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Uncertain how the following step was accomplished. I'm working through a book example that aims to find the first two nonzero terms of the Laurent expansion of $f(z)=\tan(z)$, about $z=\frac{\pi}{2}$.
The substitution $z=\frac{\pi}{2}+u$ is made
$$f(z)=\frac{\sin(\frac{\pi}{2}+u)}{\cos(\frac{\pi}{2}+u)}=-\frac{\cos(u)}{\sin(u)}$$
The respective Taylor series are then utilized
$$f(z)=-\frac{(1-\frac{u^2}{2!}+ \cdots)}{(u-\frac{u^3}{3!}+ \cdots)}=-\frac{1}{u}\frac{(1-\frac{u^2}{2!}+ \cdots)}{(1-\frac{u^2}{3!}+ \cdots)}$$
Now here is where I get lost. The book then states that it is using
$$\frac{1}{1-z}=\sum_{n=0}^{\infty} z^{n} \quad for \quad|z|<1$$
To expand the denominator into the following
$$f(z)=-\frac{1}{u}(1-\frac{u^2}{2!}+ \cdots)(1+\frac{u^2}{3!}+ \cdots)$$
I cannot seem to pick up on the math going on above. Would anyone care to enlighten me?
| Here we do a geometric series expansion
\begin{align*}
\frac{1}{1-u}&=\sum_{n=0}^{\infty}u^n\\
&=1+u+u^2+u^3+\cdots \qquad\qquad\qquad\qquad\qquad |u|<1
\end{align*}
If we substitute $h(u)$ for $u$ we obtain
\begin{align*}
\frac{1}{1-h(u)}&=\sum_{n=0}^{\infty}\left(h(u)\right)^n\\
&=1+h(u)+\left(h(u)\right)^2+\left(h(u)\right)^3+\cdots\qquad\qquad |h(u)|<1
\end{align*}
In the series expansion of $f(u)=-\frac{\cos(u)}{\sin(u)}$ we are in a similar situation
\begin{align*}
f(u)&=-\frac{\left(1-\frac{u^2}{2!}+\frac{u^4}{4!}- \cdots\right)}{\left(u-\frac{u^3}{3!}+ \frac{u^5}{5!}-\cdots\right)}\\
&=-\frac{1}{u}\frac{\left(1-\frac{u^2}{2!}+ \frac{u^4}{4!}-\cdots\right)}{\left(1-\frac{u^2}{3!}+ \frac{u^4}{5!}-\cdots\right)}\\
&=-\frac{1}{u}\left(1-\frac{u^2}{2!}+\frac{u^4}{4!}-\cdots\right)\frac{1}{(1-h(u))}
\qquad\text{with}\qquad h(u)=\frac{u^2}{3!}- \frac{u^4}{5!}+\cdots\\
&=-\frac{1}{u}\left(1-\frac{u^2}{2!}+\frac{u^4}{4!}-\cdots\right)\left(1+h(u)+(h(u))^2+\cdots\right)\\
&=-\frac{1}{u}\left(1-\frac{u^2}{2!}+\frac{u^4}{4!}-\cdots\right)\\
&\qquad\cdot\left(1+\underbrace{\left(\frac{u^2}{3!}+ \frac{u^4}{5!}-\cdots\right)}_{h(u)}
+ \underbrace{\left(\frac{u^2}{3!}+ \frac{u^4}{5!}-\cdots\right)^2}_{\left(h(u)\right)^2}+\cdots\right)\\
&=-\frac{1}{u}\left(1-\frac{u^2}{2!}+\cdots\right)\left(1+\frac{u^2}{3!}+ \cdots\right)\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these
The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+\sin^2B+\sin^2C=\sin A\sin C+\sin A\sin B\sqrt3$
I am stuck here.
| If $\triangle ABC$ is equilateral, then $3x^2=x^2(\sqrt{3}+1)$, absurd.
Now, note that the condition is symmetric respect to $b$ and $c$, ie, if $(a,b,c)$ satisfies, $(a,c,b)$ too, and then $\triangle ABC$ is isosceles with $b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Proof $\sum \frac{1}{n^a}$ is convergent for a > 1 I get to the fact that $\sum_{k=1}^n \frac{1}{k^a}$ < $\frac{1}{a-1} - \frac{1}{(a-1)(n+1)^{a-1}} - \frac{1}{(n+1)^a} + 1$ and hence $\sum_{k=1}^n \frac{1}{k^a}$ is bounded. How to deduce $\sum \frac{1}{n^a}$ is convergent ?
Note that a > 1
| Recall that a series of positive terms converges if the sequence of partial sums is bounded. Consider the $n$th partial sum of the series,
$$s_n = \sum_{k = 1}^n \frac{1}{n^a}$$
For each $m \ge 0$, $\frac{1}{k^a} \le \frac{1}{2^{ma}}$ for $k = 2^m, 2^m+1,\ldots, 2^{m+1}-1$. Choose a natural number $M$ such that $2^{M+1} > n$. Since each integer $k \ge 1$ lies between $2^m$ and $2^{m+1}-1$ for a unique $m\ge 0$,
$$s_n \le \sum_{m = 0}^M \sum_{2^m \le k < 2^{m+1}} \frac{1}{k^a} \le \sum_{m = 0}^M \sum_{2^m \le k < 2^{m+1}} \frac{1}{2^{ma}} = \sum_{m = 0}^M \frac{1}{2^{(a-1)m}}\le \sum_{m = 0}^\infty \left(\frac{1}{2^{a-1}}\right)^m$$
The last series is geometric with common ratio $\frac{1}{2^{a-1}}$; since $a > 1$, $\frac{1}{2^{a-1}} < 1$ so the geometric series converges (to $\frac{2^{a-1}}{2^{a-1}-1}$). Since $n$ was arbitrary, $(s_n)$ is bounded. Thus, the series $\sum\limits_{n = 1}^\infty \frac{1}{n^a}$ converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the limit a matrix raised to $n$ when $n$ goes to infinity Let $ A $ be a $ 3\times3 $ matrix such that
$$A
\left( \begin{array}{ccc}
1 \\
2 \\
1 \end{array} \right)=\left( \begin{array}{ccc}
1 \\
2 \\
1 \end{array} \right),~~~A
\left( \begin{array}{ccc}
2 \\
2 \\
0 \end{array} \right)=\left( \begin{array}{ccc}
1 \\
1 \\
0 \end{array} \right),~~A
\left( \begin{array}{ccc}
3 \\
0 \\
6 \end{array} \right)=\left( \begin{array}{ccc}
-1 \\
0 \\
2 \end{array} \right)
$$
Find $$ \lim_{n\to\infty}A^n \left( \begin{array}{ccc}
6 \\
7 \\
0 \end{array} \right)$$
So, do I first find $ A $ by letting A =$ \left( \begin{array}{ccc}
a&b&c \\
d&e&f \\
g&h&i \end{array} \right) $ and using the given information to solve the corresponding linear equations and then solve the actual problem of finding the limit? Is there a more efficient way of doing this? Also, I am not quite sure how to find the limit so any hints would be greatly appreciated. Thanks!
| Hint:
Notice that with the given vectors, assuming no $-$ sign,
$$Au=u,Av=2v,Aw=\frac13w,$$ so that
$$A^nu=u,A^nv=2^nv,A^nw=\frac1{3^n}w.$$
Then if you decompose the fourth vectors as a linear combination of $u,v,w$, you should easily see how it is transformed by $A^n$.
If we keep the $-$ sign,
The third equation will read
$$Aw=au+bv+cw$$ for some coefficients $a,b,c$, and
$$A^2w=au+2bv+c(au+bv+cw)=a(1+c)u+b(2+c)v+c^2w,\\
A^3w=a(1+c)u+b2(2+c)v+c(a(1+c)u+b(2+c)v+c^2w)=\\
a(1+c)^2+b(2+c)^2v+c^3w.$$
You see the pattern.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
'Destruction' of $x$ in second derivative? Can you tell whether I've taken this second derivative and determined its inflection points correctly?
$$f'(x) = (1-x)e^{ x-\frac{1}{2}x^2}$$
Now for the second derivative:
$$f''(x)=(1-x)[e^{ x-\frac{1}{2}x^2}]' + (e^{ x-\frac{1}{2}x^2} [1-x]')$$
$$(1-x)(e^{x-\frac{1}{2}x^2} - e^{x-\frac{1}{2}x^2}x) + (-e^{x-\frac{1}{2}x^2})$$
Let $e^{ x-\frac{1}{2}x^2} = \psi$
$$(1-x)(\psi-\psi x) - \psi$$
$$\psi - 2\psi x + \psi x^2 - \psi$$
$$-2\psi x + \psi x^2 $$
Now determine inflection points:
$$-2\psi x + \psi x^2 =0$$
$$\psi x(-2+x)=0$$
$$\psi x = 0 \text{ or } -2 +x = 0$$
$$e^{ x-\frac{1}{2}x^2}x = 0\text{ or } x = 2 $$
$$x = 0 \text{ or } x = 2$$
When I determine the inflection point I 'destroy' $e^{ x-\frac{1}{2}x^2}x $ by dividing on both sides by it. Is this a valid operation? Or should I have factored out only $e^{ x-\frac{1}{2}x^2}$ (without $x$)? $e^{ x-\frac{1}{2}x^2}= 0 $ doesn't have a solution and $-2x+x^2 = 0$ gives the desired inflection points (which are the same as I get using the method above ($x=0 \wedge x = 2$.)
Thank you,
Cro-Magnon
| So, you arrived at the step $e^{x-\frac{1}{2}x^{2}}x=0$ and you want to solve for $x$. Dividing both sides of this equation by $e^{x-\frac{1}{2}x^{2}}$ to conclude that $x=0$ is indeed a legal operation. This is because for all real numbers $r$, $e^{r}>0$. Therefore, for any $x$, $e^{x-\frac{1}{2}x^{2}}>0$. Since you are not dividing by $0$, you are okay.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Induction proof of the identity $\cos x+\cos(2x)+\cdots+\cos (nx) = \frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}$
Prove that:$$\cos x+\cos(2x)+\cdots+\cos (nx)=\frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}.\ (1)$$
My attempt:$$\sin\left(\frac{x}{2}\right)\sum_{k=1}^{n}\cos{(kx)}$$$$=\sum_{k=1}^{n}\sin\left(\frac{x}{2}\right)\cos{(kx)} $$ (Applying $\sin x\cos y =\frac{1}{2}\left[\sin(x+y)+\sin(x-y)\right]$)
$$=\sum_{k=1}^{n}\frac{1}{2}\sin\left(\frac{x}{2}-kx\right)+\frac{1}{2}\sin\left(\frac{x}{2}+kx\right).$$Multiplying $(1)$ by $\sin\left(\frac{x}{2}\right)$ ,gives: $$\sum_{k=1}^{n}\sin\left(\frac{x}{2}\right)\cos(kx)=\sin\left(\frac{nx}{2}\right)\cos{\frac{(n+1)x}{2}}$$ $$\Leftrightarrow\sum_{k=1}^{n}\left[\sin\left(\frac{x}{2}-kx\right)+\sin\left(\frac{x}{2}+kx\right)\right]=\sin\left(-\frac{x}{2}\right)+\sin\left(\frac{x}{2}+nx\right).$$Then, by induction on $n$ and using $(\sin x\sin y)$ and $(\sin x +\sin y)$ formulas, I end in here: $$\sum_{k=1}^{n+1}\sin\left(\frac{x}{2}\right)\cos(kx)=\left[2\sin\left(\frac{nx}{2}\right)\cos\frac{(n+1)x}{2}\right]+\left[2\sin\left(\frac{x}{2}\right)\cos(n+1)x\right].$$ Any help would be appreciated, thanks!
| Here is the induction step: it comes down to proving
\begin{gather*}\frac{\sin \dfrac{nx}{2} \cos\dfrac{(n+1)x}{2}}{\sin\dfrac{x}{2}}+\cos(n+1)x=\frac{\sin\dfrac{(n+1)x}{2}\cos\dfrac{(n+2)x}{2}}{\sin(\dfrac{x}{2})}\\
\text{or}\qquad\sin\dfrac{nx}{2}\cos\dfrac{(n+1)x}{2}+\sin\dfrac{x}{2}\cos(n+1)x=\sin\dfrac{(n+1)x}{2} \cos\dfrac{(n+2)x}{2}
\end{gather*}
Now use the linearisation formulae:
\begin{cases}\displaystyle
\sin\frac{nx}{2}\cos\frac{(n+1)x}{2}=\tfrac12\biggl(\sin\frac{(2n+1)x}{2}-\sin\frac x2\biggr),\\[1ex]
\sin\dfrac{x}{2}\cos(n+1)x=\frac12\biggl(\sin\dfrac{(2n+3)x}{2}-\sin\dfrac{(2n+1)x}{2}\biggr),
\end{cases}
whence the l.h.s. is
$$\frac12\biggl(\sin\dfrac{(2n+3)x}{2}-\sin\frac x2\biggr)=\sin\frac{(n+1)x}2\,\cos\frac{(n+2)x}2$$
by the factorisation formula: $\;\sin p-\sin q=2\sin\dfrac{p-q}2\cos\dfrac{p+q}2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the cubic polynomial given linear reminders after division by quadratic polynomials?
A cubic polynomial gives remainders $(13x-2)$ and $(-1-7x)$ when divide by
$x^2-x-3$ and $x^2-2x+5$ respectively. Find the polynomial.
I have written this as:
$P(x)=(x^2-x-3)Q(x)+(13x-2)$
$P(x)=(x^2-2x+5)G(x)+(-1-7x)$
and
$P(x)=ax^3+bx^2+cx+d$
The first method I thought about is factoring $(x^2-x-3)$ and $(x^2-2x+5)$ so that I could find roots and thereby $P(root)$ would equal the reminder for the given value of root. As the two quadratic equations would give four roots i.e. four values I can substitute, I would be able to calculate all the four constant $a$, $b$, $c$ and $d$. However, the first polynomial has very ugly roots and the second one has no real roots so I guess I should find another way. What do you suggest?
| Using the Extended Euclidean Algorithm as implemented in this answer, we get
$$
\begin{array}{r}
&&1&x+6&(x-8)/53\\\hline
1&0&1&-x-6&(x^2-2x+5)/53\\
0&1&-1&x+7&(-x^2+x+3)/53\\
x^2-x-3&x^2-2x+5&x-8&53&0\\
\end{array}
$$
That is,
$$
(x+7)(x^2-2x+5)-(x+6)(x^2-x-3)=53\tag{1}
$$
We can now use the Chinese Remainder Theorem. $(1)$ tells us that
$$
\frac{x+7}{53}\,(x^2-2x+5)\equiv
\left\{\begin{array}{}
0&\pmod{x^2-2x+5}\\
1&\pmod{x^2-x-3}
\end{array}\right.\tag{2}
$$
$$
-\frac{x+6}{53}\,(x^2-x-3)\equiv
\left\{\begin{array}{}
1&\pmod{x^2-2x+5}\\
0&\pmod{x^2-x-3}
\end{array}\right.\tag{3}
$$
Add $13x-2$ times $(2)$ and $-1-7x$ times $(3)$ to get
$$
\frac1{53}\left(20x^4+99x^3-185x^2+338x-88\right)\tag{4}
$$
taking the remainder of $(4)$ mod $(x^2-x-3)(x^2-2x+5)$ gives
$$
3x^3-5x^2+6x+4\tag{5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Positive Definite Proof Method I am interested in an effecient way of showing that the $n \times n$ matrix,
\begin{pmatrix}
2&1&1& & 1\\
1&3&1& \cdots & 1\\
1&1&4& & 1\\
&\vdots& &\ddots& \vdots\\
1&1&1& \cdots & n+1
\end{pmatrix}
is positive definite.
This is an old qualifier question, so it should not require extensive brute force computation.
Sub-question: Do row operations affect positive-definiteness?
| Observe that
\begin{align}
\begin{pmatrix}
2&1&1& \cdots & 1\\
1&3&1& \cdots & 1\\
1&1&4& \cdots & 1\\
\vdots&\vdots&\vdots &\ddots& \vdots\\
1&1&1& \cdots & n+1
\end{pmatrix}
=
\begin{pmatrix}
1&1&1& \cdots & 1\\
1&1&1& \cdots & 1\\
1&1&1& \cdots & 1\\
\vdots&\vdots&\vdots &\ddots& \vdots\\
1&1&1& \cdots & 1
\end{pmatrix}
+
\begin{pmatrix}
1&&& & \\
&2&& & \\
&&3& & \\
&& &\ddots&\\
&&& & n
\end{pmatrix}
=A_1+A_2,
\end{align}
where $A_1$ and $A_2$ denote the corresponding decomposed matrices.
Now, given ${\bf x}=(x_1,x_2,\ldots,x_n)^\top\in\mathbb{R}^n$, we have
\begin{align}
{\bf x}^\top A_1{\bf x}&=
\begin{pmatrix}
\displaystyle\sum_{k=1}^nx_k
&\displaystyle\sum_{k=1}^nx_k
&\cdots
&\displaystyle\sum_{k=1}^nx_k
\end{pmatrix}
\begin{pmatrix}
x_1\\x_2\\\vdots\\x_n
\end{pmatrix}
=\left(\sum_{k=1}^nx_k\right)^2\ge 0,\\
{\bf x}^\top A_2{\bf x}&=
\begin{pmatrix}
x_1&2x_2&\cdots&nx_n
\end{pmatrix}
\begin{pmatrix}
x_1\\x_2\\\vdots\\x_n
\end{pmatrix}
=\sum_{k=1}^nk\cdot x_k^2.
\end{align}
If ${\bf x}\neq{\bf 0}$, at least one entry $x_k\ne 0$ for
some $k$, and then the value ${\bf x}^\top A_2{\bf x}$ must be positive. Thus
\begin{align}
{\bf x}^\top\begin{pmatrix}
2&1&1& \cdots & 1\\
1&3&1& \cdots & 1\\
1&1&4& \cdots & 1\\
\vdots&\vdots&\vdots &\ddots& \vdots\\
1&1&1& \cdots & n+1
\end{pmatrix}{\bf x}
&={\bf x}^\top(A_1+A_2){\bf x}\\
&={\bf x}^\top A_1{\bf x}+{\bf x}^\top A_2{\bf x}\\
&\ge{\bf x}^\top A_2{\bf x}\\
&>0,
\end{align}
and we conclude that the matrix must be positive definite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is the following solution correct? Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$
My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$
$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$
So, either $\sqrt{x^2 + 9} = 0$ or $(\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$
From the first expression, I get $x = \pm 3 i$ and from the second expression, I get nothing.
Now, notice how in the 2nd step, I could've divided both the sides by $\sqrt{x^2 + 9}$, but I didn't because I learned here that we must never do that and that we should always factor: Why one should never divide by an expression that contains a variable.
So, my question is: is the solution above correct? Would it have been any harm had I divided both the sides by $\sqrt{x^2 + 9}$?
| Try a different method.
Take $\sqrt{x^2 +1}$ on RHS Then rationalise $\sqrt{x^2+9}$ - $\sqrt{x^2 +1}$ by it's conjugate.
So your next step would be :
$\dfrac{8}{\sqrt{x^2+1}}$[$\sqrt{x^2+9}$ + $\sqrt{x^2 +1}$] = 8
So 8 gets cancelled. and next step is as follows:
$\dfrac{\sqrt{x^2+9} + \sqrt{x^2 +1}}{\sqrt{x^2+1}}$ = 1
and so we get $\dfrac{\sqrt{x^2+9}}{\sqrt{x^2+1}}$ + 1 = 1
And finally
$\dfrac{\sqrt{x^2+9}}{\sqrt{x^2+1}}$ = 0
And only Numerator becomes zero so $\sqrt{x^2+9}=0$ hence $x=\pm3i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Any hint to solve given integral $\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}$? Show that for $ab>0$ $$\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}={{2\pi}\over ab}$$
I'm not sure how to go about this. Any solutions or hints are greatly appreciated.
| $\cos^2\theta,\sin^2\theta$ have period $\pi$, so the integral is $2I$ where $I$ is the integral from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. wlog we may take $a,b$ to be positive.
We have $I=\frac{1}{a^2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sec^2\theta}{1+(\frac{b}{a})^2\tan^2\theta}\ d\theta$. Putting $x=\tan\theta$ this becomes $\frac{1}{a^2}\int_{-\infty}^{\infty}\frac{dx}{1+(\frac{b}{a})^2x^2}$. Putting $y=\frac{b}{a}x$ we get $\frac{1}{ab}\int_{-\infty}^{\infty}\frac{dy}{1+y^2}=\frac{\pi}{ab}$. [The last integral is just $\tan^{-1}y$.]
Hence the original integral is $\frac{2\pi}{ab}$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}$ Show that $$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}.$$
Any solutions or hints are greatly appreciated.
I know I can rewrite the integral as $$\int_{-\infty}^\infty {(x-1)(x-2)\over {(x^2+1)(x^2+9)}}dx.$$ but I'm not sure how to proceed.
| Split the integral into two integrals, such that you have only one "bad" point. Now:
$$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx= \int_{-\infty}^{0} {{x^2-3x+2}\over {x^4+10x^2+9}}dx + \int_{0}^{\infty}{{x^2-3x+2}\over {x^4+10x^2+9}}dx$$
Now deal with the integrals separately:
$$\int_{-\infty}^{0} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to -\infty} \int_{t}^{0} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to -\infty} \frac{1}{48}\left(9\ln\left(\frac{x^2+9}{x^2+1}\right) + 6\arctan(x) + 14\arctan(\frac x3)\right) \Biggr|_t^0 = \frac{9}{48}\ln(9) + \frac{10\pi}{48}$$
$$\int_{0}^{\infty} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to \infty} \int_{0}^{t} {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \lim_{t\to \infty} \frac{1}{48}\left(9\ln\left(\frac{x^2+9}{x^2+1}\right) + 6\arctan(x) + 14\arctan(\frac x3)\right) \Biggr|_0^t = -\frac{9}{48}\ln(9) + \frac{10\pi}{48}$$
Summing them you will get:
$$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx = \frac{9}{48}\ln(9) + \frac{10\pi}{48} -\frac{9}{48}\ln(9) + \frac{10\pi}{48} = \frac{5\pi}{12}$$
NOTE: I skipped the calculation of the integrals, but it can be easily done by partial fraction decomposition and it will be reduced to something more familiar.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Extract coefficients for a formal power series using Lagrange Inversion Formula Given $f(x)$ is a formal power series that satisfies $f(0) = 0$
$(f(x))^{3} + 2(f(x))^{2} + f(x) - x = 0$
I know that the Lagrange inversion formula states given f(u) & $\varphi(u)$ are formal power series with respect to u, and $\varphi(0) = 1$ then the following is true.
$[x^{n}](f(u(x))) = \frac{1}{n}[u^{n-1}](f'(u)\varphi(u)^{n})$
How do I find the coefficient of $x^{n}$ in $f(x)$ using Lagrange inversion formula?
| Here we use a somewhat simpler but equivalent variant of the Lagrange Inversion Formula. (See Theorem A.2 in Analytic Combinatorics by P. Flajolet and R. Sedgewick for the equivalence of the variants).
