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Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers? $\textbf{Question:}$Do there exist any three relatively prime natural numbers so that the square of each of them is divisible by the sum of the two remaining numbers?
that is pairwise relatively prime $a,b,c \in \mathbb N $($0$ not included) such that $a+b \mid c^2,b+c \mid a^2,c+a \mid b^2$
$\textbf{Stuffs I have tried:}$
I have tried a bunch of constructions like :$(p,\frac{p+1}{2},\frac{p-1}{2})$,one of them are the sum of other two,tried small values.But was unable to find any valid triple.So,I began trying to prove such triple can't exist.I tried to bound the numbers,tried to use modular arithmatic but failed again.
Any kind of hint or solution will both be appreciated.
| Well, let's see. So we have
$$\begin{cases}
a^2 = x(b+c) \\
b^2 = y(a+c) \\
c^2 = z(a+b)
\end{cases}$$
where all quantities are integers. What does this mean?
For one thing, each of $a,b,c$ is strictly greater than 1.
For another, if $a,b,c$ are all pairwise coprime (as per the problem statement), then so are $b+c, a+c, a+b$.
Now let's make one peculiar combination out of these numbers, for no reason at all:
$$(a+b+c)^2 = a^2+2ab+2ac+(b+c)^2=x(b+c)+2a(b+c)+(b+c)^2$$
So it is divisible by $b+c$, and also (following similar reasoning) by $a+b$ and $a+c$. If these three are all mutually coprime, then it is divisible by their product.
Assume that $a<b<c$. Then $b+c>{2\over3}(a+b+c)$ and $a+c>{1\over3}(a+b+c)$. So...
$$(a+b)(b+c)(a+c)\;|\;(a+b+c)^2\\
(a+b)(b+c)(a+c)\leqslant(a+b+c)^2\\
(a+b)\cdot{1\over3}(a+b+c)\cdot{2\over3}(a+b+c)<(a+b+c)^2\\
a+b < {9\over2}\\
a+b \leqslant 4$$
With $a,b>1$ this only leaves us the option of $a=b=2$, which are not coprime.
Finally, there are no such coprime triples. Q.e.d.
| {
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How does $e^x\cdot e^X$ equal $e^{x+X}$? I know that they equal each other, but when I'm trying to prove it, something doesn't match. Please mind the difference between the two equations, one is a lowercase $x$ and the other is an uppercase $x.$ I know that the formula to get $e^x$ is $\frac{x^n}{n!}$. So I apply on $e^x$and it becomes $1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdots$ and the same for $e^X$. This image shows what I'm thinking, and what happens when I multiply them on each other. But I want to go more deeper and show the full equation of the numbers shown, for example, showing $\frac{1}{6}x^3$ and others as well. If I did, there would be a total of 16 numbers shown that are arranged in order of degrees. So I try to prove it like this picture. But the problem shows when I try to prove it more. I tried proving it more, but $\frac{1}{6}(x+X)^3$ won't work. Than what should I do to make it work? Or what formula should I use?
| Perhaps you're confused about the arrangement. We should get $e^xe^x=e^{2x}$, which can be written $$e^{2x} = 1 +(2x) + \tfrac{(2x)^2}{2!} + \tfrac{(2x)^3}{3!}+\cdots$$
$$=1 + 2x + \tfrac{4x^2}{2!} + \tfrac{8x^3}{3!} + \cdots$$
So let's multiply series:
$$e^x\cdot e^x = \left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right)\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right)$$
$$= 1\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right) + x\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right) + \tfrac{x^2}{2!}{\cdot\left(1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \cdots\right)}+\cdots$$
$$= (1) + (1\cdot x + x\cdot 1) + (1\cdot \tfrac{x^2}{2!} + x\cdot x + \tfrac{x^2}{2!}\cdot 1) + \cdots $$
$$= 1 + (2)x + (\tfrac1{2!} + 1 + \tfrac1{2!})x^2 + \cdots$$
$$=1 + 2x + \tfrac4{2!}x^2 + \cdots$$
You collect the products of a fixed degree $n$ as
$$1\cdot \tfrac{x^n}{n!} + x\cdot\tfrac{x^{n-1}}{(n-1)!} + \tfrac{x^2}{2!}\cdot\tfrac{x^{n-2}}{(n-2)!} + \cdots + \tfrac{x^{n-2}}{(n-2)!}\cdot \tfrac{x^2}{2!} +\tfrac{x^{n-1}}{(n-1)!}\cdot x + \tfrac{x^{n}}{n!}\cdot 1$$
$$= (\tfrac1{n!} + \tfrac{n}{n!} + \tfrac{n(n-1)}{n!} + \tfrac{n(n-1)(n-2)}{n!} + \cdots)x^n$$
Your job is to show that this is $\tfrac{2^n}{n!}x^n$ (expand $(1+1)^n)$ using the binomial formula).
Retry:
I think you are doing it right, you just didn't collect all terms correctly.
You are really just using the distributive law, like
$$(a + b + c+\cdots)(\textrm{terms}) = a\cdot(\textrm{terms}) + b\cdot(\textrm{terms}) + c\cdot(\textrm{terms})+\cdots$$
For each of the products on the RHS, you need to look for results of the same degree (we're really looking at the exponential series here, of course).
*
*Constant terms only occur as $1\cdot 1$, in the first product on the RHS.
*Degree 1 terms occur as $1\cdot x$ or $x\cdot 1$ (in the first and second products on the RHS).
*Degree 2 terms occur as $1\cdot x^2$, $x\cdot x$, or $x^2\cdot 1$ (in the first, second, and third products on the RHS).
*Degree 3 terms occur as $1\cdot x^3$, $x\cdot x^2$, $x^2\cdot x$, or $x^3\cdot 1$ (in the 1st, 2nd, 3rd, and 4th products on the RHS).
And so on.
| {
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Using partial information to factor $x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ I wish to find exact expressions for all roots of $p(x)=x^6+3x^5+5x^4+10x^3+13x^2+4x+1.$ By observing that for the roots $x_0 \pm iy_0, x_0 \approx -0.15883609808599033632, y_0 \approx 0.27511219196092896700,$ we have that $x_0$ is the unique real root of $r(x) = x^3+12x^2+8x+1,$ I was able to prove that all roots of the original sextic can be expressed in radicals. The process is as follows:
*
*Divide $p(x+iy)$ by $r(x)$ to get $\frac{1}{8}x^3 + \frac{3}{16}x^2 + x\left(\frac{7}{32}-\frac{15y^2}{8}\right) + \left(\frac{95}{32}-\frac{15y^2}{16}\right) + \frac{R(x,y)}{p(x)}$ where $R(x,y) = A(y)x^2 + B(y)x + C(y)$ and $A(y) = 15y^4 - \frac{15y^2}{4} - \frac{201}{16}, B(y) = 15y^8 - 30y^6 + 12y^4 + \frac{75y^2}{8} - \frac{767}{32}, C(y) = -y^6+5y^4-\frac{193y^2}{16}-\frac{63}{32}.$
*The equation $R(x_0, y_0) = 0$ is a quartic in $y_0^2,$ which we can solve exactly to obtain $y_0^2$ and hence $y_0.$
*Polynomial division reduces $p(x)$ to a quartic, and now we apply the quartic formula again to find the other $4$ roots.
However, I don't want to perform the rest of the computations. Is there a cleaner way to use the observation that $r(x_0) = 0,$ perhaps in the realm of abstract algebra?
| The hint.
Use the following:
$$x^6+3x^5+5x^4+10x^3+13x^2+4x+1=\left(x^3+\frac{3}{2}x^2-2x-1\right)^2+\frac{3}{4}x^2(3x+4)^2.$$
Now, you can get all roots of the polynomial. Just solve two cubic equations.
| {
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Find $w^2$ and compute a quadratic equation satisfied by $w$ I'm given $$\zeta_7 = e^{\frac{2\pi i}{7}} = \cos\frac{2\pi}{7}+i\sin\frac{2\pi}{7}$$
and I know $$\sum_{k=0}^{6}\zeta_7^k=0 $$
I'm then given $$w=\zeta_7+\zeta_7^2+\zeta_7^4$$
and that $\zeta_7^7 = 1$.
I'm asked to find $w^2$ and I found that that's $\zeta^2_7+2\zeta_7^3+\zeta_7^4+2\zeta_7^5+2\zeta_7^6+\zeta_7^8$ so $w^2=\zeta_7^3+\zeta_7^5+\zeta_7^6-1$ and if we add w, we get $\zeta_7^0+\zeta_7+\zeta_7^2+\zeta_7^3+\zeta_7^4+\zeta_7^5+\zeta_7^6-2\zeta_7^0-w$ and that, I think, is just $-2-w$
So, if $w^2=-2-w$ then $w^2+w+2=0$ and w solves it, so that's the quadratic, right?
| Here's a slightly simpler approach:
$$w^2=(\zeta_7+\zeta_7^2+\zeta_7^4)^2=\zeta_7^2+\zeta_7^4+\zeta_7^8+2\zeta_7^3+2\zeta_7^5+2\zeta_7^6,$$
so
$$w^2+w=2\zeta_7^2+2\zeta_7^4+2\zeta_7^1+2\zeta_7^3+2\zeta_7^5+2\zeta_7^6=2(0-\zeta_7^0)=2(-1),$$
hence
$w^2+w+2=0$.
| {
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Proof that $\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0$ How should I prove
$$\forall n\in\mathbb{N}:\, \frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}=0?$$
My attempt:
$$\begin{align}\frac{e^{\frac{\pi i}{4}}}{\sqrt{2}}\left(1+e^{-(2n+1)\frac{\pi i}{2}}\right)-e^{\frac{n^2\pi i}{2}}&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-n\pi i-\frac{\pi i}{2}+\frac{\pi i}{4}}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1}{\sqrt{2}}\left(e^{\frac{\pi i}{4}}+e^{-i\left(n\pi +\frac{\pi}{4}\right)}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1}{\sqrt{2}}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}+\frac{1}{\cos\left(n\pi +\frac{\pi}{4}\right)+i\sin\left(n\pi +\frac{\pi}{4}\right)}\right)-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{\sqrt{2}}\frac{1}{\frac{1}{\sqrt{2}}(\cos n\pi -\sin n\pi )+\frac{i}{\sqrt{2}}(\cos n\pi +\sin n\pi)}-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{\cos n\pi +i\cos n\pi}-e^{\frac{n^2\pi i}{2}}\\&=\frac{1+i}{2}+\frac{1}{i^{2n}+i^{2n+1}}-i^{n^2}\end{align}$$
How should I proceed?
| I tried what @Joe suggested on the problem statement itself. I'm getting the right answer for even $n$ but the answer for odd $n$ is equal upto a sign.
| {
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Solve $x^2+3y = u^2$ and $y^2+3x=v^2$ in positive integers. The question is from the pg - 59 from ' An Introduction to Diophantine Equations ' by Titu Andreescu , Dorin Andrica , Ion Cucurezeanu.
Example 1 : Solve in positive system of equations in positive integers
$$\begin{cases} x^2+3y = u^2 \\ y^2 + 3x = v^2 \end{cases}$$ $\;\;\;\;\;\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\text{(Titu Andreescu)}$
Solution. The inequality $x^2 + 3y ≥ (x + 2)^2 , y^2 + 3x ≥ (y + 2)^2$
cannot both be true, because adding them would yield a contradiction. So at least one of the inequalities $x^2 + 3y < (x + 2)^2$ and
$y^2 + 3x < (y + 2)^2$ is true. Without loss of generality, assume
that $x^2 + 3y < (x + 2)^2$. Then $$x^2 < x^2 + 3y < (x + 2)^2 \implies
x^2 + 3y = (x+1)^2$$ or, $3y = 2x+ 1$ . We obtain $x = 3k + 1, y = 2k + 1$
for some nonnegative integer $k$ and $y^2 + 3x = 4k^2 + 13k + 4$. For
$k > 5, (2k+ 3)^2 < 4k^2 + 13k+ 4 < (2k+ 4)^2$ ; hence $y^2 + 3x$ cannot be
a perfect square. Thus we need only consider $k ∈ {\{0, 1, 2, 3, 4\}}$ . Only
$k = 0$ makes $y^2 + 3x$ a perfect square; hence the unique solution is
$$x = y = 1,\;\;\;\;\;\; u = v = 2.$$
But if we take , $$4k^2+13k + 4 = v^2$$ $$\implies k = \dfrac{-13 \pm\sqrt{105+16v^2}}{8}$$
Since $105+16v^2 = a^2 \implies 105 = (a-4v)(a+4v)$ which gives $a \in \{\pm11 , \pm13 , \pm19 ,\pm53\}$ .
Out of these , only $a \in \{ \pm13 , \pm53\}$ works which gives $k=0,5$ , And so the the answer should be $$(x,y,u,v) = (1,1,2,2)\;\;\;,(16,11,17,13)\;\;\;\;,(11,16,13,17)$$
Who is correct here?
| You are indeed correct; it is not hard to verify that your solution(s) are indeed solutions. The proof that you quote overlooks the case $k=5$, which yields your two additional solutions.
| {
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On Resolving a Fuss on Squares and Fractions over a few Inequalities Firstly, only AM-GM and C-S are to be sought.
$1.$Let $a, b, c, d$ be positive real numbers such that $a^2+b^2+c^2+d^2=4$. Show that
$${a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge 4$$
In my textbook, this problem is credited to Michael Rozenberg, but I couldn't find a solution to it by myself or on this site, so decided to ask.
I tried my best with fallacy-
$$4(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2 \Rightarrow 4 \ge a+b+c+d$$ with the constraint given. Proceeding-
$$(a+b+c+d)\left({a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a}\right) \ge (a+b+c+d)^2 $$
$$\Rightarrow {a^2\over b}+{b^2\over c}+{c^2\over d}+{d^2\over a} \ge a+b+c+d$$
That means, I need $a+b+c+d\ge 4$ to complete the proof but instead got $4 \ge a+b+c+d$ !
$2.$Let $a, b, c$ be positive real numbers such that $abc = 1$.
Show that
$${1\over b(a+b)}+{1\over c(b+c)}+{1\over a(c+a)}\ge \frac{3}{2}.$$
$3.$If a, b, c and d are positive real numbers such that $a + b + c + d = 4$. Prove that
$$ {a \over 1+b^2c}+{b \over 1+c^2d}+{c \over 1+d^2a}+{d \over 1+a^2b} \ge 2. $$
The #2 is credited to the Zhautykov Olympiad 2008 and there is no need for revealing my attempts as I've no idea what to do, Lastly,
This is a doubt, not an question that definitely has an answer, but if the doubt comes out to have an answer, nevertheless, this is a problem.
$4.$ Prove without Induction:
$$\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2} + ...+\sqrt{a_n^2+b_n^2} \ge \sqrt{(a_1+a_2+...+a_n)^2+(b_1+b_2+...+b_n)^2}$$
| The second problem.
Let $a=\frac{x}{y}$ and $b=\frac{y}{z},$ where $x$, $y$ and $z$ are positives.
Thus, by C-S and Vasc we obtain: $$\sum_{cyx}\frac{1}{b(a+b)}=\sum_{cyc}\frac{1}{\frac{y}{z}\left(\frac{x}{y}+\frac{y}{z}\right)}=\sum_{cyc}\frac{z^2}{y^2+xz}=$$
$$=\sum_{cyc}\frac{z^4}{y^2z^2+xz^3}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(x^2y^2+x^3y)}\geq\frac{(x^2+y^2+z^2)^2}{\frac{1}{3}(x^2+y^2+z^2)^2+\frac{1}{3}(x^2+y^2+z^2)^2}=\frac{3}{2}.$$
The Vasc's inequality it's the following.
$$(x^2+y^2+z^2)^2\geq3(x^3y+y^3z+z^3x).$$
There are very many proofs of this beautiful inequality.
I think, the best of them it's the following:
$$(x^2+y^2+z^2)^2-3(x^3y+y^3z+z^3x)=\frac{1}{2}\sum_{cyc}(x^2-y^2-xy-xz+2yz)^2\geq0.$$
| {
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2010 USAMO #5:Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for integers $m$ and $n$, then $m - n$ is divisible by $p$. Let $q = \frac{3p-5}{2}$ where $p$ is an odd prime, and let $S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7} + \cdots + \frac{1}{q(q+1)(q+2)}
$
Prove that if $\frac{1}{p}-2S_q = \frac{m}{n}$ for coprime integers $m$ and $n$, then $m - n$ is divisible by $p$.
My Progress till now: $$2S_q = 2\sum_{x=1}^{\frac{q+1}{3}} \frac{1}{(3x-1)(3x)(3x+1)} = \sum_{x=1}^{\frac{p-1}{2}} \left[\frac{1}{3x(3x-1)}-\frac{1}{3x(3x+1)}\right]\\
=\sum_{x=1}^{\frac{p-1}{2}} \left[ \frac{1}{3x-1} - \frac{2}{3x} +\frac{1}{3x+1}\right]\\
=\sum_{x=1}^{\frac{p-1}{2}}\left[ \frac{1}{3x-1} + \frac{1}{3x} +\frac{1}{3x+1}\right] - \sum_{x=1}^{\frac{p-1}{2}} \frac{1}{x} $$
With the help of @user10354138 , I have got $\frac{1}{p} - 2S_q = \frac{1}{p} + \frac{1}{1} - \sum_{k=\frac{p+1}{2}}^{\frac{3p-1}{2}}\frac{1}{k} = \frac{m}{n}$
But then I am stuck.
Please give me some hints rather than a solution.
Thanks in advance.
PS: I didn't post it in AOPS, because there we don't get any guidance.
| (Original) Hint: You are almost there with the simplification. Note that you are summing over $\frac1n$ from $n=2$ to $\frac{3p-1}2$ in the first. So
$$
2S_q+1=\sum_{n=(p+1)/2}^{(3p-1)/2}\frac1n
$$
If you tweak the RHS slightly, you would be summing over $\frac1n$ as $n$ runs through representative of each of the nonzero residue classes mod $p$. So ...
Addendum (2020-07-29): As discussed in the comments,
\begin{align*}
\frac1p-2S_q-1&=-\left(\sum_{n=(p+1)/2}^{p-1}\frac1n+\sum_{n=p+1}^{p+(p-1)/2}\frac1n\right)\\
&=-\sum_{i=1}^{(p-1)/2}\left(\frac1{p-i}+\frac1{p+i}\right)
\end{align*}
and now
$$
\frac1{p-i}+\frac1{p+i}=\frac{p}{(p-i)(p+i)}
$$
so the numerators are divisible by $p$ and the denominators are not. So putting everything over a common denominator, we see
$$
\frac{m-n}{n}=-\sum_{i=1}^{(p-1)/2}\frac{p}{(p-i)(p+i)}=\frac{p\times \text{some integer}}{\text{some integer not divisible by }p}.
$$
That is, every representation of $\frac{m-n}{n}$ must have more factors of $p$ in the numerator than in the denominator, hence $m-n$ is divisible by $p$.
| {
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Finding a polynomial $f(x)$ of degree 5 such that $f(x)$ is divisible by $x^3$ and $f(x)+2$ is divisible by $(x+1)^3.$ There is some polynomial $f(x)$ of degree $5$ such that both of these properties hold:
$f(x)$ is divisible by $x^3$.
$f(x)+2$ is divisible by $(x+1)^3.$
Find that polynomial.
I know that because $f(x)$ is divisible by $x^3$ our polynomial is in the form of $ax^5+bx^4+cx^3.$ However, I'm not very sure how our second condition comes into use. Any help?
| Consider the equality
$$(x - (x+1) )^5 = -1$$
and expand getting
$$x^5 - 5 x^4 (x+1) +10 x^3 (x+1)^2- 10 x^2 (x+1)^3 +5 x (x+1)^4 - (x+1)^5 = -1$$
or
$$x^5 - 5 x^4 (x+1) + 10 x^3 (x+1)^2 = -1 + (x+1)^3 Q(x)$$
Therefore the polynomial is
$$2 x^3( x^2 - 5 x(x+1) + 10 (x+1)^2)$$
| {
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Find the eccentricity of the conic $4x^2+y^2+ax+by+c=0$, if it tangent to the $x$ axis at the origin and passes through $(-1,2)$ Solving this would require three equations
(1) Tangent to x axis at origin
Substituting zeroes in all $x$ and $y$ gives $c=0$
(2) Passes through (-1,2)
$$4(1)+4-a+2b=0$$
$$-a+2b=-8$$
How do I find the third equation?
| As you mentioned, one of the points on the ellipse is (0,0). Equation of the ellipse is
$\frac{(x+\frac{a}{8})^2}{1^2}+\frac{(y+\frac{b}{2})^2}{2^2} = {(\frac{a}{8})}^2 + {(\frac{b}{4})}^2$
For ellipse equation $\frac{(x \pm h)^2}{A^2}+\frac{(y \pm k)^2}{B^2} = 1$, eccentricity of ellipse
= $\frac{\sqrt{B^2-A^2}}{B}$ (where $B \ge A$)
= $\frac{\sqrt3}{2}$ (where $A = t, B = 2t$)
Here, $t = \sqrt{{(\frac{a}{8})}^2 + {(\frac{b}{4})}^2}$
Alternatively:
Based on the fact that it is tangent to x-axis at origin, the center of the ellipse will have to be on y-axis (as it is not a slanting ellipse). That means $a = 0, c = 0, b = -4$. So equation of ellipse is $x^2 + \frac{{(y-2)}^2} {4} = 1$ with major axis (along $y$ axis) $= 4$, minor axis $= 2$.
So, eccentricity = $\frac{\sqrt3}{2}$
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Prove that $\binom{n}{0}\binom{n+1}{n} +\binom{n}{1}\binom{n}{n-1} +\binom{n}{2}\binom{n-1}{n-2} +\cdots +\binom{n}{n}\binom{1}{0} = 2^{n-1}(n+2)$
Prove the below:
$$\binom{n}{0}\cdot\binom{n+1}{n} +\binom{n}{1}\cdot\binom{n}{n-1} +\binom{n}{2}\cdot\binom{n-1}{n-2} +\cdots +\binom{n}{n}\cdot\binom{1}{0} = 2^{n-1}\cdot(n+2)$$
Attempt:
Consider $$(1+x)^{n} = \binom{n}{0} + \binom{n} {1}x + \binom{n} {2}x^² + \cdots + \binom{n} {n} x^n$$
The series in the question is the coefficient of $x^n$ in
$$\binom{n} {0}(1+x)^{n+1}+\binom{n} {1}x(1+x)^{n} +\binom{n} {2}x^2(1+x)^{n-1 }+ \cdots + \binom {n} {n} x^n (1+x)$$
Knowing the above, how can I rewrite the coefficient of $x^n$, as that is usually the key to solving such problems?
| We can easily use the fact that you already observed: $$\binom{n} {0}(1+x)^{n}+\binom{n} {1}x(1+x)^{n-1} +\binom{n} {2}x^2(1+x)^{n-2}+ \cdots + \binom {n} {n} x^n=(x+(1+x))^n$$ Hence You just need the coefficient of $x^n$ in the binomial $(x+1)(2x+1)^{n},$ which is $$2^{n-1}\binom{n} {n-1}+2^n\binom{n} {n}=2^{n-1}(n+2).$$
| {
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Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
What I Tried :- Since $x^2 + y^2 = 2x - 2y + 2$ , we have $2x - 2y + 2 + \sqrt{32}$
=> $2(x - y + 1 + 2√2)$ . From this step I am not sure how to move forward .
Also I tried to express $x^2 + y^2 + \sqrt{32} \leq S$ , so that in that way we can say that $x^2 + y^2 + \sqrt{32}$ is maximum at $S$ , but I couldn't do it .
Can anyone help me ? Some hints or suggestions to this problem will be appreciated !!
| By C-S $$x^2+y^2=2(x-y)+2\leq2\sqrt{(1^2+(-1)^2)(x^2+y^2)}+2,$$ which gives
$$x^2+y^2-2\sqrt2\sqrt{x^2+y^2}+2\leq4$$ or
$$\left(\sqrt{x^2+y^2}-\sqrt2\right)^2\leq4,$$ which gives
$$\sqrt{x^2+y^2}\leq2+\sqrt2.$$
Id est, $$x^2+y^2+\sqrt{32}\leq(2+\sqrt2)^2+4\sqrt2=6+8\sqrt2.$$ The equality occurs for $x=1+\sqrt2$ and $y=-1-\sqrt2,$ which says that we got a maximal value.
| {
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sum of terms of series
If $$F(t)=\displaystyle\sum_{n=1}^t\frac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}$$ find $F(60)$.
I tried manipulating the general term(of sequence) in the form $V(n)-V(n-1)$ to calculate the sum by cancellation but went nowhere. I also tried using the fact that $$2n+\sqrt{4n^2-1}=\frac{1}{2}{(\sqrt{2n-1}+\sqrt{2n+1})}^2$$ Could someone please give me a hint?
| With more details, we have that
$$\frac{2n+\sqrt{(4n^2-1)}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)$$
then
$$\frac{4n+\sqrt{(4n^2-1)}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)+\frac{2n}{\sqrt{2n+1}+\sqrt{2n-1}}=\\=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)+n\left({\sqrt{2n+1}-\sqrt{2n-1}}\right)=\\=\frac12(2n+1)\sqrt{2n+1}-\frac12(2n-1)\sqrt{2n-1}=\\$$
$$=\frac12\left(\sqrt{(2n+1)^3}-\sqrt{(2n-1)^3}\right)$$
| {
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How to find all solutions for : $a^3 \equiv b^3 \pmod{7^3}$, knowing that $7 \nmid ab$.
Find all integers $a$ and $b$ such that $$a^3 \equiv b^3 \pmod{7^3}\,,$$ knowing that $7 \nmid ab$.
As a try, I noticed that, since $\gcd(b, 7)=1$, there exists $x \in \mathbb{N}$ such that $b\cdot x \equiv 1 \pmod{7} \Rightarrow b^3 \cdot x^3 \equiv 1 \pmod{7^3}$. Thus $a^3\equiv b^3 \pmod{7^3} \iff (ax)^3\equiv 1\pmod{7^3}$. After this I tried to use Euler's totient function, but I do not know where I should begin.
| Given $7\nmid ab$, $a^3\equiv b^3\bmod 7^3\iff (a/b)^3\equiv1\bmod 7^3$.
Let $x\equiv a/b\bmod 7^3$. We are looking for $x$ such that $7^3|x^3-1=(x-1)(x^2+x+1)$.
Now if $7|x-1$ and $x^2+x+1$, then $7|x^2+x+1-(x+2)(x-1)=3$, a contradiction.
So $7^3|x-1$ (i.e., $x\equiv1\bmod7^3$) or $7^3|x^2+x+1$.
Now we are looking for $x$ such that $x^2+x+1\equiv0\bmod7^3$.
Note that $x\equiv 2$ and $x\equiv 4$ are the solutions to $x^2+x+1\equiv0\bmod7$.
If $x=7k+4$ is a solution to $x^2+x+1\equiv0\bmod7^2$, then $k\equiv2\bmod7$, so $x\equiv18\bmod7^2$.
