Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding the infinite Sum of a series: $\sum\frac1{n(n+1)(n+2)}$ Find the infinite Sum of the series with general term $\frac{1}{n(n+1)(n+2)}$.
I decomposed the fraction upto this $1/(2n)-1/(n+1)+1/(2n+4)$. But I find no link about cancelling terms. So how should I find the infinite Sum?
Please don't say that this is off topic or homework. if wanna say that this is homework, at least give a hint if I am going the right way?
| Notice, use partial fractions in form of difference of two terms as follows $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$$$=\sum_{n=1}^{\infty}\frac 12\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)$$
$$=\small \frac 12\lim_{n\to \infty}\left(\left(\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}\right)+\left(\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}\right)+\left(\frac{1}{3\cdot 4}-\frac{1}{4\cdot 5}\right)+\ldots+\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\right)$$
$$=\frac 12\lim_{n\to \infty}\left(\frac{1}{1\cdot 2}-\frac{1}{(n+1)(n+2)}\right)$$
$$=\frac 12\left(\frac{1}{2}-0\right)=\color{red}{\frac 14}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the sum $\sum_{k=1}^\infty \big(\frac{1}{36k^2-1}+\frac{2}{(36k^2-1)^2}\big)$ I have to evaluate the series:
$$\sum_{k=1}^\infty \left(\frac{1}{36k^2-1}+\frac{2}{(36k^2-1)^2}\right).$$
I tried using the identity $\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$ but I got stuck.
| A more general approach, proving the identity:
$$ \sum_{k\geq 1}\frac{k^2+x^2}{(k^2-x^2)^2}=-\frac{1}{2x^2}+\frac{\pi^2}{2\sin^2(\pi x)}\tag{1}$$
for any $x\in(-1,1)$ is to check that both the LHS and the RHS of $(1)$ are meromorphic functions with the same-behaving singularities at any $x\in\mathbb{Z}\setminus\{0\}$. For instance, $(1)$ is just a consequence of:
$$ \sum_{n\in\mathbb{Z}}\frac{1}{(z-n)^2}=\frac{\pi^2}{\sin^2(\pi z)}\tag{2} $$
since $\frac{1}{(k-x)^2}+\frac{1}{(k+x)^2} = 2\cdot\frac{k^2+x^2}{(k^2-x^2)^2}$. $(2)$ can be seen as a consequence of the reflection formula for the $\Gamma$ function, too, since:
$$ \sum_{n\in\mathbb{Z}}\frac{1}{(z-n)^2}= \psi'(-z)+\psi'(1+z). \tag{3}$$
If we plug $x=\frac{1}{6}$ in $(1)$, we get:
$$ \sum_{k\geq 1}\frac{36k^2+1}{(36k^2-1)^2} = \color{red}{\frac{\pi^2-9}{18}}\tag{4} $$
and it is possible to derive such identity also through Fourier-analytic methods.
Still another way is to notice that our series is just $\frac{1}{2}\cdot L(\chi,2)$ where $\chi$ is the principal Dirichlet character $\!\!\pmod{6}$, hence our series depends on $\text{Li}_2$ evaluated at the sixth roots of unity.
An elementary proof of $(4)$ comes from the inclusion-exclusion principle:
$$ \sum_{k\geq 0}\left(\frac{1}{(6k+1)^2}+\frac{1}{(6k+5)^2}\right) = \sum_{\substack{n\geq 1 \\ 2\nmid n \\ 3\nmid n}}\frac{1}{n^2} = \sum_{n\geq 1}\frac{1}{n^2}-\sum_{\substack{n\geq 1 \\ 2\mid n}}\frac{1}{n^2}-\sum_{\substack{n\geq 1 \\ 3\mid n}}\frac{1}{n^2}+\sum_{\substack{n\geq 1 \\ 6\mid n}}\frac{1}{n^2}$$
gives:
$$ \sum_{k\geq 0}\left(\frac{1}{(6k+1)^2}+\frac{1}{(6k+5)^2}\right) = \left(1-\frac{1}{4}-\frac{1}{9}+\frac{1}{36}\right) = \frac{2}{3}\cdot\zeta(2) = \color{red}{\frac{\pi^2}{9}}. \tag{5}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the solutions of the diophantine equation $(x^2-y^2)(z^2-w^2)=2xyzw$ Let $x,y,z,w$ be postive integers. Find all solutions of:
$$(x^2-y^2)(z^2-w^2)=2xyzw$$
This gives:
$$\left(\dfrac{x}{y}-\dfrac{y}{x}\right)\left(\dfrac{z}{w}-\dfrac{w}{z}\right)=2$$
$$\left(p-\dfrac{1}{p}\right)\left(q-\dfrac{1}{q}\right)=2$$
| There are NO solutions in rationals or integers.
The equation
\begin{equation*}
\left( p-\frac{1}{p} \right) \left( q-\frac{1}{q} \right) =2
\end{equation*}
is equivalent to the quadratic
\begin{equation*}
p^2 + \frac{2q}{1-q^2} p - 1=0
\end{equation*}
For this to have rational solutions the discriminant must be a rational square, so there must exist $D \in \mathbb{Q}$ such that
\begin{equation*}
D^2=q^4-q^2+1
\end{equation*}
The quartic has a rational point when $q=0$, so is birationally equivalent to an elliptic curve. The curve is
\begin{equation*}
V^2=U^3-7U^2+12U=U(U-3)(U-4)
\end{equation*}
with
\begin{equation*}
q=\frac{V}{2U-6}
\end{equation*}
Pari-gp gives $7$ finite torsion points, $(0,0)$, $(3,0)$, $(4,0)$, $(2, \pm 2)$ and $(6,\pm 6)$. Denis Simon's ellrank package give rank $0$, so the torsion points are the only rational points.
The torsion points give $q=0$, $q=\pm 1$ or $q$ undefined, all of which are unacceptable values for a positive integer solution.
| {
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Find an equation of the tangent line to $y = \cos(x)+3\sin(x)$ at $x=\pi/3$
Find an equation of the tangent line to
$$y = \cos(x)+3\sin(x)$$ at $x=\pi/3$.
This is what I have done...
Find $y$, $y= \cos(\pi/3) + 3\sin(\pi/3)$
this equals $1 + \sqrt 3/2$
Next
Find $f'(x) = \sin(\pi/3) + 3\cos(\pi/3)$
this equals $3+ 3\sqrt3/2$
Next
Plug into point slope form
$(y-1+\sqrt 3/2) = (3 +3\sqrt3/2) (x - \pi/3)$
$y = \left(\frac{3+3\sqrt 3}{2}\right)(x-\pi/3)+(1+3\sqrt 3/2)$
Am I doing something wrong? Thanks.
| While your strategy is correct, you incorrectly evaluated the sine and cosine functions at $x = \pi/3$ and incorrectly took the derivative.
Let $f(x) = \cos x + 3\sin x$. Since
$$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$
and
$$\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$
when we evaluate $f(x)$ at $\pi/3$ we obtain
$$f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) + 3\sin\left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{3\sqrt{3}}{2} = \frac{1 + 3\sqrt{3}}{2}$$
Since the derivative of $g(x) = \cos x$ is $g'(x) = -\sin x$ and the derivative of $h(x) = \sin x$ is $h'(x) = \cos x$, the derivative of $f(x)$ is
$$f'(x) = -\sin x + 3\cos x$$
Thus, the derivative of $f(x)$ evaluated at $\pi/3$ is
$$f'\left(\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) + 3\cos\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} + \frac{3}{2} = \frac{3 - \sqrt{3}}{2}$$
Hence, the equation of the tangent line is
$$y - \frac{1 + 3\sqrt{3}}{2} = \frac{3 - \sqrt{3}}{2}\left(x - \frac{\pi}{3}\right)$$
| {
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Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$.
The term inside the bracket is divisible by 6 since we have assumed that the result is true when $n=k$. If we can show that $3k^2+15k+12$ is also divisible by 6, then we are done. But how to proceed?
| Continuing from your start, let $k^3+6k^2+5k=6p$ for some $p\in\mathbb Z$.
Now,
$$(k+1)(k+2)(k+6)=(k^3+6k^2+5k)+3k^2+15k+12=6p+3k^2+15k+12=6p+3(k^2+5k+4)=6p+3(k+1)(k+4)$$
Case 1: $k$ is even.
Then, $k+4$ is also even. We can thus write $k+4=2q$ for some $q\in\mathbb Z$.
Hence,
$$6p+3(k+1)(k+4)=6p+3(k+1)(2q)=6p+6q(k+1)=6(p+q(k+1))$$
where $(p+q(k+1))\in\mathbb Z$.
Thus, the statement is true when $k$ is even.
Case 2: $k$ is odd.
Then, $k+1$ is even. We can thus write $k+1=2q$ for some $q\in\mathbb Z$.
Hence,
$$6p+3(k+1)(k+4)=6p+3(k+4)(2q)=6p+6q(k+4)=6(p+q(k+4))$$
where $(p+q(k+4))\in\mathbb Z$.
Thus, the statement is true when $k$ is odd.
| {
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The number and amount of dividers - a power of two For a positive integer $n$ is known that the sum of all divisors of that number is a power of $2$. Prove that the number of these divisors is also a power of $2$.
My work so far:
Several of these numbers I found
Let $\tau(n) -$ the number of divisors of $n$
1) $n=3; 1+3=4=2^2$ and $\tau (3)=2$
2) $n=7; 1+7=8=2^3$ and $\tau (7)=2$
From $1$ to $20$ such numbers do not have.
3) $n=21=3 \cdot 7; \tau(21)=4 (1,3,7,21)$ and $1+3+7+21=32=2^5$
4) $n=31; 1+31=32=2^5$ and $\tau(31)=2$
| The first two divisor functions of a number with prime factorization $n=\prod_ip_i^{a_i}$ are
$$
\sigma_0(n)=\prod_i(a_i+1)
$$
and
$$
\sigma_1(n)=\prod_i\left(1+p_i+\cdots+p_i^{a_i}\right)\;.
$$
Since $\sigma_0$ is clearly a power of $2$ if and only if the $a_i+1$ are, we need to show that $a_i+1$ is a power of $2$ if $1+p_i+\cdots+p_i^{a_i}$ is a power of $2$. Now if $p_i=2$, then $1+p_i+\cdots+p_i^{a_i}=2^{a_i+1}-1$, which is not a power of $2$, so we can focus on odd primes. For odd $p_i$, if $1+p_i+\cdots+p_i^{a_i}$ is a power (and hence a multiple) of $2$, then $a_i+1$ is even, and $1+p_i+\cdots+p_i^{a_i}=(1+p_i)(1+p_i^2+\cdots+p_i^{a_i-1})$. For this to be a power of $2$, both factors must be. But then we can apply the same reasoning to the factor $1+p_i^2+\cdots+p_i^{a_i-1}$ and factor out $1+p_i^2$. We an continue to factorize the entire sum like this, and it follows that $a_i+1$, the number of summands, is a power of $2$, as required.
| {
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Solving differential equation - how to find inhomogeneous solution Given the following
$\ y'= \frac{3y^2-x^2}{2xy}$
I need to tell if the equation is linear, which I think it is because:
$\ y'= \frac{3y^2-x^2}{2xy} = \frac{3y}{2x}-\frac{x}{2y}$
Now I need to solve the equation with separation of variables which is only possible with substitution. So I substitute with $\ u(x)= \frac{y}{x}$ and do the following
$\ y'= \frac{3y}{2x}-\frac{x}{2y} => y'=\frac{3y}{2x}-\frac{1}{2u(x)} = \frac{3}{2x}y-\frac{1}{2u(x)}$
so I can use separation of variables and get the homogeneous solution and the inhomogeneous. For the homogeneous I got
$\ y_h=e^cx^{3/2} = Cx^{3/2} $
but I really don't know how to get the inhomogeneous solution because I don't exactly know what to do with the substitution, I am glad for help.
| $$y'(x)=\frac{3y(x)^2-x^2}{2xy(x)}\Longleftrightarrow$$
$$y'(x)-\frac{3y(x)}{2x}=-\frac{x}{2y(x)}\Longleftrightarrow$$
$$2y(x)y'(x)-\frac{3y(x)^2}{2x}=-x\Longleftrightarrow$$
Let $r(x)=y(x)^2$, which gives $r'(x)=2y(x)y'(x)$:
$$r'(x)-\frac{3r(x)}{x}=-x\Longleftrightarrow$$
Let $v(x)=\exp\left[\int-\frac{3}{x}\space\text{d}x\right]=\frac{1}{x^3}$:
$$\frac{r'(x)}{x^3}-\frac{3r(x)}{x^4}=-\frac{1}{x^2}\Longleftrightarrow$$
Substitute $-\frac{3}{x^4}=\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^3}\right)$:
$$\frac{r'(x)}{x^3}-\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^3}\right)r(x)=-\frac{1}{x^2}\Longleftrightarrow$$
Apply the reverse product rule:
$$\frac{\text{d}}{\text{d}x}\left(\frac{r(x)}{x^3}\right)=-\frac{1}{x^2}\Longleftrightarrow$$
$$\int\frac{\text{d}}{\text{d}x}\left(\frac{r(x)}{x^3}\right)\space\text{d}x=\int-\frac{1}{x^2}\space\text{d}x\Longleftrightarrow$$
$$\frac{r(x)}{x^3}=\frac{1}{x}+\text{C}\Longleftrightarrow$$
$$r(x)=x^2\left(\text{C}x+1\right)\Longleftrightarrow$$
$$y(x)^2=x^2\left(\text{C}x+1\right)\Longleftrightarrow$$
$$y(x)=\pm\sqrt{x^2\left(\text{C}x+1\right)}\Longleftrightarrow$$
$$y(x)=\pm x\sqrt{\text{C}x+1}$$
| {
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Line $mx + ny = 3$ is normal to the hyperbola $x^2 – y^2 = 1$ If the line $mx + ny = 3$ is normal to the hyperbola $x^2 – y^2 = 1$, then evaluate $\frac{1}{m^2}+\frac{1}{n^2}$.
I compared given equation of normal to equation of normal at parametric point i.e
$x\sin\theta+y=2\tan\theta$ and obtained $\frac{1}{m^2}-\frac{1}{n^2}=\frac{4}{9}$ but can't use it to obtain $\frac{1}{m^2}+\frac{1}{n^2}$. Any suggestion?
| There must be a flaw in the text, expression $\frac{1}{m^2}+\frac{1}{n^2}$, should have been $\frac{1}{m^2}-\frac{1}{n^2}$.
Let us see why:
First of all, I confirm the expression given by @Mathematician :
$$\frac{1}{m^2}-\frac{1}{n^2}=\frac{4}{9} \ \ \ (1)$$
Proof below.
In fact, as the question is formulated, the only reasonable answer should have been "$\frac{1}{m^2}+\frac{1}{n^2}$ is this constant". But due to (1),
$$\frac{1}{m^2}+\frac{1}{n^2}=\frac{4}{9}+\frac{2}{n^2}$$
which is a variable quantity.
Proof of (1): The equation of the tangent line $(T)$ at $(x_0,y_0)$ of the hyperbola $x^2-y^2=1$ is $xx_0-yy_0=1 \ \ (2)$.
Straight line $(D_{mn})$ with equation $mx+ny=1$ must fillful two conditions:
*
*$(D_{mn})$ and $(T)$ should be orthogonal, i.e., their normal vectors $(x_0,-y_0)$ and $(m,n)$ are orthogonal, i.e., $mx_0-ny_0=0 \ \ (3)$
*$(D_{mn})$ must pass through point $(x_0,y_0)$, i.e., $mx_0+ny_0=3 \ \ (4)$
Solving the system constituted by (3) and (4) for $x_0$ and $y_0$, we get:
$$x_0=\frac{3}{2m} \ \ and \ \ y_0=\frac{3}{2n} $$
Finally using constraint $x_0^2-y_0^2=1$, we obtain condition (1).
| {
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Solve integral $ \int \frac{dx}{( \sin x + 2 \cos x )^3}$ Alright, so this one has been giving me nightmares for a couple of days now... I've tried the substitutes for $u = \tan \frac{x}{2}$ where you can substitute
sin(x), cos(x) and dx accordingly and pretty much every possibility and combination that came to mind with them and still nothing. I mostly get stuck with a polynomial of the forth or sixth degree and can't really do anything from there. I've also looked up some integral calculator programs which gave me humongous solutions without steps of how to the program got there. Any hint appreciated.
| As what randomgirl did by auxiliary angle, I am going to complete the solution for reference.
Letting $\alpha = \arctan (\frac{1}{2})$ changes the denominator into $$
\sin x+2 \cos x=\sqrt{5} \cos (x-\alpha)
$$
which convert the integral into
$$
\begin{aligned}
I &=\int \frac{d x}{(\sin x+2 \cos x)^{3}} \\
&=\int \frac{d x}{\sqrt{5}^{3} \cos ^{3}(x-\alpha)} \\
&=\frac{1}{5 \sqrt{5}} \int \sec ^{3}(x-\alpha) d x
\end{aligned}
$$
Using integration by parts yields
$$
\begin{aligned}
\int \sec ^{3} y d y &=\int \sec y d(\tan y) \\
&=\sec y \tan y-\int \tan ^{2} y \sec y d y \\
&=\sec y \tan y-\int\left(\sec ^{2} y-1\right) \sec y d y \\
\int \sec ^{3} y d y &=\frac{1}{2}(\sec y \tan y+\ln |\sec y+\tan y|)+C
\end{aligned}
$$
Plugging back gives $$
\begin{aligned}
I &=\frac{1}{10 \sqrt{5}}(\sec (x-\alpha) \tan (x-\alpha)+\ln (\sec (x-\alpha)+\tan (x-\alpha))+C\\
&=\frac{1}{10 \sqrt{5}}\left(\frac{\sin (x-\alpha)}{\cos ^{2}(x-\alpha)}-\ln \left|\frac{1+\sin (x-\alpha)}{\cos (x-\alpha)}\right|\right)+C
\end{aligned}
$$
For simplification, using
$$
\sin (x-\alpha) =\sin x \cos \alpha-\sin \alpha \cos x =\frac{1}{\sqrt{5}}(2 \sin x-\cos x)
$$
and $$\sin x+2 \cos x=\sqrt{5} \cos (x-\alpha),$$
we can conclude that
$$
\begin{array}{rl}
I & =\frac{1}{10 \sqrt{5}}\left[\frac{\frac{1}{\sqrt{5}}(2 \sin x-\cos x)}{\frac{1}{5}(\sin x+2 \cos x)^{2}}-\ln \left|\frac{\frac{1}{\sqrt{5}}(2 \sin x-\cos x)}{\left.\frac{1}{\sqrt{5}} \sin x+2 \cos x\right)}\right|\right]+C \\
& =\frac{1}{50}\left[\frac{2 \sin x-\cos x}{(\sin x+2 \cos x)^{2}}-2 \sqrt{5} \ln \left|\frac{2 \sin x-\cos x}{\sin x+2 \cos x}\right|\right]+C
\end{array}
$$
:|D Wish you enjoy the solution!
| {
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How to solve $p^n+12^2=m^2$ Find all triples $(m,n,p) \in \mathbb{N}^3$, with $p$ prime, which satisfy $$p^n+12^2=m^2$$
| I have rewritten it as
$$ p^n = (m+12)(m-12) $$
Then there exists $\ 0 \le a \le n\ $ such that $\ p^a=m-12$. It follows that
$ p^n = p^a (p^a+24) $ and then $$ p^a(p^{n-2a}-1)= 24 = 2^3 \cdot 3 $$
We have the following cases:
*
*$\; p^a=1$ and then $a=0$ and $p^n=25$, which yields $p=5$, $n=2$ and $m=13$
*$\; p^a=2^3$ and then $p=2$, $a=3$, $n=8$ and $m=20$
*$\; p^a=3$ and then $p=3$, $a=1$, $n=4$ and $m=15$
So, all the solutions are $(13,2,5)$, $(20,8,2)$ and $(15,4,3)$.
| {
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Taylor series of $\ln{\sqrt[4]{\frac{x-2}{5-x}}}$ to $o((x-x_0)^n)$ when $x_0 = 3$ Well I have tried to get it as
$$f(x) = f(x_0) + \frac{f'(x_0)(x-x_0)}{1!} + \frac{f''(x_0)(x-x_0)^2}{2!} + ... + o((x-x_0)^n)$$
and got wrong results:
First: $$f'(x) = \frac{3}{4(x-2)(5-x)}$$
Second: $$f''(x) = \frac{3(2x-7)}{4(x-2)^2(5-x)^2} , $$
Third: $$f'''(x) = \frac{3}{4}*\frac{2(x-2)(5-x) - 2(5-x)(2x-7) + 2(x-2)(2x-7)}{(x-2)^3(5-x)^3}$$
Then I have tried changing it to $$\ln{\sqrt[4]{\frac{x-2}{5-x}}} = \frac{ln{(x-2)}}{4} - \frac{ln{(5-x)}}{4}$$ and got nothing useful. So I need an advice how to start. Or at least a hint.
| First of all, set $x-3=t$ so the computations will possibly be easier; then note that $x=t+3$ and that
$$
\ln{\sqrt[4]{\frac{x-2}{5-x}}}=
\frac{1}{4}\ln(1+t)-\frac{1}{4}\ln(2-t)=
\frac{1}{4}\ln(1+t)-\frac{\ln{2}}{4}-
\frac{1}{4}\ln\left(1-\frac{t}{2}\right)
$$
Now just use the series expansion of $\ln(1+z)$ around $z=0$.
| {
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Find the laurent series for $\frac{1}{z(z-2)^2}$ centered at z=2 and specify the region in which it converges. My attempt:
$$\frac{1}{z(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{A}{z}+\frac{B}{z-2}+\frac{C}{(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{(1/4)}{z}+\frac{(-1/4)}{z-2}+\frac{(1/2)}{(z-2)^2}$$
This is where I get stuck. The general idea I know is to get each of the 3 terms above into the form $D\cdot \frac{1}{1-z}$, where $D$ is a constant. But for the first term, it would be $\frac{1}{4}\cdot\frac{1}{(0-z)}$. I can't get rid of the zero.
| HINT:
Write $\frac1z$ as
$$\frac1z=\frac{1/2}{1+(z-2)/2}$$
Then, recall the sum of a geometric series.
SPOILER ALERT: Scroll over the highlighted area to reveal the solution
$$\begin{align}\frac1z&=\frac{1}{2+(z-2)}\\\\&=\frac{1/2}{1+(z-2)/2}\\\\&=\frac12 \sum_{n=0}^\infty (-1)^n\left(\frac{z-2}{2}\right)^n\end{align}$$for $|z-2|<2$. Therefore, the Laurent series is given by $$\frac12 (z-2)^{-2}-\frac14 (z-2)^{-1}+\frac18 \sum_{n=0}^\infty (-1)^n\left(\frac{z-2}{2}\right)^n=\frac18\sum_{n=-2}^\infty (-1)^n\left(\frac{z-2}{2}\right)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the fast way to evaluate the following integral: $\int{\frac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$? I am trying to evaluate the following integral:
$$\int{\dfrac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$$
I tried the trigonometric substitution: $u = \tan(x)$. Generally, The whole integral needs two substitutions: $u = \tan(x)$ then $v = \sin(u)$. In order to get rid of trigonometric functions, one needs to know that: $$\sin(\arctan(x))=\dfrac{x}{\sqrt{x^2+1}}$$
My question is: What is the fast substitution that leads to the answer without passing by the above steps?
| For clarifying the sign of the antiderivative, I decide to integrate it by cases:
A. When $x>0,$
$$
\begin{aligned}
I &=\int \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} d x \\
&=-\frac{1}{2} \int \sqrt{1+\frac{1}{x^{2}}} d\left(1+\frac{1}{x^{2}}\right) \\
&=-\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{\frac{3}{2}}+C\\&=-\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3 x^{3}}+C
\end{aligned}
$$
B. When $x<0,$ let $y=-x$, then $$
I=\int \frac{\sqrt{y^{2}+1}}{y^{4}}(-d y)=-\int \frac{\sqrt{y^{2}+1}}{y^{4}} d y=\frac{\left(y^{2}+1\right)^{2}}{3 y^{3}}+C =-\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3 x^{3}}+C
$$
Therefore we can conclude that
$$I=-\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3 x^{3}}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the first few Legendre polynomials without using Rodrigues' formula If a polynomial is given by $$y=\color{red}{a_0\left[1-\frac{l(l+1)}{2!}x^2+\frac{l(l+1)(l-2)(l+3)}{4!}x^4-\cdots\right]}+\color{blue}{a_1\left[x-\frac{(l-1)(l+2)}{3!}x^3+\frac{(l-1)(l+2)(l-3)(l+4)}{5!}x^5-\cdots\right]}\tag{1}$$ where $l$ is a constant and $a_0,a_1$ are coefficients.
The recurrence relation is given by $$a_{n+2}=-\frac{(l-n)(l+n+1)}{(n+2)(n+1)}a_n\tag{2}$$ The objective is to find the first few Legendre polynomials $P_l(x)$ such that $P_l(1)=1$ without using Rodrigues' formula:
$$\fbox{$P_l(x)=\frac{1}{2^{l}l!}\frac{\mathrm{d}^l}{\mathrm{d}x^l}{\left(x^2-1\right)}^l$}$$
The method given in my textbook states that:
If the value of $a_0$ or $a_1$ in each polynomial is selected so
that $y = 1$ when $x = 1$, the resulting polynomials are called Legendre Polynomials, written $P_l(x)$. From $(1)$ and $(2)$ and the requirement $P_l(1) = 1$, we find the following expressions for the first few Legendre polynomials:
$\color{#180}{\quad P_0(x)=1,\quad P_1(x)=x}\quad \text{and}\quad \color{#180}{P_2(x)=\frac12(3x^2-1)}$
I don't understand how those polynomials marked $\color{#180}{\mathrm{green}}$ were obtained as I've only just started reading about Legendre polynomials and hence I'm not sure how to tackle this problem. But since it is mandatory that OP's show their efforts for questions of this nature
Here is my attempt anyway:
I substituted $l=0$ in the $\color{red}{\mathrm{red}}$ bracket to obtain $1=a_0(1)$ so $a_0=1$ and hence $P_0(x)=1$.
