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Could someone please explain the steps in the Maclauren series?
Find the limit of $$\lim_{x\to 0}{{1-(\cos x)({\cos2x})^{1/2}(\cos3x)^{1/3}}\over{x^2}}$$
Steps to solving this problem, from the book:
Step 1: $$\lim_{x\to 0}{{1-(1-\frac{x^2}{2})({1-2x^2})^{1/2}(1-\frac{9x^2}{2})^{1/3}}\over{x^2}}$$
Step 2: $$\lim_{x\to 0}{{1-(1-\frac{x^2}{2})({1-x^2})(1-\frac{3x^2}{2})}\over{x^2}}$$
Step 3: $$\lim_{x\to 0}{{1-(1-\frac{3x^2}{2})(1-\frac{3x^2}{2})}\over{x^2}}$$
Step 4: $$\lim_{x\to 0}{\frac{1-1+3x^2}{x^2}=3 }$$
Could someone please explain to me, what happened in step 2. How did we loose the roots ?
And what happened from step 2 -> step 3 I have no idea what happened there ... :(
| You should probably get rid of the book. Most calculus texts are like this. The right approach is to use the little / big o notation.
The key here is to use Taylor / Maclaurin expansions as $x\to 0$ $$ \cos x=1-\frac{x^2}{2}+o(x^2)\tag{1}$$ and $$(1+x)^n=1+nx+o(x)\tag{2}$$ where $o(f(x)) $ represents a function $g(x) $ such that $g(x) /f(x) \to 0$.
The starting step (based on expansion $(1)$ above) thus needs to be written like $$\lim_{x\to 0}\dfrac{1-\left(1-\dfrac {x^2}{2}+o(x^2)\right)\left(1-2x^2+o(x^2)\right) ^{1/2}\left(1-\dfrac{9x^2}{2}+o(x^2)\right)^{1/3}} {x^2} $$ and the next step (using $(2)$) becomes $$\lim_{x\to 0}\dfrac{1-\left(1-\dfrac {x^2}{2}+o(x^2)\right)\left(1-x^2+o(x^2)\right)\left(1-\dfrac{3x^2}{2}+o(x^2)\right)} {x^2} $$ Finally we have via multiplication $$\lim_{x\to 0}\dfrac{1-\left(1-3x^2+o(x^2)\right)} {x^2}=3 $$
A simpler approach is to use standard limits $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2},\,\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$ which are equivalent to the Taylor expansions given earlier.
The numerator of the expression under limit is of the form $1-abc$ where $a, b, c$ all tend to $1$ as $x\to 0$. We can do a split like $$1-abc=1-a+a(1-bc)$$ and thus we have the desired limit as $$\lim_{x\to 0}\frac{1-\cos x} {x^2}+\lim_{x\to 0}\cos x\cdot\frac{1-(\cos 2x)^{1/2}(\cos 3x)^{1/3}} {x^2}$$ which is same as $$\frac{1}{2}+\lim_{x\to 0}\frac{1-(\cos 2x)^{1/2}(\cos 3x)^{1/3}}{x^2}$$ Applying the same technique we get $$\frac{1}{2}+\lim_{x\to 0}\frac{1-(\cos 2x)^{1/2}}{x^2}+\lim_{x\to 0}(\cos 2x)^{1/2}\cdot\frac{1-(\cos 3x)^{1/3}}{x^2}$$ The first limit above is evaluated as $$\lim_{x\to 0}\frac{1-(\cos 2x)^{1/2}}{1-\cos 2x}\cdot\frac{1-\cos 2x}{(2x)^2}\cdot 4=\frac{1}{2}\cdot\frac{1}{2}\cdot 4=1$$ and thus desired limit is equal to $$\frac{3}{2}+\lim_{x\to 0}\frac{1-(\cos 3x)^{1/3}}{x^2}$$ The last limit is evaluated in similar manner to get $3/2$ and thus the final answer is $3$.
This technique has been used for a more complicated problem here.
| {
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Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$. Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. This implies $(yz)x = x^{-1}x = 1$. Therefore $xyz = 1 \implies yzx = 1$.
By applying the same argument to $y(zx) = 1$ one can prove that $xyz = 1 \implies yxz =1$.
However while trying to my proof online I stumbled upon the following counterexample for the proposition $xyz = 1 \implies yxz =1$. If we take $G$ to be the group of $2\times 2$ matrices and let $x = \left( \begin{array} { c c } { 1 } & { 2 } \\ { 0 } & { 2 } \end{array} \right)$, $y = \left( \begin{array} { l l } { 0 } & { 1 } \\ { 2 } & { 1 } \end{array} \right)$ and $z = \left( \begin{array} { c c } { - 1 / 2 } & { 3 / 4 } \\ { 1 } & { - 1 } \end{array} \right)$. Then $x y z = \left( \begin{array} { c c } { 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right) = 1$ but $y x z = \left( \begin{array} { c c } { 2 } & { - 2 } \\ { 5 } & { - 9 / 2 } \end{array} \right) \neq 1$.
I don't understand where my proof went wrong.
| Since $G$ is a group then if $x\in G$ , there exists $x^{-1}\in G$ such that $$x\cdot x^{-1}=x^{-1}\cdot x=e\in G$$having this result and exploiting the other properties of group we obtain $$x^{-1}\cdot (xyz)=(x^{-1}\cdot x)\cdot yz=e\cdot yz=yz=x^{-1}\cdot e=x^{-1}$$therefore $$yz\cdot x=yzx=x^{-1}\cdot x=1$$and the statement has been proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is an integer. Prove that $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer using mathematical induction.
I tried using mathematical induction but using binomial formula also it becomes little bit complicated.
Please show me your proof.
Sorry if this question was already asked. Actually i did not found it. In that case only sharing the link will be enough.
| @I like Serena has a great answer but since the OP asked for a proof by induction, I'll show what that would look like. Define
$$f(k)=\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}=\frac{15k^7 + 21k^5+70k^3-k}{105}$$
For our base case, let $k=1$. Then we have
$$f(1)=\frac{15+21+70-1}{105}=1$$
which is an integer. Now suppose $f(k)$ is an integer for some $k\geq 1$. We want to prove that $f(k+1)$ is also an integer. To that end, observe that
\begin{align}
f(k+1)&=\frac{15(k+1)^7 + 21(k+1)^5+70(k+1)^3-(k+1)}{105}\\
&=\frac{15k^7 + 105k^6+336k^5+630k^4 + 805k^3+735k^2+419k+105}{105}
\end{align}
Therefore
\begin{align}
f(k+1)-f(k)&=\frac{105k^6+315k^5+630k^4+735k^3+735k^2+420k+105}{105}\\
&=\frac{105(k^6+3k^5+6k^4+7k^3+7k^2+4k+1)}{105}\\
&= k^6+3k^5+6k^4+7k^3+7k^2+4k+1
\end{align}
Which is an integer, say $N$. Rearranging this gives $f(k+1)=f(k)+N$ and since $f(k)$ is assumed to be an integer from the induction hypothesis, $f(k+1)$ is the sum of two integers, hence an integer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$
Prove that for $a>1$,$b>1$ and $c>1$ where $a,b,c\in \mathbb{R}$
$$\frac{1}{3}(a+b+c)^2\leq a^2 + b^2 + c^2 + 2(a-b+1).$$
My attempt: it is not so clear why is $a>1$, $b>1$ and $c>1$, but I factor left side, and all put ond side to prove $\geq 0$, but it is not an easy inequality.
| Well here is a start. Set $A=a+1$ and $B=b-1$ and $C=c$ and the equation becomes $$\frac 13(A+B+C)^2\le A^2+B^2+C^2$$
Now consider $$(A-B)^2+(B-C)^2+(C-A)^2\ge 0$$ so that $$2A^2+2B^2+2C^2-2AB-2BC-2AC\ge 0$$ so that $$AB+BC+AC\le A^2+B^2+C^2$$
Now $$(A+B+C)^2=A^2+B^2+C^2+2AB+2BC+2AC\le 3(A^2+B^2+C^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving uniform continuity of $f(x) = \sqrt{1-x^2}$ on $[-1,1]$ I want to prove that $f(x) = \sqrt{1-x^2}$ is uniform continuous on the interval $[-1,1]$. Let $f(x) = \sqrt{1 - x^2}$. Then I need to show:
$\forall \epsilon > 0 \enspace \exists \delta > 0 \enspace \forall x,y \in [-1,1] \enspace |x-y|<\delta \implies |f(x)-f(y)| < \epsilon$.
Steps:
$|f(x)-f(y)| = |\sqrt{1-x^2} - \sqrt{1-y^2}|
= \bigr\lvert\frac{(\sqrt{1-x^2}-\sqrt{1-y^2})(\sqrt{1-x^2}+\sqrt{1-y^2})}{(\sqrt{1-x^2}+\sqrt{1-y^2})} \bigr\rvert = \bigr\lvert \frac{1-x^2-(1-y^2)}{\sqrt{1-x^2}+\sqrt{1-y^2}} \bigr\rvert = \bigr\lvert \frac{x^2-y^2}{\sqrt{1-x^2}+\sqrt{1-y^2}} \bigr\lvert = \frac{|x+y||x-y|}{\sqrt{1-x^2}+\sqrt{1-y^2}} < \frac{2\delta}{\sqrt{1-x^2}+\sqrt{1-y^2}} \leq \frac{2\delta}{\sqrt{2 - x^2 - y^2}} $
The problem is that I can't bound this by constant as $\sqrt{2-x^2-y^2} \rightarrow 0$ as $x,y \rightarrow1$. I feel I'm running out of tricks. How can one solve this?
| Hint:
$$|\sqrt{1-x^2} - \sqrt{1-y^2}|^2 \leqslant |\sqrt{1-x^2} - \sqrt{1-y^2}||\sqrt{1-x^2} + \sqrt{1-y^2}| = |x^2 - y^2|$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Distinct prime number solution of an equation I am trying to solve the following question:
$$\textrm{Find all distinct prime numbers } p, q \textrm{ and } r \textrm{ such that}$$
$$3p^4-5q^4-4r^2=26$$
Progress:
$$p \geq 5, q = 3, r \geq 5$$
With these constraints, I found a solution: $p = 5, q = 3, r = 19$
My Attempt:
Let $p$ be a prime number. Then
$$p^4 \equiv p^2 \equiv 1(p \geq 5), 3(p = 3), 4(p = 2) \mod{6}$$
First notice in the equation, neither $p$ nor $q$ can be $2$ as $p$, $q$ and $r$ are distinct.
$$3\times(1, 3)-5\times(1, 3)-4\times(1, 3, 4) \equiv 2 \mod{6}$$
Possible solutions would be
$$\begin{align}
3\times1-5\times3-4\times1&\equiv\\
3\times1-5\times3-4\times4&\equiv\\
3\times3-5\times3-4\times1&\equiv\\
3\times3-5\times3-4\times4&\equiv2\mod{6}
\end{align}$$
So $q \equiv 3 \mod{6} \Rightarrow q = 3$
(And $p, r$ is not $3$)
Trying $p = 2$ and $r = 2$ shows that they don't work.
Therefore:
$$p \geq 5, q = 3, r \geq 5$$
With these constraints, I found a solution: $p = 5$, $q = 3$, $r = 19$
And indeed a computer search agrees with this.
But how to proceed?
Please help, thank you for reading <3
Gareth
| $$3p^4=5q^4+4r^2+26$$
If $s\ne 5$ is an odd prime then $s^2$ ends with $1$ or $9$, and $s^4$ ends with $1.$
If $r>5$ or $r=3$ then the right side ends with $0$ or $7$ (which is impossible) so $5\mid 3p^4\implies p=5$. So the left side is odd and thus right also. But this can be only if $q=2$. Then $$ 3\cdot 5^4 -5\cdot 2^4 -26= 4r^2\implies r=...$$
If $r=2$ we have $3p^4=5q^4+42 \implies q=3 \implies p=...$
If $r=5$ we have $3p^4=5q^4+126 \implies q=3 \implies p=...$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Help with complex numbers Algebraic form of the $z^3 = (3 + i)^6$. Can someone help me to solve this? My answer is $z = 8 + 6i$, but I'm not sure that this is true.
| Whenever you have $z^3 = k$ there will be $3$ solutions.
If $z^3 = (3+i)^6 = ((3+i)^2)^3$ then one solution is obviously $z = (3+i)^2 = 9 + 6i + i^2 = 9+6i -1 = 8+6i$.
But that is only one solution.
However $w^3 = 1$ has three solutions.
If we let $w$ be one of the solutions and we let $z = w(8+6i) = w(3+i)^2$ then $z^3 = w^3((3+i)^2)^3 = 1*(3+i)^6 = (3+i)^6$.
So if the three solutions to $w^3=1$ are $1, b,c$ then the three solutions to $z^3 = (3+i)^6$ are $8 + 6i, b(8+6i), c(8+6i)$.
So we nee to solve $w^3 = 1$.
Let $w = m+ni$ (where $m,n$ are real) then $(m+ni)^3 = m^3 + 3m^2ni + 3mn^2i^2 + n^3i^3=$
$m^3 + 3m^2ni -3mn^2 - n^3i = $
$(m^3 - 3mn^2) + (3m^2n-n^3)i = $
$1 + 0i$
So $m^3 -3mn^2 =1$ and $3m^2n -n^3=0$.
$3m^2n -n^3= n(3m^2-n^2) = n(m\sqrt 3 -n)(m\sqrt 3 + n) = 0$ so
either:
1) $n = 0$ and $m^3 = 1$ so $m = 1$ and $w = 1 + 0i = 1$.
2) $m\sqrt 3 - n = 0$ and $m\sqrt 3 = n$ so $m^3 -3mn^2 = 1$ and so
$m^3 - 3m(m\sqrt 3)^2 = m^3 - 3m(3m^2) =m^3 -9m^3 = -8m^3 =1$ so $m = -\frac 12$ and $w = -\frac 12 -\frac {\sqrt 3}2 i$.
3) $m\sqrt 3+ n = 0$ so $n = -m\sqrt 3$ and so
$m^3 - 3m(-m\sqrt 3)^2=.... = 1$ so $m = -\frac 12$ and $w = -\frac 12 + \frac {\sqrt 3}2i$.
So the three solutions to $z^3 =(3+i)^6$ are:
1)$8+6i$
2)$(-\frac 12 -\frac {\sqrt 3}2 i)(8 + 6i) = -4 -4\sqrt 3i - 3i -3\sqrt 3 i^2 = (-4+3\sqrt 3) -(4\sqrt 3-3)i$.
and 3) $(-\frac 12 +\frac {\sqrt 3}2 i)(8 + 6i) = -4 +4\sqrt 3i - 3i +3\sqrt 3 i^2 = (-4-3\sqrt 3) +(4\sqrt 3-3)i$
....
However once you learn polar notation that $z = m + ni = r(\cos \theta + i\sin \theta) := re^{i\theta}$ where $r = \sqrt{m^2 + n^2}$ and $\theta$ is the solutions to $\cos \theta = \frac mr$ and $\sin \theta = \frac nr$, once you learn that these problems become MUCH easier.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I solve limits like this by property $ \left(a-b\right)\left(a+b\right)=a^2-b^2 $
I made this far:
$$
\lim _{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)
=n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}=\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)} = \frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}$$
I will appreciate every help. Thank you
| $$ \begin{align}L&=\lim_\limits{n\to \infty }\left(n^3\left(\sqrt{n^2+\sqrt{n^{4\:}+1}}-\sqrt{2}n\right)\right)\\&
=\lim_\limits{n\to\infty}n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\\&=\lim_\limits{n\to\infty}\frac{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}-n^4\sqrt{2}\right)\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}{\left(n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}\right)}\\&=\lim_\limits{n\to\infty}\frac{-n^8+n^6\sqrt{n^4+1}}{n^3\sqrt{n^2+\sqrt{n^4+1}}+n^4\sqrt{2}}\\&=\lim_\limits{n\to\infty}\dfrac{-n^5+n^3\sqrt{n^4+1}}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt 2}\\&=\lim_\limits{n\to\infty}\dfrac{-n^5+n^3\cdot n^2\sqrt{1+\dfrac1{n^4}}}{n\sqrt{1+\sqrt{1+\dfrac1{n^4}}}+n\sqrt2}\\&=\lim_\limits{n\to\infty}\dfrac{n^5\left[\sqrt{1+\dfrac1{n^4}}-1\right]}{n\left[\sqrt{1+\sqrt{1+\dfrac1{n^4}}}+\sqrt2\right]}\\&\boxed{\text{Let }n=\dfrac1m,\text{ so as }n\to\infty,m\to 0}\\&=\lim_\limits{m\to0}\dfrac{\sqrt{1+m^4}-1}{m^4\sqrt{1+\sqrt{1+m^4}}+\sqrt2}\\&\boxed{\text{As }x\approx 0,(1+x)^n\approx 1+nx}\\&=\lim_\limits{m\to 0}\dfrac{1+\dfrac12m^4-1}{m^4\sqrt{1+1+\dfrac12m^4}+\sqrt 2}\\&=\lim_\limits{m\to0}\dfrac{\dfrac12}{\sqrt{2+\dfrac12m^4}+\sqrt2}\\&=\lim_\limits{m\to0}\dfrac1{2\sqrt2\sqrt{1+\dfrac14m^4}+2\sqrt2}\\&=\lim_\limits{m\to0}\dfrac1{2\sqrt2\left(1+\dfrac18m^4\right)+2\sqrt2}\\&=\boxed{\dfrac1{4\sqrt2}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find its arc length. Let C be the graph of hypocycloid $x^{2/3} + y^{2/3} = 1$ oriented clockwise. Parametrize the curve and find its arc length.
Attempt:
let $x = \sin^{3}t, y = \cos^3({t})$
$r(t) = (\sin^{3}(t), \cos^3(t)), 0 \le t \le 2\pi$
$r'(t) = (3 \sin^2(t) \cos(t), -3 \cos^2(t) \sin(t))$
\begin{align*}
||r'(t)||
& = \sqrt{3 \sin^2(t) + \cos(t))^2 + (3 \cos^2(t) + \sin(t)^2} \\
& = \sqrt{9 \sin^4 + \cos^2(t)+ (\sin^2(t) + \cos^2(t)) } \\
& = \sqrt{9 \sin^2(t) + \cos^2(t)}
= \sqrt{(3 \sin(t) \cos(t))^2} \\
& = 3 \sin(t) \cos(t)
= \frac{3}{2} (2 \sin(t) \cos(t))
= \frac{3}{2} \sin(2t).
\end{align*}
Now arc length:
$$
L = \int_{0}^{2\pi} \frac{3}{2} \sin(2t)dt
= 4 \cdot \frac{3}{2} \cdot \frac{1}{2}\bigg[-\cos(2t) \bigg]_{0}^{2\pi}
= 3(1+1)
= 6.$$
Is this correct?
| Now $||r'(t)||=\sqrt{(3 \sin^2(t) \cos (t))^2+(-3 \cos^2(t) \sin (t))^2}=\sqrt{9\sin^4(t)\cos^2(t)+9\cos^4(t)\sin^2(t)}=\sqrt{9\sin^2(t)\cos^2(t)(\sin^2(t)+\cos^2(t)}=\sqrt{9\sin^2(t)\cos^2(t)}=|3\sin(t)\cos(t)|=|\frac{3}{2}\sin(2t)|$
$$L=\int_0^{2\pi}\frac{3}{2} |\sin(2t)| \ dt=\frac{3}{2}\int_0^{2\pi}|\sin(2t)| \ dt=\frac{3}{2}\big(\int_0^{\frac{\pi}{2}}\sin(2t) \ dt-\int_{\frac{\pi}{2}}^{\pi}\sin(2t) \ dt+\int_\pi^{\frac{3\pi}{2}}\sin(2t) \ dt-\int_{\frac{3\pi}{2}}^{2\pi}\sin(2t) \ dt\big)=\frac{3}{2}\cdot \frac{1}{2}[(-\cos(2t))_0^{ \frac{\pi}{2}}+(\cos(2t))_{\frac{\pi}{2}}^{ \pi}+(-\cos(2t))_{\pi}^{\frac{ 3\pi}{2}}+(\cos(2t))_{\frac{3\pi}{2}}^{2 \pi}]=\frac{3}{4}(1+1+1+1+1+1+1+1)=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)\,dx$
How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx = \frac43G + \frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $\Im\operatorname{Li}_2\left(2\pm\sqrt3\right)$.
Attempt
$$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx \\ =\int_0^\infty\frac{\operatorname{arcsinh}(2x)}{1+x^2} \, dx\\
=2\int_0^\infty\frac{x\cosh x}{4+\sinh^2x} \, dx\\
=2\int_0^\infty\frac{x\cosh x}{3+\cosh^2x} \, dx\\
=2\int_0^\infty\sum_{n=0}^\infty x(-3)^n\cosh^{-2n-1}(x) \, dx$$
I failed to integrate $x\cosh^{-2n-1}(x)$. Mathematica returns a hypergeometric term while integrating it.
| Here is another variation along a theme which like @DavidG approach uses Feynman's trick of differentiating under the integral sign.
Let
$$I(a) = \int_0^{\frac{\pi}{2}} \operatorname{arcsinh} (a \tan x) \, dx, \qquad a > 1.$$
We are required to find $I(2)$. We start by finding $I(1)$ first, which will be needed later on.
For $a = 1$ we have
\begin{align}
I(1) &= \int_0^{\frac{\pi}{2}} \operatorname{arcsinh} (\tan x) \, dx\\
&= \int_0^{\frac{\pi}{2}} \ln \left (\frac{1 + \sin x}{\cos x} \right ) \, dx\\
&= \int_0^{\frac{\pi}{2}} \ln (1 + \sin x) \, dx - \int_0^{\frac{\pi}{2}} \ln (\cos x) \, dx.
\end{align}
Now the first of these integrals can be found, for example, by rewriting it as
\begin{align}
\int_0^{\frac{\pi}{2}} \ln (1 + \sin x) \, dx &= \int_0^{\frac{\pi}{2}} \ln (1 + \cos x) \, dx\\
&= \int_0^{\frac{\pi}{2}} \ln \left (\frac{1}{2} \cos^2 \frac{x}{2} \right ) \, dx\\
&= \frac{\pi}{2} \ln 2 + 4 \int_0^{\frac{\pi}{4}} \ln (\cos x) \, dx
\end{align}
and then employing the Fourier series representation for $\ln (\cos x)$ found here. The final result is
$$\int_0^{\frac{\pi}{2}} \ln (1 + \sin x) \, dx = 2 \mathbf{G} - \frac{\pi}{2} \ln 2.$$
Here $\mathbf{G}$ is Catalan's constant.
The second of the integrals is very well known. Here
$$\int_0^{\frac{\pi}{2}} \ln (\cos x) \, dx = - \frac{\pi}{2} \ln 2.$$
Thus $I(1) = 2 \mathbf{G}$.
Moving to the main event, by differentiating under the integral sign with respect to $a$ we have
\begin{align}
I'(a) &= \int_0^{\frac{\pi}{2}} \frac{\tan x}{\sqrt{a^2 \tan^2 x + 1}} \, dx\\
&= \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sqrt{a^2 - (a^2 - 1) \cos^2 x}} \, dx.
\end{align}
Observe the term $(a^2 - 1)$ is positive since $a > 1$. Letting $u = \cos x$ one has
$$I'(a) = \int_0^1 \frac{du}{\sqrt{a^2 - (a^2 - 1) u^2}} = \frac{1}{\sqrt{a^2 - 1}} \sin^{-1} \left (\frac{\sqrt{a^2 - 1}}{a} \right ).$$
As we require $I(2)$ observe that
$$I(2) - I(1) = \int_1^2 I'(a) \, da.$$
Thus
$$I(2) = 2 \mathbf{G} + \int_1^2 \frac{1}{\sqrt{a^2 - 1}} \sin^{-1} \left (\frac{\sqrt{a^2 - 1}}{a} \right ) \, da.$$
Integrating by parts leads to
$$I(2) = 2 \mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - \int_1^2 \frac{\cosh^{-1} a}{a \sqrt{a^2 - 1}} \, da.$$
Now let $a = \cosh t$. This gives
$$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - \int_0^{\ln(2 + \sqrt{3})} \frac{t}{\cosh t} \, dt.$$
Then let $t = \ln y$. This gives
$$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - 2\int_1^{2 + \sqrt{3}} \frac{\ln y}{1 + y^2} \, dy.$$
Now let $y = \tan \theta$. Then we have
$$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - 2\int_{\frac{\pi}{4}}^{\frac{5\pi}{12}} \ln (\tan \theta) \, d\theta.$$
Finally, enforcing a substitution of $\theta \mapsto \dfrac{\pi}{2} - \theta$ leads to
\begin{align}
I(2) &= 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) + 2 \int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \ln (\tan \theta) \, d\theta\\
&= 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) + 2\int_0^{\frac{\pi}{4}} \ln (\tan \theta) \, d\theta - 2 \int_0^{\frac{\pi}{12}} \ln (\tan \theta) \, d\theta.
\end{align}
For the first of the integrals we have
$$\int_0^{\frac{\pi}{4}} \ln (\tan x) \, dx = -\mathbf{G}.$$
While for the second of the integrals we have
$$\int_0^{\frac{\pi}{12}} \ln (\tan x) \, dx = -\frac{2}{3} \mathbf{G}.$$
So finally
$$I(2) = 2\mathbf{G} + \frac{\pi}{3} \ln (2 + \sqrt{3}) - 2\mathbf{G} + \frac{4}{3} \mathbf{G},$$
or
$$\int_0^{\frac{\pi}{2}} \operatorname{arcsinh} (2 \tan x) \, dx = \frac{\pi}{3} \ln (2 + \sqrt{3}) + \frac{4}{3} \mathbf{G},$$
as announced.
| {
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Mixed Fractions and Multiplication (with Variables) I stumbled over this expression: $3 \frac{1}{x^3}$.
How should you interpret something like that?
While you could see that as implicitit multiplication ($3 * \frac{1}{x^3}$),
you could also argue that $3 \frac{1}{x^3}$ is a mixed fraction ($3 + \frac{1}{x^3}$).
