Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Integration Trig Substitution After making the correct trig substitution what does the integral of
$\dfrac{1}{\sqrt{9-x^2}} dx$ reduce to without solving the equation?
I reduced it down to the integral of $3\cos(\theta)d\theta$ is this correct?
| Performing trig. substitution of $x=3 \sin \theta$ (so $dx=3\cos \theta \, d\theta$) gives $$\int\frac{1}{\sqrt{9-x^2}} \, dx=\int\frac{1}{\sqrt{9-9 \sin^2 \theta}} \, (3 \cos \theta \, d\theta)=\int\frac{1}{|3 \cos \theta|} \, (3 \cos \theta \, d\theta) = \int 1 \, d\theta$$
which is equal to $\theta + C$, or $\displaystyle\sin^{-1} \left( \frac{x}{3} \right) + C$ after back-substitution.
| {
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For which $a$ does $\lvert x+1\rvert+\lvert 2-x\rvert=a^2 -1$ have exactly two solutions? If it is not a problem, I would really appreciate if someone could explain to me how to solve and graph the following equation:
For which real numbers $a$ does the equation
$$\lvert x+1\rvert +\lvert 2-x\rvert=a^2 -1$$
Have exactly two solutions?
| This solution does not use graph or distance between points. We have $4$ cases to look at:
Case 1:
If $x < -1$, then $x+1 < 0$, and $2-x > 0$, so $LHS = -x-1 + 2-x = 1 - 2x = a^2 - 1 \to x = \dfrac{2-a^2}{2}$. This means the equation either has only one solution, or no solution depending on $a$. In particular it has one solution only if $\dfrac{2-a^2}{2} < -1$ or $a<-2$ or $a>2$
Case 2:
If $-1 < x < 2$, then $x+1 > 0$, and $2-x > 0$, so $LHS = x+1 + 2-x = 3 = a^2 - 1 \to a = \pm 2$, and the equation has infinitely many solutions on this interval $(-1,2)$ or it has no solution depending on $a$.
Case 3:
If $x > 2$, then $x+1 > 0$, and $2-x < 0$, so $LHS = x+1+x-2 = 2x-1 = a^2- 1 \to x = \dfrac{a^2}{2}$ which is one solution or no solution depending on $a$. In particular it has one solution only when $\dfrac{a^2}{2} > 2$, so $a<-2$ or $a > 2$
Case 4:
If $x = -1$ or $x = 2$, then $LHS = 3 = a^2 - 1 \to a = \pm 2$. This takes us back to case 2 which gives additional solutions.
Observe that if $a < -2$ or $a > 2$, then we have one solution $x < -1$, and another solution $x > 2$. So in total there are $2$ solutions. For other values of $a$, we have either one or no solution or more than two solutions. Thus the answer is : $a < -2$ or $a > 2$.
| {
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If $a+b+c+d=4$, then $\sum\frac{1}{a+3}\le \frac{1}{abcd}$ This is somehow related to this problem but I don't have any idea about it.
Let $a$, $b$, $c$ and $d$ be positive reals such that $a+b+c+d=4$. Prove that:
$$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}+\frac{1}{d+3}\le \frac{1}{abcd}$$
Now I also tried to prove the $3$ variable version :
$$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\le \frac{1}{abc}$$ where $a,b,c>0$ with $a+b+c=3$. But I haven't been able to solve it too. Anyone can help?? Thanks a lot.
| $\sum$ will be used to denote the cyclic sum.
For the second inequality, let $t=ab+bc+ca$. Note that $t\le 3$, and also:
$$3t=(a+b+c)(ab+bc+ca)\ge 9\sqrt[3]{(abc)(a^2b^2c^2)}=9abc$$
Now, by Cauchy-Schwarz:
$$\sum\frac{2bc}{a+2}=2\sum\frac{(bc)^2}{abc+2bc}\ge \frac{2t^2}{3abc+2t}\ge \frac{2t}{3}$$
Note further that $\frac{2t}{3}-(t-1)=\frac{3-t}{3}\ge 0$ since $t\le 3$. Hence:
$$\sum\frac{2bc}{a+2}\ge ab+bc+ca-1\implies \sum (bc-\frac{2bc}{a+2})\le 1\implies \sum\frac{1}{a+2}\le\frac{1}{abc}$$
as desired.
A similar trick works for the first inequality. Let $t=abc+bcd+cda+dab$ this time. Then by repeated applications of AM-GM,
$$t=ab(c+d)+cd(a+b)\le (\frac{a+b}{2})^2(c+d)+(a+b)(\frac{c+d}{2})^2=\frac{(a+b+c+d)^2}{4}=4$$
Also, note that by AM-GM:
$$4t=(a+b+c+d)(abc+bcd+cda+dab)\ge 16\sqrt[4]{(abcd)(a^3b^3c^3d^3)}=16abcd$$
Using these estimations:
$$\sum \frac{3bcd}{a+3}=3\sum\frac{(bcd)^2}{abcd+3bcd}\ge \frac{3t^2}{4abcd+3t}\ge\frac{3t}{4}$$
And $\frac{3t}{4}-(t-1)=\frac{4-t}{4}\ge 0$, so that:
$$\sum\frac{3bcd}{a+3}\ge abc+bcd+cda+dab-1\implies \sum (bcd-\frac{3bcd}{a+3})\le 1\implies \sum\frac{1}{a+3}\le \frac{1}{abcd}$$
as desired.
Note: Using this same method, the $n$-variable inequality can also be proven.
| {
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Evaluation of $\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx$
Compute the indefinite integral
$$
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx
$$
My Attempt:
$$
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx &= \int\frac{\cos 2x}{\sin^2 x\sqrt{\cos 2x}}\sin xdx\\
&= \int\frac{2\cos^2 x-1}{(1-\cos^2 x)\sqrt{2\cos^2 x-1} }\sin x \,dx
\end{align}
$$
Let $\cos x = t$, so that $\sin x\,dx = -dt$. This changes the integral to
$$
\begin{align}
\int\frac{(2t^2-1)}{(t^2-1)\sqrt{2t^2-1}}\,dt &= \int\frac{(2t^2-2)+1}{(t^2-1)\sqrt{2t^2-1}}\,dt\\
&= 2\int\frac{dt}{\sqrt{2t^2-1}}+\int \frac{dt}{(t^2-1)\sqrt{2t^2-1}}
\end{align}
$$
How can I solve the integral from here?
| If in the last integral you substitute $t=\frac{1}{\sqrt{2}}\cosh z$, you end with:
$$ I = \frac{1}{\sqrt{2}}\int\frac{1}{\frac{\cosh^2 z}{2}-1}dz=-\operatorname{arctanh}(\sqrt{2}\tanh z)=-\operatorname{arctanh}\left(\sqrt{2-\frac{1}{t^2}}\right).$$
| {
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Calculate $\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$ I'm an eight-grader and I need help to answer this math problem.
Problem:
Calculate $$\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$$
This one is very hard for me. It seems unsolvable. How to calculate the series without using Wolfram Alpha? Please help me. Grazie!
| Informally:
You're taking the sum of the row sums of
$
\ \ \ \displaystyle{1\over 5^{\phantom 1}}
$
$
\ \ \ \displaystyle{1\over 5^{ 3}} \ \ \ \ \displaystyle{1\over 5^{ 3}}\ \ \ \ \displaystyle{1\over 5^{ 3}}
$
$
\ \ \ \displaystyle{1\over 5^{ 5}} \ \ \ \ \displaystyle{1\over 5^{ 5}}\ \ \ \ \displaystyle{1\over 5^{ 5}}\ \ \ \ \displaystyle{1\over 5^{ 5}}\ \ \ \ \displaystyle{1\over 5^{ 5}}
$
$
\ \ \ \ \ \ \ \ \ \ \ \ \ \vdots
$
Take the sum of the column sums instead.
Towards this end, note, for example, that
$$
{1\over 5^3}+{1\over 5^5}+{1\over 5^7}+\cdots\
=\ {1\over5}\Bigl( {1\over 25}+{1\over 25^2}+{1\over25^3}\cdots \Bigr)
={1\over 5}{1/25\over1-1/25}.
$$
| {
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Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$ Solve for $x$, $\tan x +\sec x = 2\cos x$ ; $−∞ < x < ∞$
$$\tan x + \sec x = 2\cos x$$
I tried changing it all to sin and cos
$$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$$
then I made it to one fraction
$$\frac{\sin x + 1}{\cos x} = 2 \cos x$$
Then I don't know where to go from there. Please help!
| $$\sec x+\tan x=2\cos x\iff \sec x-\tan x=\frac1{2\cos x}$$
Adding we get $$2\sec x=2\cos x+\frac1{2\cos x}$$
Multiplying by $2\cos x,$
$$4=4\cos^2x+1\iff2\cos^2x=\frac32\iff\cos2x=2\cos^2x-1=\frac12=\cos60^\circ$$
$$\iff 2x=360^\circ m\pm60^\circ$$
| {
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prove $\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$
Show that if $a$, $b$, $c$ and $d$ are non-negatives and $ab+bc+cd+da=1$ then:
$$\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$$
yet again it should be solved with Cauchy inequality.
thing i have done so far:
$(\sum\limits_{cyc} \frac {a^3} {b+c+d})\times(\sum\limits_{cyc} a(b+c+d)) \geq (\sum\limits_{cyc} a^2)^2$
so my problem is simplified to proving this:
$\frac {(\sum\limits_{cyc} a^2)^2} {(\sum\limits_{cyc} a(b+c+d))} \geq \frac {1} {3}$
$3 \times (\sum\limits_{cyc} a^2)^2 \geq \sum\limits_{cyc} a(b+c+d)$
$3 \times (a^2+b^2+c^2+d^2)^2 \geq 2(ab+ac+ad+bc+bd+cd)$
someone said to me if i play around with AM-GM it could be solved and i'm almost there
my idea is this right now:
prove $(a^2+b^2+c^2+d^2) \geq ab+bc+cd+da=1$ proved(with help of Jineon Baek hint)
prove $3(a^2+b^2+c^2+d^2) \geq 2(ab+ac+ad+bc+bd+cd)$ proved(with help of Jineon Baek hint)
| By Holder an AM-GM we obtain:
$$\sum_{cyc}\frac{a^3}{b+c+d}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c+d)}=\frac{1}{12}(a+b+c+d)^2\geq$$
$$\geq\frac{1}{12}\left(2\sqrt{(a+c)(b+d)}\right)^2=\frac{1}{3}(ab+bc+cd+da)=\frac{1}{3}.$$
Done!
| {
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Antiderivative of $\frac{1}{1+\sin {x} +\cos {x}}$ How do we arrive at the following integral
$$\displaystyle\int\dfrac{dx}{1+\sin {x}+\cos {x}}=\log {\left(\sin {\frac{x}{2}}+\cos {\frac{x}{2}}\right)}-\log {\left(\cos {\frac{x}{2}}\right)}+C\ ?$$
| $$1+\cos x=2\cos^2\frac x2$$
$$1+\cos x+\sin x=2\cos^2\frac x2+2\sin\frac x2\cos\frac x2=2\cos\frac x2\left(\sin\frac x2+\cos\frac x2\right)$$
$$=\sqrt2\cos\frac x2\cos\left(\frac\pi4-\frac x2\right)$$
For numerator, $$\cos\frac\pi4=\cos\left[\frac x2-\left(\frac\pi4-\frac x2\right)\right]$$
| {
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Trigonometric Arithmetic Progression If $a$, $b$, $c$ are in arithmetic progression, prove that
$$\cos A \cot\frac{A}{2} \qquad \cos B \cot \frac{B}{2} \qquad \cos C \cot\frac{C}{2}$$
are in arithmetic progression, too.
Here, $a$, $b$, $c$ represent the sides of a triangle and $A$, $B$, $C$ are the opposite angles of the triangle.
| $$\cos B\cot\frac B2=\left(1-2\sin^2\frac B2\right)\cot\frac B2=\cot\frac B2-\sin A$$
Clearly using Law of Sines, $\sin A,\sin B,\sin C$ are in AP
So, it is sufficient to show that $\displaystyle\cot\frac A2,\cot\frac B2,\cot\frac C2$ are also in AP
We have $$a+c=2b\iff \sin A+\sin C=2\sin B$$
Now using Prosthaphaeresis Formula, $$2\sin\frac{A+C}2\cos\frac{A-C}2=2(2\sin\frac B2\cos\frac B2)$$
$$\iff\cos\frac{A-C}2=2\sin\frac B2\ \ \ \ (1)$$
Again, $\displaystyle\cot\frac A2+\cot\frac C2=2\cot\frac B2$
$$\iff\frac{\sin\frac{A+C}2}{\cos\frac A2\cos\frac C2}=2\frac{\cos\frac B2}{\sin\frac B2}$$
$$\iff\sin\frac B2=2\cos\frac A2\cos\frac C2=\cos\frac{A-C}2-\cos\frac{A+C}2$$
$$\iff\sin\frac B2=\cos\frac{A-C}2-\sin\frac B2$$ which is same as $(1)$
| {
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In a triangle ABC, prove that cot(A/2)+cot(B/2)+cot(C/2) =cot(A/2)cot(B/2)cot(C/2) In a triangle ABC, prove that $\cot \left ( \frac{A}{2} \right )+\cot \left ( \frac{B}{2} \right )+\cot \left ( \frac{C}{2} \right )=\cot \left ( \frac{A}{2} \right )\times \cot \left ( \frac{B}{2} \right )\times \cot \left ( \frac{C}{2} \right )$. I tried all identities I know but I have no idea how to proceed.
| Applying $\cot(x+y)=\dfrac{\cot x\cot y-1}{\cot x+\cot y}$ twice,
$$\cot(x+y+z)=\frac{\sum\cot x\cot y-1}{\prod\cot x-\sum\cot x}$$
Here $x=\dfrac A2$ etc. so that $x+y+z=\dfrac\pi2\implies\cot(x+y+z)=?$
| {
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Prove ${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{2/3}}=\frac{3^{3/2}}{2^{4/3}5^{5/6}\pi }\Gamma^3\left(\frac13\right)$ Here is one more conjecture I discovered numerically:
$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{\small2/3}}\stackrel{\color{#808080}?}=\frac{3^{\small3/2}}{2^{\small4/3}\,5^{\small5/6}\,\pi }\Gamma^3\!\!\left(\tfrac13\right)$$
How can we prove it?
Note that $\sqrt[3]{9+4\sqrt5}=\phi^2$.
Mathematica can evaluate this integral, but gives a large expression in terms of Gauss and Appel hypergeometric functions of irrational arguments.
| I've found a generalization of your conjecture:
If $k>0$ real number, then
$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{k\pm\tfrac{4\sqrt5k}{9}\,x}\
\left(1-x^2\right)^{\small2/3}}\stackrel{?}{=}
\frac{1}{\sqrt[3]{k}}\cdot\frac{3^{\small13/6}}{2^{\small4/3}5^{\small5/6}\pi}\Gamma^{3}\left(\tfrac{1}{3}\right).$$
For $k=9$ with $+$ sign it gives back your integral.
Another family of this type:
If $k>0$ real number, then
$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{k\pm\tfrac{4k}{5}\,x}\
\left(1-x^2\right)^{\small2/3}}\stackrel{?}{=} \frac{1}{\sqrt[3]{k}}\cdot\frac{\sqrt[3]{5}}{\sqrt[3]{2}\sqrt{3}\pi}\Gamma^3\left(\tfrac{1}{3}\right).
$$
| {
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Understanding the logic behind particular derivative I have
$\frac{\partial (f(x) g(x))}{\partial x}$=$g(x) f'(x)+f(x) g'(x)$,
I need to differentate this function with respect to x.
$f(x)=(x+1) (x+2)^2 (x+3)^3 (x+4)^4$
However I do not see the logic using the product rule.
| HINT
You must do that step by step.
$f(x)=(x+1) (x+2)^2 (x+3)^3 (x+4)^4$
$$f'(x)=((x+1))' ((x+2)^2 (x+3)^3 (x+4)^4)+((x+1)) ((x+2)^2 (x+3)^3 (x+4)^4)'$$
$$((x+2)^2 (x+3)^3 (x+4)^4)'=((x+2)^2)' ((x+3)^3(x+4)^4)+((x+2)^2) ((x+3)^3(x+4)^4)'$$
$$((x+3)^3(x+4)^4)'=((x+3)^3)'((x+4)^4)+((x+3)^3)((x+4)^4)'$$
now you go from the last to the first.
| {
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Middle binomial coefficient mod 4 It is known that the middle binomial coefficient is always even. Show that $\binom{2n}{n}= 2 \mod 4$ if and only if $n$ is a power of 2.
| The power of $2$ in $\binom{2n}{n}$ is
$$\lfloor \frac{2n}{2} \rfloor +\lfloor \frac{2n}{2^2} \rfloor +\lfloor \frac{2n}{2^3} \rfloor + ... -2 \left(\lfloor \frac{n}{2} \rfloor +\lfloor \frac{n}{2^2} \rfloor +\lfloor \frac{n}{2^3} \rfloor +...\right) \\
=\lfloor \frac{2n}{2} \rfloor - \left(\lfloor \frac{n}{2} \rfloor +\lfloor \frac{n}{2^2} \rfloor +\lfloor \frac{n}{2^3} \rfloor +...\right)\\
=n - \left(\lfloor \frac{n}{2} \rfloor +\lfloor \frac{n}{2^2} \rfloor +\lfloor \frac{n}{2^3} \rfloor +...\right)$$
Now the problem asks you to prove that
$$n - \left(\lfloor \frac{n}{2} \rfloor +\lfloor \frac{n}{2^2} \rfloor +\lfloor \frac{n}{2^3} \rfloor +...\right)=1$$
if and only if $n$ is a power of $2$.
This is equivalent to
$$\lfloor \frac{n}{2} \rfloor +\lfloor \frac{n}{2^2} \rfloor +\lfloor \frac{n}{2^3} \rfloor +... =n-1$$
if and only if $n$ is a power of $2$.
Now write $n=2^k+l$ with $l <2^k$. Then,
$$n-1= \lfloor \frac{n}{2} \rfloor +\lfloor \frac{n}{2^2} \rfloor +\lfloor \frac{n}{2^3} \rfloor +...+\lfloor \frac{n}{2^k} \rfloor \leq \frac{n}{2} + \frac{n}{2^2} + \frac{n}{2^3} +...+ \frac{n}{2^k} \\
=n (1-\frac{1}{2^k})=n-1-\frac{l}{2^k}$$
This inequality implies $l=0$.
| {
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Find depth of a half-filled parabolic cross-section Given a cross-section of an object that is parabolic in shape, how do you find the depth of the object when it is "half full".
A full example given in an exam:
A long trough whose cross-section is parabolic is $1\frac{1}{2}$ metres wide at the top and $2$ metres deep. Find the depth of water when it is half-full.
| The way to do this is to find the area of the parabola as a function of $x$. Let's assume for simplicity that function is: $x = c*y^2$. This is equivalent to $y = \sqrt{\frac{x}{c}}$ Since the parabola is symmetric, we can double this to get the total width at any point $x$.
$$ w(x) = 2 \sqrt{\frac{x}{c}}$$
Now we can integrate with respect to $x$ from $0$ to $2$. But first, we need to find the constant $c$ that gives us the particular parabola that we have in this case. To do this, at the point $(0.75,2)$ solve for $c$.
$$ y= \sqrt{\frac{x}{c}} = 2 = \sqrt{\frac{3/4}{c}} \rightarrow 4 = \frac{3}{4c} \rightarrow c = \frac{3}{16} $$
Now we can do the actual integration that will give the area.
$$A(u) = \int_0^u w(x) dx = \int_0^u 2 \sqrt {\frac{16x}{3}} dx = \frac{8}{\sqrt 3} \int_0^u \sqrt x dx = \frac{8}{\sqrt 3} [\frac{2}{3} x^{\frac{3}{2}}]_0 ^ u = \frac{16}{3 \sqrt 3} u^{\frac{3}{2}}$$.
Thus, you now have the area as a function of integration distance along the length. Just plug in $2$ to get the total area, and then solve it for $1/2$ the total area.
$$A_t = \frac{16}{3 \sqrt 3} 2^{\frac{3}{2}} = \frac{32}{3}\sqrt{\frac{2}{3}}$$
Solve:
$$\frac{\frac{32}{3}\sqrt{\frac{2}{3}}}{2} = \frac{16}{3 \sqrt 3} u^{\frac{3}{2}}$$
$$\frac{16}{3}\sqrt{\frac{2}{3}} = \frac{16}{3 \sqrt 3} u^{\frac{3}{2}}$$
$$\sqrt{\frac{2}{3}} = \frac{1}{\sqrt 3} u^{\frac{3}{2}}$$
$$\sqrt{2} = u^{\frac{3}{2}}$$
$$\sqrt{2} = u^{\frac{3}{2}}$$
$$u = \sqrt[3]{2}$$
| {
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Integrate a division of polynomials Hi I have the following integral:
$$\int \frac{2x}{x^2+6x+3}\, dx$$
I made some changes like:
$$\int \dfrac{2x+6-6}{x^2+6x+3}\, dx$$
then I have:
$$\int \dfrac{2x+6}{x^2+6x+3}\, dx -\int\dfrac{6}{x^2+6x+3}\, dx$$
and thus: $$\ln(x^2+6x+3)-\int\dfrac{6}{x^2+6x+3}\, dx$$
Ok, I have decomposed $$\frac{2x}{x^2+6x+3} $$ in: $$ \frac{3+\sqrt6}{\sqrt6(x+\sqrt 6+3)} + \frac{3-\sqrt6}{\sqrt6 (-x+\sqrt6-3)}$$
How can I integrate this expressions?
| HINT:
As
$\displaystyle x^2+6x+3=(x+3)^2-(\sqrt6)^2,$
using Trigonometric substitution, set $x+3=\sqrt6\sec\theta$
or use $\#1$ of this
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Series for logarithms This is more of a challenge than a question, but I thought I'd share anyway. Prove the following identities, and prove that the pattern continues.
