Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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A teacher divides seven girls and five boys randomly into three groups of four each. What is the probability that one group has four girls? Question in title. Sorry, this is my first question and I don't know how to format.
My thinking was
$$\frac{3\binom{7}{4}\binom{8}{4}\binom{4}{4}}{\binom{12}{4}\binom{8}{4}\binom{4}{4}}.$$
The numerator comes from 3 groups to put four girls into, and $\binom{7}{4}$ ways to choose those girls. Once that group is set, there are $\binom{8}{4}$ to put kids into the second group and $\binom{4}{4}$ ways to put kids into the third group.
That gives me an answer of $7/132$ but the book I'm using says the answer is $7/33$. Can you please help me find what I'm missing? Thank you!
| Your method of solution is indeed correct. Check your arithmetic.
$$\frac{3\cdot \binom{7}{4} \binom{8}{4} \binom{4}{4}}{\binom{12}{4}\binom{8}{4}\binom{4}{4}} = \frac{3\cdot \binom{7}{4}}{\binom{12}{4}}=\frac{3\cdot 35}{495}=\frac{7}{33}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Condition for a line to intersect a curve The line $lx + my + n = 0$ intersects the curve $ax^2+ 2hxy + by^2 = 1$ at the point $P$ and $Q$. The circle on $PQ$ as diameter passes through the origin.
We have to prove that $$n^2(a^2 + b^2) = l^2+ m^2$$
I write the equation of circle as $C+\lambda L=0$
But satisfying $(0,0)$ I got $\lambda =\dfrac{1}{n}$
Now how to proceed?
| Let the circle passes through $O(0,0)$, $P(x_1,y_1)$ and $Q(x_2,y_2)$.
Now,
$$
\left \{
\begin{align*}
aX^2+2hXY+bY^2 &= 1 \\
\ell X+mY +n &= 0
\end{align*}
\right.$$
$$\implies
aX^2-2hX\left( \frac{\ell X+n}{m} \right)+
b\left( \frac{\ell X+n}{m} \right)^2=1$$
$$\left( a-\frac{2h\ell}{m}+\frac{b\ell^2}{m^2} \right)X^2+
2n\left( \frac{b\ell}{m^2}-\frac{h}{m} \right)X+\frac{bn^2}{m^2}-1=0$$
$$x_1 x_2=\frac{bn^2-m^2}{am^2-2h\ell m+bl^2}$$
Similarly,
$$\left( b-\frac{2hm}{\ell}+\frac{am^2}{\ell^2} \right)Y^2+
2n\left( \frac{am}{\ell^2}-\frac{h}{\ell} \right)Y+\frac{an^2}{\ell^2}-1=0$$
$$y_1 y_2=\frac{an^2-\ell^2}{am^2-2h\ell m+b\ell^2}$$
Since $PQ$ is the diameter, $OP \perp OQ$.
We have
$$\frac{y_1}{x_1} \times \frac{y_2}{x_2}=-1$$
$$\frac{an^2-\ell^2}{bn^2-m^2}=-1$$
$$\fbox{$n^2(a+b)=\ell^2+m^2$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Finding the number of $4$-digit numbers so no two adjacent digits are even Find the number of $4$-digit numbers with distinct digits chosen from the set $\{0,1,2,3,4,5\}$ in which no two adjacent digits are even.
| Since you have three odd and three even digits, you must use at least one even digit. But you cannot use three even digits as that would force two adjacent even digits.
a) One even digit: There are $3$ ways to choose the even digit, and $3\cdot 2\cdot 1$ ways to choose the odd digits from left to right, and $4$ ways to choose the position of the even one:
$3\cdot 3\cdot 2\cdot 1\cdot 4=72$
b) Two even digits: There are $3\cdot 2$ ways to choose the even digits from left to right, and $3\cdot 2$ ways to choose the odd digits from left to right, and ${4\choose 2}-3$ ways to choose the even positions non-adjacent:
$3\cdot 2\cdot 3\cdot 2\cdot(6-3)=108$
In total, we get
$72+108=180$.
(The above assumes that leading $0$ is allowed.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Find a positive integer $n$ (or prove that none exists) such that $\gcd(3^{n+1}-2,2^n-3) > 1$
Find a positive integer $n$ (or prove that none exists) such that $\gcd(3^{n+1}-2,2^n-3) > 1$.
The greatest common divisor here seems to be equal to one for most values (I checked up to $7000$), but I didn't see an easy way of finding an $n$ for which this it is greater than $1$. How could we go about solving this?
| This is not correct!
If $n$ is such that $\gcd(3^{n+1}-2,2^n-3)>1$, then there exists a prime $p$ such that $3^{n+1}=2\bmod p$ and $2^n=3\bmod p$. But this implies
$$3^{n+1}=(2^n)^{n+1}=2^{n^2+n}=2\bmod p,$$
hence $2^{n^2+n-1}=1\bmod p$, and hence $\varphi(p)=p-1$ divides $n^2+n-1$, no, all we can conclude is $\gcd(p-1,n^2+n-1)>1$. Considering the possibilities modulo 4 shows that this is not possible.
Elaboration on the mod 4 argument: If $p-1$ divides $n^2+n-1$ then
$$n^2+n-1=k(p-1)$$
for some integer $k$. Note that $p$ is odd since both $3^{n+1}-2$ and $2^n-3$ are odd. Thus $p=1,3\bmod 4$, so
$$k(p-1)=0,2\bmod 4.$$
On the other side, if $n$ is odd or $n=2\bmod 4$ then
$$n^2+n-1=1\bmod 4,$$
and if $n=0\bmod 4$ then
$$n^2+n-1=3\bmod 4.$$
Therefore $n^2+n-1\ne k(p-1)$, so $p-1$ does not divide $n^2+n-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve: $(y\sqrt{1-y^2})dx+(x\sqrt{1-y^2}+y)dy=0$
$$(y\sqrt{1-y^2})dx+(x\sqrt{1-y^2}+y)dy=0$$
$$\frac{\partial M}{\partial y}=\sqrt{1-y^2}-\frac{-2y^2}{2\sqrt{1-y^2}}=\frac{1-2y^2}{\sqrt{1-y^2}}$$
$$\frac{\partial N}{\partial x}= \sqrt{1-y^2}$$
$$\frac{M_{y}-N_{x}}{M}=\frac{\frac{1-2y^2}{\sqrt{1-y^2}}-\sqrt{1-y^2}}{y\sqrt{1-y^2}}=\frac{-y}{1-y^2}=h(y)$$
$$I=\exp{\left(-\int h(y) dy\right)}= \exp{\left(-\int\frac{-y}{1-y^2}dy\right)}=\exp{\left(\frac{-\ln|1-y^2|}{2}\right)}$$
Is there a way to simplify $$\exp{\left(\frac{-\ln|1-y^2|}{2}\right)}\;?$$
| $a\ln b=\ln b^a\implies -\frac{1}{2}\ln \left |1-y^2\right |=$ $\ln \frac{1}{\sqrt{1-y^2}}$ and $e^{\ln u}=u$ integrating factor is going to be $\frac{1}{\sqrt{1-y^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega\neq 1$, find ... Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}$$
I've been having trouble with this unit, need help on solving this problem.
| Hint: using $\omega^5=1$ and bringing all fractions to numerator $1$ gives:
$$
\require{cancel}
\begin{align}
\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3} & = \frac{\omega^4}{\omega^4}\frac{\omega}{1 - \omega^2} + \frac{\omega^3}{\omega^3}\frac{\omega^2}{1 - \omega^4} + \frac{\omega^2}{\omega^2}\frac{\omega^3}{1 - \omega} + \frac{\omega}{\omega}\frac{\omega^4}{1 - \omega^3} = \\
& = \cancel{\frac{1}{\omega^4-\omega}} + \bcancel{\frac{1}{\omega^3-\omega^2}}+\bcancel{\frac{1}{\omega^2-\omega^3}} + \cancel{\frac{1}{\omega-\omega^4}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Summation problems involving $k c^{-k}$ I am trying to find a better way of solving problems involving summations of the form: $\sum_{k=0}^\infty k\cdot c^{-k}$ and $\sum_{k=0}^\infty k^2 \cdot c^{-k}$
For example, finding $\sum_{k=0}^\infty k\cdot 2^{-k}$, I used the following approach:
$S = \frac{1}{2} + 2 \cdot \frac{1}{2^2} + 3 \cdot \frac{1}{2^3} + ....$
$S = \frac{1}{2}[1+2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2^2} + .....]$
$S = \frac{1}{2}[1+\sum_{k=0}^\infty (k+1)\cdot 2^{-k}]$
$\sum_{k=0}^\infty (k+1)\cdot 2^{-k} = \sum_{k=0}^\infty k\cdot 2^{-k} + \sum_{k=0}^\infty 2^{-k} = S + \sum_{k=0}^\infty 2^{-k}$
Therefore, $S = 1 + \sum_{k=0}^\infty 2^{-k} = 1 + \frac{1}{1-\frac{1}{2}} = 3$
I may have made a few calculation mistakes (I would appreciate if you could point that out), but my main concern is a simpler way to approach this kind of problems. Can I use this method to evaluate problems of the form $\sum_{k=0}^\infty k^2 \cdot c^{-k}$ ? I tried, but failed.
I would appreciate if someone could point out the most efficient way to solve this
| Hint: Observe
\begin{align}
\frac{1}{1-x} = \sum^\infty_{n=0}x^n
\end{align}
when $|x|<1$. In particular, we have that
\begin{align}
\frac{1}{(1-x)^2}=\frac{d}{dx}\frac{1}{1-x} = \sum^\infty_{n=1} nx^{n-1} \ \ \Rightarrow \ \ \frac{x}{(1-x)^2} = \sum^\infty_{n=1}nx^n.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Linear Transformation textbook problem A problem from my Linear Algebra textbook:
Let $T$ be a linear transformation from $M_{2,2}$ into $M_{2,2}$ such that
$T\begin{pmatrix}
\begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix}\end{pmatrix}
=
\begin{bmatrix} 1&-1 \\ 0&2 \end{bmatrix}$,
$T\begin{pmatrix}
\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}\end{pmatrix}
=
\begin{bmatrix} 0&2 \\ 1&1 \end{bmatrix}$,
$T\begin{pmatrix}
\begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix}\end{pmatrix}
=
\begin{bmatrix} 1&2 \\ 0&1 \end{bmatrix}$,
$T\begin{pmatrix}
\begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}\end{pmatrix}
=
\begin{bmatrix} 3&-1 \\ 1&0 \end{bmatrix}$.
Find $T\begin{pmatrix}
\begin{bmatrix} 1&3 \\ -1&4 \end{bmatrix}\end{pmatrix}$
I just need to know how to start this. I tried setting it up like this:
$\begin{bmatrix} a&b \\ c&d \end{bmatrix}$
$\begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix}$
= $\begin{bmatrix} 1&-1 \\ 0&2 \end{bmatrix}$
which gets me to here:
$\begin{bmatrix} a&0 \\ c&0 \end{bmatrix}$
= $\begin{bmatrix} 1&-1 \\ 0&2 \end{bmatrix}$
and then I'm lost...
| Use linearity:
$$T\left(\sum_{i=1}^n \alpha_i A_i\right) = \sum_{i=1}^n \alpha_i T(A_i)$$
Also, note that a linear transformation from $M_{2,2}$ to $M_{2,2}$ is not a $2\times 2$ matrix itself.
Finally, note that
$$\begin{align*}
\pmatrix{1&3\\-1&4} & = \pmatrix{1&0\\0&0} + 3 \pmatrix{0&1\\0&0} - \pmatrix{0&0\\1&0} + 4 \pmatrix{0&0\\0&1} \\
\Rightarrow T\pmatrix{1&3\\-1&4} & = T\pmatrix{1&0\\0&0} + 3T\pmatrix{0&1\\0&0} - T\pmatrix{0&0\\1&0} +4T\pmatrix{0&0\\0&1}\\
& = \pmatrix{1&-1\\0&2} + 3\pmatrix{0&2\\1&1} - \pmatrix{1&2\\0&1} + 4\pmatrix{3&-1\\1&0} \\
& = \pmatrix{12&-1\\7&4}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Concurrent lines proof for a regular 18-gon
Let $X_1 X_2 \dotsb X_{18}$ be a regular 18-gon. Show that $X_1 X_{10}$, $X_2 X_{13}$, and $X_3 X_{15}$ are concurrent.
What would be the best way to prove this? I am actually struggling understanding 'concurrent' as I am not sure how they are concurrent in a regular 18-gon
| $\qquad\qquad$
Let $A,B,C$ be the intersection point of $X_{13}X_1$ with $X_{15}X_3$, $X_{13}X_3$ with $X_{10}X_1$, $X_{13}X_2$ with $X_1X_3$ respectively.
Then, we get
$$\begin{align}\frac{AX_1}{X_{13}A}\cdot\frac{CX_3}{X_1C}\cdot\frac{BX_{13}}{X_3B}&=\frac{X_1X_3\sin 40^\circ}{X_{13}X_3\sin 20^\circ}\cdot\frac{X_3X_{13}\sin 10^\circ}{X_1X_{13}\sin 10^\circ}\cdot\frac{X_1X_{13}\sin 30^\circ}{X_1X_3\sin 70^\circ}\\\\&=\frac{\sin 40^\circ\sin 30^\circ}{\sin 20^\circ\sin 70^\circ}\\\\&=\frac{\frac 12\sin 40^\circ}{\sin 20^\circ\sin 70^\circ}\\\\&=\frac{\sin 20^\circ\cos 20^\circ}{\sin 20^\circ \sin 70^\circ}\\\\&=1\end{align}$$
The claim follows from Ceva's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A question about domains and ranges: find the domain and range of the function $f(x)=\sqrt{1-x^2}$, without any digital help If $f(x)=\sqrt{1-x^2}$, then the domain of $f$ is $[-1,1]$. Respectively, the range of $f$ is $[0,1]$.
How do I understand that the domain of $f$ is $[-1,1]$ and that the range of $f$ is $[0,1]$, without a graphing calculator?
Let us suppose that I don't know the given intervals from start. Instead, suppose that I know only that the function that I should find the domain and range of is $f(x)=\sqrt{1-x^2}$. The question will then be: 'Find the domain and range of the function $f(x)=\sqrt{1-x^2}$, without any digital help'
| Method 1:
Domain. In the real numbers $\sqrt{w}$ exists if and only if $w \ge 0$, so the domain is all real $x$ so that $1-x^2 \ge 0$ or $x^2 \le 1$ or $-\sqrt 1 \le x \le \sqrt 1$ or $-1 \le x \le 1$.
So domain is $[-1,1]$.
Range: $x$ may be anything so that $-1 \le x \le 1$. That means $x^2$ may be anything so that $0 \le x^2 \le 1$ That means $-x^2$ may be anything $-1 \le -x^2 \le 0$. Which means $1-x^2$ may be anything $0 \le 1-x^2 \le 1$ so $0 \le \sqrt{1- x^2} \le 1$.
So range is $[0,1]$.
Method 2:
The graph is $y = \sqrt{1-x^2} \ge 0$
So $y^2 = 1-x^2$
$y^2 + x^2 = 1$. That's a graph of a unit circle centered at $(0,0)$
So $y = \sqrt{1 -x^2}$ a half circle restricted to $y \ge 0$.
In this half circle, $x$ goes from $-1$ to $1$ so the domain is $[-1,1]$.
In the full circle $y$ goes from $-1$ to $1$ but we are restricting to $y\ge 0$ so the range is $[0,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Unsure how to simplify power series If I want to find the power series representation of the following function:
$$ \ln \frac{1+x}{1-x} $$
I understand that it can be written as
$$ \ln (1+x) - \ln(1-x) $$
And I understand that if I now write in the power series representations for $ln(1+x)$ and $ln(1-x)$:
$$\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n} $$
My textbook solution does an odd thing where it writes it out as
$$\sum_{n=1}^\infty \frac{x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n} $$
$$2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1} $$
I have no idea how it got from the line where I have the power series representation for $ln(1+x)$ and $ln(1-x)$ to the last two lines. If anyone could help me link my part to the textbook solution I would really appreciate it! Thank you!
| $$\begin{align}\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-x)^{n}}{n}&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{n-1}(-1)^nx^{n}}{n}\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{(-1)^{2n-1}x^{n}}{n}\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} - \sum_{n=1}^\infty \frac{-x^{n}}{n}\\
&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{n}}{n} + \sum_{n=1}^\infty \frac{x^{n}}{n}\\
&=\left[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots\right]+ \left[x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\dots\right]\\
&=2\left[\frac{x^1}{1}+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\dots\right]\\
&=2\sum_{n=1}^\infty \frac{x^{2n-1}}{2n-1}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A problem about groups of matrices I found on the web the following problem and I have no idea how to solve it.
Let $X\subseteq \mathbb C^4$ be the set defined by the equation $$x^2=y^3+z^3+w^3.$$ Let $\tau$ be the involution over $\mathbb C^4$ given by $\tau(x,y,z,w)=(-x,y,z,w).$ Let $G$ be a finite subgroup of $GL(4,\mathbb C)$ such that:
1) $g(X)\subseteq X,$ $\forall g\in G,$
2) $G$ contains the (matrix whose action is the same as the) involution $\tau.$
Prove that $\tau$ belongs to the center of $G.$
Prove that the following assertion is false: "there exists a subgroup $H$ of $G$ of index two such that $\tau\notin H.$"
| I proved the proposition as follows.
All the linear spaces of dimension one contained in $X$ are of the form
$$\left( \begin{array}{c} x \\
y \\
z\\
w
\end{array} \right) =t\cdot \left( \begin{array}{c} a \\
b \\
c\\
d
\end{array} \right)$$ with $$(ta)^2=(tx)^3+(ty)^3+(tz)^3, \quad \forall t$$ where $t$ is a complex parameter and $(a,b,c,d)\in \mathbb C^4$ is fixed. Since the previous equation is true for every $t,$ $a$ must be $0$ and $b,c,d$ such that $b^3+c^3+d^3=0.$
Since every matirx $g\in G$ takes a linear one-dimensional subspace contained in $X$ into another, if $g=(g_{i,j})_{1\leq i,j\leq 4},$ the first row of $$g\cdot \left( \begin{array}{c} 0 \\
b \\
c\\
d
\end{array} \right)=\left( \begin{array}{c} g_{1,2}b+g_{1,3}c+g_{1,4}d \\
...\\
...\\
...
\end{array} \right)$$ must be zero. Hence $g_{1,2}b+g_{1,3}c+g_{1,4}d=0$ for every
$b,c,d\in \mathbb C$ with $b^3+c^3+d^3=0.$ This implies that $g_{1,2}=g_{1,3}=g_{1,4}=0.$
Similarly, since the vector $(t,\sqrt[3]{t^2},0,0)\in X$ $\forall t\in \mathbb R$ we have that $\forall g\in G$ and $\forall t\in\mathbb R,$ $$g\cdot \left( \begin{array}{c} t \\
\sqrt[3]{t^2} \\
0\\
0
\end{array} \right) \in X$$ so that $$(tg_{11})^2=(tg_{2,1}+\sqrt[3]{t^2}g_{2,2})^3+(tg_{3,1}+\sqrt[3]{t^2}g_{3,2})^3+(tg_{4,1}+\sqrt[3]{t^2}g_{4,2})^3\quad \forall t\in \mathbb R.$$ This implies that $g_{2,1}=g_{3,1}=g_{4,1}=0.$
Hence every matrix $g\in G$ has the form $$g=\left(\begin{array}{c|ccc}
\ast & & & \\ \hline
& \ast & \ast & \ast \\
& \ast & \ast & \ast \\
& \ast & \ast & \ast \\
\end{array}\right)$$ and $\tau $ is in the center of $G.$
For the second part of the question, consider the gruop $G$ of matrices of the form
$$\left(\begin{array}{c|ccc}
\pm 1 & 0 & 0 & 0 \\
\hline
0 & \huge{A} \\
0 & & &\\
0 & & &
\end{array}\right)$$ or $$\left(\begin{array}{c|ccc}
\pm i & 0 & 0 & 0 \\
\hline
0 & \huge{-A} \\
0 & & &\\
0 & & &
\end{array}\right)$$ where $A$ is a $3\times 3$ permutation matrix. It is easily seen that all the subgroups $H$ of index 2 of this group $G$ contain the involution $\tau.$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the second derivative of $f(x)=\frac{x^2}{10+x}$ How do you find the second derivative of $f(x)=\frac{x^2}{10+x}$?
I get $f'(x)={20x+x^2}/{(10+x)^2}$
Then get stuck here:
$(10+x)^2 (20+2x) - (20x+x^2) \frac {2(10+x)}{((10+x)^2)^2}$
| f(x) = $\frac{x^2}{10+x}$
You need to use the quotient rule: $(\frac{f}{g})'=\frac{f'*g-g'*f}{g^2}$
Then f= $x^2$ take the derivative of that and you get $\rightarrow f'=2x$
$g= 10+x$, take the derivative of that and you get $\rightarrow g'= 1$
Now let us apply the quotient rule and use our following notations.
$f'(x)= \frac{(2x)*10+x - ((1)*x^2)}{(10+x)^2}$
Now let us simplify and we get $\rightarrow \frac{20x+x^2}{(10+x)^2}$
Let us get $f''(x)$, but first let us get the notations ready to substitute them.
$f'= 20x+x^2 \rightarrow f''=20+2x$ & $g'= (10+x)^2 \rightarrow g''= 2(10+x)$
Then we use the quotient rule again and we get:
$f''(x)= \frac{(2x+10)*(10+x)^2 - (20x+x^2(2(10+x))}{((10+x)^2)^2}$
We simplify and get $\rightarrow f''(x)= \frac{200}{(x+10)^3}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that: $\cos^2 20° + \cos^2 40° +\cos^2 80° = \sin^2 20° + \sin^2 40° + \sin^2 80°$ Prove that: $\cos^2 20° + \cos^2 40° +\cos^2 80° = \sin^2 20° + \sin^2 40° + \sin^2 80°$
My Attempt:
$$L.H.S=\cos^2 20° + \cos^2 40° + \cos^2 80°$$
$$=\dfrac {1+\cos 40}{2}+\dfrac {1+\cos 80}{2} + \dfrac {1+\cos 160°}{2}$$
$$=\dfrac {3+\cos 40°+\cos 80°+\cos 160°}{2}$$
I.could not solve further from here..
| We know the identity $$\cos^2 x - \sin^2 x = \cos 2x \tag {1}$$ and $$\cos A - \cos B = 2\cos \frac {A+B}{2} \cos \frac {A-B}{2} \tag {2} $$
We have $$(\cos^2 20^\circ - \sin^2 20^\circ) + (\cos^2 40^\circ - \sin^2 40^\circ) + (\cos^2 80^\circ - \sin^2 80^\circ) $$ $$= \cos 40^\circ + \cos 80^ \circ + \cos 160^ \circ $$ (using $(1)$) $$ =2\cos 60^\circ \cos 20^\circ +\cos (90^\circ + 70^\circ) $$ (using $(2)$) $$ = \cos 20^\circ - \sin 70^\circ =0$$
where we used the identity $\cos (90^\circ + x) = -\sin x $.
Hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does there exist a factorization consisting of polynomials of $x^4+3x^2+6$ So as I understand it every polynomial with real coefficients should have a factorization consisting of polynomials of degree one(In case of real roots) and degree two(complex roots).
But I have been unable to find such a factorization of the polynomial $x^4+3x^2+6$. Even using gp and Mathematicas Factor function I only get the original polynomial as an answer.
So my question is. Does there exist a factorization?
| Let $y=x^2$ then solving the quadratic $y^2+3y+6=0$ gives $y = \frac{1}{2}(-3 \pm i \sqrt{15})$.
Then solving:
$$
x^2 = \frac{1}{2}(-3 + i \sqrt{15}) \implies
x_{1,2} = \pm \frac{1}{2}\left(\sqrt{2\sqrt{6}-3}+i \sqrt{2\sqrt{6}+3}\right) \\
x^2 = \frac{1}{2}(-3 - i \sqrt{15}) \implies
x_{3,4} = \pm \frac{1}{2}\left(\sqrt{2\sqrt{6}-3}-i \sqrt{2\sqrt{6}+3}\right)
$$
Note that $x_3=\overline{x_1}$ and $x_4=\overline{x_2}$ so both polynomials $(x-x_1)(x-x_3)$ and $(x-x_2)(x-x_4)$ have real coefficients, which can be easily calculated once you know the roots. Those two polynomials are the quadratic factors of the original polynomial.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the indicated derivative Find the indicated derivative
$\frac {d}{dt}$ $\frac {(6t-5)^6}{t+9}$
I'm stuck after getting to this part
$\frac {(6t-5)^6-36(6t-5)^5(t+9)}{(t+9)^2}$
How do they get to the answer
$\frac {(6t-5)^5(30t+329)}{(t+9)^2}$
| They simply factor out $(6t-5)^5$:
$$\frac {(6t-5)^6-36(6t-5)^5(t+9)}{(t+9)^2}=\frac{(6t−5)^5(6t−5-36(t+9))}{(t+9)^2}.
$$
(Note there seems o be a sign error in your computation)
Faster with logarithmic differentiation:
$$\frac{f'(x)}{f(x)}=\frac{36}{6t-5}-\frac1{t+9}=\frac{30t+329}{(6t-5)(t+9)}, $$
so
$$f'(x)=\frac{30t+329}{(6t-5)(t+9)}\cdot f(x)=\frac{(6t-5)^5(30t+329)}{(t+9)^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Rational solutions for $\tan(\pi x)=y$ Is there a rational solution for the following equation?
$$\tan (\pi x)=y\\y\neq-1,0,1$$
I guess there is none, but I have no idea how to solve/prove it.