I think in situations with only numbers or only variables everything should be clear:
$3 \frac{1}{2} = 3 + \frac{1}{2} = 3.5$
$a \frac{b}{c} = a * \frac{b}{c}$
This should also be true: $3 \frac{b}{c} = 3 * \frac{b}{c}$.
But what do our conventions say to something like $3 \frac{1}{x^3}$ or $6 \frac{x}{3}$ or $\frac{1}{x^2}5$? Is there any written standard that you should generally follow? Multiplication or addition?
| This can be written as $$\frac{3}{1}\cdot \frac{1}{x^3}=\frac{3\cdot 1}{1\cdot x^3}=\frac{3}{x^3}$$
| {
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Closed form expression for the harmonic sum $\sum\limits_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$ I'm wondering if one could derive a closed form expression for the series
$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$$
$$\text{With } \text{ } \text{ } \text{ }H_n=\sum_{k=1}^{n}\frac{1}{k}\text{ } \text{ } \text{} \text{ } \text{ }\text{the } n^{th} \text{ harmonic number.}$$
Now, I know series involving harmonic numbers are well suited for a summation by part (or Abel's transformation) approach, but it doesn't lead anywere here, at least not in this state.
Any suggestions ?
| Here is my way of evaluating this sum also offering a different way to calculate that polylogarithmic integral.
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}$$
First let's consider the following central binomial coefficient generating function.
$$\sum _{k=1}^{\infty }\frac{x^{2k}}{k\:4^k}\binom{2k}{k}=-2\ln \left(1+\sqrt{1-x^2}\right)+2\ln \left(2\right)$$
$$-2\sum _{k=1}^{\infty }\frac{1}{k\:4^k}\binom{2k}{k}\int _0^1x^{2k-1}\ln \left(1-x\right)\:dx=4\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx$$
$$-4\ln \left(2\right)\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx$$
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}=4\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx+4\ln \left(2\right)\zeta \left(2\right)$$
$$\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx=\int _0^1\frac{\operatorname{Li}_2\left(x\right)}{x}\:dx-\int _0^1\frac{\operatorname{Li}_2\left(x\right)}{x\sqrt{1-x^2}}\:dx$$
$$=\zeta \left(3\right)-\int _0^{\frac{\pi }{2}}\csc \left(x\right)\operatorname{Li}_2\left(\sin \left(x\right)\right)\:dx=\zeta \left(3\right)-\int _0^1\frac{\operatorname{Li}_2\left(\frac{2t}{1+t^2}\right)}{t}dt$$
$$=\zeta \left(3\right)+2\int _0^1\frac{t\ln \left(t\right)\ln \left(1-t\right)}{1+t^2}\:dt-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t^2\right)}{1+t^2}\:dt-2\int _0^1\frac{\ln \left(t\right)\ln \left(1-t\right)}{t}\:dt$$
$$+\int _0^1\frac{\ln \left(t\right)\ln \left(1+t^2\right)}{t}\:dt+2\int _0^1\frac{t\ln \left(t\right)\ln \left(1-t\right)}{1+t^2}\:dt-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t^2\right)}{1+t^2}\:dt$$
$$=\zeta \left(3\right)+4\underbrace{\int _0^1\frac{t\ln \left(t\right)\ln \left(1-t\right)}{1+t^2}\:dt}_{I}+\frac{1}{4}\underbrace{\int _0^1\frac{\ln ^2\left(1+t\right)}{t}\:dt}_{\frac{1}{4}\zeta \left(3\right)}$$
$$-2\underbrace{\int _0^1\frac{\ln \left(t\right)\ln \left(1-t\right)}{t}\:dt}_{\zeta \left(3\right)}+\underbrace{\int _0^1\frac{\ln \left(t\right)\ln \left(1+t^2\right)}{t}\:dt}_{-\frac{3}{16}\zeta \left(3\right)}$$
The integral $I$ can be found evaluated elegantly in the book (Almost) Impossible Integrals, Sums, and Series through pages $\#98,\#99,\#100$, making use of it's result we have:
$$\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+\sqrt{1-x^2}\right)}{x}\:dx=\frac{23}{16}\zeta \left(3\right)-\frac{9}{4}\ln \left(2\right)\zeta \left(2\right)$$
And in the process we also proved that:
$$\int _0^1\frac{\operatorname{Li}_2\left(x\right)}{x\sqrt{1-x^2}}\:dx=-\frac{7}{16}\zeta \left(3\right)+\frac{9}{4}\ln \left(2\right)\zeta \left(2\right)$$
Collecting the results we finally have:
$$\sum _{k=1}^{\infty }\frac{H_{2k}}{k^2\:4^k}\binom{2k}{k}=\frac{23}{4}\zeta \left(3\right)-5\ln \left(2\right)\zeta \left(2\right)$$
| {
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Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+2}<2$ I wanted to prove this by mathematical induction. I have proved that it is less than two and that the sum is greater than 0 but I am stuck on how to prove it to be greater than 1. Please help!!!
I would like a conclusive proof for this. Thank you so much!
I do believe there was an answer to a question similar to this, but the answers provided there did not use mathematical induction to provide the answer.
| This is a matter of comparison testing.
First render $n+2>n+1>0, n+3>n+1>0, \text{ etc}$. Thereby
$\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{3n+2}<\dfrac{1}{n+1}+\dfrac{1}{n+1}+\dfrac{1}{n+1}+...+\dfrac{1}{n+1}$
where the right side has $2n+2=2(n+1)$ identical terms. So
$\color{blue}{\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{3n+2}<\dfrac{2(n+1)}{n+1}=2}$
The lower bound is trickier. Here we start with the following for $a>b>0$:
$\dfrac{1}{a-b}+\dfrac{1}{a+b}=\dfrac{2a}{a^2-b^2}>\dfrac{2a}{a^2}=\dfrac{2}{a}$
Apply this with $a=2n+1$ and, successively, $b=1, b=2, n=3, ..., b=n$. Thereby
$\dfrac{1}{2n}+\dfrac{1}{2n+2}>\dfrac{2}{2n+1}$
$\dfrac{1}{2n-1}+\dfrac{1}{2n+3}>\dfrac{2}{2n+1}$
$\dfrac{1}{2n-2}+\dfrac{1}{2n+4}>\dfrac{2}{2n+1}$
...
$\dfrac{1}{n+1}+\dfrac{1}{3n+2}>\dfrac{2}{2n+1}$
Add these up:
$\color{blue}{\dfrac{1}{n+1}+\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{3n+2}>2(\dfrac{1}{2n+1})+2(\dfrac{1}{2n+1})+2(\dfrac{1}{2n+1})+...+\dfrac{1}{2n+1}=n(2(\dfrac{1}{2n+1}))+\dfrac{1}{2n+1}=\dfrac{2n+1}{2n+1}=1}$.
| {
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Maximizing $ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$, such that $a^2+b^2+c^2=1 $
If $$a^2+b^2+c^2=1 $$here a,b,c are the real numbers then find the maximum value of $$ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$$
I tried to think with vectors, that is direction cosines of lines.
But then the expression is not getting simplified.
| We need to find a minimal value of $k$ for which the following inequality is true for all reals $a$, $b$ and $c$.
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2\leq k(a^2+b^2+c^2)$$ or
$$(k-41)a^2+(k-34)b^2+(k-25)c^2+24ab+40bc+30ac\geq0$$ or
$$(k-41)a^2+6(4b+5c)a+(k-34)b^2+40bc+(k-25)c^2\geq0,$$ for which we need
$$k>41$$ and
$$9(4b+5c)^2-(k-41)((k-34)b^2+40bc+(k-25)c^2)\leq0$$ or
$$(k^2-75k+1250)b^2+(k^2-66k+800)c^2+40(k-50)bc\geq0$$ or
$$(k-50)((k-25)b^2+(k-16)c^2+40bc)\geq0.$$
But $$(k-25)b^2+(k-16)c^2+40bc\geq16b^2+25c^2+40bc=(4b+5c)^2\geq0,$$ which gives $$k\geq50.$$
For $k=50$ we obtain $$9a^2+6(4b+5c)a+(4b+5c)^2\geq0$$ or
$$(3a+4b+5c)^2\geq0,$$ which gives that the equality occurs for $$3a+4b+5c=0$$ and $$a^2+b^2+c^2=1,$$
which says that $50$ is a maximal value.
| {
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Evaluate $\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $? How to calculate
$\lim\limits_{n\to \infty} \left( \cos(1/n)-\sin(1/n) \right) ^n $?
Since $\lim\limits_{n\to \infty} \frac {\cos(1/n)-\sin(1/n) }{1-1/n} = 1 $, I guess that the limit above is $\frac{1}{e}$, but since the form $(\to 1)^{\to \infty}$ is indeterminate, I don't know how to prove it formally.
Thanks!
| You may consider
*
*$\left(\frac{\cos x - \sin x}{1-x}\right)^{\frac{1}{x}}$ for $x\to 0$ and use
*$\lim_{t\to 0} (1+t)^{\frac{1}{t}}= e$
\begin{eqnarray*} \left(\frac{\cos x - \sin x}{1-x}\right)^{\frac{1}{x}}
& = & \left( \underbrace{\left(1 + \frac{\cos x - \sin x -1 +x}{1-x}\right)^{\frac{1-x}{\cos x - \sin x -1 +x}}}_{\stackrel{x\to 0}{\longrightarrow}e}\right)^{\underbrace{\frac{\cos x - \sin x -1 +x}{x-x^2}}_{\stackrel{L'Hop}{\sim}\frac{-\sin x - \cos x +1}{1-2x}\stackrel{x\to 0}{\longrightarrow}0}}\\
& \stackrel{x\to 0}{\longrightarrow} & e^0 = 1
\end{eqnarray*}
| {
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Evaluate the constant term of $P(x-1)$ if the remainder of the division of $P(x)$ by $x-3$ is $18$ and $P(x+1) = (x^2 -4)Q(x)+3ax+6$
Assume that $$P(x+1) = (x^2 -4)Q(x)+3ax+6$$
and that the remainder of the division of polynomial $P(x)$ by $x-3$ is $18$. Evaluate the constant term of polynomial $P(x-1)$.
All I could see so far is that the polynomial $P(x)$ should be quadratic because $Q(x)$ is multiplied by quadratic term, which is $x^2$.
| We are given that $P(x+1) = (x^2 -4)Q(x)+3ax+6$ $(\ast)$. Plugging $x=2$ into $(\ast)$ yields $P(3)=6a+6$. On the other hand, for every $b$, $P(b)$ is the remainder of the division of polynomial $P(x)$ by $x-b$ hence $P(3)=18$, which shows that $a=2$.
Finally, plugging $x=-2$ into $(\ast)$ yields $P(-1)=-6a+6=-6$ hence the constant term of $P(x-1)$ is $P(-1)=-6$.
| {
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How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ I am stuck on factoring out everything properly. I feel like I am combining these fractions wrong or something because I always have an extra 1.
edit: edit: I am still stuck. Math isn't working out, I am making a mess with the constant edits, I will stop editing and not touch this so people can review the question. Sorry
a) Prove that P(1) is true
$~$
$~$
$$\frac{1}{1*2} = \frac{1}{1+1} = \frac{1}{2}$$
Show that P(k+1) is true as well
$$\frac{1}{(k+1)(k+1+1)} = \frac{k+1}{k+1+1} - \frac{k}{k+1}$$
$~~$
$$ = \frac{k+1}{k+1+1} \frac{k+1}{k+1} - \frac{k}{k+1} \frac{k+1+1}{k+1+1}$$
$~~$
$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{(k+1)\bigg((k+1) - k(+1)\bigg)}{(k+1)(k+1+1)}$$
$~~$
$$ = \frac{k-k+1}{k+1+1} = \frac{1}{k+1+1} \neq \frac{1}{(k+1)(k+1+1)}$$
| Hint:
$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)} = \frac{k(k+2)+1}{(k+1)(k+2)} = \frac{(k+1)(k+1)}{(k+1)(k+2)}.$$
Regarding your calculations, note that
$$ \frac{(k+1)(k+1) - k(k+1+1)}{(k+1)(k+1+1)} = \frac{k^2+2k+1-k^2-2k}{(k+1)(k+1+1)}=?$$
| {
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Why these $2$ methods give the exact same answer for sum of squares? Consider $n = 8$, the sum of squares from $1$ through $8$ is:
$1 \times 1 + 2 \times 2 + 3
\times 3 + 4 \times 4 + 5 \times 5 + 6 \times 6 + 7 \times 7 + 8 \times 8 = 204$.
Also, equal to
$1 \times 8 + 3 \times 7 + 5 \times 6 + 7 \times 5 + 9 \times 4 + 11 \times 3 + 13 \times 2 + 15 \times 1 = 204$.
The second one logic is that I start with $1$, then I increment by $2$ each time and subtract $1$ from the second one, until I reach $1$.
For $n = 2$.
$1 \times 1 + 2 \times 2 = 4 = 1 \times 2 + 3 \times 1$.
For $n = 3$.
$1 \times 1 + 2 \times 2 + 3 \times 3 = 1 \times 3 + 3 \times 2 + 5 \times 1$
The question is, why is it supposed to be the equal the one above? I tried it with a lot of values for $n$?
| In you increase $n$ by $1$, all terms increase by $2k+1$ and there comes an extra term $2n+1$. Then $k^2+2k+1=(k+1)^2$.
| {
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Power series representation of $f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \left( \frac{1+x}{1-x} \right)$
We consider the power series:
$$ f(x) = \sum_{n=1}^{+ \infty} \frac{1}{n(n+1)(2n+1)}.x^{2n+2} $$
Prove that:
$$ f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \bigg( \frac{1+x}{1-x} \biggr) $$
Starting from the second expression I get :
$$ f(x) = 3x^2 - (x - 1)^2 \ln(1 - x) - (x + 1)^2 \ln(1 + x) $$
Using power series of $\ln(1 + x)$ and $ \ln(1 - x)$ does not lead to the result wanted.
I do not know how to proceed to get the first expression.
| Partial fraction decomposition of the coefficient yields to
$$\frac1{n(n+1)(2n+1)}=\frac1n+\frac1{n+1}-\frac4{2n+1}$$
Therefore we may write the original series as
$$\sum_{n=1}^\infty\left[\frac1{n(n+1)(2n+1)}\right]x^{2n+2}=\sum_{n=1}^\infty\left[\frac1n+\frac1{n+1}-\frac4{2n+1}\right]x^{2n+2}$$
The easiest way from hereon is to recognize well-known series representations. To be precise we can further conclude that
\begin{align*}
&(1)&&\sum_{n=1}^\infty\frac{x^{2n+2}}{n}=x^2\sum_{n=1}^\infty\frac{x^{2n}}{n}=x^2(-\log(1-x^2))\\
&(2)&&\sum_{n=1}^\infty\frac{x^{2n+2}}{n+1}=\sum_{n=2}^\infty\frac{x^{2n}}{n}=-\log(1-x^2)+x^2\\
&(3)&&-4\sum_{n=1}^\infty\frac{x^{2n+2}}{2n+1}=-2x\left[2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}\right]=-2x\left[\log\left(\frac{1+x}{1-x}\right)-x\right]
\end{align*}
Now combining these three series we can easily see that
$$\therefore~f(x)=\sum_{n=1}^\infty\left[\frac1{n(n+1)(2n+1)}\right]x^{2n+2}=3x^2-(x^2+1)\log(1-x^2)-2x\log\left(\frac{1+x}{1-x}\right)$$
Of course, as HAMIDINE SOUMARE pointed out within his answer, we have to check beforehand whether these series converge or not and therefore if we are even allowed to split up the sum like this. However, I will leave this to you.
EDIT
Important to notice is that the given logarithm series starts at $n=0$ and not at $n=1$. Thus, we have to consider the first summand extra. Moreover I realised that I made two contrary mistakes which overall did not affect the solution. To be precise my given equations $(2)$ and $(3)$ both contain an error I which I will correct now. For $(2)$ it should rather be
$$(2)~~~\sum_{n=1}^\infty\frac{x^{2n+2}}{n+1}=\sum_{n=2}^\infty\frac{x^{2n}}{n}=-\log(1-x^2)\color{red}{-x^2}$$
Hence the series expansion of $\log(1-x^2)$ is given by
$$\log(1-x^2)=-\sum_{\color{blue}{n=1}}^\infty \frac{x^{2n}}n=-x^2-\sum_{n=2}^\infty \frac{x^{2n}}n\implies \sum_{n=2}^\infty \frac{x^{2n}}n=-\log(1-x^2)-x^2$$
From hereon I could clear my doubts concerning the last sum. The sum should correctly be given by
$$(3)~~~-4\sum_{n=1}^\infty\frac{x^{2n+2}}{2n+1}=-2x\left[2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}\right]=-2x\left[\log\left(\frac{1+x}{1-x}\right)\color{red}{-2x}\right]$$
Rather easy this can justified by exploiting the series expansion of $\log\left(\frac{1+x}{1-x}\right)$, which the same as $2\arctan(x)$, given by
$$\log\left(\frac{1+x}{1-x}\right)=2\sum_{\color{blue}{n=0}}^\infty\frac{x^{2n+1}}{2n+1}=2x+2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}\implies 2\sum_{n=1}^\infty\frac{x^{2n+1}}{2n+1}=\log\left(\frac{1+x}{1-x}\right)-2x$$
Now note that by adding up these three sums we obtain a closed-form expression of the original sum, to be precise
\begin{align*}
f(x)&=\underbrace{[x^2(-\log(1-x^2))]}_{(1)}+\underbrace{[-\log(1-x^2)-x^2]}_{(2)}+\underbrace{\left[-2x\left[\log\left(\frac{1+x}{1-x}\right)-2x\right]\right]}_{(3)}\\
&=[-x^2+4x^2]+[-x^2\log(1-x^2)-\log(1-x^2)]+\left[-2x\log\left(\frac{1+x}{1-x}\right)\right]\\
&=3x^2-(x^2+1)\log(1-x^2)-2x\log\left(\frac{1+x}{1-x}\right)
\end{align*}
Which is, in fact, the desired solution this time correctly derived.
| {
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"timestamp": "2023-03-29T00:00:00",
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A sum of Fibonacci numbers Let $F_n$ be the Fibonacci numbers. I would like to prove this really messy identity:
$$\sum_{n=0}^{\infty}(-1)^n(n+1)^2\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n+2\choose 1}{2n+2 \choose 2}{2n \choose n}}=F_{k}\tag1$$
which simplifies to:
$$\sum_{n=0}^{\infty}(-1)^n\frac{5n^2F_{k}+(F_{k+3}+3F_{k+2})n+2F_{k+2}+F_{k-1}}{{2n \choose n}(4n+2)}=F_{k}$$
How can I evaluate the following sums to show this identity (1) is correct?
$$\sum_{n=0}^{\infty}(-1)^n\frac{n^2}{{2n \choose n}(2n+1)}=A$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{n}{{2n \choose n}(2n+1)}=B$$
$$\sum_{n=0}^{\infty}(-1)^n\frac{1}{{2n \choose n}(2n+1)}=C$$
| Using Wolfram for these sums, we get approximate values
\begin{align*}
C \approx 0.86082, \qquad B \approx -0.11649, \qquad A \approx -0.07897
\end{align*}
From there, you can collect coefficients of $F_r$ to prove your identity. A more concrete way to derive exact expressions is through generating functions. Define
\begin{align*}
g(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{\binom{2n}{n}(2n+1)}x^n
\end{align*}
which you can show equals
\begin{align*}
g(x) = 4 \frac{\sinh^{-1}(\sqrt{x}/2)}{\sqrt{x(x+4)}}
\end{align*}
If we define $L = x \frac{d}{dx}$ as the operator which first takes the derivative w.r.t $x$, then multiplies by $x$, we get that
\begin{align*}
C = (L^0 g)(1), \qquad B = (L^1 g)(1), \qquad A = (L^2 g)(1)
\end{align*}
or
\begin{align*}
C = 4 \frac{\sinh^{-1}(1/2)}{\sqrt{5}}, \qquad B = \frac{2}{5} - \frac{12 \sinh^{-1}(1/2)}{5\sqrt{5}}, \qquad A = \frac{4}{125}(7\sqrt{5}\sinh^{-1}(1/2) - 10)
\end{align*}
| {
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Sum of an infinite series of fractions involving multiple terms in the denominator This is the series in question:
$$S = \frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4} \ldots$$
The general term seems to be:
$$T_n=\frac{n}{1+n^2+n^4}$$
In the original question, which is from Resonance DLPD Algebra, the value of $14S$ is asked.
I haven't been able to make any progress on this question even after trying for a long time. I don't know how to handle series where the denominator contains sum of numbers in general. Can you please provide a hint?
| We are looking for: $S_n=\sum\limits_{n=1}^\infty\frac{n}{1+n^2+n^4}$
Let's realize that $\frac{n}{1+n^2+n^4}=\frac{1}{2}\big(\frac{1}{1-n+n^2}-\frac{1}{1+n+n^2}\Big)=\frac{1}{2}\big(\frac{1}{(n-\frac{1}{2})^2+\frac{3}{4}}-\frac{1}{(n+\frac{1}{2})^2+\frac{3}{4}}\Big)$
Which provide telescopic sum so $S=\frac{1}{2}$
| {
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How do I simplify $\sqrt {4(2- \sqrt{3})}$ into $\sqrt{6} - \sqrt{2}$ This might be a stupid question, but how do I get from $$\sqrt {4(2- \sqrt{3})}$$ to $$\sqrt{6} - \sqrt{2}$$ It is obvious if you squared both, they both equal $8 - 4 \sqrt{3}$, but I'm wondering how you can find the answer from the original expression.
| Suppose $8-4\sqrt 3= (a-b\sqrt 3)^2$
This gives $a^2+3b^2=8$ and $2ab=4$ so that $ab=2$
This means that $a^2b^2+3b^4=8b^2$ or $4+3b^4=8b^2$ or $$3b^4-8b^2+4=0=(3b^2-2)(b^2-2)$$
So either $b^2=2$ or $b^2=\frac 23$.
We have also $a^2b^2=4$ so that $a^2=2$ or $a^2=6$
This should suffice to identify the solution which corresponds to the principal value (non-negative real for square root of non-negative real) of the square root.
Since the two square roots in the original are both positive, we have a unique solution. But the alternate possible signs indicate that a quartic is in the background. The other solutions of the quartic correspond to alternative choices of sign for the square root.
| {
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Solving $\sin (100^\circ-x) \sin 20^\circ =\sin (80^\circ-x)\sin 80^\circ$
Solve for $x$ such that
$$\sin (100^\circ-x) \sin 20^\circ =\sin (80^\circ-x)\sin 80^\circ$$
First, I use the co-function formula:
$$\sin 80^\circ = \cos 10^\circ \tag{1}$$
Also,
$$\sin 20^\circ = 2\sin 10^\circ \cos 10^\circ \tag{2}$$
From these, I got
$$\sin(100^\circ-x)\cdot 2\sin 10^\circ =\sin (80^\circ-x) \tag{3}$$
I thought to use
$$2\sin a \sin b =\cos(a-b)-\cos(a+b) \tag{4}$$
but I'm stuck.
Help me please.
|
A geometric solution:
First draw triangle $ABC$ where $\angle BAC=20$ and $AB=AC$. Easy to find out $\angle ABC=\angle ACB=80$.
Now add point $D$ on line $AC$ such that $\angle CBD = x$, so $\angle ABD=80-x$ and $\angle BDC = 100-x$.
${sin(80-x)\over sin20}={sin(100-x)\over sin80} \implies {AD\over BD}={BC\over BD}\implies AD=BC$.
Now draw $NM$ pass through $D$ and parallel to $BC$ where $N$ is on $AB$ and $NM=AB$. Easy to see $ANM$ is congruent to $ABC$. Now $AM=AC$ and $\angle MAC=60$ so $AMC$ is equilateral. So $\angle NMC=60-20=40$. Also since $NM=AM=MC$ so $NMC$ is isosceles and $\angle MNC=70$. Note by symmetry triangles $NDC$ and $DNB$ are congruent and therefore $\angle NDB=70$. Since $ND$ and $BC$ are parallel $x=\angle DBC=\angle NDB=70$.
| {
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"question_score": "3",
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How can I prove the following matrix equality? Let $n\in\mathbb{N}$ be arbitrary. Why is the following equality true?
$$
\left(\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}
+ 3 \begin{pmatrix}4 & 6\\-2 & -3\end{pmatrix}\right)^n
= \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}
+ (4^n-1)\begin{pmatrix}4 & 6\\-2 & -3\end{pmatrix}
.
$$
Also, it might help that:
$$\begin{pmatrix}4 & 6\\\ -2 & -3\end{pmatrix}^n=\begin{pmatrix}4 & 6\\\ -2 & -3\end{pmatrix}$$
Thanks in advance!
| Just use binomial formula and the fact that $A^2 = A \Rightarrow A^k = A$
$$(I + 3A)^n = \sum_{k=0}^n \binom{n}{k}3^kA^k =I + A\sum_{k=1}^n \binom{n}{k}3^k = I + A ((1+3)^n-1) = I +(4^n-1)A$$
Note that you can apply the binomial formula since $IA = AI$.
| {
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Minimum point of $x^2+y^2$ given that $x+y=10$ How do you I approach the following question:
Find the smallest possible value of $x^2 + y^2$ given that $x + y = 10$.
I can use my common sense and deduce that the minimum value is $5^2 + 5^2 = 50$. But how do you approach this mathematically?
Thanks in advance.
| $\large \text{How about this approach:}$
$
x + y = 10, \text{ so } y = 10 - x \\
x^2 + y^2 = k, y = \sqrt{k - x^2}\\
\rightarrow 10 - x = \sqrt{k - x^2}\\
\rightarrow 100 - 20x + x^2 = k - x^2\\
\rightarrow 2x^2 - 20x + 100 = k\\
f\prime = 4x - 20\\
\text{When x is 5, } f\prime = 0\\~\\
\text{Testing values on the right and left: }\\
f(4) = 62, f(6) = 52\\~\\
\therefore \textbf{There is a minimum point at } x = 5, \textbf{ hence } y = 5.\\
5^2 + 5^2 = 50
$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What are the solutions of $x=\cot x$? Need to find intervals in which the function $y=\frac{x}{2}\cdot \cos x$ is increasing and decreasing. I tried to solve it on the way below but don't know how to continue.