If $x=7^2k+18$ is a solution to $x^2+x+1\equiv0\bmod7^3$,
then $k\equiv0\bmod7$, so $x\equiv18\bmod7^3$.
If $x=7k+2$ is a solution to $x^2+x+1\equiv0\bmod7^2$, then $k\equiv4\bmod7$, so $x\equiv30\bmod7^2$.
If $x=7^2k+30$ is a solution to $x^2+x+1\equiv0\bmod7^3$,
then $k\equiv6\bmod7$, so $x\equiv324\bmod7^3$.
(I could have argued that, if $x\equiv18$ is a solution,
then $x^4+x^2+1\equiv x+x^2+1\equiv0$, so $x^2\equiv18^2=324$ is a solution too.)
So either $a\equiv b$ or $a\equiv18b$ or $a\equiv324 b\bmod 7^3$, and then $a^3\equiv b^3\bmod 7^3$.
| {
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Singular values of matrices which preserve the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$ Let $0<a<b$, $ab=1$, and let
$$
D_{a,b}=\biggl\{(x,y) \,\biggm | \, \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1 \biggr\}
$$
be the ellipse with diameters $a,b$.
Let $A \in \operatorname{SL}_2(\mathbb R) \setminus \operatorname{SO}(2)$ and suppose that $AD_{a,b}=D_{a,b}$.
Question: Must the singular values of $A$ be $\bigl\{\frac{a}{b},\frac{b}{a}\bigr\}$?
This option is always possible since we can take $A=\begin{pmatrix} \frac{a}{b} & 0 \\\ 0 & \frac{b}{a}\end{pmatrix}R_{\pi/2}$.
Here is a (very) partial attempt:
The condition $AD_{a,b}=D_{a,b}$ implies* that $A$ is similar to an orthogonal matrix, i.e.
$A=CQC^{-1}$, where $C \in \operatorname{SL}_2(\mathbb R) , Q \in \operatorname{SO}(2)$.
$AD_{a,b}=D_{a,b}$ implies that $Q\tilde D=\tilde D$, where $\tilde D=C^{-1}D_{a,b}$.
If $Q$ is an irrational rotation (of infinite order), then $\tilde D$ must be the unit disk, which implies that the singular values of $C$ are $a,b$. This mean that $C=U\Sigma V^T$, where $\Sigma=\operatorname{diag}(\sigma_1,\sigma_2)$.
Thus,
$$
A=CQC^{-1}=U\Sigma V^T Q V\Sigma^{-1}U^T,
$$
so up to left and right multiplication by rotations,
$A=\Sigma R\Sigma^{-1}$, where $R= V^T Q V$.
I don't see how to continue from here.
I also don't know how to start analyzing the case where $Q$ is of finite order.
*You may see this answer and the comments below it.
| That's not true in general: the singular values of $A$ are by definition the square roots of the eigenvalues of $A^*A$. Set
\begin{align*}A(\theta) & = \begin{pmatrix} a& 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1/a& 0 \\ 0 & 1/b \end{pmatrix}\\
&= \begin{pmatrix} \cos\theta & -\frac ab \sin\theta \\ \frac ba \sin\theta & \cos \theta \end{pmatrix}
\end{align*}
Then it is clear that there are choices of $\theta$ such that the singular values are not $\{ a/b, b/a\}$ when $a\neq b$ (For example, note that the singular values are continuous functions (see remark) of $\theta$, and one has $\{1, 1\}$ at $\theta = 0$ and $\{ a/b, b/a\}$ at $\theta = \pi/2$).
Remark One can calculate that the eigenvalues of $A^*A$ are
$$ \frac{1}{2} \left(2\cos ^2 \theta + \left( \frac{a^2}{b^2} + \frac{b^2}{a^2} \right) \sin^2\theta \right) \pm \sqrt{\frac 14\left(2\cos ^2 \theta + \left( \frac{a^2}{b^2} + \frac{b^2}{a^2} \right) \sin^2\theta \right)^2 -1}, $$
which are clearly continuous in $\theta$.
| {
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Prove: $\int_0^{\infty} \frac{\ln{(1+x)}\arctan{(\sqrt{x})}}{4+x^2} \, \mathrm{d}x = \frac{\pi}{2} \arctan{\left(\frac{1}{2}\right)} \ln{5}$ Prove: $$\int_0^{\infty} \frac{\ln{(1+x)}\arctan{(\sqrt{x})}}{4+x^2} \, \mathrm{d}x = \frac{\pi}{2} \arctan{\left(\frac{1}{2}\right)} \ln{5}$$
This might be a repeat question (I couldnt find a question of this here). If im being honest I dont know the first step really... Maybe a clever integration by parts, substitution, differentiation under integral sign, power series, or contour? If someone could give advice.
| Let us try to collect the most interesting ideas in one simple solution.
At first,
$$I=\int\limits_0^\infty \dfrac{\ln(1+x)\arctan\sqrt x}{x^2+4}\text{ d}x
=\int\limits_0^\infty \dfrac{\ln(1+y^2)\arctan y}{y^4+4}\,2y\text{ d}y.$$
At the second, by pisco,
$$\ln(1+y^2) = \ln(1+iy) + \ln(1-iy),\\
\arctan y = \dfrac i2(\ln(1-iy) - \ln(1+iy)),$$
$$\ln(1+y^2)\arctan y = \dfrac i2(\ln^2(1-iy)-\ln^2(1+iy)).$$
Therefore,
$$I=\int\limits_{0}^\infty \dfrac{\ln^2(1+iy)-\ln^2(1-iy)}{4iy}\dfrac {8y^2\text{ d}y}{y^4+4}.\tag1$$
Taking in account Sophie Germain identity
$$y^4+4 = (y^2+2)^2 - 4y^2 = (y^2-2y+2)(y^2+2y+2),$$
easily to get
$$\int\limits_0^\infty \dfrac{y^2}{y^4+4}\text{ d}y=\dfrac\pi4,\quad
\int\limits_0^\infty \dfrac{1}{y^4+4}\text{ d}y=\dfrac\pi8,\quad
\int\limits_0^\infty \dfrac{1}{y^2+z^2}\text{ d}y=\dfrac\pi{2z}\tag2$$
(see also Wolfram Alpha integral1, integral2, integral3).
Now, applying the definite integral
$$\int\limits_0^\infty \dfrac{\ln t}{(t+a)(t+b)} = \dfrac{\ln^2a - \ln^2 b}{2a-2b},\tag3$$
which is known from the answer of user9735739 (it should be correct if $\Re a >0,\ \Re b >0$), in the form of
$$\int\limits_0^\infty\dfrac{\ln z\text{ d}z}{(z+1+iy)(z+1-iy)} = \dfrac{\ln^2(1+iy)-\ln^2(1-iy)}{4iy},$$
integral $(1)$ can be presented in the form of
$$I=\int\limits_{0}^\infty \int\limits_0^\infty \dfrac{\ln z}{(z+1)^2+y^2}\dfrac {8y^2}{y^4+4}\text{ d}z\text{ d}y= 8 \int\limits_{0}^\infty J(z) \ln z \text{ d}z,\tag4$$
where
\begin{align}
&J(z-1) = \int\limits_0^\infty \dfrac{y^2\text{ d}y}{(y^2+z^2)(y^4+4)}
= \dfrac1{z^4+4}\int\limits_0^\infty \left(\dfrac{z^2 y^2}{y^4+4}+\dfrac{4}{y^4+4}-\dfrac{z^2}{y^2+z^2}\right)\text{ d}y\\[4pt]
&=\dfrac1{z^4+4}\left(z^2\cdot\dfrac\pi4 + 4\cdot\dfrac\pi8 - z^2\cdot\dfrac\pi{2z}\right)
= \dfrac\pi4\dfrac{z^2-2z+2}{z^4+4} = \dfrac\pi{4((z+1)^2+1)},\\[4pt]
&J(z) = \dfrac\pi{4((z+2)^2+1)}.
\end{align}
Finally, applying $(3)$ once more,
$$I=2\pi\int\limits_0^\infty \dfrac{\ln z}{(z+2+i)(z+2-i)}\text{ d}z
= 2\pi\dfrac{\ln^2(2+i)-\ln^2(2-i)}{4i} = 2\pi\dfrac{(\ln5+i\operatorname{arccot} 2)^2-(\ln5-i\operatorname{arccot}2)^2}{4i}
= \color{brown}{\mathbf{\dfrac\pi2\, \arctan\left(\frac12\right)\,\ln5}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Write $\frac{\sqrt{2}}{\sqrt{2} + \sqrt{3} - \sqrt{5}} - \frac{1}{2}$ with a rational denominator. Write $\frac{\sqrt{2}}{\sqrt{2} + \sqrt{3} - \sqrt{5}} - \frac{1}{2}$ with a rational denominator.
How should I solve this question?
| Only the first term
We have $\frac{\sqrt 2}{\sqrt 2 + \sqrt 3 - \sqrt 5}.$ Do :
$$
\frac{\sqrt 2}{\sqrt 2 + \sqrt 3 - \sqrt 5}= \frac{\sqrt 2(\sqrt 2 + \sqrt 3 + \sqrt 5)}{(\sqrt 2 + \sqrt 3)^2 - 5} = \frac{2 + \sqrt 6 + \sqrt {10}}{2\sqrt 6} \\
= \frac{2\sqrt 6 + 6 + \sqrt {60}}{2\sqrt 6 \sqrt 6} = \frac{6 + 2 \sqrt 6 + \sqrt {60}}{12}= \frac{3+\sqrt 6 + \sqrt {15}}{6}
$$
So you are correct!
| {
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how to prove this inequlity $\ln(\frac{p}{q}) \le \frac{p-q}{\sqrt{pq}} (0 < q \le p)$ by means of integral inequalities I can solve it by introducing a function, but I need to prove it by applying integral inequalities such as the holder inequality or the schwarz inequality.
| You probably can recognize this is the equivalent to prove
$$
f(x) = \ln(x)\le \sqrt{x} - \frac{1}{\sqrt{x}} = g(x) \quad \text{when} x \ge 1
$$
Easy to see $f(1) = g(1)= 0$
We also have
$$
g'(x) = \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{x}^3} = \frac{1}{2}\left( \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{x}^3}\right) \ge \sqrt{ \frac{1}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}^3}} = \frac{1}{x} = f'(x)
$$
Thus $f(x)\le g(x)$ when $x \ge 1$ as a result of integral inequalities.
| {
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If roots of equation $ax^2+bx+c=0$ is $\alpha,\beta$ find the roots of equation $bcx^2+a(bc+a^2)x+ac^2=0$ in terms of $\alpha,\beta$ If roots of equation $ax^2+bx+c=0$ is $\alpha,\beta$ find the roots of equation $bcx^2+a(bc+a^2)x+ac^2=0$ in terms of $\alpha,\beta$
My Try:
We know that
$$\alpha+\beta=-\frac ba \quad \text{ and } \quad \alpha \beta=\frac ca$$
I have tried to substitute $\alpha,\beta$ into the new equation, but I have faced some trouble how to separate roots,
So far I got;
$$x^2+\bigg(\frac {1}{\alpha\beta}-\frac {1}{(\alpha+\beta)(\alpha\beta)^2}\bigg)x+\frac{1}{\alpha+\beta}=0$$
How to proceed? If you suggest a completely different method that is fine. Thanks in advance!
| The given quadratic equation has the roots $\alpha$ and $\beta$. Then we have that
$$ \alpha + \beta = -\frac{b}{a} \quad \text{ and } \quad \alpha \cdot \beta = \frac{c}{a} $$
The new quadratic $$ bc \cdot x^2 + a(bc + a^2)x + ac^2 = 0 $$
$\blacksquare~$What I have so far:
Let the roots of the new quadratic be $m, n$. Then we have that
$$ m + n = -a \cdot \frac{bc + a^2}{bc} = - a\bigg(1 + \frac{a^2}{bc}\bigg) = - a \cdot {\bigg(1 - \color{red}{ \frac{1}{\alpha \beta (\alpha + \beta)}}\bigg)} $$
and $$ mn = \frac{ac^2}{bc} = \frac{a}{b} \cdot c = - a \cdot \bigg( \frac{\alpha \beta}{ \alpha + \beta} \bigg) $$
Hence the quadratic has become $$ x^2 + a \cdot \bigg(1 - \color{red}{\frac{1}{\alpha \beta (\alpha + \beta)}}\bigg) x - a \cdot \bigg( \frac{\alpha \beta}{ \alpha + \beta} \bigg) = 0 $$
I haven't found out any way to calculate and substitute $a$ yet (it seems too that we can't determine the roots just by $\alpha$ and $\beta$).
| {
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Systems of polynomial equations involving sums of equal powers Given the following system of polynomial equations:
$$
\left\{\begin{array}{lclclcr}
x & + & y & + & z & = & 1
\\
x^{2} & + & y^{2} & + & z^{2} & = & 14
\\
x^{3} & + & y^{3} & + & z^{3} & = & 36
\end{array}\right.
$$
What is $x^{5} + y^{5} + z^{5}\ {\large ?}$ .
How should I approach this? Is there a general formula for this kind of system?
| The requested exponent does not matter that much. Each of the three is a root of the same
$$ 6 t^3 - 6 t^2 - 39 t - 31, $$
one real and two complex conjugates, but we don't need them. Each also obeys
$$ 6 t^{n+3} - 6 t^{n+2} - 39 t^{n+1} - 31 t^n, $$ so that $a_n=x^n + y^n + z^n$ is a sequence with linear recurrence
$$ a_{n+3} = a_{n+2} + \frac{13}{2} a_{n+1} + \frac{31}{6} a_n $$ or
$$ a_1 = 1, \; a_2 = 14, \; a_3 = 36, \; a_4 = \frac{793}{6}, \; $$
$$ a_5 = \frac{877}{2}, \; a_6 = \frac{17803}{12}, \; a_7 = \frac{180601}{36}, \; a_8 = \frac{1218641}{72}, \; $$
| {
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The number of three digit numbers $abc$, which satisfy $a≤b>c$ is My approach in this question is as follows,
⇒When $a=1$
$-> 1-1-(0) / 1-2-(0,1) / 1-3-(0,1,2) ... / 1-9-(0,1...,8)$ ⇒Total numbers = $1+2+3+...+9$
⇒When $a=2$
$-> 2-2-(0,1) / 2-3-(0,1,2) ... / 2-9-(0,1...,8)$ ⇒Total numbers = $2+3+...+9$
⇒When $a=3$
$-> 3-3-(0,1,2) ... / 3-9-(0,1...,8)$ ⇒Total numbers = $3+...+9$
So on counting, we get total numbers as,
$$(1+2+...9)+(2+3+...9)+...(8+9)+(9)$$
$$⇒1^2+2^2+3^2+...9^2$$
$$⇒\frac{(9)(9+1)(2*9+1)}{6}$$
$$⇒285$$
But the actual solution says the answer is,
$$\bigl(2*({9 \choose 1}+{9 \choose 2}+{9 \choose 3})\bigr)$$
$$⇒258$$
I think I have recounted cases, can anyone explain which cases have I recounted??
| The given condition should be extended to
$$1\leq a\leq b>c\geq0\ .$$
Given $b\in\{1,2,\ldots,9\}$ we therefore have $b$ choices for $a$ and $b$ choices for $c$. The number of admissible three digit numbers $abc$ therefore is
$$\sum_{b=1}^9b^2={n\cdot(n+1)(2n+1)\over6}\biggr|_{n=9}={9\cdot10\cdot 19\over 6}=285\ .$$
| {
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Minimizing a function by finding its critical points Let $f_n(x)$ equal:
$$(2^n+2) \left(2x+1-\sqrt{2x^2+2x} \right)^n-x^n-(x+1)^n-\left(3x+1-2\sqrt{2x^2+2x}\right)^n-\left(3x+2-2\sqrt{2x^2+2x}\right)^n$$
Mathematica suggests that this function has two critical points on $(0,\infty)$, namely $x_1=1$ and $x_2=1/\sqrt{2}-1/2$. This seems to be independent of $n$ (well, I checked for all $4\leq n\leq 7$; also $f_1(x),f_2(x),f_3(x)$ identically equal to zero, a fact that I don't quite understand yet).
Problem: Prove that the only critical points of $f_n(x)$, $n\geq 4$, are $x_1=1$ and $x_2=1/\sqrt{2}-1/2$.
My goal is to show that $f_n(x)\geq 0$ for all $x\geq 0$. If I can show the only critical points are $1$ and $1/\sqrt{2}-1/2$, then by noting that $f_n(1)=0$ and $f_n(1\sqrt{2}-1/2)>0$ we have proved the claim. Of course, if you can prove $f_n(x)\geq 0$ in any way, please do share your proof.
| The fact that $f_1(x),f_2(x),f_3(x)$ are identically equal to zero is simple (even if a bit tedious) to prove; just expand the expression.
Concerning $f_4(x)$ it reduces to
$$f_4(x)=4 x (x+1) \left(17 x^2+17 x+2\right)-24 \sqrt{2}\, x^{3/2} (x+1)^{3/2} (2 x+1)$$
$$f'_4(x)=8 (2 x+1) \left(17 x^2+17 x+1\right)-12 \sqrt{2}\, x^{1/2} (x+1)^{1/2} (4 x+1) (4 x+3)$$ After one squaring step, the derivative cancels when
$$(x-1)\left(4 x^2+4 x-1\right) (x+2) \left(2 x^2+2 x+1\right) =0$$ So, since $x >0$, the only acceptable solutions are
$$x_1=1 \qquad \text{and} \qquad x_2=\frac{\sqrt{2}-1}{2} $$ as you already found.
Doing the same for $f'_5(x)$, the derivative cancels for
$$(x-1) \left(4 x^2+4 x-1\right)(x+2) \left(25 x^4+50 x^3+35 x^2+10 x+4\right)=0$$ and the quartic does not show real roots.
Continuing (it starts to be tedious), for $f'_6(x)$, the derivative cancel for
$$(x-1)\left(4 x^2+4 x-1\right)(x+2) \left(2 x^2+2 x+1\right) \left(9 x^4+18 x^3+9
x^2+1\right)$$ and again the quartic does not show real roots.
It seems that the expressions which make the derivative equal to zero show a quite clear pattern. This has been tested up to $n=100$; to confirm, the values of $f'_n(x_1)$ and $f'_n(x_2)$ have been systematically checked and they are always equal to $0$.
It is clear that $f_n(x_1)=0$. So, now
$$f_n(x_2)=2^{n/2}+2^{1-\frac{n}{2}}-2^{1-n}\Big[ \left(\sqrt{2}-1\right)^n+
\left(\sqrt{2}+1\right)^n\Big]$$ which is a very fast increasing function.
As @Alexey Burdin commented, this looks to be the result of a (difficult) recurrence relation.
| {
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Obtain sum of a sequence from sum of its odd terms. I would like to compute the sum
$$
\sum_{k=1}^\infty \frac{1}{k^4}
$$
by using the Fourier series of $f(x)=|x|$ over $(-\pi,\pi)$. Coefficients $b_k$ are all $0$ because $f$ is even. Doing the integration stuff, I obtained:
$$
a_0 = \pi
$$
and
$$
a_k = \frac{2}{k^2}\bigg((-1)^k-1\bigg)
$$
for $k>0$. The Parseval's equality gives:
$$
\frac{a_0^2}{2} + \sum_{k=1}^\infty (a_k^2+b_k^2)= \frac{1}{\pi}\int_{-\pi}^{\pi}f^2dx
$$
which gives
$$
\frac{\pi^2}{2} + \sum_{k=1}^\infty \frac{4}{\pi^2k^4}(2-2(-1)^k) = \frac{2}{3}\pi^2
$$
which simplifies to
$$
\sum_{k=1}^\infty \frac{1}{k^4} - \sum_{k=1}^\infty \frac{(-1)^k}{k^4} = \frac{\pi^4}{48}
$$
which basically says:
$$
\sum_{k=0}^\infty \frac{1}{(2k+1)^4}=\frac{\pi^4}{96}
$$
any idea how to obtain the sum from there?
| You essentially have
$${\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + ... = \frac{\pi^4}{96}}$$
You want to find
$${\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ... = ?}$$
in other words, you want to add on
$${\frac{1}{2^4} + \frac{1}{4^4} + ...}$$
Factoring out a ${\frac{1}{2^4}}$ on the above yields
$${\frac{1}{2^4}\left(\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ...\right)}$$
So overall, if you call ${S=\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + ...}$ you have
$${\left(\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + ...\right) + \left(\frac{1}{2^4} + \frac{1}{4^4} + ...\right) = S}$$
$${\Rightarrow \frac{\pi^4}{96} + \frac{1}{2^4}S = S}$$
Can you now rearrange for ${S}$?
| {
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Better proof of a numerical inequality of $e^x$ The inequality is
$$ e^z \leq 1+z+\frac{z^2/2}{1-|z|/3} \text{ for } |z|<3$$
I proved it by splitting it into 3 cases: $-3<z<0$, $z=0$ and $0<z<3$.
For $z=0$, both sides equals.
The other 2 cases are done with calculus. Define $f(x)=e^x-1-x-\frac{x^2/2}{1-|x|/3}$ and then replace $|x|$ by $x$ or $-x$ accordingly. Then just check the derivatives.
But in my opinion, it is sort of brute force, so I am wondering if there is faster(smarter) way to show it.
| Note that, if $|z|<3$,\begin{align}e^z-1-z&=\frac{z^2}2+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots\\&=\frac{z^2}2\left(1+\frac z3+\frac{z^2}{3\times4}+\frac{z^3}{3\times4\times5}+\cdots\right)\\&\leqslant\frac{z^2}2\left(1+\frac{|z|}3+\frac{|z|^2}{3\times4}+\frac{|z|^3}{3\times4\times5}+\cdots\right)\\&\leqslant\frac{z^2}2\left(1+\frac{|z|}3+\frac{|z|^2}{3^2}+\frac{|z|^3}{3^3}+\cdots\right)\\&=\frac{z^2}2\cdot\frac1{1-|z|/3}.\end{align}
| {
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Show that $x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded, monotone, and find its limit Prove that $x_1 = 0, x_2 = 0, x_{n+2} = \frac{1}{3} x_{n + 1} + \frac{1}{6} x_n + 1$ is bounded and monotonic. Then find its limit.
My attempt at boundedness:
(Using induction) For the base case we have $0 \leq x_1 = 0 \leq 2$. Assume that the sequence is bounded for $n = k$. Then,
\begin{align*}
0 \leq x_k &\leq 2 \\
\vdots \\
\text{lower bound } \leq x_{k + 1} &\leq \text{upper bound}
\end{align*}
I am thrown off by the term $x_{n + 2}$ in the recursive formula and I can't see the algebra to produce the above steps without getting $x_{n + 2}$ in the expression of the upper / lower bound.
Thank you.
Update:
I have added this to the prove:
We have $0 \leq x_1 = 0 \leq 2$ and $0 \leq x_2 = 0 \leq 2$. Assume that the sequence is bounded for $k+1$,
\begin{align*}
0 &\leq x_{k + 1} \leq 2 \\
0 &\leq x_k + x_{k+1} \leq 4 \\
0 &\leq x_k + \frac{1}{3} x_{k+1} \leq 4 \\
0 &\leq \frac{1}{6} x_{k} + \frac{1}{3} x_{k+1} \leq 4 \\
0 &\leq x_{k+2} \leq 4
\end{align*}
Therefore, by the principle of mathematical induction, the sequence is bounded.
Is this valid?
| No, your argument is not valid. You show that
$$x_{k+1}\le 2\implies x_{k+2}\le 4.$$
If you apply induction, this leads to
$$x_{k+m}\le 2^{m+1}$$ which is not bounded.
But you can use
$$x_k,x_{k+1}\le2\implies x_{k+2}=\frac{x_k}{3}+\frac{x_{k+1}}6+1\le\frac23+\frac26+1=2.$$
| {
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Coefficient of $x^7y^6$ in $(xy+x+3y+3)^8$
Find the coefficient of $x^7y^6$ in $(xy+x+3y+3)^8$.
My solution:
Factor $(xy+x+3y+3)^8$ into $(x+3)^8(y+1)^8$. To get an $x^7y^6$ term, we need to find the coefficient of $x^7$ in the first factor and $y^6$ in the second factor. Using the binomial theorem, we get the coefficient of $x^7$ to be $17496$ and $y^6$ to be $28$. Multiplying the two gets us an answer of $489888.$
However, this is wrong. This is the answer key's approach:
$x^7y^6 = (xy)^6 \cdot x = (xy)^5 \cdot x^2 \cdot y$. Now, $(xy)^6\cdot x$ can be formed by choosing $6$ $xy$'s, $1$ $x$, and $1$ $3$, which can be done in $\binom{8}{6}\binom{3}{2}\binom{1}{1} = 56$ ways. $(xy)^5\cdot x^2\cdot y$ can be formed by choosing $5$ $xy$'s, $2$ $x$'s, and $1$ $3y$, which can be done in $\binom{8}{5}\binom{3}{2}\binom{1}{1} = 168$ ways. Thus the final coefficient is $3(56+168) = 672$ ways.
I completely understand their approach, but fail to understand why mine doesn't work. Don't we just calculate the number of ways to get $x^7$, and $y^6$, then multiply them?
Interestingly enough, I noticed that when you calculate the coefficient of $x^1$ (which is $x^{8-7}$) and $y^2$ (which is $y^{8-6}$), you get $3\cdot \binom{8}{1} \cdot \binom{8}{2} = 672$, which is the answer. I'm $99\%$ sure this isn't a coincidence, but why does this method work and not the other?
I know that I did not get any calculations wrong, because I double-checked everything with WolframAlpha; the error must be in my process.
Thanks in advance!
(Question from PuMaC 2017 Algebra B)
| The factor for $x^7$ is $24$ and for $y^6$ is $28$, which their multiplication is the correct answer. Probably you're just applying the binomial theorem incorrectly.
As an add-on, your method is completely correct.
| {
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Checking the analyticity of a complex function. This was a question in an assignment that demands the use of Cauchy's Integral Formula in the questions. The question goes like this:
Integrate $\displaystyle{g(z)=\frac{e^z}{ze^z-2iz}}$ over $\displaystyle{C:\ \vert z\vert=0.5}$.