EDIT: I think also that if $y$ is rational, then $x$ is not even algebraic, but this must be much harder to prove.
| Step 1: Define Chebyshev polynomials as $T_0(x) = 1$, $T_1(x) = x$, $T_{n + 1}(x) = 2x T_n(x) - T_{n - 1}(x)\ (n \geqslant 1)$. It is well-known that (See Wiki)$$
T_n(\cos θ) = \cos(nθ). \quad \forall θ \in \mathbb{R}
$$
Now define $S_n(x) = 2T_n\left( \dfrac{x}{2} \right)$, then $S_0(x) = 1$, $S_1(x) = x$, $S_{n + 1}(x) = x S_n(x) - S_{n - 1}(x)\ (n \geqslant 1)$. It is easy to prove by induction that $S_n \in \mathbb{Z}[x]$, $\deg S_n = n$ and $S_n(x)$ is monic, i.e. the leading coefficient of $S_n(x)$ is $1$.
Step 2: If $a, b \in \mathbb{Q}$ satisfy $\cos(πa) = b$, then $b \in \left\{ 0, \pm\dfrac{1}{2}, \pm1 \right\}$.
Proof: Suppose $a = \dfrac{c_1}{d_1}$, $2b = \dfrac{c_2}{d_2}$, where $c_1, c_2 \in \mathbb{Z}$, $d_1, d_2 \in \mathbb{N}_+$, and $(c_1, d_1) = (c_2, d_2) = 1$. On the one hand,$$
S_{2d_1}(2b) = 2T_{2d_1}(b) = 2T_{2d_1}\left( \cos\left( \frac{πc_1}{d_1} \right) \right) = 2\cos(2πc_1) = 1.
$$
On the other hand, suppose that $S_{2d_1}(x) = x^{2d_1} + \sum\limits_{k = 0}^{2d_1 - 1} a_k x^k$, where $a_k \in \mathbb{Z}$, then$$
d_2^{2d_1 - 1} = d_2^{2d_1 - 1} S_{2d_1}(2b) = \frac{c_2^{2d_1}}{d_2} + \sum_{k = 0}^{2d_1 - 1} a_k c_2^k d_2^{2d_1 - k - 1} \in \mathbb{Z} \Longrightarrow d_2 \mid c_2^{2d_1} \Longrightarrow d_2 = 1,
$$
i.e. $2b \in \mathbb{Z}$. Note that $|b| \leqslant 1$, thus $b \in \left\{ 0, \pm\dfrac{1}{2}, \pm1 \right\}$.
Step 3: If $x, y \in \mathbb{Q}$ satisfy $\tan(πx) = y$, then $y \in \{0, \pm1\}$.
Proof: Note that $2x \in \mathbb{Q}$ and$$
y^2 = \tan^2(πx) = \frac{1}{\cos^2(πx)} - 1 \Longrightarrow \cos(π·2x) = 2\cos^2(πx) - 1 = \frac{1 - y^2}{1 + y^2} \in \mathbb{Q},
$$
thus $\dfrac{1 - y^2}{1 + y^2} \in \left\{ 0, \pm\dfrac{1}{2}, \pm1 \right\}$.
If $\dfrac{1 - y^2}{1 + y^2} = 0$, then $y = \pm1$ and $x$ can be taken as $\pm\dfrac{1}{4}$. If $\dfrac{1 - y^2}{1 + y^2} = 1$, then $y = 0$ and $x$ can be taken as $0$. If $\dfrac{1 - y^2}{1 + y^2} = -1$, then $1 - y^2 = -1 - y^2 \Rightarrow 1 = -1$, a contradiction. If $\dfrac{1 - y^2}{1 + y^2} = \dfrac{1}{2}$, then $y^2 = \dfrac{1}{3}$, a contradiction. If $\dfrac{1 - y^2}{1 + y^2} = -\dfrac{1}{2}$, then $y^2 = 3$, a contradiction. Therefore, $y \in \{0, \pm1\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Value of $\left \lfloor{x}\right \rfloor+\left \lfloor{-x}\right \rfloor$? While reading about greatest integer function from a book, I found a question as $\left \lfloor{x}\right \rfloor+\left \lfloor{-x}\right \rfloor$ ?
I attempted it as follows:
We know:
$x-1<\left \lfloor{x}\right \rfloor< x\tag1$
Also then: $-x-1 < \left \lfloor{-x}\right
\rfloor < -x\tag2$
Adding $(1)$ & $(2)$, we get
$-2< \left \lfloor{x}\right \rfloor+\left \lfloor{-x}\right \rfloor<0$.
This is the answer which I got, but the actual answer was $\left \lfloor{x}\right \rfloor+\left \lfloor{-x}\right \rfloor= -1$. I am not getting this. Where my method has gone wrong? Please help me.
| $T=\left\lfloor x \right\rfloor + \left\lfloor { - x} \right\rfloor $
When x=n, only integer
$\underbrace {\left\lfloor x \right\rfloor }_n + \underbrace {\left\lfloor { - x} \right\rfloor }_{ - n} \Rightarrow n - n = 0$
When x=n+a, $a\in(0,1)$
$\left\lfloor {n + a} \right\rfloor + \left\lfloor { - n - 1 + 1 - a} \right\rfloor $
$a \in \left( {0,1} \right)$, therefore $a' = 1 - a \in \left( {0,1} \right)$
$\underbrace {\left\lfloor {n + a} \right\rfloor }_n + \underbrace {\left\lfloor { - n - 1 + a'} \right\rfloor }_{ - n - 1} = n - n - 1 = - 1$.
Hence when x is an integer $T=0$, else $T=-1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that: $ \frac{x^2}{2(x + 1)} < x − \ln(x + 1) < \frac{x^2}{ 2} , x > 0$ Show that: $ \frac{x^2}{2(x + 1)} < x − \ln(x + 1) < \frac{x^2}{ 2} , x > 0$
One method is considering a function $f(x) = x − \ln(1 + x) − \frac{x^2}{2}$ and showing its derivative $f'(x) < 0$.
Similarly, considering another function $g(x) = x − \ln(1 + x) − \frac{x^2}{ 2(x + 1)}$ and showing its $g'(x) > 0$.
Is there some other method that is not lengthy?
| Note that
$$ \int_0^x\frac{t}{t+1}dt=x-\ln(x+1). $$
Since $f(t)=\frac{1}{t+1}$ is decreasing in $[0,x]$, one has
$$ \frac{x^2}{2(x+1)}=\int_0^x\frac{t}{x+1}dt<\int_0^x\frac{t}{t+1}dt<\int_0^xtdt=\frac12x^2 $$
and hence
$$ \frac{x^2}{2(x+1)}<x-\ln(x+1) <\frac12x^2, x>0.$$
| {
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"url": "https://math.stackexchange.com/questions/2138549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to find the max value of $a^2+b^2$ if $x^4+ax^3+3x^2+bx+1\ge0$ Let $f(x)=x^4+ax^3+3x^2+bx+1$, where $a,b \in \mathbb{R}$. If $f(x) \ge 0$ for all $x \in \mathbb{R}$, what is the maximum possible value of $a^2+b^2$?
I don't know how to proceed. Hints or help will be appreciated.
| $$f(x) = \frac{1}{4}\bigg[4x^4+4ax^3+12x^2+4bx+4\bigg]\geq 0$$
$$f(x)= \frac{1}{4}\bigg[(2x^2+ax)^2+(bx+2)^2+(12-a^2-b^2)x^2\bigg]\geq 0$$
So $$12-(a^2+b^2)\geq 0.$$
So $$a^2+b^2\leq 12$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem with $\int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}} \text{dx}$ I wrote
\begin{eqnarray}
I
&=& \int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}} \text{dx}\\
&=&
\int \frac{\sin^{3}x}{\sin{x}\cos{x}}\text{dx}+\int \frac{\cos^{3}x}{\sin{x}\cos{x}}\text{dx}\\
&=&
\int \frac{\sin^{2}x}{\cos^2{x}}\cos{x}\text{dx}+\int \frac{\cos^{2}x}{\sin^2{x}}\sin{x}\text{dx}\\
&=&
\int \frac{\sin^{2}x}{1-\sin^2{x}}\cos{x}\text{dx}+\int \frac{\cos^{2}x}{1-\cos^2{x}}\sin{x}\text{dx}\\
&=&
\int\frac{u^2}{1-u^2}du-\int\frac{m^2}{1-m^2}dm\\
&=&
\color{blue}{\int\frac{u^2}{1-u^2}du-\int\frac{u^2}{1-u^2}du}\\
&=&
0
\end{eqnarray}
and Other
Let $x=\dfrac{\pi}{2}-t$ so
$$I=\int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}} \text{dx}=-\int \frac{\sin^{3}t+\cos^{3}t}{\sin{t}\cos{t}} \text{dt}=-I$$
but correct answer is
$$
I=\ln\left|\dfrac{1+\tan\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}\right|+\ln\left|\tan\dfrac{x}{2}\right|-\sin x+\cos x+C
$$
Question.1 Where is wrong.?
Question.2 what conditions guarantee that our changing variable in indefinite integrals doesn't change our final solutions!.
| Note that
$$\begin{align}
\int \frac{u^2}{1-u^2}\,du&=-u+\frac12\log\left|\frac{1-u}{1+u}\right|+C\\\\
&=-\sin(x)+\frac12\log\left|\frac{1-\sin(x)}{1+\sin(x)}\right|+C
\end{align}$$
and similarly for $\int \frac{m^2}{1-m^2}\,dm$ with $m=\cos(x)$, and not $\sin(x)$.
NOTE $1$:
We need to be preserve the separate identities of the transformations $u \to \sin(x)$ and $m\to \cos(x)$ throughout the analysis. That is to say, that $u$ is not simply a "dummy" integration variable inasmuch as it represents $\sin(x)$, and not $\cos(x)$.
Naturally, both are $u$ and $m$ are "dummy" variables in the sense that we could use other symbols to represent the transformation. But, it is of critical importance to distinguish the two different transformations by the corresponding pair of symbols used as new variables.
NOTE $2$:
One way to clarify things is to turn the indefinite integral into a definite one. The separate substitutions lead to distinct integration limits.
So, let's look at the integrals $\int_a^b \frac{\sin^2(x)}{1-\sin^2(x)}\cos(x)\,dx$ and $\int_a^b \frac{\cos^2(x)}{1-\cos^2(x)}\sin(x)\,dx$. Upon enforcing the proposed substitutions we arrive at
$$\begin{align}
\int_a^b \frac{\sin^2(x)}{1-\sin^2(x)}\cos(x)\,dx=\int_{\sin(a)}^{\sin(b)} \frac{u^2}{1-u^2}\,du\tag 1
\end{align}$$
and
$$\begin{align}
\int_a^b \frac{\cos^2(x)}{1-\cos^2(x)}\sin(x)\,dx=-\int_{\cos(a)}^{\cos(b)} \frac{u^2}{1-u^2}\,du \tag 2
\end{align}$$
Clearly, $(1)$ and $(2)$ do not add to zero.
NOTE $3$:
The indefinite integral (antiderivative), $F(x)$, of a function $f$, can be more clearly written $F(x)=\int_a^x f(t)\,dt+C$ (for a suitable number $a$). This can help to avoid the potential pitfall that comes from the notation $F(x)=\int f(x)\,dx$.
| {
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How could I know that $X^4+1$ is $(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1)$? I thought that $X^4+1$ was irreducible, but in fact, $$X^4+1=(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1).$$
In general, how can I have the intuition of such a factorisation if I don't know it ?
| If you're looking for real coefficients, every polynomial factors into a product of linear and quadratic terms. Writing down the coefficients, on the other hand, can be nearly impossible. Therefore, for your given polynomial, you know that it can be factored into quadratic terms.
There are deeper reasons behind the factorization in Galois theory over the rationals, but I'll go for the elementary approach - although we will need a detour through complex numbers.
To factor $x^4+1$, let's start by setting $y=x^2$. Then, $x^4+1$ becomes $y^2+1$. A quadratic can easily be factored, in this case we use the quadratic formula to get that the roots are $y=\pm i$. Hence, this factors as
$$
y^2+1=(y-i)(y+i).
$$
Since $y=x^2$, we now know that
$$
x^4+1=(x^2-i)(x^2+i).
$$
Each of these are quadratics and can be factored with the quadratic formula. In particular, $x^2-i$ has solutions $x=\pm\sqrt{i}=\pm\frac{\sqrt{2}}{2}\pm\frac{\sqrt{2}}{2}i$. Therefore,
$$
x^2-i=\left(x-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)\left(x+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right).
$$
Similarly,
$$
x^2-i=\left(x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\left(x+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right).
$$
Since we're looking for quadratics with real coefficients, we can try to pair these factors to get real coefficients after multiplying. We can test out a few cases to find that the product
$$
\left(x-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)\left(x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)=x^2-\sqrt{2}x+1.
$$
Similarly, the other pair results in $x^2+\sqrt{2}x+1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Picking teams of 2 for a country where team members and countries are distinguishable? A country has 10 diplomats, and needs to assign two of them to each of 5 countries. How many assignments are possible?
(The diplomats and the countries are distinguishable.)
Would the possibilities be as simple as $({10 \choose 2}*{8 \choose 2}*{6 \choose 2}*{4 \choose 2}*{2 \choose 2})/5!$
| Pick any two: $\dbinom{10}{2}$
Assign to any country: $\dbinom{5}{1}$
Pick any remaining two: $\dbinom{8}{2}$
Assign to any remaining country: $\dbinom{4}{1}$
Pick any remaining two: $\dbinom{6}{2}$
Assign to any remaining country: $\dbinom{3}{1}$
Pick any remaining two: $\dbinom{4}{2}$
Assign to any remaining country: $\dbinom{2}{1}$
Total count is
$\dbinom{10}{2}\dbinom{5}{1}\dbinom{8}{2}\dbinom{4}{1}\dbinom{6}{2}\dbinom{3}{1}\dbinom{4}{2}\dbinom{2}{1}$
To avoid over counting, due to symmetry, we have to divide the above result by
$5!$
Thus,
Total count is
$\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{2}$
This agrees with the answer given by @user8795
| {
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The roots of the equation $x^2+3x-1=0$ are also the roots of $x^4+ax^2+bx+c=0$
The roots of the equation $x^2+3x-1=0$ are also the roots of quartic equation $x^4+ax^2+bx+c=0$. Find $a+b+4c$.
This problem is from yesterday's Bangladesh National Math Olympiad 2017. I tried this using Vieta Root Jumping but no luck. After the contest my friend laughed at me "One doesn't simply try a 10 point problem with Vieta Root Jumping".
How to solve this problem?
| Note that if $x^2+3x-1=0$, then roots of the quadratic $x^2+3x-1$ are also roots of $$(x^2+3x-1)(x^2-px+q)$$
Which follows from polynomial long divison.
Since the coefficient of $x^3$ is $0$, we have that $p=3$. Now note that $$(x^2+3x-1)(x^2-3x+q)=x^4+(q-10)x^2+(3q+3)x-q$$
So the value of $a+b+4c=q-10+3q+3-4q=-7$. So the answer is $-7$ no matter the value of $q$.
| {
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How to solve for $x$ algebraically, given that $x^2 = 4- \sqrt{12}$ This is a question I have been chewing on for a couple days but haven't quite solved yet. The value of $x^2$ is given as $4-\sqrt{12}$ and then the result given as $±( 1-\sqrt{3})$. How would I solve this problem algebraically without prior knowledge of the answer?
Solution to a problem
| If you want to deduce without guessing:
$$ (x^2-4)^2 = 12 \Rightarrow x^4-8x^2+16=12 \Leftrightarrow x^4-8x^2+4=0$$
A quartic polynomial without odd degree terms has a unique factorization as
$$ (x^2+ax+b)(x^2-ax+b) =x^4 +(2b -a^2) x^2 + b^2$$
You find either $b=2, a=\pm \sqrt{12}$ or $b=-2,a=\pm 2$. The first doesn't give a radical of the wanted form but $x^2-2x\pm 2=0$ does, with solutions $x=1-\sqrt{3}$ or $x=-1+\sqrt{3}$ satisfying the required condition.
| {
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Let $A = \mathbb{R} \times \mathbb{R}$, and the operation $* : A \times A \rightarrow A$ on $A$. Check if associative, commutative, identity, inverse. I have this exercise:
on the set $A = \mathbb{R} \times \mathbb{R}$ is defined the operation $* : A \times A \rightarrow A$ such that
$\forall (a,x), (b,y) \in A \quad (a,x)*(b,y) = \left ( \frac{2}{5}ab , \frac{3}{4} + y + x \right )$
*
*Check if the operation is associative;
*Check if the operation is commutative;
*Check if exists the identity element of the algebric structure $(A, *)$;
*Check if exists the inverse of $(2,-3)$ in $(A,*)$.
This is what I have done for the first two points, and I think that is right:
*
*Associativity:
I have to check if the following equation holds
$\left [ (a,x) * (b,y) \right ] * (c,z) \overset{?}{=} (a,x) * \left [ (b,y) * (c,z) \right ]$
hence,
LHS:
in the following I have assigned $m = \frac{2}{5}ab, n = \frac{3}{4}+y+x$
$\begin{array}{lcl}\left [ (a,x) * (b,y) \right ] * (c,z) & = & \left ( \frac{2}{5} ab, \frac{3}{4} +y+x\right ) * (c,z) \\ & = & (m,n)*(c,z) \\ & = & \left [ \frac{2}{5}mc, \frac{3}{4} + z +n \right ] \\ & = & \left [ \frac{2}{5} \left ( \frac{2}{5}ab \right )c, \frac{3}{4} + z + \left ( \frac{3}{4} +y +x \right )\right ] \\ & = & \left [ \frac{4}{25}abc, \frac{3}{2} +z+y+x\right ] \end{array}$
RHS:
in the following I have assigned $g = \frac{2}{5}bc, h = \frac{3}{4}+z+y$
$\begin{array}{lcl}(a*x)*\left [ (b,y) * (c,z) \right ] & = & (a,x) * \left [ \frac{2}{5}bc, \frac{3}{4}+z+y \right ] \\ & = & (a,x) * (g,h) \\ & = & \left [ \frac{2}{5} ag, \frac{3}{4} + h + x \right ] \\ & = & \left [ \frac{2}{5}a \left ( \frac{2}{5}bc \right ), \frac{3}{4}+\left ( \frac{3}{4} + z + y \right ) +x\right ] \\ & = & \left [ \frac{4}{25}abc, \frac{3}{2} +z+y+x \right ] \end{array}$
the equation holds and the operation $*$ is associative.
*Commutative element:
I have to check if the following equation holds
$(a,x) * (b,y) \overset{?}{=} (b,y) * (a,x)$
LHS:
$(a,x) * (b,y) = \left ( \frac{2}{5}ab, \frac{3}{4} + y + x \right )$
RHS:
$(b,y) * (a,x) = \left ( \frac{2}{5}ba, \frac{3}{4}+x+y \right )$
the equation holds and the operation $*$ is commutative.
*Identity element:
here, I have some problems, I am not sure on what to do, I have tried this but without success:
maybe, we have to check the following equation?
$(a,x)*(I_b, I_y) \overset{?}{=} (a,x)$
hence,
$\begin{array}{lcl}(a,x)*(I_b, I_y) & = & \left [ \frac{2}{5}aI_b, \frac{3}{4} + I_y + x \right ] \end{array}$
i.e. we should find a value to $(I_b,I_y)$ such that $\left [ \frac{2}{5}aI_b, \frac{3}{4} + I_y + x \right ] = (a,x)$
and therefore I have problems in point 4.
I don't know!
Please, can you help me?
Many thanks really!
| $(I_b, I_y)=(\frac{5}{2},\frac{-3}{4})$ and inverse of $(-3,2)$ is $(\frac{-25}{12},\frac{-14}{4})$ because $(\frac{-6}{5}b,\frac{3}4+2+y)=(\frac{5}{2},\frac{-3}{4})\Rightarrow (b,y)=(\frac{-25}{12},\frac{-14}{4})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding extrema of sublevel set of the Hénon-Heiles potential function The inequality $$\frac{1}{2}(x^2+y^2)+x^2y-\frac{1}{3}y^3 \lt C$$ produces a somewhat triangular shape near the origin for $C \le \frac 16$, where $C$ is a constant.
Example when $C = 1/10$:
I am trying to find a closed form expression for the $(x,y)$ coordinate of the top of the "triangle" in terms of the constant $C$.
$y = 0$ when $C = 0$ and $y=1$ when $C = \frac 16$.
Background: This is the Henon-Heiles potential function
| Assuming that the maximum can be found when $x=0$ and using the formulas of the "casus irreducibilis" (https://en.wikipedia.org/wiki/Casus_irreducibilis) I get
$$
y=\frac{1}{2}-\cos\left(\frac{1}{3} \arccos\left(1-12C\right)+\frac{\pi}{3}\right)
$$
The details:
Setting $x=0$ results in $y^3-\frac{3}{2}y^2+3C=0$. We substitute $y=t+\frac{1}{2}$ and get the "depressed" polynomial
$$
t^3-\frac{3}{4}t-\frac{1}{4}+3C=0
$$
We set $p=-\frac{3}{4}$ and $q=-\frac{1}{4}+3C$. It can easily be shown that
$$
\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3 \leq 0\;\; \mbox{for}\;\;0\leq C\leq \frac{1}{6}
$$
so the conditions for using the trigonometric method of the "casus irreducibilis" are fulfilled. The formula is as follows:
$$
t_k = 2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{-\frac{3}{p}}\right)-k\frac{2\pi}{3}\right)\;\;\mbox{for}\;\;k=0,1,2.
$$
We are lucky, because the coefficients of the depressed polynomial allow a lot of simplifications in this formula.
Finally, we only have to consider our initial substitution $y=t+\frac{1}{2}$ and find the appropriate $k$ for the root we are looking for. It turns out that $k=1$ is the correct choice. Additionally, I have used $\cos(\alpha) = -\cos(\alpha+\pi)$ to replace the $-\frac{2\pi}{3}$ with $\frac{\pi}{3}$, which is just a matter of personal preference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
limit $ \lim_{x\to\infty} \left( (x+2017)^{1+\frac{1}{x}} \: -\: x^{1+\frac{1}{x+2017}} \right) $ Find the following limit:
$$
\lim_{x\to\infty}
\left(
(x+2017)^{1+\frac{1}{x}} \: -\: x^{1+\frac{1}{x+2017}}
\right)
$$
I tried to exchange the infinity to zero by $x:=\frac{1}{t}$ and then use $\lim_{t\to 0^+}t^t=1$, but it doesn't lead to anything, I can't avoid having $\infty-\infty$...
The answer is 2017, as graph showed.
| I generalize this problem
in two ways.
First,
replace $2017$ with $c$.
Second,
make the exponents more general,
so that the problem is
$\lim_{x\to\infty}
\left(
(x+c)^{1+\frac{a}{x}} - x^{1+\frac{b}{x+c}}
\right)
$.
It turns out that
the result is
$(a-b)\ln(x)+c+O(\frac{\ln^2(x)}{x})
$.
If $a=b$ (as in the original problem,
the limit is $c$.
Otherwise the result
goes to $+\infty$ if $a > b$
and
goes to $-\infty$ if $a < b$.
Let's look at the parts
as $x \to \infty$.
First.
$\begin{array}\\
(x+c)^{1+\frac{a}{x}}
&=(x+c)(x+c)^{\frac{a}{x}}\\
&=(x+c)x^{a/x}(1+c/x)^{\frac{a}{x}}\\
&=(x+c)e^{a\ln(x)/x}e^{a\ln(1+c/x)/x}\\
&=(x+c)(1+\frac{a\ln(x)}{x}+O(\frac{\ln^2(x)}{x^2})e^{(ac/x+O(1/x^2))/x}\\
&=(x+c)(1+\frac{a\ln(x)}{x}+O(\frac{\ln^2(x)}{x^2})e^{ac/x^2+O(1/x^3)}\\
&=(x+c)(1+\frac{a\ln(x)}{x}+O(\frac{\ln^2(x)}{x^2})(1+ac/x^2+O(1/x^3))\\
&=(x+c)(1+\frac{a\ln(x)}{x}+O(\frac{\ln^2(x)}{x^2}+O(1/x^2))\\
&=(x+c)(1+\frac{a\ln(x)}{x}+O(\frac{\ln^2(x)}{x^2}))\\
&=x+c+a\ln(x)+O(\frac{\ln^2(x)}{x} )\\
\end{array}
$
Second.
$\begin{array}\\
x^{1+\frac{b}{x+c}}
&=xe^{b\ln(x)/(x+c)}\\
\text{and}\\
\frac{b\ln(x)}{x+c}
&=\frac{b\ln(x)}{x}\frac1{1+c/x}\\
&=\frac{b\ln(x)}{x}(1-\frac{c}{x}+O(\frac1{x^2}))\\
&=\frac{b\ln(x)}{x}-\frac{bc\ln(x)}{x^2}+O(\frac{\ln(x)}{x^3})\\
\text{so}\\
e^{b\ln(x)/(x+c)}
&=1+\frac{b\ln(x)}{x}-\frac{bc\ln(x)}{x^2}+O(\frac{\ln(x)}{x^3}))
+O(\frac{\ln^2(x)}{x^2})\\
&=1+\frac{b\ln(x)}{x}
+O(\frac{\ln^2(x)}{x^2})\\
\text{so that}\\
x^{1+\frac{b}{x+c}}
&=x+b\ln(x)+O(\frac{\ln^2(x)}{x})\\
\end{array}
$
Taking their difference,
the $x$ cancels out
and we get
$(a-b)\ln(x)+c+O(\frac{\ln^2(x)}{x})
$.