$ \\
y=\frac{x}{2}\cdot \cos x,\ x\in (0,2\pi)\\
y=\frac{1}{2}\cdot x\cdot \cos x \\
{y}'=({\frac{1}{2}\cdot x\cdot \cos x})' \\
{y}'=\frac{1}{2}\cdot ({x\cdot \cos x})' \\
{y}'=\frac{1}{2}\cdot ({x}'\cdot \cos x+x\cdot (\cos x)') \\
{y}'=\frac{1}{2}\cdot(\cos x-x\cdot \sin x) \\
{y}'=0 \\
\frac{1}{2}\cdot(\cos x-x\cdot \sin x)=0 \\ $
$\hspace{1cm}$
$\\
\cos x-x\cdot \sin x=0\ /(\cos x) \\
1-x\cdot \tan x=0 \\
-x\cdot \tan x=-1 \\
x\cdot \tan x=1 \\
x=\frac{1}{\tan x} \\
x=\cot x $
| It is a transcendental equation and you will have to find the solution numerically.
| {
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What is the value of $x+y$ if x and y are co-primes and $PR=\dfrac{x}{y}$ in the diagram?
$ST$ is the perpendicular bisector of $PR$ and $SP$ is the angle
bisector of $\angle QPR$. If $QS=9cm$ and $SR=7cm$ then
$PR=\dfrac{x}{y}$ where x, y are co-primes. $x+y$=?
Source: Bangladesh Math Olympiad 2015 junior category.
I found that $PS=7$, but I can't get the value of $PR$.
|
As shown, by the angle bisector theorem,
$${(\sqrt{7^2-a^2} + \sqrt{9^2-a^2})\over 9}={(\sqrt{7^2-a^2} + \sqrt{7^2-a^2})\over 7}$$
Simplify,
$$7\sqrt{9^2-a^2}=11\sqrt{7^2-a^2}\implies 7^29^2-49a^2=7^211^2-121a^2\implies 72a^2=$$
$$7^2(11^2-9^2)=7^2\times40\implies a^2={245\over 9}$$
Therefore $PR=2\sqrt{7^2-a^2}=2\sqrt{49-{245\over 9}}=2\sqrt{{196\over 9}}={28\over 3}$
| {
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How to convert the following sum to a geometric series?
Find $$
\sum_{n = 1}^{\infty} \frac{6 - 2^{2n - 1}}{3^n}
$$
There are many ways to find that the limit is divergent, but the question explicitly states the sum must be interpreted as a geometric series. I know how to render the exponential terms to $(n-1)$; it's just the $(6 - \ldots)$ is bugging me -- I can't for the life of me find a way find to isolate the term so I get $a \cdot r^{n-1}$.
| Expanding of Ross Millikan's comment, you could try
\begin{align}
\sum_{n = 1}^{\infty} \frac{6 - 2^{2n - 1}}{3^n}
& = \sum_{n = 1}^{\infty} \frac{6}{3^n} - \sum_{n = 1}^{\infty} \frac{2^{2n - 1}}{3^n}
= 2 \cdot \sum_{n = 1}^{\infty} \left( \frac{1}{3} \right)^{n - 1} - \frac{1}{2} \sum_{n = 1}^{\infty} \left( \frac{4}{3} \right)^n \\
& = 2 \cdot \sum_{n = 0}^{\infty} \left( \frac{1}{3} \right)^{n} - \frac{1}{2} \sum_{n = 1}^{\infty} \left( \frac{4}{3} \right)^n
= 2 \cdot \frac{3}{2} - \frac{1}{2} \left( \sum_{n = 0}^{\infty} \left(\frac{4}{3} \right)^n - 1\right) \\
& = 3 + \frac{1}{2} - \frac{1}{2} \sum_{n = 0}^{\infty} \left( \frac{4}{3} \right)^n
= \frac{7}{2} - \frac{1}{2} \sum_{n = 0}^{\infty} \left( \frac{4}{3} \right)^n
= - \infty.
\end{align}
| {
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How can $\frac{4}{3} \times 3=4$ if $ \frac{4}{3}$ is $1.3$? Ok use your closest calculator, and type $\frac{4}{3}$, which is $1.3333333333$,and then multiply it with $3$ which is $3.9999999999$ but then type $\frac{4}{3} \times 3=4$ how?. How can it be $4$ if $\frac{4}{3}$ is $1.3333333333$ and when you multiply it with $3$ is $3.9999999999$.
| Rational numbers are numbers that can be expressed as $\frac{p}{q}$, where p and q are relatively prime integers. Notice that the definition does not mention the decimal expansion. The decimal representation of $\frac{4}{3}$ is defined as $1+\sum_{i=1}^{\infty}\frac{3}{10^i}$. Multiplying by 3, the decimal expansion for $\frac{4}{3}*3$ becomes $3+\sum_{i=1}^\infty\frac{9}{10^i}$. The value of an infinite sum is defined as the limit of the partial sums, or, in simpler terms, what number is approached when we sum up the first $n$ terms, followed by the first $n+1$ term, etc. Doing that here, we see the sequence is $.9$, $.99$, $.999$, etc. It should be obvious that we are getting closer and closer to $1$. Thus, $\sum_{i=1}^{\infty}\frac{3}{10^i}=1$. We now have $\frac{4}{3}*3=1+\sum_{i=1}^{\infty}\frac{3}{10^i}=1+3=4$.
| {
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Find the difference of ceiling functions For $k \geq 1, 1 \leq r \leq k, t \geq 1, x \geq 1$, is there a lower bound or upper bound on:
$$\left\lceil \dfrac{k(t+x)}{r} \right\rceil - \left\lceil \dfrac{k(t)}{r} \right\rceil$$
edit: all the variables $k,r,t,x$ are integers
| The remainder theorem says that for any two positive integers, $q$ and $n$ you can divide $n$ by $q$ and get two unique integers $d$ and $s$ so that $n = dq + s$ and $0 \le s <q$.
So let $kt = Mr +s$ and let $kx = Nr + w$.
Then $\lceil \frac {k(t+x)}r \rceil = \lceil M + N + \frac {s+w}r\rceil= M + N + \lceil \frac {s+w}r\rceil $.
And $\lceil \frac {kt}r \rceil = \lceil M + \frac sr \rceil = M +\lceil \frac sr \rceil$.
And $\lceil \frac {k(t+x)}r \rceil-\lceil \frac {kt}r \rceil =N +\lceil \frac {s+w}r\rceil- \lceil \frac sr \rceil$
Because $0 \le s,w < r$ $0 \le \frac sr < 1$ and $0 \le \frac {s+w}r < 2$.
So $\lceil \frac sr \rceil = 0$ or $1$ and $\lceil \frac {s+w}r\rceil = 0, 1$ or $2$ and $\lceil \frac {k(t+x)}r \rceil-\lceil \frac {kt}r \rceil =N +\lceil \frac {s+w}r\rceil- \lceil \frac sr \rceil= N$ or $N+1$ or $N+2$.
There is no upper bound as we can make $N$ as large as wish by making $x$ as large as we wish.
We can make $N$ as low as $0$ by making $kx < r$ and $s+w < 1$
Example: Let $r = 5; k = 2$ and $t=3$.
We can get $\lceil \frac {k(t+x)}r \rceil-\lceil \frac {kt}r \rceil = 0$ by selection $kx = 2x < 5$ and REMAINDER $2x \div 5 < 4$. So... let $x = 1$ then
$\lceil \frac {k(t+x)}r \rceil-\lceil \frac {kt}r \rceil =$
$\lceil \frac {2(3+1)}5 \rceil-\lceil \frac {2\cdot3}5 \rceil =$
$\lceil \frac 85\rceil - \lceil \frac 65 \rceil = 2 - 2 = 0$.
And we want pick $N$ as large as we we want we can get $\lceil \frac {k(t+x)}r \rceil-\lceil \frac {kt}r \rceil \ge N$ by letting $\frac {kx}r \ge N$ i.e.$x \ge \frac {rN}k$.
So if $x = \frac 52 10^{1000}= 5^{1001}2^{999}$ we get:
$\lceil \frac {k(t+x)}r \rceil-\lceil \frac {kt}r \rceil=$
$\lceil \frac {2(3+5^{1001}2^{999})}5 \rceil-\lceil \frac {2\cdot3}5 \rceil=$
$\lceil \frac 65 + 10^{1000}\rceil - \lceil \frac 65 \rceil =$
$(10^{1000} + 2) - 2 = 10^{1000}$.
| {
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$15 [\tan 2\theta + \sin 2\theta] + 8 = 0$ if ...
QUESTION: $15[\tan 2\theta + \sin 2\theta] + 8 = 0$ if...
(a) $\tan\theta = \frac{1}{2}$
(b) $\sin\theta = \frac{1}{4}$
(c) $\tan\theta = 2$
(d) $\cos\theta = \frac{1}{5}$
What I have tried:
$$\tan2\theta + \sin2\theta = -\frac{8}{15}$$
$$\frac{2\tan\theta}{1-\tan^2\theta} + \frac{2\tan\theta}{1+\tan^2\theta} = -\frac{8}{15}$$
From here onwards, I am stuck and do not know how to continue. Please help. Thank you!
| From the final equation you gave, one gets
$$\frac{4\tan\theta}{1-\tan^4\theta} = -\frac{8}{15} \quad\Rightarrow\quad
8\tan^4\theta - 60\tan\theta -8=0.$$
This quartic factors as $4(\tan\theta-2)(2\tan^3\theta + 4\tan^2\theta + 8\tan\theta + 1)$.
This shows that (c) is a solution while (a) is not. To see that (b) and (d) are not solutions, I think you'll have to substitute the appropriate values of $\tan\theta$ into the cubic.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$
$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$$
If I use a specific path I know I can use Cauchy Theorem to get a number, but how do I prove this for all paths? Thank you!
| Swapping numerator and denominator,
$$
\frac{\sin^2 y+\ln(1+x^2)}{x^2+y^2}=\frac{y^2}{x^2+y^2}\frac{\sin^2 y}{y^2}+\frac{x^2}{x^2+y^2}\frac{\ln(1+x^2)}{x^2}.
$$ It is equal to
$$
\frac{y^2}{x^2+y^2}\left(\frac{\sin^2 y}{y^2}-1\right)+\frac{x^2}{x^2+y^2}\left(\frac{\ln(1+x^2)}{x^2}-1\right)+1
$$ The first term tends to $0$ as
$$
\left|\frac{y^2}{x^2+y^2}\left(\frac{\sin^2 y}{y^2}-1\right)\right|\le \left|\frac{\sin^2 y}{y^2}-1\right|\xrightarrow{x,y\to 0} 0.
$$ Similarly for the second term
$$\left|\frac{x^2}{x^2+y^2}\left(\frac{\ln(1+x^2)}{x^2}-1\right)\right|\le \left|\frac{\ln(1+x^2)}{x^2}-1\right|\xrightarrow{x,y\to 0} 0.
$$ So, it follows
$$
\lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{\sin^2 y+\ln(1+x^2)}=\lim_{(x,y)\to (0,0)}\frac1{\frac{\sin^2 y+\ln(1+x^2)}{x^2+y^2}}=1.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum value of expression $a+b+c$
If $a,b,c$ are non negative integers such that $$2(a^3+b^3+c^3)=3(a+b+c)^2.$$
Then maximum value of $a+b+c$ is ?
My Try: Using Jensen Inequality
Let $f(x)=x^3$. Then $f''(x)>0$ for $x>0$ is convex function
So $$\frac{f(a)+f(b)+f(c)}{3}\geq f\bigg(\frac{a+b+c}{3}\bigg)$$
$$\frac{a^3+b^3+c^3}{3}\geq \bigg(\frac{a+b+c}{3}\bigg)^3\cdots (1)$$
From given condition $$\frac{a^3+b^3+c^3}{3}=\frac{(a+b+c)^2}{2}\cdots (2)$$
So we have $$\frac{(a+b+c)^2}{2}\geq \frac{(a+b+c)^3}{27}$$
$$a+b+c\leq \frac{27}{2}=13.5$$
equality hold when $a=b=c=4.5$
but $a,b,c$ are non negative integers
Could some help me to solve it, Thanks
| You can solve it also like this. By Cauchy inequality we have $$(a+b+c)(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2$$
Further: $$a^2+b^2+c^2\geq {1\over 3}(a+b+c)^2$$
so $${3\over 2}(a+b+c)^3\geq {1\over 9}(a+b+c)^4$$
and thus $$a+b+c\leq {27\over 2}$$
Since $2\mid a+b+c$ we have $a+b+c\leq 12$.
Also, since $x^3\equiv x\pmod 3$ we have $3\mid a+b+c$ so $a+b+c\in\{0,6,12\}$.
Now if we try to find out $a,b,c$ we can assume that $a\leq b\leq c$. So if $a+b+c=12$ then $$3a^3\leq a^3+b^3+c^3 = 216\implies a\leq 4$$
Now you can find by inspection if 12 is achivable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
find equation of tangent for function $f(x)$ parallel to a line I dont have a clue how to solve this kind of problem.
I dont know how to find out slope of $\ln x$.
$f(x) = \ln x$
line $p: 2x-y-3=0$
I have searched for slope of $\ln x$, I know how find out slope for one point, for example:
first I need to derivate the $\ln x$ ==> $1/x$ then I can calculate slope for every point, for example $x=5$ then it is $1/5$.
Also I know how to find out tangent parallel to $y =x^3$ if line p is $3x-y-4=0$
first step: derivate $y =x^3$ ==> $3x^2$, $y=3x-4$ ==> slope for line is $m=3$ and slope for $y =x^3$ is $3$
second step: $3x^2$=$3$ ==> $x=+-1$
third step: use $y - y0 = m*(x-x0)$
for positive one its $y-1=3*(x-1)$ because $f(1)=1^3$ ==>$ y = 3x-2$ then for negative one its $y=3x+2$.
| Equation of tangent to $y = f(x) = \ln(x)$ at $(x_1, y_1)$ is of form
$ (y - y_1) = f'(x_1)(x - x_1) $ where
$ f'(x) = \frac{d}{dx} \ln(x) = \frac{1}{x} $
$
\Rightarrow (y - y_1) = \frac{1}{x_1}(x - x_1)
$
Since Slope of $p: 2x-y-3=0$ is 2, hence,
Slope of || tangent line = Slope of $p$ = 2
$
\Rightarrow \frac{1}{x_1} = 2
\Rightarrow x_1 = \frac{1}{2}
\Rightarrow y_1 = \ln(x_1) = \ln(\frac{1}{2}) = -\ln(2)
$
$ \therefore \ $ Tangent of $f(x)$ || p is:
$
y - (-\ln(2)) = 2(x - \frac{1}{2})
\Rightarrow 2x - y - \ln(2) - 1 = 0
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x+x^2+x^3=x^4+x^5+x^6$ implies $x^4=x$ in a ring Let $(A,+,.)$ be a ring s.t. $x+x^2+x^3=x^4+x^5+x^6$ for all $x \in A$. Prove that $x^4=x$ for all $x$ in $A$. Can somebody give me some tips, please?
| Here is a variant of the solution by @Servaes that seems simpler to me:
Plugging $x=-1$ gives $2=0$ in $A$.
Therefore, $x+x^2+x^3=x^4+x^5+x^6$ is the same as $x+x^2+x^3+x^4+x^5+x^6=0$.
Multiplying that by $x-1$ gives $x^7-x=0$ and so $x^8-x^2=0$.
Now $x^8-x^2=(x^4-x)^2$.
Let $z=x^4-x$. Then $z^2=0$ and $z+z^2+z^3=z^4+z^5+z^6$ imply $z=0$, and so $x^4=x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using induction to prove an inequality for a sequence of numbers We have the sequence $d_n = \begin{cases} 1 &\text{ if } n=0 \\
\frac{n}{d_{n-1}} &\text{ if } n>0 \end{cases}$
for all natural numbers $n$.
($d_{n-1}$ is the previous number of the sequence.)
examples: $d_0 = 1$, $d_1 = 1$, $d_2 = 2$, $d_3 = \frac{3}{2}$, $d_4 = \frac{8}{3} \dots$
I have to prove using induction that $\forall n \in \mathbb N \setminus \{0\}$, $d_{2n-1}$ $\leq$ $\sqrt{2n-1}$.
so far, I've figured out the pattern that for every n greater than or equal to $2$, $d_{2n-1} = d_{2n-3} \, \frac{2n-1}{2n-2}$.
i.e. $d_5 = d_3 \, \frac{5}{4}$
In the hints section, they told me to write $d_{2k+1}$ in terms of $d_{2k-1}$ and to use the difference of squares: $(2k-1)(2k+1) = 4k^2 - 1$ for the induction step.
Any hints/tips/advice on how to solve this problem is much appreciated!
Thank you!
| Your observation $$d_{2n-1} = d_{2n-3} \, \frac{2n-1}{2n-2}$$ is homologous with $$d_{2k+1} = d_{2k-1} \, \frac{2k+1}{2k}.$$
\begin{align}
d_{2k-1} \, \frac{2k+1}{2k} &\le \sqrt{2k-1} \, \frac{\bbox[yellow, 2px]{2k+1}}{2k} \tag{induction hypothesis}\\
&\le \color{blue}{\sqrt{2k-1}} \, \frac{\bbox[yellow, 2px]{\color{blue}{\sqrt{2k+1}}}}{2k} \, \bbox[yellow, 2px]{\sqrt{2k+1}} \\
&\le \frac{\color{blue}{\sqrt{4k^2-1}}}{2k} \sqrt{2k+1} \tag{hint} \\
&\le 1 \cdot \sqrt{2k+1} = \sqrt{2k+1} \tag*{$\square$}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the rectangular function a convolution of $L^1$ functions? Do there exist functions $f,g$ in $L^1(\mathbf{R})$ such that the convolution $f \star g$ is (almost everywhere) equal to the indicator function of the interval $[0,1]$ ?
| Following the idea of Christian Remling, I have two explicit functions that have the property that $f*g = \mathbf{1}_{[-1,1]}$:
\begin{eqnarray}
f(x) &=& \frac{K_\frac{1}{4}\left(|x|\right)}{2^{1/4}\pi\Gamma\left(\frac{1}{4}\right)|x|^{1/4}}\\\\
g(x) &=& \frac{\mathrm{sgn}(1-x)}{\sqrt{|1-x|}}\,_1F_2\left(\begin{matrix}-\frac{1}{4} \\ \frac{1}{4}, \frac{3}{4}\end{matrix}; \frac{(1-x)^2}{4}\right) +\frac{\mathrm{sgn}(1+x)}{\sqrt{|1+x|}}\,_1F_2\left(\begin{matrix}-\frac{1}{4} \\ \frac{1}{4}, \frac{3}{4}\end{matrix}; \frac{(1+x)^2}{4}\right)\\ &\,&+ \frac{\Gamma\left(\frac{5}{4}\right)(1-x)}{\sqrt{2}\Gamma\left(\frac{7}{4}\right)}\,_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2}, \frac{7}{4}\end{matrix}; \frac{(1-x)^2}{4}\right)+ \frac{\Gamma\left(\frac{5}{4}\right)(1+x)}{\sqrt{2}\Gamma\left(\frac{7}{4}\right)}\,_1F_2\left(\begin{matrix}\frac{1}{2} \\ \frac{3}{2}, \frac{7}{4}\end{matrix}; \frac{(1+x)^2}{4}\right)
\end{eqnarray}
Where $K_\frac{1}{4}$ is a modified Bessel function and $\,_1F_2$ is a hypergeometric function. While these aren't the simplest functions, and $g$ is kind of a nightmare, they have the nice property that
\begin{eqnarray}
\mathcal{F}[f](k) &=& \frac{1}{\sqrt{2\pi}(1+k^2)^{1/4}} \\
\mathcal{F}[g](k) &=& 2\sqrt{2\pi}(1+k^2)^{1/4}\frac{\sin(k)}{k},
\end{eqnarray}
so that $f*g = \mathcal{F}^{-1}[\mathcal{F}[f]\cdot\mathcal{F}[g]] = \mathcal{F}^{-1}\left[2\sin(k)/k\right] = \mathbf{1}_{[-1,1]}$.
The convolution integral isn't doable analytically, but it can be evaluated numerically. Doing this in Mathematica gives
where the blue is $f$, the yellow is $g$, and the green is $f*g$. Despite their divergences, both $f$ and $g$ are in $L^1$, as the divergences scale as $|x|^{-1/2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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maximum value of $\sum ab-2abc$ If $a+b+c=1$ and $a,b,c\in(0,1)$, then what is the maximum value of $(ab+bc+ca-2abc)$?
What I've tried:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 4(ab+bc+ca)$
$(a-b)^2=a^2+b^2-2ab\geq 0$
$a^2+b^2\geq 2ab,b^2+c^2\geq 2bc,c^2+a^2\geq 2ca$
$ab+bc+ca\leq\frac14$
How do I solve it help me please.
| Σ(ab) = Σ ( a²b+b²a ) + abc and by using AM-GM to the terms under the modified sigma we get that it is always greater than 6abc and using AM - GM on a,b,c we get abc<1/27 hence the maximal value must be 7/27
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Reduction of hypergeometric function for integer parameters Consider integers $a,b,c,d>0$ and the hypergeometric function
$${}_3F_2(1-d,b,a+b;a+b+c,b+1;1)$$
I don't know much about hypergeometric functions but I understand that when the parameters are integer valued we get some simplification. Unfortunately, I can't seem to figure this out or find a good reference.
My attempt:
For $d=1$, the function evaluates to 1.
For $d=2$, we have
$$1-\frac{b(a+b)}{(b+1)(a+b+c)}$$
For $d=3$, we have
$$1-\frac{2b(a+b)}{(b+1)(a+b+c)} + \frac{b(a+b)(a+b+1)}{(b+2)(a+b+c)(a+b+c+1)}$$
For $d=4$, we have
$$1-\frac{3b(a+b)}{(b+1)(a+b+c)} + \frac{3b(a+b)(a+b+1)}{(b+2)(a+b+c)(a+b+c+1)} - \frac{b(a+b)(a+b+1)(a+b+2)}{(b+3)(a+b+c)(a+b+c+1)(a+b+c+2)}$$
I can begin to write out the expression for general integer $d\ge 1$ as
$$\sum_{k=0}^{d-1}(-1)^k\binom{d-1}{k}\cdots$$
I'm not sure how the rest proceeds.
| Wolfram functions reference is a good place to find identities like this. Here's the one for $\;_3F_2$. For this, you probably want. Unfortunately I'm not finding much in values at $z = 1$, but there's a lot to go through.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a,b,c,d\in\mathbb{Z^+}$ where $ad=b^2+bc+c^2$, prove that $a^2+b^2+c^2+d^2$ is composite
If $a,b,c,d\in\mathbb{Z^+}$ where $ad=b^2+bc+c^2$, prove that $a^2+b^2+c^2+d^2$ is composite.
My attempt so far:
$$a^2+b^2+c^2+d^2$$
$$=a^2+d^2+2ad+b^2+c^2+2bc-2ad-2bc$$
$$=(a+d)^2+(b+c)^2-2(b^2+bc+c^2)-2bc$$
$$=(a+d)^2+(b+c)^2-2(b^2+2bc+c^2)$$
$$=(a+d)^2+(b+c)^2-2(b+c)^2$$
$$=(a+d)^2-(b+c)^2$$
$$=(a+b+c+d)(a-b-c+d)$$
Now I am trying to prove that $a-b-c+d\not=1$ so I tried to assume the contradiction but I am unable to finish. Any help would be appreciated.
| Assume that $a - b - c + d = 1$. Then you'll get from your result that
$$a^2 + b^2 + c^2 + d^2 = (a+b+c+d)(a-b-c+d) = a+b+c+d \tag{1}\label{eq1}$$
Now, since $a,b,c,d\in\mathbb{Z^+}$, note that $a^2 \gt a$ if $a \gt 1$, and likewise for $b, c, d$. Thus, \eqref{eq1} can only be true is $a = b = c = d = 1$. In that case, $a^2 + b^2 + c^2 + d^2 = 4$ which is composite. Of course, if $a - b - c + d \neq 1$, then it's also composite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Picture proof that the area of a right triangle is $xy$ I stumbled on the following result by accident:
Let $A, B, C$ be the vertices of a right triangle, with opposite side lengths $a, b, c$ respectively, where $\angle C = 90^\circ$ and $a^2 + b^2 = c^2$.
Draw the incircle, and let $x, y, z$ be the length of the tangent from the vertices $A$, $B$, and $C$ respectively to the incircle. (So $a = y + z$, $b = x + z$, and $c = x + y$.)
Then the area of the triangle is $\boldsymbol{xy}$.
I can prove this algebraically,$^1$ but is there a picture proof of this fact?
What I have in mind is that we cut up the triangle $ABC$ into finitely many pieces, and rearrange them into a rectangle with sides $x$ and $y$.
