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Prove : $\sqrt{\frac{a^2+b^2+1}{a^2+c^2+1}}+\sqrt{\frac{a^2+c^2+1}{b^2+c^2+1}}+\sqrt{\frac{b^2+c^2+1}{a^2+b^2+1}}\geq 3$ with $a+b+c=1$ Claim :
Let $a\geq b\geq c\geq 0$ and $a+b+c=1$ then we have :
$$\sqrt{\frac{a^2+b^2+1}{a^2+c^2+1}}+\sqrt{\frac{a^2+c^2+1}{b^2+c^2+1}}+\sqrt{\frac{b^2+c^2+1}{a^2+b^2+1}}\geq 3$$
To prove it I try the Ji chen's theorem because we have :
$$\frac{a^2+b^2+1}{a^2+c^2+1}+\frac{a^2+c^2+1}{b^2+c^2+1}+\frac{b^2+c^2+1}{a^2+b^2+1}\geq 3$$
$$\frac{a^2+b^2+1}{a^2+c^2+1}\frac{a^2+c^2+1}{b^2+c^2+1}+\frac{a^2+b^2+1}{a^2+c^2+1}\frac{b^2+c^2+1}{a^2+b^2+1}+\frac{a^2+c^2+1}{b^2+c^2+1}\frac{b^2+c^2+1}{a^2+b^2+1}\geq 3$$
$$\frac{a^2+b^2+1}{a^2+c^2+1}\frac{a^2+c^2+1}{b^2+c^2+1}\frac{b^2+c^2+1}{a^2+b^2+1}=1$$
To prove the two first I use Karamata's inequality as we have .
Let $x,y,z,u,v,w>0$ such that $x\geq y \geq z$ and $u\geq v \geq w$ and :
$x\geq u$,
$xy\geq uv$,
$xyz\geq uvw$
Then we have :
$$x+y+z\geq u+v+w$$
Question :
Is there simpler ?
Thanks in advance !
Regards
Max
| Hint:clearly this is of form $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge 3$$ which is true by am-gm
| {
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"timestamp": "2023-03-29T00:00:00",
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Polynomial with root $α = \sqrt{2}+\sqrt{5}$ and using it to simplify $α^6$ Find a polynomial $\space P(X) \in \mathbb{Q}[X]\space$ of degree 4 such that
$$\alpha = \sqrt{2} + \sqrt{5}$$
Is a root of $P$.
Using this polynomial, find numbers $\space a, b, c, d \space$ such that
$$\alpha^{6} = a + b\alpha + c\alpha^{2} + d\alpha^{3}$$
What have I tried so far?
I know obviously that for $\alpha$ to be a root of $P$, then $(x-\alpha)$ must be part of the polynomial. Hence, $(x-\sqrt{2} - \sqrt{5})$ will be a factor of the polynomial. Where I’m getting stuck is what to do next in order to find the other factors of the polynomial such that I get values $a, b, c$ and $d$ that satisfy the equation with $\alpha^{6}$.
Any help would be greatly appreciated!
| Hint
$a^2=(\sqrt{2}+\sqrt{5})^2=2+5+2\sqrt{10}=7+2\sqrt{10}\Rightarrow a^2-7=2\sqrt{10} \Rightarrow (a^2-7)^2=40$
Edit
$(a^2-7)^2=40 \Rightarrow a^4-14a^2+49=40 \Rightarrow a^4-14a^2+9=0$
| {
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Evaluating $\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$ I was trying to evaluate
$$\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$$
I have tried taking natural logarithm first:
$\lim_{x\to0}\frac{\ln(\sin x)-\ln x}{1-\cos x}=\lim_{x\to0}\frac{\frac{\cos x}{\sin x}-\frac{1}x}{\sin x}\quad\quad\text{(L'Hopital Rule)}\\
=\lim_{x\to0}\frac{\cos x}{x\sin x}-\frac{1}{x^2}\quad\quad(\lim_{x\to0}\frac{\sin x}x=1)\\
=\lim_{x\to0}\frac{\cos x-1}{x^2}\quad\quad\quad(\lim_{x\to0}\frac{\sin x}x=1)$
and after this I eventually have the limit equaling $-\frac{1}2$, which means that the original limit should be $\frac{1}{\sqrt{e}}$.
However, I graphed $f(x)=\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$ on Desmos, and it turned out that the limit is approximately $0.7165313$, or $\frac{1}{\sqrt[3]{e}}$.
Therefore I think there's something wrong in my approach, but I couldn't find it. Any suggestions?
| Let me try an approach without Taylor series, starting with your step
$$\lim_{x \to 0} \frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{\sin x}=\lim_{x \to 0}\frac{x\cos x - \sin x}{x \sin^2 x}$$
We continue from here:
$$
\begin{align}
\lim_{x \to 0}\frac{x\cos x - \sin x}{x \sin^2 x} &\overset{\mathrm{H}}{=} \lim_{x \to 0} \ \frac{-x \sin x}{\sin^2x + 2x\sin x \cos x} \\
&= \lim_{x \to 0} \ \frac{-x}{\sin x + 2x \cos x}
\overset{\mathrm{H}}{=} \lim_{x \to 0} \ \frac{-1}{3\cos x-2\sin x}=\frac{-1}{3}
\end{align}
$$
and the result follows from there.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate the arc length of a difficult radical function I have been struggling with an arc length question, and I want to make sure I get this right. I have the function of:
\begin{align}
f(x) = \sqrt{7.2 (x-\frac {1}{7}}) - 2.023, [0.213, 0.127].
\end{align}
I have found the derivative of the function and set up my integral this way:
\begin{align}
I &= \int_{0.127}^{0.213} \sqrt{1 + \frac{12.96}{7.2x-\frac{7.2}{7}}}~dx
\end{align}
Letting A = 12.96 and simplifying:
\begin{align}
I &= \int_{0.127}^{0.213} \sqrt{\frac{7.2x-\frac{7.2}{7}+A}{7.2x-\frac{7.2}{7}}}~dx
\end{align}
$u=7.2x-\frac{7.2}{7}, du= 7dx, dx=\frac{du}{7}$:
\begin{align}
I &= \int_a^b \sqrt{\frac{{u}+A}{u}}~\frac{du}{7}
\end{align}
\begin{align}
I &= \frac{1}{7}\int_a^b \sqrt{\frac{{u}+A}{u}}~du
\end{align}
$u = C\tan^2v\\ du = 2C \tan v \sec^2 v ~ dv$
\begin{align}
I
&= \frac{1}{7}\int_{x=a}^{x=b} \sqrt{\frac{u + A}{u}}~du\\
&= \frac{1}{7}\int_{x=a}^{x=b} \sqrt{\frac{A(\tan^2 v + 1)}{A \tan^2 v}}~2A\tan v \sec^2 v ~ dv\\
&= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{\sec^2 v}{\tan^2 v}}\tan v \sec^2 v ~ dv; & \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{\frac{1}{\cos^2 v}}{\frac{\sin^2 v}{\cos^2 v}}}\frac{\sin v}{\cos^3 v}~ dv; \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{1}{\sin^2 v}}\frac{\sin v}{\cos^3 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\sin v}\frac{\sin v}{\cos^3 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\cos^3 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{\cos^4 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{(1-\sin^2(v))^2} dv~
\end{align}
This is where I am stuck. Could I make a substitution such as:
$t = \sin v\\dt=\cos v\ dt\\\frac{dt}{cos\ v}=dv$
and then:
\begin{align}
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{(1-t^2)^2} \frac{dt}{cos\ v}~\\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{(1-t^2)^2} dt~
\end{align}
which gives me an ordinary partial fractions integral.
Could I make this substitution or is it not possible because I would have 2 different variables in my integral, and if it's not possible, how else could I solve this integral?
| I amssuming you work prior to the $u$-substitution is correct.
Instead of using the substitution $u=C\tan^2(\nu)$, consider the more simple one $u=\nu^2$, which reduces your integral to
$$I=\frac{2\sqrt{A}}7\int_{\sqrt{a}}^{\sqrt{b}}\sqrt{1+\left(\frac{\nu}{\sqrt{A}}\right)^2}d\nu.$$
There are various approaches of evaluating the integral of $\sqrt{1+x^2}$. One way is to let $\nu=\sqrt{A}\text{ sinh}(s)$.
| {
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Solve the inequality $x^4-3x^2+5\ge0$
Solve $$\sqrt{x^4-3x^2+5}+\sqrt{x^4-3x^2+12}=7.$$
$D_x:\begin{cases}x^4-3x^2+5\ge0 \\x^4-3x^2+12\ge0\end{cases}.$ We can see that $x^4-3x^2+12=(x^4-3x^2+5)+7,$ so if $x^4-3x^2+5$ is non-negative, $x^4-3x^2+12$ is also non-negative (even positive). So I am trying to solve $$x^4-3x^2+5\ge0.$$Let $x^2=y,y\ge0.$ Now we have $$y^2-3y+5\ge0; D=9-4\times5<0,a=1>0$$ so the function $f(y)=y^2-3y+5>0$ for all $y$. I don't know what to do next. The solution of the inequality is indeed $x\in(-\infty;+\infty),$ but I have the restriction $y\ge0?$
| Denote:
$x^4-3x^2+5 = a - 7/2$
Then:
$x^4-3x^2+12 = a + 7/2$
Now you want to solve this one:
$\sqrt{a-7/2} + \sqrt{a + 7/2} = 7$
This is quite symmetrical and nice to work with.
Raise it to power $2$ and proceed, should be trivial from there.
At the end do a direct check to see if the values you found for $a$ give rise to valid roots for $x$. For example, if you get a solution for $a$ smaller than $7/2$, that obviously does not work i.e. does not give you any solutions for $x$.
| {
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Verification to solution an inequality and proving another. I need solution verification an inequality, that I have solved because it seems too good to be true.
But first, I attempted this but couldn't complete:
Let $a$, $b$ and $c$ be the sides of a triangle with perimeter $3$. Prove that
$$
\sum_{cyc}{\frac{a^2}{a + 2\sqrt{b} - 1}} \geqslant
\frac{ab^3 + bc^3 + ca^3 + 9abc}
{3(ab + bc + ca) - abc}
$$
Attempt:
By the constraint,
$$
\frac{ab^3 + bc^3 + ca^3 + 9abc}
{3(ab + bc + ca) - abc} =
\frac{
\sum_{cyc}{a^3b + 3a^2bc}
}{
\left(\sum_{cyc}{a^2(b + c)}\right) + 2abc
}
$$Which is by just writing $9$ as $3(a + b + c)$ and $3$ as $a + b + c$ in the $LHS$.
By $T_2$'s Lemma and then AM-GM Inequality,
$$
\sum_{cyc}{\frac{a^2}{a + 2\sqrt{b} - 1}}\geqslant \frac32
$$
Then what is left is
$$
\frac{a+b+c}2 \geqslant \frac{
\sum_{cyc}{a^3b + 3a^2bc}
}{
\left(\sum_{cyc}{a^2(b + c)}\right) + 2abc}
$$
$$
\Rightarrow\sum_{cyc}{a^3b + a^3c + a^2b^2 + a^2c^2 + 4a^2bc} \geqslant \sum_{cyc}{2ab^3 + 6a^2bc}
$$Then AM-GM again leaves us with
$$
\sum_{cyc}{a^3b}\geqslant \sum_{cyc}{ab^3}
$$Which means it's enough to prove
$$
\sum_{cyc}{a^2b - ab^2} \geqslant 0
$$But can't prove this. I have not used the fact that they are sides of triangle so maybe it is helpful somewhere. I wish alternative solutions to this inequality.
The second:
Let $x,y,z>0$ satisfy $xyz\geqslant1$. Prove that
$$
\frac {x^5 - x^2} {x^5 + y^2 + z^2} +
\frac {y^5 - y^2} {x^2 + y^5 + z^2} +
\frac {z^5 - z^2} {x^2 + y^2 + z^5} \geqslant 0
$$
I have proven the inequality but the solution seems too easy to me.
It is here:
$$
\sum_{cyc} {\frac{x^5 - x^2} {x^5 + y^2 + z^2}} \geqslant \sum_{cyc}{\frac{x^4 - x^2yz}{x^4 + y^3z +yz^3}} \geqslant \sum_{cyc}{\frac{x^4 - x^2yz}{x^4 + y^4 + z^4}} \geqslant 0
$$Which uses $$y^4 + z^4 \geqslant y^3z + yz^3 \Leftrightarrow (y - z)^2(y^2 + z^2 + yz)\geqslant 0\ \textrm{along with others}$$ and $$\sum_{cyc}{2x^4 + y^4 + z^4} \geqslant \sum_{cyc}{4x^2yz}$$Is this solution correct?
Thanks for comments and alternatives/extensions!
| The first inequality.
Since $$a+2\sqrt{b}-1=\frac{1}{3}(3a+6\sqrt{b}-a-b-c)=$$
$$=\frac{1}{3}\left(2a+2\sqrt{3b(a+b+c)}-b-c\right)>\frac{1}{3}(a+b-c)>0,$$ by AM-GM we obtain: $$\sum_{cyc}\frac{a^2}{a+2\sqrt{b}-1}\geq\sum_{cyc}\frac{a^2}{a+b+1-1}=\sum_{cyc}\frac{a^2}{a+b}$$ and it's enough to prove that:
$$\sum_{cyc}\frac{a^2}{a+b}\geq\frac{\sum\limits_{cyc}(a^3c+3abc)}{(a+b+c)(ab+ac+bc)-abc}$$ or
$$\sum_{cyc}\frac{a^2}{a+b}\geq\frac{\sum\limits_{cyc}(a^3c+3a^2bc)}{\prod\limits_{cyc}(a+b)}$$ or
$$\sum_{cyc}(a^3b+a^2b^2-2a^2bc)\geq0,$$ which is true by Rearrangement and SOS.
We'll prove the following general statement.
For positives $a$, $b$ and $c$ the triples $(a^2,b^2,c^2)$ and $(bc,ac,ab)$ have an opposite ordering.
Proof.
Since our claim is symmetric(is not changed after any permutations of $a$, $b$ and $c$),
we can assume that $a\geq b\geq c>0$.
Thus, $a^2\geq b^2\geq c^2$ and $bc\leq ac\leq ab$ and we are done.
By using of this statement and by Rearrangement we obtain: $$\sum_{cyc}a^3b=\sum_{cyc}(a^2\cdot ab)\geq \sum_{cyc}(a^2\cdot bc)=\sum_{cyc}a^2bc$$ and
$$\sum_{cyc}(a^2b^2-a^2bc)=\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$
Your solution of the second inequality is wrong because $x^4-x^2yz$ may be negative and you can not write $$\frac{x^4-x^2yz}{x^4+y^3z+yz^3}\geq\frac{x^4-x^2yz}{x^4+y^4+z^4}.$$
The second inequality we can prove by the following way.
Let $x=ka$, $y=kb$ and $z=kc$, where $k>0$ and $abc=1$.
Thus, $$k^3abc\geq1,$$ which gives $k\geq1.$
Now, by C-S we obtain: $$\sum_{cyc}\frac{x^5-x^2}{x^5+y^2+z^2}=\sum_{cyc}\frac{k^3a^5-a^2}{k^3a^5+b^2+c^2}=3+\sum_{cyc}\left(\frac{k^3a^5-a^2}{k^3a^5+b^2+c^2}-1\right)=$$
$$=3-\sum_{cyc}\frac{a^2+b^2+c^2}{k^3a^5+b^2+c^2}\geq3-\sum_{cyc}\frac{a^2+b^2+c^2}{a^5+b^2+c^2}=3-\sum_{cyc}\frac{bc(a^2+b^2+c^2)}{a^4+b^3c+bc^3}=$$
$$=3-\sum_{cyc}\frac{bc\left(1+\frac{b}{c}+\frac{c}{b}\right)(a^2+b^2+c^2)}{(a^4+b^3c+bc^3)\left(1+\frac{b}{c}+\frac{c}{b}\right)}\geq3-\sum_{cyc}\frac{bc\left(1+\frac{b}{c}+\frac{c}{b}\right)(a^2+b^2+c^2)}{(a^2+b^2+c^2)^2}=$$
$$=3-\sum_{cyc}\frac{2a^2+ab}{a^2+b^2+c^2}\geq3-\sum_{cyc}\frac{2a^2+a^2}{a^2+b^2+c^2}=0.$$
| {
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Quadrilateral with given angles
We are looking for angles x and y.
I have found the values of the following angles:
BEA = 74,
BDA = 64,
ACD = 68,
ECD = 112,
plus the relationship $x+y = 68$.
All other angles equations, from triangles or the sum of angles in the quadrilateral (360) end up in the same equation!
I have found through Geogebra that $x=18$ and $y=50$ but I can't figure out a second relationship to determine them geometrically!
Does anyone have any ideas?
Thank you!
| A proof by Sine Law goes as follows:
Considering triangles $AED, BED, ABD, ABE$:
\begin{align}
\frac {\sin y}{AD} &= \frac {\sin 48^\circ}{ED}\\
\frac {\sin x}{EB} &= \frac {\sin 38^\circ}{ED}\\
\frac {\sin 46^\circ}{AD} &= \frac {\sin 64^\circ}{BA}\\
\frac {\sin 22^\circ}{EB} &= \frac {\sin 74^\circ}{BA}\\
\end{align}
Equating $ED$ in the first two equations we have:
$$\frac {EB \sin 38^\circ}{\sin x} = \frac {AD \sin 48^\circ}{\sin y}$$
$EB$ and $AD$ can be expressed in terms of $BA$:
$$\frac {AB \sin 22^\circ \sin 38^\circ}{\sin 74^\circ \sin x} = \frac {AB \sin 46^\circ \sin48^\circ}{\sin 64^\circ \sin y}$$
Noting that $x = 68^\circ - y$,
$$\frac {\sin 22^\circ \sin 38^\circ}{\sin 74^\circ (\sin 68^\circ \cos y - \cos 68^\circ \sin y)} = \frac {\sin 22^\circ \sin 38^\circ}{\sin 74^\circ \sin (68^\circ - y)} = \frac {\sin 46^\circ \sin48^\circ}{\sin 64^\circ \sin y}$$
Rearranging:
$$(\sin 22^\circ \sin 38^\circ \sin 64^\circ + \sin 46^\circ \sin 48^\circ \sin 74^\circ \cos 68^\circ) \sin y = \sin 46^\circ \sin 48^\circ \sin 74^\circ \sin 68^\circ \cos y$$
Giving the expression for $\tan y$:
$$\frac {\sin 46^\circ \sin 48^\circ \sin 74^\circ \sin 68^\circ} {\sin 22^\circ \sin 38^\circ \sin 64^\circ + \sin 46^\circ \sin 48^\circ \sin 74^\circ \cos 68^\circ}$$
WolframAlpha says it is $50^\circ$. One could probably reduce the expression above, but right now I don't see how.
| {
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Algebra - Piecewise Function Let $p(x)$ be defined on $2 \le x \le 10$ such that$$p(x) = \begin{cases} x + 1 &\quad \lfloor x \rfloor\text{ is prime} \\ p(y) + (x + 1 - \lfloor x \rfloor) &\quad \text{otherwise} \end{cases}$$where $y$ is the greatest prime factor of $\lfloor x\rfloor.$ Express the range of $p$ in interval notation.
I do not not where to start.
I know at $x \in [2, 3) $, the range is $[3, 4)$. Should I continue with this? Like continuing with $x\in [3,4)$, then $x \in [4, 5)$ and so on?
| If $\lfloor x \rfloor \in \mathbb{P}$, then
$$ p(x) = x + 1 $$
So we cover the interval $[\lfloor x \rfloor + 1, \lfloor x \rfloor + 2)$
If $\lfloor x \rfloor \not\in \mathbb{P}$, then
$$ p(x) = p(y) + \{x\} + 1 = y + 2 + \{x\}$$
So we cover the interval $[y+2, y+3)$
The prime values of $\lfloor x \rfloor$ we take on in the domain are 2, 3, 5, 7.
So we cover the intervals [3, 4), [4, 5), [6, 7), [8, 9)
The non-prime values of $\lfloor x \rfloor$ are 4, 6, 8, 9 - they're largest prime factors are 2 and 3
So this lets us cover [4, 5) and [5, 6)
The union of all these intervals is [3, 7) union [8, 9)
We also have to consider the special case of $x = 10$, where p(10) = 6 + 1 = 7, which makes [3, 7) closed.
Therefore the range of p is $[3,7] \bigcup [8, 9)$
| {
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Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.
Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.
If $ab \mid (a+b)^²$, then $ab\mid a^2+2ab+b^2 \Longrightarrow ab\mid a^2, ab\mid 2ab$ and $ab\mid b^2$ right?
So since $(a-b)^2 = a^2-2ab+b^2$ from the assumption we have that $ab \mid a^2$ and $ab \mid b^2$. Now only remains to show that $ab \mid -2ab$ which is clearly true.
Is this valid? I'm not sure about the implication that $ab$ would divide all the terms in $a^2+2ab+b^2$.
| Not quite: in general, for $a,b,c\in\mathbb{Z}$, $a|b+c$ does not imply $a|b$ and $a|c$.
Instead, simply suppose $ab|a^2+2ab+b^2$. That is, $abk=a^2+2ab+b^2$ for some $k\in\mathbb{Z}$. The rest sort of follows like the end of your idea, if you subtract $4ab$ from both sides of the equation, then $abk-4ab=a^2-2ab+b^2$, so clearly $ab|(a-b)^2$.
| {
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Prove inequality $|a-b+c-d| \leqslant \frac{1}{16}$ Let $a,b,c,d$ be positive real numbers that fulfill two conditions:
$$a+b+c+d \leqslant 2$$$$ab+bc+cd+ad \geqslant 1$$
Prove that $|a-b+c-d|\leqslant \frac{1}{16}$
Let:
$a+c=x$ and $b+d=y$
Both $x$ and $y$ are positive.
$$x+y \leqslant 2$$$$xy \geqslant 1$$
$$-4xy \leqslant -4$$
$$(x+y)^2 \leqslant 4$$
$$(x-y)^2 \leqslant 0 \Rightarrow |a-b+c-d|=|x-y|=0\leqslant \frac{1}{16}$$
I do not really think that this is valid solution, but I can not find any mistakes.
| Your conclusion is correct. One can shorten the argument as follows:
$$
(a-b+c-d)^2 = (a+b+c+d)^2 - 4(a+c)(b+d) \\
= (a+b+c+d)^2 - 4(ab+bc+cd+da)\le 4 - 4 = 0
$$
which implies $a-b+c-d = 0$.
| {
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Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?
Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?
If you replace $\sec^2$ with $\tan^2 + 1$, it should be $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$, right?
However, it seems that it is not multiplication here : $\sec^2 (\tan^{-1} (\frac{x}{2}))$, when the $1$ becomes independent of the $\tan^2$ somehow.
Could someone explain why?
| The correct statement is $\sec^2(y)=1+\tan^2(y)$.
Now let $y=\tan^{-1}\left(\dfrac x2\right)$.
I hope this helps you see why.
| {
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"url": "https://math.stackexchange.com/questions/3884630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $\lim_{x\rightarrow0}\frac{1}{x^4}=\infty$ via $\epsilon$-$\delta$
Show that $\lim_{x\rightarrow0}\frac{1}{x^4}=\infty$.
Consider our preliminary work:
$$\begin{align}
|\frac{1}{x^4}-0|<\epsilon &\implies|x^4-0|<\frac{1}{\epsilon} \tag1\\
&\implies|x^2-0||x^2+0|<\frac{1}{\epsilon} \tag2\\
&\implies|x-0||x+0||x^2+0|<\frac{1}{\epsilon} \tag3
\end{align}$$
Choose $\delta=1$, then
$$|x-0|<\delta=1\implies -1<x<1\Rightarrow|x-0|<1 \tag4$$
Thus,
$$|x-0||x+0||x^2+0|<\frac{1}{\epsilon}\implies|x-0|<\frac{1}{\epsilon} \tag5$$ Now, let $\epsilon>0$ and choose $\delta= \min\{1,\frac{1}{\epsilon}\}$.
What should I do next in writing this proof? I'm confused how to choose $\delta$ so that $\frac{1}{\epsilon}$ becomes $\epsilon$. Also, do I need to pick multiple upper bounds since I broke my polynomial into three separate absolute values?
| Let $M$ be given. Then, take $\delta = \dfrac{1}{\sqrt[4]{M}}$. Thus,
$$0< |x| < \delta \implies |x| < \dfrac{1}{\sqrt[4]{M}} \implies |x^4| < \dfrac{1}{M} \implies |\dfrac{1}{x^4}| > M $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$ I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$
Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} - \frac{1}{m+1} \binom{n+1}{m+1} $$
to simplify the result.
| Here are more details for @Phicar's suggested approach.
Let $A(z)=\sum_{n \ge 0} a_n z^n$ be the ordinary generating function of $(a_n)$.
