Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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The given matrix has three linearly independent eigenvectors, then $x+y=0$. The question asked is ::
If the matrix
$$
A=\left(\begin{array}{lll}
0 & 0 & 1 \\
x & 1 & y \\
1 & 0 & 0
\end{array}\right)
$$
has three linearly independent eigenvectors, then show that $x+y=0$.
solving for eigenvalues from the characteristic polynomial:
$$\left|\begin{matrix}
0-\lambda & 0 & 1 \\
x & 1-\lambda & y \\
1 & 0 & 0-\lambda
\end{matrix}\right| =-λ^3+λ^2+λ-1$$
$$=-(λ-1)*(λ^2-1)=-(λ-1)*(λ-1)=-(λ-1)^2*(λ+1)$$
So eigenvalues are $λ_1=1$ and $λ_2=-1$, Independent of the values of $x$ and $y$.
Now solving for eigenvectors I got
$\left(\begin{matrix}
0 \\
1 \\
0
\end{matrix}\right)$ and $\left(\begin{matrix}
-1 \\
\frac{x-y}{2} \\
1
\end{matrix}\right)$
From here how to show that if there are three linearly independent eigenvectors, then show that $x+y=0$.
| Your eigenvectors are correct, however, there is a possibility that there are $1$ or $2$ eigenvectors corresponding to the eigenvalue $\lambda = 1$, depending on the values of $x$ and $y$.
Let us find the missing one. We need to solve the homogeneous system $(A-I)X = 0$:
$$\begin{pmatrix}-1&0&1\\x&0&y\\1&0&-1\end{pmatrix}\sim \begin{pmatrix}1&0&-1\\0&0&0\\0&0&x+y\end{pmatrix}.$$
There are $2$ cases, $x+y \neq 0$ or $x+y= 0$.
If $x+y\neq 0$, the system is of rank $2$, and therefore $(0, 1, 0)$ is the only solution (up to scalar).
If $x+y = 0$, the system is of rank $1$, and therefore we have $2$ linearly independent solutions: $(0,1,0)$ and $(1,0,1)$.
Note that the converses are true as well.
If we count the third eigenvector, corresponding to $\lambda = -1$, that you found, we can sum up all of this in the following way:
*
*$A$ has exactly $2$ linearly independent eigenvectors if and only if $A-I$ is of rank $2$, if and only if $x+y\neq 0$.
*$A$ has exactly $3$ linearly independent eigenvectors if and only if $A-I$ is of rank $1$, if and only if $x+y= 0$.
| {
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"url": "https://math.stackexchange.com/questions/3861757",
"timestamp": "2023-03-29T00:00:00",
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How to compute the limit as $x\to 3$ of a $\textit{complicated}$ product and quotient of trigonometric functions $$\lim_{x\rightarrow 3}\frac{ \tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}$$
I substituted $x-3$ for $u$ and got as far as
$$\frac{1}{6} \lim_ {u\to 0} \frac{\sin(9 \sin u)}{\sin((u+3)^(3) -27)}.$$
This is where I get stuck. Should I try a different approach?
| $$\lim_{x\rightarrow 3}\frac{ \tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}= \lim_{x\rightarrow 3}\frac{ \frac{\tan\frac{x-3}{x+3}-0}{x-3}\frac{\sin(9\sin(x-3))-0}{x-3} }{ \frac{\sin(x-3)-0}{x-3}\frac{\sin(x^3-27)-0}{x-3}}$$
Then use the definition
$$f'(3) = \lim_{x\rightarrow 3} \frac{f(x)-f(3)}{x-3}$$
| {
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Proving $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$ using definition I got a question regarding my answer of proving limit using epsilon-delta, here's the question
Prove $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$
Here's the answer I've come up so far
let $f(x) = \frac{x+1}{x-2} + x$
by algebra manipulation, we get
$|f(x) - (-1)| = |\frac{x+1}{x-2} + x +1| $ $=|\frac{x^2 - 1}{x-2}|$ $=|\frac{(x-1)(x+1)}{x-2}|$
let $|x-1| < 1$, by triangle inequality we get $|x| < 2$, then
$|x + 1| < 3$ and $|x - 2| < 1$
now, using the definiton of limit,
for every $\epsilon > 0$, there exist $\delta = min\{1, \frac{\epsilon}{3}\}$ such that
if $0 < |x - 1| < \delta$ then,
$|f(x) - (-1)| = |\frac{x+1}{x-2} + x +1|$ $=|\frac{(x-1)(x+1)}{x-2}|$ $=|\frac{1 \cdot 3}{1}|$ $< \epsilon$
Is this correct? honestly I'm not sure on getting the upper bound of $|x-2|$, so I used the assumption of $|x-1| < 1$
Any tips would help, thanks beforehand.
| Let's take $\delta<\frac{1}{2}$, then $|x-1|<\delta$ gives $|x-2|>\frac{1}{2}$ and $|x+1|=x+1<\frac{5}{2}$, so we have
$$\left|\frac{(x-1)(x+1)}{x-2}\right|<5\delta <\varepsilon$$
| {
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Minimum value of magnitude of complex expression
If $|2z_1 + \bar z_2| = 2\sqrt2$ and $|1 + 2z_1z_2 | = 3 $ then minimum value of $(|z_1|^2 + 4|z_2|^2)$, is:-
Now
$|2z_1 + \bar z_2|^2 = 8$
$(2 z_1 + \bar z_2)(2\bar z_1 + z_2) = 8$
$4z_1\bar z_1 + 2 z_1z_2 + 2\bar z_1\bar z_2 + z_2\bar z_2 = 8$
$4|z_1|^2 + |z_2|^2 + 2 z_1z_2 + 2\bar z_1\bar z_2 = 8.......(1)$
Now
$|1 + 2z_1z_2 |^2 = 9 $
$(1 + 2z_1z_2)(1 + 2\bar z_1\bar z_2) = 9 $
$1 + 4|z_1|^2|z_2|^2 + 2z_1z_2 + 2\bar z_1\bar z_2= 9........(2)$
$(1) - (2)$
$4|z_1|^2 + |z_2|^2 = 4|z_1|^2|z_2|^2$
I am not able to make any progress after this step.
| Let $|z_1|^2 = x$, $|z_2|^2 = y$. Then we have to find the minimum value of $x+4y$. Continuing from where you left off:
$$4x + y = 4xy \Rightarrow 4xy-y = 4x \Rightarrow y = \frac{4x}{4x-1}$$
and so:
$$x+4y=x + \frac{16x}{4x-1} = x + \frac{16x-4}{4x-1} + \frac{4}{4x-1} = x+4+\frac{4}{4x-1}$$
By AM-GM, $(x-\frac{1}{4})+\frac{4}{4x-1} ≥ 2 \sqrt{(x-\frac{1}{4}) \cdot \frac{4}{4x-1}} = 2$ since $x,y ≥ 0$.
Hence, the minimum value of $x + 4y = x - \frac{1}{4} + \frac{4}{4x-1} + \frac{17}{4}$ is $2 + \frac{17}{4} = \boxed{\frac{25}{4}}$.
| {
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Multiplicative inverse of complex numbers proof I recently attempted to show that the multiplicative inverse for complex numbers exists and expressed it in complex form, as follows:
Suppose $z = a + bi$ is a non-zero complex number. Show that $z$ has a multiplicative inverse and express it in the form $c + di$.
Let $z^{-1}$ denote the multiplicative inverse of Z. Then,
$$z^{-1}z = 1 = zz^{-1}$$
$$\implies z^{-1}(a+bi) = 1 = (a+bi)z^{-1}$$
So,
$$z^{-1} = \frac{1}{a+bi}$$
Multiplying the numerator and denominator by the conjugate:
$$z^{-1} = \frac{a-bi}{a^2 + b^2}$$
$$z^{-1} = \frac{a}{a^2 + b^2} - i(\frac{b}{a^2 + b^2})$$
Thus, for all non-zero complex numbers $z$, there exists a multiplicative inverse, $z^{-1}$, where $z^{-1} = \frac{a}{a^2 + b^2} - i(\frac{b}{a^2 + b^2})$
QED.
However, I was told that this proof is circular because I assumed that the inverse exists. How can I rectify this?
Responses are much appreciated.
| there's a few things wrong. One very subtle.
First of all we have utterly no idea what a number of the form $\frac 1{a+bi}$ even means. All we did was write a $1$ put a bar underneath it and write $a+bi$ under that. We can make up rules that somehow $\frac {a+bi}{c+di}$ (whatever that means) when multiplied but $\frac {e+fi}{g+hi}$ will be equal to $\frac {(a+bi)(e+fi)}{(c+di)(g+hi)}$ but that doesn't mean anything.
We have to define that $\frac 1z$ must mean a complex number $w$ so that $z \cdot w = 1$ (assuming that there is such a number, and that it is unique; neither of which we have any reason to assume). And even if we do assume there is a $w$ so that $w(a+bi) =1$ and we write it as $w=\frac 1{a+bi}$ and if there is $v$ so that $v(c+di) = 1$ so we can write $v=\frac 1{c+di}$ we have no reason to believe that $wv = \frac 1{a+bi}\cdot \frac 1{c+di}$ that that will actually equal $\frac 1{(a+bi)(c+di)}$. (Although we can prove that.)
Anyhoo.....
So long as $z=a+bi = 0 \iff a^2 + b^2 = 0$ then the does exist a $w= \frac a{a^2 + b^2} -i\frac b{a^2 + b^2}$ and it is true that $(a+bi)(\frac a{a^2 + b^2} -i\frac b{a^2 + b^2}) = (a\cdot \frac 1{a^2+b^2} + b\frac 1{a^2 + b^2}) + i(a \frac b{a^2+b^2} - b\frac a{a^2+b^2}) = \frac {a^2 + b^2}{a^2 + b^2} + i(\frac {ab}{a^2+b^2} - \frac {ab}{a^2 + b^2}) = 1$
So an inverse does exist. But we must also prove it is unique. Now the way I'd do it, I'd simbly set up an equation $(a+bi)(c+di) =1$ and solve for $c$ and $d$ and show the solution is unique.... but it's a little too late for that!
I'd say though if $(a+bi)(c+di) =1$ and $(a+bi)(e+fi)=1$ then $c+di = (c+di)\cdot 1 = (c+di)\cdot (a+bi)(e+fi)= [(c+di)(a+bi)](e+fi)= 1(e+fi) = e+fi$. So there is only one possible solution and we know $\frac a{a^2 + b^2} -i\frac b{a^2 + b^2}$ is one solution, so it is the only solution.
| {
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Asymptotic expansion of integral of airy function In this question I am given that the asymptotic expansion of the Airy function for large $z$ is given by
$$Ai(z) = \frac{1}{2}\pi^{-\frac{1}{2}}z^{-1/4}\exp\left(-\frac{2}{3}z^{\frac{3}{2}}\right)\left[1 + O\left(\frac{1}{ z^{3/2}}\right)\right]$$
and the question is to evaluate the first to terms of the integral.
$$I = \int_{0}^x Ai(\xi) \mathrm{d}\xi$$
being told $\int_{0}^\infty Ai(\xi) \mathrm{d}\xi = \frac{1}{3}$. I have attempted this under the assumption that the first two terms are going to be larger than $O\left(\frac{1}{ z^{3/2}}\right)$ (otherwise how can I obtain the first few terms). So for the leading order terms I just need to compute the first couple of terms of
$$\int_{x}^\infty \frac{1}{2}\pi^{-\frac{1}{2}}\xi^{-1/4}\exp\left(-\frac{2}{3}\xi^{\frac{3}{2}}\right) \mathrm{d}\xi. $$
I do this with integration by parts using
$$\int_{x}^\infty \frac{1}{2}\pi^{-\frac{1}{2}}\xi^{-1/4}\frac{\xi^{\frac{1}{2}}}{\xi^{\frac{1}{2}}}\exp\left(-\frac{2}{3}\xi^{\frac{3}{2}}\right) \mathrm{d}\xi.$$
I get
$$ I \approx \frac{1}{2}\pi^{-\frac{1}{2}}\exp\left(-\frac{2}{3}x^{\frac{3}{2}}\right)\left[x^{-3/4} + \frac{3}{4}x^{-\frac{9}{4}} \right],$$
which as you can see has the second term too small compared to $\frac{1}{z^{\frac{3}{2}}}$, can anyone see where I am going wrong.
| Just to be puristic (not to say pedantic)
$$Ai(z) = \frac{1}{2}\pi^{-\frac{1}{2}}z^{-1/4}\exp\left(-\frac{2}{3}z^{\frac{3}{2}}\right)\left[1 + O\left(\frac{1}{ z^{7/4}}\right)\right]$$ For the computation of
$$I=\int \frac{1}{2}\pi^{-\frac{1}{2}}z^{-1/4}\exp\left(-\frac{2}{3}z^{\frac{3}{2}}\right)\,dz$$ let $z=\left(\frac{3}{2}\right)^{2/3} t^{4/3}$ to make
$$I=\sqrt{\frac{2}{3 \pi }}\int e^{-t^2}\,dt=\frac{\text{erf}(t)}{\sqrt{6}}$$ Back to $z$ and the bounds
$$\int_0^x \frac{1}{2}\pi^{-\frac{1}{2}}z^{-1/4}\exp\left(-\frac{2}{3}z^{\frac{3}{2}}\right)\,dz=\frac{1}{\sqrt{6}}\text{erf}\left(\sqrt{\frac{2}{3}} x^{3/4}\right)$$
$\frac{1}{\sqrt{6}} \sim 0.41$ which is $20$% higher than $\frac 13$.
| {
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Using differentation to find a power series expression for a function (a) use differentiation to find a power series representation for: $$f(x) = \frac{1}{(8+x)^2}$$
What is it's radius of convergence?
Part (a):
We start off with some tricky reverse engineering of derivatives:
$\frac{1}{(8+x)^2} = \frac{d}{dx}(\frac{-1}{x+8}) = \frac{d}{dx} ((\frac{-1}{8})(\frac{1}{\frac{x}{8}+1})) = \frac{d}{dx} ((\frac{-1}{8})(\frac{1}{1-\frac{-x}{8}}))$
Great! Now we can turn $(\frac{1}{1-\frac{-x}{8}}))$ into a power series to get :
$=\frac{d}{dx} ((\frac{-1}{8})\sum_{n=0}^\infty (-1)^n (\frac{x}{8})^n)$
Now we take the derivative....
$=(\frac{-1}{8})\sum_{n=1}^\infty (-1)^n n (\frac{x}{8})^{n-1}(\frac{1}{8})$
We are starting at $n=1$ now because the term when $n=0$ becomes $0$ when we take the derivative.
Combining all the $\frac{1}{8}$ we get:
$=\sum_{n=1}^\infty (-1)^{n+1} n x^{n-1} \frac{1}{8^{n+1}}$
notice that the alternating term went up to ${n+1}$ because of the negative in the term $\frac{-1}{8}$ out in front.
But the answer wants us to start this series at $n=0$, so we must reindex. So, if we change it so $n$ starts at $0$ rather than starting at $1$, that would means we have to add $1$ to every $n$ in our expression:
$=\sum_{n=0}^\infty (-1)^{n+2} (n+1) x^{n} \frac{1}{8^{n+2}}$
But $(-1)^{n+2} = (-1)^n$ for all $n$, so this equals:
$=\sum_{n=0}^\infty (-1)^{n} (n+1) x^{n} \frac{1}{8^{n+2}}$
Applying the ratio test you will see that this converges when $|x| < 8$, so $R = 8$
Note: Before when we were using the alternating series test, it didn't matter the form of the alternating term... It could have been $(-1)^n$ or $(-1)^{n-1}) for example.. Now it matters!!
| You are correct.
Alternatively since $$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$
for $|x|<1$ then $\sum_{n=1}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}$. So we have $$\frac{1}{(8+x)^2}=\frac{1}{8^2}\frac{1}{(1-(-\frac{x}{8}))^2}$$
$$=\frac{1}{8^2}\sum_{n=1}^{\infty}n\big(-\frac{x}{8}\big)^{n-1}=\sum_{n=1}^{\infty}(-1)^{n-1}nx^{n-1}\frac{1}{8^{n+1}}$$
$$=\sum_{n=0}^{\infty}(-1)^n(n+1)x^n\frac{1}{8^{n+2}}$$
which converges for $|\frac{x}{8}|<1$ or $|x|<8$.
| {
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Solving a differential equation using an integrating factor
I'm trying the solve the following equation: $ \left\{\begin{matrix}
(x^B+y^B)(xdy-ydx)=(1+x)x^9dx \\ y(-1)=A \end{matrix}\right. $
for $A=1$ and $A=0$.
$B\in2\mathbb{N}_0+1$
My solution is the following, but I got stuck:
$(x^B+y^B)(xdy-ydx)=(1+x)*x^9dx$
$x^B*xdy-y*x^Bdx+y^B*xdy-y*y^Bdx-(1+x)*x^9dx=0$
$x^B*xdy+y^B*xdy-y*x^Bdx-y*y^Bdx-(1+x)*x^9dx=0$
$xdy(x^B+y^B)+[-y(x^B+y^B)-(1+x)*x^9]dx=0$
where:
$Q(x,y)=x(x^B+y^B)dy$
$P(x,y)=[-y(x^B+y^B)-(1+x)*x^9]$
Then I made partial derivation:
$\frac{∂Q}{∂x}:(x^B+y^B)x(Bx^{B-1})$
$\frac{∂P}{∂y}: -(x^B+y^B)-y(By^{B-1})$
And then subtract:
$\frac{∂P}{∂y}-\frac{∂Q}{∂x}=-(x^B+y^B)-y(By^{B-1})-(x^B+y^B)-x(Bx^{B-1})$
But probably I did something wrong and
I'm stuck and not sure where made I mistake.. Can you help me please?
| $$(x^B+y^B)(xdy-ydx)=(1+x)x^9dx $$
Duivide by $x^2$:
$$(x^B+y^B)d\dfrac yx=(1+x)x^7dx $$
Divode by $x^B$:
$$\left(1+\left(\dfrac y x\right)^B \right)d\dfrac yx=\dfrac {(1+x)x^7}{x^B}dx $$
Integrate.
| {
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How to denest $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$ from scratch? I have seen several questions asking for the proof of
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$
However, I want to simplify $\sqrt[3]{\sqrt[3]{2}-1}$ into the form $\sqrt[3]{a}-\sqrt[3]{b}+\sqrt[3]{c}$ and find $a, b, $ and $c$ from scratch.
How would I go about doing it?
Thanks.
| For nested cubic-roots of the form $\sqrt[3]{p \sqrt[3]{q}+r}$, there is a systematic denesting method due to Ramanujan.
If the cubic polynomial
$x^3 +a x^2 +bx +c=(x-x_1)(x-x_2)(x-x_3)$ satisfies
$$b +a c^{1/3} +3c^{2/3}=0\tag1 $$
per Ramanujan
$$\sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 }=\sqrt[3]{ 3\sqrt[3]{9c-ab }-a-6\sqrt[3]c} \tag2$$
To denest $\sqrt[3]{\sqrt[3]{2}-1}$, match it to the right-hand-side of (2)
$$a+6c^{1/3}=1,\>\>\>\>\>27(9c-ab)=2$$
which, together with (1), determines $a=-\frac13$, $b=-\frac2{27}$ and $c= \frac8{729}$, or
$$x^3 -\frac13 x^2-\frac2{27}x + \frac8{729}= (x-\frac19)(x+\frac29)(x-\frac49)$$
Then, the Ramanujan’s formula (2) yields
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$
| {
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Question regarding a series that contains logarithms $$\sum_{n=1}^\infty\bigl(3\log(n^2+1)-2\log(n^3+1)\bigr)$$
I tried the limit comparison test with $\sum_{n=1}^\infty\frac1{n^2}$ and then applied L'Hopital's Rule. What I ended up was a limit which was a real number and as the series of $\frac1{n^2}$ converges, my original series converges as well. However, I was wondering if there is another way of solving the problem, one that involves fewer computations.
| $$3\ln(n^2+1)-2\ln(n^3+1)=\ln\left( \frac{(n^2+1)^3}{(n^3+1)^2}\right)=\ln\left( \frac{n^6+3n^4+3n^2+1}{n^6+2n^3+1}\right)$$
$$=\ln\left( 1+\frac{3n^4-2n^3+3n^2}{n^6+2n^3+1}\right) \sim \frac{3n^4-2n^3+3n^2}{n^6+2n^3+1} \sim \frac{3}{n^2}$$
Therefore the series converges.
| {
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Prove that $12(ab+ba+ac) <7a^2+15b^2+18c^2$ holds for all positive numbers. Prove that $12(ab+ba+ac) < 7a^2+15b^2+18c^2$ holds for all positive numbers.
I tried completing the square, but that solution would suggest that inequality holds for all real numbers. Inequalities between means did not work for me either.
$$12(ab+ba+ac) < 7a^2+15b^2+18c^2$$
$$(2a-3b)^2+(2b-3c)^2+(2a-3c)^2+2b^2-a^2>0$$
| By AM-GM, or completing the square,
$(p-q)^2 \ge 0 \Rightarrow p^2 + q^2 \ge 2pq$
we have
$$\color{blue}{(4a^2 + 9b^2)} + \color{red}{(3a^2 + 12c^2)} + \color{green}{(6b^2 + 6c^2)} \ge \color{blue}{12ab} + \color{red}{12ca} + \color{green}{12bc}$$
with equality for $2a=3b$, $a=2c$, $b=c$ whose simultaneous solution is $(a,b,c)=(0,0,0)$
For $a,b,c > 0$, we have strict inequality.
| {
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Application of Triple Integrals Any easy way to solve this integration?
$\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}} \int_{0}^{\frac{6-3y-2x}{2}} xyz \ dzdydx$
My work
$\int \int \int_D xyz \ dV = \dfrac{1}{2}\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}} xy(z^2 )\biggr|_{0}^{\frac{6-3y-2x}{2}} \ dy \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} \int_{0}^{\dfrac{6-2x}{3}} xy(6-3y-2x)^2 \ dy \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$
The terms are expanding, any way to find a solution in easy case. please help
| The domain of integration is a tetrahedron with vertices at $(0,0,0),(3,0,0),(0,2,0),(0,0,3)$. The integrand is symmetric with respect to permuting variables. Thus we are allowed to rotate the domain of integration so that slices perpendicular to the $z$-axis are right isosceles triangles and the integral is equal to
$$\int_0^2\int_0^{(6-3z)/2}\int_{(6-3z-2y)/2}xyz\,dx\,dy\,dz$$
We move the $z$ as far out as it will go:
$$\int_0^2z\int_0^{(6-3z)/2}\int_{(6-3z-2y)/2}xy\,dx\,dy\,dz$$
Then rewrite to make the slices of right isosceles triangles clearer. $T(k)$ is the triangle with vertices at $(0,0),(k,0),(0,k)$:
$$\int_0^2z\iint_{T((6-3z)/2)}xy\,dA\,dz$$
Now let's work out what the inner integral is before we substitute $(6-3z)/2$. In other words, what is
$$\iint_{T(k)}xy\,dA=\int_0^k\int_0^{x-k}xy\,dy\,dx$$
This evaluates to
$$\int_0^k x[y^2/2]_0^{x-k}\,dx=\frac12\int_0^k x(x-k)^2\,dx=\frac12[x^4/4-(2/3)kx^3+k^2x^2/2]_0^k=k^4/24$$
So our original integral becomes
$$\frac1{24}\int_0^2z\left(\frac{6-3z}2\right)^4\,dz=\frac{81}{24×16}\int_0^2z(2-z)^4\,dz$$
$$=\frac{27}{128}[8z^2-(32/3)z^3+6z^4-(8/5)z^5+z^6/6]_0^2$$
$$=\frac{27}{128}×\frac{32}{15}=\frac9{20}$$
| {
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Finding all integer solutions of $\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12$ If we have integers $x,y,z$ such that $x,y,z \ge 3,$ find all solutions to $$\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12.$$
I was thinking of first expanding out this, and then simplifying from there, but the equation got very messy. Is there a better method?
| As an alternative to Joshua Wang's excellent answer:
By symmetry we may assume, without loss of generality, that $x\leq y$. Clearing denominators yields
$$2xz+2yz-2xy=xyz,\tag{1}$$
and a bit of rearranging then shows that
$$xy(z+2)=2(x+y)z.$$
Of course $z+2>z$ and so $xy<2(x+y)$, or equivalently
$$(x-2)(y-2)<4.$$
Because $x\leq y$ we see that $x<4$, so $x=3$ and then $y<6$. Plugging into $(1)$ and rearranging shows that
$$yz+6y-6z=0,$$
or equivalently $z=\frac{6y}{6-y}$, yielding the following three solutions for $(x,y,z)$:
$$(3,3,6),\qquad(3,4,12),\qquad(3,5,30).$$
| {
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Prove this geometric inequality Given a triangle $\triangle ABC$, let $D$ and $E$ be on points $BC$ such that $BD=DE=EC$.The line $p$ intersects $AB,AD,AE,AC$ at $K,L,M,N$ respectively. Prove that $KN ≥ 3LM$
My attempt: I think cross ratio can be used to prove it though I am not sure how.