I substituted $l=1$ in the $\color{blue}{\mathrm{blue}}$ bracket to obtain $1=a_1(x)$ so $a_1=x$ and hence $P_1(x)=x$.
I substituted $l=2$ in the $\color{red}{\mathrm{red}}$ bracket to obtain $1=a_0\left[1-\dfrac{2(2+1)}{2!}x^2\right]=a_0\left[1-3x^2\right]$ so $a_0=\dfrac{1}{1-3x^2}\ne \frac12(3x^2-1)$.
Obviously I am doing something wrong. Can anyone please explain to me how to achieve the answers properly (and without using Rodrigues' formula)?
|
We consider for even $l$ the polynomial (OPs red part)
\begin{align*}
a_0^{(l)}\left[1-\frac{l(l+1)}{2!}x^2+\frac{l(l+1)(l-2)(l+3)}{4!}x^4-\cdots\right]\tag{1}
\end{align*}
and for odd $l$ the polynomial (OPs blue part)
\begin{align*}
a_1^{(l)}\left[x-\frac{(l-1)(l+2)}{3!}x^3+\frac{(l-1)(l+2)(l-3)(l+4)}{5!}x^5-\cdots\right]\tag{2}
\end{align*}
Note, we use for convenience an upper index $(l)$ to indicate to which polynomial the index $a_0$ resp. $a_1$ belongs. (... and I suppose this helps to clarify the things ...)
Setting $l=0$ in (1) we obtain
\begin{align*}
y_0(x)=a_0^{(0)}
\end{align*}
We observe that all terms containing $x^2$ or higher powers of $x$ vanish, since they all contain the factor $l$.
$$ $$
Setting $l=1$ in (2) we obtain
\begin{align*}
y_1(x)=a_1^{(1)}x
\end{align*}
We observe that all terms containing $x^3$ or higher powers of $x$ vanish, since they all contain the factor $l-1$.
$$ $$
Similarly to the first case we obtain with $l=2$:
\begin{align*}
y_2(x)=a_0^{(2)}\left[1-\frac{2\cdot 3}{2!}x^2\right]=a_0^{(2)}\left(1-3x^2\right)
\end{align*}
since all other terms in (1) contain the factor $l-2$.
We also know that $y_l(1)=1$ for all polynomials $y_l(x)$. We obtain
\begin{align*}
y_0(1)&=a_0^{(0)}=1\\
y_1(1)&=a_1^{(1)}=1\\
y_2(1)&=-2a_0^{(2)}=1
\end{align*}
We conclude $a_0^{(2)}=-\frac{1}{2}$ and obtain finally
\begin{align*}
y_0(x)&=a_0^{(0)}=1\\
y_1(x)&=a_1^{(1)}x=x\\
y_2(x)&=a_0^{(2)}\left(1-3x^2\right)=\frac{1}{2}\left(3x^2-1\right)
\end{align*}
[2016-03-22]: Update according to OPs comment: Why can we restrict the consideration to $a_0^{(l)}$ when $l$ is even and restrict the consideration to $a_1^{(l)}$ when $l$ is odd?
The short answer is: Since the other series diverges when considering the boundary condition at $x=1$ we can set $a_1^{(l)}=0$ when $l$ is even and we can set $a_0^{(l)}=1$ when $l$ is odd.
Some details: We start with the Legendre differential equation
\begin{align*}
(1-x^2)y^{\prime\prime}-2xy^{\prime}+l(l+1)y=0\tag{3}
\end{align*}
and we want to find polynomials as solution which additionally fulfill the boundary condition
\begin{align*}
y(1)=1
\end{align*}
We make an Ansatz with generating functions
\begin{align*}
y(x)=a_0+a_1x+a_2x^2+\cdots
\end{align*}
With the help of the differential equation (3) and via comparing coefficients we obtain
\begin{align*}
y(x)&=a_0^{(l)}\left[1-\frac{l(l+1)}{2!}x^2+\frac{l(l+1)(l-2)(l+3)}{4!}x^4-\cdots\right]\\
&+a_1^{(l)}\left[x-\frac{(l-1)(l+2)}{3!}x^3+\frac{(l-1)(l+2)(l-3)(l+4)}{5!}x^5-\cdots\right]
\end{align*}
with $a_0^{l}$ and $a_1^{l}$ being two degrees of freedom.
We next apply the boundary condition $y(1)=1$.
First case: $l=2k$ even. In this case $y$ has the shape
\begin{align*}
y(x)&=a_0^{(0)}\left[1-\frac{l(l+1)}{2!}x^2-\cdots
\pm\frac{l(l+1)\cdots(l-2k+2)(l+2k-1)}{(l-2k)!}x^{l-2k}\right]\\
&+a_1^{(l)}\left[x-\frac{(l-1)(l+2)}{3!}x^3+\frac{(l-1)(l+2)(l-3)(l+4)}{5!}x^5-\cdots\right]
\end{align*}
We see the $a_0^{(l)}$ series contains only finitely many terms, since all terms having factor $l-2k$ vanish. If we now consider $y(1)=1$, it can be shown that the other $a_1^{(l)}$ series diverges for $x=1$. To overcome this, we set $a_1^{(l)}=0$.
The second case is symmetrically. Here we correspondingly set $a_0^{(l)}=0$ since the $a_0^{(l)}$ series diverges, while the $a_1{(l)}$ series is a polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Inequality using Cauchy-Schwarz Let $a,b,c\in\mathbb{R}^+$, prove that $$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\leq \sqrt{\frac{3}{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)} $$
Hi everyone, I've been trying to do this exercise but any method that I tried has failed. First I tried to use $v=(\sqrt{\frac{a}{b+c}},\sqrt{\frac{b}{c+a}},\sqrt{\frac{c}{a+b}})$ and $w=(1,1,1)$ (the most obvious) but didn't work. Then I did this $$LHS= \frac{\sqrt{a(c+a)(a+b)}+\sqrt{b(b+c)(a+b)}+\sqrt{c(b+c)(c+a)}}{\sqrt{(b+c)(c+a)(a+b)}}\leq \frac{\sqrt{(a(a+b)+b(b+c)+c(c+a))(c+a+a+b+b+c)}}{\sqrt{(b+c)(c+a)(a+b)}} = \frac{\sqrt{2((a+b+c)^2-(ab+bc+ac))(a+b+c)}}{\sqrt{(b+c)(c+a)(a+b)}}$$
Using the fact that $\sqrt{ax}+\sqrt{by}+\sqrt{cz} \leq \sqrt{(a+b+c)(x+y+z)}$ but I don't know if I'm not looking something. If someone can give me a hint I would really appreciate it
| I finally understood what tong_nor meant so I want to write it in details.
So treat this answer as a more detailed version of the tong_nor answer.
The Cauchy-Schwarz inequality gives us that:
(1) $(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2) \geq (a_1b_1 + a_2b_2 + a_3b_3)^2$
If we put here $a_i = \sqrt {a_i}$, $b_i = \sqrt {b_i}$ we get:
(2) $(a_1 + a_2 + a_3)(b_1 + b_2 + b_3) \geq (\sqrt {a_1b_1} + \sqrt{a_2b_2} + \sqrt{a_3b_3})^2$
Now in (2) we put:
$a_1 = a$, $a_2 = b$, $a_3 = c$
$b_1 = \frac{1}{b+c}$, $b_2 = \frac{1}{c+a}$, $b_3 = \frac{1}{a+b}$
So we get:
(3) $(a+b+c) (\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}) \geq (\sqrt {\frac{a}{b+c}} + \sqrt {\frac{b}{c+a}} + \sqrt {\frac{c}{a+b}})^2$
We take square root in (3) and reverse the sides:
(4) $ \sqrt {\frac{a}{b+c}} + \sqrt {\frac{b}{c+a}} + \sqrt {\frac{c}{a+b}} \leq \sqrt {(a+b+c) (\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}) } $
Now back to (2) where we put $a_i = \frac{x_i^2}{y_i}$ and $b_i = y_i$
So we get:
(5) $(\frac{x_1^2}{y_1} + \frac{x_2^2}{y_2} + \frac{x_3^2}{y_3}) (y_1+y_2+y_3)\geq (x_1+x_2 + x_3)^2$
Now in (5) we put:
$x_1 = a, x_2 = b, x_3 = c$
$y_1 = ab, y_2 = bc, y_3 = ca$
and we obtain the following:
(6) $ ( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} ) (ab+bc+ca) \geq (a+b+c)^2 $ i.e.
(7) $ ( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} ) \geq \frac{(a+b+c)^2}{(ab+bc+ca)} $
Now we will prove that:
(8) $(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}) \le \frac{3}{2}\frac{a+b+c}{ab+bc+ca}$
Let's rewrite the inequality (8). We multiply both sides by $ab+bc+ac$. Then:
$$LHS=\frac{ab+bc+ac}{b+c}+\frac{ab+bc+ac}{c+a}+\frac{ab+bc+ac}{a+b} = \frac{a(b+c)+bc}{b+c}+\frac{b(a+c)+ac}{a+c}+\frac{c(b+a)+ab}{a+b} = a+\frac{bc}{b+c}+b+\frac{ac}{c+a}+c+\frac{ab}{a+b}$$
Now we subtract $(a+b+c)$ from both sides. This shows that:
$$(8) \Leftrightarrow \frac{bc}{b+c}+\frac{ac}{a+c}+\frac{ab}{a+b}\leq \frac{1}{2}(a+b+c) $$
Taking into account that $4ab\leq (a+b)^2$ we get:
$$\frac{bc}{b+c}+\frac{ac}{a+c}+\frac{ab}{a+b}\leq \frac{(b+c)^2}{4(b+c)}+\frac{(c+a)^2}{4(c+a)}+\frac{(a+b)^2}{4(a+b)} = \frac{1}{2}(a+b+c) $$
OK, this completes the proof of (8).
So now from (4) and by using (8) and then (7) we get:
(9) $ \sqrt {\frac{a}{b+c}} + \sqrt {\frac{b}{c+a}} + \sqrt {\frac{c}{a+b}} \leq \sqrt {(a+b+c) \cdot \frac{3}{2}\frac{(a+b+c)}{ab+bc+ca}} \leq \sqrt{\frac{3}{2}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})}$
That's all we wanted to prove.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of $\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}dx$
Evaluation of $$\int\frac{(1+x^2)(2+x^2)}{(x\cos x+\sin x)^4}dx$$
$\bf{My\; Try::}$ We can write $$x\cos x+\sin x= \sqrt{1+x^2}\left\{\frac{x}{\sqrt{1+x^2}}\cdot \cos x+\frac{1}{\sqrt{1+x^2}}\cdot \sin x\right\}$$
So we get $$(x\cos x+\sin x) = \sqrt{1+x^2}\cos(x-\alpha)\;,$$ Where $\displaystyle \alpha = \tan^{-1}\left(\frac{1}{x}\right)$
So Integral $$I = \int\frac{(1+x^2)(2+x^2)}{(1+x^2)^2\cdot \cos^4 (x-\alpha)}dx = \int\frac{(2+x^2)}{(1+x^2)}\cdot \sec^4 (x-\alpha)dx$$
Now how can i solve after that,Help me
Thanks
| Your integral becomes $$I=\int { \frac { \left( 2+{ x }^{ 2 } \right) }{ \left( 1+{ x }^{ 2 } \right) } { \left( \sec { \left( x-\cot ^{ -1 }{ x } \right) } \right) }^{ 4 } } dx$$
On substituting $$\left( x-\cot ^{ -1 }{ x } \right) =t$$
diffrentiating $$dt = \frac { \left( 2+{ x }^{ 2 } \right) }{ \left( 1+{ x }^{ 2 } \right) } dx$$
Integration becomes $$I=\int { { \sec { \left( t \right) } }^{ 4 } } dt$$
Using : $\quad \quad \sec ^{ 2 }{ \left( x \right) } -\tan ^{ 2 }{ \left( x \right) } =1$
$$I=\int { { \sec { \left( t \right) } }^{ 2 } } \left( { \tan { \left( t \right) } }^{ 2 }+1 \right) dt$$
Again substituting $$\tan{ \left (t\right)} = u$$
$$du = { { \sec { \left( t \right) } }^{ 2 } }dt$$
Intrgration becomes $$I = \int { \left( { u }^{ 2 }+1 \right) du } $$
$$I=\frac { { u }^{ 3 } }{ 3 } +u+c$$
Substituting value of u
$$I = \frac { { \tan { \left( t \right) } }^{ 3 } }{ 3 } +\tan { \left( t \right) } +c$$
$$I=\frac { { \tan { \left( x-\cot ^{ -1 }{ x } \right) } }^{ 3 } }{ 3 } +\tan { \left( x-\cot ^{ -1 }{ x } \right) } +c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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How do you calculate the area of the intersection between a rectangle and a doughnut? I'm dealing with an engineering problem, involving concentric pipes, with air flowing through the outer pipe (doughnut).
I need a cross-beam to support the inner pipe, so I need to calculate how much of the outer doughnut's area will be consumed by the cross beam, in order to determine aerodynamic impact on the airflow through the outer circle of the doughnut.
The picture below illustrates what I'm trying to do.
I need to calculate the area of the gold section.
The actual height of the blue rectangle is irrelevant - it crosses both circles exactly twice, it is perpendicular to the axes, and it is centered over the vertical axis.
The intersection is fully rounded on both ends, exactly meeting the edges of both circles.
| The equations for the upper semicircles of the small pipe and large pipe are $f(x) = \sqrt{r^2 - x^2}$ and $g(x) = \sqrt{R^2 - x^2}$ respectively. The area of the cross-beam is then found by integrating $g(x) - f(x)$ over the interval $[-\frac{w}{2}, \frac{w}{2}]$:
\begin{align*}
\int_{-w/2}^{w/2} (g(x) - f(x))\,dx &= \int_{-w/2}^{w/2} g(x)\,dx - \int_{-w/2}^{w/2} f(x)\,dx \\
&= \int_{-w/2}^{w/2} \sqrt{R^2 - x^2}\,dx - \int_{-w/2}^{w/2} \sqrt{r^2 - x^2}\,dx \\
&= \left[\frac{1}{2} x \sqrt{R^2-x^2}
+\frac{1}{2}R^2\arctan\frac{x}{\sqrt{R^2-x^2}}\right]^{w/2}_{-w/2} - \left[\frac{1}{2} x \sqrt{r^2-x^2}
+\frac{1}{2}r^2\arctan\frac{x}{\sqrt{r^2-x^2}}\right]^{w/2}_{-w/2} \tag 1
\end{align*}
I should note that this is somewhat incomplete; I don't know if it's possible to simplify this expression before calculating it. That said, it should be possible to calculate it from this point.
David Quinn pointed out that $\arctan\frac{x}{\sqrt{R^2-x^2}}=\arcsin\frac xR$ (which follows from the definitions of $\sin$ and $\tan$). Using this, we can simplify equation $(1)$ as
\begin{align*}
\left[\frac{1}{2} x \sqrt{R^2-x^2}
+\frac{1}{2}R^2\arctan\frac{x}{\sqrt{R^2-x^2}}\right]^{w/2}_{-w/2} - \left[\frac{1}{2} x \sqrt{r^2-x^2}
+\frac{1}{2}r^2\arctan\frac{x}{\sqrt{r^2-x^2}}\right]^{w/2}_{-w/2} &= \left[\frac{1}{2} x \sqrt{R^2-x^2}
+\frac{1}{2}R^2\arcsin\frac{x}{r}\right]^{w/2}_{-w/2} - \left[\frac{1}{2} x \sqrt{r^2-x^2}
+\frac{1}{2}r^2\arcsin\frac{x}{r}\right]^{w/2}_{-w/2} \tag 2
\end{align*}
Evaluating each part of $(2)$ for $x = -\frac{w}{2}$ to $x = \frac{w}{2}$, we have [INCOMPLETE]
\begin{align*}
\left[\frac{1}{2} x \sqrt{R^2-x^2}
+\frac{1}{2}R^2\arcsin\frac{x}{r}\right]^{w/2}_{-w/2} - \left[\frac{1}{2} x \sqrt{r^2-x^2}
+\frac{1}{2}r^2\arcsin\frac{x}{r}\right]^{w/2}_{-w/2} &=
\frac{w}{2}\sqrt{R^2 - x^2} - \frac{w}{2}\sqrt{r^2 - x^2} (+ \ldots)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1699176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is
A. 0;
B. 3;
C. 5;
D. 1.
I don't know how to solve this.
| If $x \ge 0$ the $x^5 + 2x^3 + x^2 + 2 \ge 2$ so there are no solutions for $x \ge 0$.
If $x = -humongouseffinglargenumber$ then $x^5 + 2x^3 + x^2+2 < 0$ so somewhere between $-humongouseffinglargenumber$ and $0$ there must be an $x$ where the term is $0$. Let's try to narrow that range down a little.
As we are only looking at negative values of $x$ we want to find values where $|x|^5 + 2|x|^3 = x^2 + 2$ and one such obvious place is when $|x| = 1$ i.e. when $x = -1$ thus $x^5 + 2x^3 + x^2 + 2 = -1 + -2 + 1 + 2 = 0$.
So that's one root. We can factor out $(x - (-1)) = x + 1$ to get $x^5 + 2x^3 + x^2 + 2 = (x + 1)(x^4 - x^3 + 3x^2 -2x + 2)$. So if there are any more real roots they are real roots to $x^4 - x^3 + 3x^2 - 2x + 2 = 0$.
As we are only considering negative $x$ these are the same as solving for $x^4 + |x|^3 + 3x^2 + 2|x| + 2 = 0$. As $|x| \ge 0$, $x^4 + |x|^3 + 3x^2 + 2|x| + 2 \ge 2$ so there are no solutions.
So $x = -1$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1700615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 3
} |
Prove that $1+\frac12+\frac13+\cdots+\frac1{2^n} \ge 1+\frac{n}{2}$ Is it true that $1+\dfrac12+\dfrac13+ \dots +\dfrac{1}{2^n} \geq 1+\dfrac{n}{2}$?
If so, a proof would be great!
Thank you!
| We have
$$1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\cdots$$
$$\geq 1+\frac{1}{2} + \frac{1}{4} + \frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\cdots$$
$$=1+\frac{1}{2} + \frac{1}{2} + \frac{1}{2}+\cdots$$
$$=1+\frac{n}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Given the following relation between the angles of a triangle, find the number of possible triangles. If $A,B,C$ are the interior angles of triangle $\Delta ABC$ such that
$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2=9$, then the number of possible triangles is
$(A)0\hspace{1cm}(B)1\hspace{1cm}(C)3\hspace{1cm}(D)\infty$
$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2=\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B+2\cos B\cos C+2\cos C\cos A+\sin^2A+\sin^2B+\sin^2C+2\sin A\sin B+2\sin B\sin C+2\sin C\sin A=9$
I do not understand how I can identify the number of possible triangles from this tedious expression.
| $$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2=9$$
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B+2\cos B\cos C+2\cos C\cos A+\sin^2A+\sin^2B+\sin^2C+2\sin A\sin B+2\sin B\sin C+2\sin C\sin A=9$$
$$3+2(\cos A\cos B+\cos B\cos C+\cos C\cos A+\sin A\sin B+\sin B\sin C+\sin C\sin A)=9$$
$$\cos A\cos B+\cos B\cos C+\cos C\cos A+\sin A\sin B+\sin B\sin C+\sin C\sin A=3$$
$$\cos(A-B)+\cos(C-B)+\cos(A-C)=3$$
Then
$A+B+C=\pi$ and $A=B=C \Rightarrow A=B=C=60^{\circ}$
Answer: $(B)$ $1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The graph of the equation $x+y=x^3+y^3$ is the union of The graph of the equation $x+y=x^3+y^3$ is the union of
$(A)$line and an ellipse$(B)$line and a parabola$(C)$line and hyperbola$(D)$line and a point
I tried to factorize the given equation.
$x^3-x+y^3-y=0$
$x(x^2-1)+y(y^2-1)=0$
The answer given is a line and an ellipse but i do not understand how this equation is split into a line and an ellipse equation.
| $$x+y=x^3+y^3=(x+y)(x^2-xy+y^2)$$$$\therefore (x+y)(x^2-xy+y^2-1)=0$$Hopefully you can conclude from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding a General Solution to a Nonhomogeneous Matrix Equation I've come across a question in which I've been asked to find the general solution to the matrix equation:
$\begin{bmatrix}1 & -2 & 1\\-2 & 4 &-2\\1 & -2 & 1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}8\\-16\\8\end{bmatrix}$
by first finding a solution to a similar homogeneous equation:
$\begin{bmatrix}1 & -2 & 1\\-2 & 4 &-2\\1 & -2 & 1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}$
and then using that along with a given solution,
\begin{bmatrix}1\\-2\\3\end{bmatrix}
to find the general solution.
I've been looking around online and in textbooks for awhile and can't seem to find any information on this type of problem. Could someone provide a rundown of the methodology?
| Notice the column structure of the target matrix:
$$
\mathbf{A}=
%
\left[
\begin{array}{rrr}
1 & -2 & 1 \\
-2 & 4 & -2 \\
1 & -2 & 1 \\
\end{array}
\right]
%
=
%
\left[
\begin{array}{rrr}
c_{1} & -2 c_{1} & c_{1}
\end{array}
\right].
$$
We have one essential column. The rank plus nullity theorem reveals there will be two vectors to span the nullspace.
$$
\mathcal{N}\left( \mathbf{A} \right) = \text{span }
\left\{ \,
\left[
\begin{array}{r}
2 \\
1 \\
0
\end{array}
\right], \,
\left[
\begin{array}{r}
1 \\
0 \\
-1
\end{array}
\right] \,
\right\}
$$
For the data vector
$$
b=
\left[
\begin{array}{r}
1 \\
0 \\
-1
\end{array}
\right]
$$
The general solution for $\mathbf{A}x = b$ is
$$
x =
\left[
\begin{array}{c}
8 \\
0 \\
0
\end{array}
\right]
+
\alpha
\left[
\begin{array}{c}
2 \\
1 \\
0
\end{array}
\right]
+
\beta
\left[
\begin{array}{r}
1 \\
0 \\
-1
\end{array}
\right], \qquad \alpha, \beta \in \mathbb{C}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
There are $5$ women, $3$ men. How many ways to form a committee of $3$ with at least $1$ member of the opposite sex? I have looked through several topics for similar solutions and I have attempted an answer to the question. Unfortunately, the sample question itself does not have an answer.
From $5$ women and $3$ men, how many ways are there to form a committee of $3$ where there has to be at least $1$ member of the opposite sex.
My first attempt:
$$\binom{5}{3} \cdot \binom{3}{3} = \frac{5 \cdot 4 \cdot 3}{3 \cdot 2 \cdot 1} = 10 \cdot \frac{3!}{3!}$$
Since there are only two ways for committees with no same gender, I did $10-2 = 8$ as my final answer.
My second attempt:
First case: There has to be at least $1$ male. So that means that there are $\binom{5}{2}$ women available. $\binom{5}{2}$ is $10$. $\binom{3}{1}$ is $3$, so $10 \cdot 3 = 30$.
Second case: There has to be at least $2$ males. So that means $\binom{5}{1}$ women are available. $\binom{5}{1}$ is $5$, and $\binom{3}{2}$ is $3$. So that means there are $15 \cdot 3 = 45$ choices available.
When you add the two choices, you get $30 + 45 = 75$ choices.
Which attempt is right? Or are they both wrong? I would love explanations because the sample question has no answer listed.
| One more formulation. From the number of all possible committees, substract the number of committees with only men and the number of committees with only women. That is :
$$
\binom{5+3}{3}-\binom{5}{3}-\binom{3}{3}=56-10-1=45
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is there a simple way of proving that $\lfloor\sqrt{n}\rfloor+\lfloor\sqrt{4n+1}\rfloor = \lfloor\frac{3}{2} \lfloor \sqrt{4n+1} \rfloor\rfloor$? It appears that $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor = \left\lfloor\frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor\right\rfloor$$ for all $n \in \mathbb{N}_{0}$, although it is not obvious to me how to prove this equality. I verified that this equality holds for $n \leq 3 \times 10^6$. I discovered this equality experimentally while researching summations involving the floor function. The integer sequence $$\left( \left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor : n \in \mathbb{N}_{0} \right) = \left(1, 3, 4, 4, 6, 6, 7, 7, 7, 9, 9, 9, 10, 10, 10, 10, \ldots \right)$$ is not currently in the On-Line Encyclopedia of Integer Sequences.
It is not clear to me how to use induction to prove this equality. Also, it is not clear to me how to use 'case analysis' to prove this equality (e.g., by considering the case whereby $4n+1$ is a perfect square). Furthermore, it is not obvious to me how to use Hermite's identity to prove this equality.
