Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
A binomial inequality with factorial fractions: $\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$ Prove that $$\left(1+\frac{1}{n}\right)^n<\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{n!}$$ for $n>1 , n \in \mathbb{N}$.
| We have by the binomial identity that
\begin{align*}
\left(1 + \frac 1n \right)^n &= \sum_{k=0}^n \binom nk \frac 1{n^k}\\
&= \sum_{k=0}^n \frac{n!}{(n-k)! n^k} \cdot \frac 1{k!}\\
&= \sum_{k=0}^n \frac{n \cdot (n-1) \cdots (n-k+1)}{n \cdot n \cdots n} \cdot \frac 1{k!}\\
&\text{now the first factor is $<1$ for $k\ge 2$}\\
&< \sum_{k=0}^n \frac 1{k!}
\end{align*}
for $n \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Showing that $a^{12}\equiv 1\pmod{210}$ if $a$ is coprime to $210$ I ran into this next question:
Show that if $a$ is coprime to $210$:
$a^{12}\equiv1 \pmod{210}$.
This is not a homework question.
Thank you very much in advance,
Yaron
| HINT:
Using Carmichael Function,
$$\lambda(210)=\text{lcm}\{\lambda(2),\lambda(3),\lambda(5),\lambda(7)\}=\text{lcm}(1,2,4,6)=12$$
In fact,we can find a larger $n$ such that $a^{12}\equiv\pmod n$ for all $a$ co-prime to $n$
We need $\lambda(n)$ divides $12$
Case $1: $ If $p=2,$ we know, $\lambda(2^r)=\begin{cases} 2^{r-1} &\mbox{if } 1\le r \le 2 \\
2^{r-2} & \text{ otherwise }\end{cases}$
$2^{r-2}$ needs to divide $12,r-2\le 2\implies r \le 4$
Case $2: $ If $p$ is odd prime, $\lambda(p^r)=\phi(p^r)=p^{r-1}(p-1),$
If $r>1,$ as the only odd divisor$(>1)$ of $12$ is $3^1\implies p=3,r=2$
Else $r=1\implies p-1$ must divide $12\implies p$ can be $2,3,5,7,13$
$2,3$ have already been encountered
So, the highest value of $n$ will be $13^1\cdot7^1\cdot5^1\cdot3^2\cdot2^4$
As $a^{12}\equiv1\pmod {13\cdot7\cdot5\cdot3^2\cdot2^4}$ if $(a,13\cdot7\cdot5\cdot3^2\cdot2^4)=1,$
$a^{12}\equiv1\pmod d$ where $d$ is any divisor of $13\cdot7\cdot5\cdot3^2\cdot2^4$ and $(a,d)=1$ as $(a,13\cdot7\cdot5\cdot3^2\cdot2^4)=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/406595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find the limit of $(\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ Could you tell me how to find $\lim_{n \rightarrow \infty} (\sin \frac{1}{n} \cdot \sin \frac{2}{n} \cdot ... \cdot \sin 1)^{\frac{1}{n}}$ ?
| As $x\rightarrow 0$ , $\sin(x) \rightarrow x$
So $ \sin(\frac{1}{n}) \rightarrow \frac{1}{n}$ as $n \rightarrow \infty$
The product $ [ \sin(\frac{1}{n}) \cdot \sin(\frac{2}{n}) \cdot ... \cdot \sin(\frac{n}{n}) ] \rightarrow \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdot...\frac{n}{n} =
\frac{n!}{n^n}$
As $n! = \sqrt{2 \cdot \pi \cdot n} \cdot (\frac{n}{e})^n $
So $\frac{n!}{n^n} = \sqrt{2\pi n} \cdot (\frac{1}{e})^n $
$\lim_{ n \rightarrow \infty}$ of $[ \sin(\frac{1}{n}) \cdot \sin(\frac{2}{n}) \cdot ... \cdot \sin(\frac{n}{n}) ] ^ \frac{1}{n} \rightarrow \frac{1}{e}$
Since $\sin(\frac{m}{n})$ is always positive if $0< m \le n$ and less than $\frac{m}{n}$ .
The limit will be less than (1/e)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/406819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Find the greatest integer $N$ such that...
Find the greatest integer $N$ such that $N<\dfrac{1}{\sqrt{33+\sqrt{128}}+\sqrt{2}-8}$.
