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How find this $f^{(4)}(0)$ let
$$f(x)=\dfrac{e^x}{1-\sin{x}}$$
Find the value of
$$f^{(4)}(0)=?$$
My try: let
$$\dfrac{e^x}{1-\sin{x}}=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+\cdots$$
so
$$e^x=(1-\sin{x})(a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+\cdots)$$
since
$$e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\cdots$$
$$1-\sin{x}=1-x+\dfrac{1}{3!}x^3-\dfrac{1}{5!}x^5+\cdots$$
Follow I fell very ugly,maybe this problem have nice methods?
Thank you
| $$
\begin{array}{lcl}
\frac{1}{1-\sin(x)} &=& 1+\sin(x)+\sin^2(x)+\sin^3(x)+\sin^4(x)+O(x^5) \\
&=& 1+\bigg(x-\frac{x^3}{6}\bigg)+\bigg(x-\frac{x^3}{6}\bigg)^2+
\bigg(x-\frac{x^3}{6}\bigg)^3+\bigg(x-\frac{x^3}{6}\bigg)^4+O(x^5) \\
&=& 1+\bigg(x-\frac{x^3}{6}\bigg)+\bigg(x^2-\frac{x^4}{3}\bigg)+
\bigg(x^3\bigg)+\bigg(x^4\bigg)+O(x^5) \\
&=& 1+x+x^2+\frac{5}{6}x^3+\frac{2}{3}x^4+O(x^5)
\end{array}
$$
We deduce
$$
\begin{array}{lcl}
f(x) &=& \bigg(1+x+x^2+\frac{5}{6}x^3+\frac{2}{3}x^4\bigg)
\bigg(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\bigg) +O(x^5) \\
&=& 1+2x+\frac{5}{2}x^2+\frac{5}{2}x^3+\frac{53}{24}x^4+O(x^5) \\
\end{array}
$$
So we see that $f^{(4)}(0)=53$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the minimum of a function of two variables
Find the smallest value of $\displaystyle \sqrt{49+a^2-7{\sqrt{2}}\ a}+\sqrt{a^2+b^2-{\sqrt{2}}\ ab}+\sqrt{50+b^2-10b}\quad \quad$ for $a,b$ real and positive.
What I've done so far:
Let $F(a,b)=\displaystyle \sqrt{49+a^2-7{\sqrt{2}}\ a}+\sqrt{a^2+b^2-{\sqrt{2}}\ ab}+\sqrt{50+b^2-10b}$
Then
$$\begin{align}
\displaystyle F_a=\frac{2a-7{\sqrt{2}}\ a}{2\sqrt{49+a^2-7{\sqrt{2}}\ a}}+\frac{2a-\sqrt{2}\ b}{2\sqrt{a^2+b^2-{\sqrt{2}}\ ab}}=0\\
\displaystyle F_b=\frac{2b-{\sqrt{2}}\ a}{2\sqrt{a^2+b^2-{\sqrt{2}}\ ab}}+\frac{2b-10}{2\sqrt{50+b^2-10b}}=0
\end{align}$$
I then tried to simplify the two equations as follows:
$$\begin{align}
\left({2a-7{\sqrt{2}}\ a}\right)\left(\sqrt{a^2+b^2-{\sqrt{2}}\ ab}\right)+\left({2a-\sqrt{2}\ b}\right)\left(\sqrt{49+a^2-7{\sqrt{2}}\ a}\right)=0\\
\left({2b-{\sqrt{2}}\ a}\right)\left(\sqrt{50+b^2-10b}\right)+\left(2b-10\right)\left(\sqrt{a^2+b^2-{\sqrt{2}}\ ab}\right)=0\\
\end{align}$$
But then I'm stuck.
Do I need to include a constraint?(i.e $a$ and $b$ are positive and real); if so what will the constraint equation be?
Is there an altogether different way to do this other than using Lagrange??
| Comment to the earlier hint: Consider a triangle XOY where $|OX|=7$ and $|OY|=\sqrt{50}$. Then that expression is equal to $|XY|$. The minimum for $|XY|$ is achieved when $∠XOY=135^∘$.
Then the law of cosine give you the minimum for $|XY|$ namely the minimum value for the original expression.
| {
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"url": "https://math.stackexchange.com/questions/619130",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Explain the origin of the number of divisors and sum of divisors formulas. I know the basic formulas which are:
For a number $n = p_1^{a_1} p_2^{a_2} \cdot \ldots \cdot p_k^{a_k}$, we have
$d(n) = ( a_1 + 1 )( a_2 + 1 ) \cdot \ldots \cdot (a_k+1)$ and
$S(n) = \frac{p_1^{a_1+1} - 1}{p_1-1} \cdot \frac{p_2^{a_2+1} - 1}{p_2-1} \cdot \ldots \cdot \frac{p_k^{a_k+1} - 1}{p_k-1}$
I found some demonstrations but none close to my level of maths understanding.I want a simple description of how those formulas were found. Basically, explain everything.
| Note that $\frac{x^{n+1}-1}{ x - 1} = x^n + x^{n-1} + \dots + x + 1$. Also note that $(a_1 + \dots + a_n) \cdot (b_1 + \dots + b_m)$ is the sum of all possible combinations of $a_ib_j$. From this you should be able to figure out $S(n)$.
Note that, in the expression of $S(n)$, if instead of the divisors we could have $1$s, that would be $d(n)$. We can make them $1$ if instead of $p_i^{a_i} + \dots + p_i + 1$ you'd have $1^{a_i} + \dots + 1 + 1 = a_i+1$. Thus $d(n)$ can be explained in terms of $S(n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $p$ is prime and $p>3$ and $k,l,m,n,p\in\mathbb N$ and $p^k+p^l+p^m=n^2$, prove that $8\mid p+1$. If $k,l,m,n,p\in\mathbb N$ and a prime number $p>3$ that satisfies $$p^k+p^l+p^m=n^2$$ is chosen, prove that $8\mid p+1$.
$n^2$, when divided by $8$, gives a remainder $1$ (it can't give the remainders $0$ and $4$, because three odd numbers sum up to an odd number, which, if it is a square, always gives a remainder 1 when divided by 8).
The prime numbers $p$ give these ones:
When $p\equiv 1\pmod 8$, then $p^{2z}\equiv 1\pmod 8$ and $p^{2z+1}\equiv 1\pmod 8$.
When $p\equiv 3\pmod 8$, then $p^{2z}\equiv 1\pmod 8$ and $p^{2z+1}\equiv 3\pmod 8$.
When $p\equiv 5\pmod 8$, then $p^{2z}\equiv 1\pmod 8$ and $p^{2z+1}\equiv 5\pmod 8$.
When $p\equiv 7\pmod 8$, then $p^{2z}\equiv 1\pmod 8$ and $p^{2z+1}\equiv 7\pmod 8$.
Where $z\in\mathbb N\cup \{0\}$. We have to prove that $p\equiv 7\pmod 8$.
And another observation is that we could mark $p$ as either $3c+1$ or $3c+2$, where $c\in\mathbb N\cup\{0\}$ (we have to use the fact that $p>3$ somehow anyway). Thanks.
And I've given a tag "diophantine-equations" to this question because this seems a bit related to them. Feel free to disagree.
| When $p\equiv 1\mod 8$, we now that $P=p^k+p^l+p^m\equiv 1+1+1\equiv 3\not \equiv 1\mod 8$. When $p\equiv 5\mod 8$, we get $P\equiv 1+1+1\equiv 3\not\equiv 1\mod4$, so this won't work either. When $p\equiv 3\mod 8$, we get $3+3+3\equiv 1\mod 8$ (and the other possibilities won't work), so the only case we still have to shoot is $k\equiv l\equiv m\equiv 1\mod 2$ with $p\equiv 3\mod 8$.
Now, we know $p|n$, so $p^2|n^2$, so $p^2|P$, so $k,l,m\geq 2$ (because $p>3$). Thus $k,l,m\geq 3$, because they are odd. Now, we get $p^3|n^2$, so $p^2|n$, so $p^4|n^2$. Thus, $k,l,m\geq 5$. This will continue forever, so there aren't any solutions in this case. The only remaining case is what you have to prove, so we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I prove $F_{n+1}^2 - F_nF_{n+2} = (-1)^n$ using induction? $F_n$ refers to the $n$ term of the Fibonacci Sequence.
I think I'm supposed to prove this by induction. I already have the base case. I am at:
$\text{F}_\text{k+1}^2 - F_k\text{F}_\text{k+2} + \text{F}_\text{k+2}^2 - \text{F}_\text{k+1}\text{F}_\text{k+3} = (-1)^k +\text{F}_\text{k+2}^2 - \text{F}_\text{k+1}\text{F}_\text{k+3}$
So it's probably easier to work backwards. Multiply $-1$ in order to get $(-1)^\text{k + 1}$ and then get:
$-\text{F}_\text{k+1}^2 + F_k\text{F}_\text{k+2} - \text{F}_\text{k+2}^2 + \text{F}_\text{k+1}\text{F}_\text{k+3} = (-1)^\text{k+1} -\text{F}_\text{k+2}^2 + \text{F}_\text{k+1}\text{F}_\text{k+3}$
So then bring everything to the left side and mess around with the left side.
$-\text{F}_\text{k+1}^2 + F_k\text{F}_\text{k+2} - \text{F}_\text{k+2}^2 + \text{F}_\text{k+1}\text{F}_\text{k+3} +\text{F}_\text{k+2}^2 - \text{F}_\text{k+1}\text{F}_\text{k+3} = (-1)^\text{k + 1} $
And so somehow get this to look more like this:
$\text{F}_\text{k+1}^2 - F_k\text{F}_\text{k+2} + \text{F}_\text{k+2}^2 - \text{F}_\text{k+1}\text{F}_\text{k+3}$
...except I don't know how to do that.
Thanks
| Hint
\begin{align}
F_{n+1}^2-F_nF_{n+2} &= F_{n+1}^2-(F_{n+2}-F_{n+1})F_{n+2}=F_{n+1}^2-F_{n+2}^2+F_{n+1}F_{n+2} \\ &=F_{n+1}(F_{n+1}+F_{n+2})-F_{n+2}^2=F_{n+1}F_{n+3}-F_{n+2}^2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Largest value of the function $f(x) = \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$ Largest value of the function $f(x) = \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}\;\;,$ where $x\in \mathbb{R}$
My Try:: Let $y=f(x)=\sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$
$\displaystyle \Rightarrow y^2=(x^4-3x^2-6x+13)+(x^4-x^2+1)-2\sqrt{(x^4-3x^2-6x+13)\cdot (x^4-x^2+1)}$
$\displaystyle y^2=(2x^4-4x^2-6x+14)-2\sqrt{\{(x^2-2)^2+(x-3)^2\}\cdot\{(x^2-1)^2+\left(x\right)^2\}}$
Now using the Cauchy-Schwarz inequality for
$\displaystyle a=(x^2-2)^2+(x-3)^2$ and $\displaystyle b=(x^2-1)^2+\left(x\right)^2$
$\displaystyle \Rightarrow (a^2+b^2)\cdot(1^2+1^2)\geq(x+y)^2$ and equality hold when $\displaystyle\frac{a}{1}=\frac{b}{1}$
$\displaystyle\Rightarrow \left\{(x^2-2)^2+(x-3)^2\right\}\cdot \left\{(x^2-1)^2+\left(x\right)^2\right\}\geq \left\{(x^2-2)\cdot\left(x^2-1\right)+(x-3)\cdot x \right\}^2 = \left\{x^4-2x^2-3x+2\right\}$
and equality holds when $\displaystyle \frac{x^2-2}{x^2-1}=\frac{x-3}{x}\Rightarrow 3x^2-x-3=0$
So $\displaystyle \Rightarrow y^2\leq (2x^4-4x^2-6x+14)-2(x^4-2x^2-3x+2)=10$
So $y=f(x)\leq \sqrt{10}$ and equality holds when $\displaystyle x=\frac{1-\sqrt{37}}{6}$
If there is any geometrical method Then please explain here.
| Hint
$$f(x)=\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$$
let
$$A(x,x^2),B(3,2),C(0,1)$$
so
$$f(x)=|AB|-|AC|\le|BC|=\sqrt{10}$$
then we have
$$AB:y-1=\dfrac{1}{3}(x-0)$$
so
$$\begin{cases}
y=x^2\\
y=\dfrac{1}{3}x+1
\end{cases}$$
$$\Longrightarrow x=\dfrac{1-\sqrt{37}}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $m^{2} + n^{2} = \csc^{2}\left(\theta\right)$ $$
\mbox{If}\quad\left\lbrace%
\begin{array}{rcl}
m^{2} + m'^{2} + 2mm'\cos\left(\theta\right) & = & 1
\\
n^{2} + n'^{2} + 2nn'\cos\left(\theta\right) & = & 1
\\
mn + m'n' + (mn' + m'n)\cos\left(\theta\right) & = & 0
\end{array}\right\rbrace
\quad\mbox{then prove that}\ m^{2} + n^{2} = \csc^{2}\left(\theta\right)
$$
Don't know how go about this. Please help.
| Using the same set of notations used by user84413,
$\displaystyle ax+by+(ay+bx)\cos\theta=0\ \ \ \ (1)$
$\displaystyle a^2+b^2+2ab\cos\theta=1\iff (b+a\cos\theta)^2=1-a^2\sin^2\theta\ \ \ \ \ (2)$
Similarly, $\displaystyle x^2+y^2+2xy\cos\theta=1\iff(y+x\cos\theta)^2=1-x^2\sin^2\theta\ \ \ \ \ (3)$
Multiplying $(2),(3)$
$\displaystyle(1-a^2\sin^2\theta)(1-x^2\sin^2\theta)=\{(b+a\cos\theta)(y+x\cos\theta)\}^2$
$\displaystyle\implies 1-(a^2+x^2)\sin^2\theta+a^2x^2\sin^4\theta=\{by+\cos\theta(ay+bx)+ax\cos^2\theta\}^2$
Now putting the value of $by+(ay+bx)\cos\theta$ from $(1),$
$\displaystyle\{by+\cos\theta(ay+bx)+ax\cos^2\theta\}^2=\{(-ax)+ax\cos^2\theta\}^2=a^2x^2(-\sin^2\theta)^2=a^2x^2\sin^4\theta$
$\displaystyle\implies 1-(a^2+x^2)\sin^2\theta+a^2x^2\sin^4\theta=a^2x^2\sin^4\theta$
Now simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find max: $M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}$ For $a,b,c>0$ and $abc=1$, find the maximum of
$$M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}.$$
| Remark: Perhaps there are alternative proofs of $(ab)^{3/2} + (bc)^{3/2} + (ca)^{3/2} \ge ab + bc + ca$ without using Chebyshev sum inequality.
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$M = \sum_{\mathrm{cyc}} \frac{a(b + c + a^2)}{(b^2 + c^2 + a)(b + c + a^2)}
\le \sum_{\mathrm{cyc}} \frac{a(b + c + a^2)}{(a^{3/2} + b^{3/2} + c^{3/2})^2}$$
$$= \frac{a^3 + b^3 + c^3 + 2ab + 2bc + 2ca}{(a^{3/2} + b^{3/2} + c^{3/2})^2}$$
and
$$1 - M \ge \frac{2ab(\sqrt{ab} - 1) + 2bc(\sqrt{bc} - 1) + 2ca(\sqrt{ca} - 1)}{(a^{3/2} + b^{3/2} + c^{3/2})^2} \ge 0$$
where we have used Chebyshev sum inequality to get
\begin{align*}
&2ab(\sqrt{ab} - 1) + 2bc(\sqrt{bc} - 1) + 2ca(\sqrt{ca} - 1)\\
\ge\,& \frac13(2ab + 2bc + 2ca)(\sqrt{ab} + \sqrt{bc} + \sqrt{ca} - 3)\\
\ge\,& 0.
\end{align*}
(Note: $2ab, 2bc, 2ca$ and $\sqrt{ab}-1, \sqrt{bc}-1, \sqrt{ca} - 1$ are in a similar order.)
Also, when $a = b = c = 1$, we have $M = 1$.
Thus, the maximum of $M$ is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
Complex roots of the polynomial $bz^{2}+2az+b$ are on the unit circle I want to show that the roots of the polynomial $bz^{2}+2az+b$ ($a,b$ are real) when $\left|\frac{a}{b}\right|\leq1$ (which is equivalent to the discriminant not being positive) are on the unit circle. The roots are given by $$z_{\pm}=\frac{-a\pm\sqrt{a^{2}-b^{2}}}{b}=-\frac{a}{b}\pm\sqrt{\left(\frac{a}{b}\right)^{2}-1}$$ I tried messing around with the modulus of this to reach the required result but without success.
Following the hint setting: $$z_{+}=-\frac{a}{b}+\sqrt{\left(\frac{a}{b}\right)^{2}-1}=-\frac{a}{b}+i\sqrt{1-\left(\frac{a}{b}\right)^{2}}$$
$$z_{-}=-\frac{a}{b}-i\sqrt{1-\left(\frac{a}{b}\right)^{2}}=-\frac{a}{b}-i\sqrt{1-\left(\frac{a}{b}\right)^{2}}$$
We get that:
$$\left|z_{+}\right|^{2}=z_{+}\cdot\overline{z_{+}}=z_{+}\cdot z_{-}=\left(-\frac{a}{b}+i\sqrt{1-\left(\frac{a}{b}\right)^{2}}\right)\left(-\frac{a}{b}-i\sqrt{1-\left(\frac{a}{b}\right)^{2}}\right)$$
$$=\left(\left(\frac{a}{b}\right)^{2}-i^{2}\left(1-\left(\frac{a}{b}\right)^{2}\right)\right)=\left(\left(\frac{a}{b}\right)^{2}+\left(1-\left(\frac{a}{b}\right)^{2}\right)\right)=1$$
| Let $z = \rho (\cos \theta + i \sin \theta)$ be a root of polynomial $z^2+cz+1$, where $c\triangleq 2a/b$. Then:
$$\rho^2\sin(2\theta)+c\rho\sin\theta = 0$$
and
$$\rho^2\cos(2\theta)+c\rho\cos\theta + 1 = 0.$$
Multiplying the first relation by $\cos\theta$ and the second by $\sin \theta$, and then subtracting, we get:
$$(\rho^2-1)\sin\theta = 0.$$
Hence $z$ either has modulus $1$ or it is real (excluded by the assumption that $|c|<2$ or making $z$ trivially $1$ or $-1$ if $|c|=2$ is assumed).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/625159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
integration by trig substitution I solved the following integral by using trig substitution $(u = 3 sec \theta)$
$$
\int \frac{dx}{x^{2}+10x+16}=\int \frac{du}{u^2-9}
$$
and got the same result as in the textbook which looked for $\int \frac{du}{u^{2}-a^2}$. Below is the process and result given in the textbook.
$$
\int \frac{du}{u^2-9}=\frac{1}{6}\ln \left| \frac{u-3}{u+3} \right|=\frac{1}{6}\ln \left| \frac{x+2}{x+8} \right|
$$
If I let $f(u) = \frac{1}{u^2-9}$, I am confused when using $u = 3 sec \theta$ due to the domain of $f(u)$ and the range of $3sec \theta$ :
$$
3 \left| \sec\theta \right|\geq3
$$
but the domain of $f(u)$ is $[(-\infty, -3)\cup(-3, 3)\cup(3,\infty)]$.
In this case, how should I interpret the answer $\frac{1}{6}\ln \left| \frac{x+2}{x+8} \right|$?
I mean, I don't think I can use the answer if I want to solve definite integrals like below:
$$
\int_{-5}^{-3} \frac{dx}{x^2+10x+16}
$$
or
$$
\int_{0}^{2} \frac{du}{u^2-9}.
$$
Is my guess right?
| This may not be what you want, but do you notice the following?
$$\begin{align}\int\frac{1}{x^2+10x+16}&=\int\frac{1}{(x+5)^2-9}dx\\&=\int\frac{1}{u^2-9}du\\&=\frac 16\int\left(\frac{1}{u-3}-\frac{1}{u+3}\right)du\\&=\frac 16(\ln|u-3|-\ln|u+3|)+C\\&=\frac 16\ln\left|\frac{u-3}{u+3}\right|+C\end{align}$$
Here, I set $x+5=u.$
I think you don't need to use $u=3\sec\theta$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor's series when x goes to infinity Let $f(x) = \frac {x^3}{(x+1)^2}$. Find constants a, b, c, so that $f(x) = ax + b + \frac cx + o(\frac 1x)$ as $x$ goes to $\pm \infty$. So i know that i can't take Taylor series as $x$ goes to infinity. So i am assuming i have to make some kind of substitution. I tried making $x = \frac 1u$ but i get nowhere.
$$f(x) = \frac {x^3}{(x+1)^2} = x^3(x+1)^{-2}=\frac{1}{u^3}(1 + \frac 1u)$$
$$\lim_{u\to 0}\frac{1}{u^3}(1 + \frac 1u) = \lim_{u\to 0}\frac{1}{u^3}(1 - \frac 2u + \frac{3}{u^2} + o({u^3})) =\lim_{u\to 0} u^3 - \frac {2}{u^4} + \frac{3}{u^5} + o(\frac 1u) = \lim_{u\to 0} \frac{1}{x^3} - 2x^4 + 3x^5 + o(x^5)$$
Thanks.
| Here is how you approach the problem. Put $x=\frac{1}{t}$ and then find the Laurent series at the point $t=0$
$$ f(x) = \frac {(1/t)^3}{(1/t+1)^2}=\frac{1}{t(1+t)^2} ={t}^{-1}-2+3\,t-4\,{t}^{2}+5\,{t}^{3}+ ( {t}^{4}) .$$
Now, substitute $t=1/x$ in the above expansion. See related problem.
Note: Note that, you only need to find the Taylor series of $\frac{1}{(1+t)^2}$ at $t=0$, since
$$ \frac{1}{t(1+t)^2}= \frac{1}{t} \frac{1}{(1+t)^2} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/632595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the limit of $\left(\dfrac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n$ I'm trying to solve this limit, for which I already know the solution thanks to Wolfram|Alpha to be $\sqrt[3]{abc}$:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n:\forall a,b,c\in\mathbb{R}^+$$
As this limit is an indeterminate form of the type $1^\infty$, I've been trying to approach it by doing:
$$\lim_{n\rightarrow\infty}\left(\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{1}{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}}\right)^{\frac{3}{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}\cdot\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}=e^{\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n}$$
But now when I approach that top limit this is what I get:
$$\lim_{n\rightarrow\infty}\frac{a^\frac{1}{n}+b^\frac{1}{n}+c^\frac{1}{n}-3}{3}\cdot n=\lim_{n\rightarrow\infty}\frac{n\cdot a^{\frac{1}{n}}}{3}+\frac{n\cdot b^{\frac{1}{n}}}{3}+\frac{n\cdot c^{\frac{1}{n}}}{3}-n=\lim_{n\rightarrow\infty}\frac{n\cdot a^0}{3}+\frac{n\cdot b^0}{3}+\frac{n\cdot c^0}{3}-n=\lim_{n\rightarrow\infty}\frac{n}{3}+\frac{n}{3}+\frac{n}{3}-n=0$$
And hence the final limit should be $e^0=1$ which is clearly wrong but I honestly don't know what I did wrong, so what do you suggest me to solve this limit?
| While $\sqrt[n]a\to 1$, it is not correct to say $(n\sqrt[n]a-n)\to 0$. Actually, $\sqrt[n]{1+\epsilon}\approx 1+\frac1n\epsilon$ so $n\sqrt[n]a-n\approx a$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Identify $\sum(-1)^{n-k}2^{2k}\binom{n+k}{2k}$ Does anybody know what the following sum evaluates to?
$$
\sum_{k=0}^n{(-1)^{n-k}}2^{2k}\binom{n+k}{2k}
$$
| Generating functions to the rescue again! Defining
$$
c_n = \sum_{k=0}^n{(-1)^{n-k}}2^{2k}\binom{n+k}{2k},
$$
we have
\begin{align*}
\sum_{n=0}^\infty c_n x^n &= \sum_{n=0}^\infty x^n \sum_{k=0}^n{(-1)^{n-k}}2^{2k}\binom{n+k}{2k} \\
&= \sum_{k=0}^\infty {(-1)^{k}}2^{2k} \sum_{n=k}^\infty (-x)^n \binom{n+k}{2k} \\
&= \sum_{k=0}^\infty {(-1)^{k}}2^{2k} \frac{(-x)^{k}}{(1-(-x))^{2k+1}} \\
&= \frac1{1+x} \sum_{k=0}^\infty \bigg( \frac{4x}{(1+x)^2} \bigg)^k \\
&= \frac1{1+x} \frac1{1-4x/(1+x)^2} = \frac{1+x}{(1-x)^2};
\end{align*}
and one can verify that
$$
\frac{1+x}{(1-x)^2} = \sum_{n=0}^\infty (2n+1)x^n.
$$
It's true that the middle equality in the long calculation isn't obvious, but
$$
\sum_{n} \binom nj x^n = \frac{x^j}{(1-x)^{j+1}}
$$
is a standard formula in the generating functions biz.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\sin^5 x$ I know $\sin x$ integrates to $-\cos x$ but ive never seen $\sin^5(x)$ integrated.
would I need to expand it to $\sin x\sin x\sin x\sin x\sin x$ and then how would I complete the integration from here?
| Here's another answer:
We know from using integration by parts that
$$
\int\sin^5x dx = -\frac{1}{5}\sin^4x\cos x + \frac{4}{5}\int\sin^3x dx
$$
Applying intgration by parts to $\int\sin^3xdx$ means
$$
\int\sin^5x dx = - \frac{1}{5}\sin^4x\cos x + \frac{4}{5}(-\frac{3}{4}\cos x + \frac{1}{12}\cos 3x)
$$
$$
= - \frac{1}{5}\sin^4x\cos x - \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= -\frac{1}{5}(\sin^2x)^2\cos x- \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= -\frac{1}{5}\cos x(1 - \cos^2 x)^2- \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= -\frac{1}{5}\cos x(1 - 2\cos^2 x + \cos^4 x) - \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= -\frac{1}{5}\cos x + \frac{2}{5}\cos^3 x - \frac{1}{5}\cos^5 x- \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= -\frac{1}{5}\cos x + \frac{2}{5}(\frac{3}{4}\cos x + \frac{1}{4}\cos 3x) - \frac{1}{5}(\frac{5}{8}\cos x + \frac{5}{16}\cos 3x + \frac{1}{16}\cos 5x) - \frac{3}{5}\cos x +\frac{1}{15}\cos 3x
$$
$$
= \cos x(-\frac{1}{40} - \frac{3}{5}) + \cos 3x(\frac{1}{15} + \frac{3}{80}) - \frac{1}{80}\cos 5x
$$
$$
= -\frac{5}{8}\cos x + \frac{5}{48}\cos 3x - \frac{1}{80}\cos 5x
$$
However, given I obtained this problem from Apostol "Calculus" Volume 1 Section 5.10 Question 10 (c) (Page 221) and the problem in Apostol states
$$
\int\sin^5x dx = -\frac{5}{8}x + \frac{5}{48}\cos 3x - \frac{1}{8}\cos 5x
$$
I can only assume that I found a typographical error in Apostol's epic tome! Hope this helps :-)
| {
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Does this equation have positive integer solutions? The only solution I can find for $a^2 + b^2 + c^2 = d^2 + e^2 + f^2$ is $a=0$, $b=0$, $c=0$, $d=0$, $e=0$, $f=0$. Are there any positive integer solutions? Any where none of $a,b,c,d,e,f$ are equal?
| Solving $a^2 + b^2 + c^2 = d^2 + e^2 + f^2$ has many many (boring) positive integer solutions.
