Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Telescoping exercise with iterations? For every $x>0$, consider the sequence $(x_n)$ defined by $x_0=x$ and, for every $n\geqslant0$, $$x_{n+1} = \sqrt{x_n + \frac12}$$
Then $x_n\to x_*=\frac{1+\sqrt3}2\ne0$ hence the sequence $$S_n(x)=\sum_{k=1}^n(-1)^kx_k^4$$ diverges. Consider its Cesàro sums, defined by $$C_n(x)=\frac1n\sum_{k=1}^nS_k(x)$$
The question is to prove that $C_n(x)\to C(x)=\frac18-x^2$.
One can probably use telescoping and / or differentiation techniques.
As safety checks, note that the proposed limit $C(x)$ satisfies the relations $$C(x_*)=-\frac12x_*^4\qquad C\left(x^2-\frac12\right)=-x^4-C(x)$$
| Here is the complete proof. I had wanted to post this before but had to wait for all close voters to rescind their votes after trying to save this question.
First off -- I think the key step in the proof is made a little clearer by using the iterated function notation for the question instead of the other that has been put to use now -- so we will first start off with a rephrase as follows: Let $f(x) = \sqrt{x + \frac{1}{2}}$. Then $x_n = f^n(x)$ and we want to prove that
$$\sum_{k=1}^{\infty} (-1)^k x_k^4 \stackrel{\mathrm{pseudo}}{=} \frac{1}{8} - x^2$$
where the left is divergent but reinterpreted using Cesaro summability (hence the pseudo-equality), i.e. that
$$\lim_{n \rightarrow \infty} C_n(x) = \frac{1}{8} - x^2$$
with $$C_n(x) = \frac{1}{n} \sum_{k=1}^{n} \left(\sum_{l=1}^{k} (-1)^l x_l^4\right)$$.
To do this, first replace $x_l$ by the corresponding iterated functions $f^l(x)$:
$$
\begin{align}
C_n(x) &= \frac{1}{n} \sum_{k=1}^{n} \left(\sum_{l=1}^{k} (-1)^l [f^l(x)]^4\right)\\
&= \frac{1}{n} \sum_{k=1}^{n} \left(-[f^1(x)]^4 + [f^2(x)]^4 - \cdots + (-1)^l [f^l(x)]^4\right)\\
\end{align}
$$
Now, we note that, by definition of iterated functions that $f^l(x) = f(f^{l-1}(x))$ and this allows us to expand out $[f^l(x)]^4$ as follows:
$$
\begin{align}
[f^l(x)]^4 &= [f(f^{l-1}(x))]^4 \\
&= \sqrt{f^{l-1}(x) + \frac{1}{2}}^4 \\
&= \left(f^{l-1}(x) + \frac{1}{2}\right)^2 \\
&= [f^{l-1}(x)]^2 + f^{l-1}(x) + \frac{1}{4} \\
&= [f(f^{l-2}(x))]^2 + f^{l-1}(x) + \frac{1}{4} \\
&= \sqrt{f^{l-2}(x) + \frac{1}{2}}^2 + f^{l-1}(x) + \frac{1}{4} \\
&= f^{l-2}(x) + \frac{1}{2} + f^{l-1}(x) + \frac{1}{4} \\
&= f^{l-2}(x) + f^{l-1}(x) + \frac{3}{4}
\end{align}
$$
We now plug this back into the previous series to get
$$
\begin{align}
C_n(x) &= \frac{1}{n} \sum_{k=1}^{n} \left(-[f^1(x)]^4 + [f^2(x)]^4 - \cdots + (-1)^l [f^l(x)]^4\right)\\
&= \frac{1}{n} \sum_{k=1}^{n} \left(-[f^{-1}(x) + f^0(x) + \frac{3}{4}] + [f^0(x) + f^1(x) + \frac{3}{4}] - [f^1(x) + f^2(x) + \frac{3}{4}] + [f^2(x) + f^3(x) + \frac{3}{4}] - \cdots + (-1)^k [f^{k-2}(x) + f^{k-1}(x) + \frac{3}{4}]\right)\\
&= \frac{1}{n} \sum_{k=1}^{n} \left(-f^{-1}(x) - [f^0(x) - f^0(x)] - \frac{3}{4} + [f^1(x) - f^1(x)] + \frac{3}{4} - [f^2(x) - f^2(x)] - \frac{3}{4} + [f^3(x) - f^3(x)] + \frac{3}{4} - \cdots + (-1)^k [f^{k-1}(x) - f^{k-1}(x)] + (-1)^k \frac{3}{4}]\right)
\end{align}
$$
Thus we now have the series in telescoping form and it telescopes down to
$$\begin{align}
C_n(x) &= \frac{1}{n} \sum_{k=1}^{n} \left(-f^{-1}(x) - [k \mod 2 = 1] \frac{3}{4}\right)
\end{align}
$$
Now of course $f^{-1}(x) = x^2 - \frac{1}{2}$ so
$$
C_n(x) = \frac{1}{n} \sum_{k=1}^{n} \left(-x^2 + \frac{1}{2} - [k \mod 2 = 1] \frac{3}{4}\right)
$$
This then becomes three separate Cesaro means
$$
C_n(x) = \left(\frac{1}{n} \sum_{k=1}^{n} (-x^2)\right) + \left(\frac{1}{n} \sum_{k=1}^{n} \frac{1}{2}\right) - \left(\frac{1}{n} \sum_{k=1}^{n} [k \mod 2 = 1] \frac{3}{4}\right)
$$
Now we take the limit as $n \rightarrow \infty$. The first two means are means of identical numbers, thus they respectively equal $-x^2$ and $\frac{1}{2}$. The last one is like the mean of Grandi's series, which has limit $\frac{1}{2}$, so this has limit $\frac{3}{8}$. Thus the final Cesaro sum is
$$
\begin{align}
C(x) &= \lim_{n \rightarrow \infty} C_n(x)\\
&= -x^2 + \frac{1}{2} - \frac{3}{8}\\
&= \frac{1}{8} - x^2
\end{align}
$$.
or
$$\sum_{k=1}^{\infty} (-1)^k x_k^4 \stackrel{\mathrm{pseudo}}{=} \frac{1}{8} - x^2$$
QED.
| {
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"url": "https://math.stackexchange.com/questions/2056934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Evaluating $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$
I have to evaluate $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$
I tried several ways like replacing $x^2-y^2$ by $z$ and then solving and breaking the denominator into $(x+y)(x-y)$ and then replacing $(x-y)$ by $z$ and then solving but always I end up with complicated things which are nowhere near the answer.
How do I solve?
Thanks for any help!!
| $$\lim _{ x\rightarrow y } \frac { \sin ^{ 2 } x-\sin ^{ 2 } y }{ x^{ 2 }-y^{ 2 } } =\lim _{ x\rightarrow y } \frac { \left( \sin { x-\sin { y } } \right) \left( \sin { x+\sin { y } } \right) }{ \left( x-y \right) \left( x+y \right) } =\\ =\lim _{ x\rightarrow y } \frac { \left( 2\sin { \frac { x-y }{ 2 } } \cos { \frac { x+y }{ 2 } } \right) \left( \sin { x+\sin { y } } \right) }{ \left( x-y \right) \left( x+y \right) } =\\ =\lim _{ x\rightarrow y } \frac { \sin { \frac { x-y }{ 2 } } }{ \frac { \left( x-y \right) }{ 2 } } \frac { \cos { \frac { x+y }{ 2 } } \left( \sin { x+\sin { y } } \right) }{ \left( x+y \right) } =\\ =\cos { y } \frac { 2\sin { y } }{ 2y } =\frac { \sin { y\cos { y } } }{ y } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve $18^{{20}^{19}}$ divided by 7? I was re-practicing some number theory and I came across this question that I had previously solved and it seems like... I can't solve it anymore. My thoughts are that $18^6 \equiv 1 \text{ mod } 7$ by Euler's formula so I am looking to express the power $20^{19}$ as some $6k + r$ to simplify the system. Doing this I get:
$$20 \equiv 2 \text{ mod } 6\\ 20^2 \equiv 4\text{ mod } 6\\ 20^3 \equiv 2 \text{ mod } 6$$
...and there seems to be no end as they repeat in 2, 4, 2 cycles. I can't seem to simplify this equation any other way though.. Could someone please enlighten me and suggest ways of viewing (or improving) on my problem solving skills when attacking problems like these?
| Since $\gcd(18,7)=1$, by Euler's theorem, $18^{\phi(7)}\bmod{7}=1$.
Since $7$ is prime, $\phi(7)=7-1=6$.
Therefore $18^{6}\bmod{7}=1$.
There exists a positive integer $n$ such that:
$20^{19}=$
$6n+(20^{19}\bmod{6})=$
$6n+((20\bmod{6})^{19}\bmod{6})=$
$6n+(2^{19}\bmod{6})$
Therefore:
$18^{20^{19}}\bmod{7}=$
$18^{6n+(2^{19}\bmod{6})}\bmod{7}=$
$18^{6n}\cdot18^{(2^{19}\bmod{6})}\bmod{7}=$
$(18^{6})^{n}\cdot18^{(2^{19}\bmod{6})}\bmod{7}=$
$(\color\red{18^{6}\bmod{7}})^{n}\cdot18^{(2^{19}\bmod{6})}\bmod{7}=$
$\color\red{1}^{n}\cdot18^{(2^{19}\bmod{6})}\bmod{7}=$
$18^{(2^{19}\bmod{6})}\bmod{7}$
Let's prove by induction that if $n$ is odd then $2^{n}\bmod{6}=2$:
First, show that this is true for $n=1$:
$2^{1}\bmod{6}=2$
Second, assume that this is true for $n=2k+1$:
$2^{2k+1}\bmod{6}=2$
Third, prove that this is true for $n=2k+3$:
$2^{2k+3}\bmod{6}=$
$2^{2k+2+1}\bmod{6}=$
$2^{2+2k+1}\bmod{6}=$
$2^2(2^{2k+1})\bmod{22}=$
$2^2(\color\red{2^{2k+1}\bmod{6}})\bmod{6}=$
$2^2(\color\red{2})\bmod{6}=$
$8\bmod{6}=$
$2$
Therefore, since $19$ is odd, $2^{19}\bmod{6}=2$.
From all of the above, we can conclude:
$\begin{align}
18^{20^{19}}\bmod{7}
&= 18^{(20^{19}\bmod{6})}\bmod{7} \\
&= 18^{(2^{19}\bmod{6})}\bmod{7} \\
&= 18^{2}\bmod{7} \\
&= 324\bmod{7} \\
&= 2
\end{align}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3,xyz=4$ Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3, xyz=4$
The first thing that I notice is that it is symetric to $a,b,c$ but it can't help me .The other idea is finding the numbers but giving it to wolfram alpha gives five complex set of answers. Another idea that looks to be nice is this:
$\sum\limits_{}^{cyc}\frac{1}{xy+z-1}=\sum\limits_{}^{cyc}\frac{x}{x^2-x+4}$
Maybe it gives the answer but it is to hard to calculate. Any hints?
| Replacing $x=2-y-z$ and similar, the sum becomes:
$$
S = \sum_{cyc}\frac{1}{x+yz-1} = \sum_{cyc}\frac{1}{1-y-z+yz} = \sum_{cyc}\frac{1}{(y-1)(z-1)}=\cfrac{1}{\prod_{cyc} (x-1)} \sum_{cyc} (x-1)
$$
From the given relations $\sum_{cyc} xy = \frac{1}{2}\left((\sum_{cyc} x)^2-\sum_{cyc} x^2\right)=\cfrac{1}{2}\,$, so:
$$
\begin{align}
\sum_{cyc}(x-1) & = \sum_{cyc} x - 3 = 2 - 3 = -1 \\
\prod_{cyc}(x-1) & = xyz - \sum_{cyc} xy + \sum_{cyc} x - 1 = 4 - \cfrac{1}{2} + 2 - 1 = \cfrac{9}{2}
\end{align}
$$
Therefore $\,S=\cfrac{1}{\frac{9}{2}} \cdot (-1) = - \cfrac{2}{9}\,$.
P.S. For an alternative derivation, note that $x,y,z$ are the roots of $\,2t^3-4t^2+t-8=0$ by Vieta's formulas. The polynomial with roots $x-1,y-1,z-1$ can be obtained with the substitution $t=u+1$ which, after expanding the powers and collecting, results in $2u^3+2u^2-u-9=0\,$. Therefore $\,\sum_{cyc} (x-1) = -1\,$ and $\,\prod_{cyc} (x-1) = \cfrac{9}{2}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proving inequality using Lagrange multipliers. I started learing about Lagrange Multipliers and I got the following question:
Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}≥\frac{3}{2}$, for each $a,b,c>0$.
I'm not sure how to use Lagrange multipliers for it... any ideas?
| Let the objective function be
$$f = a(c+a)(a+b) + b(b+c)(a+b) + c(b+c)(c+a)$$
and the constraint be
$$g = (a+b)(b+c)(c+a) - k.$$
Note that if $g=0$, then
$$\frac{f}{k} = \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}.$$
Solving $\nabla f-\lambda g = 0$ and $g=0$ we find $f$ is minimized for
$a=b=c=k^{1/3}/2$, with value $3k/2$.
Thus,
$f/k \ge 3/2$,
which proves the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
p-adic number for polynomial I have three polynomials:
$$f=x^5+x^4+4x^3+3x^2+3x$$
$$g=x^5+4x^3+2x^2+3x+6$$
$$p=x^2+3$$
Question: what is $v_p(fg)$?
First, I have multiplied f and g:
$h :=f \cdot g = x^{10}+x^9+8x^8+9x^7+24x^6+29x^5+36x^4+39x^3+27x^2+18x $
Then I have divided that $h$ with $x^2+3$ to get a factor to know how often $h$ contains $x^2+3 $:
$(x^2+3) \cdot (x^6+x^5+2x^4+3x^3+3x^2+2x)$
When I divide $x^6+x^5+2x^4+3x^3+3x^2+2x$ again by $x^2+3$ then I get a rest:
$(x^4+x^3-x^2+6) \cdot (x^2+3) + 2x-18$
Does that mean that $h$ contains $x^2+3$ only once and
$v_p(f \cdot g)=1$?
Is that the correct way to find the p-adic number? Or is there a better conventional one?
I appreciate every hint.
| Because $x^2+3$ is irreducible in $\Bbb{Z}[x]$ (Eisenstein) it is the minimal polynomial of $i\sqrt3$. You can then check that
*
*$f(i\sqrt3)=0$, so $f(x)$ must be divisible by $x^2+3$. A long division shows that $f(x)/(x^2+3)=x+x^2+x^3$, and you can check this is not a multiple of $x^2+3$.
*$g(i\sqrt3)=0$, so $g(x)$ is also divisible by $x^2+3$. A long division gives a quotient that I leave to you.
Anyway, looks like $fg$ should be divisible by $(x^2+3)^2$, so you have made a mistake somewhere. As Hagen pointed out, we have for all discrete valuations
$\nu_p$ the formula $\nu_p(fg)=\nu_p(f)+\nu_p(g)$ that comes in handy.
| {
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"url": "https://math.stackexchange.com/questions/2064727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find arc length of implicit function Result of integrating I've been trying to find the arc length of the following function:
$e^y=\frac{e^x+1}{e^x-1}$
from $x = a$ to $x = b$.
I've tried designating $y$, but it doesn't seem to work. I would really appreciate your help
| We have $y = $ ln$(\frac{e^x + 1}{e^x-1})$. Then $\frac{dy}{dx} = \frac{e^x -1}{e^x + 1} * \frac{{e^{2x} - e^x -e^{2x} - e^x}}{(e^x -1)^2} = -2\frac{e^x}{(e^x -1)(e^x +1)} = -2\frac{e^x}{e^{2x} -1}$.
Now, letting $y(x)$ be our curve and $l$ its length, $l(y) = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} dx$.
Then we have now $l(y) = \int_a^b \sqrt{1 + \frac{4e^{2x}}{e^{4x} - 2e^{2x} +1}}dx=\int_a^b \sqrt{\frac{e^{4x} +2e^{2x} + 1}{e^{4x} - 2e^{2x} +1}}dx=$ $ \int_a^b \sqrt{\frac{(e^{2x} +1)^2}{(e^{2x} -1)^2}}dx = \int_a^b \frac{e^{2x} +1}{e^{2x} -1}dx$.
Now put $u = e^{2x} -1$ Then $du = 2e^{2x}dx,$ or, $dx = \frac{du}{2(1+u)}$. For limits of integration, when $x = a, u = e^{2a} -1$, and when $x=b, u = e^{2b} -1$.
Then our integral becomes $l(y) = \frac{1}{2}\int_{e^{2a} -1}^{e^{2b} -1}\frac{u+2}{u(1+u)}du = \frac{1}{2}\int_{e^{2a} -1}^{e^{2b} -1}\frac{1}{1+u} + \frac{2}{u(1+u)}du$.
Now, you can split up the integrand into two integrals. The first is quite simple, and the second is a routine application of partial fractions decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why doesn't this property work on the following composite function? Given $f(x),g(x)$ find $(f \circ g)(x)$.
$$f(x) = x + \frac 1x \\g(x) = \frac{x+1}{x+2}\\(fog)(x) = \frac{x+1}{x+2}+\frac{1}{\frac{x+1}{x+2}} = \frac{x+1}{x+2}+\frac{x+2}{x+1}$$
Using the property $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$ I now get: $$\frac{(x+1)^2+(x+2)^2}{(x+2)(x+1)} =(x+1)(x+2)$$
But that is incorrect. The correct answer is: $$\frac{2x^2+6x+5}{(x+2)(x+1)}$$ I understand how this was reached, but I'd like to know what was wrong with the application of that property? Thank you for your help.
| The property works.
$$\frac{(x+1)^2+(x+2)^2}{(x+1)(x+2)}=\frac{x^2+2x+1+x^2+4x+4}{(x+1)(x+2)}=\frac{2x^2+6x+5}{(x+1)(x+2)}$$
I believe your careless mistake is $(x+1)^2+(x+2)^2 \neq (x+1)^2(x+2)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the number of integer solutions of $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$? What is the number of non negative integer solutions of $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$ ?
I tried it as :
${x_{3}}+{x_{4}}={12} - 2{x_{1}}-2{x_{2}}$
Now, finding the solutions of ${x_{1}}+{x_{2}}$
*
*${x_{1}}+{x_{2}} = 0 => 1$ Solution
*${x_{1}}+{x_{2}} = 1 => 2$ Solutions
*${x_{1}}+{x_{2}} = 2 => 3$ Solutions
*${x_{1}}+{x_{2}} = 3 => 4$ Solutions
*${x_{1}}+{x_{2}} = 4 => 5$ Solutions
*${x_{1}}+{x_{2}} = 5 => 6$ Solutions
*${x_{1}}+{x_{2}} = 6 => 7$ Solutions
And, now respectively finding for ${x_{3}}+{x_{4}}$
*
*${x_{3}}+{x_{4}} = 12 => 13$ Solutions
*${x_{3}}+{x_{4}} = 10 => 11$ Solutions
*${x_{3}}+{x_{4}} = 8 => 9$ Solutions
*${x_{3}}+{x_{4}} = 6 => 7$ Solutions
*${x_{3}}+{x_{4}} = 4 => 5$ Solutions
*${x_{3}}+{x_{4}} = 2 => 3$ Solutions
*${x_{3}}+{x_{4}} = 0 => 1$ Solution
Then, Multiplying respective numbers
$1.13 + 2.11 + 3.9 + 4.7 + 5.5 + 6.3 + 7.1 = 140$ Solutions
I don't have answer for it. Am i right here ?
| Sorry for making this an answer - it's a bit too long for a comment on Elliot's answer. As Elliot showed we only need to find the coefficient of $x^{12}$ for $(1-x^2)^{-2}(1-x)^{-2}$. This is quite simple without computer aid with a few tricks. First notice that $1/(1-x)^2 = 1+2x+3x^2+\cdots$. Since we are only interested in the coefficient in front of $x^{12}$ we can truncate the series at $x^{12}$. We have
$$(1-x^2)^{-2}(1-x)^{-2} = (1+2x^2+3x^4+\cdots)(1+2x+3x^2+\cdots)$$
Since the first factor has only terms of even powers we can ignore the odd powers in the second factor - so we only need to compute the coefficient of $x^{12}$ in
$$(1+2x^2+3x^4+4x^6+5x^8+6x^{10}+7x^{12})(1+3x^2+5x^4+7x^6+9x^8+11x^{10}+13x^{12})$$
Now match terms so that the powers sum to 12 giving
$$
13*1+11*2+9*3+7*4+5*5+3*6+1*7=140
$$
The advantage of this manual method is that it easily generalizes to cases were 12 is replaced by another even number.
| {
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"url": "https://math.stackexchange.com/questions/2065465",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Periodicity of an inhomogeneous system of ODEs
Consider the linear $2 \times 2$-system: $x'=Ax+b(t),$ where $b \in C(\mathbb{R}, \mathbb{R}^2)$ is a $2 \pi /{\omega}$-periodic and
$$A=
\begin{pmatrix}
0 & \omega \\ -\omega & 0
\end{pmatrix}$$
for some $\omega >0.$
(a). Find the general solution (depends on two constants $c_1 , c_2 \in \mathbb{R}$).
(b). Show that the general solution is periodic if and only if
$$\int_{0}^{2 \pi/{\omega}}
\begin{pmatrix}
\cos s & -\sin s \\ \cos s & \cos s
\end{pmatrix} \cdot b(s)~ds = \begin{pmatrix}
0 \\ 0
\end{pmatrix}
$$
My approach. I've done part (a). I need help in solving part (b). Thank you for your time.
Using the solution formula for an inhomogenous linear system
$$x=X(t)c + X(t) \int_{0}^{t} X^{-1}(s)b(s)~ds,$$
we get
$$x=\begin{pmatrix}
\cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t
\end{pmatrix} \cdot \begin{pmatrix}
c_1 \\ c_2
\end{pmatrix}+\begin{pmatrix}
\cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t
\end{pmatrix} \int_{0}^{t} \begin{pmatrix}
\cos \omega s & -\sin \omega s \\ \sin \omega s & \cos \omega s \end{pmatrix}
\cdot b(s)~ds,
$$
where $t \in \left[ 0 , 2 \pi /{\omega} \right].$
| Suppose $x(t)$ is periodic. Then $x(0)=x(p)$ for some $p>0$.
Direct computation shows that $x(0)=\begin{pmatrix}c_1\\c_2\end{pmatrix}$. And
$$x(p)=\begin{pmatrix}
\cos \omega p & \sin \omega p \\ -\sin \omega p & \cos \omega p
\end{pmatrix} \cdot \begin{pmatrix}
c_1 \\ c_2
\end{pmatrix}+\begin{pmatrix}
\cos \omega p & \sin \omega p \\ -\sin \omega p & \cos \omega p
\end{pmatrix} \int_{0}^{p} \begin{pmatrix}
\cos \omega s & -\sin \omega s \\ \sin \omega s & \cos \omega s \end{pmatrix}
\cdot b(s)~ds$$
For this to be equal to $X(0)$ for arbitrary $c_1, c_2$, the second the term has to be $0$ and $\cos \omega p=1, \sin\omega p=0$. This shows that $p=\frac{2\pi}{\omega}$. (We want this to be the period, so we choose the smallest number that satisfies the conditions.)
Plugging this into the second term, you can get the desired result.
For the other direction, suppose the integral is zero. We show that $X(t+\frac{2\pi}{\omega})=X(t)$.
