Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $\sqrt{k}+\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}$ I try to prove that $\sqrt{k}+\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}$ for all integers k greater or equal to 1 but don't know how to transform this equation full of square roots in order to finish the proof.
| First, find a common denominator.
$\sqrt{k}+\frac{1}{\sqrt{k}+\sqrt{k+1}}= \frac{\sqrt{k}(\sqrt{k} + \sqrt{k+1}) + 1}{\sqrt{k} + \sqrt{k+1}}$
Next, distribute.
$\frac{k + \sqrt{k}\sqrt{k+1} + 1}{\sqrt{k} + \sqrt{k+1}}$
Now, we rearrange the terms so we have
$\frac{k + 1 + \sqrt{k}\sqrt{k+1}}{\sqrt{k} + \sqrt{k+1}}$
Now factor out of the numerator a $\sqrt{k+1}$
$\frac{\sqrt{k+1}(\sqrt{k+1} + \sqrt{k})}{\sqrt{k} + \sqrt{k+1}}$
Cancel out common terms, and now we have our answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Give a Sequence, Find Its Generating Function The Full Question
Find the generating function(closed form) of the following sequence:
$\binom{8}{1},2\binom{8}{2},3\binom{8}{3},\dots , 8\binom{8}{8}$
My Work
The open form of this generating function is:
$1\binom{8}{1}x^0+2\binom{8}{2}x^2 +3\binom{8}{3}x^3 + \cdots +8\binom{8}{8}x^8$
This can be re-written as:
$0\binom{8}{0}x^0+1\binom{8}{1}x^0+2\binom{8}{2}x^2 +3\binom{8}{3}x^3 + \cdots +8\binom{8}{8}x^8$
Which in Sigma Notation is:
$\sum_{k=0}^{8}k\binom{8}{k}x^k$
Part of this series is obviously binomial, which has a closed form of $(1+x)^8$ I don't know how to deal with the $k$ that is always being multiplied in. I thought I could maybe use the multiplication rule for sums, but it doesn't seem to apply here because we have $a_1b_1$ instead of $a_1b_n$
My Problem
How do we deal with the k that is being multiplied in there constantly? How can I get a nice closed form generating function with that $k$ getting in my way?
| Hint: Say the generating function is $ f(x)$. What do you know about $ \int \frac{f(x) - 8}{x} \, dx $?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $n$ is an odd integer, does $x+y$ divide $x^n + y^n$? I believe that the answer is yes.
Here's my thinking:
*
*$x^n + y^n -(x+y)(x^{n-1} + y^{n-1}) = -x^{n-1}y -xy^{n-1}$
*$-x^{n-1}y -xy^{n-1} - (x+y)(-x^{n-2}y -xy^{n-2}) = x^{n-2}y^2 + x^2y^{n-2}$
*So, at each step, we get $(-1)^{c}(x^{n-c}y^{c} + x^{c}y^{n-c})$ with $c$ increasing by $1$ at each step and
*Since $n$ is odd, eventually, we get: $(-1)^c(x^{c}y^{c+1} + x^{c}y^{c+1}) = (x+y)(-1)^c(x^{c}y^{c})$
Am I correct? Am I making any mistakes in my thinking?
If this is correct, is the standard formula:
$$x^n + y^n = (x+y)(x^{n-1} -x^{n-2}y -xy^{n-2} + \dots + (-1)^{\frac{n-1}{2}}x^{\frac{n-1}{2}}y^{\frac{n-1}{2}} + y^{n-1})$$
Thanks,
-Larry
Edit: Fixed the equation based on comment.
| Hint $\ {\rm mod}\ x\!+\!y\!:\ \color{#c00}{x\equiv -y} \,\Rightarrow\, \color{#c00}x^n+y^n\equiv (\color{#c00}{-y})^n+y^n\equiv 0\,$ by $\,n\,$ odd, and $ $ Congruence Rules.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $n$ unknowns with $n$ independent equations Is it always possible to solve $n$ independent equations with $n$ unknowns?
Or is it possible to solve the following 3 equations with 3 unknowns?
$$x + y + z = a$$
$$x^2 + y^2 + z^2 = b$$
$$x^3 + y^3 + z^3 = c$$
| This problem can be solved via finding roots of polynomial of $n$-th degree.
$2$ links are helpful here:
*
*Newton's Identities;
*Vieta's formulas.
Denote (according to $1$st link)
$$
p_1 = x+y+z=a;\\
p_2 = x^2+y^2+z^2=b;\\
p_3 = x^3+y^3+z^3-c;
$$
and
$$
e_1 = x+y+z;\\
e_2 = xy+xz+yz;\\
e_3 = xyz.
$$
Then
$$
e_1=p_1=a;\\
2e_2 = e_1p_1-p_2=a^2-b;\\
3e_3 = e_2p_1-e_1p_2+p_3 = \dfrac{a^2-b}{2}a-ab+c;
$$
so
$$
e_1 = a;\\
e_2 = \dfrac{a^2-b}{2};\\
e_3 = \dfrac{a^3-3ab+2c}{6}.
$$
Then (according to $2$nd link) $x,y,z$ are three roots of polynomial
$$
p(x) = x^3-e_1x^2+e_2x-e_3.
$$
Example ($a=18, b=134, c=1152$):
$$
x+y+z=18;\\
x^2+y^2+z^2=134;\\
x^3+y^3+z^3=1152;
$$
then
$$
e_1=18, \quad e_2=95, \quad e_3=150;
$$
so $x,y,z$ are the roots of polynomial
$$
p(x) = x^3-18x^2+95x-150.
$$
$(x=3,y=5,z=10)$ and their permutations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Cauchy-Euler Equidimensional DE $x^3y'''\left(x\right)+3x^2y''\left(x\right)+5xy'\left(x\right)-5y\left(x\right)=0$ I'm studying Cauchy-Euler equations from Fundamentals of Differential Equations by Nagle et. al. and I've come across the following problem (pg. 454, #9).
\begin{align}
x^3y'''\left(x\right)+3x^2y''\left(x\right)+5xy'\left(x\right)-5y\left(x\right)=0,\:\:r>0,\tag{1}
\end{align}
and this is what I have done so far. I began with the substitution of $y=x^r$ to get
\begin{align}
x^3r\left(r-1\right)\left(r-2\right)x^{r-3}+3x^2r\left(r-1\right)x^{r-2}+5xrx^{r-1}-5x^r=0\tag{2}
\end{align}
which simplifies to
\begin{align}
r^3+4r-5=0.\tag{3}
\end{align}
(3) has a factor of $\left(r-1\right)$, so taking that out we have
\begin{align}
\left(r-1\right)\left(r^2+r+5\right)=0,\tag{4}
\end{align}
and this gives us one root of $r=\left\{1\right\}$, but then to get the others I had to resort to the well-known quadratic equation, resulting in
\begin{align}
r&=\frac{-1\pm\sqrt{1-20}}{2}\\
&=\frac{-1}{2}\pm \frac{i\sqrt{19}}{2},\tag{5}
\end{align}
and therefore I have roots of
\begin{align}
r=\left\{1,\:\frac{-1}{2}+\frac{i\sqrt{19}}{2},\:\frac{-1}{2}-\frac{i\sqrt{19}}{2}\right\}.\tag{6}
\end{align}
Using this I'm going to have to use the fact that
\begin{align}
y&=x^{\alpha+i\beta}=e^{\left(\alpha+i\beta\right)\log\left(x\right)}\tag{7}\\
&=e^{\alpha\log\left(x\right)}\cos\left(\beta\log\left(x\right)\right)+ie^{\alpha\log\left(x\right)}\sin\left(\beta\log\left(x\right)\right)\tag{8}\\
&=x^\alpha\cos\left(\beta\log\left(x\right)\right)+ix^\alpha\sin\left(\beta\log\left(x\right)\right).\tag{9}
\end{align}
This therefore (should) give us a solution of
\begin{align}
y=C_1x+C_2x^{-1/2}\cos\left(\frac{\sqrt{19}}{2}\log\left(x\right)\right)+C_3x^{-1/2}\sin\left(\frac{\sqrt{19}}{2}\log\left(x\right)\right),\:\:r>0,\tag{10}
\end{align}
if I have done everything correctly.
Have I made a mistake anywhere that you can see?
| Seems absolutely correct, except for the typo pointed out by Amzoti.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
proving trigonometric identity involving 2 arguments $\cos^2\theta=\dfrac{m^2-1}{3}$, $\tan^3\dfrac{\theta}{2}=\tan\alpha$.
How to prove $$\cos^\frac{2}{3}\alpha+\sin^\frac{2}{3}\alpha=\frac{2}{m}^{2/3}. $$ I got the following
$$\tan^\frac{2}{3}\alpha=\frac{1- \sqrt\frac{m^2 - 1}{3}}{1+\sqrt\frac{m^2 - 1}{3}} $$
but i can not proceed further.
| This is one of those problems where a step that seems like it could never work somehow does. Start with your equation and add one to both sides.
$$ \tan^{\frac23}\alpha +1 = \frac{1- \sqrt{\frac{m^2-1}{3}}}{1+ \sqrt{\frac{m^2-1}{3}}}+1$$
Multiply both sides by $\cos^{\frac23}\alpha$ to get
$$ \sin^{\frac23}\alpha + \cos^{\frac23}\alpha= \frac{2*\cos^{\frac23}\alpha}{1+\sqrt{\frac{m^2-1}{3}}}.$$
This is closer, but that cosine term is still there. Unfortunate, but lets blindly chug along and see what happens. Going back to the equation you gave, we can solve for $\cos^2\alpha$. Cube both sides and substitute for $\sin^2\alpha$ to get
$$ \frac{1-\cos^2\alpha}{\cos^2\alpha} = \left(\frac{1- \sqrt{\frac{m^2-1}{3}}}{1+ \sqrt{\frac{m^2-1}{3}}} \right)^3.$$
Now split the left side into $\frac{1}{cos^2\alpha} - 1$ and add one to both sides of the equation. Next, invert and manipulate the expression slightly to get the following expression for $\cos^2\alpha$:
$$ cos^2\alpha = \frac{\left(1+ \sqrt{\frac{m^2-1}{3}}\right)^3}{\left(1+ \sqrt{\frac{m^2-1}{3}}\right)^3 + \left(1- \sqrt{\frac{m^2-1}{3}}\right)^3}$$
Now the special thing that makes this problem work is the denominator of this expression. The two terms are identical except for the minus sign. This means that the 2nd and 4th terms of the binomial expansion cancel out and we are left with
$$ \left(1+ \sqrt{\frac{m^2-1}{3}}\right)^3 + \left(1- \sqrt{\frac{m^2-1}{3}}\right)^3 = 1+1 + 3\left(\frac{m^2-1}3\right)+3\left(\frac{m^2-1}3\right) = 2m^2.$$
This is the kind of thing that only happens in homework problems. With the denominator simplified, we can take the cube root of $cos^2\alpha$ and get
$$ \cos^{\frac23}\alpha = \frac{1+ \sqrt{\frac{m^2-1}{3}}}{2^{\frac13}m^{\frac23}}.$$
Plus this back into the expression for $cos^{\frac23}\alpha + sin^{\frac23}\alpha$ and simplify to get the desired relation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Number theory problem! Prove the following using the method that relies on “Universal Generalization”. If $n$ is the product of four consecutive integers then $n+1$ is a perfect square. Domain is all natural numbers
What I got so far:
Let $a$ be an element of natural numbers selected arbitrarily
$a(a+1)(a+2)(a+3) = n$
$a(a+3)(a+1)(a+2)= n$
$(a^2+3a)(a^2+3a+2)=n$
I am stuck after this and don't know where to go. Help appreciated!
| $a(a+1)(a+2)(a+3)+1 = (a^2+3a+2)(a^2+3a)+1 = (a^2+3a)^2+ 2(a^2+3a) + 1 = (a^2+3a+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integral of sine multiplied by Bessel function with complicated argument I need a help with integral below,
$$ \int_0^\infty \sin(ax)\ J_0\left(b\sqrt{1+x^2}\right)\ \mathrm{d}x, $$
where $a,b > 0 $ and real, $J_0(x)$ is the zeroth-order of Bessel function of the first kind.
I found some integrals similar to the integral above, but I don't have any idea on how to apply it. Here are some integrals that might help.
$$ \int_0^\infty \cos(ax)\ J_0\left(b\sqrt{1+x^2}\right)\ \mathrm{d}x = \frac{\cos\sqrt{b^2-a^2}}{\sqrt{b^2-a^2}}; \mathrm{~~for~0 < a < b} $$
$$ \int_0^\infty \sin(ax)\ J_0(bx)\ \mathrm{d}x = \frac{1}{\sqrt{a^2-b^2}}; \mathrm{~~for~0 < b < a} $$
The proof of the first integral can be seen here.
| By exploiting the integral representation for the $J_0$ function:
$$ J_0(z) = \frac{1}{2\pi}\int_{0}^{2\pi}e^{iz\cos t}\,dt \tag{1}$$
we have:
$$ \int_{0}^{+\infty}\sin(ax)\, J_0(b\sqrt{1+x^2})\,dx =\frac{1}{2\pi}\int_{0}^{+\infty}\int_{0}^{2\pi}\sin(ax)e^{ib\sqrt{x^2+1}\cos t}\,dt\,dx \tag{2}$$
where:
$$ \sqrt{x^2+1}\cos t = \cos(t-t_0)-x\sin(t-t_0),\qquad x=\tan t_0 \tag{3}$$
so:
$$\begin{eqnarray*} I &=& \frac{1}{2\pi}\int_{0}^{+\infty}\int_{0}^{2\pi}\sin(ax)\,e^{ib\cos t-ibx\sin t}\,dt\,dx\\&=&\frac{1}{2\pi}\int_{0}^{2\pi}\frac{a}{a^2-b^2\sin^2 t}e^{ib\cos t}\,dt\\&=&\frac{1}{2\pi}\int_{0}^{2\pi}\frac{a}{a^2-b^2\sin^2 t}\,\cos(b\cos t)\,dt\tag{4} \end{eqnarray*}$$
but since:
$$ \frac{1}{2\pi}\int_{0}^{2\pi}\sin^{2n}(t)\cos(b\cos t)\,dt = \frac{ (2n-1)!!}{b^n}\, J_n(b) \tag{5}$$
we have:
$$\begin{eqnarray*} I &=& \frac{1}{a}\sum_{n\geq 0}\left(\frac{b}{a}\right)^{2n} \frac{1}{2\pi}\int_{0}^{2\pi}\sin^{2n}(t)\cos(b\cos t)\,dt \\&=&\frac{1}{a}\sum_{n\geq 0}\frac{(2n)!}{n!}\left(\frac{b}{2a^2}\right)^n J_n(b)\\&=&\frac{1}{a}\sum_{m=0}^{+\infty}\frac{(-1)^m}{m!}\left(\frac{b}{2}\right)^m\sum_{n\geq 0}\left(\frac{b}{2a}\right)^{2n}\frac{(2n)!}{n!(n+m)!}\\&=&\frac{1}{\sqrt{a^2-b^2}}+\frac{1}{a}\int_{0}^{1}\sum_{m\geq 1}\frac{(-1)^m}{m!(m-1)!}\left(\frac{b}{2}\right)^m(1-u)^{m-1}\sum_{n\geq 0}\binom{2n}{n}\left(\frac{b^2}{4a^2}\right)^{n}u^n\,du\\&=&
\frac{1}{\sqrt{a^2-b^2}}+\int_{0}^{1}\sum_{m\geq 1}\left(\frac{b}{2}\right)^m \frac{(-1)^m}{m!(m-1)!}\frac{(1-u)^{m-1}}{\sqrt{a^2-b^2 u}}\,du\\&=&
\frac{1}{\sqrt{a^2-b^2}}-\sqrt{\frac{b}{2}}\int_{0}^{1}\frac{J_1(\sqrt{2b(1-u)})}{\sqrt{(a^2-b^2 u)(1-u)}}\,du\\&=&\frac{1}{\sqrt{a^2-b^2}}-\sqrt{2b}\int_{0}^{1}\frac{J_1(t\sqrt{2b})}{\sqrt{(a^2-b^2)+b^2 t^2}}\,dt.\tag{6}\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1162480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\lim_{n\rightarrow \infty} \int_{0}^{1}\frac{1}{1+x^2+x^4+.........+x^{2n}}dx\;,$ Where $n\in \mathbb{N.}$ Evaluation of $\displaystyle \lim_{n\rightarrow \infty} \int_{0}^{1}\frac{1}{1+x^2+x^4+.........+x^{2n}}dx\;,$ where $n\in \mathbb{N.}$
$\bf{My\; Solution:}$ First we will simplify $\displaystyle 1+x^2+x^4+.......+x^{2n} = \frac{1-x^{2n+2}}{1-x^2}.$
Given $0< x<1, \quad\displaystyle \lim_{n\rightarrow \infty}\frac{1-x^{2n+2}}{1-x^2} = \frac{1}{1-x^2}$.
$ \therefore \displaystyle \displaystyle \lim_{n\rightarrow \infty} \int_{0}^{1}\frac{1}{1+x^2+x^4+.........+x^{2n}}dx = \int_{0}^{1}(1-x^2)dx = \frac{2}{3}.$
Is my solution is correct? If not, then how can we solve the above question?
Thanks in advance.
| We have that:
$$ I_n = \int_{0}^{1}\frac{1-x^{2}}{1-x^{2n+2}}\,dx =\frac{2}{3}+\int_{0}^{1}x^{2n+2}\frac{1-x^2}{1-x^{2n+2}}\,dx\tag{1}$$
where, over $[0,1]$:
$$0\leq \frac{1-x^2}{1-x^{2n+2}}\leq 1,\tag{2}$$
hence:
$$ \frac{2}{3}\leq I_n\leq \frac{2}{3}+\frac{1}{2n+3}\tag{3} $$
gives that the limit is $\frac{2}{3}$ as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$ Problem
Assume that $a,b\in\mathbb{R}-\{0\}$ and that $a+b\not=0$. Prove that $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$.
My Proof
Let's assume that $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$, then it follows that
$$
\begin{equation}
(a+b)^2-ab=0
\end{equation}
$$
Let $x=a+b$ and $y=ab$. Now $b=x-a$ and so $y=a(x-a)=ax-a^2$. The previous equation can be written as
$$
x^2-y=0
$$
Substituting $y=ax-a^2$ in this equation gives
$$
x^2-ax+a^2=0
$$
The discriminant of this quadratic equation (in $x$) is $-3a^2<0$ and therefore $x=a+b$ has no real solution. This means $a+b\in\mathbb{C}$ and therefore either of $a$ or $b$ or both are not real but this contradicts our assumption that $a,b$ are real numbers. Therefore by contradiction $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$.
My Question
Is my proof correct? are there any alternative proofs?
| You have $a^2+ab+b^2=0$. Treating this as a quadratic equation in which the unknown quantity is $a$ gives us this solution:
$$
a=\frac{-b\pm\sqrt{b^2 - 4b^2}}{2} = \frac{-b\pm b\sqrt{-3}}{2} = b\left(\frac{-1\pm i\sqrt 3} 2 \right).
$$
So if $b$ is any complex number except $0$ (e.g. let $b=1$) and $a$ is as given above, then
$$
\frac 1 a + \frac 1 b = \frac 1 {a+b}.
$$
But otherwise this last identity does not hold.
Alternatively, one could just seek counterexamples. For example if $a=1$ and $b=1$ then
$$
\frac 1 a + \frac 1 b = 1 + 1 =2 \quad\text{and}\quad\frac 1 {a+b} = \frac 1 2\quad\text{and}\quad 2\ne \frac 1 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
General formula for iterated cumulative sum Consider the sequence $S_0$ consisting of ones:
$$
1,1,1,1,1,1,\ldots
$$
Now compute the cumulative sum of this sequence, and call the resulting sequence $S_1$:
$$
1,2,3,4,5,6,\ldots
$$
Proceed iteratively to generate sequence $S_2$:
$$
1,3,6,10,15,21,\ldots
$$
then $S_3$:
$$
1,4,10,20,35,56,\ldots
$$
and so on.
It is well known that each sequence $S_k$ can be represented by a $k$-degree polynomial $P_k(n)$. For the above sequences the polynomials are
$$
P_0(n) = 1
$$
$$
P_1(n) = n
$$
$$
P_2(n) = \frac{n^2+n}{2}
$$
$$
P_3(n) = \frac{n^3+3n^2+2n}{6}
$$
My question: Is there a general formula for coefficients of the polynomial $P_k$? Or more generally, is there a formula to compute $S_k(n)$ as a function of $n$ and $k$? I mean a closed formula $S_k(n) = f(k,n)$ (not an iterative procedure such as the construction method I just described).
If that helps, actually I'm not interested in $S_k(n)$, but rather in the quotient $S_k(n)/S_k(n+1)$.
| The numbers in sequence $S_k$ are the binomial coefficients $\binom{m}{k}$; the $n$-th term of $S_k$ is $\binom{n-1+k}{k} = \frac{(n-1+k)!}{(n-1)!k!}$. One can prove this by using that
$$
\binom{i}{i} + \binom{i+1}{i} + \cdots + \binom{i+j}{i} = \binom{i+j+1}{i+1}
$$
for any $i,j \geq 0$.
For $k=1$ we find $P_1(n) = \binom{n-1+1}{1} = \binom{n}{1} = n$, for $k=2$ we find $P_2(n) = \binom{n-1+2}{2} = \binom{n+1}{2} = \frac{n(n+1)}{2}$, for $k=3$ we find $P_3(n) = \binom{n-1+3}{3} = \binom{n+2}{3} = \frac{n(n+1)(n+2)}{6}$, etcetera.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
I tried to compute $\int_0^1x^2\sin^2xd x$ and mathematica disagrees with me I tried to compute $\int_0^1x^2\sin^2xd x$ and got $\frac 16-\frac 38\sin2-\frac 14\cos2$. However, Wolfram Mathematica gives the result $\frac 16-\frac 18\sin2-\frac 14\cos2$, namely there is something wrong with the coefficient of $\sin$ in my answer. And here is how I computed it(If you don't have the patience to read it through, please help me do the integral and see if we get the same answer):
$$\int_0^1x^2\sin^2xd x$$
$$=\int^1_0x^2\frac {1-\cos 2x}{2}dx$$
$$=\int^1_0\frac {x^2}2 dx+\int^1_0\frac{-x^2\cos 2x}2dx$$
$$=\frac 12*\frac 13[x^3]^1_{x=1_00}+\int^1_0\frac{-x^2\cos 2x}{2}\frac{d(\sin2x)}{2\cos 2x}$$
$$=\frac 16+\int^1_0\frac{-x^2}{4}d(\sin 2x)$$
$$=\frac 16+((\frac {-x^2}{4}\sin 2x)^1_0-\int^1_0\sin 2xd(\frac {-x^2}{4}))$$
$$=\frac 16+(-\frac 14\sin 2+\frac 14\int^1_0\sin 2x*2xdx)$$
$$\frac 16-\frac 14\sin 2+\frac 12\int^1_0x\sin 2xdx$$
But $$\int^1_0x\sin2xdx=\int^1_0x\sin2x\frac{d\cos2x}{-2\sin 2x}=-\frac 12\int^1_0xd(\cos 2x)=-\frac 12((x\cos 2x)^1_0+\int^1_0-\cos 2x\frac {d(2x)}{2})=-\frac 12(\cos 2+\frac 12(\sin 2x)^1_0)=-\frac12(\cos 2+\frac 12\sin 2)$$, So $$\int^1_0x^2\sin^2xdx=\frac 16-\frac 14\sin 2+\frac 12(-\frac 12)(\cos 2+\frac 12\sin 2)=\frac 16-\frac 14 \sin 2-14\cos 2-\frac 18\sin2=\frac 16-\frac 38\sin 2-\frac 14\cos 2$$
| $$\int x^2\cos(2x)\,dx = \text{Re}\int x^2 e^{2ix}\,dx =\frac{1}{4}\text{Re}\int x^2 e^{ix}\,dx=\frac{1}{4}\text{Re}\left((2i+2x-ix^2)e^{ix}\right)$$
hence:
$$\int_{0}^{1} x^2\cos(2x)\,dx = \frac{1}{4}\left(2\cos 2+\sin 2\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Understanding how to form a definite integral from limit of an infinite sum I have to solve the limit $$\lim_{n\rightarrow\infty}\left(\frac1n\sum_{k=1}^n(\text{log}\:k)^2-\left(\frac1n\sum_{k=1}^n(\text{log}\:k)\right)^2\right)$$
I don't know how to proceed in this problem. However, the solution says that this limit is equal to the limit $$\lim_{n\rightarrow\infty}\left(\frac1n\sum_{k=1}^{n-1}\left(\text{log}\:\frac kn\right)^2-\left(\frac1n\sum_{k=1}^{n-1}\left(\text{log}\:\frac kn\right)\right)^2\right)$$
Which is equal to $$\int_0^1 (\text{log}\:x)^2dx-\left(\int_0^1 \text{log}\:x\;dx\right)^2$$
Can someone explain how these two equations were derived from the first expression? I am aware of expressing definite integral as limit of an infinite sum. But I don't see how the above conversions were made. I need the intermediate steps.
| Write
$$\log k = \log k - \log n + \log n = \log \frac{k}{n} + \log n,$$
so that
\begin{align}\left(\frac{1}{n}\sum_{k = 1}^n \log k\right)^2 &= \left(\frac{1}{n}\sum_{k = 1}^n \log\frac{k}{n} + \log n\right)^2\\
&= \left(\frac{1}{n}\sum_{k = 1}^n \log \frac{k}{n}\right)^2 + \frac{2\log n}{n}\sum_{k = 1}^n \log \frac{k}{n} + (\log n)^2 \tag{1}
\end{align}
and
\begin{align}\frac{1}{n}\sum_{k = 1}^n (\log k)^2 &= \frac{1}{n}\sum_{k = 1}^n \left[\left(\log\frac{k}{n}\right)^2 + 2\log n\log\frac{k}{n} + (\log n)^2\right]\\
&=\frac{1}{n}\sum_{k = 1}^n \left(\log \frac{k}{n}\right)^2 + 2\frac{\log n}{n}\sum_{k = 1}^n \log \frac{k}{n} + (\log n)^2 \tag{2}
\end{align}
Subtracting $(1)$ from $(2)$, we obtain
\begin{align}
& \frac{1}{n}\sum_{k = 1}^n (\log k)^2 - \left(\frac{1}{n}\sum_{k = 1}^n \log k\right)^2 \\
&= \frac{1}{n}\sum_{k = 1}^n \left(\log \frac{k}{n}\right)^2 - \left(\frac{1}{n}\sum_{k = 1}^n \log \frac{k}{n}\right)^2\\
&= \frac{1}{n}\sum_{k = 1}^{n-1} \left(\log \frac{k}{n}\right)^2 - \left(\frac{1}{n}\sum_{k = 1}^{n-1} \log \frac{k}{n}\right)^2
\end{align}
The last step follows since $\log \frac{k}{n} = 0$ when $k = n$.
