Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that if $2a^3 + 27c = 9ab,$ then the roots of $x^3 + ax^2 + bx + c = 0$ form an arithmetic sequence. I am not sure how to begin this problem. Can someone help me?
I have a hint: Let $y = x + \frac{a}{3}$ and rewrite $x^3 + ax^2 + bx + c = 0$ in terms of $y.$
How do I do this?
| Isolating $c$ in what we are given, and then plugging it into the polynomial, we have, $$x^3+ax^2+bx+c=x^3+ax^2+bx+\frac{1}{27}(9ab-2a^3)=$$
$$=\frac{1}{27}(3x+a)(9x^2+6ax+9b-2a^2),$$
which says $$x_1=-\frac{a}{3}$$ and
$$x_2+x_3=-\frac{6a}{9}=-\frac{2a}{3}=2x_1$$ and since $$x_2-x_1=x_1-x_3,$$ we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is ...
If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is $16x^2+9y^2+k_1x-36y+k_2=0$, then find $k_1$ and $k_2$
Before solving it, I noticed a problem with it. Even if we reflect the curve, the coefficient of $x^2$ is 9, and not 16 according to what the question says.
That’s all I have as doubt. I don’t need the whole answer, just need to know if the question is right or not.
| $$\frac{(x-4)^2}{16}+\frac{(y-3)}{9}=1 \implies x=4 \cos t+4,y=3 \sin t+3 ~~~(1).$$ The image $(X,Y)$ of $(x,y)$ in the line $ax+by+c=0$ is given by:
$$\frac{X-x}{a}=\frac{Y-y}{b}=-2\frac{(ax+by+c)}{a^2+b^2}$$
So we get $$\frac{X-4\cos t-4}{1}=\frac{Y-3\sin t-3}{-1}=-2\frac{4\cos t+4-3\sin t -3-2}{2}$$, we get
$$X=-4\cos t-4 +4\cos t+4+3\sin t+3+2=3 \sin t+5$$ $$ Y=4 \cos t+4-2= 4\cos t +2$$
$$\implies \sin t=\frac{X-5}{3}, ~~ \cos t=\frac{Y-2}{4}$$
Squaring and adding these two results we get the required image the ellipse (1) as
$$\frac{(X-5)^2}{9}+\frac{(Y-2)^2}{16}=1.$$
Finally, one would write the image ellipse as $$\frac{(x-5)^2}{9}+\frac{(y-2)^2}{16}=1.$$ in the same plane as (1)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Integration through partial fractions with complex roots In integrating the following: $\frac{1}{(x^2+2x+3)^2}$ I am trying to use partial fraction decomposition as follows:
$\frac{1}{(x^2+2x+3)^2} = \frac{Ax + B}{x^2+2x+3} + \frac{Cx + D}{(x^2+2x+3)^2}$
Which gives me:
$1 = (Ax + B)(x^2 +2x +3) + (Bx + C)$
And that gets me back to where I started.
$\frac{1}{(x^2+2x+3)^2}$
What am I doing wrong?
| That is already written as partial fractions.
It seems you want:
$\begin{align*}
\int \frac{d x}{(x^2 + 2 x + 3)^2}
&= \frac{1}{\sqrt{2}} \int \frac{d u}{(u^2 + 1)^2} \qquad u = \sqrt{2} (x + 1)
\end{align*}$
This last one yields to a trigonometric substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate : $I=\int_0^{\infty}\frac{\ln (1+ax+x^{2})}{1+x^2}\,dx$ Does the following integral have a closed form :
$$I=\displaystyle\int\limits_0^{\infty}\frac{\ln (1+ax+x^{2})}{1+x^2}dx$$
Where $|a|≤1$ ,
I was trying using Feynman's trick.
Define
$$I(b)=\displaystyle\int\limits_0^{\infty}\frac{\ln (b(1+x^{2})+ax)}{1+x^2}dx$$
Differentiating with respect to $b$ we get :
$$I'(b)=\displaystyle\int\limits_0^{\infty}\frac{1}{b+ax+bx^{2}}dx$$
$$=2\left(\frac{π}{2\sqrt{4b^{2}-a^{2}}}-\frac{\arctan \left(\frac{a}{\sqrt{4b^{2}-a^{2}}}\right)}{\sqrt{4b^{2}-a^{2}}}\right)$$
Known :
$$\displaystyle\int \frac{1}{\sqrt{4b^{2}-a^{2}}}db=\frac{\log (2 x + \sqrt{-a^2 + 4 x^{2}})}{2}$$
my problem in this integral :
$$\displaystyle\int\limits_0^{1}\frac{\arctan \left(\frac{a}{\sqrt{4b^{2}-a^{2}}}\right)}{\sqrt{4b^{2}-a^{2}}}db=?$$
Of course here
$$I=I(1)=I(0)+\int\limits_0^{1}I'(b)db$$
$$I(0)=\frac{π\ln a}{2}$$
I already waiting your hints or solution.
| Parameterising $2\sin a$ instead of $a$ yields the integral
$$
I(a)=\int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x,
$$
where $a\in (0, \frac{\pi}{2}). $
Differentiating $I(a)$ w.r.t. $a$ yields
$$
\begin{aligned}
I^{\prime}(a) &=\int_{0}^{\infty} \frac{2 x \cos a}{\left(1+x^{2}\right)\left(1+2 x \sin a+x^{2}\right)} d x \\
&=\cot a\int_{0}^{\infty}\left(\frac{1}{1+x^{2}}-\frac{1}{1+2 x \sin a+x^{2}}\right) d x \\
&=\cot a\left[\tan ^{-1} x-\frac{1}{\cos a} \tan ^{-1}\left(\frac{x+\sin a}{\cos a}\right)\right]_{0}^{\infty} \\
&=\cot a\left[\frac{\pi}{2}-\frac{1}{\cos a}\left(\frac{\pi}{2}-a\right)\right]
\end{aligned}
$$
Integrating $I’(a)$ back to $I(a)$, we have
$$
\begin{aligned}
I\left(a\right)- \underbrace{I(0)}_{=\pi\ln 2} &=\int_{0}^{a} \cot x\left[\frac{\pi}{2}-\frac{1}{\cos x}\left(\frac{\pi}{2}-x\right)\right] d x \\
&=\frac{\pi}{2} \underbrace{ \int_{0}^{a}\left(\cot x-\frac{1}{\sin x}\right) d x}_{= 2 \ln \left(\cos \frac{a}{2}\right)} + \underbrace{\int_{0}^{a} \frac{x}{\sin x} d x}_{K}
\end{aligned}
$$
$$
\begin{aligned}
K &=\int_{0}^{a} \frac{x}{\sin x} d x=\int_{0}^{a} x\, d\left[\ln \left(\tan \frac{x}{2}\right)\right] \\
&=\left[x\ln \left(\tan \frac{x}{2}\right)\right]_{0}^{a}-\int_{0}^{a} \ln \left(\tan \frac{x}{2}\right) d x \\
&=a \ln \left(\tan \frac{a}{2}\right)-2 \int_{0}^{\frac{a}{2}} \ln (\tan x) d x
\end{aligned}
$$
Now we can conclude that, for any $a\in (0, \frac{\pi}{2}]$,
$$
\boxed{I=I\left(a\right)=\pi \ln 2+\pi \ln \left(\cos \frac{a}{2}\right)+a \ln \left(\tan \frac{a}{2}\right) -2 \int_{0}^{\frac{a}{2}} \ln (\tan x) d x}
$$
Similarly, for any $a\in (-\frac{\pi}{2},0)$,
$$
\boxed{I=I\left(a\right)=\pi \ln 2+\pi \ln \left(\cos \frac{a}{2}\right)-a \ln \left(\tan \frac{-a}{2}\right) +2 \int_{0}^{\frac{-a}{2}} \ln (\tan x) d x}
$$
For the last integral, we may evaluate it by expressing it as a series of sine as below:
$$
\int_{0}^{x} \ln (\tan \theta) d \theta=-\sum_{k=0}^{\infty} \frac{\sin ((4 k+2) x)}{(2 k+1)^{2}}
$$
from the web.
Example 1
$$\begin{aligned}\quad \int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} d x&=
I\left(-\frac{\pi}{6}\right)\\&= \frac{\pi}{2}[\ln (2+\sqrt{3})]+\frac{\pi}{6} \ln (2+\sqrt{3})+2\left(-\frac{2}{3} G\right)\\& =\frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \end{aligned}$$
Example 2
$$
\begin{aligned}
&\int_{0}^{\infty} \frac{\ln \left(1+\sqrt{2} x+x^{2}\right)}{1+x^{2}}\\=&I\left(\frac{\pi}{4}\right) \\
=& \pi \ln \left(2 \cos \frac{\pi}{8}\right)-\frac{\pi}{4} \ln \left(\tan \frac{\pi}{8}\right)-2 \int_{0}^{\frac{\pi}{8}} \ln (\tan x) d x \\
=& \frac{\pi}{2} \ln (2+\sqrt{2})+\frac{\pi}{4} \ln (\sqrt{2}+1) -2\left[\frac{\pi}{8} \ln (\sqrt{2}-1)-\Im\left(\operatorname{Li}_{2}(i(\sqrt{2}-1))\right]\right.\\=& \pi \ln [\sqrt[4]{2}(\sqrt{2}+1)] +2 \Im\left(\operatorname{Li}_{2}(i(\sqrt{2}-1))\right.
\end{aligned}
$$
where the last integral see post
The answer is not satisfactory as the last integral is hard to evaluate!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is: $f'(x) = x \sqrt{16-x^2}$ Section 5.2
Can somebody verify this solution for me?
Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is:
$f'(x) = x \sqrt{16-x^2}$
So the idea here is we integrate $f'(x)$ to get $f(x)=Something +C$ Where $C$ is our constant of integration, and then we use the fact that $f$ must pass through the point $(0,\frac{4}{3})$ to figure out what $C$ is.
So first lets integrate $f'(x) = x \sqrt{16-x^2}$
$\int x \sqrt{16-x^2}dx$
Let $u=16-x^2$. Then $\frac{du}{dx}=-2x$ and $\frac{du}{-2x}=dx$. And so we have:
$\int x \sqrt{16-x^2}dx$
$=\int x \sqrt{u}\frac{du}{-2x}$
$=\frac{-1}{2} \int \sqrt{u}du$
$=\frac{-1}{2} \frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C$
$=\frac{-1}{3} u^{\frac{3}{2}}+C$
$=\frac{-1}{3} (16-x^2)^{\frac{3}{2}}+C$
Okay. So by defintion $f(x)=\int f'(x)dx$. Thus $f(x) = \frac{-1}{3} (16-x^2)^{\frac{3}{2}}+C$. Lets plug in the point $(0,\frac{4}{3})$ and solve for $C$
$\frac{4}{3} = \frac{-1}{3} (16-0^2)^{\frac{3}{2}}+C$
$\rightarrow \frac{4}{3} = \frac{-1}{3} (16)^{\frac{3}{2}}+C$
$\rightarrow 4= -(16)^{\frac{3}{2}}$
$\rightarrow 4=-(4^3)+ \frac{C}{3}$
$\rightarrow 4=-64+ \frac{C}{3}$
$\rightarrow \frac{68}{3}=C$
Therefore, plugging this back into $f(x)$, we get:
$f(x) = \frac{-1}{3} (16-x^2)^{\frac{3}{2}}+\frac{68}{3}$
Unfortunately, the computer says the correct solution is:
$f(x) = \frac{-1}{3} (16-x^2)^{\frac{3}{2}}+\frac{68}{3}$
Can somebody find my error? Thanks!
| The correction is : $\frac{4}{3}=\frac{-1}{3}(16)^{3/2} +C$ then $ 4=-1(16)^{3/2} +3C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The largest area of the rectangle inscribed into an acute triangle
A triangle of base $b$ and height $h$ has acute base angles. A rectangle is inscribed in the triangle with one side on the base of the triangle. Show that the largest such rectangle has base $b/2$ and height $h/2$, so that its area is one-half the area of the triangle.
So, we have a triangle as it is shown in the picture (Sorry for the awful quality). We have that $bh/2 = A$, where $A$ is the area of the triangle. We have $y/(b-x-z) = h/(b-x-z+x_1)$ from where $y = \frac{h(b-x-z)}{b-x-z+x_1}$ and $y/z = h/(x_2+z)$, from where $y = \frac{zh}{x_2+z}$. So $\frac{zh}{x_2+z} = \frac{h(b-x-z)}{b-x-z+x_1}$. From this I get $zhx = x_2h(b-x)$. Can this be useful? If so, what should be done next?
| The solution for the general case requires a bit more effort than the right triangle case. We will be using a lot of "Basic Probotionality theorem for triangles"
$$\frac{y}{h} = \frac{b_1 - x_1}{b_1}$$
Similarly,
$$\frac{y}{h} = \frac{b_2 - x_2}{b_2}$$
Giving us,
$$\frac{x_1}{b_1} = \frac{x_2}{b_2}$$
Using Compedendo dividendo,we get
$$\frac{x_1}{b_1} = \frac{x_2}{b_2} = \frac{x_1 + x_2}{b_1 + b_2} = \frac{x}{b}$$
Now,
$$\frac{y}{h} = 1 - \frac{x_1}{b_1}$$
Hence,
$$\frac{y}{h} = 1 - \frac{x}{b}$$
Multiplying with $x$ on both sides we get,
$$ xy = hx - \frac{hx^2}{b}$$
$$ A = hx - \frac{hx^2}{b}$$
$$ \frac{dA}{dx} = h - \frac{2hx}{b}$$
For maxima this has to be equal to zero.
$$ h = \frac{2hx}{b}$$
$$ x = \frac{b}{2}$$
Also,
$$y = \frac{h}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that: $\sum_{l=0}^{m}(-1)^{m+l}4^{l-1}l{m \choose l}{2l \choose l}^{-1}{m+l \choose l}=\sum_{i=1}^{m}i^2$ By observing and alteration of the OP's question we found this sum which is equivalent to the natural square number, $\sum_{i=1}^{m}i^2$
$$\sum_{l=0}^{m}(-1)^{m+l}4^{l-1}l{m \choose l}{2l \choose l}^{-1}{m+l \choose l}=\sum_{i=1}^{m}i^2\tag1$$
How can we prove $(1)?$
$$l{m \choose l}{2l \choose l}^{-1}{m+l \choose l}=\frac{m!}{l!(m-l)!}\cdot \frac{l!l!}{(2l)!}\cdot\frac{(m+l)!}{l!m!}=\frac{(m+l)!}{(m-l)!}\cdot\frac{l}{(2l)!}$$
$$\sum_{l=0}^{m}(-1)^{m+l}4^{l-1}\frac{(m+l)!}{(m-l)!}\cdot\frac{l}{(2l)!}\tag2$$
note that $$\frac{(m+l)!}{(m-l)!}\cdot\frac{l}{(2l)!}=\frac{(m+l)!(m+l+1)}{(m-l)!}\cdot\frac{l}{(2l)!(m+l+1)}=\frac{1}{B(m-l+1,2l+1)}\cdot \frac{l}{m+l+1}$$
$$\sum_{l=0}^{m}(-1)^{m+l}4^{l-1}\frac{1}{B(m-l+1,2l+1)}\cdot \frac{l}{m+l+1}\tag3$$
Where B(n,m) is the Beta function
| We seek to evaluate
$$\sum_{q=0}^m (-1)^{m+q} 4^{q-1} q {m\choose q} {2q\choose q}^{-1}
{m+q\choose q}.$$
We get from the binomial coefficients
$$\frac{m!}{(m-q)! \times q!}
\frac{q! \times q!}{(2q)!} \frac{(m+q)!}{m! \times q!}
= \frac{(m+q)!}{(m-q)! \times (2q)!}
= {m+q\choose m-q}.$$
Our sum becomes
$$\sum_{q=0}^m (-1)^{m+q} 4^{q-1} q {m+q\choose m-q}
= [z^m] (1+z)^m
\sum_{q=0}^m (-1)^{m+q} 4^{q-1} q z^q (1+z)^q.$$
The coefficient extractor enforces the range and we find
$$[z^m] (1+z)^m
\sum_{q\ge 0} (-1)^{m+q} 4^{q-1} q z^q (1+z)^q
\\ = \frac{1}{4} (-1)^{m+1} [z^m] (1+z)^m
\frac{4z(1+z)}{(1+4z(1+z))^2}
\\ = (-1)^{m+1} [z^{m-1}] (1+z)^{m+1}
\frac{1}{(1+2z)^4}.$$
This is
$$(-1)^{m+1} \mathrm{Res}_{z=0} \frac{1}{z^m}
(1+z)^{m+1} \frac{1}{(1+2z)^4}.$$
Now we put $z/(1+z) = w$ so that $z = w/(1-w)$ and $dz = 1/(1-w)^2 \;
dw$ to get
$$(-1)^{m+1} \mathrm{Res}_{w=0} \frac{1}{w^m} \frac{1}{1-w}
\frac{1}{(1+2w/(1-w))^4}\frac{1}{(1-w)^2}
\\ = (-1)^{m+1} \mathrm{Res}_{w=0} \frac{1}{w^m}
\frac{1-w}{(1+w)^4}
\\ = (-1)^{m+1}
\left( [w^{m-1}] \frac{1}{(1+w)^4}
- [w^{m-2}] \frac{1}{(1+w)^4} \right)
\\ =
[w^{m-1}] \frac{1}{(1-w)^4}
+ [w^{m-2}] \frac{1}{(1-w)^4}
= {m+2\choose 3} + {m+1\choose 3}
\\ = \frac{1}{6} m (m+1) (m+2+m-1)
= \frac{1}{6} m (m+1) (2m+1) = \sum_{q=1}^m q^2.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find positive integer solutions for $n^4(n+1)+1=7^m$ This is the statement:
Find all pairs of natural numbers $(n,m)$ that satisfies equation $$n^4(n+1)+1=7^m$$
I found one solution (2,2) and believe there is no more solutions. I'm beginner in number theory and don't know how to even start this. Can anyone help?
| Use $$n^5+n^4+1=n^5-n^2+n^4-n+n^2+n+1=(n^2+n+1)(n^3-n+1).$$
Indeed, we obtain that there are non-negative integers $p$ and $q$, for which
$$n^2+n+1=7^p$$ and $$n^3-n+1=7^q.$$
Now, if $p=0$ or $q=0$ we obtain solutions $(0,0),$ $(-1,0)$ in integer numbers.
Let $p\geq1$ and $q\geq1$.
Thus, from the first equation we obtain $n=7k+2$ or $n=7k+4$ and the last is false for the second equation.
Id est,
$$(7k+2)^2+7k+2+1=7^p$$ and
$$(7k+2)^3-(7k+2)+1=7^q$$ or
$$7k^2+5k+1=7^{p-1}$$ and $$49k^3+42k^2+11k+1=7^{q-1}.$$
Now, if $(p-1)(q-1)=0$, we obtain $k=0$ and the solution $(2,2).$
Let $p\geq2$ and $q\geq2$.
Thus, $11k+1-(5k+1)$ is divisible by $7$, which gives $k$ is divisible by $7$,
which is a contradiction because
$$7k^2+5k+1=7\left(k^2+\frac{5k}{7}\right)+1$$ is not divisible by $7$.
Thus, our equation has no another integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Please check my work! Question about cubic polynomials I need some help with this problem. Here is the link. Can you please tell me if there is an easier way to show that cubic polynomials have a real root? The question is in an analysis book from the continuity section so it has to use that. Here is the latex:
Show that a cubic equation (i.e. one of the form $ax^3 + bx^2 + cx + d = 0$ where $a\neq 0)$ has at least one real root.
Solution:
The equation has at least one root if for some $x_1<x_2$, $\enspace f(x_1) < 0$ and $f(x_2) > 0$. Then by the intermediate value theorem $f(c) = 0$ for some $x_1 < c < x_2$. $x^3$ outgrows smaller powers of $x$ so the function is negative for some large negative number and positive for some large positive number.
If $(x_n)$ is a sequence of positive terms that tends to infinity, then
$$f(x_n) = ax_n^3 + bx_n^2 + cx_n + d = x_n^3(a+ \frac{b}{x_n} + \frac{c}{x_n^2} + \frac{d}{x_n^3})$$
Now $\frac{b}{x_n}, \frac{c}{x_n^2}, \frac{d}{x_n^3}$ are sequences that tend to zero, so for any $\epsilon$ there is an $N$ such that
$$|\frac{b}{x_n}| < \epsilon/3, \quad |\frac{c}{x_n^2}| < \epsilon/3, \quad |\frac{d}{x_n^3}| < \epsilon/3$$
and for $\epsilon = a$, we have
$$|\frac{b}{x_n}| + |\frac{c}{x_n^2}| + |\frac{d}{x_n^3}| < a$$
so that, by the triangle inequality
$$|\frac{b}{x_n} + \frac{c}{x_n^2} + \frac{d}{x_n^3}| \leq |\frac{b}{x_n}| + |\frac{c}{x_n^2}| + |\frac{d}{x_n^3}| < a$$
which means
$$-a <\frac{b}{x_n} + \frac{c}{x_n^2} + \frac{d}{x_n^3} < a$$
Then for some $|k|<1$, it can be written
$$a+ \frac{b}{x_n} + \frac{c}{x_n^2} + \frac{d}{x_n^3} = a+ ka = (1+k)a$$
and
$$f(x_n) = x_n^3(1+k)a$$
for $n\geq N$. Since $x_n$ is a sequence of positive terms, $f(x_n) = k_na$ for $n\geq N$ where $k_n>0$. If $x_n$ is instead chosen as a sequence of negative terms that tends to $-\infty$, then $f(x_n) = (k_n')a$ for $n\geq N$ where $k_n'<0$. Therefore regardless of the sign of $a$ the function $f$ takes on both positive and negative values.
It seems redundant and too many steps. Is there a more simple way to solve this problem? Any feedback is appreciated. Thank you!
| Assume wlog $a=1$ by factoring $a$ out, which is doable since $a\ne0$.
Simpler inequalities can be deduced by taking $x\ge|b|+|c|+|d|+1$ so that we have
$$x+b,x+c,x+d\ge1$$
and noticing that we then have
\begin{align}f(x)&=x^3+bx^2+cx+d\\&=(x+b)x^2+cx+d\\&\ge(x+c)x+d\\&\ge x+d\\&\ge1\end{align}
and similarly that if we have $x\le-(|b|+|c|+|d|+1)$ then
\begin{align}x+b,x+c,x+d&\le-1\\-x+b,-x+c,-x+d&\ge1\end{align}
which then gives
\begin{align}f(x)&=x^3+bx^2+cx+d\\&=(x+b)x^2+cx+d\\&\le(-x+c)x+d\\&\le x+d\\&\le-1\end{align}
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$.
Question: Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $$\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$$.
My approach: Given that $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx, \forall n\in\mathbb{N}.$ Let us make the substitution $x^n=t$, then $$nI_n=\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}}.$$
Now since $0\le t\le 1\implies \frac{1}{t}\ge 1\implies \left(\frac{1}{t}\right)^{2/n}\ge 1 \implies 1+\left(\frac{1}{t}\right)^{2/n}\ge 2\implies \sqrt{1+\left(\frac{1}{t}\right)^{2/n}}\ge \sqrt 2.$
This implies that $$\frac{1}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le\frac{1}{\sqrt 2}\\ \implies \int_0^1 \frac{dt}{\sqrt{1+\left(\frac{1}{t}\right)^{2/n}}}\le \int_0^1\frac{dt}{\sqrt 2}=\frac{1}{\sqrt 2}.$$
So, as you can see, I am trying to solve the question using Sandwich theorem.
Can someone help me to proceed after this?
Also, in $$\lim_{n\to\infty}nI_n=\lim_{n\to\infty}\int_0^1\frac{dt}{\sqrt{1+t^{-2/n}}},$$ the limit and integral interchangeable?
| You can still use Sandwich theorem, if you haven't yet seen dominated convergence theorem :
Let $ n $ be a positive integer.
As you said, using the substitution $ \left\lbrace\begin{aligned}y&=x^{n}\\ \mathrm{d}y &=n x^{n-1}\,\mathrm{d}x\end{aligned}\right. $, we get : $$\int_{0}^{1}{\frac{n x^{n}}{\sqrt{1+x^{2}}}\,\mathrm{d}x}=\int_{0}^{1}{\frac{y^{\frac{1}{n}}}{\sqrt{1+y^{\frac{2}{n}}}}\,\mathrm{d}y}$$
Meaning, we have : \begin{aligned}\left|\frac{1}{\sqrt{2}}-n I_{n}\right|&=\left|\int_{0}^{1}{\left(\frac{1}{\sqrt{2}}-\frac{y^{\frac{1}{n}}}{\sqrt{1+y^{\frac{2}{n}}}}\right)\mathrm{d}y}\right| \\ &=\left|\int_{0}^{1}{\frac{\sqrt{1+y^{\frac{2}{n}}}-\sqrt{2}y^{\frac{1}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}}\,\mathrm{d}y}\right|\\ &=\int_{0}^{1}{\frac{1-y^{\frac{2}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)}\,\mathrm{d}y}\end{aligned}
Since $ \left(\forall y\in\left[0,1\right]\right),\ \sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)\geq \sqrt{2}+\sqrt{2}y^{\frac{2}{n}}\geq 1 $, we have : $ \int\limits_{0}^{1}{\frac{1-y^{\frac{2}{n}}}{\sqrt{2\left(1+y^{\frac{2}{n}}\right)}\left(\sqrt{1+y^{\frac{2}{n}}}+\sqrt{2}y^{\frac{1}{n}}\right)}\,\mathrm{d}y}\leq\int\limits_{0}^{1}{\left(1-y^{\frac{2}{n}}\right)\mathrm{d}y} $, and thus : $$ \left|\frac{1}{\sqrt{2}}-n I_{n}\right|\leq\int_{0}^{1}{\left(1-y^{\frac{2}{n}}\right)\mathrm{d}y}=\frac{2}{n+2}\underset{n\to +\infty}{\longrightarrow}0 $$
Hence : $$ \lim_{n\to +\infty}{nI_{n}}=\frac{1}{\sqrt{2}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
How to solve quadratic matrix equations of the form $A^T B A=C$? I want to solve the following matrix equation
$$A^T
\begin{pmatrix}
10 & 0 & 0 \\
0 & 20 & 0 \\
0 & 0 & 25 \\
\end{pmatrix} A =
\begin{pmatrix}
\frac{35}{2} & \frac{5 \sqrt{3}}{2} & 0 \\
\frac{5 \sqrt{3}}{2} & \frac{25}{2} & 0 \\
0 & 0 & 25 \\
\end{pmatrix}$$
In the above formula, $A^T$ is the transpose of matrix A.
At present, I don't have a good way. I only know that the reference answer of matrix A is
$$\left(\begin{array}{ccc}
\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\
0 & 0 & 1 \end{array}\right)$$
| In our case we have two given symmetric matrices $B,C$,
$$
\begin{aligned}
B
&=
\begin{bmatrix}
10 &&\\&20&\\&&25
\end{bmatrix}=B_1^2 =B_1^TB_1=B_1B_1^T\ ,\text{ where }
\\
B_1
&=
\begin{bmatrix}
\sqrt{10} &&\\&2\sqrt 5&\\&&5
\end{bmatrix}=B_1^\ ,
\\[3mm]
C &=
\begin{bmatrix}
\frac{35}{2} & \frac{5 \sqrt{3}}{2} & 0 \\
\frac{5 \sqrt{3}}{2} & \frac{25}{2} & 0 \\
0 & 0 & 25
\end{bmatrix}=C_1^2=C_1^TC_1=C_1C_1^T\ ,\text{ where }
\\
C_1
&=
\begin{bmatrix}
a & b&\\b&c&\\&&5
\end{bmatrix}=C_1^T\ ,\text{ where its entries $a,b,c$ are}
\\
a &= \frac{\sqrt{10}}4(3\sqrt 2+1)\ ,\\
b &= \frac{\sqrt{30}}4(\sqrt 2-1)\ ,\\
c &= \frac{\sqrt{10}}4(\sqrt 2+3)\ .
\end{aligned}
$$
We have to find one or all "unknown" matrices $A$, so that the matrix equation holds:
$$A^TBA=C\ .
$$
Equivalently, after successive transformations:
$$
\begin{aligned}
A^TBA &=C\ ,\\
A^TB_1^TB_1A &=C_1^TC_1\ ,\\
(C_1^{-1})^TA^TB_1^TB_1A(C_1^{-1}) &=I\ ,\\
(B_1AC_1^{-1})^T\;(B_1AC_1^{-1}) &=I\ ,\\
B_1AC_1^{-1}&\in SO(3)\ .
\end{aligned}
$$
Here, $SO(3)$ is the special orthogonal group of all matrices $X$ (with real entries, well, i suppose that that "complex" in the title is subjectively a synonym for "difficult"), such that $X^TX=I=XX^T=$. For each such $X$ we get a solution $A$ by isolating $A$ in $B_1AC_1^{-1}=X$. (So there are infinitely many solutions.) To have one solution, take the obvious $X=I$ as a particular solution.