Choose any bijection $\theta: \{a,b,c\} \to \{d,e,f\}$. Then, choose any positive integer values for $\{a,b,c\}$.
It must be true that $a^2 + b^2 +c^2 = \theta(a)^2 + \theta(b)^2 + \theta(c)^2 = d^2 + e^2 + f^2.$ As stated though, these are all pretty boring, and probably not quite what you're looking for. For less "boring" solutions, combining some known Pythagorean triples could help.
For example, $3^2 + 4^2 = 5^2$. Meanwhile $8^2 + 15^2 = 17^2$ (I think, somebody else can fact-check this triple). Thus,
$(3^2+4^2)+17^2 = 5^2 + (8^2+15^2)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving convergence of a sequence Let the following recursively defined sequence:
$a_{n+1}=\frac{1}{2} a_n +2,$
$a_1=\dfrac{1}{2}$.
Prove that $a_n$ converges to 4 by subtracting 4 from both sides.
When I do that, I get:
$2(\frac{1}{2} a_{n+1} -2)=(\frac{1}{2} a_n -2)$, so $y=2y$,
which is true only for $0$. But I'm not sure how to formally use this in a definition of convergence?
| Just a generalization of the approach for you. Note that
$$
a_n=\frac{1}{2}a_{n-1}+2
$$
So if you subtract this expression from the $a_{n+1}$ you have above, you get rid of the constant. Also denote $\Delta a_{n+1}=a_{n+1}-a_{n}$ and you get:
$$
\Delta a_{n+1}=\frac{1}{2} \Delta a_{n}=\frac{1}{2^2} \Delta a_{n-1}=\ldots =\frac{1}{2^{n-1}} \Delta a_{2}
$$
If you sum over $n$ the LHS you get a telescoping sum: $\sum_{n=1}^{N}a_{n+1}=a_{N+1}-a_1$. Since $a_2=2 \frac{1}{4}$ and $a_1 = 0.5$ you get (using geometric sum $\sum_{n=1}^{N} \frac{1}{2^{n-1}}=2(1-(\frac{1}{2})^{N+1})$
$$
a_{N+1}=0.5+(2.25-0.5) \cdot 2 \cdot \Bigg(1-(\frac{1}{2})^{N+1} \bigg)
$$
and if you take the limit as $N \to \infty$ you get $0.5+3.5=4$.
| {
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If $x,y\in\mathbb Z$, solve $x^2+xy+y^2=x^2y^2$. If $x,y\in\mathbb Z$, solve $$x^2+xy+y^2=x^2y^2$$
We could try some factorizations. $x^2+y^2=xy(xy-1)$. We may as well add $2xy$ to both sides: $(x+y)^2=xy(xy+1)$. Then we could subtract $x^2y^2$ from both sides: $(x+y+xy)(x+y-xy)=xy$
Or we could somehow use the fact that if an integer is between two consecutive integer squares, then it can't be a square itself.
I'd like to get some more help here, some other strategies if these can't be used. Thanks.
| $$ x^2+xy+y^2=x^2y^2 \quad \Longleftrightarrow \quad (1-y^2)x^2+yx+y^2=0$$
$$ \delta=y^2-4(1-y^2)y^2=y^2(4y^2-3)$$
$$4y^2-3=p^2 \quad \implies \quad y^2=1\cdots $$
| {
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How to factorize $2x^2+5x+3$? I'm doing pre-calculus course at coursera.org and I'm in trouble with this solution $$2x^2 +5x +3 = (2x+3)(x+1)$$
By trial, using ac-method I got stuck:
$$
ac = (2)(3) = 6\\
6 + ? = 5 \Rightarrow~ ? = 5 - 6 = -1
$$
Then,
$$2x^2+6x-x+3 = 2x(x+3)-x+3$$
At this point I could not get the answer, any help?
| Complete squares:
$$2x^2+5x+3 = 2(x^2+2\alpha x + \alpha ^2) + (3- 2 \alpha ^2)$$
where $\alpha = \frac{5}{4}$. Then
$$(x+\frac{5}{4})^ 2 = \frac{1}{2}\left(\frac{25}{8}-3\right) $$
$$x+\frac{5}{4} = \pm \sqrt{\frac{1}{16}} $$
$$x= \frac{-5}{4} \pm \frac{1}{4} = \frac{-5 \pm 1}{4}$$
$$ x\in \left\{-1, \frac{-3}{2}\right\}$$
Therefore,
$$2x^2+5x+3 = (x+1)(2x+3).$$
| {
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Show that the triangle which satisfy the inequality $\frac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ Show that the triangle which satisfy the inequality $\dfrac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ is right angled.
My work:
$\sin^2 A+\sin^2 B+\sin^2 C=2(\cos^2 A+\cos^2 B+\cos^2 C)$
$3(\sin^2 A+\sin^2 B+\sin^2 C)=6$
$(\sin^2 A+\sin^2 B+\sin^2 C)=2$
What to do now? Please help!
| From where you have left off,
$$\sin^2A+\sin^2B+\sin^2C=2\implies1-\cos^2A+1-(\cos^2B-\sin^2C)=2$$
$$\implies\cos^2A+\cos^2B-\sin^2C=0$$
Using $\displaystyle\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$
$$\implies\cos^2A+\cos^2B-\sin^2C=\cos^2A+\cos(B+C)\cos(B-C)$$
As $\displaystyle A+B+C=\pi, B+C=\pi-A,\cos(B+C)=\cos(\pi-A)=-\cos A,$
$$\implies\cos^2A+\cos(B+C)\cos(B-C)=\cos^2A-\cos A\cos(B-C)$$
$$=\cos A\left[\cos A -\cos(B-C)\right]$$
$$=\cos A\left[-\cos(B+C) -\cos(B-C)\right]$$
$$=\cos A\left[-2\cos B\cos C\right]$$
| {
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How prove this $f(n)\le f(n+1)$ where $f(n)=\sum_{k=1}^{n}\frac{n}{n^2+k^2}$ let $$f(n)=\sum_{k=1}^{n}\dfrac{n}{n^2+k^2}$$
prove or disprove
$$f(n)\le f(n+1)$$
this inequality is found when I deal this follow limit:
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+(k/n)^2}=\int_{0}^{1}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{4}$$
But I can't prove
$$f(n)\le f(n+1)$$
since
$$f(n+1)=\dfrac{n+1}{(n+1)^2+1}+\dfrac{n+1}{(n+1)^2+2^2}+\cdots+\dfrac{n+1}{(n+1)^2+n^2}+\dfrac{n+1}{(n+1)^2+(n+1)^2}$$
$$f(n)=\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2^2}+\cdots+\dfrac{n}{n^2+n^2}$$
so
$$f(n+1)-f(n)=\left(\dfrac{n+1}{(n+1)^2+1}-\dfrac{n}{n^2+1}\right)+\left(\dfrac{n+1}{(n+1)^2+2^2}-\dfrac{n}{n^2+2^2}\right)+\cdots+\left(\dfrac{n+1}{(n+1)^2+n^2}-\dfrac{n}{n^2+n^2}\right)+\dfrac{1}{2(n+1)}$$
so
$$f(n+1)-f(n)=\sum_{k=1}^{n}\dfrac{k^2-n^2-n}{(k^2+n^2)((n+1)^2+k^2)}+\dfrac{1}{2(n+1)}$$
This problem is my found it,can you help to solve this problem?
| Take $g(x):=\frac{1}{n^2+x^2}$ and apply the Euler-MacLaurin's summation formula.
,
Then we obtain
$$
f(n)=\frac{\pi}{4}-\frac{1}{4n}+h(n),\qquad |h(n)|\leq\frac{3^{3/2}-2}{32n^2}.
$$
From this
$$
f(n+1)-f(n)=\frac{1}{4n(n+1)}+H(n),\qquad |H(n)|\leq\frac{3^{3/2}-2}{16n^2}.
$$
It yields
$$
f(n+1)-f(n)\geq\frac{(6-3^{3/2})n-3^{3/2}+2}{16n^2(n+1)},
$$
which is positive if $n\geq4$. For $n=1,2,3$ it can be checked by hand.
| {
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On solving $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$ How do we show that there is only one solution to,$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$$
I guess it is only $x=2$.
Please help.
| A proof by induction.
Let:
$$f_n(x) = \sqrt[3]{6+\sqrt[3]{6+\ldots+\sqrt[3]{6+x}}},\ g_n(x) = \sqrt{2+\sqrt{2+\ldots+\sqrt{2+x}}}$$
With $n$ terms. Then for $n=1$ you can easily solve the cubic equation to show that $f_1=g_1 $ only at $x=2$ (over the reals).
Now assume our claim is true for $n$, i.e. that $f_n(x)=g_n(x)$ iff for $x=2$. Then for $n+1$, raise to the sixth power to get that:
$$(6+f_n(x))^2=(2+g_n(x))^3$$
Clearly, this equality is true for $x=2$ since $f_n(2)=g_n(2)$. Now, if our claim is false and this equality holds for some $x_0\neq 2$, then:
$$g_n(x_0) = (6+f_n(x_0))^{2/3}-2$$
But since $dg_n/df_n < 1$ for all $x>0$, by the mean value theorem we have a contradiction.
$$$$
Thus we have proven that for any number of $n$ the only (real) solution of $f_n(x)=g_n(x)$ is $x=2$.
| {
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Taylor Series Expansion of $\frac{1}{1+x^2}$ about $x=a$ Let $$f(x)=\frac{1}{1+x^2}$$ Consider its Taylor series expansion about a point $a\in \mathbb{R}$. What is the radius of convergence of this series??
About $x=0$ we could expand it like
$$(1+x^2)^{-1}= 1-x^2+(-1)(-2) \frac{x^2}{2!}+(-1)(-2)(-3)\frac{x^3}{3!}+\dots$$ and get the radius as $1$.
But about $x=a$, there is a slight problem with the way as above...
$$\frac{1}{1+(x-a+a)^2}=\frac{1}{1+a^2(\frac{x-a}{a}+1)^2}.$$
Here is where I am getting stuck.
| With $u=x-a$:
$$
\begin{align}
\frac1{1+(u+a)^2}
&=\frac1{1+a^2+2au+u^2}\\
&=\frac1{1+a^2}\frac1{1+\frac{2au+u^2}{1+a^2}}\\
&=\frac1{1+a^2}\left(1-\frac{2au+u^2}{1+a^2}+\left(\frac{2au+u^2}{1+a^2}\right)^2-\left(\frac{2au+u^2}{1+a^2}\right)^3+\dots\right)\\[4pt]
&=\frac1{1+a^2}-\frac{2a}{(1+a^2)^2}u+\frac{3a^2-1}{(1+a^2)^3}u^2-\frac{4a^3-4a}{(1+a^2)^4}u^3+O\left(u^4\right)
\end{align}
$$
The radius of convergence is not simple to find from this power series, but as J.J. says, the radius of convergence of the power series for an analytic function is the distance from the point of expansion to the nearest singularity. If $a\in\mathbb{R}$, this is $\sqrt{a^2+1}$.
| {
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Show quartic polynomial has no real solutions To show a lower bound for the runtime of an algorithm, I need to show that
$$
3 x^4 - \frac{64}{5} x^3 + \frac{192}{5} x^2 - \frac{192}{5} x+ 12 > 0
$$
for all real numbers $x\in \mathbb{R}$.
I followed the Wikipedia instructions and checked the paper they are based on.
With computer help it is easy to find that the discriminant is positive. Now comes the part that has me lost:
Let $a,b,c,d,e$ be the coefficients. Then, according to both the paper and Wikipedia, the quartic function has real roots if both
$$8 a c - 3 b^2 <0\text{ and }64 a^3 e - 16 a^2 c^2 + 16 a b^2 c - 16 a^2 b d - 3 b^4 <0$$
and it has no real roots if both are positive. However, the case where
$$8 a c - 3 b^2 >0 \text{ and } 64 a^3 e - 16 a^2 c^2 + 16 a b^2 c - 16 a^2 b d - 3 b^4 <0$$ is not accounted for. Unfortunately, this is the case for my polynomial.
I appreciate any help, including hints for writing such a proof for a paper.
| this equal to:
$15x^4-6x^3+192x^2-192x+60>0$
$15x^4-6x^3+192x^2-192x+60=(3x^2-8.702x+25.0905)(5x^2-6.83x+2.3709)+.5112$
$3x^2-8.702x+25.0905>0,5x^2-6.83x+2.3709>0 \implies \\15x^4-6x^3+192x^2-192x+60>0.5 >0$
$25.0905=\dfrac{14439589717}{57550000},2.3709=\dfrac{272893639}{11510000}$
| {
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Finding all positive integers $x,y,z$ that satisfy $3^x - 5^y = z^2$
Find all positive integers $x,y,z$ that satisfy:
$$3^x - 5^y = z^2.$$
I think that $(x,y,z)= (2,1,2)$ will be the only solution. But how to prove that?
| Notice that:
$$\begin{align}3^x - 5^y &\equiv (-1)^{x} - 1 \mod 4\\
z^2 &\equiv (-1)^x - 1 \mod{4}\\
&\equiv \begin{cases} 0 & \text{$x$ even,}\\ 2 & \text{$x$ odd}\end{cases} \end{align}$$
Since $2$ is a quadratic non-residue modulo $4$, $x$ must be even $\implies x = 2a$ for positive integer $a$. Furthermore, $4| z^2 \implies 2|z \implies z = 2b$ for positive integer b.
Hence, we get:
$$\begin{align}3^{2a} - 5^y &= (2b)^2\\
5^y &= 3^{2a} - (2b)^2\\
5^y &= (3^a + 2b)(3^a - 2b)\end{align}$$
If $(3^a - 2b) > 1$, we notice that :
$$\begin{align}
&a,b > 0, a \in \mathbb{Z}\\
\implies &3^a + 2b > 3^a - 2b \ge 1\\
\implies &3^a + 2b > 1\end{align}$$
But $3^a + 2b$ is a perfect power of 5, implying that $5|(3^a + 2b)$. Hence, we have :
$$\begin{align}
3^a + 2b &\equiv 0 \mod 5\\
3^a - 2b &\equiv -4b \mod 5\\
3^a - 2b &\equiv b \mod 5\end{align}$$
If $(3^a - 2b)$ is a perfect power of $5$ (excluding 1), then from above, we get $b = 5k$ for positive integer $k$. But if $b = 5k$, then:
$$\begin{align}3^a + 2b &= 3^a + 10k\\
&\equiv 3^a \mod {5}\\
&\not\equiv 0 \mod{5}\end{align}$$
Hence, $(3^a - 2b)$ cannot be greater than $1$ $\implies 3^{\frac{x}{2}} - z = 1 \implies 3^x = (1 + z)^2$, which has infinitely many solutions, one of which is $x = 2, z = 2 \implies y = 1$. Now, we just have to find a way to show that there are no other solutions.
| {
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Finding the volume inside an elyptical cylinder and a sphere I'm trying to find the volume bounded by a sphere and an elyptical cylinder. The sphere is given by $x^2+y^2+z^2=1$ and the elyptical cylinder by $2x^2+y^2-2x=0$.
My first attempt with spherical coordinates didn't go well, now I did some work with cylindrical coordinates but I'm not sure if I have the right answer.
Playing with the equation for the cylinder I found the canonical form of the ellipse: $\displaystyle\frac{\left(x-\frac{1}{2}\right)^2}{\left(\frac{1}{2}\right)^2}+ \displaystyle\frac{y^2}{\left(\frac{1}{\sqrt{2}}\right)^2}=1$.
And this can be se clearer setting $z=0$ and looking the ellipse drawn on the $xy$-plane
Which seems to involve some kind of ugly looking function if I were to describe the ellipse in polar coordinates, which in some way I have to.
Now to find the volume I changed to cylindrical coordinates
$$x=\rho\cos\theta, y=\rho\sin\theta,z=z$$
The equation for the ellipse states $x^2+y^2-2x=0$ then $2\rho^2\cos^2\theta+\rho^2\sin^2\theta-\rho\cos\theta$ from which I have $\rho^2(2\cos^2\theta+\sin^2\theta)-\rho\cos\theta\implies \rho(\theta)=\displaystyle\frac{\cos\theta}{2\cos^2\theta+\sin^2\theta} = \displaystyle\frac{\cos\theta}{1+\cos^2\theta}$ with $0\leq\theta\leq\pi$.
Now, for $z$ considering the sphere $x^2+y^2+z^2=1$ follows $\rho^2+z^2=1\implies -\sqrt{1-\rho^2}\leq z \leq \sqrt{1-\rho^2}$.
The volume seems to be given by $\displaystyle\int_0^{\pi}\displaystyle\int_0^{\frac{\cos\theta}{1+\cos^2\theta}}\rho\displaystyle\int_{-\sqrt{1-\rho^2}}^{\sqrt{1-\rho^2}}\;dzd\rho d\theta$.
For me here is when this become tricky, I'm going to calculate this integrals one at a time.
(1)$\displaystyle\int_{-\sqrt{1-\rho^2}}^{\sqrt{1-\rho^2}}\;dz = 2\sqrt{1-\rho^2}$
(2)$\displaystyle\int_0^{\frac{\cos\theta}{1+\cos^2\theta}}\rho (2\sqrt{1-\rho^2})d\rho = \displaystyle\int_0^{\frac{\cos\theta}{1+\cos^2\theta}}2\rho\sqrt{1-\rho^2} d\rho = -\displaystyle\frac{2}{3}(1-\rho^2)^{3/2}|_{\rho=0}^{\rho=\frac{\cos\theta}{1+\cos^2\theta}} = \left[1-\left(\displaystyle\frac{\cos\theta}{1+\cos^2\theta}\right)^2\right]^{3/2} + \displaystyle\frac{2}{3}$
$=\left[\displaystyle\frac{\cos^2\theta}{1+2\cos^2\theta+\cos^2\theta}\right]^{3/2} + \displaystyle\frac{2}{3}$
(3) Stuck. I don't know what could I do to integrate $\left[1-\left(\displaystyle\frac{\cos\theta}{1+\cos^2\theta}\right)^2\right]^{3/2}$
For the term $\left[1-\left(\displaystyle\frac{\cos\theta}{1+\cos^2\theta}\right)^2\right]$ I considered using partial fractions as follows:
$\left(\displaystyle\frac{\cos\theta}{1+\cos^2\theta}\right)^2
=\displaystyle\frac{1+\cos^2\theta+\cos^4\theta}{(1+\cos^2\theta)^2} = 1 + \displaystyle\frac{1}{(1+\cos^2\theta)^2} + \displaystyle\frac{1}{1+\cos^2\theta}$
But this would make the last integral harder because I'd have to integrate $\displaystyle\int_0^{\pi} \left[1 + \displaystyle\frac{1}{(1+\cos^2\theta)^2} + \displaystyle\frac{1}{1+\cos^2\theta}\right]^{3/2}\;d\theta$
Any ideas?.
|
The equation for the ellipse states $x^2+y^2-2x=0$ (...) from which I have $\rho(\theta)= \displaystyle\frac{\cos\theta}{1+\cos^2\theta}$ with $0\leq\theta\leq\pi$.
The equation of the ellipse is
\begin{equation*}
2x^{2}+y^{2}-2x=0
\end{equation*}
from which I've obtained
\begin{equation*}
\rho (\theta)=\frac{2\cos \theta }{\cos ^{2}\theta +1},\qquad \text{with }-\pi/2 \leq
\theta <\pi/2,
\end{equation*}
because the tangent to the ellipse at $(x,y)=(0,0)$ is the vertical line $x=0$. With these corrections the volume integral becomes
\begin{eqnarray*}
V &=&\int_{-\pi /2}^{\pi /2}\left( \int_{0}^{\frac{2\cos \theta }{\cos
^{2}\theta +1}}2\rho \sqrt{1-\rho ^{2}}d\rho \right) \,d\theta \\
&=&-\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left. \left( 1-\rho ^{2}\right)
^{3/2}\right\vert _{0}^{\frac{2\cos \theta }{\cos ^{2}\theta +1}}\,d\theta
\\
&=&-\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( 1-\left( \frac{2\cos \theta }{
\cos ^{2}\theta +1}\right) ^{2}\right) ^{3/2}-1\,\ d\theta
\end{eqnarray*}
Simplifying, we have that
\begin{eqnarray*}
V &=&\frac{2}{3}\pi -\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( \left( \frac{
\sin ^{2}\theta }{\cos ^{2}\theta +1}\right) ^{2}\right) ^{3/2}\,d\theta \\
&=&\frac{2}{3}\pi -\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( \frac{\sin
^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta \\
&=&\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin ^{2}\theta
}{\cos ^{2}\theta +1}\right) ^{3}\,d\theta ,
\end{eqnarray*}
because the integrand is an even function. The remaining integral can be
converted into a rational function of $t=\tan \frac{\theta }{2}$ by the
Weirstrass substitution $t=\tan \frac{
\theta }{2}$, which I then evaluated in SWP (Scientific Work
Place):
\begin{eqnarray*}
V &=&\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin
^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta ,\qquad t=\tan \frac{
\theta }{2} \\
&=&\frac{2}{3}\pi -\frac{64}{3}\int_{0}^{1}\frac{t^{6}}{\left(
1+t^{4}\right) ^{3}\left( 1+t^{2}\right) }\,dt \\
&=&\cdots \\
&=&\frac{4}{3}\pi -\frac{7}{12}\sqrt{2}\pi =\frac{16-7\sqrt{2}}{12}\pi .
\end{eqnarray*}
ADDED. The integrand in $t$ can be expanded into partial fractions as follows
\begin{equation*}
\frac{t^{6}}{\left( 1+t^{4}\right) ^{3}\left( 1+t^{2}\right) }=\frac{1}{8}
\frac{t^{2}-1}{1+t^{4}}+\frac{1}{4}\frac{3+t^{2}}{\left( 1+t^{4}\right) ^{2}}
-\frac{1}{2}\frac{1+t^{2}}{\left( 1+t^{4}\right) ^{3}}-\frac{1}{8\left(
1+t^{2}\right) }.