Let $a=KL,b=LM,c=MN$
We have to proove $a+b+c\geq 3z$, i.e. $a+b\geq 2c$
Since $R(K,N;L,M)=R(B,C;D,E)$ so $R(K,N;L,M)= \frac{1}{2}:\frac{1}{2}$ thus $4ac=(a+b)(b+c)$.
I am not able to move further from here.
| Another way.
Let $KL=a$, $LM=x$ and $MN=b$.
Thus, since $$\frac{KM}{KN}:\frac{LM}{LN}=\frac{BE}{BC}:\frac{DE}{DC},$$ we obtain:
$$\frac{(a+x)(b+x)}{(a+b+x)x}=\frac{4}{3}$$ or
$$x^2+(a+b)x-3ab=0$$ or
$$x=\frac{-a-b+\sqrt{a^2+14ab+b^2}}{2}$$ and we need to prove that
$$\frac{-a-b+\sqrt{a^2+14ab+b^2}}{2}+a+b\geq3\cdot\frac{-a-b+\sqrt{a^2+14ab+b^2}}{2}$$ or
$$2(a+b)\geq\sqrt{a^2+14ab+b^2},$$
which is true by AM-GM:
$$\sqrt{a^2+14ab+b^2}=\sqrt{(a+b)^2+12ab}\leq\sqrt{(a+b)^2+12\left(\frac{a+b}{2}\right)^2}=2(a+b).$$
| {
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Is the process of determining whether a singular is regular or not same for the homogenous and non-homogenous ODEs? I have this second-order non-homogeneous ODE:
$$0.5x^5A''(x)+\left(2x^4+x^3\right)A'(x)+\left(x^3+x^2+x-1\right)A(x)=-1$$
I have determined that $x=0$ is a singular point and divided both sides by $0.5x^5$. What I am now trying to do is determine whether it is regular or not.
I am aware of the process with homogenous ODEs, about coefficients being divisible by $(x-x_0)$. But I am wondering, does the same property retain with non-homogeneous ODEs? I have sought the answer on the internet but all examples are of homogeneous DEs.
| $0.5x^5A''(x)+(2x^4+x^3)A'(x)+(x^3+x^2+x-1)A(x)=-1$
$A''(x)+\left(\dfrac{4}{x}+\dfrac{2}{x^2}\right)A'(x)+\left(\dfrac{2}{x^2}+\dfrac{2}{x^3}+\dfrac{2}{x^4}-\dfrac{2}{x^5}\right)A(x)=-\dfrac{2}{x^5}$
$\lim\limits_{x\to 0}\left(x\left(\dfrac{4}{x}+\dfrac{2}{x^2}\right)\right)=\lim\limits_{x\to 0}\left(4+\dfrac{2}{x}\right)=\infty$
$\lim\limits_{x\to 0}\left(x^2\left(\dfrac{2}{x^2}+\dfrac{2}{x^3}+\dfrac{2}{x^4}-\dfrac{2}{x^5}\right)\right)=\lim\limits_{x\to 0}\left(2+\dfrac{2}{x}+\dfrac{2}{x^2}-\dfrac{2}{x^3}\right)=\infty$
$\therefore$ the finite singular point $x=0$ is irregular.
That's not the end. We should also check the singularities at infinity. Because these also act as one of the key points of distinguishing the ODE type.
Let $t=\dfrac{1}{x}$ ,
Then $\dfrac{dA}{dx}=\dfrac{dA}{dt}\dfrac{dt}{dx}=-\dfrac{1}{x^2}\dfrac{dA}{dt}=-t^2\dfrac{dA}{dt}$
$\dfrac{d^2A}{dx^2}=\dfrac{d}{dx}\left(-t^2\dfrac{dA}{dt}\right)=\dfrac{d}{dt}\left(-t^2\dfrac{dA}{dt}\right)\dfrac{dt}{dx}=\left(-t^2\dfrac{d^2A}{dt^2}-2t\dfrac{dA}{dt}\right)\left(-\dfrac{1}{x^2}\right)=\left(-t^2\dfrac{d^2A}{dt^2}-2t\dfrac{dA}{dt}\right)(-t^2)=t^4\dfrac{d^2A}{dt^2}+2t^3\dfrac{dA}{dt}$
$\therefore t^4\dfrac{d^2A}{dt^2}+2t^3\dfrac{dA}{dt}-(4t+2t^2)t^2\dfrac{dA}{dt}+(2t^2+2t^3+2t^4-2t^5)A=-2t^5$
$t^4\dfrac{d^2A}{dt^2}-(2t^2+2t)t^2\dfrac{dA}{dt}-(2t^5-2t^4-2t^3-2t^2)A=-2t^5$
$\dfrac{d^2A}{dt^2}-\left(2+\dfrac{2}{t}\right)\dfrac{dA}{dt}-\left(2t-2-\dfrac{2}{t}-\dfrac{2}{t^2}\right)A=-2t$
$\lim\limits_{t\to 0}\left(-t\left(2+\dfrac{2}{t}\right)\right)=\lim\limits_{t\to 0}(-2t-2)=-2$
$\lim\limits_{t\to 0}\left(-t^2\left(2t-2-\dfrac{2}{t}-\dfrac{2}{t^2}\right)\right)=\lim\limits_{t\to 0}(-2t^3+2t^2+2t+2)=2$
$\therefore$ the singularities at $x=\pm\infty$ are regular.
| {
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Can this problem have solutions for $N>3$:Find three $(N=3)$ positive integers a, b and c such that: $a+b+c=k^2$, $a+b=t^2$, $b+c=m^2$ and $a+c=n^2$ Can this question have solutions for N>3:
Find three ($N =3$) positive integers a, b and c such that: $a+b+c=k^2$, $a+b=x^2$, $b+c=m^2$ and $a+c=n^2$; $x, m, n∈\mathbb N$
Solution:
Let $a+b+c=(x+1)^2=x^2+2x+1$
If $a+b=x^2$ then third number $c=2x+1$.
If $b+c=(x-1)^2=x^2-2x+1=m^2$ then :
$a=4x$, so $b=x^2-4x$
Now due to statement we must have $a+c=6x+1=n^2$
$n^2=6x+1$ can have infinite solutions such as:
$(x, n^2)=(20, 121=11^2), (60, 361=19^2), (140, 841=29^2)\cdot\cdot\cdot$
Which give:
$(a, b, c, k)=(80, 320, 41, 21), (240, 3360, 121, 61), (560, 19040, 281, 141),\cdot\cdot\cdot$
I tried to solve this problem for $N=4$, four numbers but no success. Now I have two questions:
1-Does this problem have solutions for N>3?
2-Any idea for better method?
| This answer is only for the case $N=4$. I fail to see how this method could help resolve the cases for larger $N$.
Consider the equations
$$a + b = u^2, c + d = v^2$$
$$a + c = w^2, b + d = x^2$$
$$a + d = y^2, b + c = z^2$$
$$a + b + c + d = k^2$$
Now the first three pairs of equations give
$$u^2 + v^2 = w^2 + x^2 = y^2 + z^2 = k^2$$
so a reasonable approach would be to investigate Pythagorean triples.
Not all Pythagorean triples generate positive integers solutions, but a solution could be generated if the pairs $(u,v), (w,x), (y,z)$ are "reasonably close". Consider
$$165^2+280^2 = 195^2+260^2 = 253^2+204^2 = 325^2$$
Take note of the parity as well (or multiply it by $2$ so you don't need to worry about it)
Solving the equations give
$$a = 11817, b = 15408, c = 26208, d = 52192$$
and more solutions can be generated this way.
| {
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Find z s.t. $z+\frac{1}{z}$ is real. Is my solution good? I must find all z s.t. $z+\frac{1}{z}$ is real. I know that $z = a + bi$ is real when the Imaginary part is 0. So, there we go:
$$z+\frac{1}{z}= \frac{z^2+1}{z}=\frac{(a+bi)^2+1}{a+bi}=\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)}$$
$$\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)} = \frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}$$
$$\frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}=\frac{a^3 +2a^2bi-b^2a+a-ba^2i+2ab^2+b^3i-bi}{(a^2+b^2)}$$
$$2a^2b-ba^2+b^3-b=0$$
$$a^2b+b^3-b=0$$
$$b_1=0$$
$$a^2+b^2-1=0$$
$$b^2=1-a^2$$
$$b_2=\sqrt{1-a^2}$$
$$b_3=-\sqrt{1-a^2}$$
so, the solutions are:
$$z_1=a+0i$$
$$z_1=a+\sqrt{1-a^2}i$$
$$z_1=a-\sqrt{1-a^2}i$$
I am not sure if that should be done like I did it.
| If $z$ is real, you know what happens.
Now if $z$ is not real, the vectors $z$, $1/z$, and $z+1/z$ form a triangle. You would like the complex angle of $z+1/z$ to be $0$.
The vectors $z$ and $1/z$ have complex angles that are negative of each other. So now this triangle has two equal angles with $z+1/z$ representing the edge between them. It's an isosceles triangle, and it follows that the magnitude of $z$ equals that of $1/z$.
So the only $z$ for which this can hold are the $z$ on the unit circle. And you can then verify that for all such complex numbers, the sum is real: $e^{it}+e^{-it}=2\cos(t)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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finding $x^6+y^6$ given $x+y$ and $xy$ Here is the question
If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.
This is from a previous timed competition, so fastest answers are the best answers.
I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$
. The only other way I can think of is solving for x and y, but that wouldn't be too quick. Any help?
| Your sum is$$\begin{align}\sum_\pm(2\pm\sqrt{2})^6&=2(2^6+\sqrt{2}^6+15(2^4\sqrt{2}^2+2^2\sqrt{2}^4))\\&=2(64+8+15(2\sqrt{2})^2(2^2+2))\\&=2(72+15\cdot8\cdot6)\\&=1584.\end{align}$$
| {
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Determine where the range of a function is positive Consider the quadratic equation $y = -\dfrac{3}{4}x^2 + 4x - 4$, with roots at $x = \tfrac{4}{3}$ and $x = 4$. I'd like to find the values for $x$ where $y > 0$; that is where
$$
-\dfrac{3}{4}x^2 + 4x - 4 > 0.
$$
Rewritting as
$$
(x - \tfrac{4}{3})(x - 4) > 0,
$$
the above inequality is true when
A) both $(x - \tfrac{4}{3})$ and $(x - 4)$ are positive
or
B) both $(x - \tfrac{4}{3})$ and $(x - 4)$ are negative.
Case A - both positive:
$x - \tfrac{4}{3} > 0$ => $x > 4/3$ and $x - 4 > 0$ => $x > 4$. This is true when $x > 4$.
Case B - both negative:
$x - \tfrac{4}{3} < 0$ => $x < 4/3$ and $x - 4 < 0$ => $x < 4$. This is true when $x < \tfrac{4}{3}$.
Therefore the solution is $(-\infty, \tfrac{4}{3}) \cup (4,\infty)$. Yet graphing this equation shows the positive range as $(4/3,4)$. The range $(-\infty, \tfrac{4}{3}) \cup (4,\infty)$ actually satisfies the equation $y = \dfrac{3}{4}x^2 - 4x + 4$.
Where have I run afoul?
| $$-\dfrac{3}{4}x^2 + 4x - 4 > 0$$ can't be written as$$(x-4/3)(x-4) > 0$$ but as $$-\frac34(x-4/3)(x-4) > 0$$which gives$$(x-4/3)(x-4)\color{red}<0$$
| {
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Is this a valid linear transformation? (1) Which of the following is a linear transformation?
$ \quad $ (a) $ T ( \begin{pmatrix} x \\ y \end{pmatrix} )= \begin{pmatrix} 0 \\ x \\y \end{pmatrix}$, $ \quad $ (b) $ T ( \begin{pmatrix} x \\ y \\ z \end{pmatrix} ) = \begin{pmatrix} y \\ x \end{pmatrix}$ $ \quad $ (c) $ T (\begin{pmatrix} x \\ y \end{pmatrix} )= \begin{pmatrix} x^2 \\ y^2 \\ 0 \end{pmatrix}$
$ \quad $ (d) $ T (\begin{pmatrix} x \\ y \end{pmatrix}) = \begin{pmatrix} 1 \\ x \\y \end{pmatrix}$ $ \quad $ (e) $ T ( \begin{pmatrix}\ x \\ y \\ z \end{pmatrix} ) = x \begin{pmatrix} 1 \\ 1 \\ z \end{pmatrix}$
I am new to this topic, am I correct in saying that only (a) is a linear transformation?
(2) $ T (\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}) = \begin{pmatrix} 2 \\ 2 \\2 \end{pmatrix}$ , $ \quad $ $ T (\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}) = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}$, $ \quad $ $ T (\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}) = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ $ \quad $
$ \quad $$ M = \begin{pmatrix}1 & 1 & 0 \\ 2 & 1 & 1 \\ 3 & 2 & -2 \end{pmatrix}. $
$ \quad $ Then, $ T (\begin{pmatrix} x \\ y \\ z \end{pmatrix}) = \begin{pmatrix} 2 & 1 & 0 \\ 2 & -1 & 0 \\ 2 & -1 & 0 \end{pmatrix} M^{-1} \begin{pmatrix} x \\ y \\ z \end{pmatrix} $ ?
| I think you need to understand the definition of linear transformation. Note that:
Let $V$ and $W$ vector spaces over field $F$. Let $T: V\to W$ a function, so $T$ is said linear transformation if, and only if,
$1.$ $\forall v_{1},v_{2}\in V: T(v_{1}+v_{2})=T(v_{1})+T(v_{2})$.
$2.$ $\forall \alpha \in F, \forall v \in V: T(\alpha v)=\alpha T(v)$.
Now, you can prove in this sense that
\begin{eqnarray}
T \text{is a linear transformation} \iff \forall \alpha, \beta\in F, \forall v_{1},v_{2} \in V: T(\alpha v_{1}+\beta v_{2})=\alpha T(v_{1})+\beta T(v_{2})
\end{eqnarray}
Now, for your question you need to see if $T$ is a linear transformation, so:
b) $T$ Is a linear transformation, because
\begin{eqnarray}
T\left(\alpha \begin{pmatrix} x \\ y \\ z \end{pmatrix} +\beta \begin{pmatrix} a \\ b \\ c \end{pmatrix} \right)
&=& T\begin{pmatrix} \alpha x+\beta a \\ \alpha y +\beta b \\ \alpha z +\beta c \end{pmatrix}\\
&=&\begin{pmatrix} \alpha y +\beta b \\ \alpha x +\beta a \end{pmatrix}\\
&=&\alpha \begin{pmatrix} y \\ x \end{pmatrix} +\beta \begin{pmatrix} b \\ a \end{pmatrix}\\
&=&\alpha T\begin{pmatrix} x \\ y \\ z \end{pmatrix} +\beta T\begin{pmatrix} a \\ b \\ c \end{pmatrix}
\end{eqnarray}
a) Also $T$ is linear transformation (I think you can do this part).
c) $T4 is not linear transformation (I think you can do this part).
d) $T$ is not linear transformation, because
\begin{eqnarray}
T\left( \alpha \begin{pmatrix} x \\ y \end{pmatrix} +\beta\begin{pmatrix} a \\ b \end{pmatrix} \right)&=& T\begin{pmatrix} \alpha x+\beta a \\ \alpha y +\beta b \end{pmatrix}\\
&=&\begin{pmatrix} 1 \\ \alpha x+\beta a \\ \alpha y +\beta b \end{pmatrix}\\
&\not=& \alpha \begin{pmatrix} 1 \\ x \\ y \end{pmatrix}+\beta \begin{pmatrix} 1 \\ a \\ b \end{pmatrix}\\
&=& \alpha T \begin{pmatrix} x\\ y \end{pmatrix} +\beta T\begin{pmatrix} a \\ b \end{pmatrix}
\end{eqnarray}
e) $T$ is not linear transformation ( I think you can do this part).
| {
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$\frac{a^{2}-1}{b+1}+\frac{b^{2}-1}{a+1}$ an integer $\Rightarrow \frac{a^{2}-1}{b+1}$ and $\frac{b^{2}-1}{a+1}$ are integers. If $\frac{a^{2}-1}{b+1}+\frac{b^{2}-1}{a+1}$ is an integer, prove that also $\frac{a^{2}-1}{b+1}$ and $\frac{b^{2}-1}{a+1}$ are integers.
By doing the math, I get $\frac{(a-1)(a+1)^{2}+(b-1)(b+1)^{2}}{(a+1)(b+1)}$ is an integer which means that $(a+1)(b+1)$ divides $(a-1)(a+1)^{2}+(b-1)(b+1)^{2}$.
but I don't know how to continue.
By the way, this is not homework or anything. I just found it in a book.
| The problem states that, for some natural numbers $a$ and $b$, there's an integer $i$ where
$$\frac{a^2 - 1}{b + 1} + \frac{b^2 - 1}{a + 1} = i \tag{1}\label{eq1A}$$
If the first term is an integer, say $\frac{a^2 - 1}{b + 1} = j$, then $\frac{b^2 - 1}{a + 1} = i - j$ is also an integer. Doing the same thing with the second term instead shows that if either term is an integer, the other one is as well.
Assume neither left side term in \eqref{eq1A} is an integer. As you did, combining the terms using a common denominator gives
$$\frac{(a - 1)(a + 1)^{2} + (b - 1)(b + 1)^{2}}{(a + 1)(b + 1)} = i \tag{2}\label{eq2A}$$
Since $a + 1 \mid (a - 1)(a + 1)^{2}$, this means $a + 1 \mid (b - 1)(b + 1)^{2} = (b + 1)(b^2 - 1)$. However, since it's assumed $a + 1 \not\mid b^2 - 1$, this means there's a prime $p$ where, using the $p$-adic order function, we have
$$\nu_{p}(a + 1) \gt \nu_{p}(b^2 - 1) \tag{3}\label{eq3A}$$
but
$$\nu_{p}(a + 1) \le \nu_{p}((b + 1)(b^2 - 1)) \tag{4}\label{eq4A}$$
This means $p \mid b + 1$. Define
$$\nu_{p}(a + 1) = k_1 \tag{5}\label{eq5A}$$
$$\nu_{p}(b + 1) = k_2 \tag{6}\label{eq6A}$$
If $p \neq 2$, then $p \not\mid b - 1$ and $p \not\mid a - 1$. Thus, \eqref{eq3A} gives
$$k_1 \gt k_2 \tag{7}\label{eq7A}$$
Since $\nu_{p}((a - 1)(a + 1)^2) = 2k_1$ and $\nu_{p}((b - 1)(b + 1)^2) = 2k_2$, this means with the numerator of \eqref{eq2A},
$$\nu_{p}((a - 1)(a + 1)^{2} + (b - 1)(b + 1)^{2}) = 2k_2 \tag{8}\label{eq8A}$$
However, with the denominator,
$$\nu_{p}((a + 1)(b + 1)) = k_1 + k_2 \gt 2k_2 \tag{9}\label{eq9A}$$
Thus, the fraction can't be an integer. Since it is an integer, this means the assumption $p \neq 2$ must be incorrect, i.e., $p = 2$. Thus, $a$ and $b$ are odd, with $p \mid a - 1$ and $p \mid b - 1$. Define
$$\nu_{p}(a - 1) = m_1 \tag{10}\label{eq10A}$$
$$\nu_{p}(b - 1) = m_2 \tag{11}\label{eq11A}$$
Using \eqref{eq6A} and \eqref{eq11A} in \eqref{eq3A} gives
$$k_1 \gt k_2 + m_2 \tag{12}\label{eq12A}$$
Note the same arguments as above could have been used to check $b + 1$ not dividing into $a^2 - 1$, with it also reaching the conclusion the prime involved must be $2$. Thus, this would give \eqref{eq12A} but with the $1$ and $2$ indices switched around, i.e.,
$$k_2 \gt k_1 + m_1 \tag{13}\label{eq13A}$$
Combining \eqref{eq12A} and \eqref{eq13A} gives
$$k_1 \gt k_2 + m_2 \gt (k_1 + m_1) + m_2 \implies 0 \gt m_1 + m_2 \tag{14}\label{eq14A}$$
This is not possible since $m_1$ and $m_2$ must be positive integers. Thus, the original assumption of the two left terms in \eqref{eq1A} not being integers must be incorrect, i.e., they are both integers.
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Stuck solving this PDE problem I need to solve $$uu_x+yu_y=x, u(x,1)=2x$$ i tried with characteristics method but got stuck solving the ODEs $$\frac{dx}{dt}=u ~,~\frac{dy}{dt}=y~,~\frac{du}{dt}=x$$ with initial values $x(0)=s,y(0)=1,u(0)=2s$. From the second one i get $y=e^t$, but how to solve the first two ODEs? I tried $$\frac{dx}{du}=\frac{u}{x} \implies x^2-u^2=k_1$$ where $k_1=-3s^2$ which gives $u=\sqrt{x^2+3s^2}$. Similarly $$\frac{dy}{y}=\frac{dx}{\sqrt{x^2+3s^2}} \implies y=k_2(x+\sqrt{x^2+3s^2})$$ And $k_2$ can be found out as $k_2=\frac{1}{3s}$, giving me $y=\frac{x+\sqrt{x^2+3s^2}}{3s}$. I'm not sure how to proceed further and solve for $u$?
| I think it is easier if we write the ODEs in the parametrisation invariant form of the Lagrange-Charpit equations
$$\frac{dx}{u} = \frac{dy}{y} = \frac{du}{x}$$
You correctly identified the first characteristic curve
$$\frac{dx}{u} = \frac{du}{x} \implies u^{2} - x^{2} = C_{2}$$
Now, we can use componendo dividendo on the first and third fractions to get
\begin{align}
\frac{dx}{u} = \frac{du}{x} &= \frac{du+dx}{u+x} \\
&= \frac{d(u+x)}{u+x} \\
&= \frac{dy}{y} \\
\end{align}
and so solving the last equality and using the functional relationship $C_{2} = f(C_{1})$ yields
\begin{align}
\ln(u+x) &= \ln y + C_{1} \\
\implies \frac{u+x}{y} &= C_{1} \\
\implies u^{2} - x^{2} &= f(C_{1}) \\
&= f \left(\frac{u+x}{y}\right)
\end{align}
which you can check solves the PDE. Now you can apply the data $u(x,1) = 2x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $x$ such that $ 2^{16^x} = 16^{2^x} $. Find $x$ such that $ 2^{16^x} = 16^{2^x}.$
I am a bit confused, what will happen when we expand $16^{2^x}$, will we get $4^{2^{2^x}}$ or $4^{2^{x+1}}$ ?
| See that $2^{16^x} = 16^{2^x}$ is $2^{2^{4x}} = 2^{2^{x+2}}$
then $4x = x + 2$ and get $x = 2/3$.
| {
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Prove $\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} <0$, where $a,b,c$ are roots of $x^3 +3x^2-24x +1 = 0$ I am working on this problem:
$a, b$ and $c$ are the roots of $x^3 +3x^2-24x +1 = 0$. Prove that
$$S = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$
After I used Vietta's Formula,expanded $S^3$ and used many algebraic tricks I have
$S[S^2 -3(a'b' +a'c +b'c')] = 0$, wich leads to $S = 0 $ or $ S^2 -3(a'b' +a'c +b'c') = 0$, where $a' =\sqrt[3]{a}$, $b' =\sqrt[3]{b}$ and $c' =\sqrt[3]{c}$
So the final step is to prove the impossibility of the second case by proving that
$$a'b' +a'c +b'c' < 0$$
(which I verified numerically). It's here that I have trouble.
I also noticed that I didn't use the fact that $ab+ac+bc = -24$ (given by Vietta's Formula), so I guess it should be used here.
| Rewrite the inequality as
$$\sqrt[3]{abc\over a}+\sqrt[3]{abc\over b}+\sqrt[3]{abc\over c}\lt0$$
but now note that $abc=-1$, so this simplifies to
$$\sqrt[3]{1\over a}+\sqrt[3]{1\over b}+\sqrt[3]{1\over c}\gt0$$
Next, note that $1/a$, $1/b$, and $1/c$ are roots of $x^3-24x^2+3x+1=0$. Calling this polynomial $P$, we see that $P(-1)\lt0$, $P(0)\gt0$, and $P(1)\lt0$, so its three roots lie in $(-1,0)$, $(0,1)$, and $(1,\infty)$. Their cube roots lie in the same intervals, and so the sum of the cube roots is clearly positive, since the one in $(1,\infty)$ more than offsets the one in $(-1,0)$.
| {
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Proof by $\epsilon - \delta$ definition that $\lim_{x \to 1} \frac{x^2-1}{x^3-1} = \frac{2}{3}$ $$\lim_{x \to 1} \frac{x^2-1}{x^3-1} = \frac{2}{3}$$
I want to know if my proof is correct. If not, please show me how to fix it, if possible.
Let $\epsilon > 0$.
