Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Another way to compute $\lim\limits_{x\to+\infty}x^2\log\left(\frac{x^2+1}{x^2+3}\right)$ I need to compute as a title a limit with $x\to+\infty$. The only way I found to obtain a result is to use the L'Hôpital's rule:
$$\lim\limits_{x\to+\infty}x^2\log\bigg(\frac{x^2+1}{x^2+3}\bigg)=\lim\limits_{t\to 0}\frac{1}{t^2}\log\bigg(\frac{1+t^2}{1+3t^2}\bigg)\stackrel{H}{=}\lim\limits_{t\to 0}\frac{1}{2t}\bigg(\frac{2t}{1+t^2}-\frac{6t}{1+3t^2}\bigg)=\lim\limits_{t\to 0}\bigg(\frac{1}{1+t^2}-\frac{3}{1+3t^2}\bigg)=-2$$
It seems the result is the difference between the two functions inside the log, but how this can be possible? Is there another way to compute these type of limits? I mean a calculus way without using theorems.
| If it is OK for you to calculate directly a derivative then you may proceed as follows:
*
*Set $x^2 = \frac{1}{t}$ and consider $t\to 0^+$
So, you get
\begin{eqnarray*}x^2\log\bigg(\dfrac{x^2+1}{x^2+3}\bigg)
& \stackrel{x^2 = \frac{1}{t}}{=} & \frac{\log \frac{1+t}{1+3t}}{t}\\
& = &\frac{\log (1+t) - 0}{t} - \frac{\log (1+3t) - 0}{t} \\
& \stackrel{t \to 0^+}{\longrightarrow} & \left. \frac{d}{dt}\log(1+t) \right|_{t=0} - \left. \frac{d}{dt}\log(1+3t) \right|_{t=0} \\
& = & \frac{1}{1+0} - \frac{3}{1+3 \cdot 0} = \boxed{-2}
\end{eqnarray*}
Another way is using
*
*$(1-y)^{\frac{1}{y}}\stackrel{y \to 0}{\longrightarrow} e^{-1}$
So,
\begin{eqnarray*}x^2\log\bigg(\dfrac{x^2+1}{x^2+3}\bigg)
& \stackrel{x^2 = \frac{1}{t}}{=} & \frac{\log \frac{1+t}{1+3t}}{t}\\
& = & \log \left(1 - \frac{2t}{1+3t}\right)^{\frac{1}{t}} \\
& = & \log \left( \left(1 - \frac{2t}{1+3t}\right)^{\frac{1+3t}{2t}}\right)^{\frac{2t}{1+3t}\cdot \frac{1}{t}} \\
& \stackrel{t \to 0^+}{\longrightarrow} & \log \left(e^{-1}\right)^2 = \boxed{-2}\\
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find minimum of function and minimum point
We have a function $$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})}$$ the task to find minimum of this function and point where is minimum.
I tried to do by mark $t=x+\frac{1}{x}$ and then simplify this function and after that find derivative etc., but it is such a long solution. The professor, in the class, said that there is very short way to do this exercise, maybe somebody know it?
| Why is that substitution bad? We get $${t^6-(t^6-6t^4+9t^2-2)\over t^3+(t^3-3t)} = {6t^4-9t^2+2\over 2t^3-3t}$$
$$ = 3t+{2\over 2t^3-3t}$$
Derivative of that is $$3-{12t^2-6\over (2t^3-3t)^2}$$
So we want to solve $$4t^6-12t^4+9t^2= 4t^2-2$$i.e.
$$4t^6-12t^4+5t^2+ 2=0$$
Write $s=t^2$, so we have $$4s^3-12s^2+5s+2=0$$
We see that we have zeroes in following intervals $(-1,0)$, $(0,1)$ and $(2,3)$. Since $$s = (x+{1\over x})^2= x^2+2+{1\over x^2} \geq 2+2 =4$$ we see that this equation has no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129231",
"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{x\to0^+}x(\lfloor \frac{1}{x}\rfloor+\lfloor \frac{2}{x}\rfloor+\cdots+\lfloor \frac{k}{x}\rfloor), \, k \in \mathbb N$. $$M:=x\left(\left\lfloor \frac{1}{x} \right\rfloor+\left\lfloor \frac{2}{x}\right\rfloor+\cdots+\left\lfloor \frac{k}{x}\right\rfloor\right),\, k \in \mathbb N.$$
Using $\lfloor y \rfloor=y-\{y\}$,
$$M=x\sum_{i=1}^{k}\left(\frac{i}{x}-\left\{\frac{i}{x}\right\}\right)=\frac{k(k+1)}{2}-x \sum_{i=1}^{k}\left\{\frac{i}{x}\right\}$$
Since $0\leq\{y\}<1$, the coefficient of $x$ is finite and thus, $$\lim_{x\to0^+}M=\frac{k(k+1)}{2}$$
Is this correct?
| Another way is, because $x$ is approaching $0^+$, i.e. it becomes very small and is positive, then $\exists n\in \mathbb{N}$ s.t. $n\leq \frac{1}{x}<n+1$ or $n=\left\lfloor \frac{1}{x} \right\rfloor$ then $\forall i\in\mathbb{N}$:
$$i\cdot n \leq \frac{i}{x}<i\cdot(n+1) \Rightarrow i\cdot n \leq \left\lfloor \frac{i}{x} \right\rfloor<i\cdot(n+1) \tag{1}$$
and
$$n\left(\sum\limits_{i=1}^k i\right)\leq\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor <(n+1)\left(\sum\limits_{i=1}^k i\right) \iff \\
n\cdot \frac{k(k+1)}{2}\leq\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor <(n+1)\cdot \frac{k(k+1)}{2} \tag{2}$$
but $n\leq \frac{1}{x}<n+1 \Rightarrow \frac{1}{n}\geq x \geq \frac{1}{n+1}$
and $(2)$ becomes
$$ \frac{n}{n+1}\cdot \frac{k(k+1)}{2} <
x\cdot n\cdot \frac{k(k+1)}{2}\leq \\
x\left(\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor\right) < \\
x\cdot(n+1)\cdot \frac{k(k+1)}{2} \leq
\frac{(n+1)}{n}\cdot \frac{k(k+1)}{2}$$
or
$$\frac{n}{n+1}\cdot \frac{k(k+1)}{2} <
x\left(\sum\limits_{i=1}^k\left\lfloor \frac{i}{x} \right\rfloor\right) <
\frac{(n+1)}{n}\cdot \frac{k(k+1)}{2} \tag{3}$$
Now, $\lim\limits_{x\rightarrow0^+} \equiv \lim\limits_{n\rightarrow\infty}$ and the result follows from $(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the modulus and argument of complex numbers My problem:
Find the magnitude and argument of the following.
$$\frac{3+4i}{1-i} + \frac{2-i}{2+3i}$$
My solution so far:
\begin{align}
\frac{3+4i}{1-i} + \frac{2-i}{2+3i} &= \frac{3+4i}{1-i} \times \frac{1+i}{1+i} + \frac{2-i}{2+3i} \times \frac{2-3i}{2-3i}\\
&=\frac{3+3i+4i+4(-1)}{1+1} + \frac{4-6i-2i+3(-1)}{4+9(-1)}\\
&= \frac{3+7i-4}{2} + \frac{4-8i-3}{13} \\
&= \frac{-1+7i}{2} + \frac{1-8i}{13}\\
&= \frac{-13+91i+2-16i}{26}
\end{align}
Thus the complex number is given by
$$ z = \frac{-11+75i}{26}= \frac{-11}{26} + \frac {75}{26}i $$
Then, in order to find the modulus, I computed
$$|z| =\sqrt{x\cdot x + y\cdot y}$$
where $x = -11/26$ and $y = 75/26 $. Then,
$$|z| = \sqrt{\frac{-121}{676} + \frac {5625}{676}} = \sqrt{\frac{5746}{676}}$$
This is where I stopped.
How do I find the argument of $z$ from here?
| in your solution, you find, correctly, $a$ and $b$ such that $\frac{3+4i}{1-i} + \frac{2-i}{2+3i} = a+bi$
but you need to find the modulus and the argument of the number. That is, you need to find $r>0$ and $\theta\in[0,2\pi)$ such that
$$\frac{3+4i}{1-i} + \frac{2-i}{2+3i} = r(\cos \theta + i\sin \theta).$$
There is a simple way to converting between the standard $a+bi$ format and the latter polar format.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A mistake on computing $\int \frac{dx}{\sqrt{x+1}+1}$ I have to find $\int \frac{dx}{\sqrt{x+1}+1}$. This was my attempt, which is wrong and I cannot find where exactly is the mistake.
First I write $\frac{1}{\sqrt{x+1}+1}=\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}}{x}-\frac{1}{x}$, therefore $\int \frac{dx}{\sqrt{x+1}+1}=\int \frac{\sqrt{x+1}}{x}dx-\log\left (x\right )$, so I only have to deal with $\int \frac{\sqrt{x+1}}{x}dx$.
Setting $u=\sqrt{x+1}$, we obtain $du=\frac{1}{2\sqrt{x+1}}dx$, therefore $dx=2udu$ and $x=u^2-1$, so
$\int\frac{\sqrt{x+1}}{x}dx=\int\frac{2u^2}{u^2-1}du=\int \left(2+\frac{2}{u^2-1}\right)du=2\sqrt{x+1}+\int \left(\frac{1}{u-1}-\frac{1}{u+1}\right)du=$
$=2\sqrt{x+1}+\log \left (\sqrt{x+1}-1\right )-\log \left (\sqrt{x+1}+1\right )=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )$
But then $\int \frac{dx}{\sqrt{x+1}+1}=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )-\log\left (x\right )=2\sqrt{x+1}+\log\left ( \frac{x+2-2\sqrt{x+1}}{x^2} \right )$, which is not what I am supposed to obtain. The actual answer is $2\sqrt{x+1}-2\log\left ( 1+\sqrt{x+1} \right )$.
Where is my mistake?
(I already know a correct way to solve it, I just want to know where I am committing a mistake).
| Note that
$$
\frac{x+2-2\sqrt{x+1}}{x^2}
=\frac
{(x+2)^2-4(x+1)}
{x+2+2\sqrt{x+1}}
\cdot\frac 1{x^2}\ ,
$$
and this has nothing to do with
$$
\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \cdot \frac 1x
=
\frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{(\sqrt{x+1}+1)(\sqrt{x+1}+1)} \cdot \frac 1x
=
\frac 1{(\sqrt{x+1}+1)^2}\ ,
$$
which leads to the solution. So the last step is the bad one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
| Here's an analytical answer with greater emphasis on reasoning than anything specific to probability which might lend greater insight into the problem.
Consider if there was only one coin. The probability to have an even number of heads is $1\over2$, since there are two possible outcomes and we are only interested in one of them.
Now, let there be $N \gt 1$ coin tosses. The $N^{\text{th}}$ coin toss will either give a heads or tails. If the $N-1$ tosses resulted in an even number of heads the probability of the $N$ coin tosses resulting in an even number of heads is $1\over2$ since the $N^{\text{th}}$ coin toss will add either $0$ or $1$ to the number of heads from the $N-1$ tosses, and we are only concerned with the parity of the count.
The conclusion is the same for the $N-1$ coin tosses resulting in an odd number of heads.
This approach is valid as this reasoning applies to all possible values of $N$ in our given domain.
$\therefore$ The probability of $N$ coin tosses resulting in an even number of heads is $1\over2$, with $N \in \mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
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"answer_id": 7
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Hypergeometric representation of Fresnel $S(x)$ I am trying to find a representation for the Fresnel integral
$$S(x)=\int_0^x\sin\frac{\pi t^2}{2}\,\mathrm dt$$
Then with $$\sin x=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
We have
$$S(x)=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!}\int_0^x\left(\frac{\pi t^2}2\right)^{2n+1}\mathrm dt$$
$$S(x)=\sum_{n\geq0}\frac{(-1)^n}{(2n+1)!}\left(\frac\pi2\right)^{2n+1}\frac{x^{4n+3}}{4n+3}$$
$$S(x)=\frac{\pi x^3}{2}\sum_{n\geq0}\frac{1}{(2n+1)!(4n+3)}\left[-\frac{\pi^2x^4}{4}\right]^{n}$$
Then setting $$S_n=\frac{1}{(2n+1)!(4n+3)}\left[-\frac{\pi^2x^4}{4}\right]^{n}$$
We have $$\frac{S_{n+1}}{S_n}=\frac{-\pi^2 x^4}{16(n+1)}\frac{n+3/4}{(n+7/4)(n+3/2)}$$
Hence we have $$S(x)=\frac{\pi x^3}2\,_1F_2\left(\frac34;\frac32,\frac74;-\frac{\pi^2x^4}{16}\right)$$
But the Wolfram Functions site Says that
$$S(x)=\frac{\pi x^3}{\color{red}6}\,_1F_2\left(\frac34;\frac32,\frac74;-\frac{\pi^2x^4}{16}\right)$$
Where did that extra $1/3$ come from? Thanks.
| Note that the recurrence relation for $(S_n)_{n \in \mathbb{N}_0}$ implies
$$ S_n = \frac{\Gamma\left(n + \frac{3}{4}\right)}{\Gamma\left(\frac{3}{4}\right) }\frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(n + \frac{3}{2}\right)} \frac{\Gamma\left(\frac{7}{4}\right)}{\Gamma\left(n + \frac{7}{4}\right)} \frac{\left(-\frac{\pi^2 x^4}{16}\right)^n}{n!} \color{red}{S_0}$$
and we have $S_0 = \frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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At how many points is this piecewise function discontinuous?
$$f(x) =
\begin{cases}
\frac{3x}{x^2-16} & \text{when $x\leq 2$} \\[2ex]
\frac{x-2}{x^2-5x+6} & \text{when $x\gt 2$}
\end{cases}
$$
At what points is this piecewise function discontinuous?
My solution:
$x=-4, 4, 2, 3$ are the points at which the denominator is zero, but $x=4\not\leq2$ and $x=2\not\gt2$. Therefore, the points are $x=-4$ and $x=3$. But the answer is given as $3$.
What am I doing wrong?
| You are correct about $x=-4$ and $x=3$, but you also need to check $x=2$ since that's the point where you jump from one function to the other.
For the piecewise function to be continuous at $x=2$ you need $$lim_{x \to 2} f(x) = f(2).$$ Since $f(x)$ is defined differently from the left and the right of $x=2$ you will need to instead use each one sided limit.
$$lim_{x \to 2^-} f(x) = f(2)$$
$$lim_{x \to 2^-} \frac{3x}{x^2-16} = \frac{3(2)}{(2)^2-16}$$
$$\frac{3(2)}{(2)^2-16} = \frac{3(2)}{(2)^2-16}$$
$$\frac{6}{-12}=\frac{6}{-12}$$
$$-\frac{1}{2}=-\frac{1}{2}$$
So this one works out. Now we also need to check the right sided limit.
$$lim_{x \to 2^+} f(x) = f(2)$$
$$lim_{x \to 2^+} \frac{x-2}{x^2-5x+6} = \frac{3(2)}{(2)^2-16}$$
$$lim_{x \to 2^+} \frac{x-2}{(x-2)(x-3)} = -\frac{1}{2}$$
$$lim_{x \to 2^+} \frac{1}{x-3} = -\frac{1}{2}$$
$$\frac{1}{(2)-3} = -\frac{1}{2}$$
$$\frac{1}{-1} = -\frac{1}{2}$$
$$-1=-\frac{1}{2}$$
Clearly this is not true, $-1 \neq -\frac{1}{2}$. Therefore the two sided limit doesn't exist and so we also have a discontinuity at $x=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the inequality $x^2 - 3 > 0$ For the inequality $x^2 - 3 > 0$, we have
\begin{align} x^2 - 3 & = (x+\sqrt 3)(x- \sqrt 3) > 0 \end{align}
Therefore,
\begin{align} x > -\sqrt 3 \end{align}
and
\begin{align} x > \sqrt 3 \end{align}
But, this is clearly wrong as we should get $x < -\sqrt 3$ and $x > \sqrt 3$ as the two intervals. What have I done wrong?
| First treat the inequality as if it were to be an equation: $ x^{2} - 3 = 0 $. Thus, $$ x^{2} = 3 $$ and solving for $x$, $$x =\pm \sqrt{3}$$
This makes $x = -\sqrt{3}$ and $x = \sqrt{3}$ the "roots" of that equation.
Now to solve the $inequality$ $ x^{2} - 3 > 0 $, keep in mind the aforementioned "roots" into the following cases:
Case #1: When $x < -\sqrt{3}$, the inequality holds true since $x^{2} - 3$ is positive.
Case #2: $-\sqrt{3} < x < \sqrt{3}$ would make the inequality false because $x^{2} - 3$ is negative.
Case #3: When $x > \sqrt{3}$, the inequality holds true since $x^{2} - 3$ is also positive.
Thus the solutions to the inequality are $x < -\sqrt{3} \quad \textbf{or} \quad x > \sqrt{3}$.
The solutions expressed in interval notation: $(-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Detemine the unit digit of a number Find the unit digit of the number:
$$3^{7005} \times 6^{8000}$$
My turn:
$$3^{7005}\times 6^{8000}=3^{7005}\times 3^{8000} \times2^{8000}$$ $$3^{13005} \times 2^{8000}$$
But i could not go on ?
| We want the unit digit of the number $3^{7005} \cdot6^{8000}$. This is equivalent to asking for the residue of the number $3^{7005} \cdot6^{8000}$ modulo $10$. Hence, we want some number $n$ with $0 \leq n \leq 9$ such that
$$
n \equiv 3^{7005} \cdot6^{8000} \pmod{10}
$$
Note that $3^4 \equiv 1 \pmod{10}$ and $6^n \equiv 6 \pmod{10}$ for all $n\in\mathbb{N}$. With this result, we have
\begin{align}
3^{7005} \cdot6^{8000} &= 3^{7004}\cdot 3^{1} \cdot 6^{8000} \\
&= \left(3^{4}\right)^{1751}\cdot 3^{1} \cdot 6^{8000}\\
&\equiv1^{1751}\cdot 3^{1} \cdot 6^{1} \pmod{10} \\
&= 3 \cdot 6 \\
&\equiv 8 \pmod{10} \\
\end{align}
Hence, the unit digit of $3^{7005} \cdot6^{8000}$ is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140090",
"timestamp": "2023-03-29T00:00:00",
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Find $k=constant$ such that $f(a,\,b,\,c\,)=\frac{3a+2b}{\sqrt{5a^2-ab+b^2}}+\frac{3b+2c}{\sqrt{5b^2-bc+c^2}}\leqq f(a,k\,a+\sqrt[3]{abc}-k\,c,c\,)$ Give $3$ positve numbers $a,\,b,\,c$ such that $abc= 1$ , prove:
$$f\left ( a,\,b,\,c \right )= \frac{3\,a+ 2\,b}{\sqrt{5\,a^{\,2}- ab+ b^{\,2}}}+ \frac{3\,b+ 2\,c}{\sqrt{5\,b^{\,2}- bc+ c^{\,2}}}\leqq f\left ( a,\,1,\,c \right )$$
Because the inequality is homogeneous, so we can write:
$$f\left ( a,\,b,\,c \right )= \frac{3\,a+ 2\,b}{\sqrt{5\,a^{\,2}- ab+ b^{\,2}}}+ \frac{3\,b+ 2\,c}{\sqrt{5\,b^{\,2}- bc+ c^{\,2}}}\leqq f\left ( a,\,\sqrt[3\,]{abc},\,c \right )$$
I tried to break the square root by using:
$$\sqrt{A}- B= \frac{A- B^{\,2}}{\sqrt{A}+ B}= \frac{A- B^{\,2}}{\frac{A- C^{\,2}}{\sqrt{A}+ C}+ B+ C}= ...$$
This is hard, I solved easier problem, we have:
$$\sqrt{5\,a^{\,2}- ab+ b^{\,2}}- \left ( 3\,a+ 2\,b \right )\sqrt{\frac{19}{40}}= \frac{\left ( 23\,a- 8\,b \right )^{\,2}}{140\left [ \sqrt{5\,a^{\,2}- ab+ b^{\,2}}+ \left ( 3\,a+ 2\,b \right )\sqrt{\frac{19}{40}} \right ]}$$
$$\sqrt{5\,b^{\,2}- bc+ c^{\,2}}- \left ( 3\,b+ 2\,c \right )\sqrt{\frac{19}{40}}= \frac{\left ( 23\,b- 8\,c \right )^{\,2}}{140\left [ \sqrt{5\,b^{\,2}- bc+ c^{\,2}}+ \left ( 3\,b+ 2\,c \right )\sqrt{\frac{19}{40}} \right ]}$$
Let $g\left ( a, b \right )= \sqrt{5\,a^{\,2}- ab+ b^{\,2}}+ \left ( 3\,a+ 2\,b \right )\sqrt{\frac{19}{40}}$ . I tried to prove:
$$\left ( 8\,c- 23\,b \right )\left \{ g\left ( \frac{b^{\,2}}{c},\,b \right )- \left [ \sqrt{5\,b^{\,2}- bc+ c^{\,2}}+ \left ( 3\,b+ 2\,c \right )\sqrt{\frac{19}{40}} \right ] \right \}\geqq 0$$
But without success, I think: $0\leqq f\left ( a,\,\sqrt[3\,]{abc},\,c \right )- f\left ( a,\,b,\,c \right )= \left ( \sqrt[3\,]{abc}- b \right )A\Leftrightarrow b^{\,2}\geqq ac\Leftrightarrow a\leqq \frac{b^{2}}{c}\Leftrightarrow g\left ( a, b \right )\leqq g\left ( \frac{b^{2}}{c}, b \right )$
Edit:
I find $k= constant$ such that: $f(\,a,\,b,\,c\,)\leqq f(\,a,\,k\,a+ \sqrt[3\,]{abc}- k\,c,\,c\,)$. Thank you a real lot!
| The inequality in your title is incorrect. There is no $k$ for which this is true.
Let us set $a = c$ in your inequality. Then we have
$$ f(a,b,a) \leq f(a, (a^2b)^{1/3}, a).$$
This implies
$$f(a,b,a) \leq f(a, (a^2b)^{1/3}, a) \leq f(a, (a^8 b)^{1/9}, a) \leq f(a, (a^{26} b)^{1/27}, a) \leq ... \leq f(a, a, a).$$
This implies that, when $a$ is fixed, $f(a,b,a)$ is maximized at $b = a$.
However, plugging in $a=1$, it seems that while the point $b=a$ is indeed an inflection point of $f(a,b,a)$, the point $b=a$ is a local minimum and not a local maximum, a contradiction. See the plot below.
NOTE: the original question was whether the inequality was correct for $k=1$. Here is my answer to that.
Your inequality is incorrect:
$$
f(1,3,1/3) = 4.169
$$
and
$$
f(1,1,1/3) = 3.914.
$$
However, your inequality does seem to hold for $a \leq 1/2$, so you should look and see for what $(a,b,c)$ you actually need it to work.
ADDED NOTE: it's actually not true for $a \leq 1/2$, as plotting the function $f(a, b, a)$ for $a=\frac{1}{2}$ shows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$ via partial fractions $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$
= $ \int \frac{Ax+B}{(x+1)} + \frac{Cx+B}{(x+1)^2} + \frac{Dx+E}{x+2}$
= $\int (Ax+B)(x+1)(x+2) + (Cx+B)(x+2) + (Dx+E)(x+1)^2$
= $ \int Ax^3 + 3Ax^2 + 2Ax + Bx^2 + 3Bx + 2B + Cx^2 + 2Cx + Bx + 2B + Dx^3 + 2Dx^2 + Dx + Ex^3 + 2Ex^2 + E$
= $ \int (A + D + E)x^3 + (3A + B + C + 2D + 2E)x^2 + (2A + 2C + 4B + D)x + (4B + E)$
Turn into matrix, find reduced row echelon form to solve system of equations:
$\begin{bmatrix}
0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 0 \\
3 & 1 & 1 & 2 & 2 & 1 \\
2 & 4 & 2 & 1 & 0 & 1 \\
0 & 4 & 0 & 0 & 1 & 1 \\
\end{bmatrix}$
This is where things go wrong, apparently this doesn't reduce down properly :( and I have been relying on RREF to solve systems of equations everytime up until now.
I am also confused where the $x^2 + x + 1$ is supposed to go exactly. I know I put it into the system of equations later but until then, I feel like I kinda just ignored it and left it out of all my work up until then (is that okay?).
Other note: I recognize that this is a proper rational fraction so no long division is nesecary and that this has irreducible factors that are repeated so that is why I split them up into the partial fractions up above in that manner. Did I miss any intermittent steps that made the RREF turn out wrong? I am not sure where I went wrong thus far either
| For a partial fraction decomposition, one only requires a polynomial on the resulting fraction numerators to be one less than those of the corresponding denominators. In this case the resulting fractions should be
$$\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}$$
which simplifies the equations significantly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Can the value of the Riemann zeta function at $n=2$ be derived from the Wallis formula for $\pi$? It is well known that the Riemann zeta function, defined for all positive integers $n>1$ by
$$
\zeta(n) = \sum_{m=1}^{\infty} m^{-n}
$$
takes the value $\displaystyle \frac{\pi^2}{6}$ at $n=2$. On the other hand, a product by Wallis converges to $\displaystyle \frac{\pi}{2}$:
$$
\prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {\pi }{2}}.
$$
There exist many proofs of $\zeta(2)=\displaystyle \frac{\pi^2}{6}$, but I have found none that involves the Wallis product.
*
*Can $\zeta(2)$ be derived from the Wallis product?