Note: The diagonalization (if possible or not) is a process to put a linear transformation in a suitable form by using a base change. In our case, it is a quadratic form to be transformed by a base change. This is different.
https://en.wikipedia.org/wiki/Quadratic_form#Equivalence_of_forms
Note that the first object, the linear map given by a matrix $A$, transforms via a base change by $S$ as
$$A\to S^{-1}AS\ ,$$
the second object, the quadratic form given by a symmetric $A$, transforms as $$A\to S^TAS\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate $ \int_0^1{\frac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx$
Given a function $f(x)$ that is differentiable on $\left[0; 1\right]$
satsifies: $$ f(1) = 1 $$ $$ f(x)f(1-x) = e^{x^2 - x} $$
Calculate: $$ \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx $$
Attempt number 1:
Using integration by parts, we have:
\begin{align}
\int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx &= \left(2x^3 - 3x^2\right)\Big|_0^1 - \int_0^1{\dfrac{\left(6x^2 - 6x\right)f(x) - \left(2x^3 - 3x^2\right)f'(x)}{\left[f(x)\right]^2}}f(x)\,dx\\
&= \left(2x^3 - 3x^2\right)\Big|_0^1 - \int_0^1{\left(6x^2 - 6x\right)}\,dx + \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx
\end{align}
This gives me equation $0 = 0$, in which I can't do anything.
Attempt number 2:
Express $f(x)$ in terms of $f(1-x)$:
$$ f(x) = \dfrac{e^{x^2-x}}{f(1-x)} $$
This implies that:
$$ f'(x) = \dfrac{2xf(1-x)e^{x^2-x} - f'(1-x)e^{x^2-x}}{\left[f(1-x)\right]^2} $$
Subtitute in, we have:
\begin{align}
\int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx &= \int_0^1{\dfrac{\left(2x^3-3x^2\right)\left(2xf(1-x) - f'(1-x)\right)}{f(1 - x)}}\,dx\\
&= \int_0^1{2x\left(2x^3-3x^2\right)}\,dx - \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(1-x)}{f(1 - x)}}\,dx\\
&= -\dfrac{7}{10} - \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(1-x)}{f(1 - x)}}\,dx
\end{align}
Then, I tried to turn $1 - x$ into $x$ in the last integral but failed to come up with anything useful.
I would like to know whether there is another way to solve this problem or how my second attempt could have been done.
Thanks in advance
| From the given functional equation, we obtain that $f(0)=f(1)=1.$
Now setting $I=\int_0^1 \frac{(2x^3-3x^2)f'(x)}{f(x)}\mathrm dx$ and integrating by parts gives $$I=(2x^3-3x^2)\int \mathrm d\left(\log f(x)\right)-\int(6x^2-6x)\log f(x)\mathrm dx.$$ Substituting the limits gives that $$I=-\int(6x^2-6x)\log f(x)\mathrm dx.$$ Now from the functional equation we get $\log f(x)+\log f(1-x)=x^2-x.$ Substituting for $\log f(x)$ and splitting the integral now gives that $$I=-6\int_0^1(x^2-x)^2\mathrm dx+6\int(x^2-x)\log f(1-x)\mathrm dx=-1/5+6\int(x^2-x)\log f(x)\mathrm dx,$$ by using the property $\int_a^bg(x)\mathrm dx=\int_a^bg(a+b-x)\mathrm dx$ and simplifying. Integrating the second part of $I$ by parts now gives that $$I=-1/5+6\log f(x)\int \mathrm d\left(\frac{2x^3-3x^2}{6}\right)-\int_0^1\frac{(2x^3-3x^2)f'(x)}{f(x)}\mathrm d x=-1/5-I,$$ giving us that $$I=-1/10.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that : $m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$ Let $m_{a},m_{b},m_{c}$ be the lengths of the medians and $a,b,c$ be the lengths of the sides of a given triangle , Prove the inequality :
$$m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$$
Where :
$s : \operatorname{Semiperimeter}$
$R : \operatorname{circumradius}$
I know the relation :
$$m_{a}^{2}=\frac{2(b^{2}+c^{2})-a^{2}}{4}$$
But when I multiple together I dont get simple formulas!
So, I need help finding a solution.
Thanks!
| In the standard notation we need to prove that:
$$\frac{1}{8}\sqrt{\prod_{cyc}(2a^2+2b^2-c^2)}\leq\frac{1}{2}\cdot\frac{abc}{4S}\cdot\frac{(a+b+c)^2}{4}$$ or
$$a^2b^2c^2(a+b+c)^3\geq\prod_{cyc}(2a^2+2b^2-c^2)\prod_{cyc}(a+b-c).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $$\prod_{cyc}(2a^2+2b^2-c^2)=\prod_{cyc}(2(a^2+b^2+c^2)-3c^2)=$$
$$=8(9u^2-6v^2)^3-12(9u^2-6v^2)^3+18(9u^2-6v^2)(9v^4-6uw^3)-27w^6=$$
$$=27(-w^6+2(3u^2-2v^2)(9v^4-6uw^3)-4(3u^2-2v^2)^3).$$
Also, $$\prod_{cyc}(a+b-c)=\prod_{cyc}(3u-2c)=27u^3-54u^3+36uv^2-8w^3=$$
$$=-8w^3-27u^3+36uv^2.$$
Thus, we need to prove that $f(w^3)\geq0,$ where
$$f(w^3)=u^3w^6-(-w^6+2(3u^2-2v^2)(9v^4-6uw^3)-4(3u^2-2v^2)^3)(-8w^3-27u^3+36uv^2).$$
But $$f''(w^3)=2u^3-2(-2w^3+2(3u^2-2v^2)(-6u))(-8)+$$
$$-(-8w^3-27u^3+36uv^2)(-2)=-4(157u^3-114uv^2+12w^3)<0,$$ which says that $f$ is a concave function.
Thus, it's enough to prove our inequality for an extreme value of $w^3$, which happens in the following cases.
*
*$w^3\rightarrow0^+$.
in this case the inequality is obvious;
*$\prod\limits_{cyc}(a+b-c)\rightarrow0^+$.
It's obvious again;
*Two variables are equal.
Since our inequality is symmetric and homogeneous, it's enough to assume $b=c=1$.
Thus, $0<a<2$ and we need to prove that
$$a^2(a+2)^3\geq(2a^2+1)^2(4-a^2)a^2(2-a)$$ or
$$(a+2)^2\geq(2a^2+1)^2(2-a)^2$$ or
$$a+2\geq(2a^2+1)(2-a)$$ or
$$a(a-1)^2\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Quadratic programming: KKT Optimality conditions I am struggling with an exercice with the following quadratic program:
$$min:x_{1}x_{2} + x^{2}_{1} + \frac{3}{2}x^{2}_{2} + 2x^{2}_{3} + 2x_{1} + x_{2} + 3x_{3}$$
subject to
$$x_{1} + x_{2} + x_{3} = 1$$
$$x_{1} − x_{2} = 0$$
$$x_{1}, x_{2}, x_{3} ≥ 0$$
Is the quadratic objective function convex? Show that $x∗ = (\frac{1}{2}, \frac{1}{2},0)$ is an optimal solution to this problem by finding vectors $y$ and $s$ that satisfy the optimality conditions jointly with $x∗$.
I already showed that the objective function is convex by proving the positive semidefiniteness of the symmetric matrix $Q$.
My idea for the second part of the question would be to compute the Karush Kuhn Tucker conditions and to check if I place the values for $x*$ in it, that there exists some vectors $y$ and $s$ that fulfill the conditions.
For the Lagrange function I got:
$$min_x:L(x,y,s)=\frac{1}{2}x^TQx+c^Tx-y^T(Ax-b)-s^Tx$$
The first derivative is then:
$$L(x,y,s)'=Qx+c-A^Ty-s$$
How do I proceed with the KKT of the function above?
| Let $x_1=x+2=a$. Thus, $c=1-2a$ and
$$x_{1}x_{2} + x^{2}_{1} + \frac{3}{2}x^{2}_{2} + 2x^{2}_{3} + 2x_{1} + x_{2} + 3x_{3}=$$
$$=\frac{23}{2}a^2-11a+5=\frac{23}{2}\left(a^2-\frac{22}{23}a+\frac{10}{23}\right)=$$
$$=\frac{23}{2}\left(a-\frac{11}{23}\right)^2+5-\frac{121}{46}\geq\frac{109}{46}.$$
The equality occurs for $x_1=x_2=\frac{11}{23}$ and $x_3=\frac{1}{23},$ which says that $\frac{109}{46}$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3607659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Suggestions for $ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} $? I'm trying to evaluate $$
\lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2}
$$
over the domain $x>0$, $y>0$.
============
My attempt:
$f(x,x^2)\to +\infty$; so if the limit exists it must be $+\infty$.
I tried to evaluate the limits "near" $(x,x)$ where, I thought, there's may be some problems:
$f(x, x-x^2)\to +\infty$.
Then I convinced myself the limit could be $+\infty$:
since $f(x,y)>0$ over the domain, I had to find such $g(x,y)$ that:
1. $f(x,y) \ge g(x,y)$
2. $ \lim_{(x,y)\to (0,0)} g(x,y)\to +\infty $
$$
f(x,y)=\frac{x-\sqrt{xy}}{x^2-y^2}=\frac{x-\sqrt{xy}+y-y}{x^2-y^2}=\frac{x-\sqrt{xy}+y}{x^2-y^2}-\frac{y}{x^2-y^2}=\frac{\sqrt{\left(x-\sqrt{xy}+y\right)^2}}{x^2-y^2}-\frac{y}{x^2-y^2}
$$
Where the last step follows by $(x-\sqrt{xy}+y) \ge 0$ with $x>0$, $y>0$. $$
\frac{\sqrt{\left(x-\sqrt{xy}+y\right)^2}}{x^2-y^2}-\frac{y}{x^2-y^2} = \frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2}.
$$ From $\left[3\left(x-\sqrt{xy}+y\right)^2\right] \ge \left[x+xy+y^2\right]$, for every $(x,y)$ with $(x>y)$: $$
\frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2} \ge \frac{\sqrt{x^2+xy+y^2}}{\sqrt{3}(x^2-y^2)}+\frac{y}{y^2-x^2}.
$$ From here I observated that $ \left[\lim_{(x,y)\to (0,0)} g(x,y)\to +\infty \right]$ and eventually $ \left[\lim_{(x,y)\to (0,0)} f(x,y)\to +\infty \right]$ for $(x>y)$.
I thought that for $(y>x)$, the inequality was formally equivalent when I replace $(x)$ with $(y)$ and viceversa: $$
\frac{\sqrt{3\left(x-\sqrt{xy}+y\right)^2}}{\sqrt{3}(x^2-y^2)}-\frac{y}{x^2-y^2} \ge \frac{\sqrt{x^2+xy+y^2}}{\sqrt{3}(y^2-x^2)}+\frac{x}{x^2-y^2}.
$$
However I could see, through an online grapher, that it is false!!
So I remained without any chance to conclude the limit.
============
Is there anybody who knows why the last inequality isn't correct?
And also, has anybody some hints to evaluate the limit?
| In order to disprove your conjecture that $\lim_{(x, y) \to (0, 0)} f(x, y) = +\infty$, take the limit along the curve $(x, y) = (t^5, t)$ as $t \to 0^+$. Then we have:
$$f(t^5, t) = \frac{t^5 - t^3}{t^{10} - t^2} = \frac{t(1-t^2)}{1-t^8}$$
and from the last expression, we see that $f(t^5, t) \to 0$ as $t \to 0^+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find delta for the limit I'm having difficulty to solve this problem:
I know that ${\displaystyle \lim_{x\to a} f(x) = L}$ means for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 <|x-a| < \delta$.
I need to find $\delta$ when $\varepsilon = 0.001$ for ${\displaystyle \lim_{x \to -1} \frac{1}{\sqrt{x^2+1}} = \frac{1}{\sqrt 2}.}$
I've started as this:
$$ \left
|\frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}\right|
= \left|\left(\frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}\right)\frac{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)}{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)} \right| = \left| \frac{\frac{1}{x^2+1} - \frac 12}{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)}
\right| < \epsilon.$$
But I do not know how can I finish solving this to find x.
| Firstable you can make $|1+x| < 1 \implies |x| -1 =|x| - |1| \le |1+x| < 1$. So:
$|x| < 2$. Now call the expression just before the $\epsilon$ in your work above $D$, then $D \le \dfrac{|1-x^2|}{2\cdot \frac{1}{\sqrt{2}}} \le |1-x^2|= |1+x||1-x|\le |1+x|(1+|x|) < 3|1+x|< \epsilon $ if $1+x| < \frac{\epsilon}{3}$. Thus choose $\delta = \min(1,\frac{\epsilon}{3})$,and you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove $\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}$ Let $1>a>0$, $1>b>0$, $1>c>0$ and $a+b+c=1$. Prove that
$$
\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.
$$
I saw the following solution. Let $x=\frac{2}{1-a^2}$, $y=\frac{2}{1-b^2}$, $z=\frac{2}{1-c^2}$, then, using AM-GM inequality, we get
$$
x+y+z-3\ge 3 \sqrt[3]{xyz}-3=\frac{27}{4}-3=\frac{15}{4}\; for \; x=y=z=\frac{9}{4}.
$$
Is it correct?
| Hint: You can solve by showing $${1\over 1-x^2} \geq {27\over 32}(x+1)$$ for $x\in (0,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find m+n. If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find $m+n$.
I did and I got
$$(x+2)^3(x-2)^{3+m}(x+1)^n(x^2-x+1)^n(x-3)^m=0$$, so I find $3+3+m+n+2n+m=18\implies 2m+3n=12$, the answer is m+n=5. What I have to do now?
| Note that $m$ and $n$ have to be non-negative integers, and perhaps it is implied in the question that $m$ and $n$ are positive as well. The only positive solution to $3n+2m=12$ is $n=2,m=3$, so $m+n=5$.
If zero powers are allowed, then we also have $n=4,m=0$ and $n=0,m=6$, for sums of $4$ and $6$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Seeking alternative methods for $\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$ I've solved this one by first tackling,
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
But i'd like to know other ways to solve it since the way i did it was a bit lengthy and not that straightforward.
| My approach.
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
$$=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx\:+\int _1^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
Let $\displaystyle x=\frac{1}{t}$ for the $2$nd integral.
$$\int _0^1\frac{\ln \left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:+\int _0^1\frac{\ln \left(t^2-t+1\right)}{t-1}\:dt-2\int _0^1\frac{\ln \left(t\right)}{t-1}\:dt$$
$$=\int _0^1\left(\frac{1}{t\left(1-t\right)}+\frac{1}{t-1}\right)\ln \left(t^2-t+1\right)\:dt\:-2\sum _{k=0}^{\infty }\frac{1}{\left(k+1\right)^2}$$
$$=\int _0^1\frac{\ln \left(t^2-t+1\right)}{t}\:dt\:-\frac{\pi ^2}{3}$$
$$\int _0^1\frac{\ln \left(t^3+1\right)}{t}\:dt-\int _0^1\frac{\ln \left(t+1\right)}{t}\:dt-\frac{\pi ^2}{3}$$
$$\int _0^1\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}t^{3k-1}\:dt-\int _0^1\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}t^{k-1}\:dt-\frac{\pi ^2}{3}$$
$$\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{3k^2}+\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+2}}{k^2}-\frac{\pi ^2}{3}$$
$$\frac{\pi ^2}{36}-\frac{\pi ^2}{12}-\frac{\pi ^2}{3}=-\frac{7\pi ^2}{18}$$
So,
$$\boxed{\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx=-\frac{7\pi ^2}{18}}$$
In order to find the desired integral i used this previous expression.
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx=\int _0^1\frac{\ln \left(t^2-t+1\right)}{t}\:dt\:-\frac{\pi ^2}{3}$$
And let $\displaystyle t=1-u$
$$-\frac{7\pi ^2}{18}=-\int _0^1\frac{\ln \left(u^2-u+1\right)}{u-1}\:du\:-\frac{\pi ^2}{3}$$
$$\boxed{\int _0^1\frac{\ln \left(u^2-u+1\right)}{u-1}\:du\:=\frac{\pi ^2}{18}}$$
Notice that on the $3$rd line all we have to do is to put in the result we just found and we'll be done.
$$-\frac{7\pi ^2}{18}=\int _0^1\frac{\ln \:\left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:+\frac{\pi ^2}{18}-\frac{\pi ^2}{3}$$
And finally.
$$\boxed{\int _0^1\frac{\ln \:\left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:=-\frac{\pi ^2}{9}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
} |
How to show the divergence of the improper integral $\int_{0}^{\infty} \frac{dx}{1+x^{2}\sin^{2}(x)}$? How can I show the given improper integral $\int_{0}^{\infty} \frac{dx}{1+x^{2}\sin^{2}(x)}$ is divergent?
Approach:
\begin{align*}
x^{2}\sin^{2}(x) \le x^2 \\
1+x^{2}\sin^{2}(x) \le 1+ x^2 \\
\frac{1}{1+x^{2}\sin^{2}(x)} \ge \frac{1}{1+x^{2}}
\end{align*}
but this inequality is not useful as $\int_{0}^{\infty}\frac{1}{1+x^{2}}dx$ is convergent.
| Let $ n $ be a positive integer, we have : \begin{aligned} \int_{0}^{n\pi}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}&=\sum_{k=0}^{n-1}{\int_{k\pi}^{\left(k+1\right)\pi}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}}\\ &=\sum_{k=0}^{n-1}{\int_{0}^{\pi}{\frac{\mathrm{d}x}{1+\left(x+k\pi\right)^{2}\sin^{2}{x}}}}\\ &\geq\sum_{k=0}^{n-1}{\int_{0}^{\pi}{\frac{\mathrm{d}x}{1+\left(\pi+k\pi\right)^{2}\sin^{2}{x}}}}=\sum_{k=1}^{n}{\int_{0}^{\pi}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}} \end{aligned}
Now, notice that $ \left(\forall k\in\mathbb{N}^{*}\right) : $ \begin{aligned}\int_{0}^{\pi}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}=\int_{0}^{\frac{\pi}{2}}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}+\int_{\frac{\pi}{2}}^{\pi}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}\end{aligned}
Let's substitute $ \small\left\lbrace\begin{aligned}y&=\pi -x\\ \mathrm{d}x&=-\,\mathrm{d}x\end{aligned}\right. $, to get : $$ \int_{0}^{\pi}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}=2\int_{0}^{\frac{\pi}{2}}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}} $$
Let's use another substitution $ \small\left\lbrace\begin{aligned}u&=\tan{x}\\ \mathrm{d}x&=\frac{\mathrm{d}u}{1+u^{2}}\end{aligned}\right. $, to get : \begin{aligned}\int_{0}^{\frac{\pi}{2}}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}&=\int_{0}^{+\infty}{\frac{\mathrm{d}u}{\left(1+k^{2}\pi^{2}\frac{u^{2}}{1+u^{2}}\right)\left(1+u^{2}\right)}}\\ &=\int_{0}^{+\infty}{\frac{\mathrm{d}u}{1+\left(1+k^{2}\pi^{2}\right)u^{2}}}\\ &=\frac{1}{\sqrt{1+k^{2}\pi^{2}}}\int_{0}^{+\infty}{\frac{\sqrt{1+k^{2}\pi^{2}}}{1+\left(\sqrt{1+k^{2}\pi^{2}}u\right)^{2}}\,\mathrm{d}u}\\ &=\frac{1}{\sqrt{1+k^{2}\pi^{2}}}\left[\arctan{\left(\sqrt{1+k^{2}\pi^{2}}u\right)}\right]_{0}^{+\infty}\\ &=\frac{\pi}{2\sqrt{1+k^{2}\pi^{2}}}\end{aligned}
Thus, $$ \int_{0}^{\pi}{\frac{\mathrm{d}x}{1+k^{2}\pi^{2}\sin^{2}{x}}}=\frac{\pi}{\sqrt{1+k^{2}\pi^{2}}} $$
Hence, $$ \int_{0}^{n\pi}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}\geq\pi\sum_{k=1}^{n}{\frac{1}{\sqrt{1+k^{2}\pi^{2}}}}=\sum_{k=1}^{n}{\frac{1}{\sqrt{\frac{1}{\pi^{2}}+k^{2}}}}\geq\sum_{k=1}^{n}{\frac{1}{k}} $$
Which means : $$ \lim_{n\to +\infty}{\int_{0}^{n\pi}}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}=+\infty $$
In other words : $$ \int_{0}^{+\infty}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}\textrm{ DV} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Expressions for system of equations in a neighborhood of the origin, $x' = y+y^2 - 2xy + x^2$, $y'=x+y^2 - 2xy + x^2.$ Do you guys agree with my solution to the following problem? Please provide feedback if possible, thanks!
Find expressions for the local stable and local unstable manifolds for the following system of equations in a neighborhood of the origin, $$x' = y+y^2 - 2xy + x^2$$ $$y'=x+y^2 - 2xy + x^2.$$
$\textbf{Solution:}$ Subtracting $x' = y+y^2 - 2xy + x^2$ from $y'=x+y^2 - 2xy + x^2$ gives us $y'-x' = x-y$ implies the following $$ x'+x=y'+y. \hspace{35pt} (1)$$ Integrating $x' = y+y^2 - 2xy + x^2$ with respect to $y$ and $y'=x+y^2 - 2xy + x^2$ with respect to $x$ gives us $$x = \frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y \hspace{35pt} (2)$$ $$y=\frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3}. \hspace{35pt} (3)$$
Applying (2) and (3) to (1) gives us $$\frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y + y +y^2 -2xy+x^2$$ $$=\frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3} + x + y^2 -2xy + x^2$$ $$\implies \frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y + y = \frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3} + x. \hspace{35pt}(4)$$
So, equation (4) denotes a function which if we replace $y$ with $x$ equation will be the same throughout. Thus, if a function of the form $$f(x,y) = \frac{t^2}{2} + \frac{t^3}{3} - t^3 + t^2 + t \text{ where } t = x, y$$ implies $$f(x,y) = \frac{t^2}{2} + \frac{t^3}{3} + t. \hspace{35pt} (5)$$
So equations (1), (4), and (5) define the stable and unstable points around the origin. So, $f'(x,y) >0$ as $$f'(x,y) = t + t^2 + 1 = (t+\frac{1}{2})^2 + \frac{3}{4}.$$ Therefore, it will be unstable and we are done.
| As $\{x^2,y^2, xy\}$ go to zero more quickly by one order degree than $\{x,y\}$ we have the near zero the dynamical system behaves as
$$
\dot x = y\\
\dot y = x
$$
The Jacobian at the origin is
$$
J = \left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
$$
characterizing a saddle point which is unstable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show $\sqrt{r_2^2-x^2}-\sqrt{r_1^2-x^2}\geq r_2-r_1$ for vertical chord distance between circles Let $0<r_1<r_2$.
Consider two circles centered at the same point, one with radius $r_1$ and the other with radius $r_2$. According to all of the pictures I have drawn, each vertical line from the smaller circle to the larger circle has length at least $r_2-r_1$. To prove this in general I think it suffices to show:
$\sqrt{r_2^2-x^2}-\sqrt{r_1^2-x^2}\geq r_2-r_1$ for all $x\in [0,r_1]$.
Is this true? How to prove it? Is there a simple geometry argument?
| Given $r_2 > r_1$ and $x\in [0,r_1]$, we have
$$r_1 \ge \sqrt{r_1^2-x^2}$$
$$ 2r_1(r_2-r_1) \ge 2(r_2-r_1)\sqrt{r_1^2-x^2}$$
$$ (r_2-r_1)^2 + 2r_1(r_2-r_1) + r_1^2-x^2 \ge (r_2-r_1)^2 + 2(r_2-r_1)\sqrt{r_1^2-x^2}+ (r_1^2-x^2) $$
$$ (r_2-r_1+r_1)^2-x^2 \ge \left[(r_2-r_1) + \sqrt{r_1^2-x^2}\right]^2$$
$$ \sqrt{r_2^2-x^2} \ge (r_2-r_1) + \sqrt{r_1^2-x^2}$$
$$\sqrt{r_2^2-x^2}-\sqrt{r_1^2-x^2}\ge r_2-r_1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
simplify $\sqrt[3]{x \sqrt[3]{ x \sqrt[3]{x ...}} }$ -- if $x$ is negative? Was given this recreational problem: simplify $$\sqrt[3]{x \sqrt[3]{ x \sqrt[3]{x ...}} }$$
The solution isn't hard. Let $y = \sqrt[3]{x \sqrt[3]{ x \sqrt[3]{x ...}} }$, then $y^3 = xy$, $y=\sqrt{x}$
The problem didn't specify, but if $x$ is negative, would this work?
| Let $y=\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\dots}}}$
So,
$y=\sqrt[3]{xy}$
Now raise both side to the power $3$, we get:
$y^3=xy$
Now, $y\ne0$ because $x$ is negative as mentioned in the problem statement, we can divide both sides by $y$, we get:
$y^2=x$
However $y^2=x$ has no real solution since $x$ is negative. Hence the given expression does not have a closed form. [THIS IS THE REQUIRED ANSWER].
When $x$ is positive, then the equation $y^2=x$ reduces to $y=\sqrt{x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
System of congruences where $\gcd(m, n)\ne1$ I have to solve this system of congruences:
$$
\begin{cases}
x^2+2x+2\equiv 0\pmod{10}\\
7x\equiv 20\pmod{22}
\end{cases}
$$
after some calculations
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 2\pmod{5}\\
x\equiv 0\pmod{2}\\
x\equiv 6\pmod{2}\\
x\equiv 6\pmod{11}\\
\end{cases}
$$
since $x\equiv 6\pmod{2}$ and $x\equiv 0\pmod{2}\\$ are equal, we get:
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 2\pmod{5}\\
x\equiv 0\pmod{2}\\
x\equiv 6\pmod{11}\\
\end{cases}
$$
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 6\pmod{11}\\
\end{cases}\implies x\equiv 46\pmod{55}
$$
$$
\begin{cases}
x\equiv 0\pmod{2}\\
x\equiv 2\pmod{5}\\
\end{cases}\implies x\equiv 2\pmod{10}
$$
but, $\gcd(55,10)\ne1$, so I cannot apply the Chinese theorem. What have I done wrong?
| Solving the linear congruence gives $x\equiv 6\bmod 22$. Write $x=22n+6$ and substitute into the quadratic congruence:
$$
(22n+6)^2 +2(22n+6)+2 \equiv 0\bmod 10
$$
$$
\Rightarrow 4n^2+4n+6+4n+2+2 \equiv 0\bmod 10
$$
$$
\Rightarrow 4n^2+8n \equiv 0\bmod 10
$$Everything is even, so let's cancel 2:
$$
\Rightarrow 2n^2+4n\equiv 0 \bmod 5
$$
$$
\Rightarrow 2n^2\equiv n \bmod 5
$$
$$
\Rightarrow n^2\equiv 3n \bmod 5
$$So, $n$ is either $0$ or $3$ mod $5$. Together, these give the solutions $x=\{6,72\}\bmod 110$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Linear Algebra: Looking at points M on a circle. Consider the points
$
M=\left\{
\begin{pmatrix}
0\\
2
\end{pmatrix}
,
\begin{pmatrix}
2\\
2
\end{pmatrix}
,
\begin{pmatrix}
1\\
0
\end{pmatrix}
,
\begin{pmatrix}
1\\
2
\end{pmatrix}
\right\}
$
in $\mathbb{R}^2$.
The points on a circle with radius r and center in $(a,b)^T\in\mathbb{R}^2$ is described as the solutions to the equation $(X-a)^2+(Y-b)^2=r^2$. This equation is equivalent with $2aX+2bY+(r^2-a^2-b^2)=X^2+Y^2$
Set
$$
\begin{pmatrix}
2a\\
2b\\
r^2-a^2-b^2
\end{pmatrix}
\in \mathbb{R}^3
$$
I have to show that the points in $M$ is on the circle if and only if $v$ is the solution to the linear equation system
$$
\begin{pmatrix}
0 & 2 & 1\\
2 & 2 & 1\\
1 & 0 & 1\\
1 & 2 & 1
\end{pmatrix}
\cdot x
=
\begin{pmatrix}
4\\
8\\
1\\
5
\end{pmatrix}
$$
I have already shown the way where we assume $M$ is on the circle and show that $v$ is then the solution to the linear equation system. However, i'm having trouble the other way around. How do i show that $M$ is on the circle if $v$ is the solution to the linear equation system? I have tried inserting the vector
$$
\begin{pmatrix}
2a\\
2b\\
r^2-a^2-b^2
\end{pmatrix}
$$
in the equation instead of x, however, I don't see how this is supposed to help me..
| Take a generic circle $\mathcal C_{a,b,r}=\{(X,Y)\in \mathbb R^2 \ | \ 2aX+2bY+(r^2-a^2-b^2) = X^2+Y^2\}$.