What I tried:
First, I noted that $\displaystyle{g(z)=\frac{e^z}{z\left(e^z-2i\right)}}$ so I could use the Cauchy's Integral Formula with $z_o=0$ and $\displaystyle{f(z)=\frac{e^z}{e^z-2i}}$. To do so, I have to make sure that $f(z)$ is analytic. So, I proceed like this:
As $z=x+iy$, so
$$\begin{align}
f(z)&=\frac{e^{x+iy}}{e^{x+iy}-2i}\\
&=\frac{e^x \cos y+ie^x\sin y}{e^x \cos y+ie^x\sin y-2i}\\
&=1+\frac{2i}{e^x \cos y+ie^x\sin y-2i}\\
&=1+\frac{2i\left(e^x \cos y-ie^x\sin y+2i\right)}{\left(e^x \cos y+ie^x\sin y-2i\right)\left(e^x \cos y-ie^x\sin y+2i\right)}\\
&=1+\frac{2ie^x\cos y+2e^x\sin y-4}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}\\
&=\left[1+\frac{2e^x\sin y-4}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}\right]+i\left[\frac{2e^x\cos y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}\right]\\
&=u+iv
\end{align}$$
Hence
$$u=1+\frac{2e^x\sin y-4}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}\ \ \ \ \ \text{and}\ \ \ \ \ v=\frac{2e^x\cos y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}$$
Now according to WolframAlpha,
$$
\begin{align}
\frac{\partial u}{\partial x}&=\frac{2e^x\sin y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}-\frac{\left(2e^x\sin y-4\right)\left(2e^x\sin y\left(e^x\sin y-2\right)+2e^{2x}\cos^2y\right)}{\left(e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2\right)^2}\\
\frac{\partial u}{\partial y}&=\frac{2e^x\cos y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}-\frac{\left(2e^x\sin y-4\right)\left(2e^x\cos y\left(e^x\sin y-2\right)-2e^{2x}\sin y\cos y\right)}{\left(e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2\right)^2}\\
\frac{\partial v}{\partial x}&=\frac{2e^x\cos y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}-\frac{2e^x\cos y\left(2e^x\sin y\left(e^x\sin y-2\right)+2e^{2x}\cos^2 y\right)}{\left(e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2\right)^2}\\
\frac{\partial v}{\partial y}&=-\frac{2e^x\sin y}{e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2}-\frac{2e^x\cos y\left(2e^x\cos y\left(e^x\sin y-2\right)-2e^{2x}\sin y\cos y\right)}{\left(e^{2x}\cos^2y+\left(e^x\sin y-2\right)^2\right)^2}
\end{align}
$$
Links for WolframAlpha: du/dx du/dy dv/dx dv/dy
Now clearly, these values for $\displaystyle\frac{\partial u}{\partial x}$, $\displaystyle\frac{\partial u}{\partial y}$, $\displaystyle\frac{\partial v}{\partial x}$ and $\displaystyle\frac{\partial v}{\partial y}$ do not satisfy the Cauchy-Riemann equations.
Did I do anything wrong during the solution? If not, what method should be used to solve the required integral?
Thanks for the attention.
| If you consider $f$ defined in $U=D(0,r)$ where $\frac{1}{2}<r<\text{ln}(2)$ then $f(z)=\frac{e^z}{e^z-2i}$ is well-defined. This is true because $e^z-2i=0 \Rightarrow |e^z|=|2i| \iff e^{Re(z)}=2$ which is impossible in $U$, therefore $f$ is analytic once is the quotient of two analytic functions. Notice that $\frac{1}{2}<r$ implies $C\subset U$.
Finally by Cauchy formula, $\int_{C} \frac{e^z}{z(e^z-2i)} dz=\int_{C} \frac{f(z)}{z} dz=2\pi if(0)\text{ind}(0,C)=\frac{2\pi i}{1-2i}$ considering that $C$ goes counterclockwise.
| {
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Change this integral $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$. It is given that $f(a+b+1 -x) = f(x)$ where $a$ and $b$ are positive real numbers then $\frac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$ is equal to
*
*$\int_{a-1}^{b-1} f(x) dx$
*$\int_{a+1}^{b+1} f(x+1) dx$
*$\int_{a-1}^{b-1} f(x+1) dx$
*$\int_{a+1}^{b+1} f(x) dx$
I tried using integration by parts along with this property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ and the one given in the question but I'm unable to get anything. It's that extra $x$ and the fraction $\frac{1}{a+b}$ which is making everything quite uneasy.
| Your question is,
$$\dfrac{1}{a+b} \int_{a}^{b} x \left[ f(x) + f(x+1) \right] dx$$
Now, in second term i.e. $f(x+1)$, put $x+1$ as $u$ and use $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ and $f(a+b+1 -x) = f(x)$.
You will get $$\dfrac{1}{a+b} \int_{a}^{b} (a+b) \left[ f(x) \right] dx$$. Now substitute $x\rightarrow x+1$, you'll get option 3.
| {
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How to find this $\prod_{k=-\infty}^{+\infty}\frac{x^2+(4k+1-y)^2}{x^2+(4k+3-y)^2}$ let $x,y$ be real numbers,show that
$$\prod_{k=-\infty}^{+\infty}\dfrac{x^2+(4k+1-y)^2}{x^2+(4k+3-y)^2}=\dfrac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}$$
This is a question that a physicist asked me. He said that he could prove the equation in a physical way, but not in a mathematical way. I thought about it for a while and might need to use the following conclusion to solve it, but it didn't work out
$$\prod_{k=1}^{+\infty}\left(1+\dfrac{1}{k^2}\right)=\dfrac{\sinh{\pi/2}}{\pi/2}$$？
| Firstly, define the complex number $z=x+yi$. Then, we are asked to prove that
$$
\prod_{k\in\mathbb{Z}}\left|\frac{z-(4k+1)i}{z-(4k+3)i}\right|^2=\frac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}.
$$
Now, let's rewrite the right-hand side of this equality:
$$
\frac{1+e^{\pi x}-2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}{1+e^{\pi x}+2e^{\frac{\pi}{2}x}\sin{\dfrac{\pi}{2}y}}
=
\frac{\left(e^{\frac{\pi}{2}x}-\sin\dfrac{\pi}{2}y\right)^2+\cos^2\dfrac{\pi}{2}y}{\left(e^{\frac{\pi}{2}x}+\sin\dfrac{\pi}{2}y\right)^2+\cos^2\dfrac{\pi}{2}y}
=
\left|\frac{e^{\frac{\pi}{2}x}+i\left(\cos\dfrac{\pi}{2}y+i\sin\dfrac{\pi}{2}y\right)}{e^{\frac{\pi}{2}x}-i\left(\cos\dfrac{\pi}{2}y+i\sin\dfrac{\pi}{2}y\right)}\right|^2=
\\
=\left|\frac{e^{\frac{\pi}{2}x}+ie^{i\frac{\pi}{2}y}}{e^{\frac{\pi}{2}x}-ie^{i\frac{\pi}{2}y}}\right|^2=\left|\frac{e^{\frac{\pi}{2}(x-iy)}+i}{e^{\frac{\pi}{2}(x-iy)}-i}\right|^2
=\left|\frac{e^{\frac{\pi}{2}\overline{z}}+i}{e^{\frac{\pi}{2}\overline{z}}-i}\right|^2=\left|\frac{e^{\frac{\pi}{2}z}-i}{e^{\frac{\pi}{2}z}+i}\right|^2.
$$
Thus, we need to prove that
$$
\prod_{k\in\mathbb{Z}}\left|\frac{z-(4k+1)i}{z-(4k+3)i}\right|=\left|\frac{e^{\frac{\pi}{2}z}-i}{e^{\frac{\pi}{2}z}+i}\right|.
$$
Let $w=z-i$, then the last equality can be rewritten as
$$
\prod_{k\in\mathbb{Z}}\left|\frac{w-4ki}{w-(4k+2)i}\right|=\left|\frac{e^{\frac{\pi}{2}w}-1}{e^{\frac{\pi}{2}w}+1}\right|.
$$
Update. As David mentioned in the comments, in order to compute the product $\prod_{k\in\mathbb{Z}}\frac{t-4k}{t-(4k+2)}$ we can use the following formula:
$$
\frac{\sin\pi z}{\pi z}=\prod_{k=1}^{\infty}\left(1-\frac{z^2}{k^2}\right),
$$
or
$$
\sin\pi z=\pi z\prod_{k\in\mathbb{Z}\backslash\{0\}}\frac{k-z}{k}.
$$
For $z=t/4$ and $z=(t-2)/4$ we have
$$
\sin\frac{\pi t}{4}=\frac{\pi t}{4}\prod_{k\in\mathbb{Z}\backslash\{0\}}\frac{4k-t}{4k}
$$
and
$$
\sin\frac{\pi (t-2)}{4}=\frac{\pi (t-2)}{4}\prod_{k\in\mathbb{Z}\backslash\{0\}}\frac{4k+2-t}{4k},
$$
respectively. Hence,
$$
\frac{\sin\pi t/4}{\sin\pi (t-2)/4}=\prod_{k\in\mathbb{Z}}\frac{t-4k}{t-(4k+2)}.
$$
Therefore,
$$
\prod_{k\in\mathbb{Z}}\left|\frac{w-4ki}{w-(4k+2)i}\right|=[t=w/i]=\prod_{k\in\mathbb{Z}}\left|\frac{t-4k}{t-(4k+2)}\right|=\left|\frac{\sin\pi t/4}{\sin\pi (t-2)/4}\right|=
\\
=\left|\frac{e^{i\pi t/4}-e^{-i\pi t/4}}{e^{i\pi (t-2)/4}-e^{-i\pi (t-2)/4}}\right|=\left|\frac{e^{i\pi t/4}-e^{-i\pi t/4}}{-ie^{i\pi t/4}-ie^{-i\pi t/4}}\right|=\left|\frac{e^{\pi w/2}-1}{e^{\pi w/2}+1}\right|,
$$
so we're done.
| {
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solutions to $\frac{1}a + \frac{1}b + \frac{1}c = \frac{1}{2018}$
Find, with proof, all ordered triplets of positive integers $(a,b,c)$ so that $\dfrac{1}a + \dfrac{1}b + \dfrac{1}c = \dfrac{1}{2018}.$
In general, if $d$ is a positive integer, then $(a,b,c) = (3d,3d,3d),(d,2d,6d),(d,6d,2d), (2d,d,6d),(2d,6d,d),(6d,d,2d),$ and $(6d,2d,d)$ are all ordered triplets of positive integers such that $\dfrac{1}a + \dfrac{1}b + \dfrac{1}c = \dfrac{1}d.$ However, I'm unsure how to find all such triplets. I tried manipulating the equation as follows:
\begin{align}
&1 + \dfrac{a}b + \dfrac{a}c = \dfrac{a}{2018}\\
&\dfrac{a}b + \dfrac{a}c = \dfrac{a-2018}{2018}\\
&\dfrac{ab}{c } = \dfrac{b(a-2018)-2018a}{2018}\\
&c[b(a-2018)-2018a]-2018ab = 0\\
&c[(a-2018)(b-2018)-2018^2]-2018ab = 0\\
&c[(a-2018)(b-2018)]-2018(ab-2018(a+b)+2018^2+2018(a+b)-2018^2)-2018^2 c =0 \\
&(c-2018)(a-2018)(b-2018)-2018^2(a+b+c)+2018^3 = 0\\
&(c-2018)(b-2018)(a-2018) = 2018^2(a+b+c) - 2018^3,
\end{align}
but I don't know whether this is useful.
Source (from comments): I based this off a contest problem. I came up with it by myself. The question I based it on was the Putnam 2018 question A1.
| Partial answer. $\,$ Let $d=\text{gcd}(a,b,c)$ and, thus, $a=dx, b=dy, c=dz$ - where $\text{gcd}(x,y,z)=1$. If we assume that $\text{gcd}(x,y)=\text{gcd}(y,z)=\text{gcd}(z,x)=1$ then $\text{gcd}(xyz, xy+yz+zx)=1$ $$\frac1a+\frac1b+\frac1c=\frac1{2018}\iff \frac{abc}{ab+bc+ca}=\frac{dxyz}{xy+yz+zx}=2018$$ This implies that $xy+yz+zx\mid d\iff d=(xy+yz+zx)\cdot k, k\in\mathbb Z$. Hence, the equation becomes $$k\cdot xyz=2018$$
Once you have the integer solutions to this equation - and this should not take long, since $2018=2\cdot 1009$ -, use the values to obtain back the solutions $$(a,b,c)\equiv (kx\cdot (xy+yz+zx),ky\cdot (xy+yz+zx), kz\cdot (xy+yz+zx) )$$
Observation. Due to symmetry, you "only" have to consider the case $\text{gcd}(x,y)>1$.
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} |
obtaining a simplified expression for the coefficient of $x^n$
I was trying to find the coefficient of $x^n$ in the expansion of $(1+x)^{-2}(1-2x)^{-2},$ denoted $[x^n]\{(1+x)^{-2}(1-2x)^{-2}\}$. Using the negative binomial theorem, I know that it is equal to
$$
\begin{split}
\sum_{j=0}^n &([x^j](1+x)^{-2})([x^{n-j}](1-2x)^{-2}) \\
&= \sum_{j=0}^n {j+1\choose 1}(-1)^j{n-j+1\choose 1}2^{n-j} \\
&= \sum_{j=0}^n (j+1)(n-j+1)(-1)^j2^{n-j}.
\end{split}
$$
However, I was wondering if there was a way to simplify this expression even further?
| As suggested by @AnginaSeng, you can apply partial fraction decomposition:
\begin{align}
\frac{1}{(1+x)^2(1-2x)^2}
&=\frac{1/9}{(1+x)^2}+\frac{4/27}{1+x}+\frac{4/9}{(1-2x)^2}+\frac{8/27}{1-2x}\\
&=\frac{1}{9}\sum_{n \ge 0}\binom{n+1}{1}(-x)^n+\frac{4}{27}\sum_{n\ge 0} (-x)^n+\frac{4}{9}\sum_{n \ge 0} \binom{n+1}{1}(2x)^n+\frac{8}{27}\sum_{n\ge 0} (2x)^n\\
&=\sum_{n \ge 0}\left(\frac{1}{9}\binom{n+1}{1}(-1)^n+\frac{4}{27}(-1)^n+\frac{4}{9}\binom{n+1}{1}2^n+\frac{8}{27} 2^n\right) x^n\\
&=\sum_{n \ge 0}\left(\color{blue}{\frac{(3n+7)(-1)^n+(12n+20)2^n}{27}}\right) x^n
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is this proof of $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ incomplete? So, for any angle $\alpha$ :
$$\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = \dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha} = \dfrac{\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}{\dfrac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}= \dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}$$
Now, $\cos\alpha = \cos\Big(2\cdot\dfrac{\alpha}{2}\Big) = \dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$
Now, using the componendo and dividendo rule, we get :
$$\dfrac{\cos\alpha+1}{\cos\alpha-1} = \dfrac{2}{-2\tan^2\dfrac{\alpha}{2}} = \dfrac{-1}{\tan^2\dfrac{\alpha}{2}} \implies \tan^2\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{1+\cos\alpha}$$
$$\implies \tan^2\dfrac{\alpha}{2} = \dfrac{(1-\cos\alpha)(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)} = \Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)^2$$
$$\implies \Bigg|\tan\Big(\dfrac{\alpha}{2}\Big)\Bigg| = \Bigg|\dfrac{1-\cos\alpha}{\sin\alpha}\Bigg|$$
Now, only if $\mathrm{sign}\Big(\tan\dfrac{\alpha}{2}\Big) = \mathrm{sign}\Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)$ is true, we can say that $\tan\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{\sin\alpha}$
So, I think that without proving that, the proof will be incomplete but my Math textbook doesn't prove it.
So, is it necessary to prove it? If not, why not?
Thanks!
| Your comment is correct. You can only get the final equality by proving that $\tan\left(\frac{\alpha}{2}\right)$ and $\frac{1-\cos \alpha}{\sin\alpha}$ have the same sign.
But this is not complicated to prove. $\tan\left(\frac{\alpha}{2}\right)$ is positive if and only if $\frac{\alpha}{2} \in (k\pi, k\pi +\frac{\pi}{2})$. Like $\sin \alpha$ while $1- \cos \alpha$ is always non negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
The limit and asymptotic analysis of $a_n^2 - n$ from $a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}$ I came up with the following question which is the follow up of
How to prove that for $a_{n+1}=\frac{a_n}{n} + \frac{n}{a_n}$ , we have $\lfloor a_n^2 \rfloor = n$?
Problem: Let $a_1 = 1,\quad a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n},\quad n\ge 1$.
*
*Prove that $\lim_{n\to \infty} (a_n^2 - n) = \frac{1}{2}$;
*Give the asymptotic analysis of $a_n^2 - n - \frac{1}{2}$.
Edit (2021/02/16) I also posted in https://mathoverflow.net/questions/384047/asymptotic-analysis-of-x-n1-fracx-nn2-fracn2x-n-2
For 1), I use the mathematical induction to prove the claim
$$n + \frac{1}{2} - \frac{2}{n} < a_n^2 < n + \frac{1}{2} + \frac{13}{4n} + \frac{13}{8n^2} + \frac{157}{16n^3}, \quad n \ge 22. \tag{1}$$
However, we need to verify it for $n = 22$ (a computer is required).
Are there simpler solutions?
$\color{blue}{\textbf{Edit}}$ 2021/02/15:
For 1), there is a solution in [1] (I know it from @haidangel's post The variation of a Ukrainian Olympiad problem: 10982). The authors proved that
$\frac{n^2}{n-1/2} \le a_n^2 \le \frac{(n-1/2)^2}{n-3/2}$ for all $n\ge 3$.
[1] Yuming Chen, Olaf Krafft and Martin Schaefer, “Variation of a Ukrainian Olympiad Problem: 10982”,
The American Mathematical Monthly, Vol. 111, No. 7 (Aug. - Sep., 2004), pp. 631-632
For 2), I have no idea currently. I want to find something like: for example, for the recurrence relation
$b_0 = 1, b_{n+1} = b_n + \frac{1}{b_n}, n\ge 0$,
we have $b_n \sim \sqrt{2n} + \frac{\sqrt{2}}{8\sqrt{n}}\ln n + o(\frac{\ln n}{\sqrt{n}})$. (Thank @Diger for pointing out the mistake. See the comment.)
About how to construct the claim (1): I want to find $d_n, c_n$ such that, for sufficiently large $n$,
$$n + \frac{1}{2} - d_n < a_n^2 < n + \frac{1}{2} + c_n.$$
To use the the mathematical induction, we need
$$a_{n+1}^2 = \frac{a_n^2}{n^2} + \frac{n^2}{a_n^2} + 2
< \frac{n + \frac{1}{2} + c_n}{n^2} + \frac{n^2}{n + \frac{1}{2} - d_n} + 2
< n + 1 + \frac{1}{2} + c_{n+1},$$
$$a_{n+1}^2 = \frac{a_n^2}{n^2} + \frac{n^2}{a_n^2} + 2
> \frac{n + \frac{1}{2} - d_n}{n^2} + \frac{n^2}{n + \frac{1}{2} + c_n} + 2
> n + 1 + \frac{1}{2} - d_{n+1}$$
which results in
$$c_{n+1} - \frac{c_n}{n^2} > \frac{n + \frac{1}{2}}{n^2} + \frac{n^2}{n + \frac{1}{2} - d_n} + \frac{1}{2} - n,$$
$$c_n < \frac{n^2}{n - \frac{1}{2} - d_{n+1} - \frac{n + \frac{1}{2} - d_n}{n^2}} - n - \frac{1}{2}.$$
We first choose $d_n$, then determine $c_n$.
For example, $d_n = \frac{2}{n}$ and $c_n = \frac{13}{4n} + \frac{13}{8n^2} + \frac{157}{16n^3}$.
| Since $a_1=1>0$ it is clear that $a_n>0$ for all $n$. Squaring gives $$a_{n+1}^2 = \frac{a_n^2}{n^2} + \frac{n^2}{a_n^2} + 2$$ and defining $a_n^2=nb_n$, this recurrence becomes $$b_{n+1}=\frac{b_n}{n(n+1)} + \frac{n}{(n+1)b_n} + \frac{2}{n+1}$$
with $b_1=1$. Now suppose $$1\leq b_n \leq 1+\frac{1}{n}+ \frac{2}{n^2}$$
which is true for $b_2=2$, $b_3=4/3$ and $b_4=(13/12)^2$ and continue inductively, i.e.
$$b_{n+1}\geq \frac{1}{n(n+1)} + \frac{n}{(n+1)(1+1/n+2/n^2)} + \frac{2}{n+1} = 1 + \frac{3n+2}{n(n+1)(n^2+n+2)}\geq 1$$
and also $$b_{n+1}\leq \frac{1+1/n+2/n^2}{n(n+1)} + \frac{n}{n+1} + \frac{2}{n+1} \\ = 1 + \frac{1}{n+1} + \frac{2}{(n+1)^2} - \frac{1-2/n-3/n^2-2/n^3}{(n+1)^2} \leq 1 + \frac{1}{n+1} + \frac{2}{(n+1)^2}$$
whenever $n\geq 4$. Taking the limit on both sides it follows $$\lim_{n\rightarrow \infty} b_n = 1 \, .$$ Next we formally write $b_n$ as an asymptotic expansion $$b_n = 1 + \sum_{k=1}^\infty \frac{c_k}{n^k}$$ and insert it into $$(n+1)b_{n+1}b_n - \frac{b_n^2}{n} - 2b_n - n = 0$$ which gives after some extensive algebra
$$0 = 2c_1 - 1 \\
+ \sum_{m=1}^\infty \frac{1}{n^m} \left\{ \sum_{k=0}^m (c_{m-k} + c_{m+1-k}) \sum_{l=0}^k \binom{-l}{k-l} c_l + \sum_{l=0}^{m+1} \binom{-l}{m+1-l} c_l - \sum_{k=0}^{m-1} c_k c_{m-1-k} - 2c_m \right\} \, .$$
Setting the coefficient of each power $n^{-m}$ to zero, iteratively gives a linear equation for $c_{m+1}$ ($m=1,2,3,...$) in terms of $c_0=1$ and $c_k$ ($k=1,2,...,m$). The coefficient of $n^0$ already gives $c_1=1/2$.
The first higher coefficients read: $c_2=5/8, c_3=13/16, c_4=155/128, c_5=505/256$. The denominator seems to follow a power of $2$ pattern.
| {
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"url": "https://math.stackexchange.com/questions/3801405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Proving $(a^2 + 1)(b ^2 + 1)(c ^2 + 1) ≥ 2(ab + bc + ca)$ where $a,b,c$ are real numbers. The inequality above seems very compelling for the pqr-method.
So this was my attempt-
$$ LHS = (a^2 + 1)(b^2 + 1)(c^2 + 1) = 1 + a^2 + b^2 + c^2 + a^2b^2 + b^2c^2 + c^2a^2 + a^2b^2c^2 $$
Now substituting $p = a+b+c$ , $q = ab+bc+ca$ and $r = abc$.
$$ LHS = 1 + p^2 - 2q + q^2 - 2pr + r^2 \geq 2q \Rightarrow 1 + p^2 + q^2 + r^2 \geq 4q + 2pr $$
It's quite well-known that $p^2\geq 3q$ and $q^2\geq 3pr$. So,
$$ 1 + 3q + 3pr + r^2 \geq 4q + 2pr \Rightarrow 1 + pr + r^2 \geq q $$
But I don't know how to prove it. It can also be seen that $a\ge b\ge c$, but I can't exploit symmetry.
Any help is thankfully welcome.
| Because $2(ab + bc + ca) \leqslant 2(|a||b| + |b||c| + |c||a|)$ and
$$(a^2 + 1)(b ^2 + 1)(c ^2 + 1) = (|a|^2 + 1)(|b| ^2 + 1)(|c| ^2 + 1),$$
so we need to prove the inequality when $ a,\,b,\,c \geqslant 0.$
Indeed, easy to check $3t^2 \geqslant 3t-1.$ Now, using the AM-GM we have
$$(a^2 + 1)(b ^2 + 1)(c ^2 + 1) \geqslant a^2 + b^2 + c^2 + 1 + a^2b^2 + b^2c^2 + c^2a^2$$
$$ \geqslant a^2+b^2+c^2+1+3\sqrt[3]{(abc)^4}$$
$$ \geqslant a^2+b^2+c^2+3\sqrt[3]{(abc)^2}.$$
Therefore we will show that
$$a^2+b^2+c^2+3\sqrt[3]{(abc)^2} \geqslant 2(ab+bc+ca).$$
Which is very known (here, here).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find all integers such that polynomial $x^4+n$ is reducible in $\mathbb{Z}[x]$ In my studies of abstract algebra and polynomials, I have the following question:
We are asked to find all integers $n$ such that the polynomial $x^4+n$ is reducible in $\mathbb{Z}[x]$.
The only thing I can think of is Eisenstein's criterion but that just tells me for $n$ that have a prime factor of power 1, it is irreducible. How would I handle the rest? I thank all helpers.
| Let's assume the polynomial is reducible. Then we can see that it must have a quadratic factor - for if it had a linear factor, it must be in form $x-p$, but then $p$ is a root and so is the $-p$ (since the polynomial has only even exponents), and so $(x-p)(x+p)=x^2-p^2$ is a quadratic factor.
So assume a generic factorization $x^4+n=(x^2+ax+b)(x^2+cx+d)$. Comparing the coefficients, we have $a+c=0$, $ac+b+d=0$, $ad+bc=0$ and $bd=n$. Hence $c=-a$, and so $0=ad+bc=ad-ab=a(d-b)$. So either $a=0$, or $d=b$.
If $a=0$, then $c=0$ and $b+d=0$. Thus we have $n=-b^2$ and $x^4+n=(x^2-b)(x^2+b)$.
If $d=b$, then $-a^2+2b=0$, and so $a=2k$ is even. But then $-4k^2+2b=0$, so $-2k^2+b=0$, and so $b=2l$ is even. So, $-k^2+l=0$, thus $a=2k$, $b=2k^2=d$. Thus we have $n=bd=4k^4$ and $x^4+n=(x^2+2kx+2k^2)(x^2-2kx+2k^2)$.
So we have found that if $x^4+n$ is reducible, then necessarily $n=-k^2$ or $n=4k^4$, and in both cases we have found these factorizations, so it is also sufficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to show the concavity of a function with an undefined point? Look at this function: $f(x)=\left\{\begin{array}{ll}
\frac{x(1-x^2)}{1-x^3}\\
2/3
\end{array}\right.$. Here $2/3=\lim_{x\to1}\frac{x(1-x^2)}{1-x^3}$. I can show that the second derivative of $\frac{x(1-x^2)}{1-x^3}$ is non-positive for $0\leq x<1$ and $x>1$; however, $\frac{x(1-x^2)}{1-x^3}$ is undefined at $x=1$. So in this case, how to show $f(x)$ is concave rigorously? (I think a point won't change the concavity of a function, but I don't know how to show it.)
| If $x\neq1$, then
$$f(x)=\frac{x(1-x^2)}{1-x^3}=\frac{x(1-x)(1+x)}{(1-x)(1+x+x^2)}=\frac{x(1+x)}{1+x+x^2}.$$
And plugging in $x=1$ into $\dfrac{x(1+x)}{1+x+x^2}$ also yields $2/3$.
This means that the given function is in fact $f(x)=\dfrac{x(1+x)}{1+x+x^2}$ for all real numbers. Now you can the second derivative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)? I am trying to evaluate the following limit:
$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$
Approach #1
$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sqrt{x+2} + \frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}}\cdot\frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{\sqrt{x^2-2x}-\sqrt{2x-4}} = \\ \sqrt{x+2} + \frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{x-2} $
But I still end up at the indefinite form $\frac12 + \frac{0}{\infty}$
My Approach #2
$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot\frac{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}= \frac1{\sqrt{x-2}}\frac{x^2-x+2\sqrt{2x}-2}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}$
Which also seems to be a dead end.