Write out $x(t+\frac{2\pi}{\omega})$, noting that $\cos(\omega (t+2\pi/\omega))=\cos \omega t$ and $\sin(\omega (t+2\pi/\omega))=\sin \omega t$, we have
$$x(t+\frac{2\pi}{\omega})=\begin{pmatrix}
\cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t
\end{pmatrix} \cdot \begin{pmatrix}
c_1 \\ c_2
\end{pmatrix}\\
\quad +\begin{pmatrix}
\cos \omega t & \sin \omega t \\ -\sin \omega t & \cos \omega t
\end{pmatrix} \int_{0}^{t+\frac{2\pi}{\omega}} \begin{pmatrix}
\cos \omega s & -\sin \omega s \\ \sin \omega s & \cos \omega s \end{pmatrix}
\cdot b(s)~ds\\
=x(t)+\int_{t}^{t+\frac{2\pi}{\omega}} \begin{pmatrix}
\cos \omega s & -\sin \omega s \\ \sin \omega s & \cos \omega s \end{pmatrix}
\cdot b(s)~ds$$
Now since the integrand is of period $2\pi/\omega$, this integral is equal to
$$\int_{0}^{2 \pi/{\omega}}
\begin{pmatrix}
\cos\omega s & -\sin\omega s \\ \cos\omega s & \cos\omega s
\end{pmatrix} \cdot b(s)~ds = \begin{pmatrix}
0 \\ 0
\end{pmatrix}.$$
This shows the other direction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate the determinant of a $4 \times 4$ matrix with multiple variables? What is the determinant: $$ \begin{vmatrix}1& a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\1 & d & d^2 &d^4 \end{vmatrix} $$
Someone gave me the following hint
Replace $d$ by a variable $x$; make use of the fact that the sum of the roots of a fourth-degree polynomial is equal to the coefficient of $x^3$
but I didn't get that.
| $$\text{Det}(A)=\begin{vmatrix}1& a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\1 & d & d^2 &d^4 \end{vmatrix}=\begin{vmatrix}1& a & a^2 & a^4 \\ 0 & b-a & b^2-a^2 & b^4-a^4 \\ 0 & c-a & c^2-a^2 & c^4-a^4 \\0 & d-a & d^2-a^2 &d^4-a^4 \end{vmatrix}=\begin{vmatrix} b-a & b^2-a^2 & b^4-a^4 \\ c-a & c^2-a^2 & c^4-a^4 \\ d-a & d^2-a^2 &d^4-a^4 \end{vmatrix}$$
thus
$$\text{Det}(A)=(b-a)(c-a)(d-a)\begin{vmatrix} 1 & b+a &(b+a)(b^2+a^2) \\ 1 & c+a & (c+a)(c^2+a^2) \\ 1 & d+a &(d+a)(d^2+a^2) \end{vmatrix}$$
As a result
$$\text{Det}(A)=(b-a)(c-a)(d-a)\begin{vmatrix} 1 & b &(b+a)(b^2+a^2) \\ 0 & c-b & (c+a)(c^2+a^2)-(b+a)(b^2+a^2) \\ 0 & d-b &(d+a)(d^2+a^2)-(b+a)(b^2+a^2) \end{vmatrix}$$
therefore
$$\text{Det}(A)=(b-a)(c-a)(d-a)\begin{vmatrix} c-b & (c+a)(c^2+a^2)-(b+a)(b^2+a^2) \\ d-b &(d+a)(d^2+a^2)-(b+a)(b^2+a^2) \end{vmatrix}$$
In other words
$$\text{Det}(A)=(b-a)(c-a)(d-a)\begin{vmatrix} c-b & (c^3-b^3)+a(c^2-b^2)+a^2(c-b) \\ d-b &(d^3-b^3)+a(d^2-b^2)+a^2(d-b) \end{vmatrix}$$
or
$$\text{Det}(A)=(b-a)(c-a)(d-a)(c-b)(d-b)\begin{vmatrix} 1 & (c^2+bc+b^2)+a(c+b)+a^2 \\ 1 &(d^2+bd+b^2)+a(d+b)+a^2 \end{vmatrix}$$
Finally we have
$$\text{Det}(A)=(b-a)(c-a)(d-a)(c-b)(d-b)\begin{vmatrix} 1 & (c^2+bc+b^2)+a(c+b)+a^2 \\ 0 &(d^2-c^2)+b(d-c)+a(d-c) \end{vmatrix}\\
\quad=(b-a)(c-a)(d-a)(c-b)(d-b)[(d^2-c^2)+b(d-c)+a(d-c)]\\
$$
or
$$\text{Det}(A)=(b-a)(c-a)(d-a)(c-b)(d-b)(d-c)(a+b+c+d)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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A.P. terms in a Quadratic equation.
The terms $a,b,c$ of quadratic equation $ax^{2}+bx+c=0$ are in A.P. and positive. Let this equation have integral root $\alpha,\ \beta$. Then find the value of $\alpha+ \beta + \alpha \cdot \beta$ ?
please point where I'm wrong:
Let common difference be $d$
$\implies \alpha+ \beta + \alpha \cdot \beta=\dfrac{c-b}{a}=\dfrac{d}{a} \implies a|d \ \ \ \ \longrightarrow \ \ \ \ \ \because (b=a+d$, $c=a+2d)$
Also
, $ax^{2}+(a+d)x+(a+2d)=0$.
$\implies$ $\alpha,\ \beta=\dfrac{-(a+d) \pm \sqrt{(a+d)^{2}-4\cdot a \cdot (a+2d)}}{2a}$.
For this to be integer $\sqrt{(a+d)^{2}-4\cdot a \cdot (a+2d)}$ must be perfect square.
$\implies$ ${(a+d)^{2}-4\cdot a \cdot (a+2d)}=p^{2}$ for some $p$.
$\implies -3a^{2}+d^{2}-6ad=p^{2}$
$\implies -3a^{2}+a^{2}q^{2}-6a^{2}q=p^{2}$ $\because$ $a|d \implies aq=d$ for some $q$.
$\implies a^{2}(-3+q^{2}-6q)=p^{2}$
$\implies -3+q^{2}-6q\ $ has to be perfect square. By trial $q=7$
But I need to get this without trial, please help.
| We have, $$p(x)=ax^2+bx+c=ax^2+(a+p)x+(a+2p)$$
and thus, $$t+r+r\cdot t=-\frac{a+p}{a}+\frac{a+2p}{a}=\frac{p}{a}$$
then $p=a\cdot k$, $k=(t+r+t\cdot r) \in \Bbb Z$.
Then our polynomial becomes,$$p(x)=ax^2+a(1+k)x+a(1+2k)$$
$$t=\frac{-(1+k) \pm \sqrt{k^2-6k-3}}{2} \quad...(1)$$
So
$$k^2-6k-3=q^2 \Rightarrow (k-3)^2-12=q^2 \Rightarrow (k+q-3)(k-q-3)=12$$
and split $12$ as a product of two integer and find all possible values for $k$.
Take for example:
\begin{cases}
k+q-3=6 \Rightarrow k+q=9 \\
k-q-3=2 \Rightarrow k-q=5
\end{cases}
Adding up both equations we get $2k=14 \Rightarrow k=7$
And back to $(1)$, we get $(t,r)=(-3,-5)$ or $(t,r)=(-5,-3)$.
| {
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Multiplicative inverse of $2x + 3 + I$ in $\mathbb{Z}_5[x]/I $? In $\mathbb{Z}_5[x]$, let $ I = \langle x^2 + x + 2\rangle$. Find the multiplicative inverse of $2x + 3 + I$ in $\mathbb{Z}_5[x]/I$.
The answer is supposed to be $3x + 1 + I$. But I don't understand how you find this. I divided $x^2 + x + 2$ by $2x+3$ to get $$ x^2 + x + 2 = (2x + 3)(3x + 1) + 4.$$ But I'm not sure how this will help me. I know that the elements of $\mathbb{Z}_5[x]/I$ are of the form $ax + b$, with $a, b \in \mathbb{Z}_5$. So I want $$ 1 + I = (ax + b)(2x+3) + I = (2ax^2 + (3a+2b)x + 3b) + I. $$ I want to find the coefficients $a, b$ from this. We must have $3b = 1$, and the coefficients by $x$ and $x^2$ must vanish. Since $\mathbb{Z}_5$ is a field, the relation $3b = 1$ means $b$ is the multiplicative inverse of $3$. So $b = 2$. From the coefficients of $x$, I get also $3a = -2b$. I multiply this with $2$ to get $a = 6a = -4b = -8 = 2$.
So I would say the inverse is $2x + 2$, in contradiction with the answer at the back of my book. So where is the mistake in my reasoning, and how to find correct answer?
| Your original approach was on the right track. By dividing, you found that
$$
x^2 + x + 2 = (2x + 3)(3x + 1) + 4 \, .
$$
Rearranging, we have
\begin{align*}
(2x + 3)(3x + 1) = -4 + x^2 + x + 2 = 1 + x^2 + x + 2 \equiv 1 \pmod{x^2 + x + 2}
\end{align*}
so $3x + 1$ is the inverse of $2x+3$.
In general, if you have the equality $a f + b g= u$ where $u$ is a unit, you can multiply through by $u^{-1}$ to get a linear combination that yields $1$: $u^{-1} a f + u^{-1} b g = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the GCD of $10$ integers
Find the greatest common divisor of the following ten integers:
\begin{align}
& 2000^3+3\cdot2000^2+2\cdot2000,\ 2001^3+3\cdot2001^2+2\cdot2001,\ \dots, \\[5pt]
& \dots, \ 2008^3+3\cdot2008^2+2\cdot2008,\ 2009^3+3\cdot2009^2+2\cdot2009.
\end{align}
How are these numbers related to each other? Is there a formula for this?
| You've got $n(n+1)(n+2)$ for $n=2000,\ldots,2009$. You want divisors that they all share in common.
Given three consecutive integers, exactly one of them is divisible by $3$; therefore their product is divisible by $3$. That is true of all $10$ of the products, so $3$ is a common divisor. Among the ten products, there are some, such as $2000\times2001\times2002$ in which $9$ is not a divisor. Therefore the highest power of $3$ that divides all ten products is $3$ itself.
If $n$ is even then so is $n+2$, and one of them is divisible by $4$ and the other is not, so the product is divisible by $8$. But if $n$ is odd, then so is $n+2$, and $n+1$ is even. Among the cases where $n+1$ is even, we have $n+1=2002$, which is not divisble by $4$, so $4$ is not a common divisor of all ten products. The highest power of $2$ that divides all ten products is therefore $2$ itself.
The next prime number is $5$. Some of these products are not divisible by $5$; for example $2001\times2002\times 2003$ is not. So $5$ is not a common divisor.
$7$ divides $2002$ and $2009$ and nothing between those, so $2003\times2004\times2005$ is not divisible by $7$.
The next prime is $11$. This one divides $2002$ but nothing else until you get up to $2013$, which is not included, so $11$ does not divide $2003\times2004\times 2005$. In a similar way you can rule out all larger primes as common divisors.
Therefore the g.c.d. is $2\times3=6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $\dfrac{\mathrm dx(t)}{\mathrm dt}=(2x(t)+6)(-2t^5+4t^4+2t^2)$ if $x(0)=-3$? How do I solve the following differential equation?
$$\frac{\mathrm dx(t)}{\mathrm dt}=(2x(t)+6)(-2t^5+4t^4+2t^2),\quad x(0)=-3$$
I've searched on Wolfram Alpha, and it shows: expand the $(2x(t)+6)(-2t^5+4t^4+2t^2)$, distribute $-2t^5+4t^4+2t^2$ over $6+2x(t)$. It's totally incomprehensible.
Thanks.
| $\dfrac{dx}{dt}=(2x+6)(-2t^5+4t^4+2t^2)$
$\implies \displaystyle \int \dfrac{dx}{2x+6}=\int -2t^5+4t^4+2t^2\,dt$
$\implies \dfrac{1}{2}\ln|2x+6|=-\dfrac{1}{3}t^6+\dfrac{4}{5}t^5+\dfrac{2}{3}t^3+\ln C_1$
$\implies \ln|2x+6|=-\dfrac{2}{3}t^6+\dfrac{8}{5}t^5+\dfrac{4}{3}t^3+\ln C$
$\implies \ln\left|\dfrac{2x+6}{C}\right|=-\dfrac{2}{3}t^6+\dfrac{8}{5}t^5+\dfrac{4}{3}t^3$
$\implies 2x+6=Ce^{-\frac{2}{3}t^6+\frac{8}{5}t^5+\frac{4}{3}t^3}$
$\implies x=\dfrac{1}{2}[Ce^{-\frac{2}{3}t^6+\frac{8}{5}t^5+\frac{4}{3}t^3}-6]$
Using $x(0)=-3$
$\implies -3=\dfrac{1}{2}\left(C-6\right)$
$\implies C=0$
Hence the solution:
$x(t)=-3$
| {
"language": "en",
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How can I isolate $x$ in this equation? Simultaneous equations - $7^{x-1} = 49 \hspace{1 mm} (7^{y})$ --------- (i)
$\hspace{46 mm} 3^x+3^y = 84$ ---------- (ii)
The first I make x subject of equation in this way:
$\Longrightarrow 7^{x-1} = 49 \hspace{1 mm} (7^{y})$
$\Longrightarrow 7 \times 7^x = 7^2 \times 7^y$
$\Longrightarrow 7^x = 7^{y+1}$
$\Longrightarrow x = y + 1$
The second I tried with logarithms but could not separate x in simple form:
$\Longrightarrow 3^x + 3^y = 84$
$\Longrightarrow 3^x = 84 - 3^y$
$\Longrightarrow \lg3^x = \lg (84 - 3^y)$
$\Longrightarrow x \lg3 = \lg(84 - 3^y)$
$\Longrightarrow x = \dfrac{\lg(84 - 3^y)}{\lg3}$
How can I simplify this further?
| First multiply equation $(i)$ by $7$ to get rid of the confusing $-1$ in the exponent and obtain $$7^x=7^{y+3}$$
Then for the second part, note that $$3^{y+3}+3^y=3^y\left(3^3+3^0\right)$$
These two key steps should make a solution rather easy.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a+b+c = 6$ and $a$, $b$, $c$ are nonnegative then $a^2+b^2+c^2 \geq 12$ Let $a,b,c$ be three positive real numbers such that $a+b+c = 6$. Prove that $a^2+b^2+c^2 \geq 12$.
I tried using the AM-GM inequality to solve the same, however I wasn't able to make any considerable progress.
| We need to prove that
$$a^2+b^2+c^2\geq12\left(\frac{a+b+c}{6}\right)^2$$ or
$$\sum\limits_{cyc}(a-b)^2\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Diophantine Equation $x^2y^2-4x-4y=z^2$ Is it true that the Diophantine equation
$$
x^2y^2-4x-4y=z^2\; ; \; x,y>0
$$
has the only solutions $(x,y)=(2,2),(2,3),(1,5),(3,2),(5,1)$?
What is the general solution in all integers?
| Note that the perfect squares next to $x^2y^2$ are
$$x^2y^2 \pm 2xy +1.$$
Therefore if $(x , y, z)$ with $x, y > 0$ is an integral point on $z^2 = x^2y^2 - 4(x + y)$, then
$$-4(x + y) \leq 1 - 2xy.$$
It follows that any solution must be of the form $(1, y)$, $(2, y)$, $(x, 1)$, $(x, 2)$, $(x, 3)$, $(x, 4)$, $(x, 5)$ or $(x,6)$.
We can use a similar argument to see which of those forms yield solutions of $x^2y^2 - 4(x + y) = z^2$.
In the family $(1, y)$ we have
$$z^2 = (y - 2)^2 - 8$$
yielding only the point $(x, y) = (1, 5)$.
In the family $(2, y)$ we have
$$z^2 = (2y - 1)^2 - 9$$
yielding the points $(x, y) = (2, 2)$ and $(2,3)$.
In the families $(x, 1)$ and $(x ,2)$ we only have the points $(5, 1)$, $(2,2)$ and $(3, 2)$ by symmetry in the equation.
In the family $(x , 3)$ we have
$$z^2 = (3x - 1)^2 + 2x - 13$$
yielding the point $(x, y) = (2,3)$.
In the family $(x, 4)$ we have
$$z^2 = (4x - 1)^2 + 4x - 17$$
but there is no $x$ for which this can happen.
In the family $(x , 5)$ we have
$$z^2 = (5x - 1)^2 + 6x - 21$$
yielding the point $(x, y) = (1,5)$.
In the family $(x, 6)$ we have
$$z^2 = (6x - 1)^2 + 8x - 25$$
but there is no $x$ for which this can happen.
So the list of solutions in your question is indeed complete for $x,y > 0$.
Over all integers you have infinitely many solutions. Here's an outline of the (hopefully) complete scenario.
When $x, y < 0$, we can adapt the method above to obtain no solution $(x,y)$ besides the infinite families $(-1, -t, \pm(t - 2))$ and $(-t, -1, \pm(t - 2))$.
When $x = 0$ we have the integral points $(0, -t^2, 2t)$, $t \in \mathbf{Z}$. Similarly, when $y = 0$ we have the solutions $(-t^2, 0, 2t)$.
When $x$ and $y$ have opposite signs, we can regard this problem as trying to find the solutions to
$$z^2 = a^2b^2 -4a + 4b, \quad a, b > 0.$$
When $-4a + 4b = 0$, we obtain the family of solutions $(x, y, z) = (-t, t, t^2)$, $t \in \mathbf{Z}$.
When $-4a + 4b > 0$, we want $4b - 4a \geq 2ab + 1$ but that can never happen.
When $-4a + 4b < 0$, we want $4b - 4a \leq -2ab + 1$, in which case $b = 1$ and the equation becomes $z^2 = (a - 2)^2$. This corresponds to integral points $(x,y,z) = (2 + t, -1, t)$, $t \in \mathbf{Z}$, $t \geq 0$.
Summing up, these should be all solutions:
*
*when $x, y > 0$: here.
*$(t, -t, \pm t^2)$, for $t \in \mathbf{Z}$.
*$(-t^2, 0, 2t)$ for $t \in \mathbf{Z}$.
*$(0,-t^2, 2t)$ for $t \in \mathbf{Z}$.
*$(2 + t, -1, t)$ for $t \in \mathbf{Z}$.
*$(-1, 2 + t, t)$ for $t \in \mathbf{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076952",
"timestamp": "2023-03-29T00:00:00",
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solving logarithmic equation $\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x}$ Solve for $x$
$$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x} $$
| Assume that $\log$ is the natural logarithm and $x > 1$ (Otherwise, it would not make sense the power in LHS). Since $\log \sqrt{x} = \dfrac{\log x}{2}$, we can simplify the equation as $$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x} \implies \left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 3)} = 1$$
$$\implies \left(\log^2x + 2\log x - 3\right)\log\left(\dfrac{\log x}{2}\right) = 0.$$
Thus, $$\log\left(\dfrac{\log x}{2}\right) = 0 \implies x = e^{2}.$$ Or,
$$\log^2x + 2\log x - 3 = 0 \implies \log x = -1\pm2 \implies x = e ~~\text{and}~~x = e^{-3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$. I have an inequality problem which is as follow:
If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$.
I am not so good in inequalities. So, please give me some hints so that I can proceed.
Thanks.
| From given equality we have $c=\sqrt {a^2+b^2-ab}$, then you can plug this in your inequality and you get $(a-b)(b-\sqrt{a^2+b^2-ab})$, now you have 2 cases to look at:
First: $a>b, a=b+x$ with this your first term is positive, so we have to show second one is negative, thus we have $b-\sqrt{b^2+bx+x^2};\sqrt{b^2+bx+x^2} > \sqrt {b^2}$, it is obvious from here that the product is negative.
Second: $a<b,a=b-x$, now we know that first term is negative, so we have to show that second one is positive, again we plug in and get $b-\sqrt{b^2-bx-x^2};\sqrt{b^2-bx-x^2} < \sqrt {b^2}$, so we know that this one is positive, so the product is again negative.
And finally if $a=b$ then their product is $0$ because $a-b=0$.
| {
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Prove that: $\csc 50° + \cot 100° = \cot 25° -\csc 100°$ Prove that: $$\csc 50° + \cot 100° = \cot 25° -\csc 100°$$
My Attempt:
$$L.H.S=\csc 50° + \cot 100°=\frac {1}{\sin 50°} + \frac {\cos 100°}{\sin 100°}$$
$$=\frac {1}{\sin 50°} + \frac {\cos 100°}{2\sin 50°\cos 50°}$$
$$=\frac {2\cos 50° + \cos 100°}{2\sin 50°\cos 50°}$$.
Now, what should I do?
| This would work better may be :
$$\csc 50° + \cot 100° = \cot 25° -\csc 100°$$
$$\csc 50° + \cot 100° +\csc 100° = \cot 25° $$
$$\frac{1}{\sin 50°} + \frac{\cos 100°}{\sin 100°} +\frac{1}{\sin 100°} = \cot 25° $$
$$\frac{1}{\sin 50°} + \frac{\cos 100° + 1}{\sin 100°} = \cot 25° $$
$$\frac{1}{\sin 50°} + \frac{2\cos^2 50° }{2\sin 50°\cos 50°} = \cot 25° $$
$$\frac{1}{\sin 50°} + \frac{\cos 50° }{\sin 50°} = \cot 25° $$
$$\frac{1+\cos 50° }{\sin 50°} = \cot 25° $$
$$\frac{2\cos^2 25° }{2\sin 25°\cos 25°} = \cot 25° $$
Done right?
| {
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How to find $3^{3^{3^{\dots}}}\pmod{100}$? I can show that $3^{3^{3^n}}\equiv7\pmod{10}$ since
$3^1\equiv3\pmod{10}$
$3^2\equiv9\pmod{10}$
$3^3\equiv7\pmod{10}$
$3^4\equiv1\pmod{10}$
Thus, it reduces to $3^{(3^{3^n}\mod4)}$. I can then notice that
$3^1\equiv3\pmod4$
$3^2\equiv1\pmod4$
Reducing it down to $3^{(3^{(3^n\mod2)}\mod4)}=3^{(3^1\mod4)}=3^3\equiv7\pmod{10}$
However, this is tedious and not capable of solving the following problem:
$$3^{3^{3^{\dots}}}\pmod{100}$$
where the power tower keeps going up until the value becomes fixed for all further power towers. How would I take this problem?
To clarify a bit, we take the exponents at the top and work our way down. For example,
$$3^{3^3}=3^{27}\ne(3^3)^3$$
For clearer notation, $3^{3^{3^{\dots}}}=3\uparrow(3\uparrow(3\uparrow(\dots(3\uparrow n)\dots)))$, and we take as many $\uparrow$'s required such that for all $n,k\in\mathbb N$ we have
$$3\uparrow(3\uparrow(3\uparrow(\dots(3\uparrow n)\dots)))\equiv3\uparrow(3\uparrow(3\uparrow(\dots(3\uparrow k)\dots)))\pmod{100}$$
In Knuth's up arrow notation.
| As $3\equiv-1\pmod4,3^{2n+1}\equiv-1\equiv3$ for any integer $n\ge0$
So, $3^{3^{3^{\cdots}}}$ can be written as $\displaystyle3^{4k+3}$ or $\displaystyle3^{3^{(4b+3)}}$
Now,$\displaystyle3^{4k+3}=27(10-1)^{2k}=27(1-10)^{2k}$
and $\displaystyle(1-10)^{2k}\equiv1-\binom{2k}110\pmod{10^2}\equiv1-20k$
So, it is sufficient to find $4k+3=3^{(4b+3)}\pmod5$
$\displaystyle4k+3=3^{(4b+3)}=3^3(3^4)^b\equiv2\cdot1^b\pmod5\equiv2$
$\displaystyle\implies4k\equiv-1\pmod5\equiv4\implies k\equiv1$ as $(4,5)=1$
$\displaystyle\implies(1-10)^{2k}\equiv1-20\cdot1\equiv81\pmod{100}$
$\displaystyle\implies3^{3^{(4b+3)}}\equiv27\cdot81\pmod{100}\equiv?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Taylor theorical question: solving an inequality Prove by using Taylor formula that in a neighbourhood of zero this inequality holds:
$\cosh x \le 12 \frac{x^2+2}{24-x^4}$
Please explicitate everything you do :)
Thanks in advance
| The hint pretty much tells you everything you need to do: take the Taylor expansion of the difference of the two sides, until you see a non-zero term. Then check the sign of that term near $0$.
In particular, the theorem that we will use is the following:
Theorem. Let $f,g$ be two functions defined on a (pointed) neighborhood of zero, and such that $g$ does not cancel there. Then, if $f(x) = g(x) + o(g(x))$ when $x\to 0$, $f$ and $g$ keep the same sign on a neighborhood of $0$.
Proof. We have $\frac{f(x)}{g(x)} = \frac{g(x)+o(g(x))}{g(x)} = 1+o(1) \xrightarrow[x\to 0]{} 1$, and thus there exists $\delta > 0$ such that $\frac{f(x)}{g(x)} \in \left(\frac{1}{2},\frac{3}{2}\right)$ for all $x\in(\delta,\delta)$. $\qquad\square$
Now, we will invoke this theorem at the end: we will prove that $12\cdot \frac{2+x^2}{24-x^4} - \cosh x = \alpha x^c + o(x^c)$ for some $\alpha\neq 0,c>0$ (by performing a Taylor expansion of both sides of the inequality until the point where they differ) and be quite happy to notice that $c$ is even, and $\alpha > 0$. Then, we will "set" $g(x) = \alpha x^c$ and conclude.
