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How can I find the limit of the following sequence $\sin ^2 (\pi \sqrt{n^2 + n})$? How can I find the limit of the following sequence:
$$\sin ^2 (\pi \sqrt{n^2 + n})$$
I feel that I will use the identity $$\sin ^2 (\pi \sqrt{n^2 + n}) = \frac{1}{2}(1- \cos(2 \pi \sqrt{n^2 + n})), $$
But then what? how can I deal with the limit of $\cos (2 \pi \sqrt{n^2 + n})$? I know that $\cos (n\pi) = (-1)^n$, if $n$ is a positive integer but then what?
| You can check $\sin^2(\pi\sqrt{n^2 + n})=\sin^2(\pi\sqrt{n^2 + n}-\pi n)$.
So $$\sin^2(\pi\sqrt{n^2 + n})=\sin^2 \pi\frac{n}{\sqrt{n^2 + n}+n}\to \sin^2\frac{\pi}{2}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving equation with fraction I don't understand how to get from the second to the third step in this equation:
$ - \frac { \sqrt { 2 - x ^ { 2 } } - x \left( \frac { - x } { \sqrt { 2 - x ^ { 2 } } } \right) } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \sqrt { 2 - x ^ { 2 } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { \frac { 2 - x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } + \frac { x ^ { 2 } } { \sqrt { 2 - x ^ { 2 } } } } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 2 } } = - \frac { 2 } { \left( \sqrt { 2 - x ^ { 2 } } \right) ^ { 3 } } $
Why can we just add $ 2 - x ^ { 2 } $ in the numerator?
Step 1 to step 2, as well as step 3 to step 4 is clear to me.
| It's simply that for $a>0$, $$\sqrt{a} = \sqrt{a}\cdot\underbrace{\left(\frac{\sqrt{a}}{\sqrt{a}}\right)}_1$$
$$=\frac{\sqrt{a}\cdot\sqrt{a}}{\sqrt{a}}$$
$$=\frac{(\sqrt{a})^2}{\sqrt{a}}$$
$$=\frac{a}{\sqrt{a}}$$
In your case, "$a$" is $2-x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Maximum value of $ab+bc+ca$ given that $a+2b+c=4$ Question:
Find the maximum value of $ab+bc+ca$ from the equation,
$$a+2b+c=4$$
My method:
I tried making quadratic equation(in $b$) and then putting discriminant greater than or equal to $0$. It doesn't help as it yields a value greater than the answer.
Thanks in advance for the solution.
| Without using calculus: Substituting $c=4-2b-a$, we get $$ab+bc+ca=ab+(a+b)(4-2b-a)=(4(a+b)-(a+b)^2)-b^2$$ and since $f(x)=4x-x^2=4-(x-2)^2$ has maximum at $(2,4)$, substituting $x:=a+b$ gives $$ab+bc+ca\le4-b^2\le4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Matrix eigenvalues Consider the matrix $$A_n=\begin{bmatrix}
a & b & 0 & 0 & 0 & \dots & 0 & 0 & 0 \\
c & a & b & 0 & 0 & \dots & 0 & 0 & 0 \\
0 & c & a & b & 0 & \dots & 0 & 0 & 0 \\
0 & 0 & c & a & b & \dots & 0 & 0 & 0 \\
0 & 0 & 0 & c & a & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & 0 & \dots & a & b & 0 \\
0 & 0 & 0 & 0 & 0 & \dots & c & a & b \\
0 & 0 & 0 & 0 & 0 & \dots & 0 & c & a
\end{bmatrix}_{n\times n}$$
The matrix with $a=2$ and $b=c=-1$ is encountered in finite difference discretization of $u_{xx}.$
(a) If $D_n = \det(A_n),$ show that $D_n = aD_{n-1}-bcD_{n-2}.$
(b) Solve the recurrence analytically to obtain $D_n$ as a function of $n.$ (and ofcourse $D_n$ will also depend on $a, b, c.$)
(c) Obtain the eigenvalues of $A_n.$ (Hint: Replace $a$ by $a-\lambda$)$$ $$
$$ $$(a)Part can be shown easily by just simple Laplace expansion.
(b)We see that $D_0=1, D_1=a$.
Let $D_n=r^n$ be a solution of the recurrence relation \begin{equation}
D_n=aD_{n-1}-bcD_{n-2}
\end{equation}
Then characteristic equation corresponding to (1) \begin{alignat*}{3}
&\quad & r^n-ar^{n-1}+bcr^{n-2} &=0
\\&\implies &r^2-ar+bc &=0
\\&\implies &r_1=\tfrac{a-\sqrt{a^2-4bc}}{2}, r_2 &=\tfrac{a+\sqrt{a^2-4bc}}{2}
\end{alignat*}$ $
Case 1: $a^2-4bc=0$
$r_1=r_2=\frac{a}{2}$
General solution of (1) :
$D_n=(C_1+nC_2)(\frac{a}{2})^n$, where $C_1$ and $C_2$ are arbitrary constants.
For $n=0$, we get $C_1=D_0=1$.
For $n=1$, we get $(C_1+C_2)\frac{a}{2}=D_1=a\implies C_2=1$
Hence $D_n=(1+n)(\frac{a}{2})^n$
$$ $$Case 2: $a^2-4bc\neq0$
General solution of (1) :
$D_n=C_1r_1^n+C_2r_2^n$, with where $C_1$ and $C_2$ are arbitrary constants.
For $n=0$, we get $C_1+C_2=D_0=1$
For $n=1$, we get $C_1r_1+C_2r_2=D_1=a\implies (C_1+C_2)\frac{a}{2}+(C_2-C_1)\frac{\sqrt{a^2-4bc}}{2}=a
\implies 2C_2-1=\frac{a}{\sqrt{a^2-4bc}}
\implies C_2=\frac{r_2}{\sqrt{a^2-4bc}}$
$\therefore C_1=\frac{-r_1}{\sqrt{a^2-4bc}}$
Hence $D_n=\frac{r_2^{n+1}-r_1^{n+1}}{\sqrt{a^2-4bc}}=\frac{1}{2^{n+1}\sqrt{a^2-4bc}}[(a+\sqrt{a^2-4bc})^{n+1}-(a-\sqrt{a^2-4bc})^{n+1}]$ $$------------------------------------$$
I have done this far, but I'm stuck now.
Is there any simpler expression for $D_n$?
How to obtain eigenvalues, if we consider replacing $a$ by $a-\lambda$?
| You got questions (a) and (b) already. For (c) the eigenvalues, you need the characteristic equation $\det (A_n - \lambda I) = 0$. This is the same as $D_n = \det (A_n) = 0$, if in there $a$ is replaced by $a-\lambda$. From your result,
$$
0 = D_n({\rm a \; replaced}) =\frac{1}{2^{n+1}\sqrt{(a-\lambda)^2-4bc}}[(a-\lambda+\sqrt{(a-\lambda)^2-4bc})^{n+1}-(a-\lambda-\sqrt{(a-\lambda)^2-4bc})^{n+1}]$$
i.e. for $(a-\lambda)^2-4bc \ne 0$ (denominator $\ne 0$) we have
$$
(a-\lambda+\sqrt{(a-\lambda)^2-4bc})^{n+1}=(a-\lambda-\sqrt{(a-\lambda)^2-4bc})^{n+1}$$
or (be careful to obtain all the roots in $\sqrt[n+1]{1}$)
$$
a-\lambda+\sqrt{(a-\lambda)^2-4bc}=(a-\lambda-\sqrt{(a-\lambda)^2-4bc})\exp{(2\pi i k/(n+1))}$$
for $k = 0,1,\cdots,n$. Indexing the $\lambda_k$ with $k$, you get the results.
E.g.
$\lambda_0 = a \pm 2 \sqrt{bc}$ but that contradicts the above condition $(a-\lambda)^2-4bc \ne 0$.
Since $k=0$ is excluded, the general result is $\lambda_k = a \pm 2 \sqrt{bc} \cos(\frac{\pi k}{n+1})$ for $k = 1,2,\cdots,n$. Since $\cos(x) = -\cos(\pi -x)$, one of the two signs in $\pm$ actually suffices: $\lambda_k = a - 2 \sqrt{bc} \cos(\frac{\pi k}{n+1}) = a + 2 \sqrt{bc} \cos(\pi - \frac{\pi k}{n+1}) \\= a + 2 \sqrt{bc} \cos(\frac{\pi (n+1-k)}{n+1}) =a + 2 \sqrt{bc} \cos(\frac{\pi m}{n+1}) $
where $1 \le m = n+1-k \le n$, so the results with the positive sign are reproduced with the same range of the counting variable $m$.
We show the general result by plugging $\lambda_k = a + 2 \sqrt{bc} \cos(\frac{\pi k}{n+1})$ (plugging $\lambda_k = a - 2 \sqrt{bc} \cos(\frac{\pi k}{n+1})$ works as well ) in the determining equation for the eigenvalues. Indeed
$$
2 \sqrt{bc} [\cos(\frac{\pi k}{n+1})+\sqrt{\cos^2(\frac{\pi k}{n+1})-1}]=2 \sqrt{bc} [\cos(\frac{\pi k}{n+1})-\sqrt{\cos^2(\frac{\pi k}{n+1})-1}]\exp{(2\pi i k/(n+1))}$$
or
$$
\cos(\frac{\pi k}{n+1}) + i \sin(\frac{\pi k}{n+1})=[\cos(\frac{\pi k}{n+1}) - i \sin(\frac{\pi k}{n+1})]\exp{(2\pi i k/(n+1))}$$
or
$$
\exp{(\pi i k/(n+1))}=\exp{(-\pi i k/(n+1))}\exp{(2\pi i k/(n+1))}$$
which is an identity.
By the way, technically, what you have here is a tridiagonal Toeplitz matrix, where references can be found easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\[1ex]
\lim_{n\to\infty}\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3} = 0 \\[1ex]
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{0}
\end{align}
$$
Then, after realizing $\frac{1}{0}$ might not be a plausible limit, I tried to multiply out the variable with the largest exponent in both the dividend and the divisor.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \lim_{n\to\infty}\frac{n^3(1 - \frac{3}{n^3})}{n^2(2 + \frac{1}{n} - \frac{1}{n^2})} =
\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\
\lim_{n\to\infty}2 + \frac{1}{n} - \frac{1}{n^2} = 2 \\
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{2} \\
\lim_{n\to\infty}n = \infty
\end{align}
$$
So, my questions about this problem:
*
*Could $\frac{1}{0}$ be a valid limit?
*Does $\infty\cdot\frac{1}{2}$
equal to $\infty$?
*In conclusion, what is the limit of the sequence
above? $\infty?$
Thank you!
| You have done everything.I just want to represent it graphically.Then you will answer your own question.
Now,you tell me.what are you seeing here?
And ofcourse the answer of your your questions is simply "yes".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 6
} |
Showing that $ \lim_{x \to \infty} x(1/2)^x = 0$ Can someone explain to me why $$ \lim\limits_{x \to \infty} x\bigg(\frac{1}{2}\bigg)^x = 0$$
Is it because the $\big(\frac{1}{2} \big)^x$ goes towards zero as $ x $ approaches $\infty$, and anything multiplied by $0 $ included $\infty$ is $0$ ?
Or does this kind of question require using l'hopital's rule because it is in the form (0*$\infty$)?
I thought it could be solved this way. Please let me know if it is correct:
$$ \lim\limits_{x \to \infty} \bigg(\frac{x}{\frac{1}{2^{-x}}}\bigg)$$
L'hopitalize the above and get:
$$ \lim\limits_{x \to \infty} \bigg(\frac{1}{x(2^{x-1})}\bigg) = 0$$
| Let
$a_x = \dfrac{x}{2^x}$.
Then
$\dfrac{a_{x+1}}{a_x}
=\dfrac{\dfrac{x+1}{2^{x+1}}}{\dfrac{x}{2^x}}
=\dfrac{1+1/x}{2}
$.
Therefore,
for $x > 3$,
$\dfrac{a_{x+1}}{a_x}
=\dfrac{1+1/x}{2}
\lt\dfrac{1+1/3}{2}
=\dfrac46
=\dfrac23
$.
Therefore,
for $x > 3$,
$\begin{array}\\
\dfrac{a_x}{a_3}
&=\prod_{y=3}^{x-1}\dfrac{a_{y+1}}{a_y}\\
&\lt\prod_{y=3}^{x-1}\dfrac23\\
&=(2/3)^{x-3}\\
\text{so}\\
a_x
&\lt a_3(2/3)^3(2/3)^x\\
&= (3/8)(8/27)(2/3)^x\\
&= (1/9)(2/3)^x\\
\end{array}
$
To show
in an elementary way
that $c^x
\to 0$
for any $0 < c < 1$,
let
$c = 1/(1+d)$
where $d > 0$.
Note:
What follows is not original.
Then,
$d = 1/c-1$
so,
by Bernoulli's inequality,
$(1+d)^x \ge 1+xd > xd$
so
$c^x
=1/(1+d)^x
\lt 1/(xd)
=1/(x(1/c-1))
=c/(x(1-c))
\to 0$
as $x \to \infty$.
With $c = 2/3$,
$(2/3)^x
\lt (2/3)/(x(1/3))
=2/x
$
so that
$a_x \lt (1/9)(2/x)
=2/(9x)
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Help with Mellin-Barnes Integral (product of two Hypergeometrics) I am trying to prove that
$$\int_0^1 \frac{dz}{z^2} z^{h}\cdot {}_{2}F_{1}(h,h;2h;z) \cdot {}_{2}F_{1}\left(\frac{1+2a}{2},\frac{1-2a}{2};1;\frac{z-1}{z}\right) = -\frac{\Gamma(2h)}{\Gamma{(h)}^2} \cdot \left(\frac{1}{a^2-(h-1/2)^2} \right).$$
To begin I first use Mellin-Barnes representation of Hypergeometric representation:
$$ {}_{2}F_{1}(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\int_{-i\infty}^{i\infty}ds\frac{\Gamma(-s)\Gamma(a+s)\Gamma(b+s)}{\Gamma(c+s)}(-z)^s$$
The integral over the z variable is (done in Mathematica):
$$\int_0^1 \frac{z^h(-z)^s(1-z)^t}{z^2z^t}=(-1)^s\frac{\Gamma(-1+h+s-t)\Gamma(1+t)}{\Gamma(h+s)} $$
Plugging this back in I get,
$$\frac{\Gamma(2h)}{\Gamma(h)^2\Gamma(1/2+a)\Gamma(1/2-a)}\int\int ds dt \frac{\Gamma(-1+h+s-t)\Gamma(-s)\Gamma(h+s)\Gamma(-t)\Gamma(1/2+a+t)\Gamma(1/2-a+t)}{\Gamma(2h+s)}(-1)^s$$
where s and t are the Mellin-Barnes variables.
Then I use Barnes lemma to integrate over the t-variable to obtain:
$$\frac{\Gamma(2h)}{\Gamma(h)^2}\int_{-i\infty}^{i\infty}\frac{\Gamma(-s)\Gamma(-1/2+h+s+a)\Gamma(-1/2+h+s-a)}{\Gamma(2h+s)} \, ds$$
I dont see how I will get to the final result from here.
Any help appreciated.
Thanks
| The hypergeometric function in contour form is
$${}_{2}F_{1}(a, b; c; x) = \frac{1}{2 \pi i} \, \int_{-i \infty}^{i \infty} \frac{\Gamma(-s) \Gamma(a+s) \Gamma(b+s)}{\Gamma(c + s)} \, ds.$$
The integral in question becomes:
\begin{align}
I &= \int_{0}^{1} {}_{2}F_{1}(p,p;2p;t) \cdot {}_{2}F_{1}\left(\frac{1+2a}{2},\frac{1-2a}{2};1;\frac{t-1}{t}\right) \, t^{p-2} \, dt \\
&= \left(\frac{1}{2 \pi i}\right)^2 \, \int \int \frac{\Gamma(-u) \Gamma^2(p+u)}{\Gamma(2p + u)} \cdot \frac{\Gamma(-s) \Gamma(\frac{1}{2} + a + s) \Gamma(\frac{1}{2} - a + s)}{\Gamma(s + 1)} \, J \, du \, ds,
\end{align}
where
\begin{align}
J &=\int_{0}^{1} t^{p-2} \, (-t)^{u} \, \left(-1 + \frac{1}{t}\right)^{s} \, dt = (-i)^{u} \, B(s+1, p+u-s-1) \\
&= (-1)^{u} \, \frac{\Gamma(s+1) \Gamma(p+u-s-1)}{\Gamma(p+u)}.
\end{align}
Now, by use of Barnes' Lemma (first) given by
$$\frac{1}{2 \pi i} \, \int_{-i \infty}^{i \infty} \Gamma(a+s) \Gamma(b+s) \Gamma(c -s) \Gamma(d -s) \, ds = \frac{\Gamma(a+c) \Gamma(a+d) \Gamma(b+c) \Gamma(b+d)}{\Gamma(a+b+c+d)},$$
with $d = 0$, $c = p+u-1$, $a = 1/2 +a$ and $b = 1/2 -a$, then
$$ \frac{1}{2 \pi i} \, \int_{- i \infty}^{i \infty} \Gamma(-s) \Gamma(\frac{1}{2} + a + s) \Gamma(\frac{1}{2} - a + s) \Gamma(p+u-s-1) \, ds = \frac{\Gamma(a + \frac{1}{2}) \Gamma(p+a-\frac{1}{2} + u) \Gamma(p - a - \frac{1}{2} +u)}{\Gamma(p+u)} $$
and
\begin{align}
I &= \left(\frac{1}{2 \pi i}\right) \, \int \frac{(-1)^u \Gamma(-u) \Gamma(p+u)}{\Gamma(2p + u)} \, du \\
& \hspace{5mm} \cdot \frac{1}{2 \pi i} \, \int_{- i \infty}^{i \infty} \Gamma(-s) \Gamma(\frac{1}{2} + a + s) \Gamma(\frac{1}{2} - a + s) \Gamma(p+u-s-1) \, ds \\
&= \frac{\Gamma(a + \frac{1}{2})}{2 \pi i} \, \int_{- i \infty}^{i \infty} \frac{\Gamma(-u) \Gamma(p-a-\frac{1}{2} + u) \Gamma(p +a-\frac{1}{2} +u)}{\Gamma(2 p + u)} \, (-1)^{u} \, du \\
&= \Gamma\left(a + \frac{1}{2}\right) \, {}_{2}F_{1}\left(p+a-\frac{1}{2}, p-a-\frac{1}{2}; 2p ; 1\right) \\
&= \frac{\Gamma(2p) \Gamma\left(a+\frac{1}{2}\right)}{\Gamma\left(p+a+\frac{1}{2}\right) \Gamma\left(p-a+\frac{1}{2}\right)}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Definite integral of $x\sin^n x$ from $0$ to $\pi/2$ How to find
\begin{equation*}
\int_0^{\pi/2} x\sin^n x dx
\end{equation*}
where $n$ is a positive integer? I tried $y=x-\pi/4$ and that gives
\begin{equation*}
\frac{1}{2^{n/2}}\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} (\sin y+\cos y)^n dy+\frac{1}{2^{n/2}}\int_{-\pi/4}^{\pi/4} y(\sin y+\cos y)^n dy.
\end{equation*}
Although some terms may be canceled, the summation of the others seems terrible.
Another method is integration by parts. Here I found the recurrence formula for $\sin^n x$:
Integrating $\int \sin^n{x} \ dx$
So the first step is to have
\begin{equation*}
\left.\left(-\frac{1}{n}x\cos x\sin^{n-1} x+\frac{n-1}{n}x\int \sin^{n-2} xdx\right)\right|_0^{\pi/2}-\int_0^{\pi/2}\left(-\frac{1}{n}\cos x\sin^{n-1} x+\frac{n-1}{n}\int_0^x \sin^{n-2} t dt\right)dx
\end{equation*}
the evaluations at $0$ and $\pi/2$ and the integral of $\cos x\sin^{n-1}x$ seem OK, but what about the second term? Any better method for this problem?
| Do the same thing: let the integral be $I_n$,
$$
\begin{align*}
&\phantom{=}\int_0^{\pi/2} x \sin(x)^n \mathrm dx = -\int_0^{\pi/2} x \sin(x)^{n-1}\mathrm d(\cos(x)) \\
&=\left. x \sin(x)^{n-1} \cos(x) \right|_{\pi/2}^0 - \int_{\pi/2}^0 \cos(x) (\sin(x)^{n-1} + (n-1)x \sin(x)^{n-2} \cos(x))\mathrm d x \\
&= 0 + \int_0^{\pi/2} \sin(x)^{n-1}\mathrm d (\sin(x)) +(n-1) \int_0^{\pi/2} x\sin(x)^{n-2}(1 - \sin(x)^2)\mathrm dx\\
&= \frac 1n + (n-1) I_{n-2} - (n-1)I_n ,
\end{align*}
$$
i.e.
$$
\fbox{$nI_n =\dfrac 1n + (n-1)I_{n-2}, $}
$$
with
$$
I_1 = 1,\quad I_2 = \frac 14 + \frac {\pi^2} {16},
$$
since
\begin{align*}
I_2&=\int_0^{\pi/2} x\sin(x)^2\mathrm dx \\
&= \left. x \sin(x) \cos(x)\right|_{\pi/2}^0 + \int_0^{\pi/2} \cos(x) (\sin(x)+ x\cos(x)) \mathrm dx \\
&= 0 + \frac 12 + \int_0^{\pi/2} x(1 - \sin(x)^2) \mathrm dx\\
&= \frac 12 + \frac {\pi^2} 8 - I_2\\
&\implies I_2 = \frac 14 + \frac {\pi^2}{16}.
\end{align*}
To go further, we are going to solve the recursive relation, but seems gruesome. So I let you take it from here.
UPDATE
According to wiki, we have a systematic method to solve this. First consider the terms with even index,
Let $J_{n} = I_{2n}$, then
$$
J_n = \frac 1{4n^2} + \frac {2n-1}{2n}J_{n-1},
$$
then divide both side of the equation by $(2n-1)!!/(2n)!!$:
$$
\frac {J_n}{\dfrac {(2n-1)!!}{(2n)!!}} = \frac 1 {4n^2} \frac {(2n)!!}{(2n-1)!!}+ \frac {J_{n-1}}{\dfrac {(2n-3)!!}{(2n-2)!!}},
$$
now let $P_n = J_n (2n)!!/(2n-1)!!$, then
$$
P_n = \frac 1{2n} \frac {(2n-2)!!}{(2n-1)!!} + P_{n-1},
$$
then
$$
P_n = \sum_2^n \frac 1{2k}\frac {(2k-2)!!}{(2k-1)!!} + P_1,
$$
which is pretty much for deriving a concrete number of $I_n$.
| {
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"url": "https://math.stackexchange.com/questions/2917725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$
$$\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$$
I tried $2$ or $3$ trigonometric transformations but it did not work. One of them is as follows
$$\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}$$ after that I am not able to figure out what to do. If I use double angle formula then
$$\frac{\frac{\sin2x}{4}}{(\sin x+\cos x)\left(1-\frac{\sin2x}{2}\right)}$$ again i am clueless
| Using identities
$$\sin x+\cos x=\sqrt{2}\cos(x-\dfrac{\pi}{4})$$
$$\sin x\cos x=\cos^2(x-\dfrac{\pi}{4})-\dfrac12$$
and then substitution $\dfrac{\pi}{4}-x=u$ gives
\begin{align}
I
&= \int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx \\
&= \int_{0}^{\frac{\pi}{4}}\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}dx \\
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{4}}\frac{(2\cos^2u-1)^2}{\cos u(3-2\cos^2u)}\ du \\
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\pi}{4}}\frac{(2\cos^2u-1)^2}{\cos^2u(3-2\cos^2u)}\cos u\ du
\end{align}
now let $\sin u=w$ therefore
\begin{align}
I
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\sqrt{2}}{2}} \frac{(2w^2-1)^2}{(1-w^2)(2w^2+1)}\ dw \\
&= \dfrac{1}{2\sqrt{2}}\int_{0}^{\frac{\sqrt{2}}{2}} -2+\frac13\frac{1}{1-w^2}+\frac83\frac{1}{2w^2+1}\ dw \\
&= \color{blue}{-\dfrac12+\dfrac{\pi}{6}+\dfrac{\sqrt{2}}{12}\operatorname{arccoth}\sqrt{2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Sum of series $\frac{ 4}{10}+\frac{4\cdot 7}{10\cdot 20}+\frac{4\cdot 7 \cdot 10}{10\cdot 20 \cdot 30}+\cdots \cdots$
Sum of the series
$$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots $$
Sum of series $\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots$
But i want to solve it using Beta ganmma function.