Any ideas on how to evaluate this?
| Since the limit is defined for $x\to 2^+$, let $y^2=x-2\to 0$ then
$$\lim _{x\to 2}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)=\lim _{y\to 0}\left(\frac{\sqrt{y^2(y^2+4)}+\sqrt{y^2+2}-\sqrt{2}}{y}\right)=$$
$$\lim _{y\to 0}\left(\frac{\sqrt{y^2(y^2+4)}+\sqrt{y^2+2}-\sqrt{2}}{y}\right)=\lim _{y\to 0}\frac{\sqrt{y^2(y^2+4)}}{y}+\lim _{y\to 0}\frac{\sqrt{y^2+2}-\sqrt{2}}{y}=$$
$$=\lim _{y\to 0}\sqrt{(y^2+4)}+\lim _{y\to 0}\frac{\sqrt{y^2+2}-\sqrt{2}}{y}=2+0=2$$
indeed by definition of derivative
$$f(y)=\sqrt{y^2+2} \implies f'(y)=\frac{2y}{2\sqrt{y^2+2}}$$
$$\lim _{y\to 0}\frac{\sqrt{y^2+2}-\sqrt{2}}{y}=f'(0)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum of series without differentiation Given a series $\sum_{i > 0}\frac{i^2}{z^i}$, and $\sum_{i > 0}\frac{i}{z^i} = \frac{z}{(z - 1)^2}$ I need to find the sum
My method does not require differentiation but there is a difficulty.
Let $S = \frac{1^2}{z} + \frac{2^2}{z^2} + \frac{3^2}{z^3} + ... + \frac{i^2}{z^i}$
Let $zS = 1 + \frac{2^2}{z} + \frac{3^2}{z^2} + ... + \frac{i^2}{z^{i - 1}}$
Thus, $zS - S = 1 + \frac{2^2 - 1^2}{z} + \frac{3^2 - 2^2}{z^2} + ... + \frac{i^2 - (i - 1)^2}{z^{i - 1}} - \frac{i^2}{z^i}$
Thus, $(z - 1)S = 1 + \frac{3}{z} + \frac{5}{z^2} + ... + \frac{2i - 1}{z^{i - 1}} - \frac{i^2}{z^i}$
My question is how can I proceed? The numerator of each term is not 1 so I cannot use any formula to calculate the sum.
| You have this
$(z - 1)S
= 1 + \frac{3}{z} + \frac{5}{z^2} + ... + \frac{2i - 1}{z^{i - 1}} - \frac{i^2}{z^i}
$
or, in summation notation,
$(z - 1)S
= \sum_{k=0}^{i-1} \dfrac{2k+1}{z^k}- \frac{i^2}{z^i}
$.
We can now split this into sums
we already know:
$\begin{array}\\
(z - 1)S
&= \sum_{k=0}^{i-1} \dfrac{2k+1}{z^k}- \dfrac{i^2}{z^i}\\
&= \sum_{k=0}^{i-1} \dfrac{2k}{z^k}+\sum_{k=0}^{i-1} \dfrac{1}{z^k}- \dfrac{i^2}{z^i}\\
&= 2\sum_{k=0}^{i-1} \dfrac{k}{z^k}+\sum_{k=0}^{i-1} \dfrac{1}{z^k}- \dfrac{i^2}{z^i}\\
\end{array}
$
You can now plug in
the known summations.
More generally,
if
$S_m(z)
=\sum_{k=0}^{\infty} \dfrac{k^m}{z^k}
$,
then
$S_0(z)
=\sum_{k=0}^{\infty} \dfrac{1}{z^k}
=\dfrac{1}{1-1/z}
=\dfrac{z}{z-1}
$
and,
for $m \ge 1$,
$S_m(z)
=\sum_{k=1}^{\infty} \dfrac{k^m}{z^k}
$,
$zS_m(z)
=\sum_{k=1}^{\infty} \dfrac{k^m}{z^{k-1}}
=\sum_{k=0}^{\infty} \dfrac{(k+1)^m}{z^{k}}
$
so
$\begin{array}\\
(z-1)S_m(z)
&= zS_m(z)-S_m(z)\\
&=\sum_{k=0}^{\infty} \dfrac{(k+1)^m}{z^{k}}-\sum_{k=0}^{\infty} \dfrac{k^m}{z^k}\\
&=\sum_{k=0}^{\infty} \dfrac{(k+1)^m-k^m}{z^{k}}\\
&=\sum_{k=0}^{\infty} \dfrac{\sum_{j=0}^m \binom{m}{j}k^j-k^m}{z^{k}}\\
&=\sum_{k=0}^{\infty} \dfrac{\sum_{j=0}^{m-1} \binom{m}{j}k^j}{z^{k}}\\
&=\sum_{j=0}^{m-1} \binom{m}{j}\sum_{k=0}^{\infty} \dfrac{k^j}{z^{k}}\\
&=\sum_{j=0}^{m-1} \binom{m}{j}S_j(z)\\
\end{array}
$
so
$S_m(z)
=\dfrac1{z-1}\sum_{j=0}^{m-1} \binom{m}{j}S_j(z)
$
so that each
$S_m(z)
$
can be gotten in terms
of the
$S_j(z)
$
for $j < m$.
In your case,
the $2$ and the $1$ comes from
$(k+1)^2-k^2
=2k+1
$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Show that $x_{n+1}=x_n(2-ax_n)$ converges and find the limit
Let $a>0$ and $x_0\in I=\left (\frac{1}{2a}, \frac{3}{2a}\right )$. Show that the sequence $(x_n)$, $n\geq 0$, $$x_{n+1}=x_n(2-ax_n), \quad n \geq 0$$ converges. Which is the limit? Hint: Consider $\phi (x)=x(2-ax)$ and show that $\phi (I)\subset \left [\frac{3}{4a}, \frac{1}{a}\right ]$.
So that I understand that correctly, with the hint we want to show that $\phi(x)$ is bounded and monotonic, which means that the sequence converges?
We have the following: \begin{align*}\frac{1}{2a}<x< \frac{3}{2a} &\Rightarrow a\cdot \frac{1}{2a}<ax<a\cdot \frac{3}{2a}\Rightarrow \frac{1}{2}<ax< \frac{3}{2} \Rightarrow -\frac{3}{2}<-ax< -\frac{1}{2}\\ & \Rightarrow 2-\frac{3}{2}<2-ax< 2-\frac{1}{2}\Rightarrow \frac{1}{2}<2-ax< \frac{3}{2} \\ & \Rightarrow x\cdot \frac{1}{2}<x(2-ax)< x\cdot \frac{3}{2}\\ & \Rightarrow \frac{1}{2a}\cdot \frac{1}{2}<x\cdot \frac{1}{2}<x(2-ax)< x\cdot \frac{3}{2}< \frac{3}{2a}\cdot \frac{3}{2}\\ & \Rightarrow \frac{1}{4a}<x(2-ax)< \frac{9}{4a}\end{align*}
That is not the interval that we want to get. Do we have to do that maybe using derivatives?
| Complete the square on the right side to find
$$
1-ax_{n+1}=(1-ax_n)^2
$$
so that the convergence of $y_n=1-ax_n$ is very easy to discuss (subsequence of a geometric sequence).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\cos ( 2 \sin^{-1}( \frac{5}{ 13} )) $ The following problem is from the $8$th edition of the book Calculus, by James Stewart. It is problem number $9$ in section $6.6$.
Problem:
Find an exact value for the expression:
$$ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } $$
Answer:
\begin{align*}
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - \sin^2{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
%
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - 2 \sin^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
%
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - 2 \left( \frac{25}{13^2} \right) \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
%
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - \left( \frac{50}{13^2} \right) \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
\end{align*}
\begin{align*}
\cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
1 - \sin^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } \\
\cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &= 1 - \frac{25}{169} = \frac{169 - 25}{169} \\
\cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \frac{144}{169} \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - \left( \frac{50}{13^2} \right) \left( \frac{144}{169} \right) } \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ \frac{13^4 - 50(144)}{13^4} } \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ \frac{21361}{13^4} } \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \frac{ \sqrt{ 21361 } } { 169 }
\end{align*}
The book's answer is $\frac{119}{169}$ and SciLab matches the book. Where did I go wrong?
| In the second line there is a mistake!
$$\sin^22\alpha=4\sin^2\alpha\cos^2\alpha.$$
Now, $$\cos2\arcsin\frac{5}{13}=\sqrt{1-4\left(\frac{5}{13}\right)^2\left(\frac{12}{13}\right)^2}=\frac{119}{169}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $Q=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3}$ For $a,b,c\geqslant 0.$ Prove: $$\text{Q}=\frac{a^{3}+b^{3}+c^{3}}{(a+b)(b+c)(c+a)}+k\cdot \frac{(ab+bc+ca)}{(a+b+c)^{2}}\geqslant \frac{3}{8}+\frac{k}{3},$$
where $k={\frac {27}{8}}+\frac{9\sqrt{3}}{4}$ is a Root Of $64{k}^{2}-432k-243=0.$
By computer (Maple) I found this inequality is equivalent to$:$ $$\sum \Big[ a+b+ ( 1-\frac{\sqrt {3}}{2} ) c\Big] \Big[ 2(a+b)- ( 1+\sqrt {3} ) c \Big] ^{2} ( \,a-b \,) ^{2} \geqslant 0$$
But I hope for alternative proof$?$
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality it's a linear inequality of $w^3$, which says that it's enough to prove our inequality for an extreme value of $w^3$, which by $uvw$ happens in the following cases.
*
*$w^3=0$.
Let $c=0$, $b=1$ and $a^2+1=2ua$.
Thus, $u\geq1$ and we need to prove here that
$$\frac{a^2-a+1}{a}+\frac{ka}{(a+1)^2}\geq\frac{3}{8}+\frac{k}{3}$$ or
$$2u-1+\frac{k}{2u+2}\geq\frac{3}{8}+\frac{k}{3}$$ or
$$4u^2-\left(\frac{2k}{3}-\frac{5}{4}\right)u+\frac{k}{3}-\frac{5}{4}\geq0,$$ for which it's enough to prove that
$$\left(\frac{2k}{3}-\frac{5}{4}\right)^2-16\left(\frac{k}{3}-\frac{5}{4}\right)\leq0,$$ which is true for $k=\frac{27}{8}+\frac{9\sqrt{3}}{4}.$
*Two variables are equal.
Let $b=c=1$.
Thus, we need to prove that:
$$\frac{a^3+2}{2(a+1)^2}+\frac{k(2a+1)}{(a+2)^2}\geq\frac{3}{8}+\frac{k}{3}$$
or
$$\frac{4a^3-3a^2-6a+5}{8(a+1)^2}\geq\frac{k(a-1)^2}{3(a+2)^2}$$ or
$$(4a+5)(a+2)^2\geq(9+6\sqrt3)(a+1)^2$$ or
$$(2a+4-\sqrt3)(2a+1-\sqrt3)^2\geq0$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find numerical matrices to remove the same coefficients on the main diameter $M=\left(
\begin{array}{cccc}
\frac{1}{4}m_{11} & m_{12} & ... & m_{1n} \\
m_{12} & \frac{1}{4}m_{22} & ... & m_{2n} \\
\colon & \colon & & \colon\\
m_{1n} & ... & & \frac{1}{4}m_{nn}
\end{array}
\right)$
$M$ is a Symmetric matrix That is, M has the $\frac{1}{4}$ coefficient along its main diagonal. How can I find the $M_1$ matrix?
$M_1=\left(
\begin{array}{cccc}
m_{11} & m_{12} & ... & m_{1n} \\
m_{12} & m_{22} & ... & m_{2n} \\
\colon & \colon & & \colon\\
m_{1n} & ... & & m_{nn}
\end{array}
\right)$
Can it be done by simple transformations?? by multiply from left and right by the specific matrix.
| Consider
$$M=\left(
\begin{array}{cccc}
\frac{1}{4} & \frac{1}{4} \\
\frac{1}{4} & \frac{1}{4}
\end{array}
\right)$$
For this matrix, you have
$$M_1=\left(
\begin{array}{cccc}
1 & \frac{1}{4} \\
\frac{1}{4} & 1
\end{array}
\right)$$
If there existed $P$, $Q$ such that $PMQ=M_1$, then taking the determinant, you would get, because $\mathrm{det}(M)=0$, that
$$\mathrm{det}(M_1)=0$$
which is not true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the probability of rolling 2 before a second odd? Consider a game of dice:
*
*You win if you roll $2$.
*You lose if you roll two odds ( need not be consecutive ).
*If you roll a $4$ or $6$, you keep playing as you have neither lost nor won.
Eg: $1$, $3$ is a loss. $1$, $6$, $4$, $3$ is a loss.
What's the probability of winning?
The answer is $7/16$.
My approach:
Suppose you roll $2$ on the first go. You win and probability of rolling $2$ is $1/6$.
Suppose you roll evens other than $2$ before you roll $2$. The probability of rolling an even other than $2$ is $1/3$. So for the probability of this event, we have,
$$\small
(1/3)(1/6) + (1/3)(1/3)(1/6) + (1/3)(1/3)(1/3)(1/6) + ... = (1/6)[(1/3)/(1-1/3)] = 1/12
$$
Now suppose we roll evens and one odd before rolling $2$. Probability of rolling an odd is $1/2$. So the probability of this event is $(1/2)(1/6)\left[1 + (1/3) + (1/3)(1/3) + \cdots + \cdots\right] = 1/8$
So required probability is $1/6 + 1/12 + 1/8 = 3/8$.
Obviously, my approach is wrong but I don't understand why. Please help me figure this out. Thanks :)
| Your approach is right, but you haven´t calculated correctly the probability of rolling evens and one odd before rolling 2, since you´re supposing you always get the odd in the first roll.
If you get a 2 in the second roll you had (odd, 2) with probability of $\frac{1}{2}*\frac{1}{6}$
Getting 2 in the third roll might be either (odd, even, 2) or (even,odd, two), with probability of $\frac{1}{2}*\frac{1}{3}*\frac{1}{6} + \frac{1}{3}*\frac{1}{2}*\frac{1}{6} = 2*(\frac{1}{2}*\frac{1}{3}*\frac{1}{6})$
In the fourth roll you have (odd,even,even,2) , (even,odd,even,2) or (even,even,odd,2) , so the probability is $3*(\frac{1}{2}*\frac{1}{3}*\frac{1}{3}*\frac{1}{6})$
So the real probability of rolling evens and one odd before rolling 2 is
$\frac{1}{2}*\frac{1}{6}*[1+2*\frac{1}{3}+3*\frac{1}{3}*\frac{1}{3}+...]$
Let´s call $S=1+2*\frac{1}{3}+3*\frac{1}{3}*\frac{1}{3}+...$ and $R=1+\frac{1}{3}+\frac{1}{3}*\frac{1}{3}+...=\frac{3}{2}$
Now you can realize that $S-R=\frac{1}{3}*S$ so $S=\frac{3}{2}*R=\frac{9}{4}$
Then, the probability of rolling evens and one odd before rolling 2 is
$\frac{1}{2}*\frac{1}{6}*\frac{9}{4}=\frac{3}{16}$
Finally, we have $\frac{1}{6}+\frac{1}{12}+\frac{3}{16}=\frac{7}{16}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stuck on proof of $\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$ for $n \geq 3$ using induction I'm relatively familiar with induction, I'm just stuck on this step. I am currently taking Introduction to Abstract Math, and have taken Calculus I and II.
$P(n)$ is
$$\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$$
Prove P(n) is true for all $n, \{n \in \mathbb{Z}^+ | \: n \geq 3 \}$
Basis Step
Show that $P(3)$ is true.
$$\sum_{k=1}^{3} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{3}$$
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2}\leq \frac{7}{4} - \frac{1}{3}$$
$$\frac{49}{36} \leq \frac{17}{12}$$
$\frac{49}{36} \leq \frac{17}{12}$ is true, therefore $P(3)$ is true.
Induction Step
Assume $P(n)$ is true for some $n, \{n \in \mathbb{Z}^+ | \: n \geq 3 \}$. Show that $P(n+1)$ holds true, with $P(n+1)$ being defined as $\sum_{k=1}^{n+1} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n+1}$.
$$\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$$
$$\sum_{k=1}^{n} \frac{1}{k^2} + \frac{1}{(n+1)^2} \leq \frac{7}{4} - \frac{1}{n} + \frac{1}{(n+1)^2}$$
That's where I got stuck. What would be the next step?
| You know that
$$\sum_{k=1}^n\frac1{k^2}+\frac1{(n+1)^2}\le\frac74-\frac1n+\frac1{(n+1)^2}\,,\tag{1}$$
and you want to show that lefthand side of $(1)$ is at most $\frac74-\frac1{n+1}$; the most straightforward way to do this is to show that
$$\frac74-\frac1n+\frac1{(n+1)^2}\le\frac74-\frac1{n+1}\,,$$
which amounts to showing that
$$\frac1{(n+1)^2}\le\frac1n-\frac1{n+1}\,.$$
And since $\frac1n-\frac1{n+1}=\frac1{n(n+1)}$, this is clear.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a_k$ be the coefficient of $x^k$ in the expansion of :- $(x+1) + (x+1)^2 + (x+1)^3 + (x+1)^4 + ... + (x+1)^{99}$ . Find $[\frac{a_4}{a_3}]$ Let $a_k$ be the coefficient of $x^k$ in the expansion of :- $(x+1) + (x+1)^2 + (x+1)^3 + (x+1)^4 + ... + (x+1)^{99}$ . Find $[\frac{a_4}{a_3}].$
What I Tried :- Honestly there's nothing I can do to start here . The problem is that I don't know how to find the coefficient of any term of any expression (assuming the term is there in the expression) . Once I know that type of method, this problem will be straight-forward for me .
So can someone say me some type of method ?
| You can use the formula for the sum of a geometric progression and then the binomial theorem to find that your sum is
$$
\frac{{1 - (x + 1)^{100} }}{{1 - (x + 1)}} - 1 = \frac{{(x + 1)^{100} - 1}}{x} - 1 = \left( {\sum\limits_{k = 1}^{100} {\binom{100}{k}x^{k - 1} } } \right) - 1 \\ = 99 + \sum\limits_{k = 2}^{100} {\binom{100}{k}x^{k - 1} } = 99 + \sum\limits_{k = 1}^{99} {\binom{100}{k+1}x^{k} }.
$$
Thus
$$
\frac{{a_4 }}{{a_3 }} = \frac{\binom{100}{5}}{\binom{100}{4}} = \frac{{96}}{5} = 19.2.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$; find my mistake solve :$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$
My try:
$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}\\\frac{y'}{y^{2}}+\frac{2}{y(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\\left(t\frac{1-x}{1+x}\right)'=\left(x-1\right)\frac{1-x}{1+x}\\\left(t\frac{1-x}{1+x}\right)=\frac{1}{2}(7+6x-x^{2}-8log(1+x))+c\\t=\frac{(1+x)(c+1/2(7+6x-x^{2}-8log(1+x))))}{1-x}\\y=-\frac{1-x}{(1+x)(c+1/2(7+6x-x^{2}-8log(1+x))))}\\\\\\\\\\$
| You made a mistake for the integrating factor. Since $\int-\frac{2}{x^2-1}dx=\ln(\frac{x+1}{1-x})+C$, your $4$th line should be:
$$(t\frac{1+x}{1-x})'=(x-1)\frac{1+x}{1-x}=-(1+x)$$
$$(t\frac{1+x}{1-x})=-x-\frac{x^2}{2}+c$$
$$t=\frac{(1-x)(c-x-\frac{x^2}{2})}{1+x}=\frac{(x-1)(c_{1}+x(x+2))}{2(1+x)}$$
where $c_{1}=-2c.$ Thus:
$$y=-\frac{2(1+x)}{(x-1)(c_{1}+x(x+2))}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why isn't $x^2+x+1$ a factor of $x^{12}+x^6+1$? When we solve the equation
$$x^{12}+x^6+1=0$$ we obtain $2$ solutions that also satisfy $$x^2+x+1=0$$
namely $-\frac{1}{2}\pm i\frac{\sqrt3}{2}$.
Shouldn't this imply that $x^2+x+1$ is a factor of $x^{12}+x^6+1$? However, the fully factorsied form of
$x^{12}+x^6+1$ is $$(x^6-x^3+1)(x^6+x^3+1)$$
The reason I think that it should be a factor is that when we have, for a function $f(x)$, (here a polynomial,) $f(a)=0$ then we know that $(x-a)$ is a factor of $f(x)$.
If we multiply $(x-(-\frac{1}{2}+ i\frac{\sqrt3}{2}))$ by $(x-(-\frac{1}{2}- i\frac{\sqrt3}{2}))$ which are both factors of $x^{12}+x^6+1$ we get $x^2+x+1$. It seems to me that this should apply to the polynomials above. I have a feeling that my logic is specious, but I'm not entirely certain why; I think the answer to my problem may be that $x^2+x+1$ is factor of $x^{12}+x^6+1$, but only if the other factor has complex coefficients, but I'm not sure. Thank you for your help.
EDIT
Oh wow, I'm so sorry for making such a stupid mistake everyone, thanks for correcting me (see numerous comments and answers). I was looking at a few polynomials simultaneously and got mixed up. Thanks for your help.
| I just want to note a technique based on my comment to the original post where a factor of $x^2+x+1$ can be shown longhand for $x^{2n}+x^n+1$ where $n$ is not a multiple of $3$.
Suppose $n=3m+1$ then $$x^{6m+2}+x^{3m+1}+1=x^{6m+2}-x^2+x^{3m+1}-x+x^2+x+1=$$$$=x^2\left(x^{6m}-1\right)+x\left(x^{3m}-1\right)+x^2+x+1$$ and the two bracketed terms are divisible by $x^3-1=(x-1)(x^2+x+1)$
With $n=3m+2$ we have similarly $$x^{6m+4}+x^{3m+2}+1=x^{6m+4}-x+x^{3m+2}-x^2+x^2+x+1=$$$$=x\left(x^{6m+3}-1\right)+x^2\left(x^{3m}-1\right)+x^2+x+1$$ and a factor $x^3-1$ is there for the two bracketed terms.
The various other methods are more efficient, but this is direct.
| {
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Find all the integer pairs $(x, y)$ which satisfy the equation $x^5-y^5=16xy$ I just came across the following question:
Find all the integer pairs $(x, y)$ which satisfy the equation $x^5-y^5=16xy$
I solved it as follows:
$x=y=0$ obvious solution. If $xy\neq0$, let $d=gcd(x, y)$ and we write $x=da$, $y=db$, $a, b\in \Bbb{Z}$ with $(a, b)=1$. Then, the given equation is:
$$d^3a^5-d^3b^5=16ab$$
So, by the above equation, $a$ divides $d^3b^5$ and hence $a$ divides $d^3$. Similarly $b$ divides $d^3$. Since $(a, b)=1$ we have that $ab$ divides $d^3$, so $d^3=abr$ with $r\in \Bbb{Z}$. Then the above equation becomes $abra^5-abrb^5=16ab$, so $r(a^5-b^5)=16$.
Hence, the difference $a^5-b^5$ must divide $16$. If $|(a^5-b^5)|\le2$ we have that $(x, y)=(-2, 2)$ is a solution. Otherwise $$|a^5-b^5|=|(x+1)^5-b^5|\ge |(x+1)^5-x^5|=|5x^4+10x^3+10x^2+5x+1|\ge31$$ which is impossible.
So only solutions are $(x, y)=(0, 0)$ or $(-2, 2)$.
I believe that this solution is not at all intuitive nor simple. Could you please post a more intuitive and simple solution where you are explaining your intuition on every step?
| The following is neither intuitive nor simple, but it does give a different approach to the proof.
If $xy\not=0$, let $p$ be an odd prime and write $x=p^ru$ and $y=p^sv$ with $p\not\mid uv$. From $p^{5r}u^5-p^{5s}v^5=16uvp^{r+s}$, we see we cannot have $r=s\not=0$, so we either have $5r=r+s$ or $5s=r+s$. This means that we can write $x$ and $y$ in the form $x=2^aA^4B$ and $y=2^bAB^4$ with $A$ and $B$ relatively prime odd numbers. But we now have $2^{5a}A^{20}B^5-2^{5b}A^5B^{20}=2^{a+b+4}A^5B^5$, from which we obtain
$$2^{5a}A^{15}-2^{5b}B^{15}=2^{a+b+4}$$
so we must now have $a=b$ (since otherwise the left hand side factors into a power of $2$ times an odd number not equal to $1$), which implies $2^{5a}(A^{15}+B^{15})=2^{2a+4}$, or
$$2^{3a}(A^{15}-B^{15})=2^4$$
The only $15$th powers of odd numbers that differ by a small power of $2$ come from $A=1$ and $B=-1$, so the only solution with $xy\not=0$ is $x=-2$ and $y=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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Prove $(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^{-1}$ is an algebraic integer Let $b = (4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^{-1}$, then $1 = b(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})$, and $1 - 4b = b(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})$.
Therefore $1 - 12b + 48b^2 - 64b^3 = (1 - 4b)^3 = b^3(3 \cdot 3^{1/3} + 2 \cdot 3^{2/3})^3$
$= b^3(81 + 18 \cdot 3^{2/3} \cdot 3^{2/3} + 12 \cdot 3^{1/3} \cdot 3^{4/3} + 72) = b^3(153 + 18 \cdot 3 \cdot 3^{1/3} + 18 \cdot 2 \cdot 3^{2/3})$
$= 18b^3(4 + 3 \cdot 3^{1/3} + 2 \cdot 3^{2/3}) + 81b^3 = 18b^3(b^{-1}) + 81b^3= 18b^2 + 81b^3$. Therefore $b$ is a root of $1 - 12b + 48b^2 - 64b^3 - 18b^2 - 81b^3 = - 145b^3 + 30b^2 - 12b + 1$
My problem is $145b^3 - 30b^2 + 12b - 1$ is not monic and is also not reducible, since if it were reducible being of degree $3$ we could factor out a root, which we can't by the rational roots theorem,
Any advice on what I am doing wrong?
| I think, there is a mistake in your computations.
We obtain:
$$4b-1+3b\sqrt[3]3+2b\sqrt[3]9=0$$ or
$$(4b-1)^3+81b^3+72b^3-3(4b-1)\cdot18b^2=0$$ or $$b^3+6b^2+12b-1=0.$$
| {
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Application of AM-GM inequality to specific contest problem Suppose that $x,y\in [0,1]$. Prove that $\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\leq \frac{2}{\sqrt{1+xy}}.$
I suppose that this problem can be solved by some application of AM-GM inequality. I was trying to do the following: since $xy\leq \frac{x^2+y^2}{2}$ then $\frac{2}{\sqrt{1+xy}}\geq \frac{2}{\sqrt{1+x^2/2+y^2/2}}$. But the inequality $\frac{2}{\sqrt{1+x^2/2+y^2/2}}\geq \frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}$ is obviously false. So I guess we have to use something which is non-trivial.
Would be grateful if someone can show the solution.