I have also considered trying to prove that $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor \leq \left\lfloor\frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor\right\rfloor$$ and $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor \geq \left\lfloor\frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor\right\rfloor,$$ but I have thus far only managed to prove that $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor \leq \left\lfloor\frac{1}{2} \left\lfloor 3 \sqrt{4n+1} \right\rfloor\right\rfloor$$ and $$\left\lfloor \frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor \right\rfloor \leq \left\lfloor \sqrt{4n+1} \right\rfloor + \left\lfloor \sqrt{n + \frac{1}{4}} \right\rfloor.$$
I would be grateful to receive feedback regarding techniques which may be used to prove that $$\left\lfloor\sqrt{n}\right\rfloor+\left\lfloor\sqrt{4n+1}\right\rfloor = \left\lfloor\frac{3}{2} \left\lfloor \sqrt{4n+1} \right\rfloor\right\rfloor$$ for $n \in \mathbb{N}_{0}$, and I would be grateful to see an explicit proof of this equality.
| Following on from your comment, suppose that
$$\left\lfloor\sqrt{n}\right\rfloor<\left\lfloor\sqrt{n+\frac14}\right\rfloor\;,$$
and let $m=\left\lfloor\sqrt{n+\frac14}\right\rfloor$. Then
$$\sqrt{n}<m\le\sqrt{n+\frac14}\;,$$
so
$$n<m^2\le n+\frac14\;.$$
Since $n$ and $m^2$ are integers, this is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1709105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Complex numbers- elementary question How can I calculate the argument of the complex number $z= (\frac{1}{2}+ \frac{i\sqrt{3}}{2}) \cdot (1+i)$?
I always get $\tan^{-1}(-2-\sqrt{3})$, but the book answer is $7 \pi/12$.
| I) Tedious: $z= (\frac{1}{2}+ \frac{3^{1/2}i}{2}) * (1+i)= (1/2 - \sqrt{3}/2) + (1/2 + \sqrt 3/2) i = $. Arg($z$) = $\arctan (1+ \sqrt{3})/(1-\sqrt{3}) = -5\pi/12 = 7\pi/12$.
(To be fair, I made data entry errors the first six times I tried to enter this into a calculator.)
II) Easier (albeit it notation intimidating): $\arg (1+i) = \pi/4$ so $i+1 = r*e^{\frac{\pi}4 i}$. ($r = \sqrt 2$ but we don't give a toss.) $\arg (1/2 + i\sqrt{3}/2) = \pi/3$ so $1/2 + i\sqrt{3}/2 = s*e^{\frac{\pi}3 i}$. ($s = 1$ but we don't give a toss.)
So $z= (\frac{1}{2}+ \frac{3^{1/2}i}{2}) * (1+i) = r*e^{\frac{\pi}4 i}*s*e^{\frac{\pi}3 i} = rse^{(\pi/3 + \pi/4)i}$ so $\arg(z) = \pi/3 + \pi/4 = 7\pi/12$.
====
Post Script: $\frac {1 + \sqrt 3}{1 - \sqrt 3}= \frac{(1 + \sqrt 3)^2}{1 - 3} = \frac {4 + 2\sqrt 3}{-2} = -2 - \sqrt 3$ so your first answer was absolutely correct. Why did you doubt yourself? Why did you assume $\tan^{-1}(-2 - \sqrt{3}) \ne 7\pi/12$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplifying the integral $\int\frac{dx}{(3 + 2\sin x - \cos x)}$ by an easy approach $$I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}$$
If $$\tan\left(\frac{x}{2}\right)=u$$
or $$x=2\cdot\tan^{-1}(u)$$
Then,
$$\sin{x}=\dfrac{2u}{1+u^2}$$
$$\cos{x}=\dfrac{1-u^2}{1+u^2}$$
$$dx=\dfrac{2}{1+u^2}$$
Substitute $$\tan\left(\dfrac{x}{2}\right)=u$$
Let us simplify the integrand before integrating
$$\dfrac{1}{3+2\sin{x}-\cos{x}}$$
$$=\dfrac{1}{3+2\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}}$$
$$=\dfrac{1}{3+\frac{4u-1+u^2}{1+u^2}}$$
$$=\dfrac{1}{\frac{4u-1+u^2+3+3u^2}{1+u^2}}$$
$$=\dfrac{1+u^2}{4u^2+4u+2}$$
$$=\dfrac{1+u^2}{(2u+1)^2+1}$$
$$I=\displaystyle\int\dfrac{1+u^2}{(2u+1)^2+1}\cdot\dfrac{2}{1+u^2}\ du$$
$$=\displaystyle\int\dfrac{1}{(2u+1)^2+1}\ 2\,du$$
Now,
Take : $$v=2u+1$$
Therefore, $$dv=2\,du$$
$$I=\displaystyle\int\dfrac{1}{v^2+1}\ dv$$
$$I=\tan^{-1}(v)$$
Substitute everything back
$$I=\tan^{-1}(2u+1)$$
$$I=\tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)$$
$$\boxed{\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)} = \tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)+C}$$
I know that my approach is also not so difficult but still I think there must be an relatively easy approach to this integral. I have tried many different things using trigonometric identities but nothing seems to bring to the solution easily. Kindly help me out.
| With $a=\tan^{-1}\frac1{\sqrt5}$ and $t=\frac12(x+a)$
\begin{align}
\int\frac{dx}{3 + 2\sin x - \cos x}
& = \int \frac{dx}{3-\sqrt5\cos(x+a)}\\
&= \int \frac{2d(\tan^2t)}{(3-\sqrt5)+(3+\sqrt5)\tan^2t}=\tan^{-1} \frac{2\tan t}{3-\sqrt5} +C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find $\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$$
What I have done is to take the $\ln$
$$e^{\lim_{x \to 0}\ln\left(\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}\right)}$$
$$y={\lim_{x \to 0}{\frac{1}{x^2}}\cdot \ln\frac{\cos x}{\cos(2x)}}$$
Applying l'hopital to get
$$y={\lim_{x \to 0}{\frac{-2}{x^3}}\cdot \frac{\cos(2x)}{\cos x}\cdot \frac{-\sin x\cos(2x)+2\sin(2x)\cos x}{(\cos(2x)^2)}}={\lim_{x \to 0}\frac{-2}{x^3\cdot \cos x}\cdot \frac{-\sin x\cos(2x)+2\sin(2x)\cos x}{(\cos(2x))}}$$
Should I apply l'hopital again?
| We can proceed using the expansion $\cos(x)=1-\frac12 x^2+O(x^4)$.
First note that
$$\frac{\cos(x)}{\cos(2x)}=1+\frac32 x^2+O(x^4)$$
so that the limit of interest becomes
$$\begin{align}
\lim_{x\to \infty}\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}&=\lim_{x\to \infty}\left(\left(1+\frac{3/2}{1/x^2}\right)^{1/x^2}\left(1+O(x^4)\right)^{1/x^2}\right) \tag 1\\\\
\end{align}$$
Second, recalling that the limit definition of the exponential function is given by
$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n$$
we find that
$$\lim_{x\to \infty} \left(1+\frac{3/2}{1/x^2}\right)^{1/x^2}=e^{\frac32} \tag 2$$
Third, using the expansion for the logarithm function $\log(1+x)=O(x)$, we find that
$$\begin{align}
\lim_{x\to \infty}\left(1+O(x^4)\right)^{1/x^2}&=\lim_{x\to \infty}e^{\frac{1}{x^2}\log\left(1+O(x^4)\right)}\\\\
&=\lim_{x\to \infty}e^{O(x^2)}\\\\
&=1 \tag 3
\end{align}$$
Putting $(1)-(3)$ together yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}=e^{3/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find polynomials $f(x),g(x)$ such that $f(x)p(x) + g(x)q(x) = 2x^2 + 6x +2$ Let $$p(x) = x^4 + 7x^3 + 14x^2 + 7x+1 $$$$q(x) = x^4 +10x^3 + 23x^2 + 10x+1$$
Find polynomials $f(x),g(x)$ with rational coefficients such that $$f(x)p(x) + g(x)q(x) = 2x^2 + 6x +2$$
I totally have no idea to solve this problem... Please help me to do it. Thank you!!
| You can use Grobner basis to solve this,
$f1=x^4+7x^3+14x^2+7x+1;$,
$f2=x^4+10x^3+23x^2+10x+1;$
$f3=(f2-f1)/3=x^2+3x^2+x$
$f4=-3(f2-(x+7)f3)=x^2+3x+1$
Since,
$2x^2+6x+2 = 2*f4$
Now just replace it as $f4->f3->f2$ till everything is in terms of $f1$ and $f2$,
Answer: $2/3(-(x+4)f2+(x+7)f1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim_{x \to 3} \sqrt{x+1} = 2$ Prove that $\displaystyle\lim_{x \to 3} \sqrt{x+1} = 2$
Attempt:
$0 < |x - 3| < \delta \Rightarrow |\sqrt{x+1} - 2| < \epsilon$
Well $|\sqrt{x+1} - 2| = |(\sqrt{x+1} - 2) \cdot \displaystyle\frac{\sqrt{x+1} + 2}{\sqrt{x+1}+2}| = |\frac{x-3}{\sqrt{x+1} + 2}| = |x-3| \cdot \frac{1}{|\sqrt{x+1}+2|}$
Here, the second term is the nuisance so maybe do something like the following:
Suppose we want $|x-3| < 1$. Then, $2 < x < 4$.
So $|x-3| \cdot \displaystyle\frac{1}{\sqrt{x+1}+2} < \frac{1}{\sqrt{3}+2} \cdot |x-3|$
Looks like we want $\delta = \min((\sqrt{3}+2)\epsilon, 1)$
Then $|\sqrt{x+1} - 2| = |x-3| \cdot \displaystyle\frac{1}{\sqrt{x+1}+2} < \frac{1}{\sqrt{3}+2} \cdot |x-3| < \frac{1}{\sqrt{3}+2} \cdot (\sqrt{3} + 2)\epsilon = \epsilon$
| Your derivation is correct (I believe, it looks right but I didn't check every detail), but you are going for too much. To prove the limit statement, you don't need to identify specifically the largest $\delta$ that works for each $\epsilon$. You just need to prove there is some positive $\delta$ that will work. And you only need to prove it for "small" $\epsilon$ (it automatically follows for larger values). So, assuming $\epsilon < 1$, $\delta = \epsilon$ would work, because the nuisance second term is less than 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
tricky system of trigonometric equations I am not very fresh in math, but I need to solve this system:
\begin{gather}
A\sin(x-y)+B\sin(z-y)=C\\
A\cos(x-y)+B\cos(z-y)=D
\end{gather}
where $A,B,C,D$ and $x$ are given.
I tried to expand and combine the bracket terms and I suppose that there are some tricky substitutions to get it out, but I am lost!
Thank you all!
| Let $x-y=U, z-y=V$. Then
$$A\sin U+B \sin V=C$$
$$A\cos U+B \cos V=D$$
Square both equations:
\begin{align}
A^2\sin^2 U + B^2 \sin^2 V + 2 AB \sin U \sin V&=C^2\\
A^2\cos^2 U + B^2 \cos^2 V + 2AB \cos U \cos V &=D^2
\end{align}
Add to get
\begin{align}
A^2(\sin^2 U + \cos^2 U) + B^2 (\sin^2 V + cos^2 V) + 2 AB (\sin U \sin V + \cos U \cos V) &=C^2 + D^2\\
A^2 + B^2 + 2 AB (\sin U \sin V + \cos U \cos V) &=C^2 + D^2\\
2 AB (\sin U \sin V + \cos U \cos V) &=C^2 + D^2 - A^2 - B^2\\
\sin U \sin V + \cos U \cos V &=\frac{C^2 + D^2 - A^2 - B^2}{2AB}\\
\cos (U-V) &=\frac{C^2 + D^2 - A^2 - B^2}{2AB}\\
(U-V) &=\cos^{-1} \frac{C^2 + D^2 - A^2 - B^2}{2AB}\\
\end{align}
Since $U - V = (x-y) - (z-y) = x - z$, this gives you
\begin{align}
z &=x - \cos^{-1} \frac{C^2 + D^2 - A^2 - B^2}{2AB}.
\end{align}
Now you can plug in $z$ and $x$ in either of your first two equations to find $y$.
Of course, this all depends on $AB \ne 0$.
If either $A$ or $B$ is zero, then you can solve the first equation directly to find $U$ or $V$, and work from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$ While I know that $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^{2}}{6}$$
But trying to evaluate this has left me stumped
$$\sum_{n=1}^{\infty} \frac{1}{n^2+1}$$
I evaluated it through wolfram alpha, it gave me $\frac{1}{2}(\pi\coth(\pi)-1)$.
What would be a good way to start evaluating this series?
| This sum is harder, and perhaps less natural, to find than the sum for $\sum \frac{1}{n^2}$. We use Fourier series of $e^{x}$ with respect to $\{e^{i nx} / 2\pi\}$and apply Parseval's theorem.
Note that we can use this method of evaluation to find $\sum \frac{1}{n^2}$, too! We do the same for the (simpler) function $f(x) = x$. We could also directly evaluate the Fourier series for $f(x) = (\pi - |x|)^2$ on $(-\pi, \pi)$ at $x = 0$ to conclude the same (this function aso gives us the sum $\sum \frac{1}{n^4} = \frac{\pi^4}{90}$). (Source of these examples to Rudin's Principles of Mathematical Analysis)
We begin by calculating $\displaystyle \int_{-\pi}^{\pi} |f(x)|^2 \, dx = \int_{-\pi}^{\pi} e^{2x} \, dx = \frac{e^{2\pi} - e^{-2\pi}}{2}$.
Moving on to the Fourier coefficients, we have $c_n = \displaystyle \dfrac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^{-inx+x} \, dx = \dfrac{1}{\sqrt{2\pi}}\left( \int_{-\pi}^{\pi} e^x \cos(nx)\,dx - i\int_{-\pi}^{\pi} e^x \sin(nx)\,dx\right).$ Integrating the first integral with parts $f = \cos(nx)$ and $e^x\,dx = dg$ we get $$\left.e^x \cos(nx)\right|_{-\pi}^{\pi} + n\int_{-\pi}^{\pi} e^x \sin(nx)\,dx$$
We integrate by parts again and combine the resulting "like terms" (integrals) to get $$(n^2+1) \int_{-\pi}^{\pi} e^x \cos(nx) \, dx = e^x (\cos(nx) + n \sin(nx))|_{-\pi}^{\pi} = (-1)^n (e^{\pi} - e^{-\pi})$$
and so $$\frac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^x \cos(nx) = \frac{(-1)^n}{\sqrt{2\pi}} \frac{e^{\pi} - e^{-\pi}}{n^2 + 1}.$$
Similarly, we get $$\dfrac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^x \sin(nx) \, dx = -\frac{n(-1)^n}{\sqrt{2\pi}} \frac{e^{\pi} - e^{-\pi}}{n^2 + 1}.$$
Hence $$|c_n|^2 = \frac{(e^\pi - e^{-\pi})^2 + n^2 (e^{\pi} - e^{-\pi})^2}{2\pi(n^2+1)^2} = \frac{1}{2\pi}\frac{(e^{\pi}-e^{-\pi})^2}{n^2+1}$$ since $|a+bi|^2 = a^2 + b^2$.
Putting it all together using Parseval's formula we have
\begin{align} \sum_{n=-\infty}^{\infty} |c_n|^2 &= \int_{-\pi}^{\pi} |f(x)|^2 \, dx \\
\frac{1}{2\pi} \sum_{n=-\infty}^{\infty} \frac{(e^\pi - e^{-\pi})^2}{n^2+1} &= \frac{e^{2\pi}-e^{-2\pi}}{2}\\
\sum_{n=-\infty}^{\infty} \frac{1}{n^2 + 1} &= \pi\frac{e^{2\pi} - e^{-2\pi}}{(e^\pi - e^{-\pi})^2}\\
2 \sum_{n=1}^{\infty} \frac{1}{n^2+ 1} &= \pi \frac{e^\pi + e^{-\pi}}{e^{\pi} - e^{-\pi}} - 1\\
\sum_{n=1}^{\infty} \frac{1}{n^2 + 1} &= \frac{\pi \coth \pi - 1}{2}
\end{align}
where we use $e^{2\pi} - e^{-2\pi} = (e^{\pi} - e^{-\pi})(e^{\pi}+e^{-\pi})$ and $c_n = c_{-n}$ in the third line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 5
} |
Find the roots of quadratic polynomial given one root of another quadratic polynomial? if $a,b,c$ are Real numbers and $1$ is a root of equation $ax^2+bx+c=0$ then curve $y = 4ax^2+3bx+2c$ , (a is not zero) intersects $x$ axis at how many points?
I get a relation $a+b+c = 0$ I tried substituting this in given polynomial and tried finding its discriminant but cannot figure out how to find whether it is positive, negative or zero? can someone help?
| If $1$ is a root of $ax^2 + bx + c = 0$ Then
\begin{align}
ax^2 + bx + c
&= a(x-1)(x - \lambda) \\
&= ax^2 - a(1 + \lambda)x + a\lambda \\
\end{align}
So $b = -a(1+\lambda)$ and $c = a\lambda$.
The descriminant of $4ax^2 + 3bx + 2c$ is
\begin{align}
9b^2 - 32ac
&= 9a^2(1+\lambda)^2 -32a^2\lambda \\
&= a^2(9\lambda^2-14\lambda+9) \\
&= 9a^2\left[\left(x- \dfrac 79 \right)^2 + \dfrac{32}{9}\right]
\end{align}
Which is positive.
Hence the polynomial has two real roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Steps to solve $\lim_{n \to \infty} (\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} ) = 4 $ $$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} \right) = 4 $$
I think $\sqrt{n^4+1} - \sqrt{n^4-1}$ is approaching to zero, but it is not correct. What steps can evaluate above limit to 4?
| Using jim's answer, by Taylor $$\sqrt{n^4+1}=n^2+\frac{1}{2 n^2}-\frac{1}{8 n^6}+O\left(\frac{1}{n^{9}}\right)$$ $$\sqrt{n^4-1}=n^2-\frac{1}{2 n^2}-\frac{1}{8 n^6}+O\left(\frac{1}{n^{9}}\right)$$ $$\sqrt{n^4+1}-\sqrt{n^4-1}=\frac{1}{n^2}+O\left(\frac{1}{n^{9}}\right)$$ Now, multiplying by $(2n+1)^2$, you get for the whole expression $$4+\frac{4}{n}+\frac{1}{n^2}+O\left(\frac{1}{n^7}\right)$$ which shows the limit and how it is approached.
Let us try with $n=10$; the exact value is $\approx 4.4100000055$ while the above approximation gives $4.41$. This does not seem to be too bad being so far from $\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Establish the inequality $\frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} \geq (1-\frac{2\Omega}{N})^{\sqrt{N}}$ Establish the inequaity
$$ \frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} \geq (1-\frac{2\Omega}{N})^{\sqrt{N}}$$ where $N > \Omega > \sqrt{N}$ and $N$ is a perfect-square.
My attempt:
$$ \frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} = \frac{(N-\sqrt{N})(N-\sqrt{N}-1)\cdots (N-\sqrt{N}-\Omega+1)}{N(N-1)(N-2)\cdots (N-\Omega+1)}$$
Therefore, by manipulating the fraction we obtain:
$$ \frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} = (1-\frac{\sqrt{N}}{N})(1-\frac{\sqrt{N}}{N-1})\cdots (1-\frac{\sqrt{N}}{N-\Omega+1})$$
Since $\Omega > \sqrt{N}$, we will have:
$$(1-\frac{\sqrt{N}}{N})(1-\frac{\sqrt{N}}{N-1})\cdots (1-\frac{\sqrt{N}}{N-\Omega+1}) \geq (1-\frac{\Omega}{N})(1-\frac{\Omega}{N-1})\cdots (1-\frac{\Omega}{N-\Omega+1})$$
Which in turn gives:
$$ \frac{{N - \sqrt{N}} \choose \Omega}{N \choose \Omega} = (1-\frac{\Omega}{N})(1-\frac{\Omega}{N-1})\cdots (1-\frac{\Omega}{N-\Omega+1}) \geq (1-\frac{\Omega}{N-\Omega+1})^{\Omega}$$
If we also assume that $\Omega < \frac{N}{2}$, we get:
$$(1-\frac{\Omega}{N-\Omega+1})^{\Omega}=(\frac{N-2\Omega+1}{N-\Omega+1})^{\Omega}\geq (\frac{N-2\Omega}{N})^{\Omega} \geq (1-\frac{2\Omega}{N})^{\Omega}$$
But since $0<1-\frac{2\Omega}{N}<1$, exponentiation by $\sqrt{N} < \Omega$ will reverse the inequality and it won't finish the proof.
| According to the comments after the question, we assume $\sqrt{N}<Q<\frac{N}{2}$.
As you started,
$$
\dfrac{\binom{N-\sqrt{N}}{\Omega}}{\binom{N}{\Omega}}
= \prod_{k=0}^{\Omega-1} \frac{N-\sqrt{N}-k}{N-k}
= \prod_{k=0}^{\Omega-1} \left(1-\dfrac{\sqrt{N}}{N-k}\right) >
\\
> \left(1-\frac{\sqrt{N}}{N-Q}\right)^{\Omega}
> \left(1-\frac{\sqrt{N}}{N-N/2}\right)^{\Omega}
= \left(1-\frac2{\sqrt{N}}\right)^{\Omega}.
$$
By Bernoulli,
$$
\left(1-\frac2{\sqrt{N}}\right)^{\frac{\Omega}{\sqrt{N}}}
> 1 -\frac2{\sqrt{N}} \cdot \frac{\Omega}{\sqrt{N}} = 1 -\frac{2\Omega}{N},
$$
so
$$
\left(1-\frac2{\sqrt{N}}\right)^{\Omega}
> \left(1 -\frac{2\Omega}{N}\right)^{\sqrt{N}}.
$$
$\textbf{Update:}$ I realised that it can be done simpler, without using Bernoulli:
$$
\dfrac{\binom{N-\sqrt{N}}{\Omega}}{\binom{N}{\Omega}}
= \prod_{k=0}^{\sqrt{N}-1} \frac{N-\Omega-k}{N-k}
= \prod_{k=0}^{\sqrt{N}-1} \left(1-\dfrac{\Omega}{N-k}\right) >
\left(1-\frac{\Omega}{N-\sqrt{N}}\right)^{\sqrt{N}} >
\left(1-\frac{2\Omega}{N}\right)^{\sqrt{N}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Summation $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+....$ I came across a question today...
Q. The sum $\dfrac{3}{1^2}+\dfrac{5}{1^2+2^2}+\dfrac{7}{1^2+2^2+3^2}+....$ upto $11$ terms is?
Okay, I think it can be written as
$$\sum_{r=1}^{11}\dfrac{2r+1}{1^2+2^2+...+r^2}$$
I can't see how to simplify it.
Another way I tried is:
$$\dfrac{2^2-1^2}{1^2}+\dfrac{3^2-2^2}{1^2+2^2}+\dfrac{4^2-3^2}{1^2+2^2+3^2}+....$$
I simplified it to
$$-1+2^2(\dfrac{1}{1^2}-\dfrac{1}{1^2+2^2})+3^2(\dfrac{1}{1^2+2^2}-\dfrac{1}{1^2+2^2+3^2})+....$$
$$\Rightarrow -1+\dfrac{2^4}{1^2(1^2+2^2)}+\dfrac{3^4}{(1^2+2^2)(1^2+2^2+3^2)}+....$$
I can't see any way after it too... :(
Can you please help?
| After the help provided by @lab bhattacharjee, I would like to complete the answer.
$$6\sum_{r=1}^{11}\left(\dfrac{r+1-r}{r(r+1)}\right)$$
$$=6\sum_{r=1}^{11}\left(\dfrac{1}{r}-\dfrac{1}{r+1}\right)$$
$$=6\left[ \left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+....+\left(\dfrac{1}{11}-\dfrac{1}{12}\right) \right]$$
$$=6\left(\dfrac{1}{1}-\dfrac{1}{12}\right)$$
$$=\dfrac{11}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Unable to derive reason/formula for permutation problem What is the probability of $n$ preceding $1$ and $n$ preceding $2$ when we randomly
select a permutation of ${1, 2, . . . , n}$ where $n ≥ 4$?
I wrote out examples of n! when n equals some number like $3$ and $4$, and I see that the answer should be $\frac{1}{3}$, but I'm unable to understand why this is the answer through a more formulaic way.
| The result holds for all $n \geq 3$.
What matters here is that of the $3! = 6$ permutations of $1, 2, n$, two of them have $n$ appearing before both $1$ and $2$. The placement of the other $n - 3$ numbers in the sequence has no effect on whether $n$ appears before both $1$ and $2$. Hence, the probability that in a permutation of $1, 2, \ldots, n$ that $n$ appears before both $1$ and $2$ is $p = 1/3$, as you concluded.