\end{equation*}
ADDED 2. WolframAlpha computation of $V =\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta$ confirms the result above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/652334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Center of Mass, Multivariable Calculus I have a solid with the bounds $z=2x^2+2y^2$ where $z=c$ and this solid has a uniform density of B. I need to find the mass and the center of mass of this solid. I know how to find a normal center of mass, but I do not know how to set up an integral for this problem, but I think it involves change of coordinates (Also, assume c>0). Thanks.
| The mass of the solid is defined as
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV,$$
that is, the integral of body density at each point over the volume of the body. In this case we have $\rho \equiv B$ which is constant, therefore the mass will be a multiple of the body's volume:
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV = B \iiint\limits_{\mathcal{B}} \, dV.$$
This is a paraboloid and its volume can be found using the cylindrical coordinate substition. We find $z = 2(x^2+y^2) = 2r^2$ and the limits for $z$ will be $2r^2 \leq z \leq \sqrt{c/2}$. This was found equating $z=2r^2 = c$, therefore $r = \sqrt{c/2}$.
$$
\begin{align}
\iiint\limits_{\mathcal{B}} \, dV & = \int_0^{2 \pi} \hspace{-5pt} \int_0^{\sqrt{c/2}} \hspace{-5pt} \int_{2r^2}^c r \, dz \, dr \, d \theta \\
& = 2 \pi \int_0^{\sqrt{c/2}} cr - 2r^3 \, dr \\
& = 2 \pi \left( \frac{cr^2}{2} - \frac{r^4}{2} \right) \Bigg\vert_0^{\sqrt{c/2}} \\
& = \pi \left( c \cdot \frac{c}{2} - \frac{c^2}{4} \right) \\
& = \frac{c^2 \pi}{4}.
\end{align}
$$
Hence $M = (Bc^2 \pi)/4$.
Coordinates for center of mass are defined as
$$
\begin{align}
\overline{x} & = \frac{1}{M} \iiint x \rho \, dV \\
\overline{y} & = \frac{1}{M} \iiint y \rho \, dV \\
\overline{z} & = \frac{1}{M} \iiint z \rho \, dV.
\end{align}
$$
In physicist notation I've seen this written as
$$\mathbf{R} = \frac{1}{M} \iiint \rho \, \mathbf{r} \, dV.$$
It is interesting to note the following: the paraboloid is symmetric around $z$ axis. This means that the center of mass must be in the $z$ axis, for the $\overline{x}$ and $\overline{y}$ will cancel (if you don't believe this, write out the integral explicitly: you will have to integrate $\cos \theta$ and $\sin \theta$ over $[0,2 \pi]$, which is zero).
Therefore it is left for us to compute the $z$ coordinate. Leaving out the density out for a second (since it is uniform), we have
$$
\begin{align}
\iiint\limits_{\mathcal{B}} z \, dV & = \int_0^{2\pi} \hspace{-5pt} \int_0^{\sqrt{c/2}} \hspace{-5pt} \int_{2r^2}^c z r \, dz \, dr \, d \theta \\
& = 2 \pi \int_0^{\sqrt{c/2}} \frac{r}{2} \left( c^2 - 4r^4 \right) \, dr \\
& = \pi \int_0^{\sqrt{c/2}} c^2 r - 4r^5 \, dr \\
& = \pi \left( \frac{(cr)^2}{2} - \frac{2r^6}{3} \right) \Bigg\vert_0^{\sqrt{c/2}} \\
& = \pi \left( \frac{c^2}{2} \cdot \frac{c}{2} - \frac{2}{3} \cdot \frac{c^3}{8} \right) \\
& = \pi \left( \frac{c^3}{4} - \frac{c^3}{12} \right) \\
& = \pi \left( \frac{c^3}{6} \right) \\
& = \frac{c^3 \pi}{6}.
\end{align}
$$
Finally
$$\overline{z} = \frac{B c^3 \pi}{6} \cdot \frac{4}{Bc^2 \pi} = \frac{2c}{3}.$$
Therefore
$$M = \frac{Bc^2 \pi}{4} \text{ and } \mathbf{R} = (\overline{x}, \overline{y}, \overline{z}) = \left( 0, 0, \frac{2c}{3} \right).$$
Hope this helps and best wishes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/653903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the value of A, B and C in the identities. $6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$
Find the value of A, B and C.
I started it like this:
$6x^3 -11x^2 + 6x + 5 \equiv (Ax-1)(Bx - 1)(x - 1) + c$
Solving the right hand side:
$ (ABx^2 - Ax - Bx + 1)(x - 1) + C$
$ ABx^3 - ABx^2 - Ax^2 + Ax - Bx^2 + Bx + x - 1 + C$
$ABx^3 - (AB + A + B)x^2 + (A + B + 1)x - 1 + C$
Comparing the coefficients:
$AB = 6$
$A = \frac6 B$
$AB + A + B = 11$
Then substitute the value of A in the above equation...is this right? Is there any error?
| Equating x coefficients you will get A+B=5.You already have AB=6.Solving you will get A=2 or 3.This will be the easier approach.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Sum $\sum_{k=0}^{2013}2^ka_{k}$ let real sequence $a_{0},a_{1},a_{2},\cdots,a_{n}$,such
$$a_{0}=2013,a_{n}=-\dfrac{2013}{n}\sum_{k=0}^{n-1}a_{k},n\ge 1$$
How find this sum
$$\sum_{k=0}^{2013}2^ka_{k}$$
My idea: since
$$-na_{n}=2013(a_{0}+a_{1}+a_{2}+\cdots+a_{n-1})\cdots\cdots(1)$$
so
$$-(n+1)a_{n+1}=2013(a_{0}+a_{1}+\cdots+a_{n})\cdots\cdots (2)$$
then
$(2)-(1)$,we have
$$na_{n}-(n+1)a_{n+1}=2013a_{n}$$
then
$$(n+1)a_{n+1}=(2013-n)a_{n}$$
then
$$\dfrac{a_{n+1}}{a_{n}}=\dfrac{2013-n}{n+1}$$
so
$$\dfrac{a_{n}}{a_{n-1}}\cdot\dfrac{a_{n-1}}{a_{n-2}}\cdots\dfrac{a_{1}}{a_{0}}=\cdots$$
so
$$\dfrac{a_{2013}}{a_{0}}=\dfrac{1}{2013}\cdot\dfrac{2}{2012}\cdots\dfrac{2013}{1}=1?$$
then How can find this sum?
| Method 1 (generating function)
Let $a(z) = \sum\limits_{k=0}^\infty a_k z^k$, notice
$$a(z)\left(\frac{z}{1-z}\right) = \left(\sum_{k=0}^\infty a_k\right)\left( \sum_{\ell=1}^\infty z^\ell\right) = \sum_{n=1}^\infty\left(\sum_{k=0}^{n-1}a_k\right)z^n$$
and
$$\left(z\frac{d}{dz}\right)a(z) = \left(z\frac{d}{dz}\right)\sum_{n=0}^\infty a_nz^n
= \sum_{n=1}^\infty na_n z^n$$
The equality $a_{n}=-\dfrac{2013}{n}\sum_{k=0}^{n-1}a_{k},\,n\ge 1$ implies
$$\frac{da(z)}{dz} = -2013\frac{a(z)}{1-z}
\quad\iff\quad \frac{d}{dz} \log a(z) = 2013\frac{d}{dz}\log(1-z)$$
and hence
$$a(z) = a_0(1-z)^{2013} = 2013(1-z)^{2013}$$
Since this is a polynomial with degree 2013, we get
$$\sum_{k=0}^{2013} a_k 2^k = a(2) = 2013 (1-2)^{2013} = -2013$$
Method 2 (more elementary, appropriate for middle school students)
Let $b_n = \sum\limits_{k=0}^n a_k$, we have $b_0 = 2013$ and for $n > 0$,
$$n(b_n-b_{n-1}) = -2013 b_{n-1}\quad\iff\quad b_n = -\frac{2013 -n}{n}b_{n-1}$$
This implies
$$b_n = (-1)^n \frac{\prod\limits_{k=1}^n (2013-k)}{n!} b_0
= (-1)^n \frac{2013!}{n!(2013-n-1)!}
= (-1)^n \binom{2013}{n} (2013-n)$$
Notice $$b_{n-1} = (-1)^{n-1}\frac{2013!}{(n-1)!(2013-n)!} = (-)^{n-1} \binom{2013}{n} n,$$
we obtain
$$a_n = b_n - b_{n-1} = (-1)^n 2013\binom{2013}{n}$$
Using binomial theorem, we can evaluate the desired sum as
$$\sum_{k=0}^{2013} a_k 2^k = 2013 \sum_{k=0}^{2013} \binom{2013}{k}(-2)^k = 2013 (1-2)^{2013} = -2013$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Why is it trivial that $\left(1+\frac{2\ln3}{3}\right)^{-3/2}\leq\frac{2}{3}$? Can someone tell me why $$\left(1+\dfrac{2\ln3}{3}\right)^{-3/2}\leq\dfrac{2}{3}$$ is trivial because for me its not and I will need to do the calculation to see it.
| You have that
\begin{align*}
\left(1+\frac{2\ln3}{3}\right)^{-\frac{3}{2}} &= e^{ -\frac{3}{2}\ln\left(1+\frac{2}{3}\ln3\right) } \\
&=e^{ \frac{3}{2}\ln\frac{1}{1+\frac{2}{3}\ln3} } \\
&\leq e^{ \frac{3}{2}\ln\left( 1-\frac{1}{2}\cdot\frac{2}{3}\ln3 \right) } \\
&\leq e^{ -\frac{3}{2}\frac{1}{3}\ln 3 } = e^{ -\frac{1}{2}\ln 3 } = \frac{1}{\sqrt{3}} \\
&\leq \frac{2}{3}
\end{align*}
where was used the "fact" that $\frac{1}{1+\frac{2}{3}\ln3}\leq 1-\frac{1}{2}\cdot\frac{2}{3}\ln3$ (proving it is not hard) and $\ln(1+x)\leq x$ (for $x>-1$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Solving $\sqrt{3\cos^2 x - \sin 2x} = - \sin x$ Please, can you suggest something for solving this equation: I have to find the solutions included in interval $\left[3\pi/2, 2\pi\right]$:
$$\sqrt{3\cos^2 x - \sin 2x} = - \sin x$$
This is what I did:
$$\begin{array}{crcl}
\Longrightarrow & 3\cos^2 x - \sin 2x &=& \sin^2 x \\
\Longrightarrow &3\left(1-\sin^2 x\right)-\sin 2x &=& \sin^2 x \\
\Longrightarrow & 4\sin^2 x + \sin 2x - 3 &=& 0 \\
\Longrightarrow &2\left(1-\cos 2x\right)+\sin 2x - 3 &=& 0\\
\Longrightarrow &-2\cos 2x + \sin 2x &=& 1\end{array}$$
So, what's next?! Thank you in advance!
| HINT:
We have $$1-\sin2x+2\cos2x=0$$
Using Double-Angle Formulas, $$1-\frac{2t}{1+t^2}+2\frac{1-t^2}{1+t^2}=0$$ where $t=\tan x$
Solve the Quadratic Equation in $\tan x$
As $x\in\left[\frac{3\pi}2,2\pi\right], t=\tan x<0$
| {
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"answer_count": 4,
"answer_id": 1
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Converting rectangular equation to polar equation How do I convert the following rectangular equation to a polar equation:
$$
(x- 3)^2 + (y+2)^2 = 1
$$
I was able to simplify it to the following:
$$
r^2 - 6r\cos(\theta) + 4r\sin(\theta) + 12 = 0
$$
but I am at a loss as to how to simplify further...
I'm trying to get r on one side and the answer on the other. In other words, r = ?
| Given that you're trying to solve for $r$, threat the equation $r^2-6r\cos\theta+4r\sin\theta+12=0$ as a quadratic equation, where $r$ is the only variable and everything else is treated as constants.
But before we begin solving for $r$, it would be best to take the terms $-6r\cos\theta$ and $4r\sin\theta$ and factor out the $r$. This will help make the actual solving part a little easier.
$r^2-6r\cos\theta+4r\sin\theta+12=0$
$r^2+\left(4\sin\theta-6\cos\theta\right)r+12=0$
Note that in addition to factoring out $r$, I also reversed the order of the 2 terms so that the positive term is followed by the negative term, though this step isn't necessarily mandatory.
From here we can either use Completing the Square or Quadratic Formula to solve the equation for $r$. I'll be using Completing the Square.
$r^2+\left(4\sin\theta-6\cos\theta\right)r+12=0$
$r^2+\left(4\sin\theta-6\cos\theta\right)r=-12$
$r^2+\left(4\sin\theta-6\cos\theta\right)r+\left(2\sin\theta-3\cos\theta\right)^2=-12+\left(2\sin\theta-3\cos\theta\right)^2$
$\left(r+2\sin\theta-3\cos\theta\right)^2=\left(2\sin\theta-3\cos\theta\right)^2-12$
$r+2\sin\theta-3\cos\theta=\pm\sqrt{\left(2\sin\theta-3\cos\theta\right)^2-12}$
$r=-2\sin\theta+3\cos\theta\pm\sqrt{\left(2\sin\theta-3\cos\theta\right)^2-12}$
$r=3\cos\theta-2\sin\theta\pm\sqrt{\left(2\sin\theta-3\cos\theta\right)^2-12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/656668",
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"source": "stackexchange",
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Help with Polar coordinates and the length of the curve. I have a test coming up today and I was going over our past midterms and this question came up. I tried it but its not working, please any hints or solution in how to do it will be really helpful.
Question:
Consider the polar curve $r = min(1 - cos(\theta), 1 + sin(\theta))$ and $\theta \in [0, \frac{3\pi}{2}]$
a) Sketch the curve. TO sketch the curve you will need to rewrite r as a case define function
$r = 1 - cos(\theta), \theta \in [a, b]$ and $1 + sin(\theta), \theta \in [c, d]$
with $a, b, c, d$ to be determined.
b) Find the length of the curve. In the final answer you don't need to evaluate the values of sin or cos.
My attempt:
So what I did was I graphed the polar curve of $1 - cos(\theta)$ and $1 + sin(\theta)$ I drew it on the same graph.
I can see that there is a point of intersection and that a and d maybe be 0. But I am not understanding the question properly. My friend told me that I need to use $tan(\theta) = -1$ but I am really confused.
and b is followed by a, please any hints or solutions would be really appreciated.
Thank you
| Both of the polar curves in this problem are cardioids: $ \ r = 1 - \cos \theta \ $ has its symmetry axis along the $ x-$ axis with the cusp opening in the positive $ x-$ direction and $ \ r = 1 + \sin \theta \ $ has its symmetry axis along the $ y-$ axis, its cusp opening in the negative $ y-$ direction.
The points of interest in dealing with these polar curves are the values of $ \ \theta \ $ in the "principal circle" ( $ 0 \le \theta < 2 \pi $ ) at which $ \ r = 0 \ $ and at which the two curves intersect. The "horizontal" cardioid meets the origin for
$$ 1 \ - \ \cos \theta \ = \ 0 \ \ \Rightarrow \ \ \cos \theta \ = \ 1 \ \ \Rightarrow \ \ \theta \ = \ 0 \ \ \ \text{and} $$
and the "vertical" cardioid, for
$$ 1 \ + \ \sin \theta \ = \ 0 \ \ \Rightarrow \ \ \sin \theta \ = \ -1 \ \ \Rightarrow \ \ \theta \ = \ \frac{3 \pi}{2} \ \ . $$
These curves intersect where
$$ 1 \ - \ \cos \theta \ = \ 1 \ + \ \sin \theta \ \ \Rightarrow \ \ \sin \theta \ = \ - \cos \theta \ \ \Rightarrow \ \ \tan \theta \ = \ -1 \ \ \Rightarrow \ \ \theta \ = \ \frac{3 \pi}{4} \ , \ \frac{7 \pi}{4} \ \ . $$
(Dividing through by cosine is "safe" because it does not equal zero at any intersection point.)
Since we wish to use the portions of each curve closest to the origin (the specified "minimum radius" condition), our polar curve will consist of two parts meeting at the intersection point in the second quadrant:
$$ r(\theta) = \left\{ \begin{array}{rcl}
1 \ - \ \cos \theta & \mbox{for}
& 0 \ \le \ \theta \ \le \ \frac{3 \pi}{4} \ \ \text{ [blue arc] } \\ 1 \ + \ \sin \theta & \mbox{for}
& \frac{3 \pi}{4} \ \le \ \theta \ \le \ \frac{3 \pi}{2} \ \ \text{ [red arc] }
\end{array}\right. $$
We wish to find the arclength of this piecewise-defined curve. Since it is symmetrical about the line $ \ \theta = \frac{3 \pi}{4} \ $ , we can integrate one-half of the curve and double the result. As a reminder, the polar form of the infinitesimal arclength element is given by
$$ ds^2 \ = \ dx^2 \ + \ dy^2 \ = \ dr^2 \ + \ r^2 \ d\theta^2 \ \ . $$
The total arclength of our curve is then
$$ s \ = \ 2 \ \int_0^{3 \pi / 4} \ \sqrt{ \ \left(\frac{dr}{d\theta} \right)^2 \ + \ [ \ r(\theta) \ ]^2} \ \ d\theta \ \ = \ \ 2 \ \int_0^{3 \pi / 4} \ \sqrt{ \ ( \ \sin \theta \ )^2 \ + \ [ \ 1 \ - \ \cos \theta \ ]^2} \ \ d\theta $$
$$ = \ \ 2 \ \int_0^{3 \pi / 4} \ \sqrt{ \ 2 \ - \ 2 \ \cos \theta } \ \ d\theta \ \ = \ \ 2 \sqrt{2} \ \int_0^{3 \pi / 4} \ \sqrt{ \ 1 \ - \ \cos \theta } \ \ d\theta $$
$$ = \ \ 2 \sqrt{2} \ \int_0^{3 \pi / 4} \ \sqrt{ \ 1 \ - \ \cos \theta } \ \cdot \ \frac{\sqrt{ \ 1 \ + \ \cos \theta }}{\sqrt{ \ 1 \ + \ \cos \theta }}\ \ d\theta \ \ = \ \ 2 \sqrt{2} \ \int_0^{3 \pi / 4} \ \frac{\sqrt{ \sin^2 \theta }}{\sqrt{ \ 1 \ + \ \cos \theta }}\ \ d\theta$$
$$ = \ \ 2 \sqrt{2} \ \int_0^{3 \pi / 4} \ \frac{ | \sin \theta | }{\sqrt{ \ 1 \ + \ \cos \theta }}\ \ d\theta \ = \ 2 \sqrt{2} \ \int_0^{3 \pi / 4} \ \frac{ \sin \theta}{\sqrt{ \ 1 \ + \ \cos \theta }}\ \ d\theta \ \ , $$
for which we have employed the "conjugate factor method" to make the integral tractable; the absolute value brackets in the numerator may be dropped since the calculation will be carried out entirely "above" the $ \ y-$ axis. Using the substitution $ \ u \ = \ 1 + \cos \theta \ , \ du \ = \ -\sin \theta \ d\theta \ $ , we find
$$ \rightarrow \ \ 2 \sqrt{2} \ \int_2^{1 - (\sqrt{2} / 2)} \ \frac{ - du}{\sqrt{ u}} \ \ = \ \ 2 \sqrt{2} \ \cdot \ 2 \ \sqrt{u} \ \vert^2_{1 - (\sqrt{2} / 2)} $$
$$ = \ 4 \sqrt{2} \ \left[ \ \sqrt{2} \ - \ \sqrt{1 - (\frac{\sqrt{2} }{ 2})} \ \ \right] \ = \ 8 \ \left[ \ 1 \ - \ \frac{\sqrt{2 - \sqrt{2}} }{ 2} \ \right] \ \approx \ 4.939 $$
or
$$ \rightarrow \ \ \ 4 \sqrt{2} \ \cdot \ \sqrt{1 \ + \ \cos \theta} \ \vert^0_{3 \pi / 4} \ = \ \ 4 \sqrt{2} \ \cdot \ \left( \ \sqrt{2} \ - \ \sqrt{1 \ + \ \cos \frac{3 \pi}{4}} \ \right) \ \ . $$
EDIT [4/19] -- As a check, we see that this curve is slightly smaller in arclength than the circumference of a circle of diameter $ \ 1 \ + \ \frac{\sqrt{2}}{2} \ $ , which is $ \ \frac{2 + \sqrt{2}}{2} \pi \ \approx \ 5.36 \ $ , so we may have some confidence in our result above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/657341",
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Proof for maths induction prove: $$ \sum_{r=1}^{n}[r^{2}+1](r!)=n[(n+1)!]$$ for all $n \in N$
prove $n=1$, $(1^2+1)(1!)$ = $1[(1+1)!]$
assume true for $n=k$, $(k^2+1)$$(k!)$= $k$$[(k+1)!]$
I got to here : $[k^{2}+1](k!)+[(k+1)^2+1](k+1)! = (k+1)[(k+2)!]$
| If
$$ \sum_{r=0}^{n}(r^{2}+1)r!=n(n+1)!$$
then
$$ \sum_{r=0}^{n+1}(r^{2}+1)r!=\sum_{r=0}^{n}(r^{2}+1)r!+((n+1)^2+1)(n+1)!=$$
$$=n(n+1)!+((n+1)^2+1)(n+1)!=$$
$$=(n+1)!(n+((n+1)^2+1))=$$
$$=(n+1)!(n+n^2+2n+1+1)=$$
$$=(n+1)!(n^2+3n+2)=$$
$$=(n+1)!(n+2)(n+1)=$$
$$=(n+1)(n+2)!$$
| {
"language": "en",
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"source": "stackexchange",
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Find $\frac{a^3}{a^6 + 1}$ given $a$ is a root of a quadratic equation
If $a$ is a root of the equation $x^2 - 3x + 1 = 0$, then find the value of $\frac{a^3}{a^6 + 1}$.
So, I figured we can use the quadratic formula, and formed the following equation:
$$a=\frac{-(-3)+\sqrt{9-4}}{2(1)}\implies a=\frac{3+\sqrt5}2$$
But what I am thinking is, if I begin to find the required value, it will take me hours. And I believe, there must be some shortcut to this question. I had tried to solve this question with the manual process but it took me a lot of squares (one was $2012^2$!)
Can someone please help me.
Thank you.
| $a^2=3a-1$ then
$$\frac{a^3}{a^6 + 1}
\\=\frac{a^3}{(a^2+1)(a^4-a^2 + 1)}
\\=\frac{a^3}{(3a)(9a^2-6a+1-3a+1+1)}
\\=\frac{a^2}{(3)(9a^2-9a+3)}
\\=\frac{a^2}{(9)(3a^2-3a+1)}
\\=\frac{a^2}{(9)(9a-3-3a+1)}
\\=\frac{a^2}{(9)(6a-2)}
\\=\frac{3a-1}{(18)(3a-1)}
\\=\frac{1}{18}
$$
| {
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"source": "stackexchange",
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Quadratic residue of $-1$ in composite modulus It is true for each odd prime number p that if $x^2\equiv-1 \pmod p$ then $p\equiv1\pmod 4$
I've observed that it should be true for all composite integers, whose prime factors are congruent to $1$ modulo $4$. However I couldn't find any remark on the internet whether it's true.
In other words, if it is correct, how do we prove that $x^2\equiv-1\pmod n$ has a solution if and only if $n=\prod_{p|n}p$ such that $p_i=4k_i+1$
P.S: My level is pretty elementary.
| For an odd $n$, there are $x$ with $x^2 \equiv -1 \pmod{n}$ if and only if all prime divisors of $n$ are of the form $p = 4m+1$.
The necessity follows from $x^2 \equiv -1 \pmod{n} \Rightarrow x^2 \equiv -1 \pmod{d}$ for all divisors $d$ of $n$, in particular its prime divisors.
The sufficiency follows from the Chinese Remainder Theorem, and the fact that for any prime $p$ and $k \geqslant 1$ the group of units in $\mathbb{Z}/(p^k)$ is cyclic.
Since the group is cyclic and its order is a multiple of $4$ for $p \equiv 1 \pmod{4}$, it contains elements of order $4$, and these satisfy $x^2 \equiv -1 \pmod{p^k}$ (since $$x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1) \equiv 0 \pmod{p^k},$$ and neither $x-1$ nor $x+1$ can be divisible by $p$).