Lagrange Inversion Formula:
Let $g(x), f(x)\in x\mathbb{C}[x]$ be inverses: $g(f(x))=x$. If $g(x)=\frac{x}{\phi(x)}$ and
$f(x)=x\phi\left(f(x)\right)$, then
\begin{align*}
[x^n]f(x)=\frac{1}{n}\left[x^{n-1}\right]\left(\phi(x)\right)^n\tag{1}
\end{align*}
The functional relation
\begin{align*}
(f(x))^{3} + 2(f(x))^{2} + f(x) - x = 0
\end{align*}
can be written as $g\left(f(x)\right)=x$ with
\begin{align*}
g(x)&=x^3+2x^2+x\\
&=x(1+x)^2
\end{align*}
We can write $$\phi(x)=\frac{x}{g(x)}=\frac{1}{(1+x)^2}$$
and obtain
\begin{align*}
[x^n]f(x)&=\frac{1}{n}\left[x^{n-1}\right]\left(\phi(x)\right)^n\tag{2}\\
&=\frac{1}{n}\left[x^{n-1}\right]\frac{1}{(1+x)^{2n}}\\
&=\frac{1}{n}\binom{-2n}{n-1}\tag{3}\\
&=\frac{(-1)^{n-1}}{n}\binom{3n-2}{n-1}
\end{align*}
Comment:
*
*In (2) we apply the Lagrange Inversion Formula (1)
*In (3) we use the binomial identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q
\end{align*}
| {
"language": "en",
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Inequality, only one solution from algebra I recently came along the following problem:
$$f(x) = {4-x^2 \over 4-\sqrt{x}}$$
Solve for:$$f(x) ≥ 1$$
My Attempt
Now I know that one of the restrictions on the domain is $x≥0$, thus one of the solutions is $0≤x≤1$. There is still one more solution according to the graph, which is $x>16$.
Graph
Where did I go wrong?
| As already mentioned in the comments, you must also account for the case where $4- \sqrt{x} < 0$, which results in the inequality sign being flipped.
The way to do this in general is to first get $0$ on one side and simplify/factor on the other:
\begin{align}
\frac{4-x^2}{4-\sqrt x} &\ge 1\\[0.3cm]
\frac{4-x^2}{4-\sqrt x} -1 &\ge 0\\[0.3cm]
\frac{4-x^2}{4-\sqrt x} - \frac{4-\sqrt x}{4-\sqrt x} &\ge 0\\[0.3cm]
\frac{4-x^2 - (4 - \sqrt x)}{4 - \sqrt x} &\ge 0\\[0.3cm]
\frac{4-x^2 - 4 + \sqrt x}{4 - \sqrt x} &\ge 0\\[0.3cm]
\frac{\sqrt x - x^2}{4 - \sqrt x} & \ge 0\\[0.3cm]
\frac{\sqrt x (1 - x^{3/2})}{4 - \sqrt x} & \ge 0
\end{align}
Therefore solving the original equality is equivalent to solving this inequality. Now we find all the values of $x$ where the numerator or denominator of the LHS is zero. So we have three equations:
\begin{align}
\sqrt x &= 0\\
1 - x^{3/2} &= 0\\
4 - \sqrt{x} &= 0
\end{align}
The first one gives us $x = 0$. The second gives us $x = 1$. The third gives us $x = 16$. Now we use these values of $x$ to split the real number line into intervals:
$$ (0, 1) \qquad (1, 16) \qquad (16, +\infty) $$
Note that the LHS of our "new" inequality above cannot change sign within each of these intervals. So we need only pick one point from each interval to determine the sign of the LHS. Also note that we don't have $(-\infty, 0)$ because that's outside of the domain.
In $(0,1)$, let's pick $x = 1/2$. We'll plug this in to the LHS above and see what its sign is. We don't care about the value, only the sign:
$$\frac{\sqrt{1/2}(1 - [1/2]^{3/2})}{4 - \sqrt{1/2}} > 0$$
Proceed similarly in $(1,16)$ and $(16,+\infty)$ to see that the LHS is negative in $(1,16)$ and positive in $(16,+\infty)$.
Finally, note that the LHS is equal to zero when $x = 0$ and when $x = 1$. Therefore the final answer is $[0,1] \cup (16, +\infty)$.
| {
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How to find continuities with square root? I don't understand how to find $$\frac{4-x^2}{3-\sqrt{x^2+5}}$$
The book says to multiply the equation by $\frac{3 + \sqrt {x^2+5}}{3 + \sqrt {x^2+5}}$. I don't understand where that comes from. It says the multiplication simplifies to "$3 + \sqrt {x^2+5}$" - I don't see how that's possible. Is the book wrong?
| $$\frac{4-x^2}{3-\sqrt{x^2+5}}=\frac{4-x^2}{3-\sqrt{x^2+5}}\cdot\frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}}=\frac{(4-x^2)(3+\sqrt{x^2+5})}{3^2-(x^2+5)}$$ $$=\frac{(4-x^2)(3+\sqrt{x^2+5})}{4-x^2}=3+\sqrt{x^2+5}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\frac{a+c}{b+d}$ lies between $\frac{a}{b}$ and $\frac{c}{d}$ (for positive $a$, $b$, $c$, $d$) I am looking for proof that, if you take any two different fractions and add the numerators together then the denominators together, the answer will always be a fraction that lies between the two original fractions.
Would be grateful for any suggestions!
| Suppose we have positive $a,b,c,d$ with $\frac{a}{b}\ge \frac{c}{d}$. Then multiplying through by $bd$ we get $ad\ge bc$. Adding $ab$ to both sides we get $a(b+d)\ge b(a+c)$. Dividing by $b(b+d)$, we get $\frac{a}{b}\ge\frac{a+c}{b+d}$.
Similarly, add $cd$ to both sides of $ad\ge bc$ to get $d(a+c)\ge c(b+d)$. Dividing by $d(b+d)$ we get $\frac{a+c}{b+d}\ge \frac{c}{d}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum $k$
Find the minimum $k$, which $\exists a,b,c>0$, satisfies
$$ \frac{kabc}{a+b+c}\geq (a+b)^2+(a+b+4c)^2$$
My Progress
With the help of Mathematica, I found that when $k=100$, we can take $a=1,b=1,c=1/2$. And I'm pretty sure that $k=100$ is the answer, but I couldn't prove it.
| Let $a=b=2$ and $c=1$.
Hence, $k\geq100$.
We'll prove that $100$ it's an answer.
Indeed, let there are positives $a$, $b$ and $c$ for which $ \frac{kabc}{a+b+c}\geq (a+b)^2+(a+b+4c)^2$ and $k<100$.
But it's impossible because we'll prove now that $ \frac{100abc}{a+b+c}\leq (a+b)^2+(a+b+4c)^2$.
Let $c=(a+b)x$.
Hence, by AM-GM $\frac{100abc}{a+b+c}\leq\frac{25(a+b)^2c}{a+b+c}=\frac{25(a+b)^2x}{x+1}$.
Thus, it remains to prove that $\frac{25(a+b)^2x}{x+1}\leq(a+b)^2+(a+b+4(a+b)x)^2$ or
$\frac{25x}{x+1}\leq1+(1+4x)^2$, which is AM-GM again or
$(4x-1)^2(x+2)\geq0$. Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\arcsin(\sqrt{1-x^2}) +\arccos(x) = \text{arccot} \left(\frac{\sqrt{1-x^2}}{x}\right) - \arcsin( x)$ If we have to find the solutions of equation
$$\arcsin(\sqrt{1-x^2}) +\arccos(x) = \text{arccot} \left(\frac{\sqrt{1-x^2}}{x}\right) - \arcsin( x)$$
Using a triangle I rewrite it as
$$2 \arctan \left(\frac{\sqrt{1-x^2}}{x}\right)= 0$$
So this equation is satisfied when $x=\pm 1$
But I saw that $x=-1/2$ is also satisfying , then where I have missed the case .
I am totally stuck , how to find it .
| Straightaway the problem reduces to $$\text{arccot}\dfrac{\sqrt{1-x^2}}x=\dfrac\pi2+\arcsin\sqrt{1-x^2}$$
As $\sqrt{1-x^2}\ge0,$ using the definition of Principal Values
$0\le\arcsin\sqrt{1-x^2}\le\dfrac\pi2$
and consequently, $\dfrac\pi2\le\text{arccot}\dfrac{\sqrt{1-x^2}}x\le\pi$
$\implies x\not>0$ but $x\ne0,$
let $-x=y>0$
$$\implies\text{arccot}\dfrac{\sqrt{1-y^2}}{-y}=\dfrac\pi2+\arcsin\sqrt{1-y^2}$$
$$\iff\dfrac\pi2-\arctan\dfrac{\sqrt{1-y^2}}{-y}=\dfrac\pi2+\arccos y$$
As $\arctan(-u)=-\arctan u,$
$$\arctan\dfrac{\sqrt{1-y^2}}y=\arccos y$$
Now as $y>0$ and let $\arccos y=v\implies\cos v=y$ and $0\le v<\dfrac\pi2$
and $\dfrac{\sqrt{1-y^2}}y=\tan v\implies\arctan\dfrac{\sqrt{1-y^2}}y=v=\arccos y$ as $0\le v<\dfrac\pi2$
So, we need $y>0\iff x<0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$ - I keep getting imaginary numbers $$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$$
My attempt
$\sqrt{x+938^2} + \sqrt{x + 140^2} = 1116$
$(\sqrt{x+938^2} + \sqrt{x + 140^2})^2 = (1116)^2$
$x+938^2 + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} + x + 140^2 = 1116^2$
$2x + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} = 1116^2 - 938^2 - 140^2$
$x + \sqrt{x^2 + 2(938^2 + 140^2)x+(938*140)^2} = 1116^2 - 938^2 - 140^2$
At this point trying to solve for x inside the sqrt in the quadratic gives me an imaginary number. How is it possible to solve this?
| After the first line of your attempt, multiply both sides by $\sqrt{x+938^2}-\sqrt{x+140^2}$
$$x+938^2-(x+140^2)=1116(\sqrt{x+938^2}-\sqrt{x+140^2})$$
Divide both sides by $1116$.
$$\dfrac{938^2-140^2}{1116}=\sqrt{x+938^2}-\sqrt{x+140^2}$$
Taken with the previous equation
$$\sqrt{x+938^2}+\sqrt{x+140^2}=1116$$
we can add both equations to solve for $\sqrt{x+938^2}$ or subtract to solve for $\sqrt{x+140^2}$. Then simply square both sides and solve for $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find $\frac{\mathrm{d}y}{\mathrm{d}x}$ when both number in front and exponent have fractions? I'm not sure how to solve this: $\frac{5}{9}x^\frac{2}{3}$. I applied the product rule and have $\frac{2}{3}\frac{5}{9}x^{-\frac{1}{3}}$.
$\frac{30}{9}x^{-\frac{1}{3}}$, then $\frac{9}{30}x^{\frac{1}{3}}$.
This isn't the answer in the book, though. Is the procedure for getting rid of a negative in the exponent to flip the number in front of $x$?
| The comments have already pointed out where you have misunderstood this question.
However, as an answer to your question:
\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x}\Bigg(\frac{5}{9}x^\frac{2}{3} \Bigg)&= \frac{5}{9}\frac{\mathrm{d}}{\mathrm{d}x}(x^\frac{2}{3})\\
&= \frac{5}{9}\Bigg(\frac{2}{3}\Bigg)x^{\frac{2}{3}-1}\\
&= \frac{10}{27}x^{-\frac{1}{3}}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the simple continued fractions for both $\pm \frac{39}{25}$... Find the simple continued fractions for both $\pm \frac{39}{25}$?
*
*So far for $\frac{39}{25}$ I have:
$39 = 1 \times 25 + 14 $
$ 25 = 1\times 14 + 11 $
$14 = 1 \times 11 + 3$
$11 = 3 \times 3 + 2$
$3 = 1 \times 2 + 1$
$2 = 2 \times 1 + 0$
Giving us [1 ; 1, 1, 3, 1, 2 ]
*
*Then for $ - \frac{39}{25}$ I have:
$ -39 = 1 \times -25 - 14 $
$ - 25 = 1 \times -14 -11 $
$ -14 = 1 \times -11 -3 $
$ -11 = 3 \times -3 -2$
$-3 = 1 \times - 2 - 1 $
$-2 = 2 \times -1 + 0 $
Giving us [1; 1,1,3,1,2]
-I think I have this problem figured out but could someone please confirm or suggest changes. Help is very much appreciated!
| I believe that for $-\frac{39}{25}$ you should first rewrite it as $-2+\frac{11}{25}$. Then
$11=0\times25+11$
$25=2\times11+3$
$11=3\times3+2$
$3=1\times2+1$
$2=2\times1+0$
This gives $-\frac{39}{25}=[-2+0;2,3,1,2]=[-2;2,3,1,2]$
Note that recently there has developed a way of representing negative continued fractions in the form
\begin{equation}
x=a_0-\frac{1}{a_1-\frac{1}{a_2-\frac{1}{\ddots}}}
\end{equation}
In this form one takes the ceiling of each improper fraction.
See the thesis of Alex Eustis.
This approach to negative continued fractions is new to me but if I understand it correctly, then
\begin{equation}
-\frac{39}{25}=-1-\frac{1}{2-\frac{1}{5-\frac{1}{3}}}=[1;2,5,3]_-
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
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Die that never rolls the same number consecutively Suppose we have a "magic" die $[1-6]$ that never rolls the same number consecutively.
That means you will never find the same number repeated in a row.
Now let's suppose that we roll this die $1000$ times.
How can I find the PDF, expected number of times and variance of getting a specific value?
| let $P_{k,n}$ be the probability to get $k$ times the result "1" in $n$ throws, and let $$P_{k,n} = p_{k,n} + q_{k,n}$$ where $p,q$ are the corresponding probabilities where you add the constraint that you finish with a 1 (for $p$) or anything but 1 (for $q$).
It is easy to get the equations (for $n \ge 1$) :
$$5p_{k,n} = q_{k-1,n-1}$$ and $$5q_{k,n} = 5p_{k,n-1} + 4 q_{k,n-1}$$
from which you get that for $n \ge 2$, $$5q_{k,n} = q_{k-1,n-2} + 4q_{k,n-1}$$
Obviously, the $(p_{k,n})$ family satisfies the same linear equation, and since $P$ is the sum of $p$ and $q$, we have for $n \ge2$, that $$5P_{k,n} = P_{k-1,n-2} + 4P_{k,n-1}$$
Together with the data that $P_{0,0} = 1, P_{0,1} = \frac 16, P_{1,1} = \frac 56$ this is enough to determine all those numbers.
Let $f(x,y)$ be the formal power series $$f(x,y)=\sum P_{k,n} x^k y^n$$ From the recurrence on $P$ we immediately have that in $(5-4y-xy^2)f(x,y)$, every coefficient vanish except the coefficients of $x^0y^0, x^0y^1, x^1y^1$, and so that $f(x,y)$ is the power series of the rational fraction $$\frac {30+y+5xy}{6(5-4y-xyy)}$$
We can check that doing the partial evaluation $x=1$ we get $$f(1,y) = \frac {30+6y}{6(5-4y-yy)} = \frac{6(5+y)}{6(5+y)(1-y)} = \frac 1 {1-y} = 1 + y + y^2 + \dots$$so for each $n$, all the probabilities $P_{k,n}$ sum to $1$.
If $E_n$ is the expectation of the number of occurences of 1 in $n$ throws then $$E_n = \sum k P_{k,n}$$so that $$\sum E_n y^n = \frac {df}{dx} (1,y)$$
We compute
$$\frac{ df}{dx} = \frac {(5y)(5-4y-xyy)-(-yy)(30+y+5xy)}{6(5-4y-xyy)^2}
= \frac {y(5+y)^2}{6(5-4y-xyy)^2}$$
Then evaluating this at $x=1$ we get $$\sum E_ny^n = \frac {y(5+y)^2}{6(5+y)^2(1-y)^2} = \frac y {6(1-y)^2} = \sum \frac n6 y^n$$
Thus $E_n = n/6$ which is as we expected.
To compute the variance we need to compute the expectancy of the square of the number of occurences,
that is, $F_n = \sum k^2 P_{k,n}$.
Then $$\sum F_n y^n = \frac {d^2f}{dx^2}(1,y) + \frac {df}{dx}(1,y) =
\frac{2y^3(5+y)^2}{6(5+y)^3(1-y)^3} + \frac y{6(1-y)^2} = \frac{2y^3+y(5+y)(1-y)}{6(5+y)(1-y)^3} = \frac{5y-4y^2+y^3}{6(5+y)(1-y)^3}$$
$$\sum E_n^2 y^n = \sum \frac 1{36} n^2y^n = \frac {y(1+y)}{36(1-y)^3}$$
Computing their difference gives the variance :
$$\sum V_n y^n = \frac {30y-24y^2+6y^3-y(1+y)(5+y)} {36(5+y)(1-y)^3}
= \frac {5y(5-6y+y^2)}{36(5+y)(1-y)^3}
= \frac {5y(5-y)}{36(5+y)(1-y)^2} \sim_{y=1} \frac {20}{216(1-y)^2}$$
Doing the partial fraction decomposition gives a formula $$V_n = \frac {50}{1296} + \frac{20}{216}n - \frac {50}{1296}(-1/5)^n$$
I expect that after renormalization, your random variable (number of 1s obtained in $n$ throws) converges in law to a gaussian variable as $n \to \infty$.
| {
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Find $\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$ Find $$\sin\frac{\pi}{3}+\frac{1}{2}\sin\frac{2\pi}{3}+\frac{1}{3}\sin\frac{3\pi}{3}+\cdots$$
The general term is $\frac{1}{r}\sin\frac{r\pi}{3}$
Let $z=e^{i\frac{\pi}{3}}$
Then, $$\frac{1}{r}z^r=\frac{1}{r}e^{i\frac{r\pi}{3}}$$
I have to find the imaginary part of $$P=\sum_{r=1}^\infty \frac{1}{r}z^r$$
Let $$S=1+z+z^2+\cdots$$
Hence, $$P=\int_0^z Sdz=z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots$$
which is the required sum.
$$S=\frac{1}{1-z}$$
$$P=\int_0^z \frac{1}{1-z}dz$$
$$P=\ln\left(\frac{1}{1-z}\right)=\ln\left(\frac{1}{1-e^{i\frac{\pi}{3}}}\right)=i\frac{\pi}{3}$$
Hence, the imaginary part of $P$ is $\frac{\pi}{3}$
Is this method correct? Is there any method that does not require complex numbers?
| This can be done with Fourier series. Let $f(x)=x$ for $x\in(-\pi,\pi)$, and its periodic extension with period $2\pi$. Then, since $f(x)=-f(-x)$ we can represent it as
$$f(x)=\sum_{n=1}^{\infty}b_n\sin\left(nx\right)$$
Then we can compute
$$\begin{align}\int_{-\pi}^{\pi}f(x)\sin(nx)dx&=\int_{-\pi}^{\pi}x\sin(nx)dx=\left[-\frac xn\cos(nx)+\frac1{n^2}\sin(nx)\right]_{-\pi}^{\pi}\\
&=-\frac{2\pi}n(-1)^n=\int_{-\pi}^{\pi}\sum_{m=1}^{\infty}b_m\sin(mx)\sin(nx)dx\\
&=\sum_{m=1}^{\infty}b_m\pi\delta_{mn}=\pi b_n\end{align}$$
So
$$f(x)=\sum_{n=1}^{\infty}\frac2n(-1)^{n+1}\sin(nx)$$
We can see that
$$\sin\left(n\left(\pi-\frac{\pi}3\right)\right)=\sin(n\pi)\cos\left(n\frac{\pi}3\right)-\cos(n\pi)\sin\left(n\frac{\pi}3\right)=(-1)^{n+1}\sin\left(n\frac{\pi}3\right)$$
And finally
$$\frac{2\pi}3=\pi-\frac{\pi}3=f\left(\pi-\frac{\pi}3\right)=\sum_{n=1}^{\infty}\frac2n(-1)^{n+1}\sin\left(n\left(\pi-\frac{\pi}3\right)\right)=\sum_{n=1}^{\infty}\frac2n\sin\left(n\frac{\pi}3\right)$$
Which agrees with the result in the question.
| {
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} |
Show that $\sin(x) > \ln(x+1)$ for any $x \in (0,1)$ Show that $\sin(x) > \ln(x+1)$ when $x \in (0,1)$.
I'm expected to use the maclaurin series (taylor series when a=0)
So if i understand it correctly I need to show that:
$$\sin(x) = \lim\limits_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{(-1)^{k-1}}{(2k-1)!} \cdot x^{2k-1} > \lim\limits_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} \cdot x^k = \ln(x+1)$$
I tried to show that for any k the general term in the bigger sum is greater then the other one (the general term in the smaller sum) but its not true :(.
$$\frac{(-1)^{k-1} \cdot x^{2k-1}}{(2k-1)!} > \frac{(-1)^{k-1} \cdot x^k}{k}$$
when k is odd we get:
$$\frac{x^{k-1}}{(2k-1)!} > \frac{1}{k}$$
and this is a contradiction since :
$x^{k-1} < 1$ for any $x \in (0,1)$ and $k > 1$ and $(2k-1)! > k $
so for any $k > 1 $ its $\frac{x^{k-1}}{(2k-1)!} < \frac{1}{k}$ and if $k=1$ its $\frac{x^{k-1}}{(2k-1)!} = \frac{1}{k}$.
What am I doing wrong and how i'm supposed to prove it ?
Thanks in advance for help .
| Let $f(x)=\sin x-\ln(x+1)$. we try to show that $f(x)$ is increasing. In fact
$$ f'(x)=\cos x-\frac{1}{x+1}=\frac{(x+1)\cos x-1}{x+1}. $$
Now we show $(x+1)\cos x>1$ or $\sec x-1<x$ for $0<x<1$.
Note
\begin{eqnarray}
\sec x-1=\frac{1-\cos x}{\cos x}=\frac{2\sin^2\frac{x}{2}}{1-2\sin^2\frac{x}{2}}.
\end{eqnarray}
Since $\frac{2u}{1-2u}$ is increasing and $\sin x\le x$ for $x\in[0,\pi/2]$, one has
\begin{eqnarray}
\sec x-1=\frac{2\sin^2\frac{x}{2}}{1-2\sin^2\frac{x}{2}}\le \frac{x^2}{2-x^2}\le x^2<x.
\end{eqnarray}
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb Q$ Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb{Q}$ (by finding a nonzero polynomial $p(x)$ with coefficients in $\mathbb{Q}$ which has $\sqrt[3] 2+\sqrt 5$ as a root).