We have:
\begin{equation}
M\subseteq \mathcal C_{a,b,r} \Longleftrightarrow
\begin{cases}
2a (0) +2b (2) + (r^2-a^2-b^2) = 0^2+2^2 = 4\\
2a (2) +2b (2) + (r^2-a^2-b^2) = 2^2+2^2 = 8\\
2a (1) +2b (0) + (r^2-a^2-b^2) = 1^2+0^2 = 1\\
2a (1) +2b (2) + (r^2-a^2-b^2) = 1^2+2^2 = 5\\
\end{cases}
\end{equation}
Where the system means that the points in $M$ satisfies the equation of $\mathcal C_{a,b,r}$. Setting now
\begin{equation}
v=\left(\begin{matrix}
2a\\
2b\\
r^2-a^2-b^2
\end{matrix}\right)
\end{equation}
we can rewrite the linear system in a more suitable way:
\begin{equation}
M\subseteq \mathcal C_{a,b,r} \Longleftrightarrow
\begin{pmatrix}
0 & 2 & 1\\
2 & 2 & 1\\
1 & 0 & 1\\
1 & 2 & 1
\end{pmatrix}
\cdot v
=
\begin{pmatrix}
4\\
8\\
1\\
5
\end{pmatrix}
\end{equation}
That is your statement.
Moreover, if you try to resolve the system written above, you will discover that it does not have solution; hence it does not exist a circle contained every points of $M$.
We can see that this result is coherent: infact the points $\left\{
\begin{pmatrix}
0\\
2
\end{pmatrix}
,
\begin{pmatrix}
2\\
2
\end{pmatrix}
,
\begin{pmatrix}
1\\
2
\end{pmatrix}
\right\}
\subseteq M$ are aligned, so we had to expect this result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3628239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\min$ for$ f(x) = (x + a + b)(x + a - b)(x - a + b)(x - a - b)$ I'm trying to find $minf(x)$ for $f(x) := (x + a + b)(x + a - b)(x - a + b)(x - a - b)$, where $a, b \in \mathbb{R},$ using inequalities.
For example, i can find $maxf(x)$, using AM-GM ineq:
$$\sqrt[4]{(x + a + b)(x + a - b)(x - a + b)(x - a - b)})^4 \leq \Big (\frac{x + a + b + x + a - b+ x - a + b+ x - a - b}{4}\Big)^4 = $$
$$= x^4.$$
So $maxf(x) = x^4$.
But i don't know how to solve is for $minf(x)$, which i need to find.
Sure do i can find structure of square difference in $f(x)$ and we can rewrite our equality:
$$f(x) = (x^2 - (a + b)^2) (x^2 - (b - a)^2).$$
But i don't know what to do next.
UPD: We need to find extremum on $x$ via fixing $a, b$. I understand that $max$ is found wrong way. How can i do it correctly?
| Here is another approach:
Denote $t=a+b$ and $s=a-b$, then $f(x)$ becomes to be $(x^2-t^2)(x^2-s^2)$. Henece $f(x)$ is quintic which is very easy to work with, more specifically,
$$f(x)=x^4-(s^2+t^2)x^2+t^2s^2.$$
Now, you can make another substitution, namely $y=x^2$ to obtain a quadratic, which you know how to find a minimum of it. Don't forget to returnt to given variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$. Use a Taylor series expansion to find $f^{(80)}(27)$.
I tried the following:
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
%
f''(x)
&= \frac{1}{3}(x+3)^{-\frac{2}{3}}+\frac{1}{3}(x-27)\cdot -\frac{2}{3}(x+3)^{-\frac{5}{3}}+\frac{1}{3}(x+3)^{-\frac{2}{3}} \\
&= \frac{2}{3}(x+3)^{-\frac{2}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}(x-27)\\
%
f'''(x) &= -\frac{2^2}{3^3}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2\cdot 5}{3^3}(x+3)^{-\frac{8}{3}}
\end{align}
This is very nasty, please help me to solve in some easy way.
| Put $g(x)=(x+3)^{1/3}$. We have $f(x)=g(x)(x-27)$. It suggests that if a Taylor series of $f$ at $x_0=27$ is $\sum_{n=0}^\infty \frac {f^{(n)}(x_0)}{n!}(x-x_0)^n$ and a Taylor series of $g$ at $x_0=27$ is $\sum_{n=0}^\infty \frac {g^{(n)}(x_0)}{n!}(x-x_0)^n$ then for each $n\ge 0$ we have $$\frac {g^{(n)}(x_0)}{n!}= \frac {f^{(n+1)}(x_0)}{(n+1)!}.$$
But we don’t need the Taylor series to show the latter equality. It can be shown directly, by a simple induction or by Leibniz differentiation rule stating that
$$f^{(n+1)}(x_0)=\sum_{i=0}^{n+1}{n+1\choose i}(x-x_0)^{(i)}g^{(i)}(x)\large|_{x=x_0}=(n+1)g^{(n)}(x_0)=$$ $$(x_0+3)^{\frac 13-n}(n+1)\prod_{i=0}^{n-1} \left(\frac 13-i\right)$$
(the last equality holds for any natural $n\ge 1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Integral $\int \frac{2x^5-2x^4+2x^3+3}{2x^4-2x^3-x^2+1}dx$ as partial fraction solved using matrix equation in order to solve the integral $$\int \frac{2x^5-2x^4+2x^3+3}{2x^4-2x^3-x^2+1}\mathrm dx,$$
the expression inside the integral can be expressed as
$$(2x^5-2x^4+2x^3+3/2x^4-2x^3-x^2+1)= x+(A/(x-1))+(B/(x-1)^2)+(Cx+D/2x^2+2x+1)$$
from here I have been ask to set up the system of linear simultaneous equations that are needed to be solved to calculate the integral, by utilising MX=Z, where M is the coefficient matrix, X is the solution vector containing the coefficients, and Z is the RHS of the matrix equation
I have attempt to do this by factoring the LHS denominator to the form of
$$(x-1)^2(2x^2+2x+1)$$ and then multiplied both sides by this giving the resulting equation
$$2x^5-2x^4+2x^3+3 = 2x^5-2x^4+2x^3+3+A(x-1)(2x^2+2x+1)+B(2x^2+2x+1)+CxD(x-1)^2$$
if I subtract $(2x^5-2x^4+2x^3+3)$ from LHS I am left with
$$LHS = 3x^3-x+3$$
I have tried expanding out the rest of the RHS and then collected like terms and to try and set up four equations
$3=2A+CD$
$0=2B-2CD$
$-1=-A+2B+CD$
$3=-A+B$
however, there is no solution.
any help would be appreciated
| Decompose the simplified integrand as,
$$\frac{3x^3-x+3}{(x-1)^2(2x^2+2x+1)}=\frac A{x-1}+\frac B{(x-1)^2}+\frac{Cx+D}{2x^2+2x+1}$$
The coefficients $A$, $B$, $C$ and $D$ are obtained successively as follows. Multiply both sides by $(x-1)^2(2x^2+2x+1)$
$$3x^3-x+3=A(x-1)(2x^2+2x+1)+B(2x^2+2x+1)+(Cx+D)(x-1)^2\tag 1$$
First, set $x=1$ to get $B=1$, which is plugged into (1) to get
$$(x+1)(3x-2)=A(2x^2+2x+1)+(Cx+D)(x-1)\tag 2$$
Then, set $x=1$ again to get $A=\frac25$, which is plugged into (2) to get
$$\frac15(11x+12) = Cx+D$$
which yields $C=\frac{11}5$ and $D=\frac{12}5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$a$ red balls and $b$ blue balls in an urn. Probability when a game lasts forever. In an urn there are $a$ indistinguishable red balls and $b$ indistinguishable blue balls. Every round, you take one ball in random from the urn. If this ball is blue, game over. If this ball is red, you put it back into the urn and put an extra indistinguishable red ball in (so the urn has an extra red ball than before). Let $E_{a,b}$ be the expected number of total rounds of games.
Suppose we know the total number of balls $N = a+b$ but don't know the exact number of $a,b$. Let $a$ follow an uniform a prioir distribution in set $\{ 1, ..., N-1 \}$. On which round, if you draw a red ball, you have are more than 90% certain that $E_{a,b}$ is infinity?
I guess the first step would be trying to derive $E_{a,b}$ in some way. $E_{a,b} = \frac{a}{b+a} (E_{a+1,b} + 1) + \frac{b}{a+b}$. Then $(a+b)E_{a,b} = a E_{a+1,b} + a + b$, so $E_{a+1,b} = \frac{a+b}{a}E_{a,b} - \frac{a+b}{a}$. Obviously $E_{0,b} = 1$, but this looks wrong..
| This calculation of $E_{a,b}$ will get you started. On the $n-$th round, the urn contins $a+n-1$ red balls and $b$ blue balls. The probability that the $n-$ th round is the last is $\frac{b}{a+b+n-1}$. The only way of playing the $n-$th round is by surviving all the previous rounds.Thus the probability that the game ends in exactly $n$ rounds is
$$E_{a,b}=\frac{a}{a+b}.\frac{a+1}{a+b+1}...\frac{a+n-2}{a+b+n-2}.\frac{b}{a+b+n-1}$$ $$=\frac{(a+n-2)!(a+b-1)!b}{(a-1)!(a+b+n-1)!}$$The expected number of rounds is $$\sum_{n=1}^{\infty}\frac{(a+n-2)!(a+b-1)!bn}{(a-1)!(a+b+n-1)!}$$ $$=\frac{b(a+b-1)!}{(a-1)!}\sum_{n=1}^{\infty}\frac{(a+n-2)!n}{(a+b+n-1)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3632281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimize $|a-1|^3+|b-1|^3$ with constant product $ab=s$ Let $0<s$, and define
$$ F(s):=\min_{a,b \in \mathbb{R}^+,ab=s} \left(|a-1|^3+|b-1|^3\right). $$
I would like to find proofs for the claim
$$
F(s)=\begin{cases}
1 - 3 s - 2s^{3/2}=F\big(a(s),b(s)\big), &\text{ if } 0<s\le1/9, \\
2 + 6 s - 2(3 + s)s^{1/2}=F(\sqrt s,\sqrt s) &\text{ if } 1/9\le s<1,
\end{cases}
$$
where $a(s),b(s)$ are uniquely defined by the equation $a+b=1-\sqrt{s},ab=s$.
(Actually I am more interested in the exact value of $F(s)$, and less in the minima points themselves, but I thought that this additional information might be helpful to find other proofs as well).
I have a proof which I present below, but I wonder if there is an easier way to prove this.
Also, is there a math software that can solve such a problem? (I am rather ignorant on that stuff, unfortunately).
Edit:
I think that maybe one can prove this without differentiation, but I am not sure. The idea is to rewrite the symmetric polynomial $(a-1)^3+(b-1)^3$, as a polynomial in $a+b,ab$, and proceed from there, but I am not sure it really works.
My proof:
First, suppose that $s \le 1$. Then the minimum is obtained at a point $(a,b)$ where both $a,b$ are not greater than $1$. Indeed, if $a>1$ (and so $b <s \le 1$), we can replace $a$ by $1$ and $b$ by $s$ to get the same product, but now both numbers are closer to $1$ then before. In fact, a symmetric argument shows that if $s \ge 1$, then both $a,b \ge 1$.
In any case, the signs of $a-1,b-1$ are identical.
Expressing the constraint as $g(a,b)=ab-s=0$, and using Lagrange's multipliers, there exist a $\lambda$ such that
$$
(a-1)^2=\lambda b, (b-1)^2=\lambda a.
\tag{1}$$
(Here we used the assumption the fact that the signs of $a-1,b-1$ are identical).
Subtracting these equations, we get
$$
(a-b)(a+b-2)=-\lambda(a-b).
$$
So, one candidate is $a=b=\sqrt{s}$. If $a \neq b$, then
$$
a+b=2-\lambda, ab =s \tag{2}.
$$
Thus, $a,b$ are the solutions of the quadratic
$$ x^2+(\lambda-2)x+s=0$$
Say that $a \le b$. Then
$$ a=\frac{2-\lambda-\sqrt{c}}{2}, b=\frac{2-\lambda+\sqrt{c}}{2}, \, \, \, \text{where } \, \, c=(2-\lambda)^2-4s.$$
Plugging this into $(a-1)^2=\lambda b$ from equation $(1)$, we get
$$
(\lambda+\sqrt c)^2=\lambda (4-2\lambda+2\sqrt c),
$$
which simplifies into
$$ 3\lambda^2-4\lambda=-c=4s-(2-\lambda)^2.$$
Further simplification gives
$$ (\lambda-1)^2=s \Rightarrow 1-\lambda=\pm \sqrt s.$$
Thus, by equation $(2)$, $a+b=1\pm \sqrt s$.
Comment: We can immediately see that this cannot happen when $s>1$. Applying the AM-GM inequality for $a,b$ implies that
$$ a+b=1+\sqrt s \Rightarrow s \le 1, \,\,\,a+b=1-\sqrt s \Rightarrow s \le \frac{1}{9}$$
Now, if $a+b=1 + \sqrt s$, the it is easy to deduce that $b \ge 1$. (recall we assumed before that $a \le b$). As commented at the beginning, the optimum point must be obtained where $a,b$ are both not greater than $1$. So, the only possible option is $b=1$, and then $a=\sqrt s$, which implies $s=ab=\sqrt s$ so $s=1$ and $F(1)=0$, $a=b=1$.
Thus, we are left with the option $a+b=1-\sqrt{s}$.
Solving explicitly the quadratic given by $a+b=1-\sqrt{s},ab=s$, we get explicit expressions $a(s),b(s)$. Then direct calculation gives
$$ F\big(a(s),b(s)\big)=1 - 3 s - 2s^{3/2}.$$
The quadratic for $a(s),b(s)$ is
$$ x^2-(1-\sqrt s)x+s=0. \tag{3}$$
It has real solutions exactly when $(1-\sqrt s)^2 \ge 4s$, or (since $s>0$), $\sqrt s \le \frac{1}{3}$. (Equivalently, this can be seen from the AM-GM inequality for $a,b$ as above.)
Now,
$F(\sqrt s,\sqrt s)=2 + 6 s - 2(3 + s)s^{1/2}$, and all is left to do is to verify that
$$ F(a(s),b(s)) \le F(\sqrt s,\sqrt s),$$ in the regime $s \le \frac{1}{9}$, where $a(s),b(s)$ exist, as solutions to the quadratic $(3)$.
Direct computation shows that
$$ F(\sqrt s,\sqrt s)-F(a(s),b(s))=(1-3\sqrt s)^2 \ge 0$$
(and equality happens only at $s=\frac{1}{9}$).
This finishes the proof.
| For $0 < s < 1$, we have
\begin{align}
F(s) &= \min_{a, b > 0;\ ab = s}\ |a-1|^3 + |b-1|^3 \\
&= \min_{a\ge b > 0;\ ab = s}\ |a-1|^3 + |b-1|^3\tag{1}\\
&= \min_{0 < b \le \sqrt{s}}\ \left|\frac{s}{b} - 1\right|^3 + (1-b)^3\\
&= \min_{s \le b \le \sqrt{s}}\ \left(1 - \frac{s}{b}\right)^3 + (1-b)^3. \tag{2}
\end{align}
Explanation: (1) holds due to symmetry.
(2) holds since $|\frac{s}{b} - 1|^3 + (1-b)^3$ is strictly decreasing on $0 < b \le s$,
and hence the minimizer occurs on the interval $s \le b \le \sqrt{s}$.
Let us solve (2).
The minimum may be achieved at points on $(s, \sqrt{s})$ with zero derivative,
or at the interval endpoints $s, \sqrt{s}$.
Let $g(b) = (1 - \frac{s}{b})^3 + (1-b)^3$.
We have $g(s) = (1-s)^3$ and $g(\sqrt{s}) = 2(1-\sqrt{s})^3$. We have
$$g'(b) = \frac{3s}{b^2}\Big(1 - \frac{s}{b}\Big)^2 - 3(1-b)^2.$$
Thus, we have, for $s < b < \sqrt{s}$,
\begin{align}
g'(b) = 0 \quad &\Longleftrightarrow \quad \frac{\sqrt{s}}{b}\Big(1 - \frac{s}{b}\Big) = 1-b\\
&\Longleftrightarrow \quad (b - \sqrt{s})(b^2 - (1-\sqrt{s})b + s) = 0, \\
&\Longleftrightarrow \quad b^2 - (1-\sqrt{s})b + s = 0. \tag{3}
\end{align}
We split into two cases:
1) $0 < s < \frac{1}{9}$: The equation (3) has exactly one real root on $(s, \sqrt{s})$, i.e.,
$b_1 = \frac{1-\sqrt{s}}{2} - \frac{1}{2}\sqrt{-3s-2\sqrt{s} + 1}$.
We have (some details are given later)
$$g(b_1) = 1 - 3s - 2s^{3/2}. \tag{4}$$
It is easy to prove that $g(b_1) \le g(s) $ and $g(b_1)\le g(\sqrt{s})$.
Thus, $F(s) = g(b_1) = 1 - 3s - 2s^{3/2}$.
2) $\frac{1}{9} \le s < 1$: The equation (3) has no real root on $(s, \sqrt{s})$.
Thus, $F(s) = \min(g(s), g(\sqrt{s})) = g(\sqrt{s}) = 2 - 6\sqrt{s} + 6s - 2s^{3/2}$.
We are done.
$\phantom{2}$
Some details about (4):
From $b_1^2 - (1-\sqrt{s})b_1 + s = 0$, we have
$s = (1-\sqrt{s})b_1 - b_1^2$ and $\frac{s}{b_1} = 1-\sqrt{s} - b_1$.
Thus,
\begin{align}
g(b_1) &= (1 - \frac{s}{b_1})^3 + (1-b_1)^3 \\
&= (\sqrt{s} + b_1)^3 + (1-b_1)^3\\
&= (3\sqrt{s}+3)b_1^2 - (3-3s)b_1 + s^{3/2} + 1\\
&= 3(1+\sqrt{s})[b_1^2 - (1-\sqrt{s})b_1] + s^{3/2} + 1\\
&= 3(1+\sqrt{s})\cdot (-s) + s^{3/2} + 1\\
&= 1 - 3s - 2 s^{3/2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Quartic polynomials of a complex variable I want to answer the following:
The equation
$$
a \bigg(z + \frac{1}{z}\bigg)^2
+ b \bigg(z + \frac{1}{z}\bigg)
+ c = 0
$$
has four solutions. Find which quartics can be put in this form after apply a linear change of vars.
| The crucial step is the substitution:
$u = z + \frac{1}{z}$
Then, we obtain the quadratic:
$au^{2} + bu + c = 0$
By the Quadratic Formula:
$u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Let the roots of the above quadratic be $s$ and $t$. Then, for $s$:
$z + \frac{1}{z} = s$
$z^{2} - sz + 1 = 0$
By the Quadratic Formula:
$z = \frac{s\pm\sqrt{s^{2} - 4}}{2}$
By a similar argument for $t$:
$z = \frac{t\pm\sqrt{t^{2} - 4}}{2}$
$\boxed{z = \frac{s+\sqrt{s^{2} - 4}}{2},\frac{s-\sqrt{s^{2} - 4}}{2},\frac{t+\sqrt{t^{2} - 4}}{2}, \frac{t-\sqrt{t^{2} - 4}}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
The probability that a binomial coefficient is divisible by 12. I do not know how to solve the following question:
What is the probability that $12$ divides $\binom{n}{12}$ for a randomly chosen natural number n?
Context:
I set this as a question in a problem set. I was trying to write these problems from memory while setting the questions.
Anyway, now I am unable to solve this variant. Essentially $12^5$ divides $12!$ and thus the question reduces to
Find the values of natural $n$ for which $12^6$ divides $n(n-1)(n-2)\cdots (n-11)$.
We can see that $12^5$ divides product of any consecutive $12$ numbers, so we need an additional $12$ from the product. When does this happen?
| $$\binom{n}{12}=\frac{n(n-1)(n-2)\ldots(n-11)}{12!}=\frac{n(n-1)(n-2)\ldots(n-11)}{2^{10}\cdot 3^5\cdot 5^2\cdot 7\cdot 11}$$
Any $12$ consecutive integers include $4$ consecutive factors of $3$, at least one of which must be a multiple of $9$, and six consecutive multiples of $2$. Three of these must be multiples of $4$, and at least one must be a multiple of $8$, so their product has at least $10$ factors of $2$. We want to know when the product has at least $6$ factors of $3$ and at least $12$ factors of $2$.
The $12$ integers include two multiples of $9$ if $n$ is congruent to $0,1$, or $2$ modulo $9$, and they include a multiple of $27$ if $n$ is congruent to $0,1,2,\ldots,10$, or $11$ modulo $27$; thus, their product has at least $6$ factors of $3$ if and only if $n$ is congruent to one of the following $15$ numbers modulo $27$:
$$0,1,2,3,4,5,6,7,8,9,10,11,18,19,20$$
If the $12$ integers include two multiples of $8$, one of those must be a multiple of $16$, and their product will then have at least $12$ factors of $2$. If they include only one multiple of $8$, it must be a multiple of $32$ if the product is to have at least $12$ factors of $2$. They include two multiples of $8$ if $n\equiv 0,1,2,3\pmod8$, and they include a multiple of $32$ if $n$ is congruent to $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, or $11$ modulo $32$. Thus, their product has at least $12$ factors of $2$ if and only if $n$ is congruent to one of the following $20$ numbers modulo $32$:
$$0,1,2,3,4,5,6,7,8,9,10,11,16,17,18,19,24,25,26,27$$
$\binom{n}{12}$ is a multiple of $12$ if and only if $n$ satisfies both of these conditions, meaning that it is in one of $15\cdot20=300$ residue classes modulo $27\cdot32=864$. As noted in the comments, the expression randomly chosen natural number is at best ill-defined, but we can at least say that the natural (or asymptotic) density of the set of natural numbers $n$ such that $12\mid\binom{n}{12}$ is
$$\frac{300}{864}=\frac{25}{72}\;,$$
and this corresponds reasonably well to the intuitive notion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence. I need to show that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence, where $x_1 = 2$ and n = 1,2,3,....
I tried to show this with induction, where since $x_1= 2$ then $x_2 = 1.5$, hence the base case satisfies.
Assuming that $x_n \geq x_{n+1}$ , I need to prove that $x_{n+1} \geq x_{n+2}$.
I started with the statement $x_n \geq x_{n+1} \Rightarrow \frac{1}{x_n} \leq \frac{1}{x_{n+1}}$
Also, $x_n \geq x_{n+1}$ then $\frac{1}{2}(x_n + \frac{2}{x_{n+1}}) \geq \frac{1}{2}(x_{n+1} + \frac{2}{x_{n+1}})$
$= \frac{1}{2}(x_n + \frac{2}{x_{n+1}} - \frac{2}{x_n} + \frac{2}{x_n}) \geq x_{n+2} $
$= x_{n+1} - \frac{1}{x_n} + \frac{1}{x_{n+1}} \geq x_{n+2} $
since $- \frac{1}{x_n} + \frac{1}{x_{n+1}} \geq 0$
then I cannot say that $ x_{n+1} \geq x_{n+2} $
How do I proceed forward?
Edit: Since $x_1 = 2$, hence $1 \leq x_1 \leq 2$ holds for base case.
If I assume that $1 \leq x_n \leq 2 $ and show that $1 \leq x_{n+1} \leq 2 $ then by induction we can say that $1 \leq x_n \leq 2 \:\forall n$
I have shown this as:
$x_{n+1} = \frac{x_n}{2} + \frac{1}{x_n}, \: now \: since\:1 \leq x_n\leq2\:\Rightarrow x_{n+1} \leq \frac{1}{2}(2 + \frac{2}{1}) = 2$
Also by AM GM inequality, $x_n \geq \sqrt{2} \: \: \forall n$
Now all that I have to show is that $x_n \geq x_{n+1} \: \forall n?$
| At first one could show that $f(x)\geq 2$ for all $x>0$, where $f(x):=\frac12\left(x+\frac2{x}\right)$. This follows immediately from $f(x)^2-4= \frac14\left(x-\frac2{x}\right)^2\geq0$. But then $f(x)\leq x$ is equivalent to $\frac2{x}\leq x$ which is always true when $x^2\geq 2$. But this holde true for $x_0=2$ by assumption, and thus by definition of $x_n$ for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Domain of $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$ $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$
I'm stuck in this one because this can be rewritten as:
$$f(x,y) = \ln(16-x^2-y^2) + \ln(x^2+y^2-4)$$
Yet, the domain of the given function is $\{(16-x^2-y^2>0)\land(x^2+y^2-4>0)\} \lor \{(16-x^2-y^2<0)\land (x^2+y^2-4<0)\}$. But the domain of the rewritten on is only $\{(16-x^2-y^2>0)\land(x^2+y^2-4>0)\}$. Which one is the correct one and why does this happen?
| When does $log(xy) = log(x) + log(y)$ hold?
Only when $x$ and $y$ are both greater than zero! Does that solve your doubt?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove that for any integers $a,b,c,$ there exists a positive integer $n$ such that the number $n^3+an^2+bn+c$ is not a perfect square.
Question: Prove that for any integers $a,b,c,$ there exists a positive integer $n$ such that the number $n^3+an^2+bn+c$ is not a perfect square.
Solution: Let $f:\mathbb{N}\to\mathbb{Z}$ be such that $$f(n)=n^3+an^2+bn+c, \forall n\in\mathbb{N}.$$
Also assume for the sake of contradiction that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$.
We have $f(1)=1+a+b+c, f(2)=8+4a+2b+c, f(3)=27+9a+3b+c$ and $f(4)=64+16a+4b+c$.
Now since $f(4)$ is a perfect square $\implies f(4)\equiv 0,1\pmod 4\implies c\equiv 0,1\pmod 4.$
First let that $c\equiv 0 \pmod 4$. Then $f(2)\equiv 0\pmod 4\implies 2b\equiv 0\pmod 4\implies b\equiv 0,2\pmod 4.$
Also we have $f(1)\equiv 0 \pmod 4$. Now we have $b+c\equiv 0,2 \pmod 4\implies 1+b+c\equiv -1,1\pmod 4.$ Thus we have $a\equiv -1,1\pmod 4$.
Also $f(3)\equiv 0\pmod 4$.
Now we have $f(3)-2f(2)+f(1)\equiv 0 \pmod 4\implies 12+2a\equiv 0\pmod 4\implies 2a \equiv 0\pmod 4 \implies a\equiv 0,2\pmod 4.$ But we have $a\equiv -1,1\pmod 4$, which is a contradiction. Thus it is not true that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$ when $c\equiv 0 \pmod 4$.
A similar analysis for $c\equiv 1\pmod 4$ will lead to a contradiction. Thus it is not true that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$ when $c\equiv 1\pmod 4$.
Hence it is not true that $f(n)$ is a perfect square $\forall n\in\mathbb{N}$ in any case, i.e., $\exists n\in\mathbb{N}$ such that $f(n)$ is not a perfect square.
Is there any better way to solve this problem?
| Let $a,b,c \in \mathbb Z$, and let $f(n)=n^3+an^2+bn+c$, $n \in \mathbb N$. We show that at least one of $f(1)$, $f(2)$, $f(3)$, $f(4)$ is not a perfect square. We use the fact that $m^2 \equiv 0\:\text{or}\:1\pmod{4}$ for $m \in \mathbb Z$.
Suppose $f(n)$ is a perfect square, $n \in \{1,2,3,4\}$. We note that
$$ \begin{eqnarray*} f(1) \equiv a+b+c+1\pmod{4}, \\
f(2) \equiv 2b+c \pmod{4}, \\
f(3) \equiv a+3b+c+3 \pmod{4}, \\
f(4) \equiv c \pmod{4}. \end{eqnarray*} $$
Since $f(3)-f(1)$, $f(4)-f(2)$ are both even, each must be divisible by $4$. But then $4$ must divide both $2b$ and $2(b+1)$. This is impossible.
Therefore, at least one of $f(1)$, $f(2)$, $f(3)$, $f(4)$ must be a non-square, as claimed. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
In how many different ways can you prove that $\sin^2x + \cos^2x = 1$ The standard proof of the identity $\sin^2x + \cos^2x = 1$ (the one that is taught in schools) is as follows: from pythagoras theorem, we have (where $h$ is hypotenuse, $b$ is base and $p$ is perpendicular)
$$h^2 = p^2 + b^2$$
dividing by $h^2$ on both sides:
$$1 = \frac{p^2}{h^2}+\frac{b^2}{h^2}$$
since $\sin x = \frac ph$ and $\cos x = \frac bh$,
$$1 = \sin^2x+\cos^2x$$
Are there any more innovative ways of proving this common identity?
| A few proofs I came up with are:
Proof 1: This uses real analysis. Let $f(x) = \sin^2x + \cos^2x \implies f'(x) = 2\sin x \cos x - 2\sin x \cos x = 0$. Thus, $f(x)$ is a constant function. To find the value of $f(x)$, it is sufficient to find $f(c)$, where c is any convenient constant (say 0). Therefore, $f(x) = f(0) = 1$
Proof 2: This uses Euler's representation of complex numbers. We can represent a complex number in the form $z = re^{ix}$, where $r$ is the modulus and $z$ is the argument. If we take the number $z = e^{ix} = \cos x + i\sin x$ (De Moivre's theorem), then $|z| = 1 = \sqrt{\cos^2x + \sin^2x} \implies \cos^2x + \sin^2x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Generating function for the sequence $(a_k) = (1,-1,2,-2,3,-3,4,-4,...).$ I'm trying to find the generating function for $(a_k)=(1,-1,2,-2,3,-3,4,-4,...)$. I know the answer is $$A(x) = \frac{1}{1-x} \cdot \frac{1}{(x+1)^2}$$ but I can't figure out how to get there.