Note:
Wolfy confirms the difference as
$2017 + 2017 \frac{1 + 2 log(x)}{x} + O((1/x)^2)
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate weighted stirling number sum - $\sum_{i=0}^n k*c(n,k)$ How to evaluate
$$\sum_{k=0}^n k*c(n,k)$$
Where $c(n,k)=|s(n,k)|$ is the stirling number of the first kind without a sign.
| We seek to verify that
$$\sum_{k=0}^n k {n\brack k} = {n+1\brack 2}.$$
This is
$$n! [z^n] \sum_{k=1}^n k
\frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k.$$
Here we can extend to infinity due to the coefficient extractor and the
fact that $\log\frac{1}{1-z} = z +\cdots$:
$$n! [z^n] \log\frac{1}{1-z} \sum_{k\ge 1}
\frac{1}{(k-1)!} \left(\log\frac{1}{1-z}\right)^{k-1}
\\ = n! [z^n] \log\frac{1}{1-z}
\exp\log\frac{1}{1-z}
= n! [z^n] \frac{1}{1-z} \log\frac{1}{1-z}
= n! \times H_n.$$
On the other hand we have
$${n+1\brack 2} = (n+1)! [z^{n+1}] \frac{1}{2}
\left(\log\frac{1}{1-z}\right)^2
\\ = (n+1)! \frac{1}{2}
\sum_{q=1}^n \frac{1}{q} \frac{1}{n+1-q}
\\ = n! \frac{1}{2}
\sum_{q=1}^n \left(\frac{1}{q} + \frac{1}{n+1-q}\right)
= n! \times H_n$$
and we have the claim. We can prove that $\exp\log\frac{1}{1-z} =
\frac{1}{1-z}$ using the combinatorial class $\mathcal{P}$ of
permutations which are sets of cycles:
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{P} = \textsc{SET}(\textsc{CYC}(\mathcal{Z}))$$
as presented in Analytic Combinatorics by Flajolet amd Sedgewick.
| {
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"url": "https://math.stackexchange.com/questions/2149187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
How do you find the derivative of $x\sqrt{x}$ with nonstandard analysis? Find the derivative of $x\sqrt{x}$. I'm doing this without power rules etc. , what I know beforehand are that
$$f'(x)=st\left(\frac{f(x+\Delta x)-f(x)}{\Delta x}\right)$$ and that
$$y'=st\left(\frac{\Delta y}{\Delta x}\right).$$
The correct answer is $y'=\frac{3}{2}\sqrt{x}$.
There are different approaches I have tried with without success. I have tried and tried, and do not know what to do to solve the problem. For example,
I do not know if it really is $\neq \frac{1}{(x+\Delta x)^{\frac{3}{2}}+x^{\frac{3}{2}}}$. But if I use that value for $\Delta y$, then I just get that $y'=\frac{1}{2x^{\frac{3}{2}}}$, and
$\frac{1}{2x^{\frac{3}{2}}} \neq \frac{3}{2} \sqrt{x}.$
PS the methods I am using are taught by Jerome H. Keisler in his book Elementary calculus: an infinitesimal approach.
Please help me,
Andreas
| $$f(x) = \sqrt{x^3}$$
$$\require{cancel}f(x+h) - f(x) = \sqrt{(x+h)^3} - \sqrt{x^3} = {(x+h)^3 - x^3\over \sqrt{(x+h)^3} + \sqrt{x^3}} = {\cancel{x^3} + 3x^2h + 3h^2x + h^3 - \cancel{x^3}\over \sqrt{(x+h)^3} + \sqrt{x^3}} = {3x^2h + 3h^2x + h^3 \over \sqrt{(x+h)^3} + \sqrt{x^3}}$$
Where $h\in \mathbb{^*R}$ is a infinitesimal quantity.
$${f(x+h) - f(x)\over h} = {3x^2 + 3hx + h^2 \over \sqrt{(x+h)^3} + \sqrt{x^3}}$$
Now from the first chapter the book, $st(x + \varepsilon) = x$, $st(x \varepsilon) = 0$ and $st(x) = x$ where $\varepsilon \in \mathbb{^*R}$ is a infinitesimal number and $x$ is a real number.
Using this to take the standard part,
$$st\left({3x^2 + 3hx + h^2 \over \sqrt{(x+h)^3} + \sqrt{x^3}}\right) = {st(3x^2) + st(3hx) + st(h^2) \over st\left(\sqrt{(x+h)^3}\right) + st\left(\sqrt{x^3}\right)} = {3x^2 \over 2\sqrt{x^3}} = {3\sqrt{x} \over 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find Minima and Maxima of $ y = \frac{x^2-3x+2}{x^2+2x+1}$ $$ y = \frac{x^2-3x+2}{x^2+2x+1}$$
I guess I made some mistakes cause after taking the first derivative and simlifying I have
$$y = \frac{2x^3-4x^2+5}{(x+1)^2}$$
but then numerator has complex roots. which should not be, IMO
| Well, you can use five rules:
*
*Quotient rule:
$$\frac{\text{d}}{\text{d}x}\left(\frac{\text{u}}{\text{v}}\right)=\frac{\text{v}\cdot\frac{\text{d}\text{u}}{\text{d}x}-\text{u}\cdot\frac{\text{d}\text{v}}{\text{d}x}}{\text{v}^2}\tag1$$
*Differentiate the sum term by term and factor out constants.
*The power rule:
$$\frac{\text{d}}{\text{d}x}\left(x^\text{n}\right)=\text{n}x^{\text{n}-1}\tag2$$
*The derivative of $x$ is $1$.
*The derivative of $1$ is $0$:
So, using the quotient rule:
$$\text{y}'\left(x\right)=\frac{\left(x^2+2x+1\right)\cdot\frac{\text{d}}{\text{d}x}\left(x^2-3x+2\right)-\left(x^2-3x+2\right)\cdot\frac{\text{d}}{\text{d}x}\left(x^2+2x+1\right)}{\left(x^2+2x+1\right)^2}\tag3$$
Now, we can use the second rule to get:
*
*$$\frac{\text{d}}{\text{d}x}\left(x^2-3x+2\right)=\frac{\text{d}}{\text{d}x}\left(x^2\right)-3\cdot\frac{\text{d}}{\text{d}x}\left(x\right)+2\cdot\frac{\text{d}}{\text{d}x}\left(1\right)\tag4$$
*$$\frac{\text{d}}{\text{d}x}\left(x^2+2x+1\right)=\frac{\text{d}}{\text{d}x}\left(x^2\right)+2\cdot\frac{\text{d}}{\text{d}x}\left(x\right)+\frac{\text{d}}{\text{d}x}\left(1\right)\tag5$$
So, we end up with:
$$\text{y}'\left(x\right)=\frac{5x-7}{\left(1+x\right)^3}=0\space\Longleftrightarrow\space x=\frac{7}{5}\tag6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For $a^3+b^3+c^3=3$ prove that $a^4b+b^4c+c^4a\leq3$
Let $a$, $b$ and $c$ be non-negative numbers such that $a^3+b^3+c^3=3$. Prove that:
$$a^4b+b^4c+c^4a\leq3$$
This inequality similar to the following.
Let $a$, $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2=3$. Prove that:
$$a^3b+b^3c+c^3a\leq3,$$
which follows from the following identity.
$$(a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)=\frac{1}{2}\sum_{cyc}(a^2-b^2-ab-ac+2bc)^2.$$
I tried Rearrangement.
Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, $$a^4b+b^4c+c^4a=a^3\cdot ab+b^3\cdot bc+c^3\cdot ca\leq x^3\cdot xy+y^3\cdot xz+z^3\cdot yz=$$
$$=y(x^4+y^2xz+z^4)$$
and I don't see what is the rest.
Thank you!
| Since $(a,b,c)$ is restricted to a compact set, it follows that $F_2 := a^4b +b^4 c + c^4 a$ has a maximal value, which must either be at an endpoint (i.e. one of $a,b,c$ is $0$, but these cases are easily eliminated) or at a point where the derivatives vanish.
Let $F_1 := a^3+b^3+c^3$. If we keep $c$ fixed and equate $F_1$ to $3$, then $b$ is a function of $a$, and we can use this to differentiate $F_2$ with respect to $a$. Equating that derivative to $0$ gives an equation. We can do the same with the roles of $a,b,c$ rotated and we get three equations:
$E = \{-a^6+4a^3b^3-4a^2b^3c+b^2c^4 = 0, 4a^3bc^2-4a^3c^3-a^2b^4+c^6 = 0, a^4c^2-4ab^2c^3-b^6+4b^3c^3 = 0\}$
To maximize $F_2$ under $F_1 = 3$, solve the equations $E \bigcup \{F_1 = 3\}$ (I need a computer for this step). These equations have 84 solutions. Most of them are complex. The only real solutions are $\{a,b,c\} = \{-0.26829, 0.80474, 1.3569\}$ and $\{a,b,c\}=1$. The only non-negative solution of $E \bigcup \{F_1 = 3\}$ is $a=b=c=1$. So that is where $F_2$ takes its maximum.
I have to ask: where did the problem come from, and was it formulated in a way that indicates that there should also be an answer that can be found with human-only computation? One way to reduce the computation would be to solve $E \bigcup \{F_1=3, F_2=3\}$ (only solution is $a=b=c=1$) (unlike my first method, it would be possible to produce a human-verifyable proof for this) and then consider derivatives at this point to show that this is a local maximum of $F_2$. That would produce a proof with computations short enough to be human-verifyable, though it still wouldn't be a nice proof.
| {
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"url": "https://math.stackexchange.com/questions/2154608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove the inequality, fractions. $\{ a,b,c \in\Bbb R_+\ \}$ If $\frac {1}{ab} +\frac {1}{bc} + \frac {1}{ac} = 3$ then prove the inequality:
$ab + bc + ac \ge 3$
How I started
*
*$ab + bc + ac \ge \frac {1}{ab} +\frac {1}{bc} + \frac {1}{ac}$
*$ab- \frac {1}{ab} + bc-\frac {1}{bc} + ac- \frac {1}{ac} \ge 0$
But now I don't know what to do.
| from the condition we have $$a+b+c=3abc$$ and by AM-GM we get $$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$$ plugging the condition in this inequlity we get $$abc\geq1$$ we will prove the inequality:
By AM-GM we have $$ab+bc+ca\geq 3\sqrt[3]{(abc)^2}\geq 3 $$ if $$abc\geq 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{x \rightarrow \infty}(\frac{x}{x^2+1}\cdot e^{x})$ I tried:
$\lim_{x \rightarrow \infty}(\frac{x}{x^2+1}\cdot e^{x}) = \frac{1}{x^2+1} \cdot e^xx = 0 \cdot \infty = 0$
But this is wrong. What did I do wrong? How do I solve this?
| You're incorrect, as $0 \times \infty \neq 0$. It is indeterminate.
Note that as $e^t \ge t+1$ for all $t$, we have that $e^{\frac{x}{2}} \ge \frac{x}{2}+1$. Squaring both sides, we have that $$e^{x} \ge \frac{x^2}{4}+x+1>\frac{x^2}{4}+\frac{1}{4} $$
So, note that $$\frac{x}{x^2+1}e^x > \frac{x}{x^2+1} \times \left( \frac{x^2}{4}+\frac{1}{4}\right)=\frac{x}{4}$$
So $$\lim_{x \rightarrow \infty}\left(\frac{x}{x^2+1}\cdot e^{x}\right) \ge \lim_{x \to \infty}\frac{x}{4}$$
However, you can see that the limit on the right tends to infinity. So your limit will also tend to infinity.
| {
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"url": "https://math.stackexchange.com/questions/2155677",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Trigonometric-substitutions for $\int \frac{x}{\sqrt{x^2+x+1}}\,dx$ Working integrals involving trigonometric substitutions, I evaluate
$$\int \frac{x}{\sqrt{x^2+x+1}}\,dx$$
and I am not sure what I am doing wrong.
What I do is change the denominator to $x^2+x+1=(x+1)^2-1$ then let $v=x+1$ and so $dv=dx$ such that I get
$$\int \frac {v-1}{\sqrt{v^2-1}}\,dv$$
Then, I set $v=a\sec\theta$ such that $dv=a\sec\theta\tan\theta\,d\theta$ and $a=1$ and I get,
$$\int \frac {(\sec\theta-1)}{\sqrt{\sec^2\theta-1}}\,\sec\theta\tan\theta\,d\theta$$
$$=\;\int \sec^2\theta-\sec\theta\,d\theta$$
$$\tan\theta - \ln|\sec\theta+\tan\theta|+C$$
Getting back in terms of the initial variable $x$ I get,
$$=\;\sqrt{v^2-1}-\ln|v+\sqrt{v^2-1}|+C$$
$$=\; \sqrt{x^2+x+1}-\ln|x+1+\sqrt{x^2+x+1}|+C$$
Which is incorrect. What I should get is
$$=\; \sqrt{x^2+x+1}-\frac 12\ln|x+\frac 12+\sqrt{x^2+x+1}|+C$$
The fact that my answer is so close and has the correct form leads me to believe that my substitutions might be correct and that I am making silly arithmetic mistakes somewhere.
All help is appreciated.
| Your mistake is -
$(x+1)^2-1=x^2+2x+1-1$
$=x^2+2x \ne x^2+x+1$
It should be -
$x^2+x+\frac 14 +1-\frac 14$
$=(x+\frac 12)^2 -\frac 34$
$=(x+\frac 12)^2 -(\frac {\sqrt3}2)^2$
| {
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"url": "https://math.stackexchange.com/questions/2157264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does one show that $\int_{0}^{1}{1\over 1+\phi x^4}\cdot{\mathrm dx\over \sqrt{1-x^2}}={\pi\over 2\sqrt{2}}?$ Consider
$$\int_{0}^{1}{1\over 1+\phi x^4}\cdot{\mathrm dx\over \sqrt{1-x^2}}={\pi\over 2\sqrt{2}}\tag1$$
$\phi$;Golden ratio
An attempt:
$x=\sin{y}$ then $\mathrm dx=\cos{y}\mathrm dy$
$(1)$ becomes
$$\int_{0}^{\pi/2}{\mathrm dy\over 1+\phi \sin^4{y}}\tag2$$
Apply Geometric series to $(2)$,
$$\int_{0}^{\pi/2}(1-\phi\sin^4{y}+\phi^2\sin^8{y}-\phi^3\sin^{12}{y}+\cdots)\mathrm dy\tag3$$
$${\pi\over 2}-\int_{0}^{\pi/2}(\phi\sin^4{y}-\phi^2\sin^8{y}+\phi^3\sin^{12}{y}-\cdots)\mathrm dy\tag4$$
Power of sine seem difficult to deal with
How else can we tackle $(1)?$
| A further substitution $\tan y=x$ makes the integral into,
$\displaystyle I = \dfrac{1}{\phi+1}\int\limits_0^\infty \dfrac{x^2+1}{x^4+\dfrac{2}{\phi^2}x^2+\dfrac{1}{\phi^2}}\; dx$
Now let us consider the integrals of the form ,
$\displaystyle F(a,b)= \int\limits_0^\infty \dfrac{x^2+1}{(x^2+a^2)(x^2+b^2)}\; dx$
As we can see that $\displaystyle \dfrac{x^2+1}{(x^2+a^2)(x^2+b^2)} = \dfrac{a^2-1}{(x^2+a^2)(a^2-b^2)}+\dfrac{1-b^2}{(x^2+b^2)(a^2-b^2)}$
Integrating and simplifying we have,
$\displaystyle F(a,b)=\pi\left(\dfrac{1+ab}{2ab(a+b)}\right)$
Now that $\displaystyle x^4+\dfrac{2}{\phi^2}x^2+\dfrac{1}{\phi^2} = \left(x^2+\dfrac{1}{\phi^2}\left(1+i\sqrt{\phi}\right)\right)\left(x^2+\dfrac{1}{\phi^2}\left(1-i\sqrt{\phi}\right)\right)$
We can easily see that our integral equals $\displaystyle I =\dfrac{1}{\phi+1}F\left(\sqrt{\dfrac{1}{\phi^2}\left(1+i\sqrt{\phi}\right)},\sqrt{\dfrac{1}{\phi^2}\left(1-i\sqrt{\phi}\right)}\right)$
Putting the value in we can see that $I=\dfrac{\pi}{2\sqrt{2}}$
The simplification becomes much easier using $\phi^2=\phi+1$ and I've checked it on paper but since it's too much nested so I'm avoiding it in latex.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How can I show that this complicated expression with square and cube roots reduces to the value 7? How would I go about reducing the complicated-looking expression $$ \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}} $$ to show that it is equal to 7?
I came across the complicated expression while calculating the single real solution for the cubic equation $$ z^3 - 33 z - 112 = 0 $$
Using that cubic equation, I can show that $ z = 7 $ satisfies it and that a standard way of solving cubic equations shows that there is only a single real root and that its value is equal to the complicated expression, which is a very roundabout way of proving that the complicated expression is equal to 7. But what if I didn't know about the cubic equation? Is there a more direct way of reducing the complicated expression to a simpler one?
| Using $a=\sqrt[3]{19\sqrt5+56}$ and $b=\frac{11}{\sqrt[3]{19\sqrt5+56}}$ and $u=a+b$, we have
$$
\begin{align}
a^3+b^3
&=19\sqrt5+56+\frac{1331}{19\sqrt5+56}\\
&=\frac{6272+2128\sqrt5}{56+19\sqrt5}\\[6pt]
&=112
\end{align}
$$
Thus,
$$
\begin{align}
&112\\
&=a^3+b^3\\
&=(a+b)\left(a^2-ab+b^2\right)\\
&=(a+b)\left((a+b)^2-3ab\right)\\
&=u(u^2-33)
\end{align}
$$
Noting that for $x\lt\sqrt{11}$,
$$
\begin{align}
x^3-33x-112
&\le22\sqrt{11}-112\\
&\lt0
\end{align}
$$
and for $x\ge\sqrt{11}$, $x^3-33x-112$ is increasing, we see that
$x^3-33x-112=0$ has only one real solution, and that is $x=7$.
Therefore, $\sqrt[3]{19\sqrt5+56}+\frac{11}{\sqrt[3]{19\sqrt5+56}}=a+b=u=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find series representation of a function $ y=\frac{1}{1+x+x^{2}+x^{3}} $ Let $y = \frac{1}{1+x+x^{2}+x^{3}}$.
And I want to find the series representation of this function.
I've noticed that I can rewrite this like $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{1+x}*\frac{1}{1+x^{2}}$ or I can rewite this like a sum of two fractions $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{2(1+x)}+\frac{1}{2(1+x^{2})}$.
I know the series representations of these functions $(1+x)^{n}$ and $\frac{1}{1-x}$, and I guess I should use one of them. But the result looks weird. Can you help me, please, may be I can't see some easy way to solve it.
| $$f(x)=\frac{1}{1+x+x^2+x^3}=\frac{1-x}{1-x^4}=\sum_{n\geq 0}\chi(n)\,x^n $$
where $\chi(n)$ equals $1$ if $n\equiv 0\pmod{4}$, $-1$ if $n\equiv 1\pmod{4}$ and zero otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding maximum of $x+y$ Let x and y be real numbers satisfying $9x^{2} + 16y^{2} = 1$. Then $x + y$ is maximum when
a. $y = \frac{9x}{16}$
b. $y = −\frac{9x}{16}$
c. $y = \frac{4x}{3}$
d. $y = −\frac{4x}{3}$
| HINT:
Let $x+y=c\iff x=c-y$
$$1=9x^2+16y^2=9(c-y)^2+16y^2\iff25y^2-18cy+9c^2-1=0$$
As $y$ is real, the discriminant must be $\ge0$
$$\implies(18c)^2\ge4\cdot25(9c^2-1)\iff144c^2\le25\iff-5\le12c\le5$$
For maximum $c=x+y=\dfrac5{12}$
consequently, $y=\dfrac{18c}{2\cdot25}=\dfrac{9c}{25}\iff x=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 1
} |
Integrate $\int \cos^n x dx$ I know how to solve it with reduction formula, but is there concrete answer for this integral (without other integrals like in reduction formula)?
WolframAlpha gives me expression with $F_1(\frac{1}{2}, \frac{n+1}{2};\frac{n+3}{2}; cos^2(x))$, but I don't know really what it is. I want to know if there is a simpler solution for this without other integrals like in reduction formula and difficult things like $F_1(\frac{1}{2}, \frac{n+1}{2};\frac{n+3}{2}; cos^2(x))$?
| Using $z=e^{ix}$ we have
$$i2^n\int \cos^nx\,dx=\int(z+z^{-1})^n\frac{dz}{z}=\int\sum_{k=0}^n\binom nkz^{2k-n-1}dz=\sum_{k=0}^n\binom nk\frac{z^{2k-n}}{2k-n},$$ with a special case when $2k=n$ (giving a term in $\log z$). We can regroup the terms in pairs of opposite exponents.
For odd $n$, this is
$$2i\sum_{k=(n+1)/2}^n\binom nk\frac{\sin(2k-n)x}{2k-n},$$
and for even $n$,
$$2i\sum_{k=n/2+1}^n\binom nk\frac{\sin(2k-n)x}{2k-n}+\binom n{n/2}ix.$$
For concrete examples,
*
*$n=3$,
$$\int\frac{(z+z^{-1})^3}{2^3}\frac{dz}{iz}=\int\frac{z^2+3z^0+3z^{-2}+z^{-4}}{8i}dz=\frac1{8i}\left(\frac{z^3-z^{-3}}3+3(z-z^{-1})+6\log z\right)\\
=\frac{\sin3x}{12}+\frac{3\sin x}{4}.$$
*
*$n=4$,
$$\int\frac{(z+z^{-1})^4}{2^4}\frac{dz}{iz}=\int\frac{z^3+4z+6z^{-1}+4z^{-3}+z^{-5}}{16i}dz=\frac1{16i}\left(\frac{z^4-z^{-4}}4+2(z^2-z^{-2})+6\log z\right)\\
=\frac{\sin4x}{32}+\frac{\sin2x}{4}+\frac38x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the value of limit when $k\in(-1,1)$ My question arises from How find this $\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)$.
$\\$
1.If $k\in(1,+\infty),$ then $$\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)=\frac{k}{12};$$
2. If $k=1,$ then $$\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)=0;$$
3.If $k\in(-\infty,-1),$ then
$$\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)=+\infty.$$
When $k\in (-1,1),$ I guess we could get $$\lim_{n\to\infty}n^2\left(\frac{1^k+2^k+\cdots+n^k}{n^{k+1}}-\frac{1}{k+1}-\frac{1}{2n}\right)=-\infty.$$
Euler-Maclaurin Summation Formula is at work in this case,but I need more convenient method to solve the problem.
Can anyone give me any hints on how to start it? Any help will be appreciated.
| For $k \in (-1,0]$, we have a very simple argument.
$$\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} = \frac{1}{n}\sum_{m = 1}^n \biggl(\frac{m}{n}\biggr)^k$$
is a Riemann sum for the integral $\int_0^1 x^k\,dx$ for the equidistant partition $0 < \frac{1}{n} < \dotsc < \frac{n-1}{n} < 1$ with the right endpoints of the partition intervals chosen as the points where the integrand is evaluated. Since $x \mapsto x^k$ is non-increasing for $k \leqslant 0$, it follows that
$$\frac{1}{n}\sum_{m = 1}^n \biggl(\frac{m}{n}\biggr)^k \leqslant \int_0^1 x^k\,dx = \frac{1}{k+1}$$
(with strict inequality for $k < 0$). Hence
$$\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} - \frac{1}{k+1} - \frac{1}{2n} \leqslant -\frac{1}{2n}$$
and
$$n^2\biggl(\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} - \frac{1}{k+1} - \frac{1}{2n}\biggr) \leqslant -\frac{n}{2},$$
from which we immediately see that the limit is $-\infty$.
For $k \in (0,1)$, the argument is a bit more complex. We first note that
$$\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} - \frac{1}{2n} = \frac{1}{n}\sum_{m = 1}^n \frac{1}{2}\Biggl(\biggl(\frac{m-1}{n}\biggr)^k + \biggl(\frac{m}{n}\biggr)^k\Biggr)$$
is a trapezium sum for the integral $\int_0^1 x^k\,dx$.