\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)=\ln2
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{2}{3n+3}\right)=\ln3
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{4n+1}+\frac{1}{4n+2}+\frac{1}{4n+3}-\frac{3}{4n+4}\right)=\ln4
\end{equation*}\begin{equation*}
\mathrm{etc.}
\end{equation*}
By the way, a good notation for discussing this problem can be found here. The problem is to prove that:
$[\overline{1,-1}]=\ln2,\;\;$ $[\overline{1,1,-2}]=\ln3,\;\;$ $[\overline{1,1,1,-3}]=\ln4,\;\;$ $[\overline{1,1,1,1,-4}]=\ln5,\;\;$ etc.
| Let $m\geq 2$. Put:
$$S_m=\sum_{n\geq 0}(\frac{1}{mn+1}+\cdots+\frac{1}{mn+m-1}-\frac{m-1}{mn+m}) $$
As $\displaystyle \frac{1}{mn+r}=\int_0^1x^{mn+r-1}dx$, We have
$$S_m=\int_0^{1}\frac{1+x+\cdots+x^{m-2}-(m-1)x^{m-1}}{1-x^m} dx$$
But
$$1+x+\cdots+x^{m-2}-(m-1)x^{m-1}=(1-x^{m-1})+\cdots+(x^{m-2}-x^{m-1})$$
Hence
$$1+x+\cdots+x^{m-2}-(m-1)x^{m-1}=[(1-x)(1+x+\cdots x^{m-2})]+\cdots+[x^{m-2}(1-x)]$$
and
$$[(1+x+\cdots x^{m-2})]+\cdots+[x^{m-2}]=1+2x+\cdots+(m-1)x^{m-2}$$
Thus
$$S_m=\int_0^1\frac{\sum_{k=0}^{m-2}(k+1)x^{k}}{1+x+\cdots+x^{m-1}}dx=\int_0^1\frac{P^{\prime}(x)}{P(x)}dx=[\log(1+x+\cdots+x^{m-1})]_0^1=\log(m)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 0
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prove $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$
If $a,b,c$ are positive real numbers and $a+b+c=1$,Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$$
Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have done so far:
Using Cauchy inequalities I can write:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})(b+c+a) \geq (a+b+c)^2$$
$a+b+c=1$,So:$$(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})\geq 1$$
| Homogenization gives $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{3(a^2+b^2+c^2)}{a+b+c}$$
We'll prove a stronger one $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge 6\cdot \frac{a^2+b^2+c^2}{a+b+c}-(a+b+c)$$
$$\Leftrightarrow (a+b+c)\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge 6(a^2+b^2+c^2)-(a+b+c)^2$$
$$\Leftrightarrow (a-b)^2\cdot \frac{a-b+c}{b}+(b-c)^2\cdot \frac{a+b-c}{c}+(c-a)^2\cdot \frac{b+c-a}{a}\ge 0$$
$$\Leftrightarrow S=S_c(a-b)^2+S_a(b-c)^2+S_b(c-a)^2\ge 0$$Since $S_a+S_b=\frac{(a-c)^2+b(a+c)}{ac}\ge 0$
and $S_aS_b+S_bS_c+S_cS_a=\frac{[\sum_{cyc} a^3+3abc-\sum_{cyc} a^2(b+c)]+3abc}{abc}>0$
$$\implies S=\frac{(aS_b+bS_a-cS_a-cS_b)^2+(a-b)^2(S_aS_b+S_bS_c+S_cS_a)}{S_a+S_b}\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Why is $f(n) =\frac{n(n+1)(n+2)}{(n+3)}$ in $O(n^2)$? Let:
$$f(n) = n(n+1)(n+2)/(n+3)$$
Therefore :
$$f∈O(n^2)$$
However, I don't understand how it could be $n^2$, shouldn't it be $n^3$? If I expand the top we get $$n^3 + 3n^2 + 2n$$ and the biggest is $n^3$ not $n^2$.
| $$f(n)=\frac{n(n+1)(n+2)}{n+3}=\frac{(n^2+n)(n+2)}{n+3}=\frac{n^3+2n^2+n^2+2n}{n+3}=\frac{n^3+3n^2+2n}{n+3} \\ =n^2-\frac{6}{n+3}+2$$
Let $f(n)=O(n^2)$.Then, $\exists c>0 \text{ and } n_0 \geq 1 \text{ such that } \forall n \geq n_0: \\ f(n) \leq cn^2 \Rightarrow n^2-\frac{6}{n+3}+2 \leq cn^2 \Rightarrow c \geq 1+\frac{2}{n^2}-\frac{6}{n^2(n+3)}$
We could pick for example $c=1$ and $n_0=1$.
Therefore,we can find such $c,n_0$,therefore:
$$f(n)=O(n^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/883972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Prove $\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$ Question:
Prove or disprove
$$I=\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$$
I know
\begin{align}\frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=&\left(1+\dfrac{1}{2^{2^n}}\right)\left(1+\dfrac{1}{2^{2^n}+2}\right)\\&\cdots\left(1+\dfrac{1}{2^{
2^n+1}}\right)\end{align}
so
$$\lim_{n\to\infty}\left(1+\dfrac{1}{2^{2^n}}\right)\left(1+\dfrac{1}{2^{2^n}+2}\right)\cdots\left(1+\dfrac{1}{2^{
2^n+1}}\right)=\sqrt{2}\ ?$$
I feel this result is very surprising. This problem comes from Chris's sis.
and I use wolfram,limit wofl can't find it
I often use this theta function and is this true? Thank you.
| $$\log \left(1+\tfrac{1}{2^{2^n}}\right)\left(1+\tfrac{1}{2^{2^n}+2}\right)\cdots\left(1+\tfrac{1}{2^{
2^n+1}}\right)
= \log \left(1+\tfrac{1}{2^{2^n}}\right)+
\log \left(1+\tfrac{1}{2^{2^n}+2}\right)+\cdots + \log\left(1+\tfrac{1}{2^{
2^n+1}}\right)$$
Expanding on Alex' idea let $t = 2^{2^n}$ which is getting very large. Notice we get only even numbers:
$$ \log \left(1+\tfrac{1}{t}\right)+
\log \left(1+\tfrac{1}{t+2}\right)+\cdots + \log\left(1+\tfrac{1}{2t}\right) \approx
\frac{1}{t} + \frac{1}{t+2}+\cdots + \frac{1}{2t}
$$
We get a Riemann sum, but only half of the terms:
$$
\frac{1}{t}\big[1 + \frac{1}{1+\tfrac{2}{t}}+\cdots + \frac{1}{2}\big] \approx \frac{1}{2} \int_1^2 \frac{dt}{t} = \frac{\log 2}{2} = \log \sqrt{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Trigonometric identity, simplifying an expression to $(1-\sin^2 a\cos^2a)/(2+\sin^2a\cos^2a)$ Question:
$$\left(\frac{1}{\sec^2A-\cos^2A}+\frac{1}{\csc^2A -\sin^2A}\right)\sin^2A\cos^2A=\frac{1-\sin^2A \cos^2A}{2+\sin^2A\ \cos^2A}$$
Prove L.H.S. = R.H.S.
My Efforts:
$$=\left(\frac{1}{\frac{1}{\cos^2 A}-\cos^2A}+\frac{1}{\frac{1}{\sin^2A} -\sin^2A}\right)\sin^2A\cos^2A$$
$$=\left(\frac{\cos^2A}{1-\cos^4A}+\frac{\sin^2A}{1-\sin^4A }\right)\sin^2A\cos^2A$$
I thought i will reach to R.H.S. by what i was doing, but now i am stuck I don't how to proceed further or is there a way to reach R.H.S. with what I am doing.
| $$\left(\frac{\cos^2A}{1-\cos^4A}+\frac{\sin^2A}{1-\sin^4A }\right)\sin^2A\cos^2A$$
$$=\left(\frac{\cos^2A}{(1+\cos^2A)(1-\cos^2A)}+\frac{\sin^2A}{(1+\sin^2A)(1-\sin^2A) }\right)\sin^2A\cos^2A$$
$$=\left(\frac{\cos^2A}{(1+\cos^2A)\sin^2A}+\frac{\sin^2A}{(1+\sin^2A)\cos^2A}\right)\sin^2A\cos^2A$$
$$=\frac{\cos^4A}{1+\cos^2A}+\frac{\sin^4A}{1+\sin^2A}=\frac{\color{red}{\cos^4A}+\cos^4A\sin^2A\color{red}{+\sin^4A}+\sin^4A\cos^2A}{(1+\cos^2A)(1+\sin^2A)}$$
Here, the numerator will be
$$\color{red}{(\cos^2A+\sin^2A)^2-2\cos^2A\sin^2A}+\cos^2A\sin^2A(\cos^2A+\sin^2A)=1-\cos^2A\sin^2A.$$
On the other hand, the denominator will be
$$1+\sin^2A+\cos^2A+\cos^2A\sin^2A=2+\cos^2A\sin^2A.$$
(Maybe I wrote too much...)
| {
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"url": "https://math.stackexchange.com/questions/886342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Does every integer $n > 2$ have a "reciprocating" pseudoprime $a$? Does every integer $n > 2$ have a "reciprocating" pseudoprime $a$ such that $\gcd(a, n) = 1$ and satisfies both $a^{n - 1} \equiv 1 \mod n$ and $n^{a - 1} \equiv 1 \mod a$? (And of course allowing for the idea of even pseudoprimes).
From Fermat's little theorem it follows all odd primes have such a corresponding number.
| For any composite $n$, take $a = \dfrac{n^n - 1}{n - 1}$. It is easy to prove that $a$ is also composite and $(a, n) = 1$. Note that$$
a = \frac{n^n - 1}{n - 1} = n^{n - 1} + \cdots + n + 1 \equiv 1 \pmod{n} \Longrightarrow a^{n - 1} \equiv 1 \pmod{n}.
$$
Also,$$
n^n = (n - 1)a + 1 \equiv 1 \pmod{a} \Longrightarrow n^{a - 1} = (n^n)^{\frac{n^{n - 1} - 1}{n - 1}} \equiv 1 \pmod{a}.
$$
Thus $a = \dfrac{n^n - 1}{n - 1}$ satisfies given conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/887070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trigonometric simplification for limits: $\lim_{x\to 0} \frac{1-\cos^3 x}{x\sin2x}$ Have to evaluate this limit, but trigonometry part is :(
$$\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}.$$
Had written the denominator as $2x\sin x\cos x$, no idea what to do next. Please help...
| Otherwise using:
$$a^3-b^3=(a-b)(a^2+ab+b^2)$$
$$\eqalign{
\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}
&=\lim_{x\to 0} \dfrac{(1-\cos x)(1+\cos x+\cos^2x)}{2x\sin x\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{(1-\cos x)}{x\sin x\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{\frac{(1-\cos x)}x}{x\frac{\sin x}x\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{(1-\cos x)}{x^2\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{(1-\cos x)}{x^2\cos x}.\frac{1+\cos x}{1+\cos x}\\
&=\frac32\lim_{x\to 0} \dfrac{1}{\cos x(1+\cos x)}.\left(\frac{\sin x}{x}\right)^2\\
&=\frac32\times\frac1{1\times2}\times1^2=\frac34.}$$
Using Taylor series:
$$\cos x =1-x^2/2+O(x^4)$$
$$\sin x =x-x^3/6+O(x^5)$$
$$(1+x)^n=1+nx+O(x^2)$$
So,
$$\eqalign{
\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}
&=\lim_{x\to 0} \dfrac{1-(1-x^2/2+O(x^4))^3}{x(2x+O(x^3))}\\
&=\lim_{x\to 0} \dfrac{1-(1-3x^2/2+O(x^3))}{(2x^2+O(x^3))}\\
&=\lim_{x\to 0} \dfrac{3x^2/2+O(x^3)}{2x^2+O(x^3)}=\frac34.}$$
Using L-Hospital Rule:
$$\eqalign{\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}&
=\lim_{x\to 0} \dfrac{3\cos^2 x.\sin x}{\sin2x+2x\cos2x}\\
&=\lim_{x\to 0} \dfrac{3\sin x}{\sin2x+2x\cos2x}\\
&=\lim_{x\to 0} \dfrac{3\cos x}{2\cos2x+2\cos2x-4x\sin2x}\\
&=\frac{3\times1}{2\times1+2\times1-4\times0}=\frac34.}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why does $\lfloor\frac{n}{x}\rfloor$ have at most $2\sqrt{n}$ values when $x=1,2,\dots n$? The question is very short and clear:
Why does $\lfloor\frac{n}{x}\rfloor$ (floor of $\frac nx$) have at most $2\sqrt{n}$ values when $x = 1, 2,\dots, n $?
I saw this statement at tutorial of 449A at codeforces, I really want to know how we get that, and since the amount of values should be integer, why $2\sqrt{n}$?
And I don't know the specific category of this problem so I just tag it as elementary number theory.
| Over the range $1 \le x \le \sqrt{n}$, $x$ can take on only $\sqrt{n}$ distinct integer values. Thus, $\left\lfloor\dfrac{n}{x}\right\rfloor$ can only take on $\sqrt{n}$ distinct values, one for each distinct value of $x$.
Over the range $\sqrt{n} \le x \le n$, we have that $1 \le \dfrac{n}{x} \le \sqrt{n}$. Thus, $\left\lfloor\dfrac{n}{x}\right\rfloor$ can only take on $\sqrt{n}$ distinct values, one for each integer between $1$ and $\sqrt{n}$.
This adds up to $2\sqrt{n}$ values over the entire range $1 \le x \le n$.
Note that this is an upper bound and not an exact number. For instance, if $n = 6$, then we have at most $2\sqrt{6} \approx 4.89$ values. Indeed, $\lfloor\frac{6}{1}\rfloor = 6$, $\lfloor\frac{6}{2}\rfloor = 3$, $\lfloor\frac{6}{3}\rfloor = 2$, $\lfloor\frac{6}{4}\rfloor = \lfloor\frac{6}{5}\rfloor = \lfloor\frac{6}{6}\rfloor = 1$, so we have $4$ distinct values.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many different positive integer factors does have? How many different positive integer factors does $(2^7)(3^4)(7^3)(23^5)$ have?
Do we have to do any combinations between the powers here?
| Note the following:
*
*$\gcd(2,3)=1$
*$\gcd(2,7)=1$
*$\gcd(2,23)=1$
*$\gcd(3,7)=1$
*$\gcd(3,23)=1$
*$\gcd(7,23)=1$
Therefore, the number of different positive integer divisors of $(2^7)(3^4)(7^3)(23^5)$ is:
$$(7+1)\cdot(4+1)\cdot(3+1)\cdot(5+1)=960$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/891005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating a limit. What makes the equality right? I'm reading a proof of a limit calculation. The limit is:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x}$$
where $a,b>0$.
The aother claims that:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x} =
\exp\left( \lim\limits_{x\to 0}\frac{\frac{a^x+b^x}{2} - 1}{x} \right)$$
How come?
Update:
Of course,
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x} =
\exp\left(\lim\limits_{x\to 0} \frac{\ln\left( \frac{a^x+b^x}{2} \right)}{x} \right)$$
But how to proceed to reach the auther's expression?
| If we try with
$$
\lim_{x\to 0} \frac{\log(a^x+b^x)-\log 2}{x}
$$
and apply l'Hôpital's theorem, we get
$$
\lim_{x\to 0}\frac{a^x\log a+b^x\log b}{a^x+b^x}=\frac{\log a+\log b}{2}=
\log\sqrt{ab}.
$$
It's just the derivative of $x\mapsto (a^x+b^x)/2$ at $0$, of course.
However,
$$
\lim_{x\to0}\frac{\log(1+x)}{x}=1
$$
so that
$$
\lim_{x\to 0} \frac{\log\dfrac{a^x+b^x}{2}}{x}=
\lim_{x\to 0} \frac{\log\dfrac{a^x+b^x}{2}}{\dfrac{a^x+b^x}{2}-1}
\frac{\dfrac{a^x+b^x}{2}-1}{x}
$$
and the limit of the first factor is $1$. I don't think it's a real simplification.
It may be worth noting that the function
$$
\mu_{a,b}(x)=\begin{cases}
\left(\dfrac{a^x+b^x}{2}\right)^{1/x} & \text{if $x\ne0$}\\[2ex]
\sqrt{ab} & \text{if $x=0$}
\end{cases}
$$
for $a,b>0$ is quite interesting, because it's increasing, $\mu_{a,b}(-1)$ is the harmonic mean, $\mu_{a,b}(0)$ is the geometric mean, $\mu_{a,b}(1)$ is the arithmetic mean and
$$
\lim_{x\to-\infty}\mu_{a,b}(x)=\min(a,b),\qquad
\lim_{x\to\infty}\mu_{a,b}(x)=\max(a,b).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Complex analysis integral (maybe using residue theory?) Trying to compute the following integral:
$$
\int_{-\pi}^{\pi} \frac{d\theta}{1+\sin^2(\theta)}.
$$
This is for a complex analysis course, so I'm trying to find a way to use residue theory of something of that nature to solve the problem. I can't think of a substitution that will easily allow me to use residue theory.
| You can solve the integral also without the help of complex analysis. For instance
\begin{align*}
\int\dfrac{1}{1+\sin^2(x)}\, dx &= \int\dfrac{1}{1+\sin^2(x)}\dfrac{\frac{1}{\sin^2(x)}}{\frac{1}{\sin^2(x)}} \ dx \\
&=\int\dfrac{\frac{1}{\sin^2(x)}}{\frac{1}{\sin^2(x)}+1}\, dx \\
&=\int\dfrac{\frac{1}{\sin^2(x)}}{2+\cot^2(x)}\, dx,\qquad \text{since }\dfrac{1}{\sin^2(x)}=1+\dfrac{\cos^2(x)}{\sin^2(x)} \\
&=\int\dfrac{\frac{1}{\sin^2(x)}}{2+u^2}\cdot \left(-\sin^2(x)\right)\, du,\qquad u=\cot(u), du = -\dfrac{1}{\sin^2(x)}dx\\
&=-\int\dfrac{1}{2+u^2}\, du \\
&=-\frac12\int\dfrac{1}{(\frac{u^2}{2}+1)}\, du \\
&=-\frac{1}{\sqrt{2}}\int\dfrac{1}{s^2+1}\,ds, \qquad s=\frac{u}{\sqrt{2}},ds = \frac{du}{\sqrt{2}}\\
&=-\frac{1}{\sqrt{2}}\arctan\left(\dfrac{\cot(x)}{\sqrt{2}}\right)=:f(x)
\end{align*}
And evaluating $f(x)$ from $-\pi$ to $\pi$ gives you $\sqrt{2}\pi$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find a real numbers $a,b$ such $a^n+b^n$ is rational Question:
prove or disprove :there exsit real numbers $a,b$ such follow two condition:
(1):$a+b$ is irrational
(2): for any postive integer $n\ge 2$, then $a^n+b^n$ is rational.
I have know if
$n=2k$ case is true,because I let $a=\sqrt{2}+1,b=\sqrt{2}-1$,so
$$a^{2k}+b^{2k}=(\sqrt{2}+1)^{2k}+(\sqrt{2}-1)^{2k}\in Q$$
But for $n=2k+1$,I can't find a example.(if you can't find,can you prove when$n=2k+1$,there can't exsit?) Thank you for help
| Theorem. There are no real numbers $a,b$ such that
$$\hbox{$a+b$ is irrational}$$
and
$$\hbox{$a^2+b^2$, $a^3+b^3$, $a^4+b^4$ and $a^6+b^6$ are rational.}$$
Proof. Suppose that there are such $a,b$; clearly they are non-zero. Then
$$2a^2b^2=(a^2+b^2)^2-(a^4+b^4)$$
and
$$2a^3b^3=(a^3+b^3)^2-(a^6+b^6)$$
are rational, and so is
$$ab=\frac{2a^3b^3}{2a^2b^2}\ .$$
Therefore
$$(a+b)^2=(a^2+b^2)+2ab$$
is rational, and since
$$(a+b)^3=(a^3+b^3)+3ab(a+b)$$
we have
$$a+b=\frac{a^3+b^3}{(a+b)^2-3ab}$$
which is rational (note that with real $a,b$, the denominator of this last fraction cannot be zero).
| {
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"timestamp": "2023-03-29T00:00:00",
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area formed by a box and line suppose we have a box defined by coordinates $(1,1)$, $(-1,1)$, $(-1,-1)$, $(1,-1)$.
Suppose, a line $y=m(x+b)$ crosses the box with $m>0$ and $b>0$. What is the area of left upper triangle. Assume that the line crosses the box.
Thank you very much.