More formally, there are $n!$ permutations of $1, 2, \ldots, n$. Let's count the number of permutations in which $n$ appears before both $1$ and $2$. We choose three of the $n$ positions for the numbers $1$, $2$, and $n$. Since $n$ appears before both $1$ and $2$, it must go in the leftmost of these three slots. We then have $2!$ ways of arranging $1$ and $2$ in the remaining selected spots. The remaining $n - 3$ numbers can be arranged in the remaining $n - 3$ open slots in $(n - 3)!$ ways. Thus, the probability that $n$ appears before both $1$ and $2$ is
$$p = \frac{\binom{n}{3} \cdot 2! \cdot (n - 3)!}{n!} = \frac{\frac{n(n - 1)(n - 2)}{3 \cdot 2} \cdot 2 \cdot (n - 3)!}{n!} = \frac{\frac{n!}{3}}{n!} = \frac{1}{3}$$
To connect the two arguments given above, note that we could also count the number of possible permutations of $1, 2, \ldots, n$ in the following manner. We choose $3$ of the $n$ slots for $1$, $2$, and $n$. We can arrange them in these slots in $3!$ ways, then arrange the remaining $n - 3$ numbers in the $n - 3$ open slots in $(n - 3)!$ ways. Hence, the probability that $n$ appears before both $1$ and $2$ is
$$p = \frac{\binom{n}{3} \cdot 2! \cdot (n - 3)!}{\binom{n}{3} \cdot 3! \cdot (n - 3)!} = \frac{2}{6} = \frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How can I solve this using prime factors? I'm stuck with this problem: $2^x \cdot 3^3 \cdot 26^y = 39^z$ for $x, y, z \in \mathbb{N}$.
I know that there isn't a natural solution for the equation, but I need to "prove" it using prime factors. I'm stuck here:
$$
2^x \cdot 3^3 \cdot (2\cdot 13)^y = (3\cdot 13)^z \\
\implies 2^x \cdot 3^3 \cdot 2^y \cdot 13^y = 3^z \cdot 13^z \\
\implies 2^{x+y} \cdot 3^3 \cdot 13^y = 3^z \cdot 13^z
$$
Could anyone give me a hint on this problem? (Sorry for the formatting)
| Assuming "natural solutions" means positive integers:
$2 \times 27 \times 26 = 702 = 2^2 \times 3^3 \times 13$ while $39 = 3 \times 13$. Therefore $x = y = z = 1$ is not a solution. In fact, no solution will work because the prime factorization of $2^x 3^3 26^y$ will always have that pesky oddly even prime while the factorization of $39^z$ will not, specifically, as $2^{x + y}$ as you have already figured.
Try seeing $39^z$ as $2^0 \times 3^z \times 13^z$. But since $x, y$ are positive integers, $x + y > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Integration by Parts with a logarithm Im not sure how i go about integrating a function being:
$$\int^{\sqrt{2}}_{1} r^{3}\log({r^2}) dr$$
I assume you use parts but it keeps making me go round in circles.
Any help would be appreciated.
My attempt was:
$$u=log(r^{2}) \rightarrow u'=\frac{2}{r},$$ $$v=\frac{r^4}{4} \rightarrow v'=r^{3}$$
so using the equation $uv- \int uv'$ i get:
$$\frac{r^{4}}{4}log(r^{2})- \int log(r^{2})r^{3}$$
| $$\int_{1}^{\sqrt{2}}r^3\ln(r^2)\space\text{d}r=$$
Integrate by parts, $\int f\space\text{d}g=fg-\int g\space\text{d}f$ where:
$$f=\ln(r),\text{d}g=r^3\space\text{d}r,\text{d}f=\frac{1}{r}\space\text{d}r,g=\frac{r^4}{4}$$
$$\left[\frac{r^4\ln(r)}{2}\right]_{1}^{\sqrt{2}}-\frac{1}{2}\int_{1}^{\sqrt{2}}r^3\space\text{d}r=$$
Use $\int y^b\space\text{d}y=\frac{y^{b+1}}{b+1}+\text{C}$:
$$\left[\frac{r^4\ln(r)}{2}\right]_{1}^{\sqrt{2}}-\frac{1}{2}\left[\frac{r^4}{4}\right]_{1}^{\sqrt{2}}=$$
$$\frac{1}{2}\left[r^4\ln(r)\right]_{1}^{\sqrt{2}}-\frac{1}{8}\left[r^4\right]_{1}^{\sqrt{2}}=$$
$$\frac{1}{2}\left(\left(\sqrt{2}\right)^4\ln(\sqrt{2})-1^4\ln(1)\right)-\frac{1}{8}\left(\left(\sqrt{2}\right)^4-1^4\right)=$$
$$\frac{1}{2}\left(4\cdot\ln(\sqrt{2})-0\right)-\frac{1}{8}\left(4-1\right)=$$
$$2\ln(\sqrt{2})-\frac{3}{8}=$$
$$\ln(2)-\frac{3}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Find the sum of series $\sum\limits_{n=2}^{+\infty}\frac{(-1)^n}{n^2+n-2}$. Using the power series $\sum\limits_{n=2}^{+\infty}\frac{1}{n^2+n-2}x^{n}$, the first derivative gives $\sum\limits_{n=2}^{+\infty}\frac{n}{n^2+n-2}x^{n-1}$ where the sum doesn't exist.
Integration of function $\frac{1}{n^2+n-2}x^n$ given $\frac{1}{(n+1)(n^2+n-2)}x^{n+1}$, and after forming the series, $\sum\limits_{n=2}^{+\infty}\frac{1}{(n+1)(n^2+n-2)}x^{n+1}$. After multiple integrations, I didn't found any simplifications.
| Decompose $n^2+n-2$ as $(n+2)(n-1)$ and then do: $$\frac{1}{n^2+n-2}=\frac{A}{n+2}+\frac{B}{n-1} \Rightarrow B=-A=\frac{1}{3}$$ And from this you can make the sum (just putting 1/3 out of the same as it is a common factor): $$\frac{1}{3}\left(\sum_{n=2}^{\infty}(-1)^n \left(\frac{1}{n-1} - \frac{1}{n+2}\right)\right) $$ Now, as you can guess, the sum of the sum of two fractions is the just the sum of the first plus the sum of the second. Consider the first one: $$\sum_{n=2}^{\infty}\frac{(-1)^n }{n-1}$$ now put n=2, the first sum, we begin with a one in the denominator and a plus one in the denominator, so let's do this: $$\sum_{n=2}^{\infty}\frac{(-1)^n }{n-1} = \sum_{n=2}^{\infty} (-1) \cdot\frac{(-1)^{n-1}}{n-1} = (-1) \sum_{n=2}^{\infty} \cdot\frac{(-1)^{n-1}}{n-1} = -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}$$ Notice that the last equality is true (plug the first values and see it for yourself) Now, I did the same to the second one and obtained the sum: $$\sum_{n=2}^{\infty}\frac{(-1)^n }{n+2} = \frac{1}{4} - \frac{1}{5} +... = \left(-1+\frac{1}{2} -\frac{1}{3} +\frac{1}{4} -\frac{1}{5} \right) + \left(1-\frac{1}{2}+\frac{1}{3} \right) = $$ $$= \sum_{n=1}^{\infty} \frac{(-1)^n}{n} +1 -\frac{1}{2} + \frac{1}{3} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} +\frac{5}{6}$$
Now putting it all together we get:
$$\frac{1}{3}\left(-\frac{5}{6} -2\sum_{n=1}^{\infty}\frac{(-1)^n}{n} \right)$$
Now that last sum is simply $-\log(2)$ therefore the total sum is: $$\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2+n-2} = \frac{1}{3}\left(2\log(2)-\frac{5}{6}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simple differential equation $\frac{x}{y}=\frac{y'}{x+1}$ $$\frac{x}{y}=\frac{y'}{x+1}$$
The solution to this is very easy, we just multiply both sides by
$$(x+1)y\ \mathrm dx$$
Then we have
$$(x^2+x)\ \mathrm dx=y\ \mathrm dy$$
Then we integrate both sides to get solution
$$\frac{x^3}{3}+ \frac{x^2}{2}=\frac{y^2}{2} + C, C\in \mathbb{R}$$
But I don't understand how can we divide by $y$ and $x+1$. What if $y=0$ or $x=-1$? What happens then?
| From the form of the DE given at the start it is implied that $x=−1$ and $y=0$ are excluded from the domain and range respectively. If you had an an initial condition this has more impact on your question.
Suppose you had the initial condition that $y\left(\frac{3}{2}\right)=\frac{3}{2}$. This initial condition will result in $c=0$.
So the solution to the DE is $\frac{x^3}{3}+\frac{x^2}{2}=\frac{y^2}{2}$. We must exclude $x=-1$ and $y=0$ from possible solutions.
When $x=-1$ then $\frac{-1}{3}+\frac{1}{2}=\frac{y^2}{2}\implies y=\pm\frac{1}{\sqrt{3}}$.
When $y=0$ then $\frac{x^3}{3}+\frac{x^2}{2}=0\implies x=0,x=-\frac{3}{2}$.
So these values would need to be excluded from your domain/range.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Median BM of triangle ABC two results Given
*
*Calculate the measure of the median $\overline{BM}$ of ABC triangle, given A (-6.1); B (-5,7) and C (2,5)
I get this result:
$Xm = \frac{Xc - Xa}{2} + Xa$
$Xm = \frac{2-(-6)}{2} + (-6) = 4 - 6 = -2$
$Ym = \frac{(Yc - Ya)}{2} + Ya$
$Ym = \frac{5-1}{2} + 1 = 2 + 1 = 3$
$M(-2, 3)$
$d_{BM}^2 = (-5-(-2))^2 + (7-3)^2 = 9 + 16 = \sqrt{25} = 5$
and someone else get:
$Xm = \frac{-5+2}{2} = \frac{-3}{2}$
$Ym = \frac{7+5}{2} = \frac{12}{2} = 6$
$...$
So what's correct ?
| First, both you and someone else seem to be using $B(-5,7)$, not $B(5,7)$.
The "someone else" appears to be computing the midpoint between some other point $B(-5,7)$ and $C(2,5)$, and you are computing the midpoint between $A(-6,1)$ and $C(2,5)$.
If you are really trying to compute the length $BM$ where $M$ is the midpoint of $AC$, the side opposite to $B$, then your answer is correct.
We don't have the complete "someone else" answer, so it is unclear what distance that person is computing, but the $M$ that person is computed is the midpoint of $BC$, not the midpoint of $AC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Define the image of the function $f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$ Question
Define the range of the function:
$$f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$$
Answer attempt
I assume that the domain for $f(x)$ is in ${\rm I\!R}$.
If we draw a triangle with the sides: $1, x$ and $\sqrt{1+x^2}$, with the angle $\theta$ opposite of $x$, we get:
$$\sin{\frac{x}{\sqrt{1+x^2}}} \rightarrow$$ $$\theta = \arcsin{\frac{x}{\sqrt{1+x^2}}}$$
$$\tan{\frac{x}{1}} \rightarrow$$ $$\theta = \arctan{x}$$
This means that:
$$\arctan x = \arcsin{\frac{x}{\sqrt{1+x^2}}}$$
The original expression can thus be written:
$$2 \arctan x + \arcsin \frac {2x}{1+x^2} = 2 \arcsin \frac{x}{\sqrt{1+x^2}} + \arcsin \frac {2x}{1+x^2}$$
We know that $\arcsin$ has the domain $-1 \leq x \leq 1$ and the range $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$
If we start with: $\arcsin \frac {2x}{1+x^2}$
$x=1 \rightarrow \arcsin{\frac{2}{2}} = \arcsin 1 = \frac{\pi}{2}$
$x=-1 \rightarrow \arcsin{\frac{-2}{2}} = \arcsin -1 = -\frac{\pi}{2}$
$-1 \leq x \leq 1 \rightarrow -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
And analyzing: $2\arcsin \frac{x}{\sqrt{1+x^2}}$
$x=1 \rightarrow 2\arcsin{\frac{1}{\sqrt{2}}} \rightarrow \sin{\theta} = \frac{1}{\sqrt{2}} \rightarrow \theta = \frac{\pi}{4}$
$x=-1 \rightarrow 2\arcsin{-\frac{1}{\sqrt{2}}} \rightarrow \sin{\theta} = -\frac{1}{\sqrt{2}} \rightarrow \theta = -\frac{\pi}{4}$
$-1 \leq x \leq 1 \rightarrow -2(\frac{\pi}{4}) \leq y \leq 2(\frac{\pi}{4}) \rightarrow -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
The range of $f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$ is therefore $-\pi \leq y \leq \pi$
I wonder if you feel that my reasoning is correct? Is there some better way to solve this problem?
| Since $-1\le 2x/(1+x^2)\le 1$ for every $x\in\mathbb{R}$, the domain of $f$ is indeed the whole real axis. Moreover
$$
\lim_{x\to-\infty}f(x)=-\pi,\qquad
\lim_{x\to\infty}f(x)=\pi
$$
Let's compute the derivative:
$$
f'(x)=\frac{2}{1+x^2}+\frac{1}{\sqrt{1-\frac{4x^2}{(1+x^2)^2}}}
\frac{2(1+x^2)-4x^2}{(1+x^2)^2}
$$
A simple computation brings this to
$$
f'(x)=\frac{2}{1+x^2}\left(1+\frac{1-x^2}{|1-x^2|}\right)
$$
(which is valid for $x\ne\pm1$).
First step.
$$
\sqrt{1-\frac{4x^2}{(1+x^2)^2}}=
\sqrt{\frac{(1+x^2)^2-4x^2}{(1+x^2)^2}}=
\sqrt{\frac{1-2x^2+x^4}{(1+x^2)^2}}=
\sqrt{\frac{(1-x^2)^2}{(1+x^2)^2}}=\frac{|1-x^2|}{1+x^2}
$$
Second step
$$
\frac{2(1+x^2)-4x^2}{(1+x^2)^2}=2\frac{1-x^2}{(1+x^2)^2}
$$
Third step
$$
f'(x)=\frac{2}{1+x^2}+2\frac{1+x^2}{|1-x^2|}\frac{1-x^2}{(1+x^2)^2}=
\frac{2}{1+x^2}\left(1+\frac{1-x^2}{|1-x^2|}\right)
$$
Fourth step
If $1-x^2<0$ (that is, $|x|>1$), we have $\dfrac{1-x^2}{|1-x^2|}=-1$
Fifth step
If $1-x^2>0$ (that is, $|x|<1$), we have $\dfrac{1-x^2}{|1-x^2|}=1$
Hence we have
$$
f'(x)=\begin{cases}
0 & \text{if $x<-1$}\\[4px]
\dfrac{4}{1+x^2} & \text{if $-1<x<1$}\\[4px]
0 & \text{if $x>1$}
\end{cases}
$$
Therefore we have
$$
f(x)=\begin{cases}
-\pi & \text{if $x<-1$}\\[4px]
4\arctan x & \text{if $-1\le x\le 1$}\\[4px]
\pi & \text{if $x>1$}
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute $\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$ I have to compute $$\frac{1^2 t}{1!}+\frac{2^2 t^2}{3!}+\frac{3^2 t^3}{5!}+\frac{4^2 t^4}{7!}+\ldots+\frac{n^2 t^n}{(2n-1)!}+\ldots$$ I know that $\sinh t$ can be represented as a series1, but for that I require only odd powers $t^{2n-1}$, but I have no idea how to get them or what to do with the even part. Can anybody give me a hint how to start?
1 The expansion is $$\sinh x = x + \frac {x^3} {3!} + \frac {x^5} {5!} + \frac {x^7} {7!} +\cdots = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}.$$
| \begin{align*}
\sqrt{t} \sinh \sqrt{t} &= \sum_{n=1}^{\infty} \frac{t^{n}}{(2n-1)!} \\
(\sqrt{t} \sinh \sqrt{t})' &=\sum_{n=1}^{\infty} \frac{nt^{n-1}}{(2n-1)!} \\
t(\sqrt{t} \sinh \sqrt{t})' &=\sum_{n=1}^{\infty} \frac{nt^{n}}{(2n-1)!} \\
[t(\sqrt{t} \sinh \sqrt{t})']' &=
\sum_{n=1}^{\infty} \frac{n^{2}t^{n-1}}{(2n-1)!} \\
t[t(\sqrt{t} \sinh \sqrt{t})']' &=
\sum_{n=1}^{\infty} \frac{n^{2}t^{n}}{(2n-1)!} \\
\sum_{n=1}^{\infty} \frac{n^{2}t^{n}}{(2n-1)!} &=
t\left[t
\left(
\frac{\sinh \sqrt{t}}{2\sqrt{t}}+
\sqrt{t} \cosh \sqrt{t} \times \frac{1}{2\sqrt{t}}
\right)
\right]' \\
&=t\left(
\frac{\sqrt{t}}{2}\sinh \sqrt{t}+\frac{t}{2} \cosh \sqrt{t}
\right)' \\
&=t\left(
\frac{1}{4\sqrt{t}}\sinh \sqrt{t}+
\frac{\sqrt{t}}{2}\cosh \sqrt{t} \times \frac{1}{2\sqrt{t}}+
\frac{1}{2} \cosh \sqrt{t}+
\frac{t}{2} \sinh \sqrt{t} \times \frac{1}{2\sqrt{t}}
\right) \\
&= \frac{(t+1)\sqrt{t}}{4} \sinh \sqrt{t}+
\frac{3t}{4} \cosh \sqrt{t}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Prove $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$ Prove that $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$
$$\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{\frac{2}{\cos\theta} +\frac{3\sin\theta}{\cos\theta}+5\sin\theta-7\cos\theta+5}{\frac{2\sin\theta}{\cos\theta} +\frac{3}{\cos\theta}+5\cos\theta+7\sin\theta+8}$$
$$=\frac{2+3\sin\theta+5\sin\theta\cos\theta-7\cos^2\theta+5\cos\theta}{2\sin\theta+3+5\cos^2\theta+7\sin\theta\cos\theta+8\cos\theta}$$
I am stuck here.I tried to factorize numerator and denominator but does not succeed.
| It is possible using the Rule $ P = n p, Q = n q $, $n$ any real number
and recognizing
$$ \frac{1-\cos\theta}{\sin\theta}= \frac{\sin\theta}{1+\cos\theta}\tag{0} ,$$
$$ \frac {u}{v} = \frac {p}{q} = \frac {P}{Q}=\frac {u \pm P}{v \pm Q}, $$ and entirely avoid all trigonometric calculations.
Multiply numerator and denominator by 7 and add
$$\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}=\frac{7-7 \cos\theta}{7 \sin\theta}\tag{1}$$
$$ F = \frac{2\sec\theta +3\tan\theta+5\sin\theta-2}{2\tan\theta +3\sec\theta+5\cos\theta+8} \tag{2} $$
Multiply numerator and denominator by 5 and subtract
$$\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{\sin\theta}{1+\cos\theta}=\frac{5 \sin\theta}{5+5\cos\theta} \tag{3} $$
$$ F=\frac{2\sec\theta +3\tan\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta +7\sin\theta+3} \tag{4}$$
Repeating the Rule by subtracting on (2) and (4)
$$ F=\frac{5 \sin \theta + 7\cos\theta-7}{5\cos\theta -7\sin\theta +5}\tag{5} $$
Repeating the rule by subtracting (2) from (3)
$$ F =\frac{5 \sin \theta + 7\cos\theta-7}{5\cos\theta -7\sin\theta +5}\tag{6} $$
which are identical.
It will be appreciated that it is simple algebra and there is no trig at all after (0) and probably that is how the problem had been set.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding $\int \frac{dx}{a+b \cos x}$ without Weierstrass substitution. I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ without using Weierstrass substitution, which is the usual technique.
When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely.
But I remember that the technique I saw was a nice way of evaluating these even when $a,b\neq 1$.
| Another way to get to the same point as C. Dubussy got to is the following:
\begin{align*}
\frac{1}{a + b \cos x} &= \frac{1}{a \left (\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} \right ) + b \left (\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} \right )}\\
&= \frac{1}{(a - b) \sin^2 \frac{x}{2} + (a + b) \cos^2 \frac{x}{2}}\\
&= \frac{\sec^2 \frac{x}{2}}{(a + b) + (a - b) \tan^2 \frac{x}{2}},
\end{align*}
and then make the substitution of $t = \tan \frac{x}{2}$ in the integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
} |
Equation arrangement. How to arrange $x=\frac{1}{2} \ln({1+\frac{10}{x}})$ to $x=\frac{10}{e^{2x}-1}$ ?
Can anyone give me step by step explanation? Thanks in advance.
| $x=\frac{1}{2} \ln(1+ \frac{10}{x}) \implies 2x= \ln(1+ \frac{10}{x}) \implies
e^{2x} = 1+ \frac{10}{x} \implies e^{2x} - 1 =\frac{10}{x} \implies x= \frac{10}{ e^{2x} - 1}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
| The partial sum decomposition of the term of your serie is : $\frac{\frac{1}{3}}{n}-\frac{\frac{1}{3}}{n+3}$. You recognise telescoping series : $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}=\sum_{n=1}^{\infty} \frac{\frac{1}{3}}{n}-\frac{\frac{1}{3}}{n+3}=\frac{\frac{1}{3}}{1}+\frac{\frac{1}{3}}{2}+\frac{\frac{1}{3}}{3}=\frac{11}{18}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Solve the equation $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$ Solve the following equation: $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$
Unfortunately I have no idea.
| First notice that $x$ must be in $[-1,1]$. Then rewrite $$\sqrt{1-x}-2x\sqrt{1-x^2} =2x^2-1$$ and square : $$(1-x)+4x^2(1-x^2)-4x(1-x)\sqrt{1+x}=4x^4-4x^2+1.$$ This can be simplified $$4x(x-1)\sqrt{1+x}=8x^4-8x^2+x=x(8x^3-8x+1).$$ Symplify by $x$ and square another time to get $$16(x+1)(x-1)^2=64x^6+64x^2+1+16x^3-16x-128x^4.$$ Finally this leads to the equation $$64x^6-128x^4+80x^2-15 = 0.$$ Set $y=x^2$ and solve $$64y^3-128y^2+80y-15=0.$$ An "obvious solution" is $\frac{3}{4}$. After that you will have to solve a simple second-degree equation.
Don't forget to discuss the solutions at the end. If no mistakes, only one solution should be in $[-1,1]$ and satisfy the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Branches of $\log(z^2)$ Suppose $z=x+iy$, then $r=\sqrt{x^2+y^2}$ and $\theta_p=\tan^{-1}\left(\frac y x\right)$, so we have
$$z=re^{i\theta},\theta=\theta_p+2n\pi$$
Now, $z^2=x^2-y^2+i(2xy)$, so $r'=x^2+y^2$ and $\theta_p'=\tan^{-1}\left(\frac{2xy}{x^2-y^2}\right)$, so we have
$$z^2=(x^2+y^2)e^{i\theta'},\theta'=\theta_p'+2n\pi$$
Hence,
$$\log(z)=\log(r)+i\theta_p+2n\pi i$$
$$\log(z^2)=\log(r^2)+i\theta_p'+2n\pi i\tag 1$$
but in my homework solution,
$$\log(z^2)=\log(r^2)+i2\theta_p+2n\pi i\tag 2$$
So from $(1)$ and $(2)$ we have,
$$\tan^{-1}\left(\frac{2xy}{x^2-y^2}\right)=2\tan^{-1}\left(\frac y x\right)$$
But I don't think that this equality should be true?
| $$\begin{align*}&\color{red}{\frac{\partial}{\partial x}\left(\arctan\frac{2xy}{x^2-y^2}\right)}=\frac{2y(x^2-y^2)-4x^2y}{(x^2-y^2)^2}\frac1{1+\frac{4x^2y^2}{(x^2-y^2)^2}}=\frac{-2y}{(x^2+y^2)}\\
&\color{green}{\frac{\partial}{\partial x}\left(2\arctan\frac yx\right)}=\frac{-2\frac y{x^2}}{1+\frac{y^2}{x^2}}=\frac{-2y}{x^2+y^2}\end{align*}$$
Fill in details.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Examine convergence of $\sum_{n=1}^{\infty}(\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2})$ How to examine convergence of $\sum_{n=1}^{\infty}(\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2})$ for $a, b, c> 0$ using Taylor's theorem?
| You have, from Taylor expansion of $e^u$ around $0$ to second order,
$$\begin{align}
\sqrt[n]{a} - \frac{\sqrt[n]{b}+\sqrt[n]{c}}{2} &=
e^{\frac{1}{n}\ln a} - \frac{1}{2}\left(e^{\frac{1}{n}\ln c}+e^{\frac{1}{n}\ln b}\right)\\ &=
1+\frac{1}{n}\ln a + \frac{1}{2n^2}\ln ^2 a \\&\qquad- \frac{1}{2}\left(2+\frac{1}{n}\ln b + \frac{1}{2n^2}\ln ^2 b+\frac{1}{n}\ln c + \frac{1}{2n^2}\ln ^2 c\right) + o\left(\frac{1}{n^2}\right) \\
&= \frac{1}{2n}(2\ln a - (\ln b + \ln c)) + \frac{1}{4n^2}(2\ln ^2 a - ( \ln ^2 b+\ln ^2 c)) + o\left(\frac{1}{n^2}\right).
\end{align}$$
If $(2\ln a - (\ln b + \ln c)) \neq 0$, can you conclude (by theorems of comparisons)? And if $(2\ln a - (\ln b + \ln c)) = 0$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Maclaurin Expansion for $e^{e^{z}}$ at $z=0$ I need to find terms up to degree $5$ of $e^{e^{z}}$ at $z=0$.
I tried letting $\omega = e^{z} \approx 1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\cdots$, and then substituting these first few terms into the Taylor series expansion for $e^{z}$ as follows:
$e^{\omega} = \sum_{n=0}^{\infty}\frac{\omega^{n}}{n!} = 1 + \omega + \frac{\omega^{2}}{2!}+\frac{\omega^{3}}{3!}+\cdots = 1 + (1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots) + \frac{1}{2}\left(1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots \right)^{2}+ \frac{1}{3!}\left( 1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots\right)^{3}+\cdots$
But, then when I try to multiply this all out (using Maple), I find that going even further out than I have here, the terms of degree $\leq 5$ do not terminate. So, then I tried putting the command into Maple, and it came out to be something long and horrible with $e$ coefficients for each term.