I first tried letting $a=\sqrt[3]{2} +\sqrt{5}$ and then square both sides. But I keep on going into a loop by continuing to square it over and over again. Then I tried $a^3=(\sqrt[3]{2} +\sqrt{5})^3$. Just can't seem to get rid of the radicals.
| As a vector space over $\mathbb Q$, $\mathbb Q(\sqrt[3]{2},\sqrt 5)$ has a basis $\{1,\sqrt 5,\sqrt[3]{2},\sqrt[3]{4},\sqrt[3]{2}\cdot\sqrt 5,\sqrt[3]{4}\cdot\sqrt 5\}$. Now consider the linear map $\phi:\mathbb Q(\sqrt[3]{2},\sqrt 5)\to\mathbb Q(\sqrt[3]{2},\sqrt 5)$ given by $\phi(y)=(\sqrt[3]{2}+\sqrt 5)y$. With respect to this basis, its matrix is given by
$$
\begin{bmatrix}
\phi(1)&\phi(\sqrt 5)&\phi(\sqrt[3]{2})&\phi(\sqrt[3]{4})&\phi(\sqrt[3]{2}\cdot\sqrt 5)&\phi(\sqrt[3]{4}\cdot\sqrt 5)
\end{bmatrix}
$$
which comes out to
$$
A=\begin{bmatrix}
0&5&0&2&0&0\\
1&0&0&0&0&2\\
1&0&0&0&5&0\\
0&0&1&0&0&5\\
0&1&1&0&0&0\\
0&0&0&1&1&0
\end{bmatrix}.
$$
The characteristic polynomial $f(x)=\det(xI-A)$ has $\sqrt[3]{2}+\sqrt 5$ as a root, since $\sqrt[3]{2}+\sqrt 5$ is an eigenvalue of $\phi$. This turns out to be
$$
\det\begin{bmatrix}
x&-5&0&-2&0&0\\
-1&x&0&0&0&-2\\
-1&0&x&0&-5&0\\
0&0&-1&x&0&-5\\
0&-1&-1&0&x&0\\
0&0&0&-1&-1&x
\end{bmatrix}=x^6-15x^4-4x^3+75x^2-60x-121.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding all real roots of the equation $(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$
Find all real roots of the equation
$$(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$$
I tried squaring the equation, but the degree of the equation became too high and unmanageable. I also tried substitutions, but it didn't work out correctly. This question was in my weekly class worksheet as were this and this question which I previously asked.
Any help will be appreciated.
Thanks.
| We have
$$x^2+7x+12-(x+1) \sqrt{x+2} - (x+6)\sqrt{x+7}=0$$
Multiplying the both sides by $3$ gives
$$3x^2+21x+36-3(x+1) \sqrt{x+2} - 3(x+6)\sqrt{x+7}=0\tag1$$
Now since
$$\begin{align}&3x^2+21x+36\\&=x^2+5x+4+x^2+13x+42+x^2+3x-10\\&=(x+1)(x+4)+(x+6)(x+7)+(x-2)(x+5)\end{align}$$
we have, from $(1)$,
$$(x+1)(x+4)+(x+6)(x+7)+(x-2)(x+5)-3(x+1) \sqrt{x+2} - 3(x+6)\sqrt{x+7}=0$$
Rearranging terms
$$(x+1)(x+4-3\sqrt{x+2})+(x+6)\sqrt{x+7}\ (\sqrt{x+7}-3)+(x-2)(x+5)=0,$$
i.e.
$$(x+1)\cdot\frac{(x-2)(x+1)}{x+4+3\sqrt{x+2}}+\frac{(x+6)\sqrt{x+7}\ (x-2)}{\sqrt{x+7}+3}+(x-2)(x+5)=0$$
and so
$$(x-2)f(x)=0$$
where
$$f(x)=\frac{(x+1)^2}{x+4+3\sqrt{x+2}}+\frac{(x+6)\sqrt{x+7}}{\sqrt{x+7}+3}+(x+5)$$
We know that $f(x)$ is positive because of $x\ge -2$.
Thus, $\color{red}{x=2}$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Show $3\cos 2x + 1 = 4\cos^2 x - 2\sin^2 x$ Show $3\cos 2x + 1 = 4\cos^2 x - 2\sin^2 x$
Using the formula $\cos 2x = \cos x - \sin^2 x$
I can say $3\cos 2x + 1 = 3(\cos^2 x - \sin^2 x) + 1$
$\Rightarrow 3\cos x^2 - 3\sin^2 x + 1$
But from there I don't see how I can get the answer.
| I continue from where you left off.
$$ 3\cos^2 x - 3\sin^2 x + 1 = 4\cos^2 x -\cos^2 x - \sin^2 x - 2\sin^2 x + 1 $$
$$ = 4\cos^2 x - 2\sin^2 x \boxed{-\cos^2 x - \sin^2 x + 1 } $$
$$ = 4\cos^2 x - 2\sin^2 x. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximum of function containing two variables $x$ and $y$
If $x+y+\sqrt{2x^2+2xy+3y^2} = k(\bf{Const.})\;,$ Then $\max(x^2y)\;,$ Where $x,y\geq 0$
$\bf{My\; Try::}$ Let $x^2y=z\;$ Then we get $$x+\frac{x^2}{z}+\sqrt{2x^2+\frac{2z}{x}+\frac{3z^2}{x^4}} = k$$
Now How can I solve it, Help me
Thanks
| Assuming $y=a x$ ($a\geq0$), the constraint leads to $$x=\frac{k}{\sqrt{3 a^2+2 a+2}+a+1}$$ and $$x^2y=\frac{a k^3}{\left(\sqrt{3 a^2+2 a+2}+a+1\right)^3}$$ Now, computing $$\frac{d}{da}x^2y=-\frac{(2 a-1) \left(3 a+\sqrt{a (3 a+2)+2}+2\right)}{\sqrt{a (3 a+2)+2}
\left(a+\sqrt{a (3 a+2)+2}+1\right)^4}k^3$$ which cancels if $a=\frac 12$.
All of this makes a maximum value equal to $$\frac{4 k^3}{\left(3+\sqrt{15}\right)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{\cos\theta\cos\delta}{\cos^2\alpha}+\frac{\sin\theta\sin\delta}{\sin^2\alpha}+1=0$
If $$\frac{\cos\theta}{\cos\alpha}+\frac{\sin\theta}{\sin\alpha}=\frac{\cos\delta}{\cos\alpha}+\frac{\sin\delta}{\sin\alpha}=1,$$ where $\theta$ and $\delta$ do not differ by an even multiple of $\pi$, then prove that $$\frac{\cos\theta\cos\delta}{\cos^2\alpha}+\frac{\sin\theta\sin\delta}{\sin^2\alpha}+1=0.$$
$$\frac{\cos\theta}{\cos\alpha}+\frac{\sin\theta}{\sin\alpha}=\frac{\cos\delta}{\cos\alpha}+\frac{\sin\delta}{\sin\alpha}=1$$
$$\frac{\cos\theta \sin\alpha+\sin\theta\cos\alpha}{\sin\alpha\cos\alpha}=\frac{\cos\delta \sin\alpha+\sin\delta\cos\alpha}{\sin\alpha\cos\alpha}=1$$
We need to prove that $$\frac{\cos\theta\cos\delta}{\cos^2\alpha}+\frac{\sin\theta\sin\delta}{\sin^2\alpha}+1=0$$
$$\implies\frac{\cos\theta\cos\delta-\cos^2\alpha\cos(\theta+\delta)}{\sin^2\alpha\cos^2\alpha}+1=0$$
I am stuck here. Please help.
| For future readers, I here combine the comments by @H.Potter and @lab bhattacharjee into an answer.
Let $\frac{\cos \theta}{\cos\alpha} $ be denoted as $(1)$,
$\frac{\sin \theta}{\sin \alpha} $ be denoted as $(2)$,
$\frac{\cos \delta}{\cos\alpha} $ be denoted as $(3)$, and
$\frac{\sin \delta}{\sin\alpha} $ be denoted as $(4)$.
As given in the question, $$(1) +(2) = (3) +(4) = 1$$
Mupltiplying $\big((1) + (2) \big)$ by $\big((3) + (4)\big)$ :
$$\left(\frac{\cos \theta}{\cos\alpha} + \frac{\sin \theta}{\sin \alpha}\right)\left(\frac{\cos \delta}{\cos\alpha} + \frac{\sin \delta}{\sin\alpha}\right) =1 \times 1$$
Expanding the left hand side :
$$\frac{\cos \theta \, \cos \delta}{\cos^2\alpha} + \frac{\sin \theta \, \sin \delta}{\sin^2\alpha} + \frac {2\sin \theta \cos \delta }{\sin\alpha \cos \alpha} = 1 \,\,\,(♣)$$
As you can hopefully see, the first two fractions are the ones as required in the final answer/"hence proved step". So, we just need to make the third fraction equal to $2$ so that when the $1$ of $RHS$ comes to $LHS$, it completes the proof.
Now , let us focus on the original statement :
$(1) + (2) = (3)+(4) = 1$
As OP has already mentioned in the "Question Details" section :
$$\frac {\cos \theta \sin \alpha + \sin \theta \cos \alpha} {\sin \alpha \cos \alpha} = \frac {\cos \delta \sin \alpha + \sin \delta \cos \alpha}{\sin \alpha \cos \alpha} = 1 $$
$$\implies \frac {\sin (\theta + \alpha)}{\sin \alpha \cos \alpha} = \frac {\sin (\alpha + \delta)}{\sin \alpha \cos \alpha} = 1$$
Now you can equate $\sin (\theta + \alpha)$ and $\sin (\alpha + \delta)$, giving
$\theta + \alpha = (2n+1)\pi - (\alpha + \delta)$.
Hopefully, this is enough to conclude that $\large \frac {2\sin \theta \cos \delta}{\sin \alpha \cos \alpha} = 2. $
Plug this back in (♣) to complete the proof.
Hope this helps :).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding lim${_{n \rightarrow \infty}}\left( \frac{n^3}{2^n} \right)$ For a class of mine we were given a midterm review; however, I just cannot figure out how to finish this one:
Find the limit $$\lim_{n \rightarrow \infty}\left( \dfrac{n^3}{2^n} \right)$$
My attempt so far:
Let $s_n=\dfrac{n^3}{2^n}$.
Note that $2^n=(1+1)^n$. Thus by the Binomial Theorem we have that,
$
(1+1)^n=\sum_{k=0}^{n} {n \choose k}(1)^{n-k}(1)^n
$
Evaluating some of the first couple I get the following terms:
$1+n+\dfrac{n(n-1)}{2}+\dfrac{n(n-1)(n-2)}{6}+\dfrac{n(n-1)(n-2)(n-3)}{24}+...$
I have noticed that for $n<4$, $n^3 \geq 2^n$. However, it seems that when $n\geq 4, n^3 < 2^n$. Thus, would it be possible to make an argument that for $n \geq 4, s_{4}>s_{5}>s_{6}>...$? Therefore, by evaluating for $n=4$ and using some algebra you get the following:
$s_{4}=\dfrac{24n^3}{24+24n+4n(n-1)(n-2)+n(n-1)(n-2)(n-3)} \leq \dfrac{24n^3}{n^4} = \dfrac{24}{n}$
and we know that lim$_{n \rightarrow \infty} \left( \dfrac{1}{n} \right)=0$ and lim$_{n \rightarrow \infty} \left( -\dfrac{1}{n} \right)=0$
Therefore, since $\dfrac{1}{n} \geq s_{4} > s_{5} >s_{6} > ...s_{n}\geq-\dfrac{1}{n}$, by the Squeeze Theorem $s_{n}$ converges to zero as well?
| Thats a long answer.Quick trick:$0 < \dfrac{n^3}{2^n} < \dfrac{1}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluate the double integral.
To find:
$$\iint_R(x^2-xy)dA$$
enclosed by
$$y=x, y=3x-x^2,$$
$$x=3x-x^2,$$
$$x=0, x=2,$$
$$y=0, y=2,$$
$$R=((x,y): 0\le x\le 2), (x \le y \le 3x-x^2)$$
$$\int_0^2\int_x^{3x-x^2}(x^2-xy)\,dy\,dx$$
$$\int_0^2 \left. \left(x^2y-x\frac{y^2}{2}\right) \right|_x^{3x-x^2}\,dx$$
where to from here?
| With the change of variables $u=x$ and $v=y-x$, that is $x=u$ and $y=v-u$, we have the Jacobian $\left|\frac{\partial(x,y)}{\partial (u,v)}\right|=1$; from $0\le x\le 2$ we have $0\le u\le 2$ and from $x\le y\le 3x-x^2$, we have $0\le y-x\le 2x-x^2$ we have $0\le v\le 2u-u^2$. The integral becomes
\begin{align}
I&=\int_0^2\int_0^{(2-u)u} (-uv)\,\mathrm dv\,\mathrm du\\
&=\int_0^2 (-u)\,\frac{u^2(2-u)^2}{2} \mathrm du=-\frac{1}{2}\int_0^2(4u^3-4u^4+u^5)\,\mathrm du\\
&=\left.-\frac{1}{2}\left(4u^4-\frac{4u^5}{5}+\frac{u^6}{6}\right)\right|_0^2=-\frac{2^4}{2}\left(4-\frac{4\cdot 2}{5}+\frac{2^2}{6}\right)=-\frac{8}{15}
\end{align}
Check with WolframAlpha
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Maximum value of the sum of absolute values of cubic polynomial coefficients $a,b,c,d$
If $p(x) = ax^3+bx^2+cx+d$ and $|p(x)|\leq 1\forall |x|\leq 1$, what is the $\max$ value of $|a|+|b|+|c|+|d|$?
My try:
*
*Put $x=0$, we get $p(0)=d$,
*Similarly put $x=1$, we get $p(1)=a+b+c+d$,
*similarly put $x=-1$, we get $p(-1)=-a+b-c+d$,
*similarly put $\displaystyle x=\frac{1}{2}$, we get $\displaystyle p\left(\frac{1}{2}\right)=\frac{a}{8}+\frac{b}{4}+\frac{c}{2}+d$
So, given that $|p(x)|\leq 1\forall |x|\leq 1$, we get $|d|\leq 1$.
Similarly $$\displaystyle |b|=\left|\frac{p(1)+p(-1)}{2}-p(0)\right|\leq \left|\frac{p(1)}{2}\right|+\left|\frac{p(1)}{2}\right|+|p(0)|\leq 2$$
Now I do not understand how can I calculate the $\max$ of $|a|$ and $|c|$.
| Let $p(1)=u,$ $p(-1)=v$, $p\left(\frac{1}{2}\right)=w$ and $p\left(-\frac{1}{2}\right)=t$.
Thus, we have the following system:
$$a+b+c+d=u,$$
$$-a+b-c+d=v,$$
$$\frac{a}{8}+\frac{b}{4}+\frac{c}{2}+d=w$$ and
$$-\frac{a}{8}+\frac{b}{4}-\frac{c}{2}+d=t,$$ which gives
$$a=\frac{2u-2v-4w+4t}{3},$$
$$b=\frac{2u+2v-2w-2t}{3},$$
$$c=\frac{-u+v+8w-8t}{6}$$ and
$$d=\frac{-u-v+4w+4t}{6}.$$
Now, $$a+b+c+d=u\leq1,$$
$$a+b+c-d=\frac{4u+v-4w-4t}{3}\leq\frac{13}{3},$$
$$a+b-c+d=\frac{4u-v-8w+8t}{3}\leq7,$$
$$a+b-c-d=\frac{5u-12w+4t}{3}\leq7,$$
$$a-b+c+d=\frac{-u-4v+4w+4t}{3}\leq\frac{13}{3},$$
$$a-b+c-d=-v\leq1,$$
$$a-b-c+d=\frac{-5v-4w+12t}{3}\leq7$$ and
$$a-b-c-d=\frac{u-4v-8w+8t}{3}\leq7,$$
which gives
$$|a|+|b|+|c|+|d|\leq7.$$
But for $p(x)=4x^3-3x$ the equality occurs, which says that $7$ is a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof of $1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=\frac{F_nF_{n+1}^2+(-1)^{n+1}[F_{n-1}+(-1)^{n+1}]}{2}$ Fibonacci series
$F_0=0$, $F_1=1$; $F_{n+1}=F_n+F_{n-1}$
This is a well known identity
$1^2+1^2+2^2+3^2+5^2+\cdots +F_n^2=F_nF_{n+1}$
I was curious and look every websites for a closed form of
$1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=S_n$, but couldn't find one, so I went and look for it.
It takes me a lot of time through experimental mathematics and deduced a closed form for the cube of the Fibonacci series.
$$1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=\frac{F_nF_{n+1}^2+(-1)^{n+1}[F_{n-1}+(-1)^{n+1}]}{2}$$
I need someone to verify it by proof. I try next to determine for the 4th power but fail, so I just wonder is there a closed form for nth power Fibonacci series.
| This can be proved by induction. It holds for $n=1$. Denoting the right-hand side by $\sigma_n$, we have
\begin{align}
\sigma_n+F_{n+1}^3-\sigma_{n+1}&=\frac{F_nF_{n+1}^2+(-1)^{n+1}F_{n-1}+1}2+F_{n+1}^3-\frac{F_{n+1}F_{n+2}^2+(-1)^{n+2}F_n+1}2
\\
&=
\frac{2F_{n+1}^3+F_nF_{n+1}^2-F_{n+1}F_{n+2}^2+(-1)^{n+1}F_{n+1}}2
\\
&=
\frac{2F_{n+1}^3+F_nF_{n+1}^2-F_{n+1}(F_n+F_{n+1})^2+(-1)^{n+1}F_{n+1}}2
\\
&=
\frac{F_{n+1}^3-F_nF_{n+1}^2-F_{n+1}F_n^2+(-1)^{n+1}F_{n+1}}2\;.
\end{align}
That this is zero can be confirmed by substituting the explicit form
$$
F_n=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}\;.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
LU decomposition using the LU factorization Algorithm with $l_{ii}=1$
For this matrix, I got $U_{31}=1$, but the answer says $U_{31}=1/2$. Since the first two elements of the third row are 0, I think this two answers actually are equivalent. I am just wondering if both of them are correct? Since it is not in the Row Reduced echelon form, I suppose the answer is not unique. Thanks!
| We have:
$$A = \begin{bmatrix}1&-1&0\\2 & 2 & 3 \\ -1 & 3 & 2\end{bmatrix} = LU=\begin{bmatrix}1 & 0 & 0\\l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1\end{bmatrix} \begin{bmatrix}u_{11} & u_{12} & u_{13}\\0 & u_{22} & u_{23} \\ 0 & 0 & u_{33}\end{bmatrix}$$
The individual calculations in order they are performed are:
*
*$u_{11} = 1, u_{12} = -1, u_{13} = 0$
*$l_{21} u_{11} = 2 \implies l_{21} = 2, ~l_{21}u_{12}+u_{22} = 2 \implies u_{22} = 4, ~l_{21}u_{13} + u_{23} = 3 \implies u_{23} = 3$
*$l_{31}u_{11} = -1 \implies l_{31} = -1, ~ l_{31} u_{12} + l_{32}u_{22} = 3 \implies l_{32} = \dfrac{1}{2}$
*$l_{31}u_{13} + l_{32}u_{23} + u_{33} = 0 +\dfrac{3}{2} + u_{33} = 2 \implies u_{33} = \dfrac{1}{2}$
This results in:
$$A = \begin{bmatrix}1&-1&0\\2 & 2 & 3 \\ -1 & 3 & 2\end{bmatrix} = LU = \begin{bmatrix}1 & 0 & 0\\2 & 1 & 0 \\ -1 & \dfrac{1}{2} & 1\end{bmatrix} \begin{bmatrix}1 & -1 & 0\\0 & 4 & 3 \\ 0 & 0 & \dfrac{1}{2}\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the area of the square using co-ordinates Given a square $ABCD$ such that the vertex $A$ is on the $x$-axis and the vertex $B$ is on the $y$-axis. The coordinates of vertex $C$ are $(u,v)$. Find the area of square in terms of $u$ and $v$ only.
What I have done
Let the coordinate of $A$ be $(x,0)$ and $B$ be $(0,y)$. Also let the side of the square be a units.
$2a^2=AC^2=(x-u)^2+v^2$
$a^2$ is the required area so if we write $x$ in terms of $u$ and $v$ then the job will be done.
Now from here I thought of two ways either using trigonometry or using rotation of axes but here none of them will work because some angles will be involved and we require just $u$ and $v$ and nothing else in the expression of the area of square.
So how to do it? Please help.
| Let $A(x;0), B(0,y), C(u,v)$, and $AB=a$. Then $$S=a^2$$.
$$\begin{cases}
AB=a=\sqrt{x^2+y^2}
\\
BC=a=\sqrt{u^2+(v-y)^2}{}
\\
AC=\sqrt2a=\sqrt{(u-x)^2+v^2}
\end{cases}$$
$$\begin{cases}
a^2=x^2+y^2 (1)
\\
a^2=u^2+(v-y)^2 (2)
\\
2a^2=(u-x)^2+v^2 (3)
\end{cases}$$
(3)+(2) $3a^2=u^2-2ux+x^2+v^2+u^2+v^2-2vy+y^2=2u^2-2ux+2v^2-2vy+(x^2+y^2)=$
$=2u^2-2ux+2v^2-2vy+a^2 \Rightarrow a^2=u^2-ux+v^2-vy$
$$\begin{cases}
a^2=x^2+y^2 (1)
\\
a^2=u^2-ux+v^2-vy (4)
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove: $\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$
Prove the trigonometric identity
$$\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$$
I've searched high and low on the net and cannot find identities where there is $+$ or $- 1$'s in the equation. Any help is appreciated.
Edit after reviewing comments.
$\cos^2 {x} (\sec {x} -1)(\sec {x} +1) = (1 - \cos {x})(1 + \cos {x})\quad$
$\cos^2 {x} (\sec^2 {x}- 1) = 1 - \cos^2 {x}\quad$
$\cos^2 {x}·tan^2 {x} = \sin^2 {x} \quad$
$\cos^2 {x} ·\frac{\sin^2 {x}}{\cos^2 {x}} = \sin^2 {x} \quad$
$\frac{\sin^2 {x}}{\cos^2 {x} }=\frac{\sin^2 {x}}{\cos^2 {x} }\quad$
How's that? I tried the $\sec {x }= \frac1{\cos {x}}\quad$ method. However I was unsuccessful.
| Hint:
$$c(\frac1c-1)c(\frac1c+1)$$
| {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Integral of the function $\frac{\cos ^2 x}{1+\tan x}$ Evaluate $$\int \frac{\cos ^2 x}{1+\tan x}dx$$ I tried converting in double angle and making the derivative of the denominator in the numerator. But, it didn't work out. Some help please. Thanks.
| Let $$I = \int\frac{\cos^2 x}{1+\tan x}dx = \int\frac{\cos^3 x}{\sin x+\cos x}dx = \frac{1}{2}\int\frac{2\cos^3 x}{\sin x+\cos x}dx$$
So we get $$I = \frac{1}{2}\left[\int\frac{\left(\cos^3 x+\sin^3 x\right)+(\cos^3 x-\sin^3 x)}{\sin x+\cos x}\right]dx$$
So we get $$I = \frac{1}{4}\int (2-\sin 2x)dx+\frac{1}{4}\int \frac{(2+\sin 2x)(\cos x-\sin x)}{\cos x+\sin x}dx$$
Now In SEcond Integral Put $(\cos x+\sin x)=t\;,$ Then $(\cos x-\sin x)dx = dt$
and $1+\sin 2x=t^2\Rightarrow \sin 2x = t^2-1$
So Our Second Integral Convert into $\displaystyle \frac{1}{4}\int\frac{t^2+1}{t}dt$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$ Problem: If $3^x +3^y +3^z=9^{13}$.Find value of $x+y+z$.
Solution: $3^x +3^y +3^z=9^{13}$
$3^x +3^y +3^z=3^{26}$
I am unable to continue from here.
Any assistance is appreciated.