$\left| \dfrac{x^2-1}{x^3-1}-\dfrac23 \right|=\left| \dfrac{3x^2-3-2x^3+2}{3x^3-3} \right|=\left| \dfrac{-2x^3+3x^2-1}{3x^3-3}\right|=\left|\dfrac{(x-1)^2(2x+1)}{3(x-1)(x^2+x+1)} \right|=\dfrac{|x-1||2x+1|}{3|x^2+x+1|}$
Assume $\delta < 1$. Then $|x-1|<1$ implies:
(i) $-2 < 0 < x < 2 \implies |x| < 2$
(ii) $-5<-3<2x+1<5 \implies |2x+1|<5$
(iii) $-3<-1<x+1<3 \implies |x+1|<3$
(iv) $|x^2+x+1| = |x(x+1)+1| \le |x||x+1|+1 < 2.3+1 = 7 $
So, make $\delta = \min \{1, \dfrac{21\epsilon}{5} \}$. Then $|x-1|<\delta \implies \dfrac{|x-1||2x+1|}{3|x^2+x+1|}<\epsilon$.
| Here it is another way to solve it for the sake of curiosity.
To begin with, notice that
\begin{align*}
\frac{|x-1||2x+1|}{3|x^{2} + x + 1|} = \frac{|x-1||(2x - 2) + 3|}{3|x^{2} + x + 1|} \leq \frac{2|x-1|^{2} + 3|x-1|}{3|x^{2}+x+1|}
\end{align*}
Since we have that
\begin{align*}
x^{2} + x + 1 = \left(x^{2} + x + \frac{1}{4}\right) - \frac{1}{4} + 1 = \left(x + \frac{1}{2}\right)^{2} + \frac{3}{4} \geq \frac{3}{4} > 0 \Rightarrow 0 < \frac{1}{x^{2}+x+1} \leq\frac{4}{3}
\end{align*}
we can conclude that
\begin{align*}
\frac{|x-1||2x+1|}{3|x^{2} + x + 1|} \leq \frac{8|x-1|^{2} + 12|x-1|}{9} \leq \frac{8\delta^{2} + 12\delta}{9} = \varepsilon
\end{align*}
Thus for every $\varepsilon > 0$ there corresponds a $\displaystyle\delta = \frac{-3 + 3\sqrt{1+2\varepsilon}}{4} > 0$ such that
\begin{align*}
0 < |x - 1| \leq \delta \Rightarrow \left|\frac{x^{2}-1}{x^{3}-1} - \frac{2}{3}\right| \leq \varepsilon
\end{align*}
and we are done.
Hopefully this helps!
| {
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Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$ I need to compute a limit:
$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$
I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.
$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$
I've calculated several values of this function, and it seems to have a limit of $1$.
| For small positive angles, the inequalities ${3\over4}\theta\le\sin\theta\le\theta$ apply. (The fraction $3/4$ is somewhat arbitrary; any fraction less than $1$ will do.) Letting $\theta=\sqrt x$ and noting that $-1\le\sin(1/x)\le1$ for all $x\not=0$, we have
$${3\over2}\sqrt x-\sqrt x\le2\sin\sqrt x+\sqrt x\sin(1/x)\le2\sqrt x+\sqrt x$$
which simplifies to
$${1\over2}\sqrt x\le2\sin\sqrt x+\sqrt x\sin(1/x)\le3\sqrt x$$
If we take for granted the limit $x^x\to1$ as $x\to0$ (easily proved with L'Hopital applied to $\ln x\over1/x$), we have $\lim_{x\to0^+}(\sqrt x/2)^x=\lim_{x\to0^+}(3\sqrt x)^x=1$ so the Squeeze Theorem tells us
$$\lim_{x\to0^+}(\sin\sqrt x+\sqrt x\sin(1/x))^x=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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proof with induction $2^n + 4 > n^2 + 2n$ prove w/ full induction : $2^n + 4 > n^2 + 2n$.
Let's skip the other steps for a moment, I'm just getting stuck with the proof
the claim is: $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$
and now the proof $$2^{n+1} + 4 > 2 * 2^n + 4$$ $$> 2 * (n^2 + 2n)$$ $$> 2n^2 + 4n$$
i am stuck here
| The claim is $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$
Assuming it is true for $n$, that is $2^n + 4 > n^2 + 2n\to 2^n>n^2+2n-4$ we have
$$2^{n+1} + 4=2\cdot 2^n+4>2\left(n^2+2n-4\right)+4=2n^2+4n-4=(n^2+2n+1)+(2n+2)+n^2-6$$
$$>(n+1)^2+2(n+1);\;n>2$$
Proved
| {
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Solving $2^{16^x} = 16^{2^x}$
Find $x$ if $2^{16^x} = 16^{2^x}$.
If $x = 0$ we have $2=16$.
$8$ times or $(2^3)$ difference and if $x = 1$ we have $65536=256$.
$256$ or $2^8$ difference. I can't see anything useful in figuring this out.
I have a feeling $x$ must be a fraction. I know only know about logarithm. I haven't learned anything about the number $e$ or $ln$ yet.
| $16^{2^x}=(2^4)^{2^x}=2^{4\cdot 2^x}=2^{2^2\cdot 2^x}=2^{2^{x+2}}$. So
$$2^{16^x}=16^{2^x}\iff 2^{16^x}=2^{2^{x+2}}\iff$$ $$\iff 16^x=2^{x+2}\iff 2^{4x}=2^{x+2} \iff 4x=x+2 \iff x=\frac{2}{3}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution for $y(x^2+y^2)dx + x(3x^2-5y^2)dy = 0$. I am teaching assistant of an elementary course of ODE and a student asked me to solve the following equation:
$$y(x^2+y^2)dx + x(3x^2-5y^2)dy = 0$$
The standard procedure to solve it consists in multiplying the equation by some function $\mu = \mu(x,y)$. Once we do this we must solve
$$-8\mu(x,y)\cdot (x^2-y^2) +\partial_y\mu(x,y)\cdot \left(yx^2 + y^3\right) - \partial_x\mu(x,y)\left(3x^3-5xy^2\right) = 0.$$
In general one assumes that $\mu$ is constant in one of each variable, but this can not be the case here.
I tried lots of manipulations, but could not handle it. I do appreciate any hints.
| $$y(x^2+y^2)dx + x(3x^2-5y^2)dy = 0$$
Multiply by $y^2$:
$$y^3x^2dx+y^5dx + x^3dy^3-xdy^5 = 0$$
$$(y^3x^2dx + x^3dy^3)+(y^5dx-xdy^5) = 0$$
Divide by $x^2$:
$$y^3dx + xdy^3-d\left (\dfrac {y^5}x \right) = 0$$
$$d(xy^3)-d\left (\dfrac {y^5}x \right) = 0$$
Integrate:
$$xy^3-\dfrac {y^5}x = K$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I find this root? How can the third root of $\sqrt{x} - 2x + x^a$ be expressed in terms of a when $a>1.5$? (Obviously there is always one at 1 and 0)
| Beside the trivial $x=0$ and $x=1$, the third solution of the equation
$$\sqrt{x} - 2x + x^a=0$$ is close to $\frac 14$ as soon as $a >3$.
So, we can have approximations using Taylor series. For example
$$0=4^{-a}+\left(4^{1-a} a-1\right) \left(x-\frac{1}{4}\right)-2^{-2 a} \left(-8 a^2+8
a+4^{a}\right)
\left(x-\frac{1}{4}\right)^2+O\left(\left(x-\frac{1}{4}\right)^3\right)$$ which is a quadratic in $\left(x-\frac{1}{4}\right)$.
Using only the first term (this would be equivalent to the first iterate of Newton method
$$x_1=\frac{1}{4}+\frac{1}{4^a-4 a}$$ is quite decent. For $a=3$, it would give $x=\frac 7{26}=0.269231$ while the "exact" solution is $0.269143$.
But we can continue with series reversion and have
$$x_2=\frac 14+\frac{1}{4^a-4 a}+\frac{8 (a-1) a-4^a}{\left(4^a-4 a\right)^3}+\cdots$$
For $a=3$, this gives $x=\frac{2365}{8788}=0.269117$
Another one which is not bad
$$x_3=\frac 14+\frac{4^a-4 a}{8 a \left(a-4^a+1\right)+4^a+4^{2a}}$$
For the case where $a$ is close to $1.5^+$, the third solution is close to $1$ but, using the same method, it can be approximated as
$$x=1-\frac{4(2a-3)}{4 (a-1) a-1}-\frac{8 (2 a-3)^3 \left(4 a^2-6 a-1\right)}{3 (4 (a-1) a-1)^3}-$$
$$\frac{4 (2 a-3)^3(320 a^6-1728 a^5+3440 a^4-2784 a^3+452 a^2+300 a+27 )}{9 (4 (a-1) a-1)^5}$$
For example, for $a=2$, the above formula gives $x=\frac{6639}{16807}=0.395014$ while the exact solution is $x=\frac{3-\sqrt{5}}{2} =0.381966$.
| {
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The Volterra-Lotka model (predator prey equations) -- Linearization ODE
Q: Find all fixed points of the equation, linearize the equation, substitute the origin point $(0, 0)$ into it and solve the linear version of Volterra-Lotka model.
The system looks like this (where $a,b,c,g,y,x_{0}$ are constants):
\begin{align}
\frac{dx}{dt}&=ax-gx^2-by\left(x-x_0\right)\\
\frac{dy}{dt}&=-cy+dy\left(x-x_0\right)
\end{align}
My take:
Critical point: $$
\begin{pmatrix}0\\ 0\end{pmatrix}
$$
Jacobian Matrix:
$$
\begin{pmatrix}a&bx_0\\ 0&-c-dx_0\end{pmatrix}
$$
$$
P(\lambda) = -(a-\lambda)(c+dx_0 + \lambda) =0 \implies \begin{cases}
\lambda_1 = a \\
\lambda_2 = -c-dx_0
\end{cases}
$$
I get two matrices:
For V1:
$$
\begin{pmatrix}0&bx_0\\ 0&-c-dx_0-a\end{pmatrix}
$$
For V2:
$$
\begin{pmatrix}a+c+dx_{0}&bx_{0}\\ \:\:0&0\end{pmatrix}
$$
Eigenvector 1 :
$$
\begin{pmatrix}1\\ 0\end{pmatrix}
$$
Eigenvector 2 :
$$
\begin{pmatrix}-bx_0\\ a+c+dx_0\end{pmatrix}
$$
So the solution is :
$$
C_1\cdot e^{at}\begin{pmatrix}0\\ 1\end{pmatrix}\:+\:C_2\cdot e^{-t\left(c+dx_0\right)}\cdot \begin{pmatrix}-bx_0\\ a+c+dx_0\end{pmatrix}
$$
I don't think that that's the solution - maybe I have an error with Eigenvector 2 ?
We didn't really went over this material so I'm having a hard time.
| I made a few edits to your question for things that looked like typos, because they didn't show up in your final answer. In particular, you had
$$\begin{pmatrix}-\color{red}{3}x_0\\ a+c+dx_0\end{pmatrix}$$
instead of
$$\begin{pmatrix}-\color{red}{b}x_0\\ a+c+dx_0\end{pmatrix}$$ and you had $$\begin{pmatrix}a+c+d\color{red}{g}&b\:\color{red}{g}\\ \:\:0&0\end{pmatrix}$$ instead of $$\begin{pmatrix}a+c+d\color{red}{x_{0}}&b\color{red}{x_{0}}\\ \:\:0&0\end{pmatrix}.$$
There's also a typo in your final answer, in particular you flipped your first eigenvector, so it should be
$$\begin{pmatrix}x\\y\end{pmatrix} = C_1e^{at}\color{red}{\begin{pmatrix}1\\ 0\end{pmatrix}}+C_2 e^{-t\left(c+dx_0\right)}\begin{pmatrix}-bx_0\\ a+c+dx_0\end{pmatrix}.$$
This is indeed the correct answer, and you can verify that it satisfies your linearized system of ODEs:
\begin{align}
\frac{dx}{dt} &= ax + bx_{0}y\\
\frac{dy}{dt} &= -(c+dx_{0})y.
\end{align}
Differentiating your solution
\begin{align}
\frac{dx}{dt} &=C_{1}ae^{at} + C_{2}bx_{0}(c+dx_{0})e^{-(c+dx_{0})t}\\
&= C_{1}ae^{at} + C_{2}bx_{0}(c+dx_{0})e^{-(c+dx_{0})t} + C_{2}bx_{0}ae^{-(c+dx_{0})t} - C_{2}bx_{0}ae^{-(c+dx_{0})t}\\
&=a\left(C_{1}e^{at} - C_{2}bx_{0}e^{-(c+dx_{0})t}\right) + bx_{0}C_{2}(a + c + dx_{0})e^{-(c+dx_{0})t}\\
&=ax + bx_{0}y\\
\frac{dy}{dt} &= -(c+dx_{0})C_{2}(a+c+dx_{0})e^{-(c+dx_{0})t}\\
&=-(c+dx_{0})y.
\end{align}
| {
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To solve an arctan sum $\sum_{n=0}^{\infty} \arctan\frac{\sqrt3a_n(a_n-1)}{(a_n+1)(2a_n^2-3a_n+2)}$ I was suggested (without any background setting) to prove a numerically checked result, which is dramatically refined
$$
\sum_{n=0}^{\infty} \arctan\frac{\sqrt3a_n(a_n-1)}{(a_n+1)(2a_n^2-3a_n+2)} = \frac{\pi}{12}
$$
where $a_n=(1+\sqrt3)^{4^n}$, with $4^n$ on shoulder.
Usually we deal with arctan series using telescoping or gamma function technique, but this $a_n$ holds a geometrical growth and I have no idea how to do the reduction. Thanks in advance for any suggestion.
| Let $b_n = \frac{\sqrt{3}}{1 + 2a_n}$. Since $a_{n+1} = a_n^4$, we have
$b_{n+1} = \frac{\sqrt{3}}{1 + 2a_{n+1}} = \frac{\sqrt{3}}{1 + 2a_n^4}$.
With help of a CAS, one can verify
$$\frac{\frac{\sqrt{3}}{1+2x} - \frac{\sqrt{3}}{1+2x^4}}{1 + \frac{\sqrt{3}}{1+2x}\frac{\sqrt{3}}{1+2x^4}}
= \frac{\sqrt{3}x(x-1)}{(x+1)(2x^2-3x+2)}
$$
Substitute $x$ by $a_n$, $x^4$ by $a_{n+1}$ and sum over $n$. The sum
at hand becomes
$$\begin{align}\sum_{n=0}^\infty \tan^{-1}\frac{b_n - b_{n+1}}{1 + b_n b_{n+1}}
&= \sum_{n=0}^\infty\left(\tan^{-1}b_n - \tan^{-1}b_{n+1}\right)
= \tan^{-1}b_0\\ &= \tan^{-1}\frac{\sqrt{3}}{3+2\sqrt{3}}
= \tan^{-1}\frac{\frac12}{1+\frac{\sqrt{3}}{2}}
= \tan^{-1}\frac{\sin\frac{\pi}{6}}{1+\cos\frac{\pi}{6}}\\
&= \frac12\cdot \frac{\pi}{6} = \frac{\pi}{12}
\end{align}
$$
Update
In case anyone wonder how the magic $b_n$ is chosen. The basic idea is to rewrite the summands into the form $\tan^{-1}\frac{b_n - b_{n+1}}{1 + b_n b_{n+1}}$ so that one can use addition formula of tangent to turn the sum into a telescoping one. Since the summands have the form $\tan^{-1}\sqrt{3}(\cdots)$, one expect $b_n$ equals to $\sqrt{3}$ multiply something simple in $a_n$.
We know the series sum to $\tan^{-1}b_0$. Since $\tan\frac{\pi}{12} = \frac{\sqrt{3}}{3+2\sqrt{3}}$ and $a_0 = 1+\sqrt{3}$. The simplest candidate of $b_n$ comes to my mind is $b_n = \frac{\sqrt{3}}{1 + 2a_n}$. It turns out this works.
| {
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Finding a polynomial with integer coefficients that has $\sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$ as a root I'm looking for an (ideally) elementary way to find a polynomial with integer coefficients that has $$\sqrt[100]{\sqrt{3} + \sqrt{2}} + \sqrt[100]{\sqrt{3} - \sqrt{2}}$$ as a root.
What I have so far is that we can write $\sqrt 3 - \sqrt 2$ as $1/(\sqrt 3 + \sqrt 2)$ and then we can quite easily check that both $\sqrt 3 + \sqrt 2$ and $\sqrt 3 - \sqrt 2$ solve $$x^4-10x^2+1=0.$$
So then we have that $\sqrt[100]{\sqrt{3} + \sqrt{2}}$ and $\sqrt[100]{\sqrt{3} - \sqrt{2}}$ both solve $$x^{400}-10x^{200}+1=0$$
but I am not sure how to find an integer-coefficient polynomial with the sum of those two values as a root. It sort of looks like Vieta formulas could somehow be relevant, but I may be missing something obvious here.
| Hint
Denote
$$\begin{cases}
a &= \sqrt[100]{\sqrt 3 + \sqrt 2}\\
b &= \sqrt[100]{\sqrt 3 - \sqrt 2}
\end{cases}$$
and notice that $ab=1$. We have $a^{100} + \frac{1}{a^{100}} = 2\sqrt 3$.
Now you can write
$$a^{100} + \frac{1}{a^{100}} = \left(a + \frac{1}{a}\right)^{100} - (p(a)+p(1/a))$$ where $p(x)$ is a polynomial of degree at most equal to $99$ with integer coefficients. You can repeat this process by induction to finally write $a^{100} + \frac{1}{a^{100}} = q(a+1/a)$ where $q$ is a polynomial of degree $100$ with integer coefficients.
At the end you get
$$q(a+b) =2 \sqrt 3$$ which implies that $a+b$ can't be a rational number as $\sqrt 3$ isn't. And that $a+b$ is a root of the polynomial with integer coefficients $$q^2(x) - 12=0$$ as desired.
| {
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Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
My Try
$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$
I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!
| Abbreviating $\sin x$ and $\cos x$ to $s$ and $c$, and writing $\sin2x=2sc$ and $\cos2x=2c^2-1$, the equation becomes
$$6sc+4(2c^2-1)-2c+6s-6=0$$
Factoring out a $2$ and grouping terms, we rewrite this as
$$3s(c+1)+4c^2-c-5=0$$
But $4c^2-c-5=(4c-5)(c+1)$, so we have two solutions in $s$ and $c$:
$$c=-1\quad\text{and}\quad3s+4c=5$$
Reverting to $\sin x$ and $\cos x$ and recognizing the Pythagorean triple $3^2+4^2=5^2$, these become
$$\cos x=-1\quad\text{and}\quad\sin(x+\theta)=1$$
where $\sin\theta=4/5$ and $\cos\theta=3/5$, e.g. $\theta=\arcsin(4/5)$. So the solution set is
$$\{\pi+2n\pi\mid n\in\mathbb{Z}\}\cup\{\pi/2-\arcsin(4/5)+2n\pi\mid n\in\mathbb{Z} \}$$
Remark: The abbreviations $s$ and $c$ are merely a convenience. Once I get all the sines and cosines down to $\sin x$ and $\cos x$, I find it tedious to write the functions out in full over and over again.
| {
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Prove arithmetic formula for the number of bits in $j=2^k-1$ If $k$ is an integer with $0\le k<155$, and $j=2^k-1$, then it holds:
$$k\,=\,5\left(\left\lfloor\frac j{31((j\bmod31)+1)}\right\rfloor\bmod 31\right)-\left\lfloor\frac{12}{(j\bmod31)+3}\right\rfloor+4$$
Why does this equation hold for such a large interval of $k$? How can we prove that?
This is a shortened version of more complex formula# posted by Hallvard B. Furuseth in a comp.c post on 2003-11-03, which reportedly works to $k$ of at least $3\cdot10^{10}$.
Update (not part of the bountied question): per this source, the same author gave that formula working for $0\le k<2040$:
$$k\,=\,8\left(\left\lfloor\frac j{255((j\bmod255)+1)}\right\rfloor\bmod 255\right)-\left\lfloor\frac{86}{(j\bmod255)+12}\right\rfloor+7$$
Try it online with Python or with Wolfram Mathematica! The code computes the upper limits for both formulas.
# Full formula, for reference (not in the above code or part of the bountied question):
$$\begin{align}k\,=\,&30\left(\left\lfloor\frac j{((j\bmod(2^{30}-1))+1)\,(2^{30}-1)}\right\rfloor\bmod (2^{30}-1)\right)+\\&5\left(\left\lfloor\frac{j\bmod(2^{30}-1)}{31((j\bmod31)+1)}\right\rfloor\bmod 31\right)-\left\lfloor\frac{12}{(j\bmod31)+3}\right\rfloor+4\end{align}$$
| There exist integers $m,r$ such that
$$k=5m+r,\qquad 0\le m\le 30,\qquad 0\le r\le 4$$
Since we have
$$j=2^k-1=2^{5m+r}-1=(2^5)^m\cdot 2^r-1\equiv 1^m\cdot 2^r-1\equiv 2^r-1\pmod{31}$$
there exists an integer $a$ such that $2^{5m+r}-1=31a+2^r-1$ which can be written as $$a=2^r(2^{5-r}a+1-2^{5m})$$ This implies that $a$ is divisible by $2^r$, so there exists an integer $b$ such that $j=2^{5m+r}-1=31\cdot 2^rb+2^r-1$.
We have $j\bmod 31=2^r-1$ and $$b\bmod 31=\frac{32^m-1}{31}\bmod 31=(32^{m-1}+32^{m-2}+\cdots +32+1)\bmod 31=m$$
Using these, we get
$$\begin{align}&5\left(\left\lfloor \frac{j}{31((j \bmod 31)+1)}\right\rfloor \bmod 31\right)-\left\lfloor\frac{12}{(j \bmod 31)+3}\right\rfloor+4
\\\\&=5\left(\left\lfloor \frac{31\cdot 2^rb+2^r-1}{31\cdot 2^r}\right\rfloor \bmod 31\right)\underbrace{-\left\lfloor\frac{12}{2^r+2}\right\rfloor+4}_{=\ r}
\\\\&=5\left(\left\lfloor b+\frac{2^r-1}{31\cdot 2^r}\right\rfloor \bmod 31\right)+r
\\\\&=5\left(b \bmod 31\right)+r
\\\\&=5m+r
\\\\&=k\end{align}$$
| {
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Proving a specific eigenvalue of a 6x6 Matrix. Currently, I am looking at Markov diagrams and their associated transition matrices. I am trying to prove that the transition matrix has eigenvalue $\lambda=1$. I am aware that to find the eigenvalues of a matrix we use:
$$\det (A - \lambda I_{6}) = 0$$
and then each $\lambda$ gives an eigenvalue of the matrix. But this seems like a very lengthy process for a $6 \times 6$ matrix.
I'm also aware that every stochastic matrix has the eigenvalue of $\lambda=1$ but I would like to learn how to work this out for this specific matrix.
$$
\begin{bmatrix}
0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & 0 \\
\frac{1}{2} & 0 & 0 & \frac{1}{3} & 0 & 0 \\
\frac{1}{2} & 0 & 0 & \frac{1}{3}& \frac{1}{2}& 0 \\
0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & \frac{1}{2} \\
0 & 0 & \frac{1}{3}& 0 & 0 & \frac{1}{2} \\
0 & 0 & 0 & \frac{1}{3} & \frac{1}{2} & 0 \\
\end{bmatrix}
$$
Any help would be greately appreciated.
edit
Having using the eigenvector of $\large[1\ 1\ 1\ 1\ 1\ 1\ \large]$
to show that $\lambda =1$ is an eigenvalue I now want to find a stable distribution.
So I am going to labe my horizontal and vertical entries M, E, C, S, W, N respectively. and will set $\large[M\ E\ C\ S\ W\ N \large]$ to be a vector to multiply my transition matrix by.
So i get:
$1/2E+1/3C = M$
$1/2M+1/3S=E$
$1/2M+1/3S+1/2W=C$
$1/2E+1/3C+1/2N=S$
$1/3C+1/2N=W$
$1/3S+1/2W=N$
So Im wondering if what I'm doing is right and whether my next steps should be to try and work out what the values of M E C S W N are?
| HINT
If you are looking for a specific eigenvalue, compute the matrix $B = A - \lambda I$, and show that $\det(B) = 0$.
If you can guess the corresponding eigenvector $\vec{x}$, it becomes even easier. Multiply matrices to show that
$$
A \vec{x} = \lambda \vec{x}
$$
and you are done.
In your case, however, it is even easier. The sum of all entries in each column is $1$, which makes you matrix stochastic. For such a matrix, the vector of all $1$'s is a left eigenvector, in other words,
$$
\begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix}
\begin{bmatrix}
0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & 0 \\
\frac{1}{2} & 0 & 0 & \frac{1}{3} & 0 & 0 \\
\frac{1}{2} & 0 & 0 & \frac{1}{3}& \frac{1}{2}& 0 \\
0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & \frac{1}{2} \\
0 & 0 & \frac{1}{3}& 0 & 0 & \frac{1}{2} \\
0 & 0 & 0 & \frac{1}{3} & \frac{1}{2} & 0 \\
\end{bmatrix}
= \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix}
$$
UPDATE
Typically to find the eigenvectors and eigenvalues of a matrix $A$, first solve $\det(A-\lambda I)=0$ and then when you get the eigenvalues, plug them into $(A - \lambda I) \vec{x}=\vec{0}$ and solve for each $\vec{x}$ separately.
| {
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Use of definition of limit to prove $\lim_{x→2} \dfrac{x}{x^2-2}=1$ I know that by definition I have to prove that $$\lim_{x→2}\dfrac{x}{x^2−2}=1⟺∀ϵ>0,∃δ>0,\, 0<|x−2|<δ⟹\left|\frac{x}{x^2−2}−1\right|<ϵ.$$
I have:
$\left|\dfrac{-x^2+x+2}{x^2-2}\right|= \left|\dfrac{-(x^2-x-2)}{x^2-2}\right|= \left|\dfrac{-(x^2-x+1-3)}{x^2-2}\right|=$
$\left|\dfrac{-(x-1)^2+3}{x^2-2}\right|$.