I must be missing something here.
Could somebody please tell me how to finish this problem? I would ask for a complete solution, not hints or leading questions. I've been working on this for hours and am beyond frustrated. The most edifying thing for me at this point would be to see the correct, full way it's supposed to be done, work backwards, take it apart, and then be able to apply it to other situations.
Thank you.
| Your approach works, but the Taylor series for the outer function should not necessarily be centered at $0$. Rather, it should be centered at the constant term of the Taylor series of the inner function.
In your example, the outer series should be centered at $x=1$. Expanding $e^x$ in a Taylor series around this point gives
$$ e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3 + \frac{e}{24}(x-1)^4 + \frac{e}{120}(x-1)^5 + \cdots $$
Now substitute in the Taylor series for $e^z$ centered at $z=0$ and you get
$$ e + e \left(z+\frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \frac{1}{120}z^5 + \cdots \right) + \frac{e}{2} \left(z+\frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \frac{1}{120}z^5 + \cdots \right)^2 + \frac{e}{6} \left(z+\frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \frac{1}{120}z^5 + \cdots \right)^3 \\ + \frac{e}{24} \left(z+\frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \frac{1}{120}z^5 + \cdots \right)^4 + \frac{e}{120} \left(z+\frac{1}{2}z^2 + \frac{1}{6}z^3 + \frac{1}{24}z^4 + \frac{1}{120}z^5 + \cdots \right)^5 + \cdots $$
As you expand each exponentiated series, you will only find finitely many terms of each given degree. Doing this and combining like terms, you get your series
$$ e + ez + \left(\frac{e}{2} + \frac{e}{2} \right) z^2 + \left(\frac{e}{6} + e + \frac{e}{6} \right) z^3 + \left(\frac{e}{24} + \frac{e}{6} + \frac{e}{8} + \frac{e}{4} + \frac{e}{24} \right) z^4 + \left( \frac{e}{120} + \frac{e}{12} + \frac{e}{24} + \frac{e}{12} + \frac{e}{8} + \frac{e}{12} + \frac{e}{120} \right) z^5 \cdots $$
which simplifies to
$$ e + ez + ez^2 + \frac{5e}{6} z^3 + \frac{5e}{8} z^4 + \frac{13e}{30} z^5 + \cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Convergence of $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ I need to prove $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges. The root test is inconclusive, so I check in W.A., $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges by comparison test, but I don't know what series to compare. I've tried to compare with $\sum \frac{1}{4n+1} - \frac{1}{4n + 3}$ because this serie converges to $\frac{\pi}{4}$, but we have exactly the opposite inaquality $\frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} > \frac{1}{4n + 1} - \frac{1}{4n + 3}$.
Can you help me?
| Hint $$\frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} = \frac{\sqrt{4n + 3}-\sqrt{4n + 1}}{\sqrt{4n + 3}\sqrt{4n + 1}}=\frac{2}{\sqrt{4n + 3}\sqrt{4n + 1}(\sqrt{4n + 3}+\sqrt{4n + 1})}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Epsilon delta proof with fractions, given what delta must be Choose $\delta = \min (1, \frac{\epsilon}{10})$
is the following statement true?
$$0 < |x − 1| < δ\text{ implies that }\left|\frac{x^2+3x}{x^2+1} − 2\right| < ε$$
Okay so this is what I have so far
$$\left|\frac{x^2+3x}{x^2+1} − 2\right| = |x-1|\frac{|-x+2|}{|x^2+1|}$$
I then chose $\delta = \frac{1}{2}$ (because if I chose $\delta = 1$, $x$ would come out as $2$ which would give me $|x-1|\cdot 0 < \epsilon$ which doesn't tell me much?)
So if $\delta = \frac{1}{2}$, then
$\frac{1}{2}< x < \frac{3}{2}$ (because it's a fraction I used $x > \frac{1}{2}$)
$$|x-1|\cdot\frac{\left|-\frac{1}{2}+2\right|}{\left|(\frac{1}{2})^2+2\right|}=\frac{2}{3}|x-1|< \epsilon$$
$$|x-1|< \frac{3}{2}\epsilon$$
I'm just so confused because I don't know how to relate the value for $\delta$ (which I found to be $\frac{3}{2}\epsilon$ to the delta they're making me choose of $\min (1, \frac{\epsilon}{9})$. I'm also not sure if I chose the correct value for $x$ as it is a fraction.
| As $|x-1|<1$ for all possible $\delta$ values then $$|x-2|< 2\ \text{and} \ |x^2+1|>1$$ So then we have
$$|x-1|\frac{|-x+2|}{|x^2+1|}< \frac{2|x-1|}{1} = 2|x-1|$$
If $\epsilon\geq10$ then remembering $|x-1|<1$ as $\delta =1$ in this case $$|x-1|\frac{|-x+2|}{|x^2+1|}<2\cdot 1<10\leq\epsilon$$ as desired. If $\epsilon<10$ then remembering $|x-1|<\delta=\frac{\epsilon}{10}$ we have
$$|x-1|\frac{|-x+2|}{|x^2+1|}<2\cdot \frac{\epsilon}{10}=\epsilon/5<\epsilon$$
and you are finished. Let me know if you have questions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to plot correct best fit line? I'm working on a software where I'm plotting graphs and finding best fit line.
I have used Least-Square Method and linear regression technique with y = mx + c
My problem is that when most of the X values of graph are equal (not all) at that time best fit line is not proper but when there is good variation it seems correct.
Following pictures are for reference, Orange line is graph and Green is best fit (please don't consider other grey lines)
1. points with variation
2. most X values are equal
What is the problem here I'm not getting? please help.
| The problem has horrific conditioning.
What is the slope$-$intercept equation for a vertical line?
The value for the slope is swinging between $\pm \infty$. This mean a very small perturbation in your data can cause an enormous change in the results.
The trial function is
$$
y(x) = a_{0} + a_{1} x.
$$
The data will consist of a sequence of measurements $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}.$
The linear system is
$$
\begin{align}
\mathbf{A} a &= y \\
%
\left[ \begin{array}{cc}
\mathbf{1} & x
\end{array} \right]
%
\left[ \begin{array}{c}
a_{0} \\ a_{1}
\end{array} \right]
%
&=
\left[ \begin{array}{c}
y
\end{array} \right]
%
\end{align}
$$
The normal equations are
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} a &= \mathbf{A}^{*} y \\
%
\left[ \begin{array}{cc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
x \cdot \mathbf{1} & x \cdot x
\end{array} \right]
%
\left[ \begin{array}{c}
a_{0} \\ a_{1}
\end{array} \right]
%
&=
\left[ \begin{array}{c}
\mathbf{1} \cdot y \\
x \cdot y
\end{array} \right].
%
\end{align}
$$
The solution for the normal equations is
$$
\begin{align}
a &= \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} y \\
%
&=
%
\left(
\det \left( \mathbf{A}^{*} \mathbf{A} \right)
\right)^{-1}
%
\left[ \begin{array}{rr}
x \cdot x & -\mathbf{1} \cdot x \\
-\mathbf{1} \cdot x & \mathbf{1} \cdot \mathbf{1}
\end{array} \right]
%
\left[ \begin{array}{c}
\mathbf{1} \cdot y \\
x \cdot y
\end{array} \right].
%
\end{align}
$$
Isolate the slope
$$
a_{1} = \frac{
\left( \mathbf{1} \cdot \mathbf{1} \right)
\left( x \cdot y \right) -
\left( \mathbf{1} \cdot x \right)
\left( \mathbf{1} \cdot y \right)}
{
\left( \mathbf{1} \cdot \mathbf{1} \right)
\left( x \cdot x \right) -
\left( \mathbf{1} \cdot x \right)^{2}
}
$$
Consider the case where $\hat{x}_{k} = \bar{x} + \epsilon_{k}$, where $\epsilon_{k}<<\bar{x}$ for all $k$. The perturbations have mean $0$. The terms in the denominator have this behavior:
$$
\begin{align}
\left( \mathbf{1} \cdot \mathbf{1} \right)
\left( \hat{x} \cdot \hat{x} \right) -
\left( \mathbf{1} \cdot \hat{x} \right)^{2}
&\to
m\left( \sum x_{k}^{2} + \sum x_{k} \epsilon_{k} \right)
-
\sum x_{k}^{2} + 2 m \bar{x}^{2} \\
&\to
2 m \sum x_{k} \epsilon_{k} - 2 \sum x_{k} \sum \epsilon_{k} \\
&\to
0
\end{align}
$$
@Wouter offers a path to salvation by rotating the coordinate system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Bag of 24 distinct objects, four colors, six objects per color, select three (probability). Could someone confirm my combinatorics solutions for this question?
Question:
A bag holds 24 different objects, of which 6 are orange, 6 are white, 6 are yellow, and 6 are red. If a juggler selects three objects to juggle, what is the probability that:
a) All three objects are the same color?
b) All three objects are different colors?
Part (a) Solution:
*
*There are a total of 24 distinct objects and three of them are to be selected, so the sample space is $C(24,3)$.
*Since all three objects to be picked are to be of the same color, one out of four colors must be selected, $C(4,1)$.
*Now three objects must be selected from the six available objects of that color, $C(6,3)$.
Part (a) Answer: $$P = \frac{C(4,1) \cdot C(6,3)}{C(24,3)}$$
Part (b) Solution:
*
*There are a total of 24 distinct objects and three of them are to be selected, so the sample space is $C(24,3)$.
*Three colors must be picked and there are four choices for the first color, $C(4,1)$.
*There are six balls of that color to choose from and only one must be picked, $C(6,1)$.
*There are three choices for the second color, $C(3,1)$, and six balls to choose from, $C(6,1)$.
*There are two choices for the third color, $(2,1)$ and six balls to choose from, $C(6,1)$.
Part (b) Answer: $$P = \frac{C(4,1) \cdot C(6,1) + C(3,1) \cdot C(6,1) + C(2,1) \cdot C(6,1)}{C(24,3)}$$
I'm definitely unsure about the (b) solution.
|
What is the probability that all three objects are the same color.
Your answer is correct.
Alternatively, we choose a ball. The probability that the second ball is the same color as the first is $\frac{5}{23}$. The probability that the third ball is the same color as the first two is $\frac{4}{22}$. Hence, the probability is
$$p = 1 \cdot \frac{5}{23} \cdot \frac{4}{22}$$
What is the probability that all three objects are different colors.
Your denominator is correct. We choose three of the four colors, then choose one object from each of the selected colors. Hence, the probability is
$$p = \frac{\dbinom{4}{3}\dbinom{6}{1}^3}{\dbinom{24}{3}}$$
Alternatively, we choose a ball. The probability that the second ball we select is of a different color than the first is $\frac{18}{23}$. The probability that the third ball we select is of a different color than the first two is $\frac{12}{22}$. Hence, the probability is
$$p = 1 \cdot \frac{18}{23} \cdot \frac{12}{22}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}}-\frac{1}{S_n}$ for partials sums of a divergent series Let $(a_n)$ be a sequence of non-negative numbers such that $a_1 > 0$ and $\sum a_n$ diverges. Let $S_n = \sum_{k=1}^n a_k$. Prove that, for all $n \geq 2$,
$$\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}}-\frac{1}{S_n}$$
How would I start this proof? I've just been staring at it and am very stuck. All i know so far is that $S_n-S_{n-1}=a_n$. Where does the inequality come from?
| You can prove this result by induction. First, in the case $n=2$,
\begin{align*}
\frac{a_2}{S^2_{2}} \leq \frac{1}{S_{1}}-\frac{1}{S_{2}},
\end{align*}
which follows that $\frac{1}{S_{1}}-\frac{1}{S_{2}}= \frac{S_2-S_1}{S_1S_{2}}=\frac{a_2}{S_1S_{2}}$ and $S_2 \geq S_1$.
Suppose that \begin{align*}
\frac{a_n}{S^2_{n}} \leq \frac{1}{S_{n-1}}-\frac{1}{S_{n}}.
\end{align*}
Then
\begin{align*}
\frac{a_{n+1}}{S^2_{n+1}} = \frac{a_{n+1}}{(S_{n}+a_{n+1})^2} \leq \frac{a_{n+1}}{S_{n}S_{n+1}}=\frac{1}{S_{n}}-\frac{1}{S_{n+1}}.
\end{align*}
The proof is completed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$
Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that
$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$$
I tried various methods. But, couldn't solve it. It'd be great if anyone can help.
| The original post was to show that
$$\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge
\sqrt{a}+\sqrt{b}+\sqrt{c}+3.
$$
Observe that
\begin{align}
\frac{a}{b}+\frac{c}{a}&\ge2\sqrt{\frac{c}{b}}=2c\sqrt{a},\\
\frac{b}{a}+\frac{c}{b}&\ge2\sqrt{\frac{c}{a}}=2c\sqrt{b},\\
\frac{c}{a}+\frac{c}{b}&\ge2\sqrt{\frac{c^2}{ab}}=2c\sqrt{c}.
\end{align}
and similarly
\begin{align}
&\frac{a}{c}+\frac{b}{a}\ge2b\sqrt{a},
&\frac{b}{a}+\frac{b}{c}\ge2b\sqrt{b},
&&\frac{c}{a}+\frac{b}{c}\ge2b\sqrt{c};\\
&\frac{a}{b}+\frac{a}{c}\ge2a\sqrt{a},
&\frac{b}{c}+\frac{a}{b}\ge2a\sqrt{b},
&&\frac{c}{b}+\frac{a}{c}\ge2a\sqrt{c}.
\end{align}
So, by combining them, and use the fact that $a+b+c\ge 3\sqrt[3]{abc}=3$, we have
\begin{align}
3\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)
&\ge 2(a+b+c)\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\\
&\ge 6\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right).
\end{align}
That is,
$$\frac{1}{2}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}.\tag{1}$$ Next, we also have
\begin{align}
\frac{1}{2}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)
&= \frac{1}{2}\left[\left(\frac{b}{a}+\frac{a}{b}\right)
+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\right]\\
&\ge \frac{1}{2}(2+2+2)\\
&=3.\tag{2}
\end{align}
Hence the result follows by combining $(1)$ and $(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's rule, that is,
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x} = \lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 }$$
since $\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \frac {1 - 1}{ 1- 1} = \frac00$. Thus, we use the L'Hopital's rule again. that is,
$$\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \lim_{x\rightarrow 0}\frac{-\sin x}{2 \sec^2 x \sec x \tan x}$$
since $\lim_{x\rightarrow 0} \frac{-\sin x}{2 \sec^2 x \sec x \tan x}= \frac{0}{2(1)(0)} = \frac00$. Thus, it always goes to zero by zero. But the answer for this question is $\frac{-1}2$. How is that?!! there must be something missing that either I forget or misunderstand.
| Notice that $\tan x = \sin x \sec x$; so we'll have:
$$\lim_{x \to 0} \frac{-\sin x}{2\sec^2x\sec x \tan x} = \lim_{x \to 0} \frac{-\sin x}{2\sec^2x\sec^2 x \sin x} = \lim_{x \to 0} \frac{-\sin x}{2\sec^4x\sin x} = \lim_{x \to 0} \frac{-1}{2\sec^4x} = \cdots$$
Hope you can go from here. :x
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Without using modular equivalence, show that: $\gcd(4n^2+1,24)=1$ Without using modular equivalence, show that: $\gcd(4n^2+1,24)=1$
Let $d=\gcd(4n^2+1,24)$ then we have:
$$d|24n^2+6,24n^2\ \Rightarrow\ d|6\ \Rightarrow\ d|6n^2,4n^2+1\ \Rightarrow\ d|12n^2,12n^2+3\ \Rightarrow\ d|3\ \Rightarrow\ d=1\ or\ 3$$
Using modular equivalence it's very easy to show that $d$ can't be 3,but how can I show it WITHOUT using modular equivalence???
| Theorem. $4n^2 + 1$ is not a multiple of $3$.
Proof. Since $n^2 = (-n)^2$, we need only prove this for
$n \in \mathbb W = \{0, 1, 2, \dots\}$.
Let $T = \{k\in \mathbb W : 3 | 4k^2 + 1\}$.
We need to show that $T$ is the empty set.
If we assume that $T$ is not the empty set, then it must contain a smallest member, say $s$. (This is called "The Well-Ordering Principle".) Then $s$ is the smallest member of $\mathbb W$ such that $3 \mid 4s^2 + 1$.
Note that
\begin{align}
4\cdot 0^2 + 1 &= 1 \\
4\cdot 1^2 + 1 &= 5 \\
4\cdot 2^2 + 1 &= 17 \\
\end{align}
So $s-3 \in \mathbb W$ and $s-3 \not \in T$. But if $3 \mid 4s^2 + 1$ and
$3 \not \mid 4(s-3)^2 + 1$, then
\begin{align}
3 &\not \mid (4s^2 + 1) - (4(s-3)^2 + 1) \\
3 &\not \mid 4(s^2 - (s-3)^2)\\
3 &\not \mid 4(6s - 9)\\
\end{align}
But clearly $3 \mid 4(6s - 9)$. So by contradiction, $T$ is the empty set.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum of the function $f(x_1,x_2) = (x_1 + x_2 + 3)^2 + (3x_1 + x_2 - 1)^2 + (2x_1 + x_2 + 1)^2$ Find the minimum of the function $f(x_1,x_2) = (x_1 + x_2 + 3)^2 + (3x_1 + x_2 - 1)^2 + (2x_1 + x_2 + 1)^2$.
We've been instructed to solve this by writing $f(x_1,x_2)$ as $||Ax - b||^2$, and $x = [x_1,x_2]^T$.
I'm not quite sure how to solve this, this is $Ax - b$ in my opinion:
$
\begin{pmatrix}
1 & 1 \\
3 & 1 \\
2 & 1 \\
\end{pmatrix}
$
$(x_1,x_2)
^T - (-3,-1,1)$
How do I continue from here?
| $$
\begin{align}
&\nabla\left[(x_1+x_2+3)^2+(3x_1+x_2-1)^2+(2x_1+x_2+1)^2\right]\\
&=2(x_1+x_2+3)\begin{bmatrix}1\\1\end{bmatrix}
+2(3x_1+x_2-1)\begin{bmatrix}3\\1\end{bmatrix}
+2(2x_1+x_2+1)\begin{bmatrix}2\\1\end{bmatrix}\\
&=\begin{bmatrix}28x_1+12x_2+4\\12x_1+6x_2+6\end{bmatrix}
\end{align}
$$
This vanishes when $x_1=2$ and $x_2=-5$. Plugging this point into the function gives
$$
0^2+0^2+0^2=0
$$
Knowing this, we can write
$$
\begin{align}
&(x_1+x_2+3)^2+(3x_1+x_2-1)^2+(2x_1+x_2+1)^2\\
&=((x_1-2)+(x_2+5))^2+(3(x_1-2)+(x_2+5))^2+(2(x_1-2)+(x_2+5))^2\\
&=14(x_1-2)^2+12(x_1-2)(x_2+5)+3(x_2+5)^2\\
&=2(x_1-2)^2+3(2(x_1-2)+(x_2+5))^2\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solution to exponential congruence Is there a clever solution to the congruence without going through all the values of x up to 58?$$2^x \equiv 43\pmod{59}$$
Can I somehow use the fact that $2^4 \equiv -43\pmod{59}$ ?
| We know that $2$ is a quadratic non-residue of $59$ from a theorem that states that $2$ is a quadratic residue of $p$, where $p$ is a prime if and only if $\frac{p^2-1}{8}$ is even. Obviously $\frac{59^2-1}{8}$ is odd.
Also, there is another theorem stating that $a$ is a quadratic residue of $p$ (prime) if and only if $a^{\frac{p-1}{2}} \equiv 1 \pmod p$. Otherwise, $a^{\frac{p-1}{2}} \equiv -1 \pmod p$. (this is known as the Euler's Criterion)
Then, just set $a=2$ and $p=59$ and we have $2^{29} \equiv -1 \pmod{59}$. Therefore $2^4\times2^{29} \equiv(-43)(-1) \equiv 43 \pmod {59}$ i.e. $2^{33} \equiv 43 \pmod{59}$.
Are we done yet? Nop!
We want to find every such $x$.
Let y be another solution. Then:
$43 \times 2^{y-33} \equiv 2^{33} \times 2^{y-33} \equiv 2^{y} \equiv 43 \pmod{59}$. Note that this works even if $y$ is smaller than $33$ -there is no such solution but there is no need to prove it.
So by cancelation ($\gcd(43,59)=1$ so we can do this) we end up with $2^{y-33} \equiv 1 \pmod{59}$. $(1)$
There is another known result, stating that the minimum solution of the equation $a^x \equiv 1 \pmod{n}$ for some $a,n$ in positive integers divides all the others. Let $d$ be the minimum solution of $2^z \equiv 1 \pmod{43}$.
Also, by Fermat Little Theorem $2^{58} \equiv 1 \pmod{59}$. By the result mentioned before, $z|58=2\times29$. Since $2^1\equiv 1 \pmod{59}$, $2^2 \equiv 4 \pmod{59}$ and $2^{29} \equiv -1 \pmod{59}$ (proved before), we have $z=58$.
Note also that from $(1)$ we have that $y$ is a solution of our equation if and only if $58|y-33$.
Therefore, the solutions are given by $x=33+58t$ where $t$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Finding out a limit using Taylor series. So the limit is the following:
$$\lim_{x \to 0}{\frac{x^2-\frac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}}$$
Expansions for $\sin(x)$ and $\cos(x)$ are given:
$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} + o(x^{2n})$$
$$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-...+(-1)^n\frac{x^{2n}}{(2n)!}+o(x^{2n+1})$$
Here is what I tried:
$$\lim_{x->0}{\frac{x^2-\frac{x^6}{2}-x^2(1-\frac{x^4}{2}+\frac{x^8}{4!}+o(x^{2*5}))}{x^{10}+o(x^{10*2})}}=\lim_{x->0}{\frac{-\frac{x^{10}}{4!}-o(x^{12})}{x^{10}+o(x^{20})}}$$
This is where I am stuck. I figured that the problem occurs when expending $o()$. What am I missing here?
| Another way is to note that $\frac{\sin x^{10}}{x^{10}} \to_x 1$, so you get your $x^{10}$ in the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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$a,b$ are positive integers . Prove that if $k=\frac{a^2+b^2}{a+b+1}$ is an integer then $k=5$. This is the original question that was posted and answered. For the question the OP intended to ask, please see the edit history.
$a,b$ are positive integers. Prove that if $k=\frac{a^2+b^2}{a+b+1}$ is an integer then $k=5$.
I tried to see the whole expression as a quadratic of $a$ but that is not helping much.
| In this case, if $k=5,$ then we would have $70$ as the sum of two integer squares, which it is not.
$$ a^2 + b^2 = 5a + 5b + 5, $$
$$ 4a^2 + 4 b^2 = 20a + 20 b + 20, $$
$$ 4 a^2 - 20a + 4 b^2 - 20 b = 20, $$
$$ 4 a^2 - 20a + 25 + 4 b^2 - 20 b + 25 = 20 + 25 + 25 = 70, $$
$$ (2a-5)^2 + (2b-5)^2 = 70. $$
However, $70 = 2 \cdot 5 \cdot 7$ is not the sum of two integer squares.
A superficially similar problem that does lead to $5$ is
$$ \frac{a^2 + b^2}{ab-1} = k, $$
see Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers? and my three or so answers there. I did, with help, eventually settle the whole thing.
If the version I have guessed is really the original problem, then it fits under https://en.wikipedia.org/wiki/Vieta_jumping
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A Chinese Exam Question which is.....quite hard Let $f(x)=x^2-2x-3$, and $x_n$ be some sequence.
$x_1=2$, $x_n =$ the $x$ coordinate of the point of intersection of the $x$ axis and the line joining $P(4,5)$ and $Q_n(x_n, f(x_n))$.
Find an expression for $x_n$
Find $\lim \limits_{n\to\infty} Q_n(x_n,f(x_n))$.
| I suppose you mean $x_1=2$ and $x_{n+1}=$ the $x$ coordinate of the point of intersection of the $x$ axis and the line joining $P(4,5)$ and $Q_n(x_n,f(x_n))$.
Now the equation of the line joining $Q_n(x_n,f(x_n))$ and $P(4,5)$ is
$$\frac{y-5}{x-4}=\frac{f(x_n)-5}{x_n-4}=\frac{x_n^2-2x_n-3-5}{x_n-4}$$
Setting $y=0$ to find $x_n+1$, we have
$$x_{n+1}-4=-\frac{5}{x_n^2-2x_n-8}(x_n-4)$$
Define $z_n=x_n-4$, we have $z_1=-2$ and
$$z_{n+1}=-\frac{5}{z_n+6}$$
Now, if $-2\le z_n\le-1$, one can show that $-2 \le z_{n+1} \le -1$ as well.
In particular, the $z_n$ sequence is bounded from above by $-1$.
Now consider
$$z_{n+1}-z_n=-\frac{5}{z_n+6}-z_n$$
$$=-\frac{(z_n+1)(z_n+5)}{z_n+6}$$
Since $-2 \le z_n \le -1$, we have
$$z_{n+1}-z_n \ge 0$$
So the $z_n$ sequence is non-decreasing and bounded from above and hence limit exists. We can then take the limit of both sides of
$$z_{n+1}=-\frac{5}{z_n+6}$$
to get
$$z=-\frac{5}{z+6}$$
which implies $z=-1$ ($z=-5$ rejected).
