Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Let $n$ be a positive integer such that $\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n} = \frac{4}{11}}$ Let $n$ be a positive integer such that
$$\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n} = \frac{4}{11}}$$
then
$$\displaystyle{\frac{2+3+\cdots+2n}{4+5+\cdots+4n}} = \frac{m}{p}.$$
The question further "Is $m+p$ a prime?"
|
$k+(k+1)+...+kn=\frac{k}{2}(n+1)(kn-k+1)$ (just the half of the sum of the first and the last terms times the number of the terms)$=\frac{k}{2}(kn^2+n-(k-1))$
Now, from the first equation we get $n$: $$33(3n^2+n-2)=20(5n^2+n-4)$$ $$n^2-13n-14=0$$ $$(n-14)(n+1)=0$$ $$n>0\Rightarrow n=14$$
And from the second equation we get $m$: $$m=n\frac{2n^2+n-1}{8n^2+2n-6}=\frac{189}{53}=3.566...$$
So, $m+n$ is not prime, because it is not even integer...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/121558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
A probability game involving dice Consider two players A and B.
Player A rolls a fair, six-sided die $m$ times and notes the highest number on the upper face out of all of the rolls.
Player B rolls the same die $n$ times and notes the highest number on the upper face out of all of the rolls. (It's clear that as $m$ and $n$ increase, the score of 6 becomes more likely).
If $m>n$, what is the probability player A will win in terms of $m$ and $n$? Note that if the numbers are equal, player B wins.
(I know the total number of outcomes is $6^{m+n}$, but I need help counting the winning games).
|
The following ingredients are enough to determine the probabilities.
$1.$ If you toss $n$ times, the probability that the highest number is $\le k$ is $\frac{k^n}{6^n}$.
$2.$ If you toss $n$ times, the probability that the highest number is equal to $k$ is $\frac{k^n -(k-1)^n}{6^n}$.
In ($1$) and ($2$), $k$ ranges from $1$ to $6$.
Added: We do the cooking, in case the list of ingredients was not enough. Assume that $m\ge 1$ and $n \ge 1$.
Player A wins if (i) her highest number is a $6$, and B's number is $\le 5$ or (ii) A's highest number is $5$, and B's is $\le 4$, or (iii) A's highest number is $4$, and B's is $\le 3$, or (iv) A's highest number is $3$, and B's is $\le 2$, or (v) A's highest number is $2$, and B's is $\le 1$.
The probability of (i) is
$$\frac{6^m-5^m}{6^m}\cdot \frac{5^n}{6^n}.$$
The probability of (ii) is
$$\frac{5^m-4^m}{6^m}\cdot \frac{4^n}{6^n}.$$
The probability of (iii) is
$$\frac{4^m-3^m}{6^m}\cdot \frac{3^n}{6^n}.$$
The probability of (iv) is
$$\frac{3^m-2^m}{6^m}\cdot \frac{2^n}{6^n}.$$
The probability of (v) is
$$\frac{2^m-1^m}{6^m}\cdot \frac{1^n}{6^n}.$$
Add up.
There are various other ways to compute the probability that A wins. If we look at the probabilities that we are adding, we can see that there are opportunities to simplify the expression for the sum. If we were dealing with $d$-sided dice, then simplification might be worthwhile, but for the case $d=6$ it probably isn't.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/122813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Let $a,b \in {\mathbb{Z_+}}$ such that $a|b^2, b^3|a^4, a^5|b^6, b^7|a^8 \cdots$, Prove $a=b$ Let $a,b \in {\mathbb{Z_+}}$ such that $$a|b^2, b^3|a^4, a^5|b^6, b^7|a^8 \cdots$$
Prove $a=b$
|
If $a > b$ then $\frac{a}{b}>1$ and hence there is an $n$ such that $\left(\frac{a}{b}\right)^n > b$, thus $a^n > b^{n+1}$. This contradicts $a^{4k+1} | b^{4k+2}$. The case $a<b$ works in just the same way.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/123584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
}
|
Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$
*
*$\displaystyle
0\cdot \binom{n}{0} + 1\cdot \binom{n}{1} + 2\binom{n}{2}+\cdots+(n-1)\cdot \binom{n}{n-1}+n\cdot \binom{n}{n}$
*$\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}$
How do you find the sum of these and prove it by induction? Can someone help me get through this?
|
1)
$$S_n=\sum_{k=0}^nk\frac{n!}{k!(n-k)!}=n\sum_{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}=n\sum_{k=0}^{n-1}\binom{n-1}k=n2^{n-1}.$$
Then the inductive proof by $S_0=0,S_n-S_{n-1}=T_n$, is a little tedious:
$$S_n-S_{n-1}
=\sum_{k=0}^nk\binom nk-\sum_{k=0}^{n-1}k\binom {n-1}k=n\binom nn+\sum_{k=0}^{n-1}k\left(\binom nk-\binom{n-1}k\right)\\
=n+\sum_{k=1}^{n-1}k\binom{n-1}{k-1}
=n+\sum_{k=1}^{n-1}(k-1)\binom{n-1}{k-1}+\sum_{k=1}^{n-1}\binom{n-1}{k-1}\\
=n+\sum_{k=1}^{n-1}(k-1)\binom{n-1}{k-1}+\sum_{k=1}^{n-1}\binom{n-1}{k-1}\\
=n+\sum_{k=0}^{n-2}k\binom{n-1}k+\sum_{k=0}^{n-2}\binom{n-1}{k}\\
=n+S_{n-1}-(n-1)+2^{n-1}-1=S_{n-1}+2^{n-1}.$$
And on the other hand
$$T_n=S_n-S_{n-1}=n2^{n-1}-(n-1)2^{n-2}=(n-1)2^{n-2}+2^{n-1}$$
2)
As $\dfrac1{k(k+1)}=\dfrac1k-\dfrac1{k+1}$, the sum telescopes as
$$S_n=\frac1{n+1}-1.$$
This time, the proof by induction is trivial:
$$S_0=0,$$
$$T_n=S_n-S_{n-1}=\frac1{n+1}-1-\frac1n+1=\frac1{n(n+1)}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/123655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
}
|
Limit: How to Conclude I have difficulty to conclude this limit ....; place of my attempts and results, can anyone help?
tanks in advance
$$\lim_{x\to +a}\, \left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right),\quad a=+\infty,\,\,\,0,\,\,\,\,-\infty$$
1):$\,\,\,{a=+\infty}$
$$\lim_{x\to +\infty}\left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right)$$
$$\sim\lim_{x\to +\infty}\, 1+6\left( \frac{\sin^x x}{x^{2x}} \cdot\frac{\log 10^x }{x}\right)=\lim_{x\to +\infty}\, 1+6\left( \frac{\sin^x x}{x^{2x}} \right)$$
$$\text{observing}\,\,\,\,\,\left|\frac{\sin x}{x^2} \right|<\frac{1}{x^2}, \text{so} \left|\left(\frac{\sin x}{x^2}\right)^x\right|<\frac{1}{x^{2x}}\to 0\,\,\,\, x\to+\infty:$$
$$\text{infact we have} \lim_{x\to+\infty} \frac{1}{x^{2x}}=\lim_{x\to +\infty} e^{-2x\log x}=0$$
$$=\lim_{x\to +\infty}\, 1+6\left( \frac{\sin^x x}{x^{2x}} \right)=1+6\cdot0=1
$$
2):$\,\,\,{a=0}$
$$\lim_{x\to 0}\, \left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right) \sim\lim_{x\to 0}\, 1+\frac{6\log2}{x}\left(\frac{ x}{x^2}\right)^x$$
$$\sim\lim_{x\to0}1+\frac{6\log2}{x}\left(\frac{1}{x}\right)^x$$
$$=\lim_{x\to 0} 1+\frac{6\log2}{x}e^{ x\ln \left(\frac{1}{x}\right)}\to \lim_{x\to 0}\, x\ln \left(\frac{1}{x}\right)=\lim_{x\to 0}\, \frac{\ln \left(\frac{1}{x }\right)}{\frac{1}{x}}=0\to e^0=1$$
$$=\lim_{x\to 0}\,1+\frac{6\log2}{x}\cdot1=+\infty$$
3):$\,\,\,{a=-\infty}$
let $x=-t,\,\,\,$if$\,\,\,\ x\to -\infty\,\,\,$we have$ \,\,\,\ t\to+\infty,\,\,\,$ and so :
$$\lim_{t\to +\infty}\, \left(1+6\left(\frac{\sin(-t)}{t^2}\right)^{-t}\cdot\frac{\log(1+10^{-t})}{-t}\right)$$
$$=\lim_{t\to +\infty}\, \left(1-6\left(-\frac{\sin t }{t^2}\right)^{-t}\cdot\frac{\log\left(1+\frac{1}{10^t}\right)}{t}\right)$$
$$=1-6\cdot\lim_{t\to +\infty}\, \left[\left(-\frac{t^2}{\sin t }\right)^{t}\cdot\frac{\log\left(1+\frac{1}{10^t}\right)}{t}\right]$$
$$\stackrel{(\bf T)}{=}1-6\cdot\lim_{t\to +\infty}\, \left[\left(-\frac{t^2}{\sin t }\right)^{t}\cdot \left(\frac{1}{10^t}-\frac{1}{2\cdot10^{2t}}\cdot\frac{1}{t}\right) \right]$$
.... but here i'm lost ....
|
$\lim_{x\to +a}\, \left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right),\quad a=+\infty,\,\,\,0,\,\,\,\,-\infty$
When $a=-\infty$,prove as below
take $x=2k\pi-\frac{\pi}{2}$,($k<0,k\in Z$) ,$(\frac{\sin x}{x^2})^x=(\frac{-1}{(2k\pi-\frac{\pi}{2})^2})^{2k\pi-\frac{\pi}{2}}$,when $k \to -\infty$,
$-1^{2k\pi-\frac{\pi}{2}}$ may not real number.so there is no limit ,when $x \to -\infty$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/125466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
}
|
Rearranging a formula, transpose for A2 - I'm lost Given the formula:
$$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
Transpose for $A_2$
I have tried this problem four times and got a different answer every time, none of which are the answer provided in the book. I would very much appreciate if someone could show me how to do this step by step.
The answer from the book is:
$$ A_2=\sqrt\frac{A_1^2q^2}{2A_1^2gh+q^2} $$
The closest I can get is the following:
$$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
$$ \frac{q^2}{A_1^2} = \frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
Invert:
$$ \frac{A_1^2}{q^2} = \frac{(\frac{A_1}{A_2})^2-1}{2gh} $$
Multiply both sides by $2gh$:
$$ 2gh\frac{A_1^2}{q^2} = (\frac{A_1}{A_2})^2-1 $$
$$ \frac{2ghA_1^2}{q^2} = (\frac{A_1}{A_2})^2-1 $$
Add 1 to both sides and re-arrange:
$$ \frac{A_1^2}{A_2^2} = \frac{2ghA_1^2}{q^2} +1 $$
Invert again:
$$ \frac{A_2^2}{A_1^2} = \frac{q^2}{2ghA_1^2} +1 $$
Multiply by $A_1^2$:
$$ A_2^2 = \frac{A_1^2q^2}{2ghA_1^2} +1 $$
Get the square root:
$$ A_2 = \sqrt{\frac{A_1^2q^2}{2ghA_1^2}+1} $$
I cannot see where the $q^2$ on the bottom of the textbook answer comes from.
|
$$\frac{q^2}{A_1^2} = \frac{2gh}{\left(\frac{A_1}{A_2}\right)^2-1} $$
$$\Leftrightarrow \frac{q^2}{A_1^2}\left(\left(\frac{A_1}{A_2}\right)^2-1\right) = 2gh $$
$$\Leftrightarrow \frac{q^2}{A_2^2}-\frac{q^2}{A_1^2} = 2gh $$
$$\Leftrightarrow \frac{q^2}{A_2^2}= 2gh+\frac{q^2}{A_1^2} $$
$$\Leftrightarrow \sqrt{\frac{q^2}{2gh+\frac{q^2}{A_1^2}}}= \pm A_2 $$
Maybe not the fastest way, but step by step how I did it. Of course this answer can be brought into several equivalent forms.
multiplying the fraction inside the square root with $\frac{1/q^2}{1/q^2}$ gives
$$ \sqrt{\frac{1}{\frac{2gh}{q^2}+\frac{1}{A_1^2}}}= \pm A_2 $$
now with $\frac{A_1^2}{A_1^2}$ to get
$$ \sqrt{\frac{A_1^2}{\frac{2ghA_1^2}{q^2}+1}}= \pm A_2 $$
same expansion with $A_1^2$ starting from my first result gives
$$ \frac{A_1q}{\sqrt{2ghA_1^2+q^2}}= \sqrt{\frac{A_1^2q^2}{2ghA_1^2+q^2}}= \pm A_2 $$
You can keep going as long as you want.... (incidently your closest solution that you have posted does not work due to a wrong invertion as already mentioned in the comments)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/125842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
}
|
Expressing a triple integral as an iterated integral in 6 different ways How can I express the triple integral (for volume) as an iterated integral in six different ways, where the solid I would like the triple integral for is bounded by the following surfaces?
$$z=0,\; x=0,\; y=2,\; z=y-2x$$
Additional advice appreciated.
|
You want to integrate over:
$z\le y-2x$
$0\le z$
$0\le x$
$y\le 2$
If you sketch a picture, you'll see that this is a tetrahedron with vertices $(0,0,0)$, $(0,2,0)$, $(1,0,2)$, $(1,2,0)$.
Obviously, you have 6 possible permutations of variables $x$, $y$, $z$.
Let's have a look at the range of these variables. We have $y\ge 2x+z \ge 0$, hence
$$0\le y \le 2.$$
We also have $x\le \frac{y-z}2 \le \frac{2-0}2=1$, thus $$0\le x\le 1.$$
From $z\le y-2x\le 2-0 = 2$ we have $0\le z\le 2$.
All you have to do now is to choose some order. E.g. we can start with $x$. Then express possible range for $y$ using $x$. And then find in what range $z$ will be, if $x$ and $y$ are given. In this way we get:
$$\begin{align}
0 &\le x \le 1\\
2x &\le y \le 1\\
0 &\le z \le y-2x
\end{align}$$
For example, we have used $y\ge 2x+z \ge 2x$ in the second line. (Since $z\ge 0$.)
If we choose different ordering $x$, $z$, $y$ we get $z\le y-2x \le 2-2x$, i.e.
$$\begin{align}
0 &\le x \le 1\\
0 &\le z \le 2-2x\\
2x+z &\le y \le 2
\end{align}$$
We can try to start with $y$:
$$\begin{align}
0 &\le y \le 2\\
0 &\le z \le y\\
0 &\le x \le \frac{y-z}2
\end{align}$$
I guess you'll be able to do the remaining three possibilities yourself.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/128914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to find $\int_0^{2\pi} |a\cos(x)+b\sin(x)|dx$, where $a^2+b^2=1$ I need help to solve this integral:
$$\int_0^{2\pi} |a\cos(x)+b\sin(x)|dx$$ where $a^2+b^2=1$. I hope someone is able to help me.
|
Hint: Let $\phi$ be a number such that $a=\sin\phi$ and $b=\cos\phi$. You want to integrate $|\sin(x+\phi)|$. Maybe make the natural change of variable, or let the geometry guide you to the answer.
Remark: The "trick" above has a number of uses. The idea works even if $a^2+b^2\ne 1$. For suppose that $a^2+b^2\ne 0$. Note that
$$a\cos x+b\sin x=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\right).$$
Let $\phi$ be a number such that $a/\sqrt{a^2+b^2}=\sin \phi$ and $b/\sqrt{a^2+b^2}=\cos\phi$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/129462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
}
|
Finding $\lim_{x\to\infty} \frac{(2x-5)^4}{(2x^2+1)(3x^2-2)}$ Finding $$\lim_{x\to\infty} \frac{(2x-5)^4}{(2x^2+1)(3x^2-2)}$$
Do I multiply top & bottom by $\frac{1}{x^2}$ or $\frac{1}{x^4}$? How do I distribute them into the numerator tho?
In the answer given, I think with some typos:
$$... = \lim_{x\to\infty} \frac{(2-5/x)^\color{red}2}{(2+1/x^2)(3-2/x^2)} = \frac{2^{\color{red}4}}{2\cdot 3} = \frac{8}{3}$$
Either way, how do I distribute the $\frac{1}{x^?}$ into the numerator?
|
The dominant term is $x^4$ (Imagine that you expanded both the numerator and denominator. What would the highest power of $x$ be?). So, you could either multiply numerator and denominator by $1/x^4$ (see below), or factor as follows:
$$\eqalign{
{(2x-5)^4 \over (2x^2+1)(3x^2-2)}
&={ \bigl(x(2-{5\over x})\bigr)^4\over x^2 (2+{1\over x^2}) \cdot x^2(3-{2\over x^2 } ) }\cr
&={x^4 (2-{5\over x})^4 \over x^2(2+{1\over x^2})\cdot x^2(3-{2\over x^2 })}\cr
&={(2-{5\over x})^4\over (2+{1\over x^2})(3-{2\over x^2})}.
}
$$
The limit as $x$ tends to infinity is $$ {2^4\over 2\cdot 3}=16/6 =8/3.$$
The $(2-5/x)^2$ is probably a typo..
Using the other approach, where you multiply numerator and denominator by $1/x^4$:
In the numerator, to distribute $1/x^4$ over $(2x-5)^4$:
$$\textstyle
{1\over x^4}(2x-5)^4 = \bigl( {1\over x} (2x-5) \bigr)^4=(2-{5\over x})^4.
$$
(just using $a^nb^n=(ab)^n$ here).
In the denominator, you could do the following:
$$\textstyle
{1\over x^4}(2x^2+1)(3x^2-2)
={1\over x^2}(2x^2+1) {1\over x^2}(3x^2-2)= (2+{1\over x^2})(3-{2\over x^2})
$$
(or just expand $(2x^2+1)(3x^2-2)$ first and then distribute the $1/x^4$ across).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/131940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
|
$$n^2+1\equiv n^2+5n+6\mod 5$$ and $$n^2+5n+6=(n+2)(n+3).$$
Then
$$(n-1)n(n+1)(n+2)(n+3)$$ is divisible by $5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/132210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 21,
"answer_id": 6
}
|
Finding the linear combinations of two vectors I am studying for my finals and I'm trying to answer the following question:
Consider the following two vectors in $\mathbb{R}^3$: $a=(1,2,3)$ and $b=(2,3,1)$. Decide
whether it is possible to express the vector $c=(2,4,5)$ as a linear
combination of $a$ and $b$.
I have used the following row operations to find what $C_1$,$C_2$ and $C_3$ are equal to:
\begin{align*}
\left[ \begin{array}{ccc|c} 1 & 2 &2 & 2\\
2 & 3 & 4 & 4 \\
3 & 1 & 5 & 5 \end{array}\right]
&\overset{R_2 = 2R_1 -R_1}{\Longrightarrow}
\left[ \begin{array}{ccc|c}
1 & 2 &2 &2 \\
0 & 1 & 0 & 0 \\
3 & 1 & 5 & 5
\end{array}\right]
\overset{R_3=3R_1-R_3}{\Longrightarrow}
\left[\begin{array}{ccc|c}
1 &2 & 2 & 2 \\
0 &1 & 0 & 0 \\
0 & 5 &1 & 1
\end{array} \right]
\\
&\overset{R_3=5R_2+R_3}{\Longrightarrow}
\left[\begin{array}{ccc|c}
1 & 2 & 2 & 2 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1
\end{array}\right]
\end{align*}
Is the following argument correct?
By looking at the last row of the matrix, we see that it says $0 = 1$, which is impossible, and the system of equations therefore has no solutions. This means that we can not find the values for $C_1$ and $C_2$, and so $c$ can not be written as a linear combination of $a$ and $b$.
Thanks in advance!
|
Note that your original linear system appears to be completely wrong for the purposes of solving the problem. What you're after is constants $c_1$ and $c_2$ such that
$$
c_1 \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \\ 5 \end{bmatrix},
$$
or, equivalently,
$$
\begin{bmatrix}
1 & 2 \\
2 & 3 \\
3 & 1
\end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \\ 5 \end{bmatrix}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/134360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
}
|
$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2}\,dx$ $$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2}\,dx$$
Tried substitution ($u = \cos\frac{x}{2}$), but I get $-\frac{\cos^3\frac{x}{2}}{3}$ ($-\frac{2}{3}$) instead of the correct answer, which is $1\frac{1}{3}$
|
$u = \cos \dfrac{x}{2} \Rightarrow du = - \dfrac{1}{2}\sin\dfrac{x}{2} \cdot dx$
Then your integrand becomes $-2 u^2 du$.
Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/135963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Integrate $\int{2^{2x}} dx$ Integrate $$\int{2^{2x}} dx$$
How do I do it, at first, I thought I treat 2 as $e$ and I will get something like $\dfrac{1}{2} 2^{2x}$, but according to WolframAlpha its supposed to be $\dfrac{4^x}{\lg{4}}$
|
$$ I = \int 2^{2x} \mathrm {d}x \tag{1}$$
Let
$$ \begin{align*}
y &=2^{2x} \hspace{5pt} \\
\Rightarrow \ln y &= 2x \ln 2
\end{align*}
$$
Differentiate both sides
$$
\begin{align*}
\frac{1}{y} \frac{dy}{dx} &= 2 \ln 2 \\
\frac{dy}{y}&=
2 \ln 2 \hspace{4pt}dx\\
&= \ln 2^2 \hspace{4pt} dx
\end{align*}
$$
$$
\begin{align*}
dx &= \frac{dy}{y \hspace{4pt} \ln 2^2}
\end{align*}
$$
Substitute for $x$ and $dx$ in $(1)$
$$
\begin{align*}
I &= \int y \frac{dy}{y \hspace{4pt} \ln 2^2}\\
&= \int \frac{1}{\ln 2^2} \mathrm{d}y = \frac{1}{\ln 2^2} \int \mathrm{d}y = \frac{1}{\ln 4} \int \mathrm{d}y \\
&= \frac{y}{\ln 4} + C \hspace{14pt} (\textit{But } y=2^{2x})\\
&= \frac{2^{2x}}{\ln 4} + C
\end{align*}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/136264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Discrete random variable summation problem. $X$ is a discrete random variable taking on the values $X = 1,3,3^2,3^3,\dots,3^m$ and $f(x) = P(X=x)=\dfrac c x$ for a constant $c$. Find $c$.
Solution:
Since $P(X)=1$, we know that $\dfrac c x=1$, so $c=x$. To find x, we have $x = \sum_0^m 3^m$. Since this series summation diverges to infinity, $c = \infty$.
This is a fascinating problem, however, something doesn't seem right ... in other words, how can $x = \infty$? Is the first statement $c=x$ incorrect?
|
The probability that $X=1$ is $\dfrac{c}{1}$, the probability that $X=3$ is $\dfrac{c}{3}$, and so on. It follows that
$$\frac{c}{1}+\frac{c}{3}+\frac{c}{3^2}+\cdots +\frac{c}{3^m}=1.$$
We want to find the sum of the finite geometric series
$1+x+x^2+\cdots+x^m$.
where $x=1/3$. Let
$$f(m)=1+x+x^2+\cdots+x^m.$$
Note that
$$xf(m)=x+x^2+x^3+\cdots+x^{m+1}.$$
Subtract. We get
$$(1-x)f(m)=1-x^{m+1},$$
and therefore
$$f(m)=\frac{1-x^{m+1}}{1-x}.$$
Finally, put $x=1/3$ and note that $c=1/f(m)$.
Remark: If instead of stopping at $3^m$, we continue forever, then instead of the finite sum of the answer, we have an infinite sum. But since $(1/3)^{n+1}$ approaches $0$ as $n\to\infty$, the infinite geometric series has sum $1/(1-x)=3/2$, and therefore we get $c=2/3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/138717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
System of equations - what are the values of the coefficients of a quadratic parabola? In the process of relearning the mathematical basics I'm stumbling over this problem:
A quadratic parabola $y = ax^2 + bx + c $ goes through the points A(1/2), B(3/7) and C(-1/1). What are the values of the coefficients $a$, $b$ and $c$?
This is a problem presented in the section about "Systems of Equalities", but I don't have the slightest idea, how to use the coordinates to calculate the values of the coefficients.
