Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
What is $\operatorname{ord}_{22}(5^6)$?
Find $\operatorname{ord}_{22}(5^6)$.
So, basically we want to find:$$\operatorname*{arg\,min}_k 5^{6k} \equiv 1\pmod {22}$$
I found that $5^5 \equiv 1\pmod {22}$ so I know that for $k=5$ we have $5^{6k} \equiv 1 \pmod {22}$. Therefore, $\text{ord}(5^6) \le 5$.
I guess that I could proceed with the somewhat tedious checking:
*
*$k=4$: $(5^6)^4 \equiv (5^4)^6 \equiv 5^4 \equiv 9 \pmod {22}$
*$k=3$ $(5^6)^3 \equiv (5^3)^6 \equiv 5^3 \equiv 15 \pmod {22}$
*$k=2$ $(5^6)^2 \equiv (5^2)^6 \equiv 5^2 \equiv 3 \pmod {22}$
*$k=1$ $(5^6)^1 \equiv (5^1)^6 \equiv 5 \pmod {22}$
So $\text{ord}_{22}(5^6) = 5$.
Questions:
*
*Is that what I'm expected to do? Or is there a simpler way? (I had to use a calculator or to tediously calculate it myself)
*Why can we get rid of the exponent when we do modular arithmetic? (I actually used it along the proof) i.e. $$x \equiv a \pmod m \iff x^b \equiv a \pmod m$$
| First observe that
\begin{eqnarray*}
5^6 \equiv 5 \pmod{22}.
\end{eqnarray*}
So we only need to calculate powers of $5$ modulo $22$, they are $5,3,15,9,1$ (by tedious calculation), the minimal value of $k$ is $\color{red}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\lim\limits_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$ In preparation for finals, I am trying to calculate $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ with proof.
Here is my approach/what I have done so far:
If we can find a dominating function, we have $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x = \int\limits_0^\infty \lim_{n\rightarrow \infty} \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ by the Dominated Convergence Theorem. If we let $f_{n} = \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)}$, then $f_{n}(x)$ converges to $0$ for all $x > 0$, which implies the limit is equal to 0 because the Dominated Convergence Theorem only requires a.e. convergence (so not having convergence at $x = 0$ is no issue). Operating under the assumption that dominating function exists, is this correct?
As far as finding a dominating function is concerned, we have
$$
|f_{n}| = \left| \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \right| = \frac{|n^2 \sin(n/x)|}{n^3x + x(1 + x^3)} \leq \frac{n^2}{n^3x + x(1 + x^3)},
$$
which is where I get stuck. The two directions that seemed the most clear from here was to either
$$
\frac{n^2}{n^3x + x(1 + x^3)} \leq\frac{n^2}{n^3x} = \frac{1}{x} \quad \text{or} \quad \frac{n^2}{n^3x + x(1 + x^3)} \leq \frac{n^2}{x(1 + x^3)}.
$$
The former is not integrable and I cannot seem to grapple with the $n^2$ on the latter and sufficiently bound it. So my main question is how can I bound $|f_{n}|$?
| Without DCT: In absolute value the integrand is bounded above by
$$\frac{n^2 (x/n)}{n^3x + x(1+x^3)} = \frac{n}{n^3 + (1+x^3)} \le \frac{n}{n^3 + x^3}.$$
Let $x = ny$ to see that
$$\int_0^\infty \frac{n}{n^3 + x^3}\, dx = \frac{1}{n}\int_0^\infty \frac{1}{1 + y^3}\, dy.$$
Since the last integral converges, the integral in question is bounded above by a constant times $1/n,$ hence converges to $0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Explanation of this hint involving the summation of cosine and sine I have this question
Let $n \ge 2$ be an integer. Prove that $$\sum_{k=0}^{n-1}\cos(\frac{2k\pi}{n}) = 0 = \sum_{k=0}^{n-1}\sin(\frac{2k\pi}{n})$$
I was given the hint to
Set $z = \cos(\frac{2\pi}{n}) + i\sin(\frac{2\pi}{n})$, so $z^n=1$. Now write this as $(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)=0$ and go from there...
I get that $(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)=0$ is just an expansion and rearrangement of $z^n=1$, but what I don't understand is, why set $z = \cos(\frac{2\pi}n) + i\sin(\frac{2\pi}{n})$ and how do we know that $z^n=1$
| When you set $z:=\text{cis}\left( \dfrac{2\pi}{n} \right)$ you win all the solutions to $z^n-1=0$, in factorization, using one of moivre law's ($\text{cis}^m(x)=\text{cis}(m.x)$),
$$ (z-1)\cdot \sum_{k=0}^{n-1} \text{cis}\left( \dfrac{2\pi}{n} \cdot k\right) = 0 + 0.\text{i} $$
If $z \neq 1$, we have that crazy sum are $0$. But the real part of this sum is $0$ and the imaginary part also is $0$.
The real part of $\displaystyle \sum_{k=0}^{n-1} \text{cis}\left( \dfrac{2\pi}{n} \cdot k\right) $ are $\displaystyle \sum_{k=0}^{n-1} \cos\left( \dfrac{2\pi}{n} \cdot k\right) = 0$ and imaginary part are $\displaystyle \sum_{k=0}^{n-1} \sin\left( \dfrac{2\pi}{n} \cdot k\right) = 0$. $\Box$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Extrapolating $\pi$ using Taylor expansion Let $P_n$ be a polygon inscribed in a circle with diameter $1$. Each side of the polygon has length $l_n=\sin(\pi /n)$ and the circumference of $P_n=nl_n$.
With
$$P_n= \pi - \frac{\pi^3}{3!}\frac{1}{n^2}+ \frac{\pi^5}{5!}\frac{1}{n^4}- \frac{\pi^7}{7!}\frac{1}{n^6}+\dots$$
We have $P_2=2,~P_3=\frac{3\sqrt{3}}{2},~P_4=2\sqrt{2},P_6=3$.
Extrapolate using the values $P_2$,...,$P_4$ and give estimates and the error for $\pi$.
$$P_2= \pi - \frac{\pi^3}{3!}\frac{1}{2^2}+ \frac{\pi^5}{5!}\frac{1}{2^4}- \frac{\pi^7}{7!}\frac{1}{2^6}+O(h^8)$$
$$P_3= \pi - \frac{\pi^3}{3!}\frac{1}{3^2}+ \frac{\pi^5}{5!}\frac{1}{3^4}- \frac{\pi^7}{7!}\frac{1}{3^6}+O(h^8)$$
$$P_4= \pi - \frac{\pi^3}{3!}\frac{1}{4^2}+ \frac{\pi^5}{5!}\frac{1}{4^4}- \frac{\pi^7}{7!}\frac{1}{4^6}+O(h^8)$$
$$P_6= \pi - \frac{\pi^3}{3!}\frac{1}{6^2}+ \frac{\pi^5}{5!}\frac{1}{6^4}- \frac{\pi^7}{7!}\frac{1}{6^6}+O(h^8)$$
To remove the first error term I extrapolarted $\frac{4P_4-P_2}{3}$ and $\frac{4P_6-P_3}{3}$, and got approximations with errors $1.17838\%$ and $0.00762\%$ respectively. How do I extrapolate $P_2$ with $P_3$ and further to eliminate the 2nd and 3rd order terms?
| Denote by $a,b,c,d$ the coefficents for $P_2,P_3,P_4,P_6$.
The errors cancel if
$$\left(
\begin{array}{ccc}
\frac{1}{2^2} & \frac{1}{2^4} & \frac{1}{2^6} \\
\frac{1}{3^2} & \frac{1}{3^4} & \frac{1}{3^6} \\
\frac{1}{4^2} & \frac{1}{4^4} & \frac{1}{4^6} \\
\frac{1}{6^2} & \frac{1}{6^4} & \frac{1}{6^6} \\
\end{array}
\right)^T.\left(
\begin{array}{c}
a \\
b \\
c \\
d \\
\end{array}
\right)=\left(
\begin{array}{c}
0 \\
0 \\
0 \\
\end{array}
\right)$$
Letting $a=1$ the equation becomes
$$\left(
\begin{array}{ccc}
\frac{1}{3^2} & \frac{1}{3^4} & \frac{1}{3^6} \\
\frac{1}{4^2} & \frac{1}{4^4} & \frac{1}{4^6} \\
\frac{1}{6^2} & \frac{1}{6^4} & \frac{1}{6^6} \\
\end{array}
\right)^T.\left(
\begin{array}{c}
b \\
c \\
d \\
\end{array}
\right)=\left(
\begin{array}{c}
-\frac{1}{2^2} \\
-\frac{1}{2^4} \\
-\frac{1}{2^6} \\
\end{array}
\right)\quad\Rightarrow\quad\left(
\begin{array}{c}
b \\
c \\
d \\
\end{array}
\right)=\left(
\begin{array}{c}
-\frac{162}{7} \\
\frac{512}{7} \\
-81 \\
\end{array}
\right)$$
The linear combination of $P$s gives $\pi$ times $a+b+c+d$, so
$\pi\approx\frac{aP_2+bP_3+cP_4+dP_6}{a+b+c+d}=3.14158884937$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Coefficient of $x^3$ in $(1-2x+3x^2-4x^3)^{1/2}$ The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$
I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?
| Expand $(1+u)^{\tfrac12}$ up to order $3$:
$$(1+u)^{\tfrac12}=1+\frac12 u-\frac18u^2+\frac1{16}u^3+o(u^3),$$
and compose with $u=-2x+3x^2-4x^3$:
*
*$u^2=4x^2-12x^3+o(x^3)$,
*$u^3=u^2\cdot u=-8x^3+o(x^3)$.
One finally obtains$$1-x+x^2-x^3+o(x^3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2555399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ Find the sum of the series
$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac19\right)+\frac1{11}\cdot\left(\frac19+\frac1{13}\right)+\cdots$$
$$=\frac13\cdot\left(\frac65\right)+\frac17\cdot \left(\frac{14}{45}\right)+\frac1{11}\cdot\left(\frac{22}{117}\right)+\cdots$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{45}\right)+\left(\frac1{117}\right)+\cdots\right)$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{5\cdot9}\right)+\left(\frac1{9\cdot13}\right)+\cdots\right)$$
It is here that I am stuck. The answer should be $\frac12$ but I don't see how to get it. Any suggestions?
Also, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills?
| \begin{align*}
\sum_{n=1}\dfrac{1}{(2n-1)(2n+1)}&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\
&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{2n-1}-\dfrac{1}{2(n+1)-1}\right)\\
&=\dfrac{1}{2}\sum_{n=1}\left(\dfrac{1}{f(n)}-\dfrac{1}{f(n+1)}\right)\\
&=\dfrac{1}{2}\dfrac{1}{f(1)},
\end{align*}
where $f(n)=2n-1$, and note that $f(n)^{-1}\rightarrow 0$ as $n\rightarrow\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solve the Integral: $\int{\sqrt{8y-x^2}dx}$ I am trying to solve the following integral:
$$\int{\sqrt{8y-x^2}}dx$$
For which I don't understand clearly how to work with root values.
What I've tried:
$$
\int{\sqrt{8y-x^2}dx} = \int{(8y-x^2)^{\frac{1}{2}}dx} \\
= \frac{(8y-x^2)^{\frac{3}{2}}}{\frac{3}{2}}\int{(8y-x^2)dx} \\
= \frac{(8y-x^2)^{\frac{3}{2}}}{\frac{3}{2}}(-\frac{x^3}{3}) \\
= -\frac{2}{3}(8y-x^2)^{\frac{3}{2}}(\frac{x^3}{3}) \\
= -\frac{2}{9}(8y-x^2)^{\frac{3}{2}}x^3
$$
Is that correct?
| this isn't correct, Substitute $$x=\sqrt{8y}\sin(t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions?
Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$
I have tried rationalizing but there is no pattern that I can observe.
Edit:
So we forget about the $x$ that is multiplied to both the functions and try to work with the expression $(x^2+1)^{1/2}-(x^3+1)^{1/3}.$ Thus we have, $$(x^2+1)^{1/2}-(x^3+1)^{1/3}=\frac{((x^2+1)^{1/2}-(x^3+1)^{1/3})((x^2+1)^{1/2}+(x^3+1)^{1/3})}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}$$
$$=\frac{x^2+1-(x^3+1)^{2/3}}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}=\frac{(x^2+1)^2-(x^3+1)^{4/3}}{(x^2+1)^{3/2}+(x^2+1)(x^3+1)^{1/3}+x^3+1+(x^3+1)^{2/3}(x^2+1)}=??$$
| You can do it with Taylor expansions:
$x\left(1+x^{2}\right)^{1/2}-x\left(1+x^{3}\right)^{1/3}=
x^{2}\left(1+\frac{1}{x^{2}}\right)^{1/2}-x^{2}\left(1+\frac{1}{x^{3}}\right)^{1/3}
=x^{2}\left(1+\frac{1}{2}\cdot\frac{1}{x^{2}}+o\left(\frac{1}{x^{2}}\right)\right)-x^{2}\left(1+\frac{1}{3}\cdot\frac{1}{x^{3}}+o\left(\frac{1}{x^{3}}\right)\right)
=\frac{1}{2}-\frac{1}{3x}+o\left(1\right)
\to\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Splitting the square root of complex function into real and imaginary parts I have these functions below: $$\sqrt{(x+iy)^2-a^2}$$
$$\frac{b(x+iy)}{\sqrt{(x+iy)^2-a^2}}$$
How do I split these to get the real and imaginary parts of these functions?
If anyone could help me out, that really would be helpful!!!!
It will help me with the method of manufactured solutions for my fracture problem that has the Westergaard solution (as the analytical solution)!!!
Thank you very much,
Mousumi
| Basically we need a complex number $A+ib$ such that its square equals $(x+iy)^2-a^2=(x^2-a^2-y^2)+i(2xy)$.
We have: $$(A+ib)^2=(A^2-b^2)+i(2Ab)=(x^2-a^2-y^2)+i(2xy)$$ giving us: $A^2-b^2=x^2-a^2-y^2$ and $Ab=xy$.
Now, we will get, $$A^2+b^2 = \sqrt{(A^2+b^2)^2} = \sqrt{(A^2-b^2)^2+4A^2b^2}=\sqrt{(x^2-a^2-y^2)^2+4x^2y^2}$$
Thus, $A^2=\frac{1}{2}[(x^2-a^2-y^2)+\sqrt{(x^2-a^2-y^2)^2+4x^2y^2}]$ and $b^2=\frac{1}{2}[\sqrt{(x^2-a^2-y^2)^2+4x^2y^2}-(x^2-a^2-y^2)]$.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How many distinct numbers between $10$ and $1000$ can be formed from the digits $2$, $3$, $4$, $5$, $6$, and $0$ without repetition of any digit? Here is my attempt:
If $0$ is not to be any digit of the number, then there are $5 \times 4 \times 3 = 60 = ^5P_2$ possibilities for a 3-digit number, whereas there are $5 \times 4 = 20 = ^5P_3$ possibilities for a 2-digit number. So there are $60 + 20 = 80$ possible numbers between $10$ and $1000$ inclusive formed from the digits $2, 3, 4, 5, 6$.
If $0$ is to be in the unit place, then there are $5 \times 4 = 20$ possibilities for a 3-digit number, and there are $5$ possibilities for a $2$-digit number. So there are a total of $20 + 5 = 25$ numbers formed from these digits with $0$ in the unit place.
If $0$ is to be in the tenth place, then there are no 2-digit numbers, and $5 \times 4 = 20$ 3-digit numbers, totaling to $0 + 20 = 20$ numbers.
Thus there are a total of $80 + 25 + 20 = 125$ numbers.
Is this calculation correct?
If not, then where lies the problem?
| First we deal with all two digit numbers ($11 - 99$). There are two numbers to pick, and tens place cannot be $0$. Hence number of ways are $5\cdot 5$.
Now three digit numbers ($100-999$). Again we cannot have $0$ at hundreds place, so we have $5\cdot 5 \cdot 4$ ways.
In all we have $5^3$ numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Domain of $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$ How do you find the domain of the function $y=\arcsin\left({2x\sqrt{1-x^{2}}}\right)$
I know that the domain of $\arcsin$ function is $[-1,1]$
So, $-1\le{2x\sqrt{1-x^{2}}}\le1$ probably?
or maybe $0\le{2x\sqrt{1-x^{2}}}\le1$ , since $\sqrt{1-x^{2}}\ge0$ ?
EDIT: So many people have answered that the domain would be $[-1,1]$ but my book says that its $[\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}].$
Can anyone explain how are those restrictions made in the given formulas?
| In order to compute $\arcsin(2x\sqrt{1-x^2})$, you need that
$$
\bigl|2x\sqrt{1-x^2}\,\bigr|\le 1
$$
which becomes
$$
4x^2(1-x^2)\le 1
$$
and therefore
$$
4x^4-4x^2+1\ge0
$$
Since the left-hand side is a square, the inequality is satisfied for every $x$, provided $1-x^2\le1$. This is equivalent to $-1\le x\le 1$.
You are misreading the assignment, which asks you to prove that, for $|x|\le1/\sqrt{2}$, you have
$$
2\arcsin x=\arcsin\bigl(2x\sqrt{1-x^2}\,\bigr)
$$
Note that this doesn't hold, for instance, when $x=1$, because the left-hand side is $\pi$, whereas the right-hand side is $0$.
Set $\alpha=\arcsin x$. Since we're assuming $|x|\le 1/\sqrt{2}$, we have $-\pi/4\le\alpha\le\pi/4$. Therefore $\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-x^2}$. Then
$$
2x\sqrt{1-x^2}=2\sin\alpha\cos\alpha=\sin2\alpha
$$
Since $-\pi/2\le 2\alpha\le \pi/2$, we have
$$
\arcsin\bigl(2x\sqrt{1-x^2}\,\bigr)=
\arcsin\sin2\alpha=2\alpha=2\arcsin x
$$
as required. The other questions are similar.
With more advanced tools. Let $f(x)=\arcsin\bigl(2x\sqrt{1-x^2}\,\bigr)$. Then
$$
f'(x)=\frac{1}{\sqrt{1-4x^2(1-x^2)}}
\left(2\sqrt{1-x^2}+2x\frac{-x}{\sqrt{1-x^2}}\right)
=\frac{1}{|1-2x^2|}\frac{2(1-2x^2)}{\sqrt{1-x^2}}
$$
This coincides with the derivative of $g(x)=2\arcsin x$ only for $1-2x^2>0$, that is, $-1/\sqrt{2}<x<1/\sqrt{2}$.
Since $f(0)=g(0)$, the two functions are equal only on the interval $[-1/\sqrt{2},1/\sqrt{2}]$ (at the extremes by continuity).
You can also say that
$$
\arcsin\bigl(2x\sqrt{1-x^2}\,\bigr)=
\begin{cases}
c_1-2\arcsin x & -1\le x<-1/\sqrt{2} \\[4px]
2\arcsin x & -1/\sqrt{2}\le x\le 1/\sqrt{2} \\[4px]
c_2-2\arcsin x & 1/\sqrt{2}<x\le1
\end{cases}
$$
and you can easily determine the constants $c_1$ and $c_2$ by using continuity or by using that $f(-1)=0$ and $f(1)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd The question:
Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.
Hence, prove $7 | 6^n + 8^n \iff n ~$ is odd
I realise that this is a proof by induction, and this is what I have so far:
\begin{align}
f(n) & = 6^n + 8^n \\
& = (2\cdot 3)^n + (2^3)^n \\
& = 2^n \cdot 3^n + 2^{3n} \\
\end{align}
\begin{align}
\text{Assume} ~ f(k) & = 6^k + 8^k \\
& = 2^k \cdot 3^k + 2^{3k} \\
f(k+1) & = 2^{k+1} \cdot 3^{k+1} + 2^{3(k+1)} \\
\end{align}
Where do I go from here?
| This is my first attempt at answering my own question. I realise the question is of a low quality, but I'm focusing on the question-and-answer aspect.
There are a few issues with the OP's attempt. Firstly, the base step was skipped. Secondly, and crucially, $f(k+1)$ is no longer odd, so it cannot be used in the inductive step.
Problem: Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.
Solution: Let $f(n)$ denote the statement
\begin{align}
f(n) & = 6^n + 8^n \\
\end{align}
Base Step ($n=1$):
\begin{align}
f(1) & = 6^1+8^1 \\
& = 6+8 \\
& = 14 \\
\end{align}
$7 | 14 \implies f(1)$ holds.
Inductive Step $f(k) \to f(k+2)$: Fix some $k$, where $k$ is odd. Assume that
$$f(k) = 6^k + 8^k$$
holds true. To be proved is that
$$f(k+2) = 6^{k+2} + 8^{k+2}$$
follows. Beginning with the left-hand side of $f(k+2)$,
\begin{align}
6^{k+2} + 8^{k+2} & = 6^k \cdot6^2 + 8^k \cdot 8^2 \\
& = 6^k \cdot 36 + 8^k \cdot 64 \\
& = 6^k + 6^k \cdot 35 + 8^k + 8^k \cdot63 \\
& = 6^k + 8^k + 6^k \cdot 35 + 8^k \cdot63 \\
& = f(k) + 7(6^k\cdot5+8^k \cdot9)
\end{align}
by the inductive assumption, $7 | f(k)$. Clearly, $7 | (6^k\cdot5+8^k \cdot9)$, thereby showing $f(k+2)$ is true, completing the inductive step.
Conclusion: By mathematical induction, it is proved that $6^n+8^n$ is divisible by $7$ iff $n$ is odd. $\Box$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.,
Is
$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4$$ true $\forall $ $n \in \mathbb{N}$
| Notice $\sum_{k=1}^n k = \frac{n(n+1)}{2}$. For the identity at hand,
$$\sum_{k=1}^n k^5 + \sum_{k=1}^n k^7 \stackrel{?}{=} 2 \left(\sum_{k=1}^n k\right)^4$$
If one compute the difference of successive terms in RHS, we find
$$\begin{align}{\rm RHS}_n - {\rm RHS}_{n-1}
&= 2\left(\frac{n(n+1)}{2}\right)^4-2\left(\frac{n(n-1)}{2}\right)^4\\
&= \frac{n^4}{8}\left((n+1)^4 - (n-1)^4\right)
= \frac{n^4}{8}\left(8n^3 + 8n\right) = n^7 + n^5\end{align}$$
This clearly equals to ${\rm LHS}_n - {\rm LHS}_{n-1}$.
