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The number of divisors of a number whose sum of divisors is a perfect square Let $n$ denote a non-prime whose sum of divisors is a perfect square.
I have noticed a few surprising facts on the number of divisors of $n$:
*
*It is either prime or semi-prime or $27$ in all cases
*It is even only when $n=9$ or $n=2401$ (see table below)
A few examples:
Number | List of divisors | Sum of divisors | Number of divisors
--------|----------------------|-----------------|--------------------
9 | 1, 3 | 4 | 2
--------|----------------------|-----------------|--------------------
12 | 1, 2, 3, 4, 6 | 16 | 5
--------|----------------------|-----------------|--------------------
15 | 1, 3, 5 | 9 | 3
--------|----------------------|-----------------|--------------------
24 | 1, 2, 3, 4, 6, 8, 12 | 36 | 7
--------|----------------------|-----------------|--------------------
2401 | 1, 7, 49, 343 | 400 | 4
I have asserted this up to $1$ million:
*
*$1$ case where the number of divisors is $2$
*$1$ case where the number of divisors is $4$
*$4$ cases where the number of divisors is $27$
*$2514$ cases where the number of divisors is an odd prime
*$165$ cases where the number of divisors is an odd semi-prime
Is any proof or related-research with regards to any of these observations?
| The only numbers with exact $2$ proper divisors are the numbers of the form $p^2$, where
p is a prime.
The proper divisors are $1$ and $p$ in this case, and $p+1$ with $p$ prime
can only be a perfect square for $p=3$.
This follows from the equation $p=a^2-1=(a-1)(a+1)$. If $a>2$ , then $p$ cannot be a prime.
So, there is only $1$ case of
$2$ divisors.
For the case of $4$ divisors, we have to find all primes $p$, such that
$p^3+p^2+p+1=(p+1)(p^2+1)$ is a perfect square.
Suppose, $q$ is a divisor of $p+1$ and $p^2+1$, so we have $p\equiv -1\ (\ mod\ q\ )$ and
$p^2\equiv -1\ (\ mod\ q\ )$.
Since we also have $p^2\equiv 1\ (\ mod\ q\ )$, we
can conclude $q=2$.
The case $gcd(p+1,p^2+1)=1$ would imply, that $p+1$ is a square, which is only
possible for $p=3$, as already mentioned, but $3^2+1=10$ is not a square.
So, we can conclude that
$$p+1=2a^2\ \ \ \ \ \ \ p^2+1=2b^2$$
with $gcd(a,b)=1$
It seems that only $p=7$ solves these equations. If there is another solution, it must
contain more than $100\ 000$ digits which I checked examining the solutions of the
equation $x^2-2y^2=-1$
The number of proper divisors is even only for squares. I checked them and
found two more examples for an even number :
$$35713^2=1275418369$$
has $8$ proper divisors.
$$102851^2=10578328201$$
has $14$ proper divisors.
Furthermore, I found an example with $3$ distinct prime factors :
$195534000$ has $399=3\times 7 \times 19$ proper divisors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Determine the limit of a series, involving trigonometric functions: $\sum \frac{\sin(nx)}{n^3}$ and $\frac{\cos(nx)}{n^2}$ I have $$\sum^\infty_{n=1} \frac{\sin(nx)}{n^3}.$$
I did prove convergence:
$0<\theta<1$
$$\left|\frac{\sin((n+1)x)n^3}{(n+1)^3\sin(nx)}\right|< \left|\frac{n^3}{(n+1)^3}\right|<\theta$$
Now I want to determine the limit. I found a similar proof but I need help understanding it; it goes like this. :
$$ F(x):=\sum^\infty_{n=1} \frac{\cos(nx)}{n^2}$$
As for this series we have uniform convergence. The series of derivatives: $$-\sum^\infty_{n=1} \frac{\sin(nx)}{n}$$ converges for every $\delta >0$ on the interval $[\delta, 2\pi-\delta]$ uniform against $\frac{x-\pi}{2}$
so, for every $x \in]0,2\pi[$ : $\displaystyle F'(x) = \frac{x-\pi}{2}$$\displaystyle F(x) = \left(\frac{x-\pi}{2}\right)^2+c,c\in \mathbb{R}$.
To determine the constant we calculate:
$$ \int^{2\pi}_0F(x)dx=\int^{2\pi}_0\left(\frac{x-\pi}{2}\right)^2dx+\int^{2\pi}_0cdx=\frac{\pi^3}{6}+2\pi c$$
(Question: Why can we do this do get the constant?)
Because $\int^{2\pi}_0cos(nx)dx= 0 \forall n≥1$ we have:
$$\int^{2\pi}_0F(x)dx = \sum^\infty_{n=1}\int^{2\pi}_0\frac{\cos(nx)}{n^2}=0,$$ so $c = -\frac{\pi^2}{12}$. (Question: How does he get to that term $\frac{\pi^2}{12}$?) With that we have proven, that
$$\sum^\infty_{n=1} \frac{\cos(nx)}{n^2}=\left(\frac{x-\pi}{2}\right)^2-\frac{\pi^2}{12}$$
If you can explain one of the questions about this proof, or if you know how to calculate the limit in my situation above, it would be cool if you leave a quick post here, thanks!
| $$\sum^\infty_{n=1}\frac{sin(nx)}{n^3}= ?$$
We know that $$\sum^\infty_{n=1}\frac{sin(nx)}{n}= \frac{x-\pi}{2}$$.
$$(\frac{sin(nx)}{n^3})''= (\frac{cos(nx)}{n^2})'=-\frac{sin(nx)}{n}$$
$$-\sum^\infty_{n=1}\frac{sin(nx)}{n}= -(\frac{x-\pi}{2})$$
$$ - \int \frac{x-\pi}{2}= -\int \frac{x}{2}-\frac{\pi}{2} = \frac{x^2}{4}+\frac{\pi x}{2}+c$$
$$\int\frac{x^2}{4}+\frac{\pi x}{2}+c=\frac{\pi x^2}{4}+\frac{x^3}{12}+xc$$
We can also determine the constant. We know that $-f'(0) = 0 + c$.
$$-\sum^\infty_{n=1} \frac{cos(n0)}{n^2}= - \sum^\infty_{n=1} \frac{1}{n^2} = -\frac{\pi^2} {6} = 0+c$$
$$-\int\frac{\pi^2} {6} = -\frac{x\pi^2} {6}$$ So
$$\sum^\infty_{n=1}\frac{sin(nx)}{n^3}= \frac{\pi x^2}{4}+\frac{x^3}{12}+(-\frac{x\pi^2}{6}) = \frac{x\pi^2}{6}-\frac{\pi x^2}{4}+\frac{x^3}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the integral of $\frac{\sqrt{x^2-49}}{x^3}$ I used trig substitution and got
$\displaystyle \int \dfrac{7\tan \theta}{343\sec ^3\theta}d\theta$
Then simplified to sin and cos functions, using U substitution with a final answer of:
$\dfrac{-7}{3x^3}+C$
Which section did I go wrong in. Any help would be appreciate!
| $$\int \frac{\sqrt{x^2-49}}{x^3}$$
$$=\int \frac{7\sec \theta\tan\theta\sqrt{49\sec^2 \theta-49}}{\sec^3 \theta}$$
$$=\int \frac{49\tan^2 \theta}{7^3\sec^2 \theta}$$
$$=\frac{1}{7}\int {\sin^2 \theta}$$
Can you find $\int \sin^2 \theta$?
Hint: $\sin^2 x= \dfrac 12 (1-\cos 2\theta)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $f^{(n)}(1)$ where $f(x)={1\over x(2-x)}$. Find $f^{(n)}(1)$ where $f(x)={1\over x(2-x)}$.
What I did so far:
$f(x)=(x(2-x))^{-1}$. $f'(x)=-(x(2-x))^{-2}[2-2x]$ $f''(x)=2(x(2-x))^{-3}[2-2x]^2+2(x(2-x))^{-2}$. It confuses me a lot. I know I have to determine where I have an expression multiplied by $2-2x$ and where not, and what is its form. I would appreciate your help.
| Notice you're asked to find $f^{(n)}(1)$, not $f^{(n)}(x)$. A series expansion around $a=1$ is a nice way to do this without taking any derivatives. If you can find a power series representation of the form $f(x) = \sum_{n=0}^\infty c_n (x-a)^n$ with positive radius of convergence, then $c_n = \frac{f^{(n)}(a)}{n!}$. Therefore $f^{(n)}(a) = c_n n!$.
Following the partial fractions hint we have
\begin{align*}
\frac{1}{x(2-x)} &= \frac{1}{2} \cdot\frac{1}{x} - \frac{1}{2} \cdot\frac{1}{x-2} \\
&= \frac{1}{2} \cdot \frac{1}{1+(x-1)} + \frac{1}{2}\cdot\frac{1}{1 - (x-1)}
\end{align*}
Now we write the fractions as sums of geometric series. Since
\begin{align*}
\frac{1}{1-u} &= \sum_{n=0}^\infty u^n \\
\frac{1}{1+u} = \frac{1}{1-(-u)} &= \sum_{n=0}^\infty(-u)^n = \sum_{n=0}^\infty(-1)^nu^n \\
\end{align*}
(when $|u|< 1$) we have
\begin{align*}
\frac{1}{2} \cdot \frac{1}{1+(x-1)} + \frac{1}{2}\cdot\frac{1}{1 - (x-1)}
&= \frac{1}{2} \sum_{n=0}^\infty (-1)^n(x-1)^n + \frac{1}{2}\sum_{n=0}^\infty (x-1)^n \\
&= \sum_{n=0}^\infty\left(\frac{1 + (-1)^n}{2}\right) (x-1)^n
\end{align*}
By the initial remark we have
$$
f^{(n)}(1) = n! \left(\frac{1 + (-1)^n}{2}\right)
$$
The factor on the right can be simplified if we know the parity of $n$. If $n$ is odd $(-1)^n = -1$ and the numerator is zero. If $n$ is even $(-1)^n =1$, the numerator is $2$, and the quotient is $1$. Therefore
$$
f^{(n)}(1) = \begin{cases} n! & \text{$n$ is even} \\ 0 & \text{$n$ is odd} \end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to use the initial condition in this DE: $x^2\frac{dy}{dx}=\frac{4x^2-x-2}{\left(x+1\right)\left(y+1\right)}$ Here is my proposed solution:
\begin{align}
x^2\frac{dy}{dx}&=\frac{4x^2-x-2}{\left(x+1\right)\left(y+1\right)}\tag{1}\\
\implies & \left(y+1\right)\frac{dy}{dx}=\frac{4x^2-x-2}{x^2\left(x+1\right)}\tag{2}\\
\implies & \int\frac{dy}{dx}\left(y+1\right)\:dx=\int\frac{4x^2-x-2}{x^2\left(x+1\right)}\:dx\tag{3}\\
\implies & \frac{y^2}{2}+y=\int\alpha\left(x\right)\:dx,\tag{4}
\end{align}
and here I'm letting $\alpha\left(x\right)$ denote the partial fraction decomposition of the integrand in the RHS, which is
\begin{align}
\alpha\left(x\right)=\frac{4x^2-x-2}{x^2\left(x+1\right)}=\frac{1}{x}-\frac{2}{x^2}+\frac{3}{x+1},\tag{5}
\end{align}
and this gives me
\begin{align}
\frac{y^2}{2}+y=\log\left|x\right|+\frac{2}{x}+3\log\left|x+1\right|+C.\tag{6}
\end{align}
Okay, so no problems there. But now the IC is such that $y\left(1\right)=1.$ I don't know what to do with these two numbers because this is an implicit answer and I cannot separate the $y$'s.
Thank you for your time,
| $y(1)=1$ implies (by plugging 1 into both $x$ and $y$)
$$\frac12+1=\log 1+\frac21+3\log 2+C.$$
Now, solve for $C$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integers $x$ such that $x^2+3x+24$ is a perfect square. Find all integers $x$ such that $x^2+3x+24$ is a perfect square.
My attempt:
$x^2+3x+24=k^2$
$3(x+8)=(k+x)(k-x)$
Now, do I find solution treating cases? But that doesn't seem very easy. Please help.
| Best way is....
$$x^2 + 3x + 24 = k^2 \Rightarrow x^2 + 3x + 24-k^2 = 0\\$$
$$\triangle = b^2 - 4ac = 3^2 - 4(1)(24-k^2) = 9 - 96 + 4k^2 = 4k^2 - 87 = n^2\\$$
$$\to 4k^2 - n^2 = (2k-n)(2k+n) = 87 \\$$
$$\to 2k+n = 29, 2k-n = 3 \to 4k = 32 \to k = 8$$. Can you finish it?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128670",
"timestamp": "2023-03-29T00:00:00",
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How to prove this: $a^4+b^4+2 \ge 4ab$? How to prove this: $a^4+b^4+2 \ge 4ab$?
$a$ and $b$ are reals.
| Inequalities like this can most often be proved or simplified by using the Arithmetic-Geometric mean inequality.
First apply AM-GM in $a^4$ and $b^4$:
$$\frac{a^4+b^4}{2} \ge \sqrt{a^4b^4}$$
$$a^4+b^4 \ge 2a^2b^2$$
Now apply AM-GM in $2a^2b^2$ and $2$:
$$\frac{2a^2b^2+2}{2} \ge \sqrt{4a^2b^2}$$
$$2a^2b^2+2 \ge 4ab$$
I suppose the result is quite obvious now. This answer was, of course, posted with the assumption that $a$ and $b$ are reals.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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I need the proving of $x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$ I could find the role of Striling Numbers in the natural logarithm function as follows
$$x\log(x)=(\frac{x-1}{x})+\frac{3}{2!}(\frac{x-1}{x})^2+\frac{11}{3!}(\frac{x-1}{x})^3+...\frac{S_{n}}{n!}(\frac{x-1}{x})^n$$
Where $S_n$= absolute Striling Numbers of first kind (0,1,3,11,50,274....)
this series numerically checked without any problem, but I need the proving. Any help?
| we will need the fact that $$n!H_n = n!\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n} \right) = S(n,2)\\ \text{ where } S(n,2) \text{ is the Stirlings number of the first kind}$$
let me make a change of variable $u= \dfrac{x-1}{x}, x = \dfrac{1}{1-u}.$ then
$\begin{align} x\ln x &= \dfrac{1}{1-u} \ln \left(\dfrac{1}{1-u}\right) = -\dfrac{1}{1-u} \ln (1-u)\\
&=\left(1+u+u^2 + \cdots\right)\left(u+\frac{u^2}{2}+\frac{u^3}{3} + \cdots\right)\\
&=u+(\frac{1}{2}+1)u^2+(\frac{1}{3}+\frac{1}{2}+1)u^3+(\frac{1}{4}+\frac{1}{3} + \frac{1}{2}+1)u^4+\cdots\\
&=u+H_2u^2+H_3u^3+H_4u^4+\cdots\\
&=u+\frac{S(2,2)}{2!}u^2+\frac{S(3,2)}{3!}u^3+\frac{S(4,2)}{4!}u^4+\cdots
+\frac{S(n,2)}{n!}u^n+\cdots\\
\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
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What to do after finding products for coefficients in generating functions? The problem is as follows:
$\text{Determine the coef. of } x^{10} \text{ in } (x^3 + x^5 + x^6)(x^4 + x^5 + x^7)(1+x^5+x^{10}+x^{15}+...)$
With this, there are three ways to get $x^{10}$.
1) $x^3\cdot x^7\cdot1$
2) $x^5\cdot x^5\cdot1$
3) $x^6\cdot x^4\cdot 1$
But then what? I don't understand where to go on from here.
| $$\sum_i x_i \sum_j y_j \sum_k z_k = \sum_{i,j,k} x_i y_j z_k$$
Suppose we wanted to multiply $(x_1 + x_2)(y_1 + y_2 + y_3)$. We take every combination of $x_iy_j$ and sum them together. In every possible way, we choose one term from the first bracket, either $x_1$ or $x_2$ and then multiply it by one term from the second bracket, either $y_1$, $y_2$ or $y_3$. Let's look at a picture.
So how do we multiply $(x^3 + x^5 + x^6) (x^4 + x^5 + x^7) (1 + x^5 + x^{10})$. Doing the same as above, we will get $27$ terms, which is unwieldy. Instead let's just pick out the ones that contribute to $x^{10}$.
$$
(x^3 + x^5 + x^6) (x^4 + x^5 + x^7) (1 + x^5 + x^{10}) \\ \; \\
\begin{array}
( &= &... &+& (x^3) (x^7)(1) &+& ... &+& (x^5)(x^5)(1) &+& ... &+& (x^6)(x^4)(1) &+& ... \\
&= &... &+& x^{10} &+& ... &+& x^{10} &+& ... &+& x^{10} &+& ...\\
&=& ... &+& 3x^{10} &+& ...\\
\end{array}
$$
There is nothing fancy about this calculation; all we are doing is standard multiplication.
You linked this post. Let's take a look at the product.
$$ (\;)_1 \;\; (\;)_2\;\; (\;)_3\;\; (\;)_4\;\; (\;)_5 = \\
\left(x^{31}\right) \left(1-x^{16}\right) \left(1-x^{15}\right) (1-x^{26}) \left(\sum_{i\geq 0} {k+2 \choose k} x^k \right)
$$
Now we want to find the coefficient of $x^{52}$. I want you to carry it out yourself so you understand. Here is your first exercise: How can we choose one term from each $()_i$ such that their combined product results in $x^{52}$ (ignoring coefficients)? You should find that there are only three ways to do this.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $x,y,z>0,xyz=1$. Prove that $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\ge \frac34$ Let $x,y,z>0$ and $xyz=1$. Prove that $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{y^3}{(1+x)(1+z)}+\dfrac{z^3}{(1+x)(1+y)}\ge \dfrac34$
My attempt:
Since it is given that $xyz=1$, I tried substituting $x=\dfrac{a}{b},y=\dfrac{b}{c},z=\dfrac{c}{a}$. But the expansion looked really ugly and I didn't think I could make out anything out of it.
So, I made another attempt, if each element was greater than $\dfrac{1}{4}$, we could have a solution, so, treating that way, I get $4x^3\ge 1+x+y+xy, 4y^3\ge 1+x+z+xz, 4z^3\ge 1+x+y+xy$. Using AM-GM I get an equality.
So, please help. Thank you.
| Edit1:
another approach is to use $\dfrac{x^2}{a}+\dfrac{y^2}{b} \ge \dfrac{(x+y)^2}{a+b},x+y+z\ge3$
LHS $\ge \dfrac{(x^2+y^2+z^2)^2}{\sum_{cyc} {(1+y)(1+z)}{x}}\ge\dfrac{(x+y+z)^4}{9(x+y+z+2(xy+yz+xz)+3)}\ge \dfrac{(x+y+z)^4}{9(x+x+z)+6(x+y+z)^2+27}\ge\dfrac{(x+y+z)^4}{9(x+y+z)+6(x+y+z)^2+9(x+y+z)}=\dfrac{(x+y+z)^3}{18+6(x+y+z)}\ge\dfrac{(x+y+z)^3}{6(x+x+z)+6(x+y+z)}=\dfrac{(x+y+z)^2}{12}\ge \dfrac{3^2}{12}=\dfrac{3}{4} $
all "=" hold when $x=y=z=1$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Is there a closed-form of $\sum_{n=1}^{\infty }\frac{\cos^2(n)}{n^2}$ Is there a closed-form of $$\sum_{n=1}^{\infty }\frac{\cos^2(n)}{n^2}$$
| Since over the interval $(0,\pi)$ we have:
$$\frac{\pi-x}{2}=\sum_{n\geq 1}\frac{\sin(nx)}{n}\tag{1}$$
over the same interval we have also:
$$\frac{2\pi x-x^2}{4} = \sum_{n\geq 1}\frac{1-\cos(nx)}{n^2}\tag{2} $$
or:
$$\frac{2\pi x-x^2}{8} = \sum_{n\geq 1}\frac{\sin^2\left(n\frac{x}{2}\right)}{n^2}\tag{3} $$
so by setting $x=2$ we get:
$$\sum_{n\geq 1}\frac{\sin^2 n}{n^2}=\frac{\pi-1}{2}, \qquad \sum_{n\geq 1}\frac{\cos^2 n}{n^2}=\color{red}{\frac{\pi^2-3\pi+3}{6}}.\tag{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $ \sum_{n=2}^{+\infty}$ $\frac{1}{n^{3}-n}$ Find:
$$ \sum_{n=2}^{+\infty}\frac{1}{n^{3}-n}$$
I tried to resolve into partial fractions to see if there are some cancellations ,but that did not helped me .How do i do this ?Thanks
| Since $n^3 - n = n(n^2 - 1) = n(n-1)(n+1)$, $$\frac{1}{n^3 - n} = \frac{1}{n}\frac{1}{(n-1)(n+1)} = \frac{1}{2n}\left(\frac{1}{n-1} - \frac{1}{n+1}\right) = \frac{1}{2(n-1)n} - \frac{1}{2n(n+1)}.$$ Now $\sum_{n = 2}^\infty 1/(n^3 - n)$ telescopes to $1/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Derivative of $\frac { y }{ x } +\frac { x }{ y } =2y$ with respect to $x$ $$\frac { y }{ x } +\frac { x }{ y } =2y$$
Steps I took:
$$\frac { d }{ dx } \left[yx^{ -1 }1+xy^{ -1 }\right]=\frac { d }{ dx } [2y]$$
$$\frac { dy }{ dx } \left(\frac { 1 }{ x } \right)+(y)\left(-\frac { 1 }{ x^{ 2 } } \right)+(1)\left(\frac { 1 }{ y } \right)+(x)\left(-\frac { 1 }{ y^{ 2 } } \right)\frac { dy }{ dx } =(2)\frac { dy }{ dx } $$
$$-\frac { y }{ x^{ 2 } } +\frac { 1 }{ y } =(2)\frac { dy }{ dx } -\left(\frac { 1 }{ x } \right)\frac { dy }{ dx } +\left(\frac { x }{ y^{ 2 } } \right)\frac { dy }{ dx } $$
$$-\frac { y }{ x^{ 2 } } +\frac { 1 }{ y } =\left(2-\frac { 1 }{ x } +\frac { x }{ y^{ 2 } } \right)\frac { dy }{ dx } $$
$$\frac { -\frac { y }{ x^{ 2 } } +\frac { 1 }{ y } }{ \left(2-\frac { 1 }{ x } +\frac { x }{ y^{ 2 } } \right) } =\frac { dy }{ dx } $$
At this point I get stuck because once I simplify the result of the last step I took, the answer is not what it should be. I think that I am making a careless mistake somewhere but I cannot seem to find it. Hints only, please. The direct answer does nothing for me.
Actual answer:
$$\frac { d y}{ dx } =\frac { y(y^{ 2 }-x^{ 2 }) }{ x(y^{ 2 }-x^{ 2 }-2xy^{ 2 }) } $$
| it will be a whole lot easier to turn $$ \frac { y }{ x } +\frac { x }{ y } =2y$$ into $$ x^2 + y^2 = 2xy^2$$ and difference as $$2x\,dx + 2y\, dy = 4xy \, dy +2y^2 \, dx $$ which can then be simplifies as $$\frac{dy}{dx} = \frac{y^2 - x}{y(1-2x)}\, \text{ and subject to the onstraint } \frac { y }{ x } +\frac { x }{ y } =2y.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the inverse of a matrix with an unknown variable. I am trying to Express the inverse of the following matrix (assuming it exists) as a matrix containing expressions in terms of k.