$^1$ Using $x = \frac{b+c-a}{2}$ and $y = \frac{a+c-b}{2}$, we get $xy = \frac14\left(c - (a-b)\right)\left(c + (a-b)\right) = \frac14\left(c^2 - a^2 - b^2 + 2ab\right)$. From $c^2 = a^2 + b^2$ this reduces to $\frac14 (2ab) = \frac12 ab$, which is the area of the triangle.
| Let A be the area of the original triangle cut up in the obvious way as on the picture. Then the area of the rectangle is $A_r = 2 A = ab = xy + A$. Thus $xy = A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the Length of $QP$ Given two circles with radii $8$ and $6$ units with centers $A$ and $B$ such that $AB=12$ If $P$ is mid point of $QR$
Find Length of $QP$
My try:
I assumed $A(0,0)$ and $B(12,0)$
So the equations of circles are:
$$x^2+y^2=64$$
$$(x-12)^2+y^2=36$$
Solving above equations we get:
$$P\left(\frac{43}{6}, \frac{\sqrt{455}}{6}\right)$$
Then we can assume
$$Q(8\cos\alpha, 8\sin \alpha)$$
$$R(12+6\cos\beta,6\sin\beta)$$
Using mid point formula we get:
$$8\cos\alpha+6\cos\beta=\frac{7}{3}$$
$$8\sin\alpha+6\sin\beta=\frac{\sqrt{455}}{3}$$
Any idea here?
|
You don't need any trigonometry or analytical geometry. The following equations are quite obvious (Pythagora):
$$p^2+\frac{x^2}{4}=8^2=64\tag{1}$$
$$q^2+\frac{x^2}{4}=6^2=36\tag{2}$$
$$(p-q)^2+x^2=12^2=144\tag{3}$$
Introduce substitution: $y=\frac{x^2}{4}$ and you get:
$$p=\sqrt{64-y}\tag{4}$$
$$q=\sqrt{36-y}\tag{5}$$
$$(p-q)^2+4y=144\tag{6}$$
Replace (4) and (5) into (6):
$$(\sqrt{64-y}-\sqrt{36-y})^2+4y=144$$
$$100-2y-2\sqrt{(64-y)(36-y)}+4y=144$$
$$2y-44 = 2\sqrt{(64-y)(36-y)}$$
$$y-22=\sqrt{(64-y)(36-y)}$$
$$(y-22)^2=(64-y)(36-y)$$
$$(y-22)^2-(64-y)(36-y)=0$$
$$56y-1820=0$$
$$y=\frac{x^2}{4}=\frac{1820}{56}=\frac{65}{2}$$
$$x=\sqrt{130}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum value of $PA+PB$ is If $P(x,y,z)$ lie on line $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}$ and $A(5,3,4)$ and $B(1,-1,2)$ . Then minimum value of $PA+PB$ is
what i try
let $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}=\lambda$
Then $P(2\lambda-2,2\lambda-7,\lambda+2)$
$PA+PB=\sqrt{(2\lambda-7)^2+(2\lambda-10)^2+(\lambda-2)^2}+\sqrt{(2\lambda-3)^2+(2\lambda-8)^2+\lambda^2)}$
$PA+PB=\sqrt{9\lambda^2-72\lambda+153}+\sqrt{
9\lambda^2-36\lambda+45}$
how do i minimize it b3cause derivative method is very tedious help me please
| By Minkowski we obtain:
$$\sqrt{9\lambda^2-72\lambda+153}+\sqrt{
9\lambda^2-36\lambda+45}=$$
$$=3\left(\sqrt{(4-\lambda)^2+1}+\sqrt{(\lambda-2)^2+1}\right)\geq$$
$$\geq3\sqrt{(4-2)^2+(1+1)^2}=6\sqrt2.$$
The equality occurs for $\lambda=3,$ which says that we got a minimal value.
| {
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"url": "https://math.stackexchange.com/questions/3143636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution to the definite integral $\int_0^{\infty}x^{-\alpha} e^{-\beta/x}\,dx$ Is there a explicit solution to this definite integral, with $\alpha>0$ and $\beta>0$ :
$$\int_0^{\infty}x^{-\alpha} e^{-\beta/x}\,dx$$
If yes, what is it?
| The method prescribed by egreg is far superior to what I'm going to put forward.
Here we will address your integral:
\begin{equation}
I\left(\alpha, \beta \right) = \int_0^\infty x^{-\alpha} e^{-\beta/x}\:dx\nonumber
\end{equation}
We now employ Fubini's Theorem and take the Laplace Transform with respect to $\beta$:
\begin{align}
\mathscr{L}_{\beta \rightarrow s}\left[ I\left(\alpha, \beta \right) \right] &= \int_0^\infty x^{-\alpha} \frac{1}{s - -\frac{1}{x}}\:dx = \int_0^\infty \frac{x^{1 - \alpha}}{xs + 1}\:dx = \frac{1}{s} \int_0^\infty \frac{x^{1 - \alpha}}{x + s^{-1}}\:dx
\end{align}
We now use the solution that I derived here:
\begin{equation}
\int_0^\infty \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right)
\end{equation}
Thus,
\begin{align}
\mathscr{L}_{\beta \rightarrow s}\left[ I\left(\alpha, \beta \right) \right] &= \frac{1}{s} \int_0^\infty \frac{x^{1 - \alpha}}{x + s^{-1}}\:dx = \frac{1}{s} \cdot \frac{1}{1} \cdot \left(s^{-1}\right)^{\frac{1 - \alpha + 1 }{1} - 1}\cdot B\left(1 - \frac{1 - \alpha + 1}{1}, \frac{1 - \alpha + 1}{1} \right)\nonumber \\
&= s^{\alpha - 2} B\left(\alpha - 1, 2 - \alpha\right)
\end{align}
Where $B(\cdot, \cdot)$ is the Beta Function. We now employ the relationship between the Beta and Gamma function being:
\begin{equation}
B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)}
\end{equation}
Thus,
\begin{align}
\mathscr{L}_{\beta \rightarrow s}\left[ I\left(\alpha, \beta \right) \right] &= s^{\alpha - 2} B\left(\alpha - 1, 2 - \alpha\right) = s^{\alpha - 2} \frac{\Gamma\left(\alpha - 1\right)\Gamma(2 - \alpha)}{\Gamma\left( \alpha - 1 + 2 - \alpha\right)} = s^{\alpha - 2}\Gamma\left(\alpha - 1\right)\Gamma(2 - \alpha)
\end{align}
We now take the inverse Laplace Transform with respect to $s$:
\begin{align}
I\left(\alpha, \beta \right)= \mathscr{L}_{s \rightarrow \beta}^{-1}\left[s^{\alpha - 2}\Gamma\left(\alpha - 1\right)\Gamma(2 - \alpha) \right] = \frac{\beta^{1 - \alpha}}{\Gamma(2 - \alpha) }\cdot\Gamma\left(\alpha - 1\right)\Gamma(2 - \alpha) =\beta^{1 - \alpha}\Gamma\left(\alpha - 1\right)
\end{align}
And thus,
\begin{equation}
I\left(\alpha, \beta \right) = \beta^{1 - \alpha}\Gamma\left(\alpha - 1\right) \nonumber
\end{equation}
| {
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Find digits $a,b$ such that $7ab + 4ba = 1a21$. I have to find all the digits $a$ and $b$ such that $7ab + 4ba = 1a21$.
Note: there is no multiplication, those are three decimal numbers.
I put this equality this way: $$7\cdot100 + a \cdot 10 + b + 4 \cdot 100 + b \cdot 10 + a = 1 \cdot 1000 + a \cdot 100 + 2 \cdot 10 + 1$$
$$1\cdot 1000 + (a-11)\cdot 100+(2-a-b)+1-a-b=0$$
So, our number will be something like this: $1(a-11)(2-a-b)(1-a-b)=0$. This is contradiction since the first digit is $1$, and on the right we have $0$.
Does that mean there are no $a$ and $b$ to satisfy this equality?
Alternatively, I did the following: since the last digit in $1a21$ is $1$, it has to be $a+b=L1$. $a$ and $b$ are digits, so there are couple of cases: $a=0$ and $b=1$, $a=1$ and $b=0$, $a=5$ and $b=6$, $a=6$ and $b=5$... None of those cases satisfy what we want, so there are no $a$ and $b$.
Are those two ways to do it the right ones? I don't have any solutions therefore I'm not sure.
Thank you.
| Another approach to solve such equations is to write it as an algebraic equation, instead of with base-10 digits. We get
$$
700 + 10a + b + 400 + 10b + a = 1000 + 100a + 21
$$
which simplifies to:
$$
11(a + b) + 1100 = 1021 + 100a
$$
or
$$
11b + 79 = 89a.
$$
Now we use the fact that $a$ and $b$ are between $0$ and $9$. In particular, $11b$ is at most $99$, so $11b + 79$ is at most $99 + 79 = 178$, which equal to two times $89$. So $a$ can be at most $2$.
*
*If $a = 2$, then $11b + 79 = 178$, so $b = 9$.
*If $a = 1$, then $11b = 10$, so this doesn't work.
*Finally, if $a = 0$, then $11b + 79 = 0$, so this doesn't work either.
Thus the only solution is $a = 2, b = 9$. We have
$$
\boxed{7\underline{29} + 4\underline{92} = 1\underline{2}21}.
$$
| {
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"answer_id": 4
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Number of possible unique patterns of $5$ points on an $8\times 8$ grid. I'm after a specific answer and workings/formula for the number of possible unique patterns that $5$ points can have on an $8\times 8$ grid. The pattern must be unique and not match another pattern when it's rotated or mirrored. Can the workings be explained in laymans terms? Thanks
| We solve the case of the points going into the slots between the grid,
so there are $64$ slots. We put at most one point into one slot. We
use the Polya Enumeration Theorem (PET), which requires the cycle
index of the action of the symmetries on the slots. Supposing the grid
is $n\times n$ with $n$ even we get for the rotations, first, the
identity:
$$a_1^{n^2}.$$
Next, two rotations by $90$ degrees and $270$ degrees:
$$2 a_4^{n^2/4}.$$
The rotation by $180$ degrees:
$$a_2^{n^2/2}.$$
For reflections there is the horizontal and the vertical reflection:
$$2 a_2^{n^2/2}$$
and the reflection in the two diagonals:
$$2 a_1^n a_2^{(n^2-n)/2}.$$
We get the cycle index
$$Z(G) = \frac{1}{8}
\left(a_1^{n^2} + 2 a_4^{n^2/4} + a_2^{n^2/2}
+ 2 a_2^{n^2/2} + 2 a_1^n a_2^{(n^2-n)/2}\right)$$
or
$$\bbox[5px,border:2px solid #00A000]{Z(G) = \frac{1}{8}
\left(a_1^{n^2} + 2 a_4^{n^2/4}
+ 3 a_2^{n^2/2} + 2 a_1^n a_2^{(n^2-n)/2}\right).}$$
For an odd number of points call is $m$ we thus have
$$[z^m] Z(G; 1+z) \\=
[z^m] \frac{1}{8}
\left((1+z)^{n^2} + 2 (1+z^4)^{n^2/4}
+ 3 (1+z^2)^{n^2/2} + 2 (1+z)^n (1+z^2)^{(n^2-n)/2}\right)
\\ = [z^m] \frac{1}{8}
\left((1+z)^{n^2} + 2 (1+z)^n (1+z^2)^{(n^2-n)/2}\right).$$
This is
$$\bbox[5px,border:2px solid #00A000]{\frac{1}{8} {n^2\choose m}
+ \frac{1}{4} \sum_{k=0}^{n/2-1}
{n\choose 2k+1} {(n^2-n)/2\choose (m-2k-1)/2}}$$
where $n$ is even and $m$ is odd.
We get for $n=8$ and $m=5$ the answer
$$\bbox[5px,border:2px solid #00A000]{954226.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all polynomials with coefficients from set $\{-1,1\}$ and which have all their roots real.
Find all polynomials with coefficients from
set $\{-1,1\}$ and which have all their roots real.
What I have tried:
Assume polynomial is $a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots +a_{0}=0$ and $a_{i}\in \{-1,1\}$ for $i=0,1,2,3,4,\cdots ,n$. Let its roots be $x=r_{i}$ for $i=1,2,3,\cdots n$. Then
$\begin{aligned}\sum r_{i}=-\frac{a_{n-1}}{a_{n}}=\pm 1\\\\
\mathop{\sum}_{i<j}r_{i}r_{j}=+\frac{a_{n-2}}{a_{n}}=\pm 1\\\\
\mathop{\sum}_{i<j<k}r_{i}r_{j}r_{k}=+\frac{a_{n-2}}{a_{n}}=\pm 1\\\\
\prod r_{i}=(-1)^n\frac{a_{0}}{a_{n}}\end{aligned}$
How do I solve it?
| WLOG, let $a_n = 1$. Also, let $r_1,\ldots,r_n$ be the roots of the polynomial.
As you noted, $\displaystyle\sum_{k = 1}^{n}r_k = -a_{n-1}$, $\displaystyle\sum_{1 \le k < \ell \le n}r_kr_{\ell} = a_{n-2}$, and $\displaystyle\prod_{k = 1}^{n}r_k = (-1)^na_0$. Hence,
$$\displaystyle\sum_{k = 1}^{n}r_k^2 = \left(\sum_{k = 1}^{n}r_k\right)^2-2\left(\sum_{1 \le k < \ell \le n}r_kr_{\ell}\right) = a_{n-1}^2-2a_{n-2} = 1-2a_{n-2} = \begin{cases}3 & \text{if} \ a_{n-2} = -1 \\ -1 & \text{if} \ a_{n-2} = 1 \end{cases}.$$
Since all the roots are real, $\displaystyle\sum_{k = 1}^{n}r_k^2 \ge 0$. Hence, we must have $a_{n-2} = -1$ and thus $\displaystyle\sum_{k = 1}^{n}r_k^2 = 3$.
Trivially, we then have $\displaystyle\sum_{k = 1}^{n}|r_k|^2 = \sum_{k = 1}^{n}r_k^2= 3$, as well as $\displaystyle\prod_{k = 1}^{n}|r_k| = \left|\prod_{k = 1}^{n}r_k\right| = \left|(-1)^na_0\right| = 1$.
So by the RMS-GM inequality, we have $$\displaystyle 1 = \left(\prod_{k = 1}^{n}|r_k|\right)^{1/n} \le \left(\dfrac{1}{n}\sum_{k = 1}^{n}|r_k|^2\right)^{1/2} = \sqrt{\dfrac{3}{n}}.$$
Therefore, we must have $n \le 3$. It is easy to check that the only solutions are:
$x-1$, $x+1$, $x^2-x-1$, $x^2+x-1$, $x^3-x^2-x+1$, $x^3+x^2-x-1$, and their negatives.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find minimum of $4(a^3 + b^3 + c^3) + 15abc$ subject to $a + b + c = 2$
$a$, $b$ and $c$ are three sides of a triangle such that $a + b + c = 2$. Calculate the minimum value of $$\large P = 4(a^3 + b^3 + c^3) + 15abc$$
Every task asking for finding the minimum value of an expression containing the product of all of the variables scares me.
Here what I've done.
Using the AM-GM inequality and the Schur's inequality, we have that
$$a^3 + b^3 + c^3 \ge 3abc \implies P \ge \dfrac{9}{2}(a^3 + b^3 + c^3 + 3abc)$$
$$\ge \dfrac{9}{2}[ab(a + b) + bc(b + c) + ca(c + a)] = \dfrac{9}{2}[ab(2 - c) + bc(2 - a) + ca(2 - b)]$$
$$\ge \dfrac{9}{2}[2(ab + bc + ca) - 3abc] \ge \dfrac{27}{2}[2\sqrt[\frac{3}{2}]{abc} - abc]$$
Let $abc = m \implies m \le \left(\dfrac{a + b + c}{3}\right)^3 = \dfrac{8}{27}$
The problem becomes
Find the minimum value of $P' = 2\sqrt[\frac{3}{2}]{m} - m$ when $ 0 < m \le \dfrac{8}{27}$.
which is invalid because there isn't a minimum with the given condition.
| I do not know if you had to use AM-GM but the problem is quite simple using pure algebra.
Considering
$$ P = 4(a^3 + b^3 + c^3) + 15abc \qquad \text{with} \qquad a+b+c=2$$ eliminate $c$ from the constaint to get
$$P=3 a^2 (8-9 b)-3 a (b-2) (9 b-8)+8 (3 (b-2) b+4)$$ Now
$$\frac{\partial P}{\partial a}=6 a (8-9 b)-3 (b-2) (9 b-8)=0 \implies a=\frac{2-b} 2$$
Reusing the constaint, this gives $c=a$ and then $a=b=c=\frac 23$.
Plug in $P$ and get the result.
Edit
Just as @KaiKoike commented, there is a mistake above
$$\frac{\partial P}{\partial a}=-3 (9 b-8) (2 a+b-2)$$
$$\frac{\partial P}{\partial b}=-3 (9 a-8) ( a+2b-2)$$
So, we can have the solutions
$$a=\frac 23 \qquad b=c=\frac 89\implies P=\frac{10720}{729}=14.7051$$
$$a=b=c=\frac 23 \implies P=\frac{10720}{729}=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Parametrizing the square spiral Related to this question concerning number spirals I have another one, more specific.
While it is rather easy to arrange the natural numbers along an Archimedean spiral by
$$x(n) = \sqrt{n}\cos(2\pi\sqrt{n})$$
$$y(n) = \sqrt{n}\sin(2\pi\sqrt{n})$$
it's much more difficult to arrange them along a square spiral by a closed formula. It's very easy to construct the square spiral algorithmically ("go along straight lines and always turn to the right if possible") but I'm totally stuck how the functions $x(n),y(n)$ would look like as formulaic expressions. The only thing I'm quite sure about is that they will make use of the square root function - but how are the "radii" and the turns coded?
This is how the two spirals look like (the Archimedean spiral being appropriately scaled and rotated to align the square numbers):
And here you can see them being morphed into each other.
Can anyone give me a hint (or the solution)?
| [This answer is inspired by user SMM's answer. Thanks for it.]
Consider "piecewise linear approximations" of the sine and cosine function, periodically defined on the unit interval, i.e. $x \in [0,1]$.
Let
$$\boxed{\cos_\bigcirc(x) = \cos(2\pi x)\\\sin_\bigcirc(x) = \sin(2\pi x)}$$
and compare this to
$$\boxed{\cos_\square(x) = \begin{cases}
+1 & \text{ for } \frac{0}{8} \leq x \leq \frac{1}{8} \\
+2 - 8x & \text{ for } \frac{1}{8} \leq x \leq \frac{3}{8} \\
-1 & \text{ for } \frac{3}{8} \leq x \leq \frac{5}{8} \\
-6 + 8x & \text{ for } \frac{5}{8} \leq x \leq \frac{7}{8} \\
+1 & \text{ for } \frac{7}{8} \leq x \leq \frac{8}{8} \\
\end{cases}
\\ \\\sin_\square(x) = \begin{cases}
+0 + 8x & \text{ for } \frac{0}{8} \leq x \leq \frac{1}{8} \\
+1 & \text{ for } \frac{1}{8} \leq x \leq \frac{3}{8} \\
+4 - 8x & \text{ for } \frac{3}{8} \leq x \leq \frac{5}{8} \\
-1 & \text{ for } \frac{5}{8} \leq x \leq \frac{7}{8} \\
-8 + 8x & \text{ for } \frac{7}{8} \leq x \leq \frac{8}{8} \\
\end{cases}}$$
or written more readable:
$$\cos_\square(x) = \begin{cases}
+1 & \text{ for } 0 \leq 8x \leq 1 \\
+2 - 8x & \text{ for } 1 \leq 8x \leq 3 \\
-1 & \text{ for } 3 \leq 8x \leq 5 \\
-6 + 8x & \text{ for } 5 \leq 8x \leq7 \\
+1 & \text{ for }7 \leq 8x \leq 8 \\
\end{cases}
\\ \\\sin_\square(x) = \begin{cases}
+0 + 8x & \text{ for } 0 \leq 8x \leq 1 \\
+1 & \text{ for } 1 \leq 8x \leq 3 \\
+4 - 8x & \text{ for } 3 \leq 8x \leq 5 \\
-1 & \text{ for } 5 \leq 8x \leq7 \\
-8 + 8x & \text{ for }7 \leq 8x \leq 8 \\
\end{cases}$$
These are the plots:
$\cos_\square$ and $\sin_\square$ are especially well suited to arrange numbers on a square with integer coordinates around the origin with uniform distance $1$ along the square.
This only works for multiples of $8$ with $8n = (2n+1)^2 - (2n-1)^2$. In this case the positions of the $8n$ numbers $k = 0, 1, \dots, 8n-1$ are given by
$$\boxed{x^{(n)}_\square(k) = n\cos_\square(\frac{k}{8n})\\ \\y^{(n)}_\square(k) = n\sin_\square(\frac{k}{8n})}$$
Compare this to the positions of $8n$ numbers on a circle around the origin with uniform distance $\frac{2\pi}{8}$ along the circle:
$$\boxed{x^{(n)}_\bigcirc(k) = n\cos_\bigcirc(\frac{k}{8n})\\
y^{(n)}_\bigcirc(k) = n\sin_\bigcirc(\frac{k}{8n})}$$
For the (circular) Archimedean spiral we have
$$x_\bigcirc(k) = -\frac{\sqrt{k}}{2}\cos_\bigcirc(\frac{\sqrt{k}}{2}-\frac{1}{8}) $$
$$y_\bigcirc(k) = -\frac{\sqrt{k}}{2}\sin_\bigcirc(\frac{\sqrt{k}}{2}-\frac{1}{8})$$
Written for the sake of comparison with the square spiral:
$$\boxed{x_\bigcirc(k) = - x_\bigcirc^{(\sqrt{k}/2)}(2k-\frac{1}{8})\\
y_\bigcirc(k) = -y_\bigcirc^{(\sqrt{k}/2)}(2k-\frac{1}{8})}$$
Note that the factors $-1$, $\frac{1}{2}$ and the phase $\frac{1}{8}$ (which corresponds to $\frac{\pi}{4}$) where chosen to align the Archimedean with the square spiral, especially the square numbers.
The formulas
$$x_\bigcirc(k) = \sqrt{k}\cos_\bigcirc(\sqrt{k})$$
$$y_\bigcirc(k) = \sqrt{k}\sin_\bigcirc(\sqrt{k})$$
would give an Archimedean spiral as well.
This is for the square spiral. Let $k'$ be the greatest odd perfect square smaller than $k$. Let $\hat{k} = (\sqrt{k'}-1)/2$. Let $x_\square(0) = 0 $ and $y_\square(0) = 0 $ and for $k > 0$
$$\boxed{x_\square(k) = x_\square^{(\hat k)}(k - k' - \hat k + 1) \\
y_\square(k) = y_\square^{(\hat k)}(k - k' - \hat k + 1)} $$
Note that $k - k' - \hat k + 1$ being negative doesn't pose a problem since $\cos_\square$ and $\sin_\square$ are periodic in both directions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Solve the following system of equations - (3)
Solve the following system of equations:
$$\large
\left\{
\begin{align*}
3x^2 + xy - 4x + 2y - 2 = 0\\
x(x + 1) + y(y + 1) = 4
\end{align*}
\right.
$$
I tried writing the first equation as $(x - 2)(3x + y - 10) = -18$, but it didn't help.
| Solving the first equation for $y$ we get $$y=\frac{-3x^2+4x+2}{2+x}$$ for $$x\neq -2$$
plugging this in the second equation we get after simplifications
$$(5x+4)(x-1)^3=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Reaching upon $9=1$ while solving $x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
$x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
Substituting $y=x+45^{\circ}$, we get
$$3\tan{(y-60^{\circ})}=\tan{(y-30^{\circ})}$$
$$3\frac{\tan y - \sqrt3}{1+\sqrt3\tan y}=\frac{\tan y - 1/\sqrt3}{1+1/\sqrt3\cdot\tan y}$$
$$3(\tan ^2 y-3)=3\tan ^2-1$$
$$9=1$$
The solution provided by the book $x=n\pi + \pi/4$ fits, so why did i get $9=1$?
| You got $$0\cdot\tan^2{y}+9=1.$$ Id est, you got that $\tan{y}$ does not exist.
Thus, $$y=\frac{\pi}{2}+\pi n,$$ where $n\in\mathbb{Z},$ or
$$x+45^{\circ}=\frac{\pi}{2}+\pi n,$$ which gives $$x=\pi n+\frac{\pi}{4}.$$
We did not get a contradiction in Math!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that $y = Ax^2+Bx+C$ satisfies this equation. I am doing an extra credit problem for college. I don't expect anyone to solve it for me, but I would really appreciate being given some hints.
The problem:
$y'' + y' -2y = x^2$, find $A, B$, & and $C$ such that the function $y = Ax^2+Bx+C$ satisfies this equation.
I understand how to find derivatives if I know the function, but this is stumping me.
| Consider $z^2+z-2=(z+2)(z-1)=-2(1+\frac{1}{2}z)(1-z)$. Its inverse (as a formal power series) is
$$
-\frac{1}{2}(1+z+z^2+\dotsb)\left(1-\frac{1}{2}z+\frac{1}{4}z^2+\dotsb\right)
=-\dfrac{1}{2}-\dfrac{1}{4}z-\dfrac{3}{8}z^2+\dotsb
$$
Interpret $z$ as the differentiation operator and “multiply” by $x^2$ (that's why we can discard higher order terms):
$$
-\dfrac{1}{2}x^2-\dfrac{1}{4}2x-\dfrac{3}{8}2=-\dfrac{1}{2}x^2-\dfrac{1}{2}x-\dfrac{3}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$.
Solve the equation: $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$$
Here's what I've done.
Let $\sqrt{x + 2} = a$ and $\sqrt{3 - x} = b$
$\implies
\left\{
\begin{align}
a^2 + b^2 &= 5\\
a^2 - b^2 &= 2x - 1
\end{align}
\right.$.
We have that $a - b = (x - 3)^2 \implies a + b = \dfrac{a^2 - b^2}{a + b} = \dfrac{2x - 1}{(x - 3)^2}$.
$\left\{
\begin{align}
a = \dfrac{(a + b) + (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 106x + 80}{2(x - 3)^2}\\
b = \dfrac{(a + b) - (a - b)}{2} = \dfrac{x^4 - 12x^3 + 54x^2 - 110x + 82}{2(x - 3)^2}
\end{align}
\right.$
| After squaring one times we obtain
$$-2\sqrt{x+2}\sqrt{3-x}=(x^2-6x+9)^2-5$$
squaring again we obtain
$$ \left( x-2 \right) \left( {x}^{7}-22\,{x}^{6}+208\,{x}^{5}-1096\,{x}
^{4}+3468\,{x}^{3}-6552\,{x}^{2}+6772\,x-2876 \right)
=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Dudeney’s solutions to haberdasher's problem exact measures of sections What is the IG length if the side of the square is 1? I wonder if it is half of the square side. The triangle below represents the haberdasher's problem.
version 2
version 1 (added after edit, here the question is about JL)
The images are taken from here.
Edit. I posted two versions of the Dudeney triangle to square dissection. Original question was intended to concern the proper Dudeney dissection. It turned out that the version 2 is a dissection of equilateral triangle to rectangle, not a square. Though the rectangle is almost a square. Special thanks to SMM for revealing the truth.
| Denote by $a$ the side of the triangle. Since the areas of the triangle and the square are the same, we have $\frac{a^2\sqrt 3}{4}=1$, so $a=\frac{2}{\sqrt[4]3}$.