As a result,
$${\rm LHS}_n - {\rm RHS}_{n} = {\rm LHS}_{n-1} - {\rm RHS}_{n-1}$$
and the expression ${\rm LHS}_n - {\rm RHS}_{n}$ is independent of $n$. Since this difference vanishes at $n = 0$, we can conclude
${\rm LHS}_n = {\rm RHS}_n$ for all $n$ and hence establishes the identity at hand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
How can I calculate $\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$ without L'Hôpital's rule? How can I calculate following limit without L'Hôpital's rule
$$\lim_{x \to 0}\frac {\cos x- \sqrt {\cos 2x}×\sqrt[3] {\cos 3x}}{x^2}$$
I tried L'Hôpital's rule and I found the result $2$.
| Write
$$\lim_{x\to0}\dfrac{\cos x-1+1-\sqrt{\cos 2x}+\sqrt{\cos 2x}-\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}=$$
$$\lim_{x\to0}\dfrac{\cos x-1}{x^2}+\lim_{x\to0}\dfrac{1-\sqrt{\cos 2x}}{x^2}+\lim_{x\to0}\sqrt{\cos 2x}\lim_{x\to0}\dfrac{1-\sqrt[3]{\cos 3x}}{x^2}$$
and we have
\begin{align}
&\lim_{x\to0}\dfrac{\cos x-1}{x^2}=\lim_{x\to0}\dfrac{-2\sin^2\frac{x}{2}}{x^2}=-\dfrac{1}{2} \\
&\lim_{x\to0}\dfrac{1-\sqrt{\cos 2x}}{x^2}=\lim_{x\to0}\dfrac{1-\cos 2x}{x^2(1+\sqrt{\cos 2x})}=\lim_{x\to0}\dfrac{2\sin^2x}{x^2(1+\sqrt{\cos 2x})}=1 \\
&\lim_{x\to0}\sqrt{\cos 2x}=1 \\
&\lim_{x\to0}\dfrac{1-\sqrt[3]{\cos 3x}}{x^2}=\lim_{x\to0}\dfrac{1-\cos3x}{x^2(1+\sqrt[3]{\cos 3x}+\sqrt[3]{\cos^23x})}=\dfrac{3}{2}
\end{align}
the answer is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Find $A^n$, if $n$ is Natural number Find eqation for:$\ A^n$ when:
$$A =\left( \begin{array}{cc}
a & 1 & 0\\
0 & a & 1\\
0 & 0 & a\end{array} \right)$$
I calculated $\ A^2$ $\ A^3$ and compared entries:
At the end I've got something like:
$$A^n =\left( \begin{array}{cc}
a^n & na^{n-1} & ???\\
0 & a^n & na^{n-1}\\
0 & 0 & a^n\end{array} \right)$$
So I'm struggling with estimating value of the last entry.
From my calculations I've got respectively: $\ 0 $ $\ 1 $ $\ 3a $. How I can find the equation for this sequence?
| $$A^2 = \left(
\begin{array}{ccc}
a^2 & 2 a & 1 \\
0 & a^2 & 2 a \\
0 & 0 & a^2 \\
\end{array}
\right)$$
$$A^3 = \left(
\begin{array}{ccc}
a^3 & 3 a^2 & 3 a \\
0 & a^3 & 3 a^2 \\
0 & 0 & a^3 \\
\end{array}
\right)$$
$$A^4 = \left(
\begin{array}{ccc}
a^4 & 4 a^3 & 6 a^2 \\
0 & a^4 & 4 a^3 \\
0 & 0 & a^4 \\
\end{array}
\right)$$
$$A^5 = \left(
\begin{array}{ccc}
a^5 & 5 a^4 & 10 a^3 \\
0 & a^5 & 5 a^4 \\
0 & 0 & a^5 \\
\end{array}
\right)$$
Now the sequence in the top right element:
$$1, 3, 6, 10, ...$$
Doesn't remember you about the binomial expansion? :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find the absolute value of the difference of two variables? The problem is as follows:
Let $x$ and $y$ integers which satisfy the following equations:
$$x+y-\sqrt{xy}=7$$
$$x^2+y^2+xy=133$$
Find the value of $\;|x-y|.$
I'm stuck on this problem due the fact that there appears a square root of $xy$ and the squares of both $x$ and $y$, hence the system cannot be solved using the regular methods. Moreover I don't know how to approach the absolute value.
The answer which would help me the most is one which addresses some theoretical basis about absolute value and steps which would led me to find $x$ and $y$.
| Let $a=x+y,b=\sqrt{xy}$. The two given equations can be rewritten as $a-b=7$ and $a^2-b^2=133$. As $a^2-b^2=(a+b)(a-b)$, $a+b=133/7=19$, so $a=(19+7)/2=13$ and $b=(19-7)/2=6$. Thus, $x+y=13$ and $xy=36$.
$$|x-y|^2=(x+y)^2-4xy=13^2-4(36)=13^2-12^2=13+12=25=5^2,$$
so $|x-y|=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Compare sum of radicals I am stuck in a difficult question:
Compare $18$ and
$$
A=\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}
$$
without using calculator.
Thank you for all solution.
| Lets be methodical about this. Consider finding a nice upper bound for
$\sqrt a + \sqrt b$.
$$(\sqrt a + \sqrt b)^2 = a + b + 2\sqrt{ab}$$
So maybe we should replace $ab$ with the smallest $n$ such that $ab \le n^2$
Then $(\sqrt a + \sqrt b)^2 < a + b + 2n$. So
$$\sqrt a + \sqrt b < \sqrt{a + b + 2n}$$
We note that $7 \times 32 = 224 < 225 = 15^2$ and
$11 \times 40 = 440 < 441 = 21^2$
Then
$$\sqrt 7 + \sqrt{32} < \sqrt{7 + 32 + 2\times 15} = \sqrt{69}$$
And
$$\sqrt{11} + \sqrt{40} < \sqrt{11 + 40 + 2\times 21} = \sqrt{93}$$
So
$$\sqrt{69} + \sqrt{93} < \sqrt{69 + 93 + 2 \times 81} = 18$$
Hence $$\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40} < 18$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Finding a specific base for matrices space I need to find a base that includes only matrices with Rank=1 for the following space: $$V=\left \{ \right.(\begin{smallmatrix}
x-y &2x+3y+3z \\
-14x-7y-21z & -8x+8y
\end{smallmatrix}\bigr):x,y,z\in\mathbb{R}\left. \right \}$$
What I did is:
1) Extracted $x,y,x$ to find the $span$: $$V=sp\left \{ x\begin{pmatrix}
1 &2 \\
-14&8
\end{pmatrix},y\begin{pmatrix}
-1 &1 \\
-7 &8
\end{pmatrix},z\begin{pmatrix}
0 &1 \\
-7 &0
\end{pmatrix} \right \}$$
2) Find the base of this $span$ using Gauss-Elimination: $$\Rightarrow V=sp\left \{ \bigl(\begin{smallmatrix}
1 &0 \\
0 &-8
\end{smallmatrix}\bigr),\bigl(\begin{smallmatrix}
0 &1 \\
-7 &0
\end{smallmatrix}\bigr) \right \}$$
The base I found includes matrices with Rank=2 therefore doesn't comply with the question terms.
I would be really glad to hear some insights about this one!
| A good start is to see which matrices out of
$$
e_1=\begin{pmatrix}
1&0\\0&0
\end{pmatrix},~
e_2=\begin{pmatrix}
0&1\\0&0
\end{pmatrix},~
e_3=\begin{pmatrix}
0&0\\
1&0
\end{pmatrix}
\text{ and }
e_4=\begin{pmatrix}
0&0\\0&1
\end{pmatrix}
$$
are contained in $V$.
You can directly check that $e_1,e_4\notin V$. But we still get
$$
e_2=\begin{pmatrix}
x-y &2x+3y+3z \\
-14x-7y-21z & -8x+8y
\end{pmatrix}
\text{ where }x=y=-z=\frac12
$$
and
$$
e_3=\begin{pmatrix}
x-y &2x+3y+3z \\
-14x-7y-21z & -8x+8y
\end{pmatrix}
\text{ where }x=y=-\frac35z=\frac1{14}
$$
From your first step, we deduce $\dim V=3$ and the last one is a little difficult to find. We observe, that a matrix with rank $1$ has to have the form
$$
\begin{pmatrix}a & b\\\lambda a & \lambda b\end{pmatrix}.
$$
for some $a,b,\lambda\in\mathbb R$. Now we compare and get
$$
\begin{pmatrix}a & b\\\lambda a & \lambda b\end{pmatrix}
=\begin{pmatrix}
x-y &2x+3y+3z \\
-14x-7y-21z & -8x+8y
\end{pmatrix}
\Leftrightarrow \begin{matrix}(i)~ a=x-y\\(ii)~ b=2x+3y+3z\\(iii)~\lambda a=-14x-7y-21z\\(iv)~\lambda b=-8x+8y\end{matrix}
$$
Here starts the tricky part. If $b=0$, we can use $(i)$ and $(iv)$ to deduce that $a=0$ and we get the null matrix, which is bad. But for $b\neq 0$, we get $\lambda=-8\frac{a}b$ and consider
$$
\begin{pmatrix}a & b\\-8\frac{a^2}b & -8a\end{pmatrix}
=\begin{pmatrix}
x-y &2x+3y+3z \\
-14x-7y-21z & -8x+8y
\end{pmatrix}\\
\Leftrightarrow\begin{pmatrix}x-y\\2x+2y+3z\\-14x-7y-21z\\-8x+8y\end{pmatrix}=\begin{pmatrix}a\\b\\-8\frac{a^2}b\\-8a\end{pmatrix}\\
\Leftrightarrow \begin{pmatrix}1 & -1 & 0\\2 & 3 & 3\\-14& -7 & -21\end{pmatrix}\cdot\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}a\\b\\-8\frac{a^2}b\end{pmatrix}
$$
You see that this linear equation has always a solution. To simplify it, we choose $a=b=1$ and solve the system to get
$$
\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}
1 &1/2 &1/14 \\
0&1/2&1/14\\
-2/3&-1/2&-5/42
\end{pmatrix}\cdot\begin{pmatrix}1\\1\\-8\end{pmatrix}=\begin{pmatrix}13/14\\-1/14\\-3/14\end{pmatrix}
$$
Next, we get
$$
f:=\begin{pmatrix}1&1\\-8&-8\end{pmatrix}=\begin{pmatrix}
x-y &2x+3y+3z \\
-14x-7y-21z & -8x+8y
\end{pmatrix}
$$
where $x=\frac{13}{14}$,$y=-\frac1{14}$ and $z=-\frac3{14}$. Finally, we see that $\{e_2,e_3,f\}\subseteq V$ is a linearly independent set of matrices with rank $1$ and the base of $V$, since $\dim V=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Triangular numbers is relating to Pythagoras triples Let
$$T_n:=1,3,6,10,15,...$$
are triangular numbers, $T_n={n(n+1)\over 2}.$
Pythagoras triples $(5,12,13),(7,24,25),(9,40,41),(11,60,61)$ and so on,...
Observe that $S_n=T_n+2T_{n+1}+T_{n+2}$ for $n\ge1$
$S_1=13$
$S_2=25$
$S_3=41$ and so on, ...
Can anyone explain how does this formula $S_n$ in term of triangular is related to Pythagoras triple numbers?
| Pythagorean triples can be gnerated by $(m^2-n^2, 2nm , n^2+m^2)$ and if $m=n+1$ you will obtain your sequence.
Note that your sequence $S_n$ are the sum of two consecutive squares. $S_=n^2+(n+1)^2$. Also
\begin{eqnarray*}
S_n&=& \frac{n(n+1)}{2}+ 2\frac{(n+1)(n+2)}{2}+\frac{(n+2)(n+3)}{2}= 2n^2+6n+5 \\&=&(n+1)^2+(n+2)^2
\end{eqnarray*}
again the sum of two consecutive squares (offset by $1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the limit of $\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)$ I have to prove that
$$\Big[\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)\Big] \to \frac 12$$
I know that $$\log\Big(1+\frac 1{n^2}\Big)+ \log\Big(1+\frac 2{n^2}\Big)+...+ \log\Big(1+\frac n{n^2}\Big)= \\ =\log\Big(\frac {n^2+1}{n^2}\Big)+ \log\Big(\frac {n^2+2}{n^2}\Big)+...+ \log\Big(\frac {n^2+n}{n^2}\Big)=\\ =\log\Big(n^2+1\Big)+ \log\Big(n^2+2\Big)+...+ \log\Big(n^2+n\Big)-2n\log n$$
Doing some manipulation we get:
$$\lim_{n \to +\infty}\log\Big(\frac{(n^2+n)!}{n^2!n^{2n}}\Big)=\frac 12$$
How can I prove this limit?
|
I thought it might be instructive to present an approach that does not rely on calculus, but rather uses the squeeze theorem, a set of inequalities that can be obtained with pre-calculus tools only, and the values of the sums $\sum_{k=1}^n k$ and $\sum_{k=1}^n k^2$. To that end we proceed.
TOOL $1$: Elementary Inequality
In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1\tag 1$$
TOOL $2$: Simple Lower Bound for $\frac{1}{1+x}$
Noting that $1-x^2\le 1$, it is trivial to see that
$$1-x\le \frac{1}{1+x}\tag 2$$
for $x>-1$.
TOOL $3$: Values of Sums of Integers and Squared Integers
It is easy to show, using induction for example, or as I showed in THIS ANSWER using elementary analysis, that
$$\sum_{k=1}^n k=\frac{n(n+1)}{2}\tag 3$$
and
$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}\tag 4$$
Applying the inequalities in $(1)$ with $x=\frac{k}{n^2}$, we find that
$$\begin{align}
\frac{\frac{k}{n^2}}{1+\frac{k}{n^2}}\le \log\left(1+\frac k{n^2}\right)\le \frac{k}{n^2}\tag 5
\end{align}$$
Next, applying the inequality $(2)$ to the left side of $(5)$ with $x=\frac{k}{n^2}$ reveals
$$\begin{align}
\frac{k}{n^2}-\frac{k^2}{n^4}\le \log\left(1+\frac k{n^2}\right)\le \frac{k}{n^2}\tag 6
\end{align}$$
Summing the terms in $(6)$ from $k=1$ to $k=n$ yields
$$\sum_{k=1}^n\left(\frac{k}{n^2}-\frac{k^2}{n^4}\right)\le \sum_{k=1}^n \log\left(1+\frac k{n^2}\right)\le \sum_{k=1}^n \frac{k}{n^2}\tag 7$$
whereupon applying $(3)$ and $(4)$ to $(7)$ we find that
$$\frac{n(n+1)}{2n^2}-\frac{n(n+1)(2n+1)}{6n^4}\le \sum_{k=1}^n \log\left(1+\frac k{n^2}\right)\le \frac{n(n+1)}{2n^2}\tag 8$$
Finally, applying the squeeze theorem to $(8)$, we obtain the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\log\left(1+\frac k{n^2}\right)=\frac12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Constant of integration change So, sometimes the constant of integration changes, and it confuses me a bit when and why it does. So for example, we have a simple antiderivative such as $$\int \frac{1}{x} dx $$ and we know that the result is $$\log|x| + C$$ and the domain is $$x\in\mathbb R \backslash \{0\} $$ If we want to show all the solutions, we need to do something like $$\begin{cases}
\log x+C_1 & x>0\\
\log(-x)+C_2 & x<0
\end{cases}$$
Do we need to do change the constant every time there is a gap in the domain or is it just when the expression changes? For example, $$ \int \frac {x^5} {x^2-1} dx$$ which is $$ \frac {1} {2} \log |x^2-1| + \frac {x^4} {4} + \frac {x^2}{2} + C$$ and the domain is $$x \in \mathbb R \backslash \{-1,1\}$$ When we want to write all the solutions, is it something like $$ \begin{cases} \ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_1 & x>1\\ \frac {1} {2}\log (-x^2+1) + \frac {x^4} {4} + \frac {x^2}{2} + C_2 & -1<x<1\\ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_3 & x<-1 \end{cases}$$ or is it that since the the first and last expression are the same they only have one constant associated? Meaning, the solutions are actually $$ \begin{cases} \ \frac {1} {2} \log (x^2-1) + \frac {x^4} {4} + \frac {x^2}{2} + C_1 & x \in \mathbb R \backslash [-1, 1]\\ \frac {1} {2}\log (-x^2+1) + \frac {x^4} {4} + \frac {x^2}{2} + C_2 & -1<x<1\\ \end{cases}$$
| To expound a bit on Catalin Zara's answer:
Your answer should give all possible antiderivatives; i.e., if you plug in any particular combination of constants for the generic constants $C_1, C_2, \dots$, you get a function which, when you differentiate, gives you the integrand. So the answer with three constants is more general than the answer with two constants; and you can check that if $C_1$ and $C_3$ are different, you still get a function with derivative $\frac{x^5}{x^2-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
If $f(x) $ is a continuous function with a given identity, how do I find $f(\sqrt{3})$?
If $f(x) $ Is a continuous function $ \forall\ x \in R$ and satisfies
$x^2+x \{f(x)\} - 3 = \sqrt{3} \ f(x) \ \forall\ x \in R$
Find $f(\sqrt{3})$.
$\{ t\}$ is the fractional part of $t$.
--
My attempt: I substituted $\sqrt{3}$ in that and could only conclude that $0 \leq f(\sqrt{3}) < 1$
| By substituting $\sqrt{3}$ in there, you get that $f(x) - \{f(x)\} = 0$, which implies that $0 \leq a < 1$ as you concluded.
Now suppose $f(\sqrt{3}) = a, 0 < a < 1$. Then, by continuity, $f$ satisfies the equation $x^2 + xf(x) - 3 = \sqrt{3} f(x)$ in a neighborhood of $x = \sqrt{3}$, so that $$f(x) = \frac{x^2 - 3}{\sqrt{3} - x} = -(\sqrt{3} + x),$$
which implies that $f(\sqrt{3}) = - 2\sqrt{3} < 0$ (a contradiction because $a > 0$). Therefore, we must have $f(\sqrt{3}) =0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that for every $n \ge 0$ the term $2^{2n+1} - 9n^2 + 3n -2$ is divisible by $54$ What I get is $4 \cdot 54k + 27(n(n-1)) - 2$.
I understand first two terms but $-2$ is giving me a headache. I suppose to be true every term should be divisible by $54$. Or did I make wrong conclusion?
P. S: Is there mathjax tutorial?
| (Alternative approach, without induction.) The following uses the notation $\,\mathcal{M}n\,$ to denote any integer multiple of $\,n\,$. The obvious rules apply $\,2 \cdot \mathcal{M}3 = \mathcal{M}6, \,\mathcal{M}3 \cdot \mathcal{M}3 = \mathcal{M}9\,$ etc.
*
*$\,4^k-1 = \mathcal{M}3\,$ for all $\,k \ge 0\,$, therefore:
$$\require{cancel}
\begin{align}
4^n - 1 &= (4-1)\cdot(4^{n-1}+\cdots+4+1) \\
&=3 \cdot \big((4^{n-1}-1)+ \cdots + (4-1) + (\cancel{1} - \cancel{1}) + n\big) \tag{1}\\
&= 3 \cdot (\mathcal{M}3 + n) \\
&= \mathcal{M}9 + 3n
\end{align}
$$
*
*Using the previous result back in $\,(1)\,$:
$$
\begin{align}
\color{blue}{2 \cdot (4^n - 1)} &=6 \cdot \big((\mathcal{M}9+3(n-1))+ (\mathcal{M}9+3(n-2))+ \cdots + (\mathcal{M}9+3 \cdot 1) + n\big)\\
&= \mathcal{M}54 + 18\big((n-1)+(n-2)+\cdots+1 \big) + 6n \\
&= \mathcal{M}54 + 18 \cdot \frac{n(n-1)}{2}+6n \\
&= \color{blue}{\mathcal{M}54 + 9 n^2 - 3n} \tag{2}
\end{align}
$$
It then follows straight from $\,(2)\,$ that: $$2^{2n+1} - 9n^2 + 3n -2 = \color{blue}{2 \cdot (4^n -1)} -9n^2+3n=\color{blue}{\mathcal{M}54 + \cancel{9 n^2} - \bcancel{3n}} - \cancel{9n^2} + \bcancel{3n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Is there an infinite number of numbers like $1600$? My reputation is at this moment at $1600$.
I did some experimenting with $1600$ and obtained the following:
Evidently, it is a perfect square $1600=40^2$
Also, it is a hypothenuse of a Pythagorean integer-triple triangle $1600=40^2=32^2+24^2$.
Also, it can be written as the sum of four non-zero squares $1600=20^2+20^2+20^2+20^2$
So, $1600$ is a perfect square, a hypothenuse of a Pythagorean integer-triple (so can be written as a sum of two non-zero squares), and a sum of four non-zero squares.
Is there an infinite number of numbers like $1600$? Can you find some more?
Edit 1: It is also a sum of $4$ non-zero positive cubes: $1600=8^3+8^3+8^3+4^3$
Edit 2: It is also a sum of powers from $1$ to $4$, as we see $1600=7^1+9^2+6^3+6^4$
| Yes. We know that $5^2=3^2+4^2$ and, if we consider $k\in\mathbb{N}$ and a semejant triangle for this with scale $2k$ then $(6k)^2+(8k)^2=(10k)^2$.
The number $(10k)^2$ is our candidate. In fact, its is a square number, its a hypothenuse of a pythagorean integer-triple.
Also, $$(10k)^2=(2\cdot 5k)^2=4\cdot(5k)^2=(5k)^2+(5k)^2+(5k)^2+(5k)^2$$
is the sum of four square numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Interesting Olympiad Style Problem about Invariance Problem: The following operations are permitted with the quadratic polynomial $ax^2 +bx +c:$ (a) switch $a$ and $c$, (b) replace $x$ by $x + t$ where $t$ is any real. By repeating these operations, can you transform $x^2 − x − 2$ into $x^2 − x − 1?$
My Attempt: Notice that the sum of coefficients $S\equiv a+b+c\pmod{t}$ is invariant. This is clear if we switch $a$ and $c.$ If we replace $x$ with $x+t$ then we have $ax^2+(2at+b)x+(at^2+bt+c)$ and so $S\equiv a+2at+b+at^2+bt+c\equiv a+b+c\pmod{t}.$ Now for $x^2-x-2$ we have $S\equiv -2\pmod{t}$ and at the end we want $S\equiv -1\pmod{t}$, which is impossible. I am not sure whether this is correct because $t\in \mathbb{R}.$ So any inputs will be much appreciated.
| I don't understand. Is it $S = a+b+c\pmod{t}$ or $S= a+b+c$. Because I don't understand how you get $S\equiv a+b+c\pmod{t}$ in the second case.