Then $z A'(z)=\sum_{n \ge 0} n a_n z^n$, and the recurrence relation implies that
\begin{align}
z A'(z)
&= \sum_{n \ge 5} \left((n-4)a_{n-1} + 12n H_n\right) z^n \\
&= z \sum_{n \ge 5} (n-1) a_{n-1} z^{n-1} - 3 z \sum_{n \ge 5} a_{n-1} z^{n-1} + 12z \sum_{n \ge 5} n H_n z^{n-1} \\
&= z \cdot z A'(z) - 3z A(z) + 12z\left(\frac{d}{dz}\left(\frac{-\log(1-z)}{1-z}\right) -\sum_{n=1}^4 n H_n z^{n-1}\right) \\
&= z^2 A'(z) - 3z A(z) + 12z\left(\frac{1-\log(1-z)}{(1-z)^2} -1-3z-\frac{11}{2}z^2-\frac{25}{3}z^4\right)
\end{align}
The resulting solution for $A(z)$ is messy, so there must be a better way.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration of $\int_{0}^{1} \frac{x^k}{1-x} dx$ Solving this using integration by parts and by letting $u = x^k$ and $dv = (1-x)^{-1}$, it follows:
$$\int_{0}^{1}\frac{x^k}{1-x} dx = -x^{k}\ln(1-x) + k\int_{0}^{1} \ln(1-x)x^{k-1} dx$$
It is known that
$$\ln(1-x) = (1-x) - (1-x)\ln(1-x)$$
But doesn't this make the integration more complicated. Can I obtain a recursive formula for such an integration?
| Let $$ \mathcal{I} = \int_0^1 \frac{x^k}{1-x} \, \mathrm{d}x $$ Now using integration by parts we get
\begin{align*}
\mathcal{I} &= -x^k\, \ln(1-x) \bigg{|}_0^1 + \int_0^1 k \, x^{k-1} \, \ln(1-x) \, \mathrm{d}x \\
&= 1 + \int_0^1 k \, x^{k-1} \left( \sum_{n=1}^\infty - \frac{x^n}{n} \right) \, \mathrm{d}x \\
&= 1 - \sum_{n=1}^\infty \frac{k}{n} \int_0^1 x^{k+n-1} \, \mathrm{d}x \\
&= 1 - \sum_{n=1}^\infty \frac{k}{n} \cdot \frac{1}{k+n} \\
&= 1 - \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{k+n} \right) \hspace{5mm} (1)
\end{align*}
Now remember Harmonic Series
(here I am using i and m because we have used n and k in our integral)
$$ \mathrm{H}_m = \sum_{i=1}^{m} \frac{1}{i} $$
We can write $\mathrm{H}_m$ in infinite series representation, see
$$ \mathrm{H}_m = \sum_{i=1}^{\infty} \left( \frac{1}{i} - \frac{1}{i+m} \right) $$
Using this we can clearly see our (1) is
$$ \boxed{ \mathcal{I} = 1 - \mathrm{H}_k} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that three numbers form an arithmetic progression
The numbers $a,b$ and $c$ form an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.
We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{ } \forall \text{ }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.
| $(a^2+ac+c^2)-(a^2+ab+b^2)=a(c-b)+c^2-b^2=(c-b)(a+b+c)\\(b^2+bc+c^2)-(a^2+ac+c^2)=b^2-a^2+(b-a)c=(b-a)(a+b+c)$
Are they equal?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the inequality $x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$. I want to prove $$x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$$
for $x>0$ real number.
I tried supposing that $$x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 \geq 0$$ holds to get a contradiction but I couldn't find the solution.
Can you help me proceed?
| Hint: Simplify that LHS to $\displaystyle -\frac{2x}{1+\sqrt x}< 0$, which is obvious for $x>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898803",
"timestamp": "2023-03-29T00:00:00",
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} |
Calculate a function with the use of a development point and geometric series Assuming I have this function :
$$ f(x) = \frac{1 + x^3}{2-x} $$
The question was to calculate this with the use of the geometric series and the development point $x_{0} = 0$.
Well setting first the function to a geometric series is quite clear here .
I got $ f(x)=$ $\sum_{n=1}^\infty \frac{x^n}{2^{n+1}}(1+x^3) $ as a result . What's the meaning of calculating an infinite series here ? By calculating does this mean that I can choose to calculate the first terms of the series (For example from n=0 to n=3 and then $ '+\ldots '$ ) ?
Plus why do I need to use the development point $x_{0}$ . I can already calculate the function with the geometric series only . Won't that be enough or is there another idea to implement this?
| We are looking for a representation
\begin{align*}
f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n=\sum_{n=0}^\infty a_nx^n\tag{1}
\end{align*}
where the series expansion is at the point $x_0=0$.
We obtain
\begin{align*}
\color{blue}{f(x)}&\color{blue}{=\frac{1+x^3}{2-x}}
=\frac{1}{2}\cdot\frac{1+x^3}{1-\frac{1}{2}x}\\
&=\frac{1}{2}\sum_{n=0}^\infty\frac{1}{2^{n}}x^n(1+x^3)\\
&=\sum_{n=0}^\infty\frac{1}{2^{n+1}}x^n+\sum_{n=0}^\infty\frac{1}{2^{n+1}}x^{n+3}\\
&=\sum_{n=0}^\infty\frac{1}{2^{n+1}}x^n+\sum_{n=3}^\infty\frac{1}{2^{n-2}}x^{n}\\
&\,\,\color{blue}{=\frac{1}{2}+\frac{1}{4}x+\frac{1}{8}x^2+\sum_{n=3}^\infty\frac{9}{2^{n+1}}x^n}
\end{align*}
with the last line being a representation according to (1).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $f(x)=\frac{1}{1+x}$ is uniformly continuous on $(-\infty,-\frac{3}{2})$
Using the $\epsilon,\delta$ definition of uniform continuity, prove that $f(x)=\frac{1}{1+x}$ is uniformly continuous on $(-\infty,-\frac{3}{2})$.
Attempt:
We must show that for any $\epsilon>0$ we can find a $\delta > 0$, such that if $|x-a|<\delta$ and $|\frac{1}{1+x}-\frac{1}{1+a}|<\epsilon$ for all $a \in (-\infty,-\frac{3}{2})$.
$\quad$Scratchwork for $\delta$:.
$\begin{align}
\quad \quad \Bigg|\frac{1}{1+x}-\frac{1}{1+a} \Bigg|=\Bigg|\frac{1+a-1\color{blue}{-}x}{(1+x)(1+a)}\Bigg|&=\fbox{|$\frac{1}{(1+x)(1+a)}|$}|x-a|\\&\color{blue}{<}4|x-a|<\epsilon \\ &\Rightarrow|x-a|<\frac{\epsilon}{4}
\end{align}$
$\quad \quad $ **In order to bound $\Bigg|\frac{1}{(1+x)(1+a)}\Bigg|$, I used the fact that $x,a<-\frac{3}{2}$.
$\quad \quad $Which would result in $\frac{1}{(1+x)(1+a)}<\frac{1}{(1-\frac{3}{2})(1-\frac{3}{2})}=\frac{1}{\frac{1}{4}}=4$.
$\quad$Choose $\delta=\frac{\epsilon}{4}\\$
$\quad$Proof:
Let $\epsilon>0$ be given and let $\delta=\frac{\epsilon}{4}$. If $|x-a|<\delta$ we have:
$\begin{align}
\quad \quad \Bigg|\frac{1}{1+x}-\frac{1}{1+a}\Bigg|&= \Bigg|\frac{1}{(1+x)(1+a)}\Bigg||x-a|\\&\leq4|x-a|\\ &<4\delta=4\frac{\epsilon}{4}=\epsilon.
\end{align}$
$\rule{18cm}{.03cm}$
Is it okay to use the fact that that $x,a<-\frac{3}{2}$, in order to place bounds on the function in order to reach a $\delta$? I am not sure if this technique holds for intervals involving infinity.
| Since $x$ and $a$ are both assumed to satisfy the inequalities $x,a < -\frac{3}{2}$ then sure, you may use those inequalities in your proof.
The proof looks fine. Your scratchwork has a mistaken inequality and a mistaken implication sign, although that seems not to have affected the proof.
| {
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Given X+Y+Z = 0, find $\frac{(X^3+Y^3+Z^3)}{XYZ}$ The result can be found using the equation :
$(X^3+Y^3+Z^3) - 3XYZ = (X+Y+Z)(X^2 - XY +Y^2 - YZ +Z^2 - XZ)$
Since X+Y+Z = 0, the right side of the equation is equal to 0. Therefore $X^3+Y^3+Z^3 = 3XYZ$ and the answer to the problem is 3.
However what if we calculate $X^3+Y^3+Z^3$ as $(X+Y+Z)^3 - 3X^2Y - 3Y^2X - 3X^2Z - 3z^2X - 3Y^2Z - 3Z^2Y$. $(X+Y+Z)^3 = 0$, so $X^3+Y^3+Z^3$ can be replaced with $- 3X^2Y - 3Y^2X - 3X^2Z - 3Z^2X - 3Y^2Z - 3Z^2Y$.
$\frac{- 3X^2Y - 3Y^2X - 3X^2Z - 3Z^2X - 3Y^2Z - 3Z^2Y}{XYZ}$ = $\frac{-3X - 3Y}{Z} - \frac{-3X-3Z}{Y} - \frac{-3Y-3Z}{X}$
If X+Y+Z = 0, consequenty 3X+3Y+3Z = 0. Our expression can be rewritten as $\frac{-3Z}{Z} + \frac{-3Y}{Y} +\frac{-3X}{X}$, so the answer is -9.
Could you please tell me which way of solving this problem is right and why
| In your second set of calculations, where you're subtracting terms from $(X + Y + Z)^3$, you're missing a $-6XYZ$ term. Also, going from $3X + 3Y + 3Z = 0$ gives $-3X - 3Y = 3Z$, not $-3Z$, with this sign error occurring for the other $2$ terms as well. When you include the $-6XYZ$ term, you'll get an additional $\frac{-6XYZ}{XYZ} = -6$ value, and correcting for the second mistake with the term sign values, it becomes $9$ instead of $-9$.
Thus, the end result becomes $9 - 6 = 3$, as you got originally. As such, both methods give the same end result, as expected.
| {
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"timestamp": "2023-03-29T00:00:00",
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Product Series Formula through Induction For the series:
$(1+\frac{1}{1})^1$$(1+\frac{1}{2})^2$$.....$ $(1+\frac{1}{n})^n$ I have the formula $\frac{(n+1)^n}{n!}$ for n$\in$ $\Bbb N$
I used induction to try and solve but I'm stuck at trying to prove it for n+1 since
$(1+\frac{1}{1})^1$$(1+\frac{1}{2})^2$$.....$$(1+\frac{1}{n})^n$$(1+\frac{1}{n+1})^{n+1}$ = $\frac{(n+1)^n}{n!}$$(1+\frac{1}{1})^{n+1}$ = $\frac{(n+1)^n}{n!}$$\sum_{k=0}^n \binom{n}{k}(\frac{1}{n+1})^k$ using the binomial theorem
but I can't tell properly how to get $\frac{(n+2)^{n+1}}{(n+1)!}$
Any hints? because I swear I'm not seeing something.
Thank you!
| Hint: We have
\begin{eqnarray*}
\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right)^2 \cdots \left( 1+\frac{1}{n} \right)^n = \frac{(n+1)^{n}}{n!}
\end{eqnarray*}
So
\begin{eqnarray*}
\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right)^2 \cdots \left( 1+\frac{1}{n} \right)^n \color{red}{\left( 1+\frac{1}{n+1} \right)^{n+1}}&=& \frac{(n+1)^{n}}{n!} \color{red}{\left(\frac{n+2}{n+1} \right)^{n+1}}\\
&=& \cdots
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solving a system of cubic equations with real roots Consider the following system of equations:
\begin{align}
\frac{1}{8} (\alpha +2 x)^2 \left((\alpha +2 x)^2-12\right)+\\ +\frac{4}{9} y^2 \left((\alpha
+2 x)^2+2\right)+\frac{32 y^4}{81}&=0\\
\alpha ^3-6 \alpha +8 x^3-4 \alpha x^2-2 \left(\alpha ^2+2\right) x&=0
\end{align}
I would like to eliminate the $\alpha$ value in order to have an equation of the form $f(x,y)=0$.
Using Cardano's formula or Mathematica, I end up with imaginary values. However if I solve them numerically I obtain a nice shape $f(x,y)=0$
Due to the numerical plots I obtain I am thinking that there must be a nice (analytical) solution for these equations. How could I derive such an expression for $f(x,y)=0$?
| The second equation write
$$\alpha^3-2x\alpha^2-2(3+2x^2)\alpha+4x(2 x-1)=0$$ The discriminant is
$$\Delta=32 \left(128 x^4+18 x^2+27\right) > 0 \qquad \forall x$$
Using the trigonometric method for three real roots leads to
$$\alpha_k=\frac{2}{3} \left(x+ \sqrt{2(8 x^2+9)} \cos \left(\frac{1}{3} \left(2 \pi k-\cos
^{-1}\left(\frac{\sqrt{2} x \left(27-16 x^2\right)}{\left(8 x^2+9\right)^{3/2}}\right)\right)\right)\right)$$ with $k=0,1,2$.
Replace in the first equation to have $f(x,y)=0$ for each value of $k$. This would lead to
$$c_0+c_1\cos(A)+c_2\cos(2A)+c_3\cos(3A)+c_4\cos(4A)=0$$ with
$$c_0=32 \left(70 x^4+3 x^2 \left(4 y^2+9\right)+y^4\right)+27 \left(8 y^2-9\right)$$
$$c_1=128 \sqrt{2} x \sqrt{8 x^2+9} \left(7 x^2+y^2\right)$$
$$c_2=2 \left(8 x^2+9\right) \left(112 x^2+8 y^2-9\right)$$
$$c_3=16 \sqrt{2} x \left(8 x^2+9\right)^{3/2}$$
$$c_4=\left(8 x^2+9\right)^2$$
$$A=\frac{1}{3} \left(2 \pi k-\cos
^{-1}\left(\frac{\sqrt{2} x \left(27-16 x^2\right)}{\left(8 x^2+9\right)^{3/2}}\right)\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many students in the school?
If they put the rows of 13, there are 8 students left; if they put the rows of 15, there are 3 students left and if they put the rows of 17, there are 9 students left. How many students are there given that the total students are < 5000?
The following congruences are: $$x\equiv 8 \pmod {13} \\ x\equiv 3 \pmod {15} \\ x\equiv 9 \pmod {17}$$
I'm still new to this, how can I apply Chinese Remainder Theorem to these congruences to find the total students?
| Let $N$ be the number of students. Let $M = 13\times 15 \times 17 = 3315$
$\qquad\ M_1 = 15 \times 17 = \color{#c00}{255},\,\ \ y_1 = 255^{-1} \equiv \ \ 8^{-1}\equiv \ \ \color{#c00}{5}\ \pmod{13} $
$\qquad\ M_2 = 13 \times 17 = \color{#0a0}{221},\,\ \ y_2 = 221^{-1} \equiv 11^{-1} \equiv \color{#0a0}{11}\pmod{15} $
$\qquad\ M_3 = 13 \times 15 = \color{#90f}{195},\,\ \ y_3 = 195^{-1} \equiv\ \ 8^{-1}\equiv \color{#90f}{15} \pmod{17} $
Thus $\ N \equiv 8 \times \color{#c00}{255 \times 5}\, +\, 3 \times \color{#0a0}{221 \times 11}\, +\, 9 \times \color{#90f}{195 \times 15}\,\equiv\, 723\pmod{3315} $
So the number of students $\, N = 723 + 3315 \times k $
With $\,N < 5000,\ N = 4038\,$ with $\,k = 1$
You can read more here (Vietnamese wiki page on CRT)
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of all possible valuse of $\frac{a}{b}+\frac{c}{d}$? If a,b,c,d are real numbers and $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=17$ and $\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}=20$, then find the sum of all possible valuse of $\frac{a}{b}$+$\frac{c}{d}$ ?
I tried this problem for a while but made no progress. I don't know how $\frac{a}{b}+\frac{c}{d}$ can take only certain values. The answer was given to be $17$. Can someone help me with this?
| \begin{align}
17 &= \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \\
&= \frac{a}{b}+\frac{c}{d}+\frac{b}{c}+\frac{d}{a} \\
&= \color{red}{\frac{ad+bc}{bd}}+\color{blue}{\frac{ab+cd}{ac}} \\
&= \color{red}{x}+\color{blue}{y} \\
20 &= \frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b} \\
&= \frac{a}{c}+\frac{b}{d}+\frac{d}{b}+\frac{c}{a} \\
&= (ad+bc) \left( \frac{1}{cd}+\frac{1}{ab} \right) \\
&= \frac{(ad+bc)(ab+cd)}{abcd} \\
&= \color{red}{\frac{ad+bc}{bd}} \times \color{blue}{\frac{ab+cd}{ac}} \\
&= \color{red}{x} \color{blue}{y} \\
\end{align}
Now, $$20=x(17-x) \implies x^2-17x+20=0$$
Note that $x$ and $y$ are conjugate pair and also symmetric in roles.
| {
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Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$ I found the following exercise reading my calculus notes:
If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$
I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.
| By A.M.$\geq$G.M.$$(x^2+1)(y^2+1)(z^2+1)=x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2+x^2+y^2+z^2+1\geq8\times \sqrt[8]{(x^2y^2z^2)\times(x^2y^2)\times(y^2z^2)\times(z^2x^2)\times(x^2)\times(y^2)\times(z^2)\times1}=8xyz$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do prove $12^n - 4^n - 3^n +1$ is divisible by 6 using mathematical induction, where n is integral? So this question is very challenging because normally the bases of the exponents are the same. There are too many different bases for me to successfully subtitue in the assumption (when $n=k$)
I was hoping someone out there will have a super smart elegant solution to this!
Base step: test when n = 1 ...
Assume true for $n = 1$
ie . $12^k - 4^k - 3^k +1 = 6M$, where $m$ is an integer
RTP: also true for $n = k+1$
ie. $12^{k+1} - 4^{k+1} - 3^{k+1} +1 = 6N$ where $N$ is an integer
LHS: $12^{k+1} - 4^{k+1} - 3^{k+1} +1$
$= 12( 4^k + 3^k - 1 + 6M) - 4^{k}(4) - 3^{k}(3) +1$ (from assumption)
$= 6(12M) + 12(4^k) + 12(3^k) -12 -4^{k}(4) - 3^{k}(3) +1$
Here is where I break down and go around in circles.
| Suppose $6 | 12^{n}-4^{n}-3^{n}+1$ for some $n$.
Then $6| 12(12^{n}-4^{n}-3^{n}+1) = 12^{n+1}-12\cdot4^{n}-12\cdot3^{n}+12=$
$12^{n+1}-3\cdot4^{n+1}-4\cdot3^{n+1}+12.$
Since $3^{m} \equiv 3 \pmod 6$, and $4^{m} \equiv 4 \pmod 6$ for every $m$, we have that
$2\cdot4^{n+1} + 3\cdot3^{n+1}-11 \equiv2+3+1\equiv0\pmod 6.$
Hence, $6|(12^{n+1}-3\cdot4^{n+1}-4\cdot3^{n+1}+12)+(2\cdot4^{n+1} + 3\cdot3^{n+1}-11) = 12^{n+1}-4^{n+1}-3^{n+1}+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Limit with parameter $p$ Find $ p $ that will make limit convergent
$$ \lim _ { x \to \infty } \left( \left( n + \left( \frac { x ^ 2 + 2 } { x ^ 2 - x + 1 } \right) ^ { 2 x + 1 } \right) \cdot \sin \left( \frac { x \pi } 2 \right) \right) $$
I know how to solve most of the basic limits. In this problem I don't know if I'm allowed to separate it and make $ 3 $ or $ 4 $ small ones. I thought about making $ p \sin.. + \sin*(big bracket)$ and then separating again because $ p $ is parameter and can go in front of limit and I wanted to separate second one so I can find big limit (it's $e^2$ or something but that doesn't matter).
| $$ \lim _ { x \to \infty } \left( \left( n + \left( \frac { x ^ 2 + 2 } { x ^ 2 - x + 1 } \right) ^ { 2 x + 1 } \right) \cdot \sin \left( \frac { x \pi } 2 \right) \right) $$
since as mentioned the SIN is diverging because it keeps oscillating. Therefore, the first part has to converge to 0.
We have that the first part converges to $(n+e^2)$ Therefore your n should be $-e^2$.
Detailing the Calculation
$$\lim\limits_{x\to\infty}\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}$$
$$=\lim\limits_{x\to\infty}\exp\left(\ln\left(\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}\right)\right)$$
$$=\exp\left(\lim\limits_{x\to\infty}\ln\left(\left(\frac{x^2+2}{x^2-x+1}\right)^{2x+1}\right)\right)$$
$$\left(\exp\left(\left(1+2x\right)\ln\left(\frac{1+\frac{2}{x^{2}}}{1-\frac{1}{x}+\frac{1}{x^{2}}}\right)\right)\right)$$
Here we have to use the definition of $ln(1+x)=x$ if x goes to 0 and subtract the numerator from the denominator after applying the small transformation.
$$\exp\left(\left(2x+1\right)\left(\frac{1}{x^{2}}+\frac{1}{x}\right)\right)$$
$$\exp\left(\frac{2x}{x^{2}}+\frac{2x}{x}\right)$$
$$exp(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How does $f^{(3)} (0) \geq 3\ $?
Let $f(x)$ be a real polynomial of degree $4.$ Suppose $f(-1) = 0,f(0) = 0, f(1) = 1$ and $f^{(1)} (0) = 0,$ where $f^{(k)} (a)$ is the value of $k^{\text {th}}$ derivative of $f(x)$ at $x = a.$ Which of the following statements are true?
$(1)$ There exists $a \in (-1,1)$ such that $f^{(3)} (a) \geq 3$
$(2)$ $f^{(3)} (a) \geq 3,$ for all $a \in (-1,1)$
$(3)$ $0 \lt f^{(3)} (0) \leq 2$
$(4)$ $f^{(3)} (0) \geq 3$
My attempt $:$ Since $f$ is a real polynomial of degree $4$ having $-1$ and $0$ as roots it follows that there exist $p,q,r \in \Bbb R$ with $p \neq 0$ such that $$f(x) = x(x+1)(px^2 + qx + r).$$ By the given hypothesis $f(1) = 1$ which yields $p + q+ r = \frac {1} {2}.$ Now by expanding $f(x)$ we get $$f(x) = p x^4 + q x^3 + (p + r) x^2 + qx + r.$$ So $$f^{(1)} (x) = 4 p x^3 + 3 q x^2 + 2 (p + r) x + q.$$ Since $f^{(1)} (0) = 0$ it follows that $q = 0.$ Hence we have $p + r = \frac {1} {2}.$ Therefore $f^{(1)} (x) = 4px^3 + x.$ Thus $f^{(3)} (x) = 24 p x.$ So $f^{(3)} (0) = 0,$ which violates $(3)$ and $(4).$ Also if we take $p = \frac {1} {24}$ then for all $a \in (-1,1)$ we have $f^{(3)} (a) \lt 1,$ which violates $(1)$ and $(2)$ as well. So neither of the given options happens to be true. But the answer key to this question suggests that $(1)$ and $(4)$ are the only correct options. Am I doing something wrong here? Can anybody please check my solution?
Thanks for your time.
Source $:$ CSIR NTA NET DECEMBER $2019.$
| The conditions $f(0)=0$ and $f'(0)=0$ tell us that $f(x)=mx^4+ax^3+bx^2$ for some $a,b\in\Bbb R$. Furthermore, $f(-1)=0$ implies $b-a=-m$ and $f(1)=1$ implies $a+b=1-m$. Thus the polynomial is $$f(x)=mx^4+\frac12x^3+\left(\frac12-m\right)x^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is this $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ irreducible in $\mathbb Q[x]$? Is this $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ irreducible in $\mathbb Q[x]$? Can you help please?
| Let $f(x)=x^6+x^5+x^4+x^3+x^2+x+1$. Define the new polynomial $f(x+1)=x^6+7x^5+21x^4+35x^2+21x+7$, using the binomial theorem expand $f(x+1)$. This polynomial is irreducible over $\mathbb{Q}$ by Eisenstein criterion, for $p=7$
If $f(x)$ were reducible over the rationals then let $p(x), q(x)\in \mathbb{Q}[x]$ such that $$f(x)=p(x)q(x)$$ where $\deg(p)< 6$ and $\deg(q)<6$ We would then have $$f(x+1)=p(x+1)q(x+1)$$ but that's impossible! Hence $x^6+x^5+x^4+x^3+x^2+x+1$ is irreducible over $\mathbb{Q}[x]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The orthogonal projection of the vector b is not equal to $AA^+b$ Given matrix $A:= \begin{pmatrix}1&2\\3&6\end{pmatrix}$
I am trying to solve this task:
... compute $P:=AA^+$. Finally, evaluate $Pb$ with
$b:=\begin{pmatrix}2\\4\end{pmatrix}$, and show that $Pb$ is indeed
the orthogonal projection of the vector b onto the subspace by the
column vectors of the matrix A.