*If so, can the general formula for $\zeta(2n)$ by derived in a similar manner?
| There are several proofs for the sine product, e.g. look for “sine product formula proof”.
You can also take a look here, there is a link to the old proof of Euler.
For a proof of the Wallis product it’s enough to look e.g. here.
It’s $~\displaystyle \frac{\pi}{2} = \frac{\pi x}{\sin(\pi x)}|_{x=\frac{1}{2}} = \Gamma(1-x)\Gamma(1+x)|_{x=\frac{1}{2}} = \prod\limits_{n=1}^\infty \frac{2n}{2n-1}\frac{2n}{2n+1}~$ .
The connection to the Riemann zeta function can be seen by using the logarithm.
$\displaystyle \ln\frac{\pi x}{\sin(\pi x)} = \ln(\Gamma(1-x)\Gamma(1+x)) = \sum\limits_{n=1}^\infty\frac{x^{2n}}{n}\zeta(2n)$
The last equation comes from $~\displaystyle \ln\prod\limits_{k=1}^n (1-zb_k)^{a_k} = \sum\limits_{k=1}^\infty\frac{z^k}{k}\sum\limits_{v=1}^n a_v b_v^k$
with $~\displaystyle (a_v;b_v;n;z):=\left(1;\frac{1}{v^2};\infty;x^2\right)~$ .
To get the formula for $~\zeta(2n)~$ one should use the Bernoulli numbers $\,B_k\,$,
traditionelly introduced by $\displaystyle \frac{x}{e^x-1} = \sum\limits_{k=0}^\infty \frac{x^k}{k!}B_k$ .
With the first derivation of $~\displaystyle \ln\frac{\pi x}{\sin(\pi x)}~$ follows:
$\displaystyle \frac{1 -\pi x\cot(\pi x)}{x} =\frac{d}{dx}\ln\frac{\pi x}{\sin(\pi x)} = \frac{d}{dx}\sum\limits_{n=1}^\infty\frac{x^{2n}}{n}\zeta(2n) = 2\sum\limits_{n=1}^\infty x^{2n-1}\zeta(2n)$
with $\enspace\displaystyle \cot(z) = \frac{\cos z}{\sin z} = i\frac{ e^{iz}+e^{-iz} }{ e^{iz}-e^{-iz} } = i\frac{ e^{iz}-e^{-iz}+2e^{-iz} }{ e^{iz}-e^{-iz} } = i\left(1+\frac{1}{iz}\frac{i2z}{e^{i2z}-1}\right)$
$\hspace{2.3cm}\displaystyle = i\left(1+\frac{1}{iz}\sum\limits_{k=0}^\infty \frac{(i2z)^k}{k!}B_k\right) = \frac{1}{z}+\frac{1}{z}\sum\limits_{k=1}^\infty (-1)^k\frac{(2z)^{2k}}{(2k)!}B_{2k}$
We get $\displaystyle \enspace\frac{1 -\pi x\cot(\pi x)}{x} = \sum\limits_{k=1}^\infty (-1)^{k-1}\frac{(2\pi)^{2k}x^{2k-1}}{(2k)!}B_{2k}$
and comparing the coefficients of $~x^{2k-1}~$ it follows $~\displaystyle \zeta(2n)=(-1)^{k-1}\frac{(2\pi)^{2k}}{2(2k)!}B_{2k}~$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluate $\int \frac{x+4}{x^2 + 2x + 5}$ I am having issues with this integral. I am not sure if it is irreducible or not. I can't use the quadratic formula, but I can rearrange the integral to be $\int \frac{x+4} {(x^2 + 2x + 1) + 4}$, but I don't know how to deal with the $+4$.
Here is my work treating the quadratic equation as irreducible with no repeating factors.
$\int \frac{x+4}{x^2 + 2x + 5}$
= $\frac {Ax + B} {x^2 + 2x + 5} $
= $Ax+B(x^2 + 2x + 5)$
= $Ax^3 + 2Ax^2 + 5Ax + Bx^2 + 2Bx + 5B$
= $Ax^3 + (2A + 2B)x^2 + (5A + 2B)x + 5B$
However I get stuck trying to solve my system of equations. This leads me to believe that I did the partial fractions improperly.
$\begin{bmatrix}
1 & 0 & 0 \\
2 & 2 & 0 \\
5 & 2 & 1 \\
0 & 5 & 4 \\
\end{bmatrix}$
| We will integrate first of all change quadratic $(x+1)^2+4$ and after fraction of $x+1+3$ and divide by quadratic and we can integrate it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can this integral be solved? I am trying to solve (by hand) the integral:$$\int_0^∞\frac1x\left(\frac{(x+a)^2}{((x+a)^2+b)^{3/2}}-\frac{(x-a)^2}{((x-a)^2+b)^{3/2}}\right)\,\mathrm dx,$$
where $a$ and $b$ are real and positive. Mathematica actually generates the answer, but I am stumped on how to solve it by hand and quite curious. Answer:
$$\frac{2a\left(\sqrt{a^2+b}-a\cdot\operatorname{arctanh}\left(\frac{a}{\sqrt{a^2+b}}\right)\right)}{\left(a^2+b\right)^{3/2}}.$$
Is there a (reasonably) nice way to solve something like this? Any tips on how to proceed?
| Let $I$ denote the integral. By a simple substitution, we find that
$$ I
= \mathrm{PV}\!\int_{-\infty}^{\infty} \frac{1}{x-a} \frac{x^2}{(x^2+b)^{3/2}} \, \mathrm{d}x
= \operatorname{Re}\left( \int_{-\infty + i\epsilon}^{\infty+i\epsilon} \frac{1}{z-a} \frac{z^2}{(z^2+b)^{3/2}} \, \mathrm{d}z \right), $$
where the path of integration in the last step denotes any straight line $\operatorname{Im}(z)=\epsilon \in (0, \sqrt{b})$ oriented to the right, and the principal branch cut is adopted for $(z^2+b)^{3/2}$. Taking integration by parts,
$$ I
= - \operatorname{Re}\left( \int_{-\infty + i\epsilon}^{\infty+i\epsilon} \frac{a}{(z-a)^2} \frac{1}{\sqrt{z^2+b}} \, \mathrm{d}z \right). $$
We notice that its branch-cut in the upper-half plane is $i[\sqrt{b},\infty)$. So, by deforming the contour to wrap this branch-cut, we have
$$ I
= \operatorname{Re}\left( \int_{\sqrt{b}}^{\infty} \frac{2a}{(y+ia)^2\sqrt{y^2-b}} \, \mathrm{d}y \right)
= \int_{\sqrt{b}}^{\infty} \frac{2a(y^2-a^2)}{(y^2+a^2)^2\sqrt{y^2-b}} \, \mathrm{d}y. $$
As for further simplification, substitute $p = b/a^2$ and $y = \sqrt{\frac{b}{1-s}}$. Then
\begin{align*}
I
&= \frac{1}{a} \int_{0}^{1} \frac{p-1+s}{(p+1-s)^2\sqrt{s}} \, \mathrm{d}s \\
&= \frac{1}{a} \left[ \frac{2p\sqrt{s}}{(p+1)(p+1-s)} - \frac{2}{(p+1)^{3/2}}\operatorname{arctanh}\left(\sqrt{\frac{s}{p+1}}\right) \right]_{0}^{1} \\
&= \frac{2}{a} \frac{\sqrt{p+1} - \operatorname{arctanh}\left(\sqrt{\frac{1}{p+1}}\right)}{(p+1)^{3/2}}.
\end{align*}
Plugging $p = b/a^2$ back shows that this expression coincides with the answer provided in OP, and so, the desired equality is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Question about partial fractions with irreducible quadratic factors Given this rational function: $$\frac{-4x^4-2x^3-26x^2-8x-44}{(x+1)(x^2 +3)^2}$$
The decomposition would look like this: $$\frac{A}{x+1} + \frac{Bx+C}{(x^2+3)} + \frac{Dx+E}{(x^2+3)^2}$$
And the final answer would be: $$\frac{-4}{x+1} - \frac{2}{(x^2+3)} - \frac{2}{(x^2+3)^2}$$
But, if you were to set it up like this: $$\frac{A}{x+1} + \frac{B}{(x^2+3)} + \frac{C}{(x^2+3)^2}$$
You end up with the same answer:
$$\frac{-4}{x+1} - \frac{2}{(x^2+3)} - \frac{2}{(x^2+3)^2}$$
I want to know why this is.
I thought that for irreducible quadratic factors, you needed a linear term in the numerator. Is this purely coincidence that it worked with constant terms in the numerator?
| The five functions
$$f_1(x)=\frac{1}{x+1},\; f_2(x)=\frac{x}{x^2+3}, \; f_3(x)=\frac{1}{x^2+3},\; f_4(x)=\frac{x}{(x^2+3)^2}, \; f_5(x)=\frac{1}{(x^2+3)^2}.$$
is a basis which linearly generates all rational functions of the form
$$\frac{P(x)}{(x+1)(x^2 +3)^2}$$
where $P$ is ANY polynomial of degree $\leq 4$.
In your case, we simply have that in the vector representation of the given function the coordinates related to $f_2$ and $f_4$ are zero. So, yes, this is a purely coincidence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Integrate $\int \frac{dx}{\sqrt{x^2-9}}$ by trig substitution $$
\begin{align}
x = 3\sec\theta, dx &= 3\sec\theta\tan\theta d\theta\\\\
\int \frac{dx}{\sqrt{x^2-9}} &= \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{(3\sec\theta)^2 - 3^2}} \\\\
& = \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{3^2(\sec^2\theta -1)}} \\\\
&= \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{3^2\tan^2\theta}} = \int \sec\theta\\\\
&= \ln|\sec\theta + \tan\theta| + C = \ln| \frac{x}{3} + \frac{\sqrt{x^2-9}}{3}|
\end{align}
$$
However, wolphram alpha says the answer is $\ln |x+ \sqrt{x^2-9}$
I am wondering how did it get rid of the 3 in the denominator? This is pretty much how I got my answer:
$$
x = 3\sec\theta \\
\frac{x}{3} = \sec\theta \\
\frac{\sqrt{x^2-9}}{3} = \tan\theta
$$
| Another approach:
$$I=\int\frac{dx}{\sqrt{x^2-9}}=\frac{1}{3}\int\frac{dx}{\sqrt{\left(\frac{x}{3}\right)^2-1}}$$
Let $u=\frac{x}{3}\implies3du=dx\implies$
$$I=\int\frac{du}{\sqrt{u^2-1}}=\text{arccosh}(u)+C=\text{arccosh}\left(\frac{x}{3}\right)+C$$
But recall the identity for $x\in[1,\infty)$ $$\text{arccosh}(x)=\ln\left(x+\sqrt{x^2-1}\right)$$
So $$\text{arccosh}\left(\frac{x}{3}\right)+C=\ln\left(\frac{x}{3}+\sqrt{\frac{x^2}{9}-1}\right)$$
Let $C=-\ln(3)$ and we have the Wolfram approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $\sin(x) - \cos(x) = 1/3$ then determine $\sin(x)\cos(x)$
If
$$\sin(x) - \cos(x) = \frac{1}{3}$$
then determine
$$\sin(x)\cos(x)$$
I know that the expected solution is squaring both sides of equation and solving it this way:
\begin{gather}
\sin^2(x)+\cos^2(x)= 1 \\[4px]
(\sin(x) - \cos(x))^2 = \left(\frac{1}{3}\right)^2 \\[4px]
\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) =\frac{1}{9} \\[4px]
-2\sin(x)\cos(x)=\frac{1}{9} -\sin^2(x)-\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +\sin^2(x)+\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +1\\[4px]
2\sin(x)\cos(x)=\frac{8}{9} \\[4px]
\sin(x)\cos(x)=\frac{4}{9}
\end{gather}
But assume I haven't noticed that I can solve it by squaring both sides in the first place. I can't figure it out how to solve it any other way.
| You can use $$\sin x = \frac{2t}{1+t^2}; \cos x = \frac {1-t^2}{1+t^2}$$
where $t=\tan (x/2)$, but you just have to know that some value of $t$ works.
Then you get a quadratic to solve for $t$. And once you know $t$ you can solve the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differentiate $11x^5 + x^4y + xy^5=18$ I am not sure how to differentiate $11x^5 + x^4y + xy^5=18$. I have a little bit of experience with implicit differentiation, but I'm not sure how to handle terms where both variables are multiplied together.
I have tried
$$\frac{d}{dx}(11x^5 + x^4y+xy^5) = \frac{d}{dx}(18)$$
$$\frac{d}{dx}(11x^5)+\frac{d}{dx}(x^4y) + \frac{d}{dx}(xy^5)=0$$
differentiating each term
$$\frac{d}{dx} (11x^5)=11\frac{d}{dy}(5x^4) = 55x^4$$
$$\frac{d}{dx} (x^4y) = [4x^3 \cdot y] + [1 \cdot x^4] = 4yx^3+x^4$$
$$\frac{d}{dx}(xy^5) = [1 \cdot y^5] + [x \cdot 5y^4] = y^5 + 5xy^4$$
finding $\frac{dy}{dx}$
$$\frac{dy}{dx}=\frac{-x}{y} = \frac{-[4yx^3+x^4] + [y^5+5xy^4]}{55x^4}$$
According to the website I'm using, "WebWork", this is wrong.
| This is how you could do it:
$11x^5+x^4y+xy^5=18$
Use the $\frac{d}{dx}$ operator (not $\frac{dy}{dx}$)
$55x^4+\frac{d}{dx}(x^4y+xy^5)=0$
Now, use the product rule on each part in the brackets:
$55x^4+y\frac{d}{dx}x^4+x^4\frac{dy}{dx}+y^5\frac{d}{dx}x+x\cdot4y^4\frac{dy}{dx}=0$
$55x^4+y\cdot3x^3+x^4\frac{dy}{dx}+y^5+4xy^4\frac{dy}{dx}=0$
$55x^4+3x^3y+y^5+\frac{dy}{dx}(x^4+4xy^4)=0$
$\frac{dy}{dx}(x^4+4xy^4)=-55x^4-3x^3y-y^5$
So, finally,
$\frac{dy}{dx}=-\frac{55x^4+3x^3y+y^5}{x^4+4xy^4}$
You were just using the operators wrong, I think
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
double integration with a domain Given that $D = \{ 1 \leq x+y \leq 2 , y+1 \leq x \leq 3y+1 \}$ and we want to evaluate $\int \int \limits_D \frac{x+y-1}{y^2 } e^{\frac{x-1}{y}}$ and used the substation $u=x+y$ and $v = \frac{x-1}{y}$ which the answer in the book use it, and get to the integral $\int \limits_{1}^{2} du \int \limits_{1}^{3} v e^v dv = 3e^3-e$ the problem is that the answer in the book is $e^3-e$, is my answer correct ? and if not, how to solve the question correctly ?
| The notations:
$$\begin{cases}u=x+y\\ v = \frac{x-1}{y} \end{cases} \Rightarrow \begin{cases} x=\frac{1+uv}{1+v}\\ y=\frac{u-1}{1+v}\end{cases}$$
The Jacobian:
$$\begin{vmatrix}x_u& x_v\\ y_u&y_v\end{vmatrix}=\begin{vmatrix}\frac{v}{1+v}& \frac{u-1}{(1+v)^2}\\ \frac1{1+v}&\frac{1-u}{(1+v)^2}\end{vmatrix}=\frac{1-u}{(1+v)^2}$$
So:
$$\int \int \limits_D \frac{x+y-1}{y^2 } e^{\frac{x-1}{y}}=\int \limits_{1}^{3} \int \limits_{1}^{2} \frac{u-1}{\left(\frac{u-1}{1+v}\right)^2}e^v\cdot \frac{1-u}{(1+v)^2}dudv = \\
\int \limits_{1}^{3} \int \limits_{1}^{2} e^vdudv =\int \limits_{1}^{3} e^vdv=e^3-e.$$
Also note:
$$\int \limits_{1}^{2} du \int \limits_{1}^{3} v e^v dv = \int \limits_{1}^{2} \left[ve^v|_1^3-\int \limits_{1}^{3} e^vdv\right] du= \int \limits_{1}^{2} 2e^3 du=2e^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove the inequality $\sum x+6\ge 2(\sum\sqrt{xy}) $ Let $x;y;z\in R^+$ such that $x+y+z+2=xyz$. Prove that $$x+y+z+6\ge 2(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}) $$
This inequality is not homogeneous and look at the condition i thought that
i would substitute the variables $x;y;z$ such as:
+)If i need to solve $x^2+y^2+z^2+2xyz=1$, i will let $x=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}$
+)If i need to solve $xy+yz+xz+xyz=4$,i will let $x=\frac{2a}{b+c}$
but failed. Please explain for me how can i get this substitution (if have a solution by substitution)
I also tried to solve it by $u,v,w$.Let $\sum_{cyc} x=3u;\sum_{cyc} xy;\Pi_{cyc}a=w^3(3u+2=w^3;u,v,w>0 )$ then $u\le w^3-3u$ or $4u\le w^3$ but stuck (I am really bad at $uvw$)
| The condition gives
$$\sum_{cyc}\frac{1}{x+1}=1.$$
Now, let $x=\frac{b+c}{a}$ and $y=\frac{a+c}{b},$ where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{a+b}{c}$ and we need to prove that:
$$\sum_{cyc}\frac{b+c}{a}+6\geq2\sum_{cyc}\sqrt{\frac{(b+c)(a+c)}{ab}}$$ or
$$\sum_{cyc}(a^2b+a^2c+2abc)\geq2\sum_{cyc}a\sqrt{bc(a+b)(a+c)},$$
which is true by AM-GM.
Indeed,
$$2\sum_{cyc}a\sqrt{bc(a+b)(a+c)}=2\sum_{cyc}a\sqrt{(ac+bc)(ab+bc)}\leq$$
$$\leq\sum_{cyc}a(ac+bc+ab+bc)=\sum_{cyc}(a^2b+a^2c+2abc).$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.
The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives
$$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$
If I could show that $\left(1 + \frac{1}{x-1} \right)^2 > e^\frac{1}{x} $ for $x>1$, then $f(x)$ would be increassing, and since $ \lim_{x \to \infty} f(x) = 0$ this would mean that $f(x) < 0$ for $x>1$. However, proving that inequality is very similar to the first one, and still involves bounding above $e^\frac{1}{x}$.
The other thing I tried is considering $f(x) = e^{x-1} - (x-1) - 1$ which is increasing for $x > 1$. Then $f(\frac{1}{x})$ is decreasing, so $e^{\frac{1}{x}-1} - \frac{1}{x}$ is decreasing; However this also does not seem to help too much.
Any ideas?
| $$e^{\frac{1}{x}}<\frac{x}{x-1}$$
$$e^{\frac{1}{x}}<\frac{1}{1-\frac{1}{x}}$$
$$e^{-\frac{1}{x}}<\frac{1}{1+\frac{1}{x}}$$
$$\frac{1}{e^{\frac{1}{x}}}<\frac{1}{1+\frac{1}{x}}$$
$$\frac{1}{1+\frac{1}{x}+\frac{1}{2x^2}+...}<\frac{1}{1+\frac{1}{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
Prove that $\frac{a + b + c}{b - a} > 3$ with $ab < b^2 < 4ca$.
$a$, $b$, $c$ are three positives such that $ab < b^2 < 4ca$. Prove that $$\large \dfrac{a + b + c}{b - a} > 3.$$
I can't think of a way to get around this problem. Although I can see that based on the given condition, $ax^2 + bx + c = 0$ has no roots, which adds almost no information whatsoever.
If you have written an answer below, thanks for that!
| Assuming $f(x) = ax^2+bx+c>0$ for all real $x$ and
$a>0$ and $(b-a)>0$
put $x=-2,$ We get
$f(-2)
=4a-2b+c> 0\Rightarrow 2a+c>2(b-a)$
$\Rightarrow \displaystyle \frac{a+b+c}{b-a}=1+\frac{2a+c}{b-a}>1+2=3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Continued fraction of $\frac{1+\sqrt{5}}{2}$ I am trying to get a better understanding of continued fractions (CF) and was watching a view tutorial clips e.g. this here and looking through some stackexchange articles. Than found this article, where someone was able to get the CF of $\frac{1+\sqrt{5}}{2}$. Unfortunately he didn't post his walktrough. So I know now how I would get the CF of just $\sqrt{5}$, as you can see below, however I am not sure how to handle the whole term.
$$Int(\sqrt{5}) = 2$$
$$\sqrt{5} = 2 -2 + \sqrt{5}$$
$$= 2 + \frac{(-2 + \sqrt{5})*(2 + \sqrt{5})}{(2 + \sqrt{5})}$$
$$= 2 + \frac{(-4 - 2\sqrt{5} + 2\sqrt{5} + 5)}{2 + \sqrt{5}}$$
$$= 2 + \frac{1}{2 + \sqrt{5}}$$
$$= 2 + \frac{1}{2 + 2 + \frac{1}{2 + \sqrt{5}}} = 2 + \frac{1}{4 + \frac{1}{2 + \sqrt{5}}}$$
$$= 2 + \frac{1}{4 + \frac{1}{2 + 2 + \frac{1}{2 + \sqrt{5}}}} = 2 + \frac{1}{4 + \frac{1}{4 + \frac{1}{2 + \sqrt{5}}}}$$
$$= 2 + \frac{1}{4 + \frac{1}{4 + \frac{1}{4 + \ddots}}}$$
Can someone please help me.
| $$
\rho = \frac{1+\sqrt{5}}2
$$
$$2 < \sqrt{5} < 3 \implies \left\lfloor \rho \right\rfloor = 1$$
$$\rho = 1 - 1 + \frac{1+\sqrt{5}}2$$
$$= 1 + \frac{\sqrt{5}-1}2$$
$$= 1 + \frac{( \sqrt{5}-1)*(1 + \sqrt{5})}{2(1 + \sqrt{5})}$$
$$= 1 + \frac{(5 + \sqrt{5} - \sqrt{5} -1 )}{2(1 + \sqrt{5})}$$
$$= 1 + \frac{4}{2(1 + \sqrt{5})}$$
$$= 1 + \frac{2}{(1 + \sqrt{5})} = 1 + \frac{1}{\rho}$$
$$= 1 + \frac{1}{1 + \frac{1}{\rho}}$$
$$= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{\rho}}}$$
$$= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \ddots}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fewest number of marbles in a bag such that drawing the probability of drawing 2 blue marbles is $\frac{1}{6}$.
Two marbles are randomly selected without replacement from a bag
containing blue and green marbles. Probability of drawing both blue is
$\frac{1}{6}$. If three marbles are drawn, then the probability all
three are blue is $\frac{1}{21}$. What is the fewest number of marbles
that must have been in the bag before any were drawn?
I use the criteria to make two equations:
$$\frac{x}{y}\cdot\frac{x-1}{y-1}=\frac{1}{6}$$
for drawing 2 marbles$$\frac{x}{y}\cdot\frac{x-1}{y-1}\cdot\frac{x-2}{y-2}=\frac{1}{21}$$
Where $x$ is the number of blue marbles and $y$ is the total number of marbles.
Simplifying these equations gives:
$$6x^2-6x=y^2-y$$ for the first equation and $$21x^3-63x^2+42x=y^3-3y^2+2y$$
I can't seem to solve this set of equations.
Is my approach correct? If so, how should I continue? If not, how would one solve this problem?
Thanks! Your help is appreciated!
Max0815
| Divide the second equation by the first:
$$\frac{x-2}{y-2} = \frac{6}{21} = \frac{2}{7}$$
Which when simplified gives:
$$7x = 2y + 10$$
What are the smallest integer values of x and y that solve this equation? Well, observe that a possible solution is $x = 2, y = 2$, but this is clearly wrong since $x = y$ in this case. Since $x$ cannot be odd, we go on to the next number, which is 4. Luckily, we can observe that a solution will be given by: $$x = 4, y = 9$$
which happens to be the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Series sum with factorial notation The sum of series
$\displaystyle 1+\frac{1}{1!}\cdot \frac{1}{4}+\frac{1\cdot 3}{2!}\cdot \frac{1}{4^2}+\frac{1\cdot 3 \cdot 5}{3!}\cdot \frac{1}{4^3}+\cdots $
what i try:
$$1+\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{1\cdot 3\cdot \cdots (2r-1)}{r!\cdot 4^r}$$
$$1+\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{(2r)!}{r!\prod^{n}_{r=1}(2r)!}$$
How do i solve it Help me please
|
The series under consideration is
\begin{align*}
\color{blue}{1+\sum_{r=1}^\infty\frac{(2r-1)!!}{r!4^r}}&=1+\sum_{r=1}^\infty\frac{(2r)!}{r!4^r(2r)!!}\\
&=1+\sum_{r=1}^\infty\frac{(2r)!}{r!r!}\left(\frac{1}{8}\right)^r\\
&=\sum_{r=0}^\infty\binom{2r}{r}\left(\frac{1}{8}\right)^r\\
&=\left.\frac{1}{\sqrt{1-4z}}\right|_{z=1/8}\tag{1}\\
&=\frac{1}{\sqrt{1-\frac{1}{2}}}\\
&\,\,\color{blue}{=\sqrt{2}}
\end{align*}
In (1) we use the ordinary generating function of the central binomial coefficients evaluated at $z=1/8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Any quicker method to find matrix exponent for a scaled permutation matrix? I am trying to find the eigen value of matrix $e^A$ such that the matrix A is given as
$$
\begin{bmatrix}
a & 0 & 0\\
0 & 0 & a\\
0 & a & 0
\end{bmatrix}
$$
I found $A^2 = \begin{bmatrix} a^2 & 0 & 0 \\ 0 & a^2 & 0 \\ 0 & 0 & a^2 \end{bmatrix}$, $A^3 = \begin{bmatrix} a^3 & 0 & 0 \\ 0 & 0 & a^3 \\ 0 & a^3 & 0 \end{bmatrix}$ and so forth...