I have spent probably 2-3 hours and did not get it.
| We have
$$\left(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\right)^2 = \frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{2}{\sqrt{(1+x^2)(1+y^2)}}$$
Using the AM-GM we have
$$\frac{2}{\sqrt{(1+x^2)(1+y^2)}} \leqslant \frac{1}{1+x^2}+\frac{1}{1+y^2}.$$
Therefore, we need to prove
$$\frac{1}{1+x^2}+\frac{1}{1+y^2}\leqslant \frac{2}{1+xy},$$
equivalent to
$$\frac{(xy-1)(x-y)^2}{(1+x^2)(1+y^2+1)(1+xy)} \leqslant 0.$$
Which is true.
| {
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A function inequality about $e^x$ and $\ln x$ If for any $x \in (1,+\infty)$，there is the inequality:
$$x^{-3} e^{x}-a \ln x \geq x+1$$
Find the value range of $a$ .
And I tried constructing the function $f(x)=x^{-3} e^{x}-a \ln x - x-1$ and deriving it, but to no avail.
| $$\max{a}=\min_{x>1}\frac{\frac{e^x}{x^3}-x-1}{\ln{x}}.$$
Let $g(x)=\frac{\frac{e^x}{x^3}-x-1}{\ln{x}}.$
Thus, for $e^x=x^3$ or $x=3\ln{x}$ we obtain: $$g'(x)=\frac{x^4+x^3+e^x(x\ln{x}-3\ln{x}-1)-x^4\ln{x}}{x^4\ln^2x}=$$
$$=\frac{x^4+x^3+x^3(x\ln{x}-3\ln{x}-1)-x^4\ln{x}}{x^4\ln^2x}=\frac{x^4-3x^3\ln{x}}{x^4\ln^2x}=0.$$
Also, for $e^x=x^3$ we obtain: $$\frac{\frac{e^x}{x^3}-x-1}{\ln{x}}=-\frac{x
}{\ln{x}}=-3.$$
Now, we'll prove that $$\frac{\frac{e^x}{x^3}-x-1}{\ln{x}}\geq-3$$ or $$\frac{e^x}{x^3}-x-1+3\ln{x}\geq0,$$ where the minimum occurs for $x=x_0$, where $x_0>1$ is a root of the equation: $e^x=x^3$.
Indeed, let $h(x)=\frac{e^x}{x^3}-x-1+3\ln{x}.$
Thus, $$h'(x)=\frac{e^x}{x^3}-\frac{3e^x}{x^4}-1+\frac{3}{x}=\frac{(x-3)(e^x-x^3)}{x^4}.$$
Now, the equation $e^x=x^3$ or $x=3\ln{x}$ has two roots maximum because $\ln$ is a concave function.
But $3\ln3-3>0$, $3\ln1-1<$ and $3\ln5-5<0$ which says that the equation $e^x=x^3$ has two roots $1<x_0<3$ and $x_1>3$ exactly and easy to see that $x_{max}=3$ and $x_{min_1}=x_0$, $x_{min_2}=x_1.$
Thus, $$h(x)\geq h(x_0)=h(x_1)=0$$ and we got the answer:
$$(-\infty,-3]$$
| {
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Prove that $83$ divides $p$ .
Let $p$ and $q$ be the positive integers such that $1 - \frac{1}{2} + \frac{1}{3} - ... - \frac{1}{54} + \frac{1}{55} = \frac{p}{q}$ . Prove that $83$ divides $p$ .
What I Tried :- To me it does not actually seem like an easy problem for me. (I guess I couldn't figure out the idea)
This expression I found is same as $$\displaystyle\sum_{n=1}^{55} \frac{(-1)^{n+1}}{n}$$
And calculating this in Wolfram Alpha gives the answer to be this in the simplest form :- $$\displaystyle\sum_{n=1}^{55} \frac{(-1)^{n+1}}{n}=\frac{115328583812490186710549}{164249358725037825439200}$$
Surprisingly, I found that $83$ really divides the numerator of the fraction. Now I got puzzled, how will you show this without using Wolfram Alpha or a Calculator?
Can Anyone Help?
| Given expression is
$$
1 - \frac{1}{2} + \frac{1}{3} - ... - \frac{1}{54} + \frac{1}{55} \\
= 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{55} - 2(\frac{1}{2} + \frac{1}{4} + ... + \frac{1}{54} ) \\
= \frac{1}{28} + \frac{1}{29} ... + \frac{1}{55} \\
= (\frac{1}{28} + \frac{1}{55}) + (\frac{1}{29} + \frac{1}{54}) + \ldots \\
= 83 (\frac{1}{28\cdot55} + \frac{1}{29\cdot54} + \ldots) = \frac{p}{q}
$$
from which it follows.
| {
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Proving that for any three integers $a,b,c$ there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer Prove that for any three integers $a,b,c$ there exists a positive integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer.
In order to solve this problem I have tried looking at the expression under the radical modulo n. Thus I want to find n such that c is a quadratic non-residue modulo n. For example, if c = 2 (mod 3), since 2 is a non-residue mod 3 we may take n to be 3 and thus the expression is never a perfect square. I need a way to do this for arbitrary c, which I could not find on my own.
| John Omielan has already provided a nice answer using $\text{mod}\ 4$.
Here is another approach using $\text{mod}\ 4$.
Let $f(n):=n^3+an^2+bn+c$.
Let us prove that at least one of $f(1),f(2),f(3),f(4)$ is not a square number.
Proof :
Let us consider in $\text{mod}\ 4$.
Suppose that $f(1),f(2),f(3),f(4)$ are square numbers.
Then, we have $f(i)\equiv 0,1$ for $i=1,2,3,4$.
Since $f(2)+f(4)\equiv 2b+2c$, we see that $f(2)+f(4)$ is even. So, we have $f(2)\equiv f(4)$.
Case 1 : $f(2)\equiv f(4)\equiv 0$
Then $2b+c\equiv c\equiv 0$ implies $b\equiv 0,2$ and $c\equiv 0$. If $b\equiv c\equiv 0$, then $f(1)\equiv 1+a\equiv 0,1$ implies $a\equiv 3,0$ for which $f(3)\equiv -1+a\equiv 2,3$, a contradiction. If $b\equiv 2$ and $c\equiv 0$, then $f(1)\equiv a-1\equiv 0,1$ implies $a\equiv 1,2$ for which $f(3)\equiv a+1\equiv 2,3$, a contradiction.
Case 2 : $f(2)\equiv f(4)\equiv 1$
Then $2b+c\equiv c\equiv 1$ implies $b\equiv 0,2$ and $c\equiv 1$. If $b\equiv 0$ and $c\equiv 1$, then $f(1)\equiv a+2\equiv 0,1$ implies $a\equiv 2,3$ for which $f(3)\equiv a\equiv 2,3$, a contradiction. If $b\equiv 2$ and $c\equiv 1$, then $f(1)\equiv a\equiv 0,1$ for which $f(3)\equiv a-2\equiv 2,3$, a contradiction.
So, we see that at least one of $f(1),f(2),f(3),f(4)$ is not a square number.
| {
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Minimum value of $f(x,y,z) = x^z + y^z - (xy)^{\frac{z}{4}}, x > 0, y > 0, z > 0$ $$f(x,y,z) = x^{z}+y^{z}-(xy)^{\frac{z}{4}}$$
for all real positive numbers x, y, z
Does anyone have a clue to find the minimum value of $f(x,y,z)$?
I honestly don't know where to start the solution, I just come up with $AM \geq GM$
$\frac{x^z + y^z}{2} \geq \sqrt{{x}^{z}{y}^{z}} \\ x^z + y^z \geq 2{x}^{\frac{z}{2}}{y}^{\frac{z}{2}}$
With the equality holds if and only if $x^z = y^z$
$x^{z}+y^{z}-(xy)^{\frac{z}{4}} \\ \geq 2{x}^{\frac{z}{2}}{y}^{\frac{z}{2}} - (xy) ^{\frac{z}{4}} \\ = (xy)^{\frac{z}{4}}(2(xy)^{\frac{z}{4}} - 1)$
Set x^z = y^z for minimum value
$(x^{\frac{z}{2}})(2x^{\frac{z}{2}} - 1)$
From here, I set the function $\leq$ 0.
Since x > 0, It's obvious that $x^{\frac{z}{2}}$ can't be $\leq$ 0
$2x^{\frac{z}{2}} - 1 \leq 0 \\ (\sqrt{2} \cdot {x}^{\frac{z}{4}} + 1)(\sqrt{2} \cdot {x}^{\frac{z}{4}} - 1) \leq 0 \\ -\frac{1}{\sqrt{2}} \leq x^{\frac{z}{4}} \leq \frac{1}{\sqrt{2}}$
Since x > 0
$0 < x^{\frac{z}{4}} \leq \frac{1}{\sqrt{2}}$
I don't know what to do after this, I probably did a wrong method to solve the problem. Does anyone have a hint to solve it?
| For a minimum $f_x=zx^{z-1}-(z/4)y^{z/4}x^{z/4-1}=0$ so $x^{3z/4}-y^{z/4}/4=0$.
As $f(x,y,z)=f(y,x,z)$ we also have $y^{3z/4}-x^{z/4}/4=0$ and equating yields $(4x^{3z/4})^3-x^{z/4}/4=0$. Thus $256x^{2z}-1=0$ which gives $x^z=y^z=1/16$ as $x,y,z>0$.
Hence the minimum value is $1/16+1/16-(1/16\cdot1/16)^{1/4}=-1/8$.
| {
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Roots and point of inflections Let $b$ and $c$ be the roots of a four degree polynomial. Also $x=b$ and $x=c$ are the real points of inflection of this four degree polynomial. If the other two roots of the polynomial be $a$ and $d$ where $a<b<c<d$ then prove that $$\int_{a}^{d}f(x)dx=0$$
My Attempt
I framed the equation $f(x)=k\left(x^4-10x^3+24x^2+5x-20\right)=k\left(x^2-5x+4\right)\left(x^2-5x-5\right)$ where $1$ and $4$ are the points of inflection as well as the roots and on integrating $f(x)$ I obtained $$\int_{\frac{5-3\sqrt{5}}{2}}^{\frac{5+3\sqrt{5}}{2}}f(x)dx=0$$But there must be a general way out.
One observation for the given polynomial is that $b+c=a+d$. Beyond this I am not able to generalize beyond this.
| If $f$ has degree $4$, $f''$ has degree $2$, so if $b$ and $c$ are inflection points (and thus roots of $f''$), we must have $f''(x) = k (x-b)(x-c)$ for some constant $k$. By integrating twice, we get $f$ as a polynomial of degree $4$ with coefficients of $x^1$ and $x^0$ arbitrary. But those coefficients will be determined by the requirement that $f(b)=f(c)=0$. So, given $b$ and $c$, $f$ is determined up to the constant factor $k$.
Now if $f(x)$ satisfies the conditions with inflection points $b$ and $c$, $f(A x + B)$ will satisfy them with inflection points $(b-B)/A$ and $(c-B)/A$. The other roots are simlarly transformed, and
$$\int_{(a-B)/A}^{(d-B)/A} f(Ax+B)\; dx = \frac{1}{A} \int_a^d f(t)\; dt$$
The conclusion is that if it's true for one particular example, it's true in general. You might as well take something conveniently symmetric such as $b=-1$, $c=1$.
| {
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Simplify $(1+\sqrt{3}) \cdot \sqrt{2-\sqrt{3}}$ Can someone help me simplify $(1+\sqrt{3})\times\sqrt{2-\sqrt{3}}$?
The end result is $\sqrt{2}$, however, I honestly do not know how to get there using my current skills.
I asked a teacher/tutor and he proposed setting the expression equal to X and working backwards, squaring both sides so that:
$$X^2 = (1 + \sqrt{3})^2 \cdot (2 - \sqrt{3}) =(4+2\sqrt{3})(2-\sqrt{3}) =8-2\sqrt{3}^2 = 2 \require{cancel}$$
$$\Rightarrow X = \sqrt{2}$$
My main question is:
*
*What are the steps to simplify this without setting equal to X?
*I tried watching a youtube video, but had no success - is difference of squares applicable here?
Thanks!
| In general, $\sqrt{2 - \sqrt3}$ can be denested using the equation $$\sqrt{X\pm Y} = \sqrt{\frac{X + \sqrt{X^2-Y^2}}{2}} \pm \sqrt{\frac{X - \sqrt{X^2-Y^2}}{2}}.$$ (source)
Setting $X = 2$ and $Y = \sqrt3$ gives $X^2-Y^2=1$ and therefore $$\sqrt{2-\sqrt3} = \sqrt{\frac{2+1}{2}} - \sqrt{\frac{2-1}{2}} = \frac{\sqrt3 - 1}{\sqrt 2}.$$
Multiplying by $\sqrt3 + 1$ turns the numerator into $3-1 = 2$, and $\frac{2}{\sqrt2} = \sqrt2$.
This is longer than bringing $\sqrt3+1$ into the radical in this particular case, but works more generally.
| {
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$ z,w\in\mathbb{C},|z|=|w|=R\gt0$. Show that $\left(\frac{z+w}{R^2+zw}\right)^2+\left(\vcenter{\frac{z-w}{R^2-zw}}\right)^2\ge\frac1{R^2}$ Let $ z, w \in \mathbb{C} $ be such that $ |z| = |w| = R > 0 $. Show that
$ \left(\frac{z + w}{R^2 + zw}\right)^2 + \left(\frac{z - w}{R^2 - zw}\right)^2 \geq \frac{1}{R^2} $
Well, i could only proof that
$ u=\left(\frac{z + w}{R^2 + zw}\right) $ , $ v = \left(\frac{z - w}{R^2 - zw}\right) \in \mathbb{R} $
By showing that
$ u = \overline{u}$ and $ v = \overline{v}$
Which leads to
$ u^2 + v^2 = |u|^2 + |v|^2 $
But i can't find a way to compute $|u|$ or $|v|$
| We must assume that $zw \ne \pm R^2$ because the left-hand side is undefined otherwise. Then we can assume that $R=1$ because of the homogeneity of the inequality, this simplifies the calculation a bit.
You already noticed that the terms on the left-hand side are real numbers, so that
$$
\left(\frac{z + w}{1 + zw}\right)^2 +
\left(\frac{z - w}{1 - zw}\right)^2 =
\left|\frac{z + w}{1 + zw}\right|^2 +
\left|\frac{z - w}{1 - zw}\right|^2 \, .
$$
Now one can apply the identity $|a+b|^2 = |a|^2 + |b|^2 + 2 \operatorname{Re}(\bar a b)$ multiple times.
Our expression becomes
$$
\frac{2+2\operatorname{Re}(\bar z w)}{2+2\operatorname{Re}( z w)} +
\frac{2-2\operatorname{Re}(\bar z w)}{2-2\operatorname{Re}( z w)} \, .
$$
To simplify the calculation further we can set $s = \operatorname{Re}(\bar z w)$ and $t = \operatorname{Re}( z w)$. Then $|s|\le 1$ and $|t| < 1$ and our expression is equal to
$$
\frac{1+s}{1+t} + \frac{1-s}{1-t} = \frac{2-2st}{1-t^2} = 1 + \frac{t^2-2st+1}{1-t^2} > 1
$$
because $t^2-2st+1 = (t-s)^2 + (1-s^2) > 0$.
The inequality is strict and sharp: For $z=w \ne \pm 1$ we have
$$
\left|\frac{2z}{1 + z^2}\right|^2 + 0^2 = \frac{4}{|1+z^2|^2} \to 1
$$
for $z \to 1$.
| {
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Convergence of $\sum_{k=1}^n\frac{(-1)^k}{\sqrt{k}}$ Does $\sum_{k=1}^n\frac{(-1)^k}{\sqrt{k}}$ converge?
My attempt:
$$\begin{aligned}
\sum_{k=1}^{2n}\frac{(-1)^k}{\sqrt{k}}&= \sum_{k=1, 3, ..., 2n-1}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\\
&=\sum_{k=1, 3,..., 2n-1}\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt{k})}
\end{aligned}$$
$$\begin{aligned}
|S_{2n}-S_{2m}|&=\sum_{k=2m+1, 2m+3,..., 2n-1}\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt{k})}\\
&<\sum_{k=2m+1, 2m+3, ..., 2n-1}\frac{1}{2k\sqrt{k}}\\
&< ?
\end{aligned}$$
| Your idea is fine but we need to consider
$$
\sum_{k=1}^{2n+1}\frac{(-1)^k}{\sqrt{k}} =-1+ \sum_{k=2}^{2n+1}\frac{(-1)^k}{\sqrt{k}}=-1+\sum_{k=1}^n\left(\frac{1}{\sqrt{2k}}-\frac{1}{\sqrt{2k+1}}\right)
$$
and since
$$\frac{1}{\sqrt{2k}}-\frac{1}{\sqrt{2k+1}}=\frac{\sqrt{2k+1}-\sqrt{2k}}{\sqrt{2k}\sqrt{2k+1}}=\frac{1}{\sqrt{2k}\sqrt{2k+1}(\sqrt{2k+1}+\sqrt{2k})}\le\frac{1}{4k\sqrt{2k}}$$
we can conclude that the given series converges also by p-test.
| {
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Show that the recursive $(g(n) = 2g(n−1) + 1, g(0) = 1)$ is equal to $ g(n) = 2^{n + 1} - 1$ How do I show that the recursive definition below is equivalent to the explicit one?
$$g(n) = 2g(n−1) + 1, g(0)=1$$
$$g(n) = 2^{n + 1} - 1$$
| By back substitution for the recurrence relation:
$$g(n)=2g(n-1)+1$$
$$g(n-1)=2g(n-2)+1$$
$$g(n-2)=2g(n-3)+1$$
$$\cdot\cdot\cdot$$
$$g(1)=2g(0)+1=2+1=3$$
So we have $$g(n)=2g(n-1)+1=2[2g(n-2)+1]+1$$
$$=2^2g(n-2)+2+1$$
$$=2^2[2g(n-3)+1]+2+1$$
$$=2^3g(n-3)+2^2+2+1$$
$$=\cdot\cdot\cdot $$
$$=2^{k}g(n-k)+2^{k-1}+2^{k-2}+...+2+1$$
$$=\cdot\cdot\cdot $$
$$=2^{n}g(n-n)+2^{n-1}+2^{n-2}+...+2+1$$
$$=2^{n}g(0)+\sum_{i=0}^{n-1}2^{i}$$
$$=2^{n}+\frac{2^{n}-1}{2-1}$$
$$=2^{n+1}-1.$$
| {
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Proving inequality: $\sum_{i=1}^n \left(a_i^7+a_i^5\right) \geq 2(\sum_{i=1}^n a_i^3)^2$
Let $a_i$ be distinct positive integers; prove that $$(a_1^7+a_2^7+\cdots + a_n^7)+(a_1^5+a_2^5+\cdots +a_n^5)\ge 2(a_1^3+a_2^3+\cdots + a_n^3)^2$$
I tried using some well known inequalities; obviously, since non homogenous and no obvious function, I don't expect either of AGM, Muirhead, CS, Jensen, Karamata, etc. should work, though I might be woefully wrong. After a while of experimentation I realized that this problem would likely be solved by either some tricky manipulations or a very obscure named inequality. Any helps? Thanks!
| We can use induction here.
For $n=1$ it's true by AM-GM.
Now, let $$\sum_{k=1}^n(a_k^7+a_k^5)\geq\left(\sum_{k=1}^na_k^3\right)^2.$$
We'll prove that:
$$\sum_{k=1}^{n+1}(a_k^7+a_k^5)\geq\left(\sum_{k=1}^{n+1}a_k^3\right)^2.$$
Indeed, let $a_{n+1}=a=\max\limits_k\{a_k\}$.
Thus, $$\sum_{k=1}^{n+1}(a_k^7+a_k^5)\geq \left(\sum_{k=1}^na_k^3\right)^2+a^7+a^5$$ and it's enough to prove that
$$a^7+a^5\geq2a^6+4a^3\sum_{k=1}^na_k^3$$ or
$$a^4+a^2\geq2a^3+4\sum_{k=1}^na_k^3.$$
Now, since $a_k\leq a-n-1+k,$ it's enough to prove that
$$\sum_{k=1}^n(a-k)^3\leq\frac{1}{4}(a^4-2a^3+a^2)$$ or
$$na^3-\frac{3n(n+1)}{2}a^2+\frac{3n(n+1)(2n+1)}{6}a-\frac{n^2(n+1)^2}{4}\leq\frac{1}{4}(a^4-2a^3+a^2)$$ or
$$(a-n)^2(a-n-1)^2\geq0$$ and we are done!
As David Cheng said, we can prove the last inequality by the following simpler way.
$$\sum_{k=1}^n(a-k)^3\leq\sum_{k=1}^{a-1}k^3=\frac{a^2(a-1)^2}{4}=\frac{1}{4}(a^4-2a^3+a^2).$$
| {
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Examples of vector fields directed towards origin I'm asked to state a vector field equation directed radially in towards origin, would $-\langle x,y\rangle $ suffice, or do I need to divide it by its magnitude r i.e. $-\left\langle \frac{x}{\sqrt {x^2+y^2}},\frac{y}{\sqrt {x^2+y^2}}\right\rangle $ or should I divide it by $r^2$?
| Any vector field of the form
$$
F(\mathbf r) = -f(|\mathbf r|)\, \mathbf r = -f(r)\, \mathbf r \ ; \qquad f(r) > 0
$$
is directed towards the origin, where $f$ is function of one argument, and its argument here is the distance $r$ from the origin, which is the length of $|\mathbf r|$. For example, in $\mathbb R^2$ we have
$$
r = \sqrt{x^2 +y^2}\ ;\quad \mathbf r = (x,y)^T
$$
one can take
\begin{align}
f(r) &= \cos^2(r^2) = \cos^2(x^2 +y^2)\\
f(r) &= \ln^2(r^2-3r) = \ln^2(x^2 +y^2-3\sqrt{x^2 +y^2}) \\
f(r) &= \frac{1}{r^3} = (x^2 +y^2)^{-\frac{3}{2}} \\
\end{align}
and $F(\mathbf r)$ above will be as required
| {
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Dummit and Foote 12.2.16: Determining all $2 \times 2$ matrices with entries from $\mathbb F _{19}$ of order $2$ This is exercise 12.2.16 of Abstract Algebra by Dummit and Foote.
Show that $x^5-1 =(x-1)(x^2-4x+1)(x^2+5x+1)$ in $\mathbb F_{19}[x]$. Use this to determine, up to similarity, all $2 \times 2$ matrices with entries from $\mathbb F_{19}$ of (multiplicative order) $5$.
*
*First of all, the multiplicative group of $\mathbb F_{19}$ is $C_{18}$, the cyclic group of order 18, with generator $2$, so I presume they mean that the matrices are of order $5$.
*Secondly, I prove the decomposition result as stated and then see that due to Cayley-Hamilton we know that any $2 \times 2$ matrix $A$ with minimal polynomial that divides $x^5-1$ must also satisfy $A^5-I=0 \implies A^5= I$.
The candidate polynomials are:
$x-1$, $x^2-4x+1$ and $x^2+5x+1$.
This question is similar to what they do on page 487, but I am not quite sure how one goes from these candidates to the list of permissible invariant factors, and after that, the matrices.
| We looks at the divisors of $x^5-1$ and use the decomposition as given. Notice for a $2 \times 2$ matrix it is only possible to get $2$ blocks of size $1$ or $1$ block of size $2\times 2$.
Recall that for a polynomial of the form $b_0 +b_1x +x^2$, we get the following companion matrix according to page 475:
$$ \begin{pmatrix}
0 & -b_0 \\
1 & -b_1
\end{pmatrix}$$
A $1\times 1$ block simply has the value of $-b_0$ for $b_0 + x$.
*
*$x-1$, $x-1$ give us:
$$\begin{pmatrix}
-(-1) & 0 \\
0 & -(-1)
\end{pmatrix}=\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}=I.$$
This matrix is of course not of order $5$, but of order $1$, therefore it satisfies $I^5=I$.
*$x^2-4x+1$ gives us:
$$A= \begin{pmatrix}
0 & -1 \\
1 & 4
\end{pmatrix}\equiv \begin{pmatrix}
0 & 18 \\
1 & 4
\end{pmatrix} \bmod {19}$$
*$x^2+5x+1$ gives us:
$$B=\begin{pmatrix}
0 & -1 \\
1 & -5
\end{pmatrix}\equiv \begin{pmatrix}
0 & 18 \\
1 & 14
\end{pmatrix} \bmod {19}$$
One can easily verify that
$$ A^5 \equiv B^5 \equiv 1$$
Since $5$ ia prime and the matrices are nonidentity, this is the order.
| {
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Simplifying $\frac{3 - 8}{\sqrt[3]{9} + 2\sqrt[3]{3} + 4} - \sqrt[3]{3}$. How to find the value?
Find the meaning of the expression
$$\frac{3 - 8}{\sqrt[3]{9} + 2\sqrt[3]{3} + 4} - \sqrt[3]{3}$$
What I tried:
$$\tag{1} \frac{-5}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2} - \frac{(\sqrt[3]{3})(\sqrt[3]{3^2}) + (\sqrt[3]{3})(2\sqrt[3]{3}) + 4(\sqrt[3]{3})}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2}$$
$$\tag{2} =\frac{-5}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2} - \frac{3 + 6 + 4(\sqrt[3]{3})}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2}$$
$$\tag{3} =\frac{-5 - 3 - 6 - 4(\sqrt[3]{3})}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2} $$
$$\tag{4} =\frac{-14 - 4(\sqrt[3]{3})}{\sqrt[3]{3^2} + 2\sqrt[3]{3} + 2^2}$$
| Note that in your computation from (1) to (2), it should be
$$(\sqrt[3]{3})(2\sqrt[3]{3})=2\sqrt[3]{3^2}\qquad (\text{it is not $6$)}.$$
It easier to follow the algebraic manipulations if we let $a=\sqrt[3]{3}$:
$$\begin{align}\frac{3 - 8}{\sqrt[3]{9} + 2\sqrt[3]{3} + 4} - \sqrt[3]{3}&=\frac{a^3-8}{a^2+2a+4}-a=\frac{a^3-8-a(a^2+2a+4)}{a^2+2a+4}\\&=
-\frac{2(a^2+2a+4)}{a^2+2a+4}=-2.\end{align}$$
| {
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Let a, b, c be ints. $\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b}$ is an int, show that each of $\frac{ab}{c}, \frac{bc}{a}, \frac{ac}{b}$ is an int. Let a, b, c $\in \mathbb{Z}$ . If $\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b}$ is an integer, prove that each of $\frac{ab}{c}, \ \frac{bc}{a}, \ \frac{ac}{b}$ is an integer.