How can I solve this problem with a system of equalities?
|
First use the fact that the curve passes through the point $(1,2)$. That says that when $x=1$, we have $y=2$. So substitute $x=1$ in the equation $y=ax^2+bx+c$. We get
$$2=a+b+c.$$
Similarly, because $(3,7)$ is on the curve, we have
$$7=9a+3b+c.$$
And finally, the third point tells us that
$$1=a-b+c.$$
We now have $3$ linear equations in the $3$ unknowns $a$, $b$ and $c$.
From the two equations $9a+3b+c=7$ and $a+b+c=2$, we obtain, by subtraction, that
$8a+2b=5$.
From the two equations $9a+3b+c=7$ and $a-b+c=1$, we obtain, by subtraction, that $8a+4b=6$. (I deliberately did not subtract the third from the first, that would have made things too easy!)
We have "eliminated" $c$, and we have two equations in the variable $a$ and $b$. Now we will "eliminate" $b$, which again is not the clever thing to do.
So recall we have $8a+2b=5$ and $8a+4b=6$. Multiply both sides of the first equation by $2$. We get $16a+4b=10$. By subtraction, using $8a+4b=6$, we get $8a=4$, and therefore $a=1/2$. Then from $8a+2b=5$ we get $4+2b=5$ and therefore $b=1/2$. Finally, from $a+b+c=2$ we get that $c=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/145470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Solving $a = b^2 + 2b^2(1 - b)$ for $b$ My algebra is very rusty, it's been about 15years since I studied, and I was stumped recently when trying to rearrange this formula;
$$a = b^2 + 2b^2(1 - b)$$
to give $b$ in terms of $a$.
Can someone show me step by step working please :)
I remember 'change side, change sign' but it all gets very confusing very quickly.
Thanks!
So,
$$2b^2(1-b)\Rightarrow 2b^2 - 2b^3$$
Giving
$$a = b^2 + 2b^2 - 2b^3$$
$$a = 3b^2 - 2b^3$$
$$a + 2b^3 - 3b^2 = 0$$
is this right so far?
|
Your last equation is correct.
$$\begin{equation*}
a+2b^{3}-3b^{2}=0
\end{equation*}\tag{0}$$
This equation is a cubic equation in $b$, which can be solved algebraically by the Cardano's cubic formula (see e.g. this PlanetMath entry). We can write it as
$$\begin{equation*}
b^{3}-\frac{3}{2}b^{2}+\frac{a}{2}=0.
\end{equation*}\tag{1}$$
*
*This equation has in general $3$ complex roots. First we make the change of variables $$b=t+\frac{1}{2}.\tag{1a}$$ Since $$\begin{equation*}
\left( t+\frac{1}{2}\right) ^{3}-\frac{3}{2}\left( t+\frac{1}{2}\right) ^{2}+\frac{a}{2}=t^{3}-\frac{3}{4}t-\frac{1}{4}+\frac{1}{2}a,
\end{equation*}\tag{2}$$ we get the reduced cubic equation $$\begin{equation*}
t^{3}-\frac{3}{4}t-\frac{1}{4}+\frac{1}{2}a=0.\tag{3}
\end{equation*}$$
*The Cardano's method to solve $(3)$ is to write the variable $t$ as a sum of two variables $$t=u+v.\tag{3a}$$ Since
$$\begin{eqnarray*}
&&\left( u+v\right) ^{3}-\frac{3}{4}\left( u+v\right) -\frac{1}{4}+\frac{1}{2}a \\
&=&\left( u^{3}+v^{3}-\frac{1}{4}+\frac{1}{2}a\right) +\left( 3uv-\frac{3}{4}\right) \left( u+v\right) =0
\end{eqnarray*}\tag{4}$$
every solution of the following system is a solution of the reduced cubic $(3)$ as well. $$
\begin{eqnarray*}
&&\left\{
\begin{array}{c}
u^{3}+v^{3}-\frac{1}{4}+\frac{1}{2}a=0 \\
3uv-\frac{3}{4}=0
\end{array}
\right. \\
&\Leftrightarrow &\left\{
\begin{array}{c}
u^{3}+v^{3}=\frac{1}{4}-\frac{1}{2}a \\
u^{3}v^{3}=\frac{1}{64}
\end{array}
\right. \\
&\Leftrightarrow &\left\{
\begin{array}{c}
u^{3}+\frac{1}{64u^{3}}-\left( \frac{1}{4}-\frac{1}{2}a\right) =0 \\
v^{3}=\frac{1}{64u^{3}}
\end{array}
\right. \\
&\Leftrightarrow &\left\{
\begin{array}{c}
\left( u^{3}\right) ^{2}-\left( \frac{1}{4}-\frac{1}{2}a\right) u^{3}+\frac{1
}{64}=0 \\
v^{3}=\frac{1}{64u^{3}}.
\end{array}
\right.
\end{eqnarray*}\tag{5}$$
*If we write $$Y=u^{3},\tag{5a}$$ we get the following quadratic equation
$$\begin{equation*}
Y^{2}-\left( \frac{1}{4}-\frac{1}{2}a\right) Y+\frac{1}{64}=0,
\end{equation*}\tag{6}$$ whose solutions are $$\begin{eqnarray*}
Y_{+} &=&\frac{1}{8}-\frac{1}{4}a+\frac{1}{4}\sqrt{-a+a^{2}} \\
Y_{-} &=&\frac{1}{8}-\frac{1}{4}a-\frac{1}{4}\sqrt{-a+a^{2}}.
\end{eqnarray*}\tag{7}$$
*So, without loss of generality, the solutions of the system $(5)$ are
$$\begin{equation*}
u=Y_{+}^{1/3},\qquad v=\frac{1}{4Y_{+}^{1/3}}=Y_{-}^{1/3}.
\end{equation*}\tag{8}$$
*Combining the above results we obtain the following formula: $$\begin{eqnarray*}
b &=&t+\frac{1}{2}=u+v+\frac{1}{2} \\
&=&\left( \frac{1}{8}-\frac{1}{4}a+\frac{1}{4}\sqrt{a^{2}-a}\right)
^{1/3}+\left( \frac{1}{8}-\frac{1}{4}a-\frac{1}{4}\sqrt{a^{2}-a}\right)
^{1/3}+\frac{1}{2}.\tag{9}
\end{eqnarray*}$$
*The $3$ complex (or real) values of $(9)$ give all the solutions of $(0)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/145814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
Evaluating $\int_{-20}^{20}\sqrt{2+t^2}\,dt$ I have this integral:
$$\int_{-20}^{20}\sqrt{2+t^2}\,dt$$
I tried solving it many times but without success.
The end result is this:
$$2\left( 10\sqrt{402}+\mathop{\mathrm{arcsinh}}(10\sqrt{2})\right).$$
I can't seem to get this end result. I got a few wrong ones but cant find this one. Perhaps it's wrong? Could anyone confirm it?
I tried Sage, and it calculates it correctly, but not with steps.
|
Another way integrate $\sqrt{2 + t^2}$ is to use the Euler substitution $u = \sqrt{2+t^2} + t$. Subtracting $t$ and squaring gives $t = \frac{1}{2}(u - \frac{2}{u})$. This implies
$$\sqrt{2+t^2} = u - t = \frac{1}{2}(u + \frac{2}{u})$$
and
$$dt = \frac{1}{2}(1 + \frac{2}{u^2})\,du$$
Thus
\begin{align*}
\int \sqrt{2+t^2}\,dt
&= \int \frac{1}{4}(u + \frac{4}{u} + \frac{4}{u^3})\,du \\
&= \frac{1}{8}u^2 - \frac{1}{2u^2} + \ln{|u|} + C \\
&= \frac{1}{8}(u - \frac{2}{u})(u + \frac{2}{u}) + \ln{|u|} + C \\
&= \frac{1}{2}t\sqrt{2 + t^2} + \ln(t + \sqrt{2 + t^2}) + C
\end{align*}
Applying this antiderivative to your integral, you get $$20\sqrt{402} + \ln(20 + \sqrt{402}) - \ln(-20 + \sqrt{402}) = 2(10\sqrt{402} + \ln(10\sqrt{2} + \sqrt{201}))$$
You can check that $\ln(10\sqrt{2} + \sqrt{201}) = \operatorname{arcsinh}(10\sqrt{2})$ so that this solution is the same as the one in your question.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/147788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Integrating $\int_{a}^{+\infty} y \exp(-by)/(1-\exp(-cy)) \ dy $ Can any one help me calculate this integral :
$$\int_{a}^{+\infty} \frac{y\ \exp{(-by)}} {1-\exp{(-cy)}} \ dy $$
a, b & c are real constant numbers, b & c > 0
|
Assuming $a>0$, you can write the integrand as $\sum_{n=0}^\infty y \exp((-b-nc)y)$ and the integral becomes the convergent series
$$\sum_{n=0}^\infty e^{-ab-acn} \frac{ab+acn+1}{(b+nc)^2} $$
According to Maple this can be written using a hypergeometric function:
$$ \frac{ab+1}{b^2 e^{ab}}\
{\mbox{$_4$F$_3$}(1,{\frac {b}{c}},{\frac {b}{c}},{\frac {1+ \left( c+b \right) a}{ac}};\,{\frac {c+b}{c}},{\frac {c+b}{c}},{\frac {ab+1}{ac}};\, {{\rm e}^{-ac}} )}
$$
However, we can do somewhat better: first write $\dfrac{ab+acn+1}{(b+nc)^2} = \dfrac{a}{b+nc}+\dfrac{1}{(b+nc)^2}$. Now
$$ \sum_{n=0}^\infty e^{-ab-acn} \left( \dfrac{a}{b+nc} + \dfrac{1}{(b+nc)^2}\right) = {\rm e}^{-ab} \left( \dfrac{a}{c} {\rm LerchPhi} \left( {{\rm e}^{-ac}},1,{\frac {b}{c}} \right)
+ \dfrac{1}{c^2} {\rm LerchPhi} \left( {{\rm e}^{-ac}},2,{\frac {b}{c}} \right)\right)
$$
where ${\rm LerchPhi}(t,m,v) = \sum_{n=0}^\infty \dfrac{t^n}{(v+n)^m}$.
In the special case $a=0$ the series becomes
$$ \sum_{n=0}^\infty \frac{1}{(b+nc)^2} = \Psi \left( 1,{\frac {b}{c}} \right) {c}^{-2}$$
where $\Psi(1,t) = \dfrac{d}{dt} \Psi(t) $ and $\Psi(t) = \dfrac{d}{dt} \ln \Gamma(t)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/147962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding square roots of $\sqrt 3 +3i$ I was reading an example, where it is calculating the square roots of $\sqrt 3 +3i$.
$w=\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)\\=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$
Let $z^2=w \Rightarrow r^2(\cos(2\theta)+i\sin(2\theta))=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$.
But how did they get from $\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$?
And can one just 'let $z^2=w$' as above?
Edit:
$w=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=z^2\\ \Rightarrow z=\sqrt{2\sqrt 3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})\\ \Rightarrow \sqrt{2\sqrt 3}\frac{\sqrt 3}{2} +i \sqrt{2\sqrt 3} \frac{1}{2}$
|
The general procedure (for square roots) goes as follows.
We want the square roots of $a+ib$ (where at least one of $a$ or $b$ is non-zero).
Note that
$$a+ib=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}+i\frac{b}{\sqrt{a^2+b^2}}i\right).\tag{$1$}$$
For brevity, write $x=\frac{a}{\sqrt{a^2+b^2}}$ and $y=\frac{b}{\sqrt{a^2+b^2}}$.
Since $x^2+y^2=1$, there is a unique $\theta$ in the interval $[0,2\pi)$ such that $x=\cos\theta$ and $y=\sin\theta$.
In your example, $x=\frac{1}{2}$ and $y=\frac{\sqrt{3}}{2}$, so, from "special angles" we recognize that $\theta=\pi/3$.
Now the square roots of $\cos\theta+i\sin\theta $ are $\pm(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})$, and from this we find that the square roots of $a+bi$ are
$$\pm\sqrt[4]{a^2+b^2}\left(\cos\tfrac{\theta}{2}+i\sin\tfrac{\theta}{2}\right).$$
Added: In language close to the language used in the post, let $w=a+bi$. We want to solve the equation $z^2=w$. Let $z=r(\cos\phi+i\sin\phi)$, where $r$ is positive. Then $z^2=r^2(\cos 2\phi+i\sin 2\phi)$. Comparing with the expression $(1)$, we find that $r^2=\sqrt{a^2+b^2}$ and therefore $r=\sqrt[4]{a^2+b^2}$.
Also, we want $\cos2\phi +i\sin 2\phi=\cos\theta+i\sin\theta$. Obviously $\phi=\frac{\theta}{2}$ works. But also, more subtly, so does $\phi=\frac{\theta}{2}+\pi$, which gives you the negative of the square root picked out by $\phi=\frac{\theta}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/148871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
calculate the volume of a zone a and b are positive numbers.
Let W be a zone between the surface $$z=2ax + 2by$$ and the parabolic $$z=x^2 + y^2.$$
I need to show that $$\mu(W)=1/2*\pi*R^4$$
I'm not really sure how to begin this.
Would appreciate your help.
|
Find the intersection between the plane and paraboloid:
$$
z = x^2 + y^2 = 2ax + 2by
$$
Or
\begin{align*}
x^2 - 2ax + y^2 - 2by &= 0 \\
(x^2 -2ax + a^2) + (y^2 - 2by + b^2) &= a^2 + b^2 \\
(x - a)^2 + (y - b)^2 &= a^2 + b^2
\end{align*}
In the $xy$ plane, this is a disk centered at $(a, b)$ with $R = \sqrt{a^2 + b^2}$:
$$
\mathcal{A} = \{(x, y) \in \mathbb{R^2} : (x - a)^2 + (y - b)^2 \le a^2 + b^2
\}
$$
Here is a plot of the plane, paraboloid and their intersection. Notice how the plane is on top of the paraboloid inside the intersection:
The volume of the region between a surface define by a function $z = f(x, y)$ and part of the $xy$ plane can be calculated via double integrals.
For the region defined in this question, we need to calculate the volume of the region below the plane, the one below the paraboloid, and then subtract to get the volume we want:
\begin{align*}
V &= \iint_{\mathcal{A}} ((2ax + 2by) - (x^2 + y^2)) \, dxdy
\end{align*}
Since the domain of integration is a disk, we can use polar coordinates to evaluate the integral:
\begin{align*}
x &= a + r\cos\theta \\
y &= b + r\sin\theta
\end{align*}
The corresponding Jacobian is:
$$
\frac{\partial(x, y)}{\partial(r, \theta)} = r
$$
Plug to get:
\begin{align*}
V &= \int_0^{2\pi} \int_0^{\sqrt{a^2+b^2}} \left((2a(a + r\cos\theta) + 2b(b + r \sin\theta)) - \left((a + r\cos\theta)^2 + (b + r\sin\theta )^2\right)\right) r \, drd\theta
\end{align*}
This iterated integral takes a bit of work to evaluate, but it should be straightforward. Evaluate to find that:
$$
V = \frac{1}{2} \pi (a^2 +b^2)^2
$$
You can also evaluate the integral without switching to polar coordinates. Notice that as $y$ changes in the range:
$$
y \in \left[b - \sqrt{a^2+b^2}, b + \sqrt{a^2+b^2}\right]
$$
$x$ changes in the range:
$$
x \in \left[a - \sqrt{a^2+b^2-(y-b)^2}, a + \sqrt{a^2+b^2-(y-b)^2}\right]
$$
Use these as the bounds of the double integral, and you will arrive at the same result. The integral will be a bit more difficult to work with, however.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/149645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to solve this equation, when the unknown variable just disappears? This problem
$\sqrt{1-x^2} + \sqrt{3+x^2} = 2$
has the solution $x = 1$ and $x = -1$.
However, I always get stuck like this:
*
*$1-x^2 + 3+x^2 = 4$
*$4 = 4$
How do I isolate that darn unknown?
|
Square both sides of original equation: $$1-x^2+3+x^2+2\sqrt{(1-x^2)(3+x^2)}=4\Longrightarrow\sqrt{(1-x^2)(3+x^2)}=0\Longrightarrow x =\pm1$$ since $\,3+x^2=0\,$has no real solutions
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/149843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
limit of power of fraction of sums of sines Find the following limit:
$$\lim_{n\to\infty} \left(\frac{{\sin\frac{2}{2n}+\sin\frac{4}{2n}+\cdot \cdot \cdot+\sin\frac{2n}{2n}}}{{\sin\frac{1}{2n}+\sin\frac{3}{2n}+\cdot \cdot \cdot+\sin\frac{2n-1}{2n}}}\right)^{n}$$
I thought of some $\sin(x)$ approximation formula, but it doesn't seem to work.
|
$$\sum_{k=1}^n \sin \left(a + (k-1)d \right) = \dfrac{\sin(dn/2) \sin(a+d(n-1)/2)}{\sin(d/2)}$$
In your case, for the numerator $a = \dfrac{2}{2n}$ and $d = \dfrac{2}{2n}$. Hence, the numerator is $$ \dfrac{\sin(1/2) \sin(2/2n+2/2n \times (n-1)/2)}{\sin(1/2n)} = \dfrac{\sin(1/2) \sin((n+1)/(2n))}{\sin(1/2n)}$$
Similarly, the denominator gives $$\dfrac{\sin^2(1/2)}{\sin(1/2n)}$$
Hence, $$\dfrac{\dfrac{\sin(1/2) \sin((n+1)/(2n))}{\sin(1/2n)}}{\dfrac{\sin^2(1/2)}{\sin(1/2n)}} = \dfrac{\sin((n+1)/(2n))}{\sin(1/2)} = \dfrac{\sin(1/2) \cos(1/2n) + \cos(1/2) \sin(1/2n)}{\sin(1/2)}$$
The series expansion at $n= \infty$ gives us $$1 + \dfrac{\cot(1/2)}{2n} - \dfrac1{8n^2} + \mathcal{O}(1/n^3)$$
Hence, the desired limit is $$\lim_{n \rightarrow \infty} \left(\dfrac{\sin((n+1)/(2n))}{\sin(1/2)} \right)^n = \lim_{n \rightarrow \infty} \left(1 + \dfrac{\cot(1/2)}{2n} - \dfrac1{8n^2} + \mathcal{O}(1/n^3) \right)^n\\ = \exp \left(\dfrac12 \cot\left( \dfrac12 \right) \right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/151216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $? How can I simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $ to rectangular form $z = a+bi$?
(Note: Wolfram Alpha says the answer is $z=-8$. My professor says the answer is $z=\pm8$.)
I've tried to figure this out for a couple hours now, but I'm getting nowhere.
Any help is much appreciated!
|
So this can be written as $$4^{3/2} \cdot (-\frac{1}{2} + \frac{\sqrt{3}}{2} i)^{3/2} = 8 \cdot (\cos\frac{2\pi}{3} + i\cdot \sin\frac{2\pi}{3})^{3/2}$$
and $4^{3/2} =8$. Use De moivre now. And you can also pull out $-4$ and get going. Hence your $z = \pm{8}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/152989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
Summation of $ \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots$ till $n$ terms What is the pattern in the following?
*
*Sum to $n$ terms of the series: $$ \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots$$
|
Hint:
Write it as $(1-{1\over2})+(1-{1\over4})+(1-{1\over 8})+\cdots+(1-{1\over 2^n}).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/153302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Solving $x^4-y^4=z^2$ I have a question
show that $x^4- y^4 = z^2$ has no nontrivial solution where $x$, $y$ and $z$ are nonzero integers
I tried infinite descent to find solution but I could not find it. square of a number in mod 4 is 1 or 0 I also tried to use but got nothing.
Can you help?
thanks
|
Suppose $z^2=y^4-x^4$ with $xyz\not=0$ for the smallest possible value of $y^4$. First we rewrite the
equation as $y^4=x^4+z^2$ so that $\{z,x^2,y^2\}$ is a Pythagorean triple.
It must be primitive, since if some prime $p$ divides $\gcd(x^2,y^2)$, then
$p\,|\,y^2$ implies $p\,|\,y$ which gives $p^4\,|\,y^4$.
Similarly, $p^4\,|\,x^4$, so $p^4\,|\,z^2$. This implies $p^2\,|\,z$, so that
$\left({z/p^2}\right)^2=\left({y/p}\right)^4-\left({x/p}\right)^4$ is a smaller solution.
The Pythagorean triple $z,x^2,y^2$ is primitive and there are two cases:
If $x$ is even, then for some $m>n$, $(m,n)=1$,
and $m\not\equiv n \pmod2$ we have
$$ z=m^2-n^2,\quad x^2=2mn,\quad y^2=m^2+n^2.$$
Since $m,n$ have opposite parity, we can let $o$ denote the odd number and $e$ the even number among $\{m,n\}$.
The primitive Pythagorean triple $\{n,m,y\}$ gives $$o=t^2-s^2,\quad e=2st,\quad y=t^2+s^2,$$ for some
$t>s$, $(t,s)=1$, and $t\not\equiv s\pmod2$. The formula for $x^2$ now
gives $$(x/2)^2=ts(t^2-s^2)$$ which expresses the product of three relatively prime numbers as a square.
That means all three of them are squares: $s=u^2$, $t=v^2$, and $t^2-s^2=w^2$.
In other words, $v^4-u^4=w^2$ is another
solution to our equation, and it is smaller, since $v^4<t^2+s^2=y\leq y^4$.
If $x$ is odd, then for some $m>n$, $(m,n)=1$,
and $m\not\equiv n\pmod2$ we have
$$ x^2=m^2-n^2,\quad z=2mn,\quad y^2=m^2+n^2.$$
In this case $m^4-n^4=(xy)^2$ is a smaller solution, since $m^4<(m^2+n^2)^2=y^4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/153546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
}
|
Diophantine equation $a^2+b^2=c^2+d^2$ I was reasonably certain I've seen this before, but I was wondering how to solve the Diophantine equation
$$a^2+b^2=c^2+d^2$$
I tried a web search and found nothing on this one. I'm trying to avoid another library trip to a less than local library (maybe I should have taken better notes on that chapter...).
I'm not quite sure how to handle this one. The only thing I can figure out with this equation, if I remember correctly, is that the sum on either side may only contain prime factors of 2 or odd primes congruent to 1 mod 4. And if I don't want a and b equal to c and d, the sum can't be prime as I believe a prime congruent to 1 mod 4 can be represented as the sum of 2 squares in exactly one way. But that doesn't give me any insight into actually solving this problem.
|
This equation is quite symmetrical so formulas making too much can be written: So for the equation:
$X^2+Y^2=Z^2+R^2$
solution:
$X=a(p^2+s^2)$
$Y=b(p^2+s^2)$
$Z=a(p^2-s^2)+2psb$
$R=2psa+(s^2-p^2)b$
solution:
$X=p^2-2(a-2b)ps+(2a^2-4ab+3b^2)s^2$
$Y=2p^2-4(a-b)ps+(4a^2-6ab+2b^2)s^2$
$Z=2p^2-2(a-2b)ps+2(b^2-a^2)s^2$
$R=p^2-2(3a-2b)ps+(4a^2-8ab+3b^2)s^2$
solution:
$X=p^2+2(a-2b)ps+(10a^2-4ab-5b^2)s^2$
$Y=2p^2+4(a+b)ps+(20a^2-14ab+2b^2)s^2$
$Z=-2p^2+2(a-2b)ps+(22a^2-16ab-2b^2)s^2$
$R=p^2+2(7a-2b)ps+(4a^2+8ab-5b^2)s^2$
solution:
$X=2(a+b)p^2+2(a+b)ps+(5a-4b)s^2$
$Y=2((2a-b)p^2+2(a+b)ps+(5a-b)s^2)$
$Z=2((a+b)p^2+(7a-2b)ps+(a+b)s^2)$
$R=2(b-2a)p^2+2(a+b)ps+(11a-4b)s^2$
solution:
$X=2(b-a)p^2+2(a-b)ps-as^2$
$Y=2((b-2a)p^2+2(a-b)ps+(b-a)s^2)$
$Z=2((b-a)p^2+(3a-2b)ps-(a-b)s^2)$
$R=2(b-2a)p^2+2(a-b)ps+as^2$
solution:
$X=(p^2-s^2)b^2+a^2s^2$
$Y=b^2(p-s)^2-2abs^2+a^2s^2$
$Z=b^2(p-s)^2+2abps-a^2s^2$
$R=s^2(a-b)^2+2abps-p^2b^2$
number $a,b,p,s$ integers and sets us, and may be of any sign.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/153603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 9,
"answer_id": 2
}
|
Polynomials in Fourier trigonometric series I'm successively integrating $x^{n} \cos{k x}$ for increasing values of positive integer n. I'm finding:
$\frac{\sin{kx}}{k}$,
$\frac{\cos{kx}}{k^2}+\frac{x\sin{kx}}{k}$,
$\frac{2 x \cos{kx}}{k^2}+\frac{\left(-2+k^2 x^2\right)sin{kx}}{k^3}$,
$\frac{3 \left(-2+k^2 x^2\right) \cos{kx}}{k^4}+\frac{x \left(-6+k^2 x^2\right) \sin(kx)}{k^3}$
Is there a name for the sequence of polynomials: $x$, $2x$, $k^2x^2-2$, $3(k^2x^2-2)$, $x(k^2x^2-6)$ ... ?