Anyway, just calculate the discriminant and show that it doesn't change:
Mark new polynomial with $a'x^2+b'x+c'$
Case 1. If we change only we get from $ax^2+bx+c$ this $cx^2+bx+a$ so $a'=c$, $b'=b$ and $c'=a$ so
$$D' = b'^2 -4a'c' = b^2-4ac = D$$
Case 2. If we replace $x$ with $x+t$ we get $$a(x+t)^2+b(x+t)+c = ax^2+(2at+b)x +at^2+bt+c$$ so $a' = a$, $b' = 2at+b$ and $c'=at^2+bt+c$ so $$D' = (2at+b)^2-4a(at^2+bt+c) = 4a^2t^2+4abt+b^2 -4at^2-4abt-4ac = D$$
So since the discriminant at begining is different from the end it is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\int_0^\infty\frac1{x^x}\, dx<2$
Prove that $$\int_0^\infty\frac1{x^x}\, dx<2.$$
Note: This inequality is rather tight. The integral approximates to $1.9955$.
Integration by parts is out of the question. If we let $f(x)=\dfrac1{x^x}$ and $g'(x)=1$ then $f'(x)=-x^{-x}(\ln x + 1)$ by implicit differentiation and $g(x)=x$. The integral $\int f'(x)g(x)\, dx$ looks even harder to evaluate.
Expressing the left-hand side as a Frullani integral $$\int_0^\infty\frac{f(ax)-f(bx)}x\, dx=(f(0)-f(\infty))\ln\frac ba$$ means that $f(ax)-f(bx)=x^{1-x}$. However, I can't seem to find a continuous function $f$ that satisfies the functional equation. Is there such a function?
(For context, user371838's post below proves sophomore's dream which I also asked about originally.)
| Remarks: Here is an alternative proof.
I used the same bounds for this question
Improper integral inequality including the golden ratio and the Sophomore's dream
We will use the following auxiliary results (Facts 1-2).
Fact 1: $x^{-x} \le \frac{3 - x}{x^2 - x + 2}$ for all $x \in [1, 2]$.
(RHS is the Pade $(1, 2)$ approximation of $x^{-x}$ at $x = 1$.)
Fact 2: $x^{-x} \le a^{-x} \mathrm{e}^{-x + a}$ for all $x, a > 0$.
(Proof: Taking logarithm on both sides, letting $u = \frac{a}{x} > 0$,
it is equivalent to $\ln u \le u - 1$ which is true (easy).)
We have (Sophomore's dream)
$$I_1 := \int_0^1 x^{-x} \mathrm{d} x
= \sum_{n = 0}^\infty \frac{1}{(n + 1)^{n + 1}}
\le 1 + \frac{1}{2^2} + \frac{1}{3^3} + \sum_{n=3}^\infty \frac{1}{4^{n + 1}} = \frac{2233}{1728}.$$
Using Fact 1, we have
$$I_2 := \int_1^2 x^{-x}\mathrm{d} x
\le \int_1^2 \frac{3 - x}{x^2 - x + 2}\mathrm{d} x = \frac{5}{\sqrt 7}\arctan \frac{\sqrt 7}{5} - \frac12\ln 2.$$
Using Fact 2, we have
$$I_3 := \int_2^{5/2} x^{-x}\,\mathrm{d} x
\le \int_2^{5/2} 2^{-x}\mathrm{e}^{-x + 2}\,\mathrm{d} x = \frac{2 - \sqrt{2\mathrm{e}^{-1}}}{8 + 8\ln 2},$$
and
$$I_4 :=
\int_{5/2}^3 x^{-x}\,\mathrm{d} x
\le \int_{5/2}^3 (5/2)^{-x}\mathrm{e}^{-x + 5/2}\,\mathrm{d} x = \frac{4\sqrt{10} - 8\sqrt{\mathrm{e}^{-1}}}{125\ln \frac{5}{2} + 125}, $$
and
$$I_5 :=
\int_3^\infty x^{-x}\, \mathrm{d} x
\le \int_3^\infty 3^{-x} \mathrm{e}^{-x + 3}\, \mathrm{d} x = \frac{1}{27\ln 3 + 27}.$$
Thus, we have
\begin{align*}
&\int_0^\infty x^{-x} \mathrm{d} x\\
=\, & I_1 + I_2 + I_3 + I_4 + I_5\\
\le\,& \frac{2233}{1728}
+ \frac{5}{\sqrt 7}\arctan \frac{\sqrt 7}{5} - \frac12\ln 2
+ \frac{2 - \sqrt{2\mathrm{e}^{-1}}}{8 + 8\ln 2} + \frac{4\sqrt{10} - 8\sqrt{\mathrm{e}^{-1}}}{125\ln \frac{5}{2} + 125}
+ \frac{1}{27\ln 3 + 27}\\
<\,& 2.
\end{align*}
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 0
} |
Find the remainder when $\frac{13!}{7!}$ is divided by $17$. How to find the remainder when $\frac{13!}{7!}$ is divided by $17$?
I started with Wilson's Theorem which states that every prime $p$ divides $(p-1)!+1$. That is $(p-1)!=-1 mod(p)$.
$16!=-1 mod(17)$
Kindly help me how to get to $\frac{13!}{7!}$.
| By Wilson's theorem $16!\equiv -1\pmod{17}$ and since $17$ is a prime of the form $8k+1$ and the square roots of $-1$ in $\mathbb{F}_{17}^*$ are $4$ and $13$, $8!\equiv 13\pmod{17}$ and $7!\equiv 8\pmod{17}$. By using the notation $\frac{1}{a}$ for the inverse of $a$,
$$ \frac{13!}{7!}\equiv\frac{16!}{(-1)(-2)(-3)8}\equiv\frac{1}{48}\equiv\frac{1}{-3}\equiv -6\equiv\color{red}{11}\pmod{17}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Error while integrating reciprocal of irreducible quadratic $\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$ I am trying to derive a formula for indefinite integral of reciprocal of irreducible real quadratic.
$$\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$$, $b^2 - 4ac \lt 0$. According to WolframAlpa it should come out as: $$\frac{2\arctan{\frac{2ax+b}{\sqrt{4ac - b^2}}}}{\sqrt{4ac - b^2}} + C$$ Instead I get $$\frac{\arctan{\frac{x + \frac{b}{2a}}{\sqrt{\frac{c}{a} - (\frac{b}{2a})^2}}}}{a(\frac{c}{a} - (\frac{b}{2a})^2)^\frac{3}{2}} + C$$ which, no matter how I try to simplify it, doesn't give the right answer. Here are my steps, what am I doing wrong?
1) Reduce to a monic quadratic and complete the square: $$\int{\frac{1}{ax^2 + bx + c}}{dx} = \frac{1}{a}\int{\frac{1}{x^2 + \frac{b}{a}x + \frac{c}{a}}}{dx} = \frac{1}{a}\int{\frac{1}{(x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2}}{dx}$$
3) Substitute $x = u - \frac{b}{2a}$, $u = x + \frac{b}{2a}$, $k = \frac{c}{a} - (\frac{b}{2a})^2$: $$\frac{1}{a}\int{\frac{1}{(x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2}}{dx} = \frac{1}{a}\int{\frac{1}{u^2 + k}}{du}$$
4) Substitute $u = v\sqrt{k}$, $v = \frac{u}{\sqrt{k}}$, $du = \frac{1}{\sqrt{k}}\hphantom ddv$:
$$\frac{1}{a}\int{\frac{1}{u^2 + k}}{du} = \frac{1}{a}\int{\frac{\frac{1}{\sqrt{k}}}{kv^2 + k}}{dv} = \frac{1}{a}\int{\frac{1}{k\sqrt{k}(v^2 + 1)}}{dv} = \frac{1}{ak^\frac{3}{2}}\int{\frac{1}{v^2 + 1}}{dv}$$
5) Use common integral $$\frac{1}{ak^\frac{3}{2}}\int{\frac{1}{v^2 + 1}}{dv} = \frac{\arctan{v}}{ak^\frac{3}{2}} + C$$
6) Substitute back $$\frac{\arctan{v}}{ak^\frac{3}{2}} + C = \frac{\arctan{\frac{u}{\sqrt{k}}}}{ak^\frac{3}{2}} + C = \frac{\arctan{\frac{x + \frac{b}{2a}}{\sqrt{\frac{c}{a} - (\frac{b}{2a})^2}}}}{a(\frac{c}{a} - (\frac{b}{2a})^2)^\frac{3}{2}} + C$$
| Everything is fine till you write $du=\frac{1}{\sqrt{k}}dv$. It is instead, $du=\sqrt{k}dv$. Aside from that, your solution method is great.
Also, for the argument to the arctan function, note that $\frac{x+\frac{b}{2a}}{\sqrt{\frac{c}{a}-\left(\frac{b}{2a}\right)^2}}=\frac{2a\left(x+\frac{b}{2a}\right)}{2a\sqrt{\frac{c}{a}-\left(\frac{b}{2a}\right)^2}}$, and simplifying this will allow the "inside" of the tangent function to have the appropriate form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\lim_{x \to 2} x^3 = 8$ by using epsilon-delta Prove that $$\lim_{x \to 2} x^3 = 8$$
My attempt,
Given $\epsilon>0$, $\exists \space \delta>0$ such that if $$|x^3-8|<\epsilon \space \text{if} \space 0<|x-2|<\delta$$
$$|(x-2)(x^2+2x+4)|<\epsilon$$
I'm stuck here. Hope someone could continue the solution and explain it for me. Thanks in advance.
| Note that
$$x^2 + 2x + 4 = (x-2)^2 + 6(x-2) + 12$$
so
\begin{align}|(x-2)(x^2 + 2x + 4)| &= |(x-2)((x-2)^2 + 6(x-2) + 12)|\\
&\le |x-2|(|x-2|^2 + 6|x-2| + 12)
< \delta(\delta^2 + 6 \delta + 12)
\end{align}
For $\varepsilon > 0$ you would take $\delta < \min\left\{\frac\varepsilon{19}, 1\right\}$ because then $|x-2| < \delta$ implies:
$$|x^3 - 8| \le \delta(\delta^2 + 6\delta + 12) < \frac\varepsilon{19} (1 + 6 + 12) = \varepsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How many positive integers from 1-1000 have 5 divisors?
How many positive integers from $1-1000$ have $5$ divisors?
Any answers would be greatly appreciated. If you have any questions, I will edit for clarification.
| (Filling in some details, on the off chance that other answers are unfamiliar to potential readers!)
Consider the prime factorization of $75 = 3^1 \cdot 5^2$. How many factors does $75$ have? Well, in forming a factor of $75$, we have to choose how many $3$s to include - there is one available, but we could also choose none; so, the number of choices is $1+1=2$ - and how many $5$s to include - there are two available, but we could also choose none; so, the number of choices is $2+1=3$. The aforementioned $2$ choices and $3$ choices yield $2 \cdot 3 = 6$ combinations using the prime factors, and these are all the factors of $75$:
$$3^0 \cdot 5^0, 3^0 \cdot 5^1, 3^0 \cdot 5^2, 3^1 \cdot 5^0, 3^1 \cdot 5^1, 3^1 \cdot 5^2$$
More generally, a number with prime factorization $p^a \cdot q^b$ for distinct primes $p$ and $q$ has $(a+1)(b+1)$ factors, since you can form these factors by making one of $a+1$ choices for how many $p$s to include (ranging from $0$ to all $a$) and one of $b+1$ choices for how many $q$s to include (ranging from $0$ to all $b$). Even more generally, you arrive at the formula in Bernard's answer.
So: In one's investigation of $5$, it becomes clear that this can arise as the number of factors if, and only if, it came from a number of the form $p^4$ for a prime $p$. All that remains now is to observe how many primes to the fourth power are present between $1$ and $1000$; this is precisely what is done in Mohammad Riazi-Kermani's answer: just $2^4 = 16$, $3^4 = 81$, and $5^4 = 625$. Already we have $6^4 = 1296 > 1000$, so the three listed numbers are exhaustive.
As a follow up exercise, you may wish to count how many factors $4^4$ and $6^4$ have.
(Which would you guess has more? Etc.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solving $\frac{1}{|x+1|} < \frac{1}{2x}$ Solving $\frac{1}{|x+1|} < \frac{1}{2x}$
I'm having trouble with this inequality. If it was $\frac{1}{|x+1|} < \frac{1}{2}$, then:
If $x+1>0, x\neq0$, then
$\frac{1}{(x+1)} < \frac{1}{2} \Rightarrow x+1 > 2 \Rightarrow x>1$
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2} \Rightarrow -(x+1) > 2 \Rightarrow x+1<-2 \Rightarrow x<-3$
So the solution is $ x \in (-\infty,-3) \cup (1,\infty)$
But when solving $\frac{1}{|x+1|} < \frac{1}{2x}$,
If $x+1>0, x\neq0$, then
$\frac{1}{x+1} < \frac{1}{2x} \Rightarrow x+1 > 2x \Rightarrow x<1 $
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2x} \Rightarrow -(x+1) > 2x \Rightarrow x+1 < -2x \Rightarrow x<-\frac{1}{3}$
But the solution should be $x \in (0,1)$.
I can see that there can't be negative values of $x$ in the inequality, because the left side would be positive and it can't be less than a negative number. But shouldn't this appear on my calculations?
| Note that $x$ cannot be negative as $|x+1|$ is always nonnegative.
With that mind, observe that $|x+1| = x+1, \forall x > 0$. Therefore
$$ \frac{1}{x+1} < \frac{1}{2x} $$
which results in $x < 1$ as you have done.
So the final result is $x \in (0,1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Factoring the polynomial $3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$ I'm trying to factor the following polynomial:
$$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$$
What I've done:
$$3(x^2 -1)^3 + 7(x^2-1)^2 + 4(x^2 -1)$$
Then I set $p=x^2 -1$ so the polynomial is:
$$3p^3 + 7p^2 + 4p$$
Therefore: $$p(3p^2 + 7p + 4)$$
I apply Cross Multiplication Method: $$p(p+3)(p+4)$$
I substitute $p$ with $x^2-1$:
$$(x^2-1)(x^2-1+3)(x^2-1+4)$$
$$(x-1)(x+1)(x^2-2)(x^2-3)$$
I don't know if I've done something wrong or if I have to proceed further and how. The result has to be: $x^2(3x^2+1)(x+1)(x-1)$. Can you help me? Thanks.
| at "Cross Multiplication" it should be $$ p(3p+4)(p+1) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
For what positive numbers $a,b$ is the series $\sum_{k=0}^{\infty} a^{\frac{1}{b}+\frac{1}{b+1}+...+\frac{1}{b+k}}$ convergent? Last one for tonight. For what positive numbers $a,b$ is the series $$\sum_{k=0}^{\infty} a^{\frac{1}{b}+\frac{1}{b+1}+...+\frac{1}{b+k}}$$ convergent?
I'm defeated by this one. I don't even know how to begin.
| Because$$\def\d{\mathrm{d}}
\sum_{j = 0}^k \frac{1}{b + j} > \int_0^{k + 1} \frac{\d x}{b + x} = \ln(b + k + 1) - \ln b,
$$
for $a \geqslant 1$,$$
a^{\sum\limits_{j = 0}^k \frac{1}{b + j}} \geqslant a^{\ln(b + k + 1) - \ln b} \geqslant a^{\ln(b + 1) - \ln b},
$$
which implies $a^{\sum\limits_{j = 0}^k \frac{1}{b + j}} \not\to 0 \ (k \to \infty)$. Therefore the series is divergent.
For $0 < a < \mathrm{e}^{-1}$,\begin{align*}
\sum_{k = 0}^\infty a^{\sum\limits_{j = 0}^k \frac{1}{b + j}} &\leqslant \sum_{k = 0}^\infty a^{\ln(b + k + 1) - \ln b} = a^{-\ln b} \sum_{k = 0}^\infty a^{\ln(b + k + 1)}\\
&= a^{-\ln b} \sum_{k = 0}^\infty (\mathrm{e}^{\ln a})^{\ln(b + k + 1)} = a^{-\ln b} \sum_{k = 0}^\infty (\mathrm{e}^{\ln(b + k + 1)})^{\ln a} \\
&= a^{-\ln b} \sum_{k = 0}^\infty (b + k + 1)^{\ln a} < +\infty.
\end{align*}
For $\mathrm{e}^{-1} \leqslant a < 1$, because$$
\sum_{j = 0}^k \frac{1}{b + j} = \frac{1}{b} + \sum_{j = 1}^k \frac{1}{b + j} \leqslant \frac{1}{b} + \int_0^k \frac{\d x}{b + x} = \frac{1}{b} + \ln(b + k) - \ln b,
$$
then analogously,\begin{align*}
\sum_{k = 0}^\infty a^{\sum\limits_{j = 0}^k \frac{1}{b + j}} &\geqslant \sum_{k = 0}^\infty a^{\frac{1}{b} + \ln(b + k) - \ln b} = a^{\frac{1}{b} - \ln b} \sum_{k = 0}^\infty (b + k)^{\ln a} \geqslant a^{\frac{1}{b} - \ln b} \sum_{k = 0}^\infty \frac{1}{b + k} = +\infty.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
In the figure, how do I find the value of the variable?
According to the problem, I have to find the value of the variable in simplest radical form. I began with the equation 18√9 = 6 (1/3)x, so x = 9√3. Now, I know we have to find the area of the base in terms of x. But how do I find the height; do I use Pythagorean Theorem?
| \begin{gathered}
V\ =\ \frac{Bh}{3} \ \ \ where\ B\ is\ the\ base\ area\ of\ the\ equilateral\ triangle\ of\ length\ x\\
\\
B\ =\ \frac{1}{2} \ ( base\ triangle\ length)( base\ triangle\ height) \ =\ \ \frac{1}{2} \ ( x)\left(\frac{\sqrt{3} x}{2} \ \right) \ =\ \frac{\sqrt{3} x^{2}}{4}\\
\\
V\ =\ \frac{1}{3} \ *\ \frac{\sqrt{3} x^{2}}{4} \ *\ 6\ \ =\ \frac{6\sqrt{3} x^{2}}{12} \ =\ \frac{\sqrt{3} x^{2}}{2}\\
\ \ \\
\ 18\sqrt{3} \ =\ \frac{\sqrt{3} x^{2}}{2}\\
\\
36\ =\ x^{2} \ \ \ \ \ \ \ \ \ so\ \ \ \ \ \ \ \ x\ =\ 6\\
\end{gathered}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
If $\cos3A + \cos3B + \cos3C = 1$ in a triangle, find one of its length I would like to solve the following problem.
In $\triangle ABC, AC = 10, BC = 13$. If $\cos3A + \cos3B + \cos3C = 1$, compute the length of $AB$.
I thought that I could apply the Law of Cosines. Using the fact that $A+B+C=\pi$, I attempted to build the equation up from there.
What I got was that $$\cos3A+\cos3B-\cos(3A+3B)=1$$
Expanding, I got $$\cos3A+\cos3B-\cos3A\cos3B-\sin3A\sin3B=1$$
Now, it's possible that I'd be able to factor it somehow by rewriting it all in terms of cosine and arrive at the answer, but is there a better way to solve the problem? Thanks!
| We have $$2\cos\frac{3A+3B}{2}\cos\frac{3A-3B}{2}-2\sin^2\frac{3C}{2}=0$$ or
$$\cos\left(\frac{3\pi}{2}-\frac{3C}{2}\right)\cos\frac{3A-3B}{2}-\sin^2\frac{3C}{2}=0$$ or
$$\sin\frac{3C}{2}\left(-\cos\frac{3A-3B}{2}-\sin\frac{3C}{2}\right)=0$$ or
$$\sin\frac{3C}{2}\left(-\cos\frac{3A-3B}{2}+\cos\frac{3A+3B}{2}\right)=0$$ or
$$\sin\frac{3C}{2}\sin\frac{3C}{2}\sin\frac{3C}{2}=0$$ or
$$\prod_{cyc}\left(3\sin\frac{A}{2}-4\sin^3\frac{A}{2}\right)=0$$ or
$$\prod_{cyc}\left(3-4\sin^2\frac{A}{2}\right)=0$$ or
$$\prod_{cyc}\left(3-2(1-\cos{A})\right)=0$$ or
$$\prod_{cyc}\left(1+\frac{b^2+c^2-a^2}{bc}\right)=0$$ or
$$\prod_{cyc}(b^2+c^2+ab-a^2)=0.$$
Can you end now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find the largest real $k$ such that: $(a+b+c)^2(ab+bc+ca)\geq k(a^2b^2+b^2c^2+c^2a^2)$
Find the largest real $k$ such that for every non negative real numbers $a,b,c$ : $$(a+b+c)^2(ab+bc+ca)\geq k(a^2b^2+b^2c^2+c^2a^2)$$
I expanded the LHS but the problem got more complicated and no progress...
| For $c=0$ and $a=b=1$ we obtain $4\geq k$.
We'll prove that $4$ is a maximal value.
Indeed, by Muirhead we obtain:
$$(a+b+c)^2(ab+ac+bc)=\sum_{cyc}(a^2+2ab)\sum_{cyc}ab=$$
$$=\sum_{cyc}(a^3b+a^3c+a^2bc+2a^2b^2+4a^2bc)=\sum_{cyc}(a^3b+a^3c+2a^2b^2+5a^2bc)\geq$$
$$\geq\sum_{cyc}(4a^2b^2+5a^2bc)\geq4(a^2b^2+a^2c^2+b^2c^2)$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}=(m^2+2) \sqrt{m^2-1}$ Im trying to get from this expression into:
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}$$
this expression:
$$(m^2+2) \sqrt{m^2-1}$$
someone know how to do it?
i tried it for hours and can't get from the first expression into the second expression.
please explain to me step by step. I'm newbie.
| Simply note that
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}=\frac{(m^2 + 2)(m^2-1)^2}{(m^2-1)^{3/2}}=(m^2 + 2)(m^2-1)^{2-\frac32}=(m^2 + 2)(m^2-1)^{-\frac12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$2^{x-3} + \frac {15}{2^{3-x}} = 256$ $$2^{x-3} + \frac {15}{2^{3-x}} = 256$$
*
*Find the unknown $x$.