Partial Answer:
the line intersects a box at $(x,y)=(\frac{1}{m}-b,1)$ and $(x,y)=(-1,m(b-1))$
Area is formed by $(-1-(\frac{1}{m}-b)) \cdot (1-m(b-1))$ ,correct?
| The box that is defined by coordinates $(1,1), (-1,1), (-1,-1), (1,-1)$ is the following:
To draw also the line $y=m(x+b) $ :
For $x=-1: y=m(-1+b)$
For $y=1: x=\frac{1}{m}-b$
Let's suppose that the line intersects the box as followed:
So we want to find the area of the yellow region.
$A(-1,1), B(\frac{1}{m}-b,1), C(-1, m(-1+b))$
$(AB)=\sqrt{(1-1)^2+(\frac{1}{m}-b+1)^2}=\frac{1}{m}-b+1$
$(AC)=\sqrt{(m(-1+b)-1)^2+(-1+1)^2}=m(-1+b)-1$
$$\text{ Area }=\frac{1}{2} (AB)(AC)=\frac{1}{2} \left (\frac{1}{m}-b+1 \right ) \left (m(-1+b)-1 \right )$$
| {
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How prove that $ x+y+z>4$ for $ a+b+c=4$ and $ ax+by+cz=xyz$? Given positive reals $ a,b,c,x,y,z$ such that $ a+b+c=4$ and $ ax+by+cz=xyz$.
How prove that $ x+y+z>4$?
| Suppose there exist $a, b, c, x, y, z > 0$ such that
$$a+b+c=4,$$
$$ax+by+cz = xyz,$$
$$x+y+z \leq 4.$$
Observe that we have
$$axyz + bxyz + cxyz = 4xyz,$$
so that
$$axyz + bxyz + cxyz = 4ax + 4by + 4cz,$$
and hence
$$ax(yz-4) + by(xz-4) + cz(xy-4) = 0.$$
Since
$x+y+z \leq 4$
and
$a, b, c, x, y, z > 0,$
what follow are the inequalities
$$xy, yz, xz < 4,$$
whence
$$ax(yz-4) + by(xz - 4) + cz(xy-4) < 0;$$
a contradiction, qed.
| {
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Do the medians (or other cevians) form all the triangles? I want to know whether set of medians of all triangles, or some other class of cevians, can form the set of all the triangles?
For example, in the case of altitudes, $(4,7,10)$ is an counterexample. While $(4,7,10)$ can form a triangle, there is no triangle with altitudes $(4,7,10)$.
| From the Wikipedia article on Median, the lengths of the medians in terms of the sidelengths are:
$m_a = \dfrac{1}{2}\sqrt{2b^2+2c^2-a^2}$
$m_b = \dfrac{1}{2}\sqrt{2c^2+2a^2-b^2}$
$m_c = \dfrac{1}{2}\sqrt{2a^2+2b^2-c^2}$
Solving for $(m_a,m_b,m_c)$ in terms of $(a,b,c)$ yields:
$a = \dfrac{2}{3}\sqrt{2m_b^2+2m_c^2-m_a^2}$
$b = \dfrac{2}{3}\sqrt{2m_c^2+2m_a^2-m_b^2}$
$c = \dfrac{2}{3}\sqrt{2m_a^2+2m_b^2-m_c^2}$
So for any desired median lengths $(m_a,m_b,m_c)$, we can find sidelengths $(a,b,c)$ such that the triangle with side lengths $(a,b,c)$ has medians of length $(m_a,m_b,m_c)$.
The area of a triangle is given by:
$T = \sqrt{s(s-a)(s-b)(s-c)} = \dfrac{4}{3} \sqrt{\sigma (\sigma - m_a)(\sigma - m_b)(\sigma - m_c)}$,
where $s = \dfrac{a+b+c}{2}$ and $\sigma = \dfrac{m_a+m_b+m_c}{2}$.
From this, we can see that $(a,b,c)$ satisfies the triangle inequality iff $(m_a,m_b,m_c)$ satisfies the triangle inequality. (If either triple did not satisfy the inequality, the corresponding quantity under the radical would be negative).
| {
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Is this real number an integer? Is this real number :
$$\Big(2+\frac{10}{9}\sqrt{3}\Big)^{1/3}+\Big(2-\frac{10}{9}\sqrt{3}\Big)^{1/3}$$
an integer ?
I've tried different factorization, but nothing seems to work.
| Let $a = \sqrt[3]{2+\frac{10}{9}\sqrt{3}}$ and $b = \sqrt[3]{2-\frac{10}{9}\sqrt{3}}$. Then:
$a^3+b^3 = (2+\frac{10}{9}\sqrt{3}) + (2-\frac{10}{9}\sqrt{3}) = 4$
$ab = \sqrt[3]{(2+\frac{10}{9}\sqrt{3})(2-\frac{10}{9}\sqrt{3})} = \sqrt[3]{4-\frac{10^2}{9^2}\cdot 3} = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}$.
Now, let $x = a+b$. Then, $x^3 = (a+b)^3 = (a^3+b^3)+3ab(a+b) = 4+2x$.
Therefore, $x$ is a real root of $x^3-2x-4 = 0$. Since $x^3-2x-4 = (x-2)(x^2+2x+2)$, it follows that the only real root of $x^3-2x-4 = 0$ is $x = 2$.
| {
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Finding the sum to infinity Question:
Find the sum to infinity for the following series
$$1, -\frac{1}{2}, \frac{1}{2^2}, -\frac{1}{2^3},\cdots$$
What would be the technique used to find such a sum?
| $(1+\frac{1}{2^2}+\frac{1}{2^4}+...)-(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...)$
Both the brackets above contain GP of $n$ terms with $r=\frac{1}{2^2}=\frac{1}{4}$
$\Large(\frac{(\frac{1}{4})^n-1}{-3/4})-(\frac{1}{2}\frac{(\frac{1}{4})^n-1}{-3/4})$
$\Large\frac{\frac{1}{2}(\frac{1}{4})^n-\frac{1}{2}}{\frac{-3}{4}}=\frac{(\frac{1}{4})^n-1}{\frac{-3}{2}}=\frac{0-1}{\frac{-3}{2}}=\frac{2}{3}$
| {
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Counting the roots of a polynomial over a finite field Let $\mathbb{F}_{11}$ be the field of 11 elements and let $\mathcal{K}$ be the splitting field of $x^{3} - 1$ over $\mathbb{F}_{11}$. How many roots does $(x^{2} - 3)(x^{3} - 3)$ have in $\mathcal{K}$?
First, note that $x^{3} - 1 = (x- 1)(x^{2} + x + 1)$ and the quadratic factor is irreducible over
$\mathbb{F}_{11}$ so that $\mathcal{K} = \mathbb{F}_{11}/(x^{2} + x + 1) \cong \mathbb{F}_{11^{2}}$.
Note that $3^{6} = 3$ in $\mathbb{F}_{11^{2}}$ so that $(x^{2} - 3)(x^{3} - 3) = (x - 27)(x+ 27)(x-9)(x^{2} + 9x + 81)$, so that the number of roots of the given polynomial in $\mathbb{F}_{11^{2}}$ is three ( I am not even sure if the quadratic factor in this decomposition does not actually have a root).
Is there a general method for determining the roots of say $(x^{n} - a)$ over a finite field of $p^{m}$ elements?
I was thinking: Find the smallest positive integer k for which $a^{k} = 1$, then $a = x^{n} \implies 1 = a^{k} = x^{kn}$, which is an equation in the cyclic group of cardinality $p^{m} -1$, hence the number of such solutions has to be the $\gcd(kn, p^{m} -1)$. This does not seem correct.
| First, note that $x^{3} - 1 = (x- 1)(x^{2} + x + 1)$ and the quadratic factor is irreducible over
$\mathbb{F}_{11}$ so that $\mathcal{K} = \mathbb{F}_{11}/(x^{2} + x + 1) \cong \mathbb{F}_{11^{2}}$.
Note that $3^{6} = 3$ in $\mathbb{F}_{11^{2}}$ so that $(x^{2} - 3)(x^{3} - 3) = (x - 27)(x+ 27)(x-9)(x^{2} + 9x + 81)$.
As suggested in the comments, $p(x) = x^{2} + 9x + 81$ splits into linear factors because $p(9x) = 81(x^{2} + x + 1) = 81(x - \zeta_{3})(x - \overline{\zeta_{3}})$.
So that $p(x) = 9(x - 9\zeta_{3})(x - 9\overline{\zeta_{3}})$ in $\mathcal{K} = \mathbb{F}_{11}(\zeta_{3})$ Therefore, $(x^{2} - 3)(x^{3} - 3)$ has five roots in $\mathcal{K}$.
| {
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What is the first step to solving $\cos3x - \sin x = \sqrt{3}(\cos x - \sin 3x)$? My calculus BC teacher has given us some trig "review".
$$\cos3x - \sin x = \sqrt{3}(\cos x - \sin 3x).$$
How do I get rewrite the cos3x and sin3x? Do I just use sum and difference, because it makes everything really complicated.
can someone teach me how to format this correctly... sorry I am a new user
| $\cos3x - \sin x = \sqrt{3}(\cos x - \sin 3x)$
$\implies \cos3x+\sqrt3\sin3x=\sqrt3\cos x+\sin x$
Now divide both sides by 2 to get :
$\frac{1}{2}\cos3x+\frac{\sqrt3}{2}\sin3x=\frac{\sqrt3}{2}\cos x+\frac{1}{2}\sin x$
$\implies \cos \frac{\pi}{3} \cos 3x + \sin \frac{\pi}{3} \sin 3x=\cos \frac{\pi}{6} \cos x + \sin \frac{\pi}{6} \sin x$
$\implies \cos(3x-\frac{\pi}{3})=\cos(x-\frac{\pi}{6})$
$\implies 3x-\frac{\pi}{3}=\pi k \pm (x-\frac{\pi}{6}), k\in\mathbb{Z}$
$\implies x=\frac{\pi}{2} \left( k+\frac{1}{6}\right) OR $ $ $ $ x=\frac{\pi}{4} \left( k+\frac{1}{2}\right)$
| {
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Calculate $\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS$ where $a^2+b^2+c^2<1$.
Let $a^2+b^2+c^2 < 1$ and $S^2$ be unit sphere in $R^3$. Calculate $$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS$$
Let $(x,y,z)=(\cos\theta \cos\phi,\cos\theta \sin\phi, \sin\theta)$. By definition,
$$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS\\
=\int_{0}^{2\pi}\int_{0}^{\pi}\frac{1}{\sqrt{(\cos\theta \cos\phi-a)^2+(\cos\theta \sin\phi-b)^2+(\sin\theta-c)^2}}\sin\theta d\theta d\phi$$
It is too complicate.
This is calculus exam problem that I took yesterday. Is there any good idea?
| Let $(x,y,z)=(a+\cos\theta \cos\phi,b+\cos\theta \sin\phi,c+ \sin\theta)$
$$\int_{S^2}\frac{1}{\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}dS\\
=\int_0^{2\pi}\int_0^{\pi}\frac{1}{\sqrt{(\cos\theta\cos\phi)^2+(\cos\theta \sin\phi)^2+(\sin\theta)^2}}\sin\theta d\theta d\phi=\int_0^{2\pi}\int_0^{\pi}\sin\theta d\theta d\phi\\=4\pi$$
| {
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Prove $\sum\limits_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le\frac{3}{2}$ if $x+y+z=1$
if $x,y,z$ are positive real numbers and $x+y+z=1$ Prove:$$\sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le\frac{3}{2}$$
where $\sum_{cyc}$ denotes sums over cyclic permutations of the symbols $x,y,z$.
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done so far: Using AM-GM $$xy+z \ge 2$$ $$\sqrt{xy+z} \ge \sqrt2$$
So manipulating this leads to $$\sum_{cyc}\frac{\sqrt{xy}}{\sqrt{xy+z}} \le \sum_{cyc}\frac{\sqrt{xy}}{\sqrt2}$$
I stuck here.I'm thinking about applying Cauchy-Schwartz.Also I have not used the assumption $x+y+z=1$.Any hint is appreciated.
| $$\sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le\frac{3}{2} $$
$$ \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}} = \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+1-x-y}} = \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{(1-x)(1-z)}} = \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{(y+z)(x+z)}} $$
And now, by AM-GM
$$ \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{(y+z)(x+z)}} \le \sum_{cyc} \frac{x}{2(x+z)}+\frac{y}{2(y+z)} = \frac{x}{2(x+z)}+\frac{y}{2(y+z)} + \frac{y}{2(y+x)}+\frac{z}{2(z+x)}+ \frac{z}{2(z+x)}+\frac{x}{2(x+y)} = \frac{3}{2} \Box $$
| {
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If $\frac{5x}{2x^2+5x+1}=\frac13$ then the value of $x+\frac{1}{2x}$ is If $\frac{5x}{2x^2+5x+1}=\frac13$ then the value of $x+\frac{1}{2x}$ is
note $5x=1$
$2x^2+5x+1=3$
or
$15x=2x^2+5x+1$
so $2x^2-10x+1$
so $\frac{5+\sqrt{23}}2$ or $\frac{5-\sqrt{23}}2$
| HINT:
As $x\ne0,$
$$3=\frac{5x}{2x^2+5x+1}=\frac{5/2}{x+5/2+1/2x}$$
$$\iff\frac{x+5/2+1/2x}{5/2}=\frac13$$
Hope you can take it from here
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Factorization of $z^4 +1 = (z^2 - \sqrt 2z+1)(z^2 + \sqrt 2 z+1)$ for complex $z$ How can I get this equation from LHS to RHS by using the four roots of $z^4 +1 = 0$ are $z=\pm\sqrt{\pm i}$
$$z^4 +1 = (z^2 - \sqrt2 z+1)(z^2 + \sqrt2 z+1)$$
| You don't need to know the roots of $P(z)=z^4+1$ for the factorization: $$z^4+1=z^4+2z^2+1-2z^2=(z^2+1)^2-(\sqrt{2}z)^2=(z^2-\sqrt{2}z+1)(z^2+\sqrt{2}z+1)$$
If you want to use that the roots of $P$ are $\omega=e^{i\frac{\pi}{4}}$, $i\omega$, $-\omega$ and $-i\omega$ (I tend to avoid to use the notation $\sqrt{z}$ for numbers that are not nonnegative real numbers), note that you have two pairs of conjugated complex numbers (namely $\omega$ and $-i\omega$, and $i\omega$ and $-\omega$) that will combine into a polynomial of degree $2$ with real coefficients. By combine, I mean $(z-a)(z-\overline{a})$ is a polynomial with real coefficients.
| {
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Proving determinants using properties of determinants $$\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & b^2+ca & b^3\\
1 & c^2+ab & c^3
\end{vmatrix}
= (a-b)(b-c)(c-a)(a^2+b^2+c^2)$$
we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calculation to apply so was hoping for an hint.
Edit: just tried the problem and here is how I have done it
$$\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & -(a^2+b^2)+c(a-b) & -(a^3-b^3)\\
1 & c^2-a^2-b(c-a) & c^3-a^3
\end{vmatrix}$$
then
$$\begin{vmatrix}
1 & a^2+bc & a^3\\
0 & -((a-b)(a+b))+c(a-b) & -((a-b)(a^2+ab+b^2))\\
0 & (c-a)(c+a)-b(c-a) & (c-a)(c^2+ca+a^2)
\end{vmatrix}$$
then
$$(a-b)(c-a)\begin{vmatrix}
1 & a^2+bc & a^3\\
0 & a+b+c & -(a^2+ab+b^2)\\
0 & a-b+c & (c^2+ca+a^2)
\end{vmatrix}$$
then
$$(a-b)(c-a)\begin{vmatrix}
a+b+c & -(a^2+ab+b^2)\\
a-b+c & (c^2+ca+a^2)
\end{vmatrix}$$
then
$$(a-b)(c-a) * ( (a+b+c)(c^2+ca+a^2)-(a-b+c)(-(a^2+ab+b^2)) )$$
Here I am Confused on how to multiply them and get the answer
note: typed because I own a very bad handwriting
Thank you every one for your help
| The first trick is to get as much zeroes as you can in the first row. That makes multiplication easier.
$\begin{vmatrix} 1 & a^2+bc & a^3 \\ 1 & b^2+ca & b^3\\ 1 & c^2+ab &c^3 \end{vmatrix}$
subtracting second row from first row and third row from second row:
$(a-b)(b-c)\begin{vmatrix} 0 & a+b-c & a^2+ab+b^2 \\ 0 & b+c-a & b^2+bc+c^2\\ 1 & c^2+ab &c^3 \end{vmatrix}$
Subtracting second row from first row:
$-(a-b)(b-c)(c-a)\begin{vmatrix} 0 & 2 & a+b+c \\ 0 & b+c-a & b^2+bc+c^2\\ 1 & c^2+ab &c^3 \end{vmatrix}$
exchanging row and column:
$(a-b)(b-c)(c-a)\begin{vmatrix} 0 & 0 & 1 \\ 2 & b+c-a & c^2+ab\\ a+b+c & b^2+bc+c^2 &c^3 \end{vmatrix}$
Now take determinant and get the result.
| {
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Finding the value of $\frac{\cos^4\beta}{\cos^2\alpha} + \frac{\sin^4\beta}{\sin^2\alpha}$. Trigonometry
$\dfrac{\cos^4 \alpha}{\cos^2 \beta}+ \dfrac{\sin^4\alpha}{\sin^2\beta} = 1$
then the value of
$\dfrac{\cos^4\beta}{\cos^2\alpha}+ \dfrac{\sin^4\beta}{\sin^2\alpha}$ is?
NOTE: can somebody help me
$\cos^2\alpha \left(\frac{\cos^2 \alpha}{\cos^2 \beta}\right)+ \sin^2\alpha \left(\frac{\sin^2 \alpha}{\sin^2\beta}\right)$
| $\cos^2 \alpha \left(\frac{\cos^2\alpha}{\cos^2 \beta}\right) + \sin^2 \alpha \left(\frac{\sin^2 \alpha}{\sin^2 \beta}\right)=1$.
It can be noticed that $\frac{\cos^2\alpha}{\cos^2 \beta}$ and $\frac{\sin^2 \alpha}{\sin^2 \beta}$ must be $1$ (Why?).
This tells us that $\cos^2 \alpha = \cos^2 \beta$ and $\sin^2 \alpha = \sin^2\beta$.
Plug those two results in the required expression. It will then become a well-known expression with a well-known value.
| {
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How to describe 4-8+12-16+20-24+28 using summation ($\Sigma$) notation? Can anyone find the ∑ summation for this please?
4-8+12-16+20-24+28
It seems to be going up by steps of 4, but I can't seem to get how I should write it down, since it uses both + and -
| $$4-8+12-16+20-24+28=4(1-2+3-4+5-6+7)=4((-1)^{1+1}\cdot 1+(-1)^{2+1}\cdot 2+(-1)^{3+1}\cdot 3+(-1)^{4+1} \cdot 4+(-1)^{5+1} \cdot 5+(-1)^{6+1} \cdot 6+(-1)^{7+1} \cdot 7)=4 \sum_{i=1}^{7}(-1)^{i+1}i=4 \left ( \sum_{i=0}^{3} (2i+1)-\sum_{i=1}^{3} 2i\right )=4 \left( 2 \sum_{i=0}^{3} i +\sum_{i=0}^{3} 1-2\sum_{i=1}^{3} i\right )=4 \left( 2 \sum_{i=1}^{3} i +\sum_{i=0}^{3} 1-2\sum_{i=1}^{3} i\right ) \\ =4 \sum_{i=0}^{3} 1=4 \cdot (3+1)=4 \cdot 4=16$$
| {
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Sum of N terms of a series Perhaps this is a silly question but how can I simplify this to find the sum of $N$ terms?
$$\sum_{n=1}^N \frac{2n - 1}{n(n+1)(n+2)}$$
I know how to find the sum of cubes and squares etc but how with the denominator like that?
Any help would be appreciated thanks.
The questions in the book have all involved finding things like:
$$\sum_{n=1}^N {(3n^3+n^2+n)}$$
This is the only one with the cubes and squares in the denominator
| Checking if there is a pair of constants $(A,B)$ such that
$$\frac{2n-1}{n(n+1)(n+2)}=A\cdot\color{red}{\frac{1}{(n+1)(n+2)}}+B\cdot\color{blue}{\frac{1}{n(n+1)}}$$
gives us $(A,B)=(5/2,-1/2)$.
Hence, since we can have
$$\frac{2n-1}{n(n+1)(n+2)}=\frac{5}{2}\color{red}{\left(\frac{1}{n+1}-\frac{1}{n+2}\right)}-\frac 12\color{blue}{\left(\frac{1}{n}-\frac{1}{n+1}\right)},$$
we have
$$\sum_{n=1}^{N}\frac{2n-1}{n(n+1)(n+2)}=\frac{5}{2}\sum_{n=1}^{N}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)-\frac 12\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+1}\right)$$
$$=\frac{5}{2}\left\{\left(\frac 12-\frac 13\right)+\left(\frac 13-\frac 14\right)+\cdots+\left(\frac 1N-\frac{1}{N+1}\right)+\left(\frac{1}{N+1}-\frac{1}{N+2}\right)\right\}$$$$-\frac 12\left\{\left(\frac 11-\frac 12\right)+\left(\frac 12-\frac 13\right)+\cdots+\left(\frac{1}{N-1}-\frac{1}{N}\right)+\left(\frac 1N-\frac{1}{N+1}\right)\right\}$$
$$=\frac 52\left(\frac 12-\frac{1}{N+2}\right)-\frac 12\left(\frac 11-\frac{1}{N+1}\right)=\frac{N(3N+1)}{4(N+1)(N+2)}.$$
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Integrate $\int \frac{6x-15}{x^2 +8x+101}dx$ I've been trying this one for days and I can't seem to get it. Any ideas?