Here is more:
$\frac{\sin{kx}}{k}$
$\frac{\cos{kx}}{k^2}+\frac{x \sin{kx}}{k}$
$\frac{2 x \cos{kx}}{k^2}+\frac{\left(-2+k^2 x^2\right) \sin{kx}}{k^3}$
$\frac{3 \left(-2+k^2 x^2\right) \cos{kx}}{k^4}+\frac{x \left(-6+k^2 x^2\right) \sin{kx}}{k^3}$
$\frac{4 x \left(-6+k^2 x^2\right) \cos{kx}}{k^4}+\frac{\left(24-12 k^2 x^2+k^4 x^4\right) \sin{kx}}{k^5}$
$\frac{5 \left(24-12 k^2 x^2+k^4 x^4\right) \cos{kx}}{k^6}+\frac{x \left(120-20 k^2 x^2+k^4 x^4\right) \sin{kx}}{k^5}$
$\frac{6 x \left(120-20 k^2 x^2+k^4 x^4\right) \cos{kx}}{k^6}+\frac{\left(-720+360 k^2 x^2-30 k^4 x^4+k^6 x^6\right) \sin{kx}}{k^7}$
|
You could also consider the exponential generating function
$$ \sum_{n=0}^\infty \dfrac{t^n}{n!} \int_0^x s^n e^{iks}\ ds = \int_0^x e^{(t+ik)s}\ ds = \dfrac{e^{(t+ik)x} - 1}{t+ik}$$
This is the product of $e^{(t+ik)x}-1 = -1 + \sum_{n=0}^\infty \dfrac{t^n x^n}{n!} e^{ikx}$ and $\dfrac{1}{t+ik} = \sum_{n=0}^\infty (-1)^n \dfrac{t^n}{(ik)^{n+1}}$, so
$$\int_0^x s^n e^{iks}\ ds = n! \left(\dfrac{1}{(-ik)^{n+1}} + \sum_{j=0}^n e^{ikx} \dfrac{(-1)^{n-j}}{j! (ik)^{n-j+1}} x^j\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/153795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
tough series involving digamma I ran across a series that is rather challenging. For kicks I ran it through Maple and it gave me a conglomeration involving digamma. Mathematica gave a solution in terms of Lerch Transcendent, which was worse yet. Perhaps residues would be a better method?.
But, it is $$\sum_{k=1}^{\infty}\frac{(-1)^{k}(k+1)}{(2k+1)^{2}-a^{2}}.$$
The answer Maple spit out was:
$$\frac{a+1}{16a}\left[\psi\left(\frac{3}{4}-\frac{a}{4}\right)-\psi\left(\frac{-a}{4}+\frac{1}{4}\right)\right]+\frac{a-1}{16a}\left[\psi\left(\frac{3}{4}+\frac{a}{4}\right)-\psi\left(\frac{1}{4}+\frac{a}{4}\right)\right]+\frac{1}{a^{2}-1}.$$
Is it possible to actually get to something like this by using $\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{k+a}\right]=\gamma+\psi(a+1)?$
I tried, but to no avail. But, then again, maybe it is too cumbersome.
i.e. I tried expanding it into
$\frac{k+1}{(2k+1)^{2}-a^{2}}=\frac{-1}{4(a-2k-1)(2k+1)}-\frac{1}{4(a-2k-1)}+\frac{1}{4(a+2k+1)(2k+1)}+\frac{1}{4(a+2k+1)}$
then using $\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{k-\frac{1}{4}-\frac{a}{4}}\right]=\psi\left(\frac{3}{4}-\frac{a}{4}\right)$ and so on, but it did not appear to be anywhere close to the given series.
On another point, can it be done using residues?. By using $$\frac{\pi csc(\pi z)(z+1)}{(2z+1)^{2}-a^{2}}.$$
This gave me a residue at $\frac{a-1}{2} and \frac{-(a+1)}{2}$ of
$\frac{-\pi}{a-1}sec(a\pi/2)$ and $\frac{\pi}{a+1}sec(\pi a/2)$
Taking the negative sum of the residues, it is $\frac{2\pi}{(a-1)(a+1)}sec(a\pi/2)$
By subbing in k=0 into the series, it gives $\frac{-1}{a^{2}-1}$.
I try adding them up and finding the sum, but it does not appear to work out.
Any suggestions?. Perhaps there is another method I am not trying?. There probably is. Thanks a million.
|
First, group the consecutive oscillating terms together:
$$\sum_{k=1}^{\infty}\frac{(-1)^{k}(k+1)}{(2k+1)^{2}-a^{2}}=\sum_{k=0}^\infty \left(\frac{2k+1}{(4k+1)^2-a^2}-\frac{2k+2}{(4k+3)^2-a^2}\right)-\frac{2(0)+1}{(2(0)+1)^2-a^2}$$
Next, invoke partial fraction decomposition and solve for coefficients:
$$
\frac{2k+1}{(4k+1)^2-a^2} = \frac{a+1}{16a}\frac{1}{k+\frac{1-a}{4}}+\frac{a-1}{16a}\frac{1}{k+\frac{1+a}{4}},$$
and similarly
$$\frac{2k+2}{(4k+3)^2-a^2}=\frac{a+1}{16a}\frac{1}{k+\frac{3-a}{4}}+\frac{a-1}{16a}\frac{1}{k+\frac{3+a}{4}}.$$
Hence we are left with
$$\frac{a+1}{16a}\sum_{k=0}^\infty \left(\frac{1}{k+\frac{1-a}{4}}-\frac{1}{k+\frac{3-a}{4}}\right)+\frac{a-1}{16a}\sum_{k=0}^\infty\left(\frac{1}{k+\frac{1+a}{4}}-\frac{1}{k+\frac{3+a}{4}}\right)+\frac{1}{a^2-1}$$
$$=\begin{array}{c} \frac{a+1}{16a}\sum_{k=0}^\infty \left(\left(\frac{1}{k+1}-\frac{1}{k+\frac{3-a}{4}}\right)-\left(\frac{1}{k+1}-\frac{1}{k+\frac{1-a}{4}}\right)\right) \\ +\frac{a-1}{16a}\sum_{k=0}^\infty\left(\left(\frac{1}{k+1}-\frac{1}{k+\frac{3+a}{4}}\right)-\left(\frac{1}{k+1}-\frac{1}{k+\frac{1+a}{4}}\right)\right)+\frac{1}{a^2-1} \end{array}$$
$$=\frac{a+1}{16a}\left[\psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{1-a}{4}\right)\right]+\frac{a-1}{16a}\left[\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{1+a}{4}\right)\right]+\frac{1}{a^{2}-1}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/153845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
}
|
Simplify this expression with nested radical signs My question is-
Simplify:
$$\frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}$$
|
Let's follow this elementary way to work out things:
$$\frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}= \frac1{\sqrt{({\sqrt{7}-\sqrt{5}})^2}}-\frac2{\sqrt{({\sqrt{7}+\sqrt{3}})^2}}-\frac1{\sqrt{({\sqrt{5}+\sqrt{3}})^2}}=\frac1{({\sqrt{7}-\sqrt{5}})}-\frac2{({\sqrt{7}+\sqrt{3}})}-\frac1{({\sqrt{5}+\sqrt{3}})}=$$
$$\frac{\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{2} =\sqrt3.$$
The proof is complete.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/155583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
}
|
What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 x}{ \cos x}dx\\
&=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\
&=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\
&=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\
&=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\
&=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\
&=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\
&=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\
&=\ln|\sec x| + \frac{\cos 2x}{4} + C
\end{align}$$
This is the wrong answer, I have went through and back and it all seems correct to me.
|
A simple $u$-substitution will work even better, IMO. ${\cos^2\theta\tan^3\theta }$ yields ${\frac {\sin^3}{\cos\theta}}$ So starting from there $${\int\frac{\sin^2\theta}{\cos\theta}\sin\theta d\theta}$$ then $${\int\frac{1-\cos^2\theta}{\cos\theta}\sin\theta d\theta}$$ Let
$u={\cos \theta}$ then ${du=-\sin\theta d\theta}$ $${-\int\frac {1-u^2}{u}du}$$ let ${v=1-u^2 }$ ${dv=-2u}$
let ${dw=1/u}$, ${w=\ln u}$
thus $${\ln u(1-u^2)+\int 2u\ln u}\,du$$
then
let ${v=\ln u}$, ${dv=u^{-1}}$
let ${dw=2u}$, ${w=u^2}$
then
$$\begin{align*}
{\ln u u^2-\int \frac{u^2}{u}\,du}\\{\ln u u^2-\int u\,du}\\
{\ln u(1-u^2)+\ln u u^2-1/2u^2}
\end{align*}$$
then
$$\begin{align*}{\ln u(1-u^2+u^2)-1/2u^2}\\
{-(\ln u-1/2u^2)}
\end{align*}
$$
substituting back yields $${-\ln\cos\theta+1/2\cos^2\theta+c}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/155829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 6
}
|
Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$ Compute
$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$$
|
Note
\begin{align}
&\int_0^1\frac{\ln (x+1)}{x^2+1}dx
=\int_0^1\underset{x\to\frac{1-x}{1+x}}{\frac{\ln \frac{x+1}{\sqrt{x^2+1}}}{x^2+1}}dx
+ \int_0^1\frac{\ln \sqrt{x^2+1}}{x^2+1}dx\\
&= \int_0^1\frac{\ln \frac{\sqrt2}{\sqrt{x^2+1}}}{x^2+1}dx
+ \int_0^1\frac{\ln \sqrt{x^2+1}}{x^2+1}dx = \frac{\ln2}2\int_0^1\frac{dx}{x^2+1}=\frac\pi8\ln2
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/155941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "102",
"answer_count": 8,
"answer_id": 4
}
|
Prove that $\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$ I need to prove that for any real number $a>1$ and $b>1$ the following inequality is true:
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}\geq8$$
|
$$\frac{{a}^{2}}{b-1}+\frac{{b}^{2}}{a-1}=\frac{a^3-a^2+b^3-b^2}{(a-1)(b-1)}=\frac{(a+b)(a^2+b^2-ab)-(a^2+b^2)}{(a-1)(b-1)}\ge\frac{(a+b)ab-(a^2+b^2)}{(a-1)(b-1)}=\frac{a^2(b-1)+b^2(a-1)}{(a-1)(b-1)}=\frac{a^2}{a-1}+\frac{b^2}{b-1}=\frac{a^2-1+1}{a-1}+\frac{b^2-1+1}{b-1}=a+1+\frac{1}{a-1}+b+1+\frac{1}{b-1}=a-1+\frac{1}{a-1}+2+b-1+\frac{1}{b-1}+2\ge2+2+2+2=8$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/157688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 1
}
|
Linear transformation for projection of a point on a line This is what my textbook wants me to do:
The matrix of the linear transformation $P_L$ that projects $\mathbb{R}^2$ on de straight line $l \leftrightarrow y = mx$ is:
\begin{pmatrix}
\frac{1}{1+m^2} & \frac{m}{1+m^2} \\
\frac{m}{1+m^2} & \frac{m^2}{1+m^2} \\
\end{pmatrix}
And I get this picture, which probably is their to inspire me:
Now, I tried to tackle this with the perpendicular line between A and C, but I got nowhere:
\begin{align*}
A &= (x_A, y_A) \\
&\Downarrow \\
AC \leftrightarrow y - y_A &= \frac{-1}{m}(x - x_A) \\
&\Downarrow \\
C &\leftrightarrow \begin{cases}
y = \frac{-x}{m} + \frac{x_A}{m} + y_A \\
y = mx \\
\end{cases}
\end{align*}
But that's were I am stuck, I can't get it to the form
\begin{align*}
\begin{bmatrix}
x_C \\
y_C
\end{bmatrix} &= \begin{bmatrix}
\frac{1}{1+m^2} & \frac{m}{1+m^2} \\
\frac{m}{1+m^2} & \frac{m^2}{1+m^2} \\
\end{bmatrix}
\begin{bmatrix}
x_A \\
y_A
\end{bmatrix}
\end{align*}
I am probably not doing it right. All tips are greatly appreciated.
|
You are off to a great start!
If you substitute your second equation into your first, you find $$mx=-\frac{x}{m}+\frac{x_A}{m}+y_A,$$ so $$\frac{1+m^2}{m}x=\frac{x_A}{m}+y_A,$$ and so $$x=\frac{1}{1+m^2}x_A+\frac{m}{1+m^2}y_A,\tag{$\star$}$$ which corresponds to what your first row should be.
Once you've found that, use $(\star)$ to substitute into your second equation, and you readily see that $$y=\frac{m}{1+m^2}x_A+\frac{m^2}{1+m^2}y_A,$$ which corresponds to the second row.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/159603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Plotting a complex argument arc I am having trouble sketching a complex argument arc
$$ \text{Sketch the following on an arcand diagram:}\\ \arg\left(\frac{w+1}{w}\right)=\frac{\pi}{6}$$
I've tried to devise a method on my own looking at questions and answers but it has failed me on this specific question so I require some help.
What I have been doing so far is I was writing $$\arg\left(\frac{x+iy+1}{x+iy}\right)$$
$$x+1=0\\y=0\\y=0 \\ x=0$$
Getting two points (-1,0) and (0,0). So I take the two points and connect them with an arc.
However the solution of the exercise was this:
However my answer was exactly the opposite on the y-axis. Can anyone show me a method and the logic behind the drawing of the arc, and the arc of any other similar question?
|
Taking $w = x+iy$, we get that $$\text{Arg} \left( \dfrac{1+w}w\right) = \text{Arg} \left( 1+\dfrac1w\right) = \text{Arg} \left( 1+\dfrac{x-iy}{x^2+y^2}\right) = \text{Arg} \left( \dfrac{x^2+y^2+x}{x^2+y^2} - i \dfrac{y}{x^2+y^2}\right)$$
Hence, we need $$\tan(\pi/6) = -\dfrac{y}{x^2+y^2+x}$$
$$x^2 + y^2 + x + y\sqrt{3} = 0 \text{ i.e. } \left(x + \dfrac12 \right)^2 + \left(y + \dfrac{\sqrt{3}}2 \right)^2 = 1$$
However, note that we have the argument in the first quadrant and hence $$\dfrac{x^2+y^2+x}{x^2+y^2} >0 \text{ & } -\dfrac{y}{x^2+y^2} > 0$$
This gives us that $y < 0$.
Hence, the curve we are interested in is $$\left(x + \dfrac12 \right)^2 + \left(y + \dfrac{\sqrt{3}}2 \right)^2 = 1 \text{ with } y< 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/160900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
How to get $ \cot(\theta/2)$ from $ \frac {\sin \theta} {1 - \cos \theta} $? According to wolfram alpha, $\dfrac {\sin \theta} {1 - \cos \theta} = \cot \left(\dfrac{\theta}{2} \right)$.
But how would you get to $\cot \left(\dfrac{\theta}{2} \right)$ if you're given $\dfrac {\sin \theta} {1 - \cos \theta}$?
|
It may not seem obvious at first, but substitute $\theta = 2\varphi$ first because we want to work with integer multiples of the angle rather than fractional multiples.
$$\begin{array}{rcl}
\dfrac{\sin \theta}{1 - \cos \theta} &=& \dfrac{\sin 2\varphi}{1 - \cos 2\varphi}
\end{array}$$
Then we use double angle formula.
$$\begin{array}{rcl}\dfrac{\sin \theta}{1 - \cos \theta} &=& \dfrac{2\sin \varphi\cos\varphi}{1-\left(1-2\sin^2 \varphi\right)}\\
&=& \dfrac{2\sin\varphi\cos\varphi}{2\sin^2\varphi}\\
&=& \dfrac{\color{blue}{2\sin \varphi} \cos \varphi}{\color{blue}{2\sin\varphi} \sin \varphi}\\
&=& \dfrac{\cos \varphi}{\sin\varphi}\\
&=& \cot \varphi
\end{array}$$
Recall that $\theta = 2\varphi$, so dividing by 2 on both sides gives $\varphi = \dfrac{\theta}{2}$.
We can conclude that
$$\begin{array}{rcl}\dfrac{\sin \theta}{1 - \cos\theta} &=& \cot \varphi\\
&=& \boxed{\cot \left(\dfrac{\theta}{2}\right)}
\end{array}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/160982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
}
|
How do I differentiate this? (logarithm & chain) I keep getting wrong results when trying to differentiate this:
${\partial \over \partial x} \ln{(x - \sqrt{x^2+a^2})}$
Thanks for hints!
|
An other method: $y=\ln |x- \sqrt{x^2+a^2}|$ gives :
$$\begin{align*} e^y &= x-\sqrt{x^2+a^2}\\ (e^y-x)^2 &=x^2+a^2 \\ 2(e^y-x) \left(e^y\frac{dy}{dx} -1\right) &=2x \\ e^y\frac{dy}{dx}&= 1+\frac{x}{e^y-x}=\frac{e^y}{e^y-x}\end{align*}$$
Since : $e^y-x=-\sqrt{x^2+a^2}$, this gives :
$$\frac{dy}{dx}= - \frac{1}{\sqrt{x^2+a^2}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/161039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
If $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $, then either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ For $a, b = 1, 2, 3, \cdots$, let $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $. Then prove that either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ holds.
|
$c\geq a\Rightarrow 1/2a+1/2b\leq 1/a\Rightarrow1/2b\leq1/2a\Rightarrow 1/2a+1/2b\geq1/b\Rightarrow c\leq b$. Similarly do when $c\leq a$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/162096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Parametric equations and area I am not sure how to work this one out.
I am suppose to find the area of this parametric equation.
$$y = b\sin\theta, x = a\cos\theta$$
$$0 \leq 0 \leq 2\pi$$
I set up the equation in the memorized formula.
$$\int_0^{2\pi} \sqrt{1 + \left(\frac{b\cos\theta}{-a\sin\theta}\right)^2}d\theta$$
$$\int_0^{2\pi} \sqrt{1 + \frac{b^2\cos^2\theta}{a^2\sin^2\theta}}d\theta$$
$$\int_0^{2\pi} \sqrt{1 + \frac{b^2}{a^2}\cdot \csc^2\theta}d\theta$$
From here I am at a loss of what to do, I tried some trig subsitution but I do not think that will work and it only seems to complicate the problem.
|
Note that $\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=\cos^2\theta + \sin^2\theta=1.$ We then recognize that this describes an ellipse, which has area $\pi a b$.
EDIT: How to find the area of the ellipse.
There are many many proofs of this, but the easiest one you might find in a single-variable calculus course is as follows.
We will find the area of the ellipse in the first quadrant and quadruple it (we assume $x,y\geq0$ for what follows). Solve for $y$ in terms of $x$.
$$
1 = \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2
$$
$$
\iff \left(\frac{y}{b}\right)^2= 1-\left(\frac{x}{a}\right)^2
$$
$$
\iff \frac{y}{b} = \sqrt{1-\left(\frac{x}{a}\right)^2}
$$
$$
y=b\sqrt{1-\left(\frac{x}{a}\right)^2}.
$$
Then we have
$$
\frac{A}{4} = \int_0^ab\sqrt{1-\left(\frac{x}{a}\right)^2}dx.
$$
Substitute $u = \frac{x}{a}$ so $dx=a\ du$ and the limits go from $u=0$ to $u=1$,
$$
\frac{A}{4} = ab\int_0^1\sqrt{1-u^2}du.
$$
At this point you could interpret the integral on the right as the the area of a quarter of a circle of radius 1, $\pi/4$. If not, we continue by making the trigonometric substitution $u = \sin t$, so $\sqrt{1-u^2}=\cos t$ and $du=\cos t\ dt$ where $t$ goes from $0$ to $\pi/2$,
$$
\frac{A}{4} = ab\int_0^{\pi/2} \cos^2t dt.
$$
Now we use the formula $\cos^2t = \frac{1}{2}(1+\cos(2t))$,
$$
\frac{A}{4} = \frac{ab}{2}\int_0^{\pi/2}(1+\cos(2t))dt
$$
$$
=\frac{ab}{2}\left(t+\frac{1}{2}\sin(2t) \right)_0^{\pi/2}
$$
$$
=\frac{ab}{2}\left(\left(\frac{\pi}{2}+0\right) - \left(0 + 0\right) \right)
$$
$$
\frac{\pi a b}{4}.
$$
Rearranging then gives us $A= \pi a b$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/163014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Factoring multivariate polynomial I'm trying to factor
$$x^3+x^2y-x^2+2xy+y^2-2x-2y \in \mathbb{Q}[x,y].$$
The hint for the exercise is to use the recursive multivariate polynomial form. So I'm using $\mathbb{Q}[x][y]$:
$$ x^3 + x^2(y-1) + x^1(y-2) + x^0(y^2-2y) $$
At this point I am stuck. Are their any general techniques to do this by hand?
|
Let’s go back to the original polynomial,
$$x^3+x^2y-x^2+2xy+y^2-2x-2y\;.\tag{1}$$
That $2xy$ looks a lot like the middle term of $(x+y)^2$, and the $-2x-2y$ can certainly be written nicely in terms of $x+y$, so let’s try something along those lines. $(x+y)^2=x^2+2xy+y^2$, and we have the $2xy+y^2$, but instead of $x^2$, we have $-x^2$. In other words, we have $(x+y)^2-2x^2$. That’s not entirely promising, but let’s see where it goes. We can rewrite $(1)$ as
$$\begin{align*}
(x+y)^2&-\;2x^2-2(x+y)+x^3+x^2y\\\\
&=(x+y)^2-2(x+y)+x^2(x+y-2)\\\\
&=(x+y-2)(x+y)+x^2(x+y-2)\\\\
&=(x+y-2)(x^2+x+y)\;.
\end{align*}$$
Added: Alternatively, following the hint, interchange the rôles of $x$ and $y$, rewriting $(1)$ as $$y^2+(x^2+2x-2)y+(x^3-x^2-2x)\;.\tag{2}$$
Remember that linear factors of a polynomial correspond to zeroes. Set $(2)$ equal to $0$ and use the quadratic formula to solve for $y$:
$$\begin{align*}
&(x^2+2x-2)^2-4(x^3-x^2-2x)\\
&\qquad=x^4+4x^3-8x+4-4x^3+4x^2+8x\\
&\qquad=x^4+4x^2+4\\
&\qquad=(x^2+2)^2\;,
\end{align*}$$
so $$y=\frac{-x^2-2x+2\pm(x^2+2)}2\;,$$
and $y=-x^2-x$ or $y=-x+2$. Thus, $(2)$ factors as $$\Big(y-(-x^2-x)\Big)\Big(y-(-x+2)\Big)=(y+x^2+x)(y+x-2)\;.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/163417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find the value of $(a^3 + b^3 + c^3)/(abc)$ if $a/b + b/c + c/a = 1$.
Find the value of $$\frac{a^3+b^3+c^3}{abc}\qquad\text{ if }\quad \frac ab + \frac bc + \frac ca = 1.$$
I tried using Cauchy's inequality but it was of no help. Please guide me.
$a, b, c$ are real.
|
Let
$$\begin{eqnarray}
u&=&a+b+c\\
v&=&a^3+b^3+c^3\\
s&=&a^2b + b^2 c + c^2 a\\
t &=& a \, b\, c
\end{eqnarray}$$
These are related by
$$u^3 = v + 3 s + t $$
Further, we are given
$$ \frac ab + \frac bc + \frac ca = \frac st =1$$
so $$u^3 = v + 4 t$$
And our target is
$$\frac{a^3+b^3+c^3}{abc}=\frac vt = \frac{u^3}{t}-4$$
This cannot be a single value, as we have a remaining degree of freedom.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/164623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
$X = \log_{12} 18$ and $Y= \log_{24} 54$. Find $XY + 5(X - Y)$ $X = \log_{12} 18$ and $Y= \log_{24} 54$. Find $XY + 5(X - Y)$
I changed the bases to 10, then performed manual addition/multiplication but it didn't yield me any result except for long terms. Please show me the way.