$\left[ {\begin{array}{cc}
-4 & -4 & -4 \\
4 & 9 & 4 \\
k & -1 & 0
\end{array} } \right]$
Now I know that it is completely possible to solve the equation by using the conventional method for finding the inverse of a matrix. However, since there is an unknown variable, this became overly complicated to solve. Is there any other method that simplifies everything?
| The adjugate matrix is a way to explicitly calculate an inverse matrix from the matrix. It is most definitely not the most efficient way to do it but it is explicit. The adjugate matrix is the transpose of the cofactor matrix. In your case, we get for the adjugate matrix
$$\begin{pmatrix}
\left|\begin{matrix} 9 & 4 \\ -1 & 0\end{matrix}\right| & -\left|\begin{matrix} -4 & -4 \\ -1 & 0\end{matrix}\right| & \left|\begin{matrix} -4 & -4 \\ 9 & 4 \end{matrix}\right| \\
-\left|
\begin{matrix}4 & 4 \\ k & 0\end{matrix}\right| & \left|\begin{matrix} -4 & -4 \\ k & 0\end{matrix}\right| & -\left|\begin{matrix} -4 & -4 \\ 4 & 4\end{matrix}\right| \\
\left|\begin{matrix} 4 & 9 \\ k & -1\end{matrix}\right| & -\left|\begin{matrix} -4 & -4 \\ k & -1\end{matrix}\right| & \left|\begin{matrix} -4 & -4 \\ 4 & 9\end{matrix}\right|
\end{pmatrix}$$
Computing these we get
$$\begin{pmatrix} 4 & 4 & 20 \\ 4k & 4k & 0 \\ -4-9k & -4k-4 & -20\end{pmatrix}.$$
Finally, to get the inverse we need only to compute the determinant. Computing the determinant gives
$$k\left|\begin{matrix} -4 & -4 \\ 9 & 4\end{matrix}\right| - (-1)\left|\begin{matrix} -4 & -4 \\ 4 & 4\end{matrix}\right| + 0\left|\begin{matrix} -4 & -4 \\ 4 & 9\end{matrix}\right| = 20 k.$$
Thus for an inverse we have
$$\frac{1}{20k}\begin{pmatrix} 4 & 4 & 20 \\ 4k & 4k & 0 \\ -4-9k & -4k-4 & -20\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve the differential equation $y'-xy^2 = 2xy$ I get it to the form $\left | \dfrac{y}{y+2} \right |=e^{x^2}e^{2C}$ but I'm not sure how to get rid of the absolute value and then solve for y. I've heard the absolute value can be ignored in differential equations. Is this true?
| We have
$$y' - x{y^2} = 2xy \Rightarrow y' - 2xy = x{y^2} \Rightarrow y'.{y^{ - 2}} - 2x{y^{ - 1}} = x,\,\,\left( {y \ne 0} \right).$$
Put $z=y^{-1}$. Then
$$\begin{gathered}
z' + 2xz = - x \Rightarrow \frac{d}{{dx}}\left( {z{e^{{x^2}}}} \right) = - x{e^{{x^2}}} \\
\Rightarrow z = {e^{ - {x^2}}}\int {\left( { - x{e^{{x^2}}}} \right)dx} = - \frac{1}{2}{e^{ - {x^2}}}\int {{e^{{x^2}}}d\left( {{x^2}} \right)} \hfill \\
= - \frac{1}{2}{e^{ - {x^2}}}\left( {{e^{{x^2}}} + C} \right) = - \frac{1}
{2} - \frac{C}{2}{e^{ - {x^2}}}. \hfill \\
\end{gathered} .$$
Finally,
$$y = {z^{ - 1}} = \frac{1}{z} = \frac{1}{{ - \frac{1}{2} - \frac{C}{2}{e^{ - {x^2}}}}} = - \frac{2}{{1 + C{e^{ - {x^2}}}}}.$$
Note: $y=0$ is also a solution of the equation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to compute this limits given these conditions.
if $f(1)=1$ and $f'(x)=\frac{1}{x^2+[f(x)]^2}$ then compute $\lim\limits_{x\to+\infty}f(x)$
i tried to write it was
$$\frac{dy}{dx}=\frac{1}{x^2+y^2}\\
(x^2+y^2)\frac{dy}{dx}=1\\
(x^2+y^2)dy=dx$$
by the help
$$\begin{align}
f(x)&\le1+\int_1^x\frac{dt}{1+t^2}\\
&\le1+\arctan t\bigg|_1^x\\
&\le1+\arctan x-\arctan 1\\
&\le1+\arctan x-\frac{\pi}{4}
\end{align}$$
so
$$\begin{align}
\lim\limits_{x\to+\infty}f(x)&\le\lim\limits_{x\to+\infty}1+\arctan x-\frac{\pi}{4}\\
&\le1+\frac{\pi}{2}-\frac{\pi}{4}\\
&\le1+\frac{\pi}{4}=\frac{4+\pi}{4}
\end{align}$$
| From the other answer it is clear that $f(x)$ is strictly increasing in $[1,\infty)$ and is bounded above by $A=1+\dfrac{\pi}{4}$ and hence $\lim\limits_{x\to\infty}f(x)=L$ exists and we have $1<L\leq A$ and $f(x)< L\leq A$. Next we can see that
\begin{align}f'(x)=\frac{1}{x^{2}+\{f(x)\}^{2}}&>\frac{1}{x^{2}+L^{2}}\notag\\
\Rightarrow \int_{1}^{x}f'(t)\,dt&>\int_{1}^{x}\frac{dt}{t^{2}+L^{2}}\notag\\
\Rightarrow f(x)&>1+\frac{1}{L}\tan^{-1}\left(\frac{x}{L}\right)-\frac{1}{L}\tan^{-1}\left(\frac{1}{L}\right)\notag\\
\Rightarrow L&\geq 1+\frac{\pi}{2L}-\frac{1}{L}\tan^{-1}\left(\frac{1}{L}\right)=1+\frac{\tan^{-1}L}{L}\notag\end{align}
Hence we need to find the least number $L \in(1,A]$ which satisfies the above inequality. From online graphing calculator it looks like the desired value of $L$ is around $1.6268\dots$
Another take on this problem is as follows. Let $g(x)=g_{1}(x)=1+\dfrac{\arctan x}{x}$ and let $g_{n}(x)$ denote $n^{\text{th}}$ iterate of $g(x)$. Then $A=g_{1}(1)$ and we can easily show that $$g_{2n}(1)\leq L\leq g_{2n+1}(1)$$ for all $n$. It follows that $L$ is given as the unique root of $g(x)=x$ which is around $1.6268\dots$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How should you prove product rules by induction? For example:
$$\prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n}$$
For every $n$ greater than or equal to $2$
my approach for this was that I need to prove that:
$$ \left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)=\frac{n+1+1}{2(n+1)}$$
is this the right approach? Because when i try and work out the algebra i keep on hitting a wall.
\begin{align}
\left(1-\frac{1}{n^2}\right)\left(1-\frac{1}{(n+1)^2}\right)&=1-\frac 1{(1-n)^2}-\frac 1{n^2}-\frac 1{n^2(n+1)} \\
&=\frac{n^2}{(n+1)^2}-\frac 1{(n+1)^2} \\
&=\frac{n^2-1}{(n+1)^2}-\frac 1{n^2}+\frac 1{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)}{n^2(n+1)^2}-\frac{(n+1)^2}{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)-(n+1)^2}{n^2(n+1)^2}+\frac 1{n^2(n+1)^2} \\
&=\frac{n^2(n^2-1)-(n+1)^2+1}{n^2(n+1)^2}
\end{align}
| Hint:
$$P_{n-1}\left(1-\frac1{n^2}\right)=\frac{(n-1)+1}{2(n-1)}\left(1-\frac1{n^2}\right)=\frac{n+1}{2n}=P_n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1154218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Un-Simplifying a fraction, i.e. computing partial fraction decomposition $\frac{3x^2+17x}{x^3+3x^2+-6x-8}$
I need to find the value of C in the form of
$\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{x+4}$
which is based on the fraction give at the top.
I can get so far to do the following:
$A(x^2+2x-8) + B(x^2+5x+4) + C(x^2-x-2) = 3x^2+17x$
No clue on my next step or even if this is the right step.
| Compare coefficients of both sides polynomials. for example by comparing coefficient next you $x^2$ you will get $A+B+C = 3$. Do the same for $x^1$ and $x^0$ and solve system of 3 linear equations.
The other way is to put arbitrary 3 values of $x$, for example $-1$, $0$ and $1$, and again you will get system of 3 linear aquations with variables $A$, $B$ and $C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does one show that for $k \in \mathbb{Z_+},3\mid2^{2^k} +5$ and $7\mid2^{2^k} + 3, \forall \space k$ odd.
For $k \in \mathbb{Z_+},3\mid2^{2^k} +5$ and $7\mid2^{2^k} + 3, \forall \space k$ odd.
Firstly,
$k \geq 1$
I can see induction is the best idea:
Show for $k=1$:
$2^{2^1} + 5 = 9 , 2^{2^1} + 3 = 7$
Assume for $k = \mu$
so: $3\mid2^{2^\mu} + 5 , \space 7\mid2^{2^\mu} + 3$
Show for $\mu +2$
Now can anyone give me a hint to go from here? My problem is being able to show that $2^{2^{\mu+2}}$ is divisible by 3, I can't think of how to begin how to show this.
| You have already shown that the base cases hold.
Assume $3\mid 2^{2^k}+5$. Then $2^{2^k}\equiv 1$ mod $3$. Hence:
$$2^{2^{k+1}}=2^{2^k*2}=\left(2^{2^k}\right)^2\equiv 1 \text{ mod } 3$$
Hence $3\mid 2^{2^{k+1}}+5$.
In the same way:
Assume $7\mid 2^{2^{k}}+3$. Then $2^{2^{k}}\equiv 4$ mod $7$. Hence:
$$2^{2^{k+2}}=\left(2^{2^k}\right)^4\equiv 4^4 \text{ mod } 7$$
And since $4^4=256=36*7+4$, we see that $256\equiv 4\text{ mod }7$. So $7\mid 2^{2^{k+2}}+3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate derivative of $f(x) = \frac{1}{1-2\cos^2x}$? $$f(x) = \frac{1}{1-2\cos^2x}$$
The result of $f'(x)$ should be equals
$$f'(x) = \frac{-4\cos x\sin x}{(1-2\cos^2x)^2}$$
I'm trying to do it in this way but my result is wrong.
$$f'(x) = \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} =
\frac {1-2\cos^2x-(1-(2\cos^2x)')}{(1-2\cos^2x)^2} = $$
$$=\frac {-2\cos^2x + 2(2\cos x(\cos x)')}{(1-2\cos^2x)^2} =
\frac {-2\cos^2x+2(-2\sin x\cos x)}{(1-2\cos^2x)^2} = $$
$$\frac {-2\cos^2x-4\sin x\cos x}{(1-2\cos^2x)^2}$$
| $f(x) = \frac{1}{1-2cos^2x}$
The result of $f'(x)$ should be equal $f'(x) = \frac{-4cosxsinx}{(1-2cos^2x)^2}$
$$
f'(x) =
\frac {0-1(1-2cos^2x)'} {(1-2cos^2x)^2} =
\frac {-4sinxcosx} {(1-2cos^2x)^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute the summation of a binomial coefficient/ show the following is true $\sum\limits_{k=0}^n \left(2k+1\right) \dbinom{n}{k} = 2^n\left(n+1\right)$.
I know that you have to use the binomial coefficient, but I'm not sure how to manipulate the original summation to make the binomial coefficient useful.
| The Binomial Theorem states that $$(x + y)^n = \sum_{k = 0}^{n} x^{n - k}y^k$$ If we set $x = y = 1$, we obtain
$$2^n = (1 + 1)^n = \sum_{k = 0}^n \binom{n}{k}1^{m - k}1^k = \sum_{k = 0}^{n} \binom{n}{k}$$
We will also make use of Pascal's Identity $$\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}$$
Here is a proof by induction.
Let $P(n)$ be the statement $$\sum_{k = 0}^n (2k + 1)\binom{n}{k} = 2^n(n + 1)$$
Let $n = 0$. Then
$$\sum_{k = 0}^{0} (2k + 1)\binom{0}{k} = (2 \cdot 0 + 1)\binom{0}{0} = 1 \cdot 1 = 1 = 1(0 + 1) = 2^0(0 + 1)$$
so $P(0)$ holds.
Hence, we may assume $P(m)$ holds for some non-negative integer $m$. Let $n = m + 1$. Then
\begin{align*}
\sum_{k = 0}^{m + 1}(2k + 1)\binom{m + 1}{k} & = 1 + \sum_{k = 1}^{m} (2k + 1)\color{red}{\binom{m + 1}{k}} + 2m + 3\\
& = 2m + 4 + \sum_{k = 1}^{m} (2k + 1)\color{red}{\left[\binom{m}{k} + \binom{m}{k - 1}\right]}\\
& = 2m + 4 + \sum_{k = 1}^{m} (2k + 1)\binom{m}{k} + \sum_{k = 1}^{m} (2k + 1)\binom{m}{k - 1}\\
& = 2m + 4 - 1 + \sum_{k = 0}^{m} (2k + 1)\binom{m}{k} + \sum_{k = 1}^{m} [2(k - 1) + 3]\binom{m}{k - 1}\\
& = 2m + 3 + \sum_{k = 0}^{m} (2k + 1)\binom{m}{k} + \sum_{k = 0}^{m - 1} (2k + 3)\binom{m}{k}\\
& = 2m + 3 + \sum_{k = 0}^{m} (2k + 1)\binom{m}{k} + \sum_{k = 0}^{m} (2k + 3)\binom{m}{k} - (2m + 3)\\
& = \sum_{k = 0}^{m} (2k + 1)\binom{m}{k} + \sum_{k = 0}^{m} (2k + 1)\binom{m}{k} + \sum_{k = 0}^{m} 2\binom{m}{k}\\
& = 2\color{blue}{\sum_{k = 0}^{m} (2k + 1)\binom{m}{k}} + 2\color{green}{\sum_{k = 0}^m \binom{m}{k}}\\
& = 2 \cdot \color{blue}{2^m(m + 1)} + 2 \cdot \color{green}{2^m}\\
& = 2^{m + 1}(m + 1) + 2^{m + 1}\\
& = 2^{m + 1}[(m + 1) + 1]
\end{align*}
where I have used red to highlight Pascal's Identity, blue to highlight the induction hypothesis, and green to highlight the use of the fact that $$2^m = \sum_{k = 0}^{m} \binom{m}{k}$$
Since $P(0)$ holds and $P(m) \Rightarrow P(m + 1)$ for each non-negative integer $m$, $P(n)$ holds for each non-negative integer $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1164366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of $\frac{e^{-x^2}}{\sqrt{1-x^2}}$ I am stuck at an integral $$\int_0^{\frac{1}{3}}\frac{e^{-x^2}}{\sqrt{1-x^2}}dx$$
My attempt is substitute the $x=\sin t$, however there may be no primitive function of $e^{-\sin^2 t}$.
So does this integral has a definitive value? If does, how can we solve it? Thank you!
| Approach $1$:
$\int_0^\frac{1}{3}\dfrac{e^{-x^2}}{\sqrt{1-x^2}}dx$
$=\int_0^{\sin^{-1}\frac{1}{3}}\dfrac{e^{-\sin^2t}}{\sqrt{1-\sin^2t}}d(\sin t)$
$=\int_0^{\sin^{-1}\frac{1}{3}}e^{-\sin^2t}~dt$
$=\int_0^{\sin^{-1}\frac{1}{3}}\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n}t}{n!}dt$
$=\int_0^{\sin^{-1}\frac{1}{3}}\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\sin^{2n}t}{n!}\right)dt$
For $n$ is any natural number,
$\int\sin^{2n}t~dt=\dfrac{(2n)!t}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
$\therefore\int_0^{\sin^{-1}\frac{1}{3}}\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\sin^{2n}t}{n!}\right)dt$
$=\left[t+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!t}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^3(2k-1)!}\right]_0^{\sin^{-1}\frac{1}{3}}$
$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!t}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^3(2k-1)!}\right]_0^{\sin^{-1}\frac{1}{3}}$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!\sin^{-1}\dfrac{1}{3}}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{2\sqrt2(-1)^n(2n)!((k-1)!)^2}{4^{n-k+1}9^k(n!)^3(2k-1)!}$
Approach $2$:
$\int_0^\frac{1}{3}\dfrac{e^{-x^2}}{\sqrt{1-x^2}}dx$
$=\int_0^\frac{1}{9}\dfrac{e^{-x}}{\sqrt{1-x}}d(\sqrt{x})$
$=\dfrac{1}{2}\int_0^\frac{1}{9}\dfrac{e^{-x}}{\sqrt{x}\sqrt{1-x}}dx$
$=\dfrac{1}{2}\int_0^1\dfrac{e^{-\frac{x}{9}}}{\sqrt{\dfrac{x}{9}}\sqrt{1-\dfrac{x}{9}}}d\left(\dfrac{x}{9}\right)$
$=\dfrac{1}{6}\int_0^1\dfrac{e^{-\frac{x}{9}}}{\sqrt{x}\sqrt{1-\dfrac{x}{9}}}dx$
$=\dfrac{1}{3}\Phi_1\left(\dfrac{1}{2},\dfrac{1}{2},\dfrac{3}{2};\dfrac{1}{9},-\dfrac{1}{9}\right)$ (according to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions)
| {
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"timestamp": "2023-03-29T00:00:00",
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Getting a specific formula for a sequence. If:
$$a_0 = \frac{5}{2}, a_k = a_{k-1}^{2} - 2$$ for $k \ge 1$.
How do I get a general formula for $a_k$? With induction proof. I even tried calculating $a_1, a_2 ...$:
$$a_0 = \frac{5}{2}$$
$$a_1 = \frac{17}{4}$$
$$a_2 = \frac{273}{16}$$
$$a_3 = \frac{74017}{256}$$
I will treat the numerator and denominator seperately.
I see that for the denominator.
$$d = 2^{2^k}$$
Now to the numerator:
I cant get it.
I tried, $$2^{2^{k}} + 1$$ but it doesnt work for $k=2$.
But it is shifted $1$, meaning for $n=2$, I got the value of $n=1$.
| $$a_0=\frac{5}{2}$$
$$a_1=\left (\frac{5}{2}\right )^2-2=\frac{17}{4}$$
$$a_2=\left (\left (\frac{5}{2}\right )^2-2\right )^2-2=\left (\frac{17}{4}\right )^2-2=\frac{257}{16}$$
As you can see, you miscalculated the $a_2$ term
and because of that, the next term is also messed up as well.
$$a_3=\frac{65537}{256}$$
Note that the numerator is a directly next term of corresponding denominator term (plus one)
Therefore, (almost as you predicted), $$a_k=\frac{2^{2^{k+1}}+1}{2^{2^k}}$$
Proof
We can see that
$P(1)$ is true as $$\frac{2^{2}+1}{2^1}=\frac{5}{2}$$
Now, let $P(n)$ be true.
We now have to prove that $P(n+1)$ is true.
$P(n)$ is $$a_{n-1}=\frac{2^{2^n}+1}{2^{2^{n-1}}}$$
$P(n+1)$ is $$a_n=\frac{2^{2^{n+1}}+1}{2^{2^n}}$$
We know that $$a_n=a_{n-1}^2-2$$
If $P(n+1)$ is true, the following will also be true
$$\frac{2^{2^{n+1}}+1}{2^{2^n}}=\left (\frac{2^{2^{n}}+1}{2^{2^{n-1}}}\right )^2-2$$
$$\implies \frac{2^{2^{n+1}}+1}{2^{2^n}}=\frac{(2^{2^{n}}+1)^2}{2^{2^{n}}}-2$$
$$\implies 2^{2^{n+1}}+1=(2^{2^{n}}+1)^2-2^{2^{n}+1}$$
$$\implies 2^{2^{n+1}}+1=2^{2^{n+1}}+2^{2^n+1}+1-2^{2^{n}+1}$$
$$\implies 0=0$$
We can see that $P(n+1)$ is true. Then by induction, QED.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I demonstrate that $x-x^9$ is divisible by 30? How can I demonstrate that $x-x^9$ is divisible by $30$ whenever $x$ is an integer?
I know that $$x-x^9=x(1-x^8)=x(1-x^4)(1+x^4)=x(1-x^2)(1+x^2)(1+x^4)$$
but I don't know how to demonstrate that this number is divisible by $30$.
| You have to prove that $x-x^9$ is divisible by $2$, $3$ and $5$.
*
*$x\equiv x^9\pmod{2}$ is obvious, isn't it?
*By Fermat's little theorem, $x^3\equiv x\pmod{3}$, so $x^9=(x^3)^3\equiv x^3\equiv x\pmod{3}$
*By Fermat's little theorem, $x^5\equiv x\pmod{5}$, so $x^9=x^4x^5\equiv x^4 x\equiv x^5\equiv x\pmod{5}$.
| {
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Factoring Polynomials in Fields I always have problems to factorize polynomials that have no linear factors any more. For example ($x^5-1$) in $\mathbb{F}_{19}$. It's easy to find the root 1 and to split it. ($x^5-1$) = ($x-1$) * ($x^4$+$x^3$+$x^2$+x+1).
I think the last part must split into two irreducible polynomials with degree 2.
($x^2$+ ax+ b) ($x^2$+ Cx+ d). I expanded it and compared the coefficients to find
values for a,b,c,d. But it wasn't solvable.
Is this approach correct or a there any other procedures or tricks to solve such a problem ? Thank you.
| Your suspicion is correct: $(x^5 - 1)$ does split into a linear factor and two quadratic ones. Your approach even works out!
$(x^2 + ax + b)(x^2 + bx + c) = x^4 + (a+c)x^3 + (b + ac + d)x^2 + (ad+bc)x + bd$
So we have
$bd = 1$, $a + c = 1$, $ad+bc = 1$, $b + ac + d = 1$.
Immediately we derive
$d = \frac{1}{b}$, $c = 1-a$.
We then substitute in $ad+bc = 1$ to obtain
$a = \frac{1-b}{\frac{1}{b}-b}$
This alerts us that we have to be careful to check $b \in \{1,18\}$ separately.
If $b$ happens to be $1$, then we get that $1 = b + ac + d = 1 + ac + 1 = 2+ac$, and so $1 + ac = 0$. As it happens, this equation and $c = 1-a$ are satisfied when $a$ is $5$, yielding $c$ as $-4$. This gives our two irreducibles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $n^3+2$ is not divisible by $9$ for any integer $n$ How to prove that $n^3+2$ is not divisible by $9$?
| The cubes modulo $9$ are $0$, $1$ and $8$; now
$$
0+2\equiv2,\quad 1+2\equiv3,\quad 8+2\equiv1\pmod{3}
$$
Computations are easy, because $(n+3)^3=n^3+9n^2+27n+27\equiv n^3\pmod{3}$, so we just need to do $0^3$, $1^3$ and $2^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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solving cubic diophantine equation Can someone show me how to find all solutions in positive integers to the diophantine equation: $$x^3 + y^3 = 35$$ I know how to do it algebraically, but I want to know how you solve it in number theory.
My Algebraic Approach:
$x^3 + y^3 = (x + y)\cdot (x^2 - xy + y^2) = 35$.
The only integer factors of $35$ are $(1, 35)$ or $(5, 7)$. There are no integers $x$ and $y$ that add to $1$, so $x + y = 5$ or $7$.
Using $x + y = 5$ we get that $y = 5 - x$, so $y^3 = 125 - 75x + 15x^2 - x^3$ so
$x^3 + y^3 = 15x^2 - 75x + 125 = 35$
or
$$15x^2 - 75x + 90 = 0$$
$$x^2 - 5x + 6 = 0$$
$$(x - 3)(x - 2) = 0$$
So $x = 2$ or $x = 3$
Thus $y = 3$ or $y = 2$ respectively.
Using $x + y = 7$ we get that $y = 7 - x$, so $y^3 = 343 - 147x + 21x^2 - x^3$ so
$$x^3 + y^3 = 21x^2 - 147x + 343 = 35$$
$$21x^2 - 147x + 308 = 0$$
$$3x^2 - 21x + 44 = 0$$
or
$$x = \frac{21 \pm \sqrt{-87}}{6}$$
Since x is complex, this can't be a solution.
