y^5} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{y^3 y^2} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{u * y^2} = \frac{x^6}{2} + C $
I'm not sure what to do next. How do I get rid of that $ y^2 $? Or did I do something wrong?
| Your equation is separable and you can rewrite it as $$\frac{dy}{y^3}=\frac{3}{4}x^5 dx$$ So, integrating both sides gives $$-\frac{1}{2 y^2}=\frac{x^6}{8}+C_1$$ which, after simplifications, write $$y^2=-\frac{4}{x^6+C_2}$$ that is to say $$y=\pm\frac{2}{\sqrt{C_3 -x^6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Evaluation of $\sum_{k=0}^n{n\choose k}^2u^k$ I am trying to evaluate the finite sum
\begin{equation}
f(u)=\sum_{k=0}^n{n\choose k}^2u^k,\quad 0<u\le1
\end{equation}
In an first attempt, I think of the generating function
\begin{equation}
(1+x)^n(u+x)^n = \sum_{k\ge0}{n\choose k}x^k\sum_{k\ge0}{n\choose k}u^kx^{n-k}=\cdots+x^n\sum_{k\ge0}{n\choose k}^2u^k+\cdots
\end{equation}
which means that $f(u)$ is the coefficients of the term $x^n$.
Expanding $(1+x)^n(u+x)^n$ into $[1+(1+u)x+x^2]^n$, and using the multinomial expansion, I get an expression for the coefficient of $x^n$. However, such an expression is more complicated than $f(u)$. It seems I am making the problem even more difficult.
Can someone help me find the value of $f(u)$.
Thank you.
| The Legendre polynomial $P_n$ happens to satisfy
$$P_n(x) = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k}^2 (x - 1)^{k}(x + 1)^{n-k}.$$
Hence
$$(1-x)^n P_n\left(\frac{x-1}{x+1}\right) = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k}^2 (2x)^{k} 2^{n-k} =\sum_{k=0}^n \binom{n}{k}^2 x^k.$$
To prove the earlier equation, note that Rodrigues' formula gives
$$P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \left((x^2 - 1)^n\right).$$
Hence
\begin{align*}
2^n P_n(x)
&= \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} \left[\frac{d^k}{dx^k} (x - 1)^n\right] \left[\frac{d^{n-k}}{dx^{n-k}} (x + 1)^n\right] \\
&= \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} \left[k! \binom{n}{k} (x - 1)^{n-k}\right] \left[(n-k)! \binom{n}{n-k} (x + 1)^k \right] \\
&= \sum_{k=0}^n \binom{n}{k}^2 (x - 1)^{n-k} (x + 1)^k \\
&= \sum_{k=0}^n \binom{n}{k}^2 (x - 1)^{k} (x + 1)^{n-k}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/912804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
The sequence $(a_0,a_1,a_2,\cdots,)$ satisfies $ a_{n+1}=a_n+2a_{n−1}$. What is $a_5$? Assume that the sequence $(a_0,a_1,a_2,\cdots,)$ satisfies the recurrence $\displaystyle a_{n+1}=a_n+2a_{n−1}$. We know that $a_0=4$ and $a_2=13$. What is $a_5$?
I got $a_1=5, a_3=23, a_4=49, a_5=95$
| If you want to calculate the $n$-th term, here is how:
$a_{n+1} - a_n - 2a_{n-1} = 0 \to x^2 - x - 2 = 0 \to (x-2)(x+1) = 0 \to x = -1, 2$. Thus the
$a_n = A(-1)^n + B2^n$. We have that: $a_0 = 4 \to a_1 = a_2 - 2a_0 = 13 - 2(4) = 5$. So:
$5 = a_1 = A(-1)^1 + B2^1 = -A + 2B$, and
$13 = a_2 = A(-1)^2 + B2^2 = A + 4B$.
Adding these equations we find $B = 3$, and solve for $A = 13 - 4(3) = 1$. Thus:
$a_n = (-1)^n + 3\cdot 2^n$.
Check that $a_5 = (-1)^5 + 3\cdot 2^5 = -1 + 96 = 95$ !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/916379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A closed-form of product the gamma functions containing $\pi$ and $\phi$ Playing with gamma functions by randomly inputting numbers to Wolfram Alpha, I got the following beautiful result
\begin{equation}
\frac{\Gamma\left(\frac{3}{10}\right)\Gamma\left(\frac{4}{10}\right)}{\Gamma\left(\frac{2}{10}\right)}=\frac{\sqrt[\large5]{4}\cdot\sqrt{\pi}}{\phi}
\end{equation}
where $\phi$ is golden ratio.