When $x\to 0$, we have (this one is standard)
$$
\cosh x = 1+\frac{x^2}{2}+\frac{x^4}{24}+\frac{x^6}{720}+o(x^6)
$$
and
$$\begin{align}
12\cdot \frac{2+x^2}{24-x^4} &=
\frac{1+\frac{x^2}{2}}{1-\frac{x^4}{24}}
= \left(1+\frac{x^2}{2}\right)\left(1+\frac{x^4}{24}+O(x^8)\right)\\
&= 1+\frac{x^2}{2}+\frac{x^4}{24}+\frac{x^6}{48}+O(x^8)\\
&= 1+\frac{x^2}{2}+\frac{x^4}{24}+\frac{x^6}{48}+o(x^6)
\end{align}$$
and thus
$$
12\cdot \frac{2+x^2}{24-x^4} - \cosh x = \frac{7}{360}x^6 + o(x^6).
$$
Since $\frac{7}{360}>0$, the difference is positive on a neighborhood of $0$, as it is equivalent to $\frac{7}{360}x^6$. This shows that on that very same neighborhood of $0$, the inequality $$
12\cdot \frac{2+x^2}{24-x^4} > \cosh x$$ holds by the above theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate $\frac{x}{e^x - 1}$ w.r.t. x? A friend of mine and I wanted to solve the following indefinite integral but got stuck:
$$
\int \frac{x}{e^x - 1} dx.
$$
My approach:
Let
$$
I = \int \frac{x}{e^x - 1} dx.\\
\implies I = x\int \frac{dx}{e^x - 1} - \int \left ( \int \frac{dx}{e^x - 1} \right ) dx.
$$
Now, let $I_2 = \int \frac{dx}{e^x - 1}$. Also, let $z = e^x \implies dx = {dz \over z}$. Then,
$$
I_2 = \int \frac{dz}{z(z-1)} \\
\implies I_2 = \int {dz \over {z - 1}} - \int {dz \over z} \\
\implies I_2 = \ln (e^x - 1) - x.
$$
Substituting the value of $I_2$ in $I$, we get,
$$
I = x[\ln (e^x - 1) - x] + {x^2 \over 2} - \int \ln (e^x - 1) dx.
$$
I got stuck right here. Is it possible to proceed further?
| \begin{equation}
I = \int \frac{x}{\mathrm{e}^{x}-1} dx = -I_{1} = -\int \frac{x}{1-\mathrm{e}^{x}} dx
\end{equation}
Integrate by parts
\begin{align}
I_{1} &= \int \frac{x}{1-\mathrm{e}^{x}} dx \\
\tag{1}
&= x^{2} - x \ln(1-\mathrm{e}^{x}) - \int x dx + \int \ln(1-\mathrm{e}^{x}) dx \\
\tag{2}
&= x^{2} - x \ln(1-\mathrm{e}^{x}) - \frac{x^{2}}{2} - \mathrm{Li}_{2}(\mathrm{e}^{x})
\end{align}
and thus
\begin{equation}
\int \frac{x}{\mathrm{e}^{x}-1} dx = x \ln(1-\mathrm{e}^{x}) - \frac{x^{2}}{2} + \mathrm{Li}_{2}(\mathrm{e}^{x})
\end{equation}
*
*Let $u=\mathrm{e}^{x}$
\begin{align}
\int \frac{1}{1-\mathrm{e}^{x}} dx &= \int \frac{1}{u(1-u)} du \\
&= \int \left( \frac{1}{u} + \frac{1}{1-u} \right) du \\
&= \ln \frac{u}{1-u} \\
&= x - \ln(1-\mathrm{e}^{x})
\end{align}
*Let $u=\mathrm{e}^{x}$
\begin{align}
\int \ln(1-\mathrm{e}^{x}) dx &= \int \frac{\ln(1-u)}{u} du \\
&= -\mathrm{Li}_{2}(u) \\
&= -\mathrm{Li}_{2}(\mathrm{e}^{x})
\end{align}
where
\begin{equation}
\mathrm{Li}_{2}(z) = -\int\limits_{0}^{z} \frac{\ln(1-x)}{x} dx
\end{equation}
is the dilogarithm function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solutions for $a^2+b^2+c^2=d^2$ I have to find an infinite set for $a,b,c,d$ such that $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb N$
I have found one possible set which is $x,2x,2x,3x$ and $x\in\mathbb N$
But I want another infinite sets and how to reach them because I have found my solution by hit and trial .
| As you noticed, given a solution you can multiply it by any natural greater than $1$ to get another solution, so we can concentrate on solutions that do not have a common divisor. You can just keep finding solutions by hand. A more systematic approach is to write $a^2+b^2=d^2-c^2=(d+c)(d-c)$ The two terms on the right are either both odd or both even. You can just choose $a,b$ with one odd and one even and $a \gt b$, then factor $a^2+b^2$. So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$ and a new solution is $4,1,8,9$. You can always use the factor with $1$ giving $a,b,\frac 12(a^2+b^2-1),\frac12(a^2+b^2+1)$ as a fundamental solution whenever $a,b$ are of opposite parity. If $a^2+b^2$ is not prime or the square of a prime there will be other solutions. So for $a=8, b=1$ we have $a^2+b^2=65=5 \cdot 13$, and we find $8,1,4,9$ as a solution.So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$. Unfortunately here it give a permutation of the solution we already have. If we choose $a=9,b=2$ we get $a^2+b^2=85=5 \cdot 17$ and find the solution $9,2,6,11$ as well as $9,2,42,43$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Formula for calculating moduli of roots of complex numbers from determinants - not working Determinants are used here as scaling factors for the areas of the polygons generated by connecting the individual roots of $z$.
*
*The roots of complex numbers are given by the formula $$z_k=\sqrt[n]{r}\Big(\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}\Big)$$
*Two consecutive roots ($k=0$ and $k=1$) can be represented as
$$\begin{bmatrix}\sqrt[n]{r}(\cos\frac{\theta}{n})&\sqrt[n]{r}(\cos\frac{\theta+2\pi}{n}) \\ \sqrt[n]{r}(\sin\frac{\theta}{n})&\sqrt[n]{r}(\sin\frac{\theta+2\pi}{n})\end{bmatrix}$$
*And the determinant of that, after some algebraic manipulation, is given by $$\text{det}=\sqrt[n]{r^2}\Big(\sin\frac{2\pi}{n}\Big)$$
*After excluding the cases where $\displaystyle\sin\frac{2\pi}{n}$ is $0$, I arrived with a final formula of
$$r=\Big(\frac{\text{det}}{\sin\frac{2\pi}{n}}\Big)^{\frac{n}{2}}$$
However, the values generated by this formula are incorrect. Here is an example with $n=4$, $\text{det}=3$; the resultant area is $162$ instead of $6$).
The area $a$ for a polygon with $n=4$ and $r=1$ is $4(\frac{r^2}{2})=4(\frac{1}{2})=2$.
The area $A$ for the same polygon is then expected to be $a\times\text{det}=2\times3=6$. Instead I get the following:
$$r=\Big(\frac{\text{det}}{\sin\frac{2\pi}{n}}\Big)^{\frac{n}{2}}=\Big(\frac{3}{\sin\frac{2\pi}{4}}\Big)^{\frac{4}{2}}=9$$
$$A=4(\frac{r^2}{2})=4(\frac{9^2}{2})=162\neq6$$
At which step did I make an error? Are my assumptions correct in the first place?
Edit: My biggest concern right now is whether complex numbers are allowed to be represented as in step 2.
| Let me write some words, from your statements. Maybe it works:
Suppose all vertices of a regular polygon are on a circle with radius $r$.
(1). The n-th roots of 1 as a complex number are given by the formula $$z_k=\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}$$
(2). All of these roots lie on circle $|z|=r$, so following points are the vertices of regular polygon
$$w_k=r\Big(\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}\Big)$$
(3). Every two consecutive points (points $k$ and $k+1$) with origin as triangle third vertex, swap an area $A_k$ equal to
$$A_k=\frac12\text{det}\begin{bmatrix}r\cos\frac{\theta+2k\pi}{n}&r\cos\frac{\theta+2(k+1)\pi}{n}\\r\sin\frac{\theta+2k\pi}{n}&r\sin\frac{\theta+2(k+1)\pi}{n}\end{bmatrix}$$
(4). And the determinant of that, after some algebraic manipulation, gives us
$$A_k=\frac12r^2\sin\frac{2\pi}{n}$$
(5). After adding all areas, the total area of polygon is
$$A=\sum_{k=0}^{n-1}A_k=\frac{nr^2}{2}\sin\frac{2\pi}{n}$$
(6). If our process were done in true way, We must have
$$\lim_{n\to\infty}A=\pi r^2$$
is the area o circle with radius $r$, because as $n\to\infty$, the polygon will convert to circle.
For example with $n=4$, the area of square is
$$A=\frac{4r^2}{2}\sin\frac{2\pi}{4}=2r^2$$
With $n=6$, the area is
$$A=\frac{6r^2}{2}\sin\frac{2\pi}{6}=\frac{3\sqrt{3}}{2}r^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Strategy on factoring quartic polynomials: cross method. I found a method to factorize quartic polynomials which I don't understand how it works.
It is presented like this:
Cross method
This methodology lets to factorize ordered and completed 4th
polynomials of the form:
$$F(x) = ax^4+bx^3+cx^2+dx+e$$
Rules:
*
*Factor the extreme terms with cross method to get a squared term (generally different from the squared term of the original
polynomial).
*Get $\Delta$ from the difference of the squared term of the polynomial and the term of the first step, and replace the result in
the original polynomial.
*Then, verify the binary combinations as double cross factoring.
I used for different exercises and it works, but I don't understand the basis for this method.
Can anyone explain the reasons?
P.D.
Example
$$x^4+2x^3+3x^2+2x-3$$
*
*Factoring the extremes terms; $x^4$ and $-3$:
$$(x^2+3)(x^2-1)$$
The result of the cross method is $3x^2-x^2=2x^2$
*Calculating $\Delta$:
$$\Delta=\text{(original squared term)}-\text{(step one squared term)}= (3x^2)-(2x^2)=x^2$$
And replaced it in the original polynomial:
$$x^4+2x^3+x^2+2x-3$$
*Using the cross method for second and fourth term:
For second term: $(x^2+x)(x^2+x)$ is $2x^3$
For fourth term: $(x+3)(x-1)$ is $2x$
So, arranging the terms, the two factors are:
$$(x^2+x+3)(x^2+x-1)$$
| I might be misunderstanding, but this just looks like concealed expanding $$(x^2+ax+b)(x^2+cx+d)\tag{1}$$ and comparing coefficients.
*
*Factoring extremes
Well, what we want is to choose $b$ and $d$ such that $(x^2+b)(x^2+d)$ matches leading and constant factor. What it amounts to is finding $b$ and $d$ such that $bd = -3$. Let's say we were lucky and chose the winning pair $b = -1,\ d=3.$
*Calculating $\Delta$
Why is this important? Well, $(x^2-1)(x^2+3) = x^4+2x^2-3$, so we are missing one $x^2$: $\Delta = 3x^2 - 2x^2$.
*Covering for the missing $\Delta$
There is only one place where $\Delta$ can come from, that is from $a$ and $c$ - we must have $ac x^2 = \Delta$. Why? Well, if you expand $(1)$, you will find that $x^2$ term must be equal to $b + d + ac$. We found $b+d$ in step 1 and 2, so $acx^2$ is equal to $\Delta$. In this case $ac = 1$. There are two possibilities in $\Bbb Z$, either $a=c=1$, or $a=c=-1$. It is easy to decide which one, since it must fit the other terms as well, namely $x^3$ term is equal to $a + c$ and the $x$ term is equal to $ad + bc = 3a - c$. Thus, $a = c = 1$.
*Collecting terms
We simply write $(x^2+x-1)(x^2+x+3) = x^4+2x^3+3x^2+2x-3$ and feel the magic all around us.
Now, let us do the same thing, only transparently:
$$(x^2+ax+b)(x^2+cx+d) = x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$
so we need to solve the following system in $\Bbb Z$:
\begin{array}{c r}\begin{align}
a+c&=2\\
b+d+ac&=3\\
ad+bc&=2\\
bd &= -3
\end{align} &\tag{2}\end{array}
There are two possibilities, either $b = 1$ and $d = -3$, or $b = -1$ and $d = 3$. We can see that former leads to no solution, so we will pursuit the latter, the system becomes:
\begin{align}
a+c&=2\\
ac&=1\\
3a-c&=2
\end{align}
The first and the third equation form linear system, which has unique solution $a = c = 1$, luckily, consistent with the second equation. Thus, we now know for certain that $$(x^2+x-1)(x^2+x+3) = x^4+2x^3+3x^2+2x-3.$$
Addendum:
There is a legitimate question why we would consider $(1)$ in the first case. Now, we are trying to factor a quartic over $\Bbb Z$. Let's say $p(x) = x^4 +2x^3+3x^2+2x-3$ and assume that $p = fg$. We have that $\deg p = \deg f + \deg g$, so there are two cases: $4=1+3$ or $4=2+2$. The first case would imply that $p$ has integer root and we can use rational root theorem to rule that out. Thus, only case $4=2+2$ remains, which leads to considering $(1)$.
Finally, if the system $(2)$ had no solutions in $\Bbb Z$, we would conclude that $p$ is irreducible over $\Bbb Z$.
| {
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"url": "https://math.stackexchange.com/questions/2090587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Inequality with square roots : $\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}+\frac{\sqrt{a+b}}{c}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$ Let $a, b, c$ be positive real numbers.
Prove that $\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}+\frac{\sqrt{a+b}}{c}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$
Please check if my work is correct or not.
$\displaystyle\sum_{cyc}\frac{\sqrt{b+c}}{a}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$
It's sufficient to show that $\displaystyle\sum_{cyc}\frac{(b+c)\sqrt{(a+b)(c+a)}}{a}\geq 4(a+b+c)$
By C-S,
$\displaystyle\sum_{cyc}\frac{(b+c)\sqrt{(a+b)(c+a)}}{a}\geq\displaystyle\sum_{cyc}\frac{(b+c)(a+\sqrt{bc})}{a}$
$ RHS = \displaystyle\sum_{cyc}(b+c) + \sum_{cyc}\frac{(b+c)\sqrt{bc}}{a} = \displaystyle\sum_{cyc}(b+c) + \sum_{cyc}\frac{2\sqrt{bc}\sqrt{bc}}{a} = 2\sum_{cyc}a + \sum_{cyc}\frac{bc}{a} = 4\sum_{cyc}a$
We're done.
| I think this inequality is not so strong. See my solution here:
https://artofproblemsolving.com/community/c6h1446900p8274628
and here:
https://artofproblemsolving.com/community/c6h1447863p8284378
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Question about inverse function $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ This was from iranian university entrance exam .Suppose $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ find $f^{-1}(x)+f^{-1}(\frac{1}{x}),x \neq 0$.
It is easy to find $f^{-1}$ and solve this like below ...
$$y=\frac{1}{2}(x+\sqrt{x^2+4}) \\(2y-x)^2=(\sqrt{x^2+4})^2\\4y^2+x^2-4xy=x^2+4\\4y^2-4=4xy\\x=\frac{y^2-1}{y}=y-\frac{1}{y} \\ \to f^{-1}(x)=x-\frac{1}{x}\\
f^{-1}(x)+f^{-1}(\frac{1}{x})=x-\frac{1}{x}+(\frac{1}{x}-\frac{1}{\frac{1}{x}})=0\\$$ And now my question is ...
is there an other method to solve this question ?
I was thinking about $f(x)f(-x)=1$ but I can't go anymore ...
Any hint ,or other Idea ?
Thanks in advanced.
| The short version:
$$f(x)f(-x)=1\implies f^{-1}(f(x))+f^{-1}\left(\frac1{f(x)}\right)=f^{-1}(f(x))+f^{-1}\left(f(-x)\right)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Calculate $\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx$ $$\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx$$
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$$z=\alpha(x)=e^{ix}$$
$$\sin(x)=\frac{z^2-1}{2zi}$$
$$2+\sin(x)=\frac{4zi+z^2-1}{2zi}$$
$$\frac{1}{2+\sin(x)}=\frac{2zi}{4zi+z^2-1}$$
$$\int_\alpha \frac{2zi}{4zi+z^2-1} \ \ \frac{1}{zi}$$
$$4zi+z^2-1=0$$
$$z_1=\frac{-4i+2\sqrt{3}i}{2}$$
$$z_2=\frac{-4i-2\sqrt{3}i}{2}$$
$$\rvert z_1 \rvert <1$$
$$\rvert z_2 \rvert >1$$
Using Residue theorem:
$$\int_\alpha \frac{2}{4zi+z^2-1}=2 \pi i \ \lim_{z\rightarrow-2i+\sqrt{3}i} \ \frac{2}{z+2i+\sqrt{3}i}=\frac{2 \pi}{\sqrt{3}}$$
Is it correct?
Thanks!
| Yes, absolutely correct but you may try alternative approach by converting $\sin\space x$ into $\tan\space (x/2)$ $$\sin(x) = \frac{2\cdot \tan(x/2)}{1+\tan^2(x/2)}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Find indefinite integral $\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$ $$\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$$
$$\frac {1}{x^2\sqrt{1-x^2}}+2\int \frac{1}{x^3\sqrt{1-x^2}}\, \mathrm{d}x$$
I try to integrate by parts method, but its doesnt want to be solved. I try to substitue $x=\sin u \mathrm{d}x=\cos u$ but failed somewhere
| $\int \frac{\arcsin(x)}{x^2}dx=\int \arcsin(x) d(-\frac{1}{x})=-\frac{1}{x}\arcsin(x)+\int \frac{dx}{x \sqrt{1-x^2}}= -\frac{1}{x}\arcsin(x)+\int \frac{dx}{x^2 \sqrt{\frac{1}{x^2}-1}}=-\frac{1}{x}\arcsin(x)+\int \frac{d(-\frac{1}{x})}{\sqrt{\frac{1}{x^2}-1}}=-\frac{1}{x}\arcsin(x)-\int \frac{dy}{\sqrt{y^2-1}}=-\frac{1}{x}\arcsin(x)-\ln(y+\sqrt{y^2-1})=-\frac{1}{x}\arcsin(x)-\ln(\frac{1}{x}+\sqrt{(1/x)^2-1})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Prove $\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\frac{\pi ^{2}}{48}+\frac{1}{4}\ln^22$. How to prove the following series,
$$\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\frac{\pi ^{2}}{48}+\frac{1}{4}\ln^22$$
I know a formula which might be usful.
$$\sum_{j=1}^{n}\frac{\left ( -1 \right )^{j-1}}{j}=\ln 2+\left ( -1 \right )^{n-1}\int_{0}^{1}\frac{x^{n}}{1+x}\mathrm{d}x$$
any hint will be appreciate.
| Yes! As you mentioned, it's a very useful formula.
Use
$$\sum_{j=1}^{n}\frac{\left ( -1 \right )^{j-1}}{j}=\ln 2+\left ( -1 \right )^{n-1}\int_{0}^{1}\frac{x^{n}}{1+x}\, \mathrm{d}x$$
we get
\begin{align*}
&\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\ln 2\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}-\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\int_{0}^{1}\frac{x^{2k}}{1+x}\, \mathrm{d}x\\
&=\ln^22+\int_{0}^{1}\frac{1}{1+x}\sum_{k=1}^{\infty }\frac{\left ( -x^{2} \right )^{k}}{k}\, \mathrm{d}x=\ln^22-\int_{0}^{1}\frac{\ln\left ( 1+x^{2} \right )}{1+x}\, \mathrm{d}x
\end{align*}
let
$$f\left ( t \right )=\int_{0}^{1}\frac{\ln\left ( 1+tx^{2} \right )}{1+x}\,\mathrm{d}x$$
then
\begin{align*}
f{}'\left ( t \right ) &=\int_{0}^{1}\frac{x^{2}}{\left ( 1+x \right )\left ( 1+tx^{2} \right )}\,\mathrm{d}x \\
&=\frac{1}{t+1}\int_{0}^{1}\frac{x-1}{1+tx^{2}}\,\mathrm{d}x+\frac{1}{t+1}\int_{0}^{1}\frac{\mathrm{d}x}{x+1} \\
&=\frac{1}{t+1}\left [ \frac{1}{2t}\ln\left ( 1+tx^{2} \right )-\frac{1}{\sqrt{t}}\arctan\left ( \sqrt{t}x \right )+\ln\left ( x+1 \right ) \right ]_{0}^{1} \\
&=\frac{1}{t+1}\left [ \frac{1}{2t}\ln\left ( 1+t \right )-\frac{1}{\sqrt{t}}\arctan\left ( \sqrt{t} \right )+\ln 2 \right ]
\end{align*}
Integrate back
$$\Rightarrow f\left ( t \right )=\frac{1}{2}\left [ -\mathrm{Li}_{2}\left ( -t \right )-\frac{1}{2}\ln^{2}\left ( t+1 \right ) \right ]-\arctan^{2}\sqrt{t}+\ln 2\ln\left ( t+1 \right )$$
then let $t=1$ and use $\displaystyle \mathrm{Li}_{2}\left ( -1 \right )=-\frac{\pi ^{2}}{12}$, you will get the answer as wanted.
EDIT:
$$\mathrm{Li}_{2}\left ( -1 \right )=\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k^{2}}=\sum_{k=1}^{\infty }\frac{1}{k^{2}}-2\sum_{k=1}^{\infty }\frac{1}{\left ( 2n-1 \right )^{2}}=\frac{\pi ^{2}}{6}-2\cdot \frac{\pi ^{2}}{8}=-\frac{\pi ^{2}}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
What is the minimum value of $\dfrac {9x^2\sin^2 x+4}{x\sin x}$
Question: What is the minimum value of $$\dfrac {9x^2\sin^2x+4}{x\sin x}\tag1$$For $0<x<\pi$.
I solved this, but somewhere, I did something wrong. My work is as follows:
First, set $y=x\sin x$. Therefore,
$$\begin{align*}\dfrac {9y^2+4}{y}=\dfrac 4y+9y\tag2\end{align*}$$
And by the AM-GM Inequality, we have$$\begin{align*}\dfrac {9y+\dfrac 4y}2 & \geq\sqrt{\dfrac 4y\cdot 9y}\tag3\\9y+\dfrac 4y & \geq12\tag4\end{align*}$$
$(4)$ is minimized only when the RHS is equal to zero. So$$9y+\dfrac 4y=0\implies 9y^2=-4\implies y^2=-\dfrac 49\tag5$$
However, if I square root both sides, I get an imaginary value. Where did I go wrong? The book says the solution is $12$ (received by plugging in $x=\dfrac 23$).
| Hint: you can show it as $$\frac { 9x^{ 2 }\sin ^{ 2 } x+4 }{ { x\sin x } } =\frac { \left( 3x\sin { x } -2 \right) ^{ 2 }+12x\sin { x } }{ x\sin x } =\frac { \left( 3x\sin { x } -2 \right) ^{ 2 } }{ x\sin x } +12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Given the quadratic equation $p(x^2 +9)= -5qx$ has two equal roots, find the ratio of $p:q$. Hence, solve the quadratic equation Given the quadratic equation $p(x^2 +9)= -5qx$ has two equal roots, find the ratio of p:q. Hence, solve the quadratic equation
so this is what i got so far :
$$px^2+9p+5qx=0$$
$$(5q)^2 - 4(p)(9p)=0$$
$$25q^2 - 36p^2 =0$$
$$(5^2 q^2) - (6^2 p^2)=0$$
$$5^2 q^2 = 6^2 p^2$$
$$5q =6p$$
I might be wrong
what to do next?
| Answer 1 -
When we have equal roots then discriminate equals to zero. As you are solving.