Next we note that for a twice continuously differentiable $f \colon [a,b] \to \mathbb{R}$, with $\mu := \frac{a+b}{2}$, we have
\begin{align}
\frac{1}{2} \int_a^b (x-a)(b-x)f''(x)\,dx
&= \frac{1}{2}(x-a)(b-x)f'(x)\biggr\rvert_a^b + \int_a^b (x-\mu)f'(x)\,dx \\
&= \int_a^b (x-\mu)f'(x)\,dx \\
&= (x-\mu)f(x)\biggr\rvert_a^b - \int_a^b f(x)\,dx \\
&= \frac{b-a}{2}\bigl(f(a) + f(b)\bigr) - \int_a^b f(x)\,dx,
\end{align}
and by the mean value theorem of integration,
$$\frac{1}{2} \int_a^b (x-a)(b-x)f''(x)\,dx = \frac{f''(\xi)}{2}\int_a^b (x-a)(b-x)\,dx = \frac{(b-a)^3}{12} f''(\xi)$$
for some $\xi \in (a,b)$. Although $f \colon x \mapsto x^k$ is for $k \in (0,1)$ not differentiable at $0$, these formulae remain correct also with $a = 0$ for that function. Thus
$$\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} - \frac{1}{k+1} - \frac{1}{2n} = \frac{k(k-1)}{12 n^3}\sum_{m = 1}^n \biggl(\frac{\eta_{n,m}}{n}\biggr)^{k-2}$$
for some $\eta_{n,m} \in (m-1,m)$. Then
$$n^2\biggl(\frac{1^k + 2^k + \dotsc + n^k}{n^{k+1}} - \frac{1}{k+1} - \frac{1}{2n}\biggr) = n^{1-k}\frac{k(k-1)}{12} \sum_{m = 1}^n \frac{1}{\eta_{n,m}^{2-k}} \to -\infty$$
since $1-k > 0$, $\frac{k(k-1)}{12} < 0$, and
$$\sum_{m = 1}^n \frac{1}{\eta_{n,m}^{2-k}} \geqslant \frac{1}{\eta_{n,1}^{2-k}} > 1$$
for all $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x \to 0}\lfloor\frac{\tan^{98}x - \sin^{98} x}{x^{100}}\rfloor=?$ fine the limit :
$$\lim_{x \to 0}\lfloor\frac{\tan^{98}x - \sin^{98} x}{x^{100}}\rfloor=?$$
We denote the floor funtion by $\lfloor x\rfloor$.
My try:
\begin{align}
\lim_{x \to 0}\frac{\tan^{n}x - \sin^{n} x}{x^{n + 2}} &= \lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}}\cdot \sum_{i = 0}^{n - 1}\frac{\tan^{n - 1 - i}x}{x^{n - 1 - i}}\cdot\frac{\sin ^{i}x}{x^{i}}\\
&= \frac{1}{2}\cdot\sum_{i = 0}^{n - 1}1\\
&= \frac{n}{2}
\end{align}
So:
\begin{align}
\lim_{x \to 0}\frac{\tan^{98}x - \sin^{98} x}{x^{100}}&=49\\
\lim_{x \to 0}\left\lfloor\frac{\tan^{98}x - \sin^{98} x}{x^{100}}\right\rfloor&=49
\end{align}
is this correct?
| Using the Taylor series
$$
\begin{aligned}
\tan x&=x+\frac13x^3+\frac2{15}x^5+\cdots,\\
\sin x&=x-\frac16x^3+\frac1{120}x^5+\cdots
\end{aligned}
$$
and the binomial formula, we arrive at
$$
\tan^{98}x=x^{98}+\frac{98}3x^{100}+[\frac19\cdot\binom{98}2+\frac{98\cdot2}{15}]x^{102}+\cdots
$$
and
$$
\sin^{98}x=x^{98}-\frac{98}6x^{100}+[\frac1{36}\cdot\binom{98}2+\frac{98}{120}]x^{102}+\cdots.
$$
Therefore
$$
\tan^{98}x-\sin^{98}x=49x^{100}+\frac{1225}3x^{102}+\cdots,
$$
and
$$
\frac{\tan^{98}x-\sin^{98}x}{x^{100}}=49+\frac{1225}3x^2+\cdots.
$$
When $x$ is close enough to zero, that quadratic term dominates the (possibly negative) cut-off terms. Therefore, when $|x|$ is small enough we have
$$
49\le \frac{\tan^{98}x-\sin^{98}x}{x^{100}}<50.
$$
The answer is thus $49$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the GCD of the following values in $Z[i]$ I'm trying to find the GCD of $4+22i$ and $17+i$ in $Z[i]$. I started by using Euclid's and did the following steps:
$4 + 22i = (17 + i) \cdot i + (5 + 5i)$
$17 + i = (5 + 5i) \cdot (1 - i) + (7 + i)$
Now when I try to divide $5 + 5i$ by $7 + i$ I get
$\frac{5 + 5i}{7 + i} = \frac{5 + 5i}{7 + i} \cdot \frac{7 - i}{7 - i} = \frac{40}{50} + \frac{30i}{50}$
I'm not sure where to go from here (or if I've made a mistake along the way) because this would seem to give a quotient of $0$ which will cause the algorithm to loop.
| Write $z=4+22i$ and $w=17+i$. Now,
$$\frac{z}{w}=\frac{4+22i}{17+i}\cdot \frac{17-i}{17-i}=\frac{9}{29}+\frac{37}{29}i.$$ The closest integers to $\frac{9}{29}$ and $\frac{37}{29}$ are $0$ and $1$, respectively. Write $q_1=0+1i=i.$
Then
$$r_1=(4+22i)-(17+i)(i)=5+5i.$$
Now,
$$\frac{17+i}{r_1}=\frac{17+i}{5+5i}\cdot \frac{5-5i}{5-5i}=\frac{9}{5}+\frac{-8}{5}i.$$ The closest integers to $\frac{9}{5}$ and $\frac{-8}{5}$ are $2$ and $-2$, respectively. Write $q_2=2-2i$. Then
$$r_2=(17+i)-r_1q_2=(17+i)-(5+5i)(2-2i)=-3+i.$$
Now,
$$\frac{r_1}{r_2}=\frac{5+5i}{-3+i}\cdot\frac{-3-i}{-3-i}=-1-2i=q_3\in\Bbb{Z}[i].$$
We have shown that
$$\begin{align}
z&=wq_1+r_1\\
w&=r_1q_2+r_2\\
r_1&=r_2q_3+0.
\end{align}$$
The last nonzero remainder is $r_2$ and hence (using the euclidean algorithm) the required $\gcd$ is $r_2=-3+i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $abc=1$ so $\sum\limits_{cyc}\frac{7-6a}{2+a^2}\geq1$
Let $a$, $b$ and $c$ be real numbers such that $abc=1$. Prove that:
$$\frac{7-6a}{2+a^2}+\frac{7-6b}{2+b^2}+\frac{7-6c}{2+c^2}\geq1$$
The equality occurs also for $a=b=2$ and $c=\frac{1}{4}$.
This inequality is a similar to the very many contest's inequalities, but nothing helps.
At least, I don't see how we can prove it.
An example of my trying.
We need to prove that
$$\sum_{cyc}\frac{7-6a}{2+a^2}\geq1$$ or
$$\sum_{cyc}\left(\frac{7-6a}{2+a^2}+1\right)\geq4$$ or
$$\sum_{cyc}\frac{(a-3)^2}{2+a^2}\geq4.$$
By C-S $$\sum_{cyc}\frac{(a-3)^2}{2+a^2}=\sum_{cyc}\frac{(a-3)^2(a+k)^2}{(2+a^2)(a+k)}\geq\frac{\left(\sum\limits_{cyc}(a-3)(a+k)\right)^2}{\sum\limits_{cyc}(2+a^2)(a+k)^2}$$
Now we'll find a value of $k$, for which the equality in the last inequality occurs for $a=b=2$ and $c=\frac{1}{2}$.
Since in all equality case we have
$$\frac{a-3}{(2+a^2)(a+k)}=\frac{b-3}{(2+b^2)(b+k)}=\frac{c-3}{(2+c^2)(a+k)},$$
we obtain:
$$\frac{2-3}{(2+2^2)(2+k)}=\frac{\frac{1}{4}-3}{(2+\left(\frac{1}{4}\right)^2)(\frac{1}{4}+k)},$$
which gives $k=-\frac{9}{4}$.
Thus, it remains to prove that
$$\left(\sum\limits_{cyc}(a-3)(4a-9)\right)^2\geq4\sum_{cyc}(2+a^2)(4a-9)^2,$$
which is wrong for $a=4$ and $b=c=\frac{1}{2}$.
Any hint?
Thank you!
| There is also the following.
Let two variables be negative.
Let $a<0$, $b<0$,$a'=-a$ and $b'=-b$.
Thus, $a'>0$, $b'>0$, $a'b'c=1$ and
$$\sum_{cyc}\frac{7-6a}{2+a^2}=\frac{7+6a'}{2+a'^2}+\frac{7+6b'}{2+b'^2}+\frac{7-6c}{2+c^2}>$$
$$>\frac{7-6a'}{2+a'^2}+\frac{7-6b'}{2+b'^2}+\frac{7-6c}{2+c^2},$$ which says that it's enough to prove our inequality for positive variables.
Now, in the starting post we saw that it's enough to prove that
$$\sum_{cyc}\frac{(a-3)^2}{2+a^2}\geq4,$$ which after substitution $a=\frac{1}{x}$, $b=\frac{1}{y}$ and $c=\frac{1}{z}$ gives
$$\sum_{cyc}\frac{(3x-1)^2}{2x^2+1}\geq4.$$
Now, by C-S
$$\sum_{cyc}\frac{(3x-1)^2}{2x^2+1}\geq\frac{\left(\sum\limits_{cyc}(3x-1)\right)^2}{\sum\limits_{cyc}(2x^2+1)}.$$
Thus, it's enough to prove that:
$$\frac{\left(\sum\limits_{cyc}(3x-1)\right)^2}{\sum\limits_{cyc}(2x^2+1)}\geq4$$ or
$$9(x+y+z-1)^2\geq8(a^2+b^2+c^2)+12.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that
$$9(3u-w)^2\geq8(9u^2-6v^2)+12w^2,$$ which is a linear inequality of $v^2,$ which says that it's enough to prove the last inequality for an extreme value of $v^2$, which happens for equality case of two variables.
Let $y=x$ and $z=\frac{1}{x^2}.$
Id est, we need to prove that
$$9\left(2x+\frac{1}{x^2}-3\right)^2\geq8\left(2x^2+\frac{1}{x^4}\right)+12$$ or
$$(2x-1)^2(x-1)^2(x+1)(5x+1)\geq0$$ and we are done!
| {
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"url": "https://math.stackexchange.com/questions/2183109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $\sum_{n=1}^\infty a_n<\frac{1}{2}(f(1)-f(2))$. Let $f:[1,\infty)\rightarrow(0,\infty)$ be a twice differentiable decreasing function such that $f''(x)$ is positive for $x\in(1,\infty)$. For each positive integer $n$, let $a_n$ denote the area of the region bounded by the graph of $f$ and the line segment joining the points $(n,f(n)$ and $(n+1,f(n+1))$. Show that
$\sum_{n=1}^\infty a_n<\frac{1}{2}(f(1)-f(2))$.
My attempt: I can see that
$a_n=f(n+1)+\frac{1}{2}(f(n)-f(n+1))-\int_n^{n+1}f(n)$
$=\frac{1}{2}(f(n)+f(n+1))-\int_n^{n+1}f(n)$
$<\frac{1}{2}(f(n)+f(n+1))-f(n+1)$
which is somewhat helpful in the infinite sum as many terms cancel, but I'm not getting the right bound. How can I use the second derivative being negative?
| On each interval $[n,n+2]$, since $f$ is convex, $f$ is above its tangent line at $n+1$ (see property 5 here), hence for all $x\in [n,n+2]$, $$ f(x) \geq f(n+1)+ f'(n+1)(x-(n+1))$$
Integrating this inequality over $x$,
$$\int_{n}^{n+2}f(x) dx \geq \int_{n}^{n+2} [f(n+1)+ f'(n+1)(x-(n+1))]dx$$
which rewrites $$\int_n^{n+2} f(x)dx \geq 2f(n+1)$$
Remember that $$a_n=\frac{1}{2}(f(n)+f(n+1))-\int_n^{n+1}f(n)$$
Hence $$\begin{align} a_n+a_{n+1} &= \frac{f(n)+f(n+1)}2 + \frac{f(n+1)+f(n+2)}2 - \int_{n}^{n+2} f(x)dx\\
&\leq \frac{f(n)+f(n+1)}2 + \frac{f(n+1)+f(n+2)}2 - 2f(n+1) \\
&=\frac{f(n)-f(n+1)}2 - \frac{f(n+1)-f(n+2)}2
\end{align} $$
Since $f$ is decreasing and positive, it has a finite limit at $\infty$ hence $\frac{f(n+1)-f(n+2)}2$ converges to $0$.
Summing all these inequalities, the sums on the right telescope, yielding $$\sum_{n=1}^\infty (a_n+a_{n+1})\leq \frac{f(1)-f(2)}2$$
that is $$\left( 2\sum_{n=1}^\infty a_n \right) -a_1 \leq \frac{f(1)-f(2)}2$$
$$\sum_{n=1}^\infty a_n \leq \frac{f(1)-f(2)}4 + \frac{a_1}2$$
The bound you found in your question yields $\displaystyle a_1\leq \frac{f(1)-f(2)}2$ and we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find all points $(x,y)$ on the graph of $f(x)$ with tangent lines passing through a certain point Find all points $(x,y)$ on the graph of $f(x) = x^2$ with tangent lines passing through the point $(3,8)$.
My attempt:
$f'(x) = x^2$
I substituted a random point into $f'(x)$ to get the gradient of the tangent, so $f'(0) = 0$.
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$
$$ m = \frac{8-0}{3-0} = 8/3$$
$$y-y_1 = m(x-x_1)$$
Now I substituted the point $(3,8)$:
$$ y - 8 = \frac{8}{3}(x-3)$$
$$y = \frac{8}{3}x$$
$$ y = x^2$$
Solving simultaneously we get $x = 0$ and $y = 0$ or $x = 8/3 $ and $ y = 64/9$
So we have the points $(0;0)$ and $(\frac{8}{3} ; \frac{64}{9})$. However, I am not so confident with my answer. Is this correct?
| $y = x^2\\
y' = 2x$
let $(x_0,y_0)$ be a point on the curve, and the line tangent to that point:
$(y-y_0) = 2x_0(x-x_0)\\
y-x_0^2 = 2x_0x-2x_0^2\\
y = 2x_0x - x_0^2$
Goes through the point $(3,8)$
$8 = 6x_0 - x_0^2\\
(x_0 - 2)(x_0 -4) = 0$
$(2,4), (4,16)$ are points on the curve where the line tangent to those points goes through the point $(3,8)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How can we show that $\int_{0}^{2\pi}{x\over \phi-\cos^2(x)}\mathrm dx=2\pi^2?$ We have the integral
$$\int_{0}^{2\pi}{x\over \phi-\cos^2(x)}\mathrm dx=2\pi^2\tag1$$
$\phi$; Golden ratio
What method do we employ to prove $(1)$?
An attempt:
If we use $u=\phi-\cos^2 x$ then $\int_{\phi-1}^{\phi-1}(...)du=0$
Another way, $(1)$ becomes
$${1\over 2\sqrt{\phi}}\int_{0}^{2\pi}\left({x\over \sqrt{\phi}-\cos x}-{x\over \sqrt{\phi}+\cos x} \right)\mathrm dx\tag2$$
We got a hint by using
$z=\tan{x\over 2}$, $dx={2dz\over 1+z^2}$, $\cos x={1-z^2\over 1+z^2}$ and $\sin x={2z\over 1+z^2}$
Let try on
$$\int_{0}^{2\pi}{x\over \sqrt{\phi}+\cos x}\mathrm dx=4\int_{0}^{0}{\tan^{-1} z\over (1+z^2)\sqrt{\phi}+1-z^2}\cdot (1-z^2)\mathrm dz=0?\tag3$$
Doesn't seem to work.
How else can we prove $(1)$?
| A purely real approach.
Consider the integral
$$C(a,b)=\int_0^{\pi}\frac{\mathrm dx}{a+b\cos(x)^2}$$
It has already been shown by other users that your integral is given by $2\pi C(\phi,-1)$.
We commence with the substitution $t=\tan\frac{x}2$:
$$C(a,b)=2\int_0^\infty \frac1{a+b\left[\frac{1-t^2}{1+t^2}\right]^2}\frac{\mathrm dt}{1+t^2}$$
$$C(a,b)=2\int_0^\infty\frac{1+t^2}{a(t^2+1)^2+b(t^2-1)^2}\mathrm dt$$
$$C(a,b)=2\int_0^\infty\frac{1+t^2}{(a+b)t^4+2(a-b)t^2+a+b}\mathrm dt$$
$$C(a,b)=\frac2{a+b}\int_0^\infty\frac{1+t^2}{t^4+2gt^2+1}\mathrm dt$$
Where $g=\frac{a-b}{a+b}$. We may split up the integral:
$$C(a,b)=\frac2{a+b}\int_0^\infty\frac{\mathrm dt}{t^4+2gt^2+1}+\frac2{a+b}\int_0^\infty\frac{t^2\mathrm dt}{t^4+2gt^2+1}$$
Then we pay attention to $$N_{n}(s)=\int_0^\infty \frac{x^{2n}}{x^4+2sx^2+1}\mathrm dx$$
With the substitution $x=1/u$, we have
$$N_n(s)=-\int_\infty^0 \frac{\frac1{u^{2n}}}{\frac1{u^4}+\frac{2s}{u^2}+1}\frac{\mathrm du}{u^2}$$
$$N_n(s)=\int_0^\infty \frac{u^{2-2n}\mathrm du}{u^4+2su^2+1}$$
$$N_n(s)=N_{1-n}(s)$$
So we have
$$C(a,b)=\frac4{a+b}\int_0^\infty\frac{\mathrm dt}{t^4+2gt^2+1}$$
For this remaining integral, we recall that
$$\begin{align}
\frac1{x^4+(2-c^2)x^2+1}&=\frac1{(x^2+cx+1)(x^2-cx+1)}\\
&=\frac1{4c}\frac{2x+c}{x^2+cx+1}-\frac1{4c}\frac{2x-c}{x^2-cx+1}+\frac14\frac1{x^2+cx+1}+\frac14\frac1{x^2-cx+1}
\end{align}$$
So we have that
$$K(c)=\int_0^\infty \frac{\mathrm dx}{x^4+(2-c^2)x^2+1}$$
becomes
$$K(c)=\frac1{4c}\int_0^\infty\frac{2x+c}{x^2+cx+1}\mathrm dx-\frac1{4c}\int_0^\infty\frac{2x-c}{x^2-cx+1}\mathrm dx\\
+\frac14\int_0^\infty \frac{\mathrm dx}{x^2+cx+1}+\frac14\int_0^\infty \frac{\mathrm dx}{x^2-cx+1}$$
The first two integrals are easy, and end up vanishing, so we end up with
$$K(c)=\frac14I(1,c,1)+\frac14I(1,-c,1)$$
Where
$$I(a,b,c)=\int_0^\infty \frac{\mathrm dx}{ax^2+bx+c}=\frac2{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}}\bigg]_0^\infty\\
=\frac\pi{\sqrt{4ac-b^2}}-\frac2{\sqrt{4ac-b^2}}\arctan\frac{b}{\sqrt{4ac-b^2}}$$
Hence
$$K(c)=\frac\pi{2\sqrt{4-c^2}}$$
So we see that
$$C(a,b)=\frac4{a+b}K\left(2\sqrt{\frac{b}{a+b}}\right)=\frac\pi{\sqrt{a^2+ab}}$$
And since, by definition, $\phi^2-\phi=1$, we have that
$$C(\phi,-1)=\pi$$
Which gives, as desired,
$$\int_0^{2\pi}\frac{x\mathrm dx}{\phi-\cos(x)^2}=2\pi^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2186007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 3
} |
Finding taylor series in Real Analysis In the following question I am trying to find the Taylor series for,
$$f(x)=x^3\sin(x^2)$$
centered at $x=0$.
My first thought was to find the Taylor series for $\sin x$ then derive it to get my series I wanted but I am not quite sure how to do that.
I worked out the Taylor series for $f(x)=\sin(x)$ centered at $x=0$ and got,
$f(x)=\sin x, f(0)=0$
$f'(x)=\cos x, f'(0)=1$
$f"(x)=-\sin x, f''(0)=0$
$f'''(x)=-\cos x, f'''(0)=-1$
$f''''(x)=\sin x, f''''(0)=0$
So the Taylor series for $\sin x$ centered at $x=0$ is,
$\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...$
But how do I use this to find the Taylor series for $f(x)=x^3\sin(x^2)$
| We have
$$\sin x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...\tag 1$$
This series converges for all $x$ (just use the Ratio Test to verify it). Thus, replacing $x$ by $x^2$ in $(1)$ we get
$$\sin (x^2)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2{(2n+1)}}}{(2n+1)!}=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\frac{x^{14}}{7!}...\tag 2$$ Then multiply $x^3$ to $(2)$ we get
$$x^3\sin (x^2)=\sum_{n=0}^{\infty}(-1)^nx^3\frac{x^{2{(2n+1)}}}{(2n+1)!}=x^5-\frac{x^9}{3!}+\frac{x^{13}}{5!}-\frac{x^{17}}{7!}...\tag 3$$ Hence,
$$x^3\sin (x^2)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n+5}}{(2n+1)!}=x^5-\frac{x^9}{3!}+\frac{x^{13}}{5!}-\frac{x^{17}}{7!}...\tag 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Derivative of $f(x) = y^4$ $(x^3 \cdot y^4)' = ?$
$a = x^3$ , $b = y^4$
$$(x^3 \cdot y^4)'= a'\cdot b + a\cdot b' = 3x^2 \cdot y^4 + x^3 \cdot (b')$$
Not sure what the derivative of $y^4$ would be, $(4y^3)$? Since the derivative of $(x\cdot y) = 1\cdot y + x \cdot y'$ and the derivative of $y' = y'$ instead of following the power rule.
| by the product and chain rule we get $$(x^3y(x)^4)'=3x^2\cdot y(x)^4+x^3\cdot 4y(x)^3\cdot y'(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the second derivative implicitely
$$\text{Having the equation }\sqrt{y}+3xy=4 \text{ find } \frac{d^2y}{dx^2} := y''$$
Ok so when I work it out implicitly I get,
$$y'=\frac{-6y^{3/2}}{1+6x\sqrt{y}}$$
Continuing on I get,
$$y''=\frac{90y^2(1+6x\sqrt{y})-108xy^{3/2}}{(1+6x\sqrt{y})^3}$$
Which is incorrect I believe. Can someone please tell me if my $y'$ is correct? And where should I go from there?
| If you have the equation (considering that $y = y(x)$ and that $y' := dy/dx$)
$$y^{1/2} + 3xy = 4 \implies \frac{1}{2y^{1/2}}y' + 3y + 3xy' = 0 \implies y' = \frac{-6y^{3/2}}{1 + 6xy^{1/2}} $$
Where I differentiate both sides of the equation and isolated $y'$. So your answer for $y'$ is correct.
Now lets calculate the second derivative $y''$
$$y'' \stackrel{\text{Leibnitz rule}}{=} \frac{1}{(1 + 6xy^{1/2})^2}\left((1 + 6xy^{1/2})(-9y^{1/2}y') - (-6y^{3/2})\left(6y^{1/2} + \frac{3xy'}{y^{1/2}}\right)\right) \implies $$
$$y'' = \frac{1}{(1 + 6xy^{1/2})^2}\left((1 + 6xy^{1/2})(-9y^{1/2}y') - \left(\frac{-6y^{3/2}}{y^{1/2}}\right)\left(6y + 3xy'\right)\right) \implies $$
$$y'' = \frac{1}{(1 + 6xy^{1/2})^2}\left((1 + 6xy^{1/2})(-9y^{1/2}y') - \left(-6y\right)\left(\frac{6yy'}{y'} + 3xy'\right)\right) \implies $$
$$y'' = \frac{y'}{(1 + 6xy^{1/2})^2}\left(-9y^{1/2} -54xy + 36\frac{y^2}{y'} +18xy \right) \implies$$
$$y'' = \frac{-6y^{3/2}}{(1 + 6xy^{1/2})^3}\left(-9y^{1/2} - 36xy + 36y^2\frac{(1+6xy^{1/2})}{-6y^{3/2}}\right) \implies $$
$$y'' = \frac{-6y^{3/2}}{(1 + 6xy^{1/2})^3}\left(-9y^{1/2} - 36xy -6y^{1/2}(1+6xy^{1/2}) \right) \implies$$
$$y'' = \frac{-6y^{3/2}}{(1 + 6xy^{1/2})^3}\left(-9y^{1/2} - 36xy -6y^{1/2} -36xy \right) \implies$$
$$y'' = \frac{-6y^{3/2}}{(1 + 6xy^{1/2})^3}\left(-15y^{1/2} - 72xy\right) \implies$$
Which leads us to the answer
$$y'' = \frac{90y^2+432xy^{5/2}}{(1 + 6xy^{1/2})^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why is $ \int_{1}^{9} \frac{dx}{x(\ln(x)^{8/5})}=\infty$? I get that it’s an improper integral.
Why is the part $\frac {5}{3\ln (t)^{3/5}}$ equal to $+\infty$?