My Process follows
Let $$S = \frac{1}{10}\bigg[\frac{ 4}{1}+\frac{4\cdot 7}{1\cdot 2}+\frac{4\cdot 7 \cdot 10}{1\cdot 2 \cdot 3}+\cdots \cdots \bigg]$$
Let $\displaystyle a_{n} = \prod^{n}_{k=1}(3k+1)=3^n \prod^{n}_{k=1}\bigg(k+\frac{1}{3}\bigg)$
$\displaystyle =3^n\Gamma\bigg(n+\frac{4}{3}\bigg)\cdot \frac{1}{\Gamma \bigg(\frac{4}{3}\bigg)}$
And Let $\displaystyle b_{n} = \prod^{n}_{k=1}k = \Gamma (n+1)$
So $\displaystyle S =\frac{1}{10} \sum^{\infty}_{n=1}3^n \frac{\Gamma \bigg(n+\frac{4}{3}\bigg)\cdot \Gamma \bigg(\frac{2}{3}\bigg)}{\Gamma \bigg(\frac{2}{3}\bigg)\Gamma \bigg(\frac{4}{3}\bigg)\cdot \Gamma(n+1)}$
using $\displaystyle \Gamma(x)\cdot \Gamma(1-x) = \frac{\pi}{\sin (\pi x)}$ and
$\displaystyle \frac{\Gamma (a) \Gamma(b)}{\Gamma(a+b)} =B(a,b)= \int^{1}_{0}x^{a-1}(1-x)^{b-1}dx$
So $\displaystyle S = \frac{1}{10}\sum^{\infty}_{n=1}3^n\int^{1}_{0}x^{n+\frac{1}{3}}(1-x)^{-\frac{1}{3}}dx$
$\displaystyle =\frac{1}{10}\int^{1}_{0}\frac{3x}{1-3x}\cdot \bigg(\frac{x}{1-x}\bigg)^{\frac{1}{3}}dx$
Now put $\displaystyle \frac{x}{1-x} = t\Rightarrow x = \frac{t}{1+t} = 1-\frac{1}{1+t}$
So $\displaystyle S = \frac{3}{10}\int^{\infty}_{0}\frac{t^{\frac{4}{3}}}{(2t-1)(1+t)^2}dt$
Could some Help me to solve it, Thanks
although it has solved Here
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{{4 \over 10} + {4 \cdot 7 \over 10 \cdot 20} +
{4 \cdot 7 \cdot 10 \over 10 \cdot 20 \cdot 30} + \cdots =
\sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{3k + 1} \over
\prod_{k = 1}^{n}10k}:\ {\LARGE ?}}$.
\begin{align}
&\bbox[10px,#ffd]{\ds{\sum_{n = 1}^{\infty}
{\prod_{k = 1}^{n}\pars{3k + 1} \over \prod_{k = 1}^{n}10k}}} =
\sum_{n = 1}^{\infty}
{3^{n}\prod_{k = 1}^{n}\pars{k + 1/3} \over 10^{n}n!} =
\sum_{n = 1}^{\infty}
{\pars{3/10}^{n} \over n!}\,\pars{4 \over 3}^{\large\overline{n}}
\\[5mm] = &\
\sum_{n = 1}^{\infty}
{\pars{3/10}^{n} \over n!}\,{\Gamma\pars{4/3 + n} \over \Gamma\pars{4/3}} =
\sum_{n = 1}^{\infty}
\pars{3 \over 10}^{n}\,{\pars{n + 1/3}! \over n!\pars{1/3}!}
\\[5mm] = &\
\sum_{n = 1}^{\infty}\pars{3 \over 10}^{n}\,{n + 1/3 \choose n} =
\sum_{n = 1}^{\infty}\pars{3 \over 10}^{n}
\bracks{{-4/3 \choose n}\pars{-1}^{n}}
\\[5mm] = &\
\sum_{n = 1}^{\infty}
{-4/3 \choose n}\pars{-\,{3 \over 10}}^{n} =
\bracks{1 + \pars{-\,{3 \over 10}}}^{-4/3} - 1
\\[5mm] = &\
\bbx{\pars{10 \over 7}^{4/3} - 1} \approx 0.6089
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove formula $\operatorname{arctanh} x = \frac12\,\ln \left(\frac{1+x}{1-x}\right)$ Problem
Prove formula $\operatorname{arctanh} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)$
Attempt to solve
To start off with definition of functions $\sinh(x)$ and $\cosh(x)$
$$ \cosh(x)=\frac{e^x+e^{-x}}{2} $$
$$ \sinh(x) = \frac{e^x-e^{-x}}{2} $$
Hyperbolic tangent is defined as:
$$ \tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{e^x-e^{-x}}{e^{x}+e^{-x}} $$
Notation $\text{arctanh}(x)$ means area tangent which is inverse of hyperbolic tangent.
$$ \operatorname{arctanh}(x)=\tanh^{-1}(x) $$
Trying to invert the $\tanh(x)$ we get:
$$\begin{align}
\frac{e^x-e^{-x}}{e^x+e^{-x}} &= y &\implies\\
e^x-e^{-x}&=y(e^x+e^{-x}) &\implies\\
e^x-e^{-x}&=ye^{x}+ye^{-x} &\implies\\
e^x(1-y)&=e^{-x}(1+y) &\implies\\
\ln(e^x(1-y)) &= \ln(e^{-x}(1+y)) &\implies\\
\ln(e^x)+\ln(1-y) &= \ln(e^{-x})+\ln(1+y) &\implies\\
x + \ln(1-y) &= -x + \ln(1+y) &\implies\\
2x &= \ln(1+y)-\ln(1-y)&\implies\\
x &= \frac{1}{2} \ln \frac{1+y}{1-y}
\end{align}
$$
By switching variables we get:
$$ \implies \operatorname{arctanh}(y) = \frac{1}{2} \ln \left(\frac{1+y}{1-y}\right) $$
| Just to add to the responses already here. You can also show this by solving this integral two different ways, one with trigonometric substitutions and the other with partial fraction decomposition.
$$ \int \frac {dx}{x^2+1} = \int \frac {dx}{x^2+1}$$
$\qquad x \mapsto \tan u$
$$ \int\frac{dx}{(1+ix)(1-ix)}= \int \frac {\sec^2 u}{1+ \tan^2u}du$$
$$ \int \Big(\frac{\frac 12}{1+ix}+\frac{\frac 12}{1-ix}\Big)dx\ = \int du$$
$$\frac 1{2i}\big(\ln(1+ix)- \ln(1-ix)\big)=u+C$$
$$\arctan x=\frac i2\ln \Big( \frac{1-ix}{1+ix} \Big)$$
Plug in $x=ix.$
$$\arctan(ix)=\frac i2 \ln\Big(\frac{1+x}{1-x} \Big)$$
Now use the identity$ \space\operatorname {arctanh}(x)=-i\arctan(ix).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Let $f(x) = 2x + 3$ and $g(x) = ax + b$. If $f(g(x)) = x$ for all $x$, determine $a$ and $b$ This is what I know:
$f(x) = 2x + 3$
$g(x) = ax+b$
$f(g(x)) = 2(ax+b) + 3 = x$
I am just not too sure what to do next to find $a$ and $b$.
| As mentioned in the comments, we have $f(g(x)) = 2ax + 2b + 3 = x$. This is only true if $2a = 1$ and $2b + 3 = 0$, otherwise the equality wouldn't hold. Solving those, we get that $a = \dfrac{1}{2}$ and $b = -\dfrac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all the positive integers k for which $7 \times 2^k+1$ is a perfect square Find all the positive integers $k$ for which $7 \times 2^k+1$ is a perfect square.
The only value of $k$ I can find is $5$. I am not sure how to find every single one or the proof, I simply used trial and error.
| \begin{align}
a^2 &= 7*2^k + 1 \\
\Rightarrow (a^2 - 1) = (a - 1)(a+1) &= 7*2^k \\
\text{For } k > 2: (\frac{a-1 }{2}) (\frac{a+1}{2}) &= 7*2^{k-2} \\
\text{Let } d = (\frac{a-1 }{2}) &\Rightarrow d + 1 = (\frac{a+1}{2}).\\
\Rightarrow d(d+1) &= 7*2^n (n = k - 2\in \mathbb{N})\\
gcd(d,d+1) = 1 &\Rightarrow (d,d+1) = (7,2^n) \text{ or } (2^n,7) \\
\text{ Only answer possible is } n = 3 &\Leftrightarrow k = 5 \text{ if } k > 2.\\
k = 1,2 \text{ do not satisfy, check individually. }
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I find the equation of a tangent to a hyperbola whose centre is (h,k)? Given that $\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$ is equation a hyperbola,
I have to find its tangent at the point $\left(-2,\frac{14}{3} \right)$.
I know about the equations $c^2=(am)^2-b^2$ and $\frac{xx1}{a^2} - \frac{yy1}{b^2} = 1$ but cant figure how to apply those here since the centre is not at origin.
| Your hyperbola is $\frac{(x- 3)^2}{9}- \frac{(y- 2)^2}{4}= 1$ and you want to find the tangent line to it "at (-2, 14/3)". The first thing I would do is check to make sure that point is on the hyperbola. With x= -2, $(x- 3)^2= 25$ and $\frac{(x- 3)^2}{9}= \frac{25}{9}$. With $y= \frac{14}{3}$, $(y- 2)^2= \frac{64}{9}$ and $\frac{y- 14/3}{4}= \frac{16}{9}$. Yes, a bit to my surprise, $\frac{(x- 3)^2}{9}- \frac{(y- 2)^2}{4}= \frac{25}{9}- \frac{16}{9}= \frac{9}{9}= 1$ and the point is on the hyperbola!
Now, do you know what a "tangent line" is and how to find the slope of a tangent line without simply plugging into formulas? Using "implicit differentiation" to differentiate both sides with respect to x, we have $\frac{2}{9}(x- 3)- \frac{1}{2}(y- 2)y'= 0$. In particular, at x= -2, y= 14/3, $\frac{2}{9}(-5)- \frac{1}{2}\frac{8}{3}y'= -\frac{10}{9}-\frac{4}{3}y'= 0$ so $y'= -\left(\frac{3}{4}\right)\left(\frac{10}{9}\right)= -\frac{5}{6}$. The tangent line is $y= -\frac{5}{6}\left(x+ 2\right)+ \frac{14}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find a Jordan form of a non-diagonazable matrix I am trying to find Jordan's form of this matrix:
\begin{pmatrix}
0 & 1 & 0\\
-4& 4& 0\\
-2 & 1& 2
\end{pmatrix}
The only eigenvalue $r$ is 2 and therefore the simplest eigenvector $v_{1}$ is (0, 0, 1)
To get the other two independent vectors (generalized eigenvectors) for the P matrix we need to compute $(A-rI)v_{2}=v_{1}$ and $(A-rI)v_{3}=v_{2}$ .
However, in the calculation for 2 I get the following system which has no solution:
$$ \left[
\begin{array}{ccc|c}
-2 & 0& 0&0 \\
0& 1& 0&0 \\
0& 0& 0 & 1
\end{array}
\right] $$
Any Idea on what I am doing wrong? If the eigenvalue is 2, when we substract it the lowest row will always contain all zeros...
| I like this method for hand calculations: first, calling your matrix $A,$ let
$$ B = A - 2 I $$
$$
B =
\left(
\begin{array}{ccc}
-2&1&0 \\
-4&2&0 \\
-2&1&0
\end{array}
\right)
$$
A basis for the genuine eigenvectors is given by the convenient
$$
E =
\left(
\begin{array}{cc}
1&0 \\
2&0 \\
0&1
\end{array}
\right)
$$
We may or may not be using these eigenvectors in the form shown. We know that the characteristic equation for $A$ is just showing $B^3 = 0.$ However, the minimal polynomial for $A$ is $B^2 = 0,$ which you can check easily enough.
We are going to make a matrix $R$ with columns $u,v,w;$ on the far right, we take $w$ as any vector with $B^2 w = 0$ (automatic) but $Bw \neq 0.$ Then $v = Bw$ will be a genuine eigenvector. Finally, we will choose an independent eigenvalue $u.$
I like ones and zeros, I choose
$$
w =
\left(
\begin{array}{c}
0 \\
1 \\
0
\end{array}
\right)
$$
Then from $v = B w$ we get
$$
v =
\left(
\begin{array}{c}
1 \\
2 \\
1
\end{array}
\right)
$$
This $v$ is a genuine eigenvector, it is the sum of the two columns of my $E.$
At last, we get to choose some $u$ eigenvector that is not a multiple of $v,$ I choose
$$
u =
\left(
\begin{array}{c}
0 \\
0 \\
1
\end{array}
\right)
$$
$$
R =
\left(
\begin{array}{ccc}
0&1&0 \\
0&2&1 \\
1&1&0
\end{array}
\right)
$$
Next we find $R^{-1} $ and $J = R^{-1}A R,$ which will be the Jordan form if we did it correctly. A piece of luck , because of choosing ones and zeros, the determinant of $R$ is small, actually $1,$ and we calculate
$$
R^{-1} =
\left(
\begin{array}{ccc}
-1&0&1 \\
1&0&0 \\
-2&1&0
\end{array}
\right)
$$
and
$$
J = R^{-1} A R =
\left(
\begin{array}{ccc}
2&0&0 \\
0&2&1 \\
0&0&2
\end{array}
\right)
$$
$$ $$
$$ R^{-1} A R = J $$
$$
\left(
\begin{array}{ccc}
-1&0&1 \\
1&0&0 \\
-2&1&0
\end{array}
\right)
\left(
\begin{array}{ccc}
0&1&0 \\
-4&4&0 \\
-2&1&2
\end{array}
\right)
\left(
\begin{array}{ccc}
0&1&0 \\
0&2&1 \\
1&1&0
\end{array}
\right) =
\left(
\begin{array}{ccc}
2&0&0 \\
0&2&1 \\
0&0&2
\end{array}
\right)
$$
$$ $$
$$ R J R^{-1} = A $$
$$
\left(
\begin{array}{ccc}
0&1&0 \\
0&2&1 \\
1&1&0
\end{array}
\right)
\left(
\begin{array}{ccc}
2&0&0 \\
0&2&1 \\
0&0&2
\end{array}
\right)
\left(
\begin{array}{ccc}
-1&0&1 \\
1&0&0 \\
-2&1&0
\end{array}
\right) =
\left(
\begin{array}{ccc}
0&1&0 \\
-4&4&0 \\
-2&1&2
\end{array}
\right)
$$
COMMENT: the line above "as any vector with $B^2 w = 0$ (automatic)" may appear silly. If, however, you were given a 5 by 5 matrix $A$ with characteristic polynomial $(x - 5)^3 (x-7)^2$ and minimal polynomial
$(x - 5)^2 (x-7),$ the demand for a vector $w$ with $(A - 5I)^2 w = 0$ but $(A - 5I) w \neq 0$ would make some sense, as the eigenvalue $7$ is ignored in this condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solution to this Differential Equation $f''(x)=f(x)f'(x)$ needed I came up with this differential equation and I don't know how to solve it.
$$f''(x)=f(x)f'(x)$$
I attempted to solve it several times, but they were all fruitless. Wolfram Alpha says that the solution is
$$f(x)=\sqrt{2a} \tan\left({\frac{\sqrt{2a}}{2} \cdot (x+b)}\right),$$ where $a$ and $b$ are constants.
How does one get this solution?
| Let us consider your differential equation:
$$f''(x)=f(x)\cdot f'(x)$$
Integrate with respect to $x$ on both sides. Recognize that $df'(x)=f''(x)\ dx$ and $df(x)=f'(x)\ dx$:
$$\int f''(x)\ dx=\int f(x)\cdot f'(x)\ dx\rightarrow \int df'(x)=\int f(x)\ df(x).$$
It follows that
$$f'(x)=\frac{(f(x))^2}{2}+a=\frac{(f(x))^2+2\cdot a}{2}.$$
Divide by $(f(x))^2+2\cdot a$ on both sides:
$$\frac{f'(x)}{(f(x))^2+2\cdot a}=\frac{1}{2}.$$
Integrate with respect to $x$ on both sides. Recognize that $df(x)=f'(x)\ dx$:
$$\int \frac{f'(x)\ dx}{(f(x))^2+2\cdot a}=\int \frac{dx}{2}\rightarrow \int \frac{df(x)}{(f(x))^2+2\cdot a}=\frac{x}{2}+b=\frac{x+2\cdot b}{2}.$$
Redefine $2\cdot b$ as $b$, since it is a constant:
$$\int \frac{df(x)}{(f(x))^2+2\cdot a}=\frac{x+b}{2}.$$
Let $f(x)=\sqrt{2\cdot a}\cdot \tan(s)$ such that $df(x)=\sqrt{2\cdot a}\cdot (\tan^2(s)+1)\ ds$:
$$\int \frac{ds}{\sqrt{2\cdot a}}=\frac{x+b}{2}\rightarrow \frac{s}{\sqrt{2\cdot a}}=\frac{x+b}{2}.$$
Isolate $s$ and let $s=\arctan \left (\frac{f(x)}{\sqrt{2\cdot a}}\right)$:
$$s=\frac{\sqrt{2\cdot a}}{2}\cdot (x+b)\rightarrow \arctan \left (\frac{f(x)}{\sqrt{2\cdot a}}\right)=\frac{\sqrt{2\cdot a}}{2}\cdot (x+b).$$
Therefore, an expression for your function $f(x)$ can be written as
$$f(x)=\sqrt{2\cdot a}\cdot \tan \left (\frac{\sqrt{2\cdot a}}{2}\cdot (x+b)\right).$$
The derived expression is equivalent to what you found with Wolfram Alpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Minimum of $\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n$
I would like to find the minimum of
$$f(x)=\left(\frac{1+\sin^2x}{\sin^2x}\right)^n+\left(\frac{1+\cos^2x}{\cos^2x}\right)^n,$$
where $n$ is a natural number.
I know there is possible by derivate, but
$$f'(x)=n \left(\left(\cos ^2(x)+1\right) \sec ^2(x)\right)^{n-1} \left(2 \left(\cos ^2(x)+1\right)\tan (x) \sec ^2(x)-2 \tan (x)\right)+n \left(\left(\sin ^2(x)+1\right) \csc^2(x)\right)^{n-1} \left(2 \cot (x)-2 \left(\sin ^2(x)+1\right) \cot (x) \csc^2(x)\right).$$
I think this is not the best way.
| Using A.M $\geq G.M.$ on the first and second inequalities yields
$$
\begin{aligned}
f(x) &=\left(\csc ^{2} x+1\right)^{n}+\left(\sec ^{2} x+1\right)^{n} \\
& \geqslant 2 \sqrt{\left[\left(\csc ^{2} x+1\right)\left(\sec ^{2} x+1\right)\right]^{n}} \\
&=2\left[\left(2+\cot ^{2} x\right)\left(2+\tan ^{2} x\right)\right]^{\frac{n}{2}} \\
&=2\left(4+2 \cot ^{2} x+2 \tan ^{2} x+1\right)^{\frac{n}{2}} \\
& \geqslant 2\left(4+2 \cdot 2 \sqrt{\cot ^{2} x \tan ^{2} x}+1\right)^{\frac{n}{2}} \\
&=2\cdot 9^{\frac{n}{2}} \\
&=2\cdot 3^{n}
\end{aligned}
$$
$\therefore f(x)$ attains its minimum value $2 \cdot 3^{n}$ when $$\cot ^{2} x=\tan ^{2} x \textrm{ and } \csc^2 x=sec^2 x\Leftrightarrow x=\pm \frac{\pi}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Why isn't this approach in solving $x^2+x+1=0$ valid? There is this question in which the real roots of the quadratic equation have to be found:
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x \neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 \neq 0$
Therefore, it is legal to divide each term by $x$:
$x + 1 + \frac{1}{x} = 0$
$x = -1 - \frac{1}{x}$
Now, substitute $x$ into the original equation and solve:
$x^2 + (-1-\frac{1}{x}) + 1 = 0$
$x^2-\frac{1}{x} = 0$
$x^3 = 1$
$x = 1$
to get $x = 1$. Clearly this isn't the right answer. But why? Thanks.
| Transformations you apply to an equation may introduce alien solutions.
Taking an extreme example,
$$x=0\implies 0=0$$ which is satisfied by all $x$ !
So you may apply transformations, but validate the solutions using the original equation.
In your example, you establish
$$x^2+x+1=0\implies x^3-1=0.$$
But as $$x^3-1=0=(x-1)(x^2+x+1),$$ nothing is wrong if you ignore the first factor.
| {
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Prove that there exists a number $x$ such that $x^3 = 6$ I want to show that there exists a real number $x$ such that $x^3 = 6$. Here is what I have so far. $\\$
Let $S = \{x \mid x \in \mathbb{R}, x \geq 0, x^3 < 6\}$. By this definition, $S$ is nonempty since $0 \in S$, and also $S$ is bounded above since $2^3 = 8 > 6$. Thus, by the Completeness Axiom, $S$ has a least upper bound; call it $b$. We will show that $b^3 = 6$ (and hence, there exists a real number such that $x^3 = 6$) by showing that we cannot have $b^3 > 6$ or $b^3 < 6$.
First, for the sake of contradiction, suppose we had $b^3 > 6$. Then, we will show that we can choose a suitably small positive number $\epsilon$ such that $b - \epsilon$ is also an upper bound for $S$, which contradicts $b$ being the least upper bound. But, I'm not sure about how to find $\epsilon$. I tried expanding:
$$(b - \epsilon)^3 = b^3 - 3b^2\epsilon + 3b\epsilon^2 - \epsilon^3,$$
and from here, I think I'm supposed to use greater-than equalities to try and come up with $\epsilon$, but I'm not really sure how to do that. Any help is appreciated.
| Note that$$b^3 - 3b^2\varepsilon + 3b\varepsilon^2 - \varepsilon^3>6\iff b^3-6>3b^2\varepsilon-3b\varepsilon^2+\varepsilon^3.$$Now, take $\varepsilon\in\left(0,1\right)$ such that$$\varepsilon<\dfrac{b^3-6}{6b^2\varepsilon}\tag1$$and that$$\varepsilon<\dfrac{b^3-6}2.\tag2$$Then\begin{align}3b^2\varepsilon+\varepsilon^3&<3b^2\varepsilon+\varepsilon\text{ (because $\varepsilon<1$)}\\&<\frac{b^3-6}2+\frac{b^3-6}2\text{ (by $(1)$ and $(2)$)}\\&=b^3-6.\end{align}Therefore$$3b^2\varepsilon-3b\varepsilon^2+\varepsilon^3<3b^2\varepsilon+\varepsilon^3<b^3-6.$$
| {
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Solve $\sin^{3}x+\cos^{3}x=1$
Solve for $x\\ \sin^{3}x+\cos^{3}x=1$
$\sin^{3}x+\cos^{3}x=1\\(\sin x+\cos x)(\sin^{2}x-\sin x\cdot\cos x+\cos^{2}x)=1\\(\sin x+\cos x)(1-\sin x\cdot\cos x)=1$
What should I do next?
| Hint $$\sin^3(x)+\cos^3(x)=\sin^2(x)+\cos^2(x).$$
| {
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Solve $7^x+x^4+47=y^2$
Solve $$7^x+x^4+47=y^2$$ where $x, y \in \mathbb{N}^*$
If $x$ is odd then the left term is congruent with $3$ mod $4$ so it couldn't be a perfect square, so we deduce that $x=2a$ and the relation becomes $$49^a+16a^4+47=y^2$$ and it is easy to see that the left term is divisible by $16$ so we obtain that $y=4b$, so we have to find $a$ and $b$ such that $$49^a+16a^4+47=16b^2$$From this point I was completely stuck. I think that there are no solutions but how can I prove it?
| When $x$ is odd then $7^x+x^4+47\equiv 3(\mod 4)$. So, it is not possible $y^2\equiv 3(\mod 4)$.
Let $x=2k$, then for $k\geq 4$
$$(7^k)^2<7^{2k}+(2k)^4+47<(7^k+1)^2$$
This mean we have $k\leq 3$.
if try to $k=1,2,3$ only $x=4$ is a solution.
| {
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Using trigonometry to prove $\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}=\frac{4abc}{\left(1-a^2\right)\left(1-b^2\right)\left(1-c^2\right)}.$
For the numbers $a,b,c$ with $ab+ac+bc=1$, prove that
$$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}=\frac{4abc}{\left(1-a^2\right)\left(1-b^2\right)\left(1-c^2\right)}.$$
This is a question from the trigonometric section of my textbook. But is it possible to use trigonometry here?
Edit:
Using the Dinesh Shankar's hint with $a=\tan\frac{x}{2}$, $b=\tan\frac{y}{2}$ and $c=\tan\frac{z}{2}$, the equation becomes
$$\tan x+\tan y+\tan z=\tan x\cdot\tan y\cdot\tan z.$$
But $z=\pi-(x+y)$, then
$$\tan x + \tan y + \tan \left(\pi-(x+y)\right)=\tan x\cdot\tan y\cdot\tan\left(\pi-(x+y)\right),$$
which implies
$$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\cdot\tan y}.$$
Therefore, the original equation is also valid.
Please feel free to give another solution.
| Hint:
You probably know that if $x+y+z=\pi$, then
$$\tan\frac{x}{2}\cdot\tan\frac{y}{2}+\tan\frac{x}{2}\cdot\tan\frac{z}{2}+\tan\frac{y}{2}\cdot\tan\frac{z}{2}=1.$$
Now, since $ab+ac+bc=1$, use $a=\tan\frac{x}{2}$....
I hope this help you.
| {
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Prove this problem using Mean - Value theorem let $m$ and $n$ are in $\mathbb{N}$ with $m>n$ . Use the Mean Value Theorem, to prove that $(1+x)^\frac{m}{n}\ge 1+\frac{m}{n}x, x \ge -1$
since $(1+x)^\frac{m}{n}$ is continuous on $[-1,x]$ and differentiable on $(-1,x)$ so by mean value theorem $\frac{f(x)-f(-1)}{x+1}=f'(c)$ for $-1\le c\le x$
then $\frac{(1+x)^{\frac{m}{n}}}{x+1}=\frac{m}{n}(1+c)^{\frac{m}{n}-1}$
form here how to we processed
| Let $f(x) = (1+x)^{\frac mn}$. Then
$$f'(x) = \frac mn (1+x)^{\frac mn -1} =\frac mn (1+x)^{\frac {m-n}n} $$
$f'(x)$ is increasing for $x\ge -1$, and in particular, $f'(0) = \frac mn$.
Case 1: $x > 0$. Consider $f(0)$ and $f(x)$. By mean value theorem, there exists a $c\in (0, x)$ that satisfies
$$\begin{align*}
f'(c) &= \frac{f(x)-f(0)}{x-0}\\
&= \frac{(1+x)^{\frac mn}-1}{x}\\
\end{align*}$$
Since $f'(x)$ is increasing for $x > -1$,
$$\begin{align*}
f'(c) &\ge f'(0)\\
\frac{(1+x)^{\frac mn}-1}{x} &\ge \frac mn\\
(1+x)^{\frac mn} &\ge 1 + \frac mn x
\end{align*}$$
Case 2: $-1 \le x < 0$. Consider $f(x)$ and $f(0)$, and there exists a $c\in(x, 0)$ that satisfies
$$\begin{align*}
f'(c) &= \frac{f(0)-f(x)}{0-x}\\
&= \frac{1-(1+x)^{\frac mn}}{-x}\\
\end{align*}$$
Since $f'(x)$ is increasing for $x > -1$,
$$\begin{align*}
f'(c) &\le f'(0)\\
\frac{1-(1+x)^{\frac mn}}{-x} &\le \frac mn\\
1-(1+x)^{\frac mn} &\le -\frac mn x\\
(1+x)^{\frac mn} &\ge 1 + \frac mn x
\end{align*}$$
Case 3: $x=0$ is trivial.
| {
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Partial sum of divergent series I am trying to find the nth partial sum of this series:
$S(n) = 2(n+1)^2$
I found the answer on WolframAlpha:
$\sum_{n=0}^m (1+2n)^2 =\frac{1}{3}(m+1)(2m+1)(2m+3)$
How can I calculate that sum, without any software?
| $\sum_\limits{i=0}^n 2(i + 1)^2 = 2\sum_\limits{i=1}^{n+1} i^2$
Which gets to the meat of the question, what is $\sum_\limits{i=1}^n i^2$?
There are a few ways to do this. I think that this one is intuitive.
In the first triangle, the sum of $i^{th}$ row equals $i^2$
The next two triangles are identical to the first but rotated 120 degrees in each direction.
Adding corresponding entries we get a triangle with $2n+1$ in every entry. What is the $n^{th}$ triangular number?
$3\sum_\limits{i=1}^n i^2 = (2n+1)\frac {n(n+1)}{2}\\
\sum_\limits{i=1}^n i^2 = \frac {n(n+1)(2n+1)}{6}$
To find: $\sum_\limits{i=1}^{n+1} i^2 $, sub $n+1$ in for $n$ in the formula above.