My attempt:
We know that $x^y . x^b = x^{y+b}$.
$$2^x . 2^{-3} + 15. 2^{-3+x} = 2^8$$
and
$$2^x . 2^{-3} + 15. 2^{-3} . 2^x = 2^8$$
From here, we get
$$2^x + 15 = 2^8$$
However, I'm stuck at here and waiting for your kindest helps.
Thank you.
| $2^{x-3} + 15\cdot 2^{x-3} = 256;$
$2^{x-3}(1+15) =256;$
$2^{x-3} =256/(16)= 16;$
$x-3=4$ , by inspection.
$x=7.$
Note:
$\dfrac{1}{2^{3-x}}= 2^{x-3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
limit with summation and product
Given $L=\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\bigg(\frac{1}{r!}\prod^{r}_{i=1}\left(\frac{i}{2}+\frac{1}{3}\right)\bigg)$. then $\lfloor L \rfloor$ is
Try: $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{1}{r!}\bigg[\left(\frac{1}{2}+\frac{1}{3}\right)\cdot \left(\frac{2}{2}+\frac{1}{3}\right)\cdots \cdots \left(\frac{r}{2}+\frac{1}{3}\right)\bigg]$$
could some help me to solve it, thanks
| For any $|a| < 1$,$$
\sum_{n = 1}^\infty \frac{1}{n!} \prod_{k = 1}^n (ak + b) = \sum_{n = 1}^\infty \frac{(-a)^n}{n!} \prod_{k = 1}^n \left( -\frac{b}{a} - k \right) = \sum_{n = 1}^\infty \frac{(-a)^n}{n!} \binom{-\frac{b}{a} - 1}{n}.
$$
Note that $|a| < 1$, by the generalized binomial theorem,$$
(1 - a)^{-\frac{b}{a} - 1} = \sum_{n = 0}^\infty \frac{(-a)^n}{n!} \binom{-\frac{b}{a} - 1}{n},
$$
thus$$
\sum_{n = 1}^\infty \frac{1}{n!} \prod_{k = 1}^n (ak + b) = \sum_{n = 0}^\infty \frac{(-a)^n}{n!} \binom{-\frac{b}{a} - 1}{n} - 1 = (1 - a)^{-\frac{b}{a} - 1} - 1.
$$
In this question, take $\displaystyle a = \frac{1}{2}$ and $\displaystyle b = \frac{1}{3}$, then$$
L = \sum_{n = 1}^\infty \frac{1}{n!} \prod_{k = 1}^n \left(\frac{k}{2} + \frac{1}{3}\right) = 2^{\frac{5}{3}} - 1,
$$
and $[L] = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Summation. What does is evaluate to? What is $\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}$ if $a_{n+2}=a_{n+1}+a_{n}$ and $a_{1}=a_{2}=1$?
| I found a nice answer, you guys might be curious to see it as well:
Let S be the sum.
Now $S=\frac{1}{16}+\frac{1}{64}+\frac{2}{256}+\frac{3}{1024}+\frac{5}{4096}$...
Multiply S by 4 to get that
$4S = \frac{1}{4}+\frac{1}{16}+\frac{2}{64}+\frac{3}{256}+\frac{5}{1024}$...
$3S = \frac{1}{4}+\frac{1}{64}+\frac{1}{256}+\frac{2}{1024}+\frac{3}{4096}$...
$3S = \frac{1}{4}+\frac{1}{4}S$
The rest is simplification... $S=\frac{1}{11}$.
Thanks for all the help guys!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\int \frac{1}{x^2+x+1} \, dx$ $$ \int \frac{1}{x^2+x+1}\, dx = \int \frac{1}{(x+\frac 1 2)^{2} + \frac 3 4}\, dx $$
Substitute $x+\frac 1 2 = u$, $dx = du$:
$$\int \frac 4 3 \frac{1}{\left(\frac{2u}{\sqrt 3}\right)^2+1} \, du = \frac 4 3 \int \frac{1}{\left(\frac{2u}{\sqrt 3}\right)^2+1} \, du$$
Substitute: $s = \frac{2u}{\sqrt 3}$, $du = \frac{\sqrt{3}}{2}ds$
$$\frac 4 3 \frac{\sqrt 3}{2}\int \frac{1}{s^2+1} \, ds$$
After multiplying and substituting back, we get the solution:
$$\frac{2}{\sqrt 3}\arctan\left(\frac{2x+1}{\sqrt 3}\right)$$
In a book I have however, this integral is evaluated as $\sqrt 3 \arctan\left(\frac{2x+1}{\sqrt 3}\right)$.
Which solution is right, and if the book's, how did the authors arrived to it?
| What you did is fine. The book, on the other hand…
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction: $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ The whole problem has been translated from German, so apologies if I made any mistakes. Thank you for taking the time to help!
So this is a problem from my math book
Prove that $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ for all $n\ \varepsilon \ N$
and I have come as far as this:
*
*let $n_0 = 1$. Then $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = (-1)^{1+1}1^2 = 1 = \frac{(-1)^{1+1}1(1+1)}{2} = 1 $
*For one $n\ \varepsilon \ N$ $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ is true.
*Now we have to show that $\sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \frac{(-1)^{(n+1)+1}(n+1)((n+1)+1)}{2} = \sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \frac{(-1)^{(n+2)}(n+1)((n+2)}{2}$ for all $n\ \varepsilon \ N$
From here, I did this: $\sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \sum\limits_{k=1}^{(n)} (-1)^{k+1}k^2 + (-1)^{(n+2)}(n+1)^2 = \frac{(-1)^{n+1}n(n+1)}{2} + (-1)^{(n+2)}(n+1)^2 = \frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2$
at which point I got stuck.
A look at the solutions at the back of the book told me that the next step is to factor (n+1) and $(-1)^{(n+2)}$, which resulted into this:
$\frac{(-1)^{n+2}(n+1)((-1)n+2(n+1))}{2} = \frac{(-1)^{n+2}(n+1)(-n+2n+2)}{2} = \frac{(-1)^{n+2}(n+1)(n+2)}{2}$
Which makes sense, but here is my question: How on earth do I get from $\frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2$ to $\frac{(-1)^{n+2}(n+1)((-1)n+2(n+1))}{2}$? I have been trying for quite a while and I just can't seem to figure it out. Any help is greatly appreciated!
| From here
$$= \frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2
=(-1)^{(n+2)}\left(\frac{-n(n+1)}{2} +\frac{2(n+1)^2)}2\right)
=(-1)^{(n+2)}\left(\frac{(n+1)(n+2)}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}$ for $a, b, c > 0$ with $abc = 1$
Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq\frac{3}{2}.$$
I tried homogenization and the BW (https://artofproblemsolving.com/community/c6h522084),
but it does not work.
Indeed, let $a=\frac{x}{y}$, $b=\frac{y}{z}$, where $x$, $y$ and $z$ are positives.
Hence, $c=\frac{z}{x}$ and we need to prove that $$\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}\leq\frac{3}{2},$$
which has a problem around $(x,y,z)=(7,5,6)$.
For these values $$\frac{3}{2}-\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}=0.0075...$$
I tried also TL, uvw, C-S, Lagrange multipliers and more, but without success.
Also, Vasc's Theorems don't help.
Also, the following method does not help here. Find the maximum of the expression
Because the inequality $\frac{x}{x^{11}+1}\leq\frac{3(a^9+1)}{4(a^{18}+a^9+1)}$ is wrong.
| Define
$$
f(a,\lambda) = -\frac{a}{a^{11}+1} + \lambda \log(a) + \frac{1}{2}
$$
Then, for any choice of $\lambda$,
$$
f(a,\lambda) + f(b,\lambda) + f(c,\lambda) = -\frac{a}{a^{11}+1} -\frac{b}{b^{11}+1} -\frac{c}{c^{11}+1} + \frac{3}{2}
$$
and we need to show that this is $\ge 0$.
It suffices to show that, for some $\lambda^*$ and for all $a$, $f(a, \lambda^*) \ge 0$.
Clearly, for any lambda, $f(a=1,\lambda) = 0$. In order to keep $f(a,\lambda) $ positive for $a >1$ and $a <1$, we demand
$$
0 = \frac{d f(a,\lambda)}{d a}|_{a=1}
$$
which results in $\lambda^* = - \frac94$. We therefore investigate
$$
f(a,\lambda^*) = -\frac{a}{a^{11}+1} -\frac{9}{4} \log(a) + \frac{1}{2}
$$
By inspection, we have that $f(a,\lambda^*) \ge 0$ for $a\in (0, 1.1]$. So the inequality is obeyed at least for $a,b,c < 1.1$, and it remains to be shown that the inequality is obeyed outside this specification.
This gives rise to three cases:
case 1: $a,b,c > 1.1$. This is not possible since $abc = 1$.
case 2: $a < 1.1$ ; $b,c > 1.1$. Now observe two facts:
*
*By inspection, $ \frac{a}{a^{11}+1} < 0.75$ for any $a$.
*For $b > 1.1$, $ \frac{b}{b^{11}+1} \le \frac{1.1}{1.1^{11}+1} \simeq 0.2855$ since $ \frac{b}{b^{11}+1}$ is falling for $b > 1.1$.
Hence, in case 2, $ \frac{a}{a^{11}+1} + \frac{b}{b^{11}+1}+ \frac{c}{c^{11}+1} < 0.75 + 2\cdot 0.2855 = 1.3210 < \frac32$ which proves case 2.
case 3: $a,b < 1.1$ ; $c > 1.1$. Here $abc = 1$ requires $a\cdot b =1/c < 1.1^{-1} = 0.909$. Also note that, for some given $c$, $1/(1.1 c) <a<1.1$ in order to observe $a,b < 1.1$. Following case 2, we have that $f(c) = \frac{c}{c^{11}+1} $ is falling with $c$. These conditions could be further exploited (this has not yet been pursued in the comments).
As Martin R. poined out, the maximum will be attained at a point where at least two out of $a,b,c$ equal. In this case, this would be $a=b$. So we can consider proving
$$
g(a) = \frac32 - \frac{2 a}{a^{11}+1} - \frac{a^{-2}}{a^{-22}+1} \ge 0
$$
for $a < 1/\sqrt{1.1} \simeq 0.9535$.
Note that in this range, the minimum of $g(a)$ occurs at $a^*\simeq 0.8385$ and has a value of $g(a^*) \simeq 0.00525$. Other than this inspection of the function $g(a)$, I couldn't offer a better proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 0
} |
Complex integral involving Cauchy integral formula Where $C$ is the circle $|z|=\frac{3}{2}$, evaluate the following integral using the Cauchy integral formula.
$$\int_{C}\frac{e^{z}}{(z^2+1)(z^{2}-4)}dz$$
Clearly the simple poles at $z=\pm i$ are the ones that are inside the circle $C$, and the simple poles at $z=\pm 2$ are outside the circle.
I use partial fractions to simplify $$\frac{e^{z}}{z^{2}-4}$$ to get it equal to $$\frac{e^{z}}{4(z-2)}-\frac{e^{z}}{4(z+2)},$$ then define $$f(z)=\frac{e^{z}}{4},$$ which gives us $$\frac{f(z)}{z-2}-\frac{f(z)}{z+2},$$ then by the C.I.F I get that
$$\int_{C}\frac{e^{z}}{(z^2+1)(z^{2}-4)}dz=\frac{\pi i}{2}[e^{i}-e^{-i}].$$
Could someone tell me if this is correct, thanks!
| Write the integral as a sum of integrals around $\{i,-i\}$. These integrals are easily evaluated using the Cauchy integral formula:
$$
\frac{1}{2\pi i}\oint_{|z-i|=1/2}\frac{1}{(z-i)}\frac{e^z}{(z+i)(z^2-4)}dz
= \frac{e^i}{(i+i)(i^2-4)} \\
\frac{1}{2\pi i}\oint_{|z+i|=1/2}\frac{1}{(z+i)}\frac{e^z}{(z-i)(z^2-4)}dz=\frac{e^{-i}}{(-i-i)((-i)^2-4)} \\
% \frac{1}{2\pi i}\oint_{|z-2|=1/2}\frac{1}{(z-2)}\frac{e^z}{(z+2)(z^2+1)}dz=\frac{e^2}{(2+2)(2^2+1)} \\
% \frac{1}{2\pi i}\oint_{|z+2|=1/2}\frac{1}{(z+2)}\frac{e^z}{(z-2)(z^2+1)}dz=\frac{e^{-2}}{(-2-2)((-2)^2+1)}
$$
The answer is the sum of these two integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2606994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Identify $\mathbb{Z}[x]/(x^2-3,2x+4)$. I need help to identify $\mathbb{Z}[x]/(x^2-3,2x+4)$.
I've been solving such problems in an approach like:
$$
2(x^2-3)=2x^2-6, x(2x+4)=2x^2+4x \\
(2x^2+4x)-(2x^2-6)=4x+6, 2(2x+4)=4x+8 \\
(4x+8)-(4x+6)=2
$$
What shall I do next, please? Thank you.
Simon
| First divide $x^2-3$ by $x+2$, dividing by a linear polynomial is the same as evaluating at the root of the linear term, so $x^2-3=4-3=1$. Thus $x^2-3=q(x)(x+2)+1$ for some $q(x)$. Multiplying by 2, and rearranging, we get $2(x^2-3)-q(x)(2x+4)=2$. Hence $2\in (x^2-3,2x+4)$, so since $2\mid 2x+4$, we have
$$(x^2-3,2x+4)=(x^2-3,2x+4,2)=(x^2-3,2).$$
Then $\newcommand{\ZZ}{\mathbb{Z}}\ZZ[x]/(x^2-3,2)\cong \ZZ_2[x]/(x^2-3),$
since the kernel of the natural map $\ZZ[x]\to \ZZ_2[x]/(x^2-3)$ is $(2,x^2-3)$.
Then $x^2-3\equiv x^2+1=x^2+1^2\equiv (x+1)^2\pmod{2}$, so
$$\ZZ_2[x]/(x^2-3)\cong \ZZ_2[x]/((x+1)^2).$$
Then by a change of variables, $$\ZZ_2[x]/((x+1)^2)\cong \ZZ_2[x]/(x^2).$$
I'm pretty sure it's impossible to simplify this further. The ring has four elements, $0,1,x,x+1$, only 0 and 1 are idempotents, since $x$ is nilpotent, so $x+1$ is a unit. Thus the ring is not a product of smaller rings.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2608636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $f\left(x\right)=$ $\sum_{n=0}^{\infty}$ $\frac{x^{3n}}{\left(3n\right)!}$ then prove $f''\left(x\right)+f'\left(x\right)+f\left(x\right)=e^{x}$
QuestionIf $f\left(x\right)=$$\sum_{n=0}^{\infty}$$\frac{x^{3n}}{\left(3n\right)!}$
then prove that
$f''\left(x\right)+f'\left(x\right)+f\left(x\right)=e^{x}$
My Approach$f\left(x\right)=$$\sum_{n=0}^{\infty}$$\frac{x^{3n}}{\left(3n\right)!}$
$f'\left(x\right)=$$\sum_{n=1}^{\infty}$$\frac{x^{3n-1}}{\left(3n-1\right)!}$
=$\sum_{n=0}^{\infty}$$\frac{x^{3n+2}}{\left(3n+2\right)!}$
$f''\left(x\right)=$$\sum_{n=2}^{\infty}$$\frac{x^{3n-2}}{\left(3n-2\right)!}$
=$\sum_{n=0}^{\infty}$$\frac{x^{3n+4}}{\left(3n+4\right)!}$
$f''\left(x\right)+f'\left(x\right)+f\left(x\right)=\sum_{n=0}^{\infty}\frac{x^{3n}}{\left(3n\right)!}+\sum_{n=2}^{\infty}\frac{x^{3n+2}}{\left(3n+2\right)!}+\sum_{n=0}^{\infty}\frac{x^{3n+4}}{\left(3n+4\right)!}$
Now i am unable to prove it to be equal to $e$$^{x}$
| Note that we can write: $$f(x) = 1 + \frac{x^3}{3!} + \frac{x^6}{6!} + \ldots $$ and $$f'(x) = \sum_{n=0}^{\infty} \frac{x^{3n+2}}{(3n+2)!}$$ $$ = \frac{x^2}{2!} + \frac{x^5}{5!} + \ldots$$ $$f''(x) = \sum_{n=0}^{\infty} \frac{x^{3n+1}}{(3n+1)!} $$ $$= x + \frac{x^4}{4!} +\ldots$$
Add them up to get $e^x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computing the series$\sum\limits_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} $ Show convergence of
$\begin{align}
\sum_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} &= \sum_{k=1}^{\infty}{\frac{3^{k-1}}{5^k}}+ \sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} \\
&= \sum_{k=1}^{\infty}{\frac{1}{3} \cdot \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\
&= \frac{1}{3} \cdot \sum_{k=1}^{\infty}{ \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\
\end{align}$
The first part of the sum converges because it is the geometric series with $q= \frac{3}{5}, 1> \left| \frac{3}{5} \right|$.
$$\sum_{n=1}^{\infty}{\left(\frac{2i}{5} \right)^k}$$
Question: Why does that series diverge (WolframAlpha)?
I mean, if $q=\frac{2i}{5}, |q|<1$ then it should be the geometric series and thus converge?
| by geometrics sum wfor any complex such that: $|z|<1$ we have
Therefore
$$\sum_{k=1}^{\infty} z^k= \lim_{n\to\infty}\sum_{k=1}^{n} z^k =\lim_{n\to\infty} \frac{z(1-z^{n+1})}{1-z} = \frac{z}{1-z} $$
since $\lim_{n\to\infty} z^n =0 $ since $|z|<1$.
Hence
\begin{align}
\sum_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} &= \sum_{k=1}^{\infty}{\frac{3^{k-1}}{5^k}}+ \sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} \\
&= \sum_{k=1}^{\infty}{\frac{1}{3} \cdot \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\
&= \frac{1}{3} \cdot \sum_{k=1}^{\infty}{ \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\&= \frac{1}{3}\frac{\frac{3}{5}}{1-\frac{3}{5}}+ \frac{\frac{2i}{5}}{1-\frac{2i}{5}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$=\frac{n^7}{12}$$
How can I reduce the numerator?
| You wish to derive a formula for the sum $\sum_{k=1}^n k^6$ using "high-school methods".
Consider $$\frac{x^7}7-\frac{(x-1)^7}7=x^6-3x^5+5x^4+\cdots\\\frac{x^6}6-\frac{(x-1)^6}6=x^5-\cdots\\\cdots$$
So, defining $$I_n:=\frac{n^7}7+\alpha_1\frac{n^6}6+\alpha_2\frac{n^5}5+\cdots+\alpha_6 n+\alpha_7$$ for some constants $\alpha_i$, you can eliminate all the other factors other than $n^6$. So you end up with $$I_n-I_{n-1}=n^6\\\implies \sum_{k=1}^n k^6=I_n-I_0=\frac{n^7}7+\text{lower order terms}$$
This is all that matters to taking your limit, since all the lower order terms go to zero when divided by the $n^7$ you have on the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find the minimum value of $|x+1|+2|x-5|+|2x-7|+|\frac{x-11}{2}|$
Find the minimum value of $|x+1|+2|x-5|+|2x-7|+|\frac{x-11}{2}|$.
I have no idea how to approach this question. However, I managed to solve it using a rather childish approach. I change this equation by multiplying $2$, getting $2|x+1|+4|x-5|+4|x-3.5|+|x-11|$. Now, I have a number line, and I mark out $-1$, $3.5$, $5$ and $11$. Here, I shall use this analogy which may seem childish. There are $2$ houses at $-1$, $4$ houses at $3.5$, $4$ houses at $5$, and $1$ house at $11$.
Here, I introduce a train station. We need to place it at an optimum place to minimise the distance from every house to it. If it is between $3.5$ and $5$, observe that it needs to move left to please more people(easier to board train). If it is placed between $-1$ and $3.5$, it needs to move right to please more people. Hence, the optimum place to put the train station would be at $3.5$, hence I have $x=3.5$, which is correct. You guys can draw a diagram to understand better. In addition, I noted that this seemed rather similar to some programming problems.
Back to the question, is there a much more elegant way to this problem?
| Case 1:$x\le -1\\f(x)=-x-1+10-2x+7-2x+5.5-0.5x=21.5-5.5x$
Case 2:$-1\le x\le 3.5\\f(x)=x+1+10-2x+7-2x+5.5-0.5x=23.5-3.5x$
Case 3:$3.5\le x\le 5\\f(x)=x+1+10-2x-7+2x+5.5-0.5x=9.5+0.5x$
Case 4:$5\le x\le 11\\f(x)=x+1-10+2x-7+2x+5.5-0.5x=-10.5+4.5x$
Case 5:$11\le x\\f(x)=x+1-10+2x-7+2x+0.5x-5.5=-21.5+5.5x$
so we have: $$\min_{x\in\Bbb R}f(x)=\min_{x\le-1}f(x)+\min_{-1\le x\le 3.5}f(x)+\min_{3.5\le x\le 5}f(x)+\min_{5\le x\le 11}f(x)+\min_{x\ge 11}f(x)=\min\lbrace{27,11.25,11.25,12,39}\rbrace=11.25$$
here is a sketch:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Geometry Problem: Find the area of $ABCD$ square
My Problem is:
The $E-$ point on the $AC$ diagonal is marked on the $ABCD$ square.If $BE= 13$, $CE=17$ find the square area.
My way:
$$a=\frac{k+17}{\sqrt2}$$
$$13^2=k^2+\frac {(k+17)^2}{2}-\frac {2k(k+17)}{\sqrt2}×\frac{\sqrt2}{2} \Rightarrow k^2+289=338 \Rightarrow k^2=49 \Rightarrow k=7$$
$$S_{ABCD}=\frac {(k+17)^2}{2}=\frac {(7+17)^2}{2}=288 $$
It looks ugly solution.
Is this solution correct?
| Let $F$ be the projection of $E$ on $BC$.