The way I did it is this: first, I rewrote the biggest square root as $\sqrt{1+2*16+8\sqrt{2}}$. Then I made the substitution $x=\sqrt{2}$, so this became $\sqrt{16x^2+8x+1}$, which factors to $\sqrt{(4x+1)^2}$ which is equal to $4x+1$. Then I substituted back: $\dfrac{1}{4\sqrt{2}+1+\sqrt {2}-8}$, or $\dfrac 1{5 \sqrt{2}-7}$. This is where I have a problem. I am able to find the answer because I know that $1.4<\sqrt 2 <1.5$. However, is there any way to find this answer without knowing this? Thanks!
| Picking up from where you left off, we have:
$$ \begin{align*}
N &= \left\lfloor \dfrac{1}{5\sqrt{2}-7} \right\rfloor\\
&= \left\lfloor \dfrac{1}{5\sqrt{2}-7} \cdot \dfrac{5\sqrt{2}+7}{5\sqrt{2}+7} \right\rfloor\\
&= \left\lfloor \dfrac{5\sqrt{2}+7}{50-49} \right\rfloor\\
&= \lfloor 5\sqrt{2}+7 \rfloor\\
&= \lfloor \sqrt{50}+7 \rfloor\\
&= \lfloor \sqrt{50} \rfloor+7 \\
&= \sqrt{49}+7 \\
&= 14 \\
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/407062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $\;x+y+z =8 ; \;\;\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}=10 $ Solve the problem
\begin{cases}x+y+z =8 \\ \\
\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}=10
\end{cases}
with $(x,y,z) \in \mathbb R^3$
I have already solved it, but I'd like to see others creative solutions and before all, share this funny problem with the community.
| Since $\sqrt{x^2+1}+\sqrt{y^2+4}+\sqrt{z^2+9}\ge \sqrt{(x+y+z)^2+(1+2+3)^2}= 10$ by the Minkowski inequality, now check when the equal sign hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/407696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving trigonometric identity with condition. Problem : If $\sin\theta +\sin^2\theta +\sin^3\theta=1$ Then prove $\cos^6\theta -4\cos^4\theta +8\cos^2\theta =4$
My working :
As $\sin\theta +\sin^2\theta +\sin^3\theta=1 \Rightarrow \sin\theta +\sin^3\theta = \cos^2\theta$
Now the given equation : $\cos^6\theta -4\cos^4\theta +8\cos^2\theta$ can be written as
$(\sin\theta +\sin^3\theta)^3-4(\sin\theta+\sin^3\theta )^2+8(\sin\theta +\sin^3\theta)$
= $\sin^3\theta +\sin^6\theta +3\sin^5\theta +3\sin^7\theta -4\sin^2\theta -4\sin^6\theta -8\sin^4\theta + 8\sin\theta + 8\sin^3\theta$
But I think this is not the right way of doing this...Please suggest other alternative.. Thanks...
| Let $x = \sin \theta$. We are given that $x^3 + x^2 + x - 1 = 0$.
We want to show that $(1-x^2)^3 -4(1-x^2)^2 +8(1-x^2) = 4$. Expanding and comparing terms, this is equivalent to
$$ x^6 + x^4 + 3x^2 - 1 = 0. $$
This is true because $$x^6 + x^4 + 3x^2 - 1 = (x^3 + x^2 + x -1 ) ( x^3 - x^2 + x +1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding indefinite integral by partial fractions $$\displaystyle \int{dx\over{x(x^4-1)}}$$
Can this integral be calculated using the Partial Fractions method.
| HINT:
We need to use Partial Fraction Decomposition
Method $1:$
As $x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1),$
$$\text{Put }\frac1{x(x^4-1)}=\frac Ax+\frac B{x-1}+\frac C{x+1}+\frac {Dx+E}{x^2+1}$$
Method $2:$
$$I=\int \frac1{x(x^4-1)}dx=\int \frac{xdx}{x^2(x^4-1)} $$
Putting $x^2=y,2xdx=dy,$
$$I=\frac12\int \frac{dy}{y(y^2-1)}$$
$$\text{ Now, put }\frac1{y(y^2-1)}=\frac A y+\frac B{y-1}+\frac C{y+1}$$
Method $3:$
$$I=\int \frac1{x(x^4-1)}dx=\int \frac{x^3dx}{x^4(x^4-1)} $$
Putting $x^4=z,4x^3dx=dz,$
$$I=\frac14\int \frac{dz}{z(z-1)}$$
$$\text{ Now, put }\frac1{z(z-1)}=\frac Az+\frac B{z-1}$$
$$\text{ or by observation, }\frac1{z(z-1)}=\frac{z-(z-1)}{z(z-1)}=\frac1{z-1}-\frac1z$$
Observe that the last method is susceptible to generalization.