Since
$$\frac{1}{n}\sum_{k = 1}^{n-1} \log \frac{k}{n}$$
is a sequence of Riemann sums for $\log x$ over $[0,1]$, and
$$\frac{1}{n}\sum_{k = 1}^{n-1} \left(\log \frac{k}{n}\right)^2$$
is a sequence of Riemann sums for $(\log x)^2$ over $[0,1]$, we have
$$\frac{1}{n}\sum_{k = 1}^{n-1} \log \frac{k}{n} \to \int_0^1 \log x\, dx$$
and
$$\frac{1}{n}\sum_{k = 1}^{n-1} \left(\log \frac{k}{n}\right)^2 \to \int_0^1 (\log x)^2\, dx,$$
whence the third step follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}=\frac{5}{36}-\frac{\log(2)}{6}$ Proving $$\sum_{n=1}^{\infty }\frac{0.5^n}{(n)(n+1)(n+2)(n+3)}=\frac{5}{36}-\frac{\log(2)}{6}$$
| Another approach. From:
$$\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{6}\frac{\Gamma(n)\Gamma(4)}{\Gamma(n+4)}=\frac{1}{6}\int_{0}^{1} u^{n-1}(1-u)^3\,du $$
it follows that:
$$\sum_{n\geq 1}\frac{1}{2^n n(n+1)(n+2)(n+3)}=\frac{1}{6}\int_{0}^{1}\frac{(1-u)^3}{2-u}\,du $$
and:
$$ \int_{0}^{1}\frac{(1-u)^3}{2-u}\,du = \int_{0}^{1}\frac{x^3}{1+x}\,du=-\log 2+\int_{0}^{1}(1-x+x^2)\,dx=-\log 2+\frac{5}{6}, $$
hence:
$$\sum_{n\geq 1}\frac{1}{2^n n(n+1)(n+2)(n+3)}=\color{red}{-\frac{\log 2}{6}+\frac{5}{36}}$$
as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
What is the indefinite integral $\int \sqrt{1-2\sqrt{x-x^2}} \ \mathrm dx$? What is
$$\int \sqrt{1-2\sqrt{x-x^2}} \ \mathrm dx$$
I have tried substituting everything and it doesn't seem to be working. Substituting trigonometry doesn't seem to work either.
| The primitive is quite hard to write down, but we can notice, by symmetry, that:
$$\int_{0}^{1}\sqrt{1-2\sqrt{x-x^2}}\,dx = 2\int_{0}^{1/2}\sqrt{1-2\sqrt{x-x^2}}\,dx = 2\int_{0}^{1/2}\sqrt{1-2\sqrt{\frac{1}{4}-x^2}}\,dx$$
hence:
$$\int_{0}^{1}\sqrt{1-2\sqrt{x-x^2}}\,dx=\int_{0}^{1}\sqrt{1-\sqrt{1-x^2}}\,dx=\int_{0}^{\pi/2}\cos\theta\sqrt{1-\cos\theta}\,d\theta$$
so:
$$\int_{0}^{1}\sqrt{1-2\sqrt{x-x^2}}\,dx = \sqrt{2}\int_{0}^{\pi/2}\cos\theta \sin\frac{\theta}{2}\,d\theta = \color{red}{\frac{2}{3}(2-\sqrt{2})}.$$
By following the same steps for the indefinite integral, we have:
$$\int_{0}^{x} \sqrt{1-2\sqrt{z-z^2}}\,dz = \frac{2}{3}+\frac{2(2x-1)\left(1-\sqrt{x-x^2}\right)}{3\sqrt{1-2\sqrt{x-x^2}}}$$
for any $x\in\left[0,\frac{1}{2}\right]$. In order to compute the integral for $x\in\left[\frac{1}{2},1\right]$ it is sufficient to notice that the integrand function is symmetric around $z=\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Prove that $g(x)=\frac{\ln(S_n (x))}{\ln(S_{n-1}(x))}$ is increasing in $x$, where $S_{n}(x)=\sum_{m=0}^{n}\frac{x^m}{m!}$ I want to prove that the function
$$g(x)=\frac{\ln(S_n (x))}{\ln(S_{n-1}(x))},\,x >0$$ is increasing in $x$ for all $n$, where
$ S_n(x)= \sum_{m=0}^{n}\frac{x^m}{m!}$. Differentiating gives something messy that I have not been able to prove it is non-negative. I have also been trying to find an appropriate $h(x)$ increasing and proving that $ h(g(x))$ is increasing $\ln(\cdot)$ seems a good candidate.
By plotting I am almost convinced the statement is true.
|
Note: Here is a first step towards a complete answer. The point I like to address is the discrete log-concavity of the function $S_n$ for fixed $x>0$. This means $S_n$ fulfils
\begin{align*}
S_{n-1}^2(x)\geq S_n(x)S_{n-2}(x)\qquad\qquad x>0, n\geq 2\tag{1}
\end{align*}
A corresponding example can be found e.g. in the slides about Log-convexity and Log-concavity by Dmitry Karp. You might have a look at the section about Bessel functions where the inequality (1) for these functions is stated and some sharper results are presented afterwards.
Please note that in order to show $g(x)$ is increasing an even sharper inequality than (1) has to be proved.
We start as @Arashium did. Using the differential operator $D_x$ we obtain
\begin{align*}
D_xS_n(x)=D_x\sum_{m=1}^{n}\frac{x^{m-1}}{(m-1)!}=\sum_{m=0}^{n-1}\frac{x^m}{m!}=S_{n-1}(x)
\end{align*}
and by omitting the argument $x$ we get
\begin{align*}
D_x\ln(S_n)&=\frac{1}{S_n}D_x(S_n)=\frac{S_{n-1}}{S_n}\\
\end{align*}
and
\begin{align*}
D_xg&=D_x\frac{\ln(S_n)}{\ln(S_{n-1})}
=\frac{\ln(S_{n-1})D_x\ln(S_n)-\ln(S_n)D_x\ln(S_{n-1})}{\ln^2(S_{n-1})}\\
&=\frac{1}{\ln^2(S_{n-1})}\left(\frac{S_{n-1}}{S_n}\ln(S_{n-1})-\frac{S_{n-2}}{S_{n-1}}\ln(S_n)\right)\tag{2}
\end{align*}
In order to show that the function $g$ is increasing we need $D_xg\geq 0$.
Therefore the following is to prove according to (2)
\begin{align*}
S_{n-1}^2\geq S_{n-2}S_n\frac{\ln(S_n)}{\ln(S_{n-1})}\tag{3}
\end{align*}
Here's the proof of the weaker inequality (1) showing that $S_n(x)$ is discrete log-concave for fixed $x>0$.
\begin{align*}
S_{n-1}(x)^2&-S_{n-2}(x)S_n(x)=\\
&=\left(S_{n-2}(x)+\frac{x^{n-1}}{(n-1)!}\right)^2-S_{n-2}(x)\left(S_{n-2}(x)+\frac{x^{n-1}}{(n-1)!}+\frac{x^{n}}{n!}\right)\\
&=S_{n-2}(x)\left(\frac{x^{n-1}}{(n-1)!}-\frac{x^{n}}{n!}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}S_{n-2}(x)\left(1-\frac{x}{n}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}\right)\left(1-\frac{x}{n}\right)+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}-\frac{1}{n}\sum_{m=0}^{n-2}\frac{x^{m+1}}{m!}\right)
+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(\sum_{m=0}^{n-2}\frac{x^m}{m!}-\frac{1}{n}\sum_{m=1}^{n-1}\frac{x^{m}}{(m-1)!}\right)
+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(1+\sum_{m=1}^{n-2}\left(\frac{1}{m}-\frac{1}{n}\right)\frac{x^m}{(m-1)!}-\frac{x^{n-1}}{n(n-2)!}\right)
+\frac{x^{2n-2}}{(n-1)!^2}\\
&=\frac{x^{n-1}}{(n-1)!}\left(1+\sum_{m=1}^{n-2}\left(\frac{1}{m}-\frac{1}{n}\right)\frac{x^m}{(m-1)!}\right)\\
&\qquad\qquad+\frac{x^{2n-2}}{(n-1)!(n-2)!}\left(\frac{1}{n-1}-\frac{1}{n}\right)\tag{4}\\
&>0\\
\end{align*}
From the line (4) it's obvious that the inequality is valid and we may therefore conclude that $S_n$ is discrete log-concave.
The challenge is of course to sharpen the inequality (4) in order to obtain (3).
Note: Since the function $\ln$ is monotonically increasing we obtain due to the log-concavity of $S_n$
\begin{align*}
\ln\left(S_{n-1}^2(x)\right)&\geq\ln\left(S_n(x)S_{n-2}(x)\right)\\
2\ln\left(S_{n-1}(x)\right)&\geq\ln\left(S_n(x)\right)+\ln\left(S_{n-2}(x)\right)\\
\end{align*}
which may of some use for further calculations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 3
} |
Trigonometry problem on product of trig functions. Questions:
*
*$\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=?$
*$\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}=?$
*$\tan\dfrac{\pi}{7}\tan\dfrac{2\pi}{7}\tan\dfrac{3\pi}{7}=?$
Approach:
If $x=\dfrac{\pi}{7}\implies 7x=\pi\implies 4x=\pi-3x\implies \sin 4x = \sin 3x$.Upon expansion, $$8cos^3x-4cos²x-4cosx+1=0$$
"$\cos \dfrac{\pi}{7}, \cos \dfrac{3\pi}{7}, \cos \dfrac{5\pi}{7}$will satisfy this cubic."
Note: I'm not able to understand the above line. Why do these values satisfy this cubic and how did we get these values?
These are therefore the roots of the cubic, and their product is:
$$\cos\dfrac{\pi}{7}\cos\dfrac{3\pi}{7}\cos\dfrac{5\pi}{7}=\dfrac{-1}{8}$$
Finally,
$\cos\dfrac{5\pi}{7} = \cos\left(\pi-\dfrac{2\pi}{7}\right)=-\cos \dfrac{2\pi}{7}$. Replacing this in (1) gives $$\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=\dfrac{1}{8}$$
How can we apply this concept to find the values of
*$\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}?$
*$\tan\dfrac{\pi}{7}\tan\dfrac{2\pi}{7}\tan\dfrac{3\pi}{7}?$
I'm not able to form the equations for these two problems. Please Help. .
| As $\sin\left[7\left(\pm\dfrac{u\pi}7\right)\right]=0$ for any integer $u$
If $\sin7x=0,7x=n\pi$ where $n$ is any integer
$\implies x=\dfrac{n\pi}7$ where $n=0,\pm1,\pm2,\pm3\pmod7$
For $f(c)=8c^3-4c^2-4c+1=0$
$f'(c)=24c^2-8c-4=4(6c^2-2c-1)$
As $f'(c)=0,f(c)=0$ don't have any root in common, $f(c)=0$ does not have Repeated Root.
As $\cos(-A)=\cos A,\cos\left(\pm\dfrac{r\pi}7\right),r=1,2,3$ will be the roots of $f(c)=0$
For the Sine product, I would recommend this
and use $\sin(\pi-B)=\sin B$
For tangent Product, I would recommend method applied in
Sum of tangent functions where arguments are in specific arithmetic series and Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$
to find the roots of $t^6-21t^4+35t^2-7=0--->(2)$ are $\tan\frac{r\pi}7$ where $r=1,2,3,4,5,6$
and use $\tan(\pi-C)=-\tan C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How many teams of $5$ players out of $15$ girls and $10$ boys can be formed with at least $2$ boys and $2$ girls [with complement]
How many teams of $5$ players out of 15 girls and 10 boys can be formed with at least 2 boys and 2 girls? The solution has to be with complement.
This is related to: How many ways to assemble a team of 5 out of 15 girls and 10 boys with limitations?
I think I'm doing something wrong with complement, is it treated like the negation of the given conditional?
In this case it would be the general case without any limitations $\binom {25} 5$, minus a team with only boys: $\binom {10} 5$, or a team with only girls: $\binom {15} 5$ or a team with one boy and four girls: $\binom{15} 4 \binom{10}1$ or a team with with one girl and four boys: $\binom {15}1\binom {10} 4$.
The total is: $$\binom {25} 5-\left(\binom {10} 5+\binom {15} 5+\binom{15} 4 \binom{10}1+\binom {15}1\binom {10} 4\right) $$
| That looks fine. It's the exactly right way to count the complement.
By the Rule of Complements: the measure of an event ($A$) is the measure of the total space ($S$) minus the measure of the complement ($A^\complement$); and vice versa. $$\lvert A\rvert = \lvert S\rvert - \lvert A^\complement\rvert \\ \lvert A^\complement \rvert = \lvert S \rvert - \lvert A \rvert$$
The total space, or general case, is counted as you did: the ways to select any $5$ students from all $25$.
The complement is partitioned as you did: counting ways to select $5$ from $10$ boys, or $5$ from $15$ girls, or $4$ boys and $1$ girl, or $1$ boy and $4$ girls.
$$\dbinom{25}{5} -\left(\binom{10}{5}+\binom{15}{5} + \binom{10}{4}\binom{15}{1}+\binom{10}{1}\binom{15}{4} \right)$$
Which, if you wish to verify, will equal the ways to select $3$ boys and $2$ girls or $2$ boys and $3$ girls.
$$\binom{10}{3}\binom{15}{2}+\binom{10}{2}\binom{15}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Orthogonal Diagonalization of a matrix I am having problem in diagonalization of \begin{bmatrix}-1&-1&-4\\-1&-4&-1\\-4&-1&-1\end{bmatrix}
This is symmetric so it must be orthogonally diagonalizable. The eigen values are $-6,-3,3$. I did all the computations and found an orthogonal matrix $P$ which is \begin{bmatrix}1/\sqrt3&1/\sqrt6&-1/\sqrt2\\1/\sqrt3&-2/\sqrt6&0\\1/\sqrt3&1/\sqrt6&1/\sqrt2\end{bmatrix}
But some how $P^TAP$ is not the diagonal matrix? Can someone please find a mistake for me?
| Note: Since $P^T = P^{-1}$ for an orthogonal $P$, the equality $P^T A P = D$ is
the same as $P^{-1} A P = D$.
Your answer is perfectly correct, just not in a simplified form.
We have:
$$P^T A P = \begin{bmatrix}
-6 & 0 & 0 \\
\frac{2 \left(-\sqrt{\frac{2}{3}}-\frac{1}{\sqrt{6}}\right)}{\sqrt{3}}+\sqrt{2} & \sqrt{\frac{2}{3}} \left(-\sqrt{\frac{2}{3}}-\frac{1}{\sqrt{6}}\right)-2 & 0 \\
\frac{-\frac{1}{\sqrt{2}}+2 \sqrt{2}}{\sqrt{3}}+\frac{\frac{1}{\sqrt{2}}-2 \sqrt{2}}{\sqrt{3}} & \frac{-\frac{1}{\sqrt{2}}+2 \sqrt{2}}{\sqrt{6}}+\frac{\frac{1}{\sqrt{2}}-2 \sqrt{2}}{\sqrt{6}} & \frac{-\frac{1}{\sqrt{2}}+2 \sqrt{2}}{\sqrt{2}}-\frac{\frac{1}{\sqrt{2}}-2 \sqrt{2}}{\sqrt{2}} \\
\end{bmatrix}$$
For example, simplifying the bottom rightmost value yields:
$$\frac{-\frac{1}{\sqrt{2}}+2 \sqrt{2}}{\sqrt{2}}-\frac{\frac{1}{\sqrt{2}}-2 \sqrt{2}}{\sqrt{2}} = \dfrac{3}{2} + \dfrac{3}{2} = 3$$
If we simplify each value in the matrix, it reduces to:
$$P^T A P = \begin{bmatrix}
-6 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & 3 \\
\end{bmatrix}$$
This is obviously a diagonal matrix $D$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Difference of two variables raised to the $k$. Let $k \ge 2$ be an integer and let $0 < y < x$ where $x, y$ are integers. Show that $x^k-y^k > 2$.
I believe I have proved this statement but want to make sure it is sound.
Proof: Since $0 < y < x$ and $x$ and $y$ are integers, $x \ge 2$ and $y \ge 1$ and $x^k>y^k$. So $x^k-y^k > 0$. Thus $0 < 2^k-1^k \le x^k-y^k.$ Since $k$ is at least $2$, $3 = 2^2-1^2 \le 2^k-1^k \le x^k-y^k $. Hence, $x^k-y^k > 2$.
| The only part of your proof that I find is not fully explained is why $2^k-1^k\leq x^k-y^k$.
Another way to see it is that
$$
x^k-y^k=(x-y)(x^{k-1}+x^{k-2}y+...+x y^{k-2}+y^{k-1})
$$
where $(x-y)\geq 1$ and $(x^{k-1}+x^{k-2}y+...+x y^{k-2}+y^{k-1})$ consists of $k$ terms all greater than or equal to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1184405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
General solution of the equation $\sin^{2015}(x)+\cos^{2015}(x) = 1\; \forall \; x\in \mathbb{R}$
Calculation of General solution of the equation $\sin^{2015}(x)+\cos^{2015}(x) = 1\; \forall \;x\in \mathbb{R}$.
$\bf{My\; Try:}$ We can Write the equation as $$\sin^{2015}(x)+\cos^{2015}(x)\leq \sin^2(x)+\cos^2(x)=1.$$
And equality hold when $$\sin(x)=1$$ and $$\cos (x) = 0.$$
So we get $$\displaystyle x=n\pi+(-1)^n\cdot \frac{\pi}{2}$$ and $$\displaystyle x=2n\pi\pm \frac{\pi}{2}.$$
Now How can I calculate common solution of $\sin (x) = 0$ and $\cos (x) = 0$.
Please help me. Thanks.
| $\sin^{2015}(x) + \cos^{2015}(x) = 1 = \sin^2(x) + \cos^2(x)$
or, $\left[\sin^2(x) \left(\sin^{2013}(x)-1 \right) \right]+ \left[ \cos^2(x) \left( \cos^{2013}(x)-1 \right) \right]= 0$
Since $\sin^2(x) \geq 0$ and $\cos^2(x) \geq 0$, this is possible only if both
$\left[\sin^2(x) \left(\sin^{2013}(x)-1 \right) \right]= 0$
and
$\left[ \cos^2(x) \left( \cos^{2013}(x)-1 \right) \right]= 0$
That is possible if either
*
*$\sin^2(x) = 0$ and $\left( \cos^{2013}(x)-1 \right) = 0$
$\implies \sin(x) = 0$ and $\cos(x) = 1 \implies x = 2n \pi$
or
*$\cos^2(x) = 0$ and $\left(\sin^{2013}(x)-1 \right) = 0$
$\implies \cos(x) = 0$ and $\sin(x) = 1 \implies x = (4n+1)\frac{\pi}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Differentiate. $y = (1 − \sec x) / \tan x$ I have no idea what I have done wrong. Please criticise.
Differentiate $y = \frac{1 - \sec x}{\tan x}$.
\begin{align*}
\frac{d}{dx} \left[ \frac{1 - \sec x}{\tan x} \right]
&= \frac{\tan x \frac{d}{dx} [ 1 - \sec x ] - (1 - \sec x) \frac{d}{dx} \tan x }{(\tan x)^2} \\
&= \frac{\tan x (- \sec x \tan x) - [(1 - \sec x) (-\csc^2 x)]}{(\tan x)^2} \\
&= \frac{-\tan^2 x \sec x + \csc^2 x - \sec x \csc^2 x}{(\tan x)^2}
\end{align*}
| The derivative of $\tan x$ is $\sec^2 x$. You wrote that it is $-\csc^2x$, in the second line. You can proceed like: $$\frac{\tan x(-\sec x \tan x) - (1-\sec x)\sec^2x}{\tan^2x} = -\sec x - \frac{(1-\sec x)}{\sin^2 x} = \cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proof of an integral identity involving $\pi$ and e In the "Surprising Identities" post from a while back, Vladimir Reshetnikov offered the following identity[1]:
$$\int_{0}^{\infty} dx \frac{1}{1 + x^2} \frac{1}{1 + x^{\pi}} = \int_{0}^{\infty}dx \frac{1}{1 + x^2} \frac{1}{1 + x^{e}} \ \ .$$
Where can I find a proof of this?
*
*https://math.stackexchange.com/a/693459/29360
| Observe the integral
$$\int_0^{\infty} \frac{1}{1+x^2}\frac{1}{1+x^s}dx$$for any exponent $s$ for which the integral converges. Then, split the integral $I$ as
$$I= \int_0^{\infty} \frac{1}{1+x^2}\frac{1}{1+x^s}dx=\int_0^{1} \frac{1}{1+x^2}\frac{1}{1+x^s}dx+\int_1^{\infty} \frac{1}{1+x^2}\frac{1}{1+x^s}dx$$
In the second integral, make the substitution $x=\frac{1}{y}$, $dx=-\frac{1}{y^2}dy$, and note that the limits of integration transform from $(1,\infty)$ to $(1,0)$. Thus, we can write
$$I=\int_0^{1} \frac{1}{1+x^2}\frac{1}{1+x^s}dx+\int_1^{0} \frac{1}{1+y^{-2}}\frac{1}{1+y^{-s}} \left(-\frac{1}{y^2}\right) dy$$
In the last integral, absorbing the negative sign by interchanging the order of integration, multiplying numerator and denominator by $y^2y^s$ and simplifying, and changing the dummy integration variable to x yields
$$I=\int_0^{1} \frac{1}{1+x^2}\frac{1}{1+x^s}dx+\int_0^{1} \frac{1}{1+x^{2}}\frac{x^s}{1+x^{s}} dx$$
which after recombining the integrals reveals that
$$I=\int_0^{1} \frac{1}{1+x^2}\frac{1+x^s}{1+x^{s}}dx=\int_0^{1} \frac{1}{1+x^2}dx=\frac{\pi}{4}$$which obviously is independent of $s$!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Does $\sum_{n=1}^{\infty}\ln(n\sin(\frac{1}{n}))$ converge? I must determine whether the following series converges: $$\sum_{n=1}^{\infty}\ln\left(n\sin\left(\frac{1}{n}\right)\right)$$
I know that in general, I must use the limit comparison test, but I cannot find an expression to which I can compare it. For instance, I have tried the usual process:
For $n$ large, we have that $\lim_{n\to \infty}n\sin\frac1n=1$, and so, $\ln(1)=0$. This fails the divergence test, but it cannot be concluded automatically that the series is convergent either. How may I proceed here? Any help would be appreciated.
| Using Taylor series, $$\sin x\sim x-\frac{x^3}{3!}+o(x^3)$$$$\ln(1+x)\sim x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$$
When $n\rightarrow\infty$, $$n\sin\frac{1}{n}\sim n(\frac{1}{n}-\frac{1}{3!n^3}+o(\frac{1}{n^3}))=1-\frac{1}{6n^2}+o(\frac{1}{n^2})$$
Thus, $$\ln(n\sin\frac{1}{n})\sim -\frac{1}{6n^2}+o(\frac{1}{n^2})-\frac{\left(-\frac{1}{6n^2}+o(\frac{1}{n^2})\right)^2}{2}+o\left(-\frac{1}{6n^2}+o(\frac{1}{n^2}\right)=-\frac{1}{6n^2}+o(\frac{1}{n^2})$$
As $\sum\dfrac{1}{n^2}$ is convergent, so is $\sum\ln(n\sin\dfrac{1}{n})$.