$A^+=\frac{1}{50}
\begin{pmatrix}1&3\\2&6\end{pmatrix}$
$P=AA^+$
$P=
\begin{pmatrix}1&2\\3&6\end{pmatrix}
\frac{1}{50}
\begin{pmatrix}1&3\\2&6\end{pmatrix}$
$P=
\frac{1}{10}
\begin{pmatrix}1&3\\3&9\end{pmatrix}$
$Pb=
\frac{1}{10}
\begin{pmatrix}1&3\\3&9\end{pmatrix}
\begin{pmatrix}2\\4\end{pmatrix}
=\frac{7}{5}\begin{pmatrix}1\\3\end{pmatrix}$
$proj_A(b)=\frac{u_1b}{u_1u_1}u_1+\frac{u_2b}{u_2u_2}u_2$
$proj_A(b)=\frac{\begin{pmatrix}1\\3\end{pmatrix}\begin{pmatrix}2\\4\end{pmatrix}}{\begin{pmatrix}1\\3\end{pmatrix}\begin{pmatrix}1\\3\end{pmatrix}}\begin{pmatrix}1\\3\end{pmatrix}+\frac{\begin{pmatrix}2\\6\end{pmatrix}\begin{pmatrix}2\\4\end{pmatrix}}{\begin{pmatrix}2\\6\end{pmatrix}\begin{pmatrix}2\\6\end{pmatrix}}\begin{pmatrix}2\\6\end{pmatrix}$
$proj_A(b)=\frac{14}{10}\begin{pmatrix}1\\3\end{pmatrix}+\frac{28}{40}\begin{pmatrix}2\\6\end{pmatrix}$
$proj_A(b)=\frac{14}{10}\begin{pmatrix}1\\3\end{pmatrix}+\frac{14}{10}\begin{pmatrix}1\\3\end{pmatrix}$
$proj_A(b)=\frac{28}{10}\begin{pmatrix}1\\3\end{pmatrix}=\frac{14}{5}\begin{pmatrix}1\\3\end{pmatrix}$
$Pb$ is not equal $proj_A(b)$. What is wrong here? Why are not they equal?
| Your solution for $P$ and $Pb$ is correct.
You can simply verify that $Pb$ is in the column space of $A$ and that $b-Pb$ is orthogonal to this column space, which is now the span of $\pmatrix{1\\3}$.
The other method works only if $u_1,u_2$ are orthogonal, and in that case it produces the projection to the plane spanned by $u_1$ and $u_2$, which only makes sense in 3 or more dimensional spaces.
Here the columns of $A$ are not orthogonal, rather they are parallel, so the subspace we project on is one dimensional, a line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3932078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to solve the homogeneous system of a differential equation I want to solve the system $\dot{x}=Ax$ where $ A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right).$ I found the eigenvalues of $A$ as $\lambda_1=4,\lambda_2=-1$ with correspongind eigenvectors $v_1=(2,3)^\intercal,v_2 =(1,-1)^\intercal $ respectively. Then, the solution of the system will be
$$x(t)=e^{At}x_0$$
so that I obtained $$x(t)=\left( \begin{matrix} \frac{3}{5}e^{-t} + \frac{2}{5}e^{4t} & -\frac{2}{5}e^{-t} + \frac{2}{5}e^{4t} \\ \frac{3}{5}e^{4t} + -\frac{3}{5}e^{-t} & \frac{3}{5}e^{4t} + \frac{2}{5}e^{-t} \end{matrix} \right) x_0$$
where $x_0$ is the initial condition. Now, if I say that $x_0 =(c_1,c_2) $ for some constants $c_1,c_2$ I obtain a sum which is very mixed as you can guess from the above matrix. (Note that I verified the matrix via Wolfram).
However, Paul's Online Notes and an online solver says that the solution can be found as
$$x(t)=c_1 e^{-t} (-1,1)^\intercal+ c_2 e^{4t}(2,3)^\intercal$$
that is not equal to the solution that I found. So, what is the problem with my solution?
Final note: The second solution takes the first eigenvector as $-v_1$.
| $$x(t)=\left( \begin{matrix} \frac{3}{5}e^{-t} + \frac{2}{5}e^{4t} & -\frac{2}{5}e^{-t} + \frac{2}{5}e^{4t} \\ \frac{3}{5}e^{4t} + -\frac{3}{5}e^{-t} & \frac{3}{5}e^{4t} + \frac{2}{5}e^{-t} \end{matrix} \right) x_0$$
$$x(t)=\left( \begin{matrix} \frac{1}{5}e^{-t}(3c_1-2c_2) + \frac{2}{5}e^{4t} (c_1+c_2) \\ - \frac{1}{5}e^{-t}(3c_1-2c_2) + \frac{3}{5}e^{4t} (c_1+c_2) \end{matrix} \right) $$
$$x(t)=\left( \begin{matrix} e^{-t}k_1 + {2}e^{4t} k_2 \\ - e^{-t}k_1 + {3}e^{4t} k_2 \end{matrix} \right) $$
This agree with the answer you posted.
$$x(t)=k_1e^{-t}\left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) +k_2e^{4t}\left( \begin{matrix}2 \\ 3\end{matrix} \right) $$
Where :$$k_1= \frac {3c_1-2c_2 }5, k_2= \frac {c_1+c_2}5 $$
There is nothing wrong with your solution.
And for initial conditions $(x_0)$ you have:
$$x(0)=k_1\left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) +k_2\left( \begin{matrix}2 \\ 3\end{matrix} \right)$$
You have the same formula as the one in Paul's notes.
found the eigenvalues of $A$ as $λ_1=4,λ_2=−1$ with correspongind eigenvectors $v_1=(2,3)^t,v_2=(1,−1)^t$ respectively. Then, the solution of the system will be:
$$x(t)=c_1e^{4t}v_1+c_2e^{-t}v_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find recurrence relation for strings of length n over {A,B,C}(none of the strings should have consecutive As) Assume that an is number of strings of length n over A B C and none of strings have consecutive As
I have no idea how to find a relation for this
| For a recurrence relation, define $u_n$ to be the number of strings of length $n$ with letters $A,B,C$ such that there are no consecutive $A$s, and additionally, the string must end in $A$. Define $v_n$ to be the number of strings of length $n$ with letters $A,B,C$ such that there are no consecutive $A$s, and additionally, the string must NOT end in $A$. Hence, we are to find $u_n+v_n$. We provide a recursive relation:
To find $u_{n+1}$, we know that the last letter has to be $A$, and the first $n$ letters have to be forming a string of $A,B,C$ which does not end in $A$, and has no consecutive $A$s. Hence, $u_{n+1}=v_n$.
Next, to find $v_{n+1}$, we know that the first $n$ letters give us $u_n+v_n$ choices and the last letter can either be $B$ or $C$, giving us $2$ choices. Hence, $v_{n+1}=2(u_n+v_n)$.
Now, we can substitute $u_n=v_{n-1}$ in the recurrence for $v_{n+1}$ to yield $v_{n+1}=2(v_n+v_{n-1})$. After solving this recurrence, we can conclude that the answer is $u_n+v_n=v_n+v_{n-1}$.
This recurrence can be solved using the standard technique. The quadratic $x^2=2x+2$ has roots:
$$\frac{2 \pm \sqrt{(2^2+4 \cdot 2 \cdot 1)}}{2}=1 \pm \sqrt{3}$$
So, we can write $v_n=P(1+\sqrt{3})^n+Q(1-\sqrt{3})^n$. This has to hold good for $n=1,2$. Since $v_1=2$ amd $v_2=6$, we have:
$$\begin{cases}
(P+Q)+\sqrt{3}(P-Q)=2 \\
4(P+Q)+2\sqrt{3}(P-Q) = 6
\end{cases} \implies (P,Q)=\bigg(\frac{1+\sqrt{3}}{2},\frac{1-\sqrt{3}}{2} \bigg)$$
Hence, $v_n=\frac{1}{2}((1+\sqrt{3})^{n+1}+(1-\sqrt{3})^{n+1})$, and thus:
$$v_{n+1}+v_n=\frac{1}{2}((1+\sqrt{3})^{n+1}+(1-\sqrt{3})^{n+1})+\frac{1}{2}((1+\sqrt{3})^{n}+(1-\sqrt{3})^{n})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $\int_0^\infty\frac{\ln x}{x^3 - 1} \, dx = \frac{4 \pi^2}{27}$ I realise this question was asked here, but I'm not able to work with any of the answers. The hint given by my professor is
Integrate around the boundary of an indented sector of aperture $\frac{2 \pi}{3}$
but when I try that I can't figure out how to deal with the (divergent) integral along the radial line at angle $2 \pi / 3$. My issue with the accepted answer is that it uses the residue theorem where it doesn't apply, at least as we've learned it, since
$$z \mapsto \frac{\log^2z}{z^3 - 1}$$
has non-isolated singularities on the closed region bounded by the proposed contour (due to branch cuts), and I am not sure how to relate the integral along the real axis to one over a contour modified to avoid the branch cut.
For a fixed $\varepsilon > 0$, and for any $\delta \in (0, 1 - \varepsilon)$, we could let $\log_{-\delta / 2}$ be the branch of the logarithmic function with a cut along the ray $\operatorname{arg}z = -\delta / 2$ and define a contour which goes along the positive real axis from $\varepsilon$ to $1 - \delta$, a semicircle in the upper half plane of radius $\delta$ around $1$, the positive real axis from $1 + \delta$ to $2$, an arc of radius $2$ around $0$ with central angle $2 \pi - \delta$, the ray $\operatorname{arg}z = 2 \pi - \delta$ from $r = 2$ to $r = \varepsilon$, and finally an arc of radius $\varepsilon$ around $0$ back to $\varepsilon$. But then, for example, I don't know how to calculate the limit of integral along the arc of radius $\varepsilon$
$$\lim_{\delta \to 0}\int_0^{2 \pi - \delta}\frac{\log_{-\delta / 2}^2(\varepsilon e^{i \theta})}{\varepsilon^3 e^{3 i \theta} - 1} \varepsilon i e^{i \theta} \, d\theta.$$
If I instead try to first use the substitution $x = e^u$ on the real integral and then compute a contour integral, I still get a divergent integral that I don't know how to handle, this time along the top of an indented rectangle.
| Here we discuss two ways of tackling the integral using contour integral. I added Solution 1 to help you implement the hint suggested by the professor. However, I personally recommend you to jump directly to Solution 2.
Solution 1. Consider
$$ f(z) = \frac{\log z}{z^3 - 1}, $$
where $\log(\cdot)$ is the principal complex logarithm. Now we integrate $f$ along the boundary of the indented sector of opening $2\pi/3$:
Here, the radius of the larger arc $\Gamma_2$ (resp. smaller arc $\Gamma_1$) is $R$ (resp. $\epsilon$) and $0 < \epsilon < 1 < R$. Then it is easy to check that
$$ \left| \int_{\Gamma_1} f(z) \, \mathrm{d}z \right| \leq C \epsilon ( C + \log (1/\epsilon) )
\qquad\text{and}\qquad
\left| \int_{\Gamma_2} f(z) \, \mathrm{d}z \right| \leq \frac{C(C + \log R)}{R^2} $$
for $C = 2\pi/3$, and so, the integrals along these curves vanish as $\epsilon \to 0^+$ and $R \to \infty$. So by the residue theorem,
\begin{align*}
\int_{L_1} f(z) \, \mathrm{d}z + \int_{L_2} f(z) \, \mathrm{d}z + \int_{L_3} f(z) \, \mathrm{d}z + \int_{\gamma} f(z) \, \mathrm{d}z = o(1)
\end{align*}
as $\epsilon \to 0^+$ and $R\to\infty$. However, using the fact that $\omega = e^{2\pi i/3}$ is a simple pole of $f(z)$, the function $(z - \omega)f(z)$ is analytic at $z = \omega$. So
\begin{align*}
\lim_{\epsilon \to 0^+} \int_{\gamma} f(z) \, \mathrm{d}z
&= \lim_{\epsilon \to 0^+} i\int_{2\pi/3}^{-\pi/3} \epsilon e^{i\theta} f(\omega + \epsilon e^{i\theta}) \, \mathrm{d}\theta \tag{$z=\omega+\epsilon e^{i\theta}$} \\
&= -i \pi \lim_{z \to \omega} (z - \omega)f(z)
= -i \pi \mathop{\mathrm{Res}}_{z = \omega} f(z) \\
&= \frac{2\pi^2}{9} \omega.
\end{align*}
Moreover,
\begin{align*}
\int_{L_2} f(z) \, \mathrm{d}z
&= -\omega \int_{1+\epsilon}^{R} f(\omega x) \, \mathrm{d}x
= -\omega \int_{1+\epsilon}^{R} \left( f(x) + \frac{2\pi i}{3} \cdot \frac{1}{x^3 - 1} \right) \, \mathrm{d}x
\end{align*}
and likewise
\begin{align*}
\int_{L_3} f(z) \, \mathrm{d}z
= -\omega \int_{\epsilon}^{1-\epsilon} \left( f(x) + \frac{2\pi i}{3} \cdot \frac{1}{x^3 - 1} \right) \, \mathrm{d}x.
\end{align*}
Combining altogether and using that $f(z)$ is analytic at $z = 1$,
\begin{align*}
(1 - \omega) \int_{\epsilon}^{R} f(x) \, \mathrm{d}x
- \frac{2\pi i \omega}{3} \left( \int_{\epsilon}^{1-\epsilon} \frac{1}{x^3 - 1} \, \mathrm{d}x + \int_{1+\epsilon}^{R} \frac{1}{x^3 - 1} \, \mathrm{d}x \right)
+ \frac{2\pi^2}{9} \omega = o(1).
\end{align*}
Letting $\epsilon \to 0^+$ and $R \to \infty$,
\begin{align*}
(1 - \omega) \int_{0}^{\infty} f(x) \, \mathrm{d}x
- \frac{2\pi i \omega}{3} \left( \mathop{\mathrm{PV}}\! \int_{0}^{\infty} \frac{1}{x^3 - 1} \, \mathrm{d}x \right)
+ \frac{2\pi^2}{9} \omega = 0.
\end{align*}
By noting that
\begin{align*}
\mathop{\mathrm{PV}}\! \int_{0}^{\infty} \frac{1}{x^3 - 1} \, \mathrm{d}x
&=-\frac{\pi}{3\sqrt{3}},
\end{align*}
we end up with
\begin{align*}
\int_{0}^{\infty} f(x) \, \mathrm{d}x
= \frac{\omega}{1 - \omega} \left( -\frac{\pi}{3\sqrt{3}} \cdot \frac{2\pi i}{3}
- \frac{2\pi^2}{9} \right)
= \frac{4\pi^2}{27}.
\end{align*}
Solution 2. Here is a more elegant solution. Let $\operatorname{Log}$ denote the complex logarithm chosen so that its argument lies between $0$ and $2\pi$. (Note: Using the principal complex logarithm, this can be realized by $\operatorname{Log}(z) = i\pi + \log(-z)$.) Then consider
$$ g(z) = \frac{(\operatorname{Log}(z) - 2\pi i)\operatorname{Log}(z)}{z^3 - 1}. $$
Then it is not hard to see that, for $x > 0$,
\begin{align*}
g(x + i0^+) := \lim_{\epsilon \to 0^+} g(x + i\epsilon)
&= \frac{(\log x - 2\pi i)\log x}{x^3 - 1} \\
g(x - i0^+) := \lim_{\epsilon \to 0^+} g(x - i\epsilon)
&= \frac{(\log x + 2\pi i)\log x}{x^3 - 1}.
\end{align*}
So by using the keyhole contour,
$$ \int_{0}^{\infty} \bigl( g(x + i0^+) - g(x - i0^+) \bigr) \, \mathrm{d}x
= 2\pi i \biggl( \mathop{\mathrm{Res}}_{z=e^{2\pi i/3}} g(z) + \mathop{\mathrm{Res}}_{z=e^{4\pi i/3}} g(z) \biggr) $$
Now the left-hand side is
$$ (-4\pi i) \int_{0}^{\infty} \frac{\log x}{x^3 - 1} \, \mathrm{d}x $$
and the right-hand side is
$$ 2\pi i \biggl( \frac{\bigl(\frac{2\pi i}{3} - 2\pi i \bigr)\bigl( \frac{2\pi i}{3} \bigr)}{3 e^{4\pi i}} + \frac{\bigl(\frac{4\pi i}{3} - 2\pi i \bigr)\bigl( \frac{4\pi i}{3} \bigr)}{3 e^{8\pi i}} \biggr) = 2\pi i \left( -\frac{8\pi^2}{27} \right).$$
Therefore the answer is again $\frac{4\pi^2}{27}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is my method of solving equation correct? The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$
using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$
$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$
Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$
$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$
Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$
Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$
However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.
So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!
| Spoiler alert: this is the solution.
Let's take $ a = \sqrt[5]{16+\sqrt x} $ and $ b = \sqrt[5]{16-\sqrt x} $ . Note that because $ \sqrt x \ge 0 $, $ a $ is always positive (technically greater than or equal to $ \sqrt[5]{16} $), but $ b $ can be positive, zero, or negative. We have:
$$ a + b = 2 $$
$$ (a + b)^5 = 32 $$
We also have:
$$ a^5 = 16 + \sqrt x $$
$$ b^5 = 16 - \sqrt x $$
$$ a^5 + b^5 = 32 $$
Aha! Let's see where this takes us:
$$ (a + b)^5 = a^5 + b^5 $$
$$ a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 = a^5 + b^5 $$
$$ 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 = 0 $$
For this sum to be zero, either all of its terms should be zero, or some of them should be positive and some others negative. The former case happens if $ b $ is zero (we noted earlier that $ a $ is always positive, so it can't be zero). The latter case can only happen if $ b $ is negative, because again as we noted earlier, $ a $ is always positive and cannot make negative terms. Let's keep that in mind and continue.
$$ 5ab(a^3 + 2a^2b + 2ab^2 + b^3) = 0 $$
The expression in the parentheses looks a bit like $ (a+b)^3 $, but not exactly! Let's add $ a^2b + ab^2 - a^2b - ab^2 $ to it:
$$ 5ab[(a+b)^3 - a^2b - ab^2] = 0 $$
$$ 5ab[(a+b)^3 - ab(a+b)] = 0 $$
$$ 5ab(a+b)[(a+b)^2-ab] = 0 $$
And recalling that $ a+b = 2 $:
$$ 10ab[4-ab]=0 $$
Now, for this equality to hold, we must have either $ ab = 4 $ or $ ab = 0 $.
But $ ab = 4 $ is not possible, because if it were so, considering that $ a $ is positive, $ b $ should necessarily be positive as well (otherwise $ ab $ would not be positive). And if both $ a $ and $ b $ were positive, then the sum $ 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 $ would necessarily be positive, whereas it should be zero.
So we are left with the conclusion that $ ab = 0 $. As we mentioned before, $ a $ is positive and not zero. So:
$$ b = 0 $$
$$ \sqrt[5]{16-\sqrt x} = 0 $$
$$ 16 - \sqrt x = 0 $$
$$ x = 16^2 = 256 $$
| {
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} |
Double root in Trigonometric equation?
Consider the function $ f( x) = \sin^2 x + \sin x$ in the domain $ [0, 2 \pi]$, find the subset of the domain where $f \geq 0 $
Factoring by treating as quadratic in $ \sin$:
$$ f = ( \sin x + 1 ) ( \sin x)$$
Hence, the roots are $ \sin x = \{ 0, -1 \}$, hence the roots in $x$ are $ \{ 0, \pi , 2 \pi, \frac{3 \pi}{2} \}$. Let us partition the interval using the roots :
$$ \left[ 0, \pi \right] \cup \left[ \pi, \frac{ 3 \pi}{2} \right] \cup \left[ \frac{ 3 \pi}{2} , 2 \pi \right]$$
From regular algebra, we know that a 'root' corresponds to a sign flip, hence since the function is negative in the second interval of a domain partition, it must be positive in the third ... but actually it's not! i.e: $f \leq 0 $ for $ x \in [ \frac{ 3 \pi}{2} , 2 \pi ]$ but there is a problem, there was only a single root at $ x= \frac{ 3 \pi}{2}$.
So, how did the sign not flip here even when it crossed a root? That is what property of the trigonometric function is which causes this?
| Your factorization $ \ (\sin x)·(1 + \sin x) \ $ will also explain how the sign of $ \ f(x) \ $ behaves. Since $ \ (1 + \sin x) \ $ is non-negative for all values of $ \ x \ \ , $ the sign of $ \ f(x) \ $ is entirely determined by the sign of the factor $ \ (\sin x) \ \ . $ As this is negative in quadrants III and IV, $ \ f(x) \ $ is negative "to either side" of its zero at $ \ x \ = \ \frac{3 \pi}{2} \ \ . $
We can also consider the symmetry of the terms $ \ \sin^2 x \ + \ \sin x \ \ . $ Using the "angle-addition" formula for cosine, we note that
$$ \cos \left(x \ - \ \frac{3 \pi}{2} \right) \ \ = \ \ \cos x · \cos \frac{3 \pi}{2} \ - \ \sin x · \sin \frac{3 \pi}{2} $$ $$ = \ \ \cos x · 0 \ - \ \sin x · (-1) \ \ = \ \ \sin x \ \ . $$
Each term in the expression for $ \ f(x) \ $ has "even symmetry" about the value $ \ x \ = \ \frac{3 \pi}{2} \ \ , $ so their sum also possesses this symmetry. Thus, there is no sign-change for $ \ f(x) \ $ on "passing through" $ \ x \ = \ \frac{3 \pi}{2} \ \ . $
| {
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What is the $n$th term of the sequence $1, 1, 4, 1, 4, 9, 1, 4, 9, 16, 1 ...$ So there exist a lot of similar questions which ask the $n$th term of
$$1^2 , (1^2 + 2^2) , (1^2 + 2^2 + 3^2) ... $$
Which have a simple answer as
$$T_n = n^3/3 + n^2/2 + n/6$$
But I want to know what will be the $n$th term if we consider each number as a single term. I tried making a pyramid of terms and got this
$1^2\\
1^2~2^2\\
1^2~2^2~3^2\\
1^2~2^2~3^2~4^2\\
...$
I got the $n$th term as$$T_n= (n\mod(m(m-1)/2)^2$$
Where $m$ is the row number.
But now I am stuck, is there a way to eliminate $m$ and write the general term only in terms of $n$?
| The number of terms in $x$ rows is $1+2+...+x=x(x+1)/2$. The row number $m$ of $n^{\text{th}}$ term is thus the smallest positive integer solution of $x(x+1)/2\ge n$. The roots of $x^2+x-2n=0$ are $\frac{-1\pm\sqrt{8n+1}}2$. First, we reject the negative root. Next, we take the ceiling value of the positive root as it may not be a whole number.
Thus,$$m=\left\lceil\frac{-1+\sqrt{8n+1}}2\right\rceil$$
As a side note, you can ditch the mod and use regular subtraction in your formula.
| {
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} |
Basic Geometric Series Question-Stuck I'm studying Calc 2 and I have a basic series question.
A geometric series $\sum _{n=0}^{\infty }\:Ar^n$ is convergent if |r|<1 and the sum equals $\frac{a}{1-r}$ if the series is convergent.
Question: $\sum _{n=1}^{\infty }\:\frac{5}{\pi ^n}=-\frac{5}{\pi }+\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$
and $\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$ totals $\frac{5}{1-\frac{1}{\pi }}=\frac{5\pi \:}{\pi \:-1}$.
Therefore, $\sum _{n=1}^{\infty }\:\frac{5}{\pi ^n}=-\frac{5}{\pi }+\sum _{n=0}^{\infty \:}\:\frac{5}{\pi ^n}$ = $\frac{5\pi \:}{\pi \:-1}\:-\frac{5}{\pi }$.
I know this is wrong per my textbook, but after an hour working on this problem I cannot figure out my error.
| You did almost everything right. You took a sum from 1 to $\infty$ and added a $n=0$ term, and then subtracted it (your $\frac{-5}{\pi}$ term).
But you wrote it wrong, because
$$
\frac{5}{\pi^0} = \frac{5}{1} = 5.
$$
Nice work otherwise!