Answer 2 -
We have,
5q = 6p
$\frac{p}{q} = \frac56$
Or
You can also write
$(5^2 q^2) - (6^2 p^2) = 0$
As $(5q)^2 - (6p)^2 = 0$
$(5q - 6p)(5q + 6p) = 0$
Either 5q - 6p = 0 or 5q + 6p = 0
$\frac{p}{q} = \frac56$ or $\frac{p}{q} = \frac{-5}{6}$
Answer 3 -
For equal roots we have each root equals to $\frac{-b}{2a}$
So x = $\frac{-5q}{2p}$
= $\frac{-5}{2} \times \frac{6}{5}$
x = -3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral of $\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$ I was asked to find the following integral:
$$\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$
What I tried to replace $\sqrt{1-\frac{1}{x^2}}$ with $u$ so that: $$du=\frac{dx}{x^3\sqrt{1-\frac{1}{x^2}}} \Rightarrow du*x=\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$
And:$$x=\sqrt{\frac{1}{1-u^2}}$$
And we can replace:
$$\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}=\int\frac{du}{\sqrt{1-u^2}}=\arctan(u)+C=\arctan(\sqrt{1-\frac{1}{x^2}})+C$$
The problem is, when that result is derived we don't get the original expression. I just can't find my mistake, so some help would be appreciated.
| Substitute $\text{u}=\frac{1}{x}$:
$$\mathcal{I}\left(x\right)=\int\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}\space\text{d}x=-\int\frac{1}{\sqrt{1-\text{u}^2}}\space\text{d}\text{u}$$
Now, use:
$$\frac{\text{d}\arcsin\left(x\right)}{\text{d}x}=\frac{1}{\sqrt{1-x^2}}$$
So, we get:
$$\mathcal{I}\left(x\right)=\text{C}-\arcsin\left(\frac{1}{x}\right)=\text{C}-\text{arccsc}\left(x\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Mathematica gives: $\int_{0}^{\infty}{\cos(x^n)-\cos(x^{2n})\over x}\cdot{\ln{x}}\mathrm dx={12\gamma^2-\pi^2\over 2(4n)^2}$ How do we show that the given result by Mathematica is correct?
$$\int_{0}^{\infty}{\cos(x^n)-\cos(x^{2n})\over x}\cdot{\ln{x}}\mathrm dx={12\gamma^2-\pi^2\over 2(4n)^2}\tag1$$
$n>0$
Where $\gamma=0.577216...$
I would try substitution, because it may help to simplify the problem into a manage integral to deal with.
$u=x^n$
$du=nx^{n-1}dx.$
$${1\over n}\int_{0}^{\infty}{\cos(u)-\cos(u^2)\over u^{1\over n}}\cdot{\ln{u^{1\over n}}}{\mathrm dx\over u^{n-1\over n}}={12\gamma^2-\pi^2\over 2(4n)^2}$$
Simplified to
$${1\over n^2}\int_{0}^{\infty}{\cos(u)-\cos(u^2)\over u}\cdot{\ln{u}}\mathrm du={12\gamma^2-\pi^2\over 2(4n)^2}$$
We can remove $\ln{u}$ by doing another substitution
$v=\ln{u}$
$udv=du$
$${1\over n^2}\int_{-\infty}^{\infty}{\cos(e^v)-\cos(e^{2v})\over e^v}\cdot{v}\cdot{e^v}\mathrm du={12\gamma^2-\pi^2\over 2(4n)^2}$$
Then we finally simplified to
$$={1\over n^2}\int_{-\infty}^{\infty}v\cos(e^v)\mathrm dv -{1\over n^2}\int_{-\infty}^{\infty}v\cos(e^{2v})\mathrm dv$$
At this stage I would apply integration by parts but it seems to show a problem for me to do. So I need some help. Thank you.
| Define
$$\mathcal{I}=\int_{0}^{\infty }\frac{\cos x-\cos x^2}{x}\, \ln x\, \mathrm{d}x$$
and
$$\mathcal{I}\left ( \alpha \right )=\int_{0}^{\infty }x^{\alpha-1}\left ( \cos x-\cos x^{2} \right )\mathrm{d}x$$
and $\mathcal{I}\left(0\right)=-\dfrac{\gamma}{2}
$, see a proof here.
Now let's see the integral below
$$\int_{0}^{\infty}x^{a-1}\cos\left(x^{b}\right)\,\mathrm{d}x=\frac{1}{b}\cos\left(\frac{\pi
a}{2b}\right)\Gamma \left(\frac{a}{b} \right), \;\ b>a, \;\ a>1$$
Proof:
Let $\displaystyle r = a/b \in (0, 1)$. Then
\begin{align*} &\int_{0}^{\infty} x^{a-1} \cos(x^{b}) \, \mathrm{d}x = \frac{1}{b} \int_{0}^{\infty} \frac{\cos t}{t^{1-r}} \, \mathrm{d}t = \frac{1}{b \Gamma(1-r)} \int_{0}^{\infty} \left( \int_{0}^{\infty} u^{-r} e^{-tu} \, \mathrm{d}u \right) \cos t \, \mathrm{d}t \\ &= \frac{1}{b \Gamma(1-r)} \int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-ut} \cos t \, \mathrm{d}t \right) u^{-r} \, \mathrm{d}u = \frac{1}{b \Gamma(1-r)} \int_{0}^{\infty} \frac{u^{1-r}}{u^{2} + 1} \, \mathrm{d}u \\ &= \frac{1}{b \Gamma(1-r)} \int_{0}^{\frac{\pi}{2}} \tan^{1-r} \theta\,\mathrm{d}\theta \end{align*}
Simplifying using the beta function identity, we have
\begin{align*} \int_{0}^{\infty} x^{a-1} \cos(x^{b}) \, \mathrm{d}x &= \frac{\Gamma\left(1-\dfrac{r}{2}\right)\Gamma\left(\dfrac{r}{2}\right)}{2b \Gamma(1-r)\Gamma(r)} \, \Gamma(r)= \frac{ \sin (\pi r)}{2b \sin \left(\dfrac{\pi r}{2}\right)} \, \Gamma(r) = \frac{1}{b} \Gamma(r) \cos \left( \frac{\pi r}{2} \right)\\
&=\frac{1}{b}\cos\left(\frac{\pi a}{2b}\right)\Gamma \left(\frac{a}{b} \right) \end{align*}
Hence
$$\mathcal{I}\left ( \alpha \right )=\Gamma \left ( a \right )\cos\left ( \frac{a\pi }{2} \right )-\frac{1}{2}\Gamma \left ( \frac{a}{2} \right )\cos\left ( \frac{a\pi }{4} \right )$$
Using
$$\Gamma \left ( x \right )\sim \frac{1}{x}-\gamma +\frac{6\gamma
^{2}+\pi ^{2}}{12}x+o(x)$$
Proof:
\begin{align*}
\lim_{x\to 0}x\Gamma(x)=\lim_{x\to 0}\Gamma(1+x)=1
\end{align*}
\begin{align*}
\lim_{x\to 0}\Gamma(x)-\frac{1}{x}&=\lim_{x\to 0}\frac{x\Gamma(x)-1}{x}\\
&=\lim_{x\to 0}\Gamma(x+1)\psi (x+1)=-\gamma\\
\end{align*}
\begin{align*}
\lim_{x\to 0}\frac{1}{x}\left(\Gamma(x)-\frac{1}{x}+\gamma\right)&=\lim_{x\to 0}\frac{\Gamma(x+1)-1+\gamma x}{x^2}\\
&=\lim_{x\to 0}\frac{\Gamma(x+1)\psi(x+1)+\gamma }{2x}\\
&=\lim_{x\to 0}\frac{\Gamma(x+1)\psi^2(x+1)+\psi'(x+1)\Gamma(x+1)}{2}\\
&=\frac{1}{2}\left(\gamma^2+\frac{\pi^2}{6}\right)
\end{align*}
So, we have
\begin{align*}
\mathcal{I}&=\mathcal{I}'\left(0\right)\\
&=\lim_{a\rightarrow 0}\frac{\mathcal{I}\left(a\right)-\mathcal{I}\left(0\right)}{a-0}\\
&=\frac{-\dfrac{1}{2}\cos\dfrac{a\pi}{4}\Gamma\left(\dfrac{a}{2}\right)+\cos\dfrac{a\pi}{2}\Gamma\left(a\right)+\dfrac{\gamma}{2}}{a}\\
&=\lim_{a\rightarrow 0}\frac{-\dfrac{1}{2}\left(1-\dfrac{a^2\pi^2}{32}+o\left(a^2\right)\right)\left(\dfrac{2}{a}-\gamma +\dfrac{1}{12}\left(6\gamma^2+\pi^2\right)\dfrac{a}{2}+o\left(a\right)\right)+\left(1-\dfrac{a^2\pi^2}{8}+o\left(a^2\right)\right)\left(\dfrac{1}{a}-\gamma +\dfrac{1}{12}\left(6\gamma^2+\pi^2\right)a+o\left(a\right)\right)+\dfrac{\gamma}{2}}{a}\\
&=\lim_{a\rightarrow 0}\frac{\dfrac{a\pi^2}{32}-\dfrac{a}{48}\left(6\gamma^2+\pi^2\right)-\dfrac{a\pi^2}{8}+\dfrac{a}{12}\left(6\gamma^2+\pi^2\right)+o\left(a\right)}{a}\\
&=\frac{3\gamma^2}{8}-\frac{\pi^2}{32}
\end{align*}
Hence
$$\int_{0}^{\infty}{\cos x^n-\cos x^{2n}\over x}\, {\ln{x}}\, \mathrm dx={1\over n^2}\int_{0}^{\infty}{\cos x-\cos x^2\over x}\,{\ln x}\, \mathrm dx={12\gamma^2-\pi^2\over 2(4n)^2}$$
| {
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"url": "https://math.stackexchange.com/questions/2100775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Prove $2^{3n}3^n | (4n)!$ Algebraically
Prove that $2^{3n}3^n | (4n)!$
I can make a combinational prove.
Consider the set $a_1a_1a_1a_1a_2a_2a_2a_2\cdots a_na_na_na_n$ this has permutations equal to
$$\frac{(4n)!}{4!^n} \\ = \frac{(4n)!}{(2^33)^n} = \frac{(4n)!}{2^{3n}3^n}$$
Since this is a integer number thus $2^{3n}3^n | (4n)!$
But I am looking for a Algebraic prove, how to Algebraically Prove it?
| In the integers from $1$ to $4n$, there will be $2n$ even numbers, of which half ($n$) will be divisible by $4$.
Also in those integers, there will be at least $n$ numbers divisible by $3$.
Since $(4n)!$ multiplies all such integers together, we can see that $2^{3n}$ divides $(4n)!$, as does $3^n$. And as these two numbers are coprime, we can conclude that their product $2^{3n}3^n$ also divides $(4n)!$
By induction:
Base case: $n=1, (4n)! = 4! = 24$ and $2^3\cdot 3^1 = 24 \mid 24$ as required
Hypothesis $2^{3k}\cdot 3^k \mid (4k)!$
Note that among any $4$ consecutive numbers we must have a multiple of $3$, a multiple of $4$ and a separate multiple of $2$
$\implies 2^3\cdot 3 \mid \big[(4k+1)\cdot (4k+2)\cdot (4k+1)\cdot (4k+4)\big]$
and since $(4(k+1))! = (4k)!\cdot (4k+1)\cdot (4k+2)\cdot (4k+1)\cdot (4k+4)$
and $2^{3k}\cdot 3^k \cdot 2^3\cdot 3 \mid (4k)!\cdot (4k+1)\cdot (4k+2)\cdot (4k+1)\cdot (4k+4)$
$\implies 2^{3(k+1)}\cdot 3^{k+1} \mid (4(k+1))! $
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Known that $a + b + c = 0$ and $a^3 + b^3 + c^3 = 27$. Find ABC!
Known that
$$a + b + c = 0$$
$$a^3 + b^3 + c^3 = 27$$
What is the value of $abc?$
A.) 1
B.) 0
C.) 7
D.) 8
E.) 10
My Work:
$$a + b = -c$$
$$a + c = -b$$
$$b + c = -a$$
Then
$$(a + b + c)(a + b + c)(a + b + c) = 0$$
Expanded into:
$$a^3 + b^3 + c^3 + 3a^2b+3b^2c+3a^2c+3ab^2 + 3bc^2+ 3ac^2+6abc = 0$$
Putting the Values:
$$27+3a^2(-a)+3b^2(-b)+3c^2(-c)+6abc = 0$$
$$27 -3(a^3 + b^3 + c^3) = -6abc$$
$$27-3(27) = -6abc$$
$$-54=-6abc$$
$$abc = 9$$
$9$ wasn't an option in the question. Am I missing something?
| No, the question is wrong, and your answer is right. However, it might be easier by recalling the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
From $a+b+c=0$, we have $abc=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I continue solving this polynomial division? I had this equation as a bonus question on a quiz today. I got to a certain extend and was completely unsure how to continue.
*$$(10x^4-18x^3-94x^2-8x+2) \div (10x+2)$$
I figured the easiest first step would be to divide. (Actually, I assumed it was factorable, but I have no idea how to go about that)
$$\frac{(10x^4-18x^3-94x^2-8x+2)} {(10x+2)}$$
$$\frac{10x^4}{10x+2} + \frac{-18x^3}{10x+2} + \frac{-94x^2}{10x+2} + \frac{-8x}{10x+2} + \frac{2}{10x+2}$$
Simplifying some more (multiplying by $10x+2$)
$$10x^4-18x^3-94x^2-8x-2 = 0$$
But what do I do from here? Do I factor? Do do something else?
If I do factor How do I go about that? If not, what do I do instead?
| One idea would be to guess that $10x+2$ is a factor.
We suppose
$$10x^4 - 18x^3 -94x^2 -8x +2 = (10x+2)(ax^3+bx^2+cx+d).$$
What does this tell us? We immediately know by considering the $x^4$ term, that $a=1$. By considering the $x^3$ term we have $-18 = 10b +2a = 10b+2$. This tells us that $b=-2$. The $x^2$ term tells us that $-94=2b+10c=10c-4$, so $c=-9$. The $x$ term tells us that $-8=2c+10d=-18+10d$ so $d=1$.
What have we shown? We have shown that if $10x+2$ divides your polynomial, then
$$10x^4-18x^3-94x^2-8x+2 = (10x+2)(x^3-2x^2-9x+1).$$
To show that this holds we need to factorise it out and check! I'll leave that to you.
Hope this helps.
| {
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"url": "https://math.stackexchange.com/questions/2105033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\lim_{n\rightarrow \infty }\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\mathrm dt=\frac{1}{4}$ How to prove
$$\lim_{n\rightarrow \infty }\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\mathrm dt=\frac{1}{4}$$
I used $x\to \dfrac{1}{t}$ but it didn't work.
Any hint?Thank you.
| It's easy to see that when $0<t\leq 1$, we have $1-2t^2< \cos2t<1$, hence
$$\frac{1}{4t^2}-\frac{1}{2}<\frac{\cos2t}{4t^2}< \frac{1}{4t^2}$$
so
$$\frac{1}{n}\int_{\frac{1}{n}}^{1}\left ( \frac{1}{4t^2}-\frac{1}{2} \right )\mathrm{d}t<\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\, \mathrm{d}t<\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{1}{4t^2}\, \mathrm{d}t$$
$$\Rightarrow \frac{1}{4}-\frac{3}{4n}+\frac{1}{2n^2}<\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\, \mathrm{d}t<\frac{1}{4}-\frac{1}{4n}$$
Now take the limit we will get the answer as wanted.
| {
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"url": "https://math.stackexchange.com/questions/2105469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Write the quadratic function in the form $g(x)= a(x-h)^2 + k$ Write the quadratic function in the form $g(x)= a(x-h)^2 + k$
Then, give the vertex of its graph.
$g(x)= 2x^2-16x+35$
~
I tried, but still lost:
$(2x^2-16x)+35$
$2(x^2-4^2-16)+35$
$2(x^2-4^2-32+35$
| I'm not too sure what you did, one of your brackets is missing, but I believe that your error lies in $2(x^2-4^2-16)+35$.
When you complete the square for $x^2+bx$, you take half of $b$ and then square that. However, starting with $x^2+16x$, you subtracted $16$, and then "turned" the $8x$ into a $4^2$. Not possible!
Instead, we're supposed to have$$\begin{align*}2x^2-16x+35 & =2(x^2-8x)+35\\ & =2\left\{x^2-8x+\left(\dfrac {8}2\right)^2-\left(\dfrac 82\right)^2\right\}+35\\ & =2\left(x^2-8x+16\right)+35-32\\ & =2(x-4)^2+3\end{align*}$$
Thus, the vertex of $g(x)$ is $(4,3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the following indefinite integral Calculate: $$\int \frac{(2x^{2}+x+\frac{1}{2})\cos2x+(6x^{2}-7x+\frac{13}{2})\sin2x}{\sqrt{(x^{2}-x+1)^{3}}}dx.$$
I have no idea how to start it.
| $$\int \frac{(2x^{2}+x+\frac{1}{2})\cos2x+(6x^{2}-7x+\frac{13}{2})\sin2x}{\sqrt{(x^{2}-x+1)^{3}}}dx \\=\int \frac{\cos2x}{\sqrt{x^2-x+1}}\Big(2+\frac{3x-\frac{3}{2}}{x^2-x+1}\Big)+ \frac{\sin2x}{\sqrt{x^2-x+1}}\Big(6+\frac{-x+\frac{1}{2}}{x^2-x+1}\Big)dx \\=\int\Bigg(\Big(\frac{2\cos2x}{\sqrt{x^2-x+1}}+\frac{6\sin2x}{\sqrt{x^2-x+1}}\Big)+\Big(\frac{\cos2x}{\sqrt{x^2-x+1}}.\frac{3x-\frac{3}{2}}{x^2-x+1}+\frac{\sin2x}{\sqrt{x^2-x+1}}.\frac{-x+\frac{1}{2}}{x^2-x+1}\Big)\Bigg)dx\\=\int\Bigg(\frac{2(\cos2x+3\sin2x)}{\sqrt{x^2-x+1}}+\frac{1}{2}.\frac{(3\cos2x-\sin2x)}{\sqrt{x^2-x+1}}.\frac{2x-1}{(x^2-x+1)}\Bigg)dx\\=\int \frac{d}{dx}\Bigg(\frac{-(3\cos2x-\sin2x)}{\sqrt{x^2-x+1}}\Bigg)dx\\=\frac{-(3\cos2x-\sin2x)}{\sqrt{x^2-x+1}}$$
| {
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Solve $y'=-\frac{x+y-2}{x-y+4}$
$$y'=-\frac{x+y-2}{x-y+4}$$
I have used the substation:
$x=u-1$
$y=v+3$
And got $$\frac{dv}{du}=\frac{-1-\frac{v}{u}}{1-\frac{v}{u}}$$
$z=\frac{v}{u}$
And got:
$\frac{du}{u}\frac{1-z}{z^2-2z-1}$
$$u=e^{\frac{-z+2z+1}{4}+C}$$
What should I do next?
| You could approach it this way:
$$
\frac{dy}{dx} = -\frac{x + y - 2}{x - y + 4} \\
\implies x\ dy - y\ dy + 4\ dy = -x\ dx - y \ dx - 2 \ dx \\
\implies (x \ dy + y \ dx) - x \ dx - 2 \ dx - y \ dy = 0 \\
\implies \int d(xy) - \int x \ dx - 2\int dx - \int y \ dy = 0 \\
\implies x^2 + y^2 - 2xy + 4x + c = 0
$$
which is a parabola.
| {
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Find $x^2+y^2$ if $x^2-\frac{2}{x}=3y^2$ and $y^2-\frac{11}{y}=3x^2$
Find $x^2+y^2$ if $x^2-\frac{2}{x}=3y^2$ and $y^2-\frac{11}{y}=3x^2$.
My try:
$$\frac{y^2-\frac{11}{y}}{3}-\frac{2}{x}=3y^2$$
then?
| Assume that $x,y\in\mathbb{R}$ and $a,b\in\mathbb{R}$ satisfy $$x^2-\frac{a}{x}=3y^2\text{ and }y^2-\frac{b}{y}=3x^2\,.$$
Then, taking $z:=x+\text{i}y$, we have
$$z^3=\left(x^3-3xy^2\right)-\text{i}\left(y^3-3x^2y\right)=a-b\text{i}\,.$$
Thus,
$$x^2+y^2=|z|^2=\sqrt[3]{\left|z^3\right|^2}=\sqrt[3]{|a-b\text{i}|^2}=\sqrt[3]{a^2+b^2}\,.$$
Caveat: This solution only works if $x$ and $y$ are real numbers. If they can be complex numbers, dxiv's first solution is better. In fact, his solution shows that $x^2+y^2$ can be any of the three cubic roots of $a^2+b^2$.
| {
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Algebraic Fractions grade 7 I am a student and I am having difficulty with answering this question. I keep getting the answer wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.
Write as a single fraction in its simplest Form.
$$\frac{3}{2x+1}+\frac{8}{2x^2-7x-4}$$
I factorised both fractions:
$$\frac{3}{(2x+1)} \qquad \frac{8}{(2x+1)(x+4)}$$
And got rid of $(2x+1)$. I don't know what to do next.
Kind Regards
| The standard approach as I understand it is:
$$\begin{align}
\frac{3}{2x+1}&+\frac{8}{2x^2-7x-4} \\
&=\frac{3}{2x+1}+\frac{8}{(2x+1)(x-4)} \tag{factorise denom.}\\
&= \frac{3(x-4)}{(2x+1)(x-4)}+\frac{8}{(2x+1)(x-4)} \tag{common denom.}\\
&= \frac{3(x-4)+8}{(2x+1)(x-4)} \tag{common frac.}\\
&= \frac{3x-12+8}{(2x+1)(x-4)} \tag{multiply out}\\
&= \frac{3x-4}{(2x+1)(x-4)} \tag{result}\\
\end{align}$$
| {
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Problem on Trigonometry If $ \sin x + \cos x = a$ , then find:
(i) $ \sin^6 x + \cos^6 x$
(ii) $ \lvert \sin x – \cos x \rvert $
I tried using the identity $a^3 + b^3$ for (i) but that didn't work and I ended up with an expression that was of no help. I have absolutely no idea on how to go about the problem. Any kind of help would be appreciated.
| Given $\displaystyle \sin x+\cos x = a\Rightarrow (\sin x+\cos x)^2 = a^2\Rightarrow \sin x \cos x = \frac{a^2-1}{2}$
Using $(\sin ^2 x+\cos^2 x)^3 = (\sin^6 x+\cos^6 x)+3(\sin x\cos x)^2(\sin^2 x+\cos^2 x)$
So $\displaystyle 1=\sin^6 x+\cos^6 x+3\left(\frac{a^2-1}{2}\right)$
$\displaystyle \bullet\; |\sin x-\cos x|^2 = (\sin x-\cos x)^2=1-2\sin x\cos x = 1-\frac{2(a^2-1)}{2}$
| {
"language": "en",
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If $a^b = b^a$ and $a=2b$ then find the value of $a^2+b^2$ If $a^b = b^a$ and $a=2b$ then find the value of $a^2+b^2$
My Attempt,
$$a^b = b^a$$
$$a^b =b^{2b}$$
$$a^b =b^b.b^b$$.
Now, what should I do further?
| You didn't specify what are $a,b$ ... Integers ? positive real numbers ? ...
I will assume that $a,b$ are positive real numbers, such that $a^b=b^a$ and $a=2b$.
So we have $(2b)^b=b^{2b}$, which simplifies to $2^b=b^b$, hence $b=2$ and $a=4$.
| {
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Find the general solution of $(x + \sin{x} +\sin{y})\ \textrm{d}{x} + \cos{y}\ \textrm{d}{y}=0$ I have tried to make this equation seperable, homogeneous, exact and use integrating factors and nothing seems to be working. Could you tell me what form to use?
| if you multply through by $e^x$ you have an exact diff eq.
$(x e^x + e^x \sin x + e^x \sin y)dx + (e^x \cos y) dy = 0\\
\nabla (x e^x-e^x + \frac 12 e^x \sin x - \frac 12 e^x \cos x + e^x \sin y) = (x e^x + e^x \sin x + e^x \sin y)dx + (e^x \cos y) dy\\
x e^x-e^x + \frac 12 e^x \sin x - \frac 12 e^x \cos x + e^x \sin y = C\\
-x + 1 - \frac 12 \sin x + \frac 12 \cos x +Ce^{-x} = \sin y\\
y = \arcsin( 1-x - \frac 12 \sin x + \frac 12 \cos x +Ce^{-x})$
| {
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Two consecutive numbers removed from first $n$ natural numbers If two consecutive numbers are removed from first $n$ natural numbers, and the arithmetic mean of remaining numbers is$ \frac {105}{4}$, find $n $.
Let $k $ and $k+1$ be two removed numbers, then we have :
$$ 1+2+3+\cdots +n = 2k+1+\frac{105n-210}{4}. $$
Rearranging I got quadratic as below
$$ 2n^2-103n+206-8k=0.$$
So:
$$ 4n=103 \pm \sqrt {8961+64k}. $$
How do I proceed from here?
| The sum of the first n natural number is $\frac{n(n+1)}{2}$. If the arithmetic mean of these remaining n-2 numbers is 105/4 then their sum is $\frac{(n- 2)105}{4}$. The difference in sums is the sum of those two missing numbers: $\frac{n(n+1)}{2}- \frac{105(n- 2)}{4}= \frac{2n^2+ 2n- 105n+ 210}{4}= \frac{2n^2- 103n+ 210}{4}$. Calling the two consecutive numbers that were removed "k" and "k+ 1", then we must have $k+ (k+ 1)= 2k+ 1= \frac{2n^2- 103n+ 210}{4}$. $2n^2- 102n+ 210= 8k+ 4$ so $2n^2- 102n+ 206- 8k= 0$. Solve for n using the quadratic formula.
| {
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Any hints on how to show that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $ I am trying to prove that $$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $$
My attempt was to suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10}$, and show that it was absurd.