$\sum_\limits{i=0}^n 2(i + 1)^2 = \frac {(n+1)(n+2)(2n+3)}{3}$
Another approach is to assume that $S_n$ can be expressed as a degree $3$ polynomial. This should seem plausible
$S(n) = a_0 + a_1 n + a_2 n^2 + a_3n^3\\
S(n+1) = S(n) + 2(n+2)^2\\
S(n+1) - S_n = 2(n+2)^2\\
S(n+1) = a_0 + a_1 (n+1) + a_2 (n+1)^2 + a_3(n+1)^3\\
a_0 + a_1 n+a_1 + a_2 n^2 + 2a_2n+a_21 + a_3n^3 + 3a_3n^2 + 3a_3n + 1\\
S(n+1) - S(n) = (a_1 + a_2 + a_3) + (2a_2 + 3a_3) n + 3a_3 n^2 = 2n^2 + 4n + 2$
giving a system of equations:
$a_1 + a_2 + a_3 = 2\\
2a_2 + 3a_3 = 4\\
3a_3 = 1\\
a_0 = S(0)$
| {
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Integral $\int\frac{\sqrt{4x^2-1}}{x^3}dx$ using trig identity substitution!
$$\int \frac{\sqrt{4x^2-1}}{x^3}\ dx$$
So, make the substitution
$ x = \sqrt{a \sec \theta}$, which simplifies to $a \tan \theta$.
$2x = \sqrt{1} \sec \theta$,
$ d\theta = \dfrac{\sqrt{1}\sec\theta\tan\theta}{2}$
$\int \dfrac{\sqrt{1}\tan\theta}{(\sqrt{1}\sec\theta)^3} d\theta$
Am I making the correct substitutions here? Substituting $d\theta$ a quantity of $(\sqrt{1}\sec\theta)$ will cancel from the denominator. Somewhere along the line I need to use the identity $\sin(2\theta)=2\sin(\theta)\cos(\theta).$
| With the substitution $x=\frac {\sec \theta }{2}$ you get $dx = \frac {\sec \theta \tan \theta }{2} d\theta $ and the integral changes to $$ \int \frac {4\tan^2 \theta \sec \theta }{ \sec ^3 \theta } d\theta =4 \int \frac {\tan^2 \theta }{ \sec ^2 \theta } d\theta =4 \int \sin ^2 \theta d\theta$$
Now you can use the double angle equality which you mentioned.
| {
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Finding value of $ \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}$
Finding value of $\displaystyle \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{4k^2}{4k^2-1}$
Try: $$\lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot \frac{2k}{2k+1} = \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{2k}{2k-1}\cdot \prod^{n}_{k=1}\frac{2k}{2k+1}$$
$$\lim_{n\rightarrow \infty}\frac{(2\cdot 4 \cdot 6\cdots 2n)^2}{1\cdot 2\cdot 3\cdots 2n}\times \frac{(2\cdot 4 \cdot 6 \cdots 2n)^2}{1\cdot 2\cdot 3\cdots 2n}$$
$$\lim_{n\rightarrow \infty}\frac{(2\cdot 4\cdot 6\cdot 2n)^4}{(1\cdot 2\cdot 3\cdots 2n)^2} = \lim_{n\rightarrow \infty}2^{4n}\cdot \frac{(n!)^4}{(2n!)^2}$$
Did not find any clue how to solve from that point
Could some help me to solve it, Thanks
| You may find the following approach useful which avoids Stirling'approximation.
Let $$a_n=\int_{0}^{\pi/2}\sin^nx\,dx\tag{1}$$ and using integration by parts we have $$a_n=\left.-\sin^{n-1}x\cos x\right|_{x=0}^{x=\pi/2}+(n-1)\int_{0}^{\pi/2}\sin^{n-2}x\cos^2x\,dx$$ and the last integral can be written as $a_{n-2}-a_n$ via the identity $\cos^2x=1-\sin^2x$ and thus we get the recurrence relation $$a_n=\frac{n-1}{n}a_{n-2}\tag{2}$$ Using the above relation repeatedly we get $$a_{2n}=\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\dots\frac{1}{2}a_0\tag{3}$$ and $$a_{2n+1}=\frac{2n}{2n+1}\cdot \frac{2n-2}{2n-1}\dots \frac{2}{3}a_1\tag{4}$$ Noting that $a_0=\pi/2,a_1=1$ we have via $(3),(4)$ $$\frac{a_{2n}}{a_{2n+1}}=\frac{\pi} {2}\prod_{k=1}^{n}\frac{4k^2-1}{4k^2}$$ The LHS of the above equation tends to $1$ as shown later in this answer and hence the product in your question evaluates to $\pi/2$.
It is easy to observe that $$a_{2n+1}\leq a_{2n}\leq a_{2n-1}$$ and hence $$1\leq \frac{a_{2n}}{a_{2n+1}}\leq \frac{a_{2n-1}}{a_{2n+1}}=\frac{2n+1}{2n}$$ (via equation $(2)$). By squeeze theorem we see that $a_{2n}/a_{2n+1}\to 1$ as $n\to\infty $.
Note: For those not well acquainted with Stirling's formula, the argument in this answer is crucial in one of the proofs of Stirling's formula.
| {
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Integrate $\int \frac {dx}{\sqrt {(x-a)(x-b)}}$ where $b>a$ Integrate: $\displaystyle\int \dfrac {dx}{\sqrt { (x-a)(x-b)}}$ where $b>a$
My Attempt:
$$\int \dfrac {dx}{\sqrt {(x-a)(x-b)}}$$
Put $x-a=t^2$
$$dx=2t\,dt$$
Now,
\begin{align}
&=\int \dfrac {2t\,dt}{\sqrt {t^2(a+t^2-b)}}\\
&=\int \dfrac {2\,dt}{\sqrt {a-b+t^2}}
\end{align}
| Alternatively, you can use an Euler substitution to rationalize the integrand.
Option 1 Change variable to $t$, where $$\sqrt{(x - a) (x - b)} = x + t.$$
rearranging gives $$x = \frac{ab - t^2}{2 t + (a + b)},$$
and substituting gives $$\int \frac{dx}{\sqrt{(x - a) (x - b)}} = \int \frac{dt}{t + \frac{1}{2}(a + b)} .$$
Option 2 Change variable to $u$, where $$\sqrt{(x - a) (x - b)} = (x - a) u.$$ Rearranging gives
$$x = \frac{a u^2 - b}{u^2 - 1},$$
and substituting gives $$-2 \int \frac{du}{u^2 - 1} .$$
| {
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Convert complex number to polar coordinates Problem
Compute when $x \in \mathbb{C}$:
$$ x^2-4ix-5-i=0 $$
and express output in polar coordinates
Attempt to solve
Solving this equation with quadratic formula:
$$ x=\frac{4i \pm \sqrt{(-4i)^2-4\cdot (-5-i)}}{2} $$
$$x= \frac{4i \pm \sqrt{4(i+1)}}{2} $$
$$ x = \frac{4i \pm 2\sqrt{i+1}}{2} $$
$$ x = 2i \pm \sqrt{i+1} $$
I can transform cartesian complex numbers to polar with eulers formula:
when $z \in \mathbb{C}$
$$ z=re^{i\theta} $$
then:
$$ r=|z|=\sqrt{(\text{Re(z)})^2+(\text{Im(z)})^2} $$
$$ \text{arg}(x)=\theta = \arctan{\frac{\text{Im}(z)}{\text{Re}(z)}} $$
Plugging in values after this computation would give us our complex in number in $(r,\theta)$ polar coordinates from $(\text{Re},\text{Im})$ cartesian coordinates.
Only problem is how do i convert complex number of form
$$ z=2i+\sqrt{i+1} $$
to polar since i don't know how to separate this into imaginary and real parts. How do you compute $\text{Re}(z)$ and $\text{Im}(z)$
| Let $a,b\in\mathbb{R}$ so that $$\sqrt{i+1} = a+bi$$
$$ i+1 = a^2 -b^2 +2abi $$
Equating real and imaginary parts, we have
$$2ab = 1$$
$$a^2 -b^2 = 1$$
Now we solve for $(a,b)$.
$$
\begin{align*}
b &= \frac{1}{2a}\\\\
\implies \,\,\, a^2 - \left(\frac{1}{2a}\right)^2 &= 1 \\\\
a^2 &= 1 + \frac{1}{4a^2}\\\\
4a^4 &= 4a^2 + 1\\\\
4a^4 - 4a^2 -1 &= 0 \\\\
\end{align*}
$$
This is a quadratic in $a^2$ (it's also a quadratic in $2a^2$, if you prefer!), so we use the quadratic formula:
$$a^2 = \frac{4 \pm \sqrt{16-4(4)(-1)}}{2(4)}$$
$$a^2 = \frac{1 \pm \sqrt{2}}{2}$$
Here we note that $a$ is real, so $a^2>0$, and we discard the negative case:
$$a^2 = \frac{1 + \sqrt{2}}{2}$$
$$a = \pm \sqrt{\frac{1 + \sqrt{2}}{2}}$$
$$ b = \frac{1}{2a} = \pm \sqrt{\frac{\sqrt{2}-1}{2}}$$
This gives what you can call the principal root:
$$\sqrt{i+1} = \sqrt{\frac{1 + \sqrt{2}}{2}} + i\sqrt{\frac{\sqrt{2}-1}{2}} $$
As well as the negation of it:
$$-\sqrt{i+1} = -\sqrt{\frac{1 + \sqrt{2}}{2}} + i\left(-\sqrt{\frac{\sqrt{2}-1}{2}}\right) $$
Finally, substituting either of these into your expression $$z=2i \pm \sqrt{i+1}$$ will give you $\text{Re}(z)$ and $\text{Im}(z)$.
At that point, as you noted in your question, conversion to polar coordinates is straightforward.
| {
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Limit of a sequence when the algebraic limit theorem 'breaks down' Background
There is the very well-known technique of computing the limit of a sequence by taking limit on both sides whenever recurrence relation arises. For example, $b_n = \frac{\alpha^n}{n!} $ for $0 < \alpha < 1$.
Question
But now there is this sequence such that taking limit on both sides will yield meaningless results:
$$a_n = \frac{1}{3}a_{n-1} + \frac{2}{3}a_{n-2} \; \; \;\text{for any
integer} \; n \geq 3, $$ and $a_1 = 0$, $a_2 = 1$.
The question is how this limit should be evaluated? (The existence of the limit is already proved by proving the sequence is a Cauchy sequence.)
My attempt
It can be observed that
$$ a_n - a_{n-1} = \frac{-2}{3} a_{n-1} + \frac{2}{3} a_{n-2} =
\frac{-2}{3} (a_{n-1} - a_{n-2})$$
Then the two sides are symmetrical and hence
$$ a_n - a_{n-1}= (\frac{-2}{3})^{n-1}(a_1-a_0) =
(\frac{-2}{3})^{n-1}$$
But again taking limit on both sides yields $ 0 = 0$
| $$a_n = \frac13a_{n-1} + \frac23a_{n-2}$$
The characteristic equation is $x^2-\frac13x-\frac23=0$.
$$3x^2-x-2=0$$
$$(3x+2)(x-1)=0$$
$$x=-\frac23,1$$
$$a_n = \alpha \left( -\frac23\right)^n+\beta$$
We have $a_1=0$ and $a_2=1$,
$$0=\alpha\left( -\frac23\right)+\beta$$
$$1=\alpha\left( -\frac23\right)^2+\beta$$
$$1=\alpha\left( -\frac23\right)\left( -\frac53\right)$$
$$\alpha=\frac9{10}, \beta=\frac23\alpha=\frac23\cdot\frac{9}{10}=\frac35$$
$$a_n = \frac{9}{10}\left( -\frac23\right)^n+\frac35$$
$$\lim_{n \to \infty}a_n = \frac35$$
| {
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A solution for $f(ax+b)=f(x)+1$ Let $a,b$ be two constant real numbers with $a\neq 0$. Can anyone give a special solution of the functional equation $f(ax+b)=f(x)+1$, where $f:\mathbb{R}\rightarrow \mathbb{R}$?
Note. It is a type of the Abel functional equations, and if $a=1$, then
$f(x)=[\frac{x}{b}]$ is an its solution.
| For fixed $a\in\mathbb{R}\setminus\{1\}$ and $b\in\mathbb{R}$, let now consider a function $f:\mathbb{R}\to\mathbb{R}$ which satisfies the functional equation
$$f(ax+b)=f(x)+1\text{ for all }x\in\mathbb{R}\setminus\left\{\frac{b}{1-a}\right\}\,.\tag{#}$$
Firstly, we assume that $a=0$. Then, we see that $f(x)=f(b)-1$ for any $x\in\mathbb{R}\setminus\{b\}$. Thus, all functions $f:\mathbb{R}\to\mathbb{R}$ with the condition (#) are of the form
$$f(x)=\begin{cases}c&\text{if }x=b\,,\\c-1&\text{if }x\neq b\,.\end{cases}$$
Secondly, we assume that $a>0$. Write $I^+:=\left(\dfrac{b}{1-a},+\infty\right)$ and $I^-:=\left(-\infty,\dfrac{b}{1-a}\right)$. For $x\in I^+$, we can see that
$$x-\frac{b}{1-a}=a^t\text{ or }t=\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}$$
for some $t\in\mathbb{R}$. Thus, if $g_+(t):=f\left(a^t+\dfrac{b}{1-a}\right)$ for each $t\in\mathbb{R}$, then
$$\begin{align}g_+(t+1)&=f\left(a^{t+1}+\frac{b}{1-a}\right)=f\Biggl(a\left(a^t+\frac{b}{1-a}\right)+b\Biggr)\\&=f\left(a^t+\frac{b}{1-a}\right)+1=g_+(t)+1\,.\end{align}$$
Therefore, if $h_+(t):=g_+(t)-t$, then $h_+:\mathbb{R}\to\mathbb{R}$ is periodic with period $1$. That is,
$$f(x)=h_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\text{ for all }x\in I^+\,.$$
We obtain a similar result for $x\in I^-$. Thus,
$$f(x)=\begin{cases}
h_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln(a)}&\text{if }x>\frac{b}{1-a}\,,\\
c&\text{if }x=\frac{b}{1-a}\,,\\
h_-\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln(a)}\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln(a)}&\text{if }x<\frac{b}{1-a}\,,
\end{cases}$$
where $h_+,h_-:\mathbb{R}\to\mathbb{R}$ are periodic functions with period $1$, and $c\in\mathbb{R}$ is an arbitrary constant.
Finally, we are dealing with the case $a<0$. We rule out the case $a=-1$, since there does not exist a solution $f$ with the said property. This is because of the contradiction below when $a=-1$:
$$f(x)=f\big(-(-x+b)+b\big)=f(-x+b)+1=\big(f(x)+1\big)+1=f(x)+2$$
for all $x\neq \dfrac{b}{2}$. From now on, we assume that $a\neq -1$.
Note that
$$\begin{align}f\big(a^2x+(a+1)b\big)&=f\big(a(ax+b)+b\big)=f(ax+b)+1\\&=\big(f(x)+1\big)+1=f(x)+2\end{align}$$
for all $x\neq \dfrac{b}{1-a}$. Let $A:=a^2$, $B:=(a+1)b$, and $\phi(t):=\dfrac{1}{2}\,f(t)$ for all $t\in\mathbb{R}$. Then,
$$\phi(Ax+B)=\phi(x)+1$$
for every $x\neq \dfrac{b}{1-a}=\dfrac{B}{1-A}$. Since $A>0$ and $A\neq 1$, we have by the previous section of this answer that
$$\phi(x)=\begin{cases}
\eta_+\left(\frac{\ln\left(x-\frac{B}{1-A}\right)}{\ln(A)}\right)+\frac{\ln\left(x-\frac{B}{1-A}\right)}{\ln(A)}&\text{if }x>\frac{B}{1-A}\,,\\
C&\text{if }x=\frac{B}{1-A}\,,\\
\eta_-\left(\frac{\ln\left(\frac{B}{1-A}-x\right)}{\ln(A)}\right)+\frac{\ln\left(\frac{B}{1-A}-x\right)}{\ln(A)}&\text{if }x<\frac{B}{1-A}\,,
\end{cases}$$
where $\eta_+,\eta_-:\mathbb{R}\to\mathbb{R}$ are periodic functions with period $1$, and $C\in\mathbb{R}$ is an arbitrary constant. Therefore,
$$f(x)=2\,\phi(x)=\begin{cases}
2\,\eta_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{2\,\ln|a|}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}&\text{if }x>\frac{b}{1-a}\,,\\
c&\text{if }x=\frac{b}{1-a}\,,\\
2\,\eta_-\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{2\,\ln|a|}\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln|a|}&\text{if }x<\frac{b}{1-a}\,,
\end{cases}$$
where $c:=2C$.
Recall that $f(ax+b)=f(x)+1$ for $x\neq \dfrac{b}{1-a}$. For $x>\dfrac{b}{1-a}$, we have
$$f(ax+b)=f(x)+1=2\,\eta_+\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{2\,\ln|a|}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}+1\,.\tag{1}$$
Because $ax+b<\dfrac{b}{1-a}$ when $x>\dfrac{b}{1-a}$, we conclude that
$$f(ax+b)=2\,\eta_-\left(\frac{\ln\left(\frac{b}{1-a}-(ax+b)\right)}{2\,\ln|a|}\right)+\frac{\ln\left(\frac{b}{1-a}-(ax+b)\right)}{\ln|a|}\,.$$
Since $\dfrac{b}{1-a}-(ax+b)=|a|\,\left(x-\dfrac{b}{1-a}\right)$, we obtain
$$f(ax+b)=2\,\eta_-\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{2\,\ln|a|}+\frac{1}{2}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}+1\,.\tag{2}$$
Equating (1) and (2), we conclude that
$$\eta_+\left(t\right)=\eta_-\left(t+\frac{1}{2}\right)\text{ for each }t\in\mathbb{R}\,.$$
Let $h:\mathbb{R}\to\mathbb{R}$ be given by
$$h(t)=2\,\eta_+\left(\frac{t}{2}\right)\text{ for every }t\in\mathbb{R}\,.$$
Ergo, $h$ is periodic with period $2$ and
$$h(t-1)=2\,\eta_-\left(\frac{t}{2}\right)\text{ for every }t\in\mathbb{R}\,.$$
Hence,
$$f(x)=\begin{cases}
h\left(\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}\right)+\frac{\ln\left(x-\frac{b}{1-a}\right)}{\ln|a|}&\text{if }x>\frac{b}{1-a}\,,\\
c&\text{if }x=\frac{b}{1-a}\,,\\
h\left(\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln|a|}-1\right)+\frac{\ln\left(\frac{b}{1-a}-x\right)}{\ln|a|}&\text{if }x<\frac{b}{1-a}\,,
\end{cases}$$
for some real constant $c$ and for some periodic function $h:\mathbb{R}\to\mathbb{R}$ with period $2$.
It is not difficult to prove that the three results are indeed solutions to (#). I shall omit the proof of this part as an exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6} \in \mathbb{Z}$.
Induction step: We want to show that it holds for $n=k+1$.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
| $$F=\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{n(n-1)(2n-1)}{6}$$
You can see that for any n(odd or even) the numerator is always a multiple of 6, so the sum of fractions is an integer; in fact we may have:
*
*$n=6k ⇒ F=6m/6=m $
*$n=6k+1 ⇒ F=(6k+1)(6k+1-1)(12k+2-1)=6m/6=m$
*$n=6k+2 ⇒ F=(6k+2)(6k+2-1)(12k+4-1)=6t/6=t$
*$n=6k+3 ⇒ F=(6k+3)(6k+3-1)(12k+6-1)=6s/6=s$
*$n=6k+4 ⇒ F=(6k+4)(6k+4-1)(12k+8-1)=6v/6=v$
*$n=6k+5 ⇒ F=(6k+5)(6k+5-1)(12k+10-1)=6u/6=u$
Where m,t,s,u and v are integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 2
} |
Square root of two positive integers less than or equal to the sum of both integers direct proof Please help with this problem.
If x and y positive integers, show:
$$2\sqrt{xy} \le x + y $$
| Observe that $(\sqrt{x} - \sqrt{y})^2 = x - 2\sqrt{xy} + y \geq 0$.
Rationale: For any real numbers a,b, $(a - b)^2 = a^2 - 2ab + b^2 \geq 0 \implies a^2 + b^2 \geq 2ab$. Since $x, y > 0$, $\sqrt{x}, \sqrt{y}$ are real numbers. Thus, if we set $a = \sqrt{x}$ and $b =\sqrt{y}$, we obtain $$(\sqrt{x} - \sqrt{y})^2 = x - 2\sqrt{xy} + y \geq 0 \iff x + y \geq 2\sqrt{xy}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Standard line element example
In Euclidean three-space, we can define paraboloidal coordinates $(u,v,\phi)$ via
\begin{align*}
x = uv\cos\phi,\quad y = uv\sin\phi,\quad z = \frac{1}{2}(u^2-v^2)
\end{align*}
Find $ds^2 = dx^2 + dy^2 + dz^2$
So I just want people to check my working and answer for this question. I have that
\begin{align*}
dx &= v\cos\phi\,du + u\cos\phi\,dv - uv\sin\phi\,d\phi\\
dy &= v\sin\phi\,du + u\sin\phi\,dv + uv\cos\phi\,d\phi\\
dz &= u\,du - \frac{1}{2}v^2 + \frac{1}{2}u^2 - v\,dv\\
\\
dx^2 &= du^2v^2\cos^2\phi + 2du\,dv\,uv\cos^2\phi - 2du\,d\phi\,uv^2\sin\phi\cos\phi\\
&+ dv^2u^2\cos^2\phi - 2dv\,d\phi\,u^2v\sin\phi\cos\phi + d\phi^2u^2v^2\sin^2\phi.\\
\\
dy^2 &= du^2v^2\sin^2\phi + 2du\,dv\,uv\sin^2\phi + 2du\,d\phi\,uv^2\sin\phi\cos\phi\\
&+ dv^2u^2\sin^2\phi + 2dv\,d\phi\,u^2v\sin\phi\cos\phi + d\phi^2u^2v^2\cos^2\phi.\\
\\
dz^2 &= du\,u^3 - du\,uv^2 + \frac{(u^2-v^2)^2}{4}\\
&- dv\,u^2v + dv\,v^3 + du^2u^2 - 2du\,dv\,uv + dv^2v^2.\\
\\
\Rightarrow ds^2 &= du^2v^2\cos^2\phi + 2du^2uv\cos^2\phi + du^2u^2\cos^2\phi\\
&+ d\phi^2u^2v^2\sin^2\phi + du^2v^2\sin^2\phi + 2du^2uv\sin^2\phi\\
&+ du^2u^2\sin^2\phi + d\phi^2u^2v^2\cos^2\phi + du^2u^2 - du\,uv^2\\
&+ du\,u^3 - 2du\,dv\,uv + dv^2v^2 + dv\,v^3 - dv\,u^2v + \frac{(u^2-v^2)^2}{4}.\\
\\
\Rightarrow ds^2 &= (2u^2 + 2uv + v^2)du^2 + v^2dv^2 + u^2v^2d\phi^2\\
&- du\,uv^2 + du\,u^3 - 2du\,dv\,uv + dv\,v^3 - dv\,u^2v + \frac{(u^2-v^2)^2}{4}
\end{align*}
I don't understand why I have these extra terms!? I know this will take a while but just think that I have had to type set it into Latex as well :). So if anyone has the time to check this answer I would be very appreciative.
| \begin{align*}
dz &= u\,du - v\,dv\\
\\
dz^2 &= u^2du^2 - 2uv\,du\,dv + v^2dv^2\\
\\
\Rightarrow ds^2 &= du^2v^2\cos^2\phi + 2du\,dv\,uv\cos^2\phi - 2du\,d\phi\,uv^2\cos^2\phi\\
&+ dv^2u^2\cos^2\phi - 2dv\,d\phi\,u^2v\sin\phi\cos\phi + d\phi^2u^2v^2\sin^2\phi\\
&+ du^2v^2\sin^2\phi + 2du\,dv\,uv\sin^2\phi + 2du\,d\phi\,uv^2\sin\phi\cos\phi\\
&+ dv^2y^2\sin^2\phi + 2dv\,d\phi\,u^2v\sin\phi\cos\phi + d\phi^2u^2v^2\cos^2\phi\\
&+ du^2u^2 + dv^2v^2 - 2du\,dv\,uv\\
\\
\Rightarrow ds^2 &= (u^2 + v^2)du^2 + (u^2 + v^2)dv^2 + (u^2v^2)d\phi^2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Dividing polynomial $f(x)$ by $x-3$ and $x+6$ leaves respective remainders $7$ and $22$. What's the remainder upon dividing by $(x-3)(x+6)$? If I have a polynomial $f(x)$ and is divided by $(x- 3)$ and $(x + 6)$ the respective remainders are $7$ and $22$, what is the remainder when $f(x)$ is divided by $(x-3)(x + 6)$?
I tried it by doing:
$$f(x) =(x-3)(x+6)q(x) + ax+b $$
And, $a$ and $b$ comes out to be $-\dfrac53$ and $12$ respectively.
But I'm not sure how to solve any further.
And kindly explain exactly how it's done
| $f(x)=(x-3)a(x)+7\Rightarrow f(3)=7$
$f(x)=(x+6)b(x)+22\Rightarrow f(-6)=22$
If you can write $f$ is of the form $f(x) =(x-3)(x+6)q(x) + ax+b$.
Solution is so easy:
$$f(3)=3a+b=7$$
$$f(-6)=-6a+b=22$$
Hence, $a=-\dfrac53$ and $b=12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition. Problem
Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition.
Note:
The problem asks us to prove that, no matter $x \to +\infty$ or $x \to -\infty$, the limit is $\infty$,which may be $+\infty$ or $-\infty.$
Proof
$\forall M>0$,$\exists X=\max(1,M+1)>0, \forall|x|>X$：
\begin{align*}
\left|\frac{x^3+1}{x^2+1}\right|&=\left|x-\frac{x-1}{x^2+1}\right|\\&\geq |x|-\left|\frac{x-1}{x^2+1}\right|\\&\geq |x|-\frac{|x|+1}{x^2+1}\\&\geq |x|-\frac{x^2+1}{x^2+1}\\&=|x|-1\\&>X-1\\&\geq M.
\end{align*}
Please verify the proof above.
| Possibly correct but unreadable.
Consider
$$
\frac{x^3+1}{x^2+1}=x-\frac{x-1}{x^2+1}>x
$$
whenever $x>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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$2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$ if and only if $a,b,c$ are the lengths of a right angled triangle Here is a problem from the book: Everything connected to Pithagoras
It is well known that in every right angled triangle $ABC$, $a^2+b^2=c^2$. Howevee, there are some more complicated equations as well. Here is one:
Prove that $$2(a^8+b^8+c^8)=(a^4+b^4+c^4)^2$$ if and only if $a,b,c$ are the lengths of a right angled triangle.
I can’t prove it. I tried to write $a_4=x$ and so on, but only algebra didn’t help for me. Aldo it is interesting because in this problem it doesn’t matter which side is the hypotenuse...
Please help!
| Bear with me as I ramble a bit ...
