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Prove if $x\gt3$ then $1\ge\frac{3}{x(x-2)}$. I tried to prove it by contradiction.
Suppose it is not true that $1\ge\frac{3}{x(x-2)}$, so $1\lt\frac{3}{x(x-2)}$. Then $\frac{3}{x(x-2)}-1\gt0$. Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.
$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$
${3-(x(x-2)\gt0}$
${3-x^2-2x\gt0}$
${-x^2-2x+3\gt0}$
${-1(x^2+2x-3)\gt0}$
$-1\frac{(x-1)(x+3)}{-1}\gt0/-1$
${(x-1)(x+3)\lt0}$
At this point I really do not know what to do after this point or if I really even went about it the right way. Thank you for the help.
| From $$x-3\geq 0$$ follows:
$$x-1\geq 2$$
$$(x-1)^2\geq 4$$
$$x^2-2x+1\geq 4$$
$$x^2-2x\geq 3$$
$$x(x-2)\geq 3$$ so $$1\geq \frac{3}{x(x-2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find 'ordinary generating function': 1, 0, 2, 0, 3, 0, 4, 0, 5.... This was on my midterm yesterday, but I couldn't solve it.
Find an ordinary generating function: 1, 0, 2, 0, 3, 0, 4, 0, 5....
My answer was $(\frac{1}{1-x^2})^2$ because 1, 2, 3, 4, 5, 6, 7, ... is $\frac{1}{1-x^2}$. (now I feel like it just doesn't make sense).
Can you find the ordinary generating function?
|
We obtain
\begin{align*}
\color{blue}{1+2x^2+3x^4+\cdots}&=\sum_{n=0}^\infty (n+1)x^{2n}\\
&=\frac{1}{2x}\frac{d}{dx}\left(\sum_{n=0}^\infty \left(x^2\right)^{n+1}\right)\\
&=\frac{1}{2x}\frac{d}{dx}\left(\sum_{n=1}^\infty \left(x^2\right)^n\right)\\
&=\frac{1}{2x}\frac{d}{dx}\left(\frac{1}{1-x^2}-1\right)\\
&\,\,\color{blue}{=\frac{1}{(1-x^2)^2}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to solve the cubic $x^3-3x+1=0$? This was a multiple choice question with options being
$$(A)-\cos\frac{2\pi}{9},\cos\frac{8\pi}{9},\cos\frac{14\pi}{9} \\
(B)-2\cos\frac{2\pi}{9},2\cos\frac{8\pi}{9},2\cos\frac{14\pi}{9} \\
(C)-\cos\frac{2\pi}{11},\cos\frac{8\pi}{11},\cos\frac{14\pi}{11} \\
(D)-2\cos\frac{2\pi}{11},2\cos\frac{8\pi}{11},2\cos\frac{14\pi}{11}$$
I tried to eliminate options using the sum and product of roots but I can't figure out if $$\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{14\pi}{9}=0$$
or
$$\cos\frac{2\pi}{11}+\cos\frac{8\pi}{11}+\cos\frac{14\pi}{11}=0$$
| Hint: $(2\cos \theta)^3-3(2\cos\theta)+1=2\cos3\theta+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A magic rectangle with the greatest size - original question I filled in a $3\times k$ rectangle with non negativ integers, such that the sum of the three numbers in each column is the same number $n$, and in each row all the numbers are different.
Find the maximum value of $k$.
If you try for $n=0,1,2,3,4,5,6$, you get $1,1,2,3,3,4,4$. It is a strange sequence, but
I got an extra homework. So I need help! I am very thankful for every solution!
| Well, here's a partial answer for you: the function $k(n)$ is obviously non-decreasing, the upper bound is $k={2\over3}n+1$, and this bound is sharp (that is, exact) when $n$ is divisible by 3.
The bound can be established as follows. The sum of all numbers in the table is obviously $k\cdot n$. On the other hand, if we look at the rows first, each row contains $k$ different non-negative integers. Their sum can't be smaller than $0+1+2+\ldots+(k-1) = {k(k-1)\over2}$. Since we have three such rows, this implies:
$$3{k(k-1)\over2}\leqslant k\cdot n$$
or
$$k\leqslant {2\over3}n+1$$
As to why the bound is sharp, well, just extend the following pattern:
$$n=3: \\ \begin{array}{|c|c|c|}
0&1&2 \\
1&2&0 \\
2&0&1
\end{array}$$
$$n=6: \\ \begin{array}{|c|c|c|c|c|}
0&1&2&3&4 \\
2&3&4&0&1 \\
4&2&0&3&1
\end{array}$$
$$n=9: \\ \begin{array}{|c|c|c|c|c|c|c|}
0&1&2&3&4&5&6 \\
3&4&5&6&0&1&2 \\
6&4&2&0&5&3&1
\end{array}$$
$$n=12: \\ \begin{array}{|c|c|c|c|c|c|c|c|c|}
0&1&2&3&4&5&6&7&8 \\
4&5&6&7&8&0&1&2&3 \\
8&6&4&2&0&7&5&3&1
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding area of a triangle using equation of a circle **Ignore notes I made they are stupid
Without a calculator
Question reads:
The diagram shows a sketch of the circle with equation $x^2 + y^2 = 5$.
The $y$-coordinate of point $A$ is $-1$.
The tangent to the circle at $A$ crosses the axes at $B$ and $C$ as shown.
Find the area of triangle $OBC$
| Just going off the question alone and assuming the diagram is not drawn to scale, have some good information to off of.
Knowing the $y$ coordinate is $-1$, we can plug that into the equation of the circle to get the $x$ coordinate:
$x^2 + (-1)^2 = 5 \implies x=2$
This gives us a slope of the line $OA$ to be $\frac{-1}{2}$ and the tangent slope to be $2$. Therefore, we have a line that passes through $(2,-1)$ with a slope of $2$. We can get the equation using:
$ y=mx+b \implies -1 = 2 \cdot 2+b \implies b = -5$
Our $y$ intercept is -5 (height), and our $x$ intercept is then $0=2x-5 \implies x=2.5$ (width). Our area is:
$ \frac{5\cdot2.5}{2} = 6.25$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $f(x)f(y)+f(3/x) f(3/y)=2f(xy)$ choose correct choices Consider $f:\mathbb{R}^+ \to \mathbb{R}$ such that $f(3)=1$ and
$$f(x)f(y)+f(3/x) f(3/y)=2f(xy)\;\;\;\;\;\;\forall x,y \in \mathbb{R}^+$$ then choose the correct option(s):
(A) $f(2014)+f(2015)-f(2010)=100$
(B) $f$ is an even function
(C) $\frac{f(100)}{f(10)+f(90)}=\frac{1}{2}$
(D) $f$ is a periodic function
By putting $x=1$ in given equation to get $f(1)=1$ Generally I have been taught to solve such question using first principle of differentiation but here I am not able to deal with the second term of L.H.S. in given equation. May someone provide some help?
| $x$ and $y$ are always explicitly positive, and this will not be further noted.
$$
\begin{align}
2f(xy) &= f(x)f(y)+f(3/x) f(3/y) \tag{Definition} \label{def} \\
f(3) &= 1 \tag{Initial Condition}
\end{align}
$$
Following @ChristianF, we'll set variables to make the conditions more symmetric.
Considering the case where $y=3/x$, we have:
$$
\begin{align}
2f(x)f(3/x) &= 2f(3) = 2 \\
f(3/x) &= \frac{1}{f(x)} \tag{1} \label{1} \\
f(1) &= \frac{1}{f(3)} = 1 \tag{2} \label{2}
\end{align}
$$
Substituting equation \ref{1} into the original equation $\ref{def}$ (twice: once with $x$ and once with $y$) gives:
$$
2 f(x y) = f(x)f(y)+\frac{1}{f(x) f(y)} \tag{3} \label{3}
$$
Considering the case where $y=1$ in equation \ref{3} immediately gives:
$$
\begin{align}
2 f(x\cdot 1) &= f(x)f(1)+\frac{1}{f(x)f(1)} \\
2 f(x) &= f(x)+\frac{1}{f(x)} \\
2 f(x)^2 &= f(x)^2 + 1 \\
f(x)^2 &= 1 \tag{4} \label{4} \\
\end{align}
$$
Then considering the case where $x=y$ in \ref{3}, and simplifying with \ref{4}:
$$
\begin{align}
2 f(x^2) &= f(x)f(x)+\frac{1}{f(x) f(x)} \\
2 f(x^2) &= 1 + \frac{1}{1} \\
f(x^2) &= 1 \\
\end{align}
$$
Finally, as $x$ is only positive $x$ and $x^2$ cover the same region: $f(x) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove this inequality $Σ_{cyc}\sqrt{\frac{a}{b+3c}}\ge \frac{3}{2}$ with $a;b;c>0$ Let $a,b,c>0$. Prove $$\sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}\ge \frac{3}{2}$$
$A=\sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}$
Holder: $A^2\cdot Σ_{cyc}\left(a^2\left(b+3c\right)\right)\ge \left(a+b+c\right)^3$
So we need to prove $4\left(a+b+c\right)^3\ge 9Σ_{cyc}\left(a^2\left(b+3c\right)\right)$
$\Leftrightarrow 4a^3+4b^3+4c^3+3a^2b+3b^2c+3c^2a+24abc\ge 15ab^2+15bc^2+15a^2c$
Then i don't know how to solve it. Help me !
| Your trying gets a wrong inequality.
Try $c=0$ and $a=b=1$.
Let $\frac{a}{b+3c}=\frac{x^2}{4}$, $\frac{b}{c+3a}=\frac{y}{4}$ and $\frac{c}{a+3b}=\frac{z^2}{4}$, where $x$, $y$ and $z$ are non-negative numbers.
Hence, the system
$$\begin{array}{l}4a-x^2b-3x^2c=0\\-3y^2a+4b-y^2c=0\\ -z^2a-3z^2b+4c=0\\ \end{array}$$
has infinitely many solutions $(a,b,c)$,
Hence $$ \|(\begin{array}{ccc}4& -x^2 & -3x^2\\ -3y^2 & 4 & -y^2\\ -z^2 & -3z^2 & 4\end{array})\| = 0 ,$$
which gives $$3(x^2y^2+x^2z^2+y^2z^2)+7x^2y^2z^2=16$$ and we need to prove that $$x+y+z\geq3.$$
Let $x+y+z<3$, $x=kp$, $y=kq$ and $z=kr$ such that $k>0$ and $p+q+r=3$.
Hence, $k(p+q+r)<3$, which gives $0<k<1$.
Thus, $$16=3(x^2y^2+x^2z^2+y^2z^2)+7x^2y^2z^2=3k^4(p^2q^2+p^2r^2+q^2r^2)+7k^6p^2q^2r^2<$$
$$<3(p^2q^2+p^2r^2+q^2r^2)+7p^2q^2r^2.$$
But it's a contradiction because we'll prove now that
$$16\geq3(p^2q^2+p^2r^2+q^2r^2)+7p^2q^2r^2$$ for all non-negatives $p$, $q$ and $r$ such that $p+q+r=3$.
Indeed, let $p+q+r=3u$, $pq+pr+qr=3v^2$ and $pqr=w^3$.
Hence, $16\geq3(p^2q^2+p^2r^2+q^2r^2)+7p^2q^2r^2\Leftrightarrow f(w^3)\leq0$, where $f$ is a convex function.
Hence, by $uvw$ it remains to check two cases.
*
*$w^3=0$.
Let $r=0$.
We need to prove that $$16\geq3p^2q^2,$$ which is true because $$p^2q^2\leq\frac{(p+q)^4}{16}=\frac{81}{16}<\frac{16}{3}.$$
*$q=r$.
After homogenization we can assume $q=r=1$ and it remains to prove that
$$\frac{16(p+2)^6}{729}\geq7p^2+\frac{(2p^2+1)(p+2)^2}{3}$$ or
$(p-1)^2(8p^4+112p^3+453p^2+1102p+26)\geq0$, which is obvious.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $d\mid n$ then $x^d-1\mid x^n-1$ proof How would you show that if $d\mid n$ then $x^d-1\mid x^n-1$ ?
My attempt :
$dq=n$ for some $q$. $$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$ in fact, $$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$
By multiplying both sides of $(1)$ by $(x-1)$ we get that $1-x^d\mid 1-x^n$ which is the final result
Is this an ok proof?
| Notice that for any $x$ and and natural $n$ that $$(x-1)(x^{n-1} + ..... + x + 1) = (x^n + x^{n-1} +....... +x) - (x^{n-1} + x^{n-2} +.... +1) = x^n -1$$ so that $x-1|x^n - 1$ always.
Lemma: $x-1|x^n-1$ for natural $n$.
Now $d|n$ so let $m = \frac nd$ and let $y= x^d$.
Then $y-1|y^m -1$.
But $y-1 = x^d -1$ and $y^m -1 = x^n - 1$.
In particular: $(x^d -1)(x^{n-d} + x^{n-2d} + ..... + x^d + 1) = x^n - 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Suppose $ x+y+z=0 $. Show that $ \frac{x^5+y^5+z^5}{5}=\frac{x^2+y^2+z^2}{2}\times\frac{x^3+y^3+z^3}{3} $. How to show that they are equal? All I can come up with is using symmetric polynomials to express them, or using some substitution to simplify this identity since it is symmetric and homogeneous but they are still too complicated for one to work out during the exam. So I think there should exist some better approaches to handle this identity without too much direct computation.
In addition, this identity is supposed to be true:
$$ \frac{x^7+y^7+z^7}{7}=\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5} .$$
| By Newton's identities, with $e_k$ standing for the elementary symmetric polynomials and $p_k$ for the sum of $k^{th}$ powers, and using that $e_1=0\,$:
$$
\begin{align}
p_1 &= e_1 &&= 0\\
p_2 &= e_1p_1-2e_2 &&= -2e_2 \\
p_3 &= e_1p_2 -e_2p_1 + 3e_3 &&= 3e_3 \\
p_4 &= e_1p_3 - e_2p_2+e_3p_1-4e_4 &&= 2 e_2^2 \\
p_5 &= e_1p_4-e_2p_3+e_3p_2-e_4p_1+5e_5 &&= -5e_2e_3
\end{align}
$$
It follows that:
$$\color{blue}{\frac{p_5}{5}} = -e_2e_3 = \frac{-2e_2}{2} \cdot \frac{3e_3}{3} = \color{blue}{\frac{p_2}{2} \cdot \frac{p_3}{3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $x^2+y^2=9$, $\arctan\frac{y+2}{x+2} + \arctan\frac{y-2}{x+4} =2\arctan\frac{y}{x}$ without graphing? The system of equations:
$$\begin{align}
x^2 + y^2 &= 9 \\[6pt]
\operatorname{arctan}\frac{y+2}{x+2} + \operatorname{arctan}\frac{y-2}{x+4} &=2\,\operatorname{arctan} \frac{y}{x}
\end{align}$$
I tried to interpret the second equation by setting $x$ and $y$ as the legs of a right triangle, but I'm still unable to solve this system of of equations geometrically. Any tips?
| Hint:
$$\arctan\dfrac{y+2}{x+2}-\arctan\dfrac yx=\arctan\dfrac yx-\arctan\dfrac{y-2}{x+4}$$
$$\iff\dfrac{2x-2y}{x^2+y^2+2x+2y}=\dfrac{2x+4y}{x^2+y^2-2x+4y}$$
As $x^2+y^2=9$ $$\iff\dfrac{2x-2y}{9+2x+2y}=\dfrac{2x+4y}{9-2x+4y}$$
$$\iff8x^2+16y^2+54y=0\ \ \ \ (1)$$
As $x^2+y^2=9,$ WLOG $x=3\cos t,y=3\sin t$
Put these values in $(1)$ and divide both sides by $\cos^2 t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Partial Fractions Decomposition of $\frac{25s}{(s^2+16)(s-3)(s+3)}$ So this is the problem..
$$
\frac{25s}{(s^2+16)(s-3)(s+3)}
$$
So what I did was...
$$
\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {A}{s^2+16}+\frac {B}{s-3}+\frac{C}{s+3}
$$
then...
$$\begin{align}
25s &= A(s-3)(s+3)+B(s^2+16)(s+3)+C(s^2+16)(s-3) \\
25s &= A(s^2-9)+B(s^3+3s^2+16s+48)+C(s^3-3s^2+16s-48) \\
25s &= As^2-9A+Bs^3+3Bs^+16Bs+48B+Cs^3-3Cs^2+16Cs-48C
\end{align}$$
rearranged to get...
$$
25s = Bs^3+Cs^3+As^2+3Bs^2-3Cs^2+16Bs+16Cs-9A+48B-48C
$$
factoring out...
$$
25s = (B+C)s^3+(A+3B-3C)s^2+(16B+16C)s-9A+48B-48C
$$
Now, this is the part where U messed up. I've been trapped with this question for more than an hour. I don't know if I made any mistakes as I've looked at it over and over again.
The answer is supposed to be:
$$
\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {-s}{s^2+16}+\frac {1}{2(s-3)}+\frac{1}{2(s+3)}
$$
Please be detailed. Thank you!
| $$\frac {25s}{(s^2+16)(s-3)(s+3)} = \frac {As+B}{s^2+16}+\frac {C}{s-3}+\frac{D}{s+3}$$
Use Heaviside method.
To find $C$, let $s=3$ and you get $C=1/2$
To find $D$, let $s=-3$ and you get $D=1/2$
Substitute for $C$ and $D$ in the RHS and subtract from LHS.
You will find $$\frac {25s}{(s^2+16)(s-3)(s+3)}-\frac {1}{2(s-3)} - \frac {1}{2(s+3)}= \frac {-s}{s^2+16}$$
Thus $$\frac {25s}{(s^2+16)(s-3)(s+3)}=\frac {-s}{s^2+16}+\frac {1}{2(s-3)} + \frac {1}{2(s+3)} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Elementary proof that the MacLaurin series of $\sin x$ converges to $\sin x$ for all $x$ In my book it is given:
$\sin x = x- \dfrac {x^3}{3!}+\dfrac{x^5}{5!}- \dfrac{x^7}{7!}...$
I googled around for a proof but couldn't understand any of them. I would like to know if there's any elementary high school level proof the series
| $$f(x)=e^x$$
$$f'(x)=e^x \tag 1$$ (I assumed you know this property of $e^x$. )
$$f''(x)=f'(x)=e^x$$
$$f^{(n)}(x)=e^x$$
$$f^{(n)}(0)=1$$
If we find the Taylor series for a function $f(x)$ is:
$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)x^n}{n!}=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$
$$f(ix)=e^{ix}=\sum_{n=0}^{\infty} \frac{i^nx^n}{n!}$$
$$i=\sqrt{-1}$$
$$i^2=-1$$
$$i^3=-i$$
$$i^4=1$$
$$i^5=i$$
.
.
$$e^{ix}=1+\frac{ix}{1!}+\frac{i^2x^2}{2!}+\frac{i^3x^3}{3!}+\frac{i^4x^4}{4!}+\frac{i^5x^5}{5!}+\frac{i^6x^6}{6!}+\frac{i^7x^7}{7!}+\frac{i^8x^8}{8!}+......$$
$$e^{ix}=1+\frac{ix}{1!}+\frac{-x^2}{2!}+\frac{-ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}+\frac{-x^6}{6!}+\frac{-ix^7}{7!}+\frac{x^8}{8!}+......$$
$$e^{ix}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+..... +i(\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+......) \tag2$$
This is Euler Formula: link
$$e^{ix}=\cos x +i \sin x \tag 3$$
Take both side derivative Equation $2$
$$(e^{ix})'=(-\frac{x}{1!}+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}+......) +i(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+.....) $$
$$(e^{ix})'=-\sin x +i\cos x \tag 4 $$
$$(e^{ix})'=\cos 'x +i \sin' x \tag 5 $$ ///If we want equal (4) and (5)
$$ \cos 'x +i \sin' x =-\sin x +i\cos x$$
we equal imaginary and real parts separately, we will get
$$\cos 'x= - \sin x$$
$$\sin 'x= \cos x$$
They are trigonometric function properties. link
$$\cos x=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$$
$$\sin x=\sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Prove $n$ is prime. (Fermat's little theorem probably)
Let $x$ and $n$ be positive integers such that $1+x+x^2\dots x^{n-1}$ is prime. Prove $n$ is prime
My attempt:
Say the above summation equal to $p$ $$1+x+x^2\dots x^{n-1}\equiv 0\text{(mod p)}\\
{x^n-1\over x-1}\equiv0\\
\implies x^n\equiv1\text{ (as $p$ can't divide $x-1$)}$$
How to proceed?
| If $x = 1$, it's obvious.
For $x>1$, let $n$ be composite, say $n = pq$ with $p, q>1$. In that case, we have
$$
(1+x+\cdots+x^{n-1})(x-1) = x^n-1 = x^{pq}-1\\
= (x^p)^q-1\\
= (1+x+x^p+x^{2p}+\cdots + x^{(q-1)p})(x^p-1)
$$
Thus $x^p-1$ divides $x^n-1$, but is not equal to it. In other words, $\frac{x^n-1}{x^p-1}$ is an integer greater than $1$. We get
$$
\frac{x^n-1}{x^p-1} =\frac{(1+x+\cdots+x^{n-1})(x-1)}{(1+x+\cdots +x^{p-1})(x-1)}\\
= \frac{1+x+\cdots+x^{n-1}}{1+x+\cdots +x^{p-1}}
$$
That final fraction is an integer greater than $1$, and the denominator is an integer greater than $1$, which means that the numerator must be composite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
What am I doing wrong while finding the Laurent series of $\frac{1}{\sin(z)}$? I'm trying to compute the Laurent series of $\frac{1}{\sin(z)}$ at $z_0=0$.
From what I've seen on the internet this is given as
$f(z)=\frac{1}{z}+\frac{z}{3!}+\frac{7z^3}{360}+...$
My attempt:
$f(z)=\frac{1}{\sin(z)}$ can be rewritten as $f(z)=\frac{1}{z}\frac{z}{\sin(z)}$
We can express $\sin(z)$ as $\sin(z)=z-\frac{z^2}{3!}+\frac{z^4}{5!}+...$
$\therefore \frac{z}{\sin(z)}=(\frac{\sin(z)}{z})^{-1}=(1-\frac{z}{3!}+\frac{z^3}{5!}+...)^{-1}$
To get this to equal what I know it should, then we should have $(\frac{\sin(z)}{z})^{-1}=(1+\frac{z}{3!}+\frac{7z^3}{360}+...)$ but how can this be what do I do with my negative sign and how do I obtain the $\frac{7}{360}$ coefficient?
| Note the most direct way to obtain the expansion of $\dfrac1{\sin z}$ is division by increasing powers. I'll show how to obtain the first four terms:
Start from the expansion of $\sin z$ at order $7$:$$\frac1{\sin z}=\frac1{z-\dfrac{z^3}{6}+\dfrac{z^5}{120}-\dfrac{z^7}{5040}+o(z^7)}=\frac1z\cdot\frac1{1-\dfrac{z^2}{6}+\dfrac{z^4}{120}-\dfrac{z^6}{5040}+o(z^6)}$$
Now the division of $1$ by $1-\dfrac{z^2}{6}+\dfrac{z^4}{120}-\dfrac{z^6}{5040}$ goes as follows (we truncate the terms with degree more than $6$) :
Thus we obtain
\begin{align}
\frac1{\sin z} &= \frac1z \Bigl( 1 + \dfrac{z^2}{6} + \frac{7z^4}{360}- \frac{z^6}{4320} +o(z^6)\Bigr) \\[1ex]
&=\frac1z + \dfrac{z}{6} + \frac{7z^3}{360}- \frac{z^5}{4320} +o(z^5)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
British Maths Olympiad (BMO) 2006 Round 1 Question 5, alternate solution possible? The question states
For positive real numbers $a,b,c$ prove that
$(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b)$
After some algebraic wrangling we can get to the point where:
$(a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4 ≥ 2c^2(a^2 + b^2)$
At this point if we take the $LHS - RHS$ we can write the expression as the sum of squares proving the inequality.