Edited
$9^{13} =3^{26}$
$=3^{25} (3)$
$=3^{25} (1+1+1)$
$=3^{25} + 3^{25} + 3^{25}$
So $x+y+z =75$
| Assume $x\geq y \geq z$. Then,
$$
3^{26}=9^{13}=3^x+3^y+3^z\leq 3(3^x)=3^{x+1}
$$
and so $x\geq 25$. Obviously, $x<26$ and so we must have $x=25$. But then
$$
2\times 3^{25}=3^{26}-3^x=3^y+3^z\leq 3^x+3^x=2\times 3^{25}.
$$
It must be the case that $y=x$ and $z=x$. We conclude that $x=y=z=25$ and their sum is $75$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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I have a problem when I go to calculate $\lim_{x\to\infty}\left( \frac {2x+a}{2x+a-1}\right)^{x}.$ The limit:
$\lim_{x\rightarrow\infty}
\left( {\frac {2\,x+a}{2\,x+a-1}} \right) ^{x}$
I make this:
$\left( {\frac {2\,x+a}{2\,x+a-1}} \right) ^{x}$=${{\rm e}^{{\it x\ln} \left( {\frac {2\,x+a}{2\,x+a-1}} \right) }}$
Then:
${{\rm e}^{{\it \lim_{x\rightarrow\infty}
x\ln} \left( {\frac {2\,x+a}{2\,x+a-1}} \right) }}$ = $0\cdot\infty$
Note: I cannot use L'Hopital
| One may recall that, as $u \to 0$, by using the Taylor series expansion,
$$
\log(1+u)=u+O(u^2)
$$ one gets, as $x \to \infty$,
$$
\begin{align}
\log\left(\frac{2x+a}{2x+a-1}\right)&=\log\left(1+\frac1{2x+a-1}\right)
\\\\&=\frac1{2x+a-1}+O\left(\frac1{(2x+a-1)^2}\right)
\\\\&=\frac1{2x+a-1}+O\left(\frac1{x^2}\right)
\end{align}
$$ thus, as $x \to \infty$,
$$
x\log\left(\frac{2x+a}{2x+a-1}\right)=\frac{x}{2x+a-1}+O\left(\frac1{x}\right) \to \color{blue}{\frac12}
$$ and, as $x \to \infty$,
$$
\left(\frac{2x+a}{2x+a-1}\right)^x=e^{x\log\left(\frac{2x+a}{2x+a-1}\right)} \to \color{blue}{\sqrt{e}.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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The elements and operations of the field $C = \Bbb R[x] / \langle x^2 + 1 \rangle$
$$C = \Bbb R[x] / \langle x^2 + 1 \rangle = \{[a + b x_{x^2 + 1}]\}$$
I know $C$ is a field since it has complex roots $(x+i)(x-i)$ and is irreducible over the reals, also since deg is $2$.
How would I find elements of this field? To find the sum and products of two elements: $[a + b x_{x^2 + 1}]$ and $[c + d x_{x^2 + 1}]$.
| Any element of $\Bbb R[x]$ has the form $$p(x) + \langle x^2 + 1 \rangle,$$
and by polynomial long division we can find a unique representative of this element of the form
$$a + b x$$
Since any two representatives of a given element differ by a polynomial of the form $(x^2 + 1) q(x)$, which if $q \neq 0$ has degree $\geq 2$, all of the elements represented by the linear polynomials $a + b x$ are distinct. Thus the elements (without duplication) are precisely $a + b x + \langle x^2 + 1 \rangle$, $a, b \in \Bbb R$. This field turns out to be the complex numbers, $\Bbb C$, and for convenience, we usually write the class $x + \langle x^2 + 1 \rangle$ as $i$, and hence write the elements as
$$a + b i .$$
If we add two elements, we get
$$(a + b i) + (c + d i) = (a + c) + (b + d) i .$$
If we multiply two elements, we get
$(a + b i) (c + d i) = (a c) + (a d + b c) i + (b d) i^2.$
Now, $i^2 = (x + \langle x^2 + 1 \rangle)^2 = x^2 + \langle x^2 + 1 \rangle$, which has linear representative $$x^2 - (x^2 + 1) = -1,$$ so that $i^2 = -1$ (by construction we identify $i$ with the root of the generating polynomial $x^2 + 1$ in the splitting field $C$). Thus,
$$(a + b i) (c + d i) = (a c - b d) + (a d + b c) i .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1782164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the range of $y = \sqrt{x} + \sqrt{3 -x}$ I have the function $y = \sqrt{x} + \sqrt{3 -x}$. The range in wolfram is $y \in\mathbb R: \sqrt{3} \leq y \leq \sqrt{6}$
(solution after correction of @mathlove)
$\sqrt{x} + \sqrt{3 -x} = y$
$$
\begin{cases}
x \geq 0\\
x \leq 3
\end{cases}
$$
then
$(\sqrt{x} + \sqrt{3 -x})^2 = y^2$
$x+3-x+2\sqrt{x(3-x)}=y^2$
$2\sqrt{x(3-x)}=y^2-3$
irrational equation, therefore:
$$
\begin{cases}
y^2-3 \geq 0\\
4(3x-x^2)=(y^2-3)^2
\end{cases}
$$
The $y^2 \geq 3$ is verified when $y \leq -\sqrt{3}$ or $y \geq \sqrt{3}$
regarding the second element of the system $4(3x-x^2)=(y^2-3)^2$
$4(3x-x^2)=(y^2-3)^2$
$12x-4x^2=y^4+9-6y^2$
$12x-4x^2-y^4-9+6y^2 = 0$
$12x-4x^2-y^4-9+6y^2 = 0$
$4x^2-12x+y^4-6y^2+9 = 0$
the quadratic equation is verified when the discriminant is $\geq 0$, then
$b^2 - 4ac = (-12)^2-16(y^4-6y^2+9) \geq 0$
$144-16y^4+96y^2-144 \geq 0$
$16y^4-96y^2 \leq 0$
change $t=y^2$ and $t^2=y^4$
$16t^2-96t \leq 0$
the inequality is verified when $t_1 \leq t \leq t_2$ because the discriminant in t is $\geq 0$
$t(16t^2-96) = 0$
then
$t_1 = 0$ and $t_2 = 6$ but $t=y^2$ and
$\sqrt(0) \leq y \leq \sqrt{6}$
$-\sqrt{6} \leq y \leq \sqrt{6}$
Finally, the solutions in $y$ in the system are:
$y\leq -\sqrt{3}$ or $y\geq \sqrt{3}$
and
$-\sqrt{6} \leq y \leq \sqrt{6}$
the solutions in $y$ (=range) of the system is:
$\sqrt{3} \leq y \leq \sqrt{6}$ because the function in the domain is satisfy only for $0 \leq x \leq 3$
|
$4x^2-12x+y^4-6y^2+9 = 0$
the quadratic equation is verified when the discriminant is $>= 0$, then
$b^2 - 4ac = (-12)^2-4(y^4-6y^2+9) >= 0$
This is the part where you have an error :
$$b^2-4ac=(-12)^2-4\cdot 4(y^4-6y^2+9)\ge 0$$
This is equivalent to $-\sqrt 6\le y\le \sqrt 6$.
With $y^2-3\ge 0$ and $y\ge 0$, the range is $\sqrt 3\le y\le\sqrt 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+x^2} = \frac{1}{x^2} + \frac{2}{x} - \frac{1}{x+1}$$
Therefore:
$$\int \frac{x^4+1}{x^3+x^2}\,dx = \int \frac{dx}{x^2} + \int \dfrac{2\,dx}{x} - \int \frac{dx}{x+1} = -\frac{1}{x} +2\log \vert x\vert - \log \vert x+1 \vert + C$$
The problem is I was supposed to find: $$\int \frac{x^4+1}{x^3+x^2}\,dx = \frac{x^2}{2} - x - \frac{1}{2} - \log \vert x \vert + 2 \log \vert x+1 \vert + C$$
Where is my mistake?
| You need to combine polynomial division and partial fractions.
$$
\begin{align}
\frac{x^4+1}{x^3+x^2}
&=x-1+\frac{x^2+1}{x^2(x+1)}\tag{1}\\
&=x-1+\frac2{x+1}-\frac1x+\frac1{x^2}\tag{2}
\end{align}
$$
Explanation:
$(1)$: polynomial division
$(2)$: partial fractions
| {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
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Consider the curve with implicit equation... I have the last question on my assignment which I cannot answer and was wondering if any of you could help me.
The two part question reads
Consider the curve with implicit equation $x^\frac{2}{3} + y^\frac{2}{3} = 4$. By using implicit differentiation show that:
a) $1 + \left(\frac{dy}{dx}\right)^2 = \frac{4}{x^\frac{2}{3}}$ , and
b) Using the result from (a), show that the arc length of the curve represented by the equation $x^\frac{2}{3} + y^\frac{2}{3} = 4$ over the interval x=1 to x=8 is 9 units.
So far I've managed to find the value of $\frac{dy}{dx}$
$\begin{aligned}
\frac{2}{3}x^\frac{-1}{3} + \frac{2}{3}y^\frac{-1}{3}\frac{dy}{dx} = 0\\
\\
\frac{2}{3}y^\frac{-1}{3}\frac{dy}{dx} = -\frac{2}{3}x^\frac{-1}{3}\\
\\
\frac{dy}{dx} = -\frac{\frac{2}{3}x^\frac{-1}{3}}{\frac{2}{3}y^\frac{-1}{3}}\\
\\
\frac{dy}{dx} = -\frac{x^\frac{-1}{3}}{y^\frac{-1}{3}}\\
\\
\frac{dy}{dx} = -\frac{y^\frac{1}{3}}{x^\frac{1}{3}}\\
\\
\frac{dy}{dx} = -^3\sqrt\frac{y}{x}\\
\end{aligned}$
I'm not sure how to proceed from this point to prove part (a) and complete part (b). If anyone could shed some light would be most appreciated.
Milo
| You are almost done. To complete (a),
\begin{align}
1+\left(\frac{dy}{dx}\right)^2 &= 1 + \frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}\\
&= 1 + \frac{4-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}\\
&=\frac{4}{x^{\frac{2}{3}}}.
\end{align}
The arc length of $y=f(x)$ from $(a,f(a))$ to $(b,f(b))$ is given by
$$
L=\int_a^b \sqrt{1+(f'(x))^2} dx.
$$
Therefore, you can get $L$ by substituting $1+(f'(x))^2=4x^{-\frac{2}{3}}$ from (a), and $a=1$ and $b=8$ from initial conditions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate $\lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)^{-1}$? $$f (x) = \frac{1}{x^2} - \frac{1}{\sin^2 x}$$
Find limit of $\dfrac1{f(x)}$ as $x\to0$.
| As already noted:
$$\frac1{f(x)}=\frac{x^2\sin^2x}{\sin^2x-x^2}=\color{blue}{\frac {x\sin^2x}{\sin x-x}}\cdot\color{purple}{\frac x{\sin x+x}}\;\color{red}{(**)}$$
Now, using l'Hospital's rule:
$$\begin{align*}&\lim_{x\to0}\color{blue}{\frac{x\sin^2x}{\sin x-x}}\stackrel{\text{l'H}}=\lim_{x\to0}\frac{\sin^2x+x\sin2x}{\cos x-1}\stackrel{\text{l'H}}=\lim_{x\to0}2\,\frac{\sin2x+x\cos2x}{-\sin x}\stackrel{\text{l'H}}=\\{}\\
&=\lim_{x\to0}2\frac{4\cos2x-2x\sin2x}{-\cos x}=2\frac{4}{-1}=-6\;,\;\;\;\;\text{whereas}\\{}\\
&\lim_{x\to0}\color{purple}{\frac x{\sin x+x}}=\lim_{x\to0}\frac1{\frac{\sin x}x+1}=\frac1{1+1}=\frac12\end{align*}$$
Thus, we finally get the limit is
$$\color{red}{(**)}=-6\cdot\frac12=-3$$
| {
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"url": "https://math.stackexchange.com/questions/1786681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $ \int \frac{1}{1 + x^3}dx $ $$ \int \frac{1}{1 + x^3}dx $$
Attempt:
I added and subtracted $x^3$ in the numerator but after a little solving I can't get through.
| Does this help?
$\dfrac{1}{x^3+1} = \dfrac{1}{(x+1)(x^2-x+1)} = \dfrac13\left(\dfrac{1}{x+1} - \dfrac{x-2}{x^2-x+1}\right)$
Can you integrate it from there?
The integral of the second part is a little tricky. Note that:
$\dfrac13\left(\dfrac{x-2}{x^2-x+1}\right) = \dfrac16\left(\dfrac{2x-4}{x^2-x+1}\right)= \dfrac16\left(\dfrac{2x-1}{x^2-x+1} - \dfrac{3}{x^2-x+1}\right)$
$=\dfrac16\left(\dfrac{2x-1}{x^2-x+1} - \dfrac{3}{\left(x-\frac12\right)^2 + \frac34}\right)$
That first integral is a $u$-substitution, and the second uses the arctangent function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\sin^2(10 ^\circ)-\sin^2(20^\circ)-\sin^2(40^\circ)=-\frac{1}{2}$ identity 10 degrees
$$\sin^2(10^\circ)-\sin^2(20^\circ)-\sin^2(40^\circ)=-\frac{1}{2}$$
$$\cos^2(10^\circ)-\cos^2(20^\circ)-\cos^2(40^\circ)=-\frac{1}{2}$$
Why are they both have same answer?
The only time they have same answer is at 45 degrees right?
$\sin(45^\circ)=\cos(45^\circ)=\frac{1}{\sqrt2}$
Can somebody provide me an explanation please?
Also how to prove these two identities
I know all others $15^\circ, 30^\circ, 45^\circ, 60^\circ$, etc, but can't seem to prove these.
| Let me show the first identity. By the double-angle identity:
\begin{align}
& \sin^2(10^\circ) - \sin^2(20^\circ) - \sin^2(40^\circ) \\
= & \frac{1 - \cos(20^\circ)}{2} - \frac{1 - \cos(40^\circ)}{2} - \frac{1 - \cos(80^\circ)}{2} \\
= & -\frac{1}{2} - \frac{1}{2}(\cos(20^\circ) - \cos(40^\circ) - \cos(80^\circ)).\\
\end{align}
So to show the result, it suffices to show that $\cos(20^\circ) - \cos(40^\circ) - \cos(80^\circ) = 0$. Indeed,
\begin{align}
& \cos(80^\circ) = \cos(60^\circ + 20^\circ) = \frac{1}{2}\cos(20^\circ) - \frac{\sqrt{3}}{2}\sin(20^\circ), \\
& \cos(40^\circ) = \cos(60^\circ - 20^\circ) = \frac{1}{2}\cos(20^\circ) +
\frac{\sqrt{3}}{2}\sin(20^\circ).
\end{align}
Adding these two equations gives that
$\cos(20^\circ) = \cos(80^\circ) + \cos(40^\circ)$, thus the result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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First Order PDE, How to Deal With This Boundary Condition? 1. The problem statement
Find a solution of $$\frac{1}{x^2}\frac{\partial u(x,y)}{\partial
x}+\frac{1}{y^3}\frac{\partial u(x,y)}{\partial y}=0$$ Which satisfies
the condition $\frac{\partial u(x,y)}{\partial x}\big |_{y=0}=x^3$ for
all $x$.
2.The attempt at a solution
I get the following general solution for the PDE:
$$u(x,y)=f(\frac{x^3}{3}-\frac{y^4}{4}):=f(z)$$
For some function f which satisfies the BC. I'm not too sure how to deal with this type of boundary condition. But here is an attempt:
$\partial_x u=\frac{\text{d}f}{\text{d}z}\frac{\partial z}{\partial x}=\frac{\text{d}f}{\text{d}z} x^2$. Thus $\frac{\text{d}f}{\text{d}z} x^2=x^3\implies \frac{\text{d}f}{\text{d}z} =x\implies f(z)=zx$
At $y=0,~ z=\frac{x^3}{3}\implies x=(3z)^{\frac{1}{3}}$. So $f(z)=(3z)^{\frac{1}{3}}z$.
Changing variables back gives:
$u(x,y)=(3(\frac{x^3}{3}-\frac{y^4}{4}))^{\frac{1}{3}}(\frac{x^3}{3}-\frac{y^4}{4})=3^{\frac{1}{3}}(\frac{x^3}{3}-\frac{y^4}{4})^{\frac{4}{3}}$
This is a solution to the PDE but doesn't satisfy the given boundary condition, it is off by a factor 3/4 and i'm not sure why.
| If $u (x,y) = f \left(\frac{x^3}{3} - \frac{y^4}{4}\right)$, then the boundary condition gives us $f ' \left(\frac{x^3}{3}\right) = x$. Hence, we have
$$f ' (z) = \sqrt[3]{3z}$$
Integrating, we obtain
$$f (z) = \frac{3\sqrt[3]{3}}{4} \, z^{\frac{4}{3}}$$
and, thus,
$$u (x,y) = \frac{3\sqrt[3]{3}}{4} \left(\frac{x^3}{3} - \frac{y^4}{4}\right)^{\frac{4}{3}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Solve $\begin{cases}x\equiv-4\pmod {17}\\ x\equiv 3\pmod{23} \end{cases}$
Solve $$\begin{cases}x\equiv-4\pmod {17}\\
x\equiv 3\pmod{23}
\end{cases}$$
My attempt:
$$\gcd (17,23)=1$$
so using the Chinese remainder theorem
there is a solution modulo $17\times 23=391$
$$x=-4+17t\\
\Longrightarrow-4+17t\equiv 3\pmod{23}\\
\Longrightarrow -4+17t\equiv 3+23k\\
k,t\in \mathbb Z$$
I am stuck here
| Once you obtain the equivalence $-4 + 17t \equiv 3 \pmod{23}$, you can add $4$ to each side of the equivalence to obtain
$$17t \equiv 7 \pmod{23}$$
Since $\gcd(17, 23) = 1$, this equivalence has a solution. To solve it, we can apply the extended Euclidean algorithm to find the multiplicative inverse of $17$ modulo $23$, then multiply both sides of the equivalence $17t \equiv 7 \pmod{23}$ by that inverse.
\begin{align*}
23 & = 1 \cdot 17 + 6\\
17 & = 2 \cdot 6 + 5\\
6 & = 1 \cdot 5 + 1\\
5 & = 5 \cdot 1
\end{align*}
Now, we work backwards to solve for $1$ in terms of $17$ and $23$.
\begin{align*}
1 & = 6 - 5\\
& = 6 - (17 - 2 \cdot 6)\\
& = 3 \cdot 6 - 17\\
& = 3(23 - 17) - 17\\
& = 3 \cdot 23 - 4 \cdot 17
\end{align*}
Thus, $-4 \cdot 17 = 1 - 3 \cdot 23 \implies -4 \cdot 17 \equiv 1 \pmod{23}$. Multiplying both sides of the equivalence
$$17t \equiv 7 \pmod{23}$$
by $-4$ yields
\begin{align*}
-4 \cdot 17t & \equiv -4 \cdot 7 \pmod{23}\\
t & \equiv -28 \pmod{23}\\
t & \equiv -28 + 2 \cdot 23 \pmod{23}\\
t & \equiv 18 \pmod{23}
\end{align*}
Hence, $t = 18 + 23n$ for some integer $n$. Substituting this value for $t$ into the expression $x = -4 + 17t$ yields
\begin{align*}
x & = -4 + 17(18 + 23n)\\
& = -4 + 306 + 17 \cdot 23n\\
& = 302 + 391n
\end{align*}
Thus, $x \equiv 302 \pmod{391}$, the result Andre Nicolas obtained in the comments by making an astute observation.
Check: Observe that
\begin{align*}
302 & = 17 \cdot 18 - 4 \implies 302 \equiv -4 \pmod{17}\\
302 & = 13 \cdot 23 + 3 \implies 302 \equiv 3 \pmod{23}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $P$ is an integer polynomial with $P(1)=P(2)=0$, then some coefficient is less than $-1$
Let $P (x)$ be a polynomial with integer coefficients. It is known that the numbers $1$ and $2$ are its roots.
Prove that there exists a coefficient that is less than $-1$.
My work so far:
Let $P(x)=a_nx^n+...+ax+a_0$. $P(1)=P(2)=0$. Prove that exists $a_i<-1$.
Suppose $\forall a_i\ge-1$. Let $S$ be the sum of positive factors. If $P(1)=0$ then $a_n+..+a_1+a_0=0$, then number of negative factors equally $S$.
I need help here.
| We assume that $deg P(x) \geq 2$ and that $a_n \neq 0$. Since $P(1) = P(2) = 0$, we have $a_n2^n + a_{n - 1}2^{n-1} + .. + a_12 + a_0 = 0$ and $a_n + a_{n - 1} + .. + a_1 + a_0 = 0$. Together, we have $(a_{n - 1} + .. + a_1 + a_0)2^n = a_{n - 1}2^{n - 1} + .. + a_12 + a_0$. If we assume that $a_k \geq -1$ for $k \leq n - 1$, then $a_{n - 1}2^{n - 1} + .. + 2a_1 + a_0 > -2^n$. Therefore $(a_{n - 1} + .. + a_1 + a_0)2^n > -2^n$; moreover, $a_{n-1} + .. + a_1 + a_0 > -1$ which implies that $a_n < 1$. Since $a_n \neq 0$, we have $a_n \leq -1$. However, if $a_n = -1$, then $a_{n-1} + .. + a_1 + a_0 = 1$ with $a_k \geq -1$ for $k \leq n - 1$ (as well as $n - 1 \geq 1$ because of degree of $P(x)$) and $a_{n-1}2^{n-1} + .. + a_12 + a_0 = 2^n$ which leads to a contradiction, just observe the two cases for $n \geq 3$: (1) $a_{n-1} = 2$ and $a_0 = -1$; (2) $a_{n - 1} = 3$ and $a_{n-2} = a_{n-3} = -1$.
| {
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"url": "https://math.stackexchange.com/questions/1797990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum of $(ab+cd)(ac+bd)(ad+bc)$ Let $a,b,c,d\ge 0$ satisfy $a+b+c+d=4$. Find the maximum value of $(ab+cd)(ac+bd)(ad+bc)$.
When all of the variables are $1$, the value is $8$. Using the AM-GM inequality gives $$(ab+cd)(ac+bd)(ad+bc)\leq\left(\frac{ab+cd+ac+bd+ad+bc}{3}\right)^3.$$
Can we upper bound the right-hand side in terms of $a+b+c+d$?
| We have
$$ab+ac+ad+bc+bd+cd=\frac12\bigl((a+b+c+d)^2-(a^2+b^2+c^2+d^2)\bigr) $$
and for $a+b+c+d=4$ we have $a^2+b^2+c^2+d^2\ge4$ (why?) so that
$$ab+ac+ad+bc+bd+cd\le\frac{16-4}{2}= 6 $$
and you indeed obtain $\le 8$ for the original inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$. Problem :
Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$.
My approach :
Since :
$(a^3+b^3)=(a+b)^3-3ab(a+b)$
$\Rightarrow z^3+\frac{1}{z^3}=(z+\frac{1}{z})^3-3(z+\frac{1}{z})$
Now I don't know further about this problem whether or not this will help here. Please guide to solve this problem will be of great help, thanks.
| You have $(z+\frac{1}{z})^3-3(z+\frac{1}{z})=z^3+\frac{1}{z^3}$, and $\left|(z+\frac{1}{z})^3-3(z+\frac{1}{z})\right|=\left|z+\frac{1}{z}\right|\cdot\left|(z+\frac{1}{z})^2-3\right|$.