The relation in question resembles a way of writing Heron's formula for the area of a triangle with sides $x$, $y$, $z$:
$$16\;|\triangle xyz|^2 = \left(x^2+y^2+z^2\right)^2-2\left(x^4+y^4+z^4\right) \tag{1}$$ Now, some (most?) people think of Heron's formula as more like
$$|\triangle xyz|^2 = s(s-x)(s-y)(s-z) \tag{2}$$
where $s = (x+y+z)/2$ is the semi-perimeter. I personally prefer not to bother introducing an extra value, so I write
$$16\;|\triangle xyz|^2 = (x+y+z)(-x+y+z)(x-y+z)(x+y-z) \tag{3}$$
What's the point of all this? Well, it actually has nothing to do with triangle areas. My point is simply that my familiarity with various forms of Heron formula allows me to immediately realize that expressions that look like $(1)$ factor into expressions that look like $(3)$. (So, I no longer need to go through the trouble of expanding the product, combining terms, and attempting to factor. It's second-nature to me now, and should be to any olympiad contender. :) The same must be true of the relation in question:
$$\begin{align}
0 &= \left(a^4+b^4+c^4\right)^2-2\left(a^8+b^8+c^8\right) \\[4pt]
&= \left(a^2+b^2+c^2\right)\left(-a^2+b^2+c^2\right)\left(a^2-b^2+c^2\right)\left(a^2+b^2-c^2\right)
\end{align} \tag{4}$$
Assuming this relation holds, we see that one of the factors must vanish. Obviously, first never does. Whichever of the latter three does corresponds to a Pythagorean relation in $a$, $b$, $c$, and therefore a right triangle. Conversely, a right triangle admits a Pythagorean relation, which causes a $(4)$ to hold. $\square$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{x \to 4} \frac{x^4-4^x}{x-4}$, where is my mistake? Once again, I am not interested in the answer. But rather, where is/are my mistake(s)? Perhaps the solution route is hopeless:
Question is: evaluate $\lim_{x \to 4} \frac{x^4 -4^x}{x-4}$.
My workings are:
Let $y=x-4$. Then when $x \to 4$, we have that $y \to 0$. Thus:
$$\lim_{y \to 0} \frac{(y+4)^4 - 4^{y+4}}{y} = \\ = \lim_{y \to 0}\frac{(y+4)^4}{y} - \lim_{y \to 0} \frac{4^{(y+4)}}{y} $$
And this step is not allowed from the get go, as I am deducting infinities, which is indeterminate. What I should have done though:
$$4^4 \lim_{y \to 0} \frac{(1+y/4)^4-1+(4^y-1)}{y} = \\ 4^4 \lim_{y \to 0} \left( \frac{(1+y/4)^4-1}{\frac{y}{4}4} - \frac{4^y-1}{y} \right) = \\
=4^4\left(\frac{1}{4} \cdot 4 - \ln 4 \right) = 256(1-\ln 4)$$
| The following step is not allowed
$$\lim_{x \to 4} \frac{x^4-4^x}{x-4}=\lim_{y \to 0} \frac{(y+4)^4-4^{y+4}}{y}\color{red}{=\lim_{y \to 0} \frac{(y+4)^4}{y}-\lim_{y \to 0} \frac{4^{y+4}}{y}}$$
Refer also to the related
*
*Analyzing limits problem Calculus (tell me where I'm wrong).
*Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Expressiong $\frac{t+2}{t^3+3}$ in the form $a_o+a_1t+...+a_4t^4$, where $t$ is a root of $x^5+2x+2$ Expressing $\frac{t+2}{t^3+3}$ in the form $a_o+a_1t+...+a_4t^4$, where $t$ is a root of $x^5+2x+2$.
So i can deal with the numerator, but how do I get rid of the denomiator to get it into the correct form? Thanks in advance!
| Using the Euclidean algorithm for computing $\gcd(x^3+3,x^5+2x+2)$, we get
$$
367=(10 x^4 - 31 x^3 - 14 x^2 - 30 x + 113)(x^3+3)+(-10 x^2 + 31 x + 14)(x^5+2x+2)
$$
and so
$$
367=(10 t^4 - 31 t^3 - 14 t^2 - 30 t + 113)(t^3+3)
$$
Thus,
$$
\begin{align}
367\frac{t+2}{t^3+3}
&=(t+2)(10 t^4 - 31 t^3 - 14 t^2 - 30 t + 113)\\
&=10(t^5+2t+2)+(-11 t^4 - 76 t^3 - 58 t^2 + 33 t + 206)\\
&=-11 t^4 - 76 t^3 - 58 t^2 + 33 t + 206
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $(x \mod 7) - (x \mod 8) = 5$? I'm trying to solve $(x \mod 7) - (x \mod 8) = 5$ but no idea where to start. Help appreciated!
| If CRT = Chinese Remainder Theorem is known then we can reformulate it as
$$\begin{align}
x&\equiv a\!+\!5\!\!\pmod{7},\ \ \ \overbrace{ 0\le a\!+\!5 \le 6}^{\Large \iff a\ =\ 0,1}\\
x&\equiv a\quad\pmod{8},\ \ \ 0\le a\le 7\end{align}\qquad \qquad\qquad $$
Solve the $\,a=0\,$ case $\,x\equiv (5,0)\bmod (7,8),\,$ then $\,x\!+\!1\equiv (6,1)\,$ yields the $\,a=1\,$ case.
W/o $ $ CRT: $\,\ \overbrace{ x-7j}^{\large x\bmod 7}-(\overbrace{x-8k)}^{\large x\bmod 8} = 5\iff 8k - 7j = 5\,\Rightarrow\, j \equiv \color{#c00}{5}\pmod{\!8},\ k\equiv \color{#c00}{5}\pmod{\!7}$
Working in the intial range $\, 0\le x\le 55\ $ and enforcing the remainder bounds
$\qquad\quad \begin{align} 0\le \overbrace{x-7\cdot \color{#c00}5}^{\large x\bmod 7} \le 6\iff 35\le x \le \color{#0a0}{41}\\[.5em]
0\le \underbrace{x-8\cdot \color{#c00}5}_{\large x\bmod 8} \le 7\iff \color{#0a0}{40}\le x \le 47\end{align}$
Therefore $\, x\equiv \color{#0a0}{40},\color{#0a0}{41}\pmod{56}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the exact values of the square roots of $z=1+i$ are... Show that the exact values of the square roots of $z=1+i$ are...
$w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}$
$w_1=-\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{-1+\sqrt{2}}{2}}$
My attempt
Let $z=1+i\in \mathbb{C}$. Then
$r=|z|=\sqrt{2}$
Moreover,
$\theta=\tan^{-1}(1)=\frac{\pi}{4}$
Then, the polar form of $z$ is
$z=\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$
Let $w\in\mathbb{C}$ such that $w^2=z$
Then
$w_k=\sqrt{\sqrt{2}}(\cos(\frac{\frac{\pi}{4}+2k\pi}{2})+i\sin(\frac{\frac{\pi}{4}+2k\pi}{2})$ for $k=0,1$.
Then the square roots, are:
$w_0=\sqrt[4]2(\cos\frac{\pi}{8}+i\sin\frac{\pi}{8})=1,84+0,76i.$
$w_1=\sqrt[4]2(\cos\frac{9\pi}{8}+i\sin\frac{9\pi}{8})=-1,84-0,76i.$
But, here i'm stuck because the $w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}\not = 1,84+0,76i$
| Setting $$\sqrt{1+i}=A+Bi$$ then $$1+i=A^2-B^2+2ABi$$ and we have to solve
$$A^2-B^2=1$$
$$2AB=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Maximise $(x+1)\sqrt{1-x^2}$ without calculus Problem
Maximise $f:[-1,1]\rightarrow \mathbb{R}$, with $f(x)=(1+x)\sqrt{1-x^2}$
With calculus, this problem would be easily solved by setting $f'(x)=0$ and obtaining $x=\frac{1}{2}$, then checking that $f''(\frac{1}{2})<0$ to obtain the final answer of $f(\frac{1}{2})=\frac{3\sqrt{3}}{4}$
The motivation behind this function comes from maximising the area of an inscribed triangle in the unit circle, for anyone that is curious.
My Attempt
$$f(x)=(1+x)\sqrt{1-x^2}=\sqrt{(1-x^2)(1+x)^2}=\sqrt 3 \sqrt{(1-x^2)\frac{(1+x)^2}{3}}$$
By the AM-GM Inequality, $\sqrt{ab}\leq \frac{a+b}{2}$, with equality iff $a=b$
This means that
$$\sqrt 3 \sqrt{ab} \leq \frac{\sqrt 3}{2}(a+b)$$
Substituting $a=1-x^2, b=\frac{(1+x)^2}{3}$,
$$f(x)=\sqrt 3 \sqrt{(1-x^2)\frac{(1+x)^2}{3}} \leq \frac{\sqrt 3}{2} \left((1-x^2)+\frac{(1+x)^2}{3}\right)$$
$$=\frac{\sqrt 3}{2} \left(\frac{4}{3} -\frac{2}{3} x^2 + \frac{2}{3} x\right)$$
$$=-\frac{\sqrt 3}{2}\frac{2}{3}(x^2-x-2)$$
$$=-\frac{\sqrt 3}{3}\left(\left(x-\frac{1}{2}\right)^2-\frac{9}{4}\right)$$
$$\leq -\frac{\sqrt 3}{3}\left(-\frac{9}{4}\right)=\frac{3\sqrt 3}{4}$$
Both inequalities have equality when $x=\frac{1}{2}$
Hence, $f(x)$ is maximum at $\frac{3\sqrt 3}{4}$ when $x=\frac{1}{2}$
However, this solution is (rather obviously I think) heavily reverse-engineered, with the two inequalities carefully manipulated to give identical equality conditions of $x=\frac{1}{2}$. Is there some better or more "natural" way to find the minimum point, perhaps with better uses of AM-GM or other inequalities like Jensen's inequality?
| You can solve the problem using geometry.
$(x+1)\sqrt{1-x^2}$ is the area of triangle
$$(-1, 0), (x, \sqrt{1-x^2}), (x, -\sqrt{1-x^2}).$$
This unit inscribed triangle has maximal area if and only if it is a equilateral triangle.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Proving that $\sum\limits_{n=1}^∞\frac{a_n}{s_n^2}$ converges
Let $a_n>0\;$ ($n=1,2,...\,$) with $\sum\limits_{n=1}^\infty a_n$ divergent and $s_n =\sum\limits_{k=1}^na_k$. For all $n \ge 2$, prove that $\sum\limits_{n=1}^∞\dfrac{a_n}{s_n^2}$ converges.
Proof: For all $n \ge 2$, we have $\dfrac{a_n}{s_n^2} \le \dfrac1{s_{n-1}} -\dfrac1{s_n}$ and $\sum\limits_{n=2}^{\infty} \dfrac{a_n}{s_n^2} \le \sum\limits_{n=2}^{k} \left(\dfrac{1}{s_{n-1}} - \dfrac{1}{s_n} \right)$.
Now $\sum\limits_{n=2}^k \left (\dfrac{1}{s_{n-1}} - \dfrac{1}{s_n} \right) = \dfrac{1}{s_1} - \dfrac{1}{s_k}$ converges to $\dfrac{1}{s_1}$. This follows because $\sum\limits_{n=1}^\infty a_n$ diverges and $\dfrac{1}{s_n} \to 0$ as $ n \to \infty$. Thus, by the comparison test, $\sum\limits_{n=1}^{\infty} \dfrac{a_n}{s_n^2}$ converges.
Is this proof correct?
$\dfrac{a_n}{s_n^2} \le \dfrac1{s_{n-1}} -\dfrac1{s_n}$
Proof: Let $n \le 2$
$ s_{n-1} \le s_{n}$ $\Leftrightarrow \frac{1}{s_{n}} \le \frac{1}{s_{n-1}} $
$ \Leftrightarrow \frac{1}{s_{n^2}} \le \frac{1}{s_{n}s_{n-1}}$
$ \Leftrightarrow \frac{a_{n}}{s_{n^2}} \le \frac{a_{n}}{s_{n}s_{n-1}} = \frac{s_{n} - s_{n-1}}{s_{n}s_{n-1}}$
$\Leftrightarrow \frac{a_{n}}{s_{n^2}} \le \frac{1}{s_{n-1}} - \frac{1}{s_{n}} $
| Your proof is correct if $a_n$ is nonnegative since it then follows that
$$\frac{a_n}{S_n^2} \leqslant \frac{a_n}{S_nS_{n-1}}= \frac{S_n - S_{n-1}}{S_nS_{n-1}}= \frac{1}{S_{n-1}}- \frac{1}{S_n}$$
and the RHS has a telescoping sum.
You also need to make it clear that $1/S_n \to 0$ as $n \to \infty$ to argue that the telescoping sum converges to $1/S_1$.
Without nonnegativity it is another story.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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} |
Let $a, b, c \in \mathbb{R^+}$ and $abc=8$ Prove that $\frac {ab+4}{a+2} + \frac {bc+4}{b+2} + \frac {ca+4}{c+2} \ge 6$
Let $a, b, c \in \mathbb{R^+}$ and $abc=8$ Prove that $$\frac {ab+4}{a+2} + \frac {bc+4}{b+2} + \frac {ca+4}{c+2} \ge 6$$
I have attempted multiple times in this question and the only method that I can think of is by adding up the fractions and expanding, but took too long and I eventually got stuck. Is there any other method I can use to help me in this question?
| Let $a=\frac{2y}{x}$ and $b=\frac{2z}{y}$, where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{2x}{z}$ and by AM-GM we obtain:
$$\sum_{cyc}\frac{ab+4}{a+2}=\sum_{cyc}\frac{\frac{4z}{x}+4}{\frac{2y}{x}+2}=2\sum_{cyc}\frac{z+x}{x+y}\geq6\sqrt[3]{\prod_{cyc}\frac{z+x}{x+y}}=6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2962974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
How to prove that $3^{x^2+x} (x+1)^{-x} \Gamma (x+1)\ge 1$ for $x>0$? Let
$$f(x)=3^{x^2+x} (x+1)^{-x} \Gamma (x+1).$$
Drawing a picture with any computer algebra system,
it is obviously that $f(x) \ge 1$ on $[0,\infty)$.
But How can we prove this? If we take derivative, then we get
$$
\frac{\mathrm d}{\mathrm dx}\log(f(x))=-\frac{x}{x+1}+(2 x+1) \log (3)-\log (x+1)+\psi(x+1),
$$
where $\psi(\cdot)$ is the digamma function.
Drawing a picture again, we see that this is positive and increasing
But again, how can we prove this?
Okay, I have a proof now for $x \in (0,1)$.
We can expand $\log(f(x))$ by this formula to get
$$
\log(f(x))=
\underset{t=2}{\overset{\infty }{\sum }}\frac{(-x)^t ((t-1) \zeta (t)-t)}{(t-1) t}+x^2 (3 \log )+x (3 \log -\gamma ).
$$
Thus it suffices to show that is decreasing for $t \ge 2$.
$$
\left|\frac{((t-1) \zeta (t)-t)}{(t-1) t}\right|
$$
This can be proved using this paper.
| Here's a proof for $x > 1$.
If $c > 1$,
since
$x! > \sqrt{2\pi x}(x/e)^x$
for $x > 1$,
if $x > 1 $ then
$\begin{array}\\
f(x)
&=c^{x^2+x} (x+1)^{-x} \Gamma (x+1)\\
&=c^{x^2+x} (x+1)^{-x} x!\\
&>c^{x^2+x} (x+1)^{-x} \sqrt{2\pi x}(x/e)^x\\
&=\sqrt{2\pi x}\left(c^{x+1} \dfrac{x}{e(x+1)}\right)^x\\
&>\sqrt{2\pi x}\left( \dfrac{c^2x}{e(x+1)}\right)^x\\
&>\sqrt{2\pi x}\left( \dfrac{c^2}{2e}\right)^x
\qquad\text{since } x/(x+1) > 1/2 \text{ for } x > 1\\
\end{array}
$
Therefore,
if $c^2 > 2e$,
or $c > 2.34 > \sqrt{2e}
$,
$f(x) \gt \sqrt{2\pi x}$
for $x > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What methods can be used to solve $ \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx $ I'm seeking methods to solve the following definite integral:
$$ I = \int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx $$
| The method I took was:
First make the substitution $t = \tan(x)$
$$ I = \int_{0}^{\infty} \frac{\arctan(t)}{t\left(1 + t^2\right)} \:dt $$
Now, let
$$ I\left(\omega\right) = \int_{0}^{\infty} \frac{\arctan(\omega t)}{t\left(1 + t^2\right)} \:dt $$
Thus,
\begin{align}
\frac{dI}{d\omega} &= \int_{0}^{\infty} \frac{t}{t\left(1 + t^2\right)\left(1 + \omega^2t^2\right)} \:dt \\
&= \int_{0}^{\infty} \frac{1}{\left(1 + t^2\right)\left(1 + \omega^2t^2\right)} \\
&= \frac{1}{\omega^2 - 1} \int_{0}^{\infty}\left[\frac{\omega^2}{\left(1 + \omega^2t^2\right)} - \frac{1}{\left(1 + t^2\right)}\right]dt \\
&= \frac{1}{\omega^2 - 1} \left[\omega\arctan(\omega t) - \arctan(t) \right]_{0}^{\infty} \\
&= \frac{1}{\omega^2 - 1} \left[\omega\frac{\pi}{2} - \frac{\pi}{2}\right]\\
&= \frac{1}{\omega + 1}\frac{\pi}{2}
\end{align}
Hence,
$$ I(\omega) = \int \frac{1}{\omega + 1}\frac{\pi}{2}\:d\omega = \frac{\pi}{2}\ln|\omega + 1| + C$$
Setting $\omega = 0$ we find:
$$I(0) = C = \int_{0}^{\infty} \frac{\arctan(0 \cdot t)}{t\left(1 + t^2\right)}\:dt = 0 $$
Thus,
$$ I(\omega) = \frac{\pi}{2}\ln|\omega + 1| $$
And finally,
$$I(1) = \int_{0}^{\infty} \frac{\arctan(t)}{t\left(1 + t^2\right)} \:dt =\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = \frac{\pi}{2}\ln(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded
Let $n \in \mathbb N$ and:
$$
x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1}
$$
Prove $x_n$ is bounded sequence.
Start with $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}}\right)
\end{align}
$$
From here:
$$
\sqrt[^3]{1 + {1\over n^3}} \gt 1 \\
\sqrt{1 - {1\over n^2}} \lt 1
$$
Therefore:
$$
\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}} \gt 0
$$
Which means $x_n \gt 0$.
Consider the following inequality:
$$
\sqrt[^3]{n^3 + 1} \le \sqrt{n^2 + 1} \implies \\
\implies x_n < \sqrt{n^2 + 1} - \sqrt{n^2 - 1}
$$
Or:
$$
x_n < \frac{(\sqrt{n^2 + 1} - \sqrt{n^2 - 1})(\sqrt{n^2 + 1} + \sqrt{n^2 - 1})}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} = \\
= \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} <2
$$
Also $x_n \gt0$ so finally:
$$
0 < x_n <2
$$
Have i done it the right way?
| Your prove is fine but a lot more work than necessary.
As $n \ge 1$ we have
$n = \sqrt[3]{n^3} < \sqrt[3]{n^3 + 1} < \sqrt[3]{n^3 + 3n^2 + 3n + 1} = \sqrt[3]{(n+1)^3} = n+1$
and
$n = \sqrt{n^2} > \sqrt{n^2 -1 } = \sqrt{n^2 - 2 + 1} \ge \sqrt{n^2 - 2n + 1} = \sqrt{(n-1)^2} = n-1$.
So $0 = n - n < \sqrt[3]{n^3 + 1} - \sqrt{n^2 -1} < (n+1) - (n-1) = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Which grows at a faster rate $\sqrt {n!}$ vs $(\sqrt {n})!$ when $n \rightarrow \infty$? Which grows at a faster rate $\sqrt {n!}$ vs $(\sqrt {n})!$ ? How to solve such type of questions considering $n \rightarrow \infty$?
| As alluded to in the comments, $(\sqrt{n})!$ doesn't make sense, so I'm going to compare the growth of $n!$ to the growth of $\sqrt{(n^2)!}$. Or, equivalently, compare $(n!)^2$ to $(n^2)!$.
Let $a_n = \frac{(n!)^2}{(n^2)!}$. Then
\begin{align*}
\frac{a_{n+1}}{a_n} &= \frac{((n+1)!)^2 \div (n!)^2}{((n+1)^2)! \div(n^2)!} \\
&= \frac{(n+1)^2}{(n + 1)^2 ((n + 1)^2 - 1)((n + 1)^2 - 2) \ldots(n+1)} \\
&= \frac{1}{((n + 1)^2 - 1)((n + 1)^2 - 2) \ldots(n+1)}.
\end{align*}
So, the ratio between $(n!)^2$ and $(n^2)!$ very quickly approaches $0$, telling you that $(n^2)!$ grows much faster than $(n!)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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is there are any short cut method to find the determinant of A? Find the determinant of A
$$A=\left(\begin{matrix}
x^1 & x^2 & x^3 \\
x^8 & x^9 & x^4 \\
x^7 & x^6 & x^5 \\
\end{matrix}\right)$$
My attempts : By doing a laplace expansion along the first column i can calculate,but it is a long process My question is that is there are any short cut method to find the determinant of A?
Thanks u
| \begin{align*}
\det(A) =\begin{vmatrix}
x^1 & x^2 & x^3 \\
x^8 & x^9 & x^4 \\
x^7 & x^6 & x^5 \\
\end{vmatrix}
&=\color{red}{x^5}\begin{vmatrix}
x^1 & x^2 & x^3 \\
x^8 & x^9 & x^4 \\
\color{red}{x^2} & \color{red}{x^1} & \color{red}{1}
\end{vmatrix}\\
&=x^5\cdot x^4 \cdot x^1\begin{vmatrix}
1 & x & x^2 \\
x^4 & x^5 & 1 \\
x^2 & x & 1 \\
\end{vmatrix}\\
&=x^{10}\cdot \color{red}{x^1}\begin{vmatrix}
\color{blue}{1} & \color{red}{1} & x^2 \\
\color{blue}{x^4} & \color{red}{x^4} & 1 \\
\color{blue}{x^2} & \color{red}{1} & 1
\end{vmatrix}\\
&=x^{11}\begin{vmatrix}
\color{blue}{0} & 1 & x^2 \\
\color{blue}{0} & x^4 & 1 \\
\color{blue}{x^2-1} & 1 & 1
\end{vmatrix}\\
&=x^{11}(x^2-1)(1-x^6).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the value of an integer $a$ such that $ a^2 +6a +1 $ is a perfect square. I was able to solve this but it required me using hit and trial at one step. I was wondering if i could find a more solid method to solve it.
p.s. this is the first time im asking a question here so sorry if i couldn't construct the question properly.
this is my solution -
$a^2 + 6a + 1 = k^2$
$(a+3)^2 = k^2 + 8$
$k^2 + 8 = m^2 $ where [$m=a+3$]
$(m-k)(m+k) = 8$ ... here by inspection $m= 3$ and $k =1$
hence, $a = 0, -6$.
| hint
$$a^2+6a+1=b^2$$
$$\iff a^2+6a+1-b^2=0$$
the reduced discriminant is
$$\Delta'=9-1+b^2=b^2+8$$
$$a=-3\pm \sqrt{b^2+8}$$
thus
$$b^2+8=c^2$$
and
$$(c+b)(c-b)=8$$
$$=4×2=-4×(-2)$$
$$c+b=\pm 4,\;\; c-b=\pm 2$$
gives $$\;\; b=\pm 1$$
and in all cases,
$$a=0 \text{ or } a=-6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that a sequence $a_n: n\in\mathbb{N}$ is (not) monotonic, bounded and converging $$a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$
$(0\in\mathbb{N})$
Monotonicity:
To prove, that a sequence is monotonic, I can use the following inequalities:
\begin{align}
a_n \leq a_{n+1}; a_n < a_{n+1}\\
a_n \geq a_{n+1}; a_n > a_{n+1}
\end{align}
I inserted some $n$'s to get an idea on how the sequence is going to look like.
I got:
\begin{align}
a_0&=3\\
a_1&=1\\
a_2&=\frac{7}{9}\approx 0.\overline{7}\\
a_3&=\frac{3}{4}=0.75
\end{align}
Assumption: The sequence is monotonic for $\forall n\in \mathbb{N}$
Therefore, I show that
\begin{align}
a_n \leq a_{n+1}; a_n < a_{n+1}\\
a_n \geq a_{n+1}; a_n > a_{n+1}
\end{align}
I am having problems when trying to prove the inequalities above:
\begin{align}
& a_n \geq a_{n+1}\Longleftrightarrow \left|\frac{a_{n+1}}{a_n}\right |\leq 1\\
& = \left|\dfrac{\dfrac{(n+1)^2+3}{(n+2)^2}}{\dfrac{n^2+3}{(n+1)^2}}\right|\\
& = \frac{4 + 10 n + 9 n^2 + 4 n^3 + n^4}{12 + 12 n + 7 n^2 + 4 n^3 + n^4}\\
& = \cdots \text{ not sure what steps I could do now}
\end{align}
Boundedness:
The upper bound with $a_n<s_o;\; s_o \in \mathbb{N}$ is obviously the first number of $\mathbb{N}$:
\begin{align}
a_0=s_o&=\frac{0^2+3}{(0+1)^2}\\
&=3
\end{align}
The lower bound $a_n>s_u;\; s_u \in \mathbb{N}$
$s_u$ should be $1$, because ${n^2+3}$ will expand similar to ${n^2+2n+1}$ when approaching infinity. I don't know how to prove that formally.
Convergence
Assumption (s.a) $\lim_{ n \to \infty} a_n =1$
Let $\varepsilon$ contain some value, so that $\forall \varepsilon > 0\, \exists N\in\mathbb{N}\, \forall n\ge N: |a_n-a| < \varepsilon$:
\begin{align}
\mid a_n -a\mid&=\left|\frac{n^2+3}{(n+1)^2}-1\right|\\
&= \left|\frac{n^2+3}{(n+1)^2}-\left(\frac{n+1}{n+1}\right)^2\right|\\
&= \left|\frac{n^2+3-(n+1)^2}{(n+1)^2}\right|\\
&= \left|\frac{n^2+3-(n^2+2n+1)}{(n+1)^2}\right|\\
&= \left|\frac{2-2n}{(n+1)^2}\right|\\
&= \cdots \text{(how to go on?)}
\end{align}
| Hint: $$a_{n+1}-a_n=2\,{\frac {{n}^{2}-n-4}{ \left( n+2 \right) ^{2} \left( n+1 \right) ^{
2}}}
$$
Second hint: $$a_n=\frac{n^2(1+\frac{3}{n^2})}{n^2(1+\frac{2}{n}+\frac{1}{n^2})}$$ this tends to $1$ for $n$ tends to infinity
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Sum the first $n$ terms of the series $1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots$ The question
Sum the first $n$ terms of the series:
$$ 1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots. $$
This was asked under the heading using method of difference and the answer given was
$$ S_n = \frac{1}{10}n(n+1)(n+2)(n+3)(2n+3). $$
My approach
First, I get $$ U_n=n(n+2)(n+1)^2. $$
Then I tried to make $U_n = V_n - V_{n-1}$
in order to get $S_n = V_n - V_0$.