So, I don't know how to continue.
| Two points: 1) It's often much easier to deal with $x\to 0$ than $x\to c\neq 0$; and 2) you may want to avoid fractions whenever you can.
So let's denote $d=x-2$ and prove $\lim_{d\to 0} \dfrac{d+2}{(d+2)^2-2} = 1$.
$$
\left| \dfrac{d+2}{(d+2)^2-2} - 1 \right|=\left| \dfrac{d+2-d^2-4d-2}{d^2+4d+2} \right| = \left| d \right| \left| \dfrac{d+3}{d^2+d+2} \right|
$$
If $|d|<1, |d+3| \le |d| + 3<4$. Also note $|d^2+d+2|=\left(d+\dfrac{1}{2}\right)^2+\dfrac{7}{4} > 1$.
Therefore, if $|d| < \delta = \min(1, \dfrac{1}{4} \varepsilon )$
$$\left| d \right| \left| \dfrac{d+3}{d^2+d+2} \right| < \dfrac{1}{4} \varepsilon \cdot \dfrac{4}{1} = \varepsilon.$$
| {
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Limit $\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$ Determine whether or not the limit below exists.
$$\lim_{x\rightarrow0}\frac{1}{x}\int_{0}^{\sin x}\sin \frac{1}{t} \cos t^{2}\mathrm{d}t$$
I tried to use the Mean value theorem integrals to prove the limit exists, but it does not exist for $\lim\limits_{x\rightarrow0}\sin \frac{1}{x}$. So I guessed the limit does not exist and used the Cauchy principle to prove it, but I failed.
Any idea will be helpful.
| Since $\cos t^2$ is monotone decreasing for $t$ sufficiently close to $0$, by the second mean value theorem for integrals there exists $\xi_x \in (0, \sin x)$ such that
$$\int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = \cos(0) \int_0^{\xi_x}\sin \frac{1}{t} \, dt = \int_0^{\xi_x}\sin \frac{1}{t} \, dt$$
Taking $g(t) = t^2 \cos \frac{1}{t}$ for $t > 0$ and $g(0) = 0$, we have $g’(0) =0$ and for $t>0$,
$$g'(t) = 2t \cos \frac{1}{t} + \sin \frac{1}{t},$$ and, using the FTC,
$$\int_0^{\xi_x}\sin \frac{1}{t} \, dt = \int_0^{\xi_x} g'(t) \, dt-\int_0^{\xi_x}2t \cos \frac{1}{t} \, dt=\xi_x^2\cos \frac{1}{\xi_x} - \int_0^{\xi_x}2t \cos \frac{1}{t} \, dt$$
Thus,
$$\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = \xi_x \frac{\xi_x}{x}\cos \frac{1}{\xi_x} - \frac{1}{x} \int_0^{\xi_x}2t \cos \frac{1}{t} \, dt$$
We can apply the mean value theorem to the integral on the RHS (since the integrand is continuous) to find $\theta_x \in (0,\xi_x)$ such that
$$\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = \xi_x \frac{\xi_x}{x}\cos \frac{1}{\xi_x} - \frac{\xi_x}{x} 2\theta_x \cos \frac{1}{\theta_x} $$
Since $\xi_x/x < 1$, we have
$$\left|\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt\right| \leqslant \xi_x +2 \theta_x$$
Since $\xi_x , \theta_x \to 0$ as $x \to 0+$, we get
$$\lim_{x \to 0+}\frac{1}{x} \int_0^{\sin x} \sin \frac{1}{t} \cos t^2 \, dt = 0$$
Similarly we can show that the limit as $x \to 0-$ is $0$ as well.
| {
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Question regarding surjectivity of a nonzero function satisfying $ f \big( a + 2 f ( a ) f ( b ) \big) = f ( a ) + 2 a f ( b ) $ I was practicing that how to prove surjectivity of functions in a functional equation and I came across the following question.
Let $ f : \mathbb R \to \mathbb R $ be a function such that
$$ f \big( a + 2 f ( a ) f ( b ) \big) = f ( a ) + 2 a f ( b ) $$
for all $ a , b \in \mathbb R $. Also, suppose that $ f $ is not the all-zero function. Prove that $ f $ is surjective.
I really don't have any idea about how to prove it. A pedantic and easiest possible proof will be highly appreciated and would be helpful.
| Consider a function $ f : \mathbb R \to \mathbb R $ such that
$$ f \big( a + 2 f ( a ) f ( b ) \big) = f ( a ) + 2 a f ( b ) \tag 0 \label 0 $$
for all $ a , b \in \mathbb R $ which is not the constant zero function. Setting $ a = b = - \frac 1 2 $ in \eqref{0} gives
$$ f \left( - \frac 1 2 + 2 f \left( - \frac 1 2 \right) ^ 2 \right) = 0 \text . \tag 1 \label 1 $$
For any $ a \in \mathbb R $ with $ f ( a ) = 0 $, \eqref{0} gives $ a f ( b ) = 0 $. In case $ a \ne 0 $, that shows that $ f $ is the constant zero function, contrary to the assumption. Thus such $ a $ must be equal to $ 0 $, which by \eqref{1} not only shows that $ f ( 0 ) = 0 $, but also $ - \frac 1 2 + 2 f \left( - \frac 1 2 \right) ^ 2 = 0 $, or equivalently $ f \left( - \frac 1 2 \right) = - \frac k 2 $ where $ k ^ 2 = 1 $, i.e. $ k \in \{ - 1 , 1 \} $. Now, setting $ b = - \frac 1 2 $ in \eqref{0} shows that
$$ f \big( a - k f ( a ) \big) = f ( a ) - k a \text , \tag 2 \label 2 $$
while plugging $ a = - \frac 1 2 $ in \eqref{0} gives
$$ f \left( - \frac 1 2 - k f ( b ) \right) = - \frac k 2 - f ( b ) \text . \tag 3 \label 3 $$
By setting $ a = \frac 1 2 $ in \eqref{2} we have $ f \Big( \frac 1 2 - k f \left( \frac 1 2 \right) \Big) = f \left( \frac 1 2 \right) - \frac k 2 $, which shows that letting $ b = \frac 1 2 - k f \left( \frac 1 2 \right) $ in \eqref{3} we get
$$ f \Bigg( - k f \left( \frac 1 2 \right) \Bigg) = - f \left( \frac 1 2 \right) \text . \tag 4 \label 4 $$
By plugging $ a = \frac 1 2 $ and $ b = - k f \left( \frac 1 2 \right) $ in \eqref{0} and using \eqref{4} we get $ f \Big( \frac 1 2 - 2 f \left( \frac 1 2 \right) \Big) = 0 $, which means that we must have $ f \left( \frac 1 2 \right) ^ 2 = \frac 1 4 $, or equivalently $ f \left( \frac 1 2 \right) = \pm \frac k 2 $. But by \eqref{4}, $ f \left( \frac 1 2 \right) = - \frac k 2 $ leads to $ f \left( \frac 1 2 \right) = - f \left( \frac 1 2 \right) $, contradicting $ f \left( \frac 1 2 \right) ^ 2 = \frac 1 4 $. Thus we have $ f \left( \frac 1 2 \right) = \frac k 2 $.
Now, assume that for some $ c \in \mathbb R $ we have $ f ( c ) = - k c $. By \eqref{3} we have $ f \left( - \frac 1 2 + c \right) = k \left( - \frac 1 2 + c \right) $. Then, setting $ x = \frac 1 2 $ and $ y = - \frac 1 2 + c $ in \eqref{0}, we get $ f ( c ) = k c $, which means that we must have $ c = 0 $. By \eqref{2}, we know that for any $ a \in \mathbb R $, letting $ c = a - k f ( a ) $ we have $ f ( c ) = - k c $, and therefore we must have $ k f ( a ) = a $, or equivalently $ f ( a ) = k a $. Hence, $ f $ has to either be the identity function or its opposite. It's straightforward to verify that those are both surjective solutions.
\eqref{0} is a variant of the functional equation
$$ f \big( a + f ( a ) f ( b ) \big) = f ( a ) + a f ( b ) \text , $$
which has appeared in The American Mathematical Monthly, with an official answer here. This question has been asked many times on AoPS, and you can for example find solutions here, here, here, here and here. The official answer and several others first prove surjectivity and then use that to characterize the solutions of the equation. The answer I gave above is similar to the one in the last linked page. Not only it doesn't appeal to surjectivity first, but also it's shorter and simpler than all others.
Wu Wei Chao, and The University of Louisiana at Lafayette Math Club, Two Functional Equations: 11053, The American Mathematical Monthly 112 (2005), no. 7, 661--662.
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Calculus Integral Question
Question: Given the point (a, b) in the first quadrant, find the downward-opening parabola that passed through the point (a, b) and the origin such that the area under the parabola is a minimum.
I've tried to solve it. But the answer can't be right, since it leads to a upward-opening parabola.
Here's what I've done so far.
Step One:
Let $$y = cx^2 + dx + e$$ be the equation for the parabola.
Because the parabola has a root at the origin, $$e = 0$$ The function for parabola can be written as $$y = cx^2 + dx$$
Step Two:
We are given that the parabola passes through the point (a, b) in the first quadrant. Plugging the numbers into the function above, $$b = c(a)^2 + d(a)$$ $$\frac{b-a^2c}{a}=d $$
Therefore, $$y=cx^2+\frac{b-a^2c}{a}x$$
Step Three:
Next, to find the area under the parabola, use the integral: $$\int_0^n(cx^2+\frac{b-a^2c}{a}x)dx$$where n is the second root of the parabola (first root is the origin).
For n, $$0 = cx^2+\frac{b-a^2c}{a}x$$ $$0 = cx+\frac{b-a^2c}{a}$$ $$x=-\frac{b-a^2c}{ac} = n$$
Now, substitute the expression for n, $$\int_0^{-\frac{b-a^2c}{ac}} (cx^2+\frac{b-a^2c}{a}x)dx$$ which equals $$\frac{c}{3}\bigl(-\frac{b-a^2c}{ac}\bigr)^3+\frac{b-a^2c}{2a}\bigl(-\frac{b-a^2c}{ac}\bigr)^2$$ which equals $$\frac{b-a^2c}{6a^3c^2}$$
It can be concluded that the area under the parabola has the function $$A=\frac{b-a^2c}{6a^3c^2}$$
Step Four:
To find the minimum area -- as requested, take the derivative of A. $$A'=\frac{6a^3c^2(-a)^2-(b-a^2c)(12a^3c)}{(6a^3c^2)^2}$$ $$=\frac{a^2c-2b}{6a^3c^3}$$ Find its root: $$0=\frac{a^2c-2b}{6a^3c^3}$$ $$0=a^2c-2b$$ $$c=\frac{2b}{a^2}$$
Since (a, b) is in the first quadrant, c is a positive number.
Can you tell me where have I gone wrong?
Thanks!
| Please note
$\displaystyle A = \frac{c}{3}\bigl(-\frac{b-a^2c}{ac}\bigr)^3+\frac{b-a^2c}{2a}\bigl(-\frac{b-a^2c}{ac}\bigr)^2 = \frac{(b - a^2 c)^3} {6 a^3 c^2}$
Taking derivatative wrt. $c$ and equating to zero,
$\displaystyle A' = 0 = \frac{2(b-a^2c)^3 + 3ca^2(b-a^2c)^2}{6a^3c^3}$
This gives us $\displaystyle c = \frac{b}{a^2}, c = - \frac{2b}{a^2}$
| {
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Approximation of $\arcsin(\sqrt{1-x^2})$ by $2\arcsin(\sqrt{\frac{1-x}2})$ I stumbled upon an use of $2\arcsin(\sqrt{\frac{1-x}2})$ to approximate $\arcsin(\sqrt{1-x^2})$. The latter is equal to $\frac\pi 2-\arcsin(x)$ when $0\le x\le 1$ and is useful if you have a good approximation of $\arcsin$ around $0$ and want to use it for values near $1$. On the flip side the starting approximation needs to cover $[-\frac1{\sqrt2},\frac1{\sqrt2}]$ in order to use the previous fact to approximate $\arcsin$ on the entire domain.
However if we use the fact that $\arcsin(\sqrt{1-x^2})\approx2\arcsin(\sqrt{\frac{1-x}2})$, our starting approximation only needs to cover $[-0.5, 0.5]$. Empirically this approximation is extremely good, but I'm wondering how would one come up with this kind of approximation and are there any insights to explain why the two are so close?
| Let $\theta = \arcsin \sqrt{1-x^2}$. Since the argument of $\arcsin$ is non-negative, consider the range $0\le \theta\le \frac \pi2$.
$$\begin{align*}
\sin\theta &= \sqrt{1 - x^2}\\
\sin^2\theta &= 1 - x^2\\
x &= \cos \theta\\
\end{align*}$$
Then using half-angle formula,
$$\begin{align*}
1-2\sin^2\frac\theta2&= \cos\theta\\
\sin^2\frac\theta2 &= \frac{1-\cos \theta}2\\
\sin\frac\theta 2&= \sqrt{\frac{1-\cos \theta}2}\\
&= \sqrt{\frac{1-x}2}\\
\frac\theta2 &= \arcsin \sqrt{\frac{1-x}2}\\
\theta &= 2\arcsin \sqrt{\frac{1-x}2}
\end{align*}$$
| {
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Determine for which $a$ it holds that $\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=0$ Let $a>0$.
Determine for which values of $a$ it holds that:
$$\lim_{(x,y)\to(0,0)}\frac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=0$$
The solution indicated is $a>3$.
Passing in polar coordinates, I obtain:
$$\frac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=\frac{\rho^{a-3}|\cos\theta-\sin\theta|^{a-1}}{(\cos\theta+\sin\theta)}$$
If $a > 3$, when $\rho \to 0$, the fraction goes to $0$, and this does not depend on $\theta$.
However, this is not enough for the limit to exist and it must exist a $g(\rho)$ such that:
$$\frac{\rho^{a-3}|\cos\theta-\sin\theta|^{a-1}}{\cos\theta+\sin\theta}\leqslant g(\rho)\to 0 $$
But
$$\frac{|\cos\theta-\sin\theta|^{a-1}}{\cos\theta+\sin\theta}$$
is not limited, because, for example, when $\theta \to \pm \frac{3}{4}\pi$, it goes to $\mp \infty$.
Where is the mistake?
Thanks in advance.
|
Determine for which values of $\;a\;$ it holds that:
$$\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=0\;.$$
By letting $\;X=x-y\;$ and $\;Y=x+y\;,\;$ it follows that
$\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}\;.$
First case : $\;a\le1\;.$
$\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2}{Y|X|^{1-a}\sqrt{X^2+Y^2}}\;,$
but the last limit does not exist because we get different results when we calculate the limits of the restrictions of the function to two different subsets of its domain, indeed
$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y>0\end{align}}\dfrac{\sqrt2}{Y|X|^{1-a}\sqrt{X^2+Y^2}}=$
$=\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y>0\end{align}}\dfrac1{Y|X|^{1-a}}$$\cdot\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y>0\end{align}}\dfrac{\sqrt2}{\sqrt{X^2+Y^2}}=$
$=+\infty\cdot(+\infty)=+\infty\;,$
$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y<0\end{align}}\dfrac{\sqrt2}{Y|X|^{1-a}\sqrt{X^2+Y^2}}=$
$=\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y<0\end{align}}\dfrac1{Y|X|^{1-a}}$$\cdot\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y<0\end{align}}\dfrac{\sqrt2}{\sqrt{X^2+Y^2}}=$
$=-\infty\cdot(+\infty)=-\infty\;.$
Second case : $\;a>1\;.$
The limit $\;\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}\;$ does not exist because we get different results when we calculate the limits of the restrictions of the function to two different subsets of its domain, indeed
$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&X=0\end{align}}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}=$$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&X=0\end{align}}\dfrac0{Y\big|Y\big|}=0\;,$
$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y=|X|^{a-1}\end{align}}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}=$
$=\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y=|X|^{a-1}\end{align}}\dfrac{\sqrt2}{\sqrt{X^2+X^{2(a-1)}}}=+\infty\;.$
Conclusion :
In any case, for any $\;a\in\mathbb{R}\;,$ the limit
$\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}$
does not exist.
| {
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Compute difficult integral $\int \frac{dx}{2 + x + \sqrt{1 - x^2}}$ To solve the integral
$$I = \int \frac{dx}{2 + x + \sqrt{1 - x^2}}$$
I have tried several things, such as $t = \arcsin x$, because $\cos(\arcsin x) = \sqrt{1 - x^2}$. If I am not wrong, we can conclude with this variable change
$$
I = \int \frac{\cos t\,dt}{2 + \sin t + \cos t}
$$
but if it were correct, how could I go on?
| Use Euler substitution
$$tx-1=\sqrt{1-x^2}$$
$$t=\frac{1+\sqrt{1-x^2}}{x}$$
$$x=\frac{2t}{t^2+1}$$
$$dx=\frac{2(1-t^2)}{(t^2+1)^2}dt$$
$$\int \frac{dx}{2+x+\sqrt{1-x^2}}=\int\frac{1}{2+(t+1)\frac{2t}{t^2+1}-1}\frac{2(1-t^2)}{(t^2+1)^2}dt=\int\frac{2(1-t^2)dt}{(3t^2+2t+1)(t^2+1)}$$$$=-\int\frac{t+1}{t^2+1}dt+\frac{1}{2}\int\frac{6t+2}{3t^2+2t+1}dt+\frac{2}{3}\int\frac{dt}{(t+\frac{1}{3})^2+\frac{2}{9}}$$$$=-\frac{1}{2}\ln(t^2+1)-\arctan(t)+\frac{1}{2}\ln(3t^2+2t+1)+\frac{2}{3}\frac{1}{\frac{2}{9}}\arctan\left(\frac{t+\frac{1}{3}}{\frac{2}{9}}\right)+C$$
$$=\frac{1}{2}\ln\left(\frac{3t^2+2t+1}{t^2+1}\right)-\arctan(t)+3\arctan\left(\frac{9t+3}{2}\right)+C$$
where $t=\frac{1+\sqrt{1-x^2}}{x}$
| {
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Solving $\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$ for $x$ Suppose I have an expression as follows
$$\sqrt{x^2+ay^2} - \sqrt{x^2+y^2} = z$$
How would I go about rearranging this to deduce a value for x?
Thanks for the help.
| Bring $\sqrt{x^2+y^2}$ to the right
$\sqrt{x^2+ay^2}=z+\sqrt{x^2+y^2}$
Square both sides
$x^2+ay^2= z^2+2z\sqrt{x^2+y^2}+x^2+y^2$
Get rid of $x^2$ on both sides, group together the $y^2$ terms and bring $z^2$ to the left
$(a-1)y^2-z^2=2z\sqrt{x^2+y^2}$
Square both sides again to get rid of the $\sqrt{}$ sign on the right
$\left((a-1)y^2-z^2\right)^2=4z^2(x^2+y^2)$
...at least that’s how Stewart Calculus does it.
| {
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How to show that if $a^2+b^2+c^2=1$ then $\frac{-1}{2}\le ab+ac+bc \le 1$ How to show that if $a^2+b^2+c^2=1$ then:
$$\frac{-1}{2}\le ab+ac+bc \le 1$$
From the assumption I see that:
$$-\frac{1}{2}\le\frac{\left(a+b+c\right)^{2}-1}{2}=\frac{\left(a+b+c\right)^{2}-\left(a^{2}+b^{2}+c^{2}\right)}{2}=ab+ac+bc $$
On the other hand $a^2=1-(b^2+c^2)<1$,then same does hold for $b^2$ and $c^2$ which shows that $-1<a<1$ and so :$$-3<a+b+c<3$$ $$\implies ab+ac+bc =\frac{\left(a+b+c\right)^{2}-1}{2}<4$$
But it's does not what I've been asked to prove,so how can I show that?
| Hint:
For the 2 inequalities you want to prove, expand the left hand sides of:
\begin{align}
(a+b+c)^2&\geq 0,\\
(a-b)^2+(b-c)^2+(c-a)^2&\geq 0.
\end{align}
And remember to use $a^2+b^2+c^2=1$.
| {
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Can't find the inverse of Laplace Transform. Find $\mathcal{L}^{-1}{(F(s))}$, if given $F(s)=\dfrac{2}{s(s^2+4)}$.
I have tried as below.
To find inverse of Laplace transform, I want to make partial fraction as below.
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+4}=\dfrac{(A+B)s^2+Cs+4A}{s(s^2+4)}.
\end{align*}
After that, we have system of linear equation
\begin{align*}
A+B&=0\\
C&=0\\
4A&=2.
\end{align*}
Thus we have $A=2$, $B=-2$, and $C=0$.
Now, substituting $A, B, C$ and we have
\begin{align*}
\dfrac{2}{s(s^2+4)}=\dfrac{2}{s}+\dfrac{-2s}{s^2+4}.
\end{align*}
But the fact is
\begin{align*}
\dfrac{2}{s(s^2+4)}\neq \dfrac{2}{s}+\dfrac{-2s}{s^2+4} = \dfrac{8}{s^2+4}.
\end{align*}
I'm stuck here. I can't make a partial fraction for $F(s)$ and I can't find inverse of Laplace transform for $F(s)$.
Anyone can give me hint to give me hint for this problem?
| Note that $$\frac{2}{s(s^{2}+4)}=\frac{1}{2s}-\frac{s}{2(s^{2}+4)}$$
So, you have $$\mathcal{L}^{-1}\left\{\frac{2}{s(s^{2}+2)} \right\}=\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{s} \right\}-\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{s}{s^{2}+4} \right\}$$
Finally, you can use $$\mathcal{L}^{-1}\left\{ \frac{s}{s^{2}+4}\right\}=\cos(2t)$$
and $$\mathcal{L}^{-1}\left\{ \frac{1}{s}\right\}=1$$
| {
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How to read and execute $\sum_{1 \leq \ell
How to read and execute this sum?
$$\sum_{1 \leq \ell <m<n} \frac{1}{5^{\ell}3^{m}2^{n}}$$
I am having trouble to understand where is my error.
The question does not say, but I am assuming that $\ell$ starts at $1$, $m$ at $2$, and so $n$ at $3$.
This is essential a product of pg:
$$\sum_{1 \leq \ell<m<n} \frac{1}{5^{\ell}3^{m}2^{n}} = \sum_{\ell=1}\frac{1}{5^{\ell}}\sum_{m=2}\frac{1}{3^{m}}\sum_{n=3}\frac{1}{2^{n}} = \frac{1/5}{1-1/5}\frac{1/9}{1-1/3}\frac{1/8}{1-1/2}$$
But this does not agree with the answer :/
| *
*Assuming $n$ is not fixed:
Rewrite as triple sum
$$\sum_{1 \leq \ell <m<n} \frac{1}{5^{\ell}3^{m}2^{n}} = \sum_{\ell=1}^\infty\sum_{m=\ell+1}^\infty\sum_{n=m+1}^\infty\frac{1}{5^{\ell}3^{m}2^{n}} =\sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \sum_{n=m+1}^\infty \frac{1}{2^n} $$
Then, this should be obvious in terms of how to read and execute.
Indeed, since each of them are Geometric Progressions, we have
$$\begin{align}\sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \sum_{n=m+1}^\infty \frac{1}{2^n} &= \sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \left(\frac{1}{2^m}\right) \\&= \sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{6^m} \\&= \sum_{\ell=1}^\infty \frac{1}{5^\ell}\left(\frac{1}{5\cdot 6^\ell}\right) \\& = \frac{1}{5}\sum_{\ell=1}^\infty \frac{1}{30^\ell} \\&= \frac{1}{5}\cdot\frac{1}{29} \\&= \frac{1}{145}\end{align} $$
*Assuming $n$ is fixed:
The sum will be a finite double sum and you can exclude $1/2^n$ from the sum(treating that as a constant). Then, again use the GP formula to calculate each $\ell$-sum and $m$-sum.
| {
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Show that if $m$, $n$, $m^2+n^3$, $m^3+n^2$ are all prime, then $(m,n)=(2,3)$ or $(3,2)$ if $m$, $n$, $m^2+n^3$, $m^3+n^2$ are all prime
then $(m,n)=(2,3)$ or $(3,2)$
I found out that one of the $m$ and $n$ must be two since $n^2+m^3$ will be even if $m$ and $n$ have the same parity.
However, how to prove that if $4+n^3$ and $8+n^2$ are both prime, then $n=3$?
| If $n \neq 3$ and is also prime, then $n \equiv 1 \pmod{3} \implies 8 + n^2 \equiv 0 \pmod{3}$, or $n \equiv 2 \pmod{3} \implies 4 + n^3 \equiv 0 \pmod{3}$. Since both $4 + n^3$ and $8 + n^2$ are greater than $3$, this means at least one of them can't be prime.
| {
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Prove relation about area of triangle
In the below picture, angle A is obtuse, AD is a median.