Hence if $\left|z+\frac{1}{z}\right|>2$, then $\left|z^3+\frac{1}{z^3}\right|>2\cdot(4-3)=2$. Contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the maximum and minimum values of $x-\sin2x+\frac{1}{3}\sin 3x$ in $[-\pi,\pi]$ Find the maximum and minimum values of $x-\sin2x+\frac{1}{3}\sin 3x$ in $[-\pi,\pi]$.
Let $f(x)=x-\sin2x+\frac{1}{3}\sin 3x$
$f'(x)=1-2\cos2x+\cos3x$
Put $f'(x)=0$
$1-2\cos2x+\cos3x=0$ gives $2\sin^2x-\cos2x+\cos3x=0$
I am stuck here.I cannot find the critical values and the maximum and minimum values of the function.Please help me.
| Continuing from $1-2\cos 2x+\cos 3x=0$:
$1-2\cos 2x+\cos 3x=0$
$1-2\left(2\cos ^2x-1\right)+\cos \left(x+2x\right)=0$
$1-2\left(2\cos ^2x-1\right)+\cos x\cos 2x-\sin x\sin 2x=0$
$1-2\left(2\cos ^2x-1\right)+\cos x\left(2\cos ^2x-1\right)-\sin x\left(2\sin x\cos x\right)=0$
$1-2\left(2\cos ^2x-1\right)+\cos x\left(2\cos ^2x-1\right)-2\sin ^2x\cos x=0$
$1-2\left(2\cos ^2x-1\right)+\cos x\left(2\cos ^2x-1\right)-2\left(1-\cos ^2x\right)\cos x=0$
$1-4\cos ^2x+2+2\cos ^3x-\cos x-2\cos x+2\cos ^3x=0$
$4\cos ^3x-4\cos ^2x-3\cos x+3=0$
$4\cos ^2x\left(\cos x-1\right)-3\left(\cos x-1\right)=0$
$\left(4\cos ^2x-3\right)\left(\cos x-1\right)=0$
$\left(\cos ^2x-\frac34\right)\left(\cos x-1\right)=0$
$\left(\cos x-\frac{\sqrt3}2\right)\left(\cos x+\frac{\sqrt3}2\right)\left(\cos x-1\right)=0$
$\cos x=\frac{\sqrt3}2$ or $-\frac{\sqrt3}2$ or $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximize system of linear equations Suppose you have the system
$$
\begin{bmatrix}
4 & 3\\
1 & 7\\
5 & 9\\
2 & 4\\
\end{bmatrix}
\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix}b_1\\b_2\\b_3\\b_4\end{bmatrix}
$$
How could one find scalars $x$ and $y$ such that $b_1+b_2+b_3+b_4$ is maximized? I can see the math behind it as I just took Theory of Linear Algebra, but am stuck on the question.
EDIT: it must be that $0\leq x\leq 1$, $0\leq y\leq 1$, and $x+y=1$.
| First note
$$
b_1 + b_2 + b_3 + b_4
= (4 + 1 + 5 + 2) x + (3 + 7 + 9 + 4)
= 12 x + 23 y
$$
then we can formulate it as linear program
$$
\begin{array}{rr}
\max & c^\top u \\
\text{w.r.t.} & A u = b \\
& B u \le d \\
& u \ge 0
\end{array}
$$
for cost vector $c = (12, 23)^\top$, vector of unknowns $u = (x, y)^\top$ and constraints
$$
A =
\begin{pmatrix}
1 & 1
\end{pmatrix}
\quad
b =
\begin{pmatrix}
1
\end{pmatrix}
\\
B =
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\quad
d =
\begin{pmatrix}
1 \\
1
\end{pmatrix}
$$
As a problem with only $2$ unknowns, it can be solved graphically:
(Large version)
The blue lines and half spaces show the constraints $u \ge 0$ and $B u \le d$. The red line is the constraint $A u = b$.
The yellow lines show the cost isolines, the lowest one is $c = 0$, in the middle $c = 10$ and above $c = 20$.
The set of feasible solutions $u$ consists of the red line clamped to the box $[0,1]^2$.
The maximum is determined by the highest possible cost isoline, thus $12 x + 23 y = c$ which intersects the point $(0, 1)^\top$, thus $12 \cdot 0 + 23 \cdot 1 = 23 = c$.
This gives the maximal solution $u = (0, 1)^\top$ with $c^\top u = 23$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Recurrence relation $a_r+6a_{r-1}+9a_{r-2}=3$, then find $a_{20}$
Consider the recurrence relation $a_r+6a_{r-1}+9a_{r-2}=3$, given that $a_0=0, a_1=1$. Let $a_{20}=x\times10^9$, then the value of $x$ is______ .
My attempt:
$a_r=3-6a_{r-1}-9a_{r-2}$
I calculated manually, I get $a_{20}=-6465079410=-6.46\times10^9$, so $x=-6.46$
Can you explain in formal way, please?
| Let $f(z)=\sum_{n=0}^\infty a_n z^n$. Multiplying both sides of the recurrence by $z^n$ and summing over $n$ yields
$$\sum_{n=2}^\infty a_n z^n +6\sum_{n=2}^\infty a_{n-1}z^n +9\sum_{n=2}^\infty a_{n-2}z^n = 3\sum_{n=2}^\infty z^n, $$
or
$$f(z)-(a_0+a_1z)+6z(f(z)-a_0) + 9z^2f(z)=\frac{3z^2}{1-z}. $$ Substituting $a_0$ and $a_1$ and isolating $f(z)$, we have $$f(z)(1+6z+9z^2) = \frac{3z^2}{1-z}+z, $$
and hence
$$f(z) = \frac z{(1+3z)^2}+\frac{3z^2}{(1-z)(1+3z)^2}. $$ Partial fraction decomposition yields
$$f(z) = \frac3{16}\left(\frac1{1-z}\right)-\frac5{48}\left(\frac1{1+3z}\right)-\frac1{12}\left(\frac1{(1+3z)^2}\right). $$
Manipulating the geometric series $$\sum_{n=0}^\infty z^n=\frac1{1-z} $$ we find that $$f(z) = \sum_{n=0}^\infty \frac3{16}z^n - \sum_{n=0}^\infty\frac5{48}(-3)^n z^n-\sum_{n=0}^\infty \frac1{12}(n+1)(-3)^nz^n.$$
Simplifying, we have
$$f(z) = \sum_{n=0}^\infty\left(\frac3{16}-\frac5{48}(-1)^n3^{n}-\frac1{12}(n+1)(-1)^n3^{n} \right)z^n, $$
and so we conclude that $$a_n=\frac3{16}-\frac5{48}(-1)^n3^{n}-\frac1{12}(n+1)(-1)^n3^{n}. $$
In particular,
$$a_{20} = \frac{3}{16}-\frac{5}{48} (-1)^{20} 3^{20}-\frac{21}{12} (-1)^{20} 3^{20}=-6465079410. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this equation using logarithms? I have to solve for all real values of $x$.
$(5+2\sqrt6)^{x^2-3}+(5-2\sqrt6)^{x^2-3}=10$
I tried to take $\log_{10}$ on both sides but could not do this.
How do I do this?Thanks for any hint or answer!!
| Hint.
note that $$(5-2\sqrt{6})=\frac{1}{5+2\sqrt{6}}$$
and use the substitution
$$
\left(5+2\sqrt{6}\right)^{x^2-3x}=y
$$
so the equation becomes:
$$
y+\frac{1}{y}=10
$$
that becomes a second degree equation (multiply by $y \ne 0$). Solve this equation and you can find the final solution (without logarithms).
$$
y^2-10y+1=0 \quad \Rightarrow \quad y=5\pm \sqrt{25-1}=5\pm 2\sqrt{6}
$$
so you have the two solutions:
$$
\left(5+2\sqrt{6}\right)^{x^2-3x}=5+ 2\sqrt{6} $$
$$\left(5+2\sqrt{6}\right)^{x^2-3x}=5- 2\sqrt{6}=\left(5+ 2\sqrt{6} \right)^{-1}
$$
so:
$$
x^2-3x= 1 \quad \lor \quad x^2-3x= -1
$$
Now, can you find $x$?
| {
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"url": "https://math.stackexchange.com/questions/1803586",
"timestamp": "2023-03-29T00:00:00",
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Limit of ($\sqrt{x^2+8x}-\sqrt{x^2+7x}$) as $x$ approaches infinity I've been stuck on this one problem for 3 days now, I don't know how to proceed. Any help would be appreciated.
The problem is asking for the $$\lim_{x\to\infty} (\sqrt{x^2+8x}-\sqrt{x^2+7x}) $$
Every time I attempt this problem, I can never get rid of infinity in the numerator without making the denominator zero. How can I do this problem?
| Another way to do it.
Considering that $x\to\infty$ $$\sqrt{x^2+8x}-\sqrt{x^2+7x}=x\left(\sqrt{1+\frac 8x}-\sqrt{1+\frac 7x} \right)$$ Now, using Taylor for small $y$ $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ make $y=\frac 8x$ in the first radical and $y=\frac 7x$ in the second radical to get $$\sqrt{x^2+8x}-\sqrt{x^2+7x}=x\left(\left(1+\frac{4}{x}-\frac{8}{x^2}+O\left(\frac{1}{x^3}\right) \right)-\left(1+\frac{7}{2 x}-\frac{49}{8 x^2}+O\left(\frac{1}{x^3}\right)\right)\right)$$ After simplification $$\sqrt{x^2+8x}-\sqrt{x^2+7x}=\frac{1}{2}-\frac{15}{8 x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and how it is approached.
Edit
Making the problem more general as Marty Cohen did in his answer, doing the same, we should get
$$\sqrt{x^2+ax+c}-\sqrt{x^2+bx+d}=\frac{a-b}{2}+\frac{-a^2+4 b+c^2-4 d}{8 x}+O\left(\frac{1}{x^2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
My solution: Consider
$$x^2 + y^2 + z^2 < 2(xy + yz + xz)$$
Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$
Hence
$$(x+y+z)^2-2(xy+yz+xz) < 2(xy + yz + xz)$$
$$ (x+y+z)^2 - 4(xy+yz+xz) < 0 $$
As $x,y,z$ are all greater than zero as a side length of a triangle can not be negative. And because $(x+y+z)^2 >0$ for all real $x,y,z$ therefore the whole expression is less than zero. $Q.E.D$
Am I correct? Or could I be more "rigorous" I am a highschool student and getting into proof so any tips would be appreciated as well :)
| Since $x,y,z$ are sides of a triangle, we have
$|x-y| < z $
Squaring both sides we get
$(x-y)^2=x^2 +y^2 -2xy < z^2$
Setting up similar inequalities and adding, we get the desired result.
Your proof doesn't work because
$(x+y+z)^2 \geq 0$ doesn't imply that
$(x+y+z)^2 -4(xy+yz+zx) <0$, even when $x,y,z>0$
A simple counter-example is $(x,y,z)=(1,1,5)$
Obviously, this triple doesn't form a triangle, but contradicts the validity of the claim that $(x+y+z)^2 -4(xy+yz+zx) <0$, when $x,y,z>0$, which is the flaw in the given argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $a^2-b^2+c^2$ Let $a$, $b$, and $c$ be real numbers such that $a − 7b + 8c = 4$ and $8a + 4b − c = 7$. What is the
value of $a^2-b^2+c^2$ ?
| Given $a + 8c= 7b + 4$ and $8a - c = -4b + 7$
Now Squaring both equation, We get
\begin{align}a^2 + 64c^2 + 16ac&=49b^2 + 16 + 56b\tag{1} \\
64a^2 + c^2 - 16ac&=16b^2 + 49 - 56b\tag{2}\end{align}
Summing these two equations gives
$$ 65a^2 + 65c^2 = 65b^2 + 65 \Rightarrow a^2 - b^2 + c^2 = 1 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How should I try to evaluate the integral $\int_a^b \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ I've tried to evaluate $\displaystyle\int_{-r}^r \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ on my own, but I have encountered a problem I cannot get around.
The indefinite integral $\sqrt{\frac{r^2}{r^2-x^2}} \sqrt{r^2-x^2} \tan ^{-1}\left(\frac{x}{\sqrt{r^2-x^2}}\right) + C$ is undefined at both limits of integration
(the fractions $\frac{r^2}{r^2-r^2} = \frac{r^2}0 = \text{indetermined}$ and $\frac{({-r})^2}{r^2-r^2} = \frac{r^2}0 = \text{indetermined}$)
so I really don't know what to do. Apparently, the value of this integral should be $\pi r.$ Any help would be appreciated.
| hint...The integrand is equivalent to $$\frac {r}{\sqrt{r^2-x^2}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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A Sine and Inverse Sine integral A demonstration of methods
While reviewing an old text book an integral containing sines and sine inverse was encountered, namely,
$$\int_{0}^{\pi/2} \int_{0}^{\pi/2} \sin(x) \, \sin^{-1}(\sin(x) \, \sin(y)) \, dx \, dy = \frac{\pi^{2}}{4} - \frac{\pi}{2}.$$
One can express $\sin^{-1}(z)$ as a series and integrate, but are there other methods nearly as efficient?
| As mickep has observed
\begin{equation*}
I_{1} = \int_0^{\pi/2}\int_0^{\pi/2}\sin(x)\arcsin(\sin(x)\sin(y))\,dxdy = \dfrac{1}{2} \int_{0}^{\pi/2}\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\dfrac{\cos^{2}x}{\sin x}\, dx.
\end{equation*}
We intend to finish the solution without using series expansion. Instead we will use complex integration and Cauchy integral theorem.
After integration by parts we have
\begin{gather*}
I_{1} = \dfrac{1}{2}\int_{0}^{\pi/2}\left(\dfrac{1}{\sin x}-\sin x\right)\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\, dx = \dfrac{1}{2}\left[\left(\ln\left(\tan\dfrac{x}{2}\right)+\cos x\right)\ln\left(\dfrac{1+\sin x}{1-\sin x}\right)\right]_{0}^{\pi/2}\\
- \int_{0}^{\pi/2}\left(\ln\left(\tan\dfrac{x}{2}\right)+\cos x\right)\dfrac{1}{\cos x}\, dx = - \underbrace{\int_{0}^{\pi/2}\dfrac{\ln\left(\tan\dfrac{x}{2}\right)}{\cos x}\, dx}_{I_{2}} - \dfrac{\pi}{2}.
\end{gather*}
Finally we will prove that $I_2 = -\dfrac{\pi^{2}}{4}.$ Since $\displaystyle \tan^{2}\dfrac{x}{2} = \dfrac{1-\cos x}{1+\cos x}$ we get
\begin{gather*}
I_{2}= \dfrac{1}{2}\int_{0}^{\pi/2}\dfrac{1}{\cos x}\ln\left(\dfrac{1-\cos x}{1+\cos x}\right)\, dx = \dfrac{1}{2}\int_{0}^{\pi/2}\dfrac{1}{\sin x}\ln\left(\dfrac{1-\sin x}{1+\sin x}\right)\, dx = \left[t = \tan\dfrac{x}{2}\right]\\
= \int_{0}^{1}\dfrac{1}{t}\ln\left(\dfrac{1-t}{1+t}\right)\, dt = \left[\ln(t)\ln\left(\dfrac{1-t}{1+t}\right)\right]_{0}^{1} - 2\int_{0}^{1}\dfrac{\ln(t)}{t^{2}-1}\, dt.
\end{gather*}
But
\begin{equation*}
\int_{0}^{1}\dfrac{\ln(t)}{t^{2}-1}\, dt = \int_{1}^{\infty}\dfrac{\ln(t)}{t^{2}-1}\, dt .
\end{equation*}
Thus $
\displaystyle I_{2} = -\int_{0}^{\infty}\dfrac{\ln(t)}{t^{2}-1}\, dt.$ If $\log $ is the principal branch of the logarithm then the function $ f(z) = \dfrac{\log(z)}{z^{2}-1}$ has a removable singularity at $z=1$. If
we integrate $f(z)$ ''around quadrant 1'' and use Cauchy integral theorem we get
\begin{equation*}
\int_{0}^{\infty}\dfrac{\ln(x)}{x^{2}-1}\, dx - \int_{0}^{\infty}\dfrac{\ln(|iy|)+i\pi/2}{(iy)^{2}-1}i\, dy = 0.
\end{equation*}
Consequently
\begin{equation*}
-I_{2} + i\int_{0}^{\infty}\dfrac{\ln(y)}{y^{2}+1}\, dy -\dfrac{\pi}{2}\underbrace{\int_{0}^{\infty}\dfrac{1}{y^{2}+1}\, dy}_{ = \pi/2} = 0.\tag{1}
\end{equation*}
From the real part in (1) we finally get $I_2 = -\dfrac{\pi^2}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a_{n+1} \leq \left(1-\frac{2}{n+1}\right)a_n + b\left(\frac{2}{n+1}\right)^2$ then $a_n \leq \frac{4b}{n+1}$ A sequence $(a_n)$ satisfy $a_{n+1} \leq (1-\gamma_n)a_n + \frac{\beta R^2}{2}\gamma_n^2$ where $\gamma_n = \frac{2}{n+1}$, $\beta$ and $R$ are constant.
How to verify that
$$a_n \leq \frac{2\beta R^2}{n+1},$$ note that it is easy to show $a_2\leq\frac{\beta R^2}{2}$
| It is easy to show that $a_2 \leq \frac{\beta R^2}{2}$
We use induction to prove the inequality. Suppose that for $i = 2\ldots n$, we have $a_i \leq \frac{2\beta R^2}{i+1}$, then
$$a_{n+1}\leq (1-\gamma_n)a_n + \frac{\beta R^2}{2}\gamma_n^2 = \frac{n-1}{n+1}a_n+\frac{2\beta R^2}{(n+1)^2} \leq \frac{2(n-1)\beta R^2}{(n+1)^2}+\frac{2\beta R^2}{(n+1)^2}=\frac{2n\beta R^2}{(n+1)^2}<\frac{2\beta R^2}{n+2}$$
The last inequality simply based on $\frac{n}{(n+1)^2}\leq \frac{1}{n+2}\Leftrightarrow n^2+2n \leq n^2 + 2n + 1$
So we can conclude the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $ord_x(2) = 20$ Given that the (multiplicative) order of $2$ mod $x$ is $20$, how can I work out what $x$ is?
| You need $$2^k\cdot2^{20-k}=1$$ for $k=1,2,...,20$. In other words $2^k$ has as inverse $2^{20-k}$ Since $2$ is a primitive root modulo $11$ we examine if this is verified for $x=11$. We have in the field $\mathbb F_{11}$ $$\begin{cases}2^1=2\\2^2=4\\2^3=8\\2^4=5\\2^5=10\\2^6=9\\2^7=7\\2^8=3\\2^9=6\\2^{10}=1\end{cases}$$ and we can see the condition $2^k\cdot2^{20-k}=1$ is verified. So $\color{red}{x=11}$
Is there another solution? I will come back (some trouble to write in English and I could less the first answer as many times).
NOTE.-This is an answer to the question as it was formulated before it was changed: this question was $2^{20}\equiv 1\pmod x$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the summation of this infinite series: $\sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!}(1 - \frac{2}{k})$ I've been trying to figure out the following sum for a while now:
$$\sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!}\left(1 - \frac{2}{k}\right)$$
I'm pretty sure that this doesn't evaluate to $0$.
As $k$ increases the term tends to $0$, but the first few terms add up to give a non-zero number.
I'm just having trouble figuring out how to find that number. Any help would be appreciated.
| Let's try splitting up the series (since if it converges to a finite number, it will be absolutely convergent, since all terms are positive).
$$\sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!}(1 - \frac{2}{k}) = \sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!} - 2 \sum_{k=1}^{\infty} \frac{1}{(k+1)(k)(k-1)!} \\ = \sum_{k=1}^{\infty} \frac{k}{(k+1)(k)(k-1)!} - 2 \sum_{k=1}^{\infty} \frac{1}{(k+1)!} \\ = \sum_{k=1}^{\infty} \frac{k}{(k+1)!} - 2 \left[\sum_{k=0}^{\infty} \frac{1}{(k+2)!} + \frac{1}{0!} + \frac{1}{1!} -2 \right] = \\ \sum_{k=0}^{\infty} \frac{k+1}{(k+2)!} -2\left[ \sum_{k=0}^{\infty}\frac{1^k}{k!} -2 \right] = \sum_{k=0}^{\infty}\frac{k}{(k+2)!} + \sum_{k=0}^{\infty} \frac{1}{(k+2)!} - 2[e^1 -2]$$
Using the same technique as before:
$$ = \sum_{k=0}^{\infty}\frac{k}{(k+2)!} + [e^1 -2] - 2[e^1 -2] \\= \sum_{k=1}^{\infty} \frac{ k }{(k+2)!} -[e^1 -2] \\ = \sum_{k=1}^{\infty} \frac{ k }{(k+2)!} + \sum_{k=0}^{\infty} \frac{2}{(k+2)!} - \sum_{k=0}^{\infty} \frac{2}{(k+2)!} -[e^1 -2] \\ = \sum_{k=0}^{\infty} \frac{k+2}{(k+2)!} - 2 \sum_{k=0}^{\infty} \frac{1}{(k+2)!} - [e^1 -2]$$
Using a second time our computation that $\sum_{k=0}^{\infty} \frac{1}{(k+2)!} = e^1 -2$ we get $$= \sum_{k=0}^{\infty} \frac{k+2}{(k+2)!} - 3[e^1 -2] \\ = \sum_{k=0}^{\infty} \frac{1}{(k+1)!} - 3[e^1 -2] \\ = \sum_{k=0}^{\infty}\frac{1}{(k+1)!} + 1 -1 -3[e^1-2] \\ = \sum_{k=0}^{\infty} \frac{1}{k!} -1 - 3[e -2] \\ = e-1 -3e +6 \\ \\= -2e+5 $$
| {
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Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot \sin^2 xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot (1-\cos^2 x)dx$$
$$I_{n} =I_{n-1}-\int_{0}^{\frac{\pi}{2}}\cos x\cdot \sin^{2n-2}\cdot \cos xdx$$
Now Using Integration by parts, We get $$I_{n} = I_{n-1}-\frac{I_{n}}{2n-1}\Rightarrow I_{n} = \frac{2n-1}{2n}I_{n-1}$$
Now Using Recursively, We get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}I_{n-2} =\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}I_{n-3}$$
So we get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}\cdot \frac{2n-7}{2n-6}\cdot \cdot \cdot \cdot \cdot \cdot \cdot\cdot \frac{3}{2}I_{0}$$
and we get $\displaystyle I_{0} = \frac{\pi}{2}$
So we get $$I_{n} = \frac{(2n)!}{4^n\cdot n!\cdot n!}\cdot \frac{\pi}{2}$$
Now I did not understand How can I calculate value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Help Required, Thanks.
| Just completing Olivier's answer:
$$ I=\int_{0}^{\pi/2}\frac{4\,dx}{4-\sin^2 x}=\int_{0}^{\pi/2}\frac{4\,dx}{4-\cos^2 x}=\int_{0}^{+\infty}\frac{dt}{1+t^2}\cdot\frac{4}{4-\frac{1}{1+t^2}}$$
gives:
$$ S = \sum_{n\geq 0}\frac{1}{16^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{+\infty}\frac{4}{4t^2+3} = \color{red}{\frac{2}{\sqrt{3}}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How does $\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$ relate to $\sqrt{x^2+y^2+z^2}$? Another possible 'mean' for three positive real numbers $x,y,z$ is made of pairwise quadratic means:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}$$
By QM-AM inequality it is greater than or equal to arithmetic mean of the three numbers:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \frac{x+y+z}{3}$$
Now what is its relationship to the quadratic mean of the three numbers:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \text{ ? } \sqrt{\frac{x^2+y^2+z^2}{3}}$$
Using the obvious inequality $x^2+y^2+z^2 \geq x^2+y^2$, we obtain the following relationship:
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \sqrt{\frac{2}{3}} \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} $$
Can we make this bound any tighter? How?
| If $\sqrt{x^2+y^2}=c$, $\sqrt{x^2+z^2}=b$ and $\sqrt{y^2+z^2}=a$ so we can use
$\sqrt{3(a^2+b^2+c^2)}\geq a+b+c$.
| {
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Extending the ordered sequence of 'three-number means' beyond AM, GM and HM I want to create an ordered sequence of various 'three-number means' with as many different elements in it as possible.