Set the coordinate system as follows: $A=(0,0)$, $C=(a,0)$ and $B=(\frac{a}{2},\frac{a\sqrt 3}{2})$. Then $D=\frac{A+B}{2}=(\frac{a}{4},\frac{a\sqrt 3}{4})$ and $E=\frac{C+B}{2}=(\frac{3a}{4},\frac{a\sqrt 3}{4})$. Further, by projecting $D$ and $E$, $F=(\frac{a}{4},0)$ and $G=(\frac{3a}{4},0)$.
We have $\vec{FE}=(\frac{a}{2},\frac{a\sqrt 3}{4})$, and $I=F+\alpha\vec{FE}=(\frac{a}{4}+\alpha\frac{a}{2},\alpha\frac{a\sqrt 3}{4})$ for some $\alpha$. Then $\vec{GI}= (\alpha\frac{a}{2}-\frac{a}{2},\alpha\frac{a\sqrt 3}{4})$. Since $FE\perp GI$ we have $\vec{FE}\cdot\vec{GI}=0$, so $\frac{a}{2}(\alpha\frac{a}{2}-\frac{a}{2})+ \frac{a\sqrt 3}{4}\alpha\frac{a\sqrt 3}{4}=0$, i.e. $\alpha-1+\frac{3}{4}\alpha=0$, where from $\alpha=\frac{4}{7}$.
So $\vec{GI}=(\frac{2a}{7}-\frac{a}{2},\frac{a\sqrt 3}{7})$. Therefore $GI^2=(\frac{2a}{7}-\frac{a}{2})^2+(\frac{a\sqrt 3}{7})^2= \frac{4a^2}{49}-\frac{2a^2}{7}+\frac{a^2}{4}+\frac{3a^2}{49}= \frac{a^2}{4}-\frac{a^2}{7}=\frac{3a^2}{28}$. Thus $GI=a\frac{\sqrt 3}{2\sqrt 7}= \frac{\sqrt[4]{3}}{\sqrt 7}$, and this is not $\frac{1}{2}$.
Edit. After Aretino posted his solution, which is correct, I started to wonder where is the mistake in mine. And there is no mistake. Mistake is in Version 2 of the construction. Namely, the rectangle obtained by Version 2 is not a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Examining an inequality involving exponential functions and hyperbolic cosine Let $a,b$ be real numbers with $0 < a < b$.
Problem: I would like to prove/disprove that $$ \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b} \leq \cosh(x \log{2})
$$ is true for all real $x \geq 0$.
Approach:
*
*I defined a function $f: \mathbb{R} \to \mathbb{R}$ with $$ f(x) = \cosh(x \log{2}) - \frac{a \cdot 2^x+ b \cdot 2^{-x}}{a+b}.
$$ In order to show the inequality, it suffices to show $f(x) \geq 0$ for all $x \geq 0$.
*I tried to plot the function for some chosen parameters like $a=1$ and $b=2$. In all those cases the function was non-negative, so I suppose that this inequality is true.
*It is $f(0) = 1-1 = 0 \geq 0$.
*Now $f$ is differentiable, so I computed
$$ f'(x) = \frac{\log{2}}{2(a+b)} \big(a (2^x - 2^{-x} - 2^{x+1}) + b(2^x-2^{-x}+2^{-x+1} ) \big). $$
*I would be done if I could show that $f'(x) \geq 0$ for all $x \geq 0$, so $f$ is monotonically increasing and we get our desired result. However, I can not see how this can be shown.
Could you please help me with this problem? That would be nice, thank you in advance!
| Using
$$
\frac{a}{a+b} = \frac 12 \left( 1 - \frac{b-a}{a+b}\right) \, , \quad
\frac{b}{a+b} = \frac 12 \left( 1 + \frac{b-a}{a+b}\right)
$$
we have
$$
\begin{aligned}
\frac{a \cdot 2^x + b \cdot 2^{-x}}{a+b} &= \left(\frac{2^x + 2^{-x}}{2}\right) - \frac{b-a}{a+b} \left(\frac{2^x - 2^{-x}}{2}\right) \\
&= \cosh(x \log 2) - \frac{b-a}{a+b} \sinh(x \log 2) \\
&\le \cosh(x \log 2) \, .
\end{aligned}
$$
An alternative argument: Fix $x \ge 0$. For $0 \le t \le \frac 12$ we have
$$
t \cdot e^x + (1-t) \cdot e^{-x} \le \max(e^{-x}, \frac{2^x + 2^{-x}}{2}) = \frac{2^x + 2^{-x}}{2}
$$
because the left-hand side is a linear function of $t$ which attains its maximum at a boundary point of the interval $[0, \frac 12]$. Setting $t = \frac{a}{a+b}$ gives the desired conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Number of ways to express sum.
Consider three sets of cards colored Blue, Red and Yellow. Each set has cards numbered $1-10$. The $4$ remaining cards are all indistinguishable cards numbered $0$.
Card numbered $i$ has the value of $2^i$. How many ways are there to choose a group of cards that sums up to $2016$.
I'm having a problem with creating a generating function for this problem, any help would be appreciated.
So far I have $f(x) = (x^1 + x^2 + x^3 + x^4)\cdot(1 + x^{2^{1}} + x^{2^{2}} + x^{2^{3}} ... + x^{2^{10}})^3$ but I'm not really sure that's even correct.
| We can represent the number of selecting zero up to four indistinguishible cards which have value $2^0=1$ as
\begin{align*}
1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}
\end{align*}
We can represent the number of selecting zero up to three distinguishible cards numbered with value $2^i$ as
\begin{align*}
\left(1+x^{2^i}\right)^3\qquad 1\leq i\leq 10
\end{align*}
The corresponding generating function is
\begin{align*}
&\frac{1-x^5}{1-x}\left(1+x^{2^{1}}\right)^3\left(1+x^{2^{2}}\right)^3\left(1+x^{2^{3}}\right)^3\cdots\left(1+x^{2^{10}}\right)^3\\
&\qquad=\frac{1-x^5}{1-x}\left(\sum_{j=0}^{2^{10}-1}x^{2j}\right)^3\\
&\qquad=\frac{1-x^5}{1-x}\left(\frac{1-x^{2^{11}}}{1-x^2}\right)^3\tag{1}
\end{align*}
in accordance with the generating function given by the answer from @RossMillikan.
Next we have to extract the coefficient of $x^{2016}$ from (1). It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series.
We obtain from (1)
\begin{align*}
\color{blue}{[x^{2016}]}&\color{blue}{\frac{1-x^5}{1-x}\left(\frac{1-x^{2^{11}}}{1-x^2}\right)^3}\\
&=[x^{2016}]\frac{1-x^5}{1-x}\left(\frac{1}{1-x^2}\right)^3\tag{2}\\
&=\left([x^{2016}]-[x^{2011}]\right)\sum_{j=0}^\infty x^k\sum_{j=0}^\infty\binom{-3}{j}(-x^2)^j\tag{3}\\
&=\left(\sum_{k=0}^{2016}[x^k]-\sum_{k=0}^{2011}[x^k]\right)\sum_{j=0}^\infty\binom{j+2}{2}x^{2j}\tag{4}\\
&=\left(\sum_{k=0}^{1008}[x^{2k}]-\sum_{k=0}^{1005}[x^{2k}]\right)\sum_{j=0}^\infty\binom{j+2}{2}x^{2j}\tag{5}\\
&=\binom{1008}{2}+\binom{1009}{2}+\binom{1010}{2}\tag{6}\\
&\,\,\color{blue}{=1\,525\,609}
\end{align*}
Comment:
*
*In (2) we omit terms in the numerator with powers of $x^{2^{11}}$ since they do not contribute to the coefficient of $x^{2016}$.
*In (3) we apply $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and do a geometric and binomial series expansion.
*In (4) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.
*In (5) we skip coefficients of odd powers since they do not contribute.
*In (6) we select the coefficients accordingly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the remainder when the polynomial $1+x^2+x^4+x^6+....+x^{22}$ is divided by $1+x+x^2+x^3+...+x^{11}$ Find the remainder when the polynomial $$1+x^2+x^4+x^6+....+x^{22}$$ is divided by $$1+x+x^2+x^3+...+x^{11}$$
$1+x^2+x^4+x^6+....+x^{22}=\frac{x^{24}-1}{x^2-1}$
$1+x+x^2+x^3+...+x^{11}=\frac{x^{12}-1}{x-1}$
Now$$\frac{1+x^2+x^4+x^6+....+x^{22}}{1+x+x^2+x^3+...+x^{11}}=\frac{x^{12}+1}{x+1}$$
Dont know how to proceed from here
| Assume $\deg Q(x)=11$ and $\deg R(x)\le10$, with
\begin{align*}
(1+x^2+\cdots+x^{22})=(1+x+\cdots+x^{11})Q(x)+R(x).\tag 1
\end{align*}
Multiply $(x^2-1)$ on both sides,
\begin{align*}
\color{blue}{(x^{12}-1)}(x^{12}+1)&=\color{blue}{(x^{12}-1)}(x+1)Q(x)+(x^2-1)R(x)
\end{align*}
implies $\color{blue}{(x^{12}-1)}\mid(x^2-1)R(x)$. Since $\deg (x^2-1)R(x)\le12$, we have
\begin{align*}
R(x)=k\cdot\frac{x^{12}-1}{x^2-1}=k(1+x^2+\cdots +x^{10}),\qquad k\in\Bbb R. \tag 2
\end{align*}
Put $(2)$ into $(1)$ and substitute $x=-1$,
$$12=0+k(6)\implies k=2.$$
$$\therefore R(x)=2(1+x^2+\cdots +x^{10}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solutions of sin(x) = cos(x) I know that the solutions to the equation $\sin(x) = \cos(x)$ are :
$ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°)
However when I try to solve it algebraically I get the following :
$$ \sin x = \cos x$$
$$ \sin^2 x = \cos^2 x$$
$$ \sin^2 = 1 - \sin^2 x$$
$$ 2\sin^2 x = 1$$
$$ sin^2 x = \frac{1}{2}$$
$$ \sqrt {sin^2 x} = \sqrt{\frac{1}{2}}$$
$$ \sqrt {sin^2 x} = \sqrt{\frac{1}{2}}$$
$$ \sin x= \lvert\frac{1}{\sqrt2}\rvert$$
$$ \sin x= \frac{\sqrt2}{2} ; \sin x= -\frac{\sqrt2}{2}$$
So if I look for all the values of $x$ that solve the above I should get not only $ x= \frac{\pi}{4}$ (45°) ; $ x= \frac{5 \pi}{4}$ (225°) but also $ x= \frac{3\pi}{4}$ (135°) ; $ x= \frac{7 \pi}{4}$ (315°).
What am I doing wrong?
| A better alternative would be to do the following: $$\begin{aligned}\sin x&=\cos x\\\tan x&=1\\x&=\arctan 1\\ x&=\dfrac{\pi}{4}+\pi n, n\in \mathbb{Z}\end{aligned}$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 1
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Solve $\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-4x}}{x}$ without L'Hospital's Rule. I need to solve $\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-4x}}{x}$ without using L'Hospital's Rule. Using that rule I found the equation becomes $\lim_{x \to 0}(\frac{1}{\sqrt{1+2x}} - \frac{2}{\sqrt{1-4x}}) = \frac{-1}{\sqrt{1}}$.
However, I'm not sure how to solve this without L'Hospital's Rule. The only tool I know of is multiply both numerator and denominator by a conjugate, but multiplying by $\frac{\sqrt{1+2x} + \sqrt{1-4x}}{\sqrt{1+2x} + \sqrt{1-4x}}$ doesn't seem to get me the same answer.
$$\lim_{x \to 0}\frac{\sqrt{1+2x} - \sqrt{1-4x}}{x} \cdot \frac{\sqrt{1+2x} + \sqrt{1-4x}}{\sqrt{1+2x} + \sqrt{1-4x}}$$
$$= \lim_{x \to 0} \frac{1+2x-(1-4x)}{x(\sqrt{1+2x} + \sqrt{1-4x})}$$
$$= \lim_{x \to 0} \frac{6x}{x(\sqrt{1+2x} + \sqrt{1-4x})}$$
$$= \lim_{x \to 0} \frac{6}{\sqrt{1+2x} + \sqrt{1-4x}}$$
$$=\frac{6}{\sqrt{1} + \sqrt{1}}$$
But $\frac{6}{\sqrt{1} + \sqrt{1}} \neq \frac{-1}{\sqrt{1}}$.
| $$\frac{\sqrt{1+2x}-\sqrt{1-4x}}{x} =\frac{1+x + o(x)-1 + 2x + o(x)}{x} = 3 + o(1) $$
So the limit is $3$.
| {
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"source": "stackexchange",
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"answer_count": 2,
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} |
Find length of the arc of $y^2=x$ I am embarrassingly stuck on this example. My textbook provides the answer (picture below) and the steps, but I am unable to follow the math. I have been stuck for more then an hour.
Find the length of the arc of the parabola $y^2=x$ from (0,0) to (1,1).
Since $x=y^2$, we have $\frac{dx}{dy} = 2y$.
$$\begin{align}
L &= \int_{0}^{1} \sqrt{1+(\frac{dx}{dy})^2}dy \\
&= \int_{0}^{1} \sqrt{1+(2y)^2}dy
\end{align}$$
Here is where I am getting stuck. I am supposed to make a trigonometric where $y = \frac{1}{2}\tan\theta$, but based off the trig substitution equation I have, $\sqrt{a^2+y^2} \rightarrow y = \tan\theta$ I am getting confused how I am supposed to get $y = \frac{1}{2}\tan\theta$. I keep thinking it should be $ y = \frac{1}{4}\tan\theta$.
Here is where I get my $y$ value from:
$$\sqrt{a^2 + y^2} = \sqrt{1+ 2^2 y^2} \\
= \sqrt{\frac{1}{4} + y^2}$$ Which gives me $y = \frac{1}{4} tan\theta$
Here is a picture of the example in the textbook. I am not able to follow the math :(
| If you have a sum under the square root sign ($\sqrt{a^2+x^2}=\sqrt{\frac{1}{a^2}\left(1+a^2x^2\right)}=\frac{1}{a}\sqrt{1+a^2x^2},\ a>0$), your trigonometric substitution would be $x=\frac{1}{a}\tan{\theta}$:
$$
\int\sqrt{1+4y^2}\,dy=\int\sqrt{1+2^2y^2}\,dy\rightarrow y=
\frac{1}{2}\tan{\theta}, dy=\frac{dy}{d\theta}d\theta=
\frac{1}{2}\sec^2\theta\,d\theta\\
\int\sqrt{1+4\left(\frac{1}{2}\tan{\theta}\right)^2}\frac{1}{2}\sec^2\theta\,d\theta=
\frac{1}{2}\int\sqrt{1+4\cdot\frac{1}{4}\tan^2{\theta}}\sec^2\theta\,d\theta=\\
\frac{1}{2}\int\sqrt{1+\tan^2{\theta}}\sec^2\theta\,d\theta=
\frac{1}{2}\int\sqrt{\sec^2\theta}\sec^2\theta\,d\theta=\\
\frac{1}{2}\int|\sec{\theta}|\sec^2\theta\,d\theta=
\frac{1}{2}\int\sec^3\theta\,d\theta.
$$
Here, $\sec\theta>0$, so we can drop the absolute value sign. That's because we're considering only an interval on which the secant function is positive (the interval of choice is typically $0<\theta<\frac{\pi}{2}$).
| {
"language": "en",
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"source": "stackexchange",
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} |
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
| How about using Cramer's Rule? Define $\Delta_x=\left[\begin{matrix}36 & 2 \\ 64 & 4\end{matrix}\right]$, $\Delta_y=\left[\begin{matrix}3 & 36\\ 5 & 64\end{matrix}\right]$
and $\Delta_0=\left[\begin{matrix}3 & 2\\ 5 &4\end{matrix}\right]$.
Now computation is trivial as you have: $x=\dfrac{\det\Delta_x}{\det\Delta_0}$ and $y=\dfrac{\det\Delta_y}{\det\Delta_0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 7
} |
What is $\int_{0}^{\pi} (a^2+1+2a \cos\theta)^{\frac{p-1}{2}}d\theta$? I have tried the problem using $\theta\rightarrow\theta/2$, then taking $z=e^{i\theta}$ we get $\frac{1}{2i}\int_{|z|=1}\frac{(a^2+1+a\sqrt{z}+\frac{a}{\sqrt{z}})^{\frac{p-1}{2}}}{z}dz$.
| With the change of variables $t = 2 \arccos \sqrt u$,
$$\int_0^\pi (a^2 + 2 a \cos t + 1)^{(p - 1)/2} dt = \\
|a - 1|^{p - 1} \int_0^1 u^{-1/2} (1 - u)^{-1/2}
\left( 1 + \frac {4 a} {(a - 1)^2} u \right)^{(p - 1)/2} du = \\
\pi |a - 1|^{p - 1} {_2 \hspace{-1px} F_{\hspace{-1px} 1}} {\left(
\frac 1 2, \frac {1 - p} 2; 1; -\frac {4 a} {(a - 1)^2} \right)},
\quad a, p \in \mathbb R, \; a^2 \neq 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the video, the best I can do with this is relying on expansion formulas and substitution. As trivial a problem this is, the numerous trinomials and binomials with mixed terms makes it very, very tedious.
What is the quickest/shortest approach to this problem (meaning it doesn't need to be solved algebraically)? You don't have to type the entire solution out, I think if I'm given a good hint then I can take it from there.
| Let's start with the basic symmetric expressions: $ab+bc+ca$ and $abc$. You can refer to giannispapav's answer for details, which shows that
$$ab+bc+ca = -1/2, abc = 1/6.$$
With that, Vieta's formulas implies that $a,b,c$ satisfy:
$$ x^3 -x^2 - x/2 -1/6=0,\tag{1}$$
Or
$$x^3 = x^2 + x/2 + 1/6.$$
That means, for $x$ equals $a,b,c$,
$$x^4 = x^3 + x^2/2 + x/6,$$
and
$$x^5 = x^4 + x^3/2 + x^2/6.$$
Adding the two equations above, we have
$$x^5 = \frac32x^3 + \frac23x^2 + \frac16x.$$
Now replace $x$ as $a,b,c$ and add them all up, we have
$$a^5+b^5+c^5 = \frac32(a^3+b^3+c^3) + \frac23(a^2+b^2+c^2) + \frac16(a+b+c).$$
Note: if you feels that
$$a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$$
is too complicated to verify, then Vieta's formulas is the way to go. That is, replace $a,b, c$ in Equation $1$ and add them up, where $1/6$ is indeed $abc$ as in Vieta's formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve for $x^2 + 7x +1 = 3n(x^2 + x +1), n \in \Bbb{Z}$ Solve for $x$ and $n$ in the equation
$$ \dfrac{x^2 + 7x +1}{ x^2 + x +1}= 3n,\qquad n \in \Bbb{Z}.$$
The original problem was a trigonometric equation but solved it till I got stuck here. I am a high school student who is self studying and preparing for college entrance exam. Original problem was a trigonometric equation which came from 'cengage exam crack'.
[After Mark's and Jyrki's hints in the comments, I got the answer and posted it below.]
Original problem:
| $3n
=\dfrac{x^2 + 7x +1}{ x^2 + x +1}
$
so
$3nx^2+3nx+3n
=x^2+7x+1
$
so
$(3n-1)x^2+(3n-7)x+3n-1
=0
$
or
$\begin{array}\\
x
&=\dfrac{-3n+7\pm \sqrt{(3n-7)^2-4(3n-1)^2}}{2(3n-1)}\\
&=\dfrac{-3n+7\pm \sqrt{(3n-7-2(3n-1))(3n-7+2(3n-1))}}{2(3n-1)}\\
&=\dfrac{-3n+7\pm \sqrt{(-3n-5)(9n-9)}}{2(3n-1)}\\
&=\dfrac{-3n+7\pm 3\sqrt{(-3n-5)(n-1)}}{2(3n-1)}\\
\end{array}
$
If nothing is specified about $x$,
this is as far as we can go.
If $x$ has to be real,
then
$(3n+5)(n-1) \le 0$
so that
$-\dfrac53 \le n \le 1$.
Since $n$ is an integer,
$n = -1, 0, 1$.
If $x$ is supposed to be rational,
then
$(3n-7)^2-4(3n-1)^2
=m^2
$.
From the preceding result,
$n=1
\implies
m^2
=4^2-4(2)^2
=0
$,
$n=0
\implies
m^2
=7^2-4(1)^2
=45
$,
$n=-1
\implies
m^2
=10^2-4(-4)^2
=36
$.
Therefore the only rational solutions
with integer $n$ are
$n=1, 0, -1$
for which
$n=1
\implies x
=\dfrac{-3n+7\pm 3\sqrt{(-3n-5)(n-1)}}{2(3n-1)}
=\dfrac{4}{4}
=1
$,
$n=0
\implies x
=\dfrac{-3n+7\pm 3\sqrt{(-3n-5)(n-1)}}{2(3n-1)}
=\dfrac{7\pm 3\sqrt{5}}{-2}
$,
and
$n=-1
\implies x
=\dfrac{-3n+7\pm 3\sqrt{(-3n-5)(n-1)}}{2(3n-1)}
=\dfrac{-10\pm 4}{-8}
=\dfrac{-14, -6}{-8}
=\dfrac74, \dfrac52
$.
| {
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"url": "https://math.stackexchange.com/questions/3184575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Very indeterminate form: $\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x \longrightarrow (\infty-\infty)^{\infty}$ Here is problem:
$$\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x$$
The solution I presented in the picture below was made by a Mathematics Teacher
I tried to solve this Limit without using derivative (L'hospital) and Big O notation. Although I get the answer, I don't know if the technique I'm using definitely correct.
And here is my method:
$$\begin{align*}\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x&=\lim_{x \to \infty} \left(\frac {2x}{\sqrt{x^2+2x+3} +\sqrt{x^2+3}}\right)^x\\&=\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}\end{align*}$$
Then, I define a new function here
$$y(x)=\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1$$
We have
$$\begin{align*}
\lim _{x\to\infty} y(x)&=\lim_{x \to \infty}\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1\\
&=\lim_{x \to \infty}(\sqrt{x^2+2x+3}-(x+1))+(\sqrt{x^2+3}-x)\\
&=\lim_{x \to \infty}\frac{2}{\sqrt{x^2+2x+3}+x+1}+ \lim_{x \to \infty}\frac{3}{\sqrt{x^2+3}+x}\\
&=0.
\end{align*}$$
This implies that
$$\lim_{x \to \infty}\frac{2x}{y(x)+1}=\infty $$
Therefore,
$$\begin{align*}
\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}&=\lim_{x \to\infty} \frac{1}{ \left(\frac{y(x)+2x+1}{2x} \right)^x}\\
&=\lim_{x \to\infty} \frac{1}{ \left(1+\frac{y(x)+1}{2x} \right)^x}\\
&=\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}\\
&
\end{align*}$$
Here, we define two functions:
$$f(x)=\left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}},\quad
g(x)=\frac{y(x)+1}{2}.
$$
We deduce that,
$$
\lim_{x\to\infty} f(x)=e>0,\quad \lim_{x\to\infty} g(x)=\frac 12>0.
$$
Thus, the limit $\lim_{x\to\infty} f(x)^{g(x)} $ exists and is finite.
Finally we get,
$$\begin{align*}
\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}
&=\frac{1}{\lim_{x \to \infty}\left( \left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}\right)}\\
&=\frac{1}{\left(\lim_{x\to\infty} \left( 1+\frac{1}{\frac{2x}{y(x)+1}} \right)^{\frac{2x}{y(x)+1}}\right)^{ \lim_{x\to\infty} \frac{y(x)+1}{2}}}\\
&=\frac {1}{e^{\frac12}}=\frac{\sqrt e}{e}.\\&&
\end{align*}$$
Is the method I use correct?
I have received criticisms against my work. What can I do to make the method I use, rigorous? What are the points I missed in the method?
Thank you!
| The proof can be accelerated, using binomial Maclaurin series in the form of
$$\sqrt{x^2+2x+3} = (x+1)\sqrt{1+\dfrac2{(x+1)^2}} = (x+1)\left(1 + \dfrac1{(x+1)^2}+O\left(x^{-4}\right)\right)$$
$$ = x+1+\dfrac1x+O\left(x^{-2}\right),$$
$$\sqrt{x^2+3} = x\left(1+\dfrac3{2x^2}+O(x^{-4})\right) = x + \dfrac3{2x}+O(x^{-3}).$$
Then
$$\ln L = \ln \lim\limits_{x\to\infty} \left(\sqrt{x^2+2x+3}-\sqrt{x^2+3}\right)^x
= \lim\limits_{x\to\infty} x\ln\left(1-\frac1{2x}+O\left(x^{-2}\right)\right)$$
$$= \lim\limits_{x\to\infty} x\left(-\frac1{2x}+O\left(x^{-2}\right)\right) = -\frac12,$$
$$L=e^{\Large^{-\frac12}}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Representing a function as a power series and finding the coefficients of the terms? My prblem asks me to find the coefficients of the first five terms of the power series representation of the function:
$$f(x)=\frac{10}{17+x}$$
$$f(x)=10*\frac{\frac{1}{17}}{1-\frac{-x}{17}}$$
$$\sum_{n=0}^{\infty}10*\frac{1}{17}*(\frac{-x}{17})^n$$
$$\sum_{n=0}^{\infty}\frac{10}{17}*\frac{(-1)^nx^n}{17^n}$$
First five terms are:
$$\frac{10}{17} + \frac{10}{17}*\frac{-x}{17} + \frac{10}{17}*\frac{x^2}{17^2} + \frac{10}{17}*\frac{-x^3}{17^3} + \frac{10}{17}*\frac{x^4}{17^4}$$
I entered $0$ for the first coefficient. Marked me correct.
I put $\frac{10}{17}$ for the second coefficient. Marked me correct.
I put $\frac{10}{17}$ for the third coefficient. Marked me wrong.
I'm not sure what I'm doing wrong. Looking at the series again, the coefficient should be $-\frac{10}{17}$ for the second coefficient, shouldn't it?