$$J=\int\frac{dx}{x(x^n-a)}=\int\frac{x^{n-1}dx}{x^n(x^n-a)}$$
Putting $x^n=u,nx^{n-1}dx=du,$
$$J=\frac1n\int \frac{du}{ u(u-a)}$$
$$\text{ and }\frac1{u(u-a)}=\frac1a\cdot\frac{u-(u-a)}{u(u-a)}=\frac1a\left(\frac1{u-a}-\frac1u\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simple generalized integral The integral to compute is $\displaystyle\int_0^\infty \frac{1}{3+x^2} \, \mathrm dx$.
I know how to compute the indefinite integral of this function its gives me
$$\frac{\sqrt{3}}{3} \arctan\left(\frac{x}{\sqrt{3}}\right).$$
When i compute the definite integral it now gives me :
$$\frac{\sqrt{3}}{3} \left(\lim\limits_{x \to \infty }\arctan\left(\frac{x}{\sqrt{3}}\right)-\arctan(0)\right).$$
Then i don't understand why my teacher writes that arctan(0)=0 because it also can be equal to π, and more stranger he found i don't know how that $\lim\limits_{x \to \infty }\arctan\left(\frac{x}{\sqrt{3}}\right)=\pi/2$. Thank you for help !
EDIT : I only need how to compute the limit now.
| The way i would approach this problem is .
$$ \int_0^\infty \frac{1}{3+x^2}dx $$
I would divide the $$ x^2+3 $$ by 3 to factor out the constants.
$$ \int_0^\infty \frac{1}{3(\frac{x^2}{3})+1}dx $$
then we can substitute our function with
$ u^2=\frac{x^2}{3}$
$u= \frac{x}{\sqrt{3}} $
The integral of
$$ \int_0^\infty \frac{1}{3(\frac{x^2}{3})+1}dx $$
$du= \frac{1}{\sqrt{3}}dx$
When we substitute we get
$$ \frac{1}{\sqrt{3}}\int_0^\infty \frac{1}{{u^2}+1}dx $$
the integral would result to
$$ \frac{1}{\sqrt{3}}\int_0^\infty \frac{1}{{u^2}+1}dx =\frac{1}{\sqrt{3}}arctan(u)|_0^\infty $$
Then we substitute back $u= \frac{x}{\sqrt{3}} $
This would give us the step before the final answer.
$$\frac{1}{\sqrt{3}}arctan(\frac{x}{\sqrt{3}})|_0^\infty $$
then we can evaluate the value and the limit
$$\frac{1}{\sqrt{3}}\left(\lim\limits_{a \to \infty }\arctan\left(\frac{a}{\sqrt{3}}\right)-\arctan(0)\right)= (\frac{pi}{2\sqrt{3}}) $$
because as "a" approaches infinity arctan(a/sqrt(3)) = Pi/2 and arctan(0/sqrt(3)) = 0
$$ \frac{1}{\sqrt{3}}( \frac{pi}{2} - 0 ) = (\frac{pi}{2\sqrt{3}}) $$
Thats the final correct answer and evaluation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$1=2$ | Continued fraction fallacy It's easy to check that for any natural $n$
$$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$
Now,
$$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots
=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},$$
$$2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots
=\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.$$
Since the right hand sides are the same, hence $1=2$.
| This is of the same type as the following
$$0=(1-1)+(1-1)+(1-1)+\ldots=1+(-1+1)+(-1+1)+(-1+1)+\ldots = 1 \; .$$
Did's example is even simpler and closer to yours in spirit.