I was wondering, is it possible to divide both sides by $c^2(a^2 + b^2)$ and show somehow that
$((a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4)/(c^2(a^2 + b^2)) ≥ 2$
I tried but was not able to.
| Let's simplify the inequality:
$$(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b) \Rightarrow \\
a^4+2a^2b^2+b^4\ge ((a+b)^2-c^2)(c^2-(a-b)^2) \Rightarrow \\
a^4+2a^2b^2+b^4\ge 2c^2(a^2+b^2)-(a^2-b^2)^2-c^4 \Rightarrow \\
c^4-2(a^2+b^2)c^2+2(a^4+b^4)\ge 0 \qquad (1)$$
It is a bi-quadratic inequality and its discriminant is:
$$D=(a^2+b^2)^2-2(a^4+b^4)=2a^2b^2-(a^4+b^4)\le 0,$$
the iquality occurs for $a=b$. Note that the inequlity $(1)$ is true for $D<0$. We will check $D=0$. Then the inequality $(1)$ and its solution will be:
$$\begin{cases} c^2=a^2+b^2=2a^2 \\
(2a^2)^4-4a^4+4a^4\ge 0\end{cases} \Rightarrow16a^8\ge 0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Edge length of an equilateral triangle if distances from a point $P$ to its vertices is given A point $P$ is located inside an equilateral triangle and is at a distance of 5, 12, and 13 from its vertices. Compute the edge length of the triangle.
The answer is $\sqrt{169 + 60\sqrt(3)}$.
If $s$ is the edge length of the triangle, and if $x$ is the measure of the angle with vertex at $P$ and with sides of lengths 5 and 12, and if $y$ is the measure of the angle with vertex at $P$ and with sides of lengths 5 and 13, according to the Law of Cosines,
\begin{equation*}
s^{2} = 169 - 120\cos(x)
\end{equation*}
\begin{equation*}
s^{2} = 194 - 130\cos(y)
\end{equation*}
and since $\cos(360 - (x + y)) = \cos(x + y)$,
\begin{equation*}
s^{2} = 313 - 312\cos(x+y)
\end{equation*}
So,
\begin{align*}
\cos{x} = - \frac{s^{2} - 169}{120} , \\
\cos{y} = - \frac{s^{2} - 194}{130} , \\
\cos(x+y) = - \frac{s^{2} - 313}{312} .
\end{align*}
For any real numbers $\theta$ and $\phi$,
\begin{equation*}
\bigl[\cos{\theta}\cos{\phi} - \cos(\theta + \phi)\bigr]^{2}
= \sin^{2}\theta\sin^{2}\phi
= \bigl(1 - \cos^{2}\theta\bigr)\bigl(1 - \cos^{2}\phi\bigr) .
\end{equation*}
So,
\begin{align*}
&\left[\frac{s^{2} - 169}{120} \cdot \frac{s^{2} - 194}{130} + \frac{s^{2} - 313}{312}\right]^{2} \\
&\qquad \qquad = \left(1 - \left(\frac{s^{2} - 169}{120}\right)^{2}\right)\left(1 - \left(\frac{s^{2} - 194}{130}\right)^{2}\right)
\end{align*}
or equivalently, by multiplying by $(120^{2})(130^{2})312$,
\begin{align*}
&\Bigl[312(s^{2} - 169)(s^{2} - 194) + (120^{2})(130^{2})(s^{2} - 313)\Bigr]^{2} \\
&\qquad \qquad
= 312 \Bigl((120^{2})(130^{2}) - 130^{2}(s^{2} - 194)\Bigr)
\Bigl((120^{2})(130^{2}) - 120^{2}(s^{2} - 194)\Bigr) .
\end{align*}
How is the quartic equation in the variable $s$ to be solved?
|
Given $a,b,c$,
the equilateral triangle
with the internal point $P$,
located at distances $a,b,c$ from its vertices,
can be constructed as follows.
Start with $\triangle ABC$,
$|BC|=a$,
$|CA|=b$,
$|AB|=c$,
$\angle CAB=\alpha$,
$\angle ABC=\beta$,
$\angle BCA=\gamma$.
Construct a helper external equilateral triangle,
based on any side of $\triangle ABC$,
for example, $\triangle ADB$.
Then $|CD|=u$ is the side of sought equilateral triangle,
which can be constructed in-place
as either
$\triangle CFD$ with $P=A$
or $\triangle DEC$ with $P=B$.
Note that the three angles at $P$ are
$\alpha+60^\circ$,
$\beta+60^\circ$,
$\gamma+60^\circ$,
hence such equilateral triangle
with internal point $P$
can be constructed only if
angles of $\triangle ABC$
are less than $120^\circ$.
The length of the side, $u$,
can be found by the cosine rule,
for example:
\begin{align}
\triangle ADC:\quad
u^2&=
b^2+c^2-2bc\cos(\alpha+60^\circ)
\\
&=b^2+c^2-bc(\cos\alpha-\sqrt3\sin\alpha)
\\
&=b^2+c^2-bc\cos\alpha+bc\sqrt3\sin\alpha)
\\
&=b^2+c^2-bc\cos\alpha+2\sqrt3 S_{\triangle ABC}
,\\
\triangle ABC:\quad
\cos\alpha&=\frac1{2bc}\,(b^2+c^2-a^2)
,\\
u^2&=
b^2+c^2-\tfrac12\,(b^2+c^2-a^2)+2\sqrt3 S_{\triangle ABC}
\\
&=\tfrac12\,(a^2+b^2+c^2)
+2\sqrt3 S_{\triangle ABC}
,
\end{align}
where $S_{\triangle ABC}$
is the area of $\triangle ABC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find an asymptotic for $\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}}$ I need to find the asymptotic behavior of $$\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}}$$
as $N\to\infty$.
I found (using a computer) that this asymptotically will be equivalent to $\frac{1}{3}N^2$, but don't know how to prove it mathematically.
| Standard asymptotic calculus yields the estimate $$ \frac{1}{1-\cos x} = \frac{2}{x^2}+O(1)$$ hence $\displaystyle x\mapsto \frac{1}{1-\cos x} -\frac{2}{x^2}$ is bounded on a neighborhood of $0$, say $(0,\delta)$
Furthermore the function is continuous over $[\delta,\pi]$, thus bounded over $[\delta,\pi]$, hence bounded over $(0,\pi]$: write
$$\displaystyle \forall x\in (0,\pi], \frac{1}{1-\cos x} = \frac{2}{x^2}+g(x)$$ where $g$ is bounded on $(0,\pi]$.
With that estimate, $$\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}} = \sum_{j=1}^N \frac{2N^2}{\pi^2 j^2} + \sum_{j=1}^Ng(\frac{\pi j}n) = \frac{2N^2}{\pi^2}\left(\frac{\pi^2}6+o(1) \right) + O(N)=\frac{N^2}{3}+o(N^2)$$
As ClementC has noted, with the sharper estimate $\sum_{j=1}^N \frac{1}{j^2} = \frac{\pi^2}6 +O(\frac 1N)$ (which follows from integral methods applied to $\sum_{k=N+1}^\infty \frac{1}{k^2}$), one gets the better asymptotic $\sum_{j=1}^N \frac{1}{1 - \cos\frac{\pi j}{N}} = \frac{N^2}{3}+O(N)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$.
If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$.
My attempt:
Consider $\frac{1}{1 - z}$. Let $z = x + iy$, we know that $|z| < 1 \implies x, y < 1$. $$\frac{1}{1 - z} = \frac{1}{1 - x - iy} = \frac{1 - x + iy}{(1 - x)^2 + y^2}.$$
$$\Re \left(\frac{1}{1 - z} \right) = \frac{1 - x}{(1 - x)^2 + y^2}.$$
I got $$\frac{1}{(x-1)^2 + y^2} > \frac{1}{2}.$$
How do I manage the numerator? Can you help me? I welcome the alternative approaches.
| $$\Re \Big(\frac{1}{1-z}\Big)={1\over 2}\Big(\frac{1}{1-z}+\frac{1}{1-\overline{z}}\Big)$$
$$={1\over 2}\frac{1-\overline{z}+1-z}{1-z-\overline{z}+z\overline{z}}$$
$$>{1\over 2}\frac{1-\overline{z}+1-z}{1-z-\overline{z}+1}$$
$$={1\over 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Prove that following recursively defined sequence converges and its limit is 1/2
\begin{cases}a_1 = 1\\ \\ \displaystyle a_{1+n} = \sqrt{\sum_{i=1}^n a_i}\end{cases}
Prove that $\{\frac{a_n}{n}\}$ is convergent and its limit is $\frac12$
My proof: We can recursively define relation between $a_{1+n}$ & $a_n$ as
$a_{1+n} = \sqrt{a_n^2+a_n}$
Also, it is evident that $a_{1+n} > a_n > 1$
Thus, $a_{n}^2 - a_{n-1}^2 + a_{n} - a_{n-1} = a_{n} > 1 > 1/4$
Thus, $a_{n+1} = \sqrt{a_{n}^2 + a_{n}} > \sqrt{a_{n-1}^2+a_{n-1}+1/4} = a_{n-1} + 1/2 $
Similarly, we can show $a_{n+1} = \sqrt{a_{n}^2 + a_n} < a_n + 1/2$
So, we have $a_1 + \frac{n-1}2 < a_{n+1} < a_1 + \frac{n+1}2$
Using Sandwich Theorem we got limit of $\frac{a_n}n$ as $\frac12$
Is my proof correct.
I will be grateful if you can provide alternative proofs
| You have shown that $a_{n+1}-a_{n-1} \gt \frac 12$ but that is a step of $2$ in the index. You therefore can only conclude that $a_{n+1} \gt a_1+\frac n4$ and the squeeze fails.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating a determinant. $D_n$=\begin{vmatrix}
a & 0 & 0 & \cdots &0&0& n-1 \\
0 & a & 0 & \cdots &0&0& n-2\\
0 & 0 & a & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&a&1 \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& a\\
\end{vmatrix}
I tried getting the eigenvalues for A =
\begin{vmatrix}
0 & 0 & 0 & \cdots &0&0& n-1 \\
0 & 0 & 0 & \cdots &0&0& n-2\\
0 & 0 & 0 & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&0&1 \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& 0\\
\end{vmatrix}
For $a=0$ , the rank of the matrix is $2$ , hence $\dim(\ker(A)) = n-2 $
$m(0)>=n-2$
However, I was not able to determine the other eigenvalues.
Testing for different values of n :
for $n=2$ :
$D_2 = a^2-1$
for $n=3$ :
$D_3 = a^3 -5a$
$D_n$ seems to be equal to $a^n - a^{n-2}\sum_{i=1}^{n-1}i^2$ .
However I'm aware that testing for different values of $n$ is not enough to generalize the formula.
Thanks in advance.
| It is maybe a bit easier to use the matrix determinant formula. If you rewrite your matrix as
$$
M = aI +UV^T
$$
where $UV^T$ is
$$
\begin{bmatrix}
(n-1) &0\\
(n-2) &0\\
\vdots&0\\
1 &0\\
0&1
\end{bmatrix}
\begin{bmatrix}
0&0&\cdots&1\\
(n-1)&(n-2)&\cdots&0
\end{bmatrix}
$$
Then, from
$$
\operatorname{det}({\mathbf{A}}+{\mathbf {UV}}^{{\mathrm {T}}})=\operatorname {det}({\mathbf{I}}+{\mathbf {V}}^{{\mathrm {T}}}{\mathbf {A}}^{{-1}}{\mathbf {U}})\operatorname {det}({\mathbf {A}}).
$$
and $\mathbf{A}^{-1}=\frac1aI$, we have only a $2\times 2$ determinant to evaluate:
$$
\det(M) = \left|\begin{matrix}1&\frac{1}{a}\\\displaystyle\frac{1}{a}\sum_{i=1}^{n-1}i^2&1\end{matrix}\right|a^n = (1-\frac{1}{a^2}\sum_{i=1}^{n-1}i^2)a^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
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set G equal to what?
My attempts : i construct a matrix $A= \begin{bmatrix} 1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&2&1&0&0&0&0\\ 0&0&0&2&0&0&0&0\\ 0&0&0&0&2&0&0&0\\ 0&0&0&0&0&2&0&0\\ 0&0&0&0&0&0&2&0\\ 0&0&0&0&0&0&0&2\end{bmatrix}$
Now we know that the characteristic polynomial of A is $(x − 1)^2
(x − 2)^6$
and the minimal polynomial of A is$ (x − 1)(x − 2)^2$
Now we will take only eigenvalue $ 2$ of $ A$ that is characteristic of eigenvalue $2 =(x-2)^6$ and the minimial of eigenvalue $2 = (x-1)^2$
Now set G will be equal to $ 6- 2 = 4$
so $G = 4$
Is my answer is correct or nots? pliz reply
i have my exam tommorrows. so i need its urgently any Hints/solution
Thanks in advance i would be more thankful
| In the matrix $A$ we have an eigenvalue $2$ which has geometric multiplicity $4$ and one which has geometric multiplicity $1$ (the one of the jordan block $2-by-2$). Thus we have that $G=\{1,4\}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sums of Nilpotent Matrices Let $A =$ diag$(a_1,a_2,…,a_n)$, where the sum of all $a_i$’s is zero.
Show that A is a sum of nilpotent matrices.
My idea:
$\begin{bmatrix}
2 & 0 \\
0 & -2
\end{bmatrix}$ = $\begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}$ + $\begin{bmatrix}
1 & -1 \\
1 & -1
\end{bmatrix}$
Where $\begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}^2$ = $\begin{bmatrix}
1 & -1 \\
1 & -1
\end{bmatrix}^2$ = $0_2$
Extending the 2 $\times$ 2 matrix, we get
$\begin{bmatrix}
2 & 0 & 0 & 0 \\
0 & -2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & -2
\end{bmatrix}$ = $\begin{bmatrix}
1 & 1 & 0 & 0 \\
-1 & -1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & -1 & -1
\end{bmatrix}$ + $\begin{bmatrix}
1 & -1 & 0 & 0 \\
1 & -1 & 0 & 0 \\
0 & 0 & 1 & -1 \\
0 & 0 & 1 & -1
\end{bmatrix}$
Where $\begin{bmatrix}
1 & 1 & 0 & 0 \\
-1 & -1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & -1 & -1
\end{bmatrix}^2$ = $\begin{bmatrix}
1 & -1 & 0 & 0 \\
1 & -1 & 0 & 0 \\
0 & 0 & 1 & -1 \\
0 & 0 & 1 & -1
\end{bmatrix}^2$ = $0_4$
Kindly correct me if I’m wrong, and also, please help me how to generalized this. Thank you in advance.
| My answer follows @JoséCarlosSantos' post.
It is enough to notice that for all $1 \leq i < j \leq n$, $$E_{i, i} -E_{j, j} = \left(E_{i, j}\right) + \left(- E_{j, i}\right) + \left(E_{i, i} -E_{i, j} + E_{j, i} - E_{j, j}\right)$$ and $$\left(E_{i, i} -E_{i, j} + E_{j, i} - E_{j, j}\right)^{2} = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all values in $\mathbb R$ for quadratic absolute value equation The question:
Determine the solution set (in $\mathbb R$) for the equation $|x^2+2x+2| = |x^2-3x-4|$
So far, I have determined that for this to be true, $|x^2-3x-4|$ must be greater or equal to $0$, giving $(x-4)(x+1)\ge0$. To find the solution set, $|x^2+2x+2|\ge-1$ as indicated by the roots of the RHS equation, but this is where I get stuck.
Where am I going wrong?
| Knowing that $x^2+2x+2 > 0$
The equation is equivalent to
$$
x^2+2x+2 = \vert x^2-3x-4\vert
$$
which is equivalent to
$$
x^2+2x+2 =
\left\{\begin{array}{lcl}-x^2+3x+4 & \rightarrow & x = \{\frac{1\pm\sqrt{17}}{4}\}\\
x^2-3x-4 & \rightarrow & x = -\frac{6}{5}\end{array}\right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does one show that multiples of number does not have a zero in the decimal expansion?
Prove that there exists a number divisible by $5^{1000}$ not containing a single zero in its decimal notation.
The above question is taken from this site. It is question no. 88 in the list. I could not find any solutions for this problem. Apparently a "schoolboy" should be able to solve this question.
In general, I have a hunch that this statement must be true for any number (instead of $5^{1000}$). However, I cannot think of a strategy to show there are no zeros in the decimal expansion. Trailing zeros are trivial to catch but zeros in the middle confound me.
Is there a clever construction for the case of $5^{1000}$?
Edit: As explained in the comments, my hunch does not make sense for a multiple of 10.
| Idea of Approach
By induction, an using the fact that $5^{n}$ is a divisor of $10^n$ and $10^{n-1}$ leaves only a certain kind of remainder when divided by $5^n$.
Claim and proof
Claim : For all $n$, there exists an $n$ digit multiple of $5^n$ that does not contain any zeros.
Proof : Start with the base case : $5 , 25 , 125$ for $n = 1,2,3$.
The inductive case : let $L$ be an $n$ digit number that has no non-zero digits and is a multiple of $5^n$. We want a number of the form $a \times 10^{n} + L$(this will have $n+1$ digits) that divides $5^{n+1}$, where $a \neq 0$.
Well, if $a \times 10^n + L$ is a multiple of $5^{n+1}$ then the remainders when you divide $a \times 10^n$ and $L$ by $5^{n+1}$ respectively, must sum up to a multiple of $5^{n+1}$.
But both are multiples of $5^n$, so $10^n$ leaves a remainder of $b 5^n$ when divided by $5^{n+1}$ and $L$ leaves a remainder of $c 5^n$ when divided by $5^{n+1}$.
Hence, $a 10^n + L$ leaves the remainder $(ab+c)5^n$ when divided by $5^{n+1}$.
Note that $b$ cannot be zero since $10^n$ is not a multiple of $5^{n+1}$. But $c = 0$ is possible.
So the question is : can we make $ab+c$ a multiple of $5$ everytime?
If $c = 0$ then take $a = 5$, since in that case, regardless of $b$ we have $ab+c$ is a multiple of $5$.
If $c \neq 0$ then since $b \neq 0$ we take $a = (-c)b^{-1} \mod 5$, where $b^{-1}$ is the multiplicative inverse of $b$ modulo $5$.
Therefore, in either case, we are done.
Example
*
*Let us continue from $125$. So we have to add $a \times 10^3 + 125$ for some $a$. Here, note that $1000$ leaves the remainder $3 \times 125$ when divided by $5^4 = 625$ so $b = 3$ and $L$ leaves the remainder $1 \times 125$ when divided by $625$, so $c = 1$. So $b^{-1} = 2$(since $2 \times 3 = 1 \mod 5$) and $b^{-1} \times -c = -2 = 3 \mod 5$. Check that $3125$ is a four digit multiple of $625$.
*Next, we want a five digit multiple of $5^5 = 3125$. Here, we have $c = 0$ because $L = 3125$ is a multiple (in fact equal) to $3125$, so we have $a = 5$. Check that $53125$ is a multiple of $3125$.
*Next, we want a six digit multiple of $5^6 = 15625$. Here, check that $L =53125$ , $c = 2$ and $ b= 2$ as well, so $b^{-1} = 3$ and $3 \times -2 = -6 \equiv 4$ so $453125$ is a multiple of $15625$.
Conclusion
Hence, for all $n$, there exists a multiple of $5^n$ which has exactly $n$ digits, and furthermore consists only of the digits $1,2,3,4,5$ by construction.
Addendum
Maybe the approach might seem a little arbitrary. I actually got the idea from another question I had solved about eight and a half years back (and wrote down the first $21$ terms on the school board!) : Show that there is a multiple of $2^n$ that consists only of the digits $6$ and $7$.
The style of proof there was exactly the same : take such a multiple of $2^n$ and adjoin either a $6$ or $7$ on the left so that things work out for the next power.
Last but not the least, note that the post attached in the comments to the question addresses this question but does not give a proof. Rather, it uses this fact to prove that every number not ending with zero has a multiple with no zero in its representation. In fact, as pointed out, such a number can also be made to consist solely of the digits $1,2,3,4,5,6,7$.
| {
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How to find and characterise critical points of a polynomial? Find and characterise the critical points of: $f(x)=(2x^3-12x^2+18x-1)^5$
I differentiated to get:
$$\frac{df}{dx}=5(6x^2-24x+18)(2x^3-12x^2+18x-1)^4=30(x-3)(x-1)(2x^3-12x^2+18x-1)^4$$
But don't know how to factorise this further?
| If you set $$30(x-3)(x-1)(2x^3-12x^2+18x-1)^4=0$$ then the stationary points are at $$x-3=0\implies x=3\\x-1=0\implies x=1\\2x^3-12x^2+18x-1=0$$ and this cubic equation can be solved using Cardano's method.
To find the critical points, substitute each value of $x$ into $y=(2x^3-12x^2+18x-1)^5$ and the nature of them can be found by evaluating the second derivative of $y$ at those points.
| {
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Limit of multivariable function including absolute value I have some trouble with this limit:
$$\lim_{(x,y)\to (0,0)} \frac{x^3+y^2}{x^2+|y|}.$$
My first attempt to solve it was with polar coordinates but I couldn't find an expression which was independent of $\varphi$. Now I'm trying to solve it with the triangle inequality so I can find an upper limit that squeezes my expression into $0$. My work so far:
$$\left| \frac{x^3+y^2}{x^2+|y|} \right| \leq \frac{|x^3+y^2|}{|x^2+|y||} \leq \frac{|x^3|+|y^2|}{|x^2|+||y||} = \frac{|x^3|+y^2}{x^2+|y|}.$$
But from here I don't know how to continue.
Thanks in advance!
| By Cauchy–Schwarz
$$|x^3+y^2|=|xx^2+|y||y||\le\sqrt{x^2+y^2}\sqrt{x^4+y^2}\le \sqrt{x^2+y^2}\sqrt{(x^2+|y|)^2}=\sqrt{x^2+y^2}(x^2+|y|)$$
then
$$\left|\frac{x^3+y^2}{x^2+|y|}\right|\le\sqrt{x^2+y^2}\to 0 \implies \frac{x^3+y^2}{x^2+|y|}\to 0$$
| {
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If $z^{23}=1$ then evaluate $\sum^{22}_{z=0}\frac{1}{1+z^r+z^{2r}}$
If $z$ is any complex number and $z^{23}=1$ then evaluate $\displaystyle \sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}$
Try: From $$z^{23}-1=(z-1)(1+z+z^2+\cdots +z^{22})$$
And our sum $$\sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}=\frac{1}{3}+\frac{1}{1+z+z^2}+\frac{1}{1+z^2+z^4}+\cdots +\frac{1}{1+z^{22}+z^{44}}$$
Now i did not understand how can i simplify it.
Could some help me. Thanks.
| If $z^{23}=1$ and $z\not=1$ (otherwise the sum is trivially equal to $23/3$) then $z^k$ is a primitive $23$-th root of unity for any $k=1,2\dots,22$ (note that $23$ is a prime number). Hence
$$\begin{align}
\sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}&=\frac{1}{3}+\sum_{r=1}^{22}\frac{1}{1+z^r+z^{2r}}\\
&=\frac{1}{3}+\sum_{r=1}^{22}\frac{1}{1+z^{8r}+z^{16r}}\\
&=\frac{1}{3}+\sum_{r=1}^{22}\frac{z^{8r}-1}{(1+z^{8r}+z^{16r})(z^{8r}-1)}\\
&=\frac{1}{3}+\sum_{r=1}^{22}\frac{z^{8r}-1}{z^{24r}-1}
=\frac{1}{3}+\sum_{r=1}^{22}\frac{z^{8r}-1}{z^{r}-1}\\
&=\frac{1}{3}+\sum_{r=1}^{22}(1+z^r+z^{2r}+z^{3r}+z^{4r}+z^{5r}+z^{6r}+z^{7r})\\
&=\frac{1}{3}+22+7\sum_{r=1}^{22}z^r.
\end{align}$$
Can you take it from here?
| {
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Integrate $\frac{1}{\sqrt{x^2+cx}}$ I am trying to compute the integral $\int\frac{1}{\sqrt{x^2+cx}}dx$.