So $(x, y) = (2, 3)$ or $(x, y) = (3, 2)$
| You know that $x$ and $y$ cannot both be greater than or equal to $3$, because this would give you too large of a sum. And $x$ and $y$ must both be less than $4$ by similar reasoning. So in this case, that's probably the best way to approach the problem: Just try $1,2,3$ for $x$ and $y$, where they are not both $3$. And note you can assume $x \leq y$ by symmetry, and modify your answer after the fact based on symmetry and the solutions you get.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluating the sum $ \sum_{n = 1}^{44} {\sin^{2}}(n^{\circ}) ~ {\cos^{2}}(n^{\circ}) $. I want to find the sum
$$
{\sin^{2}}(1^{\circ}) ~ {\cos^{2}}(1^{\circ}) +
{\sin^{2}}(2^{\circ}) ~ {\cos^{2}}(2^{\circ}) +
{\sin^{2}}(3^{\circ}) ~ {\cos^{2}}(3^{\circ}) + \cdots +
{\sin^{2}}(44^{\circ}) ~ {\cos^{2}}(44^{\circ}).
$$
I thought of using the identity $ \sin(2 x) = 2 \sin(x) \cos(x) $, so
$$
[2 \sin(1^{\circ}) \cos(1^{\circ})]^{2} +
[2 \sin(2^{\circ}) \cos(2^{\circ})]^{2} + \cdots +
[2 \sin(44^{\circ}) \cos(44^{\circ})]^{2}
= 4 y.
$$
By the identity above, I get
$$
{\sin^{2}}(2^{\circ}) + {sin^{2}}(4^{\circ}) + \cdots + {\sin^{2}}(88^{\circ}) = 4 y,
$$
but then I don’t know how to simplify this or if it could be done in other simpler ways.
Thanks.
| Note that $\sin^2 \theta=\frac{1-\cos 2\theta}{2}$. As there are $44$ terms in the sum, this means that
$$
4y=\frac{44}{2}-\frac{1}{2}\left(\cos 4^\circ + \cos 8^\circ + \dots + \cos 176^\circ \right) \, .
$$
Now, since $\cos(180^\circ -\theta)=-\cos \theta$, the terms in parentheses will cancel in pairs:
\begin{align}
\cos 4^\circ + \cos 176^\circ&= 0 \\
\cos 8^\circ + \cos 172^\circ&= 0 \\
\cdots \\
\cos 88^\circ + \cos 92^\circ&= 0
\end{align}
So the parenthesized sum vanishes, meaning that $4y=\frac{44}{2}=22$, and thus $y=\frac{11}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\lim\limits_{x \to \infty } \frac{5 + \sqrt{x^2+5}}{x-6}$ I don't know how to evaluate
$$\lim_{x \to \infty } \frac{5 + \sqrt{x^2+5}}{x-6}$$
*
*It is $\frac{\infty}{\infty}$
*I have tried to multiply by $5 - \sqrt{x^2+5}$, but I reach a wrong result.
*I have tried to take the derivative of $5 + \sqrt{x^2+5}$ and $x-6$, but this also doesn't help.
Probably, I should use another method. Which method do I use?
| Hint:
$$\frac{5+\sqrt{x^2+5}}{x-6}=\frac{\frac{5}{x}+\sqrt{\frac{x^2}{x^2}+\frac{5}{x^2}}}{\frac{x}{x}-\frac{6}{x}}=\frac{\frac{5}{x}+\sqrt{1+\frac{5}{x^2}}}{1-\frac{6}{x}}$$
$$\lim_{x\to\infty}\frac{\frac{5}{x}+\sqrt{1+\frac{5}{x^2}}}{1-\frac{6}{x}}=\frac{0+\sqrt{1+0}}{1-0}=1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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critical points of the function Find the critical points of the function
$$f(x,y)=(4x-x^2)\cos y$$
first let's determinate Partial derivatives:
$$\dfrac{\partial f}{\partial x}(x,y)=(4-2x)\cos y$$
$$\dfrac{\partial f}{\partial y}(x,y)=-\sin y(4x-x^2) $$
To find the critical points, we solve:
\begin{cases}\dfrac{\partial f}{\partial x}(x,y)=0 & \\
\\
\dfrac{\partial f}{\partial y}(x,y)=0 & \end{cases}
\begin{cases}
(4-2x)\cos y=0 & \\
\\
-\sin y(4x-x^2)=0 & \end{cases}
\begin{cases}
(4-2x)\cos y=0 & \\
\\
(x^2-4x)\sin y=0 & \end{cases}
which gives
$ \begin{cases}
4-2x=0 & \\
\\
x^2-4x=0 & \end{cases} \textrm{or}
\begin{cases}
4-2x=0 & \\
\\
\sin y=0 & \end{cases}
\textrm{or}
\begin{cases}
\cos y=0 & \\
\\
x^2-4x=0 & \end{cases}
\textrm{or}
\begin{cases}
\cos y=0 & \\
\\
\sin y=0 & \end{cases} $
let's treat each case individually
*
*$ \begin{cases}
4-2x=0 & \\
\\
x^2-4x=0 & \end{cases} \implies \begin{cases}
x=2 & \\
\\
x=0 \textrm{ or } x=4 & \end{cases} $ thus The system does not have solution
*$ \begin{cases}
4-2x=0 & \\
\\
\sin y=0 & \end{cases} \implies \begin{cases}
x=2 & \\
\\
\ y=k\pi,\ k \in \mathbb{Z} & \end{cases} \implies (2,k\pi ) $
*$\begin{cases}
\cos y=0 & \\
\\
x^2-4x=0 & \end{cases}\implies \begin{cases}
y=\dfrac{\pi}{2}+k\pi, \ k \in \mathbb{Z} & \\
\\
x=0 \textrm{ or } x=4 & \end{cases}\implies (0,\dfrac{\pi}{2}+k\pi) \textrm{ or } (4,\dfrac{\pi}{2}+k\pi)$
*$\begin{cases}
\cos y=0 & \\
\\
\sin y=0 & \end{cases} \implies$ thus The system does not have solution
The function has three critical points $$(2,k\pi ), (0,\dfrac{\pi}{2}+k\pi) \textrm{ and }(4,\dfrac{\pi}{2}+k\pi)$$
*
*Am i right ?
| 1)$(4-2x)\cos y= 0$: The possible Solutions are $x=2,y= \frac{\pi}{2} + \pi n$. $n$ is arbitrary natural number.
2)$(x^2-4x)\sin y = 0$: Possible Solutions: $x=0,x=4,y=\pi n$.
Your critical Points are all possible $x,y$ which satisfy both equations. For example, $(x=2,y= \pi(n+ \frac{1}{2}))$ is not a critical Point, because it satisfies 1) but not 2). All critical Points are the points $(x,y)$ where $x$ satisfies either 1) or 2) and $y$ satisfies the other equation (i.e. if $x$ satisfies 1), then $y$ satisfies 2) and if $x$ satisfies 2), then $y$ satisfies 1)).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Volume of a Solid, $x^2 - y^2 = a^2$ The question is
Find the volume of a solid rotated around the y axis, bounded by the given curves: $$x^2 - y^2 = a^2$$
$$x = a + h$$
I am lost by the number of variables in this question and the question does not tell me what kind of variables they are, only that they are both greater than 0.
|
The region that is to be revolved about the $ \ y-$axis [marked in gold in the graph above] is bounded by the "vertical" hyperbola $ \ x^2 - y^2 \ = \ a^2 \ $ and the vertical line $ \ x \ = \ a + h \ \ , \ $ with $ \ a \ $ and $ \ h \ $ being unspecified constants. What is asked for in the problem then is a formula for the volume of the solid of revolution expressed in terms of those constants.
The solid "generated" will be within a cylinder of radius $ \ a + h \ $ and a height $ \ \mathcal{H} \ $ which we have yet to determine from the intersections of the bounding curves and outside of the hyperboloid of one sheet identified by Narasimham. These intersections are found from
$$ x^2 \ - \ y^2 \ \ = \ \ a^2 \ \ \rightarrow \ \ y^2 \ \ = \ \ x^2 \ - \ a^2 $$ $$ \Rightarrow \ \ \mathcal{H}^2 \ \ = \ \ (a + h)^2 \ - \ a^2 \ \ = \ \ 2ah \ + \ h^2 \ \ , $$
so the top and bottom of the solid are found at $ \ y \ = \ \pm \mathcal{H} \ = \ \pm \sqrt{2ah \ + \ h^2} \ \ , \ $ respectively.
This suggests one relatively quick method to obtain the volume (which might be called the "Cavalieri approach"): use a familiar formula to calculate the volume of the cylinder, then calculate the volume "inside" the hyperboloid by integrating the "vertical stack of disks" it is comprised of, and "subtract away" that volume. The volume of the cylinder is simply $ \ V_{cyl} \ = \ \pi·(a + h)^2 · (\ \mathcal{H} - [\mathcal{-H}] \ ) $ $ = \ \pi·(a + h)^2 · 2·\sqrt{2ah \ + \ h^2} \ \ . $ The disks making up the solid hyperboloid have a "radius-squared" as a function of $ \ y \ $ given by $ \ x^2 \ = \ a^2 + y^2 \ \ $ and "infinitesimal thicknesses" of $ \ dy \ \ . \ $ Because the "stack" is symmetrical about the plane $ \ y \ = \ 0 \ \ , \ $ its volume is given by
$$ V_{hyp} \ \ = \ \ 2 \ \int_0^{\mathcal{H}} \ \pi · x^2 \ \ dy \ \ = \ \ 2 \pi \ \int_0^{\mathcal{H}} \ (a^2 + y^2) \ \ dy \ \ = \ \ 2 \pi \ · \ \left( \ a^2·y \ + \ \frac13 y^3 \ \right)|_0^\mathcal{H}$$
$$ = \ \ 2 \pi \ · \ ( \ a^2·\mathcal{H} \ + \ \frac13 \mathcal{H}^3 \ ) \ \ = \ \ 2 \pi \mathcal{H} \ · \ ( \ a^2 \ + \ \frac13 \mathcal{H}^2 \ ) \ \ . $$
We can now determine the volume of the specified solid of revolution as
$$ \mathcal{V} \ \ = \ \ V_{cyl} \ - \ V_{hyp} \ \ = \ \ 2 \pi \mathcal{H} \ · \ \left[ \ (a + h)^2 \ - \ \left( \ a^2 \ + \ \frac13 \mathcal{H}^2 \ \right) \ \right] \ \ $$
$$ = \ \ 2 \pi \mathcal{H} \ · \ \left[ \ (a^2 + 2ah + h^2) \ - \ \left( \ a^2 \ + \ \frac13 · (2ah \ + \ h^2) \ \right) \ \right] \ \ $$
$$ = \ \ 2 \pi \mathcal{H} \ · \ \frac23 · (2ah \ + \ h^2) \ \ = \ \ \frac{4 \pi}{3} · \sqrt{2ah \ + \ h^2} · (2ah \ + \ h^2) \ \ = \ \ \frac{4 \pi}{3} · (2ah \ + \ h^2)^{3/2} \ \ . $$
$ \rule{15 cm}{0.5 pt} $
This problem is also amenable to either of the volume integration methods taught for dealing with solids of revolution.
If we apply the "cylindrical shells" method, the radii of the shells "run" from $ \ x \ = \ a \ $ to $ \ x \ = \ a + h \ \ , \ $ with "heights" given by $ \ y \ - \ (-y) \ = \ 2·\sqrt{x^2 - a^2} \ \ . $ The volume integral is then
$$ \mathcal{V} \ \ = \ \ \int_a^{a+h} \ 2 \pi \ · \ x \ · \ 2·\sqrt{x^2 - a^2} \ \ dx $$
[making the substitution $ \ u \ = \ x^2 - a^2 \ \Rightarrow \ du \ = \ 2x \ dx \ \ ] $
$$ \rightarrow \ 4 \pi \ \int \ u^{1/2} \ \ \left(\frac12 \ du\right) \ \ = \ \ 2 \pi \ · \ \frac{u^{3/2}}{3/2} \ \ = \ \ \frac{4 \pi}{3} \ · \ u^{3/2} $$
[we'll evaluate the definite integral after "back-substitution"]
$$ \rightarrow \ \frac{4 \pi}{3} \ · \ (x^2 - a^2)^{3/2}|_a^{a+h} \ \ = \ \ \frac{4 \pi}{3} \ · \ [ \ ( \ [a+h]^2 - a^2 \ )^{3/2} \ - \ ( \ a^2 - a^2 \ )^{3/2} \ ] $$
$$ = \ \ \frac{4 \pi}{3} \ · \ [ \ ( \ [a^2 + 2ah + h^2] - a^2 \ )^{3/2} \ - \ 0 \ ] \ \ = \ \ \frac{4 \pi}{3} · ( 2ah + h^2 )^{3/2} \ \ . $$
The "disk/washer" method calculation resembles the "Cavalieri approach" we used above, except that the washers lie between the hyperboloid and the cylinder: the "outer radii" are the cylinder's radius $ \ a + h \ \ , \ $ and the "inner radii-squared" are $ \ x^2 \ = \ a^2 + y^2 \ \ . \ $ The volume integral for this "stack of washers" is
$$ \mathcal{V} \ \ = \ \ \int_{-\mathcal{H}}^{\mathcal{H}} \ [ \ \pi · (a + h)^2 \ - \ \pi · (a^2 + y^2) \ ] \ \ dy \ \ = \ \ 2 \pi \ \int_0^{\mathcal{H}} \ ( \ a^2 + 2ah + h^2 - a^2 - y^2 \ ) \ \ dy $$
[again exploiting the symmetry of the "stack" about $ \ y \ = \ 0 \ \ ] $
$$ = \ \ 2 \pi · \left[ \ ( 2ah + h^2 )·y \ - \ \frac13·y^3 \ \right]|_0^{\mathcal{H}} \ \ = \ \ 2 \pi · \left[ \ ( 2ah + h^2 )·\mathcal{H} \ - \ \frac13·\mathcal{H}^3 \ \right] $$
$$ = \ \ 2 \pi · \left[ \ ( 2ah + h^2 )·( 2ah + h^2 )^{1/2} \ - \ \frac13·( 2ah + h^2 )^{3/2} \ \right] \ \ = \ \ \frac{4 \pi}{3}·(2ah + h^2 )^{3/2} \ \ . $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Global maxima of $f(x,y)=x^2y$ restricted to D Let $f(x,y) = x^2y$ and $D = \{(x,y): y\geq0 \land 2x^2+y^2 \leq a\}$ with $a>0$.
I need to find $a$ such that the global maxima of $f$ restricted to $D$ is $\frac{1}{8}$.
I found, using Lagrange multipliers, that $a$ = $\frac{3}{4}$. With that value, $f(\frac{1}{2}, \frac{1}{2}) = f(-\frac{1}{2}, \frac{1}{2}) = \frac{1}{8}$, and those points are local maxima.
Tha part where I am struggling with is to prove that those are global maxima.
The restriction now states that $2x^2+y^2 \leq \frac{3}{4}$, but from there I can't get a close enough bound for $x^2y$. What else can I try?
Thank you.
| i am wondering if you can do this. we need to find the maximum of $k$ so that $$y = \frac k{x^2} \tag 1$$ touches the ellipse $$2x^2 + y^2 = a \tag 2$$ in the first quadrant.
for $(1)$ an $(2)$ to touch, $$2x^2 + \frac{k^2}{x^4} = a $$ must have a double root. that is, with $x^2 = u$ the function $g(u) = 2u^3 - au^2 + k^2$ and its derivative $g'(u) = 6u^2 - 2au = 0$ must have a positive common root $u = a/3.$ so that
$$0 = g(a/3) = \frac 2{27}a^3 - \frac 19a^3 + k^2 = 0 \to k = \left(\frac a 3\right)^{3/2} \text{ or } a = 3k^{2/3}$$ which agrees with the max pair found
$a = \frac 34, k = \frac1 8.$
| {
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"timestamp": "2023-03-29T00:00:00",
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$a\frac{d^{3}x}{dt^{3}}+b\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=0$, on what condition of k that we have $\lim_{t\rightarrow\infty}x(t)=0$ $a\frac{d^{3}x}{dt^{3}}+b\frac{d^{2}x}{dt^{2}}+c\frac{dx}{dt}+kx=0$, $a,b,c$ are constant, and on what condition of k that we have $\lim_{t\rightarrow\infty}x(t)=0$
I completely have no idea how to solve this problem, can anyone could help me? Thanks very much!
What I did:
If we let $a=b=c=1$, then we have $r^{3}+r^{2}+r+k=0,x(t)=C_{1}e^{r_{1}t}+C_{2}e^{r_{2}t}+C_{3}e^{r_{3}t}$, since $\lim_{t\rightarrow\infty}x(t)=0$, we get $r_{1}<0,r_{2}<0,r_{3}<0$. Assume that $r^{3}+r^{2}+r+k=(r+r_{1})(r+r_{2})(r+r_{3})$, then expand it we get $r_{1}+r_{2}+r_{3}=1,r_{3}(r_{1}+r_{2})+r_{1}r_{2}=1,r_{1}r_{2}r_{3}=k$, then how can I get the condition on k? I can just get $k>0$
| The characteristic polynomial is
$$ ar^3 + br^2 + cr + k = 0 $$
Vieta's formulas give us
$$ \left\{ \begin{matrix}
r_1 + r_2 + r_3 = -\frac{b}{a} \\
r_1r_2 + r_2r_3 + r_1r_3 = \frac{c}{a} \\
r_1r_2r_3 = \frac{k}{a}
\end{matrix} \right. $$
where $r_1, r_2, r_3$ are the 3 roots, which can be either all real or 1 real and 2 complex conjugates. In the case of 2 complex roots, their sum is a real number and their product is a positive real number.
In order for the limit to be $0$, the 3 roots must all have negative real parts, so $\frac{b}{a} > 0$. $\frac{c}{a} > 0$ and $\frac{k}{a} < 0$
All 3 conditions are needed, so you have to know more about $a, b, c$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why doesn't the ratio test work to evaluate this series $$\sum_{k=1}^{\infty}\ln\frac{k+5}{k+4}$$
$$
\frac{\ln\frac{k+6}{k+5}}{\ln\frac{k+5}{k+4}}
=\cdots
$$
(ln(k+6/k+5)/ln(k+5/k+4))
ln((k+6/k+5)*(k+4/k+5))
simplify
lim as k approaches infinity of ln(kˆ2 +10k +24/ kˆ2 +10k+25)= ln(1) = 0.
why doesn't the series converge?
| Concerning the ratio test, if you consider $$a_k=\ln\frac{k+5}{k+4}$$ then $$\frac{a_{k+1}}{a_k}=\frac{\log \left(\frac{k+6}{k+5}\right)}{\log \left(\frac{k+5}{k+4}\right)}=\frac{\log \left(1+\frac{1}{k+5}\right)}{\log \left(1+\frac{1}{k+4}\right)}$$ Now, use that, for small values of $y$, $$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+O\left(y^4\right)$$ Make $y=\frac{1}{k+5}$ for the numerator, $y=\frac{1}{k+4}$ for the denominator, use the long division to arrive to $$\frac{a_{k+1}}{a_k}=1-\frac{1}{k}+\frac{11}{2 k^2}+O\left(\left(\frac{1}{k}\right)^3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178969",
"timestamp": "2023-03-29T00:00:00",
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How to show $n^5 + 29 n$ is divisible by $30$ Show that $n^5 + 29 n$ is divisible by $30$.
Attempt:
$n^4 ≡ 1 \pmod 5$ By Fermat Little Theorem
| Modulo $6$, the expression is $n^5 + (-1)n = n^5 -n$.
Modulo $5$, $n^5 \equiv n$ by Fermat's Little Theorem, so $n^5 - n \equiv 0 \pmod 5$.
Since $5$ and $6$ are coprime $n^5 + 29n \equiv 0 \pmod {30}$.
| {
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"url": "https://math.stackexchange.com/questions/1180122",
"timestamp": "2023-03-29T00:00:00",
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Fourier sine series expansion
The function $f(x)$ is defined as $$f(x)=1\qquad0<x<\pi$$
Sketch the odd extension and show that the Fourier sine series expansion is
$$f(x)~\frac4\pi\sum_{n=1}^\infty\frac{\sin((2n-1)x)}{2n-1}$$
In this question, $f(x)=1$, $a_0$ is obviously $2$ and $a_m$ and $b_m$ are zero when plugged into Fourier series equation. Could you please explain why the Fourier series can be expanded in such this form?
| I don't know how you got $a_0 = 2$ and $A_n, B_n = 0$ so I'll post the solution.
You have
$$\begin{align}
f(x) &= 1 \\
&= \sum_{n=1}^{\infty} B_n \sin\bigg(\frac{n\pi x}{L}\bigg) \ \ \text{(because we're doing the odd extension)} \\
&= \sum_{n=1}^{\infty} B_n \sin(nx) \ \ \text{(with $L = \pi$)} \\
\end{align}$$
Integrating over our domain and using orthogonality, we find
$$\begin{align}
\int_{0}^{\pi} f(x)\sin(mx)dx &= \int_{0}^{\pi} \sin(mx)dx \ \ (1)\\
&= \sum_{n = 1}^{\infty} B_n \int_{o}^{\pi} \sin(nx)\sin(mx) dx \\
&= \sum_{n = 1}^{\infty} \frac{B_n}{2} \int_{-\pi}^{\pi} \sin(nx)\sin(mx) dx \ \ \text{(as an odd function $\times$ odd function is an even function)} \\
&= \frac{B_m \pi}{2},\ \ n = m \ \ (2)
\end{align} $$
Equating $(1)$ and $(2)$, we find
$$\begin{align}
\frac{B_m \pi}{2} &= \int_{0}^{\pi} \sin(mx)dx \\
\implies B_m &= \frac{2}{\pi} \int_{0}^{\pi} \sin(mx)dx \\
&= \frac{2}{\pi} \bigg[\frac{-\cos(mx)}{m} \bigg]_{0}^{\pi} \\
&= \frac{-2}{m \pi}\bigg[(-1)^{m} - 1 \bigg] \\
&= \begin{cases}
0 & m = \text{even} \\
\frac{4}{m \pi} & m = \text{odd} \\
\end{cases}
\end{align}$$
Hence we should set $m = 2j - 1$ for $j \ge 1$ to keep only those cases that are non-zero.
Hence, our solution is given by
$$\begin{align}
f(x) &= \sum_{n=1}^{\infty} B_n \sin(nx) \\
&= \sum_{m=1}^{\infty} B_m \sin(mx) \ \ \text{(using $(2)$)} \\
&= \sum_{m=1}^{\infty} \frac{4}{m \pi} \sin(mx) \\
&= \frac{4}{\pi} \sum_{m=1}^{\infty} \frac{\sin(mx)}{m} \\
&= \frac{4}{\pi} \sum_{j=1}^{\infty} \frac{\sin((2j - 1)x)}{2j - 1} \\
\end{align}$$
EDIT
$$\begin{align}
f(x) &= 1 \\
&= \sum_{n=1}^{\infty} B_n \sin(nx) \\
\end{align}$$
Multiply both sides by $\sin(mx)$ for orthogonality
$$\begin{align}
\implies f(x)\sin(mx) &= \sin(mx) \sum_{n=1}^{\infty} B_n \sin(nx) \\
&= \sum_{n=1}^{\infty} B_n \sin(nx)\sin(mx) \\
\end{align}$$
where we can take the $\sin(mx)$ term inside the series because we aren't summing over $m$ so it can be thought of almost like a constant.
Integrate both sides over $[0, \pi]$
$$\begin{align}
\implies \int_{0}^{\pi} f(x)\sin(mx) dx &= \int_{0}^{\pi} \bigg( \sum_{n=1}^{\infty} B_n \sin(nx)\sin(mx) \bigg) dx \\
&= \sum_{n=1}^{\infty} B_n \int_{0}^{\pi} \sin(nx)\sin(mx) dx \\
\end{align}$$
Replace $f(x)$ with $1$
$$\begin{align}
\implies \int_{0}^{\pi} f(x)\sin(mx) dx &= \int_{0}^{\pi} \sin(mx) dx \\ &= \int_{0}^{\pi} \bigg( \sum_{n=1}^{\infty} B_n \sin(nx)\sin(mx) \bigg) dx \\
&= \sum_{n=1}^{\infty} B_n \int_{0}^{\pi} \sin(nx)\sin(mx) dx \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Norm in the cyclotomic integers
Let $\alpha \in \mathbb{Z}[\zeta_3]$, where $\zeta_3=e^{2\pi i/3}$ is a cube root of unity. So $\alpha=x+y\zeta_3$ for $x,y\in\mathbb{Z}$. Show that the norm $N(\alpha)$ can be written as $\frac{a^2+3b^2}{4}$ for integers $a,b$ of the same parity.