Here's what I did:
Suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10} $, then $2 - \sqrt{3} \geq \frac{2\sqrt{3}}{10}$ thus:
$$\frac{20 -10 \sqrt{3}}{10} \geq \frac{2 \sqrt{3}}{10} \implies 20 -\sqrt{300} \geq 2 \sqrt{3} \implies 10 - \sqrt{\frac{300}{4}}\geq \sqrt{3} \implies 10 - \sqrt{75} \geq \sqrt{3} \implies \frac{10}{\sqrt{3}}- \sqrt{25} \geq 1 \implies \frac{10}{\sqrt{3}}- 5 \geq 1$$
Now I elevate everything to the square:
$$\frac{100}{3} - \frac{100}{\sqrt{3}} + 25 \geq 1 \implies \frac{100}{3} - \frac{100 \sqrt{3}}{3} + \frac{75}{3} \geq 1 \implies \frac{100(1-\sqrt{3}) +75}{3} \geq \frac{3}{3}$$
All is left to show is that $$100(1-\sqrt{3}) +75 \geq 3$$
And here, I don't know where to go, and it doesn't seem as I simplified the problem.
Any hints of help will be appreciated
| If you multiply the numerator and denominator by $2+\sqrt3$, you get $\dfrac1{4\sqrt3+6}$. Since $\sqrt3>1$
$$4\sqrt3+6>4(1)+6$$
$$4\sqrt3+6>10$$
$$\frac1{10}>\frac1{4\sqrt3+6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the number of $4$-digit numbers so no two adjacent digits are even Find the number of $4$-digit numbers with distinct digits chosen from the set $\{0,1,2,3,4,5\}$ in which no two adjacent digits are even.
| Even digits $\{0,2,4\}$
Odd digits $\{1,3,4\}$
Numbers of the format: Even Odd Even Odd
$2 \times 3 \times 2 \times 2=24$
Numbers of the format: **Odd Even Odd Even
$3 \times 3 \times 2 \times 2=36$
Numbers of the format: Odd Odd Odd Even
$3 \times 2 \times 1 \times 3=18$
Numbers of the format: Odd Odd Even Odd
$18$
Numbers of the format: Odd Even Odd Odd
$18$
Numbers of the format: Even Odd Odd Odd
$2 \times 3 \times 2 \times 1=12$
**Numbers of the format: Even Odd Odd Even **
$2 \times 3 \times 2 \times 2=24$
Required count:
$24+36+18+18+18+12+24=150$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate$\sum_{n=0}^\infty\binom{3n}{n}x^n$ The question is:
Evaluate $$\sum_{n=0}^\infty\binom{3n}{n}x^n$$
After applying a few numbers as $x$ in Wolfram Alpha, I guess that the answer is probably:
$$2\sqrt{\frac1{4-27x}}\cos\left( \frac13\sin^{-1}\frac{3\sqrt{3x}}{2} \right)$$
that I can never prove. (Interestingly the above becomes simply $2\cos\frac{\pi}9$ when $x=\frac19$.)
(*) To give you the background, the motivation that led me to this question is the Algebra problem #$10$ in the Harvard-MIT Math Test in Feb. 2008, that concludes to:
$$\sum_{n=0}^\infty\binom{2n}{n}x^n=\frac1{\sqrt{1-4x}}$$
And then I thought about what if it was $3n$ instead of $2n$.
| This is too long for a comment.
Considering $$f_k=\sum_{n=0}^\infty\binom{kn}{n}x^n$$ where $k$ is a positive integer,we could notice nice patterns using the generalized hypergeometric function
$$f_4=\, _3F_2\left(\frac{1}{4},\frac{2}{4},\frac{3}{4};\frac{1}{3},\frac{2}{3};\frac{4^4
}{3^3}x\right)$$
$$f_5=\,
_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{4},\frac{2}{
4},\frac{3}{4};\frac{5^5 }{4^4}x\right)$$
$$f_6=\,
_5F_4\left(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{1}{
5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{6^6 }{5^5}x\right)$$ As you already noticed, this only simplifies for $k=2$ and $k=3$.
| {
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Solve: $(y\sqrt{1-y^2})dx+(x\sqrt{1-y^2}+y)dy=0$
$$(y\sqrt{1-y^2})dx+(x\sqrt{1-y^2}+y)dy=0$$
$$\frac{\partial M}{\partial y}=\sqrt{1-y^2}-\frac{-2y^2}{2\sqrt{1-y^2}}=\frac{1-2y^2}{\sqrt{1-y^2}}$$
$$\frac{\partial N}{\partial x}= \sqrt{1-y^2}$$
$$\frac{M_{y}-N_{x}}{M}=\frac{\frac{1-2y^2}{\sqrt{1-y^2}}-\sqrt{1-y^2}}{y\sqrt{1-y^2}}=\frac{-y}{1-y^2}=h(y)$$
$$I=\exp{\left(-\int h(y) dy\right)}= \exp{\left(-\int\frac{-y}{1-y^2}dy\right)}=\exp{\left(\frac{-\ln|1-y^2|}{2}\right)}$$
Is there a way to simplify $$\exp{\left(\frac{-\ln|1-y^2|}{2}\right)}\;?$$
| Apart from your question you can solve this equation by this way as well
$$(y\sqrt { 1-y^{ 2 } } )dx+(x\sqrt { 1-y^{ 2 } } +y)dy=0\\ ydx+xdy+\frac { ydy }{ \sqrt { 1-{ y }^{ 2 } } } =0\\ d\left( xy \right) -\frac { 1 }{ 2 } d\left( \sqrt { 1-{ y }^{ 2 } } \right) =0\\ d\left( xy-\frac { \sqrt { 1-{ y }^{ 2 } } }{ 2 } \right) =0\\ xy-\frac { \sqrt { 1-{ y }^{ 2 } } }{ 2 } =C\\ \\ $$
| {
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Finding $u$ for $u_t+uu_x=f$ For a river flowing on a slope, the resistive force is $R = av^2$ and gravitational force, $F = bh$. The flow speed adjusts itself to $av^2=bh \implies v^2=\frac{b}{a}h$. Given the mass conservation law,
$$h_t + (hv)_x = r.$$
If $v^2=\frac{b}{a}h$,
$$v^2=\frac{b}{a}h\rightarrow (v^2)_t =\frac{b}{a} h_t\rightarrow h_t =\frac{a}{b}(v^2)_t$$
$$(hv)_x=(\frac{a}{b}v^3)_x=\frac{a}{b}3v^2v_x$$
Thus,
$$h_t+(hv)_x=\frac{a}{b}(v^2)_t+\frac{a}{b}3v^2v_x=r$$
$$(v^2)_t+3v^2v_x=r\frac{b}{a}$$
With $u=3\frac{b}{a}h=3v^2$ write,
$$\frac{u_t}{3}+u\frac{(u^{\frac{1}{2}})_x}{3}=r\frac{b}{a}$$
$$\frac{u_t}{3}+u\frac{\frac{1}{2}u_x}{3u^{\frac{1}{2}}}=r\frac{b}{a}$$
$$\frac{u_t}{3}+\frac{u^{\frac{1}{2}}u_x}{6}=r\frac{b}{a}$$
I am trying to get the PDE in the form $u_t+uu_x=f$ (inviscid Burger's equation) but am out of tricks.
| $$\frac{u_t}{3}+\frac{u^{\frac{1}{2}}u_x}{6}=r\frac{b}{a} \qquad (1)$$
Supposing $a$ , $b$ and $r$ are constants.
Solving thanks to the method of characteristics :
The system of differential equations for the characteristics curves is :
$$3dt=6\frac{dx}{ u^{\frac{1}{2} }}=\frac{a}{br}du$$
From $\quad 3dt=\frac{a}{br}du\quad$ the equation of a first family of characteristic cuves is :
$$\frac{a}{br}u-3t=c_1$$
From $\quad 6\frac{dx}{u^{\frac{1}{2}}}=\frac{a}{br}du \quad\to\quad 6\frac{br}{a}dx=u^{\frac{1}{2}}du\quad$ the equation of a second family of characteristic cuves is :
$$6\frac{br}{a}x-\frac{2}{3}u^{\frac{3}{2}}=c_2$$
The general solution of the PDE $(1)$ is obtained on the form of an implicit equation :
$$\Phi\left(\frac{a}{br}u-3t \; , \; 6\frac{br}{a}x-\frac{2}{3}u^{\frac{3}{2}}\right)=0$$
where $\Phi$ is any differentiable function of two variables.
Other equivalent forms to express the general solution are :
$$\frac{a}{br}u-3t =F\left( 6\frac{br}{a}x-\frac{2}{3}u^{\frac{3}{2}}\right)$$
$$6\frac{br}{a}x-\frac{2}{3}u^{\frac{3}{2}}=G\left( \frac{a}{br}u-3t \right)$$
where $F$ and $G$ are any differentiable functions, but related ( one is the inverse of the other).
The possibility (or not) to reduce the implicit equation to an explicite equation $u(x,t)$ depends on the kind of function $\Phi$ (or $F$ or $G$).
Unfortunately the boundary conditions are missing in the wording of the question. Without them, one cannot determine the function $\Phi$ (or $F$ or $G$). So, we will not go further.
| {
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Solve the system ${x}^{2}+ \left( y-1 \right) ^{2}=4,z^{4}+y{z}^{2}+xz+1=0. $ Find all real solutions of the system of equation
\begin{cases}
{x}^{2}+ \left( y-1 \right) ^{2}=4,\\{z}^{4}+y{z}^{2}+xz+1=0.
\end{cases}
| let $t = y-1$
$x^2+t^2 = 4$
$tz^2 + xz = -z^4 - z^2 - 1$
Solutions (x,t) exist if and only if $\frac{z^4 +z^2 +1}{\sqrt{z^4 +z^2}} \le 2$ (geometrically)
Let $k = z^4 + z^2$
$\frac{k +1}{sqrt(k)} \le 2$ equals $k^2 +2k +1 \le 4k$ equals $k^2 -2k +1 \le 0$ equals k = 1;
So, $z^4 + z^2 = 1$
| {
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Range of a 1-2 function $$f(x)=\frac x {x^2+1}$$ I want to find range of $f(x)$ and I do like below . If someone has different Idea please Hint me .
Thanks in advanced .
This is 1-1 function $\\f(x)=\dfrac{ax+b}{cx+d}\\$,
This is 2-2 function $\\f(x)=\dfrac{ax^2+bx+c}{a'x^2+b'x+c'}\\$,
This is 1-2 function $\\f(x)=\dfrac{ax+b}{a'x^2+b'x+c'}\\$
| Method 1
$$y=\dfrac{x}{x^2+1} \to yx^2-x+y=0 \\\Delta \geq 0 \to 1-4(y)(y) \geq0 \\y^2 \leq \dfrac14 \\ y \in \left[-\dfrac12,\dfrac12\right]$$
Method 2
It is well known that$$\left|x+\frac1x\right| \geq 2 \to y=\dfrac{x}{x^2+1} \to \dfrac1y=\dfrac{x^2+1}{x}=x+\dfrac1x \\\to |\dfrac1y|\geq 2 \\|y| \leq \dfrac12 \space y \in\left[-\dfrac12,\dfrac12\right]$$
Method 3
$$y=\frac{x}{x^2+1} \to y'=\frac{1-x^2}{(x^2+1)^2}=0 \to x=\pm1 \\ \begin{cases}x=1 & f(1)=\frac12 \\x=-1 & f(1)=-\dfrac12\\
x\to -\infty & \lim_{x \rightarrow -\infty}f(x) \to 0\\x\to -\infty & \lim_{x \rightarrow +\infty}f(x) \to 0\end{cases} \to y \in\left[-\frac12, \dfrac12 \right] $$
Method 4
By substituting $x=\tan \alpha$ we have
$$y=\dfrac x {x^2+1}=y=\frac{\tan \alpha}{\tan^2 \alpha+1}=\frac{\dfrac{\sin \alpha}{\cos \alpha}}{\left(\dfrac{\sin \alpha}{\cos \alpha}\right)^2+1}=\\=\sin \alpha \cdot \cos \alpha=\frac22 \sin \alpha \cdot\cos \alpha =\frac12 \sin 2\alpha \\ -1 \leq \sin 2\alpha \leq 1 \\ \to -\frac12 \leq \frac12 \sin 2\alpha \leq \frac12 \to y \in \left[-\dfrac12,\dfrac12\right]$$
| {
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Cubic system of equations modulo $n$. What would be the best method to solve the systems of equations? If solving it, please show how to solve for $x$ and $y$ because that's where it's confusing. Thanks.
$5y^3+4x^2-y^2+10x-y$ $=$ $1$ $\pmod {31}$
$2y^3+13x^3+7y^2-x^2+5x-2$ $=$ $1$ $\pmod {31}$
| Systematically eliminating $y$ by starting with the cubic term and then moving on to lower powers of $y$ leads to the equation:
$$16 x^9+9 x^8+10 x^7+14 x^6+28 x^5+28 x^4+19 x^3+13 x^2+2 x+26=0 \bmod 31$$
and
$$y = \frac{15 x^3+11 x^2+29 x+9}{x^3+6 x^2+12 x+15}\bmod 31$$
where the denominator is to be interpreted as the multiplicative inverse of the polynomial $\bmod 31$ there, which can be rewritten as a polynomial (because it doesn't have linear factors), but we don't need to do that. Then we can solve for $x$ using trial and error, we find that $x=11\bmod 31$, substituting in the equation for $y$ yields $y = 16\bmod 31$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution of $\lim_{x\to 0} \frac{(x \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $ using Mclaurin series The problem is
$$\lim_{x\to 0} \frac{(x \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $$
I know that
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}... $$
got this
$$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15}...$$
and suppose to use taylor expansion for $\ln {(1 + x)}$
Tried to expand $\sqrt{1 + \sin x}$ and $\ln{\sqrt{(1 + x^2)}}$ but it seems to be messy. Am I on the right direction?
| $$\ln\sqrt{1+x^2}=\dfrac{\ln(1+x^2)}2=\dfrac12\left(x^2-\dfrac{x^4}2+\dfrac{x^6}3-\cdots\right)$$
Now $1+\sin x=\left(\cos\dfrac x2+\sin\dfrac x2\right)^2$
Now $\cos\dfrac x2+\sin\dfrac x2=\sqrt2\sin\left(\dfrac x2+\dfrac \pi4\right)>0$ for $x\to0$
$\implies\sqrt{1+\sin x}=\cos\dfrac x2+\sin\dfrac x2$
and $\cos\dfrac x2=1-\dfrac{(x/2)^2}{2!}+\dfrac{(x/2)^2}{4!}-\cdots$
$\sin\dfrac x2=x/2-\dfrac{(x/2)^3}{3!}+\dfrac{(x/2)^5}{5!}-\cdots$
Dord it help?
| {
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A question about domains and ranges: find the domain and range of the function $f(x)=\sqrt{1-x^2}$, without any digital help If $f(x)=\sqrt{1-x^2}$, then the domain of $f$ is $[-1,1]$. Respectively, the range of $f$ is $[0,1]$.
How do I understand that the domain of $f$ is $[-1,1]$ and that the range of $f$ is $[0,1]$, without a graphing calculator?
Let us suppose that I don't know the given intervals from start. Instead, suppose that I know only that the function that I should find the domain and range of is $f(x)=\sqrt{1-x^2}$. The question will then be: 'Find the domain and range of the function $f(x)=\sqrt{1-x^2}$, without any digital help'
| One way to determine the domain and range of $f(x) = \sqrt{1 - x^2}$ is to graph the function. If $y = \sqrt{1 - x^2}$, then $y \geq 0$. Moreover, \begin{align*}
y & = \sqrt{1 - x^2}\\
y^2 & = 1 - x^2\\
x^2 + y^2 & = 1
\end{align*}
which is the equation of the unit circle, that is, the equation of the circle with center at the origin and radius $1$.
The restriction that $y \geq 0$ means that we obtain the upper semi-circle.
From the graph, we can see that $f(x) = \sqrt{1 - x^2}$ has domain $[-1, 1]$ and range $[0, 1]$.
If we did not know how to graph the function, we would start by observing that for $f(x) = \sqrt{1 - x^2}$ to be real-valued, $1 - x^2 \geq 0$. Solving the inequality yields
\begin{align*}
1 - x^2 \geq 0\\
1 \geq x^2\\
1 \geq |x|\\
|x| \leq 1
\end{align*}
which implies that $-1 \leq x \leq 1$, from which we obtain the domain $[-1, 1]$.
We further observe that if $x$ is in the domain, $1 - x^2 \leq 1$, with equality holding only at $0$. Moreover, the square of a number in the domain is at most $1$, so $1 - x^2 \geq 1 - 1 = 0$. Since a polynomial function is continuous, $1 - x^2$ assumes every value in the interval $[0, 1]$, so we have $0 \leq 1 - x^2 \leq 1$. Taking square roots yields $0 \leq \sqrt{1 - x^2} \leq 1$, so the range is $[0, 1]$.
That said, it helps to recognize that the graph of $f(x) = \sqrt{r^2 - x^2}$ is the upper half of a circle with radius $r$ and center at the origin.
| {
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"url": "https://math.stackexchange.com/questions/2130188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the second derivative of $f(x)=\frac{x^2}{10+x}$ How do you find the second derivative of $f(x)=\frac{x^2}{10+x}$?
I get $f'(x)={20x+x^2}/{(10+x)^2}$
Then get stuck here:
$(10+x)^2 (20+2x) - (20x+x^2) \frac {2(10+x)}{((10+x)^2)^2}$
| First make the Euclidean division:
$$x^2=(x-10)(x+10)+100, \enspace\text{whence}\quad\frac{x^2}{x+10}=x-10+\frac{100}{x+10}$$
Thus
\begin{align}
f'(x)&=1-\frac{100}{(x+10)^2},\\
f''(x)&=\frac{200}{(x+10)^3}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the sum of $100$ terms of the sequence $a_{k} = \frac{1}{k} - \frac{1}{k+1}$? In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence?
The answer given is $\frac{100}{101}$, but I am not sure how.
So far I am have plugged in the values of $k's$ and have the following values $$\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30},....$$
The numerator makes sense to me as it's just $1$ and $100 \times 1 = 100$, however, I am not sure about the denominator.
| Such a sum is called telescop sum. We have
$$\sum_{k=1}^{100} \left( \frac{1}{k}- \frac{1}{k+1} \right) \stackrel{\star}{=}1 - \frac{1}{101}.$$
In $(\star)$ we split the sum into two and make an index shift.
| {
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Solving a polynomial in an easier manner I need to solve the following polynomial: $$(12x-1)(6x-1)(4x-1)(3x-1)=15$$
I tried do the multiplication and ended up with an even worse expression. There are a few other questions like this one on the list of exercises I'm trying to solve, and I just wanted an easier method, without having to do the "brutal" work.
| Hint: write it as:
$$12\cdot 6\cdot 4 \cdot 3 \cdot \left(x-\frac{1}{12}\right)\left(x-\frac{1}{6}\right)\left(x-\frac{1}{4}\right)\left(x-\frac{1}{3}\right)=15$$
Note that the symmetric terms have equal sums $\left(x-\frac{1}{12}\right)+\left(x-\frac{1}{3}\right)=\left(x-\frac{1}{6}\right)+\left(x-\frac{1}{4}\right)$ $=2x - \frac{5}{12}$. This suggests the substitution $y=x-\frac{5}{24}\,$ which "shifts" the center of symmetry to $0\,$:
$$
12\cdot 6\cdot 4 \cdot 3 \cdot \left(y+\frac{1}{8}\right)\left(y+\frac{1}{24}\right)\left(y-\frac{1}{24}\right)\left(y-\frac{1}{8}\right)=15 \\[5px]
\iff \quad 12\cdot 6\cdot 4 \cdot 3 \cdot \left(y^2-\frac{1}{8^2}\right)\left(y^2-\frac{1}{24^2}\right) = 15
$$
The latter is a biquadratic which can be solved for $y^2\,$, which then gives $y\,$, then $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\cos(\pi/7)$ is root of equation $8x^3-4x^2-4x+1=0$
Prove that $\cos\theta$ is root of equation $8x^3-4x^2-4x+1=0$, given $\theta=\frac{\pi}{7}$.
I put $\cos\theta$ in equation, but couldn't show the left-hand side to be zero.
| We can use theory of equations (and some disguised Galois theory).
Given equation can be rewritten as $(2x)^3-(2x)^2 -2( 2x)+1=0$. So we need to show $2\cos \theta $ is a root of $x^3-x^2-2x+1=0$. We will do this by showing there is a cubic equation with integer coefficients satisfied by $2\cos\theta$, and determine the other two roots, and from that reconstruct the equation.
First let us write $\pi/7 = 2\pi/14=\theta$. Let $\alpha=e^{2\pi i/14}$, a primitive 14th root of unity.
Clearly $-\alpha$ is a primitive $7$th root of unity. The latter is any solution of $x^6+x^5+x^4+x^3+x^2+x+1=0$. This tells us that $\alpha$ is a root of $x^6-x^5+x^4-x^3+x^2-x+1=0$.
Now $\alpha+\bar\alpha = 2\cos\theta=\alpha+\alpha^{13}$. We will compute the polynomials satisfied by this number.
The Galois conjugates (the other roots) of $\alpha+\alpha^{13}$ are $\alpha^3+\alpha^{11},\alpha^5+\alpha^9$ (pair numbers less than 14 and coprime to it such that they add up to 14, and sum the corresponding powers of $\alpha$.)
It is a cubic with roots $a,b,c$ where $a=\alpha+\alpha^{13}, b=\alpha^3+\alpha^{11}, c=\alpha^5+\alpha^9$. So we need to calculate $a+b+c, ab+bc+ca$ and $abc$. So it boils down to showing $a+b+c=1, ab+bc+ca=-1,
abc=-1$. Now this is a routine verification using the fact that $\alpha^{14}=1$,
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this system of equations in Lagrange Multiplier problem Find the maximum and minimum values of ${x^{2} + y^{2} + z^{2}}$ subject to the conditions ${\frac{x^{2}}{4} + \frac{y^{2}}{5} + \frac{z^{2}}{25} = 1}$ and ${x + y - z = 0}$.
Using Lagrange multiplier method, I got following equations:
$$ {2x = \frac{\lambda_{1} x}{2} + \lambda_{2}}$$
$$ {2y = \frac{2 \lambda_{1} y}{5} + \lambda_{2}}$$
$$ {2z = \frac{2 \lambda_{1} z}{25} - \lambda_{2}}$$
$${\frac{x^{2}}{4} + \frac{y^{2}}{5} + \frac{z^{2}}{25} = 1}$$
$${x + y - z = 0}$$
I'm stuck after this. I've tried to solve this system of equations to get critical point many times. Any help will be greatly appreciated. Also is there any other way to approach this problem?
| With the Lagrange multipliers
$$
L(x,y,z,\lambda,\mu) = x^2+y^2+z^2+\lambda\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}-1\right)+\mu(x+y-z)
$$
The stationary points are located at
$$
\nabla L = 0 =
\left\{
\begin{array}{l}
\frac{2\lambda x}{a}+\mu +2 x \\
\frac{2 \lambda y}{b}+\mu +2 y \\
\frac{2 \lambda z}{c}-\mu +2 z \\
\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}-1 \\
x+y-z \\
\end{array}
\right.
$$
taking the first three rows we have
$$
\left(\frac{2\lambda x}{a}+\mu +2 x\right)x+\left(\frac{2 \lambda y}{b}+\mu +2 y\right)y+\left(\frac{2 \lambda z}{c}-\mu +2 z\right)z=2\lambda+2(x^2+y^2+z^2)=0
$$
and also
$$
\cases{
x = -\frac{\mu}{2+2\lambda/a}\\
y = -\frac{\mu}{2+2\lambda/b}\\
z = \frac{\mu}{2+2\lambda/c}\\
}
$$
or
$$
\mu\left(\frac{1}{2+2\lambda/a}+\frac{1}{2+2\lambda/b}+\frac{1}{2+2\lambda/c}\right)=0
$$
Solving for $\lambda$ we have the values for $x^2+y^2+z^2$ at the stationary points.
| {
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"timestamp": "2023-03-29T00:00:00",
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Various methods to find value of $\sin 18^\circ$ To find value of $\sin 18^\circ$. Now my textbook gives a proof in which it takes $\theta=18^\circ$ and then multiply it out by 5 and write again as sum of $2\theta+3\theta$ and then taking sin on both sides forms a quadratic equation.