I tried a bunch of stuff for the second coefficient too. $0$, $-1$, $1$, $-10$, $10$, $\frac{10}{17}$, $-\frac{10}{17}$, $\frac{100}{17^2}$, $-\frac{100}{17^2}$, $\frac{10}{17^3}$, $-\frac{10}{17^3}$ (those last few are from different ways of writing the power series I tried). Nothing is working. Where am I going wrong?
Edit: to clarify a bit, the question wants me to find the coefficients of $x^0$, $x$, $x^2$, $x^3$ and $x^4$ in the series.
| You allready found $$f(x)=\frac{10}{17} + \frac{10}{17}\frac{-x}{17} + \frac{10}{17}\frac{x^2}{17^2} + \frac{10}{17}*\frac{-x^3}{17^3} + \frac{10}{17}\frac{x^4}{17^4}+...$$
$$=\frac{10}{17} - \frac{10}{17^2}x + \frac{10}{17^3}x^2 - \frac{10}{17^4}x^3 + \frac{10}{17^5}x^5 \mp ...$$
The general coefficient $a_n$ is given by
$$a_n=(-1)^n\dfrac{10}{17^{n+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$f: \mathbb{R} \to \mathbb{R},\space\space\space f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,\space\space$find $f(3)$ in terms of $f(0)$.
$f: \mathbb{R} \to \mathbb{R},\space\space\space\space f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2,\space\space\space\space$
Find $f(3)$ in terms of $f(0)$.
My approach:
$$f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$$
$$\Rightarrow f(\frac{x}{2})-2f(\frac{x}{4})+f(\frac{x}{8})=\frac{x^2}{4}$$
$$\Rightarrow f(\frac{x}{4})-2f(\frac{x}{8})+f(\frac{x}{16})=\frac{x^2}{16}$$
$$\vdots$$
$$\Rightarrow f(\frac{x}{2^{n-1}})-2f(\frac{x}{2^n})+f(\frac{x}{2^{n+1}})=\frac{x^2}{4^{n-1}}$$
Summing up these when n $\rightarrow$ $\infty$
$$f(x)-f(\frac{x}{2})=x^2\cdot\frac{4}{3}$$
I am unable to proceed further.
| I'll go ahead and assume that $f$ is continuous at $0$.
For any $x \in \mathbb{R}$, let $P(x)$ denote the property that $f(x) - 2f(\tfrac{x}{2}) + f(\tfrac{x}{4}) = x^2$. Then, for any $x$, we have the following:
\begin{align}
P(x)&: \quad f(x) - 2f(\tfrac{x}{2}) + f(\tfrac{x}{4}) = x^2
\\
P(\tfrac{x}{2})&: \quad f(\tfrac{x}{2}) - 2f(\tfrac{x}{4}) + f(\tfrac{x}{8}) = \tfrac{1}{4}x^2
\\
& \vdots
\\
P(\tfrac{x}{2^{n-1}})&: \quad f(\tfrac{x}{2^{n-1}}) - 2f(\tfrac{x}{2^n}) + f(\tfrac{x}{2^{n+1}}) = \tfrac{1}{4^{n-1}}x^2
\end{align}
Adding these up gives us $$f(x)-f(\tfrac{x}{2})-f(\tfrac{x}{2^n})+f(\tfrac{x}{2^{n+1}}) = \tfrac{4}{3}(1-\tfrac{1}{4^n})x^2$$
If we take the limit as $n \to \infty$ and use the assumption that $f$ is continuous at $0$, we get $$f(x)-f(\tfrac{x}{2})-f(0)+f(0) = \tfrac{4}{3}(1-0)x^2$$ i.e. $$f(x)-f(\tfrac{x}{2}) = \tfrac{4}{3}x^2.$$
Now, we can repeat the same trick. For any $x \in \mathbb{R}$, let $Q(x)$ denote the property that $f(x)-f(\tfrac{x}{2}) = \tfrac{4}{3}x^2$. Then, for any $x$, we have the following:
\begin{align}
Q(x)&: \quad f(x) - f(\tfrac{x}{2}) = \tfrac{4}{3}x^2
\\
Q(\tfrac{x}{2})&: \quad f(\tfrac{x}{2}) - f(\tfrac{x}{4}) = \tfrac{1}{3}x^2
\\
& \vdots
\\
Q(\tfrac{x}{2^{n-1}})&: \quad f(\tfrac{x}{2^{n-1}}) - f(\tfrac{x}{2^n}) = \tfrac{1}{3 \cdot 4^{n-2}}x^2
\end{align}
Adding these up gives us $$f(x)-f(\tfrac{x}{2^n}) = \tfrac{16}{9}(1-\tfrac{1}{4^n})x^2$$
If we take the limit as $n \to \infty$ and use the assumption that $f$ is continuous at $0$, we get $$f(x)-f(0) = \tfrac{16}{9}x^2.$$ Hence, $f(x) = \tfrac{16}{9}x^2+f(0)$. It is easy to check that any function in the form $f(x) = \tfrac{16}{9}x^2+C$ for some constant $C$ satisfies the given functional equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The smallest value of the expression: $(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$ lies in which of the following intervals? Question:
Let $a_1,a_2,a_3,a_4\in\Bbb R$, such that $a_1+a_2+a_3+a_4 = 0$ and $a_1^2+a_2^2+a_3^2+a_4^2 = 1$.
Then the smallest value of the expression, $$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$$
lies in the interval:
*
*$(0,1.5)$
*$(1.5,2.5)$
*$(2.5,3)$
*$(3,3.5)$
My Attempt:
Here,
$$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2\gt0\qquad\forall a_i\in\Bbb R$$
Now:
$$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2 = 2(1)-2(a_1a_2+a_2a_3+a_3a_4+a_4a_1)$$
since,$$(a_1-a_2)^2\gt0\implies a_1^2+a_2^2 \gt 2a_1a_2$$
$$\implies 2(a_1^2+a_2^2+a_3^2+a_4^2)\gt 2(a_1a_2+a_2a_3+a_3a_4+a_4a_1)$$
$$\implies a_1a_2+a_2a_3+a_3a_4+a_4a_1\lt 1$$
Therefore,
$$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2\lt1$$
Therefore the answer (according to me) should be option $(1)$, or the interval: $(0,1.5)$,
however, the answer given is option $(2)$, or the interval: $(1.5,2.5)$.
Can someone tell me where my error is (or) a better method to solve this problem?
| Hint: Your sum simplifies to $$2-2(a_1a_3+a_2a_3+a_3a_4+a_4a_1)=2-2(a_2(a_1+a_3)+a_4(a_1+a_3))=2-2(-(a_2+a_4)^2)$$ and $$2+2(a_2+a_4)^2\geq 2$$. The minimum is equal to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3201420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\left(\sqrt{\sqrt{2}-4-x}\right)+x^{\frac{1}{4}}=2^{\frac{-1}{4}}$ Solve the Equation in real/Complex numbers:
Solve $$\left(\sqrt{\sqrt{2}-4-x}\right)+x^{\frac{1}{4}}=2^{\frac{-1}{4}}$$
My try:
Letting $x=t^4$ we get
We get
$$\left(\sqrt{\sqrt{2}-4-t^4}\right)+|t|=2^{\frac{-1}{4}}$$
Then:
$$\left(\sqrt{\sqrt{2}-4-t^4}\right)=2^{\frac{-1}{4}}-|t|$$
Squaring we get:
$$\sqrt{2}-4-t^4=t^2+\frac{1}{\sqrt{2}}-|t|2^{\frac{3}{4}}$$
$\implies$
we get
$$t^4+t^2+4=\frac{1}{\sqrt{2}}+|t|2^{\frac{3}{4}}$$
Any clue here?
| As said in comments, I seriously wonder it the problem is correctly stated.
This equation shows two complex roots. Using Newton method, we have the awful iterates
$$\left(
\begin{array}{cc}
n & x_n \\
0 & +1.00000+1.00000\, i \\
1 & -3.58944+1.23135\, i \\
2 & -0.75253+0.37882\, i \\
3 & -1.89384+0.00982\, i \\
4 & -1.89949+0.01512\, i \\
5 & -1.89949+0.01510\, i
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
n & x_n \\
0 & +1.00000-1.00000\, i \\
1 & -3.58944-1.23135\, i \\
2 & -0.75253-0.37882\, i \\
3 & -1.89384-0.00982\,i \\
4 & -1.89949-0.01512\, i \\
5 & -1.89949-0.01510\, i
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Real roots of the equation $\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$
Find the set of real roots of the equation$$\log_{(5x+4)}(2x+3)^3-\log_{(2x+3)}(10x^2+23x+12)=1$$
My Attempt
$$
2x+3>0, 5x+4>0, 2x+3,5x+4\neq1\implies x>-4/5\;\&\;x\neq -1\;\&\;x\neq -3/5
$$
$$
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)(2x+3)=1\\
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)-\log_{(2x+3)}(2x+3)=1\\
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)-1=1\\
3\log_{(5x+4)}(2x+3)-\log_{(2x+3)}(5x+4)=2\\
3\log_{(5x+4)}(2x+3)-\frac{1}{\log_{(5x+4)}(2x+3)}=2
$$
Set $y=\log_{(5x+4)}(2x+3)$
$$
3y^2-2y-1=0\implies y=\log_{(5x+4)}(2x+3)=1,\frac{-1}{3}
$$
Case 1:
$$
\log_{(5x+4)}(2x+3)=\frac{\log(5x+4)}{\log(2x+3)}=1\implies\log(5x+4)=\log(2x+3) \implies \boxed{x=\frac{-1}{3}}
$$
Case 2:
$$
\log_{(5x+4)}(2x+3)=\frac{\log(2x+3)}{\log(5x+4)}=\frac{-1}{3}\implies \color{red} ?
$$
My reference says $x=\frac{-1}{3}$ is the only solution. How do I prove it ?
Possible Solution
$$
y=\log_{(5x+4)}(2x+3)=\frac{-1}{3}<0\:,\quad x>-4/5=-0.8
$$
Case 1: $5x+4>1\implies x>-3/5=-0.6$
$$
0<2x+3<1\implies -3/2<x<-1\implies-1.5<x<-1\\
\text{Not Possible !}
$$
Case 2: $0<5x+4<1\implies -4/5<x<-3/5\implies-0.8<x<-0.6$
$$
2x+3>1\implies x>-1\\
\implies \boxed{x\in(-0.8,-0.6)}
$$
It seems like there could be a solution for "case 2" in my attempt ?
| Since $10x^2+23x+12=(2x+3)(5x+4)$, my suggestion is to set $t=\log_{(2x+3)}(5x+4)$, so that
$$
\log_{(5x+4)}(2x+3)=\frac{1}{t}
$$
and the equation becomes
$$
\frac{3}{t}-1-t=1
$$
so $t^2+2t-3=0$ and $t=1$ or $t=-3$.
The first case yields $5x+4=2x+3$, hence $x=-1/3$.
The second case yields $5x+4=(2x+3)^{-3}$. If we consider the two functions $f(x)=5x+4$ and $g(x)=(2x+3)^{-3}$, we can see that
$$
f(-2)=-6,\quad g(-2)=-1,\qquad
f(-1)=-1,\quad g(-1)=1,\qquad
f(0)=4,\quad g(0)=1/27
$$
Hence the equation $f(x)=g(x)$ has two roots in the interval $(-2,0)$. For the one in the interval $(-2,-1)$, we have $f(x)=g(x)<0$, so this is not a solution.
For the one in the interval $(-1,0)$, we have $f(x)=g(x)>0$ and not $1$. So the solution is good for the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculate $\sum_{n=1}^{\infty}\left(\frac{1}{3}\frac{2}{5}\cdots\frac{n}{2n+1}\frac{1}{n+1}\right)$
Denote $a_n=\frac{1}{3}\cdot\frac{2}{5}\cdots\frac{n}{2n+1}\cdot\frac{1}{n+1}$.
Please prove
$$\sum_{n=1}^{\infty}a_n=\frac{\pi^2}{8}-1$$
This is the answer given by my friend. He also used the integral $\int_{-\pi}^{\pi}\cos^kx\,dx$ along with much calculation in which the answer came out almost accidentally.
I want to see other ways to deal with it. Feel free to use the integral above. Thanks in advance.
Edit:This may be a little long
(I)$\int_{0}^{\pi}\sin^{2n+1}xdx=\frac{2^n\dot n!}{(2n+1)!!}$
(II)$\sum_{n=1}^\infty \frac{x^{n+1}}{n+1}=-x-ln(1-x)$
(III)$-\int_{-1}^1\frac{ln(\frac{1+x^2}{2})}{1-x^2}dx=-\int_{-1}^1\frac{ln(1-\frac{1-x^2}{2})}{1-x^2}dx=-\int_{-1}^1\frac{ln(1-t(1-x^2))}{1-x^2}\big|_{t=0}^{1/2}dx=\int_{-1}^1\int_{0}^{1/2}\frac{dtdx}{1-t(1-x^2)}=\int_{0}^{1/2}\int_{-1}^1\frac{dxdt}{1-t(1-x^2)}=2\int_0^{1/2}\frac{\arctan \sqrt{\frac{t}{1-t}}}{\sqrt{t(1-t)}}dt=4\int_0^{1/2}\arctan \sqrt{\frac{t}{1-t}}d\arctan\sqrt{\frac{t}{1-t}}=4\cdot\frac{1}{2}\cdot(\frac{\pi}{4})^2=\frac{\pi^2}{8}$
(IV)$\sum_{n=1}^{\infty}\left(\frac{1}{3}\frac{2}{5}\cdots\frac{n}{2n+1}\frac{1}{n+1}\right)=\sum_{n=1}^{\infty}\frac{n!}{(2n+1)!!}\frac{1}{n+1}=\sum_{n=1}^{\infty}\frac{\int_{0}^{\pi}\sin^{2n+1}xdx}{2^{n+1}}\frac{1}{n+1}=\int_{0}^{\pi}\sum_{n=1}^{\infty}\frac{\sin^{2n+1}x}{2^{n+1}}\frac{dx}{n+1}=\int_{0}^{\pi}\frac{1}{\sin x}\sum_{n=1}^{\infty}(\frac{\sin^{2}x}{2})^{n+1}\frac{dx}{n+1}=\int_0^{\pi}\frac{-\frac{\sin^{2}x}{2}-ln(1-\frac{\sin^{2}x}{2})}{\sin x}dx=-1+\int_0^{\pi}\frac{-ln(1-\frac{\sin^{2}x}{2})}{\sin x}dx=-1+\int_0^{\pi}\frac{ln(1-\frac{\sin^{2}x}{2})}{\sin^2 x}d\cos x=-1+\int_0^{\pi}\frac{ln(\frac{1+\cos^{2}x}{2})}{1-\cos^2 x}d\cos{x}=-1+\frac{\pi^2}{8}$
(Sorry for ugly typing style.)
| Using the Factorial aswell as the Double Factorial we may rewrite $a_n$ as
$$a_n=\frac13\frac25\cdots\frac n{2n+1}\frac1{n+1}=\frac{n!}{(2n+1)!!}\frac1{n+1}$$
Expressing the utilized Double Factorial in terms of the Factorial this can be further reduced
$$a_n=\frac1{(2n+1)!!}\frac{n!}{n+1}=\frac{2^{n+1}(n+1)!}{(2n+2)!}\frac{n!}{n+1}=\frac{2^n}{(n+1)(2n+1)\binom{2n}n}$$
Now, the sum comes into play. Plugging our new gained expression within the summation followed by an index shift gives us the following
\begin{align*}
\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{2^n}{(n+1)(2n+1)\binom{2n}n}&=\sum_{n=2}^\infty\frac{2^{n-1}}{n(2n-1)\binom{2n-2}{n-1}}\\
&=\sum_{n=2}^\infty \frac{2^n}{(2n)(2n-1)\frac{(2n-2)!}{(n-1)!^2}}\\
&=\sum_{n=2}^\infty \frac{2^n}{n^2\frac{(2n)!}{n!^2}}\\
&=\sum_{n=2}^\infty\frac{2^n}{n^2\binom{2n}n}
\end{align*}
The last sum is of a truly remarkable structure, to be precise the famous series expansion of the squared inverse sine function! Applying this knowledge we finally obtain
$$\sum_{n=2}^\infty\frac{2^n}{n^2\binom{2n}n}=2\arcsin^2\left(\frac{\sqrt2}2\right)-1=\frac{\pi^2}8-1$$
$$\therefore~\sum_{n=1}^\infty a_n~=~\frac{\pi^2}8-1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Writing $3.8473221018630726$ in the form $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$. I attempted the following question on Brilliant which has to do with finding roots of a cubic polynomial. I was successful in finding what the only real root is but I am facing a problem rewriting the root in the sought expression.
The equation $x^3-3x^2-3x-1=0$ has exactly one real solution that can be written in the form $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$. What is the value of $a+b+c$?
I've found the value of $x$ to be equal to $3.8473221018630726$ by the method of depressing the cubic. Any hints to proceed are appreciated.
Edit:
Would the fact that $x=\dfrac{2}{\sqrt[3]{4}}+\sqrt[3]{4}+1=\dfrac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1\approx3.8473221018630726$ help somewhere in determining $a, b$ and $c$?
|
Would the fact that $x=\frac{2}{\sqrt[3]{4}}+\sqrt[3]{4}+1=\frac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1\approx3.8473221018630726$ help somewhere in determining $a, b$ and $c$?
Yes, it would very much. If $\frac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1$ is the root you're after, then
$$
\frac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1 = \sqrt[3]{\frac82} + \sqrt[3]{2} + \sqrt[3]{1}
$$
and you have the solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do i integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$? Integrate $\int \frac{dx}{(2x+3)\sqrt{(x^2+3x+2})}$
I put $x^2+3x+2=t,$ and notice that $2x+3 dx=dt$, but the $dx$ is above! Please help me!
| Let $ t = \sqrt{x^2+3x+2}$, $dt = \frac{2x+3}{2\sqrt{x^2+ 3x +2}}dx$, $dx = \frac{2\sqrt{x^2+ 3x +2}}{2x+3}dt$
$I = \int{ \frac{2\sqrt{x^2+ 3x +2}}{2x+3}. \frac{1}{{(2x+3)}.{\sqrt{x^2+ 3x +2}}}dt} =\int{ \frac{2}{2x+3}. \frac{1}{{2x+3}}dt} =\int{ \frac{2}{4x^2+6x+9}dt} = 2\int{\frac{1}{4t^2 + 1}} $(on solving x in terms of t)
$I = 2.\frac{1}{2}.tan^{-1}(2t)+ c = tan^{-1}(2t)+c= tan^{-1}(2\sqrt{x^2+3x+2})+c = sec^{-1}(2x+3) +c = cos^{-1}(\frac{1}{2x+3})+c = \pi/2 - sin^{-1}(\frac{1}{2x+3}) +c = -sin^{-1}(\frac{1}{2x+3}) +c'$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$1+\sqrt{2}$ seems special: its powers quickly approach integers $ \alpha = 1+\sqrt{2}$ has some interesting properties. One of which is that if you look at powers
\begin{align*}
\alpha^2 &= 3+2\sqrt{2} \approx 5.8284 \\
\alpha^3 & \approx 14.0710 \\
\alpha^4 & \approx 33.9705 \\
& \vdots \\
\alpha^{10} & \approx 6725.9998 \\
\alpha^{11} & \approx 16238.0000 \\
\vdots &
\end{align*}
(values truncated, not rounded) then the sequence $\{ \alpha^n \}$ is "almost" an unbounded integer sequence. What is going on here? Is this property secondary to $\alpha$ being a unit in $\mathbb{Z}[\sqrt{2}]$? Can $\alpha$ be interpreted as a type of approximate eigenvalue? Does this behavior have to do with the continued fraction representation?
Other numbers such as $1+\sqrt{3}$ have a similar property, but they don't appear to "become integer" as quickly.
| You could observe the pattern if you would calculate the powers:
$$\begin{align}\alpha^1&=(1+\sqrt{2})^1=1+1\sqrt{2}\\
a^2&=(1+\sqrt{2})^2=3+2\sqrt{2}\\
a^3&=(1+\sqrt{2})^3=7+5\sqrt{2}\\
a^4&=(1+\sqrt{2})^4=17+12\sqrt{2}\\
\vdots
\end{align}$$
So, basically, the second terms are important. The coefficients $1,2,5,12,...$ form the recurrence relation OEIS A000129:
$$a_1=1,a_2=2,a_n=2a_{n-1}+a_{n-2},$$
Its solution is:
$$a_n=\frac1{2\sqrt{2}}\left[(1+\sqrt{2})^n-(1-\sqrt{2})^n\right]$$
Hence:
$$a_n\sqrt{2}=\frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2}=\frac{(1+\sqrt{2})^n+(1-\sqrt{2})^n-2(1-\sqrt{2})^n}{2}=\\
\frac{(1+\sqrt{2})^n+(1-\sqrt{2})^n}{2}-(1-\sqrt{2})^n$$
The first term is an integer, because its numerator is an even number and the second term approaches $0$ for large $n$, because $|1-\sqrt{2}|<1$, which were also stated in other answers and comments.
Hence, you can generalize for $\alpha=1+\sqrt{k}$:
$$a_n\sqrt{k}=\frac{(1+\sqrt{k})^n+(1-\sqrt{k})^n}{2}-(1-\sqrt{k})^n$$
So, it depends on how fast the second term $(1-\sqrt{k})^n$ approaches $0$.
| {
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When recreating the quadratic formula by completing the square of $ax^2+bx+c=0$ I cannot shorten the right hand side I am attempting to derive the quadratic formula by completing the square on the generic generic expression:
$$ax^2+bc+c=0$$
I'm struggling with the right hand side of the equation which, for the step I'm on I know should be $\frac{b^2-4ac}{4a^2}$. However, I arrive at $\frac{b^2a-4a^2c}{4a^3}$
Here's my working:
(Approach copied largely from textbook)
Start with:
$ax^2+bx+c=0$
Move constant term to the right:
$ax^2+bx=-c$
Divide by $a$ to ensure leading coefficient is 1:
$x^2+\frac{b}{a}x=-\frac{c}{a}$
Calculate the amount needed to complete the square and add to both sides:
$(\frac{1}{2}*\frac{b}{a})^2$ = $(\frac{b}{2a})^2$ = $\frac{b^2}{4a^2}$
Now add this to both sides:
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}+-\frac{c}{a}$
Write the left side as a perfect square:
$(x^2+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}$
Simplify the right hand side by finding a common denominator:
This is where I'm tripping up
$\frac{b^2}{4a^2}-\frac{c}{a}$
The common denominator will be the product of the denominators so $4a^3$
This doesn't "feel" right and I suspect I should be looking for a "least common denominator" but I don't know what that would be given the existence of the radical.
Rewriting using the common denominator $4a^3$ I multiply the numerator and denominator of left side of the minus sign by just $a$. I then multiple the numerator and denominator on the right side of the minus sign by $4a^2$:
$\frac{b^2a}{4a^3}-\frac{4a^2c}{4a^3}$ = $\frac{b^2a-4a^2c}{4a^3}$
How can I arrive at $\frac{b^2-4ac}{4a^2}$?
I know that I'm not done yet after figuring out the above, but it's this in between step I'm tripping up on.
| Note that
$$
\frac{c}{a}=\frac{4ac}{4a^2}
$$
whence
$$
\frac{b^2}{4a^2}-\frac{c}{a}=\frac{b^2-4ac}{4a^2}
$$
In your approach
$$
\frac{b^2a-4a^2c}{4a^3}=\frac{b^2-4ac}{4a^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Integrate $S = \int_{0}^{1} 2\pi (y+y^3) \sqrt{1+(1+3y^2)^2}dy$ $S = \int_{0}^{1} 2\pi (y+y^3) \sqrt{1+(1+3y^2)^2}dy$
$$
\begin{align}
S &= \int_{0}^{1} 2\pi (y+y^3) \sqrt{1+(1+3y^2)^2}dy \\
&= 2\pi \int_{0}^{1} (y+y^3) \sqrt{1+(1+3y^2)^2}dy \\
&= 2\pi \int_{0}^{2} \sqrt{1+u^2}du ~~~~~~\textrm{substitution with $u = y + y^3$, $du = 1 + 3y^2$} \\
&= 2\pi \int_{0}^{2} \sec^3\theta d\theta = \frac{2\pi}{2} \bigg(\sec\theta \tan\theta + \ln|\sec\theta + \tan\theta| \bigg)\\
&= \pi \bigg[ \sec\theta \tan\theta + ln|\sec\theta + \tan\theta|\bigg]_{0}^{2} \\
&= \pi \bigg[ \sec(\sqrt{1+u}) u + ln|\sec(\sqrt{1+u}) + u|\bigg]_{0}^{2} \\
&= \pi \bigg[ \sec(\sqrt{11}) \cdot 11 + ln|\sec(\sqrt{11}) + 11|\bigg] - \pi\bigg[ \sec(\sqrt{1}) \cdot 0 + ln|\sec(\sqrt{1}) + 0|\bigg]\\
&\approx 21.918\pi
\end{align}
$$
However the approximation per Wolfram Alpha and my textbook both agree on a value of 13.5134.
| We see that
$$S/2\pi=\int_0^1\sqrt{1+(1+3x^2)^2}\,xdx+\int_0^1x^2\sqrt{1+(1+3x^2)^2}\,xdx$$
Then set $t=1+3x^2\Rightarrow xdx=\frac16dt$ in both integrals to get
$$S/2\pi=\frac16\int_1^4 \frac{t-1}{3}\sqrt{1+t^2}\,dt+\frac16\int_1^4\sqrt{1+t^2}\,dt\\
=\frac1{18}\int_1^4t\sqrt{1+t^2}\,dt+\frac19\int_1^4\sqrt{1+t^2}\,dt$$
In the first integral, sub $u=t^2$, and in the second integral, use a trig sub. Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve Shonk Sequences? A Shonk sequence is a sequence of positive integers in which
*
*each term after the first is greater than the previous term, and
*the product of all the terms is a perfect square
For example: 2, 6, 27 is a Shonk sequence since 6>2 and27>6 and 2*6*27 = 324 or 18^2
a. If 12, x, 24 is a Shonk sequence, what is the value of x?
b. If 28, y, z, 65 is a Shonk sequence, what are the values of y and z?
c. Determine the length of the longest Shonk sequence, each of whose terms is an integer between 1 and 12, inclusive. Your solution should include an example of this longest length, as well as justification as to why no longer sequence is possible.
d. A sequence of four terms a,b,c,d is a super-duper-Shonkolistic sequence (SDSS) exactly when a,b,c,d and a,b,c and b,c,d are a Shonk sequence. Determine the number of pairs (m, n) such that m, 1176, n, 48400 is an SDSS.