In working with an infinite number of operations, you have to be very careful about how you are performing them. Normally, one uses some kind of limit, but then what you really do is define a sequence of finite but ever increasing number of operations. Changing something in the order of these operations will change the entire limit. Or in your case, you hide away the fact that in each term of your sequence, the last operation is subtracting a different number, $1$ in the first case, $(n+2)/(n+1)$ in the second.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "89",
"answer_count": 7,
"answer_id": 3
} |
Prove $BA - A^2B^2 = I_n$. I have a problem with this. Actually, still don't have the right way to start :/
Problem :
Let $A$ and $B$ be $n \times n$ complex matrices such that $AB - B^2A^2 = I_n$.
Prove that if $A^3 + B^3 = 0$, then $BA - A^2B^2 = I_n$.
Thanks for any help.
| By the two given conditions, we see that
\begin{align*}
A(BA-A^2B^2) &= ABA-A^3B^2\\
&=(I+B^2A^2)A-A^3B^2\\
&=A+B^2A^3-A^3B^2\\
&=A-B^5+B^5\\
&=A.
\end{align*}
Therefore, if we can prove that $A$ is invertible, we are done.
Suppose the contrary. Then there exists a nonzero vector $v$ such that $Av=0$. We now prove by mathematical induction that for $k\ge1$, $AB^kv = a_kB^{k-1}v$ for some $a_k\neq0$. The base case is easy: $ABv=(I+B^2A^2)v=v$. Now, for $k\ge1$,
\begin{align*}
AB^{k+1}v
&= (AB)B^kv\\
&= (B^2A^2+I)B^kv\\
&= B^2A(AB^kv) + B^kv\\
&= B^2A(a_kB^{k-1}v) + B^kv\\
&= a_kB^2(AB^{k-1}v) + B^kv\\
&= \begin{cases}
0+B^kv &\text{ when } k=1\\
a_k B^2 (a_{k-1}B^{k-2}v) + B^kv &\text{ when } k\ge2
\end{cases}\\
&= \begin{cases}
B^kv &\text{ when } k=1\\
(a_{k-1}a_k+1) B^kv &\text{ when } k\ge2
\end{cases}.
\end{align*}
In short, if we define $a_0=0,\ a_1=1$ and $a_k=a_{k-2}a_{k-1}+1$ for $k\ge2$, we have
\begin{align*}
Av &= 0,\tag{1}\\
AB^kv &= a_kB^{k-1}v\ \text{ for }\ k\ge1\tag{2}
\end{align*}
with $a_k\neq0$ for $k\ge1$.
Let $m\ge0$ be the largest integer such that $v,Bv,B^2v,\ldots,B^mv$ are linearly independent, and let $B^{m+1}v=\sum_{k=0}^m c_kB^kv$. By $(1)$ and $(2)$,
$$
a_{m+1}B^mv = AB^{m+1}v
= A \sum_{k=0}^m c_k B^k v
= \sum_{k=\color{red}{1}}^m c_k AB^k v
= \sum_{k=1}^m c_ka_k B^{k-1} v.\tag{3}
$$
Yet this is impossible because $v,Bv,B^2v,\ldots,B^mv$ are linearly independent and $a_{m+1}\neq0$. Hence the assumption that $v$ is nonzero cannot be true and $A$ is invertible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
} |
How to solve this integral $\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx$? I will be grateful if you would write me a solution procedure for this integral
$$\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx. $$
I am sure that an antiderivative is
$$\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right).$$
Now if I put $+\infty $ instead of $x$ I get
\begin{align*}
\left[ \frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right) \right]^{\infty }_{0}
&= \frac{1}{4} \left( \frac{\infty}{\infty}-\infty \right)-\frac{1}{4} \left( \frac{2 \log 1}{1}-\log 1 \right) \\
&= \frac{1}{4} \left( \frac{\infty}{\infty}-\infty \right).
\end{align*}
As you can see, it is useless. Can you help me please? Thanks
Can I use this solution below?
Let $$I=\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right).$$
Now if I calculate the limit of I i get:
$$\lim_{x\to\infty}I=0$$
So the final result is
\begin{align*}
\left[ \frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right) \right]^{\infty }_{0}=0
\end{align*}
| Related problems: (I). Recalling the Mellin transform
$$ F(s) = \int_{0}^{\infty} x^{s-1}f(x) dx \implies F'(s) = \int_{0}^{\infty} x^{s-1}\ln(x)f(x) dx$$
So, taking $f(x)=\frac{1}{(1+x^2)^2}$ and finding its Mellin transform
$$F(s) = \frac{1}{4}\,{\frac { \left( 2-s \right) \pi }{\sin \left( \frac{\pi s}{2}
\right) }}.$$
Now, differentiating and taking the limit as $s\to 2$ gives the desired result
$$ \lim_{s\to 2} F'(s) = 0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaulate $\lim_{x\to\infty}\frac{3x^2-36x+12}{5x^2+113x-2}$ Question: Find the limit, $$\lim_{x\to\infty}\frac{3x^2-36x+12}{5x^2+113x-2}$$
The limit should be $\frac{3}{5}$ since when $x$ approaches infinity since $\frac{3*\infty^2}{5*\infty^2}$ and infinity squared cancels out. Am I correct?