Hope this can help you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
Find the point $T(a, b)$ on the curve $y = x^2 + 1$ whose tangent passes through the point $P(1, 0)$ from the left. Find the point $T(a, b)$ on the curve $y = x^2 + 1$ whose tangent passes through the point $P(1, 0)$ from the left.
$$
\\ \begin{align}
\\
\\ f(x) &= x^2 + 1
\\ f(a) &= a^2 + 1 = b
\\
\\ T(a, f(a))
\\
\\ f'(x) &= 2x
\\ f'(a) &= 2a = m_{tan}
\\
\\ x_1 &= 1
\\ y_1 &= 0
\\
\\ P(x_1, y_1)
\\
\\ y_1 - b &= m_{tan}(x_1 - a)
\\ 0 - a^2 - 1 &= 2a(1 - a)
\\ -a^2 - 1 &= 2a - 2a^2
\\ a^2 - 2a - 1 &= 0
\\ (a - 1)(a - 1) &= 0
\\ a &= 1
\\ b &= 1^2 + 1 = 2
\\
\\ T(1, 2)
\\ P(1, 0)
\\ \end{align}
$$
At this point, I see that the line that goes through the points $T$ and $P$ would have to have a vertical slope.
I've made an error somewhere along the line, but I don't quite see it. Perhaps I'm going about this all wrong?
| the tangent to $y = x^2 + 1$ at $(a, a^2 + 1$ is $$y - a^2 - 1 = 2a(x-a)$$ if the point $x = 1, y = 0$ is on it, then $$-a^2 - 1 = 2a(1-a) $$ this is a quadratic equation, you can find two values for $a$ by solving it.
you get $$a^2 - 2a = 1 \to (a-1)^2=2 \to a = 1 \pm \sqrt 2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sum of numbers in a grouping question A person grouped numbers in the following way:
$$\left \{ 1 \right \},\left \{ 3,5 \right \},\left \{ 7,9,11 \right \},\left \{ 13,15,17,19 \right \},...$$
What is the sum of the numbers in the $9$th grouping?
I got the answer as $729$, by inspection. I wanted to try it again with a more mathematically method so here is my progress.
I noticed that the sum of the first $n$ groups containing odd numbers is $\frac{1}{2}n(n+1)$. I also know that the sum of the first $k$ odd terms is $k^{2}$. So I said to get the sum of the terms in the $n$th group, I must add sum of the numbers in the $n$th group and subtract it from the sum of the first $(n-1)$ groups. That is:
$$ [\frac{1}{2}n(n+1)]^{2} - [\frac{1}{2}n(1+(n-1))]^{2}$$
$$= \frac{1}{2}n^{3}+\frac{1}{4}n^{2}$$.
When I plugged in $n = 9$, I didn't get $729$. Where did I go wrong?
| Use the fact that arithmetic sum of odd numbers is a square:
$$\sum_{k=1}^{n} (2k-1) = n^2$$
The sum of a grouping will therefore be the difference of two squares. Which squares? Those are also arithmetic sequence $1,2,3,4...n$ which is $a_n = n(n+1)/2$. So it should be:
${a_{n+1}}^2 - {a_n}^2 = ((n+1)(n+2)/2)^2 - (n(n+1)/2)^2 = (n+1)^2((n+2)^2-n^2)/4 = (n+1)^2(2n+2)2/4 = (n+1)^2(n+1) = (n+1)^3$ which is $9^3 = 729$ for $n = 8$. So it seems to me you are right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Sum of factorial fractions Find the sum $$\sum\limits_{a=0}^{\infty}\sum\limits_{b=0}^{\infty}\sum\limits_{c=0}^{\infty}\frac{1}{(a+b+c)!}$$ I tried making something like a geometric series but couldn't. Then I couldn't think of anything.
| Consider an integer $m$ where $a+b+c=m$ $\lbrace m,a,b,c\in \Bbb N_0\rbrace$.
The number of possible combinations of $a, b, c$ is given by
$$\begin{align}
N(m)&=\sum_{a=0}^{m}\sum_{b=0}^{m-a}1\\
&=\sum_{a=0}^{m}(m-a+1)\\
&=\sum_{a=1}^{m+1}a\quad \text{ or } {a\choose 1}\\
&={m+2\choose 2}
\end{align}$$
The above can also be derived using the "stars and bars" method to count the number of ways to partition a set with $m$ elements into $n$ groups (here $n=3$), which is given as
$$N(m)=\binom {m+n-1}{n-1}\Biggr|_{n=3}=\binom {m+2}2$$
.
Hence
$$\begin{align}
\sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\sum_{c=0}^{\infty}\frac 1{(a+b+c)!}
&=\sum_{m=0}^\infty \frac {N(m)}{m!}\\
&=\sum_{m=0}^\infty \frac {1}{m!}{m+2\choose 2}\\
&=\sum_{m=0}^\infty \frac {1}{m!}\frac{(m+2)(m+1)}2\\
&=\frac 12 \sum_{m=0}^\infty \frac {m^2+3m+2}{m!}\\
&=\frac 12 \sum_{m=0}^\infty \frac {m(m-1)+4m+2}{m!}\\
&=\frac 12 \sum_{m=0}^\infty \frac 1{(m-2)!}+\frac 4{(m-1)!}+\frac 2{m!}\\
&=\frac 72 \sum_{m=0}^\infty \frac 1{m!}\\
&=\frac 72e\qquad \blacksquare
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
If $\sin^{-1}\frac{2a}{1+a^2}-\cos^{-1}\frac{1-b^2}{1+b^2}=\tan^{-1}\frac{2x}{1-x^2}$ then what is value of x? If $\sin^{-1}\frac{2a}{1+a^2}-\cos^{-1}\frac{1-b^2}{1+b^2}=\tan^{-1}\frac{2x}{1-x^2}$ then what is value of x?
Solution
$\tan^{-1}x=\tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab}$
x=$\frac{a-b}{1+ab}$
i have a doubt what is $\tan^{-1}a$ and $\tan^{-1}b$ and how come $\tan^{-1}x=\tan^{-1}a-\tan^{-1}b$?
| Hint:
Assume a=tan(y) and similarly for x and b .
Solve the identity and replace y with $tan^{-1}$b .
You will get your expression.
Use this ,this and this from here for help with those identities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\int \frac{dx}{x^2 \sqrt{x-1}}$? I need to integrate$$\int \dfrac{dx}{x^2 \sqrt{x-1}}.$$
I've tried everything from substitutions ($\sqrt{x-1}=u$) to integration by parts but I cannot get anywhere. Please help.
| Substitute $u=\sqrt{x-1}$. Then $2du=\dfrac{1}{\sqrt{x-1}}dx$ and $x=u^2+1$.
Therefore $$\int \dfrac{1}{x^2 \sqrt{x-1}}dx=\int \dfrac{2}{(u^2+1)^2}du.$$
Put $U=\dfrac{1}{u^{2}+1}$ and $dV=du$. Then integration by parts,
$$\int \dfrac{1}{(u^2+1)^2}du=\dfrac{u}{u^{2}+1}+\int \dfrac{2u^2}{(u^2+1)^2}du\\=\dfrac{u}{u^{2}+1}+2\int \left( \dfrac{u^{2}+1}{(u^{2}+1)^2}-\dfrac{1}{(u^{2}+1)^2}\right)du\\
=\dfrac{u}{u^{2}+1}+2\int \left( \dfrac{1}{(u^{2}+1)}-\dfrac{1}{(u^{2}+1)^2}\right)du $$
Hence $$\int \dfrac{1}{(u^2+1)^2}du=\dfrac{u}{2(u^{2}+1)}+\dfrac{1}{2}\int \dfrac{1}{(u^{2}+1)}du=\dfrac{u}{2(u^{2}+1)}+\dfrac{1}{2}\tan^{-1}(u).$$
Therefore $$\int \dfrac{1}{x^2 \sqrt{x-1}}dx=\dfrac{\sqrt{x-1}}{x}+\tan^{-1}(\sqrt{x-1}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
If $f$ is differentiable at $4$ and $f^\prime(4) = 5$, evaluate the following limit: $\lim_{x \to 4} \frac{f(x)-f(4)}{\sqrt{x} - 2}$. If $f$ is differentiable at $4$ and $f^\prime(4) = 5$, evaluate the following limit: $\lim_{x \to 4} \frac{f(x)-f(4)}{\sqrt{x} - 2}$.
\begin{align*}
\\ h &= \sqrt{x} - 2
\\
\\ h \to 0 &\text{ as } x \to 4
\\
\\ \lim_{x \to 4} \frac{f(x)-f(4)}{\sqrt{x} - 2} &= \lim_{h \to 0} \frac{f(x)-f(4)}{h}
\\ \lim_{h \to 0} \frac{f(4-h)-f(x)}{-h} &\implies \lim_{h \to 0} \frac{f(4-h)-f(4)}{-h} = f^\prime(4) = 5
\end{align*}
| you can do this. $$f(4 + h) = f(4) + 5h+\cdots, \sqrt{4+h} = (4+h)^{1/2} = 4^{1/2} + \frac12 4^{-1/2}h+\cdots \\\to f(4+h) - f(4) = 5h+\cdots, \sqrt{4+h} = 2 +\frac 14 h+\cdots \\
\to \frac{f(4+h) - f(4)}{\sqrt{4+h} - 2 }= 20+\cdots$$ so the limit is $20.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Summation over a product of binomial coefficients Question: I can't figure out why the following equality is true
$\sum_\limits{k=a-b-c}^{d} (-1)^k \binom{d}{k}\binom{k+b+c}{a} = (-1)^d \binom{b+c}{a-d} $
How can this be shown? (In the book it just says that the summation is accomplished by elementary means.)
Attempt:
I tried setting $d=a-b-c+n$ so there will be $n$ terms in the summation. Using the briefer notation $x=a-b-c$, I get the LHS
\begin{align}
\sum_\limits{k=x}^{x+n} (-1)^k \binom{d}{k}\binom{k+b+c}{a} = &
(-1)^{x}\binom{x+n}{x}\binom{a}{a} + (-1)^{x+1}\binom{x+n}{x+1}\binom{a+1}{a} +
\\ & \dots + (-1)^{x+i}\binom{x+n}{x+i}\binom{a+i}{a} + \dots + \\ & (-1)^{x+n-1}\binom{x+n-1}{x+n}\binom{a+n-1}{a} \\
= & \sum_\limits{i=0}^{n-1} (-1)^{x+i} \binom{x+n}{x+i}\binom{a+i}{a}
\end{align}
and then I get stuck. I can't recognize it as being of the form of some of the normal products or summations of binomial coefficents.
I can verify the equation if I put in numbers or just make n=1 or n=2 or similar. So I thought of using induction to prove it. However with the $n$ in $\binom{x+n}{x+i}$ part, I am not sure that if it is possible. I thought of maybe using the relation
$\binom{n}{k} k = n \binom{n-1}{k-1}$ but I can't see that making sense.
| Since $b$ and $c$ always appear together, the solution can be simplified visually by putting $n=b+c$.
$$\begin{align}
\sum_{k=a-b-c}^{d}(-1)^k {d\choose k}{k+b+c\choose a}&=
\sum_{k=a-n}^{d}(-1)^k {d\choose k}{k+n\choose a}
\qquad \qquad \quad \;\;\text{putting $n=b+c$}\\
&=\sum_{k=a-n}^{d}(-1)^k {d\choose k}{k+n\choose k+n-a}\\
&=\sum_{k=a-n}^{d}(-1)^{k+k+n-a} {d\choose k}{-a-1\choose k+n-a}
\qquad \text{using upper negation}\\
&=(-1)^{n-a}\sum_{k=a-n}^{d} {d\choose d-k}{-a-1\choose k+n-a}\\
&=(-1)^{n-a}{d-a-1\choose d+n-a}
\qquad \qquad \qquad \quad \text{using Vandermonde}\\
&=(-1)^{n-a+(d+n-a)}{n\choose d+n-a}
\qquad \qquad \;\text{using upper negation}\\
&=(-1)^d {n\choose a-d}\\
&=(-1)^d {b+c\choose a-d}\qquad \blacksquare
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
reduction formulae for $\int_0^1 (x^{\frac{n+2}2}\sqrt{1-x})dx\,\!$ how do I prove that
$\int_0^1 (x^{\frac{n+2}2}\sqrt{1-x})dx\,\!={\frac{n+2}{n+5}}\int_0^1 (x^{\frac{n}2}\sqrt{1-x})dx\,\!$
I couldn't find which functions u and v' to use for the intgeration by parts.
Help please.
| Recall that when $\Re(a),\Re(b)>0$ $$\int_0^1x^{a-1}(1-x)^{b-1}=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$thus
\begin{align}
\int_0^1x^{\frac{n+2}{2}}\sqrt{1-x}dx&=\int_0^1x^{\frac{n}{2}+2-1}(1-x)^{\frac32-1}dx\\
&=\frac{\Gamma(\frac{n}{2}+2)\Gamma(\frac32)}{\Gamma(\frac{n}{2}+2+\frac32)}\\
&=\frac{(\frac{n}{2}+1)\Gamma(\frac{n}{2}+1)\Gamma(\frac32)}{(\frac{n}{2}+1+\frac32)\Gamma(\frac{n}{2}+1+\frac32)}\\
&=\frac{n+2}{n+5}\frac{\Gamma(\frac{n}{2}+1)\Gamma(\frac32)}{\Gamma(\frac{n}{2}+1+\frac32)}\\
&=\frac{n+2}{n+5} \int_0^1 x^{\frac n2+1-1}(1-x)^{\frac32-1}dx\\
&=\frac{n+2}{n+5} \int_0^1 x^{\frac n2}\sqrt{1-x}dx\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Ball and urn method (counting problems)
How many ordered triples $(a, b, c)$ of positive integers exist with the property that $abc = 500$?
Since, $500 = 2^2 5^3$
I believe this can be solved using Ball and Urn
let $a = 2^{x_1}5^{y_1}$ and then the subscripts increase by $1$ for $b, c$.
There is:
$$x_1 + x_2 + x_3 = 2$$
$$y_1 + y_2 + y_3 = 3$$
We count each seperately.
There are 3 "bins," $x_1, x_2, x_3$.
Then what are the objects? Im confused?
| Expanding on robjohn's answer.
The 'bins' (or urns) are the factors $a, b, c$.
The number of 'balls' are counted by the exponents of $2$. Since $500= 2^25^3$ there are two 'balls' to distribute among the three bins ($a,b,c$) to determine what exponent of $2$ each might receive.
$$\begin{array}{lll}a & b & c \\ \hline
2^2 & 2^0 & 2^0 & \ast\ast \vert\vert
\\ 2^1 & 2^1 & 2^0 & \ast\vert\ast \vert
\\ 2^1 & 2^0 & 2^1 & \ast\vert\vert\ast
\\ 2^0 & 2^2 & 2^0 & \vert\ast\ast\vert
\\ 2^0 & 2^1 & 2^1 & \vert\ast\vert\ast
\\ 2^0 & 2^0 & 2^2 & \vert\vert\ast\ast
\end{array}\qquad \dbinom{2+2}{2,2}=\dfrac{4!}{2!2!} 6$$
And similarly we count the ways the exponents of $5$ may be assigned by counting the permutations of $3$ stars and $2$ bars.$$\dbinom{3+2}{3,2} = \dfrac{5!}{3!2!}=10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Question in Differential Equation. (Don't know how to proceed) By using the substitution $y=vx$, find the particular solution of the differential equation $$2xy\frac{dy}{dx}=y^2-x^2$$, given that y=4 when x=2. Express in terms of x.
My attempt,
$$y=xv$$
$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$
$$2x^2(x\frac{dv}{dx}+v)-x^2+x^2v^2$$
$$2x^2(x\frac{dv}{dx}+v)v=x^2(v^2-1)$$
$$\frac{dv}{dx}=\frac{-v^2-1}{2xv}$$
$$\frac{2\frac{dv}{dx}v}{-v^2-1}=\frac{1}{x}$$
$$\int \frac{2\frac{dv}{dx}v}{-v^2-1}=\int \frac{1}{x}dx$$
$$-ln (v^2+1)=\ln (x)+c_1$$
How do I simplify this?
| exponentiating $$\ln (v^2 + 1) = -\ln x + \ln C, $$ you get
$$ v^2 + 1 = \frac C x \to \frac{y^2 }{x^2} + 1=\frac C x \to y^2 = Cx - x^2\to y =\sqrt{Cx - x^2}$$ use the initial condition $y= 4, x = 2$ implies $4 = \sqrt{2C - 4}$ gives you $C = 10.$ therefore the solution is $$ y=\sqrt{10x - x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Related Trigonometric Rates A port and a radar station are $3 \text{ mi}$ apart on a straight shore running east and west. A ship leaves the port at noon traveling at a rate of $13 \frac{\text{mi}}{\text{hr}}$. Find the rate of change of the tracking angle $\theta$ between the shore and the line between the radar station and the ship at $12:30 \text{ PM}$, assuming the ship maintains its speed and course.
The following graphic represents this scenario:
$\hspace{1.95in}$
Let $\dfrac{ds}{dt} = 13 \frac{\text{mi}}{\text{hr}}$.
The ship is given to move from $A \to C$ at $\dfrac{ds}{dt}$ departing at noon and arriving at $12:30 \text{ PM}$ such that:
$$s = \dfrac{13}{2}$$
Let the segment $\overline{CB} = x$.
By the law of cosines:
$$x = \sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}$$
Then apply the law of sines to find that:
$$\dfrac{\sin{\left( \frac{\pi}{4} \right)}}{x} = \dfrac{\sin{\theta}}{s}$$
Through substitution and manipulation of the equality its found that:
$$\dfrac{s \cdot \sin{\left( \frac{\pi}{4} \right)}}{\sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}} = \sin{\theta}$$
Take the derivative with respect to time of both sides to find that:
$$
\\ \begin{align*}
\\ \frac{d}{dt}\left[ \arcsin \left( \dfrac{s \cdot \sin{\left( \frac{\pi}{4} \right)}}{\sqrt{s^2 + 3^2 - (2)(s)(3)\cos{\left( \frac{\pi}{4} \right)}}} \right) \right] &= \frac{d}{dt}\left[ \theta \right]
\\
\\ \dfrac{\dfrac{1}{\sqrt{2}\sqrt{s^2-3\sqrt{2}s+9}} - \dfrac{s^2}{\sqrt{2}{(s^2-3\sqrt{2}s+9)}^{3/2}}}{\sqrt{1-\dfrac{s^2}{2(s^2-3\sqrt{2}s+9)}}}\cdot\frac{ds}{dt} &= \frac{d\theta}{dt}
\\ \end{align*}
$$
Which Wolfram evaluates to be $-4.5193 \frac{\text{rad}}{\text{hr}}$ as seen here.
This doesn't appear to be correct, however.
I'm uncertain of the error I've made, so if you could explicitly state at what point I've made an error in my approach, that would be very helpful. All that I'm certain of is that this is not the expected answer, but my approach appears proper to me (though it is clearly not).
| let the distance between the ship and the radar station be $y.$ so we have $$y^2 = s^2 + 3^2 - 2 \times 3 \times s\cos 45^\circ = s^2 + 9 -3\sqrt 2s$$ differencing gives $$2y\frac{dy}{dt} = 2s\frac{ds}{st} - 3\sqrt 2 \frac{ds}{dt} \tag 1$$ now we put the data $$\frac{ds}{dt} = 13, s = 6.5, y = \sqrt{6.5^2 + 9 - 3\sqrt 2 \times 6.5}= 4.865$$
putting all these in $(1),$ we get $$2\times 4.865\frac{dy}{dt}=13\times 13-3\sqrt 2 \times 13=113.845\to \frac{dy}{dt} = 11.700\, \text{ mile/hour} $$
$\bf edit:$ if you need $\frac{d\theta}{dt},$ we can use $$\frac{s}{\sin \theta} = \frac{y}{\sin 45^\circ}\to\sin 45^\circ \frac{ds}{dt} = y\cos \theta \frac{d\theta}{dt} + \sin \theta \frac{dy}{dt} \tag 2$$ we will first compute $\theta.$
we have $$\cos \theta = \frac{-\frac{6.5}{\sqrt 2} + 3}{y} = -0.328, \sin \theta = \frac{6.5}{\sqrt 2 y}=0.944$$
now subbing $$13\frac{\sqrt 2}2=4.865\times (-0.328)\frac{d\theta}{dt}+0.944\times 11.700 $$
i get $$\frac{d\theta}{dt} = 1.165\, radian /hour = 66.787^\circ/hour$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of a ratio For a positive integer $n$, let $a_n, b_n, c_n, d_n $ be positive integers such that
$$\left(1+\sqrt 2+\sqrt 3\right)^n=a_n+b_n \sqrt 2+c_n\sqrt 3+d_n\sqrt 6$$
Then for large $ n$, find the limit of the expression
$$\frac{a_n^2+b_n^2+c_n^2+d_n^2}{(a_n+b_n+c_n+d_n)^2}$$ I tried some binomial theorem for the expansion, but no progress.
| Please check my simplification work to get $I$:
$$A=(1+\sqrt{2}+\sqrt{3})^n=a_{n}+b_{n}\sqrt{2}+c_{n}\sqrt{3}+d_{n}\sqrt{6}$$
$$B=(1-\sqrt{2}+\sqrt{3})^n=a_{n}-b_{n}\sqrt{2}+c_{n}\sqrt{3}-d_{n}\sqrt{6}$$
$$C=(1+\sqrt{2}-\sqrt{3})^n=a_{n}+b_{n}\sqrt{2}-c_{n}\sqrt{3}-d_{n}\sqrt{6}$$
$$D=(1-\sqrt{2}-\sqrt{3})^n=a_{n}-b_{n}\sqrt{2}-c_{n}\sqrt{3}+d_{n}\sqrt{6}$$
then we have
$$4a_{n}=A^n+B^n+C^n+D^n,$$
$$4\sqrt{2}b_{n}=A^n-B^n+C^n-D^n,$$
$$4\sqrt{3}c_{n}=A^n+B^n-C^n-D^n,$$
$$ \sqrt{6}d_{n}=A^n-B^n-C^n+D^n,$$
then it is easy to find
$$x=\lim_{n\to\infty}\dfrac{b_{n}}{a_{n}}=\lim_{n\to\infty}\dfrac{\dfrac{1}{4\sqrt{2}}(A^n-B^n+C^n-D^n)}{\dfrac{1}{4}(A^n+B^n+C^n+D^n}=\dfrac{\sqrt{2}}{2}$$
and similarly
$$y=\lim_{n\to\infty}\dfrac{c_{n}}{a_{n}}=\dfrac{1}{\sqrt{3}},z=\lim_{n\to\infty}\dfrac{d_{n}}{a_{n}}=\dfrac{1}{\sqrt{6}}$$
So
$$I=\lim_{n\to\infty}\dfrac{a^2_{n}+b^2_{n}+c^2_{n}+d^2_{n}}{(a_{n}+b_{n}+c_{n}+d_{n})^2}
=\dfrac{1+x^2+y^2+z^2}{(1+x+y+z)^2}=\dfrac{72}{(6+3\sqrt{2}+2\sqrt{3}+\sqrt{6})^2}$$
and note
$$6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}=3\sqrt{2}(1+\sqrt{2})+\sqrt{6}(1+\sqrt{2})=(1+\sqrt{2})(3\sqrt{2}+\sqrt{6})=\sqrt{6}(1+\sqrt{2})^2$$
so
$$I=\dfrac{12}{(1+\sqrt{2})^4}=12(\sqrt{2}-1)^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Continued fraction manipulation I have the following continued fraction
$$ \frac{1}{a_1x+}\;\;\frac{1}{b_1+}\;\;\frac{1}{a_2x+}\;\;\frac{1}{b_2} $$
The paper I am reading then converts this to the following continued z-fraction but does not show any work
$$ \frac{\frac{1}{a_1x}}{1+}\;\;\frac{\frac{1}{b_1a_1x}}{1+}\;\;\frac{\frac{1}{a_2x}}{1+}\;\;\frac{\frac{1}{a_2b_2x}}{1+0} $$
Is this relationship correct? When I try a variety of numbers, the relationship seems to hold for large values of $x$ but becomes increasingly inaccurate as $x$ becomes smaller ($x<1$).