$CE=17$ implies $EF=FC=\frac{17}{\sqrt{2}}$. $BE=13$ and the Pythagorean theorem imply
$$ BF = \sqrt{13^2-\frac{17^2}{2}}=\frac{7}{\sqrt{2}} $$
hence
$$ BC=BF+FC = \frac{17+7}{\sqrt{2}} = 12\sqrt{2} $$
and the area of $ABCD$ is clearly $288$. You do not really need the cosine theorem, just the trick behind one of its proofs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve the integral $\int \frac{4+12x}{({1-2x-3x^2})^\frac{2}{3}}\,dx$ $$\int \frac{4+12x}{({1-2x-3x^2})^\frac{1}{3}}\,dx$$
So first I tried setting $u = 1-2x-3x^2, du = -2-6x\,dx$
Which gives
$$-2\int u^\frac{-1}{3}du = \frac{-6 u^\frac{2}{3}}{2} =-3u^\frac{2}{3}$$
And ultimately:
$$-3(1-2x-3x^2)^\frac{2}{3}$$
For some odd reason when I take the derivative I'm getting something completely different, what's wrong with my substitution?
| Let $u=1-2x-3x^2$, then $\mathrm{d}u=-\left(6x+2\right)\mathrm{d}x$
$$
\begin{align}
\int\frac{4+12x}{(1-2x-3x^2)^{1/3}}\,\mathrm{d}x
&=\int\frac{-2\,\mathrm{d}u}{u^{1/3}}\\
&=-3u^{2/3}+C\\[3pt]
&=-3\left(1-2x-3x^2\right)^{2/3}+C
\end{align}
$$
Looking at the answer in the question, this is pretty much exactly what was done. So the question is about what happened when the derivative was taken.
$$
\begin{align}
-3\frac{\mathrm{d}}{\mathrm{d}x}\left(1-2x-3x^2\right)^{2/3}
&=-2\left(1-2x-3x^2\right)^{-1/3}\frac{\mathrm{d}}{\mathrm{d}x}\left(1-2x-3x^2\right)\\
&=\frac{4+12x}{\left(1-2x-3x^2\right)^{1/3}}
\end{align}
$$
It's hard to tell what went wrong when you computed the derivative, since that wasn't shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
is there away to give $= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}$ question is there away to give
$$= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}$$
in just reals?
I am thinkting maybe use some clever algebra raise it to like 3 and magically imaginary terms dissapear. Or expressed the complex numbers in polar form but idk
Background (might delete if this is not really the question)
Let $$ m(x)=x^3-15x-10 $$
Find the roots of $m(x)$ using the cubic formula and show that they are all real
Using Howell complex approach
Lets change variables $$ \begin{aligned}
&z^3-15z-10
\\=&z^3+0z^2-15z-10
\end{aligned}$$
so, general form is denoted as
$$ z^3+a_2z^2+a_1z+a_0=0$$
so $$\begin{aligned}
a_2&=0
\\a_1&=-15
\\a_0&=-10
\end{aligned}$$
we get by substituting that $z=x-a_2/3$. $a_2=0$ its already depressed cubic but so theses steps are unnecesary for this one problem but will use them for other problems
$$ x^3+bx+c$$
where $$\begin{aligned}
b&=a_1-\frac{a_2^2}{3}
=-15-\frac{0^2}{3}=-15
\\c&=\frac{-a_1a_2}{3}+\frac{2}{27}a_2^3+a_0
=\frac{-(-15)(0)}{3}+\frac{2}{27}(0)^3+-10=-10
\end{aligned} $$
our depressed cubic is of $$ x^3-15x-10 $$
Ferro-Tartaglia Formula
$$ x=\sqrt[3]{\frac{-c}{2} + \sqrt{\frac{c^2}{4} +\frac{b^3}{27}}}\
+ \sqrt[3]{\frac{-c}{2} - \sqrt{\frac{c^2}{4} -\frac{b^3}{27}}}\ $$
in our case $$
\begin{aligned}
x&=\sqrt[3]{\frac{10}{2} + \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\
+ \sqrt[3]{\frac{10}{2} - \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\
\\&= \sqrt[3]{5 + \sqrt{25 -125}}
+ \sqrt[3]{5- \sqrt{25 -125}}
\\ &= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}
\end{aligned}
$$
| When there are $3$ real roots, Cardano-Tartaglia's formulæ can't be applied, and we need a trigonometric solution:
Setting $x=A\cos \theta\;$ ($A>0,\; 0\le \theta <\pi$), the equation becomes
$$A^3\cos^3\theta -15A \cos\theta=10.$$
We'll try to write the l.h.s. as $\;B\cos 3\theta=10\;$ to obtain an easy-to-solve trigonometric equation. For this , we need to satisfy the proportionality equation
$$\frac{A^3}4=\frac{15A}{3}\iff A^2=20\iff A=2\sqrt 5.$$
So the equation becomes
$$10\sqrt 5\cos 3\theta=10\iff\cos 3\theta=\frac{\sqrt 5}5.$$
The general solutions are
$$3\theta\equiv \pm\arccos\Bigl(\frac{\sqrt 5}5\Bigr)\mod 2\pi\iff \theta\equiv\pm\frac13\arccos\Bigl(\frac{\sqrt 5}5\Bigr)\mod\frac{2\pi}{3}. $$
One checks the three solutions in the interval $[0,\pi]$ are
$$ \theta= \frac13\arccos\Bigl(\frac{\sqrt 5}5\Bigr),\quad \theta=\frac13\arccos\Bigl(\frac{\sqrt 5}5\Bigr)+\frac{2\pi}3,\theta =-\frac13\arccos\Bigl(\frac{\sqrt 5}5\Bigr)+\frac{4\pi}3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the value of $\frac{a+b+c}{d+e+f}$ Given real numbers $a, b, c, d, e, f$, such that:
$a^2 + b^2 + c^2 = 25$
$d^2 + e^2 + f^2 = 36$
$ad + be + cf = 30$
What is the value of $\frac{a+b+c}{d+e+f}$?
I've tried combining equations in several ways but haven't gotten very far. Any hints would be appreciated.
| Hint: by the Cauchy-Schwarz inequality:
$$
900 = 30^2 = (ad + be + cf )^2 \le (a^2 + b^2 + c^2 )(d^2 + e^2 + f^2) = 25 \cdot 36 = 900
$$
Equality occurs when $a,b,c$ and $d,e,f$ are proportional.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Trigonometry problem, Evaluate: $\frac {1}{\sin 18°}$ My problem is,
Evaluate: $$\frac {1}{\sin 18°}$$
I tried to do something myself.
It is obvious,
$$\cos 18°= \sin 72°$$
I accept $\left\{18°=x \right\}$ for convenience and here, $\sin (x)>0$
$$\cos (x)=\sin (4x)$$
$$\cos (x)=2× \sin(2x) \cos (2x)$$
$$\cos (x)=2× 2\sin(x) \cos (x)×(1-2\sin^2(x)), \cos(x)>0$$
$$8\sin^3(x)-4\sin(x)+1=0$$
$$(2\sin(x)-1)(4\sin^2(x)+2\sin (x)-1)=0$$
$$4\sin^2(x)+2\sin (x)-1=0, \sin(x)≠\frac 12$$
$$4t^2+2t-1=0$$
$$t_{1,2}=\frac {-1±\sqrt 5}{4}$$
$$t=\frac {\sqrt5-1}{4} ,t>0$$
$$\sin 18°=\frac {\sqrt5-1}{4} .$$
Finally, $$\frac {1}{\sin 18°}=\frac {4}{\sqrt5-1}=\sqrt5+1$$
Is this way correct and is there a better/elegant way to do it? As always, it was an ugly solution.
Thank you!
| Note that $\forall x\in \mathbb{R}$
$$\sin x= \sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots.
$$
which converges “very fast for small” x.
In this case $x=\frac{\pi}{10}$ so we have that
$$ \sin \left(\frac{\pi}{10}\right) \approx \frac{\pi}{10} - \frac{\pi^3}{6000} + \frac{\pi^5}{10^5\cdot 5!}\approx 0.30901705...$$
while the exact value is
$$\sin \left( \frac{\pi}{10} \right)=0.30901699...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is wrong. How can I solve this without using the binomial theorem?
| by part method
$$\int(x^2 + 1)^7x^3dx=x^2*\frac{(x^2+1)^8}{2}-\int(x^2 + 1)^8xdx$$
$$x^2*\frac{(x^2+1)^8}{2}-\frac{1}{18}(x^2+1)^9+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Find all possible values of $ d_{17}$ knowing that $ (d_7)^2 + (d_{15})^2= (d_{16})^2$
Let $ d_1,d_2 \ldots, d_r$ be the positive divisors of $ n$. Namely,
$$ 1=d_1<d_2< \ldots <d_r=n$$
Now if $ (d_7)^2 + (d_{15})^2= (d_{16})^2$ find all possible values of $ d_{17}$
I don't know which tools should I use here. I know so far that all the prime divisors satisfy the condition $d^2\le \frac{n}2$
Patently this tells us that must be always some relationships between divisors of a number $n$ satisfying Pythagorian relation such as $ (d_7)^2 + (d_{15})^2= (d_{16})^2$.
Any hint or help is welcome
| Hint:
It is well-known, that if $a^2+b^2=c^2$ then $60|abc$ so $d_2=2,d_3=3,d_4=4,d_5=5,d_6=6$ and $d_7=7,8,9$ or $10$
Further Hint
If $d_7=7$ then $d_{16}-d_{15}=1$, hence $d_{16} + d_{15}=49 \implies d_{16}=25,d_{15}=24$ then $4200|n$ and $d_8=8$ and...
$10,12,14,15,20,21,24|n$ \to $d_{15} \leq 24$ but it means ,that $d_{9},d_{10},...,d_{15}=10,12,14,15,20,21,24$ and so $9,13 \not | n$ so $d_{17}=28$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Minimum distance from the points of the function $\frac{1}{4xy}$ to the point $(0, 0, 0)$ I am trying to find the minimum distance from the points of the function $\large{\frac{1}{4xy}}$ to the point $(0, 0, 0)$.
This appears to be a problem of Lagrange in which my condition: $C(x,y,z) = z - \frac{1}{4xy} = 0$, and my function would be $f(x,y,z) = \sqrt{x^2+y^2+z^2}$ or if i'm correct, it would be the same as the minimum value I get from using thee function as $f(x,y,z) = x^2+y^2+z^2$.
If I do this, I would then have:
$$
f(x,y,z,\lambda) = x^2 + y^2 + z^2 + \lambda(z-\frac{1}{4xy}) \\
= x^2 + y^2 + z^2 + \lambda z-\frac{1}{4xy} \lambda
$$
Then findind the partial derivatives ($\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}, \frac{\partial f}{\partial \lambda}$) and solving for the values of $x, y, z, \lambda$ and the minimum value I get at the end would be my answer.
Would that be the correct solution?
| Note that by substitution of the constraint we need to minimize
$$f(x,y)=\sqrt{x^2+y^2+z^2}=\sqrt{x^2+y^2+\frac{1}{16x^2y^2}}$$
and we don't need Lagrange's method.
In this case AM-GM is the most effective method, indeed
$$\frac{x^2+y^2+\frac{1}{16x^2y^2} }{3}\ge\sqrt[3]{x^2\cdot y^2\cdot \frac{1}{16x^2y^2}}\iff x^2+y^2+\frac{1}{16x^2y^2}\ge3\sqrt[3]{\frac1{16}}=\frac32\sqrt[3]{\frac1{2}}$$
and equality holds when
$$x^2=y^2=\frac{1}{16x^2y^2}\implies x=\pm\sqrt[6]\frac1{16}=\pm\sqrt[3]\frac14 \quad y=\pm x$$
Thus the minimum distance is
$$d_{min}=\sqrt{\frac32 \sqrt[3]{\frac1{2}} } =\sqrt{\frac32}\sqrt[6]{\frac1{2}}$$
attained for
$$P_1=\left(\sqrt[3]\frac14,\sqrt[3]\frac14,\frac14\sqrt[3]16\right)\quad P_2=\left(\sqrt[3]\frac14,-\sqrt[3]\frac14,-\frac14\sqrt[3]16\right)$$
$$P_3=\left( -\sqrt[3]\frac14,\sqrt[3]\frac14,-\frac14\sqrt[3]16\right)\quad P_4=\left( -\sqrt[3]\frac14,-\sqrt[3]\frac14,\frac14\sqrt[3]16\right)$$
As an alternative, since $f(x,x)>0$ its minimum coincide with the ninimum of
$$g(x,y)=f^2(x,y)=x^2+y^2+\frac{1}{16x^2y^2}$$
thus for the critical points we have
$$g_x=2 x - \frac1{8 x^3 y^2}=0$$
$$g_y=2 y - \frac1{8 x^2 y^3}=0$$
from which we obtain
$$x=\pm\sqrt[6]\frac1{16}=\pm\sqrt[3]\frac14 \quad y=\pm x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is there any trig identity which can be used for $\int \sin^4x$? What is $\int \sin^4x$?
I tried to use the identity $\sin^2x = \frac{1-\cos 2x}{2}$, but I'm stuck.
| Using the substitution above:
$\sin^4 x = (\frac {1-\cos 2x}2)^2 = \frac 14 - \frac 12 \cos 2x + \frac 14\cos^2 2x $
$\cos^2 2x = \frac {1+\cos 4x}{2}$
Or, using a little complex analysis
$\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\
\sin^4 x = \left(\frac {e^{ix} - e^{-ix}}{2i}\right)^4\\
\frac {e^{4ix} - 4e^{3ix}e^{-ix} + 6e^{2ix}e^{-2ix} - 4e^{ix}e^{-3ix} + 4e{-4ix})}{16}\\
\frac {(e^{4ix} + e^{-4ix})- 4(e^{2ix}+2e^{-2ix}) +6}{16}\\
\frac 18 \cos 4x - \frac 12 \cos 2x + \frac {3}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
limit without expansion I was solving this limit instead of using sum of expansion i did this and got zero but answer is 1/5 by expansion why this is wrong
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots + n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n^5}$$
$$= \lim_{n\rightarrow \infty } \frac{( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n}$$
$$= \frac{(0 + 0 +0 \ldots + 1 )}{n} = 0$$ this is how i got ,
| You can’t do this step
$$\lim_{n\rightarrow \infty } \frac{ \frac{1^4}{n^4} + \frac{2^4}{n^4} +...+1 }{n}\color{red}{=\frac{0 + 0 +0 +...+1 }{n}= 0} $$
since you have infinitely many terms which tend to 0 and their sum not necessarily is $0$.
Think to $\sum_1^{\infty} \frac1k$ which diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Limits connected to a right angled triangle Given a right angled triangle with sides $1,x$ and hypotenuse $y$. Let $\theta$ be the angle contained by side $1$ and hypotenuse. Then evaluate the following limits:
*$\lim_{\theta\to\pi/2}\sqrt{y}-\sqrt{x}$
*$\lim_{\theta\to\pi/2}y-x$
*$\lim_{\theta\to\pi/2}y^2-x^2$
*$\lim_{\theta\to\pi/2}y^3-x^3$
I was able to evaluate the third one, using $y=\sec{\theta}$ and $x=\tan{\theta}$.
EDIT:
I also solved the first first and second one. Only second and fourth are is unsolved.
How to evaluate others?
| First note that$$y^3-x^3=\dfrac{y^6-x^6}{x^3+y^3}=\dfrac{3y^4-3y^2+1}{x^3+y^3}\ge\dfrac{3y^4-3y^2+1}{2y^3}=1.5y-1.5\dfrac{1}{y}+\dfrac{1}{y^3}$$Also $\cos\theta=\dfrac{1}{y}\to 0^+$ leads to $y\to \infty$ and makes the limit $\infty$ either.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Origin Triangle Tetrahedron Volume I have a problem that goes like this:
Triangle $ ABC$, with sides of length $ 5$, $ 6$, and $ 7$, has one vertex on the positive $ x$-axis, one on the positive $ y$-axis, and one on the positive $ z$-axis. Let $ O$ be the origin. What is the volume of tetrahedron $ OABC$?
I really can't wrap my head around this problem. I thought I could find the area of the triangle base by using Heron's Formula, and then multiply that by $1/3$ of the height, but I've got no way of finding the height.
The posted solution for this question was very confusing.
It says:
wlog use $A(a,0,0), B(0,b,0), C(0,0,c)$
(i) wlog set $ 5 = \sqrt{a^2+b^2} \qquad 6 = \sqrt{b^2+c^2} \qquad 7 = \sqrt{c^2+a^2}$
(ii) Square for $ 25 = a^2+b^2 \qquad 36 = b^2+c^2 \qquad 49 = c^2+a^2$
(iii) Adding and dividing by 2 gives $ a^2+b^2+c^2 = 55$
(iv) Subtracting each element of (ii) from (iii) gives $ a^2 = 19 \qquad b^2 = 6 \qquad c^2 = 30$
(v) Multiplying gives $ a^2b^2c^2 = 19 \cdot 6 \cdot 30 = 19 \cdot 5 \cdot 6^2$
(vi) Squareroot and divide by 6 to get $ V = \frac{1}{6} abc = \sqrt{95} \implies \boxed{\text{C}}$
What I don't understand about this solution is why they are using these $a, b$, and $c$ coordinates and taking the square roots of two coordinates added together and setting them equal to the triangle's side lengths.
Perhaps I just don't properly understand how tetrahedrons and $x$-$y$-$z$ coordinates work. I'd really appreciate some help wrapping my head around this problem.
Thanks :)
| Because $A$ is on the $x$-axis, there exists a number $a$ such that the coordinates of $A$ are $(a,0,0)$. Similarly, there exist numbers $b$ and $c$ such that the coordinates of $B$ and $C$ are $(0,b,0)$ and $(0,0,c)$, respectively.
We know that $AB,BC,$ and $CA$ are $5,6,7$ in some order. We may assume $AB=5,BC=6,$ and $CA=7$, because otherwise we could relabel the tetrahedron so that this is the case.
Note that $\triangle OAB$ is a right triangle, so $OA^2+OB^2=AB^2$. But $OA=a,OB=b,$ and $AB=5$, so the equation becomes $a^2+b^2=25$.
Is the rest of the solution now clear?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Probability of being dealt a bridge hand with... Hey Guys I'd like to know if this question can be tackled this way.
What is the probability of being dealt a bridge hand with exactly 4 honour cards and exactly 4 cards from the 5 through 10?
I wrote that having in mind that the 10 is not included as it wasn't specified.
$$\frac{\binom{20}{4}\binom{20}{4}\binom{12}{5}}{\binom{52}{13}}$$
I wasn't really sure about choosing between the two fours. For example knowing which one is what. For that matter I wanted to go with the multinomial way.
$$\frac{\binom{40}{4,4}\binom{12}{5}}{\binom{52}{13}}$$
Any clarification would be appreciated.
| We split the bridge cards into $4$ mutually exclusive sets:
*
*$A$ is the set of $10$s; it consists of $4$ cards.
*$B$ is the set of honour cards which are not $10$s; it consists of $16$ cards.
*$C$ is the set of cards with number $5$ through $9$; it consists of $20$ cards.
*$D$ is the set of cards which are not in any of the sets above; it consists of $52-4-16-20=12$ cards.
Consider that a honour card is a card in the set $A$ or $B$, and a card with number $5$ through $10$ is a card in the set $A$ or $C$. That means we pick exactly $4$ cards in $A\cup B$ and exactly $4$ cards in $A\cup C$. That implies the number of cards we pick from $B$ and $C$ are the same.
Now we split cases on the number of members of $A$ picked:
*
*Case 1: $0$ in $A$, so $4$ in $B$, $4$ in $C$ and $5$ in $D$. There are $\binom{4}{0}\binom{16}{4}\binom{20}{4}\binom{12}{5}$ possibilities.
*Case 2: $1$ in $A$, so $3$ in $B$, $3$ in $C$ and $6$ in $D$. There are $\binom{4}{1}\binom{16}{3}\binom{20}{3}\binom{12}{6}$ possibilities.
*Case 3: $2$ in $A$, so $2$ in $B$, $2$ in $C$ and $7$ in $D$. There are $\binom{4}{2}\binom{16}{2}\binom{20}{2}\binom{12}{7}$ possibilities.
*Case 4: $3$ in $A$, so $1$ in $B$, $1$ in $C$ and $8$ in $D$. There are $\binom{4}{3}\binom{16}{1}\binom{20}{1}\binom{12}{8}$ possibilities.
*Case 5: $4$ in $A$, so $0$ in $B$, $0$ in $C$ and $9$ in $D$. There are $\binom{4}{4}\binom{16}{0}\binom{20}{0}\binom{12}{9}$ possibilities.
Now we add those numbers and divide by $\binom{52}{13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ How do we integrate the following?
$\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ given that $\cos 2x \gt 0$
I tried to simplify this, but I cannot seem to proceed further than the below form:
$\int{\frac{\sec2x}{\sqrt{2}}dx + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos2x}dx}}$
$\implies \frac{1}{2\sqrt2}\log |\sec 2x + \tan 2x| + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos^2x-\sin^2x}dx} + C$
The answer that I'm supposed to get is:
$\frac{x}{\sqrt2}+C$
Please help, thanks!
| Using $\cos2x=1-2\sin^2x=2\cos^2x-1,$
$$I=\dfrac{\sin^4x+\cos^4x}{\sqrt{1+\cos4x}}=\dfrac{(1-\cos2x)^2+(1+\cos2x)^2}{4\sqrt2|\cos2x|}=\dfrac{1+\cos^22x}{2\sqrt2|\cos2x|}$$
For $\cos2x>0,$
$$2\sqrt2I=\sec2x+\cos2x$$
Now use Integral of the secant function
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Solve congruence with large exponents I'm trying to solve:
$$ x^{1477} \equiv 54 \mod 97 $$
Applying Euler-Fermat gives:
$$
x^{1477} = x^{15\cdot 96 + 37} = x^{37}\cdot x^{{96}^{15}} \equiv x^{37}\cdot 1^{15} = x^{37} \mod 97
$$
So instead of solving $ x^{1477} \equiv 54 \mod 97 $ one can solve $ x^{37} \equiv 54 \mod 97 $. How could I proceed from here? I know about the Chinese Remainer Theorem and Euler-Fermat.
| If $x^{37} \equiv 54 \pmod {97}$ then $x^{37\cdot 24} \equiv 54^{24} \equiv 1 \pmod {97}$ because the order of $54 \pmod {97}$ is $24$.
This means the order of $x$ is a common divisor of $96$ and $37 \cdot 24$.
Therefore the order of $x$ is a divisor of $24$, and so $x^{24} \equiv 1 \pmod {97}$.
But if $k$ is a divisor of $24$ then $x^k \equiv 1 \pmod {97}$ implies $1 \equiv x^{37k} \equiv 54^k \pmod {97}$ so $24$ divides $k$.
Therefore the order of $x$ is $24$.
That means $x^{12} \equiv -1 \pmod {97}$.