$\int \frac{6x-15}{x^2 +8x+101}dx$
Thanks,
Nick
| $\displaystyle\int\dfrac{6x-15}{x^2+8x+101} dx=\int\dfrac{6x-15}{(x+4)^2+85} dx$.
Now let $u=x+4$, so $x=u-4$ and $dx=du$ to get
$\displaystyle\int\frac{6u-39}{u^2+85} du=3\int\frac{2u}{u^2+85} du-39\int\frac{1}{u^2+85} du$
$=3\ln(u^2+85)-39\left(\frac{1}{\sqrt{85}}\arctan\frac{u}{\sqrt{85}}\right)+C$.
(Now substitute back for u.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor polynomial of $\frac{1}{2-x}$ Can someone show how to find the Taylor polynomial of $\frac{1}{2-x}$?
I tried this: $\frac{1}{2-x}=\frac{1}{1-(x-1)}$ and then use that $ \ T_n(\frac{1}{1-x})=1+x+\dots +x^n.$ But this gives $1+(x-1)+\dots +(x-1)^n$ which is the wrong answer. Why?
| We want to find $a_0, a_1, a_2,\ldots$ so that $$\frac1{2-x} = a_0 + a_1x + a_2x^2 + \ldots.$$
Multiply both sides by $2-x$: $$
\begin{align}
1 & = (2-x)a_0 + (2-x)a_1x + (2-x)a_2x^2 + \cdots\\
& = (2a_0-a_0x)+(2a_1x-a_1x^2) + (2a_2x^2 - a_2x^3) + \cdots \\
\end{align}$$
Combine like terms:
$$ 1 = 2a_0 + (2a_1-a_0)x + (2a_2 -a_1)x^2 + \cdots$$
Equate coefficients on both sides:
$$\begin{align}
2a_0 & = 1 \\
2a_1 - a_0 & = 0 \\
2a_2 - a_1 & = 0 \\
& \vdots
\end{align}$$
Now just solve the equations starting with the top one:
$$\begin{align}
a_0 &= \frac12 \\
a_1 &= \frac14\\
a_2 &= \frac18 \\
& \vdots
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of the following series I'm stuck with this series:$$\frac12 + \frac{1\cdot 4}{2 \cdot 5} + \frac{1\cdot 4 \cdot 7}{2 \cdot 5 \cdot8} + \frac{1 \cdot 4 \cdot 7 \cdot 10}{2 \cdot 5 \cdot 8 \cdot 11}+\ldots$$
I cant even find the $n$-th th term here...
I have to prove that $$\sum_{k=1}^n A_k = \frac12 \left[ \frac{4\cdot 7 \cdot 10 \cdot \ldots \cdot(3n+1)}{2 \cdot 5 \cdot 8 \cdot\ldots\cdot(3n -1)}\ -1 \right] $$
Here are the instructions given in the question: write the $A_n$ term of this series and express $A_{n+1}$ using $A_n$
find $C$ and $D$ such that $f(n) = (Cn + D)A_{n+1}$ and $f(n) - f(n -1) = A_n$
and after that i have to come up with that above answer... P.S :- Im stuck in the step of writing the $A_n$ term. I don't know how to write a $n^{th}$ term for a series like this. I think I could manage if someone could help show me how to write the $n^{th}$ term.
| Notice that the $n$-th term of the sum is given by:
$$A_n=\prod_{j=1}^{n}\frac{3j-2}{3j-1}.$$
Let now $B_n$ be:
$$ B_n = \prod_{j=1}^{n}\frac{3j+1}{3j-1}=(3n+1)\prod_{j=1}^{n}\frac{3j-2}{3j-1}=(3n+1)A_n=(3n-1)A_n+2A_n.$$
Since:
$$ (3n-1)A_n = (3n-2)A_{n-1} = B_{n-1}$$
it happens that:
$$ B_n = B_{n-1} + 2 A_n = B_{n-2} + 2 A_{n-1} + 2 A_n = \ldots =B_1 + 2\sum_{k=1}^{n}A_k.$$
Since $B_1 = 2$, the last identity gives:
$$\sum_{k=1}^{n}A_k = \frac{B_n-2}{2}.$$
| {
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Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we have $x^3-y^3\ge 0$ or $(x-y)(x^2+xy+y^2)\ge 0$. Since $x\ge y$ we can divide by $x-y$ to get $x^2+xy+y^2\ge 0$.
Is it right?
Thanks for your help!
| One (simple) way:
let t = x2 + y2 + xy
then t - xy ≥ 0 since it is the sum of two real squares x2 + y2
and t + xy ≥ 0 since it is the square of the real (x + y) [since (x + y)2 = x2 + y2 + 2xy]
adding these, we get 2t ≥ 0, therefore t ≥ 0
| {
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The expression $1 + x^2 +(-T_px+y)^2 +z^2$ is bounded below by a constant multiple of $(1+x^2+y^2+z^2)$ Suppose $T_p > 0$. Is there a simply way to show that
$1 + x^2 +(-T_px+y)^2 +z^2 \geq C (1+x^2+y^2+z^2)$, for all $(x,y,z) \in \mathbb R^3$, where $C>0$.
| We can show that for a certain positive constant $C$
$$ x^2+(-T_p x+y)^2+z^2+w^2 \geq C(x^2+y^2+z^2+w^2) \tag{1} $$
holds for any $(x,y,z,w)\in\mathbb{R}^4$. We just need that the eigenvalues of the symmetric matrix
$$ M=\left(\begin{array}{cccc}T_p^2+1-C & -T_p & 0 & 0 \\ -T_p & 1-C & 0 & 0 \\ 0 & 0 & 1-C & 0 \\ 0 & 0 & 0 & 1-C\end{array} \right) $$
are non-negative. This is equivalent to asking that:
$$ C\in(0,1],\qquad (T_p^2+1-C)(1-C)-T_p^2 \geq 0 \tag{2} $$
hence
$$ C = \frac{2+T_p^2-T_p\sqrt{4+T_p^2}}{2}\in (0,1]\tag{3}$$
is an effective choice.
| {
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How to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$. I am trying to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$ and I have worked this problem a couple times and keep getting the same answer. So I will show my process and please point my errors out.
$$a=\sqrt{7}\hspace{10pt} x=\sqrt{7}\sin\theta\hspace{10pt}dx=\sqrt{7}\cos\theta d\theta \hspace{7pt}7-x^2=7\cos^2\theta$$
$$\int\frac{7\sin^2\theta \sqrt{7}\cos\theta d\theta}{\sqrt{7\cos^2\theta}}$$
$$7\int\sin^2\theta d\theta\rightarrow \frac{7}{2}\int1-\cos(2\theta)d\theta$$
$$\frac{7}{2}\left(\theta-\frac{\sin(2\theta)}{2} \right) \rightarrow \frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{2\sin\theta\cos\theta}{2} \right)$$
$$\frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{x}{\sqrt{7-x^2}}\frac{\sqrt{7-x^2}}{\sqrt{7}} \right)$$
$$\frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{x}{\sqrt{7}} \right)$$
But apparently the answer is
$$\frac{1}{2}\left(7\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-x\sqrt{7-x^2} \right)$$
So how do I go about getting that answer? Thanks for all the help in advance.
| $$\frac{x^2}{\sqrt{7-x^2}}=\frac{7-(7-x^2)}{\sqrt{7-x^2}}=7\cdot\frac1{\sqrt{(\sqrt7)^2-x^2}}-\sqrt{(\sqrt7)^2-x^2}$$
Now, $$\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa$$
and $$\int\sqrt{a^2-x^2}dx=x\frac{\sqrt{a^2-x^2}}2+\frac{a^2}2\arcsin\frac xa$$
Reference : http://www.sosmath.com/tables/integral/integ13/integ13.html
| {
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Does the sequence converge, and to what? We have a sequence $\{a_n\}$
$$a_0 = 0$$
$$a_{n+1} = \frac{a_{n}}{2} + 1$$
Does it converge? And to what?
| $$a_{0}=0\\a_{n+1}=\frac{a_{n}}{2}+1\\a_{1}=\frac{a_{0}}{2}+1=1\\a_{2}=\frac{a_{2}}{2}+1=1+\frac{1}{2}\\a_{3}=\frac{\frac{3}{2}}{2}+1=1+\frac{3}{4}\\a_{4}=\frac{\frac{7}{4}}{2}+1=1+\frac{7}{8}\\a_{5}=\frac{\frac{15}{8}}{2}+1=1+\frac{15}{16}\\a_{n}=\frac{2^{n-1}-1}{2^{n}}+1\\...\\a_{n}<2\\so\\it-converge
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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On a recursive sequence (exercise 8.9 Apostol) The exercise states: show convergence of the sequence ${a_n}$ knowing that:
$$|a_n| \le 2, \ \ \ |a_{n+2}-a_{n+1}| \le \frac{1}{8}|a_{n+1}^2 - a_{n}^2|.$$
The solution states:
$$|a_{n+2}-a_{n+1}| \le \frac{1}{8}|a_{n+1}^2 - a_{n}^2| = \frac{1}{8}|a_{n+1} - a_{n}||a_{n+1} + a_{n}| \le \frac{1}{2}|a_{n+1} - a_{n}| $$
since $|a_n| \le 2$ we know that $|a_{n+1} - a_{n}| \le (\frac{1}{2})^{n-1}$ ...
This last step is not very clear to me, could somebody explain it to me.
Thanks in advance
| Since $-2\leq a_n\leq 2$
$$|a_{n+2}-a_{n+1}| \le \frac{1}{8}|a_{n+1}^2 - a_{n}^2| = \frac{1}{8}|a_{n+1} - a_{n}||a_{n+1} + a_{n}| \leq \frac{1}{8}|a_{n+1} - a_{n}||4|= \frac{1}{2}|a_{n+1} - a_{n}| $$
Next, call $b_n=a_{n+1}-a_n$ and rewrite the above.
| {
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For the following system to be consistent, what must k not be equal to? $6x - 4y + 4z = 5$
$9x - 6y + kz = -4$
$12x - 8y = -10$
Originally I just multiplied the first row by $\frac{3}{2}$ and subtracted it from the second, which gives you a value of $6$ for the answer. However, this is not the correct answer. Any idea what I am doing wrong?
|
A system of linear equations is called inconsistent if it has no solutions. A
system which has a solution is called consistent.
Now the augmented matrix
$
\left[\begin{array}{rrr|r}
6 & -4 & 4 & 5 \\
9 & -6 & k & -4 \\
12 & -8 & 0 & -10
\end{array}\right]$
Replace $R_3 $ by $R_3-2R_1$ and $R_2$ by $ R_2 - \frac{3}{2}R_1$
$
\left[\begin{array}{rrr|r}
6 & -4 & 4 & 5 \\
0 & 0 & k-6 & -\frac{23}{2} \\
0 & 0 & -8 & -20
\end{array}\right]$
Replace $R_3$ by
Clearly, $z=\frac{20}{8}=\frac{5}{2}$ and
$(k-6)z=-\frac{23}{2}\implies k-6=-\frac{23}{5}\implies k=-\frac{23}{5} +6=\frac{7}{5}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Lucas Numbers Proof $L_n = \alpha^n + \beta^n$ Proof by Induction:
Lucas numbers are recursively defined as:
$L_n = L_{n-1} + L_{n-2}$ where $L_1 = 1$ and $ L_2 = 3 $for $n \ge 3$
Show that: $L_n = \alpha^n + \beta^n$ for $\alpha = {1+\sqrt{5}\over2}$ and $\beta = {1-\sqrt{5}\over2}$
Is this the correct approach?
base case:
n = 3
$L_3 = L_2 + L_1 = 4$
$L_3 = \alpha^3 + \beta^3 = 4$
inductive hypothesis:
Assume true: $L_k = L_{k-1} + L_{k-2} = \alpha^k + \beta^k$ for $k\ge3$
inductive step:
$L_{k+1} = L_k + L_{k-1} = \alpha^{k+1} + \beta^{k+1}$
$(\alpha^k + \beta^k) + L_{k-1} = \alpha^{k+1} + \beta^{k+1}$
$L_{k-1} = (\alpha^{k+1} - \alpha^k) + (\beta^{k+1} - \beta^k) = \alpha^{k-1} + \beta^{k-1}$
| Note that for any $\alpha$, $\beta$, $k$, $$(\alpha^k + \beta^k)(\alpha + \beta) = \alpha^{k+1} + \alpha \beta^k + \alpha^k \beta + \beta^{k+1} = \alpha^{k+1} + \beta^{k+1} + \alpha \beta(\alpha^{k-1} + \beta^{k-1}).$$
Now, if $\alpha = (1+\sqrt{5})/2$, $\beta = (1-\sqrt{5})/2$, what are the sum $\alpha + \beta$ and product $\alpha\beta$, respectively?
This should suggest an explicit proof by induction. For the inductive step, we claim that there exists at least one $n$ such that $L_k = \alpha^k + \beta^k$ for all $1 \le k \le n$. Then in particular, $L_n = \alpha^n + \beta^n$ and $L_{n-1} = \alpha^{n-1} + \beta^{n-1}$. But the identity we established above tells us that $L_{n+1} = L_n + L_{n-1} = \alpha^{n+1} + \beta^{n+1}$, completing the inductive step.
| {
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Prove combination identity $ \sum_{k=0}^n {2k \choose k} {2n-2k \choose n-k} = 4^n $
I tried with mathematical induction only to fail.
Is this formula related to some special function like Beta, Gamma, etc?
| Consider the generating function method. It is seen that for
\begin{align}
S_{n} = \sum_{k=0}^{n} \binom{2k}{k} \binom{2n-2k}{n-k}
\end{align}
yields
\begin{align}
\sum_{n=0}^{\infty} S_{n} t^{n} &= \sum_{n=0}^{\infty} \sum_{k=0}^{n} \binom{2k}{k} \binom{2n-2k}{n-k} t^{n} \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \binom{2k}{k} \binom{2n}{n} t^{n+k} \\
&= \left( \sum_{k=0}^{\infty} \binom{2k}{k} t^{k} \right)^{2} \\
&= \left( \frac{1}{\sqrt{1-4t}} \right)^{2} \\
&= \frac{1}{1-4t} \\
&= \sum_{n=0}^{\infty} 4^{n} \, t^{n}.
\end{align}
Equating both sides yields
\begin{align}
\sum_{k=0}^{n} \binom{2k}{k} \binom{2n-2k}{n-k} = 4^{n}.
\end{align}
| {
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$[\mathbb{Q}(\sqrt [3] {2}+\sqrt {5}):\mathbb{Q}]$ What is $[\mathbb{Q}(\sqrt [3] {2}+\sqrt {5}):\mathbb{Q}]$? A straight forward way will be to just set $x=\sqrt [3] {2}+\sqrt {5}$, take powers and reach at a polynomial in $x$ and show the polynomial is irreducible. But the method is very tedious and seems to be hopeless. Any idea?
| Denote $\alpha = \sqrt[3]{2}+ \sqrt{5}$.
Let $\zeta = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$.$\ $
The product
\begin{eqnarray}
P(X) =\prod_{0\le k \le 2, 0 \le l \le 1} ( X - (\sqrt[3]{2}\cdot \zeta ^k + \sqrt{5}\cdot(-1)^l)\ )
\end{eqnarray}
is a polynomial with integer coefficients. We calculate:
\begin{eqnarray}
P(X) = X^6- 15 X^4 - 4 X^3 + 75 X^2- 60 X -121
\end{eqnarray}
$P(X)$ is irreducible $\mod\!\! 7$ and therefore irreducible over $\mathbb{Q}$. It follows that
$$[\mathbb{Q}(\alpha) \colon \mathbb{Q}] = 6$$
We can check by direct computation that
\begin{eqnarray}
\frac{-2275 + 7482 \alpha - 390 \alpha^2 - 1000 \alpha^3 +
9 \alpha^4 + 60 \alpha^5}{4054} = \sqrt{5}
\end{eqnarray}
and therefore
\begin{eqnarray}
\alpha - \frac{-2275 + 7482 \alpha - 390 \alpha^2 - 1000 \alpha^3 +
9 \alpha^4 + 60 \alpha^5}{4054} = \sqrt[3]{2}
\end{eqnarray}
We conclude that
$$\mathbb{Q}(\alpha) = \mathbb{Q}( \sqrt[3]{2}, \sqrt{5})
$$
| {
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Do we simplify the Proof by Contradiction? Prove the following by contradiction:
Suppose $a,b\in\mathbb{Z}$. If $4|\left(a^2+b^2\right)$, then $a$ and $b$ are not both odd (in other words, $a$ and $b$ are even)
So, I did this:
Assume $a$ and $b$ are odd
Let $a={2k+1}$
Let $b={2l+1}$
$4|\left((2k+1\right)^2+\left(2l+1\right))^2$
$\left(2k+1\right)*\left(2k+1\right)=4k^2+4k+1$
$\left(2l+1\right)*\left(2l+1\right)=4l^2+4l+1$
$4|\left(\left(4k^2+4k+1\right)+\left(4l^2+4l+1\right)\right)$
For this, do we still, simplify it
Or, what do you do after that??
| Because
$$
a^2+b^2=(a-b)^2+2ab,
$$
if $a^2+b^2$ is even, then $a$ and $b$ must share parity. But if $a$ and $b$ are both odd, then $(a-b)^2$ is divisible by $4$ whereas $2ab$ is not, so it must be that $a$ and $b$ are both even.
| {
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If $\int \frac{f(x)}{x^2(x+1)^3}\hspace{1mm}dx$ is rational and $f$ is quadratic with $f(0)=1$, then find $f'(0)$
If $\int \dfrac{f(x)}{x^2(x+1)^3}\hspace{1mm}dx$ is a rational function, and $f$ is quadratic function, such that $f(0)=1$. Then Find $f'(0)$
This looks like an interesting problem with an elegant solution.
But I cannot figure out how to approach this.
Can anyone help
| Let $f(x):=ax^2+bx+1$
$$\int \frac{ax^2+bx+1}{x^2(x+1)^3}dx=\int\left(\frac{a-b+1}{(x+1)^3}+\frac{2-b}{(x+1)^2}+\frac{b-3}{x}+\frac{3-b}{x+1}+\frac1{x^2}\right)dx$$
So $b=3$
Now $f'(x)=2ax+b,f'(0)=b=3$
| {
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Is $5^2x^3-x^5 = x^3(x-5)(x+5)$ or $-x^3(5-x)(5+x)$ Geogebra's Factor function says that
$5^2x^3-x^5$
is
$-x^3(x-5)(x+5)$
but from what I do, it is positive, $x^3(5+x)(5-x)$
Note the x isnt in the same position
Am I wrong?
| Look:
$$5^2x^3-x^5=x^3(5^2-x^2)=x^3(5-x)(5+x)x^3(-1)(x-5)(x+5)=-x^3(x-5)(x+5).$$
| {
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Calculate $\int_0^1\frac{\log^2x\log(1+x^2)}{1-x^2}dx$ Evaluating $$\int_0^1\frac{\log^2(x)\log(1+x^2)}{1-x^2}dx$$
I found $- \dfrac{\pi^4}{32}+2G^2+\dfrac74 ζ(3)\log2 $ where $G$ is the Catalan's constant.
| \begin{align}J&=\int_0^1\frac{\log^2(x)\log(1+x^2)}{1-x^2}dx\\
&\overset{\text{IBP}}=\left[\left(\int_0^x\frac{\ln^2 t}{1-t^2}dt\right)\ln(1+x^2)\right]_0^1 -\int_0^1 \frac{\color{red}{2x}}{1+x^2}\left(\int_0^x\frac{\ln^2 t}{1-t^2}dt\right)dx\\
&=\frac{7}{4}\zeta(3)\ln 2-\int_0^1\int_0^1 \frac{2x^2\ln^2(tx)}{(1+x^2)(1-t^2x^2)}dtdx\\
&=\frac{7}{4}\zeta(3)\ln 2+2\int_0^1\int_0^1 \left(\frac{\ln^2(tx)}{(1+t^2)(1+x^2)}-\frac{\ln^2(tx)}{(1+t^2)(1-t^2x^2)}\right)dtdx\\
&=\frac{7}{4}\zeta(3)\ln 2+4\int_0^1\int_0^1 \int_0^1\frac{\ln^2 x}{(1+t^2)(1+x^2)}dtdx+4\left(\int_0^1\frac{\ln x}{1+x^2}dx\right)^2-\\&2\int_0^1\int_0^1\frac{\ln^2(tx)}{(1+t^2)(1-t^2x^2)}dtdx\\
&=\frac{7}{4}\zeta(3)\ln 2+4\times\frac{1}{16}\pi^3\times \frac{1}{4}\pi+4\text{G}^2-\int_0^1 \frac{2}{t(1+t^2)}\left(\int_0^t\frac{\ln^2 u}{1-u^2}du\right)dt\\
&\overset{\text{IBP}}=\frac{7}{4}\zeta(3)\ln 2+\frac{1}{16}\pi^4+4\text{G}^2-\left[\left(2\ln t-\ln(1+t^2)\right)\left(\int_0^t\frac{\ln^2 u}{1-u^2}du\right)\right]_0^1+\\&\int_0^1 \frac{\left(2\ln t-\ln(1+t^2)\right)\ln^2 t}{1-t^2}dt\\
&=\frac{7}{4}\zeta(3)\ln 2+\frac{1}{16}\pi^4+4\text{G}^2+\frac{7}{4}\zeta(3)\ln 2+2\int_0^1 \frac{\ln^3 t}{1-t^2}dt-J\\
J&=\boxed{\frac{7}{4}\zeta(3)\ln 2+2\text{G}^2-\frac{1}{32}\pi^4}
\end{align}
NB: i assume that,
\begin{align}\int_0^1 \frac{\ln t}{1+t^2}dt&=-\text{G}\\
\int_0^1 \frac{\ln^2 t}{1+t^2}dt&=\frac{\pi^3}{16}\\
\int_0^1 \frac{\ln^2 t}{1-t^2}dt&=\frac{7}{4}\zeta(3)\\
\int_0^1 \frac{\ln^3 t}{1-t^2}dt&=-\frac{\pi^4}{16}
\end{align}
Addendum: typo fixed. Thanks Sewer Keeper.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/941994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove or disprove $ p^{r+s}\mid q^{ke} - 1 \iff p^s \mid k$. Let $p$ be an odd prime and $q$ be a power of prime. Suppose $e := \min\{\, e \in \mathbb{N} : p \mid q^e - 1 \,\}$ exists. Put $r := \nu_p(q^e - 1)$ (that is, $p^r \mid q^e - 1$ and $p^{r+1} \nmid q^e - 1$). What I want to prove is the following:
$ \forall k \in \mathbb{N},~ \forall s \in \mathbb{N}_0,~p^{r+s} \mid q^{ke} - 1 \iff p^s \mid k.$
I feel that this is true (but perhaps some upper bounds may needed for $k$ and $s$). How can I prove or disprove this statement?