So that I can find, $e^A$ by substituting above matrices in its power series expansion
\begin{equation}
e^A = 1 + A + \frac{1}{2!} A^2+\dots
\end{equation}
which gives
\begin{equation}
e^A =
\begin{bmatrix}
1+a+\frac{1}{2}a^2+\frac{1}{6}a^3+\dots & 0 & 0 \\
0 & 1+\frac{1}{2}a^2+\frac{1}{24}a^4+\dots & a+\frac{1}{6}a^3+\frac{1}{120}a^5\dots \\
0 & a+\frac{1}{6}a^3+\frac{1}{120}a^5\dots & 1+\frac{1}{2}a^2+\frac{1}{24}a^4+\dots
\end{bmatrix}
\end{equation}
which can be written as
\begin{equation}
e^A =
\begin{bmatrix}
e^a & 0 & 0 \\
0 & \frac{e^a}{2}+\frac{e^{-a}}{2} & \frac{e^a}{2}-\frac{e^{-a}}{2} \\
0 & \frac{e^a}{2}-\frac{e^{-a}}{2} & \frac{e^a}{2}+\frac{e^{-a}}{2}
\end{bmatrix}
\end{equation}
I am able to solve this, but I am wondering if there is a quicker way or a trick that I can use to find $e^A$ quickly that this method?
| By diagonalization,
$$\exp\begin{pmatrix} 0 & a \\ a & 0 \end{pmatrix} = \begin{pmatrix} \cosh(a) & \sinh(a) \\ \sinh(a) & \cosh(a)\end{pmatrix} $$
is just De Moivre/Euler's formula $e^{i\alpha}=\cos(\alpha)+i\sin(\alpha)$ written in a equivalent way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim\limits_{n\to\infty}{[x]+[2^2x]+\dots +[n^2x]\over n^3}$ Find the limit, $$\lim_{n\to\infty}{[x]+[2^2x]+\dots +[n^2x]\over n^3}$$
Where $[.]$ denotes the integral part of $.$?
Efforts: If $x$ is integer,
$$\lim_{n\to\infty}{[x]+[2^2x]+\dots +[n^2x]\over n^3}$$ $$\lim_{n\to\infty}{x(1+2^2+\dots +n^2)\over n^3}$$
We know $\sum\limits_{i=1}^ni^2=(n+1)(2n+1)n/6$
Therefore we get that limit is equal to $x/3$.
How to solve it for non integral values.
Thanks in advance.
| Applying Stolz–Cesàro theorem with
$a_n=\sum\limits_{k=1}^n [x\cdot k^2]$ and
$b_n=n^3$ (which is strictly monotone and divergent)
$$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=
\frac{[x(n+1)^2]}{(n+1)^3-n^3}=
\frac{[x(n+1)^2]}{(n+1)^2+(n+1)n+n^2}$$
and we have (using inequality @Winther suggested in the comments $x-1\leq[x]<x$ and squeeze theorem)
$$\frac{x(n+1)^2-1}{3(n+1)^2}\leq
\frac{[x(n+1)^2]}{3(n+1)^2}<
\frac{a_{n+1}-a_n}{b_{n+1}-b_n}<
\frac{[x(n+1)^2]}{3n^2}<
\frac{x(n+1)^2}{3n^2}$$
as a result
$$\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\frac{x}{3}$$
and finally
$$\lim\limits_{n\rightarrow\infty}\frac{a_n}{b_n}=
\lim\limits_{n\rightarrow\infty}\frac{\sum\limits_{k=1}^n [x\cdot k^2]}{n^3}=
\frac{x}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Showing that the locus of point $N$ is $x^2+y^2=a^2$
Question: A point $P(a\cos\theta,b\sin\theta)$ lies on an ellipse with equation $$\varepsilon:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$The tangent to the ellipse $\varepsilon$ at point $P$ is perpendicular to a straight line $l$ which has passed through its focus and intersected at point $N$. Show that the equation of locus of point $N$ is $x^2+y^2=a^2$.
Suppose the coordinate point of the focus is $F(c,0)$. To find the point $N$ parametrically, I have to deal with the following simultaneous equations.
$$\begin{cases}
y-b \sin \theta=-\dfrac{b}{a} \cot \theta \, (x- a \cos \theta)\\y=\dfrac{b}{a}\tan \theta \,(x-c)
\end{cases}$$
where the first equation is the equation of tangent line at point $P$, and the second equation represents the perpendicular line $l$ that passes through the focus of the ellipse $\varepsilon$ and point $N$.
Solving the above equation for $x$ and $y$ in terms of $a$, $b$ and $c$ is quite complicated and outsmarted because it also involves a new variable $\theta$. Any pretty way to deal with it?
| Denote $P$ by $(x_P, y_P)$ instead, then $\dfrac{x_P^2}{a^2} + \dfrac{y_P^2}{b^2} = 1$ and the tangent line at $P$ is$$
t_P: \frac{x_P x}{a^2} + \frac{y_P y}{b^2} = 1. \tag{1}
$$
Because $l$ is perpendicular to $t_P$, then $l$ is given by$$
l: -\frac{y_P x}{b^2} + \frac{x_P y}{a^2} = C_P,
$$
where $C_P$ is a constant depending on $P$. Given that $l$ passes through $(c, 0)$, thus $C_P = -\dfrac{c y_P }{b^2}$ and$$
l: -\frac{y_P x}{b^2} + \frac{x_P y}{a^2} = -\frac{c y_P}{b^2}. \tag{2}
$$
Now, $(1)^2 + (2)^2$ yields$$
1 + \frac{c^2 y_P^2}{b^4} = \left( -\frac{y_P x}{b^2} + \frac{x_P y}{a^2} \right)^2 + \left( -\frac{y_P x}{b^2} + \frac{x_P y}{a^2} \right)^2 = \left( \frac{x_P^2}{a^4} + \frac{y_P^2}{b^4} \right) (x^2 + y^2). \tag{3}
$$
Since $x_P^2 = a^2 \left( 1 - \dfrac{y_P^2}{b^2} \right)$ and $a^2 - b^2 = c^2$, then$$
\frac{x_P^2}{a^4} + \frac{y_P^2}{b^4} = \frac{1}{a^2} \left( 1 - \frac{y_P^2}{b^2} \right) + \frac{y_P^2}{b^4} = \frac{1}{a^2} \left( 1 - \frac{y_P^2}{b^2} + \frac{a^2 y_P^2}{b^4} \right) = \frac{1}{a^2} \left( 1 + \frac{c^2 y_P^2}{b^4} \right)
$$
and (3) becomes$$
x^2 + y^2 = a^2.
$$
| {
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"url": "https://math.stackexchange.com/questions/3161926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Shortcut for value of $f(1)$ where $f(x) = \int e^x \left(\arctan x + \frac {2x}{(1+x^2)^2}\right)\,dx$ If $$f(x) = \int e^x \biggr(\arctan x + \frac {2x}{(1+x^2)^2}\biggr)\,dx$$ and $f(0)=0$ then value of $f(1)$ is?
This is actually a Joint Entrance Examination question so I have to do it in two minutes. Is there a shortcut to finding this result quickly? It seems very complicated. The answer is $e(\pi/4-(1/2)). $
| Actually there is a formula $$\int e^x (g (x)+g'(x))\,dx = e^x\cdot g (x)+c.$$
Now for $$\int e^x \biggr(\arctan x + \frac {2x}{(1+x^2)^2}\biggr)\,dx $$, do the following manipulation:
$$\int e^x \biggr(\arctan x + \frac {2x}{(1+x^2)^2}\biggr)\,dx =\int e^x \biggr(\arctan x - \frac {1}{1+x^2}+\frac {1}{1+x^2}+\frac {2x}{(1+x^2)^2}\biggr)\,dx. $$
Note that $$\biggr(\arctan x - \frac {1}{1+x^2}\biggr)'=\frac {1}{1+x^2}+\frac {2x}{(1+x^2)^2}. $$
Then by the above formula $$\int e^x \biggr(\arctan x + \frac {2x}{(1+x^2)^2}\biggr)\,dx=e^x \biggr(\arctan x - \frac {1}{1+x^2}\biggr)+c.$$
So $$f (1)=\biggr[e^x \biggr(\arctan x - \frac {1}{1+x^2}\biggr)\biggr]_0^1=\frac {e\pi}{4}-\frac {e}{2}+1. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$
They all are positive terms so arithmetic mean is greater than equal to geometric mean.
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq 3( \sec^6 x \csc^6 x \sec^6 x\csc^6 x)^\frac{1}{3} $$
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x \geq 3( \sec x \csc)^4 $$
$$
\sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq
\frac{3 * 2^4}{\sin ^4 2x} $$
Clearly least value is 48, but something is wrong here, as the answer is 80, if I use other methods.
| Alternatively:
$$\sec^6 x +\csc^6 x + \sec^6 x\csc^6 x=\frac{\sin^6x+\cos^6x+1}{\sin^6 x\cos^6x}=\\
\frac{2^6[(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)+1]}{\sin^62x}=\\
\frac{64(2-\frac34\sin^22x)}{(\sin^22x)^3}=\frac{128-48(1-\cos^22x)}{(\sin^22x)^3}=\\
\frac{80+48\cos^22x}{(\sin^22x)^3}\ge 80.$$
equality occurs for $\sin2x=\pm1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Upper Bound for Polynomial Using Evenly Spaced Points Suppose that $x \in [0,1]$ and the points $x_1, x_2,\ldots x_n$ are evenly spaced in the interval $[0,1]$. I am trying to find a tight bound for the maximum of:
$$(x - x_1)(x-x_2)\cdots(x-x_n) $$
I realize that the above expression will be smaller in absolute value than the absolute value of the smallest term $(x-x_i)$ for $ 1\le i \le n$ since each term will is less than $1$. Thus, if there are $n$ equally spaced points, each distanced $\frac{1}{n-1}$ apart, then smallest term will be bounded above by $\frac{1}{2(n-1)}$.
I am wondering if there is a way to get a tighter bound than the one I have found above.
| Let $n\geq3$ and let $x_i:=\frac{i}{n-1}$ so that we want to determine the maximum of the polynomial
$$f_n(x):=\prod_{i=1}^n(x-x_i)=\prod_{i=1}^n\left(x-\frac{1-i}{n-1}\right),$$
on the interval $[0,1]$. If $n$ is odd then for all $x\in(x_1,x_2)$ and $i>2$ we have
$$\frac{1-i}{n-1}<x-x_i<\frac{2-i}{n-1}<0,$$
from which it follows that
$$(x-x_1)(x-x_2)\prod_{i=3}^n\frac{2-i}{n-1}
<f_n(x)
<(x-x_1)(x-x_2)\prod_{i=3}^n\frac{1-i}{n-1}.$$
The maximum of $(x-x_1)(x-x_2)$ on the interval $(x_1,x_2)$ is at the midpoint $\frac{x_1+x_2}{2}=\frac{1}{2(n-1)}$, yielding the following bounds for the maximum $M_n$ of $f_n$ on the interval $(x_1,x_2)$:
$$M_n
>\left(\tfrac{1}{2(n-1)}-x_1\right)\left(\tfrac{1}{2(n-1)}-x_2\right)\prod_{i=3}^n\frac{2-i}{n-1}
=\frac14\frac{(n-2)!}{(n-1)^n}.
$$
$$M_n
<\left(\tfrac{1}{2(n-1)}-x_1\right)\left(\tfrac{1}{2(n-1)}-x_2\right)\prod_{i=3}^n\frac{1-i}{n-1}
=\frac14\frac{(n-1)!}{(n-1)^n},$$
Similarly, if $n$ is even then for all $x\in(x_2,x_3)$ and $i>3$ we have
$$(x-x_1)(x-x_2)(x-x_3)\prod_{i=4}^n\frac{3-i}{n-1}
<f_n(x)
<(x-x_1)(x-x_2)(x-x_3)\prod_{i=4}^n\frac{2-i}{n-1},$$
and some basic algebra shows that the maximum of $(x-x_1)(x-x_2)(x-x_3)$ is at $x=\frac{1+\sqrt{3}}{3(n-1)}$, so
$$M_n>\left(\tfrac{1+\sqrt{3}}{3(n-1)}-x_1\right)
\left(\tfrac{1+\sqrt{3}}{3(n-1)}-x_2\right)
\left(\tfrac{1+\sqrt{3}}{3(n-1)}-x_3\right)\prod_{i=4}^n\frac{3-i}{n-1}
=2\frac{\sqrt{3}}{9}\frac{(n-3)!}{(n-1)^n},$$
$$M_n<\left(\tfrac{1+\sqrt{3}}{3(n-1)}-x_1\right)
\left(\tfrac{1+\sqrt{3}}{3(n-1)}-x_2\right)
\left(\tfrac{1+\sqrt{3}}{3(n-1)}-x_3\right)\prod_{i=4}^n\frac{2-i}{n-1}
=2\frac{\sqrt{3}}{9}\frac{(n-2)!}{(n-1)^n}.$$
Proof that the maximum is in $(x_1,x_2)$ if $n$ is odd, and in $(x_2,x_3)$ if $n$ is even:
(A bit messy, might clean up later)
It it not hard to see that for $x\in(x_k,x_{k+1})$
we have $\operatorname{sgn}f(x)=(-1)^{n-k}$. So the maximum is in some interval $(x_k,x_{k+1})$ with $k\equiv n\pmod{2}$.
The symmetry in the product shows that for all $x\in(0,1]$ we have
$$f\left(x-\frac{1}{n-1}\right)
=\frac{x-x_n}{x-x_1}f\left(x\right)
=(1-x^{-1})f(x),$$
and of course for $x\in[\tfrac12,1]$ we have $|1-x^{-1}|\leq1$, so the maximum value of $f$ on the interval $[\tfrac12,1]$ is assumed on the interval $(x_{n-2},x_{n-1})$. It is clear that
$$f(1-x)=(-1)^nf(x),$$
for all $x\in[0,1]$, from which it follows that the maximum value of $f$ on the interval $[0,1]$ is assumed on $(x_1,x_2)$ if $n$ is odd, and on $(x_2,x_3)$ if $n$ is even.
The polynomial $f$ of degree $n$ has precisely $n$ distinct roots in the interval $[0,1]$. It follows that the polynomial $f'$ has precisely one root in each interval $(x_k,x_{k+1})$ for $k\in\{1,\ldots,n-1\}$. The maximum of $f$ is then assumed at the unique root of $f'$ in the interval $(x_k,x_{k+1})$, where $k=1$ if $n$ is odd and $k=2$ if $n$ is even.
| {
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"timestamp": "2023-03-29T00:00:00",
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$28 + p \sqrt{3} = (q + 2 \sqrt{3})^2$, where p and q are integers, find the values of p and of q. This is how much I have solved so far,
$$28 + p \sqrt{3} = (q + 2 \sqrt{3})^2$$
$$ \implies28 + p \sqrt{3} = q^2 + (4 \sqrt{3})q + 12$$
$$\implies p \sqrt{3} = q^2 + (4 \sqrt{3})q - 16$$
$$\implies p = (q^2 + (4 \sqrt{3})q - 16)/(\sqrt{3})$$
which gives $p$ as ${-8.755604237, 1.827401007}$
But since $p$ is supposed to be an integer (and $q$) this would be the wrong answer to my question.
How could the answers be in such a way that $p$ and $q$ are integers?
| Just do it.
$28 + p\sqrt{3}=(q+2\sqrt 3)^2 = q^2 +4q\sqrt 3 + 12$
$16 - q^2 = \sqrt 3(4q - p)$.
$16-q^2$ is an integer. So is $4q -p$. So $\sqrt 3\times (4q-p)$ is an integer.
Let $\sqrt 3(4q-p) = M$ then our first thought is $\sqrt 3 = \frac M{4q-p}$. But thats impossible as $\sqrt 3$ is irrational.
The only way this makes sense is if $4q - p = 0$ so $p=4q$.
So $16 -q^2 = \sqrt 3(4q -p) = 0$ so $q^2 = 16$ and $q = \pm 4$ and $p = \pm 16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$x^2 + y^2+xy = 1$ , then find the minimum of $x^3 y + xy^3 +4$ x and y belongs to real numbers. $ x^2 + y^2+xy = 1 $. then find the minimum value of $x^3 y + xy^3 +4$.
I assume $ x = r \sin (w)$ and $ y = r\cos(w) $. $ x^3 y + xy^3 +4 = L $ which give me $ \frac{2}{3} \le r^2 \le 2 $ I am stuck after that.Its my Humble request to help me after that.
| Lagrangian multiplier for constrained function
$$L(x,y)= F(x,y)-\lambda G(x,y) $$
set in the order given, partially differentiating with respect to both variables
$$ \dfrac{F_x}{F_y}=\dfrac{G_x}{G_y}= \lambda $$
$$\dfrac{2x+y}{2y+x}=\dfrac{3x^2y+x^3}{x^3+3y^2} $$
Cross multiplying to simplify we get two conditions
$$ (x^2-y^2)(x^2+y^2-xy)=0 $$
Minimal value verifiable by sign of second partial derivatives of $L(x,y)$..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Proving that $g(x)$ defined as $x^2$ on the rationals and $x^4$ on irrationals, is discontinuous at $2$ Define $g: \mathbb{R} \to \mathbb{R}$ by $$g(x) =\begin{cases} x^2 & \text{if } x \text{ is rational}, \\\\ x^4 & \text{if } x \text{ is irrational}. \end{cases}$$ Prove that $g$ is discontinuous at $x = 2$.
Solution Attempt:
Assume instead that $g$ is continuous at 2. Then since $g(2)= 4$, $\;\;\displaystyle \lim_{x\to2}g(x)=4$;
so taking $\epsilon=1$, there is a $\delta>0$ such that if $0<|x-2|<\delta$, then $\big|g(x)-4\big|<1$.
Therefore if $x$ is irrational and $2-\delta<x<2$, then
$\big|x^4-4\big|<1$
$-1 <x^4-4<1\implies 3<x^4<5\implies x^4>3 \implies x = 3^\frac{1}{4} \implies x < 2$ Contradiction
| Rather than $x^4 > 3$, instead conclude that $x^4 < 5$, or equivalently,
$$-\sqrt[4]{5} < x < \sqrt[4]{5}.$$
If you had chosen instead an irrational $x$ such that
$$2 - \min\{\delta, 2 - \sqrt[4]{5}\} < x < 2,$$
then you'd have $x > 2 - (2 - \sqrt[4]{5}) = \sqrt[4]{5}$, a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Computation of the Laplacian form of $g_{\sigma, \eta}$ in Huisken and Sinestrari's paper I'm trying understand how to compute evolution equation for $g_{\sigma, \eta}$ defined on page $3$ of this article and I'm reading the Lectures on Mean Curvature Flows by Xi-Ping Zhu, which gives some details about the computation, I could understand all the computations, except for the computation of the Laplacian, which is
\begin{align*}
\Delta g_{\sigma,\eta} &= \frac{1}{H^{2 - \sigma}} \left[ \Delta |A|^2 - \sigma (1+\eta) H \Delta H + (1 + \eta) \sigma (\sigma - 3) |\nabla H|^2 \right]\\
&- 2 (2 - \sigma) \frac{\nabla_i H}{H} \cdot \nabla_i g_{\sigma,\eta} + \frac{(2 - \sigma)}{H^{4 - \sigma}} \left[ |A|^2 |\nabla H|^2 (\sigma - 1) - H |A|^2 \Delta H \right]
\end{align*}
according to the book (the author use '$\cdot$' to represent the inner product ), but my computation differ from the above by the terms with $(1 + \eta)$. I would like to know how I can obtain the Laplacian correctly.
This is what I did
$\triangle g_{\sigma, \eta} = \nabla_i \nabla_i g_{\sigma,\eta}$
$= \{ \left[ \triangle |A|^2 -2(1+\eta)(|\nabla H|^2 + H\triangle H) \right] H^{2- \sigma} - (\nabla_i |A|^2 - 2(1 + \eta) H \nabla_i H) [ (2 - \sigma) H^{1 - \sigma} \nabla_i H ] \} \frac{1}{(H^{2 - \sigma})^2} - (2 - \sigma) \{ \left( \frac{|A|^2 - (1 + \eta) H^2}{H^{3 - \sigma}} \right) \triangle H \left[ (\nabla_i |A|^2 - (1 + \eta) 2H \nabla_i H) H^{3 - \sigma} - (|A|^2 - (1 + \eta)H^2) (3- \sigma) H^{2 - \sigma} \nabla_i H \right]$
$\frac{\nabla_i H}{(H^{3 - \sigma})^2} \}$
$= \frac{\triangle |A|^2}{H^{\sigma - 2}} - \frac{\sigma(1 + \eta) H \triangle H}{H^{\sigma - 2}} - (\sigma^2 - 3\sigma + 4) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}} -2 \frac{(2 - \sigma)}{H^{2 - \sigma}} \left\langle \nabla_i |A|^2, \frac{\nabla_i H}{H} \right\rangle$
$- \frac{(2 - \sigma) |A|^2 H \triangle H}{H^{4 - \sigma}} + \frac{(2 - \sigma)(3 - \sigma) |A|^2 |\nabla H|^2}{H^{4 - \sigma}} (\star)$
Observe that
$-2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle = - \frac{2(2 - \sigma) \left\langle \nabla_i H, \nabla_i |A|^2 \right\rangle}{H^{3 - \sigma}} + 4(2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}}$
$\frac{2(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}} - \frac{2(2 - \sigma)^2 (1 + \eta) H^2 |\nabla H|^2}{H^{4 - \sigma}}$
$= -\frac{2(2 - \sigma) \left\langle \nabla_i H, \nabla_i |A|^2 \right\rangle}{H^{3 - \sigma}} + 2 \sigma (2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}} + 2 \frac{(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}},$
therefore
$-\frac{2(2 - \sigma)}{H^{2 - \sigma}} \left\langle \frac{\nabla_i H}{H}, \nabla_i |A|^2 \right\rangle = -2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle - 2 \sigma (2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}} - 2 \frac{(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}}$
Substituting this in $(\star)$,
$\triangle g_{\sigma,\eta} = \frac{\triangle |A|^2}{H^{\sigma - 2}} - \frac{\sigma(1 + \eta) H \triangle H}{H^{\sigma - 2}} - (-\sigma^2 + \sigma + 4) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}} -2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle$
$+ \frac{(2 - \sigma)}{H^{4 - \sigma}} \left[ |A|^2 |\nabla H|^2 (\sigma - 1) - H |A|^2 \triangle H \right].$
Thanks in advance!
| Computing in normal coordinates,
\begin{align*}
\Delta g_{\sigma, \eta} &= \nabla_i \nabla_i g_{\sigma,\eta}\\
&= \nabla_i \left[ \frac{(\nabla_i |A|^2 - (1 + \eta) 2 H \nabla_i H) H^{2 - \sigma} - (|A|^2 - (1 + \eta) H^2) (2 - \sigma) H^{1 - \sigma} \nabla_i H}{(H^{2 - \sigma})^2} \right]\\
&= \nabla_i \left[ \frac{(\nabla_i |A|^2 - 2 (1 + \eta) H \nabla_i H}{H^{2 - \sigma}} - (2 - \sigma) \frac{(|A|^2 - (1 + \eta) H^2) \nabla_i H}{H^{3 - \sigma}} \right]\\
&= \left[ (\Delta |A|^2 - 2 (1 + \eta) (H \Delta H + |\nabla H|^2)) H^{2 - \sigma} - (\nabla_i |A|^2 - 2 (1 + \eta) H \nabla_i H) \right.\\
&\left. (2 - \sigma) H^{1 - \sigma} \nabla_i H \right] \frac{1}{(H^{2 - \sigma})^2}\\
&- (2 - \sigma) \left[ (\left\langle \nabla_i |A|^2, \nabla_i H \right\rangle + |A|^2 \Delta H - (1 + \eta) (2 H |\nabla H|^2 + H \Delta H)) H^{3 - \sigma} \right.\\
&\left. - (|A|^2 \nabla_i H - (1+ \eta) H^2 \nabla_i H) (3 - \sigma) H^{2 - \sigma} \nabla_i H \right] \frac{1}{(H^{3 - \sigma})^2}\\
&= \left[ (\Delta |A|^2 - 2 (1 + \eta) (H \Delta H + |\nabla H|^2)) H^{2 - \sigma} - (\left\langle \nabla_i |A|^2, \nabla_i H \right\rangle - 2 (1 + \eta) H |\nabla H|^2) \right.\\
&\left. (2 - \sigma) H^{1 - \sigma} \right] \frac{1}{(H^{2 - \sigma})^2}\\
&- (2 - \sigma) \left[ (\left\langle \nabla_i |A|^2, \nabla_i H \right\rangle + |A|^2 \Delta H - (1 + \eta) (2 H |\nabla H|^2 + H \Delta H)) H^{3 - \sigma} \right.\\
&\left. - (|A|^2 |\nabla H|^2 - (1+ \eta) H^2 |\nabla H|^2) (3 - \sigma) H^{2 - \sigma} \right] \frac{1}{(H^{3 - \sigma})^2}\\
&= \left[ \Delta |A|^2 - \sigma (1 + \eta) H \Delta H + (- \sigma^2 + \sigma) (1 + \eta) |\nabla H|^2 \right] \frac{1}{(H^{2 - \sigma})^2} + \frac{2 \left\langle \nabla_i |A|^2, \nabla_i H \right\rangle}{H^{3 - \sigma}}\\
&+ \frac{(2 - \sigma)}{H^{4 - \sigma}} \left[ |A|^2 |\nabla H|^2 (3 - \sigma) - H |A|^2 \Delta H \right] (\star)
\end{align*}
Observe that
\begin{align*}
-2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle &= - \frac{2(2 - \sigma) \left\langle \nabla_i H, \nabla_i |A|^2 \right\rangle}{H^{3 - \sigma}} + 4(2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}}\\
&+ \frac{2(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}} - \frac{2(2 - \sigma)^2 (1 + \eta) H^2 |\nabla H|^2}{H^{4 - \sigma}}\\
&= -\frac{2(2 - \sigma) \left\langle \nabla_i H, \nabla_i |A|^2 \right\rangle}{H^{3 - \sigma}} + 2 \sigma (2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}}\\
&+ 2 \frac{(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}},
\end{align*}
i.e.,
\begin{align*}
-\frac{2(2 - \sigma) \left\langle \nabla_i H, \nabla_i |A|^2 \right\rangle}{H^{3 - \sigma}} &= -2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle - 2 \sigma (2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}}\\
&- 2 \frac{(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}}
\end{align*}
Substituting this in $(\star)$,
\begin{align*}
\Delta g_{\sigma,\eta} &= \frac{\triangle |A|^2}{H^{\sigma - 2}} - \frac{\sigma(1 + \eta) H \triangle H}{H^{\sigma - 2}} + \sigma (\sigma - 3) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}} -2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle\\
&+ \frac{(2 - \sigma)}{H^{4 - \sigma}} \left[ |A|^2 |\nabla H|^2 (\sigma - 1) - H |A|^2 \triangle H \right].