Could anyone here please help me to prove it by hand? I mean without using table for the specific values of $\Gamma(x)$ except for $\Gamma\left(\frac{1}{2}\right)$. As usual, preferably with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
| In this answer, we derive Gauss's Multiplication Formula:
$$
\prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right)
=\sqrt{n2^{n-1}\pi^{n-1}}\frac{\Gamma(nx)}{n^{nx}}\tag{1}
$$
In this answer, we derive Euler's Reflection Formula:
$$
\Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)\tag{2}
$$
Multiply $(2)$ by $\frac{\Gamma\left(\frac8{10}\right)}{\Gamma\left(\frac8{10}\right)}$ to get
$$
\begin{align}
\frac{\Gamma\left(\frac3{10}\right)\Gamma\left(\frac4{10}\right)}{\Gamma\left(\frac2{10}\right)}
&=\frac{\color{#C00000}{\Gamma\left(\frac3{10}\right)}\Gamma\left(\frac4{10}\right)\color{#C00000}{\Gamma\left(\frac8{10}\right)}}{\color{#00A000}{\Gamma\left(\frac2{10}\right)\Gamma\left(\frac8{10}\right)}}\tag{3a}\\
&=\frac{\color{#C00000}{2\sqrt{\pi}\frac{\Gamma\left(\frac35\right)}{2^{3/5}}}\Gamma\left(\frac25\right)}{\color{#00A000}{\pi\csc\left(\frac\pi5\right)}}\tag{3b}\\
&=\frac{2^{2/5}\sqrt\pi\,\pi\csc\left(\frac{2\pi}5\right)}{\pi\csc\left(\frac\pi5\right)}\tag{3c}\\
&=\frac{2^{2/5}\sqrt\pi}{2\cos\left(\frac\pi5\right)}\tag{3d}
\end{align}
$$
Explanation:
$\mathrm{(3a)}$: multiply numerator and denominator by $\Gamma\left(\frac8{10}\right)$
$\mathrm{(3b)}$: in red, apply $(1)$ with $x=\frac3{10}$ and $n=2$; in green, apply $(2)$ with $x=\frac15$
$\mathrm{(3c)}$: apply $(2)$ with $x=\frac25$
$\mathrm{(3d)}$: $\sin(2x)=2\sin(x)\cos(x)$
We can use the identity $\cos(5x)=16\cos^5(x)-20\cos^3(x)+5\cos(x)$ to get
$$
\begin{align}
-1&=16\cos^5\left(\frac\pi5\right)-20\cos^3\left(\frac\pi5\right)+5\cos\left(\frac\pi5\right)\\
0&=16\cos^5\left(\frac\pi5\right)-20\cos^3\left(\frac\pi5\right)+5\cos\left(\frac\pi5\right)+1\\
&=\left[4\cos^2\left(\frac\pi5\right)-2\cos\left(\frac\pi5\right)-1\right]^2\left[\cos\left(\frac\pi5\right)+1\right]\tag{4}
\end{align}
$$
which implies, since $\cos\left(\frac\pi5\right)\gt0$, that
$$
2\cos\left(\frac\pi5\right)=\phi\tag{5}
$$
Combining $(3)$ and $(5)$ yields
$$
\frac{\Gamma\left(\frac3{10}\right)\Gamma\left(\frac4{10}\right)}{\Gamma\left(\frac2{10}\right)}
=\frac{2^{2/5}\sqrt\pi}{\phi}\tag{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/916853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we have $x^3-y^3\ge 0$ or $(x-y)(x^2+xy+y^2)\ge 0$. Since $x\ge y$ we can divide by $x-y$ to get $x^2+xy+y^2\ge 0$.
Is it right?