Are there any other elegant ways to find its value?
Thanks
| $$S=\cos A+\cos3A=2\cos A\cos2A=\dfrac{(2\cos A\sin A)\cos2A}{\sin A}=\cdots=\dfrac{\sin4A}{2\sin A}$$
Now if $S=\dfrac12,\sin4A=\sin A$ with $\sin A\ne0$
either $4A=m360^\circ+A\iff A=120^\circ m$ with $3\nmid m$ as $\sin A\ne0$ $\implies A \equiv\pm120^\circ\pmod{360^\circ}$
or $4A=(2m+1)180^\circ-A\iff A=(2m+1)36^\circ$ with $m\not\equiv2\pmod5$ as $\sin A\ne0$
$\implies A \equiv\pm36^\circ,\pm108^\circ\pmod{360^\circ}$
All the four cases reduce to $$\cos36^\circ+\cos108^\circ=\dfrac12$$
Now use $\cos36^\circ=1-2\sin^218^\circ$
and $\cos108^\circ=\cos(90+18)^\circ=-\sin18^\circ$ to form a Quadratic Equation in $\sin18^\circ$
Use the fact that $\sin18^\circ>0$
See also : Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
| {
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"url": "https://math.stackexchange.com/questions/2140356",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\sum_{r=0}^{10}\frac{\binom{10}{r}}{\binom{50}{30+r}}$ The question is to evaluate $$\sum_{r=0}^{10}\frac{\binom{10}{r}}{\binom{50}{30+r}}$$I tried rewriting the expression as $$\sum_{r=0}^{10}\frac{\binom{20-r}{10}\binom{30+r}{r}}{\binom{50}{30}\binom{20}{10}}$$ I am facing trouble on how to further simplify the numerator.Any ideas?Thanks.
| $$\begin{align}
\sum_{r=0}^{10}\frac {\displaystyle\binom {10}r}{\displaystyle\binom{50}{30+r}}
&=\frac {\color{blue}{\displaystyle\sum_{r=0}^{10}\displaystyle\binom {20-r}{10}\binom {30+r}{30}}}{\color{}{\displaystyle\binom {50}{20}\binom {20}{10}}}\\\\
&=\frac {\color{blue}{\displaystyle\binom {51}{41}}}{\color{green}{\displaystyle\binom {50}{20}\binom {20}{10}}}
&&\scriptsize\color{blue}{\text{using }\sum_{r=0}^{a-b}\binom {a-r}b\binom {c+r}d=\binom {a+c+1}{b+d+1}}\\\\
&=\frac {\displaystyle\binom {51}{10}}{\color{green}{\displaystyle\binom {50}{10}\binom {40}{10}}}
&&\scriptsize \color{green}{\text{using }\color{green}{\binom ab\binom bc=\binom ac\binom {a-c}{b-c}}}
\\\\
&= \frac {51\cdot 10!\;30!}{41!}\\\\
&=\color{red}{\frac 3{2\; 044\; 357\; 744}}\\\\
\end{align}$$
See also wolframalpha confirmation here and here.
NB: The first line makes use of the following:
$$\begin{align}
\binom {20}{10}\binom {10}r&=\binom {20}r\binom {20-r}{10}
&&\scriptsize\text{using }\binom ab\binom bc=\binom ac\binom {a-c}{b-c}\\
\color{orange}{\binom {50}{20}}\binom {20}{10}\binom {10}r&=\color{orange}{\binom {50}{20}}\binom {20}r\binom {20-r}{10}
&&\scriptsize\text{multiplying by }\binom {50}{20}\\
\color{purple}{\binom{50}{10}\binom{40}{10}}\binom{10}r
&=\binom {50}{20}\binom {20}r\binom{20-r}{10}
&&\scriptsize\text{using }\binom ab\binom bc=\binom ac\binom {a-c}{b-c}\\
&=\color{green}{\binom{50}{30+r}\binom{30+r}{30}}\binom{20-r}{10}
&&\scriptsize\text{using }\binom ab\binom {a-b}c=\binom a{b+c}\binom {b+c}{b}\\
\frac{\displaystyle\binom{10}r}{\displaystyle\binom{50}{30+r}}&=
\frac {\displaystyle\binom{20-r}{10}\binom{30+r}{30}}{\displaystyle\binom {50}{10}\binom {40}{10}}
\end{align}$$
| {
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How to solve for $x$ algebraically, given that $x^2 = 4- \sqrt{12}$ This is a question I have been chewing on for a couple days but haven't quite solved yet. The value of $x^2$ is given as $4-\sqrt{12}$ and then the result given as $±( 1-\sqrt{3})$. How would I solve this problem algebraically without prior knowledge of the answer?
Solution to a problem
| From $x^2=4-2\sqrt3$, we can take the square root to obtain$$x=\pm\sqrt{4-2\sqrt3}\tag1$$
However, we can factor $(1)$ into$$x^2=4-2\sqrt3\iff (x-1+\sqrt3)(x+1-\sqrt3)=0\tag2$$
Which implies the roots are $x=\pm\sqrt3\mp1$. To prove $(2)$, we can employ a general formula for $\sqrt{X\pm Y}$ for $X,Y\in\mathbb{R}$ and $X>Y$:
General Denesting Identity:$$\sqrt{X\pm Y}=\sqrt{\dfrac {X+\sqrt{X^2-Y^2}}{2}}\pm\sqrt{\dfrac {X-\sqrt{X^2-Y^2}}2}\tag3$$
Here, we have $X=4,Y=2\sqrt3$. Hence,$$\sqrt{4-2\sqrt3}=\sqrt{\dfrac {4+\sqrt{16-12}}2}-\sqrt{\dfrac {4-\sqrt{16-12}}2}=\sqrt3-1\tag4$$
Therefore, $(1)$ can be expressed as$$x=\pm\sqrt{4-2\sqrt3}=\pm(\sqrt3-1)\tag5$$
Can you continue from here?
Another way is to assume$$\sqrt{4-2\sqrt{3}}=a-\sqrt b\iff 4-2\sqrt3=a^2+b-2a\sqrt b\tag6$$And solve the resulting system that follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) > 10^{2009}$
Show that the number $ 3^{{4}^{5}} {+} 4^{{5}^{6}}$ can be expressed as the product of two integers greater than $ 10^{2009}$
By Sophie Germain:
$ 3^{{4}^{5}} {+} 4^{{5}^{6}}=\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)\cdot \left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)$
Now $\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)>\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)$
So need to prove $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)>10^{2009}$.
Please help with this, and I'm very enthusiast to know your method to solve above problem.
| $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) = \left(2^{\frac{5^6}{2}}-3^{2^8}\right)^2 > (2^{7811}-3^{256})^2$
$2^8>3^5$ so $(2^{7811}-3^{256})^2 > (2^{7811}-2^{410})^2 >(2^{7810})^2 = 2^{15620}$
$2^{10}>10^3$ so $2^{15620}>10^{4686}>10^{2009}$
| {
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"url": "https://math.stackexchange.com/questions/2143979",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Limit of $\lim_{x\to0}\frac{f(x)-\sqrt{x+9}}{x}$? How to find this limit?
For $|f(x)-3|\le x^2$,
$$\lim_{x\to0}\frac{f(x)-\sqrt{x+9}}{x}$$
Can I make $f(x)=x^2+3$, and then
$$\lim_{x\to0}\frac{x^2+3-\sqrt{x+9}}{x}$$
Using l'Hopital's,
$$\lim_{x\to0}\frac{2x-\frac{1}{2\sqrt{x+9}}}{1} = -\frac{1}{6}$$
I'm not confident about this answer. How do I do this problem?
| Note that $ -x^2 \leq (f(x) - 3) \leq x^2$, implying that $ 3-x^2 \leq f(x) \leq 3+x^2$ for all $x$. Hence, fixing $x$, we have that:
$$
\frac{3-x^2-\sqrt{x+9}}{x} \leq \frac{f(x) - \sqrt{x+9}}{x} \leq \frac{3+x^2 - \sqrt{x+9}}{x}
$$
Now, take the limit as $x \to 0$ on both sides of this inequality. Both the left and right sides converge to $\frac {-1}6$ as $x \to 0$ (you can do l'hopital for this).
Hence, the answer is $\frac{-1}{6}$ by the squeeze theorem, but not because you assumed $f$ to be a certain function, only using what was given in a clever manner.
| {
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If $a^2+b^2+c^2=1$ so $\sum\limits_{cyc}\frac{1}{(1-ab)^2}\leq\frac{27}{4}$
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that:
$$\frac{1}{(1-ab)^2}+\frac{1}{(1-ac)^2}+\frac{1}{(1-bc)^2}\leq\frac{27}{4}$$
This inequality is stronger than $\sum\limits_{cyc}\frac{1}{1-ab}\leq\frac{9}{2}$ with the same condition,
which we can prove by AM-GM and C-S:
$$\sum\limits_{cyc}\frac{1}{1-ab}=3+\sum\limits_{cyc}\left(\frac{1}{1-ab}-1\right)=3+\sum\limits_{cyc}\frac{ab}{1-ab}\leq$$
$$\leq3+\sum\limits_{cyc}\frac{(a+b)^2}{2(2a^2+2b^2+2c^2-2ab)}\leq3+\sum\limits_{cyc}\frac{(a+b)^2}{2(a^2+b^2+2c^2)}\leq$$
$$\leq3+\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)=\frac{9}{2},$$
but for the starting inequality this idea does not work.
By the way, I have a proof of the following inequality.
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=3$. Prove that:
$$\sum\limits_{cyc}\frac{1}{(4-ab)^2}\leq\frac{1}{3}$$
(we can prove it by SOS and uvw).
This inequality is weaker and it not comforting.
We can assume of course that all variables are non-negatives.
Thank you!
| My second proof:
Fact 1: $\frac{1}{(1-x)^2} \le \frac{729}{40}x^4 + \frac{243}{40}x^2 + \frac{27}{20}$ for all $x\in [0, 1/2]$.
(Note: $\frac{729}{40}x^4 + \frac{243}{40}x^2 + \frac{27}{20} - \frac{1}{(1-x)^2}
= \frac{(81x^4 - 108x^3 + 27x^2 - 24x + 14)(1-3x)^2}{40(1-x)^2} \ge 0$.)
Noting that $ab, bc, ca \le 1/2$, by Fact 1, it suffices to prove that
$$\frac{729}{40}((ab)^4 + (bc)^4 + (ca)^4) + \frac{243}{40}((ab)^2 + (bc)^2 + (ca)^2) + \frac{81}{20}
\le \frac{27}{4}$$
or
$$27(a^4b^4 + b^4c^4 + c^4a^4) + 9(a^2b^2 + b^2c^2 + c^2a^2) \le 4.$$
Letting $x = a^2, y = b^2, z = c^2$,
we have $x + y + z = 1$. We need to prove that
$$27(x^2y^2 + y^2z^2 + z^2x^2) + 9(xy + yz + zx) \le 4. \tag{1}$$
We use pqr method.
Let $p = x + y + z = 1, q = xy + yz + zx, r = xyz$. Using $p^2 \ge 3q$, we have $q \le 1/3$.
The desired inequality (1) is written as
$$27(q^2 - 2pr) + 9q \le 4.$$
We split into two cases.
Case 1: $1/4 < q \le 1/3$
Degree three Schur yields
$r \ge \frac{4pq - p^3}{9} = \frac{4q - 1}{9}$. We have
$$27(q^2 - 2pr) + 9q - 4
\le 27\left(q^2 - 2p\cdot \frac{4q - 1}{9}\right) + 9q - 4
= (9q - 2)(3q - 1) \le 0.$$
Case 2: $0 \le q \le 1/4$
We have
$$27(q^2 - 2pr) + 9q - 4 \le 27q^2 + 9q - 4 < 0.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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} |
$A,B$ coefficients for maximal convergence
In order to approximate the solution to the equation $f(x) = x^2-2$ we use the following iteration:
$$x_{n+1} = x_n + A\frac{x_n^2 -2}{x^n} + B \frac{x_n^2-2}{x_n^3}$$
Find $A,B$ for maximal rate of convergence.
So first we can easily see that it is a proper function since if we apply $\sqrt 2$ for $x_n$ we indeed get $x_{n+1} = \sqrt 2$.
Now, I've understood that in order to get a maximal rate of convergence one should check when the derivatives are vanished. Is that the case here? What exactly the criteria?
I'd be glad for guidance.
Thanks!
|
By varying the coefficients $A$ and $B$, we can achieve some mathematical convenience, this is called method of undetermined coefficients.
Let $x_{n}=\sqrt{2}+e_{n}$, then
\begin{align*}
x_{n+1} &= x_{n}+A\frac{x_{n}^{2}-2}{x_{n}}+B\frac{x_{n}^{2}-2}{x_{n}^{3}} \\
&= \sqrt{2}+e_{n}+A\frac{(\sqrt{2}+e_{n})^{2}-2}{\sqrt{2}+e_{n}}+
B\frac{(\sqrt{2}+e_{n})^2-2}{(\sqrt{2}+e_{n})^3} \\
&= \sqrt{2}+e_{n}+A(\sqrt{2}+e_{n})+\frac{B-2A}{\sqrt{2}+e_{n}}-
\frac{2B}{(\sqrt{2}+e_{n})^{3}} \\
&= \sqrt{2}+e_{n}+A(\sqrt{2}+e_{n})+(B-2A)
\left(
\frac{1}{\sqrt{2}}-\frac{e_{n}}{2}+
\frac{e_{n}^{2}}{2\sqrt{2}}-\frac{e_{n}^{3}}{4}+\ldots
\right) \\
& \quad -2B
\left(
\frac{1}{2\sqrt{2}}-\frac{3e_{n}}{4}+
\frac{3e_{n}^{2}}{2\sqrt{2}}-\frac{5e_{n}^{3}}{4}+\ldots
\right) \\
&= \sqrt{2}+(1+2A+B)e_{n}+\frac{2A+5B}{2\sqrt{2}}e_{n}^{2}+
\frac{2A+9B}{4}e_{n}^{3}+\ldots
\end{align*}
To optimize the rate of convergence, we make the linear and quadratic terms in $e_{n}$ vanished:
$$
\left \{
\begin{align*}
1+2A+B &= 0 \\
2A+5B &= 0
\end{align*}
\right.$$
On solving, we have
$$(A,B)=\left( -\frac{5}{8},\frac{1}{4} \right)$$
Substitute back into $x_{n+1}$, we obtain cubic convergence:
$$e_{n+1} \approx \frac{e_{n}^3}{4}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a^2+b^2+c^2+d^2=4$ what is the range of $a^3+b^3+c^3+d^3$? I tried to used AM-GM but could not do.Setting any one as $2$ or $-2$ and the others $0$ we get the range as $[-8,8]$ but what is the formal way to do this?
| It's enough to prove that $a^3+b^3+c^3+d^3\leq8$
for non-negatives $a$, $b$, $c$ and $d$ such that $a^2+b^2+c^2+d^2=4$ or
it's enough to prove that $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}+d^{\frac{3}{2}}\leq8$
for non-negatives $a$, $b$, $c$ and $d$ such that $a+b+c+d=4$.
Let $f(x)=x^{\frac{3}{2}}$.
Hence, $f$ is a convex function and $(4,0,0,0)\succ(a,b,c,d)$.
Thus, by Karamata
$$\sum_{cyc}a^{\frac{3}{2}}\leq f(4)+3f(0)=8$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sign of integral of $\frac{ 2 ^{\frac{it}{2/3}} \Gamma ( \frac{it +1}{2/3}) }{ 2 ^{\frac{it}{1.5}} \Gamma ( \frac{it +1}{1.5}) } \frac{1}{(a+it)^k}$ Can we determine the sign of the following function
\begin{align}
f(a,k)=\int_{-\infty}^\infty \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right) }{ 2 ^{\frac{it}{3/2}} \Gamma \left( \frac{it +1}{3/2}\right) } \frac{1}{(a+it)^k} dt,
\end{align}
where $a\neq 0$ and $k \ge 1$ is some positive integer.
The conjecture is that the sign of the integral is equal to
\begin{align}
{\rm sign } (f(a,k))={\rm sign}(a)^k.
\end{align}
Perhpas the following limit can be usefull. By using a method in this question it is not difficult to see that
\begin{align}
\left | \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right) }{ 2 ^{\frac{it}{3/2}} \Gamma \left( \frac{it +1}{3/2}\right) } \right| \to O( e^{- (\frac{3}{2}-\frac{2}{3}) t}) \text{ as } t \to \infty.
\end{align}
Thanks
| Not quite an answer, but a good start. Let's look at
$$f(a,k)=\int_{-\infty}^\infty \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right) }{ 2 ^{\frac{it}{3/2}} \Gamma \left( \frac{it +1}{3/2}\right) } \frac{1}{(a+it)^k} dt$$
which can be rewritten as
$$f(a,k)= (-i) \int_{-i\infty}^{i\infty} 2 ^{\frac{5z}{6}}\frac{ \Gamma \left( \frac{3(z +1)}{2}\right) }{ \Gamma \left( \frac{2(z +1)}{3}\right) } \frac{1}{(a+z)^k} dz$$
Observe that $\Gamma \left( \frac{3(z +1)}{2}\right)$ has poles in the left half plane when $\frac{3(z+1)}{2} = n$ and $ n = 0, -1, -2, -3,....$, so when $z = \frac{2}{3}n - 1$ we have a pole. The principal part is
$$\frac{(-1)^n}{n!(\frac{3(z+1)}{2} +n)} = \frac{2}{3}\frac{(-1)^n}{n!(z + 1 + \frac{2}{3}n)}$$
Take a semicircle contour that grows in the left half plane. A simple exercise in Mellin transforms gives that, if $a < 0$
$$f(a,k) = \frac{4\pi}{3}\sum_{n=0}^\infty \frac{(-1)^n2^{-\frac{5}{6}(1+\frac{2}{3}n)}}{n!\Gamma(-\frac{4}{9}n)(a-1-\frac{2}{3}n)^k}$$
Now showing that $\text{sign}(f(a,k)) = \text{sign}(a)^k$ involves talking about this series. Note that some of the terms disappear if $n = 0 \,\mod 9$ (because the Gamma function on the bottom vanishes there). Not sure how you would really approach this, but probably discussing that this series oscillates wildly has something to do with it. It seems obvious though that if $a<0$ then $\text{sign}(f(a,k))^k = \text{sign}(a)^k$.
EDIT: If $a > 0$ there's another term added because $\frac{1}{(a+z)^k}$ has a pole in the left half plane. This needs to be handled in cases, because when $a = -\frac{2}{3}n -1$ the residues get all wonky. I'll leave it to you to find that extra term, which isn't too hard to get at. It just involves taking the $k$'th derivative of the rest of the integrand (something I'm not in the mood to do).
PS: I may have screwed up some arithmetic; the amount of fractions I just crossed out and rearranged in my head had me blurry eyed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Show $9 \mid a^3 + b^3 + c^3 \Rightarrow 3 \mid a \text{ or } 3 \mid b \text{ or } 3 \mid c$ Let $a, b,c \in \mathbb{Z}: 9 \mid a^3 + b^3 + c^3$
I need to show that $3 \mid a \text{ or } 3 \mid b \text{ or } 3 \mid c$
I don´t have any clue how should I approach this problem. Could someone give my hints or name some theorems that should I use.
Thank you in advance
| If $x\not\equiv 0\pmod{3}$, then $x^3\equiv \pm 1\pmod{9}$, since
$ (3x\pm 1)^3 = 9X\pm 1$.
The sum of three numbers in $\{-1,+1\}$ cannot be $\equiv 0\pmod{9}$, hence $a^3+b^3+c^3\equiv 0\pmod{9}$ implies that at least one number among $a,b,c$ is a multiple of three.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate integral with sin and cos $$\int _0^{3\pi }\:\frac{\mathrm{d}x}{\sin ^4\left(x\right)+\cos ^4\left(x\right)}=?$$
I have used that $\cos^2 \left(x\right)=\frac{1+\cos \left(2x\right)}{2}$ and $\sin^2 \left(x\right)=\frac{1-\cos \left(2x\right)}{2}$, so:
$$\int _0^{3\pi }\:\frac{\mathrm{d}x}{\sin ^4\left(x\right)+\cos ^4\left(x\right)}=\int _0^{3\pi }\:\frac{\mathrm{d}x}{3+\cos \left(4x\right)}.$$
Now I know that $\cos(4x)$ is periodic with $T=\pi/2$ and that it is an even function. How can I use that so I can say that
$$\int _0^{3\pi }\:\frac{\mathrm{d}x}{3+\cos \left(4x\right)} = 48\int _0^{\frac{\pi }{4}}\:\frac{\mathrm{d}x}{3+\cos \left(4x\right)}$$ and what would be the purpose ? (I am asking that because the last integral was a hint and I don't know how to get to that form)
| Through the substitution $x=\frac{t}{2}$ and the periodicity of the $\cos^2$ function we have
$$ \int_{0}^{3\pi}\frac{dx}{3+\cos(4x)} = \frac{1}{2}\int_{0}^{6\pi}\frac{dt}{3+\frac{2\cos^2(t)-1}{2}} = 6\int_{0}^{\pi/2}\frac{dt}{\frac{5}{2}+\cos^2(t)} $$
and at last is is enough to set $t=\arctan u$, so that $\cos^2\arctan(u)=\frac{1}{1+u^2}$ and $dt=\frac{du}{1+u^2}$ leads to $\color{red}{\large\frac{3\pi}{2\sqrt{2}}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
value of trigonometric product Finding $\displaystyle \tan \left(\frac{\pi}{14}\right)\cdot \tan \left(\frac{3\pi}{14}\right)\cdot \tan \left(\frac{5\pi}{14}\right) $
assume $\displaystyle \frac{\pi}{14} = \theta$ then $\tan (14\theta) = 0$
wan,t be able to go further, some help me
| We have $\tan \frac {\pi}{14} = \cot (\frac {\pi}{2}-\frac {\pi}{14}) = \cot \frac {3\pi}{7} $. Similarly, we have, $\tan \frac {3\pi}{14} = \cot \frac {2\pi}{7} $ and $\tan \frac {5\pi}{14} = \cot \frac {\pi}{7} $.
Now, our expression simplifies to evaluating $$P = \tan \frac {\pi}{14} \tan \frac {3\pi}{14} \tan \frac {5\pi}{14}$$ $$ = \cot \frac {\pi}{7}\cot \frac {2\pi}{7}\cot \frac {3\pi}{7} $$ $$= \frac {1}{\tan \frac {\pi}{7}\tan \frac {2\pi}{7}\tan \frac {3\pi}{7}} \tag {1}$$
If we let $\theta = \pi/7$, then we get, $7\theta = 4\theta + 3\theta = \pi \Rightarrow 4\theta = \pi -3\theta $. Thus, $\tan 4\theta = \tan (\pi -3\theta) = -\tan 3\theta $. So, we get, $$\frac {4\tan \theta - 4\tan^3 \theta}{1-6\tan^2 \theta + \tan^4 \theta} = -\frac {3\tan \theta -\tan^3 \theta}{1-3\tan^2 \theta} $$
To proceed, note that you will get a sixth degree equation whose roots are $\tan \frac {k\pi}{7} , k\in \{1,2,\cdots,6\} $. Also, notice that $\tan^2 \frac {k\pi}{7} = \tan^2 \frac {(7-k)\pi}{7}, k \in \{1,2,3\} $. Keep $x^2=y $, and $P $ will turn out to be reciprocal of the square root of the product of the roots in the new equation in terms of $y $ (why?). The answer is thus $\boxed {\displaystyle \frac {1}{\sqrt {7}}} $.