I've tried to solve this problem but still got no solution. All I think is divisibility and GCD
$\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b} \\ = \frac{a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2}}{abc}$
Note that $2a^{2}bc + 2ab^{2}c + 2abc^{2}$ is divisible by abc. Put those in, we get:
$\frac{a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2} + 2a^{2}bc + 2ab^{2}c + 2abc^{2}}{abc} \\ = \frac{(ab + bc + ac)^{2}}{abc}$
Because it's an integer, thus $abc \mid (ab + bc + ac)^{2}$
Assume $GCD(ab + bc + ac, abc) = d$, then $ab + bc + ac = dk_1$ and $abc = dk_2$ for an integer d where $GCD(k_1, k_2) = 1$
$\frac{(ab + bc + ac)^{2}}{abc} = \frac{d^{2}{k_1}^{2}}{dk_2} = \frac{d{k_1}^2}{k_2}$
Because $GCD(k_1, k_2) = 1$, thus the only possibility is $k_2 \mid d$. Let d = $k_{2}p$ where p is an integer, thus it implies that $abc = dk_2 = {k_2}^{2}p$
I got stuck here, I probably used the wrong method to solve this problem, does anyone know how to solve this?
| You can use divisibility as I show here. First, let
$$\frac{ab}{c} = \frac{d_1}{e_1} \tag{1}\label{eq1A}$$
$$\frac{bc}{a} = \frac{d_2}{e_2} \tag{2}\label{eq2A}$$
$$\frac{ac}{b} = \frac{d_3}{e_3} \tag{3}\label{eq3A}$$
where each fraction $\frac{d_i}{e_i}$ for $1 \le i \le 3$ is in lowest terms, i.e., $\gcd(d_i, e_i) = 1$. Since the sum of these fractions is an integer, say $n$, we have
$$\begin{equation}\begin{aligned}
\frac{d_1}{e_1} + \frac{d_2}{e_2} + \frac{d_3}{e_3} & = n \\
d_1(e_2)(e_3) + d_2(e_1)(e_3) + d_3(e_1)(e_2) & = n(e_1)(e_2)(e_3)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Consider one of the fractions in the first $3$ equations to not be an integer, say WLOG in \eqref{eq1A}, then there exists a prime $p \mid e_1$, so $p \not\mid d_1$. Using the $p$-adic order function, i.e., which gives the highest power of $p$ which divides a given value, we have
$$\nu_p(e_1) \gt 0 \implies \nu_p(c) \gt \nu_p(a) + \nu_p(b) \tag{5}\label{eq5A}$$
If $p \not\mid e_2$ and $p \not\mid e_3$, then in \eqref{eq4A} on the left side, $p$ doesn't divide the first term, but it divides the second & third terms, plus it divides the right side term, which is not possible. Thus, $p \mid e_2$ and/or $p \mid e_3$, say WLOG we have $p \mid e_2$. This gives
$$\nu_p(e_2) \gt 0 \implies \nu_p(a) \gt \nu_p(b) + \nu_p(c) \tag{6}\label{eq6A}$$
Substituting this into \eqref{eq5A} gives
$$\begin{equation}\begin{aligned}
\nu_p(c) & \gt (\nu_p(b) + \nu_p(c)) + \nu_p(b) \\
\nu_p(c) & \gt 2\nu_p(b) + \nu_p(c) \\
0 & \gt 2\nu_p(b)
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
which is not possible since $\nu_p(b) \ge 0$. Thus, the original assumption that one of the fractions in \eqref{eq1A}, \eqref{eq2A} or \eqref{eq3A} is not an integer must be false, i.e., they are actually all integers instead.
| {
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Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;
*
*Using calculators is not allowed,
*The student have $72$ seconds on average to answer one question.
PROBLEM:
Compare $a=(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $b=(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$.
CHOICES:
A) $a>b$
B) $a<b$
C) $a=b$
D) Given information is not enough
Using algebra to evaluate each expression is easy, and the correct choice is $A$, but that will take a long time.
Any suggestion to solve this problem in a short time? THANKS.
| If $a<b$ then
$$(a+x)(b-x)$$
is increasing in x for $0\leq x \leq \frac{b-a}{2}$.
Using this $(2+1/2)*(5+1/5)$ is larger than $(2+1/5)*(5+1/2)$ and $(3+1/3)*(4+1/4)$ is larger than $(3+1/4)*(4+1/3)$.
Intuitively, the square maximises the area over all rectangles with the same circumference. To maximise a product where the factors have a fixed sum, we must try to get the factors as close as possible.
| {
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Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, $x < 0, a = 1$ The textbook answer is $\frac {1}{3}$, I went through all the steps, but couldn't interpret it.
Below were my steps.
Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, x < 0, a = 1
First I tried to find the inverse, rearrange the equation to $$x = y - \frac{2}{y} \implies x = \frac{y^{2}-2}{y}$$
Rearrange, $xy=y^{2}-2 \implies y^{2}-xy-2 = 0$
Then, I tried to use the equation $\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$, and I got:
$y = \frac{x \pm \sqrt{x^{2}+8}}{2}=f^{-1}$
Next, I tried to find $(f^{-1}){'}$, for both $+$ and $-$
$\frac{1}{2} \cdot (x' + ((x^{2}+8)^{\frac{1}{2}})')$
Apply chain rule:
$((x^{2}+8)^{\frac{1}{2}})'$ = $\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x$
Then do the same for -
And I got $(f^{-1})'$ = $\begin{cases} \frac{1}{2}\cdot(1+\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x) \\ \frac{1}{2}\cdot(1-\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x) \end{cases}$
When a = 1,
$(f^{-1})'=\begin{cases} \frac{\sqrt{9}+1}{2\sqrt{9}} = \frac{2}{3} \\ \frac{\sqrt{9}-1}{2\sqrt{9}} = \frac{1}{3} \end{cases}$
I need some help. It says when $x < 0$, but both of my answers are positive. How do I interpret it?
| The domain of $f$ must be the range of $f^{-1}$, i.e. $f^{-1}$ must only output negative values for any value of $x$. So, you should reject the solution $y=\frac{x+\sqrt{x^2+8}}{2}$ and notice $$\frac{x-\sqrt{x^2+8}}{2} \lt \frac{x-\sqrt{x^2}}{2} \le 0 $$ which satisfies our requirement and the corresponding answer is $\frac 13$.
| {
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Intersection of median and segment between two sides of a triangle
In triangle $ABC,$ $M$ is the midpoint of $\overline{BC},$ $AB=12,$ and $AC=16.$ Points $E$ and $F$ are taken on $\overline{AC}$ and $\overline{AB},$ respectively, and $\overline{EF}$ and $\overline{AM}$ intersect at $G.$ If $AE=2AF,$ then what is $EG/GF?$
This seemingly-easy problem (at least for my standards) is driving me crazy. I tried an analytical approach:
WLOG, assume that $\triangle{ABC}$ is right. (There is no specific angle measures.) Fix the points on the cartesian plane such that $A = (0, 0), B = (0, 12), \text{ and } C = (0, 16).$ Then $M$ is at $(8, 6)$, and the equation of line $AM$ is $y=\frac{3}{4}x.$ Next, let $E = (4, 0) \text{ and } F = (0, 8).$ The equation of line $EF$ is $y=-2x+8.$ Therefore, we have the system of equations $$y=\frac{3}{4}x$$$$y=-2x+8$$
Solving gets $$x=\frac{32}{11} \text{ and } y=\frac{24}{11}.$$
Therefore, the ratio of $EG$ to $GF$ is just $\frac{\frac{32}{11}}{4-\frac{32}{11}} = \frac{8}{3}.$
However, my approach is incorrect. Can anyone point out any flaws and present a solution to the correct answer? I also tried using mass points to no avail.
TIA!
| The key step is the ratio lemma. Let $b$ and $c$ be the measures of angles $BAM$ and $MAC$, respectively, and say $BM=m=MC$. Also, say $AF=\ell$ and $AE=2\ell$. Then, applying the ratio lemma to triangle $ABC$ at $A$, we have $\frac{\sin b}{\sin c}=\frac{m/12}{m/16}=\frac{4}{3}$. Then, we apply the ratio lemma to triangle $AEF$ at $A$ where $FG=x$ and $GE=y$ (we want $x/y$), so $\frac{4}{3}=\frac{\sin b}{\sin c}=\frac{x/\ell}{y/2\ell}=2\frac{x}{y}$. Thus $\frac{x}{y}=\frac{2}{3}$ (unless I've made a computation mistake).
| {
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Hint to prove $\sin^4(x) + \cos^4(x) = \frac{3 + \cos(4x)}{4}$ Could go from LHS to RHS by adding zero but I need to know how to do this WITHOUT knowing the half-angle formula. So from RHS to LHS, you an expand $\cos4x$ twice. I get as close as
$$\frac{ \cos^4x + \sin^4x + 3(1 - 2\sin^2x\cos^2x)}{4}$$
| \begin{align*}
\frac{3+\cos4x}{4}&=\frac{3+2\cos^22x-1}{4}=\frac{(\cos^2x-\sin^2x)^2+1}{2}\\
&=\dfrac{\sin^4x+\cos^4x+(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}{2}\\
&=\sin^4x+\cos^4x
\end{align*}
| {
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Prove a bijection between $\mathbb{N}^2$ and $\mathbb{N}$.
Prove that the function
$$f(m,n)=\frac{1}{2}\left(m^2+2 m n+n^2+m+3 n\right)$$
is a bijection between $\mathbb{N}^2$ and $\mathbb{N}$.
The problem arose in a series problem. I have to show that for each couple $(m,n)$ we get a different natural number and that all natural numbers are got applying $f$.
Below an example of what happens for $m,n$ from $0$ to $6$
$$
\begin{array}{ccccccc}
0 & 2 & 5 & 9 & 14 & 20 & 27 &\ldots\\
1 & 4 & 8 & 13 & 19 & 26 & 34 &\ldots\\
3 & 7 & 12 & 18 & 25 & 33 & 42 &\ldots\\
6 & 11 & 17 & 24 & 32 & 41 & 51 &\ldots\\
10 & 16 & 23 & 31 & 40 & 50 & 61 &\ldots\\
15 & 22 & 30 & 39 & 49 & 60 & 72 &\ldots\\
21 & 29 & 38 & 48 & 59 & 71 & 84 &\ldots\\
\ldots\\
\end{array}
$$
| Here is a rigorous proof, but first we rewrite $f$:
$$f(m,n) = \frac12(m^2+2mn+n^2 + m + 3n) = \frac12((m+n)(m+n+1)+2n)$$
$\Large \textbf{Injectivity}$
Suppose we have $f(m,n) = f(a,b)$. Then $(m+n)(m+n+1)+2n = (a+b)(a+b+1)+2b$.
First, suppose $m+n\ne a+b$. WLOG suppose $m+n > a+b$. Then:
\begin{align}(m+n)(m+n+1)+2n &\ge (a+b+1)(a+b+2)\\&=(a+b)(a+b+1)+2a+2b+2 \\&>(a+b)(a+b+1)+2b \\&= (m+n)(m+n+1)+2n\end{align}
which is a contradiction. Hence $m+n=a+b$.
Using this fact we have $2n=2b$, and hence $(m,n) = (a,b)$.
$\Large \textbf{Surjectivity}$
Your table provides a great insight: $f(m,0)$ are precisely the triangular numbers, and $f(m-1, n+1) = 1+f(m,n)$ for $m > 0$.
We can prove this by:
$$f(m,0) = \frac12(m^2+m) = T_m$$
\begin{align}f(m-1,n+1) &= \frac12((m-1+n+1)(m-1+n+1+1)+2(n+1))\\&=\frac12((m+n)(m+n+1)+2n)+1\\&=f(m,n)+1\end{align}
Now take any $x\in \mathbb N$. We can find a triangular number $T_k = \frac{k(k+1)}2$ such that $T_k \le x < T_{k+1}$.
Intuitively this $k$ would be $m+n$, and we need to shift over by $x-T_k$ numbers.
That is, notice that:
$$f(k-x+T_k, x-T_k) = \frac12((k)(k+1)+2(x-T_k))= T_k+x-T_k=x$$
This shows surjectivity.
| {
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Best way to evaluate $\lim_{n \rightarrow \infty} |\frac{(3(n+1)+4)(4^{n+1}+5)(5^n+3)}{(5^{n+1}+3)(3n+4)(4^n+5)}|$ Best way to evaluate $\lim_{n \rightarrow \infty} |\frac{(3(n+1)+4)(4^{n+1}+5)(5^n+3)}{(5^{n+1}+3)(3n+4)(4^n+5)}|$
Can you all show me some different ways to evaluate this limit? I was thinking of multiplying the numerator and denominator by something.. maybe $\frac{1}{5^{n+1}}$... I don't know. I'm sure there are a lot of ways to evalute this and you guys will show me a lot of slick awesome ways!!
As you may have guessed, i'm in the midst of employing the ratio test for a series.
| We need to factor out the leading terms to obtain
$$\frac{(3(n+1)+4)(4^{n+1}+5)(5^n+3)}{(5^{n+1}+3)(3n+4)(4^n+5)}
=\frac{(n+1)\cdot 4^{n+1} \cdot 5^n}{5^{n+1} \cdot n \cdot 4^n}\frac{\left(3+\frac4{n+1}\right)\left(1+\frac5{4^{n+1}}\right)\left(1+\frac3{5^n}\right)}{\left(1+\frac3{5^{n+1}}\right)\left(3+\frac4n\right)\left(1+\frac 5 {4^n}\right)}$$
with
$$\frac{(n+1)\cdot 4^{n+1} \cdot 5^n}{5^{n+1} \cdot n \cdot 4^n}=\frac{4}{5}\frac{n+1}n \to \frac45$$
and
$$\frac{\left(3+\frac4{n+1}\right)\left(1+\frac5{4^{n+1}}\right)\left(1+\frac3{5^n}\right)}{\left(1+\frac3{5^{n+1}}\right)\left(3+\frac4n\right)\left(1+\frac 5 {4^n}\right)} \to1$$
| {
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Where is the mistake in my evaluation of $\int\frac{x}{x^2+2x+3}\,dx$? Here is how I did it:
First, write $$\int\frac{x}{x^2+2x+3}\,dx=\int\frac{2x+2-x-2}{x^2+2x+3}\,dx=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx.$$
Now consider the integral in the minuend. Letting $u=x^2+2x+3$, one finds $du=(2x+2)\,dx$, and so $$\int\frac{2x+2}{x^2+2x+3}\,dx=\int\frac{du}{u}=\ln{|x^2+2x+3|}.$$
Next consider the other integral. Put $t\sqrt{2}=x+1$. Then $dx=\sqrt 2\,dt$. Now
\begin{align*}
\int\frac{x+2}{x^2+2x+3}\,dx&=\int\frac{(x+1)+1}{(x+1)^2+2}\,dx\\
&=\int\frac{t\sqrt 2+1}{2t^2+2}\sqrt 2\,dt\\
&=\frac{1}{2}\int\frac{2t+\sqrt 2}{t^2+1}\,dt\\
&=\frac{1}{2}\left(\int\frac{2t}{t^2+1}\,dt+\sqrt 2\int\frac{1}{t^2+1}\,dt\right)\\
&=\frac{1}{2}\left(\ln{|t^2+1|}+\sqrt 2\arctan t\right)
\end{align*}
and hence this is equal to
$$\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+1}{2}+1\right|}+\sqrt 2\arctan\frac{x+1}{\sqrt 2}\right)=\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+3}{2}\right|}+\sqrt 2\arctan\frac{x+1}{\sqrt 2}\right)$$
therefore
\begin{align*}
\int\frac{x}{x^2+2x+3}\,dx&=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx\\
&=\ln{|x^2+2x+3|}-\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+3}{2}\right|}+\sqrt{2}\arctan\frac{x+1}{\sqrt 2}\right)+C.
\end{align*}
apparently, the correct answer is $\frac{(\ln|x^2+2x+3|)}{2}-\frac{\sqrt{2}\arctan{\frac{(x+1)}{\sqrt{2}}}}{2}+C.$ what went wrong?
| Answer :
$\int_{}^{} \frac{x}{x^2 +2x+3}dx=\frac{1}{2}\int_{}^{} \frac{2x+2-2}{x^2 +2x+3} dx =\frac{1}{2}([ln(|x^2 +2x+3|] +c_1) - \frac{1}{2}\int_{}^{} \frac{2}{x^2 +2x+1 +2} dx= \frac{1}{2}([ln(|x^2 +2x+3|] +c_1) - \frac{1}{2}\int_{}^{} \frac{2}{(x+1)^2 +2 } dx =\frac{1}{2}([ln(|x^2 +2x+3|]+c_1) -\frac{1}{2}\int_{}^{} \frac{1}{(\frac{x+1}{\sqrt{2}})^2 +1} dx=\frac{1}{2}([ln(|x^2 +2x+3|]+c_1) - \frac{\sqrt{2}}{2}([arct(\frac{x+1}{\sqrt{2}})] +c_2) $
We put $C=\frac{1}{2}(c_1-\sqrt{2}c_2 )$.
| {
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} |
Find $f$ and $g$ by trial and error and a rigorous proof for showing that $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}]$ Here is the question I am trying to solve:
Find polynomials $f(x), g(x) \in \mathbb{Q}[x]$ such that $\sqrt{2} = f(\sqrt{2} + \sqrt{3})$ and $\sqrt{3} = g(\sqrt{2} + \sqrt{3}).$ Deduce the equality of fields: $\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}].$
And I found the following solution online:
a formula for $f$ and $g$
$\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \mathbb{Q}[\sqrt{2},\sqrt{3}]$
My questions are:
1- How did we know that $f$ and $g$ look like that? what are the trials that lead to this?
2- Is there a more rigorous way of proving that $\mathbb{Q}[\sqrt{2} + \sqrt{3}] \subseteq \mathbb{Q}[\sqrt{2},\sqrt{3}]$? or what is written is enough?
3- the stated reason for showing that $ \mathbb{Q}[\sqrt{2},\sqrt{3}] \subseteq \mathbb{Q}[\sqrt{2} + \sqrt{3}]$ is not that much clear for me ......could anyone explain it in details for me please?
Note: $F[\alpha]$ is a symbol for subring while $F(\alpha)$ is a symbol for a subfield for any $\alpha$
| $(\sqrt 2 + \sqrt 3)^k$ will be a linear combination of $\sqrt 2$, $\sqrt 3$, and $\sqrt 6$ so any polynomial $f(\sqrt 2 + \sqrt 3)$ will yield result of $a\sqrt 6 + b \sqrt 2 +c \sqrt 3 + d$ so we need a polynomial where the yielded values are $a=c=d=0$ and $b = 1$ (and for $g$, $a=b=d=0; c=1$). And, for simplicity, we want the least power.
Now if we have a power of $k$ and $k+1$ coefficients, $e_k$ we will end up with the $4$ variables, $a,b,c,d$ in $a\sqrt 6 + b \sqrt 2 + c\sqrt 3 + d$ being linear combinations of $e_j$. That is we will have $4$ equations of combinations of $e_j$ equaling each of $a,b,c,d$. If we have fewer than $4$ coefficients, $e_j$, we must have linear dependence which are likely to result in inconsistancies. If we have more than $4$ coefficients we will have superflous solutions.
Best guess is to try a polyinomial of $k+1 = 4$ or of power $k = 3$.
A power of $3$ with $mx^3 + nx^2 + px + q$ will yield $m(2\sqrt 2 + 6\sqrt 3 + 9\sqrt 2+3\sqrt 3) + n(5 + 2\sqrt 6) + p(\sqrt 2 + \sqrt 3) + q$ so we'd need:
$2m +9m+p =11m + p = 1; 6m + 3m+p =9m + p= 0; 2n = 0; 5n+q = 0$ which is a far more promising set of equations.
$n=q=0$ and $11m + p = 1$ and $9m + p =0$. So $m= \frac 12$ and $p=-\frac 92$.
So $f(x) = \frac 12x^3 -\frac 92x$ will yeild $f(\sqrt 2+ \sqrt 3) = \sqrt 2$.
TO find $g$ is much the same but we must solve $11m + p = 0$ and $9m+p = 1$ so $m=-\frac 12$ and $p=\frac {11}2$.
ANd $g(x) = -\frac 12x^3 + \frac {11}2 x$ will yeild $f(\sqrt 2 + \sqrt 3) = \sqrt 3$.
====
Note we could have tried powers of $k < 3$ but the $4$ equations and fewer than $4$ unknown all lead to inconsistencies.
$k = 0$ and $f(x) = e_0$ requires $e_0 = \sqrt 2 \in \mathbb Q$. Impossible. $k = 1$ and $f(x) = e_1x + e_0$ requires $e_1 = 1; e_1=0;e_0=0$. Impossible.
$k = 2$ and $f(x)=e_2x^2 + e_1x + e_0$ when evaluated for $x=\sqrt 2 + \sqrt 3$ would yield
$e_2(5+2 \sqrt 6) + e_1(\sqrt 2 +\sqrt 3) +e_0$ which to equal $\sqrt 2$ would require $2e_2 = 0; e_1=1; e_1=0; 5e_1+e_0 = 0$ which is of course impossible ($4$ equations and $3$ unknowns requires linear dependence, and in this case inconsistently so).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$n\times n$ real matrix with characteristic polynomial $p$, where $p$ is an irreducible polynomial in $\mathbb{R}[X]$. Is there a $2\times 2$ real matrix which has $x^2+x+1$ as its characteristic polynomial?
Edit: For any irreducible polynomial of degree n in $\mathbb{R}[X]$, is there a matrix $A$ in $M_n(\mathbb{R})$ for some $n\in\mathbb{N}$, whose characteristic polynomial is the irreducible polynomial chosen.
| The polynomial $x^2+x+1$ has the zeros
$$
\frac{-1 \pm \sqrt{ 1^2 - 4(1)(1) } }{2 (1) } = \frac{-1 \pm \sqrt{-3 } }{2} = \frac{-1 \pm \iota \sqrt{3} }{2}.
$$
So we can write
$$
x^2 + x + 1 = \left( x - \frac{-1 + \iota \sqrt{3} }{2} \right) \left( x - \frac{-1 - \iota \sqrt{3} }{2} \right) = \left( \frac{-1 + \iota \sqrt{3} }{2} - x \right) \left( \frac{-1 - \iota \sqrt{3} }{2} - x \right).
$$
Thus if we put
$$
A := \left[ \begin{matrix} \frac{-1 + \iota \sqrt{3} }{2} & a \\ b & \frac{-1 - \iota \sqrt{3} }{2} \end{matrix} \right],
$$
where $a$ and $b$ are any complex numbers such that $ab = 0$, then we note that
\begin{align}
& \ \ \ A - x I_2 \\
&= \left[ \begin{matrix} \frac{-1 + \iota \sqrt{3} }{2} & a \\ b & \frac{-1 - \iota \sqrt{3} }{2} \end{matrix} \right] - x \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \\
&= \left[ \begin{matrix} \frac{-1 + \iota \sqrt{3} }{2} - x & a \\ b & \frac{-1 - \iota \sqrt{3} }{2} - x \end{matrix} \right],
\end{align}
and so
\begin{align}
& \ \ \ \det \left( A - x I_2 \right) \\
&= \left( \frac{-1 + \iota \sqrt{3} }{2} - x \right) \left( \frac{-1 - \iota \sqrt{3} }{2} - x \right) - ab \\
&= \left( \frac{-1 + \iota \sqrt{3} }{2} - x \right) \left( \frac{-1 - \iota \sqrt{3} }{2} - x \right) \\
&= x^2+x+1,
\end{align}
as required.
Hope this helps.
PS:
Let
$$
A := \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right]
$$
be such a matrix, where $a, b, c, d \in \mathbb{R}$. Then we have
$$
\det \left(A - x I_2 \right) = x^2 + x + 1.
$$
But we note that
$$
A - x I_2 = \left[ \begin{matrix} a-x & b \\ c & d-x \end{matrix} \right],
$$
and thus
\begin{align}
\det \left( A - x I_2 \right) &= (a-x)(d-x) - bc \\
&= (x-a) (x-d) - bc \\
&= x^2 - (a+d)x + (ad-bc) \\
&= x^2 - \mbox{tr} (A) x + \det (A).
\end{align}
Thus we must have
$$
a + d = -1, \qquad \qquad \mbox{ and } \qquad \qquad ad-bc = 1.
$$
So we have
$$
a (-a-1) - bc = 1,
$$
that is,
$$
-a^2 -a - bc = 1,
$$
and so
$$
bc = -a^2 - a - 1 = - \left( a^2 + a + 1 \right),
$$
and since $a$ is real, $a^2 + a + 1$ cannot be zero, which implies that $b \neq 0$ and $c \neq 0$.
Thus $A$ is a matrix of the form
$$
A = \left[ \begin{matrix} a & b \\ -\frac{a^2+a+1}{b} & -a-1 \end{matrix} \right],
$$
where $a, b \in \mathbb{R}$ such that $b \neq 0$.
Conversely, if
$$
A = \left[ \begin{matrix} a & b \\ -\frac{a^2+a+1}{b} & -a-1 \end{matrix} \right],
$$
where $a, b \in \mathbb{R}$ such that $b \neq 0$, then we obtain
$$
A - x I_2 = \left[ \begin{matrix} a -x & b \\ -\frac{a^2+a+1}{b} & -a-1 -x \end{matrix} \right],
$$
and so
\begin{align}
\det \left( A - x I_2 \right) &= (a-x) (-a-1-x) - b \left( - \frac{a^2+a+1}{b} \right) \\
&= (x-a)(x+a+1) + a^2+a+1 \\
&= (x-a)(x+a) + (x-a) + a^2 + a + 1 \\
&= x^2 - a^2 + x - a + a^2 + a + 1 \\
&= x^2 + x + 1.
\end{align}
Thus the set of real matrices with the characteristic polynomial $x^2 + x + 1$ is
$$
\left\{ \, \left[ \begin{matrix} a & b \\ -\frac{a^2+a+1}{b} & -a-1 \end{matrix} \right] : a, b \in \mathbb{R}, b \neq 0 \right\}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to compute the limit as $x\to 3$ of a $\textit{complicated}$ product and quotient of trigonometric functions $$\lim_{x\rightarrow 3}\frac{ \tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}$$
I substituted $x-3$ for $u$ and got as far as
$$\frac{1}{6} \lim_ {u\to 0} \frac{\sin(9 \sin u)}{\sin((u+3)^(3) -27)}.$$
This is where I get stuck. Should I try a different approach?
| You can use the following trick:
As $\sin(x)\sim x$ for $x\to0$, in a product you can replace the sine (tangent) of an argument that tends to zero by the argument itself. In your case,
$$\frac{ \tan\left(\dfrac{x-3}{x+3}\right)\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}\to \frac{ \dfrac{x-3}{x+3}\sin(9(x-3)) }{ (x-3)(x^3-27))}\to \frac{ \dfrac{9(x-3)}{x+3}}{ (x-3)(x^3-27))}
\\\to \frac9{(x+3)(x^2+3x+9)}
\to\frac9{6\cdot27}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How to find steady-state probabilities in a queuing system? Queuing system set
$$\begin{matrix}\text{intake intensity} & \lambda=26 \\
\text{service channels} & m=4\\
\text{service intensity} & \mu=8\\
\text{maximum queue size} &n=18\end{matrix}$$
It is required to
Draw up a graph of a Markov process, write down the Kolmogorov system
of equations and find the steady-state probabilities.
I understand what the queuing system graph looks like
But how can we find the probabilities?