So $54 \equiv x^{37} \equiv x^{1 + 3\cdot 12} \equiv x\left(x^{12}\right)^3\equiv x(-1)^3 \equiv -x \pmod {97}$
Finally, we obtain $x \equiv -54 \pmod {97}$, or $x \equiv 43$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $x+\frac1{4x} ≥ 1$ for $x>0$
Let $x$ be a real number such that $x > 0$. Prove that x +$\frac {1} {4x} ≥ 1$.
Not really sure on the correct way to approach it/is valid and could use some help.
Answer:
Proof Strategy: Proof by cases:
*
*$x = 1$
*$x > 1$
*$x < 1$
--
Case 1: $x = 1$
$x+\frac1{4x}$
$= (1) + 1/(4(1))$
$= 1.25 \ge 1$
Case is true
Case 2: $x > 1$, in this case $x = 2$.
$x+\frac1{4x} $
$= (2) + 1/(4(2))$
$= 2.125 \ge 1$
Case is true
Case 3: $x < 1$, in this case $x = 0.5$.
$= (0.5) + 1/(4(0.5))$
$= 1 \ge 1$
Case is true
Since all possible cases were satisfied therefore $x+\frac1{4x} \ge 1 $ when $x > 0$.
| For fun:
Let $x>0,$ real. Multiply by $4x$ :
$4x^2-4x +1 \ge 0.$
Need to show that above inequality is true for $x \gt 0$.
$4x^2 -4x +1 = 4(x^2 -x) +1 = 4(x-1/2)^2 \ge 0$
(why?).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
integral of $\cos (x+y+z)$ over a sphere Let $r>0$ and $S_r=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2=r^2\}$.
Show that
$$\int_{S_r}\cos x\cos y\cos z\ dS=\int_{S_r}\cos (x+y+z)\ dS$$
and find the value of the integrals.
I know that we can parametrize $S_r$ and try to solve the integrals by hand or try to expand the integrands using trigonometric identities, but I had no success in doing so.
I am asking for a hint on how to do this. Moreover, is there some way to handle with integrals over spheres that involve "cosines and sines" (before parametrization). What I mean by this is that when the integrand does not involve "cosines and sines", parametrization often helps but in a situation like the above I don't know what to do.
| As a partial answer I can show the first equation is valid:
On the upper surface of the sphere,
$$\vec r=\langle x,y,z\rangle=\langle x,y,\sqrt{r^2-x^2-y^2}\rangle$$
So
$$d\vec r=\langle 1,0,\frac{-x}{\sqrt{r^2-x^2-y^2}}\rangle dx+\langle 0,1,\frac{-y}{\sqrt{r^2-x^2-y^2}}\rangle dx$$
And then
$$\begin{align}d\vec A&=\pm\langle 1,0,\frac{-x}{\sqrt{r^2-x^2-y^2}}\rangle dx\times\langle 0,1,\frac{-y}{\sqrt{r^2-x^2-y^2}}\rangle dx\\
&=\pm\langle\frac{x}{\sqrt{r^2-x^2-y^2}},\frac{y}{\sqrt{r^2-x^2-y^2}},1\rangle dx\,dy\\
&=\langle\frac{x}{\sqrt{r^2-x^2-y^2}},\frac{y}{\sqrt{r^2-x^2-y^2}},1\rangle dx\,dy\end{align}$$
Because the outward normal points up here. Then
$$d^2A=\left|\left|d^2\vec A\right|\right|=\frac{r\,dx\,dy}{\sqrt{r^2-x^2-y^2}}$$
On the lower surface of the sphere,
$$\vec r=\langle x,y,z\rangle=\langle x,y,-\sqrt{r^2-x^2-y^2}\rangle$$
And we can work out
$$d^2A=\left|\left|d^2\vec A\right|\right|=\frac{r\,dx\,dy}{\sqrt{r^2-x^2-y^2}}$$
Just as it was on the upper surface.
Start with the right hand side:
$$\begin{align}\int\int_{S_r}\cos(x+y+z)d^2A&=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\left[\cos\left(x+y-\sqrt{r^2-x^2-y^2}\right)\right.\\
&+\left.\cos\left(x+y+\sqrt{r^2-x^2-y^2}\right)\right]\frac{r\,dy}{\sqrt{r^2-x^2-y^2}}dx\end{align}$$
The outer $(x)$ integral has a domain that is a circle parallel to the $yz$-plane and the inner integral just needs the two points that have the same $x$ and $y$. Now we can expand the cosines using the angle sum formula to get
$$\begin{align}\int\int_{S_r}\cos(x+y+z)d^2A&=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}2\cos(x+y)\cos\left(\sqrt{r^2-x^2-y^2}\right)\frac{r\,dy}{\sqrt{r^2-x^2-y^2}}dx\\
&=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\cos(x+y)\cos\left(-\sqrt{r^2-x^2-y^2}\right)\frac{r\,dy}{\sqrt{r^2-x^2-y^2}}dx\\
&+\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\cos(x+y)\cos\left(\sqrt{r^2-x^2-y^2}\right)\frac{r\,dy}{\sqrt{r^2-x^2-y^2}}dx\\
&=\int\int_{S_r}\cos(x+y)\cos z\,d^2A\end{align}$$
Let's set up the same integral with $x$ outer, $z$ middle and $y$ inner:
$$\begin{align}\int\int_{S_r}\cos(x+y)\cos z\,d^2A&=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\left[\cos\left(x-\sqrt{r^2-x^2-z^2}\right)\cos z\right.\\
&+\left.\cos\left(x+\sqrt{r^2-x^2-z^2}\right)\cos z\right]\frac{r\,dz}{\sqrt{r^2-x^2-z^2}}dx\\
&=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}2\cos x\cos\left(\sqrt{r^2-x^2-z^2}\right)\cos z\frac{r\,dz}{\sqrt{r^2-x^2-z^2}}dx\\
&=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\cos x\cos\left(-\sqrt{r^2-x^2-z^2}\right)\cos z\frac{r\,dz}{\sqrt{r^2-x^2-z^2}}dx\\
&+\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\cos x\cos\left(\sqrt{r^2-x^2-z^2}\right)\cos z\frac{r\,dz}{\sqrt{r^2-x^2-z^2}}dx\\
&=\int\int_{S_r}\cos x\cos y\cos z\,d^2A\end{align}$$
EDIT: Now we have to actually finish up the integral. We start out with the right hand side and transform to
$$\begin{align}u&=\frac1{\sqrt3}(x+y+z)\\
v&=\frac1{\sqrt6}(-2x+y+z)\\
w&=\frac1{\sqrt2}(z-y)\end{align}$$
This is just a rigid rotation about the origin, so
$$\begin{align}\int\int_{S_r}\cos(x+y+x)d^2A&=\int\int_{S_r}\cos\left(\sqrt3u\right)d^2A\\
&=\int_{-r}^r\int_{-\sqrt{r^2-u^2}}^{\sqrt{r^2-u^2}}2\cos\left(\sqrt3u\right)\frac{r\,dv}{\sqrt{r^2-u^2-v^2}}du\\
&=\int_{-r}^r\int_{-\pi/2}^{\pi/2}2\cos\left(\sqrt3u\right)\frac{r\sqrt{r^2-u^2}\cos\theta\,d\theta}{\sqrt{r^2-u^2}\cos\theta}du\\
&=2\pi r\int_{-r}^r\cos\left(\sqrt3u\right)du\\
&=\frac{2\pi r}{\sqrt3}\left.\sin\left(\sqrt3u\right)\right|_{-r}^r\\
&=\frac{4\pi r}{\sqrt3}\sin\left(\sqrt3r\right)\end{align}$$
EDIT: Pity that factor of $\frac r{\sqrt{r^2-u^2-v^2}}$ in the area element destroyed my beautiful solution with the Bessel functions and left us with this prosaic answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
P.M.F and expected value/expected payout The question is:
Let a random variable X be the number of days that a certain patient needs to be in the hospital. Suppose that X has the p.m.f.
$$\displaystyle f(x) = \frac{5 - x}{10}, \quad x = 1, 2, 3, 4 $$
If the patient is to receive 166 dollars from an insurance company for each of the first two days in the hospital and $118 for each day after the first two days, what is the expected payment for the hospitalization?
The given answer is 312.8
I got: 284.
How I got this:
$$E(X) = 166(1(\frac{5-1}{10}) + 2(\frac{5-2}{10})) + 118(3(\frac{5-3}{10}) + 4(\frac{5-4}{10})$$
$$E(X) = 166(0.4 + 0.6) + 118(0.6 + 0.4) $$
$$E(X) = 166 + 118 = 284 $$
What am I doing wrong?
|
What am I doing wrong?
$$E(X) = 166( 1(\frac{5-1}{10}) + 2(\frac{5-2}{10})) + \underbrace{118( 3(\frac{5-3}{10}) + 4(\frac{5-4}{10}))}$$
If you spend three or four days in the hospital, you pay \$166 for each of the first two days and \$118 for the remainder. Not $118 for each of the days.
$$E(X) = 166( 1(\frac{5-1}{10}) + 2(\frac{5-2}{10})) + (166\cdot 2+118)(\frac{5-3}{10}) + (166\cdot 2+118\cdot 2)(\frac{5-4}{10})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the distance between two parallel tangents of a rational function A Curve C has equation $y=\frac{2x-5}{x-1}$
a) A line y = mx+c is tangent to the curve. Find a condition for c in terms of m.
This can be solved relatively easily
$mx + c = \frac{2x-5}{x-1}$
$mx^2 + (c-m-2)x -c+5=0$
take discriminant as 0, as there is only one point at which the tangent intersects the curve
$c^2 +(2m-4)c +m^2 -16m+4=0$
By completing the square for c and m separately
$(c+m-2)^2-(m-2)^2+(m-8)^2-64+4=0$
$(c+m-2)^2 = 12m$
$c = 2-m\pm\sqrt{12m}$
b) Two parallel lines are each tangents to the curve, touching at points P(p,q) and R(r,s) with p<1
I have plotted this graph using desmos and attached an image of it
P can be taken to be a point in the top-left part of the graph and R a point in the bottom-right section of the graph.
If the distance $PR$ is $d$, find $d^2$ in terms of $m$, the gradient of each of the lines.
No calculus is required for this part.
$d^2=(p-r)^2 +(q-s)^2$
the tangents intersecting P and Q are given by:
$y = mx - mp + q$
$y = mx - mr +s$
$PR$ is perpendicular to the tangents
Therefore $grad_{PR} = -\frac{1}{m}$
Using the original equation:
$q = \frac{2p-5}{p-1}$
and $s = \frac{2r-5}{r-1}$
$q-s = \frac{3(p-r)}{(p-1)(r-1)}$
also take the leftmost tangent as $y_1$ and the other as $y_2$
$y_1 = mx - mp + \frac{2p-5}{p-1}$
$y_2 = mx - mr + \frac{2r-5}{r-1}$
$grad_{PR}=-\frac{1}{m} =\frac{p-r}{q-s} = \frac{(p-1)(r-1)}{3}$
$m = -\frac{3}{(p-1)(r-1)}$
$(q-s)^2 = \frac{9(p-r)^2}{(p-1)^2(r-1)^2}$
$(p-r)^2=\frac{(p-r)^2(p-1)^2(r-1)^2}{(p-1)^2(r-1)^2}$
$d^2 = \frac{(p-r)^2[9+(p-1)^2(r-1)^2]}{(p-1)^2(r-1)^2}$
$(p-1)^2(r-1)^2 = \frac{9}{m^2}$
$d^2 = (p-r)^2(m^2+1)$
using the conclusion from part a
$y_1 = mx + 2 - m - \sqrt{12m}$
$y_2 = mx+ 2- m + \sqrt{12m}$
Hence $ 2 - m - \sqrt{12m} = - mp + \frac{2p-5}{p-1}$
and $ 2- m + \sqrt{12m} =- mr + \frac{2r-5}{r-1} $
I still need to find $p-r$ in order to express $d^2$ in terms of $m$
I've been working on this problem for ages so any help would be much appreciated
The answer is
$d^2 = 4(\frac{3}{m}+3m)$
however, this is not particularly useful without a solution to it
c) By differentiating $d^2$ with respect to $m$, or otherwise, find the shortest distance between the two branches of the curve.
This part evidently does require calculus.
Any suggestions as to how I should answer this question would also be helpful.
| Given $y=\frac{2x-5}{x-1}$, the two tangents must have same slope:
$$m=y'(p)=\frac{3}{(p-1)^2}=y'(r)=\frac{3}{(r-1)^2} \Rightarrow p=-\sqrt{\frac{3}{m}}+1; r=\sqrt{\frac{3}{m}}+1 \ \ \ (*)$$
The squared distance between the tangents is
$$d^2=(p-r)^2+(q-s)^2=(p-r)^2+\left(\frac{2p-5}{p-1}-\frac{2r-5}{r-1}\right)^2.$$
Now, it remains to plug (*) in here.
Note: $p<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far:
(1) For n = 1:
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if divisible by $27$: $189$ mod $27$ = $0$
As no remainder is left, the base case is divisible by $27$.
(2) Assume $n = k$, then $2^{5k + 1} + 5^{k + 2} = 27k$
(3) Prove that this is true for n = k + 1:
$$2^{5(k + 1) + 1} + 5^{(k + 1) + 2} $$
$$= 2^{5k + 5 + 1} + 5^{k + 1 + 2} $$
$$ = 32 * 2^{5k + 1} + 5 * 5^{k + 2}$$
$$= ? $$
I know I am supposed to factor out 27 somehow, I just cant seem to figure it out. Any help would be appreciated.
| \begin{eqnarray*}
=\color{red}{27} \times 2^{5n+1}+5 \color{red}{(2^{5n+1}+5^{n+1})}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Find all $f$ that satisfies $f:\mathbb{R}\rightarrow\mathbb{R};f(x+y)+f(x)f(y)=(1+x)f(y)+(1+y)f(x)+f(xy)$ Find all $f$ that satisfies:
$1, ~f:\mathbb{R}\rightarrow\mathbb{R};\\
2,\forall x,y\in\mathbb{R},f(x+y)+f(x)f(y)=(1+x)f(y)+(1+y)f(x)+f(xy);
$
Maybe we can prove it's derivable or it's a linear function.
Any idea?
| The only solutions are $f(x)=0$, $f(x)=3x$, and $f(x)=x(x+1)$. It is easy to verify these all work; let me now prove there are no other solutions.
First note that setting $y=0$ gives $$f(x)f(0)=(2+x)f(0).$$ So either $f(0)=0$ or $f(x)=2+x$ for all $x$. Since $f(x)=2+x$ does not satisfy the functional equation, we have $f(0)=0$.
Now let $c=f(1)$. Setting $y=1$, we get $$f(x+1)=(3-c)f(x)+c(x+1).$$ If $c=3$, this tells us $f(x)=3x$ for all $x\in\mathbb{R}$.
Now suppose $c\neq 3$. The solutions $f(x)=0$ and $f(x)=x(x+1)$ correspond to $c=0$ and $c=2$, so let us first show these are the only possible values of $c$.
Setting $x=-1$ in the recurrence above, we get $$0=(3-c)f(-1)$$ and so $f(-1)=0$ since $c\neq 3$. Setting $y=-1$ in the original equation then gives $$f(x-1)=f(-x).$$ Setting $x=2$ gives $f(-2)=f(1)=c$. But evaluating $f(-2)$ using the recurrence, we get $$f(-2)=\frac{c}{3-c}.$$ Thus $$c=\frac{c}{3-c}$$ which implies $c=0$ or $c=2$.
Now suppose $c=0$. In that case, our recurrence is $f(x+1)=3f(x)$. We also have $f(-x)=f(x-1)=f(x)/3$. But then $f(x)=f(-(-x))=f(x)/9$, so $f(x)=0$ for all $x$.
This leaves the case $c=2$. In this case, the recurrence is $f(x+1)=f(x)+2x+2$. We also then have $f(-x)=f(x-1)=f(x)-2x$. Applying the original equation with $y=2$, using $f(x+2)=f((x+1)+1)=f(x)+4x+6$ and $f(2)=6$, gives $$f(2x)=4f(x)-2x.$$ Applying the original equation with $y=x$ then gives $$f(x^2)=f(2x)+f(x)^2-2(1+x)f(x)=f(x)^2-2xf(x)+2f(x)-2x.$$ On the other hand, applying the original equation with $y=-x$ gives $$f(x)f(-x)=(1+x)f(-x)+(1-x)f(-x)+f(-x^2).$$ Substituting $f(-x)=f(x)-2x$ and $f(-x^2)=f(x^2)-2x^2$ and solving for $f(x^2)$ gives $$f(x^2)=f(x)^2-2xf(x)-2f(x)+4x^2+2x.$$ Comparing our two equations for $f(x^2)$, we see that $4f(x)=4x^2+4x$ and so $f(x)=x(x+1)$ for all $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can verify this through expansion, but if I were a mathematician without access to CAS, how might I notice that this is the case?
I suppose what I'm asking is how one should "see" that $P$ can be simplified to $(1 + x + x^2 + x^3 + x^4 + x^5)^2$? Is it a multinomial thing (which seems a bit too complicated for someone to "notice"), or is there something simpler about the polynomial that one could use to factor it?
| The coefficient of $x^n$ is the number of integer compositions of $n$ into two parts where every part has length between $0$ and $5$. You can find the coefficients with a stars and bars approach. For every $n$, draw $n$ stars in a line, and see how many places it is possible to insert one bar such that there are between $0$ and $5$ stars on either side.
The reason this works is that in the expanded form, every instance of $x^n$ results from multiplying one of $\{x^0, x^1, x^2, x^3, x^4, x^5\}$ and another one of $\{x^0, x^1, x^2, x^3, x^4, x^5\}$. Thus to find the coefficient of $x^n$, it suffices to count the number of ways to get $n$ from a sum of two members of $\{0,1,2,3,4,5\}$, where the order of the sum matters.
I do realize that my and other answers address expanding the polynomial and not factoring it. I don't think there's such an easy trick to factoring it, and I think that seeing it can be factored just comes with experience and exposure to this kind of thing. (I mean, factoring an 11-digit square number isn't really easy, so why should factoring a degree-10 square polynomial be?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 7
} |
Deriving the closed form of $M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)$ I have the sequence, let $M_0=1$
$$M_{n+1} =\frac{2^{2n+1}}{M_n}\left(\sqrt{1+ 2^{-2n}M^2_{n}}-1\right)$$
Which I would like first to study the convergence and fine the closed form.
I failed to show that $M_n$ is bounded and monotone.
This could be easy if have the explicit expression of it.
Question:
Is there a closed a form of this sequence? does anyone has an idea?
FYI In the book it is mentioned that this sequence is used to approximate the area of the unit circle. may be some else has a more clever explanation to this connection
| Define $a_n = \dfrac{M_n}{2^n} \ (n \in \mathbb{N}_+)$, then$$
a_{n + 1} = \frac{\sqrt{1 + a_n^2} - 1}{a_n} > 0.
$$
Define $b_n = \sqrt{1 + a_n^2} \ (n \in \mathbb{N}_+)$, then$$
b_n > 1 \Longrightarrow a_n = \sqrt{b_n^2 - 1},
$$
and\begin{align*}
&\mathrel{\phantom{\Longrightarrow}}\sqrt{b_{n + 1}^2 - 1} = \frac{b_n - 1}{\sqrt{b_n^2 - 1}} = \sqrt{\frac{b_n - 1}{b_n + 1}}\\
&\Longrightarrow b_{n + 1}^2 = \frac{b_n - 1}{b_n + 1} + 1 = \frac{2b_n}{b_n + 1}.
\end{align*}
Define $c_n = \dfrac{1}{b_n} \ (n \in \mathbb{N}_+)$, then$$
c_{n + 1}^2 = \frac{1 + c_n}{2}.
$$
Because $c_1 = \dfrac{2}{\sqrt{5}} < 1$, denote $θ = \arccos \dfrac{2}{\sqrt{5}}$, then by induction on $n$, there is $0 < c_n < 1$. Define $θ_n = \arccos c_n \ (n \in \mathbb{N}_+)$, then$$
\cos θ_{n + 1} = \sqrt{\frac{1 + \cos θ_n}{2}} = \cos \frac{θ_n}{2} \Longrightarrow θ_{n + 1} = \frac{θ_n}{2}.
$$
Therefore, $θ_n = \dfrac{θ}{2^{n - 1}} \ (n \in \mathbb{N}_+)$, which implies$$
M_n = 2^n \tan \frac{θ}{2^{n - 1}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Proving that $\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 } $
Let $x,y,z\geq 1$ and $x+y+z=6$. Then $$\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 }. $$
I tried to use Cauchy- Schwartz inequality but it doesn't work.
| Also, the Tangent Line method works: $$\sum_{cyc}\frac{1}{3x^2+1}=\sum_{cyc}\left(\frac{1}{3x^2+1}-\frac{1}{13}+\frac{12}{169}(x-2)\right)+\frac{3}{13}=$$
$$=\sum_{cyc}\frac{3(x-2)^2(12x+11)}{169(3x^2+1)}+\frac{3}{13}\geq\frac{3}{13}>\frac{3}{16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove that $(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}$
If $x,y,z$ are real and $x^2+y^2+z^2=1$, prove that$$(x-y)(y-z)(z-x) \leq \frac{1}{\sqrt{2}}.$$
Equality is achieved in some strange cases: For example, if $x = -\dfrac{1}{\sqrt{2}}$, $y = 0$ and $z = \dfrac{1}{\sqrt{2}}$, then $(x-y)(y-z)(z-x)=\dfrac{1}{\sqrt{2}}$.
Note that the claim is obvious in the case when $x \geq y \geq z$ (since $(x-y)(y-z)(z-x)\leq 0$ in this case). But the inequality is not symmetric in $x, y, z$ (only cyclic). Thus, you cannot WLOG assume that $x \geq y \geq z$. (But you can WLOG assume that $x \leq y \leq z$ because of the previous observation.)
| If $\prod\limits_{cyc}(x-y)<0$ then it's obvious.
But for $\prod\limits_{cyc}(x-y)\geq0$ it's enough to prove that
$$(x^2+y^2+z^2)^3\geq2(x-y)^2(x-z)^2(y-z)^2.$$
Now, let $x\leq y\leq z$, $y=x+u$ and $z=x+v$.