So far I've got $12$ ($8$ unusual ones are highlighted):
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{ \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}} \geq $$
$$\geq \color{blue}{\frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}} } \geq \frac{x+y+z}{3} \geq $$
$$ \geq \color{blue}{ \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} } \geq \color{blue}{\sqrt{\frac{xy+yz+zx}{3}}} \geq $$
$$\geq \color{blue}{\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{3}} \geq \color{blue}{\sqrt{\frac{x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}}{3}}} \geq $$
$$ \geq \sqrt[3]{xyz} \geq \color{blue}{ \frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}} \geq $$
$$ \geq\color{blue}{ 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}}} \geq \frac{3xyz}{xy+yz+zx}$$
All the inequalities here are proven (note that $x,y,z>0$).
The rules are as follows - only explicit expressions allowed (no AGM for example) and these expressions should contain only addition, multiplication, division, squares, cubes, square and cube roots.
For any new mean we need to find a place between a pair of existing means, so we can keep our sequence precisely ordered.
I partially placed several other means, for example:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \\ \geq \color{blue}{ \frac{\sqrt[6]{(x^2+y^2)(y^2+z^2)(z^2+x^2)}}{\sqrt{2}}} \geq \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2}$$
$$\sqrt{\frac{xy+yz+zx}{3}} \geq \color{blue}{\frac{xy+yz+zx}{x+y+z}} \geq \sqrt[3]{xyz}$$
But I can't place them exactly among other means so far.
| Suppose $f(x,y,z)$ is a function that is:
*
*symmetric, i.e. $f(x,y,z)=f(y,x,z)=f(x,z,y)$ and so on,
*positive, i.e. $f(x,y,z)\ge0$ for all $x,y,z\ge0$, and
*has power-law scaling with exponent $p$, that is, $f(ax,ay,az)=a^pf(x,y,z)$ for all $a\ge0$.
Suppose $g(x,y,z)$ is a similar function but with exponent $q\ne p$. Then
$$m(x,y,z)=\left(\frac{f(x,y,z)\,g(1,1,1)}{f(1,1,1)\,g(x,y,z)}\right)^{q/p}$$
is a "three-number mean".
(Edit: The OP hasn't provided any criteria for what counts as a "mean", but for one thing we probably want the convex hull property $m(x,y,z)\in[\min(x,y,z),\max(x,y,z)]$, which the above function need not satisfy in general. So it probably shouldn't be called a mean until we figure out additional conditions to impose on $f$ and $g$ to guarantee this.)
If $f(x,y,z)$ is a polynomial, it must be a linear combination of monomial symmetric polynomials of degree $p$, and for positivity it is sufficient (though not necessary) that the linear combination is convex.
Almost all of the means in your list can be constructed this way, with $f$ and $g$ being convex combinations of monomial symmetric polynomials either in $x,y,z$ or in $\sqrt x,\sqrt y,\sqrt z$. The sole exception is the numerator of the second example, $\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$, which is not a polynomial but is still symmetric, positive, and scales with exponent $p=1$.
In particular, suppose we take $p=2$ and $q=0$. The two monomial symmetric polynomials of degree $2$ are $x^2+y^2+z^2$ and $xy+yz+zx$, so $f(x,y,z)$ is a multiple of $\alpha(x^2+y^2+z^2) + (1-\alpha)(xy+yz+zx)$ for some $\alpha\in[0,1]$. Therefore our mean is
$$m(x,y,z) = \sqrt{\frac{\alpha(x^2+y^2+z^2) + (1-\alpha)(xy+yz+zx)}3}.$$
It includes the following special cases which exist in your sequence:
$$\begin{align}
\alpha &= 1 \implies & m(x,y,z) &= \sqrt{\frac{x^2+y^2+z^2}3}, \\
\alpha &= \tfrac12 \implies & m(x,y,z) &= \frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2\sqrt3}, \\
\alpha &= \tfrac13 \implies & m(x,y,z) &= \frac{x+y+z}3, \\
\alpha &= 0 \implies & m(x,y,z) &= \sqrt{\frac{xy+yz+zx}3}. \\
\end{align}$$
In this way, using different $p$ and $q$, one can construct arbitrarily many continuous families of "three-number means".
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int\frac{\sin x}{\sqrt{1-\sin x}}dx=?$ Calculate this integral $\displaystyle\int\dfrac{\sin x}{\sqrt{1-\sin x}}dx=?$
Effort;
$1-\sin x=t^2\Rightarrow \sin x=1-t^2\Rightarrow \cos x=\sqrt{2t^2-t^4}$
$1-\sin x=t^2\Rightarrow-\cos x dx=2tdt\Rightarrow dx=\frac{2t}{\sqrt{t^4-2t^2}}dt$
$\displaystyle\int\frac{1-t^2}{t}\cdot\frac{2t}{\sqrt{t^4-2t^2}}dt=2\int\frac{1-t^2}{\sqrt{t^4-2t^2}}dt$
$\ = 2\displaystyle\int\frac{1}{\sqrt{t^4-2t^2}}dt-2\displaystyle\int\frac{t}{\sqrt{t^2-2t}}dt$
$\ = 2\displaystyle\int t^{-1}(t^2-2)^{-\frac{1}{2}}dt-2\displaystyle\int t(t^2-2t)^{-\frac{1}{2}}dt$
But after that I don't know how to continue.
| HINT:
$$\dfrac{\sin x}{\sqrt{1-\sin x}}=-\sqrt{1-\sin x}+\dfrac1{\sqrt{1-\sin x}}$$
$$1-\sin x=1-\cos\left(\dfrac\pi2-x\right)=2\sin^2\left(\dfrac\pi4-\dfrac x2\right)$$
Now for real $a,$ $$\sqrt{a^2}=|a|=\begin{cases}+a &\mbox{if } a\ge0 \\-a & \mbox{if } a<0 \end{cases}$$
| {
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Showing that $\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$ Integrate
$$I=\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$$
Substitution $x=\sqrt{\tan(u)}\rightarrow dx={\sec^2(u)\over 2\sqrt{\tan(u)}}du$
$x=1\rightarrow u={\pi\over 4}$
$x=0\rightarrow u=0$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{1-\tan^2(u)}\over 1+\tan^2(u)}\cdot {\sec^2(u)\over \sqrt{\tan(u)}}du$$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{1-\tan^2(u)\over \tan(u)}}du$$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{\cot(u)-\tan(u)}}du$$
Recall
$$\cot(u)-\tan(u)={\cos^2(u)-\sin^2(u)\over \sin(u)\cos(u)}=2\cot(2u)$$
Substitute back into I
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{2\cot(2u)}}du$$
$$I={\sqrt2\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{\cot(2u)}}du$$
Well I know that $$\int{\cos(2u)\over\sin(2u)}du={1\over 2}\ln(\sin(2u))+C$$ but
$$\int\sqrt{{\cos(2u)\over\sin(2u)}}du$$ I have not idea, so can anyone please give a hand? Thank.
| On the path of Achille Hui,
$\displaystyle J=\int_0^{\infty} \dfrac{x^2}{x^4+4}dx$
Perform the change of variable $y=\dfrac{x}{\sqrt{2}}$,
$\displaystyle J=\dfrac{\sqrt{2}}{2}\int_0^{\infty} \dfrac{x^2}{x^4+1}dx$
1)
Perform the change of variable $y=x^4$,
$\begin{align} J&=\dfrac{\sqrt{2}}{8}\int_0^{\infty} \dfrac{x^{-\tfrac{1}{4}}}{1+x}dx\\
&=\dfrac{\sqrt{2}}{8}\text{B}\left(\dfrac{3}{4},\dfrac{1}{4}\right)\\
&=\dfrac{\sqrt{2}}{8}\dfrac{\Gamma\left(\dfrac{3}{4}\right)\Gamma\left(\dfrac{1}{4}\right)}{\Gamma\left(1\right)}\\
&=\dfrac{\sqrt{2}}{8}\dfrac{\pi}{\sin\left(\tfrac{\pi}{4}\right)}\\
&=\dfrac{\sqrt{2}}{8}\dfrac{\pi}{\tfrac{\sqrt{2}}{2}}\\
&=\boxed{\dfrac{\pi}{4}}
\end{align}$
2) alternatively
$\displaystyle J=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty} \dfrac{x^2}{1+x^4}dx$
Perform the change of variable $y=\dfrac{1}{x}$,
$\displaystyle J=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty} \dfrac{1}{1+x^4}dx$
therefore,
$\displaystyle 2J=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty}\dfrac{1+\tfrac{1}{x^2}}{x^2+\tfrac{1}{x^2}}dx$
Observe that,
$x^2 +\dfrac{1}{x^2}=\left(x-\dfrac{1}{x}\right)^2+2$
Perform the change of variable $y=x-\dfrac{1}{x}$,
$\begin{align} 2J&=\dfrac{\sqrt{2}}{4}\int_{-\infty}^{+\infty} \dfrac{1}{x^2+2}dx\\
&=\dfrac{\sqrt{2}}{4}\left[\frac{\arctan\left(\frac{x}{\sqrt{2}}\right)}{\sqrt{2}}\right]_{-\infty}^{+\infty}\\
&=\dfrac{\pi}{2}
\end{align}$
therefore,
$\boxed{J=\dfrac{\pi}{4}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x$ and $y$ are positive numbers less than $20$ for which $x+y+xy=76$, what is $x+y$? What is a simple way to solve this problem? I can do it by trying $x$ and $y$, starting from $1$. That does not look like the best way.
If $x$ and $y$ are positive numbers less than $20$ for which $x+y+xy=76$, what is $x+y$?
| \begin{align}xy+x+y&=76\\
x(y+1)+y&=76\\
x(y+1)+y+1&=77\\
(x+1)(y+1)&=77
\end{align}
As $1,7,11,77$ are the only positive divisors of $77$, $(x,y)$ should be $(6,10)$ or $(10,6)$. Thus, $x+y=16$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving by induction that $\sum\limits_{i=1}^n\frac{1}{n+i}=\sum\limits_{i=1}^n\left(\frac1{2i-1}-\frac1{2i}\right)$ I have a homework problem to prove the following via induction:
$$\sum_{i=1}^n \frac{1}{n+i} = \sum_{i=1}^n \left(\frac{1}{2i-1} - \frac{1}{2i}\right) $$
The base case is true.
So far I've done the following:
$$\sum_{i=1}^{k+1} \frac{1}{(k+1)+i}$$
$$a_{k+1} = \frac{1}{(k+1)+(k+1)} = \frac{1}{2k +2} $$
$$s_k + a_{k+1} = \frac{1}{2k +2} + \sum_{i=1}^k (\frac{1}{2k-1} - \frac{1}{2k}) $$
$$s_{k+1} = \sum_{i=1}^{k+1} (\frac{1}{2(k+1)-1} - \frac{1}{2(k+1)}) $$
$$ \frac{1}{2k +2} + \sum_{i=1}^k (\frac{1}{2k-1} - \frac{1}{2k}) = \sum_{i=1}^{k+1} (\frac{1}{2(k+1)-1} - \frac{1}{2(k+1)}) $$
$$ \frac{1}{2k +2} + \sum_{i=1}^k (\frac{1}{2k-1} - \frac{1}{2k}) = \sum_{i=1}^{k+1} (\frac{1}{2k-1} - \frac{1}{2k+2)}) $$
I'm having a hard time figuring out how to proceed from here. Anyone have any hints or tips for search terms? Or corrections to what I've started above?
| Write the left hand side as $f(n)$.
Note that $$f(n+1)-f(n)=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$$
That's the key step - that $f(n+1)$ adds two terms to the right of the sum, but subtracts one term from the left.
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the values of $a_0,a_1,...,a_{10}$ if $\cos^{10} {\theta}=\sum_{k=0}^{10}a_k\cos {k\theta }$? I was thinking of doing the following:
Let $A=a_0+a_1\cos {\theta }+a_2\cos {2\theta }+...+a_{10}\cos {10\theta }$
and
$B=a_1\sin {\theta }+a_2\sin {2\theta }+...+a_{10}\sin {10\theta }$
Then, $S=A+iB=a_0+a_1(\cos {\theta}+i\sin {\theta})+a_2(\cos {2\theta}+i\sin {2\theta})+...+a_{10}(\cos {10\theta}+i\sin {10\theta})$
$=a_0+a_1e^{i\theta}+a_2e^{2i\theta}+...+a_{10}e^{10i\theta}$
If I could evaluate $S$ then my required sum would be $\Re (S)$.
| Let $x=\cos {\theta}+i\sin {\theta}$ whence, $x+\frac{1}{x}=2\cos {\theta}$ and $x^n+\frac{1}{x^n}=2\cos {n\theta}$
$\therefore (2\cos {\theta})^{10}=(x+\frac{1}{x})^{10}$
$=(x^{10}+\frac{1}{x^{10}})+10(x^8+\frac{1}{x^8})+45(x^6+\frac{1}{x^6})+120(x^4+\frac{1}{x^4})+210(x^2+\frac{1}{x^2})+252$
$=2\cos {10\theta}+10\times 2\cos {8\theta}+45\times 2\cos {6\theta}+120\times 2\cos {4\theta}+210\times 2\cos {2\theta}+2\times 126$
$\Rightarrow \cos^{10}{\theta}=\frac{1}{2^9}\left [\cos {10\theta}+10\cos {8\theta}+45\cos {6\theta}+120\cos {4\theta}+210\cos {2\theta}+126 \right]$
So, by the problem, $a_0=\frac{126}{2^9},a_2=\frac{210}{2^9},a_4=\frac{120}{2^9},a_6=\frac{45}{2^9},a_8=\frac{10}{2^9},a_{10}=\frac{1}{2^9}$
and $a_1=a_3=a_5=a_7=a_9=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Show that: $97|2^{48}-1$ Show that: $97|2^{48}-1$
My work:
$$\begin{align}
2^{96}&\equiv{1}\pmod{97}\\
\implies (2^{48}-1)(2^{48}+1)&=97k\\
\implies (2^{24}-1)(2^{24}+1)(2^{48}+1) &=97k\\
\implies (2^{12}-1)(2^{12}+1)(2^{24}+1)(2^{48}+1)&=97k\\
\implies (2^6-1)(2^6+1)(2^{12}+1)(2^{24}+1)(2^{48}+1) &=97k
\end{align}$$
None of the terms on LHS seem to be divisible by 97!!
Direct calculation shows that: $97\mid 2^{24}+1$ , but how to find it mathematically (of course not using calculator)?
| $2^{48} - 1 = ( (2^{24})^2 - 1) = (2^{24} - 1)(2^{24}+1) = ( (2^{12})^2 - 1)(2^{24}+1) = (2^{12} - 1)(2^{12} + 1)(2^{24} + 1). $
Now $2^6 = 64 $ thus $2^{12} = 64^2 = 4096 = 22 \mod 97$.
Therefore $2^{24} = (2^{12})^2 = 22^2 = 484 = 96 \mod 97$.
Hence $2^{24} + 1 = 96 +1 = 97 = 0 \mod 97$.
Therefore $97$ divides $2^{24} + 1$ so $97$ divides $2^{48} - 1$ as well.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Integration by Parts on $\sqrt{1+x^2}dx$ I have been asked to show that $$\int\sqrt{1+x^2}dx=\frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\int\frac{1}{{\sqrt{1+x^2}}}dx$$
I'm aware of how to do this with trig substitution, but the question is specifically regarding integration by parts, and the only examples I can find keep using trig subs.
So far I have integrated twice. The first time was with $u=\sqrt{1+x^2},v'=1$. Resulting in the following:
$$\int\sqrt{1+x^2}dx=x\sqrt{1+x^2}-\int\frac{x}{\sqrt{1+x^2}}dx$$
The first term is almost correct. Integrating this new integral, with $u=\frac{x}{\sqrt{1+x^2}}, v'=1$. Gave me the following:
$$\int\sqrt{1+x^2}dx=x\sqrt{1+x^2}-\frac{x^2}{\sqrt{1+x^2}}+\frac{1}{2}\int\frac{x}{\sqrt{1+x^2}}dx$$
From here I'm pretty much stuck. I can't seem to figure out how I am supposed to get rid of the middle term. The final integral can be put into the right form if I divide by $x$, but then the first term loses it's $x$ scalar and I still can't drop the middle term.
I feel as though I've made a mistake on one of the integrations, but I've been looking over it for that long that the whole thing is blurry.
| Alternative method, but not really so different from the other answer:
\begin{align}
\int\sqrt{1+x^2}\,dx
&=
\int\frac{1+x^2}{\sqrt{1+x^2}}\,dx
\\[6px]
&=\int\frac{1}{\sqrt{1+x^2}}\,dx+
\int x\frac{x}{\sqrt{1+x^2}}\,dx
\qquad \biggl(u=x, v'=\frac{x}{\sqrt{1+x^2}}\biggr)
\\[6px]
&=\int\frac{1}{\sqrt{1+x^2}}\,dx+x\sqrt{1+x^2}-\int\sqrt{1+x^2}\,dx
\end{align}
Therefore
$$
2\int\sqrt{1+x^2}\,dx
=
x\sqrt{1+x^2}+\int\frac{1}{\sqrt{1+x^2}}\,dx
$$
| {
"language": "en",
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"answer_count": 2,
"answer_id": 1
} |
3D Perspective projection I have this following question to answer, however I am not sure how I should combine my calculation into one final answer.
Suppose the Centre of Projection in a viewing space is at an offset
$(0, 0, -5)$ from $(0,0,0)$, and the view plane is the $UV$ plane containing
$c$. Find the transform matrix for the perspective projection, and give
the projected Word Coordinate point $(10,-20,-10)$ on the view plane.
So this is the transformation matrix: with $1/d=0.2$. ($1/5$)
$$
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1/d & 1
\end{pmatrix}
$$
Because the view plane is placed at $z=0$ we use the following similar triangles:
$$
x' = x/(z/d+1), \quad
y' = y/(z/d+1), \quad
z'=z/(z/d+1), \quad
w=z/d
$$
My question is, what values do I use for the $x$,$y$,$z$. If using the project world coordinates $x'$, $y'$, $z'$ will be undefined due to the $0$. Once I have the values for $x'$, $y'$, $z'$ do I multiple them by the transformation matrix?
Thanks for any help
| Let me try based on the theories described in the arXiv.org article Unified Framework of Elementary Geometric Transformation Representation.
A so-called perspective projection in this specific problem can be realized by a central projection defined in definition 3.2 (pages 8-9) and formulated in equation (3.1) or (see definitions and formulations in line No. 1 of Table 1 in page 9 of the arXiv.org article Unified Framework of Elementary Geometric Transformation Representation.
As we know from the problem, the projection center $C=(0,0,-5)$ and view plane is $xOy$ axis plane of the 3D world coordinate system. So for the central projection in homogenous matrix form to be found, the homogenous coordinate of its projection center $(s)=(0,0,-5,1)^T$ in column vector convention, and that of its interface plane $z=0$ which is written into $(\pi)=(0,0,1,0)^T$, are all readily available. So the homogeneous matrix of the desired central projection per definition 3.2 (pages 8-9) and equation (3.1) (or definition and formulation in line No. 1 of Table 1 in page 9) of the arXiv.org article Unified Framework of Elementary Geometric Transformation Representation can be obtained easily:
$$T=\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right]
-\dfrac{\left[
\begin{array}{c}
0 \\
0 \\
-5 \\
1 \\
\end{array}
\right]\cdot \left[
\begin{array}{cccc}
0 & 0 & 1 & 0 \\
\end{array}
\right]}{\left[
\begin{array}{cccc}
0 & 0 & -5 & 1 \\
\end{array}
\right]\cdot \left[
\begin{array}{c}
0 \\
0 \\
1 \\
0 \\
\end{array}
\right]}=\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & \dfrac{1}{5} & 1 \\
\end{array}
\right]$$
It is easy to apply the obtained central projection to the given $(10,-20,-10)$ of which a column vector homogenous coordinate is :
$$X= \left[
\begin{array}{c}
10 \\
-20 \\
-10 \\
1 \\
\end{array}
\right]$$
then
$$X'=\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & \dfrac{1}{5} & 1 \\
\end{array}
\right]\cdot \left[
\begin{array}{c}
10 \\
-20 \\
-10 \\
1 \\
\end{array}
\right]=\left[
\begin{array}{c}
10 \\
-20 \\
0 \\
\color{red}{-1} \\
\end{array}
\right]\approx \color{red}{({\text{nonzeor scalar } -1})} \left[
\begin{array}{c}
-10 \\
20 \\
0 \\
\color{red}1 \\
\end{array}
\right]$$
So the final projected point has the 3D Euclidean coordinate of $\color{red}{(-10,20,0)}$.
Similarly for your point $(10,-10,10)$ you will obtain
$$\left[
\begin{array}{c}
10 \\
-10 \\
0 \\
\color{red}3 \\
\end{array}
\right]\approx \color{red}{({\text{nonzeor scalar } 3})} \left[
\begin{array}{c}
\dfrac{10}{3} \\
-\dfrac{10}{3} \\
0 \\
\color{red}1 \\
\end{array}
\right]$$
the projected 3D Euclidean coordinate of which should be $\left(\dfrac{10}{3}, -\dfrac{10}{3},
0\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I thought maybe it's better to consider odd and even cases for $n$ but there's no unified rule here, because for example: $8^2\equiv{4}\pmod{10}$ and $8^4\equiv{6}\pmod{10}$. Any ideas??
| Our sum is odd, so all we need to do is to compute it modulo $5$.
Note that the congruence class of $k^n$ modulo $5$ is the same as the congruence class of $k^{n+4}$ modulo $5$. This is obvious if $k$ is divisible by $5$. And if $k$ is not divisible by $5$ then $k^4\equiv 1\pmod{5}$.
So to find the last digit for any $n$, it is enough to know the last digit for $n=1,2,3,4$. Now you need to compute the last digit for $n=1,2,3,4$, and you will know the situation for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Asymptotes of $(x(t),y(t)) = \bigg(\frac{1+t^2}{2+t^3}, \frac{t}{2+t^3}\bigg)$, collinear points, ... Consider the curve:
\begin{equation}
(x(t),y(t)) = \bigg(\frac{1+t^2}{2+t^3}, \frac{t}{2+t^3}\bigg)
\end{equation}
Question 1: What are his asymptotes?
Answer: In projective space: $[(2+t^3,1+t^2,t)]$. The intersection with the line at infinity ($x_0=0$), is $t= -(2)^{\tfrac{1}{3}}$. But how to check that this gives an asymptote and not a parabolic direction?
Question 2: Give a necessary condition such that $t_1,t_2,t_3$ gives collinear points.