Example:
If $p = 3,~q = 5$ then $e = 2,~r = 1$.
*
*$5^{1 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 0}$
*$5^{2 \cdot 2} - 1 = 2^4 \cdot 3^{1 + 0} \cdot 13$
*$5^{3 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 1} \cdot 7 \cdot 31$
*$5^{4 \cdot 2} - 1 = 2^5 \cdot 3^{1 + 0} \cdot 13 \cdot 313$
*$5^{5 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 0} \cdot 11 \cdot 71 \cdot 521$
My attempt:
In case $(\Leftarrow)$,
$$\begin{align}
q^{ke} - 1 &= ((q^e - 1) + 1)^k - 1 \\
&= (q^e - 1)^k + k(q^e - 1)^{k - 1} + \dotsb + \binom{k}{2}(q^e - 1)^2 + k(q^e - 1).
\end{align}$$
I can see that the last term is divisible by $p^{r + s}$ from the assumption. But is it true for the former terms?
In case $(\Rightarrow)$, we may start by introducing the quotient and the remainder $k = ap^s + b~(0 \le b < p^s)$ and calculate similar expansion as before.
| Suppose that $p$ divides $q^n-1$. Let $t=q^n$. We have that $p \mid (t-1)$, and we want to show that $\nu_p (t^k-1) = \nu_p (t-1) + \nu_p (k)$. We may assume that $k$ is prime, as the statement follows from this case by induction on the number of prime factors.
What we need to check is that $\nu_p (\frac{t^k-1}{t-1}) = \delta_{k,p}$. This is straightforward when $k=2$ ($t+1$ cannot have an odd factor in common with $t-1$), so we will further assume that $k$ is an odd prime.
We compute the quantity $\frac{t^k-1}{t-1}=t^{k-1} + \cdots + 1$ modulo $p^2$. Since $p$ divides $t^{\frac{j-1}{2}}-1$ for all odd $j$, we have $(t^{\frac{j-1}{2}}-1)^2 \equiv 0 \pmod{p^2}$, or $t^{j-1} + 1 \equiv 2t^{\frac{j-1}{2}} \pmod{p^2}$.
We have $t^{k-1} + \cdots + 1 = (t^{k-1} + 1) + t(t^{k-3} + 1) + \cdots + t^{\frac{k-3}{2}} (t^2+1) + t^{\frac{k-1}{2}}$. By the above formula, taken modulo $p^2$ this is $2t^{\frac{k-1}{2}} + 2t \cdot t^{\frac{k-3}{2}} + \cdots + 2t^{\frac{k-3}{2}} \cdot t + t^{\frac{k-1}{2}} = kt^{\frac{k-1}{2}}$.
Since $(t,p)=1$, this last fact tells us exactly what we want.
I believe that we can use a similar proof to show that, if $4$ divides $q^n-1$, then $\nu_2 (q^{kn}-1) = \nu_2 (q^n-1) + \nu_2 (k)$. This reflects the fact that the $p$-adic exponential converges on $p\mathbb{Z}_p$ for $p\neq 2$, but the $2$-adic exponential only converges on $4\mathbb{Z}_2$.
Note that there are strong parallels here to the Fibonacci sequence, which also has the property that if $p\mid F_n$, then $\nu_p (F_{kn}) = \nu_p(F_n) + \nu_p (k)$ (with the same caveat for $p=2$). The proof is similar, and uses the Lucas numbers (or $p$-adic analysis, take your pick).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Proof that $|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$ Any hints on how I can prove the inequality:
$$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0$$
Thank you.
| Case 1: If $x = y$, both sides equal zero.
Case 2: If $x \neq y$, both sides of the inequality are positive, so we can square both sides of
$$|\sqrt{x} - \sqrt{y}| \leq \sqrt{|x - y|}$$
to obtain
$$|\sqrt{x} - \sqrt{y}|^2 \leq |x - y| = |(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})| = |\sqrt{x} + \sqrt{y}||\sqrt{x} - \sqrt{y}|$$
Divide both sides of the inequality by $|\sqrt{x} - \sqrt{y}|$ to obtain
$$|\sqrt{x} - \sqrt{y}| \leq |\sqrt{x} + \sqrt{y}|$$
Since both sides of the inequality are positive, we can square both sides to obtain
$$x - 2\sqrt{xy} + y \leq x + 2\sqrt{xy} + y$$
Cancelling $x + y$ yields
$$-2\sqrt{xy} \leq 2\sqrt{xy}$$
Dividing both sides of the inequality by $-2\sqrt{xy}$ yields
$$0 \geq -1$$
Since the steps are reversible, the original inequality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/946804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 6
} |
Is the sequence $x_n = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} +\dots + \frac{\sin n}{2^n}$ Cauchy? $$x_{n} = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} + ... + \frac{\sin n}{2^n}$$
I came across this sequence while studying, and while it is convergent, I'm curious as to whether or not it is a Cauchy sequence. If so, what's the proof?
| You could be interested to notice that quite simple manipulations of the sines of multiple angles lead to $$x_n=\sum_{i=1}^n\frac{\sin (i)}{2^i}= \frac{2^{-n} \left(\sin (n)-2 \sin (n+1)+2^{n+1} \sin (1)\right)}{5-4 \cos (1)}$$ Similarly $$x_n=\sum_{i=1}^n\frac{\sin (i)}{3^i}=\frac{3^{-n} \left(\sin (n)-3 \sin (n+1)+3^{n+1} \sin (1)\right)}{10 - 6 \cos (1)}$$ $$x_n=\sum_{i=1}^n\frac{\sin (i)}{4^i}=\frac{4^{-n} \left(\sin (n)-4 \sin (n+1)+4^{n+1} \sin (1)\right)}{17-8 \cos (1)}$$
If fact, for any value of $k$
$$x_n(k)=\sum_{i=1}^n\frac{\sin (i)}{k^i}=\frac{k^{-n} \left(\sin (n)-k \sin (n+1)+k^{n+1} \sin (1)\right)}{1+k^2-2 k \cos (1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
How can I find $\lim_{n\to\infty}\int_0^\infty\frac{n\cos^2(x/n)}{n+x^4}dx$? I am trying to find the value of this integral:
$\displaystyle{\lim_{n\to\infty}\int_0^\infty\frac{n\cos^2(x/n)}{n+x^4}dx}$.
The integrand tends to 1 as $n$ goes to infinity. So if some convergence theorem holds, the integral would tend to infinity(not Lebesgue dominated convergence theorem, since if there were an integrable function dominating the integrand, there would be a contradiction). So I guess the value of this integral should be infinity, but I am not sure.
| By a direct integration, we may obtain more than the value of the desired limit.
We have, as $n$ tends to $+\infty$:
$$
\int_0^{+\infty}n\:\frac{\cos^2(x/n)}{n+x^4}{\rm d}x =\color{#008B8B}{\frac{\pi\sqrt{2}}{4}} \color{#006699}{\:n^{1/4}}+\color{#008B8B}{\frac{\pi\sqrt{2}}{4}}\color{#006699}{\frac{1}{n^{5/4}}}+\color{#008B8B}{\frac{\pi}{3}}\color{#006699}{\frac{1}{n^2}}+\mathcal{O}\left(\color{#006699}{\frac{1}{n^{11/4}}}\right) \tag1
$$
Of course the desired limit is $+\infty$.
One may recall that we have, using inverse Laplace transform, the standard result:
$$
\int_0^{+\infty}\frac{\cos^2(ax)}{b^4+x^4}{\rm d}x =\frac{\pi +e^{-\sqrt{2} a b} \pi \left(\cos\left(\sqrt{2} a b\right)+\sin\left(\sqrt{2} a b\right)\right)}{4 \sqrt{2} \:b^3}\quad a>0,\,b>0.
$$
Then putting $a:=\dfrac 1n$ and $b:=n^{1/4}$, we get
$$
\int_0^{+\infty}\frac{\cos^2(x/n)}{n+x^4}{\rm d}x =\frac{\pi +e^{-\Large \frac{\sqrt{2}}{n^{3/4}}} \pi \left(\cos\left(\Large \frac{\sqrt{2}}{n^{3/4}}\right)+\sin\left(\Large \frac{\sqrt{2}}{n^{3/4}}\right)\right)}{4 \sqrt{2} \:n^{3/4}}.
$$ Now, as $x$ is near $0$, we apply the standard Taylor series expansions:
$$
\begin{align}
e^{-x} & =1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+\mathcal{O}\left(x^5\right) \\
\cos x & =1-\frac{x^2}{2!}+\frac{x^4}{4!}+\mathcal{O}\left(x^5\right) \\
\sin x & =x-\frac{x^3}{3!}+\mathcal{O}\left(x^5\right)
\end{align}
$$
with $\displaystyle x:=\frac{\sqrt{2}}{n^{3/4}} $ giving $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Find the order of the elements in the given groups I have to find the order of the following elements in the given groups:
*
*$(1 \ \ 2 \ \ 3) \ (1 \ \ 2\ \ 4) \text{ in } S_5$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 4 & 1 & 2 & 5
\end{pmatrix}=\begin{pmatrix}
1 & 3
\end{pmatrix}\begin{pmatrix}
2 & 4
\end{pmatrix}$$ order $=\operatorname{lcm}(2,2)=2$
Is it correct?
*
*$a^{45}$ in $G=\langle a\mid a^{140}=1\rangle$
$$\operatorname{ord}(a^{45})=\frac{140}{\gcd(45, 140)}=\frac{140}{5}=28$$
Is it right?
*
*$\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} \text{ in } GL_2(\mathbb{R})$
$$A=\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} \\ A^2=\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} =\begin{bmatrix}
-1 & 0\\
0 & -1
\end{bmatrix} \\ A^3=\begin{bmatrix}
-1 & 0\\
0 & -1
\end{bmatrix} \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} =\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix} $$
But how can I find the order?
| Your first two questions are indeed correct.
Note that your calculation of $A^3$ is "off":
$$A^3= A^2 A=\begin{bmatrix}
-1 & 0\\
0 & -1
\end{bmatrix} \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} =\begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}$$
And $A^4 = A^2A^2 = -I\cdot -I = I$. Hence the order of $A = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving terms are rational using Mathematical Induction
I was able to do the first part of the question, in the second part (Proof by Induction),
I showed it holds for $n=1$:
Then I Assumed its true for $n=k$.
$$\begin{align}I_{k+1}=\frac{e^2}{2}-\frac{(k+1)}{2}I_k\end{align}$$
Then for $n=k+1$:
$$\begin{align}I_{k+2}=\frac{e^2}{2}-\frac{(k+1)+1}{2}I_{k+1}=\frac{e^2}{2}-(\frac{(k+1)+1}{2})(A_{k+1} e^2+B_{k+1})\end{align}$$
I don't know how to proceed.
| You wrote you showed it holds for $n=1$.
Assume that $I_k=A_ke^2+B_k$ where $A_k,B_k$ are rational numbers.
Then, $$\begin{align}I_{k+1}&=\frac 12e^2-\frac 12(k+1)I_k\\&=\frac 12e^2-\frac 12(k+1)(A_ke^2+B_k)\\&=\left(\frac{1}{2}-\frac{(k+1)A_k}{2}\right)e^2+\left(-\frac{(k+1)B_k}{2}\right).\end{align}$$
Since $$\frac{1}{2}-\frac{(k+1)A_k}{2},\ \ -\frac{(k+1)B_k}{2}$$
are rational number, $I_{k+1}$ is of the form $A_{k+1}e^2+B_{k+1}$ where $A_{k+1},B_{k+1}$ are rational numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949843",
"timestamp": "2023-03-29T00:00:00",
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Show that the equation Ax=x can be rewritten as (A-I)x = 0 and use this result to solve Ax=x for x. Given matrix A = \begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix}
and x = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}
Answer: Given $Ax = x$. Subtract from x from both sides to get $Ax - x = 0$. We know $x = I \cdot x$ , where $I$ is the identity matrix.
Simplify $Ax - x = 0$ to get $x(A - I) = 0$.
$$\begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} \cdot \left(\begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \right) = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} \cdot \begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$
Do I need to write the system of linear equations?
| One has
\begin{equation}
A-I
=
\left(\begin{matrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{matrix}\right)
-
\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)
=
\left(\begin{matrix}1 & 1 & 2 \\ 2 & 1 & -2 \\ 3 & 1 & 0\end{matrix}\right)
\end{equation}
With the rule of Sarrus
(see http://en.wikipedia.org/wiki/Rule_of_Sarrus), one gets
\begin{equation}
det(A-I) = 0 + (-6) + (+4) - (-2) - 0 - (+6) = -6
\end{equation}
Thus one has $x=0$.$\Box$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/950626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:
*
*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ and $z_2=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}$
*How do I proceed?
| Hint: $z^2+z+1=0$ implies that $0=(z-1)(z^2+z+1)=z^3-1$, so $z^3=1$. Also $z^{10}=z^9\cdot z$ and $z^{-10}=z^2\cdot z^{-12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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$\lim_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}$ May I know how can I calculate the following expression?
$$
\lim\limits_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}
$$
From calculator, the answer is $-\sqrt{6}$ , my approach is by using dividing numerator and denominator by using $x$. Which is,
$$
\lim\limits_{x\to-\infty}\frac{\frac{\sqrt{6x^2 - 2}}{x}}{\frac{x+2}{x}}
=
\lim\limits_{x\to-\infty}\frac{{\sqrt{\frac{6x^2 - 2}{x^2}}}}{\frac{x+2}{x}}=\sqrt{6}
$$
My answer is $\sqrt{6}$, is my working wrong or there are actually another approach? Thank you.
| $$\lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{x+2}$$
First note that as $x\to -\infty$, then $x+2\lt 0$ and
$$ x+2=-|x+2|=-\sqrt{(x+2)^2} $$
Therefore
$$ \lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{x+2} = \lim_{x\to -\infty} \frac{\sqrt{6x^2-2}}{-\sqrt{(x+2)^2}} $$
$$ = -\lim_{x\to -\infty} \sqrt{\frac{6x^2-2}{x^2+4x+4}} = -\lim_{x\to -\infty} \sqrt{\frac{6-\frac{2}{x^2}}{1+\frac{4}{x}+\frac{4}{x^2}}} $$
$$ =-\sqrt{\frac{6-0}{1+0+0}}=-\sqrt{6} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Chinese Remainder Theorem RSA I want to solve the following modular quadratic equation:
$x^2 \equiv 188 \pmod {437}$ using the fact that $437$ can be factorized by the primes as: $19⋅23$.
So far I have done:
$$x^2 \equiv 188 \pmod{19} \Rightarrow x^2 \equiv 17 \pmod {19}\\
x^2 \equiv 188 \pmod {23} \Rightarrow x^2 \equiv 4 \pmod {23}$$
$$x^2 \equiv 7 \pmod{ 19} \Rightarrow x \equiv \pm 6 \pmod {19}\\
x^2 \equiv 4 \pmod {23} \Rightarrow x \equiv \pm 2 \pmod {23}$$
And the last step. But I do not understand how to combine:
$$x \equiv 6 \pmod {19} \text{ and } x \equiv 2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}\\
x \equiv -6 \pmod {19} \text{ and } x \equiv 2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod{ 437}\\
x \equiv 6 \pmod {19} \text{ and } x \equiv -2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}\\
x \equiv -6 \pmod {19} \text{ and } x \equiv -2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}$$
Please help me fill the "insert-her".
| You can find the solution using this calculator just fill in $$ x=6 \mod 19$$ and $$
x=2 \mod 23
$$
Your solution is $25$ and indeed $188 \equiv 25^2 \pmod {437} $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How is $\lim_{x \to 0}\frac{\sqrt {2-x}-\sqrt {2}}{x} = \frac{-1}{2\sqrt{2}}$? How is
$$\displaystyle\lim_{x \to 0}\frac{\sqrt {2-x}-\sqrt {2}}{x} = \frac{-1}{2\sqrt{2}}$$
computable?
| Multiplying$$\frac{\sqrt{2-x}-\sqrt 2}{x}$$
by$$\frac{\sqrt{2-x}+\sqrt 2}{\sqrt{2-x}+\sqrt 2}$$
gives us
$$\begin{align}\lim_{x\to 0}\frac{\sqrt{2-x}-\sqrt 2}{x}&=\lim_{x\to 0}\frac{(\sqrt{2-x}-\sqrt 2)(\sqrt{2-x}+\sqrt 2)}{x(\sqrt{2-x}+\sqrt 2)}\\&=\lim_{x\to 0}\frac{(2-x)-2}{x(\sqrt{2-x}+\sqrt 2)}\\&=\lim_{x\to 0}\frac{-1}{\sqrt{2-x}+\sqrt 2}\\&=\frac{-1}{\sqrt 2+\sqrt 2}\\&=-\frac{1}{2\sqrt 2}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Confused by the solution of $x^3+bx^2+cx+d=0$ From $x^3 + bx^2 + cx + d = 0$, we have $(x-x_1)(x-x_2)(x-x_3)=0$ for some roots $x_1$, $x_2$ and $x_3$.
Expanding this second expression gives us
$$x^3 + \left(x_1+x_2+x_3\right)x^2 + \left(x_1x_2 + x_1 x_3 + x_2 x_3\right) x + x_1 x_2 x_3 = 0$$
and comparing coefficients gives
*
*$x_1+x_2+x_3 = b$
*$x_1 x_2 + x_1 x_3 + x_2 x_3 = c$
*$x_1 x_2 x_3 = d$
From this we can see that
*
*$x_2 x_3 = d/x_1$ (where $x_1 \neq 0$)
*$x_2 + x_3 = b - x_1$
*$x_1 \left(x_2 + x_3\right) + x_2 x_3 = c$
This gives us
$$x_1 \left(b-x_1\right) + \frac{d}{x_1} = c \implies -x_1^3 + bx_1^2 - cx_1 + d = 0
\implies x_1^3 - bx_1^2 + cx_1 -d = 0$$
But since $x_1$ is a root of the polynomial, we know that
$$x_1^3 = bx_1^2 + cx_1 + d = 0$$
and differencing the two gives
$$bx_1^2 + d = 0$$
so we have $x_1 = \pm \sqrt{-d/b}$, but that’s not always true. Why?
| If $$x^3+bx^2+cx+d=(x-x_1)(x-x_2)(x-x_3),$$
then
$$x_1+x_2+x_3=-b$$
$$x_1x_2x_3=-d$$
instead of
$$x_1+x_2+x_3=b$$
$$x_1x_2x_3=d$$
because
$$(x-x_1)(x-x_2)(x-x_3)$$
$$=x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_2x_3+x_3x_1)x-x_1x_2x_3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the inverse of $f(x) = x\sqrt{2+x^2}$ For a function $f(x) = x\sqrt{2+x^2}$ find out if it's bijective and if so, find its inverse.
The function is surjective because $x^2 > 0:\forall x\in\mathbb{R}$. I'm having difficulties proving that the function is injective.
I tried the following: $$f(x) = f(y) \iff x\sqrt{2+x^2} = y\sqrt{2+y^2} \\ x^2(2+x^2) = y^2(2+y^2) \\ 2x^2+x^4 = 2y^2 + y^4$$
And well, I think I just made it more difficult for myself. I couldn't find any thread on MSE where one tried tackling such a specific problem.