Thus, we need to prove that
$$(3x^2+2(u+v)x+u^2+v^2)^3\geq2(u-v)^2u^2v^2$$ or
$$3x^2+2(u+v)x+u^2+v^2-\sqrt[3]{2(u-v)^2u^2v^2}\geq0,$$
for which it's enough to prove that
$$(u+v)^2-3\left(u^2+v^2-\sqrt[3]{2(u-v)^2u^2v^2}\right)\leq0$$ or
$$2(u^2-uv+v^2)\geq3\sqrt[3]{2(u-v)^2u^2v^2},$$ which is true by AM-GM:
$$2(u^2-uv+v^2)=2(u-v)^2+uv+uv\geq3\sqrt[3]{2(u-v)^2u^2v^2}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Finding the area of a region of circle Consider the following shape. It is requested to find the area of the shaded region. The radius of the circles and the overlapping half circles is 1 unit.
The region is irregular and I couldn't find a way to find it.
| Rotate the image, and place the origin of the Cartesian system at the center of the main circle. The other two circles will be centered at $(\pm 1/ \sqrt{2},-1/ \sqrt{2})$.
Obviously, it's enough to find half of the area for $x>0$.
For the green circle for $y>0$ we have:
$$y=\sqrt{1-x^2}$$
While for the right blue circle:
$$y =\sqrt{1-(x-1/ \sqrt{2})^2}-1/ \sqrt{2}$$
While polar coordinates are a good way to go, the problem can be solved as well in Cartesian.
We need to find the intersection point for $y>0$ of the two circles. Let's work with the equations:
$$y+1/ \sqrt{2} =\sqrt{1-x^2+\sqrt{2} x-1/2}=\sqrt{y^2+\sqrt{2} x-1/2}$$
Squaring:
$$y^2+\sqrt{2}y+1/2 =y^2+\sqrt{2} x-1/2$$
$$\sqrt{2}y =\sqrt{2} x-1$$
$$y =x-1/\sqrt{2}$$
From the second equation again:
$$y+1/ \sqrt{2} =\sqrt{1-y^2}$$
$$y^2+\sqrt{2}y+1/2 =1-y^2$$
$$2y^2+\sqrt{2}y-1/2 =0$$
$$y_0=\frac{\sqrt{3}-1}{2 \sqrt{2}}$$
$$x_0=\frac{\sqrt{3}+1}{2 \sqrt{2}}$$
Now all we need to do is integrate from $x=0$ to $x_0$ (subtracting the result for the blue circle from the green one):
$$S/2=\int_0^{x_0} \sqrt{1-x^2} dx-\int_0^{x_0}(\sqrt{1-(x-1/ \sqrt{2})^2}-1/ \sqrt{2}) dx=$$
$$=x_0/ \sqrt{2}+(x_0 \sqrt{1-x_0^2}+\arcsin x_0 )/2-(y_0 \sqrt{1-y_0^2}+\arcsin y_0 )/2+$$
$$+(-1/ \sqrt{2} \sqrt{1-1/2}+\arcsin (-1/ \sqrt{2}) )/2$$
$$S=\frac{\sqrt{3}+1}{2}+\frac{1}{4}+\frac{5 \pi}{12}-\frac{1}{4}-\frac{ \pi}{12}-\frac{1}{2}-\frac{ \pi}{4}$$
$$S=\frac{6 \sqrt{3}+\pi}{12}$$
Same result as Allawonder obtained.
This method is tedious (I have made liberal use of Wolfram Alpha), but we didn't need to think much.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding $\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$ Finding
$$\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$$
Try: I have tried it to convert it into $\displaystyle \left(x+\frac{1}{x}\right)=t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)$ but nothing happen.
I felt that it must be in Elliptical Integral of first kind, second kind or in third kind.
Can someone explain me how to write in elliptical form
| The other answers to this question use complex numbers. Here is a "real-only" answer.
Let us first work out
$$\int_y^\infty\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx\qquad y\ge0$$
"But wait a minute, isn't this always infinite?" Yes, but the integral from $0$ to $y$ (the generally implied bounds of the indefinite integral) can be obtained by a simple subtraction once we work out the antiderivative, so there's no problem.
The denominator quartic can be written as $(x^2+z^2)(x^2+\overline z^2)$ where $z=e^{i\pi/6}$. Thus Byrd and Friedman 225.09 applies:
$$\int_y^\infty\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=\frac12\int_0^{u_1}\left(\frac{1+\operatorname{cn}u}{1-\operatorname{cn}u}-1\right)\,du=\int_0^{u_1}\frac{\operatorname{cn}u}{1-\operatorname{cn}u}\,du$$
where $u_1=F(\varphi,m)$, $m=\frac14$ is the parameter and $\operatorname{cn}u_1=\cos\varphi=\frac{y^2-1}{y^2+1}$. B&F 361.61 provides the solution to this second integral:
$$\int_0^{u_1}\frac{\operatorname{cn}u}{1-\operatorname{cn}u}\,du=-E(\varphi,m)-\frac{\operatorname{sn}u_1\operatorname{dn}u_1}{1-\operatorname{cn}u_1}$$
We can work out $\operatorname{sn}u_1$ and $\operatorname{dn}u_1$ using the known value of $\operatorname{cn}u_1$ and the fundamental relations between Jacobian elliptic functions, yielding
$$\int_y^\infty\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=-E\left(\varphi,\frac14\right)-\frac{y\sqrt{y^4+y^2+1}}{y^2+1}=J(y)$$
The integral from $0$ to $y$ is then $J(0)-J(y)$ and $J(0)=-2E(1/4)$ (a complete integral), so
$$\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=-2E\left(\frac14\right)+E\left(\cos^{-1}\frac{y^2-1}{y^2+1},\frac14\right)+\frac{y\sqrt{y^4+y^2+1}}{y^2+1}$$
The quasiperiodicity of $E$ allows us to absorb $2E(1/4)$. Applying a few trigonometric identities then allows us to simplify the amplitude, in the process extending the domain of validity to all real numbers:
$$\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=E\left(\cos^{-1}\frac{y^2-1}{y^2+1}-\pi,\frac14\right)+\frac{y\sqrt{y^4+y^2+1}}{y^2+1}$$
$$\color{red}{\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=\frac{y\sqrt{y^4+y^2+1}}{y^2+1}-E\left(2\tan^{-1}y,\frac14\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2651254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Cosine of a matrix I came across this question, asked in a competitive exam. It is as follows.
Given a matrix $M = \begin{bmatrix}2&1\\1&2\end{bmatrix}$ what is the value of $cos(πM/6)$?
I've tried series expansion but I think there is an alternative way doing it, any help is appreciated.
Options given are
\begin{bmatrix}1/2&1\\1&1/2\end{bmatrix} \begin{bmatrix}\sqrt3/4&-\sqrt3/4\\-\sqrt3/4&\sqrt3/4\end{bmatrix} \begin{bmatrix}\sqrt3/4&\sqrt3/4\\\sqrt3/4&\sqrt3/4\end{bmatrix} \begin{bmatrix}1/2&\sqrt3/2\\\sqrt3/2&1/2\end{bmatrix}
| $$M=A+B$$
where
$$A=2I$$
and
$$B=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$$
We have
$$[A,B]=0$$
and
$$B^2=I$$
So
$$\exp\left(i\frac{\pi}{6}M\right)=\exp\left(i\frac{\pi}{6}2I\right)\exp\left(i\frac{\pi}{6}B\right)$$
$$=\exp\left(i\frac{\pi}{3}I\right)\left(I+i\frac{\pi}{6}B+\cdots\right)$$
$$=\exp\left(i\frac{\pi}{3}\right)I\left(I\cos(\pi/6)+iB\sin(\pi/6)\right)$$
$$=\exp(i\pi/3)\cos(\pi/6)I+i\exp(i\pi/3)\sin(\pi/6)B$$
$$=\left(\frac{\sqrt{3}}{4}+i\frac{1}{4}\right)I+\left(\frac{1}{4}i-\frac{\sqrt{3}}{4}\right)B$$
Similary
$$\exp\left(-i\frac{\pi}{6}M\right)=\left(\frac{\sqrt{3}}{4}-i\frac{1}{4}\right)I+\left(-\frac{1}{4}i-\frac{\sqrt{3}}{4}\right)B$$
Hence
$$\cos\left(i\frac{\pi}{6}M\right)=\frac{\sqrt{3}}{4}(I-B)$$
$$=\left(\begin{array}{cc}\frac{\sqrt{3}}{4}&-\frac{\sqrt{3}}{4}\\-\frac{\sqrt{3}}{4}&\frac{\sqrt{3}}{4}\end{array}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Repeated linear factors in partial fractions I have a question about the following partial fraction:
$$\frac{x^4+2x^3+6x^2+20x+6}{x^3+2x^2+x}$$
After long division you get:
$$x+\frac{5x^2+20x+6}{x^3+2x^2+x}$$
So the factored form of the denominator is
$$x(x+1)^2$$
So
$$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$$
Why is the denominator under $C$ not simply $x+1$? It is $x$ times $(x+1)^2$ and not $(x+1)^3$
| Explanation 1: If you have only $(x+1)$, your denominators are too weak:
$$\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+1} = \frac{A(x+1)}{x(x+1)}+\frac{Bx}{x(x+1)}+\frac{Cx}{x(x+1)}$$
Thus, you your denominator is quadratic whereas the desired denominator is $x(x+1)^2$, cubic. I mean, $\frac{C}{x+1}$ is redundant. Why? Just let $D=B+C$. Then we have $\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+1} = \frac{A}{x}+\frac{D}{x+1}$.
Explanation 2: Would you do this?
$$\frac{1}{x^2} = \frac{A}{x} + \frac{B}{x \ \text{and not} \ x^2 ?}$$
Explanation 3: Actually, some texts suggest
$$\frac{5x^2+20x+6}{x(x+1)^2} = \frac{\text{something}}{x}+\frac{\text{something}}{(x+1)^2}$$
where $\text{something}$ is an arbitrary polynomial with degree 1 lower than the denominator, i.e.
$$\frac{5x^2+20x+6}{x(x+1)^2} = \frac{D}{x}+\frac{Ex+F}{(x+1)^2}$$
This is actually equivalent to what your text suggests! How?
$$\frac{Ex+F}{(x+1)^2} = \frac{Ex+F+E-E}{(x+1)^2}$$
$$= \frac{E(x+1)+F-E}{(x+1)^2} = \frac{E}{x+1} + \frac{F-E}{(x+1)^2}$$
where $E=B$ and $F-E = C$
If you're confused about equivalence or of why there's a squared, I think it's best to just play safe:
Partial fractions, if I understand right, is pretty much an ansatz in basic calculus, iirc (the word 'ansatz' is used here, but I guess different from how I use it).
Instead of trying to understand, I just go on the safe side and do:
$$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{(x+1)^2}$$
Similarly,
$$\frac{5x^2+20x+6}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{(x)^2}$$
$$\frac{5x^2+20x+6}{x(x+1)^3}=\frac{A}{x}+\frac{B}{x+1}+\frac{Cx+D}{(x+1)^2}+\frac{Ex^2+Dx+F}{(x+1)^3}$$
Maybe there's some rule about how something is redundant, but I don't care (if I start caring, I may end up with your question) because I'm covering everything: Whenever there's an exponent in the denominator, I do as many fractions at that exponent. The additional coefficients will end up as zero anyway.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How can I show $(x^2+1, y^2+1)$ is not maximal in $\mathbb R[x,y]$? Is there a geometric interpretation for this?
How can I show $(x^2+1, y^2+1)$ is not maximal in $\mathbb R[x,y]$?
I know I can mod out the ideal one piece at a time and show $\mathbb C[x]/(x^2+1)$ is not a field since $(x^2+1)$ is not maximal in $\mathbb C[x]$, but is there another way of showing this?
| Let $I = (x^2+1,y^2+1)$, let $J=(x^2+1,x-y)$, and let $K=(x^2+1,x+y)$.
If $1 \in J$, then $a(x^2+1)+b(x-y)=1$, for some $a,b \in \mathbb{R}[x,y]$. But then, letting $y=x$, we get $a(x,x)(x^2+1)=1$, contradiction, since a nonzero multiple of $x^2+1$ must have degree at least $2$.
If $1 \in K$, then $c(x^2+1)+d(x+y)=1$, for some $c,d \in \mathbb{R}[x,y]$. But then, letting $y=-x$, we get $c(x,-x)(x^2+1)=1$, contradiction, since a nonzero multiple of $x^2+1$ must have degree at least $2$.
Thus, $J$ and $K$ are proper ideals.
Since $x-y \in J$, we get $(x+y)(x-y) \in J$, hence $x^2-y^2\in J$. Then since $y^2+1 = (x^2 + 1) - (x^2 - y^2)$, we get that $y^2 + 1 \in J$. It follows that $I \subseteq J$.
Since $x+y \in K$, we get $(x+y)(x-y) \in K$, hence $x^2-y^2\in K$. Then since $y^2+1 = (x^2 + 1) - (x^2 - y^2)$, we get that $y^2 + 1 \in K$. It follows that $I \subseteq K$.
Suppose $I$ is maximal. Then we must have $J=I$ and $K=I$.
From $J=I$, we get $x-y\in I$, and from $K=I$, we get $x+y\in I$.
From $x-y \in I$ and $x+y \in I$, we get $x \in I$.
But then, from $x \in I$ and $x^2+1\in I$, we get $1 \in I$, contradiction.
It follows that $I$ is not maximal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Expected number of turns in dice throwing I generated a transition probability matrix for a scenario where I throw five dice and set aside those dice that are sixes. Then, I throw the remaining dice and again set aside the sixes - then I repeat this procedure until I get all the sixes. $X_n$ here represents the number of dices that are sixes after n rolls.
$$\begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} & \frac{1250}{6^5} & \frac{250}{6^5} & \frac{25}{6^5} & \frac{1}{6^5}\\\ 0 & \frac{625}{6^4} & \frac{500}{6^4} & \frac{150}{6^4} & \frac{20}{6^4} & \frac{1}{6^4} \\\ 0& 0 & \frac{125}{6^3}& \frac{75}{6^3}& \frac{15}{6^3} & \frac{1}{6^3} \\\ 0 & 0& 0& \frac{25}{6^2}& \frac{10}{6^2}& \frac{1}{6^2}& \\ 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6} \end{pmatrix}$$
I want to figure out how many turns it takes for me on average to get all sixes.
I'm not even sure where to start with this problem. Is it a right approach to write a program where I calculate $P^n$ and see when the 6th column all equals to 1?
Any pointers would be greatly appreciated.
| Yet another way:
Let's call rolling all five dice a "turn". Let $X$ be the number of the first turn on which all five dice have rolled at least one six. If $X>n$ then we have at least one die which has not rolled a six by turn $n$. So
$$\begin{align}
E(X) &= \sum_{n>0} P(X > n) \tag{1} \\
&= \sum_{n=0}^{\infty} \{1 - [1-(5/6)^n)]^5 \} \\
&= \sum_{n=0}^{\infty} \left( 1 - \sum_{i=0}^5 (-1)^i \binom{5}{i} (5/6)^{ni} \right) \tag{2}\\
&= \sum_{i=1}^5 (-1)^{i+1} \binom{5}{i} \sum_{n=0}^{\infty} (5/6)^{ni} \\
&= \sum_{i=1}^5 (-1)^{i+1} \binom{5}{i} \frac{1}{1-(5/6)^i} \tag{3}\\
&= 13.02366
\end{align}$$
$(1)$ is true for any discrete random variable which only takes on non-negative values.
$(2)$ is by the binomial theorem.
$(3)$ is by the formula for the sum of an infinite geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding value of product of Cosines
Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$
My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\sin \frac{\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
So we have $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
Could some help me to solve it, Thanks in Advanced
| Hint:
Using $\cos2x=1-2\sin^2x,$
$$\sin3x=3\sin x-4\sin^3x=\cdots=\sin x(1+2\cos2x)$$
Put $2x=\dfrac\pi{20},\dfrac{3\pi}{20},\dfrac{9\pi}{20},\dfrac{27\pi}{20}$ one by one to recognize the Telescoping nature.
Finally use $\sin(2m\pi+y)=\sin y$ where $m$ is any integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Inequality with positive reals For each $x,y,z>0$, define
$$
f(x,y,z)=\frac{3x^2(y+z)+2xyz}{(x+y)(y+z)(z+x)}.
$$
Fix also $a,b,c,d>0$ with $a\le b$. How can we prove "reasonably" the following inequality?
$$
\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)} \le \frac{a+c}{b+c}.
$$
| The expression
$$\frac{f(a,b,d)+f(c,b,d)}{f(b,a,d)+f(c,a,d)}
\le \frac{a+c}{b+c}$$ is equivalent to
$$(b+c)(f(a,b,d)+f(c,b,d))\le(a+c)(f(b,a,d)+f(c,a,d))$$ The $LHS$ is equal to
$$\frac{3a^2(b+c)}{(a+b)(a+d)}+\frac{2adb^2+2abcd}{(a+b)(a+d)(b+d)}+\frac{3c^2}{c+d}+\frac{2bcd}{(b+d)(c+d)}$$
The $RHS$ is equal to
$$\frac{3b^2(a+c)}{(a+b)(b+d)}+\frac{2bda^2+2abcd}{(a+b)(a+d)(b+d)}+\frac{3c^2}{c+d}+\frac{2acd}{(a+d)(c+d)}$$ so we have to prove that
$$\frac{3a^2(b+c)}{(a+b)(a+d)}+\frac{2adb^2}{(a+b)(a+d)(b+d)}+\frac{2bcd}{(b+d)(c+d)}$$ is less or equal than
$$\frac{3b^2(a+c)}{(a+b)(b+d)}+\frac{2bda^2}{(a+b)(a+d)(b+d)}+\frac{2acd}{(a+d)(c+d)}$$ Multiplying by $(a+b)(a+d)(b+d)(c+d)$ one has respectively
$$(b+d)(c+d)(b+c)(3a^2)+(c+d)(2adb^2)+(a+b)(a+d)(2bcd)$$ and
$$(a+d)(a+c)(c+d)(3b^2)+(c+d)(2bda^2)+(a+b)(b+d)(2acd)$$
What precedes becomes after simplification $$a^2(b+c)(b+d)\le b^2(a+c)(a+d)$$ where $a,b,c,d\gt 0$ and $a\le b$.
Finally we have the evident inequality
$$\color{red}{0\le ab(b-a)(c+d)+(b^2-a^2)cd}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $|z-10|=3|z-2$| is the equation of a circle with radius $3$ and center $1$.
Prove that $|z-10|=3|z-2$| is the equation of a circle with radius $3$ and center $1$.
I tried the following solution, with no result:
Considering $z=x+yi$,
$$|x+yi-10|=3|x+yi-2|$$
$$|(x-10)+yi|=3|(x-2)+yi|$$
$$\sqrt{(x-10)^2+y^2}=3\sqrt{(x-2)^2+y^2}$$
power of $2$:
$$(x-10)^2+y^2=9((x-2)^2+y^2)$$
after simplification, we will have:
$$8x^2+8y^2-16x-64=0$$
or: $$x^2+y^2-2x-8=0$$
I tried $z=x-yi$, as well. But there is no way I can come up with the intended result, that is, radius of $3$ and center of $1$. Any help would be appreciated.
| You are almost at end of your exercise. Just notice that
$$0=x^2+y^2-2x-8=(x-1)^2+y^2-9=|z-1|^2-3^2\Leftrightarrow |z-1|=3$$
where $z=x+iy$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find $\lim_\limits{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$
How to find $$\lim_{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$$
My attempt:
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n+4\right)}=\lim _{n\to \infty }\left(\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n\right)}+\left(1-\frac{5}{n^4}\right)^4\right)=\lim _{n\to \infty }\left(\left(\left(1-\frac{5}{n^4}\right)^{n}\right)^{4*2018}+\left(1-\frac{5}{n^4}\right)^4\right)$$
so we get :
*
*$\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^4=1$
let:
$b_n=\frac{-5}{n^3}$
$\lim _{n\to \infty }\frac{-5}{n^3}=0$
*
*$\lim _{n\to \infty }\left(\left(1-\frac{5}{n^4}\right)^n\right)^{4\cdot 2018}=\lim \:_{n\to \:\infty \:}\left(\left(1+\frac{b_n}{n}\right)^n\right)^{4\cdot \:2018}=\left(e^0\right)^{4\cdot 2018}=1$
so the answer is :
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=1+1=2$$
Is this answer correct if not how can I find the limit ?
thanks.
| The result seems to me weird. You wrote $\displaystyle \left(2018n+1\right)^4=4\left(2018n+1\right)$ which is false. And you when an expression is at power $n$, you cannot say the argument tends to $0$ hence it tends to $0^n$.
I would rather write
$$
\left(1-\frac{5}{n^4}\right)^{\left(2018n+1\right)^4}=e^{\left(2018n+1\right)^4\ln\left(1-\frac{5}{n^4}\right)}
$$
Then
$$
\ln\left(1-\frac{5}{n^4}\right)\underset{(+\infty)}{=}-\frac{5}{n^4}+o\left(\frac{1}{n^4}\right)
$$
and
$$ \left(2018n+1\right)^4=2018^4n^4+o\left(n^4\right)=16583822760976n^4+o\left(n^4\right)$$
Finally
$$
\left(1-\frac{5}{n^4}\right)^{\left(2018n+1\right)^4}\underset{n \rightarrow +\infty}{\rightarrow}e^{-82919113804880}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find all $z$ s.t $|z|=1$ and $|z^2 + \overline{z}^2|=1$ Here's my attempt. Let $z=x+i\ y$, then $$z^2=x^2-y^2+i\ 2xy$$ and $$\bar z^2=x^2-y^2-i\ 2xy$$
Then, $$z^2+\bar z^2=2x^2-2y^2$$ so $$1=|z^2+\bar z^2|=\sqrt{(2x^2)^2+(-2y^2)^2}$$
Simpliflying the expression above, we get $$1=4x^4+4y^4$$
which gives us $$\frac14=x^4+y^4$$. I am stuck here. Is it wrong approach? is there an easier one?
| Since $w+\overline{w}\in \mathbb{R}$ for each $w\in \mathbb{C}$
we have $z^2+ \overline{z}^2 =1$ and from $|z|=1$ we have $z\overline{z}=1$.