Thanks for your help!
| Note that we can prove it by $$x^2+y^2+xy=\left(x+\frac y2\right)^2+\frac{3}{4}y^2\ge 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/920605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove $3(\sin x-\cos x)^4 + 6(\sin x+ \cos x)^2 + 4(\sin^6 x + \cos^6 x) -13 = 0$
Q) Prove that $3(\sin \theta-\cos \theta)^4 + 6(\sin \theta+ \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) -13 = 0$
Source: Trigonometric Functions, Page 5.9, Mathematics XI - R.D. Sharma
My Attempt::
For writing convenience, let $\color{red}{s} = \sin \theta$ and $\color{blue}{c} = \cos \theta$
$
\begin{align}
\text{To Prove }&:\space 3(\color{red}{s}-\color{blue}{c})^4 + 6(\color{red}{s}+\color{blue}{c})^2 + 4(\color{red}{s^6} + \color{blue}{c^6}) - 13 = 0\\
\equiv \text{TP }&:\space 3(\color{red}{s}-\color{blue}{c})^4 + 6(\color{red}{s}+\color{blue}{c})^2 + 4(\color{red}{s^6} + \color{blue}{c^6}) = 13
\end{align}
$
$$
\require{cancel}
\begin{align}
\text{LHS }
&= 3\left[(\color{red}{s}-\color{blue}{c})^2\right]^2
+ 6(\color{red}{s} +\color{blue}{c})^2
+ 4(\color{red}{s^6} + \color{blue}{c^6})
\\
&= 3(\color{limegreen}{s^2 + c^2} - 2\color{red}{s}\color{blue}{c})^2
+ 6(\color{limegreen}{s^2 + c^2} + 2\color{red}{s}\color{blue}{c})
+ 4\left[(\color{red}{s^2})^3 + (\color{blue}{c^2})^3\right]
\\
&= 3(1 - 2\color{red}{s}\color{blue}{c})^2 + 6(1+2\color{red}{s}\color{blue}{c})
+ 4(\color{limegreen}{s^2 + c^2})(\color{red}{s^4}
- \color{red}{s^2}\color{blue}{c^2}
+ \color{blue}{c^4})
\\
&= 3(1 - 4\color{red}{s}\color{blue}{c} + 4\color{red}{s^2}\color{blue}{c^2}) + 6 (1+2\color{red}{s}\color{blue}{c}) + 4(\color{red}{s^4} - \color{red}{s^2} \color{blue}{c^2}+\color{blue}{c^4})
\\
&= 3
\cancel{- 12\color{red}{s}\color{blue}{c}}
+12\color{red}{s^2}\color{blue}{c^2}
+ 6
\cancel{+ 12\color{red}{s}\color{blue}{c}} + 4\color{red}{s^4}
-4\color{red}{s^2}\color{blue}{c^2} + 4\color{blue}{c^4}\\
&= 4\color{red}{s^4} + 8\color{red}{s^2}\color{blue}{c^2} + 4\color{blue}{c^4} + 9\\
&= 4(\color{red}{s^4} + 2\color{red}{s^2}\color{blue}{c^2} + \color{blue}{c^4}) + 9\\
&= 4(\color{limegreen}{s^2 + c^2})^2 + 9\\
&= 4 + 9 = 13 = \text{RHS} \tag{Q.E.D.}
\end{align}
$$
Thanks to @mathlove, I found and corrected the mistake in my attempt.
$\Huge\color{lightgrey}{☺}$ Although, a quicker alternate way will always be nice.
| Alternative Simplification of $$\displaystyle \sin^6(x)+\cos^6(x)$$
Using $$\sin^2(x)+\cos^2(x)=1\Rightarrow \underbrace{\sin^2 x}_{a}+\underbrace{\cos^2 x}_{b}+\underbrace{(-1)}_{c}=0$$
Now Using The Formula, If $${a+b+c=0\;,}$$ Then $$a^2+b^2+c^3=3abc.$$
So $$\left(\sin^2x\right)^3+\left(\cos^2x\right)^3+(-1)^3=3\cdot \left(\sin^2x\right)\cdot \left(\cos^2x\right)\cdot (-1)$$
So $$\displaystyle \sin^6 x+\cos^6 x=1-3\sin^2 x\cdot \cos^2 x$$
| {
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"url": "https://math.stackexchange.com/questions/921345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 3
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How to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$. I am trying to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$ and I have worked this problem a couple times and keep getting the same answer. So I will show my process and please point my errors out.
$$a=\sqrt{7}\hspace{10pt} x=\sqrt{7}\sin\theta\hspace{10pt}dx=\sqrt{7}\cos\theta d\theta \hspace{7pt}7-x^2=7\cos^2\theta$$
$$\int\frac{7\sin^2\theta \sqrt{7}\cos\theta d\theta}{\sqrt{7\cos^2\theta}}$$
$$7\int\sin^2\theta d\theta\rightarrow \frac{7}{2}\int1-\cos(2\theta)d\theta$$
$$\frac{7}{2}\left(\theta-\frac{\sin(2\theta)}{2} \right) \rightarrow \frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{2\sin\theta\cos\theta}{2} \right)$$
$$\frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{x}{\sqrt{7-x^2}}\frac{\sqrt{7-x^2}}{\sqrt{7}} \right)$$
$$\frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{x}{\sqrt{7}} \right)$$
But apparently the answer is
$$\frac{1}{2}\left(7\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-x\sqrt{7-x^2} \right)$$
So how do I go about getting that answer? Thanks for all the help in advance.