I don’t understand that part.
| Let $u=\ln x\implies du=\dfrac{dx}{x}$
$$\begin{equation}\begin{split}\displaystyle \int_1^9\dfrac{1}{x(\ln x)^{\frac{8}{5}}}\mathrm{dx}&=\int_0^{\ln 9}\dfrac{1}{u^{\frac{8}{5}}}\mathrm{du}\\&=\dfrac{u^{-\frac{3}{5}}}{-\frac{3}{5}}\bigg|_0^{\ln 9}\\&=-\dfrac{5}{3}u^{-\frac{3}{5}}\bigg|_0^{\ln 9}\\&=-\dfrac{5}{3(\ln 9)^{\frac{3}{5}}}+\lim_\limits{b\to0}\dfrac{5}{3b^{\frac{5}{3}}}\\&=+\infty\end{split}\end{equation}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The equations $x^3+5x^2+px+q=0$and $x^3+7x^2+px+r=0$ have 2 common roots, then find the third root of both equations
The equations $x^3+5x^2+px+q=0$and $x^3+7x^2+px+r=0$ have 2 common roots, then find the third root of both equations
From the first equation we can say, $\alpha\beta+\beta\gamma+\gamma\alpha=p/1=p$. Similarly from the second equation we know, $\alpha\beta+\beta\delta+\delta\alpha=p/1=p$
Hence,
$\alpha\beta+\beta\delta+\delta\alpha=\alpha\beta+\beta\gamma+\gamma\alpha$
$\delta(\beta+\alpha)=\gamma(\beta+\alpha)$
$\delta=\gamma$
Hence the third root of both equations should be equal, but $\alpha+\beta+\gamma=-5$ and $\alpha+\beta+\delta=-7$. Now, where did I go wrong?
| $$
P(x) = x^3+5x^2+p x+q = (x-x_1)b(x)\\
Q(x) = x^3+7x^2+p x+r = (x-x_2)b(x)
$$
then
$$
Q(x)-P(x) = (x_1-x_2)b(x) = 2x^2+r-q
$$
and finally
$$
(x-x_1) = \frac{P(x)}{b(x)} = (x_1-x_2)\frac{x^3+5x^2+p x+q}{2x^2+r-q}\\
(x-x_2) = \frac{Q(x)}{b(x)} = (x_1-x_2)\frac{x^3+7x^2+p x+r}{2x^2+r-q}\
$$
so from any of those equations, equating to $0$ the polynomial coefficients we have
$$
2-x_1+x_2 = 0\\
5(x_1-x_2)+2x_1 = 0
$$
or
$$
2-x_1+x_2 = 0\\
7(x_1-x_2)+2x_2 = 0
$$
giving
$$
x_1 = -5,\ \ x_2 = -7
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$ let $$f(x)=\frac{1}{1+2\cos x}$$
prove that :
$$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$$
My Try :
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{4\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{6\pi}{7})}$$
$$L=\frac{1}{1+2\cos (\frac{6\pi}{7})}+\frac{1}{1+2\cos (\frac{4\pi}{7})}+\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
what now ?
| $f(2x)=\dfrac1{1+2\cos2x}=\dfrac1{1+2(1-2\sin^2x)}=\dfrac{\sin x}{\sin3x}$ for $\sin x\ne0$
$x=\dfrac\pi7\implies f\left(\dfrac{2\pi}7\right)=\dfrac{\sin\dfrac{\pi}7}{\sin\dfrac{3\pi}7}=2\cos\dfrac{3\pi}7$ as $\sin\dfrac{\pi}7=\sin\left(\pi-
\dfrac\pi7\right)=2\sin\dfrac{3\pi}7\cos\dfrac{3\pi}7$
$x=\dfrac{2\pi}7\implies f\left(\dfrac{4\pi}7\right)=\dfrac{\sin\dfrac{2\pi}7}{\sin\dfrac{6\pi}7}=2\cos\dfrac{\pi}7$ as $\sin\dfrac{6\pi}7=\sin\left(\pi-\dfrac{6\pi}7\right)$
$x=\dfrac{3\pi}7\implies f\left(\dfrac{6\pi}7\right)=\dfrac{\sin\dfrac{3\pi}7}{\sin\dfrac{9\pi}7}=\dfrac{-\sin\left(\pi+\dfrac{3\pi}7\right)}{-\sin\left(2\pi-\dfrac{5\pi}7\right)}=2\cos\dfrac{5\pi}7$
So, we need $S=2\cos\dfrac{\pi}7+2\cos\dfrac{3\pi}7+2\cos\dfrac{5\pi}7$
Like $\sum \cos$ when angles are in arithmetic progression,
$\implies\sin\dfrac\pi7\cdot S=2\cos\dfrac{\pi}7\cdot\sin\dfrac\pi7+2\cos\dfrac{3\pi}7\cdot\sin\dfrac\pi7+2\cos\dfrac{5\pi}7\cdot\sin\dfrac\pi7$
Using Werner Formulas,
$\sin\dfrac\pi7\cdot S=\sin\dfrac{6\pi}7=\cdots=\sin\dfrac\pi7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to find max integer value of $6\sin(x)-8\cos(x)$ without using derivative I have $6\sin(x)-8\cos(x)$ and want to find its maximum value. If we use derivative and assuming that it is equal to 0, we get $6\cos(x)+8\sin(x)=0$ which implies that $\tan(x)=-\cfrac{3}{4}$ and $x = 143^\circ$ and we can conclude that the maximum value is equal to $10$ . How can I find the maximum integer value of $6\sin(x)-8\cos(x)$ without using derivative?
| Write
$$f = 6 \sin(x) - 8 \cos(x) = \sqrt{6^2 + 8^2} (\frac {6}{\sqrt{6^2 + 8^2}} \sin(x) + \frac {8}{\sqrt{6^2 + 8^2}} \cos(x) ) $$
and now let
$$\frac {6}{\sqrt{6^2 + 8^2}} = \sin(\alpha)$$
then (remember $\sin(z)^2 + \cos(z)^2 = 1$ for any $z$)
$$\frac {8}{\sqrt{6^2 + 8^2}} = \cos(\alpha)$$
And we can write
$$f = 10 \;(\;\sin(\alpha)\sin(x) - \cos(\alpha) \cos(x)\; ) = - 10 \cos(x+\alpha) $$
where we have used the trigonometric addition theorem.
Hence the maximum is obviously $10$.
EDIT
As a bonus let us determine the values of $x$ where $f$ has its maximum (and its minimum).
From the previous result we have to look for the extrema of $\cos(z)$ which leads us to
$$f\to -10: x =2 n \pi - \alpha $$
$$f\to +10: x=(2 n+1) \pi - \alpha$$
Where
$$\alpha = \arctan(\frac{3}{4})\;\simeq 0.6435010.643501 $$
and $n$ integer.
The extrema closest to the origin are at
$$x_{min} = - \alpha\;\simeq -0.6435010.643501 $$
$$x_{max} = \pi - \alpha \; \simeq 2.498091544796$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial fractioning an exponential function $\sum\limits_{n=1}^{\infty}\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}$ I saw this and immediately thought it was a telescoping series, so I tried to partial fraction it. $$\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}=\frac{A}{3^{n+1}-2^{n+1}}+\frac{B}{3^n-2^n}$$
\begin{align}
6^n &= A(3^n-2^n)+B(3^{n+1}-2^{n+1})\\
6^n &=3^n(A+3B)+2^n(-A-2B)
\end{align}
This is where I got stuck, how can the sum of these 2 be $6^n$ so I figured one way it could be true was if
$$A+3B=\frac{2^n}{2}$$
$$-A-2B=\frac{3^n}{2}$$
Because then I will just be doing $\frac{6^n}{2}+\frac{6^n}{2}=6^n$
Solving this system gave me
\begin{align}
A&=-\frac{3^{n+1}+2^{n+1}}{2}\\ B&=\frac{3^n+2^n}{2}
\end{align}
As terrible as it looks, I got that it did telescope to $\frac{5}{2}$ however it actually came out to $2$ with the partial fraction of $$\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$$
So 2 questions,
Why am I wrong? and
What would be the correct approach to get this answer?
| Is there a transcription problem? A check on your work shows
$$
\frac{6^n}{\left(3^{n+1}-2^{n+1}\right) \left(3^n-2^n\right)} =
\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}},
$$
and
$$
\sum _{n=1}^{\infty } \frac{6^n}{\left(3^{n+1}-2^{n+1}\right) \left(3^n-2^n\right)} = 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to factor $r^6 -3r^4 +3r^2 - 1 = 0$ I tried two ways to factor $r^6 -3r^4 +3r^2 - 1 = 0$
When I factor $r^4$ out of $r^6 -3r^4$:
$r^4(r^2-3)+(3r^2-1) = 0$
When I factor $r^2$ out of $r^6 + 3r^2$:
$r^2(r^4+3)-(3r^4 + 1) = 0$
For both methods, I'm stuck at an equation that isn't factorable. But Wolfram Alpha says $r^6 -3r^4 +3r^2 - 1 = (r-1)^3(r+1)^3$. How do you get this?
| Option 1
Notice that if you make the substitution $x = r^2$ to make it $x^3-3x^2+3x-1$, the coefficients are (1, -3, 3, -1), which is exactly the values on the 4th row of Pascal's triangle, with alternating signs. So it must be something like $(x \pm 1)^3$. The last coefficient is $-1$, so there must be an odd number of $(x-1)$ terms, and it doesn't take long to realise it must be $(x-1)^3=(r^2-1)^3=(r-1)^3(r+1)^3$.
Option 2
Using the rational roots theorem, we note that both $r=1$ and $r=-1$ are roots. Divide the whole thing by $(r^2-1)$ and we get $r^4-2r^2+1$. Applying rational roots again and we see that $r=1$ and $r=-1$ are still roots. Divide again by $(r^2-1)$ and we're left with $r^2-1$, so we know that the whole thing is $(r^2-1)^3=(r-1)^3(r+1)^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
why doubling the number in a contingency table changes the p-value? I am doing a statistics problem, which is testing if the evaluation of a person is independent of the person's sex.
I am given a contingency table, I calculated the expected value for each entry and calculated the chi-square value then I got a p-value.
Then the question asked me to do the same thing after doubling all entries in the contingency table, I got a p-value smaller than the one I got before. Why does this happen?
Can anyone give me an explanation for the difference? Thx!
| $$
\begin{array}{ccc}
\begin{array}{cc}
1 & 2 \\ 3 & 4 \\ 5 & 5
\end{array} & \text{Now multiply all entries by 10 } \longrightarrow &
\begin{array}{cc}
10 & 20 \\ 30 & 40 \\ 50 & 50
\end{array}
\end{array}
$$
In the first table, the chi-square test statistic for the null hypothesis of independence of rows and columns is
\begin{align}
& \sum \frac{(\text{observed} - \text{expected})^2}{\text{expected}} \\[10pt]
= {} & \frac{(1-1.35)^2}{1.35} + \frac{(2-1.65)^2}{1.65} + \frac{(3-3.15)^2}{3.15} + \frac{(4 - 3.85)^2}{3.85} + \frac{(5-4.5)^2}{4.5} + \frac{(5-5.5)^2}{5.5}
\end{align}
In the second table, it is
$$
\frac{(10-13.5)^2}{13.5} + \frac{(20-16.5)^2}{16.5} + \frac{(30-31.5)^2}{31.5} + \frac{(40 - 38.5)^2}{38.5} + \frac{(50-45)^2}{45} + \frac{(50-55)^2}{55}
$$
Observe two things:
*
*We multiplied each numerator by $10^2$ and each denominator by $10$, thereby multiplying the entire expression by $10^2/10 = 10.$ Thus the value of the chi-square test statistic is $10$ times as big.
*If the pattern of deviation from independence of rows and columns persists as sampling continues until we have $10$ times as many observations as we had before, then we have $10$ times as much evidence of the non-independent distribution, so it makes sense that the evidence against independence is stronger.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$3\mid a^3+b^3+c^3$ if only if $3\mid a+b+c $
Prove the following equivalence: $3\mid (a^3 + b^3 + c^3) $ if and only if $3\mid (a + b + c) $.
My try:
I know $a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc$, but I can't seem to proceed from here.
Thanks all!
| Fact is that, $n^3−n=(n−1)n(n+1)$ being a product of three consecutive integers is a multiple of $3$.
Hence $3$ divides $(a^3−a)+(b^3−b)+(c^3−c)$ for any triple of integers $a,b,c$. Rearranging the terms, we have that $3$ dividing $(a^3+b^3+c^3)−(a+b+c)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Last three digits of $\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$ Find the last three digits of $$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$$
I don't know how to continue my work:
$$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}=\sqrt{2^{2(2016)}+2\cdot (3\cdot 2)^{2016}+3^{2(2016)}}=\sqrt{2^{4032}+2\cdot 3^{2016}\cdot 2^{2016}+3^{4032}}=\sqrt{2^{4032}+2^{2017}\cdot 3^{2016}+3^{4032}}=?$$
| The presence of the squares $4$ and $9$ here is a big hint to help you find that the expression $E := \sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}} = \sqrt{(2^{2016}+3^{2016})^2} = 2^{2016}+3^{2016}$.
Then finding hte last three digits of this expression is a matter of finding the remainders $2^{2016}\bmod 1000$ and $3^{2016}\bmod 1000$ and then adding.
The Carmichael function $\lambda$ gives us the maximum cyclic order of exponentiation and here $\lambda(1000) = {\rm lcm}(\lambda(2^3), \lambda(5^3)) = {\rm lcm}(2, 100) = 100 $. Any smaller cycle will divide this value. We could also use Euler's totient $\phi(1000)=400$ to similar effect. So this gives us $a^{2016} \equiv a^{16} \bmod 1000$ (noting $16\ge 3$, the greatest prime exponent in $1000$) and we can calculate this easily enough for both cases:
$ 2^{16}\equiv 1024\cdot 64 \equiv 24.64 \equiv 1536 \equiv 536 \bmod 1000$
$ 3^{16} \equiv 81^4 \equiv 6561^2 \equiv 561^2 \equiv 314721 \equiv 721 \bmod 1000$
And then we can add to get $E\equiv 257 \bmod 1000$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
How many roots does the polynomial $acx^4 + b(a + c)x^3 + (a^2 + b^2 + c^2)x^2 + b(a + c)x + ac$ have Let a, b and c be real numbers. Then the fourth degree
polynomial in $x$, $acx^4 + b(a + c)x^3 + (a^2 + b^2 + c^2)x^2 + b(a + c)x + ac$
(a) Has four complex (non-real) roots
(b) Has either four real roots or four complex roots
(c) Has two real roots and two complex roots
(d) Has four real roots
I have no clue as to how to approach it, a hint would suffice
| HINT: write your equation in the form
$$(x(ax+b)+c)(a+x(b+cx))=0$$ and solve it
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove my matrix's powers? Being bored I played around with matrices I stumbled upon this matrix.
$$T = \left[\begin{array}{ll|cc}
0&2&0&0\\
\frac{1}{2}&0&0&0\\
\hline 1&0&1&0\\
1&1&0&1
\end{array}\right]$$
What is $T^{2^k}$ for $\cases{k\in \mathbb N\\k>1}$? I have my suspicions what it should be, but how can we prove it?
| Well, let
$$
T = \begin{pmatrix}A & 0 \\ B & I\end{pmatrix}, \quad A = \begin{pmatrix}0 & 2 \\ \frac{1}{2} & 0\end{pmatrix}, \quad B = \begin{pmatrix}1 & 0 \\ 1 & 1\end{pmatrix},
$$
and $I$ is a unit matrix. One may see (by simple induction) that
$$
T^n = \begin{pmatrix}A^n & 0 \\ B\sum_{i=0}^{n-1}A^i & I\end{pmatrix}.
$$
Also one may note that $A^2 = I$. Thus $A^{2^k} = I$ and
$$
B\sum_{i=0}^{2^k-1}A^i = B\cdot2^{k-1}(A + I) = 2^{k-1}\begin{pmatrix}1 & 2 \\ \frac{3}{2} & 3\end{pmatrix}.
$$
So the full answer is
$$
T^{2^k} = \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 2^{k-1} & 2^k & 1 & 0 \\
3\cdot 2^{k-2} & 3\cdot 2^{k-1} & 0 & 1\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is there something special about $x^2-x-1$? I was working on proofs by induction and had to produce numerous proofs about the properties the Fibonacci sequence as exercises.
It was pointed out to me while proving that $ F_n = \frac 1{\sqrt 5} \left(\phi^n-\left(-\frac 1\phi\right)^n\right)= F_{n-1}+F_{n-2}$, that the roots of the second degree polynomial
$$y = x^2-x-1$$
are the golden ratio $\phi = \frac {1+\sqrt 5}2 = \lim_{n \to \infty} \frac {F_{n+1}}{F_n}$ and its negative inverse $\left(-\frac 1\phi\right)$.
I never realized that before and now I wonder whether there is something special about the polynomial $y=x^2-x-1$ appart from the fact that its positive root is the golden ratio $\phi$?
| Perhaps what is more interseting than: $y = x^2 - x - 1$
Is this equation: $x = \frac 1x + 1$
Is there some number that equals its reciprocal makes plus 1?
I think the decimal expansion (rounded off at 3 decimals) makes it is easier to follow some of the implications.
$\frac {1}{1.618} = 0.618\\
(1.618)(0.681) = 1\\
(1.618)^2 = 1 + 1.618\\
(0.618)^2 = 1-0.618$
As it related to the Fibonacci sequence. We start with the conjecture that $n$ gets large $\frac {F_{n+1}}{F_{n}}$ approaches some constant.
$\frac {F_{n+1}}{F_{n}} = \frac {F_{n}}{F_{n-1}}$
And for the Fibonacci sequence $F_{n+1} = F_n + F_{n-1}$
$\frac {F_{n} + F_{n-1}}{F_{n}} = \frac {F_{n}}{F_{n-1}}\\
1 + \frac {F_n}{F_{n-1}} = \frac {F_{n}}{F_{n-1}}$
and we say $\frac {F_{n}}{F_{n-1}}$ equals $x$ and we are back to
$1 + \frac {1}{x} = x$
One more:
The continued fraction.
$x = \dfrac {1}{1+\frac {1}{1+\frac{1}{1+\cdots}}}\\
x = \frac {1}{1+x}\\
x^2 + x - 1 = 0$
The roots of which are the negative of
$x^2 - x - 1 = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Pseudo Inverse of product of Matrices Let $A$ and $B$ are two matrices where $A \in \mathbb{R}^{m\times p}$ and $B \in \mathbb{R}^{p\times n}$ and both $A$ and $B$ are full rank matrices Now I really want to know in what cases
$(AB)^+ = B^+A^+$ ,where $A^+$ is Moore-Penrose Pseudo-inverse of $A$
Here $m,p$ and $n$ are in any order like $m<p<d$ or $m>p<d$ etc.
| Wikipedia
Example
Define matrices
$$
\mathbf{A} =
\left[
\begin{array}{cc}
0 & 1 \\
3 & 2 \\
0 & 2 \\
\end{array}
\right]
, \quad
%
\mathbf{B} =
\left[
\begin{array}{ccc}
0 & 3 & 0 \\
1 & 2 & 2 \\
\end{array}
\right]
, \quad
%
\mathbf{C} = \mathbf{A} \mathbf{B} =
\left[
\begin{array}{rrr}
1 & 2 & 2 \\
2 & 13 & 4 \\
2 & 4 & 4 \\
\end{array}
\right]
$$
$$
\mathbf{A}^{\dagger} = \frac{1}{15}
\left[
\begin{array}{rrr}
-2 & 5 & -4 \\
3 & 0 & 6 \\
\end{array}
\right]
, \quad
%
\mathbf{B}^{\dagger} = \frac{1}{15}
\left[
\begin{array}{rr}
-2 & 3 \\
5 & 0 \\
-4 & 6 \\
\end{array}
\right]
, \quad
%
\mathbf{C}^{\dagger} = \frac{1}{225}
\left[
\begin{array}{rrr}
13 & -10 & 26 \\
-10 & 25 & -20 \\
26 & -20 & 52 \\
\end{array}
\right]
$$
Test premise
Does $\mathbf{C}^{\dagger} = \mathbf{B}^{\dagger}\mathbf{A}^{\dagger}$?
$$
\begin{align}
\mathbf{B}^{\dagger}\mathbf{A}^{\dagger} &= \frac{1}{15}
\left[
\begin{array}{rr}
-2 & 3 \\
5 & 0 \\
-4 & 6 \\
\end{array}
\right]
\frac{1}{15}
\left[
\begin{array}{rrr}
-2 & 5 & -4 \\
3 & 0 & 6 \\
\end{array}
\right] \\[3pt]
%%
& =
%%
\frac{1}{225}
\left[
\begin{array}{rrr}
13 & -10 & 26 \\
-10 & 25 & -20 \\
26 & -20 & 52 \\
\end{array}
\right] \\
&= \mathbf{C}^{\dagger}
\end{align}
$$
Conclusion
$$\therefore \qquad \left(\mathbf{A}\mathbf{B}\right)^{\dagger} = \mathbf{B}^{\dagger}\mathbf{A}^{\dagger}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
$(a) \sum_{n=2}^{\infty} \frac{3}{n^2+2n}$ Determine whether it converges and find sum Determine whether the following are convergent or divergent. If it is
convergent find its sum. Make sure to fully justify all of your work.
$$(a) \sum_{n=2}^\infty \frac{3}{n^2+3n}$$
Solution (My attempt):
Let $f(n) = a_n$
On $[2,\infty), f(x) = \frac 3 {x^2+3x} > 0$
$$f'(x) = 3\left(\frac{-(2x+3)}{(x^2+3x)^2}\right) < 0, \forall x \in [2,\infty)$$
Therefore the hypothesis of the integral test is met.
Consider
$$\lim_{A\to\infty} \int_2^A \frac 3 {(x)(x+3)} \, dx$$
$$\lim_{A\to\infty} \int_{2}^{A} \left( \frac{1}{x} - \frac{1}{x+3} \right)dx \text{ By pfd}$$
$$\lim_{A\to\infty} \left( \ln \right(\frac A {A+3}\left) - \ln \right(\frac 2 5 \left)\right) $$
$$= \ln 1 - \ln \left(\frac{2}{5}\right)$$
Therefore by Integral test the original series converges.
I was never thought how to find the sum of this type of series. The only way I know how to find the sum of a series is if its geometric. Does anyone know how too find the sum?
| To compute the sum observe
\begin{align}
\frac{1}{n(n+2)} = \frac{1}{2n}-\frac{1}{2(n+2)}
\end{align}
which means
\begin{align}
\sum^\infty_{n=2}\frac{3}{n^2+2n} =&\ \frac{3}{2}\sum^\infty_{n=2}\left( \frac{1}{n}-\frac{1}{n+2}\right)\\
=&\ \frac{3}{2}\sum^\infty_{n=2}\left(\frac{1}{n}-\frac{1}{n+1}\right)+\frac{3}{2}\sum^\infty_{n=2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\\
=&\ \frac{1}{2}+\frac{3}{4}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Proving that there are no other integer solutions The integer solutions for this equation:
$m ^ 2 = 2 \cdot 3 ^ n - 5$
are these four couples: $(n; m): (1;-1) (1;1) (3;7) (3;-7)$.
How can I prove there are no other solutions?
| Here is an approach.
Since $m$ is odd, let $m = 2k + 1$
We have $4k^2 + 4k + 6 = 2*3^n \implies 2k^2 + 2k + 3 = 3^n \implies 2k(k+1) = 3(3^{n-1} - 1)$
So $3$ has to divide $k$ or $k+1$ and $3$ cannot be a factor of both $k$ and $k+1$.
If $3$ divides $k$, then $k$ divides $3 \implies k = 3 \implies m = 7, n = 3$
If $3$ divides $k$ and $k+1$ divides $3 \implies k = 0, m = 1, n = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding a closed formula for a generating function involving catalan numbers For $n$ $\ge$ $1$, let the Catalan number $C_n$ be defined to be the number of ways of partitioning a convex $(n+2)$-gon into $n$ triangles by using diagonals that do not cross one another (except perhaps at their ends). By convention set $C_0$ $=$ $1$.
Find a recurrence relation expressing $C_n$ in terms of the Catalan numbers of smaller index.
I have found this to be $C_{n+1}$ $=$ $\frac{4n+2}{n+2}C_n.$
I want to use this to find a closed formula for the generating function $$g(x) = \sum_0^{\infty}C_nx^n.$$
Please help me..
| Here is another two-step approach to derive the generating function of the Catalan numbers from the recurrence relation
\begin{align*}
C_{n+1}&=\frac{4n+2}{n+2}C_n\qquad\qquad n\geq 0\tag{1}\\
C_0&=1
\end{align*}
First step: $C_n$
We can iteratively apply the recurrence relation (1) and obtain
\begin{align*}
C_n&=\frac{4n-2}{n+1}\cdot C_{n-1}\\
&=\frac{4n-2}{n+1}\cdot\frac{4n-6}{n}\cdot C_{n-2}\\
&=\frac{4n-2}{n+1}\cdot\frac{4n-6}{n}\cdots\frac{2}{1}\cdot C_0\\
&=\frac{2^n}{(n+1)!}\cdot (2n-1)(2n-3)\cdots 3\cdot 1\tag{2}\\
&=\frac{2^n(2n-1)!!}{(n+1)!}\tag{3}\\
&=\frac{2^n}{(n+1)!}\cdot\frac{(2n)!}{2^nn!}\tag{4}\\
&=\frac{1}{n+1}\binom{2n}{n}
\end{align*}
Comment:
*
*In (2) we factor out $2^n$ and use $C_0=1$.
*In (3) we use the double factorial as compact notation.
\begin{align*}
(2n)!!&=2n\cdot(2n-2)\cdots 4\cdot 2\\
(2n-1)!!&=(2n-1)\cdot(2n-3)\cdots 3\cdot 1\\
\end{align*}
*In (4) we use
\begin{align*}
(2n)!=(2n)!!(2n-1)!!\qquad\text{and}\qquad (2n)!!=2^nn!