Once you have solutions $x_i$ modulo each prime power $p_i^{k_i}$ dividing $n$, the Chinese Remainder Theorem asserts the existence of $x$ with
$$x \equiv x_i \pmod{p_i^{k_i}},\quad 1\leqslant i \leqslant r,$$
where $n = \prod\limits_{i=1}^r p_i^{k_i}$, and such an $x$ satisfies $x^2 \equiv -1\pmod{n}$. Since for each prime power $p_i^{k_i}$ there are exactly $2$ solutions, there are $2^r$ solutions of $x^2\equiv -1\pmod{n}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The equation $3^n+4^m=5^k$ in positive integer numbers Please help me to prove that the equation $3^n + 4^m = 5^k$
where $n$, $m$, $k$ are positive integer numbers has only the solution $n=m=k=2$.
I know how to prove it for $n=m=k$.
If $3^x + 4^x = 5^x$ then $(3/4)^x + 1 = (5/4)^x$,
and this equation has at most one solution
since the function $(3/4)^x + 1$ decreases and the function $(5/4)^x$ increases
for all real $x$. So the solution $x = 2$ is unique.
Thank you very much in advance!
| Considering the equation modulo 3 and 4 separately, we see that $2 | n \implies n = 2a$ and $2 | k \implies k = 2c$ for some positive integers $a, c$. Rewriting, we get
$$\begin{align}
3^{2a} + 4^m &= 5^{2c}\\
2^{2m} &= (5^c)^2 - (3^a)^2\\
2^{2m} &= (5^c + 3^a)(5^c - 3^a)\end{align}$$
We thus allow $5^c + 3^a = 2^y$ and $5^c - 3^a = 2^x$ for non-negative integers $x, y$ such that $x + y = 2m$. Summing both up, we get:
$$2\cdot5^c = 2^x + 2^y$$
If $x = 0$, then $2\cdot5^c = 1 + 2^y\implies y = 0 \implies m = 0$. But the question states that $m > 0$, so the case where $x = 0$ is impossible $\implies x > 0 \implies y > 0$.
This allows us to divide throughout by $2$ to get:
$$5^c = 2^{x - 1} + 2^{y - 1}$$
Taking both sides modulo $2$, we find that the above is impossible for $x-1 > 0$. Hence, $x - 1 = 0$, or $x = 1$. From this, we get:
$$\begin{align}5^c = 1 + 2^{y - 1}\\
5^c - 2^{y - 1} = 1\end{align}$$
This smells of Catalan's Conjecture (now promoted to a theorem), which implies (in this context) that either $c$ or $y - 1$ must be equal to $1$.
If $y - 1 = 1$, then $5^c - 2^1 = 1 \implies 5^c = 3$, which has no solution.
If $c = 1$, then $k = 2c = 2$ and $5 - 2^{y - 1} = 1 \implies 2^{y - 1} = 4 \implies y = 3$. But we deduced that $x = 1$, so by substituting back into the relation $x + y = 2m$, we get $m = 2$, which after substituting back into the original equation gives us directly $n = 2$.
We therefore conclude that there is only one solution, that is, $(n, m, k)$ = $(2, 2, 2)$.
Perhaps the above approach could somehow be used to solve the mess I made over here at this thread.
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove $\measuredangle CDE=2\measuredangle ABE$ In rectangular $ABCD$,and $E\in AC$,such
$$BE=\sqrt{2}\cdot AE$$
show that
$$\measuredangle CDE=2\measuredangle ABE$$
My try: let $$AB=a,AD=b,\dfrac{AE}{AC}=k,$$
then
$$AE=k\sqrt{a^2+b^2},BE=k\sqrt{2(a^2+b^2)}$$
I know have this nice relsut
$$AE^2+EC^2=BE^2+ED^2$$
then
$$ED^2=k^2\cdot AC^2+(1-k)^2\cdot AC^2-2k^2\cdot AC^2=(1-2k)AC^2$$
so
$$\cos{\measuredangle EBD}=\dfrac{AB^2+BE^2-AE^2}{2AB\cdot BE}=\dfrac{a^2+k^2(a^2+b^2)}{2a\cdot k\sqrt{2(a^2+b^2)}}$$
$$\cos{\measuredangle CDE}=\dfrac{DC^2+DE^2-EC^2}{2 DC\cdot DE}=\dfrac{a^2+(1-2k)(a^2+b^2)-(1-k)^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}=\dfrac{a^2-k^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}$$
we will prove
$$2\cos^2{\measuredangle ABE}-1=\cos{\measuredangle CDE}$$
$$\Longleftrightarrow 2\dfrac{((k^2+1)a^2+b^2)^2}{8a^2k^2(a^2+b^2)}-1=\dfrac{a^2-k^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}$$
But I can't
Thank you for you
| Probably not the best way to do it, but maybe you can simplify it.
Let $F \in AB$ such that $AB\perp EF$ and let $G \in CD$ such that $CD \perp EG$.
Let $|AE|=x\Rightarrow |BE|=x\sqrt2$
Let $|EG|=y$
Finally let $\measuredangle ABE=\alpha$, $\measuredangle BAC = \measuredangle ACD = \beta$ and $\measuredangle ADC=\gamma$
$\sin\alpha=\Large\frac{|FE|}{x\sqrt2}$ $\Rightarrow |FE|=\sin\alpha\cdot x\sqrt2$
Likewise, $|BF|=|CG|=\cos\alpha\cdot x\sqrt2$ and $|AF|=|GD|=\cos\beta\cdot x$
$\sin\beta=\Large\frac{|EF|}{|AE|}$$=\sin\alpha\cdot \sqrt2$
$\tan\beta = \Large\frac{y}{\cos\alpha\cdot x\sqrt2}$=$\Large\frac{\sin\beta}{\cos\beta}$ $\Rightarrow y=\Large\frac{\sin\beta\cdot \cos\alpha\cdot x\sqrt2}{\cos\beta}$
Since $\cos\beta=\sqrt{1-\sin^2\beta}=\sqrt{1-2\sin^2\alpha}$ and $\sin\beta=\sin\alpha\cdot \sqrt2$ we can plug these in. So,
$y=\Large\frac{\sin\alpha\cdot \cos\alpha\cdot 2x}{\sqrt{1-2\sin^2\alpha}}$
$\tan\gamma=\Large\frac{y}{\cos\beta\cdot x}$=$\Large\frac{y}{\sqrt{1-2\sin^2\alpha}\cdot x}$
Plug $y$ in
$\tan\gamma=\Large\frac{\sin\alpha\cdot \cos\alpha\cdot 2x}{(1-2\sin^2\alpha)\cdot x}$=$\Large\frac{2\sin\alpha\cos\alpha}{1-2\sin^2\alpha}$=$\Large\frac{\sin2\alpha}{\cos2\alpha}$=$\tan2\alpha$
Hence, $\gamma=2\alpha \Rightarrow \measuredangle CDE=2\measuredangle ABE$
| {
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"source": "stackexchange",
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If $|x + 1| + |x - 3| = 6$, solve for $x$ $|x + 1| + |x - 3| = 6$. Solve for X.
So I know when you have a problem like this: |x| = 6, you solve by doing x=6 and x=-6. That doesn't help us much in the above example.
You can also solve problems of this fashion by negative the variable portion of the equation. For ex:
$|x+3| = 6. $
You can solve it either by doing:
$x+3 = 6$ or
$x+3 = -6$
$x= 3,-9$
OR
$(x+3) = 6$ so $x=3$
or
$-(x+3) = 6$
$-x -3 =6$
$-x = 9$
$x = -9$
So back to the original problem of $|x + 1| + |x - 3| = 6$
if $X>3$, then $x+1 + x-3 = 6$
$2x=8$
$x=4$
or if $X<3$
$x+1 + x-3 = -6$
$2x=-4$
$x=-2$
Is that correct? Is that all I need to do? I also feel like I got the "critical points" of the equation wrong.
| Divide the domain to
(1) $x \le -1$
(2) $-1 \le x \le 3$
(3) $x \ge 3$
solve the equation in each domain
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear algebra : eigenvalues of an integral operator on polynomials Consider the linear transformation
$$
T : \left\{
\begin{array}{ccc}
\mathbb{R}_n[X] & \to & \mathbb{R}_n[X] \\
P & \mapsto & \int_0^1 (X + t)^n\,P(t)\,dt
\end{array}\right.
$$
where $\mathbb{R}_n[X]$ denotes the vector space of polynomials with real coefficients and degree $\leqslant n$.
NB: $T$ is self-adjoint with respect to the $L^2$-inner product $\langle P, Q \rangle = \int_0^1 P(t)\, Q(t) \,dt$.
The question is: what are the eigenvalues of $T$?
Edit: This question was originally posted by someone on the French maths forum les-mathematiques.net (here). I do not know if there is a reason to think that it is actually possible to find an explicit expression for the eigenvalues.
| Using the adhoc convention $\frac{x}{0}=x$, Newton’s binomial formula yields
$$
T(X^j)=\sum_{i=0}^{n} \frac{\binom{n}{i}}{n+1+j-i}X^i \tag{1}
$$
If we denote by ${\cal B}_n=(1,X,X^2, \ldots, X^{n})$ the canonical
basis of ${\mathbb R}_n[X]$, the matrix $A$ of $T$ relatively to ${\cal B}_n$ is
$A=(a_{ij})_{1\leq i,j \leq n+1}$ where
$$
a_{ij}=\frac{\binom{n}{i-1}}{n+1+j-i} \tag{2}
$$
For example, when $n=4$ we have
$$
A=\left(\begin{matrix}
\frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} \\
\ & & & & \\
1 & \frac{4}{5} & \frac{2}{3} & \frac{4}{7} & \frac{1}{2} \\
\ & & & & \\
2 & \frac{3}{2} & \frac{6}{5} & 1 & \frac{6}{7} \\
\ & & & & \\
2 & \frac{4}{3} & 1 & \frac{4}{5} & \frac{2}{3} \\
\ & & & & \\
1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\
\end{matrix}\right)
$$
The characteristic polynomial of this $A$ is
$$\chi_A=X^5 - \frac{16}{5}X^4 - \frac{251}{450}X^3 + \frac{2977}{330750}X^2 -
\frac{19}{1890000}X + \frac{1}{2778300000}$$
a polynomial whose Galois group over $\mathbb Q$ is ${\mathfrak S}_5$ and
whose roots are therefore not expressible by radicals.
Thus I concur with Christopher A.Wong’s comment that it is unlikely that
a simple closed-form solution exists.
| {
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Is it possible to solve $i^2+i+1\equiv 0\pmod{2^p-1}$ in general? While looking at the Mersenne numbers (for prime $p$, the number $2^p-1$), I noticed that only certain of them had any solution to the modular equation $i^2+i+1\equiv 0\pmod{2^p-1}$, e.g., $p=3,5,7,13,17,19,31,37,...$ but not $p=11,23,29,...$.
I have gotten as far as realizing that if $k$ is a solution then so is $-k-1$.
Questions:
*
*Is it possible to solve this modular equation (assuming there is a solution) in general given these restrictions?
*Suppose that for a given $p$ there is a solution $k$ such that $k^2+k+1\equiv 0\pmod{2^p-1}$; then $k\cdot(k+1)\equiv -1\pmod{2^p-1}$. Then (as noted in a comment) $k^3\equiv -(k+1)^3\equiv 1\pmod{2^p-1}$. Is it possible to conclusively prove whether $k$ or $k+1$ is a quadratic non-residue from this information?
| Just as an alternate view of the particular solutions to this modular equation, if $$k^2+k+1\equiv 0\pmod q$$ then $$k^2\equiv -k-1\pmod q$$ so that solution is a quadratic residue $\mod q.$ Taking note of the comments, we have $$(k^2+k+1)(k-1)=k^3-1\implies k^3\equiv 1\pmod q$$ Since we also have $$k(k+1)\equiv -1\pmod q$$ then $$k^3(k+1)^3\equiv -1\pmod q$$ or $k^3\equiv -(k+1)^3\pmod q.$ Multiplying both sides by $k$ yields $$k^4\equiv k\equiv -k\cdot (k+1)\cdot(k+1)^2\equiv (k+1)^2\pmod q$$ thus showing that $k$ is also a quadratic residue.
Based on this success, consider whether $\exists m:m^2\equiv k+1\pmod q$ or $\exists n:n^2\equiv -k\pmod q.$ In the first case, note that $$m^2\equiv k+1\implies (k+1)^2m^2\equiv (k+1)^3\equiv -1\pmod q$$ and in the second case note that $$n^2\equiv -k\implies k^2n^2\equiv -k^3\equiv -1\pmod q$$ so the quadratic residue status of $k+1$ and $-k$ is the same as that of $-1.$ The real modulus is $2^p-1$ which is of the form $4s+3,$ which does not have $-1$ as a quadratic residue.
Taking this one step further (and using the solution $m^2\equiv -3\pmod q$ mentioned in the comments and the other answer), we have Lagrange's formula for the value of $m$ for $m^2=(2i+1)^2\equiv -3\pmod {2^p-1}$ which is $m\equiv \pm (-3)^{2^p-1+1\over 4}=\pm 3^{2^{p-2}}$. This is also obvious since $3$ is a quadratic non-residue $\pmod {2^p-1}$ and thus $3^{2^{p-1}-1}\equiv -1\pmod {2^p-1}$. Now we have that $i=2^{-1}\cdot(\pm 3^{2^{p-2}}-1)$.
The futility of attempting to use the solutions of $i^2+i+1\equiv 0\pmod{2^p-1}$ to reduce the computing time required for $3^{2^{p-1}-1}\pmod{2^p-1}$ is now readily apparent.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$ I need to find a clear formula (without summation) for the following sum:
$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$
Well, the first few elements look like this:
$1,1,1,2,2,2,2,2,3,3,3,...$
In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s etc.
Still I have absolutely no idea how to generalize it for $n$ first terms...
|
The following is valid for $n\geq 1$
\begin{align*}
\sum_{k=1}^{n}\lfloor\sqrt{k}\rfloor
=n\lfloor\sqrt{n}\rfloor
-\frac{1}{3}\lfloor\sqrt{n}\rfloor^3-\frac{1}{2}\lfloor\sqrt{n}\rfloor^2+\frac{5}{6}\lfloor\sqrt{n}\rfloor
\end{align*}
For convenient calculations we consider two aspects:
*
*We introduce a variable $a=\lfloor\sqrt{n}\rfloor$. So, we have
$$a\leq \sqrt{n} < a+1$$
*We use the Iverson Bracket notation, so we can replace the expression $\lfloor x\rfloor$ by
$$\lfloor x\rfloor=\sum_{j\geq 0}[1\leq j \leq x]$$
Special case: $n=a^2,a=\lfloor\sqrt{n}\rfloor$
We start the calculation by conveniently assuming $n=a^2$. We obtain
\begin{align*}
\sum_{k=1}^{n}\lfloor\sqrt{k}\rfloor&=\sum_{k=1}^n\sum_{j\geq 0}[1\leq j \leq \sqrt{k}][0\leq k \leq a^2]\\
&=\sum_{j=1}^{a}\sum_{k=1}^{n}[j^2\leq k \leq a^2]\\
&=\sum_{j=1}^{a}(a^2-j^2+1)\\
&=a^3-\frac{1}{6}a(a+1)(2a+1)+a\\
&=\frac{2}{3}a^3-\frac{1}{2}a^2+\frac{5}{6}a\tag{1}
\end{align*}
General case: $n\geq a^2,a=\lfloor\sqrt{n}\rfloor$
In the general case we let again $a=\lfloor\sqrt{n}\rfloor$ and have additionally to consider the terms for
$a^2< k \leq n$.
They are all equal to $a$, so they sum up to $$(n-a^2)a.$$ Adding this to (1) we get the general formula with $a=\lfloor \sqrt{n}\rfloor$
\begin{align*}
\sum_{k=1}^{n}\lfloor\sqrt{k}\rfloor&=na-\frac{1}{3}a^3-\frac{1}{2}a^2+\frac{5}{6}a\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
Note: This approach can be found in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik
Note: I've also added here an answer for the general case $\sum_{k=1}^n\lfloor\sqrt[p]{k}\rfloor$ with $p\geq 1$.
| {
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Is the following scheme for generating $p_n=(1/3)^n$ stable or not. $p_n=(5/6)p_{n-1}-(1/6)p_{n-2}$. Is the scheme for generating $p_n=(1/3)^n$ stable or not?
$$p_n= \frac{5}{6} p_{n-1} - \frac{1}{6}p_{n-2}.$$
| From your recursion, we have
$$
\left[\begin{array}{c}p_n\\p_{n-1}\end{array}\right]
=\left[\begin{array}{c}5/6&-1/6\\1&0\end{array}\right]
\left[\begin{array}{c}p_{n-1}\\p_{n-2}\end{array}\right]
$$
The eigenvalues of the matrix are
$$
\frac12\quad\text{for }\left[\begin{array}{c}1/2\\1\end{array}\right]
$$
and
$$
\frac13\quad\text{for }\left[\begin{array}{c}1/3\\1\end{array}\right]
$$
Since
$$
\left[\begin{array}{c}1/2&1/3\\1&1\end{array}\right]^{-1}
\left[\begin{array}{c}p_1\\p_0\end{array}\right]
=\left[\begin{array}{c}6p_1-2p_0\\-6p_1+3p_0\end{array}\right]
$$
we get
$$
\left[\begin{array}{c}p_n\\p_{n-1}\end{array}\right]
=\dfrac1{2^{n-1}}\left[\begin{array}{c}1/2\\1\end{array}\right](6p_1-2p_0)
+\dfrac1{3^{n-1}}\left[\begin{array}{c}1/3\\1\end{array}\right](-6p_1+3p_0)
$$
Thus, unless you start with a multiple of $\left[\begin{array}{c}1/3\\1\end{array}\right]$, the system will tend to
$$
\left[\begin{array}{c}p_n\\p_{n-1}\end{array}\right]
\to\dfrac1{2^{n-1}}\left[\begin{array}{c}1/2\\1\end{array}\right](6p_1-2p_0)
$$
Another approach:
Your recursion has the characteristic equation
$$
0=x^2-\frac56x+\frac16=\left(x-\frac12\right)\left(x-\frac13\right)
$$
Thus, the solutions are of the form
$$
\begin{align}
p_n
&=\frac{a}{2^n}+\frac{b}{3^n}\\
&=\frac{-2p_0+6p_1}{2^n}+\frac{3p_0-6p_1}{3^n}
\end{align}
$$
Note that if $p_0=3p_1$, then $p_n$ grows like $3^{-n}$; otherwise, it grows like $2^{-n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/672727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrals involving reciprocal square root of a quartic For two integration below, what is the ratio of them?
$$I_1 = \int_0^1\frac{dt}{\sqrt{1-t^4}}$$ $$I_2 = \int_0^1 \frac{dt}{\sqrt{1+t^4}}$$
What is the ration $\frac{I_1}{I_2}$?
I do not have any thoughts for solving this question so could anyone give me some hints?
Thank you!!
| Hint:
$$I_1(x)=\int_x^1\frac{dt}{\sqrt{1-t^4}}=\frac{1}{\sqrt{2}}\int_{0}^{\arccos x}\frac{du}{\sqrt{1-\frac{1}{2}\sin^2u}}$$
$$I_2(x)=\int_0^x \frac{dt}{\sqrt{1+t^4}}=\frac{1}{2}\int_{0}^{\arccos\frac{1-x^2}{1+x^2}}\frac{du}{\sqrt{1-\frac{1}{2}\sin^2u}}$$
and since $$\arccos 0 = \arccos \frac{1-1^2}{1+1^2}$$
$$\frac{I_1(0)}{I_2(1)}=\sqrt{2}$$
From the hint the substitutions should be pretty straightforward.
Added due to the comment below:
The first one: let $t=\cos u$, $u=\arccos t$.
$$\frac{dt}{\sqrt{1-t^4}}=\frac{-\sin udu}{\sqrt{1-\cos^4 u}}=\frac{-\sin udu}{\sqrt{\sin^2 u(1+\cos^2 u)}}$$
and you get the RHS almost immediately.
The second one: let $t=\tan\frac{u}{2}$, $t^2=\frac{1}{\cos^2 \frac{u}{2}}-1=\frac{1-\cos u}{1+\cos u}$, $u=\arccos \frac{1-t^2}{1+t^2}$. Plug it in, and get the RHS also (almost) immediately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/673373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
If a and b are odd integers, then $8\mid (a^2-b^2)$ Prove: If $a$ and $b$ are odd integers, then $8\mid (a^2-b^2)$
So far I have: $$8\mid \left((2n+1)^2-(2m+1)^2\right)\Longleftrightarrow 8\mid \left(4n^2+4n+1-4m^2-4m-1\right) $$
Is this right so far? Where do I go from here? Step by step explanation please!
| $$4n^2+4n+1-4m^2-4m-1=4n^2+4n-4m^2-4m=4(n^2-m^2+n-m) \\
4[(n-m)(n+m)+(n-m)]=4(n-m)(n+m+1)$$
Now
$$(n-m)+(n+m+1)=2n+1$$
is odd, thus one of $(n-m)$ or $(n+m+1)$ must be even.
P.S. A simpler solution is the following
$$a^2-1=(a-1)(a+1)$$
is the product of two consecutive even numbers. Thus one is divisible by $4$ and the other is even. This shows that $8 |a^2-1$. Same way $8|b^2-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/675856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
If $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$, then $a^3+b^3+c^3=$ If $a,b,c\in \mathbb{R}$ and $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$ and $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31$. Then $a^3+b^3+c^3 = $
$\bf{My\; Trial\; Solution::}$ Given $a^2+b^2+c^2 = 23$ and
$a+b+c = 7\Rightarrow (a+b+c)^2 = 49\Rightarrow (a^2+b^2+c^2)+2(ab+bc+ca) = 49$
So $23+2(ab+bc+ca) = 49\Rightarrow (ab+bc+ca) = 13$
Now from $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31\Rightarrow \frac{(a+1)\cdot (b+1)+(b+1)\cdot (c+1)+(c+a)\cdot (a+1)}{(a+1)(b+1)(c+1)} = 31$
So $\displaystyle \frac{(ab+bc+ca)+2(a+b+c)+3}{1+(a+b+c)+(ab+bc+ca)+abc} = 31\Rightarrow \frac{13+2\cdot 7+3}{1+7+13+abc} = 31$
So $\displaystyle \frac{30}{21+abc} = 31\Rightarrow 21\times 31+31(abc) = 30\Rightarrow (abc) = \frac{30-21\times 31}{31}=-\frac{621}{31}$
Now How can I calculate $a^3+b^3+c^3$
Is there is any better method by which we can calculate $abc$
Help me
Thanks
| Note that $a,b,c$ are the roots of the equation $$x^3-(a+b+c)x^2+(bc+ac+ab)x-abc=0$$ which we know to be $$x^3-7x^2+13x-\frac {621}{31}=0 \dots (A)$$
Add the three equations for $a, b, c$ to obtain
$$(a^3+b^3+c^3)-7(a^2+b^2+c^2)+13(a+b+c)-\frac {3\cdot 621}{31}=0$$
Note that if we define $P_n=a^n+b^n+c^n$ we can multiply equation $A$ by $x^n$ before substituting $a,b,c$ and we get $$P_{n+3}-7P_{n+2}+13P_{n+1}-\frac {621}{31}P_n=0$$ which is a recurrence relation for the sums of higher powers. It works with negative powers too, provided the roots are all non-zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/677184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find all complex values for $a$ where there's no solution for the non homogeneous system
Find all complex values for $a$ where there's no solution for the non homogeneous system
$$\left\{ {\begin{array}{*{20}{c}}
{x + (1 - a)y + z = 1} \\
{2x - y + z = 3} \\
{3x - ay + (a - 1)z = 4} \\
\end{array}} \right.$$
After moving to a matrix representation and reduction, I have:
$$\left( {\begin{array}{*{20}{c}}
1 & {1 - a} & 1 & | & 1 \\
0 & {2a - 3} & { - 1} & | & 1 \\
0 & 0 & {a - 1} & | & 0 \\
\end{array}} \right)$$
Now, I don't see an $a$ that answering this request, and I double-checked my reduction process.