Because this is a homogeneous ergodic process, we can solve it using the Kolmogorov equation by equating the derivative to 0.
$$\dot{p}_i(t)=\sum\limits_j p_j(t)\lambda_{ji}(t)-p_i(t)\sum\limits_j\lambda_{ij}(t),\qquad (i=1,2,3...)\\
\frac{\partial p_i(t)}{\partial t}=\dot{p}_i(t)=0$$
Then we get
$$\sum\limits_j p_j\lambda_{ji}-p_i\sum\limits_j\lambda_{ij}=0\\$$
In the tutorial I found the so called system of equations for n = 3
$$\begin{equation*}
\begin{cases}
p_2\lambda_{21}+p_3\lambda_{31}-p_1(\lambda_{12}+\lambda_{13})=0\\
p_1\lambda_{12}+p_3\lambda_{32}-p_2(\lambda_{21}+\lambda_{23})=0\\
p_1\lambda_{13}+p_2\lambda_{23}-p_3(\lambda_{31}+\lambda_{32})=0
\end{cases}
\end{equation*}$$
$$M=\begin{pmatrix}
-(\lambda_{12}+\lambda_{13}) & \lambda_{21} & \lambda_{31} \\
\lambda_{12} & -(\lambda_{21}+\lambda_{23}) & \lambda_{32} \\
\lambda_{13} & \lambda_{23} & -(\lambda_{31}+\lambda_{32})
\end{pmatrix}, X=\begin{pmatrix}p_1\\p_2\\p_3\end{pmatrix}, B=\begin{pmatrix}0\\0\\0\end{pmatrix}$$
$$M = \Lambda^T-D\text{, where } D = diag(\sum\nolimits_j\lambda_{1j},\sum\nolimits_j\lambda_{2j},\sum\nolimits_j\lambda_{3j})\\ \text{-
Diagonal matrix of the sums of the rows of the $\Lambda$ transition intensities matrix}$$
The solution will only give the proportions between $p_i$ we need to replace any string with a string of ones
$$\begin{equation*}
\begin{cases}
p_2\lambda_{21}+p_3\lambda_{31}-p_1(\lambda_{12}+\lambda_{13})=0\\
p_1\lambda_{12}+p_3\lambda_{32}-p_2(\lambda_{21}+\lambda_{23})=0\\
\textbf{p}_1+\textbf{p}_2+\textbf{p}_3=\textbf{1}
\end{cases}
\end{equation*}$$
$$M_*=\begin{pmatrix}
-(\lambda_{12}+\lambda_{13}) & \lambda_{21} & \lambda_{31} \\
\lambda_{12} & -(\lambda_{21}+\lambda_{23}) & \lambda_{32} \\
\textbf{1} & \textbf{1} & \textbf{1}
\end{pmatrix}, B=\begin{pmatrix}0\\0\\\textbf{1}\end{pmatrix}$$
$$M_*\times X = B$$
Whence multiplying both sides on the left by the inverse matrix we get
$$X = M^{-1}_*\times B$$
But how to compose this system in our case? How many rows and columns do we have? And what are different lambdas equal to? We essentially have the same value and everywhere will be the same number or not?
| As I understand it, the values of the matrix are derived from the queuing system graph.
Namely we have $m+n+1$ rows and $m+n+1$ columns since we also take cell with the zero state
$$ \Lambda = \begin{pmatrix} 0 & \lambda & 0 & 0 & \dots & 0 & 0\\
\mu & 0 & \lambda & 0 & \dots & 0 & 0 \\
0 & 2\mu & 0 & \lambda & \dots & 0 & 0\\
0 & 0 & 3\mu & 0 & \dots & 0 & 0\\
& & & & \dots & & \\
0 & 0 &0 & m\mu & 0 & \lambda & 0\\
0 & 0 &0 &0 & m\mu & 0 & \lambda \\
0 & 0 &0 &0 & 0 & m\mu & 0
\end{pmatrix} $$
In our case
$$ \Lambda_{22\times 22} = \begin{pmatrix} 0 & 26 & 0 & 0 & \dots & 0 & 0\\
1\cdot 8 & 0 & 26 & 0 & \dots & 0 & 0 \\
0 & 2\cdot 8 & 0 & 26 & \dots & 0 & 0\\
0 & 0 & 3\cdot 8 & 0 & \dots & 0 & 0\\
& & & & \dots & & \\
0 & 0 &0 & 4\cdot 8 & 0 & 26 & 0\\
0 & 0 &0 &0 & 4\cdot 8 & 0 & 26 \\
0 & 0 &0 &0 & 0 & 4\cdot 8 & 0
\end{pmatrix} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3868772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Possible ways of choosing 7 courses out of 20 with the constraint that at least 1 course has to be a stat course To fulfill the requirements for a certain degree, a student can choose to take any $7$ out of a list of $20$ courses, with the constraint that at least $1$ of the $7$ courses must be a statistics course. Suppose that $5$ of the $20$ courses are statistics courses.
(a) How many choices are there for which 7 courses to take?
The solution I believe is : $$\binom{5}{1}\binom{15}{6} + \binom{5}{2}\binom{15}{5} + \binom{5}{3}\binom{15}{4} + \binom{5}{4}\binom{15}{3} + \binom{5}{5}\binom{15}{2}$$
But the solution seems to be $$\binom{20}{7} - \binom{15}{7}$$
I evaluated the two expressions and they give different numerical values.
I want to know where am I going wrong.
The question was earlier asked in the following link:
Take seven courses out of 20 with requirement
| The answers are equal and both correct.
The relationship between the answers is explained by Vandermonde's Identity
$$\binom{m+n}{r}=\sum\limits_{k=0}^r\binom{m}{k}\binom{n}{r-k}$$
In your case, that is $$\binom{20}{7}=\binom{5}{0}\binom{15}{7}+\binom{5}{1}\binom{15}{6}+\binom{5}{2}\binom{15}{5}+\dots+\binom{5}{5}\binom{15}{2}\color{grey}{+\binom{5}{6}\binom{15}{1}+\binom{5}{7}\binom{15}{0}}$$
The terms at the end were zero and so can be discarded. Then, by subtracting the $\binom{5}{0}\binom{15}{7}$ term from both sides and evaluating $\binom{5}{0}$ as $1$ you are left with $$\binom{20}{7}-\binom{15}{7}=\binom{5}{1}\binom{15}{6}+\dots+\binom{5}{5}\binom{15}{2}$$
Arguably, the expression on the left is easier to work with since there are less terms and less arithmetic involved, but both are equally correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problem to find exact solution with steepest descent I have a problem that is formulated as this:
$$\begin{matrix}\min\\x \in \mathbb{R}^2\end{matrix} f(\mathbf{x}) := (2 x_1^2 - x_2^2)^2 + 3x_1^2-x_2$$
The task is: Perform one iteration using the steepest descent algorithm when $\mathbf{x}_0 = (1/2, 5/4)^T$.
And I get the solution to be:
$\mathbf{x}_1 = (1/2, 3/4)^T$
but it should be:
$\mathbf{x}_1 = (1/2, 1)^T$
This is how I solved it:
*
*$\mathbf{p}_k = -\nabla f(\mathbf{x}_k)$
*$\mathbf{x}_{k+1} = \mathbf{x}_k + \alpha_k \cdot \mathbf{p}_k$
Since nothing is said about choosing $\alpha$ I set it to $\alpha = 1$.
$\nabla f(\mathbf{x}) = (8x_1(2x_1^2-x_2) + 6x_1, -2(2x_1^2-x_2)-1)^T = (16x_1^3 - 8x_1x_2 + 6x_1, -4x_1^2+2x_2-1)^T$
$\nabla f(\mathbf{x}_0) = (0, 1/2)^T, \mathbf{p}_k = (0, -1/2)^T$
$\mathbf{x}_1 = \mathbf{x}_0 + 1 \cdot (0, -1/2)^T = (1/2, 3/4)$
I think it depends on that I picked my $\alpha$ to be 1 but it gets correct when it's 1/2.
So why should $\alpha = 1/2$? Should I use Armijo step rule to find out, or how?
| Let $x_1=x$ and $x_2=y$ and $y^2\leq\frac{3}{4}$.
Thus, by AM-GM
$$(2x^2-y^2)^2+3x^2-y=4x^4-4x^2y^2+3x^2+y^4-y=$$
$$\geq x^2(4x^2+3-4y^2)+y^4-y\geq y^4-y=$$
$$=y^4+\frac{3}{4\sqrt[3]4}-y-\frac{3}{4\sqrt[3]4}\geq4\sqrt[4]{y^4\left(\frac{1}{4\sqrt[3]4}\right)^3}-y-\frac{3}{4\sqrt[3]4}=$$
$$=|y|-y-\frac{3}{4\sqrt[3]4}\geq-\frac{3}{4\sqrt[3]4}.$$
The equality occurs for $x=0$ and $y=\frac{1}{\sqrt[3]4}.$
Now, let $y^2\geq\frac{3}{4}$.
Thus, $$(2x^2-y^2)^2+3x^2-y=$$
$$=4x^4-(4y^2-3)x^2+\left(y^2-\frac{3}{4}\right)^2+y^4-\left(y^2-\frac{3}{4}\right)^2-y\geq$$
$$\geq y^4-\left(y^2-\frac{3}{4}\right)^2-y=\frac{3}{2}y^2-y-\frac{9}{16}\geq\frac{3}{4}+\frac{1}{2}y^2-y-\frac{9}{16}\geq$$
$$\geq\frac{3}{4}-\frac{1}{2}-\frac{9}{16}=-\frac{5}{16}>-\frac{3}{4\sqrt[3]4},$$
which says that $-\frac{3}{4\sqrt[3]4}$ it's a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Asymptotic expansion of an integral as $x \to 0^{+}$ Find the asymptotic expansion of $$F(x) := \int_{x}^{1} \frac{1}{t \sqrt{1+t^2}} \ dt, \text{ as } x \to 0^{+}$$
I tried expanding $\frac{1}{ \sqrt{1+t^2}} = 1 - \frac{t^2}{2} + \frac{3 t^{4}}{8} + \cdot \cdot \cdot$
The integral is then :$$F(x) := \int_{x}^{1} \frac{1}{t \sqrt{1+t^2}} \ dt = \bigg[\frac{1}{t} - \frac{t}{2} + \frac{3t^{3}}{8} \cdot \cdot \cdot \bigg] dt$$
I observed that the first term goes to infinity as $x \to 0^{+}$.
Can someone point out if I am going in the wrong direction?
| HINT
I would start by making the substitution $t = \sinh(u)$. Thus we get
\begin{align*}
\int\frac{\mathrm{d}t}{t\sqrt{1+t^{2}}} = \int\frac{\cosh(u)}{\sinh(u)\cosh(u)}\mathrm{d}u = \int\frac{\mathrm{d}u}{\sinh(u)} = \int\frac{\sinh(u)}{\sinh^{2}(u)}\mathrm{d}u = \int\frac{\sinh(u)}{\cosh^{2}(u)-1}\mathrm{d}u
\end{align*}
Now you can make the substitution $v = \cosh(u)$, which leads to
\begin{align*}
\int\frac{\sinh(u)}{\cosh^{2}(u) - 1}\mathrm{d}u = \int\frac{\mathrm{d}v}{v^{2} - 1} = \frac{1}{2}\int\left(\frac{1}{v-1} - \frac{1}{v+1}\right)\mathrm{d}v
\end{align*}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Where did I go wrong in applying the factor theorem?
Given that $x + 1$ and $x - 3$ are two of the four factors of the expression $x^4 + px^3 + 5x^2 + 5x + q$, find the values of $p$ and $q$.
I tried to answer this question using the factor theorem but got the answer wrong:
$$ \text{Let } f(x) = x^4 + px^3 + 5x^2 + 5x + q $$
$$ \text{Since } x + 1 \text{ and } x - 3 \text{ are factors of } f(x), \text{ then } f(-1) = 0 \text{ and } f(3) = 0, \text{ i.e.} $$
\begin{align}
(-1)^4 + p(-1)^3 + 5(-1)^2 + 5(-1) + q &= 0 \color{red}{\leftarrow (1)} \\
(3)^4 + p(3)^3 + 5(3)^2 + 5(3) + q &= 0 \color{blue}{\leftarrow (2)}
\end{align}
$$ \text{From } \color{red}{(1)}: $$
\begin{align}
(-1)^4 + p(-1)^3 + 5(-1)^2 + 5(-1) + q &= 0 \\
1 + p(-1) + 5(1) + (-5) + q &= 0 \\
1 - p + 5 - 5 + q &= 0 \\
1 - p + q &= 0 \\
q &= p - 1 \color{limegreen}{\leftarrow (3)}
\end{align}
$$ \text{From } \color{blue}{(2)}: $$
\begin{align}
(3)^4 + p(3)^3 + 5(3)^2 + 5(3) + q &= 0 \\
81 + 27p + 45 + 15 + q &= 0 \\
27p + q + 60 + 81 &= 0 \\
27p + q + 141 &= 0 \\
q &= -27p - 144 \color{orange}{\leftarrow (4)}
\end{align}
$$ \color{orange}{(4)} + \color{limegreen}{(3)}: $$
\begin{align}
-27p - 144 &= p - 1 \\
-27p - p &= 144 - 1 \\
-28p &= 143 \\
28p &= -143 \\
p &= -\frac{143}{28} \\
\therefore p &= -5\frac{3}{28} \color{mediumpurple}{\leftarrow (5)}
\end{align}
$$ \text{Substitute } \color{mediumpurple}{(5)} \text{ into } \color{limegreen}{(3)}: $$
\begin{align}
q &= -5\frac{3}{28} - 1 \\
\therefore q &= -6\frac{3}{28}
\end{align}
The answers were $ p = -5, q = -6 $.
Where did I go wrong?
| Here is a way to see where you went wrong by comparison. Let
\begin{align*}
f(x) & = x^4+px^3 +5x^2+5x+q \\
g(x) & = (x+1)(x-3)(x-a)(x-b)
\end{align*}
Then write down the polynomial $f-g=0$. The coefficients must be all zero, i.e., we have
$$
a + b + p + 2=0,\; ab + 2a + 2b - 8=0,\; - 2ab + 3a + 3b - 5=0, q-3ab=0.
$$
Solve for $a$ and $b$, and then put $p=-a-b-2$, $q=3ab$.
| {
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how would I find integer solutions for $y^2 = mx + b$? For example; $$y^2 = 30x + 1$$
For which one answer is; $$11^2 = 30(4) + 1 = 121$$
WA kindly gave me 4 answers;
$$x = 2 (15 n^2 - 29 n + 14), y = 29 - 30 n, n \in Z$$
$$x = 2 (15 n^2 - 19 n + 6), y = 19 - 30 n, n \in Z$$
$$x = 2 (15 n^2 - 11 n + 2), y = 11 - 30 n, n \in Z$$
$$x = 2 (15 n^2 - n), y = 1 - 30 n, n \in Z$$
Please help me learn how to get from my example to those 4 answers.
I'm more interested in a general solution in terms of $m$ and $b$. $30$ and $1$ are just random, convenient values for example. Though, I can work backward from example values to the general solution.
I recognize the two coefficients of $n$ being $1$ and $29$ are trivially $b$ and $m-b$, respectively.
I'm particularly interested in whether there's an algebraic calculation that yields the coefficients of $11$ and $19$, or whether it's a search function. I.e., what if $m$ were very large?
| $y^2 = mx +b$ so $x=\frac {y^2-b}m$ needs to be an integer.
I don't know of any general way to do it but you could solve $y^2 = b\pmod m$ (which may or may not have tricks; but brute force we can test $m$ values). $\alpha$ is a such a solution then $\alpha + mk$ will be solutions.
For example for $y^2 = 6m + 7$ we must have $y^2 \equiv 7\equiv 1\pmod 6$ and and so $y^2-1\equiv (y-1)(y+1)\equiv 0\pmod 6$ so $y\equiv 1, -1$ and $y-1\equiv 2,3$ while $y+1\equiv 3,2$ but the latter are impossible so $y=6k\pm 1$ will be a solution.
Less neat example $y^2 = 7m + 12$ we must have $y^2 \equiv 12\equiv 5\pmod 7$. Brute force $0^2,(\pm1)^2,(\pm 2)^2,(\pm 3)^2\equiv 0,1,4,2$ so that has no solutions. But for $y^2 = 7m + 11$ we'd have $y=7m \pm 2$ always a solution.
Verification $(7m\pm 2)^2 = 49m^2 \pm 28m + 4= 7(7m^2 \pm 4m-1) + 11$. So $x=7m^2\pm 4m-1; y=7m\pm 2$ is a solution to $y^2 = 7x +11$
And for $k=0....6$ $(7m+k)^2 = 49m^2 +14mk + k^2 = 7(7m^2+2mk) +k^2$ but for $k^2\ne 7j + 12$ for any $j$ so there are no integer solutions to $y^2=7x + 12$.
| {
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Why is $r^3+4t^3+2s^3-6rts$ non-zero (unless $r=s=t=0$)? When solving How to find the multiplicative inverse of a polynomial? I created a $3 \times 3$ linear system with determinant $r^3+4t^3+2s^3-6rts, \text{ where } r, t, s \in \Bbb Q$. Basic field theory tells me this determinant must be non-zero (unless $r=s=t=0$), but there must be a more direct way to see that. Can anyone explain why this quantity must be non-zero in all but the trivial case? Thanks.
| Let $x = r$, $y =\sqrt[3]{4}t$ and $z = \sqrt[3]{2}s$
$$r^3 + 4t^3 +2s^3 - 6rts = \left(r + t\sqrt[3]{4} + s\sqrt[3]{2}\right)\left(x^2 + y^2 + z^2 - xy -yz - zx\right)$$
$r + t\sqrt[3]{4} + s\sqrt[3]{2} = 0 \iff r = s = t = 0$
$$x^2 + y^2 + z^2 - xy - yz - zx = \frac{1}{2}(x-y)^2 + \frac{1}{2}(y-z)^2 + \frac{1}{2}(z-x)^2$$
so $x^2 + y^2 + z^2 - xy - yz - zx = 0 \iff r = s = t = 0$
| {
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| We have
$$3 = b^2+bc+c^2 = \frac{3(b+c)^2+(b-c)^2}{4} \geqslant \frac{3}{4}(b+c)^2.$$
Therefore $(b+c)^2 \leqslant 4,$ so $b+c \leqslant 2.$
| {
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How do I calculate the definite integral $\int_{-1}^1\frac{\sqrt{1-x^2}}{1+x^2}dx$ using complex variables? I have tried solving the integral $\oint_{C}\frac{\sqrt{1-z^2}}{1+z^2}dz$ using the upper semi-circle contour; I am getting the poles $z=\pm i$.
Only $z=i$ exists within the contour and I have evaluated the residue at $z=i$ as follows:
$Res(z=i)=\lim_{z\to i} (z-i)\frac{\sqrt{1-z^2}}{1+z^2}=\frac{1}{\sqrt{2}i}$
Using residue theorem, I am ending up with $\oint_{C}\frac{\sqrt{1-z^2}}{1+z^2}dz = \sqrt{2}\pi$.
I am confused as to how to proceed from this point.
| Let $I$ be given by
$$I=\int_{-1}^1\frac{\sqrt{1-x^2}}{1+x^2}\,dx$$
Next, let $C$ be the classical dog bone contour around $[-1,1]$ in the complex plane. It is straightforward to show that
$$\oint_C \frac{\sqrt{1-z^2}}{1+z^2}\,dz=-2I$$
since $C$ is traversed in the counter-clockwise direction.
In the following analysis, we cut the plane with branch cuts along the real axis from the branch points at $\pm 1$ to $-\infty$. These two branch cut coalesce as a branch cut from $-1$ to $1$. We ensure that the chosen branches are taken such that on the real axis above the branch cut $[-1,1]$, we have $\sqrt{1-x^2}\ge 0$.
Using the Residue Theorem, we have for $R>1$
$$\begin{align}
\oint_C \frac{\sqrt{1-z^2}}{1+z^2}\,dz&=\oint_{|z|=R}\frac{\sqrt{1-z^2}}{1+z^2}\,dz-2\pi i\,\text{Res}\left(\frac{\sqrt{1-z^2}}{1+z^2}, z=\pm 1\right)\\\\
&=\oint_{|z|=R}\frac{\sqrt{1-z^2}}{1+z^2}\,dz-2\pi i \left(\frac{\sqrt{2}}{2i}+\frac{-\sqrt{2}}{-2i}\right)\tag1
\end{align}$$
Since $R$ is arbitrary, we can let $R\to \infty$. Proceeding accordingly we have
$$\begin{align}
\lim_{R\to \infty}\oint_{|z|=R }\frac{\sqrt{1-z^2}}{1+z^2}\,dz&=\lim_{R\to \infty}\int_0^{2\pi} \frac{\sqrt{1-R^2e^{i2\phi}}}{1+R^2e^{i2\phi}}\,iRe^{i\phi}\,d\phi\\\\
&=2\pi\tag2
\end{align}$$
NOTE:
In arriving at $(2)$, we wrote $\sqrt{1-R^2e^{i2\phi}}=-i\sqrt{R^2e^{i2\phi}-1}$, which is consistent with the chosen branch.
Finally, using $(2)$ in $(1)$ we find that
$$\oint_C \frac{\sqrt{1-z^2}}{1+z^2}\,dz =2\pi(1-\sqrt {2}) \tag3$$
whereupon dividing $(3)$ by $-2$ yields the coveted result
$$I=2\pi (\sqrt 2-1)$$
And we are done!
| {
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Prove by Induction. For $n \in \mathbb{N}$, $10|(9^{n+1}+7^{2n})$. So far, this is what I have, but I'm confused as to how to 1. remove the 7 from inside the brackets to be able to substitute 10k and 2. make the whole thing divisible by 10 so that I can prove it.
Basic Step: Let $n = 1$. Therefore,
$$
9^{1+1} + 7^{2 \cdot 1} = 9^2 + 7^2 = 130
$$
Therefore, $10|(9^{n+1}+7^{2n})$, is true.
Inductive Step: There exists an integer $k$ such that $(9^{n+1}+7^{2n}) = 10k$. Let $P(k) = 9^{k+1} + 7^{2k}$. Therefore,
$$
P(k+1) = 9^{(k+1)+1} + 7^{2(k+1)} = 9^{k+2} + 7^{2k+2}
$$
$$
P(k+1) = (9^1 \cdot 9^{k + 1}) + (7^2 \cdot 7^{2k})
$$
$$P(k+1) = (2+7)(9^{k+1}) + (7^2 \cdot 7^{2k}) = (2 \cdot 9^{k+1}) + (7 \cdot 9^{k+1}) + (7 \cdot 7 \cdot 7^{2k})
$$
$$
P(k+1) = (2 \cdot 9^{k+1}) + 7(9^{k+1} + 7 \cdot 7^{2k})
$$
| Can you show that $$9^{k+2}+7^{2k+2}=9\cdot9^{k+1}+49\cdot7^{2k}=9(9^{k+1}+7^{2k})+40\cdot7^{2k}$$ is divisible by $10$?
| {
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Find maximize of $P=\frac{x\sqrt{yz}}{\sqrt{x^2+1}\sqrt[4]{\left(y^2+4\right)\left(z^2+9\right)}}$ Let $x,y,z\in \mathbb{R^+}$ such that $6x+3y+2z=xyz$. Find maximize of $$P=\frac{x\sqrt{yz}}{\sqrt{x^2+1}\sqrt[4]{\left(y^2+4\right)\left(z^2+9\right)}}$$
We will prove $$P\le \sqrt{\frac{16}{27}}$$
$$\Leftrightarrow \frac{1}{\sqrt{1+\frac{1}{x^2}}\sqrt[4]{\left(1+\frac{4}{y^2}\right)\left(1+\frac{9}{z^2}\right)}}\le \sqrt{\frac{16}{27}}$$
$$\Leftrightarrow \frac{1}{\sqrt[4]{\frac{\left(2x+y\right)^3\left(3y+2z\right)^2\left(3x+z\right)^3}{x^6y^5z^5}}}\le \sqrt{\frac{16}{27}}$$
$$\Leftrightarrow \left(2x+y\right)^3\left(3y+2z\right)^2\left(3x+z\right)^3\left(xyz\right)^4\ge \frac{729}{256}x^6y^5z^5\left(6x+3y+2z\right)^4$$
I don't know how to solve it now. Help me.
| Also, $uvw$ helps.
Indeed, in my previous solution it's enough to prove that:
$$64(2a+b)^3(2a+c)^3(b+c)^2\geq729(2a+b+c)^4a^2bc.$$
Now, let $b+c=2u$ and $bc=v^2$, where $v>0$.
Thus, by AM-GM $u\geq v$ and we need to prove that $f(u)\geq0,$ where
$$f(u)=16u^2(4a^2+4ua+v^2)^3-729(a+u)^4a^2v^2.$$
But, $$f'(u)=32u(4a^2+4ua+v^2)^3+192u^2a(4a^2+4ua+v^2)^2-4\cdot729(a+u)^3a^2v^2=$$
$$=4(512ua^6+9(256u^2-81v^2)a^5+3u(1024u^2-601v^2)a^4+5u^2(256u^2-207v^2)a^3+$$
$$+3uv^2(13u^2+32v^2)a^2+144u^2v^4a+8uv^6)>0,$$
which says that $f$ increases.
Thus, it's enough to prove our inequality for a minimal value of $u$, which happens for $u=v$,
id est, for $b=c$.
Since our inequality is homogeneous, it's enough to assume that $b=c=1$ and we need to prove that:
$$16(4a^2+4a+1)^3\geq729(a+1)^4a^2$$ or
$$4(2a+1)^3\geq27(a+1)^2a,$$ which is true by AM-GM:
$$4(2a+1)^3=4\left(a+2\cdot\frac{a+1}{2}\right)^3\geq4\left(3\sqrt[3]{a\left(\frac{a+1}{2}\right)^2}\right)^3=27(a+1)^2a.$$
| {
"language": "en",
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Simplifying the derivative of $\sin(\cos^2 x)\cos(\sin^2 x)$ Differentiating the term
$$\sin(\cos^2 x)\cos(\sin^2 x)$$
leads me through the chain and product rule to
$$-\sin(2x)\cos(\cos^2 x)\cos(\sin^2 x)+(-\sin(\sin^2 x)\sin(2x)\sin(\cos^2 x))$$
where the derivative of $\sin^2 x$ equals to
$$\frac{d}{dx} \sin^2 x = 2\sin x \frac{d}{dx} \sin x = 2\sin x \cos x = \sin 2x$$
and $\frac{d}{dx} cos^2 x$ to $-\sin 2x$ respectively.