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
McLaurin expansion of $(a^2+z^2)^{-\frac{1}{2}}$
Show that $$(a^2+z^2)^{-\frac{1}{2}}=\frac{1}{a}+\sum_\limits{n=1}^{\infty}\frac{(-1)^n(2n-1)!!)}{a^{2n+1}(2n)!!}z^{2n},\:|z|<|a|,a>0$$
$$(a^2+z^2)^{-\frac{1}{2}}=\frac{1}{a}(1+\frac{z^2}{a^2})^{-\frac{1}{2}}=\frac{1}{a}(1+\frac{z^2}{a^2})^{-\frac{1}{2}}=\frac{1}{a}+\frac{1}{a}\sum_\limits{n=1}^{\infty}{-\frac{1}{2}\choose n}(\frac{z^2}{a^2})^n$$
Now my problem lies in the simplification of ${-\frac{1}{2}\choose n}$
$${-\frac{1}{2}\choose n}=\frac{-\frac{1}{2}!}{n!(-n-\frac{1}{2})!}=\frac{(-1)^n(\frac{1}{2}(\frac{1}{2}+1)...(\frac{1}{2}+n-1))}{n!}$$
Now I am stuck and I do not know how to simplify the expression any longer.
Question:
Can someone help me solve the exercise?
Thanks in advance!
| Hint. This might be useful
\begin{align}
\frac{(2n-1)!!}{(2n)!!} &=\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n}\\\\
&=\frac{1\cdot \color{red}{2}\cdot 3\cdot \color{red}{4} \cdot 5\cdot\color{red}{6}\cdots(2n-1)\cdot \color{red}{2n}}{(2\cdot 4\cdot 6\cdots (2n))^\color{red}{2}}\\\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^\color{red}{2}}\\\\
& =\frac{(2n)!}{2^{2n} (n!)^2 }.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
maximum value of $\sum (a-b)^2$
If $a^2+b^2+c^2=5$ and $a,b,c \in \mathbb{R},$ find the maximum
value of $(a-b)^2+(b-c)^2+(c-a)^2$.
My Try: $(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ac)$
$$=10-2(ab+bc+ac)$$
Now this implies $ab+bc+ac\le 5$. So
$$(a-b)^2+(b-c)^2+(c-a)^2 = 10-2(ab+bc+ac) \geq 20.$$
Could some help me to find max of $(a-b)^2+(b-c)^2+(c-a)^2$? Thanks.
| Here's a more visual approach; for a point $(a,b,c)\in\Bbb{R}^3$ the condition that
$$a^2+b^2+c^2=5,$$
is equivalent to being on the sphere of radius $\sqrt{5}$ centered at the origin, and
$$(a-b)^2+(b-c)^2+(c-a)^2=||(a,b,c)-(b,c,a)||^2,$$
where the point $(b,c,a)\in\Bbb{R}^3$ is obtained from $(a,b,c)$ by a rotation of one third of a full turn around the line spanned by $(1,1,1)$. Then clearly the distance is maximal precisely when $(a,b,c)$ is on the equator w.r.t. the axis of rotation, and as the following picture shows;
by elementary geometry the distance between $(a,b,c)$ and $(b,c,a)$ is $\sqrt{3}$ times the radius, which is $\sqrt{5}$. Hence the desired maximum is $(\sqrt{3}\times\sqrt{5})^2=15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Earth's surface area
Here we are trying to calculate the earth's surface area via geodetic coordinates:
$x=(Rp(\lambda)+h)\sin (\lambda)\cos(\phi)$
$y=(Rp(\lambda)+h)\sin (\lambda)\sin(\phi)$
$x=((1-e^2)Rp(\lambda)+h)\cos (\lambda)$
where $p(\lambda)=\frac 1 {\sqrt{1-e^2\cos ^2 \lambda}}$
We first compute the flat metric $ds^2$ in these coordinates:
$ds^2=dh^2+(h+(1-e^2)Rp^3)^2d\lambda ^2+\sin ^2\lambda (h+Rp)^2d\phi ^2$
Then we compute induced metric on the surface by putting $h=0$
$ds_*^2=R^2p^2((1-e^2)^2p^4d\lambda ^2+\sin ^2\lambda d\phi ^2)$
where the determinant of this metric is $g_*=R^4p^8(1-e^2)^2\sin ^2\lambda$
We found that $A=\int_0^{2\pi}\int_0^{\pi}\sqrt{g_*} d\lambda d\phi=2\pi R^2(1-e^2) [\frac{1}{1-e^2}+\frac{arctanh(e)}{e}]=2\pi R^2[1+ (\frac{1}{e}-e)arctanh(e)]$
Now from here my professor wrote that
$e=0 \implies A=4\pi R ^2$ and $e\approx 0 \implies A\approx 4\pi R ^2 - \frac 4 3 \pi R^2 e^2 + O(e^4)$, and I didn't understand how he got these.
Also from here how do we get that for the earth: $A_{ellipsoid}=5.10\times 10^8 km^2$ and $A_{sphere}=5.12\times 10^8 km^2$
| Consider $$Z=1+\left(\frac{1}{e}-e\right) \tanh ^{-1}(e)$$ Now, by Taylor around $e=0$
$$\tanh ^{-1}(e)=e+\frac{e^3}{3}+\frac{e^5}{5}+O\left(e^7\right)$$
$$Z=1+\left(\frac{1}{e}-e\right)\left(e+\frac{e^3}{3}+\frac{e^5}{5}+O\left(e^7\right)\right)=2-\frac{2 e^2}{3}-\frac{2 e^4}{15}+O\left(e^6\right)=2-\frac{2 e^2}{3}+O\left(e^4\right)$$
Then
$$A=2\pi R^2 \left(2-\frac{2 e^2}{3}+O\left(e^4\right)\right)=4\pi R^2- \frac{4 \pi e^2}{3} R^2+O\left(e^4\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $f \in L^p(0,\infty)$ and $g(x) = \int_0^\infty \frac{1}{x+y} f(y) \, dy$, show $\lim\limits_{x \to \infty} g(x) = 0$
For $x \in (0, \infty)$, let:
\begin{align*}
g(x) &= \int_0^\infty \frac{1}{x+y} f(y) \, dy \\
\end{align*}
Show that $g(x) \mathop{\longrightarrow}\limits_{x \to \infty} 0$ for $f \in L^p(0,\infty)$, $1 \le p \le \infty$
I'm struggling with the $p = \infty$ case, but I believe I have a solution for $1 \le p < \infty$:
Let $h(x,y) = \frac{1}{x+y}$ such that:
\begin{align*}
g(x) &= \int_0^\infty \frac{1}{x+y} f(y) \, dy \\
g(x) &= \int_0^\infty f(y) h(x,y) \, dy \\
g(x) &= \lVert f \cdot h(x) \rVert_1 \\
\end{align*}
From there we can use Holder's Inequality with $q = p/(p-1)$:
\begin{align*}
g(x) = \lVert f \cdot h(x) \rVert_1 &\le \lVert f \rVert_p \lVert h(x) \rVert_q \\
&= \lVert f \rVert_p \left( \int_0^\infty h(x)^q \, dy \right)^{1/q} \\
&= \lVert f \rVert_p \left( \int_0^\infty (x+y)^{-q} \, dy \right)^{1/q} \\
\end{align*}
Substitute $z=x+y$:
\begin{align*}
g(x) &= \lVert f \rVert_p \left( \int_x^\infty z^{-q} \, dz \right)^{1/q} \\
&= \lVert f \rVert_p \left( \frac{1}{q-1} \cdot x^{1-q} \right)^{1/q} \\
&= \lVert f \rVert_p \left( \frac{1}{q-1} \right)^{1/q} \cdot x^{1/q-1} \\
&= \lVert f \rVert_p \left( p-1 \right)^{1-1/p} \cdot x^{-1/p} \\
\end{align*}
Take the limit of both sides:
\begin{align*}
\lim\limits_{x \to \infty} g(x) &= \lim\limits_{x \to \infty} \lVert f \rVert_p \left( p-1 \right)^{1-1/p} \cdot x^{-1/p} \\
&= \lVert f \rVert_p \left( p-1 \right)^{1-1/p} \cdot \lim\limits_{x \to \infty} x^{-1/p} \\
&= 0 \\
\end{align*}
If $p=\infty$, then $q=1$, and $\lVert h(x) \rVert_1$ does not converge, and this proof does not work. Is there a different proof that works for that case?
| The result is not true for $p =\infty$. In fact if $f \equiv 1$ then $g(x)=\infty$ for all $x$ even though $f \in L^{\infty}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can't solve system of linear equations (that need simplification first) I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics", and am stuck on a problem at the end of the "Linear Equations" chapter.
I've a system of two linear equations:
$$\frac{3x+2}{4} - \frac{x+2y}{2} = \frac{x-3}{12}$$
$$\frac{2y+1}{5} + \frac{x-3y}{4} = \frac{3x+1}{10}$$
So, these two first need to be simplified. I assume that the LSM for the first one (for 2, 4 and 12) is 12, so we have:
$$12\frac{3x+2}{4} - 12\frac{x+2y}{2} = 12\frac{x-3}{12}$$
Simplifying further, we have:
$$3(3x+2) - 6(x+2y) = (x-3)$$
$$9x + 6 - 6x - 12 y = x - 3$$
$$9x -6x -x - 12y = -3 +6$$
Finally, we get our first simplified linear equation:
$$2x - 12y = 3$$
Now, onto the second one. The LSM of 5, 4 and 10 is 20, so we have:
$$20\frac{2y+1}{5} + 20\frac{x-3y}{4} = 20\frac{3x+1}{10}$$
Simplifying further, we have:
$$4(2y+1) + 5(x-3y) = 2(3x + 1)$$
$$8y + 4 + 5x - 15y = 6x + 2$$
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation:
$$-x -7y = 2$$
Now we can solve our system of linear equations:
$$2x - 12y = 3$$
$$-x -7y = 4$$
Multiplying the second one by 2:
$$2x - 12y = 3$$
$$-2x -14y = 4$$
Now, we add the two equations, and get:
$$-26y = 7$$
Solving for $y$, we get:
$$y = -\frac{7}{26}$$
which I'm fairly certain is not a correct answer.
Can anyone see where I'm going wrong here?
| The first mistake I see is here:
$$9x + 6 - 6x - 12 y = x - 3$$ $$9x -6x -x - 12y = -3 +6$$
The second line should read
$$9x-6x-x-12y=-3-6$$
you forgot to invert the sign when moving $6$ to the other side of the $=$ sign.
You were equally sloppy with the second equation, when you did this:
$$5x - 6x + 8y - 15y = 2 - 4$$
We get our second simplified linear equation: $$-x -7y = 2$$
In fact, $2-4$ is not equal to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Given three non-negative numbers $a,b,c$ so that $a+b+c=3$. Prove that $\prod\limits_{cyc}(\!2+ a^{2}\!)+abc\geqq 28$ .
Given three non-negative numbers $a,\!b,\!c$ so that $a+\!b+\!c\!=\!3$. Prove that $\prod\limits_{cyc}(\!2+ a^{2}\!)+ abc\geqq 28$.
Let $a+ b+ c= 3u= 3, ab+ bc+ ca= \frac{3u}{X}, abc= \frac{u^{3}}{wX}$ so $1\leqq X\leqq w$, so
$$\therefore\frac{(16X^{2}- 24X+ 18)w^{2}- 11X+ 1}{w^{2}X^{2}}= \frac{F(w)}{w^{2}X^{2}}\geqq 0$$
$$\because F(w)= (16X^{2}- 24X+ 18)w^{2}- 11X+ 1\geqq 0$$
By discriminant and sos: $(16X^{2}- 24X+ 18)w^{2}- 11X+ 1\geqq (16X^{2}- 24X+ 18)(w- X)^{2}$ .
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$\prod_{cyc}(2u^2+a^2)+u^3w^3\geq28u^6$$ or $f(w^3)\geq0,$ where
$$f(w^3)=w^6-11u^3w^3+16u^6-24u^4v^2+18u^2v^4.$$
But $$f'(w^3)=2w^3-11u^3\leq0,$$ which says that it's enough to prove our inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$.
Id est, we need to prove that
$$(2+a^2)^2(2+(3-2a)^2)+a^2(3-2a)\geq28$$ or
$$(a-1)^2(4a^4-4a^3+15a^2-16a+16)\geq0,$$ which is obvious.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Maclaurin Series Expansion of $\ln(1+\sin x)$ Hello can anyone help with this question
Show that the Maclaurin series of the function
$$\ln(1+\sin x)$$
up to the term in $x^4$ is
$$x-x^2/2 + x^3/6 - x^4/12 + \ldots$$
So I know the expansion for $\ln(1+x)= x - x^2 + x^3/3 +\dots$ and that of $\sin x= x - x^3/3!+x^5/5!-\dots$ hence I tried by substituting the first two terms of $\sin x$ into the expansion of $\ln(1+x)$ to get $\ln(1+x-x^3/6)$ up to the $x^4$ term of the expansion of $\ln(1+x)$. But I got stucked with the algebra so I will value the help anyone can provide.
| What you tried is, however, an interesting attempt to solve the given task but not quite right. It is in fact (and please don't ask me why) correct up a few terms if you finish what you tried. Anyway, this is not the standard way of finding a MacLaurin Series of a given function.
Recall, a MacLaurin Series Expansion is a Taylor Series Expansion centered at $0$. By Taylor's Theorem we know that the series expansion is then given by
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n\tag1$$
Since you are only asked to find the expansion up to the $x^4$-term we only need to compute the first four derivatives and evaluate them at $0$. Thus, we obtain
\begin{align*}
&f(x)=\ln(1+\sin x),&&f(0)=\ln(1+0)=0\\
&f^{(1)}(x)=\frac{\cos x}{1+\sin x},&&f^{(1)}(0)=\frac1{1+0}=1\\
&f^{(2)}(x)=-\frac1{1+\sin x},&&f^{(2)}(0)=-\frac1{1+0}=-1\\
&f^{(3)}(x)=\frac{\cos x}{(1+\sin x)^2},&&f^{(3)}(0)=\frac1{(1+0)^2}=1\\
&f^{(4)}(x)=-\frac{1+\sin x+\cos^2x}{(1+\sin x)^3},&&f^{(4)}(0)=-\frac{1+0+1}{(1+0)^3}=-2
\end{align*}
Plugging these values in $(1)$ we obtain
\begin{align*}
\ln(1+\sin x)&=f(0)+f^{(1)}(0)x+\frac{f^{(2)}(0)}{2}x^2+\frac{f^{(3)}(0)}{6}x^3+\frac{f^{(4)}(0)}{24}x^4+\cdots\\
&=0+1\cdot x-\frac12x^2+\frac16x^3-\frac2{24}x^4+\cdots\\
&=x-\frac{x^2}2+\frac{x^3}6-\frac{x^4}{12}+\cdots
\end{align*}
$$\therefore~\ln(1+\sin x)~=~x-\frac{x^2}2+\frac{x^3}6-\frac{x^4}{12}+\cdots$$
In a similiar way you can obtain the MacLaurin Series Expansions for $sin x$ or $\ln(1+x)$. Just substituting one into another isn't the afterall the exspected way to do this but rather computing the derivatives at $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Have I found the correct Laurent series expansion? I am supposed to find the Laurent series expansion for
$$f(z) = \frac{z+1}{z^2-4}$$ in the region $1<|z+1|<3$.
My solution:
$$w=z+1 \Leftrightarrow z=w-1 \Rightarrow f(z) = \frac{w-1+1}{(w-1)^2-4} = \frac{w}{(w+1)(w-3)}$$
Partial fractions: $$\frac{w}{(w+1)(w-3)} = \frac{1}{4}\frac{1}{w+1} + \frac{3}{4}\frac{1}{w-3}$$
We have:
\begin{eqnarray}\frac{1}{4}\frac{1}{w+1} + \frac{3}{4}\frac{1}{w-3} &=& \frac{1}{4}\frac{1}{w} \bigg(\frac{1}{1+\frac{1}{w}}\bigg) + \frac{3}{4}\frac{1}{w}\bigg(\frac{1}{1-\frac{3}{w}}\bigg)\\ &=& \frac{1}{4}\frac{1}{w} \sum_{n=0}^\infty \bigg[\frac{(-1)}{w}\bigg]^n + \frac{3}{4}\frac{1}{w} \sum_{n=0}^\infty \bigg[\frac{3}{w}\bigg]^n\\ &=& \frac{1}{4} \sum_{n=0}^\infty \frac{(-1)^n}{w^{n+1}} + \frac{3}{4} \sum_{n=0}^\infty \frac{3^n}{w^{n+1}}\\ &=& \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{w^{n}} + \frac{3}{4} \sum_{n=1}^\infty \frac{3^{n-1}}{w^n}\\ &=& \frac{1}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(z+1)^n} + \frac{3}{4} \sum_{n=1}^\infty \frac{3^{n-1}}{(z+1)^{n}}\end{eqnarray}
for the region $1<|z+1|<3$. It feels like there is something wrong here. Any help is greatly appreciated.
| Seems like the second sum is good on $\mid z+1\mid\gt3$. You should factor out $3$ instead of $w$ (in the denominator).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Taylor series of functions with matrix input Today my professor said something interesting, replace $x$ in $f(x)$ with matrix \begin{bmatrix}x&1\\0&x\end{bmatrix}
Then $$f(\begin{bmatrix}x&1\\0&x\end{bmatrix}) = f(x)\begin{bmatrix}1&0\\0&1\end{bmatrix} + f^{'}(x)\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$
and asked us to find the higher order terms ($f^{''}(x), f^{'''}(x) ....$) and extend it to multi variable functions.
I didn't understand how he came up with this.
After some googling, this seems similar to matrix exponential from lie groups.
| I believe that, as mentioned by mathreadler in his comment on the question itself, that there is indeed a typo and that the correct formula may be written
$f \left (\begin{bmatrix} x & 1 \\ 0 & x \end{bmatrix} \right ) = f(x)I + f'(x)N, \tag 0$
with $I$ and $N$ explained in what follows.
I assume $f(x)$ is represented by a Taylor series about $x = 0$:
$f(x) = f(0) + f'(0)x + \dfrac{1}{2}f''(0)x^2 + \ldots = \displaystyle \sum_0^\infty \dfrac{1}{n!} f^{(n)}(0) x^n; \tag 1$
we have
$\begin{bmatrix} x & 1 \\ 0 & x \end{bmatrix}^n = \left ( \begin{bmatrix} x & 0 \\ 0 & x \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right )^n = \left ( x\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right )^n; \tag 2$
setting
$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \tag 3$
and
$N = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \; N^2 = 0, \tag 4$
we write
$\begin{bmatrix} x & 1 \\ 0 & x \end{bmatrix}^n = (xI + N)^n; \tag 5$
since $IN = NI$, (5) may be subject to the ordinary binomial expansion, and since $N^2 = 0$, the terms containing the powers of $N$ greater than the second vanish; thus
$\begin{bmatrix} x & 1 \\ 0 & x \end{bmatrix}^n = (xI + N)^n = x^nI + nx^{n - 1}N; \tag 6$
if we substitute this into (1) we obtain
$f \left (\begin{bmatrix} x & 1 \\ 0 & x \end{bmatrix} \right ) = f(0)I + f'(0)(xI + N) + \dfrac{1}{2}f''(0) (x^2 I + 2xN) + \ldots$
$= (f(0) + f'(0)x + \dfrac{1}{2} f''(0)x^2 + \ldots)I + (f'(0) + f''(0)x + \ldots)N$
$= \displaystyle \sum_0^\infty \dfrac{1}{n!} f^{(n)}(0)(xI + N)^n = \sum_0^\infty \dfrac{1}{n!}f^{(n)}(0)(x^nI + nx^{n - 1}N)$
$= \displaystyle \sum_0^\infty \dfrac{1}{n!} f^{(n)}(0)x^nI + \sum_1^\infty \dfrac{1}{n!} f^{(n)}(0)nx^{n - 1}N$
$= \left ( \displaystyle \sum_0^\infty \dfrac{1}{n!} f^{(n)}(0)x^n \right ) I + \left ( \displaystyle \sum_1^\infty \dfrac{1}{(n - 1)!} f^{(n)}(0)x^{n - 1} \right )N; \tag7$
we observe that the coefficient of $I$ is the Taylor series of $f(x)$ and that of $N$ is the Talylor series of $f'(x)$; thus
$f \left (\begin{bmatrix} x & 1 \\ 0 & x \end{bmatrix} \right ) = f(x)I + f'(x)N. \tag 8$
This "expansion" is in fact exact on any inteval containing $0$ on which the Taylor series for $f(x)$ and $f'(x)$ converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int _0^{\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx$ The function
$$f\left(z\right)=\frac{z^6}{\left(z^4+a^4\right)^2}$$
Has the following poles of order 2:
$$ z(k)=a \exp\left( \frac{\left(2k+1\right)}4 i\pi \right)$$
$f$ is even, therefore: $$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx =\frac{1}{2}\int _{-\infty }^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx$$
$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=i\pi \sum _k\:Res\left(f,\:z\left(k\right)\right)$$
$$Res\left(f,\:z\left(k\right)\right)=\lim _{z\to z\left(k\right)}\left(\frac{1}{\left(2-1\right)!}\left(\frac{d}{dz}\right)^{2-1}\frac{z^6\left(z-z\left(k\right)\right)^2}{\left(z^4+a^4\right)^2}\right)$$
$$z^4+a^4=z^4-z_k^4\implies\dfrac{z^6(z-z_k)^2}{(z^4+a^4)^2}=\dfrac{z^6}{(z^3+z_k z^2+z_k^2 z+z_k^3)^2}$$
$$Res\left(f,\:z_k\right)=\lim _{z\to \:z_k}\left(\frac{d}{dz}\left(\frac{z^6}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^2}\right)\right)$$
$$Res\left(f,\:z_k\right)=\frac{2z_kz^5\left(z^2+2z_kz+3z_k^2\right)}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^3}=\frac{2z_k^6\cdot 6z_k^2}{\left(4z_k^3\right)^3}$$
$$Res\left(f,\:z_k\right)=\frac{12z_k^8}{64z_k^9}=\frac{3}{16z_k}$$
$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi }{16a}\sum _{k=0}^n\:e^{-\frac{\left(2k+1\right)}{4}i\pi }$$
We consider only the residues within the upper half plane, that is to say those corresponding to $k=0$ and $k=1$.
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(e^{-\frac{i\pi }{4}\:\:}+e^{-\frac{3i\pi \:}{4}\:\:}\right)$$
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(\frac{\sqrt{2}}{2}\:-i\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)$$
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3\pi \sqrt{2}\:}{16a}$$
| If you want an approach that doesn't require as much differentiation, first use partial fractions to write $\dfrac{z^3}{z^4+a^4}=\sum_{k=0}^3\dfrac{c_k}{z-z(k)}$ so $$\frac{z^6}{(z^4+a^4)^2}=\sum_k\frac{c_k^2}{(z-z(k))^2}+2\sum_{k<l}\frac{c_kc_l}{(z-z(k))(z-z(l))}\\=\sum_k\frac{c_k^2}{(z-z(k))^2}+2\sum_{k<l}\frac{c_kc_l}{z(k)-z(l)}\left(\frac{1}{z-z(k)}-\frac{1}{z-z(l)}\right).$$The first sum doesn't contribute to $\int_{\Bbb R}\dfrac{x^6 dx}{(x^4+a^4)^2}$, but some of the latter sum's terms do, viz. $$\oint\frac{dz}{(z-w)^{n+1}}=2\pi i\delta_{n0}$$for enclosed $w$. Hence$$\int_0^\infty\frac{x^6 dx}{(x^4+a^4)^2}=\pi i\sum_{k<l}\frac{c_kc_l}{z(k)-z(l)}\left([k\in S]-[l\in S]\right),$$where $\{z(k)|k\in S\}$ is the set of residues your contour encloses and $[]$ is the Iverson bracket, i.e. $[k\in S]$ is $1$ if $k\in S$ or $0$ otherwise. If you're experienced with Beta functions, you should try calculating the integral separately with $x=a\tan^{1/2}t$ to double-check you get the same answer twice. (I get $\frac{3\pi}{8a\sqrt{2}}$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Laplace Transforms with Non-repeated Irreducible quadratic factors I'm solving the following differential equation with a Laplace Transform:
$$x″+ 9x = \cos(t) + \delta(t-\pi)$$
The initial conditions are that x(0) and x'(0) are equal to 0.