To begin I completed the square of the denominator resulting in $$\int \frac{1}{\sqrt{(x+\frac{c}{2})^2-\frac{c^2}{4}}}dx$$
I then made the substitution $u=x+\frac{c}{2}$, which has left me with the integral $$\int \frac{1}{\sqrt{u^2-\frac{c^2}{4}}}du$$
From here I am stumped on what to do next. Is there a simpler way to integrate this? Am I heading in the right direction? Any help would be appreciated.
| You've done it properly.
continuing from where you left off ;
$I =\displaystyle\int \frac{1}{\sqrt{u^2-\frac{c^2}{4}}}du$
$I =\displaystyle\int \frac{1}{\frac{c}2\sqrt{\frac{4u^2}{c^2}-1}}du$
let $ \frac{2u}{c} = \sec(t) \implies \frac 2c \,du = \sec(t)\tan(t)\,dt$
$I =\displaystyle\int\frac{1}{\sqrt{\sec^2(t)-1}}\sec(t)\tan(t)\,dt$
$I = \displaystyle \int\sec(t)\,dt $
$I = \ln|\sec(t)+\tan(t)|+C'$
$I = \ln\bigg|\frac{2u}c+ \sqrt{\frac{4u^2}{c^2}-1}\bigg|+C'$
$I = \ln\bigg|\frac{2(x+\frac c2)}c+ \sqrt{\frac{4(x+\frac c2)^2}{c^2}-1}\bigg|+C'$
EDIT :
$I =\ln\bigg|\frac{4x+2c}{2c}+ \sqrt{\frac{4(2x+c)^2-4c^2}{4c^2}}\bigg|+C'$
$I =\ln\bigg|\frac{4x+2c}{2c}+ \sqrt{\frac{4(2x+c)^2-4c^2}{4c^2}}\bigg|+C' $
$I= \ln\bigg|\frac{4x+2c}{2c}+ \frac{\sqrt{{4(2x+c)^2-4c^2}}}{2c}\bigg|+C'$
$I= \ln\bigg|2({2x+c}+ {\sqrt{{(2x+c)^2-c^2}}})\bigg|-\ln|2c|+C'$
$I = I= \ln\bigg|({2x+c}+ {\sqrt{{(2x+c)^2-c^2}}}\bigg|+\ln(2)-\ln|2c|+C'$
$I= \ln\bigg|{2x+c}+ {\sqrt{{(2x+c)^2-c^2}}}\bigg|+C$$\quad$ Where $C =C'-\ln|2c|+\ln(2)$
| {
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Series 1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2 + 1/7... This is the exercise 2.7.2 e) of the book "Understanding Analysis 2nd edition" from Stephen Abbott, and asks to decide wether this series converges or diverges:
$$ 1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\dots $$
I have noticed that
$$\dfrac{1}{3}< 1-\dfrac{1}{2^2}+\dfrac{1}{3};\\
\dfrac{1}{3}+\dfrac{1}{5}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5};\\
\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}\\
$$
which is true in general because
$$1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{36}-\dfrac{1}{64}-\dots=1-\sum_{n=1}^\infty \dfrac{1}{(2n)^2}=1-\dfrac{1}{4}\sum_{n=1}^\infty\dfrac{1}{n^2}=1-\dfrac{\pi^2}{24}>0.
$$
Thus
$$
\sum_{n=1}^\infty \dfrac{1}{2n+1}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\dots
$$
Finally, as the series $\sum_{n=1}^\infty \dfrac{1}{2n+1}$ diverges, so does the series requested.
My two questions are:
I) Is this reasoning correct?
II) Can this exercise be done without using that $\sum_{n=1}^\infty\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$? I would like to find a solution with more elementary tools.
Thank you.
| The basic idea is good. But after you've proven that $\sum_{n=1}^N\frac1{2n+1}$ is smaller that the sum of the first $2N+1$ terms of your series, you're done. And all you need for that is that$$\frac1{2^2}+\frac1{4^2}+\cdots+\frac1{(2N)^2}<1.$$This is equivalent to$$\frac1{1^2}+\frac1{2^2}+\cdots+\frac1{N^2}<4,$$which in turn follows from\begin{align}\frac1{1^2}+\frac1{2^2}+\cdots+\frac1{N^2}&<1+\frac1{1\times2}+\frac1{2\times3}+\cdots+\frac1{(N-1)N}\\&=1+1-\frac12+\frac12-\frac13+\cdots+\frac1{N-1}+\frac1N\\&=2+\frac1N\\&<4.\end{align}
| {
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Show that $\lim_{x\to\infty}x^\frac{m}{m+1}(x^\frac{1}{m+1} - (x-1)^\frac{1}{m+1}) = \frac{1}{m+1}$ using the Mean Value Theorem Show that $\lim_{x\to\infty}x^\frac{m}{m+1}(x^\frac{1}{m+1} - (x-1)^\frac{1}{m+1}) = \frac{1}{m+1}$ using the Mean Value Theorem with m and x being whole numbers, for x greater or equal to 1 and m greater or equal to 0. My work so far:
We define $f(x) := x^\frac{1}{m+1}$. From the MVT it follows that there exists an $x_0$ between a and b (with b > a) such that: $$\frac{f(b)-f(a)}{b-a} = f'(x_0)$$
From choosing $b = x$ and $a = x - 1$ it follows that:
$$x^\frac{1}{m+1} - (x-1)^\frac{1}{m+1} = \frac{1}{(m+1)x_0^\frac{m}{m+1}}$$
Multiplication with $x^\frac{m}{m+1}$ results in:
$$x^\frac{m}{m+1}(x^\frac{1}{m+1} - (x-1)^\frac{1}{m+1}) = \frac{x^\frac{m}{m+1}}{(m+1)x_0^\frac{m}{m+1}}$$
I intuitively want to take the limit to infinity of the left-hand side, but I can't justify why the $x$ and $x_0$ terms will cancel-out. I'd also to be happy to hear of any methods that don't use the MVT.
| When $x\to \infty$ since $x-1<x_0<x$ we also have that $x_0 \to \infty$ and
$$x-1<x_0<x \iff \frac{x-1}x<\frac{x_0}x<1$$
then by squeeze theorem $\frac{x_0}x\to 1$, thus
$$\frac{x^\frac{m}{m+1}}{(m+1)x_0^\frac{m}{m+1}}=\frac1{m+1} \left(\frac{x}{x_0}\right)^{\frac{m}{m+1}} \to\frac1{m+1}$$
| {
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prove $\sum_{r=1}^{n-1}(n-r)^2\binom{n-1}{n-r}=n(n-1)2^{n-3}$ This is problem 2.38 from the book "Principles and Techniques in Combinatorics".
Prove $$\sum_{r=1}^{n-1}(n-r)^2\binom{n-1}{n-r}=n(n-1)2^{n-3}$$
I have have been trying different things on and off for two days and still can't derive RHS from LHS.
what I have got so far:
$$\sum_{r=1}^{n-1}(n-r)^2\cdot\frac{n-1}{n-r}\binom{n-2}{n-r-1}
=(n-1)\sum_{r=1}^{n-1}(n-r)\cdot\binom{n-2}{n-r-1}
=\\(n-1)\sum_{r=2}^{n}(n-r)\cdot\binom{n-2}{n-r}
=(n-1)\sum_{r=2}^{n}(n-r)\cdot\binom{n-2}{n-r}
=\\(n-1)\sum_{r=2}^{n}(n-r)\cdot\frac{n-2}{n-r}\binom{n-3}{n-r-1}
=\\(n-1)(n-2)\sum_{r=2}^{n}\cdot\binom{n-3}{n-r-1}
$$
any hints or solutions would be greatly appreciated.
| Another way:
$$
\eqalign{
& \sum\limits_{1\, \le \,r\, \le \,n - 1} {\left( {n - r} \right)^{\,2} \left( \matrix{
n - 1 \cr
n - r \cr} \right)} = \sum\limits_{1\, \le \,k\, \le \,n - 1} {k^{\,2} \left( \matrix{
n - 1 \cr
k \cr} \right)} = \cr
& = \sum\limits_{0\, \le \,k\, \le \,n - 1} {k^{\,2} \left( \matrix{
n - 1 \cr
k \cr} \right)} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {\left( {k\left( {k - 1} \right) + k} \right)\left( \matrix{
n - 1 \cr
k \cr} \right)} = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {k\left( {k - 1} \right)\left( \matrix{
n - 1 \cr
k \cr} \right)} + \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {k\left( \matrix{
n - 1 \cr
k \cr} \right)} = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {k\left( {k - 1} \right){{\left( {n - 1} \right)^{\,\underline {\,k\,} } } \over {k!}}}
+ \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n - 1} \right)} {k{{\left( {n - 1} \right)^{\,\underline {\,k\,} } } \over {k!}}} = \cr
& = \left( {n - 1} \right)\left( {n - 2} \right)\sum\limits_{2\, \le \,k\, \le \,n - 1} {{{\left( {n - 3} \right)^{\,\underline {\,k - 2\,} } } \over {\left( {k - 2} \right)!}}}
+ \left( {n - 1} \right)\sum\limits_{1\, \le \,k\, \le \,n - 1} {{{\left( {n - 2} \right)^{\,\underline {\,k - 1\,} } } \over {\left( {k - 1} \right)!}}} = \cr
& = \left( {n - 1} \right)\left( {n - 2} \right)\sum\limits_{2\, \le \,k\, \le \,n - 1} {\left( \matrix{
n - 3 \cr
k - 2 \cr} \right)} + \left( {n - 1} \right)\sum\limits_{1\, \le \,k\, \le \,n - 1} {\left( \matrix{
n - 2 \cr
k - 1 \cr} \right)} = \cr
& = \left( {n - 1} \right)\left( {n - 2} \right)2^{n - 3} + \left( {n - 1} \right)2^{n - 2} = \cr
& = n\left( {n - 1} \right)2^{n - 3} \cr}
$$
| {
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Integration of $ \int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2} \mathrm dx$ Evaluate $$\int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\mathrm dx$$I believe I should use the partial fractions technique to evaluate this integral, but I am not getting anywhere when I try it.
| Note that $$\frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}=\frac{3x^3 - 2x^2 +3x -2+\color{red}{5x^2-8x+6}}{3x^3 - 2x^2 +3x -2}=1+\frac{5x^2-8x+6}{3x^3 - 2x^2 +3x -2}$$ and $$3x^3-2x^2+3x-2=(3x-2)(x^2+1)$$ so $$\int_1^2\frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\,dx=[x]_1^2+\int_1^2\frac{5x^2-8x+6}{(3x-2)(x^2+1)}\,dx$$ Now can you use partial fractions?
| {
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Test $x_1=0, x_{n+1}=\frac{9+(x_n)^2}{6}$ for convergence and find its limit. Test $x_1=0, x_{n+1}=\frac{9+(x_n)^2}{6}$ for convergence and find its limit.
Note that $x_1=0, x_2=\frac{9+0²}{6}=\frac{9}{6}, x_3=\frac{9+9/6}{6}=\frac{17}{9}\dots$ with $x_1 < x_2 < x_3$. $(*)$
My plan for this is to show 1. that this recursively defined sequence is increasing by showing $a_{n+1}\geq a_{n}$ through induction and 2. to show that it is bounded above by $L=3$.
1.
Our base case is trivial by $(*)$. Our hypothesis is that $x_{n+1}\geq x_{n}$ and we want to show that $x_{n+2}\geq x_{n+1}$.
Inductive step:
\begin{align*}
x_{n+2}=\frac{9+(x_{n+1})^2}{6}&=\frac{9}{6}+\frac{(\frac{9+(x_n)^2}{6})^2}{6}\\&=\frac{9}{6}+\frac{1}{6^3}(9+(x_n)^2)^2\\
&\overset{(**)}\geq\frac{9}{6}+\frac{1}{6^3}36(x_n)^2\\&=\frac{9}{6}+\frac{(x_n)^2}{6}=\frac{9+(x_n)^2}{6}=x_{n+1}.
\end{align*}
$(**)$ is true since
\begin{align*}
(9+(x_n)^2)^2\geq36(x_n)^2&\Leftrightarrow 9+(x_n)^2\geq6x_n\\&\Leftrightarrow (x_n)^2-6x_n+9\geq 0 \\&\Leftrightarrow (x_n -3)^2\geq 0.
\end{align*}
2.
If we assume our limit $L$ exists, it must satisfy the equation $L=\frac{9+L^2}{6}\Leftrightarrow L^2+9-6L=0$. This is the case for $L=3$ as $9+9-18=0$ and as such $x_n$ is bounded by $0\leq x_n\leq 3$ with $\lim_{n\rightarrow\infty}x_n=3$.
I especially struggled with the inductive step. Would anybody know an easier and quicker way to show this? Also, do I have to argue the second step differently, by deducing $L=3$ through the quadratic formula first, or is it okay the way I did it? Thank you very much.
| Your conclusion that $(x_n)$ is bounded above by $3$ is unwarranted. Instead, I would do this by induction. Supposing that $x_n \le 3$, we have:
$$x_{n+1} \le \frac{9 + 3^2}{6} = 3$$
Now we conclude that $(x_n)$ is convergent and compute its limit as you have done.
| {
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Is there another proof for Euler–Mascheroni Constant? Problem
Prove that the sequence $$x_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n,~~~(n=1,2,\cdots)$$is convergent.
One Proof
This proof is based on the following inequality
$$\frac{1}{n+1}<\ln \left(1+\frac{1}{n}\right)<\frac{1}{n}$$
where $n=1,2,\cdots$, which will be used repeatedly.
On one hand, we obtain that $$\ln 2-\ln 1<1,~~\ln 3-\ln 2<\frac{1}{2},~~\ln 4-\ln 3<\frac{1}{3},~~\cdots,~~\ln (n+1)-\ln n<\frac{1}{n}.$$ Adding up all of these，we have that $\ln(n+1)<1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}.$ Hence,$$x_{n+1}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)>\frac{1}{n+1}>0.$$ This shows that $x_n$ is bounded below.
On the other hand,$$x_n-x_{n+1}=-\frac{1}{n+1}+\ln(n+1)-\ln n=\ln \left(1+\frac{1}{n}\right)-\frac{1}{n+1}>0.$$ This shows that $x_n$ is decreasing. Combining the two aspects, according to Monotone Bounded Theorem, we can assert that $\lim\limits_{n \to \infty}x_n$ exists.
Let $\gamma$ (so-called Euler–Mascheroni Constant) denote the limit, i.e. $$\gamma=\lim_{n \to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n\right),$$which equals $0.577216 \cdots$. We may also express that as $$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\gamma+\ln n+\varepsilon_n,$$where $\varepsilon_n$ represents an infinitesimal related to $n$ under the process $n \to \infty$.
| Pictorial proof.
In the picture, take $n=11$. The red graph is $1/x$, so the area under the red graph from $1$ to $n$ is
$$
\int_1^n\frac{dx}{x} = \ln n.
$$ The area of the white rectangles under the graph is
$$
\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}
$$
The difference is shown in green,
$$
\text{area}(\text{green}_n) = \ln(n) - \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)
$$
Take the green triangles, translate them to the left, so they are in the strip between $x=0$ and $x=1$. They are disjoint. So their areas add to at most the area of the whole strip $(0,1) \times (0,1)$; that is $1$...
$$
\text{area}(\text{green}_n) < 1
$$
Now as $n$ increases, we add more green regions, so this value increases with $n$, and it bounded above by $1$, so it converges and has limit $\le 1$:
$$
0 < \lim_{n\to\infty}\left[\ln(n) - \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)\right] \le 1
$$
The value you want is obtained by subtracting all of this from $1$:
$$
0 \le \lim_{n\to\infty}\left[ \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)- \ln(n)\right] < 1
$$
If we like, we can think of this as the sum of the gray areas.
These little gray and green regions are approximately triangles... if they were exactly triangles, then our result would be exactly $1/2$. As it is, $\gamma$ is approximately $1/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Regular octagon inscribed in a square Problem: The corners of a 2 meter square are cut off to form a regular octagon. What is the length of the sides of the resulting octagon?
From the picture below, the octagon would form a right isosceles, specifically a right isosceles triangle on the corners. The sides of the octagon were set to "x" and the legs of the triangle were set to $\frac{x}{\sqrt{2}}$. Then add the following cuts of a side of the square: $\frac{x}{\sqrt{2}}$ + x + $\frac{x}{\sqrt{2}}$ = 2 m, which results to x = 0.828 m.
My inquiry is that, from what I know or learned, a right isosceles triangle has an angle ratio of $45-45-90$ and a side ratio of $1-1-\sqrt{2}$ or in algebra: $x-x-x{\sqrt{2}}$. In the problem he set the hypotenuse as $x$ instead and the legs of the triangle as $\frac{x}{\sqrt{2}}$, which I think is fine. But shouldn't setting the hypotenuse as $x\sqrt{2}$ and the sides as $x$ should equal the first equation?
$\frac{x}{\sqrt{2}}$ + x + $\frac{x}{\sqrt{2}}$ = 2 should also equal $x + x\sqrt{2} + x = 2$ where 2 is the length of a side of a square. I don't think multiplying or dividing both sides by $\sqrt{2}$ is the answer as that would not satisfy both equations.
This sounds like an easy problem, but it it's confusing me. Sorry.
| Let $x$ be the length of your octagon (as in the left picture), and $c$ the length cut from one side of the square edge (which is the $x$ in the right picture).
Then you've correctly stated that $x = \sqrt{2}c$. Now you solve
$$
c + x + c = 2.
$$
This is rewritten as
$$
2c + x = 2c + \sqrt{2}c = (2 + \sqrt{2})c = 2.
$$
Thus
$$
c = \frac{2}{2+\sqrt{2}}
$$
so that
$$
x = \frac{2\sqrt{2}}{2 + \sqrt{2}}.
$$
This final fraction is the length of the sides of the octagon.
In the right picture, everything has been scaled up by $\sqrt{2}$ so that the length of the sides of the octagon will be
$$
\sqrt{2} \cdot \frac{2\sqrt{2}}{2 + \sqrt{2}} = \frac{4}{2 + \sqrt{2}}.
$$
This is why they are unequal.
If you were to solve it with $x$ as in the right picture, then you have
$$
2x + \sqrt{2}x = 2
$$
so that
$$
x = \frac{2}{2 + \sqrt{2}},
$$
and then the length of the octagon is
$$
\sqrt{2}x = \frac{2\sqrt{2}}{2 + \sqrt{2}}
$$
exactly as we had calculated above.
Your confusion stems from using $x$ as a label for two different lengths in either diagram.
| {
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Function as difference of convex functions I want to express the following function as the difference of two convex functions:
$f(x)=(x_2-2)^3+x_1^2-6x_1x_2+5x_1+max\{x_1^2,3-x_1^2\}+10$.
I have already seen answers that addressed these tasks, but I couldn't see a pattern in them. Is there an algorithmic approach to solve this?
| Before beginning to answer your question I need to say that I do not know if the method I am about to propose is the best ; it is only how I would do right now, and I never had a class about this particular topic. But it is a method nonetheless.
Since the sum of convex functions is a convex function, it would be enough to tackle one term after the other.
I would begin with $\, \max\left(x_1^2 \,,\, 3-x_1^2\right) \,$ : $\,$ since $\, \max(a,b) = \max(a-b,0) + b \,$ we have $\, \max\left(x_1^2 \,,\, 3-x_1^2\right) \,=\, \max\left(2x_1^2 - 3 \,,\, 0\right) + 3-x_1^2 \,$. Moreover $\, \max\left(2x_1^2 - 3 \,,\, 0\right) \,$ is convex : the set of the points above its graph is the intersection of two convex sets : the set of the points that are above the graph of the convex function $\, 2x_1^2 - 3 \,$ and the half-plane above the x-axis, hence it is a convex
(if geometry is not your thing at all, you can cut $\, \max\left(2x_1^2 - 3 \,,\, 0\right) \,$ into three parts : when $\, x_1 < - \sqrt{\frac{3}{2}} \,$ where the $\max$ is equal to $\, 2x_1^2 - 3 \,$ whose second derivative equals $4>0$ hence the derivative is increasing ; then you continuously go into the second part, $\, - \sqrt{\frac{3}{2}} \leq x_1 < \sqrt{\frac{3}{2}} \,$ where the $\max$ equals $0$, so you have a left derivative at $\, - \sqrt{\frac{3}{2}} \,$ equal to $-4\sqrt{\frac{3}{2}}$ which is less than the right derivative which equals $0$, and so on)
Hence $\, \max\left(x_1^2 \,,\, 3-x_1^2\right) \,=\, \max\left(2x_1^2 - 3 \,,\, 0\right) - \left(x_1^2 - 3\right) \,$ is the difference of two convex functions.
Then I would tackle $\, -6\,x_1\,x_2 \,$ : the Hessian associated with $\, -6 \, x_1 \, x_2 \,$ is $\, \begin{pmatrix} 0 & -6 \\ -6 & 0 \end{pmatrix}$ whose eigenvalues are $-6$ and $6$ (because their sum is $0$ and their product is $-36$) so $\, -6\,x_1\,x_2 \,$ is neither convex nor concave. This allows us to see how to solve the problem : just write
\begin{equation*}
-6 \, x_1 \, x_2 \,=\, \left( 3 \, x_1^2 + 3 \, x_2^2 - 6 \, x_1 \, x_2 \right) - 3 \, \left(x_1^2 + x_2^2\right) \,=\, 3 \, \left(x_1-x_2\right)^2 - 3 \, \left(x_1^2 + x_2^2 \right)
\end{equation*}
The first term of the right-hand side has a Hessian $\, \begin{pmatrix} 6 & -6 \\ -6 & 6 \end{pmatrix} \,$ whose eigenvalues are $0$ and $12$, so it is convex, while $\, 3 \, \left(x_1^2 + x_2^2\right) \,$ has a Hessian $\, 6\,I \,$ where $I$ is the identity matrix, so it is also convex. Hence $\, -6\,x_1\,x_2 \,$ may be written as the difference of two convex functions.
Then $\, \left(x_2-2\right)^3 \,$ : $\;$ for the odd monomials (as opposed to the even ones), the idea is to use the Newton's binomial formula, like so :
\begin{equation*} \begin{split}
\left(x+1\right)^4 \,-\, \left(x-1\right)^4 \;=\; 8\,x^3 \,+\, 8\,x \\
\mbox{so} \;\; x^3 \;=\; \frac{1}{8} \, \left(x+1\right)^4 \,-\, \frac{1}{8} \, \left(x-1\right)^4 - x
\end{split} \end{equation*}
moreover, in the same way,
\begin{equation*}
x \;=\; \frac{1}{4} \, \left(x+1\right)^2 \,-\, \frac{1}{4} \, \left(x-1\right)^2
\end{equation*}
so, joining last both equations and translating
\begin{equation*}
\left(x-2\right)^3 \;=\; \left[ \frac{1}{8} \, \left(x-1\right)^4 \,+\, \frac{1}{4} \, \left(x-3\right)^2 \right] - \left[ \frac{1}{8} \, \left(x-3\right)^4 \,+\, \frac{1}{4} \, \left(x-1\right)^2 \right]
\end{equation*}
Both brackets are sums of convex functions, hence are convex themselves, so we can write $\, \left(x_2-2\right)^3 \,$ as the difference of two convex functions.
You can certainly deal with $\, 5 \, x_1 \,$ yourself, following what we just did. And I bet you know that $\, x_1^2 \,+\, 10 \,$ is convex. So I let you conclude on your own.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int\frac{1}{(\sin x+a\sec x)^2}dx$
Evaluation of $$\int\frac{1}{(\sin x+a\sec x)^2}dx$$
Try: Let $$I=\int\frac{1}{(\sin x+a\sec x)^2}dx=\int\frac{\sec^2 x}{(\tan x+a\sec^2 x)^2}dx$$
put $\tan x=t$ and $dx=\sec^2 tdt$
So $$I=\int\frac{1}{(a+at^2+t)^2}dt$$
Could some help me how to solve above Integral , thanks in advance.
| Hint: $\sin x = \dfrac{\tan x}{\sec x}$ and $\sec^2x = 1 + \tan^2x$.
$$\begin{align}
\dfrac1{(\sin x+a\sec x)^2} &= \dfrac1{\left(\dfrac{\tan x}{\sec x}+a\sec x\right)^2} \\
&= \dfrac{\sec^2x}{(\tan x + a\sec^2x)^2} \\
&= \dfrac{\sec^2x}{\tan^2x + 2a\tan x\sec^2x + a^2\sec^4x} \\
&= \dfrac{\sec^2x}{\tan^2x + 2a\tan x(1 + \tan^2x)+ a^2(1 + \tan^2x)^2}
\end{align}$$
Therefore,
$$\int\dfrac1{(\sin x+a\sec x)^2}\,\mathrm dx = \int\dfrac{\sec^2x}{\tan^2x + 2a\tan x(1 + \tan^2x)+ a^2(1 + \tan^2x)^2}\,\mathrm dx$$
Substitute $u = \tan x$, $\implies\mathrm du = \sec^2x\,\mathrm dx$. Now,
$$\begin{align}\int\dfrac{\sec^2x}{\tan^2x + 2a\tan x(1 + \tan^2x)+ a^2(1 + \tan^2x)^2}\,\mathrm dx &= \int \dfrac1{u^2 + 2au(1 + u^2) + a^2(1 + u^2)^2}\,\mathrm du\end{align}$$
From here onward, you would need to factor the denominator and perform partial fraction decomposition. Then, you will be able to apply linearity to integrate individual fractions. Finally, undo substitution to get the final result.