We are also given the relation $AB^2 = AF*AC$.
We want to prove that area of triangle $(ABC) = AB*AD$.
What I have tried:
Area of triangle is $A = \frac {1}{2}*AC*BF$.
Replacing AC from the given relation $AB^2 = AF*AC$, we get $A = \frac {1}{2}*\frac{AB^2*BF}{AF}$.
Also from Pythagoras, we replace $AB^2$ with $BF^2+AF^2$.
But I can't see how to involve AD and prove the required relation!
Any help please?
| $BE$ parallel to $DA$, $E$ on the line $CA$
Let $AB=c$, $AC=b$, $BC=2a$ , $AD=d$
we have $BE=2d$
Given $AF = \frac{c^2}{b}$
We get $EF=b-\frac{c^2}{b}$, $CF=b+\frac{c^2}{b}$
using Pythagoras, express $BF^2$ in two different ways
$4d^2-\left(b-\frac{c^2}{b}\right)^2=4a^2-\left(b+\frac{c^2}{b}\right)^2$
it follows from this that $d^2+c^2=a^2$
and so $BAD$ is a right angle, area $ABD = \frac{1}{2} cd$ and therefore area $ABC=cd$
| {
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Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$. I am working on inequality problems for mathematical olympiads. I have come across this problem:
Find the maximum and minimum value of $2x^2 + y^2 + z^2$ subject to $x + y + z = 10$, with $x,y,z \ge 0 $
I found the minimum using Cauchy-Schwarz: $$100 = \left(\frac 1 {\sqrt 2} \left(\sqrt 2 x\right) + y + z\right)^2 \le \left(\left(1/\sqrt 2\right)^2 + 1^2 + 1^2\right) \left(\left(\sqrt 2 x\right)^2 + y^2 + z^2\right)=2.5\left(2x^2 + y^2 + z^2\right)$$
so $2x^2 + y^2 + z^2 \ge 40$ with equality if $x=2$ and $y=z=4$.
I am not sure how to find the maximum though. Could I have a hint? Ideally I would like to do it without using calculus, though if it is easy to do with calculus feel free to share the method.
| For a calculus-based method that delivers both solutions, you want a Lagrange multiplier. Let: \begin{align*}f\left(x,y,z\right) &= 2x^2 + y^2 + z^2 \\ g\left(x,y,z\right) &= x + y + z - 10\end{align*} The condition for an extremum is $$\vec{\nabla} f\left(x,y,z\right) = \lambda \vec{\nabla} g\left(x,y,z\right)$$ Calculating this out, we have a system with four equations and four unknowns: \begin{align*}4x &= \lambda \\ 2y &= \lambda \\ 2z &= \lambda \\ 0 &= x+y+z-10\end{align*}
Solving for $\lambda$ first is easiest: $$0 = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{2} - 10 \hspace{2.00cm} \rightarrow \hspace{2.00cm} \lambda = 8$$ With $\lambda=8$ figured out, we know one extreme point to be: $$x = 2 \hspace{2.54cm} y = 4 \hspace{2.54cm} z = 4$$
So what about the maximum? The best hint, which undoubtedly drives the other solutions to this, is the factor of $2$ in front of $x^2$, meaning this term grows faster than its counterparts. It only makes sense to try $y=z=0$, so that $x=10$ is evident from inspection. This makes the whole $\lambda$-apparatus somewhat useless, but at least the form is there in case the problem was trickier than it looked.
| {
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've found (I'm sorry, I forgot where I got it):
Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.
Now since I love to punish myself, I tried to find a harder proof as such:
We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2) +2t} + 1
$$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$
Now how do I prove the sextic polynomial inequality above (which is true)?
| An alternative way of proving, is going into parametric equations. The cubic equation is a Follium of Descartes, and can be represented by $(\frac{\sqrt{2}t}{1+t^3},\frac{\sqrt{2}t^2}{1+t^3})$. With respect to the 1st quadrant we must now show that all points on the unit circle fall "outside" (that is on the "right" side) of the Follium. Note that the point$P(\sqrt{2}/2,\sqrt{2}/2)$ coincides with both curves, hence the need for the equality symbol. When we use the distance formula on the parametric representation (in squared form), we get $d^2=\frac{2t^2+2t^4}{(1+t^3)^2}$. We can optimize $d^2$ and show that for $t=1$, a local maximum is achieved for the distance. You need to apply the quotient rule, which is annoying algebra. The numerator will be $(1+t^3)(-4t^6-8t^4+8t^3+4t)$ and when you set this equal to zero, $t=-1$ doesn't yield anything productive (what's going on with $t=-1$?). But the sixth degree polynomial inside has a root $t=0$ (Origin) and $t=1$ (Rational Zero Theorem!) and thus is divisible by $(t-1)$, which (after factoring out $-4t$) gives $t^4+t^3+3t^2+t+1$ which has no real roots. It shouldn't be to hard to show that for $t=1$ a maximum distance (Point P) in the first quadrant is achieved. Here you have all the ingredients to set up the proof which is what I want you to do.
| {
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Find constants in system of equations In the below system of equations, find the values of constants $q_1, q_2, q_3 \in \mathbb R$ so that the system has no solution.
$$\begin{cases}
2x&+y-z&+2s &= q_1\\
3x&+y-2z&+s &= q_2\\
x&+y+3z&-s &= q_3
\end{cases}$$
In order to eliminate some of the unknowns, we add the 1st with the 3rd and subtract the 2nd, so we get:
$y+4z = q1+q3-q2$
By similar more ways (multiplying, adding and subtracting) we get some more such equations.
I don't see how this system can have no solution; plus that we are one equation short (3 equations - 4 unknowns).
Any assistance is much appreciated.
EDIT:
(Apologies but I am not familiar with linear algebra!!)
What I believe I must do is the following:
Reduce the unknowns by one (don't see how), so as to have a 3x3 system. Calculate the determinant, for which it must be $D \neq 0$.
Then calculate $D_x, D_y, D_z$ assuming we have eliminated s.
Then, for the system NOT to have a solution, it must be $D=0$ and $D_x, D,y, D_z \neq 0$.
Any further help?
| The system can be written as$$\begin{pmatrix}
2& 1 &-1& 2\\
3&1 &-2& 1\\
1&1& 3&-1\end{pmatrix} \begin{pmatrix}
x\\ y \\ z \\ s\end{pmatrix} = \begin{pmatrix}
q_1\\ q_2 \\ q_3\end{pmatrix}.$$ Now since we have that $$\mathrm{rank}\begin{pmatrix}
2& 1 &-1& 2\\
3&1 &-2& 1\\
1&1& 3&-1\end{pmatrix} =3$$ the system must be consistent, independently of the values of $q_1,q_2,q_3$. (Obviously it is $$\mathrm{rank}\begin{pmatrix}
2& 1 &-1& 2\\
3&1 &-2& 1\\
1&1& 3&-1\end{pmatrix} =\mathrm{rank}\begin{pmatrix}
2& 1 &-1& 2 &q_1\\
3&1 &-2& 1 & q_2\\
1&1& 3&-1 & q_3\end{pmatrix}=3.)$$
EDIT
$$\begin{cases}
2x&+y&-z&+2s &= q_1\\
3x&+y&-2z&+s &= q_2\\
x&+y&+3z&-s &= q_3
\end{cases} \iff \begin{cases}
2x&+y&-z &= q_1-2s\\
3x&+y&-2z&= q_2-s\\
x&+y&+3z&= q_3+s
\end{cases} $$
$$\iff \begin{cases}
2x&+y&-z &= q_1-2s\\
x&&-z&= q_2-q_1+s\\
-x&&+4z&= q_3-q_1+3s
\end{cases}$$
$$\iff \begin{cases}
2x&+y&-z &= q_1-2s\\
x&&-z&= q_2-q_1+s\\
&&3z&= q_3+q_2-2q_1+4s
\end{cases}$$
From the last equation we get
$$z=\dfrac{q_3+q_2-2q_1+4s}{3}.$$
Now $$x=z+q_2-q_1+s=\dfrac{q_3+4q_2-5q_1+7s}{3}.$$
Finally we get $$y=z-2x+q_1-2s=\dfrac{-q_3-7q_2+11q_1-16s}{3}.$$ That is, the system is consistent independently of the values of $q_1,q_2,q_3.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Power-mean Inequality: Prove that for positive reals $a$, $b$, $c$ we have $3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$.
Taiwan Quiz 2014:
Prove that for positive reals $a$, $b$, $c$ we have
$$3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$
Applying Power-mean inequality (weighted) in $$a_1={abc} , a_2={\frac{a^3+b^3+c^3}{3}}$$ with $r=1$, $s=\dfrac{1}{3}$ and weights $\dfrac{8}{9}, \dfrac{1}{9}$ respectively, we get
\begin{align}
P(1) &\ge P\left(\frac{1}{3} \right) \\
\frac{8}{9}\cdot {abc} +\frac{1}{9}\cdot {\frac{a^3+b^3+c^3}{3}} & \ge
\left(
\frac{8}{9}\sqrt[3]{abc}+
\frac{1}{9}\sqrt[3]{\frac{a^3+b^3+c^3}{3}}
\right)^3 \tag{1}
\end{align}
In the book it says this simplifies into
$$a^3 +b^3 +c^3 +24abc \le (a+b+c)^3$$
which is true by $AM \ge GM$, we get on $2$ terms
$$8abc\le(a+b)(b+c)(c+a) \tag{2}$$
Can someone explain me how the simplification happened from $(1)$ to $(2)$?
| From (1), we have
$$27(24abc + a^3 + b^3 + c^3) \ge \left(8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}\right)^3.$$
Thus, it suffices to prove that
$$[3(a+b+c)]^3 \ge 27(24abc + a^3 + b^3 + c^3)$$
or
$$(a+b+c)^3 \ge 24abc + a^3 + b^3 + c^3.$$
| {
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Question regarding big O notation. Let $z\in \mathbb{C}$. I want to show that $\frac{1}{z+n}+\frac{1}{z-n}=O(\frac{1}{n^2})$.
By definition, I need to show that for some constant $M$ and $N$, we have $\mid\frac{1}{z+n}+\frac{1}{z-n}\mid=\mid\frac{2z}{z^2-n^2}\mid\leq \frac{M}{n^2}$ for every $n\geq N$. I need to change this inequality and treat $z$ as a constant. Is this the right approach?
| $$\frac1{z+n}+\frac1{z-n}=\frac{z+n+z-n}{(z+n)(z-n)}=\frac{2z}{z^2-n^2}$$
$$\left|\frac{2z}{z^2-n^2}\right|=2|z|\frac{1}{|z^2-n^2|}$$
now we can pick apart this second term:
$$|z^2-n^2|=|n^2[(z/n)^2-1]|=n^2|(z/n)^2-1|$$
so:
$$\left|\frac{2z}{z^2-n^2}\right|=\frac{2}{n^2}\left|\frac{z}{(z/n)^2-1}\right|$$
| {
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Convergence/Divergence proof for Infinite products Exercise 2.4.10 is about infinite products. I'd like someone to verify, if my convergence/divergence proof is technically correct and rigorous.
A close relative of the infinite series is the infinite product.
\begin{align*}
\prod_{n=1}^{\infty} b_n = b_1 b_2 b_3 \cdots
\end{align*}
which is understood in terms of its sequence of partial products.
\begin{align*}
p_m = \prod_{n=1}^{m} b_n = b_1 b_2 b_3 \cdots b_m
\end{align*}
Consider the special class of infinite products of the form
\begin{align*}
\prod_{n=1}^{\infty}(1+a_n) = (1+a_1)(1+a_2)(1+a_3)\cdots
\end{align*}
where $a_n > 0$.
(a) Find an explicit formula for the sequence of the partial products in the case where $a_n = 1/n$ and decide whether the sequence converges. Write out the first few terms in the sequence of partial products in the case where $a_n = 1/n^2$ and make a conjecture about the convergence of this sequence.
(b) Show, in general, that the sequence of partial products converges if and only if $\sum_{n=1}^{\infty}a_n$ converges. (The inequality $1+x \le 3^x$ for positive $x$ will be useful in one direction.)
Proof.
(a) If $a_n = 1/n$, the partial product
\begin{align*}
p_m &:= \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{m}\right)\\
&= 1 + \sum_{m} \frac{1}{m} + \sum_{m < n} \frac{1}{m \cdot n} + \sum_{m < n < p} \frac{1}{m \cdot n \cdot p} + \ldots
\end{align*}
Since $\sum 1/m$ is the harmonic series, which is well-known to be divergent, this sequence of partial products is divergent.
Consider the case of $a_n = 1/n^2$. We are going to use the inequality $1 + x^2 \le e^{x^2}$. We have:
\begin{align*}
p_m &= \left(1+\frac{1}{1^2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{3^2}\right)\cdots\left(1+\frac{1}{m^2}\right)\\
&\le e^{\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots + \frac{1}{m^2}}\\
&\le e^{\frac{1}{1\cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \ldots + \frac{1}{m\cdot (m+1)}}\\
&\le e^{\left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3}-\frac{1}{4}\right) + \ldots + \left(\frac{1}{m} - \frac{1}{m+1}\right)}\\
&= e^{1 - \frac{1}{m+1}}\\
&\le e
\end{align*}
Moreover, $p_1 = 2$, $p_2 = (2)(5/4)$, $p_3 = (2)(5/4)(10/9)$, $p_4 = (2)(5/4)(10/9)(17/16)$. So, $(p_m)$ is monotonic increasing and bounded by $e$. Consequently, by the Montone Convergence Theorem, $(p_m)$ is convergent.
(b) ($\longrightarrow$) direction.
The sequence of partial products,
\begin{align*}
p_n &= (1+a_1)(1+a_2)\cdots(1+a_n)\\
&\le e^{a_1}\cdot e^{a_2}\cdots e^{a_n}\\
&\le \exp \left({\sum_{k=1}^{n}a_k}\right)
\end{align*}
If the sequence of partial products is convergent and therefore bounded, the sequence of partial sums $\sum_{k=1}^{n} a_k$ must be bounded and convergent.
| For (a) what you have done is correct.
For (b) your argument is not valid. Note that $\sum \ln (1+a_n) <\infty$. This implies that $\ln (1+a_n) \to 0$ so $a_n \to 0$. Now, there exists $\delta >0$ such that $\ln (1+x) \geq \frac 1 2 x$ for $0<x <\delta$ (because $\lim_{x \to 0}\frac {ln (1+x)} x=1$). Hence, $a_n <2 \ln (1+a_n)$ for $n$ sufficiently large. This proves that $\sum a_n <\infty$.
The converse part follows similarly by looking at $\ln (1+a_n)$ and noting that $\ln (1+x) \leq x$ for all $x >0$.
| {
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Evaluate $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$ This integral
$$\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
is fabricated based on an interesting point which I may post later.
The question is: How would you do it?
| At first, we must check whether integrand function is bounded or not. [More specifically, if it is not bounded, we must approach this problem as improper integral..] Since both numerator and denominator are continuous at [0,1], we enough to check whether denominator has 0-value or not. Since $3\left(x-\dfrac{1}{2}\right)^{2}<3x^{2}-3x+1$, we can conclude that denominator does NOT have 0 value. So if we can get indefinite integral of integrand, then we can also calculate the definite integral.
Let's manipulate integrand function so that we can get indefinite integral. At first, I want to divide numerator into denominator.
$3x^{3}+3x^{2}-3x+1$
$=3(x^{3}+1)-3+(3x^{2}-3x+1)$
$=3(x+1)(x^{2}-x+1)-3+(3x^{2}-3x+1)$
$=(x+1)(3x^{2}-3x+3)-3+(3x^{2}-3x+1)$
$=(x+1)(3x^{2}-3x+1)+2(x+1)-3+(3x^{2}-3x+1)$
$=(x+2)(3x^{2}-3x+1)+(2x-1)$
From this calculation, we can manipulate integrand as $x+2+\dfrac{2x-1}{3(x^{2}-x)+1}$.
Indefinite integral of $x+2$ is $\dfrac{x^{2}}{2}+2x$, so we only need to compute indefinite integral of the last term. To integrate it, I will first substitute $x^{2}-x$ as u, so that $u'(x)=2x-1$. So it can be changed as $\dfrac{1}{3} \dfrac{1}{u+\dfrac{1}{3}}$ and its indefinite integral is $\dfrac{1}{3} \log{\left(u+\dfrac{1}{3}\right)}$. Since u(1)=u(0) in this situation, we only need to calculate $\left[\dfrac{x^{2}}{2}+2x \right]^{1}_{0}=\dfrac{5}{2}$.
| {
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How to solve $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$ without L'Hopital? $\lim_{n \to \infty}\frac{1}{\sqrt[3]{n^3+n+1}-\sqrt{n^2-n+2}}$
$\lim_{n \to \infty}\frac{1}{\sqrt[6]{(n^3+n+1)^2}-\sqrt[6]{(n^2-n+2)^3}}$ but because this limit is still the type of $\frac{1}{\infty-\infty}$ I tried to do this:
$\lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{(n^3+n+1)^2-(n^2-n+2)^3} = \lim_{n \to \infty}\frac{\sqrt[6]{(n^3+n+1)^2}+\sqrt[6]{(n^2-n+2)^3}}{3n^5-7n^4+15n^3-17n^2+14n-7}$ I'm totally stuck here. I would divide the fraction by $3n^5$ and then the solution is $0$. Not the correct answer.
Did I miss something?
| For large $n$, $(1+n^{-2}+n^{-3})^{1/3}\in 1+\tfrac13n^{-2}+O(n^{-2})\subseteq 1+o(n^{-1})$, so$$\begin{align}\frac{n^{-1}}{(1+n^{-2}+n^{-3})^{1/3}-(1-n^{-1}+2n^{-2})^{1/2}}&\in\frac{n^{-1}}{1+o(n^{-1})-1+\tfrac12n^{-1}+o(n^{-1})}\\&=\frac{n^{-1}}{\tfrac12n^{-1}+o(n^{-1})}\\&\stackrel{n\to\infty}{\sim}2.\end{align}$$
| {
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Solving $\sqrt{a-\sqrt{a+x}} = x$ Solve the equation
$$\sqrt{a-\sqrt{a+x}} = x$$
My approach:
Tried shifting the variables into different options, but couldn't get anything out of it. So, please help.
| Let $\sqrt{a+x}=y\ge0\implies x=y^2-a$
$$y^2-a=\sqrt{a-y}\ge0\implies y^2\ge a$$
$$(y^2-a)^2=a-y\iff a^2-(2y^2+1)a+y^4-y=0$$
$$a=\dfrac{2y^2+1\pm\sqrt{(2y^2+1)^2-4(y^4-y)}}2=\dfrac{2y^2+1\pm(2y+1)}2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$
This solution isn't true. Where am I wrong?
EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.
| Let $x=\sin(u)$, then $dx=\cos(u)du$. So
$$\begin{align}\int x\sqrt{1-x^2}\arcsin(x)dx&=\int u\sin(u)\cos^2(u)du\\
&=\int u\sin(u)du-\int u\sin^3(u)du\end{align}$$
| {
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Upper bound of non-negative concave function on interval
Let $f(x)$ be a non-negative and upper convex (concave) function
defined on the interval $[a,b]$. Suppose
$f\left(\frac{a+b}{2}\right)\leq2$. Show that $f(x)\leq4$ for all
$x\in[a,b]$.
How can I show the statement above? Note that we do not assume $f$ being continuous or differentiable.
I can't come up with any trick rather than either to use a non-negativity like
$$f(x)\leq f(x) + f\left(\frac{a+b}{2}\right)\leq f(x) + 2f\left(\frac{a+b}{2}\right)\leq f(x) + 4$$
or to use a subadditivity along the lines of
$$f\left(x+\frac{a+b}{2}\right)\leq f(x) + f\left(\frac{a+b}{2}\right)\leq f(x) + 2$$
which doesn't lead to the desired upper bound. A complete solution or at least a hint will be appreciated.
| In the case $\frac{a+b}{2} \le x \le b$ we can use the concavity condition for $a < \frac{a+b}{2} < x$, which gives
$$
2 \ge f(\frac{a+b}{2}) \ge \frac{x-(a+b)/2}{x-a}f(a) + \frac{(a+b)/2-a}{x-a} f(x) \, .
$$
Now use that $f(a) \ge 0$ and $x-a \le b-a$.
Graphically: Let $l$ be the line joining $(a, f(a))$ and $(\frac{a+b}{2}, f(\frac{a+b}{2}))$. Then $f(x) \le l(x) \le 4$ for $\frac{a+b}{2} \le x \le b$.
The case $a \le x \le \frac{a+b}{2}$ works similarly, or can be reduced to the first case because of the symmetry of the problem.
| {
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Find constants $a$, $b$, $c$, $d$ and $e$ such that $\cos4x=a\sin^4x+b\sin^3x+c\sin^2x+d\sin x+e$ for all angles $x$ Basically, write $\cos4x$ as a polynomial in $\sin x$.
I've tried the double angles theorem and $\cos2x=\cos^2x-\sin^2x$. I'm still having trouble right now though.
Please help! Thanks!
| If you prefer we can use complex numbers (yes!).
Let $z=\cos x+i\sin x$. Using de Moivre's theorem:
$$\begin{align} z^4&=\cos 4x+i\sin 4x=(\cos x+i\sin x)^4\\&=\cos^4x+4\cos x(i\sin x)+6\cos^2x(i\sin x)^2+4\cos x(i\sin x)^3+(i\sin x)^4\end{align}$$
Since we're only concerned with the real part of this expression, $\cos 4x$, we see that
$$\cos4x=\cos^4x-6\cos^2x\sin^2x+\sin^4x$$
and now use $\cos^2x=1-\sin^2x$ repeatedly.
| {
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What is wrong with my 'proof' of $i=1$? So I recently learnt about $i$ and I can't wrap my head around the concept of $i^2=-1$ or that $\sqrt{-1}$ can even exist. Today I was thinking about $i$ again and thought of a "proof" that $i=1$.
$$i^4=1 \text{ and } 1^4=1 \text{ so } i^4=1^4$$
$$i^4=1^4 \to \sqrt[4]{i^4}=\sqrt[4]{1^4} \to i=1$$
But if $i=1$, then $i^2 \neq -1$. So I think I must have messed up something in the proof. Can someone point out where this went wrong? I know you can $\text{"prove" }1=2$ by accidentally dividing by $0$ and I suspect something similar is happening.
For anyone else having trouble with complex numbers @mrsamy commented this link and I found it quite helpful: https://www.math.toronto.edu/mathnet/answers/imaginary.html
| The thing is that if $m= n$ then we can conclude $f(m) = f(n)$.
But if $f(m) = f(n)$ we can not conclude $m = n$ unless $f$ is one to one.
If $f$ is not one to one is perfectly possible to have $m \ne n$ but $f(m) = f(n)$.
A simple example is $f(x) = x^2 - 3x +6 $ If $x=2$ we get $f(2) = 4 - 6 + 6 = 4$. And if $x=1$ we get $f(1)=1-3 + 6 = 4$. So $f(1) = f(2) =4$ but $1\ne 2$.
An even easier example is $f(x) = x^4$ and $f(-1) = f(1) = f(i) = f(-i)$ will give us $1^4 = 1$ and $(-1)^4 = 1$ and $i^4 = (i^2)^2 =(-1)^2 =1$ and $(-i)^4 = ((-i)^2)^2=((-1)^2i^2)^2 = (1*(-1))^2 = 1$.
the problem is when sloppy teachers give to impressionable students this INCORRECT definition:
THIS IS WRONG: $\sqrt[k]{m}$ is equal to the $x$ so that $x^k = m$.
The problem is that there are $k$ different $x$ that give $x^k = m$ and so that isn't actually a definition of a single value.
Ex. $\sqrt{16}$ iss the $x$ so that $x^2= 16$. Well $4^2 = 16$ and $(-4)^2 = 16$. So which is it? is $\sqrt{16} = 4$ of is $\sqrt{16} = -4$.
Well, the answer is we define that $\sqrt{m}$ is the positive value $x$ so that $x^2 = m$.
But two things come about from this.
One $\sqrt{x^2} \ne x$. That is just wrong. $\sqrt{x^2} =|x|$ because we don't know that $x$ is positive.
And 2) $i^2 = 1$ is a property of $i$ but that does not mean $\sqrt{-1} = i$. Because $i$ is neither positive nor negative.
Anyway.....
tl;dr
There are four values, all different of $x$ so that $x^4 = 1$. They are $1^4 =1; (-1)^4 = 1; (-i)^4=1$
$\sqrt[4]{x^4} \ne x$. IF $x^4$ is a positive real number then $\sqrt[4]{x^4} = |x|$ and indeed $|1| = |-1| = |i| = |-i|$ but $\sqrt[4]{x^4} \ne x$.
And if $x^4$ is not a positive real number we don't actually have a definition for $\sqrt[4]{x^4}$.
| {
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Given $\frac{A}{4B}+\frac{2a^3A}{B^3}-\frac{a^2A}{B^3}-\frac{2a^2A^2}{B^3}=0$, what does $\frac{a}{B}$ tend to as $\frac{A}{B}\to \infty$? In this problem, all of $A, a, B$ can vary.