After after the transform, we obtain:
$$F(S) = \frac{S}{(S^2+9)[(S^2+1)+9]} + \frac{e^{-\pi\cdot t}}{S^2+9}$$
The problem is that there are two unique irreducible quadratic factors in the denominator of that first term: therefore I cannot seem to find a way to simplify this to find equivalent components which would make the inverse transform easier. How do I find the inverse of that first term?
$$\frac{S}{(S^2+9)[(S^2+1)+9]}$$
The answer includes simple cos(t) and cos(3t) terms with coefficients of $\frac{1}{8}$ and $\frac{-1}{8}$ respectively. What is the process of going from this complex laplace transform to something so simple?
| Alternative Method
$$
F(s)=\underbrace{\frac{s}{(s^2+1)(s^2+9)}}_{\mathbf{(I)}}+\underbrace{\frac{e^{-\pi s}}{s^2+9}}_{\mathbf{(II)}}
$$
Alternative (and fast) method for $\mathbf{(I)}$:
We can also decompose $\mathbf{(I)}$ using partial fractions by the following way:
$$
\frac{s}{(s^2+1)(s^2+9)}\equiv\frac{Es+F}{s^2+1}+\frac{Gs+H}{s^2+9}
$$
Finding the constants of partial fractions decomposition
$$
\frac{s}{(s^2+1)(s^2+9)}\equiv\frac{(Es+F)\cdot(s^2+9)+(Gs+H)\cdot(s^2+1)}{(s^2+1)(s^2+9)}
$$
Now, working only with the numerators:
$$s\equiv (E+G)\cdot s^3+(F+H)\cdot s^2+(9E+G)\cdot s+(9F+H)$$
Thus, we can find $E$, $F$, $G$ and $H$ solving the following system of equations:
$$\mbox{$
\begin{cases}
\begin{alignat}{1}
E+G&=0
\\F+H&=0
\\9E+G&=1
\\9F+H&=0
\end{alignat}
\end{cases}
$}
\implies
\mbox{$
\begin{cases}
\begin{alignat}{1}
E&=1/8
\\F&=0
\\G&=-1/8
\\H&=0
\end{alignat}
\end{cases}
$}
$$
Thus, we get:
$$
\frac{s}{(s^2+1)(s^2+9)}\equiv \frac{s}{8\cdot(s^2+1)}-\frac{s}{8\cdot(s^2+9)}
$$
Now we can apply the Inverse Laplace Transform:
Remember that the Laplace transform of cosine function is:
$$
\cos(\omega t) \iff \frac{s}{s^2+\omega^2}
$$
So, we have:
$$
\begin{alignat}{1}
\mathscr{L}^{-1}\left[\frac{s}{8\cdot(s^2+1)}-\frac{s}{8\cdot(s^2+9)}\right]&=\frac 1 8 \,\mathscr{L}^{-1}\left[\frac{s}{s^2+1^2}\right]-\frac 1 8 \,\mathscr{L}^{-1}\left[\frac{s}{s^2+3^2}\right]
\\&=\frac{1}{8}\left[cos(t)-cos(3t)\right]
\end{alignat}
$$
Same answer here
$$
\bbox[5px,border:1.1px solid black] {
x(t)=\frac{1}{8}\left[cos(t)-cos(3t)\right]-\frac 1 3\sin(3t)\cdot\operatorname{H}(t-\pi)
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convert $\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$ into an algebraic function
Convert trigonometric function into algebraic function.
$$\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$$
My approach is as follows:
$sin2\theta$ is to be calculated
$$\tan^{-1}x=\gamma \tag{1}$$
Hence
$$\begin{align}
\cos\theta&=\cot2\gamma \tag{2}\\[4pt]
\cos\theta&=\frac{\cot^2\gamma-1}{2\cot\gamma} \tag{3}\\[4pt]
\cos\theta&=\frac{1-x^2}{2x} \tag{4}
\end{align}$$
The value of $\sin\theta$ is coming in negative.
| I firstly apologise about how this is written, I really ought to learn how to type this stuff properly. Secondly, if there's any mistakes please do say!
$$\tan(2x) = 2\tan(x)/1-\tan^2(x)$$
so $$\cot(2x) = (1-\tan^2(x))/2\tan(x)$$
swapping $x = \arctan(x)$
$$\cot(2\arctan(x)) = (1-x^2)/2x$$
Okay, that bit is done,
Let $(1-x^2)/2x = Y$ for simplicity...
So we want to evaluate
$$\sin(2\arccos(Y))$$
Well, $$\sin(2x)= 2\sin(x)\cos(x)$$
this gives us
$$2\sin(\arccos(Y))(Y)$$
So we just want to now evaluate $$2\sin(\arccos(Y))$$
Well $$\sin^2(x)+\cos^2(x)=1$$
so,
$$\sin(x) = \sqrt {(1 - \cos^2(x)})$$
$$\sin(\arccos(y)) = \sqrt {(1-Y^2)}$$
$$2\sin(\arccos(y)) = \sqrt{2(1-Y^2)}$$
So the answer is $$2Y \sqrt {(1-Y^2)}$$
where $Y=(1-x^2)/2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Equation solution set of $f^{-1}(x)=f(x)$?
let $f(x)=-(x+2)^3-2 $ . Now what is the equation solution set of $f^{-1}(x)=f(x)$ ?
My try : I know that if a function $f$ is increasing in the domain $D$ on
which its inverse function $f^{-1}$ exists, and a lies in
$D$, then $f (a) = f^{-1}(a)$ if and only if $f(a) = a$
Now :
$$f'(x)=-3(x+2)^2 \leq 0$$ So function $f$ is steady decreasing and $-f(x)$ is steady increasing. Thus these equations have the same solution :
$$f(x)=f^{-1}(x)\\-f(x)=-f^{-1}(x)\\-f(x)=x \to f(x)=-x$$
$$-(x+2)^3-2=-x \\ x = -3.7$$
It is right ?
| No, if you have $f(x) = f^{-1}(x)$, then $f(f(x)) = x$, but $f(f(-3.7)) = 2.913 \ne -3.7$, so it's wrong. Since $f$ is monotone, $f(x)=f^{-1}(x)$ iff $f(f(x)) = x$. Let $$g(x) := f(f(x))-x = -[(-(x+2)^3-2)+2]^3-2 - x = (x+2)^9 - 2 - x = (x+2)((x+2)^8-1)$$
Therefore, $f(f(x)) = x$ iff $g(x) = 0$ iff $x = -2$ or $(x+2)^8 = 1$, so $(x+2)^4 = (x+2)^2 = 1$, so $x + 2 = \pm1$ and $x = -3$ or $x = -1$.
Hence, $g(x)$ has only three real roots $-3,-2,-1$.
Checking: I'll use Julia to speed up the checking work.
julia> f(x) = -(x+2)^3-2
f (generic function with 1 method)
julia> f.(-3:-1)
3-element Array{Int64,1}:
-1
-2
-3
The . in f.() means to apply user-defined function f element-wise on the range -3:-1. For $x = -3$, it's the first element in the argument for f, so f(x) will be shown in the first element in the results (-1). Then search for -3 in the results: it occurs at the last element, so it corresponds to the last element in the argument for f (-1). The other two elements can be checked in the same way.
julia> f.(-10:10)
21-element Array{Int64,1}:
510
341
214
123
62
25
6
-1
-2
-3
-10
-29
-66
-127
-218
-345
-514
-731
-1002
-1333
-1730
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrate $\int \frac {\sin (2x)}{(\sin x+\cos x)^2}\,dx$ Integrate
$$\int \frac {\sin (2x)}{(\sin x+\cos x)^2} \,dx$$
My Attempt:
$$=\int \frac {\sin (2x)}{(\sin x + \cos x)^2} \,dx$$
$$=\int \frac {2\sin x \cos x}{(\sin x+ \cos x)^2} \,dx$$
Dividing the numerator and denominator by $\cos^2 x$
$$=\int \frac {2\tan x}{(1+\tan x)^2} \,dx$$
| $$I=\int\frac{\sin(2x)}{(\sin(x)+\cos(x))^2}dx=\int\frac{2\sin(x)\cos(x)}{\sin^2(x)+2\sin(x)\cos(x)+\cos^2(x)}dx\tag{1}$$ $$=\int\frac{2\sin(x)\cos(x)+1-1}{2\sin(x)\cos(x)+1}dx=\int dx-\int\frac{dx}{\sin(2x)+1}=x-\int\frac{1-\sin(2x)}{\cos^2(2x)}dx\tag{2}$$ $$=x-\int\sec^2(2x)dx+\int\frac{\sin(2x)}{\cos^2(2x)}dx\tag{3}$$
Enforce the substitution $u=\cos(2x)$ on the second integral so that $du=-2\sin(2x)dx$.
$(1):\text{Recall}\space\sin(2x)=2\sin(x)\cos(x)\space\text{and}\space(a+b)^2=a^2+2ab+b^2$
$(2):$ $\frac{\sin(2x)}{\sin(2x)+1}=\frac{\left(\sin(2x)+1\right)-1}{\sin(2x)+1}=\frac{\sin(2x)+1}{\sin(2x)+1}-\frac{1}{\sin(2x)+1}=1-\frac{1}{1+\sin(2x)}\cdot\frac{1-\sin(2x)}{1-\sin(2x)}$
$(3):$ For $\int\sec^2(2x)dx$, let $t=2x\implies dx=\frac{dt}{2}\implies\int \sec^2(2x)dx=\frac{1}{2}\int\sec^2(t)dt$
Then $$I=x-\int \sec^2(2x)dx-\frac{1}{2}\int\frac{du}{u^2}=\boxed{x-\frac{1}{2}\tan(2x)+\frac{1}{2}\sec(2x)+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
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Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Question:
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.
Following from the question, I tried:
Let $N = \sqrt{3} + \sqrt{7} + \sqrt{21}$. Then,
$$
\begin{align}
N+1 &= 1+\sqrt{3} + \sqrt{7} + \sqrt{21}\\
&= 1+\sqrt{3} + \sqrt{7} + \sqrt{3}\sqrt{7}\\
&= (1+\sqrt{3})(1+\sqrt{7}).
\end{align}
$$
Using the above stated fact, $\sqrt{3}$ and $\sqrt{7}$ are irrational. Also, sum of a rational and irrational number is always irrational, so $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational. Similarly, if we prove that $N+1$ is irrational, $N$ will also be proved to be irrational.
But, how do I prove that product of $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational.
| Suppose $(1+\sqrt3)(1+\sqrt7)=p/q$ for some $p,q\in\Bbb Z^+$. Then we have that $$q(1+\sqrt3)=\frac p{1+\sqrt7}=\frac{p(1-\sqrt7)}{-6}\implies p\sqrt7-6q\sqrt3=p+6q\ne0\tag1$$ This implies that $$p\sqrt7+6q\sqrt3=\frac{(p\sqrt7+6q\sqrt3)(p\sqrt7-6q\sqrt3)}{p\sqrt7-6q\sqrt3}=\frac{7p^2-108q^2}{p+6q}\tag2$$ Adding $(1)$ and $(2)$ together gives $$2p\sqrt7=p+6q+\frac{7p^2-108q^2}{p+6q}\implies\sqrt7\in\Bbb Q$$ which is a contradiction. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 0
} |
Finding $\lim_{n\to\infty}\sqrt n\int_1^n \frac{\arctan (x/\sqrt n)}{x^2+n}\,dx$ Let $I_n = \int_{1}^{n} \frac{\arctan \frac{x}{\sqrt n}}{x^{2}+n}\,dx$, for $n \geq 1$. How can I show that $\lim_{n \to \infty} \sqrt nI_n=\frac{\pi}{8}$ ?
I've tried to first solve the integral but I'm really stuck, I couldn't find an ideea to start from, any help or suggestions would be truly appreciated!
| Let $\tan^{-1}\frac{x}{\sqrt n} = t$
$\frac{dx}{ 1+ \frac{x}{\sqrt n^2}} \frac{1}{\sqrt n} = dt$
$\frac{dx}{x^2 + n} = \frac{1}{\sqrt n}dt$
So, $I_n = \int^n_{x=1} \ \frac{1}{\sqrt n} tdt = \frac{1}{\sqrt n}\big[\frac{t^2}{2}\big]^n_{x=1}$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[\tan^{-1}(\frac{x}{\sqrt n}))^2\big]^n_{x=1}$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[\tan^{-1}\sqrt n)^2 - (\tan^{-1}\frac{1}{\sqrt n})^2 \big]$
For $x>0$, $\tan^{-1}\frac{1}{x} = \cot^{-1}(x)$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[(\tan^{-1}\sqrt n)^2 - (cot^{-1}\sqrt n)^2 \big]$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[(\tan^{-1}\sqrt n)^2 - (\frac{\pi}{2}-\tan^{-1}\sqrt n)^2 \big]$
$I_n = \frac{1}{2}\frac{1}{\sqrt n}\big[ - \frac{\pi^2}{4}+2\frac{\pi}{2}\tan^{-1}\sqrt n \big]$
$\sqrt n I_n = \frac{1}{2}\big[ - \frac{\pi^2}{4}+\pi \tan^{-1}\sqrt n \big]$
$\lim_{n \to \infty}\sqrt n I_n = \frac{1}{2}\lim_{n \to \infty}\big[ - \frac{\pi^2}{4}+\pi \tan^{-1}\sqrt n \big]$
$\lim_{n \to \infty}\sqrt n I_n = \frac{1}{2}\big[ - \frac{\pi^2}{4}+\pi .\frac{\pi}{2} \big] = \frac{1}{2}[\frac{\pi^2}{4}] = \frac{\pi^2}{8}$
Thus, $$\lim_{n \to \infty}\sqrt n I_n = \frac{\pi^2}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3218641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do you solve $\frac{x-1}{\sqrt{x}+2}=\frac{5}{2}$? I solved it using the quadratic formula and it went like:
\begin{gather}
\frac{x-1}{\sqrt{x}+2}=\frac{5}{2} \\
2(x-1)=5(\sqrt{x}+2) \\
2x-2=5\sqrt{x}+10 \\
2x-12=5\sqrt{x} \\
4x^2+144-48x=25x \tag*{(squaring both sides)}\\
4x^2-73x+144=0
\end{gather}
Then do the usual stuff and the solutions are $16$ and $9/4$. However $9/4$ doesn't work when you apply it to the initial equation. Is that solution correct, if not, where is the mistake?
| If you rewrite the equation as
$$
2(x-1)=5(\sqrt{x}+2)
$$
and then as
$$
2x-12=5\sqrt{x}
$$
you can immedately notice that a necessary condition for a solution is $x\ge6$, because the right-hand side is positive.
With this side condition, you can square: $4x^2-48x+144=25x$, which indeed has the roots $16$ and $9/4$, but the latter fails to satisfy the condition.
Indeed, if you substitute it in the left-hand side you get $9/2-12=-15/2$, while the right-hand side is $15/2$.
A different strategy could be to set $t=\sqrt{x}+2$, with the condition $t\ge2$, because $\sqrt{x}\ge0$. Then we have $x=t^2-4t+4$ and the equation becomes
$$
t^2-4t+3=\frac{5}{2}t
$$
or
$$
2t^2-13t+6=0
$$
that has roots $6$ and $1/2$. Only the first one is good, whence $\sqrt{x}=4$ and $x=16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3218914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Change of variables for double integral Problem: Using the change of variables $$x=\sqrt2u-\sqrt\frac{2}{3}v,y=\sqrt2u+\sqrt\frac{2}{3}v$$
Calculate the double integral $$\iint_Rx^4-2x^3y+3x^2y^2-2xy^3+y^4dA$$ where $R$ is the region bound by $x^2-xy+y^2=2$.
My work so far: The Jacobian is fairly trivial: $\frac{4}{\sqrt3}$
The region $R$ becomes $$(\sqrt2u-\sqrt\frac{2}{3}v)^2-(\sqrt2u-\sqrt\frac{2}{3}v)(\sqrt2u+\sqrt\frac{2}{3}v)+(\sqrt2u+\sqrt\frac{2}{3}v)^2=2$$$$\rightarrow2u^2+2v^2=2$$$$\rightarrow u^2+v^2=1$$
which is the unit disk. And after a lengthy calculation, I believe $$x^4-2x^3y+3x^2y^2-2xy^3+y^4$$$$\rightarrow \frac{8}{3}(-3u^4+6u^2v^2+v^2)$$
So we have
$$\iint_S\frac{8}{3}(-3u^4+6u^2v^2+v^2)\frac{4}{\sqrt3} \ du \ dv$$
where $S$ is the unit disk.
Now what to do? I tried proceeding using polar coordinates, but didn't find it an easy integral to compute. Am I just very bad at polar coordinates, or is there another way? Or have I miscalculated somewhere?
| We can save a lot of pain and get the correct answer by completing the square:
\begin{align}
x^4-2x^3 y+3x^2y^2-2xy^3+y^4
&= (x-y)^4 + 2x^3 y -3x^2 y^2 +2x y^3
\end{align}
Pull a common factor $xy$ out of the leftover terms and complete the square again:
\begin{align}
&= (x-y)^4 + xy(2x^2 -3x y +2 y^2) \\
&= (x-y)^4 + xy(\sqrt{2}x-\sqrt{2}y)^2 + (xy)^2\\
&= (x-y)^4 + 2 xy(x-y)^2 + (xy)^2
\end{align}
This is a sum of squares:
\begin{align}
&= ((x-y)^2 + xy)^2
\end{align}
Substituting in terms of $u$ and $v$ is now easy:
\begin{align}
&= ((-2\sqrt{\frac{2}{3}}v)^2 + 2u^2 -\frac{2}{3}v^2)^2\\
&= 4(v^2 + u^2)^2
\end{align}
and now polar coordinates is easy to do.
Moral of the story: always look for repeated opportunities to complete the square when we have funny polynomials like the one you linked (ie. looks almost exactly like the polynomials appearing in the binomial theorem). Or maybe just in any case in multivariable calculus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding A solution to a matrix equation I have the following problem in a past exam that involves matrix equations:
Suppose that $X$is a $2 \times2$ matrix satisfying:
$X^{2} = 6X +I$
a) Find the values of $\alpha$ and $\beta$ such that $X^{3} = \alpha X + \beta I$
b) Hence find a solution of the matrix equation
$X^{3} = \begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} X + \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix}$
I need a hint for part (b) as I am unable to use the result for part (a).
For part a the values for $\alpha $ and $\beta$ are 37 and 6. In particular, subbing in the values I have found in part (a), gives me the relation
$X^{3} = 37X + 6I$, however I fail to see how I can apply this to part b, as there is a matrix instead of a real number
| Suppose $X=\begin{bmatrix} a & b \\ c &d \end{bmatrix}$.
It is given that $X^{3} = \begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} X + \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix}$.
Therefore,$\begin{bmatrix} 36 &-1\\ 2&35\end{bmatrix} \times \begin{bmatrix} a & b \\ c &d \end{bmatrix}+ \begin{bmatrix} 9 &8\\ 2&10\end{bmatrix} = \begin{bmatrix} 36a-c+9 & 36b-d+8 \\ 2a+35c+2 & 2b+35d+10 \end{bmatrix}$.
Since, $X^3=37X+6I$, we have $X^3= \begin{bmatrix} 37a+6 &37b\\ 37c&37d+6\end{bmatrix}$
Comparing both $X^3$'s. [4 variables and 4 distinct eqations]
We get, $37a+6-36a+c-9=0\Rightarrow \boxed{c+a=3}$,
$37b-36b+d-8=0\Rightarrow \boxed{d+b=8}$,
$37c-2a-35c-2=0\Rightarrow \boxed{c-a=1}$, and
$37d+6-2b-35d-10=0\Rightarrow \boxed{d-b=2}$.
On solving these simultaneous equations we get $\bf{\boxed{a=1,b=3,c=2,d=5}}$.
Therefore, our matrix $X$ is $\begin{bmatrix} 1 & 3 \\ 2&5\end{bmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\int_0^{\frac{\pi}{2}}(\cos t)^{\frac{5}{2}}{\rm d}t$? Let $\sqrt{\cos t}=:u$，then $t=\arccos u^2$，${\rm d}t=-\dfrac{2u}{\sqrt{1-u^4}}{\rm d}u$. Therefore
\begin{align*}
\int_0^{\frac{\pi}{2}}(\cos t)^{\frac{5}{2}}{\rm d}t&=2\int_0^{1}\frac{u^6}{\sqrt{1-u^4}}{\rm d}u,
\end{align*}
which seems to be the elliptic integral...
Edit
\begin{align*}
\int_0^{\frac{\pi}{2}}(\cos t)^{\frac{5}{2}}{\rm d}t&=\int_0^{\frac{\pi}{2}}\sin^0 t \cos^{\frac{5}{2}}t{\rm d}t\\
&=\frac{1}{2}B\left(\frac{1}{2},\frac{7}{4}\right)\\
&=\frac{1}{2}\cdot\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{7}{4}\right)}{\Gamma\left(\frac{9}{4}\right)}\\
&=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma\left(\frac{7}{4}\right)}{\Gamma\left(\frac{9}{4}\right)}.\\
\end{align*}
| Solution
Consider integrating by part. One can obtain
\begin{align*}
\int_0^{\frac{\pi}{2}}\cos ^{\frac{5}{2}}t{\rm d}t&=\int_0^{\frac{\pi}{2}}\cos^{\frac{3}{2}}t{\rm d}(\sin t)\\
&=\left[\cos^{\frac{3}{2}}t\sin t\right]_0^{\frac{\pi}{2}}+\frac{3}{2}\int_0^{\frac{\pi}{2}}\cos^{\frac{1}{2}}t\sin^2 t{\rm d}t\\
&=0+\frac{3}{2}\int_0^{\frac{\pi}{2}}\cos^{\frac{1}{2}}t(1-\cos^2 t){\rm d}t\\
&=\frac{3}{2}\int_0^{\frac{\pi}{2}}\cos ^{\frac{1}{2}}t{\rm d}t-\frac{3}{2}\int_0^{\frac{\pi}{2}}\cos ^{\frac{5}{2}}t{\rm d}t.
\end{align*}
Thus
$$\int_0^{\frac{\pi}{2}}\cos ^{\frac{5}{2}}t{\rm d}t=\frac{3}{5}\int_0^{\frac{\pi}{2}}\cos ^{\frac{1}{2}}t{\rm d}t=\frac{3}{5}\int_0^{\frac{\pi}{2}}\sin ^{\frac{1}{2}}t{\rm d}t.$$
Let $u:=\sin^2 t$. Then ${\rm d} t=\dfrac{{\rm d}u}{2\sin t\cos t}.$ Hence
\begin{align*}
\int_0^{\frac{\pi}{2}}\sin ^{\frac{1}{2}}t{\rm d}t&=\int_0^1\frac{u^{\frac{1}{4}}}{2u^{\frac{1}{2}}(1-u)^{\frac{1}{2}}}{\rm d}u\\
&=\frac{1}{2}\int_0^1 u^{-1/4}(1-u)^{-1/2}{\rm d}u\\
&=\frac{1}{2}{\rm B}\left(\frac{3}{4},\frac{1}{2}\right)\\
&=\frac{\Gamma(3/4)\Gamma(1/2)}{2\Gamma(5/4)}\\
&=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma(3/4)}{\Gamma(5/4)}.
\end{align*}
It follows that
$$\int_0^{\frac{\pi}{2}}\cos ^{\frac{5}{2}}t{\rm d}t=\frac{3\sqrt{\pi}}{10}\cdot\frac{\Gamma(3/4)}{\Gamma(5/2)}\approx 0.71888\cdots$$
| {
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"url": "https://math.stackexchange.com/questions/3222948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral Bounds of Part of an Ellipse
I'm looking through my notes, and I've completed this questions integral, getting $ -144 \int cos^4(\theta)$ and due to the fact that $x=4cos(\theta)$ I calculated that the lower bound is $cos^-1(\frac{2\sqrt{3}}{4})=\frac{\pi}{6}$. The lecture notes confirm that this is correct however when I try the upper bound of the integral I get two different answers regardless of how I use $x=4cos(\theta)$ or $y=6sin(\theta)$ (the two ways in which he curve is parametererised).
The upper bound that the lecture notes specify is $\frac{4\pi}{3}$
| You need to solve both equations at once.
$$
-2 = 4 \cos\theta\qquad\qquad
-3 \sqrt 3 = 6 \sin\theta
$$
If you try to use inverse functions on the first, you get an angle in the second quadrant: $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$. If you try to use inverse functions on the second, you get an angle in the fourth quadrant: $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = - \frac{\pi}{3}$. Neither of these is the angle you are looking for: the point is in the third quadrant.