Any help is greatly appreciated.
Thanks
| It works like this:
$$\frac{1}{a_1x+\frac{1}{b_1+\frac{1}{a_2x+\frac{1}{b_2}}}}=\frac{\frac{1}{a_1x}}{1+\frac{\frac{1}{a_1x}}{b_1+\frac{1}{a_2x+\frac{1}{b_2}}}}$$
because we multiply the top and bottom of the outermost fraction by $\frac{1}{a_1x}$. Now, isolate the fraction from the denominator above:
$$\frac{\frac{1}{a_1x}}{b_1+\frac{1}{a_2x+\frac{1}{b_2}}}=\frac{\frac{1}{b_1a_1x}}{1+\frac{\frac{1}{b_1}}{a_2x+\frac{1}{b_2}}}$$
because we multiply the top and bottom by $\frac{1}{b_1}$. Moving along, we're going to multiply the top and bottom of that fraction in the denominator we just looked at by $\frac{1}{a_2x}$:
$$\frac{\frac{1}{b_1}}{a_2x+\frac{1}{b_2}}=\frac{\frac{1}{b_1a_2x}}{1+\frac{\frac{1}{a_2x}}{b_2}}$$
Finally, we take the fraction in that last denominator, and multiply it, top and bottom, by $\frac{1}{b_2}$:
$$\frac{\frac{1}{a_2x}}{b_2}=\frac{\frac{1}{a_2b_2x}}{1}=\frac{\frac{1}{a_2b_2x}}{1+0}$$
In the notation from your question, this calculation is giving us:
$$ \frac{\frac{1}{a_1x}}{1+}\;\;\frac{\frac{1}{b_1a_1x}}{1+}\;\;\frac{\frac{1}{b_1a_2x}}{1+}\;\;\frac{\frac{1}{a_2b_2x}}{1+0} $$
which differs from what you posted in having a $b_1$ term in the denominator of the third numerator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\log (2-x)$ in powers of x Find $\log (2-x)$ in powers of $x$.
I know this is how I've gotten so fare,
$$\frac{1}{1-x}=\sum _{n=0}^{\infty } x^n$$
given that $\left| x\right| <1$.
Now
$$\begin{align}\frac{1}{2-x}&=\frac{1}{2} \sum _{n=0}^{\infty } \left(\frac{x}{2}\right)^n \end{align}$$
I therefore try integrating
$$\begin{align} \int \frac{1}{2-x} \, dx&=-\log(2 - x)\\
\int \frac{1}{2} \left(\frac{x}{2}\right)^n dx&=
\frac{2^{-n-1} x^{n+1}}{n+1}\end{align}$$
Now multiplying both sides with $-1$ i still not get the answer which is
$$\begin{align}\ln(2-x)&=\ln(2)-\sum_{n=1}^{+\infty}\frac{x^n}{2^nn} & -2\leq x<2\end{align}$$
Could someone please show me the steps?
| We have that $$\frac{1}{1-x}=\sum _{n=0}^{\infty } x^n$$
given that $\left| x\right| <1$.
Since $$\dfrac{d}{dx} \log(2-x) =-\dfrac{1}{2-x} $$
we have to transform $-\dfrac{1}{2-x}$ somehow into $\dfrac{1}{1-x}$.
$$-\dfrac{1}{2-x}=-\dfrac{1}{2(1-\frac{x}{2})}=-\dfrac{1}{2}\cdot \dfrac{1}{1-\frac{x}{2}}= -\dfrac{1}{2}\cdot \sum _{n=0}^{\infty } \left(\frac{x}{2}\right)^n$$
Because the series $$\sum _{n=0}^{\infty } \left(\frac{x}{2}\right)^n$$ is uniformly convergent for $|\frac{x}{2}|<1 \Leftrightarrow x\in (-2,2)$ we can integrate the following
$$-\dfrac{1}{2-x}=-\dfrac{1}{2}\cdot \sum _{n=0}^{\infty } \left(\frac{x}{2}\right)^n$$
and we'll get that
$$ \int_0^x \left(-\dfrac{1}{2-t}\right)dt = -\dfrac{1}{2}\cdot \sum _{n=0}^{\infty }\int_0^x \left(\frac{t}{2}\right)^n dt $$
After the integration we get:
$$\log(2-x)-\log2=-\sum _{n=0}^{\infty } \dfrac{x^{n+1}}{(n+1)\cdot 2^{n+1}}$$
Let's say that $k=n+1$ then we have that:
$$\log(2-x)=\log2-\sum _{k=1}^{\infty } \dfrac{x^{k}}{k\cdot 2^{k}}$$
Or
$$\log(2-x)=\log2-\sum _{n=1}^{\infty } \dfrac{x^{n}}{n\cdot 2^{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplifying An Inverse Tan Function I would like to know how this equality holds.
$$
\tan^{-1} \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)} = \tan^{-1} \frac{1}{2n-1} - \tan^{-1} \frac{1}{2n+1}.$$
I was told to use the double angle formula for $\tan \theta$ but I can't seem to show this.
Thank you.
For some context, I was asked to show
$$
\sum_{n=1}^\infty \tan^{-1} \frac{1}{2n^2} = \frac{\pi}{4}
$$
| Observe that $\tan(x-y)=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}$. Then, letting $x=\arctan(\frac{1}{2n+1})$ and $y=\arctan(\frac{1}{2n-1})$, we see that
$$\begin{align}
\tan(x-y)&=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\\
&=\frac{\frac{1}{2n+1}-\frac{1}{2n-1}}{1+\frac{1}{2n+1}\frac{1}{2n-1}}\\
&=\frac{(2n-1)-(2n+1)}{1+(2n-1)(2n+1)}
\end{align}$$
Taking the inverse tangent of both sides of this last expression reveals
$$\begin{align}
x-y&=\arctan\left(\frac{1}{2n+1}\right)-\arctan\left(\frac{1}{2n-1}\right)\\
&=\arctan\left(\frac{(2n-1)-(2n+1)}{1+(2n-1)(2n+1)}\right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
The bisector of $\angle BAC$ of triangle $\Delta ABC$ cuts $BC$ at $D$ The bisector of $\angle BAC$ oF triangle $\Delta ABC$ cuts $BC$ at $D$ and circumcircle of triangle at $E$. if
$$AD=5 \text{ cm} ,\ DE=3 \text{ cm},\ AC=4 \text{ cm}, $$
then what is the length of $AB$?
|
Let $\angle ACB = \phi$. We have
\begin{align*}
\angle EBC &= \theta\\
\angle ABE &= 180^\circ -\theta - \phi
\end{align*}
Since the triangles $ABE$ and $ABC$ have the same circum circle, we have
\begin{align}
\frac{8}{\sin(180^\circ - \theta - \phi)} = 2R = \frac{c}{\sin \phi} \label{eq1}
\end{align}
Also, in triangle $ACD$, $\angle ADC = 180^\circ - \theta - \phi$. Thus,
\begin{align}
\frac{5}{\sin \phi} = \frac{4}{\sin(180^\circ - \theta - \phi)} \label{eq2}
\end{align}
From the above two equations, we get
\begin{align}
\frac{8}{c} = \frac{4}{5}
\end{align}
and hence $c=10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Infinitely many $n$ such that $n^2+1$ is squarefree How can we show that there are infinitely many positive integers $n$ such that $n^2+1$ is square free? I was thinking we could split it into cases of squares of primes congruent to either 1 or 3 (mod 4).
Even stronger: could we prove that the set of numbers positive integers $m$ such that $m^2+1$ is not squarefree is a set of measure zero?
EDIT: could I get an elementary solution to the first question?
| Let $S(n)$ be the number of squareful numbers of the form $k^2+1$ less than or equal to $n^2+1$. We can bound it from above by counting every time a number is divisible by the square of a prime:
$S(n) \le \sum_{1 \le p \le n}{\sum_{1 \le k \le n, p^2 \vert k^2+1}{1}}$
Since $k^2+1$ is never divisible by $4$ or $9$, this becomes:
$S(n) \le \sum_{5 \le p \le n}{\sum_{1 \le k \le n, p^2 \vert k^2+1}{1}}$
$-1$ has at most two square roots mod $p^2$, so we can write this as:
$S(n) \le \sum_{5 \le p \le n}{(\frac{2 \cdot n}{p^2}+2)}$
And we can separate the constant term to get:
$S(n) \le 2 \cdot \pi(n) + \sum_{5 \le p \le n}{\frac{2 \cdot n}{p^2}}$
Assuming $n \gt 120$:
$S(n) \le \frac{2 \cdot \phi(120) \cdot n}{120} + 2 \cdot n \cdot \sum_{5 \le p \le n}{\frac{1}{p^2}}$
Simplifying and ignoring that the sum is limited to primes:
$S(n) \le \frac{8 \cdot n}{15} + 2 \cdot n \cdot (\frac{\pi^2}{6} - \frac{1}{16} - \frac{1}{9} - \frac{1}{4} - 1)$
Calculating and rounding up:
$S(n) \le 0.54 \cdot n + 0.45 \cdot n$
$S(n) \le 0.99 \cdot n$
So the density of squarefree numbers of the form $k^2+1$ is at least $0.01$, and therefore there are an infinite number of them. I really wasn't expecting to end up with such a narrow margin! It is possible to do better with some obvious improvements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
combinatorics summation problem My problem is following:
$$\binom{n}{r} + \binom{n+1}{r+1} + \binom{n+2}{r+2} + \dots + \binom{n+M}{r+M}$$
how can we reduce it to a more short solution
Here $\dbinom{n}{r} = \dfrac{n!}{r! (n-r)!}$ and thus same as regular.Please help me in solving the above expression
| This question can be treated using basic complex variables, which is
an instructive exercise.
Suppose we seek to compute
$$\sum_{q=0}^M {n+q\choose r+q}$$
with $n\ge r.$
Introduce the integral representation
$${n+q\choose r+q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+q}}{z^{r+q+1}} \; dz.$$
This gives for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r+1}}
\sum_{q=0}^M \frac{(1+z)^q}{z^q}\; dz.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r+1}}
\frac{(1+z)^{M+1}/z^{M+1}-1}{(1+z)/z-1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r}}
\frac{(1+z)^{M+1}/z^{M+1}-1}{(1+z)-z} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r}}
\left((1+z)^{M+1}/z^{M+1}-1\right) \; dz.$$
This has two pieces, the first is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+M+1}}{z^{r+M+1}} \; dz
= {n+M+1\choose r+M}$$
and the second is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r}} \; dz
= {n\choose r-1}.$$
Combine these two to get
$${n+M+1\choose r+M} - {n\choose r-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1218912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
The maximum value of expression $ \sqrt{\sin^2x+ 2a^2} - \sqrt{-1 -\cos^2x+ 2a^2} $ If $a,x\in\Bbb R$, what is the maximum value of the expression
$ \sqrt{\sin^2x+ 2a^2} - \sqrt{-1 -\cos^2x+ 2a^2} $?
I tried to differentiate but it became messy.
| Let $$ f(x) = \sqrt{\sin^2x+ 2a^2} - \sqrt{2a^2-1-\cos^2 x} $$
Now for $\max$ of $f(x)\;,$ Put $$2a^2-1-\cos^2 x=0\Rightarrow 2a^2=1+\cos^2 x$$
So $$f_{\max} = \sqrt{\sin^2 x+1+\cos^2 x} = \sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the integral $\int \frac{\left(\sin\left(2 \:x\right)\right)^2}{\left(\sin^3x+\cos^3x\right)^2}$ That's decided a similar example. This I do not know how to solve. Help me please.
My integral:
$$\int \frac{\left(sin\left(2 \:x\right)\right)^2}{\left(sin^3x+cos^3x\right)^2}$$
| $\displaystyle\int\frac{(\sin 2x)^2}{(\sin^{3}x+\cos^{3}x)^2}dx=\int\frac{4\sin^2x\cos^2x}{(\sin^{3}x+\cos^{3}x)^2}dx=\int\frac{4\tan^2x\sec^2x}{(\tan^3x+1)^2}dx$
$\hspace{1.8 in}$(after dividing by $\cos^6x$ on the top and bottom).
Now let $u=\tan^3x+1, du=3\tan^2x\sec^2x\;dx$ to get
$\displaystyle\frac{4}{3}\int\frac{1}{u^2}\;du=-\frac{4}{3}\left(\tan^3x+1\right)^{-1}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$f \colon \mathbb{R} \rightarrow \mathbb{R}$, $f$ is continuous , and $f(x+1)+f(x)=x^2$ I would appreciate if somebody could help me with the following problem:
Find $f(x)$ ($f(x)$ is not Polynomial function), given that:
$f \colon \mathbb{R} \rightarrow \mathbb{R}$, $f$ is continuous, and $f(x+1)+f(x)=x^2$
I tried but couldn't get it that way.
| It might be helpful to first find $f:\Bbb N \to \Bbb R$, then extend the result.
We can solve the recurrence relation $f(n) + f(n+1) = n^2$ via generating functions: define $F(x) = \sum_{n=0}^\infty f(n) x^n$. Then
$$
f(n) + f(n+1) = n^2\\
\sum_{n=0}^\infty [f(n) + f(n+1)]x^n = \sum_{n=0}^\infty n^2 x^n \\
\sum_{n=0}^\infty f(n)x^n +
\sum_{n=0}^\infty f(n+1)x^n =
\sum_{n=0}^\infty n^2 x^n\\
\sum_{n=0}^\infty f(n)x^n +
\sum_{n=1}^\infty f(n)x^{n-1} =
\sum_{n=0}^\infty n^2 x^n\\
F(x) +
\frac 1x \left(F(x) - f(0)\right) =
\frac{x(x+1)}{(x-1)^3}\\
xF(x) +
F(x) - f(0) =
\frac{x^2(x+1)}{(x-1)^3}\\
(x+1)F(x) =
\frac{x^2(x+1)}{(x-1)^3} + f(0)\\
F(x) =
\frac{x^2}{(x-1)^3} + \frac{f(0)}{x+1}\\
\sum_{n=0}^\infty f(n) x^n =
\sum_{n=0}^\infty \frac{n(n+1)}{2} x^n +
f(0)\sum_{n=0}^\infty (-1)^n x^n
$$
So setting $f(0) = a$ for any $a$, we find
$$
f(n) = \frac{n(n+1)}{2} + a(-1)^n =
\frac{n(n+1)}{2} + a\cos(\pi n)
$$
if a solution to the problem over $\Bbb R$ exists, it must be an extension of this function for some value of $a$.
In fact, $f(x) = \frac{x(x+1)}{2} + a\cos(\pi x)$ seems to work for any $a \in \Bbb R$.
A more general solution (taken from Salomo's work) is
$$
f(x) = \frac{x(x+1)}{2} + g(x)
$$
where $g(x)$ satisfies $g(x+1) + g(x) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
An Eigen Value of $\tiny \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$ is : Let $a,b,c,d$ be distinct non zero real numbers with $a+b=c+d.$ Then, an eigen value of the matrix $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$ is :
$(i)~a+c ~~(ii)~a+b~~(iii)~a-b~~(iv)~(b-d$
Attempt:
If $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$, then :
$A - \lambda I = \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix}$
$\det ( A- \lambda I) = \det \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix} $
Solving $\det ( A - \lambda I) = 0$ results in a third degree equation which is difficult to solve.
Could anyone tell me of an easier method to solve it. Thank you very much for your help in this regard.
| $a+b=c+d$ implies that we should add up the first two columns of the matrix using a vector of the form $\begin{pmatrix}
1\\
1\\
t
\end{pmatrix}$.
You can see that if $A$ is the matrix, then $A\begin{pmatrix}
1\\
1\\
t
\end{pmatrix}$=$\begin{pmatrix}
a+b+t\\
c+d+t\\
0
\end{pmatrix}$.
Therefore, the suitable choice is of course $t=0$, and so the eigenvalue is $a+b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
What is $\tan \alpha$ if $\sin \alpha + \cos \alpha = \frac{\sqrt{3}-1}{2}$ and $\alpha \in (90^\circ,135^\circ)$ I am stuck with this problem. Any ideas on how to solve this?
| The addition theorem for the sine function gives:
$$
\sin(\alpha + \pi/4) = \sin(\alpha) \cos(\pi/4) + \cos(\alpha) \sin(\pi/4)
$$
The triangle
$$
A=(0,0), B=(1,0), C=(1,1)
$$
gives:
$$
\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}
$$
So we have
$$
\frac{\sqrt{3}-1}{2} = \sin(\alpha) + \cos(\alpha) = \sqrt{2} \sin(\alpha + \pi/4)
$$
Inverting gives
\begin{align}
\alpha &= 2\pi k + \arcsin\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right) - \pi/4 \vee \\
\alpha &= 2\pi k - \arcsin\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right) - (\pi/4 + \pi)
\end{align}
Using a computer algebra system, one finds
$$
\arcsin\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right) = \pi/12
$$
which gives
\begin{align}
\alpha &= 2\pi k + \pi/12 - \pi/4 = 2\pi k - \pi/6 = k\cdot 360^\circ - 30^\circ\vee \\
\alpha &= 2\pi k - \pi/12 - (\pi/4 + \pi) = 2\pi k - 4\pi/3 = k\cdot 360^\circ - 240^\circ
\end{align}
and then $\alpha = 120^\circ \in (90^\circ, 135^\circ)$.
which then gives $\tan 120^\circ = -\sqrt{3}$.
$$
\tan(120^\circ)
= \frac{\sin(120^\circ)}{\cos(120^\circ)}
= -\frac{\cos(30^\circ)}{\sin(30^\circ)} = -\frac{\sqrt{3}/2}{1/2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
Can this be shown: $\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\dots}}} = \sqrt a$? $$\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}=\sqrt{a}$$
Just for fun. I would like to read the proof of this if it exists. Any references would be appreciated.
| Let $\displaystyle b=\sqrt[3]{a\cdot\ \underbrace{\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\sqrt[3]{a\cdots}}}}}}}}}_{\text{This is $b$, which is allegedly $\sqrt a$.}}}$.
Then $\displaystyle b = \sqrt[3]{ab}$, so $b^3 = ab$, and then $b^2 = a$. Thus $b = \sqrt a$.
If we put $\sqrt{a}$ in place of the expression said to be $b=\sqrt{a}$, we get $\sqrt[3]{a\sqrt{a}}$. And it is easy to see that that is indeed $\sqrt a$.
As to convergence: let $g(x) = (ax)^{1/3}$. The question is the behavior of the sequence
$$
a,\ g(a),\ g(g(a)),\ g(g(g(a))),\ \ldots\ .
$$
For $x$ between $\sqrt{a}$ and $a$, we have $0<g'(x)<1/3$, so $g$ is a contraction and thus has a unique attractive fixed point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
What did I do wrong trying to find this limit? In another question, a user asked to find:
$$\lim_{x\to 0} \frac{\exp(x^2)-\cos(x)}{\sin(x)^2}$$
I thought I could use pure trigonometric identities to find the limit. Apparently I was mistaken, but I can't find my mistake. Here's what I did:
Knowing $\sin(x)^2 = \frac{1-\cos(2x)}{2}$:
$$\frac{\exp(x^2)-\cos(x)}{\sin(x)^2} = \frac{\exp(x^2)-\cos(x)}{\frac{1-\cos(2x)}{2}} = \frac{\exp(x^2)-\cos(x)}{1} \cdot \frac{2}{1-\cos(2x)}$$
$$ = \frac{2(\exp(x^2)-\cos(x))}{1-\cos(2x)}$$
And knowing $\cos(2x) = 2\cos(x)^2-1$:
$$\frac{2(\exp(x^2)-\cos(x))}{1-\cos(2x)} = \frac{2(\exp(x^2)-\cos(x))}{1-(2\cos(x)^2-1)} = \frac{2(\exp(x^2)-\cos(x))}{-2(\cos(x)^2)}$$
$$ = -1 \cdot \frac{\exp(x^2)-\cos(x)}{\cos(x)^2} = -1 \cdot \left( \frac{\exp(x^2)}{cos(x)^2} -\frac{\cos(x)}{\cos(x)^2} \right)
= -1 \cdot \left( \frac{\exp(x^2)}{cos(x)^2} -\frac{1}{cos(x)} \right) $$
And we can plug in $x=0$ since we don't run the risk of an indeterminate form:
$$ -1 \cdot \left( \frac{\exp(0^2)}{cos(0)^2} -\frac{1}{cos(0)} \right) = -1 \cdot \left( \frac{1}{1^2} -\frac{1}{1} \right) = -1 \cdot 0 = 0$$
Apparently the correct answer is $3/2$. Where did I go wrong?
| The wrong step is $1-(2\cos^2x-1)=2-2\cos^2x\neq -2\cos^2x$.
Notice that you are simply itroducing double angle, then removing it, so it's equivalent to use the identity $\cos^2x +\sin^2x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim\limits_{x \to 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}}$ How do I evaluate
$$\lim_{x \to 0}\left(\frac{\sin x}{x}\right)^{\dfrac{1}{1-\cos x}}\ ?$$
I tried using the fact that $\left(\frac{\sin x}{x}\right)^{\frac{1}{1-\cos x}} = \exp\left(\ln\bigg(\frac{\sin x}{x}\right)\frac{1}{1-\cos x}\bigg)$ and then I am already stuck.
| One may recall that, as $x \to 0$, we have
$$
\cos x =1-\frac {x^2}{2}+\mathcal{O}(x^3), \quad
\sin x =x-\frac{x^3}{6}+\mathcal{O}(x^4), \quad
\ln (1+x)=x+\mathcal{O}(x^2),
$$ giving
$$
\begin{align}
\frac{1}{1-\cos x}&=\frac 2{x^2}+\mathcal{O}(1)\\\\
\ln \left(\frac{\sin x}x\right)& =-\frac{x^2}{6}+\mathcal{O}(x^3).
\end{align}
$$ Then, as $x \to 0$,
$$
\begin{align}
\left(\frac{\sin x}x\right)^{\small \dfrac{1}{1-\cos x}}&=e^{\Large \frac{1}{1-\cos x}\ln \left(\frac{\sin x}x\right)}\\\\
& =e^{\large \left(\frac 2{x^2}+\mathcal{O}(1)\right)\left(-\frac{x^2}{6}+\mathcal{O}(x^3)\right)}\\\\
& =e^{\Large -\frac 13+\mathcal{O}(x)}
\end{align}
$$ and the desired limit is $\displaystyle e^{-\large \frac 13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Proving that for Fibonacci numbers $a_n \lt (\frac {1 + \sqrt 5} 2)^n$ for $n \ge 1$ I'd like to prove that for Fibonacci numbers $a_n \lt \left(\frac {1 + \sqrt 5} 2\right)^n$ for $n \ge 1$. I suppose it needs induction so, after verifying the trivial case $n=1$, the inductive step needs to be justified: $$a_{n+1} \lt \left(\frac {1 + \sqrt 5} 2\right)^{n+1}.$$
Assume $$a_n \lt \left(\frac {1 + \sqrt 5} 2\right)^n.$$
I can only think of multiplying both sides by $\left(\frac {1 + \sqrt 5} 2\right)$, which yields $$a_n\left(\frac {1 + \sqrt 5} 2\right) \lt \left(\frac {1 + \sqrt 5} 2\right)^{n+1}$$
and then somehow show that $a_{n+1} \lt a_n\left(\frac {1 + \sqrt 5} 2\right)$.