First I broke down the sequence, $$(a_k) = 1-x+2x^2-2x^3+3x^4-3x^5+4x^6-4x^7+...$$ and tried working backwards by looking at each Taylor series separately, $$\frac{1}{1-x} = 1+x+x^2+x^3+... $$
$$\frac{1}{(x+1)^2}=1-2x+3x^2-4x^3+5x^4-... $$
The last thing I attempted was factoring,
$$(a_k)=(1-x+ 2x^2-2x^3+3x^4-3x^5+4x^6-4x^7+...$$
$$=(0-x+2x^2-2x^3+3x^4-3x^5+...)+1$$
$$=(-1+2x-2x^2+3x^3-3x^4+...)x + 1 $$
But from there I'm stuck. Would love some help!
| This is
$$\sum_{i=1}^\infty i(x^2)^{i-1}-x\sum_{i=1}^\infty i(x^2)^{i-1}$$
which is equal to
$$\frac1{(1-x^2)^2}-\frac x{(1-x^2)^2} = \frac{1-x}{(1-x)^2(1+x)^2}=\frac{1}{(1-x)(1+x)^2}$$
Here I have used the fact that $\sum_{i=1}^\infty{ix^{i-1}}$ is the derivative of a geometric series and used the substitution $x\mapsto x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha^3 = -2\alpha^2+5\alpha -1$.
And, $(\alpha^2+2\alpha-4)^3+2(\alpha^2+2\alpha-4)^2-5(\alpha^2+2\alpha-4)+1=0$ gives $\alpha^6+6\alpha^5+2\alpha^4-13\alpha^2+54\alpha-11=0$
Additionally, $\alpha^6 = (-2\alpha^2+5\alpha -1)^2 = 4\alpha^4-20\alpha^3+29\alpha^2-10\alpha+1$
I do not know how to move forward from here. I tried setting $f(x)$ equal to the product of the roots and expanding that out to a 14-term polynomial with $\alpha$ and $\beta$ coefficients but that seems unproductive.
| Here is another, slightly more sophisticated, take:
The other root is $\beta= \gamma ^2+2\gamma-4$, where $\gamma = \alpha^2+2\alpha-4$.
Expanding $\beta$ in terms of $\alpha$ and reducing it mod $\alpha^3+2\alpha^2-5\alpha+1$ gives $\beta=-\alpha^2 - 3 \alpha + 2$. Note that $\beta=g(\gamma)=g(g(\alpha))$, where $g(x)=x^2+2x-4$.
(This is because $\mathbb Q(\alpha)$ must be the splitting field of $x^3+2x^2-5x+1$ since it already contains two roots and so must contain the third. The Galois group is cyclic of order $3$ and so the roots are $\alpha$, $g(\alpha)$, $g^2(\alpha)$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Using definition of limit to prove that $\lim\limits_{x \to 1}\frac{2-x}{4-x}=\frac{1}{3}$ I hit a block when discovering a negative $\delta$. This is how:
I need to show that$$\forall \epsilon>0 \; \exists \delta>0 \text{ s.t. } \mid x-1 \mid < \delta \Rightarrow \Bigl| \frac{2-x}{4-x}-\frac{1}{3} \Bigr| < \epsilon$$
To find such a $\delta$: $$\left| \frac{3(2-x)-(4-x)}{3(4-x)} \right|=\left|\frac{2-2x}{3(4-x)} \right|=\left| \frac{2(x-1)}{3(4-x)} \right|.$$
Since $|x-1|<\delta$, consider $\frac{2}{3}\Bigl|\frac{1}{4-x} \Bigr|$ further for $\delta=1$, $$-1<x-1<1,\ -4<x-4<-2 \Rightarrow \frac{1}{x-4}<-\frac{1}{4}.$$
How can I get an appropriate positive $\delta$?
| If $\delta < 1$ then $|x-1| < \delta$ implies
$$x-1 \in [-1,1] \implies x-4 \in [-4,-2] \implies |x-4| \in [2,4] \implies \frac1{|x-4|} \in \left[\frac14,\frac12\right]$$
so $\frac1{|x-4|} \le \frac12$.
Therefore we have
$$\left|\frac{2(x-1)}{3(4-x)}\right| = \frac23 \cdot \frac1{|x-4|} \cdot |x-1| < \frac23 \cdot \frac12 \cdot \delta = \frac{\delta}3 < \delta.$$
We want this to be $\le \varepsilon$ so we can take $\delta = \min\{\varepsilon, 1\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3648178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving two binomial identities I would like to show that
\begin{align}
&\sum_{j=n-k}^n\binom nj(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&\qquad=\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&\qquad=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k-1}k(-x)^k
\end{align}
I feel I exhausted all identities/properties of binomials without success. Mathematica says it is true, but how to show it?
| First Identity
$$
\begin{align}
&\sum_{j=n-k}^n\binom{n}{j}(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&=\sum_{j=n-k}^n\binom{n}{j}(1-x)^{n-j-1}x^{j-1}[j(1-x)-(n-j)x]\tag1\\
&=\sum_{j=n-k}^nn\binom{n-1}{j-1}(1-x)^{n-j}x^{j-1}-\sum_{j=n-k}^nn\binom{n-1}{j}(1-x)^{n-j-1}x^{j}\tag2\\
&=\sum_{j=n-k-1}^{n-1}n\binom{n-1}{j}(1-x)^{n-j-1}x^{j}-\sum_{j=n-k}^nn\binom{n-1}{j}(1-x)^{n-j-1}x^{j}\tag3\\[3pt]
&=n\binom{n-1}{n-k-1}(1-x)^{k}x^{n-k-1}\tag4\\[9pt]
&=(n-k)\binom{n}{n-k}(1-x)^{k}x^{n-k-1}\tag5
\end{align}
$$
Explanation:
$(1)$: rewrite $j-nx=j(1-x)-(n-j)x$
$(2)$: $\binom{n}{j}j=\binom{n-1}{j-1}n$ and $\binom{n}{j}(n-j)=\binom{n-1}{j}n$
$(3)$: substitute $j\mapsto j+1$ in the left sum
$(4)$: combine the cancelling terms
$(5)$: $\binom{n-1}{n-k-1}n=\binom{n}{n-k}(n-k)$
Second Identity
$$
\begin{align}
&\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom{n}{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&=\sum_{k=0}^{n-1}(-1)^k\binom{n-1}k\binom{n-1}{n-k-1}(1-x)^kx^{n-k-1}\tag6\\
&=\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}(-1)^k\binom{n-1}{k}\binom{n-1}{n-k-1}\binom{k}{j}(-x)^{k-j}x^{n-k-1}\tag7\\
&=\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}(-1)^j\binom{n-1}j\binom{n-1}{n-k-1}\binom{n-j-1}{k-j}x^{n-j-1}\tag8\\
&=\sum_{j=0}^{n-1}(-1)^j\binom{n-1}{j}\binom{2n-j-2}{n-j-1}x^{n-j-1}\tag9\\
&=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{n+k-1}{k}(-x)^k\tag{10}
\end{align}
$$
Explanation:
$\phantom{1}(6)$: $\binom{n}{n-k}\frac{n-k}n=\binom{n-1}{n-k-1}$
$\phantom{1}(7)$: Binomial Theorem: $(1-x)^k=\sum_{j=0}^{n-1}\binom{k}{j}(-x)^{k-j}$
$\phantom{1}(8)$: $\binom{n-1}{k}\binom{k}{j}=\binom{n-1}{j}\binom{n-j-1}{k-j}$
$\phantom{1}(9)$: Vandermonde Identity: $\sum_{k=0}^{n-1}\binom{n-1}{n-k-1}\binom{n-j-1}{k-j}=\binom{2n-j-2}{n-j-1}$
$(10)$: substitute $j\mapsto n-k-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3650275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Is there a way to show that $x^2=1$ is the only solution of $x^2+3=k^2$ for $x,k\in\mathbb{N}$ The question i'm trying to solve is: The sum of all natural numbers a such that $a^2-16a+67$ is a perfect square is (a)10, (b)12, (c)16, (4)22.
$a^2-16a+67$
= $a^2-16a+64+3$
=$(a-8)^2+3$
This is a perfect square when $(a-8)^2=1$ and so $a=7,9$
So the answer is $7+9 = 16$, which agrees with the answer given .
But i don't know how to show that these are the only possible values of a, or rather that $x^2=1$ is the only solution for this $x^2+3=k^2$ is the only value that works, can i get any tips/directions? Thanks to everyone in advance.
| If $k,x \in \Bbb N$ and $3=k^2-x^2$
then $k^2>x^2\,$ so $k>x\,$ so $k\ge x+1\,$ so $k^2\ge (x+1)^2$
so $3=k^2-x^2\ge (x+1)^2-x^2=2x+1$
so $3\ge 2x+1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A function divisible by $p$ I want to ask about what is the intuition for making functions like these
*
*$4^n+2$ is divisible by $3$
*$ 2^{4n+2}+3^{n+2}$ is divisible by $13$
And if so, how can I make my own ones? Thank you all for your time.
| There is a simple explanation using the binomial theorem:
$$
4^n+2 = (3+1)^n+2 = 3a + 1 + 2
$$
and
$$\small
2^{4n+2}+3^{n+2}
=4 \cdot 16^n + 9 \cdot 3^n
=4 \cdot (13+3)^n + 9 \cdot 3^n
=13a + 4 \cdot 3^n + 9 \cdot 3^n
=13a + 13 \cdot 3^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Olympiad inequalities. I am trying to solve this problem but unable to. Can someone please give some hint or help.
I have to use holder inequality.
For $a,b,c$ positive real numbers prove.
$ \frac{a^6}{b^2+c^2} + \frac{b^6}{c^2+a^2} + \frac{c^6}{a^2+b^2} \ge \frac{abc(a+b+c)}{2}$
| By Cauchy-Schwarz inequality (Titu's lemma or Sedrakyan's inequality specifically):
$$F=\frac{a^6}{b^2+c^2}+\frac{b^6}{c^2+a^2}+\frac{c^6}{a^2+b^2} \ge
\frac{(a^3+b^3+c^3)^2}{2(a^2+b^2+c^2)}$$
Next use, $$(a^3+b^3+c^3) \ge \frac {(a^2+b^2+c^2)(a+b+c)}{3}~~\text{(Tchebechev)}$$
So,
$$F \ge \frac{1}{18} \ \underbrace{(a^2+b^2+c^2)}_\text{$AM \ge GM$} \ \underbrace{(a+b+c)}_\text{$AM \ge GM $} (a+b+c)$$
$$ F \ge \frac{1}{18} \left(3(abc)^{\frac{2}{3}}\cdot 3(abc)^{\frac{1}{3}}\cdot(a+b+c)\right) = \frac{abc(a+b+c)}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3660230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Derivative of the inverse of a parametric function with respect to the parameter Given a function
$y = f(x)$
its inverse with respect to the argument x is
$x = F(y) = f^{-1}(y)$
Now, suppose that function has a parameter p
$y = f(x;p)$
whose inverse is
$x = F(y;p) = f^{-1}(y;p)$
How do I compute the parametric derivative of the inverse function
$\partial{F}(y;p)/\partial{p}$
using the easily computable parametric derivative of the original function
$\partial{f}(x;p)/\partial{p}$
where p is actually a vector of parameters. This does not fit the mold of the inverse function theorem as usually presented. It seems too simple to just reciprocate each of the partial derivatives. Could it be something like
$\frac{\partial{F(y;p)}}{\partial{p}}=
-\frac{\partial{f(x;p)}}{\partial{p}} /
\frac{\partial{f(x;p)}}{\partial{x}}
$
?
@Narasimham suggested that I start with a simple problem, then generalize. A simplified version of the problem is
$y = x + k x^3$
with parametric derivative
$d y / d k = x^3$
This is a cubic equation, so luckily there is an analytic inverse -- actually three three of them, but we are only interested in the positive real one.
$x = \frac{2 \sqrt[3]{3} k-\sqrt[3]{2} \left(\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2 y\right)^{2/3}}{6^{2/3} k
\sqrt[3]{\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2 y}}$
Its parametric derivative is
$dx / dk = \frac{-36\ 3^{5/6} k^3 y^2+9 \sqrt[3]{2} \sqrt{3} k^2 y^2 \left(\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2 y\right)^{2/3}+2
\sqrt[3]{2} \sqrt{3} k \left(\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2 y\right)^{2/3}+12 \sqrt[3]{3} k y \sqrt{k^3 \left(27 k
y^2+4\right)}-3 \sqrt[3]{2} y \sqrt{k^3 \left(27 k y^2+4\right)} \left(\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2
y\right)^{2/3}-4\ 3^{5/6} k^2}{6^{2/3} \sqrt{k^3 \left(27 k y^2+4\right)} \left(\sqrt{3} \sqrt{k^3 \left(27 k y^2+4\right)}-9 k^2
y\right)^{4/3}}$
But this gives me no insight. Substituting
$y \to x + k x^3$
doesn't simplify things nor provide any insight into extending this to a nonic polynomial. It seems like I could evaluate this expression from the $dy/dk$ derivative.
| We have the following relations
\begin{align}
y &= f(x; p), \tag{1}\\
x &= f^{-1}(y; p). \tag{2}
\end{align}
By taking derivative of (1) with respect to $p$ and by noting (2), we have
$$0 = \frac{\partial f}{\partial x} \frac{\partial x}{\partial p} + \frac{\partial f}{\partial p}$$
and hence
$$\frac{\partial x}{\partial p} = \frac{\partial f^{-1}(y; p)}{\partial p} = - \frac{\partial f}{\partial p}\cdot \frac{1}{\frac{\partial f}{\partial x}}. \tag{3}$$
Let us see three examples.
Example 1: $y = x + p^2$, its inverse $x = y - p^2$, $\frac{\partial x}{\partial p} = -2p$,
$\frac{\partial f}{\partial p} = 2p$, $\frac{\partial f}{\partial x} = 1$, (3) is valid.
Example 2: $y = x^p$ ($p > 0$, $x > 0$), its inverse $x = y^{1/p}$,
$\frac{\partial x}{\partial p} = y^{1/p}\cdot \frac{-1}{p^2}\ln y$,
$\frac{\partial f}{\partial p} = x^p \ln x$,
$\frac{\partial f}{\partial x} = px^{p-1}$,
$$- \frac{\partial f}{\partial p}\cdot \frac{1}{\frac{\partial f}{\partial x}}
= - x^p \ln x \cdot \frac{1}{px^{p-1}} = - y \cdot \frac{1}{p}\ln y \cdot \frac{1}{p y^{(p-1)/p}}
= - y^{1/p}\cdot \frac{-1}{p^2}\ln y,$$
(3) is valid.
Example 3: $y = x^2 + x(p^2 + p)$ ($p > 0$, $x > 0$), its inverse
$x = - \frac{1}{2}p^2 - \frac{1}{2}p + \frac{1}{2}\sqrt{p^4 + 2p^3 + p^2 + 4y}$, $\cdots$, (3) is valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\frac{d^{100}}{dx^{100}}(\frac{1+x}{\sqrt{1-x}})$ Calculate $$\frac{d^{100}}{dx^{100}}\left(\frac{1+x}{\sqrt{1-x}}\right).$$
I gathered that I can use Leibniz's formula, so the differentiation can be represented by the following sum:
$$ \sum^{100}_{r=0} \binom{100}{r}\left[\frac{d^{100-r}}{dx^{100-r}}(1+x)\right]\left[ \frac{d^r}{dx^r}(1-x)^{-\frac{1}{2}} \right].$$
Since we know that $ \frac{d^{2}}{dx^2}(1+x) = 0 $, we can simplify the above sum to the following:
$$ 0+\ldots+\binom{100}{98}\left[\frac{d^{2}}{dx^2}(1+x)\right] \left[ \frac{d^{98}}{dx^{98}}(1-x)^{-\frac{1}{2}} \right]
+\binom{100}{99}\left[\frac{d}{dx}(1+x)\right] \left[ \frac{d^{99}}{dx^{99}}(1-x)^{-\frac{1}{2}} \right] +\binom{100}{100}(1+x) \left[ \frac{d^{100}}{dx^{100}}(1-x)^{-\frac{1}{2}} \right], $$
which then gives us,
$$ 100\left[ \frac{d^{99}}{dx^{99}}(1-x)^{-\frac{1}{2}} \right] + (1+x)\left[ \frac{d^{100}}{dx^{100}}(1-x)^{-\frac{1}{2}} \right].$$
This is where I am stucked. I'm not sure resolve those differentials.
| Note
$$\frac{1}{\sqrt{1-x}}=\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^n}{2^n n!}$$
and hence
$$\frac{1+x}{\sqrt{1-x}}=(1+x)\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^n}{2^n n!}=\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^n}{2^n n!}+\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^{n+1}}{2^n n!}.$$
So
$$ \frac{d^{100}}{dx^{100}}\frac{1+x}{\sqrt{1-x}}=\cdots. $$
Here
$$ (2n-1)!!=1\cdot3\cdot5\cdots(2n-1). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove this algebraic version of the sine law? How to solve the following problem from Hall and Knight's Higher Algebra?
Suppose that
\begin{align}
a&=zb+yc,\tag{1}\\
b&=xc+za,\tag{2}\\
c&=ya+xb.\tag{3}
\end{align}
Prove that
$$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}.\tag{4}$$
(I suppose that $x,y,z$ are real numbers whose moduli are not equal to $1$.)
I discovered this problem from chapter 3 of Prelude to Mathematics by W. W. Sawyer. Sawyer thought that this problem arose from the sine law: let $a,b,c$ be respectively the lengths of the edges opposite to three vertices $A,B,C$ of a triangle. Define $x=\cos A$ and define $y,z$ analogously. Now equalities $(1)-(3)$ simply relate $a,b$ and $c$ to each other by the cosines of the angles and $(4)$ is just a rewrite of the sine law
$$
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.
$$
However, the algebraic version $(4)$ looks more general. For example, it does not state that $a,b,c$ must be positive or that they must satisfy the triangle inequality.
Sawyer wrote that this isn't a hard problem, but he didn't provide any solution. I can prove $(4)$ using linear algebra. Suppose that $(a,b,c)\ne(0,0,0)$ (otherwise $(4)$ is obvious). Rewrite $(1)-(3)$ in the form of $M\mathbf a=0$:
$$\begin{bmatrix}-1&z&y\\ z&-1&x\\ y&x&-1\end{bmatrix}\begin{bmatrix}a\\ b\\ c\end{bmatrix}=0.$$
Since $x^2,y^2,z^2\ne1$, $M$ has rank $2$ and $D=\operatorname{adj}(M)$ has rank $1$. Hence all columns of $D$ are parallel to $(a,b,c)^T$ and $\frac{d_{11}}{d_{21}}=\frac{d_{12}}{d_{22}}=\frac{a}{b}$. Since $M$ is symmetric, $D$ is symmetric too. Therefore $\frac{1-x^2}{1-y^2}=\frac{d_{11}}{d_{22}}=\frac{d_{11}d_{12}}{d_{21}d_{22}}=\frac{a^2}{b^2}$, i.e. $\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}$.
As this problem comes from Hall and Knight's book, I think there should be a more elementary solution. Any ideas?
| We can write
I) $z=\frac{a-yc}{b}$ from (1)
Now, multiplying $y$ on both sides of (3) we get $yc=y^2a+xyb$.
So, from I) we get $\frac{a-ay^2}{b}-xy=z$.....(1')
Similarly from equation (2) we get
II) $z=\frac{b-xc}{a}$ from (2)
Now, multiplying $x$ on both sides of (3) we get $xc=xya+x^2b$.
And we get from (2) $\frac{b-bx^2}{a}-xy=z$.....(2')
From (1') and (2') we get,
$\frac{a-ay^2}{b}=\frac{b-bx^2}{a} \rightarrow \frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}$.
Similarly, $\frac{a^2}{1-x^2}=\frac{c^2}{1-z^2}$
So, ultimately we have proved
$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}$
Suppose, $b=0$, then $\frac{a}{c}=y=\frac{c}{a}$ or $y=1$. So, $\frac{b^2}{1-y^2}$ will be undefined.
If also $c=0$ then $a=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 1
} |
If $a^2 + b^2 + c^2$ is divisible by $16$, then show that$ a^3 + b^3 + c^3$ is divisible by $64$. Where $a, b, c \in \mathbb{Z}$. If $a^2 + b^2 + c^2$ is divisible by $16$, then show that $a^3 + b^3 + c^3$ is divisible by $64$; where $a, b, c \in \mathbb{Z}$.
I began by proving that if $(a^3+b^3+c^3) -(a^2+b^2+c^2)$ is divisible $16$, then $a^3+b^3+c^3$ will be divisible by $64$. That is equal to $a^3-a^2+b^3-b^2+c^3-c^2$ $= a(a^2-a)+b(b^2-b)+c(c^2-c)$.
But this doesn't give a clear path for the question.
Any help?
| I don't see any way offhand to continue with what you started to finish the proof.
Instead, note that perfect squares have remainders of only $0$, $1$, $4$ or $9$ when divided by $16$. As such, if $a^2$ had a remainder of $1$, the other $2$ would need to have remainders which sum to $15$. This is not possible as the only options are $0$ plus $1$, $4$ or $9$, $1 + 1 = 2$, $1 + 4 = 5$, $1 + 9 = 10$, $4 + 4 = 8$, $4 + 9 = 13$ and $9 + 9 = 18 \equiv 2 \pmod{16}$. Similarly, if $a^2$ had a remainder or $4$ or $9$, the other $2$ also can't add up to give a multiple of $16$.
This means that all of $a^2$, $b^2$ and $c^2$ must have a remainder of $0$ when divided by $16$, i.e., each of $a$, $b$ and $c$ are multiples of $4$. Thus, each of $a^3$, $b^3$ and $c^3$ must be a multiple of $4^3$, i.e., $64$, so their sum is as well.
Update: As Will Jagy's comment indicates, since perfect squares have remainders of only $0$ or $1$ when divided by $4$, this means $a^2 + b^2 + c^2$ can be a perfect square only if $a$, $b$ and $c$ are all even. If you factor out $4$, then for $\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 + \left(\frac{c}{2}\right)^2$ to be a multiple of $4$ again requires each half value to be even. This means each of $a$, $b$ and $c$ has to be a multiple of $4$ which, as before, results in $a^3 + b^3 + c^3$ being a multiple of $64$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that the value of $\Delta$ is an integer for the given determinant
$\Delta=\begin{vmatrix} \sqrt 6& 2i& 3+\sqrt 6 \\ \sqrt{12}&\sqrt 3 +\sqrt8i &3\sqrt 2 +\sqrt 6i \\ \sqrt{18} &\sqrt 2+ \sqrt {12}i &\sqrt {27}+2i \end{vmatrix}$
taking $\sqrt 6$ out from the first column and performing the following operations
$$R_2\rightarrow R_2 -\sqrt 2 R_1$$
$$R_3 \rightarrow R_3-\sqrt 3 R_1$$
$$\sqrt 6\begin {vmatrix} 1&2i&3+\sqrt 6 \\\ 0&\sqrt 3&\sqrt 6i-2\sqrt 3 \\\ 0&\sqrt 2&2i-3\sqrt 2 \end {vmatrix}$$
What should I do next?
| The obvious way is to simply find the determinant right away. But the fact that you have asked this must mean that must not be allowed. So to prove this without finding the determinant at all:
Take a factor of $\sqrt3$ from $R_2$ and $\sqrt2$ from $R_3$.
$$\Delta=6\begin{vmatrix} 1 & 2i & 3+\sqrt6 \\
0 & 1 & \sqrt2 i-2\\
0 &1 & \sqrt2 i-3 \end{vmatrix}$$
Now perform $R_2 \to R_2-R_3$
$$=6\begin{vmatrix} 1 & 2i & 3+\sqrt6\\
0 &0 & 1\\
0&1& \sqrt2 i -3 \end{vmatrix}$$
Now $C_2 \to C_2- 2iC_1$, $C_3 \to C_3 -(3+\sqrt6)C_1$
$$=6\begin{vmatrix} 1 &0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & \sqrt2 i-3 \end{vmatrix}$$
Finally : $C_3 \to C_3 - (\sqrt2 i - 3)C_2 $
$$\Delta = 6\begin{vmatrix} 1 & 0 & 0 \\ 0 & 0& 1 \\ 0&1&0 \end{vmatrix}$$
Now this determinant will obviously be a integral value (as it will be a linear combination of product of integers)
NOTE: I am trying to get rid of the non-integral terms using the 1 0 0 columns.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3680882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$ For $a,b,c>0$$,$ prove$:$ $$(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$$
My proof by S-S method$,$ see here.
Another proof by $pqr$ method$:$
Let $p=a+b+c,\,q=ab+bc+ca,\, r=abc.$ This inequality equivalent to$:$ $${p}^{6}-4\,{p}^{4}q+8\,{p}^{3}r+27\,{r}^{2} \geqq 0$$
Or$:$ $${\frac { \left( {p}^{4}-5\,{p}^{2}q+6\,pr+4\,{q}^{2} \right)
\left( 7\,{p}^{4}+45\,{p}^{2}q+54\,pr-36\,{q}^{2} \right) }{12{p}^{2}}}
+\,{\frac { \left( {p}^{2}-3\,q \right) \left( 5\,{p}^{2}-3\,q
\right) \left( {p}^{2}-4\,q \right) ^{2}}{12{p}^{2}}} \geqq 0$$
Which is obvious because $p^2 \geqq 3q,\, p^4 -5p^2 q+6pr+4q^2 \geqq 0 \,(\text{Schur degree 4})$
I hope for another proof (without $uvw$!). Thanks for a real lot!
PS$:$ You can get $pqr$'s form more faster by using Maple$,$ see here.
| Geometric approach :
Let $a,b,c$ be the side of an triangle $ABC$ then your inequality is :
$$T\leq\frac{3\sqrt{3}abc}{4(a+b+c)}$$
Or $$4\sqrt{3}T\leq \frac{9abc}{a+b+c}$$
Where $T$ is the area of the triangle $ABC$
For a proof see (found on Wikipedia):
Posamentier, Alfred S. and Lehmann, Ingmar. The Secrets of Triangles, Prometheus Books, 2012.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $AD\cdot BD \cdot CD \leq \dfrac{32}{27}$ where $ABC$ is a triangle of circumradius 1 and $D\in (BC)$.
Let triangle $ABC$ of circumradius $1$ and $D$ a point on side $(BC)$.
Prove that $$AD\cdot BD\cdot CD\leq \dfrac{32}{27}.$$
My idea. By letting $\alpha = \dfrac{BD}{BC}$ (of course $0<\alpha <1$) we get $BD=BC\cdot \alpha
, \enspace CD=BC\cdot(1-\alpha)\tag{1}$ and also
$$\overrightarrow{AD}=(1-\alpha)\cdot\overrightarrow{AB}+\alpha\cdot \overrightarrow{AC}.$$
By squaring this relation we have that $$AD^2=AB^2(1-\alpha)+AC^2\alpha+BC^2(\alpha^2-\alpha). \tag{$2$}$$
By law of sines we also have $AB=2\sin C$, $AC=2\sin B$ and $BC=2\sin A$.
Now combining with $(1)$ and $(2)$ we may rewrite the desired inequality as follows:
$$((1-\alpha)\sin^2C+\alpha\sin^2B+(\alpha^2-\alpha)\sin^2A)\cdot\alpha^2(1-\alpha)^2\sin^4A\leq \dfrac{2^4}{27^2}.$$
This is where I got stuck. Maybe we could also use the fact that $\sin A=\sin (\pi -B-C)=-\sin(B+C)=-(\sin B\cos C+\sin C\cos B)$ to get rid of $\sin A$?
Thank you in advance!
|
Let $M$ the second intersection of $AD$ and the circumcircle of triangle $ABC$. By power of a point, $BD \cdot DC=AD\cdot DM$. Let $x=AD$ and $y=DM$. The inequality can be rewritten as $$x^2y\leq \dfrac{32}{27}.$$
It is clear that $$(x-2y)^2(4x+y)\geq 0$$ which is equivalent to $$x^2y\leq (x+y)^3\cdot \dfrac{4}{27}$$ and combining with $x+y\leq 2$ (which is true as $x+y=AM$ is not greater than the diameter of the circle) we get the conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$x-\sin(x) \geq \dfrac{x^3}{(x+\pi)^2}$ Let $x \geq 0.$ I need to prove that $x-\sin(x)\geq\dfrac{x^3}{(\pi+x)^2}.$
I tried the derivative, of $f(x)=x-\sin(x)-\dfrac{x^3}{(\pi+x)^2}$ which is $1-\cos(x)-\dfrac{x^2(x+3\pi)}{(\pi+x)^3},$ but it has a complicated formula.
Any ideas, hints?