I think this is supposed to be a somewhat straightforward arithmetic-heavy problem, but I'm having trouble with it.
Since $\zeta_3^2=-1-\zeta_3$, I was able to show that $N(\alpha)=x^3+y^3$, but I don't see how this reduces to what I need.
| The norm $x^2+y^2-xy$ can be simply computed via square-completion procedure to get
$4(x^2+y^2-xy)$=$(x^2+2xy+y^2)+3(x^2-2xy+y^2)$
. =$(x+y)^2+3(x-y)^2$
And also since
$x^2+y^2-xy=x^2+(x^2-2xy+y^2)-(x^2-xy)$
. =$x^2+(x-y)^2-x(x-y)$
The we also have
$4(x^2+y^2-xy)=(2x-y)^2+3y^2$
Either way, it shows that the norm can be written in the form $a^2+3b^2$, by dividing the above expressions $4$ depending on whether$ x$ and $y$ are of the same parity or not. Use the first expression when $x$ and $y$ are of the same parity and use the second expression when $x$ and $y$ are of different parity, assuming $y$ is even and $x$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Intersection between two planes and a line? What is the coordinates of the point where the planes: $3x-2y+z-6=0$ and $x+y-2z-8=0$ and the line: $(x, y, z) = (1, 1, -1) + t(5, 1, -1)$ intersects with eachother?
I've tried letting the line where the two planes intersect eachother be equal to the given line, this results in no solutions.
I have tried inserting the lines x, y and z values into the planes equations, this too, results in no solutions.
According to the answer sheet the correct solution is: $\frac{1}{2}(7,3,-3)$
|
I get $(3.5,1.5,-1.5)^T$ as well.
The yellow line is the intersection of the two planes, the big red dot the intersection with the given line (black).
$$
E_1: 3x - 2y + z = 6 \wedge E_2: x + y - 2z = 8
$$
Solving for $z$ gives
$$
z = 6 - 3x + 2y = -4 + x/2 + y/2 \\
10 = 7/2 x - 3/2 y \iff y = 7/3 x - 20/3
$$
E.g.choosing $x=3$ and $x=4$ then $a = (3, 1/3, 6 - 9 + 2/3)^T = (3, 1/3, -7/3)^T$ and
$b = (4, 8/3, 6-12+16/3)^T = (4, 8/3, -2/3)^T$ are part of the intersection $E_1 \cap E_2$.
From this we generate the line
\begin{align}
g &: (3, 1/3,-7/3)^T + ((4, 8/3,-2/3)^T - (3, 1/3,-7/3)^T) s \\
&= (3, 1/3,-7/3)^T + (1, 7/3, 5/3)^T s
\end{align}
and intersect it with the given line
$$
f: (1,1,-1)^T + (5,1,-1)^T t
$$
$g(s) = f(t)$, which gives the system:
$$
\left(
\begin{matrix}
1 & -5 \\
7/3 & -1 \\
5/3 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
s \\
t
\end{matrix}
\right) =
\left(
\begin{matrix}
-2 \\
2/3 \\
4/3
\end{matrix}
\right)
$$
This has the solution $(s, t) = (1/2, 1/2)$.
Using for example $t=1/2$ with $f$, we get
$$
P = (3.5, 1.5, -1.5)^T
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
square root / factor problem $(A/B)^{13} - (B/A)^{13}$ Let
$A=\sqrt{13+\sqrt{1}}+\sqrt{13+\sqrt{2}}+\sqrt{13+\sqrt{3}}+\cdots+\sqrt{13+\sqrt{168}}$ and
$B=\sqrt{13-\sqrt{1}}+\sqrt{13-\sqrt{2}}+\sqrt{13-\sqrt{3}}+\cdots+\sqrt{13-\sqrt{168}}$.
Evaluate $(\frac{A}{B})^{13}-(\frac{B}{A})^{13}$.
By Calculator, I have $\frac{A}{B}=\sqrt{2}+1$ and $\frac{B}{A}=\sqrt{2}-1$.
But, I don't know how. Has someone any idea about this.
| This took me some time to solve. Here you go:
First, we find this:
$$\begin{aligned}
(\sqrt{13+\sqrt{a}}-\sqrt{13-\sqrt{a}})^2
&=13+\sqrt{a}+13-\sqrt{a}-2\sqrt{13+\sqrt{a}}\sqrt{13-\sqrt{a}}\\
&=2(13-\sqrt{169-a})
\end{aligned}$$
So,
$$\sqrt{13+\sqrt{a}}-\sqrt{13-\sqrt{a}}=\sqrt{2}\sqrt{13-\sqrt{169-a}}$$
By what we have, we write,
$A-B=\sqrt{2}B$
What I used here is the fact that I've summed over all $a$ from $1$ to $168$, and that summing with $\sqrt{169-a}$ is the same as summing with $\sqrt{a}$ in this question.
Now, we have $A=(1+\sqrt{2})B$
Thus, $\frac{A}{B}=1+\sqrt{2}$ and $\frac{B}{A}=\sqrt{2}-1$
We just calculate $(\frac{A}{B})^{13}$ and $(\frac{B}{A})^{13}$ which I believe is okay to be done using calculator. Else, comment so I can edit my answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
Prove $3a^4-4a^3b+b^4\ge0$ . $(a,b\in\mathbb{R})$ $$3a^4-4a^3b+b^4\ge0\ \ (a,b\in\mathbb{R})$$
We must factorize $3a^4-4a^3b+b^4\ge0$ and get an expression with an even power like $(x+y)^2$ and say an expression with an even power can not have a negative value in $\mathbb{R}$.
But I don't know how to factorize it since it is not in the shape of any standard formula.
| $$\color{red}{3a^4}-4a^3b+b^4=\color{red}{4a^4-a^4}-4a^3b+b^4=4a^3(a-b)+b^4-a^4$$
writing $$b^4-a^4=(b-a)(b+a)(b^2+a^2)$$ and factoring out $(a-b)$ gives
$$4a^3(a-b)+(b-a)(b+a)(b^2+a^2)=(a-b)\color{blue}{(4a^3-(a^3+ab^2+ba^2+b^3\big)})$$
Consider two cases.
1)$a\ge b$. Then the blue color expression is $\ge 0$. Because
$$a^3+ab^2+ba^2+b^3\le a^3+a^3+a^3=4a^3$$
2)$a\le b$. The proof is Similar to the case 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integration question on $\int \frac{x}{x^2-10x+50} \, dx$ How would I integrate
$$\int \frac{x}{x^2-10x+50} \, dx$$
I am not sure on how to start the problem
| write the integral as
$$\int\frac{x}{x^2 - 10x +50}dx = \int \left(\frac{2x-10}{2(x^2-10x+50)} + \frac{5}{x^2-10x+50}\right)dx$$
Let $u = x^2 -10x +50, \; \text{then } du = 2x-10$
$$\frac{1}{2}\int\frac{1}{u}du + 5\int \frac{1}{(x-5)^2 + 25}dx$$
$$\frac{\log (u)}{2}+5\int\frac{1}{\frac{(x-5)^2}{25}+1}dx$$
$$\frac{1}{2}\log(x^2-10x+50) + \arctan\left(\frac{x-5}{5}\right) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that the complex expression is real Let $|z_1|=\dots=|z_n|=1$ on the complex plane.
Prove that:
$$
\left(1+\frac{z_2}{z_1}\right)
\left(1+\frac{z_3}{z_2}\right)
\dots
\left(1+\frac{z_n}{z_{n-1}}\right)
\left(1+\frac{z_1}{z_n}\right)
\in\mathbb{R}
$$
I have tried induction and writing every "subexpression" as $(1+e^{i(\theta_n-\theta_{n-1})})$.
Any ideas?
| Use induction:
for $n=2$, we have
$$\big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_1}{z_2} \big)= 2 + \frac{z_2}{z_1} + \frac{z_1}{z_2} \in\mathbb{R} $$
since $ \dfrac{z_2}{z_1} = \exp\big( i \theta \big)$ for some $\theta\in\mathbb{R}$ while $\dfrac{z_1}{z_2} = \exp\big( - i\theta \big)$ and $ \exp\big( i\theta \big) + \exp\big( - i\theta \big) = 2 \cos\theta$ is real.
Assume the result holds true for $k = n \geq 2$, that is,
$$\big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_3}{z_2} \big)\ldots \big( 1 + \frac{z_n}{z_{n-1}} \big) \cdot \big( 1 + \frac{z_1}{z_n} \big)\in\mathbb{R} , $$
then for $k = n+1$, we have
\begin{align}
& \qquad \big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_3}{z_2} \big)\ldots \big( 1 + \frac{z_n}{z_{n-1}} \big) \cdot \big( 1 + \frac{z_{n+1}}{z_n} \big) \big( 1 + \frac{z_1}{z_{n+1}} \big) \\
& = \left[ \big( 1 + \frac{z_2}{z_1} \big) \cdot \big( 1 + \frac{z_3}{z_2} \big)\ldots \big( 1 + \frac{z_n}{z_{n-1}} \big) \cdot \big( 1 + \frac{z_1}{z_n} \big) \right]\times \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big) \big( 1 + \dfrac{z_1}{z_{n+1}} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big)}
\end{align}
now it suffices to show $ \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big) \big( 1 + \dfrac{z_1}{z_{n+1}} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big)}$ is a real number.
Clearly, $ \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big) \big( 1 + \dfrac{z_1}{z_{n+1}} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big)} = \frac{ \big( 1 + \dfrac{z_{n+1}}{z_n} \big) \big( 1 + \dfrac{z_1}{z_{n+1}} \big) \big( 1 + \dfrac{z_n}{z_1} \big)}{ \big( 1 + \dfrac{z_1}{z_n} \big) \big( 1 + \dfrac{z_n}{z_1} \big)}$ is real, since the denominator is real by step-1 and the nominator is real by some easy calculation. Q.E.D
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1184048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Linear operator from matrix to general presentaition Let $T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ be a linear operator defined by:
$T(0,1,1)=(2,-1,1), T(2,-1,0)=(1,1,0), T(-1,0,0)=(1,-1,1)$.
I need to find $T(x,y,z)$.
First of, i have found $[T]_{A}^{B}$ where A is the origin vectors and B is the image vectors that A is sent to.
I computed it to be
$$[T]_A^B=\begin{pmatrix} 1 & 0 & 2 \\
2 & - 1& 2 \\
2 & -3 & 3 \end{pmatrix}$$
But now after I have that matrix I am kind of lost. Is that even the right way to start?
| There are lots of ways to solve this problem. Here is a relatively straightforward srategy.
We are given that $T:\Bbb R^3\to\Bbb R^3$ is the linear map satisfying
\begin{align*}
T(0,1,1)&=(2,-1,1) & T(2,-1,0) &=(1,1,0) & T(-1,0,0)&=(1,-1,1)
\end{align*}
We wish to find a formula for $T(\vec x)$. To do so, we begin by solving the system
\begin{array}{rclclcl}
(1,0,0) & = & a_{11} (0,1,1) & + & a_{12} (2,-1,0) & + & a_{13} (-1,0,0)\\
(0,1,0) & = & a_{21} (0,1,1) & + & a_{22} (2,-1,0) & + & a_{23} (-1,0,0)\\
(0,0,1) & = & a_{31} (0,1,1) & + & a_{32} (2,-1,0) & + & a_{33} (-1,0,0)
\end{array}
which is equivalent to solving $AB = I$ where
\begin{align*}
A &=
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}
&
B
&=
\begin{bmatrix}
0 & 1 & 1 \\
2 & -1 & 0 \\
-1 & 0 & 0
\end{bmatrix}
&
I &=
\begin{bmatrix}
1 & 0 & 0\\ 0&1&0\\0&0&1
\end{bmatrix}
\end{align*}
But multiplying $AB=I$ on the left by $B^{-1}$ gives $A=B^{-1}$ and one computes
$$
B^{-1}=\begin{bmatrix} 0&0&-1\\0&-1&-2\\1&1&2\end{bmatrix}
$$
This gives
\begin{align*}
(1,0,0) &= -1\cdot(-1,0,0) \\
(0,1,0) &= -1\cdot (2,-1,0)+(-2)\cdot (-1,0,0) \\
(0,0,1) &= 1\cdot (0,1,1) +1\cdot(2,-1,0)+2\cdot(-1,0,0)
\end{align*}
Finally, we may compute
\begin{align*}
T(x,y,z)
&= T(x\cdot(1,0,0)+y\cdot(0,1,0)+z\cdot(0,0,1)) \\
&= x\cdot T(1,0,0)+y\cdot T(0,1,0)+z\cdot T(0,0,1) \\
&= ?
\end{align*}
Can you finish the computation?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that for all positive integers n, $\text{hcf}(6n + 8, 2n + 3) = 1$ To show that for all positive integers $n$, $\operatorname{hcf}(6n + 8, 2n + 3) = 1$
Can I just show that since n is a positive integer, so $6n + 8 > 4n + 5$.
When we apply long division
$6n + 8 = 1(4n + 5) + (2n +3)$
$4n + 5 = 1(2n + 3) + (2n +2)$
$2n + 3 = 1(2n + 2) + 1$
Therefore the highest common factor of $6n+8$ and $4n+5$ is $1.$
| Yes, your proof is simple and it works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality involving exponential partial sums Consider the exponential partial sums $E_n(x) = \sum_{i=0}^n \frac{x^i}{i!}$.
I want to prove that for all $x \ge 0$:
$$2 \frac {E_{n-1}(x)} {E_n(x)} \ge \frac {E_{n}(x)} {E_{n+1}(x)} + \frac {E_{n-2}(x)} {E_{n-1}(x)}$$
My approach so far
First observe that $E_{n-1}(x) = E_{n}(x) - \frac {x^n}{n!}$.
So the inequality becomes:
$$2 \frac {E_{n}(x) - \frac {x^n}{n!}} {E_n(x)} \ge \frac {E_{n+1}(x) - \frac {x^{n+1}}{(n+1)!}} {E_{n+1}(x)} + \frac {E_{n-1}(x) - \frac {x^{n-1}}{(n-1)!}} {E_{n-1}(x)}$$
which leads to
$$2 \frac {\frac {x^n}{n!}} {E_n(x)} \le \frac {\frac {x^{n+1}}{(n+1)!}} {E_{n+1}(x)} + \frac {\frac {x^{n-1}}{(n-1)!}} {E_{n-1}(x)}$$
So all we need to show is that $\frac {x^n} {n! E_n(x)}$ is convex in $n$. Unfortunately, I didn't have much luck going forward. A good direction could be to use the fact that $n! E_n(x) = e^x \Gamma(n+1,x)$, where $\Gamma(n+1,x) = \int_x^\infty t^n e^{-t} \textrm{dt}$ is the incomplete gamma function. I feel that this way, I will be able to prove the inequality analytically without working painfully with factorials and large sums. So it suffices to show that the following is convex as a function of $n$:
$$\frac {x^n} {\int_x^\infty t^n e^{-t} \textrm{dt}}$$
Any ideas on how to continue? Unfortunately, derivatives of the incomplete gamma function with respect to $n$ are not as nice as those with respect to $x$.
| Following thelionkingrafiki's reduction, we only need to show that, for $x > 0$ and $n \ge 1$,
$$
\frac{a_{n-1}}{E_{n-1}}
+\frac{a_{n+1}}{E_{n+1}}
-2\frac{a_{n}}{E_n}
> 0,
$$
where $a_{n} = x^n/n!$ and $E_n = \sum_{k=0}^n a_k$.
It turns out the inverse $E_n/a_n$ is easier to handle, so we shall define
$$
y_n
\equiv \frac{E_n}{a_n} - 1.
$$
and it can be shown by direct expansion that our statement is equivalent to
$$
(1 + 2 \, y_{n-1} - y_n)(y_{n-1} + y_{n+1} - 2 \, y_n)
<
2 \, (y_n - y_{n-1})^2.
$$
Now, by expanding the sum we have
\begin{align}
y_n &=
\frac{n}{x} + \frac{n(n-1)}{x^2} + \frac{n(n-1)(n-2)}{x^3} + \cdots + \frac{n!}{x^n} \\
y_n - y_{n-1} &=
\frac{1}{x} + \frac{2\,(n-1)}{x^2} + \frac{3\,(n-1)(n-2)}{x^3} + \cdots + \frac{n!}{x^n} \\
&=
\frac{1}{x}\left(
1 \, b_1 + 2 \, b_2 + 3 \, b_3 + \cdots + n \, b_n
\right),
\end{align}
where, $b_1 = 1$, and
$$
b_k = \frac{ (n-1)\cdots (n - k + 1) } { x^{k+1} },
$$
for $k \ge 2$.
Thus,
\begin{align}
y_{n-1} + y_{n+1} - 2 \, y_n
&=
\frac{2}{x^2} + \frac{3\cdot 2 \, (n-1)}{x^3} + \frac{4\cdot 3 \,(n-1)(n-2)}{x^4} + \cdots + \frac{(n+1)!}{x^{n+1}}\\
&\le
\frac{2}{x^2}\left(
1 + \frac{2^2 \, (n-1)}{x} + \frac{3^2 \,(n-1)(n-2)}{x^2} + \cdots + \frac{n^2 (n-1)!}{x^{n-1}} \right)\\
&=
\frac{2}{x^2}\left(
1^2 \, b_1 + 2^2 \, b_2 + 3^2 \, b_2 + \cdots + n^2 \, b_n
\right) \\
1 + 2 \, y_{n-1} - y_n
&=
1 + \frac{n-2}{x} + \frac{(n-1)(n-4)}{x^2} + \frac{(n-1)(n-2)(n-6)}{x^3} + \cdots - \frac{n!}{x^{n}}\\
&<
1 + \frac{n-1}{x} + \frac{(n-1)(n-2)}{x^2} + \frac{(n-1)(n-2)(n-3)}{x^3} + \cdots + \frac{(n-1)!}{x^{n-1}}\\
&=
b_1 + b_2 + b_3 + \cdots + b_n.
\end{align}
Finally, by the Cauchy-Schwarz inequality, we have
\begin{align}
(1 + 2 \, y_{n-1} - y_n)(y_{n-1} + y_{n+1} - 2 \, y_n)
&<
\frac{2}{x^2}
\left( \sum_{k=1}^{n} k^2 \, b_k \right)
\left( \sum_{k=1}^{n} b_k \right)
\\
&\le
\frac{2}{x^2}
\left( \sum_{k=1}^{n} k \, b_k \right)^2 \\
&=
2 \, \left(y_n - y_{n-1} \right)^2.
\end{align}
Q. E. D.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this inequality with $xyz\le 1$
If $x,y,z\ge 0$ and $\color{red}{xyz\le 1}$, show that
$$\color{blue}{\dfrac{x^2-x+1}{x^2+y^2+1}+\dfrac{y^2-y+1}{y^2+z^2+1}
+\dfrac{z^2-z+1}{z^2+x^2+1}\ge 1}.$$
| This inequality admits an SOS (Sum of Squares) representation
\begin{align}
&\frac{x^2-x+1}{x^2+y^2+1} + \frac{y^2-y+1}{y^2+z^2+1} + \frac{z^2-z+1}{z^2+x^2+1} - 1 \\
=\ & \frac{1}{(x^2+y^2+1)(y^2+z^2+1)(z^2+x^2+1)}\Big[(1-xyz)(f_1^2+f_2^2+f_3^2) + \frac{1}{12}u^TQu\Big]
\end{align}
where $f_1, f_2, f_3$ are polynomials with integer coefficients,
$u = [1, x, y, z, xy, xz, yz, x^2, y^2, z^2, xyz, x^2y, xz^2, y^2z]^T$
and $Q$ is a positive semi-definite (PSD) $14\times 14$ constant integer matrix.
The desired result follows.
Remarks:
1) Since the problem is tagged as contest-math but my proof is not an human proof, I do not give $f_1, f_2, f_3$ and $Q$ currently
and hope to see nice proofs.
2) Mathematica Resolve can prove the inequality although we do not see the step-by-step solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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integration of definite integral involving sinx and cos x Evaluate $\int_0^{\pi}\frac{dx}{a^2\cos^2x +b^2 \sin^2x}$
I got numerator $\sec^2 x$ and denominator $b^2 ( a^2/b^2 + \tan^2x)$.
I made substitution $u= \tan x$. That way $\sec^2 x$ got cancelled and the answer was of form $1/ab$ ($\tan^{-1} (bu/a)$)
And then if I put limits answer is $0$ but answer is wrong. Where did I go wrong?
| $$\begin{eqnarray*}\color{red}{I}&=&\int_{0}^{\pi}\frac{dx}{a^2\cos^2 x+b^2\sin^2 x}=2\int_{0}^{\pi/2}\frac{dx}{a^2\cos^2 x+b^2\sin^2 x}\\&=&2\int_{0}^{+\infty}\frac{dt}{(1+t^2)\left(a^2\frac{1}{1+t^2}+b^2\frac{t^2}{t^2+1}\right)}=2\int_{0}^{+\infty}\frac{dt}{a^2+b^2 t^2}\\&=&\frac{2}{ab}\int_{0}^{+\infty}\frac{du}{1+u^2}=\color{red}{\frac{\pi}{ab}}.\end{eqnarray*}$$
Here we replaced $x$ with $\arctan t$, then $t$ with $\frac{a}{b}\,u$.
| {
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Posed in regional mathematics Olympiad 2005 Let a, b, c be three positive real numbers such that $a+b+c = 1$.
Let $\Delta= \min( a^{3} + a^{2}bc, b^{3}+ab^{2}c, c^{3}+abc^{2} )$.
Prove that the roots of the equation $x^{2} + x + 4 \Delta = 0$ are real.
The last line is equivalent to $\Delta \leq \frac{1}{16} $. So I tried to prove by contradiction. I assumed all of them are $ > \frac{1}{16} $and tried to draw a contradiction with the fact $a+ b+ c=1$ but failed
| First, we see that the roots of the polynimium are real iff:
$$1-16\Delta \ge 0 \Leftrightarrow \Delta \le \frac{1}{16}$$
Assume without loss of gennerality that $a \le b \le c$. Then:
$$\Delta = a^3 + a^2bc = a^2(a+bc)$$
$bc$ is biggest when $b=c=\frac{1-a}{2}$, so:
$$\Delta\le a^2\left(a+\left(\frac{1-a}{2}\right)^2\right)=\frac{a^4+2a^3+a^2}{4}$$
Notice that this expression grows monotonically with $a$, but $a\le \frac13$, so it's maximum is at $a=\frac13$. But then:
$$\Delta\le\frac{\frac{1}{81}+\frac{2}{27}+\frac{1}{9}}{4}=\frac{4}{81} < \frac{1}{16}$$
| {
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Build the graph of a function with absolute value. The function is:
And my idea of graphic (i did it using two graphs and deleting some parts)
Is it correct?
| Your choice of scale makes the graph a bit difficult to interpret, but it appears you correctly graphed $y = x^2 + 3$ when $|x| > 1$. However, you incorrectly graphed $y = |x + 2| + |x - 1|$ in the interval $[-1, 1]$.