\end{align*}
Second step: $g(x)= \sum_{n=0}^{\infty}C_nx^n$
We use the binomial identity
\begin{align*}
\binom{2n}{n}=(-4)^n\binom{-\frac{1}{2}}{n}
\end{align*}
(see e.g. formula (1.9) in H.W. Goulds Combinatorial Identities, vol. I) and obtain
\begin{align*}
C_n&=\frac{1}{n+1}\binom{2n}{n}=\frac{(-4)^n}{n+1}\binom{-\frac{1}{2}}{n}
=2(-4)^{n}\binom{\frac{1}{2}}{n+1}\\
&=-\frac{1}{2}(-4)^{n+1}\binom{\frac{1}{2}}{n+1}\tag{5}
\end{align*}
With the representation of $C_n$ stated in (5) we are well prepared to derive the generating function $g(x)$ by a binomial series expansion.
We obtain
\begin{align*}
\sum_{n=0}^\infty C_nx^n&=\sum_{n=0}^\infty-\frac{1}{2}(-4)^{n+1}\binom{\frac{1}{2}}{n+1}x^n\\
&=-\frac{1}{2x}\sum_{n=0}^\infty\binom{\frac{1}{2}}{n+1}(-4x)^{n+1}\\
&=-\frac{1}{2x}\sum_{n=1}^\infty\binom{\frac{1}{2}}{n}(-4x)^{n}\\
&=-\frac{1}{2x}\left((1-4x)^{\frac{1}{2}}-1\right)\\
&=\frac{1}{2x}\left(1-\sqrt{1-4x}\right)
\end{align*}
| {
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Proof that ${2x\over 2x-1}={A(x)\over A(x-1)}$ for every integer $x\ge2$, where $A(x)=\sum\limits_{n=0}^\infty\prod\limits_{k=0}^n\frac{x-k}{x+k}$
Let $x\ge2$ denote an integer. Consider:
$$A=1+{x-1\over x+1}+{(x-1)(x-2)\over (x+1)(x+2)}+{(x-1)(x-2)(x-3)\over (x+1)(x+2)(x+3)}+\cdots\tag1$$
and
$$B=1+{x-2\over x}+{(x-2)(x-3)\over x(x+1)}+{(x-2)(x-3)(x-4)\over x(x+1)(x+2)}+\cdots\tag2$$
How does one show that $${2x\over 2x-1}={A\over B}\ ?$$
An attempt: consider $(x)^n=x(x+1)\cdots(x+n-1)$ and $(x)_n=x(x-1)\cdots(x-(n-1))$. One can re-write $(1)$ as
$$A=x+{(x)_2\over (x)^2}+{(x)_3\over (x)^3}+{(x)_4\over (x)^4}+\cdots\tag3$$
and $(2)$ as
$$x+x(x-1)B=x+(x)_2+{(x)_3\over x}+{(x)_4\over (x)^2}+{(x)_5\over (x)^3}+\cdots\tag4$$
but I am not sure how to continue.
| Frist
$$
\prod^{n}_{k=0}\frac{x-k}{x+k}=\prod^{n}_{k=1}\frac{x-k}{x+k}=(-1)^n\frac{\prod^{n}_{k=1}(-x+1+k)}{\prod^{n}_{k=1}(x+1+k)}=(-1)^n\frac{(-x+1)_n}{(x+1)_n},
$$
where $(a)_n:=a(a+1)(a+2)\ldots(a+n-1)$. Hence
$$
A(x)=\sum^{\infty}_{n=0}(-1)^n\frac{(1-x)_n}{(x+1)_n}=\sum^{\infty}_{n=0}\frac{(1-x)_n(1)_n}{(x+1)_n}\frac{(-1)^n}{n!},
$$
since $(1)_n=n!$, for $n=0,1,2,\ldots$. Hence
$$
A(x)={}_2F_{1}\left(1-x,1;x+1;-1\right) \tag 1
$$
But from [Leb] Chapter 9, pg. 240 relation (9.1.6) it is known that
$$
{}_2F_1\left(a,b;c;z\right)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int^{1}_{0}t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a}dt,
$$
where $Re(c)>Re(b)$, $|\textrm{arg}(1-z)|<\pi$.
Hence whenever $x>0$:
$$
A(x)=\frac{\Gamma(x+1)}{\Gamma(x)}\int^{1}_{0}(1-t)^{x-1}(1+t)^{x-1}dt=x\int^{1}_{0}(1-t^2)^{x-1}dt=
$$
$$
=\frac{x}{2}\int^{1}_{0}(1-w)^{x-1}w^{-1/2}dw
=\frac{x}{2}B\left(x,\frac{1}{2}\right)=\frac{x}{2}\frac{\Gamma(x)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(x+\frac{1}{2}\right)}=\frac{1}{2}\frac{\Gamma(x+1)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(x+\frac{1}{2}\right)}\tag 2
$$
Where we have make the change of variable $x\rightarrow\sqrt{w}$ and use the identity (see [Leb] Chapter 1, pg.13-14)
$$
\int^{1}_{0}t^{a_1-1}(1-t)^{b_1-1}dt=\frac{\Gamma(a_1)\Gamma(b_1)}{\Gamma(a_1+b_1)},
$$
where $Re(a_1)>0$, $Re(b_1)>0$. Hence
$$
\frac{A(x)}{A(x-1)}=\frac{\Gamma(x+1)\Gamma\left(x-\frac{1}{2}\right)}{\Gamma(x)\Gamma\left(x+\frac{1}{2}\right)}=\frac{x\Gamma\left(x-\frac{1}{2}\right)}{\left(x-\frac{1}{2}\right)\Gamma\left(x-\frac{1}{2}\right)}=\frac{2x}{2x-1}\tag 3
$$
where we have used the identity $\Gamma(x+1)=x\Gamma(x)$, $x>0$.
Hence we have proved that
$$
\frac{A(x)}{A(x-1)}=\frac{2x}{2x-1},
$$
$x$ real greater than 1.
References
[Leb] N.N. Lebedev "Special Functions and their Applications". Dover Publications, Inc. New York (1972)
| {
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How to solve an equation of the form $f(x)=f(a)$ for a fixed real a. I got stuck on this question: find all solutions $x$ for $a\in R$:
$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{(a^2-a+1)^3}{a^2(a-1)^2}$$
I see that if we simplify we get:
$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{[(x-{\frac 12})^2+{\frac 34}]^3}{[(x-{\frac 12})^2-{\frac 14}]^2}$$
From the expression $(x-{\frac 12})^2$, I see that if $x=x_1$ is a solution, then $x=1-x_1$ is also a solution. But in the solution to this exercise, it was stated that $x=\frac{1}{x_1}$ must also be a solution, and I don't see how.
[EDIT]
Ok, thx for the help guys. What do you think of this solution (doesn't involve any above precalculus math, and needs no long calculations)?
From the above we know that if $x_1=a$ is a solution, then $x_2=1-a$ is also a solution.
Also, from here:
$$\require{cancel}\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{\cancel{x^3}(x+{\frac 1x}-1)^3}{\cancel{x^3}(x+{\frac 1x}-2)}$$
in the expression $x+{\frac 1x}$ we see that if $x=x_1$ is a solution, then $x=\frac{1}{x_1}$ is also a solution, so $x_3=\frac{1}{a}$.
With these two rules we can now keep generating roots until we have 6 total.
If $x=x_2$ is a solution, then $x=\frac{1}{x_2}$ is also a solution, so $x_4=\frac{1}{1-a}$.
If $x=x_3$ is a solution, then $x=1-x_3$ is also a solution, so $x_5=\frac{a-1}{a}$.
Finally, if $x=x_5$ is a solution, then $x=\frac{1}{x_5}$ is also a solution, so $x_6=\frac{a}{a-1}$
The 6 obtained values are distinct, so they cover all the roots.
[EDIT2]
I guess this is answered. No sure whose particular answer to actually select as the right one since they're all correct, so I'll just leave it like this.
| factorizing the given equation and cancel the denominators we get
$$(ax-a-x)(ax-x+1)(x-1+a)(-x+a)(ax-1)(ax-a+1)=0$$
| {
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Solving the differential equation $y'= 2 xy + 4 x$ using power series. This differential equation can be solved using a Power Series Method:
$$f'(x) = 2 x f(x) + 4 x$$
I found $f'(x)$ and substituted it back into the equation but i do not understand where to go from there. How would i be able to find the power series which satisfies this differential equation?
| Assume
\begin{align}
f(x) = \sum^\infty_{n=0}a_n x^n
\end{align}
then substitute the expression back to the differential equation yields
\begin{align}
\sum^\infty_{n=1}a_n nx^{n-1}- 2\sum^\infty_{n=0} a_nx^{n+1}-4x=0.
\end{align}
Rearranging terms will lead to
\begin{align}
&\sum^\infty_{n=0}(n+1)a_{n+1} x^n -2\sum^\infty_{n=1} a_{n-1}x^n-4x\\
&= a_1+(2a_2-2a_0-4)x+\sum^\infty_{n=2}\left\{(n+1)a_{n+1}-2a_{n-1} \right\}x^n =0.
\end{align}
Hence we have that
\begin{align}
a_1 =0, \ \ a_2=a_0+2, \ \ a_{n+1} = \frac{2a_{n-1}}{n+1}.
\end{align}
In particular, if $n=2k$ then we have that
\begin{align}
a_{2k+1} = \frac{2}{2k+1}a_{2k-1} = \frac{2^2}{(2k+1)(2(k-1)+1)}a_{2k-3} = \frac{2^{k}}{(2k+1)\ldots (2+1)} a_1 =0
\end{align}
for all $k\geq 1$. If $n=2k+1$, then we have
\begin{align}
a_{2(k+1)}= \frac{1}{k+1}a_{2k}=\frac{1}{(k+1)k}a_{2k-2} = \frac{1}{(k+1)!}a_2
\end{align}
Thus, we see that
\begin{align}
f(x) = a_0+(a_0+2)x^2+(a_0+2)\sum^\infty_{k=2}\frac{x^{2k}}{k!} = -2+(a_0+2)\sum^\infty_{k=0}\frac{x^{2k}}{k!} =-2+(a_0+2)e^{x^2}
\end{align}
is the solution.
| {
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Solve for $x$ in $\frac{x^2+12x+36}{x^2+4x-12}=2$ $\frac{x^2+12x+36}{x^2+4x-12}=2$
I factored top and bottom to $\frac{(x+6)(x+6)}{(x+6)(x-2)}=2$
and eliminated the common factor (x+6) for $\frac{x+6}{x-2}=2$
then $x+6=2(x-2)$ and $x+6=2x-4$ and $6+4=2x-x$ and finally $x = 10$
Plugging this back into the original fraction proves true
My instructor however in her exam prep test says
$\frac{x^2+12x+36}{x^2+4x-12}=2$
$$x^2+12x+36 = 2(x^2+4x-12)$$
$$x^2+12x+36 = 2x^2+8x-24$$
$$0 =2x^2-x^2 +8x -12x -24 -36$$
$$0 =x^2-4x -60$$
which she then factors to $(x-10)(x+6)$ saying the answer to the problem is $x=10$ or $x=-6$
But if you plug -6 back into the original fraction you get $0=2$. Where is the problem here? Thanks
| You're right, for (almost) the very reason that you say: $-6$ is not a solution because the left-hand side is not defined when $x=-6$.
| {
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Prove that $3(a^5b+b^5c+c^5a)\geq(a^2c+b^2a+c^2b)^2$
Let $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ be sides-lengths of a triangle. Prove that:
$$3(a^5b+b^5c+c^5a)\geq(a^2c+b^2a+c^2b)^2$$
I tried to apply the way like the proof of the following inequality, but without success.
Let $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ be sides-lengths of a triangle. Prove that:
$$2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq3+\frac{a}{c}+\frac{b}{a}+\frac{c}{b}.$$
Proof.
We need to prove that
$$\sum_{cyc}(2a^2c-a^2b-abc)\geq0$$ or
$$\sum_{cyc}(a^3-abc)-\sum_{cyc}(b^3-2ab^2+a^2b)\geq0$$ or
$$\frac{a+b+c}{2}\sum_{cyc}(a-b)^2-\sum_{cyc}b(a-b)^2\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+c-b)\geq0.$$
Lemma.
Let $x+y+z\geq0$ and $xy+xz+yz\geq0$. Prove that:
$$(b-c)^2x+(a-c)^2y+(b-c)^2z\geq0.$$
Proof of the lemma.
Let $x+y\geq0$.
If $x+y=0$ then $xy+xz+yz=-x^2\geq0$,
which gives $x=y=0$ and $z\geq0$, which says that
$(b-c)^2x+(a-c)^2y+(a-b)^2z\geq0$ is true.
Thus, we can assume that $x+y>0$ and we need to prove that
$$(b-c)^2x+(a-b+b-c)^2y+(a-b)^2z\geq0$$ or
$$(x+y)(b-c)^2+2y(a-b)(b-c)+(y+z)(a-b)^2\geq0,$$
for which it's enough to prove that
$$y^2-(x+y)(y+z)\leq0$$ or
$$xy+xz+yz\geq0,$$
which ends a proof of the lemma.
Now, $\sum\limits_{cyc}(a+c-b)=a+b+c>0$ and
$$\sum_{cyc}(a+b-c)(a+c-b)=\sum_{cyc}(2ab-a^2)=$$
$$=(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{b}+\sqrt{c}-\sqrt{a})>0,$$
which ends the proof by the lemma.
Thank you!
| Remark: Since no complete solution is available, I give the solution by the Buffalo Way. Hope to see nice solutions.
Let $a = (u+v)^2, b = (v+w)^2, c = (w+u)^2$ for $u, v, w > 0$ (Ravi Substitution). It suffices to prove that
$g(u,v,w) \ge 0$ where $g(u,v,w)$ is a homogeneous polynomial of degree $12$.
WLOG, assume that $w = \min(u, v, w)$. Let $v = w + s$ and $u = w + t$ for $s, t\ge 0$.
It suffices to prove that
$$q_{10}w^{10} + q_9w^9 + q_8w^8 + q_7w^7 + q_6w^6 + q_5w^5 + q_4w^4 + q_3w^3 + q_2w^2 + q_1w + q_0 \ge 0.$$
It is not hard to prove that $q_{10}, q_9, \cdots, q_0\ge 0$.
I do not give the expressions of $q_i, \forall i$. If someone uses the Computer Algebra Systems (CAS), it is an easy job.
| {
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Value of n for which f(n) = $\,n^2 + 9n + 30\,$ is a perfect square. I attempted this by setting $f(n) = \,m^2.\,$
So $\,n^2 + 9n + 30 = m ^2\,$.
Then $\,9(n + 10/3) = (m + n)(m - n)\,$.
So $m = 10/3$ and $n = -17/3$ which is incorrect.
| If $n>0$ then
By comparison we have
$$ (n+4)^2 <n^2+9n+30 < (n+6)^2$$
Therefore
$$n^2+9n+30=(n+5)^2$$
Thus
$$n=5$$
If $n\leq0$, similarly
$$(n+5)^2<n^2+9n+30< (n+4)^2$$ for all $n<-14$
Now we only need to consider the case $-14\leq n \leq 0$
Manually, it leads to $n=-14$, $n=-7$ and $n=-2$
| {
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How many positive integer solutions does the equation $a+b+c=100$ have if we require $aHow many positive integer solutions does the equation $a+b+c=100$ have if we require $a<b<c$?
I know how to solve the problem if it was just $a+b+c=100$ but the fact it has the restriction $a<b<c$ is throwing me off.
How would I solve this?
| Since $a$ is the smallest, the largest number it can be is $32$, so $a$ ranges from $1$ to $32$. The remaining sum $b+c$ must be equal to $100 - a$ and since $b$ is smaller than $c$ then $b$ ranges from $a+1$ to $\lfloor \frac{100-a}{2}\rfloor $. So for a given $a$ there are $\lfloor \frac{100-a}{2}\rfloor - a -1$ different $b$s one can choose. Therefore the possible permutations are
$$\sum_{a=1}^{32} \lfloor \frac{100-a}{2}\rfloor - a -1$$
$$=\sum_{a=1}^{32} \lfloor \frac{100-a}{2}\rfloor - \sum_{a=1}^{32}a -32 $$
$$ = 2\sum_{a=34}^{49}a - \sum_{a=1}^{32}a -32$$
$$=49(50) - 34(33) - \frac{32(33)}{2} -32= 768$$
EDIT:
Following the comments of Vik, it is quite clear that since we are including the case $b = a +1$ the expression should change to
$$\sum_{a=1}^{32} \lfloor \frac{100-a}{2}\rfloor - a $$
However as Vik also mentioned this is an over estimate as it would also give us the combination where we have $b = c= (100-a)/2$ the case where $a$ is even. Since $b$ is strictly less than $c$, we need to remove $16$ off of our expression as there are 16 cases of when $a$ is even. Therefore, we get $$\sum_{a=1}^{32}( \lfloor \frac{100-a}{2}\rfloor - a) - 16= 784$$
| {
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The roots of quadratic equation $3x^2 - 5x - k = 0$ are $m/3$ and $m + 3$. Find the values of $m$ and $k$ Can someone please help me solve this equation. Please put in your working as well so I can see how to do this because this equation has been driving me crazy for days!
| Quadratic equation factorization using values of roots:
$ax^2+bx+c=a(x-x_1)(x-x_2)$
Where $x_1$ and $x_2$ are the roots.
So we get:
$3x^2-5x-k=3(x-\frac{m}{3})(x-m-3)$
$3x^2-5x-k=(3x-m)(x-m-3)$
$3x^2-5x-k=3x^2-3xm-9x-mx+m^2+3m$
$3x^2-5x-k=3x^2-4xm-9x+m^2+3m$
$-5x-k=-4xm-9x+m^2+3m$
$5x+k=4xm+9x-m^2-3m$
$5x+k=(4m+9)x+(-m^2-3m)$
Now by comparing terms, we get:
$\begin{cases}
4m+9=5
\\
-m^2-3m=k
\end{cases}$
$1)\quad 4m+9=5 \\
\hspace{9mm} 4m=-4 \\
\hspace{9mm} m=-1$
$2)\quad -m^2-3m=k \\
\hspace{9mm} -(-1)^2-3\times(-1)=k \\
\hspace{9mm} -1+3=k \\
\hspace{9mm} k=2$
So the answer is:
$\begin{cases}
m=-1 \\
k=2
\end{cases}$
| {
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How can I prove/disprove that if n is an integer which is not divisible by 3, then $n^3+n^2+2n+1 ≡ 2 \pmod 3$? By doing a few examples, it seems as if the statement is true. However, I'm having trouble proving (or disproving) it. I know that the numerical value of some $N$ can be written as $n_{0}10^{0}+n_{1}10^{1}+n_{2}10^{2}+\dotsb$ and $10\equiv 1 \pmod{3}$, but this just shows the divisibility rule for $3$. Can I use this somehow?
| You have to check every case:
$1)$ If $n\equiv0 \pmod 3 \to n^3+n^2+2n+1\equiv 1\pmod 3$
$2)$ If $n\equiv 1 \pmod 3 \to n^3+n^2+2n+1\equiv 1+1+2+1\equiv 2\pmod 3$
$3)$ If $n\equiv -1 \pmod 3 \to n^3+n^2+2n+1\equiv -1+1-2+1 \equiv 2\pmod 3$
So if you want $n^3+n^2+2n+1\equiv 2\pmod 3$ you must have $n\equiv \pm 1 \pmod 3$
| {
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Solving the recurrence relation with repeated roots (2 in $a_g$ and same root in $a_p$) $$a_{n+2} - 6a_{n+1} + 9a_n = 3^n$$
Solving for this recurrence can be done by the following:
1) solve for the recurrence relation $a_g$ such that: $a_{n+2} - 6a_{n+1} + 9a_n = 0$
then,
2) $a_h$ such that: $a_{n+2} - 6a_{n+1} + 9a_n = 3^n$
then,
3) $a_n = a_g + a_p$
Solving for the $a_g$ I have the solution: $C_1(3)^n + C_2n(3)^n$, where the extra $n$ is due to the repeated root.
Now, solving for $a_p$ I use the trial solution $a_n = A(3^n)\cdot(n^2)$, where $n^2$ is due to the third root.
This is the part I'm having trouble with now: when i substitute for $a_{n+2}$ is it $A\cdot3^{n+2}\cdot(n+2)^2$ and then similarly I would substitute for $6a_{n+1}$ which should be $6A\cdot3^{n+1}\cdot(n+1)^2$ and $9a_n$ should be $9A\cdot 3^n \cdot n^2$
Then, I need to solve:
$$(A\cdot3^{n+2}\cdot(n+2)^2) - (6A\cdot 3^{n+1} \cdot (n+1)^2) + (9A\cdot 3^n \cdot n^2) = 3^n$$
Would this be the correct approach to solve for $a_p$?
| Writing, as you did,
$$(A\cdot3^{n+2}\cdot(n+2)^2) - (6A\cdot 3^{n+1} \cdot (n+1)^2) + (9A\cdot 3^n \cdot n^2) = 3^n$$ Expanding the lhs, you should end with $$2 A\, 3^{n+2}=3^n\implies 18 A=1\implies A=\frac 1 {18}$$ But, as said in comments, there is a faster way to solve the problem.
| {
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Double binomial series simplification Could someone please check that this double binomial series (with a delta function) simplification makes sense:
$$\sum_{m_J = -N/2}^{N/2} \sum_{m_K = -N/2}^{N/2} \dbinom{N}{N/2 +m_J}^{1/2}\dbinom{N}{N/2 + m_K}^{1/2} \delta_{m_K, m_J +1} = \bigg[\sum_{m_J = - N/2}^{N/2}\dbinom{N}{N/2 +m_J}^{1/2}\bigg]\bigg[ \sum_{m_J = - N/2-1}^{N/2} \dbinom{N}{N/2 + m_J + 1}^{1/2} \bigg] = \bigg[\sum_{m_J = - N/2}^{N/2}\dbinom{N}{N/2 +m_J}^{1/2}\bigg]^{2}$$
Please advise on the correct simplification if there are any inconsistencies.
Thanks for the assistance.
| We consider the case $N=2M$
We obtain
\begin{align*}
\sum_{m_J=-M}^M&\sum_{m_K=-M}^M\binom{2M}{M+m_J}^\frac{1}{2}\binom{2M}{M+m_K}^\frac{1}{2}\delta_{m_K,m_J+1}\\
&=\sum_{m_J=-M}^M\sum_{{-M\leq m_K\leq M}\atop{m_K=m_J+1}}\binom{2M}{M+m_J}^\frac{1}{2}\binom{2M}{M+m_K}^\frac{1}{2}\\
&=\sum_{m_J=-M}^{M-1}\binom{2M}{M+m_J}^\frac{1}{2}\binom{2M}{M+m_J+1}^\frac{1}{2}
\end{align*}
The inner sum reduces to one summand with $m_K=m_J+1$. Since $\binom{2M}{2M+1}=0$ the upper limit of the outer sum can be set to $M-1$.
The calculation $N=2M+1$ follows similarly.
[Add-On 2017-04-10]: We consider according to OPs comment
\begin{align*}
\frac{1}{2^N}\sum_{m_J=-N/2}^{N/2}&\sum_{m_K=-N/2}^{N/2}\binom{N}{N/2+m_J}^\frac{1}{2}\binom{N}{N/2+m_K}^\frac{1}{2}\delta_{m_K,m_J}\\
&=\frac{1}{2^N}\sum_{m_J=-N/2}^{N/2}\binom{N}{N/2+m_J}^\frac{1}{2}\sum_{{-N/2\leq m_K\leq N/2}\atop{m_K=m_J}}\binom{N}{N/2+m_K}^\frac{1}{2}\\
&=\frac{1}{2^N}\sum_{m_J=-N/2}^{N/2}\binom{N}{N/2+m_J}^\frac{1}{2}\binom{N}{N/2+m_J}^\frac{1}{2}\\
&=\frac{1}{2^N}\sum_{m_J=-N/2}^{N/2}\binom{N}{N/2+m_J}\\
&=\frac{1}{2^N}\sum_{m_J=0}^{N}\binom{N}{m_J}\\
&=1
\end{align*}
| {
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Solution of differential equation which is quadratic in $\frac{dy}{dx}$ Consider the following differential equation
$$ x\frac{dy}{dx} + y = x^4(\frac{dy}{dx})^2$$
I used the quadratic formula and got $\frac{dy}{dx} = \frac{x \pm \sqrt {x^2 + 4yx^4}}{2x^4}$. Now how to proceed or is there any other method?
Edit:
I proceeded further this way
$\frac{dy}{dx} = \frac{1 \pm \sqrt {1 +4yx^2}}{2x^3}$
Let $1 + 4yx^4 = t^2$
then $8yx + 4x^2\frac{dy}{dx} = 2t\frac{dt}{dx}$
multiplying throughout with $x$ I got $4yx^3 +2x^3\frac{dy}{dx} = tx\frac{dt}{dx}$
now putting the value of $2x^3\frac{dy}{dx}$ and $t^2$ in 1st equation I got
$tx\frac{dt}{dx} - t^2 = \pm t$
which gave 2 solutions $$t = 0$$ $$and$$ $$x\frac{dt}{dx} = t \pm 1$$
Using $t = 0$ I got one solution $$1 + 4x^2y = 0$$ and the other by solving the differential equation $x\frac{dt}{dx} = t \pm 1$ as $$c^2x^2 + 4y^2x^2 = 4cyx^2 + 2c$$
Now the answer to this question which I have got is $$4x^2y + 1 = 0$$ $$and$$ $$xy - c^2x + c = 0$$
Why I couldn't get the second solution right?
| The equation $\quad x\frac{dt}{dx}=t\pm 1\quad$ is correct.