All I'm getting is $$\frac{\lg 18\lg54 + 5 \lg18\lg24 - 5\lg54\lg12}{\lg12\lg24} $$
|
Let $I=\dfrac{\log 18}{\log 12}\cdot\dfrac{\log 54}{\log 24}+5 \left( \dfrac{\log 18}{\log 12}-\dfrac{\log 54}{\log 24} \right)$. Also, let $\log 3=x $ and $\log 2=y$.
Then,
$$I= \frac{\log 3^2\cdot2}{\log 2^2\cdot 3}.\frac{\log 3^3 \cdot 2}{\log 2^3\cdot 3}+5\left(\frac{\log 3^2\cdot2}{\log 2^2\cdot3}-\frac{\log 3^3\cdot2}{\log 2^3\cdot3}\right)= \frac{2x+y}{2y+x}\cdot\frac{3x+y}{3y+x}+5 \left( \frac{2x+y}{2y+x}-\frac{3x+y}{3y+x} \right)$$
$$=\frac{6x^2+5xy+y^2+10x^2+35xy+15y^2-15x^2-35xy-10y^2}{(2y+x)(3y+x)}=\frac{x^2+5xy+6y^2}{x^2+5xy+6y^2}=1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/165667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
All integer solutions for $x^4-y^4=15$ I'm trying to find all the integer solutions for $x^4-y^4=15$.
I know that the options are $x^2-y^2=5, x^2+y^2=3$, or $x^2-y^2=1, x^2+y^2=15$, or $x^2-y^2=15, x^2+y^2=1$, and the last one $x^2-y^2=3, x^2+y^2=5$.
Only the last one is valid. $x^2+y^2=15$ is not solvable since the primes which have residue $3$ modulo $4$ is not of an equal power.
One particular solution for $x^2-y^2=3, x^2+y^2=5$, is $x_0=2, y_o=1$. How do I get to an expression of a general solution for this system?
Thanks!
|
$$(x^2-y^2)+(x^2+y^2)=2x^2=3+5=8\Rightarrow x^2=4\Rightarrow x=2,-2$$
$$(x^2+y^2)-(x^2-y^2)=2y^2=5-3=2\Rightarrow y^2=1\Rightarrow y=1,-1$$
Substitute the values to check that these are indeed the soloutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/166070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
}
|
Complex part of a contour integration not using contour integration A propos of a user's comment on this question, quoting Feynman to the effect that some integrals are only possible using contour integration, I wonder what the simplest example of such an integral might be. In particular, he spoke of integrals that were the complex part of a solution, divorced from the rest of the solution.
For me, something like
$$I = \int_0^\infty \frac{\ln x}{x^2+1}dx, $$
seems hard; it falls out as the complex part of contour integration using the residue theorem in the problem,
$$J = \int_0^\infty \frac{(\ln x)^2}{x^2+1}dx . $$ That is, we can use contour integration to evaluate J and end up with something like
$J + iI = \frac{\pi^3}{8}$ so we conclude $I = 0.$
Perhaps it is hard but I don't know that it can't be done without complex analysis [edit: Robjohn has shown it can be done without complex analysis]. So I would like to see one good example of the sort of thing Feynman might have had in mind.
Here is the original quote: "So Paul puts up this tremendous damn integral he had obtained by starting out with a complex function that he knew the answer to, taking out the real part of it and leaving only the complex part. He had unwrapped it so it was only possible by contour integration!"
In terms of this quote, do we have an example of what Paul did?
Thanks.
|
A change of variables yields
$$
\begin{align}
\int_0^\infty\frac{\log(x)^2}{1+x^2}\,\mathrm{d}x
&=\int_{-\infty}^\infty\frac{t^2}{1+e^{-2t}}e^-t\,\mathrm{d}t\\
&=2\int_0^\infty\frac{t^2}{1+e^{-2t}}e^{-t}\,\mathrm{d}t\\
&=2\int_{-\infty}^\infty t^2\left(e^{-t}-e^{-3t}+e^{-5t}-e^{-7t}+\dots\right)\,\mathrm{d}t\\
&=4\left(1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\dots\right)\tag{1}
\end{align}
$$
Define
$$
\xi(n)=\sum_{k=0}^\infty(-1)^k\frac{1}{(2k+1)^n}\tag{2}
$$
and consider the generating function of $\xi(n)$ for odd $n$:
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^\infty(-1)^k\frac{z^{2n+1}}{(2k+1)^{2n+1}}
&=\sum_{k=0}^\infty(-1)^k\frac{\frac{z}{2k+1}}{1-\left(\raise{1pt}{\frac{z}{2k+1}}\right)^2}\\
&=\frac{z}{2}\sum_{k=0}^\infty(-1)^k\left(\frac{1}{z+2k+1}-\frac{1}{z-2k-1}\right)\\
&=\frac{z}{2}\sum_{k=-\infty}^\infty\left(\frac{1}{z+4k+1}-\frac{1}{z+4k-1}\right)\\
&=\frac{z}{8}\sum_{k=-\infty}^\infty\left(\frac{1}{k+\frac{z+1}{4}}-\frac{1}{k+\frac{z-1}{4}}\right)\\
&=\frac{z}{8}\left(\pi\cot\left(\pi\frac{z+1}{4}\right)-\pi\cot\left(\pi\frac{z-1}{4}\right)\right)\\
&=\frac{\pi z}{4}\sec\left(\frac{\pi z}{2}\right)\\
&=\frac{\pi z}{4}+\frac{\pi^3z^3}{32}+\frac{5\pi^5z^5}{1536}+\dots\tag{3}
\end{align}
$$
where we used $(7)$ from this answer.
Putting together $(1)$, $(2)$, and $(3)$ yields
$$
\int_0^\infty\frac{\log(x)^2}{1+x^2}\,\mathrm{d}x=\frac{\pi^3}{8}\tag{4}
$$
Here is another method using contour integration.
In order to define $\log(z)$, consider the path $\gamma$ that runs from $0$ to $\infty$ just north of the real axis, circles the complex plane counter-clockwise, then returns from $\infty$ to $0$ just south of the real axis.
Adding the residues at $z=i$ and $z=-i$ gives
$$
\begin{align}
\int_\gamma\frac{\log(z)^3}{1+z^2}\mathrm{d}z
&=2\pi i\left(\frac{\left(\frac\pi2i\right)^3}{2i}+\frac{\left(\frac{3\pi}2i\right)^3}{-2i}\right)\\
&=\frac{13\pi^4}{4}i\tag{5}
\end{align}
$$
Computing the integral along the real axis yields
$$
\begin{align}
\int_\gamma\frac{\log(z)^3}{1+z^2}\mathrm{d}z
&=\int_0^\infty\frac{\log(x)^3-(\log(x)+2\pi i)^3}{1+x^2}\mathrm{d}x\\
&=\int_0^\infty\frac{-3\log(x)^22\pi i+\color{#C00000}{3\log(x)4\pi^2}+\color{#00A000}{8\pi^3i}}{1+x^2}\mathrm{d}x\\
&=-6\pi i\int_0^\infty\frac{\log(x)^2}{1+x^2}\mathrm{d}x+\color{#C00000}{0}+\color{#00A000}{4\pi^4i}\tag{6}
\end{align}
$$
Equating the real and imaginary parts of $(5)$ and $(6)$ gives
$$
\int_0^\infty\frac{\log(x)}{1+x^2}\mathrm{d}x=0\tag{7}
$$
and
$$
-6\pi i\int_0^\infty\frac{\log(x)^2}{1+x^2}\mathrm{d}x=-\frac{3\pi^4}{4}i\tag{8}
$$
which yields
$$
\int_0^\infty\frac{\log(x)^2}{1+x^2}\mathrm{d}x=\frac{\pi^3}{8}\tag{9}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/167304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
}
|
Finding the area of triangle if length of medians are given My question is:
In triangle ABC ,length of median from vertex A is $13$ , length of
median from vertex B is $14$ , length of median from vertex C is $15$.
Compute the area of triangle ABC.
|
One strategy, probably not optimal, is to find the lengths of the sides, and then use Heron's Formula.
Take a triangle $XYZ$, with sides $x$, $y$, $z$ as usual, and let $m$ be the length of the median from $Z$ to the side $XY$, which has length $z$.
Let $P$ be the midpoint of $XY$. We have divided our triangle into two triangles, $ZPX$ and $ZPY$. Let $\theta=\angle ZPX$ and $\phi=\angle ZPY$. Note that $\theta+\phi=\pi$. By the Cosine Law, we have
$$y^2=m^2+\left(\frac{z}{2}\right)^2-mz\cos\theta.$$
Also by the Cosine Law, we have
$$x^2=m^2+\left(\frac{z}{2}\right)^2-mz\cos\phi.$$
But $\cos\phi=-\cos\theta$. Add. The awkward cosine terms cancel, and we get
$$x^2+y^2=2m^2+2\left(\frac{z}{2}\right)^2,$$
which yields the result
$$m^2=\frac{1}{4}\left(2x^2+2y^2-z^2\right).$$
The rest is downhill. We go back to the triangle of the question. The three lengths of the medians give us three linear equations in the squares of the side lengths. If the sides of our triangle are $a$, $b$, and $c$, we get
$$4\cdot 13^2=2a^2+2b^2-c^2;\quad 4\cdot 14^2=2b^2+2c^2-a^2;\quad 4\cdot 15^2=2c^2+2a^2-b^2.$$
Solve. A simple way is to use the symmetry of the equations by first of all "adding" the above three equations to first find $a^2+b^2+c^2$.
Now that we have the sides, we can use Heron's Formula.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/168701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Finding all primes $p$ such that $\frac{(11^{p-1}-1)}{p}$ is a perfect square How to Find all primes $p$ such that $\dfrac{(11^{p-1}-1)}{p}$ is a perfect square
|
Let $\frac{a^{p-1}-1}{p}=b^2$ for some integer b.
If p=2, $a=2b^2+1$(11 is not of the form).
If p>2 & prime ,so must be odd=(2k+1) (say)
$(a^k+1)(a^k-1)=b^2(2k+1)$
Now, if a is odd (like 11), $a^k±1$ is even => b is even =2d(say).
$\frac{a^k+1}{2}\frac{a^k-1}{2}=d^2(2k+1)$
Now, $(\frac{a^k+1}{2},\frac{a^k-1}{2})$=1
So, either $\frac{a^k+1}{2}=m^2p$ and $\frac{a^k-1}{2}=n^2$ or vice versa.
Now, $n^2$≡0,1,4,9,5,3(mod 11) =>$11∤(2n^2-1)$
In other case,
If $11^k=2n^2+1$
n must be even=2m (say)=> $2n^2+1≡1(mod\ 8)$.
Now, 11≡3(mod 8) =>$11^2≡1(mod\ 8)$=>$11^{2r}≡1(mod\ 8)$ and $11^{2r+1}≡3(mod\ 8)$
So, k must even=2r(say).
The part problem becomes $(11^r)^2-2n^2=1$ which is well known Pell equation.
The minimum solution in natural number for $A^2-2B^2=1$ is (3,2).
So, A is $\sum_{s≥2t≥0}sC_{2t}3^{s-2t}(2\sqrt2)^{2t}$ where s is a natural number, can this be a power of 11?
Find here
An observation:
$a^k≡-1(mod\ p)=>a^{2k}≡1(mod\ p) $
Let $d=ord_pa$ => d|2k,but d∤k and d∤2 as k is even
=> a(=11) must be a primitive root of p to admit any solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/169935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
}
|
Induction Proof: $5^n + 5 < 5^{n+1}$ I am trying to prove for all natural $n$ that:
$$5^n + 5 < 5^{n+1}$$
I did the basic step with $n=1$ and inequality holds, I am now at the induction step:
$$5^{k+1} + 5 < 5^{k+2}$$
and I have no idea how to proceed from here. Can someone give me a clue?
|
Inequality holds for $n=1$.
For $n=k$
$$5^k + 5 < 5^{k+1}$$
We can write this as
$$5^{k+1} - (5^k + 5) = p$$
where $p$ is a positive integer.
$$5*5^k - 5^k - 5 = p$$
$$4*5^k - 5 = p$$
$$5^k=\frac{p+5}{4}$$
Now we have to prove it is true for $n = k+1$. So if $(5^{k+2} - (5^{k+1}+5))$ is positive then inequality holds true for $n=k+1$.
$$5^{k+2} - (5^{k+1}+5)$$
$$25*5^k - (5*5^{k}+5)$$
$$25*5^k - 5*5^{k}-5$$
$$20*5^k - 5$$
substitute value of $5^k$,
$$20*(\frac{p+5}{4}) - 5$$
$$5*(p+5)-5$$
$$5p+25-5$$
$$5p+20$$
$$5(p+4)$$
As $p$ is a positive integer, so $5(p+4)$ is also positive integer.
Thus, we can say that
$$5^{k+1}+5 < 5^{k+2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/171193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
How to solve this Pell's equation $x^{2} - 991y^{2} = 1 $ How to solve the following Pell's equation?
$$x^{2} - 991y^{2} = 1 $$
where $(x, y)$ are naturals.
The answer is $$x = 379,516,400,906,811,930,638,014,896,080$$
$$y = 12,055,735,790,331,359,447,442,538,767$$
I can't think of any way apart from brute force. Please help.
Also, what is the general way of solving any Pell's equation?
I read the wiki article on it but couldn't get any general method to solve it.
|
$$x^2 - 991y^2 = 1$$ -------(1)
Assuming $y \not= 0$ divide both sides of the equation by $y^2$
i.e. $$(x/y)^2 - 991 = (1/y)^2$$
i.e. $(x/y)^2 - (1/y)^2 = 991$
i.e. $(x/y - 1/y)(x/y + 1/y) = (1)(991)$
i.e. $(x/y - 1/y)(x/y + 1/y) = (496 - 495)(496 + 495)$
i.e. $x/y = 496$ and $1/y = 495$
i.e. $x = (496/495)$ and $y = 1/495$
The equation represents a hyperbola of the form $$x^2/a^2 - y^2/b^2 = 1$$
i.e. the equation is $$x^2/(1^2) - y^2/(\sqrt {1/991})^2 = 1$$
foci of this hyperbola are the co ordinates $(\pm c, 0)$
Here $c = \sqrt{a^2 + b^2}$
The asymptotes are $y = \pm (b/a)x$
i.e. The asymptotes are $y = \pm (1/\sqrt{991})x$
There exists a $CONJUGATE$ $HYPERBOLA$ of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$, for the $HYPERBOLA$ $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
Here the equation of the $conjugate$ $hyperbola$ is $x^2 - 991(y^2) = -1$
$$x^2 - 991y^2 = -1$$ -------(2)
Now solve equations (1) and (2) for $x$ and $y$
i.e. $(2)(x^2) - (2)(991)(y^2) = 0$
i.e. $(x^2) - (991)(y^2) = 0$
i.e. $(x^2) = (991)(y^2)$
i.e. $\frac{x^2}{y^2} = 991$
i.e. $\left(\frac{x^2}{y^2} - (\sqrt{991})^2\right) = 0$
i.e. $\left(\frac{x}{y} - \sqrt{991}\right)\left(\frac{x}{y} + \sqrt{991}\right) = 0$
i.e. $(\frac{x}{y} = \pm\sqrt{991})$
i.e. $\frac{y}{x} = \pm\frac{1}{\sqrt{991}}$ , the equations of the asymptotes to the hyperbola
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/171803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
}
|
Proof of $\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}$ Numerically it seems to be true that
$$
\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}.
$$
Any ideas how to prove this?
|
Let's start out with the following relation:
$$\int_0^{\infty} \frac{\sin x}{\sqrt{x}} e^{-a x} dx = \frac{2}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{1+(a+x^2)^2} dx \tag1$$
Proof of the relation $(1)$
$$\int_0^{\infty} \frac{\sin x}{\sqrt{x}} e^{-a x} dx=$$
Notice that $\displaystyle \frac{1}{\sqrt x}= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} e^{-xt^2} dt$ and have that
$$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \sin x e^{-ax}\left(\int_{0}^{\infty} e^{-xt^2} dt\right) dx=$$
$$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \left(\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dt\right) dx=$$
Change the integration order
$$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \left(\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dx\right) dt=$$
Now let's recollect the formula
$$ \int e^{\alpha x} \sin (\beta x) \ dx = \frac{e^{\alpha x}(-\beta (\cos (\beta x) + \alpha \sin(\beta x)))}{{\alpha}^2+{\beta}^2}$$
Hence
$$\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dx=-\frac{e^{-(a+t^2)x}((a+t^2)\sin x + \cos x)}{1+(a+t^2)^2}\bigg|_{0}^{\infty}=\frac{1}{1+(a+t^2)^2}$$
Then
$$\frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \left(\int_{0}^{\infty}\sin x e^{-(a+t^2)x} dx\right) dt=\frac{2}{\sqrt{\pi}} \int_{0}^{\infty}\frac{1}{1+(a+t^2)^2} \ dt.$$
End of the relation $(1)$ proof.
Based upon the above relation we get that
$$\int_0^{\infty} \frac{\sin x}{\sqrt{x}} dx=$$
$$ \lim_{a\to0+} \int_0^{\infty} \frac{\sin x}{\sqrt{x}} e^{-a x} dx =$$
$$ \lim_{a\to0+} \frac{2}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{1+(a+x^2)^2} dx=$$
$$\frac{2}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{1+x^4} dx \tag2$$
For the last integral we may change the variable and everything gets reduced to computing beta function
Change the variable
$$x=\left(\frac{t}{1-t}\right)^{\frac{1}{4}}$$
Then
$$\int_0^\infty \frac{1}{1+x^4} \ dx = \int_0^1 \frac{1}{4} (1-t)^{\frac{3}{4}-1} t^{\frac{1}{4}-1} \mathrm{d} t = \frac{1}{4} \operatorname{B}\left(\frac{1}{4}, \frac{3}{4}\right) = \frac{1}{4} \sqrt{2} \pi \tag3$$
Finally, from $(2)$ and $(3)$ we obtain the desired result
$$\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}.$$
Q.E.D.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/171970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 8,
"answer_id": 1
}
|
Evaluating $\int \frac{l\sin x+m\cos x}{(a\sin x+b\cos x)^2}dx$ How do I integrate this expression:
$$\int \frac{l\sin x+m\cos x}{(a\sin x+b\cos x)^2}dx$$.I got this in a book.I do not know how to evaluate integrals of this type.
|
One uses trigonometric substitution: $t = \tan\left(\frac{x}{2}\right)$. Then
$$
\sin(x) = \frac{2t}{1+t^2} \quad \cos(x) = \frac{1-t^2}{1+t^2} \quad \mathrm{d} x = \frac{2}{1+t^2} \mathrm{d} t
$$
Hence:
$$\begin{eqnarray}
\int \frac{ \ell \sin(x) + m \cos(x)}{(a \sin(x)+ b \cos(x))^2} \mathrm{d}x &=& \int \frac{ \ell \frac{2 t}{1-t^2} + m \frac{1-t^2}{1+t^2}}{\left(a \frac{2 t}{1+t^2}+ b \frac{1-t^2}{1+t^2}\right)^2} \frac{2}{1+t^2}\mathrm{d}t \\
&=& \int \frac{2 \ell t + m (1-t^2)}{\left(2 a t + b (1-t^2)\right)^2} 2 \mathrm{d}t
\end{eqnarray}$$
The resulting rational function can be integrated using partial fraction decomposition of the integrand, for example.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/172698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
Find the remainder for $(x^5-x^4+x^3+2x^2-x+4)/(x^3+1)$
Find the remainder for $\dfrac{x^5-x^4+x^3+2x^2-x+4}{x^3+1}$
I know exactly how to synthetically divide in the format of: $(x\pm a)$. But not $(x^n\pm a)$ (with an exponent). So if anyone can tell me if anything changes or if the steps are the exact same just with the remainder as $\dfrac{remainder}{x^n\pm a}$. And please do not solve the problem for me.
Edit:
I think I've reached an answer. Long division confused me so I used expanded synthetic division and got: $x^2-x+1+\left(\dfrac{x^2+3}{x^3+1}\right)$. Can anyone verify my answer please.
|
All you have to do is do long division; alternatively, you can subtract appropriate multiples of $x^3+1$ and replace the dividend by the difference until you get a term that is of degree strictly smaller than $3$. (In other words, do polynomial long division analytically instead of synthetically).
For instance, suppose we were trying to find the remainder of dividing
$$2x^5 - 3x^2 + 1$$
by
$$x^2-2.$$
This is the same as the remainder of dividing
$$2x^5 - 3x^2 + 1 - p(x)(x^2-2)$$
by $p(x)$, for any polynomial $p(x)$.
Multiplying $x^2-2$ by $2x^3$ we get $2x^5 - 4x^3$. We can subtract this from $2x^5 - 3x^2+1$ we get
$$2x^5 - 3x^2 + 1 - (2x^5-4x^3) = 4x^3 - 3x^2 + 1.$$
So the remainder when dividing $2x^5-3x^2+1$ by $x^2-2$ is the same as the remainder when dividing $4x^3-3x^2+1$ by $x^2-2$.
Now, multiplying $x^2-2$ by $4x$ we get $4x^3 - 8x$; subtracting it from $4x^3-3x^2+1$ we get
$$4x^3-3x^2+1 - (4x^3 - 8x) = -3x^2 + 8x + 1.$$
So the remainder of dividing $4x^3-3x^2+1$ by $x^2-2$ is the same as the remainder of dividing $-3x^2+8x+1$ by $x^2-2$. Multiplying $x^2-2$ by $-3$ we get $-3x^2+6$; subtracting it from $-3x^2+8x+1$ we get:
$$-3x^2+8x+1 -(-3x^2+6) = 8x -5.$$
So the remainder of dividing $-3x^2+8x+1$ by $x^2-2$ is the same as the remainder of dividing $8x-5$ by $x^2-2$. But the remainder of dividing $8x-5$ by $x^2-2$ is just $8x-5$ (already of degree smaller than $2$).
So the remainder in the original division is $8x-5$.
(This is just, as I noted above, polynomial long division done in the discourse manner rather than synthetically)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/173082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Solve the following inequality $x^2+x+1\gt 0$
Solve the following inequality $x^2+x+1\gt 0$
I understand how to solve inequalities and what the graphs look like. Usually the first step is to set this as in equation and then find the zeros. But for this one when I used the quadratic formula my two answers were:
$$\dfrac{-1+i\sqrt{3}}{2}$$
$$\dfrac{-1-i\sqrt{3}}{2}$$
I do not know what to do after this, or if I messed up in any way. Or, if there is another way to solve this problem. Any hints help. Please do not solve this problem in any way for me. Thanks!
|
Complete the square: $x^2+x+1=\left(x+\frac12\right)^2+\frac34$. For what values of $x$ is this positive?
Added: Here’s an explanation of completing the square. Suppose that you have a quadratic $x^2+ax+b$. You want to write this in the form $(x+c)^2+d$ for some constants $c$ and $d$. We know that $(x+c)^2+d=x^2+2cx+(c^2+d)$, and we want this to be identically equal to $x^2+ax+b$. That is, we want $x^2+2cx+(c^2+d)$ and $x^2+ax+b$ to be the same polynomial. Clearly this means that we must have $2c=a$ and $c^2+d=b$. In particular, we must have $c=\frac{a}2$. I could also solve for $d$, but in practice it’s easier to work it out each time than it is to use a formula.
Knowing now that $c=\frac{a}2$, I write $\left(x+\frac{a}2\right)^2$ as a first approximation to $x^2+ax+b$, and then I multiply it out to get $x^2+ax+\frac{a^2}4$. This approximation gives me the right $x^2$ and $x$ terms, but in general it gives me the wrong constant term, because $\frac{a^2}4$ is rarely equal to $b$. Therefore I have to adjust my approximation $\left(x+\frac{a}2\right)^2$. I do so by subtracting $\frac{a^2}4$ and adding $b$:
$$\left(x+\frac{a}2\right)^2-\frac{a^2}4+b=x^2+ax+b\;,$$
as desired. In the case of the quadratic $x^2+x+1$, $a=1$, so $c=\frac{a}2=\frac12$, and my first approximation was $\left(x+\frac12\right)^2$. This has a constant term of $\frac14$ instead of the desired $1$, so I knew that I had to add another $\frac34$.