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the inequality, fractions. $\{ a,b,c \in\Bbb R_+\ \}$ If $\frac {1}{ab} +\frac {1}{bc} + \frac {1}{ac} = 3$ then prove the inequality:
$ab + bc + ac \ge 3$
How I started
*
*$ab + bc + ac \ge \frac {1}{ab} +\frac {1}{bc} + \frac {1}{ac}$
*$ab- \frac {1}{ab} + bc-\frac {1}{bc} + ac- \frac {1}{ac} \ge 0$
But now I don't know what to do.
| We need to prove that
$$ab+ac+bc\geq\frac{9}{\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}}$$ or
$$(ab+ac+bc)(a+b+c)\geq9abc$$
which is just AM-GM:
$$(ab+ac+bc)(a+b+c)\geq3\sqrt[3]{a^2b^2c^2}\cdot3\sqrt[3]{abc}=9abc$$
Or by C-S:
$$(ab+ac+bc)\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)\geq(1+1+1)^2=9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does one show that $\int_{0}^{1}{1\over 1+\phi x^4}\cdot{\mathrm dx\over \sqrt{1-x^2}}={\pi\over 2\sqrt{2}}?$ Consider
$$\int_{0}^{1}{1\over 1+\phi x^4}\cdot{\mathrm dx\over \sqrt{1-x^2}}={\pi\over 2\sqrt{2}}\tag1$$
$\phi$;Golden ratio
An attempt:
$x=\sin{y}$ then $\mathrm dx=\cos{y}\mathrm dy$
$(1)$ becomes
$$\int_{0}^{\pi/2}{\mathrm dy\over 1+\phi \sin^4{y}}\tag2$$
Apply Geometric series to $(2)$,
$$\int_{0}^{\pi/2}(1-\phi\sin^4{y}+\phi^2\sin^8{y}-\phi^3\sin^{12}{y}+\cdots)\mathrm dy\tag3$$
$${\pi\over 2}-\int_{0}^{\pi/2}(\phi\sin^4{y}-\phi^2\sin^8{y}+\phi^3\sin^{12}{y}-\cdots)\mathrm dy\tag4$$
Power of sine seem difficult to deal with
How else can we tackle $(1)?$
| On the path of Aditya Narayan Sharma.
Define for $a\geq 0$,
$\displaystyle F(a)=\int_0^1 \dfrac{1}{(1+ax^4)\sqrt{1-x^2}}dx$
Perform the change of variable $y=\sin x$,
$\displaystyle F(a)=\int_0^{\tfrac{\pi}{2}} \dfrac{1}{1+a(\sin x)^4}dx$
$(\sin x)^2=\dfrac{1}{1+\dfrac{1}{(\tan x)^2}}$
Perform the change of variable $y=\tan x$,
$\begin{align}\displaystyle F(a)&=\int_0^{+\infty} \frac{1}{(1+x^2)\left(1+a\left(\tfrac{1}{1+\frac{1}{x^2}}\right)^2\right)}dx\\
&=\int_0^{+\infty} \dfrac{1+x^2}{(1+a)x^4+2x^2+1}dx\\
\end{align}$
Perform the change of variable $y=x\sqrt[4]{1+a}$,
$\displaystyle F(a)=\dfrac{1}{\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{1+\tfrac{x^2}{\sqrt{1+a}}}{x^4+\tfrac{2}{\sqrt{1+a}}x^2+1}dx$
Perform the change of variable $y=\dfrac{1}{x}$,
$\displaystyle F(a)=\dfrac{1}{\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{x^2+\tfrac{1}{\sqrt{1+a}}}{x^4+\tfrac{2}{\sqrt{1+a}}x^2+1}dx$
Therefore,
$\begin{align}
\displaystyle F(a)&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{x^2+1}{x^4+\tfrac{2}{\sqrt{1+a}}x^2+1}dx\\
&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{\tfrac{1}{x^2}+1}{x^2+\tfrac{1}{x^2}+\tfrac{2}{\sqrt{1+a}}}dx\\
\end{align}$
Perform the change of variable $y=x-\dfrac{1}{x}$,
$\begin{align}
\displaystyle F(a)&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\int_{-\infty}^{+\infty} \dfrac{1}{x^2+2+\tfrac{2}{\sqrt{1+a}}}dx\\
&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\left[\dfrac{\sqrt{1+a}}{\sqrt{2}\sqrt{1+a+\sqrt{1+a}}}\arctan\left(\dfrac{x\sqrt{1+a}}{\sqrt{2}\sqrt{1+a+\sqrt{1+a}}}\right)\right]_{-\infty}^{+\infty}\\
&=\frac{\sqrt{1+\frac{1}{\sqrt{1+a}}}}{2\sqrt{2}\sqrt[4]{1+a}}\pi\\
&=\boxed{\frac{\sqrt{1+\sqrt{1+a}}}{2\sqrt{2}\sqrt{1+a}}\pi}
\end{align}$
Since $\sqrt{1+\phi}=\phi$ then,
$\begin{align}\displaystyle F(\phi)&=\frac{\sqrt{1+\sqrt{1+\phi}}}{2\sqrt{2}\sqrt{1+\phi}}\pi\\
&=\frac{\sqrt{1+\phi}}{2\sqrt{2}\phi}\pi\\
&=\dfrac{\phi}{2\sqrt{2}\phi}\pi\\
&=\boxed{\dfrac{1}{2\sqrt{2}}\pi}\\
\end{align}$
| {
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"url": "https://math.stackexchange.com/questions/2157531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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find the limit: $\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=?$ fine the limit:
$$\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=?$$
my try :
$$\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=\sin ^2(\lim_{ n \to \infty }\pi\sqrt{n^2+n})\\\sin^2(\infty)=$$Does not exist
is it right ?
| $$\begin{align}\lim_{n\rightarrow \infty}\sin^2\left(\pi\sqrt{n^2+n}\right) &=\lim_{n\rightarrow \infty}\sin^2\left(n\pi-\pi\sqrt{n^2+n}\right)\\
&=\lim_{n\rightarrow \infty}\sin^2\bigg(\frac{-\pi^2 n}{n\pi+\pi\sqrt{n^2+n}}\bigg)\end{align}$$
So $$\lim_{n\rightarrow \infty}\sin^2\left(\pi\sqrt{n^2+n}\right) = 1$$
Because:$$\displaystyle \star \; \lim_{n\rightarrow \infty}\frac{\pi^2 n}{n\pi+\pi\sqrt{n^2+n}} = -\lim_{n\rightarrow \infty}\frac{\pi}{1+\sqrt{1+\frac{1}{n}}} = -\frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem on indefinite integration $$ f(x) = \int\frac{x^2}{(1+x^2)(1 + \sqrt{1+x^2})}dx$$
where $f(0)=0$, find $f(1)$.
Despite many tries by taking $$\sqrt{1+x^2}=t $$ or
$$ 1+\sqrt{1+x^2}$$ as $t$, I could not get it solved.
| Hint:
$$I = \int \frac{x^2}{(1+x^2)(1+\sqrt{1+x^2})} \mathrm{d}x = \frac{x^2(\sqrt{1+x^2}-1)}{(1+x^2)(1+\sqrt{1+x^2})(\sqrt{1+x^2}-1)} \mathrm{d}x =\int \frac{\sqrt{x^2+1}-1}{1+x^2}\mathrm{d}x = \int \frac{1}{\sqrt{1+x^2}}-\frac{1}{1+x^2}\mathrm{d}x$$
Hope you can take it from here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove equality concerning differentiable function Let $f(x)$ be a function that is n-times differentiable. Prove the following equality(without L'Hospital):
$$(x^{n-1}\cdot f(\frac{1}{x}) )^{(n)}=\frac{(-1)^{n}}{x^{n+1}}\cdot f^{(n)}(\frac{1}{x})\\$$
So, I wanted to prove it by induction, for $n=0$ the statement is obviously true. So let's assume it's true for some $n \in \mathbb{N_0}$
$\\$
Now we need to check for $n+1$ but I'm not sure how to proceed.
$$(x^{n}\cdot f(\frac{1}{x}) )^{(n+1)}$$
I thought about taking the first derivative of the product, then separately observe the n-th derivative of the two expressions but haven't reached any useful conclusions. Any kind of hint would be appreciated, thanks in advance!
| Prove by induction. Let $P_n$ be the claim for the specific value of $n$
Base Case: $$ x^{-1} \cdot f \left( \frac{1}{x} \right) = \frac{1}{x} f \left( \frac{1}{x} \right) $$
$\therefore P_0$
Inductive Step: Assume $P_k$ holds for some whole number $k$, for the sake of induction. By $P_k$ we have:
\begin{align}
\frac{d^k}{dx^k} \left[ x^{k-1} \cdot f \left( \frac{1}{x} \right) \right] &= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right)
\end{align}
Now we note that:
\begin{align}
\frac{d^{k+1}}{dx^{k+1}} \left[ x^{k} \cdot f \left( \frac{1}{x} \right) \right] &= \frac{d^{k}}{dx^{k}} \frac{d}{dx} \left[ x \cdot x^{k-1} \cdot f \left( \frac{1}{x} \right) \right] \\
&= \frac{d^{k}}{dx^{k}}\left[x^{k-1} \cdot f \left( \frac{1}{x} \right) + x \frac{d}{dx} \left[ x^{k-1} \cdot f \left( \frac{1}{x} \right) \right] \right] \\
&= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) +\frac{d^{k}}{dx^{k}}\left[x \frac{d}{dx} \left[ x^{k-1} \cdot f \left( \frac{1}{x} \right) \right] \right] \\
&= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) +\frac{d^{k}}{dx^{k}}\left[x \left[ (k-1) x^{k-2} \cdot f \left( \frac{1}{x} \right) + x^{k-1}\cdot f' \left( \frac{1}{x} \right) \cdot (-x^{-2}) \right] \right] \\
&= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) +\frac{d^{k}}{dx^{k}}\left[(k-1) x^{k-1} \cdot f \left( \frac{1}{x} \right) - x^{k-2}\cdot f' \left( \frac{1}{x} \right) \right] \\
&= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) + (k-1)\frac{d^{k}}{dx^{k}} \left[x^{k-1} \cdot f \left( \frac{1}{x} \right) \right] - \frac{d^{k}}{dx^{k}} \left[ x^{k-2}\cdot f' \left( \frac{1}{x} \right) \right] \\
&= \frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) + (k-1)\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{d^{k}}{dx^{k}} \left[ x^{k-2}\cdot f' \left( \frac{1}{x} \right) \right] \\
&= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{d^{k}}{dx^{k}} \left[ x^{k-2}\cdot f' \left( \frac{1}{x} \right) \right] \\
&= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{d^{k}}{dx^{k}} \left[ x^{k-1}\cdot \frac{1}{x} f' \left( \frac{1}{x} \right) \right] \\
&= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{d^{k}}{dx^{k}} \left[ x^{k-1}\cdot g\left( \frac{1}{x} \right) \right] \end{align}
where
$$ g(x) = xf'(x) $$
So
\begin{align}
\frac{d^{k+1}}{dx^{k+1}} \left[ x^{k} \cdot f \left( \frac{1}{x} \right) \right] &= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{d^{k}}{dx^{k}} \left[ x^{k-1}\cdot g\left( \frac{1}{x} \right) \right] \\
&= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{(-1)^k}{x^{k+1}} \cdot g^{(k)} \left( \frac{1}{x} \right)
\end{align}
By the General Leibniz Rule:
\begin{align*}
\frac{d^k}{dx^k}[g(x)] &= \sum_{m = 0}^k \binom{k}{m}\frac{d^m}{dx^m}[x] \frac{d^{k-m}}{dx^{k-m}}[f'(x)] \\
&= x f^{(k+1)}(x) + k f^{(k)}(x)
\end{align*}
So
\begin{align*}
\frac{d^{k+1}}{dx^{k+1}} \left[ x^{k} \cdot f \left( \frac{1}{x} \right) \right] &= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{(-1)^k}{x^{k+1}} \cdot g^{(k)} \left( \frac{1}{x} \right) \\
&= k\frac{(-1)^k}{x^{k+1}} \cdot f^{(k)} \left( \frac{1}{x} \right) - \frac{(-1)^k}{x^{k+1}} \cdot \left[ \frac{1}{x} f^{(k+1)}\left(\frac{1}{x} \right) + k f^{(k)} \left(\frac{1}{x} \right) \right] \\
&= - \frac{(-1)^k}{x^{k+1}} \cdot \frac{1}{x} f^{(k+1)}\left(\frac{1}{x} \right) \\
&= \frac{(-1)^{k+1}}{x^{k+2}} \cdot f^{(k+1)}\left(\frac{1}{x} \right)
\end{align*}
$\therefore P_k \implies P_{k+1} $
$\therefore P_0 \wedge (P_k \implies P_{k+1}) $
$\therefore P_n \quad \forall n \in \mathbb{W} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I show that this complicated expression with square and cube roots reduces to the value 7? How would I go about reducing the complicated-looking expression $$ \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}} $$ to show that it is equal to 7?
I came across the complicated expression while calculating the single real solution for the cubic equation $$ z^3 - 33 z - 112 = 0 $$
Using that cubic equation, I can show that $ z = 7 $ satisfies it and that a standard way of solving cubic equations shows that there is only a single real root and that its value is equal to the complicated expression, which is a very roundabout way of proving that the complicated expression is equal to 7. But what if I didn't know about the cubic equation? Is there a more direct way of reducing the complicated expression to a simpler one?
|
But what if I didn't know about the cubic equation?
As long as you have a reasonable suspicion that the expression might be rational, the polynomial it satisfies can be derived rather easily. Let:
$$
x = \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}}
$$
Then using $(a+b)^3=a^3+b^3+3ab(a+b)\,$ and $(56+19\sqrt{5})(56-19\sqrt{5})=1331=11^3\,$:
$$
\require{cancel}
\begin{align}
x^3 & = 19\sqrt{5} + 56 + \frac{11^3}{19\sqrt{5} + 56} + 3 \cdot \cancel{\sqrt[3]{19\sqrt{5} + 56}} \cdot \frac{11}{\cancel{\sqrt[3]{19\sqrt{5} + 56}}} \cdot x \\
& = 56+ \cancel{19\sqrt{5}} + 56 - \cancel{19\sqrt{5}} + 33x \\
& = 112 + 33x
\end{align}
$$
By the rational root theorem, the equation $x^3-33x-112=0$ can only have rational roots that are integer divisors of $112=2^4\cdot 7\,$. Obviously $\pm 1$ do not satisfy the equation, and even roots can be eliminated by comparing the powers of $2$ between the terms, which leaves $\pm 7\,$ to try.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find series representation of a function $ y=\frac{1}{1+x+x^{2}+x^{3}} $ Let $y = \frac{1}{1+x+x^{2}+x^{3}}$.
And I want to find the series representation of this function.
I've noticed that I can rewrite this like $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{1+x}*\frac{1}{1+x^{2}}$ or I can rewite this like a sum of two fractions $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{2(1+x)}+\frac{1}{2(1+x^{2})}$.
I know the series representations of these functions $(1+x)^{n}$ and $\frac{1}{1-x}$, and I guess I should use one of them. But the result looks weird. Can you help me, please, may be I can't see some easy way to solve it.
| Note $$\frac{1}{1 + x + \cdots + x^{n-1}} = \frac{1-x}{1-x^n}.$$ Since we know $$\frac{1}{1-z} = 1 + z + z^2 + \cdots = \sum_{k=0}^\infty z^k, \quad |z| < 1,$$ we have $$\frac{1-x}{1-x^n} = (1-x)\sum_{k=0}^\infty (x^n)^k = (1-x)(1+x^n + x^{2n} + x^{3n} + \cdots).$$ We conclude that the desired series expansion is $$1 - x + x^n - x^{n+1} + x^{2n} - x^{2n+1} + x^{3n} - x^{3n+1} + \cdots,$$ for integers $n > 1$. Your case is $n = 4$.
It is also worth pointing out that because of the restriction $|z| < 1$ for the geometric series expansion we used above, the given series expansion does not converge even when the original function is well-defined. This is addressed by performing the series expansion about $z = \infty$ rather than $z = 0$. We do this by writing $$\frac{1}{1-z} = \frac{z^{-1}}{z^{-1} - 1} = -z^{-1} \sum_{k=0}^\infty z^{-k} = -\sum_{k=1}^\infty z^{-k}, \quad |z| > 1,$$ hence $$\frac{1-x}{1-x^n} = -(1-x) \sum_{k=1}^\infty (x^n)^{-k} = x^{-n+1} - x^{-n} + x^{-2n+1} - x^{-2n} + x^{-3n+1} - x^{-3n} + \cdots, \quad |x| > 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to rewrite $2n^3+9n^2+13n+6$ as a polynomial in terms of $n+1$? I'm working on an induction proof for proving that the sum of the squares of the first n natural numbers is equal to: $$\frac{2n^3 +3n^2 + n}6$$ I understand how to get to the answer starting in the form of: $$\frac{n(n+1)(2n+1)}6$$ which just seems to require simpler manipulations.
I've solved for the series of: $$1^2 + 2^2 + \space...\space+ n^2 +(n+1)^2 = \frac{2n^3+9n^2+13n+6}6$$
I know that using synthetic division I get the factors as $(n+1), (n+2), (2n+3)$ or $$\frac{(n+1)*(n+2)*(2n+3)}6$$ I just can't figure out how to manipulate the factors from here to give me: $$\frac{2(n+1)^3+3(n+1)^2+(n+1)}6$$
How would I manipulate the equation to get it in terms of $(n+1)$ ?
| $(n + 1)(n + 2)(2n + 3) = (n + 1)((n + 1) + 1)(2(n + 1) + 1) = ((n + 1)^2 + (n+ 1))(2(n + 1) + 1) = 2((n + 1)^2 + (n + 1))(n + 1) + (n + 1)^2 + (n + 1)= 2(n + 1)^3 + 3(n + 1)^2 + (n + 1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Improper integral $\int\limits_0^\infty x\exp (-x-x^2)\,\text dx$ Integrate $\int\limits_0^\infty
x\exp (-x-x^2)\,\text dx$
Hint: Use $\int\limits_0^\infty
\exp (-x-x^2)\,\text{d}x = 0.4965$
I don't know how to use this hint in solving the integration. Help!
| $I= \int xe^{-x-x^2}dx = \frac{1}{2} \left[ \int (2x-1+1)e^{-x-x^2}dx\right]=\frac{1}{2} \left[ \int (2x+1)e^{-x-x^2}dx - \int e^{-x-x^2}dx \right]$
Observe that $\frac{d}{dx}(-x-x^2)=-(1+2x)$
Thus the previous expression simplifies to $\frac{1}{2} \left[ -e^{-x-x^2} - \int e^{-x-x^2}dx \right]$
On applying the limits we get $\int_0^{\infty} xe^{-x-x^2}dx=\frac{1}{2} \left[ -e^{-x-x^2} - \int e^{-x-x^2}dx \right]_0^{\infty} = \frac{1}{2} \left[0-(-1) - 0.4965\right] = \frac{1}{2} \left[ 0.5035 \right]=0.25175$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding maximum of $x+y$ Let x and y be real numbers satisfying $9x^{2} + 16y^{2} = 1$. Then $x + y$ is maximum when
a. $y = \frac{9x}{16}$
b. $y = −\frac{9x}{16}$
c. $y = \frac{4x}{3}$
d. $y = −\frac{4x}{3}$
| HINT:
WLOG $3x=\cos t,4y=\sin t$
$$x+y=\dfrac{\cos t}3+\dfrac{\sin t}4=\dfrac{4\sin t +3\cos t}{12}$$
Now $(4\sin t +3\cos t)^2+(3\sin t -4\cos t)^2=3^2+4^2$
$\implies (4\sin t +3\cos t)^2\le25\iff-5\le4\sin t +3\cos t\le5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
} |
Find the equation of a circle tangent to a circle and x-axis, with center on a certain line. Find the equation of the circle, which is tangent to the circle $x^2 + y^2=4$ and $x-axis$, with center on the line $x=4$.
So far I have:$$(x-h)^2 + (y-k)^2=r^2$$
Since the center is on the line $x=4$, then the center is $(4;k)$.
$\therefore ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(x-4)^2 + (y-k)^2=r^2$
And because the circle touches the $x-axis$, $r=k$.
$\therefore~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(x-4)^2 + (y-k)^2=k^2$
Now, I am struggling to use $x^2 + y^2=4$ to find $k$.
Please help.
| let $$B(xB,yB)$$ the common Point of the two circles then we have
$$xB^2+yB^2=4$$
$$(xB-4)^2+(yB-yM)^2=yM^2$$
solving this System we get
$$\frac{10-4xB)^2}{yM}=yB$$
from here we get the equation
$$xB^2+\left(\frac{10-4xB}{y_M}\right)^2=4$$
the dicriminant of this equation must be Zero, thus we have
$$yM^2(yM^2-9)=0$$ and we get $$yM=\pm 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$ I am studying the book Equations and Inequalities by Herman et al, and am stuck on the following exercise:
Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$
The hint says to consider $(n+2)(n+1)S$.
Doing so, I get:
$(n+2)(n+1)S = \frac{1}{2}{n+2 \choose 0} + \frac{1}{3}{n+2 \choose 1} + \cdots + \frac{1}{n+2}{n+2 \choose n}$,
but I don't see how this is any easier to solve.
I wonder if somebody might help by expanding on the given hint, and maybe offering their suggestions as to how to use it properly.
Thanks.
| Hint : Consider the function $$f(x) = \sum_{k=2}^{n+2}\frac{x^k}{k}\binom{n+2}{k}.$$
Then, compute $f'(x)$ and express it in a closed form, using the binomial expansion theorem. The value you wanna find is $f(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Stokes theorem problem involving sketch The question is Let $S$ be that part of the surface of the paraboloid $z=x^2+y^2$ between the planes $z=1$ and $z=4$. Using suitable diagrams, show how Stokes' theorem may be applied to this surface in order to write
$\iint_s \nabla \times V.\hat ndS$
as the sum of two line integrals. I want to show clearly the direction of integration along the two curves assuming that the z components of $\hat n$ are positive.
part (b)
given $\vec{V}=x^3j+z^3k$ I want to evaluate both the surface intergal
$\iint_s\nabla\times V.\hat n dS$ and the sum of the line intergrals $\int_{C_{1}} V.dR +\int_{C_{2}} V.dR$ where $C_1$ and $C_2$ are two curves bounding S, hence verifying stokes' theorem for this case.
My questions are:
1) How does one find the upper and lower limits of the intergrals clearly one of the integrals should be z=4 and z=1 but how can one calculate the ones for x?
2) Do we need to calculate $\hat n$ and if so how? is it just $\frac{\nabla V}{|\nabla V|}=\hat n$
3) What is the diagram supposed to look like because I'm finding it very hard to believe my diagram is correct?
4) Could someone show me the working for when evaluating the surface intergral specifically
update
would we substitute the values of $z$ into $z=x^2+y^2$ to find the curves $C_1$ and $C_2$?.
| We take the following parametrisation
$${\varphi ^{ - 1}}\left( {u,v} \right) = \left( {u,v,{u^2} + {v^2}} \right)$$
That is
$$\begin{gathered}
x = u, \hfill \\
y = v, \hfill \\
z = {u^2} + {v^2} = {x^2} + {y^2} \hfill \\
\end{gathered} $$
From the given parametrisation we get a base of tangent-vectors
like
$$\begin{gathered}
{\xi _x} = {\partial _x}{\varphi ^{ - 1}}\left( {x,y} \right) = \left( {1,0,2x} \right) \hfill \\
{\xi _y} = {\partial _y}{\varphi ^{ - 1}}\left( {x,y} \right) = \left( {0,1,2y} \right) \hfill \\
\end{gathered} $$
And here is the normal and unit-normal vector
$$n = {\xi _x} \times {\xi _y} = \left( { - 2x, - 2y,1} \right)$$
with length
$$\left\| n \right\| = \left\| {{\xi _x} \times {\xi _y}} \right\| = \sqrt {4\left( {{x^2} + {y^2}} \right) + 1} $$
so
$${n_o} = \frac{1}{{\left\| n \right\|}}n = \frac{1}{{\sqrt {4\left( {{x^2} + {y^2}} \right) + 1} }}\left( { - 2x, - 2y,1} \right)$$
From given vectorfield
$$V = \left( {0,{x^3},{z^3}} \right)$$
we get the curl
$$\nabla \times V = \left| {\begin{array}{*{20}{c}}
{{\partial _x}}&0&i \\
{{\partial _y}}&{{x^3}}&j \\
{{\partial _z}}&{{z^3}}&k
\end{array}} \right| = \left( {0,0,3{x^2}} \right)$$
Together with unit-normal vector, the dot-product is
$$\left( {\nabla \times V} \right) \cdot {n_o} = \frac{{3{x^2}}}{{\left\| n \right\|}} = \frac{{3{x^2}}}{{\sqrt {4\left( {{x^2} + {y^2}} \right) + 1} }}$$
And now with
$$dS = \left\| n \right\|dxdy$$
we have
$$\left( {\nabla \times V} \right) \cdot {n_o}dS = \frac{{3{x^2}}}{{\left\| n \right\|}}\left\| n \right\|dxdy = 3{x^2}dxdy$$.
We are now going to integrate this.