EDIT:
followed by @DonAntonio suggestion:
defining $a= x + yi$
Developing the determinant by first column:
$$\begin{array}{l}
\left| {\begin{array}{*{20}{c}}
{2a - 3} & { - 1} \\
0 & {a - 1} \\
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{2(x + yi) - 3} & { - 1} \\
0 & {x + yi - 1} \\
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{2x - 3 + 2yi} & { - 1} \\
0 & {x - 1 + yi} \\
\end{array}} \right| = \\
(2x - 3 + 2yi) \cdot (x - 1 + yi) = 2{x^2} - 2x + 2xyi - 3x + 3 - 3yi + 2xyi - 2yi + 2{y^2}{i^2} \\
= 2{x^2} - 5x - 5yi + 4xyi - 2{y^2} \\
\end{array}$$
Comparing the result to $0$ we get:
$$\left\{ {\begin{array}{*{20}{c}}
{2{x^2} - 5x - 2{y^2} = 0} \\
{ - 5y + 4xy = 0} \\
\end{array}} \right.$$
Am I doing it right?
| If you rref you get this
$ [1 \ 0 \ 0 \ \frac{3a - 4}{2a - 3} \\ \ 0 \ 1 \ 0 \ \frac{1}{2a - 3} \\ \ 0 \ 0 \ 1 \ \ \ \ \ \ 0 \ \ ] $
So $$ 2a - 3 \ne 0 $$
$$ a \ne \frac{3}{2}$$
Now you still need to check a = 1 because if your calculations are correct , (I didn't check them) to get a 1 in the last row you have to divide by $a - 1$. If a = 1 you are dividing by 0
I checked a = 1 and it works. Putting a = 1 leads to the system
$[1 \ 0 \ 0 \ 1 \\ 0 \ 1 \ 0 -1 \\ 0 \ 0 \ 1 \ 0 \ ]$
And this works by substituting a = 1 , x = 1 , y = -1 z = 0 in the OP.
UPDATE: I got it to this
$[1 \ (1-a) \ \ 1 \ 1 \\ 0 \ (2a - 3) \ \ -1 \ \ 1 \\ 0 \ \ \ 0 \ (a-3) \ \ 0 \ \ ] $
So even if a = 3 it will not produce 'no solution' because of 0 in last entry of row 3 , we would then have 0/0 which leads to a system with a row of 0's
I invite you to check a = 3 in the OP.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/677256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying long fractions How would I go about simplifying long fractions, such as the likes of this:
$((8+\frac{3}{4}) + (3\frac{2}{3}))$ / $((4+\frac{2}{5}) - (1\frac{7}{8}))$
The correct answer is ($4 + \frac{278}{303}$)
I'm not really sure how to approach this problem,
regards.
| Combine the numerator into an improper fraction:
$$\begin{align}8 + \frac{3}{4} + 3\frac{2}{3} &= 8 + 3 + \frac{3}{4} + \frac{2}{3} \\
&= \frac{132}{12} + \frac{9}{12} + \frac{8}{12} \\
&= \frac{149}{12}\end{align}$$
Do the same for the denominator:
$$\begin{align}4 + \frac{2}{5} - 1\frac{7}{8} &= 4 - 1 + \frac{2}{5} - \frac{7}{8}\\
&=\frac{120}{30} + \frac{16}{40} - \frac{35}{40} \\
&= \frac{101}{40}\end{align}$$
Then divide both:
$$\begin{align}\frac{\frac{149}{12}}{\frac{101}{40}} &= \frac{149}{101}\cdot\frac{40}{12}\\
&= \frac{149}{101}\cdot\frac{10}{3}\\
&=\frac{1490}{303}\\
&=4 + \frac{278}{303}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/678286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to use $x + \frac{1}{x} = 7$ to compute $x^2 + \frac{1}{x^2}$. I am not sure how to approach this question:
You know that $x + \frac{1}{x} = 7$. Compute $x^2 + \frac{1}{x^2}$.
I have tried adding $x + \frac{1}{x}$ to get $\frac{x^2 +1}{x}$ but can't see if this was useful or not.
I need help in getting started.
| Thank you to Yiyuan Lee and awllower who really answered the question.
I am only adding this for completeness, and to remind myself of the solution.
We are given $x + \frac{1}{x} =7$ and want to use this to find $x^2 + \frac{1}{x^2}$.
Using the fact $$\left(x +\frac{1}{x}\right)^2 = x^2 +\frac{1}{x^2} +2$$
and rewriting this as:
$$\left(x +\frac{1}{x}\right)^2 -2 = x^2 +\frac{1}{x^2}$$
We can substitute $x + \frac{1}{x} =7$ to get:
$$(7)^2 -2 = x^2 +\frac{1}{x^2}$$
So $$x^2 +\frac{1}{x^2} = 47$$
The same method can be used to calculate $x^3 + \frac{1}{x^3}$:
$$\left(x +\frac{1}{x}\right)^3 = x^3 +\frac{1}{x^3} +3x +\frac{3}{x}$$
Noticing that $3x +\frac{3}{x}$ = $3(x+ \frac{1}{x})$, which means $3x +\frac{3}{x} = 3.(7) = 21$
Then it is possible to write:
$$\left(x +\frac{1}{x}\right)^3 -21 = x^3 + \frac{1}{x^3}$$
$$(7)^3 -21 = x^3 + \frac{1}{x^3}$$
$$x^3 + \frac{1}{x^3} = 343 -21$$
$$x^3 + \frac{1}{x^3} = 322$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/679402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
How to compute the value of $1 \cdot \frac{2}3 \cdot \frac 45 \cdots$, as $n$ goes to infinity? It's supposed to be $\dfrac 12\sqrt{\dfrac \pi n}$.
| From the Wallis's product in the following form ( Advanced Calculus by Angus Taylor, formula (20.7-9))
\begin{equation*}
\frac{\pi }{2}=\lim_{n\rightarrow \infty }\left( \frac{2\cdot 4\cdots (2n)}{
1\cdot 3\cdots (2n-1)}\right) ^{2}\frac{1}{2n+1}\tag{1}
\end{equation*}
follows that the limit in the title is zero.
ADDED. The formula $(1)$ is a consequence of the following double inequality
\begin{equation*}
\frac{2n}{2n+1}\frac{\pi }{2}<\left( \frac{2\cdot 4\cdots (2n)}{1\cdot
3\cdots (2n-1)}\right) ^{2}\frac{1}{2n+1}<\frac{\pi }{2},\tag{2}
\end{equation*}
which can be derived integrating over the interval $]0,\pi/2[$
$$\sin^{2n+1}\theta<\sin^{2n}\theta<\sin^{2n-1}\theta.\tag{3}$$
Inequality $(2)$ can be rewritten as
\begin{equation*}
\frac{\sqrt{n}}{2n+1}\sqrt{\pi }<\frac{2\cdot 4\cdots (2n)}{1\cdot 3\cdots
(2n-1)\left( 2n+1\right) }<\sqrt{\frac{\pi }{2\left( 2n+1\right) }}<\frac{1}{
2}\sqrt{\frac{\pi }{n}},\tag{4}
\end{equation*}
thus proving the asymptotic expression you indicate in the question body. Is this what you mean?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/680062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Factoring the quintic polynomial $x^5+4x^3+x^2+4=0$ I am trying to factor
$$x^5+4x^3+x^2+4=0$$
I've used Ruffini's rule to get
$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$
But I don't know what to do next.
The solution is $(x+1) (x^2+4) (x^2-x+1) = 0$. I've tried using the completing square method but with no result. Could you give me hints?
| The answers actually
$$
(x^2 + 4)(x^3 + 1)\\
x^5 + 4x^3 + x^2 + 4\\
(x^5 + 4x^3)(x^2 + 4)\\
x^3(x^2 + 4) 1(x^2 + 4)\\
(x^2 + 4)(x^3 + 1)
$$
(Sorry, too lazy to actually change them to exponents)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/681059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Evaluating a trigonometric limit What is the limit as $x$ approaches $0$ of: $$\frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{1+x-\cos x}?$$
We cannot use L'Hôpital's rule or anything advanced like Taylor series. I reduced it to, by considering the numerator's conjugate:
$$\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{1+x-\cos x}$$
But I cannot go further. Please help.
EDIT: Thinking carefully, I think I can simplify further:
$$\frac{1}{2}\lim_{x \to 0}\frac{\frac{\tan x - \sin x}{x}}{\frac{1-\cos x}{x}+1}=\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{x}$$
And perhaps the solution:
$$\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{x}=\frac{1}{2}(\lim_{x \to 0}\frac{\tan x}{x} - 1)=\frac{1}{2}(\lim_{x \to 0}\frac{\sin x}{\cos x \cdot x}) - 1=\frac{1}{2}(1- 1)=0$$
Is this correct?
| $$\frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{1+x-\cos x}\sim_0\frac{\sqrt{1+x+\frac{x^3}3} - \sqrt{1+x-\frac{x^3}6}}{x+\frac{x^2}2}\sim_0\frac{\frac{1}{2}\left(\frac{x^3}3+\frac{x^3}6\right)}{x}\sim_0\frac14 x^2$$
so the desired limit is $0$.
Edit Without Taylor series the calculus is tedious: let
$$f(x)=\sqrt{1+\tan x} - \sqrt{1+\sin x}$$
and
$$g(x)=1+x-\cos x$$
then
$$\lim_{x\to0}\frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{1+x-\cos x}=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}\frac{x-0}{g(x)-g(0)}=\frac{f'(0)}{g'(0)}$$
and with painful calculus we find that $f'(0)=0$ and easily $g'(0)=1$ so we conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/681774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Dimension and its consequences Find a basis for $S=[\mathrm{(1,2,3),(3,4,7),(5,-2,3)}]\subseteq \Bbb R^3$ and give the dimension.
Then, putting all the vectors as the columns of a new matrix:
$A=\begin{bmatrix}
1 & 3 & 5 \\
2 & 4 & -2 \\
3 & 7 & 3
\end{bmatrix}$
By row reducing:
$A_R=\begin{bmatrix}
1 & 0 & -13 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}$
Let $C=(c_1,c_2, c_3) \in \Bbb R^3$
$A_R\times C =0 \implies \begin{cases} c_3 = {1\over{13}}c_1 \\ c_2 = {-6\over{13}}c_1 \end{cases}$
Then $S$ is not linearly independent, so I remove the vector $(5,-2,3)$ from $S$
Then now $S=\mathrm{gen}[(1,2,3),(3,4,7)]$
Now $(1,2,3)$ and $(3,4,7)$ are linearly independent and $\dim(S)=2$
So, if $S$ is a basis for $W\subseteq \Bbb R^3$ and $\dim(S)=2 \implies W= \Bbb R^2$
Then, I want to check if $gen(S)=W$.
Let $B=\begin{bmatrix}
1 & 3 \\
2 & 4 \\
3 & 7
\end{bmatrix}
$ which reduced form is
$ \begin{bmatrix}
1 & 0 \\
0 & 1 \\
0 & 0
\end{bmatrix}
$, $k_1,k_2 \in \Bbb R$
$k_1 \times \begin{bmatrix}1\\0\\0 \end{bmatrix} + k_2 \times \begin{bmatrix} 0 \\ 1 \\0 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_ 2 \\ x_3 \end{bmatrix} \implies \begin{cases} k_1=x_1 \\ k_2 = x_2 \\ x_3 = 0 \end{cases}$
So, finally, my question:
This not generates $\Bbb R^3$, but $\mathrm{gen}(S)=\Bbb R^2$?
| By putting the vectors as the columns of the matrix, the leading entries in the row echelon form indicate columns of the original matrix that form a basis of its column space. In this case, we have:
$$
\overbrace{\begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & -2 \\ 3 & 7 & 3 \\ \end{bmatrix}}^A
\xrightarrow{\text{row operations}}
\overbrace{\begin{bmatrix} 1 & 0 & -13 \\ 0 & 1 & 6 \\ 0 & 0 & 0 \\ \end{bmatrix}}^{A_R}
$$
(there's a bug in the $A_R$ in the question). This implies the linear combination $$\color{blue}{13}C_1\color{red}{-6}C_2+\color{purple}{1}C_3=\mathbf{0}$$ of the columns of either $A_R$. The same linear combination applies for the columns of $A$:
$$\color{blue}{13}\begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix}\color{red}{-6}\begin{bmatrix} 3 \\ 4 \\ 7 \\ \end{bmatrix}+\color{purple}{1}\begin{bmatrix} 5 \\ -2 \\ 3 \\ \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}.$$
We see that $A_R$ has rank $2$, so its column space will be $2$-dimensional, and will be $$\mathrm{span}\{(1,2,3),(3,4,7),(5,-2,3)\}=\mathrm{span}\{(1,2,3),(3,4,7)\}.$$ So $$\{(1,2,3),(3,4,7)\}$$ is a basis (as is $\{(1,2,3),(5,-2,3)\}$ and $\{(3,4,7),(5,-2,3)\}$).
Note: $\{(1,0,0),(0,1,0)\}$ is not a basis: $\mathrm{span}\{(1,0,0),(0,1,0)\}$ does not even contain $(1,2,3)$.
We may also find a basis by putting the vectors as the rows of the matrix:
$$\begin{bmatrix} 1 & 2 & 3 \\ 3 & 4 & 7 \\ 5 & -2 & 3 \\ \end{bmatrix}
\xrightarrow{\text{row operations}}
\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}$$
which gives the basis $\{(1,2,3),(0,1,1)\}$ formed by the non-zero rows in the row echelon form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove continuity for cubic root using epsilon-delta I am trying to prove that a function is continuous at a point a using the $\epsilon$-$\delta$ theorem. I managed to find a $\delta$ in this case $|2x^2+1 - (2a^2+1)| < \epsilon$. But I have a hard time when the function under consideration is $f(x) = \sqrt[3]{x}$. That is, I want to have if $|x-a|<\delta$, then $|\sqrt[3]{x}-\sqrt[3]{a}| < \epsilon$. Any suggestions?
| $$|\sqrt[3]{x}-\sqrt[3]{a}| = |\sqrt[3]{x}-\sqrt[3]{a}| \times \frac {| x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |}{|x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |} = \frac {|x - a|}{|x^{\frac 2 3} + \sqrt[3]{a}\sqrt[3]{x} + a^{\frac 2 3} |} \le \frac {| x - a |}{| {ax} |^{\frac 1 3}}$$
The final inequality is due to the fact that $x^{\frac 2 3} + a^{\frac 2 3} \ge 0$.
Let us assume $a \neq 0$. Then We can bound $|x|$ as follows.
Say $|x - a| \lt |a| $ then $ |x| \lt 2|a| \implies |ax| \lt 2|a|^2 \implies \frac {1}{2|a|^2} \lt \frac {1} {|ax|}$.
Therefore $|\sqrt[3]{x}-\sqrt[3]{a}| \lt \frac {| x - a |}{| {a} |^{\frac 2 3}}$ as long as $|x - a| \lt |a|$ and $a \neq 0$. So if we pick $\delta = \text {Min} \{ |a|, \epsilon|a|^{\frac 2 3} \}$ then $|x - a| \lt \delta \implies |\sqrt[3]{x}-\sqrt[3]{a}| \lt \epsilon$. This proves the function is continuous everywhere except at $0$. To get rid of the case when $a = 0$ just pick $\delta = \epsilon^{3} $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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N-th power of matrix Find the formula for the n-th power of this matrix.
$$
\pmatrix{1&1\\1&0}
$$
Well $f^2 = \pmatrix{2&1\\1&1}$ and $f^3 = \pmatrix{3&2\\2&1}$ and $f^4 = \pmatrix{5&3\\3&2}$
and $f^5 = \pmatrix{8&5\\5&3}$...
I can't seem to find the pattern here. Would someone mind helping me
| Hint: For $F = \pmatrix{1&1\\1&0}$, show (perhaps inductively) that
$$
F^n = F^{n-1} + F^{n-2}
$$
Inductive step:
$$
F^{n+1} = F \cdot F^n = F(F^{n-1} + F^{n-2}) = F^n + F^{n-1}
$$
For the base case, note that $F^0 = I$.
| {
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How many "Two pair" poker hands are there in a standard deck? From the solutions to our midterm, I know the answer of this question to be:
$$\binom{13}{2} \times \binom{4}{2} \times \binom{4}{2} \times \binom{11}{1} \times \binom{4}{1}$$
However, I initially thought it would be:
$$\binom{13}{1} \times \binom{4}{2} \times \binom{12}{1} \times \binom{4}{2} \times \binom{11}{1} \times \binom{4}{1}$$
I think I see that my answer would could KKQQ5 and QQKK5 as two separate hands, but wouldn't the solution read 5QQKK and QQKK5 as separate hands as well?
| The difference between yours and the solution is $\binom{13}{2}$ versus $\binom{13}{1} \times \binom{12}{1}$. If you write them out you will see that $2 \binom{13}{2} = \binom{13}{1} \times \binom{12}{1}$, the 2 double counting as you have noticed (that KKQQ5 and QQKK5 are essentially the same).
(Note that $\binom{13}{2}$, the number of ways to choose $2$ pairs from 13 possible is unordered. Meanwhile, $\binom{13}{1} \times \binom{12}{1}$ involves order which we have to remove)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maclaurin series of $\frac{1}{1+\sin x}$ Find the terms through degree four of the Maclaurin series of $f(x)$.
$$f(x) = \frac{1}{1+\sin x}$$
My work:
The Maclaurin series for $\sin x$ up to degree $4$ is $x - \frac{x^3}{6} + \frac{x^5}{120}$
The Maclaurin series for $\frac{1}{1+x}$ up to degree $4$ is $1 - x + x^2 - x^3 + x^4$
I substituted $x - \frac{x^3}{6} + \frac{x^5}{120}$ for $x$ in $1 - x + x^2 - x^3 + x^4$
Did I do this right?
Plugging this into WolframAlpha, I get this: http://goo.gl/SKddyh
Which doesn't seem like the answer in the text: $1-x+x^2-\frac{5x^3}{6}+\frac{2x^4}{3}$
| Your answer is correct, but there is a somewhat simpler method. You are looking for a polynomial $P=a+bx+cx^2+dx^3+ex^4$ such that $P\times(1+x-\frac{x^3}6)\equiv 1\pmod{x^5}$. This gives (by comparing coefficients of $x^0,x^1,\ldots,x^4$) the equations
$$
\begin{align}a&=1\\a+b&=0\\b+c&=0\\-\tfrac16a+c+d&=0\\-\tfrac16b+d+e&=0,
\end{align}
$$
which you can solve straight away as $a=1$, $b=-1$, $c=1$, $d=-\tfrac56$, $e=\tfrac23$.
| {
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Solving $6ab+a+b-36pq-19p-13q=7$ where $a,b,p,q \in \mathbb{N}$, $a,b,p,q \neq 0$ Is there an efficient way to find solutions to the equation:
$6ab+a+b-36pq-19p-13q=7$ where $a,b,p,q \in \mathbb{N}$ and $a,b,p,q \neq 0$
If the equation has no solutions, how could you prove that, that is the case?
Edit: The equation can in fact be expressed in the form;
$\frac{(6a+1)(6b+1)}{6}- \frac{1}{6} - \frac{(36p+13)(36q+19)}{36} - \frac{247}{36} =7$
which is the same as
$6(6a+1)(6b+1)-(36p+13)(36q+19)=11$
but I am still left wondering how to move forwards.
| Partial answer:
You can factor:
$$\frac{1}{6} (6a + 1)(6b + 1) = 36(p + \frac{13}{36})(q + \frac{19}{36}) + \frac{11}{36}$$
$$6(6a+1)(6b+1) = (36p + 13)(36q + 19) + 11$$
Note that $RHS \equiv LHS \equiv 6 \pmod {36}$. Some trial and error with $p, q$ yields $(5, 7, 1, 4), (2, 32, 1, 8), (3, 27, 1, 10), (9, 11, 1, 12)$. With $p = 2$, the first solution is $(15, 5, 2, 5)$. Solutions certainly exist.
| {
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Prove, that two equations are equivalent EDIT: Missed something very important! Sorry! We have $x^4+1=2(2x-1)^{1/4}$ not $x^4+1=2\sqrt{2x-1}$. One friend of mine told me that the equation $x^4+1=2(2x-1)^{1/4}$, where $x\geq \frac{1}{2}$
is equivalent to $$x^4+1=2x$$. How did he obtain this? The equation $x^4+1=2(2x-1)^{1/4}$ is equivalent to $\frac{x^4+1}{2}=\sqrt{2x-1}$ and we can consider two functions: One is $f:[\frac{1}{2},\infty)\rightarrow[0,\infty)$, $f(x)=(2x-1)^{1/4}$ and $g:[0,\infty)\rightarrow [\frac{1}{2},\infty), g(x)=\frac{x^4+1}{2}$. $ g $is the inverse of$ f$ so they can only intersect on the line $y=x$. Then we can consider the equation $x^4+1=2x$
Is that enough?
| What you can say is that $$x^4 +1 = 2\sqrt{2x - 1}\tag{1}$$ $$x^4 + 1 = 2x\tag{2}$$
share one solution: when $x = 1$, both equations are satisfied.
But each equation has a second solution not shared by the other.
*
*Real solutions to $(1)$ are $x = 1,\;x\approx 0.68682$.
*Real solutions to $(2)$ are $\;x = 1,\;\text{ and }\;x \approx
0.54369.$
Hence, equations $(1)$ and $(2)$ cannot be equivalent.
Note also that $$2x = 2\sqrt{2x-1} \iff x = 1.$$ The equality holds if and only if $x = 1$, and nowhere else.
We can see each of $f(x) = x^4 + 1, \;g(x) = 2\sqrt{2x - 1},\;h(x) = 2x$ in the following graph (refer to top graph only):
| {
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Solving for $\sum_{n = 1}^{\infty} \frac{n^3}{8^n}$? I was trying to solve $ \displaystyle \sum_{n = 1}^{\infty} \frac{n^3}{8^n}$ and I found a way to solve it and I want if there are generalizations for, say, $\displaystyle \sum_{n=1}^{\infty} \frac{n^k}{a^n}$ in terms of $k$ and $a$. I would also like to know if there is a better way to solve it. Here's how I did it:
First I decomposed the series into the following sums:
$S_1 = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$
$S_2 = \frac{7}{64} + \frac{7}{512} + \dots = \frac{\frac{7}{64}}{\frac{7}{8}}$
$S_3 = \frac{19}{512} + \frac{19}{4096} + \dots = \frac{\frac{19}{512}}{\frac{7}{8}}$
And deduced that the sum can be written as $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{3n^2 - 3n + 1}{8^n}$
$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ is easy to evaluate -- it's $\frac{1}{7} $by geometric series
$\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ can be evaluated in a whole host of ways to get an answer of $\frac{8}{49}$.
It remains to evaluate $\displaystyle \sum_{n = 1}^{\infty} \frac{n^2}{8^n}$, for which I took a similar approach as the cubics by decomposing it into many sums:
$T_1 = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$
$T_2 = \frac{3}{64} + \frac{3}{512} + \dots = \frac{\frac{3}{64}}{\frac{7}{8}}$
And so forth, coming to the conclusion that it is equal to $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{2n-1}{8^n}$
Now, I used this information and the above values for $\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ and $\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ to get the sum as $\frac{776}{2401}$, which is confirmed by WA.
So, I would like to reiterate here: Is there a simpler way to compute this sum, and are there any known generalizations for this problem given an arbitrary $a$ in the denominator and arbitrary $k$ as the exponent in the numerator?
| $$\sum {n\choose k}x^n$$ is easy to sum, using the binomial theorem. Then if you can express $n^k$ in terms of ${n\choose0},{n\choose1},\dots,{n\choose k}$, you can get a formula for $\sum n^kx^n$. Expressing powers of $n$ in terms of those binomial coefficients can be done using Stirling numbers, which I invite you to look up.
| {
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"timestamp": "2023-03-29T00:00:00",
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Different solutions of $x+y+z=10$ where $x$, $y$, $z$ are all positive integers and $x, y, z \leq 10$ The number of solutions to the equation $x+y+z=10$ where $x,y,z$ are positive integers, is given by ${k−1 \choose n−1}$, where in this case $k=10,n=3$, giving us ${9 \choose 2} = 36$
Now we have
$x + y + z = 10$ with $x, y, z \leq 10$ (where $x,y,z$ are positive integers and can be the same)
What are the different methods by which we can solve this?
| One way is to consider this as a multiset of 3 types of elements, 7 in total (as there is at least 1 of each type, substract that). So the number of solutions is:
$$
\left(\!\!\binom{3}{7} \!\! \right) = \binom{3 + 7 - 1}{7} = \binom{9}{7} = 36
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this fraction even possible to put into partial fractions? I'm to integrate $\int\frac{x}{x^2+6x+13}dx$
But I'm finding it impossible to do anything with it.
Since $x^2+6x+13$ is an irreducible quadratic factor, and the only factor it means that the partial fraction should be in the form...
$\frac{Ax+B}{x^2+6x+13}$
Hence $x = (Ax+B)(x^2+6x+13)$
$=>x= Ax^3+(6A+B)x^2+(13A+6B)x+13B$
Equating coefficients mean that A =0, 6A+B = 0, 13A+6B = 1 and 13B = 0.