But I really don't know how can I figure this out.
| \begin{align*}
\sum_{k=1}^n k(k+2)(k+1)^2&=\sum_{k=1}^n (k+3)(k+2)(k+1)k-2\sum_{k=1}^n (k+2)(k+1)k\\
&=24 \sum_{k=1}^n \binom{k+3}{4}-12 \sum_{k=1}^n \binom{k+2}{3}\\
&=24\binom{n+4}{5}-12\binom{n+3}{4}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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What is the smallest integer greater than 1 such that $\frac12$ of it is a perfect square and $\frac15$ of it is a perfect fifth power?
What is the smallest integer greater than 1 such that $\frac12$ of it is a perfect square and $\frac15$ of it is a perfect fifth power?
I have tried multiplying every perfect square (up to 400 by two and checking if it is a perfect 5th power, but still nothing. I don't know what to do at this point.
| Hint: Let the required number be x:
$\frac{1}{2}x= A^2$
$\frac{1}{5}x= B^5$
$\frac{1}{2}x+\frac{1}{5}x =A^2+B^5$
$\frac{5x+2x}{10}=A^2+B^5$
$7x=10(A^2+B^5)$
⇒ $x=10k$; $k ∈ N $.
So x is a power of 10.
The smallest 5th power of 10 is $10^5$ so the number must be $5\times 10^5=500000$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
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Find $\cos(\alpha+\beta)$ if $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$
If $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$, then prove that $\cos(\alpha+\beta)=\dfrac{a^2-b^2}{a^2+b^2}$
My Attempt
$$
b\sin x=c-a\cos x\implies b^2(1-\cos^2x)=c^2+a^2\cos^2x-2ac\cos x\\
(a^2+b^2)\cos^2x-2ac\cos x+(c^2-b^2)=0\\
\implies\cos^2x-\frac{2ac}{a^2+b^2}\cos x+\frac{c^2-b^2}{a^2+b^2}=0
$$
$$
a\cos\alpha+b\sin\alpha=c\implies a\cos^2\alpha\cos\beta+b\sin\alpha\cos\alpha\cos\beta=c\cos\alpha\cos\beta\\
a\cos\beta+b\sin\beta=c\implies a\sin\alpha\sin\beta\cos\beta+b\sin\alpha\sin^2\beta=c\sin\alpha\sin\beta\\
c\cos(\alpha+\beta)=a\cos\beta+a\sin\alpha\cos\beta.(\sin\beta-\sin\alpha)+b\sin\alpha+b\sin\alpha\cos\beta(\cos\alpha-\cos\beta)\\
$$
I think its getting complicated to solve now. What is the simplest way to solve this kind of problems?
| Guide: $c= a\cos \alpha +b\sin \alpha = a\cos \beta + b\sin \beta \implies a(\cos \alpha -\cos \beta) = b(\sin \beta-\sin \alpha)\implies \dfrac{a^2}{b^2}=\dfrac{(\sin \alpha - \sin \beta)^2}{(\cos \alpha - \cos \beta)^2}=m\implies RHS = \dfrac{m-1}{m+1}=...LHS$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Seeking methods to solve $ I = \int_{0}^{\infty} \frac{\sin(kx)}{x\left(x^2 + 1\right)} \:dx$ I am currently working on an definite integral that requires the following definite integral to be evaluated.
$$ I = \int_{0}^{\infty} \frac{\sin(kx)}{x\left(x^2 + 1\right)} \:dx$$
Where $k \in \mathbb{R}^{+}$
I was wondering what methods can be employed to solve this integral?
| The method I took was:
Let
$$ I(t) = \int_{0}^{\infty} \frac{\sin(kxt)}{x\left(x^2 + 1\right)} \:dx$$
Take the Laplace Transform
\begin{align}
\mathscr{L} \left[I(t) \right]&= \int_{0}^{\infty} \frac{\mathscr{L} \left[\sin(kxt)\right]}{x\left(x^2 + 1\right)} \:dx \\
&= \int_{0}^{\infty} \frac{kx}{\left(k^2x^2 + s^2\right)x\left(x^2 + 1\right)} \:dx \\
&= \int_{0}^{\infty} \frac{k}{\left(k^2x^2 + s^2\right)\left(x^2 + 1\right)} \:dx \\
&= \int_{0}^{\infty} \left[\frac{k^3}{\left(k^2 - s^2\right)\left(k^2x^2 + s^2 \right)} - \frac{k}{\left(k^2 - s^2\right)\left(x^2 + 1\right)}\right] \:dx \\
&= \left[\frac{k^3}{\left(k^2 - s^2\right)}\frac{\arctan\left(kx\right)}{ks} - \frac{k}{\left(k^2 - s^2\right)}\arctan(x)\right]_{0}^{\infty} \\
&= \frac{k^2}{s\left(k^2 - s^2\right)}\frac{\pi}{2} - \frac{k}{\left(k^2 - s^2\right)}\frac{\pi}{2} \\
&= \frac{k}{k^2 - s^2}\left[ \frac{k}{s} - 1 \right]\frac{\pi}{2} \\
&= \frac{k}{s\left(k + s\right)}\frac{\pi}{2}
\end{align}
And thus,
$$ I(t) = \mathscr{L}^{-1}\left[\frac{k}{s\left(k + s\right)}\frac{\pi}{2} \right] = \left[1 - e^{-kt} \right]\frac{\pi}{2}$$
Lastly,
$$ I = I(1) = \left[1 - e^{-k} \right]\frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
On existence of positive integer solution of $\binom{x+y}{2}=ax+by$ How can I prove this?
Prove that for any two positive integers $a,b$ there are two positive integers $x,y$ satisfying the following equation:
$$\binom{x+y}{2}=ax+by$$
My idea was that $\binom{x+y}{2}=\dfrac{x+2y-1}{2}+\dfrac{y(y-1)}{2}$ and choose $x,y$, such that $2a=x+2y-1, 2b=y(y-1)$, but using this idea, $x,y$ won’t be always positive.
| If $a=b$ then let $(x,y)=(a,a+1)$.
Otherwise, w.l.g. suppose $a>b$ and let $x+y=2t(a-b)$ for some positive integer $t$. Then
$$t(a-b)\Big(2t(a-b)-1\Big) =ax+by=(a-b)x+2bt(a-b)$$
$\text{Therefore } x=t\Big(2t(a-b)-1\Big)-2bt=t\Big(2t(a-b)-(2b+1)\Big)$.
$x$ will be a positive integer providing $t>\frac{2b+1}{2(a-b)}.$
$y=2t(a-b)-x$ will be a positive integer providing $2(a-b)>2t(a-b)-(2b+1)$ i.e. $\frac{2b+1}{2(a-b)}>t-1$.
$\frac{2b+1}{2(a-b)}$ is positive but not an integer and so, precisely as required, there is a positive integer $t$ such that
$$t>\frac{2b+1}{2(a-b)}>t-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2984918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Basic questions about pythagorean triples and "n-lets" I've had some difficulties finding answer to the two following questions:
1) Given one of natural numbers $a,b$ where $b$ is even and $a^2+b^2=c^2$ is there only one such a pythagorean triple?
2)How about a sum of $n$ squares of natural numbers that is equal to a square of a natural number? Given one of the summed squares is there only one such a pythagorean "n-let"?
I know that $a$ and $c$ are odd and I am aware of Euclid's formula but I don't know how to use it. I have no idea how to tackle 2). I only know that for exemple of $n=4$ there may be or not such two summed squares that would be lengths of the legs of a right triangle. Perhaps this is trivial but I don't know what to do.
Edit: 3) How about a given $c$ instead of $a,b$ in 1)?
| To construct n-lets, we can begin with this function to find values of $(m,n)$ for Euclid's formula:
$$n=\sqrt{m^2-A}\text{ where m varies from }\lceil\sqrt{A}\rceil\text{ to }\frac{A+1}{2}$$
This will let us find a triple with a matching side A, if it exists, for any $m$ that yields a positive integer $n$.
Let's begin with $(3,4,5)$ and find a triple to match the hypotenuse. In this case,
$$\lceil\sqrt{A}\rceil=\frac{A+1}{2}=3\text{ and } \sqrt{3^2-5}=2\text{ so we have }(3,2)$$
$$A=3^2-2^2=5\qquad B=2*3*2=12\qquad C=3^2+2^2=13$$
$\text{ The n-let that follows is }\\3^2+4^2+12^3=13^2\\\text{ and continuing with this process, we can get}$
$$3^2+4^2+12^2+84^2+3612^2=3613^2$$
or
$$3^2+4^2+12^2+84^2+132^2=157^2$$
Here we have two triples that match the hypotenuse of $(13,84,85)$ and here is how we found them
$$m =\lceil\sqrt{85}\rceil=10\text{ to }\frac{85+1}{2}=43$$
In the loop from $10$ to $43$, we find $(11,6)$ and $(43,42)$ with integer $n$.
$$A=11^2-6^2=85\qquad B=2*11*6=132\qquad C=11^2+6^2=157$$
$$A=43^2-42^2=85\qquad B=2*43*42=3612\qquad C=43^2+42^2=3613$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2985610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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What is the remainder when $4^{10}+6^{10}$ is divided by $25$? Without using calculator, how to decide? Must go with last two digits of $4^{10}+6^{10}$, can tell the last digit is $2$. How to tell the tenth digit of the sum?
Thanks!
| The last digit of $4^x$ is $4,6,4,6,4,6\dots \implies 4^{10}$ has last digit $6$
The last digit of $6^x$ is always $6 \implies 6^{10}$ has last digit $6$
Therefore $4^{10} + 6^{10}$ has last digit $2$.
Any multiple of $25$ has last digit $0$ or $5\implies
25$ cannot be a divisor!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
For which polynomials $p(x)$ is $p(p(x))+p(x)$=$x^4+3x^2+3$, for all $x \in \mathbb{R}$ For which polynomials $p(x)$ is $p(p(x))+p(x)$=$x^4+3x^2+3$, for all $x \in \mathbb{R}$
Since the power of the right hand side is 4, $p(x)$ has to be 2.
So I assumed a solution of: $p(x)=ax^2+bx+c$ and then i put it in $p(p(x))+p(x)$ and got:
$a(ax^2+bc+c)^2+b(ax^2+bx+c)+c+ax^2+bx+c=x^4+3x^2+3$
To find the coefficients I tried $x=0$ and got
$ac^2+bc+2c=3$. But here I'm stuck, how do I go from here?
| Identifying the coefficient of $x^4$ one gets $a=1$ and then the coefficient of $x^3$ you get $b=0$ hence $c=1$ or $c=-3$. Then check that $x^2+1$ is indeed a solution while $x^2-3$ is not.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find a general solution for $\int_{0}^{\infty} \sin\left(x^n\right)\:dx$ So, I was recently working on the Sine Fresnal integral and was curious whether we could generalise for any Real Number, i.e.
$$I = \int_{0}^{\infty} \sin\left(x^n\right)\:dx$$
I have formed a solution that I'm uncomfortable with and was hoping for qualified eyes to have a look over.
So, the approach I took was to employ Complex Numbers (I forget the name(s) of the theorem that allows this).
But
$$\sin\left(x^n\right) = \Im\left[-e^{-ix^n}\right]$$
And so, n
$$ I = \int_{0}^{\infty} \sin\left(x^n\right)\:dx = \Im\left[\int_{0}^{\infty} -e^{-ix^n}\:dx \right]= -\Im\left[\int_{0}^{\infty} e^{-\left(i^{\frac{1}{n}}x\right)^{n}}\:dx \right]$$
Applying a change of variable $u = i^{\frac{1}{n}}x$ we arrive at:
\begin{align}
I &= -\Im\left[i^{-\frac{1}{n}}\int_{0}^{\infty} e^{-u^{n}}\:du \right] \\
&= -\Im\left[i^{-\frac{1}{n}}\frac{\Gamma\left(\frac{1}{n}\right)}{n} \right]\\
&= \sin\left(\frac{\pi}{2n}\right)\frac{\Gamma\left(\frac{1}{n}\right)}{n}
\end{align}
My area of concern is in the substitution. As $i^{-\frac{1}{n}} \in \mathbb{C}$, I believe the limits of the integral should have been from $0$ to $i^{-\frac{1}{n}}\infty$. Is that correct or not?
I'm also struggling with bounds on $n$ for convergence. Is this expression valid for all $n\in\mathbb{R}$
Any guidance would be greatly appreciated
| Here is an alternative approach that avoids complex numbers and series altogether. To get round these two obstacles I will use a property of the Laplace transform.
Let
$$I = \int_0^\infty \sin (x^n) \, dx, \qquad n > 1.$$
We begin by enforcing a substitution of $x \mapsto x^{1/n}$. This gives
$$I = \frac{1}{n} \int_0^\infty \frac{\sin x}{x^{1 - 1/n}} \, dx.$$
The following useful property (does this result have a name? It would be so much nicer if it did!) for the Laplace transform will be used:
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$
Noting that
$$\mathcal{L} \{\sin x\}(t) = \frac{1}{1 + t^2},$$
and
$$\mathcal{L}^{-1} \left \{\frac{1}{x^{1-1/n}} \right \} (t)= \frac{1}{\Gamma (1 - \frac{1}{n})} \mathcal{L}^{-1} \left \{\frac{\Gamma (1 - \frac{1}{n})}{x^{1-1/n}} \right \} (t) = \frac{t^{-1/n}}{\Gamma (1 - \frac{1}{n})},$$
then
\begin{align}
I &= \frac{1}{n} \int_0^\infty \sin x \cdot \frac{1}{x^{1 - \frac{1}{n}}} \, dx\\
&= \frac{1}{n} \int_0^\infty \mathcal{L} \{\sin x\} (t) \cdot \mathcal{L}^{-1} \left \{\frac{1}{x^{1 - \frac{1}{n}}} \right \} (t) \, dt\\
&= \frac{1}{n\Gamma (1 - \frac{1}{n})} \int_0^\infty \frac{t^{-1/n}}{1 + t^2} \, dt.
\end{align}
Enforcing a substitution of $t \mapsto \sqrt{t}$ yields
\begin{align}
I &= \frac{1}{2 n \Gamma \left (1 - \frac{1}{n} \right )} \int_0^\infty \frac{t^{-\frac{1}{2} - \frac{1}{2n}}}{t + 1} \, dt\\
&= \frac{1}{2 n \Gamma \left (1 - \frac{1}{n} \right )} \operatorname{B} \left (\frac{1}{2} - \frac{1}{2n}, \frac{1}{2} + \frac{1}{2n} \right )\\
&= \frac{1}{2 n \Gamma \left (1 - \frac{1}{n} \right )} \Gamma \left (\frac{1}{2} - \frac{1}{2n} \right ) \Gamma \left (\frac{1}{2} + \frac{1}{2n} \right ). \tag1
\end{align}
Applying Euler's reflexion formula we have
$$\Gamma \left (\frac{1}{2} - \frac{1}{2n} \right ) \Gamma \left (\frac{1}{2} + \frac{1}{2n} \right ) = \frac{\pi}{\sin \left (\frac{\pi}{2n} + \frac{\pi}{2} \right )} = \frac{\pi}{\cos \left (\frac{\pi}{2n} \right )},$$
and
$$\Gamma \left (1 - \frac{1}{n} \right ) = \frac{\pi}{\sin \left (\frac{\pi}{n} \right ) \Gamma \left (\frac{1}{n} \right )}.$$
So (1) becomes
$$I = \frac{\sin (\frac{\pi}{n} ) \Gamma (\frac{1}{n})}{2n \cos (\frac{\pi}{2n} )},$$
or
$$I = \sin \left (\frac{\pi}{2n} \right ) \frac{\Gamma \left (\frac{1}{n} \right )}{n}, \qquad n > 1$$
where in the last line the double angle formula for sine has been used.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\big(1+\sin^2x\big)\\
&=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\
&=
\end{align}
I don't think its getting anywhere with my attempt, so how do I solve it ?
Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?
Note: The solution given in my reference is $4$.
| Let $t=\sin x$ and solve the cubic $$t^3+t^2+t=1$$
Wolfram Alpha gives the real solution as
$$t=(1/3) (-1 - 2/(17 + 3 \sqrt {33})^{1/3} + (17 + 3 \sqrt{33})^{1/3})$$
Plug the real solution of the above to get $$(1-t^2)^3 -4(1-t^2)^2+8(1-t^2) =4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Hints about the limit $\lim_{x \to \infty} ((1+x^2)/(x+x^2))^{2x}$ without l'Hôpital's rule? I've tried to evaluate $\lim_{x \to \infty} \left(\frac {1+x^2}{x+x^2}\right)^{2x}$ as $$\lim_{x \to \infty} \left(\left(\frac {1+ \frac{1}{x^2}}{1+ \frac{1}{x}}\right)^{x}\right)^{2}$$
So the denominator goes to $e^2$, but I don't know how to solve the numerator, because of the $x^2$. Any hint?
Thanks in advance!
| HINT
We have
$$\left(\frac {1+x^2}{x+x^2}\right)^{2x}=\left(\frac {x+x^2+1-x}{x+x^2}\right)^{2x}=\left[\left(1+\frac {1-x}{x+x^2}\right)^{\frac {x+x^2}{1-x}}\right]^{\frac {2x(1-x)}{x+x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Taylor series of functions Consider the Taylor series of the function
$$\frac{2e^x}{e^{2x}+1} = \sum_{n=0}^{\infty} \frac{E_n}{n!} x^n$$
Prove that $E_0 = 1, E_{2n-1} = 0$ and, for $n \ge 1$, $$E_{2n} = - \sum_{l=0}^{n-1} C_{2l}^{2n} {E_{2l}}$$
The hint given was to consider $\cosh(x)$.
I understand that the function is actually the reciprocal of $\cosh(x)$, however, I have no idea how to proceed.
| We use the Taylor series expansion of $\cosh x$ which you can find e.g. here.
\begin{align*}
\cosh x = \frac{e^{2x}+1}{2e^x} = \sum_{l=0}^\infty \frac{x^{2l}}{(2l)!}
\end{align*}
We obtain
\begin{align*}
\color{blue}{1}&=\left(\sum_{k=0}^\infty\frac{E_k}{k!}x^k\right)\frac{e^{2x}+1}{2e^x}\tag{1}\\
&=\left(\sum_{k=0}^\infty\frac{E_k}{k!}x^k\right)\left(\sum_{l=0}^\infty\frac{x^{2l}}{(2l)!}\right)\\
&=\left(\sum_{k=0}^\infty\frac{E_k}{k!}x^k\right)\left(\sum_{l=0}^\infty\frac{1+(-1)^l}{2}\frac{x^l}{l!}\right)\\
&=\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{E_k}{k!}\cdot\frac{1+(-1)^l}{2l!}\right)x^n\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}\frac{1+(-1)^{n-k}}{2}E_k\right)\frac{x^n}{n!}}\tag{2}
\end{align*}
Coefficient comparison of (1) and (2) gives
\begin{align*}
E_0=1\quad \text{and}\quad E_n=\sum_{k=0}^n\binom{n}{k}\frac{1+(-1)^{n-k}}{2}E_k\qquad\qquad n\geq 1\tag{3}
\end{align*}
From (3) we get for even index
\begin{align*}
\color{blue}{E_{2n}}&=-\sum_{k=0}^{2n-1}\binom{2n}{k}\frac{1+(-1)^{k}}{2}E_k\\
&\,\,\color{blue}{=-\sum_{k=0}^{n-1}\binom{2n}{2k}E_{2k}\qquad\qquad\qquad\qquad\qquad n\geq 1}
\end{align*}
Since
\begin{align*}
\frac{2e^{-x}}{e^{-2x}+1}=\frac{2e^x}{e^{2x}+1}\\
\end{align*}
we see the function $\frac{2e^x}{e^{2x}+1}$ is even and we conclude $\color{blue}{E_{2n-1}=0, n\geq 1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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coloring with a dihedral group $D_n$ with n prime I need to find out how many different colorings you can make with 2 colors in a dihedral group $D_n$ with $n$ prime and $m$ black and $p-m$ white beads. So first I compute the cycle index:
The cycle index of a dihedral group with $n$ prime (odd) is equal to:
$$Z(D_n) = \frac{1}{2}(\frac{1}{n}a_1^n + \frac{(n-1)}{n}a_n + a_1a_2^\frac{n-1}{2})$$
Now I fill in:
$$a_1 = (b+w), a_2 = (b^2 + w^2), a_n = (b^n + w^n)$$
After that, I find the number before the $b^mw^{p-m}$ and that is the amount of different colorings with $m$ black and $p-m$ white beads. But is there a general formule to find that number?
| Cycle index.
$$Z(D_p) = \frac{1}{2p}
\left(a_1^{p} + (p-1) a_p + p a_1 a_2^{(p-1)/2}\right)$$
We are interested in
$$[B^m W^{p-m}] Z(D_p; B+W).$$
This has three components.
First component.
$$[B^m W^{p-m}] \frac{1}{2p} (B+W)^p
= \frac{1}{2p} {p\choose m}.$$
Second component.
$$[B^m W^{p-m}] \frac{p-1}{2p} (B^p+W^p).$$
This is using an Iverson bracket:
$$\frac{p-1}{2p} [[m=0 \lor m=p]].$$
Third component.
$$[B^m W^{p-m}] \frac{1}{2} (B+W) (B^2+W^2)^{(p-1)/2}.$$
Now with $p$ prime we cannot have both $m$ and $p-m$ even, or both odd,
so one is odd and the other one even. Supposing that $m$ is odd we get
$$[B^{m-1} W^{p-m}] \frac{1}{2} (B^2+W^2)^{(p-1)/2}
\\ = [B^{(m-1)/2} W^{(p-m)/2}] \frac{1}{2} (B+W)^{(p-1)/2}
= \frac{1}{2} {(p-1)/2 \choose (m-1)/2}.$$
Alternatively, if $p-m$ is odd we get
$$[B^{m} W^{p-m-1}] \frac{1}{2} (B^2+W^2)^{(p-1)/2}
\\ = [B^{m/2} W^{(p-m-1)/2}] \frac{1}{2} (B+W)^{(p-1)/2}
= \frac{1}{2} {(p-1)/2 \choose m/2}.$$
Closed form.
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2p} {p\choose m}
+ \frac{p-1}{2p} [[m=0 \lor m=p]]
+ \frac{1}{2} {(p-1)/2 \choose (m-[[m \;\text{odd}]])/2}.}$$
Sanity check.
With a monochrome coloring we should get one as the answer, and
we find for $m=0$ ($B^0 W^p = W^p$)
$$\frac{1}{2p} {p\choose 0} + \frac{p-1}{2p}
+ \frac{1}{2} {(p-1)/2 \choose 0}
= \frac{p}{2p} + \frac{1}{2} = 1.$$
Similarly we get for $m=p$ ($B^p W^0 = B^p$)
$$\frac{1}{2p} {p\choose p} + \frac{p-1}{2p}
+ \frac{1}{2} {(p-1)/2 \choose (p-1)/2}
= \frac{p}{2p} + \frac{1}{2} = 1.$$
The sanity check goes through. Another sanity check is $m=1$ or
$m=p-1$ which should give one coloring as well. We find
$$\frac{1}{2p} {p\choose 1}
+ \frac{1}{2} {(p-1)/2\choose 0} = 1$$
and
$$\frac{1}{2p} {p\choose p-1}
+ \frac{1}{2} {(p-1)/2\choose (p-1)/2} = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998245",
"timestamp": "2023-03-29T00:00:00",
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Solving $\int_{0}^{\infty} \frac{\sin(x)}{x^3}dx$ In my attempt to solve the this improper integral, I employed a well known improper integral (part of the Borwein family of integrals):
$$ \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\sin\left(\frac{x}{5}\right)}{\left(\frac{x}{1}\right)\left(\frac{x}{3}\right)\left(\frac{x}{5}\right)} \: dx = \frac{\pi}{2}$$
To begin with, I made a simple rearrangement
$$ \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\sin\left(\frac{x}{5}\right)}{x^3} \: dx = \frac{\pi}{30}$$
From here I used the Sine/Cosine Identities
$$ \int_{0}^{\infty} \frac{\frac{1}{4}\left(-\sin\left(\frac{7}{15}x\right)+ \sin\left(\frac{13}{15}x\right) + \sin\left(\frac{17}{15}x\right) -\sin\left(\frac{23}{15}x\right) \right)}{x^3} \: dx = \frac{\pi}{30}$$
Which when expanded becomes
$$ -\int_{0}^{\infty} \frac{\sin\left(\frac{7}{15}x\right)}{x^3}\:dx + \int_{0}^{\infty} \frac{\sin\left(\frac{13}{15}x\right)}{x^3}\:dx +
\int_{0}^{\infty} \frac{\sin\left(\frac{17}{15}x\right)}{x^3}\:dx -
\int_{0}^{\infty} \frac{\sin\left(\frac{23}{15}x\right)}{x^3}\:dx
= \frac{2\pi}{15}$$
Using the property
$$\int_{0}^{\infty}\frac{\sin(ax)}{x^3}\:dx = a^2 \int_{0}^{\infty}\frac{\sin(x)}{x^3}\:dx$$
We can reduce our expression to
$$\left[ -\left(\frac{7}{15}\right)^2 + \left(\frac{13}{15}\right)^2 + \left(\frac{17}{15}\right)^2 - \left(\frac{23}{15}\right)^2\right] \int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = \frac{2\pi}{15}$$
Which simplifies to
$$ -\frac{120}{15^2}\int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = \frac{2\pi}{15}$$
And from which we arrive at
$$\int_{0}^{\infty} \frac{\sin(x)}{x^3}\:dx = -\frac{\pi}{4}$$
Is this correct? I'm not sure but when I plug into Wolframalpha it keeps timing out...
|
$$-\int_{0}^{\infty} \frac{\sin\left(\frac{7}{15}x\right)}{x^3}\:dx + \int_{0}^{\infty} \frac{\sin\left(\frac{13}{15}x\right)}{x^3}\:dx +
\int_{0}^{\infty} \frac{\sin\left(\frac{17}{15}x\right)}{x^3}\:dx -
\int_{0}^{\infty} \frac{\sin\left(\frac{23}{15}x\right)}{x^3}\:dx
= \frac{2\pi}{15}$$
You cannot expand the integrals since they are not convergent.