So $$(z+\overline{z})^2= z^2+2z\overline{z} +\overline{z}^2 =3$$
and thus we have $z+\overline{z} =\pm \sqrt{3}$
So, since $\overline{z} = {1\over z} $ we have $z^2\pm \sqrt{3}z+1=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find the limit of $\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$ I was trying to solve the following task but I stumbled across something I do not understand:
Calculate:
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
my attempt was to factorize n^2 out of the squareroot:
$$$$
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
\begin{align}
\\a_n & = \frac{1}{n^2-\sqrt{n^4+4n^2+n}} \\
& =\frac{1}{n^2-\sqrt{n^4\left(1+\frac{4}{n^2}+\frac{1}{n^3}\right)}} \\
& =\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}} \\
\end{align}
Therefor, I thought that:
$$\lim_{n \to \infty}\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}}$$
$$ = \lim_{n \to \infty}\frac{1}{\left(n^2-n^2\right)\sqrt{1}} = \infty$$
I also tried a different way where I got to the result of $-\dfrac{1}{2}$.
I am not going to show that method here but it starts with using the 3rd binomial formula. Then, having the squareroot at the top of the fraction, I factorized $n^2$ and it all worked.
Why does this method like shown above not work?
I am very happy for any help.
P.S. This is not the only example where this kind of getting to a solution does not work for me. Are there cases where I am not allowed to factorize something?
| HINT
Use
$$\frac{1}{n^2-\sqrt{n^4+4n^2+n}}\frac{n^2+\sqrt{n^4+4n^2+n}}{n^2+\sqrt{n^4+4n^2+n}}=\frac{n^2+\sqrt{n^4+4n^2+n}}{-4n^2-n}=\frac{1+\sqrt{1+4/n^2+1/n^3}}{-4-1/n}$$
In your attempt the following
$$n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}=n^2\cdot \sqrt1$$
is uncorrect, indeed by binomial series expansion
$$\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}=1+\frac{2}{n^2}+o(1/n^2)$$
from here you can conclude by your attempt.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
What is the value of $\sin^2 (\frac{\pi}{10}) \sin^2 (\frac{3\pi}{10})$?
PROBLEM
$$ \prod_{i=0}^4 \left(1 + \cos \left(\frac{(2k+1)\pi}{10}\right)\right)$$
My Try
$$
\left(1 + \cos \frac{\pi}{10}\right)
\left(1 + \cos \frac{9\pi}{10}\right)
\left(1 + \cos \frac{7\pi}{10}\right)
\left(1 + \cos \frac{3\pi}{10}\right)
= \sin^2 \left(\frac{\pi}{10}\right)
\sin^2 \left(\frac{3\pi}{10}\right)
$$
I am not able to proceed further.
Please help me.
| use that $$\sin\left(\frac{\pi}{10}\right)=\frac{\sqrt{5}-1}{4}$$
$$\sin\left(\frac{3\pi}{10}\right)=\frac{\sqrt{5}+1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
For real numbers $x,y$ satisfy the condition $x^2+5y^2-4xy+2x-8y+1=0$. Find the Maximum and Minimum Value of the expression $A=3x-2y$. For real numbers $x,y$ satisfy the condition $x^2+5y^2-4xy+2x-8y+1=0$.Find the Maximum and Minimum Value of the expression $A=3x-2y$.
| Using secondary school "tools", you can replace
$$
x = \left( {A + 2y} \right)/3
$$
into the quadric equation, to get, after some simple simplifications
$$
25y^{\,2} - \left( {8A + 60} \right)y + \left( {8A + 60} \right)^{\,2} = 0
$$
whose solution is
$$
\eqalign{
& y = {{2\left( {2A + 15} \right) \pm \sqrt {\left( {8A + 60} \right)^{\,2} - \left( {10A + 30} \right)^{\,2} } } \over {25}} = \cr
& = {{2\left( {2A + 15} \right) \pm \sqrt {\left( {18A + 90} \right)\left( { - 2A + 30} \right)} } \over {25}} = \cr
& = {{2\left( {2A + 15} \right) \pm 3\sqrt {\left( {2A + 10} \right)\left( { - 2A + 30} \right)} } \over {25}} \cr}
$$
For the solutions to be real, we must have
$$
0 \le \left( {2A + 10} \right)\left( { - 2A + 30} \right)\quad \Rightarrow \quad - 5 \le A \le 15
$$
corresponding to the $(x,y)$ values
$$
\left\{ \matrix{
A = - 5 \hfill \cr
y = 10/25 = 2/5 \hfill \cr
x = \left( { - 5 + 4/5} \right)/3 = - 7/5 \hfill \cr} \right.\quad \left\{ \matrix{
A = 15 \hfill \cr
y = 90/25 = 18/5 \hfill \cr
x = \left( {15 + 36/5} \right)/3 = 37/5 \hfill \cr} \right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Here is what I have done so far:
\begin{align}
\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}
&=
\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)\\
&=\lim_{x\to 0} \left(\frac{3^x-1}{x}\frac{x}{4^x+2^x-2}+\frac{2^x-1}{x}\frac{x}{4^x+2^x-2}\right)\\
&=\ln 3\lim_{x\to 0} \frac{x}{4^x+2^x-2}+\ln 2\lim_{x\to 0} \frac{x}{4^x+2^x-2}
\end{align}
| \begin{align}\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}&=\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)
\\&=\lim_{x\to 0} \left(\frac{3^x-1}{x}\frac{x}{4^x+2^x-2}+\frac{2^x-1}{x}\frac{x}{4^x+2^x-2}\right)
\\&=\ln 3\lim_{x\to 0} \frac{x}{4^x+2^x-2}+\ln 2\lim_{x\to 0} \frac{x}{4^x+2^x-2} \\&= (\ln 3 + \ln 2) \lim_{x\to 0} \frac{x}{4^x+2^x-2}
\\&= (\ln 3 + \ln 2) \lim_{x\to 0} \frac{1}{\frac{4^x-1}{x}+\frac{2^x-1}{x}} \\
&=\frac{\ln 3 + \ln 2}{\ln 4 + \ln 2} \\
&= \frac{\ln 6}{\ln 8}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solving this system of equations without a CAS: $4=a+c$, $5=d+ac+b$, $4=ad+bc$, $4=bd$ Solving this system of equations :
$$4=a+c$$ $$5=d+ac+b$$ $$4=ad+bc$$ $$4=bd$$
Anyone knows how to solve this without a CAS?
| We obtain $$c=4-a$$ and $$d=\frac{4}{b},$$ which gives
$$\frac{4}{b}+b+a(4-a)=5$$ and
$$a\cdot\frac{4}{b}+b(4-a)=4.$$
From the last equation we obtain:
$$a=\frac{4b^2-4b}{b^2-4},$$ which gives
$$\frac{4}{b}+b+\frac{4b^2-4b}{b^2-4}\left(4-\frac{4b^2-4b}{b^2-4}\right)=5$$ or
$$\frac{b^2-5b+4}{b}+\frac{16b(b-4)(b-1)}{(b^2-4)^2}=0$$ or
$$(b-1)(b-4)\left(\frac{1}{b}+\frac{16b}{(b^2-4)^2}\right)=0$$ or
$$(b-1)(b-4)=0$$ because $$(b^2-4)^2+16b^2>0.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
For what values $s$ does $\int_{0}^{\infty} \frac{\sin x}{x^s}dx$ converge? For what values $s$ does $\int_{0}^{\infty} \frac{\sin x}{x^s}dx$ converge?
Solution says $0<s<2$. Any ideas?
| Let us split the integral into two parts.
\begin{align}
\int^\infty_0 \frac{\sin x}{x^s}\ dx = \int^\pi_0 \frac{\sin x}{x^s}\ dx + \int^\infty_\pi \frac{\sin x}{x^s}\ dx
\end{align}
where
\begin{align}
\int^\pi_0 \frac{\sin x}{x^s}\ dx=\int^\pi_0 \frac{\operatorname{sinc} x}{x^{s-1}}\ dx< \infty \ \ \ \text{ if and only if }\ \ \ s<2.
\end{align}
Hence it remains to show that
\begin{align}
\int^\infty_\pi \frac{\sin x}{x^s}\ dx<\infty
\end{align}
for all $0<s$. Note that
\begin{align}
\int^\infty_\pi \frac{\sin x}{x^s}\ dx =&\ \sum^\infty_{n=1} \int^{(n+1)\pi}_{n\pi} \frac{\sin x}{x^s}\ dx
=\ \sum^\infty_{n=1} \int^\pi_0 \frac{\sin(n\pi +y)}{(n\pi+y)^s}\ dy\\
=&\ \sum^\infty_{n=0} (-1)^n \int^\pi_0 \frac{\sin y}{(n\pi +y)^s}\ dy
=: \sum^\infty_{n=1}(-1)^n a_n
\end{align}
where $a_n \geq a_{n+1}$ and $\lim_{n\rightarrow \infty} a_n = 0$. Hence by the alternating series test, we see that the integral converges in the sense of Riemann.
Hence it follows $0<s<2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Monotonic recursive sequence with recursive term in denominator: $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ I am trying to show that $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ is a monotonic decreasing sequence for $n \ge 2$. Currently, my approach using induction is stuck because of the $\mathbf{s_n}$ term appearing in the denominator.
What I have so far:
Calculating a few terms: $s_1 = 1$, $s_2 = 2$, $s_3 = \frac{7}{4} < s_2$.
For this to be monotonic decreasing, we have to show $s_{k+2} < s_{k+1}$.
For the base case, we assume for $k$ we have $s_{k+1} < s_k.$
The inductive step: to show $s_{k+2} < s_{k+1}$, i.e. to show $$\color{grey}{s_{k+2} = }\frac{1}{2} \left(s_{k+1} + \frac{3}{s_{k+1}}\right) < \frac{1}{2} \left(s_{k} + \frac{3}{s_{k}}\right) \color{grey}{= {s_{k+1}}_.}$$
In order to show $\dfrac{1}{2} \left(s_{k+1} + \frac{3}{s_{k+1}}\right) < \dfrac{1}{2} \left(s_{k} + \frac{3}{s_{k}}\right)$, I am stuck:
Given, $s_{k+1} < s_k$, I cannot then say $s_{k+1} + \mathbf{\frac{3}{s_{k+1}}} < s_k + \mathbf{\frac{3}{s_{k}}}$, because the $s_k, s_{k+1}$ terms which appears in the denominator may reverse the inequality. Any pointers on how to proceed further?
Disclaimer: I am revising real analysis on my own from the Kenneth Ross book, this is not strictly homework.
| The hint.
Use for all $n\geq2$ $$s_{n+1}-\sqrt3=\frac{(s_n-\sqrt3)^2}{2s_n}>0$$ and
$$s_{n+1}-s_n=\frac{3-s_n^2}{2s_n}<0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$? When two digit numbers in base $5$ are multiplied the result is $4103_5$. What are the numbers in base $5$?
Well given by two digit numbers in base $5$ I tried out the multiplication and tried to simplify.
$(ab_5)(cd_5)=4103_5$ or $(5a+b)(5c+d)=4*5^3+1*5^2+0*5^1+3$
Then get that $5^2ac+5ad+5cb+bd=5^3*4+5^2+3$
I notice that $bd=3$ so b=1,3 and $d=1,3$. Now this is where I'm stuck I don't see a clear way to get what the numbers should be without a lot of trial and error.
| Continuing along your route, where:
$$4103=(10a+b)(10c+d)$$
(notice all values are represented in base $5$ here)
All values will be in base $5$ throughout.
Notice that $40^2=3100$, which is less than $4103$, so $a$ and $c$ are both $4$:
$$4103=(40+b)(40+d)=3100+40(b+d)+bd$$
Now, notice that $\left|{b-d}\right|=2$ in order for the last digit in the product to be a $3$. Since the order of factors is optional, and we know the first digits are the same, we will set $d=b+2$:
$$4103=3100+40(2b+2)+b(b+2)$$
$$4103=3100+(130b+130)+(b^2+2b)$$
$$4103=3230+132b+b^2$$
$$0=b^2+132b-323$$
$$0=(b-2)(b+134)$$
$$ b=\{2,-134\}$$
Since the digit must be between $0$ and $4$ inclusive, only $2$ can be valid. For $d$:
$$d=b+2=2+2=4$$
Finally, since $a=4$; $b=2$; $c=4$; $d=4$, the factors are $\{42_5, 44_5\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2}\\
&=x(x-1)(x^2+1)+\dfrac{1}{2}.
\end{align*}
Is there any way to solve this question?
| For $0 \leqslant x \leqslant 1$,$$
x(x - 1)(x^2 + 1) \geqslant \left(-\frac{1}{4}\right)(x^2 + 1) > \left(-\frac{1}{4}\right) \times 2 = -\frac{1}{2}.
$$
For $x < 0$ or $x > 1$,$$
x(x - 1)(x^2 + 1) > 0 > -\frac{1}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Solving $4^{32^x} = 16^{8^x}$
Solve for $x$: $4^{32^x} = 16^{8^x}$.
So I have tried using log (so $x\log{4^{32}} = x\log{16^8}$), but that wasn't very helpful to me, and some random guessing gave me an answer of $1/2$, but I was wondering how I could be more mathematical…
| Since$$
4^{32^x} = 2^{2 \cdot 32^x} = 2^{2 \cdot 2^{5x}} = 2^{2^{5x + 1}}
$$
and$$
16^{8^x} = 2^{4 \cdot 8^x} = 2^{4 \cdot 2^{3x}} = 2^{2^{3x + 2}},
$$
then$$
4^{32^x} = 16^{8^x} \Longleftrightarrow 2^{2^{5x + 1}} = 2^{2^{3x + 2}} \Longleftrightarrow 2^{5x + 1} = 2^{3x + 2}\\\Longleftrightarrow 5x + 1 = 3x + 2 \Longleftrightarrow x = \frac{1}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the value of a 5th-root expression. Simplify and find the value of the expression:
$$\sqrt[5]{\frac{123+\sqrt{15125}}{2}}+\sqrt[5]{\frac{123-\sqrt{15125}}{2}}.$$
I tried to rationalise it.
It was of no use..
| $$\frac{123+\sqrt{15125}}2=\frac{123+55\sqrt5}2=\left(\frac{1+\sqrt5}2\right)^{10}=\left(\frac{3+\sqrt5}2\right)^5.$$
Likewise
$$\frac{123-\sqrt{15125}}2=\left(\frac{3-\sqrt5}2\right)^5.$$
So
$$\sqrt[5]{\frac{123+\sqrt{15125}}2}+\sqrt[5]{\frac{123-\sqrt{15125}}2}
=\frac{3+\sqrt5}2+\frac{3-\sqrt5}2=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Unable to use Chinese Remainder Theorem on a question. So I had recently learned about Chinese Remainder theorem, and I got a Question to solve:
Find X if-
$X\equiv 1\bmod 2\\X\equiv 2 \bmod 3\\X\equiv 3 \bmod 4\tag*{}$
So I went on to find the solution modulus, $N=24$ & $N_1=12\\ N_2 = 8\\ N_3 = 6\tag*{}$
And now I just had to find the three inverse modulus, but for the first one
$12y_1\equiv 1 \bmod 2\tag*{}$ there's no $y_1$ satisfying the equation. I was stuck on this part. Please Highlight what I am doing wrong.
| $x \equiv 1 (\mod 2) \Rightarrow x-1$ divisible by $2$ $\Rightarrow x-1+2=x+1$ is divisible by $2$.
$x \equiv 2 (\mod 3) \Rightarrow x-2$ divisible by $3$ $\Rightarrow x-2+3=x+1$ is divisible by $3$.
$x \equiv 3 (\mod 4) \Rightarrow x-3$ divisible by $4$ $\Rightarrow x-3+4=x+1$ is divisible by $4$.
Because $x+1$ is divisible by $2;3;4$, it is also divisible by the least common multiple of them, which is $12$ (but not neccesarily the product of them or $24$, because these numbers are not relatively prime)
We can imply that $x \equiv -1 $(mod $12)$ or $x \equiv 11 $(mod $12)$, which means $x$ can be expressed in the form $x=12k+11$ for non-negative integers $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove the identity $ \sum\limits_{k=0}^{n-1} (-1)^{k} \binom{n}{k} (n-k) \frac{(a+k-2)!}{(a+k-n)!}=0.$ Prove the identity
$$
\sum_{k=0}^{n-1} (-1)^{k} \binom{n}{k} (n-k) \frac{(a+k-2)!}{(a+k-n)!}=0
$$
I want to reduce
$$
\frac{(a+k-2)!}{(a+k-n)!}=(2-n)\binom{a+k-n}{a+k-2}^{-1}=(2-n) (a+k-2) \int_0^1 z^{a+k-2}(1-z)^{2-n} dz,
$$
and then try integrate it
but the binomial coefficient
$$
\binom{a+k-n}{a+k-2}=\frac{(a+k-n)!}{(2-n)! (a+k-2)!},
$$
is undefined for $n>2$ however the sum seems correct for any $n.$
Any ideas?
|
We obtain for integral $n>0$ and $a\in\mathbb{C}\setminus\{n-1,n-2,n-3,\ldots\}$
\begin{align*}
\color{blue}{\sum_{k=0}^{n-1}}&\color{blue}{(-1)^k\binom{n}{k}(n-k)\frac{(a+k-2)!}{(a+k-n)!}}\\
&=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}k\frac{(a+n-k-2)!}{(a-k)!}\tag{1}\\
&=(-1)^nn(n-2)!\sum_{k=1}^n(-1)^k\binom{n-1}{k-1}\binom{a+n-k-2}{a-k}\tag{2}\\
&=(-1)^{n+1}n(n-2)!\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{a+n-k-3}{a-k-1}\tag{3}\\
&=(-1)^{n+1}n(n-2)!\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{-n+1}{a-k+1}\tag{4}\\
&=(-1)^{n+1}n(n-2)!\sum_{k=0}^{n-1}\binom{n-1}{k}[z^{a-k+1}](1+z)^{-n+1}\tag{5}\\
&=(-1)^{n+1}n(n-2)^{-n+1}\sum_{k=0}^{n-1}\binom{n-1}{k}z^k\tag{6}\\
&=(-1)^{n+1}n(n-2)^{-n+1}(1+z)^{n-1}\tag{7}\\
&=(-1)^{n+1}n(n-2)![z^{a+1}]1\\
&\,\,\color{blue}{=0}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we set the upper limit of the sum to $n$ (i.e. adding zero) and change the order of summation by $k\to n-k$.
*In (2) we apply the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$ and use binomial coefficients.
*In (3) we shift the index to start with $k=0$.
*In (4) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^{q}$.
*In (5) we use the coefficient of operator $[z^p]$ to denote the coefficient of a series. This way we can write
$[z^p](1+z)^q=\binom{q}{p}$.
*In (6) we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (7) we apply the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove that perspective projection of circle is ellipse How to prove that perspective projection of a circle is an ellipse?
I start with the parametric equation of circle and ellipse:
Circle:
$x = r\cos t$
$y = r\sin t$
Ellipse:
$x = a\cos(t)$
$y = b\sin(t)$
Then I have perspective projection matrix:
$$
\begin{pmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & 1 \\
\end{pmatrix}
$$
And obtain projected coordinates (affine transform for simplicity):
$$
\begin{pmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
0 & 0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
1 \\
\end{pmatrix}
=
\begin{pmatrix}
u \\
v \\
1 \\
\end{pmatrix}
$$
If we convert this to equations:
$u = a_{11}x+a_{12}y+a_{13}$
$v = a_{21}x+a_{22}y+a_{23}$
Substituting $x,y$ with parametric equations:
$a\cos t = a_{11}\,r\, \cos t + a_{12}\, r \,\sin t + a_{13}$
$b \sin t = a_{21}\,r \,\cos t + a_{22}\, r \,\sin t + a_{23}$
What to do next? I can't see how these equations can be equal.
| As noted in the answer of @rfabbri ( +1), the perspective projection of a circle is not always an ellipse, but it is in general a conic section: an ellipse, a parabola, or a hyperbola.
If you want a matrix that transform the circle in an ellipse by projection, than note that the general equation of a conic:
$$
Ax^2+Bxy+Cy^2+Dx+Ey+F=0
$$
can be written in the form (see here):
$$
\begin{pmatrix}
x&y&1
\end{pmatrix}
\begin{pmatrix}
A&\frac{B}{2}&\frac{D}{2}\\
\frac{B}{2}&C&\frac{E}{2}\\
\frac{D}{2}&\frac{E}{2}&F\\
\end{pmatrix}
\begin{pmatrix}
x\\y\\1
\end{pmatrix}=0
$$
so, for $A=C=1$ and $B,D,E=0$ and $F=-r^2$ we have the circle center at the origin and radius $r$, that becomes an ellipse if the matrix is such that
$B^2-4AC <0$ and the conic is not degenerate, i.e. the determinat of the matrix is not null.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2674874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ (prove by induction) I'm having some difficulty proving by induction the following statement.
$$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$
I have shown that $\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$ holds for $n=1$ (equals $\frac{1}{20}$) , but I am getting stuck on the induction step.
As far as I know I have to show $$\sum_{i=0}^n \frac{1}{(i+3)(i+4)} = \frac{n}{4(n+4)}$$
implies
$$\sum_{i=0}^{n+1} \frac{1}{(i+3)(i+4)} = \frac{n+1}{4(n+5)}$$
To do this I think I should add the number $\frac{1}{(n+4)(n+5)}$ to $\frac{n}{4(n+4)}$ and see if it gives $\frac{n+1}{4(n+5)}$ , if I am not mistaken.
When trying to do that however I get stuck. I have:
$$\frac{n}{4(n+4)} +\frac{1}{(n+4)(n+5)} = \frac{n(n+4)(n+5)}{4(n+4)^2(n+5)} + \frac{4(n+4)}{4(n+4)^2(n+5)} = \frac{n(n+4)(n+5)+4(n+4)}{4(n+4)^2(n+5)} = \frac{n(n+5)+4}{4(n+4)(n+5)}$$
However beyond this point I don't know how to reach $\frac{n+1}{4(n+5)}$ I always just end up at the starting point of that calculation.
So I think that either my approach must be wrong or I am missing some trick how to simplify $$\frac{(n(n+5)+4}{4(n+4)(n+5)}$$
I would be very grateful for any help, as this is a task on a preparation sheet for the next exam and I don't know anyone, that has a correct solution.
| $$\dfrac{n(n+5)+4}{4(n+4)(n+5)}=\dfrac{n^2+5n+4}{4(n+4)(n+5)}=\dfrac{(n+1)(n+4)}{4(n+4)(n+5)}$$
This simplifies to $\dfrac {n+1}{4(n+5)}$ and you want to substitute $n$ for $n+1$ for $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.