The only remaining issue is what to do when the coefficient of $x^2$ isn’t $1$, i.e., when we’re dealing with $ax^2+bx+c$ with $a\ne 1$. The easiest approach is to factor out the $a$ to get
$$a\left(x^2+\frac{b}ax+\frac{c}a\right)\;.$$
Then complete the square on $x^2+\frac{b}ax+\frac{c}a$ to get
$$a\left(\left(x+\frac{b}{2a}\right)^2+\text{some constant}\right)\;.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/173140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Laurent series of $f(z)=\frac{1}{z(z-1)(z-2)}.$ Consider $$f(z)=\frac{1}{z(z-1)(z-2)}.$$
I want to determine the Laurent series in the point $z_0=0$ on $0<|z|<1$.
Partial decomposition yields:
$$f(z)=\frac{1}{z(z-1)(z-2)}=(1/2)\cdot (1/z) - (1/(z-1)) + (1/2)(1/(z-2)).$$
Is the general strategy now, to try to use the geometric series?
$(1/(z-1))=-(1/(1-z))=-\sum_{k=0}^\infty z^k$
$\displaystyle 1/(z-2)=-(1/2)\frac{1}{1-\frac{z}{2}}=-(1/2)\cdot\sum_{k=0}^\infty (z/2)^k$
So $f(z)=(1/2)\cdot (1/z)+(1/2)\cdot\sum_{k=0}^\infty z^k-(1/4)\cdot\sum_{k=0}^\infty (z/2)^k$ (*)
Some questions:
1) What is the difference between a Laurent and a Taylor series? I don't get it. It seems you calculate them the same.
2) Why didn't we write (1/z) as a series expansion too?
3) What makes the end result (*) to be a Laurent series?
|
Partial fractions and geometric series give
$$
\begin{align}
\frac1{(1-x)(2-x)}
&=\frac1{1-x}-\frac1{2-x}\\
&=\frac1{1-x}-\frac12\frac1{1-x/2}\\
&=(1+x+x^2+x^3+x^4+\dots)\\
&-\left(\frac12+\frac14x+\frac18x^2+\frac1{16}x^3+\frac1{32x^4}+\dots\right)\\
&=\frac12+\frac34x+\frac78x^2+\frac{15}{16}x^3+\frac{31}{32}x^4+\dots
\end{align}
$$
Thus, the Laurent series for $\frac1{x(x-1)(x-2)}$ at $x=0$ is
$$
\frac1{x(x-1)(x-2)}=\frac1{2x}+\frac34+\frac78x+\frac{15}{16}x^2+\frac{31}{32}x^3+\dots
$$
We could also expand the series at $x=1$. Let $y=x-1$ and then
$$
\begin{align}
\frac1{x(x-1)(x-2)}
&=\frac1{(y+1)y(y-1)}\\
&=-\frac1y\frac1{1-y^2}\\
&=-\frac1y-y-y^3-y^5-y^7-\dots\\
&=-\frac1{x-1}-(x-1)-(x-1)^3-(x-1)^5-(x-1)^7-\dots
\end{align}
$$
*
*The Laurent series is much like the Taylor series except terms of negative degree are allowed.
*We don't expand $\frac1x$ since there is no power series for $\frac1x$ at $0$ other than $\frac1x$.
*Definition. It is a power series at a point, $x_0$ which can have both positive and negative powers of $x-x_0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/174452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solve $x^2 (a - bx) {d^2y \over dx^2} - x (5a - 4bx) {dy \over dx} + 3(2a - bx)y = 6a^2$ What do I substitute here
$$ x^2 (a - bx) {d^2y \over dx^2} - x (5a - 4bx) {dy \over dx} + 3(2a - bx)y = 6a^2 $$ to get it into this form?
$$ u^2 {dv^2 \over du^2} + P_1 u {dv \over du} + P_2 v = F(u) $$
The solution (according to answer sheet):
$y(a-bx) = Ax^2 + Bx^3 + C$
|
I will post a partial answer, as I might have made a sign error or been taken astray at some point. The method amounts to finding some sort of integrating factor for the equation. It seems reasonable to start with the change of the dependent variable:
$$u=a-bx$$
$$x=-\frac{1}{b}\left(u-a\right)$$
$$\frac{dy}{dx}=-b\frac{dy}{du}$$
$$\frac{d^{2}y}{dx^{2}}=-b^{2}\frac{d^{2}y}{du^{2}}$$
Rewrite the equation:
$$\left(u-a\right)^{2}u\frac{d^{2}y}{du}+\left(u-a\right)\left(4u+a\right)\frac{dy}{du}+3\left(u+a\right)y=6a^{3}$$
Split the $\frac{dy}{du}$ term and regroup:
$$\left(u-a\right)^{2}u\frac{d^{2}y}{du}+\left(u-a\right)\left(u+a\right)\frac{dy}{du}+3\left(u-a\right)u\frac{dy}{du}+3\left(u+a\right)y=6a^{3}$$
$$\left(u-a\right)\left[\left(u-a\right)u\frac{d^{2}y}{du^{2}}+\left(u+a\right)\frac{dy}{du}\right]+3\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]=6a^{3}$$
Differentiating the second square bracket:
$$\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]=\left(2u-a\right)\frac{dy}{du}+\left(u-a\right)u\frac{d^{2}y}{du^{2}}+y+\left(u+a\right)y\qquad(1)$$
We can express the first one as follows:
$$\left(u-a\right)u\frac{d^{2}y}{du^{2}}+\left(u+a\right)\frac{dy}{du}=\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]-\left(2u-a\right)\frac{dy}{du}-y$$
Inserting in the equation:
$$\left(u-a\right)\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]-\left(u-a\right)\left[\left(2u-a\right)\frac{dy}{du}+y\right]+3\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]=6a^{3}$$
$$\left(u-a\right)\frac{d}{du}\left[\left(u-a\right)u\frac{dy}{du}+\left(u+a\right)y\right]+\left(u^{2}-a^{2}\right)\frac{dy}{du}+2ay=6a^{3}$$
Now using the product rule we obtain:
$$\left(u-a\right)\frac{d}{du}\left[\left(u-a\right)u\frac{dy}
{du}+\left(u+a\right)y\right]=\frac{d}{du}\left[\left(u-a\right)^{2}u\frac{dy}{du}+\left(u^{2}-a^{2}\right)y\right]-\left(u-a\right)u\frac{dy}{du}-\left(u+a\right)y$$
The original equation is then expressed as follows:
$$\frac{d}{du}\left[\left(u-a\right)^{2}u\frac{dy}{du}+\left(u^{2}-a^{2}\right)y\right]+\left(u-a\right)\left[a\frac{dy}{du}-y\right]=6a^{3}$$
Now it might have made sense in (1) to look for $y$ in some specific form, say $y=rs$ and then impose suitable conditions on the factors.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/176220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Prove $\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$ if $A+B+C=180$ degrees I most humbly beseech help for this question.
If $A+B+C=180$ degrees, then prove
$$
\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)
$$
I am not sure what trig identity I should use to begin this problem.
|
Let's take the Right Hand Side.
$$2 \sin B \cos C = \sin(B+C) + \sin(B-C) = \sin A + \sin(B-C)$$
Therefore, $2 \sin A \sin B \cos C = \sin^2 A + \sin A \sin(B-C)$
Now, $$2 \sin A \sin (B-C) = \cos (A - B +C) - \cos (A + B -C) = \cos 2C - \cos 2B = 2 \sin^2 B - 2 \sin ^2 C$$ Cancelling the $2$ gives us therefore, that $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$.
To finish, we already established $2 \sin A \sin B \cos C = \sin^2 A = \sin A \sin (B-C)$. Now, since $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$, therefore, $2 \sin A \sin B \cos C = \sin^2 A + \sin^2 B - \sin^2 C$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/177208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
}
|
Which of the following statements are false Let$$A=\frac{1}{3}\begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix}$$
Which of the following statements are false?
(a) $A$ has only one real eigenvalue.
(b) $\operatorname{Rank}(A) = \operatorname{Trace}(A)$.
(c) Determinant of $A$ equals the determinant of $A^n$ for each integer $n > 1$.
I proceed in this way:
(a) let $\lambda$ be the eigenvalue of $A$. So,$[A-\lambda I]=\begin{pmatrix} 2/3-\lambda & -1/3 & -1/3 \\ -1/3 & 2/3-\lambda & -1/3 \\ -1/3 & -1/3 & 2/3-\lambda \end{pmatrix}=\begin{pmatrix} -\lambda & -\lambda & -\lambda \\ -1/3 & 2/3-\lambda & -1/3 \\ -1/3 & -1/3 & 2/3-\lambda \end{pmatrix}$$[R_1'\to R_1+R_2+R_3]$$$=\begin{pmatrix} -\lambda & 0 & 0 \\ -1/3 & 1-\lambda & 0 \\ -1/3 & 0 & 1-\lambda \end{pmatrix}[C_2'\to C_2-C_1, C_3'\to C_3-C_1]$$ Hence we get $$|A-\lambda I|=0$$$=>\lambda(1-\lambda)^2=0$ $=>\lambda= $ $0$ or $1$ $=>$There are two real eigenvalue.
(b) Note that $|A|=0$ and $$\begin{vmatrix} 2/3 & -1/3 \\ -1/3 & 2/3 \end{vmatrix}=5/9 \not= 0$$ So , $\operatorname{Rank}(A) =2$ and $\operatorname{Trace}(A)=\frac{1}{3}(2+2+2)=2$ Hence $\operatorname{Rank}(A) = \operatorname{Trace}(A)$
(c) $A^2=AA=A$ and $A^3=A^2A=AA=A$ so determinant of $A$ equals the determinant of $A^n$ for each integer $n > 1$. So only first statement is wrong.
|
You are totally right, except the third part. You do not have $A^2=A$, but instead you have that $det(A^n)=(det(A))^n=0^n=0=det(A)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/178268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality $\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$? Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality
$$\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$$?
|
Although the function $f(x)=\sqrt{x}/(1-x)$ is not convex on (0,1), its tangent at $x=1/3$ lower bounds the
function and passes through the origin.
That is, for $0\leq x\leq 1$, we have
$${\sqrt{x}\over 1-x}\geq {3\sqrt{3}\over2}\, x.$$
Plugging in $a,b,c$ and adding gives
$${\sqrt{a}\over 1-a}+{\sqrt{b}\over 1-b}+{\sqrt{c}\over 1-c}\geq {3\sqrt{3}\over2}.$$
Added reference: Exercise 8.1 on page 131 of The Cauchy-Schwarz Master Class by J. Michael Steele asks
you to prove that for $p\geq 1$, and positive $a,b,c$,
$${a^p\over b+c}+{b^p\over a+c}+{c^p\over a+b}\geq {1\over 2}\,3^{2-p}\,(a+b+c)^{p-1}.\tag1$$
He notes that for $p=1$ this reduces to Nesbitt's inequality.
The inequality (1) fails for $0<p<p_c$, where $p_c={3\log(2)-2\log(3)\over \log(2)-\log(3)}=.29048$
by looking at $a=b=1/2$ and $c$ close to zero. But it holds again for $p=0$ by Jensen's inequality .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/180937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
}
|
Inequality $a+b+c \geqslant abc +2$ Assuming $a,b,c \in (0, \infty)$, we need to prove that:
$$a+b+c \geqslant a b c+2 \quad \text{if} \quad ab+bc+ca=3$$
Can you give me an idea, please? This inequality seem to be known, but I didn't manage to solve it.
|
For $x,y,z \geq 0 $, $ f(x,y,z) = x+y+z $ given that, $xy+yz+xz =3 = \phi(x, y, z)$
$$ \nabla f = \lambda \nabla \phi $$
$$ 1 = \lambda (y+z) \hspace {2 cm} (1) $$
$$ 1 = \lambda (x+z) \hspace {2 cm} (2) $$
$$ 1 = \lambda (x+y) \hspace {2 cm} (3) $$
$$ xy+yz+xz =3 \hspace{2 cm} (4)$$
Solveing $(1), (2), (3), \text{ and } (4)$ we get, $x=y=z=1$, so the minimum value of $f(x, y, z) = x+y+z$ under the constraint $xy+yz+xz = 3$ is $ 1 + 1 + 1 = 3$
Again $g(x, y, z) = xyz + 2$
$$ \nabla g = \lambda \nabla \phi $$
$$ yz = \lambda (y+z) \hspace {2 cm} (5) $$
$$ xz = \lambda (x+z) \hspace {2 cm} (6) $$
$$ xy = \lambda (x+y) \hspace {2 cm} (7) $$
$$ xy+yz+xz =3 \hspace{2 cm} (4)$$
Solving these we get $x=y=z=1$, the maximum value of $g(x, y, z) = 1\cdot 1 \cdot 1 + 2 = 3 $
Since, $ \text{ min }( f) = \text{ max}(g) $, we have $a+b+c \geq abc+2$
I hope there is a better method!!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/181121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
}
|
How to evaluate the integral $\iint_C \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy$ I would like to evaluate
$$\iint\limits_C \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy$$
where $C$ is the first quadrant, i.e.
$$\int\limits_0^\infty\int\limits_0^\infty \sin (x^2+y^2)\,\mathrm dx\,\mathrm dy$$
where
$$\int\limits_0^\infty \sin x^2\, dx=\int\limits_0^\infty \cos x^2\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$$
|
$$\begin{array}{c l}\int_0^\infty\int_0^\infty \sin(x^2+y^2)dxdy & = \int_0^\infty\int_0^\infty \sin(x^2)\cos(y^2)+\cos(x^2)\sin(y^2)dxdy \\[10pt] & =\int_0^\infty\sin(x^2)dx\int_0^\infty\cos(y^2)dy \\ & \qquad~~~+\int_0^\infty\cos(x^2)dx\int_0^\infty\sin(y^2)dy \\[10pt] & = 2\left(\frac{1}{2}\sqrt{\frac{\pi}{2}}\right)^2=\frac{\pi}{4}. \end{array}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/181910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
}
|
Getting square root of negative in completing the square problem I try to solve the equation $f(x) = 7x - 11 - 2x^2 = 0$ for $x$, but run into troubles. I've gone through it over and over again as well as similar problems, but can't find what I'm doing wrong.
$$f(x) = 7x - 11 - 2x^2 = 0$$
$$\iff x^2 - \frac{7}{2}x + \frac{11}{2} = 0 $$
$$\iff \left(x + \frac{7}{4}\right)^2 = \left(\frac{7}{4}\right)^2 - \frac{11}{2}$$
$$\iff x + \frac{7}{4} = \pm \sqrt{\left(\frac{7}{4}\right)^2 - \frac{11}{2}}$$
$$\iff x = -\frac{7}{4} \pm \sqrt{\frac{49}{16} - \frac{88}{16}}$$
$$\iff x = -\frac{7}{4} \pm \sqrt{\frac{-39}{16}}$$
I should be able to continue but I'm stuck (seeing as it's a negative number). What am I doing wrong?
|
You did everything fine but your quadratic equation has no real solutions, which you could have found out way more easily had you first calculated the equation's discriminant:
$$\Delta:=b^2-4ac=7^2-4(-2)(-11)=49-88=-39<0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/183731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to combine ratios? If $a:b$ is $2:5$, and $c:d$ is $5:2$, and $d:b$ is $3:2$, what is the ratio $a:c$? How would I go about solving this math problem?
if the ratio of $a:b$ is $2:5$ the ratio of $c:d$ is $5:2$ and the ratio of $d:b$ is $3:2$, what is the ratio of $a:c$?
I got $\frac{a}{c} = \frac{2}{5}$ but that is not a correct answer.
|
These ratios are just simple equations. For example $a:b=2:5$ is
$$a= \frac{2}{5}b$$
No need for confusing tricks here. Just substitutions :
$$ a = \frac{2}{5}b = \frac{2}{5}\frac{2}{3} d = \frac{2}{5}\frac{2}{3}\frac{2}{5} c = \frac{8}{75} c$$
So that
$$ a:c = 8:75 $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/184669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Proof of $(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$ For positive real numbers $a$, $b$ and $c$, how do we prove that:
$$(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$$
|
Just by sum of square and Schur's inequality:
$$
3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)-(ab+bc+ca)^3\\
=
\sum_{cyc} (a^{3}b+a^{3}c+a^{2}bc)(b-c)^2+3abc\sum_{cyc}a(a-b)(a-c) \geq 0
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/186575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$
Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$
how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$
we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$
now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$
|
Define a sequence $\{a_n\}_{n \geq 1}$ such that $a_1 = \sqrt 2$ and $a_{n+1} = \sqrt{2 + a_n}$. It should be clear that $x$ is the limit of this sequence as $n$ goes to infinity, i.e:
$$x = \lim_{n \rightarrow \infty} a_n $$
To prove that this limit exists, it is sufficient to show that the sequence is bounded above, and monotonically increasing. Both of these facts may be proved by induction.
The fact that $a_n$ is bounded above follows since $a_1 < 2$ and $a_{n+1} = \sqrt{2+a_n}$ which is less than $2$ if $a_n$ is, because then $a_{n+1} < \sqrt{2+2} = 2$.
To prove that $a_n$ is monotonically increasing note that $a_1 > a_2$ and $a_{n+1} > \sqrt{2+a_{n-1}} = a_n$, assuming of course that $a_n > a_{n-1}$.
Every monotonically increasing sequence that is bounded from above converges, so the other solutions are justified in concluding that $x=2$.
Edit: As pointed out by did, the argument above only applies when $a_1 < 2$ as there is no reason to choose $a_1 = \sqrt{2}$ specifically, as $a_1$ occurs in the "$\dots$" portion of the nested radicals.
If $a_1>2$ we may show that the sequence $\{a_n\}$ is bounded below by 2 and monotonically decreasing, by a similar argument. Finally, it is simple to show that $x$ converges if $a_1 = 2$. Thus, $x$ converges to $2$ regardless of the starting value $a_1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/186652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
}
|
Equation of line and its points
In the xy coordinate system if (a,b) and (a+3,b+k) are two points on the line defined by equation (the equation is kind of faded in text , but it seems to be like x=3y-7) then k =
A)9 , B)3 , C)1 , D)1 (Ans is 1)
Any suggestions on how that answer was calculated ?
|
The slope of the line through $(a,b)$ and $(a+3, b+k)$ is $\frac{b+k-b}{a+3-a}$, which is $\frac{k}{3}$.
The slope of the line $x=3y-7$ is $\frac{1}{3}$. This is because the equation can be rewritten as $3y=x+7$, and then in standard slope-intercept form as $y=\frac{1}{3}x+\frac{7}{3}$.
These slopes are equal $\frac{k}{3}$ and $\frac{1}{3}$ are equal.
Another way: Because $(a,b)$ is on the line, we have $a=3b-7$. Because $(a+3,b+k)$ is on the line, we have $a+3=3(b+k)-7$, that is, $a+3=3b+3k-7$.
Since $a=3b-7$, we conclude that $3=3k$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/187762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Does $i^i$ and $i^{1\over e}$ have more than one root in $[0, 2 \pi]$ How to find all roots if power contains imaginary or irrational power of complex number? How do I find all roots of the following complex numbers?
$$(1 + i)^i, (1 + i)^e, (1 + i)^{ i\over e}$$
EDIT:: Find the value of $i^i$ and $i^{1\over e}$ in $[0, 4\pi]$?
|
Let $e^z=N≠0$ ,So, $N=e^z=e^z\cdot e^{2n\pi i}$ as $e^{2n\pi i}=1$ for any integer $n$.
$N=e^{z+2n\pi i}\implies Log N=z+2n\pi i$
If $n=0$ ,we get the principal value $z=\log N$, so $LogN=\log N+2n\pi i$
Let $A=(a+ib)^{x+iy}$
$Log A=(x+iy)Log(a+ib)$
Let $a=r\cos \theta$ and $b=r\sin \theta\implies a+ib=r(\cos \theta+i\sin \theta)=re^{i\theta}$
So, $r=\sqrt{a^2+b^2}$ and $\theta=\tan ^{-1}\frac{b}{a}$
then $Log(a+ib)=\frac{1}{2}\log (a^2+b^2)+i(2k\pi+\tan ^{-1}\frac{b}{a})$ where $k$ is any integer.
So, $Log A=(x+iy)Log(a+ib)=(x+iy)(\frac{1}{2}\log (a^2+b^2)+i(2k\pi+\tan ^{-1}\frac{b}{a}))$
So, $Log A=\frac{1}{2}x\log (a^2+b^2)-y(2k\pi+\tan ^{-1}\frac{b}{a})+i(\frac{1}{2}y\log (a^2+b^2)+x((2k\pi+\tan ^{-1}\frac{b}{a}))$
So, $$A=(a+ib)^{x+iy}=e^{\frac{1}{2}x\log (a^2+b^2)-y(2k\pi+\tan ^{-1}\frac{b}{a})}\cdot e^{i({\frac{1}{2}y\log (a^2+b^2)+x(2k\pi+\tan ^{-1}\frac{b}{a}})}$$
$$=e^{\frac{1}{2}x\log (a^2+b^2)-y(2k\pi+\tan ^{-1}\frac{b}{a})} cis(({\frac{1}{2}y\log (a^2+b^2)+x(2k\pi+\tan ^{-1}\frac{b}{a}})) $$ where $cis(P)=\cos P+i\sin P$
(1)For $(1+i)^i$ $a=b=y=1$ and $x=0$
So, $r^2=a^2+b^2=2$ and $\sin \theta=\cos \theta=\frac{1}{\sqrt2}\implies \theta=\frac{\pi}{4}$
$(1+i)^i=e^{-(2k\pi+\frac{\pi}{4})}(\cos (\frac{1}{2}\log 2)+i\sin (\frac{1}{2}\log 2))$ where $k$ is any integer.
(2)For $(1+i)^e, a=b=1$ and $x=e,y=0$
So, $r^2=a^2+b^2=2$ and $\sin \theta=\cos \theta=\frac{1}{\sqrt2}\implies \theta=\frac{\pi}{4}$
$(1+i)^e=e^{\frac{1}{2}e\log 2}\cdot cis(e(2k\pi+\frac{\pi}{4}))$ where $k$ is any integer.
$(1+i)^e=2^{(\frac{e}{2})}\cdot cis(e(2k\pi+\frac{\pi}{4}))$
(3) For $(1+i)^{\frac{i}{e}}, a=b=1$ and $x=0,y=\frac{1}{e}$
$(1+i)^{\frac{i}{e}}=e^{-\frac{1}{e}(2k\pi+\frac{\pi}{4})}\cdot cis(\frac{1}{2e}\log 2)$ where $k$ is any integer.
(4) For $i^i,a=0,b=1,x=0,y=1\implies r^2=1$ and $\sin \theta=1$ and $\cos \theta=0\implies \theta=\frac{\pi}{2}$
So, the general value of $i^i=e^{-(2k\pi+\frac{\pi}{2})}$ where $k$ is any integer.
Now by the given condition, $0 ≤e^{-(2k\pi+\frac{\pi}{2})} ≤4\pi$
$0 ≤e^{-(2k\pi+\frac{\pi}{2})}$ is true for all real finite $k$.
So, we need check $ -(2k\pi+\frac{\pi}{2})≤\ln _e{4\pi}$ .
or $2k\pi+\frac{\pi}{2}≥-\ln _e{4\pi}$
Without using calculator, we can observe that this inequality is satisfied by all integer $k≥0$
(5) For $i^{\frac{1}{e}}, a=0,b=1,x=\frac{1}{e},y=0$
$\implies r^2=1$ and $\sin \theta=1$ and $\cos \theta=0\implies \theta=\frac{\pi}{2}$
$i^{\frac{1}{e}}=e^{\frac{1}{2e}\log 2}\cdot cis(\frac{1}{e}(2k\pi+\frac{\pi}{2}))$
can be further simplified to $2^{\frac{1}{2e}}\cdot cis(\frac{1}{e}(2k\pi+\frac{\pi}{2}))$ where $k$ is any integer.
It does have an imaginary part, so can not have any pure real value in any specified range.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/189703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
}
|
Find the angle between two bivariate functions.. Find the angle between $x^2+y^2=8$ and $xy=4$ at the intersection points.
I have thought in find the angle between the tangent lines of each function at the intersection point, but i don´t know how to do it
|
Let the intersection point be $(a,b)$, so $a^2+b^2=8$ and $ab=4≠0$
$$\implies \frac{a^2+b^2}{ab}=\frac{4}{2}\implies (a-b)^2=0\implies a=b≠0$$
So,$a=±2$ as $ab=4$.
$$x^2+y^2=4\implies 2x+2y\frac{dy}{dx}=0\implies (\frac{dy}{dx})_{x=y}=-1$$
$$xy=4\implies x\frac{dy}{dx}+y=0\implies (\frac{dy}{dx})_{x=y}=-1 $$
So, at $(a,a)$ the gradients of the two given curves are same.