Moreover, given that $\int_a^b f(x)+g(x)dx$ converges,
$\int_a^b f(x)+g(x)dx=\int_a^b f(x)dx+\int_a^b g(x)dx$ only if $\int_a^b f(x)dx$ and $\int_a^b g(x)dx$ converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
If $n$ is an integer , find all the possible values for $(8n+6,6n+3)$ I have got 2 questions which I could not solve:
1) if $n$ is an integer , find all the possible values for $(8n+6,6n+3)$
2)if $n$ is an integer, find all possible values of $(2n^2+3n+5,n^2+n+1)$
| Let $d=\gcd(8n+6,6n+3)$, then $$d\mid 8n+6$$
$$d\mid 6n+3$$
so $$d\mid 6(8n+6)-8(6n+3)= 12$$
so $d\in \{1,2,3,4,6,12\}$ Since $6n+3$ is odd $d$ can not be $2,4,6$ or $12$ so $d=1$ or $d=3$ (which is realised at $n=3k$ for some integer $k$)
For second one:
Let $d=\gcd(2n^2+3n+5,n^2 + n+1)$, then $$d\mid 2n^2+3n+5$$
$$d\mid n^2+n+1$$
so $$d\mid 2n^2+3n+5-2(n^2 + n+1) =n+3$$
then $$d\mid (n^2+n+1)-(n^2-9)-(n+3)=7$$
So $d=1$ which is ok or $d=7$ which is realised if $n=7k+4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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System of equations with three variables Characterize all triples $(a,b,c)$ of positive real numbers such that
$$ a^2-ab+bc = b^2-bc+ca = c^2-ca+ab. $$
This is the equality case of the so-called Vasc inequality. I think the answer is that $a=b=c$ or $a:b:c = \sin^2(4\pi/7) : \sin^2(2\pi/7) : \sin^2(\pi/7)$ and cyclic equality cases. I'm mostly interested if there is a way to derive this reasonably, other than just magically guessing the solutions and then doing degree-counting. This was left as a "good exercise" in Mildorf that has bothered me for years.
| If $c=0$ then $a^2-ab=b^2=ab$, which gives $a=b=c=0.$
Let $abc\neq0$ and $a=xb$.
Thus, from the first equation we obtain:
$$a^2-ab-b^2=(a-2b)c.$$
If $a=2b$ so $a=b=c=0$, which is impossible here.
Thus, $c=\frac{a^2-ab-b^2}{a-2b}$ and from the second equation we obtain:
$$b^2+\frac{(a-b)(a^2-ab-b^2)}{a-2b}=\frac{(a^2-ab-b^2)^2}{(a-2b)^2}-\frac{(a^2-ab-b^2)a}{a-2b}+ab$$ or
$$1+\frac{(x-1)(x^2-x-1)}{x-2}=\frac{(x^2-x-1)^2}{(x-2)^2}-\frac{(x^2-x-1)x}{x-2}+x$$ or
$$(x-1)(x^3-5x^2+6x-1)=0,$$ which gives $x=1$ and $a=b=c$ or
$$x^3-5x^2+6x-1=0.$$
Now, easy to show that $\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}$, $\frac{\sin^2\frac{\pi}{7}}{\sin^2\frac{3\pi}{7}}$ and $\frac{\sin^2\frac{3\pi}{7}}{\sin^2\frac{2\pi}{7}}$ they are roots of the last equation.
For example:
$$\left(\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}\right)^3-5\left(\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}\right)^2+6\cdot\frac{\sin^2\frac{2\pi}{7}}{\sin^2\frac{\pi}{7}}-1=$$
$$=\left(4\cos^2\frac{\pi}{7}\right)^3-5\left(4\cos^2\frac{\pi}{7}\right)^2+6\left(4\cos^2\frac{\pi}{7}\right)-1=$$
$$=\left(2+2\cos\frac{2\pi}{7}\right)^3-5\left(2+2\cos\frac{2\pi}{7}\right)^2+6\left(2+2\cos\frac{2\pi}{7}\right)-1=$$
$$=8\cos^3\frac{2\pi}{7}+4\cos^2\frac{2\pi}{7}-4\cos\frac{2\pi}{7}-1=$$
$$=2\left(4\cos^3\frac{2\pi}{7}-3\cos\frac{2\pi}{7}\right)+6\cos\frac{2\pi}{7}+2+2\cos\frac{4\pi}{7}-4\cos\frac{2\pi}{7}-1=$$
$$=2\cos\frac{2\pi}{7}+2\cos\frac{4\pi}{7}+2\cos\frac{6\pi}{7}+1=$$
$$=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{\sin\frac{\pi}{7}}+1=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{\sin\frac{\pi}{7}}+1=0.$$
Since we have no another roots, we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given that $X \sim \operatorname{Binomial}(n,p)$, Find $\mathbb{E}[X(X-1)(X-2)(X-3)]$
Given that $X \sim \operatorname{Binomial}(n,p)$, Find $\mathbb{E}[X(X-1)(X-2)(X-3)]$.
It is suggested that I can transform it into
\begin{align}
\mathbb{E}[X(X-1)(X-2)(X-3)]
&=\sum_{k=0}^n k(k-1)(k-2)\mathbb{P}\{X=k\}\\
&=\sum_{k=3}^{n+3} (k-3)(k-4)(k-5)\mathbb{P}\{X=k-3\}\\
&=\sum_{k=0}^n i(i-1)(i-2)\mathbb{P}\{X=i\}
\end{align}
But then I just have no idea about how can i do it. I suspect that it needs something similar to this post but the steps are quite different from this one.
Please help.
| Start as suggested, and write down what the probability mass function (pmf) of the Binomial actually is:
$$\begin{align*}
\mathbb{E}[X(X-1)(X-2)(X-3)]
&= \sum_{k=0}^n k(k-1)(k-2)(k-3)\mathbb{P}\{X=k\}\\
&= \sum_{k=4}^n k(k-1)(k-2)(k-3)\mathbb{P}\{X=k\}\\
&= \sum_{k=4}^n k(k-1)(k-2)(k-3)\binom{n}{k}p^k(1-p)^{n-k}\\
&= \sum_{k=4}^n k(k-1)(k-2)(k-3)\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\\
&= \sum_{k=4}^n \frac{n!}{(k-4)!(n-k)!}p^k(1-p)^{n-k}\\
&= \sum_{k=4}^n \frac{n(n-1)(n-2)(n-3)(n-4)!}{(k-4)!((n-4)-(k-4))!}p^k(1-p)^{n-k}\\
&= n(n-1)(n-2)(n-3)p^4\sum_{k=4}^n \binom{n-4}{k-4}p^{k-4}(1-p)^{(n-4)-(k-4)}\\
&= n(n-1)(n-2)(n-3)p^4\sum_{\ell=0}^n \binom{n-4}{\ell}p^{\ell}(1-p)^{(n-4)-\ell}\\
&= \boxed{n(n-1)(n-2)(n-3)p^4}
\end{align*}$$
since $\sum_{\ell=0}^n \binom{n-4}{\ell}p^{\ell}(1-p)^{(n-4)-\ell}=1$, recognizing the sum of probabilities for a Binomial with parameters $n-4$ and $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does this specific SO(4) matrix have to be block-diagonal? So I have a specific real $4\times4$ matrix $\mathbf{P}$ given by
\begin{align}
\mathbf{P}=
\begin{pmatrix}
p_{11} & -p_{21} & p_{13} &-p_{23}\\ p_{21} & p_{11} & p_{23} & p_{13}\\p_{31} & -p_{41}& p_{33} & -p_{43}\\p_{41} & p_{31} & p_{43} & p_{33}.
\end{pmatrix},
\end{align}
and I'm confident that if this matrix is in $SO(4)$ then it must be block-diagonal OR anti-block-diagonal i.e. if $\mathbf{P}\in SO(4)$ then either $p_{11}=p_{21}=p_{33}=p_{43}=0$ or $p_{13}=p_{23}=p_{31}=p_{41}=0$, but I can't seem to show this...
I've tried picking one column and generating a set of orthonormal vectors to fill the matrix, but this just gives examples where it works and I'd like to show it generally. Is this possible? Am I just missing something trivial here?
| Skip to the end for the big reveal, or read through this for the "how I got there" version.
Let's rewrite that as
\begin{align}
\mathbf{P}=
\begin{pmatrix}
a & -b & p &-q\\
b & a & q & p\\
c & -d& r & -s\\
d & c & s & r.
\end{pmatrix},
\end{align}
Orthogonality of the first and third and first and 4th columns tells us that
\begin{align}
ap + bq + cr + ds &= 0\\
-aq + bp - cs + dr &= 0\\
\end{align}
Cross-multiply to get
\begin{align}
apq + bq^2 + crq + dsq &= 0 \\
-apq + bp^2 - cps + dpr &= 0\\
\end{align}
and sum to get
\begin{align}
b(p^2 + q^2) + crq - cps + dsq + dpr &= 0 \\
\end{align}
Doing the same for columns 2 against 3 and 4 gives
\begin{align}
a(p^2 + q^2) - drq + csq + dps + cpr &= 0 \\
\end{align}
Let's factor those to get
\begin{align}
b(p^2 + q^2) + c(rq - ps) + d(sq + pr) &= 0 \\
a(p^2 + q^2) + c(sq + pr) - d(rq - ps) &= 0 \\
\end{align}
Looking at the dot product between rows 2 and 3, we see that
$$
rq - ps = ad - bc
$$
and similarly, for rows 2 and 4, we get
$$
qs + pr = - (ac + bd)
$$
so
\begin{align}
b(p^2 + q^2) + c(ad-bc) - d(ac + bd) &= 0 \\
a(p^2 + q^2) - c(ac + bd) - d(ad-bc) &= 0 \\
\end{align}
which simplifies to
\begin{align}
b(p^2 + q^2) - bc^2 - bd^2 &= 0 \\
a(p^2 + q^2) - ac^2 - ad^2 &= 0 \\
\end{align}
which become
\begin{align}
b(p^2 + q^2 - c^2 - d^2) &= 0 \\
a(p^2 + q^2 - c^2 - d^2) &= 0 \\
\end{align}
We conclude that either
(1) $p^2 + q^2 = c^2 + d^2$ or
(2) $a = b = 0$.
In the second case, we have that the squared norm of the first row is $p^2 + q^2$, which must be $1$, and the squared norm of the first column is $c^2 + d^2$, which must also be $1$, hence $p^2 + q^2 = c^2 + d^2$.
In other words, in all cases, $p^2 + q^2 = c^2 + d^2$.
By looking at the squared norms of the first row and the 4th column, we find that
$$
a^2 + b^2 = r^2 + s^2
$$
as well.
But nothing else obvious seems to jump out...
...and so I began to wonder if it was actually true, and came up with this:
\begin{align}
\mathbf{P}=
\begin{pmatrix}
\frac{1}{2} & -\frac{1}{2} & 0 &-s\\
\frac{1}{2} & \frac{1}{2} & s & 0\\
\frac{1}{2} & -\frac{1}{2}& 0 & s\\
\frac{1}{2} & \frac{1}{2} & -s & 0
\end{pmatrix},
\end{align}
where $s = \frac{1}{\sqrt{2}}$.
That seems to be a counterexample to your conjecture. So I guess the answer to your question is "Yes, you are just missing something trivial." :)
| {
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"url": "https://math.stackexchange.com/questions/3007826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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What is $\lim_{x \to 3} (3^{x-2}-3)/(x-3)(x+5)$ without l'Hôpital's rule? I'm trying to solve the limit $\lim_{x \to 3} \frac{3^{x-2}-3}{(x-3)(x+5)}$
but I don't know how to proceed: $\lim_{x \to 3} \frac{1}{x+5}$ $\lim_{x \to 3} \frac{3^{x-2}-3}{x-3}$ = $1\over8$ $\lim_{x \to 3} \frac{\frac{1}{9}(3^{x}-27)}{x-3}$
Any hints? Thanks in advance.
| Perhaps use definition of $3^x$ ... namely $3^x = e^{x\log 3}$. Instaed of $x \to 3$ write $y=x-3$ and do $y \to 0$.
$$
\lim_{x \to 3} \frac{3^{x-2}-3}{(x-3)(x+5)} = \lim_{y \to 0}\frac{3^{y+1}-3}{y(y+8)}
=3 \lim_{y \to 0}\frac{3^{y}-1}{y(y+8)}
\\
3^y = \exp(y\log 3) = 1 + y\log 3 + o(y)
\\
3^y-1 = y\log 3 + o(y)
\\
y(y+8) = 8y+o(y)
\\
\frac{1}{y(y+8)} = y^{-1}\frac{1}{8}+o(y^{-1})
\\
\frac{3^y-1}{y(y+8)} = \frac{\log 3}{8} + o(y)
\\
3\frac{3^y-1}{y(y+8)} = \frac{3\log 3}{8} + o(1)
\\
3 \lim_{y \to 0}\frac{3^{y}-1}{y(y+8)} = \frac{3\log 3}{8}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Maclaurin series for $\arctan^{2}(x)$ I have a question here that requires me to find the Maclaurin series expansion of
$\arctan^{2}(x)$. Now I know how to find it for $\arctan(x)$, by taking the derivative, expanding it into a series, and integrating back (given x is in the interval of uniform convergence), But applying that here leaves me with
$$\frac{df}{dx}=2\arctan(x)\frac{1}{(x^2+1)}$$ I am not sure If I can pick out parts of the product (like $\arctan(x)$) and differentiate, expand , then integrate them. Even if I did, how would I go about multiplying two series? Is there an easier way to do this? Thanks.
| We can try to obtain the series in the following way:
$$f(x)=\arctan^2 x=x^2 \int_0^1 \int_0^1 \frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$
It's easier to consider:
$$g(x)=\int_0^1 \int_0^1 \frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$
Let's use partial fractions:
$$\frac{1}{(1+x^2u^2)(1+x^2v^2)}=\frac{u^2}{(u^2-v^2)(1+x^2u^2)}-\frac{v^2}{(u^2-v^2)(1+x^2v^2)}$$
We obtain a sum of two singular integrals, which however, can both be formally expanded into a series:
$$g(x)=\int_0^1 \int_0^1 \left(\frac{u^2}{(u^2-v^2)(1+x^2u^2)}-\frac{v^2}{(u^2-v^2)(1+x^2v^2)} \right) du ~dv=$$
$$g(x)=\sum_{n=0}^\infty (-1)^n x^{2n} \int_0^1 \int_0^1 \left(\frac{u^{2n+2}}{u^2-v^2}-\frac{v^{2n+2}}{u^2-v^2} \right) du ~dv$$
Now:
$$g(x)=\sum_{n=0}^\infty (-1)^n x^{2n} \int_0^1 \int_0^1 \frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} du ~dv$$
Obviously, every integral is finite now, and we can write:
$$\int_0^1 \int_0^1 \frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du=2 \int_0^1 \int_0^u \frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du= \\ = 2 \sum_{k=0}^\infty \int_0^1 \int_0^u u^{2n} \left(1-\frac{v^{2n+2}}{u^{2n+2}} \right) \frac{v^{2k}}{u^{2k}} dv ~du =2 \sum_{k=0}^\infty \int_0^1 \int_0^1 u^{2n+1} \left(1-t^{2n+2} \right) t^{2k} dt ~du= \\ = 2 \sum_{k=0}^\infty \int_0^1 \left(\frac{1}{2k+1}-\frac{1}{2k+2n+3} \right) u^{2n+1}~du= 2\sum_{k=0}^\infty \frac{1}{(2k+1)(2k+2n+3)}$$
So we get:
$$g(x)=\frac{1}{2} \sum_{n=0}^\infty (-1)^n x^{2n} \sum_{k=0}^\infty \frac{1}{(k+\frac{1}{2})(k+n+\frac{3}{2})}$$
The inner series converges for all $n$, and we can formally represent it as a difference of two divergent harmonic series, which, after some manipulations, should give us the same result as robjohn obtained.
I suppose, a kind of closed form for the general term can also be given in terms of digamma function:
$$g(x)=\frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{\psi \left(n+\frac32 \right)-\psi \left(\frac12 \right)}{n+1} x^{2n}$$
Which makes:
$$\arctan^2 x=\frac{x^2}{2} \sum_{n=0}^\infty (-1)^n \frac{\psi \left(n+\frac32 \right)-\psi \left(\frac12 \right)}{n+1} x^{2n}$$
Which is essentially the same as robjohn's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the area of the surface formed by revolving the given curve about $(i)x$-axis and $(i)y$-axis
Q:Find the area of the surface formed by revolving the given curve about $(i)x-axis$ and $(i)y-axis$
$$x=a\cos\theta ,y=b\sin\theta,0\le\theta\le2\pi$$
About $x-$axis is, $S=2\pi\int_0^{2\pi}b\sin\theta \sqrt{a^2(\sin\theta)^2+b^2(\cos\theta)^2} d\theta$
About $y-$axis is, $S=2\pi\int_0^{2\pi}a\cos\theta \sqrt{a^2(\sin\theta)^2+b^2(\cos\theta)^2} d\theta$from now i get stuck.I can't not figure out the integral part.Any hints or solution will be appreciated.Thanks in advance.
| The limits of integration need some correction. While finding the surface area about the $x$-axis, $x$ ranges from $-a$ to $a\implies\theta$ ranges from $\pi\rightarrow 2\pi$, not $0\rightarrow 2\pi$. For the surface area about the $y$-axis, $\theta$ ranges from $-\pi/2 \rightarrow +\pi/2$, or from $3\pi/2\rightarrow 2\pi$ and $0\rightarrow\pi/2$.
For the surface area about $x$-axis, take $t=\cos\theta \implies dt=-\sin\theta\ d\theta$
$S_x=2\pi\int_\pi^{2\pi}|b\sin\theta| \sqrt{a^2\sin^2\theta+b^2\cos^2\theta}\ d\theta\\ \ \ \ \ =2\pi b\int_\pi^{2\pi}|\sin\theta| \sqrt{a^2(1-\cos^2\theta)+b^2\cos^2\theta}\ d\theta\\\\ \ \ \ \ =2\pi b\int_\pi^{2\pi}(-\sin\theta) \sqrt{a^2+(b^2-a^2)\cos^2\theta}\ d\theta\\\\ \ \ \ \ =2\pi b\int_{-1}^{1}\sqrt{a^2+(b^2-a^2)t^2}\ dt\\\\ \ \ \ \ =4\pi b\int_0^{1}\sqrt{a^2+(b^2-a^2)t^2}\ dt\\$
Depending on the sign of $(b^2-a^2)$, this integral can take either of the standard forms $\int \sqrt{a^2-x^2}\ dx$ or $\int \sqrt{a^2+x^2}\ dx$.
For the surface area about $y$-axis, because we have $\cos\theta\ d\theta$ outside the square root, take $t=\sin\theta$, and try to get the argument of the square root in terms of $\sin\theta$ alone, this time by substituting for $\cos^2\theta$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof that $\frac{1+x^2}{n^2} \geq 1-e^{-x^2/n^2}$ for all $x,n \in \mathbb{R}$ I'm looking for a simple proof that $\frac{1+x^2}{n^2} \geq 1-e^{-x^2/n^2}$ for all $x,n \in \mathbb{R}$.
My first attempt was to express the exponential as a Taylor series:
$$\frac{1+x^2}{n^2} \geq \frac{x^2}{n^2}-\frac{1}{2!}\frac{x^4}{n^4}+\frac{1}{3!}\frac{x^6}{n^6}- \, ... \, .$$
Obviously
$$\frac{1+x^2}{n^2} \geq \frac{x^2}{n^2},$$
so if I can show
$$-\frac{1}{2!}\frac{x^4}{n^4}+\frac{1}{3!}\frac{x^6}{n^6}- \, ... <0,$$
then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.
| Set $y=x^2/n^2$. Then you want to show that
$$
\frac{1}{n^2}+y\ge 1-e^{-y}
$$
Note that $y\ge0$. A standard process is to consider
$$
f(y)=\frac{1}{n^2}+y-1+e^{-y}
$$
and note that $f(0)=1/n^2>0$. Also
$$
f'(y)=1-e^{-y}=\frac{e^y-1}{e^y}>0
$$
for $y>0$. Therefore the function $f$ is strictly increasing over $[0,\infty)$ and so
$$
f(y)>0
$$
for $y\ge0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3019372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Values of $x$ satisfying $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$
For what values of $x$ between $0$ and $\pi$ does the inequality $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$ hold?
My Attempt
$$
\sin x\cos x\cdot(\cos^2x-\sin^2x)=\frac{1}{2}\cdot\sin2x\cdot\cos2x=\frac{1}{4}\cdot\sin4x>0\implies\sin4x>0\\
x\in(0,\pi)\implies4x\in(0,4\pi)\\
4x\in(0,\pi)\cup(2\pi,3\pi)\implies x\in\Big(0,\frac{\pi}{4}\Big)\cup\Big(\frac{\pi}{2},\frac{3\pi}{4}\Big)
$$
But, my reference gives the solution, $x\in\Big(0,\dfrac{\pi}{4}\Big)\cup\Big(\dfrac{3\pi}{4},\pi\Big)$, where am I going wrong with my attempt?
| As an alternative for a full solution we can consider two cases
*
*$\sin x \cos x >0$ that is $x\in(0,\pi/2)\cup(\pi,3\pi/2)$
$$\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x \iff\cos^2x>\sin^2x \iff2\sin^2 x<1$$
$$-\frac{\sqrt 2}2<\sin x<0 \,\land\, 0<\sin x<\frac{\sqrt 2}2 \iff \color{red}{x\in(0,\pi/4)}\cup(\pi,5\pi/4)$$
*
*$\sin x \cos x <0$ that is $x\in(\pi/2,\pi)\cup(3\pi/2,2\pi)$
$$\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x \iff\cos^2x<\sin^2x \iff2\sin^2 x>1$$
$$-1<\sin x<-\frac{\sqrt 2}2\,\land\, \frac{\sqrt 2}2<\sin x <1 \iff \color{red}{x\in(\pi/2,3\pi/4)}\cup(3\pi/2,7\pi/4)$$
and then your solution is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What are the complex roots of $x^3-1$? What are the complex roots of $x^3-1$?
Work I've done so far:
I've set $x = a + bi$. Since $x^3-1=0$, I set $x^3 = (a+bi)^3=1$.
This gives me the following:
(1) $(-ab^2 + a^3) + (2ab^2 + 2a^2b + a^2b - b^3)i$
Which means that I set $(-ab^2 + a^3) = a(a^2-b^2)= 1$ which is also equivalent to
(2) $a(a-b)(a+b)=(a-b)(a^2+ab)=1$.
I also set
(3) $(2ab^2 + 2a^2b + a^2b - b^3) = 0$.
I simplify (3) to
(4) $2b(a^2 + ab) + (a^2 -b^2)b = 0 $
which gives me
(5) $\frac{2b}{a-b} + \frac{b}{a} = 0$ using (2).
Then I get
(6) $\frac{2b}{a-b} = -\frac{b}{a}$. Then I get that $3a=b$. Plugging into (2) I get
(7) $a(a^2 - (3a)^2)=1 = a(a^2 -9a^2) = -8a^3$. So that $a= \frac{-1}{2}$. Now I get that $b= \frac{3}{2}$, which would give me $\frac{1}{2} + \frac{3}{2}i$. But on Wolfram, the imaginary component is close to $.9$. Where am I going wrong?
| Hint:
The simpler way is to factorize:
$$
x^3-1=(x-1)(x^2+x+1)
$$
can you find all the roots?
Anyway, your algebra is wrong because:
$$
(a+ib)^3=a^3+3a^2(ib)+3a(ib)^2+(ib)^3=a^3-3ab^2+i(3a^2b-b^3)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why is $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$ not correct? $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$
Apparently, the 2nd step is illegal here. Probably because for $x=-\infty$ I'd get $(+\infty-\infty)$ which is not possible. I see why this wouldn't be possible, I'm not sure if it really is the cause which makes that equation illegal though.
But now, apparently, I could do this:
$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x=\lim_{x\to -\infty}\frac{[\sqrt{x^2+5x+3}+x][\sqrt{x^2+5x+3}-x]}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{x^2+5x+3-x^2}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5x+3}{\sqrt{x^2+5x+3}-x}=\lim_{x\to -\infty}\frac{5+3/x}{\sqrt{1+5/x+3/x^2}-1}=-5/2$
which gives me the correct result.
But in the 3th step I used $x^2-x^2=0$, how is that legal?
Also, in the 2nd step I implicitly used:
$-x\sqrt{x^2+5x+3}+x\sqrt{x^2+5x+3}=0$
Which also seems to be fine, but why?
| The first step is illegal, not the second. This is because $\sqrt{x^2 + 5x + 3} \not\equiv -x$.
Also addressing your comment:
The idea behind $\lim_{x \to -\infty} \sqrt{x^2 + 5x + 3} = \dots = \lim_{x \to -\infty} −x$ was, that $x^2$ dominantes $5x+3$. I also thought if that argument is wrong here, if $\lim_{x \to -\infty} x − x = 0$ is legal and everything of the same "type"; then I'm okay again. Then I have to think about why the domination argument doesn't work as I thought it does here.
You are right in thinking that $x^2$ dominates $5x+3$, so that $\color{blue}{\sqrt{x^2 + 5x + 3}} \sim \color{red}{-x}$ for large negative $x$. Indeed there is no problem when computing their ratio:
$$\lim_{x \to -\infty} \frac{\color{blue}{\sqrt{x^2 + 5x + 3}}}{\color{red}{-x}} = 1$$
However, because you are computing the difference between $\color{blue}{\sqrt{x^2 + 5x + 3}}$ and $\color{red}{-x}$, such an approximation is not fine enough. You need more terms of the series expansion:
$$
\begin{align*}
\color{blue}{\sqrt{x^2 + 5x + 3}}
&= -x \sqrt{1 + \frac{5}{x} + \frac{3}{x^2}} \\
&\sim -x \left(1 + \frac{1}{2} \cdot \frac{5}{x} + O\left(\frac{1}{x^2}\right) \right) \\
&= \color{red}{-x} - \frac{5}{2} + O\left(\frac{1}{x}\right)
\end{align*}
$$
Then may you conclude that the required limit is $-5/2$.
By analogy, an approximation such as $\color{blue}{x + 1} \sim \color{red}{x}$ is perfectly fine for computing the ratio limit $\lim_{x \to -\infty} (\color{blue}{x + 1}) / (\color{red}{x})$. But if you are computing the difference limit, then you may not discard the $1$ in $\color{blue}{x + 1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
} |
Number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$
Find the number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$
\begin{align}
2\cos(\pi\sqrt{x-4})&.\cos(\pi\sqrt{x})=2\\\implies\cos\Big[\pi(\sqrt{x-4}+\sqrt{x})\Big]&+\cos\Big[\pi(\sqrt{x-4}-\sqrt{x})\Big]=2\\
\implies\cos\Big[\pi(\sqrt{x-4}+\sqrt{x})\Big]=1\quad&\&\quad\cos\Big[\pi(\sqrt{x-4}-\sqrt{x})\Big]=1\\
\pi(\sqrt{x-4}+\sqrt{x})=2n\pi\quad&\&\quad\pi(\sqrt{x-4}-\sqrt{x})=2m\pi\\
\sqrt{x-4}+\sqrt{x}=2n\quad&\&\quad\sqrt{x-4}-\sqrt{x}=2m\\
2\sqrt{x}=2(n-m)\quad&\&\quad2\sqrt{x-4}=2(n+m)\\
\sqrt{x}=n-m\quad&\&\quad\sqrt{x-4}=n+m\quad\&\quad x\geq4
\end{align}
How do I properly find the solutions ?