Solution appreciated, thanks in advance.
| Hint:
To find the inverse let $y= f^{-1}(x)$ then since $f(f^{-1}(x)) = x$ we have
$$f(f^{-1}(x)) = x = y\sqrt{2+y^2}$$
which gives
$$x^2 = y^2(2+y^2)\to y^4 + 2y^2 - x^2 = 0$$
This is a quadratic equation in $y^2$: $(y^2)^2 + 2y^2 - x^2 = 0$ for which you can use the standard formula on. To determine which root to pick notice that $f^{-1}(0)=0$ and $\text{sign}(f^{-1}(x)) = \text{sign}(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
general way to get formula as multiplication. We assume:
$$ 1^n + 2^n + 3^n + .. + k^n$$
where k and n are natural numbers.
Are there a general way to get it as multiplication?
For example:
$$ 1^3 + 2^3 + 3^3 + .. + k^3 = \binom {k+1} 2 ^2 $$
| Suppose that for $0\le j\lt m$, there is a polynomial $P_j$, with degree $j+1$, so that
$$
\sum_{k=0}^nk^j=P_j(n)\tag{1}
$$
The binomial theorem says that
$$
(k+1)^{m+1}-k^{m+1}=\sum_{j=0}^m\binom{m+1}{j}k^j\tag{2}
$$
Sum $(2)$ in $k$ using $(1)$:
$$
\begin{align}
(n+1)^{m+1}
&=\sum_{k=0}^n\sum_{j=0}^m\binom{m+1}{j}k^j\\
&=\sum_{k=0}^n\left[(m+1)k^m+\sum_{j=0}^{m-1}\binom{m+1}{j}k^j\right]\\
&=(m+1)\sum_{k=0}^nk^m+\sum_{j=0}^{m-1}\binom{m+1}{j}P_j(n)\tag{3}
\end{align}
$$
Solving $(3)$ for $\sum\limits_{k=0}^nk^m$
$$
\begin{align}
\sum_{k=0}^nk^m
&=\frac1{m+1}\left[(n+1)^{m+1}-\sum_{j=0}^{m-1}\binom{m+1}{j}P_j(n)\right]\\
&=P_m(n)\tag{4}
\end{align}
$$
That is, $(1)$ is true for $j=m$.
Thus, we have shown that if $(1)$ is true for $0\le j\lt m$, then $(1)$ is also true for $j=m$. Since $P_0(n)=n+1$, we have that $(1)$ is true for $j=0$. Therefore, by induction, $(1)$ is true for all $j\ge0$.
Notice that $(4)$ also gives us a formula to compute $P_m$ from the $P_j$ for $j\lt m$.
$$
\begin{align}
P_1(n)
&=\frac12\left[\vphantom{\binom{2}{0}}\right.(n+1)^2-\binom{2}{0}\overbrace{(n+1)\vphantom{\frac{n^2+n}2}}^{P_0(n)}\left.\vphantom{\binom{2}{0}}\right]\\[4pt]
&=\frac{n^2+n}2\\
P_2(n)
&=\frac13\left[\vphantom{\binom{2}{0}}\right.(n+1)^3-\binom{3}{0}\overbrace{(n+1)\vphantom{\frac{n^2+n}2}}^{P_0(n)}-\binom{3}{1}\overbrace{\frac{n^2+n}2}^{P_1(n)}\left.\vphantom{\binom{2}{0}}\right]\\[4pt]
&=\frac{2n^3+3n^2+n}6\\
P_3(n)
&=\frac14\left[\vphantom{\binom{2}{0}}\right.(n+1)^4-\binom{4}{0}\overbrace{(n+1)\vphantom{\frac{n^2+n}2}}^{P_0(n)}-\binom{4}{1}\overbrace{\frac{n^2+n}2}^{P_1(n)}-\binom{4}{2}\overbrace{\frac{2n^3+3n^2+n}6}^{P_2(n)}\left.\vphantom{\binom{2}{0}}\right]\\[4pt]
&=\frac{n^4+2n^3+n^2}4
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}
\end{align}
I know the final answer is $\frac{9}{2}$. After multiplying by the conjugate, I see where the $9$ in the numerator comes from. I just can't remember how I solved the rest of the problem.
| Expanding the rightmost term, we get:
$$
\frac{(x^2+2x) - (x^2-7x)}{\sqrt{x^2+2x} + \sqrt{x^2-7x}} = \frac{9x}{\sqrt{x^2+2x} + \sqrt{x^2-7x}}
$$
now just divide the numerator and denominator by $x = \sqrt{x^2}$:
$$
\frac{9}{\sqrt{1+\frac{2}{x}} + \sqrt{1-\frac{7}{x}}}.
$$
We have that $\sqrt{1+\frac{2}{x}} + \sqrt{1-\frac{7}{x}} \rightarrow 2$ as $x \rightarrow \infty$, so the final answer is $9/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
$\Sigma_{k=m+1}^{\infty}\frac{1}{10^k} \leq \frac{1}{10^m}$ Is it true that $\Sigma_{k=m+1}^{\infty}\frac{1}{10^k} \leq \frac{1}{10^m}$ for $m \in \mathbb{N}$?
| $\Sigma_{k=m+1}^{\infty}\frac{1}{10^k}=\frac{1}{10^{m+1}}+\frac{1}{10^{m+2}}+\frac{1}{10^{m+3}}+.... =\frac{1}{10^m}(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...)\leq \frac{1}{10^m}\cdot1=\frac{1}{10^m}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Proving $R +r\le h_{max} $ If $R$ is the circumradius , $r$ the inradius and $ h_{max}$ is the largest altitude of acute angled triangle $ABC$, then prove that $$R +r\le h_{max}. $$
I tried this using Euler's inequality but I did not succeed.
| By assuming $a\leq b\leq c$ we have that the greatest altitude is $h_a$ and:
$$ R = \frac{abc}{4\Delta},\quad r=\frac{2\Delta}{a+b+c},\quad h_a=\frac{2\Delta}{a} $$
so, by Heron's formula, we have to prove that:
$$ 2abc + (-a+b+c)(a-b+c)(a+b-c) \leq \frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{a} $$
or:
$$2a^2 bc\leq(b+c)(-a+b+c)(a-b+c)(a+b-c)\tag{1}$$
where $2a^2\leq(b+c)(b+c-a)$ is trivial (since $c\geq b\geq a$) and
$$ bc \leq (a-b+c)(a+b-c) = a^2-(b-c)^2 $$
is equivalent to:
$$ b^2+c^2-bc \leq a^2 \tag{2}$$
that follows from the cosine theorem, since $\widehat{A}\leq\frac{\pi}{3}$ (otherwise, $a$ cannot be the shortest side).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Closed-form of $\sum_{k=1}^{\infty}\arctan(1/k^3)$ Wolfram said that
$$\sum_{k=1}^{\infty}\arctan\left(\frac{1}{k^2}\right)=\arctan\left(\frac{1-\cot\left(\frac{\pi}{\sqrt 2}\right)\tanh\left(\frac{\pi}{\sqrt 2}\right)}{1+\cot\left(\frac{\pi}{\sqrt 2}\right)\tanh\left(\frac{\pi}{\sqrt 2}\right)}\right).$$
I was wondering about a closed-form of
$$S = \sum_{k=1}^{\infty}\arctan\left(\frac{1}{k^3}\right).$$
A numerical approximation of $S$ is
$$S \approx 0.986791652613071125515794193830247643724471031136456434\dots$$
Is there a closed-form of $S$?
| Notice for any positive number $x$, we have
$$\tan^{-1}(x) = \frac{1}{2i}\log\left(\frac{1+ix}{1-ix}\right) = \Im\log(1 + ix)$$
We can rewrite the sum at hand as
$$\sum_{k=1}^\infty \tan^{-1}\frac{1}{k^3} = \Im\left[ \sum_{k=1}^\infty \log\left(1 + \frac{i}{k^3}\right) \right]
$$
For each $k$, we have the factorization
$$1 + \frac{i}{k^3} = 1 - \left(\frac{i}{k}\right)^3 = \prod_{j=0}^2 \left(1 - \frac{i\omega^j}{k}\right)$$
where $\omega = e^{i2\pi/3}$ is a primitive cubic root of unity.
Recall the infinite product expansion of gamma function
$$\frac{1}{\Gamma(z)} = z e^{\gamma z}\prod_{k=1}^\infty \left(1 + \frac{z}{k}\right) e^{-\frac{z}{k}}
\quad\implies\quad
\prod_{k=1}^\infty \left(1+\frac{z}{k}\right)e^{-\frac{z}{k}} = \frac{e^{-\gamma z}}{\Gamma(1+z)}
$$
If we replace $z$ by $-i \omega^j$ for $j = 0,1,2$ and taking the product, we get
$$\begin{align}
\prod_{k=1}^\infty\left(1 + \frac{i}{k^3}\right)
&= \prod_{k=1}^\infty\prod_{j=0}^2\left[\left(1 - \frac{i\omega^j}{k}\right)e^{i\omega^j/k}
\right]
= \prod_{j=0}^2\prod_{k=1}^\infty\left[\left(1 - \frac{i\omega^j}{k}\right)e^{i\omega^j/k}
\right]\\
& = \frac{e^{-\gamma( -i\sum\limits_{j=0}^2 \omega^j )}}{\prod\limits_{j=0}^2\Gamma(1 - i\omega^j)}
= \frac{1}{
\Gamma(1-i)
\Gamma\left(1 + \frac{\sqrt{3}}{2} + \frac{i}{2}\right)
\Gamma\left(1 - \frac{\sqrt{3}}{2} + \frac{i}{2}\right)}
\\
\end{align}
$$
Taking logarithm on both sides, we will get something close to what we want.
The catch is the $\log$ function is not holomorphic over the whole complex plane.
In general, the sum of logarithms is equal to the log of the product only up
to some multiples of $2\pi$. i.e.
$$ \sum_{k=1}^\infty \log\left( 1 + \frac{i}{k^3}\right)
= \log\left[ \prod_{k=1}^\infty \left( 1 + \frac{i}{k^3}\right) \right] + 2N\pi$$
for some unknown integer $N$.
Instead of determining what $N$ is, we will solve this problem in a different manner.
We use the fact in the sum of the log, the imaginary part for the $k \ge 2$ terms are
small enough.
If we remove the $k = 1$ term from the sum, the sum of log will be equal to the log of product.
At the end, we have
$$\begin{align}
\sum_{k=1}^\infty \tan^{-1}\frac{1}{k^3}
&= \frac{\pi}{4} + \sum_{k=2}^\infty \tan^{-1}\frac{1}{k^3}
= \frac{\pi}{4} + \Im\left[\sum_{k=2}^\infty\log\left(1 + \frac{i}{k^3}\right)\right]\\
&= \frac{\pi}{4} + \Im\log\left[\prod_{k=2}^\infty\left(1 + \frac{i}{k^3}\right)\right]\\
&= \frac{\pi}{4} - \Im\log\left[(1+i)
\Gamma(1-i)
\Gamma\left(1 + \frac{\sqrt{3}}{2} + \frac{i}{2}\right)
\Gamma\left(1 - \frac{\sqrt{3}}{2} + \frac{i}{2}\right)
\right]\\
&=
\frac{\pi}{4} - \Im\log\left[(-1+i)
\Gamma(-i)
\Gamma\left( \frac{\sqrt{3}}{2} + \frac{i}{2}\right)
\Gamma\left( - \frac{\sqrt{3}}{2} + \frac{i}{2}\right)
\right]\\
\end{align}$$
According to WA, this is approximately
$$0.986791652613071125515794193830247643724471031136456434028974\ldots$$
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/968098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Help with using Master Theorem on $T(n)=9T(n/3) + \Theta(n^2/\operatorname{lg}(n))$ I want to use the Master theorem to solve the following recurrence.
$$T(n)=9T(n/3) + \Theta(n^2/\operatorname{lg}(n))$$
We can easily see that $a=9$ and $b=3$ and $f(n) = n^2/\operatorname{lg}(n)$.
I am very new to using this theorem, and am hoping for some help or pointers at this stage.
Is there a way to eyeball which 'case' to use or should we go through each of them (that actually takes me a pretty long time).
Thanks in advance.
| By way of enrichment we solve another closely related recurrence that
admits an exact solution. Suppose we have $T(0)=0$ and $T(1)=T(2)=1$
and for $n\ge 3$
$$T(n) = 9 T(\lfloor n/3 \rfloor) +
\frac{n^2}{\lfloor \log_3 n \rfloor}.$$
Furthermore let the base three representation of $n$ be
$$n = \sum_{k=0}^{\lfloor \log_3 n \rfloor} d_k 3^k.$$
Then we can unroll the recurrence to obtain the following exact
formula for $n\ge 3$
$$T(n) =
9^{\lfloor \log_3 n \rfloor} +
\sum_{j=0}^{\lfloor \log_3 n \rfloor-1}
9^j \frac{1}{\lfloor \log_3 n \rfloor-j} \left(
\sum_{k=j}^{\lfloor \log_3 n \rfloor} d_k 3^{k-j}
\right)^2
\\ = 9^{\lfloor \log_3 n \rfloor} +
\sum_{j=0}^{\lfloor \log_3 n \rfloor-1}
\frac{1}{\lfloor \log_3 n \rfloor-j} \left(
\sum_{k=j}^{\lfloor \log_3 n \rfloor} d_k 3^k\right)^2.$$
Now to get an upper bound consider a string of two digits which yields
$$T(n) \le
9^{\lfloor \log_3 n \rfloor} +
\sum_{j=0}^{\lfloor \log_3 n \rfloor-1}
\frac{1}{\lfloor \log_3 n \rfloor-j} \left(
2 \times \sum_{k=j}^{\lfloor \log_3 n \rfloor} 3^k\right)^2
\\ =
9^{\lfloor \log_3 n \rfloor} +
\sum_{j=0}^{\lfloor \log_3 n \rfloor-1}
\frac{1}{\lfloor \log_3 n \rfloor-j} \left(
3^{\lfloor \log_3 n \rfloor+1} - 3^j\right)^2
\\ = 9^{\lfloor \log_3 n \rfloor} +
9\times 3^{2\lfloor \log_3 n \rfloor}
\sum_{j=0}^{\lfloor \log_3 n \rfloor-1}
\frac{1}{\lfloor \log_3 n \rfloor-j}
\left(1 - 3^{j-(\lfloor \log_3 n \rfloor+1)}\right)^2
\\ = 9^{\lfloor \log_3 n \rfloor} +
9\times 3^{2\lfloor \log_3 n \rfloor}
\sum_{j=1}^{\lfloor \log_3 n \rfloor}
\frac{1}{j}
\left(1 - 3^{-j-1}\right)^2
\\ = 9^{\lfloor \log_3 n \rfloor} +
9\times 3^{2\lfloor \log_3 n \rfloor}
\sum_{j=1}^{\lfloor \log_3 n \rfloor}
\frac{1}{j}\left(
1 - \frac{2}{3} 3^{-j} + \frac{1}{9} 3^{-2j}\right).$$
Note that this bound is attained and cannot be improved. It is
asymptotic to
$$9^{\lfloor \log_3 n \rfloor} +
9\times 3^{2\lfloor \log_3 n \rfloor}
\left(H_{\lfloor \log_3 n \rfloor} +
\frac{2}{3}\log(2/3) -\frac{1}{9} \log(8/9)\right)
\\ = 9^{\lfloor \log_3 n \rfloor} +
9\times 3^{2\lfloor \log_3 n \rfloor}
\left(H_{\lfloor \log_3 n \rfloor} +
\frac{1}{3}\log 2 - \frac{4}{9}\log 3\right).$$
The lower bound is for the case of a one digit followed by a string of
zeros and yields
$$T(n) \ge
9^{\lfloor \log_3 n \rfloor} +
\sum_{j=0}^{\lfloor \log_3 n \rfloor-1}
\frac{1}{\lfloor \log_3 n \rfloor-j} \left(
3^{\lfloor \log_3 n \rfloor}\right)^2
\\ = 9^{\lfloor \log_3 n \rfloor} +
3^{2\lfloor \log_3 n \rfloor}
\sum_{j=0}^{\lfloor \log_3 n \rfloor-1}
\frac{1}{\lfloor \log_3 n \rfloor-j}
= 9^{\lfloor \log_3 n \rfloor} +
3^{2\lfloor \log_3 n \rfloor}
H_{\lfloor \log_3 n \rfloor}.$$
The lower bound too is attained.
Joining the dominant terms of the upper and the lower bound we obtain
the asymptotics
$$\color{#0A0}{3^{2\lfloor \log_3 n \rfloor} H_{\lfloor \log_3 n \rfloor}
\in \Theta\left(\log\log_3(n) \times 3^{2 \log_3 n}\right)
= \Theta\left(\log\log n \times n^2\right)}.$$
These are in agreement with what the Master theorem would produce.
Addendum. Wikipedia lists the above as an inadmissible case for the Master Theorem. However Akra-Bazzi applies and the reader is invited to do this calculation.
| {
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"url": "https://math.stackexchange.com/questions/972275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Difference of inverse squares Given that the positive number $a$ is the difference of inverse squares:
$$a = \frac{1}{n^2} - \frac{1}{m^2}, m, n \in \mathbb{N},$$
could it well be that the $pa$ is also a difference of inverse squares , when p - some natural number ?
| If
$$a = \frac{1}{n^2} - \frac{1}{m^2},$$
where $m, n \in \mathbb{N}$ and $\mathbb{N}$ is the set of natural numbers, then
$$2a = \frac{2}{n^2} - \frac{2}{m^2}.$$
The OP is asking for solutions to the Diophantine equation
$$2a = \frac{2}{n^2} - \frac{2}{m^2} = \frac{1}{r^2} - \frac{1}{s^2},$$
where $r, s \in \mathbb{N}$.
This reduces to
$$2{r^2}{s^2}(m + n)(m - n) = {m^2}{n^2}(r + s)(r - s).$$
Consequently, we have:
$$2{r^2}{s^2} \mid {m^2}{n^2}(r + s)(r - s).$$
Assuming $\gcd(r, s) = 1$, then we have either:
(1) $2 \mid m$, or
(2) $2 \mid n$, or
(3) $2{r^2}{s^2} \mid {m^2}{n^2}$, or
(4) $2 \mid (r + s)(r - s)$.
Perhaps this could shed some light into Antony's answer.
Update - Antony gave the following solutions:
$$n \hspace{0.1in} m \hspace{0.1in} r \hspace{0.1in} s$$
$$5 \hspace{0.1in} 7 \hspace{0.1in} 5 \hspace{0.1in} 35$$
$$5 \hspace{0.1in} 10 \hspace{0.1in} 4 \hspace{0.1in} 20$$
$$6 \hspace{0.1in} 10 \hspace{0.1in} 5 \hspace{0.1in} 15$$
$$9 \hspace{0.1in} 11 \hspace{0.1in} 11 \hspace{0.1in} 99$$
$$10 \hspace{0.1in} 14 \hspace{0.1in} 10 \hspace{0.1in} 70$$
$$5 \hspace{0.1in} 15 \hspace{0.1in} 3 \hspace{0.1in} 5$$
Notice that all the known solutions (so far) satisfy:
(4) $2 \mid (r + s)(r - s)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Evaluating Telescopic Sum $ \sum\frac{n}{1+n^2+n^4} $ How to evaluate following
$$ \sum_{n=1}^{\infty}\frac{n}{1+n^2+n^4}$$
I posted my way as an answer, Is there another Interesting approach to evaluate this sum of series?
| I did it this way
$x$th Term of this sum can be written as
$$T_x=\frac{x}{1+x^2+x^4}=\left(\frac {1}{2\cdot(x^2-x+1)}\right)-\left(\frac{1}{2\cdot(x^2+x+1)}\right)$$
Similarly $(x+1)$th Term can be written as
$$T_{x+1}=\frac{x+1}{1+(x+1)^2+(x+1)^4}=\left(
\frac{1}{2\cdot(x^2+x+1)}\right)-\left(\frac{1}{2\cdot(x^2+3x+3)}\right)$$
So,
$$\begin{align}
\sum_{n=1}^{\infty}\frac{n}{1+n^2+n^4}&=\frac{1}{3}+\frac{2}{21}+\frac{3}{91}+\cdots+\frac{1}{\infty}\\
&=
\left(\frac{1}{2}-\frac{1}{6}\right)+
\left(\frac{1}{6}-\frac{1}{14}\right)+
\left(\frac{1}{14}-\frac{1}{26}\right)+
\cdots
\left(\frac{1}{\infty}-\frac{1}{\infty}\right)\\
&=\frac{1}{2}+
\left(-\frac{1}{6}+\frac{1}{6}\right)+
\left(-\frac{1}{14}+\frac{1}{14}\right)+
\cdots
\left(\frac{1}{\infty}-\frac{1}{\infty}\right)\\
&=\frac{1}{2}+
\left(0\right)+
\left(0\right)+
\cdots
\left(0\right)\\
&=\frac{1}{2}
\end{align}$$
Hence,
$$\sum_{n=1}^{\infty}\frac{n}{1+n^2+n^4}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Quadratic equation $4x^2+4x=7$ using quadratic formula Solve using quadratic formula. $4x^2+4x=7$
So $4x^2+4x-7=0$
$A=4$ $b=4$ $c=-7$
$$x=\frac{-4\pm\sqrt{(4)^2-4(4)(-7)}}{2(4)}=\frac{-4\pm\sqrt{16+112}}{8}=\frac{-4\pm\sqrt{128}}{8}$$
What's next?
| I can't see what you have written but it should look like:
$$\begin{align}
4x^2+4x-7&=0
\\ \Rightarrow x_{\pm}&=\frac{-4\pm\sqrt{(4)^2-4(4)(-7)}}{2(4)}
\\ &=\frac{-4\pm{\sqrt{16+112}}}{8}
\\&=\frac{-4\pm\sqrt{128}}{8}
\\ &=\frac{-4+\sqrt{128}}{8}\,\,\,\text{ or }\frac{-4-\sqrt{128}}{8}
\\ &=\frac{-4+\sqrt{8^2\cdot 2}}{8}\,\,\,\text{ or }\frac{-4-\sqrt{8^2\cdot 2}}{8}
\\ &=\frac{-4+\sqrt{8^2}\cdot \sqrt{2}}{8}\,\,\,\text{ or }\frac{-4-\sqrt{8^2}\cdot \sqrt{2}}{8}
\\ &=\frac{-4+8\cdot \sqrt{2}}{8}\,\,\,\text{ or }\frac{-4-8\cdot \sqrt{2}}{8}
\\ &=-\frac{1}{2}+\sqrt{2}\,\,\,\text{ or }-\frac{1}{2}-\sqrt{2}.