My attempt was to let $\frac{A}{B}=C$ and $\frac{a}{B}=c$, which gives $ \frac{C}{4}+2Ac^3-c^2C-2Ac^2C=0$.
Now $C\to \infty $ means either $A\to \infty $ or $B\to 0 $. At this point I'm not sure at all how to proceed.
| my attempt, starting from your attempt:
$$\frac{C}{4} + 2Ac^3 - c^2C-2Ac^2C = 0$$
divide by $c^2$:
$$\frac{C}{4c^2} + 2Ac - C-2AC = 0$$
$$2Ac = -\frac{C}{4c^2} + C + 2AC$$
$$2Ac = C(-\frac{1}{4c^2} + 1 + 2A)$$
and isolate $C$:
$$\frac{2Ac}{-\frac{1}{4c^2} + 1 + 2A} = C$$
as we know, $C$ approaches infinity. Therefore, the LHS of the equation should too. Assuming $2Ac$ does not approach infinity, the following can be deduced:
$$-\frac{1}{4c^2} + 1 + 2A \rightarrow 0$$
$$-1 + 4c^2 + 8Ac^2 \rightarrow 0$$
$$ c^2(4 + 8A) = 1$$
and therefore:
$$c = \sqrt{\frac{1}{4+8A}}$$
| {
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Find all solutions of congruence $3x^2−2x+9≡0\pmod {35}$ Find all solutions of congruence $3x^2 - 2x + 9 ≡ 0 \bmod 35$:
Attempt:
\begin{align}
3x^2 - 2x + 9 &\equiv 0 \bmod 35\tag{* 3} \\
9x^2-6x+27 &\equiv 0 \bmod 35 \tag{- 26} \\
(3x-1)^2 &\equiv -26 \bmod 35 \\
\\
-26 + 35 &= 9 = 3^2 \\
\\
\iff (3x-1)^2 &\equiv 3^2 \bmod 35 \\
\iff (3x-1-3)*(3x-1+3) &\equiv 0 \bmod 35 \\
\\
\implies \underbrace{3x - 4 \equiv 0 \bmod 35}_{(a)}
&\lor \underbrace{3x + 2 \equiv 0 \bmod 35}_{(b)}\\
\end{align}
Case $(a){:}\; 3x - 4 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv 4 \bmod 35$ $\Rightarrow 3x \equiv 39 \bmod 35$
$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 13 \bmod 35$
Case $(b){:}\; 3x +2 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv -2 \bmod 35$ $\Rightarrow 3x \equiv 33 \bmod 35$
$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 11 \bmod 35$
So $x = 13$ or $x = 11$.
Is it correct that way?
| There is a direct way to solve modulo $35$ with the traditional quadratics method.
$\Delta^2=b^2-4ac=2^2-4\times3\times 9=-104\equiv 1\pmod {35}$
However as you can see it requires you to know how to solve $\Delta^2\equiv 1\pmod{35}$, but let assume you have a precomputed table of the squares then:
$\Delta\in\{1,6,29,34\}=\{-1,1,-6,6\}$
Then use $x=(-b\pm\Delta)\times (2a)^{-1}\pmod {35}$
$(2a)^{-1}\equiv 6^{-1}\equiv 6\pmod{35}$ and we do not have to worry about the $\pm$ in the formula since $\Delta$ already covers additive inverses.
Therefore for each $4$ values of $\Delta$ we get same number values of $x\in\{6,11,13,18\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to show nonassociativity of the positive rationals under a binary operation defined in terms of max and min? Consider $\mathbb{Q}^+$ with the usual $\leq$ relation and the binary operation $\circ$ defined as:
$$p \circ q = max(p,q) + \frac{1}{2} min(p,q)$$
A book that I'm reading states that the operation $\circ$ is not associative with respect to $\mathbb{Q}^+$, with verification left to the reader. I have tried to verify this by picking random triples $(p,q,r)$ of positive rational numbers as examples, but in each case I end up with the result that $p \circ (q \circ r) = (p \circ q) \circ r$.
For example, take $p=4.5, q=18.1, r=7$:
\begin{align}
4.5 \circ (18.1 \circ 7) &= 4.5 \circ (max(18.1,7) + \frac{1}{2} min(18.1,7))\\
&= 4.5 \circ (18.1 + 3.5)\\
&= 4.5 \circ 21.6\\
&= max(4.5,21.6) + \frac{1}{2} min(4.5,21.6)\\
&= 21.6 + \frac{1}{2} 4.5\\
&= 23.85
\end{align}
and then:
\begin{align}
(4.5 \circ 18.1) \circ 7 &= (max(4.5,18.1) + \frac{1}{2} min(4.5,18.1)) \circ 7\\
&= (18.1 + \frac{1}{2} 4.5) \circ 7\\
&= 20.35 \circ 7\\
&= max(20.35,7) + \frac{1}{2} min(20.35,7)\\
&= 20.35 + \frac{1}{2}7\\
&= 23.85
\end{align}
Am I not picking the right triples of positive rationals for a counterexample, or am I misunderstanding somehow?
| Select $q<p<p+q/2<r$, for example, $q=2,p=3,r=5$.
We get $(p\circ q)\circ r=r+p/2+q/4$ and $p\circ(q\circ r)=r+p/2+q/2$.
| {
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"url": "https://math.stackexchange.com/questions/3997348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving definite integral from -1 to 1 Please help me with this. I can't prove the result. Tried integral by parts or notations, nothing working
$$\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx$$
| Let $t=-x$, then
\begin{align*}
\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx
&=\int_{1}^{-1}{\frac{t^2}{e^{-t}+1}}(-dt)\\
&=\int_{-1}^{1}{\frac{t^2}{e^{-t}+1}}dt\\
&=\int_{-1}^{1}{\frac{x^2e^x}{e^x+1}}dx.
\end{align*}
\begin{align*}
\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx
&=\frac{1}{2}\left(\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx
+\int_{-1}^{1}{\frac{x^2e^x}{e^x+1}}dx\right)\\
&=\frac{1}{2}\int_{-1}^{1}x^2dx=\int_{0}^{1}x^2dx\\
&=\frac{1}{3}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$ We have a parabola and a circle with the following equations and their graph placed at the end of my question.
Parabola: $y^2 = 4x -4$
Circle: $(x-2)^2 + y^2 = 9$
My goal was to calculate their intersection points so I substituted $y^2$ from the parabola equation into the circle equation and I got
$(x-2)^2 + (4x-4)=9 \implies x^2 - 4x + 4 + (4x - 4) = 9 \implies x^2 = 9 \implies x = \pm3$
$x=3$ is the only correct solution but why is $x=-3$ produced as an extra invalid solution?
What is the exact mathematical explanation behind this? Why substituting one equation into the other has produced extra answers?
update
When I calculate $x$ from the parabola equation and substitute it in the circle equation, I don't get any extra answers for $y$:
$y^2=4x-4 \implies y^2 +4 = 4x \implies x = \frac{y^2}{4} + 1$
$(x-2)^2 +(4x-4)=9 \implies ((\frac{y^2}{4} + 1) - 2)^2 + (4x - 4)=9 \implies y^4 +8y - 128 = 0 \implies y^2=8,-16$
$y^2 = -16$ cannot be true so $y^2 = 8 \implies y=\pm 2\sqrt{2}$ and these are correct answers for $y$.
2nd update
I made a mistake in the calculation in the previous update although the final solutions where correct. I write the correct calculation:
$(x-2)^2 +y^2=9 \implies
((\frac{y^2}{4} + 1) - 2)^2 + y^2=9 \implies
(\frac{y^2}{4} - 1)^2 + y^2=9 \implies
(\frac{y^4}{16} - \frac{y^2}{2} + 1) + y^2=9 \implies
\frac{y^4}{16} + \frac{y^2}{2} + 1=9 \implies
(\frac{y^2}{4} + 1)^2=9 \implies
(\frac{y^2}{4} + 1)=\pm3 \implies
\frac{y^2}{4} =2,-4 \implies
y^2=8,-16$
| I suppose that we are working on the real number system and not consider complex numbers but I think my reasoning works in complex numbers too. We want to solve a system of equations containing two equations. I move all terms of each equation to one side and name them $A(x,y)$ and $B(x,y)$
$\begin{cases}
A(x,y)=y^2 -4x +4=0\\
B(x,y)=y^2 +(x-2)^2-9=0\\
\end{cases}$
We want to find tuples like $(x', y')$ such that $A(x',y')=0$ and $B(x',y')=0$ simultaneously.
If we calculate $y^2$ from $A(x,y)=0$ and substitute it into $B(x,y)$ we arrive at a third equation $C(x)=0$
$C(x) = 4x-4 +(x-2)^2-9 = 0 \implies C(x) = x^2 -9=0$
And when we solve $C(x) = 0$ we get $x^2=9 \implies x=\pm 3$ but any point $(-3,y)$ with $y\in\mathbb{R}$ does not satisfy $\begin{cases}
A(x,y)=0\\
B(x,y)=0\\
\end{cases}$
and the solving procedure has produced extraneous solutions. The reason is this the line of reasoning is not reversible.
$\begin{cases}
A(x,y)=0\\
B(x,y)=0\\
\end{cases} \overset{1}{\implies} C(x)=0 \overset{2}{\iff} x=\pm 3$
The $\overset{1}{\implies}$ is not reversible. In this case when there exists $(x_0,y_0)$ as a solution to the system we will have:
$\begin{cases}
A(x_0,y_0)=0\\
B(x_0,y_0)=0\\
\end{cases} \overset{3}\implies C(x_0)=0 \overset{4}\iff x_0=\pm 3$
but if there exits $(x_1,y_1)$ as a solution to $C(x) = 0$ we will have
$C(x_1)=0 \;\not\!\!\!\implies \begin{cases}
A(x_1,y_1)=0\\
B(x_1,y_1)=0\\
\end{cases}$
When we combine the equations we loose information and cannot retrive the system $\begin{cases}
A(x_1,y_1)=0\\
B(x_1,y_1)=0\\
\end{cases}$ from $C(x)=0$
so every solution to the system is a solution to $C(x)=0$ but we cannot say every solution to $C(x)=0$ must be a solution to the system and extraneous solutions might have been produced.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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} |
for which real values of $a$ the limit is finite and different from $0$ $\lim_{ x \to 0 } \frac{ \cos (\pi \cdot \frac{1-\cos ax}{x^2})}{x^2}$ I have to find for which real values of $a$ this limit is finite and different from $0$:
$$\lim_{ x \to 0 } \frac{ \cos (\pi \frac{1-\cos ax}{x^2})}{x^2}$$
I've tried to change the form of cos but with no success.
According to my book the final result should be for $a^2=2k+1 (k \in N)$ for which $\lim=\frac{(-1)^k (2k+1)^2 \pi}{24}$.
| We start from
$$\cos(ax)=1-\frac{a^2x^2}{2}+\frac{a^4x^4}{4!}(1+\epsilon(x))$$
to get
$$\frac{1-\cos(ax)}{x^2}=\frac{a^2}{2}-\frac{a^4x^2}{24}(1+\epsilon(x))$$
If The limit is finite, then necessarily
$$\cos(\pi.\frac{a^2}{2})=0$$
or
$$a^2=1+2k , k\ge 0$$
In this case
$$A=\pi(\frac{a^2}{2}-\frac{a^4x^2}{24}(1+\epsilon(x))=$$
$$\frac{\pi}{2}+k\pi-\frac{a^4x^2\pi}{24}(1+\epsilon(x))$$
and
$$\cos(A)=-(-1)^k\sin(\frac{a^4x^2\pi}{24}(1+\epsilon(x)($$
$$\sim -(-1)^k\frac{a^2x^2\pi}{24}$$
the limit will be
$$-(-1)^k\frac{a^4\pi}{24}=-(-1)^k\frac{(1+2k)^2\pi}{24}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Missing factor equal to 0 after factoring
The equation $x^{10}+(13x-1)^{10}=0$ has $10$ complex roots, $r_i$ for integers $1\le i\le5$. Find $\sum^5_{i=1}\frac{1}{r_i\overline{r_i}}.$
My solution was simply to factor the given polynomial into $(r^2+(13r-1)^2)(r^8-\dots+(13r-1)^8)=0.$ We suppose the left factor is $0,$ such that $170r^2-26r+1=0.$ By Vieta's, the product of the roots if $1/170.$ Since this product is the product of the conjugate pairs, we can multiply $170\times5$ to find our answer, and $850$ turns out to be correct.
However, how come we don't need to account for the right factor here? We need not consider it if it were never equal to $0$, so how do we prove that?
| We solve equation $x^{10}+(13x-1)^{10}=0$.
$$\left(13-\frac{1}{x}\right)^{10}=-1$$
$$13-\frac{1}{x_k}=\cos\frac{(2k+1)\pi}{10}+i\sin\frac{(2k+1)\pi}{10},$$
$$\frac{1}{x_k}=13-\cos\frac{(2k+1)\pi}{10}-i\sin\frac{(2k+1)\pi}{10},\quad k=0\ldots9.$$
Then
$$\frac{1}{x_k\bar x_k}=170-26\cos\frac{(2k+1)\pi}{10}$$
$$\sum_{k=0}^4\frac{1}{x_k\bar x_k}=850-26\left(\cos\frac{\pi}{10}+\cos\frac{3\pi}{10}+\cos\frac{7\pi}{10}+\cos\frac{9\pi}{10}\right)=850$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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calculating $\lim\limits _{x\to0}\left(\frac{1}{x^{2}}-\frac{e^{x}}{\left(e^{x}-1\right)^{2}}\right)$ using taylor polynomials I am trying to calculate the limit $$\lim\limits _{x\to0}\left(\frac{1}{x^{2}}-\frac{e^{x}}{\left(e^{x}-1\right)^{2}}\right)$$ using taylor polynomials. I tried using L'Hôpital's rule, but it was tedious. the final answer should be $\frac{1}{12}$.
I tried taylor many times but couldn't make it work. here is my best attempt:
$$\lim\limits _{x\to0}\left(\frac{1}{x^{2}}-\frac{e^{x}}{\left(e^{x}-1\right)^{2}}\right)=\lim\limits _{x\to0}\left(\frac{\left(e^{x}-1\right)^{2}-x^{2}e^{x}}{x^{2}}\right)=\lim\limits _{x\to0}\left(\frac{\left(1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+R_{3}\left(x\right)-1\right)^{2}-x^{2}\left(1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+R_{3}\left(x\right)\right)}{x^{2}}\right)=\lim\limits _{x\to0}\left(\frac{\frac{1}{3}R_{3}\left(x\right)x^{4}+R_{3}^{2}\left(x\right)x^{2}+2x^{2}R_{3}\left(x\right)+\frac{x^{6}}{36}+\frac{x^{4}}{12}}{x^{2}}\right)=\lim\limits _{x\to0}\left(\frac{R_{3}\left(x\right)\left(\frac{1}{3}x^{4}+R_{3}\left(x\right)x^{2}+2x^{2}\right)}{x^{2}}\right)=\lim\limits _{x\to0}\left(R_{3}\left(x\right)\left(\frac{1}{3}x^{2}+R_{3}\left(x\right)+2\right)\right)$$
the number $\frac{1}{12}$ appears here but then goes away since $x^4$ goes to zero. There must be a simpler way then opening the brackets here, and making all those tedious calculations.
| You have $(e^x-1)^2-x^2e^x=e^{2x}-(x^2+2)e^x+1$ and the first $5$ terms of the Taylor series of this function are simply $\frac{x^4}{12}$. And the first $5$ terms of the Taylor series of $x^2(e^x-1)^2$ are simply $x^4$. Therefore your limit is indeed $\frac1{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4001672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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From the system of equations prove $a+b+c=0$
$a,b,c$ are distinct real numbers and $x,y$ are also real numbers. we have these equations:
$${ \begin{cases}{a^3+ax+y=0} \\ {b^3+bx+y=0} \\
{c^3+cx+y=0}\end{cases} }$$ Prove $a+b+c=0$
I added all the equations together and get:
$$a^3+b^3+c^3+(a+b+c)x+3y=0$$
It is similar to Euler identity (because we have $a^3+b^3+c^3$). if $a+b+c=0$ then from Euler identity we can conclude $a^3+b^3+c^3=3abc$. and equation change to:
$$3abc+3y=0$$
But it seems that doesn't work.
| Subtract consecutive equations to get$$a^3-b^3+(a-b)x=0\iff a^2+b^2+ab+x=0~~\because a\ne b\\b^3-c^3+(b-c)x=0\iff b^2+c^2+bc+x=0~~\because b\ne c$$
Subtract these equations to get$$a^2-c^2+b(a-c)=0\iff a+b+c=0~~\because a\ne c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding an equation of a parabola I was asked to show that a$\pm\frac{cx}{b}=\sqrt{x^2+y^2}$ is a parabola if $b=c$
$$\text{If }b=c\rightarrow a\pm x=\sqrt{x^2+y^2}\\
\left(a\pm x\right)^2=x^2+y^2\\
\left(a\pm x\right)^2-x^2=y^2\\
\pm\sqrt{a^2+2ax+x^2-x^2}=y \text{ or } \pm\sqrt{a^2-2ax+x^2-x^2}=y\\
\pm\sqrt{a^2\pm2ax}=y$$
Can I go from here to a standard form of a parabola? How?
How can I show that if $b>c$ it's an ellipse and if $b<c$ it's a hyperbola?
*Edit: I am also attaching my work for $b>c$. $$\text{If } b>c\rightarrow a\pm\frac{cx}{b}=\sqrt{x^2+y^2}\\
\left(a\pm\frac{cx}{b}\right)^2=x^2+y^2\\
a^2\pm2a\frac{cx}{b}+\frac{\left(cx\right)^2}{b^2}=x^2+y^2\\
\pm2\frac{cx}{ab}+\frac{\left(cx\right)^2}{\left(ab\right)^2}=\frac{x^2}{a^2}+\frac{y^2}{a^2}\\
\frac{cx}{ab}\left(\pm2+\frac{cx}{ab}\right)=\frac{x^2}{a^2}+\frac{y^2}{a^2}\\
\pm2+\frac{cx}{ab}=\frac{b}{cx}\left(\frac{x^2}{a}+\frac{y^2}{a}\right)\\
\pm2=\frac{b}{cx}\left(\frac{x^2}{a}+\frac{y^2}{a}\right)-\frac{cx}{ab}\\
\pm2=\frac{b^2x^2+b^2y^2}{abcx}-\frac{\left(cx\right)^2}{abcx}\\
\pm2=\frac{b^2x^2+b^2y^2-c^2x^2}{abcx}\\
\pm1=\frac{\left(b^2-c^2\right)x^2+b^2y^2}{abcx}$$
| All conic sections follow the following general form
$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $, with $A, B, C, D, E, F$ all real and $A,B,C$ non-zero.
(see: https://en.wikipedia.org/wiki/Conic_section#cite_ref-Protter_1970_326_15-0):
The discriminant $B^2 - 4AC$ determines ellipse, circle, parabola, hyperbola, etc.
(1) if $B^2 - 4AC = 0$, the equation is a parabola.
(2) if $B^2 - 4AC < 0$, the equation is an ellipse, and so on.
From $\left(a\pm \frac{cx}{b}\right)^2=x^2 + y^2$,
$\big(\frac{c^2 - b^2}{b^2}\big)x^2 - y^2 \pm2\frac{ac}{b}x + a^2 = 0$,
Then, $A = \big(\frac{c^2 - b^2}{b^2}\big)$, $B = 0$, $C = -1$, $D =\pm2\frac{ac}{b}$, $E = 0$, $F = a^2$.
If $b = c$, then $B^2 - 4AC = 0$, so it is a parabola.
For the ellipse, the discriminant is negative if $b > c$, and similarly for hyperbola.
Alternatively,
$ y^2 = a^2 \pm 2ax$ from your last step,
then, use a substitution $x = x' - \frac{a}{2}$:
$ y^2 = a^2 \pm 2a\big(x' - \frac{a}{2}\big) = \pm 2ax'$, which is an equation of parabola.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Quadratic equation in trigonometric form $\sqrt 3 a\cos x + 2b\sin x = c$ Let a,b,c be three non-zero real number such that the equation, $\sqrt 3 a\cos x + 2b\sin x = c$, $x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$, has two distict roots $\alpha$ and $\beta $ where $\alpha +\beta=\frac{\pi}{6}$. Then find the value of $\frac{b}{a}$.
My approach is as follow
$\sqrt 3 a\cos x + 2b\sin x = c$
$3{a^2}{\cos ^2}x = {c^2} + 4{b^2}{\sin ^2}x - 4bc\sin x$
$ {\sin ^2}x\left( {4{b^2} + 3{a^2}} \right) - 4bc\sin x + {c^2} - 3{a^2} = 0$
Roots are real when $16{b^2}{c^2} + 4\left( {4{b^2} + 3{a^2}} \right)\left( {{c^2} - 3{a^2}} \right) > 0$
$\sin \alpha + \sin \beta = \frac{{4bc}}{{4{b^2} + 3{a^2}}};\sin \alpha \sin \beta = \frac{{{c^2} - 3{a^2}}}{{4{b^2} + 3{a^2}}}$, how I will proceed from here
| let $\alpha =A,\beta =B$
as $$c-c=0$$ $$\sqrt{3}a \cos A+2b\sin A-\sqrt{3}a\cos B-2b\sin B=0$$
$$\sqrt{3}a \cos A-\sqrt{3}a\cos B+ 2b\sin A-2b\sin B=0$$
$$-2\sqrt{3}a \sin\left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)+2b\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)=0$$ Can you end it now? :cancel $\sin (\frac{A-B}{2})$
| {
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"timestamp": "2023-03-29T00:00:00",
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USA MO 1980's inequality Let $x;y;z$ be real numbers and $x;y;z \in [0;1]$
Prove: $\dfrac{x}{y+z+1}+\dfrac{y}{z+x+1}+\dfrac{z}{x+y+1} \leq 1-(1-x)(1-y)(1-z)$
The official solution uses calculus, so I can't understand it.
I hope someone have a "simpler solution" (for middle-school student). Thank you.
| Show $ \ \dfrac{x}{y+z+1}+\dfrac{y}{z+x+1}+\dfrac{z}{x+y+1} \leq 1-(1-x)(1-y)(1-z)$
WLOG, $x \geq y \geq z$
Then $ \ \dfrac{y}{z+x+1} \leq \dfrac{y}{y+z+1} \ $ and $ \ \dfrac{z}{x+y+1} \leq \dfrac{z}{y+z+1}$
So LHS $\leq \dfrac{x+y+z}{y+z+1}$ and hence it suffices to show,
$(1-x)(1-y)(1-z) \leq 1 - \dfrac{x+y+z}{y+z+1} = \dfrac{1-x}{y+z+1}$
or it suffices to show $(1-y)(1-z) \leq \dfrac{1}{(1+y)(1+z)} \ $
[as $ \ \dfrac{1}{(1+y)(1+z)} \leq \dfrac{1}{y+z+1}$]
or to show that $(1-y^2)(1-z^2) \leq 1 \ $ which is obviously true.
| {
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"url": "https://math.stackexchange.com/questions/4013985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How to determine the convergence of the series$\sum_{n=1}^\infty (n^\frac{1}{n^2+1}-1)$? I am currently working on some analysis exercise and was struggling to prove that the series converges $$\sum_{n=1}^\infty (n^\frac{1}{n^2+1}-1)$$
I have tried several tricks such as converting $n^\frac{1}{n^2+1}$ to $e^{logn\frac{1}{n^2+1}}$ and try to apply ratio test and root test to $\sum_{n=1}^\infty( e^{logn\frac{1}{n^2+1}}-1)$ but couldn't succeed.
Also, I tried to upper bound the terms by $n^\frac{1}{n^2}-1$ and try to use a comparison test, but couldn't make it work either, I calculated the sum with Mathematica and it showed me it does converge to approximately $0.87357$, but I just couldn't find a way to prove it.
| Comparison can work: let's try to show $\frac{1}{x^p}\ge x^\frac{1}{x^2+1}-1$ for some $1<p$.
$$\frac{1}{x^p}+1\ge x^\frac{1}{x^2+1}$$
$$\log\left(\frac{1}{x^p}+1\right)\ge\frac{\log(x)}{x^2+1}$$
Then by the taylor series for the logarithm we know $\log(1+x)>x-\frac{x^2}{2}$ for $x\in(0,2)$, so we have
$$\frac{1}{x^p}-\frac{1}{2x^{2p}}\ge\frac{\log(x)}{x^2+1}$$
$$\frac{(2x^p-1)(x^2+1)}{2x^{2p}}\ge\log(x)$$
Next, $\sqrt{x}\ge\log(x)$ so it's enough to find $p$ so that
$$\frac{(2x^p-1)(x^2+1)}{2x^{2p}}\ge\sqrt{x}$$
However, notice that the limit of the lhs as $x\to\infty$ is a simple power of $x$: $x^{2-p}$. Hence if $2-p>\frac{1}{2}$, there will be some finite $N$ after which $x^{2-p}\ge\sqrt{x}$. The sum up to $N$ converges because it is a finite sum, and the sum after $N$ converges because this inequality shows the terms after $N$ are bounded by the convergent series $\frac{1}{x^p}$. Any $1<p<\frac{3}{2}$ will work, like say $p=\frac{4}{3}$.
| {
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A quite difficult limit: $\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x}$ I tried this limit with Taylor series. However, proceeding in this manner I obtained a series over another series. Furthermore, I do not know how to continue as long as we have $x$ tends to $0$. Were it to be $x$ tends to infinite, this limit would be simpler.