There's another way to deal with a series starting at $1$. You write
\begin{align}
S
&= \sum_{n=1}^\infty \frac5{\pi^n} \\
\end{align}
and then you let $k = n-1$ in that expression, or $n = k + 1$ (which si the same thing), and rewrite the sum in terms of $k$. As $n$ goes from $1$ to $\infty$, $k$ goes from $0$ to $\infty$. So you get
\begin{align}
S
&= \sum_{n=1}^\infty \frac5{\pi^n} \\
&= \sum_{k=0}^\infty \frac5{\pi^{k+1}} \\
\end{align}
Now that's not actually in geometric-series form, but a little algebra -- $\pi^{k+1} = \pi \cdot \pi^k$, is enough to get you there:
\begin{align}
S
&= \sum_{n=1}^\infty \frac5{\pi^n} \\
&= \sum_{k=0}^\infty \frac5{\pi^{k+1}} \\
&= \sum_{k=0}^\infty \frac5{\pi\cdot\pi^{k+1}} \\
&= \sum_{k=0}^\infty \frac5{\pi}\cdot\frac{1}{\pi^{k}} \\
&= \frac{5}{\pi}\sum_{k=0}^\infty \frac{1}{\pi^{k}} \\
&= \frac{5}{\pi}\sum_{k=0}^\infty \left(\frac{1}{\pi}\right)^{k} \\
\end{align}
and that's a geometric series with $A = 1$ and $r = \frac1\pi$.
| {
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Show that for $a,b>0$ and $n \in \mathbb N^+$ and $a \ne b$ $\left[a-\left(n+1\right)\left(a-b\right)\right]a^{n}Show that for $a,b>0$ and $n \in \mathbb N^+$ and $a \ne b$ the following does hold:
$$\left[a-\left(n+1\right)\left(a-b\right)\right]a^{n}<b^{n+1}$$
I tried to simplify the inequality to $$na^n(b-a)<b(b^n-a^n)$$
$$na^n(b-a)<b(b-a)(\sum_{k=0}^{n-1}b^{k}a^{n-1-k})$$
How to continue? (Notice that I don't know if $b-a$ is necessarily positive).
| We want to prove that, for any $n\in\mathbb{N}^+$ and $a,b>0;\;a\ne b$ the following inequality holds
$$a^n (a-(n+1) (a-b))-b^{n+1}<0$$
Proof by induction. It is true for $n=1$
$$a^1 (a-(1+1) (a-b))-b^{1+1}=-(a-b)^2<0$$
Now suppose it is true for $n$ that is
$$P(n)=-n a^{n+1}+b a^n+b n a^n-b^{n+1}<0\tag{1}$$
and let's prove it for $(n+1)$
$$P(n+1)=2 b a^{n+1}+b n a^{n+1}-n a^{n+2}-a^{n+2}-b^{n+2}$$
write the quotient and perform the long division wrt $b$
$$\frac{P(n+1)}{P(n)}=\frac{2 b a^{n+1}+b n a^{n+1}-n a^{n+2}-a^{n+2}-b^{n+2}}{-n a^{n+1}+b a^n+b n a^n-b^{n+1}}=\\=\frac{b(-n a^{n+1}+b a^n+b n a^n-b^{n+1})+(-2 b a^{n+1}-2 b n a^{n+1}+n a^{n+2}+a^{n+2}+b^2 a^n+b^2 n a^n)}{-n a^{n+1}+b a^n+b n a^n-b^{n+1}}$$
$$b+\frac{a^n \left(a^2 n+a^2-2 a b n-2 a b+b^2 n+b^2\right)}{n a^{n+1}-b n a^n-b a^n+b^{n+1}}=b+\frac{a^n (n+1) (a-b)^2}{n a^{n+1}-b n a^n-b a^n+b^{n+1}}$$
as the result is positive and the denominator is negative, the numerator must be negative, that is $P(n+1)<0$
| {
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Solve $\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$ Solve the equation,
$$\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$$
We have $$y^3\frac{d^2y}{dx^2}+y^4=1$$
I tried using change of dependent variable
Let $z=y^3\frac{dy}{dx}$
Then we get
$$y^3\frac{d^2y}{dx^2}+3y^2\left(\frac{dy}{dx}\right)^2=\frac{dz}{dx}$$
But i could not get an equation completely involving $z,x$
| Another Way to look at it
$$\frac{d^2y}{d^2x} = \frac{1}{y^3} - y => \frac{d^2y}{d^2x} = \frac{1-y^4}{y^3} =>$$
$$\frac{y^3}{1-y^4} d^2y = d^2x $$
Now this can be double integrated using the integral of a division formula
| {
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How do this question? A discrete random variable $X$ can only take the value of 1, 2, 3 and 4, with the probabilities $P(X=1)=P(X=3)$ and $P(X=2)=P(X=4)=2P(X=1)$.
A random sample of size 3 is chosen from $X$ with replacement.
(a) given that $X_1,\ X_2,\text{ and }X_3$ are the three independent random values of $X$, find the probability that exactly one of these values is more than 3. $\left[ans:\dfrac{4}{9}\right]$
(b) find the probability that the sample mean is greater than $\dfrac{4}{3}$. $\left[ans:\dfrac{209}{216}\right]$
I do part (a) like this,
$\Sigma P(X=x)=1$
$P(X=1)+P(X=2)+P(X=3)+P(X=4)=1$
$P(X=1)+2P(X=1)+P(X=1)+2P(X=1)=1$
$6P(X=1)=1$
$P(X=1)=\dfrac{1}{6}$
$X$
$1$
$2$
$3$
$4$
$P(X=x)$
$\dfrac{1}{6}$
$\dfrac{2}{6}$
$\dfrac{1}{6}$
$\dfrac{2}{6}$
$\dfrac{\left(\dfrac{1}{6}+\dfrac{2}{6}+\dfrac{1}{6}\right)^2 \cdot \dfrac{2}{6} \cdot 3!}{2!}=\dfrac{4}{9}$
My part (b),
$E(X)=\Sigma xP(X=x)=\dfrac{8}{3}$
$E(X^2)=\Sigma x^2P(X=x)=\dfrac{25}{3}$
$Var(X)=\sigma^2=E(X^2)-[E(X)]^2=\dfrac{17}{3}$
$\overline X \sim \mathcal{N} \left(\dfrac{8}{3},\dfrac{\dfrac{17}{3}}{3}\right)$
$\overline X \sim \mathcal{N} \left(\dfrac{8}{3},\dfrac{17}{9}\right)$
$P(X>\dfrac{4}{3})=P\left(\dfrac{\overline X-\dfrac{8}{3}}{\sqrt{\dfrac{17}{3}}}>\dfrac{\dfrac{4}{3}-\dfrac{8}{3}}{\sqrt{\dfrac{17}{3}}}\right)$
$=P(Z>-0.5601) $
$=1-P(Z<-0.5601) $
$=1-0.2877 $
$=0.7123 \neq \dfrac{209}{216}$
Can anyone check my work?
| Your calculations for (a) is correct. You choosing three events from
event
probability
$\{X \le 3 \}$
$2/3$
$\{X > 3 \}$
$1/3$
You need $2!$ in the denominator as events with two '$\le3$' like $\{X_1\le3, X_2\le3, X_3>3\}$ are double-counted by $3!$.
Your error in (b) comes from $\bar{X} \sim \mathcal{N}(E[X], \frac13 \mathrm{Var}(X))$. Why does the sample mean $\bar{X}$ follow the normal distribution?
As @sudeep5221 points out, the required probability is
$$\begin{aligned}
P(\bar{X} > 4/3) &= P(X_1+X_2+X_3 > 4) \\
&= 1 - P(X_1+X_2+X_3 \le 4) \\
&= 1 - P(X_1=X_2=X_3=1) - P(X_1+X_2+X_3 = 4) \\
&= 1 - (1/6)^3 - 3 \cdot (1/6)^2 (2/6) \\
&= \frac{6^3 - 1 - 3 \cdot 1 \cdot 2}{6^3} \\
&= \frac{209}{216}
\end{aligned}$$
| {
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Evaluate $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$ for $k=3$ $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$
How can I think this for k=3 ?
Here is my idea: $\prod _{i=n}^{6}\frac{i}{i+1}=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}$
then
$\frac{3}{14}\sum _{n=1}^{3}\frac{{\left(-4\right)}^{n+2}}{{\left(n-2\right)}^{2}}=\frac{3}{14}.\left(\frac{{\left(-4\right)}^{3}}{{\left(1-2\right)}^{2}}+\frac{{\left(-4\right)}^{5}}{{\left(3-2\right)}^{2}}\right)=-233,1428571$
Is my idea correct?
| That's a telescoping product so
$\begin{array}\\
\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\dfrac{i}{i+1}
&=\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\dfrac{n}{2k+1}\\
\end{array}
$
but then there is a problem at $n=2$
when there is a division by zero.
I'll change that to $n+2$
and we get
$\begin{array}\\
\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n+2{)}^{2}}\prod _{i=n}^{2k}\dfrac{i}{i+1}
&=\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n+2{)}^{2}}\dfrac{n}{2k+1}\\
&=\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}(n-2)}{n^2}\\
&=\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}n}{n^2}-\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}2}{n^2}\\
&=\dfrac1{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}}{n}-\dfrac{2}{2k+1}\sum _{n=3}^{k+2}\dfrac{{\left(-4\right)}^{n}}{n^2}\\
\end{array}
$
and these can be handeled
by the usual methods.
| {
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Why $\omega\left[\begin{pmatrix} 0 &Q\\ R &0 \end{pmatrix}\right]\leq\omega\left[\begin{pmatrix} P &Q\\ R &S \end{pmatrix}\right]$? Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all operators on $E$. The numerical radius of an operator $T\in\mathcal{L}(E)$ is given by
$$\omega(T)=\sup_{\|x\|=1}|\langle Tx, x\rangle|.$$
Let $P,Q,R,S\in\mathcal{L}(E)$. I want to prove that
$\omega\left[\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix}\right]\leq\omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right].$
Here $\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix},\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\in \mathcal{L}(E\oplus E)$.
My attempt: One can remark that
$$
\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix} = \frac{1}{2} \begin{pmatrix}
P &Q\\
R &S
\end{pmatrix} + \frac{1}{2} \begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix},
$$
This implies that
$$
\omega\left[\begin{pmatrix}
0 &Q\\
R &0
\end{pmatrix}\right] \leq \frac{1}{2} \omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right] + \frac{1}{2} \omega\left[\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}\right].
$$
But I'm facing difficulties to prove that
$$\omega\left[\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}\right]=\omega\left[\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}\right].$$
Let $T=\begin{pmatrix}
P &Q\\
R &S
\end{pmatrix}$. I want to find an unitary operator $U$ such that
$$U^*TU=\begin{pmatrix}
-P &Q\\
R &-S
\end{pmatrix}.$$
In this case we get the desired result since
$$\omega(U^*TU)=\omega(T).$$
| Note that $\omega(T) = \omega(-T)$. Thus it suffices to find a unitary operator such that satisfies
$$U^*TU=\begin{pmatrix}
P & -Q\\
-R &S
\end{pmatrix}.$$
But for this we can take
$$U=\begin{pmatrix}
I & 0\\
0 & -I
\end{pmatrix}.$$
| {
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Numbers of the form $\sum_{n=0}^{N} 2^{a_n} 3^n$ Just out of curiosity.
Does the numbers of the form
\begin{align}
\sum_{n=0}^{N} 2^{a_n} 3^n \text{,}
\end{align}
with $a_n \ge 0$ and $N \ge 0$, cover all integers not divisible by 3?
These numbers can never be divisible by three.
It is also easy to discover integers with multiple representations (e.g., $4 = 2^2 3^0 = 2^0 3^0 + 2^0 3^1$).
I am however stuck to prove that they cover all integers not divisible by 3.
Second question: The numbers of the form
\begin{align}
\sum_{n=0}^{N} 5^{a_n} 3^n \text{,}
\end{align}
with $a_n \ge 0$ and $N \ge 0$, are numbers congruent to $\{1, 4, 5, 8\} \pmod{12}$.
These are the numbers not divisible by 3 and not $\{ 2, 3 \} \pmod{4}$.
Many other such forms may be considered.
What is the name of such numbers (numbers of such a form)?
| If $x \le 8$, then all numbers not divisible by $3$ have the desired form as one can check by hand (e.g. $1=1$, $2=2$, $4=4$, $5=2+3$, $7=1+2 \cdot 3$, $8 = 8$).
If $x > 8$, and $x \equiv 1,2,4,5,7,8 \bmod 9$, then let $m = 4,8,1,2,4,2$ respectively, so $m$ is a power of $2$ and $x \equiv m \bmod 3$ but $x \not\equiv m \bmod 9$. Now write
$$x = m + 3y$$
Now, by construction, $y<x$ is not divisible by $3$. By induction, it has the required form, which implies that $x$ also has the required form. Indeed from this proof you see that one can take the power of $2$ to be in the set $\{1,2,4,8\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Jacobian linearization of trigonometric functions If I have to linearize a nonlinear trigonometric system around the origin $(0,0)$:
$$\dot{x_1} = x_2$$
$$\dot{x_2} = \cos(x_1)$$
I can apply the small angle approximation to find the matrix A:
$$ A = \pmatrix{0&1\\1&0} $$
However, if I apply Jacobian linearization and take the partial derivatives of $\dot{x}$, I get:
$$ A = \pmatrix{0&1\\-\sin(x_1)&0}$$
Evaluated at the origin, I get:
$$ A = \pmatrix{0&1\\0&0} $$
Seeing as how the two different linearization techniques yield different results, is one more valid than the other?
| Your small angle approximation is wrong and your Jacobian linearization is incomplete.
If $x_1 \approx 0$ then $\cos(x_1) \approx 1$ and so the system
$$
\begin{align}
\dot{x_1} &= x_2 \\
\dot{x_2} &= \cos(x_1) \\
\end{align}
$$
behaves close to $(0,0)$ approximately like
$$
\begin{align}
\dot{x_1} &= x_2 \\
\dot{x_2} &= 1 \\
\end{align}
$$
If you put this in matrix form:
$$
\begin{pmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} + \begin{pmatrix}
0 \\
1
\end{pmatrix}
$$
This is the "small angle approximation" and as you can see this is essentially the same as the Jacobian linearization. Remember that the Jacobian linearization of a nonlinear system $\dot{x} = f(x)$ at $x_0$ is
$$
\dot{x} = f(x_0) + A(x - x_0)
$$
where $A$ is the Jacobian matrix of $f$ at $x_0$. If you insert your system and your point you get:
$$
\begin{pmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
-\sin(0) & 0
\end{pmatrix} \begin{pmatrix}
x_1 - 0 \\
x_2 - 0
\end{pmatrix} + \begin{pmatrix}
0 \\
\cos(0)
\end{pmatrix}
$$
which is just:
$$
\begin{pmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{pmatrix} = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} \begin{pmatrix}
x_1 \\
x_2
\end{pmatrix} + \begin{pmatrix}
0 \\
1
\end{pmatrix}
$$
so the same as the small angle approximation.
Your mistake was that you neglected the fact that you linearized at a non-equilibrium point. Usually we linearize at an equilibrium and then $f(x_0) = 0$, so the constant term can be neglected.
But you linearize at $(0, 0)$ which is not an equilibrium point of your system because $\dot{x_2} \neq 0$ at $x_1 = 0$, so the system is not at rest at the origin.
| {
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Derive half angle formula $ \cos\frac x 2= \sqrt{\frac {1 + \cos x} {2}}$
How can the cosine of half angle formula
$$ \cos(\frac x 2) = \sqrt{\frac {1 + \cos x} {2}}$$
can be derived using the bisector angle theorem $ OA:OB = AM:MB$ for the $ \triangle AOM $?
Can anybody explain this to me?
|
Beside $\frac{AM}{BM} = \frac{OA}{OB}$ per the bisector angle theorem, $MN= \frac{BM}{OM}AM$ holds from similar triangles. Substitute them below
\begin{align}
\cos^2\frac x2= \frac{ON}{OA} \frac{OB}{OM}
&=\left( \frac{OM}{OA} + \frac{MN}{OA} \right) \frac{OB}{OM}
= \frac{OB}{OA} + \frac{\frac{AM}{BM}}{\frac{OA}{OB}} \cdot \frac{BM^2}{OM^2}\\
&= {\cos x}+ 1\cdot\sin^2\frac x2 = {\cos x}+ 1-\cos^2\frac x2
\end{align}
which yields
$$ \cos^2\frac x2 = \frac{1+\cos x}2$$
| {
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Prove that $\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\leq\frac14(a+b+c+3)$ if $abc=1$ I found the following exercise in a problem book (with no solutions):
Given $a,b,c>0$ such that $abc=1$ prove that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq\frac{a+b+c+3}{4}$$
I tried AM-GM for the fraction on the LHS but got stuck from there.
| let $p=a+b+c,q=ab+bc+ca,r=abc$
Now it is easy to prove the following as $r=1$
*
*$pq\ge 9r=9$
*$p^2\ge 3q$
Now we have to prove $$4(3+2p+q)\le (p+3)(2+p+q)$$ or $$p^2+pq\ge 6+3p+q$$ $$p^2+9\ge 6+3p+q$$ using $q\le p^2/3$ $$\frac{2p^2}{3}-3p+3\ge 0$$ which is true as $p\ge 3$
| {
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Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$ I am trying to solve the following problem:
Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$
the given hint to the problem is:
"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, use the uniqueness part of the division algorithm."
If I take $a = x^2 - x +1$ I have
$$(x^2 + x +1)^n = (a + 2x)^n = a^n + \binom{n}{1}a^{n-1} 2x+ \binom{n}{2}a^{n-2} (2x)^2 + \dots + (2x)^n$$
but how do I proceed further?
| Hint: every term but the last is divisible by $a$, so your remainder is congruent to $(2x)^n$. But that's not quite enough either, because that's still too 'large' if $n\gt 1$. You need to figure out how to divide $(2x)^n$ by $x^2-x+1$. The easiest way (IMHO) to do that is this: modulo $x^2-x+1$, we can just say that $x^2\equiv x-1$. This means that every polynomial is equivalent to one of the form $ax+b$; if you have $(2x)^n\equiv a_nx+b_n$, then you can multiply the RHS by $2x$ and use the equivalence to find $a_{n+1}$ and $b_{n+1}$ in terms of $a_n$ and $b_n$.
| {
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Find the highest natural number which is divisible by $30$ and has exactly $30$ different positive divisors.
Find the highest natural number which is divisible by $30$ and have exactly $30$ different positive divisors.
What I Tried: I am not sure about any specific approach to this problem. Of course as the number is divisible by $30$ , some of the factors of the number would be $1,2,3,5,6,10,20,30$ , but that only makes $8$ divisors and there will be $12$ more which I have no idea of .
Can anyone help me?
| Because it must be a multiple of $30 = 2 \cdot 3 \cdot 5$, then the number must be of the form $$N = 2^{\alpha+1} \cdot 3^{\beta+1} \cdot 5^{\gamma+1} \cdot M $$ where $M$ is $1$ or a product of primes that are greater than $5$ and $\alpha, \beta$, and $\gamma$ are non negative integers.
Let $d(n)$ represent the number of different positive divisors of $n$. Then
$$d(N) = (\alpha+2)(\beta +2)(\gamma + 2)d(M)$$
and $d(N)$ is a multiple of $30 = 2 \cdot 3 \cdot 5$. It follows that
$\{\alpha+2, \beta+2, \gamma+2\} = \{2,3,5\}$ and $d(M)=1$. The largest value of $N$ will occur when $\alpha = 0, \beta = 1$, and $\gamma = 3$. The value of $N$ is therefore
$$N = 2^1 \cdot 3^2 \cdot 5^4 = 11250$$
| {
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How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.
| $f(x)=\frac{1-\cos x}{x^2}$ is an even function, so we only look at $x \in [0,1]$. It's easy to prove $f(x)$ is continuous and differentiable at $x=0$, and
$$f'(x) = \frac{x \sin x + 2 \cos x - 2}{x^3}$$
Now $$x \sin x + 2 \cos x - 2 \le 2 \tan \frac x2 \cdot \sin x - 4 \sin^2 \frac x2\\= 2 \tan \frac x2 \cdot 2 \sin \frac x2 \cos \frac x2 - 4 \sin^2 \frac x2=0, \forall x \in [0,1]$$
Hence $f(x) \ge f(1) = 1 - \cos 1\approx .46 > \frac 13, 1-\cos x \ge (1-\cos 1) x^2 > \frac 13 x^2.\blacksquare$
| {
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Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$ Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.
We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\sqrt{\dfrac{\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha}{\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha}}=\sqrt{\dfrac{(\sin\alpha+\cos\alpha)^2}{(\sin\alpha-\cos\alpha)^2}}=\sqrt{\left(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right)^2}.$
Using the fact that $\sqrt{a^2}=|a|$ the given expression is equal to $\left|\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right|.$ I think that in the inverval $\left(45^\circ;90^\circ\right) \sin\alpha>\cos\alpha$ but how can I prove that? What to do next?
| After your 3rd line, you are actually done, without realizing it; you simply went down the wrong path.
The RHS
$$= \frac{\sin(a) + \cos(a)}{\sin(a) - \cos(a)}.$$
Edit
See the comments following this answer.
A case can be made that my analyis is flawed, since I didn't bother to prove that the denominator above is always positive in the interval $(45^\circ, 90^\circ)$.
Addendum
Responding to
You can prove what the OP wants using pure geometry. Only using the definition of cosine and sine.
Okay: Imagine that you have a unit circle (i.e. of radius $= 1$) centered at the origin, with any point $(x,y)$ that is on the unit circle representing $(\cos[a],\sin[a])$, where $a$ is the angle formed by the two line segments $\overline{(0,0),(1,0)}$ and $\overline{(0,0),(x,y)}$.
Clearly, at $a = 45^\circ$, you have that $(x,y) = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),$ since in a $45^\circ - 45^\circ - 90^\circ$ triangle, the two legs are equal.
Further, you can see geometrically, that for $a$ equal to any angle in $(45^\circ,90^\circ)$, for the corresponding point on the unit circle, the corresponding $y$ coordinate will have increased from $\frac{1}{\sqrt{2}}$, and the corresponding $x$ coordinate will have decreased from $\frac{1}{\sqrt{2}}$.
Therefore, since the $y$ coordinate represents $\sin(a)$ and the $x$ coordinate represents $\cos(a)$, you have that
$\sin(a) > \frac{1}{\sqrt{2}} > \cos(a),$ for $a$ equal to any angle in $(45^\circ,90^\circ)$.
A variation on the above argument is to notice that for $a$ equal to any angle in $(45^\circ,90^\circ)$, the slope of the line segment $\overline{(0,0),(x,y)}$ is $> 1$, which (alternatively) implies that $y = \sin(a) > x = \cos(a).$
| {
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Finding the maximizer Let be $ A\in \mathbb{R}^{2,2} $ a symmetrical and positive definite matrix with different eigenvalues $ \lambda_1>\lambda_2>0 $. Further we have the rayleigh quotient $$ \max_{x\neq 0}\frac{\overline{x}^T\cdot A\cdot x}{\overline{x}^T\cdot x}=\max_{\|x\|_2=1}\overline{x}^T\cdot A\cdot x $$.
For how many points $ x\in \mathbb{R}^2 $ the maximum $$ \max_{\|x\|_2=1}\overline{x}^T\cdot A\cdot x $$ is reached?
My idea:
Let $$ A:=\begin{pmatrix}a&b\\b&c\end{pmatrix},\quad x:=\begin{pmatrix}x_1\\x_2\end{pmatrix} ,\quad \|x\|_2^2=1=x_1^2+x_2^2$$
and define
$$ F(x)=x^T\cdot A\cdot x=(x_1, x_2)\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}x_1\\x_2\end{pmatrix}\\=(x_1, \sqrt{1-x_1^2})\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}x_1\\\sqrt{1-x_1^2}\end{pmatrix}=:F(x_1) $$.
Derivative:
$$ \nabla F(x_1)=\left(1,\frac{-x_1}{\sqrt{1-x_1^2}}\right)\cdot \begin{pmatrix}a&b\\b&c\end{pmatrix} \cdot \begin{pmatrix}1\\\frac{-x_1}{\sqrt{1-x_1^2}}\end{pmatrix}\stackrel{!}{=}0 $$.
From here I get stuck and I don't see how can I use the information about $ A $, the eigenvalues and the rayleigh qoutient.
| Note that the set $\|x\|=1$ is compact, so a $\max$ exists.
Solve $\max_{ \|x\|^2 =1 } {1 \over 2}x^T A x$ using Lagrange multipliers to get
$Ax = \lambda x$, so we see that the Lagrange multiplier must be an eigenvalue.
Hence $\max_{ \|x\|^2 =1 } x^T A x = \lambda_\max$ and in the given example,
since the eigenvalues are distinct we see that $x$ must be an eigenvector and hence the two solutions are $\pm x$.
| {
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Least residue power modulo May someone confirm that the least residue of $44^8$ modulo $7$ is $4$ please?
And that $2^2 \equiv 4\, (\!\!\!\mod 7)$?
Thanks in advance.
| It is correct: $44\equiv 2\mod 7$, so $44^8\equiv 2^8\mod 7$. Now $2^3\equiv 1\mod 7$, so $2^8\equiv 2^{8\bmod 3}=2^2\mod 7=4$, since $0\le 4<7$.
| {
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Misunderstanding on Taylor series expansion Sometimes, I try to find the series expansion of some random functions but I am usually wrong, especially when using already known formulas such as $$\frac{1}{1-X} = 1 + X + X^2 + X^3 + ... \quad (1)$$
Now imagine I want to find the series expansion of $\frac{1}{2+x^2}$, here is what I do :
$$\frac{1}{2+x^2}=\frac{1}{1-(-x^2-1)}=1+(-x^2-1)+(-x^2-1)^3+... \quad (2)$$
But the result is not the same as the results in the internet that finds : $$\frac{1}{2+x^2}= \frac{1}{2}- \frac{1}{4}x^{2}+\frac{1}{8}x^{4}- \frac{1}{16}x^{6}$$.