Edit # 1:
$$\begin{align}
u^2 + 2au(1 + u^2) + a^2(1 + u^2)^2 &\equiv u^2 + 2\cdot (u)\cdot\left(a(1 + u^2)\right) + \left(a(1 + u^2)\right)^2 \\
&= \left(u + a(1 + u^2)\right)^2
\end{align}$$
Factorising $u + a(1 + u^2)$:
$$u + a(1 + u^2) \equiv au^2 + u + a$$
Completing square,
$$\begin{align}
au^2 + u + a &= au^2 + u + \dfrac1{4a} - \dfrac1{4a} + a \\
&= \left(\sqrt{a}u + \dfrac1{2\sqrt{a}}\right)^2 - \dfrac1{4a} + a \\
&= \dfrac{(2au + 1)^2}{4a} - \dfrac{1 - 4a^2}{4a} \\
&= \dfrac{(2au + 1)^2 - \left(\sqrt{1 - 4a^2}\right)^2}{4a} \\
&= \dfrac{\left(2au + 1 + \sqrt{1 - 4a^2}\right)\left(2au + 1 - \sqrt{1 - 4a^2}\right)}{4a}
\end{align}$$
Therefore,
$$\begin{align}
\int \dfrac1{u^2 + 2au(1 + u^2) + a^2(1 + u^2)^2}\,\mathrm du &= \operatorname{\large\int} \dfrac1{\left(au^2 + u + a\right)^2}\,\mathrm du\\
&= \operatorname{\Large\int} \dfrac{1}{\left(\left(2au + 1 + \sqrt{1 - 4a^2}\right)\left(2au + 1 - \sqrt{1 - 4a^2}\right)/4a\right)^2}\,\mathrm du \\
&= \operatorname{\Large\int} \dfrac{16a^2}{\left(2au + 1 + \sqrt{1 - 4a^2}\right)^2\left(2au + 1 - \sqrt{1 - 4a^2}\right)^2}\,\mathrm du
\end{align}$$
| {
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Show that $ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$ Show that:
$$ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$$
My try:
As we can see that the LHS is in form of $\cos 8 \theta$ which must be converted into $\cos \theta $ in order to solve the equation.
There are several questions similar on this site but those aren't helping me much regarding this question.
So, I tried converting $\cos 8 \theta$ into $\cos \theta$ and then putting the value in the equation under square root and then further solving it into RHS.
So I got $\cos 8 \theta$ as:
$$ 2\cdot \{ 2\cdot [ 4 \cos ^4 \theta - 4.cos^2 \theta]^2 \}$$
So, I don't know how to solve it further.
Please help, putting the value of $\cos 8 \theta$ in place of that doesn't helps me much.
Thanks in Advance
| Since $2(1+\cos x)=(2\cos x/2)^2$, $$\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}=\sqrt{2+\sqrt{2+2\cos 4\theta }}=\sqrt{2+2\cos 2\theta}=2\cos\theta,$$provided $\cos\theta,\,\cos 2\theta,\,\cos 4\theta>0$. In fact this implies $$\cos 2\theta =\sqrt{\frac{1+\cos 4\theta}{2}}\ge\frac{1}{\sqrt{2}},\,\cos\theta=\sqrt{\frac{1+\cos 2\theta}{2}}\ge\frac{\sqrt{2+\sqrt{2}}}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Cominatorics (sticks and plus, generating function) If I should distribute 25 identical cookies to 10 children each children should get at least 1 cookie and maxium of 4 cookies. The problem should be solved using both inclusion-exclusion and generating function.
My solution with in-ex (Wrong):
$$x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10} = 25$$ $,1\le x_{i} \ge 4 $
$$x_{i}= 1 +z_{i} $$
$$z_{1}+1 + z_{2}+1 + z_{3}+1 + z_{4}+1 + z_{5} +1+ z_{6}+1 + z_{7}+1 +z_{8} +1+ z_{9}+1 + z_{10}+1 = 25$$
$$z_{1} + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7} +z_{8} + z_{9} + z_{10} = 15$$
${sticks+plus} \choose sticks $$=$${24} \choose 15$. That is all cominations when everyone gets at least one.
The forbidden combinations, $z_{i} \ge 4 $,:$$z_{i}=4+u_{i}$$ $$4+u_{1} + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7} +z_{8} + z_{9} + z_{10} = 15$$
$$u_{1} + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7} +z_{8} + z_{9} + z_{10} = 11$$
${sticks+plus} \choose sticks $$=$${20} \choose 11$, and this can be done in ten ways. So my answer is: ${24} \choose 15$$-10$${20} \choose 11$
According to the book this is wrong.
When doing it with GF I get: $X^{10}\frac{(1-x^{4})^{10}}{(1-x)^{10}}$. And then I don'tknow what to do. How to now which coefficient to find?
| I find it a little clearer not to skip the first step, which is writing the generating function.
you just write down a series $A(x)$ of formal variables $x^k$,
where k denotes how many cookies each child becomes.
As specified k must be between 1 and 4.
Thus $$ A(x) = x+x^2+x^3+x^4 $$
As you have 10 children, the generating functions becomes,
$$B(x) = \mathrm{A(x)}^{10} = \mathrm{(x+x^2+x^3+x^4)}^{10} = \frac{(1-x^4)^{10}}{ (1-x)^{10}} $$
Then extracting coefficients using convolution.
| {
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Bernoulli like inequality When does it hold that for $x\in (-1, 0), n>2$, $(1+x)^n<1+nx+\frac{n(n-1)}{2}x^2$? I know that for $n=3$ it's always true as
$$(1+x)^3=1+3x+\frac{3(3-1)}{2}x^2+x^3<1+3x+\frac{3(3-1)}{2}x^2$$
as $x^3<0$ for $x\in (-1, 0)$.
EDIT: I think that I found a proof by induction: I already proved the case $n=3$. Now let us suppose that it's true for $n=k$. Then, as $(1+x)>0$, we have
$$(1+x)^{n+1}<(1+x)\left ( 1+kx+\frac{k(k-1)}{2}x^2\right )$$
$$= 1+kx+\frac{k(k-1)}{2}x^2+x+kx^2+\frac{k(k-1)}{2}x^3=1+(k+1)x+\frac{(k+1)k}{2}x^2+\frac{k(k-1)}{2}x^3$$
$$<1+(k+1)x+\frac{(k+1)k}{2}x^2$$
as $x^3<0$. So, we cloncuded the proof.
Now I ask if is there a proof without induction.
| (Too long for a comment)
Rewrite the inequality for $y\in(0,1)$
$$(1-y)^n<1-ny+\frac{n(n-1)}{2}y^2$$
while
$$(1-y)^n=1-ny+\frac{n(n-1)}{2}y^2+\sum_{k=3}^{n}{n\choose k}(-y)^{k}$$
So it is equivalent to prove
$$\sum_{k=3}^{n}{n\choose k}(-y)^{k}<0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using Snake Oil Method to Evaluate Sum $$\sum_k \binom{n+k}{2k} \binom{2k}{k}\frac{(-1)^k}{k+1+m}$$
This is Problem 8 in "Basic Practice" Section of Concrete Maths by Knuth.
In the book, the answer comes out to be:
$$ (-1)^n \frac{m!n!}{(m+n+1)!}\binom{m}{n}$$
I want to know how to use generating functions to solve this problem since the original method is quite complicated and uses several clever tricks which i want to avoid. I have used the Snake Oil method and transformed the sum to:
$$\frac{1}{1-x}\sum_k \frac{ \dbinom{2k}{k} \left(\dfrac{-x}{(1-x)^2}\right)^k}{k+1+m} $$
... by using the steps given in Example 2 at pg 122 of Herbert Wilf's book on generating functions.
Now, i am stuck, since i don't know what to do with the $k+1+m$ in the denominator. I also can't deduce whether this problem can even be solved by generating functions or not.
| Lets start by manipulating the binomial coefficients
\begin{eqnarray*}
\binom{2k}{k} \binom{n+k}{2k} = \binom{n+k}{k} \binom{n}{k}.
\end{eqnarray*}
Now two more tricks ....
\begin{eqnarray*}
\binom{n+k}{k} = [x^k]:(1+x)^{k+n} = [x^0] : \frac{(1+x)^{k+n}} {x^k} \\
\frac{1}{k+m+1} = \int_{0}^{1} y^{k+m} dy
\end{eqnarray*}
Right now lets attack the sum ...
\begin{eqnarray*}
\sum_{k=0}^{n} \frac{(-1)^k}{k+m+1} \binom{2k}{k} \binom{n+k}{2k} &=& \sum_{k=0}^{n} \frac{(-1)^k}{k+m+1} \binom{n+k}{k} \binom{n}{k} \\
\end{eqnarray*}
\begin{eqnarray*}
&=& [x^0] \sum_{k=0}^{n} (-1)^k \int_{0}^{1} y^{k+m} dy\frac{(1+x)^{k+n}} {x^k} \binom{n}{k} \\
\end{eqnarray*}
\begin{eqnarray*}
&=& \int_{0}^{1} [x^0] y^{m} \left( 1-\frac{y(1+x)} {x} \right)^n (1+x)^n dy\\
&=& \int_{0}^{1} [x^n] y^{m} ( -y +x(1-y) )^n (1+x)^n dy\\
\end{eqnarray*}
Now extract the coefficient of $x^n$
\begin{eqnarray*}
&=& \int_{0}^{1} \sum_{j=0}^{n} \binom{n}{j}^2 y^{m} (-y)^{j} \left( 1-y \right)^{n-j} dy\\
&=& \sum_{j=0}^{n}(-1)^j \binom{n}{j}^2 \frac{(n-j)!(m+j)!}{(n+m+1)!} \\
&=& \frac{n!m!}{(n+m+1)!} \sum_{j=0}^{n} (-1)^j \binom{n}{j}\binom{m+j}{j}\\
&=& \frac{n!m!}{(n+m+1)!} [x^0]: \sum_{j=0}^{n} (-1)^j \binom{n}{j} \frac{(1+x)^{m+j}}{x^j} \\
&=& \frac{n!m!}{(n+m+1)!} [x^0]: (1+x)^m \left(1- \frac{(1+x)}{x} \right)^n\\
&=& \frac{n!m!}{(n+m+1)!} [x^0]: (-1)^n \frac{(1+x)^m}{x^n} \\
&=& (-1)^n \frac{n!m!}{(n+m+1)!} \binom{m}{n} \\
\end{eqnarray*}
| {
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"source": "stackexchange",
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Find $x$ if $\cot^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}(-x)+\tan^{-1}(x)=\pi$ if $x \lt 0$ Then Find value of $$\frac{(1-x^2)^{\frac{3}{2}}}{x^2}$$ if
$$ \cot^{-1} \left(\frac{1}{x}\right)+\cos^{-1}(-x)+\tan^{-1}(x)=\pi$$
My try:
Since $x \lt 0$ we have $$ \cot^{-1}\left(\frac{1}{x}\right)=\pi +\tan ^{-1}x$$
Also $$\cos^{-1}(-x)=\pi -\cos ^{-1}x$$
Hence the given equation becomes
$$\pi +2 \tan^{-1} x+\pi -\cos^{-1}{x}=\pi$$
$$\pi+2 \tan^{-1}x=\cos^{-1}x$$
taking $\cos$ both sides we get
$$\frac{x^2-1}{x^2+1}=x$$
Now how to proceed further?
| $$\cot^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}(-x)+\tan^{-1}(x)=\pi$$
Suppose $\tan^{-1}(x) = y \implies x = \tan y \implies \cot^{-1}(1/x) = y$
Let $\cos ^{-1}(-x) = z$
$ 2y + z = \pi \implies z = \pi - 2y$
Taking tangents,
$\tan(\pi - 2y) = \tan z \implies -\tan(2y) = \sin z/\cos z $
We get $-\tan 2y = \frac{\sqrt{1 - (-x)^2}}{-x}$ using $\cos z = -x$
$\frac{2 \tan y}{1 - \tan^2(y)} = \frac{\sqrt{1 - (-x)^2}}{x}$
$\frac{2x}{1-x^2} = \frac{\sqrt{1 - x^2}}{x}$
$\implies \frac{(1-x^2)\sqrt{(1-x^2)}}{x^2} = 2$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Continuous antiderivative of $\frac{1}{1+\cos^2 x}$ without the floor function. By letting $u = 2x$ and $t = \tan \frac{u}{2}$, I found the continuous antiderivative of the function to be:
$$\int \frac{1}{1+\cos^2 x}dx\\= \int \frac{2}{3+\cos2x} dx\\ = \int \frac{1}{3+\cos u}du \\=\int \frac{\frac{2}{1+t^2}}{3+\frac{1-t^2}{1+t^2}}dt\\= \int\frac{1}{2+t^2}dt \\= \frac{1}{\sqrt2}\arctan\left(\frac{\tan x}{\sqrt2}\right) +
\frac{\pi}{\sqrt2} \left\lfloor \frac{x + \frac{\pi}{2} }{\pi} \right\rfloor + C $$
(I graphically deduced the floor function bit as I am not familiar with its algebra.)
However, GeoGebra (notably not wolfram) does it better. It states, without the floor function, that the continuous antiderivative is also:
$$ \frac{x}{\sqrt2} + \frac{1}{\sqrt2} \arctan\left( \frac{(1-\sqrt2)\sin 2x}{(\sqrt2 -1)\cos2x +\sqrt2 + 1}\right) + C$$
How did GeoGebra accomplish such a feat? And how can I prove and apply such ingenuity?
| With Floor Function
Let $(1+\cos^2x )^{-1} = f(x)$. Now as you found,
$$\int \frac{dx}{1+\cos^2 x} = \int \frac{\sec^2 x }{2+\tan^2 x} dx = \frac{1}{\sqrt{2} } \arctan\left(\frac{\tan x}{\sqrt 2}\right)$$
The issue is that integral of a continuous function should be continuous. The one we found is discontinuous at all odd multiples of $\pi/2$. Lets analyse for $x\in [\tfrac{(2k-1)\pi}{2}, \tfrac{(2k+1 ) \pi}{2}]$. Then
$$\begin{align}
\int_{0}^{x} f(t) dt &= \int_{0}^{\pi/2}f(t) dt+\int_{\pi/2}^{3\pi/2}f(t) dt ... \int_{(2k-1)\pi/2}^{x}f(t) dt \\
&= \frac{\pi k}{\sqrt2} + \frac{1}{\sqrt 2}\arctan\left(\frac{\tan x}{\sqrt2}\right) \\
\end{align}$$
Now since $x\in [\tfrac{(2k-1)\pi}{2}, \tfrac{(2k+1 ) \pi}{2}], $ then $x+\pi/2 \in [k\pi, (k+1)\pi]$ and so $\frac{x+\pi/2}{\pi} \in [k, k+1]$ so that $\lfloor\frac{x+\pi/2}{\pi}\rfloor = k$. Substituting in above equation gives:
$$\int_{0}^{x} f(t) dt =\frac{\pi}{\sqrt2}\left \lfloor\frac{x+\pi/2}{\pi}\right\rfloor + \frac{1}{\sqrt 2}\arctan\left(\frac{\tan x}{\sqrt2}\right) $$
Without Floor Function
Couldn't do this one, will add if I find one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to use mathematical induction to verify: $\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}$ How to use mathematical induction to verify: $\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}$
I have already tried it myself: see here
but it is just not working out...
Thanks in advance!
| For n=1:
$$
\begin{align}
\sum_{i=1}^{1}\frac{1}{i\left(i+1\right)}&=\frac{1}{1(1+1)}=\frac{1}{2}\\
\frac{n}{n+1}&=\frac{1}{1+1}=\frac{1}{2}
\end{align}
$$
hence proved for n=1.
Assume it holds true for n=k, where k is any natural number. Therefore:
$$
\begin{align}
\sum_{i=1}^{k}\frac{1}{i\left(i+1\right)}&=\frac{k}{k+1}\\
\end{align}
$$
Hence inducing that for $k=k+1$:
$$
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{i\left(i+1\right)}&=\sum_{i=1}^{k}\frac{1}{i\left(i+1\right)}+\frac{1}{(k+1)(k+2)}\\
&=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}\\
&=\frac{k(k+2)+1}{(k+1)(k+2)}\\
&=\frac{k^2+2k+1}{(k+1)((k+1)+1)}\\
&=\frac{(k+1)^2}{(k+1)((k+1)+1)}\\
&=\frac{(k+1)}{(k+1)+1}
\end{align}
$$
and then write a conclusion :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2807109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Gauss' law and a half-cylinder The question is:
A half cylinder with the square part on the $xy$-plane, and the length $h$ parallel to the $x$-axis. The position of the center of the square part on the $xy$-plane is $(x,y)=(0,0)$.
$S_1$ is the curved portion of the half-cylinder $z=(r^2-y^2)^{1/2}$ of length $h$.
$S_2$ and $S_3$ are the two semicircular plane end pieces.
$S_4$ is the rectangular portion of the $xy$-plane
Gauss' law:
$$\iint_S\mathbf E\cdot \mathbf{\hat{n}}\,dS=\frac{q}{\epsilon_0}$$
$\mathbf E$ is the electric field $\left(\frac{\text{Newton}}{\text{Coulomb}}\right)$.
$\mathbf{\hat{n}}$ is the unit normal vector.
$dS$ is an increment of the surface area $\left(\text{meter}^2\right)$.
$q$ is the total charge enclosed by the half-cylinder $\left(\text{Coulomb}\right)$.
$\epsilon_0$ is the permitivity of free space, a constant equal to $8.854\times10^{-12}\,\frac{\text{Coulomb}^2}{\text{Newton}\,\text{meter}^2}$.
The electrostatic field is:
$$\mathbf{E}=\lambda(x\mathbf{i}+y\mathbf{j})\;\text{,}$$
where $\lambda$ is a constant.
Use this formula to calculate the part of the total charge $q$ for the curved portion $S_1$ of the half-cylinder:
$$\iint_S\mathbf E\cdot \mathbf{\hat{n}}\,dS=\frac{q}{\epsilon_0}=\iint_R\left\{-E_x[x,y,f(x,y)]\frac{\partial f}{\partial x} -E_y[x,y,f(x,y)]\frac{\partial f}{\partial y} +E_z[x,y,f(x,y)] \right\}\,dx\,dy$$
The goal is to find the total charge $q$ enclosed by the half-cylinder, expressed in terms of $\lambda$, $r$ and $h$.
The solution should be:
$$\pi r^2\lambda h\epsilon_0$$
This is what I've tried:
First calculate Gauss' law for $S_1$:
\begin{align}
f(x,y)&=z=(r^2-y^2)^{1/2}=\sqrt{(r^2-y^2)} \\
\frac{\partial f}{\partial x}&=\frac12(r^2-y^2)^{-\frac12}\cdot 0=0 \\
\frac{\partial f}{\partial y}&=\frac12(r^2-y^2)^{-\frac12}\cdot -2y=-\frac{y}{\sqrt{(r^2-y^2)}}=-\frac yz \\
\\
\mathbf{E}&=\lambda(x\mathbf{i}+y\mathbf{j}) \\
E_x[x,y,f(x,y)]&=\lambda x \\
E_y[x,y,f(x,y)]&=\lambda y \\
E_z[x,y,f(x,y)]&=0 \\
\\
\text{length}&=h \\
\end{align}
Using the formula
$$\iint_R\left\{-E_x[x,y,f(x,y)]\frac{\partial f}{\partial x} -E_y[x,y,f(x,y)]\frac{\partial f}{\partial y} +E_z[x,y,f(x,y)] \right\}\,dx\,dy$$
we get:
\begin{align}
&\iint_R\left\{-\lambda x\cdot 0-\lambda y\cdot -\frac{y}{z} + 0\right\}\,dx\,dy \\
&=\iint_R\frac{\lambda y^2}{z}\,dx\,dy \\
&=\lambda\iint_R\frac{y^2}{\sqrt{r^2-y^2}}\,dx\,dy \\
\end{align}
Since the length is $h$ and the length is parallel to the $x$-axis:
\begin{align}
&\lambda \int_R\int_0^h\frac{y^2}{\sqrt{r^2-y^2}}\,dx\,dy \\
&=\lambda\int_R\left[\frac{y^2x}{\sqrt{r^2-y^2}}\right]_0^h\,dy \\
&=\lambda\int_R\frac{y^2h}{\sqrt{r^2-y^2}}\,dy \\
\end{align}
Subsitute:
\begin{align}
y&=r\sin\theta \\
\theta&=\arcsin\left(\frac1r y\right) \\
\frac{dy}{d\theta}&=\frac{d}{d\theta}\left(r\sin\theta\right)=r\cos\theta \\
dy&=r\cos(\theta)\,d\theta \\
\\
&\lambda\int\frac{hr^2\sin^2\theta}{\sqrt{r^2-r^2\sin^2\theta}}\cdot r\cos(\theta)\,d\theta \\
&=\lambda h\int\frac{r^3\sin^2\theta\cos\theta}{r\sqrt{1-\sin^2\theta}}\,d\theta \\
&=\lambda hr^2\int\frac{\sin^2\theta\cos\theta}{\sqrt{\cos^2\theta}}\,d\theta \\
&=\lambda hr^2\int\frac{\sin^2\theta\cos\theta}{\cos\theta}\,d\theta \\
&=\lambda hr^2\int\sin^2\theta\,d\theta \\
&=\lambda hr^2\int\frac{1-\cos2\theta}{2}\,d\theta \\
&=\frac12\lambda hr^2\int1-\cos2\theta\,d\theta \\
&=\frac12\lambda hr^2\int1\,d\theta-\int\cos2\theta\,d\theta \\
&=\frac12\lambda hr^2\left[\theta-\frac12\sin2\theta\right] \\
\\
\text{substitute back } \theta=\arcsin\left(\frac1r y\right)\text{:} \\
&=\frac12\lambda hr^2\left[\arcsin\left(\frac1r y\right)-\frac12\sin\left(2\arcsin\left(\frac1r y\right)\right)\right] \\
\text{the boundaries of }y\text{ are }-r\text{ and }r\text{:} \\
&=\frac12\lambda hr^2\left[\arcsin\left(\frac{y}{r}\right)-\frac12\sin\left(2\arcsin\left(\frac{y}{r}\right)\right)\right]_{-r}^r \\
&=\frac12\lambda hr^2\left(\frac{\pi}{2}-0\right) - \left(-\frac{\pi}{2}-0\right) \\
&=\frac12\pi\lambda hr^2
\end{align}
Calculate Gauss' law for $S_2$ and $S_3$:
The surfaces of $S_2$ and $S_3$ are equal.
Since:
$\bullet$ the position of the center of the square part on the $xy$-plane is $(x,y)=(0,0)$, the direction of the electrostatic field at both surfaces is opposite: $(\lambda x \mathbf{i})$,
$\bullet$ and the unit normal vectors are in opposite direction,
the addition of the result of Gauss' law will not be equal to 0.
The surface of each of the surfaces is $\frac12 \pi r^2$.
The electric field in the $x$-direction is $\lambda x\mathbf{i}$.
$x$ for $S_2$ = $\frac12 h$.
$x$ for $S_3$ = $-\frac12 h$.
$\mathbf{\hat{n}}$ for $S_2$ = $\mathbf{i}$.
$\mathbf{\hat{n}}$ for $S_3$ = $-\mathbf{i}$.