Obviously this doesn't hold. Since 13A+6B apparently = 1 despite A and B both coming out as zero.
The other way I did it is
Let x = 0, hence 0 = 13B, therefore B = 0.
Let x = 1, hence 1 = A + 6A + 13A = 20A, therefore A = $\frac{1}{20}$
But that means $\int\frac{x}{x^2+6x+13}dx$ = $\int\frac{x}{20(x^2+6x+13)}dx$, which isn't possible unless x is always 0. And it doesn't help me integrate it.
Quite clearly I've done something wrong something along the lines, but I have no idea what?
Any pointers?
| Hints:
$$\frac x{x^2+6x+13}=\frac12\frac{2x+6}{x^2+6x+13}-\frac3{(x+3)^2+4}=$$
$$\frac12\frac{(x^2+6x+13)'}{x^2+6x+13}-\frac32\frac{\left(\frac{x+3}2\right)'}{1+\left(\frac{x+3}2\right)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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a quadratic equation for two unknown number find values of $p$ such that the equation $4x^2 + 3px - 2p = 0$ has? below are a few choices of the value p:
a) 2 real roots
b) 1 real roots
c) no roots or complex roots
so far i did for a) 2 real roots
$$(3p)^2 - 4(4)(-2p) > 0\\
9p^2 +32p > 0\\
9\left( p^2+\frac{32p}{9} \right) > 0\\
p^2+\frac{32p}{9} > 0\\
p^2 +\frac{32p}{9} + \left(\frac{32}{18}\right)^2 > 0\\
\left(p + \frac{32}{18} \right) ^2 > 0 \\
p + \frac{32}{18} > 0\\
p > -\frac{32}{18} $$
| $D=(3p)^2+4\cdot4\cdot2p=9p^2+32p=p(9p+32)$
if $D>0$ then there is 2(different) real roots
$D=9p^2+32p=p(9p+32)>0$
$p\in(-\infty;-32/9)\cup(0;+\infty)$
if $D=0$ then there is 1(or 2 equal) real root
$p=0$ and $p=-32/9$
if $D<0$ then there is no roots or 2 complex roots
$p\in(-32/9;0)$
| {
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If $\,x>1$, then $\lim\limits_{n\rightarrow\infty}\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}=x$. How can I prove that
$$
\lim_{n\rightarrow\infty}\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}=x,
$$
whenever $x>1$. Here $\left\lfloor \cdot\right\rfloor$ denotes the floor function,
or the integer part function.
The integer part $\lfloor z\rfloor$ of $z$ is the largest integer, which does not exceed $z$.
Thanks for your answer.
| Since $y-1< \lfloor y\rfloor\le y$, for every $y\in\mathbb R$, then
$$
\frac{x^{n+1}-1}{x^{n}}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}< \frac{x^{n+1}}{x^n-1},
$$
and hence
$$
x-\frac{1}{x^n}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}<x+\frac{x}{x^n-1},
$$
or
$$
-\frac{1}{x^n}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}-x<\frac{x}{x^n-1}.
$$
Since both
$
-\frac{1}{x^n},\,\frac{x}{x^n-1}\to 0, \quad\text{as}\quad n\to\infty,
$
then $\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}\to x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How does $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$? Good afternoon my wonderful friends!
Whenever I do this equation I set it up using the difference of two cubes, which is as follows:
$(a+b)^3 = (a+b)(a^2-ab+b^2) = a^3 + b^3$
Whenever I try to use this formula I always get:
$(x+h) x^2 - xh + h^2$
Simplify:
$x^3 - x^2h + h^2x + x^2h - xh^2 + h^3$
Simplify:
x^3 + h^3$
I don't understand were the three's come from in the final answer: $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$
| $$(a+b)^3 \neq a^3 + b^3$$
for example for $a=1$ and $b=1$ you get $(a+b)^3=(1+1)^3=2^3=8 \neq 2 = 1^3 + 1^3 = a^3 + b^3$
To evaluate $(a+b)^n$ one can use the property of the Pascal's triangle
in your case:
$$(a+b)^3 = (a+b)*(a+b)*(a+b)=(a^2 + ab + ab + b^2) * (a+b)=$$
$$(a^2 + 2ab + b^2)* (a+b) = a^3 + a^2b + 2a^2b + 2ab^2 + b^2a+ b^3 = $$
$$1a^2 + 3a^2b + 3ab^2 + 1b^3$$
You can look at the Pascal's triangle in line 4 (where you have $n = 3$) and see what the coefficient of your expanded polynomial are. You will see that are exactly $1,3,3,1$
I.e. for $n=4$, line $5$ We have the coefficients $1,4,6,4,1$ and indeed $(a+b)^4 = 1a^4 + 4 a^3b + 6a^2b^2 + 4ab^3 + 1b^4$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to algebraically prove $\binom{n+m}{2} = nm + \binom{n}{2} + \binom{m}{2}$? Need help trying to prove this problem algebraically.
$$\binom{n+m}{2} = nm + \binom{n}{2} + \binom{m}{2}$$
The farthest I've got is simplifying the RHS to $$nm + \frac{n(n-1)}{2!} + \frac{m(m-1)}{2!}$$
but not sure what to do after that.
| $${n+m\choose2}=mn+{n\choose2}+{m\choose2}$$
Without loss of generality we will prove this identity by induction on $m$.
Notice that for $m=2$
\begin{align}{n+2\choose2}&=\frac{(n+2)!}{n!2!}=\frac{(n+2)(n+1)n!}{n(n-1)(n-2)!2!}=\frac{(n+2)(n+1)}{n(n-1)}\cdot{n\choose2}\\&=\frac{n(n-1)+4n+2}{n(n-1)}\cdot{n\choose2}={n\choose2}+\frac{4n+2}{n(n-1)}\cdot{n\choose2}={n\choose2}+\frac{4n+2}{2}\\&={n\choose2}+1+2n={n\choose2}+{2\choose2}+2n\end{align}
for $m=3$
\begin{align}{n+3\choose2}&=\frac{n+3}{n+1}{n+2\choose2}=\frac{n+3}{n+1}({n\choose2}+{2\choose2}+2n)=(1+\frac{2}{n+1})({n\choose2}+{2\choose2}+2n)\\&={n\choose2}+{2\choose2}+2n+\frac{n(n-1)}{n+1}+\frac{2}{n+1}+\frac{4n}{n+1}\\&={n\choose2}+{2\choose2}+2n+\frac{(n+1)(n+2)}{n+1}={n\choose2}+{2\choose2}+3n+2\\&={n\choose2}+{3\choose2}+3n
\end{align}
Assume identity holds true for $m=k$. Let $m=k+1$ then
\begin{align}{n+k+1\choose2}&=\frac{(n+k+1)!}{(n+k-1)!2!}=\frac{(n+k+1)(n+k)!}{(n+k-1)(n+k-2)!2!}\\&=\frac{n+k+1}{n+k-1}{n+k\choose2}\\&=\frac{n+k+1}{n+k-1}(nk+{n\choose2}+{k\choose2})\\&=(1+\frac{2}{n+k-1})(nk+{n\choose2}+{k\choose2})\\&=nk+{n\choose2}+{k\choose2}+\frac{2nk}{n+k-1}+\frac{n(n-1)}{n+k-1}+\frac{k(k-1)}{n+k-1}\\
&=nk+{n\choose2}+{k\choose2}+\frac{(n+k)(n+k-1)}{n+k-1}\\
&=nk+{n\choose2}+{k\choose2}+(n+k)\\
&=n(k+1)+{n\choose2}+{k\choose2}+k\\
&=n(k+1)+{n\choose2}+\frac{k!}{(k-2)!2!}+k\\
&=n(k+1)+{n\choose2}+\frac{k!+2k(k-2)!}{(k-2)!2!}\\
&=n(k+1)+{n\choose2}+\frac{(k+1)k(k-2)!}{(k-2)!2!}\\
&=n(k+1)+{n\choose2}+\frac{(k+1)!}{(k-1)!2!}\\
&=n(k+1)+{n\choose2}+{k+1\choose2}
\end{align}
| {
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Rationalizing mixed denominators? How would I rationalize the following Fraction?
$$ \frac {2}{5-\sqrt2+\sqrt3}$$
I have considered the idea of multiplying by the same radicals, but the 5 prevents that.
| $$\begin{align}\frac{2}{5-\sqrt2+\sqrt3}\left(\frac{5+\sqrt2-\sqrt3}{5+\sqrt2-\sqrt3}\right)&=\frac{10+2\sqrt2-2\sqrt3}{20+2\sqrt6}\\
&=\frac{10+2\sqrt2-2\sqrt3}{20+2\sqrt6}\left(\frac{20-2\sqrt6}{20-2\sqrt6}\right)\\
&=\frac{200-20\sqrt6+40\sqrt2-4\sqrt{12}-40\sqrt3+4\sqrt{18}}{400-24}\\
&=\frac{200-20\sqrt6+52\sqrt2-48\sqrt3}{376}\\
&=\frac{50-5\sqrt6+13\sqrt2-12\sqrt3}{94}\\
&\approx.37609
\end{align}$$
You can check you answer here
| {
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Derivatives of Logarithmic function Determine $f'(x)$ for $f(x) = ln(x + \sqrt{x^2 + 1})$
My handbook has the answer as $\displaystyle\frac{1}{\sqrt{x^2 + 1}}$ with no steps on how they got there.
I tried to get there, but somewhere I am getting things wrong. This is what I have
\begin{align}
f'(x) & = \displaystyle\frac{1}{x + \sqrt{x^2 + 1}}\cdot\tfrac{d}{dx}(x + \sqrt{x^2 + 1})\\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} \cdot\tfrac{d}{dx}(x + \sqrt{x^2 + 1})}{x + \sqrt{x^2 + 1}}\\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} \cdot (1 + \tfrac{d}{dx}(x^2 + 1)^{\tfrac{1}{2}})}{x + \sqrt{x^2 + 1}} \\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} \cdot (1 + \tfrac{1}{2}(\tfrac{d}{dx}x^2 + 0)^{-\tfrac{1}{2}})}{x + \sqrt{x^2 + 1}} \\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} \cdot (1 + \tfrac{1}{2}(2x)^{-\tfrac{1}{2}})}{x + \sqrt{x^2 + 1}} \\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} \cdot (1 + \tfrac{1}{2\sqrt{2x}})}{x + \sqrt{x^2 + 1}} \\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} + \tfrac{2\sqrt{2x}}{x + \sqrt{x^2 + 1}}}{x + \sqrt{x^2 + 1}} \\
\end{align}
Here I gave up, I have a mistake, but I can't figure out where
| Why, in your second line, did you divide the derivative by $$x + \sqrt{x^2 + 1}\;\;?$$
Your first step is correct:
$$f'(x)=\displaystyle\frac{1}{x + \sqrt{x^2 + 1}}\cdot\tfrac{d}{dx}(x + \sqrt{x^2 + 1})$$
But when we take the derivative of $x + \sqrt{x^2 + 1} = x + (x^2 + 1)^{1/2},$ we get
$$f'(x) = \dfrac 1{x+\sqrt{x^2 + 1}}\cdot \underbrace{\left(1 + \require{cancel}\dfrac{\cancel{2}x}{\cancel{2}(x^2 + 1)^{1/2}}\right)}_{\frac d{dx} \left(x + (x^2 + 1)^{1/2}\right)}$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/714599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is $f(x)=\sum_{k\in\mathbb N}\frac1k\sin\frac x{2^k}$ bounded?
$$f(x)=\sum_{k\in\mathbb N}\frac1k\sin\frac x{2^k}$$Is this function bounded?
So obviously this converges because $|\frac1k\sin\frac x{2^k}|<|\frac x{2^k}|$ and $\sum\frac x{2^k}$ converges by the integral test.
Now I need to show that there exists a $N$ for all $y\in\text{range}\,f$ such that $|y|<N$.
So I think:
Let $a_k=\frac1k$ and $b_k=\sin\frac x{2^k}$. From the Schwarz inequality we get $|\sum a_kb_k|\leq\sqrt{\sum|a_k|^2\sum|b_k|^2}$. Since $a_k>0$ for all $k$ we have $a_k=|a_k|$ and thus $|a_k|^2=\frac1{k^2}$. $\sum\frac1{k^2}=\frac{\pi^2}6$ from the Reimann-Zeta. Now all we have left to prove is that $\sum|b_k|^2$ is bounded as well.
| The function is unbounded. In fact, take $x_m = \frac{2^{3m}}7 2\pi$; then the sequence $f(x_m)$ tends to infinity. To see this, note that
$$
\sin\Big( \frac n72\pi \Big) \approx \begin{cases}
0.781831, &\text{if }n\equiv1\pmod 7, \\
0.974928, &\text{if }n\equiv2\pmod 7, \\
-0.433884, &\text{if }n\equiv4\pmod 7.
\end{cases}
$$
Therefore
\begin{align*}
f(x_m) &= \sum_{j=1}^m \bigg( \frac1{3j-2}\sin\Big(\frac{2^{3m-(3j-2)}}7 2\pi\Big) + \frac1{3j-1}\sin\Big(\frac{2^{3m-(3j-1)}}7 2\pi\Big) + \frac1{3j}\sin\Big(\frac{2^{3m-3j}}7 2\pi\Big) \bigg) \\
&\qquad{}+ \sum_{k=3m+1}^\infty \frac1k\sin\Big( \frac\pi{7\cdot 2^{k-3m-1}} \Big) \\
&\approx \sum_{j=1}^m \bigg( \frac{0.974928}{3j-2} - \frac{0.433884}{3j-1} + \frac{0.781831}{3j} \bigg) + \sum_{k=3m+1}^\infty \frac1k\sin\Big( \frac\pi{7\cdot 2^{k-3m-1}} \Big) \\
&> \sum_{j=1}^m \frac{0.44}j + \sum_{k=3m+1}^\infty 0 > 0.44 \ln m.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/714921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Evaluate the limit $\lim\limits_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\frac1x}$ Evaluate
$$
\displaystyle\lim_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\displaystyle\frac{1}{x}}
$$
And actually I have my answer and just need someone to verify this for me since I haven't done something like this for a long time.
First, to deal with the pesky $1/x$, I take the natural log inside the limit:
\begin{align}
\lim_{x\to0+}\ln\left(\frac{3^x+5^x}{2}\right)^{\displaystyle\frac{1}{x}}
&= \lim_{x\to0+}\frac{1}{x}\ln\left(\frac{3^x+5^x}{2}\right)\\
&= \lim_{x\to0+}\frac{\ln(3^x+5^x)-\ln2}{x}\\
&= \lim_{x\to0+}\frac{3^x\ln3+5^x\ln5}{3^x+5^x}......L'Hopital's \;Rule\\
&=\frac{\ln3+\ln5}{2}\\
&=\frac{1}{2}\ln3+\frac{1}{2}\ln5
\end{align}
And since what we calculated was the limit the of the natural log, the final answer would be $\displaystyle e^{\frac{1}{2}ln3+\frac{1}{2}ln5}=e^{\sqrt{3}+\sqrt{5}}$. Please tell me if I did this correctly, thanks.
| Everything is good until you try to simplify $e^{\frac{1}{2}\ln 3 +\frac{1}{2}\ln 5}$. By the rules of logarithms, $\frac{1}{2}\ln 3 + \frac{1}{2}\ln 5 = \ln \sqrt{3} + \ln \sqrt{5} = \ln (\sqrt{3}\cdot \sqrt{5}) = \ln \sqrt{15}$. So the final answer is $e^{\ln \sqrt{15}} = \sqrt{15}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $n^7+7$ can never be a perfect square. Prove that for a positive integer $n$, $n^7+7$ cannot be a perfect square.
I managed to show that $n \equiv 5 \pmod{8}$ or $n \equiv 9 \pmod{16}$. But nothing came from that so I presume another approach is needed???
Thanks for any help.
| probably easier than previous solutions:
Suppose that $n^7+7=x^2, x \in \mathbb{N}$ has a solution. Then we must have that $x^2+11^2=x^2+121=n^7+128=n^7+2^7$. Now $n+2|n^7+2^7=x^2+11^2$. If $11|n+2$, then $\nu_{11}(n^7+2^7)=\nu_{11}(n+2)$. This also means that $11|x^2+11^2 \implies 11|x$. Let $x=11y$, then we get $\nu_{11}(n+2)=\nu_{11}(n^7+2^7)=\nu_{11}(x^2+11^2)=2+\nu_{11}(y^2+1)=2$ since $-1$ is a quadratic non residue modulo $11$( because $11 \equiv 3 \pmod{4}$). If $11\not|n+2$ then obviously $\nu_{11}(n+2)=0$. in any case $\nu_{11}(n+2)$ is even. Consider any prime $p$ apart from $11$ dividing $n+2$, we must have that $x^2 \equiv -11^2 \pmod{p} \implies (x \cdot (11^{-1}))^2 \equiv -1 \pmod{p} \implies p \equiv 1 \pmod{4}$ because $-1$ is a quadratic residue for only those primes which are $\equiv 1 \pmod{4}$. Hence we get $$n+2 \equiv 11^{2k}\equiv 1 \pmod{4} \implies n \equiv -1 \pmod{4}$$ where $k=0,1$. But then $x^2=n^7+7 \equiv 7-1 \equiv 2 \pmod{4}$ which is clearly false. Hence we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/721834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Given six numbers $x,y,z,a,b,c$ which satisfy the following relations. Express $x+y+z$ in terms of $a,b,c$ Given six numbers $x,y,z,a,b,c$ which satisfy the following relations
$y^2+yz+z^2=a^2$
$z^2+zx+x^2=b^2$
$x^2+xy+y^2=c^2$
Express $x+y+z$ in terms of $a,b,c$
My attempt:
$\dfrac{(y^3-z^3)}{(y-z)}=a^2,\dfrac{(z^3-x^3)}{(z-x)}=b^2,\dfrac{(x^3-y^3)}{(x-y)}=c^2$
But this step didn't seem to help.
So I tried adding and expressing $k(a+b+c)^2$ but some other unmanagable elements came along. I could not think of any other way to approach the problem. Please help.
| Let $A=a^2,B=b^2,C=c^2$. We have
$$
\begin{array}{lcl}
B-A &=& (x-y)(x+y+z) \\
C-A &=& (x-z)(x+y+z) \\
C-B &=& (y-z)(x+y+z) \\
\end{array}\tag{1}
$$
Then, if we put $D=(B-A)^2+(C-A)^2+(C-B)^2$, we deduce
$$
\begin{array}
D
&=& (x+y+z)^2\Bigg((x-y)^2+(x-z)^2+(y-z)^2\Bigg) \\
&=& 2(x+y+z)^2\Bigg(x^2+y^2+z^2-xy-xz-yz\Bigg) \\
&=& 2(x+y+z)^2\Bigg(
\big(2(x^2+y^2+z^2)+xy+xz+yz\big)-\big(x^2+y^2+z^2+2(xy+xz+yz)\big)
\Bigg) \\
&=& 2(x+y+z)^2\Bigg(
\big(A+B+C\big)-\big((x+y+z)^2\big)
\Bigg) \\
\end{array}\tag{2}
$$
So the sum $w=x+y+z$ satisfies
$$
2w^4-2(A+B+C)w^2+(B-A)^2+(C-A)^2+(C-B)^2=0 \tag{3}
$$
If we simplify by 2 we obtain
$$
w^4-(a^2+b^2+c^2)w^2+(a^4+b^4+c^4-((ab)^2+(ac)^2+(bc)^2)=0 \tag{4}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Bessel function integral How to solve the integral for $J_1{(2x\sin{\frac{\theta}{2}})}$ at $[0,\pi]$? If solving by Matlab, please provide me the source. Thank you!
| A simple analytical result is derived from the series representation of $J_1$:
$$J_1(z) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (k+1)!} \left ( \frac{z}{2}\right )^{2 k+1} $$
as well as the integral
$$\int_0^{\pi/2} du \, \sin^{2 k+1}{u} = \frac{2^{2 k}}{\displaystyle (2 k+1) \binom{2 k}{k}}$$
Thus
$$\begin{align}\int_0^{\pi} d\theta \, J_1\left (2 x \sin{\frac{\theta}{2}} \right ) &= \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (k+1)!} x^{2 k+1} \int_0^{\pi} d\theta \, \sin^{2 k+1} {\frac{\theta}{2}} \\ &= 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{k! (k+1)!} x^{2 k+1} \int_0^{\pi/2} du \, \sin^{2 k+1}{u} \\ &= \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)(2 k+1)} \frac{(2 x)^{2 k+1}}{(2 k)!}\\ &= \frac1{x}\sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+2)!} (2 x)^{2 k+2}\\ &= \frac{1-\cos{2 x}}{x}\end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/723892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help with discrete math proof? I am having trouble proving the following:
If $x\in R$ and $x > 0$, then $x^4+1 \geq x^3+x$.
Work: I tried to rearrange the equation as $x^4-x^3-x+1 \geq 1$, but that does not really help. I also tried proof by cases where case 1 would be that x is irrational and case 2 would be that x is rational. However, that has not got me far either. I am not really sure how to approach this problem.
| As $x\in \mathbb{R}$ and $x>0$, then we will analyze the sign of the expression
$$x^4-x^3-x+1. $$
Indeed,
$$x^4-x^3-x+1=x^3(x-1)-1(x-1)=(x-1)(x^3-1) $$
Now, as $x^3-1=(x-1)(x^2+x+1)$, then
$$x^4-x^3-x+1=(x-1)(x-1)(x^2+x+1) =(x-1)^2(x^2+x+1).$$
Note that $(x-1)^2(x^2+x+1)\geq 0$ because $x>0$ (it can be zero if $x=1>0$).
Therefore, it follows that $x^4-x^3-x+1\geq 0$, then $x^4+1\geq x^3+x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Factorising a complex polynomial over $\mathbb{C}$ I'm given $f(z)=z^6-1$ to factorise over $\mathbb{C}$.
My working is as follows up to the point I don't understand:
$f(-1)=0$ and $f(1)=0$
So $(z+1)$ and $(z-1)$ are factors
$(z+1)(z-1)=z^2-1$
$(z^2-1)(z^4+pz^3+qz^2+rz+s)=z^6-1$
$z^6+pz^5+qz^4-z^4+rz^3-pz^3+sz^2-qz^2-rz-s=z^6-1$
$s=1$ because $-s=-1$
$q=1$ because $qz^4-z^4=0$
$p=0$ and $r=0$ because $pz^5-rz=0$
$(z^2-1)(z^4+z^2+1)=z^6-1$
If $z^4+z^2+1=0$
Let $y=z^2$
$y^2+y+1=0$
$y=$$-1\pm\sqrt3i\over2$
$z^2=$$-1\pm\sqrt3i\over2$
Now here's where I get stuck. In the answers section of my textbook the remaining 4 factors are listed as:
$z+{\frac{1}{2}\pm\frac{\sqrt3}{2}i},$
$z-{\frac{1}{2}\pm\frac{\sqrt3}{2}i}$
but I have:
$z=\pm\sqrt{-\frac12+\frac{\sqrt3}{2}i},$
$z=\pm\sqrt{-\frac12-\frac{\sqrt3}{2}i}$
and I don't know if I've made a mistake somewhere, or if I just don't know how to link what I have so far to the answers given by the textbook.
| This solution refers to the original post: $z^6 + 1$. We have: $z^3 + i$ and $z^3 - i$ are factors. And $f(z) = z^3 + i$ has $f(i) = i^3 + i = -i + i = 0$. So $z - i$ is a factor, and you can easily find that: $z^3 + i = (z - i)(z^2 + iz - 1)$. Let $g(z) = z^3 - i$. Then $g(-i) = 0$. So $z + i$ is a factor of $g(z)$, and you can also find that: $z^3 - i = (z + i)(z^2 - iz - 1)$. And from this you can use quadratic formula to solve for the remaining zeroes, and find all factors and roots.