| $\displaystyle\frac{3x^2-36x+12}{5x^2+113x-2}=\displaystyle\frac{3-36\frac{1}{x}+12\frac{1}{x^2}}{5+113\frac{1}{x}-2\frac{1}{x^2}}$(Dividing by numerator and denominator by $x^2$ )
As $\lim_{x\to \infty }(3-36\frac{1}{x}+12\frac{1}{x^2})=3$ and $\lim_{x\to \infty }(5+113\frac{1}{x}-2\frac{1}{x^2})=5$ using the fact that $\lim_{x\to \infty } \frac{1}{x}=\lim_{x\to \infty }\frac{1}{x^2}=0$
So $\displaystyle \lim_{x\to \infty }\frac{3x^2-36x+12}{5x^2+113x-2}=\frac{\lim_{x\to \infty }(3-36\frac{1}{x}+12\frac{1}{x^2})}{\lim_{x\to \infty }(5+113\frac{1}{x}-2\frac{1}{x^2})}=\frac{3}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/421238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. Is my induction solution correct? Show using mathematical induction that $3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. I'm not sure whether what I did at the last is valid?
Basis step:
for all non-negative integers
$$P(n) = 3^{3n+1} < 2^{5n+6} $$
$$P(0) = 3^{3(0) + 1} = 3 < 64 = 2^{5(0) + 6}$$
$$P(0) = T$$
Inductive Step:
Assume: $3^{3k+1} < 2^{5k+6}$
Show: $3^{3(k+1)+1} < 2^{5(k+1)+6}$
$$ 3^{3(k+1)+1} = 3^{3k+4} = 3^3 \cdot 3^{3k+1}$$
By inductive hypothesis~
$$3^3 \cdot 3^{3k+1} < 3^3 \cdot 2^{5n+6} $$
This is the part where I'm not sure if you can do this in induction but it seems logically correct.
$$3^3 \cdot 2^{5n+6} = 27 \cdot 2^{5n+6}$$
$$2^{5(k+1)+6} = 2^{5k+5+6}= 2^5 \cdot 2^{5n+6} = 32 \cdot 2^{5n+6}$$
I'm not sure whether it should be $\le$ or $<$ but I used '$<$' for $3^3 \cdot 2^{5n+6}<2^{5(k+1)+6} $
Therefore:
$$3^{3(k+1)+1} < 3^3 \cdot 2^{5n+6}<2^{5(k+1)+6} $$
| It’s fine. You can compress it to a single chain of inequalities:
$$3^{3(k+1)+1}=3^3\cdot3^{3k+1}=27\cdot3^{3k+1}<32\cdot3^{3k+1}<32\cdot2^{5k+6}=2^5\cdot2^{5k+6}=2^{5(k+1)+6}\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/421362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the derivative of a relational problem I am self studying some calculus and I have gotten really stuck! I thought I had the right idea but I keep getting the answer totally wrong. I am sure I am missing something important. Here is the problem:
For the equation $6x^{\frac{1}{2}}+12y^{-\frac{1}{2}} = 3xy$, find an equation of the tangent line at the point $(1, 4)$.