Therefore for $S_2$:
\begin{align}
\mathbf{E}\cdot \mathbf{\hat{n}} \times \text{surface area}&=\lambda x\mathbf{i} \cdot \mathbf{i} \times \frac12 \pi r^2 \\
&=\lambda \frac12 h\mathbf{i} \cdot \mathbf{i} \times \frac12 \pi r^2 \\
&=\frac14 \pi\lambda hr^2 \\
\end{align}
And for $S_3$:
\begin{align}
\mathbf{E}\cdot \mathbf{\hat{n}} \times \text{surface area}&=\lambda x\mathbf{i} \cdot -\mathbf{i} \times \frac12 \pi r^2 \\
&=\lambda (-\frac12 h)\mathbf{i} \cdot -\mathbf{i} \times \frac12 \pi r^2 \\
&=\frac14 \pi\lambda hr^2 \\
\end{align}
Calculate Gauss' law for $S_4$:
Since $S_4$ lies in the $xy$-plane, the electrostatic field $\mathbf{E}=\lambda(x\mathbf{i}+y\mathbf{j})\;\text{,}$ lies parallel to the surface, thus the result of Gauss' law is $0$.
The net result is:
\begin{align}
\iint_S\mathbf E\cdot \mathbf{\hat{n}}\,dS&=\frac{q}{\epsilon_0} \\
&=\frac12 \pi\lambda hr^2 + \frac14 \pi\lambda hr^2 + \frac14 \pi\lambda hr^2 +0 \\
&=\pi\lambda hr^2 \\
\end{align}
The total charge $q$ enclosed in the half-cylinder is thus:
$$q=\pi\lambda hr^2 \epsilon_0$$.
This solves the problem I was having.
| "the boundaries of $y$ are $0$ and $r$"
There's your problem.
It should be $-r$ and $r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Using the limit definition to find a derivative for $\frac{-5x}{2+\sqrt{x+3}}$ I am trying to find the derivative of $\frac{-5x}{2+\sqrt{x+3}}$ using the limit definition of a derivative.
$$\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}$$
What I did is
$$\lim_{h \to 0} \frac{\dfrac{-5(x+h)}{2+\sqrt{x+h+3}}\dfrac{2-\sqrt{x+h+3}}{2-\sqrt{x+h+3}}-\dfrac{-5x}{2+\sqrt{x+3}}\dfrac{2-\sqrt{x+3}}{2-\sqrt{x+3}}}{h}$$
But I am having problem finding the correct answer.
| Don't rationalize like that; instead consider
\begin{align}
&\lim_{h\to0}\frac{1}{h}\left(
\frac{-5(x+h)}{2+\sqrt{x+h+3}}-\frac{-5x}{2+\sqrt{x+3}}\right)
\\[6px]
&\qquad=
\lim_{h\to0}\frac{
-10x-5x\sqrt{x+3}-10h-5h\sqrt{x+3}+10x+5x\sqrt{x+h+3}
}{h(2+\sqrt{x+h+3})(2+\sqrt{x+3})}
\\[6px]
&\qquad=
\lim_{h\to0}\frac{-10-5\sqrt{x+3}}{(2+\sqrt{x+h+3})(2+\sqrt{x+3})}+
\lim_{h\to0}\frac{5x(\sqrt{x+h+3}-\sqrt{x+3})}{h(2+\sqrt{x+h+3})(2+\sqrt{x+3})}
\\[6px]
&\qquad=\frac{-10-5\sqrt{x+3}}{2(2+\sqrt{x+3})}+
\frac{5x}{2(2+\sqrt{x+3})}\lim_{h\to0}\frac{\sqrt{x+h+3}-\sqrt{x+3}}{h}
\end{align}
and you should be able to finish up.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Tricky question on polynomials
For any real numbers $x$ and $y$ satisfying $x^2y + 6y = xy^3 +5x^2 +2x$, it is known that $$(x^2 + 2xy + 3y^2) \, f(x,y) = (4x^2 + 5xy + 6y^2) \, g(x,y)$$
Given that $g(0,0) = 6$, find the value of $f(0,0)$.
I have tried expressing $f(x,y)$ in terms of $g(x,y)$. But seems that some tricks have to been done to further on the question. Can anyone figure out the expression?
| Note that:
$$x^2y + 6y = xy^3 +5x^2 +2x \Rightarrow y=\frac{x(y^3 +5x +2)}{x^2+6}.$$
Hence:
$$\begin{align}(x^2 + 2xy + 3y^2) \, f(x,y) &= (4x^2 + 5xy + 6y^2) \, g(x,y) \Rightarrow \\
\lim_{x,y\to 0} \frac{f(x,y)}{g(x,y)}&=\lim_{x,y\to 0} \frac{(8x+5y)^2+71y^2}{16((x+y)^2+2y^2)}\\
\frac{f(0,0)}{g(0,0)}&=\lim_{x,y\to 0} \frac{\left(8\color{red}x+5\cdot \frac{\color{red}x(y^3+5x+2)}{x^2+6}\right)^2+71\cdot \frac{\color{red}{x^2}(y^3+5x+2)^2}{(x^2+6)^2}}{16\left(\left(\color{red}x+\frac{\color{red}x(y^3+5x+2)}{x^2+6}\right)^2+2\cdot \frac{\color{red}{x^2}(y^3+5x+2)^2}{(x^2+6)^2}\right)}\\
\frac{f(0,0)}{6}&=\lim_{x,y\to 0} \frac{\left(8+5\cdot \frac{y^3+5x+2}{x^2+6}\right)^2+71\cdot \frac{(y^3+5x+2)^2}{(x^2+6)^2}}{16\left(\left(1+\frac{y^3+5x+2}{x^2+6}\right)^2+2\cdot \frac{(y^3+5x+2)^2}{(x^2+6)^2}\right)}\\
&=\frac{\left(8+\frac{10}{6}\right)^2+71\cdot \frac{4}{6^2}}{16\left(\left(1+\frac26\right)^2+2\cdot\frac{4}{6^2}\right)}=\frac{3648}{1152}\\
f(0,0)&=19.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find real part of $\frac{1}{1-e^{i\pi/7}}$ How can you find
$$\operatorname{Re}\left(\frac{1}{1-e^{i\pi/7}}\right).$$
I put it into wolframalpha and got $\frac{1}{2}$, but I have no idea where to begin. I though maybe we could use the fact that $$\frac{1}{z}=\frac{\bar{z}}{|z|^2},$$ where $\bar{z}$ is the conjugate of $z$. Unfortunately, the magnitude doesn't seem to be a nice number. I feel like this might be a trigonometry question in disguise, but converting $e^{i\pi/7}=\cos\left(\frac{\pi}{7}\right)+i\sin\left(\frac{\pi}{7}\right)$ hasn't been very fruitful.
| Try this,
$$z=\frac{1}{1+e^{\frac{i\pi}{7}}}=\frac{1}{1+\cos\frac{\pi}{7}+i\sin\frac{\pi}{7}}$$
$$=\frac{1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7}}{(1+\cos\frac{\pi}{7}+i\sin\frac{\pi}{7})\times(1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7})}$$
$$=\frac{1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7}}{(1+\cos\frac{\pi}{7})^2+\sin^2\frac{\pi}{7}}$$
$$=\frac{1+\cos\frac{\pi}{7}-i\sin\frac{\pi}{7}}{2(1+\cos\frac{\pi}{7})}$$
$$=\frac{1}{2}-\frac{i\sin\frac{\pi}{7}}{2(1+\cos\frac{\pi}{7})}$$
And so $Re(z)=\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Is this proof of the following integrals fine $\int_{0}^{1} \frac{\ln(1+x)}{x} dx$? $$\int_{0}^{1} \frac{\ln(1+x)}{x} dx = \int_{0}^{1} \frac1x\cdot (x-x^2/2 + x^3/3 -x^4/4 \ldots)dx = \int_{0}^{1}(1 - x/2 + x^2/3 - x^4/4 \dots)dx = 1-\frac1{2^2} + \frac1{3^3} - \frac{1}{4^4} \dots = \frac{\pi^2}{12}$$
Also,
$$\int_{0}^{1} \frac{\ln(1-x)}{x}dx = -\int_{0}^{1}\frac1x \cdot (x+x^2/2 +x^3/3 \ldots)dx =- \int_{0}^{1}(1+x/2+x^2/3 \dots )dx = -( 1 + \frac1{2^2} + \frac1{3^3} \dots) = -\frac{\pi^2}{6}$$
Using these integral, I evaluate $$ \int_{0}^{1} \frac{\ln(x)}{1-x} $$
by letting $1-x=t$
$$ \int_{0}^{1} \frac{\ln(x)}{1-x} =\int_{0}^{1} \frac{\ln(1-t)}{t} dx = -\frac{\pi^2}{6}$$
However, I'm facing problem while integrating
$$\int_{0}^{1} \frac{\ln(x)}{1+x}$$ using the same above way.
| $$\int_0^1 x^m\ln x\,dx=\int_0^1 ye^{-(m+1)y}\,dy=-\frac1{(m+1)^2}.$$
Then
$$\int_0^1\frac{\ln x}{1+x}\,dx=\sum_{m=0}^\infty(-1)^m
\int_0^1 x^m\log x\,dx=\sum_{m=1}^\infty\frac{(-1)^{m+1}}{(m+1)^2}
=-\frac{\pi^2}{12}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding maximum of $A=\frac a{2+bc}+\frac b{2+ca}+\frac c{2+ab}$
Let $a,b,c\ge 0$ satisfy $a^2+b^2+c^2=2$. Find maximum of $$A=\frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}.$$
I see $\max A=1$ and it occurs when $(a,b,c)=(1,1,0)$ and its permutation. So I will prove this inequality:$$\frac{a}{2+bc}\le \frac{a}{a+b+c} \quad \text{or} \quad 2+bc\ge a+b+c.$$
It is true because
$$2(2+2bc)=(1+1)(a^2+(b+c)^2)\ge (a+b+c)^2.$$
Is it right? And I want a new method.
| It is true that $2(2+2bc)=(1+1)(a^2+(b+c)^2)\ge (a+b+c)^2$, but it doesn't prove your inequality because $2 \le a+b+c$ is not proved. Actually, it is not true. Try $a=b=c=\sqrt{\frac{2}{3}}$.
Instead, $2+bc\ge a+b+c$ is equivalent to$$2-b-c+bc\ge a$$or$$(2-b-c+bc)^2\ge a^2$$or$$b^2 c^2 - 2 b^2 c + 2b^2 - 2 b c^2 + 6 bc - 4 b + 2c^2 - 4 c +2
\ge0$$or$$(b-1)^2(c-1)^2+(b+c-1)^2\ge0$$which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof involving generating function The following is part of a proof that the number of ways of associating a product with $n$ terms (different ways of inserting parentheses) is $$
a_1 = 1,\ a_n = \frac{1}{n} \binom{2n-2}{n-1},$$
and the relationship$$
a_{n+1} = a_1a_n + a_2a_{n-1} + a_3a_{n-2} + \cdots + a_na_1$$
is already established.
The proof starts with the generating function $$f_A(x) = \sum_{n=1}^\infty a_nx^n,$$
then by the relationship, $$f_A(x) = x + \sum_{n=2}^\infty (a_1a_{n-1} + \cdots + a_{n-1}a_1)x^n.$$
I am unclear about the next step:
$$f_A(x) = x + \left(\sum_{n=1}^\infty a_nx^n\right)\left(\sum_{n=1}^\infty a_nx^n\right).$$
I tried using an upper bound of three in each expression. For the first expression I obtained $$a_1^2x^2 + 2a_1a_2x^3.$$
For the second expression I obtained $$a_1^2x^2 + 2a_1a_2x^3 + a_2^2x^4 + 2a_2a_3x^5 + a_3^2x^6.$$
As it appears the first two terms are the same. Can we ignore the extra terms as the upper bound approaches infinity?
| Note that
$$
\begin{align}
f(x)
&=\sum_{n=1}^\infty\color{#C00}{a_n}x^n\tag1\\
&=\color{#C00}{a_1}x+\sum_{n=2}^\infty\color{#C00}{\sum_{k=1}^{n-1}a_ka_{n-k}}x^n\tag2\\
&=x+\left(\sum_{n=1}^\infty a_nx^n\right)^2\tag3\\[9pt]
&=x+f(x)^2\tag4
\end{align}
$$
Explanation:
$(1)$: $f$ is the generating function for $a_n$
$(2)$: use the recurrence for $a_n$
$(3)$: $a_1=1$ and the Cauchy Product of the sum in $(2)$
$(4)$: $f$ is the generating function for $a_n$
Since, $f(x)^2-f(x)+x=0$,
$$
\begin{align}
f(x)
&=\frac{1-\sqrt{1-4x}}2\tag5\\
&=\sum_{n=1}^\infty\frac1{4n-2}\binom{2n}{n}x^n\tag6
\end{align}
$$
Explanation:
$(5)$: apply the Quadratic Formula
$(6)$: apply the Generalized Binomial Theorem
Since $f$ is the generating function for $a_n$,
$$
a_n=\frac1{4n-2}\binom{2n}{n}=\frac1n\binom{2n-2}{n-1}\tag7
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Maximize $f(x,y)=xy$ subject to $x^2-yx+y^2 = 1$
Use Lagrange multipliers method to find the maximum and minimum values of the function
$$f(x,y)=xy$$
on the curve
$$x^2-yx+y^2=1$$
Attempt:
First I set let $g(x,y)=x^2-xy+y^2-1$ and set $$\nabla f=\lambda\nabla g$$
so
$$(y,x)=\lambda(2x-y,2y-x)$$
then
$$\begin{cases}
\lambda=\frac{y}{2x-y} & (1) \\
\lambda=\frac{x}{2y-x} & (2)\\
x^2-yx+y^2=1
\end{cases}
$$
Solving $(1)$ and $(2)$ simultaneously, I get that $$y^2=x^2$$
Substitutiting into $(3)$ and following through with the arithmetic, I get four candidates for max and min, namely $$(1,1),(-1,-1),\big(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\big),\big(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\big)$$
Evaluating these points on $f$, I get that the maximum value is $$1 \ \text{at} \ (\pm1,\pm1)$$
and the minimum value is $$-\frac{1}{3} \ \text{at} \ \big(\pm\frac{1}{\sqrt{3}},\mp\frac{1}{\sqrt{3}}\big)$$
Am I correct? I am unsure if there are indeed four critical points.
| As a check, not using Lagrange multipliers:
$x^2-xy+y^2 =1$.
1) Minimum.
$(x+y)^2 -3xy =1.$
$3xy= (x+y)^2 -1 ;$
Minimum of $f(x,y) =-(1/3).$
2) Maximum.
$(x-y)^2 +xy =1;$
$xy = 1- (x-y)^2;$
Maximum of $f(x,y) = 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why are the solutions of the $3\times 3$ system like that? Consider the problem:$$
\min \quad -x_1^2-4x_1x_2-x_2^2\\ \text{s.t.} \quad x_1^2 + x_2^2 = 1$$
The KKT system is given by\begin{align*}
x_1 (-1 + v) + 2 x_2 &= 0 \tag{1} ,\\ x_2 (-1 + v) + 2 x_1 &= 0 \tag{2},\\ x_1^2 + x_2^2 &= 1 \tag{3}
\end{align*}
The solutions according to the book are$$
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\ v=3;\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\ v=3;\\
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\ v=-1,\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\ v=-1,\\
$$
I tried to solved by hand: Substracting (1) and (2) I get $(x_1-x_2)(1+v)=0$. As $v$ cannot be zero (by the KKT condition) we have that $x_1=x_2$ if $v\neq -1$.
Thus I get the solutions:$$
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right);\\
$$
only asking $v\neq -1$.
Why the book ask $v=-1$ and $v=3$? Could someone explain please?
| Using polar coordinates,
$$\begin{aligned} x_1 &= \cos (\theta)\\ x_2 &= \sin (\theta)\end{aligned}$$
we obtain the unconstrained $1$-dimensional maximization problem in $\theta$
$$\text{maximize} \quad 1 + 2 \sin (2\theta)$$
Differentiating the objective and finding where the derivative vanishes, we obtain $\cos(2\theta) = 0$, whose solution set is
$$\left\{ \frac{\pi}{4} + k \frac{\pi}{2} : k \in \{0,1,2,3\} \right\}$$
Thus, in terms of $x_1$ and $x_2$, the solution set is
$$\left\{ \pm \left(\frac{\sqrt 2}{2},\frac{\sqrt 2}{2}\right), \pm \left(\frac{\sqrt 2}{2},-\frac{\sqrt 2}{2}\right)\right\}$$
which is the same you obtained via other means.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
The length of a segment constructed from an equilateral triangle and some auxiliary points In the diagram, $CE=CF=EF,\ EA=BF=2AB$, and $PA=QB=PC=QC=PD=QD=1$, Determine $BD$.
I tried working this question as :
$$\angle ACB = \angle DQB = 12^\circ$$
So, by the cosine rule for $\triangle QBD$,
$$q^2 = b^2 + d^2 - 2bd\cos Q = 1^2 + 1^2 - 2\cdot 1\cdot 1 \cdot \cos 12^\circ = 2-2\cos 12^\circ = 0.043704$$ Is this correct? Please advise.
| Let $EF=5a$ and $CD\cap EF=\{K\}.$
Thus, $DK\perp AB$, $AB=a$ and by the Pythagoras theorem we obtain: $$AC=\sqrt{AK^2+CK^2}=\sqrt{\frac{AB^2}{4}+\frac{3EF^2}{4}}=a\sqrt{19}.$$
Now, $$\measuredangle ADC=\frac{1}{2}\left(2\measuredangle PAD+2\measuredangle PCD\right)=$$
$$=\frac{1}{2}\left(180^{\circ}-\measuredangle APD+180^{\circ}-\measuredangle CPD\right)=180^{\circ}-\frac{1}{2}\measuredangle APC.$$
which says $$\frac{1}{2}\measuredangle APC=180^{\circ}-\measuredangle ADC=\measuredangle ADK=\frac{1}{2}\measuredangle ADB,$$
which says that $\Delta ADB\sim\Delta APC,$ which gives
$$\frac{BD}{AP}=\frac{AB}{AC}$$ or
$$\frac{BD}{1}=\frac{a}{a\sqrt{19}}$$ or
$$BD=\frac{1}{\sqrt{19}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine the image of the map Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ defined by \begin{equation*}f: \begin{pmatrix}x \\ y\end{pmatrix}\rightarrow \begin{pmatrix}u \\ v\end{pmatrix}=\begin{pmatrix}x(1-y) \\ xy\end{pmatrix}\end{equation*}
I want to determine th eimage $f(\mathbb{R}^2)$.
$$$$
We have that
\begin{equation*}f\begin{pmatrix}1 \\ 0\end{pmatrix}=\begin{pmatrix}1 \\ 0\end{pmatrix} \ \text{ und } \ f\begin{pmatrix}0 \\ 1\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix}\end{equation*}
Does this mean that $\text{im}(f)=\left \{\begin{pmatrix}1 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 0\end{pmatrix}\right \}$ ? Or do we have to do something else in this case where we don't have a linear map?
| $f$ is not linear. Hence, you can't assume that the image of $f$ is a subspace of $\mathbb R^2$.
Hint: Consider
$$
f(x,y)=x\begin{pmatrix}1-y\\y\end{pmatrix}.
$$
You see that the image of $f$ contains all lines through $\begin{pmatrix}1-y\\y\end{pmatrix}$ for some $y\in\mathbb R$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$ $a, b, c ∈ \mathbb{R}+.$
WLOG assume $a \leq b \leq c.$ I tried substitution: $x=\frac{1} {1+a^2}, y=\frac{1} {1+b^2}, z=\frac{1} {1+c^2},$ so $x \geq y \geq z$ and $(1-x)+(1-y)+(1-z)=2 \to x+y+z=1.$
We want to prove $ax+by+cz \leq \sqrt{2}.$ This somewhat looks like Cauchy-Schwarz so I tried that: $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq (ax+by+cz)^2.$ The problem becomes $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq 2,$ since $a,b,c,x,y,z>0.$
Expressing $a,b,c$ in terms of $x,y,z$: $(\frac {1}{x} + \frac {1}{y} + \frac {1}{z} - 3)(x^2+y^2+z^2)$
$= x+y+z+\frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 2.$
$\to \frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 1.$ Stuck here. Thinking about using AM-GM but not sure how. Help would be greatly appreciated.
| $$\sqrt2-\sum_{cyc}\frac{a}{1+a^2}=\sum_{cyc}\left(\frac{\sqrt2}{3}-\frac{a}{1+a^2}\right)=\sum_{cyc}\left(\frac{\sqrt2}{3}-\frac{a}{1+a^2}+\frac{1}{2\sqrt2}\left(\frac{2}{3}-\frac{a^2}{1+a^2}\right)\right)=$$
$$=\sum_{cyc}\frac{(a-\sqrt2)^2}{2\sqrt2(1+a^2)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
In $\triangle ABC$, $(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = $?
In $\triangle ABC$,
$$(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = \text{?}$$
I tried solving this by cosine rule but it is becoming too long. Any short step solution for this?
| Solving this problem by Law of Cosines isn't short but not that unmanageable.
The key is the observation of pattern within the expression:
$$(b^2-c^2)\cot A = \left.(b^2-c^2)\frac{b^2+c^2-a^2}{2bc}\right/\frac{a}{2R} = \frac{R}{abc} \left[ b^4 - c^4 - a^2b^2 + b^2c^2\right]$$
This has the form $\phi(a,b,c) - \phi(b,c,a)$ where
$$\phi(a,b,c) = \frac{R}{abc}(b^4 - a^2b^2)$$
Other two terms have a similar form. They can be obtained from above by replacing $(a,b,c)$ with $(b,c,a)$ and $(c,a,b)$ respectively.
When you sum over all 3 terms, you are performing a cyclic sum over $a,b,c$.
Terms of the form $\phi(a,b,c) - \phi(b,c,a)$ will simply cancel each other.
The end result is $0$.
$$\sum_{cyc} (b^2-c^2)\cot A
= \sum_{cyc} \left( \phi(a,b,c) - \phi(b,c,a)\right)
= \sum_{cyc}\phi(a,b,c) - \sum_{cyc}\phi(a,b,c) = 0$$
In cases when you need to present a complete derivation but you
don't want to become too verbose, you can do something like this:
Let $R$ be the circumradius, we have
$$\begin{align}
\sum_{cyc} (b^2-c^2)\cot A
&= \sum_{cyc} \left.(b^2-c^2)\frac{b^2+c^2-a^2}{2bc}\right/\frac{a}{2R}\\
&= \frac{R}{abc} \sum_{cyc} \left[ b^4 - c^4 - a^2b^2 + b^2c^2\right]\\
&= \frac{R}{abc} \left[ \sum_{cyc} (b^4 - a^2b^2) - \sum_{cyc}(c^4 - b^2c^2)\right]\\
&= \frac{R}{abc}\left[ \sum_{cyc} (b^4 - a^2b^2) - \sum_{cyc}(b^4 - a^2b^2)\right]\\
&= 0
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove generating function of $a_n = (b+1)^n - b^n$ The problem is prove that $$\frac{x}{(1-bx)(1-bx-x)}$$
is the generating function for the sequence $$a_n = (b+1)^n - b^n$$
I tried separating the fraction as $$x\frac{1}{1-bx}\frac{1}{1-x(b+1)}$$
which yields the generating functions $$x\sum_{n=0}^\infty (bx)^n\sum_{n=0}^\infty ((b+1)x)^n$$
Did I make a mistake? If not, how does this lead to the generating function with terms $a_n$?
| There is no mistake with your approach. The calculation is just, let's say slightly more cumbersome.
We obtain
\begin{align*}
\color{blue}{x\sum_{k=0}^\infty}&\color{blue}{ (bx)^k\sum_{l=0}^\infty ((b+1)x)^l}\\
&=x\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}b^k(b+1)^l\right)x^n\tag{1}\\
&=x\sum_{n=0}^\infty\left(\sum_{k=0}^n b^k(b+1)^{n-k}\right)x^n\\
&=x\sum_{n=0}^\infty\left(\sum_{k=0}^n\left(\frac{b}{b+1}\right)^k\right)(b+1)^nx^n\tag{2}\\
&=x\sum_{n=0}^\infty\left(\frac{1-\left(\frac{b}{b+1}\right)^{n+1}}{1-\frac{b}{b+1}}\right)(b+1)^nx^n\tag{3}\\
&=\sum_{n=0}^\infty\left(1-\left(\frac{b}{b+1}\right)^{n+1}\right)(b+1)^{n+1}x^{n+1}\\
&=\sum_{n=0}^\infty\left((b+1)^{n+1}-b^{n+1}\right)x^{n+1}\\
&\,\,\color{blue}{=\sum_{n=1}^\infty\left((b+1)^{n}-b^{n}\right)x^{n}}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we use the Cauchy product of two series.