We know if the angle between the two curves is $\theta$, then $\tan\theta=\frac{g_1-g_2}{1+g_1g_2}$ where $g_i$ s are the gradient of the intersecting curves.
So , the angle between the given two curves at $(a,a)$, i.e., at $(2,2)$ and $(-2,-2)$, is
$$\tan^{-1}\left(\frac{-1-(-1)}{1+(-1)(-1)}\right)=\tan^{-1}(0)=0 \text{ or }\pi$$
Alternatively,
we are given $x^2+y^2=8$ and $xy=4$. From the latter, $x=\frac{4}{y}$
$$\left(\frac{4}{y}\right)^2+y^2=8\implies y^4-8y^2+16=0\implies (y^2-4)^2=0\implies y=±2$$
$$\tag 1 y=2\implies x=2$$ and $$\tag 2 y=-2\implies x=-2$$
So, $x=y$ at the points of intersection.
Observe that the equation of the common tangents are $x+y=a+a=2a$ i.e., $x+y=±4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/192282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prove $ 1 + 2 + 4 + 8 + \dots = -1$
Possible Duplicate:
Infinity = -1 paradox
I was told by a friend that $1 + 2 + 4 + 8 + \dots$ equaled negative one. When I asked for an explanation, he said:
Do I have to?
Okay so, Let $x = 1+2+4+8+\dots$
$2x-x=x$
$2(1+2+4+8+\dots) - (1+2+4+8+\dots) = (1+2+4+8+\dots)$
Therefore, $(2+4+8+16+\dots) + (-1-2-4-8+\dots) = (1+2+4+8+\dots)$. Now $-2$ and $2$, $-4$ and $4$, $-8$ and $8$ and so on, cancel out, and the only thing left is $-1$.
Therefore, $1+2+4+8+\dots = -1$.
I feel that this conclusion is not right, but I cannot express it. Can anyone tell if this proof is wrong, and if it is, how it is wrong?
|
Look at Calculus textbook, undergraduate level. When we treat infinite sum, we cannot change the order to compute.
Example. $1-1+1-\cdots$
$$(1-1)+(1-1)+\cdots=0+0+\cdots=0$$
$$1+(-1+1)+(-1+1)+\cdots=1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/193872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 0
}
|
Calculation of a strange series Is it possible to find an expression for:
$$S(N)=\sum_{k=0}^{+\infty}\frac{1}{\sum_{n=0}^{N}k^n}?$$
For $N=1$ we have
$$S(1) = \displaystyle\sum_{k=0}^{+\infty}\frac{1}{1 + k} = \displaystyle\sum_{k=1}^{+\infty}\frac{1}{k}$$
which is the (divergent) harmonic series. Thus, $S (1) = \infty$.
For $N=2$ this sum is:
$$S(2)=\sum_{k=0}^{+\infty}\frac{1}{1+k+k^2}$$
which can be expressed as:
$$S(2)=-1+\frac{1}{3}\sqrt 3 \pi \tanh(\frac{1}{2}\pi\sqrt 3)\approx 0.798$$
For $N=3$ we have:
$$S(3)=\frac{1}{4}\Psi(I)+\frac{1}{4I}\Psi(I)-\frac{1}{4I}\pi\coth(\pi)+\frac{1}{4}\pi\coth(\pi)+\frac{1/}{4}\Psi(1+I)-\frac{1}{4I}\Psi(1+I)-\frac{1}{2}+\frac{1}{2}\gamma \approx 0.374$$
|
Let $T(N) = S(N-1)$. Then
$$ \begin{align*}T(n)
&= 1 + \frac{1}{n} + \sum_{k=2}^{\infty} \frac{1}{k^{n-1}+k^{n-2}+\cdots+k+1} \\
&= 1 + \frac{1}{n} + \sum_{k=2}^{\infty} \frac{k - 1}{k^n - 1} \\
&= 1 + \frac{1}{n} + \sum_{k=2}^{\infty} \frac{1}{n} \sum_{l=1}^{n-1} \frac{\omega_l (\omega_l - 1)}{k - \omega_l} \\
&= 1 + \frac{1}{n} + \sum_{k=0}^{\infty} \frac{1}{n} \sum_{l=1}^{n-1} \frac{\omega_l (\omega_l - 1)}{k + 2 - \omega_l},
\end{align*}$$
where $\omega_l = \exp\left(\tfrac{2\pi l i}{n}\right)$. Since
$$ \frac{1}{n} \sum_{l=0}^{n-1} \omega_l (\omega_l - 1) = 0, $$
we may write
$$ \begin{align*}T(n)
&= 1 + \frac{1}{n} + \sum_{k=0}^{\infty} \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \left( \frac{1}{k + 2 - \omega_l} - \frac{1}{k+1} \right) \\
&= 1 + \frac{1}{n} + \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \sum_{k=0}^{\infty} \left( \frac{1}{k + 2 - \omega_l} - \frac{1}{k+1} \right) \\
&= 1 + \frac{1}{n} - \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \left( \gamma + \psi_0 (2 - \omega_l) \right) \\
&= 1 + \frac{1}{n} - \frac{1}{n} \sum_{l=1}^{n-1} \omega_l (\omega_l - 1) \psi_0 (2 - \omega_l).
\end{align*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/194096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
}
|
Estimation of a polynomial I'm currently reading the following paper: http://arxiv.org/abs/1209.0612 and got stuck on Proposition 3.1 (2).
The claim translated to polynomials is the following:
Assume $n\geq 3, c\geq 1, d\geq 1$ are natural numbers such that $c²+d²-(n-1)cd<0$. Show that $(n³-n+1)c²+(n+1)d²-(n²+n-1)cd>1$.
Anyone an idea to solve this?
|
Let $c,d$ be positive real numbers and $n>1$ (esp., $n^3-n+1>0$).
By the arithmetic-geometric inequality
$$(n^3-n+1)c^2+(n+1)d^2\ge 2\cdot\sqrt{(n^3-n+1)(n+1)}\cdot c d.$$
One checks by multiplying out that
$$4(n^3-n+1)(n+1)=(n^2+n-1)^2+3+3n^2(n^2-1)+2n(n^2+1),$$
hence $2\cdot\sqrt{(n^3-n+1)(n+1)}>n^2+n-1$ and finally
$$(n^3-n+1)c^2+(n+1)d^2>(n^2+n-1) cd. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/195261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$ How to prove this inequality
$$\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}} $$
for $a,b,c,d\gt0$?
Thanks
|
Here is my proof for positive variables from 1979.
Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$, where $v>0$.
Hence, $a$, $b$, $c$ and $d$ are positive roots of the following equation:
$(x-a)(x-b)(x-c)(x-d)=0$ or $x^4-4ux^3+6v^2x^2-4w^3x+t^4=0$.
Hence, by Rolle the equation $(x^4-4ux^3+6v^2x^2-4w^3x+t^4)'=0$ or
$x^3-3ux^2+3v^2x-w^3=0$ has three positive roots.
Since we can replace $x$ at $\frac{1}{x}$, we get that the equation
$w^3x^3-3v^2x^2+3ux-1=0$ has three positive roots.
Now by the Rolle's theorem again we get that the equation
$(w^3x^3-3v^2x^2+3ux-1)'=0$ or $w^3x^2-2v^2x+u=0$ has two real roots,
which says that $v^4\geq uw^3$.
By the Rolle's theorem again the equation $(x^3-3ux^2+3v^2x-w^3)'=0$
or $x^2-2ux+v^2=0$ has two positive roots, which says that $u^2\geq v^2$ or $u\geq v$.
The last inequality is also $\sum\limits_{sym}(a-b)^2\geq0$.
Id est, $v^4\geq uw^3\geq vw^3$, which gives $v\geq w$ and we are done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/197955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
$(1+i)$ to the power $n$
Possible Duplicate:
Complex number: calculate $(1 + i)^n$.
I came across a difficult problem which I would like to ask you about:
Compute
$ (1+i)^n $ for $ n \in \mathbb{Z}$
My ideas so far were to write out what this expression gives for $n=1,2,\ldots,8$, but I see no pattern such that I can come up with a closed formula.
Then I remember that one can write any complex number $a+bi$ like:
$$(a+bi)=\sqrt{a^2+b^2} \cdot \left( \frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{a^2+b^2}}\cdot i\right)$$
and $\frac{a}{\sqrt{a^2+b^2}} = \cos(\phi)$ and $\frac{b}{\sqrt{a^2+b^2}} = \sin(\phi)$ where $\phi$ is $\arctan{\frac{b}{a}} $
So it becomes,
$(a+bi)=\sqrt{a^2+b^2} \cdot ( \cos(\phi)+\sin(\phi)\cdot i)$ Taking this entire thing to the power $n$ using De Moivre
$$(a+bi)^n=(\sqrt{a^2+b^2})^n \cdot ( \cos(n\phi)+\sin(n\phi)\cdot i)$$
Substituting my $a=1$ and $b=1$
$(1+i)^n=(\sqrt{2})^n \cdot ( \cos(n\cdot\frac{\pi}{4})+\sin(n\cdot\frac{\pi}{4})\cdot i)$
$\phi$ is 45 degrees hence $\frac{\pi}{4}$
But now I don't know how to continue further and I would really appreciate any help! Again, Im looking for a closed formula depending on n.
Best regards
|
You have a closed form already if you do just a little more work:
$$\begin{align*}
(1+i)^n&=\left(\sqrt{2}\right)^n\left(\cos\frac{n\pi}4+i\sin\frac{n\pi}4\right)\\
&=2^{n/2}\left(\cos\frac{n\pi}4+i\sin\frac{n\pi}4\right)\\
&=\begin{cases}
2^{n/2},&\text{if }n\equiv 0\pmod 8\\
2^{n/2}\left(\frac{\sqrt2}2+\frac{\sqrt2}2i\right),&\text{if }n\equiv 1\pmod 8\\
2^{n/2}i,&\text{if }n\equiv 2\pmod 8\\
2^{n/2}\left(\frac{\sqrt2}2-\frac{\sqrt2}2i\right),&\text{if }n\equiv 3\pmod 8\\
-2^{n/2},&\text{if }n\equiv 4\pmod 8\\
-2^{n/2}\left(\frac{\sqrt2}2+\frac{\sqrt2}2i\right),&\text{if }n\equiv 5\pmod 8\\
-2^{n/2}i,&\text{if }n\equiv 6\pmod 8\\
-2^{n/2}\left(\frac{\sqrt2}2-\frac{\sqrt2}2i\right),&\text{if }n\equiv 7\pmod 8\;.
\end{cases}\\
\end{align*}$$
A function defined by cases is still a closed form. You can do better, though, if you allow the exponential form of the complex number: $1+i=\sqrt2 e^{i\pi/4}$, so
$$(1+i)^n=2^{n/2}e^{in\pi/4}\;.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/199004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
the least possible value for :$ \lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor $ If we know that for every $a,b,c>0$ ,how we can find the least possible value for :
$$ \lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor $$
|
The expression
$$
E := \frac {a + b} c + \frac {b + c} a + \frac {c + a} b
$$
is homogeneous, so we can assume $a + b + c = 1$. Then it becomes
$$
E = \frac 1 a + \frac 1 b + \frac 1 c - 3
$$
By CS inequality we get
$$
\frac 1 a + \frac 1 b + \frac 1 c \geq \frac {(1 + 1 + 1)^2} {a + b + c} = 9
$$
So $E \geq 6$. By the following relation
$$
\left\lfloor \frac {a + b} c \right\rfloor > \frac {a + b} c - 1
$$
(and the other ones got by cyclic permutations of variables), we have
$$
E' := \left\lfloor \frac {a + b} c \right\rfloor + \left\lfloor \frac {b + c} a \right\rfloor + \left\lfloor \frac {c + a} b \right\rfloor > E - 3 \geq 6 - 3 = 3
$$
Being $E'$ an integer, the above inequality is equivalent to
$$
E' \geq 4
$$
To conclude, let's note that for $a = b = 4$ and $c = 3$, $E' = 4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/200289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
}
|
A inequality proposed at Zhautykov Olympiad 2008 An inequality proposed at Zhautykov Olympiad 2008.
Let be $a,b,c >0$ with $abc=1$. Prove that:
$$\sum_{\mathrm{cyc}}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$
Set $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.
Our inequality becomes:
$$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \frac{3}{2}.$$
Now we use that: $z^2+x^2 \geq 2zx.$
$$\sum_{\mathrm{cyc}}{\frac{z^2}{zx+y^2}} \geq \sum_{\mathrm{cyc}}{\frac{2z^2}{z^2+x^2+2y^2}} \geq \frac{3}{2}.$$
Now applying Cauchy-Schwarz we obtain the desired result.
What I wrote can be found on this link: mateforum.
But now, I don't know how to apply Cauchy-Schwarz.
Thanks:)
|
Let $a=\frac{x}{y}$ and $b=\frac{y}{z}$, where $x$, $y$ and $z$ be positives.
Hence, since $abc=1$, we get $c=\frac{z}{x}$ and by C-S we obtain:
$$\sum_{cyc}\frac{1}{(a+b)b}=\sum_{cyc}\frac{z^2}{xz+y^2}=\sum_{cyc}\frac{z^4}{xz^3+z^2y^2}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(x^2y^2+x^3y)}.$$
Thus, it remains to prove that
$$2(x^2+y^2+z^2)^2\geq3\sum\limits_{cyc}(x^2y^2+x^3y)$$ or
$$\sum_{cyc}(2x^4-3x^3y+x^2y^2)\geq0$$ or
$$\sum_{cyc}(x-y)x^2(2x-y)\geq0$$ or
$$\sum_{cyc}\left((x-y)x^2(2x-y)-\frac{x^4-y^4}{4}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2(7x^2+2xy+y^2)\geq0.$$
Done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/202053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
}
|
Help me proof the identity $\sin^2\alpha+\cos^2\alpha=1$ Prove the identity $\sin^2\alpha+\cos^2\alpha=1$. Thanks
|
$$\sin\alpha=\frac{a}{c}\Rightarrow\sin^2 \alpha=\frac{a^2}{c^2}$$
$$\cos\alpha=\frac{b}{c}\Rightarrow\cos^2\alpha=\frac{b^2}{c^2}$$
$$\sin^2\alpha+\cos^2\alpha=\frac{a^2}{c^2}+\frac{b^2}{c^2}=\frac{a^2+b^2}{c^2}=\frac{c^2}{c^2}=1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/203473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Ideas on the ways to integrate $\int \tan^2( x)\sec^3( x) dx$ I would proceed by thus , let $y = [\sec (x)]^2 $
then
$$dy = 2 \cdot \sec(x) \cdot \sec(x) \cdot \tan(x) \cdot dx = 2 \cdot ( \sec (x))^2 \cdot \tan(x) \cdot dx $$
so,
$$
2 \tan^2(x) \sec^2 (x) dx = \sec(x) \cdot \tan(x) \cdot dy = y(y-1)^\frac{1}{2} \cdot dy
$$
since $$\sec(x) = y^{\frac{1}{2}}$$ and by considering
positive square roots only $\tan y = ( \sec^2(x) - 1)^{1/2} = (y - 1)^{1/2}$. Thus the substitution $y = \sec^2 x$ yields
$$
2 \int \tan^2 (x) \sec^3(x) dx = \int (y(y - 1) )^{1/2} dy
$$
and this later form can be reduced to the standard form $\int(z^2 - a^2)^{1/2} dz$ since
$$
y(y-1)=(y-(1/2))^2 - (1/2)^2 .
$$
What are the other ways to integrate this expression, except for the substitution
$\tan^2(x)^2=\sec (x)^2 -1$ which gives
$$
\sec^5(x) - \sec^3 (x)
$$
in the integrand which I quite don't like.
|
Let you want to solve $\int R\big(\sin(x),\cos(x)\big)dx$ and you know that $$R\big(\sin(x),-\cos(x)\big)\equiv -R\big(\sin(x),\cos(x)\big)$$ then you can take $\sin(x)=t$ for a good substitution. We have here $$\int \tan^2(x)\sec^3(x)dx=\int \frac{\sin^2(x)dx}{\cos^5(x)}$$ and we can see the above statement is true for the last integrand. By taking $\sin(x)=t$, we have $$\int\frac{t^2}{(1-t^2)^3}dt$$ which can be solve by fractions method.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/203861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Vector Project onto Subspace So the question is:
Let S be the subspace of $\mathbb{R}^3$ spanned by the vectors $ u_2 = \begin{pmatrix} \frac{2}{3}\\\frac{2}{3}\\\frac{1}{3}\end{pmatrix} u_3 = \begin{pmatrix} \frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}\\0\end{pmatrix}$. Let $x=(1,2,2)^T$. Find the projection p of x onto S. Show that $(p-x)\perp u_2$ and $(p-x)\perp u_2$.
I understand how to show that they are perpendicular and I actually found the answer for the projection. It's:
$\begin{pmatrix} \frac{23}{18}\\\frac{41}{18}\\\frac{8}{9} \end{pmatrix}$
The problem is, i have no idea why i am doing what I am doing, I just followed my notes. Can someone explain why I was supposed to do:
$(xu_2)u_2 + (xu_3)u_3$
To find p.
|
Let $\{v_1,\dotsc,v_m\}$ be a basis for a subspace $V$ of $\Bbb R^n$. The matrix of the orthogonal projection onto $V$ is
$$
P=A\left(A^\top A\right)^{-1}A^\top
$$
where $A$ is the matrix whose columns are $\{v_1,\dotsc,v_m\}$. A nice discussion of this can be found in these notes.
In our case we have
$$
A=
\left[\begin{array}{rr}
\frac{2}{3} & \frac{1}{2} \, \sqrt{2} \\
\frac{2}{3} & -\frac{1}{2} \, \sqrt{2} \\
\frac{1}{3} & 0
\end{array}\right]
$$
Thus
$$
P =
\left[\begin{array}{rrr}
\frac{17}{18} & -\frac{1}{18} & \frac{2}{9} \\
-\frac{1}{18} & \frac{17}{18} & \frac{2}{9} \\
\frac{2}{9} & \frac{2}{9} & \frac{1}{9}
\end{array}\right]
$$
Now, the projection of $v=(1,2,2)$ is
$$
Pv=
\left(\frac{23}{18},\,\frac{41}{18},\,\frac{8}{9}\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/211403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Solve the dirichlet pde with the given conditions. Solve the Dirichelt problem:
$$\nabla^2u(x,y)=0$$
$$0\le x\le3,0<y<7$$
$$$u(x,0)=0, u(x,7)=sin((\pi)x/3$$ $$9\le x\le3$$
$$u(0,y)=u(3,y)=0$$
$$0\le y\le7$$
Using separation of variables I found that $X(0)=0$ and $X(3)=0$ because $u(0,y)=u(3,y)=0$. Also, $u(x,0)=0$ so $Y(0)=0$ and $u(x,7)=sin(\pi x/3)$. I found that $\lambda=n^2$ (this came from previous calculations from a different problem) and $X_n=sin(\pi x/3)$. I am having trouble with the $Y''-n^2Y=0$. The solution for Y would be $Y=ae^{-n y}
+be^{n y}$. If you plug $0$ in for y then you would get $a+b=0$ so $a=-b$ or vice versa. Is this right? I get the purpose of this is to find $Y_n$ so you can combine that with $X_n$ to get the solution of $u(x,t)$. Am I on the right track?
|
Although your approach is not completely wrong, your approach is not the best.
Note that for $\nabla^2u(x,y)=0$ with conditions of the types $u(x,0)$ , $u(x,7)$ , $u(0,y)$ and $u(3,y)$ , according to http://eqworld.ipmnet.ru/en/solutions/lpde/lpde301.pdf#page=2 , we have special consideration:
$u(x,y)=\sum\limits_{n=1}^\infty A(n)\sinh\dfrac{n\pi(3-x)}{7}\sin\dfrac{n\pi y}{7}+\sum\limits_{n=1}^\infty B(n)\sinh\dfrac{n\pi x}{7}\sin\dfrac{n\pi y}{7}+\sum\limits_{n=1}^\infty C(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi(7-y)}{3}+\sum\limits_{n=1}^\infty D(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi y}{3}$
$u(0,y)=0$ :
$\sum\limits_{n=1}^\infty A(n)\sinh\dfrac{3n\pi}{7}\sin\dfrac{n\pi y}{7}=0$
$A(n)=0$
$\therefore u(x,y)=\sum\limits_{n=1}^\infty B(n)\sinh\dfrac{n\pi x}{7}\sin\dfrac{n\pi y}{7}+\sum\limits_{n=1}^\infty C(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi(7-y)}{3}+\sum\limits_{n=1}^\infty D(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi y}{3}$
$u(3,y)=0$ :
$\sum\limits_{n=1}^\infty B(n)\sinh\dfrac{3n\pi}{7}\sin\dfrac{n\pi y}{7}=0$
$B(n)=0$
$\therefore u(x,y)=\sum\limits_{n=1}^\infty C(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi(7-y)}{3}+\sum\limits_{n=1}^\infty D(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi y}{3}$
$u(x,0)=0$ :
$\sum\limits_{n=1}^\infty C(n)\sinh\dfrac{7n\pi}{3}\sin\dfrac{n\pi x}{3}=0$
$C(n)=0$
$\therefore u(x,y)=\sum\limits_{n=1}^\infty D(n)\sin\dfrac{n\pi x}{3}\sinh\dfrac{n\pi y}{3}$
$u(x,7)=\sin\dfrac{\pi x}{3}$ :
$\sum\limits_{n=1}^\infty D(n)\sinh\dfrac{7n\pi}{3}\sin\dfrac{n\pi x}{3}=\sin\dfrac{\pi x}{3}$
$D(n)=\begin{cases}\text{csch}\dfrac{7\pi}{3}&\text{when}~n=1\\0&\text{when}~n\neq1\end{cases}$
$\therefore u(x,y)=\text{csch}\dfrac{7\pi}{3}\sin\dfrac{\pi x}{3}\sinh\dfrac{\pi y}{3}$
Note that this solution suitable for $x,y\in\mathbb{C}$ , not only suitable for $0\leq x\leq3$ and $0<y<7$ .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/214376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Elementary Question about limits Prove
$$
\lim_{x \to 0}\frac{x^3 - \sin^3x}{x - \ln{(1+x)} - 1 + \cos x} = 0
$$
Obviously, Using l'Hôpitals rule, we can evaluate this limit. But, taking derivatives of such functions is such a mess. Anyone sees a trick to do this problem faster? any ideas?
|
We have $x^3-\sin^3x=(x-\sin x)(x^2+x\sin x+\sin^2x)\approx\frac{x^3}{6}(3x^2)=\displaystyle\frac{x^5}{2}$ as $x\to 0$.
Hence
\begin{equation*}
\begin{array}{lll}
\displaystyle\lim_{x \to 0}\frac{x^3 - \sin^3x}{x - \ln{(1+x)} - 1 + \cos x} &=&\displaystyle\lim_{x\to 0}\frac{x^5}{2(x - \ln{(1+x)} - 1 + \cos x)}\\
&\overset{H}{=}&\displaystyle\lim_{x \to 0}\frac{5x^4}{2(1-\frac{1}{1+x}-\sin x)}\\
&\overset{H}{=}&\displaystyle\lim_{x \to 0}\frac{10x^3}{\frac{1}{(1+x)^2}-\cos x}\\
&\overset{H}{=}&\displaystyle\lim_{x \to 0}\frac{30x^2}{\frac{-2}{(1+x)^3}+\sin x}\\
&=&\displaystyle\frac{0}{\frac{-2}{(1+0)^3}+0}=0.