Edit: sorry, there was a mistake in the derivative, I corrected it.
| Put $x=z-\pi$; it suffices to show the inequality for for $z\geq \pi$.
$$
z-\pi -\sin(z-\pi)\geq \frac{(z-\pi)^3}{z^2}
$$
$$
z-\pi +\sin(z)\geq z -3\pi+\frac{3\pi^2}{z} -\frac{\pi^3}{z^2}
$$
$$
2\pi +\sin(z)\geq \frac{3\pi^2}{z} -\frac{\pi^3}{z^2}
$$Note that $\sin(z)\geq -1$. If we set
$$
2\pi- 1=\frac{3\pi^2}{z} -\frac{\pi^3}{z^2},
$$the last solution is $z=\frac{3 π^2 + π^{3/2} \sqrt{4 + π}}{4 π - 2}(:=z_0)\approx 4.21$. But on $(\pi,z_0)$, we can use the approximation $\sin(z)> (\pi-z)-\frac{(\pi-z)^3}{7}$, whence
$$
2\pi +\sin(z) - \frac{3\pi^2}{z} +\frac{\pi^3}{z^2}
$$
$$
> 2\pi + (\pi-z)-\frac{(\pi-z)^3}{7}-\frac{3\pi^2}{z} +\frac{\pi^3}{z^2}
$$
$$
=\frac{(z-\pi )^3 \left(z^2-7\right)}{7 z^2}>0
$$
So there are no solutions on $(\pi,z_0)$, we don't actually have equality at $z_0$, and we know there are no solutions for $z>z_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Sigmoid Function Typically sigmoid function is calculated as
1/(1 + exp(-x))
I see sometimes it is calculated as
1 - 1/(1 + exp(x))
or even
exp(x)/(1 + exp(x))
Could you clarify the difference please?
| $$
\frac
{1}{1+\exp(-x)}
=
\frac
{1}{1+e^{-x}}
=
\frac
{1}{1+e^{-x}}
\cdot
\frac
{e^x}{e^x}
=
\frac
{e^x}{e^x+1}
=
\frac
{e^x}{1+e^x}
=
\frac
{\exp(x)}{1+\exp{x}}
$$
and also
$$
1-\frac
{1}{1+\exp(x)}
=
1-\frac
{1}{1+e^{x}}
=
\frac{1+e^x}{1+e^x}
-
\frac{1}{1+e^x}
=
\frac{(1+e^x)-1}{1+e^x}
=
\frac
{e^x}{1+e^x}
=
\frac
{\exp(x)}{1+\exp{x}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integer solution of $a+b+c=15$ with restrictions on $a, b$ and $c$ I'm checking this problem:
Find the integer solutions of $a+b+c=15$ if $a$ is multiple of 3, $b$ is less than 10 and $c$ is multiple of 2. With $a, b, c ≥ 0$
We can make a series of polynomials with combinatorics. The polynomials would be:
$$P_a(x)=1+x^3+x^6+x^9+\dots$$
$$P_b(x) = 1 +x+x^2+x^3+\dots + x^9 = \frac{1-x^{10}}{1-x}$$
$$P_c(x) = 1+x^2 + x^4 + x^6 + \dots$$
I want to know if my $P_a(x)$ and $P_c(x)$ are stated correctly and how can they be simplified so that I get to $x^{15}$ in an easier way.
Big thanks for your help.
| Can't this simply be done by making use of the fact that $b + c = 15 - a$ is divisible by $3$ to conclude that $15-a$ can only equal one of $0,$ $3,$ $6,$ $9,$ $12,$ $15$? In the lists below, keep in mind that the values are nonnegative and $b = 0,\,1,\,\ldots,\,9$ and $c$ is even. Thus, for example, if $b+c = 9,$ then $c$ can only be $0,\,2,\,4,\,6,\,8$ because $c \geq 0$ is even and $b \leq 9.$ Each of these $5$ values of $c$ in this example is also easily seen correspond to a unique solution for $(a,b,c).$
$a=15$ and $b+c=0$
There is only $1$ solution, namely $(b,c) = (0,0).$
$a=12$ and $b+c=3$
There are $2$ solutions, namely $(b,c) = (1,2),\;(3,0).$
$a=9$ and $b+c=6$
There are $4$ solutions, namely $(b,c) = (0,6),\;(2,4),\;(4,2),\;(6,0).$
$a=6$ and $b+c=9$
There are $5$ solutions, namely $(b,c) = (1,8),\;(3,6),\;(5,4),\;(7,2),\,(9,0).$
$a=3$ and $b+c=12$
There are $5$ solutions, namely $(b,c) = (0,12),\;(2,10),\;(4,8),\;(6,6),\,(8,4).$ (No more, because we must have $b \leq 9.)$
$a=0$ and $b+c=15$
There are $5$ solutions, namely $(b,c) = (1,14),\;(3,12),\;(5,10),\;(7,8),\,(9,6).$
Therefore, altogether there are $1+2+4+5+5+5 = 22$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3685534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding $\lim_{(x,y) \to (0,0)} \frac{x^5y^5}{|x|^9 + |y|^{11}}$ Setting $y = mx^k$ suggests the limit evaluates to $0$, and as far as the techniques I know, dismisses the possibility of its non-existence. But I'm having trouble using the Squeeze Theorem to prove that it does exist.
$$\lim_{(x,y) \to (0,0)} \frac{x^5y^5}{|x|^9 + |y|^{11}}$$
Major Edit: $(x,y) \to (0,0)$ not $(x,y) \to \infty$
| By weighted AM-GM inequality, we get
\begin{align*}
\lvert x \rvert^9 + \lvert y \rvert^{11}
&= \frac{11}{20} \cdot \frac{20}{11} \lvert x \rvert^9 + \frac{9}{20} \cdot \frac{20}{9} \lvert y \rvert^{11} \\
&\geq \left(\frac{20}{11} \lvert x \rvert^9\right)^{11/20} \left(\frac{20}{9} \lvert y \rvert^{11}\right)^{9/20} \\
&= \text{const} \cdot \lvert x \rvert^{99/20} \lvert y \rvert^{99/20}.
\end{align*}
Hence
\begin{align*}
\frac{\lvert x \rvert^5 \lvert y \rvert^5}{\lvert x \rvert^9 + \lvert y \rvert^{11}}
\leq \text{const} \cdot \lvert x \rvert^{1/20} \lvert y \rvert^{1/20},
\end{align*}
which shows that
$$
\lim_{(x, y) \to (0, 0)} \frac{x^5 y^5}{\lvert x \rvert^9 + \lvert y \rvert^{11}} = 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Modular calculation high exponent? I want to show, that $5^{96}\equiv -1 \pmod{193}$, without using the formula for quadratic residue.
So far I have :
$5^{96}\equiv 5^{4\cdot24} \equiv 625^{24}\equiv 46^{24}\equiv 186^{12}\equiv -7^{12}\equiv 7^{12}\equiv 7^{3\cdot4}\equiv 150^4\equiv -43^4\equiv 43^4\equiv 112^2\equiv -81^2\equiv 3^8\\\
$
I think Euler's totient doesn't help, as $\varphi(193)=192>96=\frac{192}{2}$ or can I write this ?
$5^{96}\equiv 5^{192-96} \equiv 5^{-96}\equiv(5^{-1})^{96}\equiv 116^{96}\equiv -77^{96} \pmod{193} $
What am I doing wrong ? What am I missing ? Thanks in advance.
| Another idea: since $96$ is a multiple of $3$ and you know $(5^{96})^2\equiv 1$, you have $(5^{32})^3\pm 1\equiv 0$. Factor this using the familiar factorization for the sum or difference of cubes:
$5^{96}\pm1=(5^{32}\pm1)×(5^{64}\mp5^{32}+1)$
Now square $5$ six times to get
$5^2\equiv 25, 5^4\equiv 46, 5^8\equiv 186\equiv -7, 5^{16}\equiv 49, 5^{32}\equiv 85, 5^{64}\equiv 84$.
Then $5^{64}-5^{32}+1\equiv 0$ and this is a factor of $5^{96}+1$, forcing $5^{96}\equiv -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help with complex contour integral involving a logarithm and a square root I've been trying to do the following integral using the 'keyhole' contour: $\int_0^{\infty} \frac{\ln(x)}{\sqrt{x}(1+x^3)}d x.$ I know the result is $-\frac{2\pi^2}{3\sqrt{3}},$ yet when I use the Residue Theorem the result amounts to something else. I chose the branch cut on the real positive axis. I calculated the residues: $\frac{\pi}{9} e^{i5\pi/3}, \frac{\pi}{3}, - \frac{5\pi}{9} e^{i\pi/3}$. Then I applied the theorem: the integral over the contour is the sum of the residues times $2\pi i$. It is pretty clear that the integrals over the arcs of the keyhole approach zero as their radii approach infinity and zero, so all that's left are the integrals over the segments. The integrals over the segments, as the keyhole 'closes' converge to $\int_{r}^{R} \frac{\ln(x)}{\sqrt{x}(1+x^3)}d x$ and $\int_{r}^{R} \frac{\ln(x)+2\pi i}{\sqrt{x}(1+x^3)}d x.$ So we should have $2\pi i \big[ \frac{\pi}{9} e^{i5\pi/3}+ \frac{\pi}{3} - \frac{5\pi}{9} e^{i\pi/3}] = 2\int_{0}^{\infty} \frac{\ln(x)}{\sqrt{x}(1+x^3)}dx + 2\pi i\int_{r}^{R} \frac{dx}{\sqrt{x}(1+x^3)}.$ However when I solve for the integrals on the right I don't get the right value. Where have I made mistakes? Please help!
| We seek to compute using contour integration the integral
$$J = \int_0^\infty \frac{\log{x}}{\sqrt{x}(x^3+1)} \; dx$$
We work with
$$f(z) = \frac{\mathrm{Log}(z)}
{\exp(\mathrm{Log}(z)/2) (z^3+1)}$$
where $\mathrm{Log}(z)$ is the branch with argument in $[0,2\pi).$ We
use a keyhole contour wih radius $R$ and the slot on the positive real
axis. Let $\Gamma_0$ be the segment on the real axis up to $R$,
$\Gamma_1$ the big circle of radius $R$, $\Gamma_2$ the segment below
the real axis coming in from $R$ and finally $\Gamma_3$ the small
circle of radius $\epsilon$ enclosing the origin. We then have with
$$\rho_k = \exp(\pi i/3+ 2\pi i k/3)$$
that
$$\left(\int_{\Gamma_0}
+ \int_{\Gamma_1}+ \int_{\Gamma_2}+ \int_{\Gamma_3}\right) f(z) \; dz
= 2\pi i \times
\sum_{k=0}^2 \mathrm{Res}_{z=\rho_k} f(z).$$
The poles are all simple and we get for the residues
$$\frac{\pi i/3 + 2\pi i k/3}
{\exp(\pi i/6 + \pi i k/3) 3\rho_k^2}
= - \frac{1}{3} \frac{(\pi i/3 + 2\pi i k/3)\rho_k}
{\exp(\pi i/6 + \pi i k/3)}
\\ = - \frac{1}{3} (\pi i/3 + 2\pi i k/3)
\exp(\pi i/6 + \pi i k/3)
\\ = - \frac{1}{9} (\pi i + 2\pi i k)
\exp(\pi i/6 + \pi i k/3)
.$$
Evaluating these we obtain
$$\begin{array}{rll}
\alpha_0 & = & \mathrm{Res}_{z=\rho_0} f(z)
= -\frac{1}{9} \pi i \exp(\pi i/6) \\
\alpha_1 & = & \mathrm{Res}_{z=\rho_1} f(z)
= -\frac{1}{3} \pi i \exp(\pi i/2)
= \frac{\pi}{3} \\
\alpha_2 & = & \mathrm{Res}_{z=\rho_2} f(z)
= -\frac{5}{9} \pi i \exp(5 \pi i/6).
\end{array}$$
Now observe that in the limit for $\Gamma_0$ and $\Gamma_2$
$$\int_{\Gamma_0} f(z) \; dz = J$$
and
$$\int_{\Gamma_2} f(z) \; dz =
\int_{\infty}^0 \exp(-\pi i)
\frac{\log x + 2\pi i}{\sqrt{x}(x^3+1)} \; dx
\\ = \int_0^\infty
\frac{\log x + 2\pi i}{\sqrt{x}(x^3+1)} \; dx
\\ = J + 2\pi i \int_0^\infty \frac{1}{\sqrt{x}(x^3+1)} \; dx
= J + 2\pi i K,$$
where $K$ is a real number, as is of course, $J.$
For the large circle of radius $R$ we have by the ML-estimate on
$\Gamma_1$ the bound $\lim_{R\rightarrow\infty} 2\pi R\times
\frac{\log R + 2\pi}{\sqrt{R}(R^3-1)} = 0,$ so this vanishes. For the
small circle $\Gamma_3$ of radius $\epsilon$ we find
$\lim_{\epsilon\rightarrow 0} 2\pi \epsilon \times
\frac{|\log\epsilon|+2\pi}{\sqrt{\epsilon}(1-\epsilon^3)} = 0,$ and
this too vanishes.
The conclusion is that
$$2J + 2\pi i K
= 2\pi i\times (\alpha_0+\alpha_1+\alpha_2).$$
or
$$J = \frac{1}{2}
\Re\left(2\pi i\times
(\alpha_0+\alpha_1+\alpha_2)\right)
= - \pi
\Im\left(\alpha_0+\alpha_1+\alpha_2\right)
\\ = \frac{\pi^2}{9} (\cos(\pi/6) + 5 \cos(5 \pi/6))
= \frac{\pi^2}{9}
\left(\frac{\sqrt{3}}{2}-5\frac{\sqrt{3}}{2}\right)
= - \frac{\pi^2}{9} 2\sqrt{3}.$$
This at last yields the closed form
$$\bbox[5px,border:2px solid #00A000]{
J = - \frac{2\pi^2}{3\sqrt{3}}.}$$
We also get $K$ as a bonus integral as in
$$K= \frac{1}{2\pi} \Im(2\pi i \times (\alpha_0+\alpha_1+\alpha_2))
= \Re(\alpha_0+\alpha_1+\alpha_2)
\\ = \frac{\pi}{3}
+ \frac{1}{9} \pi \sin(\pi / 6)
+ \frac{5}{9} \pi \sin(5\pi / 6)
= \frac{\pi}{3}
+ \frac{1}{9} \pi \frac{1}{2}
+ \frac{5}{9} \pi \frac{1}{2}.$$
We get the closed form
$$\bbox[5px,border:2px solid #00A000]{
K = \frac{2}{3} \pi.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
integral of $x^{1/n}\sin x$ how to prove $\lim_{n \to \infty} \int_0^{\pi/2} x^{1/n}\sin x dx = 1$? I've tried it by using squeeze theorm. It worked on right side by putting $\pi/2$ into $x$, but i have no idea for left side
| Here's an elementary approach (without using any measure theory):
Since $\int_0^\frac{\pi}{2} \sin(x)dx=1$, we have that
$$
\left| \int_0^\frac{\pi}{2} x^\frac{1}{n}\sin(x)dx-1\right|=\left|\int_0^\frac{\pi}{2}\sin(x)\left(x^\frac{1}{n}-1\right)dx\right|\le \int_0^\frac{\pi}{2}\left|x^\frac{1}{n}-1\right|dx
$$
were the last inequality follows from $\left|\int f\right|\le \int |f|$ and $|\sin(x)|\le 1$.
Since $x^\frac{1}{n}\ge 1 \iff x\ge 1$, we can write
$$
\int_0^\frac{\pi}{2}\left|x^\frac{1}{n}-1\right|dx =\int_0^1 1-x^\frac{1}{n}dx+\int_1^\frac{\pi}{2}x^\frac{1}{n}-1dx
$$
$$
=1-\left[\frac{x^{\frac{1}{n}+1}}{\frac{1}{n}+1} \right]_0^1+\left[\frac{x^{\frac{1}{n}+1}}{\frac{1}{n}+1} \right]_1^\frac{\pi}{2}-\frac{\pi}{2}+1
$$
$$
=1-\frac{1}{\frac{1}{n}+1}+\frac{\left(\frac{\pi}{2}\right)^{\frac{1}{n}+1}}{\frac{1}{n}+1}-\frac{1}{\frac{1}{n}+1}-\frac{\pi}{2}+1
$$
Letting $n\to\infty$, we see that the last expression tends to $0$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is this series convergent? $1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + ...$ Is this series convergent? How to prove?
$$1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{8} \ (8 \times 1/8) + \frac{1}{16} + ...$$
It's equal to $\{1 + 1 + 1 + 1 + ...\}$, when I take the $(n+1)^{th}$ number and $n^{th}$ number, the ratio of them can be $\frac{1}{2}$ or $1$. I don't think this is a convergent series, but what is it?
Thanks!
| Observe that the $(2^n - 1)$th partial sum is given by $$s_{2^n - 1} = 1 + 2 \cdot \frac 1 2 + 4 \cdot \frac 1 4 + \cdots + 2^{n - 1} \cdot \frac 1 {2^{n - 1}} = \underbrace{1 + 1 + 1 + \cdots + 1}_{n \text{ summands}} = n.$$ Consequently, we have that $\lim_{n \to \infty} s_{2^n - 1} = \lim_{n \to \infty} n = \infty,$ and the series diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Leading behavior of a logarithmic integral The integral
$$\int_0^1 du \frac{\ln u}{u^2+b^2}$$
is governed by the divergence at $u=0$. I'm interested in the leading behavior of the integral, and it seems reasonable to expand around $u=0$ (as it is typically done in Laplace's method for instance), but what should be done next?
| Here's my attempt. First we start off by integrating by parts.
\begin{equation}
\int_0^1 \frac{\ln u}{u^2+b^2} du = \frac{1}{b} \arctan\left( \frac{u}{b} \right) \ln u \bigg |_0^1 - \frac{1}{b} \int_0^1 \frac{\arctan\left(\frac{u}{b} \right)}{u} du.
\end{equation}
Assuming $b\neq0$, we can evaluate the limit
\begin{equation}
\lim_{u\rightarrow 1} \arctan\left( \frac{u}{b} \right) \ln u =0, \\
\lim_{u \rightarrow 0} \arctan\left( \frac{u}{b} \right) \ln u \text{ - undefined}.
\end{equation}
Applying L'Hôpital's rule to the second limit
\begin{equation}
\lim_{u \rightarrow 0} \frac{1}{b} \arctan\left( \frac{u}{b} \right) \ln u = \lim_{u \rightarrow 0} \frac{1}{b} \frac{\ln u}{\frac{1}{\arctan\left(\frac{u}{b} \right)}} =\frac{1}{b} \lim_{u \rightarrow 0} \frac{ \frac{1}{u}}{-\left(\frac{1}{\arctan\left(\frac{u}{b} \right)} \right)^2 \frac{b}{u^2+b^2} }\\
= -\frac{1}{b^2} \lim_{u\rightarrow 0} \frac{\arctan \left( \frac{u}{b} \right)^2 (u^2+b^2)}{u}=0,
\end{equation}
since $\arctan(x) \sim x$, for small $x$.
We see that
\begin{equation}
\int_0^1 \frac{\ln u}{u^2+b^2} du = - \frac{1}{b} \int_0^1 \frac{\arctan\left( \frac{u}{b}\right)}{u} du.
\end{equation}
Since you asked for a power series, taylor expanding gives us the following result:
\begin{equation}
-\frac{1}{b} \int_0^1 \frac{\arctan\left( \frac{u}{b} \right)}{u} = -\frac{1}{b}\int_0^1 \frac{1}{u}\left( \left(\frac{u}{b}\right) - \frac{1}{3}\left(\frac{u}{b}\right)^3 + \frac{1}{5} \left(\frac{u}{b}\right)^5 + O(u^7) \right) \\
=-\frac{1}{b^2}\int_0^1 \left(1 - \frac{1}{3b^2} u^2 + \frac{1}{5b^4} u^4 + O(u^6) \right)\\
= - \frac{1}{b^2} \left(1 - \frac{1}{3^2 b^2} + \frac{1}{5^2 b^4} + O\left(\frac{1}{b^6} \right) \right)
= - \frac{1}{b^2} \sum_{i=0}^{\infty} \frac{(-1)^{i}}{(2i+1)^{2i} b^{2i}}
\end{equation}
Note that the $b\rightarrow0$ divergence makes sense, since the initial integral is also divergent for $b=0$. I suspect that this could probably be reexpressed in terms of the zeta function for the case of $b=1$.
I hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
I saw steps in an equation that added a 1 and a -1. Why is this okay? $(x^2+x)-(x+1)$=
$(x \cdot x + x \cdot 1)+(-1)\cdot(x+1)$
Does this have to be a multiplication by +1 and -1? I'm not sure why the - in the middle changed to a + when the (-1) appeared.
(Sorry I don't know the notation for the dot)
| It might be easier to keep track of everything if you use a number instead of the variable $x$. If we substitute $x = 3$, the equation becomes
$$
(3^2+3)-(3+1) = (3 \cdot 3 + 3 \cdot 1)+(-1)\cdot(3+1).
$$
It's easy to see that $(3^2 + 3)$ is the same as $(3 \cdot 3 + 3 \cdot 1)$. $3 \cdot 3$ and $3^2$ are both just $9$, and multiplying $3 \cdot 1$ just leaves us with $3$. So, $3^2 + 3$ and $3 \cdot 3 + 3\cdot 1$ are both just $9+3 = 12$.
Now, on the left side we're supposed to take $12$ and subtract $3 + 1$. On the right side, we're supposed to take $12$ and add $(-1)\cdot (3+1)$. In other words, on one side we subtract $4$, and on the other we add $-4$.
Can you see why these operations lead to the same result?
The point now is that there's nothing special about the number $3$ here; this would work if we plugged anything else in. In other words, it is always true that $x^2 = x \cdot x$, that $x \cdot 1 = x$, and that adding $(-1)\cdot (x+1)$ is the same as subtracting $x+1$. So, it is indeed true that
$$
(x^2 + x) - (x+1) = (x\cdot x + x\cdot 1) + (-1) \cdot (x+1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$F(x , y)=\cos ^{y}\left(\frac{\pi}{x}\right)+\cos ^{y}\left(\frac{3 \pi}{x}\right)+\cos ^{y}\left(\frac{5 \pi}{x}\right)$ I am stuck with this problem , may you help me , it is from Spanish math olympiad ....
Given
$F(x , y)=\cos ^{y}\left(\frac{\pi}{x}\right)+\cos ^{y}\left(\frac{3 \pi}{x}\right)+\cos ^{y}\left(\frac{5 \pi}{x}\right)$
Calculate
$M=F(7 , 2)+F(7 , 3)+F(7 , 5)-F(7 , 6)$
a) $37 / 32$
b) $7 / 4$
c) $19 / 16$
d) $53 / 32$
e)$41 / 32$
| First let's define
$$p_n:=\left(2\cos\frac\pi7\right)^n+\left(2\cos\frac{5\pi}7\right)^n+\left(2\cos\frac{3\pi}7\right)^n\quad(n\in\mathbb{Z}) $$
$$p_0=3,\quad p_1=1,\quad p_2=5$$
$$\forall n\in\mathbb{Z}c\quad p_{n+3}-p_{n+2}-2p_{n+1}+p_n=0$$
Then your sum is :
$$M=\dfrac{p_2}{4}+\dfrac{p_3}{8}+\dfrac{p_5}{32}-\dfrac{p_6}{64} $$
So
$$ M= \dfrac{5}{4}+\dfrac{4}{8}+\dfrac{16}{32}-\dfrac{38}{64}=\dfrac{106}{64}=\dfrac{53}{32} $$
Nota bene:
The expression of $p_n$ can be deduced to the fact that :
$$x^3-\frac{x^2}2-\frac x2+\frac18=0$$
has for roots $$\cos\frac\pi7,\quad-\cos\frac{2\pi}7,\quad\cos\frac{3\pi}7$$
and using the relation between coefficient of your polynom and the roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $| P ( A )+2 I |$ Given $A=\left[\begin{array}{ccc}4 & 1 & 0 \\ 2 & -1 & 1 \\ 0 & 2 & 1\end{array}\right]$
And
$P(x)=x^{3}-3 x+4$
Calculate $| P ( A )+2 I |$
a) $1648$
b) $1624$
c) $1528$
d) $1728$
e) $1696$
My try
$P(A)+2I=A^{3}-3 A+4+2I$
$P(A)+2I=A^{3}-3 A+4I+2I$
$P(A)+2I=A^{3}-3 A+6I$
= ${\left[\begin{array}{ccc}4 & 1 & 0 \\ 2 & -1 & 1 \\ 0 & 2 & 1\end{array}\right] }^{3} - 3{ \left[\begin{array}{ccc}4 & 1 & 0 \\ 2 & -1 & 1 \\ 0 & 2 & 1\end{array}\right]}+6I$
$P(A)+2I={\left[\begin{array}{ccc}72 & 14 & 4 \\ 28 & 10 & 2 \\ 16 & 4 & 6\end{array}\right] }$
$| P ( A )+2 I |$= $
\text{Det}(\left(
\begin{array}{ccc}
72 & 14 & 4 \\
28 & 10 & 2 \\
16 & 4 & 6
\end{array}
\right) )=1648$
Is there a short method for solving this problem since "cubing" $A$ is a bit time consuming .... thank you
| It's not a lot easier, but here is the calculation using the characteristic polynomial and eigenvalues of $A$. We calculate the characteristic polynomial
$$
|xI - A| = x^3 - 4x^2 - 5x + 14 = (x-2)(x-6x + 7).
$$
By the Cayley Hamilton theorem,
$$
P(A) + 2I = A^3 - 3A + 6I =
A^3 - 3A + 6I - 0
\\=
(A^3 - 3A + 6I) - (A^3 - 4A^2 - 5A + 14 I)
\\= 4A^2 + 2A - 8I = 2(2A^2 + A - 4I).
$$
Thus, we want to calculate the determinant of $P(A) + 2I = 2(2A^2 + A - 4I)$. Let $q(x) = 2(2x^2 + x - 4)$.
We find that the eigenvalues (the zeros of the characteristic polynomial) are $\lambda_1=-2, \lambda_2 = 3 + \sqrt{2}, \lambda_3 = 3 - \sqrt{2}$. The eigenvalues of $q(A)$ are equal to $q(\lambda_j)$ for $j = 1,2,3$, i.e.
$$
q(\lambda_1) = 4, \quad q(\lambda_2) = 42 + 26\sqrt{2}, \quad q(\lambda_3) = 42 - 26\sqrt{2}.
$$
The determinant of $q(A)$ is the product of these eigenvalues of $q(A)$, namely
$$
2^2 (2\cdot 21 + 2 \cdot 13\sqrt{2})(2\cdot 21 - 2 \cdot 13\sqrt{2}) = 2^4(21^2 - 2\cdot 13^2) = 16 \cdot 103 = 1648.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Maximum ellipse inscribed in Witch of Agnesi curve An ellipse with variable $(2a,2b)$ axes parallel to the $(x,y)$ coordinate axes is inscribed inside fixed curve of equation.
$$ y=\pm\dfrac{1}{1+x^2}$$
Show that maximum ellipse area occurs when it touches the curve at its inflection point.
I am looking to generalizing a variable ellipse contact point with a curve having an inflection, like in the recent Bell Curve post. My intuition needs to be validated or disproved later using simple methods of differential calculus.
| Take $ f $ an even function and consider $ \mathcal E $ the ellipse that touches $ f $ in $ (c, f(c)) $. Suppose it has equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ (and thus an area of $ ab $). You get, from $ (c, f(c)) \in \mathcal E $ and from $ f $ tangent to $ \mathcal E $,
$$ \frac{c^2}{a^2} + \frac{f(c)^2}{b^2} = 1 $$
$$ \frac c{a^2} + \frac{f(c)f'(c)}{b^2} = 0 $$
Solving this, you find $ a^2 = c^2 - \frac{c f(c)}{f'(c)} $ and $ b^2 = f(c)^2 - cf(c)f'(c) $. Thus you want to find the maximum of
$$ a^2b^2 = \left(c^2 - \frac{c f(c)}{f'(c)}\right)\left(f(c)^2 - cf(c)f'(c)\right) = 2c^2f(c)^2 - c^3f(c)f'(c) - \frac{cf(c)^3}{f'(c)} $$
Deriving with respect to $ c $, you get
$$ \begin{eqnarray}
\frac{\mathrm da^2b^2}{\mathrm dc}
& = & 4cf(c)^2 + 4c^2f(c)f'(c) - 3c^2f(c)f'(c) - c^3f'(c)^2 - c^3f(c)f''(c) - \frac{(f(c)^3 + 3cf(c)^2f'(c))f'(c) - cf(c)^3f''(c)}{f'(c)^2} \\
& = & \frac{cf(c)^2f'(c)^2 + c^2f(c)f'(c)^3 - c^3f'(c)^4 - c^3f(c)f'(c)^2f''(c) - f(c)^3f'(c) + cf(c)^3f''(c)}{f'(c)^2}
\end{eqnarray} $$
There's no connection between this derivative vanishing and $ f''(c) = 0 $, so your conjecture is wrong.