| I suppose that the error is when you worked $$\frac{2\sin\theta\cos\theta}{2}=\sin\theta\cos\theta$$ You had $\sin\theta=\frac{x}{\sqrt 7}$, so $\cos^2\theta=1-\frac{x^2}{ 7}$, $cos \theta=\sqrt{1-\frac{x^2}{7}}$ and then $$\sin\theta\cos\theta=\frac{x}{\sqrt 7}\sqrt{1-\frac{x^2}{7}}=\frac{1}{7}x\sqrt{7-{x^2}} $$
I am sure that you can take from here.
By the way, it could have been faster getting rid of the radical in denominator, setting $x=\sqrt{7-u^2}$, $dx=-\frac{u}{\sqrt{7-u^2}}$ so the integrand reduces to $-\sqrt{7-u^2}$ from which you start the substitution (just as you did).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does the sequence converge, and to what? We have a sequence $\{a_n\}$
$$a_0 = 0$$
$$a_{n+1} = \frac{a_{n}}{2} + 1$$
Does it converge? And to what?
| Claim: For $n \geq 0$, $$a_{n}=\frac{2^{n}-1}{2^{n-1}}$$ We can show this inductively.
Begin with $n=0$ we have $$a_{0}=0=\frac{2^0-1}{2^{-1}} \ \ \checkmark$$ Now let's suppose the result holds for all $k$ where $0<k\leq n$. By definition we have $a_{k+1}=\frac{a_{k}}{2}+1$, and by our induction hypothesis we have $a_{k}=\frac{2^{k}-1}{2^{k-1}}$ Now plug $a_{k}$ into our expression for $a_{k+1}$ and we will get $$a_{k+1}=\frac{\left(\frac{2^{k}-1}{2^{k-1}}\right)}{2}+1$$ $$=\frac{2^{k}-1}{2^{k}}+1$$ $$=\frac{2^{k}-1}{2^{k}}+\frac{2^{k}}{2^{k}}$$ $$=\frac{2\cdot2^{k}-1}{2^{k}}$$ $$=\frac{2^{k+1}-1}{2^{k}}$$ We now know this relationship holds for all $k \in \mathbb{N}$.You can do a little algebra and rewrite this as $a_{n}=2-\frac{1}{2^n}$. Taking the limit of this expression should make it clear that $lim_{n \rightarrow \infty}(2-\frac{1}{2^n})=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Volume of Solid Revolution region bounded by $$y=x$$ and $$y=x^2$$
a) find the volume of the solid of revolution formed by revolving R region about the line $x=2$.
b) find the volume of the solid of revolution formed by revolving R region about the line $y=b$ where $b > 1$ is a constant
c) Find any values of a for which the volume of solid in part b is same as volume of solid in part a. Approximate to two decimal places and may use a calculator
I got $\frac \pi2$ for my answer for a but I feel like thats wrong. I integrated with respect to y and found the bounds as 0 to 1 but after calculating I get $\frac \pi2$ for an answer. Please explain in detail how to do these parts and let me know what I did wrong in part a (if its wrong).
| we have that the region is bounded by $y=x$ and $y=x^2$, finding the intersections
$\left.\begin{matrix}y=x\\y=x^2\end{matrix}\right\} x=x^2\iff x=0\vee x=1$
so, them intercepts on $(0,0)$ and $(1,1)$, then:
a) then the volume of revolution about the line $x=2$, we have that $x=y,x=\sqrt{y}$, then.