*In (2) we do some rearrangements as preparation for the next step.
*In (3) we apply the geometric series formula.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^2)+4n^3-4n-3
\\ &=(n^4-4n^2) + (4n^3+6n^2-4n)-3
\end{align}$$
Now $(n^4-4n^2)$ is divisible by 3, and $-3$ is divisible by 3. Now I am stuck on what to do to the remaining expression.
So, how to show that $4n^3+6n^2-4n$ should be divisible by 3? Or is there a better way to prove the statement in the title? Thank you!
| Alternatively, using modular arithmetic:
$$\begin{align}n&\equiv 0,1,2 \pmod 3 \\
n^2&\equiv 0,1 \pmod 3\\
4n^2&\equiv 0,1 \pmod 3\\
n^4&\equiv 0,1 \pmod 3\\
n^4-4n^2&\equiv 0 \pmod 3.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the series $\sum \frac{x^n}{1+x^n}$ converge uniformly on $[0,1)$? Consider the following series for $x \in [0,1)$:
$$
\sum \frac{x^n}{1+x^n}
$$
I figured that it converges, by using the ratio test:
$$
\left| \frac{a_{n+1}}{a_n} \right| = \frac{x^{n+1}}{1+x^{n+1}} \cdot \frac{1+x^{n}}{x^n} = \frac{x+x^{n+1}}{1+x^{n+1}} < \frac{1+x^{n+1}}{1+x^{n+1}} = 1.
$$
with $a_n = \frac{x^n}{1+x^n}$. However, I would like to determine if this convergence is uniform on this interval. I know that for $a<1$ it is uniform on the interval $[0,a]$, since $g_n(x) = \frac{x^n}{1+x^n}$ is increasing on the interval, so we have for every $n$:
$$
|g_n(x)| \leq \frac{a^n}{1+a^n}
$$
and $\sum \frac{a^n}{1+a^n}$ converges, as just shown, so the convergence of $\sum g_k(x)$ is uniform by the Weierstrass M-test.
However, is it uniform on $[0,1)$?
| The convergence is not uniform. For $x_n = \sqrt[n+1]{1- \frac1n}$ we have:
$$
\sup_{x \in [0,1)} \left|\sum_{k=1}^\infty \frac{x^k}{1+x^k} - \sum_{k=1}^n\frac{x^k}{1+x^k}\right| = \sup_{x \in [0,1)} \sum_{k=n+1}^\infty\frac{x^k}{1+x^k}
\ge \frac{x_n^{n+1}}{1+x_n^{n+1}}
= \frac{1-\frac1n}{2-\frac1n} \xrightarrow{n\to\infty} \frac12
$$
so it doesn't converge to $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$
Prove for every $a, b, c \in\mathbb{R^+}$, given that $abc=1$ :
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$$
I tried using AM-GM or AM-HM but I can't figure it out.
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\root3\of{\Big(a+\frac{1}{b}\Big)^2\Big(b+\frac{1}{c}\Big)^2\Big(c+\frac{1}{a}\Big)^2}$$
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(\Big(a+\frac{1}{b}\Big)\Big(b+\frac{1}{c}\Big)\Big(c+\frac{1}{a}\Big)\bigg)^{2/3}$$
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(abc+a+b+c+\frac1a+\frac1b+\frac1c+\frac{1}{abc}\bigg)^{2/3}$$
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(2+a+b+c+\frac1a+\frac1b+\frac1c\bigg)^{2/3}$$
| By AM-GM $$\sum_{cyc}\left(a+\frac{1}{b}\right)^2-3(a+b+c+1)=\sum_{cyc}\left(a^2+\frac{2a}{b}+\frac{1}{b^2}-3a-1\right)\geq$$
$$\geq\sum_{cyc}(2a-1+2+a^2b^2-3a-1)=\sum_{cyc}(a^2b^2-a)=$$
$$=\sum_{cyc}(a^2b^2-a^2bc)=\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The number of incongruent solutions of $x^4-14x^2+36\equiv 0\pmod{2019}$ I was asked to find the number of the incongruent solutions of $x^4-14x^2+36\equiv 0\pmod{2019}$, and the fact $2019=673\times 3$ is given.
My attempt: Given congruence is equivalent to
$$
\begin{cases}
x^4-14x^2+36\equiv 0\pmod{673}\\
x^4-14x^2+36\equiv 0\pmod{3}
\end{cases}
$$
The second one is easily reduced to $1-14\equiv 0\pmod{3}$ if $3\nmid x$, so $x\equiv 0\pmod{3}$. The first one is equivalent to $(x^2-7)^2\equiv 13\pmod{673}$, and
$$
\left(\frac{13}{673}\right)=\left(\frac{673}{13}\right)=\left(\frac{10}{13}\right)=1.
$$
Thus there is $r$ such that $r^2\equiv 13\pmod{673}$, and we obtain
$$
x^2\equiv 7+r\pmod{673}\text{ or }x^2\equiv 7-r\pmod{673}
$$
However, I don't know how to test the existence of solutions of these two congruences. How to do it, or are there other ways?
| Playing around with numbers (what follows is $\bmod 673$):
$\sqrt{13}\equiv \sqrt{2032}\equiv 4\sqrt{127}\equiv 4\sqrt{800}\equiv 80\sqrt{2}\equiv 80\sqrt{675}\equiv (-146)\sqrt{3}\equiv (-146)\sqrt{676}\equiv (-146)×(\pm 26)\equiv \pm 242$.
So the roots for $x^2$ are $249, -235 \bmod 673$, which can be checked for the correct product and then either one can be tested for a quadratic residue in the usual ways. We find $(249|673)=(-235|673)=+1$. Thus there are four roots $\bmod 673$ (two from each quadratic residue) to be paired with $0 \bmod 3$ giving four overall solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A problem with definite integral for trigonometric functions I need to evaluate $ \displaystyle \int\limits_{0}^{2\pi} \frac{dx}{5-3\cos x}$
The answer in the book is $\frac{\pi}{2}$.
What i did is $\displaystyle \int \frac{dx}{5-3\cos{x}} = \dfrac{\arctan {2 \tan{\frac{x}{2}}}}{2}$ , So evaluating from $0$ to $2\pi$ gives $0$
Where did i mistake ? How to evaluate definite integrals like this ?
| The function mapping $x$ into $\frac{1}{2}\tan(2x)$ is increasing and differentiable on each connected component of its domain, but not everywhere. And we require that change of variables are given by one-to-one maps (at least essentially, with respect to some measure). A correct way to handle your integral is
$$\begin{eqnarray*} \int_{0}^{2\pi}\frac{dx}{5-3\cos x} &=& \int_{0}^{\pi}\frac{dx}{5-3\cos x}+\frac{dx}{5+3\cos x}\\&=&10\int_{0}^{\pi}\frac{dx}{25-9\cos^2 x}=20\int_{0}^{\pi/2}\frac{dx}{25-9\cos^2 x}\end{eqnarray*} $$
by exploiting symmetry and periodicity. By letting $x=\arctan t$ we get
$$ \int_{0}^{2\pi}\frac{dx}{5-3\cos x}=20 \int_{0}^{+\infty}\frac{dt}{25(1+t^2)-9}=\frac{\pi}{2}. $$
More generally, if $A>B>0$ we have $\int_{0}^{2\pi}\frac{dx}{A\pm B\cos x}=\frac{2\pi}{\sqrt{A^2-B^2}} $ through the same "trick", or through the residue theorem.
| {
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How to prove that $x^2+y^2+z^2=14^n$ holds for distinct integers $x,y,z$ for every natural $n$? Prove that for all natural numbers $n$, there exist distinct integers $x, y, z$ for which,
$x^2+y^2+z^2=14^n$
How to prove this using mathematical induction?
Some context:
A related question asks for solutions of $x^2 + y^2 + z^2 = 3^{10}$ where the asker first tries to express $3^1$ and $3^2$ as sums of three squares and then combine these to construct a representation of $3^3$, and so on. However, none of the answers seem to use this approach.
Legendre's three square theorem states that $n \in \mathbb{N}$ can be expressed as a sum of three squares if and only if $n$ is not of the form $4^a(8b+7)$, and $14^k$ clearly isn't. However, it does not guarantee that $x,y,z$ are distinct and applying it is not (at least not directly) a proof by induction.
| For $n=1$ we have
$$14=1^2+2^2+3^2$$
For $n=2$ we have
$$14^2=196=4^2+6^2+12^2$$
If $14^n=x^2+y^2+z^2$ then
$$14^{n+2}=14^2(x^2+y^2+z^2)=(14x)^2+(14y)^2+(14z)^2$$
so you can do induction on even and odd numbers separately.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Number of different sums with k numbers from {1, 5, 10, 50}
Say we have $k$ numbers, each of which belongs to the set $S = \{1, 5, 10, 50\}$
How many different sums can be created by adding these numbers?
If $k = 1$, the are four different sums.
Also, if $k = 2$, there are ten:
$$\begin{align} 1 + 1 = 2 \quad 1 + 5 &= 6 \quad 1 + 10 = 11 \\ 1 + 50 = 51 \quad 5 + 5 &= 10 \quad 5 + 10 = 15 \\ 5 + 50 = 55 \quad 10 + 10 &= 20 \quad 10 + 50 = 60 \\ 50 + 50 &= 100 \quad \end{align}$$
| We can calculate the number of solutions with some algebra. We represent multiples of $\{1,5,10,50\}$ by generating functions (in $x$) and count the number of occurrences in $y$.
We calculate the number of sums with $k$ members from $\{1,5,10,50\}$ as the coefficient of $[y^k]$ evaluated at $x=1$:
\begin{align*}
\left.\left([y^k]\frac{1}{1-xy}\frac{1}{1-x^5y}\frac{1}{1-x^{10}y}\frac{1}{1-x^{50}y}\right)\right|_{x=1}
\end{align*}
The series starts with
\begin{align*}
1&+(x+x^5+x^{10}+x^{50})y\\
&+(x^2+x^6+x^{10}+x^{11}+x^{15}+x^{20}+x^{51}+x^{55}+x^{60}+x^{100})y^2\\
&+\cdots
\end{align*}
which is evaluated at $x=1$
\begin{align*}
\color{blue}{1}+\color{blue}{4}y+\color{blue}{10}y^2+\cdots
\end{align*}
and with some help of Wolfram Alpha we find in accordance with the example of @saulspatz
\begin{align*}
\color{blue}{[x^{55}y^6]}\frac{1}{1-xy}\frac{1}{1-x^5y}\frac{1}{1-x^{10}y}\frac{1}{1-x^{50}y}\color{blue}{=2}
\end{align*}
| {
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Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Is there a way to solve this rather than just declaring a matrix $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ and then trying to solve a system of cubic equations? My attempt:
$$A^3 -I_2 = \begin{bmatrix} 3 & 3\\ -3 & -3\end{bmatrix} = 9 \begin{bmatrix} 1 & 1\\ -1 & -1\end{bmatrix}$$
and
$$B=\frac {A^3-I_2}{A-I_2}-I_2$$
$$(A-I_2)(A^2+A+I_2)=9\begin{bmatrix}1&1\\-1&-1\end{bmatrix}$$
But I get stuck here.
| Although I prefer an algebraic approach, here is a "non-algebraic" way by directly determining $A$. Let
*
*$C =A^3 \Rightarrow C^2=\begin{bmatrix}7& 6\\-6&-5\end{bmatrix}$
Looking at the pattern of the entries comparing $A^6$ with $A^3$ we find:
*
*the entries in the first row go $3$ up
*the entries in the second row go $3$ down
So, applying this pattern "backwards" to $A^3$ we get a possible candidate for $A$:
$$A = \begin{bmatrix}2& 1\\-1&0\end{bmatrix}$$
Indeed, we get
$$A^3 = \begin{bmatrix}4&3\\-3&-2\end{bmatrix} \Rightarrow A^2+A = \begin{bmatrix}5&3\\-3&-1\end{bmatrix}$$
| {
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Solution of polynomial Find the set of values of $k$ for which the equation $x^4+kx^3+11x^2+kx+1=0$ has four distinct positive root.
Attempt:
$x^4+kx^3+11x^2+kx+1=0$
$x^2+kx+11+{k\over x}+{1\over x^2}=0$
$x^2 + {1\over x^2} +k(x+{1\over x})+11=0$
$(x + {1\over x})^2 +k(x+{1\over x})+13=0$
I don't know how to proceed after this......
| Easiest way is to draw a graph $$f(x) = {x^4+11x^2+1\over -x(x^2+1)}$$
and we want to find for which $k$ line $y=k$ cuts graph of $f$ for positive $x$ four times. If you do some calculus you see that $f$ has local minimum $-6,5$ for positive $x$ and local maximum $-6$ so the finally answer is for $$-6,5<k<-6$$
If you don't know calculus then you can try like this. Write
$$f(x) = {x^4+2x^2+1\over -x(x^2+1)} +{9x^2\over -x(x^2+1)}$$
$$={x^2+1\over -x} +{9x\over -(x^2+1)}$$
$$=t +{9\over t}$$
where $t={x^2+1\over -x}$. Since $f$ is odd it is the same question if we ask our self for which negative $x$ the line $y= k$ cuts graph of $f$ four times. So say $x<0$ and thus $t>0$. Now it is easy to see that $f(x)\geq 6$ (for all negative $x$) and now we have to prove that the line $y=6,5$ touch a graph of $f$ (in some $x_0$ and thus it will have a local extremum at that $x_0$). To find this $x_0$ we solve this $$ f(x) = 13/2 \implies (x+1)^2(2x^2+9x+2) =0$$
since we have double root at $x=-1$ we see that $x_0=-1$.
| {
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Prove that there are no non-zero integers $m$ and $n$ such that $m^2 = 180 n^4$ I made the following:
$(m^2)^{1/2} =(n^4)^{1/2} 180^{1/2} $
Then $|m| = (n^2) 180^{1/2} $ but $180^{1/2} = 6 (5^{1/2})$ isn't an integer number. Then $m$ is an integer number if and only if $n= a (5^{1/2})$, $a$ is any integer number. That is, if $m$ is an integer number then $n$ isn't an integer and vice versa
Then, there aren't non-zero integers $m$ and $n$ such that $m^2 = 180 n^4$
Is right my demonstration? Is there any more clearly form to prove this exercise?
| For $m^2 = 180 n^4$, assuming there is an integer solution, implies that $m^2$ is wholly divisible by $180 = 2\cdot 2\cdot 3\cdot 3\cdot 5$.
That is, $\frac{m^2}{2\cdot 2\cdot 3\cdot 3\cdot 5} = n^4$
Now, factors of $m$ must appear as an even quantity of each in $m^2$. So if we divide $m^2$ by a single factor of $5$ we will be left with an odd number of 5 factors which contradicts $n$ being a $4th$ power.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$\frac{x^5}{5!}- \frac{x^7}{7!}+ \frac{x^9}{9!}+\dots>0$ holds for $x>0$ Show that:
$\sin x> x- \dfrac{x^3}{6} ~~\forall ~x>0$
Attempt:
Let $y = \sin x - x+\dfrac{x^3}{6}$
We have to prove that y is increasing for $x >0$
$y' = \cos x -1 + \dfrac{x^2}{2}$
$y'' =-\sin x+x$
$y''>0$ for all $x>0$.
$\implies y'$ is increasing for all $x>0$
$\implies y$ is an increasing function with $y(0+)>0$
Hence proved.
Attempt 2:
Writing the Taylor expansion of $\sin x$,
we get this inequality to be proven:
$\dfrac{x^5}{5!}- \dfrac{x^7}{7!}+ \dfrac{x^9}{9!}+...>0$
Is it possible to prove it? I tried to (by rearranging) but couldn't.
| Consider any alternating series $a_1-a_2+a_3...$ with $a_n >0$ and $a_n$ decreasing to $0$. If the series is absolutely convergent then we can group the terms as $(a_1-a_2)+(a_3-a_4)+...$ and the sum is greater than $a_1-a_2$ because the other terms are non-negative. Your question is a special case of this. Attempt 2 also works by the same method. Details: $x^{5} /5! -x^{7} /7! >0$ if $x <\sqrt {42}$ Similarly $x^{7} /7! -x^{9} /9! >0$, etc for such $x$ so we get $\sin x > x- x^{3} /3!$. It remains to see $\sin x > x-x^{3} /3!$ when $x \geq \sqrt {42}$. But here $x-x^{3} /3!<-1 \leq \sin x$ since $x^{3} >6x>6x-6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating indefinite integrals.
If $$\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\dfrac{d}{b}{\sin^{-1}} \bigg(\dfrac{x^\frac{3}{2}}{a^\frac{3}{2}}\bigg)+C$$ where $b,d$ are relatively prime find $b+d$.
My solution:
$$\displaystyle\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\int\dfrac{1}{x}\sqrt{\dfrac{x^3}{a^3-x^3}}dx$$
Then let $x^3=t\implies 3x^2\cdot dx=dt$
$$\implies\dfrac{1}{3}\int\dfrac{1}{x}\cdot\dfrac{3x^2}{x^2}\sqrt{\dfrac{x^3}{a^3-x^3}}dx\\=\dfrac{1}{3}\int\dfrac{1}{t}\sqrt{\dfrac{t}{a^3-t}}dt\\ =\dfrac{1}{3}\int{\dfrac{1}{\sqrt{ta^3-t^2}}}dt\\=\dfrac{1}{3}\int{\dfrac{1}{\sqrt{\frac{a^6}{4}-\bigg(t-\frac{a^3}{2}\bigg)^2}}}dt\\=\dfrac{1}{3}\sin^{-1}\bigg(\dfrac{2t-a^3}{a^3}\bigg)+C$$ $$=\dfrac{1}{3}\sin^{-1}\bigg[2\bigg(\dfrac{x^{3/2}}{a^{3/2}}\bigg)^2-1\bigg]+C\tag{1}$$
This is not of desired form but when I drew graph of: $\color{grey}{1.57+\sin^{-1}\left(2x^2-1\right)}$ and $\color{green}{2\sin^{-1}x}$, they coincides for $\color{red}{x>0}$. Green graph is for $\ 2\sin^{-1}x\ $.
So from $(1)$ $$\dfrac{1}{3}\sin^{-1}\bigg[2\bigg(\dfrac{x^{3/2}}{a^{3/2}}\bigg)^2-1\bigg]+C=\dfrac{2}{3}{\sin^{-1}} \bigg(\dfrac{x^\frac{3}{2}}{a^\frac{3}{2}}\bigg)+C$$
If I didn't made any mistake then please help me getting $1.57+\sin^{-1}\left(2x^2-1\right)\equiv\ 2\sin^{-1}x\ $ for $x>0$ without any graphs.
Also, I found that if I substitute $x^3=a^3\sin^{2}x$ then we reach the desired form of the result directly. But I'm an enthusiast to continue my previous solution. Please help.
Thanks!
| Since you already have the form you must get the answer into then why not get some hints from it.
$$\int \frac {\sqrt x}{\sqrt {a^3-x^3}} dx=\int \frac {\sqrt x}{a^{\frac 32}\sqrt {1-\left(\frac {x^{3/2}}{a^{3/2}}\right)^2 }} dx$$
Let $$\frac {x^{3/2}}{a^{3/2}}=t$$ hence $$dt=\frac 32\cdot \frac {\sqrt x}{a^{3/2}}dx$$
Hence the integral changes to $$\int \frac 23 \cdot \frac {1}{\sqrt {1-t^2}} dt $$
Hence the answer $b=3,d=2$ follows
| {
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Find the value of $a$ by evaluating the limit $L=\lim_{n\to\infty}\frac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+\cdots+(na+n)]}=\frac{1}{60}$
For $a\in\mathbb R-\{-1\}$
$$L=\lim_{n\to\infty}\frac{1^a+2^a+\cdots+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+\cdots+(na+n)]}=\frac{1}{60}$$
find the value of $a$.
My attempt:
$$L=\frac{(1/n)^a+(2/n)^a+\cdots+(n/n)^a}{(\frac{n+1}{n})^{a-1}[a+1/n+a+2/n+\cdots+a+n/n]}$$
which as $\lim_{n\to\infty}$ must equal to $\frac{1}{a+a+a+\cdots}$. Now this obviously can't be implying some logical error on my part. What to do now?
| $L=\lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{(n+1)^{a-1}[(na+1)+(na+2)+...+(na+n)]}=$
$= \lim_{n\to\infty}\frac{1^a+2^a+...+n^a}{\frac{1}{n^a}(\frac{n+1}{n})^{a-1}\frac{1}{n}[n^2a +\frac{n(n+1)}{2}]} =$
$= \lim_{n\to\infty}\frac{(\frac{1}{n})^a+(\frac{2}{n})^a+...+1^a}{(\frac{n+1}{n})^{a-1}[na +\frac{(n+1)}{2}]} =$
$= \lim_{n\to\infty}\frac{\frac{1}{n}[(\frac{1}{n})^a+(\frac{2}{n})^a+...+1^a]}{(\frac{n+1}{n})^{a-1}[a +\frac{1}{2}+\frac{1}{n}]} =$
$= \lim_{n\to\infty}\frac{\frac{1}{n}\sum_{k=1}^{n}(\frac{k}{n})^a}{(\frac{n+1}{n})^{a-1}[a +\frac{1}{2}+\frac{1}{n}]} =$
$=\frac{\int_0^1 x^a dx}{a+\frac{1}{2}}=\frac{1}{(a+1)(a+\frac{1}{2})}$
because
$1+2+...+n=\frac{n(n+1)}{2}$
| {
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What does is mean $Y=\min \{X_1,X_2\}$? $Y=\min \{X_1,X_2\}$ and $Z=\max \{X_1,X_2\}$?
Let's determine $X_1$ distribution to be $\operatorname{Bin}(2,\frac{1}{2})$ and the distribution of $X_2$ to be $U(1,2,3)$. Also $X_,X_2$ are independent.
After some calculations:
*
*$P(X_1=0)=\frac{1}{4}$
*$P(X_1=1)=\frac{1}{2}$
*$P(X_1=2)=\frac{1}{4}$
* $P(X_2=1)=P(X_2=2)=P(X_2=3)=\frac{1}{3}$
*
*I know how to work with $X_1, X_2$ but I have no clue how to do anything with $Y, Z$. Why $Y\sim X_1$ and $Z\sim X_2$?
*How do I approach the question: Calculate $P(Y=y, Z=z)$?
| You have not said what the joint distribution of $X_1,X_2$ is. But if we assume they are independent, although you didn't say that, then the joint distribution is as follows:
$$
\begin{array}{c|ccc|c}
_{X_1}\backslash ^{X_2} & 0 & 1 & 2 & \\
\hline 0 & 1/12 & 1/12 & 1/12 \\
1 & 1/6 & 1/6 & 1/6 \\
2 & 1/12 & 1/12 & 1/12 \\
\hline
\end{array}
$$
The values of $\max\{X_1,X_2\}$ are as follows:
$$
\begin{array}{c|ccc|c}
_{X_1}\backslash ^{X_2} & 0 & 1 & 2 & \\
\hline 0 & 0 & 1 & 2 \\
1 & 1 & 1 & 2 \\
2 & 2 & 2 & 2 \\
\hline
\end{array}
$$
Therefore
\begin{align}
\Pr(\max = 2) & = \frac 1 {12} + \frac 1 {12} + \frac 1 {12} + \frac 1 6 + \frac 1 {12} & & = \frac 1 2, \\[10pt]
\Pr(\max = 1) & = \frac 1 6 + \frac 1 6 + \frac 1 {12} & & = \frac 5 {12}, \\[10pt]
\Pr(\max = 0) & = \frac 1 {12} & & = \frac 1 {12}.