Answer: I put it in a determinant, but then it's difficult to find a condition. Are there better ways to do this?
Question 3: Show that the curve has a 'turning point' (point with multiplicity two where the two tangent lines are the same)
Answer: I have no idea.
| To the third question: Entering into M2
R=QQ[x,y]
S=QQ[t]
KS=frac S
f=map(KS,R,{(1+t^2)/(2+t^3),t/(2+t^3)})
ker f
we see $2x^3-6x y^2+5y^3-x^2-2x y+2y^2+y=0$ is the implicit equation. The partial derivatives
$6x^2-6y^2-2x-2y,-12xy+15y^2-2x+4y+1$
have a common zero at $(x,y)=(2/3,1/3)$ which is on the curve. This is your singular point ($t=1$). Moving this point to the origin; the tangent cone is the lowest degree term of $2(x+2/3)^3-6(x+2/3) (y+1/3)^2+5(y+1/3)^3-(x+2/3)^2-2(x+2/3) (y+1/3)+2(y+1/3)^2+(y+1/3)=5 y^3-6 x y^2+3y^2-6x y+2x^3+3x^2$ which is $3(y^2-2xy+x^2)=3(y-x)^2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
given $f(x)$ is continuous and $\int_0^{x^2}f(t)dt = x+sin\frac{\pi}{2}x$,solve $f(1)=?$ given $f(x)$ is continuous and $\int_0^{x^2}f(t)dt = x+\sin\frac{\pi}{2}x$,solve $f(1)=?$
we know
$$[\int_0^{x^2}f(t)dt]^\prime= f(x^2) \cdot2x$$
and
$$[x+\sin\frac{\pi}{2}x]^\prime = 1+ \frac{\pi}{2}\cos\frac{\pi}{2}x$$
therefore
$$f(x^2) \cdot2x = 1+ \frac{\pi}{2}\cos\frac{\pi}{2}x$$
substitute $x$ by $x=1$, we have:
$$f(1) = \frac{1}{2}$$
however by substitute $x$ by $x = -1$ we have:
$$f(1) = -\frac{1}{2}$$
more strangely , if I substitute $x$ by $x = 0 $ we have:
$$ f(x^2)\cdot 2x = 0$$ while
$$ 1 + \frac{\pi}{2} \cos\frac{\pi}{2}x \neq 0$$
what's wrong with it?
Thank you!
| The issue is that the identity can only be valid for $x\geq 0$ or $x\leq 0$ but not both. The reason is simple: The LHS of the desired relation is even in $x$, but the RHS is odd! So the formula can't hold for all $x$, and we should limit ourselves to one particular sign.
Regarding the behavior at zero, the conclusion that $f(x^2)\cdot 2x\to 0$ as $x\to 0$ assumes that $f(0)$ exists. But if we rearrange the results shown in the OP to solve for $f(x^2)$, we have $$f(x^2)=\frac{1}{2x}\left(1+\frac{\pi}{2}\sin\frac{\pi}{2}x\right)$$ which diverges as $x\to 0$. More precisely, $f(x^2)\approx 1/(2x)$ for small $x$ and so $f(x^2)\cdot 2x\to 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
expansion of a generating function I found this formula in a book
$$\sqrt{1-4x}=1-2\sum_{n=1}^{\infty}\frac{1}{n} {{2n-2}\choose {n-1}} x^n$$
How can I prove that?
| Hint. One may apply the Taylor series expansion to $f(x):=\sqrt{1-4x}$,
$$
f(x)=f(0)+\sum_{n=1}^\infty\frac{f^{(n)}(0)}{n!}x^n,\quad |x|<\frac14. \tag1
$$
Then one may prove by induction that
$$
\begin{align}
(\sqrt{u})'=& \frac12\cdot u^{1/2-1}
\\(\sqrt{u})^{(2)}=& \frac12\left(\frac12-1 \right)\cdot u^{1/2-2}
\\(\sqrt{u})^{(3)}=& \frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdot u^{1/2-3}
\\\cdots =& \cdots
\\(\sqrt{u})^{(n)}=& \frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdots \left(\frac12-n+1 \right)\cdot u^{1/2-n}
\end{align}
$$ that is
$$
\left.(\sqrt{u})^{(n)}\right|_{u=0}= \frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdots \left(\frac12-n+1 \right) \tag2
$$
From $(2)$ we deduce by the chain rule that, for $n\geq1$,
$$
\begin{align}
\frac{f^{(n)}(0)}{n!}=&\frac1{n!}\left.(\sqrt{1-4x})^{(n)}\right|_{x=0}
\\\\= &\frac{(-4)^n}{n!}\cdot\frac12\left(\frac12-1 \right)\left(\frac12-2 \right)\cdots \left(\frac12-n+1 \right)
\\\\= &\frac{(-4)^n}{n!}\cdot\frac{1\times(-1)\times(-3)\cdots \times(-(2n-3))}{2^n}
\\\\= &2\cdot\frac{(-1)^n}{n!}\cdot\frac{(-1)^{n-1} 1\times2\times3\cdots \times(2n-3)\times(2n-2)}{1\times2\times3\cdots \times(n-1)}
\\\\= &-\frac2n\cdot\frac{(2n-2)!}{((n-1)!)^2}
\\\\= &-\frac2n\binom{2n-2}{n-1}
\end{align}
$$ which inserted in $(1)$ gives the announced identity.
| {
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"source": "stackexchange",
"question_score": "2",
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Find $k$ if Sum of all Trigonometric ratios is $7$ Given $0< A< 90^{\circ}$ and $$\sin A+\cos A+\tan A+\sec A+\operatorname{cosec} A+\cot A=7$$
and if $\sin A$ and $\cos A$ are roots of $4x^2+3x+k=0$
Find the value of $k$
sum of the roots is $$\sin A+\cos A=\frac{-3}{4}$$
Squaring both sides we get
$$1+2\sin A\cos A=\frac{9}{16}$$ $\implies$
$$\sin A\cos A=\frac{-7}{32}$$
But Product of roots is
$$\sin A\cos A=\frac{k}{4}$$
so $$\frac{k}{4}=\frac{-7}{32}$$ Hence
$$k=\frac{-7}{8}$$
But it is also given that
$$\sin A+\cos A+\tan A+\sec A+\operatorname{cosec}A+\cot A=7$$ $\implies$
$$\sin A+\cos A+\frac{1}{\sin A\cos A}+\frac{\sin A+\cos A}{\sin A\cos A}=7$$ substituting sum of the roots and product of the roots we get
$$\frac{-3}{4}+\frac{4}{k}+\frac{-3}{k}=7$$ so
$$\frac{1}{k}=\frac{31}{4}$$ so
$$k=\frac{4}{31}$$
The two values are mismatching?
| You're right. This equation is not right.
From this post: If $\sin \theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$, then $\sin 2\theta$ is a root of $x^2 -44x +36=0$ My own bonafide attempt.
it shows that $\sin x \cos x$ is a root of $x^2 -44x +36=0$
=> $\sin x \cos x$ = $22-8\sqrt{7}$ (Another value greater than 1 is ignored).
However, $(\sin x + \cos x)^2=1+2\sin x \cos x = 45-16 \sqrt{7} \neq 9/16$
Therefore there seems to be a problem in the equation given.
| {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Given that $\tan 2x+\tan x=0$, show that $\tan x=0$
Using the Trigonometric Addition Formulae,
\begin{align}
\tan 2x & = \frac{2\tan x}{1-\tan ^2 x} \\
\Rightarrow \frac{2\tan x}{1-\tan ^2 x}+\tan x & = 0 \\
\ 2\tan x+\tan x(1-\tan ^2 x) & = 0 \\
2+1-\tan ^2 x & = 0 \\
\tan ^2 x & = 3
\end{align}
This is as far as I can get, and when I look at the Mark Scheme no other Trignometric Identities have been used. Thanks
| Hint. As written, the assertion is not correct since
$$
\tan\left(\frac{2\pi}3\right)+\tan\left(\frac{\pi}3\right)=0
$$ but $$\tan\left(\frac{\pi}3\right)=\sqrt{3}\color{red}{\neq}0.$$
There is a mistake in your reasoning, starting as you did you obtain
$$
\left(\frac{2\color{blue}{\tan x}}{1-\tan ^2 x}+\color{blue}{\tan x}\right)=0
$$ which one may rewrite as
$$
\color{blue}{\tan x} \times\left(\frac{2}{1-\tan ^2 x}+1\right)=0
$$ or
$$
\tan x \times\left(\frac{\color{red}{3-\tan^2 x}}{1-\tan ^2 x}\right)=0
$$ giving
$$
\tan x=0 \quad \text{or}\quad\color{red}{\tan^2 x=3}.
$$
that is explicitly
$$
x=\pm k\pi \quad \text{or}\quad x=\pm \dfrac{\pi}3\pm k\pi
$$ $k=0,1,2,\ldots$.
| {
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"question_score": "2",
"answer_count": 5,
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Prove by induction the particular inequality $\left(1.3\right)^n \ge 1 + \left(0.3\right)n$ for every $n \in \mathbb N$ $\left(1.3\right)^n \ge 1 + \left(0.3\right)n$ for every $n \in \mathbb N$
Not sure where I'm going wrong in my Algebra, but I assume it's because I'm adding an extra term. Is the extra term unnecessary because it's not a summation I'm trying to prove?
Starting the proof:
$n=1 \Rightarrow \left(1.3\right)^{1} \ge 1 + \left(0.3\right)(1) \Rightarrow 1.3 \ge 1.3$
$n=k \Rightarrow \left(1.3\right)^{k} \ge 1 + \left(0.3\right)(k) \Rightarrow 1.3 \ge 1.3$
$n=k+1 \Rightarrow \left(1.3\right)^{k}+\left(1.3\right)^{k+1} \ge 1 + \left(0.3\right)(k+1) \Rightarrow 1.3 \ge 1.3$
After trying some Algebra I end up with this:
$= 1 + \left(0.3\right)k+\left(1.3\right)\left(1.3\right)^{k} \ge 1 + \left(0.3\right)k + .3$
Any hints?
| Let's prove Bernoulli's inequality by induction. Assume $x\ge -1$.
Base step $n=1$:
$$(1+x)^1 = 1+1\cdot x.$$
Now induction step. Assume $(1+x)^n \ge 1+nx$. Then
\begin{align*} (1+x)^{n+1} &= (1+x)^n (1+x) \\
& \ge (1+nx) (1+x)\\
& = 1+(n+1)x +nx^2\\
&\ge 1+(n+1)x.
\end{align*}
Note that the first inequality holds because of the induction assumption and since $x\ge -1$.
| {
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Bound for the sum of the divisors of a number
Let us denote by $s(n) = \sum_{d|n} d$ the sum of divisors of a natural number $n$ ($1$ and $n$ included). If $n$ has at most $5$ distinct prime divisors, prove that $s(n) < \dfrac{77}{16}n$. Also prove that there exists a natural number $n$ for which $s(n) > \dfrac{76}{16}n$ holds.
Attempt:
Let $n = p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}p_4^{\alpha_4}p_5^{\alpha_5}$ where the $p_i$ are primes. The sum of the divisors is $$(1+p_1+\cdots+p_1^{\alpha_1})(1+p_2+\cdots+p_2^{\alpha_2})\cdots(1+p_5+\cdots+p_5^{\alpha_5})=\frac{(p_1^{\alpha_1+1}-1)(p_2^{\alpha_2+1}-1)\cdots(p_5^{\alpha_5+1}-1)}{(p_1-1)(p_2-1)\cdots(p_5-1)}.$$ Then we see that $\dfrac{s(n)}{n}=\prod_{i=1}^5 \dfrac{p_i^{\alpha_i+1}-1}{p_i^{\alpha_i}(p_i-1)}$. But then $$\dfrac{p_i^{\alpha_i+1}-1}{p^\alpha_i(p_i-1)} = \dfrac{p_i-\dfrac{1}{p_i^{\alpha_i}}}{p_i-1}<\dfrac{p_i}{p_i-1}=1+\dfrac{1}{p_i-1} = \dfrac{p_i}{p_i-1}.$$ Since the last expression is decreasing for all $p_i$, we have
$$\prod_{i=1}^5 \frac{p_i^{\alpha_i+1}}{p_i^{\alpha_i}(p_i-1)} < \prod_{i=1}^5 \dfrac{p_i}{p_i-1} \le \dfrac{2}{1}\cdot \dfrac{3}{2}\cdot \dfrac{5}{3}\cdot \dfrac{7}{6}\cdot \dfrac{11}{10}=\dfrac{77}{16}.$$
If we have less than $5$ primes, notice that by repeating the same process we arrive at smaller bounds. If $n = 1$ then clearly our bound holds.
How do we find an $n$ such that $s(n) > \dfrac{76}{16}n$?
| Let $n = \prod_i^6 p_i^k$.
You've pointed out that $s(n)/n = \prod_i^6 \frac{p_i - \frac 1{p_i^k}}{p_i -1} < \prod_i^6 \frac{p_i }{p_i -1}=77*13/16*12$.
But we can make $k$ arbitrarily large so for any $\epsilon > 0$ in particular any epsilon $77/16 < 77*13/16*12 - \epsilon$ we can find $k$ so that $77*13/16*12 >s(n)/n = \prod_i^6 \frac{p_i - \frac 1{p_i^k}}{p_i -1} > 77*13/16*12 - \epsilon$.
I figure, but haven't varified, that as
$\prod_i^6\frac{p_i}{p_i - 1} = 77*13/16*12 = 1001/192=5.213$
$\prod_i^6\frac{1 + p_i}{p_i} = 32256/30030= 1.0741$
$\log_{1.0741} 5.213 = 23.01$
that $k = 24$ ought to be large enough the $n = \prod_i^6 p_i^{24}$ will satisfy.
Not sure if my reasoning for $k =24$ is valid but if not some larger $k$ will do.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Integral of $\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx$
Find the integral of the following:
$$\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx$$
Do set $u=x^3+3x^2+1$?
So, $du=(3x^2+6x)dx$?
And, $x^2+2x={u-1-x^2\over x}$?
So then,
$$\int{({u-1-x^2\over x})\over \sqrt u} du$$
This seems very complicated, is there any easier way to do this? or if this is correct, could anyone show me how to go on with this? Thanks!
| You have $du=3(x^2+2x)dx$, so your integral become :
$$\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx=\int{1\over 3\sqrt{u}} du=\frac{1}{3}\int {1\over \sqrt{u}} du=\frac{2}{3}\sqrt u$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n}{\left(n+1\right)!} = 1-\frac{1}{\left(n+1\right)!}$
So I proved the base case where $n=1$ and got $\frac{1}{2}$
Then since $n=k$ implies $n=k+1$ I setup the problem like so:
$\frac{k}{(k+1)!}+\frac{(k+1)}{(k+2)!}=1-\frac{1}{(k+2)!}$
After trying to simplify it I got the following:
$\frac{k(k+2)!+(k+1)(k+1)!}{(k+1)!(k+2)!}=1-\frac{1}{(k+2)!}$
However, I'm having trouble simplifying it to match the RHS. Hints?
| You can simply write it like this also this solution doesn't need induction:
$=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n+1-1}{(n+1)!}$
$=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{n!}-\frac{1}{(n+1)!}$
$=1-\frac{1}{(n+1)!}$
$$solved!$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\frac{1}{1+x^2}\right) \cdot \frac{1}{x} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x(1+x^2)} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1+x^2-x^2}{x(1+x^2)} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x}+ \frac{1}{2}\int^{\infty}_{0} \frac{2x}{1+x^2} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 -\left[ \ln x -\frac{1}{2}\cdot \ln(1+x^2) \right]^\infty_0 $$
$$-2I=\left[ \frac{\ln x}{1+x^2}\right]^\infty_0 -\left[\ln \left (\frac{x}{\sqrt{1+x^2}} \right) \right]^\infty_0 $$
How can I evaluate the last limits ?
| HINT:
Let $x=\tan y$
$$\implies2J=\int_0^{\pi/2}\sin2y\cdot\ln(\tan y)dy$$
Now use $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
$$\implies2I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b\left(f(x)+f(a+b-x)\right)dx$$
| {
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With $y_n$ a sequence of real numbers, prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges Let $y_n$ be a sequence of real numbers. Prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges.
Let us suppose that $y_{n}$ goes to a limit $L$. Then for all $\varepsilon >0$, for sufficiently large $n$, $|y_n -L| < \varepsilon$ .
But $|y_n - L| < 2|x_n-\frac{L}{3}|+|x_n - \frac{L}{3}|$
But then how to proceed?
| We have
$$
\begin{aligned}
x_2 &= \frac 12 y_2 - \frac 12 x_1 \\
x_3 &= \frac 12 y_3 - \frac 12 x_2 = \frac 12 y_3 + \frac 14 y_2 - \frac 14 x_1 \\
x_4 &= \frac 12 y_4 - \frac 12 x_3 = \frac 12 y_4 + \frac 14 y_3 - \frac 18 y_2 + \frac 18 x_1 \\
\ldots
\end{aligned}
$$
and generally for $n\ge 2$
$$ \begin{aligned}
x_n &=
\frac 12 y_n - \frac 14 y_{n-1} + \frac 18 y_{n-2}
- \ldots + \frac{(-1)^n}{2^{n-1}} y_2 - \frac{(-1)^n}{2^{n-1}} x_1 \\
&= \sum_{k=2}^{n} \frac{(-1)^{n-k}}{2^{n-k+1}}y_{k} - \frac{(-1)^n}{2^{n-1}} x_1 \\
&= L \sum_{k=2}^{n} \frac{(-1)^{n-k}}{2^{n-k+1}}
+ \sum_{k=2}^{n} \frac{(-1)^{n-k}}{2^{n-k+1}} \bigl(y_{k} - L \bigr)
- \frac{(-1)^n}{2^{n-1}} x_1 \\
&=:A_n + B_n + C_n
\end{aligned}
$$
Now
$$
A_n = L \sum_{j=0}^{n-2} \frac{(-1)^{j}}{2^{j+1}} \to \frac L3
$$
and
$$
C_n = - \frac{(-1)^n}{2^{n-1}} x_1 \to 0
$$
for $n \to \infty$.
For all $\varepsilon > 0$ then is a $N$ such that
$\lvert y_n - L \rvert < \varepsilon$ for $n \ge N$.
Then
$$
B_n = \left( \sum_{k=2}^{N-1} + \sum_{k=N}^{n} \right) \frac{(-1)^{n-k}}{2^{n-k+1}} \bigl(y_{k} - L \bigr)
$$
The first (finite) sum converges to zero,
and the absolute value of the second sum can be estimated above by
$$
\varepsilon \sum_{k=0}^\infty \frac{1}{2^{j+1}}
= \varepsilon \, .
$$
It follows that that $B_n \to 0$ and therefore
$$
\lim_{n \to \infty} x_n = \lim_{n \to \infty} A_n
+ \lim_{n \to \infty} B_n + \lim_{n \to \infty} C_n = \frac L3 + 0 + 0
= \frac L3\, .
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove ${2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n\ge1}{k^2\over k^2+x^2}dx={n\over 2n-1}$ Prove
$$I={2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n\ge1}{k^2\over k^2+x^2}dx={n\over 2n-1}\tag1$$
Expand out $(1)$
$$I={2n!\over \pi}\int_{0}^{\infty}{1\over (1+x^2)(2^2+x^2)(3^2+x^2)\cdots(n^2+x^2)}dx\tag2$$
Noticing
$${1\over (1+x^2)(4+x^2)}={1\over 3}\left({1\over 1+x^2}-{1\over 4+x^2}\right)$$
$${1\over (1+x^2)(4+x^2)(9+x^2)}={1\over 2x^2-12x+30}\left({1\over 1+x^2}-{2\over 4+x^2}+{1\over 9+x^2}\right)$$
$${1\over (1+x^2)(4+x^2)(9+x^2)(16+x^2)}={1\over -60x^2+300}\left({1\over 1+x^2}-{3\over 4+x^2}+{3\over 9+x^2}-{1\over 16+x^2)}\right)$$
and so on...,
k is a polynomial function of x, we have
(I just realised that if k is a function of x, then what follow from (3) are all wrong!)
$${1\over (1+x^2)(2^2+x^2)\cdots(n^2+x^2)}=k\left({{n-1\choose 0}\over 1+x^2}-{{n-1\choose 1}\over 2^2+x^2}+{{n-1\choose 2}\over 3^2+x^2}-\cdots{{n-1\choose n-1}\over n^2+x^2}\right)\tag3$$
Recall
$$\int_{0}^{\infty}{1\over a^2+x^2}dx={\pi\over 2a}\tag4$$
Sub $(3)$ into $(2)$ and applying $(4)$ hence we have
$$I={2n!k\over \pi}\cdot{\pi\over 2}\left[{{n-1\choose 0}\over 1}-{{n-1\choose 1}\over 2}+{{n-1\choose 2}\over 3}-\cdots-{{n-1\choose n-1}\over n}\right]\tag5$$
$$I={n!k}\left[{{n-1\choose 0}\over 1}-{{n-1\choose 1}\over 2}+{{n-1\choose 2}\over 3}-\cdots-{{n-1\choose n-1}\over n}\right]\tag6$$
How can we get from $(6)$ to $I={n\over 2n-1}$?
Can anyone produce another method less lengthy than this method above to tackle Integral (1)?
I have saw some authors using the residue theorem to tackle another simple case like the above (1) but I don't know how to apply it.
| I dont know if the follow help you. put $$I_n={2\over \pi}\int_{0}^{\infty}\prod_{k=1}^{n}{k^2\over k^2+x^2}dx$$ By induction we have:
$$I_1={2\over \pi}\int_{0}^{\infty}{1\over 1+x^2}dx={2\over \pi}{ \pi\over2}=1$$
Suppose that it true for $I_n$. Now
\begin{align}I_{n+1}&={2\over \pi}\int_{0}^{\infty}\left(\prod_{k=1}^{n}{k^2\over k^2+x^2}\right){(n+1)^2\over (n+1)^2+x^2}dx\\
&={2\over \pi}\int_{0}^{\infty}\left(\prod_{k=1}^{n}{k^2\over k^2+x^2}\right)dx-{2\over \pi}\int_{0}^{\infty}\left(\prod_{k=1}^{n}{k^2\over k^2+x^2}\right){x^2\over (n+1)^2+x^2}dx\\
&=I_n-{2\over \pi}\int_{0}^{\infty}\left(\prod_{k=1}^{n}{(kx)^2\over k^2+x^2}\right){1\over (n+1)^2+x^2}dx
\end{align}
and so we are able to finish the proof if we can prove that
$$\int_{0}^{\infty}\left(\prod_{k=1}^{n}{(kx)^2\over k^2+x^2}\right){1\over (n+1)^2+x^2}dx={\pi\over 2(2n-1)(2n+1)}$$
| {
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Converting from base $2$ to base 3
Transform the binary expansion $y = 0.110110110\ldots$ into a ternary expansion.
We are given that $y = 0.110110110\ldots_2$ and thus $1000_2y = 110_2+y \implies y = \frac{6}{7}$. Then we see that $\frac{6}{7}=0.857142\ldots$. How do I convert this to base $3$?
| Notice that this is $0.\overline{110}$, so it corresponds to $\frac{110_2}{1000_2-1}=\frac 6 7$.
This is how repeating decimals in different bases work: The repeating part can be written as the part that repeats over the difference between the power of the base and $1$.