Hence,
$$\lim_{n\rightarrow \infty}x_n=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can we show that $ \sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4 ?$ We proposed this sum, but we are lacking in knowledge of this area of maths and we would ask if any of the authors would be willing to show us step by step how to go about proving this sum.
$$
\sum_{n=0}^{\infty}\frac{2^nn[n(\pi^3+1)+\pi^2](n^2+n-1)}{(2n+1)(2n+3){2n \choose n}}=1+\pi+\pi^2+\pi^3+\pi^4
$$
| Generally the following sums may help:
$$
\sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}}=4 \left(\frac{1}{4-x}+\frac{\sqrt{x} \arcsin\left(\frac{\sqrt{x}}{2}\right)}{(4-x)^{3/2}}\right)
$$
$$
\sum_{n=0}^\infty\frac{nx^n}{\binom{2n}{n}}=\frac{\partial}{\partial x}\sum_{n=0}^\infty\frac{x^{n+1}}{\binom{2n}{n}}-\sum_{n=0}^\infty\frac{x^{n}}{\binom{2n}{n}}
$$
$$
\sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}(2n+1)}=\frac{4 \arcsin \left(\frac{\sqrt{x}}{2}\right)}{\sqrt{(4-x) x}}
$$
$$
\sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}(2n+3)}=-\frac{4 \left(2 \sqrt{(4-x) x}+(x-8) \sin ^{-1}\left(\frac{\sqrt{x}}{2}\right)\right)}{\sqrt{4-x} x^{3/2}}$$
Derivation of the third line from the first is as follows:
$$
\sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}}=f(x)
$$
$$
C+\sum_{n=0}^\infty\frac{x^{n}}{\binom{2n}{n}(2n+3)}=\frac{1}{x^3}\int\sum_{n=0}^\infty\frac{x^{n+2}}{\binom{2n}{n}}\mathrm{d}x=\frac{1}{x^3}\int f(x)x^2\mathrm{d}x
$$
Where $C$ should be matched to the value at $x=0$+ there are some convergence criteria that should be checked. Also
$$
\frac{n[n(y^3+1)+y^2](n^2+n-1)}{(2n+1)(2n+3)}=\frac{1}{2}\frac{n[n(y^3+1)+y^2](n^2+n-1)}{(2n+1)}-\frac{1}{2}\frac{n[n(y^3+1)+y^2](n^2+n-1)}{(2n+3)}
$$
And then:
$$
\frac{1}{2}\frac{n[n(y^3+1)+y^2](n^2+n-1)}{(2n+1)}=\frac{n^3 \left(8 y^3+8\right)+n^2 \left(-4 y^3+8 y^2-4\right)+n \left(-2 y^3-4 y^2-2\right)+3 y^3-2 y^2+3}{32}+\frac{-9y^3+6y^2-9}{2n+1}
$$
Applying these, and a few more similarly obtained formulas, may help. There might be a better approach, with less calculation. (and there may be some mistakes). Finally plug in $x=2$ and $y=\pi$
| {
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"question_score": "13",
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Let $f(x)=x^5+x^2+1$ with $x_1,x_2,x_3,x_4,x_5$ as zeros and ... Let $f(x)=x^5+x^2+1$ with $x_1,x_2,x_3,x_4,x_5$ as zeros and $g(x)=x^2-2.$ Show that $$g (x_1)g (x_2)g (x_3)g (x_4)g (x_5)-30g(x_1x_2x_3x_4x_5)=7$$.
I found this question in a local question paper. And I have no idea how to solve it...
Please Help.
| It is known that $$x^2+1+x^5=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)\tag1$$
Comparing the constants of each side, we have $-1=x_1x_2x_3x_4x_5$.
Now, puting in $-x$ in $\text{(1)}$, we have $$x^2+1-x^5=-(x+x_1)(x+x_2)(x+x_3)(x+x_4)(x+x_5)\tag 2$$Multipling $\text{(1)}$ and $\text{(2)}$, we have $$h(x)=(x^2+1)^2-x^{10}=(x_1^2-x^2)(x_2^2-x^2)(x_3^2-x^2)(x_4^2-x^2)(x_5^2-x^2)$$
Now put in $x=\sqrt{2}$. We have $$g (x_1)g (x_2)g (x_3)g (x_4)g (x_5)-30g(x_1x_2x_3x_4x_5)=h(\sqrt{2})+30=9-32+30=7$$
| {
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"source": "stackexchange",
"question_score": "2",
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Evaluation of $\lim_{n\rightarrow \infty}\frac{1}{n^{2m}}\left[(n^2+1^2)^{m}(n^2+2^2)^m(n^2+3^2)^m..............(2n^2)^{m}\right]^{\frac{1}{n}}$
Evaluation of $$\lim_{n\rightarrow \infty}\frac{1}{n^{2m}}\left[(n^2+1^2)^{m}(n^2+2^2)^m(n^2+3^2)^m..............(2n^2)^{m}\right]^{\frac{1}{n}}$$
$\bf{My\; Try::}$ Let $$L=\lim_{n\rightarrow \infty}\frac{n^{\frac{2m}{n}}}{n^{2m}}\left[\left(\frac{n^2+1^2}{n^2}\right)^m\cdot \left(\frac{n^2+2^2}{n^2}\right)^m\cdot ..........\left(\frac{n^2+n^2}{n^2}\right)^m\right]^{\frac{1}{n}}$$
Altgough I know that we use Riemann Sum of Integral, But I did not understand How can
I Adjust into $\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{r=1}f\left(\frac{r}{n}\right)\cdot \frac{1}{n}$
Help Required, Thanks
| If we set
$$ \color{blue}{L}=\lim_{n\to +\infty}\left[\left(1+\frac{1^2}{n^2}\right)\cdot\left(1+\frac{2^2}{n^2}\right)\cdot\ldots\cdot\left(1+\frac{n^2}{n^2}\right)\right]^{1/n}\tag{1}$$
the original limit is just $L^m$. But:
$$ \log L = \lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left[1+\left(\frac{k}{n}\right)^2\right]=\int_{0}^{1}\log(1+x^2)\,dx\tag{2}$$
by Riemann sums, hence it is enough to evaluate the last integral. By integration by parts:
$$ \int_{0}^{1}\log(1+x^2)\,dx = \left. x\log(1+x^2)\right|_{0}^{1}-\int_{0}^{1}\frac{2x^2\,dx}{1+x^2}=\log(2)-2+\frac{\pi}{2}, \tag{3}$$
hence:
$$ \color{blue}{L} = \color{red}{2\cdot\exp\left(\frac{\pi}{2}-2\right)}.\tag{4}$$
| {
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Under certain conditions $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}\Rightarrow \{a,b,c\}=\{a',b',c'\}$ Let $a,b,c,a',b',c'\in \mathbb{Z}_{\geq 1}$ be such that
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<1,\quad \frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}<1.
$$
Suppose
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a'}+\frac{1}{b'}+\frac{1}{c'}.
$$
Is it true that $\{a,b,c\}=\{a',b',c'\}$? If it is not, it should be easy to give a counterexample, but I am not able to find one.
| For the more general case when $a\neq b\neq c$ and $a'\neq b'\neq c'$, the answer is also no.
In particular, let $\frac1a +\frac1b = \frac{1}{a'}$, and $\frac{1}{c} = \frac{1}{b'} +\frac{1}{c'}$. Then $a' = \frac{ab}{a + b}$, and $c = \frac{b' c' }{b'+ c'}$. We'd want these to be integers.
A simple way of making $\frac{ab}{a + b}$ an integer is by setting one to $a = k+1$ and $b = k (k + 1)$. In this way, for $k = 2$, we can have $\frac{1}{3} + \frac{1}{6} = \frac{1}{2}$; and for $k = 4$, $\frac{1}{5} + \frac{1}{20} = \frac{1}{4}$. We have a counter-example:
$$\frac{1}{3} + \frac{1}{6} + \frac{1}{4} = \frac{1}{5} + \frac{1}{20} + \frac{1}{2} = \frac{3}{4}$$
| {
"language": "en",
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$A^2+B^2=AB$ and $BA-AB$ is non-singular The question is:
Are there square matrices $A,B$ over $\mathbb{C}$ s.t. $A^2+B^2=AB$ and $BA-AB$ is non-singular?
From $A^2+B^2=AB$ one could obtain $A^3+B^3=0$. Can we get something from this?
Edit:
$$A^2+B^2=AB\implies\\
A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\
A^3+(A^2+B^2)B=A^2B \implies \\
A^3+A^2B+B^3=A^2B \implies \\
A^3+B^3=0
$$
| We consider the equations
(1) $A^2+B^2=AB$,
(2) $A^2+B^2=2AB$.
A couple $(A,B)$, solution of (2), is simultaneously triangularizable, and, moreover, when $n=2$, $AB=BA$.
Eq. (1) has not the second property: indeed $$A=\begin{pmatrix}0&-1\\1&-1\end{pmatrix},\ B=\begin{pmatrix}\dfrac{1-i\sqrt{3}}{2}&0\\\dfrac{3+i\sqrt{3}}{2}&-1\end{pmatrix}$$ is a solution s.t. $AB\not= BA$. Note that $\mathrm{spectrum}(A)=\{e^{\pm 2i\pi/3}\}$ and $A^3=I_2$, $B^3=-I_2$; moreover $(AB-BA)^2=0$. I think that Eq. (1) has the first property, or at least, $AB-BA$ is nilpotent, but I don't know how to prove it. We see, in the following, that, a solution, in general position, satisfies $AB=BA$.
Proposition. We consider the equation in the unknown $X$: $X^2-AX+A^2=0$. Let $\mathrm{spectrum}(A)=(\lambda_i)_i$ denote the complete spectrum of $A$. If, for every $i\not= j$ $\lambda_i^3\not=\lambda_j^3$, then any solution $X$ satisfies $AX=XA$.
Proof. Since $A$ is diagonalizable we may assume $A=\mathrm{diag}(\lambda_i)$. Thus $A^3=\mathrm{diag}(\lambda_i^3)$ has distinct eigenvalues; since $X$ and $A^3$ commute, $X$ is a diagonal matrix and we are done.
EDIT. There are solutions of $A^3+B^3=0$ s.t. $AB-BA$ is invertible; for example, choose $A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},\ B=\begin{pmatrix}0&0&0\\1&0&0\\0&-1&0\end{pmatrix}$; of course $A^2+B^2\not= AB$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How to Find $ \lim\limits_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$. Can someone help me with this limit? I'm working on it for hours and cant figure it out.
$$ \lim_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$$
I started transforming to the form
$ \lim_{x\to 0} e^{ {\frac{\ln \left(\frac {\tan x}{x} \right)}{x^2}} }$
and applied the l'Hopital rule (since indeterminated $\frac00$), getting:
$$ \lim_{x\to 0} \left( \frac{2x-\sin 2x }{2x^2\sin 2x} \right)$$
From here, I try continue with various forms of trigonometric substitutions, appling the l'Hopital rule again and again, but no luck for me. Can someone help me?
| In the same spirit as other answers, consider $$A=\left(\frac {\tan (x) }{x} \right)^{\frac{1}{x^2}}$$ Take logarithms $$\log(A)=\frac{1}{x^2}\log\left(\frac {\tan (x) }{x} \right)$$ Now, consider Taylor series $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ $$\frac {\tan (x) }{x}=1+\frac{x^2}{3}+\frac{2 x^4}{15}+O\left(x^6\right)$$ $$\log\left(\frac {\tan (x) }{x} \right)=\log\left(1+\frac{x^2}{3}+\frac{2 x^4}{15}+O\left(x^6\right)\right)$$ Now use $\log(1+y)=y-\frac{1}{2}y^2+O\left(y^3\right)$ and replace $y$ by $\frac{x^2}{3}+\frac{2 x^4}{15}$ to get $$\log(A)=\frac{1}{x^2} \left(\frac{x^2}{3}+\frac{7 x^4}{90}+O\left(x^5\right)\right)=\frac{1}{3}+\frac{7 x^2}{90}+O\left(x^3\right)$$ which shows the limit and also how it is approached.
You can continue using $A=e^{\log(A)}$ and still using Taylor arrive to $$A=\sqrt[3]{e}+\frac{7}{90} \sqrt[3]{e} x^2+O\left(x^3\right)$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 10,
"answer_id": 7
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If $p>5$ is prime, $2p+1$ is a prime, $\frac{4p+1}{3}$ is prime, $8p+1$ is prime, Then $p \equiv 29 (mod \; 30)$ Assume that $p>5$ is prime, $2p+1$ is a prime, $\frac{4p+1}{3}$ is prime, $8p+1$ is prime. Then I want to prove that $p \equiv 29 (mod \; 30).$
First of all I have to show that $4p+1$ is a multiple of 3 if $2p+1$ is prime and $p > 5.$
Any help will be appreciated.
Thank you so much.
| We have $p\equiv \pm 1\pmod{3}$. But if $p\equiv 1\pmod{3}$ then $2p+1$ is divisible by $3$, so $p\equiv -1\pmod{3}$, and therefore $p\equiv 5\pmod{6}$.
Also, $p$ is congruent to one of $1,2,3,4$ modulo $5$. Since $2p+1$ is prime, we cannot have $p\equiv 2\pmod{5}$. Since $\frac{4p+1}{3}$ is prime, $p$ cannot be congruent to $1$ modulo $5$. And since $8p+1$ is prime, $p$ cannot be congruent to $3$ modulo $5$. So $p\equiv 4\pmod{5}$.
Since $p\equiv 5\pmod{6}$ and $p\equiv 4\pmod{5}$, we have $p\equiv 29\pmod{30}$.
| {
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"source": "stackexchange",
"question_score": "2",
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Prove that $a^2b+b^2c+c^2a \leqslant 3$ for $a,b,c >0$ with $a^ab^bc^c=1$
Let $a,b,c >0$ and $a^ab^bc^c=1$. Prove that
$$a^2b+b^2c+c^2a \leqslant 3.$$
I don't even know what to do with the condition $a^ab^bc^c=1$. At first I think $x^x>1$, but I was wrong. This inequality is true, following by the verification from Mathematica
| We shall prove the following inequalities:
If $x+y+z=3$ and $x,y,z>0$, then
$$\tag{1}x^2y+y^2z+z^2x+xyz\le 4\label{1},$$
$$\tag{2}3x^xy^yz^z+xyz\ge 4\label{2},$$
i.e.,
$$\tag{*}x^2y+y^2z+z^2x\le 3x^xy^yz^z\label{*}.$$
Accept $\eqref{*}$ for a moment. Homogenizing the inequality by substitution
$$x=\frac{3a}{a+b+c},~y=\frac{3b}{a+b+c},~z=\frac{3c}{a+b+c},$$
we have $\forall a,b,c>0$,
$$\tag{**}a^2b+b^2c+c^2a\le 3(a^ab^bc^c)^{3/(a+b+c)}.\label{**}$$
If $a^ab^bc^c=1$, then we get the original inequality as desired.
Proof of $\eqref{1}$:
Note the LHS is cyclic, we cannot assume a specific order like $x\le
y\le z$, but we can assume WLOG that $y$ is in the middle (neither the minimum nor the maximum), then
$$z(y-x)(y-z)\le 0\\
\Rightarrow x^2y+y^2z+z^2x+xyz\le y(x^2+2xz+z^2)=\frac{1}{2}(2y)(x+z)^2\le\frac{1}{2}\left(\frac{2(x+y+z)}{3}\right)^3=4,$$
by AM-GM. Equality holds when $(x,y,z)=(1,1,1),(2,1,0)$ along with the cyclic permutations.
Proof of $\eqref{2}$:
WLOG we assume $x=\min(x,y,z)$ and consider the following two cases:
*
*$x\ge 1/3$, then $y,z\ge 1/3$. We note that the function
$$f(t)=\left(t+\frac{1}{3}\right)\ln t$$
is convex for $t\ge 1/3$, as
$$f''(t)=\frac{3t-1}{3t^2}\ge 0.$$
By Jensen,
$$
\begin{align}
&~~~~~~~x^{x+1/3}y^{y+1/3}z^{z+1/3}=\exp\left\{f(x)+f(y)+f(z)\right\}
\ge\exp\left\{3f\left(\frac{x+y+z}{3}\right)\right\}=\exp\left\{3f(1)\right\}=1,\\
&\Rightarrow x^xy^yz^z\ge (xyz)^{-1/3},\\
&\Rightarrow 3x^xy^yz^z+xyz\ge 3(xyz)^{-1/3}+xyz\ge 4\sqrt[4]{(xyz)^{-1}(xyz)}=4,
\end{align}
$$
where we have applied AM-GM for the last line. Equality holds when $x=y=z=1$.
*
*$0<x\le1/3$. We note that the function
$$g(t)=t\ln t$$
is convex for $t>0$, as
$$g''(t)=\frac{1}{t}> 0.$$
Again by Jensen, we have
$$
\begin{align}
&~~~~~~~y^yz^z=\exp\left\{g(y)+g(z)\right\}\ge\exp\left\{2g\left(\frac{y+z}{2}\right)\right\}
=\exp\left\{2g\left(\frac{3-x}{2}\right)\right\},\\
&\Rightarrow x^xy^yz^z\ge\exp\left\{g(x)+2g\left(\frac{3-x}{2}\right)\right\}
\ge\exp\left\{g\left(\frac{1}{3}\right)+2g\left(\frac{4}{3}\right)\right\}
=\frac{32}{27}\sqrt[3]{2}>\frac{4}{3},\\
&\Rightarrow 3x^xy^yz^z+xyz>4+xyz>4.
\end{align}
$$
The second line follows from that $h(t):=g(t)+2g((3-t)/2)$ is monotonically
decreasing for $t\in(0,1)$, so $h(x)\ge h(1/3)$ as $x\le 1/3$. Proof of monotonicity:
$$h'(t)=\ln\left(\frac{2t}{3-t}\right)<0\iff 0<t<1.$$
Post-mortem:
*
*Despite the flow of the proof, the starting point is the homogeneous version \eqref{**};
*\eqref{1} is due to Vasile, which allows us to strengthen the inequality into a
symmetric one as \eqref{2} (I am not able to find the originial post on artofproblemsolving, but see e.g.,
here). It seems that converting \eqref{*} into other symmetric form such as $x^3+y^3+z^3$ is too strong for it to hold;
*For \eqref{2}, the case of $x\le 1/3$ is unfortunately necessary, as
$x^xy^yz^z\ge(xyz)^{-1/3}$ does NOT always hold, e.g., check $x\to0,y-z\to 0$. Is there any better estimate of $x^xy^yz^z$ so we can avoid the derivatives? Another follow-up question is to determine the smallest $k$ such that $kx^xy^yz^z+xyz\ge k+1$ holds given the constraints. It definitely fails for $k\le 1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Evaluation of $\cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$
Evaluation of $$\lim_{n\rightarrow \infty}\left(\tan \frac{\pi}{2n}\cdot \tan \frac{2\pi}{2n}\cdot \tan \frac{\pi}{3n}\cdot ...............\tan \frac{(n-1)\pi}{2n}\right)^{\frac{1}{n}} = $$ without using Limit as a sum.
$\bf{My\; Try::}$ Using the formula $$\displaystyle \sin\left(\frac{\pi}{n}\right)\cdot \sin\left(\frac{2\pi}{n} \right)....\sin\left(\frac{(n-1)\pi}{n}\right) = \frac{n}{2^{n-1}}$$
Replace $n\rightarrow 2n$
$$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(2n-1)\pi}{2n}\right) = \frac{2n}{2^{2n-1}}$$
Now How can I calculate $$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right)$$
and also How can I calculate $$\displaystyle \cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$$
Help required, Thanks
| First we have
\begin{align}
\lim_{n\to \infty}\frac1{n}\ln{\left(\tan\frac{\pi}{2n}.\tan\frac{2\pi}{2n}\cdots\tan\frac{(n-1)\pi}{2n}\right)}
&=\lim_{n\to \infty}\frac1{n}\sum\limits_{k=0}^{n-1}\ln{\left(\tan\frac{k\pi}{2n}\right)}
\\
&=\frac{2}{\pi}\int_0^{\pi/2}\ln{\tan{x}}\:dx
\\
&=\frac{2}{\pi}\left(\int_0^{\pi/2}\ln\sin(x)dx -\int_0^{\pi/2}\ln\cos(x)dx\right)
\\
&=0
\end{align}
The last step is through $y=\pi/2-x$.
Hence
$$
\lim_{n\to\infty}\left(\tan\frac{\pi}{2n}.\tan\frac{2\pi}{2n}\cdots\tan\frac{(n-1)\pi}{2n}\right)^{\frac1{n}}=e^0=1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Splitting fractions with a linear denominator: $\frac{2x-1}{x+2}$ How can $$\frac{2x-1}{x+2}$$ be split to give $$A-\frac{B}{x+2}$$
where $A$ and $B$ are integers?
The solution is $$2-\frac{5}{x+2}.$$
| $\dfrac{2x-1}{x+2}:$
\begin{array}{ccccccc}
&&&& 2 \\
&&&& -- & -- & --\\
x & + & 2 & | & 2x & - & 1 \\
&&&&2x & + & 4 \\
&&&& -- & -- & --\\
&&&&&&-5
\end{array}
So $\dfrac{2x-1}{x+2} = 2 + \dfrac{-5}{x+2} = 2 - \dfrac{5}{x+2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to differentiate this fraction $\frac{2}{x^2+3^3}$? $\frac{2}{(x^2+3)^3}$.
I have ${dy}/{dx}$ x 2 x ${x^2+3^3}$ - 2 x ${dy}/{dx}$ x ${x^2+3^3}$ over $({x^2+3)^6}$
And then simplifying to $-12x^5 + 36x^2$ over $({x^2+3)^6}$
I'm not sure if this is right.
| You can use the quotient rule or the product rule. Let's use the quotient rule for this:
$$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'g-g'f}{g^2}$$
In your case we have: $f(x)=2 \implies f'(x)=0$ and $g(x)=x^2+3^3 \implies g'(x)=2x$
It follows:
$$\implies \frac{f'g-g'f}{g^2}=\frac{0\cdot (x^2+3^3)-2x\cdot2}{(x^2+3^3)^2}=\frac{-4x}{(x^2+3^3)^2}$$
The other approach would be to use the product rule:
$$\frac{d}{dx}(f(x)\cdot g(x))=f'g+g'f$$
In your case we have $$\frac{2}{x^2+3^3}=2(x^2+3^3)^{-1}$$
The corresponding functions and derivatives are: $f(x)=2 \implies f'(x)=0$ and $g(x)=(x^2+3^3)^{-1} \implies g'(x)=-2x(x^2+3^3)^{-2}$ (I used the chain rule here)
$$\implies f'g+g'f=0 \cdot (x^2+3^3)^{-1}-2x(x^2+3^3)^{-2}\cdot 2=\frac{-4x}{(x^2+3^3)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Trigonometric identities involved $\sin(\alpha)$, $\cos(\alpha)$, $\tan(\alpha)$ $\alpha$ is valid between $-90<\alpha<90$ Degrees
Show that,
$$ \frac{\tan^6(\alpha)-\tan^4(\alpha)+2}{\tan^6(\alpha)-2\tan^2(\alpha)+4 }\cdot\cos^2(\alpha) =
\frac{\sin^6(\alpha)+\sin^4(\alpha)-2}{\sin^6(\alpha)-2\sin^2(\alpha)-4 } $$
$$
\frac{\sin^6(\alpha)+\sin^4(\alpha)-2}{\sin^6(\alpha)-2\sin^2(\alpha)-4 } =\frac{(\sin^2(\alpha)-1)(\sin^4(\alpha)+2\sin(\alpha)+2)} {(\sin^2(\alpha)-4)(\sin^4(\alpha)+2\sin(\alpha)+2)} $$
$$
\frac{\sin^6(\alpha)+\sin^4(\alpha)-2}{\sin^6(\alpha)-2\sin^2(\alpha)-4 } =\frac{\sin^2(\alpha)-1} {\sin^2(\alpha)-4} $$
Can this expression be simplify more further?
$$ \frac{\tan^6(\alpha)-\tan^4(\alpha)+2}{\tan^6(\alpha)-2\tan^2(\alpha)+4 }\cdot\cos^2(\alpha)$$
please give a hand here can't seem to do it
| So, for the LH fraction:
$$
{{x^{\,6} - x^{\,4} + 2} \over {x^{\,6} - 2x^{\,2} + 4}} = {{y^{\,3} - y^{\,2} + 2} \over {y^{\,3} - 2y + 4}}
$$
then by Ruffini's method
$$
= {{\left( {y + 1} \right)\left( {y^{\,2} - 2y^{\,2} + 1} \right)} \over {\left( {y + 2} \right)\left( {y^{\,2} - 2y^{\,2} + 1} \right)}} = {{x^{\,2} + 1} \over {x^{\,2} + 2}}
$$
Same method for the RH fraction, as already done, but corrected to$$
{{\sin ^{\,2} \alpha - 1} \over {\sin ^{\,2} \alpha - 2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\cos^4x - \sin^4x - \cos^2x + \sin^2x$ is always $0$? So I have a small problem here where I have to prove the following :
$$\cos^4x - \sin^4x - \cos^2x + \sin^2x = 0 $$
I know that the 2nd part is always $1$, so I need to prove that the first part also equals $1$. So how should I prove it ?