Choosing polar-coordinate-system
$$x = r\cos t,y = r\sin t$$
calculating differentials
$$\begin{gathered}
dx = \cos \left( t \right)dr - r\sin \left( t \right)dt \hfill \\
dy = \sin \left( t \right)dr + r\cos \left( t \right)dt \hfill \\
\end{gathered} $$
and surface-volume form in these coordinates
$$dxdy = \left| {\begin{array}{*{20}{c}}
{\cos \left( t \right)}&{ - r\sin \left( t \right)} \\
{\sin \left( t \right)}&{r\cos \left( t \right)}
\end{array}} \right|drdt = r \cdot drdt$$
then
$$\int_S {\left( {\nabla \times V} \right) \cdot {n_o}dS = \int\limits_R {3{x^2}dxdy = } } \int\limits_0^{2\pi } {\int\limits_1^2 {3 \cdot {r^3}{{\cos }^2}\left( t \right)} } \cdot drdt$$
and
$$\int\limits_0^{2\pi } {\int\limits_1^2 {3 \cdot {r^3}{{\cos }^2}\left( t \right)} } \cdot drdt = \frac{{45}}{4}\pi $$
Next part. We are looking at
$$V \cdot \left( {dx,dy,dz} \right) = {x^3} \cdot dy + {z^3} \cdot dz$$
For the top-boundary, we take r=2 and counterclockwise orientation
$$\begin{gathered}
x\left( t \right) = 2\cos \left( t \right) \hfill \\
y\left( t \right) = 2\sin \left( t \right),dy\left( t \right) = 2\cos \left( t \right) \cdot dt \hfill \\
z\left( t \right) = 4,dz\left( t \right) = 0 \cdot dt \hfill \\
\end{gathered} $$
This leads us to
$$\int\limits_0^{2\pi } {8{{\cos }^3}\left( t \right)} \cdot 2\cos \left( t \right) \cdot dt = 16 \cdot \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)} \cdot dt$$
For the bottom boundary we take r=1 and clockwise orientation
$$\begin{gathered}
x\left( t \right) = \cos \left( {2\pi - t} \right) = \cos \left( t \right) \hfill \\
y\left( t \right) = \sin \left( {2\pi - t} \right) = - \sin \left( t \right),dy\left( t \right) = - \cos \left( t \right) \cdot dt \hfill \\
z\left( t \right) = 1,dz\left( t \right) = 0 \cdot dt \hfill \\
\end{gathered} $$
Now we get
$$\int\limits_0^{2\pi } {{{\cos }^3}\left( t \right)} \cdot \left( { - \cos \left( t \right)} \right) \cdot dt = - \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)} \cdot dt$$
Putting results together, we have
$$16 \cdot \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)} \cdot dt - \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)} \cdot dt = 15 \cdot \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)} \cdot dt$$
And
$$15 \cdot \int\limits_0^{2\pi } {{{\cos }^4}\left( t \right)} \cdot dt = \frac{{45}}{4}\pi $$
that's it.
Forgot to remark the following:
Using differential-forms, we can write:
$$\omega = {x^3} \cdot dy + {z^3} \cdot dz$$
This is a differential 1-Form.
Calculating the exterior derivative, applying $d$
this becomes
$$d\omega = 3{x^2}dx \wedge dy + 3{z^2}dz \wedge dz = 3{x^2}dx \wedge dy$$
We have proven
$$\int\limits_S {d\omega } = \int\limits_{\partial S} \omega $$
Stoke's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Factorize $x^9 - 1 $ in $\mathbb F_3[x]$
I was able to factorize $x^9-x$ in $\mathbb F_3[x]$, however, with $x^9-1$ I am not sure what to do.
My first intuition was to do $(x^3-1)(x^3+1)$ but this is clearly just the first step, if not completely misplaced.
Thanks for help!
| This is a very "hands-on" answer.
There are only three elements of $\mathbb F_3$, i.e. $0$, $1$ and $2$.
The remainder theorem tells us that if $\mathrm f(a)=0$ then $(x-a)$ is a factor of $\mathrm f(x)$.
Let's try $x=0,1,2$ and see what happens:
\begin{eqnarray*}
0^9-1 &\equiv& 2 \pmod 3 \\
1^9-1&\equiv& 0 \pmod 3 \\
2^9-1 &\equiv& 1 \pmod 3
\end{eqnarray*}
Since $\mathrm f(1) \equiv 0 \pmod 3$, it follows that $(x-1)$ is a factor of $\mathrm f(x)$.
You need to use polynomial division, and reduce modulo $3$.
$$\frac{x^9-1}{x-1} = 1+x+x^2+\cdots +x^7+x^8$$
Next, we need to use the Factor Theorem on $\mathrm{g}(x):=1+x+x^2+\cdots +x^7+x^8$.
\begin{eqnarray*}
\mathrm g(0) &\equiv& 1 \pmod 3 \\
\mathrm g(1) &\equiv& 0 \pmod 3 \\
\mathrm g(2) &\equiv& 1 \pmod 3
\end{eqnarray*}
It follows that $(x-1)$ divides $\mathrm g(x)$ modulo $3$. Using Polynomial division
\begin{eqnarray*}
\frac{\mathrm g(x)}{x-1} &=&
x^7+2x^6+3x^5+4x^4+5x^3+6x^2+7x+8+\frac{9}{x-1} \\ \\
&\equiv& x^7+2x^6+x^4+2x^3+x+2 \pmod 3
\end{eqnarray*}
Let $\mathrm h(x) := x^7+2x^6+x^4+2x^3+x+2$ and proceed as before.
We see that $\mathrm h(x) \equiv 0 \pmod 3 \iff x \equiv 1 \pmod 3$, so consider
\begin{eqnarray*}
\frac{\mathrm h(x)}{x-1} &=& x^6+3x^5+3x^4+4x^3+6x^2+6x+7+\frac{9}{x-1} \\ \\
&\equiv& x^6+x^3+1 \pmod 3
\end{eqnarray*}
Again, we see that $(x-1)$ divides $x^6+x^3+1 \pmod 3$, leaving $x^5+x^4+x^3+2x^2+2x+2$, which is divisible by $x-1$. This leaves $x^4+2x^3+2x+1$, which is divisible by $x-1$. This gives $x^3+2$. This is divisible by $x-1$ which leaves $x^2+x+1$. This is divisible by $x-1$, and that leaves $x+2$. This is congruent to $x-1$.
\begin{eqnarray*}
(x-1)^9 &=& x^9-9x^8+36x^7-84x^6+126x^5-126x^4+84x^3-36x^2+9x-1 \\
&\equiv& x^9 + 2 \pmod 3 \\
&\equiv& x^9 -1 \pmod 3
\end{eqnarray*}
| {
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"url": "https://math.stackexchange.com/questions/2176528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does this integral converge? $\int \limits^{-1}_{-\infty} \frac{x^2dx}{(4+x^3)^3}$ I'm new with this topic and I try to solve this integral
$$\int \limits^{-1}_{-\infty} \frac{x^2dx}{(4+x^3)^3}$$
There is a break point in $\displaystyle{x= -\sqrt[3]{4}\thinspace}$
I think, that I need to divide line segment into two like this:
$$\int \limits^{{-\sqrt[3]{4}\thinspace}}_{-\infty} \frac{x^2dx}{(4+x^3)^3} + \int \limits^{-1}_{{-\sqrt[3]{4}\thinspace}} \frac{x^2dx}{(4+x^3)^3}$$
But here I face double limits in the first integral. How should I deal with this?
| As $x\to -\infty$, $\frac{x^2}{(4+x^3)^3}\sim \frac{1}{x^4}$ and so there is no issue at the lower limit.
Next, note that
$$x^3+4=(x+4^{1/3})(x^2-4^{1/3}x+4^{2/3})$$
Hence, as $x\to -4^{1/3}$, $\frac{x^2}{(4+x^3)^3}\sim \frac{4^{2/3}}{432(x+4^{1/3})^3}$ and the integral diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $abc=1$ so $\sum\limits_{cyc}\frac{7-6a}{2+a^2}\geq1$
Let $a$, $b$ and $c$ be real numbers such that $abc=1$. Prove that:
$$\frac{7-6a}{2+a^2}+\frac{7-6b}{2+b^2}+\frac{7-6c}{2+c^2}\geq1$$
The equality occurs also for $a=b=2$ and $c=\frac{1}{4}$.
This inequality is a similar to the very many contest's inequalities, but nothing helps.
At least, I don't see how we can prove it.
An example of my trying.
We need to prove that
$$\sum_{cyc}\frac{7-6a}{2+a^2}\geq1$$ or
$$\sum_{cyc}\left(\frac{7-6a}{2+a^2}+1\right)\geq4$$ or
$$\sum_{cyc}\frac{(a-3)^2}{2+a^2}\geq4.$$
By C-S $$\sum_{cyc}\frac{(a-3)^2}{2+a^2}=\sum_{cyc}\frac{(a-3)^2(a+k)^2}{(2+a^2)(a+k)}\geq\frac{\left(\sum\limits_{cyc}(a-3)(a+k)\right)^2}{\sum\limits_{cyc}(2+a^2)(a+k)^2}$$
Now we'll find a value of $k$, for which the equality in the last inequality occurs for $a=b=2$ and $c=\frac{1}{2}$.
Since in all equality case we have
$$\frac{a-3}{(2+a^2)(a+k)}=\frac{b-3}{(2+b^2)(b+k)}=\frac{c-3}{(2+c^2)(a+k)},$$
we obtain:
$$\frac{2-3}{(2+2^2)(2+k)}=\frac{\frac{1}{4}-3}{(2+\left(\frac{1}{4}\right)^2)(\frac{1}{4}+k)},$$
which gives $k=-\frac{9}{4}$.
Thus, it remains to prove that
$$\left(\sum\limits_{cyc}(a-3)(4a-9)\right)^2\geq4\sum_{cyc}(2+a^2)(4a-9)^2,$$
which is wrong for $a=4$ and $b=c=\frac{1}{2}$.
Any hint?
Thank you!
| Clearly, we only need to prove the case $a, b, c > 0$.
Let $f(x) = \frac{7-6x}{x^2+2}.$
We have $f'(x) = \frac{(6x+4)(x-3)}{(x^2+2)^2}$.
Thus, $f(x)$ is strictly decreasing on $(0, 3)$
and strictly increasing on $(3,+\infty).$
Also, since $f(x) + 1 = \frac{(x-3)^2}{x^2+2}\ge 0$, we have $f(x)\ge -1.$
WLOG, assume that $c\le b\le a$. Clearly $c\le 1$.
If $c < \frac{1}{9}$, we have $f(a)+f(b)+f(c) \ge (-1)+(-1) + f(\frac{1}{9}) > 1.$
If $\frac{1}{9} \le c \le 1$, we have $1\le ab \le 9$. Denote $p = \sqrt{ab} \in [1, 3]$. We have
\begin{align*}
&\frac{7-6a}{a^2+2} + \frac{7-6b}{b^2+2} - \frac{2(7-6\sqrt{ab})}{ab + 2}\\
=\ &\frac{7-6a}{a^2+2} + \frac{7-6(\frac{p^2}{a})}{(\frac{p^2}{a})^2+2} - \frac{2(7-6p)}{p^2 + 2}\\
=\ & \frac{(a-p)^2[(7p^2+24p-14)a^2+(-6p^4+14p^3+24p^2-28p-24)a+7p^4+24p^3-14p^2] }{(a^2+2)(p^4+2a^2)(p^2+2)}\\
\ge\ & 0,
\end{align*}
since $7p^2+24p-14 > 0$, $7p^4+24p^3-14p^2 > 0$ and
\begin{align*}
&4(7p^2+24p-14)(7p^4+24p^3-14p^2) - (-6p^4+14p^3+24p^2-28p-24)^2\\
=\ & -12(3p^2+4p-6)(p^2-6p-2)(p^2+2)^2 > 0.
\end{align*}
Thus, it suffices to prove that $$ \frac{7-6c}{c^2+2} + \frac{2(7-6\sqrt{ab})}{ab + 2} \ge 1$$ or
$$\frac{7-6c}{c^2+2} + \frac{2(7-6\sqrt{\frac{1}{c}})}{\frac{1}{c} + 2} \ge 1$$ or
$$\frac{(3c+6\sqrt{c}+5)(2\sqrt{c}-1)^2(\sqrt{c}-1)^2}{(c^2+2)(2c+1)} \ge 0.$$
This completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all points $(x,y)$ on the graph of $f(x)$ with tangent lines passing through a certain point Find all points $(x,y)$ on the graph of $f(x) = x^2$ with tangent lines passing through the point $(3,8)$.
My attempt:
$f'(x) = x^2$
I substituted a random point into $f'(x)$ to get the gradient of the tangent, so $f'(0) = 0$.
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
$$ m = \frac{8-0}{3-0} = 8/3$$
$$y-y_1 = m(x-x_1)$$
Now I substituted the point $(3,8)$:
$$ y - 8 = \frac{8}{3}(x-3)$$
$$y = \frac{8}{3}x$$
$$ y = x^2$$
Solving simultaneously we get $x = 0$ and $y = 0$ or $x = 8/3 $ and $ y = 64/9$
So we have the points $(0;0)$ and $(\frac{8}{3} ; \frac{64}{9})$. However, I am not so confident with my answer. Is this correct?
| Let $(a, a^2)$ be a point on the graph of $f(x) = x^2$ the tangent line at which passes through the point $(3, 8)$. We know that the slope of the tangent line at $(a, a^2)$ is $$ \frac{\mathrm{d} f(x) }{\mathrm{d} x} \vert_{x=a} = 2a.$$
But the tangent line passes through the points $(a, a^2)$ and $(3, 8)$. So its slope is
$$\frac{a^2 - 8}{a-3}.$$
Now equating the above two expressions for the slope, we obtain
$$\frac{a^2 - 8}{a-3} = 2a.$$
which yields
$$a^2 - 6a + 8 = 0, $$
from which we obtain $a = 2$ or $a=4$.
Hence the two points are $(2, 4)$ and $(4, 16)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Derivative of $f(x) = y^4$ $(x^3 \cdot y^4)' = ?$
$a = x^3$ , $b = y^4$
$$(x^3 \cdot y^4)'= a'\cdot b + a\cdot b' = 3x^2 \cdot y^4 + x^3 \cdot (b')$$
Not sure what the derivative of $y^4$ would be, $(4y^3)$? Since the derivative of $(x\cdot y) = 1\cdot y + x \cdot y'$ and the derivative of $y' = y'$ instead of following the power rule.
| If $y$ is a function of $x$, then $y(x)^4$ is a composite function and you need to apply the chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the maximum of the value $F=x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+\cdots+x_{n-2}x_{n-1}x_{n}+x_{n-1}x_{n}x_{1}$
Let $x_{i}\ge 0$ and such $$x_{1}+x_{2}+\cdots+x_{n}=1$$
Find the maximum of the value
$$F=x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+\cdots+x_{n-2}x_{n-1}x_{n}+x_{n-1}x_{n}x_{1}$$
case $1$. when $n=3$,then
$$F=2x_{1}x_{2}x_{3}\le2\dfrac{(x_{1}+x_{2}+x_{3})^3}{27}=\dfrac{2}{27}$$
but for $n\ge 4$,I can't solve it
| I think you meant that $F=x_1x_2x_3+x_2x_3x_4+...+x_nx_1x_2$,
but I am ready to think also about your $F$.
For $n=4$ we'll prove that $abc+abd+acd+bcd\leq4$,
where $a$, $b$, $c$ and $d$ are non-negatives such that $a+b+c+d=4$,
Indeed, by AM-GM
$$abc+abd+acd+bcd=ab(c+d)+cd(a+b)\leq\left(\frac{a+b}{2}\right)^2(c+d)+\left(\frac{c+d}{2}\right)^2(a+b)=$$
$$=\frac{(a+b)(c+d)(a+b+c+d)}{4}=(a+b)(c+d)\leq\left(\frac{a+b+c+d}{2}\right)^2=4$$
Thus, $$abc+abd+acd+bcd\leq4\left(\frac{a+b+c+d}{4}\right)^3,$$
which says that $$x_1x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2\leq4\left(\frac{x_1+x_2+x_3+x_4}{4}\right)^3=\frac{1}{16}.$$
The equality occurs for $x_1=x_2=x_3=x_4=\frac{1}{4}$, which says that the answer is $\frac{1}{16}$.
For $n=5$ the answer is $\frac{1}{25}$.
Indeed, let $x_1=a$, $x_2=b$, $x_3=c$, $x_4=d$,$x_5=e$ and $a=b=c=d=e=\frac{1}{5}$.
Hence, $abc+bcd+cde+dea+eab=\frac{1}{25}$ and it's enough to prove that
$$abc+bcd+cde+dea+eab\leq\frac{1}{25}.$$
Let $a=\min\{a,b,c,d,e\}$.
Hence, $0\leq a\leq\frac{1}{5}$ and by AM-GM we obtain
$$abc+bcd+cde+dea+eab=a(b+d)(c+e)+cd(b+e-a)\leq$$
$$\leq a\left(\frac{b+c+d+e}{2}\right)^2+\left(\frac{c+d+b+e-a}{3}\right)^3=\frac{a(1-a)^2}{4}+\frac{(1-2a)^3}{27}$$
and it remains to prove that
$$\frac{a(1-a)^2}{4}+\frac{(1-2a)^3}{27}\leq\frac{1}{25},$$
which is $(5a-1)^2(5a+8)\geq0$ and we are done!
For $n\geq6$ the idea is:
$$F\leq\left(x_1+x_4+...\right)\left(x_2+x_5+...\right)\left(x_3+x_6+...\right)\leq\left(\frac{x_1+x_2+...+x_n}{3}\right)^3=\frac{1}{27}.$$
the equality occurs for $x_1=x_2=x_3=\frac{1}{3},$ which says that the answer is $\frac{1}{27}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating a finite sum using a double sum I want to evaluate the sum $S_n = \sum_{k=0}^n k 3^k $.
Try:
Since $ \sum_{j=0}^{k-1} 1 = k $, then
$$ S_n = \sum_{k=0}^n \sum_{j=0}^{k-1} 3^k $$
Since we are summing over $(0 \leq k \leq n )$ and $(0 \leq j \leq k -1 )$, then we are summing over $(0 \leq j < k \leq n )$. Thus, we can write
$$ S_n = \sum_{j=0}^n \sum_{k=j}^n 3^k = \sum_{j=1}^n \frac{ 3^{n+1} - 3^j }{2} = \frac{n 3^{n+1}}{2} - \frac{1}{2} \sum_{j=0}^n 3^j = \boxed {\frac{n 3^{n+1} }{2} - \frac{1}{2} \cdot \left( \frac{ 3^{n+1} - 1 }{2} \right) } $$
Is this a correct argument?
| Small additional info. The expression
$$ S_n = \sum_{k=0}^n \sum_{j=0}^{k-1} 3^k $$
has for $k=0$ an empty inner sum
\begin{align*}
\sum_{j=0}^{-1} 3^k=0
\end{align*}
since the upper limit of the sum is smaller than the lower limit.
So, we have effectively
\begin{align*}
\sum_{k=0}^n \sum_{j=0}^{k-1} 3^k&= \sum_{k=\color{blue}{1}}^n \sum_{j=0}^{k-1} 3^k=\sum_{k=0}^{n-1} \sum_{j=0}^{k} 3^{k+1}
\end{align*}
with range
\begin{align*}
0\leq j\leq k\leq n-1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Given $x^9 = e$ and $x^{11} = e$ prove $x = e$. Full Problem: Prove that for any element $x$ in a group $G$ that satisfies
$$x^9 = e \\
x^{11} = e,$$
where $e$ is the identity element, that $x$ itself must be $e$.
Is this as simple as showing that
*
*$x^{11} = x^{9} \cdot x^{2} = e \cdot x^2 \Rightarrow x^2 = e$
*$x^{9} = x^{2} \cdot x^{7} = e \cdot x^7 \Rightarrow x^7 = e$
*$x^{7} = x^{2} \cdot x^{5} = e \cdot x^5 \Rightarrow x^5 = e$
*$x^{5} = x^{2} \cdot x^{3} = e \cdot x^3 \Rightarrow x^3 = e$
*$x^{3} = x^{2} \cdot x = e \cdot x \Rightarrow x = e$
Therefore, $x = e$.
| Your reasoning is correct but here is a more direct argument.
Since $x^9 = e$ and $x^{11} = e$, the order of $x$ divides both $9$ and $11$. Therefore, the order of $x$ is $1$ so $x = e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Trignometric integral : $\int \frac{dx}{\sin x + \sec x}$ The integral I am trying to compute is :
$$\int \dfrac{dx}{\sin x + \sec x}$$
I have tried manipulating trignometric functions and it took me nowhere. Then finally I tried putting $\tan\dfrac{x}{2} = t$, and subsequently:
$$=\int\dfrac{1-t^2}{(1+t^2)^2-2t(t^2-1)}dt$$
I cannot see how to approach after this, I know we have to factor out two quadratics, but cant see how to.
Also, if there was another method instead of this substitution, please hint on that too! Thanks!
| As Michael Wang commented $$A=(1+t^2)^2-2t(t^2-1)=t^4-2 t^3+2 t^2+2 t+1$$ Computing the roots of the quartic equation (all are complex), you end with
$$A=\left(t^2+(\sqrt{3}-1)t-\sqrt{3}+2\right)\left(t^2-(\sqrt{3}+1) t+\sqrt{3}+2\right)$$ Partial fraction decomposition then leads to
$$\dfrac{1-t^2}{(1+t^2)^2-2t(t^2-1)}=\frac{1+t}{t^2+(\sqrt{3}-1)t-\sqrt{3}+2}+\frac{1+t}{t^2-(\sqrt{3}+1) t+\sqrt{3}+2}$$ which makes the problem workable.
Edit
Being less lazy, write $$(t^2+at+b)(t^2+ct+d)-(t^4-2 t^3+2 t^2+2 t+1)=0$$ Expand and group terms to get $$(a+c+2)t^3+ (a c+b+d-2)t^2+ (a d+b c-2)t+(b d-1)=0$$ So the equations to be solved are $$a+c+2=0\tag 1$$ $$a c+b+d-2=0\tag 2$$ $$a d+b c-2=0\tag 3$$ $$bd-1=0\tag 4$$ Using $(4)$, then $d=\frac 1b$. Using $(1)$, then $c=-2-a$. Replace in $(2)$ to get $$\frac{a}{b}-(a+2) b-2=0\implies a=\frac{2 b}{1-b}$$ Now, plug everything in $(3)$ to get $$\frac{(b^2-4b+1) \left(b^2+1\right)}{(b-1)^2 b}=0\implies b^2-4b+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$ let $$f(x)=\frac{1}{1+2\cos x}$$
prove that :
$$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$$
My Try :
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{4\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{6\pi}{7})}$$
$$L=\frac{1}{1+2\cos (\frac{6\pi}{7})}+\frac{1}{1+2\cos (\frac{4\pi}{7})}+\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
what now ?
| The minimal polynomial of $2\cos\frac{2\pi}{7}$ over $\mathbb{Q}$ is given by $x^3+x^2-2x-1$, and the conjugated roots are $2\cos\frac{4\pi}{7}$ and $2\cos\frac{6\pi}{7}$. It follows that $1+2\cos\frac{2\pi}{7}$, $1+2\cos\frac{4\pi}{7}$ and $1+2\cos\frac{6\pi}{7}$ are roots of
$$ p(x)= x^3-2x^2-x+1 $$
and by Vieta's theorem
$$\sum_{k=1}^{3}\frac{1}{1+2\cos\frac{2\pi k}{7}} = -\frac{[x^1]\,p(x)}{[x^0]\,p(x)}=\color{red}{1} $$
as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Laurent series for $f(z) = \frac{1}{(z-4)(z+2)}$ Find the Laurent series for $f(z) = \frac{1}{(z-4)(z+2)}$, valid for $2<|z|<4$.
first i did partial fraction expansion:
$\frac{1}{(z-4)(z+2)} = \frac{A}{(z-4)} + \frac{B}{(z+2)}$
The result i got : $\frac{1}{6}[\frac{1}{(z-4)} - \frac{1}{(z+2)}]$
For the annuls centered at $0$, I solved for the laurent series:
$\frac{1}{(z-4)}= \frac{1}{4}\frac{1}{-1+\frac{z}{4}} = \frac{1}{4} \sum_{n=0}^{\infty}\frac{(-1)^nz^n}{4^n}$ is the power series for $|z|<4$
What I am supposed to do about $\frac{1}{(z+2)}$ since it is not within $2<|z|<4$|?
| You need to write the term such that the region of convergence changes:
$$
\frac{1}{z+2} = \frac{1}{z} \, \frac{1}{1 + 2z^{-1}}
$$
which can be series expanded (in powers of $1/z$) for $|2z^{-1}| < 1$, i.e. for $|z| > 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194019",
"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.