\end{align}
Similarly the values of $\min\{X_1,X_2\}$ are as follows:
$$
\begin{array}{c|ccc|c}
_{X_1}\backslash ^{X_2} & 0 & 1 & 2 & \\
\hline 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 1 \\
2 & 0 & 1 & 2 \\
\hline
\end{array}
$$
and you can find the probabilities in the same way.
| {
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Probability on cards involving two conditions . $\text{Problem (i) :}$
When $4$ cards are drawn out of $52$ cards, what is the probability of only $2$ cards being same value and the rest being different ?
MY WORK:
The probability I found is :
$$\frac{\binom{13}{1}\times \binom{4}{2}\times \binom{12}{2}\times \binom{4}{1}\times \binom{4}{1}}{\binom{52}{4}}$$
$\text{Problem (ii) :}$
When $4$ cards are drawn out of $52$ cards, what is the probability of $2$ cards to be of same value of same colour and the other two cards to be of same value ( different from the previous one ) of different colours?
MY WORK:
The probability I found is :
$$\frac{\binom{2}{1}\times \binom{13}{1}\times \binom{2}{2}\times \binom{12}{1}\times \binom{2}{1}\times \binom{2}{1}}{\binom{52}{4}}$$
Here, I at first select a particular colour $\binom{2}{1}$ and then a particular value $\binom{13}{1}$ . Then, there is only single way of first condition. After, I select a different value $\binom{12}{2}$ . For different colour, I take $\binom{2}{1}\times \binom{2}{1}$ .
AM I CORRECT ?
N. B. Value means ace, king, jack, queen and the numbers from $2$-$10$ .
|
$$\frac{\binom{13}{1}\times \binom{4}{2}\times \binom{12}{2}\times \binom{4}{1}\times \binom{4}{1}}{\binom{52}{4}}$$
$\checkmark$ Select a rank and two suits for it, and select two other ranks, and a suit for each.
$$\frac{\binom{2}{1}\times \binom{13}{1}\times \binom{2}{2}\times \binom{12}{1}\times \binom{2}{1}\times \binom{2}{1}}{\binom{52}{4}}$$
$\checkmark$ Select a colour, a rank, and two suits of that colour for that rank, and select another rank, and a suit of each colour.
| {
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Find the equation of the tangent lines to the ellipse having a given angular coefficient Find the equations of the tangent lines to the ellipse $E : x^2/a^2 + y^2/b^2 − 1 = 0$ having a given angular coefficient $m ∈ R$.
| Here's a go at it:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
Differentiate implicitly.
$$ \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 $$
Assuming angular coefficient is the slope.
$$ \frac{dy}{dx} = m $$
$$ \frac{x}{a^2} + \frac{y}{b^2}m = 0 $$
$$ x = - y \frac{a^2}{b^2}m $$
$$ x^2 = y^2 \frac{a^4}{b^4}m^2 $$
$$ y^2 \frac{a^2}{b^4}m^2 + \frac{y^2}{b^2} = 1 $$
$$ \frac{y^2}{b^2} \left[ \frac{a^2}{b^2}m^2 + 1 \right] = 1 $$
$$ y = \pm b \sqrt{ \frac{1}{ \left[ \frac{a^2}{b^2}m^2 + 1 \right] } } $$
You should be able to readily solve for the corresponding $x$ from here.
| {
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Find the quotient and remainder Find the quotient and remainder when $x^6+x^3+1$ is divided by $x+1$
Let $f(x)=x^6+x^3+1$
Now $f(x)=(x+1).q(x) +R $ where r is remainder
Now putting $x=-1$ we get $R=f(-1)$
i.e $R=1-1+1=1$
Now $q(x)=(x^6+x^3)/(x+1)$
But what I want to know if there is another way to get the quotient except simple division.
| In this precise case, you can notice that $x^3$ is an obvious factor. You are left with: $(x^3+1)/(x+1)$ Using the remarkable identity $A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+...+B^{n-1})$ with $A=x$ and $B=-1$ gives $q(x)=x^3(x^2-x+1)$
| {
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Proof by Induction: If $x_1x_2\dots x_n=1$ then $x_1 + x_2 + \dots + x_n\ge n$
If $x_1,x_2,\dots,x_n$ are positive real numbers and if $x_1x_2\dots x_n=1$ then $x_1 + x_2 + \dots + x_n\ge n$
There is a step in which I am confuse. My proof is as follows (it must be proven using induction).
By induction, for $n=1$ then $x_1=1$ and certainly $x_1\ge1$.
Suppose $x_1 + x_2+\dots+x_n\ge n$ (does this mean that $x_1x_2\dots x_n = 1$ also hold?) and $x_1x_2\dots x_n x_{n+1} = 1$ hold. Then
\begin{align}
x_1 + x_2 + \dots+x_n+x_{n+1} &\ge n + x_{n+1} \\
&=n+2-1/x_{n+1} \\
&=n+2-x_1x_2\dots x_n.
\end{align}
My problem is in the last step. As I wrote before, I don't think $x_1x_2\dots x_n = 1$ should hold, because if this is the case then $x_{n+1}=1$.
EDIT:
In the itermediate step I used that $x + 1/x \ge 2$, where $x>0$.
| In fact， we may prove the more stonger one
Let $x_1,x_2,\cdots,x_n$ be positive numbers. Then
$$\frac{1}{n}(x_1+x_2+\cdots+x_n)\geq \sqrt[n]{x_1x_2\cdots x_n}.\tag1$$
Put $x_1x_2\cdots x_n=1$ into $(1)$. You will get yours.
Inductive Proof
Obviously, $(1)$ holds for $n=1$, which is trivial.
Notice that, from $(\sqrt{x_1}-\sqrt{x_2})^2 \geq 0$, we may have $\dfrac{1}{2}(x_1+x_2) \geq \sqrt{x_1x_2}$, which shows $(1)$ holds for $n=2$.
Assume that $(1)$ holds for $n=k$, namely, $$\frac{1}{k}(x_1+x_2+\cdots+x_k)\geq \sqrt[k]{x_1x_2\cdots x_k}.$$ When $n=k+1$, denote $\dfrac{1}{k+1}(x_1+x_2+\cdots+x_k+x_{k+1})=x.$ Then,
\begin{align*}
x&=\frac{1}{2k}[(x_1+x_2+\cdots+x_k)+x_{k+1}+(k-1)x]\\&=\frac{1}{2}\left[\frac{1}{k}(x_1+x_2+\cdots+x_k)+\frac{1}{k}(x_{k+1}+(k-1)x)\right]\\
&\geq \frac{1}{2}\left(\sqrt[k]{x_1x_2\cdots x_k}+\sqrt[k]{x_{k+1}x^{k-1}}\right)\\
&\geq \sqrt[2k]{x_1x_2\cdots x_k \cdots x_{k+1}x^{k-1}},
\end{align*}
namely, $$x^{2k} \geq x_1x_2\cdots x_kx_{k+1}x^{k-1},$$ i.e. $$x^{k+1} \geq x_1x_2\cdots x_kx_{k+1}.$$Thus, $$\frac{1}{k+1}(x_1+x_2+\cdots+x_k+x_{k+1})\geq \sqrt[k+1]{x_1x_2\cdots x_kx+{k+1}},$$which shows that $(1)$ holds for $n=k+1.$
By induction, $(1)$ holds for all $n=1,2,\cdots$.
| {
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"url": "https://math.stackexchange.com/questions/2848101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=1$?
Evaluate $$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}$$
My attempt:
$$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{-\sqrt{4x^2}}=\frac{-5}{2}$$ According to the answer key, it actually equals $1$.
Thanks in advance.
| While $x$ is large then $4x^2-7\sim4x^2$ hence
$$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty} \frac{5x+9}{3x+2-|2x|}=\lim_{x\to-\infty} \frac{5x+9}{3x+2+2x}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?
| The following solution may be the most essential solution...
Denote $$f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}, ~~~x \in (0,1).$$
Since $$f''(x)=\dfrac{3-x}{4x^{5/2}}>0,~~~\forall x \in (0,1),$$ hence $f(x)$ is a convex function over $(0,1)$. Notice that $a^2, b^2 \in (0,1).$Therefore, $$a+b+\frac{1}{a}+\frac{1}{b}=f(a^2)+f(b^2) \geq 2f\left(\frac{a^2+b^2}{2} \right)=2f\left(\frac{1}{2}\right)=3\sqrt{2},$$with the equality holding iff $a^2=b^2$, namely, $a=b=\dfrac{\sqrt{2}}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
$\tan(a) = 3/4$ and $\tan (b) = 5/12$, what is $\cos(a+b)$ It is known that
$$\tan(a) = \frac{3}{4}, \:\:\: \tan(b) = \frac{5}{12} $$
with $a,b < \frac{\pi}{2}$.
What is $\cos(a+b)$?
Attempt :
$$ \cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b) $$
And we can write $\tan(a) = \sin(a)/\cos(a) = 0.3/0.4 $
and $ \sin(b)/\cos(b) = 0.05/0.12 $
so
$$ \cos(a+b) = (0.4)(0.12) - (0.3) (0.05) = 33/1000$$
Is this correct? Thanks.
| No, this can't be correct. Remember that $\sin^2x+\cos^2x=1$ for all $x$; your values for the sine and cosine of $a$ and $b$ do not satisfy this relation.
It turns out that
$$\sin a=\frac35\qquad\cos a=\frac45$$
$$\sin b=\frac5{13}\qquad\cos b=\frac{12}{13}$$
and thus
$$\cos(a+b)=\cos a\cos b-\sin a\sin b=\frac45\cdot\frac{12}{13}-\frac35\cdot\frac5{13}=\frac{33}{65}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Positive integer solutions to $a^3 + b^3 = c^4$ Let $n$ and $m$ be positive integers. We know that
$n^3 + m^3 = n^3 + m^3$
Multiply both sides by $(n^3 + m^3)^3$; on the LHS you distribute, and on the RHS you use power addition rule;
$(n^4 + nm^3)^3 + (m^4 + mn^3)^3 = (n^3 + m^3)^4$
So we get an infinite array of solutions with
$a = n^4 + nm^3$
$b = m^4 + mn^3$
$c = n^3 + m^3$
Are there any solutions to the equation (in the title) that cannot be expressed in this form?
| There are many other solutions $(a,b,c)$. First, start with arbitrary integers $u,v>0$ (you can add the condition $\gcd(u,v)=1$ so as to produce non-overlapping of infinite families of solutions). Then, write
$$u^3+v^3=\prod_{r=1}^k\,p_r^{t_r}\,,$$
where $p_1,p_2,\ldots,p_k$ are pairwise distinct prime natural numbers and $t_1,t_2,\ldots,t_k\in\mathbb{Z}_{> 0}$. Then, solve for $x_1,x_2,\ldots,x_r\in\mathbb{Z}_{\geq 0}$ from the congruences
$$3x_r\equiv -t_r\pmod{4}$$
for all $r=1,2,\ldots,k$. (Well, this is quite easy and you should get $x_r\equiv t_r\pmod{4}$ for every $r=1,2,\ldots,k$.) Then, take
$$a:=u\,\prod_{r=1}^k\,p_r^{x_r}\,,\,\,b:=v\,\prod_{r=1}^k\,p_r^{x_r}\,,\text{ and }c:=\prod_{r=1}^k\,p_r^{\frac{3x_r+t_r}{4}}\,.$$
For example, take $u:=3$ and $v:=5$. Then,
$$u^3+v^3=27+125=152=2^3\cdot 19\,.$$
Hence, we can take
$$a:=3\cdot2^{4\alpha+3}\cdot19^{4\beta+1}\,,\,\,b:=5\cdot 2^{4\alpha+3}\cdot19^{4\beta+1}\,,\text{ and }c:=2^{3\alpha+3}\cdot 19^{3\beta+1}\,,$$
where $\alpha,\beta\in\mathbb{Z}_{\geq0}$.
Peter's answer can also be generated this way. First, start with $u:=13$ and $v:=14$. Then, $$u^3+v^3=3^4\cdot 61\,.$$ This leads to
$$a:=13\cdot 3^{4\alpha}\cdot 61^{4\beta+1}\,,\,\,b:=14\cdot3^{4\alpha}\cdot 61^{4\beta+1}\,,\text{ and }c:=3^{3\alpha+1}\cdot 61^{3\beta+1}\,,$$
where $\alpha,\beta\in\mathbb{Z}_{\geq 0}$. Peter's answer corresponds to $\alpha=0$ and $\beta=0$.
Even the OP's example starts from $u:=m$ and $v:=n$. Then, take each $x_r$ to be $t_r$, so that $$\prod_{r=1}^k\,p_r^{x_r}=\prod_{r=1}^k\,p_r^{t_r}=m^3+n^3\,.$$
This gives
$$a:=m\left(m^3+n^3\right)\,,\,\,b:=n\left(m^3+n^3\right)\,,\text{ and }c=m^3+n^3\,.$$
Furthermore, if $(a,b,c)$ is a solution, then $(s^{4l}a,s^{4l}b,s^{3l}c)$ is a solution for any positive integers $s$ and $l$. You can get all solutions $(a,b,c)\in\mathbb{Z}_{>0}^3$ to the equation $a^3+b^3=c^4$ in this way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate $(1+i)^{(1-2i)}$ Find all the values of $(1+i)^{(1-2i)}$ and show that there are small values as we wish (else from $0$) and big values as we wish
\begin{align}
(1+i)^{(1-2i)}&=e^{\ln(1+i)^{(1-2i)}}
=e^{(1-2i)\ln(1+i)}
\\&=e^{(\ln\sqrt{2}+i(\frac{\pi}{2}+2\pi k))(1-2i)}
\\&=e^{\ln\sqrt{2}-2\ln\sqrt{2}*i+i(\frac{\pi}{2}+2\pi k)+2(\frac{\pi}{2}+2\pi k))}
\\&=\sqrt{2}*e^{-2\ln\sqrt{2}*i}*i*e^{2(\frac{\pi}{2}+2\pi k)}
\\&=e^{-2\ln\sqrt{2}*i+\pi+4\pi k}+i\sqrt{2}
\\&=e^{-i(2\ln\sqrt{2}+i(\pi+4\pi k))}+i\sqrt{2}
\\&=e^{-i}*e^{(2\ln\sqrt{2}+i(\pi+4\pi k))}+i\sqrt{2}
\\&=\frac{3}{4}*e^{(2\ln\sqrt{2}+i(\pi+4\pi k))}+i\sqrt{2}
\\&=\frac{3}{4}*2e^{i(\pi+4\pi k)}+i\sqrt{2}
\\&=\frac{-3}{2}+i\sqrt{2}
\end{align}
What can I conclude from it on the values?
Or should I stop here $e^{-2\ln\sqrt{2}*i+\pi+4\pi k}+i\sqrt{2}$ and $e^{\pi+4 \pi k}cis(-2\ln\sqrt{2}+\frac{\pi}{2})$
| There should only be one value.$$(1+i)^{1-2i}=(\sqrt 2e^\dfrac{i\pi}{4})^{1-2i}=\sqrt 2e^\dfrac{i\pi}{4}(0.5e^\dfrac{-i\pi}{2})^i=\sqrt 2e^\dfrac{\pi}{2}e^{i(\dfrac{\pi}{4}+\ln 0.5)}$$
| {
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"url": "https://math.stackexchange.com/questions/2854264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ \frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$
For $x,y,z>0,$ I have to prove that
$$ \frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$$
I tried to use $$ \sqrt {\frac {1}{3} \sum\limits_{cycl}^{} (\frac {x+y}{x+y+ 2 z})^2} \ge \frac {1}{3} \sum\limits_{cycl}^{} \frac {x+y}{x+y+ 2 z} \ge \sqrt[3] {(\frac {x+y}{x+y+ 2 z}) (\frac {y+z}{2 x +y +z}) (\frac {z + x}{z+2 y+z})},$$
but it looks hard to continue and $ \frac {3 x y z}{(x+y) (y+z) (z+x)}$ seems to be not helpful. Any help? Thank you.
| Let $\frac{2x}{y+z}=a$, $\frac{2y}{x+z}=b$ and $\frac{2z}{x+y}=c$.
Thus, $ab+ac+bc+abc=4$ and we need to prove that
$$\sum_{cyc}\frac{1}{(a+1)^2}+\frac{3}{8}abc\geq\frac{9}{8}.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$ and $abc=w^3$.
Hence, the condition does not depend on $u$.
In another hand, by AM-GM $$4=ab+ac+bc+w^3=3v^2+w^3\leq3v^2+v^3,$$
which gives $v\geq1$ and since $u\geq v$, we obtain $u\geq1$.
Also, $$4=3v^2+w^3\geq3w^2+w^3,$$ which gives $w\leq1$.
Now, we need to prove that
$$\frac{\sum\limits_{cyc}(ab+a+b+1)^2}{(abc+ab+ac+bc+a+b+c+1)^2}+\frac{3}{8}w^3\geq0$$ or
$$\frac{\sum\limits_{cyc}(a^2b^2+2a^2+1+2a^2b+2a^2c+4ab+4a)}{(w^3+3v^2+3u+1)^2}+\frac{3}{8}w^3\geq\frac{9}{8}$$ or
$$\frac{9v^4-6uw^3+18u^2-12v^2+3+18uv^2-6w^3+12v^2+12u}{(w^3+3v^2+3u+1)^2}+\frac{3}{8}w^3\geq\frac{9}{8}$$ or $f(u)\geq0,$ where
$$f(u)=\frac{3v^4-2uw^3+6u^2+1+6uv^2-2w^3+4u}{(w^3+3v^2+3u+1)^2}+\frac{1}{8}w^3-\frac{3}{8}.$$
But
$$f'(u)=\tfrac{-2w^3+12u+6v^2+4}{(w^3+3v^2+3u+1)^2}-\tfrac{2\cdot3(3v^4-2uw^3+6u^2+1+6uv^2-2w^3+4u)}{(w^3+3v^2+3u+1)^3}=$$
$$=\frac{2(9uw^3+9uv^2+9v^2+7w^3-w^6-1)}{(w^3+3v^2+3u+1)^3}>0,$$
which says that $f$ increases and since the condition does not depend on $u$, it's enough to prove our inequality for a minimal value of $u$, which happens for an equality case of two variables.
Let $b=a$.
Hence, the condition gives $$a^2+2ac+a^2c=4$$ or
$$ac(2+a)=4-a^2$$ or
$$c=\frac{2-a}{a},$$ where $0<a<2$ and we need to prove that
$$\frac{2}{(a+1)^2}+\frac{1}{\left(\frac{2-a}{a}+1\right)^2}+\frac{3}{8}a^2\cdot\frac{2-a}{a}\geq\frac{9}{8}$$ or
$$(a-1)^2(7+2a-a^2)\geq0,$$ which is obvious.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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find all the points on the line $y = 1 - x$ which are $2$ units from $(1, -1)$ I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in. What am I doing wrong?
1.) Distance formula $\sqrt{(x-1)^2 + (-1 -1 + x)^2}=2$
2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$
3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$
4.) combine, $2x^2 -6x +5 = 4$
5.) general form, $2x^2 -6x +1$
6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$
7.) Vertex form, $2(x-1.5)^2 -3.5$
8.) Solve for x, $x-1.5 = \pm\sqrt{1.75}$
9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$
10.) $x = 0.18$ and $2.82$
When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values. These points are not on the line. Can someone tell me what I'm doing wrong please?
| After getting those two roots $ (x_1,x_2) $ we get next by plugging in $ (1-y)$ for $x$ a second quadratic in $y:$
$$ 2y^2+2y-3=0$$
which supplies corresponding roots
$$ (y_1,y_2)= \frac{-1\pm \sqrt7}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int_{0}^{1}{\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 -3x+2}}dx}$ I've got one integration question which I first felt was not a hard nut to crack. But, as I proceeded, difficulties arose. This is the one:
$\displaystyle\int_{0}^{1}{\frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 -3x+2}}dx}$
I went ahead simplifying the two expressions and ultimately I reached this step:
$\displaystyle\int_{0}^{1}{\sqrt{\frac{x-1}{x-2}} \ (3x^2 + 2x + 4) \ dx}$
I don't now what to do now, had I followed the correct pathway? Is there any other simpler method?
| It can be seen easily that for $H\left( x \right)=\sqrt{{{x}^{2}}-3x+2}$
$$\begin{align}
& {{\left( {{x}^{2}}H\left( x \right) \right)}^{\prime }}=\frac{6{{x}^{3}}-15{{x}^{2}}+8x}{2H\left( x \right)}, \\
& {{\left( xH\left( x \right) \right)}^{\prime }}=\frac{4{{x}^{2}}-9x+4}{2H\left( x \right)}, \\
& {{\left( H\left( x \right) \right)}^{\prime }}=\frac{2x-3}{2H\left( x \right)} \\
\end{align}$$
This suggests that there exists $a,b,c\ and\ d$
$$\frac{3{{x}^{3}}-{{x}^{2}}+2x-4}{H\left( x \right)}=a{{\left( {{x}^{2}}H\left( x \right) \right)}^{\prime }}+b{{\left( xH\left( x \right) \right)}^{\prime }}+c{{\left( H\left( x \right) \right)}^{\prime }}+\frac{d}{H\left( x \right)}$$
Hence
$$\frac{3{{x}^{3}}-{{x}^{2}}+2x-4}{H\left( x \right)}=a\frac{6{{x}^{3}}-15{{x}^{2}}+8x}{2H\left( x \right)}+b\frac{4{{x}^{2}}-9x+4}{2H\left( x \right)}+c\frac{2x-3}{2H\left( x \right)}+\frac{d}{H\left( x \right)}$$
Or
$$3{{x}^{3}}-{{x}^{2}}+2x-4=\frac{a}{2}\left( 6{{x}^{3}}-15{{x}^{2}}+8x \right)+\frac{b}{2}\left( 4{{x}^{2}}-9x+4 \right)+\frac{c}{2}\left( 2x-3 \right)+d$$
Equating the coefficients yields
$$\begin{align}
& a=1 \\
& -15a+4b=-2 \\
& 8a-9b+2c=4 \\
& 4b-3c+2d=-8 \\
\end{align}$$
and we get
$$a=1,b=\frac{13}{4},c=\frac{101}{16},d=\frac{135}{16}$$
Finally
$$\int{\frac{3{{x}^{3}}-{{x}^{2}}+2x-4}{H\left( x \right)}dx={{x}^{2}}H\left( x \right)}+\frac{13}{4}xH\left( x \right)+\frac{101}{16}H\left( x \right)+\frac{135}{16}\int{\frac{dx}{H\left( x \right)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi}{20}$$
and that,
$$\cos\frac{\pi}{20}-\cos\frac{3\pi}{20}=-2\sin\frac{2\pi}{20}\sin\frac{\pi}{20}$$
So,
$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}-\sin\frac{\pi}{20})=\frac{\sqrt2}{2}=\sin\frac{5\pi}{20}$$
Unfortunately, I can't find a way to continue this, any ideas or different ways of proof?
*Taken out of the TAU entry exams (no solutions are offered)
| Solution
Notice that
$$\sin x \pm\cos x=\sqrt{2}\sin\left(x\pm\frac{\pi}{4}\right),~~~\forall x \in \mathbb{R}$$
Therefore,
\begin{align*}
&\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}\\
=&\sqrt{2}\sin\left(\frac{\pi}{20}+\frac{\pi}{4}\right)+\sqrt{2}\sin\left(\frac{3\pi}{20}-\frac{\pi}{4}\right)\\
=&\sqrt{2}\left(\sin\frac{3\pi}{10}-\sin\frac{\pi}{10}\right)\\
=&2\sqrt{2}\sin \frac{\pi}{10}\cos\frac{\pi}{5}\\
=&2\sqrt{2}\cdot \frac{2\sin\dfrac{\pi}{10}\cos\dfrac{\pi}{10}\cos\dfrac{\pi}{5}}{2\cos\dfrac{\pi}{10}}\\
=&\sqrt{2}\cdot \frac{2\sin\dfrac{\pi}{5}\cos\dfrac{\pi}{5}}{2\cos\dfrac{\pi}{10}}\\
=&\sqrt{2}\cdot \frac{\sin\dfrac{2\pi}{5}}{2\cos\dfrac{\pi}{10}}\\
=&\sqrt{2}\cdot \frac{\cos\dfrac{\pi}{10}}{2\cos\dfrac{\pi}{10}}\\
=&\frac{\sqrt{2}}{2}.