I tried using the fact that $a_{n+1}=a_n + a_{n-1}$ but wasn't able to conclude anything useful. Can you suggest me a way to prove it?
| Let $a_n=F_n$ and $b_n=\left(\frac{1+\sqrt{5}}{2}\right)^n$. Since $\frac{1+\sqrt{5}}{2}$ is a root of the polynomial $t^2-t-1$, we have:
$$ a_{n+2} = a_{n+1}+a_{n}\quad\text{as well as}\quad b_{n+2}=b_{n+1}+b_n, \tag{1}$$
hence in order to prove that
$$ a_n < b_n \tag{2}$$
holds for every $n\geq 1$ it is enough to check that $(2)$ holds for $n=1,2,$ then apply induction exploiting $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $\triangle ABC\;,\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^0,$ Then find ordered pairs $\left(\angle A,\angle B\right)$
In a $\displaystyle \triangle ABC\;,\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^0,$ Then find the ordered pairs $\left(\angle A,\angle B\right)=$
$\bf{Options}::$ $(i)\; \left(45^0,75^0\right)\;\;\;\; (ii)\; \left(105^0,15^0\right)\;\;\;\; (ii) \; \left(15^0,105^0\right)\;\;\;\; (ii)\;\;\;\; (75^0\;,45^0)$
$\bf{My\; Solution::}$ Using Cosine Rule, $$\displaystyle \cos (60^0)=\frac{a^2+b^2-c^2}{2ab}\Rightarrow a^2+b^2-c^2=ab.$$
Now Divide both side by $b^2\;,$ We get $$\displaystyle \left(\frac{a}{b}\right)^2+1-\left(\frac{c}{b}\right)^2=\left(\frac{a}{b}\right)\;,$$ Where $b\neq 0$
Now put value of $$\displaystyle \frac{a}{b}=2+\sqrt{3}\;,$$ We get $$\displaystyle 7+4\sqrt{3}+1-\left(\frac{c}{b}\right)^2=2+\sqrt{3}.$$
So we get $$\displaystyle \left(\frac{c}{b}\right)^2=6+3\sqrt{3}\Rightarrow \frac{c}{b}=\sqrt{6+3\sqrt{3}}$$
Now How can i solve it, Help required, Thanks
| Start with Sine Law since you are given the quotient of sides.
Make a rough sketch of segment containing an angle $60$, to note that $A$ must be > 90 an obtuse angle anyhow, B is noted to be a narrow angle, should be an acute angle.
Arguments in degrees to avoid Latex. (I prefer such practice only when there is no degree/radian conflict anyway).
$\dfrac{\sin A}{\sin (A + 60)} = \dfrac{2 + \sqrt 3}{1}$, recognize the right hand side as $ \tan 75 . $
So, using sine supplementary angle identiity $ \sin\theta = \sin ( 180 - \theta) $:
$$\dfrac{\sin A}{\sin (A + 60)} = \dfrac{\sin 75}{\cos 75} = \dfrac{\sin 75}{\sin 15} = \dfrac{\sin 105}{ \sin 165} = \dfrac{\sin (105)}{\sin ( 105 + 60) }$$
Comparing arguments of first and last fractions we straightaway note that $ A = 105 $, $$ (A,B,C)= (105,15,60).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proof by Induction $3^n > n^3$ I am trying to prove the following, however I'm stuck at the Induction hypothesis
Prove by induction that, for all integers $n$, if $n\geq 5$, then $3^n>n^3$
What I have Done:
Base Case:
$n = 5$
$3^5 > 5^3$
$243 > 125$ so TRUE
Assume True for $n = k$
$3^k > k^3$ must be TRUE
Should be TRUE for $n = k+1$
$3^k + 3^{k+1} > k^3 + 3^{k+1}$
Im a little stuck here. Any help would be appreciated!
| By induction, you assume that for $ n = k,
3^k > k^3$ must be true. We want to show $ 3^{k+1} > (k+1)^3$.
Thus,
$3^{k+1} = 3^k\cdot 3 >3\cdot k^3$, by your induction hypothesis.
So we want to show $ 3^{k+1} > (k+1)^3$, but have $3^{k+1} > 3k^3$.
So just show $3k^3 > (k+1)^3$ for $k>5$. We have
$(k+1)^3 = k^3+3k^2+3k +1 < k^3+k^3+k^3 = 3k^3$.
Therefore,$3^{k+1} > 3k^3 > (k+1)^3 \Longrightarrow 3^{k+1} > (k+1)^3,$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Something wrong at $\int \frac{x^2}{x^2+2x+1}dx$ I have to calculate $$\int \frac{x^2}{x^2+2x+1}dx$$ and I obtain: $$\int \frac{x^2}{x^2+2x+1}dx=\frac{-x^2}{x+1}+2\left(x-\log\left(x+1\right)\right)$$
but I verify on wolfram and this is equal with: $$x-\frac{1}{x+1}-2\log\left(x+1\right)$$ where did I go wrong?
P.S: Here is all steps: $$\int \:\frac{x^2}{x^2+2x+1}dx=-\frac{x^2}{x+1}+\int \:\frac{2x}{x+1}dx=-\frac{x^2}{x+1}+2\left(x-\log\left(x+1\right)\right)$$ I am sure that is not wrong, but the form on wolfram seems easily then it.
| Your answer is equivalent, since
$$\frac{-x^2}{x+1} + 2x = \frac{2x^2 + 2x-x^2}{x+1} = \frac{x^2 + 2x + 1 - 1}{x+1} = x - \frac{1}{x+1} +1$$
and it differs by a constant $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
"$111 \dots$ upto $3^n$ digits" is divisible by $3^n$
Prove that an integer of the form "$111 \dots$ upto $3^n$ digits" is divisible by $3^n$
My attempt
For $n=1,$ $111$ is divisible by 3.
Let $T_n=111...$ upto $3^n$ digits is divisible by $3^n$.
Then $T_{n+1}=111...$ upto $3^{n+1}$ digits. I am unable to prove that Then $T_{n+1}$ is divisible by $3^n$.
| step 1. Write the number K = 111... in decimal form as 1*100 + 1*101 + 1*102 +... = $\sum_{i=0}^m10^i$
step 2. Examine K mod $3^n$ = $\sum_{i=0}^m10^i$ mod $3^n$ = $\frac{1-10^{m+1}}{1-10}$ mod $3^n$ = $\frac{10^{m+1}-1}{9}$ mod $3^n$ = $\frac{10^{3^n+1} -1}{9}$ mod $3^n$. Where the first equality follows from the formula for the sum of a finite geometric series. Clearly, 9|$10^{3^n+1} -1$ is necessary, so compute K' = $10^{3^n+1} -1$ mod (9*$3^n$)= $10^{3^n+1} -1$ mod $3^{n+2}$. If K' = 0, we're done.
step 3.
$\phi$($3^{n+2}$) = $3^{n+1}$-$3^n$ = a
Thus, by Euler's formula, $10^a$ = 1 mod $3^{n+2}$. Therefore, K' + 1 = $(10^a)^{3^n}$= 1 mod $3^{n+2}$.
So, $10^{3^n +1}$ -1 mod $3^{n+2}$ = 0.
The End
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1240655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find spectrum for matrix $A$ Let $A = \left[ \begin{array}{*{20}{c}}
0&b&0&0&0&0\\
c&0&b&0&0&0\\
0&c&0&b&0&0\\
0&0&c&0&b&0\\
0&0&0&c&0&b\\
0&0&0&0&c&0
\end{array} \right]_{n \times n}$.
Why does matrix $A$ have spectrum $\sigma (A) = \left\{ 2\sqrt {bc} \cos \left(\frac{\pi k}{n + 1}\right):k = 1,2,\ldots,n \right\}$?
| Let $\lambda$ be an eigenvalue of $A$ and $\mathbf{v}$ be the respective eigenvector. If you expand $(A - \lambda I)\mathbf{v} = \mathbf{0}$, you'll notice that the components $v_k$ ($k = 1, \ldots ,n$) of $\mathbf{v}$ satisfy the second order linear difference equation
$$\tag{1} cv_{k-1} - \lambda v_k + v_{k+1}= 0, \; k = 1, \ldots ,n,$$
with $v_0=v_{n+1}=0$.
We can write $(1)$ as
$$\tag{2} v_{k+2} - \frac{\lambda}{b} v_{k+1} + \frac{c}{b} v_{k}= 0, \; k = 0, \ldots ,n-1,$$
again with $v_0=v_{n+1}=0$.
In order to find the general solution of $(2)$, in the same spirit of one of the techniques we can use to find the general solution of a second order ordinary differential equation, we seek solutions of the form $v_k = \xi r^k$, for some constants $\xi$ and $r$. Substituting in $(2)$, we arrive at the characteristic equation
$$\tag{3} r^2 - \frac{\lambda}{b} r + \frac{c}{b}= 0, $$
which, being a quadratic equation, will have two roots, say $r_1$ and $r_2$.
It can be proven that the general solution of $(2)$ is then given by
$$ v_k =
\left\{
\begin{array}{ll}
\alpha (r_1)^k + \beta (r_2)^k &, \text{ if } r_1 \neq r_2, \\
\alpha \rho^k + \beta k \rho^k &, \text{ if } r_1 = r_2 = \rho,
\end{array}
\right.
$$
where $\alpha$ and $\beta$ are arbitrary constants.
(If you want to learn a general method of solving linear difference equations of second or higher order, you can read this answer.)
However, in our particular case, we must have $r_1 \neq r_2$, since otherwise $v_k = \alpha \rho^k + \beta k \rho^k$ and $v_0=v_{n+1}=0$ would imply that $v_k = 0$, for $k = 1, \ldots ,n$, and hence $\mathbf{v} = \mathbf{0}$, so $\mathbf{v}$ wouldn't be an eigenvector.
Thus, we know that $v_k = \alpha (r_1)^k + \beta (r_2)^k$ and $v_0=v_{n+1}=0$, which yields
$$
\left\{
\begin{array}{l}
\alpha + \beta = 0 \\
\alpha (r_1)^{n+1} + \beta (r_2)^{n+1} = 0
\end{array}
\right.,
$$
and this implies
$$\tag{4} \left( \frac{r_1}{r_2}\right)^{n+1} = 1 \Rightarrow \frac{r_1}{r_2} = \exp \left( i \frac{2 \pi j}{n+1} \right) \Rightarrow r_1 = r_2 \exp \left( i \frac{2 \pi j}{n+1} \right) ,$$
for some $1 \leq j \leq n$.
Now, applying the Viète's formulas to $(3)$, we also have
$$\tag{5}
\left\{
\begin{array}{l}
r_1 r_2 = c/b\\
r_1 + r_2 = \lambda / b
\end{array}
\right..
$$
Plugging $(4)$ in the first equation of $(5)$ we get
$$\tag{6} (r_2)^2 \exp \left( 2 \frac{i \pi j}{n+1} \right) = c/b \Rightarrow r_2 = \sqrt{c/b} \exp \left( - i \frac{\pi j}{n+1} \right),$$
and
$$\tag{7} r_1 = \sqrt{c/b} \exp \left( i \frac{\pi j}{n+1} \right).$$
Finally, by plugging $(6)$ and $(7)$ in the second equation of $(5)$, we arrive at
$$\lambda = b (r_1 + r_2) = b \sqrt{c/b} \underbrace{\left\{ \exp \left( i \frac{\pi j}{n+1} \right) + \exp \left( - i \frac{\pi j}{n+1} \right) \right\}}_{= 2 \cos \left( \frac{\pi j}{n+1} \right)} = 2 \sqrt{bc} \cos \left( \frac{\pi j}{n+1} \right).$$
Therefore, the eigenvalues of $A$ are given by
$$\lambda_j = 2 \sqrt{bc} \cos \left( \frac{\pi j}{n+1} \right), \; j=1,\ldots,n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $ \sum\limits_{k=1}^{n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$ using induction? This is as far as I get. I get stuck here because both sides to not equal each other, but I am not sure what I am doing wrong.
$$ \sum\limits_{k=1}^{n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$$
Assume: $ \sum\limits_{k=1}^{c}\frac{k}{(k+1)!} = 1 – 1/(c+1)!$
Prove: $ \sum\limits_{k=1}^{c+1}\frac{k}{(k+1)!} = 1 – 1/((c+1)+1)!$
$$\sum_{k=1}^c \frac{k}{(k+1)!} + \frac{c+1}{((c+1)+1)!} \stackrel{?}{=} 1- \frac{1}{((c+1)+1)!}$$
$$1 – \frac{1}{(c+1)!} + \frac{(c+1)}{((c+1)+1)!} \stackrel{?}{=} 1 – \frac{1}{((c+1)+1)!}$$
| Make a common denominator on your left hand side (noting that $(c+1)+1 = c+2$ and that $(c+2)! = (c+2)(c+1)!$):
\begin{eqnarray}
1 - \frac{1}{(c+1)!} + \frac{c+1}{(c+2)!} &=& 1 - \frac{c+2}{(c+1)!(c+2)} + \frac{c+1}{(c+2)!} \\
&=& 1 - \frac{c+2}{(c+2)!} + \frac{c+1}{(c+2)!}
\end{eqnarray}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1244335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Covering curved 1/8 of a circle A circle of radius 1 is given, and 8 semicircles of radius 1/2, like in this picture:
What is the radius of the smallest circle that can cover shaded area?
There was another problem involving the same picture, and this problem occurred to me while working on it. I find the formulation of this problem simple, yet it seems difficult. I used software simulation to find the desired radius, however, this does not count, a pure math solution is needed.
| Such a region is enclosed by the circular arcs defined by the parametric equations $$A: (x,y) = \left( \tfrac{1}{2} (1 + \cos \theta), \tfrac{1}{2} \sin \theta \right), \quad 0 \le \theta \le \pi,$$ $$B : (x,y) = (\cos \theta, \sin \theta), \quad -\frac{\pi}{4} \le \theta \le 0,$$ $$C : (x,y) = \left( \tfrac{1}{2}\left(\cos \tfrac{\pi}{4} + \cos \theta \right), \tfrac{1}{2} \left(-\sin \tfrac{\pi}{4} + \sin \theta \right) \right), \quad -\frac{\pi}{4} \le \theta \le \frac{3\pi}{4}.$$ It should be visually obvious that the maximum diameter of this shape must be measured from the point $(x,y) = \left(\cos \tfrac{-\pi}{4}, \sin \tfrac{-\pi}{4} \right)$, and the other endpoint must lie on the arc $A$. So it is simply a calculus problem: For a given $x$, locate $y = f(x) = \sqrt{x-x^2}$ such that $$P^2 = \left(x - \tfrac{1}{\sqrt{2}}\right)^2 + \left(y + \tfrac{1}{\sqrt{2}}\right)^2$$ is maximized. It is a simple but computationally tedious matter to compute the derivative with respect to $x$ and obtain the point $$(x,y)_{\text max} = \left( \frac{1}{2}\left(1 - \sqrt{\frac{7 - 4\sqrt{2}}{17}}\right), \sqrt{\frac{5 + 2\sqrt{2}}{34}} \right).$$ This corresponds to a distance of $$P = \frac{1}{4}(1 + \sqrt{5 - 2\sqrt{2}}).$$ Thus a lower bound for the disk of minimal radius that will enclose this shape is $P/2 \approx 0.618406$. Indeed, this lower bound is attainable, since the resulting disk has curvature strictly less than the curvature of the arc $A$, and it is not hard (but again, a bit tedious) to show that the disk does in fact cover.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Intriguing Indefinite Integral: $\int ( \frac{x^2-3x+1/3 }{x^3-x+1})^2 \mathrm{d}x$
Evaluate
$$\int \left( \frac{x^2-3x+\frac{1}{3}}{x^3-x+1}\right)^2 \mathrm{d}x$$
I tried using partial fractions but the denominator doesn't factor out nicely. I also substituted $x=\dfrac{1}{t}$ to get
$$\frac{-1}{9} \int \left(\frac{t^2-9t+3}{t^3-t^2+1}\right)^2 \, \mathrm{d}t $$
But I don't know how to solve this either.
Please Help.
Thanks in advance.
| Ostrogradsky-Hermite method. As I have written in more detail here, we can use the Ostrogradski-Hermite method to integrate a rational fraction $P(x)/Q(x)$ without decomposing it into partial fractions and without finding the multiple roots of the denominator. You can also find a simple description of this method here, subsection 4.5.2, as well as some exercises (1301-1304).
Assume that $\deg P<\deg Q$. Then, there exist polynomials $$P_{1},\quad Q_{1}=\gcd \left\{ Q,Q^{\prime }\right\} ,\quad P_{2}\quad \text{and}\quad Q_{2}=Q/Q_{1},$$ with $\deg P_{1}<\deg Q_{1}$, $\deg P_{2}<\deg Q_{2}$, such that
\begin{equation*}
\int \frac{P}{Q}\, dx=\frac{P_{1}}{Q_{1}}+\int \frac{P_{2}}{
Q_{2}}\, dx.\tag{1}
\end{equation*}
Since the integrand is
\begin{equation*}
\frac{P}{Q}=\frac{\left( x^{2}-3x+1/3\right) ^{2}}{\left(
x^{3}-x+1\right) ^{2}},\tag{2}
\end{equation*}
and
\begin{eqnarray*}
Q &=&\left( x^{3}-x+1\right) ^{2} \\
Q^{\prime } &=&2\left( x^{3}-x+1\right) \left( 3x^{2}-1\right) \\
Q_{1} &=&\gcd \left\{ Q,Q^{\prime }\right\} =x^{3}-x+1,\\
Q_{2}&=&\frac{Q}{Q_{1}}=x^{3}-x+1,
\end{eqnarray*}
we can expand the given integral as
\begin{equation*}
\int \frac{\left( x^{2}-3x+1/3\right) ^{2}}{\left( x^{3}-x+1\right) ^{2}}\, dx=
\frac{Ax^{2}+Bx+C}{x^{3}-x+1}+\int \frac{Dx^{2}+Ex+F}{x^{3}-x+1}\, dx,\tag{3}
\end{equation*}
where $A$, $B$, $C$, $D$, $E$, $F$ are constants. They can be found by
differentiating $(3)$, reducing both sides to a common denominator and equating the coefficients of like powers of $x$ of the numerators. We thus have
\begin{eqnarray*}
\frac{\left( x^{2}-3x+1/3\right) ^{2}}{\left( x^{3}-x+1\right) ^{2}} &=&
\frac{\left( 2Ax+B\right) \left( x^{3}-x+1\right) -\left( 3x^{2}-1\right)
\left( Ax^{2}+Bx+C\right) }{\left( x^{3}-x+1\right) ^{2}}\\ &&+\frac{\left(Dx^{2}+Ex+F\right)\left(x^{3}-x+1\right)}{\left( x^{3}-x+1\right) ^{2}}.
\end{eqnarray*}
Since $\left( x^{2}-3x+1/3\right) ^{2}=x^{4}-6x^{3}+(29/3)x^{2}-2x+1/9$, after some algebra we obtain
\begin{align}
x^{4}-6x^{3}+(29/3)x^{2}-2x+1/9&= Dx^{5}+\left( -A+E\right) x^{4}+\left( -2B-D+F\right) x^{3}\\
&\qquad+\left(-A-3C+D-E\right) x^{2}+\left( 2A+E-F\right) x\\
&\qquad+\left( B+C+F\right),
\end{align}
which means that the constants satisfy the following system of equations
\begin{equation*}
\left\{
\begin{array}{l}
D =0,\\ -A+E=1,\\ -2B-D+F=-6, \\
-A-3C+D-E =\frac{29}{3},\\ 2A+E-F=-2, \\ B+C+F=\frac{1}{9},
\end{array}
\right.
\end{equation*}
whose solution is
\begin{equation*}
A=-1,\, B=3,\, C=-\frac{26}{9},\, D=0,\, E=0,\, F=0.
\end{equation*}
Hence
\begin{equation*}
\int \frac{\left( x^{2}-3x+1/3\right) ^{2}}{\left( x^{3}-x+1\right) ^{2}}\, dx=
\frac{-x^{2}+3x- 26/9 }{x^{3}-x+1}+C.\tag{4}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 2
} |
Find the value of $a$, $b$ and $c$ for the given limit. Question -
Find the values of $a$, $b$ and $c$ so that $$ \lim_{x\to 0} \cfrac{ae^x - b\cos x +c e^{-x} }{x\sin x} = 2 $$
This is what I've tried yet :
For $ x\to 0 $ the numerator must also tend to zero as : $ e^x , \ \cos x, e^{-x} $ all $\to 1$ for $x \to 0$ .
Therefore, we have:
$$
a - b + c = 0 \\
\color{blue}{\text{OR}} \\ b = a + c \\
$$
Now, plugging this in the original equation:
$$\lim_{x\to 0} \cfrac{ae^x -(a+c)\cos x + ce^{-x} }{x^2 \times \frac{\sin x}{x} } = 2 $$
Which implies -
$$\lim_{x\to 0} \cfrac{a(e^x - \cos x) + c(e^{-x} - \cos x) }{x^2} =2 $$
I do get a positive wave that there is some application of expansion series of $e^x \ \& \ \cos x $ . I tried that too but that just didn't work. May be I'm not doing right simplification. Any help will be greatly appreciated.
| Since you have tagged this with "taylor-expansion", you might want to see your limit as
$$\lim_{x\to 0} \cfrac{a(1+x+\frac{x^2}{2}+\cdots) - b(1-\frac{x^2}{2}+\cdots) +c (1-x+\frac{x^2}{2}-\cdots) }{x(x-\frac{x^3}{6}+\cdots)}$$
Matching powers will give you $$\lim_{x\to 0} \cfrac{a-b+c}{x^2} + \cfrac{a-c}{x} + \cfrac{\frac{a}{2}+\frac{b}{2}+\frac{c}{2}}{1} + \cdots =2 $$ and so
$$a-b+c=0$$ $$a-c=0$$ $$\frac{a}{2}+\frac{b}{2}+\frac{c}{2}=2$$
which are three equations in three unknowns, so you can solve these.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Given $\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}$ show that $x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$ Given:
$$\dfrac{\log x}{b-c}=\dfrac{\log y}{c-a}=\dfrac{\log z}{a-b}$$
We have to show that :
$$x^{b+c-a}\cdot y^{c+a-b}\cdot z^{a+b-c} = 1$$
I made three equations using cross multiplication :
$$1.~~x^{c-a}=y^{b-c}$$
$$2.~~y^{a-b}=z^{c-a}$$
$$3.~~z^{b-c}=x^{a-b}$$
How do I proceed hereafter? If I multiply the equations, one variable goes away from exponents.
Thank you.
| If you want to use your equations, here is a method.
Multiplying the equations together, we obtain: $$x^{c-a}y^{a-b}z^{b-c}=y^{b-c}z^{c-a}x^{a-b}$$ which gives after reordering: $$x^{b+c-a}y^{c+a-b}z^{a+b-c}=x^a y^b z^c$$
Therefore it suffices to show that $x^a y^b z^c = 1$.
Your first and third equations give $y = x^{\frac {c-a}{b-c}}, z = x^{\frac{a-b}{b-c}}$. This gives us:
$$x^a y^b z^c = x^a x^{\frac {c-a}{b-c}\times b} x^{\frac{a-b}{b-c}\times c} = x^{a + \frac{bc-ba+ca-bc}{b-c}} = x^{a-a} = x^0 = 1 $$
QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 4
} |
Integral of $\frac{e^x}{5+2e^x}$ Regarding the integral of this term$\frac{e^x}{5+2e^x}=\frac{e^x}{2(\frac{5}{2}+e^x)}$
Is the answer $\frac{1}{2} \ln(\frac{5}{2} +e^x)$ or $\frac{1}{2} \ln(5+2e^x)$?
When I substitute $u= \frac{5}{2}+e^x$ I get the first one, when it's $u=5+2e^x$, then it is the second one...
| $$C + \frac 12 \ln(5+2e^x ) =C + \frac 12\ln[2(\frac 52 +e^x)]=C +\frac12\ln 2 + \frac 12 \ln\left(\frac 52 + e^x\right) $$ the two antiderivatives differ by a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the remainder when $x^{100} + 2x + 10$ is divided by $x − 11$ in $\mathbb Z_{17}[x]$ Find the remainder when $x^{100} + 2x + 10$ is divided by $x − 11$ in $\mathbb Z_{17}[x]$
I simplified
$x^{100} + 2x + 10$ to $x^{15} + 2x + 10$
and $x − 11$ to $x+6$ to be in $\mathbb Z_{17}$.