This is what I got
$$\begin{align}&
\lim _{x\to 0}\frac{\exp(x^2)-\cos x-x\sin x}{\sinh^2x} \\[6pt]
=\;&\frac{\frac12x^2+\frac{5}{8}x^4+\frac{23}{144}x^6+\frac{241}{5760}x^8+\frac{3359}{403200}x^{10}+\cdots}{{x^2+\frac{1}{3}x^4+\frac{2}{45}x^6+\frac{1}{315}x^8+\frac{2}{14175}x^{10}+\cdots}}
\end{align}$$
I can't continue.
| You may use some standard limits as follows:
\begin{eqnarray*}\frac{e^{x^2}-\cos x - x\sin x}{\sinh^2 x}
& = & \left(\frac{e^{x^2}-1}{x^2}+ \frac{1-\cos x}{x^2}-\frac{\sin x}{x}\right)\cdot \left(\frac{x}{\sinh x}\right)^2\\
& \stackrel{x\to 0}{\longrightarrow} & \left(1+\frac 12 - 1\right)\cdot 1^2 = \frac 12
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
help faulhaber's formula summation problem I'm given that $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$
and told to prove $\sum_{k=1}^n (2k-1)=n^2$
$\sum_{k=1}^n (2k-1)=2\sum_{k=1}^n k- \sum_{k=1}^n 1=2\frac{n(n+1)}{2}-n$
$=2\frac{n(n+1)}{2}-n$
$=\frac{n\left(n+1\right)\cdot \:2}{2}-n$
$=n\left(n+1\right)-n$
$=n^2+n-n$
$=n^2$
$\sum_{k=1}^n (2k-1)=n^2$
next I need to write the summation that gives the sum
of the odd numbers found in the first $n$ rows of this sequence.
$1 = a_1$
$3 + 5 = a_2$
$7 + 9 + 11 = a_3$
$13 + 15 + 17 + 19 = a_4$
$21 + 23 + 25 + 27 + 29 = a_5$
$\vdots \vdots \vdots$
I'm givin that $n=3$ should yield the same result as $\sum_{k=1}^6 (2k-1)=36$
if you sub in $n^2$ to the formula for $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ you get $\frac{n^{2}\left(n+1\right)^{2}}{2^{2}}$
at $n=3$; $\frac{3^{2}\left(3+1\right)^{2}}{2^{2}}=36$
I'm having trouble getting that formula into the correct form I believe this is the end result but I'm unsure how to prove the equality $\frac{n^{2}\left(n+1\right)^{2}}{2^{2}}=\sum_{k=1}^{n}k^{3}$
Then I need to replace that sum with an explicit formula of n
| Hint: Write the rows of your sequence this way:
$$
\begin{aligned}
a_1 &= (2\cdot1 -1)\\
a_2 &= (2\cdot2-1) + (2\cdot3-1)\\
a_3 &= (2\cdot4-1) + (2\cdot5-1) + (2\cdot6-1)\\
a_4 &= (2\cdot7-1) + (2\cdot8-1) + (2\cdot9-1) + (2\cdot10-1)
\end{aligned}
$$
When you the sum the first $n$ rows of your sequence, you are summing $2k-1$ from $k=1$ up to a final value for $k$.
For $n=1$, the final value of $k$ is $1$.
For $n=2$, the final value of $k$ is $3$.
For $n=3$, the final value of $k$ is $6$.
For $n=4$, the final value of $k$ is $10$.
Find an expression for the final value of $k$, as a function of $n$. Then plug this final value into your formula
$$
\sum_{k=1}^{\text {final}}(2k-1) = ({\text {final}})^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+) Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$
I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this using $x+2y=8$?
| $$
x + y + \frac{3}{x} + \frac{9}{2y} = \frac{1}{4}(x+2y) + \left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right)
= 2 + \left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right)
$$
And
$$
\left (\frac{3}{4}x + \frac{1}{2}y \right) + \left( \frac{3}{x} + \frac{9}{2y} \right) \geq 2 \sqrt{\left (\frac{3}{4}x + \frac{1}{2}y \right) \left( \frac{3}{x} + \frac{9}{2y} \right)}
$$ (AM-GM, equality: $\left (\frac{3}{4}x + \frac{1}{2}y \right) = \left( \frac{3}{x} + \frac{9}{2y} \right)$)
Further,
$$
\left (\frac{3}{4}x + \frac{1}{2}y \right) \left( \frac{3}{x} + \frac{9}{2y} \right) \geq \left( \frac{3}{2} + \frac{3}{2} \right)^2 = 9
$$ (Cauchy– Schwarz inequality, equality: $\frac{4}{x^2} = \frac{9}{y^2}$)
Observe that the equalities hold when $x=2$ and $y=3$, and the minimum is $2+2*(\sqrt{9}) = 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A trigonometric function comparing with $0$ Short Version
We need to solve this problem:
Prove that function $$f(x) = 2\sin\left(\frac{\pi}{x+\frac{1}{x}+1}\right)-\sin\left(\frac{\pi}{x^{2}+x+1}\right)-\sin\left(\frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}\right)$$ is always larger than to $0$ on $(0,1)$, and is always smaller than $0$ on $(-1,0)$.
Long Version
Me and my friends are working on an extended problem from another forum:
Prove that function $$f(x) = 2\sin\left(\frac{\pi}{x+\frac{1}{x}+1}\right)-\sin\left(\frac{\pi}{x^{2}+x+1}\right)-\sin\left(\frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}\right)$$ is always larger than or equal to $0$ on $(0,+\infty)$, and is always smaller than or equal to $0$ on $(-\infty,0)$.
Despite our best effort, we failed to solve it. Our initial opinion is that to show that $f'(x)$ is always smaller than or larger than $0$ before or after a constant since we are able to show that $$\lim_{x\to\pm\infty} = 0$$ However we failed.
Every approch or guidance to the problem, if mentioned, will be appreciated. Thanks in advance.
UPD: Calculating $f'(x)$ and determining whether it is positive / negetive throughout is not necessary!
OK, after long thoughts I decided to post an answer to the problem.
Actually, $f(x) = f\left(\frac{1}{x}\right)$ is a very helpful tool.
Via calculating $f'(x)$ we can discover that $f(x)$ is always positive on $(0, 1)$ and always negative on $(-1, 0)$.
Hence, let $g:(0,1)\to(1,+\infty)$ become $g(x)=\frac 1 x$, we know that if $f(x)>0$, then $f(g(x))>0$, and hence the question is proved.
Similarly, the other part can be proved.
Some questioned the inequality, well, you can check the link: https://www.desmos.com/calculator/clfhhlq5zi
Some asked for the proof of "$f(x)$ is always positive on $(0, 1)$"
Firstly, given $$
f'(x)=\frac{\pi (2 x+1) \cos \left(\frac{\pi }{x^2+x+1}\right)}{\left(x^2+x+1\right)^2}-\frac{2 \pi \left(1-\frac{1}{x^2}\right) \cos \left(\frac{\pi }{x+\frac{1}{x}+1}\right)}{\left(x+\frac{1}{x}+1\right)^2}+\frac{\pi \left(-\frac{2}{x^3}-\frac{1}{x^2}\right) \cos \left(\frac{\pi }{\frac{1}{x^2}+\frac{1}{x}+1}\right)}{\left(\frac{1}{x^2}+\frac{1}{x}+1\right)^2}
$$
Concluding from demos graph, $f'(x)$ is negative from a constant, $x_0$, and returns to $0$ at $1$. This conclusion can also be proven by pluging in $f'(x)$.
Hence, $f(1)$ is a minimum point, since $f'(1) = 0$. However, $f(1)>0$ and, also, $f(x_0)>0$, we can now conclude that $\forall x\in(0,1)$, $f(x)>0$.
This, well, is not a rigorous proof, but it seems convincing to me and I hope someone can help me finish this.
Therefore, the problem is not ended since we had a little problem when it comes to $(0,1)$.
| Edit: A slightly different approach is described as follows:
Denote $A = \frac{\pi}{x+\frac{1}{x}+1}$, $B = \frac{\pi}{x^{2}+x+1}$ and $C = \frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}$.
Clearly, $A + B + C = \pi$.
We have
\begin{align}
f(x) &= 2\sin A - \sin B - \sin C\\
&= 2\sin (B + C) - 2 \sin\frac{B+C}{2} \cdot \cos \frac{B-C}{2}\\
&= 2\sin \frac{B+C}{2} \cdot \left(2\cos \frac{B+C}{2} - \cos \frac{B-C}{2}\right).
\end{align}
Thus, if suffices to prove that
$$2\cos \frac{B+C}{2} - \cos \frac{B-C}{2} \ge 0.$$
Similarly, we may use bounds. Omitted.
Proof of $f(x) \ge 0$ on $(0, \infty)$:
Since $f(x) = f(1/x)$ for all $x > 0$, we only need to prove the case $x \ge 1$.
First we give the following auxiliary results (Facts 1 through 5). The proofs are easy and thus omitted.
Fact 1: $\sin u \le u$ for all $u \ge 0$.
Fact 2: $\sin u \ge \frac{2}{\pi}u$ for all $u \in [0, \pi/2]$.
Fact 3: $\sin u \ge \frac{\sqrt{3}}{2} + \frac{1}{2}(u - \pi/3) - \frac{\sqrt{3}}{4}(u - \pi/3)^2$ on $[0, \pi/3]$.
Fact 4: $\sin u \le \frac{\sqrt{3}}{2} + \frac{1}{2}(u - \pi/3) - \frac{\sqrt{3}}{6}(u - \pi/3)^2$ on $[0, \pi/3]$.
Fact 5: $\sin u \le \frac{\sqrt{3}}{2} + \frac{1}{2}(u - \pi/3) - \frac{\sqrt{3}}{4}(u - \pi/3)^2$ on $[\pi/3, \pi]$.
Now, we split into two cases:
*
*$x \ge 8$
It suffice to prove that
$$2\sin\left(\frac{\pi}{x+\frac{1}{x}+1}\right)-\sin\left(\frac{\pi}{x^{2}+x+1}\right)-\sin\left(
\pi - \frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}\right) \ge 0.$$
By Facts 1-2, it suffices to prove that
$$2\cdot \frac{2}{\pi} \cdot \frac{\pi}{x+\frac{1}{x}+1} - \frac{\pi}{x^{2}+x+1} - \left(\pi - \frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}\right) \ge 0$$
that is
$$\frac{(4 - \pi)x - 2\pi}{x^2+x+1} \ge 0$$
which is true.
*$1\le x < 8$
Denote $A = \frac{\pi}{x+\frac{1}{x}+1}$, $B = \frac{\pi}{x^{2}+x+1}$ and $C = \frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}$.
Note that $A \in [0, \pi/3]$, $B \in [0, \pi/3]$ and $C \in [\pi/3, \pi)$.
By Facts 3-5, it suffices to prove that
\begin{align}
&2\cdot \left(\frac{\sqrt{3}}{2} + \frac{1}{2}(A - \pi/3) - \frac{\sqrt{3}}{4}(A - \pi/3)^2\right)\\
&\quad - \left(\frac{\sqrt{3}}{2} + \frac{1}{2}(B - \pi/3) - \frac{\sqrt{3}}{6}(B - \pi/3)^2\right)\\
&\quad - \left(\frac{\sqrt{3}}{2} + \frac{1}{2}(C - \pi/3) - \frac{\sqrt{3}}{4}(C - \pi/3)^2\right) \ge 0
\end{align}
that is
$$\frac{\pi (x-1)^2\Big[(8\pi \sqrt{3} - 54)x^2 + (32\pi \sqrt{3} - 54)x + 5\pi \sqrt{3} - 54\Big]}{108(x^2+x+1)^2} \ge 0$$
which is true.
We are done.
| {
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"url": "https://math.stackexchange.com/questions/4029177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$ Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$
and by the same method we have :
$$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$
Adding them together will give us the desired inequality.
and please don't use any $\sum_{cyc}$ because I get confused with it.
| We can rewrite the LHS :
$$LHS = \frac1{1+ \frac{bc}{a^2}} + \frac1{1+\frac{ac}{b^2}} + \frac1{1+\frac{ab}{c^2}}$$
Let $u = \frac{b}{a}$, $v = \frac{a}{c}$, $w = \frac{c}{b}$. Then :
$$LHS = \frac{v}{u+v} + \frac{u}{w+u} + \frac{w}{w+v}$$
We suppose that $u \geq v$ and $u \geq w$ (other cases are similar).
If $w > v$ :
$$LHS < \frac{v}{u+v} + \frac{u}{v+u} + \frac{w}{w} = 2$$
If $w \leq v$ :
$$LHS < \frac{v}{v+v} + \frac{u}{u} + \frac{w}{w+w} = 2$$
Taking $u = n^2$, $w = n$, $v = 1$ :
$$LHS = \frac1{1+n}+\frac{n^2}{n+n^2}+\frac{n}{n+1} \to 2 \quad [n \to \infty]$$
Thus $2$ can't be improved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4029795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Close-form for triple integral $ \int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$ I am able to work out the double integral
$$\int_0^b \int_0^a \sqrt{x^2+y^2} dx dy $$
with brute-force (i.e. integrating $x$, then $y$) to arrive at the close-form result
$$\frac13ab\sqrt{a^2+b^2} +\frac16a^3\sinh^{-1}\frac ba +\frac16 b^3 \sinh^{-1}\frac ab$$
which has the expected parity between $a$ and $b$. However, it gets unwieldy to tackle the triple-integral extension $$\int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$$
this way and I am unable to slug it out. Does anyone know the corresponding close-form expression for the triple version?
| You can refer to J.M. Borwein's box integral
\begin{gather*}
\int_0^c\int_0^b\int_0^a\dfrac{1}{\sqrt{x^2+y^2+z^2}}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\
=
\begin{array}{r}
ab\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\
+ac\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right)
+bc\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\
-\frac{a^2}{2}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^2}{2}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^2}{2}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)
\end{array}\\
\\
\\
\int_0^c\int_0^b\int_0^a\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\
=
\begin{array}{r}
\frac{abc}{4}\sqrt{a^2+b^2+c^2}+\frac{ab\left(a^2+b^2\right)}{6}\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\
+\frac{ac\left(a^2+c^2\right)}{6}\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right)
+\frac{bc\left(b^2+c^2\right)}{6}\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\
-\frac{a^4}{12}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^4}{12}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^4}{12}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)
\end{array}\\
\\
\\
\int_0^c\int_0^b\int_0^a\bigg(x^2+y^2+z^2\bigg)^{\frac{3}{2}}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\
=
\begin{array}{r}
\frac{2abc\left(a^2+b^2+c^2\right)}{15}\sqrt{a^2+b^2+c^2}+\frac{ab\left(9a^4+10a^2b^2+9b^4\right)}{120}\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\
+\frac{ac\left(9a^4+10a^2c^2+9c^4\right)}{120}\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right)\\
+\frac{bc\left(9b^4+10b^2c^2+9c^4\right)}{120}\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\
-\frac{a^6}{30}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^6}{30}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^6}{30}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)
\end{array}
\end{gather*}
| {
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"url": "https://math.stackexchange.com/questions/4030756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Number of roots of $\lfloor\frac x3\rfloor=\frac x2$ How many roots does the equation $$\left\lfloor\frac x3\right\rfloor=\frac x2$$ have?
*
*$1$
*$2$
*$3$
*infinitely many
I checked that $x=0$ and $x=-2$ are the answers, so I think the answer is $(2)$. but I don't know how to solve the problem in general.
|
$$\left\lfloor \frac{x}{3} \right\rfloor = \frac{x}{2}$$
Hint Since $\left\lfloor \frac{x}{3} \right\rfloor$ is an integer, any solution $x$ must be a multiple of $2$. Now, for $x > 0$ we have $$\left\lfloor \frac{x}{3} \right\rfloor \leq \frac{x}{3} < \frac{x}{2} ,$$ and for $x \leq -6$ we have
$$\left\lfloor \frac{x}{3} \right\rfloor > \frac{x}{3} - 1 \geq \frac{x}{2} ,$$ leaving only a few possibilities to check manually.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
find all primes $p$ such that $3^p-(p+2)^2$ is a prime find all primes $p$ such that $3^p-(p+2)^2$ is also a prime.
My proof
set $3^p-(p+2)^2=q$
case $1$:$p=2 \implies q = -7 \notin \mathbb N$. so there is no solution.
case $2$: $p \geq 3$
notice that $p=3 \implies q = 2$ so there is a solution.
claim: with $p\geq 4$ there is no solution.
Proof: we gonna show that $3^p-(p+2)^2$ is always even, and we know that for $p\geq3$ there is no even prime.
so :$$q = 3^p-(p+2)^2= 3^p-p^2-4p-4$$ $3^p$ and $p^2$is odd and $-4p-4$ is even so :
$$odd + odd+even+even=even+even=even$$
therefore there is no solution with $p\geq4$.
is this a valid proof?
| Hint: As is very often the case, you don't need calculus when dealing only with integers. In stead of differentiating, you can consider the differences between consecutive values of $3^k-(k+2)^2$.
The details:
For every nonnegative integer $k$ define $f(k):=3^k-(k+2)^2$. With a bit of algebra you'll find that the first difference of this sequence equals
\begin{eqnarray*}
f(k+1)-f(k)&=&
\big(3^{k+1}-((k+1)+2)^2\big)-\big(3^k-(k+2)^2\big)\\
&=&\big(3^{k+1}-3^k\big)-\big((k+3)^2-(k+2)^2\big)\\
&=&3^k\big(3-1\big)-\big(2k+5\big)\\
&=&2\cdot3^k-2k-5.
\end{eqnarray*}
Similarly, the second difference equals
\begin{eqnarray*}
\big(2\cdot3^{k+1}-2(k+1)-5\big)-\big(2\cdot3^k-2k-5\big)
&=&\big(2\cdot3^{k+1}-2\cdot3^k\big)-\big(2(k+1)-2k\big)\\
&=&2\cdot3^k\big(3-1\big)-2\\
&=&4\cdot3^k-2.
\end{eqnarray*}
Clearly the second difference is positive for all $k\geq0$. Then looking at the first difference, for $k=2$ we have
$$2\cdot3^2-2\cdot2-5=4>0,$$
and hence the first difference is positive for all $k\ge2$. It follows that for all $p>3$ we have
$$f(p)>f(3)=2,$$
and hence $f(p)$ is not prime for any prime number $p>3$, because $f(p)$ is even and greater than $2$. For $p=2$ we have $f(2)=-7$ which is of course not prime. So $p=3$ is the unique prime number for which $f(p)$ is prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate an absolute monster integral $\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$ I want to figure out a way to evaluate $$\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$$
I tried to substitute $x = u^2$ and cancel the square root in the denominator, getting $$2\int\limits_{0}^{1} \frac{\log(1-u^2+u^4)}{1+u^2}\mathrm{d}u.$$But now I am stuck again.
| Svyatoslav showed that $$\int_{0}^{1} \frac{\log(1-x+x^{2})}{\sqrt{x}(1+x)} \, \mathrm dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx- 4 G.$$
The following is a slightly different way to show that $$\int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx = 2 \pi \log (3). $$
First notice that $$ \begin{align} \int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{\log(1+\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx+ \int_{-\infty}^{\infty} \frac{\log(1-\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx \\ &= \int_{-\infty}^{\infty} \frac{\log(1+\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx + \int_{\infty}^{-\infty} \frac{\log(1+\sqrt{3}u + u^{2})}{1+u^{2}} (- \mathrm du) \\ &= 2 \int_{-\infty}^{\infty} \frac{\log(1+\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx \end{align}$$
In a previous answer I used contour integration to show that $$I(a,b,\theta) = \int_{-\infty}^{\infty} \frac{\log \left(a^{2}+2ax \cos \theta + x^{2}\right)}{x^{2}+b^{2}} \, \mathrm dx = \frac{\pi}{b} \, \log \left(a^{2}+2ab \sin \theta +b^{2} \right), $$ where $a, b >0$ and $0 < \theta < \pi$.
Therefore, $$ \int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx = 2 I \left( 1,1, \frac{\pi}{6} \right) = 2 \pi \log(3).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove $ (\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1$ Let $\triangle ABC$ be a acute triangle. Prove that: $$(\frac{1}{cosA}-1)(\frac{1}{cosB}-1)(\frac{1}{cosC}-1) \ge 1 $$
My attempt:
$$\Leftrightarrow (1-cosA)(1-cosB)(1-cosC)\ge cosA.cosB.cosC$$
$$\Leftrightarrow 1-2cosA.cosB.cosC + cosA.cosB + cosB.cosC+cosA.cosC \ge cosA+cosB+cosC $$
$$\Leftrightarrow cos^2A+cos^2B+cos^2C + cosA.cosB + cosB.cosC+cosA.cosC \ge cosA+cosB+cosC $$
$$ 0<cos A,cos B,cosC<1$$
$$cos^2A+cos^2B+cos^2C=1-2cosA.cosB.cosC\ge 3\sqrt[3]{cosA.cosB.cosC}$$
$$\Rightarrow cosA.cosB.cosC \le \frac{1}{8}$$
And I was stuck here. Could you help me ?
| We have
$$1- \cos A = 1- \frac{b^2+c^2-a^2}{2bc} = \frac{(a+b-c)(c+a-b)}{2bc},$$
so
$$(1-\cos A)(1-\cos B)(1-\cos C) = \frac{(a+b-c)^2(b+c-a)^2(c+a-b)^2}{8a^2b^2c^2},$$
and
$$\cos A \cos B \cos C = \frac{(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)}{8a^2b^2c^2}.$$
Thefore, the original inequality become
$$(a+b-c)^2(b+c-a)^2(c+a-b)^2 \geqslant (a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2).$$
It's remain to prove that
$$(a+b-c)^2(c+a-b)^2 \geqslant (a^2+b^2-c^2)(c^2+a^2-b^2).$$
But, this is true because
$$(a+b-c)^2(c+a-b)^2 - (a^2+b^2-c^2)(c^2+a^2-b^2) = 2(b-c)^2(b^2+c^2-a^2) \geqslant 0.$$
The proof is completed.
| {
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Find $\int_0^4(g\circ f\circ g)(x)\mathrm{d}x$ where $f(x)=\sqrt[3]{x+\sqrt{x^2+1/27}}+\sqrt[3]{x-\sqrt{x^2+1/27}}$, $g(x)=x^3+x+1$
Let $$f(x)=\sqrt[3]{x+\sqrt{x^2+\frac{1}{27}}}+\sqrt[3]{x-\sqrt{x^2+\frac{1}{27}}}$$
and
$$g(x)=x^3+x+1$$
then, find $$\int_0^4(g\circ f\circ g)(x) \mathrm dx$$
My attempt:
Let $\displaystyle h(x)=\sqrt{x^2+\frac{1}{27}}$
$$(g\circ f)(x)=2x+3((2x)(x^2-[h(x)]^2))^{1/3}+(x+h(x))^{1/3}+(x-h(x))^{1/3}+1$$
Is finding $(g\circ f \circ g)(x)$ in term of $x$ necessary? Because it is quite a work to do.
| This is a nice problem.
$$f(x)=\underbrace{\sqrt[3]{x+\sqrt{x^2+\frac{1}{27}}}}_{a}+\underbrace{\sqrt[3]{x-\sqrt{x^2+\frac{1}{27}}}}_b$$
Now note that from the identity $(a+b)^3=a^3+b^3+3ab(a+b)$, and using $ab=-\dfrac{1}3$,
we have:
$$(f(x))^3=2x-f(x) \implies (f(x))^3+f(x)=2x $$
And from $g(x)=x^3+x+1$ , we have $$g(f(x))=(f(x))^3+f(x)+1=2x+1$$
Therefore $$g(f(g(x))=2g(x)+1=2x^3+2x+3$$
So all you have got is $$\int_0^4 (2x^3+2x+3)\mathrm{d}x$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve determinants with $|AB| = |A||B|$ How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.
(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & \dots & 1+x_ny_n
\end{vmatrix}$
(2) $D_2 = \begin{vmatrix}
1 & \cos(a_1-a_2) & \cos(a_1-a_3) & \dots & \cos(a_1-a_n) \\
\cos(a_1-a_2) & 1 & \cos(a_2-a_3) & \dots & \cos(a_2-a_n) \\
\cos(a_1-a_3) & \cos(a_2-a_3) & 1 & \dots & \cos(a_3-a_n) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\cos(a_1-a_n) & \cos(a_2-a_n) & \cos(a_3-a_n) & \dots & 1
\end{vmatrix}$
(3) $D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}$
(4) Let $s_k=a_1^k+a_2^k+a_3^k+a_4^k \quad (k=1,2,3,4,5,6)$,
$$D_4 = \begin{vmatrix}
4 & s_1 & s_2 & s_3 \\
s_1 & s_2 & s_3 & s_4 \\
s_2 & s_3 & s_4 & s_5 \\
s_3 & s_4 & s_5 & s_6
\end{vmatrix}$$
My Attempt:
(1) I noticed for any $i,j$,
$1+x_iy_j = \begin{bmatrix} 1 & x_i \end{bmatrix}
\begin{bmatrix} 1 \\ y_j \end{bmatrix}$.