The solutions of this linear ODE are :
$$t\pm 1=Cx$$
From $t^2=1+4yx^2 \quad\to\quad (Cx\pm 1)^2=1+4yx^2$
$$4yx^2=(Cx\pm 1)^2-1=C^2x^2\pm 2Cx$$
$$xy-\frac{C^2}{4}x\pm \frac{C}{2}=0$$
Let $c=\frac{C}{2}$
$$xy-c^2x\pm c=0$$
The solution is :
$$xy-c^2x+ c=0\quad\to\quad y=c^2-\frac{c}{x} $$
and $\quad xy-c^2x- c=0\quad\to\quad y=c^2+\frac{c}{x} \quad$ which is the same as the preceding one (just change $c$ into $-c$ ) since $c$ is any constant.
Of course, without forgetting the particular solution previously obtained for $t=0$
$$4yx^2+1=0\quad\to\quad y=-\frac{1}{4x^2}$$
NOTE :
$y=-\frac{1}{4x^2}$ is the low envelop (drawn in red) of the family of curves $y=c^2-\frac{c}{x}$ in blue ($c<0$) and in black ($c\geq 0$).
| {
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Finding the eigenvalues and a basis for the eigenspaces of a $3\times3$ matrix. For the matrix $A \in M_{3\times3}(\mathbb{R})$ below, I need to find the eigenvalues and a basis for the corresponding eigenspaces:
$$\begin{bmatrix}\
1 & -3 & 3 \\
3 & -5 & 3 \\
6 & -6 & 4 \\
\end{bmatrix}$$
I have tried to use the formula $\det(I\lambda - A) = 0$ but I ended up with equation $\lambda^3 - 12\lambda - 16 = 0$, of which I can't seem to find the solutions to.
| You can use the rationnal root theorem to find a root of your polynomial : $x$ is a rationnal root of $\lambda^3 - 12\lambda - 16\Rightarrow x$ is a divisor of $16$. You check if there is a root of $\lambda^3 - 12\lambda - 16$ in $\{\pm1,\pm2,\pm4,\pm8,\pm16\}$, and you find that $4$ is indeed a root. Thus $\lambda^3 - 12\lambda - 16=(\lambda-4)(\lambda^2+4\lambda+4)$.
Can you finish ?
The eigenvalues are $-2$ and $4$
Now will find the eigen sace associated to $4$ for example. $X=\begin{pmatrix} x \\ y\\z \end{pmatrix}$ is in this eigenspace iff :
$$AX=4X\iff \begin{pmatrix}\
1 & -3 & 3 \\
3 & -5 & 3 \\
6 & -6 & 4 \\
\end{pmatrix}\begin{pmatrix} x \\ y\\z \end{pmatrix}=4\begin{pmatrix} x \\ y\\z \end{pmatrix}\\\iff\begin{cases}x-3y+3z=4x\\3x-5y+3z=4y\\6x-6y+4z=4z\end{cases}\\\iff\begin{cases}-3x-3y+3z=0\\3x-9y+3z=0\\6x-6y=0\end{cases}\\\iff\begin{cases}x=y\\z=2x\end{cases}$$
So the eigenspace associated to $4$ is $\text{Vect}\left(\begin{pmatrix} 1 \\ 1\\2 \end{pmatrix}\right)$.
You can do the same for the eigenspace associated to $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\sum_{k=0}^{n} {3n \choose 3k}$ Evaluate $\sum_{k=0}^{n} {3n \choose 3k}$.
It's not hard to evaluate by putting roots of unity.
Therefore, I would like to see some solutions using elementary mathematical stuffs but not roots of unity.
And the answer is $\frac{1}{3}(2^{3n}+2(-1)^n)$
| If you let $f(n)$ be the sum in question, then, knowing the answer, we see that $$f(n+1)=7f(n)+8f(n-1).$$
If you can prove this recursion directly from the definition of $f(n)$ then you are done.
Now, by Vandermond's identity, we have: $$\begin{align}\binom{3n+3}{3k}=&\binom{3n-3}{3k-6}+6\binom{3n-3}{3k-5}+15\binom{3n-3}{3k-4}+20\binom{3n-3}{3k-3}\\
&+15\binom{3n-3}{3k-2}+6\binom{3n-3}{3k-1}+\binom{3n-3}{3k}
\end{align}\tag{1}$$
And:
$$\binom{3n}{3k}=\binom{3n-3}{3k-3}+3\binom{3n-3}{3k-2}+3\binom{3n-3}{3k-1}+\binom{3n-3}{3k}$$
This means that:
$$f(n)=\sum_{k=0}^{n-1} 2\binom{3n-3}{3k} + 3\sum_{k=1}^{n-1}\left(\binom{3n-3}{3k-2}+\binom{3n-3}{3k-1}\right)$$
So $$7f(n)+8f(n-1)=\sum_{k=0}^{n-1}22\binom{3n-3}{3k} + 21\sum_{k=1}^{n-1}\left(\binom{3n-3}{3k-2}+\binom{3n-3}{3k-1}\right)$$
Combine the terms of $(1)$ to get the same expression.
To complete the induction, show that the expression is true for $n=0,1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How can I simplify $\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$? This problem got me thinking hard:
$$\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$$
Considering the fact that the common base is '$3$', I decided to do this:
$$\log_3\left(\frac{(n+1)(n-1)}{(n+2)(n)}\right)$$
$$\log_3\left(\frac{n^2-n+n+1}{n^2+2n}\right)$$
$$\log_3\left(\frac{n^2 + 1}{n^2 + 2n}\right)$$
$$\log_3\left(\frac{1}{2n}\right)$$
I am not 100% sure if have simplified the problem. I need some clarification.
| Why are taking logarithm? No need of that. It's just simple Indices.
$\dfrac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}
=\dfrac{3^{n-1}\left(3^2+1\right)}{3^n\left(3^2+1\right)}
=\dfrac{3^{n-1}}{3^n}
=\dfrac{1}{3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove $8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c$
Let $a$, $b$ and $c$ be angles of an acute triangle. Prove that:
$$8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c$$
I tried am-gm, rm-gm rm-am and a couple of other inequalities but I didn't get anywhere.
| Here's a proof that accounts for all acute-angled triangles whose angles lie in $ \left(\tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$.
Note that $f(x) = \ln \sin(x) \tan(x)$ is convex over $ \left( \tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$ since $f''(x) = \dfrac{\tan^{2}(x) -2}{\sin^{2}(x)} > 0 \quad \forall x\in \left(\tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$
Hence we can apply Jensen's inequality to $f(x)$ and $a, b,c \in \left(\tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$ to get:
$f\left(\dfrac{a+b+c}{3}\right)\leq \dfrac{f(a) + f(b)+ f(c)}{3}$
$\Rightarrow \ln \dfrac{3}{2} \leq \dfrac{\ln \dfrac{\sin^{2}(a)\,\sin^{2}(b)\,\sin^{2}(c)}{\cos(a)\,\cos(b)\,\cos(c)}}{3}$
$\Rightarrow \dfrac{27}{8} \leq \dfrac{\sin^{2}(a)\,\sin^{2}(b)\,\sin^{2}(c)}{\cos(a)\,\cos(b)\,\cos(c)}$, as required.
Can someone figure out an extension of this argument that accounts for those acute-angled triangles with one angle in $ \left(0, \tan^{-1}{\sqrt{2}}\right)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $p\mid (3^n+1)$, then $p\equiv 1\pmod{3}$
Show that if $p> 2$ is a prime, $n > 1$ is odd and $p\mid (3^n+1)$,
then $p\equiv 1\pmod{3}$.
Since $n$ is odd, we have $3^{n+1} \equiv -3 \pmod{p}$ is a quadratic residue. Then I thought about using Quadratic Reciprocity but didn't see how to apply it.
We have $x^2 \equiv -3 \pmod{p}$ for some integer $x$. By Fermat's Little Theorem we have $$x^{p-1} \equiv x^2 \cdot x^{p-3} \equiv -3 \cdot x^{p-3} \equiv 1 \pmod{p}.$$ Thus, $x^{p-3} \equiv -3^{-1} \pmod{p}$.
I didn't see how to get a contradiction if $p \equiv 2 \pmod{3}$.
| This may qualify as cheating, but I will use a few known facts from Wikipedia. First of all, indeed $3^{n+1} \equiv -3 \pmod{p}$, where $n+1$ is even, thus
$$\left(\frac{-3}{p}\right)=1 \Rightarrow \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=1 \tag{1}$$
But
$$\left(\frac{-1}{p}\right)=\left\{\begin{matrix}
1, & p \equiv 1 \pmod{4} \\
-1,& p \equiv 3 \pmod{4}
\end{matrix}\right. \color{red}{\text{ and }} \left(\frac{3}{p}\right)=\left\{\begin{matrix}
1, & p \equiv 1 \text{ or } 11 \pmod{12} \\
-1, & p \equiv 5 \text{ or } 7 \pmod{12}
\end{matrix}\right.$$
Due to $(1)$, we have
$$\left\{\begin{matrix}
\left(\frac{-1}{p}\right)=1\\
\left(\frac{3}{p}\right)=1
\end{matrix}\right. \color{red}{\text{ or }} \left\{\begin{matrix}
\left(\frac{-1}{p}\right)=-1\\
\left(\frac{3}{p}\right)=-1
\end{matrix}\right. $$
Or, for the first case
$$\left\{\begin{matrix}
\left(\frac{-1}{p}\right)=1\\
\left(\frac{3}{p}\right)=1
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
p \equiv 1 \pmod{4}\\
p \equiv 1 \text{ or } 11 \pmod{12}
\end{matrix}\right. \Rightarrow p \equiv 1 \pmod{12}$$
and for the latter
$$\left\{\begin{matrix}
\left(\frac{-1}{p}\right)=-1\\
\left(\frac{3}{p}\right)=-1
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
p \equiv 3 \pmod{4}\\
p \equiv 5 \text{ or } 7 \pmod{12}
\end{matrix}\right. \Rightarrow p \equiv 7 \pmod{12}$$
As a result
$$p \equiv 1 \pmod{12} \color{red}{\text{ or }} p \equiv 7 \pmod{12} \Rightarrow \\ p \equiv 1 \pmod{3} \color{red}{\text{ or }} p \equiv 7 \pmod{3} \Rightarrow \\ p \equiv 1 \pmod{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Evaluation of $\sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$ Find the following sum
$$S= \sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$$
I am not able to make it telescopic series. Could someone help me with this?
| This can be rewritten as
$$\sum_{n=1}^\infty \left(\arctan \frac2n - \arctan \frac2{n+1}\right) = \arctan 2.$$
Of course if we mysteriously find this expression, we can verify it using the formula $\arctan x - \arctan y = \arctan \frac{x-y}{1+xy}$. But how do we find this expression in the first place?
If we guess the form of $x$ and $y$, namely, $\frac1{a n + b}$ and $\frac1{a(n+1)+b}$, then we can figure out when $\frac{x-y}{1+xy} = \frac{2}{n^2+n+4}$. This gives us $$\frac{a}{a^2 n^2+a^2 n+2 a b n+a b+b^2+1} = \frac{2}{n^2+n+4}.$$ To get a $2$ in the numerator at the same time as an $n^2$ in the denominator, we want $a = \frac12$, and then $b=0$ follows by straightforward algebra.
(Guessing the form $\frac1{an+b}$ is not too unreasonable, since we want something that goes to $0$ as $n \to \infty.$ It's not the only possibility, of course.)
Another answer is "find the pattern and check". If you compute the first few partial sums and try to stuff them into one $\arctan$ (by using the corresponding addition formula), then we get
$$\arctan \frac13, \arctan \frac47, \arctan \frac34, \arctan \frac89,\arctan 1, \arctan \frac{12}{11},\arctan \frac{7}{6}, \arctan \frac{16}{13}, \dots,$$ and that's how long it took me to spot the pattern: this is
$$\arctan \frac26,\arctan \frac47,\arctan \frac68,\arctan \frac89,\arctan \frac{10}{10},\arctan \frac{12}{11},\arctan \frac{14}{12},\arctan \frac{16}{13},$$
so the $n^{\text{th}}$ term is $\arctan \frac{2n}{n+5}$. This doesn't let us find the telescoping form, but now we can prove that this is the correct partial sum by induction, and the limit as $n \to \infty$ is $\arctan 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Dalzell-type integrals for convergents to $2\pi$
Should $\pi$ be harder than $\log(2)$?
Simple integrals relate $\pi$ to its first two convergents $3$ and $\dfrac{22}{7}$.
$$
\begin{align}
\int_{0}^{1}\frac{2x(1-x)^2}{1+x^2}dx&=\pi-3\\
\\
\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx&=\frac{22}{7}-\pi (Dalzell)
\end{align}
$$
However, the following $\dfrac{333}{106}$ and $\dfrac{355}{113}$ are more involved. Is there an integral that proves $\pi > 333/106$?
In contrast, analogous integrals that evaluate to the error between $\log(2)$ and its first four convergents are easy to find.
$$
\begin{align}
\int_0^1\frac{2x}{1+x^2}dx
&=
\log\left(2\right) \\
\\
\int_0^1\frac{(1-x)^2}{1+x^2}dx
&=
1-\log\left(2\right) \\
\\
\int_0^1\frac{x^2(1-x)^2}{1+x^2}dx
&=
\log\left(2\right)-\frac{2}{3} \\
\\
\int_0^1\frac{x^4(1-x)^2}{1+x^2}dx
&=\frac{7}{10}-\log\left(2\right) \\
\end{align}
$$
Proof that $\frac{2}{3} < \log(2) < \frac{7}{10}$
Why not $2\pi$?
Comparing the list of convergents to $\pi$
$$3,\dfrac{22}{7},\dfrac{333}{106},\dfrac{355}{113},...$$
to that of $2\pi$
$$6,\dfrac{19}{3},\dfrac{25}{4},\dfrac{44}{7},\dfrac{333}{53},\dfrac{710}{113},...$$
shows that the convergents for $2\pi$ are not simply twice the convergents for $\pi$.
The first four convergents of $\pi$ have their correspondent fractions in the list for $2\pi$, but two more approximations appear: $$\pi \approx \frac{19}{6}$$ $$\pi\approx\frac{25}{8}$$
The list of integrals for the first four convergents of $2\pi$ is simpler than that of $\pi$.
$$
\begin{align}
4\int_0^1 \dfrac{x(1-x)^2}{1+x^2}dx&=2\pi-6\\
\\
4\int_0^1 \dfrac{x^3(1-x)^2}{1+x^2}dx&=\dfrac{19}{3}-2\pi\\
\\
\dfrac{1}{2} \int_0^1 \dfrac{x(1-x)^4(1+4x+x^2)}{1+x^2}dx&=2\pi-\dfrac{25}{4}\\
\\
2\int_0^1 \dfrac{x^4(1-x)^4}{1+x^2}dx&=\dfrac{44}{7}-2\pi\\
\end{align}
$$
The main question is:
Are there integer $m$,$n$ and a rational $q$ for the third convergent of $2\pi$? $$q\int_0^1 \dfrac{x^m(1-x)^n}{1+x^2}dx=2\pi-\frac{25}{4}$$ If not, how to prove it?
and a more speculative one:
What rational multiple of $\pi$ should we focus on?
| It's not a definitive and complete answer at all and it's too lenghty for a comment. It's an empirically based answer.
Let,
$\displaystyle J(m,n)=\int_0^1 \dfrac{x^m(1-x)^n}{1+x^2}dx$
You are searching an integer relation between $J(m,n)$ and $2\pi-\dfrac{25}{4}$
Here is a program for PARI-GP to search empirically for $m,n$:
$J(m,n)={intnum(x=0,1,x^m*(1-x)^n/(1+x^2))};$
$scan(r,p)=\{$
$for(m=1,r,for(n=1,r,L=lindep([J(m,n),2*Pi-25/4]);if(abs(L[1])<p\&\&abs(L[2])<p,print(m,\text{" "},n,\text{" "},L[1],\text{" "},L[2]);return)))\}$
r=max(m,n).
p=precision for integers coefficients.
To use:
start to define the precision
\p 100
(precision to compute integrals=100)
scan(50,1000)
(i have tested this, no m,n found)
If m,n are found,
$L[1]J(m,n)+L[2]\times 2\pi=0$
$L[1],L[2]$ are integers.
If you replace 25/4 by 19/3 or by 44/7 this program finds an integer relation.
Bon courage dans tes recherches !
ADDENDUM:
To search for integer relation between $J(m,n),\pi,1$
i mean searching for integers $a,b,c$ and integers $m,n$ such that:
$aJ(m,n)+b\pi+c=0$
here is a variant of the program above:
$scanb(r,p)=\{$
$for(m=1,r,for(n=1,r,L=lindep([J(m,n),Pi,1]);if(abs(L[1])<p\&\&abs(L[2])<p\&\&abs(L[3])<p,print(m," ",n," ",L[1]," ",L[2]," ",L[3]))))
\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
MLE of uniform distribution on the interval $[\theta − \frac{1}{2}, \theta + \frac{1}{2}]$ $X_1, X_2, . . . , X_n$ are i.i.d. observations from a uniform distribution on
the interval $[\theta − \frac{1}{2}, \theta + \frac{1}{2}]$.
Show that any $\theta$ between $X_{max} − \frac{1}{2}$ and $X_{min} + \frac{1}{2}$ maximizes the likelihood, and therefore, can be taken as the MLE.
I am confused...because $f(x|\theta) = \frac{1}{(\theta+\frac{1}{2}) -(\theta -\frac{1}{2})} = 1$ for $\theta-\frac{1}{2} \le x \le\theta+\frac{1}{2}$ and 0 otherwise... Isn't the likelihood function $L(x|\theta) = 1$?
The how can it be maximized between $X_{max} − \frac{1}{2}$ and $X_{min} + \frac{1}{2}$ ?
| The likelihood function is not $L(x|\theta)$. It is $L(X_1,\ldots,X_n|\theta)$ and is considered as a function of the variable $\theta$.
Firstly rewrite pdf and express $\theta$ from the inequality $\theta-\frac{1}{2} \le x \le\theta+\frac{1}{2}$ with respect to $x$:
$$
f(x|\theta) = \begin{cases}1, & \theta-\frac{1}{2} \le x \le\theta+\frac{1}{2}\\ 0 & \text{ else}\end{cases} = \begin{cases}1, & x-\frac{1}{2} \le \theta \le x+\frac{1}{2}\\ 0 & \text{ else}\end{cases}
$$
Next find the likelihood function:
$$
L(X_1,\ldots,X_n|\theta)=f(X_1|\theta)\cdot\ldots\cdot f(X_n|\theta).$$
This product equals to $1$ iff
$$X_i-\frac{1}{2} \le \theta \le X_i+\frac{1}{2}\quad \text{ for all } i=1,2,\ldots,n$$
or, equivalently, if
$$X_{max}-\frac{1}{2} \le \theta \le X_{min}+\frac{1}{2}$$
Then
$$
L(X_1,\ldots,X_n|\theta)=\begin{cases}1, & X_{max}-\frac{1}{2} \le \theta \le X_{min}+\frac{1}{2}\\ 0 & \text{ else}\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Show this $|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\frac85$ Let $x\in R$ show that
$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{8}{5}$$
since
$$f(x)=f(x+\pi),$$it sufficient to show $x\in (0,\pi]$
| We'll prove that
$$\sin1>\frac{4}{5}$$ by hand.
Indeed, let $f(x)=\sin{x}-x+\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040},$ where $x\in\left[0,\frac{\pi}{2}\right].$
Thus, $$f'(x)=\cos{x}-1+\frac{x^2}{2}-\frac{x^4}{24}+\frac{x^6}{720};$$
$$f''(x)=-\sin{x}+x-\frac{x^3}{6}+\frac{x^5}{120};$$
$$f'''(x)=-\cos{x}+1-\frac{x^2}{2}+\frac{x^4}{24};$$
$$f''''(x)=\sin{x}-x+\frac{x^3}{6}$$ and
$$f'''''(x)=\cos{x}-1+\frac{x^2}{2}=\frac{x^2}{2}-2\sin^2\frac{x}{2}=2\left(\frac{x}{2}-\sin\frac{x}{2}\right)\left(\frac{x}{2}+\sin\frac{x}{2}\right)\geq0,$$
which since
$$f''''(0)=f'''(0)=f''(0)=f'(0)=0,$$ says that $f$ increases.
Thus, $$f(1)>0$$ or $$\sin1-1+\frac{1^3}{6}-\frac{1^5}{120}+\frac{1^7}{5040}>0$$ or
$$\sin1>\frac{4241}{5040}$$ and since
$$\frac{4241}{5040}>\frac{4}{5},$$ we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
How to integrate this : $\int \frac{x^2+\cos^2 x}{1+x^2}\csc^2xdx$ How to integrate :
$\int \frac{x^2+\cos^2x}{1+x^2}\csc^2xdx$
I am not getting any clue how to move further, request you to please provide hint, will be of great help.
Thanks..
| Let
$$\begin{align}
f(x) &= \frac{x^2+\cos^2x}{1+x^2} \csc^2x \\
&= \frac{x^2+ 1- \sin^2x}{1+x^2} \cdot \frac{1}{\sin^2x}\\
&= \frac{1}{\sin^2x} - \frac{1}{1+x^2}.
\end{align}$$
These two terms can be integrated easily.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Do the following series converge or diverge? Justify. $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ $$\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$$
Can I do this using limit comparison test?
for $n \in [1, \infty), a_n = \frac{1}{2^n - 1} \geq 0$
for $n \in [1,\infty), b_n = \frac{1}{2^n} \geq 0$
Since $$\lim_{n\to\infty} \frac{2^n}{2^n-1} = 1 \in (0,\infty)$$
$\sum_{n=1}^{\infty} \frac{1}{2^n}$. This is a geometric series with $r = \frac{1}{2} < 1$. Therefore by the geometric series this $\sum b_n$ converges. So does $\sum a_n$ by the limit comparison test.
Is this right?
| By setting boundaries:
$$\begin{array}{rcccl}
\dfrac1{2^n} &\le& \dfrac{1}{2^n - 1} &\le& \dfrac1{2^{n-1}} \\
\displaystyle \sum_{n=1}^\infty \dfrac1{2^n} &\le& \displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n - 1} &\le& \displaystyle \sum_{n=1}^\infty \dfrac1{2^{n-1}} \\
1 &\le& \displaystyle \sum_{n=1}^\infty \dfrac{1}{2^n - 1} &\le& 2
\end{array}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Three points: $A=(4,3), B=(2,5)$, and $C=(3,6)$. Find the angle between vector $AB$ and vector $AC$. Okay, so where am I going wrong? Here is step by step what I tried.
First I did this to find vectors:
$a=[A_x-B_x, A_y-B_y]$ so $a=[2,-2]$
$b=[A_x-C_x, A_y-C_y]$ so $b=[1,-3]$
Then I found the magnitudes of both vectors:
$$\|a\|= \sqrt{x^2 + y^2} \text{ so } \|a\|= 0$$
$$\|b\|=\sqrt{x^2 + y^2} \text{ so } \|b\| = \sqrt{10}$$
Then with this new information I tried the dot product:
$$\frac{a\cdot b}{\|a\|\cdot \|b\|} \text{ so } \frac{(2\cdot -2)+(1\cdot -3)}{0\cdot \sqrt{10}}$$
My calculator says "DIVIDE BY 0 Error"
So I assume I needed to try this:
$$\frac{(2 \cdot -2)+(1 \cdot -3)}{\sqrt{10}}= \frac{-7\sqrt{10}}{10} = -2.213594362$$
inverse of cos(-2.213594362)=DOMAIN ERROR
| Here are some mistakes you've made:
Note that $(-2)^2=4$, which is not equal to $-4$. Therefore, $\|\mathbf{a}\|\neq0$:
$$\|\mathbf{a}\|=\sqrt{2^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}$$
Also, note that your dot product is incorrect. For vectors $\mathbf{a}=\begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$, we should have:
$$\mathbf{a}\cdot \mathbf{b}=a_1b_1+a_2b_2$$
Here is what you seem to have done instead:
$$\mathbf{a}\cdot \mathbf{b}\neq a_1a_2+b_1b_2$$
Hence for your vectors, we should have:
$$\mathbf{a}\cdot \mathbf{b}=\begin{pmatrix} 2 \\ -2 \end{pmatrix}\cdot \begin{pmatrix} 1 \\ -3 \end{pmatrix}=2+6=8$$
The correction:
Hence, you should have:
$$\cos{\theta}=\frac{\mathbf{a}\cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|}=\frac{8}{\sqrt{8}\cdot \sqrt{10}}=\cdots$$
After simplifying, you should be able to find the value of $\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $x$ and $y$ such that $15\sin(x+y)+7\sin x+7\sin y$ is maximized Question is: Find $x$ and $y$ such that
$$15\sin(x+y)+7\sin x+7\sin y$$
is maximized.
What I tried was
$$f(x,y)=15\sin(x+y)+7\sin x+7\sin y=15(\sin x\cos y+\cos x\sin y)+7(\sin x +\sin y)$$
$$\frac{\partial f}{\partial x}=15\cos(x+y)+7\cos x=0$$
$$\frac{\partial f}{\partial y}=15\cos(x+y)+7\cos y=0$$
$$\cos x=\cos y$$
$$y=-x$$
$$\cos x=-{15\over 7}\quad ??$$
| I think, the answer is $25.6$.