We need to find where we can do this for $\frac 6 7$ in base $3$. We need to find:
$$\frac 6 7=\frac{a}{3^b-1}$$
We can do this by guessing and checking values of $b$ and seeing if $a$ is an integer. After guessing $b=1,2,3,4,5$, we get non-integer $a$s, but $b=6$ yields $a=624$, so we have:
$$\frac 6 7=\frac{624}{3^6-1}=\frac{212010_3}{1000000_3-1}$$
Thus, the answer is $0.\overline{212010}_3$.
As @BarryCipra, we can solve this without guessing using modular arithmetic:
We want $a=6\frac{3^b-1}{7}$ to be an integer. Leave $6$ out of this since there's no way it's cancelling with the $7$. Thus, $3^b-1$ needs to be a multiple of $7$, so we have: $$3^b-1 \equiv 0 \pmod 7$$ Add $1$ to both sides: $$3^b \equiv 1 \pmod 7$$ Now apply Euler's theorem to get $b=\phi(7)$ as a solution. In this case, $\phi(7)$ is the smallest possible power because $3$ is a primitive root $\pmod 7$.
Note that $\phi(d)$ won't always be the smallest power. For example, in the case of $$\frac 1 8=\frac{a}{3^b-1}$$ If we use $b=\phi(8)=4$, we get $a=10=101_3$ and thus: $$\frac 1 8=0.\overline{0101}_3$$ However, this can be simplified to: $$\frac 1 8=0.\overline{01}_3$$ meaning that $b=2$ was the actual smallest power. However, it still works out because we still find the pattern using $\phi(d)$ and we can simplify the decimal once we find the repeating pattern.
| {
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Let $A$ be a $2 \times 2$ real matrix such that $A^2 - A + (1/2)I = 0$. Prove that $A^n \to 0$ as $n \to \infty$. Question: Let $A$ be a $2 \times 2$ matrix with real entries such that $A^2 - A + (1/2)I = 0$, where $I$ is the $2 \times 2$ identity matrix and $0$ is the $2 \times 2$ zero matrix. Prove that $A^n \to 0$ as $n \to \infty$.
My attempt: Here is my idea so far. Consider $A$ as a $2 \times 2$ matrix over the field of complex numbers. Now, the polynomial
$$g(t) = t^2 - t + \frac{1}{2} $$
factors as
$$g(t) = \left(\frac{1}{2} - \frac{i}{2}\right)\left(\frac{1}{2}+\frac{i}{2}\right)$$
over $C$. This means that the minimal polynomial of $A$ over $C$ is:
$$m(t) = \left(t-(\frac{1}{2} - \frac{i}{2})\right), m(t) = \left(t-(\frac{1}{2} + \frac{i}{2})\right), \ \ \ \text{ or } \ \ \ m(t) = \left(t-(\frac{1}{2} - \frac{i}{2})\right)\left(t-(\frac{1}{2} + \frac{i}{2})\right),$$
and thus we have $A = Q D Q^{-1}$, where
$$D =\left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 - i)}&0\\
0&{\frac{1}{2}(1 - i)}
\end{array}} \right],$$
$$D = \left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 + i)}&0\\
0&{\frac{1}{2}(1 + i)}
\end{array}} \right],\ \ \ \text{ or } $$
$$D = \left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 - i)}&0\\
0&{\frac{1}{2}(1 + i)}
\end{array}} \right].$$
Now, $A^n = QD^nQ^{-1}$.
Here is where I get stuck. $D^n$ doesn't seem to be converging to $0$ as $n$ approaches infinity. In addition, I am concerned that my strategy is bad, since we are talking about a real matrix and I'm using a minimal polynomial over $C$. Is it still true that $A$ must have one of the three forms above, even if $A$ is supposed to be real?
Thanks for any help/suggestions you may be able to provide.
| You can get $A^n=A^{n-1}-1/2A^{n-2}=(A^{n-2}-1/2A^{n-3})-1/2A^{n-2}=1/2A^{n-2}-1/2A^{n-3}=-1/4A^{n-4}$. Then the conclusion is obvious by induction on k, while let n=4k+i.
| {
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Integrating $\int\frac{x^3}{\sqrt{9-x^2}}dx$ via trig substitution What I have done so far:
Substituting $$x=3\sin(t)\Rightarrow dx=3\cos(t)dt$$ converting our integral to
$$I=\int\frac{x^3}{\sqrt{9-x^2}}dx=\int \frac{27\sin^3(t) dt}{3\sqrt{\cos^2(t)}}3\cos(t)dt\\
\Rightarrow \frac{I}{27}=\int \sin^3(t)dt=-\frac{\sin^2(t)\cos(t)}{3}+\frac{2}{3}\int\sin(t)dt\\
\Rightarrow I=18\cos(t)-9\sin^2(t)\cos(t)+c$$
Which is where I'm stuck. I can substitute back in for the $\sin(t)$ terms but don't know hot to deal with the $\cos(t)$ terms and back out information about $x$.
edited: because I miswrote the problem, I am sorry.
| Substitution $x=3\sin{y} \rightarrow dx=3\cos{y}$ $dy$
Then
$27 \int \frac{\sin^3{y}}{3\cos{y}}.3\cos{y}$ $dy$
$=\frac{27}{4}\int 4\sin^3{y}$ $dy$
Use Identity
$4\sin^3{y}=3\sin{y}-\sin{3y}$
$\frac{27}{4}\int 4\sin^3{y}$ $dy$ $=\frac{27}{4}\int 3\sin{y}-\sin{3y}$ $dy$
$=\frac{27}{4}(-3\cos{y}-\frac{\cos{3y}}{3})$
From substitution,
$\cos{y}=\frac{3}{\sqrt{9-x^2}}$
$\cos{3y}=4\cos^3{y}-3\cos{y}$
$\therefore \frac{27}{4}(-3\cos{y}-\frac{\cos{3y}}{3})=\frac{27}{4}[-\frac{9}{\sqrt{9-x^2}}-\frac{1}{3}(\frac{108}{(9-x^2)^{\frac{3}{2}}}-\frac{9}{\sqrt{9-x^2}})]$
You should be able to carry forward.Good luck!
| {
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Solve $3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$ Can someone point me in the right direction how to solve this?
$3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$
I guess I have to get to logarithms of the same base. But how? What principle should I use here?
Thx
| $$3^{ x+2 }\cdot 4^{ -(x+3) }+3^{ x+4 }\cdot 4^{ -(x+3) }=\frac { 40 }{ 9 } \\ { 4 }^{ -\left( x+3 \right) }{ 3 }^{ x+2 }\left( 1+9 \right) =\frac { 40 }{ 9 } \\ { 4 }^{ -\left( x+3 \right) }{ 3 }^{ x+2 }=\frac { 4 }{ 9 } \\ { \left( \frac { 3 }{ 4 } \right) }^{ x }\frac { 9 }{ 64 } =\frac { 4 }{ 9 } \\ { \left( \frac { 3 }{ 4 } \right) }^{ x }={ \left( \frac { 4 }{ 3 } \right) }^{ 4 }\\ x=-4\\ \\ $$
| {
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Show that the vertex lies on the surface $z^2(\frac{x}{a}+\frac{y}{b})=4(x^2+y^2)$ Two cones with a common vertex pass through the curves $z^2=4ax,y=0$ and $z^2=4by,x=0.$ The plane $z=0$ meets them in two conics which intersect in four concyclic points.Show that the vertex lies on the surface $z^2(\frac{x}{a}+\frac{y}{b})=4(x^2+y^2)$
Let the common vertex of the two cones is $(\alpha,\beta,\gamma)$ and let the direction cosines of the generator line of first cone which passes through $z^2=4ax,y=0$ be $l_1,m_1,n_1$ and the direction cosines of the generator line of second cone which passes through $z^2=4ay,x=0$ be $l_2,m_2,n_2.$
The generator line of the first cone is $\frac{x-\alpha}{l_1}=\frac{y-\beta}{m_1}=\frac{z-\gamma}{n_1}.$The equation of the first cone is $(\gamma-\frac{z\beta}{y})^2=4a(\alpha-\frac{\beta x}{y})$
The generator line of the second cone is $\frac{x-\alpha}{l_2}=\frac{y-\beta}{m_2}=\frac{z-\gamma}{n_2}.$The equation of the second cone is $(\gamma-\frac{z\alpha}{x})^2=4b(\beta-\frac{\alpha y}{x})$
I am stuck here.Please help me.
|
The generator line of the first cone is $\frac{x-\alpha}{l_1}=\frac{y-\beta}{m_1}=\frac{z-\gamma}{n_1}.$
Setting $y=0$ gives
$$\frac{x-\alpha}{l_1}=\frac{0-\beta}{m_1}=\frac{z-\gamma}{n_1}\implies x=-\beta\frac{l_1}{m_1}+\alpha,\quad z=-\beta\frac{n_1}{m_1}+\gamma\tag1$$
Also,
$$\frac{l_1}{m_1}=\frac{x-\alpha}{y-\beta},\quad \frac{n_1}{m_1}=\frac{z-\gamma}{y-\beta}\tag2$$
From $(1)(2)$,
$$z^2=4ax\implies \left(-\beta\cdot \frac{z-\gamma}{y-\beta}+\gamma\right)^2=4a\left(-\beta\cdot\frac{x-\alpha}{y-\beta}+\alpha\right)\tag3$$
The generator line of the second cone is $\frac{x-\alpha}{l_2}=\frac{y-\beta}{m_2}=\frac{z-\gamma}{n_2}.$
Setting $x=0$ gives
$$\frac{0-\alpha}{l_2}=\frac{y-\beta}{m_2}=\frac{z-\gamma}{n_2}\implies y=-\alpha\frac{m_2}{l_2}+\beta,\quad z=-\alpha\frac{n_2}{l_2}+\gamma\tag4$$
Also,
$$\frac{m_2}{l_2}=\frac{y-\beta}{x-\alpha},\quad \frac{n_2}{l_2}=\frac{z-\gamma}{x-\alpha}\tag5$$
From $(4)(5)$,
$$z^2=4by\implies \left(-\alpha\cdot \frac{z-\gamma}{x-\alpha}+\gamma\right)^2=4b\left(-\alpha\cdot \frac{y-\beta}{x-\alpha}+\beta\right)\tag6$$
Setting $z=0$ in $(3)(6)$ give
$$\left(-\beta\cdot \frac{0-\gamma}{y-\beta}+\gamma\right)^2=4a\left(-\beta\cdot\frac{x-\alpha}{y-\beta}+\alpha\right)\implies y^2\gamma^2=4a(\alpha y-\beta x)(y-\beta)$$
$$\implies b\alpha (\gamma^2-4a\alpha) y^2+4ab\alpha \beta xy+4ab\alpha^2\beta y-4ab\alpha \beta^2 x=0\tag7$$
$$\left(-\alpha\cdot \frac{0-\gamma}{x-\alpha}+\gamma\right)^2=4b\left(-\alpha\cdot \frac{y-\beta}{x-\alpha}+\beta\right)\implies x^2\gamma^2=4b(x\beta-\alpha y)(x-\alpha)$$
$$\implies a\beta (\gamma^2-4b\beta) x^2+4ab\alpha\beta xy+4a\beta^2 b\alpha x-4ba\beta \alpha^2 y=0\tag8$$
The four intersection points of $(7)$ with $(8)$ satisfy
$$b\alpha (\gamma^2-4a\alpha) y^2+4ab\alpha \beta xy+4ab\alpha^2\beta y-4ab\alpha \beta^2 x+k(a\beta (\gamma^2-4b\beta) x^2+4ab\alpha\beta xy+4a\beta^2 b\alpha x-4ba\beta \alpha^2 y)=0$$
Since it represents a circle for some $k$, we have to have $k=-1$ and
$$b\alpha (\gamma^2-4a\alpha)=-a\beta (\gamma^2-4b\beta) ,$$
i.e.
$$\gamma^2\left(\frac{\alpha}{a}+\frac{\beta}{b}\right)=4(\alpha^2+\beta^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this Taylor expansion of $\frac{1}{(1+x)^2}=-1\times\displaystyle\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$? I came across this series of the Taylor Expansion-
$$\frac{1}{(1+x)^2}=1 - 2x + 3x^2 -4x^3 + \dots.=-1\times\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$$
But I have no idea how to prove this...
Thanks for any help!
| Observe that
$$\frac x{(1+x)^2}=x - 2x^2 + 3x^3 -4x^4 + \dots$$ and adding the original
$$\frac1{(1+x)^2}=1-2x+3x^2-4x^3+5x^4\cdots$$
you verify
$$\frac x{(1+x)^2}+\frac1{(1+x)^2}=1-x+x^2-x^3+x^4-\cdots=\frac1{x+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Prove that $\frac{2^a+1}{2^b-1}$ is not an integer
Let $a$ and $b$ be positive integers with $a>b>2$. Prove that $\frac{2^a+1}{2^b-1}$ is not an integer.
This is equivalent to showing there always exists some power of a prime $p$ such that $2^a+1 \not \equiv 0 \pmod{p^a}$ but $2^b-1 \equiv 0 \pmod{p^a}$. How do we prove the statement from this or is there an easier way?
| Note for this type of question is always useful to ask what how many times does the denominator easily go into the numerator, here we have $2^a/2^b = 2^{a-b}$, thus $$\frac{2^a + 1}{2^b -1} = \frac{2^a + 1 - 2^{a-b} (2^b - 1)}{2^b - 1} + 2^{a-b} = \frac{2^{a-b} + 1}{2^b - 1} + 2^{a-b}$$
Thus if $a > b$, then
$$\frac{2^a + 1}{2^b -1} = \frac{2^{a-b} + 1}{2^b - 1} + \mbox{some integer}$$
And this jumps out to me as something that can be repeated.
It is easy to show that similarly for any $x$ with $x > b$, $$\frac{2^{x} + 1}{2^b - 1} = \frac{2^{x-b} + 1}{2^b - 1} + \mbox{some integer}$$
Thus we can repeat this process, "subtracting" $b$ from the power in the numerator and get
$$\frac{2^a + 1}{2^b -1} = \frac{2^{r} + 1}{2^b - 1} + \mbox{some integer}$$
where $r < b$.
The rest is just inequalities to show that this can never be equal to an integer.
Now as $r < b$, thus $r \leq b-1$, thus $2^r \leq 2^{b-1}$.
Also $2^{b-1} < 2^b - 2$, thus $2^r + 1 < 2^b + 1$, thus $$0< \frac{2^{r} + 1}{2^b - 1} < 1$$
Thus $$\frac{2^a + 1}{2^b -1} = \mbox{non integer} + \mbox{some integer}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \cdot \frac{1}{2\sqrt{x+2}} = \frac{1}{(2x+6)\sqrt{x+2}}$$
$$g'(x) = 1$$
$$g(x) = x$$
So:
$$\bigg[\arctan(\sqrt{x+2}) \cdot x\bigg]_{-1}^{1} - \int_{-1}^{1} \frac{x}{(2x-6)\sqrt{x+2}}\ dx$$
How should I proceed with the new integral?
| $$\int \arctan(\sqrt{x+2})\, dx$$
$$=\int \arctan (\sqrt{x+2})\, d(x+2)$$
$$=2\int (\arctan(\sqrt{x+2}))(\sqrt{x+2})\, d(\sqrt{x+2})$$
$$=2\int (\arctan u)(u)\, du,$$
where $u=\sqrt{x+2}$. Now use integration by parts:
$$=2\left((\arctan u)\left(\frac{u^2}{2}\right)-\int \frac{u}{u^2+1}\, du\right)$$
$$=2\left((\arctan u)\left(\frac{u^2}{2}\right)-\frac{1}{2}\int \frac{d\left(u^2+1\right)}{u^2+1}\right)$$
$$=2\left((\arctan u)\left(\frac{u^2}{2}\right)-\frac{1}{2}\ln\left|u^2+1\right|\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + 1/β$ and $β + 1/α$. Quadratic equation question, as specified in the title.
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + \frac{1}{β}$ and $β + \frac{1}{α}$.
I gather that $α + β = -\frac{b}{a} = \frac{6}{1} = 6$ and that $αβ = \frac{c}{a} = \frac{7}{1} = 7$. Do I need to convert $α + \frac{1}{β}$ and $β + \frac{1}{α}$ into a format whereby I can sub in the values for adding together or multiplying $α$ and $β$ ? If so, how ?
| If $\alpha,\beta$ are the roots of $x^2-6x+7=0$, then all quadratic equations with roots $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$ are $$a\left(x-\left(\alpha+\frac{1}{\beta}\right)\right)\left(x-\left(\beta+\frac{1}{\alpha}\right)\right)=0,$$
where $a\in\mathbb R$, $a\neq 0$ (saying "the equation" is wrong, because there are infinitely many of them).
Using Vieta's Formulas, we get $\alpha+\beta=6$ and $\alpha\beta=7$. Use these two equalities in the following:
$$a\left(x^2-\left(\alpha+\beta+\frac{\alpha+\beta}{\alpha\beta}\right)x+\left(\alpha\beta+1+1+\frac{1}{\alpha\beta}\right)\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$
Prove that $\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$.
I tried using the partial fraction decomposition $a_j = \frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}$, but I don't see how that helps.
| More generally,
if
$x_{(n)}
=\prod_{k=0}^{n-1} (x+k)
$,
$\begin{array}\\
\dfrac1{x_{(n)}}-\dfrac1{(x+1)_{(n)}}
&=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=0}^{n-1} (x+1+k)}\\
&=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=1}^{n} (x+k)}\\
&=\dfrac1{\prod_{k=1}^{n-1} (x+k)}\left(\dfrac1{x}-\dfrac1{x+n}\right)\\
&=\dfrac1{\prod_{k=1}^{n-1} (x+k)}\left(\dfrac{n}{x(x+n)}\right)\\
&=\dfrac{n}{\prod_{k=0}^{n} (x+k)}\\
&=\dfrac{n}{x_{(n+1)}}\\
\end{array}
$
Therefore
$\begin{array}\\
\sum_{j=1}^m \dfrac{1}{j_{(n+1)}}
&=\sum_{j=1}^m \dfrac1{n}\left(\dfrac1{j_{(n)}}-\dfrac1{(j+1)_{(n)}}\right)\\
&= \dfrac1{n}\left(\dfrac1{1_{(n)}}-\dfrac1{(m+1)_{(n)}}\right)\\
\text{or}\\
\sum_{j=1}^m \dfrac{1}{\prod_{k=0}^{n} (j+k)}
&= \dfrac1{n}\left(\dfrac1{n!}-\dfrac1{\prod_{k=0}^{n-1} (m+1+k)}\right)\\
\end{array}
$
If
$n=2$,
this becomes
$\sum_{j=1}^m \dfrac{1}{\prod_{k=0}^{2} (j+k)}
= \dfrac1{2}\left(\dfrac1{2!}-\dfrac1{\prod_{k=0}^{1} (m+1+k)}\right)
$
or
$\sum_{j=1}^m \dfrac{1}{j(j+1)j+2)}
= \dfrac1{4}-\dfrac1{2(m+1)(m+2)}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer.
I'm stuck at the basis step.
If I started with $1$. I get the right hand side is $18$ which is clearly not even close. It says prove shouldn't it be always true?
| For $n=1$, since $3 + 3 \times 5^1= 3+15 = 18$, there is no problem with the righthand side being $18$ too.
Note though that the base case is $n=0$, the condition being $n$ nonnegative, and then the lefthand side is $3$ as is the righthand side, which is $3 \times (5^{0+1}-1)/4 = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
A convergent series: $\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$ I would like to find the value of:
$$\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$$
I could only see that the ratio of two consecutive terms is $\dfrac{1}{27\cos(2\theta)}$.
| Hint. One may observe that
$$
\sin^3 (a)=\frac34\sin (a)-\frac14 \sin(3a)
$$ giving
$$
3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\frac{3^{n}}4\sin\left(\frac{\pi}{3^{n+1}}\right)-\frac{3^{n-1}}4\sin\left(\frac{\pi}{3^{n}}\right)
$$
then one gets a telescoping sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find last 5 significant digits of 2017! Since there are less powers of $5$ than of $2$ and since $10 = 2 \cdot 5$, I counted the number of zeros in $2017!$:
$\left \lfloor{ \frac{2017}{5^1}}\right \rfloor + \left \lfloor{ \frac{2017}{5^2}}\right \rfloor +\left \lfloor{ \frac{2017}{5^3}}\right \rfloor +\left \lfloor{ \frac{2017}{5^4}}\right \rfloor = 502$
Now I need to find $\frac{2017!}{10^{502}} \pmod{10^5}$
Using Chinese remainder theorem I split the $\pmod{10^5}$ into $\pmod{2^5}$ and $\pmod{5^5}$. Since there are a lot more powers of $2$ left, $\frac{2017!}{10^{502}}\equiv 0 \pmod{2^5}$
I am having trouble with getting $\frac{2017!}{10^{502}} \pmod{5^5}$. What would be the best way to compute it?
Clarification: significant digits - the ones that follow up before the zeros at the right side of the number.
| I know how to do the last 3 significant digits, if that helps.
$4! = 24\\
\frac {9!}{5!} = 3024 \equiv 24\pmod {1000}$
$\frac {5n!}{5^n n!} = 24^n \pmod {1000}$
$2017! = \frac {2017!}{2015!}\frac {2015!}{5^{403}403!}\frac {5^{403}403!}{5^{80}80!}\frac {5^{80}80!}{5^{16}16!}\frac {5^{16}16!}{5^{3}3!} 3!\\$
$\frac {2017!}{10^502} \equiv 12^{502}\frac {2017!}{2015!}\frac {403!}{400!}\frac {16!}{15!}3!\pmod{1000}$
$144\cdot17\cdot16\cdot403\cdot402\cdot401\cdot16\cdot6 = 968\pmod{1000}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Solution of Diophantine equation Find all integral solutions of $x^2+1= y^2+z^2$. Actually I have to find all integral solution of $a(a+1)=b(b+1)+c(c+1)$. I reduced this in the above form I.e., $ (2a+1)^2+1= (2b+1)^2+(2c+1)^2$ .
| The case where $z^2-x^2=1-y^2=0$ is self-evident.We have $z=\pm x$ and $y=\pm 1$
Now consider the case where $(1-y^2)(z^2-x^2)\neq 0$
$$z^2-x^2=1-y^2$$
$$(z+x)(z-x)=(1+y)(1-y)$$
$$\dfrac{z+x}{1+y}=\dfrac{1-y}{z-x}
=\dfrac{p}{q}$$ with $pq\neq 0$ and $\gcd(p,q)=1$.
It follows:
$$z+x=\dfrac{p}{q}(1+y)$$
$$z-x=\dfrac{q}{p}{(1-y)}$$
Hence, in order to find all the integral solutions, the following restrictions must apply
$$1-y\equiv 0\pmod p$$
$$1+y\equiv 0\pmod q$$
In other words,by alternatively subtracting 1 from 2/ adding 1 and 2 $$qs-pr=2y$$
$$qs+pr=2$$ where $r,s$ are 2 coprime integers.
Since
$$z+x=ps$$
$$z-x=qr$$
we have:
$$z=\dfrac{1}{2}(ps+qr)$$ $$x=\dfrac{1}{2}(ps-qr)$$ $$y=\dfrac{1}{2}(qs-pr)$$where $pr,qs$ have the same parity and $qs+pr=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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