\end{align}$$
The approximate values of these may be found using a calculator.
| {
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Proof for $|1 - z| \geq 1 - |z|$ for $|z| < 1$, $z \in \mathbb{C}$ I can prove it "by picture" by drawing a picture of a circle of radius $|z|$ centered at $(0, 1)$. Then $1 - |z|$ is the length from the origin to the intersection of the circle with the x-axis (to the left). $|1 - z|$ is length from the origin to any point on the circle.
However, I'm having trouble proving this algebraically / with words (aka not proof by picture). I tried using this picture again and the triangle equality, but that didn't work out.
| let $z = a + bi$ with $a^2 + b^2 < 1$, then $|1-z| = |1 - (a+bi)| = |1-a + (-b)i| = \sqrt{(1-a)^2 + b^2}$. So we need to prove: $\sqrt{(1-a)^2 + b^2} \geq 1 - \sqrt{a^2 + b^2}$.
Squaring both sides of the above inequality we have: $(1-a)^2 + b^2 \geq 1 - 2\sqrt{a^2 + b^2} + a^2 + b^2 \iff -2a \geq -2\sqrt{a^2 + b^2} \iff \sqrt{a^2 + b^2} \geq a$ which is true since $\sqrt{a^2 + b^2} \geq |a| \geq a$.
| {
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Triplets satisfying $(a^3+b)(b^3+a)=2^c$
Find the number of triplets $(a,b,c)$ satisfying $(a^3+b)(b^3+a)=2^c$, where $a,b,c\in \mathbb{N}$
A trivial solution is $(1,1,2)$. I think there aren't any other such triplets, so I've been trying to prove that but haven't been able to do so.
How can other solutions be found?
| shaurya gupta already noted the solution $(a,b,c) = (1,1,2)$;
rogerl found another pair, $(3,5,12)$ and $(5,3,12)$, and conjectured
that there are no others. I prove that this conjecture is correct.
[The question required that $a,b,c \in \mathbb N$; this notation
"$\mathbb N$" is sometimes used for nonnegative integers,
but I assume that zero is not allowed here, else
$(a,b,c) = (0,2^r,4r)$ is a solution for every nonnegative integer.]
rogerl shows that $a,b$ must be odd.
Assume without loss of generality that $a \leq b$, and write
$$
a^3 + b = 2^d, \quad
b^3 + a = 2^e
$$
for some integers $d,e$ with $d \leq e$ and $c = d+e$.
Then $b = \lfloor 2^{e/3} \rfloor$, which makes it easy to check that
the three known solutions are the only ones with $e \leq 15$
(for each $e$, solve for $b$, recover $a = 2^e - b^3$,
and check whether $0 < a \leq b$). Thus any other solution
must have $e>15$, and therefore $d>5$ (because $b^3+a < (a^3+b)^3$).
Now $a,b$ satisfy $a^3 \equiv -b$ and $b^3 \equiv -a$ mod $2^d$.
Therefore $a^9 \equiv a \bmod 2^d$. Since $a$ is odd it follows that $a^8 \equiv 1$,
so $a \equiv \pm 1 \bmod 2^{d-3}$. Hence $a=1$, else $a \geq 2^{d-3}-1$
and $a^3 > 2^d$. But then $b=2^d-1$, and (since $d>5$) we have reached a
contradiction because $b^3+a$ is strictly between $2^{3d-1}$ and $2^{3d}$.
QED
P.S. For the record:
@ $a,b$ are odd because if a power of $2$ is the sum of two positive integers
then their 2-valuations are equal, and this would not be possible with
$a,b$ even.
@ One proof of the implication
$a^8 \equiv 1 \bmod 2^d \Longrightarrow a \equiv \pm1 \bmod 2^{d-3}$ is to
write $\pm a = 1 + 4n$ and expand $(a^8-1)/32$ as a polynomial in $n$.
| {
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An inequality, which is supposed to be simple Let $x,y,z\in\mathbb{R}$.Let $xy+yz+xz=1$.
Prove:$\displaystyle \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\leq \frac{3}{2}$
| Let $a,b,c\in(0,\pi)$ be such that $x=\cot a$, $y=\cot b$, $z=\cot c$.
Using the addition formula for cotangent, one can show that
$$ \cot(a+b+c)
= \frac{\cot a\cot b\cot c - \cot a - \cot b - \cot c}
{\cot a\cot b + \cot b\cot c + \cot c\cot a - 1}
\tag{$\ast$} $$
By hypothesis, the denominator on the RHS is 0, so (see below) the LHS is $\infty$; since $a,b,c\in (0,\pi)$, this implies $a+b+c\in\{\pi,2\pi\}$.
Case $a+b+c=\pi$: Then at most one of $a,b,c$ is greater than $\frac\pi2$; wlog, $a,b\in(0,\frac\pi2]$. Since cosine is concave on that interval, we have
\begin{align*}
\frac{x}{\sqrt{x^2+1}} + \frac{y}{\sqrt{y^2+1}} + \frac{z}{\sqrt{z^2+1}}
&= \cos a + \cos b + \cos c \\
&\le 2\cos(\tfrac{a+b}{2}) + \cos c \\
&= 2\cos(\tfrac{a+b}{2}) - \cos(a+b) \\
&= 2\cos(\tfrac{a+b}{2}) - 2\cos^2(\tfrac{a+b}{2}) + 1 \\
&= \tfrac32 - 2\big(\cos(\tfrac{a+b}{2}) - \tfrac12\big)^2 \\
&\le \tfrac32
\end{align*}
with equality iff $a=b=\frac\pi3$ (whence $c=\frac\pi3$ also), that is, $x=y=z=\frac1{\sqrt3}$.
Case $a+b+c=2\pi$: Then at most one of $a,b,c$ is less than $\frac\pi2$; wlog, $a,b\in[\frac\pi2,\pi)$. Since cosine is nonpositive in that interval, we have
$$ \frac{x}{\sqrt{x^2+1}} + \frac{y}{\sqrt{y^2+1}} + \frac{z}{\sqrt{z^2+1}}
= \cos a + \cos b + \cos c
\le \cos c
\le 1
< \tfrac32
$$
Now, one annoying detail about ($\ast$): I noted that the denominator on the RHS is zero, and inferred that the RHS, and hence the LHS, is $\infty$. But what if the numerator on the RHS is also zero? Well, suppose for contradiction that both numerator and denominator are zero. Then $xyz=x+y+z$ and $xy+yz+zx=1$, and so
\begin{align*}
(t-x)(t-y)(t-z)
&= t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz \\
&= t^3 - xyzt^2 + t - xyz \\
&= (t^2+1)(t-xyz)
\end{align*}
which is impossible because the first polynomial has three real roots (counting multiplicity) but the last has only one.
| {
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Proof of: if $x^2+y^2=2xy$ then $x=y$ I am trying to prove $x^2+y^2=2xy$ then $x=y$
What I have done is suppose $x^2+y^2=2xy$
then $x^2+y^2+(-2xy)=0\iff
x^2+(-xy)+(-xy)+y^2=0
\iff (x+(-y))\cdot x+(x+(-y))\cdot-y=0
\iff (x+(-y))^2=0$
i then square root both sides but i'm not sure if that's mathematically correct?
which gives me: $(x+(-y))=0$
so $x=y$
| This isn't true: Consider $$x = \cos \frac{\pi}{8}, \qquad y = \sin \frac{\pi}{8}. \qquad(\ast)$$ Then, substituting gives
$$x^2 - y^2 = \left(\cos \frac{\pi}{8}\right)^2 - \left(\sin \frac{\pi}{8}\right)^2 = \cos\left[2 \left( \cos \frac{\pi}{8}\right)\right] = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}.$$ On the other hand,
$$2xy = 2 \left(\sin \frac{\pi}{8}\right) \left(\cos \frac{\pi}{8}\right) = \sin\left[2\left(\frac{\pi}{8}\right)\right] = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}.$$
So, we have $x^2 - y^2 = 2xy$ but, since $\cos \theta > \sin \theta$ on $\left[0, \frac{\pi}{4}\right) \ni \frac{\pi}{8}$, we have $x \neq y$.
Note that the original equation just says that
$$\Re(z^2) = \Im(z^2)$$
for some complex number $z := x + iy$, so, we must have $\arg (z^2) = \frac{\pi}{4} + \pi k$ for some integer $k$ (or $z = 0$). Thus, the solutions $(x, y)$ to the given equation correspond exactly to the points $z := x + iy$ with $\arg z = \frac{\pi}{8} + \pi k$ for some integer $k$ (and $(0, 0)$), and so $(\ast)$ is actually the unique solution up to a common scaling of $x, y$.
Edit The condition has been changed by OP to $$x^2 \color{red}{+} y^2 = 2xy,$$ and handling this is much easier: Rearranging gives $$0 = x^2 - 2xy + y^2 = (x - y)^2,$$
so $x - y = 0$ and thus $x = y$ as desired.
| {
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Boolean Simplification of $(\overline{a+b+c})+a\cdot(b+ \overline{c})$ I'm lost, when checking my answer via truth tables, my simplified form does not match the original equation. My work, with reasoning step by step is below. Can you help me figure out where I'm wrong, and what it should be?
$(\overline{a+b+c})+a\cdot(b+ \overline{c})$ $\quad $ original equation
$(\overline{a+b+c})+a\cdot b+ a \cdot \overline{c}$ $\quad$ Distribution
$(\overline{a} \cdot \overline{b} \cdot \overline{c}) +a\cdot b+ a \cdot \overline{c}$ $\quad$ DeMorgan
$(\overline{a} \cdot \overline{b}) \cdot \overline{c} +a\cdot b+ a \cdot \overline{c}$ $\quad$ Associativity
$(\overline{a} \cdot \overline{b}) +a\cdot b \cdot \overline{c} + a \cdot \overline{c}$ $\quad$ Commutativity
$\overline{c} + a \cdot \overline{c}$ $\quad $ Complement
$\overline{c}$ $\quad $ Absorption
| Your application of what you call "commutativity" is incorrect. Variables commute only over the same operations.
Instead: $$\overline{a} \cdot \overline{b} \cdot \overline{c} +a\cdot b+ a \cdot \overline{c}= \overline{a} \cdot \overline{b} \cdot \overline{c} + a\cdot (b+\overline c) $$
$$= \overline{a} \cdot \overline{b} \cdot \overline{c} + b + \overline c$$
$$ = (\overline a \cdot \overline b + 1)\cdot \overline c + b$$
$$= 1\cdot \overline c + b$$
$$ = \overline c + b$$
| {
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Define continuity for $f(x)=\arctan(2x^3)/x^2$ at $x=0$. $$f(x)=\dfrac{\arctan(2x^3)}{x^2}.$$
*
*How are we allowed to define $f(x)$ at $x=0$ for it to be continuous there?
*Find the derivative for all $x$ real numbers.
I can't see this work out since $x=0$ is not defined in the denominator.
Thanks beforehand if anyone can explain :)
| Compute the limit $$\lim_{x\to0} f(x)=\lim_{x\to0}\frac{\arctan ( 2x^3 )}{x^2}=\lim_{x\to0}\frac{2x^3}{x^2}=0.$$ Hence we define $f(0)=0$, and $f(x)$ is continuous.
Now compute the derivative. For $x\ne 0$, we have \begin{align}
& f'(x)=\frac{{( \arctan ( 2x^3 ))^{\prime }}\cdot x^2-\arctan( 2x^3)\cdot{ (x^2)^{\prime }}}{(x^2)^2}=\frac{\frac{6x^4}{1+4x^6}-2x\arctan(2x^3)}{x^4} \\
& =\frac{6}{1+4x^2}-\frac{2\arctan ( 2x^3 )}{x^3}. \tag1
\end{align}
And $$f'(0)= \lim_{x\to0} \frac{\frac{\arctan( 2x^3)}{x^2}-0}{x-0} =\lim_{x\to0} \frac{\arctan ( 2x^3t)}{x^3} =\lim_{x\to0} \frac{2x^3}{x^3}=2.$$
Or we can simply let $x$ tends to zero in $(1)$.
| {
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Assume that 495 divides the integer $\overline{273x49y5}$ where $x,y \in \{0,1,2...9\}$. Find $x$ and $y$. So, I know that $495 = 5\times 9\times 11$. So then, if that's the case, then the number $\overline{273x49y5}$ must be divisible by $495$ if and only if it is divisible by $5$ and $9$ and $11$.
Then, I know that $5$ divides it for any number $x$ and $y$ because it only has to divide the digits number which is $5$.
To see if it divides by $9$, the sum of the digits must be divisible by $9$, and as for $11$, the alternating sum of digits must be divisible by $11$.
So, to test it by $9$, then, $2+7+3+x+4+9+y+5 = 30+x+y$ has be divisble by $9$. So, the only numbers $x$ and $y$ that make $30+x+y$ divisible by $9$ are $x= \{0,1,2,3,4,5,6\}$ and $y=\{0,1,2,3,4,5,6\}$.
But $11$ is what gets me. To make it divisible by $11$, then the alternating sums must be divisible by $11$. So then, $-2+7-3+x-4+9-y+5= 12+x-y$ must be divided by $11$. Then, the only numbers $x$ and $y$ that make it divisible by $11$ must be $x=\{0,1,2,\ldots,8\}$ and $y=\{1,2,\ldots,9\}$, if I did my math right.
Next, I assume we must then find numbers in $x$ and $y$ that satisfy the division by $9$ and $11$. So, if that is the case, then I know that since the number must be divisible by $9$ and $11$, then the numbers $x$ and $y$ must fit between $0$ to $6$ because that's $6$ is the highest number $x$ or $y$ can be in the sequence to divide the number. So, how do I check to see which numbers both satisfy division of $9$ and $11$?
| Notice $495 = 5\times 99$. It will be easier to look at everything modulo $99$.
$$0 \equiv \overline{273x49y5}
\equiv 27 + \overline{3x} + 49 + \overline{y5}
\equiv \overline{yx} + 111
\equiv \overline{yx} + 12 \pmod{99}
$$
This leads to $\overline{yx} = 87 \implies (x,y) = (7,8)$.
| {
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Cool little system of equations. Solving the system of equations for integers:
$2^a+3^b=5^b$
$3^a+6^b=9^b$
How is it done? I tried substituting the $2^a$ from the first equation into the second, and dividing the two equations by $2^a$.
| The second equation gives you $b\le a\lt2b$, because $3^b+6^b\le9^b\lt3^{2b}+6^b$ when $b\gt0$ (and the equations obviously have no solutions when $b\le0$).
By diving it by $3^b$, you get $3^{a-b}+2^b=3^b$.
It means $2^b=3^b-3^{a-b}=3^{a-b}(3^{2b-a}-1)$.
By unicity of decomposition in prime factors, we must have $a-b=0$, so $a=b$, and $2^b=3^b-1$.
This is only possible if $a=b=1$. Then the two equations are verified. Notice that the first one is not needed to find the necessary condition $a=b=1$.
| {
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Solve $\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$
Find the set of integer solutions $(x,y)$ to
$$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$$
I can't seem to multiply both sides by the expression in the denominator.
Nor can I simplify and cancel any terms. How should I it?
| Hint
$$\frac{x^2+2xy+y^2}{x^2-y^2}=\frac{(x+y)^2}{(x+y)(x-y)}=\frac{x+y}{x-y}$$
Edit
So, the initial inequality is equivalent to
$$\frac{x+y}{x-y}>x+y.$$
If $x+y>0$ it is
$$\frac{1}{x-y}>1,$$ which has no solution. (If $x-y<0$ then $\frac{1}{x-y}<0$ can't be greater than $1.$ If $x-y>0$ then $\frac{1}{x-y}\le 1$ can't be greater than $1.$)
Thus, it must be $x+y<0.$ In such a case, the inequality is equivalent to
$$\frac{1}{x-y}<1,$$ which is satisfied if $x-y<0$ or $x-y>1.$
So we have the regions $x+y<0, x-y<0$ and $x+y<0,x-y>1.$
Thus, the solution is $$\{(x,y)\in \mathbb{Z}^2:x< y<-x \:\mathrm{or} \: y+1<x<-y\}.$$
(Remark for readers just coming to this post: All the intial comments below this answer, down to the one with a little boldface, pertain to earlier versions of the answer. There's no need to read them.)
| {
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Remainder of $3^7/8$ I read here that the remainder of $\frac{ab}{c}$ is equal to the remainder of $\frac{a}{c}\frac{b}{c}$ implying that the remainder of $\frac{a^b}{c}$ is equal to the remainder of $[\frac{a}{c}]^b$. However, when I apply this here, I would get remainder of $[\frac{3}{8}]^7=3^7$ (while the correct answer is $3$). What am I doing wrong here? And if this method is not valid, are there other quick and simple methods of finding remainders like this?
| You may use congruences:
$$3^2=9\equiv1\pmod{8}$$
So
$$3^7=(3^2)^3\times 3\equiv 1\times 3\equiv3\pmod{8}$$
Remember that $a\equiv b\pmod{m}$ means that $a$ and $b$ have same remainder when divided by $m$. It's also equivalent to $m|a-b$ (you read it $m$ divides $a-b$).
And with congruences you can write that if $a\equiv b \pmod{m}$ and $c\equiv d \pmod{m}$, then, among others:
$$ac\equiv bd \pmod{m}$$
$$a+c\equiv b+d \pmod{m}$$
| {
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Factoring the quartic polynomial $x^4-2x^2+1$ Okay, I am practicing factoring for an upcoming assignment and I know that this is basic algebra, but I forgot how to attack this polynomial. Every method that I have used so far from simply guessing to using the quadratic formula to long division has failed me in replicating the answer. So either I am attacking this problem wrong up or there is another method that I just forgot.
So here it is: $x^4-2x^2+1=0\\$
The 4 roots of the polynomial are $\pm 1,\pm 1$ (it only has 2 distinct roots).
My confusion is in how to get those values.
Any help in deriving the solution would be greatly appreciated.
| $x^4-2x^2+1=0$ clearly has a root of $1$. Since $\dfrac{x^4-2x^2+1}{x-1}= x^3+x^2-x-1$ you now have $$(x-1)(x^3+x^2-x-1)=0.$$
$x^3+x^2-x-1=0$ clearly has a root of $1$. Since $\dfrac{x^3+x^2-x-1}{x-1}= x^2+2x+1$ you now have $$(x-1)(x-1)(x^2+2x+1)=0.$$
And so on.
| {
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Given $\sin{A}:\sin{B}:\sin{C}$ how to find $\sin{(2A)}:\sin{(2B)}:\sin{(2C)}$? Question:
Given $x,y,z>0$, and the positive number $k$ such
$$\dfrac{x^2}{x^2+k}+\dfrac{y^2}{y^2+k}+\dfrac{z^2}{z^2+k}=1$$
in $\Delta ABC$,
$$(1):\sin{A}:\sin{B}:\sin{C}=\dfrac{x}{x^2+k}:\dfrac{y}{y^2+k}:\dfrac{z}{z^2+k}$$
$$(2): \sin{(2A)}:\sin{(2B)}:\sin{(2C)}=\dfrac{1}{x^2+k}:\dfrac{1}{y^2+k}:\dfrac{1}{z^2+k}$$
show that: $$(1)\Longrightarrow (2)$$
this problem is from I write a paper, and this result seems interesting.