My guess is that the $X$ in (1) must verify $\lvert X \rvert < 1$ and when I take $X = (-x^2-1)$ in (2) then this condition is not verified. I also think that to find the right answer, computers do $$\frac{1}{2+x^2} = \frac{1}{2}\frac{1}{1+(\frac{x^2}{2})}$$
| The expansion $\frac{1}{1-x} = 1 + x + x^2 + x^3 + ... $ only works if $|x|<1$ as you pointed out yourself. Because $2+x^2>1$ the substitution you make is guaranteed to give you a divergent sum.
Now, by factoring out $\frac{1}{2}$ you solve that problem, since now it works whenever $|x|<\sqrt{2}$. So the substitution is fine (that is, it results in a convergent series).
| {
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Calculate $x^3 + \frac{1}{x^3}$ Question
$x^2 + \frac{1}{x^2}=34$ and $x$ is a natural number. Find the value of $x^3 + \frac{1}{x^3}$ and choose the correct answer from the following options:
*
*198
*216
*200
*186
What I have did yet
I tried to find the value of $x + \frac{1}{x}$. Here are my steps to do so:
$$x^2 + \frac{1}{x^2} = 34$$
$$\text{Since}, (x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$$
$$\Rightarrow (x+\frac{1}{x})^2-2=34$$
$$\Rightarrow (x+\frac{1}{x})^2=34+2 = 36$$
$$\Rightarrow x+\frac{1}{x}=\sqrt{36}=6$$
I have calculated the value of $x+\frac{1}{x}$ is $6$. I do not know what to do next. Any help will be appreciated. Thank you in advanced!
| Hint $(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$ also $(x+\frac{1}{x})=6$ not $36$ you seem to have typed an incorrect calculation.
| {
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Solve $e^{2z} - 2e^z + 2= 0$ So I've started by looking at
\begin{align} e^{z} = x:\\
x^2 - 2x + 2 = 0
\end{align}
Whose solutions should be: \begin{align}\ 1+i, 1-i \end{align}
Then I did the following:
\begin{align}
e^z=e^{x+iy}=e^x(\cos y+i\sin y)&=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{1}{2}+\frac{i \cdot 1}{2}\\
e^x= {\sqrt{2}} \quad \text{and}\quad y&=\frac{\pi}{4}+2\pi k\quad k \in \Bbb{Z}\\
x&= \sqrt{2} \quad \text{and} \quad y=\frac{\pi}{4}\\
\text{Also,} \quad e^x(\cos y+\sin y)&=\cos(\frac{7\pi}{4})+i\sin\frac{7\pi}{4})=\frac{1}{2}-\frac{i \cdot 1}{2}\\
e^x= \sqrt{2} \quad \text{and}\quad y&=\frac{7\pi}{4}+2\pi k\quad k \in \Bbb{Z}\\
x&= \sqrt{2} \quad \text{and} \quad y=\frac{7\pi}{4}\\
\end{align}
Are the solutions the following? \begin{align} \sqrt{2} + \frac{\pi}{4} \quad \text{and} \quad \sqrt{2} + \frac{7\pi}{4} \end{align}
Forgive the shoddy formatting as this is my first ever post here. Thanks.
| Yes.....
Once you get $e^z = 1 +i, 1-i$ you can take the following as a formula (assuming $a,b$ are real):
$a + bi= \sqrt{a^2 + b^2}e^{\arctan \frac ab i}= e^{\ln(a^2 + b^2)}e^{\arctan \frac ab i}= e^{\ln(a^2+b^2) + \arctan \frac ab i}$
[I suppose another way putting this is the principal natural log $\operatorname{Ln}(a+bi) = \ln(\sqrt {a^2 + b^2}) + \arctan \frac abi$. But as natural log is a "multivalued function" on complex numbers $\ln (a+bi) = \ln(\sqrt {a^2 + b^2}) + \arctan \frac abi + 2k\pi i$.]
so $z = \ln 2 + \arctan \pm 1 i = \ln 2 + \frac \pi 4i, \ln -\frac \pi 4 i$
.....
But as $\arg$ is congruent $\mod 2\pi i$ we can actually have an infinite number of solutions $z = \ln 2 +(\pm \frac \pi 4 + 2k\pi )i$ for any integer (positive or negative) $k$.
.....
Also you shouldn't use $x$ as a variable for two different meanings (although being in isolated steps of the proof there was no harm done).
| {
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How to find a upper bound of the set $A= \{ (1+\frac{1}{n})^n : n \in \mathbb{N}^{*} \} $ i have the set $A= \{(1+\frac{1}{n})^{n} : n \in \mathbb{N}^* \}$ and the exercise ask me to find at least $2$ upper bound least equal to $\frac{14}{5}$.
My first question is, how i know $\frac{14}{5}$ is a upper bound of $A$?
My attempt is show doesn't exist $m \in \mathbb{N}^*$ such that $(1+\frac{1}{m})^{m} = \frac{14}{5}$, but i'm stuck, for the other upper bound numbers, i could reason the same way?
Thanks for suggestions
| If you are not allowed to use the definition of $e$, then by observing that the sequence terms are bound by (refer to this)
$$
\begin{align}\left(1+\frac 1n\right)^n&=\sum_{k=0}^{n}\binom{n}{k}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots(n-k+1)}{k!}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{1}{k!}\left(1-\frac 1n\right)\left(1-\frac 2n\right)\cdots\left(1-\frac{k-1}{n}\right)\\&\lt \sum_{k=0}^{n}\frac{1}{k!}\\
&=1+\frac{1}{1}+\frac{1}{2}+\frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2}\\
&\lt 1+\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\\&=3-\frac{1}{2^{n-1}}\\&\lt 3\end{align}
$$
Now to get a stricter bound, since
$$
\frac{1}{2^1} + \frac{1}{3\cdot 2}+\frac{1}{4\cdot 3\cdot 2}+\cdots+\frac{1}{n\cdot(n-1)\cdots 2} < \frac
{1}{2} \left( 1+\frac{1}{2^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{n-1}} \right)
$$
we have
$$
\left(1+\frac 1n\right)^n < 1 + \frac{1}{2^0} + \frac
{1}{2} \left(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{n-1}} \right) < 2 + \frac{1}{2} \frac{1}{1-\frac{1}{3}} = \frac{11}{4} < \frac{14}{5}
$$
as required.
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Closed form for $\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$ How to get a closed form for the following sum:
$$\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$$
Where $F_k$ is the $k$th Fibonacci number.
I tried to get a recurrence relation,but I failed and the closed form is given by $$\frac{\left(-1\right)^{n}F_{2n+1}-\left(2n+1\right)}{5}$$
| $F_k=\frac{a^k-b^k}{\sqrt{5}}, a+b=1,ab=-1, a^2=a+1, b^2=b+1$
We will make multiple use of there relations below:
$$S=\frac{1}{5}\sum_{k=0}^{n} (-1)^k[a^{2k}+b^{2k}-2(ab)^k]~~~~(1)$$ $$S=\frac{1}{5}\sum_{k=0}^{n}(-1)^k\left[a^{2k}+b^{2k}-2(n+1)]\right]~~~~(2)$$
$$S=\frac{1}{5}\left[\frac{(-1)^{n+1}a^{2(n+1)}-1}{-a^2-1}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2-1}-2(n+1)\right]~~~~~(3)$$
$$S=\frac{1}{5}\left[\frac{(-1)^{n+1}a^{2(n+1)}-1}{-a^2+ab}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2+ab}-2(n+1)\right]~~~~~(4)$$
$$S=\frac{1}{5(a-b)}\left[\frac{(-1)^{n+1}a^{2(n+1)(a-b)}}{-a^2+ab}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2+ab}-2(n+1)(a-b)\right]~~~~(5)$$
$$=\frac{1}{5(a-b)}\left[(-1)^{n}[a^{2n+1}-b^{2n+1}]-1/a-1/b-2(n+1)(a-b)\right]$$
$$S=\frac{1}{5(a-b)}\left[(-1)^{n}[a^{2n+1}-b^{2n+1}]-(2n+1)(a-b)\right]~~~~(6)$$
$$\implies S=\frac{F_{2n+1}-(2n+1)}{5}~~~~~(7)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$? Given: $\begin{cases}x;y;z \in\Bbb R\\x+y+z=3\end{cases}$
Then what is the minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$ ?
I start with $AM-GM$
So $A \geq 8xyz$
Then I need to prove $xyz\geq 1$ right?
How to do that and if it's a wrong way, please help me to solve it !!!
| Let $A(x,y,z)=(1+x^2)(1+y^2)(1+z^2)$, defined on $\{(x,y,z)\in\mathbb{R}^3{\,\mid\,}x+y+z=3\}$.
Since $\max(|x|,|y|,|z|)\ge 3$ would yield $A > 8$, it follows $A$ has a global minimum value, $a$ say.
Since $(x,y,z)=1$ yields $A=8$, it follows that $a\le 8$.
Suppose $A(x,y,z)=a$.
Without loss of generality, assume $x\le y\le z$.
From $x+y+z=3$, it follows that $z\ge 1$.
We must have $\max(|x|,|y|,|z|)< 3$, hence $z < 3$.
Let $s=x+y$.
From $x+y+z=3$ and $1\le z < 3$, we get $0 < s \le 2$.
Let $u=(1+x^2)(1+y^2)$ and let $v=\left(1+\left({\Large{\frac{s}{2}}}\right)^2\right)^2$.
Replacing $y$ by $s-x$, we get
$$
u-v
=
\Bigl({\small{\frac{1}{16}}}\Bigr)
\!
\left((s-2x)^2+2(4-s^2)\right)
(s-2x)^2
$$
hence $u\ge v$ with equality if and only if $s=2x$.
But we can't have $u > v$, else $A(x,y,z) > A\Bigl({\Large{\frac{s}{2}}},{\Large{\frac{s}{2}}},z\Bigr)$, contrary to $A(x,y,z)=a$.
Thus $u=v$, hence $s=2x$, so $y=x$.
It follows that $a$ is the minimum value of
$$
f(x)
=
A(x,x,3-2x)
=
(1+x^2)^2\left(1+(3-2x)^2\right)
$$
where $f$ can be restricted to the interval $0\le x\le 1$.
At the endpoints we have $f(0)=10$ and $f(1)=8$.
Identically we have
$$
f'(x)
=
12(2x-1)(1+x^2)(x-1)^2
$$
which on the interval $(0,1)$ has the unique critical point $x={\large{\frac{1}{2}}}$.
Noting that $f\Bigl({\large{\frac{1}{2}}}\Bigr)={\large{\frac{125}{16}}}=7.8125 < 8$, it follows that $a={\large{\frac{125}{16}}}$, with $(x,y,z)=\Bigl({\large{\frac{1}{2}}},{\large{\frac{1}{2}}},2\Bigr)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Area bounded by $2 \leq|x+3 y|+|x-y| \leq 4$
Find the area of the region bounded by $$2 \leq|x+3 y|+|x-y| \leq 4$$
I tried taking four cases which are:
$$x+3y \geq 0, x-y \geq 0$$
$$x+3y \geq 0, x-y \leq 0$$
$$x+3y \leq 0, x-y \geq 0$$
$$x+3y \leq 0, x-y \leq 0$$
But it becomes so confusing to draw the region. Any better way ?
| Well, if you're looking for a plot:
Now, if we want to find the area of the part where $y\ge0$ and $x\ge0$, we can take a look at the following picture:
We can find the shaded area by finding:
$$3\cdot\left(\int_0^\frac{1}{2}\left(1-x\right)\space\text{d}x-\frac{1}{2}\cdot\frac{1}{2}\right)+\frac{1}{2}\cdot\frac{1}{2}+\int_1^2\left(2-x\right)\space\text{d}x=\frac{9}{8}\tag1$$
And the place where $y\ge0$ and $x\le0$, we can take a look at the following picture:
We can find the shaded area by finding:
$$3\cdot\frac{1}{2}\cdot1+\frac{1}{2}\cdot\frac{1}{2}+\int_0^\frac{1}{2}\left(1-x\right)\space\text{d}x-\frac{1}{2}\cdot\frac{1}{2}=\frac{15}{8}\tag 2$$
So, the total area is given by:
$$\mathscr{A}=2\cdot\frac{9}{8}+2\cdot\frac{15}{8}=6\tag3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding standard deviation given mean and probability of a sample (algebraically). I am trying to solve for the standard deviation of a normal distribution given the probability of a sample and mean:
$P(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp(-0.5(\frac{x-\mu}{\sigma})^2)$
So given $P(x)=a, x=b, \mu=c$ is it possible to find a closed-form solution of $\sigma$?
I tried reducing the problem to a form where it could be solved using the Lambert W function with no luck.
| $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Indeed, this case is suitable for solution in terms of the Lambert $\W$ function as follows:
\begin{align}
\frac1{\sqrt{2\pi}\sigma}\exp\left(-\frac{(b-c)^2}{2\sigma^2}\right)
&=a
\\
\frac1{2\pi\sigma^2}\exp\left(-\frac{(b-c)^2}{\sigma^2}\right)
&=a^2
\\
\frac1{\sigma^2}\exp\left(-\frac{(b-c)^2}{\sigma^2}\right)
&=2\pi a^2
\\
\frac{-(b-c)^2}{\sigma^2}\exp\left(-\frac{(b-c)^2}{\sigma^2}\right)
&=-2\pi(b-c)^2 a^2
\\
\W\left(\frac{-(b-c)^2}{\sigma^2}\exp\left(-\frac{(b-c)^2}{\sigma^2}\right)\right)
&=\W\left(-2\pi(b-c)^2 a^2\right)
\\
\frac{-(b-c)^2}{\sigma^2}
&=\W\left(-2\pi(b-c)^2 a^2\right)
\\
\sigma^2&=
\frac{(b-c)^2}{-\W\left(-2\pi(b-c)^2 a^2\right)}
\\
\sigma&=
\frac{|b-c|}{\sqrt{-\W\left(-2\pi(b-c)^2 a^2\right)}}
.
\end{align}
Or, using identity $\W(x)=x\exp(-\W(x))$, we can get
an equivalent, but more convenient form that includes the case $b=c$
without resort to the limits,
\begin{align}
\sigma&=
\frac1{a\sqrt{2\pi}}\,\exp\left(\tfrac12\,\W\left(-2\pi a^2(b-c)^2\right)\right)
.
\end{align}
Note that if $(-2\pi a^2(b-c)^2)\in(-\tfrac1{\e},0)$ there are two real solutions for $\sigma$:
\begin{align}
\sigma_0&=
\frac1{a\sqrt{2\pi}}\,\exp\left(\tfrac12\,\Wp\left(-2\pi a^2(b-c)^2\right)\right)
\\
\text{and }\quad
\sigma_1&=
\frac1{a\sqrt{2\pi}}\,\exp\left(\tfrac12\,\Wm\left(-2\pi a^2(b-c)^2\right)\right)
,
\end{align}
where $\Wp$ and $\Wm$ are the two real branches of the Lambert $\W$ function.
For example, setting $a=0.75,\ \mu=0.5$, $x=0.2$ (or $x=0.8$)
corresponds to two curves, one with
$\sigma=\sigma_0\approx0.403437$
and another
with $\sigma=\sigma_1\approx0.234237$, see image below:
$\endgroup$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4002059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
what value of K does the system have a unique solution $\begin{cases}x_1 + kx_2 - x_3 = 2\\2x_1 - x_2 + kx_3 = 5\\x_1 + 10x_2 -6x_3= 1\\
\end{cases}$
I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$
So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\2&-2&K&5\\1&10&-6&1\end{array}\right]$
I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$
and reduced it
So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\0&-1-2K&K+2&1\\0&10-K&-5&-1\end{array}\right]$
But now I am not sure how i could remove $10-K$ with $-1-2K$ any help would be appreciated
| If $k\neq -\frac12$
$$\left(
\begin{matrix}
1 & k & -1 & | & 2\\
2 & -1 & k & | & 5 \\
1 & 10 & -6 & | & 1 \\
\end{matrix}
\right) \xrightarrow[\text{$R_3=R_3-R_1$}]{\text{$R_2=R_2-2R_1$}}$$ $$\left(
\begin{matrix}
1 & k & -1 & | & 2\\
0 & -1-2k & k+2 & | & 1 \\
0 & 10-k & -5 & | & -1 \\
\end{matrix}
\right) \xrightarrow[\text{$(2k+1\neq0)$}]{R_3=R_3-(\frac{10-k}{2k+1})R_2} $$
$$\left(
\begin{matrix}
1 & k & -1 & | & 2\\
0 & -1-2k & k+2 & | & 1 \\
0 & 0 & \frac{-k^2-2k+15}{2k+1} & | & \frac{-3k+9}{2k+1} \\
\end{matrix}
\right) $$
Which has a unique solution $\iff \frac{-3k+9}{2k+1} \neq 0 \iff -3k+9\neq 0 \iff k\neq3$
If $k=-\frac12$ it is easy to show that the system has a unique solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$.
Evaluate:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$$
The only thing I can think of doing here is long division to simplify the integral down and see if I can work with some easier sections. Here's my attempt:
\begin{align}
\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx &= \int \left(\:x^4+3x^2+4x+3+\frac{18}{x+1} \right )dx \\
&= \int \:x^4dx+\int \:3x^2dx+\int \:4xdx+\int \:3dx+\int \frac{18}{x+1}dx \\
&= \frac{x^5}{5}+x^3+2x^2+3x+18\ln \left|x+1\right|+ c, c \in \mathbb{R}
\end{align}
The only issue I had is that the polynomial long division took quite some time. Is there another way to do this that is less time consuming? The reason I ask this is that, this kind of question can come in an exam where time is of the essence so anything that I can do to speed up the process will benefit me greatly.
| As substitution by parts goes nowhere It simply do what you did but skip the step of multiplying out the polynomial $ \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}=\frac {(x(x+1) + 3)(x^3+7)}{x+1}= x(x^3 +7) + \frac 3{x+1}(x^3 + 7)=$
$x(x^3 +7) +\frac 3{x+1}(x^2(x+1) -x(x+1) + x+1 + 6)=x(x^3 +7) + 3(x^2-x+1) + \frac {18}{x+1} $
Saves one or two tablets of ibuprofen.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution? In this answer, I give an elementary solution of the Diophantine equation $$y^2=3x^4+3x^2+1.$$
In this post, the related equation $$y^2=3x^4-3x^2+1 \tag{$\star$}$$ is solved (in two different answers!) using elliptic curve theory. Is there an elementary method of proving that $x=1$ is the only positive rational solution to ($\star$)? In particular, I’d love to see a method using descent.
| An elementary (and general) method
This is a little-known method which will solve many equations of this type. For this example it will show that the only positive rational solution is $p=1$. We start by letting $p=\frac{y}{x}$, where $x$ and $y$ are coprime integers. Thus we seek positive integer solutions of $$ x^4-3x^2y^2+3y^4=z^2.$$ This is one of a family of equations which can be dealt with together. (Apologies for using the letters for variables which I am used to when solving these types of equation.)
Theorem
The only positive integer solutions of any equation of either of the forms
$$ Ax^4+6x^2y^2+Cy^4=z^2,AC=-3\tag{1}$$ $$ ax^4-3x^2y^2+cy^4=z^2,ac=3\tag{2}$$
have $x=y=1$.
Proof
First note that, for either equation, we can suppose that $x,y,z$ are pairwise coprime since a common factor of any pair of $x,y,z$ would be a factor of all and cancellation can occur.
Also note that precisely one of $A$ and $C$ is divisible by $3$. Without loss of generality we can suppose that $3$ is a factor of $C$ and that $A\in \{\pm 1\}.$ Then $z^2\equiv A\pmod 3$ and so $A=1$. Similarly, we can suppose $a=1$.
An equation of form (1)
$x^4+6x^2y^2-3y^4=z^2$ can be rewritten, using completing the square, as $$\left (\frac{x^2+3y^2-z}{2}\right )\left (\frac{x^2+3y^2+z}{2}\right)=3y^4.$$
Since the two bracketed factors, $L$ and $M$ say, differ by the integer $z$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $z$ and $y$, a contradiction.
Therefore $\{L,M\}=\{au^4,cv^4\}$, where $ac=3$ and $y=uv$, with $u$ and $v$ coprime. Then $$au^4+cv^4=x^2+3y^2=x^2+3u^2v^2.$$ Therefore $au^4-3u^2v^2+cv^4=x^2$, $ac=3$, an equation of form (2).
It is important to note that the mapping $(x,y,z)\rightarrow (u,v,w)$ is invertible. Only one solution set can map to $(u,v,w)$ by this process.
An equation of form (2)
Let $u,v,x$ be a pairwise coprime solution of $ u^4-3u^2v^2+3v^4=x^2$ and let $t$ be the greatest common divisor of $v$ and $2$. Then $(U,V,W)=(\frac{2u}{t},\frac{v}{t},\frac{4x}{t^2})$ is a pairwise coprime solution of $U^4-12U^2V^2+48V^4=W^2$.
This can be rewritten, using completing the square, as $$\left (\frac{U^2-6V^2-W}{2}\right )\left (\frac{U^2-6V^2+W}{2}\right)=-3V^4.$$
The bracketed factors, $L$ and $M$, are again coprime integers.Therefore $\{L,M\}=\{aX^4,cY^4\}$, where $ac=-3$ and $V=XY$, with $X$ and $Y$ coprime. Then $$aX^4+6X^2Y^2+cY^4=U^2,ac=-3,$$ an equation of form (1). Again, this mapping of solutions is invertible.
Fermat's infinite descent
We have seen that any positive integer solution $(x,y,z)$ of an equation of form (1) leads to another positive integer solution $(X,Y,Z)$, where $$Y=\frac{y}{tuX}.$$
We have to agree with Fermat that "there cannot be a series of numbers [positive integers] smaller than any given [positive] integer we please" and therefore the above process must lead to solutions with $tuX=1$.
It is now straightforward to plug $t=u=X=1$ back into the equations to see that there is only a simple loop consisting of the solutions
$$x^4+6x^2y^2-3y^4=z^2,x=y=1,z=2$$ $$ x^4-3x^2y^2+3y^4=z^2,x=y=z=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$ What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$
But, by solving it myself the result comes out to be different. My solution is as follow:
$$\begin{aligned} Put,\, &\cot^{-1}\left(\frac{-3}{4}\right) = \theta
\\\implies &\cot(\theta) = \frac{-3}{4} =\frac{b}{p}
\\So,\,&\cos(\theta) = \frac{-3}{5}\end{aligned}$$
$$\begin{aligned}\\Then,
\\\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right) &= \sin\left(\frac{\theta}{2}\right) \\&= \sqrt{\frac{1-\cos{\theta}}{2}}
\\&= \sqrt{\frac{1+\frac{3}{5}}{2}}
\\&= \frac{2}{\sqrt{5}}
\end{aligned}$$
This solution is used by a lot of websites.
So, I got two different values of single expression but I am not sure which one is correct. Can you point out where I have done the mistake?
| You got the expression $\cot \theta= \frac{-3}{4}$
As you may know, the value of $\cot \theta$ is postive in the first and third quadrants, which means it is negative in the second and fourth quadrants.
When $\cot \theta=\frac{-3}{4}$, $\cos \theta$ can either equal $\frac{-3}{5}$ or $\frac{3}{5}$
$\cos\theta=\frac{3}{5}$ is true wrt the fourth quadrant.
If $\cos \theta=\frac{3}{5}$, $\sin \theta= \frac{1}{\sqrt{5}}$
But, we know that in the fourth quadrant $\sin \theta$ is negative.
Thus, $\sin\theta=\frac{-1}{\sqrt{5}}$
In short, both answers are correct, for their respective quadrants.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Sum of square roots inequality
For all $a, b, c, d > 0$, prove that
$$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$
The idea would be to use AM-GM, but $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a time doesn't really work as well. How can I solve this question? Any help is appreciated.
| Alternative solution using Cauchy-Schwarz, which finishes the problem off immediately. By C-S, we have:
\begin{align}
& (a+b+c+d)(1+1+1+1) \geq (\sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d} )^2 \\
& \Rightarrow 4(a+b+c+d) \geq (\sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d} )^2 \\
& \Rightarrow 2\sqrt{a+b+c+d} \geq \sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$ \sum_{i=1}^{10} x_i =5 $, and $ \sum_{i=1}^{10} x_i^2 =6.1 $ ; Find the largest value of these ten numbers
Ten non-negative numbers are such that $ \sum_{i=1}^{10} x_i =5 $,
and $ \sum_{i=1}^{10} x_i^2 =6.1 $. What is the greatest value of the
largest of these numbers?