For the case: $z^6 - 1$. We have: $z^6 - 1 = (z^3 - 1)(z^3 + 1) = (z - 1)(z^2 + z + 1)(z + 1)(z^2 - z + 1)$. And use quadratic formula for the two quadratic factors to find the zeroes and then all the factors and roots.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In Triangle, $\sin\frac A2\!+\!\sin\frac B2\!+\!\sin\frac C2\!-\!1\!=\!4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$ To prove
$$\sin\frac A 2+\sin\frac B 2+\sin\frac C 2-1=4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$$
My approach :
$$
\begin{align}
\text{L.H.S.} & = \sin\frac{A}{2}+\sin \frac{B}{2}+\sin\frac{C}{2} -1 \\[8pt]
& = 2\sin\frac{A+B}{4}\cos\frac{A-B}{4}+\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)-1 \\[8pt]
& = 2\sin\frac{\pi -C }{4}\cos\frac{A-B}{4} - 2\sin^2\left(\frac{\pi -C}{4} \right) \\[8pt]
& =2\sin\frac{\pi -C }{4}\left\{ \cos\frac{A-B}{4} - \sin\left(\frac{\pi -C}{4} \right)\right\}
\end{align}
$$
Unable to move further please help. thanks.
| I will prove,
$$\sin\frac{A}{2}+\sin \frac{B}{2}+\sin\frac{C}{2} - \sin \frac{\pi}{2}= 4\sin \frac{\pi -A}{4}\sin\frac{\pi -B}{4} \sin\frac{\pi-C}{4}$$
Let $z_1=e^{iA/2}$, $z_2=e^{iB/2}$, $z_3 = e^{iC/2}$, $z_4=e^{i\pi/2}=i$
If $z_0=e^{i\theta}$, $\sin(\theta)= \frac1{2i}\left(z_0 - \frac1{z_0}\right)$
Therefore, LHS,
$$\frac{1}{2i}\left(z_1-\frac{1}{z_1}+z_2-\frac{1}{z_2}+z_3-\frac{1}{z_3} -z_4+\frac{1}{z_4}\right)$$
Substituting, $z_4=i$,
$$\frac{1}{2i}\left(z_1-\frac{1}{z_1}+z_2-\frac{1}{z_2}+z_3-\frac{1}{z_3} -2i\right)$$
Now for the RHS,
$\sin\left(\frac{\pi-A}{4}\right) = \sin((\pi/2-A/2)/2) = \frac{1}{2i}(t-\frac{1}{t})$ where $t=\sqrt{\frac{z_4}{z_1}}$
Similarly for the other two terms. Multiply and we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two people A and B throwing dice A begins by throwing a dice until he gets $6$, then B does the same thing.
What is the probability that A throws more times than B?
I try to solve it, but I got 2 different answers:
1. We mark: $X=$How many times A throws, $Y$=How many times B throws.
$$P(X>n|Y=n)=\frac{P(X>n)\cdot P(Y=n)}{P(Y=n)}=P(X>n)\\ \sum_{k=n}^\infty\left(\frac56\right)^{k-1}\cdot \left(\frac16\right)=\left(\frac65\right)^{1-n}=\left(\frac56\right)^{n-1}$$
2. The second way is: We can say that "B throw less times than A", so let's assume that A throws the dice $n$ times. So:
$$P(Y<n)=\sum_{k=1}^{n-1}\left(\frac56\right)^{k-1}\cdot \left(\frac16\right)=1-\left(\frac56\right)^{n}$$
They are opposite, and I don't know which one of them is correct or both of them are incorrect..
Please help with this.
Thank you!
| To find the probability that A throws more times than B, we need to sum the probabilities that A throws two or more times given that B throws once, A throws three or more times given that B throws twice and so on.
So we need to evaluate
$\sum_{k=1}^{\infty}P(A_{throws} > k,B_{throws}=k)$
Now for B throwing once, and A throwing two or more times, we have as probability (using the sum to infinity of a geometric progression)
$\large \frac{1}{6}\times\sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k\left(\frac{1}{6}\right)=\left(\frac{1}{6}\right)^2\left(\frac{\frac{5}{6}}{1-\frac{5}{6}}\right)=\frac{1}{6}\left(\frac{5}{6}\right)$
For B throwing twice, and A throwing three or more times, we have as probability (using the sum to infinity of a geometric progression as before)
$\large\left(\frac{5}{6}\right) \frac{1}{6}\times\sum_{k=2}^{\infty}\left(\frac{5}{6}\right)^k\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^2\left(\frac{\left(\frac{5}{6}\right)^2}{1-\frac{5}{6}}\right)=\frac{1}{6}\left(\frac{5}{6}\right)^3$
For B throwing three times, and A four or more times we have
$\large\left(\frac{5}{6}\right)^2 \frac{1}{6}\times\sum_{k=3}^{\infty}\left(\frac{5}{6}\right)^k\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)^2\left(\frac{\left(\frac{5}{6}\right)^3}{1-\frac{5}{6}}\right)=\frac{1}{6}\left(\frac{5}{6}\right)^5$
Generalising for B throwing $m$ times and A throwing $m+1$ or more times, the probability is given by
$\large \frac{1}{6}\left(\frac{5}{6}\right)^{2m-1}$
Thus, the probability that A throws more times than B is given by the following scaled sum to infinity of a geometric progression with ratio $\left(\frac{5}{6}\right)^2$
$\large \sum_{k=1}^{\infty}P(A_{throws}>k,B_{throws}=k)=\frac{1}{6}\sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^{2k-1}=\frac{1}{6}\frac{\frac{5}{6}}{1-\left(\frac{5}{6}\right)^2}=\frac{5}{11}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
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Harder Trigonometry Identity ($\sec A+\csc A$) How do I prove:
$\sin A (1 + \tan A) + \cos A (1 + \cot A) = \sec A + \csc A$
I've tried expanding the brackets by multiplying sin A and cos A to the left hand side but to no avail. Where should I start from?
| Another possibility, just continuing to get common denominators:
$$\sin A \left(1 + {\sin A\over \cos A}\right) + \cos A\left(1 + {\cos A\over \sin A}\right) \\ = \sin A \left({\cos A + \sin A\over \cos A}\right) + \cos A \left({\sin A + \cos A\over \sin A}\right) \\ = {\sin A(\cos A + \sin A)\over \cos A} + {\cos A (\sin A + \cos A)\over \sin A}\\= {\sin^2 A(\cos A + \sin A) + \cos^2 A(\sin A + \cos A)\over \cos A \sin A} \\ = {(\sin^2 A + \cos^2 A)(\cos A + \sin A)\over \cos A \sin A} \\ = {\cos A + \sin A\over \cos A \sin A} \\ = {1\over \sin A} + {1\over \cos A}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a^2 - b^2 + c^2 - d^2 \ge (a - b + c - d)^2$ In thinking about a base case in this problem, I came up with the following question.
Given real numbers $a \ge b \ge c \ge d \ge 0$, prove that the following holds:
$a^2 - b^2 + c^2 - d^2 \ge (a - b + c - d)^2 \tag{A}$
My attempt:
After simplification, this reduces to proving the inequality:
$\underbrace{ab}_{(1)} + \underbrace{bc}_{(2)} + \underbrace{cd + da}_{(3)} \ge \underbrace{ac}_{(1)} + \underbrace{bd}_{(2)} + \underbrace{b^2 + d^2}_{(3)} \tag{B}$
I tried to attack pairs of terms individually. This gave pairs $(1)$ and $(2)$ which satisfied the $\ge$ relation, since $ab \ge ac \implies b \ge c$ true and $bc \ge bd \implies c \ge d$ true.
But then I got stuck at proving pair $(3)$ also satisfied the $\ge$ relation. That is $cd + da \ge b^2 + d^2 \tag{C}$
It turned out that $(C)$ doesn't hold in general. For example $(a, b, c, d) = (5, 4, 3, 2)$ gives $3 \cdot 2 + 2 \cdot 5 \ge 4^2 + 2^2 \implies 16 \ge 20$ false.
So, my strategy was incorrect. I would appreciate if anyone could show me the right approach for proving either (A) or (B).
Update: See this for a generalization of this problem.
| Here's a method without expanding everything:
$$\begin{align}(a^2-b^2)+(c^2-d^2)&\ge((a-b)+(c-d))^2\\
(a-b)(a+b)+(c-d)(c+d)&\ge(a-b)^2+(c-d)^2+2(a-b)(c-d)\\
(a-b)(a+b-(a-b))+(c-d)(c+d-(c-d)&\ge2(a-b)(c-d)\\
(a-b)b+(c-d)d&\ge(a-b)(c-d)\\
(a-b)(b-c+d)+(c-d)d&\ge0\end{align}$$
And that's obvious. If you want to see when equality holds, consider all $4$ cases:
*
*$a=b$ and $c=d$
*$a=b$ and $d=0$
*$b=c-d$ and $d=0$, that is $b=c$ and $d=0$
*$b=c-d$ and $c=d$, this implies $b=c=d=0$ and is already handled by the previous case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/734007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Trignometry: Find the value in terms of $ \ l \ $ If $ \ \tan x + \sec x \ = \ l \ , $
find the value of secant x in terms of $ \ l \ $ .
I do not know how to solve it but i do know that the basic trig identities are involved.
Pls help me!!!
| Simply use the definitions:
$$\tan x + \sec x = \frac{\sin x}{\cos x} + \frac{1}{\cos x} = \frac{1 + \sin x}{\cos x}.$$
So we seek $x$ such that
$$1 + \sin x = L \cos x \quad \text{or} \quad L \cos x - \sin x = 1.$$
Let $t = \tan(x/2)$. Then
$$\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) = 2\frac{\sin(x/2)}{\cos(x/2)}\cos^2\left(\frac{x}{2}\right) = 2\frac{\tan(x/2)}{\sec^2(x/2)} = \color{red}{\frac{2t}{1 + t^2}}.$$
Similarly,
$$\cos x = \cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right) = \cos^2\left(\frac{x}{2}\right)\left(1 - \frac{\sin^2(x/2)}{\cos^2(x/2)}\right) = \frac{1 - \tan^2(x/2)}{\sec^2(x/2)} = \color{red}{\frac{1 - t^2}{1 + t^2}}.$$
To obtain the expressions in $\color{red}{\text{red}}$ just draw a right triangle with angle $= x/2$.
Therefore, $L \cos x - \sin x = 1$ transforms into
$$L\frac{1 - t^2}{1 + t^2} - \frac{2t}{1 + t^2} - 1 = 0.$$
If you solve for $t$ you should get $t = -1$ or $t = (L - 1)/(L + 1)$. So
$$x = -\frac{\pi}{2} \quad \text{or} \quad x = 2\tan^{-1}\left(\frac{L - 1}{L + 1}\right).$$
Clearly, $x \ne -\pi/2$ because $\cos(-\pi/2) = 0$, so
$$x = 2\tan^{-1}\left(\frac{L - 1}{L + 1}\right), \quad L \ne -1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/736035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric Limits - solution needed: $\lim_{x\to π/2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}$ how to solve this problem? (without using l'Hopital rule)
$$\lim_{x\to π/2} \frac{1-\sin x+\cos x}{\sin 2x -\cos x}$$
thanks for helping.
| It is much simpler to put $x = \dfrac{\pi}{2} - t$ so that we have $\sin x = \cos t, \cos x = \sin t$ and $\sin 2x = \sin (\pi - 2t) = \sin 2t$. Also as $x \to \pi/2$ we have $t \to 0$. Clearly we then see that $$\begin{aligned}L &= \lim_{x \to \pi/2}\frac{1 - \sin x + \cos x}{\sin 2x - \cos x}\\
&= \lim_{t \to 0}\frac{1 - \cos t + \sin t}{\sin 2t - \sin t}\\
&= \lim_{t \to 0}\frac{1 - \cos t + \sin t}{\sin t(2\cos t - 1)}\\
&= \lim_{t \to 0}\frac{1 - \cos t + \sin t}{\sin t\cdot 1}\\
&= \lim_{t \to 0}\frac{1 - \cos t}{\sin t} + 1\\
&= \lim_{t \to 0}\frac{1 - \cos t}{t^{2}}\cdot t\cdot \frac{t}{\sin t} + 1\\
&= \frac{1}{2}\cdot 0\cdot 1 + 1 = 1\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/739645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
An almost impossible limit The following limit appeared in a qualification exam: Find the limit of
$$\lim_{x \to 0} \left( \frac{\tan (\sin (x))-\sin (\tan (x))}{x^7} \right).$$
I ended up doing it in Mathematica, is there any other way?
Thanks in advance!
| Recall that
$$\sin(x) = x-x^3/6+x^5/120-x^7/5040+O(x^9)$$
$$\tan(x)=x+x^3/3+2x^5/15+17x^7/315+O(x^9)$$
thus
$$\begin{align}
\tan(\sin(x)) &= x-x^3/6+x^5/120+17x^7/315+O(x^9)\\
&\;+\;(x-x^3/6+x^5/120+17x^7/315+O(x^9))^3/3\\
&\;+\;2(x-x^3/6+x^5/120+17x^7/315+O(x^9))^5/15\\
&\;-\;(x-x^3/6+x^5/120+17x^7/315+O(x^9))^7/5040+O(x^9)\\
&= x-x^3/6+x^5/120+17x^7/315+O(x^9)\\
&\;+\;(x\cdot x \cdot x+3x\cdot x\cdot -x^3/6+3x\cdot (-x^3/6)^2+3x\cdot x\cdot x^5/120 + O(x^9))/3\\
&\;+\;2(x\cdot x\cdot x\cdot x\cdot x + 5x\cdot x\cdot x\cdot x\cdot -x^3/6 + O(x^9))/15\\
&\;-\;(x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x + O(x^9))/5040\\
&= x + x^3/6-x^5/40-107x^7/5040 + O(x^9)
\end{align}$$
which agrees with WolframAlpha's result. Similarly we can compute
$$\sin(\tan(x))=x + x^3/6-x^5/40-55x^7/1008 + O(x^9)$$
and so
$$\begin{align}
\lim_{x\to 0}\frac{\tan(\sin(x))-\sin(\tan(x))}{x^7}
&=\lim_{x\to 0}\frac{-107x^7/5040 + 55x^7/1008+O(x^9)}{x^7}\\
&=\frac{-107}{5040}+\frac{55}{1008}=\frac{1}{30}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/741446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Discrete Math Induction: $\sum^n_{i=1} \frac1{i(i+1)}$ For $\sum^n_{i=1} \frac1{i(i+1)}$ Find a formula and proofs that it holds for all n ≥ 1.
How would I find the formula for this one that can hold for all n ≥ 1?
| Notice that:
$$\frac{1}{i} - \frac{1}{i + 1} = \frac{1}{i(i+1)}$$
Afterwards, observe the cancellations that occur in:
$$\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} + \dots - \frac{1}{n} + \frac{1}{n} - \frac{1}{n + 1}\\
= 1 - \frac{1}{n + 1}$$
Then, we have to prove this by induction (although I find it unnecessary): Here's the inductive step:
$$\begin{align}\sum_{i = 1}^{n + 1} \frac{1}{i(i+1)} &= \frac{1}{(n+1)(n+2)} + \sum_{i = 1}^{n} \frac{1}{i(i+1)}\\
&= \frac{1}{(n+1)(n+2)} + 1 - \frac{1}{n + 1}\\
&= 1 + \frac{1 - n - 2}{(n+1)(n+2)}\\
&= 1 - \frac{1}{n + 2}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/741750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Value range of parameter inequality. Assume $a,b,c>0$ satisfies :$$5c-3a \leq b \leq 4c-a$$ and $$c\ln b \geq a + c\ln c$$. Find the value range of $\frac{b}{a}$.
My approach:
because $a>0$ and I want to construct the target expression : $\frac{b}{a}$. I divide the first inequality by $a$ in the three part. and I get : $$5\frac{c}{a}-3 \leq \frac{b}{a} \leq 4 \frac{c}{a} - 1$$. In the second inequality. I also want to construct the $\frac{b}{a}$. $$c\ln b -c \ln a \geq a+c\ln c -c \ln a \Rightarrow \frac{b}{a} \geq \frac{c}{a} \exp{\frac{a}{c}}$$
But I can't go any further. It is depressed me a lot .. What is the right way to solve this ? I don't know..
| Here what I did:
We have the following inequality:
$$5c-3a\leq b\leq 4c-a.\quad\quad (E_0)$$
Since $b>0$, then I can divide by $b$:
$$\dfrac{5c-3a}{b}\leq 1\leq \dfrac{4c-a}{b}.\quad (1)$$
Further $(1)$ can be seen as:
$$\dfrac{5c-3a}{b}\leq 1\leq \dfrac{5c-3a+2a-c}{b},$$
or
$$\dfrac{5c-3a}{b}\leq 1\leq \dfrac{5c-3a}{b}+\dfrac{2a-c}{b},$$
or
$$0\leq 1-\dfrac{5c-3a}{b}\leq \dfrac{2a-c}{b}.\quad (2)$$
With $(2)$ we can conclude that: $\dfrac{2a-c}{b}\geq0$ and since $b>0$ then $2a-c\geq0$.
Finally we have: $$\dfrac{c}{a}\leq2.\quad\quad (E_1)$$
Now, back to $(E_0)$ and divide it by $a$, we get:
$$5\dfrac{c}{a}-3\leq \dfrac{b}{a}\leq 4\dfrac{c}{a}-1.\quad (3)$$
Using $(E_1)$, we have an upper bound:
$$\dfrac{b}{a}\leq 4\dfrac{c}{a}-1\leq7.$$
Now, it time to use the logarithm inequality to get the lower bound.
We have:
$$c\log b\geq a+c\log c.$$
First divide it by $c$, we get:
$$\log b\geq \dfrac{a}{c}+\log c.$$
Again we use $(E_1)$ to see that: $\log \dfrac{b}{c}\geq \dfrac{a}{c} \geq\dfrac{1}{2}>0.$ (I needed this to guarantee that $\log \dfrac{b}{c}>0$). Then, this is useful to divide by $\log \dfrac{b}{c}$ in both sides. Then,
$$\dfrac{c}{a}\geq\dfrac{1}{\log \dfrac{b}{c}}.$$
Back to $(3)$ and use the left hand side, we get:
$$-3+5\dfrac{c}{a}\geq -3+\dfrac{5}{\log \dfrac{b}{c}}.\quad (4)$$
Now divide $(E_0)$ by $c$ and apply log (we about the right hand side) you get:
$$\log \dfrac{b}{c}\leq \log(4-\dfrac{a}{c}).$$
Which is equivalent to (using $(E_1)$)
$$\log \dfrac{b}{c}\leq \log(4-\dfrac{a}{c})\leq \log\dfrac{7}{2}.$$
Now use $(4)$ to get:
$$-3+5\dfrac{c}{a}\geq -3+\dfrac{5}{\log \dfrac{b}{c}}\geq -3+\dfrac{5}{\log \dfrac{7}{2}}.\quad (E_2)$$
Conclusion:
$$-3+\dfrac{5}{\log \dfrac{7}{2}}\leq \dfrac{b}{a}\leq 7.$$
$$0.99\leq \dfrac{b}{a}\leq 7.$$
Edit: $\log$ is the natural logarithm. I calculated wrong before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/742067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Limit of $\sum_{k=0}^{\lfloor n/2 \rfloor} 2^{-2nk} \binom{n}{2k}\left(\binom{2k}{k}^n\right)$ I can see numerically that $$\lim_{n \to \infty} \sum_{k=0}^{\lfloor n/2 \rfloor} 2^{-2nk} \binom{n}{2k}\left(\binom{2k}{k}^n\right) = 1$$ but how can you prove this? Using Stirling's approximation doesn't seem to be enough.
| $$
\sum_{k=1}^{\lfloor n/2\rfloor}2^{-2nk}\binom{n}{2k}\binom{2k}{k}^{\large n}
=\sum_{k=1}^{\lfloor n/2\rfloor}\binom{n}{2k}\left[4^{-k}\binom{2k}{k}\right]^{\large n}\tag{1}
$$
The $k=0$ term is $1$ so we need to determine how the rest of the terms go to $0$. Let's look at the ratio between consecutive terms in $(1)$
$$
\begin{align}
&\frac{(n-2k)(n-2k-1)}{(2k+1)(2k+2)}\left[\frac14\frac{(2k+2)(2k+1)}{(k+1)(k+1)}\right]^{\large n}\\
&=\left(\frac{n+1}{2k+1}-1\right)\left(\frac{n+1}{2k+2}-1\right)\left[1-\frac1{2k+2}\right]^{\large n}\\[6pt]
&=\left(\frac{2k+2}{2k+1}\frac{n+1}{2k+2}-1\right)\left(\frac{n+1}{2k+2}-1\right)\frac{2k+2}{2k+1}\left[1-\frac1{2k+2}\right]^{\large n+1}\\[6pt]
&\le\left(\frac43\frac{n+1}{2k+2}-1\right)\left(\frac{n+1}{2k+2}-1\right)\frac43\left[1-\frac1{2k+2}\right]^{\large n+1}\\[6pt]
&\le\frac43\left(\frac43x-1\right)(x-1)e^{-x}\qquad\text{where }x=\frac{n+1}{2k+2}\\[6pt]
&\lt\frac25\qquad\qquad\text{for all }x\ge1\tag{2}
\end{align}
$$
Thus, the error when truncating the sum is no more than $\frac53$ times the first truncated term.
Therefore,
$$
\sum_{k=0}^{\lfloor n/2\rfloor}2^{-2nk}\binom{n}{2k}\binom{2k}{k}^{\large n}
\sim1+\binom{n}{2}(1/2)^n+O\left(n^4(3/8)^n\right)\tag{3}
$$
since $\binom{n}{4}\sim\frac{n^4}{24}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/742510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Taylor series and Maclaurin Series expansion
Hi! I am currently working on some calc2 online homework problems on Taylor series and Maclaurin series. I have tried a few different answers to this question, but I am really not sure how to go about solving this problem. If anyone can help me I would greatly appreciate it!
| Observe that
$$\frac{1}{1 - 3x} = \frac{1}{-2 - 3(x - 1)} = -\frac{1}{2 + 3(x - 1)} = -\frac{1/2}{1 + (3/2)(x - 1)}.$$
Therefore
$$\frac{1}{1 - 3x} = -\frac{1}{2}\sum_{n = 0}^\infty (-1)^n \left(\frac{3(x - 1)}{2}\right)^n = -\frac{1}{2}\sum_{n = 0}^\infty (-1)^n \left(\frac{3}{2}\right)^n (x - 1)^n, \quad \left|\frac{3(x - 1)}{2}\right| < 1.$$
Because
$$\left|\frac{3(x - 1)}{2}\right| < 1 \Longleftrightarrow \frac{1}{3} < x < \frac{5}{3},$$
and the series above diverges when $x = 1/3$ or $x = 5/3$, we conclude that the interval of convergence is $(1/3, 5/3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/742773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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question on surds i already asked this question but the answer I got did not match the one in the book $$\sqrt{ 3x }= x + \sqrt {3}$$
Give x in the form
$$A \sqrt {B} + C $$
Can you show me how this is done step by step.
The answer I have in the book is:
$$\frac {1}{2} \sqrt{3} + \frac {3}{2} $$
this is where I got stuck:
$$ \frac {x^2 +2x \sqrt{3} +3}{3x} $$
| I figured out the problem: The solution to $\color{blue}{\sqrt 3}(x) = x + \sqrt 3$ is indeed $$\frac {1}{2} \sqrt{3} + \frac {3}{2}$$
But that's not the problem you posted. In the above, only $3$ is under the radical sign. In your post, you have $\sqrt{3x}$
In the event that the problem should read: $$\sqrt 3(x) = x + \sqrt 3$$
then $$\begin{align} \sqrt 3(x) = x + \sqrt 3 & \iff (\sqrt 3 - 1)x = \sqrt 3 \\ \\ &\iff x = \dfrac {\sqrt 3}{\sqrt 3 - 1} \\ \\ &\iff x = \frac{\sqrt 3}{\sqrt 3 - 1} \cdot \frac{\sqrt 3 + 1}{\sqrt 3 + 1} = \dfrac{3 +\sqrt 3}{3 - 1} = \dfrac 32 + \dfrac {\sqrt 3}2\end{align}$$
And in the desired form, that gives you $$\frac {1}{2} \sqrt{3} + \frac {3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/743508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Several (advanced) L'Hospital problems Problems :
$$
\begin{align}
&\text{A}.\ \lim_{x\rightarrow1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}\\
&\text{B}.\ \lim_{x\rightarrow 0}\left(\cot x-\frac{1}{x}\right)\\
&\text{C}.\ \lim_{x\rightarrow0}\frac{\sin^{-1}x-x}{\tan^{-1}x-x}\\
&\text{D}.\ \lim_{x\rightarrow 1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}\\
&\text{E}.\ \lim_{x\rightarrow0}\left(a^x+b^x\right)^\frac{1}{x}
\end{align}
$$
Here are my solutions concerning A~E.
A. $$\ln y=\tan\left(\frac{\pi x}{2}\right)\ln(x-2)=\frac{\tan\left(\frac{\pi x}{2}\right)}{\frac{1}{\ln(2-x)}}$$
$$\lim_{x\rightarrow 1}\frac{\tan\left(\frac{\pi x}{2}\right)}{\frac{1}{\ln(2-x)}}=_H\lim_{x\rightarrow1}\frac{\frac{\pi}{2}\sec^2(\frac{\pi x}{2})}{x-2}=\infty$$so, $$A\rightarrow \infty$$
Is this conclusion right?