However, using the symmetry properties of sine and cosine, we see that:
\begin{align*}
\cos\left(-\frac{2\pi}{3}\right) &= \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}\\
\sin\left(-\frac{2\pi}{3}\right) &= -\sin\left(\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2}
\end{align*}
So this angle satisfies both equations. But we need to start from the lower limit $\theta=\frac{\pi}{6}$ and go counterclockwise, increasing $\theta$. So we use $-\frac{2\pi}{3} + 2\pi = \frac{4\pi}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3223714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^4}$
Evaluate $\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$
The answer stated is $\displaystyle {1 \over 6}$.
What I've tried:
$$\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{\cos (\sin x) -1+1- \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{1- \cos x}{x^4} - \frac {1-\cos (\sin x)}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{2 \sin^2(\frac {x}{2})}{x^4} - \frac {2 \sin^2(\frac {\sin x}{2})}{x^4}$$
$$=\displaystyle \lim_{x\to0} \left(\frac{\sin(\frac {x}{2})}{x} \right)^2. \left( \dfrac{1}{2x^2} \right) - \frac {2 \sin^2(\frac {\sin x}{2})}{x^4}$$
I'm not sure how I can evaluate the limit by proceeding this way. All help will be appreciated.
P.S. I'd prefer not using L'Hôpital's rule, it can get really messy.
EDIT: I should have mentioned that I would prefer if the solution does not use taylor series approximations (or any approximations) for that matter.
| For small $x$, $\frac{\cos(\sin x)-\cos x}{\sin x-x}\sim\cos^\prime x\sim -x$, while $\frac{\sin x-x}{x^4}\sim-\frac{1}{6x}$, so the limit is $\frac16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the coefficient of $x^5$ in $(1+x+x^2+x^3+x^4+x^5)^{17}$? I figured that $(1+x+x^2+x^3+x^4)^{17} = (1-x^6)^{17}*(1-x)^{-17}$ but don't know what else to do.
I would really appreciate any help
| You can pick up an $x^5$ from one term and a $1$ from every other term 17 different ways.
You can pick up an $x^4$ from one term, an $x$ from a second term in $\dfrac{17!}{15!}$ ways.
You can get $x^3$ and $x^2$ in $\dfrac{17!}{15!}$ ways.
You can get $x^3$, $x$, $x$ in $\dfrac{17!}{1!2!14!}$ ways.
You can get $x^2, x^2, x$ in $\dfrac{17!}{1!2!14!}$ ways.
You can get $x^2,x,x,x$ in $\dfrac{17!}{1!3!13!}$ ways.
And finally, you can get $x,x,x,x,x$ in $\dbinom{17}{5}$ ways.
Show that this is exhaustive, and it should give you the correct answer.
Final tally:
$$17+2\cdot \dfrac{17!}{15!} + 2\cdot \dfrac{17!}{2!14!} + \dfrac{17!}{3!13!} + \dfrac{17!}{5!12!} = 20,349$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$
$$\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$$
I have to resolve this limit, I tried factoring out x, I tried rewriting $x^3+x$ as
$x^3(1+ \frac{x}{x^3})$ and it doesn't seem to cancel. What should I do?
| \begin{align}
\lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3+x}}{x}
=&
\lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3\cdot(1+1/x^2)}}{x}
\\
=&
\lim_{x\to -\infty} \frac{\sqrt[\,3]{x^3}\cdot\sqrt[\,3]{(1+1/x^2)}}{x}
\\
=&
\lim_{x\to -\infty} \frac{x\cdot\sqrt[\,3]{(1+1/x^2)}}{x}
\\
=&
\lim_{x\to -\infty} {\sqrt[\,3]{(1+1/x^2)}}
\\
=&
{\sqrt[\,3]{\lim_{x\to -\infty} (1+1/x^2)}}
\\
=&
{\sqrt[\,3]{(1+\lim_{x\to -\infty} (1/x^2))}}
\\
=&
{\sqrt[\,3]{(1+0)}}
\\
=&
{\sqrt[\,3]{1}}
\\
=&
1
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve ODE $y'' = y^3 -y y'$ I want to solve $y'' = y^3 -y y'$ with boundary conditions $y(1) = 1/2$ and $y(2) = 1/3$ but not sure how to start. If someone could give me a complete step by step explanation, it would be greatly appreciated because I want to fully understand it.
Edit:
By the hint $(y'+\frac{y^2}{2})'=y^3$
$y' + \frac{y^2}{2} = \frac{y^4}{4} + c1$
$\frac{dy}{dx} = \frac{y^4}{4} - \frac{y^2}{2} + c1$
Integrate $\frac{dy}{\frac{y^4}{4} - \frac{y^2}{2} + c1} = dx$
Integration looks very messy.
I know that the answer is $y = \frac{1}{(x+1)}$
| The integration is not messy. After the hint, you have
$$dx=\frac {dy}{\frac {y^4}4 -\frac {y^2} 2+c_1}=4\frac {dy}{y^4 -2y^2+c_1}=4 \frac {dy}{(y^2-a)(y^2-b)}$$ Now, use partial fraction decomposition
$$\frac {1}{(y^2-a)(y^2-b)}=\frac 1{a-b}\left(\frac 1{y^2-a}-\frac 1{y^2-b} \right)$$ Just continue and, when done, apply the conditions to get $a$ and $b$ which define $c_1$ and the next constant of integration.
| {
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Prove that $2\tan^{-1}\frac{\sqrt{a^{2}- 2}}{a}= \tan^{-\,1}a\sqrt{a^{2}- 2} $ Prove
$$2\tan^{-1}\dfrac{\sqrt{a^{2}- 2}}{a}= \tan^{-\,1}a\sqrt{a^{2}- 2} \tag{a p u}$$
I find
$$a\sqrt{a^{2}- 2}= \dfrac{\dfrac{\sqrt{a^{2}- 2}}{a}+ \dfrac{\sqrt{a^{2}- 2}}{a}}{1- \dfrac{\sqrt{a^{2}- 2}}{a}\dfrac{\sqrt{a^{2}- 2}}{a}}$$
This may help! How I can use it to solve my problem? Thanks for all the interests!
| Using
$$\tan^{-1}(u) - \tan^{-1}(v) = \tan^{-1} \frac{u - v}{1 + uv}$$
where $u=\frac{\sqrt{a^2-2}}{a}$ and $v = a\sqrt{a^2- 2}$
you get
$$\tan^{-1}\frac{\sqrt{a^2-2}}{a} - \tan^{-1} a\sqrt{a^2- 2} = \tan^{-1} \frac{\frac{\sqrt{a^2-2}}{a} - a\sqrt{a^2-2}}{1+a^2-2} = -\tan^{-1}\frac{\sqrt{a^2-2}}{a} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Sub-Sum of Roots of Unity Let $\alpha$ be an algebraic integer of a cyclotomic field, and let $\theta_1, \theta_2, ..., \theta_n$ be roots of unity such that
$$\sum_{i=1}^n \theta_i = 2\alpha.$$
Does there necessarily exists a sub-sum of the $\theta_i$ that equals $\alpha$?
| No. Here we use the cyclotomic polynomial $\Phi_{105}$.
$$
\begin{align}
\Phi_{105}(x)&= x^{48} + x^{47} + x^{46} - x^{43} - x^{42} - 2 x^{41} - x^{40} - x^{39} \\ &\qquad+ x^{36} + x^{35} + x^{34} + x^{33} + x^{32} + x^{31} - x^{28} - x^{26} \\
&\qquad- x^{24} - x^{22} - x^{20} + x^{17} + x^{16} + x^{15} + x^{14} + x^{13} \\ &\qquad+ x^{12} - x^9 - x^8 - 2 x^7 - x^6 - x^5 + x^2 + x + 1.
\end{align}
$$
Note that the coefficient of $x^{41}$ and $x^7$ are $-2$. Let $\zeta=e^{2\pi i /105}$. Then $\Phi_{105}(x)$ is the minimal polynomial for $\zeta$ over integers. This means that there are no polynomials with integer coefficients of degree smaller than $48$, that has $\zeta$ as a root.
Also, the polynomial has all nonzero coefficient other than $x^{41}$, $x^7$ are $\pm 1$. If $x$ is a root of unity, then so is $-x$.
By evaluating $\Phi_{105}$ at $\zeta$, we obtain a sum of $210$-th root of unity, that equals $2(\zeta^{41}+\zeta^7)$, which will be used as $2\alpha$ in your question.
If there is such sub-sum equals $\zeta^{41}+\zeta^7$ exists, then there should be another monic polynomial over integers of degree $\leq 48$, having $\zeta$ as a root. This is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233238",
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"source": "stackexchange",
"question_score": "1",
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Help calculate the limits Help calculate the limits:
1) $$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{\sqrt {k(n-k)}}$$
2) $$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{(n-k)\ln{n}} $$
3) $$\lim_{n\to \infty}{n}^p \sin(\pi(\sqrt 2 +1)^n) $$
In the directions to this number it is written: compare $$(\sqrt 2 +1)^n$$ with the whole part $$(\sqrt 2 +1)^n + (-\sqrt 2 +1)^n $$
But I just can not understand how this will help. I got that with even n $$(\sqrt 2 +1)^n + (-\sqrt 2 +1)^n > (\sqrt 2 +1)^n $$
Аnd with odd on the contrary.
| This suppose to be a comment but it getting kinda long so I placed it here in term of an answer instead, although it's not directly answering your question but it may helps... In regard to Rick's comment, I think this is what he meant:
Suppose I have something like:
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^n \dfrac{1}{\sqrt{k^2 + n^2}} $$
Then I can rewrite this as follow:
$$ \lim_{n \rightarrow \infty} \dfrac{1}{ \sqrt{1^2 + n^2}} + \dfrac{1}{ \sqrt{2^2 + n^2}} + \dfrac{1}{ \sqrt{3^2 + n^2}} + \cdots \dfrac{1}{ \sqrt{n^2 + n^2}} $$
but note that
$$ \lim_{n \rightarrow \infty} \dfrac{1}{ \sqrt{1^2 + n^2}} + \dfrac{1}{ \sqrt{2^2 + n^2}} + \cdots + \dfrac{1}{ \sqrt{n^2 + n^2}}
= \lim_{n \rightarrow \infty} \dfrac{1}{n} \bigg( \dfrac{1}{\sqrt{ \big(\dfrac{1}{n}\big)^2 + 1 }} + \dfrac{1}{\sqrt{ \big(\dfrac{2}{n}\big)^2 + 1 }} + \cdots + \dfrac{1}{\sqrt{ \big(\dfrac{n}{n}\big)^2 + 1 }} \bigg) $$
From here you can see that it starting to look like how you would define a Riemann sum for the function $ \dfrac{1}{\sqrt{x^2 + 1}} $ from $0$ to $1$ with subdivision point $\dfrac{1}{n}, \dfrac{2}{n}, \cdots $ So now you can evaluate
$$ \lim_{n \rightarrow \infty} \sum_{k=1}^n \dfrac{1}{\sqrt{k^2 + n^2}} = \int_0^1 \dfrac{1}{\sqrt{x^2 + 1}} dx = \log \big[ x + \sqrt(x^2 + 1) \big] \bigg|_0^1 = \log(1 + \sqrt{2} ) $$
For your problem, can you think of doing something similar? Hope this help...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Checking the Monotonic function
Let $g''(x)>0$, such that $x \in \mathbb{R} $ and $f(x)=2g(2x^3–3x^2)+g(6x^2–4x^3–3)$, $x \in \mathbb{R} $,then
(A)in interval $x \in (1,\infty) $; $f(x)$ is increasing function
(B)in interval $x \in (-\infty , \frac{-1}{2}) $; $f(x)$ is decreasing function
(C) in interval $x \in (0,\infty) $; $f(x)$ is monotonic function
(D)in interval $x \in (-\infty , 0) $; $f(x)$ is non-monotonic function
My approach is as follow
as $g''(x)>0$ therefore $g'(x)$ is an increasing function.
But while substituting I end up getting $f'(x)=-3g'(x)$
| I will just give a hint.
$g''(x)>0$ implies that $g'(x)$ is increasing, meaning $g'(a)>g'(b)$ whenever $b<a$.
Now
$$
\begin{split}
f'(x)&= 2g'(2x^3-3x^2)\cdot (6x^2-6x) + g'(6x^2-4x^3-3) \cdot (12x - 12x^2) \\
&= 12x(x^2-1)g'(2x^3-3x^2) + 12x(1-x^2)g'(6x^2-4x^3-3) \\
&= 12x(x^2-1)g'(2x^3-3x^2) - 12x(x^2-1)g'(6x^2-4x^3-3) \\
&= 12x(x^2-1)\big(g'(2x^3-3x^2) - g'(6x^2-4x^3-3)\big)
\end{split}
$$
Now $12x(x^2-1)>0$ if $x>1$. But what can you say about the other term, $g'(2x^3-3x^2) - g'(6x^2-4x^3-3)$? Maybe examine the functions on the inside....perhaps plotting them on the same axes will help you (given my first comment).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me please
| Set $y=x^2+6x+7,$ and note that the resulting system $$y^2+6y+7-x=0,\,\,\,x^2+6x+7-y=0$$ is symmetric. That is, performing the transformation $(x,y)\mapsto(y,x)$ leaves the system unchanged except for a permutation. Thus, for every solution $(x,y),$ it is always the case that $(y,x)$ is also a solution, where $x,y\in\mathrm R.$
It follows that this system can only have an even number of real solutions, which must be in pairs that are reflections of each other in the line $y=x,$ and they cannot both fall on this line because the given equation is a quartic with real coefficients, and as such cannot have an odd number of real roots.
But the parabolas defined by the system above are orthogonal, congruent parabolas. Thus, they can only intersect along the line $x=y.$ This contradicts the observation above that the points of intersection cannot fall on this line. Thus, there are no real solutions for the system, and therefore to the original quartic.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Contour integration problem with sin and cos so I'm revising contour integration for an upcoming complex analysis exam. I have been asked to integrate $$\int_0^{2\pi}\frac{\sin^2x}{a+b \cos x}dx$$
I thought the sensible thing to do here would be to substitute in $z=e^{ix}$ and take a contour integral around the unit circle, call this path $\ast$ so that my integral becomes $$\frac{1}{2i}Re(\int_\ast\frac{1-z^2}{az+bz^2}dz)$$
Then, letting $f(z)=\frac{1-z^2}{az+bz^2}$, I thought the function had simple poles at $z=0$ with residue $\frac{1}{a}$ and another simple pole at $z=\frac{-a}{b}$ with residue $\frac{a}{b^2}-\frac{1}{a}$ and thus I get the that $$Re(\int_\ast\frac{1-z^2}{az+bz^2}dz)=2i(2\pi i)\frac{a}{b^2}=-4\pi(\frac{a}{b^2})$$ which is not the answer given, that is: $=\frac{2\pi}{b^2}[a-\sqrt{a^2-b^2}]$,but I can't work out why.
Any help appreciated, thank you in advance.
| $$
\begin{align}
\int_0^{2\pi}\frac{\color{#C00}{\sin^2(x)}}{\color{#090}{a+b\cos(x)}}\,\color{#00F}{\mathrm{d}x}
&=\oint\frac{\color{#C00}{-\frac{z^4-2z^2+1}{4z^2}}}{\color{#090}{\frac{bz^2+2az+b}{2z}}}\color{#00F}{\frac{\mathrm{d}z}{iz}}\\
&=-\frac1{2i}\oint\frac{z^4-2z^2+1}{bz^2+2az+b}\,\frac{\mathrm{d}z}{z^2}
\end{align}
$$
The residue of $\frac{z^4-2z^2+1}{bz^2+2az+b}\frac1{z^2}$ at $z=\frac{-a\pm\sqrt{a^2-b^2}}b$ (simple poles) is $\pm\frac{2\sqrt{a^2-b^2}}{b^2}$ and the residue at $z=0$ (degree $2$ pole) is $-\frac{2a}{b^2}$. Assuming $a\gt0$, we get $2\pi i$ times the sum of the residues inside the unit circle, $\frac{-a+\sqrt{a^2-b^2}}b$ and $0$, to be
$$
\int_0^{2\pi}\frac{\sin^2(x)}{a+b\cos(x)}\,\mathrm{d}x=2\pi\frac{a-\sqrt{a^2-b^2}}{b^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $n^{3/2} \le 2^n/n$ for all $n\geq 8$ via Induction? How to prove $n^{3/2} \le 2^n/n$ for all $n \geq 8$
My attempt:
$\textbf{Base case}$ is $n_0=8$: $\quad \left(8^{3/2}=4\right)\leq \left(32=\frac{2^8}{8}\right) \checkmark$
$\textbf{Induction hypothesis}$: $\exists n\in \mathbb{N}_{\geq 8}:(n+1)^{3/2}\leq \frac{2^{n+1}}{n+1}$
$\textbf{Inductive step}$:
$$
\begin{gather}
(n+1)^{3/2}\leq \frac{2^{n+1}}{n+1} \end{gather} \\
\iff (n+1)^{5/3}\leq 2^{n+1} \\
\iff (n+1)^{5/3}\leq 2^n \cdot 2 \quad |\uparrow ^{3/5} \\
\iff n+1\leq 2^{3n/5}\cdot 2^{3/5}\ $$
This, sadly, doesn't get me anywhere, any ideas?
| Consider function $f(x)=4^x-x^5$. Now we have to prove that $f(x)>0$ for $x\geq8$ using induction.
Let start with $f(8)=4^8-8^5=2^{16}-2^{15}\gt0$
Now, for any $x=k\geq8$ let us assume $f(k)=4^k-k^5\geq0$
Now put $x=k+1$, then $f(k+1)=4^{k+1}-(k+1)^5=4\times4^k-(k+1)^5$
$f(k+1)=[4^k-k^5]+[4^k-5k^4]+[4^k-10k^3]+[4^k-10k^2-5k^-1]$
$f(k+1)=[4^k-k^5]+[4^k-\frac{5}{k}k^5]+[4^k-\frac{10}{k^2}k^5]+[4^k-(\frac{10}{k^3}+\frac{5}{k^4}+\frac{1}{k^5})k^5]$
$f(k+1)=t_1+t_2+t_3+t_4$ (Assume all one square bracket as a single term)
since we are assuming $t_1\geq0$ and in other terms, the quotient of $k^5$ in $t_2,t_3,t_4$ is $\lt1$ which makes positive part more dominant.
Hence, all the terms are positive giving $f(k+1)\geq0$
Now, by theorem of Induction, $f(k)\geq0$ for $x\geq8$.
Hence Proved!
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find this kind of relationship: $x-\frac{1}{x}=A$ and $x+\frac{1}{x}=\sqrt{A^2+4}$? Could someone explain how to get from: $x-\frac{1}{x}=A$ to $x+\frac{1}{x}=\sqrt{A^2+4}$ ? It is one of the Algebra II tricks.
Thanks.
| Start by squaring both sides:
$$\begin{align}x-\frac{1}{x}&=A\\\left(x-\frac{1}{x}\right)^2&=A^2\\x^2-2+\frac{1}{x^2}&=A^2.\end{align}$$
Then try adding $4$ to both sides and "reversing" the processes above.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that, for an acute $\triangle ABC$, $\sin A + \sin B+\sin C\gt \cos A+\cos B+\cos C$
I need to prove or disprove that in any acute $\triangle ABC$, the following property holds:
$$\sin A + \sin B + \sin C \gt \cos A + \cos B + \cos C$$
To begin, I proved a lemma:
Lemma. An acute triangle has at most one angle which is less than or equal to $\dfrac{\pi}{4}$.
Proof:
Let there be an acute angled $\Delta ABC$ with the angles $A$ & $B \le \frac{\pi}{4}$. Then
$$ A + B \le \frac{\pi}{2}
\implies - (A + B) \ge -\frac{\pi}{2}
\implies C = \pi - (A+B) \ge \frac{\pi}{2}$$
thus contradicting that the triangle is obtuse. Hence, by contradiction, the lemma is proved. $\square$
Further, I used the identity that $\sin x - \cos x = \sqrt{2}\sin (x - \frac{\pi}{4})$ to rewrite the inequality as
$$\sin \biggr(A - \frac{\pi}{4}\biggr) + \sin \biggr(B - \frac{\pi}{4}\biggr) + \sin \biggr(C - \frac{\pi}{4}\biggr) \gt 0$$
Without loss of generality, I assumed that $A \le \frac{\pi}{4}$.
If $A = \dfrac{\pi}{4}$, then the inequality follows, since both $B$ and $C$ are strictly greater than $\dfrac{\pi}{4}$.
How do I prove the inequality if $A \lt \dfrac{\pi}{4}$?
Any help or hint will be appreciated.
| Hint: Use the formulas $$\cos(\alpha)+\cos(\beta)+\cos(\gamma)=\frac{r}{R}+1$$
$$2A=ab\sin(\gamma)=ac\sin(\beta)=ab\sin(\gamma)$$
$$A=sr=\frac{abc}{4R}$$
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
Doing this you will get
$$4A(a+b+c)>a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)$$
| {
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"url": "https://math.stackexchange.com/questions/3238915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate $\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}$ The question:\,
Calculate $$\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}.$$
Book's final solution: $\dfrac\pi 2$.
My mistaken solution: I don't see where is my mistake because my final solution is $-\dfrac{\pi i}2$:
$$\begin{align}
\text{A}:\int_{-\infty}^\infty {x^2\over (1+x^2)^2}dx &= 2\int_{0}^\infty {x^2\over (1+x^2)^2}dx \quad\text{as the function is even}\\&=
-2\left(\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,i\right)+\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,-i\right)\right)\end{align}
$$
Calculate the residues: Since $\pm i$ are poles of order $2$, then
$$\begin{align}\text{B}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,i\right)&={(2z\ln z+z^2\cdot z^{-1})(z+i)^2-z^2\ln z\cdot 2(z+i)\over (z+i)^4} \\&= \frac{(2i\ln i+i)(-4)+\ln i\cdot 4i }{16} \\ &= {1\over 16} (-8i\ln i -4i+4i\ln i)={-1\over 4}i(\ln i+1)\end{align}
$$
whereas
$$\begin{align}\text{C}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,-i\right)&={(2z\ln z+z^2\cdot z^{-1})(z-i)^2-z^2\ln z\cdot 2(z-i)\over (z-i)^4} \\ &=\frac{(-2i\ln(- i)-i)(-4)+\ln (-i)\cdot (-4i)}{16} \\ &= {1\over 16} (8i\ln (-i) +4i-4i\ln(- i))={1\over 4}i(\ln (-i)+1)\end{align}
$$
Combining $\text A$, $\text B$ and $\text C$, we get
$$
\int_{-\infty}^\infty {x^2\over (1+x^2)^2}\,dx=-2\left({i\over 4}(\ln(-i)+1)-{i\over 4}(\ln i+1)\right)={-i\over 2}\left({3\pi\over 2}-{\pi\over 2}\right)={-\pi i \over 2}
$$
Where is my mistake? Thanks in advance!
| Anyway, this integral doesn't require complex analysis: it is well-known that the antiderivatives of $\:\dfrac 1{(1+x^2)^n}\:$ can be calculated recursively. In this case, it is particularly simple:
Use integration parts to calculate the integral of $\frac1{1+x^2}$, setting
\begin{gather}
u=\frac1{1+x^2},\quad \mathrm d v=\mathrm dx,\quad\text{whence }\quad\mathrm du=\frac{-2x\,\mathrm dx}{(1+x^2)^2},\quad v=x \\[1ex]
\text{so }\qquad \frac\pi 2=\int_{0}^\infty {\mathrm dx \over 1+x^2}=\frac x{1+x^2}\biggm|_0^\infty +2 \int_{0}^\infty {x^2\over (1+x^2)^2}\,\mathrm dx =2 \int_{0}^\infty {x^2\over (1+x^2)^2}\,\mathrm dx.
\end{gather}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Approximate solution: factorial and exponentials If z= $\dbinom{200}{100}/(4^{100})$, what is the value of z?