Edit : Sorry, the equation itself was wrong. I've edited it.
|
Notice, your statement isn't true. Now we know that:
$$\cos^2(x)+\sin^2(x)=1$$
So, we can see that:
$$\cos^4(x)+\sin^4(x)-\cos^2(x)-\sin^2(x)=\cos^4(x)+\sin^4(x)-1$$
Now, prove that $\cos^4(x)+\sin^4(x)=\frac{\cos(4x)+3}{4}$:
$$\cos^4(x)+\sin^4(x)=\frac{\cos(4x)+3}{4}\Longleftrightarrow$$
$$4\left(\cos^4(x)+\sin^4(x)\right)=\cos(4x)+3\Longleftrightarrow$$
Use $\sin^4(x)=\left(\sin^2(x)\right)^2=\left(1-\cos^2(x)\right)^2$:
$$4\left(\cos^4(x)+\left(1-\cos^2(x)\right)^2\right)=\cos(4x)+3\Longleftrightarrow$$
Use $\left(1-\cos^2(x)\right)^2=1-2\cos^2(x)+\cos^4(x)$:
$$4\left(\cos^4(x)+1-2\cos^2(x)+\cos^4(x)\right)=\cos(4x)+3\Longleftrightarrow$$
$$4\left(1-2\cos^2(x)+2\cos^4(x)\right)=\cos(4x)+3\Longleftrightarrow$$
$$4-8\cos^2(x)+8\cos^4(x)=\cos(4x)+3\Longleftrightarrow$$
Use $\cos(4x)=\cos(2(2x))=2\cos^2(2x)-1$:
$$4-8\cos^2(x)+8\cos^4(x)=2\cos^2(2x)-1+3\Longleftrightarrow$$
$$4-8\cos^2(x)+8\cos^4(x)=2\cos^2(2x)+2\Longleftrightarrow$$
Use $\cos(2x)=2\cos^2(x)-1$:
$$4-8\cos^2(x)+8\cos^4(x)=2\left(2\cos^2(x)-1\right)^2+2\Longleftrightarrow$$
$$4-8\cos^2(x)+8\cos^4(x)=4-8\cos^2(x)+8\cos^4(x)$$
So:
$$\color{red}{\cos^4(x)+\sin^4(x)-\cos^2(x)-\sin^2(x)=\frac{\cos(4x)+3}{4}-1=\frac{\cos(4x)-1}{4}\ne0}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the limit of $\lim_{n\to \infty}\frac{1-2+3-4+\cdots+(-2n)}{\sqrt{n^2+1}}$? Find the $$\lim_{n\to \infty}\frac{1-2+3-4+\cdots+(-2n)}{\sqrt{n^2+1}}.$$
I thought to apply squeeze theorem so
$$\frac{1-2+3-4+\cdots+(-2n)}{\sqrt{n^2+1}}\leq \frac{1+2+3+4+\cdots+(2n)}{\sqrt{n^2+1}}=\frac{n(2n+1)}{\sqrt{n^2+1}}$$ But I am unable to find a lower bound though. How can I do this?
| You can do the following. For any $k$ we have $$1-2+3-4 + ... +(-2k) = (1-2) + (3-4) + \cdots +(2k-1 - 2k) =-k$$
| {
"language": "en",
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"question_score": "3",
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Problem solving limit infinity/infinity. I cannot solve this limit:
$$\lim_{x\rightarrow\infty}
\frac {(3x^2-4) \left( \sqrt [3]{2x^2+1}+1
\right)^2}{ (2x-1) \left( 4-\sqrt {8x^3-2}
\right)x^{3/2}}$$
I make this:
$${\frac { \left( 3\,{x}^{2}-4 \right) \left( \sqrt [3]{2\,{x}^{2}+1}+1
\right) ^{2}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) {x}^{3/2}}} =3\,{\frac {\sqrt {x} \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-
1 \right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) }}-4\,{\frac {
\left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-1 \right) \left( 4-
\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}+6\,{\frac {\sqrt {x}\sqrt [3]
{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) }}-8\,{\frac {\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1
\right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}+3\,{\frac
{\sqrt {x}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) }}-4\,{\frac {1}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{
x}^{3}-2} \right) {x}^{3/2}}}$$
then
$$\lim_{x\rightarrow\infty}3\,{\frac {\sqrt {x} \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-
1 \right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-4\,{\frac { \left( 2\,{x}^{2}+1 \right) ^{2/3}}{ \left( 2\,x-1
\right) \left( 4-\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}
$$
$$\lim_{x\rightarrow\infty}6\,{\frac {\sqrt {x}\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right)
\left( 4-\sqrt {8\,{x}^{3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-8\,{\frac {\sqrt [3]{2\,{x}^{2}+1}}{ \left( 2\,x-1 \right) \left( 4-
\sqrt {8\,{x}^{3}-2} \right) {x}^{3/2}}}
$$
$$\lim_{x\rightarrow\infty}3\,{\frac {\sqrt {x}}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{
3}-2} \right) }}
$$
$$\lim_{x\rightarrow\infty}-4\,{\frac {1}{ \left( 2\,x-1 \right) \left( 4-\sqrt {8\,{x}^{3}-2}
\right) {x}^{3/2}}}
$$
but I cannot solve this
NOTE: I cannot use L'hopital for finding this limit.
| The total power of $x$ in the numerator is $2 + 4/3 = 10/3.$ The total power of $x$ in the denominator is $1 + 3/2 + 3/2 = 4.$ The denominator wins: The limit must be $0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_1^N \frac{-3N+6t-3}{t^3(N-t+1)^4}dt$ when $N=3$ or $N=5$ Let the Cauchy product $$(\zeta(3))^2=\sum_{n=1}^\infty c_n,$$
where $$c_n=\sum_{k=1}^n\frac{1}{k^3(n-k+1)^3},$$
and $\zeta(3)$ is the Apèry constant.
Taking $f(x)=\frac{1}{x^3(N-x+1)^3}$ in Abel's identity after tedious computations using the division of polynomials, since $f'(x)=\frac{-3N+6x-3}{x^4(N-x+1)^4}$ we've
$$\sum_{n\leq N}\frac{1}{n^3(N-n+1)^3}=\frac{1}{N^2}-\int_1^N \frac{-3N+6t-3}{t^3(N-t+1)^4}dt+O \left( \frac{1}{N^6} \right). $$
Then I understand that I can't get a closed form from this way since I have an error term that I don't know how compute it, but I've asked to me if at least we can compute previous integral, seems a hard integral, but Wolfram Alpha knows how compute particular values of previous integral. Then since I would like learn more mathematics
Question. Can you explain how get a closed-form, for example for $N=3$ or $N=5$ by specializacion of the previous integral
$$\int_1^N \frac{-3N+6t-3}{t^3(N-t+1)^4}dt?$$
It is neccesary only one case $N=3$ or $5$, I say these because I see that these are posible with this online tool. I will accept one answer with good details. Very thanks much!
| $$\frac{1}{n(N-n)} = \frac{1}{N}\left(\frac{1}{n}+\frac{1}{N-n}\right) \tag{1}$$
gives, by squaring:
$$\frac{1}{n^2(N-n)^2} = \frac{1}{N^2}\left(\frac{1}{n^2}+\frac{1}{(N-n)^2}+\frac{\frac{2}{N}}{n}+\frac{\frac{2}{N}}{N-n}\right)\tag{2} $$
as well as:
$$ \frac{1}{n^3 (N-n)^3} = \frac{1}{N^3}\left(\frac{1}{n^3}+\frac{1}{(N-n)^3}+\frac{3N}{n^2(N-n)^2}\right)\tag{3}$$
hence it follows that:
$$ \sum_{k=1}^{n}\frac{1}{k^3(n+1-k)^3} = \frac{2 H_n^{(3)}}{n^3}+\frac{6 H_n^{(2)}}{n^4}+\frac{12 H_n}{n^5}\tag{4} $$
and the same strategy (partial fraction decomposition) can be applied to the integral resulting from Abel's identity. However, Abel's identity is not really needed to compute the LHS of $(4)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to write this cubic equation For an imaginary number $i=\sqrt{-1}$ ,the cubic equation $24x^3+21x^2-72x-7=0$ can be represented in the form $$\frac{((x+i)^3-(x-i)^3)}{((x+i)^3+(x-i)^3)}\frac{((2+i)^4+(2-i)^4)}{((2+i)^4-(2-i)^4)}=1$$
How can we write the cubic $141x^3+72x^2-141x-8=0$ in a similar form?
| HINT:
By applying Componendo and Dividendo repeatedly,
we find $$\dfrac{x^3-3x}{-7}=\dfrac{3x^2-1}{24}\iff24x^3+21x^2-72x-7=0$$
Now,
$$14x1^3+72x^2-141x-8=0\iff\dfrac{x^3-x}8=\dfrac{1-9x^2}{141}$$
Again, $(ax+i)^3=a^2x^3+i(3a^2x^2)+3ax(-1)-i$
Comparing with $x^3-x+i(1-9x^2)$
we need, $$\dfrac{a^3}1=\dfrac{3a}1\iff a^3=3a$$
As $a\ne0,a=\pm\sqrt3$
Taking $a=\sqrt3,$
$$\dfrac{3\sqrt3x^3-\sqrt3x}{24\sqrt3}=\dfrac{(1-9x^2)}{141}$$
$$\iff\dfrac{3\sqrt3x^3-\sqrt3x}{(1-9x^2)i}=\dfrac{-24\sqrt3i}{141}$$
Applying Componendo and Dividendo, $$\left(\dfrac{i+\sqrt3x}{i-\sqrt3x}\right)^3=\dfrac{141+24\sqrt3i}{141-24\sqrt3i}$$
I leave to you to express $$\dfrac{141+24\sqrt3i}{141-24\sqrt3i}$$
as $$\left(\dfrac{c+id}{c-id}\right)^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the closed form for $\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{x^{\pi}}{1+x^{\pi}}\cdot\frac{1}{1+x^ e}dx $? On my previou page Jack D'Aurizio offered a concise elegant prove of Vladimir Reshetnikov's identity and a closed form for it.
(1)
$$\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{1}{1+x^{\pi}}dx=\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{1}{1+x^{e}}dx=\frac{\pi}{4}$$
Here we have another imitation of Vladimir Reshetnikov's identity
(2)
$$\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{x^{\pi}}{1+x^{\pi}}\cdot\frac{1}{1+x^e}dx =\int_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{x^e}{1+x^e}\cdot\frac{1}{1+x^{\pi}}dx$$
A closed form of (2) is unknown
We ask if this identity (2) can be proven in the same way as (1) and with a closed form.
| Through the substitution $x\mapsto\frac{1}{x}$ we have:
$$ I(a,b) = \int_{0}^{+\infty}\frac{x^a}{1+x^a}\cdot\frac{1}{1+x^b}\cdot\frac{dx}{1+x^2} = \int_{0}^{+\infty}\frac{1}{1+x^a}\cdot\frac{x^b}{1+x^b}\cdot\frac{dx}{1+x^2} = I(b,a) $$
hence, by your previous question:
$$\begin{eqnarray*} I(a,b) &=& \frac{1}{2}\int_{0}^{+\infty}\frac{x^a+x^b}{(1+x^a)(1+x^b)}\cdot\frac{dx}{1+x^2}\\&=&\frac{1}{2}\left[\int_{0}^{+\infty}\left(\frac{1}{1+x^a}+\frac{1}{1+x^b}\right)\frac{dx}{1+x^2}-2\int_{0}^{+\infty}\frac{dx}{(1+x^a)(1+x^b)(1+x^2)}\right]\\&=&\frac{\pi}{4}-\int_{0}^{+\infty}\frac{dx}{(1+x^a)(1+x^b)(1+x^2)}\end{eqnarray*}$$
that, at least in principle, can be computed from partial fraction decomposition, the residue theorem and the identity:
$$ \int_{0}^{+\infty}\frac{x^b dx}{1+x^a} = \frac{\pi}{a\sin\left(\frac{\pi (b+1)}{a}\right)}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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For which $\lambda$ do we have solutions I'm trying to find for what values of $\lambda$ the following matrix has either no solutions, infinitely many or unique solutions.
$$A=\begin{pmatrix} 1 & 1 & \lambda & 1 \\ 4 & \lambda ^2 & -8 & 4 \\ \lambda & 2 & 4 & \lambda + 1\end{pmatrix}$$
I know that in general a system of equations has no solutions if there is a leading $1$ in the last column, unique if there is a leading $1$ in every column except the last and infinitely many otherwise. I'm not sure how to apply this here though.
| Because of the way your question is phrased, I will assume that $A$ is the augmented matrix of a system of 3 equations on 3 variable
If we perform row reduction:
\begin{align}
\begin{bmatrix} 1 & 1 & \lambda & 1 \\ 4 & \lambda ^2 & -8 & 4 \\ \lambda & 2 & 4 & \lambda + 1\end{bmatrix}
&\xrightarrow[R3-\lambda R1]{R2-4R1}
\begin{bmatrix} 1 & 1 & \lambda & 1 \\ 0 & \lambda ^2 -4& -8-4\lambda & 0 \\ 0 & 2-\lambda & 4-\lambda^2 & 1\end{bmatrix}\\ \ \\
&=
\begin{bmatrix} 1 & 1 & \lambda & 1 \\ 0 & (\lambda -2)(\lambda+2)& -4(\lambda+2) & 0 \\ 0 & 2-\lambda & -(\lambda-2)(\lambda+2) & 1\end{bmatrix}
.
\end{align}
*
*If $\lambda=2$, we get
$$
\begin{bmatrix} 1 & 1 & 2 & 1 \\ 0 & 0& -16 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}
$$
The fourth row now signifies the equation $0=1$, so the system has no solution.
*If $\lambda=-2$, we get
$$
\begin{bmatrix} 1 & 1 & -2 & 1 \\ 0 & 0& 0 & 0 \\ 0 & 4 & 0 & 1\end{bmatrix}
\xrightarrow{}
\begin{bmatrix} 1 & 0 & -2 & 3/4 \\ 0 & 1 & 0 & 1/4\\ 0 & 0& 0 & 0\end{bmatrix}
$$
This says that $x_2=1/4$, and $x_1=2x_3+3/4$. As we are free to choose $x_3$, the system has infinitely many solutions.
*If $\lambda$ is neither $2$ nor $-2$, we can divide the second row by $\lambda+2$ and the third one by $\lambda-2$ to get
\begin{align}
\begin{bmatrix} 1 & 1 & \lambda & 1 \\ 0 & \lambda -2& -4 & 0 \\ 0 & -1& -\lambda+2 & 1\end{bmatrix}
&\xrightarrow[R2+(\lambda+2)R3]{R1+R3}
\begin{bmatrix} 1 & 0 & 2 & 2 \\ 0 & 0& -\lambda^2 & 0 \\ 0 & -1& -\lambda+2 & 1\end{bmatrix}\\ \ \\
&\xrightarrow[]{}
\begin{bmatrix} 1 & 0 & 2 & 2 \\ 0 & 1& \lambda-2 & -1\\ 0 & 0& \lambda^2 & 0\end{bmatrix}.
\end{align}
If $\lambda\ne0$, then the system will have three leading ones and no inconsistency, so it has unique solution.
If $\lambda=0$, then again we are free to choose $x_3$ and so the system has infinitely many solutions.
In summary:
*
*If $\lambda=0$, infinitely many solutions.
*If $\lambda=-2$, infinitely many solutions.
*If $\lambda=2$, no solution.
*If $\lambda$ is not $0$ nor $2$ nor $-2$, the system has unique solution.
| {
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"question_score": "2",
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Shortest distance between two lines in 3-dimensional space Can someone explain to me how to solve this question?
Find the shortest distance between the lines
$L_1 = \left\{t \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} : t \in \mathbb{R}\right\}$ and $L_2 = \left\{s \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} + \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}: s \in \mathbb{R}\right\}$
Thanks
| The Euclidean distance between the two lines is
$$\left\|s \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} - t \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} + \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}\right\| = \left\|\begin{bmatrix} 1 & -1\\ 2 & -1\\ 3 & -1\end{bmatrix} \begin{bmatrix} s\\ t\end{bmatrix} - \begin{bmatrix} -1\\ 0\\ 0\end{bmatrix}\right\|$$
Minimizing this Euclidean distance is a least-squares problem. The mininimizer of $\| A x - b\|$ is given by the solution to the so-called normal equations $A^T A x = A^T b$. Since
$$\begin{bmatrix} 1 & -1\\ 2 & -1\\ 3 & -1\end{bmatrix}^T \begin{bmatrix} 1 & -1\\ 2 & -1\\ 3 & -1\end{bmatrix} = \begin{bmatrix} 14 & -6\\ -6 & 3\end{bmatrix}, \qquad{} \begin{bmatrix} 1 & -1\\ 2 & -1\\ 3 & -1\end{bmatrix}^T \begin{bmatrix} -1\\ 0\\ 0\end{bmatrix} = \begin{bmatrix} -1\\ 1\end{bmatrix}$$
the normal equations give us the following system of equations
$$\begin{bmatrix} 14 & -6\\ -6 & 3\end{bmatrix} \begin{bmatrix} s\\ t\end{bmatrix} = \begin{bmatrix} -1\\ 1\end{bmatrix}$$
whose solution is $\begin{bmatrix} s\\ t\end{bmatrix} = \begin{bmatrix} \frac{1}{2}\\ \frac{4}{3}\end{bmatrix}$. The minimum Euclidean distance between the two lines is, thus,
$$\left\|\frac{1}{2} \begin{bmatrix} 1\\ 2\\ 3\end{bmatrix} - \frac{4}{3} \begin{bmatrix} 1\\ 1\\ 1\end{bmatrix} + \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix}\right\| = \left\|\begin{bmatrix} \frac{1}{6}\\ -\frac{1}{3}\\ \frac{1}{6}\end{bmatrix}\right\| = \frac{1}{\sqrt{6}}$$
| {
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"question_score": "5",
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Factors of the numbers of the form $a^2+nb^2$
Let $N=a^2+nb^2$ with $\gcd(a,b) =1$ and $n \in \mathbb{Z^+}$. If $N=xy$ where $x$ and $y$ are relatively prime numbers, in what condition can $x$ and $y$ be also written in the same form as $N$ (i.e, $c^2+nd^2$) ? How can we prove it?
The above statement is true for sum of two squares (i.e, the case $n=1$), which has been proved by Fermat and Euler $($Here $ \text{is Euler's proof, see Proposition} \ 4)$ using Infinite Descent and Brahmagupta's Identity. But can it be generalized?
Any help will be appreciated.
Thanks.
| This is true when $a^2 + n b^2$ is of class number one ( well, the form class number of the discriminant $-4n$). I refer to the number of (equivalence classes of) primitive binary quadratic forms with that discriminant, $-4n.$ Even in this case, we need to be cautious when there are also imprimitive forms of that discriminant. This is all in Dickson's little book, page 91
An easy example would be $x^2 + 6 y^2,$ where the other class of this discriminant is $2 x^2 + 3 y^2.$ Now, $ 3 \cdot 11 = 33 = 9 + 6 \cdot 4.$ Neither of $3,11$ can be expressed as $x^2 + 6 y^2.$ However, in accordance with Dickson, $11 = 2 \cdot 4 + 3,$ and $3 = 2 \cdot 0 + 3.$ This example even gives a formula,
$$ (2u^2 + 3 v^2)(2x^2 + 3 y^2) = (2ux+3vy)^2 + 6 (uy-vx)^2 $$
There are always such formulas, this is Gauss composition, but no guarantee that the whole thng will be pretty. The most elegant ones happen when we have already written the forms ready for Dirichlet's presentation of composition. This can be found, for example, in the book by Cox, Primes of the Form $x^2 + n y^2.$
Dirichlet on composition: we usually write the forms of discriminant $-44$ as $x^2 + 11 y^2,$ $3 x^2 + 2 xy + 4 y^2,$ $3 x^2 - 2 xy + 4 y^2.$ However, the first is $SL_2 \mathbb Z$ equivalent to $27 x^2 + 8 xy + y^2,$ the second form to $3 x^2 + 8 xy + 9 y^2,$ the third (its "opposite" class) to
$9 x^2 + 8 xy + 3 y^2.$ And composition comes out well:
$$ (3 x^2 + 8xy + 9 y^2)(9 z^2 + 8zw + 3 w^2) = 27 X^2 + 8XY + Y^2, $$
where
$$ X =xz-yw, \; \; Y=3xw+9yz+8yw $$
This is on page 49 in the first edition of Cox, which had a typo in the formula for $X.$ Corrected in the second edition, see if I can paste in an excerpt.
| {
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Prove $(x^2y+y^2z+z^2x)\cdot \left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2} +\frac{1}{(z+x)^2}\right) \geqslant \frac94$ $x,y,z > 0$ and $x+y+z=3$, prove
$$(x^2y+y^2z+z^2x)\cdot \left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2} +\frac{1}{(z+x)^2}\right) \geqslant \frac94$$
My immediate thought is that this inequality is similar to the famous Iran inequality
$$(xy+yz+zx)\cdot \left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2} +\frac{1}{(z+x)^2}\right) \geqslant \frac94$$
then if I can prove that
$$x^2y+y^2z+z^2x \geqslant xy+yz+zx$$ for positive $x,y,z$ satistifes $x+y+z=3$ then the problem is solved.
However, it turns out that $x^2y+y^2z+z^2x$ is neither always greater or lesser than $xy+yz+zx$, so I get stuck here.
I don't like solution involved computer or numerical methods. I will down vote all of answers that showing these methods
| A full expanding gives
$$\sum\limits_{cyc}(4x^6y+5x^5y^2+x^5z^2+3x^4y^3-x^4z^3+$$
$$+2x^5yz+9x^4y^2z-7x^4z^2y-2x^3y^3z-14x^3y^2z^2)\geq0$$
and since by Muirhead and AM-GM
$$\sum\limits_{cyc}(x^5y^2+x^5z^2)\geq\sum\limits_{cyc}(x^4y^3+x^4z^3)$$
$$6x^4y^3+2y^4z^3+5z^4x^3\geq13x^3y^2z^2$$
$$19x^6y+2y^6z+10z^6x\geq31x^4z^2y$$
$$9x^5y^2+4y^5z^2+6z^5x^2\geq19x^3y^2z^2$$
which gives $4\sum\limits_{cyc}x^6y\geq4\sum\limits_{cyc}x^4z^2y$, $4\sum\limits_{cyc}x^4y^3\geq4\sum\limits_{cyc}x^3y^2z^2$ and $4\sum\limits_{cyc}x^5y^2\geq4\sum\limits_{cyc}x^3y^2z^2$,
it remains to prove that $$\sum\limits_{cyc}(2x^5yz+9x^4y^2z-3x^4z^2y-2z^3y^3z-6x^3y^2z^2)\geq0$$ or
$$\sum\limits_{cyc}(x^4+9x^3y-3x^3z-2x^2y^2-6x^2yz)\geq0$$ or
$$\sum\limits_{cyc}(2x^4+3x^3y+3x^3z-2x^2y^2-6x^2yz)\geq6\sum\limits_{cyc}(x^3z-x^3y)$$ or
$$\sum\limits_{cyc}((x^2-y^2)^2+3z(x+y)(x-y)^2)\geq6(x+y+z)(x-y)(y-z)(z-x)$$ or
$$\sum\limits_{cyc}(x-y)^2(x+y)(x+y+3z)\geq6(x+y+z)(x-y)(y-z)(z-x)$$
Since for $x\geq y\geq z$ we get $\prod\limits_{cyc}(x-y)\leq0$, we can assume $x\geq z\geq y$ and since
$(x+y+z)(x-y)(y-z)(z-x)=(x+y+z)(x-y)(x-z)(z-y)\leq(x+z)(x-z)xz$,
it remains to prove
$$\sum\limits_{cyc}(x-y)^2(x+y)(x+y+3z)\geq6(x+y+z)(x-y)(y-z)(z-x)$$
for $y\rightarrow0^+$, which gives
$$x^4-3x^3z-2x^2z^2+9xz^3+z^4\geq0$$
which is obviously true.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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write down the expression for $\sin (15°)$ using the double angle formula. show that $\sin 15^\circ=\frac {\sqrt3 -1}{2\sqrt2}$ using $\cos2A=1-2\sin^2A$
However I got $\sin 15^\circ= \sqrt{\frac {2-\sqrt 3}{4}}$ instead.
| Your form, $\sqrt{\frac{2-\sqrt3}{4}}$, is already close:
$$\begin{align*}
\sin15^\circ &=\sqrt{\frac{2-\sqrt3}{4}}\\
&= \sqrt{\frac{4-2\sqrt3}{8}}\\
&= \sqrt{\frac{1-2\sqrt3+3}{8}}\\
&= \sqrt{\frac{(1-\sqrt3)^2}{8}}\\
&= \frac{\sqrt3 -1 }{2\sqrt2}
\end{align*}$$
| {
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Finding K such that a line is perpendicular or parallel to a given plane? Known information:
Plane $P = 3x -5y + 2z =1 (R^3)$
Parametrics:
x = 5 + t
y = 2 - 2t
z = -1 - 6t
I've solved for $t$, $t = -2$. Point of intersection: $(3,6,11)$
The part causing problems is, given vectors:
A = (2, -7, -6)
B = (-4, 3, k)
Vector of $A$ and $B = AB: (-8, -21, -6k)$
Solve for $k$ such that vector $AB$ is perpendicular to the plane, and for $k$ such that $AB$ is parallel to the plane.
Thank you very much for any help. I've been pouring over notes for how to do this, with little success. Thank you.