This question was taken directly from the Grade 9 Fryer Contest for the University of Waterloo. Permission has been granted to discuss these questions online as the contests are officialy over.
I'd like to know the answers and the steps of the solution. Thanks.
| Although you don't show any work in your question text, I assume you have at least tried to answer them yourself. If you haven't, I suggest you do this first before looking at the solutions hidden below.
For question (a):
For $12, x, 24$ to be a Shonk sequence, note that $12 \times 24 = 2^5 \times 3^2$. Thus, $x$ must be $2$ times a perfect square. The only such value between $12$ and $24$ is $18$, so $x = 18$.
For question (b):
For $28, y, z, 65$ to be a Shonk sequence, $28 \times 65 = 2^2 \times 5 \times 7 \times 13$. Thus, $yz$ must be $5 \times 7 \times 13$ times a perfect square. $y$ and/or $z$ must contain a factor of $13$, with only $39 = 3 \times 13$ and $52 = 4 \times 13$ being between $28$ and $65$. As neither value has a factor of $5$ or $7$, this means the other unknown must be a multiple of $35$, with only $35$ being in the range. Therefore, $35$ is one value, and the other one must be $13$ times a perfect square, so it must be $52$, giving that $y = 35$ and $z = 52$.
For question (c):
It's easiest to consider the longest possible Shonk sequence by examining the prime factors decomposition of $12!$ to determine the minimum factors which can't be included. For the power of $2$, it is $\lfloor \frac{12}{2} \rfloor + \lfloor \frac{12}{4} \rfloor + \lfloor \frac{12}{8} \rfloor = 6 + 3 + 1 = 10$. Using similar calculations for the other prime factors gives that $12! = 2^{10} \times 3^5 \times 5^2 \times 7 \times 11$. To get a perfect square, each power must be even, so at least one factor of $3$, plus the factors of $7$ and $11$, can't be included. As no single integer from $1$ to $12$ contains $2$ or more of these factors, at least $3$ numbers must be excluded. As excluding $3, 7, 11$ satisfies the conditions, this shows that the $9$ remaining values of $1, 2, 4, 5, 6, 8, 9, 10, 12$ forms the longest possible Shank sequence.
For question (d):
For an SDSS, with $a,b,c,d$ and $a,b,c$ and $b,c,d$ being Shonk sequences means that $abcd$, $abc$ and $bcd$ are each perfect squares. As such, $\frac{abcd}{abc} = d$ and $\frac{abcd}{bcd} = a$ must also be perfect squares. In addition, then, $\frac{abc}{a} = bc$ must be a perfect square too. For $m, 1176, n, 48400$ to be a SDSS, $m$ must be a perfect square. Since $\sqrt{1176} = 34.2\ldots$, then $m$ is $i^2$ for $1 \le i \le 34$ resulting in $34$ such values. Since $1176 = 2^3 \times 3 \times 7^2$, for $1176 \times n$ to be a perfect square requires that $n$ be $6$ times a perfect square $i^2$ with $i \gt 2 \times 7 = 14$. As $\sqrt{\frac{48400}{6}} = 89.8\ldots$, then $n$ is $6 \times i^2$ for $15 \le i \le 89$ resulting in $75$ such values. In total, there are $34 \times 75 = 2550$ pairs of $(m,n)$ values which work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$ Proof that $\int_{\pi/6}^{\pi/2} \frac{x}{\sin{x}} \le \frac{\pi^2}{6}$
After few calculations I get that if I take $\frac{3}{2}x$ then after integral I get $\frac{3}{4}x^2+ C$ and
$$ \int_{\pi/6}^{\pi/2} \frac{3}{4}x^2+ C = \pi^2 / 6$$ so I should show that
$$\frac{x}{\sin{x}} \le \frac{3x}{2} $$
but last inequality is not true...
| Call the given integral as I, then $$I=\int_{\pi/6}^{\pi/2} \frac{x}{\sin x} dx, \mbox{use}~ x=2y~ ~ \mbox{ and get} ~ I= \int_{\pi/12}^{\pi/4} ~\frac{2y}{\sin 2y} 2 dy= 2 \int_{\pi/12}^{\pi/4} \frac{y}{\sin y \cos y} dy.$$Next, let $y=\tan^{-1} z$. then $$I=2 \int_{2-\sqrt{3}}^{1} \frac{\tan^{-1}z}{z} dz < 2(\sqrt{3}-1) <\frac{\pi^2}{6}~~~ (\tan^{-1}z <z, z>0, ~~~\mbox{used here}).$$
Note that $2(\sqrt{3}-1)$ is a better bound to this integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How many Taylor series terms are needed to accurately approximate $\sqrt{a+x}-\sqrt{a}$? Naive evaluation of $\sqrt{a + x} - \sqrt{a}$ when $|a| >> |x|$ suffers from catastrophic cancellation and loss of significance.
WolframAlpha gives the Taylor series for $\sqrt{a+x}-\sqrt{a}$ as:
$$\frac{x}{2 \sqrt{a}} - \frac{x^2}{8 a^{3/2}} + \frac{x^3}{16 a^{5/2}} - \frac{5 x^4}{128 a^{7/2}} + \frac{7 x^5}{256 a^{9/2}} + O(x^6)$$
which (I think) equals:
$$\sqrt{a} \left( \frac{1}{2} \left(\frac{x}{a}\right) - \frac{1}{8} \left(\frac{x}{a}\right)^2 + \frac{1}{16} \left(\frac{x}{a}\right)^3 - \frac{5}{128} \left(\frac{x}{a}\right)^4 + \frac{7}{256} \left(\frac{x}{a}\right)^5 + O\left(\left(\frac{x}{a}\right)^6\right) \right)$$
How quickly do the coeffients decrease?
How many terms are needed to reach $53$ bits of accuracy (IEEE double precision) in the result given that $10^{-300} < \left|\frac{x}{a}\right| < 1$ is known?
Alternatively, what are the threshold values of $\left|\frac{x}{a}\right|$ where the number of terms changes?
What about rounding errors, assuming each value is stored in double precision?
| To avoid cancellation error the first thing to do is to write:
$$
\sqrt{a+x}-\sqrt{a}=\frac{x}{\sqrt{a+x}+\sqrt{a}}=\sqrt{a}\frac{x}{a}
\frac{1}{1+\sqrt{1+\frac{x}{a}}}
$$
then with $y=\frac{x}{a}$ you must approximate this
$$
\sqrt{a}\frac{y}{1+\sqrt{1+y}}
$$
fonction for $y\in[10^{-300},1]$. This function has nothing pathological and IMHO can be computed in a straightforward way.
If you really want to use Taylor series for $y\sim 0$
$$
\sqrt{a}\frac{y}{1+\sqrt{1+y}}=\sqrt{a}(\frac{y}{2}-\frac{y^2}{8}+\frac{y^3}{16}-\frac{5 y^4}{128}+\frac{7
y^5}{256}+O\left(y^6\right))
$$
I assume that the series is alternating, hence the error term $e$ is majored by $|e|<\sqrt{a}\frac{7y^5}{256}$. For instance if you want $|e|<10^{-q}$ you can use the Taylors series for $0\le y \le y_*$ where $y_*$ is such that
$$\sqrt{a}\frac{7y_*^5}{256}<10^{-q}$$ which gives
$$y_*<10^{-q/5}(\frac{256}{7\sqrt{a}})^{1/5}$$
Example: with $q=5$, $a=3$
We get $y_*<0.184042$.
That means that you can use the Taylors series for $ y_*=\frac{x_*}{a}<0.184042$, hence $x_*<3\times 0.184042 \approx 0.552125$.
Let's try with $x=0.55$.
With the initial formula we find:
$$
\sqrt{a+x}-\sqrt{a}\approx 0.152094
$$
With the Taylor series, with $y=\frac{0.55}{3}$ we get
$$
\sqrt{a}(\frac{y}{2}-\frac{y^2}{8}+\frac{y^3}{16}-\frac{5 y^4}{128}+\frac{7
y^5}{256})\approx 0.152095
$$
We see that the error $|e|=|0.152094-0.152095|\approx 1.17957\times 10^{-6}$ is less that $10^{-q}=10^{-5}$ as expected
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $a$ such that $-3\sin^{2}(x)-4\sin(x)+3-a=0$ has solution I have the function $f:[0,2\pi]\rightarrow \mathbb{R},f(x)=3\cos^{2}(x)-4\sin(x)$
I need to find the values of $a$, a real parameter such that $f(x)=a$ has solution.
My try:I got $-3\sin^{2}(x)-4\sin(x)+3=a$ so $-3\sin^{2}(x)-4\sin(x)+3-a=0$
I noted $\sin(x)=t$ and I got a quadratic equation and I put the condition that the discriminant to be $\geq0$ and I got $a\leq\frac{13}{3}$ and the right answer is $[-4,\frac{13}{3}]$
Which condition I forgot?
| Consider the function
$$
f(x)=3\cos^2x-4\sin x=-3\sin^2x-4\sin x+3
$$
Then $f'(x)=-6\sin x\cos x-4\cos x=-2\cos x(3\sin x+2)$. The derivative vanishes for $x=\pi/2$, $x=3\pi/2$, $x=\arcsin(-2/3)$ and $x=\pi-\arcsin(-2/3)$.
The second derivative is
$$
f''(x)=12\sin^2x+4\sin x-6
$$
and $f''(\pi/2)=10>0$, $f(3\pi/2)=2>0$, $f''(\arcsin(-2/3))=f''(\pi-\arcsin(-2/3))=-10/3<0$.
Thus the function has points of maximum at $\arcsin(-2/3)$ and $\pi-\arcsin(-2/3)$, with
$$
f(\arcsin(-2/3))=f(\pi-\arcsin(-2/3))=-3\cdot\frac{4}{9}+\frac{8}{3}+3=\frac{13}{3}
$$
and points of minimum at $\pi/2$ and $3\pi/2$, with
$$
f(\pi/2)=-4,\qquad f(3\pi/2)=4
$$
Thus the range of the function is $[-4,13/3]$ and the equation $f(x)=a$ only has solutions for $a\in[-4,13/3]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of positive definite sequence with terms $a_{n} = \frac{a_{n - 1} + 1}{2}$ or $a_{n} = sin(a_{n - 1})$ In the sequence all terms are positive.
Each term can be either $a_{n} = \frac{a_{n - 1} + 1}{2}$ or $a_{n} = \sin{a_{n - 1}}$.
Can this sequence have a limit in the interval $(0, 1)$?
| No, it's not possible. Let us use an indirect prrof and assume that we have such a sequence, convergent to $g\in(0,1)$. Then by the definition of convergence $$ \forall \epsilon >0 \,\exists N \,\forall n>N :|a_n-g|<\epsilon$$
Let us take $\epsilon < \min\{\frac{1-g}{4},\frac{g-\sin g}{2}\}$. For $n$ big enough we have $|a_n-g|<\epsilon$ and $|a_{n+1}-g|<\epsilon$. We have two cases.
If $a_{n+1} = \frac{a_n+1}{2}$, then
$$ |a_{n+1}-g| = |\frac{a_n+1}{2} - g| = |\frac{a_n-g}{2} + \frac{1-g}{2}| \ge \big| |\frac{1-g}{2}| - |\frac{a_n-g}{2}|\big| \rightarrow^{n\rightarrow\infty} \frac{1-g}{2}$$
so for $n$ big enough $|a_{n+1}-g| > \frac{1-g}{4} \ge \epsilon$.
If $a_{n+1} = \sin a_n$, then
$$ |a_{n+1}-g| = |\sin a_n - g| \ge \big| |g-\sin g| - |\sin a_n - \sin g|\big| \rightarrow^{n\rightarrow\infty} g-\sin g $$
so for $n$ big enough $|a_{n+1}-g| > \frac{g-\sin g}{2} \ge \epsilon$.
We get a contradiction with the assumption that $|a_{n+1}-g|<\epsilon$.
In conclusion it's impossible to get a convergent series; if one element of the sequence gets close to the would-be limit, then the next one cannot be arbitratily close.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the number $3^{3^n} + 1$ has at least $2n + 1$ prime factors. For any natural $n,$ prove that $3^{3^n} + 1$ has at least $2n + 1$ prime factors.
My idea was to use induction:
*
*for $n = 1$:
$$f(1) = 3^3 + 1 = 28 = 7*2^2$$
*let it be true for $n = k$, then for $n = k + 1$:
$$f(k + 1) = 3^{3^{k + 1}} + 1 = 3^{3*3^k} + 1 = (3^{3^k} + 1)(3^{2*3^k} - 3^{3^k} + 1) = f(k)\times(3^{2*3^k} - 3^{3^k} + 1)$$
Now I have a problem: how to prove that $(3^{2*3^k} - 3^{3^k} + 1)$ is not a prime number?
Or, if it is harder than solving the original problem, please give a hint where I turned the wrong way.
| We have:
$$3^{2\cdot3^k} - 3^{3^k} + 1=(3^{3^k}+3^{\frac{3^k+1}{2}}+1)(3^{3^k}-3^{\frac{3^k+1}{2}}+1)$$
And for $k>0$ both factors are greater than one. This factorization can be deduced from the fact that $f(2)=387400807=19441\cdot19927$ and that both factors are close to each other.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
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Proof of the proposition which says that the column rank of $A$ is equal to the row rank of $A$. (Gilbert Strang's new lecture) I am watching this new lecture by Gilbert Strang.
I have the following question.
Let $A = \begin{bmatrix}2&1&3\\3&1&4\\5&7&12\\\end{bmatrix}$.
Prof. Strang showed that the column rank of $A$ is equal to the row rank of $A$.
His proof is like the following:
Because $$\begin{bmatrix}3\\4\\12\end{bmatrix} = 1 \begin{bmatrix}2\\3\\5\end{bmatrix}+ 1 \begin{bmatrix}1\\1\\7\end{bmatrix},$$
the column rank of $A$ is $2$ and
$$\begin{bmatrix}2&1&3\\3&1&4\\5&7&12\\\end{bmatrix}=\begin{bmatrix}2&1\\3&1\\5&7\end{bmatrix}\begin{bmatrix}1&0&1\\0&1&1\\\end{bmatrix}.$$
Then he says that $\begin{bmatrix}1\\0\\1\end{bmatrix}$, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ are the basis of the row space of $A$ because $\begin{bmatrix}1\\0\\1\end{bmatrix}$, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ are linearly independent and each of the vectors $\begin{bmatrix}2\\1\\3\end{bmatrix}$, $\begin{bmatrix}3\\1\\4\end{bmatrix}$, $\begin{bmatrix}5\\7\\12\end{bmatrix}$ is a linear combination of $\begin{bmatrix}1\\0\\1\end{bmatrix}$, $\begin{bmatrix}0\\1\\1\end{bmatrix}$.
Don't we need to show that $\begin{bmatrix}1\\0\\1\end{bmatrix}$, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ are in the row space of $A$?
For example, $$\begin{bmatrix}2&1&3\\3&1&4\\5&7&12\\\end{bmatrix}=\begin{bmatrix}2&1&3\\3&1&4\\5&7&12\\\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}.$$
$\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}0\\0\\1\end{bmatrix}$ are linearly independent and each of the vectors $\begin{bmatrix}2\\1\\3\end{bmatrix}$, $\begin{bmatrix}3\\1\\4\end{bmatrix}$, $\begin{bmatrix}5\\7\\12\end{bmatrix}$ is a linear combination of $\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}0\\0\\1\end{bmatrix}$. But $\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}0\\0\\1\end{bmatrix}$ are not the basis of the row space of $A$.
$\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}0\\0\\1\end{bmatrix}$ generate $\mathbb{R}^3$ which is bigger than the row space of $A$.
| We have $A=C R$ as in the lecture. the row space of $A$ is the column space of $A^T$ and $A^T=R^T C^T$ so the colums space of $A^T$ is generated by the columns of $C^T$ (using coefficients from $R^T$) and those columns are just the rows of $C$ again.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can't solve $\int_{0}^{\pi} \frac{x}{1 + \cos^2x} dx$ I tried this :-
Let $$I =\int_{0}^{\pi}\frac{x}{1 + \cos^2x}dx\tag{1}$$
then $$I = \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2(\pi-x)}dx= \int_{0}^{\pi}\frac{\pi-x}{1 + \cos^2x}dx\tag{2}$$
Adding (1) and (2), we get
$$
2I = \int_{0}^\pi\frac{\pi}{1 + \cos^2x}dx\\
= \pi\int_{0}^{\pi}\frac{1}{1 + \frac{1}{\sec^2x}}dx\\
= \pi\int_{0}^{\pi}\frac{\sec^2x}{\sec^2x + 1}dx\\
= \pi\int_{0}^{\pi}\frac{\sec^2x}{2 + \tan^2x}dx
$$
Let $\tan x = u$, then $du = \sec^2x dx$
Then,
$$\int \frac{\sec^2x}{2+\tan^2x}dx = \int \frac{du}{2 + u^2} = \frac{1}{\sqrt{2}}\tan^{-1}\frac{u}{\sqrt{2}}+c = \frac{1}{\sqrt{2}}\tan^{-1}\frac{\tan x}{\sqrt{2}}+c
$$
Therefore,
$$
2I = \frac{\pi}{\sqrt{2}}\left[\tan^{-1}\frac{\tan x}{\sqrt{2}}\right]_0^\pi\\
\Rightarrow I = \frac{\pi}{2\sqrt{2}}\left[\tan^{-1}\frac{\tan x}{\sqrt{2}}\right]_0^\pi\\= \frac{π}{2\sqrt{2}}\left[\tan^{-1}\frac{\tan \pi}{\sqrt{2}} - \tan^{-1}\frac{\tan 0}{\sqrt{2}}\right]\\=\frac{\pi}{2\sqrt{2}}[\tan^{-1}0 - \tan^{-1}0] \\= 0\\$$
But the answer given in the book is $\frac{\pi^2}{2\sqrt{2}}$
What am I doing wrong ?
| As $\tan x$ is discontinuous at $\dfrac\pi2,$ let's fold the integral in the first quadrant
$$I=\int_0^{2a}f(x)\ dx=\int_0^af(x)\ dx+\int_a^{2a}f(x)\ dx$$
Now set $2a-x=y$ in $\displaystyle J=\int_a^{2a}f(x)\ dx$
to find $\displaystyle J=\int_a^0f(2a-y)\ (-dy)=\int_0^af(2a-y)\ dy=\int_0^af(2a-x)\ dx$
$$\displaystyle I=\int_0^{2a}f(x)\ dx=\begin{cases} 2\displaystyle\int_0^af(x)\ dx &\mbox{if } f(2a-x)=f(x) \\
0& \mbox{if } f(2a-x)=-f(x)\end{cases} $$
Here $2a=\pi,f(x)=\dfrac1{1+\cos^2x}$
| {
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How do I show that $w = \frac{1}{(z-a)^n}$ is an analytic function using the Cauchy Riemann equations,where z = x + iy? I know I need to use the Cauchy Riemann equations, to test for the analyticity, but how do I break the function $w = \frac{1}{(z-a)^n}$ into real and imaginary parts?
| With
$z = x + iy, \tag 1$
$a = \sigma + i\omega, \tag 2$
$z - a = (x - \sigma) + i(y - \omega); \tag 3$
$\dfrac{1}{z - a} = \dfrac{1}{(x - \sigma) + i(y - \omega)} = \dfrac{(x - \sigma) - i(y - \omega)}{(x - \sigma)^2 + (y - \omega)^2}$
$= \dfrac{x - \sigma}{(x - \sigma)^2 + (y - \omega)^2} - i\dfrac{y - \omega}{(x - \sigma)^2 + (y - \omega)^2} = u(x, y) + iv(x, y); \tag 4$
it follows that
$u(x, y) = \dfrac{x - \sigma}{(x - \sigma)^2 + (y - \omega)^2} \tag 5$
and
$v(x, y) = - \dfrac{y - \omega}{(x - \sigma)^2 + (y - \omega)^2} \tag 6$
are the real and imaginary parts of $(z - a)^{-1}$, respectively. At this point it is a relatively easy matter to apply the Cauchy-Riemann equations to $u(x, y)$ and $v(x, y)$ directly, though the necessary algebra is a bit tedious:
writing
$u(x, y) = (x - \sigma)((x - \sigma)^2 + (y - \omega)^2)^{-1}, \tag 7$
$v(x, y) = - (y - \omega)((x - \sigma)^2 + (y - \omega)^2)^{-1}, \tag 8$
we have
$u_x(x, y) = ((x - \sigma)^2 + (y - \omega)^2)^{-1} -2(x - \sigma)^2((x - \sigma)^2 + (y - \omega)^2)^{-2}$
$ = \dfrac{1}{(x - \sigma)^2 + (y - \omega)^2} - \dfrac{2(x - \sigma)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}$
$= \dfrac{(x - \sigma)^2 + (y - \omega)^2}{((x - \sigma)^2 + (y - \omega)^2)^2} - \dfrac{2(x - \sigma)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}$
$= \dfrac{(y - \omega)^2 - (x - \sigma)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}; \tag 9$
we also have
$v_y(x, y) = -((x - \sigma)^2 + (y - \omega)^2)^{-1} + 2(y - \omega)^2((x - \sigma)^2 + (y - \omega)^2)^{-2}$
$ = -\dfrac{1}{(x - \sigma)^2 + (y - \omega)^2} + \dfrac{2(y - \omega)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}$
$= -\dfrac{(x - \sigma)^2 + (y - \omega)^2}{((x - \sigma)^2 + (y - \omega)^2)^2} + \dfrac{2(y - \omega)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}$
$= \dfrac{(y - \omega)^2 - (x - \sigma)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}; \tag{10}$
comparing (9) and (10) we see that
$u_x(x, y) = v_y(x, y); \tag{11}$
a similar calculatinon validates
$u_y(x, y) = -v_x(x, y), \tag{12}$
and thus, with of course the proviso that $z \ne a$,
$\dfrac{1}{z - a} = u(x, y) + iv(x, y) \tag{13}$
satisfies the Cauchy-Riemann equations, and hence is holomorphic.
The preceding calculations may be simplified somewhat if we express $(z - a)^{-1}$ in term of a polar coordinate system centered at $a$; that is, we write
$z = a + re^{i\theta}, \tag{14}$
whence
$\dfrac{1}{z - a} = (z - a)^{-1}$
$= (re^{i\theta})^{-1} = r^{-1}e^{-i\theta} = \dfrac{\cos \theta}{r} - i\dfrac{\sin \theta}{r} = u(r, \theta) + iv(r, \theta); \tag{15}$
the Cauchy-Riemann equations in such a polar coordinate system are given by
$u_r = \dfrac{1}{r}v_\theta, \tag{16}$
$v_r = -\dfrac{1}{r}u_\theta; \tag{17}$
now a relatively simple calculation shows that, with $u(r, \theta)$ and $v(r, \theta)$ as in (15),
$\left ( \dfrac{\cos \theta}{r} \right)_r = -\dfrac{\cos \theta}{r^2} = \dfrac{1}{r} \left ( -\dfrac{\sin \theta}{r} \right)_\theta;$
$\left ( -\dfrac{\sin \theta}{r} \right)_r = \dfrac{\sin \theta}{r^2} = -\dfrac{1}{r} \left ( \dfrac{\cos \theta}{r} \right)_\theta, \tag{18}$
again establishing that $(z - a)^{-1}$ is holomorphic away from $z = a$, though with much less effort.
Once we have that $(z - 1)^{-1}$ is a holomorphic function, we may simply invoke the well-known fact that the product of holomorphic functions is again holomorphic to affirm that
$\dfrac{1}{(z - a)^2} = \dfrac{1}{z - a} \dfrac{1}{z - a} \tag{19}$
is holomorphic, and then use a simple inductive argument, assuming that $(z - a)^{-k}$ is holomorphic for $k \in \Bbb N$, to see that
$\dfrac{1}{(z - a)^{k + 1}} = \dfrac{1}{(z - a)^k}\dfrac{1}{z - a} \tag{20}$
is also holomrphic; we conclude that $(z - a)^{-n}$ is holomorphic for all $n \in \Bbb N$. $OE\Delta$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Long division of $\frac{3x^3-x^2-13x-13}{x^2-x-6}$ I'm self-studying from Stroud & Booth's amazing textbook "Engineering Mathematics", and am on the "Partial Fractions" chapter. As part of an exercise I need to do long division of two polynomial equations.
The problem is, long division of polynomials was never explained in the textbook. The long division I need to do is this one:
$$\frac{3x^3-x^2-13x-13}{x^2-x-6}$$
The book states that the result is:
$$3x+2+\frac{7x-1}{x^2-x-6}$$
But the mechanics of the long division are never explained. Can I please get some help with this one?
| First, you see which monomial you must multiply $x^2-x-6$ by in order to get something close to $3x^2-x^2-13x-13$. Clearly, that would be $3x$: $3x(x^2-x-6)=3x^3-3x^2-18x$. Now, you subtract this from $3x^2-x^2-13x-13$:$$3x^2-x^2-13x-13-(3x^3-3x^2-18x)=2x^2+5x-13.$$Now, you start all over again. What is the monomial you must multiply $x^2-x-6$ by in order to get something close to $2x^2+5x-13$? Clearly, it is $2$: $2(x^2-x-6)=2x^2-2x-12$. Now, you subtract this from $2x^2+5x-13$:$$2x^2+5x-13-(2x^2-2x-12)=7x-1.$$Since you now have a polynomial whose degree is smaller than the degree of the polynomial that you are dividing by, you're done:
*
*the quotient is the sum of the monomials that you got: $3x+2$;
*that final polynomial ($7x-1$) is the remainder.
| {
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Evaluate the Limit Without L'Hopital Rule Let $\lim_{x\to a}\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}=L$, find the value of $\lim_{x\to a}\frac{x(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}$ for $a\neq0$.
Using L'Hopital rule I found that the answer is $a^4-aL$. My question is how to solve this problem without using L'Hopital rule.