Since $x - 1 \geq 0$ if $x \geq 1$ and $x - 1 < 0$ if $x < 1$,
\begin{align*}
|x - 1| & =
\begin{cases}
x - 1 & \text{if $x \geq 1$}\\
-(x - 1) & \text{if $x < 1$}
\end{cases}\\
& =
\begin{cases}
x - 1 & \text{if $x \geq 1$}\\
-x + 1 & \text{if $x < 1$}
\end{cases}
\end{align*}
Since $x + 2 \geq 0$ if $x \geq -2$ and $x + 2 < 0$ if $x < -2$,
\begin{align*}
|x + 2| & =
\begin{cases}
x + 2 & \text{if $x \geq -2$}\\
-(x + 2) & \text{if $x < -2$}
\end{cases}\\
& =
\begin{cases}
x - 1 & \text{if $x \geq -2$}\\
-x - 2 & \text{if $x < -2$}
\end{cases}
\end{align*}
Putting these rules together yields
\begin{align*}
|x + 2| + |x - 1| & =
\begin{cases}
x + 2 + x - 1 & \text{if $x \geq 1$}\\
x + 2 - (x - 1) & \text{if $-2 \leq x < 1$}\\
-(x + 2) - (x - 1) & \text{if $x < -2$}
\end{cases}\\
& =
\begin{cases}
2x + 1 & \text{if $x \geq 1$}\\
3 & \text{if $-2 \leq x < 1$}\\
-2x - 1 & \text{if $x < 1$}
\end{cases}
\end{align*}
If $|x| \leq 1$, then $-1 \leq x \leq 1$. If $-1 \leq x < 1$, then the rule for $-2 \leq x < 1$ applies, so $$y = |x + 2| + |x - 1| = 3$$ If $x = 1$, then the rule for $x \geq 1$ applies, so $$y = |x + 2| + |x - 1| = 2x + 1 = 2 \cdot 1 + 1 = 2 + 1 = 3$$ Thus, in the interval $[-1, 1]$, the graph of $y = |x + 2| + |x - 1|$ lies on the horizontal line $y = 3$.
The graph of the function
$$
y =
\begin{cases}
x^2 + 3 & \text{if $|x| > 1$}\\
|x + 2| + |x - 1| & \text{if $|x| \leq 1$}
\end{cases}
$$
is sketched below. While the grid lines make it somewhat difficult to tell, the circles at the points $(1, 3)$ and $(-1, 3)$ are filled since these points are on the graph while the circles at the points $(1, 4)$ and $(1, -4)$ are empty since these points are not on the graph.
| {
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Evaluate the following indefinite integral Evaluate the integral : $$\int\frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}\,dx$$
I tried through putting $x=\tan \theta$ as well as $x=\tan^2\theta$ .but I am unable to remove the square root. I also tride by putting $x+x^2+x^3=z^2$. But I could not proceed anyway...Please help...
Update :
putting $x=\frac{1-t}{1+t}$ , I get , $$\sqrt{x+x^2+x^3}=\sqrt{\frac{2-3t+t^2-t^3}{(1+t)^3}}$$
How you got $\sqrt{x+x^2+x^3}=\sqrt{(t^3+3)(1-t^2)}/(t+1)^2$ ?
| $\bf{My\; Solution::}$ Given $$\displaystyle \int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx = -\int\frac{(x^2-1)}{(x+1)^2\sqrt{x+x^2+x^3}}dx$$
We can Write it as $$\displaystyle -\int\frac{\left(1-\frac{1}{x^2}\right)}{\left(x+\frac{1}{x}+2\right)\sqrt{x+\frac{1}{x}+1}}dx$$
Now Let $$\displaystyle \left(x+\frac{1}{x}+1\right)=t^2\;,$$ Then $$\displaystyle \left(1-\frac{1}{x^2}\right)dx = 2tdt$$
So Integral Convert into $$\displaystyle - \int\frac{2t}{(t^2+1)\cdot t}dt = -\int\frac{2}{1+t^2}dt = -2\tan^{-1}(t)+\mathcal{C}$$
So Integral Convert into $$\displaystyle \int \frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}dx =-2\tan^{-1}\left(\sqrt{x+\frac{1}{x}+1}\right)+\mathcal{C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solutions of the Laplace equation How do I find solutions $u=f(r)$ of the two-dimensional Laplace equation $u_{xx}+u_{yy}=0$ that depend only on the radial coordinate $r= \sqrt{x^2+y^2}$
| As @CameronWilliams pointed out, the easiest way to do this is to switch to polar coordinates. But, here is a brute force way to proceed.
Let $f(x,y)=g(x^2+y^2)$. Then, $\frac{\partial f}{\partial x}=2xg'(x^2+y^2)$.
Taking a second partial yields $\frac{\partial^2 f}{\partial x^2}=(2x)^2g''(x^2+y^2)+2g'(x^2+y^2)$.
One can easily see then (replace $x$ with $y$) that $\frac{\partial^2 f}{\partial y^2}=(2y)^2g''(x^2+y^2)+2g'(x^2+y^2)$.
Thus,
$$\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=4(x^2+y^2)g''(x^2+y^2) + 4g'(x^2+y^2)=0$$
Then, we may write
$$\frac{g''(x^2+y^2)}{g'(x^2+y^2)}=\left(\log g'(x^2+y^2)\right)'=-\frac{1}{x^2+y^2}$$
where again the "prime" is on $x^2+y^2$. Integrating once reveals that $\log g'(x^2+y^2) = -\log(x^2+y^2) +C \Rightarrow g(x^2+y^2) = A\log (x^2+y^2) +B$.
| {
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Simplifying algebra fraction expression The below equation is after taking the derivative for a calculus problem, but when I try to simplify I don't see how. My calculator shows that
$${5(\sqrt{1+x^2}-x) \over \sqrt{1+x^2}(1+(x-\sqrt{1+x^2})^2)}$$ simplifies to
$${5 \over 2(x^2+1)}$$
and no matter how much I rearrange the equation I just don't see how anything cancels. What am I missing?
| $$\sqrt{1+x^2}(1+(x-\sqrt{1+x^2})^2) \\ =\sqrt{1+x^2}+x^2\sqrt{1+x^2}+\sqrt{1+x^2}(1+x^2)-2x(\sqrt{1+x^2})^2 \\ =2\sqrt{1+x^2}+2x^2\sqrt{1+x^2}-2x(1+x^2) \\ = (2+2x^2)\sqrt{1+x^2}-2x(1+x^2) \\ = 2(1+x^2)\left[\sqrt{1+x^2}-x\right]$$
Then cancel from the numerator and denominator to get your result.
| {
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Inequality and Induction: $\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$ I needed to prove that
$\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$, $\forall n \geq 1$ .
I've atempted by induction.
I proved the case for $n=1$ and assumed it holds for some $n$.
The left-side of the n+1 case is
$\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2}$.
Using the inductive hypothesis, i could reach the result that
$\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2} <\frac{2n+1}{\sqrt{2n+1}(2n+2)}=\frac{\sqrt{2n+1}}{(2n+2)}$.
Now, i'm wondering how should i connect it to my goal :
$\frac{1}{\sqrt{2n+2+1}}=\frac{1}{\sqrt{2n+3}} $
I know one way to prove that
$\frac{\sqrt{2n+1}}{(2n+2)}<\frac{1}{\sqrt{2n+3}} $
We just square things, then eventually reach
$ (n+1).(n+3)<(n+2)^2$
Which is easily provable because $3<4$ ....
But I was wondering if there was another way to show that ... perharps a more direct way to show that last bit ... a way that was not so direct and brute as to involve squaring both sides. A way of gradually manipulating the left-side until reaching the inequality with the right side.
Thanks in advance.
| Let $$a_n = \prod_{k=1}^n\frac{2k-1}{2k},\quad b_n=\prod_{k=1}^n\frac{2k}{2k+1}$$ then $0<a_n<b_n$ and so $a_n^2<a_nb_n=\frac{1}{2n+1}$.
| {
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Another optimization problem I am having trouble figuring out a next step in an optimization problem
the question is to find the max and min values of $f(x,y)=\frac{x+y}{2+x^2+y^2}$
I calculated $f_x$ and $f_y$ and set both them equal to zero, and the only possibility you get is x=y. I dont know how else to find it after this. But the back of the book says the answer is a max at $f(1,1)$ and a min at $f(-1,-1)$ but I dont know how?
$$f_x=\frac{-x^2-2xy+y^2+2}{(2+x^2+y^2)^2}$$
$$f_y= \frac{-y^2-2xy+x^2+2}{(2+x^2+y^2)^2}$$
Can anyone see why please?
Thankyou
| Rewriting $\displaystyle \frac{x+y}{2+x^2+y^2}=t\;\;$ gives $\;\;tx^2-x+ty^2-y+2t=0$.
This will have a solution for $x$ if $b^2-4ac=1-4t(ty^2-y+2t)\ge0$,
which gives $\;\;4t^2y^2-4ty+8t^2-1\le0$.
This will have a solution for $y$ if $b^2-4ac=16t^2-16t^2(8t^2-1)\ge0$.
Then $16t^2(2-8t^2)\ge0,\;\;$ so $8t^2\le2\implies t^2\le\frac{1}{4}\implies-\frac{1}{2}\le t\le \frac{1}{2}$.
Since $t=\frac{1}{2}\implies x^2+y^2-2x-2y+2=0\implies (x-1)^2+(y-1)^2=0\implies x=1 \text{ and } y=1$
and $t=-\frac{1}{2}\implies x^2+y^2+2x+2y+2=0\implies (x+1)^2+(y+1)^2=0\implies x=-1 \text{ and } y=-1$,
$f(1,1)=\frac{1}{2}$ is the maximum value for f and $f(-1,-1)=-\frac{1}{2}$ is the minimum value for f.
| {
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Finding a limit with two independent variables: $\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$ I must find the following limit: $$\lim_{(x,y)\to (0,0)}\frac{x^2y^2}{x^2+y^2}$$ Substituting $y=mx$ and $y=x^2$, I have found the limit to be $0$ both times, as $x \to 0$. I have thus assumed that the above limit is $0$, and will attempt to prove it. Let $\varepsilon>0$. We have that: $$\left\lvert\frac{x^2y^2}{x^2+y^2}\right\rvert=\frac{x^2y^2}{x^2+y^2}\leq\frac{(x^2+y^2)(x^2+y^2)}{x^2+y^2}=x^2+y^2$$ However, I must find $\delta>0$ such that $0<\sqrt{x^2+y^2}<\delta$, and I cannot see a way to obtain $\sqrt{x^2+y^2}$ in the above inequality to complete the proof. Am I mistaken in my process? Thank you.
| $$0<\sqrt{x^2+y^2}<\delta\iff 0<x^2+y^2<\delta^2$$
But you have already shown that $\left\lvert\frac{x^2y^2}{x^2+y^2}-0\right\rvert<x^2+y^2$ and so $$\implies\left\lvert\frac{x^2y^2}{x^2+y^2}-0\right\rvert<\delta^2$$
So choosing $\delta=\sqrt{\epsilon}$ suffices.
| {
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What method was used here to expand $\ln(z)$? On Wikipedia's entry for bilinear transform, there is this formula:
\begin{align}
s &= \frac{1}{T} \ln(z) \\[6pt]
&= \frac{2}{T} \left[\frac{z-1}{z+1} + \frac{1}{3} \left( \frac{z-1}{z+1} \right)^3 + \frac{1}{5} \left( \frac{z-1}{z+1} \right)^5 + \frac{1}{7} \left( \frac{z-1}{z+1} \right)^7 + \cdots \right] \\[6pt]
&\approx \frac{2}{T} \frac{z - 1}{z + 1} \\[6pt]
&= \frac{2}{T} \frac{1 - z^{-1}}{1 + z^{-1}}
\end{align}
What is the method that is used to expand $\ln(z)$? Taylor series? Laurent series? Some other techniques?
| $$
z = \frac{1+\frac{z-1}{z+1}}{1-\frac{z-1}{z+1}}
$$
| {
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The diophantine equation $x^2+y^2=3z^2$ I tried to solve this question but without success:
Find all the integer solutions of the equation: $x^2+y^2=3z^2$
I know that if the sum of two squares is divided by $3$ then the two numbers are divided by $3$, hence if $(x,y,z)$ is a solution then $x=3a,y=3b$. I have $3a^2+3b^2=z^2$ and that implies
$$
\left(\frac{a}{z}\right)^2+\left(\frac{b}{z}\right)^2=\frac{1}{3}
$$
so I need to find the rational solutions of the equation $u^2+v^2=\frac{1}{3}$ and I think that there are no solutions for that because $\frac{1}{3}$ doesn't have a rational root, but I dont know how to explain it.
Thanks
| Let $(x,y)=d$ and $\dfrac xX=\dfrac yY=d$
$$\implies d^2(X^2+Y^2)=3z^2\implies d|z,$$ $z=dZ$(say)
$$\implies X^2+Y^2=3Z^2$$
Now $X^2+Y^2\equiv1,2\pmod4$ as $X,Y$ both can not be even
$$Z^2\equiv0,1\pmod4\implies3Z^2\equiv0,3\pmod4$$
| {
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Prove that equation $x^6+x^5-x^4-x^3+x^2+x-1=0$ has two real roots Prove that equation
$$x^6+x^5-x^4-x^3+x^2+x-1=0$$ has two real roots
and $$x^6-x^5+x^4+x^3-x^2-x+1=0$$
has two real roots
I think that:
$$x^{4k+2}+x^{4k+1}-x^{4k}-x^{4k-1}+x^{4k-2}+x^{4k-3}-..+x^2+x-1=0$$
and
$$x^{4k+2}-x^{4k+1}+x^{4k}+x^{4k-1}-x^{4k-2}-x^{4k-3}-..+x^2+x-1=0$$
has two real roots but i don't have solution
| The first equation is dealt with well by @user140591's answer. The second equation, $x^6-x^5+x^4+x^3-x^2-x+1=0,$ has no real roots. To prove this, we first show that it has no negative roots. This can be seen by writing the equation as $$x=\frac{x^6+x^4-x^2+1}{x^4-x^2+1}\qquad\qquad\qquad\quad$$
$$=\dfrac43\frac{3(x^3-x)^2+(3x^2-1)^2+2}{(2x^2-1)^2+3}.$$In the first displayed equation, write $x^2=t$ and divide to get $$\sqrt t=t+2-\frac{4}{(2t-1)^2+3}.$$To show that the RHS exceeds the LHS for $t\geqslant0$, it is enough to show that it does so when the negative term is replaced by its least value, namely $-\frac43$, and the LHS is replaced by $\frac12t+\frac12$, which upper-bounds it. This is easily seen to be the case.
| {
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Calculation of real root values of $x$ in $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}.$
Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $
$\bf{My\; Solution::}$ Here domain of equation is $\displaystyle x\geq 1$. So squaring both sides we get
$\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$.
$\displaystyle (1-2x)^2=4(x^2-1)\Rightarrow 1+4x^2-4x=4x^2-4\Rightarrow x=\frac{5}{4}.$
But when we put $\displaystyle x = \frac{5}{4}\;,$ We get $\displaystyle \frac{3}{2}-\frac{1}{2}=2\Rightarrow 1=2.$(False.)
So we get no solution.
My Question is : Can we solve above question by using comparision of expressions?
Something like $\sqrt{x+1}<\sqrt{x-1}+\sqrt{4x-1}\; \forall x\geq 1?$
If that way possible, please help me solve it. Thanks.
| Also, for $x\ge1$ $$\sqrt{4x-1}+\sqrt{x-1}\ge\sqrt{4x-1+x-1}=\sqrt{5x-2}>\sqrt{2x}\ge\sqrt{x+1}. $$
| {
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How to prove that $ \sum_{k=1}^{n-1} \frac{1}{1-e^{2 \pi i k/n}} = \frac{n-1}{2}$? I came across the fact that
$$ \sum_{k=1}^{n-1} \frac{1}{1-e^{2 \pi i k/n}} = \frac{n-1}{2}.$$
How can we prove this identity?
| One has for $f(X)=\frac{X^n-1}{X-1}=\prod\limits_{k=1}^{n-1}(X-e^{\frac{2ik\pi}{n}})$
$$\frac{f'(X)}{f(X)}=\sum\limits_{k=1}^{n-1}\frac{1}{X-e^{\frac{2ik\pi}{n}}}$$
So we have $\sum\limits_{k=1}^{n-1}\frac{1}{1-e^{\frac{2ik\pi}{n}}}=\frac{f'(1)}{f(1)}$
Now let us compute $\frac{f'(X)}{f(X)}=\frac{(n-1)X^{n-2}+(n-2)X^{n-3}+\cdots+2X+1}{X^{n-1}+\cdots+1}$ so we have
$$\frac{f'(1)}{f(1)}=\frac{(n-1)+(n-2)+\cdots+2+1}{n}=\frac{n^2-\frac{n(n+1)}{2}}{n}=n-\frac{n+1}{2}=\frac{n-1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 0
} |
How to differentiate $y=\sqrt{\frac{1+x}{1-x}}$? I'm trying to solve this problem but I think I'm missing something.
Here's what I've done so far:
$$g(x) = \frac{1+x}{1-x}$$
$$u = 1+x$$
$$u' = 1$$
$$v = 1-x$$
$$v' = -1$$
$$g'(x) = \frac{(1-x) -(-1)(1+x)}{(1-x)^2}$$
$$g'(x) = \frac{1-x+1+x}{(1-x)^2}$$
$$g'(x) = \frac{2}{(1-x)^2}$$
$$y' = \frac{1}{2}(\frac{1+x}{1-x})^{-\frac{1}{2}}(\frac{2}{(1-x)^2})
$$
| $$
y' = \frac{d}{dx}\left(\sqrt{\frac{1+x}{1-x}}\right)
$$
Chain rule through the square root:
$$
y' = \frac{1}{2}\sqrt{\frac{1-x}{1+x}}\frac{d}{dx}\left(\frac{1+x}{1-x}\right)
$$
Quotient rule on what's left
$$
y' = \frac{1}{2}\sqrt{\frac{1-x}{1+x}}\left(\frac{1-x-(1+x)(-1)}{(1-x)^2}\right)
$$
Simplify
$$
y' = \sqrt{\frac{1-x}{1+x}}\frac{1}{(1-x)^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
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Proof of sum results I was going through some of my notes when I found both these sums with their results
$$
x^0+x^1+x^2+x^3+... = \frac{1}{1-x}, |x|<1
$$
$$
0+1+2x+3x^2+4x^3+... = \frac{1}{(1-x)^2}
$$
I tried but I was unable to prove or confirm that these results are actually correct, could anyone please help me confirm whether these work or not?
| Let $S=1+x+x^2+\dotsb$. Now, take a look at this:
$\phantom xS=1+x+x^2+x^3+\dotsb$
$xS=\phantom{1+{}}x+x^2+x^3+\dotsb$
Thus, we have:
\begin{align}
S&=1+xS\\
S-Sx&=1\\
S(1-x)&=1\\
S&=\frac1{1-x}
\end{align}
Moving on to the second one: This one is a bit trickier, but we'll use the identity $(n+2)-2(n+1)+n=0$ in a clever way.
Let $T=x+2x^2+3x^3+\dotsb$. Now:
$\phantom{(1-2x+x^2)}T=x+2x^2+3x^3+4x^4+5x^5+\dotsb$
$\phantom{(1+x^2)}-2xT=\phantom{x}-2x^2-4x^3-6x^4-8x^4-\dotsb$
$\phantom{(1-2x+{})}x^2T=\phantom{x+2x^2+3}x^3+2x^4+3x^5+\dotsb$
Add them up:
$(1-2x+x^2)T=x$
Thus, we have:
$$T=\frac x{1-2x+x^2}=\frac x{(1-x)^2}$$
EDIT: By the way, we haven't been paying much attention to the matter of convergence. All we've proven, in fact, is that where $S$ and $T$ converge, their sums are equal to what we've derived above.
We've actually implicitly assumed that $\displaystyle\lim_{N\to\infty}x^N=0$, which is only true where $-1<x<1$. This happens to be where $S$ and $T$ converge. (We should check the boundary points — that is, $x=1$ and $x=-1$ — but it's not too hard to see that the sums diverge there.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
No definite integrals of trigonometry I have big problems with the following integrals:
$$\int\frac{dx}{\sin^6 x+\cos^6x}$$
$$\int\frac{\sin^2x}{\sin x+2\cos x}dx$$
It isn't nice of me but I almost have no idea, yet I tried the trigonometric substitution $\;t=\tan\frac x2\;$ , but I obtained terrible things and can't do the rational function integral.
Perhaps there is exist some trigonometry equalities? I tried also
$$\frac1{\sin^6x+\cos^6x}=\frac{\sec^6x}{1+\tan^6x}=\frac13\frac{3\sec^2x\tan^2x}{1+\left(\tan^3\right)^2}\cdot\overbrace{\frac1{\sin^2x\cos^2x}}^{=\frac14\sin^22x}$$
and then doing parts with
$$u=\frac14\sin^22x\;\;:\;\;u'=\sin2x\cos2x=\frac12\sin4x\\{}\\v'=\frac13\frac{3\sec^2x\tan^2x}{1+\left(\tan^3\right)^2}\;\;:\;\;v=\arctan\tan^3x$$
But it is impossible to me doing the integral of $\;u'v\;$ .
Any help is greatly appreciated
| $\bf{My\; Solution}::$ Given $$\displaystyle \int\frac{1}{\sin^6 x+\cos ^6 x}dx\;,$$ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\cos^6 x\;,$
We Get $$\displaystyle \int\frac{\sec^6 x}{1+\tan^ 6 x}dx\;,$$ Now let $$\tan x=t\;,$$ Then $$\sec^2 xdx = dt\;,$$ So we get
$$\displaystyle \int\frac{(1+t^2)^2}{1+t^6}dt = \int\frac{(1+t^2)^2}{(1+t^2)\cdot (t^4-t^2+1)}dt = \int\frac{1+t^2}{t^4-t^2+1}dt$$.
Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $t^2\;,$ We get
$$\displaystyle \int\frac{\left(1+\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}-1}dt = \int\frac{\left(1+\frac{1}{t^2}\right)}{\left(t-\frac{1}{t}\right)^2+1},$$ Now Let $$\displaystyle \left(t-\frac{1}{t}\right)=u\;,$$ Then $$\displaystyle \left(1+\frac{1}{t^2}\right)dt = du$$
So We get $$\displaystyle \int\frac{1}{u^2+1}du = \tan^{-1}\left(u\right)+\mathcal{C} = \tan^{-1}\left(\frac{t^2-1}{t}\right)+\mathcal{C}$$
So we get $$\displaystyle \int\frac{1}{\sin^6 x+\cos ^6x}dx = \tan^{-1}\left(\frac{\tan^2 x-1}{\tan x}\right)+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
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Finding Eigenvalues and Eigenvectors for Leslie Matrix A Leslie Matrix is given by
$$L =\begin{pmatrix}0 & (3/2)a^2 & (3/2)a^3\\1/2 & 0 & 0\\ 0 & 1/3 & 0\end{pmatrix}\cdot$$
Find the Eigenvalues and determine the dominant eigenvalue and eigenvector. I am struggling with how to do this without a calculator, even when I use Wolfram alpha, the answers I am getting do not seem to make sense. Should I guess a root?
| $\det(L-\lambda I)=-\lambda\begin{vmatrix}-\lambda & 0\\\frac{1}{3} &-\lambda\end{vmatrix}-\frac{1}{2}\begin{vmatrix}\frac{3}{2}a^2 & \frac{3}{2}a^3\\\frac{1}{3}&-\lambda\end{vmatrix}=-\lambda^3-\frac{1}{2}(\frac{3}{2}a^2)\begin{vmatrix}1&a\\\frac{1}{3}&-\lambda\end{vmatrix}=-\lambda^3-\frac{3}{4}a^2(-\lambda-\frac{1}{3}a)$
$=-\frac{1}{4}(4\lambda^3-3a^2\lambda-a^3)=-\frac{1}{4}(\lambda-a)(4\lambda^2+a\lambda+a^2)$, so $\lambda=a$ is the dominant eigenvalue.
Reducing $\begin{bmatrix}-a &\frac{3}{2}a^2&\frac{3}{2}a^3&0\\\frac{1}{2}&-a&0&0\\0&\frac{1}{3}&-a&0\end{bmatrix}$ gives $\begin{bmatrix}1&-2a &0&0\\0&1&-3a &0\\0&-\frac{1}{2}a &\frac{3}{2}a^2 &0\end{bmatrix}$ and then
$\hspace{.6 in}\begin{bmatrix}1&-2a&0&0\\0&1&-3a&0\\0&0&0&0\end{bmatrix}$ and $\begin{bmatrix}1&0&-6a^2&0\\0&1&-3a&0\\0&0&0&0\end{bmatrix}$.