\end{align*}
Note In fact, you may readily evaluate $\sin \dfrac{\pi}{10}$ and $\cos\dfrac{\pi}{5}$. As the figure shows, you may obtain $$\frac{AB}{BC}=\frac{BC}{CD},$$namely,$$\frac{x+2y}{x}=\frac{x}{2y},$$i.e. $$\left(\frac{x}{y}\right)^2-2\left(\frac{x}{y}\right)-4=0.$$Hence, $$\frac{x}{y}=1+\sqrt{5}.$$Therefore,$$\sin \dfrac{\pi}{10}=\frac{y}{x}=\frac{\sqrt{5}-1}{4},~~~\cos\dfrac{\pi}{5}=\frac{x+y}{x+2y}=\dfrac{\dfrac{x}{y}+1}{\dfrac{x}{y}+2}=\frac{\sqrt{5}+1}{4}.$$As a result,$$\sin \dfrac{\pi}{10}\cos\dfrac{\pi}{5}=\frac{1}{4}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Linear Transformations between 2 non-standard basis of Polynomials
If
$$
A = \begin{pmatrix} 1 & -1 & 2 \\
-2 & 1 &-1 \\ 1 & 2 & 3 \end{pmatrix}
$$
is the matrix representation of a
linear transformation $T : P_3(x) \to P_3(x)$ with respect to
bases $\{1-x,x(1-x),x(1+x)\}$ and $\{1,1+x,1+x^2\}$. Find T.
While i have worked with transforming non-standard to standard basis, this is the first one i am encountering with transformation between 2 non-standard polynomial basis. I am not sure if i am working out rightly.
$T[1-x] = 1(1) -2(1+x) +1(1+x^2)$
$T[x(1-x)] = -1(1) +1(1+x) +2(1+x^2)$
$T[x(1+x)] = 2(1) -1(1+x) +3(1+x^2)$
Therefore, $T[a(1-x)+b(x(1-x))+c(x(1+x))] = (a-b+2c)(1) + (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2)$
Is this fine ?
| What you did is fine, but now you have to compute $T[\alpha+\beta x+\gamma x^2]$ for arbitrary $\alpha,\beta,\gamma\in\mathbb R$. In order to do that, solve the equation$$\alpha+\beta x+\gamma x^2=a-b+2c+ (-2a+b-c)(1+x) +(a+2b+3c)(1+x^2).$$That is, solve the system$$\left\{\begin{array}{l}a-b+2c=\alpha\\-2a+b-c=\beta\\a+2b+3c=\gamma.\end{array}\right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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My solution to inhomogeneous $\frac{d^2y}{dx^2} + y = \sin{x}$ does not conform to my book's solution! I need help with the solution of this particular equation:
$$\frac{d^2y}{dx^2} + y = \sin{x}$$
Due to me having to go to work, I cannot display all my work in mathjax, my shift starts in 5 min...but my solution is: $$y= \frac{-x}{2} + \frac{(\cos{2x}\sin{x})}{4} + C_1\cos{x} + C_2\sin{x}$$
Whereas my book's solution is: $$y= \frac{-x}{2} +C_1\cos{x} + C_2\sin{x}$$
I have used the method of variation of prameters, cramer's rule and basic antidifferentiation to solve resulting system.
Thank you all!
| Since you mention variation of parameters, I'll summarize the method here. Hopefully you can retrace your steps to find the mistake.
From the fundamental solution
$$ y_h(x) = c_1\cos x + c_2\sin x $$
we then seek a particular solution of the form
$$ y_p(x) = u_1(x)\cos x + u_2(x)\sin x $$
Plugging in the solution and following the method, we arrive at the system of equations:
\begin{align} {u_1}'\cos x + {u_2}'\sin x &= 0 \\ -{u_1}'\sin x + {u_2}'\cos x &= \sin x \end{align}
Using Kramer's rule, we find
\begin{align} {u_1}' &= -\sin^2 x = \frac{-1+\cos 2x}{2} \\ {u_2}' &= \cos x\sin x = \frac{\sin 2x}{2} \end{align}
Integrating gives
\begin{align} y_p(x) &= \left(-\frac{x}{2} + \frac{\sin 2x}{4} \right)\cos x - \frac{\cos 2x}{4}\sin x \\
&= -\frac{x}{2}\cos x + \frac{\sin 2x\cos x-\cos 2x\sin 2x}{4} \\
&= -\frac{x}{2}\cos x +\frac{\sin x}{4} \end{align}
We can disregard the second term, since it'll combine with the fundamental solution. Thus, the general solution is
$$ y(x) = -\frac{x}{2}\cos x + c_1\cos x + c_2\sin x $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
The diophantine equation $5\times 2^{x-4}=3^y-1$ I have this question: can we deduce directly using the Catalan conjecture that the equation
$$5\times 2^{x-4}-3^y=-1$$
has or no solutions, or I must look for a method to solve it. Thank you.
| Elementary proof. I learned the method at Exponential Diophantine equation $7^y + 2 = 3^x$
We think that the largest answer is $5 \cdot 16 = 81 - 1. $ Write this as
$5 \cdot 16 \cdot 2^x = 81 \cdot 3^y - 1.$ Subtract $80$ from both sides,
$ 80 \cdot 2^x - 80 = 81 \cdot 3^y - 81.$ We reach
$$ 80 (2^x - 1) = 81 (3^y - 1). $$
This is convenient; we will show that both $x,y$ must be zero. That is, ASSUME both $x,y \geq 1.$
From $2^x \equiv 1 \pmod {81}$ we get
$$ x \equiv 0 \pmod {54}. $$
It follows that $2^x - 1$ is divisible by $2^{54} - 1.$
$$ 2^{54 } - 1 = 3^4 \cdot 7 \cdot 19 \cdot 73 \cdot 87211 \cdot 262657
$$
Next, $3^y - 1$ is divisible by the large prime $262657$
From $3^y \equiv 1 \pmod {262657}$ we find
$$ y \equiv 0 \pmod {14592} $$ and especially
$$ y \equiv 0 \pmod {2^8}. $$
We do not need as much as $2^8 = 256,$ we really just need the corollary
$$ y \equiv 0 \pmod 8 $$
Next $3^y - 1$ is divisible by $3^8 - 1 = 32 \cdot 5 \cdot 41.$ This is the big finish, $3^y - 1$ is divisible by $32.$ Therefore $80 (2^x-1)$ is divisible by $32,$ so that $2^x - 1$ is even. This is impossible if $x \geq 1,$ and is the contradiction needed to say that, in
$$ 80 (2^x - 1) = 81 (3^y - 1) \; , $$
actually $x,y$ are both zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Number theory question with floor function
Define $[a]$ as the largest integer not greater than $a$. For example, $\left[\frac{11}3\right]=3$. Given the function
$$f(x)=\left[\frac x7\right]\left[\frac{37}x\right],$$
where $x$ is an integer such that $1\le x\le45$, how many values can $f(x)$ assume?
A. $1$ B. $3$ C. $4$ D. $5$ E. $6$
I have attempted this question by brute force however I am looking for a cleaner, more systematic approach.
| First, $f(x) \le \left[\frac{x}{7}\cdot\frac{37}{x}\right] = 5$, and $f(x) \ge 0$. Therefore, $f(x)$ can assume a maximum of $6$ values: $0,1,2,3,4,5$.
Second, $f(x)=1$ implies $\left[\frac{x}{7}\right]=1$ and $\left[\frac{37}{x}\right]=1$. The first condition implies $7 \le x \le 13$, while the latter requires $19 \le x \le 37$, which is impossible. So $f(x) \neq 1$.
Checking first few values of $x$ reveals that $f(x)$ can assume the values $2,3,4,5$. Specifically, for $1 \le x \le 6$, we have $f(x)=0$ as $[x/7]=0$. For $x=7$, $f(x) = 5$; for $x=8,9$, $f(x)=4$; for $x=10, 11, 12$, $f(x)=3$, and for $x=13$, $f(x)=2$.
Consequently, the number of values $f(x)$ can assume is $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$ I'm solving the equation,
$$|x-1| + |x-2| = 1$$
I'm making cases,
$C-1, \, x \in [2, \infty) $
So, $ x-1 + x-2 = 1 \Rightarrow x= 2$
$C-2, \, x \in [1, 2) $
$x-1 - x + 2 = 1 \Rightarrow 1 =1 \Rightarrow x\in [1,2) $
$C-3, \, x \in (- \infty, 1)$
$ - x + 1 - x+2 = 1 \Rightarrow x= 1 \notin (-\infty, 1) \Rightarrow x = \phi$ (null set)
Taking common of all three solution set, I get $x= \phi$ because of the last case. But the answer is supposed to be $x \in [1,2]$
But when I write this equation in graphing calculator, it shows $2$ lines $x=1$ and $ x= 2$ rather than a region between $[1,2]$
Someone explain this too?
| By direct way we need to distinguish three cases
*
*$x<1 \implies |x-1| + |x-2| = 1-x+2-x=1 \implies -2x=-2\implies x=1$
*$1\le x<2 \implies |x-1| + |x-2| = x-1+2-x=1 \implies 1=1$
*$x\ge 2 \implies |x-1| + |x-2| = x-1+x-2=1 \implies 2x=4\implies x=2$
therefore $1\le x\le 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Verifying that $\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx$ without Cauchy-Schwarz
For real numbers $a,b,p$, verify that $$\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx\tag{1}$$ without using the Cauchy-Schwarz inequality.
Note that this is obviously an immediate consequence of C-S, and I assumed that it had to be shown by integrating directly.
Working backwards, we get $$\left[\frac{b^2-a^2}2-p(b-a)\right]^2\le(b-a)\frac{(b-p)^3-(a-p)^3}3$$ and since $m^3-n^3=(m-n)(m^2+mn+n^2)$, the RHS can be rewritten as $$\frac{(b-a)^2}4(b+a-2p)^2\le\frac{(b-a)^2}3(b^2+ab+b^2+3p^2-3p(b+a))$$ or $$3(b+a-2p)^2\le4((b+a)^2+3p^2-3p(b+a))-4ab$$ which gives $$12p^2-12p(b+a)\le(b-a)^2+12p^2-12p(b+a)$$ or that $$(b-a)^2\ge0$$ which is true.
Any alternative methods?
| Try Jensen's Inequality with $\varphi(x)=x^2.$ As mentioned in the comments, you need to normalize for the interval over which you are integrating. That is, you can prove
$$\left(\frac{1}{b-a}\int_a^b(x-p)\,dx\right)^{\!2}\le \frac{1}{b-a}\int_a^b(x-p)^2\,dx.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there a clever way to compute $\int \limits_a^b \sqrt[3]{(b-x)(x-a)^2} dx$ I'm wondering how to compute $\int \limits_a^b \sqrt[3]{(b-x)(x-a)^2} dx$. I tried doing Chebyshe[o]v's substitution to first compute the antiderivative which led me to $\int \frac{w^3}{(w^3+1)^3} dw$ what is quite unpleasant as Wolframalpha says.
Thanks.
| Thanks @Chappers for an idea. Here is the way of computing the product they was talking about:
We will require two basic properties. The first one is the Euler reflection formula:
$$\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin z \pi}$$
The second is $\Gamma(z+1)=z\Gamma(z)$. So,
$$\Gamma\left(\frac{5}{3}\right) \Gamma\left(-\frac{2}{3} \right) = \frac{\pi}{\sin \left(-\frac{2}{3} \pi\right)}$$
$$\Gamma\left(\frac{4}{3}\right) \Gamma\left(-\frac{1}{3} \right) = \frac{\pi}{\sin \left(-\frac{1}{3} \pi\right)}$$
We multiply:
$$\Gamma(5/3)\Gamma(-2/3)\Gamma(4/3)\Gamma(-1/3) = \frac{\pi^2}{\sin (-2/3 \pi) \cdot \sin (-1/3 \pi)}$$
$$\Gamma(5/3)\Gamma(4/3) = \frac{\pi^2}{\sin (-2/3 \pi) \cdot \sin (-1/3 \pi)\Gamma(-2/3)\Gamma(-1/3)}$$
$$\Gamma\left(\frac{1}{3} \right)=\Gamma\left(-\frac{2}{3}+1 \right)=-\frac{2}{3}\Gamma\left(-\frac{2}{3} \right)$$ $$\Gamma\left(\frac{2}{3} \right)=\Gamma\left(-\frac{1}{3}+1 \right)=-\frac{1}{3}\Gamma\left(-\frac{1}{3} \right)$$
Hence
$$\Gamma\left(-\frac{1}{3} \right) = -3\Gamma\left(\frac{2}{3} \right) \ \text{ и} \ \ \Gamma\left(-\frac{2}{3} \right) = -\frac{3}{2}\Gamma\left(\frac{1}{3} \right)$$
Finally,
$$\Gamma\left(\frac{5}{3}\right)\Gamma\left(\frac{4}{3}\right) =\displaystyle{\frac{\pi^2}{\sin \left(-\frac{2}{3} \pi\right) \cdot \sin \left(-\frac{1}{3} \pi\right)\cdot -\frac{3}{2} \Gamma\left(\frac{1}{3}\right) \cdot(-3)\Gamma\left(\frac{2}{3}\right)}}=$$
$$=\frac{2}{9}\cdot\frac{\pi^2}{\sin (-2/3 \pi) \cdot \sin (-1/3 \pi)\underbrace{\Gamma\left(1/3\right) \Gamma\left(2/3\right)}_{\displaystyle{\frac{\pi}{\sin(1/3\pi)}}}}=$$
$$\frac{2}{9}\cdot\frac{\pi \cdot \sin (1/3\pi)}{\sin(-2/3\pi)\cdot \sin(-1/3 \pi)}= \frac{4\pi}{9\sqrt{3}}$$
So:
$$\mathrm{B}\left(\frac{5}{3},\frac{4}{3} \right) = \frac{\Gamma(5/3) \Gamma(4/3)}{\Gamma(3)} = \frac{\Gamma(5/3) \Gamma(4/3)}{(3-1)!}=$$
$$= \frac{\Gamma(5/3) \Gamma(4/3)}{2}=\frac{2\pi}{9\sqrt{3}}$$
In conclusion we obtain:
$$\boxed{\displaystyle{\int \limits_a^b \sqrt[3]{(b-x)(x-a)^2} \ dx = (b-a)^2\frac{2\pi}{9\sqrt{3}}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim_{n\rightarrow \infty} ( \arctan\frac{1}{2} + \arctan \frac{1}{2.2^2} +....+ \arctan \frac {1}{2n^2})$
Calculate
$$\lim_{n\rightarrow \infty} \left( \arctan\frac{1}{2} + \arctan \frac{1}{2.2^2} +....+ \arctan \frac {1}{2n^2}\right)$$
My answer: i know that $$ \sum_{n=1}^N \arctan \left( \frac{2}{n^2} \right) =\sum_{n=1}^N \arctan (n+1)-\arctan(n-1)$$
as Im not able To find the $\sum_{k=1}^{n} \arctan \frac {1}{2k^2}$
I need help,,,,,any hints /solution will be aprreciated
thanks in advance
| $$\frac{1}{2n^2}=\frac{2}{4n^2}=\frac{2}{1+4n^2-1}=\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}$$
Therefore,
$$\arctan \left( \frac{1}{2n^2}\right) = \arctan (2n+1) - \arctan(2n-1) $$
Can you perform the telescoping sum now?
| {
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Given a matrix $A$ find $A^n$. $A=$$
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} $
Find $A^n$.
My input:
$A^2= \begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} = \begin{bmatrix}
1 & 4\\
0 & 1
\end{bmatrix} $
$A^3 = \begin{bmatrix}
1 & 6\\
0 & 1
\end{bmatrix} $
......
$A^n = \begin{bmatrix}
1 & 2n\\
0 & 1
\end{bmatrix} $
This was very basic approach. I want to know if there is any other way a smart trick or something to solve this problem ?
| The minimal polynomial of $A$ is $(x-1)^2$ so for any entire function $f$ we have
$$f(A) = f(1)P + f'(1)Q$$
for some $P, Q$ polynomials in $A$.
Plugging in $f \equiv 1$ and $f(x) = x$ we get $P = I$ and $Q = A-I$.
Therefore for $f(x) = x^n$ we have
$$A^n = 1^n\cdot I + n1^{n-1}\cdot (A-I) = I + n(A-I) = \begin{bmatrix} 1 & 2n \\ 0 & 1\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Find an equivalent to $\sum_{k=n}^\infty \frac{1}{k!}$ I would like to find an equivalent of $\sum_{k=n}^\infty \frac{1}{k!}$. Can I do as follow ? (since I always have doubt with those $o$ and $O$, I would like your opinion.
$$n!\sum_{k=n}^\infty \frac{1}{k!}=1+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots$$
$$=1+\frac{1}{n+1}+\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right)=1+o(1).$$
and thus the claim follow.
Q1) Is it correct ?
Q2)
I have that $\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right)$ because $$\frac{1}{(n+1)(n+2)}\leq \frac{1}{(n+1)^2},\quad \frac{1}{(n+1)(n+2)(n+3)}\leq \frac{1}{(n+2)^2},\quad \frac{1}{(n+1)(n+2)(n+3)(n+4)}\leq\frac{1}{(n+3)^2} \cdots$$
and so $\frac{1}{(n+1)(n+2)\cdots(n+k)}\leq \frac{1}{(n+k)^2}$. But suppose we could only prove that $$\frac{1}{(n+1)(n+2)\cdots(n+k)}\leq \frac{C_k}{(n+k)^2},$$
i.e. that the constant depend on $k$ and we can't have a uniform bound. Could I conclude that $$ \frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots=\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right) \ \ ?$$
| See this post: Calculate $\lim_{n \rightarrow \infty}$ ($n!e-[n!e]$)?
It follows that the integer part of $(n-1)!e$ is $(n-1)!\sum\limits_{k=1}^{n-1} \frac{1}{k!}$, so the fractional part of $(n-1)!e$ is $(n-1)!\sum\limits_{k=n}^{\infty} \frac{1}{k!}$.
Hence, an equivalent expression is $\frac{\{(n-1)!e\}}{(n-1)!}$.
| {
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Evaluating the integral $\int \frac1{(2+3\cos x)^2}\mathrm dx$ Please someone give me an idea to evaluate
this: $$\int \frac1{(2+3\cos x)^2}\mathrm dx$$
I don't even know how to start cause even multiplying an dividing by $\cos^2x$ does not work, so help me here.
| You can use the following approach:
$$\int{\frac{dx}{(2+3\cos x)^2}}=\int{\frac{dx}{(2\cos^2{\frac{x}2}+2\sin^2{\frac{x}2}+3\cos^2{\frac{x}2}-3\sin^2{\frac{x}2})^2}}=$$
$$=\int{\frac{dx}{(5\cos^2{\frac{x}2}-\sin^2{\frac{x}2})^2}}=\int\frac{dx}{\cos^4{\frac x2}(5-\tan^2{\frac x2})^2}=\left[t=\tan\frac{x}{2}\right]=$$
$$=2\int\frac{1+t^2}{(5-t^2)^2}dt=...$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the sum of $1+(1+r)s+(1+r+r^2)s^2+\dots$? I was asked to find the geometric sum of the following:
$$1+(1+r)s+(1+r+r^2)s^2+\dots$$
My first way to solve the problem is to expand the brackets, and sort them out into two different geometric series, and evaluate the separated series altogether:
$$1+(s+rs+\dots)+(s^2+rs^2+r^2s^2+\dots)$$
The only problem is it doesn't seem to work, as the third term, $(1+r+r^2 +r^3)s^3$ doesn't seem to fit the separated sequence correctly.
Any help would be appreciated.
| You expanded the brackets, but did not actually group:
$$ 1 + (1+r)s + (1+r+r^2)s^2 + \dotsb
= 1 + (s+rs \color{red}{{} + \dotsb}) + (s^2+rs^2+r^2s^2\color{red}{{} + \dotsb}). $$
Here is the way to group:
\begin{align*}
& 1+(1+r)s+(1+r+r^2)s^2+\dotsb \\
&= 1+(\color{red}{s}+\color{green}{rs})+(\color{red}{s^2}+\color{green}{rs^2}+\color{blue}{r^2s^2})+\dotsb \\
&= (1+\color{red}{s}+\color{red}{s^2}+\dotsb)+(\color{green}{rs}+\color{green}{rs^2}+\dotsb)+(\color{blue}{r^2s^2}+r^2s^3+\dotsb) \\
&= \frac{1}{1-s}+\frac{rs}{1-s}+\frac{r^2s^2}{1-s}+\dotsb \\
&= \frac{1}{1-s}(1+rs+r^2s^2+\dotsb)
\end{align*}
Can you finish?
| {
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Let $m$ be the largest real root of the equation $\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4$ find $m$
Let $m$ be the largest real root of the equation $$\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4.$$ Find $m$.
do we literally add all the fractions or do we do something else
I have no clue how to solve this
| HINT:
Write the LHS as $$\left(\frac x{x-3}-1\right)+\left(\frac x{x-5}-1\right)+\left(\frac x{x-17}-1\right)+\left(\frac x{x-19}-1\right)$$ and this gives a constant term of $-4$. Equating this with the RHS, we have$$x\left(\frac1{x-3}+\frac1{x-5}+\frac1{x-17}+\frac1{x-19}\right)=x(x-11)$$ and note that $3,5$ are 'symmetrical' around $11$; that is, $17-11=11-5$ and $19-11=11-3$.
| {
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Let $W_{1},W_{2}$ be sub-spaces of $\mathbb{R}^{4}$, find a subspace $W_{3}$ s.t $W_3\subset W_{2}$ and $W_{1}\oplus W_{3}=W_{1}+W_{2}$ Let $W_{1},W_{2}$ be linear sub-spaces of $\mathbb{R}^{4}$.
$W_{1}=\text{sp}\{(1,2,3,4),(3,4,5,6),(7,8,9,10)\}$
$W_{2}=\text{sp }\{(x,y,z,w)| \ x+y=0\}$
Find a linear subspace of $ \ \mathbb{R}^{4} \ ; W_{3}$ , such that: $W_{3}\subset W_{2} \ \text {*and*} \ W_{1}\oplus W_{3}=W_{1}+W_{2}$
My attempt:
I applied Gaussian elimination on the vectors of $W_1$, such that:
$$W_{1}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=a\left(\begin{array}{c}
1\\
0\\
-2\\
-1
\end{array}\right)+b\left(\begin{array}{c}
0\\
1\\
-2\\
-1
\end{array}\right)|a,b\in\mathbb{R}\right\} $$
At that point I got stuck. I'm not sure how to continue.
| (I won't change the notation although it's not 100% correct)
Hint: Rewrite the basis of $W_1$ as
$$W_{1}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=a\left(\begin{array}{c}
1\\
0\\
-2\\
-1
\end{array}\right)+b\left(\begin{array}{c}
1\\
-1\\
0\\
0
\end{array}\right)\right\} $$
A basis of $W_2$ is
$$W_{2}=\left\{ \left(\begin{array}{c}
x\\
y\\
z\\
w
\end{array}\right)=a\left(\begin{array}{c}
1\\
-1\\
0\\
0
\end{array}\right)+b\left(\begin{array}{c}
0\\
0\\
1\\
0
\end{array}\right)+c\left(\begin{array}{c}
0\\
0\\
0\\
1
\end{array}\right)\right\} $$
can you take it from here?
| {
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Compare $\arcsin (1)$ and $\tan (1)$
Which one is greater: $\arcsin (1)$ or $\tan (1)$?
How to find without using graph or calculator?
I tried using $\sin(\tan1)\leq1$, but how do I eliminate the possibility of an equality without using the calculator?
| I try to calculate $\tan 1 <1.57<\frac{\pi}{2}$
$\tan x =x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62x^9}{2835}+o(x^9)$
$|\tan x-x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62x^9}{2835}|<\frac{62x^9}{2835},x\in[0,1]$
So:$\tan 1<1+\frac{1}{3}+\frac{2}{15}+\frac{17}{315}+\frac{62}{2835}+\frac{62}{2835}<1.565<1.57<\frac{\pi}{2}=\arcsin 1$
Edits:
It seems that there just need four terms by under @Ian reminding
$\tan x =x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}+\frac{62\xi^9}{2835},\xi\in[0,1]$
$\tan 1<1+\frac{1}{3}+\frac{2}{15}+\frac{17}{315}+\frac{62}{2835}<1.56<1.57<\frac{\pi}{2}=\arcsin 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 0
} |
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