My idea:
and if $x=3,y=4,z=18$,then $$\dfrac{9}{9+k}+\dfrac{16}{16+k}+\dfrac{324}{324+k}=1,k>0$$
then
$$k=96$$see:equation
then $$\sin{A}:\sin{B}:\sin{C}=4:5:6$$
see:2
so use sine therom
$$a:b:c=4:5:6$$
let $$a=4t,b=5t,c=6t$$
so
$$\cos{A}=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{3}{4},\cos{B}=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{9}{16},\cos{C}=\dfrac{a^2+b^2-c^2}{2ac}=\dfrac{1}{8}$$
so
$$\cos{A}:\cos{B}:\cos{C}=12:9:2$$
so
$$\sin{2A}:\sin{2B}:\sin{2C}=(4\cdot 12):(5\cdot 9):(6\cdot 2)=16:15:4$$
and other hand
$$\dfrac{1}{x^2+k}:\dfrac{1}{y^2+k}:\dfrac{1}{z^2+k}=\dfrac{1}{9+96}:\dfrac{1}{16+96}:\dfrac{1}{324+96}=16:15:4$$
see 3
second example:
let $x=\sqrt{2},y=\sqrt{5},z=\sqrt{10}$,then
we easy to find $k=10$ see 4,
and
$$\sin{A}:\sin{B}:\sin{C}=\dfrac{\sqrt{2}}{12}:\dfrac{\sqrt{5}}{15}:\dfrac{\sqrt{10}}{20}=5\sqrt{2}:4\sqrt{5}:3\sqrt{10}$$
so
$$\cos{B}=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{50+90-80}{2\times 5\sqrt{2}\times 3\sqrt{10}}=\dfrac{1}{\sqrt{5}}$$
$$\cos{A}=\dfrac{1}{\sqrt{2}},\cos{C}=\dfrac{1}{\sqrt{10}}$$
so
$$\sin{2A}:\sin{2B}:\sin{2C}=5:4:3$$
other hand
$$\sin{2A}:\sin{2B}:\sin{2C}=\dfrac{1}{x^2+k}:\dfrac{1}{y^2+k}:\dfrac{1}{z^2+k}=\dfrac{1}{12}:\dfrac{1}{15}
:\dfrac{1}{20}=5:4:3$$
so this case is also true
But I can't prove this.can you help?
| Let
$$p = \frac{x^2}{x^2 + k}, \quad q = \frac{y^2}{y^2 + k}, \quad r = \frac{z^2}{z^2 + k}.$$
Then by assumption $p + q + r = 1$.
Letting $a$, $b$ and $c$ be the sides of the triangle, we have by the sine law
$$\begin{align*}
a^2 : b^2 : c^2 &= \sin^2 A : \sin^2 B : \sin^2 C \\
&= \frac{x^2}{(x^2 + k)^2} : \frac{y^2}{(y^2 + k)^2} : \frac{z^2}{(z^2 + k)^2} \\
&= \frac{x^2}{x^2 + k}\frac{k}{x^2 + k} : \frac{y^2}{y^2 + k}\frac{k}{y^2 + k} : \frac{z^2}{z^2 + k}\frac{k}{z^2 + k} \\
&= p(1-p):q(1-q):r(1-r) \\
&= p(q + r):q(p + r): r(p + q).
\end{align*}
$$
Now by the sine and cosine laws, we have (using the symbol $\propto$ to mean "is proportional to"),
$$
\begin{align*}
\sin 2A &= 2\sin A \cos A \\
&\propto a\frac{b^2 + c^2 - a^2}{2bc} \\
&\propto a^2(b^2 + c^2 - a^2) \\
&\propto p(q+r)[q(p+r) + r(p+q) - p(q + r)] \\
&=2pqr(q+r) \\
&\propto q + r \\
&= 1 - p \\
&= \frac{k}{x^2 + k} \\
&\propto \frac{1}{x^2 + k}.
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Ellipse $3x^2-x+6xy-3y+5y^2=0$: what are the semi-major and semi-minor axes, displacement of centre, and angle of incline? Given the ellipse
$$3x^2-x+6xy-3y+5y^2=0$$
find the following:
*
*semi-major axis, $a$
*semi-minor axis, $b$
*displacement of centre from origin (or coordinates of centre of ellipse $(h,k)$)
*angle of incline $\theta$ (or $\tan\theta$).
This can of course be done systematically by equating coefficients with
$$\frac{((x-h)\cos\theta-(y-k)\sin\theta)^2}{a^2}+\frac{((x-h)\sin\theta+(y-k)\cos\theta)^2}{b^2}=1$$
which is derived by applying the rotation and translation parameters to the standard ellipse equation. Perhaps someone could complete the solution?
Also, can it be solved in any other way, and in particular, without using trigonometric
ratios?
(NB - this problem came about whilst trying to solve another problem on MSE here
How to Solve this Arithmetic Progression Question? which gave rise to the Diophantine Equation above; assuming instead that $x,y$ can take on non-integer values the equation describes a rotated and translated ellipse)
| Here's another method suggested by a friend of mine.
Keep the solution of the centre of the ellipse identified earlier as $(-\frac13,\frac12)$ .
Change the centre of the ellipse such to the origin. Equation of translated ellipse is:
$$E':\qquad 3x^2+6xy+5y^2-\frac7{12}=0\qquad \cdots (1)$$
Consider a circle at the same centre as the ellipse (i.e. the origin) with equation
$$C': \qquad x^2+y^2=r^2\qquad \qquad \qquad \qquad \cdots (2) $$
At intersection of $E'$ and $C$:
$$(1)-3\times(2):\\
6xy+2y^2=\frac7{12}-3r^2$$
Substitute in $(1)$:
$$\begin{align}
6\sqrt{r^2-y^2}y&=-\left[2y^2+\left((3r^2-\frac7{12}\right)\right]\\
36(r^2-y^2)y^2&=4y^4+4\left(3r^2-\frac7{12}\right)y^2+\left(3r^2-\frac7{12}\right)^2\\
40y^4+4\left(-6r^2-\frac7{12}\right)y^2+\left(3r^2-\frac7{12}\right)^2&=0
\end{align}$$
For tangency,
$$\begin{align}
4^2\left(6r^2+\frac7{12}\right)^2&=4(40)\left(3r^2-\frac7{12}\right)^2\\
6r^2+\frac7{12}&=\sqrt{10}\left(3r^2-\frac7{12}\right)\\
3r^2(2-\sqrt{10})&=\frac7{12}(-1-\sqrt{10})\\
r^2&=\frac7{36}\frac{\sqrt{10}+1}{\sqrt{10}-2}=\frac7{72}(4+\sqrt{10})\\
r&=\sqrt{\frac7{72}(4+\sqrt{10})}\qquad \text{(semi-major axis)}\qquad \blacksquare
\end{align}$$
Correspondingly,
$$\qquad\qquad r=\sqrt{\frac7{72}(4-\sqrt{10})}\qquad \text{(semi-minor axis)}\qquad \blacksquare$$
The incline of the ellipse can be calculated easily from values of $x,y$ obtained by substituting values of $r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
How to prove $\int_0^{2\pi} \ln(1+a^2+2a\cos x)\, dx=0$? How can I prove $\int_0^{2\pi} \ln(1+a^2+2a\cos x)\, dx=0$, where $a<1$?
Thanks.
| Let $a$ be a real number.
*
*$\color{blue}{\text{Case 1.}}$ $\quad|a|<1$
$$
\int_0^{2\pi}\log \left(1+2a\cos x+ a^2\right){\rm d}x=0
$$
Observe that
$$
\left(1+ae^{ix}\right)\left(1+ae^{-ix}\right)=1+2a\cos x+ a^2, \quad x \in [0,2\pi],
$$
and that
$$
\begin{align}
\log \left(1+2a\cos x+ a^2\right)
&=\log \left(1+ae^{ix}\right)+\log \left(1+ae^{-ix}\right)\\\\
&=-\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n e^{inx}-\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n e^{-inx}\\\\
&=-2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n \cos (nx)\\\\
\end{align}
$$
Using the normal convergence of the series $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n \cos (nx) $ as a function of $x \in [0,2\pi]$,
$$
\left|\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n \cos (nx)\right|\leq \sum_{n=1}^{\infty}\frac{|a|^n }{n}=-\log(1- |a|)<\infty,
$$
we are allowed to perform a termwise integration giving
$$
\begin{align}
\int_0^{2\pi}\log \left(1+2a\cos x+ a^2\right) {\rm d}x
=-2\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}a^n \int_0^{2\pi}\cos (nx) dx=0
\end{align}$$
due to
$$
\begin{align}
\int_0^{2\pi}\cos (nx) {\rm d}x=\left. \frac{\sin (nx)}{n}\right|_0^{2\pi}=0, \quad n=1,2,3,\ldots.
\end{align}$$
*
*$\color{blue}{\text{Case 2.}}$ $\quad|a|>1$
$$
\int_0^{2\pi}\log \left(1+2a\cos x+ a^2\right){\rm d}x=4\pi \log |a|
$$
Observe that
$$
\log \left(1+2a\cos x+ a^2\right)=2\log |a| +\log \left(1+2\cdot\frac1a\cos x+ \frac{1}{a^2}\right), \quad x \in [0,2\pi],
$$
then
$$
\int_0^{2\pi}\log \left(1+2a\cos x+ a^2\right){\rm d}x=2\log |a|\int_0^{2\pi} {\rm d}x+\int_0^{2\pi}\log \left(1+2\cdot\frac1a\cos x+ \frac{1}{a^2}\right){\rm d}x,
$$ and, applying the previous case to the last integral, we get the desired result.
Remark 1.
Here $\displaystyle \log (z)$ denotes the principal value of the logarithm defined for $z \neq 0$ by
$$ \begin{align}
\displaystyle \log (z) = \ln |z| + i \: \mathrm{arg}z, \quad -\pi <\mathrm{arg} z \leq \pi.
\end{align}
$$
Remark 2.
$$
\begin{align}
& a=-1 & \text{gives} \quad &\int_0^{2\pi}\log \left(2-2\cos x\right){\rm d}x=4\int_0^{\pi}\log \left(2\sin u\right){\rm d}u=0\\
& a=1 & \text{gives} \quad &\int_0^{2\pi}\log \left(2+2\cos x\right){\rm d}x=2\int_0^{\pi}\log (4\cos^2 u){\rm d}u=0.
\end{align}
$$
Remark 3. One may readily notice that the preceding reasoning gives
$$
\begin{align} \int_0^{\pi}\log \left(1+2a\cos x+ a^2\right){\rm d}x =
\left\{
\begin{array}{ll}
0 & \mbox{if } |a| \leq 1 \\
2\pi \log |a| & \mbox{if } |a| > 1
\end{array}
\right.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
The minimum value of $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$ Problem :
The minimum value of $$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$$
Can I use this in numerator and denominator :
The minimum value of $a +\frac{1}{a}$
Using A.M and G.M inequality :
$a +\frac{1}{a} \geq 2\sqrt{a \times \frac{1}{a}}$
$\Rightarrow a +\frac{1}{a} \geq 2$ .....(1)
By putting the minimum value of (1) in $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$
we get ; $\frac{2^6-2-2}{2^3+2}$ but I think this is wrong especially denominator as we need to find the maximum value of denominator to get the minimum value.
Please suggest ,thanks.
| Here is one way to go about it. We want to express$\left(x^6+\frac1{x^6}\right)$ in terms of $\left(x+\frac1x\right)$.
$$\left(x+\frac1x\right)^6={x}^{6}+6{x}^{4}+15{x}^{2}+20+\frac{15}{x^2}+\frac{6}{x^4}+\frac1{x^6}$$
We can remove the $x^4$ and $\frac1{x^4}$ terms by subtracting $6\left( x+\frac1x\right)^4$, which gives
$$\left(x+\frac1x\right)^6-6\left( x+\frac1x\right)^4={x}^{6}-9{x}^{2}-16-\frac{9}{x^2}+\frac1{x^6}$$
Continuing this process and using the substitution $u=x+\frac1x$ leads to
$$x^6+\frac1{x^6}=\left(x+\frac1x\right)^6-6\left(x+\frac1x\right)^4+9\left(x+\frac1x\right)^2-2=u^6-6u^4+9u^2-2$$
and similarly
$$x^3+\frac1{x^3}=\left(x+\frac1x\right)^3-3\left(x+\frac1x\right)=u^3-3u$$
Now you have the function
$$f(x)=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}=\frac{u^6-(u^6-6u^4+9u^2-2)-2}{u^3+(u^3-3u)}$$
$$=\frac{6u^4-9u^2}{2u^3-3u}=3u=3\left(x+\frac1x\right)$$
Since $x+\frac1x\geq2$, the minimum value of $f(x)$ is thus $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How do I know if two vectors with $n$ components are parallel? How do I know if two vectors with $n$ components are parallel?
For example
$$\begin{pmatrix}5\\2\\1\\3\\4\end{pmatrix} \text{, and } \begin{pmatrix}4\\1\\2\\3\\6\end{pmatrix}.$$
| Check to see if they are constant multiples of one another. In your example, if there were a $c$ such that
$$c\begin{pmatrix}
5\\2\\1\\3\\4
\end{pmatrix}=
\begin{pmatrix}
4\\1\\2\\3\\6
\end{pmatrix}$$
then by equating the first components, we see that $c$ would have to be $\frac{4}{5}$. However, when we check the second coordinate, $\frac{4}{5}$ doesn't work, since $\frac{4}{5}\cdot2\neq1$. This proves that no such $c$ exists. Since they are not constant multiples of one another, they are not parallel.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/998001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd. Prove: If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd.
I know that the $n$th row in pascal's triangle correspond to the coefficients of $(x+y)^n$:
$$\begin{align}(x+y)^n = {n \choose 0} x^n + {n \choose 1} x^{n-1} y + {n \choose 2}x^{n-2}y^2+{n \choose 3} x^{n-3} + \cdots + {n \choose n-1} xy^{n-1}+{n \choose n} y^n \end{align}.$$
I have to show that each coefficient $\begin{align} {n \choose 0},{n \choose 1},{n \choose 2},\ldots,{n \choose n-1},{n \choose n} \end{align}$ is of the form $2b+1$, but I'm not sure how to do this with $n =2^k-1$.
| Counting the Number of Factors of $\boldsymbol{2}$
In Corollary $(7)$ of this answer, it is shown that the number of factors of $p$ in $\binom{n}{k}$ is
$$
\frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}\tag{1}
$$
where $\sigma_p(n)$ us the sum of the digits in the base-$p$ representation of $n$.
In base-$2$, each digit of $n=2^k-1$ is a $1$. For $0\le k\le n$, each digit of $k$ that is a $0$ is a $1$ in $n-k$ and vice-versa. Thus, $\sigma_p(k)+\sigma_p(n-k)=\sigma_p(n)$. Thus, formula $(1)$ says that there are $0$ factors of $2$ in $\binom{n}{k}$.
That is, $\binom{n}{k}$ is odd.
Induction
We will show that for $n\ge1$,
$$
\binom{2^n}{k}\text{ is even for }0\lt k\lt2^n\tag{2}
$$
and of course $\binom{2^n}{0}=\binom{2^n}{2^n}=1$.
Note that $(2)$ is true for $n=1$. Suppose that $(2)$ is true for some $n$. Then we can use Vandermonde's Identity to get
$$
\binom{2^{n+1}}{k}=\sum_{j=0}^k\binom{2^n}{j}\binom{2^n}{k-j}\tag{3}
$$
where inductively, $\binom{2^n}{j}\binom{2^n}{k-j}$ is even except when either $j=0$ or $j=2^n$ and either $k-j=0$ or $k-j=2^n$. That gives $4$ cases:
*
*$j=k=0$: makes $\binom{2^{n+1}}{0}$ odd
*$j=0$ and $k=2^n$: paired with case 3, makes $\binom{2^{n+1}}{2^n}$ even
*$j=2^n$ and $k=2^n$: paired with case 2, makes $\binom{2^{n+1}}{2^n}$ even
*$j=2^n$ and $k=2^{n+1}$: makes $\binom{2^{n+1}}{2^{n+1}}$ odd
Thus, accounting for these $4$ special cases, $(3)$ shows that $(2)$ holds for $n+1$.
Since $\binom{2^n-1}{0}=1$ and for $0\lt k\lt2^n$
$$
\binom{2^n-1}{k}=\binom{2^n}{k}-\binom{2^n-1}{k-1}\tag{4}
$$
we get that $\binom{2^n-1}{k}$ is odd (even minus odd) for $0\le k\le2^n-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Tough inequality problem, splitting into cases Find the set of all $x$ for which $\dfrac{2x}{2x^2 + 5x + 2} > \dfrac{1}{x+1}$.
I am new to solving such problems and am not able to appropriately break up the problem into cases, so any help would be thoroughly appreciated.
| $$\dfrac{2x}{2x^2+5x+2}>\dfrac{1}{x+1}\iff\frac{2x}{2x^2+5x+2}-\frac{1}{x+1}>0 \\ \iff\frac{2x(x+1)-(2x^2+5x+2)}{(2x^2+5x+2)(x+1)}>0 \\ \iff-\frac{3x+2}{(2x+1)(x+2)(x+1)}>0 \\ \iff \frac{3x+2}{(2x+1)(x+2)(x+1)}<0$$
We can see that the numerator vanishes when $x=-\frac{2}{3}$, also the fraction is discontinuous for $x\in\{-2,-1,-\frac{1}{2}\}$. Since rational functions are continuous in their domain we must analyze what happens for $x$ in each of the intervals $(-\infty,-2),\,(-2,-1),\,(-1,-2/3),\,(-2/3,-1/2)$ and $(-1/2,\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimum of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$ What is the minimum of $$f(a,b,c):=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}$$ where $a,b,c$ are positive real numbers?
When $a=b=c$, we have $f(a,b,c)=\dfrac{3}{\sqrt{2}}\approx 2.12$
When $a=1,b=c\rightarrow\infty$, we have $f(a,b,c)\rightarrow 2$. So the minimum is at most $2$.
| If $a=b=1$ and $c\rightarrow0^+$ we get $f(a,b,c)\rightarrow2$.
We'll prove that
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq2.$$
Indeed, by AM-GM
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}=\sum_{cyc}\frac{2a}{2\sqrt{a(b+c)}}\geq\sum_{cyc}\frac{2a}{a+b+c}=2.$$
Done!
Also we can use Holder:
$$\left( \sum_{cyc}\sqrt{\frac{a}{b+c}}\right)^2\sum_{cyc}a^2(b+c)\geq(a+b+c)^3.$$
Thus, it remains to prove that
$$(a+b+c)^3\geq4\sum_{cyc}(a^2b+a^2c)$$ or
$$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$
which follows from Schur.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Poisson Integral is equal to 1
Show
$$
\int_{-\pi}^{\pi}P(r, \theta)d\theta = 1
$$
Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\big(\frac{r^2 + 1}{2r}\big)$.
Then
$$
\frac{1}{2\pi}
\int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}d\theta
= \frac{\alpha}{2\pi}
\int_{-\pi}^{\pi}\frac{1}{\cos(\theta) + \gamma}d\theta
$$
where
$$
\frac{r^2 - 1}{2r}\frac{1}{\cos(\theta) - \frac{1}{2r} - \frac{r^2}{2r}} = \frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}
$$
Next, let $z = e^{i\theta}$. Then $d\theta = \frac{-i}{z}dz$. Since $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$, $\cos(\theta) = \frac{z + z^{-1}}{2}$.
$$
= \frac{-i\alpha}{\pi}\int_C\frac{1}{z^2 + 2z\gamma + 1}dz
$$
and $C$ is the contour oriented counter clockwise with simple poles at $z = -\gamma\pm\sqrt{\gamma^2 - 1}$. Let $f(z) = \frac{1}{z^2 + 2z\gamma + 1}$. Then
$$
= \Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z)
\tag{1}
$$
The only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$.
Then
$$
2\pi i\lim_{z\to z_j}\Bigg[\big(z + \gamma - \sqrt{\gamma^2 - 1}\big)
\frac{1}{\big(z_j + \gamma + \sqrt{\gamma^2 - 1}\big)
\big(z + \gamma - \sqrt{\gamma^2 - 1}\big)}\Bigg] =
\frac{\pi i}{\sqrt{\gamma^2 - 1}}
$$
Now, we can substitute $\frac{\pi i}{\sqrt{\gamma^2 - 1}}$ for $2\pi i\sum\text{Res}$ in equation (1).
\begin{align*}
\Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z)
&= \frac{\alpha}{\sqrt{\gamma^2 - 1}}\\
&= \frac{r^2 - 1}{2r\sqrt{\frac{(r^2 + 1)^2}{4r^2} - 1}}\\
&= \frac{r^2 - 1}{\sqrt{(r^2 + 1)^2 - 4r^2}}\\
&= \frac{r^2 - 1}{\sqrt{r^4 - 2r^2 + 1}}\\
&= \frac{(r - 1)(1 + r)}{(r - 1)(r + 1)}\\
&= 1
\end{align*}
I have been unable to convince myself that the only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$. I know it is the case because if I use the other pole, the integral becomes $-1$
| For $0 \le r < 1$,
$$
\begin{align}
\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}
& =\sum_{n=0}^{\infty}r^{n}e^{in\theta}+\sum_{n=1}^{\infty}r^{n}e^{-in\theta} \\
& =\frac{1}{1-re^{i\theta}}+\left(\frac{1}{1-re^{-i\theta}}-1\right) \\
& = \frac{1}{1-re^{i\theta}}+\frac{re^{-i\theta}}{1-re^{-i\theta}} \\
& = \frac{1-re^{-i\theta}+re^{-i\theta}-r^{2}}{(1-re^{i\theta})(1-re^{-i\theta})} \\
& = \frac{1-r^{2}}{1-2r\cos\theta+r^{2}}.
\end{align}
$$
Therefore, because $\int_{-\pi}^{\pi}e^{in\theta}\,d\theta=0$ for $n \ne 0$, the following holds for $0 \le r < 1$:
$$
\int_{-\pi}^{\pi}\frac{1-r^{2}}{1-2r\cos\theta+r^{2}}\,d\theta= \sum_{n=-\infty}^{\infty}r^{|n|}\int_{-\pi}^{\pi}e^{in\theta}\,d\theta = \int_{-\pi}^{\pi}d\theta=2\pi.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
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.