Therefore $\begin{bmatrix}6a^2\\3a\\1\end{bmatrix}$ is an eigenvector for the dominant eigenvalue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Demand $z=x+y$ and $x^2/4 + y^2/5 + z^2/25 = 1$. What is the maximum value of $f(x,y,z) = x^2+y^2+z^2$?
Demand $z=x+y$ and $x^2/4 + y^2/5 + z^2/25 = 1$. What is the maximum value of $f(x,y,z) = x^2+y^2+z^2$?
I've been attempting this with Lagrange multipliers in a few different ways. However, the resulting system of equations with two lagrangians has so many variables that it becomes very complicated. Can someone show how this is to be done manually?
I also attempted turning it into two unknowns by replacing $z$ with $x+y$. However, this also led nowhere.
| If $3x=2y$, we get a value $10$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$x^2+y^2+z^2\leq10\left(\frac{x^2}{4}+\frac{y^2}{5}+\frac{z^2}{25}\right)$$ or
$$x^2+y^2+(x+y)^2\leq10\left(\frac{x^2}{4}+\frac{y^2}{5}+\frac{(x+y)^2}{25}\right)$$ or
$$(3x-2y)^2\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1230637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $2^{105} + 3^{105}$ is divisible by $7$ I know that $$\frac{(ak \pm 1)^n}{a}$$ gives remainder $a - 1$ is n is odd or $1$ is n is even.
So, I wrote $ 2^{105} + 3^{105}$ as $8^{35} + 27^{35}$ and then as $(7\cdot 1+1)^{35} + (7\cdot 4-1)^{35}$, which on division should give remainder of $6$ for each term and total remainder of 5 (12/7).
But, as per question, it should be divisible by 7, so remainder should be zero not 5. Where did I go wrong?
[note: i don't know binomial theorem or number theory.]
| $$\begin{align}
2^3 &\equiv 1 (\mod 7)\\
(2^3)^{35} &\equiv 1^{35} (\mod 7)\\
2^{105} &\equiv 1 (\mod 7)\tag1
\end{align}$$
Again,
$$\begin{align}
3^3 &\equiv -1 (\mod 7)\\
(3^3)^{35} &\equiv (-1)^{35} (\mod 7)\\
3^{105} &\equiv -1 (\mod 7)\tag2
\end{align}$$
Adding (1) and (2) we get,
$$\begin{align}
2^{105}+3^{105} &\equiv1+(-1)\space (\mod7)\\
\text{or,}\quad 2^{105}+3^{105} &\equiv0\space (\mod7)
\end{align}$$
This implies that $\space2^{105}+3^{105}$ is divisible by 7.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Is there another way to solve $\int \frac{x}{\sqrt{2x-1}}dx$? $$\int \frac{x}{\sqrt{2x-1}}dx$$
Let $u=2x-1$
$du=2dx$
$$=\frac{1}{2}\int \frac{u+1}{2\sqrt{u}}du$$
$$=\frac{1}{2}\int (\frac{\sqrt{u}}{2}+\frac{1}{2\sqrt{u}})du$$
$$=\frac{1}{4}\int \sqrt{u}du+\frac{1}{4}\int\frac{1}{\sqrt{u}}du$$
$$=\frac{u^{\frac{3}{2}}}{6}+\frac{\sqrt{u}}{2}+c$$
$$=\frac{1}{6}(2x-1)^{\frac{3}{2}}+\frac{1}{2}\sqrt{2x-1}+c$$
$$=\frac{1}{3}(x+1)\sqrt{2x-1}+c$$
Is there another way except this?
| Another method is to let $u=\sqrt{2x-1}$, so $x=\frac{1}{2}(u^2+1), dx=udu$.
Then $\displaystyle\int\frac{x}{\sqrt{2x-1}}dx=\int\frac{\frac{1}{2}(u^2+1)}{u}\cdot udu=\frac{1}{2}\int(u^2+1)du=\frac{1}{2}\left[\frac{u^3}{3}+u\right]+C$
$\hspace{.23in}=\frac{1}{6}(2x-1)^{3/2}+\frac{1}{2}(2x-1)^{1/2}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How can you derive $\sin(x) = \sin(x+2\pi)$ from the Taylor series for $\sin(x)$? \begin{eqnarray*}
\sin(x) & = & x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\\
\sin(x+2π) & = & x + 2\pi - \frac{(x+2π)^3}{3!} + \frac{(x+2π)^5}{5!} - \ldots \\
\end{eqnarray*}
Those two series must be equal, but how can you show that by only manipulating the series?
| Define $f(x)$ as the residual:
$$\sin(x+2\pi) = x + 2\pi - \frac{(x+2\pi)^3}{3!} + \frac{(x+2\pi)^5}{5!}+\ldots$$
$$\equiv x - \frac{x^3}{3!} + \frac{x^5}{5!}+\ldots + f(x)=\sin(x)+f(x)$$
Now prove it is zero:
$$f(x) = 2\pi -\frac{1}{3!}\left((2\pi)^3 + 3(2\pi)^2 x + 3(2\pi)x^2 \right)+
\frac{1}{5!}\left( (2\pi)^5 +5x (2\pi)^4+10x^2(2\pi)^3+10x^3(2\pi)^2+5(2\pi)x^4\right)+\ldots$$
Extract all powers of $x$ lower than $2$, and define yet another residual, $g(x)$:
$$f(x)=\sin(2\pi) + x\sum_{k=1}^\infty \frac{(-1)^k(2\pi)^{2k}}{(2k)!}+g(x)$$
$$=\sin(2\pi) +x(1-\cos(2\pi))+g(x)$$
The first terms are zero, and now we must prove that the remaining residual $g(x)$ is also zero, ad infinitum.
A more general continuation to follow...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of m. Problem :
If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of $m$.
My Approach:
$|x^2+4x+3|-mx+2m=0$
Case I : $x^2+4x+3-mx+2m=0$
$\Rightarrow x^2+ x (4-m) + 3+2m=0 $
Discriminant of above qudratic is
$D = (4-m)^2 -4(3+2m) \geq 0$
$D = 16+m^2-8m-12-8m$
Solving for $m$ we get the values $-8 \pm 2\sqrt{15}$
Case II :
Similarly solving for the given equation taking negative sign of modulus we get the solution
for $m =$$8 \pm 2\sqrt{15}$
Can we take all the values of m to satisfy the given condition of the problem , please suggest which value of m should be neglected in this. Thanks.
| the value of $m$ you are looking for is the slope of the tangent line to the graph $y = -x^2 - 4x _ 3, -3 < x < 1.$ it has a tangent at the $x$-coordinate $2 - \sqrt {15}$ with a slope of $-2a - 4 = -8 + 2\sqrt{15}.$ any line with a bigger slope cuts at two points and any with a positive and smaller slope will cut at four points and if the slope is zero, then it cuts at two pints $x = -1, x = -3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
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How to integrate $\frac{1}{(1+a\cos x)}$ from $-\pi$ to $\pi$ How to solve the following integration?$$\int_{-\pi}^\pi\frac{1}{1+a \cos x}$$
| So, here's how I like to think of proceeding.
We know the following identities:
$$\sin^2(x/2)=\frac{1-\cos x}{2}$$
$$\cos^2(x/2)=\frac{1+\cos x}{2}$$
Now, the function $1+a\cos u$ can be written as
$$1+a\cos u=A(1+\cos u)+B(1-\cos u)$$
where $A+B=1$ and $A-B=a$. Thus, $A=\frac12(1+a)$ and $B=\frac12(1-a)$.
Now, we can write
$$\begin{align}
\frac{1}{1+\cos u}&=\frac{1}{\frac12(1+a)(1+\cos u)+\frac12(1-a)(1-\cos u)}\\\\
&=\frac{1}{(1+a)\cos^2(u/2)+(1-a)\sin^2(u/2)}\\\\
&=\frac{1}{1+a}\frac{\sec^2(u/2)}{1+\frac{1-a}{1+a}\tan^2(u/2)}\\\\
\end{align}$$
To integrate this, we have
$$\begin{align}
\int_{-\pi}^{\pi} \frac{1}{1+\cos u}\,\, du&=\int_{-\pi}^{\pi} \frac{1}{1+a}\frac{\sec^2(u/2)}{1+\frac{1-a}{1+a}\tan^2(u/2)}\,\, du\\\\
\end{align}$$
Case 1: Assume $|a|<1$
Making the substitution $t=\sqrt{\frac{1-a}{1+a}}\tan(u/2)$ implies that $dt=\frac12 \sqrt{\frac{1-a}{1+a}}\sec^2(u/2)du$, and the limits of integration transform from $(-\pi, \pi)$ to $(-\infty,\infty)$. Thus,
$$\begin{align}
\int_{-\pi}^{\pi} \frac{1}{1+\cos u}\,\, du&=2\frac{1}{\sqrt{1-a^2)}}\int_{-\infty}^{\infty} \frac{dt}{1+t^2}\\\\
\end{align}$$
The anti-derivative of $\frac{1}{1+t^2}$ is $\arctan(t)$, which when evaluated between $-\infty$ and $+\infty$ is $\pi$.
The final result is thus $\frac{2\pi}{\sqrt{1-a^2}}$.
Case 2: Assume $a>1$
Making the substitution $t=\sqrt{\frac{a-1}{1+a}}\tan(u/2)$ implies that $dt=\frac12 \sqrt{\frac{a-1}{1+a}}\sec^2(u/2)du$, and the limits of integration transform from $(-\pi, \pi)$ to $(-\infty,\infty)$. Thus,
$$\begin{align}
\int_{-\pi}^{\pi} \frac{1}{1+\cos u}\,\, du&=2\frac{1}{\sqrt{a^2-1)}}\int_{-\infty}^{\infty} \frac{dt}{1-t^2}\\\\
\end{align}$$
This integral diverges (singularities at $\pm \infty$ and $\pm 1$), unless it is interpreted in a Cauchy Principal Value sense with
$$\int_{- \infty}^{\infty} \frac{dt}{1-t^2} =\lim_{M \to \infty} \lim_{\epsilon \to 0} \left( \int_{-M}^{-1-\epsilon} \frac{dt}{1-t^2} +\int_{-1+\epsilon}^{1-\epsilon} \frac{dt}{1-t^2} +\int_{1+\epsilon}^{M} \frac{dt}{1-t^2} \right)$$
in which case the integral is zero!
Case 3: Assume that $a<-1$
This case is similar to Case 2 - the integral diverges unless it is interpreted in a Cauchy Principal Value sense for which the result in zero.
Case 4: $|a| =1$
The integral diverges whether or not it is interpreted in a Cauchy Principal Value sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1235200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Express $\ln(3+x)$ and $\frac{1+x}{1-x}$ as Maclaurin series its probably a lot to ask.. but how can I obtain the Maclaurin series for the two functions $f(x)=\ln(3+x)$ and $g(x)=\frac{1+x}{1-x}$ ? as far as I know I cant use any commonly known series to help me with this one ? so finding the derivatives at 0:
$f(0) = \ln(3)$
$f'(x)= 1/(x+3) \implies f'(0) = 1/3$
$f''(x)= -1/(x+3)^2 \implies f''(0)= -1/ 3^2$
$f'''(x)= 2/(x+3)^3 \implies f'''(0)= 2/3^3$
$f''''(x)= -6/(x+3)^4 \implies f''''(0)= -6/3^4$
and now I guess Ill have to use the maclaurin formula:
$$f(x)=\sum \frac{f^{(n)}(0) x^n}{n!}$$
but now how do I continue from here? I've got:
$$(\ln 3) + \frac{x}{3} - \frac{x^2}{3^2 \cdot 2!} + \frac{2x^3}{3^3 \cdot 3!} - \frac{6x^4}{3^4 \cdot 4!} $$
same question goes for $g(x)$ honestly, on that one upon simplifying the numerator from $g(0)$ with the factorial when I plug it into the maclaurien series formula I just get $g(x)= 1+2x+2x^2 + 2x^3 + 2x^4 + \cdots$
| For the second function, rewrite it as
$$ g(x) = -1 + \frac{2}{1-x} $$
Then apply what you know for $\frac{1}{1-x}$, which is the geometric series
$$ -1 + 2\cdot\frac{1}{1-x} = -1 + 2\sum_{n=0}^{\infty}x^n $$
The first function is a bit tricky, note that
$$\frac{d}{dx}\ln(x+3) = \frac{1}{x+3}$$
With that in mind, we can find the series expression for the derivative and then integrate the series
$$\frac{1}{x+3} = \frac{1}{3\left(1 + \frac{x}{3}\right)} = \frac{1}{3}\cdot\frac{1}{1 - \left(-\frac{x}{3}\right)} \\= \frac{1}{3} \sum_{n=0}^{\infty} \left(-\frac{x}{3}\right)^n = \sum_{n=0}^\infty \frac{(-1)^n x^n}{3^{n+1}}$$
Integrating
$$\begin{align}\ln(x+3) &= \int \frac{1}{x+3}\,dx \\&= \int \sum_{n=0}^\infty \frac{(-1)^n x^n}{3^{n+1}} dx \\&= \sum_{n=0}^\infty \int \frac{(-1)^n x^n}{3^{n+1}}\,dx \\&= \sum_{n=0}^{\infty} \frac{(-1)^nx^{n+1}}{(n+1)3^{n+1}} \\&= \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n\cdot3^n} \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality with condition $x+y+z=xy+yz+zx$ I'm trying to prove the following inequality:
For $x,y,z\in\mathbb{R}$ with $x+y+z=xy+yz+zx$, prove that
$$
\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{z^2+1}\ge-\frac{1}{2}
$$
My approach:
After slight manipulation the inequality is equivalent to:
$$
\sum_{cyc}\frac{(x+1)^2}{x^2+1}\ge 2
$$
Now, applying CS is legitimate and it reduces the inequality to proving:
$$
s^2-10s-3\le0
$$
with $s=x+y+z=xy+yz+zx$, but I'm not quite sure if this is still true. Could anybody give me a hint in the right direction? Any help is highly appreciated.
| Here is a way to use CS to solve the inequality. First, we re-write what we want to prove as
$$\frac{(x+1)^2}{x^2+1}+\frac{(y+1)^2}{y^2+1} \ge 2-\frac{(z+1)^2}{z^2+1} = \frac{(z-1)^2}{z^2+1}$$
Now the constraint gives $z = \dfrac{x+y-xy}{x+y-1}$. Using this, we need to only show
$$\frac{(x+1)^2}{x^2+1}+\frac{(y+1)^2}{y^2+1} \ge \frac{(xy-1)^2}{(x+y-xy)^2+(x+y-1)^2}$$
Using CS on the LHS, we have
$$\left(\frac{(x+1)^2}{x^2+1}+\frac{(y+1)^2}{y^2+1}\right)\left((x^2+1)(y-1)^2+(y^2+1)(x-1)^2 \right) \ge \left((x+1)(y-1)+(y+1)(x-1) \right)^2=4(xy-1)^2$$
Thus it is sufficient to show
$$4(x+y-xy)^2+4(x+y-1)^2 \ge (x^2+1)(y-1)^2+(y^2+1)(x-1)^2$$
which reduces to showing that the following quadratic (in say $x$) is non-negative:
$$(y^2-3y+3)x^2 - (3 y^2-8 y+3)x + (3y^2-3y+1) \ge 0$$
which is easy to show as its discriminant, $-3(y^2-1)^2$ is never positive.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find whole number answers in systems of square root equations Given the following 4 equations, can you find 4 whole number answers using whole number variable inputs?
$x,y,z$ where $x>y>z$
$Eq 1 = (x^2-2xy+y^2-2xz+z^2)^{\frac{1}{2}} $
$Eq 2 = (x^2-2xy+y^2+2xz-z^2)^{\frac{1}{2}} $
$Eq 3 = (x^2+2xy-y^2+2xz-z^2)^{\frac{1}{2}}$
$Eq 4 = (x^2+2xy-y^2-2xz+z^2)^{\frac{1}{2}} $
for example, $x=1921, y=792, z=272$ yields $551$ in equation 1, but non-whole number answers for remaining 3 equations
| Do the substitution $x=c,\,$ $y=x-a,\,$ $z=x-b,\,$ and we get the system,
$$a^2+b^2-c^2 = s_1^2\tag1$$
$$a^2-b^2+c^2 = s_2^2\tag2$$
$$-a^2-b^2+3c^2 = s_3^2\tag3$$
$$-a^2+b^2+c^2 = s_4^2\tag4$$
Solving for $(1)$ as $u_1^2+u_2^2=v_1^2+v_2^2,\,$ and the other three are just quadratic forms to be made squares. Then solving simultaneously any two of the remaining three involves an elliptic curve, so there are infinitely many solutions to the $3$-out-of-$4$ system. In fact, there is a polynomial solution to $(1), (2),(4)$, a "small" particular example being,
$$x=5405,\; y=4440,\; z=22$$
However, I haven't found a solution for all four.
| {
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Probability of number of drawing cards in a scenario being equal to that in another scenario I came across the following question in a book:-
$Q.$ Cards are drawn one by one at random from a well shuffled pack of $52$ cards.
$(a)$Find the probability that exactly $n$ cards are drawn before the first king appears.
$(b)$If $N$ is the number of cards required to be drawn until $2$ aces are obtained for the first time then show that probability of $N$ being equal to $n$ is $$\frac {(n-1)(52-n)(51-n)}{50*49*17*13} \space \space where \space 2\le n\le 50 .$$
The first part is easy. It is simply that no king has been chosen in the last $n$ cards. At the $(n+1)th$ card, we have $4$ kings to choose from. Hence $$P(n)=\frac {\binom {48}{n}\binom{4}{1}}{\binom {52}{n}}=\frac {(52-n)(51-n)(50-n)(49-n)}{13*51*50*49}$$
For the second part we now that the last card drawn is the second ace (which can be chosen in $3$ ways as one ace has already been chosen) and none of the cards drawn before is a king (as $N=n$). Hence probability of drawing the aces is $$P(N)=\frac {3\binom{4}{1}\binom{44}{n-1}}{\binom{52}{n}}$$ But how do I find the probability of $n$ being equal to $N$?
| (a) Andre's comment is right about needing to include the king in your calculations.
There are:
\begin{eqnarray*}
&&\text{$\binom{48}{n}$ ways to choose the first $n$ cards} \\
&&\text{$\binom{4}{1}$ ways to choose the first king} \\
&&\text{$n!$ ways to arrange those first $n+1$ cards} \\
&&\text{$(51-n)!$ ways to arrange the last $51-n$ cards} \\
&&\text{$52!$ ways to arrange all $52$ cards.} \\
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
P(\text{$n$ cards before first king}) &=& \dfrac{\binom{48}{n}\binom{4}{1}n!(51-n)!}{52!} \\
&& \\
&=&\dfrac{(51-n)(50-n)(49-n)}{13\cdot51\cdot50\cdot49}\qquad\qquad\text{after simplifying.} \\
\end{eqnarray*}
$$\\$$
(b) Here, there are:
\begin{eqnarray*}
&&\text{$\binom{48}{n-2}$ ways to choose the first $n-2$ non-ace cards} \\
&&\text{$\binom{4}{2}$ ways to choose the first two aces} \\
&&\text{$2!(n-2)!$ ways to arrange those first $n$ cards} \\
&&\text{$(52-n)!$ ways to arrange the last $52-n$ cards} \\
&&\text{$52!$ ways to arrange all $52$ cards.} \\
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
P(\text{$n^{th}$ card is the second ace}) &=& \dfrac{\binom{48}{n-2}\binom{4}{2}2!(n-2)!(52-n)!}{52!} \\
&& \\
&=&\dfrac{(n-1)(52-n)(51-n)}{50\cdot49\cdot17\cdot13}\qquad\qquad\text{after simplifying.} \\
\end{eqnarray*}
| {
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Show that $\ln(1+x)\leq x-{1\over 2}x^2+{1\over 3}x^3$. Prove that for $x\in (0,\infty)$, $\ln(1+x)\leq x-{1\over 2}x^2+{1\over 3}x^3$.
I'm a little bit stuck, but I think I have the right idea. Any hints or solutions are greatly appreciated. Here is what I have.
If we assume that $\ln(1+x)>x-{1\over 2}x^2+{1\over 3}x^3$ and let $f(x)=\ln(1+x)-x+{1\over 2}x^2-{1\over 3}x^3>0$, then $f'(x)={1\over{1+x}}-1+x-x^2={-x^3\over 1+x}<0$ for all $x>0$. That is, $f'(x)<0$ implies that $f(x)$ decreasing which is a contradiction.
Does this make sense?
| You could convert yours into a direct proof easily. As you have, let
$$f(x) = \ln(1+x) - x + \dfrac{x^2}2 - \dfrac{x^3}3$$
We then have
$$f'(x) = \dfrac1{1+x} - 1 + x - x^2 = \dfrac{-x^3}{1+x} < 0 \text{ since }x \in (0,\infty)$$
Hence, $f(x)$ is a decreasing function, which means
$$f(x) < f(0) \implies \ln(1+x) - x + \dfrac{x^2}2 - \dfrac{x^3}3 < 0 \implies \ln(1+x) < x - \dfrac{x^2}2 + \dfrac{x^3}3$$
| {
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Basic question about modular arithmetic applied to the divisor sum function $\sigma(n)$ when $n=5p$ While studying the divisor sum function $\sigma(n)$ (as the sum of the divisors of a number) I observed that the following expression seems to be true always (1):
$\forall\ n=5p, p\in\Bbb P,\ p\gt 5,\ if\ d(5p)=4\ then\ \sigma(5p)=3,0\ (mod\ 9)$
Meaning that if the number $n=5k$ has only four divisors, so $5$ and other prime $p$ divide $n$, then the sum of the divisors is congruent to 3 or 0 modulo 9. I just observed that manually in the range $[1..400]$.
The divisors of a $5p$ number as the type above are $\{1,5,p,5p\}$
E.g.
$d(35) = \{1,5,7,35\}$
$d(55) = \{1,5,11,55\}$
And the sum of the divisors is $\sigma(5p)= 1+5+p+5p$, thus (2):
$(1+5+p+5p)\ mod\ 9 = 1 + 5 + ((p + 5p)\ mod\ 9) =$
$6 + ((p + 5p)\ mod\ 9) = 6 + (6p\ mod\ 9)$
... but from that point I am not sure how to conclude that for the case above (1) it will be $3$ or $0$.
I think that this must happen at (2) if (1) is true:
$6p\ mod\ 9 = 6 * (p\ mod\ 9) \equiv 3, 6\ (mod\ 9)$
and then is true only if:
$p\equiv 1,x?\ (mod\ 9)$
Any help or hint is very appreciated, or a counterexample if (1) is wrong and then the congruence is false, thank you!
| The divisors of 5k are $\{1, 5, k, 5k\}$ if and only if k is a prime number or if $k = 25$.
The divisors of $5 \times 25$ are $\{1, 5, 25, 125\}$. Their sum is $3 \pmod 9$.
Otherwise, you want to compute $1 + 5 + k + 5k \pmod 9 \equiv 6(1+k) \pmod 9$ for all prime numbers $k$ (Except $k=5$).
For $k = 0 \text{ to } 8 \pmod 9$, the values of $6(1+k) \pmod 9$ are $6, 3, 0, 6, 3, 0, 6, 3$.
If $k$ is a multiple of $3$, then $k \in \{3, 6, 9, 12, \dots\}$.
Except for $k = 3$, none of those numbers are prime numbers.
This means that, except for $k = 3$, k cannot be congruent to $0, 3$, or $6$ modulo $9$.
This eliminates the bolded numbers in the sequence of possible values
6, $3$, $0$, 6, $3$, $0$, 6, $3$.
The conclusion is
*
*$d(5k) = 4$ if and only if k = 25 or if k is a prime number other than 5.
*For $k = 3$ the sum of the divisors of $5 \times 3$ is $6$ modulo $9$.
*For $k = 25$ the sum of the divisors of $5 \times 25$ is $3$ modulo
$9$.
*Otherwise, k cannot be congruent to $0, 3$, or $6$ modulo $9$.