In the case of the Witch of Agnesi, the inflexion points are $ c = \pm \frac 1{\sqrt 3} $ and this doesn't correspond to the ellipse of maximal area. Indeed,
$$ a^2b^2
= 2c^2f(c)^2 - c^3f(c)f'(c) - \frac{cf(c)^3}{f'(c)}
= \frac{2c^2}{(1 + c^2)^2} + \frac{2c^4}{(1 + c^2)^3} + \frac 1{2(1 + c^2)}
= \frac{4c^2(1 + c^2) + 4c^4 + (1 + c^2)^2}{2(1 + c^2)^3}
= \frac{(1 + 3c^2)^2}{2(1 + c^2)^3} \le 1 $$
with equality iff $ c = \pm 1 $. (the last inequality is equivalent to $ \frac{c^6 + c^6 + 1}3 \ge c^4 $ which is true iff $ c^6 = 1 $ by the arithmetico-geometric inequality)
Surprisingly, the zeroes of $ f''' $ are $ 0, \pm 1 $ and correspond to the extrema of $ ab $. This is not the case in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Mistake in the proof of Binet's formula I want to prove Binet's formula $F_{k} = \dfrac{1}{\sqrt{5}}\times\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^k-\left(\dfrac{1-\sqrt{5}}{2}\right)^k\right]$ for the $k_{th}$ fibonacci number.
I did as follows -
$F_{0}=0, F_{1} = 1$ .....
$F_{k} = F_{k-1}+F_{k-2}$
$F_kx^k=xF_{k-1}x^{k-1} + x^2F_{k-2}x^{k-2}$
summing over $k$ from $2$ to $\infty$,
$g(x)-1 = x(g(x)-1)+x^2g(x)\implies g(x) = \dfrac{x}{1-x-x^2} = \dfrac{-x}{x^2+x-1}$ where $g$ is the generating function of $F_0,F_1,F_2....$
Factoring we get
$g(x) = \dfrac{-x}{\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)}
=-\left[\dfrac{\dfrac{\sqrt{5}-1}{2\sqrt{5}}\left(x-\dfrac{1+\sqrt{5}}{2}\right)+\dfrac{\sqrt{5}+1}{2\sqrt{5}}\left(x-\dfrac{1-\sqrt{5}}{2}\right)}{\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right)}\right]$
$\implies g(x) = -\left[\dfrac{\sqrt{5}-1}{2\sqrt{5}}\cdot\dfrac{1}{\left(x-\frac{1-\sqrt{5}}{2}\right)} + \dfrac{\sqrt{5}+1}{2\sqrt{5}}\cdot\dfrac{1}{\left(x-\frac{1+\sqrt{5}}{2}\right)}\right] = \left[\dfrac{\sqrt{5}-1}{2\sqrt{5}}\cdot\dfrac{1}{\left(\frac{1-\sqrt{5}}{2}-x\right)} + \dfrac{\sqrt{5}+1}{2\sqrt{5}}\cdot\dfrac{1}{\left(\frac{1+\sqrt{5}}{2}-x\right)}\right]$
$\implies g(x) = \dfrac{\sqrt{5}-1}{2\sqrt{5}}\cdot\dfrac{\frac{2}{1-\sqrt{5}}}{\left(1-\frac{2x}{1-\sqrt{5}}\right)}+\dfrac{\sqrt{5}+1}{2\sqrt{5}}\cdot\dfrac{\frac{2}{1+\sqrt{5}}}{\left(1-\frac{2x}{1+\sqrt{5}}\right)} = \dfrac{1}{\sqrt{5}} \left[\dfrac{1}{1-\frac{2x}{1+\sqrt{5}}}-\dfrac{1}{1-\frac{2x}{1-\sqrt{5}}} \right]$
$\implies g(x) = \dfrac{1}{\sqrt{5}}\left[\displaystyle\sum_{k=0}^{\infty}\left(\dfrac{2x}{1+\sqrt{5}}\right)^k-\displaystyle\sum_{k=0}^{\infty}\left(\dfrac{2x}{1-\sqrt{5}}\right)^k\right] = \dfrac{1}{\sqrt{5}}\displaystyle\sum_{k=0}^{\infty}\left[\left(\dfrac{2}{1+\sqrt{5}}\right)^k-\left(\dfrac{2}{1-\sqrt{5}}\right)^k\right]x^k$
$\\$
So I'm getting $F_k = \dfrac{1}{\sqrt{5}}\times \left[\left(\dfrac{2}{1+\sqrt{5}}\right)^k-\left(\dfrac{2}{1-\sqrt{5}}\right)^k\right]$ wheres actually $F_{k} = \dfrac{1}{\sqrt{5}}\times\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^k-\left(\dfrac{1-\sqrt{5}}{2}\right)^k\right]$.
Where did I go wrong? I cant figure it out. Any help is very appreciated. Thanks!
| Since the roots of $x^2+x-1$ are $\frac{-1\pm\sqrt{5}}{2}$, at the begining it should be
$$g(x) = \dfrac{-x}{\left(x+\frac{1+\sqrt{5}}{2}\right)\left(x+\frac{1-\sqrt{5}}{2}\right)}.$$
and therefore, at the end, we find
$$F_k = -\dfrac{1}{\sqrt{5}}\times \left[\left(\dfrac{-2}{1+\sqrt{5}}\right)^k-\left(\dfrac{-2}{1-\sqrt{5}}\right)^k\right].$$
We recover the classic Binet's formula, by noting that,
$$\frac{-2}{1+\sqrt{5}}=\frac{-2}{1+\sqrt{5}}\cdot\frac{1-\sqrt{5}}{1-\sqrt{5}}=
\frac{-2(1-\sqrt{5})}{1-5}=\frac{1-\sqrt{5}}{2}$$
and
$$\frac{-2}{1-\sqrt{5}}=\frac{-2}{1-\sqrt{5}}\cdot\frac{1+\sqrt{5}}{1+\sqrt{5}}=
\frac{-2(1+\sqrt{5})}{1-5}=\frac{1+\sqrt{5}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the values $a$ and $b$ such that the function is differentiable at $x=0$ $\mathbf{Question:}$
Find the values $a$ and $b$ such that the function is differentiable at $x=0$
$$
f(x)=
\begin{cases}
x^{2}+1 &x≥0\\
a\sin x+b\cos x & x<0\\
\end{cases}
$$
$\mathbf{Solution:}$
$f(x)$ is differentiable at $x=0$ if $f'(0)$ exists. This implies that for $f$ to be differentiable at $x=0$, the left hand limit and the right hand limit must exist and be equal.
$$
\begin{align}
\lim_{x\to 0-}f'(0) & =\lim_{x\to 0-}\frac{f(x)-f(0)}{x} \\
& =\lim_{x\to 0-}\frac{a\sin x +b\cos x-1}{x} \\
& =\lim_{x\to 0-}\frac{a\sin x}{x}+\frac{b\cos x-1}{x}=a \\
\end{align}
$$
$$
\begin{align}
\lim_{x\to 0+}f'(0) & =\lim_{x\to 0+}\frac{f(x)-f(0)}{x} \\
& =\lim_{x\to 0+}\frac{x^{2}+1-1}{x} = 0
\end{align}
$$
Therefore, $a=0$
To find $b$, we can use the fact that if $f(x)$ is differentiable at $x=0$ then, it must be continuous at $x=0$.
So if $f(x)$ is continuous, $\lim_{x \to0-}f(x) = \lim_{x \to0+}f(x)=b$
$$
\begin{align}
\lim_{x\to 0-}f(x) & =\lim_{x\to 0-}a\sin x +b\cos x \\
& = a\sin (0) + b\cos (0) = b
\end{align}
$$
$$
\begin{align}
\lim_{x \to 0+}f(x) & = \lim_{x \to0+}x^{2}+1 =1
\end{align}
$$
Therefore, $b=1$
Thus,
$
f(x)=
\begin{cases}
x^{2}+1 &x≥0\\
\cos x & x<0\\
\end{cases}
$
is differentiable at $x=0$
| $$f(0)=1$$
$$\lim_{x\to 0^+}f(x)=\lim_{x\to0^+}(a\sin(x)+b\cos(x))=b$$
$f$ is continuous at $x=0$ if $b=f(0)=1$.
$$f'(0^-)=\lim_{x\to 0^-}\frac{f(x)-f(0)}{x-0}$$
$$=\lim_{x\to0^-}x=\color{red}{0}$$
$f$ is differentiable at $x=0$ if
$$\lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=\color{red}{0}$$
or
$$\lim_{x\to0^+}\frac{a\sin(x)+\cos(x)-1}{x}=0$$
but
$$\sin(x)\sim x \; and \; \cos(x)-1\sim \frac{-x^2}{2}$$
thus
$a=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
inequality :$x^2+y^2+z^2=1$ , $(1-xy)(1-yz)(1-zx)\ge\frac{8} {27}$ and $a+b+c=1$,$a^2+b^2+c^2 +3abc \ge \frac {4}{9}$ 1.Let $x$,$y$,$z$ be nonnegetive reals such that $x^2+y^2+z^2=1$.
$(1-xy)(1-yz)(1-zx)\ge\frac{8} {27}$
*let $a$,$b$,$c$ be positive reals with $a+b+c=1$ . Prove that $a^2+b^2+c^2 +3abc \ge \frac {4}{9}$.
| *
*For the first one:
Let $p=x+y+z, q= xy+yz+zx, r=xyz$ Which are greater or equal to zero.
$x^2+y^2+z^2=p^2-2q \rightarrow p^2-2q=1 $ (1)
$\rightarrow (1-xy)(1-yz)(1-zx)=1-q+pr-r^2\geq \frac{8}{27}$ (2)
$p^3 - 4pq +9r \geq 0 \rightarrow p(p^2 -4q)+9r\geq 0 \rightarrow 9r \geq p(2q-1)$ (3)
Also $p^2 \geq 3q $ and $p^2-2q=1$ thus $2q +1 \geq 3q \rightarrow q \leq 1$ (4)
According to above and $ pq -9r \geq 0$ we have
$p\geq pq \geq 9r \rightarrow p-r \geq \frac{8}{9}p$
Thus we have $r(p − r) \geq \frac{8}{9}pr \geq \frac{8}{9}p \frac{p(2q-1)}{9}=\frac{8(2q+1)(2q-1)}{81}$ (5)
with (2) and (5) we have :
$1-q +\frac{8(2q+1)(2q-1)}{81}\geq \frac{8}{27} \rightarrow (1-q)(49-32q)\geq 0$ which is true due to (4)
*Normalize as follow: $9(a+b+c)(a^2 +b^2 +c^2)+27abc \geq 4(a+b+c)^3$
$\rightarrow 5(a^3 +b^3 +c^3) + 3abc \geq 3(ab(a+b) + bc(b+c)+ca(c+a))$
If you know Schur's inequality:
$a^3 +b^3 +c^3 + 3abc \geq (ab(a+b) + bc(b+c)+ca(c+a)) (1) $
and by Muirhead theorem we have :
$4(a^3 +b^3 +c^3) \geq 2(ab(a+b) + bc(b+c)+ca(c+a)) (2)$
by adding (1) and (2) we have the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum_{i=1}^{n}\binom{a_{i}}{i}\ge\frac{\binom{2n+1}{n}}{2^{2n+1}}\sum_{i=1}^{n}2^{a_{i}}$ let $a_{1}<a_{2}<\cdots<a_{n}=2n$ be postive integers,Prove that
$$\sum_{i=1}^{n}\binom{a_{i}}{i}\ge\dfrac{\binom{2n+1}{n}}{2^{2n+1}}\sum_{i=1}^{n}2^{a_{i}}$$
from Yongxi wang
I want use induction to prove it.
Suppose that the inequality is true for $n$ and Let us prove it for $n+1$,under the inductive hypothesis,it suffices to show that
$$\binom{a_{2n+2}}{n+1}+\dfrac{\binom{2n+1}{n}}{2^{2n+1}}\sum_{i=1}^{n}2^{a_{i}}\ge \dfrac{\binom{2n+3}{n+1}}{2^{2n+3}}\sum_{i=1}^{n+1}2^{a_{i}}$$
I fell last inequality don't right,Thanks
| The induction approach is a good choice in this case, but we have to take into account the fact that, when passing from $n$ to $n+1$, only the first terms of the sequence can be left unchanged. For given $n$, the $n^{th}$ term is necessarily equal to $2n$, whereas for $n+1$ it has no fixed value and can range from $n$ to $2n+1$, depending on the distribution of the first $n-1$ terms. A method to solve this problem is to determine the change in the LHS and that in the RHS when passing from $n$ to $n+1$, trying to identify the case in which any increase in the LHS is minimized and any increase in the RHS is maximized (and vice versa for the decreasing components). If the inequality holds even in this worst scenario, it is true.
For $n=1$, the inequality is trivially satisfied. Passing from $n$ to $n+1$, there is a new term at the end of the sequence, given by $a_{n+1}=2n+2$. So the first component of LHS change is positive and given by
$$\Delta_{1}(LHS)=\binom{2n+2}{n+1}$$
However, we must take into account the above mentioned behaviour of the $n^{th}$ term. If this term decreases when passing to $n+1$, this creates a second component of change for the LHS, which tends to decrease it. In this regard, we can note that this decreasing component is maximal when the $n^{th}$ term changes from $2n$ to $n$, implying that the first $n-1$ terms are the integers from $1$ to $n-1$. In this case, the change in LHS is $$\Delta_2(LHS)=-\binom{2n}{n}+\binom{n}{n}=- \frac{n+1}{2(2n+1)} \binom{2n+2}{n+1} +1$$
On the other hand, passing from $n$ to $n+1$, the RHS changes by two components as well. Firstly, the summation of the first $n$ terms is no longer multiplied to $\binom{2n+1}{n}/(2^{2n+1})$, but to $\binom{2n+3}{n+1}/(2^{2n+3})$. Considering that the $n^{th}$ term changes from $2n$ to $n$, the RHS changes by
$$ \left[\dfrac{\binom{2n+3}{n+1}}{2^{2n+3}} - \dfrac{\binom{2n+1}{n}}{2^{2n+1}} \right] \sum_{i=1}^{n-1}2^{a_i} + \left[\dfrac{\binom{2n+3}{n+1}}{2^{2n+3}}2^n - \dfrac{\binom{2n+1}{n}}{2^{2n+1}}2^{2n} \right] $$
which can be simplified, after some hand calculations on factorials, as
$$\left[-\frac{1}{(n+2) 2^{2n+3}}\binom{2n+2}{n+1} \right]\sum_{i=1}^{n-1}2^{a_i} + \left( \frac{2n+3}{(n+2)2^{n+3}} -\frac{1}{4}\right)\binom{2n+2}{n+1}
$$
Note that both terms are negative, i.e. this component of the change in RHS leads to a decrease. Also note that the minimal decrease occurs when the first $n-1$ terms correspond to the integers between $1$ and $n-1$, as already assumed. In this case the summation of the first $n+1$ terms is $\sum_{i=1}^{n-1}2^{i} = 2^n-2$. So the last quantity can be rewritten as
$$\Delta_1(RHS)=\left[-\frac{1}{(n+2) 2^{n+3}} +
\frac{1}{(n+2) 2^{2n+2}} \right]
\binom{2n+2}{n+1} +
\left( \frac{2n+3}{(n+2)2^{n+3}} -\frac{1}{4}\right)\binom{2n+2}{n+1}
$$
In addition, the new $a_{n+1}=2n+2$ term at the end of the sequence generates in the RHS a positive quantity equal to
$$\Delta_2(RHS)=\dfrac{\binom{2n+3}{n+1}}{2^{2n+3}} 2^{2n+2}
= \dfrac{1}{2}\binom{2n+3}{n+1}\
= \frac{(2n+3)}{2(n+2)}\binom{2n+2}{n+1}$$
We can now collect all these results by taking the main terms of the $\Delta$ components (it is not difficult to show that the effect of the other terms is negligible and tends to zero as $n\rightarrow \infty$). Thus, passing from $n$ to $n+1$, we create a difference between the LHS and the RHS given by
$$\Delta_1(LHS)+\Delta_2(LHS)-\Delta_1(RHS)-\Delta_2(RHS)\\
=\left[1- \frac{n+1}{2(2n+1)} +\frac{1}{4} - \frac{(2n+3)}{2(n+2)} \right]\binom{2n+2}{n+1}
\\=
\left[\frac{5}{4}- \frac{5 n^2 + 11 n + 5}{4 n^2 + 10 n + 4} \right]\binom{2n+2}{n+1}
$$
The term in square brackets is always positive for positive $n$, and tends to $0$ as $n\rightarrow \infty$ (here is a plot). So, even in the worst scenario, passing from $n$ to $n+1$ we create a positive difference between the LHS and the RHS. This proves the inequality of the OP.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3712767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the following equation in integers $x,y:$ $x^2+6xy+8y^2+3x+6y=2.$ Question: Solve the following equation in integers $x,y:$ $$x^2+6xy+8y^2+3x+6y=2.$$
Solution: For some $x,y\in\mathbb{Z}$ $$x^2+6xy+8y^2+3x+6y=2\\\iff x^2+2xy+4xy+8y^2+3x+6y=2\\\iff x(x+2y)+4y(x+2y)+3(x+2y)=2\\\iff(x+4y+3)(x+2y)=2.$$
Now if $(x+4y+3)(x+2y)=2$, then either $$\begin{cases} x+4y+3=1\\ x+2y=2\end{cases}\text{ or }\begin{cases} x+4y+3=2\\ x+2y=1\end{cases}\text{ or }\begin{cases} x+4y+3=-1\\ x+2y=-2\end{cases}\text{ or }\begin{cases} x+4y+3=-2\\ x+2y=-1\end{cases}.$$
We have $$\begin{cases} x+4y+3=1\\ x+2y=2\end{cases}\iff (x,y)=(6,-2), \\\begin{cases} x+4y+3=2\\ x+2y=1\end{cases}\iff (x,y)=(3,-1), \\\begin{cases} x+4y+3=-1\\ x+2y=-2\end{cases}\iff (x,y)=(0,-1),\\\begin{cases} x+4y+3=-2\\ x+2y=-1\end{cases}\iff (x,y)=(3,-2).$$
Now since, all the four pairs $(6,-2),(3,-1),(0,-1),(3,-2)$ satisfies the integer equation $(x+4y+3)(x+2y)=2$, thus we can conclude that $(x+4y+3)(x+2y)=2\iff (x,y)=(6,-2),(3,-1),(0,-1),(3,-2).$
Hence, we can conclude that the integer equation $x^2+6xy+8y^2+3x+6x=2$ is satisfied if and only if $(x,y)=(6,-2),(3,-1),(0,-1),(3,-2)$, and we are done.
Is the solution correct and rigorous enough? And, I am always confused while solving equations regarding the usage of the if and only if arguments, which I feel is very necessary in order to have a complete and rigorous solution, but I rarely find it's usage in any book while solving equations of any kind. So, is it necessary? Also, is there a better solution than this?
| Another way:
Discriminant $$=(6y+3)^2-4(8y^2+6y-2)=4y^2+12y+17=(2y+3)^2+8$$ which needs to be perfect square $=z^2$(say)
WLOG $z\ge0$
so that $$x=\dfrac{-(6y+3)\pm z}2$$
Now $$(2y+3)^2-z^2=8$$
$$\iff\dfrac{2y+3-z}2\cdot\dfrac{2y+z+3}2=2=1\cdot2=(-2)(-1)$$
As $2y+3+z\ge2y+3-z,$
$\dfrac{2y+3+z}2=2\iff\dfrac{2y+3-z}2=-1$
Or $\dfrac{2y+3+z}2=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{n\to\infty} \left(\frac{n}{1+n^2}+\frac{n}{4+n^2}+\cdots+ \frac{n}{n^2+n^2}\right)$ $$\lim_{n\to\infty} \left(\frac{n}{1+n^2}+\frac{n}{4+n^2}+\cdots+ \frac{n}{n^2+n^2}\right)$$
I will give my solution, which, to my surprise, turned out to be erroneous:
$S_n=\frac{a_1+a_n}{2}n=\frac{\frac{n}{1+n^2}+\frac{n}{n^2+n^2}}{2}n=\frac{1}{4}\frac{3n^2+1}{1+n^2}$
$\lim_{n\to\infty}S_n=\frac{3}{4}$
But the correct answer is $\frac{\pi}{4}$
I do not understand how to get the exact answer.
| Hint:
You can rewrite it as a Riemann sum:
$$\sum_{k=1}^n\frac{n}{n^2+k^2}=\sum_{k=1}^n\frac{n}{n^2\Bigl(1+\frac{k^2}{n^2}\Bigr)}=\frac1n\sum_{k=1}^n\frac{1}{1+\frac{k^2}{n^2}}.$$
Can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
what is the volume of a tilted glass I was wondering about the volume of a tilted glass (60$^o$).
Not sure, but my attempt:
$$y=\sqrt{x}, \space 0<x<9$$
$$x=y^2,\space 0<y<3$$
$$A(y)=\pi x^2=\pi y^4$$
$$V(y)=\int_0^3 \pi y^4\,dy$$
$$V=\frac{243 \pi}{5}$$
$$V(\text{adjusted})=\sin(60^o)*V=\frac{243\sqrt{3 \pi}}{10}$$
Some help to get?
Image:
| The line of the $60$ degree angle can be described by $x = \frac{1}{\sqrt{3}}y + 9 - \sqrt{3}$
The area enclosed by the curve and the line can be calculated with an integral.
$\int{}{}{\frac{1}{\sqrt{3}}y + 9 - \sqrt{3}} - y^2$
You need to find the limits of integration. For the right side, it will be 3 since the limit of the glass is 9. For the left side, it will be $-\sqrt{9-2\sqrt{3}}$.
$$\int_{-\sqrt{9-2\sqrt{3}}}^3{\frac{1}{\sqrt{3}}y + 9 - \sqrt{3}} - y^2 dy$$
Find the value of the integral and that is the two dimensional area of the glass tilted.
To find the volume for a three dimensional version you will have to evaluate an integral of the area of a circle from zero to the liquid level.
$$f_r(x) = \sqrt{x}$$
$$\int_0^{9 - \sqrt{3}} \pi f_r(x)^2 dx$$
$$\int_0^{9 - \sqrt{3}} \pi x \ dx$$
$$ \pi [\frac{x^2}{2} ]^{9 - \sqrt{3}}_0 $$
$$ \pi * \frac{(9-\sqrt{3})^2}{2} $$
$$=~ 82.974...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding Inverse of a 4*4 matrix Find the inverse of the following matrix
$$
\begin{array}{c}
{\left[\begin{array}{cccc}
c_{0} & c_{1} & c_{2} & c_{3} \\
c_{2} & c_{3} & c_{0} & c_{1} \\
c_{3} & -c_{2} & c_{1} & -c_{0} \\
c_{1} & -c_{0} & c_{3} & -c_{2}
\end{array}\right]} \\
\text { where } c_{0}=\frac{1+\sqrt{3}}{4 \sqrt{2}}, c_{1}=\frac{3+\sqrt{3}}{4 \sqrt{2}}, c_{2}=\frac{3-\sqrt{3}}{4 \sqrt{2}} \text { and } c_{3}=\frac{1-\sqrt{3}}{4 \sqrt{2}}
\end{array}
$$
I had thought of replacing the values with trigonometric values like Sin 15 , cos 15 .... But it seems very lengthy. Any shorter approach?
| The columns are pairwise orthogonal to each other.
So, with
$$A=
{\left[\begin{array}{cccc}
c_{0} & c_{1} & c_{2} & c_{3} \\
c_{2} & c_{3} & c_{0} & c_{1} \\
c_{3} & -c_{2} & c_{1} & -c_{0} \\
c_{1} & -c_{0} & c_{3} & -c_{2}
\end{array}\right]} \\
$$
You have
$$A^TA = qI_{4\times 4} \text{ with } q= \sum_{k=0}^3c_k^2 =1$$
Hence,
$$A^{-1} = A^T$$
So, the columns of matrix $A$ form an orthonormal basis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find inverse element of $1+2\alpha$ in $\mathbb{F}_9$
Let $$\mathbb{F}_9 = \frac{\mathbb{F}_3[x]}{(x^2+1)}$$ and consider $\alpha = \bar{x}$. Compute $(1+2 \alpha)^{-1}$
I think I should use the extended Euclidean algorithm: so I divide $x^2 +1 $ by $(1+2x)$:
$$x^2 + 1 = (1+2x)(2x+2)+2$$
$$(2x+2)(1+2x) + 2(x^2+1) = 1$$
Therefore, considering $\text{mod}(x^2+1) $, I have $$(2x+2)(1+2x) = 1\text{mod}(x^2+1)$$
and so $2x+2 = (1+2x)^{-1}$
Is it okay, or did I misunderstood something?
| Your answer is correct.
Note that $\alpha^2=-1$, and the conjugate of $\alpha$ is $-\alpha$,
so $\dfrac1{1+2\alpha}=\dfrac1{1+2\alpha}\dfrac{1-2\alpha}{1-2\alpha}=\dfrac{1-2\alpha}{1-4\alpha^2}=\dfrac{1-2\alpha}{1-\alpha^2}=\dfrac{1-2\alpha}2=2^{-1}-\alpha=2+2\alpha$
in $\mathbb F_9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all pairs of positive integers $(m, n)$ for which $X^m + X + 1$ divides $X^n + 1$ and pairs $(m,n)$ for which $X^m +X −1$ divides $X^n +1$?
The special case I can think of is when $n=3$, and $m=2$ for the first part. But I don't know if other cases exist. Any help would be appreciated. Thanks!
| I get that
there are no solutions
for $x^m+x+1$.
The roots of
$x^n+1$ are
$e^{(2k+1) \pi i/n}$
for $k = 0$ to $n-1$,
so
all the roots have magnitude $1$.
If $m = 1$ then
$x = -\frac12$
so
$\dfrac{(-1)^n}{2^n}+1 = 0$
which can't be.
If $m = 2$
the roots of
$x^2+x+1$
are
$x_{1, 2}
=\dfrac{-1\pm\sqrt{-3}}{2}
=e^{\pm 2\pi i/3}
$.
Then
$x_1^n
=e^{2\pi n i/3}
=-1
=e^{(2k+1)\pi i}
$
for some $k$
so
$2n/3 =2k+1
$
so
$2n = 6k+3
$
and this can't be since
the left side is even
and the right side is odd.
The same thing holds for
$x_2
=e^{-2\pi i/3}
$.
Therefore
there are no solutions for
$m = 2$.
If $m > 2$ then
for $m$ values of
$0 \le k \le n-1$,
$\begin{array}\\
0
&= x^m+x+1\\
&=e^{(2k+1)m \pi i/n}+e^{(2k+1) \pi i/n}+1\\
&=\cos((2k+1)m\pi/n)+i\sin((2k+1)m\pi/m)
+\cos((2k+1)\pi/n)+i\sin((2k+1)\pi/m)
+1\\
&=\cos((2k+1)m\pi/n)+\cos((2k+1)\pi/n)+1
+i(\sin((2k+1)m\pi/m)+\sin((2k+1)\pi/m))\\
&=\cos((2k+1)m\pi/n)+\cos((2k+1)\pi/n)+1\\
&=\cos(mr)+\cos(r)+1
\qquad\text{where }r = (2k+1)\pi/n\\
\text{and}\\
0
&=\sin((2k+1)m\pi/n)+\sin((2k+1)\pi/n)\\
&=\sin(mr)+\sin(r)\\
\end{array}
$
Therefore
$\sin(mr)
=-\sin(r)
$
so
$\sin^2(mr)
=\sin^2(r)
$
$\cos^2(mr)
=\cos^2(r)
$.
If
$\cos(mr)
=\cos(r)
$
then
$0
=\cos(mr)+\cos(r)+1
=2\cos(r)+1
$
so,
since
$0 < r < 2\pi$,
$r
= 2\pi/3,
= -2\pi/3
$.
If $r = 2\pi/3$,
$2/3
=(2k+1)/n
$
so
$2n
=3(2k+1)
=3k+3
$
but this can not be since
the left side is even
and the right side is odd.
Similarly,
If $r = -2\pi/3$,
then
$-2/3
=(2k+1)/n
$
so
$-2n
=3(2k+1)
=3k+3
$
but this also can not be since
the left side is even
and the right side is odd.
Therefore there
are no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim\limits_{n \to \infty}\sum\limits_{i=1}^n\frac{1-\cos \frac{\pi}{\sqrt{n}}}{1+\cos \frac{i\pi}{\sqrt{2n}}}$ We may have
\begin{align*}
\lim_{n \to \infty}\sum_{i=1}^n\frac{1-\cos \frac{\pi}{\sqrt{n}}}{1+\cos \frac{i\pi}{\sqrt{2n}}}&=\lim_{n \to \infty}\left(1-\cos \frac{\pi}{\sqrt{n}}\right)\sum_{i=1}^n\frac{1}{1+\cos \frac{i\pi}{\sqrt{2n}}}\\&=\lim_{n \to \infty}\frac{\pi^2}{4n}\sum_{i=1}^n\frac{1}{\cos^2 \frac{i\pi}{2\sqrt{2n}}},
\end{align*}
but how to go on?
| The limit doesn't exist (i.e. it is infinite).