It presents a lot of confusion to me, but I think the largest number maybe a integer. Please help me with procedure.
Source : Moscow Institute of Physics and Technology admission assessment for computer Science
| A General Problem
Suppose
$$
\frac{\sum_{i = 1}^n x_i}{n} = a
$$
and
$$
\frac{\sum_{i = 1}^n x_i^2}{n} = b^2
$$
Set $y_i = x_i - a$. Then (as you can check)
$$
\frac{\sum_{i = 1}^n y_i}{n} = 0
$$
So
$$
b^2 = \frac{\sum_{i = 1}^n (y_i + a)^2}{n} = \frac{\sum_{i = 1}^n y_i^2 + 2ay_i + a^2}{n} = \frac{\sum_{i = 1}^n y_i^2}{n} + 2a \frac{\sum_{i = 1}^n y_i}{n} + \frac{a^2}{n} \sum_{i = 1}^n 1
$$
Setting
$$
c^2 = \frac{\sum_{i = 1}^n y_i^2}{n}
$$
We obtain
$$
b^2 = c^2 + 2a(0) + a^2
$$
So
$$
c^2 = b^2 - a^2
$$
Observe that such a sequence $y_i$ achieves its largest possible element precisely when $x_i$ achieves its largest possible element.
The upshot of all of this is that we've reduced the problem to finding the largest possible value of a sequence of numbers $y_i$ with a mean of $0$ (i.e. $\frac{\sum_{i =1}^n y_i}{n} = 0$) and variance of $c^2$ (i.e. $\frac{\sum_{i = 1}^n y_i^2}{n} = c^2$).
Intuitively, it's clear that the maximum should occur when $y_1 = \ldots = y_{n -1} = y$. Applying the mean and variance condition and performing a straightforward calculation yields $y_n = \sqrt{n - 1} c$. It remains to check that this is the best we can do.
Suppose for contradiction $y_n > \sqrt{n - 1} c$. Then
$$
\sum_{i = 1}^{n - 1} y_i = -y_n
$$
and
$$
\sum_{i = 1}^{n - 1} y_i^2 = nc^2 - y_n^2
$$
Thus
$$
\frac{\sum_{i =1}^{n - 1} y_i}{n -1} = -\frac{y_n}{n - 1}
$$
and
$$
\frac{\sum_{i = 1}^{n - 1} y_i^2}{n - 1} = \frac{nc^2 - y_n^2}{n - 1}
$$
Set $z_i = y_i + \frac{y_n}{n - 1}$. Then
$$
0 \leq \frac{\sum_{i = 1}^{n - 1} z_i^2}{n - 1} = \frac{nc^2 - y_n^2}{n - 1} - \left(\frac{y_n}{n - 1}\right)^2 < \frac{nc^2 - (n - 1)c^2}{n -1} - \frac{(n - 1)c^2}{(n - 1)^2} = 0
$$
which is absurd. Thus $y_n = \sqrt{n - 1} c$ is the best we can do.
In terms of $x_n$, $a$, and $b$ this is
$$
x_n = a + \sqrt{(n - 1)(b^2 - a^2)}
$$
Your Problem
In your problem $a = .5$ and $b^2 = .61$. Plugging in, you find that
$$
x_n = 2.3
$$
is the best you can do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Numerically evaluate a series involving Stirling numbers of the second kind How can I numerically evaluate the following for $n$ above one million:
$$ \sum_{k>=n} \sum_{i=0}^{n-1} (-1)^i {{n-1}\choose{i}} \left( \frac{n-1-i}{n} \right)^{k-1} k (k+1)$$?
I have tried changing the order of summation and the getting closed form for the geometric series in $k$ but the terms are large (even for $n = 100$) and I run into precision problems.
| Recall that $${n \brace k}=\frac{1}{k!}\sum _{i=0}^k(-1)^i\binom{k}{i}(k-i)^n,$$
so your expression is equal to
$$\sum _{k\geq n}\frac{(n-1)!k(k+1)}{n^{k-1}}{k-1\brace n-1}=2(n-1)!\sum _{k\geq n}\frac{1}{n^{k-1}}\binom{k+1}{2}{k-1\brace n-1},$$
if $n$ is large, then $k$ is large, so ${k-1\brace n-1}\sim \frac{(n-1)^{k-1}}{(n-1)!}$ and so
$$2(n-1)!\sum _{k\geq n}\frac{(n-1)^{k-1}}{n^{k-1}(n-1)!}\binom{k+1}{2}=2\sum _{k\geq n}\left (\frac{n-1}{n}\right )^{k-1}\binom{k+1}{2}$$
now take $x=(n-1)/n,$ we complete the sum to $\infty$ and take out the finite sum getting
$$2\left (\sum _{k=0}^{\infty}\binom{k+1}{2}x^{k-1}-\sum _{k=0}^{n-1}\binom{k+1}{2}x^{k-1}\right )=2\left (\frac{1}{(1-x)^3}-\sum _{k=0}^{n-1}\binom{k+1}{2}x^{k-1}\right ),$$
recall too that taking the second derivative of $\sum _{k=1}^{n}x^{k}$ gives you
$\sum _{k=1}^{n-1}(k+1)k\cdot x^{k-1}$ but that sum is $\frac{x^{n+1}-1}{x-1}$ and so we get
$$\frac{2}{(1-x)^3}-\frac{(n + 1)n\cdot x^{n-1}}{x-1} + \frac{2 (n+1) x^n}{(x-1)^2} - \frac{2 (-1 + x^{1 + n})}{(x-1)^3},$$
taking $x=(n-1)/n$ back we get
$$2n^3+(n + 1)n^2\cdot \left (1-\frac{1}{n}\right )^{n-1} + 2 n^2(n+1) \left(1-\frac{1}{n}\right )^n - 2 n^3(-1 + \left (1-\frac{1}{n}\right )^{n+1}).$$
For $n$ as big, if I am not mistaken, your expression should be close to
$$(4-\frac{2}{e})n^3+\frac{3}{e}(n+1)n^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008869",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $x^3+2x+1=2^n$ over positive integers? I just have proved that $x^3+2x+1$ can not be divided by $2^n$ ($n>3$), for $x=8k+t$ and $t=0,1,2,3,4,6,7$, but for $x=8k+5$, I have no way to deal with this case, may someone give me some hints for this question? It seems that $(x,n)=(1,2)$ is the unique solution. Also what I know is that $x$ must divide $2^n-1$.
Note. Here both $x$ and $n$ are positive integers.
| Use the Quadratic Reciprocity - https://en.wikipedia.org/wiki/Quadratic_reciprocity (The supplementary laws using Legendre symbols):
\begin{align}
\left(\frac{-2}{p}\right) = (-1)^{\frac{p^2+4p-5}{8}}=\begin{cases}
1 & p\equiv 1,3~\text{(mod 8)}\\
-1 & p\equiv 5,7~\text{(mod 8)}
\end{cases}
\end{align}
\begin{align}
{}
\end{align}
As you mentioned, for $n\ge 3$, we must have $x\equiv 5 ~~\text{(mod 8)}$. Now note that
\begin{align}
x^3+2x+1 \equiv 1 ~~ \text{(mod 3)}.
\end{align}
You can directly check it by using Fermat's little theorem. Thus, we have \begin{align}
x^3+2x+1 = 2^n\equiv (-1)^n\equiv 1 ~~\text{(mod 3)},
\end{align}
which implies that $n$ is even. Let $n=2m$. By adding $2$ to both sides, we obtain
\begin{align}
(x+1)(x^2-x+3) = 2^{2m}+2.
\end{align}
Now consider $x^2-x+3$. Clearly, it is an odd number, and its prime factor $p$ must divide $(2^m)^2+2$. It then follows that $p=8k+1$ or $p=8k+3$ due to the quadratic reciprocity. This implies that
\begin{align}
x^2-x+3 \equiv \text{1 or 3}~~\text{(mod 8)}.
\end{align}
However, because $x\equiv 5 ~~\text{(mod 8)}$, we have
\begin{align}
x^2-x+3 \equiv 25-5+3\equiv 7~~\text{(mod 8)},
\end{align}
which is a contradiction.
| {
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"source": "stackexchange",
"question_score": "6",
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maximum and minimum value of $(a+b)(b+c)(c+d)(d+e)(e+a).$
Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$
I couldn't proceed much, however I think I got the minimum and maximum case.
For minimum, we get $-512$ with equality on $(-1,-1,-1,-1,9).$
For maximum, we get $288$ with equality on $(-1,-1,-1,4,4).$
| We can solve the problem by substitution.
Step 1: Let
\begin{align}
&A = a+b \\
&B = b+c \\
&C = c+d \\
&D = d+e \\
&E = e+a \\
\end{align}
Then the problem is equivalent to find the maximum and minimum of $ABCDE$, where $A,B,C,D,E\ge -2$ and $A+B+C+D+E = 10$.
\begin{align}
{}
\end{align}
Step 2: Consider each case separately, according to the sign of $A$, $B$, $C$, $D$, and $E$. Without loss of generality, we only need to consider cases as follows:
\begin{align}
&(1):~(A,B,C,D,E) = (+,+,+,+,+)\\
&(2):~(A,B,C,D,E) = (+,+,+,+,-)\\
&(3):~(A,B,C,D,E) = (+,+,+,-,-)\\
&(4):~(A,B,C,D,E) = (+,+,-,-,-)\\
&(5):~(A,B,C,D,E) = (+,-,-,-,-)\\
\end{align}
where + means the number is non-negative, and - means the number is non-positive.
Step 3 (Minimum): Of course, the minimum can occur in cases (2) and (4). First, consider case (2). As your intuition suggested, the minimum occurs when $E=-2$ and $A=B=C=D=3$. We can show this as follows. First, observe that for all $(A,B,C,D,E)$, we have
\begin{align}
-2ABCD \le ABCDE.
\end{align}
Next, observe that $4(ABCD)^{1/4}\le A+B+C+D = 10-E\le 12$, which implies that $ABCD\le 81$. Thus, in case (2), the minimum of $ABCD$ is $-162$. Similarly, you can find the minimum value under case (4), which is $-512$.
Step 4 (Maximum): In the same vein, the maximum can occur in cases (1), (3), and (5). And their maximum values are $32$, $784/9\approx 87.11$, and $288$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I find the domain of this sqrt log function?
Can someone help me to find the domain of this function step by steps,
I know how to find the domain for a log by making it >0 but I am unsure about the process doing it with 2 different logarithm and a square root.
$$f(x)=\sqrt{\log_{0.5}(5-x)+\log_2(2x-4)-1}$$
| Let's first define both the $\log$s, input of $\log$ should be a positive quantity.
$5 - x > 0 \implies x \in (-\infty, 5)$
$2x - 4 > 0 \implies x \in (2, \infty)$
Now we know that $\log_{a}{b} = -\log_{\frac{1}{a}}{b} = \log_{\frac{1}{a}}{\frac{1}{b}}$
$\implies\log_{0.5}(5-x) = \log_{2}{\frac{1}{5-x}}$
Now, we know that anything inside square root should be non negative.
$\implies \log_{0.5}(5-x)+\log_2(2x-4)-1 \geq 0$
$\implies \log_{2}{\frac{1}{5-x}+\log_2(2x-4)-1} \geq 0$
$\implies \log_2{\frac{2x-4}{5-x}} \geq 1$
$\implies \frac{2x-4}{5-x} \geq 2$
$\implies \frac{2x-4}{5-x} - 2 \geq 0$
$\implies \frac{4x-14}{5-x} \geq 0$
$\implies \frac{4x-14}{x-5} \leq 0$
$\implies x \in [\frac72, 5)$
Taking intersection of all 3 intervals:
$$\boxed{x \in [\frac72,5)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$ Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$
saw some similar problems like
show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true
but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.
\begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\
&= 10001
\end{align}
so $x > 100$
| Looking at your method, you've actually proven $\frac{3}{4}x^2 > 10000$ as well – just, instead of writing $\frac{2}{1} \geqslant \frac{3}{2}$, you can incorporate that $\frac{3}{4}$ here, and you'll have exactly $\frac{3}{2} = \frac{2}{1} \cdot \frac{3}{4}$. Now we can deduce that
$$ x > \sqrt{\frac{40000}{3}} \approx 115.47 > 115$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Holomorphic Function $g(z)=\sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}$ Let $g(z)=\sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}$
Prove that this function is holomorphic in $\mathbb{C} \setminus \{0\}$ and calculate $$\int_{|z|=2} g(z)\,\mathrm{d}z$$
Edit:
What I have done:
\begin{align}
\begin{split}
g(z) & = \sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}\\
& = (2-3i)\cdot \sum_{n=1}^\infty \dfrac{1}{n! z^n}\\
& = (2-3i)\cdot \sum_{n=1}^{\infty} \dfrac{z^{-n}}{n!}\\
& = (2-3i)\cdot \sum_{n=0}^{\infty} \dfrac{z^{-n}}{n!} - \dfrac{z^{-0}}{0!}\\
& = (2-3i)\cdot (e^{z^{-1}} -1)\\
& = (2-3i) e^{z^{-1}} -(2-3i) \\
& = 2e^{z^{-1}} - 3ie^{z^{-1}} +3i - 2
\end{split}
\end{align}
let $z^{-1}=z^*$ for a $z^* = x+ iy$
Then let's see $2e^{z^{*}} - 3ie^{z^{*}} +3i - 2$ is holomorphic
Proof:
$ 2e^{z^{*}} - 3ie^{z^{*}} +3i - 2$
$ = 2e^{x + iy} - 3ie^{x + iy} +3i - 2 $
$ = 2e^{x} e^{iy} - 3ie^{x}e^{iy} +3i - 2 $
$ = 2e^{x} [ \cos(y) + i \sin(y) ] - 3 ie^{x} [ \cos(y) + i \sin(y) ] +3i - 2 $
$ = 2e^{x}\cos(y) + 2 i e^{x} \sin(y) - 3 ie^{x} \cos(y) - 3 i^2 e^{x} \sin(y)+3i-2 $
$ = 2e^{x}\cos(y) + 2 i e^{x} \sin(y) - 3 ie^{x} \cos(y) + 3 e^{x} \sin(y) +3i - 2 $
$ = 2e^{x}\cos(y) + 3e^{x} \sin(y) - 2 + 2 i e^{x} \sin(y) - 3 i e^{x} \cos(y) +3i $
$ = 2e^{x}\cos(y) + 3e^{x} \sin(y) - 2 + i[2 e^{x} \sin(y) - 3 e^{x} \cos(y) +3] $
Now let $u(x,y) = 2e^{x}\cos(y) + 3e^{x} \sin(y) - 2$
And let $v(x,y) = 2 e^{x} \sin(y) - 3 e^{x} \cos(y) +3$
| It was discussed in the comments that $g(z)=(2-3i)(e^{1/z}-1)$ is holomorphic in the complex plane excluding the origin. The function $e^{1/z}$ is a composition of holomorphic functions, thus holomorphic.
The residue is $2-3i$ so the integral evaluates to $2 \pi i \cdot(2-3i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$
If
$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$
I added $x^2$ to both side of the equation:
$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem useful.
| We have$$x^2+y^2+z^2+t^2=x(y+z+t)\iff x^2-\frac34x^2+(y-\frac x2)^2+(z-\frac x2)^2+(t-\frac x2)^2=0$$ i.e.$$\left(\frac x2\right)^2+\left(y-\frac x2\right)^2+\left(z-\frac x2\right)^2+\left(t-\frac x2\right)^2=0$$
We are done
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to prove that for $n=k+1$ that,
$$7|4^{2^{k+1}}+2^{2^{k+1}}+1,\;\; \forall k\in \mathbb{N}$$
We know that if $a$ is divisible by 7 then $b$ is divisible by 7 iff $b-a$ is divisible by 7.
Then,
$$
\begin{aligned}
b-a &= 4^{2^{k+1}}+2^{2^{k+1}}+1 - (4^{2^{k}}+2^{2^{k}}+1)\\
&= 4^{2^{k+1}}+2^{2^{k+1}} - 4^{2^{k}}-2^{2^{k}}\\
&= 4^{2\cdot 2^{k}}+2^{2\cdot 2^{k}} - 4^{2^{k}}-2^{2^{k}}
\end{aligned}
$$
I get stuck here, please help me.
| $7|8\cdot 2^{2^{k}}+2\cdot 2^{2^{k}}-2\cdot 2^{2^{k}}-2^{2^{k}}$
$7| 2^{2^{k}}\cdot (8+2-2-1)$
$7| 2^{2^{k}}\cdot (7)$
I think I proved it.
| {
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"url": "https://math.stackexchange.com/questions/4016714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Maximum value of $(x−1)^2+ (y−1)^2+ (z−1)^2$ with constraint $x^2+y^2+z^2 ≤2 , z≤1$ So the problem is that I have $D(f)=\{(x,y,z), x^2+y^2+z^2 ≤2 , z≤1\}$ and I have to determine the maximum value for the function $(x−1)^2+ (y−1)^2+ (z−1)^2$ in $D$.
I'm just confused as I don't actually know if $z\le1$ counts as a constraint as well, or is it just for me to sketch the area, which is actually a part of the question.
Furthermore, I know that I have to use Lagrange multiplier method, but I honestly don't know how because $\le$ is making the question hard for me. Do I just calculate as usual and count $\le$ the same as $=$?
appreciate all the feedback
Edit: I have calculated the grad f =0 which is = $D(f)=(2(x-1), 2(y-1), 2(z-1))$ where I've got that $x=y=z= 1$ and $f(1, 1, 1)=0$. (I don't know what to do with this though). Then I calculated $L(x, y, x, λ) = (x−1)^2+(y−1)^2+(z−1)^2 +λ(x^2+y^2+z^2-2)$, then the four cases where I got the same value which is $-2λ= 2(x-1)/x = 2(y-1)/y = 2(z-1)/z$. Which means that $x=y=z$, put it in the $D$ function $x^2+ x^2+ x^2$ and ended up with $x=y=z= −+√2/√3$. I took the minus sign for the maximum distance from $(1,1,1)$. which means that the answer is $x=y=z= −√2/√3$. Is it correct?
| From the inequality: $(a+b)^2 \le 2(a^2+b^2) $, we have $$(-(x+y))^2 \le 2(x^2+y^2) \implies -(x+y) \le\sqrt{2(x^2+y^2)}$$
And by applying the constraint $x^2+y^2+z^2 ≤2$,
\begin{align}
(x−1)^2+ (y−1)^2 &= (x^2+y^2) -2(x+y)+2 \\
& \le (x^2+y^2) +2 \sqrt{2(x^2+y^2)}+2\\
& \le (2-z^2) +2 \sqrt{2(2-z^2)}+2\\
\end{align}
So, we can deduce that $$(x−1)^2+ (y−1)^2 +(z-1)^2 \le 5-2z + 2 \sqrt{2(2-z^2)}$$ with $-\sqrt{2} \le z\le 1$
Let's study the function $g(z)= -z+\sqrt{2(2-z^2)}$ in the interval $z\in(-\sqrt{2},1)$. We have $g'(z)=-1 -\frac{2z}{\sqrt{4-2z^2}}$ and reaches its maximum at $\sqrt{4-2z^2} = -2z$ or $z =-\frac{\sqrt{2}}{\sqrt{3}}$. We verify that this value is in the interval $z = -\frac{\sqrt{2}}{\sqrt{3}} \in(-\sqrt{2},1)$.
Note: we can use Cauchy Schwart inequality to find the maximum value of $g(z)$. This approach is more direct but not orthodox.
Finally, $(x−1)^2+ (y−1)^2 +(z-1)^2$ reaches the maximum value when ($z =-\frac{\sqrt{2}}{\sqrt{3}} $ and $x = y$ and $x^2+y^2 +z^2 =2$), or $x=y=z = -\frac{\sqrt{2}}{\sqrt{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find a matrix $A$ with no zero entries such that $A^3=A$ I took a standard $2 × 2$ matrix with entries $a, b, c, d$ and multiplied it out three times and tried to algebraically make it work, but that quickly turned into a algebraic mess. Is there an easier method to solve this?
| If $ad-bc=1,$ with $a,b,c,d$ non-zero, you can use:
$$\begin{pmatrix}-bc&-bd\\ac&ad\end{pmatrix}$$
For example, $a=2,b=1,c=1,d=1$ gives:
$$\begin{pmatrix}-1&-1\\2&2\end{pmatrix}$$
I got this by taking:
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}\begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
Where $B=\begin{pmatrix}0&0\\0&1\end{pmatrix}$ satisfies $B^2=B$ and hence $B^3=B.$
You can also choose $B=\begin{pmatrix}-1&0\\0&1\end{pmatrix}$ or $B=\begin{pmatrix}0&0\\0&-1\end{pmatrix}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$3^n+4^n<5^n$ for all $n>2$ I'm doing the following induction proof and wanted to know if this was valid. I think it is, but I'm seeing more complicated solutions than what I did. What I did seems much easier.
Prove that $3^n+4^n<5^n$ for all $n>2$.
When $n=3$ we get $91<125$. No problem, now assume the result is true from $k<n$, $(5^k>3^k+4^k)$ and consider $5^{k+1}=5 \times 5^k>5(3^k+4^k)=5\times 3^k + 5\times 4^k>3\times 3^k+4\times 4^k=3^{k+1}+4^{k+1}$ since $5\times 3^k>3\times 3^k$ and $5\times 4^k> 4\times 4^k$.
| Here's a method I like
for showing
$f(n) > g(n)$
for $n \ge n_0$.
*
*Show that
$f(n_0) > g(n_0)$.
*Show that,
if $n \ge n_0$
and $f(n) > g(n)$ then
$f(n+1)-f(n)
\ge g(n+1)-g(n)
$.
Then $f(n) > g(n)$
for $n \ge n-0$.
In this case,
$f(n) = 5^n,
g(n) = 3^n+4^n$.
Choose $n_0 = 3$.
$f(n_0) = 5^3=125,
g(n_0) = 3^3+4^3 = 91
\lt f(n_0)$.
So (1) holds.
If $n \ge 3$
and $f(n) > g(n)$ then
$f(n+1)-f(n)
=5^{n+1}-5^n
=4\cdot 5^n
\gt 4(3^n+4^n),\\
g(n+1)-g(n)
=2\cdot 3^n+3\cdot 4^n
\lt 3(3^n+4^n)
\lt f(n+1)-f(n)
$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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At least how many different numbers are there from $a_1,a_2,\cdots,a_{1394}$?
Consider positive integers $a_1,a_2,\ldots,a_{1394}$ so that neither
of two numbers of
$\dfrac{a_1}{a_2},\dfrac{a_2}{a_3},\cdots,\dfrac{a_{1393}}{a_{1394}}$
be equal to each other.
at least how many different numbers are there
from $a_1,a_2,\cdots,a_{1394}$?
$$1)38\quad\quad\quad\quad\quad\quad
2)45\quad\quad\quad\quad\quad\quad3)49\quad\quad\quad\quad\quad\quad4)53\quad\quad\quad\quad\quad\quad5)55$$
To solve this problem in order to have unique $\dfrac{a_m}{a_{m+1}}$ I considered $a_m$ be prime numbers. so I started form $2$ and wrote few terms of the sequence: $2,3,2,5,3,5,2$ from here I should use new number ($7$). $2,3,2,5,3,5,2,7,3,7,5,7,2$ and here should continue with $11$. I don't know if writing numbers help or not.
| You are on the right track.
What you need is to realize that essentially you are looking for the smallest $n$ such that $2{n\choose 2}\geq 1394$.
See, the first sequence $2,3,2$ has length of pairs $2=2{2\choose 2}$.
The second sequence $2,3,2,5,3,5,2$ has length of pairs $6=2{3\choose 2}$
The third sequence $2,3,2,5,3,5,2,7,3,7,5,7,2$ has length of pairs $12=2{4\choose 2}$
The result is pretty obvious because you use every pair of numbers exactly twice, once the bigger one first and once the smaller one first.
Thus $2{38\choose 2}=1406$ is the smallest $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to use telescoping series to find: $\sum_{r=1}^{n}\frac{1}{r+2}$ I am a bit confused in this one, how do i do the required modification in this case?
| As stated in the comments, $\sum_{r = 1}^{n} \frac{1}{r + 2}$ does not telescope. As for an example of a sum that does telescope, consider $\sum_{r = 1}^{n} \frac{1}{r(r + 1)}$. We can rewrite this summation as
$$\sum_{r = 1}^{n} \frac{1}{r(r + 1)} = \sum_{r = 1}^{n} \left(\frac{1}{r} - \frac{1}{r + 1}\right) = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n + 1}\right),$$
and we can pair off terms to see that the entire sum ends up being $1 - \frac{1}{n + 1} = \frac{n}{n + 1}$.
| {
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Proof by induction: Inductive step struggles Using induction to prove that:
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left( \frac{1}{2}\right)^{n} = \frac{2^{n+1}+(-1)^{n}}{3\times2^{n}}
$$
where $ n $ is a nonnegative integer.