I got stuck here and used Mathematica and got this answer:
Is there any way to get this remainder without actually diving x+6 into $x^{15} + 2x + 10$?
| $\ \ {\rm mod}\ \color{#c00}{17},\, \color{#0a0}{x\!-\!11}\!:\,\ \color{#0a0}{x\equiv 11}\,\Rightarrow\,\color{#0a0}x^{\large \color{#a0f}{100}}\equiv {\color{#0a0}{11}}^{100}\equiv \overbrace{11^4(\color{#c00}{11^{\color{#c00}16}})^6\equiv (-6)^4\color{#c00}1^6}^{\rm little\ \color{#c00}{Fermat}}\equiv 2^2\equiv \color{#a0f}{4}$
Therefore we conclude $\quad \begin{align} &{\color{#0a0}{x}}^{\large\color{#a0f}{100}}+2\color{#0a0}x+10\\ \ \ \equiv \ \ \ &\color{#a0f}4+\,2\,(\color{#0a0}{11})+10\equiv 36\equiv 2\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
finiding a limit base on the limit of $e$ How can i find: $\lim_{x\to -\infty} \left(\cfrac{2x^3+5}{2x^3+3}\right)^{3x^3+5}$
I know that $\lim_{x\to \infty} \left(\cfrac{x+1}{x}\right)^{x}=e$ but can't see how it helps.
| $$
\left(1+\frac{2}{2x^3+3}\right)^{\frac{3}{2}(2x^3+3)}\cdot \left(1+\frac{2}{2x^3+3}\right)^{-\frac{5}{2}} \to e^{2\cdot \frac{3}{2}} \cdot 1=e^3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove expression don't relate to value of $k$ Given: $$\cfrac{1 - \cos (x) + k \sin (x) }{\sin (x) + k + k \cos (x) }$$
How to prove expression don't relate to value of $k$
Solve without using $\sin(2x) = 2 \sin(x)\cos(x)$
Please help.
| $$\begin{align}\frac{1-\cos x+k\sin x}{\sin x+k(1+\cos x)}&=\frac{1-\cos x}{1-\cos x}\cdot \frac{1-\cos x+k\sin x}{\sin x+k(1+\cos x)}\\&=\frac{(1-\cos x)(1-\cos x+k\sin x)}{(1-\cos x)\sin x+k\sin^ 2x}\\&=\frac{(1-\cos x)(1-\cos x+k\sin x)}{\sin x(1-\cos x+k\sin x)}\\&=\frac{1-\cos x}{\sin x}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1258931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!...n!)^2} \right)$
Evaluating $$\lim\limits_{n\rightarrow \infty} \frac1{n^2}\ln \left( \frac{(n!)^n}{(0!1!2!...n!)^2} \right)$$
I'm not quite sure where to start in evaluating this. Some pointers, or a solution, would greatly be appreciated.
| No need to reference Barnes G-functions. These are integers after all.
$$\log{\left ( \frac{(n!)^n}{(0!1!2!...n!)^2} \right )} = n \log{n!} - 2 \sum_{k=0}^n \log{k!}$$
Now
$$\sum_{k=0}^n \log{k!} = \sum_{k=0}^n \log{\Gamma(k+1)} = \sum_{k=1}^{n+1} \log{\Gamma(k)}$$
Now use Euler-Maclurin:
$$\begin{align}\sum_{k=1}^{n+1} \log{\Gamma(k)}&= \int_0^{n+1} dx \, \log{\Gamma(x)} + \frac12 \left [\log{\Gamma(n+1)} - \log{\Gamma(1)} \right ] + o(n)\end{align}$$
Now, we use Raabe's integral, which is
$$ \int_{k}^{k+1} dx \, \log \Gamma(x) = \frac12 \log{(2 \pi)} + k \log{k}-k$$
so that
$$\begin{align} \int_{0}^{n+1} dx \, \log{\Gamma(x)} &= \sum_{k=0}^n \int_{k}^{k+1} dx \, \log{\Gamma(x)}\\ &= \frac{n+1}{2} \log (2 \pi) - \frac{n (n+1)}{2} + \sum_{k=1}^n k \log{k} \\ &= \frac{n+1}{2} \log (2 \pi) - \frac{n (n+1)}{2} + \log{(1^1 2^2 \cdots n^n)} \\ &= \frac{n+1}{2} \log (2 \pi) - \frac{n (n+1)}{2} + \log{\left [\frac{n!^n}{(n-1)!(n-2)!^2 \cdots 2^{n-2} 1^{n-1}} \right ]} \\ &= \frac{n+1}{2} \log (2 \pi) - \frac{n (n+1)}{2} + n \log{n!} - \sum_{k=1}^{n-1} \log{k!}\end{align}$$
Putting this all together, we find that
$$2 \sum_{k=1}^{n} \log{k!} = \frac{n+1}{2} \log (2 \pi) - \frac{n (n+1)}{2} + n \log{n!} + \frac{3}{2} \log{n!} +o(n) $$
so that, finally, using the fact that Stirling's approximation is equivalent to $\log{n!} = n \log{n} - n + o(n) $,
$$\lim_{n \to \infty} \frac1{n^2} \left ( n \log{n!} - 2 \sum_{k=1}^{n} \log{k!} \right ) = \frac12 $$
ADDENDUM
Raabe's integral is actually not all that hard to evaluate. Rewrite as
$$\int_k^{k+1} dx \, \log{\Gamma(x)} = \int_0^1 dx \, \log{\Gamma(x+k)} $$
$$\log{\Gamma(x+k)} = \log{\Gamma(x)} + \sum_{m=0}^{k-1} \log{(x+m)} $$
Thus,
$$\begin{align}\int_k^{k+1} dx \, \log{\Gamma(x)} &= \int_0^1 dx \, \log{\Gamma(x)} + \sum_{m=0}^{k-1} \int_0^1 dx \, \log{(x+m)}\\ &= \int_0^1 dx \, \log{\Gamma(x)} + \sum_{m=0}^{k-1} [(m+1) \log{(m+1)} - m \log{m} - (m+1)+m] \\ &= \int_0^1 dx \, \log{\Gamma(x)} + k \log{k} - k \end{align}$$
To evaluate the integral on the RHS, use the duplication formula:
$$\Gamma \left ( \frac{x}{2} \right ) \Gamma \left ( \frac{x+1}{2} \right ) = 2^{1-x} \sqrt{\pi} \Gamma(x) $$
so that
$$\log{\Gamma(x)} = \log{\left [\frac{\Gamma \left ( \frac{x}{2} \right ) \Gamma \left ( \frac{x+1}{2} \right )}{\sqrt{\pi} 2^{1-x}} \right ]} = -\frac12 \log{\pi} - \log{2} + \log{\Gamma \left ( \frac{x}{2} \right )} + \log{\Gamma \left ( \frac{x+1}{2} \right )} + x \log{2}$$
Thus,
$$\begin{align}\int_0^1 dx \, \log{\Gamma(x)} &= -\frac12 \log{(2 \pi)} + \int_0^1 dx \, \log{\Gamma \left ( \frac{x}{2} \right )} + \int_0^1 dx \, \log{\Gamma \left ( \frac{x+1}{2} \right )}\\ &= -\frac12 \log{(2 \pi)} + 2 \int_0^{1/2} dx \, \log{\Gamma(x)} + 2 \int_{1/2}^1 dx \, \log{\Gamma(x)} \\ &= -\frac12 \log{(2 \pi)} + 2 \int_0^{1} dx \, \log{\Gamma(x)} \end{align}$$
The result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
find the derivative of a function with natural log find the derivative of $f(x)=\ln(x^4)(\sqrt{5x-3})$
I just need help getting to the answer.
The first answer I got was $f(x)=(x^4)(2.5)+(5x-3)^{1/2}(4x^3)$.
| We will first use the product rule:
$$\frac{d}{dx} \ln(x^4)\sqrt{5x-3} = \ln(x^4)\cdot\frac{d}{dx}\sqrt{5x-3} + \sqrt{5x-3}\cdot\frac{d}{dx}\ln(x^4)$$
Then we use the chain rule on the first part. $u=5x-3$.
\begin{align}\ln(x^4)\frac{d}{dx}\sqrt{5x-3} & =\ln(x^4)\frac{d}{du}\sqrt{u} \frac{d}{dx}5x-3 \\ &= 5\ln(x^4) 0.5 u^{-\frac{1}{2}} \\ &=2.5 \ln(x^4)(5x-3)^{-\frac{1}{2}}\end{align}
Then we use the chain rule on the second part. $v=x^4$.
\begin{align}\sqrt{5x-3}\cdot\frac{d}{dx}\ln(x^4) &=\sqrt{5x-3}\cdot\frac{d}{dv}\ln(v)\cdot\frac{d}{dx}x^4 \\ &= \sqrt{5x-3}\cdot\frac{1}{v}\cdot4x^3 \\ &= 4x^3\cdot\sqrt{5x-3}\cdot\frac{1}{x^4} \\ &= 4 \frac{1}{x}\cdot\sqrt{5x-3}
\end{align}
So the derivative is $$2.5 \ln(x^4)(5x-3)^{-\frac{1}{2}} + 4 \frac{1}{x}\sqrt{5x-3}$$
EDIT:
You could also simplify $\ln(x^4) = 4 \ln(x)$, that would have saved one chain rule.
EDIT 2:
I think you went wrong on the product rule.
Also (I suppose you know this but I will say it): The derivative of $\ln(x)$ is $\frac{1}{x}$. This can be proven elegantly by the chain rule if you know that the derivative of $e^x$ is $e^x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Check the convergence of $a_{n+1}=\sqrt{a_n+\frac{4}{a_n}}$ where $a_1=4$ Check the convergence of $a_{n+1}=\sqrt{a_n+\frac{4}{a_n}}$ where $a_1=4$. If it converges, find its limit.
I tried to prove that the sequence is monotonically decreasing and bounded by 2, but I failed. Even though, I know that if there exists a limit, than it should satisfy the equation: $a^3-a^2=4$, and its solution is $a=2$.
| Let $x$ be a positive real. Then by arithmetic-geometric means inequality we have
$$
\frac{x+4/x}{2} \ge \sqrt{x\cdot \frac{4}{x}}=2,
$$
with equality iff $x=4/x$, i.e. $x=2$.
Then, we miss only to check that the function is monotonically decreasing starting from $a_1>2$. Take $x>2$, then
$$
x^2>x+\frac{4}{x} \text{ if and only if }x^3-x^2-4>0.
$$
Its first derivative is $3x^2-2x=3x(x-2/3)>0$, hence it is increasing for $x>2$ Therefore $a_{n+1}>a_n$ for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to factorize $x^5 -1=0$? how can I factorize $x^5 -1 =0 $ so I get a quartic equation separately. Is there any specific way of factorizing such equations to get something of a lesser power?
| Divide $x^5-1$ by $x-1$ to get
$$x^5-1=(x-1)(x^4+x^3+x^2+x+1).$$
Now note that
$$x^4+x^3+x^2+x+1=\left(x^2+\frac 12x+1\right)^2-\frac 54x^2$$
$$=\left(x^2+\frac 12x+1-\frac{\sqrt 5}{2}x\right)\left(x^2+\frac 12x+1+\frac{\sqrt 5}{2}x\right).$$
Hence, one has
$$x^5-1=(x-1)\left(x^2+\frac{1-\sqrt 5}{2}x+1\right)\left(x^2+\frac{1+\sqrt 5}{2}x+1\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
finding the value of $a$ for minimal solid Given that $y=x^2+ax-0.25$ is bounded by $x=0$ and $y=1$ and spinning around the x-axis.
What is the value of $a$ for min volume of that solid?
My attempt: $x^2+ax-0.25=1$ so $x=\frac{-a\pm \sqrt{a^2+5}}{2}$.
Then, $f(a)=\pi \int_{0}^{\frac{-a+\sqrt{a^2+5}}{2}}(1-(x^2+ax-0.25)^2)dx$
But that will lead to: $f(a)=1/480 (-225a+20a^3+8a^5+200 \sqrt{5+a^2}-8a^4 \sqrt{5+a^2})π$
And then $f'(a)=0$ will be very hard to solve.
Any ideas?
Update: wolfram gave me an answer but i wonder if it can be done without that.
Thanks.
| I think you overcomplicated the problem.
You only need to find out what is the minimum of :
$f(a)=\int_{0}^{\frac{-a+\sqrt{a^2+5}}{2}}|x^2+ax-0.25|dx$
Since the volume is going to be the rotation of this area around the axis $x$. If you minimize the area, you minimize the volume.
The function $y=x^2+ax-0.25$ has a positive zero in $x=\dfrac{-a+\sqrt{a^2+1}}{2}$, which is smaller than $\dfrac{-a+\sqrt{a^2+5}}{2}$, so that divide the integral in two domains.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the least squares solution for the system of equations $y=Ax^2+B$ Find the least squares solution for the system of equations $y = Ax^2 + B$ where
$(x, y)$ belongs to the set {$(0, 1),(1, 5),(−1, 3)$}. What is the geometric (graphical) interpretation
of the solution?
Is this done the same way as a least squares solution for $Ax=B$? If not then how do I start this?
| Constant + quadratic
The trial function is
$$y(x) = \alpha + \gamma x^2.
$$
The input vectors are
$$
x = \left[
\begin{array}{r}
0\\
1\\
-1
\end{array}
\right], \quad
y = \left[
\begin{array}{r}
1\\
5\\
3
\end{array}
\right]
$$
They imply the linear system of column vectors:
$$
\begin{align}
\mathbf{A} c &= y \\
\left[
\begin{array}{rr}
1 & x^2
\end{array}
\right]
\left[
\begin{array}{r}
\alpha \\
\gamma
\end{array}
\right]
& =
\left[
\begin{array}{rr}
y
\end{array}
\right]
\end{align}
$$
The system matrix $\mathbf{A}$ is built with the $x$ data, the data vector is constructed with the $y$ data. The task is to find the coefficient vector $c$.
Using the input data, we have
$$
\begin{align}
\mathbf{A} c &= y \\
\left[
\begin{array}{rr}
1 & 0\\
1 & 1\\
1 & 1
\end{array}
\right]
\left[
\begin{array}{r}
\alpha \\
\gamma
\end{array}
\right]
& =
\left[
\begin{array}{r}
1\\
5\\
3
\end{array}
\right]
\end{align}
$$
Use the normal equations
$$
\mathbf{A}^{*}\mathbf{A} c =\mathbf{A}^{*}y
$$
to find the least squares solution:
$$
c = \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1}\mathbf{A}^{*}y
$$
The product matrix and its inverse are
$$
\mathbf{A}^{*}\mathbf{A} =
%
\left[
\begin{array}{rr}
3 & 2\\
2 & 2
\end{array}
\right], \qquad
%
\left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} =
%
\left[
\begin{array}{rr}
1 & -1\\
-1 & \frac{3}{2}
\end{array}
\right].
$$
The solution vector is
$$
c = \left[
\begin{array}{r}
\alpha\\
\gamma
\end{array}
\right]
=
\left[
\begin{array}{r}
1\\
3
\end{array}
\right],
$$
that is, the solution function is $y(x) = 1 + 3x^2$. The error vector is
$$
r = \mathbf{A} c - y =
\left[
\begin{array}{r}
0 \\
-1 \\
1
\end{array}
\right], \qquad
r \cdot r = 2.
$$
The solution curve is plotted against the data points below.
Constant + linear
With the quadratic fit complete, turn to the linear fit. The trial function becomes
$$
y(x) = \alpha + \beta x
$$
The $x$ and $y$ vectors are unchanged. The linear system of column vectors is
$$
\begin{align}
\mathbf{A} c &= y \\
\left[
\begin{array}{rr}
1 & 0\\
1 & 1\\
1 & -1
\end{array}
\right]
\left[
\begin{array}{r}
\alpha \\
\beta
\end{array}
\right]
& =
\left[
\begin{array}{r}
1\\
5\\
3
\end{array}
\right]
\end{align}
$$
With the input data, we have
$$ \left[
\begin{array}{rr}
1 & 0\\
1 & 1\\
1 & -1
\end{array}
\right]
\left[
\begin{array}{r}
\alpha \\
\beta
\end{array}
\right]
=
\left[
\begin{array}{r}
1\\
5\\
3
\end{array}
\right]
$$
The product matrix and its inverse are
$$
\mathbf{A}^{*}\mathbf{A} =
%
\left[
\begin{array}{rr}
3 & 0 \\
0 & 2
\end{array}
\right], \qquad
\left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} = \frac{1}{6}
\left[
\begin{array}{rr}
2 & 0 \\
0 & 3
\end{array}
\right].
$$
The solution vector here is
$$c =
\left[
\begin{array}{r}
3\\
1
\end{array}
\right],
$$
so the solution function is $y(x) = 3 + x$. The error vector is
$$
r = \mathbf{A} c - y =
\left[
\begin{array}{r}
2 \\
-1 \\
1
\end{array}
\right], \qquad
r \cdot r = 6.
$$
The first solution represented a quadratic equation, the second a linear equation (see plot below).
To emphasize the difference between the quadratic solution $y_{q}$ and the linear solution $y_{l}$, they are written with terms in increasing order:
$$
\begin{matrix}
y_{q}(x) & = & \alpha_q & & & + & \gamma x^2\\
y_{l}(x) & = & \alpha_l & + & \beta x
\end{matrix}
$$
Constant + linear + quadratic
The final case is
$$
y(x) = \alpha + \beta x + \gamma x^{2}.
$$
The linear system is
$$
\begin{align}
\mathbf{A} c &= y \\
\left[
\begin{array}{rr}
1 & 0 & 0\\
1 & 1 & 1\\
1 & -1 & 1
\end{array}
\right]
\left[
\begin{array}{r}
\alpha \\
\beta \\
\gamma
\end{array}
\right]
& =
\left[
\begin{array}{r}
1\\
5\\
3
\end{array}
\right]
\end{align}
$$
The product matrix and its inverse are
$$
\mathbf{A}^{*}\mathbf{A} =
%
\left[
\begin{array}{ccc}
3 & 0 & 2 \\
0 & 2 & 0 \\
2 & 0 & 2 \\
\end{array}
\right], \qquad
\left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} = \frac{1}{2}
\left[
\begin{array}{rrr}
2 & 0 & -2 \\
0 & 1 & 0 \\
-2 & 0 & 3 \\
\end{array}
\right].
$$
The solution vector is:
$$c =
\left[
\begin{array}{r}
1\\
1\\
3
\end{array}
\right].
$$
The solution vector is $y(x) = 1 + x + 3x^{2}$. The error vector is
$$
r = \mathbf{A} c - y =
\left[
\begin{array}{r}
0 \\
0 \\
0
\end{array}
\right], \qquad
r \cdot r = 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
In how many ways can I sum integers from $1$ to $N-1$ to obtain $N$? I'm looking for the exact formula for $f(N) =$ number of ways to sum $1, 2, ..., N-1$ to obtain $N$.
$N$ is an integer $> 0$. Integers $1, 2, ..., N-1$ can be used $0$ or 1 time as an element in the sum.
For example:
$f(6) = 3$. There are 3 ways to obtain 6 from integers less than $6$:
$1+2+3$
$1+5$
$2+4$
| The power series for
$$
\prod_{k=1}^\infty\left(1+x^k\right)
$$
which is equal to
$$
\prod_{k=1}^\infty\frac1{1-x^{2k-1}}
$$
is the generating function for your number plus $1$. The beginning of the power series is
$$
\small1+x+x^2+2x^3+2x^4+3x^5+4x^6+5x^7+6x^8+8x^9+10x^{10}+12x^{11}+15x^{12}+\dots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Differential Equation (Non linear to linear differential equation) Show that the substitution $u=\frac{1}{y}$ transform the non-linear differential equation $$\frac{dy}{dx}+\frac{y}{x}=y^2\ln (x)$$ into the linear differential equation $$\frac{du}{dx}-\frac{u}{x}=-\ln (x)$$. Solve this linear differential equation, and hence obtain $y$ in terms of $x$, given that $y=\frac{1}{2}$ when $x=1$
My attempt,
$\frac{dy}{dx}+\frac{y}{x}=y^2\ln (x)$
$\frac{-\frac{dy}{dx}}{y^2}-\frac{1}{xy}=-\ln (x)$
Let $u=\frac{1}{y}$ $\frac{du}{dx}=\frac{-\frac{dy}{dx}}{y^2}$
$\frac{du}{dx}-\frac{y}{x}=-\ln (x)$
$e^{\int -\frac{1}{x}dx}=\frac{1}{x}$
$\frac{\frac{du}{dx}}{x}-\frac{u}{x^2}=-\frac{\ln (x)}{x}$
$\frac{\frac{du}{dx}}{x}+\frac{d}{dx}(\frac{1}{x})u=-\frac{\ln (x)}{x}$
$\frac{d}{dx}(\frac{u}{x})=\frac{-\ln (x)}{x}$
$\int \frac{d}{dx}(\frac{u}{x})dx=\int \frac{-\ln (x)}{x}dx$
$\frac{u}{x}=-\frac{1}{2}\ln ^2(x)+c_1$
$u=x(-\frac{1}{2}\ln ^2(x)+c_1)$
$y=\frac{1}{u}=\frac{2}{-x\ln ^2(x)+2c_1x}$
$y=\frac{2}{-x\ln ^2(x)+c_1x}$
When $x=1$, $y=\frac{1}{2}$
I got $c_1=4$
Therefore, $y=\frac{2}{-x\ln ^2(x)+4x}$
Is this working right? Is there another method to solve this question?
| $$
y' + \frac{y}{x} = y^2\ln x
$$
lets multiply by $x$ we find
$$
xy' + y = \frac{d}{dx}(xy) = xy^2\ln x
$$
let $u = xy$ we find
$$
u' = \frac{u^2}{x}\ln x
$$
thus we find
$$
\int\frac{du}{u^2} = \int \frac{\ln x}{x}dx = \int \ln x\dfrac{d}{dx}\ln x dx = \int \dfrac{d}{dx}\frac{(\ln x)^2}{2}dx = \frac{(\ln x)^2}{2} +C
$$
so we find
$$
-\frac{1}{u} = \frac{(\ln x)^2}{2} + C
$$
so we get
$$
u = \frac{-1}{ \frac{(\ln x)^2}{2} + C}
$$
thus we lead to
$$
y = \frac{1}{x}\frac{-1}{ \frac{(\ln x)^2}{2} + C} = \frac{2}{-x\ln^2x -2Cx}
$$
Is this easier, I can not say.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$a^2+b^2+c^2=(abc)^2-2\leq 6$. Proof or counter-example needed for $a,b,c\gt 0$ I was working on an inequality proof in which I need to use the following inequality to conclude.
$$\forall~a,b,c\gt 0~,~a^2+b^2+c^2=(abc)^2-2\implies a^2+b^2+c^2\leq 6$$
I can't think of any way to prove this. I only need a hint since I dislike people just giving away the answers. Thanks in advance!
If this claim is false, I'd like a counter-example. Thanks again!
| True for $a = b = c$, then $a = \sqrt{\frac{6}{3}} = \sqrt{2}$
$\sqrt{2}^2 + \sqrt{2}^2 + \sqrt{2}^2 = (\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2})^2 - 2$
$6 = 8-2$ and $6 \le 6$
$a,b,c = \sqrt{2} > 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $\int\limits_{0}^{2\pi} \frac{1}{5-3\cos(x)} \,\,dx$ I have to find $$\int_0^{2\pi} \frac{1}{5-3\cos(x)} \,\,dx$$
I tried to do it by substitution $t = \tan(\frac{x}{2})$
Then we have that $$\cos(x) = \frac{1-t^2}{1+t^2} \quad dx = \frac{2\,dt}{1+t^2}$$ but then also limits of integration are changing so we have
$$\int\limits_{t(0)}^{t(2\pi)} \frac{1}{5 - \frac{1-t^2}{1+t^2}} \cdot \frac{2dt}{1+t^2} = \int\limits_0^0 \frac{1}{5 - \frac{1-t^2}{1+t^2}} \cdot \frac{2\,dt}{1+t^2} = 0$$
I figured out that it is not correct because $\tan(\frac{\pi}{2})$ is not defined and $t(\pi) = \tan(\frac{\pi}{2})$ and $\pi \in [0, 2\pi]$. How can I "repair" that and do it right?
| While there have been several clean and different approaches shown here, I'm surprised that nobody has shown that contour integration works well also. To that end here we go....
Let $z=e^{ix}$ so that $dx=dz/iz$ and $\cos x = \frac12 (z+z^{-1})$. We have
$$\begin{align}
I&=\int_0^{2\pi} \frac{1}{5-3\cos x}dx\\\\
&=\oint_C \frac{1}{5-3\frac12 (z+z^{-1})}\frac{dz}{iz}\\\\
&=\oint_C \frac{\frac23 i}{(z-3)(z-1/3)}dz\\\\
&=2\pi i \text{Res}_{\text{in C}} \left(\frac{\frac23 i}{(z-3)(z-1/3)}\right)
\end{align}$$
where $C$ is the unit circle in the complex $z$-plane.
Note that the pole at $z=1/3$ is inside the unit circle whereas the pole at $z=3$ is outside the unit circle.
Thus, the residue is given by
$$\text{Res}_{z=1/3} \left(\frac{\frac23 i}{(z-3)(z-1/3)}\right)=\frac23 i\,\lim_{z\to 1/3}\left((z-1/3)\,\frac{1}{(z-3)(z-1/3)}\right)=-i/4$$
which after multiplying by $2\pi i$ shows that
$$I=\pi/2$$
as expected!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 2
} |
Cubic root of unity Is there anyway to solve this without substituting with the values?