Or can I simply say
$$
x=(n-m)^2=(n+m)^2-4nm=x-4-4nm\implies nm=-1\\
\implies x=\bigg[n+\frac{1}{n}\bigg]^2\in\mathbb{Z}\implies n,\frac{1}{n}\in\mathbb{Z}\\
\implies n\neq0\implies n=1,x=4\text{ is the only solution}
$$
| I suppose $x$ is real in the following and that $\cos$ is the function $\cos:\Bbb R\to\Bbb R$. (There is an other function $\cos:\Bbb C\to\Bbb C$, to use it i have to ask for the branch of the square root(s) first.)
The two $\cos$ functions in the product (evaluated at those two places) must have (in a correlated way) the value $\pm 1$. This makes things easier maybe to decide that $\sqrt x$ and $\sqrt {x-4}$ are two integers of same parity. In particular $x\ge 4$. Starting with the perfect square $2^2=4$ the distance between two perfect squares is $\ge 3^2-2^2=9-4=5$, so it is at least $5$. So the bigger perfect square $x$ (among $x,x-4$) is at most $4$. We get the solution $x=2^2=4$. (There is no $x=1^2$ or $x=0^2$ as solution, because $x-4<0$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve for the exponent of a matrix we discussed matrices in class and had the following task: Given
$$U=\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\\1 & \frac{1}{15} & 0 & 0 & 0 &0\\0 & \frac{8}{15} & \frac{3}{15} & 0 & 0 & 0\\0 & \frac{6}{15} & \frac{9}{15} & \frac{6}{15} & 0 & 0\\ 0 & 0 & \frac{3}{15} & \frac{8}{15} & \frac{10}{15} & 0\\0 & 0 & 0 & \frac{1}{15} & \frac{5}{15} & 1\end{bmatrix},
\quad \overrightarrow{s}=\begin{bmatrix}1\\0\\0\\0\\0\\0\end{bmatrix},$$
solve for $x$ such that $U^x\overrightarrow{s}=\overrightarrow{s_x}$ where the last element (Row) of $\overrightarrow{s_x}$ should be equal or greater than 0.99.
We were told that the only way to get $x$ is by inserting random numbers and "search" for it. By doing this we actually found:
$$\overrightarrow{s_{15}}=\begin{bmatrix}0\\0\\0\\0\\0,0137\\0,9863\end{bmatrix},\quad \overrightarrow{s_{16}}=\begin{bmatrix}0\\0\\0\\0\\0,0091\\0,9909\end{bmatrix}.$$
So the answer is pretty much $x=16$, but is there no way to solve for $x$ instead of inserting random numbers until you find the answer?
Thanks for any answers
| Hint :
Using this you get the diagonalization of $U=PDP^{-1}$, then $U^x = PDP^{-1} \dots PDP^{-1} = P D^x P^{-1}$ where $D$ is the matrix with the elements of $D$ elevated to power $x$. Using this you can solve for the last element being greater to $0.99$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Positive divisors of n = $2^{14} \cdot 3^9 \cdot 5^8 \cdot 7^{10} \cdot 11^3 \cdot 13^5 \cdot 37^{10}$ How do I find positive divisors of n that are perfect cubes that are multiples of 2^10 * 3^9 * 5^2 * 7^5 * 11^2 * 13^2 * 37^2
The answer is (1)(1)(2)(2)(1)(1)(3) = 12
I don't understand though because I would have done something like:
2: [(14-10)/3]+1 = 2 (taking the floor)
3: [(9-9)/3]+1 = 1
5: [(8-2)/3]+1 = 3
7: [(10-5)/3]+1 = 2
11: [(3-2)/3]+1 = 1
13: [(5-2)/3]+1 = 2
37: [(10-2)/3]+1 =3
2*1*3*2*1*2*3
| For $d$ to be a divisor of $n$, $d$ must be of the form $2^a \cdot 3^b \cdot 5^c \cdot 7^d \cdot 11^e \cdot 13^f \cdot 37^g$, where $0 \leq a \leq 14, 0 \leq b \leq 9, 0 \leq c \leq 8, 0 \leq d \leq 10, 0 \leq e\leq 3, 0\leq f \leq 5, 0 \leq g \leq 10$.
Now we want $d$ ot be a multiple of the number given, that means $d$ must be of the form $2^a \cdot 3^\color{red}{9} \cdot 5^c \cdot 7^d \cdot 11^e \cdot 13^f \cdot 37^g$, where $\color{red}{10} \leq a \leq 14, \color{red}{2} \leq c \leq 8, \color{red}{5} \leq d \leq 10, \color{red}{2} \leq e\leq 3, \color{red}{2}\leq f \leq 5, \color{red}{2} \leq g \leq 10$.
Now we want $d$ to be a cube as well. This means all powers appearing must by divisible by $3$. Thus
$$a=12, b=9, c \in \{3,6\}, d \in \{6,9\}, e=3, f=3, g \in \{3,6,9\}.$$
Thus the total number of choices we have are
$$2 \cdot 2 \cdot 3=12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a+\sqrt{a^2+1}= b+\sqrt{b^2+1}$, then $a=b$ or not? It might be a silly question but if $$a+\sqrt{a^2+1}= b+\sqrt{b^2+1},$$ then can I conclude that $a=b$? I thought about squaring both sides but I think it is wrong! Because radicals will not be removed by doing that! Can you help me with proving that $a=b$ or not?
Actually I'm going to prove that $x+\sqrt{x^2+1}$ is a $1$-$1$ function.
| Alternatively, for $a,b\in \Bbb R$,
\begin{align}a+\sqrt{a^2+1}&=b+\sqrt{b^2+1}\\&\implies \frac{1}{a+\sqrt{a^2+1}}=\frac{1}{b+\sqrt{b^2+1}}\wedge a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\\
&\implies \sqrt{a^2+1}-a=\sqrt{b^2+1}-b \wedge a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\\
&\implies \sqrt{a^2+1}-a=\sqrt{b^2+1}-b \wedge a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\\&\implies (a+\sqrt{a^2+1})-(\sqrt{a^2+1}-a)=(b+\sqrt{b^2+1})-(\sqrt{b^2+1}-b)
\\&\implies 2a=2b\implies a=b. \end{align}
It is also easy to see that $a=b\implies a+\sqrt{a^2+1}=b+\sqrt{b^2+1}$. That is,
$$a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\iff a=b.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\sum\limits_{n=0}^\infty a_nx^n$ if $a_0=3$, $a_1=5$, and $na_n=\frac23a_{n-1}-(n-1)a_{n-1}$ for every $n>1$
Assume that $a_0=3$, $a_1=5$, and, for arbitrary $n>1$ , $na_n=\frac{2}{3}a_{n-1}-(n-1)a_{n-1}$. Prove that, when $|x|<1$, the series $\sum\limits_{n=0}^\infty a_nx^n$ converges, and compute its sum.
I tried to let $\displaystyle a_n-a_1=\sum_{k=2}^{n}\frac{5-6k}{3k}a_{n-1}$ , and $\displaystyle a_n=(\frac{5-3n}{3n})a_{n-1}$
$$
a_{n-1}(\frac{5-3n}{3n}-\sum_{k=2}^{n}\frac{5-6k}{3k})=5
$$
I want to know how to continue it.
Edit: (after reading ideas by @JV.Stalker)
I made the following supplement
$$
S(x)=\sum_{n=0}^{\infty}a_nx^n\\
S'(x)=\sum_{n=1}^{\infty}na_nx^{n-1}\\
\sum_{n=2}^{\infty}na_nx^n=\sum_{n=2}^{\infty}\frac{2}{3}a_{n-1}x^n-\sum_{n=2}^{\infty}(n-1)a_{n-1}x^n\\
[xS'(x)-5x]=\frac{2}{3}x·\sum_{n=2}^{\infty}(n-1)a_{n-1}x^{n-1}-x\sum_{n=2}^{\infty}(n-1)a_{n-1}x^{n-1}\\
x[S'(x)-5]=\frac{2}{3}x(S(x)-3)-x(xS'(x))\\
S'(x)-5=S(x)-3-xS'(x)\\
(x+1)S'(x)=\frac{2}{3}S(x)+3\\
S'(x)-\frac{2}{3}\frac{1}{x+1}S(x)=\frac{3}{x+1}\\
S(x)=c(x+1)^{\frac{2}{3}}-\frac{9}{2}\\
S(0)=a_0=3\\
c=\frac{15}{2}\\
S(x)=\frac{15}{2}(x+1)^{\frac{2}{3}}-\frac{9}{2}
$$
| $na_n=\frac{5}{3}a_{n-1}-na_{n-1}$
Multiply by $x^n$ both sides and sum from $n=1$ to $\infty$
$\sum\limits_{n=1}^\infty na_n x^n=\frac{5}{3}\sum\limits_{n=1}^\infty a_{n-1}x^n-\sum\limits_{n=1}^\infty na_{n-1}x^n$
Reindex of the RHS:
$\sum\limits_{n=1}^\infty na_nx^n=\frac{5}{3}x\sum\limits_{n=0}^\infty a_{n}x^n-x\sum\limits_{n=0}^\infty (n+1)a_{n}x^n$
After sorting the eqution:
$\sum\limits_{n=1}^\infty na_nx^n+x\sum\limits_{n=0}^\infty na_{n}x^n=\frac{2}{3}x\sum\limits_{n=0}^\infty a_{n}x^n$
$(1+x)\sum\limits_{n=0}^\infty na_nx^n=\frac{2}{3}x\sum\limits_{n=0}^\infty a_{n}x^n$
Use that $nx^{n-1}=\frac{dx^n}{dx}$ and let $f(x)=\sum\limits_{n=0}^\infty a_nx^n$
We have the following differential eqution:
$(x+1)\frac{df(x)}{dx}=\frac{2}{3}f(x)$
Finally $f(x)=(x+1)^\frac{2}{3}+c$
$\sum\limits_{n=1}^\infty a_n x^n$ is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3033773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the limit of the sequence: $\lim_{n_\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$ Evaluate the limit of the sequence:
$$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}$$
My try:
Stolz-cesaro: The limit of the sequence is $\frac{\infty}{\infty}$
$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$
For our sequence:
$\lim_{n\to\infty}\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{n!}-\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})\cdot(1+\sqrt{n+1})-(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})}=\lim_{n\to\infty}\frac{\sqrt{(n-1)!}\cdot(\sqrt{n-1})}{\left((1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})\right)\cdot(\sqrt{n}+1)}$
Which got me nowhere.
| Consider:
$$
(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})
$$
Take the root from each pair of parentheses and multiply them, then:
$$
(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n}) > \sqrt{n!} \iff \\
\iff \frac{1}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})} < \frac{1}{\sqrt{n!}}
$$
Going back to original we have that:
$$
\frac{\sqrt{(n-1)!}}{(1+\sqrt{1})\cdot(1+\sqrt{2})\cdot (1+\sqrt{3})\cdots (1+\sqrt{n})} \le \frac{\sqrt{(n-1)!}}{\sqrt{n!}} = \frac{1}{\sqrt n}
$$
But the function is greater than $0$ and hence using squeeze theorem we conclude that:
$$
0 \le \lim_{n\to\infty}x_n \le \lim_{n\to\infty}\frac{1}{\sqrt n} = 0
$$
Hence the limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Average distance between point in a disc and line segment What is the average distance between a (randomly chosen) point in a disc of radius r and a line segment of length $a < 2r$ whose midpoint is at the center of the disc? ["Distance" here being the shortest distance to any point on the line segment.]
|
We devote our calculations to only the first quadrant because all the other quadrants are symmetrical in terms of calculations.
Dividing the quadrant into three regions, we will use the concept of expectation value of a variable denoted as follows $$\lt x\gt = \frac{\int xP(x)dx}{\int P(x)dx}$$
where $P(x)$ is the number of times a particular value $x$ of the quantity we seek occurs. (which we will contemplate as area of infinitesimal strips)
For region $\mathbf I$:
The distances of all the points on a vertical line are the same ($x$).
So the $N^r$ contributed to the final formula of expected value or $N^r_{\mathbf I}$ is calculated as $$\int_0^{\sqrt{R^2-\frac{a^2}{4}}}x\frac a2dx=\frac a4\Biggl(R^2-\frac{a^2}{4}\Biggr)$$
where $\frac a2dx$ is the area of the strip on which all the points are at a distance $x$ from the line of length $a\lt 2R$.
And obviously the denominator contributed $$D^r_{\mathbf I} = area(rectangle\ OABC) =\int_0^{\sqrt{R^2-\frac{a^2}{4}}}\frac a2dx=\frac a2\sqrt{R^2-\frac{a^2}{4}}$$
For region $\mathbf {II}$:
$$N^r_{\mathbf {II}}=\int_{\sqrt{R^2-\frac{a^2}{4}}}^R\ x\sqrt{R^2-x^2}dx$$
Subtituting $t=x^2$ and $dt=2xdx$, it becomes at last
$$-\frac13\biggl[(R^2-x^2)^{\frac32}\biggr]_{\sqrt{R^2-\frac{a^2}{4}}}^R=\frac{a^3}{24}$$
where $\sqrt{R^2-x^2}$ is the height of each strip.
And $D^r_{\mathbf {II}}$ = area of half the segment $\mathbf {II}$ $$= \int_{\sqrt{R^2-\frac{a^2}{4}}}^R\ \sqrt{R^2-x^2}dx$$
which can also simply be calculated by $\frac12R^2(\angle BOD)^c - area (\triangle OBC)$
$$=\frac12R^2\sin^{-1}(\frac a{2R})-\frac a4\sqrt{R^2-\frac{a^2}{4}}$$.
For region $\mathbf {III}$: ($\mathbf {My\ attempt}$)
Since the equal distances between line and collection of random points are radial in sense, we now switch to polar coordinates.
Also you'd reconcile that there is some relation between $\theta$ and the radial distance $r$ from point $A$ i.e. $max(r)$ depends on what $\theta$ line you are seeing.
So, applying $\mathit {Law \ of \ cosines}$ in $\triangle OAE$:
$$\cos(\frac{\pi}2+\theta)=\frac{r^2+\frac{a^2}4-R^2}{ra}$$
or
$$\sin(\theta)=\frac{R^2-\big(r^2+\frac{a^2}4\big)}{ra}$$
or
$$r=\frac{\sqrt{4R^2-a^2\cos^2{\theta}}-a\sin{\theta}}2$$
which is strictly decreasing in $\big[0,\frac{\pi}2\big)$.
Now, there are two integral sums that can lead us to $D^r_{\mathbf {III}}$:
$$D^r_{\mathbf {III}}=\int_0^{R-\frac a2}\int_0^{\frac{\pi}2}rd{\theta}dr+\int_{R-\frac a2}^{\sqrt{R^2-\frac{a^2}4}}\int_0^{\sin^{-1}\biggl(\frac{R^2-\big(r^2+\frac{a^2}4\big)}{ra}\biggr)}rd{\theta}dr=\frac{\pi}4\big(R-\frac a2\big)^2+\int_{R-\frac a2}^{\sqrt{R^2-\frac{a^2}4}}r\ {\sin^{-1}\Biggl(\frac{R^2-\big(r^2+\frac{a^2}4\big)}{ra}\Biggr)}dr$$
or
$$D^r_{\mathbf {III}}=\int_0^{\frac{\pi}2}\int_{R-\frac a2}^{\frac{\sqrt{4R^2-a^2\cos^2{\theta}}-a\sin{\theta}}2}rdrd\theta+\int_0^{\frac{\pi}2}\int_0^{R-\frac a2}rdrd{\theta}=\int_0^{\frac{\pi}2}\int_0^{\frac{\sqrt{4R^2-a^2\cos^2{\theta}}-a\sin{\theta}}2}rdrd\theta$$
(I prefer latter to check if we are on the right track and we are! That's what I checked through above)
which can also simply be calculated by the area of half the segment $\mathbf {III}$
$$=\frac12R^2(\angle BOA)^c - area (\triangle OBA)=\frac12R^2\cos^{-1}\Bigl(\frac a{2R}\Bigr)-\frac a4\biggl(\sqrt{R^2-\frac{a^2}4}\biggr)$$
$$\sum_{i=1,2,3} D^r_{\mathbf {i}}= area(quadrant)=\frac{\pi}4R^2$$
since $\sin^{-1}x+\cos^{-1}x=\frac{\pi}2$
Since the distance between points at radial distance of $r$ from point $A$ is the same, thus multiplying $r$ in the integrand would mean $distance\times (number\ of\ points\ having\ such\ distance)$ thing:
$$N^r_{\mathbf {III}}=\int_0^{\frac{\pi}2}\int_0^{\frac{\sqrt{4R^2-a^2\cos^2{\theta}}-a\sin{\theta}}2}r.rdrd\theta=\frac1{24}\int_0^{\frac{\pi}2}\Big({\sqrt{4R^2-a^2\cos^2{\theta}}-a\sin{\theta}}\Big)^3drd\theta$$
Average distance(as defined) of a random point from the given line is $$\frac{\underset{i=1,2,3}{\sum} N^r_{\mathbf {i}}}{\frac{\pi}4R^2}$$
| {
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Find $\lim_{n\to\infty} \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8})\ldots \cos(\frac{\pi}{2^n}) $ I already know that $$ a_n = \cos\left(\frac{\pi}{2^{n+1}}\right) = \overbrace{\frac{\sqrt{2+\sqrt{2+\ldots + \sqrt{2}}}}{2}}^{n\text{ roots}}$$
Also I know that $$\lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^n}\right) = 2
\text{ and if } a_n \xrightarrow {n\to\infty} a \text{ then } \sqrt[n]{a_1 a_2 \ldots a_n} \xrightarrow{n\to\infty} a $$
With that method I only got indeterminate form
$$ \lim_{n\to\infty} \cos\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{8}\right)\ldots \cos\left(\frac{\pi}{2^n}\right) = \Big(\frac{\sqrt[n]{a_1 a_2 \ldots a_n}}{2}\Big)^n = 1^\infty $$
Anyone knows a working solution?
| What you are trying to proof is Viete's formula. What he did was trying to compare area's of regular polygons that are inscribed in a unit circle. The area of a regular polygon with $n$ sides is given by
$$ A_n = \frac12 n \sin\left(\frac\pi n\right)$$
If you compute now the ratio between two regular polygons, one with $2^n$ sides, and one with $2^{n-1}$ sides, then you get:
$$B_n=\frac{A_{2^{n-1}}}{A_{2^n}} = \frac{2^{n-1} \sin\left(\frac{\pi}{2^{n-1}}\right)}{2^{n} \sin\left(\frac{\pi}{2^{n}}\right)} = \cos\left(\frac{\pi}{2^{n}}\right)$$
This now implies that the product the OP tries to compute is equal to
$$C_n=B_3 B_4 ... B_n = \frac{A_4}{A_8}\cdot\frac{A_8}{A_{16}}\cdot\cdots\cdot\frac{A_{2^{n-1}}}{A_{2^n}}=\frac{A_4}{A_{2^n}}$$
Sine a regular polygon with an infinite amount of sides is equivalent to a circle, you have $$\lim_{n\rightarrow\infty}A_n=\pi$$. In essence, the complete product is nothing more than comparing the size of a circle with respect to its inscribed square. Hence,
$$\prod_{n=2}^\infty\cos\left(\frac{\pi}{2^n}\right)=\lim_{n\rightarrow\infty}C_n=\frac 2 \pi$$
| {
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Extra factor of 2 when evaluating an infinite sum using fourier series and parseval's theorem. I'm asked to find the fourier series of the $2 \pi $ periodic function f(x) which is $sin(x)$ between $0$ and $\pi$ and $0$ between $\pi$ and $2\pi$
I use the complex form to proceed and get $$\frac{1}{2\pi}\int_{0}^{\pi}sin(x)e^{-ikx}dx$$. The complex coefficient $c_k$ I get as result is $\frac{-1}{\pi(k^2-1)}$ where $k=2n$ (k even) which is also correct according to WolframAlpha.
But then, I'm asked to use this result to evaluate $\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^2}$. For that, I switch to the real coefficients using $a_k=c_k+c_{-k}, b_k=i(c_k-c_{-k})$. I get: $a_k=\frac{-2}{\pi(k^2-1)}, a_0=\frac{2}{\pi}$ and $b_k$ is $0$.
So $f(x)=\frac{a_0}{2}+\sum_{k=1}^{\infty}a_kcos(kx)$
I then use Parseval's theorem to evaluate the sum we are looking for, remembering that k is even, i.e. $k=2n$ and that the function is $0$ between $\pi$ and $2\pi$:
$$\frac{1}{\pi}\int_{0}^{2\pi}|f(x)|^2 dx=\frac{a_0^2}{2}+\sum_{k=1}^{\infty}a_k^2$$ (1)
$$\frac{1}{\pi}\int_{0}^{\pi}sin^2(x)dx=\frac{\frac{2^2}{\pi^2}}{2}+\sum_{n=1}^{\infty}\frac{4}{\pi^2(4n^2-1)^2}$$ (2)
$$\frac{1}{2}-\frac{2}{\pi^2}=\sum_{n=1}^{\infty}\frac{4}{\pi^2(4n^2-1)^2}$$ (3)
So finally I get$\frac{\pi^2}{8}-\frac{1}{2}=\sum_{n=1}^{\infty}\frac{1}{(4n^2-1)^2}$ (4)
However, WolframAlpha gets $\frac{\pi^2}{16}-\frac{1}{2}$ so I must somehow have forgotten a factor of $\frac{1}{2}$ or put an extra factor of $2$ by $\frac{\pi^2}{8}$.
Logically, this missing/extra factor must have happened while I was evaluating $\frac{1}{\pi}\int_{0}^{\pi}sin^2(x)dx$ because the $-\frac{1}{2}$ at the end that came from the right side of the equality is correct according to WolframAlpha. But even when I evaluate $\frac{1}{\pi}\int_{0}^{\pi}sin^2(x)dx$ in WolframAlpha, I get the $\frac{1}{2}$ from step (3) which finally becomes $\frac{\pi^2}{8}$ and again, an factor of $\frac{1}{2}$ is missing, so I'm a bit perplex about what's wrong.
Thanks for your help !
Edit: the strange thing is that when I evaluate this sum using Parseval but with the fourier series of |sin(x)| between $0$ and $2\pi$, I get the correct result.
| I can't check your calculations since you haven't included them, but it is clear that the Fourier series you found is not the Fourier series of $f$. Your function is not even, so it cannot have a Fourier cosine series. For example,
$$ b_1 = \frac{1}{\pi} \int_0^{2\pi} f(x) \sin(x) \, dx = \frac{1}{\pi} \int_0^{\pi} \sin^2(x) \, dx = \frac{1}{2}. $$
My guess is that you haven't been careful in checking the special case when integrating the complex form (that is, $\int e^{ikx} \, dx = \frac{e^{ikx}}{ik} + C$ only when $k \neq 0$). In fact, the Fourier series of $f$ is given by
$$ \sum_{k = 1}^{\infty} \frac{2}{\pi(1-4k^2)} \cos(2kx) + \frac{1}{\pi} + \frac{1}{2} \sin(x) $$
and you'll get your missing factor from the extra $\sin$ term.
The complex coefficient $c_1$ is given by
$$ c_1 = \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-ix} \, dx = \frac{1}{2\pi} \int_0^{\pi} \frac{e^{ix} - e^{-ix}}{2i} e^{-ix} \, dx = \frac{1}{4\pi i} \int_0^{\pi} (1 - e^{-2ix}) \ dx = \frac{1}{4\pi i} \left[x - \frac{e^{-2ix}}{-2i} \right]_{x = 0}^{x = \pi} = -\frac{1}{4}i. $$
Similarly,
$$ c_{-1} = \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{ix} \, dx = \frac{1}{2\pi} \int_0^{\pi} \frac{e^{ix} - e^{-ix}}{2i} e^{ix} \, dx = \frac{1}{4\pi i} \int_0^{\pi} (e^{2ix} - 1) \ dx = \frac{1}{4\pi i} \left[\frac{e^{2ix}}{2i} - x \right]_{x = 0}^{x = \pi} = \frac{1}{4}i. $$
Hence,
$$ b_1 = i(c_1 - c_{-1}) = i(-\frac{1}{4}i - \frac{1}{4}i) = \frac{1}{2}. $$
| {
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$\lim_{n \to \infty}(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})=e^{\frac{1}{2}}$. Here is the beginning of a proof:
Suppose $0<k \leq n$,
$1+\frac{1}{n}<(1+\frac{k}{n^2})(1+\frac{n+1-k}{n^2})=1+\frac{n+1}{n^2}+\frac{k(n+1-k)}{n^4}\leq 1+\frac{1}{n}+\frac{1}{n^2}+\frac{(n+1)^2}{4n^4}$.
I'm confused by the second inequality above.
| Hint: By AM-GM Inequality
$$\frac{k+ n+1-k}{2} \geq \sqrt{k(n+1-k)}$$
| {
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Invertible matrix properties of a matrix I have here the following question:
Let $X$ be the $5 \times 5$ matrix "full of ones":
$X = \begin{pmatrix}1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 & 1\end{pmatrix}$
$(a)$ Is $X$ invertible? Explain.
$(b)$ Find a number $c$ such that $X^2=cX$.
$(c)$ Compute $(X-aI_5)(X+(a-c)I_5)$, where $a$ is a real number, $c$ is the constant from part $(b)$, $a\neq 0$, and $a\neq c$. If $M=(X-aI_5)$, what is $M^{-1}$? (You may express your answer in terms of $X,I_5,a$, and $c.$)
I already did part $(a)$ and $(b)$.
$(a)$ No. It's not invertible since there are two or more identical rows or columns so the determinant would be $0$ and hence it is not invertible.
$(b)$ $X^2 = \begin{pmatrix}5 & 5 & 5 & 5 & 5 \\ 5 & 5 & 5 & 5 & 5 \\ 5 & 5 & 5 & 5 & 5 \\ 5 & 5 & 5 & 5 & 5\\ 5 & 5 & 5 & 5 & 5 \end{pmatrix}$
Thus, $c=5$.
$(c)$ I almost got part $(c)$ too. It's the last part I can't figure out.
$=(X-aI_5)(X+(a-c)I_5)$
$=-a(a-c)I_5$
If $M=(X-aI_5)$, that would mean if I multiply both sides by $(X+(a-c)I_5)$, I get that:
$$M(X+(a-c)I_5)=(X-aI_5)(X+(a-c)I_5)$$
$$M(X+(a-c)I_5)=-a(a-c)I_5$$
I can literally see the answer in front of me, but it's not quite there. How do I proceed from here?