The options are:
a. $z<1/3$
b. $1/3<z<1/2$
c. $1/2<z<2/3$
d. $2/3<z<1$
How should I go about solving these type of problems?
| Here is an elementary approach without estimating the value of the expression:
\begin{eqnarray*} \frac{1}{4^m}\binom{2m}{m}
& = & \frac{1}{4^m}\cdot \frac{\prod_{i=1}^m 2i \cdot \prod_{i=1}^m (2i-1) }{(m!)^2} \\
& = & \frac{1}{4^m}\cdot 4^m\frac{\prod_{i=1}^m i \cdot \prod_{i=1}^m \left(i-\frac{1}{2}\right) }{\prod_{i=1}^m i \cdot \prod_{i=1}^m i} \\
& = & \prod_{i=1}^m \left( 1 - \frac{1}{2i} \right) \\
& = & \prod^m_{i=1} \frac{2i-1}{2i} \\
& \stackrel{m=100}{<} & \frac{1}{2}\cdot\frac{3}{4}\cdot \frac{5}{6} \\
& = & \frac{5}{16}\\
& < & \frac{1}{3}\\
\end{eqnarray*}
| {
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"url": "https://math.stackexchange.com/questions/3239926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Compute without calculator, $ \frac{1}{\cos^{2}(10)} + \frac{1}{\sin^{2}(20)} + \frac{1}{\sin^{2}(40)} - \frac{1}{\cos^{2}(45)} $ Compute without calculator,
$$ \frac{1}{\cos^{2}(10^{\circ})} + \frac{1}{\sin^{2}(20^{\circ})} + \frac{1}{\sin^{2}(40^{\circ})} - \frac{1}{\cos^{2}(45^{\circ})} $$
Attempt:
Let $A = \cos(10) \sin(20) \sin(40)$, then
$$ \frac{1}{\cos^{2}(10)} + \frac{1}{\sin^{2}(20)} + \frac{1}{\sin^{2}(40)} = \frac{\sin^{2}(20) \sin^{2}(40) + \cos^{2}(10) \sin^{2}(40) + \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
notice also
$$2A \sin(10) \sin(40) = \sin^{2}(20) \sin^{2}(40) $$
$$2A \cos(20) \cos(10) = \cos^{2}(10) \sin^{2}(40) $$
so we have
$$\frac{2A \sin(10) \sin(40) + 2A \cos(20) \cos(10)+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
$$ = \frac{2A \left[2 \sin(10) \sin(20) \cos(20) + \frac{\sqrt{3}}{2} + \sin(10)\sin(20) \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
$$ = \frac{2A \left[\sin(10) \sin(20) (1+ \cos(20) + \cos(20)) + \frac{\sqrt{3}}{2} \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
$$ = \frac{2A \left[\sin(10) \sin(20) (3\cos^{2}(10) - \sin^{2}(10)) + \frac{\sqrt{3}}{2} \right]+ \cos^{2}(10) \sin^{2}(20)}{A^{2}} $$
How to continue then?
| \begin{align*}
&\;\frac{1}{\cos^{2}10^{\circ}} + \frac{1}{\sin^{2}20^{\circ}} + \frac{1}{\sin^{2}40^{\circ}} - \frac{1}{\cos^{2}45^{\circ}} \\
=&\;\frac {16\sin^210^\circ\cos^220^\circ}{16\sin^210^\circ\cos^210^\circ\cos^220^\circ}+\frac{4\cos^220^\circ}{4\sin^220^\circ\cos^220^\circ}+ \frac{1}{\sin^{2}40^{\circ}} -2\\
=&\;\frac{4(1-\cos20^\circ)(1+\cos40^\circ)+2(1+\cos40^\circ)+1}{\sin^240^\circ}-2\\
=&\;\frac{7-4\cos20^\circ+6\cos40^\circ-4\cos20^\circ\cos40^\circ}{\sin^240^\circ}-2\\
=&\;\frac{7-4\cos20^\circ+6\cos40^\circ-2\cos60^\circ-2\cos20^\circ}{\sin^240^\circ}-2\\
=&\;\frac{6-6\cos20^\circ+6\cos40^\circ}{\sin^240^\circ}-2\\
=&\;\frac{6-12\sin30^\circ\sin 10^\circ}{\sin^240^\circ}-2\\
=&\;\frac{6-6\cos 80^\circ}{\frac12(1-\cos80^\circ)}-2\\
=&\;10
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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A problem Dealing with Sampling with replacement Below is a problem from the Schaum book: "Probability and Statistics". I started
the problem but I am confident that I am on the wrong approach. I am hoping somebody will tell me where I went wrong.
Problem:
An urn holds $60$ red marbles and $40$ white marbles. Two sets of $30$ marbles are drawn with replacement
from the urn, and their color is noted. What is the probability that the two sets differ by $8$ or more red marbles.
Answer:
First we note that both sets should have a mean of $0.4*30 = 7.5$ white marbles. Let $X$ be a random variable
whose value is $1$ if on a single draw the marble is red and $0$ if the marble is white.
\begin{align*}
u_x &= \frac{3}{5} \\
\sigma_x^2 &= E(X^2) - u_x^2 \\
E(X^2) &= \frac{3}{5} \\
\sigma_x^2 &= \frac{3}{5} - \left( \frac{3}{5} \right)^2 = \frac{15}{25} - \frac{9}{25} \\
\sigma_x^2 &= \frac{6}{25} \\
\sigma_s^2 &= \frac{\sigma_x^2}{n} \\
n &= 30 \\
\sigma_s^2 &= \frac{\frac{6}{25}}{30} = \frac{3}{25(15)} \\
\sigma_s^2 &= \frac{1}{125} \\
\sigma_s &= \frac{1}{5\sqrt{5}} \\
\end{align*}
The book's answer to the problem is $0.0482$ and I am confident that my work is wrong. Please tell me where I went wrong.
Thanks,
Bob
Based upon the response I received, I have been able to continue my work. However, I am stuck. Here is my updated work:
\begin{align*}
u_x &= \frac{3}{5} \\
\sigma_x^2 &= E(X^2) - u_x^2 \\
E(X^2) &= \frac{3}{5} \\
\sigma_x^2 &= \frac{3}{5} - \left( \frac{3}{5} \right)^2 = \frac{15}{25} - \frac{9}{25} \\
\sigma_x^2 &= \frac{6}{25} \\
\sigma_s^2 &= n \sigma_x^2 \\
n &= 30 \\
\sigma_s^2 &= 30 \left( \frac{6}{25} \right) = \frac{6(6)}{5} \\
\sigma_s^2 &= \frac{36}{5} \\
\sigma_s &= \frac{6}{\sqrt{5}} \\
\end{align*}
Let $SD$ be the number of standard deviations $8$ marbles is.
\begin{align*}
SD &= \frac{8}{ \frac{6}{\sqrt{5}} } = \frac{8\sqrt{5}}{6} \\
SD &= \frac{4 \sqrt{5}}{3} \doteq 2.981424 \\
\end{align*}
At this point, I am not sure what to do.
Bob
| The sample variance is $n$ times the variance of one marble. This makes $\sigma_s^2=\frac {36}{5}$. The variance of the difference of two samples is twice the variance of one, so $\frac{72}5$. The square root of this is about $3.79$, so a difference of $8$ is about $2$ standard deviations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Find minimum value of $P=17x^2+17y^2+16xy$ Let $x,y$ be positive real numbers such that $4x^2+4y^2+17xy+5x+5y\ge 1$. Find minimum value of $$P=17x^2+17y^2+16xy$$
My idea:
I see that $x=y=\frac{\sqrt 2 -1}{5}\rightarrow P=2(3-2\sqrt 2)$
So I proved $P\ge 2(3-2 \sqrt 2)(4x^2+4y^2+17xy+5x+5y)\ge2(3-2\sqrt 2)$
Is my approachment is true?
| We have
$$17\,x^{\,2}+ 17\,y^{\,2}+ 16\,xy\geqq 2(\,3- 2^{\,\frac{3}{2}}\,)$$
$$\because\,17\,x^{\,2}+ 17\,y^{\,2}+ 16\,xy- 2(\,3- 2^{\,\frac{3}{2}}\,)- (\,2- 2^{\,\frac{1}{2}}\,)(\,\underbrace{4\,x^{\,2}+ 4\,y^{\,2}+ 17\,xy+ 5\,x+ 5\,y- 1}_{\,\geqq 0}\,)\geqq 0$$
$$\because\,{\rm discriminant}[\,17\,x^{\,2}+ 17\,y^{\,2}+ 16\,xy- 2(\,3- 2^{\,\frac{3}{2}}\,)- (\,2- 2^{\,\frac{1}{2}}\,)(\,4\,x^{\,2}+ 4\,y^{\,2}+ 17\,xy+ 5\,x+ 5\,y- 1\,),\,x]= -\,\frac{126\,(\,1+ 2^{\,\frac{3}{2}}\,)(\,-\,5\,y+ 2^{\,\frac{1}{2}}- 1\,)^{\,2}}{9+ 2^{\,\frac{5}{2}}}\leqq 0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Gravity of a Circular Ring at a co-planar external point. I have a circular ring of unit mass and fixed radius R which lies in the XY plane at point $O$ with coordinates $O:(0,0)$.
I wish to find a formula for the gravitational force at a point $P: (D,0)$ which lies in the same plane as the ring and is at some variable distance D from the ring centre O.
(Note: There are many treatments of the case for a target lying on the axis of the ring. The reference so far found nearest to this co-planar case is Problems 5-12, 5-13, (no solutions given) p.127 in Classical Dynamics of Particles and Systems by Jerry B. Marion.
I expect that the formula should be of the form $F = GM*f(D)$ where $G$ is the gravitational constant, $M$ is the mass and $f$ is some function similar to the Newtonian spherical divergence function $f(D) = \frac{1}{ D^2}$ (where the factor $\frac{1}{4.\pi}$ is absorbed in the value of the constant $G$ ).
So far I have obtained an integral formula by initially modelling the ring as a series of $N$ small point masses of mass $\frac{1}{N}$ separated by angle $\delta\theta$, whose distance from target is $L$ where:
$$L^2 = (D-a)^2+b^2 = D^2-2aD+R^2 = D^2\left(1 -\frac{2a}{D} +\frac{R^2}{D^2}\right)$$
where $a (= R\cos\theta)$ and $b(=R\sin\theta)$ are the $x$ and $y$ coordinates of the point.
Due to symmetry and vector addition of forces there is no net force in the y-direction and so the effective force contribution (along $x$) for a point is given by multiplying by the cosine factor $(D-a)/L$ thus:-
$$ F = \frac{-GM}{N}\frac{1}{4\pi.L^2}\frac{D-a}{L}
= \frac{-GM}{ N} \frac{D-a}{L^3} $$
$$ F = \frac{-GM}{ N} \frac{D-R\cos\theta}{\left(D^2\left(1 -\frac{2a}{D} +\frac{R^2}{D^2}\right)\right)^{\frac{3}{2}}} $$
$$ F = \frac{-GM}{ N} \frac{D-R\cos\theta}{D^3 \left(1 -\frac{2a}{D} +\frac{R^2}{D^2} \right)^{\frac{3}{2}}} $$
$$ F = \frac{-GM}{ N} \frac{1-(R/D)\cos\theta}{D^2 \left(1 -\frac{2a}{D} +\frac{R^2}{D^2} \right)^{\frac{3}{2}}} $$
I then obtained the following integral formula for the force exerted on the target point by the ring:-
$$ F = \frac{-GM}{ D^2} \frac{1}{2\pi}\int_0^{2\pi}\frac{1-Q\cos\theta}{\left(1-2Q\cos\theta+Q^2\right)^{\frac{3}{2}}} \text{d}\theta$$
where $Q = R/D$.
$$ F = \frac{-GM}{ D^2} \frac{1}{2\pi} \frac{1}{(2Q)^{3/2}}\int_0^{2\pi}\frac{1-Q\cos\theta}
{\left(\frac{Q^2+ 1}{2Q} - \cos\theta\right)^{\frac{3}{2}}} \text{d}\theta$$
Defining $A = \frac{Q^2+ 1}{2Q}$, Wolfram Alpha gives...
$$ \int_0^{2\pi}\frac{ 1 - Q \cos x}{(A -\cos x)^{3/2}} dx $$
$$=\left[\frac{2}{(A^2-1)\sqrt{A - \cos x}}\left(A^2-1\right)Q\sqrt{\frac{A - \cos x}{A-1}} \operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)-AQ\sin x- (A-1)(AQ-1)\sqrt{\frac{A-\cos x}{A-1}}\operatorname{E}\left(\frac{x}{2}~\big|~\frac{2}{1-A}\right)
+\sin x\right]_0^{2\pi}$$
Where $E(x|m)$ is an elliptic integral of the 2nd kind with parameter $m=k^2$, and $F(x|m)$ is an elliptic integral of the 1st kind with parameter $m=k^2$.
Replacing $\cos x$ by $1$ and $\sin x$ by $0$...
$$=\frac{2}{(A^2-1)\sqrt{A -1}}*\left[(A^2-1)Q \operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)-(A-1)(AQ-1) \operatorname{E}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)\right]_0^{2\pi}$$
Cancelling $(A^2-1)$...
$$=\frac{2}{\sqrt{A -1}}\left[Q\operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)
- \frac{(AQ-1)}{A+1} \operatorname{E}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right) \right]_0^{2\pi}$$
Being unfamiliar with Elliptic Integrals, this is as far as I can comfortably go at present.
After reading the wikipedia article Elliptic Integral, proceeding tentatively, from the definitions of elliptic integrals I think that $E(x|k^2)$ and $F(x|k^2)$ both go to zero when $x$ is zero, thus...
$$=\frac{2Q}{\sqrt{A -1}}\left[\operatorname{F}\left(\pi~\big|~\frac{-2}{A-1}\right)-\frac{(AQ-1)}{AQ+Q} \operatorname{E}\left(\pi~\big|~\frac{-2}{A-1}\right)\right]$$
Next perhaps it would be helpful to reformulate the problem so that the amplitude(?) term in the elliptic integrals changes from $\pi$ to $\pi/2$, thereby making the elliptic integrals "complete" and permitting them to be expressed as power series. This reformulation could be done by modelling the gravitational effect ($Fx$ component only) of two half rings (positive $y$ and negative $y$), independently, and using respectively the angles $\theta_1$ and $\theta_2$ which both range from $0$ to $\pi/2$ but in different directions.
| In order to express the result in terms of complete elliptic integrals, it is easier to compute the gravitational potential $\phi(D)$ first. Then the (radial) field is given by $F(D) = - \phi'(D)$. Following your approach we find
$$ \phi(D) = - \frac{G M}{2 \pi D} \int \limits_0^{2\pi} \frac{\mathrm{d} \theta}{\sqrt{1 - 2 Q \cos(\theta) + Q^2}} = - \frac{G M}{\pi D} \int \limits_0^{\pi} \frac{\mathrm{d} \theta}{\sqrt{1 - 2 Q \cos(\theta) + Q^2}} \, . $$
In the last step we have used the fact that the integral from $0$ to $\pi$ and that from $\pi$ to $2\pi$ have the same value. Now we can write
$$ - \cos(\theta) = \cos(\pi - \theta) = 1 - 2 \sin^2\left(\frac{\pi - \theta}{2}\right) $$
and introduce the new integration variable $\alpha = \frac{\pi - \theta}{2}$ to obtain
$$ \phi(D) = -\frac{2 G M}{\pi D} \int \limits_0^{\pi/2} \frac{\mathrm{d} \alpha}{\sqrt{1 + 2 Q + Q^2 - 4 Q \sin^2(\alpha)}} = -\frac{2 G M}{\pi D} \frac{1}{1+Q} \int \limits_0^{\pi/2} \frac{\mathrm{d} \alpha}{\sqrt{1 - \frac{4 Q}{(1+Q)^2} \sin^2(\alpha)}} \, . $$
But this integral is just the definition of the complete elliptic integral of the first kind and (using the parameter $m = k^2$ as the argument)
$$ \phi(D) = - \frac{2 G M}{\pi D} \frac{1}{1+Q} \operatorname{K}\left(\frac{4 Q}{(1+Q)^2}\right) = - \frac{2 G M}{\pi D} \operatorname{K}(Q^2) = - \frac{2 G M}{\pi D} \operatorname{K}\left(\frac{R^2}{D^2}\right)$$
follows. The final simplification is an application of Gauss's transformation. Taking the derivative we find the field
$$ F(D) = - \frac{2 G M}{\pi(D^2 - R^2)} \operatorname{E}\left(\frac{R^2}{D^2}\right) $$
in terms of the complete elliptic integral of the second kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $ \int \frac{a^2\cos^2x+b^2\sin^2x}{a^4\cos^2x+b^4\sin^2x}\,dx$ Evaluate $$ \int \frac{a^2\cos^2x+b^2\sin^2x}{a^4\cos^2x+b^4\sin^2x}\,dx$$
I have tried Weierstrass substitution and tried to split into two integrations, but it gets really messy.
Is there a better way to approach this problem? I feel that complex numbers are the best way out, but I couldn't get anything using that as well.
| \begin{align}\int \frac{a^2\cos^2x+b^2\sin^2x}{a^4\cos^2x+b^4\sin^2x}\,dx
=& \int \frac{a^2\csc^2x+b^2\sec^2x}{a^4\csc^2x+b^4\sec^2x}\,dx\\
=& \int \frac{ d(b^2\tan x -a^2\cot x)}{(a^2+b^2)^2 +(b^2\tan x -a^2\cot x)^2} \\
=& \ \frac1{a^2+b^2}\tan^{-1}\frac{b^2\tan x -a^2\cot x}{a^2+b^2}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$. I need to show that $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$.
This happens if $(x+1)\ln\Big(1 + \frac 2 {2x+1}\Big) > 1$ so let's study the function $f(x) = (x+1)\ln\Big(1 + \frac 2 {2x+1}\Big)$
Edit
We have abandoned the previous method of only using a characterization of $e$ with inequalities since we thought it was impractical to do it. Let's return instead to analysis.
*
*$f(1) = 2 \ln(\frac 5 3) > 1$
*$f'(x) = \ln(1+ \frac 2 {2x+1}) - \frac {4(x+1)}{(4x^2+8x+3)} < 0$ on $[1,+\infty[$
*$\lim f(x) = 1$
However how do I show formally 2.?
| From $f(x) = (x+1)\log\Big(1 + \frac 2 {2x+1}\Big)$ we compute the derivative.
$$
\begin{eqnarray}
f'(x) = & (x+1)'\log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\log\Big(1 + \frac 2 {2x+1}\Big)' \\
= & (x+1)'\log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\log\Big(\frac {2x+3} {2x+1}\Big)' \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {2x+1} {2x+3} \Big( \frac{2(2x+1) - 2(2x+3)}{(2x+1)^2} \Big) \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {2x+1} {2x+3} \Big( \frac{-4}{(2x+1)^2} \Big) \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {1} {2x+3} \Big( \frac{-4}{(2x+1)} \Big) \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) - \frac {4(x+1)} {4x^2+8x+3}
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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A non-trivial solution for $\lambda_1 + 2 \lambda_2 + \cdots + (p-2)\lambda_{p-2} = 0$ in $\mathbb{Z}_p$ Let $K = \mathbb{Z}_p$, where $p$ is a prime number and $p \neq 2,3$ . Find a non-trivial solution for the system:
$$\left\{ \begin{align}
& \lambda_1 + \lambda_2 + \cdots + \lambda_{p-2} = 0 \ (\mod p) \\
& \lambda_1 + 2 \lambda_2 + \cdots + (p-2)\lambda_{p-2} = 0(\mod p) \\
\end{align} \right.,$$
where $\lambda_i \in K$, for all $i=1, \cdots, p-2$.
Comments: In the first part of the exercise I solved the system
$$\left\{ \begin{align}
& \lambda_1 + \lambda_2 + \cdots + \lambda_{p-2}+ \lambda_{p-1} = 0 (\mod p)\\ &\lambda_1 + 2 \lambda_2 + \cdots + (p-2)\lambda_{p-2} + (p-1)\lambda_{p-1}= 0
(\mod p) \\
\end{align} \right.,$$
using that
$$1+2+ \cdot + (p-1) = \frac{p(p-1)}{2}$$ and
$$1+2^2+3^2+ \cdots + (p-1)^2 = \frac{(p-1)p(2p-1)}{6}$$.
I can not see a logic look like the one in this part to solve the next one, for exemple, if $p = 5$, a solution is $\lambda_1 = 2$, $\lambda_2 = 1$ and $\lambda_3 = 2$, but I can not see a logic to the general case.
| If we have $$\lambda_2 + 2\lambda_3 + \cdots + (p-3) \lambda_{p-2} \equiv 0 \pmod{p} \tag{$\ast$}$$ then setting $\lambda_1 = -(\lambda_2 + \cdots+\lambda_{p-2})$ gives a solution $(\lambda_1, \cdots, \lambda_{p-2})$ to the described system of two equations. So, we just need to find a nontrivial solution to the single equation $(\ast)$. We can do this by choosing $\lambda_2, \cdots, \lambda_{p-3}$ arbitrarily and then taking $\lambda_{p-2}$ to be the unique residue forcing the equation to hold.
For example, choose $\lambda_2 = \cdots = \lambda_{p-3} = 1$. Then we require $$1+2+\cdots + (p-4) + (p-3) \lambda_{p-2} \equiv 0 \pmod{p}$$ $$\implies \frac{(p-4)(p-3)}{2} + (p-3) \lambda_{p-2} \equiv 0 \pmod{p}$$ $$ \implies \lambda_{p-2} = \frac{4-p}{2} \pmod{p}$$ where we have assumed $p \neq 3$ so we can divide by $p-3$. Then we can solve for $\lambda_1$ as well, yielding a solution to the original system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247614",
"timestamp": "2023-03-29T00:00:00",
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Integrating factor for $(x^2-y^2-y)dx-(x^2-y^2-x)dy=0$ I am having trouble finding the integrating factor for turning the below differential equation into an exact one (Tenenbaum and Pollard, exercise 10, problem 6). Any hints and suggestions would extremely helpful and lead me to the solution.
Solve the differential equation :
$$ (x^2-y^2-y)dx - (x^2-y^2-x)dy=0$$
My attempt:
The coefficients of $dx$ and $dy$ are not homogenous functions. Further,
$
\begin{align}
P(x,y) &= x^2 - y^2 - y \\
\frac{\partial P(x,y)}{\partial y}&=-2y-1 \\
Q(x,y) &= -(x^2-y^2-x)\\
\frac{\partial Q(x,y)}{\partial x}&=-(2x-1) \\
\therefore \frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}
\end{align}
$
The given differential equation is not exact. We have :
$\begin{align}
& \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \\
=& -2y-1+(2x-1)\\
=&(2x-2y-2)
\end{align}$.
Moreover,
$\begin{align}
& yQ-xP\\
= & -y(x^2-y^2-x)-x(x^2-y^2-y)\\
= & -x^2 y + y^3 + xy - x^3 + xy^2 + xy\\
= & y^3 - x^3 + xy - xy(x - y - 1)
\end{align}$
It doesn't look like $(\partial P / \partial y - \partial Q / \partial x)/(yQ-xP)$ will be a function of $u=xy$ alone.
Also,
$\begin{align}
& yQ+xP\\
= & -y(x^2-y^2-x)+x(x^2-y^2-y)\\
= & -x^2 y + y^3 + xy + x^3 - xy^2 - xy\\
= & y^3 + x^3 - x^2 y - x y^2 \\
= & (y + x)(y^2 + x^2 - xy) - xy(y + x) \\
= & (y + x)(y^2 + x^2 - 2xy)\\
= & (y + x)(y - x)^2
\end{align}$
It doesn't look like $y^2(\partial P / \partial y - \partial Q / \partial x)/(yQ+xP)$ will be a function of $u=x/y$ or $x^2(\partial P / \partial y - \partial Q / \partial x)/(yQ+xP)$ will be a function of $u=y/x$ alone.
| The part of $\dfrac{\partial P}{\partial y} - \dfrac{\partial Q}{\partial x} = 2(x-y-1)=2(z-1)$ shows that you may assume a new variable $z=x-y$ which which shows the denominator of integrating factor might be of the form
$$\dfrac{\partial I}{\partial x}Q - \dfrac{\partial I}{\partial y}P = P+Q=z$$
so
$$p(z)=\dfrac{2(z-1)}{z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Test for the convergence of the series $\sum _{_{n=1}}^{\infty }\frac{\frac{1}{2}+\left(-1\right)^n}{n}$ Test for the convergence of the series $\sum _{_{n=1}}^{\infty }\frac{\frac{1}{2}+\left(-1\right)^n}{n}$
My attempt
$\sum _{_{n=1}}^{\infty }\frac{\frac{1}{2}+\left(-1\right)^n}{n}=\frac{-\frac{1}{2}}{1}+\frac{\frac{3}{2}}{2}+\frac{-\frac{1}{2}}{3}+\frac{\frac{3}{2}}{4}+\frac{-\frac{1}{2}}{5}+\frac{\frac{3}{2}}{6}+...$. I tried to use Drichlet test, ratio test.
I applied limit comparison test with $a_n=\frac{1}{n}$. $\lim_{n\to \infty}\frac{\frac{\frac{1}{2}+\left(-1\right)^n}{n}}{\frac{1}{n}}=\lim_{ n \to \infty}\frac{1}{2}+\left(-1\right)^n\neq 0(\because$ sequence doesnot converges to anyone of the real number, I can prove it). So, series diverges.
| Concatenate terms in pairs, with the odd n first. $(\frac{-1}{2n}+\frac{3}{2(n+1)})=\frac{2n-1}{2n^2}\gt \frac{1}{2n}$ The sum then becomes $\gt \sum_{k=1}^\infty \frac{1}{2(1+2k)}$ which diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to prove if $3\nmid n$, then $n^2$ has a remainder $1$ when divided by $3$? Lets say:
$n^2=3k+1$ ; $n\neq3$.
I'm trying to prove this by induction, therefore:
$(n+1)^2=n^2+2n+1=3k+2n+2$
Any suggestion on how to move foward?
| If 3 does not divide $n$, that means that $n$ has remainder 1 or 2 by division with 3.
So either $n=3k+1$ or $n=3k+2$.
In both cases we have $n^2=(3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\mod 3$
or $n^2=(3k+2)^2=3(3k^2+4k)+4=3(3k^2+4k)+3\cdot 1+1\equiv 1\mod 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$ If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$
After we solve for $x$ (its a quadratic), and find that $x=1\pm i$, it's trivial to see the powers of $x$ in the complex plane, but the problem must be solved without using the complex numbers. With complex numbers i found that the result is $1$.