Also, please note: the accuracy of vector $AB$ is not 100%. IF the math is wrong, that may be part of the issue, so please note that.
| Your plane is described by
$$
(3, -5, 2) \cdot (x,y,z) = 1 \iff \\
\frac{1}{\lVert (3,-5,2) \rVert} (3, -5, 2) \cdot (x,y,z) =
\frac{1}{\lVert (3,-5,2) \rVert} \\
\frac{1}{\sqrt{38}} (3, -5, 2) \cdot (x,y,z) =
\frac{1}{\sqrt{38}} \iff \\
n \cdot u = d
$$
where $n$ is a unit normal vector and $d$ is the signed distance to the origin.
a) Assuming $AB = B - A = (-6, 10, k + 6)$ then a vector perpendicular to the plane is parallel to $n$:
$$
\alpha n = AB \iff \\
\alpha \frac{1}{\sqrt{38}} (3, -5, 2) = (-6, 10, k + 6)
$$
where $\alpha \in \mathbb{R}$ is some scalar. Using $\alpha = -2 \sqrt{38}$ we get
$$
(-6, 10, -4) = (-6, 10, k + 6) \Rightarrow \\\
-4 = k + 6 \iff \\
k = -10
$$
b) If $AB$ is parallel to the plane it is perpendicular to $n$:
$$
0
= n \cdot (-6, 10, k+6) \\
= \frac{1}{\sqrt{38}} (3, -5, 2) \cdot (-6, 10, k+6) \\
= \frac{1}{\sqrt{38}} (-18 -50 + 2k + 12) \\
= \frac{1}{\sqrt{38}} (2k - 56) \iff \\
0 = 2k - 56 \iff \\
2k = 56 \iff \\
k = 28
$$
| {
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Can anyone help me with this finite sum? I have to calculate the sum
$\displaystyle\sum_{k=1}^n \displaystyle\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1}$
We can re-write the sum as follows
$\displaystyle\sum_{k=1}^n \displaystyle\frac{3^k-1+1}{(3^{k+1}-1)(3^k-1)}$
And then we obtain
$\displaystyle\sum_{k=1}^n \displaystyle\frac{1}{3^{k+1}-1}+\displaystyle\sum_{k=1}^n \displaystyle\frac{1}{(3^{k+1}-1)(3^k-1)}$
But I don't know what to do with the last two sums. Can anyone help me with them. Thanks
| For such problem, we usually look for some telescoping structure, note that
$$\frac{1}{3^{k} - 1} - \frac{1}{3^{k + 1} - 1} = \frac{2 \times 3^k}{3^{k + 1} - 1)(3^k - 1)}.$$
So based on the decomposition you already found, the summation can be written as
$$\frac{1}{2}\sum_{k = 1}^n \left(\frac{1}{3^{k} - 1} - \frac{1}{3^{k + 1} - 1}\right) = \frac{1}{2}\left(\frac{1}{2} - \frac{1}{3^{n + 1} - 1}\right).$$
| {
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Evaluating the inverse trigonometric limit $\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}$
$$
\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}
$$
I was doing some questions on limits, I saw one in which there is $\arccos x$. I am stuck there, not able to proceed.
Can you give me some hint?
| Set $t=\arccos(2x\sqrt{1-x^2})$, so $2x\sqrt{1-x^2}=\cos t$ and, by definition, $0\le t\le\pi$. Then
$$
\sin t=\sqrt{1-\cos^2t}=\sqrt{1-4x^2+4x^4}=|2x^2-1|
$$
and so $t=\arcsin|2x^2-1|$. Thus your limit (from the right) can be written
$$
\lim_{x\to(1/\sqrt2)^+}\sqrt{2}\frac{\arcsin|2x^2-1|}{\sqrt{2}x-1}=
\lim_{x\to(1/\sqrt2)^+}\sqrt{2}
\frac{(\sqrt{2}x+1)\arcsin(2x^2-1)}{2x^2-1}
$$
which you should be able to compute.
Note then that, for $x<1/\sqrt{2}$, $|2x^2-1|=-(2x^2-1)$ and you can go similarly for the limit as $x\to(1/\sqrt{2})^-$
| {
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Factor proofs problem The coolness of an integer is equal to the integer divided by the total number of factors that it has. For example, $48$ has $10$ factors therefore, coolness $(48) = \frac { 48 }{ 10 } =\quad 4.8$
1. Provide an explanation for why coolness$(xy)$ cannot be equal to an integer if both $x$ and $y$ are different prime numbers.
My attempt: Just from a couple of trials a found that the number of factors of $xy$ seems to always be $4$. Therefore, we can note that coolness$(xy)$ can only be a whole number if $xy$ is divisible by $4$. I am pretty sure that we can’t make a multiple of $4$ by multiplying any combination of 2 prime numbers.
The above working is not valid without proofs and I am unsure of how to approach them. How can I prove that $xy$ has $4$ factors and that it is not possible to get a multiple of $4$ by multiplying two prime numbers?
2. $x$ and $y$ are different prime numbers. Identify the numbers of the form $x y^4$ which have a coolness that is equal to an
integer.
My attempt: I am not sure how to solve this one but I think listing the factors of $xy^4$ might help.
3. Prove that the square of any prime number $x$ is equal to the coolness of some integer.
My attempt: No idea other than just listing a couple of prime numbers and then squaring them to check if the result is equal to the coolness of some integer.
Please help me solve the above problems.
|
1. Provide an explanation for why coolness$(xy)$ cannot be equal to an integer if both $x$ and $y$ are different prime numbers.
You are correct in your thinking for this part:
As $x$ and $y$ are distinct primes, $xy$ has $4$ factors: $1, x, y, xy$. This means that $xy$ must be divisible by $4$. However, as $2$ is the only even prime number and cannot be used twice, this means that at least one of $x$ or $y$ must be odd. The product of a integer with an odd number will never yield a multiple of $4$, therefore coolness$(xy)$ will never equal a integer where $x$ and $y$ are distinct primes.
2. $x$ and $y$ are different prime numbers. Identify the numbers of the form $x y^4$ which have a coolness that is equal to an
integer.
$xy^4$ has $10$ factors: $1, x, y, y^2, y^3, y^4, xy, xy^2, xy^3, xy^4$. This means that for coolness$(xy^4)$ to be an integer, $xy^4$ must be divisible by $10$. Then $x$ and $y$ must multiple to a factor of $10$. The only prime numbers which serve this are $5$ and $2$. Therefore $x=2, y=5$, or $x=5, y=2$.
3. Prove that the square of any prime number $x$ is equal to the coolness of some integer.
Given the prime $x\neq 3$, the integer number which has its coolness as the square of any prime number $x$, can be written $n(x) = 3^2x^2$. It has $3^2 = 9$ factors ($1$, $3$, $x$, $3x$, $3^2x$, $3x^2$, $3^2$, $x^2$, $3^2x^2$) and
$$
\mbox{Coolness}(n(x)) = \frac{3^2x^2}{3^2} = x^2
$$
If $x=3$ then the number $n(x)=2^2 3^3 = 108$. In this case
$$
\mbox{Coolness}(108) = \frac{2^23^3}{2^2 3} = 3^2
$$
Therefore the square of any prime number $x$ is equal to the coolness of some integer.
I hope this helps you on your problems!
| {
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proof that $\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$ diverges, by comparsion I need to prove that
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$$
diverges by comparsion.
The way I did was to use
$$\frac{n}{n^2+2n+1}>\frac{n}{n^2+2n^2+n^2} = \frac{n}{4n^2} = \frac{1}{4n}$$
which diverges. Can I do that? Because $$2n+1<2n^2+n^2\implies n^2+2n+1<n^2+2n^2+n^2\implies $$
$$\frac{1}{n^2+2n+1}>\frac{1}{n^2+2n^2+n^2}$$
for large $n$.
I also tried seeing
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1} = \sum_{n=1}^{\infty}\frac{n}{(n+1)^2}$$
but I couldn't find any comparsion for that. Am I right in the first demonstration? Can you find a comparsion for the second attempt I did?
| $$\frac{n}{n^2+2n+1} = \frac{n}{(n+1)^2} = \frac{n + 1 - 1}{(n+1)^2} = \frac{1}{(n+1)} - \frac{1}{(n+1)^2}$$
Which diverges because $\frac{1}{n+1}$ does so and $\frac{-1}{(n+1)^2}$ doesn't.
| {
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Solving $nx_n=(n+2)x_{n-1} + 1$ by the telescoping method I am trying to solve this recurrence relation from a book "Problem solving through Problems" by Loren c. Larson (5.3.14 (b)) using the telescoping method.
$$x_0=0\qquad nx_n=(n+2)x_{n-1} + 1\ (n > 0)$$
Here some additional terms I computed: $x_1=1, x_2 = 5/2, x_3 = 9/2, x_4=7$.
So far I tried to sum both parts of the expression for all $x_i$ and indeed, after some telescoping I got this:
$$x_n = \frac3n(x_{n-1} + \ldots + x_1) + 1$$
Although it seems of no help to actually get the result.
| Dividing both parts of the expression by $n(n+1)(n+2)$, one gets:
$$\frac{x_n}{(n+2)(n+1)}=\frac{x_{n-1}}{(n+1)n} + \frac{1}{(n+2)(n+1)n}$$
Let us define $$y_n = \frac{x_n}{(n+2)(n+1)}$$
Then the new recurrence is easier to solve:
$$y_n = y_{n-1} + \frac{1}{(n+2)(n+1)n}\qquad y_0 = 0$$
One knows that there exists $(A,B,C)$ such that $$\frac{1}{(n+2)(n+1)n}=\frac{A}{n+2} + \frac{B}{n+1} + \frac{C}{n}$$
One easily checks that $A = 1/2$, $B = -1$, $C = 1/2$, hence
our recurrence takes the following form:
$$y_n = y_{n-1} + \frac{1/2}{n+2} - \frac{1}{n+1} + \frac{1/2}{n}$$
It follows that
$$y_1 - y_0 = \color{red}{\frac{1/2}{3}} - \frac{1}{2} + \frac{1/2}{1}$$
$$y_2 - y_1 = \color{blue}{\frac{1/2}{4}} - \color{red}{\frac{1}{3}} + \frac{1/2}{2}$$
$$y_3 - y_2 = \color{green}{\frac{1/2}{5}} - \color{blue}{\frac{1}{4}} + \color{red}{\frac{1/2}{3}}$$
$$y_4 - y_3 = \color{orange}{\frac{1/2}{6}} - \color{green}{\frac{1}{5}} + \color{blue}{\frac{1/2}{4}}$$
$$\ldots$$
$$y_{n-2} - y_{n-3} = \color{orange}{\frac{1/2}{n}} - \color{#08f}{\frac{1}{n-1}} + \color{#08A}{\frac{1/2}{n-2}}$$
$$y_{n-1} - y_{n-2} = \frac{1/2}{n+1} - \color{orange}{\frac{1}{n}} + \color{#08f}{\frac{1/2}{n-1}}$$
$$y_{n} - y_{n-1} = \frac{1/2}{n+2} - \frac{1}{n+1} + \color{orange}{\frac{1/2}{n}}$$
Now if we sum both parts of all the expressions notice how terms of the same color cancel each other. Also all $y_i$ for the exception of $y_0$ and $y_n$ will be canceled as well.
Eventually we get
$$y_n = \frac14 - \frac{1}{2(n+1)} + \frac{1}{2(n+2)}$$
Then going back to $x_n$ we get
$$x_n = \frac{n^2 + 3n}{4}$$
| {
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Number of discontinuous values We have to find the number of values of $x$ at which the function $$ f(x) = \frac{2x^5-8x^2+11}{x^4+4x^3+8x^2+8x+4}$$ is discontinuous.
I thought that since both numerator and denominator are polynomials, and so are continuous, that the function would be continuous except when the denominator is zero.
But I am unable to find points where the denominator is zero.
| Note $(x+1)^4=x^4+4x^3+6x^2+4x+1$.
Then, $$x^4+4x^3+8x^2+8x+4=(x+1)^4+2x^2+4x+3=(x+1)^4+2(x+1)^2+1$$
Thus we conclude that it is never discontinuous because $$x^4+4x^3+8x^2+8x+4=(x^2+2x+2)^2>0$$
| {
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Integral ${\large\int}_0^\infty\big(2J_0(2x)^2+2Y_0(2x)^2-J_0(x)^2-Y_0(x)^2\big)\,dx$ I'm interested in the following definite integral:
$$\int_0^\infty\big(2J_0(2x)^2-J_0(x)^2+2Y_0(2x)^2-Y_0(x)^2\big)\,dx,\tag1$$
where $J_\nu$ and $Y_\nu$ are the Bessel functions of the first and the second kind.
Mathematica evaluates this integral symbolically to $\frac{\ln2}\pi$, but the result of a numerical integration looks more like $\frac{\ln4}\pi$, so I suspect the symbolic result is incorrect.
Moreover, it looks like both components converge and make equal contribution:
$$\int_0^\infty\big(2J_0(2x)^2-J_0(x)^2\big)\,dx\stackrel?=\int_0^\infty\big(2Y_0(2x)^2-Y_0(x)^2\big)\,dx\stackrel?=\frac{\ln2}\pi,\tag2$$
but it is more difficult to check numerically, because both integrands here are oscillating (unlike $(1)$ where the integrand looks monotonic).
How can we find values of these integrals and prove them correct? Can we generalize results for values of the index $\nu$ other than $0$?
| Somewhat similar to this answer, we can use Ramanujan's master theorem to show that $$\int_{0}^{\infty}\big(2J_0(2x)^2-J_0(x)^2\big)\,dx = \frac{\ln 2}{\pi}. $$
As $x \to \infty$, $$\big(2J_0(2x)^2-J_0(x)^2\big) \sim \frac{\sin (4x)-\sin (2x)}{\pi x}. \tag{1}$$ So the integral does indeed converge (but not absolutely).
The hypergeometric representation of the square of the Bessel function of the first kind of order zero is $$J_{0}(z)^{2} = \, _1F_2\left(\frac{1}{2}; 1, 1; -z^{2} \right). $$
So for $a>0$ and $0 < s < 1$, the Mellin transform of $J_{0}(ax)^{2}$ is $$ \begin{align} \int_{0}^{\infty} x^{s-1} J_{0}(ax)^{2} \,dx &= \int_{0}^{\infty} x^{s-1} \, _1F_2\left(\frac{1}{2}; 1, 1; -(ax)^{2} \right) \, dx \\ &= \frac{1}{2a^{s}}\int_{0}^{\infty} u^{s/2-1} \, _1F_2\left(\frac{1}{2}; 1, 1; -u \right) \, du \\ &= \frac{1}{2a^{s}} \Gamma\left(\frac{s}{2} \right) \frac{\Gamma \left(\frac{1}{2}-\frac{s}{2} \right) \Gamma(1)^{2} }{\Gamma\left(\frac{1}{2}\right) \Gamma\left(1- \frac{s}{2} \right)^{2}} \\ &= \frac{1}{2a^{s}} \Gamma\left(\frac{s}{2} \right) \frac{\Gamma \left(\frac{1}{2}-\frac{s}{2} \right)}{\sqrt{\pi} \, \Gamma\left(1- \frac{s}{2} \right)^{2}} . \end{align}$$
Therefore, assuming we can bring the limit inside the integral, $$ \begin{align} \int_0^\infty\big(2J_0(2x)^2-J_0(x)^2\big)\,dx &= \frac{1}{\pi} \lim_{s \to 1^{-}} \left(\frac{1}{2^{s}}- \frac{1}{2} \right) \Gamma \left(\frac{1}{2} - \frac{s}{2} \right) \\ &= \frac{1}{\pi} \lim_{s \to 0^{-}} \left( \frac{1}{2^{s+1}} - \frac{1}{2} \right)\Gamma \left(- \frac{s}{2} \right) \\ &= \frac{1}{\pi} \lim_{s \to 0^{-}} \left(- \frac{\ln 2}{2} s + \mathcal{O}(s^{2}) \right) \left(- \frac{2}{s} + \mathcal{O}(1) \right) \\ &= \frac{\ln 2}{\pi}. \end{align}$$
The same approach seemingly also shows that $$\int_{0}^{\infty}\big(2J_v(2x)^2-J_v(x)^2\big)\,dx = \frac{\ln 2}{\pi} \, , \quad v> - \frac{1}{2}. $$
$(1)$ https://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms
| {
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If $a,b>0$ and $a+b=1\;,$ Then minumum value of $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ is If $a,b>0$ and $a+b=1\;,$ Then minumum value of $\displaystyle \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2$ is
$\bf{My\; Try::}$ Let $a=\sin^2 \theta$ and $b=\cos^2 \theta\;,$ Then We have to minimize
$$\displaystyle f(\theta) = (\sin^2 \theta+\csc^2 \theta)^2+(\cos^2 \theta+\sec ^2 \theta) = \sin^4 \theta+\csc^4 \theta+\cos^4 \theta+\sec^4 \theta+4$$
So $$f(\theta) = 1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4$$
So $$f(\theta) = 1-\frac{\sin^2 2\theta}{2}+\frac{4}{\sin^2 \theta}+4$$
Now Put $\sin^2 2\theta = t\;, t\in \left(0,1\right]$
So we get $$f(t)=5-\frac{t}{2}+\frac{4}{t}\;,$$ So we get $\displaystyle f'(t) = -\frac{1}{2}-\frac{4}{t^2}<0\;\forall t \in (0,1]$
So $$f(1)_{\min} = 5-\frac{1}{2}+4=9-\frac{1}{2}=\frac{17}{2}$$
But answer given as $$\frac{25}{2}$$
plz help me, Where I am wrong, Thanks
| The following is not true
$$\sin^4 \theta+\csc^4 \theta+\cos^4 \theta+\sec^4 \theta+4=1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4.$$
For example, let $\theta=\pi/4$. Then $\text{LHS}=1/4+4+1/4+4+4=12.5$, while $\text{RHS}=1-1/2+4+4=8.5$.
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"answer_id": 0
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Linear Regression: linear or reciprocal function? The problem is given below:
Simultaneous values of time $t$ and output $y$ from a specific sensor has been measured and is tabulated below $$\begin{array}{cc}
t & y \\
\hline
1 & 17 \\
2 & 15 \\
3 & 11 \\
4 & 10\\
5&8\\
6&7\\
7&7
\end{array} $$
Determine whether the model $y_1(t) = \beta_0 + \beta_1 t$ or $y_2(t) = \gamma_0 + \gamma_1t^{-1}$ gives the best least squares fit to the data.
by having the design matrix and observation vector
In the first model I get following beta: $\beta_0=17.71$ and $\beta_1= -1.75$
by the formula:
$\beta = (X^TX)^{-1}X^Ty$.
My question is then, how to make the design matrix for the second model. the $t^{-1}$ confuse me!
| Linear function
$$
y_{1}(t) = \beta_{0} + \beta_{1} t
$$
Linear system:
$$
\begin{align}
\mathbf{A} \beta &= y \\
%
\left[
\begin{array}{cc}
1 & 1 \\
1 & 2 \\
1 & 3 \\
1 & 4 \\
1 & 5 \\
1 & 6 \\
1 & 7 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
\beta_{0} \\
\beta_{1}
\end{array}
\right]
%
&=
%
\left[ \begin{array}{c}
17 \\
15 \\
11 \\
10 \\
8 \\
7 \\
7
\end{array}
\right]
%
\end{align}
$$
Solution:
$$
\beta_{LS} =
%
\left[
\begin{array}{c}
\beta_{0} \\
\beta_{1}
\end{array}
\right]
=
%
\frac{1}{28}
\left[
\begin{array}{r}
496 \\
-49
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
17.7143 \\
-1.753
\end{array}
\right]
%
$$
Error:
$$
r = \mathbf{A} \beta - y
\qquad \Rightarrow \qquad
r \cdot r \approx 7.67857
$$
Reciprocal function
$$
y_{2}(t) = \gamma_{0} + \frac{\gamma_{1}} {t}
$$
Linear system:
$$
\begin{align}
\mathbf{A} \beta &= \frac{1}{y} \\
%
\left[
\begin{array}{cc}
1 & 1 \\
1 & 2 \\
1 & 3 \\
1 & 4 \\
1 & 5 \\
1 & 6 \\
1 & 7 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
\gamma_{0} \\
\gamma_{1}
\end{array}
\right]
%
&=
%
\left[ \begin{array}{c}
\frac{1}{17} \\
\frac{1}{15} \\
\frac{1}{11} \\
\frac{1}{10} \\
\frac{1}{8} \\
\frac{1}{7} \\
\frac{1}{7}
\end{array}
\right]
%
\end{align}
$$
Solution:
$$
\gamma_{LS} =
%
\left[
\begin{array}{c}
\gamma_{0} \\
\gamma_{1}
\end{array}
\right]
=
%
\frac{1}{4398240}
\left[
\begin{array}{c}
181296 \\
68891
\end{array}
\right]
%
=
%
\left[
\begin{array}{c}
0.0412201 \\
0.0156633
\end{array}
\right]
%
$$
Error:
$$
r = \mathbf{A} \gamma - \frac{1}{y}
\qquad \Rightarrow \qquad
r \cdot r \approx 0.000213205
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof that $3^n | 2^{3^n} + 1$ Question:
Proof by induction that $3^n | 2^{3^n} + 1$.
Attempt: $$ 2^{3^{n+1}} + 1 = 2^{3^n} 2^3 + 1 = 2^{3^n} 2^3 + 1 + 2^3 - 2^3 =
2^3( 2^{3^n} + 1 ) + 1 -2^3$$
And the first is $3^n |$ but second I don't know how to proof that.
| Hint: you should be looking at $2^{3^{n+1}}$, not $2^{3^n+1}$. Assume, by induction that $2^{3^n} + 1 = 3^nd$, i.e., $2^{3^n} = 3^nd - 1$, and see what you can get from $2^{3^{n+1}} + 1 = 2^{3^n\cdot 3} + 1 = (2^{3^n})^3 + 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $a+b+c=0$ what is the value of $\frac{a^2}{2a^2 +bc }+\frac{b^2}{2b^2 +ca }+\frac{c^2}{2c^2 +ab }$ Let $s=\frac{a^2}{2a^2 +bc }+\frac{b^2}{2b^2 +ca }+\frac{c^2}{2c^2 +ab }$.
If we use inequality $\frac{x^2}{a}+\frac{y^2}{b} \ge \frac{(x+y)^2}{(a+b)}$ we get $s \ge 0$ as $a+b+c=0$.
Again $s \le \frac{a^2}{bc }+\frac{b^2}{ca }+\frac{c^2}{ab }=\frac{a^3}{abc }+\frac{b^3}{abc }+\frac{c^3}{abc }=3$ as $a+b+c=0$ implies $a^3+b^3+c^3=3abc$.
Thus $0 \le s \le 3$. Putting any one of $a$,$b$,$c$=0 we get $s=1$. Hence my hunch is $s=1$, but i can't prove it. Can anyone help?
| From your observation, we have
$$\sum_{cyc}\frac{a^2}{2a^2+bc}-1=\frac{abc(a+b+c)(ab+bc+ca-a^2-b^2-c^2)}{(2a^2+bc)(2b^2+ac)(2c^2+ab)}$$
Since $a+b+c=0$, we get the value of $S=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trouble with a Probability (Integral) Computation from *All of Statistics* From All of Statistics pg. 36:
Suppose that $X$ and $Y$ are independent and both have the same density $f(x) = 2x$ for $0 \le x \le 1$ and $f(x) = 0$ otherwise. Let us find $\mathbb{P}(X + Y \le 1)$. Using independence, the joint density is
$$
f(x,y) = f_X(x) f_Y(y) = 4xy \text { if } 0 \le x \le 1, 0 \le y \le 1
$$
Now
\begin{equation}
\begin{split}
\mathbb{P}(X + Y \le 1) & = \int \int_{x+y \le 1} f(x,y) dy dx \\
& = 4 \int_0^1 x \left[ \int_0^{1-x} y dy \right] dx \\
& = 4 \int_0^1 x \frac{(1-x)^2}{2} dx \\
& = \frac{1}{6}
\end{split}
\end{equation}
Question 1: Why are we integrating on $\int_0^1$ for the outside integral and on $\int_0^{1-x}$ on the inside integral?
Question 2: According to the last step, it must be that $\int_0^1 x \frac{(1-x)^2}{2} dx = 1/24$. But why is this the case? How is this integral computed?
| Question 1: The integral is over the region $x+y \leq 1$. You are also constrained to $0 \leq x \leq 1$ and $0 \leq y \leq 1$ because of the densities of $X$ and $Y$. If you draw the $1 \times 1$ box where $0 \leq x \leq 1$ and $0 \leq y \leq 1$, then draw the line $x+y=1$ and shade the part below the line where $x+y \leq 1$ (equivalently, $y \leq 1-x$), you will see the triangular region over which you are integrating. Since the order of integration is $dydx$, you must find constant limits for the $x$ variable, which can range between $0$ and $1$. Then you must find limits for $y$ in terms of $x$. For every $x$ value, $y$ is bounded below by $0$ and above by $1-x$, since $y \leq 1-x$.
Question 2:
\begin{align*}
\int_0^1 x\frac{(1-x)^2}{2}dx &= \frac{1}{2}\int_0^1 x(1-2x+x^2)dx\\
&=\frac{1}{2}\int_0^1 (x-2x^2+x^3)dx\\
&=\frac{1}{2}\left(\frac{x^2}{2}-\frac{2}{3}x^3+\frac{1}{4}x^3\right)\Bigg|_0^1\\
&=\frac{1}{2}\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)\\
&=\frac{1}{2}\left(\frac{6}{12}-\frac{8}{12}+\frac{3}{12}\right)\\
&=\frac{1}{2}\cdot \frac{1}{12}\\
&=\frac{1}{24}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.