Here's my attempt using L'Hopital.
$$
\begin{aligned}
\lim_{x\to a}\frac{a^4-(x^2-x\left | x \right |-a^2)^2}{x-a}&=L\\
\lim_{x\to a}\frac{d}{dx}(x^2-x\left | x \right |-a^2)^2&=-L...(1)\\
\end{aligned}
$$
Let the numerator equal to zero.
$$
(x^2-x\left | x \right |-a^2)^2=a^4...(2)
$$
Ergo
$$\begin{aligned}
\lim_{x\to a}\frac{(x^2-x\left | x \right |-a^2)^2-a^4\left | a \right |}{x-a}&=\lim_{x\to a}(x^2-x\left | x \right |-a^2)^2+x\cdot\frac{d}{dx}(x^2-x\left | x \right |-a^2)^2\\
&=a^4-aL
\end{aligned}
$$
| Hint: The numerator can be written as $$-x \left( \left| x \right| -x \right) \left( x \left| x \right| +2\,
{a}^{2}-{x}^{2} \right)
$$ and now distinguish the cases $$x\geq 0$$ or $$x<0$$
| {
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Prove that hat if $a,b \ge 2$ then $ab \ge a+b$ Prove that if $a,b \ge 2$ then $a+b \le ab$
so if $a \ge 2$ and $b \ge 2$ then $a-1 \ge 1$ and $b-1 \ge 1$
$(a-1)(b-1) \ge b-1$
$(a-1)(b-1) \ge 1$
$(a-1)(b-1) - 1 \ge 0$
$ab -a -b\ge 0$
$ab \ge a + b$
Thanks in advance, is this valid?
| $$ ab\ge 2a$$
$$ab\ge 2b$$
Add both sides
$$2ab \ge 2a+2b$$
$$ ab\ge a+b $$
| {
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"question_score": "2",
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If $x^2+kx+1$ is a factor of $px^5+qx^2+r$ prove that $(p^2-r^2)(p^2-r^2+qr)=q^2p^2$ if $x^2+kx+1$ is a factor of $px^5+qx^2+r$ prove that $(p^2-r^2)(p^2-r^2+qr)=q^2p^2$
My try
Since this is a factor I tired finding A, B, C & D such that,
$(x^2+kx+1)(Ax^3+Bx^2+cx+D)$ =$px^5+qx^2+r$
if I can find k in terms of p, q & r I can try to get the given condition, But when I tried this equating coefficients, I ended up with quadratic expression for k? Is there a better way to approach this question? Please Help! thanks
| Let $px^5+qx^2+r=(x^2+kx+1)(px^3+ax^2+bx+r)$
Let's compare the coefficients of $x$
$0=kr+b\iff b=-kr$
That of $x^4$
$0=a+kp\iff a=-kp$
That of $x^3$
$0=b+ka+p=-kr+k(-kp)+p$
which is on rearrangement, a quadratic equation in $k$
Compare the values of $a$ to find $k$
Coefficients of $x^2$
$q=r+kb+a=r+(-kr)(k-1)$
Replace the values of $a,b$ to form another quadratic equation in $k$
Solve the two quadratic equations for $k,k^2$
Use $k^2=(k)^2$ to eliminate $k$
| {
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Solve the equation: $x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$ Here is the equation:
$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$$
The equation looks very simple. But, it's a little misleading (for me).
$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}\Longrightarrow x(x^2-b^2)-a^2x(x^2-b^2)+b^2+a(x^2-b^2)-x^2=0\Longrightarrow (x^2-b^2)(x-a^2x+a-1)=0$$
We have $x(1-a^2)=1-a$ and $x≠±b$.
1) $a=1$, $x\in (-\infty;b) ∪(-b;b)∪(b;\infty)$
2) $a=-1$ then $x\in\emptyset$
3) $a≠±1$ and $b≠±\frac{1}{a+1}$ ,then $x=\frac{1}{a+1}$
What I want to know, is there a mistake in the solution, or is there any other point I missed?
Thank you.
| Your solution is almost right, except that 2) is not a solution because that yields $0=2$ in your equation.
Here’s a simpler method.
Adding $\frac{b^2}{b^2-x^2}$ to both sides, note that the RHS now becomes $1$. So we now need to solve $x(1-a^2)=1-a$, which factorises to $(1-a)(x(a+1)-1)=0$ (where $x \neq \pm b$).
| {
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$f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$ $f(x) = \frac{4 + x}{2 + x - x^2}$, calculate $f^{(9)}(1)$, where $f^{(9)}$ is the $9$-th derivative of $f$.
Domain of $f$ is $\mathbb{R} - \{-1, 2\}$. I've got that $f(x) = \frac{1}{1 - (-x)} + \frac{1}{1 - \frac{1}{2}x} = \sum_{n=0}^\infty ((-1)^n + 2^{-n})x^n$, but there is a problem that $\frac{1}{1 - (-x)} = \sum_{n=0}^\infty (-1)^nx^n$ is convergent only for $|x| < 1$, so not for $1$. How can I go about this?
| I've solved this using Julian Mejia's suggestion by making substitution $t = x -1$. We have $g(t) = \frac{1}{2} \cdot \left(\frac{4}{1 - t} + \frac{1}{1 + \frac{1}{2}t}\right) = \sum_{n=0}^\infty (2-(-\frac{1}{2})^{n+1})t^n$ which converges for $t \in (-1, 1)$. Now $9$-th derivative is just $(2-(-\frac{1}{2})^{10}) \cdot 9!$, which is exactly what InterstellarProbe got.
| {
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"timestamp": "2023-03-29T00:00:00",
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Undetermined coefficients in a perturbative expansion In order to familiarize myself with perturbation methods, I've been trying to derive the Lorentz transformations, given by
\begin{align*}
x \rightarrow \frac{x + vt}{\sqrt{1 - v^2}} & = (x + vt)(1 + \frac{1}{2}v^2 + \frac{3}{8}v^4 + ...) \\
& = (x + vt + \frac{1}{2}v^2x + \frac{1}{2}v^3t + \frac{3}{8}v^4x + \frac{3}{8}v^5t + ...)
\end{align*}
\begin{align*}
t \rightarrow \frac{t + vx}{\sqrt{1 - v^2}} & = (t + vx)(1 + \frac{1}{2}v^2 + \frac{3}{8}v^4 + ...) \\
& = (t + vx + \frac{1}{2}v^2t + \frac{1}{2}v^3x + \frac{3}{8}v^4t + \frac{3}{8}v^5x + ...)
\end{align*}
using a perturbative approach which entails starting with the Gallilean transformations (i.e. $x \rightarrow x + vt$, $t \rightarrow t$) and in each step adding variations $\delta x$ and $\delta t$, both of order $n$ in $v$ and linear in $x$ and $t$, such that the spacetime interval $t^2 - x^2$ is invariant with respect to the transformations up to order $n+1$ in $v$. However, I keep running into the problem that, in each step of the process, some of the coefficients in my expansions for the transformations are undetermined. For example, for $\delta x$ and $\delta t$ of order $2$ in $v$,
$$x \rightarrow x + vt + \delta x$$
$$t \rightarrow t + vx + \delta t$$
$$\delta x = v^2(a_1x + b_1t) + \mathcal{O}(v^3)$$
$$\delta t = v^2(a_2x + b_2t) + \mathcal{O}(v^3)$$
\begin{align*}
t^2 - x^2 & = (t + vx + \delta t)^2 - (x + vt + \delta x)^2 \\
& = (t + vx)^2 + 2(t + vx)\delta t - (x + vt)^2 - 2(x + vt)\delta x + \mathcal{O}(v^4) \\
& = t^2 + 2vxt + v^2x^2 + 2v^2(t + vx)(a_2x + b_2t) \\
& \qquad - x^2 - 2vxt - v^2t^2 - 2v^2(x + vt)(a_1x + b_1t) + \mathcal{O}(v^4) \\
& = t^2 + v^2x^2 + 2v^2a_2xt + 2v^2b_2t^2 \\
& \qquad - x^2 - v^2t^2 - 2v^2a_1x^2 - 2v^2b_1xt + \mathcal{O}(v^3) \\
& = t^2 - x^2 + v^2[(2b_2-1)t^2 + 2(a_2-b_1)xt + (1-2a_1)x^2] + \mathcal{O}(v^3)
\end{align*}
I end up with $a_1 = b_2 = \frac{1}{2}$ and $a_2 = b_1 \equiv c_1$, which is so far an undetermined coefficient.
I know from the Taylor expansions of the full transformations above that $c_1$ should be equal to $0$. Is there a way to mathematically determine this coefficient?
| What you're trying to do here cannot possibly work. The parameter $v$ in the Lorentz transformation is an arbitrary parameter that you can redefine at will. Your approach has just as much reason to yield the coefficients in the expansion in $2v$ or $\sin v$ as it does to yield the coefficients in the expansion in $v$, so it's not surprising that it has a lot of degrees of freedom.
| {
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"timestamp": "2023-03-29T00:00:00",
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Combinatorial proof of $\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$ and then use result to find a formula for $1^2 + 2^2 + \ldots + n^2$ a) Give a combinatorial proof that for every $n \geq r \geq 1$ that:
$$\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$$
And use (a) to concoct a formula for $1^2 + 2^2 + ... + n^2$
| Maybe not combinatorial, but maybe it helps somebody else for a proper idea.
\begin{align}
\binom{n+1}{r+1} &= \frac{n+1}{r+1}\binom{n}{r} \\
&=\binom{n}{r} + \frac{n-r}{r+1} \binom{n}{r} =\binom{n}{r} + \frac{n}{r+1} \binom{n-1}{r} =\binom{n}{r} + \binom{n}{r+1} \\
&=\binom{n}{r} + \binom{n-1}{r} + \binom{n-1}{r+1} \\
&= \dots \\
&=\sum_{i=0}^k \binom{n-i}{r} + \binom{n-k}{r+1} \, .
\end{align}
The last term vanishes for $k=n-r$. In that case use this formula for $r=2$ which immediately yields $1^2 + 2^2 + \dots + n^2$ in terms of simpler sums.
| {
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"timestamp": "2023-03-29T00:00:00",
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Angle formed by orthocenter, incenter and circumcenter of a triangle $>135^\circ$? If $H$ is the orthocenter, $I$ the incenter and $O$ the circumcenter of a triangle , the I need to show that the angle $HIO>135^\circ$
With the assumptions of $OI^2=R^2-2Rr$, $OH^2=9R^2-(a^2+b^2+c^2)$, $HI^2=2r^2-4R^2\cos A\cos B \cos C$ and $R^2\cdot8(1+\cos A\cos B\cos C)= a^2+b^2+c^2$
Applying the $cosine$ rule I got to something like $$\cos (HIO)=\frac{2r^2-2Rr+R^2\cdot 4\cos A\cos B\cos C}{2(2r^2-4R^2\cos A\cos B\cos C)(R^2-2Rr)}$$
I need to show that the LHS is between $\left[-\frac{1}{\sqrt{2}},-1\right]$
How t0 proceed?
Is there any simpler way to prove the question (without applying cosine rule) ?
| Too long for a comment.
Put $X=4R^2\cos A\cos B \cos C$.
We have to show that
$$-\frac 1{\sqrt{2}}>\cos (HIO)=\frac{HI^2+OI^2-OH^2}{2\cdot HI\cdot OI}.$$
Since it is assmed that $2\cdot HI\cdot OI>0$, we have to show that
$$OH^2- HI^2-OI^2>\sqrt{2}\cdot HI\cdot OI$$
$$2Rr-2r^2-X>\sqrt{2(2r^2-X)(R^2-2Rr)}$$
Since $HI^2>0$ and $OI^2>0$, $2r^2>X$ and $R>2r$, so $2Rr-2r^2-X>0$. Thus we have to show that
$$(2Rr-2r^2-X)^2>2(2r^2-X)(R^2-2Rr)$$
$$X^2+2XR^2+4Xr^2+4r^4>8RrX$$
The respective quadratic equation for $X$ has a discriminant $D=(R-2r)^2R(R-4r)$ and roots $X_1=4Rr-2r^2-R^2-\sqrt{D}$ and $X_2=4Rr-2r^2-R^2+\sqrt{D}$. So if $R<4r$ then $D<0$ and the inequality is proved.
Otherwise I tried to evaluate $X$ in terms of $R$ and $r$. According to exercises at p.23 of “Topics in Inequalities - Theorems and Techniques” by Hojoo Lee (version February 25, 2006), we have $X=s^2-(2R+r)^2$. Moreover, in W. J. Blundon (see Problem E1935, Amer. Math. Monthly 73 (1966), 1122) found the best possible inequalities of the form $A(R,r)\le s^2\le B(R, r)$, where $A(x, y)$ and $B(x, y)$ are real quadratic forms $\alpha x^2+\beta xy+\gamma y^2$, namely $$16Rr-5r^2\le s^2\le 4R^2+4Rr+3r^2.$$
This implies $$12Rr-4R^2-6r^2\le X\le 2r^2.$$
Unfortunately, this doesn’t help because
$$12Rr-4R^2-6r^2\le X_1\le X_2\le 2r^2.$$
| {
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Find the ratio of twelfth terms if the sum of first n terms of two AP's are $3n+8,7n+15$
The sum of first n terms of two AP's are $3n+8,7n+15$. Then the ratio of their twelfth term is:
$$
\frac{a_{12}}{A_{12}}=\frac{a+11d}{A+11D}=\frac{2a+22d}{2A+22D}=\frac{s_{23}}{S_{23}}=\frac{3(23)+8}{7(23)+15}=\frac{77}{176}=\frac{7}{16}
$$
Fine but if I do the following
$$
a_{12}=s_{12}-s_{11}=44-41
=3\\
A_{12}=S_{12}-S_{11}=99-92=7\\
\frac{a_{12}}{A_{12}}=\frac{3}{7}
$$
Why am I getting diferent results for method 1 and 2 ?. Is the question wrong because the sum of terms of AP is a quadratic expression in $n$ ?
Note: This was asked as a multiple choice question with options including both the results of method 1 and 2.
| The question should have been
The ratio of the sum of first $n$ terms of two AP's are $3n+8:7n+15$
instead of
The sum of first $n$ terms of two AP's are $3n+8,7n+15$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For every integer $n$, the quantity $n^2 + 2n \equiv 0\pmod 4$ or $n^2 + 2n \equiv 3\pmod 4$ I'm trying to prove this question using induction
So far I have
Base Case
Let $n = 1$, $(1^2 + 2)\equiv 3 \pmod 4 $
Claim holds for base case
Induction
Assume $n = k$ holds, that is $k^2 + 2k \equiv 0\pmod 4$ or $k^2 + 2k \equiv 3\pmod 4$
Let $n = k+1$ such that
$${(k+1)^2 + 2(k+1)}\equiv {k^2 + 2k + 1 + 2k + 2}\pmod 4$$
Then I substitute $k^2 + 2k$ with both $0$ or $3$ from the earlier assumption.
So we have
$$(0 + 2k + 3) \equiv (2k + 3) \pmod 4$$
Or,
$$(3 + 2k + 3) \equiv (2k + 6) \pmod 4$$
Where do I go from here?
| Separate cases for $n$ even and $n$ odd.
If $n$ is even, than it is $n=2m$.
We get: $(2m)^2+2(2m)=4m^2+4m=4(m^2+m)\equiv 0\mod 4$
If $n$ is odd, then it is $n=2m+1$.
We get: $(2m+1)^2+2(2m+1)=4m^2+4m+1+4m+2=4(m^2+2m)+3\equiv 3\mod 4$
| {
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"url": "https://math.stackexchange.com/questions/3249053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Checking a proof that the equality $ax^2+bx+c=0$ is never true for integer $a , b$ and $c$ if $ x = 2^{\frac{1}{3}} $ The solution of $ax^2+bx+c=0$ is $x=-\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}$ . Since $ x^3=2$ then after raising both sides to the third power we get $$ 2=-\frac{b^3}{8a^3}+\frac{\left(b^2-4ac\right)^{\frac{3}{2}}}{8a^3}+\frac{3b^2\sqrt{b^2-4ac}}{8a^3}-\frac{3b\left(b^2-4ac\right)}{8a^3}$$ multiplying by $ 8a^3$ and simplifying we get $$4a^3=-b^3+3abc+\left(b^2-4ac\right)^{\frac{1}{2}}\left(b^2-ac\right)$$ Now $a$ , $b$ and $c$ are integers and the left hand side is an integer and so must the right hand side be and this happens only if the term $b^2-4ac$ is a perfect square. Let $b^2-4ac = m^2$. This implies that $$x=\frac{m-b}{2a}$$ Substituting $ x =2^{\frac{1}{3}}$ and multiplying by $2$ yields $$2^{\frac{4}{3}}=\frac{m-b}{a}$$ so $2=\left(\frac{m-b}{a}\right)^{\frac{3}{4}}$ which implies that all powers of two can be written in this form. choosing $ 2^4=16$ shows that $16^{\frac{1}{3}}=\frac{h}{a}\ $ where $h=m-b$. Now all I have to do is prove that $16^{\frac{1}{3}} = 2\ \left(2^{\frac{1}{3}}\right)$ is an irrational number. Now $h=2a\left(2^{\frac{1}{3}}\right)$. the number $a$ is the product of some primes and this product must contain a $\left(2^{\frac{r}{3}}\right)$ where $ r \equiv 2 mod(3)$ but that would mean that $r$ is not divisible by $3$ which means that the term $\left(2^{\frac{r}{3}}\right)$ can't exist in the prime factorization of any integer which shows that the initial assumption ( that $a$ , $b$ and $c$ are integers) is wrong.
The same argument works for $x=-\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a}$ but the signs are different.
edit: Guys this is proof verification. please don't suggest alternative methods without mentioning why the proof is wrong.
| There is a fast way to finish. Once you have $2^\frac{4}{3} = \frac{m-b}{a}$, note that the LHS is $2 \times 2^\frac{1}{3}$. Dividing both sides by $2$ we obtain that $2^\frac{1}{3}$ is rational; this yields a contradiction in a similar method to the proof that $\sqrt 2$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $n$ is Divisible by $6$ when the Quadratics have integer roots If $m,n \in \mathbb{Z}$ and Both the Quadratics $x^2+mx-n=0$ and $x^2-mx+n=0$ has integer roots, Prove that $n$ is a Multiple of $6$
My try:
Let $n$ is not a multiple of $6$
We have
$$n=6q\pm1$$
OR
$$n=6q\pm 2$$
OR
$$n=6q\pm 3$$
Now since the Discriminants of both the quadratics should be a perfect square we have
$$m^2+4n=p^2$$
$$m^2-4n=r^2$$
So we get
$$8n=(p-r)(p+r)$$
$\implies$
$$6n+2n=(p-r)(p+r)$$
$$6n+2(6q\pm s)=(p-r)(p+r)$$
$$6a\pm2s=(p-r)(p+r)$$
where $s=1,2,3$
Any clue from here?
I already know this Question has an existing thread, but i will be happy if i can continue with my solution.
| You mention in your question that it already "has an existing thread". You should specify where this is to help avoid people duplicating efforts. However, if it's Prove that $n$ is divisible by $6$, note my solution below is somewhat different and, as you asked, I try to continue your solution.
You started with the two discriminant equations of
$$m^2+4n=p^2 \tag{1}\label{eq1}$$
$$m^2-4n=r^2 \tag{2}\label{eq2}$$
and then proceeded to take the difference of them to get
$$6a\pm2s=(p-r)(p+r) = p^2 - r^2 \tag{3}\label{eq3}$$
where $n = 6q \pm s$ and $s=1,2,3$.
One way to proceed would be to check each case to show it results in a situation which is not possible. First, start with adding $s = 1$. In case you're not familiar with it, basic modulo notation of $x \equiv y \pmod z$ means that $x - y$ is an integral multiple of $z$. For all integers $x$, $x^2 \equiv 0 \text{ or } 1 \pmod 3$. The LHS (left hand side) of \eqref{eq3} gives $6a + 2 \equiv 2 \pmod 3$. On the right side, only $p^2 \equiv 0 \pmod 3$ and $r^2 \equiv 1 \pmod 3$ matches. This causes \eqref{eq1} to become $m^2 + 1 \equiv p^2 \equiv 0 \pmod 3$, so $m^2 \equiv 2 \pmod 3$ which is not possible.
Next, with adding $s = 2$, the LHS of \eqref{eq3} gives $6a + 4 \equiv 1 \pmod 3$. On the right side, only $p^2 \equiv 1 \pmod 3$ and $r^2 \equiv 0 \pmod 3$ matches. This causes \eqref{eq2} to become $m^2 - 8 \equiv m^2 + 1 \equiv r^2 \equiv 0 \pmod 3$, so $m^2 \equiv 2 \pmod 3$ which is not possible.
You can similarly show that subtracting $s = 1, 2$ also doesn't work. Finally, consider the case where $s = 3$, which means $n = 6q + 3$ is odd. For this, I don't see any direct way to use your final equation since it's lost some important information. Instead, from \eqref{eq1}, moving $m^2$ to the other side gives
$$4n = p^2 - m^2 \tag{4}\label{eq4}$$
The LHS side is even so both $p$ and $m$ must be either odd or even. For the first case, note for all odd integers $x$ that $x^2 \equiv 1 \pmod 8$. As such, $p^2 - m^2 \equiv 0 \pmod 8$. However, $4n \equiv 0 \pmod 8$ if $n$ is even and $4n \equiv 4 \pmod 8$ if $n$ is odd, so the values only match when $n$ is even. For the case where $p$ and $m$ are both even, then consider \eqref{eq1} plus \eqref{eq2}:
$$2m^2 = p^2 + r^2 \tag{5}\label{eq5}$$
Since $m$ and $p$ are even, so is $r$. Thus, if $p = 2p_1$, $m = 2m_1$ and $r = 2r_1$, you can divide by a factor of $4$ to get
$$2m_1^2 = p_1^2 + r_1^2 \tag{6}\label{eq6}$$
Now, both $p_1$ and $r_1$ must be both either even or odd. If they're even, then $p_1^2 + r_1^2$ has a factor of $4$, so $m_1$ must be even. In this case, $p$ and $m$ must each be a multiple of $4$ so $p^2 - m^2$ is a multiple of $16$ meaning $n$ is even in \eqref{eq4}. If both $p_1$ and $r_1$ are odd, then their squares are each congruent to $1$ modulo $4$, so $m_1$ must be odd. Thus, in \eqref{eq4}, $p^2 - m^2 = 4(p_1 + m_1)(p_1 - m_1)$ must be a multiple of $16$. As such, in either case, $n$ must be even, so $s = 3$ doesn't work.
This shows that $s$ must be $0$, i.e., $n$ is a multiple of $6$. However, a somewhat simpler & more direct way to have accomplished this is to use the method above to show that $n$ must be even and then check modulo $3$. In particular, if $n \equiv 1 \pmod 3$, \eqref{eq1} gives that $m^2 + 1 \equiv p^2 \pmod 3$. Since $m^2 \equiv 1 \pmod 3$ doesn't work as this gives that $p^2 \equiv 2 \pmod 3$, then $m^2 \equiv 0 \pmod 3$. However, \eqref{eq2} then gives that $0 - 1 \equiv 2 \equiv r^2 \pmod 3$, which can't be true. Similarly, if $n \equiv 2 \pmod 3$, then $m^2 \equiv 1$ from \eqref{eq1}, but this again requires that $r^2 \equiv 2 \pmod 3$ from \eqref{eq2}. Thus, the only choice left is that $n \equiv 0 \pmod 3$. Since $n$ must have a factor of $2$ and a factor of $3$, it's a multiple of $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Calculate $\frac{a^8 + b^8}{a^8 - b^8} + \frac{a^8 - b^8}{a^8 + b^8}$ based on $\frac{a^2 + b^2}{a^2 - b^2} + \frac{a^2 - b^2}{a^2 + b^2}$.
Given that $a$ and $b$ are distinct reals,
calculate $\dfrac{a^8 + b^8}{a^8 - b^8} + \dfrac{a^8 - b^8}{a^8 + b^8}$ based on $\dfrac{a^2 + b^2}{a^2 - b^2} + \dfrac{a^2 - b^2}{a^2 + b^2}$.
Well... this problem is adapted from a recent competition... And I can't solve it. Please help me.
| Let
$$x=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}=2\frac{a^{4}+b^{4}}{a^{4}-b^{4}}.$$
Then
$$\frac{a^{4}+b^{4}}{a^{4}-b^{4}}+\frac{a^{4}-b^{4}}{a^{4}+b^{4}}=\frac{x}{2}+\frac{2}{x}.$$
We can repeat this to get
$$\frac{a^{8}+b^{8}}{a^{8}-b^{8}}+\frac{a^{8}-b^{8}}{a^{8}+b^{8}}=\frac{\frac{x}{2}+\frac{2}{x}}{2}+\frac{2}{\frac{x}{2}+\frac{2}{x}}=\frac{x}{4}+\frac{1}{x}+\frac{4x}{x^{2}+4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the maximum value of $\frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy}$ where $x, y, z > 0$ and $x^2 + y^2 + z^2 = 3xyz$.
$x$, $y$ and $z$ are positives such that $x^2 + y^2 + z^2 = 3xyz$. Calculate the maximum value of $$\large \frac{x^2}{x^4 + yz} + \frac{y^2}{y^4 + zx} + \frac{z^2}{z^4 + xy}$$
This is (obviously) adapted from a recent competition. There ought to be better solutions that the one I have provided below. So if you could, please post them.
| Also, by AM-GM and Muirhead we obtain:
$$\sum_{cyc}\frac{x^2}{x^4+yz}=\sum_{cyc}\frac{\frac{9x^4y^2z^2}{(x^2+y^2+z^2)^2}}{x^4+\frac{9x^2y^3z^3}{(x^2+y^2+z^2)^2}}=\sum_{cyc}\frac{9x^2y^2z^2}{x^2(x^2+y^2+z^2)^2+9y^3z^3}\leq$$
$$\leq\sum_{cyc}\frac{9x^2y^2z^2}{6x(x^2+y^2+z^2)\sqrt{y^3z^3}}=\frac{3\sum\limits_{cyc}x\sqrt{yz}}{2(x^2+y^2+z^2)}\leq\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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