\end{array}
\end{equation*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/214579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
proof by by directly manipulating the sum: Prove that
$$\sum_{i=0}^{n} x ^{i} = \frac{ 1-x ^{n+1} }{ 1-x }$$
to be used by directly manipulating the sum: let A be the sum, and show that xA = A + x^(n+1) -1
I don't get how its going to equal $\frac{ 1-x ^{n+1} }{ 1-x }$
$$xA=x\sum_{i=0}^n x^i=x(x^0+x^1+x^2+x^3+...)=x^1+x^2+x^3+x^4+...$$
so then i have
$$\sum_{i=0}^{n} x ^{n+1}-1$$
I'm stock on how its going to equal one to each other.
|
Write it out like this:
$$\begin{align*}
A&=x^0+\color{red}{x^1+x^2+\ldots+x^n}\\
xA&=\quad\quad\,\color{red}{x^1+x^2+\ldots+x^n}+x^{n+1}\\
A-xA&=x^0+\qquad\qquad\color{red}{0}\qquad\quad\,-x^{n+1}
\end{align*}$$
Then $(1-x)A=A-xA=x^1-x^{n+1}=1-x^{n+1}$, so $$A=\frac{1-x^{n+1}}{1-x}\;.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/215210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Prove the trigonometric identity $(35)$ Prove that
\begin{equation}
\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)= \left\{
\begin{aligned}
\sqrt{n} \space \space \text{for $n$ odd}\\
\\
\ 1 \space \space \text{for $n$ even}\\
\end{aligned}
\right.
\end{equation}
I found this identity here at$(35)$. At the moment I don't know where I should start from.
Thanks!
|
$$\tan nx=\frac{^nC_1t-^nC_3t^3+^nC_5t^5-\cdots }{^nC_0t^0-^nC_2t^2+^nC_4t^4-\cdots }$$ where $t=\tan x$
If $\tan nx=0, x=\frac {k\pi}n$ where $0\le k< n$, clearly, the roots of this $n$-degree equation are $\tan\frac{k\pi}n$
If $n$ is odd,
$^nC_n(-1)^{\frac{n-1}2}t^n+^nC_{n-2}(-1)^{\frac{n-3}2}t^{n-2}+\cdots-^nC_3t^3+^nC_1t=0$
$^nC_n(-1)^{\frac{n-1}2}t^{n-1}+^nC_{n-2}(-1)^{\frac{n-3}2}t^{n-3}+\cdots-^nC_3t^2+^nC_1=0$ if we exclude $k=0$
So, $\prod_{k=1}^{n-1}\tan \left(\frac{k \pi}{n}\right)=n(-1)^{\frac{n-1}2}$ (applying Vieta's formula)
Now, $\tan \left(\frac{(n-k) \pi}{n}\right)=\tan \left(\pi-\frac{k \pi}{n}\right)=-\tan \left(\frac{k \pi}{n}\right)$
So, there are $\frac{n-1}2$ such pairs and $\lfloor \frac{n-1}2 \rfloor=\frac{n-1}2$ as $n$ is odd.
$\prod_{k=1}^{n-1}\tan \left(\frac{k \pi}{n}\right)$
$=(-1)^{\frac{n-1}2}\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan^2 \left(\frac{k \pi}{n}\right) $
$\implies (-1)^{\frac{n-1}2}\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan^2 \left(\frac{k \pi}{n}\right)=n(-1)^{\frac{n-1}2} $
$\implies \left(\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)\right)^2=n$
$\implies \prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)=\sqrt n$ as all the angles lies in $(0,\frac \pi 2)$
If $n$ is even,
$ ^nC_1t^{n-1}-^nC_3t^{n-3}+^nC_5t^{n-5}-\cdots+^nC_{n-1}(-1)^{\frac n 2}t=0$ which has roots $\tan\frac{k\pi}n$ where $0\le k<n$ and $k\ne \frac n 2$ as $k=\frac n 2$ corresponds to $\tan \frac \pi 2(=\infty)$ which has occurred as the co-efficient of $t^n$ is $0$.
So, $ ^nC_1t^{n-2}-^nC_3t^{n-4}+^nC_5t^{n-6}-\cdots+^nC_{n-1}(-1)^{\frac n 2}=0$ if we exclude $k=0$ i.e., $(n-2)$ degree equation in $t$.
So, $$\prod_{\substack{k=1 \\ k\neq \frac{n}{2}}}^{n-1}\tan \left(\frac{k \pi}{n}\right)=-(-1)^{\frac n 2}$$
Now, $\tan \left(\frac{(n-k) \pi}{n}\right)=\tan \left(\pi-\frac{k \pi}{n}\right)=-\tan \left(\frac{k \pi}{n}\right)$
So, there are $\frac{n-2}2=(\frac n 2 -1)$ such pairs and $\lfloor \frac{n-1}2 \rfloor=\frac{n-2}2$ as $n$ is even.
$-(-1)^{\frac n 2}$
$={\displaystyle\prod_{\substack{k=1 \\ k \neq \frac{n}{2}}}^{n-1}} \tan \left(\frac{k \pi}{n}\right)$
$=(-1)^{\frac{n-2}2}\left(\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)\right)^2$
$\implies \left(\prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)\right)^2=1$
$\implies \prod_{k=1}^{\lfloor (n-1)/2 \rfloor}\tan \left(\frac{k \pi}{n}\right)=1$ as all the angles lies in $(0,\frac \pi 2)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/218766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
}
|
Differentiating $x^2 \sqrt{2x+5}-6$ How do I differentiate this function: f(x)= $x^2 \sqrt{2x+5}-6$
I had: I had $2x\sqrt{2x+5} + x^2 \dfrac{1}{2\sqrt{2x+5}}$ but the correction model said it was I had $2x\sqrt{2x+5} + x^2 \dfrac{2}{2\sqrt{2x+5}}$
|
I would use the product rule, which you seem to have tried from your comment above. Here would be the idea:
$\begin{split}
\frac{d[x^2 \sqrt{2x+5} - 6]}{dx}
&= \frac{dx^2}{dx} \sqrt{2x+5} + x^2 \frac{d(2x+5)^{\1/2}}{dx} \\
&= 2x \sqrt{2x+5} + x^2 \frac{1}{2} (2x+5)^{-1/2} \cdot 2 \\
&= 2x \sqrt{2x+5} + \frac{x^2}{\sqrt{2x+5}}
\end{split}
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/220858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
Evaluate the following integral, $\int\sqrt{4-\sqrt{x}}dx$ Evaluate the following integral,
$$\int\sqrt{4-\sqrt{x}}dx$$
$$\int \sqrt{4-\sqrt{x}}dx=\int \sqrt{2^2-(x^{1/4})^2}dx$$
Considering the common subsitution for $a^2-x^2$, let $$x^{1/4}=2\sin t$$
$$x=16\sin^4t$$$$\int dx=\int 64\sin^3t\cos t dt$$
Therefore by subsitution, we have
$$\int \sqrt{4-\sqrt{x}}dx=\int \sqrt{{4-(2\sin t)^2}}(64 \sin^3t\cos t)dt=\int \sqrt{4\cos^2 t}(64 \sin^3t\cos t) dt=\int128\cos^2t\sin^3 t dt=\int 128(\cos^2 t)*(1-cos^2t)\sin tdx=128\int \cos^2t\sin t-\cos^4 t\sin t dt=128(-1/3\cos^3 t+1/5\cos^5t)+C$$
Is there any mistake in my workings? This is a very important piece of work for me, and I was hoping SE could check it for me. Thanks!
I suspect something is very wrong but I can't dig the mistake...i.e I tested it for a definite integral and it got me a different answer.
P.S Sorry about the messy typing. I am rather new to LaTex.
|
So, substitute $u = \sqrt{x}$ and $\mathrm{d}u = \frac{1}{2 \sqrt{x}} \,\mathrm{d}x$:
$$= 2 \int \!\sqrt{4-u}\, u \, \mathrm{d}u$$
For the integrand $\sqrt{4-u}\, u$, substitute $s = 4-u$ and $\mathrm{d}s = - \mathrm{d}u$:
$$= 2 \int \!(s-4) \sqrt{s}\, \mathrm{d}s$$
Expanding the integrand $(s-4) \sqrt{s}$ gives $s^{\frac{3}{2}}-4 \sqrt{s}$:
$$= 2 \int\! (s^{\frac{3}{2}}-4 \sqrt{s})\, \mathrm{d}s$$
Integrate the sum term by term and factor out constants:
$$= 2 \int \!s^{\frac{3}{2}} \, \mathrm{d}s-8 \int\! \sqrt{s}\,\mathrm{d}s$$
The integral of $\sqrt{s}$ is $\frac{2 }{3}\,s^\frac{3}{2}$:
$$= 2 \int \!s^{\frac{3}{2}}\,\mathrm{d}s-\frac{16}{3}\,s^{\frac{3}{2}}$$
The integral of $s^\frac{3}{2}$ is $\frac{2}{5}s^\frac{5}{2}$:
$$= \frac{4}{5}s^\frac{3}{2}-\frac{16}{3} s^\frac{3}{2}+constant$$
Substitute back for $s = 4-u$:
Hope I didn't made any typos.
$$= \frac{4}{5} (4-u)^\frac{5}{2}-\frac{16}{7} (4-u)^\frac{3}{2}+constant$$
Substitute back for $u = \sqrt{x}$:
$$= \frac{4}{5} (4-\sqrt{x})^\frac{5}{2}-\frac{16}{3} (4-\sqrt{x})^\frac{3}{2}+constant$$
Factor the answer a different way:
$$= -\frac{4}{15} (4-\sqrt{x})^\frac{3}{2} \,(3 \sqrt{x}+8)+constant$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/223463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
How to find a parametric equation for the tangent line to the curve of intersection of the cylinders? How can i find a parametric equation for the tangent line to the curve of intersection of the cylinders $x^2 + y^2 = 4$ and and $x^2 + z^2 = 1$ at the point $P_0(1,\sqrt{3}, 0)$?
|
The equation of any line passing through $(1,\sqrt 3,0)$ can be written as
$\frac{x-1}a=\frac{y-\sqrt 3}b=\frac z c$ where $a^2+b^2+c^2=1$
So,$cx=az+c,cy=bz+\sqrt 3c$
Putting the values of $x,y$ in $x^2+y^2=4,$
$(az+c)^2+(bz+\sqrt 3c)^2=4c^2$
$(a^2+b^2)z^2+2zc(a+\sqrt 3 b)=0$
But this is a quadratic in $z,$, each root represent the $z$ co-ordinate of the intersection.
For tangency, both root should be same, so $\{2c(a+\sqrt 3 b)\}^2=4(a^2+b^2)0$
$\implies c(a+\sqrt 3 b)=0--->(1)$
Putting the value of $x$ in $x^2+z^2=1,$
$(az+c)^2+c^2z^2=c^2\implies (a^2+c^2)z^2+2caz=0$
So like 1st case, $(2ca)^2=4(a^2+c^2)0\implies ca=0--->(2)$
Form $(1)$ and $(2)$,
if $c=0,a^2+b^2=1\implies x=a+1,y=b+\sqrt 3,z=0 $
if $c\ne 0,a=0$ and $a+\sqrt 3 b=0\implies b=0\implies x=1,y=\sqrt 3,z=c=\pm1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/225398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
sum of a series Can
\begin{equation}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left(
2k+1\right) ^{2}+a^{2}},
\end{equation}
be summed explicitly, where $a$ is a constant real number? If $a=0,$ this
sum becomes
\begin{equation}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1}=\frac{\pi }{4}.
\end{equation}
What about for $a\neq 0$. I tried this method
\begin{eqnarray*}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left(
2k+1\right) ^{2}+a^{2}} &=&\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}%
\left[ \frac{1}{2k+1+ja}+\frac{1}{2k+1-ja}\right] \\
&=&\frac{1}{2}\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1+ja}+\frac{1}{2}%
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1-ja} \\
&=&\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}\int_{0}^{1}x^{2k+ja}dx+%
\frac{1}{2}\sum_{k\geq 0}\left( -1\right) ^{k}\int_{0}^{1}x^{2k-ja}dx \\
&=&\frac{1}{2}\int_{0}^{1}\left[ \sum_{k\geq 0}\left( -1\right) ^{k}x^{2k+ja}%
\right] dx+\frac{1}{2}\int_{0}^{1}\left[ \sum_{k\geq 0}\left( -1\right)
^{k}x^{2k-ja}\right] dx \\
&=&\frac{1}{2}\int_{0}^{1}\frac{x^{ja}}{1+x^{2}}dx+\frac{1}{2}\int_{0}^{1}%
\frac{x^{-ja}}{1+x^{2}}dx \\
&=&\frac{1}{2}\int_{0}^{1}\frac{x^{ja}+x^{-ja}}{1+x^{2}}dx
\end{eqnarray*}
but I was stucked at the last equations. Can any one give me some hint or tell me that the analytic expression doesn't exist. Thanks very much!
|
The series is summable and has the closed form formula
$$-\frac{1}{4}\,{\frac {\pi }{\sin \left( \frac{1}{2}\pi \, \left( 3 + ia \right )
\right) }} = \frac{1}{4}{\frac {\pi }{\cosh \left( \frac{\pi \,a}{2} \right) }} .$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/226308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
}
|
Definite integral involving Fresnel integrals I am seeking to evaluate
$\int_0^{\infty} f(x)/x^2 \, dx$
with
$f(x)=1-\sqrt{\pi/6} \left(\cos (x) C\left(\sqrt{\frac{6 x}{\pi }} \right)+S\left(\sqrt{\frac{6 x}{\pi }} \right) \sin
(x)\right)/\sqrt{x}$.
$C(x)$ and $S(x)$ are the Fresnel integrals. Numerical integration suggests that the integral equals $2 \pi/(3 \sqrt{3})$, which would also be desirable within the (physical) context it arose. How can this be proved?
|
First of all,
$$
C(x)=\int_{0}^{x}\cos\left(\frac{1}{2}\pi t^{2}\right)dt=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{\pi/2} \,\left (x\right )}\cos(z^{2})\, dz,
$$
and
$$
S(x)=\int_{0}^{x}\sin\left(\frac{1}{2}\pi t^{2}\right)dt=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{\pi/2} \,\left (x\right )}\sin(z^{2})\, dz.
$$
So we have
$$
C\left(\sqrt{\frac{6x}{\pi}}\right)=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{3x}}\cos(z^{2})\, dz,
$$
and
$$
S\left(\sqrt{\frac{6x}{\pi}}\right)=\sqrt{\frac{2}{\pi}}\int_{0}^{\sqrt{3x}}\sin(z^{2})\, dz.
$$
Using these we can write
$$
\begin{eqnarray*}
f(x) & = & 1-\sqrt{\pi/6}\frac{\cos(x)C\left(\sqrt{\frac{6x}{\pi}}\right)+\sin(x)S\left(\sqrt{\frac{6x}{\pi}}\right)}{\sqrt{x}} \\ & = & \frac{\int_{0}^{\sqrt{3x}}(1-\cos(x-z^{2}))\, dz}{\sqrt{3x}}
\end{eqnarray*}
$$
and
$$
\int_{0}^{\infty}\frac{f(x)}{x^{2}}\, dx=\frac{2}{\sqrt{3}}\int_{0}^{\infty}\int_{0}^{\sqrt{3x}}\frac{\sin^{2}((x-z^{2})/2)}{x^{5/2}}\, dz\, dx.
$$
Let's introduce the new variable $z=t\sqrt{x}$. Then $dz=\sqrt{x}dt$ and
$$
\begin{eqnarray*}
\int_{0}^{\infty}\int_{0}^{\sqrt{3x}}\frac{\sin^{2}((x-z^{2})/2)}{x^{5/2}}\, dz\, dx & = & \int_{0}^{\infty}\int_{0}^{\sqrt{3}}\frac{\sin^{2}(x(1-t^{2})/2)}{x^{2}}\, dt\, dx\\
& = & \frac{1}{2}\int_{0}^{\infty}\int_{0}^{\sqrt{3}}\frac{\sin^{2}(x(1-t^{2}))}{x^{2}}\, dt\, dx\\
& = & \frac{1}{2}\int_{0}^{\infty}\int_{0}^{1}\frac{\sin^{2}(x(1-t^{2}))}{x^{2}}\, dt\, dx\\
& & +\frac{1}{2}\int_{0}^{\infty}\int_{1}^{\sqrt{3}}\frac{\sin^{2}(x(t^{2}-1))}{x^{2}}\, dt\, dx\\
& = & \frac{1}{2}\int_{0}^{1}\int_{0}^{\infty}\frac{\sin^{2}(x(1-t^{2}))}{x^{2}}\, dx\, dt\\
& & +\frac{1}{2}\int_{1}^{\sqrt{3}}\int_{0}^{\infty}\frac{\sin^{2}(x(t^{2}-1))}{x^{2}}\, dx\, dt\\
& = & \int_{0}^{1}\frac{\pi}{4}(1-t^{2})\, dt+\int_{1}^{\sqrt{3}}\frac{\pi}{4}(t^{2}-1)\, dt\\
& = & \frac{\pi}{3},
\end{eqnarray*}
$$
where the formula
$$
\int_0^{\infty}\frac{\sin^2(Ax)}{x^2}\,dx=\frac{1}{2}A\pi,\quad(A>0)
$$
was applied. Your conjecture was excellent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/227641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
Prove the following relation: I must prove the relation $$\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k}=2\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}.$$
I got this far before I got stuck:
$\begin{eqnarray*}
\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k} & = & \sum_{k=0}^{n+1}\left\{\binom{n+k}{k}+\binom{n+k}{k-1}\right\}\frac1{2^k}\\
& = & \sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^n\binom{n+k}{k-1}\frac1{2^k}+\binom{2n+1}{n}\frac1{2^k}.
\end{eqnarray*}$
If I can combine the second and third terms and get something same as first term, I am done but I could not do that.
|
Let $s_n=\displaystyle\sum_{k=0}^n\binom{n+k}k\frac1{2^k}$. Then
$$\begin{align*}
s_{n+1}&=\sum_{k=0}^{n+1}\binom{n+k+1}k\frac1{2^k}\\
&=\sum_{k=0}^{n+1}\left(\binom{n+k}k+\binom{n+k}{k-1}\right)\frac1{2^k}\\
&=\binom{2n+1}{n+1}\frac1{2^{n+1}}+\sum_{k=0}^n\binom{n+k}k\frac1{2^k}+\sum_{k=0}^n\binom{n+1+k}k\frac1{2^{k+1}}\\
&=\binom{2n+1}{n+1}\frac1{2^{n+1}}+s_n+\frac12\sum_{k=0}^n\binom{n+1+k}k\frac1{2^k}\\
&=s_n+\frac12\left(s_{n+1}-\binom{2n+2}{n+1}\frac1{2^{n+1}}\right)+\binom{2n+1}{n+1}\frac1{2^{n+1}}\\
&=s_n+\frac12s_{n+1}+\binom{2n+1}{n+1}\frac1{2^{n+1}}-\binom{2n+2}{n+1}\frac1{2^{n+2}}\;,
\end{align*}$$
and therefore
$$\begin{align*}
s_{n+1}&=2s_n+\binom{2n+1}{n+1}\frac1{2^n}-\binom{2n+2}{n+1}\frac1{2^{n+1}}\\
&=2s_n+\frac1{2^{n+1}}\left(2\binom{2n+1}{n+1}-\binom{2n+2}{n+1}\right)\\
&=2s_n+\frac1{2^{n+1}}\left(2\binom{2n+1}{n+1}-\binom{2n+1}{n+1}-\binom{2n+1}n\right)\\
&=2s_n+\frac1{2^{n+1}}\left(\binom{2n+1}{n+1}-\binom{2n+1}n\right)\\
&=2s_n\;.
\end{align*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/228339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
Permutation and cycles I need to determine the missing number to fulfill the following reproduction:
$$\pi=\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&1&2&6&7&8}$$
And here is the representation of $pi$ as the product of transpositions:
$$(1 i) (1 3) (2 5) (9 8) (8 7) (7 6) (3 6) = \pi$$
The result for $i$ must be 2 but I have no clue how to get there.
I tried to start at $(3 6)$ and then check on which transposition it gets pictured on but it doesn't seem to work for me. Any ideas on how to generally solve these kind of problems?
-Freddy
|
According to the comments, you’re multiplying from right to left. If you multiply the transpositions that you know completely, you get
$$\pmatrix{6&3&7&8&9&5&2&1&4\\1&9&6&7&8&2&5&3&4}\;,$$
which we rearrange into standard form as
$$\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&2&1&6&7&8}\;.\tag{1}$$
We want to choose $i$ so that
$$(1i)\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&2&1&6&7&8}=\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&1&2&6&7&8}\;.\tag{2}$$
There are several ways to think about this.
*
*You can notice that $(1)$ and $\pi$ differ only in what they do to $5$ and $6$, and it’s pretty clear that setting $i=2$ will fix that discrepancy.
*You can notice that if you carry out the multiplication on the lefthand side of $(2)$, you’ll get $$\pmatrix{1&2&3&4&5&6&7&8&9\\\cdot&\cdot&\cdot&\cdot&\cdot&i&\cdot&\cdot&\cdot}\;,$$ so if this product is to be $\pi$, $i$ must be $2$. (Of course you should then finish the multiplication to check.)
*You can use the fact that a transposition is its own inverse: $(2)$ holds if and only if $$\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&2&1&6&7&8}=(1i)\pmatrix{1&2&3&4&5&6&7&8&9\\3&5&9&4&1&2&6&7&8}\;,$$ and the product on the right is going to have a column $\binom5i$, so again $i$ must be $2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/229015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Probability that I will get a $4$ consecutive numbers I have five dice to roll. I roll them. What is the probability that I will get a straight with exactly four consecutive numbers and not $5$?
There are three options: $1,2,3,4$ or $2,3,4,5$ or $3,4,5,6$.
I have $1,2,3,4,*$ where $*$ can be either $1/2/3/4/6$. It cannot be five. Now we do $4!*5$.
Next, I take $2,3,4,5,*$ where $*= 2,3,4,5$. We get $4!*4$ and $4!*5$ for $3/4/5/6$.
I get the probability $(4!*14)/6^5$ but there are some complexities involved. Can anyone explain this clearly? I tried $4!*5*5$ since the position of * is variable. Now, I get $(4!*70)/6^5$ which is not correct either.
Can someone explain it systematically?
|
Imagine tossing the dice one at a time, and recording the results, or equivalently labelling the dice A to E, and recording the results as a string of length $5$, result on A, result on B, and so on. There are $6^5$ possibilities, all equally likely. Now we count the favourables.
The "straight" part can be of any of type $1,2,3,4$, $2,3,4,5$, or $3,4,5,6$.
First we deal with the $2,3,4,5$. In order not to get $5$ in a row, we must avoid $1$ and $6$, so we must double up something. What we double up can be chosen in $4$ ways. For each of these ways, the smallest non-doubled number can be placed in $5$ ways, then the second smallest in $4$ ways, then the third smallest in $3$ ways. Now the doubleton falls in the remaining spaces.
That gives a total of $(4)(5)(4)(3)=240$ ways to have $2,3,4,5$ as the "straight" part.
Now we deal with $1,2,3,4$. We could have the other number be a $6$, and then the $5$ numbers can be arranged in $5!=120$ ways.
Or else we could double up one of our numbers. We have already analyzed this, and seen there are $240$ ways to do it. So there are $360$ patterns where the straight part is $1,2,3,4$.
Similarly, there are $360$ patterns where the straight part is $3,4,5,6$.
So our total is $240+360+360=960$, and the required probability is
$$\dfrac{960}{6^5}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/231122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Induction: $\sum_{k=0}^n \binom nk k^2 = n(1+n)2^{n-2}$ I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: $$0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} $$
|
Marvis has given a typically excellent answer. I'll go ahead and show you an induction-style proof, just in case you're interested.
A useful basic combinatoric fact for this induction proof is Pascal's identity: $$\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}\tag{1}$$ Another nice basic fact is $$\sum_{k=0}^n\binom{n}k=2^n\tag{2}$$ for all $n\in\Bbb N$.
For each $n$, define $$f(n)=n(1+n)2^{n-2}\qquad \text{and}\qquad g(n)=\sum_{k=0}^nk^2\binom{n+1}{k},$$ so what we're trying to show is that $f(n)=g(n)$ for all $n$. In the $n=0$ case, this is clear, yes?
For the induction step, supposing that $f(n)=g(n)$ for some $n$, we must show that $f(n+1)=g(n+1)$.