So, the matrix corresponds $D_1$ equals $\begin{bmatrix}
1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n
\end{bmatrix}_{n \times 2}
\begin{bmatrix}
1 & 1 & \dots & 1 \\
y_1 & y_2 & \dots & y_n
\end{bmatrix}_{2 \times n}$.
But it's not helpful at all. :(
(2) It's obvious that
$a_{ij} = \cos(a_i - a_j) = \cos a_i \cos a_j + \sin a_i \sin a_j$.
(3) I've no idea about this problem at all. All I came up with, is
$$D_3 = a^4 \begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
1 & -1 & -1 & 1
\end{vmatrix}$$
(4) I noticed that $a_{ij} = s_{i+j-2}$. But still, not helpful.
Plz give me some hints. Thx in advance.
| For the fourth determinant start with
$$
V=\pmatrix{1 & 1 & 1 & 1\\a_1 & a_2 & a_3 & a_4\\a_1^2 & a_2^2 & a_3^2 & a_4^2\\a_1^3 & a_2^3 & a_3^3 & a_4^3\\}
$$
and notice that
$$
\begin{pmatrix}
4 & s_1 & s_2 & s_3 \\
s_1 & s_2 & s_3 & s_4 \\
s_2 & s_3 & s_4 & s_5 \\
s_3 & s_4 & s_5 & s_6
\end{pmatrix}=VV^{T}.
$$
The determinant of $V$ is well-known to be $\prod_{i<j}(a_i-a_j)$, so $D_4=(\prod_{i<j}(a_i-a_j))^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor expansion of a gaussian integral I tried Taylor expanding the following function for small $x \ll 1$:
\begin{align*}
f(x) = \frac{1}{\sqrt{2\pi}} \int\limits^{+\infty}_{-\infty} e^{\frac{-xy^4}{24}-\frac{y^2}{2}}\,\mathrm{d}y.
\end{align*}
I already know that $f$ takes on finite values for $x \geq0$. My goal is to get a series expression for $f$ of the following form:
\begin{align*}
f_N(x) = \sum^N_{n = 0}a_n x^n.
\end{align*}
So after using the formula (for $a = 0$)
\begin{align*}
f(x) = \sum^\infty_{n = 0} \frac{f^{(n)}(a)}{n!}(x-a)^n,
\end{align*}
I get up to the $N$-th term:
\begin{align*}
f_N(x) = \sum^N_{n = 0} \frac{1}{n!}\frac{(-x)^n}{24^n}(4n-1)!!
\end{align*}
However, when plugging in a small value for $x$ (for example $x = 0.1$) in Wolfram Alpha, I get that till $N = 25$ the approximation is very good, but after around $N = 30$ the series diverges away from the true value ($\approx 0.988306$ ). So for $N \rightarrow + \infty$, the series seems to diverge.
My questions are:
How can this be ? I thought that for larger and larger $N$ the Taylor approximation would be more and more better ? (well, for small $x$ atleast)
Does this mean that the radius of convergence is equal to $0$ and that the interval of convergence is just the point $x = 0$ ? If not, then what is the radius of convergence ? If yes, doesn't this contradict the fact that $f$ is finite for $x \geq1$ ?
| This is not a gaussian integral.
$$f(x) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} e^{-\frac{xy^4}{24}-\frac{y^2}{2}} \, dy=\frac{\sqrt{3}}{\sqrt{2\pi}}\frac{e^{\frac{3}{4 x}} }{\sqrt{x}}K_{\frac{1}{4}}\left(\frac{3}{4 x}\right)$$
Using the asymptotics of the Bessel function (since $x \to 0 \implies \frac 1 x \to \infty$), you should have
$$f(x)=1-\frac{x}{8}+\frac{35 x^2}{384}-\frac{385 x^3}{3072}+\frac{25025
x^4}{98304}-\frac{1616615 x^5}{2359296}+O\left(x^6\right) $$
If you make $x=\frac 1{10}$, this truncated series will give
$$f\left(\frac{1}{10}\right)=\frac{46634068277}{47185920000}=0.9883047$$ while the exact value is
$$f\left(\frac{1}{10}\right)=e^{15/2} \sqrt{\frac{15}{\pi }} K_{\frac{1}{4}}\left(\frac{15}{2}\right)=0.9883064$$
If you want more terms, let $x=\frac{3}{4 z}$ which makes
$$f(z)=\sqrt{\frac{2}{\pi }} e^z \sqrt{z} K_{\frac{1}{4}}(z)$$ with $z \to \infty$. Have a look at this paper (equation $(1.10)$) to have
$$f(z)=\Bigg[\sqrt{\frac{2}{\pi }} e^z \sqrt{z}\Bigg]\sqrt{\frac{\pi }{2z}} e^{-z} \sum_{n=0}^\infty a_n\left({\frac 14}\right)\,z^{-n}=\sum_{n=0}^\infty a_n\left({\frac 14}\right)\,z^{-n}$$
From equation $(1.9)$
$$ a_n\left({\frac 14}\right)=(-1)^n\frac{ \cos \left(\frac{\pi }{4}\right) \Gamma
\left(n+\frac{1}{4}\right) \Gamma \left(n+\frac{3}{4}\right)}{\pi\, 2^n\, \Gamma (n+1)}$$
Make $z=\frac 3{4x}$ to recover the first expansion.
Edit
Without using the exact solution (as I did), you could have done the same work expanding first the integrand as a Taylor series around $x=0$
$$e^{-\frac{xy^4}{24}-\frac{y^2}{2}}=\sum_{n=0}^\infty\frac {(-1)^n}{24^n\,n!} e^{-\frac{y^2}{2}} y^{4 n} x^n$$ and use that
$$\int_{-\infty}^\infty e^{-\frac{y^2}{2}} y^{4 n}\,dy=2^{2 n+\frac{1}{2}} \Gamma \left(2 n+\frac{1}{2}\right)$$ This makes
$$f(x)= \frac{1}{\sqrt{\pi}}\sum_{n=0}^\infty\frac {(-1)^n}{6^n\,n!} \Gamma \left(2 n+\frac{1}{2}\right) x^n$$ ad the same result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find number of solutions to $a+b+c=0$ As titled, I need to find number of solutions to the equation $a+b+c=0$, where $a,b,c$ are integers in the range $[-k,k]$.($k$ is a positive integer) and $a$ doesn't equal to $0$.
My attempt: let $$x=a+k, y=b+k, z=c+k$$ we need to find non-negative solutions to the equation $$x+y+z=3k$$ with $$0\leq x,y,z\leq 2k.$$
Using the formula, there are $\left(\frac{3k+3-1}{3-1}\right)$=$\left(\frac{3k+2}{2}\right)$ solutions.
We then erase the cases where one of them is $>2k$ (only one of them can be $>2k$).
Let $$x'= x-(2k+1)$$ then x' is non-negative, number of solutions to the equation: $$x'+y+z=k-1$$ is:
$$\left(\frac{k+3-1}{3-1}\right)$$
But this gives a total of $3k^2-2$ solutions if we calculate $\left(\frac{3k+2}{2}\right)$-$3 \cdot \left(\frac{k+2}{2}\right)$,which is not the desired answer. And I haven't delete the case where $a=0$
What did I do wrong?
Any help will be appreciated.
| Your first binomial coefficient, $\binom{3k+2}2$, is correct, but your second, counting the solutions in non-negative integers to $x_1'+x_2+x_3=k-1$, is not: it should be
$$\binom{(k-1)+3-1}{3-1}=\binom{k+1}2\,,$$
giving you a total of
$$\begin{align*}
\binom{3k+2}2-3\binom{k+1}2&=\frac12\big((3k+2)(3k+1)-3k(k+1)\big)\\
&=3k^2+3k+1
\end{align*}$$
solutions in non-negative integers to $x+y+z=3k$. Now you just have to subtract the solutions for which $x=k$, i.e., the solutions in non-negative integers to $y+z=2k$. I’ll leave that to you: these are straightforward to count.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factoring out $4x^3+2x^2y-2xy^2-y^3$ This can be factored as follows:
$$4x^3+2x^2y-2xy^2-y^3 = (2x^2-y^2)(2x+y)$$
What is a systematic way for finding this factorization?
| \begin{gather*}
4x^{3} +2x^{2} y-2xy^{2} -y^{3}\\
=\left( 4x^{3} -2xy^{2}\right) +\left( 2x^{2} y-y^{3}\right)\\
=2x\left( 2x^{2} -y^{2}\right) +y\left( 2x^{2} -y^{2}\right)\\
=( 2x+y)\left( 2x^{2} -y^{2}\right)
\end{gather*}
Does this help?
| {
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To show that a sequence is bounded above by $3$. Let $x_1=\sqrt2$ and $x_{n+1}=\sqrt{2 x_n}$.
I would like to know if there is a way to see that this sequence is bounded above by 3 without induction.
I know how to use induction to show that it is in fact bounded by 2. I would like to not use induction though. Thanks for your time.
| OK. Let's give it a try. $x_{n} = \sqrt{2x_{n-1}}=2^{\frac{1}{2}}\cdot \sqrt{\sqrt{2x_{n-2}}}=2^{\frac{1}{2}}\cdot 2^{\frac{1}{4}}\cdot \sqrt[4]{x_{n-2}}= 2^{\frac{1}{2}+\frac{1}{4}}\cdot \sqrt[4]{\sqrt{2x_{n-3}}}= 2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}\cdot x_{n-3}^{\frac{1}{8}}= 2^{\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{n-1}}}\cdot x_1^{\frac{1}{2^{n-1}}}= 2^{\frac{1}{2}+\frac{1}{4}+\cdots + \frac{1}{2^n}}= 2^{1-\frac{1}{2^n}}< 2 < 3.$
| {
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Solving differential equation ${\rm d}y/{\rm d}x = (4x - y)^2$ Let $u = 4x - y$
${\rm d}u = 4{\rm d}x - {\rm d}y$
$4{\rm d}x - {\rm d}u = {\rm d}y$
$4{\rm d}x - {\rm d}u = (4x - y)^2$
$4{\rm d}x - {\rm d}u = u^2$
I was stuck at $-{\rm d}u = (u - 4){\rm d}x$
by separation of variables (SOV), so what's next?
| To solve
$$dy/dx = (4x-y)^2$$
is to find a function $y(x)$, that satisfies this equation.
Substitute
$$u=4x-y$$
It follows
$$\Rightarrow du/dx = 4 - dy/dx \Leftrightarrow dy/dx = 4-du/dx $$
Plug in the $u$ and the $du/dx$ in the original equation
$$\begin{align} & dy/dx = 4-du/dx = u^2 = (4x-y)^2 \\ \Rightarrow & \ du/dx = 4-u^2 \\ \Rightarrow & \ dx/du = 1/(4-u^2) \\ \Rightarrow & \ dx = du/(4-u^2)\end{align}$$
It follows
$$\begin{align}x &= \int du/(4-u^2) +C\\
&= (1/4) \cdot \int du/(1-(u/2)^2) + C\\
&= (2/4) \cdot \int da/(1-a^2) + C && a = u/2 \ \ \text{and} \ \ da/du = 1/2\\
&= (1/2) \cdot \text{arctan}(a) + C \\
&= (1/2) \cdot \text{arctan}\left(u/2\right) + C \\
&= (1/2) \cdot \text{arctan}\left((4x-y)/2\right) + C && u = 4x-y \\
\end{align}$$
This means that
$$ x = (1/2) \cdot \text{arctan}\left((4x-y)/2\right) + C$$
which has to be solved for $y$. Maybe this way:
Substract C from both sides
$$ x - C = (1/2) \cdot \text{arctan}\left((4x-y)/2\right) $$
Multiply by $2$
$$ 2(x - C) = \text{arctan}\left((4x-y)/2\right) $$
Take $\tan$ on both sides
$$ \tan(2(x - C)) = (4x-y)/2 $$
Multiply by $2$
$$ 2\tan(2(x - C)) = 4x-y $$
Rearrange
$$ y = 4x-2\tan(2(x - C)) $$
Not so sure if $ y = 4x-2\tan(2(x - C)) $ satisfies $dy/dx = (4x-y)^2$ ... ?
| {
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Prove that $\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}$ is convergent or divergent. I have the series $$\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}.$$ How can I prove the convergence or divergence of it?
I tried to use the comparison test and claim that:
$$\sum_{n=1}^{\infty}\left( 1-\frac{1}{n^2+1}\right)^{n^3}\leq \sum_{n=1}^{\infty}\left( 1+\frac{1}{n^2}\right)^{n^2}$$ but the right series divergent because the limit of the sequence is e and therefore the series is divergent.
| Hint
$$
\left( 1-\frac{1}{(n^2+1)}\right)^{n^3}= \frac{1}{\left( 1+\frac{1}{n^2}\right)^{n^3}}
$$
Now, for $n \geq 2$ you have
$$
\left( 1+\frac{1}{n^2}\right)^{n^3} \geq \binom{n^3}{2}(\frac{1}{n^2})^2=\frac{n^3-1}{2n}
$$
Therefore
$$
\sum_{n=1}^\infty \left( 1-\frac{1}{(n^2+1)}\right)^{n^3} \leq \frac{1}{2}+\sum_{n=2}^\infty \frac{2n}{n^3-1}
$$
It is trivial to show that the sum on the RHS is convergent.
| {
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Inequality question involving a cyclic expression and moduli Consider three real numbers a,b,c such that $a^2+b^2+c^2 =1$. What will be the maximum value of the expression $|a-b| + |b-c| + |c-a|$ ?
I tried two approaches. One I used the fact that $a^2+b^2+c^2 =1 = \frac{1}{2} (|a-b|^2 + |b-c|^2 + |c-a|^2) + ab + bc + ca$. Original plan was to somehow eliminate ab + bc + ca , but unable to do so - got stuck in a series of repeating expressions
Another variant was to try and assert that $a \geq b \geq c$ without loss of generality. Then
$(|a-b| + |b-c| + |c-a|)^2 \leq 4(a-c)^2$ , but I am unable to use the given fact about $a^2 + b^2 + c^2$.
I realize that I am missing something very trivial here as this seems a fairly simple problem - so any help would be greatly appreciated.
| Let's assume that $a\ge b\ge c$, then $|a-b|+|b-c|+|c-a|=2(a-c)$
And $(a-c)^2=a^2+c^2-2ac$
Because $a^2+c^2\le a^2+c^2+b^2\le1$ and $-2ac\le a^2+c^2\le a^2+c^2+b^2\le1$
We got $(a-c)^2=a^2+c^2-2ac\le 2$
So $(a-c)\le \sqrt 2$
So the answer is $2\sqrt 2$
Equality holds at $(a,b,c)=(\sqrt \frac12,-\sqrt \frac12,0)$
| {
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Area of the section enclosed by the curve Curve is formed by the intersection of the surface:
$\displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$
with the plane $2x + 2y + z = 1$
What is the area enclosed by this curve?
Eliminating z from both equations, I get an implicit equation relating $x$ and y, with terms in $x^2, y^2, xy, x,$ and $y$. But I don't know where to go from there.
| Two given surfaces are
$ \ \displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \ $ and
$ \ 2x+2y+z=1$
We need to find area enclosed by the intersection curve. The projection of the enclosed area in XY plane is ellipse
$x^2+y^2-\frac{x}{2}-\frac{y}{2}+ \frac{3xy}{2} = 0$.
As the enclosed area is in plane $z = 1- 2x - 2y$, the area element can be expressed in terms of the area element of its projection in $XY$ plane,
$dS = \sqrt{1+2^2+2^2} \ dA \implies dS = 3 \ dA$
Now I will suggest you to plot the ellipse $x^2+y^2-\frac{x}{2}-\frac{y}{2}+ \frac{3xy}{2} = 0$ in a tool and see. It is apparent that to make it easier to work, we either rotate the ellipse clockwise or anticlockwise by $\frac{\pi}{4}$. If we rotate it anticlockwise, we will get an ellipse with major axis parallel to $x-$ axis and minor axis along $y-$axis. A combination of rotation and shifting of origin will convert it into an ellipse centered at the origin and major and minor axes along coordinate axes. We can first rotate by substituting,
$x = \frac{X+Y}{\sqrt2}, y = \frac{Y-X}{\sqrt2}$. So the equation of ellipse becomes,
$\displaystyle \frac{X^2} {(\sqrt{2/7})^2} + \frac{(Y - \sqrt2 / 7)^2}{(\sqrt2 / 7)^2} = 1$
Now you can directly apply the formula for area of ellipse which is $A = \pi ab = \displaystyle \pi \frac{\sqrt2}{\sqrt7} \frac{\sqrt2}{7} = \frac{2 \pi}{7 \sqrt7}$
So the area enclosed by the curve is $\displaystyle \frac{6 \pi}{7 \sqrt7}$.
If you want to find the area by integral, use polar coordinates with the following transformation -
$X = \sqrt{\frac{2}{7}} \ r \cos\theta, Y = \frac{\sqrt2}{7} (1 + r \sin\theta)$
which transforms the ellipse into a unit circle centered at origin.
Now find the Jacobian of transformation and integrate with limits $0 \leq r \leq 1, 0 \leq \theta \leq 2\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution? Here's my system of linear equations:
$\begin{cases}
x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\
\end{cases}$
Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$
By row reduction, I obtain:
$\left[\begin{array}
{ccc|c}1&2&2&1\\0&a-2&1&2\\0&9&a-2&-1\end{array}\right]$
Unfortunately, I am stuck at this stage. I have tried swapping rows around but I didn't have much luck.
Update: I have managed to solve this with the use of the determinant.
Matrix of minors:
$\left[\begin{array}
{ccc}a^2-33&a-3&11-a\\2a-22&a-2&9\\6-2a&1&a-2\end{array}\right]$
Matrix of cofactors:
$\left[\begin{array}
{ccc}a^2-33&3-a&11-a\\22-a&9&a-2\\6-2a&-1&a-2\end{array}\right]$
Adjugate matrix:
$\left[\begin{array}
{ccc}a^2-33&22-2a&6-2a\\3-a&a-2&-1\\11-a&-9&a-2\end{array}\right]$
Det(A) = $1(a^2 - 33) + 2(3 - a) + 2(11 - a) = a^2 - 4a - 5$
$(a - 5)(a + 1) = 0$
Thank you all for your help!
| Hint: Set the determinant equal to zero and solve for $a$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\delta$-response for challenging the $\lim_{x \to 10} \frac{1}{[[x]]} = \frac{1}{10}$ with $\epsilon=\frac{1}{2}$ $\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I am self-studying Real Analysis from Stephen Abbott's Understanding Analysis. I'd like to ask if my conclusions pertaining to exercise (a), (b) of the below problem are correct.
Notation. $f(x)= [[x]]$ is the box function, the greatest integer less than or equal to $x$, for all $x \in \mathbf{R}$.
Exercise 4.2.4. Consider the reasonable but erroneous claim that
\begin{align*}
\lim_{x \to 10} \frac{1}{[[x]]} = \frac{1}{10}
\end{align*}
(a) Find the largest $\delta$ that represents a proper response to the challenge of $\epsilon = 1/2$.
(b) Find the largest $\delta$ that represents a proper response to $\epsilon = 1/50$.
(c) Find the largest $\epsilon$ challenge for which there is no suitable $\delta$ response possible.
Proof.
(a) We require
\begin{align*}
\frac{1}{10} - \frac{1}{2} &< \frac{1}{[[x]]} &< \frac{1}{10} + \frac{1}{2} \\
\frac{-4}{10} &< \frac{1}{[[x]]} &< \frac{6}{10}
\end{align*}
So, $[[x]] < \frac{-10}{4} < -2$ and $[[x]]>\frac{10}{6} > 1$. In other words, $x-10 < -12$ and $x-10>-8$. The absolute distance must satisfy the inequality $\absval{x - 10} < 8$. Thus, the largest $\delta-$response to the challenge $\epsilon=1/2$ appears to be $\delta = 8$.
(b) We require
\begin{align*}
\frac{1}{10} - \frac{1}{50} &< \frac{1}{[[x]]} &< \frac{1}{10} + \frac{1}{50} \\
\frac{4}{50} &< \frac{1}{[[x]]} &< \frac{6}{50}
\end{align*}
So, $[[x]] < \frac{50}{4} < 13$ and $[[x]]>\frac{50}{6} > 8$. In other words, $x < 13$ or $x >9$. The absolute distance must satisfy the inequality $\absval{x - 10} < 1$. Thus, the largest $\delta-$response to the challenge $\epsilon=1/50$ appears to be $\delta = 1$.
(c) We would like to have the distance
\begin{align*}
\absval{\frac{1}{[[x]]} - \frac{1}{10}} > \epsilon
\end{align*}
no matter what the open interval $(10-\delta,10+\delta)$ in which $x$ lies. Rearranging, I get:
\begin{align*}
\epsilon < \frac{\absval{[[x]] - 10}}{10 \absval{[[x]]}}
\end{align*}
I am not sure how to proceed from here. I know that, $\absval{[[x]] - 10}$. That yields,
\begin{align*}
\epsilon < \frac{\delta}{10 \absval{[[x]]}}
\end{align*}
I can further write, $[[x]] > \lceil{10 - \delta}\rceil$. But, this $\epsilon$ is dependent on the $\delta$-interval I choose.
| For part (c), we’re looking for a distance $\epsilon$ such that, when we look around x = 10, the function values aren’t all closer than $\epsilon$ to 1/10.
I feel like a more intuitive approach might be effective here. What value does the function take on to the left of x = 10? How far is that value from 1/10? That difference should be your $\epsilon$, no?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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show that $y^2 = x^5 - x +2 $ has no integer solutions I considered everything$\mod 3$:
$x,y \equiv 0 \mod 3 ,\;\;\;\; y^2 \equiv 0 \mod 3 ,\;\;\;\; x^5 \equiv 0 \equiv x \mod 3$
$x,y \equiv 1 \mod 3 ,\;\;\;\; y^2 \equiv 1 \mod 3 ,\;\;\;\; x^5 \equiv 1 \equiv x \mod 3$
$x,y \equiv 2 \mod 3 ,\;\;\;\; y^2 \equiv 1 \mod 3 ,\;\;\;\; x^5 \equiv 2 \equiv x \mod 3$
thus in particular $x^5 - x \equiv 0 \mod 3, \;\;\; \forall x \in \mathbb{Z},$ and furthermore
$x^5 - x + 2 \equiv 2 \mod 3, \;\;\; \forall x \in \mathbb{Z}$.
But, $y^2 \equiv 0,1 \mod 3$, and if we were considering that there would be an integer solution then both sides should have the same remainder modulo $3$, but LHS and RHS never have the same remainder thus there are no integer solutions?
This is my first time doing a proof of this form, I'm familiar with modular arithmetic, however I've never used it to prove that a polynomial equation didn't have integer solutions.
| You can also use $\bmod 3$:
$$y^2-2=x(x^2-1)(x^2+1)$$
RHS: $x^2-1\equiv 0\bmod 3\Rightarrow RHS \equiv 0\bmod 3$
But :
LHS: $$y^2-1-1\equiv (0\bmod 3-1\bmod 3)\equiv -1\bmod 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculation of a hard integral $\int^{1/2}_0\frac{1}{x^2-x+1} \ln\frac{1+x}{1-x}\,dx $ Calculate
$$\int^{1/2}_0\frac{1}{x^2-x+1} \cdot \ln\frac{1+x}{1-x}\,dx $$
I tried to subsitute $x=1/2-t$ so $dx=-dt$ but I just complicated more my problem.
| Substitute $t=\frac{1-x}{1+x}$
\begin{align}
&\int^{1/2}_0\frac{1}{x^2-x+1} \ln\frac{1+x}{1-x}\,dx \\
=&-2\int_{\frac13}^1 \frac{\ln t}{1+3t^2}dt
\overset{\sqrt3t\to t}= -\frac2{\sqrt3} \int_{\frac1{\sqrt3}}^{\sqrt3} \frac{\ln t-\ln\sqrt3}{1+t^2}dt\\
=& \frac{\ln 3}{\sqrt3} \int_{\frac1{\sqrt3}}^{\sqrt3} \frac{1}{1+t^2}dt = \frac{\pi\ln3}{6\sqrt3}
\end{align}
Note $\int_{\frac1{\sqrt3}}^{\sqrt3} \frac{\ln t}{1+t^2}dt \overset{t\to\frac1t} =0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem with system of equations $(x,y,z)$ System of equations:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}$$
Solution:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}\Rightarrow\begin{cases}x^2+y^2+z^2=3\\xy+yz+zx=3\end{cases}$$
$x^{2}+y^{2}+z^{2}=xy+yz+zx$
$(x+y+z)^{2}-2(xy+yz+zx)=xy+yz+zx$
$(x+y+z)^{2}-2*3=3$
$(x+y+z)^{2}=9$
What should I do next?
| Hint:
$$(x-y)^2+(y-z)^2+(z-x)^2=?$$
Alternatively, $$x^2-x(y+z)+y^2+z^2-yz=0$$
As $x$ is real, the discriminant must be $\ge0$
But $$(y+z)^2-4(y^2+z^2-yz)=-3(y-z)^2\le0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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