C.
$$=_H\lim_{x\rightarrow0}\frac{\frac{1}{\sqrt{1-x^2}}-1}{\frac{1}{1+x^2}-1}=_H\lim_{x\rightarrow0}\frac{\frac{-1}{2}\frac{1}{\sqrt{(1-x^2)^3}}(-2x)}{-\frac{2x}{(1+x^2)^2}}=\frac{1}{-2}=-\frac{1}{2}$$
| For example A,D and E
*
*put $y=\lim_{x\to a} f(x)$
*Apply ln as $\ln y=\ln\lim_{x\to a} f(x)=\lim_{x\to a} \ln f(x)$
*use L'Hopital rule
*$\ln y=l\implies y=e^l$
For A
$$y=\lim_{x\rightarrow1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}
\\ \ln y=\lim_{x\rightarrow1}\ln (2-x)^{\tan\left(\frac{\pi x}{2}\right)}
\\ \ln y=\lim_{x\rightarrow1}{\tan\left(\frac{\pi x}{2}\right)}\ln (2-x)
\\ \ln y=\lim_{x\rightarrow1}{}\frac{\ln (2-x)}{\cot\left(\frac{\pi x}{2}\right)}=\frac{0}{0}$$ Now apply L'Hopital rule
For E
$$y=\lim_{x\rightarrow0}\left(a^x+b^x\right)^\frac{1}{x}
\\ \ln y=\lim_{x\rightarrow0}\ln\left(a^x+b^x\right)^\frac{1}{x}
\\ \ln y=\lim_{x\rightarrow0}\frac{\ln\left(a^x+b^x\right)}{x}=\infty\implies y=\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/745427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Expressing a Polynomial as a sum of cube roots of integers How do you prove $x^3-3x^2-6x-4$ has a zero of the form $\sqrt[3]a+\sqrt[3]b+\sqrt[3]c$, for distinct positive integers a,b,c
| Use a classic method for solving a cubic (not having to remember the formula).
First eliminate the squared term by setting $y=x-1$ to obtain $$y^3-9y-12=0$$
Now we look for solutions of the form $y=v+w$, where $y^3=(v+w)^3=(v^3+w^3)+3vw(v+w)$ which we can rewrite as $$y^3-3vw\cdot y-(v^3+w^3)=0$$
Comparing coefficients we obtain: $3vw=9$ from which we derive $$v^3w^3=27$$ and we also have $$v^3+w^3=12$$ which means that $v^3, w^3$ are the roots of the quadratic $$z^2-12z+27=0=(z-3)(z-9)$$[or use the quadratic formula]
So $v^3=3, w^3=9$ (or the other way about). $y=v+w$, and $x=v+w+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/745609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the easiest way to integrate $y=\frac {x+4}{\sqrt{-x^2-2x+3}}$? What is the easiest way to integrate $y=\frac{x+4}{\sqrt{-x^2-2x+3}}$ ? I tried to integrate it by making numerator in form: $-2x-2$ and then pulling it under differential, but the result drastically differed from Mathematica output.
| Since $-x^2-2x+3=-(x+3)(x-1)$, both roots $x_1=-3,x_2=1$ are real. As such it is known that e.g. the Euler substitution
\begin{align}
\sqrt{-x^{2}-2x+3}&=t(x-x_2)=t(x-1)\\[2ex] &\Leftrightarrow x=\frac{t^{2}-3}{t^{2}+1}=1-\frac{4
}{1+t^{2}},\quad dx=\frac{8t}{\left( 1+t^{2}\right) ^{2}}dt, \\[2ex]
& \Leftrightarrow\sqrt{-x^{2}-2x+3}=-\frac{4t}{1+t^{2}}
\end{align}
will turn the original integral into an integral of a rational function in $t$ which can be integrated by the method of partial fractions:
\begin{eqnarray*}
I &=&\int \frac{x+4}{\sqrt{-x^{2}-2x+3}}dx = \int\frac {1-{\dfrac {4}{1+t^2}+4}}{-\dfrac {4t}{1+t^2}}\frac {8t}{(1+t^2)^2}\, dt\\[2ex]
&=&\int -2\frac{1+5t^{2}}{\left( 1+t^{2}\right) ^{2}}dt=\int \frac{8}{\left( 1+t^{2}\right) ^{2}}- \frac{10}{1+t^{2}}dt\\[2ex]
&=&\int \frac{8}{\left( 1+t^{2}\right) ^{2}}dt-\int \frac{10}{1+t^{2}}dt,\quad \text{(see note below) }\\[2ex]
&=&\left(\frac{4t}{1+t^{2}}+4\arctan t\right)-10\arctan t+C \\[2ex]
&=&\frac{4t}{1+t^{2}}-6\arctan t +C\\[2ex]
&=&-\sqrt{-x^{2}-2x+3}-6\arctan \frac{\sqrt{-x^{2}-2x+3}}{x-1}+C.
\end{eqnarray*}
Note.
*
*By the method explained here we can reduce the integration of the function $f(t)=\frac{1}{\left( 1+t^{2}\right) ^{2}}$ to the integration of $\frac{1}{1+t^{2}}$. We start by adding and subtracting $t^2$ in the numerator. The first integral is a standard integral and the second one is integrable by parts:
$$\begin{eqnarray*}
\int \frac{1}{\left( 1+t^{2}\right) ^{2}}dt &=&\int \frac{1}{1+t^{2}}dx-\int \frac{t^{2}}{\left( 1+t^{2}\right) ^{2}}dt
\\
&=&\arctan t-\int t\frac{t}{\left( 1+t^{2}\right) ^{2}}dt,
\end{eqnarray*}$$
and
$$\begin{eqnarray*}
\int t\frac{t}{\left( 1+t^{2}\right) ^{2}}dt &=&t\left( -\frac{1}{2\left(
1+t^{2}\right) }\right) +\int \frac{1}{2\left( 1+t^{2}\right) }dt \\
&=&-\frac{t}{2\left( 1+t^{2}\right) }+\frac{1}{2}\arctan t.
\end{eqnarray*}$$
We thus get
$$\int \frac{1}{\left( 1+t^{2}\right) ^{2}}dt =\frac{t}{2\left( 1+t^{2}\right) }+\frac{1}{2}\arctan t,$$
$$\int \frac{8}{\left( 1+t^{2}\right) ^{2}}dt =\frac{4t}{1+t^{2}}+4\arctan t.$$
Comments.
*
*The Euler substitutions are more general, but when the integrand is a rational function of $x$ and $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, $\sqrt{x^2-a^2}$, with $a>0$, trigonometric and hyperbolic substitutions based on the trigonometric identities $$1-\sin^2 x=\cos^2 x,\quad 1+\tan^2 x=\sec^2x, \quad \sec^2x-1=\tan^2 x$$ and on the hyperbolic identities $$1-\tanh^2 x=\operatorname{sech}^2 x,\quad 1+\sinh^2 x=\cosh^2 x, \quad\cosh^2 x-1=\sinh^2 x$$ are in general faster. For example for $\sqrt{a^2-x^2}$ you can use $x=a\sin t$ or $x=a\tanh t$.
*Note that if you use the trigonometric substitution $x=2\sin t-1$ suggested by
lab bhattacharjee, instead of the one I indicate above, the two constants of integrations are not equal.
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Why does $2^{n+1} + 2^{n+1} = 2^{n+2}$? Simple question, why does:
$2^{n+1} + 2^{n+1} = 2^{n+2}$ ?
Furthermore, why does this only work for powers of 2?
Thanks.
| Note that $2^{n+1}=2\cdot 2^n=2^n+2^n$
If you try it with $3$ you get $3^{n+1}=3\cdot 3^n=3^n+3^n+3^n$
| {
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Solutions for quartic Suppose I have an equation in the form $(x-a)^4 + (x-b)^4 = c$. What is a clever way to find all four solutions? I have tried expanding and then used long division. However, I believe a better way is to reduce to cubic, then to quadratic. I don't know how to do this.
| The solutions of $$(x-a)^4 + (x-b)^4 = c$$ are less complex that we could think. They are, after some tedious simplifications given by $$x_1=\frac{1}{2} \left(a+b-\sqrt{-2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$ $$x_2=\frac{1}{2} \left(a+b+\sqrt{-2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$ $$x_3=\frac{1}{2} \left(a+b-\sqrt{2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$
$$x_4=\frac{1}{2} \left(a+b+\sqrt{2 \sqrt{2} \sqrt{(a-b)^4+c}-3 (a-b)^2}\right)$$
| {
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the geometric explain of $t = x-\frac{a}{3}$ in the simplify of cubic equation $x^3+ax^2+bx+c=0$ Assume $$f(x) = x^3+ax^2+bx+c$$
we have $$f''(x)=2a+6x$$. we get $x = -\frac{a}{3}$
Magically, If we take the transformation: $$t = x -\left(-\frac{a}{3}\right)$$. we can transform the above equation into $$f(x)=g(t) = t^3+\left(b-\frac{a^2}{3}\right)t+\frac{2a}{27}-\frac{ab}{3}+c$$
Why ? why this transform cancellation the quadratic term? Can anyone give an Geometric explain? Thanks very much.
| Let's look at the quadratic term of $f(t - \dfrac{a}{3})$. It came from $2$ sources. One is the $2$nd term of the cubic expansion $(t - \dfrac{a}{3})^3$: $-3t^2\cdot \dfrac{a}{3}$ which is: $-at^2$, and the other term comes from the $1$st term of $a(t - \dfrac{a}{3})^2$ which is: $at^2$. So the coefficient of $t^2$ is: $-a + a = 0$. So there is no quadratic term $t^2$.
| {
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If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$
This can be easily done by calculas but is there any way to do do this by algebra
| Let $a = \sin x, b = \cos x$. Then we need to find the minimum of the function $\cos x + \sin x + \sec(x) \csc(x)$ which is $2+ \sqrt{2}$ at $x = \frac{\pi}4$.
| {
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How find this limit $\lim_{n\to \infty} \left(\frac{(2n)!}{2^n\cdot n!}\right)^{\frac{1}{n}}\cdot \cdots$ Find this limit
$$\lim_{n\to\infty}\left(\dfrac{(2n)!}{2^n\cdot n!}\right)^{\frac{1}{n}}\left(\tan{\left(\dfrac{\pi\sqrt[n+1]{(n+1)!}}{4\sqrt[n]{n!}}\right)}-1\right)$$
I know we must use this
$$n!\approx\left(\dfrac{n}{e}\right)^{n}\sqrt{2n\pi}$$
so
$$\dfrac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}\approx\dfrac{n+1}{n}\dfrac{\sqrt[n+1]{2(n+1)\pi}}{\sqrt[n]{n!}}\to 1,n\to \infty$$
But I can't .Thank you
| $$L=\lim_{n\to\infty}\left(\dfrac{(2n)!}{2^n\cdot n!}\right)^{\frac{1}{n}}\left(\tan{\left(\dfrac{\pi\sqrt[n+1]{(n+1)!}}{4\sqrt[n]{n!}}\right)}-1\right)=\lim_{n\to\infty}\left(\dfrac{(2n)!}{2^n\cdot (n!)^2}\right)^{\frac{1}{n}}\cdot\lim_{n\to\infty}\sqrt[n]{n!}\left(\tan{\left(\dfrac{\pi\sqrt[n+1]{(n+1)!}}{4\sqrt[n]{n!}}\right)}-1\right)= \lim_{n\to\infty}a_n\cdot\lim_{n\to\infty}b_n$$
1) Apply root criterion (Chauchy-d'Alembert): $$ \lim_{n\to\infty} a_n = \lim_{n\to\infty}\frac{(2n+2)!}{2^{n+1}\cdot((n+1)!)^2}\cdot\frac{2^n\cdot(n!)^2}{(2n)!} =2$$
2) Note $\dfrac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}=c_n$$$\lim_{n\to\infty}b_n= \lim_{n\to\infty}\sqrt[n]{n!}\cdot(\tan(\frac{\pi}{4}c_n)-1)= \lim_{n\to\infty}\sqrt[n]{n!}\cdot\frac{\sqrt{2}\sin\frac{\pi}{4}(c_n-1)}{\cos\frac{\pi}{4}c_n}=\lim_{n\to\infty}\sqrt[n]{n!}\cdot\frac{2\sin\frac{\pi}{4}(c_n-1)}{\frac{\pi}{4}(c_n-1)}\cdot\frac{\pi}{4}(c_n-1)=\frac{\pi}{2}\lim_{n\to\infty}\sqrt[n]{n!}\cdot(c_n-1)= \frac{\pi}{2}\cdot\lim_{n\to\infty}(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}) =\frac{\pi}{2}\cdot\frac{1}{e} = \frac{\pi}{2e}.$$
Applied $$\lim_{n\to\infty}(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}) = \frac{1}{e}$$ the Lalescu's Sequence.(See http://www.artofproblemsolving.com/blog/44744 )
Conclusion $$L= \frac{\pi}{e}.$$
| {
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Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ = 5$$
Find
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
| Very important fact of physics and harmonics and digital signal processing:
The sum of two sinusoids of equal frequency is another sinusoid of that same frequency, regardless of amplitude or phase.
Or stated mathematically:
If
$$f(x) = A_1\cos\left(\omega\,x + p_1\right) + A_2\cos\left(\omega\,x + p_2\right)$$
then
$$f(x) = A_3\cos\left(\omega\,x + p_3\right)$$
and
$$A_1 \angle p_1 + A_2 \angle p_2 = A_3 \angle p_3$$
where $M \angle \theta$ is polar notation. The same holds is $\cos$ is replaced with $\sin$.
So your problem is:
$$4\sin(\theta) + 3\cos(\theta) = 5$$
$$4\sin(\theta) + 3\sin(\theta + \pi/2) = 5$$
$$M\sin(\theta + p) = 5$$
$$\theta = \arcsin(5/M) - p$$
And you find $M$ and $p$ by adding:
$$M\angle p = 4 \angle 0 + 3 \angle \pi/2$$
$$\begin{cases} M = 5 \\ p = \arctan(3/4)\end{cases}$$
So
$$\theta = \arcsin(1) - \arctan(3/4)$$
$$\theta = \pi/2 + Z\cdot 2\pi - \arctan(3/4)$$
So now
$$y = 4\cos(\theta) - 3\sin(\theta)$$
$$y = 4\cos(\theta) + 3\cos(\theta + \pi/2)$$
$$y = 5\cos(\theta + \arctan(3/4))$$
So
$$y = 5\cos(\pi/2 + Z\cdot 2\pi - \arctan(3/4) + \arctan(3/4))$$
$$y = 5\cos(\pi/2 + Z\cdot 2\pi)$$
$$y = 0$$
| {
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How do I show that $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$ According to wolfram alpha this is true: $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$
But how do you show this? I know of no rules that works with addition inside square roots.
I noticed I could do this:
$\sqrt{24} = 2\sqrt{3}\sqrt{2}$
But I still don't see how I should show this since $\sqrt{5+2\sqrt{3}\sqrt{2}} = \sqrt{3}+\sqrt{2}$ still contains that addition
| On can easily discover the denesting using my simple radical denesting algorithm.
$\ w = 5+\sqrt{24}\,$ has norm $\,n = ww' = 5^2-24 = 1.\,$ Subtracting out $\,\sqrt{n}=1\,$ yields $\,4+\sqrt{24}.$
This has trace $\,t = 8,\,$ so dividing $\,\sqrt{t} = 2\sqrt{2}\,$ out of $\,4+\sqrt{24}=4+2\sqrt{6}\,$ yields
$$ \frac{4+2\sqrt{6}}{2\sqrt{2}}\,=\, \frac{2+\sqrt{6}}{\sqrt{2}\ } \,=\, \sqrt{2}+\sqrt{3}$$
| {
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Prove that $\int_0^{\infty}\int_0^{\infty}e^{-(x^2 + 2xy\cos(\alpha)+y^2)}\mathrm dx\mathrm dy=\frac{\alpha}{2\sin(\alpha)}$ I'm having difficulty with a question. It says
By putting $x=r\cos(\theta), y=r\sin(\theta)$, prove that $$\int_0^{\infty}\int_0^{\infty}e^{-(x^2 + 2xy\cos(\alpha)+y^2)}\mathrm dx\mathrm dy=\frac{\alpha}{2\sin(\alpha)}$$
Making the substitution we get \begin{align*}&\int_0^{\pi/2}\int_0^{\infty}re^{-r^2 (\sin(2\theta) \cos(\alpha) + 1)}dr\ d\theta \\ \\
=&\int_0^{\pi/2} \frac{1}{2(\sin(2\theta) \cos(\alpha) + 1)} d\theta\\ \\
=&\frac{\alpha}{2\sin(\alpha)} \int_0^{\pi/2} \frac{\sin(\alpha)}{2 \alpha \cos(\alpha) \sin(\theta) \cos(\theta) + \alpha} d\theta\\ \\
=&\frac{\alpha}{2\sin(\alpha)} \int_0^{\pi/2}\frac{\tan(\alpha) \sec^2(\theta)}{2 \alpha \tan(\theta) + \alpha \sec(\alpha) \sec^2(\theta)} d\theta
\end{align*}
setting $u=\tan(\theta),\ du=\sec^2(\theta)\ d\theta$
$$=\frac{\alpha}{2\sin(\alpha)}\int_0^{\infty}\frac{\tan(\alpha)\ du}{\alpha \sec(\alpha) + 2\alpha u+ \alpha \sec(\alpha) u^2}$$
Now \begin{align*}u&=\frac{\cos(\alpha)\{-2\alpha\pm\sqrt{4\alpha ^2 - 4\alpha ^2\sec^2(\alpha)}\}}{2\alpha}\\ \\
&=-\cos(\alpha) \mp i\sin(\alpha)\\
&=-e^{\pm i\alpha}
\end{align*}
So substituting in
\begin{align*}
=&\frac{\alpha}{2\sin(\alpha)}\int_0^{\infty}\frac{\tan(\alpha)\ du}{(u-e^{i\alpha})(u-e^{- i\alpha})}\\ \\
=&\frac{\alpha}{2\sin(\alpha)}\left[ \frac{\tan(\alpha)\{\log(1-ue^{i\alpha}) - \log(-u+e^{i\alpha})\}}{-1+e^{2i\alpha}} \right]^{\infty}_{u\, =\, 0}
\end{align*}
which I then can't evaluate (and certainly isn't $1$). Any help?
| \begin{align}
\frac{1}{1+\sin{(2\, \theta)}\, \cos{\alpha}} &=\frac{1}{1+2\sin{\theta}\cos{\theta}\, \cos{\alpha}}\\
&= \frac{\sec{(\theta)}^2}{\sec{(\theta)}^2+2\tan{\theta}\, \cos{\alpha}}\\
&= \frac{\sec{(\theta)}^2}{1+\tan{(\theta)}^2+2\tan{\theta}\, \cos{\alpha}}\\
&= \frac{\sec{(\theta)}^2}{\sin{(\alpha)}^2+\cos{(\alpha)}^2+\tan{(\theta)}^2+2\tan{\theta}\, \cos{\alpha}}\\
&= \frac{\sec{(\theta)}^2}{\sin{(\alpha)}^2+\left(\cos{(\alpha)}+\tan{(\theta)}\right)^2}\\
\end{align}
\begin{align}
\therefore \int \frac{1}{1+\sin{(2\, \theta)}\, \cos{\alpha}} d\theta &= \int \frac{\sec{(\theta)}^2}{\sin{(\alpha)}^2+\left(\cos{(\alpha)}+\tan{(\theta)}\right)^2}\, d\theta\\
&= \frac{1}{\sin{\alpha}}\, \arctan{\frac{\tan{\theta}+\cos{\alpha}}{\sin{\alpha}}}+C
\end{align}
Thus, the required integral is
\begin{align}
\frac{1}{2} \int_0^{\pi/2} \frac{1}{1+\sin{(2\, \theta)}\, \cos{\alpha}} d\theta &= \frac{\alpha}{2\, \sin{\alpha}}
\end{align}
| {
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How have they done the algebra here? Proof by induction
\begin{align}&4-\frac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^k\\
=&4-\frac{2(k+2)}{2^k}+\frac{k+1}{2^k}
\\
=&4-\frac{(k+1)+2}{2^{(k+1)-1}}
\end{align}
Original image
Can someone explain these steps to me please? Did the $2^{k-1}$ change to $2^k$ by multiplying numerator by 2?? Even so, if you add them when they have the common denominator, shouldn't you get $3k+5$??
| When you look at the second line notice that they have an extra $2$ in the numerator they multiplied by $1$, which can also be written as $1 = \frac{2}{2}$. So just for completeness:
\begin{align*} &4-\frac{k+2}{2^{k-1}}+(k+1)(\frac{1}{2})^k \\
& = 4-1\cdot\frac{k+2}{2^{k-1}}+(k+1)(\frac{1}{2})^k \\
& = 4-\frac{2}{2}\frac{k+2}{2^{k-1}}+(k+1)(\frac{1}{2})^k\\
&= 4-\frac{2(k+2)}{2^k}+\frac{k+1}{2^k} \end{align*}
For the numerator you get $-2(k+2)+(k+1) = -(k+3) = -((k+1)+2)$
So it becomes:
$$4-\frac{(k+1)+2}{2^{k+0}} = 4-\frac{(k+1)+2}{2^{(k+1)-1}}$$
| {
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Bombelli's wild thought of cubic equations In many books, like Visual Complex Analysis. talk about the real original of complex number. the author begin with this equation:
$$x^3=15x+4$$
Then the author use the formula $$x=\sqrt[3]{q+\sqrt{q^2-p^3}}+\sqrt[3]{q-\sqrt{q^2-p^3}}$$
to say that the equation has a root $$x = \sqrt[3]{2+11i}+\sqrt[3]{2-11i}$$
Apparently, $x=4$ is a root of the equation $x^3=15x+4$. Then the author guess $$\sqrt[3]{2+11i}+\sqrt[3]{2-11i} =4.$$ then introduction complex arithmetic. It seems very natural. but equation $x^3=15x+4$ has three different roots. why the author, or probably Bombelli guess this equal $4$, not other roots?
| He (Bombelli) probably used the Cardano formula to obtain the $\textbf{one}$ root solution you just described $$x = \sqrt[3]{2+11i}+\sqrt[3]{2-11i}$$
He then had to find the cube roots of each of the radicals above. I am not sure exactly how he found them but he may have used the following sometimes used polynomial to find each of the cube roots
$$\frac{-64a^9+(48x)a^6+((15(x)^2)-3(3y)^2)a^3+(x)^3}{-64} = 0$$ where $x$ = $2$ and $y$ = $11i$ to get
$$a^9 + \frac{3}{2}a^6 - \frac{3327}{64}a^3 + \frac{1}{8} = 0$$
This polynomial has one rational root $a$ = $2$
He could now take the rational root $a$ = $2$ he found above to find $b$ in the equation below in order to denest the two cube roots in Cardano's formula. The $2$ on the $RHS$ is the $2$ under the radical
$$ a^3+3ab^2=2$$
$$2^3 + 3(2)b^2 = 2$$
$$b=\pm \sqrt\frac{-6}{6} = \pm i$$
So using the $a$ he found and the $b$ he found he could derive the cube roots as $a + b$.
This gives the two results $$2 + i$$ and $$2 -i$$ Each is the cube root of the nested radicals in the Cardano solution.
So using the $a$ he found and the $b$ he found he could observe that this was equal to the real root of $4$ found using other methods
$$\left(2 + i\right) + \left(2 - i\right) = 4$$
| {
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How find the $AP+\frac{1}{2}BP$ minmum value An equilateral triangle $ABC$ such $$AB=BC=AC=2a>0$$
A circle $O$ is inscribed in triangle $ABC$,and the point $P$ on the circle $O$.
Find the minimum $$AP+\dfrac{1}{2}BP$$
My idea: let $$A(-a,0),B(a,0),O(0,\dfrac{\sqrt{3}}{3}a)$$
then the circle equation is $$
x^2+(y-\dfrac{\sqrt{3}a}{6})^2=\dfrac{1}{12}a^2$$
let $P(x,y)\;$ , then
$$|PA|+\dfrac{1}{2}|PB|=\sqrt{(x+a)^2+y^2}+\dfrac{1}{2}\sqrt{(x-a)^2+y^2}$$
where $$
x^2+(y-\dfrac{\sqrt{3}a}{6})^2=\dfrac{1}{12}a^2$$
then I can't.Thank you
| The solution of Omri Nissan Solan.
Let $K$ and $L$ be common points of the circle with $AB$ and $BC$ respectively
and let $BO\cap KL=\{M\}$.
Thus, our circle it's the Apollonian circle of $\Delta KLB$.
Id est, $AP+\frac{1}{2}PB=AP+PM\geq AM=\frac{a\sqrt7}{2}$.
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.