Prove that: $$\frac{1+10w^2}{1-2w} + \frac{2+17w}{2+3w} = 6$$. (Where $w$ & $w^2$ are the cubic roots of unity)
| Note that
$1+w+w^2 = 0$, as $$w(1+w+w^2) = w+w^2+w^3 = 1+w+w^2$$
and $w \ne 1$. Now we have
$$ (1-2w)(2+3w) = 2 - w - 6w^2 = 2 - w - 6(-1-w) = 8+5w $$
and
\begin{align*}
(1+10w^2)(2+3w) + (1-2w)(2+17w) &= 2+ 3w+20w^2 + 30 + 2 +13 w - 34w^2\\
&= 34 + 16w - 14w^2\\
&= 34 + 16w + 14 + 14w\\
&= 48 + 30w
\end{align*}
And therefore
$$
\frac{1+ 10w^2}{1-2w} + \frac{2+17w}{2+3w} = \frac{48+30w}{8+5w} = 6.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Using least squares regression for line of best fit Use the least square approximation to find the closest line (the line of "Best Fit") to the points:
$$(-6,-1), \quad (-2,2), \quad (1,1), \quad (7,6)$$
I'm attempting to use the least squares approximation formulation that is as follows:
$$A^TAx = A^Tb$$
However, I'm confused because I'm given four vectors. Does that mean I can use the first three vectors $(-6,-1),(-2,2),(1,1)$ to create my matrix A, and then use the last vector $(7,6)$ for the "$b$" value? I understand the process is very straight forward once proper substitution has been done, but I tried the method I described above and got the wrong answer.Thank you guys.
| The trial function is
$$
y(x) = c_{0} + c_{1} x.
$$
As noted by @user147263,you have the linear system
$$
\begin{align}
\mathbf{A} c & = y\\
%
\left[
\begin{array}{rr}
1 & -6 \\
1 & -2 \\
1 & 1 \\
1 & 7 \\
\end{array}
\right]
%
\left[
\begin{array}{r}
c_{0} \\
c_{1} \\
\end{array}
\right]
%
&=
%
\left[
\begin{array}{r}
-1 \\
2 \\
1 \\
6
\end{array}
\right].
%
\end{align}
$$
Your choice for solution is the normal equations
$$
\begin{align}
%
\mathbf{A}^{*} \mathbf{A} c &= \mathbf{A}^{*}y \\
%
\left[
\begin{array}{cc}
4 & 0 \\
0 & 90
\end{array}
\right]
%
%
\left[
\begin{array}{r}
c_{0} \\
c_{1} \\
\end{array}
\right]
%
&=
%
\left[
\begin{array}{r}
8 \\
45
\end{array}
\right] .
%
\end{align}
$$
The solution is
$$
\begin{align}
c &= \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} y \\
%
\left[
\begin{array}{r}
c_{0} \\
c_{1} \\
\end{array}
\right]
%
&=
%
\left[
\begin{array}{cc}
\frac{1}{4} & 0 \\
0 & \frac{1}{90} \\
\end{array}
\right]
%
\left[
\begin{array}{r}
8 \\
45
\end{array}
\right] \\
%
&=
%
\left[
\begin{array}{c}
2 \\
\frac{1}{2}
\end{array}
\right] .
%
\end{align}
$$
The solution function is
$$
y(x) = 2 + \frac{1}{2} x.
$$
The residual error vector is
$$
r = \mathbf{A}c - 7 =
\left[
\begin{array}{r}
0 \\
-1 \\
\frac{3}{2} \\
-\frac{1}{2}
\end{array}
\right]
$$
with a total error of $r^{2} = \frac{7}{2}.$
The solution is plotted against the data points below.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $ \tan(20^{\circ}) = p $, find $ \frac{\tan(160^{\circ}) - \tan(110^{\circ})}{1 + \tan(160^{\circ}) \tan(110^{\circ})} $.
I applied the $ \tan(A - B) $-formula to make it $\tan(50^{\circ}) $, then I split it to $ \tan(30^{\circ} + 20^{\circ}) $. My answer came out to be $ \dfrac{\sqrt[3]{p + 1}}{\sqrt[3]{- p}} $, but the actual answer is $ \dfrac{1 - p^{2}}{2 p} $.
| $$\frac{\tan 160 - \tan 110}{1+\tan 160 \tan 110} = \tan(160 - 110) = \tan 50 = \cot 40 = \frac{1}{\tan 40} \\
\tan 40 = \frac{\tan 20 + \tan 20}{1-\tan^2 20} = \frac{2p}{1-p^2} \\
\frac{\tan 160 - \tan 110}{1+\tan 160 \tan 110} = \frac{1-p^2}{2p}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finite Summation of Fractional Factorial Series Is there a closed form solution for the following series? (Without Using Gamma Function):
$$
S=\sum _{i=1}^{n-1} \frac{1}{(i+1)!}
$$
| Note that
$$n!\,e=\sum_{k=0}^\infty\frac{n!}{k!}=\sum_{k=0}^n\frac{n!}{k!}+\sum_{k=n+1}^\infty\frac{n!}{k!}$$
The first sum on the RHS is always an integer since $n\geq k$. The second sum satisfies
$$\begin{align} \sum_{k=n+1}^\infty\frac{n!}{k!}
&=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots\\
&<\frac{1}{n+1}+\frac{1}{(n+1)(n+1)}+\frac{1}{(n+1) (n+1)(n+1)}+\cdots\\
&=\sum_{k=1}^\infty\frac{1}{(n+1)^k}\\
&= \frac{1}{n}\\
&\leq1
\end{align}$$
when $n\geq1$. Hence we have
$$\lfloor n!\,e\rfloor=\sum_{k=0}^n\frac{n!}{k!}\\
\implies \frac{\lfloor n!\,e\rfloor}{n!}=\sum_{k=0}^n\frac{1}{k!}\\
\implies \frac{\lfloor n!\,e\rfloor}{n!}-2=\sum _{k=1}^{n-1} \frac{1}{(k+1)!}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $\frac {\sin^4\theta}a +\frac {\cos^4\theta}b=\frac 1 {a+b}$ P.T $\frac {\sin^8\theta}{a^3} + \frac {\cos^8\theta} {b^3} = \frac1{(a+b)^3}$ The question reads if
$$\frac {\sin^4\theta} {a} + \frac {\cos^4\theta} {b} = \frac 1 {a+b}$$
Then prove that
$$\frac {\sin^8\theta} {a^3} + \frac {\cos^8\theta} {b^3} = \frac 1 {(a+b)^3}$$
I know simply cubing the first equation would not work. I have also tried arriving at the solution using various identities like $ \sin^2\theta + \cos^2\theta = 1$ but to no avail. I would also like to know the approach one should use while solving such trigonometry questions.
I am an eleventh grader so sorry if this question is too silly.
| Let $\frac{\sin^4\theta}{a}=x$ and $\frac{\cos^4\theta}{b}=y$
Cubing $(x+y)$
$(x+y)^3=x^3+y^3+3xy(x+y)$
Substituting..
$\frac1{(a+b)^3}=\frac{\sin^{12}\theta}{a^3}+\frac{\cos^{12}\theta}{b^3}+3\frac{\sin^4\theta\cos^4\theta}{ab}(\frac{\sin^4\theta}{a}+\frac{\cos^4\theta}{b})$
Now equate this expression to $\frac {\sin^8\theta}{a^3} + \frac {\cos^8\theta} {b^3}$ And LHS=RHS. Hence proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $f(a,b) = \min\left(\max\left(a^2+b\;,b^2+a\right)\right)$
Evaluation of $f(a,b) = \min\left(\max\left(a^2+b\;,b^2+a\right)\right)\;,$ Where $a,b\in \mathbb{R}$
$\bf{My\; Try::}$ First we have to calculate $\max(a^2+b,b^2+a) = \left\{\begin{matrix}
a^2+b& \;,a^2+b>b^2+a \\\\
b^2+a& \;,a^2+b\leq b^2+a \\
\end{matrix}\right.$
Now For $\bf{I^{st}}$ case, Here $f(a,b) = a^2+b\;\;,$If $ a^2+b>b^2+a\Rightarrow (a-b)\cdot (a+b-1)>0$
Similarly For $\bf{II^{st}}$ case, $f(a,b) = b^2+a\;\;,$If $ a^2+b\leq b^2+a\Rightarrow (a-b)\cdot (a+b-1)\leq 0$
Now I did not understand how can i solve it, Help me
Thanks
| you can go on further, for example, 1st case:
sub case 1: $a \ge b, a+b \ge 1$ we need to find min$a^2+b$
$a^2+b=a^2-a+a+b \ge a^2-a+1=(a-\dfrac{1}{2})^2+\dfrac{3}{4} \ge \dfrac{3}{4} $ when $a=b=\dfrac{1}{2}$ get min.
rest cases you can do simliar thing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Check my general solution to the differential equation? Given differential equation:
$$y' = \frac{(2xy^{3}+4x)}{(x^{2}y^{2}+y^{2})}$$
This is the general solution that I got for the above differential equation:
$$\frac{1}{3} \ln{\lvert y^3+2\rvert}=\ln{\lvert x^2+1\rvert}+\ln C$$
Please check if it is correct?
And for the solution satisfying y(1)=0, this is what I got:
$$(y^3+2)^{\frac{1}{3}}=(x^2+1)+C$$
| For the sake of completion, I'll run through the solution of the differential equation. First, our original equation is the same as
$$\frac{dy}{dx} = \frac{(2x)(y^{3} + 2)}{(x^{2} + 1)(y^{2})}.$$
By separation of variables, we get
$$ \frac{y^{2}}{y^{3}+2} dy = \frac{2x}{x^{2}+1} dx.$$
Equivalently,
$$ \frac{1}{3} \frac{3y^{2}}{y^{3}+2} dy = \frac{2x}{x^{2} + 1} dx.$$
Now, we note that this
$$ \frac{1}{3} \frac{d}{dy} \ln \left| y^{3} + 2 \right| = \frac{d}{dx} \ln \left| x^{2} + 1 \right|,$$
so we integrate to get
$$ \frac{1}{3} \ln \left| y^{3} + 2 \right| = \ln \left| x^{2} + 1 \right| + C.$$
So that part of your answer is correct; I've just renamed your $``\ln C"$ to just $C$.
There is, however, a problem with your solution to the initial value problem. Most importantly, there should not be any undetermined constants! To solve the initial value problem, simply plug $x = 1$ and $y = 0$ into the general solution of the differential equation to figure out what $C$ is. That is,
$$\begin{aligned}
\frac{1}{3} \ln \left| (0)^{3} + 2 \right| &= \ln \left| (1)^{2} + 1 \right| + C \\
\frac{1}{3} \ln 2 &= \ln 2 + C \\
-\frac{2}{3} \ln 2 &= C.
\end{aligned}$$
Hence, the solution to your initial value problem is
$$\frac{1}{3} \ln \left| y^{3} + 2 \right| = \ln \left| x^{2} + 1 \right| - \frac{2}{3} \ln 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the Range of the function $f(x) = |x-6|+x^2-1$
find the Range of $f(x) = |x-6|+x^2-1$
$$ f(x) = |x-6|+x^2-1 =\left\{
\begin{array}{c}
x^2+x-7,& x>0 .....(b) \\
5,& x=0 .....(a) \\
x^2-x+5,& x<0 ......(c)
\end{array}
\right.
$$
from eq (b) i got $$f(x)= \left(x+\frac12\right)^2-\frac{29}4 \ge-\frac{29}4$$
and from eq (c) i got $$f(x)= \left(x-\frac12\right)^2+\frac{19}4 \ge\frac{19}4$$
and eq(b) tells me that it also passes through 5 and so generalize all this and found its range is $\left[-\frac{29}4 , \infty\right)$
but the graph says its range is $(5, \infty)$
| $$f'(x)=\frac{x-6}{|x-6|}+2x$$
$f'(x)=0\iff x=\frac{1}{2}$ and $f'(x)<0$ if $x<\frac{1}{2}$ and $f'(x)>0$ if $x>\frac{1}{2}$, therefore the range is $[f(\frac{1}{2}),+\infty [$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Laurent Series Expansion of $\frac{-3z^2+8z+1}{(z-2)(z^2+1)}$
Laurent Series Expansion of $\frac{-3z^2+8z+1}{(z-2)(z^2+1)}$ on the annulus $A(1,2)$
I think $A(1,2)$ denotes the set $\{z:1<|z-0|<2\}$, so it excludes the poles.
using partial fraction decomposition I got;
$\frac{-3z^2+8z+1}{(z-2)(z^2+1)}=\frac{1}{z-2}-\frac{2}{z+i}-\frac{2}{z-i}$
In general The Laurent Series of $\frac{1}{z-z_0}$ about $a$
If $|z-a|<|z_0-a|$ then $\frac{1}{z-z_0}=-\frac{1}{z_0-a}\sum\limits_{n=0}^{\infty}\left(\frac{z-a}{z_0-a}\right)^n$
If $|z-a|>|z_0-a|$ then $\frac{1}{z-z_0}=\frac{1}{z-a}\sum\limits_{n=0}^{\infty}\left(\frac{z_0-a}{z-a}\right)^n$
then I have about $0$
$\frac{1}{z-2}=-\frac{1}{2}\sum\limits_{n=0}^{\infty}\left(\frac{z}{2}\right)^n\quad$ for $|z|<2$
$\frac{2}{z+i}=-\frac{2}{z}\sum\limits_{n=0}^{\infty}\left(\frac{-i}{z}\right)^n\quad$ for $|z|>1$
$\frac{2}{z-i}=-\frac{2}{z}\sum\limits_{n=0}^{\infty}\left(\frac{i}{z}\right)^n\quad$ for $|z|>1$
Can you verify my steps, and Is then their difference or sum the laurent series on the annulus, or is there another one ?
| That's certainly a valid way to do it, however if I'm doing this right,$$\frac{2}{z+i} = \frac{2}{z} \cdot \frac 1 {1 - (-i/z)} = + \frac 2 z \sum_{n=0}^\infty (-i/z)^n$$which is the opposite overall sign from what you seem to have, and similarly for $2/(z - i)$. However it will be simpler to proceed if you realize that$$\frac 1 {z - i} + \frac 1 {z + i} = \frac {(z + i) + (z - i)}{(z - i)(z + i)} = \frac {2 z}{z^2 + 1}.$$This expands more simply on the annulus as:$$\frac{-4z}{z^2 + 1} = \frac{-4}{z} \cdot \frac{1}{1 - (-z^{-2})} = \frac{-4}{z} \left( 1 - z^{-2} + z^{-4} - z^{-6} + \dots\right)$$which shows more clearly that the negative even powers vanish while the rest of the negative series alternates in sign.
You can probably also get this your way; you'll get:$$i^n + (-i)^n = (i^n) \cdot (1 + (-1)^n) = (-1)^{n/2} \cdot 2 \operatorname{even}(n) $$ where $\operatorname{even}(n) = (1 + (-1)^n)/2$ is $1$ when $n$ is even and $0$ when $n$ is odd. Combined with the $2/z$ prefactor these appear to be identical, if you subtract them in keeping with the $-$ signs in your expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the area of the region $ABCD$. In the Figure $\square PQRS$ is a square with side $2\sqrt6$.
By joining the midpoints another square $\square WXYZ$ is formed .
Circles are drawn with $4$ vertices as the center and
radius equal to the side of the square $\square WXYZ$.
Find the area common to all the $4$ circles .
$a.)6\pi\left(\dfrac{\sqrt3-1}{2}\right)\\
b.)4\pi-3\sqrt3\\
c.)\dfrac12\left(\pi -3\sqrt3\right)\\
d.)4\pi-12\left(\sqrt3 -1\right)\quad \LARGE \color{green}{\checkmark}\\$
So i have to find the area of region $ABCD$.
I have found that $WX=WY=YZ=XZ=2\sqrt3.$
Let Region $WBA$=Region $XAD$=Region $ZCD$=Region $YBC=a$
and
Region $WAX$=Region $XDZ$=Region $ZCY$=Region $YBW=b$
and
Region $ABCD=m$
$Area(\square WXYZ)=12\\
m+4a+4b=12$.
Area of sectors.
$Area$(sector $WZY$)=$Area$(sector $XYZ$)=$Area$(sector $WXY$)=$Area$(sector $WZX$)=$m+3a+2b=3\pi$.
Now i m stucked.
I m looking for simple short way.
I have studied maths upto $12th$ grade.
| Let $[F]$ be the area of a figure $F$.
Since $\angle{WYA}=30^\circ,\angle{AYZ}=60^\circ$, one has
$$\begin{align}a+b&=[YWA]\\&=[\text{sector}\ YWA]-\left([\text{sector}\ ZYA]-[\triangle{ZYA}]\right)\\&=(2\sqrt 3)^2\pi\times\frac{30}{360}-\left((2\sqrt 3)^2\pi\times\frac{60}{360}-\frac{\sqrt 3}{4}\times(2\sqrt 3)^2\right)\\&=3\sqrt 3-\pi\end{align}$$
Hence, one has$$m=12-4(a+b)=12-4(3\sqrt 3-\pi)=4\pi+12-12\sqrt 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Express $w=f(z)=\frac{1}{(1-z)^2}$ in the form $w=u(x,y)+iv(x,y)$ I start by writing $f(z)$ as $$\frac{1}{(1-(x+iy))^2}$$
and then I expand the bottom to get $$\frac{1}{(1-2x+x^2-y^2) + i(2y-2xy)}$$
The answer says $$w=\frac{(1+x^2-2x-y^2)-(i(2xy-2y)}{(1+x^2-2x-y^2)^2+(2xy-2y)^2}$$
How do I get to this stage?
| $$
\begin{align*}
\frac{1}{(1 - (x + iy))^2}
&= \frac{1}{1-2(x+iy)+(x+iy)^2}
= \frac{1}{1+x^2-y^2-2x + i(2xy - 2y)} \\
&= \frac{1 +x^2-y^2-2x-i(2xy - 2y)}{\left[ (1+x^2-y^2-2x)+i(2xy-2y) \right]\left[ (1+x^2-y^2-2x)-i(2xy-2y) \right]} \\
&= \frac{(1+x^2-y^2-2x) - i(2xy - 2y)}{(1+x^2-y^2-2x)^2 + (2xy-2y)^2}
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluating the Surface Integral $\iint_{x^3+y^3+z^3=a^3} \frac{\bf{x}}{||\bf{x}||} \cdot d\bf{S}$ Compute the surface integral: $$\int_S({x\over \sqrt{x^2+y^2+z^2}}, {y\over \sqrt{ x^2+y^2+z^2}}, {z\over \sqrt{x^2+y^2+z^2}}) \cdot \vec n \ dS$$
where $S: x^3+y^3+z^3=a^3$
The first parametrization that came to my mind was: $r(x,y)=(x,y,(a^3-x^3-y^3)^{1/3})$ but the integral becomes very hard to compute; I also gave
$$r(u,v)=(a(\cos(u)\sin(v))^{2/3},a(\sin(u)\sin(v))^{2/3},a(\cos(v))^{2/3})$$
(I was thinking about some type of spherical transformation) but then again the integral becomes vey hard to compute. Can you please help me with this problem? I would really appreciate it :)
| This is not a complete answer. I did some progress and convert the surface integral into an improper double integral. So anyone willing to find a solution may use the final result in this post.
Your surface for $a=1$ and ratio $1:1:1$ between the $x$, $y$, and $z$ axis looks like this
$\qquad \qquad \qquad \,\,$
so it is not a closed surface and we cannot use the divergence theorem as it was mentioned in the comments too. Next, note that
$$\begin{array}{}
g(x,y,z)=x^3+y^3+z^3-a^3 \\
{\bf{F}} = \frac{\bf{x}}{\left\| {\bf{x}} \right\|}= {x\over \sqrt{x^2+y^2+z^2}}{\bf{i}} + {y\over \sqrt{ x^2+y^2+z^2}} {\bf{j}} + {z\over \sqrt{x^2+y^2+z^2}} {\bf{k}} \\
{d\bf{S}}= {1 \over {\partial{g} \over \partial z}} \nabla g \, dx dy \\
{1 \over {\partial{g} \over \partial z}}\nabla g=\frac{1}{z^2}(x^2 {\bf{i}} + y^2 {\bf{j}} + z^2 {\bf{k})} \\
{\bf{F}} \cdot {d\bf{S}} = \frac{x^3+y^3+z^3}{z^2 \sqrt{x^2+y^2+z^2}} dx dy = \frac{a^3}{z^2 \sqrt{x^2+y^2+z^2}} dx dy\\
\end{array}$$
and finally the surface integral becomes
$$\iint_{S} {\bf{F}} \cdot {d\bf{S}} = \int_{x=-\infty}^{+\infty} \int_{y=-\infty}^{+\infty} \frac{a^3}{z^2 \sqrt{x^2+y^2+z^2}}dydx \tag{*}$$
and note that $z$ is a function of $x$ and $y$ by the relation
$$z=(a^3-x^3-y^3)^{\frac{1}{3}}$$
Now, one may work on $(*)$ to obtain some result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry express $4\cos x+3\sin x$ in the form $R \cos (x+a)$. I have been asked to express $4\cos x+3\sin x$ in the form $R \cos (x+a)$.
I know that the formula to express it in that form is $a \cos x+b\sin x=R \cos (x-a)$.
But as the question is asking me to express it in where it is $(x+a)$ instead of $(x-a)$ I am unsure of what to do.
Any help is much appreciated.
| I have never seen this kind of question but here is my attempt for a solution.
Assume, a $\triangle ABC$ right angled at B, opposite angle $\angle CAB$ as $a$, and hypotenuse as AC.
Now, presume BC is 4 units and BA is 3. So, AC becomes 5 units.
That implies, $\cos a=\frac{4}{5}$, $\sin a=\frac{3}{5}$
Coming back to your original question.
$4\cos x+3\sin x=k$
Divide by 5 both sides,
$\frac{4}{5}\cos x+\frac{3}{5} \sin x=\frac{k}{5}$
which is the form of
$\cos a\times\cos x+\sin a\times\sin x=\frac{k}{5}$
$\implies \cos(a-x)=\cos(x-a)=\frac{k}{5}$
Or,
$k=5\cos(x-a)=5\cos(x+(-a))$
$a$ is approximately $0.64$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use quadratic formula to find upper and lower limits of an expression
Using quadratic formula show that $\frac{x^2-x+1}{x^2+x+1}$ lies
between $3$ and $\frac{1}{3}$ for all real values of $x$.