Through factorization, I can then simplify the term to
$$-\sin 2x\ (\cos(\cos^2 x)\cos(\sin^2 x) + \sin(\sin^2 x)\sin(\cos^2 x))$$
The problem starts here where I fail to find a simplification for the second factor which, according to Wolfram Mathematica, should lead to $\cos(\cos 2x)$ and ultimately to
$$-\sin(2x)\cos(\cos 2x)$$
How and which trigonometric identities could I apply to get to that? Is my approach right?
| Denote:
$$f(x) = \sin(\cos^2 x)\cos(\sin^2 x)$$
$$g(x) = \sin(\cos^2 x)$$
$$h(x) = \cos(\sin^2 x)$$
Thus:
$$f(x) = g(x)\cdot h(x)$$
$$g'(x) = \cos(\cos^2{x}) \cdot 2 \cos{x} \cdot (-\sin{x})$$
$$h'(x) = -\sin(\sin^2{x}) \cdot 2 \sin{x} \cdot \cos{x}$$
Now we get:
$$f'(x) = g(x)\cdot h'(x) + g'(x)\cdot h(x) $$
$$f'(x) = \sin(\cos^2 x) \cdot (-\sin(\sin^2{x}) \cdot 2 \sin{x} \cdot \cos{x}) + \\ \cos(\cos^2{x}) \cdot 2 \cos{x} \cdot (-\sin{x}) \cdot \cos(\sin^2 x) $$
$$f'(x) = -2\sin{x}\cos{x} [ \sin(\cos^2 x) \cdot \sin(\sin^2{x}) + \cos(\cos^2 x) \cdot \cos(\sin^2{x})]$$
Now we use the formula $\cos(a-b) = \sin{a}\sin{b} + \cos{a}\cos{b}$ and we get:
$$f'(x) = -2\sin{x}\cos{x} [ \cos(\cos^2 x - \sin^2{x})]$$
$$f'(x) = -2\sin{x}\cos{x} \cos{2x}$$
$$f'(x) = -\sin{2x}\cos{2x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Jordan normal form of $\left(\begin{smallmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{smallmatrix}\right)$ I want to find the Jordan normal form of $A=\begin{pmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{pmatrix}$, but somewhere I think that I make a mistake (I am quite new to the computation of the Jordan normal form).
I could find that $\lambda=3$ is $A$'s only eigenvalue and its algebraic multiplicity is $3$.
Afterwards, I solved the system $(A-3I_3)X=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ and I got that $x=\begin{pmatrix} -a-b \\ a \\ b \end{pmatrix}$ for some parametres $a, b\in \mathbb{C}$.
I picked the eigenvectors $X_1=\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix}$ and $X_2=\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$ and then I got stuck. I tried solving both the equations $(A-3I_3)X'=X_1$ and $(A-3I_3)X''=X_2$, but these have no solutions, so I can't find any generalised eigenvectors. What should I do?
| After you get the Jordan form (as psidaga's answer did)
\begin{align*}
J = \begin{pmatrix}
3 & 1 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix},
\end{align*}
in order to determine the similarity matrix $P$ such that $P^{-1}AP = J$, it is is equivalent to solve the system
\begin{align*}
& A\alpha_1 = 3\alpha_1, \tag{1} \\
& A\alpha_2 = \alpha_1 + 3\alpha_2, \tag{2} \\
& A\alpha_3 = 3\alpha_3,
\end{align*}
where $\alpha_1, \alpha_2, \alpha_3$ are three columns of $P$. As you did, you can fix $\alpha_3$ to be $(-1, 0, 1)^T$. To determine $\alpha_1$ and $\alpha_2$, note that $A - 3I_{(3)}$ is nilpotent with nilpotent index $2$, i.e., $(A - 3I_{(3)})^2 = 0$, hence any non-zero vector $\alpha_2$ that is not already an eigenvector and $\alpha_1 := (A - 3I_{(3)})\alpha_2$ would satisfy equations $(1)$ and $(2)$, since $(A - 3I_{(3)})\alpha_1 = (A - 3I_{(3)})^2\alpha_2 = 0$. For example, let $\alpha_2 = (1, 0, 0)^T$, then $\alpha_1 = (1, -2, 1)^T$. It can be easily verified that
\begin{align*}
P = \begin{pmatrix}
1 & 1 & -1 \\
-2 & 0 & 0 \\
1 & 0 & 1
\end{pmatrix}
\end{align*}
is invertible and satisfies $A = PJP^{-1}$.
Of course, $P$ is not unique, depending on what $\alpha_2$ you are choosing.
| {
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Find $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$ $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$
Can we apply Stolz-Cesaro?
$\lim\limits_{n \to \infty}\frac {\frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n}}} {{\ln(n+1)-\ln(n)}}$ =
$\lim\limits_{n \to \infty}\frac{\sqrt n - \sqrt {n+1}}{\sqrt n \sqrt{n+1}\ln(1+\frac{1}{n})}$ = $\lim\limits_{n \to \infty}\frac{1-\sqrt{1+\frac{1}{n}}}{\sqrt{n+1}\ln(1+\frac{1}{n})}$
What can I do from here?
| Option:
$n/√n < 1/√1+1/√2....+1/√n;$
$\dfrac{√n}{2 \log (√n)} < \dfrac{1/√1+1/√2+...1/√n}{\log n};$
$(1/2) \lim_{n \rightarrow \infty} \dfrac{√n}{\log √n} = $?
| {
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$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof Does anyone know hot to prove this inequality?
Having: $a, b, c \gt 0$
$$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$
I tried with the AM-GM inequality but I couldn't get any improvement. I'm on a still point and I don't know how to continue.
Furthermore, I don't know how to get rid of the known term. I tried with the AM-HM inequality but I'm still not getting any results.
Also, just some hints would be appreciated, thanks
| We can prove it without expending by SOS and the Tangent Line method.
Indeed, $$\sum_{cyc}\frac{b^2-a^2}{a+c}=\frac{1}{2}\sum_{cyc}\frac{2b^2-2a^2}{a+c}=\frac{1}{2}\sum_{cyc}\left(\frac{2b^2-2a^2}{a+c}+a-c\right)=$$
$$=\frac{1}{2}\sum_{cyc}\frac{b^2-c^2-(a^2-b^2)}{a+c}=\frac{1}{2}\sum_{cyc}\left(\frac{a^2-b^2}{c+b}-\frac{a^2-b^2}{a+c}\right)=\sum_{cyc}\frac{(a-b)^2(a+b)}{2(a+c)(b+c)}\geq0.$$
| {
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Simplifying $\cos^{-1}x +\cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$ A question has this equation: $$f(x) = \cos^{-1}x + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$$ and you're supposed to simplify it and find $f\left(\frac{2}{3}\right)$ and $f\left(\frac{1}{3}\right)$.
By taking $\cos\alpha = x$, the equation on the right can be simplified to $\cos^{-1}\left(\cos\left(\frac {\pi}{3} - \alpha\right)\right)$. Finally, you get $\frac{\pi}{3}$ as the final answer.
But the answers are $\frac{\pi}{3}$ and $2\cos^{-1}\left(\frac{1}{3}\right)-\frac{\pi}{3}$. How does it work out that way?
| When you used the replacement $\cos \alpha = x$, the expression in the bracket became
$$\frac{\cos \alpha}{2} + \frac{\sqrt{3} \sin \alpha}{2} = \cos \frac{\pi}{3} \, \cos \alpha + \sin \frac{\pi}{3} \, \sin \alpha$$
but note that this can be written as both
$$\cos \left( \frac{\pi}{3} - \alpha \right) \quad \text{or} \quad \cos \left( \alpha - \frac{\pi}{3} \right)$$
In order for $\cos^{-1} ( \cos \theta) = \theta,$ we want to avoid $\,\theta \,$ being negative.
When evaluating $f(\frac{2}{3})$, we have that $\,\cos \alpha = \frac{2}{3} \,$, so $\, 0 < \alpha < \frac{\pi}{3}$; but when evaluating $f(\frac{1}{3})$, we have that $\,\cos \alpha = \frac{1}{3} \,$, so $\, 0 < \frac{\pi}{3} < \alpha$.
Thus
$$\begin{align}
f(\frac{2}{3}) &= \alpha + \cos^{-1} \left( \cos \left[ \frac{\pi}{3} - \alpha \right] \right) \\
\\
&= \alpha + \left( \frac{\pi}{3} - \alpha \right) = \boxed{\frac{\pi}{3}}
\end{align}$$
but
$$\begin{align}
f(\frac{1}{3}) &= \alpha + \cos^{-1} \left( \cos \left[ \alpha - \frac{\pi}{3} \right] \right) \\
\\
&= \alpha + \left( \alpha - \frac{\pi}{3} \right) \\
\\
&= 2 \alpha - \frac{\pi}{3} = \boxed{2 \cos^{-1} \left( \frac{1}{3} \right) - \frac{\pi}{3}}
\end{align}$$
| {
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Integration of a Bessel function times a sine function times a polynomial I have recently faced the following expression (while trying to compute the Fourier transform of a RKKY-like potential):
\begin{align}
\int_0^{\infty} \text{d} r\; J_0 (k r) \frac{\sin \left(\alpha \sqrt{1+r^2} \right) }{(1+r^2)^2} \; r = -\frac{1}{\alpha} \int_0^{\infty} \text{d} r\; J_0 (k r) \frac{1}{(1+r^2)^{3/2}} \frac{\text{d}}{\text d r} \left[\cos \left( \alpha\sqrt{1+r^2} \right) \right]
\end{align}
Although I can solve it numerically, I would like to find a closed-form expression, if there is one.
I found some related questions here, such as,
\begin{align}
\int_0^{\infty} \text{d}r\; J_0 \left( \alpha\sqrt{x^2 +z^2} \right) \, \cos \left(\beta x \right) = \frac{\cos \left(z\sqrt{\alpha^2 -\beta^2} \right)}{\sqrt{\alpha^2 -\beta^2}}
\end{align}
but did not really helped me.
| According to this reference, Table 8.2 , entry (7):
$$\int_0^\infty \dfrac{1}{\left(1+r^2\right)^\frac{3}{2}}J_0(kr)r \; dr = e^{-k}$$
and according to Table 8.2, entry (41):
$$\int_0^\infty \dfrac{\sin\left[\alpha\left(1+r^2\right)^\frac{1}{2}\right]}{\left(1+r^2\right)^\frac{1}{2}}J_0(kr)r \; dr = \dfrac{\cos\left[\left(\alpha^2-k^2\right)^\frac{1}{2}\right]}{\left(\alpha^2-k^2\right)^\frac{1}{2}}\Pi\left(\dfrac{k}{\alpha}-\dfrac{1}{2}\right)$$
where
$$\Pi(t) = \begin{cases} 1 & -\frac{1}{2} < t< \frac{1}{2}\\ 0 & \mathrm{otherwise} \end{cases}$$
Using the convolution property of the Hankel Transform, if the convolution exists
$$\begin{align*} \int_0^\infty \dfrac{1}{\left(1+r^2\right)^\frac{3}{2}}\dfrac{\sin\left[\alpha\left(1+r^2\right)^\frac{1}{2}\right]}{\left(1+r^2\right)^\frac{1}{2}}J_0(kr)r \; dr &= \dfrac{1}{2\pi}e^{-k}**\dfrac{\cos\left[\left(\alpha^2-k^2\right)^\frac{1}{2}\right]}{\left(\alpha^2-k^2\right)^\frac{1}{2}}\Pi\left(\dfrac{k}{\alpha}-\dfrac{1}{2}\right)\\
\\
&= \dfrac{1}{2\pi}\int_0^{2\pi}\int_0^{k\cos\theta+\sqrt{k^2\cos^2\theta-(k^2-\alpha^2)}} e^{-k'}\dfrac{\cos\left[\left(\alpha^2-\left[k^2+k'^2-2kk'\cos\theta\right]\right)^\frac{1}{2}\right]}{\left(\alpha^2-\left[k^2+k'^2-2kk'\cos\theta\right]\right)^\frac{1}{2}}k' \; dk' \; d\theta\\
\\
&\mathrm{or}\\
\\
&= \dfrac{1}{2\pi}\int_0^{2\pi}\int_0^{\alpha^-} e^{-\sqrt{k^2+k'^2-2kk'\cos\theta}}\dfrac{\cos\left[\left(\alpha^2-k'^2\right)^\frac{1}{2}\right]}{\left(\alpha^2-k'^2\right)^\frac{1}{2}}k' \; dk' \; d\theta\\
\end{align*}$$
Neither of which look possible to easily evaluate symbolically. Although maybe one of them takes less effort to evaluate numerically than your original integral.
| {
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"url": "https://math.stackexchange.com/questions/3889025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+x^2y^2-2xy^2+y^2+x^2y^2-2xy^2+y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+2x^3+x^2-2x^3+4x^2-2x+x^2+2x+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+6x^2+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
And I have no idea what to do next.
Any help would be much appreciated
| From the beginning, move one of the absolute values to the other side. It will simplify the calculations:
$$|z+i|=2-|z-i|$$
Square this and you get $$|z+i|^2=(2-|z-i|)^2=4-4|z-i|+|z-i|^2$$
$$x^2+(y+1)^2=4-4|z-i|+x^2+(y-1)^2$$
$$x^2+y^2+2y+1=4-4|z-i|+y^2-2y+1$$
$$4y-4=-4|z-i|$$
$$y-1=-|z-1|$$
$$(y-1)^2=x^2+(y-1)^2$$
The only solution is $x=0$. This is independent of $y$. Next is to check if by squaring we add extra solutions. Not all $y$'s might be valid.
$$|y-1|+|y+1|=2$$
If $y\ge1$, the equation is $$2y=2$$So $y=1$. If $y\le -1$ then $$-(y-1)-(y+1)=2$$, meaning $y=-1$.In between, $$-(y-1)+y+1=2$$or $2=2$. The final answer is $z=\alpha i$, for $\alpha\in[-1,1]$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a simple, but tight lower bound for the error made when $\sum_{n=1}^{k}\frac{1}{n^2}$ is used to approximate $\frac{\pi^2}{6}$? As a math-for-fun exercise, I've recently been seeking bounds for the error $R_k$ made when using $\sum_{n=1}^{k}1/n^2$ to estimate its beautiful sum $\pi^2/6$. Applying the Comparison Test for series multiple times, I derived the following estimates:
$$\frac{1}{k+1}<\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k<\ln\left(1+\frac{1}{k}\right)<\frac{1}{2k+2}+\frac{1}{2k}$$
The lower estimate
$$\frac{\pi\coth(\pi)-1}{2}-\sum_{n=1}^{k}\frac{1}{n^2+1}<R_k$$
which I derived with WolframAlpha's help, is pretty useless. $1/(k+1)<R_k$ is good, but I want to improve it. Given how clean and tight the upper estimate
$$R_k<\ln\left(1+\frac{1}{k}\right)$$
is, I figured I could find a similar lower estimate that's just as clean and tight. After thinking for a while, I came up empty handed. I can't seem to find a positive sequence $a_n$ lying between $1/(n^2+1)$ and $1/n^2$ for which $\sum_{n=1}^{k}a_n$ has a clean expression. Any ideas or hints?
Edit: I'm not trying to prove the convergence of $\sum_{n=1}^{\infty}\frac{1}{n^2}$ nor any other series.
| Consider $$g(n) = \frac{1}{n-1/2} - \frac{1}{n+1/2}$$
Then
$$ g(n) - \frac{1}{n^2} = \frac{1}{4n^4 - n^2} > 0 \ \text{for}\ n \ge 1$$
Now $1/(4n^4 - n^2)$ is a decreasing function of $n$ for $n > 1$, so
$$\eqalign{\sum_{n=N+1}^\infty \frac{1}{n^2} &= \frac{1}{N+1/2} - \sum_{n=N+1}^\infty \frac{1}{4n^4-n^2}\cr & > \frac{1}{N+1/2} - \int_{N}^\infty \frac{dx}{4x^4 - x^2} \cr
&= \frac{1}{N+1/2} - \ln \left(\frac{2N+1}{2N-1}\right) + \frac{1}{N}}$$
while on the other side
$$ \eqalign{\sum_{n=N+1}^\infty \frac{1}{n^2} &< \frac{1}{N+1/2} - \int_{N+1}^\infty \dfrac{dx}{4x^4 - x^2}\cr
&= \frac{1}{N+1/2} - \ln\left(\frac{2N+3}{2N+1}\right)+ \frac{1}{N+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893405",
"timestamp": "2023-03-29T00:00:00",
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Number Theory : Find the group of A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ Find the group A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ ?
Polynomial Long Division we get
$\frac{x^{2}}{2}-\frac{x}{4}-\frac{11}{8}+\frac{27}{8\left(2x+1\right)}$
but how i can from here find all x such that $x \in \mathbb{Z}$ ??
i did use the hint for $y = 2x+1$ so that $x =\frac{y-1}{2}$
made the equation to be $\frac{(y-1)^3}{8}-\frac{3(y-1)}{2}+2=\frac{y^3-3y^2-9y+27}{8}$
$\frac{\frac{y^3-3y^2-9y+27}{8}}{y}=\frac{y^3-3y^2-9y+27}{8y}$
I can not understand with the help of the clues, can some 1 some formal proof ?
| Hint: Since $2x+1$ is odd for all integers $x$, we have that $\frac{x^3-3x+2}{2x+1}\in\mathbb Z$ if and only if $\frac{2(x^3-3x+2)}{2x+1}\in\mathbb Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3896747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sketching a set of $Re(z^2(1+i))$ on the Complex plane. So I am trying to sketch a set of complex numbers that is characterized by inequality: $ Re((1+i)z^2) < 0 $.
I know that if I take $ z = a + bi $ , then:
$$Re(z^2) \implies (a + bi)^2 = a^2 + 2abi - b^2 \implies Re(z^2) = a^2 - b^2$$
Similarly:
$$Re(z^2(1+i)) \implies (a^2 + 2abi - b^2)(1+i) = a^2 + 2abi - b^2 + a^2i - 2ab - b^2i \implies Re(z^2(1+i)) = a^2 - 2abi + b^2 = (a-b)^2 $$
So how am I supposed to sketch a set of complex numbers $(a-b)^2 $ on a 2D plane? Or did I do something wrong?
| When dealing with complex number, polar form is almost always more useful than breaking it into real and imaginary part.
$$
Re((1+i)z^{2})<0\implies \frac{\pi}{2}<\arg{((1+i)z^{2})}<\frac{3\pi}{2}
$$
Now we use the following
$$
\begin{align}
\arg{((1+i)z^{2})}&=\arg{(1+i)}+ 2\arg{(z)}\\
&=\frac{\pi}{4}+2\arg{(z)}
\end{align}
$$
Substitute this to the first equation to obtain
$$
\frac{\pi}{8}<\arg{(z)}<\frac{5\pi}{8}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Domain of function $\frac{1}{x + \sqrt{x+1}}$ I need to find the domain of the fuction: $$\frac{1}{x + \sqrt{x+1}}$$
I can see that the conditions for the domain are $x+\sqrt{x+1} \neq 0$ and $x + 1\geq 0$. Hence:
$$x+\sqrt{x+1} \neq 0 \quad \wedge \quad x + 1\geq 0$$
$$x \neq -\sqrt{x+1} \quad \wedge \quad x\geq -1$$
$$x^2 \neq |x+1| \quad \wedge \quad x\geq -1$$
Since $x \geq -1$ the expression $x+1$ is never negative, therefore:
$$x^2 \neq x+1 \quad \wedge \quad x\geq -1$$
$$x^2 - x - 1 \neq 0 \quad \wedge \quad x\geq -1$$
$$x \neq \frac{1\pm \sqrt{5}}{2} \quad \wedge \quad x\geq -1$$
Now both $\frac{1 + \sqrt{5}}{2}$ and $\frac{1 - \sqrt{5}}{2}$ are greater than $-1$. So why isn't my domain $$\mathbb{R} \setminus \{\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\}$$
When I plot the graph I can see that $\frac{1 + \sqrt{5}}{2}$ is part of the domain, but analitically I can't see why or derive from these equations that conclusion.
| $$x\ge-1$$ is obvious.
Then the denominator is zero when
$$x+\sqrt{x+1}=0$$ or by squaring $$x^2=x+1$$
and there is a single valid root
$$x=1-\phi\ge-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of $(\sin x + \cos x)^3 + \frac{1}{\sin^2x\cos^2x}$ Let $\sin x + \cos x = t\qquad (\vert t \vert \leq \sqrt2)$
and solve it as $t^3 + \dfrac{4}{(t^2 - 1)^2}$
Is there any easier solution to solve this $(\sin x + \cos x)^3 + \dfrac{1}{\sin^2x\cos^2x}$?
Is there any available inequality can solve it?
Please help!
| There may not be a specific inequality, but sure there is a faster way.
$$(\sin x + \cos x)^3 + \dfrac{1}{\sin^2x\cos^2x}=2\sqrt{2}\left(\sin\left(x+\frac{\pi}{4}\right)\right)^{3}+4\csc^{2}\left(2x\right)$$
Observe that only the expression having an odd power can contribute to reducing the value of the expression. That is, when $\sin(x+\pi/4)$ is the lowest.
A possible solution is $x=-\dfrac{3\pi}{4}$ which gives the global minimum as $4-2\sqrt{2}=1.1715728\dots$
Here is a graphical solution in aid of the above.
| {
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What is the remainder when $1^n + 2^n + 3^n + \ldots + 99^n$ is divided by $1 + 2 + 3 + \ldots + 99$? My first idea on how to approach was to create a polynomial expansion for the first few terms and then try to find a pattern for the rest but this became cumbersome and I don't think that this is a right approach as this would quickly become a problem that involves factorials and I have not covered modular arithmetic, is there a way to approach this problem in such a way that does not involve using factorials with mods.
| An ad-hoc solution:
$n$ odd one can group $1^n+99^n, 2^n+98^n,...50^n$ to get $S_n$ divisible by $50$ and similarly $1^n+98^n,..49^n+50^n$ to get $S_n$ divisible by $99$ so as noted above the remainder is zero.
For $n$ even we notice that modulo $3$ each group $3k-2,3k-1, k=1,..33$ gives a sum that is $2$ modulo $3$ and since there are $33$ such sums, $S_n$ is then $3$ modulo $9$ as the multiples of $3$ give zero modulo $9$ for $n \ge 2$ but there is an odd number of them $15$ that are odd so $S_n=9\times (2m+1)+q$
Modulo $5$ we have to split into $n=4k+2$ when we have $20$ sums that are $0$ modulo $5$ so $S_n$ is divisible by $25$ and then $n=4k$ when similarly we get $S_n=80=5$ modulo $25$ and since $S_n$ is always even, $S_n$ is $30$ modulo $50$
Modulo $11$ we have two cases - $n$ multiple of $10$ we get then $S_n=90=2$ modulo $11$ ($9$ sums of $10$ each) and $n$ not multiple of $10$ when $S_n$ is divisble by $11$ (easily seen for $n \le 8$ since $S_n$ has a factor of $99$ and a denominator that doesn't contain $11$ and then by periodicity since $S_n=S_{n+10}$ modulo $11$)
Putting it together we get:
$n=2k+1$ we get $0$
$n=4k+2, n\ne 10m$ the remainders modulo $50,9,11$ are $0,3,0$ so we get $1650$
$n=4k, n \ne 10m$ the remainders modulo $50,9,11$ are $30,3,0$ so we get $3630$
$n=20k+10$ the remainders modulo $50,9,11$ are $0,3,2$ so we get $750$
$n=20k$ the remainders modulo $50,9,11$ are $30,3,2$ so we get $2730$ and that's all!
| {
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"url": "https://math.stackexchange.com/questions/3902114",
"timestamp": "2023-03-29T00:00:00",
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Quadratic equation where constant term is an integral function Consider a quadratic equation ${x^2} + 2x = k + \int\limits_0^1 {\left| {t + k} \right|dt}$. Then choose the correct option(s),
(A) Roots are Real
(B) Roots are Imaginary
(C) Roots are Distinct
(D) Roots are complex number
My approach is as follow
Let $-c=k + \int\limits_0^1 {\left| {t + k} \right|dt}$
$a=1$ & $b=2$
$b^2-4ac$
$4-4(k + \int\limits_0^1 {\left| {t + k} \right|dt})$
How do I check the nature of the roots
| The discriminant $b^2 - 4ac$ is actually
$$d = 4 + 4(k + \int_{0}^{1}\left| t + k \right| dt) \tag{1}\label{eq1A}$$
Regarding calculating $k + \int_{0}^{1}\left| t + k \right| dt$, there are $3$ cases to consider:
Case $1$: $\; k \lt -1$
Since $t + k \lt 0$ for $t$ from $0$ to $1$, this gives
$$\begin{equation}\begin{aligned}
k + \int_{0}^{1}\left| t + k \right| dt & = k + \int_{0}^{1}(-t - k) dt \\
& = k + \left. \left(-\frac{t^2}{2} - kt\right)\right|_{0}^{1} \\
& = k + \left(-\frac{1}{2} - k\right) \\
& = -\frac{1}{2}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
In \eqref{eq1A}, this gives $d = 2$.
Case $2$: $\; -1 \le k \lt 0$
This requires splitting the integration into $2$ parts, the initial one for $t$ from $0$ to $-k$ where $t + k \le 0 \implies \left|t + k\right| = -t -k$ and a second part for $t$ from $-k$ to $1$ where $t + k \ge 0 \implies \left|t + k\right| = t + k$, giving
$$\begin{equation}\begin{aligned}
k + \int_{0}^{1}\left| t + k \right| dt & = k + \int_{0}^{-k}(-t - k) dt + \int_{-k}^{1}(t + k) dt \\
& = k + \left. \left(-\frac{t^2}{2} - kt\right)\right|_{0}^{-k} + \left. \left(\frac{t^2}{2} + kt\right)\right|_{-k}^{1} \\
& = k - \frac{k^2}{2} + k^2 + \frac{1}{2} + k - \frac{k^2}{2} + k^2 \\
& = 2k + k^2 + \frac{1}{2} \\
& = k^2 + 2k + 1 - 1 + \frac{1}{2} \\
& = (k + 1)^2 - \frac{1}{2}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
For $k \ge -1$, the value above is an increasing function, going from $-\frac{1}{2}$ to $\frac{1}{2}$. Thus, $d$ in \eqref{eq1A} goes from $2$ to $6$.
Case $3$: $\; k \ge 0$
This obviously gives $d \gt 0$, but doing the calculations results in
$$\begin{equation}\begin{aligned}
k + \int_{0}^{1}\left| t + k \right| dt & = k + \int_{0}^{1}(t + k) dt \\
& = k + \left. \left(\frac{t^2}{2} + kt\right)\right|_{0}^{1} \\
& = k + \left(\frac{1}{2} + k\right) \\
& = 2k + \frac{1}{2}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Thus, $d$ in \eqref{eq1A} is $6$ at $k = 0$, and increases without bound as $k$ goes up.
This shows all $3$ cases gives $d \gt 0$, so the roots are real and distinct. Since the roots are also always complex numbers, this means options (A), (C) and (D) are true, with only (B) being false since there are no imaginary roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$.
I had the following idea, we write:
$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$
Let's pretend the identity we want to prove is true, then:
$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$
We take then $k-x=j$ and solve for $x$. If the given identities are true, $x$ must be the sum of $2^2+4^2+\dots+(2n)^2$, and we have that
$$x= \frac{2n (n+1) (2 n+1)}{3} $$
We still don't know that $x=2^2+4^2+\dots+(2n)^2$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?
Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.
| Hint: notice that $$\sum(2k)^2=\sum4k^2=4\sum k^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Fractions in Questions and Answers
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