For $1\leqslant k\leqslant m_n:=\lfloor\sqrt{n/8}\rfloor$ (and $n$ large enough), let $$i_k:=\lfloor(2k-1)\sqrt{2n}\rfloor=(2k-1)\sqrt{2n}-\epsilon_k.$$ Then $$\cos^2\frac{i_k\pi}{2\sqrt{2n}}=\cos^2\left((2k-1)\frac{\pi}{2}-\frac{\epsilon_k\pi}{\sqrt{8n}}\right)=\sin^2\frac{\epsilon_k\pi}{\sqrt{8n}}\leqslant\frac{\pi^2}{8n},$$ hence $$\frac{\pi^2}{4n}\sum_{i=1}^{n}\frac{1}{\cos^2\frac{i\pi}{2\sqrt{2n}}}\geqslant\frac{\pi^2}{4n}\sum_{k=1}^{m_n}\frac{1}{\cos^2\frac{i_k\pi}{2\sqrt{2n}}}\geqslant 2m_n\underset{n\to\infty}{\longrightarrow}\infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$ For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$
By dividing the numbers between $1$ and $5^m$ as intervals of $5^k$, I was getting the following expression:
$$\binom{5^{m-k}}{2}$$ which is not only too good to be true but it turns out it is wrong. Any suggestions on how I should approach this?
Edit: After reading through @heropup's example, I am starting to realize that I may have forgotten $5^k$ term and so the following could be correct.
$$5^k \binom{5^{m-k}}{2}$$
Does that sound correct?
| Your original approach of doing casework on intervals works. The key (at least to my method) is to interpret $$\left\lfloor\frac{a}{b}\right\rfloor$$ as the number of positive multiples of $b$ that are less than or equal to $a,$ where $a$ and $b$ are positive integers. Your list of disjoint intervals that cover all of the $l$ is $$[1,5^k -1],[5^k, 2\cdot 5^k -1],[2\cdot 5^k,3\cdot 5^k -1],\ldots, [5^{m-k-1}\cdot 5^k,5^{m-k}\cdot 5^k -1].$$
In the first interval, none of the elements have a positive multiple of $5^k$ less than or equal to the element. For each integer in the $t^{\text{th}}$ interval for $t\ge 2$ the number of positive multiples of $5^k$ that are less than or equal to the element is $t-1.$ There are $5^k$ elements in each interval after first interval. So the answer is
\begin{align*}
\sum_{n=1}^{5^{m-k}-1}{5^k\cdot n} &= 5^k\cdot\sum_{n=1}^{5^{m-k}-1}{n}\\
&= 5^k \cdot \frac{(5^{m-k}-1)5^{m-k}}{2}\\
&= \frac{5^{m}\cdot (5^{m-k}-1)}{2}.
\end{align*}
This is the same as $5^k \cdot \binom{5^{m-k}}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Is there a quick (hopefully elementary) way to prove that $6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0$ has only one solution? I have the Diophantine equation $$6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0.$$ Numerical calculations suggest this has only one positive integer solution, namely $(b,c)=(2,3)$. Is there a quick way to prove or disprove that?
| Solving for $c$, we get
$$ c=-{\frac {-18\,b \pm \sqrt {24\,{b}^{6}+36\,{b}^{4}+18\,{b}^{2}-159}
}{3 (2\,{b}^{2}+1)}}$$
If there is to be an integer solution, we need $24\,{b}^{6}+36\,{b}^{4}+18\,{b}^{2}-159$ to be a square. Taking $s = b^2$, let's look for integer solutions of
$t^2 = 24 s^3 + 36 s^2 + 18 s - 159$, which is an elliptic curve.
Taking $s = x/24 - 1/2$ and $t = y/24$, this becomes $y^2 = x^3 - 93312$: note that if $s$ and $t$ are integers, then $x = 24 s + 12$ and $y = 24 t$ are also integers.
Now $y^2 = x^3 - 93312$ is an elliptic curve in Weierstrass form. According to Sage, its only integer solutions are $x = 108, y = \pm 1080$. This corresponds to $s=4$ and $t = \pm 45$. And indeed $s=4$ gives us $b = \pm 2$; $b=2$ gives us an integer solution with $c=3$, and $b=-2$ gives us an integer solution with $c=-3$. Those are the only integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Let z be a complex number such that $\frac{z-i}{z-1}$ is purely imaginary. Then the minimum value of $|z-(2+2 i)|$ Let
z be a complex number such that $\frac{z-i}{z-1}$ is purely imaginary. Then the minimum value of $|z-(2+2 i)|$ is ?
My approach:-
$$
\begin{array}{l}
\\
\left|z-z^{\prime}\right| \geqslant| \ |z|-\left|z^{\prime}\right||
\end{array}
$$
where, $z^{\prime}=2+2 i$
$$
\left|z^{\prime}\right|=\sqrt{8}=2 \sqrt{2}
$$
$Let, \quad z=x+i y$
$$
\begin{array}{l}
\frac{z-i}{z-1} \\
=\frac{\{x+i(y-1)\}\{(x-1)-i y\}}{(x-1)^{2}+(i y)^{2}} \\
=\frac{x(x-1)+y(y-1)+i(x-1)(y-1)-x y}{(x-1)^{2}-(i y)^{2}}
\end{array}
$$
${}{} \frac{z-i}{z-1}$ is purely imagivery
$$
\begin{array}{c}
\text { Hence, } x(x-1)+y(y-1)=0 \\
x(x-1)=-y(y-1) \\
x^{2}+y^{2}=x+y
\end{array}
$$
What to do next?
| A geometric approach:
$\frac{z-i}{z-1}$ is purely imaginary signifies that $$\arg\left(\frac{z-i}{z-1}\right)={\pi \over 2}+k\pi$$ with $k\in \mathbb{Z}.$
Geometrically, locus of point $z$ is the circle with diameter joining $i$ and $1,$ except these two points.
The center is ${1\over2}+i{1\over2},$ the midpoint of the diameter.
The nearest point of this circle to $2+2i$ lies on the line joining $2+2i$ and the center of the circle.
Therefore, the nearest point is $1+i,$ and the minimal distance $\sqrt2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do I evaluate $\lim_{n\to\infty} \,\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}$? I came across the following problem recently in a problem sheet aimed at high school students:
Evaluate $$\lim_{n\to\infty} \,\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}.$$
I tried to rewrite the inner sum as a Riemann sum hoping that the limit would become a definite integral, but no gain because of the extra $1/n$'s.
| We can rewrite the sum as
$$S = \lim_{n\to\infty}\sum_{k=1}^n \left(\frac{k}{n^2}\right)^{\frac{k}{n^2}}\cdot\frac{k}{n}\cdot\frac{1}{n}$$
We also have that for $1\leq k \leq n$
$$\left(\frac{1}{n}\right)^{\frac{1}{n}} \leq \left(\frac{k}{n^2}\right)^{\frac{k}{n^2}} \leq \left(\frac{1}{n^2}\right)^{\frac{1}{n^2}}$$
for $n > e$. Thus we can sandwich the original limit
$$\lim_{n\to\infty} \left(\frac{1}{n}\right)^{\frac{1}{n}} \cdot \sum_{k=1}^n \frac{k}{n}\cdot\frac{1}{n} \leq S \leq \lim_{n\to\infty} \left(\frac{1}{n^2}\right)^{\frac{1}{n^2}} \cdot \sum_{k=1}^n \frac{k}{n}\cdot\frac{1}{n}$$
which means that
$$S = \lim_{n\to\infty} \sum_{k=1}^n \frac{k}{n}\cdot\frac{1}{n} = \int_0^1x\:dx = \frac{1}{2}$$
by squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
The functional equation $\big(1 + yf(x)\big)\big(1 - yf(x + y)\big) = 1$ for $f:\mathbb R^+\to\mathbb R^+$ Functional equation from USAMO 2010 preparation session:
Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that
$\big(1 + yf(x)\big)\big(1 - yf(x + y)\big) = 1$
for all $x, y \in \mathbb R^+$, where $\mathbb R^+$ is a set of all positive real numbers.
Well I don't really see what we can do. I mean I could have plugged in some numbers but the best one ($0$) which could have caused simplifications is out of reach and so are the negative numbers. Maybe if we could take two cases:
*
*Both factors are equal to $\pm 1$.
*The factors are reciprocals of each other.
In the first case $+1$ in fact is only possible when $f(x)=0$ $ \forall x$, which isn't possible as $0$ isnt in the codomain. For $-1$ we would get $f(x+y)=-f(x)$, contradiction again.
So we must have the two factors being reciprocals. Well now what? I'm stuck here. Clearly my approach is not just unprofessional, it's bad too. How can this thing be solved?
EDIT:
After considering @Yesit'sme's comment, I retried the problem and would like to present a solution. PLEASE DO TELL ME WHETHER IT IS CORRECT OR NOT. Here we go.
Since $x,y \in \mathbb R^+$, we may without restriction assume $x,y \neq 0$.
Now from given,
$\begin{align}
\big(1 + yf(x)\big)\big(1 − yf(x + y)\big) &= 1 \\
1 − yf(x + y)&= \frac{1}{1 + yf(x)} \\
1-\frac{1}{1 + yf(x)} &= yf(x + y) \\
\frac{1+yf(x)-1}{1 + yf(x)} &= yf(x + y) \\
\frac{f(x)}{1+yf(x)} &= f(x+y)= \frac{f(y)}{1+xf(y)} \tag 1 \label 1
\end{align}$
$\forall x,y \in \mathbb R^+$.
The last equation follow from symmetry (or plugging in $y+x$ into $f$).
Now by plugging in $y=1$ we see that,
$ f(x+1) =\frac{f(x)}{1+f(x)}<f(x), \forall x \in \mathbb R^+$.
The last inequality follows from the fact that,
$\begin{align} f(x)+1 &>1 \\
1 &>\frac{1}{f(x)+1} \\
f(x)&>\frac{f(x)}{f(x)+1}=f(x+1) \end{align}$
As $f(x) \in \mathbb R^+$.
Thus $f(x)$ is in fact decreasing. We now define a new function $Q$ such that,
$f(x)=\frac{1}{Q(x)}$
where $Q$ is strictly increasing $\forall x$.
Plugging this into \eqref{1} we get,
$\begin{align} \frac{\frac{1}{Q(x)}}{1+\frac{y}{Q(x)}} &= \frac{\frac{1}{Q(y)}}{1+\frac{x}{Q(y)}} \\
\frac{1}{Q(x)+y} &= \frac{1}{Q(y)+x} \\
Q(x)+y &=Q(y)+x \\
Q(y)-y &=Q(x)-x=k \\ \end{align} $
Where $k \in \mathbb R$ is a constant. This gives,
$\begin{align}\frac{1}{f(x)} &= x+k \\
\therefore f(x) &= \frac{1}{x+k} \blacksquare. \\ \end{align} $
Plugging this into the original equation we see that the equation is satisfied and hence the solution is complete.
(I didn't show the checking part as typing out this much already took a boatload of time. Hope you understand. It does satisfy though, I have checked it by hand.)
| Since $x,y \in \mathbb R^+$, we may without restriction assume $x,y \neq 0$.
Now from given,
$ \begin{align} (1 + yf(x))(1 − yf(x + y)) &= 1 \\
1 − yf(x + y)&= \frac{1}{1 + yf(x)} \\
1-\frac{1}{1 + yf(x)} &= yf(x + y) \\
\frac{1+yf(x)-1}{1 + yf(x)} &= yf(x + y) \\
\frac{f(x)}{1+yf(x)} &= f(x+y)= \frac{f(y)}{1+xf(y)} \tag 1 \label {eqn1} \\ \end{align} $
$\forall x,y \in \mathbb R^+$.
The last equation follow from symmetry (or plugging in $y+x$ into $f$).
Since,
$f ( x + y ) = \frac { f ( x ) } { f ( x ) + y } = \frac 1 { \frac 1 { f ( x ) } + y }<\frac{1}{\frac{1}{f(x)}}=f(x)$
The inequality follows from the fact that,
$\begin{align} y &>0 \\
\frac{1}{f(x)}+y &>\frac{1}{f(x)} \\
\frac{1}{\frac{1}{f(x)}} &> \frac{1}{\frac{1}{f(x)}+y} \end{align}$
As $f(x) \in \mathbb R^+$.
Thus, $f(x)$ is in fact decreasing.
We now define a new function $Q$ such that,
$f(x)=\frac{1}{Q(x)}$
Since $f \neq 0$.
Plugging this into \eqref{eqn1} we get,
$\begin{align} \frac{\frac{1}{Q(x)}}{1+\frac{y}{Q(x)}} &= \frac{\frac{1}{Q(y)}}{1+\frac{x}{Q(y)}} \\
\frac{1}{Q(x)+y} &= \frac{1}{Q(y)+x} \\
Q(x)+y &=Q(y)+x \\
Q(y)-y &=Q(x)-x=k \\ \end{align} $
Where $k \in \mathbb R$ is a constant. This gives,
$\begin{align}\frac{1}{f(x)} &= x+k \\
\therefore f(x) &= \frac{1}{x+k} \blacksquare. \\ \end{align} $
Plugging this into the original equation we see that the equation is satisfied and hence the solution is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculating $ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$ I need to find the limit of:
$$ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$$
I've tried throught "rationalization" and completing squares... This is my work so far (i'm learning by myself limits since my teachers doesn't respond any email and they are not making lectures, just pdf's... I'm trying to do my best, help pls). Also is there any good book or suggestion to learn limits .
\begin{align*}&\lim_{b \to a} \dfrac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2} \cdot \dfrac{(a-\sqrt{a^2-b^2})}{(a-\sqrt{a^2-b^2})} \\=&
\lim_{b \to a} \dfrac{a[a^2-(a^2-b^2)]-b^2(a-\sqrt{a^2-b^2})}{a[(a^2-\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
= &\lim_{b \to a} \dfrac{-ab^2-ab^2+b^2(\sqrt{a^2-b^2})}{a[(a^2-(\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
=&\lim_{b \to a} \dfrac{-2ab^2+b^2(\sqrt{a^2-b^2})}{a[(a^2-(\sqrt{a^2-b^2})^2] -b^2 (a-\sqrt{a^2-b^2)}}\\
\end{align*}
| Let
$b = a-x$
where $x > 0$.
$\begin{array}\\
f(a, b)
&=\dfrac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}\\
&=\dfrac{a \cdot (a + \sqrt{a^2-(a-x)^2}) - (a-x)^2}{a \cdot (a - \sqrt{a^2-(a-x)^2})-(a-x)^2}\\
&=\dfrac{a(a + \sqrt{2ax-x^2}) - a^2+2ax-x^2}{a(a - \sqrt{2ax-x^2})-a^2+2ax-x^2}\\
&=\dfrac{a\sqrt{2ax-x^2} +2ax-x^2}{- a\sqrt{2ax-x^2}+2ax-x^2}\\
&=\dfrac{a\sqrt{x}\sqrt{2a-x} +2ax-x^2}{- a\sqrt{x}\sqrt{2a-x}+2ax-x^2}\\
f(a, b)+1
&=\dfrac{a\sqrt{x}\sqrt{2a-x} +2ax-x^2}{- a\sqrt{x}\sqrt{2a-x}+2ax-x^2}+1\\
&=\dfrac{a\sqrt{x}\sqrt{2a-x} +2ax-x^2- a\sqrt{x}\sqrt{2a-x}+2ax-x^2}{- a\sqrt{x}\sqrt{2a-x}+2ax-x^2}\\
&=\dfrac{4ax-2x^2}{- a\sqrt{x}\sqrt{2a-x}+2ax-x^2}\\
&=\dfrac{4a\sqrt{x}-2x^{3/2}}{- a\sqrt{2a-x}+2a\sqrt{x}-x^{3/2}}\\
&\to 0
\qquad\text{as }x \to 0\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $a+b+c=k$ and $a^2+b^2+c^2 =2k$ what is the maximum value of $k$? $a,b,c$ are real numbers and they satisfy the following equations.
$a+b+c=k$
$a^2+b^2+c^2=2k$
Find the maximum value of $k$.
I tried substituting for k in the second equation from the first and got
$a^2+b^2+c^2=2(a+b+c)$
Rearranging the terms I got
$a^2-2a+b^2-2b+c^2-2c=0$
Adding 3 to both sides we get
$a^2-2a+1+b^2-2b+1+c^2-2c+1=3$
This can be simplified to the following
$(a-1)^2+(b-1)^2+(c-1)^2=3$
Therefore,
$0\leq(a-1)^2,(b-1)^2,(c-1)^2\leq3$
From here we can deduce the range of values that a,b,c can take as
$1-\sqrt{3}\leq a,b,c\leq1+\sqrt{3}$
I don't know know if this helps to answer the question.
| Your ideas are good. You obtained $(a-1)^2 + (b-1)^2 + (c-1)^2 = 3$. That means there is a sphere around the point $(1,1,1)$ with radius $\sqrt{3}$ and the point $(a,b,c)$ is situated on that sphere.
The condition $a^2+b^2+c^2 = 2k$ tells us that the point $(a,b,c)$ lives on a sphere around the origin at distance $\sqrt{2k}$.
The point exists on both spheres simultaneously, hence ...? (what is the largest distance from the origin these spheres can intersect?)
2D analogue. The point $C$ in the picture corresponds to $(a,b)$. Then $2\sqrt{2} = \sqrt{2k}$ i.e $k=4$.
In the 3D case, we would have $2\sqrt{3} = \sqrt{2k}$, thus $k=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Bound for $\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}$. In the book Complex analysis by Bak J. & Newman J., chapter 11, talks about Sums Involving Binomial coefficients and find a bound $\frac{16}{9}\sqrt{3}$ for $|(z-1)^2(z+1)|$ in the "Example 3" on the unit circle, my way was using lagrange multipliers in this form:
We want find $\max{|(z-1)^2(z+1)|}$ ahnd have that $|z-1|^2+|z+1|^2=4$. Let be $a=|z-1|$ and $b=|z+1|$ and then the exercise is: "Maximize $f(a,b)=a^2b$ subject to $a^2+b^2=4$" then $\nabla f=\lambda\nabla g$ so $\begin{cases}2ab=\lambda(2a)\\a^2=\lambda(2b)\end{cases}$, i.e., $ab=\lambda a$ if $a(b-\lambda)=0$ therefore
i)$a=0$ or $b=\lambda$, $b^2=4$ then $b=2$ then $|z+1|=2$ or $|z-1|=0$ then $z=1$ and $a^2b=0$.
ii)$b=\lambda$, $a^2=2b^2$ then $4=a^2+b^2=3b^2$ then $b^2=4/3$ then $b=\frac{2\sqrt{3}}{3}$ then $a^2=\frac{8}{3}$ then $a=\frac{2\sqrt{2}}{\sqrt{3}}$
So $a^2b=\frac{8}{3}\frac{2\sqrt{3}}{3}=\frac{16}{9}\sqrt{3}$.
After i want to use this idea for exercise 17.b with $\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}$ but, i don´t see what should be $a$ and $b$.
Edit: this exercise says that $|\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}|\leq4^n$.
| Using the "coefficient-of" notation, we have
$$S_n:=\sum_{k=0}^{n}(-1)^k\binom{3n}{k}\binom{n}{k}=\sum_{k=0}^{n}[z^k](1-z)^{3n}\times[z^{n-k}](1+z)^n=[z^n]\big((1-z)^3(1+z)\big)^n.$$ Let $f(z)=(1-z)^3(1+z)/z$; then, by Cauchy integral formula, for any $r>0$ we have $$S_n=\frac{1}{2\pi\mathrm{i}}\oint_{|z|=r}\big(f(z)\big)^n\frac{dz}{z}\implies|S_n|\leqslant\Big(\max_{|z|=r}\big|f(z)\big|\Big)^n.$$ The key to a solution is to choose $\color{blue}{r=1/\sqrt{3}}$ (this is suggested by saddle points of $\big|f(z)\big|$, that is, the solutions $z=(-1\pm\mathrm{i}\sqrt{2})/3$ of $f'(z)=0$; alternatively, we can let $r$ be arbitrary, and minimize the result w.r.t. $r$ in the end). Using $|1+z|^2+|1-z|^2=2(1+|z|^2)$ again, we arrive at $$\text{maximize}\quad a^3 b\quad\text{subject to}\quad a^2+b^2=8/3.$$ Solving it the way you know, we find $a^2=2$, $b^2=2/3$, $a^3 b=4/\sqrt{3}$ and $\color{blue}{\max\limits_{|z|=r}\big|f(z)\big|=4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that for every integer $n$, $n^3$mod$6$=$n$mod$6$ I will use induction to prove this.
Firstly for $n=1$,
$1^3\text{mod}6=1\text{mod}6$
Now we assume that this holds for some $n=k$ and prove that if it holds for $n=k$ it should also hold for $n=k+1$.
$(k+1)^3\text{mod}6=(k^3+3k^2+3k+1)\text{mod}6$
As for $n=k$ because it holds,
$k^3\text{mod}6=k\text{mod}6$
$k^3-6p(k)=k-6q(k)$
$k^3-k=6p(k)-6q(k)$
Consider $(k^3+3k^2+3k+1)-6a(k)=6p(k)-6q(k)+k+1+3k^2+3k$
Now we just need to prove that $3k^2+3k$ is always a multiple of 6.
For $k=1$ it holds.
Assume $k=j$ holds. Then for $k=j+1$,
$3(j+1)^2+3(j+1)=3(j^2+2j+1)+3j+3=3j^2+3j+6j+6$
So for $k=j+1$ it is also a multiple. As it holds for $k=1$, then it holds for all positive integers $k$.
Then this means that
$6p(k)-6q(k)+k+1+3k^2+3k=k+1-6b(k)$
And if $(k^3+3k^2+3k+1)-6a(k)=k+1-6b(k)$ this means $(k+1)^3\text{mod}6=(k+1)\text{mod}6$
Is this correct?
| An easy approach:
$$n^3-n=(n-1)n(n+1)$$
As in any two consecutive integers, one is even, and in any three consecutive, one is multiple of three. So the product of three consecutive numbers is a multiple of $6$. $$\begin{split}n^3-n&\equiv0\pmod6\\n^3&\equiv n\pmod 6\end{split}$$ Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Remainder when divided by $7$ What would be the remainder when
$12^1 + 12^2 + 12^3 +\cdots + 12^{100}$ is divided by $7$ ?
I tried cyclic approach (pattern method), but I couldn't solve this particular question.
| As $12\equiv-2\pmod7, 12^3\equiv(-2)^3\equiv-8\equiv-1,$ord$_712=6$
$\implies12^{6k+r}\equiv12^r\equiv(-2)^r\pmod7$
$$\sum_{r=1}^{100}12^r\equiv12^1+12^2+12^3+12^4+16\sum_{r=0}^512^r\pmod7$$
$$\equiv(-2)+(-2)^2+(-2)^3+(-2)^4+2\sum_{r=0}^5(-2)^r\pmod7$$
Finally $\displaystyle\sum_{r=0}^5(-2)^r\equiv\dfrac{(-2)^6-1}{-2-1}\equiv0\pmod7$ as $(-2-1,7)=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to solve $\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$?
How can i evaluate the following integral $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=?$$
This is taken from a definite integral where $x$ varies from $0$ to $1$.
My attempt:
multiplied by conjugate
$$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}$$
$$=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{1+x-1+x}$$
$$=\int \frac{(\sqrt{1+x}+\sqrt{1-x})dx}{2x}$$
*
*if i use $x=\sin^2\theta$
$$\int \frac{(\sqrt{1+\sin^2\theta}+\cos\theta)}{2\sin^2\theta}\sin2\theta\ d\theta=\int (\sqrt{1+\sin^2\theta}+\cos\theta)\cot\theta d\theta$$
*if i use $x=\tan^2\theta$
$$\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{2\tan^2\theta}2\tan\theta\sec^2\theta d\theta\ d\theta=\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{\sin\theta\cos\theta} d\theta$$
Should I use substitution $x=\sin^2\theta$ or $x=\tan^2\theta$?. I can't decide which substitution will work further. Please help me solve this integration.
Thanks
| $\begin{aligned} I &:= \int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{\sqrt{1+x}+\sqrt{1-x}}{2 x} d x =\frac{1}{2}\left[ \underbrace{\int \frac{\sqrt{1+x}}{x}-d x}_{J}+\underbrace{\int \frac{\sqrt{1-x}}{x} d x}_{K}\right] \end{aligned}$
$$
\begin{aligned}
J &=\int \frac{\sqrt{1+x}}{x} d x \\
&=\int \frac{1+x}{x \sqrt{1+x}} d x \\
&=\int \frac{1+x}{x} d (\sqrt{1+x}) \\
&=\int\left(\frac{1}{x}+1\right) d(\sqrt{1+x}) \\
&=\int \frac{d (\sqrt{1+x})}{(\sqrt{1+x})^{2}-1}+\sqrt{1+x} \\
&=\frac{1}{2} \ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+\sqrt{1+x}+c_{1}
\end{aligned}
$$$$
K \stackrel{x\mapsto -x}{=} \sqrt{1-x}+\frac{1}{2} \ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|+c_{2}
$$
Now we can conclude that $$
I=\sqrt{1-x}+\sqrt{1+x}+\frac{1}{2} \ln \left| \frac{(\sqrt{1+x}-1)(\sqrt{1-x}-1)}{(\sqrt{1+x}+1)(\sqrt{1-x}+1)}\right|+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find every equation of the line that passes through the point $(5,13)$ "Find every equation of the line that passes through the point $(5,13)$ and passes both axis at non-negative, whole values."
Here's my attempt:
Finding first two equations, with $k=\pm1$ is fairly simple. After that, plugging in the $x=5$ and $y=13$ in the equation yields $b=13-5k$. Since the line passes the $y$ axis at $(0,b)$, $b$ has to be whole. That means $13-5k$ has to be whole, $\implies 5k \in Z$.
Only non-negative value of $k$ that passes through the $x$ axis at non-negative value is $k=1$, so for every other line $k<0$. For lines with $k<0$, $b>13$ and value of $x >=6$. $$kx+b=0$$ $$x=\frac{-b}{k}$$ $$\frac{5k-13}{k} \geq 6 $$ $$ k \leq -13 \implies b\leq 78$$
I'm not sure how to proceed from here on, I could just check for each value of $b \in (13,78]$ but that doesn't seem very efficient.
What am I missing? Is my way of doing this correct? Or is there a better way? And if my attempt is correct, how do I proceed?
| A few things to note: If $b$ is the $y$ intercept of line and $c$ is the $x$ intercept then the line goes through $(0,b), (c,0)$ and the slope is $m =-\frac bc$. As $b,c$ are positive the slope is negative.
Point $(5,13),(0,b),(c,0)$ is on the line and the line has a negative slope. So $\frac {13-b}5 =\frac {13}{c-5} < 0$. Therefore $b > 13$. $c > 5$. $b=18\iff c=18$. And $b< 18\implies c >18$ and $b>18\implies c<18$.
So we just need to test $b=14..17$ and $c >18$. And $c=6....17$ and $b > 18$. And $b = c =18$.
The equation for a line is $y=mx + b = m(x-c)$.
And as $(5,13)$ is on the line $13 = -5\frac bc + b= -\frac bc(5-c)$
Which tells us that $c=\frac {5b}{b-13}$ and that $b=\frac {13c}{c-5}$
Case 1: $b=c=18$ then $m = -1$ then $13 = -5+b$ and $13=-(5-c)$ so $b=c =18$ and the equation is $y=-x + 18$
Case 2: $13 < b < 18$ and $c > 18$. But $c =\frac {5b}{b-13}$ is a positive number and plugging in $b=14...17$ we have $b-13=1...4$ so relatively prime to $5$ so we $b-13|b$ and $1|14$ and $2\not \mid 15$ and $3\not|16$ but $4\not \mid 17$ so the only such equation is $b=14$ and $c = 5*14= 70$ and $y=-\frac 15x + 14$
Case 3: $b > 18$ and $5 < c < 18$ But we have $b=\frac {13c}{c-5}$ so we have $c-5 = 1....12$ dividing in $13c$ so $c-5|c$ and we have $1|6$ and $5|10$
so we we have $c=6;b=6*13=78$ and so $y =-13x + 78$ or $c=10; b=26$ and $y =-\frac {13}5x + 26$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
limit of the sequence $x_{n}:= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n\ldots}}}$ I was thinking what happens with the sequence $\{x_n\}_{n\in \Bbb N}$ where:
$$x_{n}:= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n\ldots}}}$$
When you look some terms, for example $x_{1}=1$, $x_{2}=\sqrt[]{2 \sqrt[]{2 \sqrt[]{2 ...}}}$, $x_{3}=\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}$, these terms and the others will be continued fractions, where each one converges.