Let $\frac{x^2-x+1}{x^2+x+1}=y$, then $\frac{x^2+1}{-x}=\frac{y+1}{y-1}$
$\therefore (y-1)x^2+(y+1)x+(y-1)=0$, if $x$ is real then
$(y+1)^2-4(y-1)(y-1)\geq 0$
$-3y^2+10y-3 \geq 0$
$(y-3)(-3y+1) \geq 0 \Rightarrow y>3 \text{ AND } y <\frac{1}{3}$
What am I doing wrong?
| you brought the equation till there
$$-3y^2+10y-3 \geq 0$$
divide it with -3 and reverse the inequality
$$y^2-\frac{10}{3}y+1\lt0$$
bring y in easy terms
$$\left( y-\frac{5}{3}\right)^2+1-\frac{25}{9}\lt0$$
$$\left( y-\frac{5}{3}\right)^2\lt\frac{16}{9}$$
taking square root note how I use the absolute function
$$\left| y-\frac{5}{3}\right|\lt\frac{4}{3}$$
then,
$$\frac{-4}{3}\lt y-\frac{5}{3}\lt\frac{4}{3}$$
adding $\frac{5}{3}$ throughout
$$\frac{1}{3}\lt y \lt{3}$$
hint
just be carful when you're playing with inequalities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
how to prove by contradiction that any distance between a curve $x^4 - x^2 + y^4 - y^2 = 0$ and the origin is less than or equal to $\sqrt{2}$ Given a closed trajectory $x^4 - x^2 + y^4 - y^2= 0$
Prove that any distance between any point on the curve and the origin does not exceed $\sqrt2$ (ie, maximum distance from the origin to the curve is $\sqrt2$)
I proved it using polar transformations but i'd rather use proof by contradiction without using polar transformation, so i tried
$$\begin{array}{l}
\left( {\exists \alpha ,\beta } \right){\rm{ }}{\alpha ^2} + {\beta ^2}>2\\
{\alpha ^4} - {\alpha ^2} + {\beta ^4} - {\beta ^2} = 0\\
\Rightarrow {\alpha ^4} + {\beta ^4} = {\alpha ^2} + {\beta ^2}\\
\Rightarrow {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} - 2{\left( {\alpha \beta } \right)^2} = {\alpha ^2} + {\beta ^2}{\rm{ }}\left( {{\rm{Completing - Square}}} \right)\\
\Rightarrow {\left( {{\alpha ^2} + {\beta ^2}} \right)^2} = {\alpha ^2} + {\beta ^2} + 2{\left( {\alpha \beta } \right)^2}\\
\Rightarrow {\left( {a + b} \right)^2} = a + b + 2ab \le 2\left( {{a^2} + {b^2}} \right){\rm{ }}\left( {{\rm{Cauchy - Schwarz - Inequality}}} \right)\\
\Rightarrow a + b \le \left( {{a^2} + {b^2}} \right) + \left( {{a^2} + {b^2}} \right) - 2ab\\
\Rightarrow a + b \le \left( {{a^2} + {b^2}} \right) + {\left( {a - b} \right)^2}\\
\Rightarrow {\left( {a - b} \right)^2} \ge \left( {a + b} \right) - \left( {{a^2} + {b^2}} \right) \ge 0\\
\Rightarrow \left( {a + b} \right) \ge \left( {{a^2} + {b^2}} \right)\\
\Rightarrow \left( {{\alpha ^2} + {\beta ^2}} \right) \ge \left( {{\alpha ^4} + {\beta ^4}} \right)\\
\Rightarrow \left( {{\alpha ^2} + {\beta ^2}} \right) - \left( {{\alpha ^4} + {\beta ^4}} \right) \ge 0
\end{array}
% MathType!MTEF!2!1!+-
% faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj
% 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x
% bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
% pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa
% aiaabeqaamaabaabaaGceaqabeaadaqadaqaaiabgoGiKiabeg7aHj
% aacYcacqaHYoGyaiaawIcacaGLPaaacaqGGaGaeqySde2aaWbaaSqa
% beaacaaIYaaaaOGaey4kaSIaeqOSdi2aaWbaaSqabeaacaaIYaaaaO
% GaaGOmaaqaaiabeg7aHnaaCaaaleqabaGaaGinaaaakiabgkHiTiab
% eg7aHnaaCaaaleqabaGaaGOmaaaakiabgUcaRiabek7aInaaCaaale
% qabaGaaGinaaaakiabgkHiTiabek7aInaaCaaaleqabaGaaGOmaaaa
% kiabg2da9iaaicdaaeaacqGHshI3cqaHXoqydaahaaWcbeqaaiaais
% daaaGccqGHRaWkcqaHYoGydaahaaWcbeqaaiaaisdaaaGccqGH9aqp
% cqaHXoqydaahaaWcbeqaaiaaikdaaaGccqGHRaWkcqaHYoGydaahaa
% WcbeqaaiaaikdaaaaakeaacqGHshI3daqadaqaaiabeg7aHnaaCaaa
% leqabaGaaGOmaaaakiabgUcaRiabek7aInaaCaaaleqabaGaaGOmaa
% aaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaa
% ikdadaqadaqaaiabeg7aHjabek7aIbGaayjkaiaawMcaamaaCaaale
% qabaGaaGOmaaaakiabg2da9iabeg7aHnaaCaaaleqabaGaaGOmaaaa
% kiabgUcaRiabek7aInaaCaaaleqabaGaaGOmaaaakiaabccadaqada
% qaaiaaboeacaqGVbGaaeyBaiaabchacaqGSbGaaeyzaiaabshacaqG
% PbGaaeOBaiaabEgacaqGTaGaae4uaiaabghacaqG1bGaaeyyaiaabk
% hacaqGLbaacaGLOaGaayzkaaaabaGaeyO0H49aaeWaaeaacqaHXoqy
% daahaaWcbeqaaiaaikdaaaGccqGHRaWkcqaHYoGydaahaaWcbeqaai
% aaikdaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGH
% 9aqpcqaHXoqydaahaaWcbeqaaiaaikdaaaGccqGHRaWkcqaHYoGyda
% ahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaWaaeWaaeaacqaHXoqy
% cqaHYoGyaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaakeaacq
% GHshI3daqadaqaaiaadggacqGHRaWkcaWGIbaacaGLOaGaayzkaaWa
% aWbaaSqabeaacaaIYaaaaOGaeyypa0JaamyyaiabgUcaRiaadkgacq
% GHRaWkcaaIYaGaamyyaiaadkgacqGHKjYOcaaIYaWaaeWaaeaacaWG
% HbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOyamaaCaaaleqaba
% GaaGOmaaaaaOGaayjkaiaawMcaaiaabccadaqadaqaaiaaboeacaqG
% HbGaaeyDaiaabogacaqGObGaaeyEaiaab2cacaqGtbGaae4yaiaabI
% gacaqG3bGaaeyyaiaabkhacaqG6bGaaeylaiaabMeacaqGUbGaaeyz
% aiaabghacaqG1bGaaeyyaiaabYgacaqGPbGaaeiDaiaabMhaaiaawI
% cacaGLPaaaaeaacqGHshI3caWGHbGaey4kaSIaamOyaiabgsMiJoaa
% bmaabaGaamyyamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadkgada
% ahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaacqGHRaWkdaqadaqa
% aiaadggadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGIbWaaWbaaS
% qabeaacaaIYaaaaaGccaGLOaGaayzkaaGaeyOeI0IaaGOmaiaadgga
% caWGIbaabaGaeyO0H4TaamyyaiabgUcaRiaadkgacqGHKjYOdaqada
% qaaiaadggadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGIbWaaWba
% aSqabeaacaaIYaaaaaGccaGLOaGaayzkaaGaey4kaSYaaeWaaeaaca
% WGHbGaeyOeI0IaamOyaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOm
% aaaaaOqaaiabgkDiEpaabmaabaGaamyyaiabgkHiTiaadkgaaiaawI
% cacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHLjYSdaqadaqaaiaa
% dggacqGHRaWkcaWGIbaacaGLOaGaayzkaaGaeyOeI0YaaeWaaeaaca
% WGHbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOyamaaCaaaleqa
% baGaaGOmaaaaaOGaayjkaiaawMcaaiabgwMiZkaaicdaaeaacqGHsh
% I3daqadaqaaiaadggacqGHRaWkcaWGIbaacaGLOaGaayzkaaGaeyyz
% Im7aaeWaaeaacaWGHbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaam
% OyamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaaqaaiabgkDi
% EpaabmaabaGaeqySde2aaWbaaSqabeaacaaIYaaaaOGaey4kaSIaeq
% OSdi2aaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaGaeyyzIm7a
% aeWaaeaacqaHXoqydaahaaWcbeqaaiaaisdaaaGccqGHRaWkcqaHYo
% GydaahaaWcbeqaaiaaisdaaaaakiaawIcacaGLPaaaaeaacqGHshI3
% daqadaqaaiabeg7aHnaaCaaaleqabaGaaGOmaaaakiabgUcaRiabek
% 7aInaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaiabgkHiTmaa
% bmaabaGaeqySde2aaWbaaSqabeaacaaI0aaaaOGaey4kaSIaeqOSdi
% 2aaWbaaSqabeaacaaI0aaaaaGccaGLOaGaayzkaaGaeyyzImRaaGim
% aaaaaa!4654!
$$
Since the last equation includes 0, I cannot show any contradiction.
I think the problem is that I did not use my hypothesis, ${\alpha ^2} + {\beta ^2}>2.
% MathType!MTEF!2!1!+-
% faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj
% 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x
% bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq
% pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa
% aiaabeqaamaabaabaaGcbaGaeqySde2aaWbaaSqabeaacaaIYaaaaO
% Gaey4kaSIaeqOSdi2aaWbaaSqabeaacaaIYaaaaOGaaGOmaaaa!34DD!
$
I think this claim can be proven using contradiction by I'm just not sure how to use my hypothesis.
| Here is a way using Power Means or QM-AM inequality:
$$\frac{a^2+b^2}2 \le \sqrt{\frac{a^4+b^4}2} = \sqrt{\frac{a^2+b^2}2} \implies \sqrt{a^2+b^2} \le \sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Infinite integrals$\int_0^{ + \infty } {\frac{1}{{\left( {x + 1} \right)\left( {{x^n} + 1} \right)}}dx} .$ How to calculate
$$\int_0^{ + \infty } {\frac{1}{{\left( {x + 1} \right)\left( {{x^n} + 1} \right)}}dx} .$$
| We have $\displaystyle\int_0^{\infty}\dfrac{dx}{(1+x)(1+x^n)}=\int_0^{\infty}\dfrac{dx}{1+x^n}-\int_0^{\infty}\dfrac{xdx}{1+x^n}$
$=\displaystyle\frac{1}{n}(B\Bigl(\frac{1}{n},1-\frac{1}{n},\Bigr)-B\Bigl(\frac{2}{n},1-\frac{2}{n},\Bigr))=\dfrac{\pi}{n\sin(\pi/n)}-\dfrac{\pi}{n\sin(2\pi/n)}$.
Second try. We have
$\displaystyle\int_0^\infty\dfrac{dx}{(1+x)(1+x^n)}$
$=\displaystyle\dfrac{1}{2}\int_0^\infty\dfrac{1}{1+x}-\dfrac{x^{n-1}}{1+x^n}dx+\sum_{k=1}^{n-1}\dfrac{(-1)^k}{2}\int_0^\infty\dfrac{x^{n-k}}{1+x^n}$
$=\displaystyle\sum_{k=1}^{n-1}\dfrac{(-1)^k}{2}\int_0^\infty\dfrac{x^{n-k}}{1+x^n}$
$=\displaystyle\sum_{k=1}^{n-1}\dfrac{(-1)^k}{2n}B\Bigl(\frac{n-k+1}{n},1-\frac{n-k+1}{n}\Bigr)$
$=\displaystyle\sum_{k=1}^{n-1}\dfrac{(-1)^k}{2n}\dfrac{\pi}{sin(\pi(n-k+1)/n)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
"Rationalizing the denominator" of $1/(a + b\sqrt[3]{2} + c\sqrt[3]{4})$? If $(a, b, c) \in \mathbb{Q}^3 \setminus \{(0, 0, 0)\}$, so that $a + b\sqrt[3]{2} + c\sqrt[3]{4}$ is a nonzero element of $\mathbb{Q}(\sqrt[3]{2})$, is there a formula for $${1\over{a + b\sqrt[3]{2} + c\sqrt[3]{4}}}$$ as a rational linear combination of $1$, $\sqrt[3]{2}$, and $\sqrt[3]{4}$?
| The following method may be practical for high school graduates:
Let $$y = x^2 + ax + b,\tag{1}$$ where $x = \sqrt[3]{z}$.
Then $$x^2 = y - ax - b.\tag{2}$$
Multiplying $(1)$ by $x$:
$$xy = z + ax^2 + bx.\tag{3}$$
Substituting $x^2$ from $(2)$ in $(3)$: $$xy = ay + (b - a^2)x - (ab - z). \tag{4}$$
Expressing $y$ from $(4)$: $$y = \frac{ux - v}{x - a},\tag{5}$$ where $u = b - a^2$ and $v = ab - z$.
Using the difference of cubes formula: $$(ux - v)(u^2x^2 + uvx + v^2) = u^3z - v^3.\tag{6}$$
Expressing $(ux - v)$ from $(6)$: $$ux - v = \frac{u^3z - v^3}{u^2x^2 + uvx + v^2}.\tag{7}$$
Substituting $(ux - v)$ from $(7)$in $(5)$: $$\frac{1}{y} = \frac{(x - a)(u^2x^2 + uvx + v^2)}{u^3z - v^3}.\tag{8}$$
Finally, opening parentheses in $(8)$: $$\frac{1}{y} = \frac{(uv - au^2)x^2 + (v^2 - auv)x + (u^2z - av^2)}{u^3z - v^3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Rolling two dice, what is the probability that two consecutive $7$s happens earlier than a $12$? Alice and Bob are playing a game involving two dice. If a sum of 12 appears, Alice wins and they stop playing. If a 7 appears twice in a row, Bob wins and they stop playing. What is the probability that Bob wins this game?
My thought was to draw a tree diagram, which I did, but I can't seem to wrap my head around the recursion that is in the problem. How do I put this into the tree? Thanks!
| Such problems are often solved by partitioning sample space on the first throw. Let $X$ be the RV denoting the sum in the first throw and $P(B)$ be the probability $B$ wins,then: $$\begin{align} P(B) &= P(B,X = 7) + P(B, X\neq 7) \\
& = P(B|X = 7) P(X = 7) + P(B|X\neq7) P(X\neq 7)\tag{a}\label{eq}\end{align}$$
Given that the first throw is 7, B wins immediately if the second throw is 7 as well. If second throw is neither 7 nor 12, then the game begins again. Hence, $$P(B|X=7) = \frac{1}{6} + \frac{29P(B)}{36}$$
For $P(B|X\neq 7)$ partition further by $X\neq 12$. $$\begin{align}P(B|X\neq 7) &= P(B, X\neq 12|X\neq 7) + P(B, X =12 |X\neq 7) \\ &= P(B| X\neq 12,X\neq 7)P(X\neq 12|X \neq 7) + P(B| X =12 ,X\neq 7)P(X = 12|X\neq 7) \end{align}$$
B can't win if first roll is $12$, i.e. $P(B| X =12 ,X\neq 7) = 0$. If first roll is neither 7 nor 12, we are back to original game. Lastly,
$$P(X\neq 12|X \neq 7) = \frac{P(X\neq 12,X \neq 7)}{P(X\neq 7)} = \frac{29}{30} $$
Thus, $P(B|X\neq 7) = \frac{29P(B)}{30}$
Put, all values in equation (\ref{eq})
$$P(B) = \left(\frac{1}{6}+ \frac{29P(B)}{36}\right)\frac{1}{6} + \frac{29P(B)}{30}\frac{5}{6}$$ Solves to $P(B) = \frac{6}{13}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
The equation of a tangent to a circle at a given point
18. Show that the equation of the tangent $PT$ at the point $P \left(\frac{1}{5}, \frac{3}{5}\right)$ on the circle
$$x^{2} + y^{2} + 8x + 10y - 8 = 0$$
is $3x + 4y - 3 = 0$.
Find the equations of the chords, each of length $4\ \sqrt{10}$, which are parallel to $PT$.
I proceeded as I've been taught:
The centre of the circle is $(-4, -5)$.
The gradient of the radius to $P$ is $\left.\frac{28}{5}\right/\frac{21}{5} = \frac{4}{3}$.
The gradient the reqd. tangent is $-\frac{3}{4}$.
$\therefore$ The equation of the tangent is:
$y + 5 = -\dfrac{3}{4}(x + 4)$
$4y + 20 = -3x - 12$
$3x + 4y + 32 = 0 \nLeftrightarrow 3x + 4y - 3 = 0$
Is this a misprint in the book, or am I losing my marbles? Since I'm practising in my spare time with nothing more than the book itself, it's hard for me to be sure. But given that I've got the coefficients of $x$ and $y$ right and I can't find an error, I'm forced to conclude that either the problem is malformed, or I have seriously misunderstood something.
| the radius at the point $P =(1/5, 3/5)$ has a slope of $\frac43.$ radius of the circle is $5$ and if half the chord is $2\sqrt{10},$ then by phytagora's theorem, it is $$\sqrt{7^2 - (2\sqrt{10})^2} = 3.$$ a point on the diameter with one endpoint $P$ is of the form $$(-4+3k, -5 + 4k) $$ equating the distance from the center $3$, we get $$k = \pm 3/5$$ the equation of the two chords that are of length $4\sqrt{10}$ are $$3x+4y = 3(-4+3k) + 4(-5+4k)=-32+25k= -17, -42. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Given $\csc\theta=-\frac53$ and $\pi<\theta<\frac32\pi$, evaluate sine ,cosine, and tangent of $2\theta$ If $\csc\theta=\frac{-5}{3}$, what is the exact value of $\tan(2\theta)$, $\sin(2\theta)$, and $\cos(2\theta)$ on the interval of $\left(\pi, \frac{3\pi}{2}\right)$?
I think I'm getting the fraction negatives wrong. I've used the sine formula($2\sin{\theta}\cos{\theta}$), the cosine formula ($2\cos^2{\theta}-1$) and the tangent formula, $\left(\frac{\sin(2\theta)}{\cos(2\theta)}\right)$
originally answering the problem I got $\sin(2\theta)=\frac{24}{25}$,
$\cos(2\theta)=\frac{7}{25}$, and $\tan(2\theta)=\frac{24}{7}$
| We are given that $\csc\theta = -\dfrac{5}{3}$ with $\pi < \theta < \dfrac{3\pi}{2}$.
Since $\csc\theta = \dfrac{1}{\sin\theta}$,
$$\sin\theta = \frac{1}{\csc\theta} = \frac{1}{-\frac{5}{3}} = -\frac{3}{5}$$
Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$ yields
\begin{align*}
\cos^2\theta & = 1 - \sin^2\theta\\
& = 1 - \left(-\frac{3}{5}\right)^2\\
& = 1 - \frac{9}{25}\\
& = \frac{16}{25}
\end{align*}
Since $\pi < \theta < \dfrac{3\pi}{2}$, $\theta$ is a third-quadrant angle, so we take the negative square root. Thus,
$$\cos\theta = -\frac{4}{5}$$
Hence,
\begin{align*}
\sin(2\theta) & = 2\sin\theta\cos\theta\\
& = 2\left(-\frac{3}{5}\right)\left(-\frac{4}{5}\right)\\
& = \frac{24}{25}\\
\cos(2\theta) & = \cos^2\theta - \sin^2\theta\\
& = \left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2\\
& = \frac{16}{25} - \frac{9}{25}\\
& = \frac{7}{25}\\
\tan(2\theta) & = \frac{\sin(2\theta)}{\cos(2\theta)}\\
& = \frac{\frac{24}{25}}{\frac{7}{25}}\\
& = \frac{24}{7}
\end{align*}
as you found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
If $\tan x=\sqrt{\frac{a}{b}}$ where a,b are positive real numbers and x is in 1st quadrant then find the value of $\sin x\sec^7x+\cos x\csc^7x$ The answer is $\frac{(a+b)^3(a^4+b^4)}{(ab)^{\frac{7}{2}}}$. I just want to now how to do it.
| you say $$\tan t = \sqrt{a/b}, 0 < t < \pi/2 \to y = \sin t = \sqrt{\frac a{a+b}}, x = \cos t = \sqrt{\frac b{a+b}}$$
then $$\begin{align}\sin t \sec^7 t + \cos t \csc ^7 t &= \frac y{x^7} + \frac x{y^7} \\
&= \frac{x^8+y^8}{(xy)^7} \\
&= \frac{(a+b)^3(a^4+b^4)}{(ab)^{7/2}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the value of $a^4+b^4+c^4$?
Consider $a,b,c$ such that $a+b+c =1, a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$. Find the value of $a^4+b^4+c^4$, if possible.
Trial: I observe that
\begin{align}
a^4+b^4+c^4 &=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)\\&=2^2-2[(ab+bc+ca)^2-2abc(a+b+c)]\\&=4-2[(ab+bc+ca)^2-2abc]
\end{align}
Then
\begin{align}
(a+b+c)^2 &=a^2+b^2+c^2+2(ab+bc+ca)\\\implies 1^2= 2+ 2(ab+bc+ca)\\\implies ab +bc+ca &=-\dfrac12
\end{align}
Then I am stuck. Please help.
| You may just use Newton's identities. If we consider
$$p(x)=(x-a)(x-b)(x-c) = x^3-e_1 x^2+e_2 x - e_3 \tag{1}$$
and define $p_n = a^n+b^n+c^n$, we have:
$$\left\{\begin{array}{rcl} e_1 &=& p_1,\\ 2e_2 &=& e_1 p_1-p_2,\\3e_3&=&e_2 p_1-e_1 p_2+p_3,\tag{2}\end{array}\right.$$
hence $e_1=1, e_2 =-\frac{1}{2},e_3 = \frac{1}{6} $. Assuming $x\in\{a,b,c\}$ we have $p(x)=0$, from which:
$$ x^4 = e_1 x^3 -e_2 x^2 +e_3 x \tag{3}$$
follows. Then summing $(3)$ over $x\in\{a,b,c\}$ we get:
$$ p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1 = 3e_1-2e_2+e_3 = 3+1+\frac{1}{6}=\color{red}{\frac{25}{6}}.\tag{4}$$
We may also notice that the discriminant of $p(x)$ is negative, hence the set $\{a,b,c\}$ is made by one real number and two (conjugated) complex numbers. That also follows from the fact that $S_1=\frac{e_1}{3},S_2=\frac{e_2}{3},S_3=e_3$ do not fulfill the Newton's inequality $S_1 S_3\leq S_2^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve trigonometric integral $\int_{-\pi/2}^{\pi/2} \frac{\sin^{2014}x}{\sin^{2014}x+\cos^{2014}x} dx $ Please help me to solve the following integral:
$$\int_{-\pi/2}^{\pi/2} \frac{\sin^{2014}x}{\sin^{2014}x+\cos^{2014}x} dx$$
I have tried a lot, but no results. I only transformed this integral to the following also not easy integral:
$$\int \frac{1}{(1+t^2)(1+t^{2014})} dt.$$
| Consider the integral
\begin{align}
I = \int_{-\pi/2}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx.
\end{align}
This may also be seen as
\begin{align}
I &= \int_{0}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx + \int_{-\pi/2}^{0} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx \\
&= \int_{0}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx - \int_{\pi/2}^{0} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx \\
&= 2 \, \int_{0}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx
\end{align}
where $x \to -x$ was made in the second integral. Now make the substitution $x = t-\pi/2$ to obtain
\begin{align}
I = 2 \, \int_{-\pi/2}^{0} \frac{\cos^{2a}t}{\sin^{2a}t + \cos^{2a}t} \, dt.
\end{align}
Now let $t \to -x$ to obtain
\begin{align}
I = 2 \, \int_{0}^{\pi/2} \frac{\cos^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx.
\end{align}
Adding the two integral expressions leads to
\begin{align}
2 I &= 2 \, \int_{0}^{\pi/2} \frac{\sin^{2a}x + \cos^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx = 2 \, \int_{0}^{\pi/2} dx = \pi.
\end{align}
It can now be stated that
\begin{align}
\int_{-\pi/2}^{\pi/2} \frac{\sin^{2a}x}{\sin^{2a}x + \cos^{2a}x} \, dx = \frac{\pi}{2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\int\frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+\zeta$ using trigonometric substitution We know that $\int\frac{1}{x^2-a^2}dx=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+\zeta$.
I tried to verify formula using trigonometric substitution and I had some problems.
Here is all my steps:
$\int\frac{1}{x^2-a^2}dx=\frac{1}{a}\int\frac{sec(\theta)\:d\theta}{tg(\theta)}=\frac{1}{a}\int{csc(\theta)}\:d\theta=-\frac{1}{a}\ln\left(\frac{x}{\sqrt{x^2-a^2}}+\frac{a}{x}\right)+\zeta\Rightarrow\theta=\sec^{-1}(\frac{x}{a})$.
How can I continue such that to get $\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|$ ?
| If I'm not mistaken, you cannot compute this integral without using partial fractions. If you make the substitution $x= a \sec(\theta)$, your integral "simplifies" to $\int \frac{d \theta}{a \sin \theta}$, from which you can write $\sin \theta = \sqrt {1-\cos^2 \theta}$ and use the substitution $y=\cos x$, which leads to the integral $-\frac{1}{a}\int \frac{dy}{1-y^2}$, which is more or less what you started with!
Thus the only (sensible?) way is to note that $$\int\frac{1}{x^2-a^2}dx = \frac{1}{2a}\left[\int\frac{1}{x-a}dx - \int\frac{1}{x+a}dx\right]$$ and the solution follows immediately as these are elementary integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate $\lim\limits_{x\to1}\frac{(1-x^{1/2})(1-x^{1/3})\cdots(1-x^{1/n})} {(1-x)^{n-1}}$ I used substitution
$t=1-x, x=1-t, t\rightarrow 0$
After the substitution:
$$\lim\limits_{t\to0}\frac{(1-(1-t)^{1/2})(1-(1-t)^{1/3})\cdots(1-(1-t)^{1/n})} {t^{n-1}}$$
How to use rationalization to get:
$$\lim\limits_{t\to0}\frac{t^{n-1}}{n! t^{n-1}}$$
Thanks for replies.
| The limit can be easily evaluated if we use the standard formula $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ for $a > 0$ and rational $n$. Putting $a = 1$ and $n = 1/k$ where $k$ is a positive integer we can see that $$\lim_{x \to 1}\frac{1 - x^{1/k}}{1 - x} = \lim_{x \to 1}\frac{x^{1/k} - 1}{x - 1} = \frac{1}{k}\tag{2}$$ Now our desired limit can be calculated as
\begin{align}
L &= \lim_{x \to 1}\frac{(1 - x^{1/2})(1 - x^{1/3})\cdots (1 - x^{1/n})}{(1 - x)^{n - 1}}\notag\\
&= \lim_{x \to 1}\frac{1 - x^{1/2}}{1 - x}\cdot \frac{1 - x^{1/3}}{1 - x}\cdots \frac{1 - x^{1/n}}{1 - x}\notag\\
&= \frac{1}{2}\cdot\frac{1}{3}\cdots\frac{1}{n} = \frac{1}{n!}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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.