*For $k \equiv 1, 4$, or $7 \pmod 9$, the sum of the divisors of $5k$ is
$6$ modulo $9$.
*For $k \equiv 2, 5$, or $8 \pmod 9$, the sum of the divisors of 5k is
$3$ modulo $9$.
| {
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Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$. Find $m$.
Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$. Find $m$.
$m$ is a 3 digit number (because this was an AIME problem).
$$m \equiv 0 \pmod{17}$$
$$m \equiv 17 \pmod{9} \equiv -1 \pmod{9}$$
Applying the chinese remainder theorem, the solution is supposed to be:
$$m = 17 \cdot 9 = 153$$
but that isnt correct.
| Lebes, from your lead and solving for $N$, we have
$$N \equiv 0 \pmod{17}\equiv 17 \pmod{17}$$
and from the digit sum condition
$$N \equiv 17 \pmod{9}.$$
Since $(17,9)=1$, we have $N\equiv17\pmod{153}$. [This uses principles from the Chinese Remainder Theorem - see the bottom for a detailed explanation]
Since this is only finding solutions where the digitsum $\equiv8\pmod{9}$, we now simply check the first few solutions to find the correct answer:
$1\times 153+17=170\longrightarrow $ digit sum $=8\qquad$Nope.
$2\times 153+17=323\longrightarrow $ digit sum $=8\qquad$Nope. [Fixed the error here]
$3\times 153+17=476\longrightarrow $ digit sum $=17\qquad$Found it!
Therefore $N=476$.
Thanks @AaronMaroja for the correction.
Explanation of the use of CRT for @Amad27:
Using the Chinese Remainder Theorem to solve $N \equiv 0 \pmod{17}$ and $N \equiv 17 \pmod{9} \equiv 8 \pmod{9}$. Breaking these down we get $a_1=0$, $a_2=8$, $m_1=17$, $ m_2=9$, $M=m_1\times m_2=153$, $M_1=M/m_1=153/17=9$, $M_2=M/m_2=153/9=17$.
Now applying CRT we get
$$M_1y_1\equiv 1\pmod{m_1}\rightarrow 9y_1\equiv 1\pmod{17},$$
and
$$M_2y_2\equiv 1\pmod{m_2}\rightarrow 17y_2\equiv 1\pmod{9}\rightarrow 8y_2\equiv 1\pmod{9}.$$
It can be seen by observation that $y_1=2$ and $y_2=8$.
Finally, combining the above information we get
$$m=a_1M_1y_1+a_2M_2y_2=0\times9\times2+8\times17\times8=1088.$$
Since $1088\equiv17\pmod{153}$, we have
$$N\equiv m\pmod{M}\equiv17\pmod{153}$$
| {
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Find area of rhombus Given the following rhombus, where points E and F divide the sides CD and BC respectively, AF = 13 and EF = 10
I think the length of the diagonal BD is two times EF = 20, but i got stuck from there.
| I am assuming that the midpoint of $CD$ is $E$ and not $F$. Let $BF=CF=CE=a$. We then have $AB=2a$. If $\angle{ECF} = \theta$, we have $\angle{ABF} = \pi-\theta$.
Applying cosine rule for triangles $ABF$ and $CEF$, we obtain
\begin{align}
\cos(\theta) & = \dfrac{a^2+a^2-10^2}{2\cdot a \cdot a}\\
\cos(\pi-\theta) & = \dfrac{a^2+(2a)^2-13^2}{2\cdot a \cdot (2a)}
\end{align}
Adding both the equations, we obtain
$$\dfrac{2a^2-10^2}{2a^2} + \dfrac{5a^2-13^2}{4a^2} = 0 \implies 1-\dfrac{50}{a^2} + \dfrac54 - \dfrac{169}{4a^2} = 0 \implies \dfrac94 - \dfrac{369}{4a^2} = 0$$
This gives us the side to be $$2a = 2\sqrt{41}$$
| {
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Use implicit differentiation to find an equation of the tangent line to the curve at the given point
Use implicit differentiation to find an equation of the tangent line to the curve $$x^2+xy+y^2=1$$ at $(1,1)$.
I am not sure how I should work this out because the given point is not on the curve.
| [1] $x^2 + xy + y^2 = 1$
differentiating implicitly by $\frac{d}{dx}$
$2x + x\frac{dy}{dx} + y + 2y \frac{dy}{dx} = 0$
Which simplifies to
[2] $\frac{dy}{dx} = \frac{-2x - y}{x + 2y}$
Now consider a tangent linear equation which passes through a point on the curve and through $(1,1)$
It has the form
[3] $y = mx + c$
To find the points that these equations [1] and [3] intersect we must solve them simultaneously.
Substituting $y$ into [1]
$x^2 + x(mx+c) + (mx+c)^2 = 1$
This simplifies to
$(m^2+m+1)x^2 + (2mc+c)x + c^2 = 1$
using the quadratic formula
[4] $x = \frac{-2mc-c \pm \sqrt{(2mc+c)^2 - 4(m^2+m+1)c^2}}{2(m^2+m+1)}$
Note that there are 2 solutions because there are two points on the curve where this will work.
Also, since we know that equation [3] passes through the point $(1,1)$
we know
$1= m1+c$
So [5] $m = 1 - c$
Se expression [4] becomes
[6] $x = \frac{-2(1 - c)c-c \pm \sqrt{(2(1 - c)c+c)^2 - 4((1 - c)^2+(1 - c)+1)c^2}}{2((1 - c)^2+(1 - c)+1)}$
Lets call these $x_1$ and $x_2$ They have a corresponding $y_1$ and $y_2$
We also know that the gradient of [3] is $m$. This will be equal to the derivative given by [2] at $(x_1, y_1)$ and $(x_2,y_2)$
ie:
[7] $m = \frac{-2x_1 - y_1}{x_1 + 2y_1} = 1 - c$
and
[8] $m = \frac{-2x_2 - y_2}{x_2 + 2y_2} = 1 - c$
Substitute [6] into [7] to get an expression for $y_1$ in terms of c
Then you can substitute $x_1$ and $y_1$ into [1] to find what $c$ is.
From here you can calculate $m$ which will give you the tangent line you are looking for.
A similar process can be used for $x_2$ and $y_2$ there are two tangent lines which will solve this problem.
I'll leave the rest for you
| {
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Find the least degree Polynomial whose one of the roots is $ \cos(12^{\circ})$ Find the least degree Polynomial with Integer Coefficients whose one of the roots is $ \cos(12^{\circ})$
My Try: we know that $$\cos(5x)=\cos^5x-10\cos^3x\sin^2x+5\cos x\sin^4x$$ Putting $x=12^{\circ}$ and Converting $\sin$ to $\cos$ we have
$$\frac{1}{2}=x^5-10x^3(1-x^2)+5x(1-2x^2+x^4)$$ $\implies$
$$32x^5-40x^3+10x-1=0$$
Is this the Least degree? Please let me know.
| Looking at function $$f(x)=32x^5-40x^3+10x-1$$ by inspection $x=\frac 12$ is a root. So $$f(x)=(2x-1)(16 x^4+8 x^3-16 x^2-8 x+1)$$ Now, you have a quartic which can be solved with radicals and integers.
| {
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If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x)$
If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x).$
My Solution:: Let $$\displaystyle y = \sin^4 x+\cos^2 x \leq \sin^2 x+\cos^2 x=1$$
And for Minimum, We take $$\displaystyle y = \sin^4 x+\cos^2 x=(1-\cos^2 x)^2+\cos^2 x$$
So $$\displaystyle y=\cos^4 x-\cos^2 x+1 = \left(\cos^2 x-\frac{1}{2}\right)^2+\frac{3}{4}\geq \frac{3}{4}$$
So We get $\displaystyle y=\sin^4 x+\cos^2 x\in \left[\frac{3}{4}\;,1\right]$
My question is How can we find Min. of $f(x)$ other then that method,
Something Like Using Inequality.,Thanks
| We have
\begin{align}
f(x) & = \sin^4(x) + \cos^2(x) = \sin^4(x) + 1 - \sin^2(x) = 1 - \sin^2(x)(1-\sin^2(x))\\
& = 1-\sin^2(x)\cos^2(x) = 1 - \dfrac{\sin^2(2x)}4
\end{align}
I trust you can finish off from here.
| {
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Proving convergence/divergence via the ratio test Consider the series
$$\sum\limits_{k=1}^\infty \frac{-3^k\cdot k!}{k^k}$$
Using the ratio test, the expression $\frac{|a_{k+1}|}{|a_k|}$ is calculated as:
$$\frac{3^{k+1}\cdot (k+1)!}{(k+1)^{k+1}}\cdot \frac{k^k}{3^k\cdot k!}=\frac{3}{(k+1)^{k}}\cdot {k^k}=3\cdot \frac{k^k}{(k+1)^{k}}=3\cdot \left(\frac{k}{k+1}\right)^k$$
How to continue?
| $$\log L=\lim_{k\rightarrow \infty }k\log(1-\frac{1}{k+1})=\lim_{k\rightarrow \infty }\frac{\log(1-\frac{1}{k+1})}{\frac{1}{k}}$$
by using Lopital rule
$$\log L=\lim_{k\rightarrow \infty }\frac{\frac{\frac{1}{(k+1)^2}}{1-\frac{1}{k+1}}}{\frac{-1}{k^2}}=\lim_{k\rightarrow \infty }\frac{-k^2}{(k^2+1)(1-\frac{1}{k+1})}=\lim_{k\rightarrow\infty }\frac{-1}{(1+\frac{1}{k^2})(1-\frac{1}{1+k})}=-1$$
hence the limit $$\frac{3}{e}>1$$
| {
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Questions concerning elements in $F = \big\{f: \{1, 2, 3\} \to \{1, 2, 3, 4, 5\}\big\}$. a) Find and simplify the number of functions $f \in F$ so that $f(1) = 4$.
My attempt: there is $1$ choice for $f(1)$, and $5$ choices for $f(2)$ and $5$ choices for $f(3)$, thus $1\cdot 5\cdot 5 = 25$ functions
b) Find and simplify the number of one-to-one functions $f \in F$ so that $f(1) \geq 4$
Attempt: since there are $2$ choices for $f(1)$, I broke it into $2$ cases, one when $f(1) = 5$ and when $f(1) = 4$.
Then there are $(1 \cdot 4 \cdot 3) + (1 \cdot 4 \cdot 3) = 24$ choices.
Is this correct?
| Yes, both solutions are correct. Your reasoning for (a) is spot-on.
For (b), you could have simplified your reasoning a little -- there are $2$ choices for $f(1)$, and then $4$ choices for $f(2)$, and $3$ choices for $f(3)$. By the rule of product, you have $2 \cdot 4 \cdot 3 = 24$ choices in total.
| {
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Formula of parabola from two points and the $y$ coordinate of the vertex The parabola has a vertical axis of symmetry. Given two points and the $y$ coordinate of the vertex, how to determine its formula?
For example:
| Formula of parabola is $y=ax^2+bx+c$. The task here is first to find the values of $a,b,c$. We know that the vertex happens at $x^*=-\frac{b}{2a}$ (right?). We plug the points we have in the formula for parabola:
\begin{align}
6&=5^2a+5b+c\\
8&=8^2a+8b+c
\end{align}
and for the vertex we have
\begin{align}
10&=(-\frac{b}{2a})^2a+b(-\frac{b}{2a})+c\\
\end{align}
From the first equation we find that $c=6-25a-5b$. This we plug in the other two equations and simplify to obtain
\begin{align}
39a+3b-2&=0\\
5+25a+5b+\frac{b^2}{4a}&=0
\end{align}
For the first equation we obtain $b=\frac23-\frac{39}{3}a$, which we plug in the second equation to obtain $$\frac{9a}{4}+\frac{1}{9a}+3=0$$ Solving this equations we obtain
\begin{align}
a&=\frac29(-3-2\sqrt{2})\\
a&=\frac29(-3+2\sqrt{2})
\end{align}
and consequently you can find two values for $b$ and two values for $c$. Hence there are two values for the unknown ($-b/2a$) you are after $$?=11-3\sqrt{2}$$ and $$?=11+3\sqrt{2}$$
| {
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How to factor the polynomial $x^4-x^2 + 1$ How do I factor this polynomial: $$x^4-x^2+1$$
The solution is: $$(x^2-x\sqrt{3}+1)(x^2+x\sqrt{3}+1)$$
Can you please explain what method is used there and what methods can I use generally for 4th or 5th degree polynomials?
| You can get it by$$x^4-x^2+1=x^4+2x^2-2x^2-x^2+1=x^4+2x^2+1-3x^2=(x^2+1)^2-(\sqrt 3x)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Evaluating trigonometric limit: $\lim_{x \to 0} \frac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2}$
Evaluate $\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $
This is what I've tried yet:
$$\begin{align} & \cfrac{x(\tan 2x - 2\tan x)}{4\sin^4 x} \\
=&\cfrac{x\left\{\left(\frac{2\tan x}{1-\tan^2 x} \right) - 2\tan x\right\}}{4\sin^4 x}\\
=& \cfrac{2x\tan x \left(\frac{\tan^2 x}{1 - \tan^2 x}\right) }{4\sin^4 x} \\
=& \cfrac{x\tan^3 x}{2\sin^4 x (1-\tan^2 x)} \\
=& \cfrac{\tan^3 x}{2x^3\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} \\
=& \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)}\end{align}$$
Taking limit of the above expression, we've :
$$\lim_{x\to 0} \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} = \lim_{x\to 0} \cfrac{\cos^2x}{2\cos 2x} = \cfrac{1}{2} $$
Firstly, is my answer right or am I doing somewhere wrong?
Secondly, this seems a comparatively longer method than expected for objective type questions. I'm seeking for a shortcut method for such type of questions. Is there any method I should've preferred?
| Ok this is a better way of doing this:
$$\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $$
$$= \lim_{x\to 0} \frac{1}{4} \frac{x}{\sin x} \frac{\tan 2x - 2\tan x}{\sin^3 x}$$
$$= \lim_{x\to 0} \frac{1}{4} \frac{\sin2x \cos x - 2\sin x \cos 2x}{\cos 2x \cos x \sin^3 x}$$
$$= \lim_{x\to 0} \frac{1}{2} \frac{\cos^2 x - \cos 2x}{\cos 2x \cos x \sin^ 2 x}$$
$$= \lim_{x\to 0} \frac{1}{2} \frac{\cos^2 x - \cos^2 x + \sin^2x}{\cos2x \cos x \sin^2 x}$$
$$ = \lim_{x\to 0} \frac{1}{2} \frac{1}{\cos 2x \cos x} = \frac{1}{2}$$
So your answer was right, but I feel like this is a faster and neater way of doing this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1263968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Why represent a complex number $a+ib$ as $[\begin{smallmatrix}a & -b\\ b & \hphantom{-}a\end{smallmatrix}]$? I am reading through John Stillwell's Naive Lie Algebra and it is claimed that all complex numbers can be represented by a $2\times 2$ matrix $\begin{bmatrix}a & -b\\ b & \hphantom{-}a\end{bmatrix}$.
But obviously $a+ib$ is quite different from $\begin{bmatrix}a & -b\\ b & \hphantom{-}a\end{bmatrix}$, as the latter being quite clumsy to use and seldom seen in any applications I am aware of. Furthermore, it complicates simple operations such as matrix multiplication whereby you have to go one extra step and extract the complex number after doing the multiplication.
Can someone explain what exactly is the difference (if there is any) between the two different representations? In what instances is a matrix representation advantageous?
| Quirky little thing, there is a real version, little 2 by 2 blocks, of the Jordan Canonical Form that goes with this. For a real matrix, eigenvalues come in conjugate pairs;
$$
\left(
\begin{array}{rrrr}
i & 1 & 0 & 0 \\
0 & i & 0 & 0 \\
0 & 0 & -i & 1 \\
0 & 0 & 0 & -i
\end{array}
\right)
$$
is similar to
$$
\left(
\begin{array}{rr|rr}
0 & 1 & 1 & 0 \\
-1 & 0 & 0 & 1 \\ \hline
0 & 0 & 0 & 1 \\
0 & 0 & -1 & 0
\end{array}
\right)
$$
Then
$$
\left(
\begin{array}{rrrrrr}
i & 1 & 0 & 0 & 0 & 0 \\
0 & i & 1 & 0 & 0 & 0 \\
0 & 0 & i & 0 & 0 & 0 \\
0 & 0 & 0 & -i & 1 & 0 \\
0 & 0 & 0 & 0 & -i & 1 \\
0 & 0 & 0 & 0 & 0 & -i
\end{array}
\right)
$$
is similar to
$$
\left(
\begin{array}{rr|rr|rr}
0 & 1 & 1 & 0 & 0 & 0 \\
-1 & 0 & 0 & 1 & 0 & 0 \\ \hline
0 & 0 & 0 & 1 & 1 & 0 \\
0 & 0 & -1 & 0 & 0 & 1 \\ \hline
0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & -1 & 0 \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 11,
"answer_id": 6
} |
Differentiate $\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$ with respect to $x$
Differentiate $$\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$$ with respect to $x$.
I started like this: Consider $$\frac {\sin x + \cos x}{\sqrt{2}}$$, substitute $\cos x$ as $\sin \left(\frac {\pi}{2} - x\right)$, and proceed with the simplification. Finally I am getting it as $\cos \left(x - \frac {\pi}{4}\right)$. After this I could not proceed. Any help would be appreciated. Thanks in advance!
| $$\frac{d(\arcsin x)}{dx}=\frac{1}{\sqrt{1-x^2}}$$ It is as simple as that. This is followed by the substitution $v(x)=\frac{\sin x+\cos x}{\sqrt{2}}$. $$ \begin{align} \frac{d}{dx}\arcsin(v(x)) & =\frac{1}{\sqrt{1-v^2(x)}}v'(x) \\
&= \frac{1}{\sqrt{1-\Biggl(\frac{\sin x+\cos x}{\sqrt{2}}\Biggl)^2}}\Biggl(\frac{\cos x-\sin x}{\sqrt{2}}\Biggl) \\
&= \frac{\sqrt{2}}{\sqrt{(\cos x-\sin x)^2}}\frac{\cos x-\sin x}{\sqrt{2}} \\
&= \frac{\cos x-\sin x}{|\cos x-\sin x|} \\
&= \frac{\cos(x+\frac{\pi}{4})}{|\cos(x+\frac{\pi}{4})|} \\ \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to solve this nonstandard system of equations? How to solve this system of equations
$$\begin{cases}
2x^2+y^2=1,\\
x^2 + y \sqrt{1-x^2}=1+(1-y)\sqrt{x}.
\end{cases}$$
I see $(0,1)$ is a root.
| (More a comment.) If we allow the non-principal square root,
$$\begin{cases}
2x^2+y^2=1,\\
x^2 + y \sqrt{1-x^2}=1\color{red}{\pm} (1-y)\sqrt{x}
\end{cases}$$
the $+$ case has one real solution, but the $-$ case has three real solutions: $(x,y)=(0,1)$ and two which surprisingly are roots of $11$-deg equations. Using,
$$x=+\sqrt{\frac{1-y^2}{2}}$$
and with $y$ as two appropriate real roots of,
$$\small49 - 301 y + 327 y^2 - 51 y^3 - 1038 y^4 + 1334 y^5 - 1066 y^6 + 850 y^7 - 259 y^8 + 279 y^9 + 3 y^{10} + y^{11}=0$$
namely $y_1 \approx -0.619107$, and $y_2 \approx 0.939251$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a).$$
Clearly, $$\left\{\det(A)\neq0\left|\begin{matrix}c\neq b\\a\neq c\\b\neq a\\a,b,c\neq 0\end{matrix}\right.\right\}\\$$
Is it sufficient to say that the matrix is invertible provided that all 4 constraints are met? Would Cramer's rule yield more explicit results for $a,b,c$ such that $\det(A)\neq0$?
| If $A$ is a square matrix, $\det(A)=\det(A^T)=\det\begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix}$
If $B$ is obtained from $A$ by adding a multiple of a row of $A$ to another row in $A$, $\det(B)=\det(A)\Rightarrow\text{let }B=\begin{pmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{pmatrix}\\\Rightarrow\det(B)=\det(A)=\begin{vmatrix}b-a&b^2-a^2\\c-a&c^2-a^2\end{vmatrix}=(c^2-a^2)(b-a)-(b^2-a^2)(c-a)\\\\=(c-a)(b-a)((c-a)-(b-a))=(c-a)(b-a)(c-b)\\\therefore\det(A)\neq0\iff\begin{cases}a\neq c\\a\neq b\\b\neq c\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Transformed pde but my answer doesn't match solution? $$\frac{d^2u}{dx^2} + \frac{d^2u}{dy^2} + \frac{du}{dx} + 2\frac{du}{dy} + 3u = 0$$
Let $u = ve^{ax + by}$ and find $a, b$ such that we can transform to the following equation
$$\frac{d^2v}{dx^2} + \frac{d^2v}{dy^2} + Av = 0$$
where $A$ is an arbitrary constant.
I used the chain rule to get the first and second derivatives in terms of $v$.
$\frac{du}{dx} = \frac{du}{dv}\frac{dv}{dx}$
Then
$v_{xx} + v_{yy} + (a+1)v_x + (b+2)v_y + 3v = 0$
and letting $a = -1$ and $b = -2$ gives
$v_{xx} + v_{yy} + Av = 0$
However I have a solution for this problem and it says, $a=-\frac{1}{2}$ and $b=-1$
Have I made a mistake or is the given solution incorrect?
| $$
u_{xx} + u_{yy} + u_x+2u_y + 3u = 0
$$
use the sub
$$
u_x = v_x\mathrm{e}^{ax+by} + au\\
u_{xx} = v_{xx}\mathrm{e}^{ax+by} + 2av_x\mathrm{e}^{ax+by} + a^2u\\
u_y = v_y\mathrm{e}^{ax+by} + bu\\
u_{yy} = v_{yy}\mathrm{e}^{ax+by} + 2bv_y\mathrm{e}^{ax+by} + b^2u\\
$$
thus we get
$$
v_{xx}\mathrm{e}^{ax+by} + 2av_x\mathrm{e}^{ax+by} + a^2u + v_{yy}\mathrm{e}^{ax+by} + 2bv_y\mathrm{e}^{ax+by} + b^2u + v_x\mathrm{e}^{ax+by} + au + 2v_y\mathrm{e}^{ax+by} + 2bu + 3u = 0
$$
we get
$$
\mathrm{e}^{ax+by}\left[v_{xx} + 2av_x + v_{yy} + 2bv_y + v_x + 2v_y\right] + \left(a^2+b^2+a + 2b+3\right)v\mathrm{e}^{ax+by} = 0
$$
combining terms
$$
\mathrm{e}^{ax+by}\left[v_{xx} + v_{yy} + 2(b+1)v_y +(2a+1)v_x + \left(a^2+b^2+a + 2b+3\right)\right] = 0
$$
this means
$$
2a+1 = 0 \implies a = -\frac{1}{2}\\
b+1 = 0 \implies b = -1.
$$
and to wrap it up
$$
a^2+b^2+a + 2b+3 = A = \left(-\frac{1}{2}\right)^2+(-1)^2 -\frac{1}{2} -2 + 3=\frac{7}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Least Square method, find vector x that minimises $ ||Ax-b||_2^2$ Given Matrix A =
| 1 0 1 |
| 1 1 2 |
| 0 -1 -1|
and b = $[1\ \ 4\ -2]^T$
find x such that $||Ax - b||_2^2$ is minimised.
I know I have to do something along the line $A^TAx = A^Tb$
got the vector $(1/3)* [4\ 7\ 0] ^T$.