I'm asking what happens with $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}$ ?. I have an idea and it is $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}=1$. My reasoning is in the fact:
$$\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}= \displaystyle {n^{\frac{1}{n}}} n^{\frac{1}{n^2}} n^{\frac{1}{n^3}}...$$
And you know that:
$${\displaystyle \frac{1}{n}> \frac{1}{n^k} \textrm{ for } n,k \in \Bbb N}$$
Then:
$$n^{\frac{1}{n}}> n^{\frac{1}{n^k}} \geq 1$$
Like $\lim\limits_{n \to \infty}n^{\frac{1}{n}}=1$ and $\lim\limits_{n \to \infty}1=1$, by the Squeeze Theorem $\lim\limits_{n \to \infty}\sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}=1$. Is this reasoning correct? What do you think about $x_{n}$? Do you think there is another way to prove it? I receive suggestions or comments. Thank you.
| An Inductive idea is:
$$x_{2}=\sqrt[]{2 \sqrt[]{2 \sqrt[]{2 ...}}}\to 2\\
x_{3}=\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}\to \sqrt[2]3\\
x_{4}=\sqrt[4]{4 \sqrt[4]{4 \sqrt[4]{4 ...}}}\to \sqrt[3]4\\
x_{5}=\sqrt[5]{5 \sqrt[5]{5 \sqrt[5]{5 ...}}}\to \sqrt[4]5\\\vdots\\
x_{n}= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}\to \sqrt[n-1]n$$and it tends to $$\sqrt[n-1]n=n^{\frac{1}{n-1}}\to 1$$
Implicit : idea to solve for example$$\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}=a\to \text{to the power of 3}\\a^3=3\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}\\a^3=3\underbrace{\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}}_{a}\\a^3=3a\underbrace{\to}_{a\neq 0}a^2=3\to a=\sqrt 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Simplifying $\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$
Simplify $$\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}\,.$$
I tried very hard but I am not being able to solve it easily I opened up everything and multiplied all of it and got the answer -2. But it took me 1 hour and I also made many silly mistakes. Is there a quicker way than brute force?
| First of all, we must have $$a\ne b\ne c$$
Take out the terms containing $b^2$ in the numerator
$$b^2\left(-\dfrac1{(a-b)(c-a)}+\dfrac1{(b-c)(a-b)}-\dfrac1{(c-a)(b-c)}\right)$$
$$=b^2\cdot\dfrac{-(b-c)+c-a-(a-b))}{(a-b)(b-c)(c-a}$$
$$=\dfrac{2b^2(c-a)}{(a-b)(b-c)(c-a)}$$
$$\implies\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$$
$$=\dfrac2{(a-b)(b-c)(c-a)}\cdot\left( b^2(c-a)+c^2(a-b)+a^2(b-c)\right)$$
Now $b^2(c-a)+c^2(a-b)+a^2(b-c)$
$=b^2(c-a)+c^2a-bc^2+a^2b-ca^2$
$=b^2(c-a)+ca(c-a)-b(c^2-a^2)$
$=(c-a)(b^2-b(c+a)+ca)$
$=\cdots$
$=-(a-b)(b-c)(c-a)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Solution verification: $ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $ This limit is not too difficult but I was just wondering if my work/solution looked good?
Thanks so much for your input!!
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $$
$$ 2 x - 6 = 2 x \left( 1 - \frac 6 { 2 x } \right) $$
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } =
\lim _ { x \to 3 } \frac { 2 x \left( 1 - \frac 6 { 2 x } \right) } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \frac { x \left( 1 - \frac 6 { 2 x } \right) } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \frac { x - 3 } { \sqrt x - \sqrt 3 } $$
By rationalizing the denominator:
$$ \frac { x - 3 } { \sqrt x - \sqrt 3 } = \sqrt x + \sqrt 3 $$
$$ 2 \cdot \lim _ { x \to 3 } \frac { x - 3 } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \left( \sqrt x + \sqrt 3 \right) $$
By plugging in $ x = 3 $:
$$ 2 \cdot \lim _ { x \to 3 } \left( \sqrt x + \sqrt 3 \right) =
2 \left( \sqrt 3 + \sqrt 3 \right) = 4 \sqrt 3 $$
| Shortly: for $x\ne\sqrt 3$,
$$\frac{2x-6}{\sqrt x-\sqrt3}=2\frac{(\sqrt x-\sqrt3)(\sqrt x+\sqrt3)}{\sqrt x-\sqrt3}=2(\sqrt x+\sqrt3)\to4\sqrt3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How can i solve $\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$ How can i solve following $$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$$
My work:
I substituted $x=\tan\theta$, $dx=\sec^2\theta d\theta $
integral becomes
$\int \dfrac{\tan^3\theta+2\tan \theta-7}{\sqrt{\tan^2\theta+1}}\ \sec^2\theta d\theta$
$\int \dfrac{\tan\theta(\tan^2\theta+1)+\tan \theta-7}{\sec\theta}\sec^2\theta d\theta$
$\int (\tan\theta(\sec^2\theta)+\tan \theta-7)\sec\theta d\theta$
$\int \tan\theta\sec^3\theta\ d\theta+\int \sec\theta \tan \theta\ d\theta-7\int \sec\theta d\theta$
$\int \tan\theta\sec^3\theta+\sec\theta -7\ln|\sec\theta+\tan\theta|+C$
I got stuck here in solving first part of above integral. I can't see the way to solve it. please help me solve it by substitution or other method. thanks
| \begin{aligned}
\frac{x^{3}+2 x-7}{\sqrt{x^{2}+1}} d x &=\int \frac{x^{3}+2 x-7}{x} d \sqrt{x^{2}+1} \\
&=\int\left(x^{2}+2-\frac{7}{x}\right) d \sqrt{x^{2}+1} \\
&=\int\left(x^{2}+1\right) d \sqrt{x^{2}+1}+\sqrt{x^{2}+1}-7 \int \frac{d \sqrt{x^{2}+1}}{x}\\
&=\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3}+\sqrt{x^{2}+1}-7 \int \frac{d \sqrt{x^{2}+1}}{\sqrt{\left(\sqrt{x^{2}+1}\right)^{2}-1}} \\
&=\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3}+\sqrt{x^{2}+1}+\frac{7}{2} \ln \left|\frac{\sqrt{x^{2}+1}+1}{\sqrt{x^{2}+1}-1}\right| +C\\
&=\frac{\sqrt{x^{2}+1}}{3}\left(x^{2}+4\right)+\frac{7}{2} \ln \left|\frac{\sqrt{x^{2}+1}+1}{\sqrt{x^{2}+1}-1}\right|+C
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Sums of powers of cosines and sines shifted by $2\pi/3$ I have stumbled across these two identities
$$
\begin{split}
\cos^2(x)+\cos^2(x+2\pi/3)+\cos^2(x+4\pi/3) &= 3/2,\\
\cos^4(x)+\cos^4(x+2\pi/3)+\cos^4(x+4\pi/3) &= 9/8.
\end{split}
$$
There is also the more intricate
$$
\begin{split}
\cos^2(x)\sin^2(x)+\cos^2(x+2\pi/3)\sin^2(x+2\pi/3)+\cos^2(x+4\pi/3)\sin^2(x+4\pi/3) &= 3/8,\\
\cos^4(x)\sin^4(x)+\cos^4(x+2\pi/3)\sin^4(x+2\pi/3)+\cos^4(x+4\pi/3)\sin^4(x+4\pi/3) &= 9/128,
\end{split}
$$
and of course the most elementary
$$
\cos(x)+\cos(x+2\pi/3)+\cos(x+4\pi/3)=0.
$$
The last identity admits a rather intuitive interpretation in terms of unitary complex numbers centered about the origin. My questions are:
*
*Do the other identities admit similar more or less intuitive interpretations as well?
*Do such identities have names?
*Not all powers and combinations produce a constant; What is the general form of the expressions that do?
Context: The first two identities came up while calculating the elastic response of a two-dimensional truss (a planar lattice of nodes connected with springs) that is invariant by rotations of order 3, in which case $x$ describes the orientation of the truss. We know that such trusses must exhibit an isotropic response and that justifies, in a rather convoluted manner, that these expressions must be constants. The other expressions I found by trial and error. I am looking for a satisfying, non-brute-force, non-too-group-theoretic, explanation.
| If $\cos3y=\cos3x$
$3y=2n\pi\pm3x$ where $n$ is any integer
$y=\dfrac{2n\pi}3+x$ where $n=0,1,2$
Again, $\cos3y=4\cos^3y-3\cos y$
So, the roots of $$4\cos^3y-3\cos y-\cos3x=0$$ are $p=\cos x,q=\cos\left(\dfrac{2\pi}3+x\right),r=\cos\left(\dfrac{4\pi}3+x\right)$
Using Vieta's formula, $$p+q+r=\dfrac04\ \ \ \ (1)\text{ and }pq+qr+rp=\dfrac{-3}4\ \ \ \ (2)\text{ and }pqr=\dfrac{\cos3x}4\ \ \ \ (3)$$
By $(1),(2)$ $$p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+rp)=?\ \ \ \ (4)$$
By $(1),(3)$ $$p^3+q^3+r^3=3pqr=?\ \ \ \ (5)$$
A little Transformation of equation
*
*Let $c=\cos^2y$
$$(\cos3x)^2=(4\cos^3y-3\cos y)^2$$
$$\implies16c^3-24c^2+9c-\cos^23x=0$$ whose roots are $p^2,q^2,r^2$
Again applying Vieta's formula, $$p^2+q^2+r^2=\dfrac{24}{16}\ \ \ \ (6)\text{ compare with }(4)$$
$$p^2q^2+q^2r^2+r^2p^2=\dfrac9{16}\ \ \ \ (7)\text{ and } p^2q^2r^2=\dfrac{\cos^23x}{16}\ \ \ \ (8)\text{ compare with }(3)$$
By $(6),(7)$ $$p^4+q^4+r^4=(p^2+q^2+r^2)^2-2(p^2q^2+q^2r^2+r^2p^2)=?\ \ \ \ (9)$$
*
*Let $s=\dfrac1{\cos y}$
$$\dfrac4{s^3}-\dfrac3s-\cos3x=0\iff(\cos3x)s^3+3s^2-4=0$$ whose roots are $\dfrac1p,\dfrac1q,\dfrac1r$
$$\implies\dfrac1p+\dfrac1q+\dfrac1r=-\dfrac3{\cos3x}=-3\sec3x\ \ \ \ (10)$$
Similarly, $$\dfrac1{pq}+\dfrac1{qr}+\dfrac1{rp}=?\ \ \ \ (11)\text{ and }\dfrac1{pqr}=?\ \ \ \ (12)$$
Finally as $\dfrac1p=\sec x$ etc.,
using $(10,11),$ $$\sec^2x+\sec^2\left(\dfrac{2\pi}3+x\right)+\sec^2\left(\dfrac{4\pi}3+x\right)=\left(\dfrac1p+\dfrac1q+\dfrac1r\right)^2-2\left(\dfrac1{pq}+\dfrac1{qr}+\dfrac1{rp}\right)=?$$
Generalization
$$\cos ny=\cos nx$$ Can this be left as an exercise?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
plane perpendicular to the straight line I have this problem to find the equation of the plane perpendicular to the straight line joining the points $(1, 3, 5)$ and $(4, 3 ,2)$ at its middle points. How would you solve this?
| The formula for it is: $(X-X_0)\cdot \vec{n}=0$ whereas $X = (x,y,z), X_0 = \left(\frac{1+4}{2}, \frac{3+3}{2}, \frac{2+5}{2}\right)=\left(\frac{5}{2}, 3, \frac{7}{2}\right)$ , and $\vec{n}= \dfrac{1}{\sqrt{3^2+0^2+(-3)^2}}\left(4-1,3-3,2-5\right)=\dfrac{1}{2\sqrt{3}}\left(3,0,-3\right)=\left(\frac{\sqrt{3}}{2}, 0,\dfrac{-\sqrt{3}}{2}\right)$. Can you complete the equation?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve $\int\limits_0^{1/\sqrt{2}} \frac{au^2}{5(1-u^2)^2}du = 1$ for $a$ Problem:
The function $f_U(u) = \frac{au^2}{5(1-u^2)^2}$ is a probability density for the random variable $U$, which is non-zero on the interval $(0, \frac{1}{\sqrt{2}})$. I am supposed to find the value $a$.
I understand that that amounts to solving the equation $\int_0^{\frac{1}{\sqrt{2}}} \frac{au^2}{5(1-u^2)^2}du = 1$ for $a$.
Hint:
The hint I am given is to use integration by parts on the left hand side of $$\int_0^{\frac{1}{\sqrt{2}}} \frac{1}{1-u^2}du = \log(1 + \sqrt2)$$
My Attempt:
I don't understand how to use this hint. I know how to integrate by parts.
If the standard expression for integrating by parts is $st\bigg\rvert_0^{\frac{1}{\sqrt{2}}} - \int_0^{\frac{1}{\sqrt{2}}}tds$, then I guess I am supposed to find the appropriate $s$ and $dt$ in the expression $\dfrac{u^2}{(1-u^2)^2}$? In particular, I am supposed to choose $s$ and $dt$ such that $tds = \dfrac{1}{1-u^2}$?
I am led to believe that this is not a particularly nasty integral, though the online integral calculators do some sort of partial fraction decomposition that looks like a mess.
Another idea I had was to do a trigonometric substitution, but I don't think that will work.
| Start by pulling the $\frac{a}{5}$ out of the integral. You know how to factor the denominator into linear factors, so perform partial fraction decomposition.
$$ \frac{u^2}{(1-u^2)^2} = \frac{1/4}{u-1} + \frac{1/4}{(u-1)^2} + \frac{-1/4}{u+1} + \frac{1/4}{(u+1)^2} \text{.} $$
The antiderivatives of these terms are, respectively,
*
*$\frac{1}{4} \ln|u-1| + C$,
*$\frac{1}{4} \frac{-1}{u-1} + C$,
*$\frac{-1}{4} \ln|u+1| + C$, and
*$\frac{1}{4} \frac{-1}{u+1} + C$.
Then your integral is
$$ \frac{a}{20}\left( \left( \ln\left|\frac{1}{\sqrt{2}} - 1\right| + \frac{-1}{\frac{1}{\sqrt{2}} - 1} - \ln\left|\frac{1}{\sqrt{2}} + 1\right| + \frac{-1}{\frac{1}{\sqrt{2}} + 1} \right) - \left( \ln\left|0 - 1\right| + \frac{-1}{0 - 1} - \ln\left|0 + 1\right| + \frac{-1}{0 + 1} \right) \right) $$
$$ = \frac{a}{20} \left( 2\sqrt{2} +\ln \frac{1-1/\sqrt{2}}{1+1/\sqrt{2}} \right) $$
$$ = \frac{a}{10} \left( \sqrt{2} +\ln (\sqrt{2}-1) \right) $$
(Using an identity and the fact that the (inverse) hyperbolic tangent is an odd function, we can see this is equivalent to other answers here.)
(For the last bit: \begin{align*}
\ln \frac{1-1/\sqrt{2}}{1+1/\sqrt{2}}
&= 2 \ln \sqrt{\frac{1-1/\sqrt{2}}{1+1/\sqrt{2}} } \\
&= 2 \ln \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} } \\
&= 2 \ln \sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)} } \\
&= 2 \ln \sqrt{\frac{(\sqrt{2}-1)^2}{2-1} } \\
&= 2 \ln \left| \sqrt{2}-1 \right| \\
&= 2 \ln ( \sqrt{2}-1 ) \text{.}
\end{align*})
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3752754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Derivative of $y = \log_{\sqrt[3]{x}}(7)$. Never dealt with a derivative of these type. My approach was $$y = \log_{\sqrt[3]{x}}(7) \iff 7 = (\sqrt[3]{x})^y.$$ Then,
$$\frac{d}{dx}(7) = \frac{d}{dx}\left(\sqrt[3]{x}\right)^y \Rightarrow (\sqrt[3]{x})^y = e^{\frac{y\ln(x)}{3}} $$
From here,
$0 = e^u\dfrac{du}{dx}$ and $u = \dfrac{y\ln(x)}{3}.$ Thus,
$$0 = \frac{du}{dx} = \frac{y}{3x} +\frac{\ln(x)}{3}\frac{dy}{dx}.$$
Which implies that $$\frac{dy}{dx}= \frac{-\log_{\sqrt[3]{x}}(7)}{x\ln(x)}.$$
Is this the correct derivative? Can I alternatively use $\log_{b}(a) = \dfrac{\ln(a)}{\ln(b)}$, with $b = \sqrt[3]{x}$ and $a=7$? In that case, I arrive at
$$\frac{dy}{dx}= \dfrac{-3\ln(7)}{x(\ln(x))^2}.$$
| $$\frac{\mathrm{d}}{\mathrm{d}x}\left(\log_{\sqrt[3]{x}}(7)\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\dfrac{3\ln\left(7\right)}{\ln\left(x\right)}\right]$$
$$=3\ln\left(7\right)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{\ln\left(x\right)}\right]$$
$$=-\frac{3\ln\left(7\right)\cdot\dfrac{1}{x}}{\ln^2\left(x\right)}$$
$$=-\frac{3\ln\left(7\right)}{x\ln^2\left(x\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluating Sum at bounds I have to find an expression in terms of n using standard results for $$\sum_{r=n+1}^{2n} r(r+1)$$
And have found the general equation
$$\sum_{r=n+1}^{2n} r(r+1) = \frac{2n^3+6n^2+4n}{6}$$
However evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n+1)^3+6(n+1)^2+4(n+1)}{6}$$
does not yield the correct answer, yet evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n)^3+6(n)^2+4(n)}{6}$$
gives the correct answer
Im at a loss here, why am I not getting the correct answer by finding the difference of the sum between the two bounds?
| Your general equation should be
$$\begin {align} \sum_{r=n+1}^{2n} r(r+1)&=\sum_{r=n+1}^{2n} (r^2+r)\\
&=\sum_{r=1}^{2n} (r^2+r)-\sum_{r=1}^{n} (r^2+r)\\
&=\frac 16\left((2n)(2n+1)(4n+1)\right)+\frac 12\left(2n(2n+1)\right)-\frac 16\left((n)(n+1)(2n+1)\right)+\frac 12\left(n(n+1)\right)\\
&=\frac 16\left(16n^3+12n^2+2n\right)+\frac 12\left(4n^2+2n)\right)-\frac 16\left(2n^3+3n^2+n\right)+\frac 12\left(n^2+n\right)\\
&=\frac 16\left(14n^3+9n^2+n\right)+\frac 12\left(3n^2+n)\right)\\
&=\frac 16\left(14n^3+18n^2+4n\right)
\end {align}$$
which does not match your result and checks with Alpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simplify $\sqrt{8-\sqrt{63}}$ I simplified the expression into $$\sqrt{8-3\cdot \sqrt{7}}$$ but my tutor said it wasn't the answer he was looking for. Can someone help me?
| if $x = \sqrt{8-\sqrt{63}},$ then $0<x<1$ and $x^2 - 8 = - \sqrt{63},$ then $x^4 - 16 x^2 + 64= 63,$ then
$$ x^4 - 16 x^2 + 1 = 0. $$ Also
$$ x^2 - 16 + \frac{1}{x^2} = 0 $$
Taking $$ u = x + \frac{1}{x} $$
we get $u^2 - 18 = 0 $ and $$ u = \sqrt {18} $$
and
$$ x = \frac{3 \sqrt 2 \pm \sqrt{14}}{2} $$
and $x<1$ gives
$$ \color{blue}{ x = \frac{3 \sqrt 2 - \sqrt{14}}{2} } $$
Let's see, I could have chosen $$ v = \frac{1}{x} - x > 0 $$
with $v^2-14 = 0,$ then $v = \sqrt{14}$ or
$$ x^2 + \sqrt{14} x - 1 = 0, $$
$$ x = \frac{- \sqrt{14} \pm \sqrt{18}}{2} $$
and $x>0$ gives
$$ \color{red}{ x = \frac{- \sqrt{14} + \sqrt{18}}{2} } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Problem with proving inequalities Question:
Prove that if $x,y,z$ are positive real numbers such that $x+y+z=a$ then $(a-x)(a-y)(a-z)>\frac8{27}a^3$ is not true.
My Approach:
$$\frac{a-x}{2}=\frac{y+z}2$$
$$\frac{a-y}{2}=\frac{x+z}2$$
$$\frac{a-z}{2}=\frac{x+y}2$$
Using $AM>GM$ we get $$\frac{x+y+z}{3}>\root 3 \of {xyz}$$ Cubing both sides and multiplying by $8$, $$\frac{8a^3}{27}>8xyz$$
Also, by $AM>GM$, $$(\frac{y+z}2)(\frac{x+z}2)(\frac{x+y}2)>8xyz$$
Now, how do I find the relation between $(\frac{y+z}2)(\frac{x+z}2)(\frac{x+y}2)$ and $\frac{8a^3}{27}$?
| Use AM-GM of three items as
$$F=[(a-x)(a-y)(a-z)]^{1/3} \le \frac{3a-(x+y+z)}{3}$$
$$\implies (a-x)(a-y)(a-z) \le \frac{8a^3}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find an inequality with more precise bound? In an other topic, i found that : $x-1\le\lfloor x\rfloor\le x$ which really easy to show.
However, i want to found an inequality with more precise bound.
Like i found that : $$\forall x\in \mathbb R\text{, }x-1\le x-1+\frac{1}{\pi}\vert\sin{\pi x}\vert\le \lfloor x\rfloor\le x -\frac{1}{\pi}\vert\sin{\pi x}\vert\le x$$
So in the end, we have a more precise inequality :$$x-1+\frac{1}{\pi}\vert\sin{\pi x}\vert\le \lfloor x\rfloor\le x -\frac{1}{\pi}\vert\sin{\pi x}\vert$$
But now i just want to know what more precise inequality you could find out !
Just two things :
*
*The boundaries must not have any floor or ceil function in them
*The inequality must work for all x in the real numbers
I am interested what we could find out !
| Let us prove that
$$x-1+\frac{1 + \frac{1}{6}(\sin \pi x)^2}{\pi}\vert\sin{\pi x}\vert\le
\lfloor x\rfloor\le x -\frac{1 + \frac{1}{6}(\sin \pi x)^2}{\pi}\vert\sin{\pi x}\vert .$$
By letting $y = \pi(x - \lfloor x\rfloor)\in [0, \pi)$ and $z = \pi (1 - (x - \lfloor x\rfloor))\in (0, \pi]$, it suffices to prove that
$$(1 + \tfrac{1}{6}(\sin y)^2)\sin y \le y$$
and
$$(1 + \tfrac{1}{6}(\sin z)^2)\sin z \le z.$$
Thus, it suffices to prove that, for $u \in [0, \pi]$,
$$(1 + \tfrac{1}{6}(\sin u)^2)\sin u \le u.$$
It is easy to prove that $\sin u \le u$ and $\sin u \le u - \frac{1}{6}u^3 + \frac{1}{120}u^5$ for $u\in [0, \pi]$.
Thus, it suffices to prove that
$$\left(1 + \tfrac{1}{6}u^2\right)(u - \tfrac{1}{6}u^3 + \tfrac{1}{120}u^5) \le u$$
or
$$\frac{1}{720}u^5(14 - u^2) \ge 0.$$
It is true.
Remark: We may find the bounds of the form
$$x-1+\frac{f(\sin \pi x)}{\pi}\vert\sin{\pi x}\vert\le \lfloor x\rfloor\le x -\frac{f(\sin \pi x)}{\pi}\vert\sin{\pi x}\vert.$$
A better one is the following:
$$x-1+\frac{\frac{3}{2 + \sqrt{1 - (\sin \pi x)^2}}}{\pi}\vert\sin{\pi x}\vert\le
\lfloor x\rfloor\le x -\frac{\frac{3}{2 + \sqrt{1 - (\sin \pi x)^2}}}{\pi}\vert\sin{\pi x}\vert .$$
Here we have used the Shafer-Fink’s inequality for $\arcsin x$:
$$\frac{3x}{2 + \sqrt{1-x^2}} \le \arcsin x \le \frac{\pi x}{2 + \sqrt{1-x^2}}, \ x\in [0, 1]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the number of $6$ digit positive integers whose sum of the digits is at least $52$?
What is the number of $6$ digit positive integers whose sum of the digits is at least $52$?
My kind of approach was:
I thought to use Multinomial Theorem concept here.
So my primary aim was to find the coefficient of $x^{52}$in {followed by $x^{53}$ and $x^{54}$[$54$ will be highest because $6\cdot 9=54$]}
$$(x^{1}+x^{2}+x^{3}+\ldots+x^{9})^{6};$$
$=\text{finding coefficient of } x^{52 } \text{ from } x^{6}\cdot (1-x^{9})^{6}\cdot (1-x)^{-6}$
$=\text{this is equivalent to finding coefficient of } x^{46 } \text{ from } (1-x^{9})^{6}\cdot (1-x)^{-6}$
$=(1-x^{9}+x^{18}-x^{27}+x^{36}-x^{45})\cdot (1-x)^{-6}$[rest term will be a waste]
This leads to:
$\binom{51}{46}-\binom{42}{37}+\binom{33}{28}-\binom{24}{19}+\binom{15}{10}-\binom{6}{1}$
But it is a wrong answer. How to use Multinomial correctly?
| You should have:
$$[x^{52}](1+x+x^2+\cdots +x^9)^6=[x^{52}]\left(\frac{1-x^{10}}{1-x}\right)^6=\\
[x^{52}](1-x^{10})^6\cdot (1-x)^{-6}=\\
[x^{52}]\sum_{i=0}^6 {6\choose i}(-x^{10})^i\cdot \sum_{j=0}^{\infty}{-6\choose j}(-x)^j=\\
[x^{52}]\sum_{i=0}^6 {6\choose i}(-x^{10})^i\cdot \sum_{j=0}^{\infty}{6+j-1\choose j}x^j=\\
\small{57\choose 52}-{6\choose 1}{47\choose 42}+{6\choose 2}{37\choose 32}-{6\choose 3}{27\choose 22}+{6\choose 4}{17\choose 12}-{6\choose 5}{7\choose 2}=21$$
Wolfram answer.
Indeed, to have the sum of $52$, the digits should be $799999$ or $889999$:
$${6\choose 1}+\frac{6!}{2!4!}=21.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many 6 digit numbers can be formed from two sets of digit? There are two sets of digit :
$ \text{set 1 :} \{~0,1,2,3,4~\}$
$ \text{set 2 :} \{~5,6,7,8,9~\}$
Now how many 6 digit number can we make by taking numbers from these two sets ? From $\text{set 1}$ repetition is permitted but from $\text{set 2}$ repetition is prohibited .
Solution :
The possible combinations are :
\begin{array}{c|cccc}
\ combo &\text{group 1}& \text{group 2} & \text{without considering 0} &\text{considering 0}&\text{result}\\
\hline
1 & 6 & \fbox0 & 5^6 & 5^6 - 5^5 & 12,500\\
2 & 5 & \fbox1 & 5 \times 5 \times 5 \times 5 \times 5 \times \fbox 5& 5 \times 5 \times 5 \times 5 \times 5 \times \fbox 5 - 5^5 & 12,500\\
3 & 4 & \fbox2 & 5 \times 5 \times 5 \times 5 \times \fbox {$5 \times 4 $}& (5 \times 5 \times 5 \times 5 \times \fbox {$5 \times 4 $})- (5 \times 5 \times 5 \times \fbox {$5 \times 4 $}) & 10,000\\
4 & 3 & \fbox3 & 5 \times 5 \times 5 \times \fbox {$5 \times 4 \times 3 $}& (5\times 5 \times 5 \times \fbox {$5 \times 4 \times 3 $})-(5 \times 5 \times \fbox{$5 \times 4 \times 3 $}) & 6000 \\
5 & 2 & \fbox4 & 5 \times 5 \times \fbox {$5 \times 4 \times 3 \times 2$} & (5 \times 5 \times \fbox {$5 \times 4 \times 3 \times 2$}) - ( 5 \times \fbox {$5 \times 4 \times 3 \times 2$}) & 2400\\
6 & 1 & \fbox5 & 5 \times \fbox {$5 \times 4 \times 3 \times 2 \times 1$} & (5 \times \fbox {$5 \times 4 \times 3 \times 2 \times 1$}) - (\fbox {$5 \times 4 \times 3 \times 2 \times 1$}) & 480\\
\end{array}
So,total number of possible combinations is $= 43880$
Is this procedure correct ?
| Yes, but I don't see you taking into account the positioning of the digits. For example, in combo 2, the digit from group 2 can be in one of the 6 places, while you let it be only in the last (it will affect also the 0 consideration).
Sanity check: assume we don't allow the usage of any digit twice. There are 9∗9∗7∗6∗5∗4=68K possible numbers like that. If you allow using some of the digits twice, you should get more options. Your answer, 43K is too small
Solution:
You can write an explicit formula in the following way: $k$ out of the $6$ digits should be from group 1. We need to choose where they will be placed - ${6 \choose k}$, and in each location choose a digit - $5^k$.
For the rest $6-k$ locations - you need to choose which digits will be placed there and in what order: ${5 \choose 6-k}(6-k)!=\tfrac{5!}{(k-1)!}$.
Now, sum over all possi$k$ to obtain
$$\sum\limits_{k=1}^6 {6 \choose k} 5^k \tfrac{5!}{(k-1)!}=495475$$
From that, we need to remove the ones that start with a $0$ (he is chosen to be at place 1):
$$\sum\limits_{k=1}^6 {5 \choose k-1} 5^{k-1} \tfrac{5!}{(k-1)!}=61870$$
Total: 433605.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.