Preforming the basis step where $ n $ is equal to 0
$$
1 = \frac{2^{1}+(-1)^{0}}{3\times2^{0}} = \frac{3}{3} = 1
$$
Now the basis step is confirmed.
Then I started the inductive step where $ n = k $ is assumed true and I needed to prove $ n = k+1 $
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left(- \frac{1}{2}\right)^{k} + \left(- \frac{1}{2}\right)^{k+1} = \frac{2^{k+1+1}+(-1)^{k+1}}{3\times2^{k+1}}
$$
Using the inductive hypothesis
$$
\frac{2^{k+1}+(-1)^{k}}{3\times2^{k}} + \left(- \frac{1}{2}\right)^{k+1} = \frac{2^{k+1+1}+(-1)^{k+1}}{3\times 2^{k+1}}
$$
After this I am struggling here trying to get around to the end. I would appreciate any guidance.
| Multplying numerator and denominator of the first term by $2$ we get $$
\frac{2^{k+1}+(-1)^{k}}{3*2^{k}} + (- \frac{1}{2})^{k+1} =\frac{2^{k+2}+2(-1)^{k}}{3*2^{k+1}} + (- \frac{1}{2})^{k+1} =\frac {2^{k+2}+2(-1)^{k}+3(-1)^{k+1}} {3*2^{k+1}}
$$ Now use the fact that $2(-1)^{k}+3(-1)^{k+1}=(-1)^{k+1}$ (which can be checked by considering the case $k$ even and $k$ odd separtely).
| {
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Find $\lim_{n \to \infty}\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$ Find the following limit $$\lim_{n \to \infty}\left(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}\right)$$
I don´t get catch a idea, I notice that
$$\left(\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\cdots + \frac{2n+1}{2^n} \right)$$ is such that
$$ (\frac{3}{2}-\frac{1}{2})=1, \, (\frac{5}{2^2}-\frac{3}{2^2})=\frac{1}{2},\, \, (\frac{7}{2^3}-\frac{5}{2^3})=\frac{1}{4}\cdots (\frac{2n+1}{2^n}-\frac{2n-1}{2^n})=\frac{1}{2^{n-1}}\text{Which converges to 0 }$$
Too I try use terms of the form $\sum_{n=1}^{\infty}\frac{2n}{2^n}$ and relatione with the orignal sum and consider the factorization and try sum this kind of terms$$\frac{1}{2}\lim_{n \to \infty }\left(1+\frac{3}{2}+\frac{5}{2^2}+ \dots +\frac{2n-1}{2^{n-1}}\right)$$.
Update:
I try use partial sum of the form
$$S_1=\frac{1}{2},S_{2}=\frac{5}{2^2},S_{3}=\frac{15}{2^3},S_{4}=\frac{37}{2^4} $$
and try find $\lim_{n \to \infty }S_{n}$ but I don´t get the term of the numerator.
Unfortunelly I don´t get nice results, I hope someone can give me a idea of how I should start.
| 1)First you need to proof that this sequence is a absolute convergent sequence.
Hence the convergence properties wont be affected by the finite basic algebraic manipulation ,such as multiplying by a real number ,or exchanging the position of elements of the S(n).
*let S=lim S(n) (n⇒+∞)
1/2 S=1/2 limS(n) (n⇒+∞)
S（n）-1/2S（n）=1/2+1/2+1/2²+...1/2ⁿ⁻¹-（2n+1）/2ⁿ⁺¹
let n⇒+∞ ，THEN $1/2S=1/2+1-0=3/2$
THUS $S=3$
| {
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Picking out unique solution of $x,y,z$ from two constraints
Let $x,y,z$ be positive real numbers in $\mathbb{R}$ satisfying the conditions:$x+y+z=12$ and $x^3 y^4z^5=(.1)600^3$. Then the value of $x^3 + y^3 +z^3=?$
This is a question from JEE mains, in it I have a doubt of how we are able to uniquely determine an $x,y,z$ triplet from just two equations(upto permutation)! The answer involves observing $x=3, y=4 $ and $z=5$ but how can we make that inference?
| For positive reals, we can use AM-GM inequality. This inequality for $12$ terms applies as
$$\frac{3\cdot (x/3)+4\cdot (y/4)+5\cdot (z/5)}{3+4+5} \ge \left(\frac{x^3}{3^3}\cdot \frac{y^4}{4^4}\cdot \frac{z^5}{5^5} \right)^{1/(3+4+5)}$$
$$\Rightarrow x^3\cdot y^4 \cdot z^5 \le 3^3 \cdot 2^8 \cdot 5^5$$
$$\Rightarrow x^3y^4z^5 \le (0.1)(600)^3$$
where equality (maximum) is achieved when each term is equal i.e.,
$$\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{12}{12}=1$$
from which one infers $x=3, y=4, z=5$.
| {
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Let $a$ be a non zero real number. Evaluate the integral $\int \frac{-7x}{x^{4}-a^{4}}dx$ I hit a wall on this question. Below are my steps
$$\int \frac{-7x}{x^{4}-a^{4}}dx=-7\int \frac{x}{x^{4}-a^{4}}dx$$
Let $u=\frac{x^2}{2}, dx = \frac{du}{x}, x^{4}=4u^{2}.$
$$-7\int \frac{1}{4u^{2}-a^{4}}du=-7\int \frac{1}{(2u+a^2)(2u-a^2)}du$$
Use partial fraction decomposition,
$$\frac{1}{(2u+a^2)(2u-a^2)}=\frac{A}{2u+a^{2}}+\frac{B}{2u-a^{2}}.$$
Solve for $A$ and $B$:
$$\begin{cases}
A=\frac{1}{-2a^{2}}
\\
B=\frac{1}{2a^{2}}
\end{cases}$$
Now $$\int \frac{1}{(2u+a^2)(2u-a^2)}du=\int \frac{1}{-2a^2(2u+a^{2})}+\int \frac{1}{2a^{2}(2u-a^{2})}$$
Factoring out $a$ yields
$$\frac{7}{2a^{2}}(\int \frac{1}{2u+a^{2}}-\int \frac{1}{2u-a^{2}})$$
Evaluate the integral and substitute $u=\frac{x^{2}}{2}$ back.
My final answer is $$\frac{7}{2a^{2}}(\log(x^2+a^2)-\log(x^2-a^2)).$$
Feedback says my answer is wrong. Where did I mess up?
| Mistake appeared in evaluating:
$$\int\frac{1}{2u\pm a^2}\,\mathrm{d}u.$$
To solve this you have to substitute $v=2u\pm a^2$, $\mathrm{d}u=1/2 \mathrm{d}v$. We get
$$\frac{1}{2}\int\frac{1}{v}\,\mathrm{d}v.$$
From now on everything should be fine.
There is also a much easier way to evaluate this integral.
Substitute $u=-ix^2/{a^2}$. The differential is $\mathrm{d}x=ia^2/x^2\,\mathrm{d}u$.
$$\int-\frac{7ia^2}{2(-a^4u^2-a^4)}\,\mathrm{d}u = \frac{7i}{2a^2}\int\frac{1}{u^2+1}\,\mathrm{d}u = \frac{7i}{2a^2}\arctan u+C.$$
Substituting back gives us
$$\frac{7i\arctan\left(-\frac{ix^2}{a^2}\right)}{2a^2}+C=\frac{7i\operatorname{arctanh}\left(\frac{x^2}{a^2}\right)}{2a^2}+C.$$
If you want to get a solution with logarithm function involved, you can do that with this identity
$$\arctan(x)= \frac{1}{2i}\log \left( \frac{x-i}{x+i}\right).$$
| {
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What is $(7^{2005}-1)/6 \pmod {1000}$? What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$
My approach:
Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.
So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000x + 16806)/6 \pmod {1000}$.
Obviously, this must be an integer, so $x=3y$ for some $y$. Then, we are trying to find $16807000\cdot 3y/6+2801 \pmod {1000} = 500y+801 \pmod {1000}$. However, this value can be $301$ or $801$, and I am not sure how to find which one is correct.
Any help is appreciated!
| By geometric sum formula we have
$$\frac{7^{2005}-1}{6} = \frac{7^{2005}-1}{7-1} = 1+7+7^2+\dots+7^{2004}$$
The sequence $1,7,7^2,\dots$ has period $20 $ modulo $1000$ (since $7^{20} \equiv 1 \pmod{1000}$).
$$1+7+7^2+\dots+7^{2004} \equiv 100(1+\dots+7^{19})+1+7+7^2+7^3+7^4 \equiv 100\cdot 0+801\equiv 801 \pmod{1000} $$
| {
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How to find the second time when a piezoelectric crystal vibrates given a cubic trigonometric equation? The problem is as follows:
In an electronics factory a quartz crystal is analyzed to get its
vibration so this frequency can be used to adjust their components.
The length that it expands from a certain charge is given by:
$6\tan\left(\frac{\pi t}{24}\right)\cdot\tan\left(\frac{\pi
t}{24}+\frac{\pi}{3}\right)\cdot\tan\left(\frac{\pi
t}{24}-\frac{\pi}{3}\right)$ micrometers.
Assuming this interval $0\leq t \leq 12$. Where $t$ is measured in
seconds. Find on which second the crystal expands $6$ micrometers?
I'm not sure how to solve this question because I'm ending with a cubic equation and I don't really know how to solve that using precalculus tools.
The thing is that all that equation is:
$6\tan\left(\frac{\pi t}{24}\right)\cdot\tan\left(\frac{\pi t}{24}+\frac{\pi}{3}\right)\cdot\tan\left(\frac{\pi t}{24}-\frac{\pi}{3}\right)=6$
Hence:
$\tan\left(\frac{\pi t}{24}\right)\cdot\tan\left(\frac{\pi t}{24}+\frac{\pi}{3}\right)\cdot\tan\left(\frac{\pi t}{24}-\frac{\pi}{3}\right)=1$
$\tan\left(\frac{\pi t}{24}\right)\cdot\left[\frac{\tan\frac{\pi t}{24}+\sqrt{3}}{1-\sqrt{3}\tan\frac{\pi t}{24}}\right]\cdot\left[\frac{\tan\frac{\pi t}{24}-\sqrt{3}}{1+\sqrt{3}\tan\frac{\pi t}{24}}\right]=1$
Then:
$\tan\left(\frac{\pi t}{24}\right)\cdot\left[\frac{\tan \frac{\pi t}{24}-3}{1-3\tan^2\frac{\pi t}{24}}\right]=1$
$\tan\left(\frac{\pi t}{24}\right)\cdot\left[\tan \frac{\pi t}{24}-3\right]=1-3\tan^2\frac{\pi t}{24}$
$\tan^3 \left(\frac{\pi t}{24}\right)-3 \tan \left(\frac{\pi t}{24}\right)=1-3\tan^2\left(\frac{\pi t}{24}\right)$
$\tan^3 \left(\frac{\pi t}{24}\right) +3\tan^2\left(\frac{\pi t}{24}\right) -3 \tan \left(\frac{\pi t}{24}\right) -1 =0$
Then I landed here, now what?
Because I'm not familiar with solving this sort of equation it would help me alot someone could guide on me what should be done next.
The key from what I could notice is that if I could get that time, I would find when it gets its second value, hence the second time. But as I mentioned. I don't know how to do that.
| You made the same mistake a few times: you neglected the $t$ in $\tan\frac{\pi t}{24}$ in some of your equations, which may be the cause of your confusion. We have the equation
$$x^3+3x^2-3x-1=0$$ where $x=\tan\frac{\pi t}{24}$. Simply by inspection, after trying a few different values for $x$ we see that $x=1$ is a solution; hence by the factor theorem
$$\begin{align}
x^3+3x^2-3x-1&\equiv(x-1)(Ax^2+Bx+C)\\
&\equiv Ax^3+(B-A)x^2+(C-B)x-C
\end{align}$$
We are now in a position to compare coefficients. We can first see that $A=1$. Also, $B-A=3$, but since $A=1$ this means that $B=4$. Similarly, we can see that the constant terms on both sides must be equal so finally we have $C=1$, and thus
$$
x^3+3x^2-3x-1\equiv(x-1)(x^2+4x+1)$$
This is now something that we can solve, as we have two factors, one of which is simply linear and hence very easy to deal with, and the other quadratic which can be dealt with by employing the quadratic formula.
I think you must have made an error: the quadratic factor is factorisable over the reals, as shown below.
$$x^2+4x+1=(x-(-2+\sqrt 3))(x-(-2-\sqrt 3))$$
I got these values by using the quadratic formula- I didn't work them out just by looking at the equation :) These are actually very nice factors in our case, as... drumroll...
$$\tan\left(\frac{7\pi}{12}\right)=-2-\sqrt3~~~~\tan\left(\frac{11\pi}{12}\right)=-2+\sqrt3$$
I hope that was helpful. If you have any questions please don't hesitate to ask :)
| {
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Alternative proof of the Fibonacci property $\sum\limits_{j=n}^{n+9}F_j=11\cdot F_{n+6}.$ If we consider the standard Fibonacci sequence$$F_0=0;\quad F_1=1;\quad F_n=F_{n-1}+F_{n-2}\quad\forall n\geq 2,$$ it is easy to proof that, given any chunk of $10$ consecutive Fibonacci, its sum is equal to $11$ times the seventh (this statement is also valid for any Fibonacci sequence startinng from any two values). That is $$\sum\limits_{j=n}^{n+9}F_j=11\cdot F_{n+6}$$ For example, lets consider the first $10$-numbers chunk of the sequence. Then $$0+1+1+2+3+5+\color{red}{8}+13+21+34=88=11\cdot\color{red}{8}.$$
I proved this by applying the recursive property several times in the following way:
\begin{gather*}\underbrace{F_n+F_{n+1}}_{F_{n+2}}+\underbrace{F_{n+2}+F_{n+3}}_{F_{n+4}}+\underbrace{F_{n+4}+F_{n+5}}_{F_{n+6}}+F_{n+6}+\underbrace{F_{n+7}}_{\underbrace{F_{n+5}}_{F_{n+3}+F_{n+4}}+F_{n+6}}+\underbrace{F_{n+8}}_{F_{n+6}+\underbrace{F_{n+7}}_{F_{n+5}+F_{n+6}}}+\underbrace{F_{n+9}}_{\underbrace{F_{n+7}}_{F_{n+5}+F_{n+6}}+\underbrace{F_{n+8}}_{F_{n+6}+F_{n+7}}}=&\\= F_{n+2}+F_{n+4}+F_{n+6}+F_{n+6}+F_{n+3}+F_{n+4}+F_{n+6}+F_{n+6}+F_{n+5}+F_{n+6}+F_{n+5}+F_{n+6}+F_{n+6}+F_{n+7}=&\\=\underbrace{F_{n+2}+F_{n+3}}_{F_{n+4}}+\underbrace{2F_{n+4}+2F_{n+5}}_{2F_{n+6}}+7F_{n+6}+\underbrace{F_{n+7}}_{F_{n+5}+F_{n+6}}=&\\=\underbrace{F_{n+4}+F_{n+5}}_{F_{n+6}}+10F_{n+6}=&\\=11F_{n+6}&\end{gather*}
This proof looks a bit naive to me compared to the strength of the statement. That is why I wanted to know if someone knows an alternative way to prove this statement. Thank you.
| After OP changes 10 to 9.
Use $F_j=\frac{a^j-b^j}{\sqrt{5}}, a+b=1, ab=-1, a^2=a+1, b^2=b+1$
$$S_n=\sum_{j=n}^{n+9} F_j= \sum \frac{a^j-b^{j}}{\sqrt{5}}$$
USE sum of finite GP, changing the index as $j-n=k$
$$S_n=\sum_{k=0}^{9} \frac{a^{n+k}-b^{n+k}}{\sqrt{5}}=\frac{1}{\sqrt{5}} \left(a^n \frac{a^{10}-1}{a-1}-b^n\frac{b^{10}-1}{b-1}\right)$$
See $a-1=-b, -1/b=a$, then
$$S_n=\frac{1}{\sqrt{5}} [(a^{n+11}-a^{n+1})-(b^{n+11}-b^n)]=F_{n+11}-F_{n+1}.$$
I hope to come back.
Edit:
Now I prove that $F_{n+11}-F_{n+1}=11F_{n+6}$
Note that $$\left(\frac{1\pm\sqrt{5}}{2}\right)^5=\frac{11\pm5\sqrt{5}}{2}$$
$$F_{n+11}-F_{n+1}-11F_{n+6}=a(a^{10}-11a^5-1)-b(b^{10}-11b^5-1)~~~~(*)$$
As roots of $x^{10}-11x^5-1=0$ are $\frac{11\pm5\sqrt{5}}{2}=a^5,b^5;$
the RHS of (*) is zero.
| {
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$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$ I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$$
applying Taylor's series.
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}=\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \frac{\sin 2x}{\cos 2x}}= \lim_{x\to 0^{+}} \frac{\ln (2 \cdot( sin x)^2)}{\ln \sin 2x - \ln \cos 2x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln \sin 2x - \ln \cos 2x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + \ln \sin x + \ln \cos x - 2\ln \cos x + 2 \ln \sin x}= \lim_{x\to 0^{+}} \frac{\ln 2+ 2 \ln sin x}{\ln 2 + 3\ln \sin x }= \lim_{x\to 0^{+}} \frac{\ln 2( sin x)^2}{\ln 2( sin x)^3}$$
$$\frac{\ln 2( sin x)^2}{\ln 2( sin x)^3} \sim \frac{\ln (2 x^2- \frac{2}{3} x^4+ o(x^4))}{\ln (2 x^3- x^5+ o(x^5))}= \frac{\ln (x^2)+ \ln(2 - \frac{2}{3} x^2+ o(x^2))}{\ln(x^3)+\ln (2 - x^2+ o(x^2))} \sim \frac{2\ln x+ \ln 2}{3\ln x+\ln 2} \sim \frac{2}{3}$$
The suggested solution in my book is $2$. can someone indicate where I made mistakes?
| Now @AdamRubinson has identified the error, note the limit is $L+2$ with$$L:=\lim_{x\to0^+}\frac{\ln\frac{1-\cos2x}{\tan^22x}}{\ln\tan2x}=\lim_{x\to0^+}\frac{\ln\tfrac12}{\ln2x}=\tfrac{-\ln2}{-\infty}=0.$$
| {
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Integrate $\frac{\log(x^2+4)}{(x^2+1)^2}$.
Using residue calculus show that
$$\int_0^{\infty}\frac{\log(x^2+4)}{(x^2+1)^2}dx=\frac{\pi}2\log 3-\frac{\pi}6.$$
I was thinking of using some keyhole or semi-circular contour here. But the problem is apart from poles at $x=-i$ and $x=i$, the logarithm has singularities when $x=\pm 4i$.
I consider $C_R$, a semicircle contour oriented clockwise with radius $R$ centered at origin. The semicircle resides in the lower half plane, so that it encloses $x=-i$ as a pole. I set
$$\int_{C_R} \frac{\log(2-ix)}{(x^2+1)^2}dx$$
But it seems like it leads to wrong answer.
| Integrating
$$
\int_\gamma\frac{\log\left(z^2+4\right)}{\left(z^2+1\right)^2}\,\mathrm{d}z\tag1
$$
along the contour
gives $2\pi i$ times the residue at $z=i$.
The integral along the curved pieces vanish as the radius of the large semi-circle grows to $\infty$.
The integral along the downward line along the right side of the imaginary axis to $2i$ (in red) is
$$
-i\int_2^\infty\frac{\log\left(x^2-4\right)+\pi i}{\left(x^2-1\right)^2}\,\mathrm{d}x\tag2
$$
The integral along the upward line along the left side of the imaginary axis from $2i$ (in green) is
$$
i\int_2^\infty\frac{\log\left(x^2-4\right)-\pi i}{\left(x^2-1\right)^2}\,\mathrm{d}x\tag3
$$
For $z=ix$ and $x\gt2$, we have one of $\log\left(z^2+4\right)=\log\left(x^2-4\right)\pm\pi i$. Integrating $\frac1{z-2i}+\frac1{z+2i}$ clockwise around $2i$ decreases $\log\left(z^2+4\right)$ by $2\pi i$. Thus, on the right side of the branch cut, we have $\log\left(x^2-4\right)+\pi i$, and on the left side, we have $\log\left(x^2-4\right)-\pi i$.
The sum of the integrals in $(2)$ and $(3)$ is
$$
2\pi\int_2^\infty\frac1{(x^2-1)^2}\,\mathrm{d}x=\frac\pi6(4-3\log(3))\tag4
$$
The integral along the real axis (in blue) is
$$
2\int_0^\infty\frac{\log(x^2+4)}{(x^2+1)^2}\,\mathrm{d}x\tag5
$$
Furthermore,
$$
2\pi i\operatorname*{Res}_{z=i}\left(\frac{\log(z^2+4)}{(z^2+1)^2}\right)=\frac\pi6(2+3\log(3))\tag6
$$
Subtracting $(4)$ from $(6)$ and dividing by $2$ gives
$$
\int_0^\infty\frac{\log(x^2+4)}{(x^2+1)^2}\,\mathrm{d}x=\frac\pi2\log(3)-\frac\pi6\tag7
$$
| {
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If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square. If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square. $\\$
Well I tried expressing $a=p_1^{a_1}.p_2^{a_2} \cdots p_k^{a_k}$ and $b = q_1^{b_1}.q_2^{b_2}\cdots q_k^{b_k}$ and $c=r_1^{c_1}.r_2^{c_2}\cdots r_k^{c_k}$ basically emphasizing on the fact that the primes which divide $a$ are different from those that divide $b$ and $c$, but I couldn't come up with anything fruitful. $\\$
Any help would be appreciated. $\\$
Thanks
EDIT:- $a,b,c$ are positive integers.
| Well $\frac {ab}{a-b}$ being an integer seems unlikely and specific.
In particular if $p$ is a prime divisor of $c$ then $p|\frac {ab}{a-b}$ so $p|ab$ so $p|a$ or $b|b$. But $p$ can't divide both as $\gcd(a,b,c) =1$.
But we have that if $p|c$ then either $p|a$ or $p|b$ but not both but then $p\not \mid a-b$ yet for any $q|a-b$ we must have $q|a$ or $q|b$ so $q$ must divide both $a$ and $b$ so....
Okay.... Let $\gcd(a,b) = d$ and with $a=a'd; b= b'd$ and as $\gcd(a,b,c) = 1$ we have $\gcd(d,c) =1$. (Also we have $\gcd(a',b') = 1$. so.....
$c = \frac {ab}{a-b} = \frac {a'b'd^2}{d(a'-b')} = \frac {a'b'}{a'-b'}d$.
But two things to note. 1) $d\not \mid c$ so we must have $d|(a'-b')$ and $\gcd(a',b') = 1$ so $a' - b'$ can't have any factors in common with $a'$ or $b'$ and so cant have any factors in common with $a'b'$[1]. So we must have $a'-b'|d$.
So we must have:
$c = a'b'\frac d{a'-b'} = a'b'$ and $a'-b' = d$ and $a-b = d(a'-b') = d^2 = \gcd(a,b)^2$.
Too find a case where this is possible we can let $a',b'$ but any two relatively prime integers. Say $a' = 7$ and $b'=3$ so $a'-b' = 4$. Multiply both by $4$ to get $a = 28$ and $b = 12$ and let $c = \frac {ab}{a-b} = a'b'$ or $c = \frac {28\cdot 12}{28-12}=\frac {28\cdot 12}{16} = 7\cdot 3=21$.
SO $\gcd(28,12, 21) = 1$ and $21 = \frac {28\cdot 12}{28-12}$.
Cute.
=====
[1] A cute, but distractingly tangential lemma: If $\gcd(m,n) =1$ then $\gcd(mn, m- n)=1$. Pf: If $p|mn$ then $p|m$ or $p|n$ but not both. So $p\not \mid m-n$.
| {
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Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare $\frac{x}{y}+\frac{y}{x} \geq 2 $ and the equality $(1+\frac{1}{x}) (1+\frac{1}{y})\leq 2$ unfortunelly not is true the last inequality and hence I can´t conclude our first inequality.
Third attempt:comparing $x^2+y^2$ and $(\sqrt{x}+\sqrt{y})^2$ but unfortunelly I don´t get bound the term $2\sqrt{xy}$ with $xy$.