Is there any way to solve this in $\Bbb R$ without entering $\Bbb C$?
| It's enough to prove that
$$x^9-(x^4+x^2+1)(x^6+x^3+1)=1$$ or
$$(x^3-1-(x^4+x^2+1))(x^6+x^3+1)=0,$$ for which it's enough to prove that
$$x^4-x^3+x^2+2=0$$ or
$$x^4-2x^3+2x^2+x^3-2x^2+2x+x^2-2x+2=0$$ or
$$(x^2-2x+2)(x^2+x+1)=0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Calculate $\int_3^4 \sqrt {x^2-3x+2} \, dx$ using Euler's substitution
Calculate $\int_3^4 \sqrt {x^2-3x+2}\, dx$ using Euler's substitution
My try:
$$\sqrt {x^2-3x+2}=x+t$$
$$x=\frac{2-t^2}{2t+3}$$
$$\sqrt {x^2-3x+2}=\frac{2-t^2}{2t+3}+t=\frac{t^2+3t+2}{2t+3}$$
$$dx=\frac{-2(t^2+3t+2)}{(2t+3)^2} dt$$
$$\int_3^4 \sqrt {x^2-3x+2}\, dx=\int_{\sqrt {2} -3}^{\sqrt {2} -4} \frac{t^2+3t+2}{2t+3}\cdot \frac{-2(t^2+3t+2)}{(2t+3)^2}\, dt=2\int_{\sqrt {2} -4}^{\sqrt {2} -3}\frac{(t^2+3t+2)^2}{(2t+3)^3}\, dt$$ However I think that I can have a mistake because Euler's substition it should make my task easier, meanwhile it still seems quite complicated and I do not know what to do next.Can you help me?P.S. I must use Euler's substitution because that's the command.
| Using the third substitution of Euler
$$\sqrt{{{x}^{2}}-3x+2}=\sqrt{\left( x-1 \right)\left( x-2 \right)}=\left( x-1 \right)t$$
$$x=\frac{2-{{t}^{2}}}{1-{{t}^{2}}}\Rightarrow dx=\frac{2t}{{{\left( {{t}^{2}}-1 \right)}^{2}}}dt$$
$$\begin{align}
& =-\int_{1/\sqrt{2}}^{\sqrt{2/3}}{\frac{2{{t}^{2}}}{{{\left( {{t}^{2}}-1 \right)}^{3}}}dt} \\
& =\left[ \frac{\left( t+{{t}^{3}} \right)}{{{\left( {{t}^{2}}-1 \right)}^{2}}}+\ln \left( \frac{1-t}{1+t} \right) \right]_{1/\sqrt{2}}^{\sqrt{2/3}} \\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral of $\frac{1}{\sqrt{2-x^2}}$ I know that the integral of $\frac{1}{\sqrt{1-x^2}} = arcsin(x)$, but what is the the integral of $\frac{1}{\sqrt{2-x^2}}$? Is this how you do it?
$$\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2}(1-\frac{x^2}{\sqrt{2}})} = \frac{1}{\sqrt{2}} \arcsin\left(\frac x{\sqrt{2}}\right)$$
| This is a simple u-substitution problem. Use $u=\frac{x}{\sqrt{2}}$:
$$
\int\frac{1}{\sqrt{2-x^2}}\,dx=
\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{1-\left(\frac{x}{\sqrt{2}}\right)^2}}\,dx=\\
\frac{\sqrt{2}}{\sqrt{2}}\int\frac{1}{\sqrt{1-\left(\frac{x}{\sqrt{2}}\right)^2}}\frac{d}{dx}\left(\frac{x}{\sqrt{2}}\right)\,dx=
\int\frac{1}{\sqrt{1-u^2}}\,du=\\
\arcsin{u}+C=\arcsin{\left(\frac{x}{\sqrt{2}}\right)}+C.
$$
| {
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"url": "https://math.stackexchange.com/questions/3251320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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New bound for Am-Gm of 2 variables Today I'm interested by the following problem :
Let $x,y>0$ then we have :
$$x+y-\sqrt{xy}\leq\exp\Big(\frac{x\ln(x)+y\ln(y)}{x+y}\Big)$$
The equality case comes when $x=y$
My proof uses derivative because for $x\geq y $ the function :
$$f(x)=x+y-\sqrt{xy}-\exp\Big(\frac{x\ln(x)+y\ln(y)}{x+y}\Big)$$
is decreasing and for $y\geq x$ the function is increasing and the maximum occurs when $x=y$
My question is : Have you an alternative proof wich doesn't use derivative ?
Thanks in advance.
| Idea for a proof :
As the inequality is homogenous we obtain an equivalent inequality like :
$$\left(\frac{1}{x}\right)^{\frac{2}{x^2+1}}+\frac{1}{x}-\frac{1}{x^2}\geq1$$
Using the Bernoulli's first approximation and some others terms in the Newton's expansion of $(1+x)^a$ we have ($0< x\leq 1$).:
$$\left(\frac{1}{x}\right)^{\frac{2}{x^{2}+1}}+\frac{1}{x}-\frac{1}{x^{2}}\geq 1+\frac{1}{6}\cdot\left(\frac{2}{x^{2}+1}-2\right)\left(\frac{2}{x^{2}+1}-1\right)\left(\frac{2}{x^{2}+1}\right)\left(\frac{1}{x}-1\right)^{3}+\left(\frac{1}{x}-1\right)\left(\frac{2}{x^{2}+1}\right)+\frac{1}{x}-\frac{1}{x^{2}}+0.5\left(\frac{2}{x^{2}+1}-1\right)\left(\frac{2}{x^{2}+1}\right)\left(\frac{1}{x}-1\right)^{2}\geq 1$$
The RHS becomes (using Wolfram alpha for the simpplification) :
$$ \frac{1}{3}\frac{(3x^6-14x^5+27x^4-28x^3+17x^2-6x+1)}{(x(x^2+1)^3)}\geq 0$$
Or :
$$\frac{1}{3}\frac{(x-1)^4 (3x^2-2x+1)}{(x(x^2+1)^3)}\geq 0$$
Wich is obvious !
Done !
For a reference of proof without calculus of the binomial theorem see :
Reference :
https://www.jstor.org/stable/2319010?seq=1
Binomial theorem proof for rational index without calculus
| {
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Find $x_1^3+x_2^3+x_3^3$ for a degree 3 polynomial. I have the polynomial $P(x) = x^3+mx^2-3x+1, m\in \mathbb{R}$. I need to find $x_1^3+x_2^3+x_3^3$ as a $m$ function.
I tried to use Viette equations: $x_1+x_2+x_3 = m, x_1x_2+x_1x_3+x_2x_3 = 3, x_1x_2x_3 = 1$ Then I expanded $(x_1+x_2+x_3)^3 = (x_1^3+x_2^3+x_3^3) + 3(x_1^2x_2+x_1^2x_3+x_1x_2^2+x_2^2x_3+x_1x_3^2+x_2x_3^2) + 6x_1x_2x_3$. After that I expanded $(x_1x_2+x_1x_3+x_2x_3)^2 = 2(x_1x_2x_3)+x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2$. I don't know if my way of doing it is the right way, and if it is can you help from this point on?
| Since $x_1,x_2,x_3$ are the roots we find that $x_1^3=3x_1-mx_1^2-1$ similarly for $x_2,x_3$ adding the three equations we find that $x_1^3+x_2^3+x_3^3=3(x_1+x_2+x_3)-m(x_1^2+x_2^2+x_3^2)-3$ we see that $(x_1+x_2+x_3)^2=x_1^2+x_2^2+x_3^2+2(x_1x_2+x_2x_3+x_1x_3)$ thus the required value is $x_1^3+x_2^3+x_3^3=3(-m)-m(((-m)^2)-2(-3))-3=-3m-m^3-6m-3=-m^3-9m-3$
| {
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How to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$. Please tell if the problem can be solved using telescoping technique or not.
If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using that. It is given that $a,b \in \mathbb{R}{+},\, a\gt b,\, n \in \mathbb{N}.$
I tried as follows, but was unsuccessful to pursue:
$a^n − b^n = a^n+\sum_{i=1}^{n-1}(a^ib^{n-i}-a^ib^{n-i})-b^n=a^n+\sum_{i=1}^{n-1}a^ib^{n-i}-\sum_{i=1}^{n-1}a^ib^{n-i}-b^n$
Edit : based on the selected answer's comment.
Writing a few terms of the series, $\sum_{i=1}^n (a^{n+1-i}b^{i-1}-a^{n-i}b^i)$ get:
For $n =5$, get the terms as:
$i=1, \,\, a^{5+1-1}b^{1-1}-a^{5-1}b^1 = a^5-a^4b.$
$i=2, \,\, a^{5-1}b^{2-1}-a^{5-2}b^2 = a^4b-a^3b^2.$
$i=3, \,\, a^{5-2}b^{3-1}-a^{5-3}b^3 = a^3b^2-a^2b^3.$
$i=4, \,\, a^{5-3}b^{4-1}-a^{5-4}b^4 = a^2b^3-a^1b^4.$
$i=5, \,\, a^{5-4}b^{3-1}-a^{5-3}b^5 = a^1b^4-b^5.$
Adding all the terms, get:
$a^5-a^4b+ a^4b-a^3b^2+a^3b^2-a^2b^3+a^2b^3-a^1b^4+a^1b^4-b^5 = a^5 - b^5$
| Hint: Use that $$\sum_{i=1}^na^{n-i}b^{i-1}=\frac{a^n-b^n}{a-b}$$ if $$a\ne b$$
| {
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Divide the numbers 1,2,…,n into two sets of equal sum. My approach:
If sum(1 to n) is divisible by 2, then it is possible to divide the set.
If n is even take 1st and last element and add to 1st set, then add 2nd and 2nd last to 2nd set...and so on, In then end to balance out both set I remove 1 from 1st set and add it to 2nd.
I don't know how to do it if n is odd.
For odd: eg: n=7
4
1 2 4 7 = 14
3
3 5 6 = 14
| A triangular number is even if and only if its argument is $\in\{0,3\}\bmod 4$.
If the argument is $\equiv 0\bmod 4$, you can specify blocks from the "outside in". For $n=8$ you put $1$ and $8$ into one block, then $2$ and $7$ into the second block, and alternate blocks until you hit the middle:
$1+2+3+4+5+6+7+8=\color{blue}{1+8}\color{brown}{+2+7}\color{blue}{+3+6}\color{brown}{+4+5}$
$=\color{blue}{1+3+6+8}\color{brown}{+2+4+5+7}$
For an argument $\equiv 3\bmod 4$, start by identifying the multiples of $m$ where the argument is $4m-1$. There will be three such values $m,2m,3m$ such that $m+2m=3m$, so put $m+2m$ in one block and $3m$ in the other:
$1+2+3+4+5+6+7+8+9+10+11=1+2\color{blue}{+3}+4+5\color{blue}{+6}+7+8\color{brown}{+9}+10+11$
The remaining terms are partitioned outside in, the same way as for a multiple of $4$ terms described above:
$\color{blue}{1+11}\color{brown}{+2+10}\color{blue}{+4+8}\color{brown}{+5+7}\color{blue}{+3+6}\color{brown}{+9}$
$=\color{blue}{1+3+4+6+8+11}\color{brown}{+2+5+7+9+10}$
| {
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I need to define the cubic equation $f(x)= ax^3 + bx^2 + cx + d$ It is given that if cubic function $f(x)$ is divided by $(x^2+4)$ and $(x+3)$, the remainders will be $7x-20$ and $-2$ respectively. Also it is said show that $f(x)$ is $x^3 + 6x^2 + 11x +4$. I tried but can i find the $a,b,c$, and $d$ without knowing them in from the start?
| Solving in pari/gp:
? chinese(Mod(7*x-20,x^2+4),Mod(-2,x+3))
%1 = Mod(3*x^2 + 7*x - 8, x^3 + 3*x^2 + 4*x + 12)
?
? Mod(a*x^3+b*x^2+c*x+d,(x^2+4)*(x+3))
%2 = Mod((-3*a + b)*x^2 + (-4*a + c)*x + (-12*a + d), x^3 + 3*x^2 + 4*x + 12)
I.e. $3=-3a+b$, $7=-4a+c$, $-8=-12a+d$.
And $\quad a=1\quad\Longrightarrow\quad b=6,c=11,d=4$.
| {
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Prove that inequality $27+7\left(ab+bc+ca\right)\le 8\left(\sum \sqrt{a+3b}\right)$ For the positive reals $a,b$ and $c$ so that $a+b+c=3$. Show that $$27+7\left(ab+bc+ca\right)\le 8\left(\sqrt{a+3b}+\sqrt{b+3c}+\sqrt{c+3a}\right)$$
That inequality has been created by Imad Zak, i think it is interesting problem.
We can see that $$\text{L.H.S}=3(a+b+c)^2+7(ab+bc+ca)$$
$$=3(a^2+b^2+c^2)+13(ab+bc+ca)$$
$$=\sum_{cyc} (a+3b)(b+3c)$$
Let $\sqrt{a+3b}=x;\sqrt{b+3c}=y$ and $z=\sqrt{c+3a}$ where $x,y,z>0$ and $x^2+y^2+z^2=12$
We prove $$8(x+y+z)\ge x^2y^2+y^2z^2+z^2x^2$$
I am stuck here. I tried to homogenize the last inequality but i only get wrong inequalities. I also used C-S or Holder but failed.
| You are actually very close. We can go on from your last inequality, with $x=2u, y=2v, z=2w$, we need to show
$$u+v+w \ge u^2v^2+v^2w^2+ w^2u^2\tag{1}$$
for $u^2+v^2+w^2=3$.
Now, the RHS of (1) is
$$\frac{(u^2+v^2+w^2)^2 - (u^4+v^4+w^4)}2 = \frac{9- (u^4+v^4+w^4)}2$$
So we need to prove
$$u^4+ v^4 + w^4 + 2(u+v+w) \ge9.$$
But this is clear since:
$$u^4 + u + u \ge 3\sqrt[3]{u^4\cdot u\cdot u} = 3u^2,$$
and so on. QED.
| {
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Exercise 3 in Section 7.11 of Apostol's Calculus (Vol. 1) little o-notation The problem is stated as:
Find the polynomial $P(x)$ of minimal degree such that
$$
\sin(x-x^2)=P(x)+o(x^6)\quad \textrm{as }x\to 0.
$$
and the answer in the book is
$$
P(x)=x-x^2-\frac{x^3}{6}+\frac{x^4}{2}-\frac{59x^5}{120}+\frac{x^6}{8}
$$
I am wondering why the answer cannot be
$P(x) = x-x^2$
Since we can use the fact that $\sin(x-x^2) = x-x^2 + o((x-x^2)^2)$ as $x \to 0$
And we know that $o((x-x^2)^2) = o(x^5) = o(x^6)$ as $x$ goes to zero.
Then why is the answer from the back of the book not equal to $x-x^2$??
| Note that one has $o(x^6)=o(x^5)$ (as $x\to 0$), but not $o(x^5)=o(x^6)$.
Also, $o((x-x^2)^2)=o(x^5)$ is incorrect: one has $\lim_{x\to 0}\frac{x^5}{(x-x^2)^2}=0$ but $\lim_{x\to 0}\frac{x^5}{x^5}=1$. In general, $o(f(x))=o(g(x))$ does NOT imply that $o(g(x))=o(f(x))$ as discussed in an answer to another question of yours.
Why the answer cannot be $x-x^2$?
It is an instructive exercise to show (using the correct answer in Apostol) that the following is FALSE:
$$
\lim_{x\to 0}\frac{\sin (x-x^2)-(x-x^2)}{x^6}= 0\tag{1}
$$
[Added.]
To see why (1) is false, you can use the solution of $P(x)$ in the book
so that
$$\begin{align}
f(x):=\frac{\sin (x-x^2)-(x-x^2)}{x^6}
&=\frac{a_3x^3+a_4x^4+a_5x^5+a_6x^6+o(x^6)}{x^6}\\
&=\frac{1}{x^3}
\left(
a_3+a_4x+a_5x^2+a_6x^3
\right)+\frac{o(x^6)}{x^6}\\
&=\frac{1}{x^3}\cdot g(x)+h(x)
\end{align}
$$
where $g(x):=a_3+a_4x+a_5x^2+a_6x^3$.
Now, note that
$$
\lim_{x\to 0}g(x)=a_3\neq 0,\quad \lim_{x\to 0}\frac{o(x^6)}{x^6}=0.
$$
So
$
\lim_{x\to 0+}f(x)=\infty
$
and (1) is thus impossible.
| {
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Modular arithmetic: 5th powers (mod 11) Doing a question about modular arithmetic and number theory and I have a mark scheme that doesn't really explain very well.
It say " The 5th powers (mod 11) are 0,1 and 10 ". I can find examples to support this, such as $11^5$ is equal to 0(mod 11), $2^5$ is equal to 10(mod 11).
How do we know that these are the only values that it can take?
| Well you can write any number as $11n,11n+1,11n+2,11n+3,...11n+10$
This basically means that a number can leave a remainder from 0 to 10 upon division by 11
So if you know the remainders for the fifth powers of all the numbers upto 11 , you can know which remainders all the numbers will leave upon division by 11.
So ,
$1^5 \equiv 1$ mod $11$
$2^5 \equiv 10$ mod $11$
$3^5 \equiv 1$ mod $11$
$4^5 \equiv 1$ mod $11$
$5^5 \equiv 1$ mod $11$
$6^5 \equiv 10$ mod $11$
$7^5 \equiv 10$ mod $11$
$8^5 \equiv 10$ mod $11$
$9^5 \equiv 1$ mod $11$
$10^5 \equiv 10$ mod $11$
So , as you can see the remainders are only 0 , 1 , 10 , so any number will only leave 0,1,10 as remainder when their fifth power is divided by 11.
Eg) $51^5$ mod $11$
$51 = 44 + 7$ mod $11$
So $51 ^5 $ will leave the remainder of $ 7^ 5$ upon division by 11 ( i.e 10 )
I hope this helps you .
| {
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Factoring out the expression I'm sorry, it must pretty basic, but...
How to factor out this expression $x^2 + 5x - 5 = 0$ ?
(The method I normally use to factor expressions doesn't work there because when we factor expression above so that format will be $(ax + b)(qx + c)$ $b$ and $c$ definitely won't be integers, and method that I use only works when $b$ and $c$ are integers)
| You can use completing square method to factorize it.
Look at:
$$x^2+5x-5=x^2+ 2x\times\frac{5}{2}+\frac{25}{4}-\frac{25}{4}-5=(x +\frac{5}{2})^2-(\sqrt\frac{45}{2})^2$$
Now use the fact that $a^2-b^2 = (a+b)(a-b)$, so:
$$x^2+ 5x -5 =(x +\frac{5}{2})^2-(\sqrt\frac{45}{2})^2 = (x+\frac{5}{2}+\sqrt\frac{45}{2})(x+\frac{5}{2}-\sqrt\frac{45}2)$$
which is your answer.
| {
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If $u=(1+\cos t)(1+\cos 2t)-\sin t\sin 2t$ and $v=\sin t(1+\cos 2t)+\sin 2t(1+\cos t)$, then $u^2+v^2=4(1+\cos t)(1+\cos 2t)$
If
$$u = (1+\cos\theta)(1+\cos2\theta) - \sin\theta \sin 2\theta \qquad v = \sin\theta (1+\cos2\theta) + \sin2\theta(1+\cos\theta)$$
then show that
$$u^2 + v^2 = 4(1+\cos\theta)(1+\cos2\theta)$$
I have simplified the values of $u$ and $v$ and got:
$$u = 2\cos\theta (1+\cos2\theta) \qquad v=\sin2\theta(1+\cos\theta)$$
Then I tried to square both $u$ and $v$ individually before doing summation.
Still could not prove the statement.
| You simplified the values incorrectly. See below.
Alternatively:
$$\begin{align}\color{blue}u&=(1+\cos\theta)(1+\cos2\theta) - \sin\theta \sin 2\theta=\\
&=1+\cos \theta +\cos 2\theta +\cos \theta \cos 2\theta-\sin\theta \sin2\theta=\\
&=1+\cos\theta+\cos2\theta+\cos3\theta=\\
&=1+\cos2\theta+2\cos\theta\cos2\theta= \ \ \ \ [\color{red}{\ne 2\cos\theta (1+\cos2\theta)}]\\
&=\color{blue}{1+\cos2\theta(1+2\cos\theta)};\\
\color{green}v&=\sin\theta (1+\cos2\theta) + \sin2\theta(1+\cos\theta)=\\
&=\sin\theta+\sin2\theta+\sin\theta\cos2\theta+\cos\theta\sin2\theta=\\
&=\sin\theta+\sin2\theta+\sin3\theta=\\
&=\sin2\theta+2\sin2\theta\cos\theta=\\
&=\color{green}{\sin2\theta(1+2\cos\theta)}; \ \ \ \ \ \ \ \ [\color{red}{\ne \sin2\theta(1+\cos\theta)}]\\
\color{blue}{u^2}+\color{green}{v^2}&=\color{blue}{1+2\cos2\theta(1+2\cos\theta)+\cos^22\theta(1+2\cos\theta)^2}+\color{green}{\sin^22\theta(1+2\cos\theta)^2}=\\
&=1+2\cos2\theta(1+2\cos\theta)+1\cdot(1+2\cos\theta)^2=\\
&=1+2\cos2\theta+4\cos\theta\cos2\theta+1+4\cos\theta+4\cos^2\theta=\\
&=2+2\cos2\theta+4\cos\theta\cos2\theta+4\cos\theta+2(\cos2\theta+1)=\\
&= 4(1+\cos\theta)(1+\cos2\theta).\end{align}$$
| {
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Dividing polynomial $f(x)$ by $(x-a)(x-b)(x-c)$
$f(x)$ is a polynomial with a degree greater than 3. When $f(x)$ is divided by $(x-a)(x-b)(x-c)$, prove remainder is
$$\frac{f(a)(x-b)(x-c)}{(a-b)(a-c)}+ \frac{f(b)(x-a)(x-c)}{(b-c)(b-a)} +\frac{f(c)(x-a)(x-b)}{(c-b)(c-a)}$$
My Try
I tried this using the conventional method,$$f(x)=Q(x)(x-a)(x-b)(x-c)+Ax^2+Bx+C$$
But then I got long answers for coefficients $A$, $B$ & $C$.
Is there a better way to solve this? Can anyone give me a hint to work this?
| Because $$f(a)=Aa^2+Ba+C,$$
$$f(b)=Ab^2+Bb+C$$ and $$f(c)=Ac^2+Bc+C.$$
Now, work with the given remainder (it's also a polynomial of the second degree).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $f(x)=\frac{x}{\sqrt{x^2+1}}$ then find $ f^{-1}(\sin x)$ let $f(x)=\dfrac{x}{\sqrt{x^2+1}}$ then find $ f^{-1}(\sin x)$
my try : $x:=\tan t$ then $f(\tan t)=\dfrac{\tan t}{1+\tan^2 t}=\sin t \ \ \ \text{or} \ \ -\sin t$
Now what ?
| I would try to find the inverse of $f$ first - this can be done by switching the $x$ and $y$ variables, as follows:
Square to get rid of the square root:
\begin{align}
y = \frac{x}{\sqrt{x^2+1}} \Rightarrow y^2&=\frac{x^2}{x^2+1} \\
&= 1-\frac{1}{x^2+1}
\end{align}
then rearrange to get $x$ to be the subject variable:
\begin{align}
1-y^2=\frac{1}{x^2+1} &\Rightarrow x^2+1=\frac{1}{1-y^2}\\
&\Rightarrow x= \sqrt{\frac{1}{1-y^2}-1}
\end{align}
So $f^{-1}(x)$ can be defined as
$$
f^{-1}(x) = \sqrt{\frac{1}{1-x^2}-1}
$$
Now plugging in $\sin(x)$ yields
\begin{align}
\sqrt{\frac{1}{1-\sin^2(x)}-1} &= \sqrt{\frac{1}{\cos^2(x)}-1}\\
&= \sqrt{\sec^2(x)-1}\\
&= \sqrt{\tan^2(x)}\\
&= \tan(x)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of infinite series with variable range between -1 to 1 Let $S$ denote the infinite sum $2 + 5x + 9x^2 + 14x^3 + 20x^4 + ....$, where $| x | < 1$, then what is the value of $S$?
I am not able to find the generalized form. What is the trick to solve it?
| HINT:
$$S=2+5x+9x^2+14x^3+20x^4+\cdots$$
$$(1-x)S=2+(5-2)x+(9-5)x^2+(14-9)x^3+(20-14)x^4+\cdots$$
Let $T=(1-x)S$
$$T(1-x)=2+(3-2)x+(4-3)x^2+(5-4)x^4+\cdots=1+\dfrac1{1-x}$$ for $|x|<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)$$
I found $A=\frac{1}{3}$ by letting $x=1$.
Now in the book they let me know that $B=\frac{2}{3}$, $C=-\frac{1}{3}$, $D=-1$ and $E=0$. But I would really like to figure out how I can find the values for $B, C, D, E$.
| Since\begin{multline}A(x^2+2)^2+(Bx+C)(x-1)(x^2+2)+(Dx+E)(x-1)=\\=Ax^4+Bx^4-Bx^3+Cx^3+4Ax^2+2Bx^2-Cx^2+Dx^2+\\-2Bx+2Cx-Dx+Ex+4A-2C-E,\end{multline}solve the system$$\left\{\begin{array}{l}A+B=1\\-B+C=-1\\4A+2B-C+D=2\\-2B+2C-D+E=-1\\4A-2C-E=2.\end{array}\right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Prove that $x^2-5y=3$ has no integer solutions. Here is my attempt:
$x^2-5y=3$ is the same as $2x^2-10y=6$, which is equivalent to $2x^2=10y+6$.
This means that 2 divides $x^2$. Since 2 is prime, then 2 divides x, or $x=2k$ for some integer $k$.
Then we have that $(2k)^2-5y=3$, or $4k^2=5y+3$. This implies that 2 divides $5y+3$, but this is a contradiction.
This looks valid to me but any feedback is greatly appreciated.
| If $x^2-5y=3$ then $x^2\equiv3\pmod 5,$
but that would contradict that,
for all integers $x$, $x^2\equiv0, 1, $ or $4\pmod5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.