However the answer is
$x = (1/3)* [4\ 7\ 0] ^T + \lambda*[-1\ -1 \ \ 1]^T $. I have no clue where does the $\lambda*[-1\ -1 \ \ 1]^T$ come from. Really appreciate for some help.
| Since $A^{T}A=\begin{bmatrix}2&1&3\\1&2&3\\3&3&6\end{bmatrix}\;\;$ and $\;\;A^{T}b=\begin{bmatrix}5\\6\\11\end{bmatrix}$, solving $A^{T}Ax=A^{T}b$ gives
$\begin{bmatrix}2&1&3&5\\1&2&3&6\\3&3&6&11\end{bmatrix}\longrightarrow\begin{bmatrix}1&0&1&\frac{4}{3}\\0&1&1&\frac{7}{3}\\0&0&0&0\end{bmatrix}$
$\;\;\;$so $x=\frac{4}{3}-t, \;\;, y=\frac{7}{3}-t,\;\;z=t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Absolute Value of Complex Integral Let $[a,b]$ be a closed real interval. Let $f:[a,b] \to \mathbb{C}$ be a continuous complex-valued function. Then $$\bigg|\int_{a}^{b} f(t)dt \ \bigg| \leq \int_{a}^{b} \bigg|f(t)\bigg| dt,$$ where the first integral is a complex integral, and the second integral is a definite real integral.
There's a neat "rotational" proof of this in D'Angelo's An Introduction to Complex Analysis and Geometry.
Question: Can this fact also be proven using the Cauchy-Schwarz Inequality? If so, some help would be nice.
Thank you...
| We first state the Cauchy-Schwarz inequality for definite integrals:
Let $u$ and $v$ be real functions which are continuous on the closed interval $[a,b].$ Then
$$\left(\int_{a}^{b} u(t)v(t)dt \right) \leq \left(\int_{a}^{b} u(t)dt \right)^2 \cdot \left(\int_{a}^{b} v(t)dt \right)^2$$
Since $F$ is complex-valued, we can write $F(t) = u(t) + iv(t),$ with $u(t), v(t)$ real-valued. Thus $|F(t)|^2 = u^2+v^2.$ Also
\begin{align}\left(\int_{a}^{b} F(t)dt \right)^2 = \left(\int_{a}^{b} u(t) + iv(t) dt \right)^2.
\end{align}
Observe that
\begin{align*}
\left(\int_{a}^{b} u ~dt \right)^2 &= \left(\int_{a}^{b} \dfrac{u}{(u^2+v^2)^{1/4}}(u^2+v^2)^{1/4}dt \right)^2 \\
&\overset{\mathrm{C-S \ ineq.}}{\leq} \left(\int_{a}^{b} \dfrac{u}{(u^2+v^2)^{1/4}}dt \right)^2\left(\int_{a}^{b} \sqrt[4]{u^2+v^2} ~dt \right)^2 \\ &= \left(\int_{a}^{b} \dfrac{u^2}{\sqrt{u^2+v^2}}dt \right)\left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right)
\end{align*}
and
\begin{align*}
\left(\int_{a}^{b} v ~dt \right)^2 &= \left(\int_{a}^{b} \dfrac{v}{(u^2+v^2)^{1/4}}(u^2+v^2)^{1/4}dt \right)^2 \\
&\overset{\mathrm{C-S \ ineq.}}{\leq} \left(\int_{a}^{b} \dfrac{v}{(u^2+v^2)^{1/4}}dt \right)^2\left(\int_{a}^{b} \sqrt[4]{u^2+v^2} ~dt \right)^2 \\ &= \left(\int_{a}^{b} \dfrac{v^2}{\sqrt{u^2+v^2}}dt \right)\left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right).
\end{align*}
So
\begin{align*}
\bigg|\int_{a}^{b} F(t) ~dt \bigg|^2 &= \bigg|\int_{a}^{b} u(t) + iv(t) ~dt \bigg|^2 \\ &= \bigg|\int_{a}^{b} u(t) ~dt + i\int_{a}^{b}v(t) ~dt \bigg|^2 \\ &= \left(\int_{a}^{b} u ~dt \right)^2 + \left(\int_{a}^{b} v ~dt \right)^2 \\ &\leq \left(\int_{a}^{b} \dfrac{u^2}{\sqrt{u^2+v^2}}dt \right)\left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right) \\ &+ \left(\int_{a}^{b} \dfrac{v^2}{\sqrt{u^2+v^2}}dt \right)\left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right) \\ &= \left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right)\left(\int_{a}^{b} \dfrac{u^2+v^2}{\sqrt{u^2+v^2}}dt \right) \\ &= \left(\int_{a}^{b} \sqrt{u^2+v^2} ~dt \right)^2 \\ &=
\left(\int_{a}^{b} \bigg|F(t)\bigg|\right)^2 dt
\end{align*}
Now, recalling that the modulus is always non-negative we can take the square root of both sides and we arrive at the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Surface area generated by revolving $r = \sqrt {\cos 2\theta}$ I've been giving a good time trying to solve this problem, I do not find a clear way to solve appreciate your help.
\begin{array}{rcl}
r& =& \sqrt{\cos 2\theta }
\end{array}
This Around to axis y and the limits of integration is 0 to π/4
P.d :
I would appreciate if anyone could recommend me some text or page in which you could learn to plot in polar and learn more about these problems.
| Parametric Breakdown
Start by dividing the function from polar form into a set of two parametric functions. That is, the horizontal and vertical component of the radial coordinate $r$:
$$
x =r \cos{\theta}\qquad\quad y=r\sin{\theta}\\
\implies \dfrac{dx}{d\theta}=\dfrac{dr}{d\theta}\cos{\theta}-r\sin{\theta} \qquad\quad \dfrac{dy}{d\theta} = \dfrac{dr}{d\theta}\sin{\theta}+r\cos{\theta}\\
$$
However the components alone don't really help, we need to find a differential element $ds$ that represents the hypotenuse formed by the two differential components (basically Pythagorean Theorem) :
\begin{align}
(ds)^2 &= (\dfrac{dx}{d\theta})^2+(\dfrac{dy}{d\theta})^2\\
& = (\dfrac{dr}{d\theta}\cos{\theta}-r\sin{\theta})^2 + (\dfrac{dr}{d\theta}\sin{\theta}+r\cos{\theta})^2\\
& = \left[(\dfrac{dr}{d\theta})^2\cos^2\theta\color{red}{-(\dfrac{dr}{d\theta})2r\cos\theta\sin\theta} +r^2\sin^2\theta\right] \space\space\space + \left[(\dfrac{dr}{d\theta})^2\sin^2\theta \color{red}{+ (\dfrac{dr}{d\theta})2r\cos\theta\sin\theta} +r^2cos^2\theta \right]\\
& =(\dfrac{dr}{d\theta})^2(\cos^2\theta + \sin^2\theta) + r^2(\cos^2\theta + \sin^2\theta)\\
&=(\dfrac{dr}{d\theta})^2+r^2
\end{align}
Hence we conclude for this step that:
$$ds = \sqrt{(\dfrac{dr}{d\theta})^2+r^2} d\theta$$
Surface of Revolution Derivation
We should notice (through derivation using frustrums not shown here) that just as in moving from the arc length of a curve to its surface area in parametric coordinates requires multiplication by $2\pi * \left(x(t) \,\mathrm{or}\,y(t)\right)$, the same applies in polar coordinates:
$$ L = \int ds $$
Since we are rotating about the y-axis, the height of each frustrum would be the $x$ component of the polar equation:
$$A = 2\pi\int \operatorname{x}(\theta)\, ds\\
\boxed{A = 2\pi\int \operatorname{x}(\theta)\,\sqrt{(\dfrac{dr}{d\theta})^2+r^2} d\theta}
$$
If you wish to try it from here yourself, don't continue reading as I will propose the calculated solution!
Calculated Solution
Recollecting variables:
$r = \sqrt{\cos2\theta}\\
\operatorname{x}(\theta)= r\cos\theta = \cos\theta\sqrt{\cos2\theta}\\
\dfrac{dr}{d\theta}= \dfrac{-2\sin2\theta}{2\sqrt{\cos2\theta}} = \dfrac{-\sin2\theta}{\sqrt{\cos2\theta}}
$
Plugging in variables and simplifying:
\begin{align}
A &= 2\pi\int_0^{\pi/4} \operatorname{x}(\theta)\,\sqrt{(\dfrac{dr}{d\theta})^2+r^2}\, d\theta\\
& = 2\pi\int_0^{\pi/4}\cos\theta\sqrt{\cos2\theta}\,\sqrt{(\dfrac{-\sin2\theta}{\sqrt{\cos2\theta}})^2+(\sqrt{\cos2\theta})^2}\, d\theta\\
& = 2\pi\int_0^{\pi/4}\cos\theta\sqrt{\cos2\theta}\,\sqrt{\dfrac{\sin^22\theta}{\cos2\theta}+\cos2\theta}\, d\theta
\end{align}
Combining the square roots and using the fact that $\sin^2x+\cos^2x=1$:
\begin{align}
&= 2\pi\int_0^{\pi/4}\cos\theta\,\sqrt{\cos2\theta \left(\dfrac{\sin^22\theta}{\cos2\theta}+\cos2\theta\right)}\, d\theta\\
&= 2\pi\int_0^{\pi/4}\cos\theta\,\sqrt{\sin^22\theta + \cos^22\theta}\, d\theta\\
&= 2\pi\int_0^{\pi/4}\cos\theta\,\sqrt{(1)}\, d\theta\\
&= 2\pi\, \left.\sin\theta\right\rvert_0^{\pi/4}\\
&= 2\pi\, \left(\dfrac{\sqrt{2}}{2}\right)\\
& = \boxed{\pi\sqrt{2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a nullspace of a matrix. I am given the following matrix $A$ and I need to find a nullspace of this matrix.
$$A =
\begin{pmatrix}
2&1&4&-1 \\
1&1&1&1 \\
1&0&3&-2 \\
-3&-2&-5&0
\end{pmatrix}$$
I have found a row reduced form of this matrix, which is:
$$A' =
\begin{pmatrix}
1&0&3&-2 \\
0&1&-2&3 \\
0&0&0&0 \\
0&0&0&0
\end{pmatrix}$$
And then I used the formula $A'x=0$, which gave me:
$$A'x =
\begin{pmatrix}
1&0&3&-2 \\
0&1&-2&3 \\
0&0&0&0 \\
0&0&0&0
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{pmatrix}=
\begin{pmatrix}
0 \\
0 \\
0 \\
0
\end{pmatrix}$$
Hence I obtained the following system of linear equations:
$$\begin{cases} x_1+3x_3-2x_4=0 \\ x_2-2x_3+3x_4=0 \end{cases}$$
So I just said that $x_3=\alpha$, $x_4=\beta$ and the nullspace is:
$$nullspace(A)=\{2\beta-3\alpha,2\alpha-3\beta,\alpha,\beta) \ | \ \alpha,\beta \in \mathbb{R}\}$$
Is my thinking correct? Thank you guys!
| Everything is right. But after finding the equations \begin{cases} x_1+3x_3-2x_4=0 \\ x_2-2x_3+3x_4=0 \end{cases} This $x_1$ and $x_2$ are pivot variables and $x_3$ and $x_4$ are free variables. The number of non zero rows is the rank of the matrix, in our case $2$ and hence nullity of the matrix is $2$. since $dim(W) = Rank + Nullity$. Now since $nullity =2$, the usual basis is $\{(1,0),(0,1)\}$. Hence Substitute $x_3=1$ and $x_4=0$ in two equations, we will get $x_1=-3$ and $x_2 =2$ and we get a point $(-3,2,1,0)$ and again substitute $x_3=0$ and $x_4=1$ in two equations, we will get $x_1 = 2$ and $x_2 =-3$ and we will get another point $(2,-3,0,1)$. These two points $(-3,2,1,0)$ and $(2,-3,0,1)$ serves as a basis for the nullspace of the matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Inequality used in the proof of Kolmogorov Strong Law of Large Numbers I'm trying show that convergence follows.
$$\sum_{k \geq 1} \frac{\sigma^2_k}{k^2} < \infty \Rightarrow \lim_{M \rightarrow \infty}\frac{1}{M^2}\sum_{k \leq M} \sigma^2_k=0.$$
Let's consider $D_k = \sum_{n \geq k} \displaystyle\frac{\sigma_n^2}{n^2}$ para $k \geq 1$ and it is noted that for $k=1$ we have to:
$$D_1 = \sum_{n \geq 1} \frac{\sigma_n^2}{n^2} < \infty \text{(by hypothesis)}.$$
or this I have proved that $\lim_{k \rightarrow \infty} D_k = \lim_{k \rightarrow \infty} \sum_{n \geq k} \frac{\sigma_n^2}{n^2} = 0$,
Now note the following:
\begin{eqnarray*}
D_2 &=& \sum_{n \geq 2} \frac{\sigma_n^2}{n^2} = \frac{\sigma_2^2}{2^2} + \frac{\sigma_3^2}{3^2} + \frac{\sigma_4^2}{4^2} \ldots\nonumber \\
D_3 &=& \sum_{n \geq 3} \frac{\sigma_n^2}{n^2} = \frac{\sigma_3^2}{3^2} + \frac{\sigma_4^2}{4^2} \ldots \nonumber\\
D_4 &=& \sum_{n \geq 4} \frac{\sigma_n^2}{n^2} = \frac{\sigma_4^2}{4^2} \ldots \nonumber\\
&\vdots& \\
\lim_{k \rightarrow \infty} D_k &=& \sum_{n \geq k} \frac{\sigma_n^2}{n^2} = 0. \nonumber
\end{eqnarray*}
then I considered a $M$ such that $M \geq 1$ and I then come to the following equation
$$\frac{1}{M^2}\sum_{k=1}^M \sigma^2_k = \frac{1}{M^2} \sum_{k=1}^M k^2\left(D_k - D_{k+1}\right).$$
I noted in a book which obtained the following inequality
$$\sum_{n \geq k} \displaystyle\frac{\sigma^2_k}{k^2} = \frac{1}{M^2} \sum_{k=1}^M k^2 \left(D_k - D_{k+1}\right)\text{(How could justify this inequality?)} \leq \frac{1}{M^2} \sum_{k=1}^M (2k - 1)D_k$$
It is easy to see that
\begin{eqnarray}
\sum_{n \geq k} \displaystyle\frac{\sigma^2_k}{k^2} &=& D_k - D_{k+1}\nonumber \\
&=& \sum_{n \geq k} \displaystyle\frac{\sigma_n^2}{n^2} - \sum_{n \geq k+1} \displaystyle\frac{\sigma_n^2}{n^2} \nonumber \\
&=& \left\{\displaystyle\frac{\sigma_k^2}{k^2} + \displaystyle\frac{\sigma_{(k+1)}^2}{(k+1)^2} + \ldots \right\} - \left\{\displaystyle\frac{\sigma_{(k+1)}^2}{(k+1)^2} + \displaystyle\frac{\sigma_{(k+2)}^2}{(k+1)^2} + \ldots \right\}.
\end{eqnarray}
where
$$\lim_{k \rightarrow \infty} \frac{1}{M^2} \sum_{k=1}^M (2k - 1)D_k = 0$$
but I don´t know to justify that step. Could you please give me a suggestion on how to justify the last step.
Thank you very much, for you help.
| \begin{align*}
\frac{1}{M^2}\sum_{k \leq M} \sigma^2_k &\le \frac{1}{M^2} \sum_{k=1}^M (2k-1) D_k \\
&=\frac{1}{M^2} \left(D_1 + 3D_2 + 5D_3 + 7D_4 + \ldots \right)\\
&=\frac{1}{M^2} \left(\sum_{n \geq 1}\frac{\sigma^2_n}{n^2} + 3\sum_{n \geq 2}\frac{\sigma^2_n}{n^2} + 5\sum_{n \geq 3}\frac{\sigma^2_n}{n^2}+7\sum_{n \geq 4}\frac{\sigma^2_n}{n^2} + \ldots\right)\\
&=\frac{1}{M^2} \left\{\left(\frac{\sigma_1^2}{1^2} + \frac{\sigma_2^2}{2^2} + \ldots \right) + 3\left(\frac{\sigma_2^2}{2^2} + \frac{\sigma_3^2}{3^2} + \ldots \right) + 5\left(\frac{\sigma_3^2}{3^2} + \frac{\sigma_4^2}{4^2} + \ldots \right) + 7\left(\frac{\sigma_4^2}{4^2} + \ldots \right) + \ldots\right\}\\
&=\frac{1}{M^2} \left\{\frac{\sigma_1^2}{1^2} + \frac{4\sigma_2^2}{2^2} + \frac{9\sigma_3^2}{3^2} + \frac{16\sigma_4^2}{4^2} + \ldots \right\}\\
&= \frac{1}{M^2} \left(\sigma_1^2 + \sigma_2^2 + \sigma_3^2 + \sigma_4^2 + \ldots\right)\\
&= \frac{1}{M^2}\sigma_1^2 + \frac{1}{M^2}\sigma_2^2 + \frac{1}{M^2}\sigma_3^2 + \frac{1}{M^2}\sigma_4^2 + \ldots
\end{align*}
Applying limit
\begin{align*}
\lim_{M \rightarrow \infty} \frac{1}{M^2}\sum_{k=1}^M \sigma^2_k &\le \lim_{M \rightarrow \infty} \left(\frac{1}{M^2}\sigma_1^2 + \frac{1}{M^2}\sigma_2^2 + \frac{1}{M^2}\sigma_3^2 + \frac{1}{M^2}\sigma_4^2 + \ldots \right) \\
&= 0 + 0 + 0 + 0 + \ldots \\
&= 0.
\end{align*}
Therefore
$$\lim_{M \rightarrow \infty} \frac{1}{M^2}\sum_{k \leq M} \sigma^2_k = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solutions to $z^3 - z^2- z =15 $ Find in the form $a+bi$, all the solutions to the equation
$$z^3 - z^2- z =15 $$
I have no idea what to do - am I meant to factor out z to get $z(z^2-z-1)=15$ or should I plug in $a+bi$ to z?
Please help!!!!!!!
| Real solution:
$$z^3 - z^2- z =15<=> $$
$$z^3-z^2-z-15=0<=>$$
$$(z-3)(z^2+2z+5)=0<=>$$
So:
$$z-3=0<=>z=3$$
Or:
$$z^2+2z+5=0<=>z^2+2z=-5<=>z^2+2z+1=-4<=>(z+1)^2=-4$$
($(z+1)^2=-4$ hasno solution since for all $z$ on the real line, $(z+1)^2$ is bigger or even to zero and $-4$ is smaller than zero)
Complex solution:
$$z^3 - z^2- z =15<=> $$
$$z^3 - z^2- z -15=0<=>$$
$$(z-3)(z^2+2z+5)=0<=>$$
So:
$$z-3=0<=>z=3$$
Or:
$$z^2+2z+5=0<=>z^2+2z=-5<=>z^2+2z+1=-4<=>(z+1)^2=-4<=>$$
(the square root of both sides)
$$z+1=2i<=>z=-1+2i$$
Or:
$$z+1=-2i<=>z=-1-2i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Number of ways to throw at most 14 with 4 dice - generating functions Determine the chance to throw at most 14 with 4 normal dice. I will set up the right generating function to determine the number of ways tot thow at most 14 with 4 normal dice and I need some help. I know that:
\begin{align}
x_1 + x_2 + x_3 + x_4 \leq 14 \qquad \text{with} \qquad 1 \leq x_i \leq 6.
\end{align}
\begin{align}
x_1 + x_2 + x_3 + x_4 + x' = 14 \qquad \text{with} \qquad 0 \leq x' \leq 10
\end{align}
So we have:
\begin{align}(x + x^2 + x^3 + x^4 + x^5 + x^6)^4 \cdot (0 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10}).
\end{align}
Does this make sense? I do not have to determine the coefficient of $x^{14}$ only the right generating function.
| As mentioned in Karl's answer, one surefire way of doing it is by adding together all coefficients of $(x+x^2+x^3+x^4+x^5+x^6)$ on the terms of power less than or equal to $14$.
You have included in your proposed solution a very nice way to get around the difficulty of having to either personally sum or word to the computer how to sum all of those coefficients by multiplying another polynomial which takes care of the summing for you. There is but one problem: as $x'$ is allowed to be zero, you are missing a term, $x^0=1$, in the right term.
The correct generating function will be:
$$(x^1+x^2+x^3+x^4+x^5+x^6)^4(0+\color{red}{x^0}+x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10})$$
With the generating function, you can plug it into your favorite CAS and read the coefficient to the power in question. In this case wolfram reads the coefficient is $721$, agreeing with Karl's earlier arithmetic.
It is common as well to see this written instead as $(x^1+x^2+x^3+x^4+x^5+x^6)^4(0+1+x+x^2+\dots)$ where the right term is an infinite series. We don't actually mind that the series doesn't converge in combinatorics as we look at partial sums anyways. This allows us to reuse the same generating function for other sums as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve the following summation $S = \dfrac{n \choose 0}{1} + \dfrac{n \choose 1}{2} +
\dfrac{n \choose 2}{3}+\dotsb+\dfrac{n \choose n}{n+1}$
| $\bf{My\; Solution::}$ We can write $$\displaystyle S = \sum_{k=0}^{n}\binom{n}{k}\cdot \frac{1}{k+1} = \sum_{k=0}^{n}\frac{n!}{k!\cdot (n-k)!}\cdot \frac{1}{k+1}$$
$$\displaystyle S = \frac{1}{n+1}\sum_{k=0}^{n}\frac{(n+1)!}{(k+1)!\cdot (n-k)!}=\frac{1}{n+1}\sum_{k=0}^{n}\binom{n+1}{k+1} = \frac{2^{n+1}-1}{n+1}.$$
Above we have used the Identity $$\displaystyle \sum_{k=0}^{n}\binom{n}{k} = 2^n$$.
So for $$\displaystyle \sum_{k=0}^{n}\binom{n+1}{k+1} = 2^{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x\rightarrow0}\frac{1-\left(\cos x\right)^{\ln\left(x+1\right)}}{x^{4}}$ Could you please check if I derive the limit correctly?
$$\lim_{x\rightarrow0}\frac{1-\left(\cos x\right)^{\ln\left(x+1\right)}}{x^{4}}=\lim_{x\rightarrow0}\frac{1-\left(O\left(1\right)\right)^{\left(O\left(1\right)\right)}}{x^{4}}=\lim_{x\rightarrow0}\frac{0}{x^{4}}=0$$
| We can proceed as follows:
\begin{align}
L &= \lim_{x \to 0}\frac{1 - (\cos x)^{\log(1 + x)}}{x^{4}}\notag\\
&= \lim_{x \to 0}\frac{1 - \exp\{\log(1 + x)\log(\cos x)\}}{x^{4}}\notag\\
&= \lim_{x \to 0}\frac{1 - \exp\{\log(1 + x)\log(\cos x)\}}{\log(1 + x)\log(\cos x)}\cdot\frac{\log(1 + x)\log(\cos x)}{x^{4}}\notag\\
&= \lim_{y \to 0}\frac{1 - e^{y}}{y}\cdot\lim_{x \to 0}\frac{\log(1 + x)}{x}\cdot\frac{\log \cos x}{x^{3}}\notag\\
&= -1\cdot 1\lim_{x \to 0}\frac{\log(1 + \cos x - 1)}{x^{3}}\notag\\
&= -\lim_{x \to 0}\frac{\log(1 + \cos x - 1)}{\cos x - 1}\cdot\frac{\cos x - 1}{x^{3}}\notag\\
&= -\lim_{z \to 0}\frac{\log(1 + z)}{z}\cdot\lim_{x \to 0}\frac{\cos^{2}x - 1}{x^{3}(\cos x + 1)}\notag\\
&= -\lim_{x \to 0}\frac{1}{1 + \cos x}\cdot\frac{-\sin^{2}x}{x^{2}}\cdot\frac{1}{x}\notag\\
&= -\frac{1}{2}\cdot (-1)\cdot\lim_{x \to 0}\frac{1}{x}\notag\\
&= \infty
\end{align}
Here $y = \log(1 + x)\log(\cos x) \to 0$ as $x \to 0$ and $z = \cos x - 1 \to 0$ as $x \to 0$. We see that effective use of standard limits is sufficient in many seemingly tough limit problems and there is no need for higher level theorems like Taylor or L'Hospital Rule.
Also as mentioned in one of the comments (to the question) we can see that had the denominator been $x^{3}$ there would not have been the factor $1/x$ in the second last line of our derivation and the limit would have been $1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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