For reference, the answer should be:
$M^{-1}=\frac{X+(a-c)I_5}{-a(a-c)}$
| Hint. You have $M(X+(a-c)I_5)=-a(a-c)I_5$. Pre-multiply by $M^{-1}$.
| {
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Calculate the limit $\lim_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)}$. We could use L'Hospital here, because both numerator as well as denominator tend towards 0, I guess. The derivative of the numerator is $$x^2\cdot \left(-\sin\left(\frac{1}{x}\right)\right) \cdot \left( -\frac{1}{x^2}\right) + 2x \cos\left(\frac{1}{x}\right)=\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right) $$ The derivative of the denominator is $\cos(x)$. So, $$\lim\limits_{x\rightarrow 0}\frac{x^2 \cos\left(\frac{1}{x}\right)}{\sin(x)} = \lim\limits_{x\rightarrow 0}\displaystyle\frac{\sin\left(\frac{1}{x}\right) + 2x \cos\left(\frac{1}{x}\right)}{\cos(x)}$$
Is that right so far?
Thanks for the help in advance.
Best Regards,
Ahmed Hossam
| Hint: Without using L’Hôpital’s Rule, note that
$$\frac{x^2\cos\big(\frac{1}{x}\big)}{\sin x} = \frac{x}{\sin x}\cdot x\cos\bigg(\frac{1}{x}\bigg)$$
and recall $\lim_\limits{x \to 0}\frac{\sin x}{x} = 1$.
| {
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Verifying that $ \prod_{j=1}^{\infty} \frac{1}{1-q^j} = \prod_{j=1}^{\infty} \frac{1}{(1-q^{2j-1})(1-q^{2j})}$ On page 165 of Chapter 13, how was the equality made from line 1 to line 2?
https://archive.org/details/NumberTheory_862/page/n173
Namely, how $$ \prod_{j=1}^{\infty} \frac{1}{1-q^j} = \prod_{j=1}^{\infty} \frac{1}{(1-q^{2j-1})(1-q^{2j})}$$
| Simply pair-off factors and use the fact that multiplication is commutative:
\begin{align*}
\prod_{j=1}^{\infty}\frac{1}{(1-q^{j})} & = \frac{1}{1-q}\frac{1}{1-q^2}\frac{1}{1-q^3}\frac{1}{1-q^4}\cdots\\
& = \left(\frac{1}{1-q}\frac{1}{1-q^2}\right)\left(\frac{1}{1-q^3}\frac{1}{1-q^4}\right)\cdots\\
& = \left(\frac{1}{1-q^{2}}\frac{1}{1-q}\right)\left(\frac{1}{1-q^4}\frac{1}{1-q^{3}}\right)\cdots\\
& = \prod_{j=1}^{\infty}\frac{1}{(1-q^{2j})(1-q^{2j-1})}.
\end{align*}
| {
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Limit Question $\lim_{x\to\infty} \sqrt{x^2+1}-x+1$ I understand the answer is 1 which kind of makes sense intuitively but I can't seem to get there. I would appreciate if someone pointed out which line of my reasoning is wrong, thanks. I tried writing all my steps
\begin{equation}
\lim_{x\to\infty} \sqrt{x^2+1}-x+1
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x^2+1 - x +1}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x^2 - x +2}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x \left( x - 1 +\frac{2}{x}\right)}{x \left( \sqrt{1+\frac{1}{x}}+1-\frac{1}{x} \right)}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x - 1 +\frac{2}{x}}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\infty - 1 + 0}{1+1-0}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\infty - 1}{2} = \infty
\end{equation}
Edit: Added correct steps for completeness, thanks for the quick answers!
\begin{equation}
\lim_{x\to\infty} \sqrt{x^2+1}-x+1
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x^2+1 - x^2+2x -1}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{2x}{\sqrt{x^2+1}+x-1}
\end{equation}
\begin{equation}
\lim_{x\to\infty} \frac{x}{x} \frac{2}{\sqrt{1+\frac{1}{x}}+1-\frac{1}{x}}
\end{equation}
\begin{equation}
\frac{2}{1+1} = 1
\end{equation}
| From here we have
$$\frac{\left( \sqrt{x^2+1}-(x-1) \right) \left( \sqrt{x^2+1}+(x-1) \right)}{\sqrt{x^2+1}+(x-1)}=\frac{(\sqrt{x^2+1})^2-(x-1)^2}{\sqrt{x^2+1}+(x-1)}=$$$$=\frac{x^2+1-x^2+2x-1}{\sqrt{x^2+1}+(x-1)}=\frac{2x}{\sqrt{x^2+1}+(x-1)}$$
| {
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How many sequences can be made with 5 digits so that the difference between any two consecutive digits is $1$? Using the digits $0$, $1$, $2$, $3$, and $4$, how many ten-digit sequences can be written so that the difference between any two consecutive digits is $1$?
I was wondering if my solution is right.
Let $a(n)$ be the number of $n$ digit sequences that end with $0$ or $4$ so that the difference between any two consecutive digits is $1$.
$b(n)$ be the number of n digit sequences that end with $1$ or $3$ so that the difference between any two consecutive digits is $1$.
$c(n)$ be the number of n digit sequences that end with $2$ so
that the difference between any two consecutive digits is $1$.
$x(n)$ be the number of n digit sequences so that the difference between any two consecutive digits is $1$.
$x(n) = a(n) + b(n) + c(n)$
$a(n) = b(n-1)$
$b(n) = a(n-1) + 2c(n-1)$
$c(n) = b(n-1)$
By substituting $a(n-1)$ and $c(n-1)$ in the formula for $b(n)$ we get $b(n) = 3b(n-2)$
We know that $b(1) = 2, b(2) = 4$.
The characteristic equation for this recursion is $x^2-3 = 0$ with have the roots $3^{1/2}$ and $-3^{1/2}$, so $b(n) = A{(3^{1/2})}^{n} + B{(-3^{1/2})}^{n}$ where $A = {(3^{1/2}+2)}/{3}$ and $B = {(2-3^{1/2})}/{3}$. I think this is an integer.
We get $x(n) = 2b(n-1) + 3b(n-2)$ and by substituting we get something.
|
here is your answer. Your approach was absolutely correct.
| {
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Proving the following quadratic inequality? Apologies if this has been asked before - I could not find a question with this exact inequality.
Basically the inequality is
$$(a+b+c)^2 \leq 3 a^2 + 3 b^2 + 3 c^2$$
Expanding it out we see that
$$(a+b+c)^2 = a^2 +b^2 + c^2 + 2ab + 2bc + 2ac$$
so I guess it is equivalent to showing that
$$ab + bc + ac \leq a^2 + b^2 + c^2$$
Which makes sense to me. But how exactly do I prove it?
We can assume WLOG that each $a,b,c > 0$ since $ab \leq |a||b|$. From here, I guess we need to show that
$$ab \leq \frac{1}{2} \left(\max(a,b)^2 + \min(a,b)^2
\right)$$
And the result follows by adding up each term. But I'm not really sure why this must hold.
| It follows immediately from Cauchy-Schwarz:
$$(a+b+c)^2 = (1\cdot a + 1 \cdot b + 1 \cdot c)^2\leq (1^2+1^2+1^2) (a^2 + b^2 + c^2)$$
| {
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Proof that $ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = \frac{-1}{2} + \frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$ Task
Proof that $ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = -\frac{1}{2} + \frac{4}{3}\sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n}$
About
Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ \sum_{n=1}^{\infty} \frac{1}{n \cdot 3^n} $ there is formula for that.
So I want to evaluate in the same way left side. So I compute that
$$ \sum_{n=2}^{\infty} \frac{2}{3^n \cdot (n^3-n)} = \sum_{n=2}^{\infty} \frac{1}{3^n \cdot n (n-1)} - \frac{1}{3^n \cdot n (n+1)} $$
I am trying to transform it to use that formula:
$$ \sum_{n=1}^{\infty} \frac{1}{n\cdot p^n} = ln\frac{p}{p-1} $$
but I still defeat.
So please tell me, there are better ways to proof that or should I consider to change my field of study?
| hint
$$\frac{2}{n^3-n}=\frac{-2}{n}+\frac{1}{n-1}+\frac{1}{n+1}$$
| {
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An exercise on the calculation of a function of operator The operator is given by
$$A=\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 4
\end{pmatrix}$$
I have to write down the operator $$B=\tan(\frac{\pi} {4}A)$$
I calculate $$\mathcal{R} (z) =\frac{1}{z\mathbb{1}-A}=\begin{pmatrix}
\frac{1}{z-1} & 0 & 0\\
\frac{1}{(z-1)^2} & \frac{1}{z-1} & 0\\
0 & 0 & \frac{1}{z-4}\end{pmatrix} $$
Now the B operator is given by:
$$B=\begin{pmatrix}
Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{z-1} & 0 & 0\\
Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{(z-1)^2} & Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{z-1} & 0\\
0 & 0 & Res_{z=4}\frac{\tan(\frac{\pi}{4}z)}{z-4}
\end{pmatrix} $$
For me the result should be
$$ B=\begin{pmatrix}
1 & 0 & 0\\
\frac{\pi}{2} & 1 & 0\\
0 & 0 & 0\end{pmatrix}$$
But the exercise gives as solution:
$$ B=\begin{pmatrix}
1 & 0 & 0\\
\frac{\pi}{4} & 1 & 0\\
0 & 0 & 1\end{pmatrix}$$
Where is the error?
Thank you and sorry for bad English
| It would appear that there’s an error in either the problem statement or the solution.
You can check your own answer by computing this via a consequence of the Cayley-Hamilton theorem: any analytic function of $A$ can be expressed as a quadratic polynomial in $A$, i.e., $\tan\left(\frac\pi4A\right) = aI+bA+cA^2$ for some unknown coefficients $a$, $b$, $c$. The eigenvalues of $A$ are obviously $1$, $1$ and $4$, so you can find these coefficients by solving the following system of linear equations $$a+b+c = \tan\frac\pi4 \\ b+2c = \frac\pi4\sec^2\frac\pi4 \\ a+4b+16 = \tan\pi$$ obtained by substituting the eigenvalues of $A$ into the above equation for $\tan\left(\frac\pi4A\right)$, as well as into its derivative in order to get a third independent equation in the case of the repeated eigenvalue. Solving these equations and computing the polynomial of $A$ produces $\pi/4$, not $\pi/2$ on the off-diagonal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Value of k to give matrix infinite, 0, 1 solutions. I have a question that goes:
For which values of the constant $k$ does the system of equations below have:
*
*a unique solution,
*no solutions at all,
*infinitely many solutions?
$$
\begin{cases}
x &- 3y & &= 6\\
x & &+ 3z &= -3\\
2x &+ ky &+ (3-k)z &= 1
\end{cases}
$$
I tried putting the system of equations into matrix in reduced row echelon form, ended up with the last line being
$$
\begin{matrix}
0 & k+6 & 3-k &| 1,
\end{matrix}
$$ which I don't think would make sense.
| You have
$$
\begin{pmatrix}
1 & -3 & 0 & 6 \\
1 & 0 & 3 & -3 \\
2 & k & 3-k & 1
\end{pmatrix}
\to
\begin{pmatrix}
1 & -3 & 0 & 6 \\
0 & 3 & 3 & -9 \\
0 & k+6 & 3-k & -11
\end{pmatrix}
\to\\
\begin{pmatrix}
1 & 0 & 3 & -3 \\
0 & 1 & 1 & -3 \\
0 & 0 & 3-k-(k+6) & -11 + 3(k+6)
\end{pmatrix}
\to \\
\begin{pmatrix}
1 & 0 & 3 & -3 \\
0 & 1 & 1 & -3 \\
0 & 0 & -3-2k& 7 + 3k
\end{pmatrix}
$$
What happens in $-3-2k=0$? Can you finish this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Understanding the solution to this polynomial estimation problem Show that there exist $K, N > 0$ such that for $x ∈\mathbb{R}$,
$$x ≥ N \implies
\frac{3x^
2 − 4x + 8}{
5x + 6}
≥ Kx.$$
Solution:
For $x ≥ 4,$ we have $4x ≤ x^
2$
so that
$$3x^
2 − 4x + 8 ≥ 3x^
2 − x^
2 = 2x^
2.$$
Similarly, for $x ≥ 1,$ we have
$$5x + 6 ≤ 11x.$$
Therefore, when $x ≥ 4,$ we have
$$\frac{3x^
2 − 4x + 8}{
5x + 6}
≥\frac{
2x^
2}{
11x}
=
\frac{2}
{11} x,$$
and so we can take $N = 4$ and
$K = \frac{2}{
11} .$
I'm struggling to follow the logical steps in this solution. I'm not sure why the inequality of $x ≥ 4$ and $x ≥ 1$ has been chosen. Can someone please break down this solution and show where the steps come from?
| Essentially the method of attack that was used in this proof is to find a quadratic function that is less than the numerator and a linear function that is greater than the denominator. Then when you divide these you get a linear function which is less than the original $\frac{3x^2-4x+8}{5x+6}$.
In layman's terms the question is asking if it is possible to find some real number $N$ such that $\frac{3x^2-4x+8}{5x+6}$ is greater than some linear function $Kx$ for all $x$ greater than or equal to $N$.
So again we are looking for a quadratic equation which is less than $3x^2-4x+8$ for some values of $x > 0$. Notice that $3x^2 + 8 \geq 3x^2 $ for all $x > 0$. Now we need to account for the $-4x$ term still in the numerator. There's actually an infinite amount of choices we could do for the next step. In the proof you presented the author realized that $x^2$ would be greater than or equal to $4x$ eventually (namely at $x = 4$). So from the inequality above we note that $3x^2 - 4x + 8 \geq 3x^2 - x^2 = 2x^2$ when $x \geq 4$ (because we subtracted something greater from the RHS than the LHS).
Now for the next step again we have an infinite amount of choices, however the author realized that $5x +6 \leq 11x$ when $x \geq 1$. Of course they could just have easily said $5x+6 \leq 12x$ or an infinite number of other choices for that matter (thus finding a different value of $K$) but they chose this. Now since $5x + 6 \leq 11x$ when $x \geq 1, $ it is certainly also true when $x \geq4.$
Now for the final step, we can obtain the equation $\frac{3x^2 - 4x + 8}{5x+6} \geq \frac{2x^2}{11x} = \frac{2}{11}x$ for $x \geq 4$ because the numerator of the LHS is greater than the numerator of the RHS and the denominator of the LHS is less than the numerator of the RHS for $x\geq4$ as we showed above.
Thus we can pick $N = 4$ and $K = \frac{2}{11}$. To make sure you understand it try and find another value of $N$ and $K$, for instance by realizing that $3x^2 - 4x + 8 \geq 3x^2 -2x^2 = x^2$ for $x \geq 4$ in the first step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\int_{x^2+y^2+z^2 \leq 1}\frac{dx\,dy\,dz}{x^2+y^2+(z-2)^2}$ I'm trying to calculate the integral $$\int_{x^2+y^2+z^2 \leq 1}\frac{dx\,dy\,dz}{x^2+y^2+(z-2)^2}.$$
I've tried in two methods:
Regular spherical coordinates, but this leads to really unfun integrals and logarithms with negative numbers inside them and other beasts I'd rather avoid.
The other method was to try shifted spherical coordinates, $x = r\sin(\theta)\cos(\phi), y=r\sin(\theta)\sin(\phi)$ but $z-2 = r\cos(\theta)$.
Now the integral is very easy but finding the limits of integration is tougher. We still have $0 < \theta < \pi$ and $0 < \phi < 2\pi$, I think, but the limits on $r$ are harder.
At the very limit of the domain of integration we have $x^2+y^2+z^2 = 1$, so $r^2+4r\cos(\theta)+4 = 1$, so we need to have $r^2+4r\cos(\theta) + 3 = 0$.
This happens when $$r = \frac{-4\cos(\theta)\pm\sqrt{16\cos^2(\theta)-12}}{2}.$$ I don't know how this helps us or if I'm going in the right direction.
The unfun way:
$\int_{x^2+y^2+z^2 \leq 1}\frac{dxdydz}{x^2+y^2+(z-2)^2} = \int_{0}^{1}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{r^2\sin(\theta)}{r^2-4r\cos(\theta)+4}d\phi d\theta dr = 2\pi \int_{0}^{1}\int_{0}^{\pi}\frac{r^2\sin(\theta)}{r^2-4r\cos(\theta)+4}d\theta dr $
Use $u-$substitution $u=\cos(\theta)$ to get:
$2\pi \int_{0}^{1}\int_{0}^{\pi}\frac{r^2\sin(\theta)}{r^2-4r\cos(\theta)+4}d\theta dr =-2\pi \int_{0}^{1}\int_{1}^{-1}\frac{r^2}{r^2-4ru+4}dudr = 2\pi\int_{0}^{1}\int_{-1}^{1}\frac{r^2}{r^2-4ru+r}dudr = 2\pi\int_{0}^{1}r^2[\frac{\ln(r^2-4ru+4)}{-4r}]_{-1}^{1}dr = -\frac{\pi}{2}\int_{0}^{1}r(\ln(r^2-4r+4) - \ln(r^2+4r+4))dr=-\pi\int_{0}^{1}r(\ln(|r-2|)-\ln(r+2))dr$
Is this the right way? seems very unpleasant...
| Consider the integral $$I(a, b) = \int_{x^2 + y^2 \le a} \frac{dx \, dy\, }{x^2 + y^2 + b}$$ By changing to polar coordinates, we compute $$I(a,b) = \int_{0}^{2\pi} \int_{0}^{\sqrt{a}} \frac{r \, dr\, d\theta}{r^2 + b} = \pi \ln(a/b + 1)$$ Your desired integral is $$\int_{-1}^{1} I(1-z^2, (z-2)^2) \, dz = \pi \int_{-1}^{1} \ln\left(\frac{1-z^2 + (z-2)^2}{(z-2)^{2}} \right)\, dz = \pi \int_{-1}^{1} \ln(5-4z) - 2\ln(|z-2|) \, dz$$ which one can compute by elementary techniques (e.g. integration by parts).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Proving the Fibonacci identity $(−1)^{m−k}(F_{m+k+1}F_{m−k−1}−F_{m+k}F_{m−k}) =F_{k}^2+F_{k+1}^2$ Prove that for two natural numbers $m$ and $k$, where $m>k$ the following identity holds:
$$(−1)^{m−k}(F_{m+k+1}F_{m−k−1}−F_{m+k}F_{m−k}) =F_{k}^2+F_{k+1}^2$$
Here the exercise comes with a hint:
The constant is $F^2 _m$, try to substitute $k=0$ in.
Okay fair enough, let us try this, this will give us:
$$ (−1)^{m}(F_{m+1}F_{m−1}−F_{m}F_{m}) =F_{0}^2+F_{1}^2$$
$$ (−1)^{m}(F_{m+1}F_{m−1}−F_{m}^2) =F_{0}^2+F_{1}^2=0^2+1^2=1$$
We can now use a identity by Cassini, namely that for $m>0=k$ we have:
$$ (−1)^{m}(F_{m+1}F_{m−1}−F_{m}^2) =(-1)^m(-1)^m=(-1)^{2m}=1 \checkmark$$
Okay great, so our identity works whenever $k=0$, but I do not see how this helps with a generalisation.
Edit/idea: Did I just write down the base case for an approach via induction on $k$ perhaps?
| Here is a proof based upon Binets formula
\begin{align*}
F_k=\frac{\varphi^k-\psi^k}{\varphi-\psi}\qquad k\geq 0\tag{1}
\end{align*}
where $\varphi=\frac{1}{2}\left(1+\sqrt{5}\right), \psi=\frac{1}{2}\left(1-\sqrt{5}\right)=-1/\varphi$.
We start with the right-hand side of OPs formula and obtain
\begin{align*}
F_k^2=\left(\frac{\varphi^k-\psi^k}{\varphi-\psi}\right)^2&=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}-2\left(\varphi\psi\right)^k+\psi^{2k}\right)\\
&=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}-2(-1)^k+\psi^{2k}\right)\tag{2}\\
\end{align*}
Putting $k\to k+1$ in (2) we get
\begin{align*}
\color{blue}{F_k^2+F_{k+1}^2}&=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}-2(-1)^k+\psi^{2k}\right)\\
&\qquad+\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k+2}-2(-1)^k+\psi^{2k+2}\right)\\
&\,\,\color{blue}{=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}\left(1+\varphi^2\right)+\psi^{2k}\left(1+\psi^2\right)\right)}\tag{3}
\end{align*}
And now the left-hand side.
We obtain
\begin{align*}
F_{m+k}F_{m-k}&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{m+k}-\psi^{m+k}\right)\left(\varphi^{m-k}-\psi^{m-k}\right)\\
&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}-\varphi^{m+k}\psi^{m+k}-\varphi^{m-k}\psi^{m+k}+\psi^{2m}\right)\\
&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}-\left(\varphi\psi\right)^{m-k}\left(\varphi^{2k}+\psi^{2k}\right)+\psi^{2m}\right)\\
&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}-(-1)^{m-k}\left(\varphi^{2k}+\psi^{2k}\right)\right)\tag{4}
\end{align*}
Replacing $k$ with $k+1$ in (4) we get
\begin{align*}
F_{m+k+1}F_{m-k-1}&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}+(-1)^{m-k}\left(\varphi^{2k+2}+\psi^{2k+2}\right)\right)\tag{5}
\end{align*}
The left-hand side of OPs formula can now be rewritten using (4) and (5) as
\begin{align*}
\color{blue}{(-1)^{m-k}}&\color{blue}{\left(F_{m+k+1}F_{m-k-1}-F_{m+k}F_{m-k}\right)}\\
&=\frac{(-1)^{m-k}}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}+(-1)^{m-k}\left(\varphi^{2k+2}+\psi^{2k+2}\right)\right)\\
&\qquad-\frac{(-1)^{m-k}}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}-(-1)^{m-k}\left(\varphi^{2k}+\psi^{2k}\right)\right)\\
&\,\,\color{blue}{=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}\left(1+\varphi^2\right)+\psi^{2k}\left(1+\psi^2\right)\right)}\tag{6}
\end{align*}
A comparison of (3) and (6) shows OPs identity is valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
In $\Delta ABC$ if $(\sqrt{3}-1)a=2b$, $A=3B$, then find $C$
In $\Delta ABC$ if $(\sqrt{3}-1)a=2b$, $A=3B$, then find $C$
My Attempt
$$
b=\frac{\sqrt{3}-1}{2}a\quad\& \quad \frac{A-B}{2}=B\quad\&\quad\frac{A+B}{2}=2B\\\frac{a-b}{a+b}=\frac{\tan\frac{A-B}{2}}{\tan\frac{A+B}{2}}\implies \frac{3-\sqrt{3}}{\sqrt{3}+1}=\frac{\tan B}{\tan 2B}=\frac{1-\tan^2B}{2}
$$
| $$2\sin B=(\sqrt3-1)\sin3B=(...)(\sin B)(3-4\sin^2B)$$
As $\sin B>0,$ $$\sin^2B=\dfrac{2-\sqrt3}4$$
$$\cos2B=1-\dfrac{2-\sqrt3}2=\cos30^\circ$$
$0<2B<360^\circ,2B=360^\circ n\pm30^\circ$ for some integer $n$
$\implies2B=30^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
&=H_{1}\left(1-\frac{1}{2}\right)+H_{2}\left(\frac{1}{2}-\frac{1}{3}\right)+H_{3}\left(\frac{1}{3}-\frac{1}{4}\right)\\
&=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{3}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{9}-\frac{1}{12}
\end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
Then,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
| Using summation by parts we have $$\sum_{n\leq N}\frac{H_{n}}{n\left(n+1\right)}=H_{N}\sum_{n\leq N}\frac{1}{n\left(n+1\right)}-\sum_{n\leq N-1}\frac{1}{n+1}\left(\sum_{k\leq n}\frac{1}{k\left(k+1\right)}\right).$$ Clearly $$\sum_{n\leq N}\frac{1}{n\left(n+1\right)}=\left(1-\frac{1}{N+1}\right)$$ then $$\sum_{n\leq N}\frac{H_{n}}{n\left(n+1\right)}=H_{N}\left(1-\frac{1}{N+1}\right)-H_{N}+\sum_{n\leq N-1}\frac{1}{\left(n+1\right)^{2}}.$$ Now, since $H_{N}\sim\log\left(N\right)$ as $N\rightarrow+\infty$, the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
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solving $\lim\limits_{x\rightarrow\infty} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}$ To investigate the convergence of a series I have to solve the folliwing limit:
\begin{equation}
\lim\limits_{x\rightarrow\infty} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}
\end{equation}
It should be $\frac{1}{2}$ but i can't quite seem to get to that solution. I've tried to factor the square root out of the quotient which was:
\begin{equation}
\lim\limits_{x\rightarrow\infty} \sqrt{\frac{((x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1})^2}{x^2(x + 1)}}
\end{equation}
Then i worked out the square in the numerator which was:
\begin{equation}
(x^2-1)^2 (x + 2)-2x^2(x^2-1)\sqrt{x+1}\sqrt{x+2}+x^4(x+1)
\end{equation}
I could then factor the terms and take and divide the numerator with the (x+1) from the denominator. Then i could expand the terms in the numerator wich became:
\begin{equation}
\lim\limits_{x\rightarrow\infty}\sqrt{\frac{(x^4+x^3-3x^2-x+2)-\sqrt{16x^8+32x^7-12x^6-20x^5+8x^4}+x^4}{x^2}}
\end{equation}
Now i can take the $16x^8$ out of the root and then i then looked at the terms with the highest exponent so i had:
\begin{equation}
\lim\limits_{x\rightarrow\infty}\sqrt{\frac{x^4-4x^4+x^4}{x^2}} =\lim\limits_{x\rightarrow\infty}\sqrt{\frac{-2x^2}{x^2}}
\end{equation}
Which could only be solved with complex numbers, so i should be wrong somewhere in my calculations, since i know that i should get $\frac{1}{2}$. I also checked my sollution with WolframAlpha which also gave $\frac{1}{2}$ so I know that the sollution i have is correct.
Would anyone know where i was wrong or how i could better solve it?
| Another way:
\begin{eqnarray*} \frac{(x^2-1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}}
& = & \frac{(x-1)(x+1) \sqrt{x + 2}-x^2\sqrt{x+1}}{x\sqrt{x + 1}} \\
& = & \frac{(x-1)\sqrt{(x+1)(x + 2)}-x^2}{x} \\
& \stackrel{x^2 = x(x-1)+x}{=} & \underbrace{\frac{x-1}{x}}_{\stackrel{x \to +\infty}{\longrightarrow}1}\underbrace{\left(\sqrt{(x+1)(x + 2)} - x\right)}_{= \frac{3x+2}{\sqrt{(x+1)(x + 2)} + x}\stackrel{x \to +\infty}{\longrightarrow}\frac{3}{2}} - 1 \\
& \stackrel{x \to +\infty}{\longrightarrow} & 1 \cdot \frac{3}{2} -1 = \frac{1}{2}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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