Observe on the one hand that $$f(n+1)=(n+1)(2+n)2^{n-1}=(n+1)2^n+n(n+1)2^{n-1}=(n+1)2^n+2f(n),$$ so by inductive hypothesis, $$2g(n)=2f(n)=f(n+1)-(n+1)2^n.\tag{3}$$ On the other hand, we have $$\begin{align}g(n+1) &= \sum_{k=0}^{n+1}k^2\binom{n+1}k\\ &=\sum_{k=1}^{n+1}k^2\binom{n+1}{k}\\ &=(n+1)^2\binom{n+1}{n+1}+\sum_{k=1}^nk^2\binom{n+1}{k}\\ &\overset{(1)}{=}(n+1)^2+\sum_{k=1}^nk^2\left[\binom{n}{k}+\binom{n}{k-1}\right]\\ &=(n+1)^2+\sum_{k=1}^nk^2\binom{n}{k}+\sum_{k=1}^nk^2\binom{n}{k-1}\\ &=(n+1)^2+\sum_{k=0}^nk^2\binom{n}{k}+\sum_{k=1}^nk^2\binom{n}{k-1}\\ &=(n+1)^2+g(n)+\sum_{k=1}^nk^2\binom{n}{k-1}\\ &=(n+1)^2+g(n)+\sum_{k=0}^{n-1}(k+1)^2\binom{n}{k}\\ &=n^2+2n+1+g(n)+\sum_{k=0}^{n-1}k^2\binom{n}{k}+2\sum_{k=0}^{n-1}k\binom{n}{k}+\sum_{k=0}^{n-1}\binom{n}{k}\\ &=(n^2+2n+1)\binom{n}{n}+g(n)+\sum_{k=0}^{n-1}k^2\binom{n}{k}+2\sum_{k=0}^{n-1}k\binom{n}{k}+\sum_{k=0}^{n-1}\binom{n}{k}\\ &=g(n)+\sum_{k=0}^nk^2\binom{n}{k}+2\sum_{k=0}^nk\binom{n}{k}+\sum_{k=0}^n\binom{n}{k}\\ &=2g(n)+2\sum_{k=0}^nk\binom{n}{k}+\sum_{k=0}^n\binom{n}{k}\\ &\overset{(3)}{=}f(n+1)-(n+1)2^n+2\sum_{k=0}^nk\binom{n}{k}+\sum_{k=0}^n\binom{n}{k}\\ &\overset{(2)}{=}f(n+1)-n2^n+2\sum_{k=0}^nk\binom{n}{k}\\ &=f(n+1)-2\left(n2^{n-1}-\sum_{k=0}^nk\binom{n}{k}\right)\end{align}$$ Hence, to see that $f(n+1)=g(n+1)$, it suffices that $$n2^{n-1}=\sum_{k=0}^nk\binom{n}{k}$$ Indeed, note that for $1\leq k\leq n$ we have $$k\binom{n}{k}=k\cdot\frac{n!}{k!(n-k)!}=n\cdot\frac{(n-1)!}{(k-1)!\bigl((n-1)-(k-1)\bigr)!}=n\binom{n-1}{k-1},\tag{4}$$ and so $$\begin{align}n2^{n-1} &\overset{(2)}{=} n\sum_{k=0}^{n-1}\binom{n-1}{k}\\ &= n\sum_{k=1}^n\binom{n-1}{k-1}\\ &= \sum_{k=0}^{n-1}n\binom{n-1}{k-1}\\ &\overset{(4)}{=} \sum_{k=1}^nk\binom{n}{k}\\ &= \sum_{k=0}^nk\binom{n}{k},\end{align}$$ as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/231596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 4
}
|
Find $2\times 2$ matrix of linear transformation $T$ Given:
\[
T\left(\begin{bmatrix}-1 \\ -2\end{bmatrix}\right) = \begin{bmatrix}17 \\ 11\end{bmatrix} \text{ and } T\left(\begin{bmatrix}2 \\ 3\end{bmatrix}\right) = \begin{bmatrix}-30 \\ -17\end{bmatrix}
\]
Find a matrix such that:
\[ T(v) = \begin{bmatrix}? & ? \\ ? & ? \end{bmatrix}\cdot v \]
I have no idea where to start... Any help will be much appreciate it.
|
Let's write $$\begin{bmatrix} a & b \\ c & d\end{bmatrix}$$ for the matrix we want to find. Then the two given equations read
\begin{align*}
-a - 2b &= 17\\
-c - 2d &= 11\\[3mm]
2a + 3b &= -30\\
2c + 3d &= -17
\end{align*}
We have to solve this system, so let's first look at the equations for $a$ and $b$
\begin{align*}
-a - 2b &= 17\\
2a + 3b &= -30
\end{align*}
Adding the first two times to the second gives
\begin{align*}
-a - 2b &= 17\\
-b &= 4
\end{align*}
Hence $b = -4$ and $a = -9$. The second system reads
\begin{align*}
-c - 2d &= 11\\
2c + 3d &= -17
\end{align*}
Adding again, we have
\begin{align*}
-c - 2d &= 11\\
-d &= 5
\end{align*}
Hence $d = -5$ and $c = -1$.
So $$ \begin{bmatrix} -9 & -4 \\ -1 & -5\end{bmatrix} $$ is the matrix we looked for.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/233412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Finding all real zeros of the polynomial $$x^5 - 5x^4 +6x^3 -30x^2 +8x - 40 = 0$$
So far I have...
$$r/s: +- 1, +- 40, +- 2, +- 20, +- 4, +- 10, +-5, +- 8$$
Only $+ 5$ works.
Then I have $$(x + 5)( ) = x^5 - 5x^4 +6x^3 -30x^2 +8x - 40$$
Then you have to use long devision between $x + 5$ and $ x^5 - 5x^4 +6x^3 -30x^2 +8x - 40$
That's where I get lost. Help?!
|
Hint:
$$x^5-5x^4+6x^3-30x^2+8x-40=x^4(x-5)+6x^2(x-5)+8(x-5)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/237389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
If $a,b,c,d$ be the roots of the biquadratic $x^4-x^3+2x^2+x+1=0$ then show that $(a^3+1)(b^3+1)(c^3+1) (d^3+1)=16$ If $a,b,c,d$ be the roots of the biquadratic $x^4-x^3+2x^2+x+1=0$ then show that $(a^3+1)(b^3+1)(c^3+1) (d^3+1)=16$
I have tried to solve the equation first and find the values of the roots but it becomes very long process. Is there any easy process?
|
$$\begin{align}
a+b+c+d &= \phantom{-}1 \\
ab+bc+cd+ac+ad+bd &= \phantom{-}2 \\
abc+bcd+abd+acd &= -1 \\
abcd &= -1
\end{align}$$
All you need to find is
$$\begin{align}
(abc)^3 + (abd)^3 + (acd)^3 + (bcd)^3 &= p \\
(ab)^3 + (ac)^3 + (ad)^3 + (bc)^3 + (bd)^3 + (cd)^3 & = q \\
a^3 + b^3 + c^3 +d^3 &= r
\end{align}$$
And you need to show
$$(abcd)^3 + p + q + r + 1 = 16$$
which is really just
$$p+q+r = 16$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/241751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Why is the determinant of a symplectic matrix 1? Suppose $A \in M_{2n}(\mathbb{R})$. and$$J=\begin{pmatrix}
0 & E_n\\
-E_n&0
\end{pmatrix}$$
where $E_n$ represents identity matrix.
if $A$ satisfies $$AJA^T=J.$$
How to figure out $$\det(A)=1~?$$
My approach:
I have tried to separate $A$ into four submartix:$$A=\begin{pmatrix}A_1&A_2 \\A_3&A_4 \end{pmatrix}$$
and I must add a assumption that $A_1$ is invertible.
by elementary transfromation:$$\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}\rightarrow \begin{pmatrix}A_1&A_2 \\ 0&A_4-A_3A_1^{-1}A_2\end{pmatrix}$$
we have:
$$\det(A)=\det(A_1)\det(A_4-A_3A_1^{-1}A_2).$$
From$$\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}\begin{pmatrix}0&E_n \\ -E_n&0\end{pmatrix}\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}^T=\begin{pmatrix}0&E_n \\ -E_n&0\end{pmatrix}.$$
we get two equalities:$$A_1A_2^T=A_2A_1^T$$ and $$A_1A_4^T-A_2A_3^T=E_n.$$
then $$\det(A)=\det(A_1(A_4-A_3A_1^{-1}A_2)^T)=\det(A_1A_4^T-A_1A_2^T(A_1^T)^{-1}A_3^T)=\det(A_1A_4^T-A_2A_1^T(A_1^T)^{-1}A_3^T)=\det(E_n)=1,$$
but I have no idea to deal with this problem when $A_1$ is not invertible.
|
There is an easy proof for real and complex case which does not require the use of Pfaffians. This proof first appeared in a Chinese text. Please see http://arxiv.org/abs/1505.04240 for the reference.
I reproduce the proof for the real case here. The approach extends to complex symplectic matrices.
Taking the determinant on both sides of $A^T J A = J$,
$$\det(A^T J A) = \det(A^T) \det(J) \det(A) = \det(J).$$
So we immediately have that $\det(A) = \pm 1$.
Then let us consider the matrix $A^TA + I.$
Since $A^TA$ is symmetric positive definite,
its eigenvalues are real and greater than $1$.Therefore its determinant, being the product of its eigenvalues, has $\det(A^TA +I) > 1$.
Now as $\det(A) \ne 0$, $A$ is invertible. Using this we may write
$$ A^TA + I = A^T( A + A^{-T}) = A^T(A + JAJ^{-1}).$$
Denote the four $N \times N$ subblocks of $A$ as follows,
$$
A = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix},
\text{ where } A_{11},A_{12},A_{21},A_{22} \in \mathbb{R}^{N \times N}.
$$
Then we compute
$$ A + JAJ^{-1} = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}
+ \begin{bmatrix} O & I_N \\ -I_N & O \end{bmatrix}
\begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}
\begin{bmatrix} O & - I_N \\ I_N & O \end{bmatrix}
= \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}
+ \begin{bmatrix} A_{22} & -A_{21} \\ -A_{12} & A_{11} \end{bmatrix}
= \begin{bmatrix} A_{11}+ A_{22} & A_{12} - A_{21} \\ - A_{12}+ A_{21} & A_{11} + A_{22} \end{bmatrix}.$$
Writing the blocks as $C := A_{11} + A_{22}$ and $D:= A_{12} - A_{21}$,
we make use of a unitary transform
$$
A + JAJ^{-1} = \begin{bmatrix} C & D \\ -D & C \end{bmatrix}
=
\frac{1}{\sqrt{2}}\begin{bmatrix} I_N & I_N \\ iI_N & -iI_N \end{bmatrix}
\begin{bmatrix} C + i D & O \\ O & C - i D \end{bmatrix}
\frac{1}{\sqrt{2}} \begin{bmatrix} I_N & -iI_N \\ I_N & iI_N \end{bmatrix}.
$$
We plug this factorization into our identity.
Note that $C,D$ are both real. This allows the complex
conjugation to commute with the determinant (as it is a polynomial of its
entries)
$$0 < 1 < \det(A^TA + I) = \det(A^T(A + JAJ^{-1})) \\
= \det(A) \det(C + i D) \det(C - iD) \\
= \det(A) \det(C + i D) \det\left(\overline{C + iD}\right)\\
= \det(A) \det(C + iD) \overline{\det(C + iD)}
= \det(A) \left\lvert \det(C + iD)\right\rvert^2.
$$
Clearly, none of the two determinants on the RHS can be zero,
so we may conclude
$\left\lvert \det(C + iD) \right\rvert^2 > 0$. Dividing this through on both sides,
we have $\det(A) > 0$, and thus $\det(A) = 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/242091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 7,
"answer_id": 0
}
|
Solve for x $\left\lfloor x \right\rfloor^3+2x^2=x^3+2\left\lfloor x\right\rfloor^2$ Solve for x
$$\left\lfloor x \right\rfloor^3+2x^2=x^3+2\left\lfloor x\right\rfloor^2$$
where $\left\lfloor t \right\rfloor$ denotes the largest integer not exceeding t
$X \in \mathbb{Z}$ is a solution. Is there other root? Thanks.
|
Write the equation as $x^3 - 2x^2 = \lfloor x \rfloor^3 - 2\lfloor x \rfloor^2$. Clearly this has solutions in $\mathbb{Z}$.
On most intervals of the form $[n, n+1]$ with $n \in \mathbb{Z}$, the function $x^3 - 2x^2$ is monotone. You will only get interesting values on the intervals where it isn't.
To find those, take the derivativve $3x^2 - 4x$ and set it to zero - you get $0$ and $\frac{4}{3}$ as critical points. So the only possible zero outside of $\mathbb{Z}$ will occur where $\lfloor x \rfloor = 1$.
Solve the equation now: $x^3 - 2x^2 = -1$ has three real solutions, one of which lies in the interval $(1, 2)$ and therefore satisfies $\lfloor x \rfloor = 1$.
This solution is $\frac{1}{2}(1 + \sqrt{5})$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/243299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Minimum Possible Number How to find the minimum possible number of length N ,which is simultaneously divisible by the single digit prime number like 2,3,5,7 ?like of length 5 minimum possible number is 10080.
|
As the lcm of $2,3,5,7$ is $210,$ we need to find the minimum multiple in $N$ digits.
So, $N$ must be $\ge 3$ to admit solution.
The minimum natural number with $N$ digits is $M=100\cdots00$ with $(N-1)$ zeros.
So, if $D=\lfloor\frac M {210}\rfloor,$ the answer will be $210(D+1)$
For example if $N=4,D=\lfloor\frac {1000} {210}\rfloor=4,$ the answer will be $210(4+1)=1050$
As pointed out by Thomas in his comment,
we can use Carmichael Function, $\lambda(21)=lcm(\lambda(3), \lambda(7))=lcm(2,6)=6$
So, $10^6\equiv1\pmod {21}\implies 10^{6m}\equiv1\pmod {21}$
$\implies 10^{6m+1}\equiv{10}\pmod {210},$ the minimum number with length $(6m+1+1)=6m+2,$ will be $10^{6m+1}+(210-10)$
$\implies 10^{6m+2}\equiv{100}\pmod {210},$ the minimum number with length $6m+3,$ will be $10^{6m+1}+(210-100)$
and so on.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/244345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Dual of a Linear Program \begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b
\end{align}
Note that here $x$ is unrestricted. I need to prove that the dual of this program is given by
\begin{align}
\max_{\lambda} \lambda^Tb \\
s.t.~\lambda^TA\leq c^T
\end{align}
But in the constraint, I always get an equality (using what I learnt)
\begin{align}
\max_{\lambda} \lambda^Tb \\
s.t.~\lambda^TA = c^T
\end{align}
Please give some explanation also.
|
\begin{align}
\min_{x} c^Tx \\
s.t.~Ax=b \\
Unrestricted
\end{align}
Take $x=x_1-x_2$
\begin{align}
\min c^T(x_1-x_2) \\
s.t.~A(x_1-x_2)=b \\
~ x_1, x_2 \ge 0
\end{align}
This is can be written as
\begin{align}
\min {\begin {pmatrix}c \\ -c \\ \end {pmatrix} }^T \begin {pmatrix} x_1 \\ x_2 \\ \end {pmatrix} \\
s.t.~\begin {pmatrix} A, & -A \end {pmatrix} \begin {pmatrix} x_1 \\ x_2 \\ \end {pmatrix}=b \\
~ x_1, x_2 \ge 0
\end{align}
Now, this is in the standard form of the linear program. Therefore, the dual can be written as
\begin{align}
\max b^T y \\
s.t.~ \begin {pmatrix}c \\ -c \\ \end {pmatrix} -
\begin {pmatrix} A^T\\ -A^T \\ \end {pmatrix} y \ge 0\\
~ y \ge 0
\end{align}
This can be simplify as
\begin{align}
\max b^T y \\
s.t.~ c-A^T y \ge 0\\
~ -c+A^T y \ge 0\\
~ y \ge 0
\end{align}
\begin{align}
\max b^T y \\
s.t.~ A^T y \le c\\
~ A^T y \ge c\\
~ y \ge 0
\end{align}
This is equivalent to
\begin{align}
\max b^T y \\
s.t.~ A^T y = c\\
~ y \ge 0
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/244875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Proving $\lim\limits_{x\to a}(a-x)\int\limits_0^x\frac{f(y)}{(a-y)^2}\,dy=f(a)$ How to prove for a continuous function $f$, the following limit holds?
$$\lim_{x\to a}\,(a-x)\int_0^x\frac{f(y)}{(a-y)^2}\,dy=f(a)$$
|
Here is general proof without using limit theorems:
First remark that
$$(x-a) \cdot \int_0^x \frac{f(a)}{(a-y)^2} = f(a)+ \frac{f(a) \cdot (x-a)}{a}$$
hence
$$\left|(a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) \right| \leq \left| (a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) + \frac{f(a) \cdot (x-a)}{a} \right| + \underbrace{\left|\frac{f(a) \cdot (x-a)}{a} \right|}_{\to 0 \, (x \to a)}$$
Now we have by the first equation
$$(a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) + \frac{f(a) \cdot (x-a)}{a} = (x-a) \cdot \int_0^x \frac{f(y)-f(a)}{(a-y)^2} \, dy $$
Let $\varepsilon>0$. Since $f$ is continuous we find $\delta>0$ such that $|f(y)-f(a)| \leq \varepsilon$ for all $y \in B(a,\delta)$. Thus
$$\left|(x-a) \cdot \int_{(0,x) \cap B(a,\delta)} \frac{f(y)-f(a)}{(a-y)^2}\right| \leq \left(1- \frac{x-a}{a} \right) \cdot \varepsilon \\
\left|(x-a) \cdot \int_{(0,x) \cap B(a,\delta)^c} \frac{f(y)-f(a)}{(a-y)^2}\right| \leq (x-a) \cdot 2 \|f\|_{\infty} \cdot \frac{x}{\delta^2} \leq (x-a) \cdot 2 \|f\|_{\infty} \cdot \frac{x}{\delta_0^2} $$
for some fixed $\delta_0>0$. Hence
$$ \left| (a-x) \cdot \int_0^x \frac{f(y)}{(a-y)^2} \, dy - f(a) + \frac{f(a) \cdot (x-a)}{a} \right| \leq (x-a) \cdot 2 \|f\|_{\infty} \cdot \frac{x}{\delta_0^2} + \left(1- \frac{x-a}{a} \right) \cdot \varepsilon \to 0 \quad (\varepsilon \to 0, x \to a)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/246401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
}
|
What am I doing wrong in calculating this determinant? I have matrix:
$$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
And I want to calculate $\det{A}$, so I have written:
$$
\begin{array}{|cccc|ccc}
1 & 2 & 3 & 4 & 1 & 2 & 3 \\
2 & 3 & 3 & 3 & 2 & 3 & 3 \\
0 & 1 & 2 & 3 & 0 & 1 & 2 \\
0 & 0 & 1 & 2 & 0 & 0 & 1
\end{array}
$$
From this I get that:
$$
\det{A} = (1 \cdot 3 \cdot 2 \cdot 2 + 2 \cdot 3 \cdot 3 \cdot 0 + 3 \cdot 3 \cdot 0 \cdot 0 + 4 \cdot 2 \cdot 1 \cdot 1) - (3 \cdot 3 \cdot 0 \cdot 2 + 2 \cdot 2 \cdot 3 \cdot 1 + 1 \cdot 3 \cdot 2 \cdot 0 + 4 \cdot 3 \cdot 1 \cdot 0) = (12 + 0 + 0 + 8) - (0 + 12 + 0 + 0) = 8
$$
But WolframAlpha is saying that it is equal 0. So my question is where am I wrong?
|
The others have pointed out what's wrong with your solution. Let's calculate the determinant now:
\begin{align*}
\det \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix} &\stackrel{r1 - \frac12(r2+r3+r4)}{=}
\det \begin{bmatrix}
0 & 0 & 0 & 0 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
= 0.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/246606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
}
|
Compute the limit Consider an equation
$$
\tan (x) = \frac{a x}{x^2+b}
$$
where $a,b \neq 0$. Plotting $\tan(x)$ and function on the RHS we can see that this equation has infinitely many positive solutions $x_{n}$ and $x_{n} \sim \pi n$ as $n \to \infty$, i.e.
$$
\lim\limits_{n \to \infty} \frac{x_{n}}{n} = \pi.
$$
But is it possible to show the latter equality analitically?
|
Let us assume $a,b>0$. The proof can be easily adapted if $a,b$ are not positive.
Consider $f(x) = \dfrac{ax}{x^2+b} - \tan(x)$. Note that $f(x)$ is nice except at $x = m \pi + \pi/2$ i.e. $$f((m \pi + \pi/2)^-) = - \infty \,\,\,\,\,\, f((m \pi + \pi/2)^+) = \infty$$
$$f'(x) = \dfrac{a(b-x^2)}{(b+x^2)^2} - \sec^2(x)$$ For large enough $x$ i.e. $x > \sqrt{ab}$, we have that $f'(x) < 0$.
Hence, for large enough $n$ within $\left( (n \pi - \pi/2), (n \pi + \pi/2) \right)$, the function is decreasing and changes sign. Hence, there is exactly one root in this interval.
Further, we have that $f(n \pi) = \dfrac{an \pi}{n^2 \pi^2+b} > 0$.
\begin{align}
f \left(n \pi + \dfrac{a}{n \pi} \right) & = \dfrac{a\left(n \pi + \dfrac{a}{n \pi} \right)}{\left(n \pi + \dfrac{a}{n \pi} \right)^2+b} - \tan\left(n \pi + \dfrac{a}{n \pi} \right)\\
& = \dfrac{a\left(n \pi + \dfrac{a}{n \pi} \right)}{\left(n \pi + \dfrac{a}{n \pi} \right)^2+b} - \tan\left(\dfrac{a}{n \pi} \right)\\
& \leq \dfrac{a\left(n \pi + \dfrac{a}{n \pi} \right)}{\left(n \pi + \dfrac{a}{n \pi} \right)^2} - \tan\left(\dfrac{a}{n \pi} \right)\\
& = \dfrac{a}{\left(n \pi + \dfrac{a}{n \pi} \right)} - \tan\left(\dfrac{a}{n \pi} \right)\\
& \leq \dfrac{a}{n \pi} - \tan\left(\dfrac{a}{n \pi} \right)\\
& \leq 0.
\end{align}
Hence, the root in the interval $\left( (n \pi - \pi/2), (n \pi + \pi/2) \right) $ in fact lies within $\left( n \pi, n \pi + \dfrac{a}{n \pi} \right)$.
Hence, we have that $$\left \vert x_{n+k} - n\pi \right \vert \leq \dfrac{a}{n \pi}$$ where $k$ is a fixed natural number and takes into account some initial ugliness in the function where there could be possibly more or less roots. Hence,
$$\left \vert \dfrac{x_{n+k}}{n} - \pi \right \vert \leq \dfrac{a}{n^2 \pi}$$
This gives your desired result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/246802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Limit $\lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $ $\displaystyle \lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $
|
$$\dfrac12 \cdot \dfrac34 \cdot \dfrac78 \cdots \dfrac{2n-1}{2n} = \left(1 - \dfrac12\right)\left(1 - \dfrac14\right)\left(1 - \dfrac16\right)\left(1 - \dfrac18\right)\cdots\left(1 - \dfrac1{2n}\right)$$
Since $$\dfrac12 + \dfrac14 + \dfrac16 + \cdots + \dfrac1{2n} + \cdots$$ diverges, the infinite product goes to $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/247360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
Differential equation problems I need your help to finish my homework. Can somebody help me?
I cannot finish all my homework especially in this problem.
*
*$\dfrac{dy}{dx} = \dfrac{y^2 -1}{x}$; my answer is $-\dfrac{1}{y} -y = \ln x + x$, is it right?
*$\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$
*$\dfrac{dy}{dx} = \dfrac{2xy + 3y^2}{x^2 + 2xy}$
|
For 2. The equation can be formed as:$$\frac{dy}{dx}=\frac{x^2\left(1+\frac{y^2}{x^2}\right)}{x^2\left(\frac{y}{x}\right)}$$
then if $x\neq0$ by taking $u=\frac{y}{x}$, we have $$\frac{dy}{dx}=\frac{\left(1+u^2\right)}{u}$$
but $u=\frac{y}{x}$ leads us to $xu=y$ and then $1+u'=y'$ where in $y'=\frac{dy}{dx}$. Now we have $$1+\frac{du}{dx}=\frac{\left(1+u^2\right)}{u}$$ This is a separable ode of first order. We have $$\frac{du}{dx}=\frac{\left(1+u^2\right)}{u}-1=\frac{1-u+u^2}{u}$$ so $$\frac{u\;du}{1-u+u^2}=dx$$ Now take an integral of both sides regarding to corresponding variable. We have $$\frac{1}{2}\ln(1-u+u^2)+\frac{\sqrt{3}}{3}\arctan\left(
\frac{\sqrt{3}}{3}(2u-1)\right)=x+C$$ put $u=\frac{y}{x}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/251600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"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.