for $0\le y\le 1\Rightarrow 2\ge\sqrt{y}\ge y$, wich implies that $(y-2)^2\ge(\sqrt{x}-2)^2$, then we gets that
$$\begin{align}V&=\pi\int_{0}^{1}\left|(y-2)^2-(\sqrt{y}-2)^2\right|dy\\
&=\pi\int_{0}^{1}(y-2)^2-(\sqrt{y}-2)^2dy\\
&=\pi\int_{0}^{1}(y^2-4y+4)-(y-4\sqrt{y}+4)dy\\
&=\pi\int_{0}^{1}y^2-4y+4-y+4\sqrt{y}-4dy\\
&=\pi\int_{0}^{1}y^2-5y+4\sqrt{y}dy\\
&=\pi\left[\frac{y^3}{3}-\frac{5y^2}{2}+\frac{8y\sqrt{y}}{3}\right]_{0}^{1}\\
&=\pi\left(\frac{1}{3}-\frac{5}{2}+\frac{8}{3}\right)\\
&=\pi\frac{2-15+16}{6}\\
&=\frac{\pi}{2}\end{align}$$
b) for this, the solid is revolving over line $y=a$, with $a>1$, then we got:
for $0\le x\le1\Rightarrow a\ge x\ge x^2$, then $(x^2-a)^2\ge(x-a)^2$, then we gets that
$$\begin{align}V&=\pi\int_{0}^{1}\left|(x-a)^2-(x^2-a)\right|dx\\
&=\pi\int-\left[(x-a)^2-(x^2-a)\right]dx\\
&=-\pi\int(x^2-2ax+a^2)-(x^4-2ax^2+a^2)dx\\
&=-\pi\int x^2-2ax+a^2-x^4+2ax^2-a^2dx\\
&=-\pi\int-x^4+(1+2a)x^2-2axdx\\
&=-\pi\left[-\frac{x^5}{5}+\frac{(1+2a)x^3}{3}-ax^2\right]_{0}^{1}\\
&=-\pi\left(-\frac{1}{5}+\frac{1+2a}{3}-a\right)\\
&=-\pi\frac{-3+5(1+2a)-15a}{15}\\
&=-\pi\frac{-3+5+10a-15a}{15}\\
&=-\pi\frac{2-5a}{15}\\
&=\pi\frac{5a-2}{15}
\end{align}$$
c) for $c$ we have
$$\frac{\pi}{2}=\pi\frac{5a-2}{15}\\
15=2(5a-2)\\
15=10a-4\\
10a=15+4=19\\
a=\frac{19}{10}=1.9
$$
| {
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"url": "https://math.stackexchange.com/questions/922907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit with trigonometric function $\lim_{x \to \frac{\pi}{4}} \frac{\tan^2(x)-1}{\cos(x)-\sin(x)}$ I have this limit, I have resolved it until a part but I'm stucked now.
$$\lim_{x \to \frac{\pi}{4}} \frac{\tan^2(x)-1}{\cos(x)-\sin(x)}$$
$$ \lim_{x \to \frac{\pi}{4}} \frac{\frac{\sin^2(x)}{\cos^2(x)}-\frac{\cos^2(x)}{\cos^2(x)}}{\cos(x)-\sin(x)}$$
$$ \lim_{x \to \frac{\pi}{4}} \frac{\frac{\sin^2(x)-\cos^2(x)}{\cos^2(x)}}{\frac{\cos(x)-\sin(x)}{1}}$$
$$\lim_{x \to \frac{\pi}{4}} \frac{ \sin^2(x)-\cos^2(x)}{\cos^2(x)(\cos(x)-\sin(x)}$$
$$\lim_{x \to \frac{\pi}{4}} \frac{\sin(x)-\cos(x)}{\cos^2(x)}$$
$$\lim_{x \to \frac{\pi}{4}} \frac{\sin(x)}{\cos^2(x)}-\frac{\cos(x)}{\cos^2(x)}$$
$$\lim_{x \to \frac{\pi}{4}} \frac{\sin(x)}{\cos^2(x)}-\frac{1}{\cos(x)}$$
And now I replace by $\frac{\pi}{4}$ and what I get is $0$.
What am I doing wrong?
| Going from line 4 to line 5, $(\sin^2x-\cos^2x)/(\cos x-\sin x)=-\sin x-\cos x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the following inequality I have the following inequality $\frac {2x^2}{x+2} < x-2$. I tried to solve it the with the following steps.
step 1 $\frac {2x^2}{x+2} < x-2$
step 2 $\frac {2x^2}{x+2} - (x-2) < 0$
step 3 $\frac {2x^2}{x+2} - \frac {(x-2)(x+2)}{1(x+2)} < 0$
step 4 $\frac {2x^2}{x+2} - \frac {x^2-2^2}{x+2} < 0$
step 5 $\frac {2x^2 - x^2 + 4}{x+2} < 0$
step 5 $\frac {x^2 + 4}{x+2} < 0$
step 6 I used character study to get the result x > 2. But this is incorrect.