Indeed, by C-S
$$15\sin(x+y)+7(\sin{x}+\sin{y})=15(\sin{x}\cos{y}+\cos{x}+\sin{y})+7(\sin{x}+\sin{y})=$$
$$=(15\cos{y}+7)\sin{x}+15\sin{y}\cos{x}+7\sin{y}\leq$$
$$\leq\sqrt{(\sin^2x+\cos^2x)\left(\left(15\cos{y}+7\right)^2+\left(15\sin{y}\right)^2\right)}+7\sin{y}=$$
$$=\sqrt{274+210\cos{y}}+7\sin{y}\leq\sqrt{274+210\cos{y}}+7|\sin{y}|\leq$$
$$=20\sqrt{0.685+0.525\cos{y}}+5.6\sqrt{1.5625\sin^2y}\leq$$
$$\leq\sqrt{(20+5.6)\left(20(0.685+0.525\cos{y})+5.6\cdot1.5625\sin^2y\right)}=$$
$$=25.6\sqrt{1-\frac{7}{512}(3-4\cos{y})^2}\leq25.6.$$
The equality occurs for $x=y=\arccos\frac{3}{5}$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Perturbation series for $x^5+\varepsilon x-1=0$ I want to find a closed form for the perturbation coefficients $a_n$ defined by the perturbative solution
$$
x(\varepsilon)=1+\sum_{n=1}^\infty a_n \varepsilon^n
$$
to the quintic equation
$$
x^5+\varepsilon x-1=0.
$$
By computing the determinant, as was suggested for this related question regarding the cubic case, we can argue that the radius of convergence of the above series must be
$$
\rho=\frac{5}{4^{4/5}}=1.64938\dots
$$
Furthermore, the Lagrange-Bürmann theorem allows us to formally write down a closed form for the coefficients, namely
$$
a_n=\frac{(-1)^n}{n!}\frac{d^{n-1}}{dx^{n-1}}\left(\frac{x}{1+x+x^2+x^3+x^4}\right)^n_{x=1}
$$
but this doesn't look very illuminating. (For the case of the cubic equation a slightly more explicit but still cumbersone rewriting was made possible by the simpler form of $a_n$).
What I would like to achieve is obtaining $\rho$ from the explicit closed expression for the $a_n$. Therefore I set out to compute some of them. Here they are:
$$
a_1=-\frac{1}{5},\
a_2=-\frac{1}{5^2},\
a_3=-\frac{1}{5^3},\
a_4=0,\\
a_5=\frac{21}{5^6},\
a_6=\frac{78}{5^7},\
a_7=\frac{187}{5^8},\
a_8=\frac{286}{5^9},\
a_{9}=0,\\
a_{10}=-\frac{9367}{5^{12}},\
a_{11}=-\frac{39767}{5^{13}},\
a_{12}=-\frac{105672}{5^{14}},\
a_{13}=-\frac{175398}{5^{15}},\
a_{14}=0.
$$
The behavior of the $a_n$ for $n=1,\ldots,30$ supports the following conjecture:
$$\boxed{
a_n = -(-1)^{\lfloor n/5\rfloor}\frac{c_n}{5^{\alpha_n}}
}
$$
where
$$
\alpha_n=\sum_{k=0}^\infty \left\lfloor \frac{n}{5^k} \right\rfloor
$$
and the $c_n$ are nonnegative integer coefficients which are not divisible by $5$ and vanish for $n=5m-1$.
I think that $c_n$ should be something of the form
$$
\frac{1}{4n+1}\binom{5n}{n}
$$
which for $n=4$ goes very near to reproducing $c_8=286$ and has a scaling similar to that exhibited by the $c_n$. This problem is also motivated by this video where it is suggested that the answer can be guessed with some effort by staring at the coefficients hard enough (27.08).
I think I need some help, however!
| Making Robert Israel's formula a bit more compact we can write
$$\boxed{
x(\varepsilon)=
\sum_{k=0}^\infty (-1)^{k+1}(2)^{4k}\sum_{j=0}^3(-1)^{\delta_{j0}}\frac{\left(\frac{2(j+1)}{5}\right)_{2k}\left(\frac{2(j+1)}{5}-\frac{1}{2}\right)_{2k}}{\left(\frac{j+5}{5}\right)_{k}\left(\frac{j+4}{5}\right)_{k}\left(\frac{j+3}{5}\right)_{k}\left(\frac{j+2}{5}\right)_{k}}\frac{\varepsilon^{5k+j}}{5^{5k+j}}
}
$$
where $(a)_m$ is the Pochhammer symbol and
$$
(a)_m = \frac{\Gamma(a+m)}{\Gamma(a)}=(a+m-1)(a+m-2)\cdots(a+1)a
$$
for positive integer $m$.
Applying Stirling's asymptotic expansion,
$$\begin{aligned}
&\lim_{k\to\infty}\left[\frac{2^{4k}}{5^{5k+j}}\frac{\left(\frac{2(j+1)}{5}\right)_{2k}\left|\left(\frac{2(j+1)}{5}-\frac{1}{2}\right)_{2k}\right|}{\left(\frac{j+5}{5}\right)_{k}\left(\frac{j+4}{5}\right)_{k}\left(\frac{j+3}{5}\right)_{k}\left(\frac{j+2}{5}\right)_{k}}\right]^{1/{5k}}\\
=&\ \lim_{k\to\infty}\frac{2^{4/5}}{5}\frac{\left(2+\frac{2(j+1)}{5k}\right)^{2/5}\left(2+\frac{2(j+1)}{5k}-\frac{1}{2k}\right)^{2/5}}{\left(1+\frac{j+5}{k}\right)^{1/5}\left(1+\frac{j+4}{k}\right)^{1/5}\left(1+\frac{j+3}{k}\right)^{1/5}\left(1+\frac{j+2}{k}\right)^{1/5}}\\
=&\ \frac{2^{4/5}}{5}\cdot 2^{2/5}\cdot 2^{2/5}\\
=&\ \frac{4^{4/5}}{5}
\end{aligned}$$
independently of $j=0,1,2,3$. Therefore by Hadamard's formula, the radius of convergence is
$$\boxed{
\rho=\frac{5}{4^{4/5}}.
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find f(x) using concept of Definite-Integral-as a limit of sum $$ f(x) = \lim_{n\to\infty} \left[\frac{n^n\left(x^2+n^2\right)\left(x^2+\frac{n^2}{4}\right)\cdots\left(x^2+\frac{n^2}{n^2}\right)}{n!\left(x^3+n^3\right)\left(x^3+\frac{n^3}{8}\right)\cdots\left(x^3+\frac{n^3}{n^3}\right)}\right]^\frac{x}{n},\qquad x>0$$
My approach:
I took $\log$ on both sides..
$$
\log(f(x)) = x \lim_{n\to\infty}\left[ \sum_{r=1}^n \frac{1}{n} \log\left(x^2+\frac{1}{\left(\frac{r}{n}\right)^2}\right) - \sum_{r=1}^n \frac{1}{n} \log\left(x^3 + \frac{1}{(\frac{r}{n})^3}\right) + \frac{1}{n} \log\left(\frac{n^n}{n!}\right)\right]$$
The first two terms inside the square brackets become..
$$
\log(f(x)) = x \left[\int_0^1 \log\left(x^2+\frac{1}{t^2}\right) dt + \int_0^1 \log\left(x^3+\frac{1}{t^3}\right) dt + \lim_{n\to\infty} \frac{1}{n} \log\left(\frac{n^n}{n!}\right)\right]$$
I was unable to solve the integrals and the last term..Any Help ??
| Rather than first using riemann sum lets simplify some things. Take out $n^2,\frac {n^2}{4},...\frac {n^2}{n^2} $ from the brackets of numerator and $n^3,\frac {n^3}{8},....\frac {n^3}{n^3} $ from brackets of denominator. This converts the function as $$(\frac {n^n.n^{2n}.\frac {1}{1.4.9...n^2}(\frac {x^2}{n^2}+1)(\frac{2^2}{n^2}x^2+1)...(\frac {n^2}{n^2}x^2+1)}{n!.n^{3n}.\frac {1}{1^3.2^3...n^3}(\frac{x^3}{n^3}+1)(\frac {2^3}{n^3}x^3+1)... (\frac {n^3}{n^3}x^3+1)})^{x/n}$$ . Now remember that $\pi _1^n \frac {n^3}{n^2}=n!$ thus the function becomes $\frac {(\frac {x^2}{n^2}+1)(\frac {4x^2}{n^2}+1).... (\frac {n^2x^2}{n^2}+1)}{(\frac {x^3}{n^3}+1)(\frac {2^3x^3}{n^3}+1).. (\frac {n^3x^3}{n^3}+1)}^{\frac {x}{n}} $ now take logs and the $f (x)=\int _0 ^1 x\log (\frac {(xr)^2+1}{(xr)^3+1})dr $ now just put $rx=k $ note that x is constant and then integration can be done by separating logs and then by parts indivdually.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find an inverse in a quotient of a Polynomial Ring I know this question is so easy that should be a first year algebra, but It has been a long time since I last studied this argument.
I have to find the inverse of $x+2$ in the ring $\mathbb{Q}[x]/\left<x^6 + x^4 + x^2 + 1\right>$.
I tried to divide $(x^6 + x^4 + x^2 + 1)+1$ for $x + 2$ but I obtained a reminder of $86$. Then I tried to solve
$$(x + 2)(ax^5+bx^4+cx^3+dx^2+ex+f)=q(x^6 + x^4 + x^2 + 1)+1$$
and obtained that q must be a complex number. So now I'm wondering if there's a better way to proceed. Can anybody give me an hint?
Thanks
| Use Horner's scheme to divide by $x+2$:
$$\begin{matrix}\\\\\times -2\end{matrix}\quad\begin{matrix}
1&0&1&0&1&0&1\\
&-2&4&-10&20&-42&84 \\
\hline
1&-2&5&-10&21&-42&\!\!\vert\; 85
\end{matrix}$$
Thus $\;x^6+x^4+x^2+1=(x^5-2x^4+5x^3-10x^2+21x-42)(x+2)+85$, so that
\begin{align}-\frac1{85}(x^5-2x^4+5x^3-10x^2+21x-42)(x+2)&=1-(x^6+x^4+x^2+1)\\&\equiv 1\mod (x^6+x^4+x^2+1).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to show $\int_{0}^{\infty}{{\gamma+\ln x\over e^x}}\cdot{1-\cos x\over x} \,\mathrm dx={1\over 2}\cdot{\pi-\ln 4\over 4}\cdot{\pi+\ln4\over 4}?$ Consider this integral $(1)$
$$\int_{0}^{\infty}\color{red}{{\gamma+\ln x\over e^x}}\cdot{1-\cos x\over x}\,\mathrm dx={1\over 2}\cdot{\pi-\ln 4\over 4}\cdot{\pi+\ln 4\over 4}\tag1$$
Recall a well-known integral for $\gamma$:
$$\int_{0}^{\infty}e^{-x}\ln x\,\mathrm dx=-\gamma.$$
Making an attempt:
I am not sure, what to do...
Recall: $\cos x={e^{ix}+e^{-ix}\over 2}$, then $(1)$ becomes
$$\int_{0}^{\infty}{\gamma+\ln x\over e^x}\cdot{2-e^{ix}-e^{-ix}\over 2x}\,\mathrm dx\tag2$$
Or using $e^{-x}$ series, then $(1)$ becomes
$$\sum_{n=0}^{\infty}{(-1)^n\over n!}\color{blue}{\int_{0}^{\infty}(\gamma+\ln x)(1-\cos x)x^{n-1}\,\mathrm dx}\tag3$$
$$\color{blue}{blue}=\int_{0}^{\infty}(\gamma+\ln x)x^{n-1}\,\mathrm dx-\int_{0}^{\infty}(\gamma+\ln x)\cos(x) x^{n-1}\,\mathrm dx=I_1-I_2\tag4$$
Indefinite integral of $$I_1={x^n\over n^2}(n\ln x+n\gamma-1)+C$$
Put in the limit and $I_1$ is diverges and $I_2$ it is a lengthy IBP and it also diverges.
How to prove (1)?
| Assuming that $a>1$,
$$ \begin{align} I(a) &= \int_{0}^{\infty} \frac{\gamma+\ln x}{e^{ax}} \frac{1-\cos x}{x} \, dx \\ &= -\int_{0}^{\infty} \frac{\gamma+\ln x}{e^{ax}} \sum_{n=1}^{\infty} \frac{(-1)^{n}x^{2n-1}}{(2n)!} \, dx \\ &= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n)!} \int_{0}^{\infty} (\gamma + \ln x) e^{-ax} x^{2n-1} \, dx \\ &= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n)!} \left(\gamma \, \frac{\Gamma(2n)}{a^{2n}} + \frac{\Gamma'(2n) -\ln(a) \Gamma(2n)}{a^{2n}}\right) \\ &=-\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n} \left(\frac{\gamma + \psi_{0}(2n)- \ln(a)}{a^{2n}} \right) \tag{1} \\ &= -\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2n} \frac{H_{2n}- \frac{1}{2n} -\ln (a) }{a^{2n}} \tag{2} \\ &= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n}\frac{H_{2n}}{a^{2n}} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{4n^{2}} \frac{1}{a^{2n}} + \ln (a) \sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n} \frac{1}{a^{2n}} \\\ &= -\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n}\frac{H_{2n}}{a^{2n}} + \, \frac{1}{4} \, \operatorname{Li}_{2} \left(- \frac{1}{a^{2}} \right) - \frac{\ln (a)}{2} \, \ln\left(1+ \frac{1}{a^{2}} \right) \tag{3}. \end{align}$$
From the ordinary generating function for the harmonic numbers, we see that $$f(z) = \sum_{n=1}^{\infty} \frac{(-1)^{n} H_{n}}{n} \, z^{n} = -\int_{0}^{z} \frac{\log(1+t)}{t(1+t)} \, dt = \operatorname{Li}_{2}(-z) + \frac{1}{2} \, \ln^{2}(1+z), \quad |z| <1. $$
Therefore,$$\begin{align} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n}\frac{H_{2n}}{a^{2n}}&= \frac{1}{2} \left[f\left(\frac{i}{a}\right)+f\left(-\frac{i}{a} \right) \right] \\ &= \frac{1}{2} \left[\operatorname{Li}_{2} \left(-\frac{i}{a}\right)+ \frac{1}{2} \, \ln^{2} \left(1+ \frac{i}{a} \right)+ \operatorname{Li}_{2} \left(\frac{i}{a}\right) + \frac{1}{2} \, \ln^{2} \left(1- \frac{i}{a} \right) \right] \\ &= \frac{1}{4} \left[\operatorname{Li}_{2} \left(-\frac{1}{a^{2}} \right) +\ln^{2} \left(1+ \frac{i}{a} \right) + \ln^{2} \left(1- \frac{i}{a} \right)\right], \tag{5} \end{align}$$
and $$\begin{align} I(a) &= -\frac{1}{4} \left[ \ln^{2} \left(1+ \frac{i}{a}\right)+ \ln^{2} \left(1- \frac{i}{a} \right)+2 \ln (a) \ln \left(1+ \frac{1}{a^{2}} \right)\right] \\ &= - \frac{1}{8} \left[ \ln^{2} \left(1+ \frac{1}{a^{2}} \right)-4\arctan^{2} \left(\frac{1}{a} \right) + 4 \ln (a) \ln \left(1+ \frac{1}{a^{2}} \right)\right]. \end{align}$$
Letting $a \downarrow 1$, we get $$I(1) = - \frac{1}{8} \, \ln^{2}(2)+\frac{1}{2} \left(\frac{\pi^{2}}{16} \right) = \frac{\pi^{2}-4 \ln^{2}(2)}{32} = \frac{\pi^{2}-\ln^{2}(4)}{32}.$$
For evaluation purposes, I assumed that $a >1$. But the result should hold for $a>0$.
$(1)$ https://en.wikipedia.org/wiki/Digamma_function
$(2)$ https://en.wikipedia.org/wiki/Digamma_function#Relation_to_harmonic_numbers
$(3)$ https://en.wikipedia.org/wiki/Polylogarithm
$(4)$ https://en.wikipedia.org/wiki/Polylogarithm#Properties
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
trignometric equation, can i move the 2cos(2x) to the other part of the equation? When solving $2\cos(2x)+\sin(x)=1/2$ can i move the first $2$ from $\cos$ to the other parcel and multiply like this: $\cos(2x)+\sin(x)=1/4$?
| No you may not. If you'd like to divide by 2, you must divide both terms on the LHS by 2. Not just your $2\cos(2x)$ term. If you divided both terms by two, you'd be left with $\cos(2x)+\frac{1}{2}\sin(x)=\frac{1}{4}$ Try playing around with the double angle formula to solve this problem.
For example:
$2\cos(2x)=2(1-2\sin^2x)$,
So, we have $2-4\sin^2x+\sin x=\frac{1}{2}$.
Multiply both sides of the equation by 2 to get rid of fractions and you're left with $4-8\sin^2x+2\sin x=1$. Let $u=\sin x$ and set equation equal to zero and you have $-8u^2+2u+3=0$. Factor and we have $(2u+1)(-4u+3)=0$. Which gives $u=-\frac{1}{2}$ and $u=\frac{3}{4}$. Substituting back in we have $\sin x=-\frac{1}{2}$ and $\sin x=\frac{3}{4}$. $\sin x=\frac{1}{2}$ at $\frac{4\pi}{3}+2\pi k$ and $\frac{5\pi}{3}+2\pi k$. $\sin x=\frac{3}{4}$ at $x=\sin^{-1}(\frac{3}{4})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve a simple equation: $x+x\sqrt{(2x+2)}=3$
$x+x\sqrt{(2x+2)}=3$
I must solve this, but I always get to a point where I don't know what to do. The answer is 1.
Here is what I did:
$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$
Then I got:
$-2x^{3}-x^{2}-6x+9=0$
| I wouldn't have gotten there is such a convoluted way (and I wouldn't have divided by $x$ without checking that $x \ne 0$ first) but that would really make a difference.
And I'd watch out for extraneous roots.
$x +x\sqrt{2x+2} = 3$
$x\sqrt{2x+2} = 3-x$ $2x+2 > 0$ so $x > -1$
$x^2(2x+2) = 9 -6x + x^2; x > -1$
$2x^3 + x^2 + 6x -9=0; x > -1$ which is basically what you have.
Rational root theoreom says that if there are any rational roots the will be of the form $\frac {\pm 1,3,9}{\pm 1,2}$.
It's pretty clear that $x = 1$ is solution so $x-1$ is a factor and so
$(2x^3 + x^2 + 6x - 9)=(x-1)(2x^2 + bx + 9)$ where $bx^2 -2x^2 = x^2$ and $9x-bx=6x$ so $b=3$ so $(x-1)(2x^2 + 3x + 9)=0$.
$x =1$ is a solution and $2x^2 + 3x + 9 = 0$ is another. But that'd imply $2x^2 = -3x - 9$. But that's imply $3x < -9$ and $x<-3$ which is impossible as $(2x + 2) > 0$.
So $x=1$ is the only real solution.
====
Another way to factor. If you KNOW $x-1$ is a factor, which we do we have:
$2x^3 + x^2 + 6x - 9=$
$2x^2(x-1) + 2x^2 + x^2 + 6x - 9 =$
$2x^2(x-1) + 3x^2 + 6x - 9=$
$2x^2(x-1) + 3x(x-1) + 3x + 6x - 9=$
$2x^2(x-1) + 3x(x-1) + 9x - 9 = $
$2x^2(x-1) + 3x(x-1) + 9(x-1) + 9 - 9=$
$2x^2(x-1) + 3x(x-1) + 9(x-1)=$
$(x-1)(2x^2 + 3x + 9) = 0$.
We can try to solve $2x^2 + 3x + 9 =0$ by the quadratic equation. But we don't actually have to. We know as $\sqrt{2x+2}$ must be real that $x \ge -1$ So $2x^2 + 3x + 9 \ge 2x^2 + 3(-1) + 9 = 2x^2 + 9 > 9 > 0$ so there is no solution that will satisfy both $2x^2 + 3x + 9 = 0$ and $\sqrt{2x+2}$ is real.
(There's actually no real solution to just $2x^2 + 3x + 9=0$ at all as the quadratic equation or completeing the square would show us.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find locus with circle and intersection of lines
$M(x_0,y_0), P(x,y)$
The equation of the circle is $x^2+y^2=25$.
MN is a chord in the circle perpendicular to the axis $x$. P is the intersection of BM and NA.
I found that
$$MB: y = \frac{y_0}{x_0+5}x+\frac{5y_0}{x_0+5}$$
$$NA: y = \frac{y_0}{5-x_0}x-\frac{5y_0}{5-x_0}$$
I need to find the locus of the points P.
| With $O=(0,0)$ as circle center, let $\theta=\angle AOM$. Then the slope of line $BM$ is
\begin{equation}
m_1=\frac{\sin\theta}{1+\cos\theta}
\end{equation}
and the slope of line $NA$ is
\begin{equation}
m_2=\frac{\sin\theta}{1-\cos\theta}
\end{equation}
So the point $P$ lies at the intersection of the lines
\begin{eqnarray}
BM:\quad y&=&\frac{\sin\theta}{1+\cos\theta}(x+5)\\
NA:\quad y&=&\frac{\sin\theta}{1-\cos\theta}(x-5)
\end{eqnarray}
which have point of intersection
\begin{equation}
P=\left(5\sec\theta,5\tan\theta\right)
\end{equation}
Since $\sec^2\theta-\tan^2\theta=1$ we have the hyperbola
\begin{equation}
\frac{x^2}{5^2}-\frac{y^2}{5^2}=1
\end{equation}
EDIT: Since the OP wants a solution not referencing trigonometry, I am adding the following additional solution:
First, give the point $M$ coordinates $(s,t)$ then find the coordinates of $P$ in terms of $s$ and $t$. Since $(s,t)$ lies on the circle we know that
$$ s^2+t^2=25$$
and that, therefore
$$ 25-t^2=s^2\tag{1} $$
We will need equation $(1)$ below.
The slope of line $BM$ is $m_1=\dfrac{t}{s+5}$ and the slope of line $NA$ is $m_2=\dfrac{t}{5-s}$ so the equations of the two lines are
$$ BM:\quad y=\frac{t}{s+5}(x+5) $$
and
$$ NA:\quad y=\frac{t}{5-s}(x-5) $$
The two lines intersect at $P=(x,y)$. Solving the two equations for $x$ and $y$ yields
$$ P=\left(\frac{25}{s},\frac{5t}{s}\right)$$
Then
\begin{eqnarray}
x^2-y^2&=&\frac{625-25t^2}{s^2}\\
&=&25\left(\frac{25-t^2}{s^2}\right) \text{ which, using equation } (1)\\
&=&25\cdot\frac{s^2}{s^2}\\
&=&25
\end{eqnarray}
Thus the equation of the locus of points $P$ is
$$ x^2-y^2=25$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos\left (\theta -\frac{\pi}{4}\right)$ If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos \left(\theta -\frac{\pi}{4}\right).$
I could not get any idea to solve. However I tried by using $\theta =0^\circ $. But could not get the answer.
| Hint -
$\tan (\pi \cos \theta) =\cot (\pi \sin \theta)$
$\tan(\pi \cos \theta) =\tan (\frac{\pi}2 - \pi \sin \theta)$
$\pi \cos \theta =\frac{\pi}2 - \pi \sin \theta$
$\cos \theta = \frac 12 - \sin \theta$
$\sin \theta + \cos \theta = \frac 12$
Now we have,
$\cos\left(\theta-\frac{\pi}{4}\right)$
$= \cos\theta \cos \frac{\pi}4 +\sin\theta \sin \frac{\pi}4$
$= \cos\theta \frac 1{\sqrt{2}} + \sin\theta \frac 1{\sqrt{2}}$
$= \frac1{\sqrt 2} \left( \cos\theta +\sin\theta \right)$
$ = \frac 1{\sqrt2} \left(\frac 12\right)$
$= \frac 1{2\sqrt2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluating $\int {\sin \theta + \cos \theta \over \sqrt{\sin 2\theta} }d\theta$
$$\int {\sin \theta + \cos \theta \over \sqrt{\sin 2\theta} }d\theta$$
After simplifying I get,
$${1\over \sqrt{2}}\int {\tan \theta + 1 \over \sqrt{\tan \theta}} d\theta$$
Substituting $u = \tan \theta$
$${1\over \sqrt{2}}\int {u + 1\over \sqrt{u}(1 + u^2)} du$$
Substituting $t^2 = u$
$${\sqrt{2}}\int {t^2 + 1\over (1 + t^4)} dt = \sqrt{2} \int {1 + 1/t^2 \over(t - 1/t)^2 + 2 }dt$$
Substituting $z = t - 1/t$
$$\sqrt{2} \int {1\over z^2 + 2 }dz = \arctan\left(\tan \theta - 1\over \sqrt{2\tan \theta} \right) + C = {\arcsin(2\sqrt{\sin2\theta} (\sin\theta - \cos \theta))\over 2} + C$$
Which is far from the right answer of $\arcsin(\sin \theta - \cos \theta) + C$.
Where did I go wrong ?
Edit :
I would like to know where I am wrong rather than knowing how to solve the problem.
| Hint -
Put $\sin\theta - \cos \theta = t$
$(\cos \theta + \sin \theta)\text{d}\theta = \text{d}t$
Also on squaring above equation,
$\sin^2\theta + \cos^2 \theta - 2 \sin \theta \cos \theta = t^2$
$1 - \sin2\theta = t^2$
$\sin2\theta = 1 - t^2$
Then given integral becomes,
$\int \frac{1}{\sqrt {1-t^2}} \text{d}t$
Hope now its easy to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.