Any hint or advice of how I should think the problem was very useful.
| Apply Cauchy-Schwarz or Buniakovsky inequality: $(xy+x+y)^2 = (x\cdot y + 1\cdot x + y\cdot 1)^2 \le (x^2+1^2+y^2)(y^2+x^2+1^2) = (x^2+y^2+1)^2\implies xy+x+y \le x^2+y^2+1$.
| {
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generalized f-mean/power mean inequalities I'd like to prove the following inequality (which seems to be true by numerics)
$$
(p-1)\frac{a^2+b^2}{2} \leq \Big(\frac{a^p+b^p}{2}\Big)^{2/p}
$$
for all $a,b\in [0,1]$ and $p\in [1,2]$. I'd eventually like a generalization when I am summing over many numbers and not just $2$, but I'd like to see how to prove this simple case first.
| By applying the Holder's inequality where $\frac{1}{m}+\frac{1}{n} = 1$
$$|x_1 y_1 +x_2y_2| \le (x_1^m +x_2^m)^{\frac{1}{m}} (y_1^m +y_2^n)^{\frac{1}{n}}$$
with $(m,n)=\left(\frac{p}{2},\frac{p}{p-2} \right)$, $(x_1,x_2) = (a^2,b^2)$ and $(y_1,y_2) = (1,1)$, we have
\begin{align}
a^2 +b^2 \le \left((a^2)^{\frac{p}{2}}+(b^2)^{\frac{p}{2}} \right)^{\frac{2}{p}} \left(1+1 \right)^{\frac{p-2}{p}} &\iff a^2 +b^2 \le \left(a^p +b^p\right)^{\frac{2}{p}} 2^{\frac{p-2}{p}} \\
&\iff \frac{a^2 +b^2}{2} \le \left(\frac{a^p +b^p}{2}\right)^{\frac{2}{p}} \tag{*}
\end{align}
And because $\frac{a^2 +b^2}{2} \ge (p-1)\frac{a^2 +b^2}{2} $ for $p \in [1,2]$, then from $(*)$ we can conclude that
$$
(p-1)\frac{a^2+b^2}{2} \leq \Big(\frac{a^p+b^p}{2}\Big)^{2/p}
$$
| {
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Expected value of the sum of two dice. Came across this question:
We roll two dice. Let $X$ be the sum of the two numbers appearing on the dice.
*
*Find the expected value of $X$.
*Find the variance of $X$.
I'm not sure how to do either, but this was my thinking for part 1:
$$E(X) = 2((1/6)^2) + \\
3(2(1/6)^2) + \\
4(2(1/6)^2 + (1/6)^2) + \\
5(2(1/6)^2 + 2(1/6)^2) + \\
6(2(1/6)^2 + 2(1/6)^2 + (1/6)^2) + \\
7(2(1/6)^2 + 2(1/6)^2 + 2(1/6)^2) + \\
8(2(1/6)^2 + 2(1/6)^2 + (1/6)^2) + \\
9(2(1/6)^2 + 2(1/6)^2) + \\
10(2(1/6)^2 + (1/6)^2) + \\
11(2(1/6)^2) + \\
12((1/6)^2)$$
The reason I multiplied some by 2 is because it could possibly switch up or permute. So, for example, for 4, the two sums that could give us 4 are (3,1) and (2,2), so I multiplied one of the probabilities by 2 because (3,1) could come as either (3,1) or (1,3) whereas (2,2) can only come in one form.
| $$E(X)=E(X_1+X_2)=E(X_1)+E(X_2)=3.5+3.5=7$$
can you calculate $V(X)$?
$$V(X_1)=E(X_1^2)-E^2(X_1)\approx 2.917$$
thus
$$V(X)=2\times 2.917=5.8\overline{3}$$
| {
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$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$
How many solutions the following equation has in real numbers
?$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$$
$1)1\quad\quad\quad\quad\quad\quad\quad2)2\quad\quad\quad\quad\quad\quad\quad3
)3\quad\quad\quad\quad\quad\quad\quad4)4\quad\quad\quad\quad\quad5)\text{zero}$
Obviously we have $x\ge1390$. starting from $x=1390$ we can see LHS is very close to $1+1+1=3$ and RHS is close to zero. for higher values of $x$ we can see right side of the equation grows much faster than left side. so it seems the equation has not answer in real numbers.
I'm not sure how to proceed mathematically. I can rewrite it as follow:
$$\left(\sqrt{\frac{x-8}{1388}}-\sqrt{\frac{x-1388}{8}}\right)+\left(\sqrt{\frac{x-7}{1389}}-\sqrt{\frac{x-1389}{7}}\right)+\left(\sqrt{\frac{x-6}{1390}}-\sqrt{\frac{x-1390}{6}}\right)=0$$
I putted fractions with similar numbers in the same parentheses so it looks better now, but don't know how to continue.
| Write $x=1396+t$. Then
$$\sqrt{1+\frac t{1388}}\:+\sqrt{1+\frac t{1389}}\:+\sqrt{1+\frac t{1390}}$$
$$=\sqrt{1+\frac t8}\:+\sqrt{1+\frac t7}\:+\sqrt{1+\frac t6}.$$
Clearly $t=0$ is one solution. Now, $t$ cannot be positive, because that would make the terms on the LHS each smaller than the terms on the RHS. Similarly, $t$ cannot be negative, for the opposite reason. Therefore $t=0$ or $x=1396$ is the unique solution.
| {
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Triangle Properties application In a triangle ABC, AD is the altitude from A. Given b>c, $\angle C = {23^o}$,$AD = \frac{{abc}}{{{b^2} - {c^2}}}$. Then $\angle B = \_\_\_{\_^o}$
My approach is as follow
$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R$
$AD = \frac{{8{R^3}sinA\sin B\sin C}}{{4{R^2}{{\sin }^2}B - 4{R^2}{{\sin }^2}C}} = \frac{{2RsinA\sin B\sin C}}{{\sin \left( {B - C} \right)\sin \left( {B + C} \right)}} = \frac{{2R\sin B\sin C}}{{\sin \left( {B - C} \right)}}$
How do I approach from here
| We can use the extended law of sines and the area of a triangle in terms of sides and circumradius formulae.
$$\bigtriangleup=\frac{1}{2}a\cdot AD \implies AD= \frac{2\triangle}{a} = \frac{{abc}}{{{b^2} - {c^2}}}=\frac{4R \bigtriangleup}{{b^2} - {c^2}}$$
$$\implies 2Ra={b^2} - {c^2}$$
$$\implies 2R(2R\cdot \sin(A))=(2R\cdot \sin(B))^2 - {(2R\cdot \sin(C))}^2$$
$$\implies \sin(A)=\sin^2(B)-\sin^2(C)$$
Can you take it from here?
$$B=C+90°$$
There is also another version of this problem here.
| {
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Why is Euler's statement $\exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$ true? Euler, in his paper Variae observationes circa series infinitas [src], makes the following statements in his Theorem 19.
$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \dots $$
Where he defines:
$$A = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots $$
$$B = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots $$
$$C = \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots $$
Question: I can't see how that statement is true.
I would appreciate replies that don't assume university level training in mathematics.
| This is just an idea, why don't you view it like this
$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \dots $$
Where $$A = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots $$
$$B = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots $$
$$C = \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots $$
But this in terms of a function also say $p(n)$ is the function of the $n$th prime
$$ \exp[ \sum_{k=1}^{\infty } ( \frac{1}{k} \cdot \sum_{n=1}^{\infty} ( \frac{1}{p(n)^k} ) )] = \sum_{n = 1}^{\infty} ( \frac{1}{n} ) $$
| {
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Probabilities in a game John plays a game with a die. The game is as follows: He rolls a fair six-sided die. If the roll is a $1$ or $2$, he gets $0$ dollars. If he rolls a $3, 4$ or $5$, he gets $5$ dollars. If he rolls a $6$, he wins $X$ dollars where $X$ is a continuous random variable that is uniform on the interval $(10, 30)$. Let $Y$ be the amount of money John wins by playing the game.
(i) Compute the cumulative distribution function (cdf) of $Y$.
(ii) What is the probability that John rolled a $6$ given that he won less than $15$ dollars?
(iii) Compute $E(Y)$.
My attempt:
(i) If $Y > 5$ then $P(Y \leq y) = P(X \leq y)$. Then $P(Y \leq 5) = P(Y = 0) + P(Y = 5) = P(roll = 1 or 2) + P(roll = 3, 4 or 5) = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}$.
Not sure if this is even correct.
(ii) $P(roll = 6 | Y < 15) = \frac{P(roll = 6 \land Y < 15)}{P(Y<15)}$
But by Bayes theorem $P(roll = 6 | Y < 15) = \frac{P(Y < 15 | roll = 6) \cdot P(roll = 6)}{P(Y<15)}$
We know $P(roll = 6) = \frac{1}{6}$.
Given that he rolled a $6$, the probability that he won less than $15$ dollars is equal to $P(X < 15) = \frac{15 - 10}{30 - 10} = \frac{1}{4}$
Lastly, $P(Y < 15) = 1 - P(Y \geq 15)$. However, since the only situation in which John wins $15$ dollars or more is if he rolls a $6$. We may infer that $P(Y \geq 15) = P(X \geq 15) = 1 - P(X < 15) = \frac{3}{4}$. So $P(Y < 15) = 1 - \frac{3}{4} = \frac{1}{4}$.
So $P(roll = 6 | Y < 15) = \frac{\frac{1}{4} \cdot \frac{1}{6}}{\frac{1}{4}} = \frac{1}{6}$
Not sure how to determine the other probabilities.
(iii) We can check $Y$ conditioned on $X = x$: $E(Y|X=x) = (0 \cdot \frac{1}{3}) + (5 \cdot \frac{1}{2}) + (x \cdot \frac{1}{6}) = \frac{5}{2} + \frac{x}{6} = \frac{x + 15}{6}$. But $E(E(Y|X)) = E(Y)$ by the law of iterated expectation.
Applying it we have $E(Y) = E(E(Y|X)) = E(\frac{X + 15}{6}) = \frac{1}{6} E(X) + \frac{15}{6}$
Since $X$ is uniform on $(10, 30)$ we know $E(X) = \frac{1}{2} (10 + 30) = 20$
Therefore $E(Y) = \frac{35}{6}$.
Is this correct? I have a feeling that my attempt for (i) is incorrect, and I am unsure about (ii) and (iii). Any assistance is much appreciated.
| In short $Y= 0\mathbf 1_{1\leq X\leq 2}+5\mathbf 1_{3\leq X\leq 5}+Z\mathbf 1_{X=6}$ where $X\sim\mathcal U\{1,2,3,4,5,6\}$ and independently $Z\sim\mathcal U(10..30)$. That means:-
$$Y=\begin{cases}0 &:& 1\leq X\leq 2\\5&:& 3\leq X\leq 5\\Z&:& X=6\\0&:&\textsf{else}\end{cases}$$
(i) The CDF will be a piecewise function. You need to evaluate the probability in each of the pieces.
So, the cumulative probability equals zero for for all $y<0$. There is a massive step in the CDF at $y=0$, which equals the probability that the die rolls $1$ or $2$; ie $\mathsf P(1\leq X\leq 2)$. There is another massive step at $y=5$, equal to the probability that the die rolls $3,4,$ or $5$. At and after $10$ there is no step, but a linear accumulation until $y=30$, when we reach total probability.
$\qquad\begin{align}\mathsf P(Y\leqslant y)&=\begin{cases}0&:&\qquad y<0\\\mathsf P(1\leq X\leq 2)&:&0\leqslant y\lt 5\\\mathsf P(1\leq X\leq 5)&:& 5\leqslant y\lt 10\\\mathsf P(1\leq X\leq 5)+\mathsf P(X=6)\mathsf P(Z\leqslant y)&:& 10\leqslant y< 30\\1&:& 30\leqslant y\end{cases}\end{align}$
(ii) You mostly have this. Under the condition that $X=6$, there $Y=Z$; and $X\perp Z$, so...
$\qquad\mathsf P(X=6\mid Y\leqslant 15)~{=\dfrac{\mathsf P(X=6, Y\leqslant 15)}{\mathsf P(Y\leqslant 15)}\\=\dfrac{\mathsf P(X=6)\mathsf P(Z\leqslant 15)}{\mathsf P(Y\leqslant 15)}}$
(iii) Expectation is Linear, and the expectation of an indicator equals the probability of the event. Also, again, $X$ and $Z$ are independent.
$\qquad\mathsf E(Y)~{=\mathsf E(0\mathbf 1_{1\leq X\leq 2}+5\mathbf 1_{3\leq X\leq 5}+Z\mathbf 1_{X=6})\\[1ex]=0\,\mathsf P(1\leq X\leq 2)+5\,\mathsf P(3\leq X\leq 5)+\mathsf E(Z)\,\mathsf P(X=6)}$
| {
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$63^{63^{63}} \mod 100$ I need to find $63^{63^{63}} \bmod 100$.
This is what I've got so far:
Since $\gcd(63,100)=1$ we can use Euler's theorem. We have $\phi (100)=40$ so $63^{40} \equiv 1 \mod 100$
Again $\gcd(16,100)=1$ and $\phi (40)=16$, that is $63^{16} \equiv 1 \mod 40$
Using this I got that $63^{63} \equiv 7 \mod 40 $ which led me to $63^{63^{63}} \equiv 63^7 \mod 100$
I'm stuck here and don't know what to do next, what could I do now?
| $(63)^7 = (60+3)^7 = (60^7 + {7\choose 1} 60^6\cdot 3 + \cdots + {7\choose 6} 60\cdot 3^6 + 3^7 $
All the terms to the left of the last 2 are equivalent to 0 modulo 100, and can be dropped.
Leaving $7\cdot 60\cdot 3^6 + 3^7\pmod {100}$
We only need to worry about the one figure for the first term since we know it will end in 0.
$3^4 = 81, 3^6\equiv 9 \pmod {10}$
$6\cdot 7 \cdot 9 \cdot 10 + 81\cdot 27\equiv 80 + 87 \equiv 67 \pmod {100}$
| {
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How to show that there is no positive rational number a such that $a^3$ = 2? I'm stuck on this one.
So far, I've tried this:
\begin{align}
a^3 &= 2 \\
a &= \left(\frac mn\right) \\
a^3 &= \left(\frac mn\right)^3 = 2 \\
m^3 &= 2n^3 \\
m &= 2p \\
m^3 &= (2p)^3
\end{align}
I'm really confused about what to do after this—the book answer says that $m^3$ becomes $m^3 = 2(4p)^3$. And $2n^3 = 2(4p)^3$. I'm super confused here and can't really understand what's going on here.
Can someone tell what exactly is going on here and how should I prove this?
| Alternative route:
Let $\alpha^3=2$, and assume $\alpha\in\Bbb Q$, then we have $$\frac{p^3}{q^3}=2\\p^3=2q^3\\p^3=q^3+q^3$$
The last line contradicts Fermat's Last Theorem.
Let $x,y,z$ be non-zero integers. Then if $x^3+y^3=z^3$ we have $xyz=0$.
Proof: We can assume that $x,y,z$ are relatively prime, since if they were not we could divide by $\gcd(x,y,z)^3$.
Reducing the equation modulo $9$, and noticing that the only cubes are $0,-1,1$ we conclude that exactly one of $x,y,z$ is divisible by $9$. Without loss of generality we may assume that $9\mid z$.
We can factor the equation is question as $$z^3=(x+y)(x^2-xy+y^2)$$
Since $9\mid z^3$ we have that $3$ divides one of the factors $x+y, x^2-xy+y^2$. In fact, $3$ divides both of them as $$x^2-xy+y^2=(x+y)^2-3xy$$ That is, $3\mid x+y\iff 3\mid x^2-xy+y^2$. Since $3^3\mid z^3$ one of these factors are divisible by $9$ which will yield a contradiction:
*
*If $9\mid x^2-xy+y^2$ then $9\mid (x+y)^2-3xy$, so $9\mid 3xy$ which is impossible, as $x,y$ are assumed not to be divisible by $3$.
*If $9\mid x+y$, let $k$ be the highest power such that $3^k\mid x+y$, $k\geq 2$. Then we have $$3^{3k}\mid (x+y)^3=x^3+y^3+3xy(x+y)$$ As $3^{3k}\mid x^3+y^3=z^3$, we have $$3^{3k}\mid 3xy(x+y)$$ This is impossible, since then $3^{2k-1}\mid xy$, and $2k-1\geq 3$.
| {
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If $3(x +7) - y(2x+9)$ is the same for all values of $x$, what number must $y$ be? If we let $x = 0$.
\begin{align*}
3(0+7)-y(2(0)+9) \\
21-9y \\
\end{align*}
Then $9y$ should always equal $21$?
Solving for $y$ finds $\frac{7}{3}$.
But $3(x+7)-\frac{7}{3}(2(x)+9)$ does not have the same result for diffrent values of $x$.
Where am I going wrong?
| If we let $x=0$ then we do indeed find that the expression equals $21-9y$. But this does not mean that $21-9y$ must equal $0$; it only means that $21-9y$ is a constant. If, as Raffaele has suggested, we let $x=1$, then we find that the expression equals $24-11y$. Hence, $21-9y$ and $24-11y$ must be equal to the same constant, and so solving the problem boils down to solving the equation $21-9y=24-11y$:
\begin{align}
21 - 9y &= 24 - 11y &&\text{Subtract $21$ from both sides}\\
-9y &= 3 - 11y &&\text{Add $11y$ to both sides}\\
2y &= 3 &&\text{Divide both sides by $2$}\\
y &= \frac{3}{2} \, .
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Can anyone tell me if my proof is right? Let $0<a<b,$ compute $\lim_{n\to \infty}{\frac{a^{n+1}{+}b^{n+1}}{a^n+b^n}}$
My proof are followings :
Assuming that $r=b-a>0,$ so $b=a+r.$ So the original formula equals to $\frac{a\cdot a^{n}{+}{(a+r)\cdot}b^{n}}{a^n+b^n}=a+\frac{r\cdot b^n}{a^n+b^n}=a+\frac{r}{1+(a/b)^n},$ so it can limit to $a+r=b.$
From my point of view, I think that the results should contain $a,$ however my result contains no $a.$ I am a self learner, I don't know wheather my result is right. Hope that someone can help me out.
| Your proof is correct. Here's another way to see why the limit is $b$:
Notice that
\begin{align*}
\frac{a^{n+1}+b^{n+1}}{a^n+b^n} &= a^n\cdot\frac{a}{a^n+b^n}+b^n\cdot\frac{b}{a^n+b^n}\\
&= \frac{1}{\frac{1}{a^n}}\cdot\frac{a}{a^n+b^n}+\frac{1}{\frac{1}{b^n}}\cdot\frac{b}{a^n+b^n}\\
&= \frac{a}{1+\left(\frac{b}{a}\right)^n}+ \frac{b}{1+\left(\frac{a}{b}\right)^n}
\end{align*}
Since $b>a$, it's easy to see that $\left(\frac{b}{a}\right)^n\to\infty$ and $\left(\frac{a}{b}\right)^n\to 0$ as $n\to\infty$, so
$$\lim\limits_{n\to\infty}\left(\frac{a}{1+\left(\frac{b}{a}\right)^n}+ \frac{b}{1+\left(\frac{a}{b}\right)^n}\right)=0+\frac{b}{1}=b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Question about basis change/transformation Let
$B=\left[ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix} 0 \\ 0\\ 1 \end{pmatrix}\right]$
$B_1=\left[ \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 0 \\ -2\end{pmatrix}, \begin{pmatrix} 2 \\ 5\\ 6 \end{pmatrix}\right]$
and the following map
$f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$, $f\left[\begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix}\right]=\begin{pmatrix} \alpha+\gamma \\ 2\alpha+2\beta+\gamma \\ 4\alpha+2\beta+\gamma \end{pmatrix}$
I need to determine $\left[f\right]_{B,B}$ and $T=[id]_{B,B_1}$
So I have an idea how to start with $\left[f\right]_{B,B}$
I need to take the first column vector and
$f\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}$
and then I need to disassemble the resulting vector $\begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}$ into = $1 \cdot\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+2\cdot \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}+4 \cdot \begin{pmatrix} 0 \\ 0\\ 1 \end{pmatrix}$.
and the same also with the second and third column vector. Is this correct?
In this case the breakdown into the basis vectors was very easy, but what if it's not the standard basis? How can I see how to disassemble in such cases? Can anyone recommend me any good .pdf with examples provided where I can read about this topic?
my second question how to determine $T=[id]_{B,B_1}$?
Do I need to invert something here?
| Yes, you are correct. In general, given a linear transformation from $V$ to $W$ vector spaces $f:V \to W$, $\alpha=\{v_1, \cdots, v_n\}$ basis of $V$ and $\beta=\{w_1, \cdots, w_n\}$ basis of $W$ to construct $[f]_{\alpha,\beta}$ you proceed like this: The $i$-th column of the matrix $[f]_{\alpha,\beta}$ is given by the coordinates of $f(v_i)$ with respect to the basis $\beta$. That is the $i$-th column of $[f]_{\alpha,\beta}$
$$
\begin{bmatrix}
c_1 \\
c_2\\
\cdot\\
\cdot\\
\cdot\\
c_m
\end{bmatrix}
$$
is such that $f(v_i)=c_1 w_1+ c_2 w_2 + \cdots + c_m w_m$. As you noticed this is not always so simple as it was in this case with the canonical basis of $\mathbb{R}^3$. But in your case with $V=W=\mathbb{R}^3$ and $\alpha=\beta=\{u_,u_2,u_3\}$ we must find $[f]_{\alpha,\alpha}$. Then we need to decompose $f(u_i)$ with respect to the basis $\{u_1,u_2,u_3\}$. Hence we must find $a_{1,i}$, $a_{2,i}$ and $a_{3,i}$ such that $f(u_i)=a_{1,i} u_1 + a_{2,i} u_2 + a_{3,i} u_3$. Now denoting the vectors as
$$
f(u_i)=\begin{pmatrix}
y_{1,i}\\
y_{2,i}\\
y_{3,i}
\end{pmatrix}
, u_i=\begin{pmatrix}
x_{1,i}\\
x_{2,i}\\
x_{3,i}
\end{pmatrix}
$$
we have that
$$
\begin{pmatrix}
y_{1,i}\\
y_{2,i}\\
y_{3,i}
\end{pmatrix}=\begin{pmatrix}
a_{1,i}x_{1,1}\\
a_{1,i}x_{2,1}\\
a_{1,i}x_{3,1}
\end{pmatrix}+\begin{pmatrix}
a_{2,i}x_{1,2}\\
a_{2,i}x_{2,2}\\
a_{2,i}x_{3,2}
\end{pmatrix}+\begin{pmatrix}
a_{3,i}x_{1,3}\\
a_{3,i}x_{2,3}\\
a_{3,i}x_{3,3}
\end{pmatrix}
$$
$$
\begin{pmatrix}
y_{1,i}\\
y_{2,i}\\
y_{3,i}
\end{pmatrix}=\begin{pmatrix}
a_{1,i}x_{1,1}+a_{2,i}x_{1,2}+a_{3,i}x_{1,3}\\
a_{1,i}x_{2,1}+a_{2,i}x_{2,2}+a_{3,i}x_{2,3}\\
a_{1,i}x_{3,1}+a_{2,i}x_{3,2}+a_{3,i}x_{3,3}
\end{pmatrix}
$$
$$
\begin{pmatrix}
y_{1,i}\\
y_{2,i}\\
y_{3,i}
\end{pmatrix}=\begin{pmatrix}
x_{1,1} && x_{1,2} && x_{1,3}\\
x_{2,1} && x_{2,2} && x_{2,3}\\
x_{3,1} && x_{3,2} && x_{3,3}
\end{pmatrix} \begin{pmatrix}
a_{1,i}\\
a_{2,i}\\
a_{3,i}
\end{pmatrix}.
$$
And in order to find $a_{1,i}$, $a_{2,i}$ and $a_{3,i}$ you solve the $i$-th linear system. Now $[id]_{B,B_1}$ is the representation of the identity transformation $id:\mathbb{R}^3 \to \mathbb{R}^3$ relative with the basis $B$ and $B_1$. The first column of the matrix is the coordinates of the identity applied to $(1, 0, 0)$ with respect to the basis $B_1$. That is we must find $a, b$ and $c$ such that
$$
\begin{pmatrix}
1\\
0\\
0
\end{pmatrix}=id \begin{pmatrix}
1\\
0\\
0
\end{pmatrix}=a\begin{pmatrix}
1\\
-2\\
0
\end{pmatrix}+b\begin{pmatrix}
1\\
0\\
-2
\end{pmatrix}+c\begin{pmatrix}
2\\
5\\
6
\end{pmatrix}
$$
$$
\begin{pmatrix}
1\\
0\\
0
\end{pmatrix}=\begin{pmatrix}
1 && 1 && 2\\
-2 && 0 && 5\\
0 && -2 && 6
\end{pmatrix} \begin{pmatrix}
a\\
b\\
c
\end{pmatrix}.
$$
Then you can apply the same procedure to $(0,1,0)$ and $(0,0,1)$ in order to find the second and third columns of $[id]_{B,B_1}$. Another way to calculate $[id]_{B,B_1}$ is to find $[id]_{B_1,B}$ and since $B$ is the canonical basis
$$
[id]_{B_1,B}=\begin{pmatrix}
1 && 1 && 2\\
-2 && 0 && 5\\
0 && -2 && 6
\end{pmatrix}.
$$
And then you can use the fact that $[id]_{B,B_1}=([id]_{B_1,B})^{-1}$ and calculate the inverse matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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