Where did I go wrong with this?. I feel that I made a mistake somewhere in step 2 but not sure what I did wrong.
Thanks!
| I would like to suggest another way: you can define two cases, one where $x>-2$, the other where $x<-2$, and multiply the initial inequality by $(x+2)$.
It is then straightforward to get the final result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Prove that $x+\frac{1}{2x}-\frac{1}{8x^3}<\sqrt{x^2+1}This is a math problem from the German Math Olympiad, but in this case I do not know where to start, probably because I do not have enough intuition regarding inequalities.
However, I tried to apply the standard AM-GM equation which didn't help, and in general I could not find a good up-/ downward estimation for the root term.
I also looked at the function plots and it seems that those three terms are very close to each other which seems to make this problem even more difficult to prove.
Any tips on how to begin? Please do not post a complete solution here.
BTW usually those problems provide an elegant solution based on non-university level knowledge, so I do not like to use an approach like a series expansion of the root term!
EDIT: Alright, the right side should've been obvious:
$$0<\frac{1}{4x^2}\Leftrightarrow \sqrt{x^2+1}<\sqrt{x^2+1+\frac{1}{4x^2}}=\sqrt{x^2+2x\frac{1}{2x}+\frac{1}{4x^2}}=x+\frac{1}{2x}$$
EDIT 2: So, I noted that squaring the left inequality (considering $x>0$ as well as $LHS>0$) gives me an inequality equivalent to $x\gt\sqrt{\frac{1}{8}}$.
Therefore, I only have to show that the Assumption $LHS\gt 0$ is implying exactly this.
I thus noted that $0=x+\frac{1}{2x}-\frac{1}{8x^3}$ has only one positive solution $\frac{\sqrt{\sqrt{3}-1}}{2}\gt\sqrt{\frac{1}{8}}$ -- but how do I show this as an inequality? I always get very confused about inequations with quadratic polynomials, especially regarding the direction of the inequality sign...
EDIT 3: Got it! So, let me state the whole proof ;)
To prove: $L:=x+\frac{1}{2x}-\frac{1}{8x^3}<\sqrt{x^2+1}=:R,\quad x>0$
a) Consider $L\leq0$. $x>0$ implies $R>0 \Rightarrow L<R$
b) Consider $L\gt0$. We get
$$
0<x+\frac{1}{2x}-\frac{1}{8x^3} \Rightarrow 0\lt x^4+\frac{1}{2}x^2-\frac{1}{8} = (x^2)^2 + 2\frac{1}{4}x^2+\frac{1}{4^2}-\frac{1}{4^2}-\frac{1}{8} = (x^2+\frac{1}{4})^2-\frac{3}{16}\\
\Leftrightarrow \frac{\sqrt{3}}{4}\lt x^2+\frac{1}{4} \Leftrightarrow x^2 \gt \frac{\sqrt{3}-1}{4}
\Leftrightarrow x \gt \sqrt{\frac{\sqrt{3}-1}{4}} \Rightarrow x\gt \sqrt{\frac{1}{8}}
$$
From that, we get
$$
\frac{1}{8x^2}\lt 1 \Leftrightarrow \frac{1}{8x^2}\frac{1}{8x^4}\lt \frac{1}{8x^4}
\Leftrightarrow (\frac{1}{8x^3})^2-\frac{1}{8x^4}\lt0 \\
\Leftrightarrow x^2+1+(\frac{1}{8x^3})^2-\frac{1}{8x^4}\lt x^2+1
$$
Noticing that $L^2=x^2+1+(\frac{1}{8x^3})^2-\frac{1}{8x^4}$ and $R^2=x^2+1$ gives you
$$
\Leftrightarrow L^2 \lt R^2 \Leftrightarrow L\lt R
$$
...quod erat demonstrandum.
Sorry it took so long, I am not used to write that much TeX ;)
BTW, can anyone leave a comment regarding the "beauty" of my proof? I tried to compose this final version as clean as possible!
| It might be helpful to note that
$$\begin{align}\sqrt{x^2+1}&=x\sqrt{1+\frac 1{x^2}}\\
&=x\left(1+\frac 1{x^2}\right)^{\frac 12}\\
&=x\left[1+\frac12\cdot \frac 1{x^2}+\frac {\frac 12\cdot -\frac 12}{1\cdot 2}\frac1{4x^4}+\cdots \right]\\
&=x+\frac 1{2x}-\frac 1{8x^3}+\cdots \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
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