Here is my work:
$$6x^{\frac{1}{2}}+12y^{-\frac{1}{2}} = 3xy$$
$$6(x^{\frac{1}{2}}+2y^{-\frac{1}{2}}) = 3xy$$
$$2(x^{\frac{1}{2}}+2y^{-\frac{1}{2}}) = xy$$
$$2x^{\frac{1}{2}}+4y^{-\frac{1}{2}} = xy$$
$$\ln(2x^{\frac{1}{2}})+\ln(4y^{-\frac{1}{2}}) = \ln(xy)$$
$$\ln2 + \ln x^{\frac{1}{2}}+\ln4 + \ln y^{-\frac{1}{2}} = \ln x + \ln y$$
$$\ln2 + \frac{1}{2}\ln x+\ln4 -\frac{1}{2} \ln y = \ln x + \ln y$$
$$\ln2 + \ln4 + \frac{1}{2}\ln x - \ln x = \ln y + \frac{1}{2} \ln y$$
$$\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x = \ln y$$
$$e^{\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x} = y$$
$$\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x = \ln y$$
$$\frac{1}{3x} = y'\frac{1}{y}$$
$$y' = \frac{y}{3x}$$
$$y' = \frac{e^{\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x}}{3x}$$
$$y - 4 = \frac{4}{3}(x-1)$$
$$y = \frac{4}{3}x-\frac{4}{3} + 4$$
$$y = \frac{4}{3}x+\frac{8}{3}$$
The Solution I found was: $y = \frac{4}{3}x+\frac{8}{3}$ but this is wrong! Can you tell me where I have gone wrong?
| In general, you've just put yourself through a lot of unnecessary work. Just hit the thing immediately with implicit differentiation:
$$6x^{\frac{1}{2}}+12y^{-\frac{1}{2}} = 3xy$$
Becomes:
$$3x^{-1/2}-6y^{-3/2} = 3xy' + 3y$$
You're told to find the tangent line at $(1, 4)$, so plug in $x=1,\;y=4$:
$$3(1)^{-1/2}-6(4)^{-3/2} = 3(1)y' + 3(4)$$
$$3-6\frac{1}{8} = 3y' + 12$$
Now solve for $y'$:
$$y' = \frac{-9-\frac{3}{4}}{3}$$
EDIT: Dang. I missed the "easy" mistake, and found a harder one instead. :o
Doing it your way:
$$\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x = \ln y$$
When you differentiated, you forgot a minus sign:
$$\color{red}- \frac{1}{3x} = \frac{y'}{y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} = \prod_{k=1}^n \frac{k}{x+k}$ and more The current issue (vol. 120, no. 6)
of the American Mathematical Monthly
has a proof by probabilistic means
that
$$\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} = \prod_{k=1}^n \frac{k}{x+k}
$$
for all $x > 0$
and all
$n \in \mathbb{N}$.
The article also mentions two other ways to prove this,
one using hypergeometric functions and
the Chu-Vandermonde formula,
and the other using the Rice integral formula
and complex contour integration.
This made me wonder if there were
more elementary ways to prove this result,
and my question is a challenge
to find the most elementary proof.
Two ideas that have occurred to me
are (1) using Lagrange interpolation
and (2) induction.
I have not yet completed a proof,
so I am putting the problem out here.
The article states with hints of proofs
the following results:
$$\sum_{k=0}^n \binom{n}{k}(-1)^k \big(\frac{x}{x+k}\big)^2
= \big(\prod_{k=1}^n \frac{k}{x+k}\big)\big(1+\sum_{k=1}^n \frac{x}{x+k}\big)
$$
and,
for $m\in \mathbb{N}$,
$$\sum_{k=0}^n \binom{n}{k}(-1)^k \big(\frac{x}{x+k}\big)^m
= \big(\prod_{k=1}^n \frac{k}{x+k}\big)
\big(1+\sum_{j=1}^{m-1} \sum_{1\le k_1 \le k_2 \le ... \le k_j \le n}
\frac{x^j}{\prod_{i=1}^j (x+k_i)}\big)
$$
What (relatively) elementary proofs
of these are there?
| $$\sum_{k=0}^n \dbinom{n}k p^k = (1+p)^n$$
$$\sum_{k=0}^n \dbinom{n}k (-1)^k = (1-1)^n=0$$
$$\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} =\sum_{k=0}^n \binom{n}{k}(-1)^k (1-\frac{k}{x+k} )= \sum_{k=0}^n \binom{n}{k}(-1)^k -\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{k}{x+k}=
$$
$$\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{x}{x+k} = -\sum_{k=0}^n \binom{n}{k}(-1)^k \frac{k}{x+k}= \sum_{k=1}^n \binom{n}{k}(-1)^{k+1} \frac{k}{x+k}
\tag1$$
$$
\prod_{k=1}^n \frac{k}{x+k}= \frac{A_1}{x+1}+\frac{A_2}{x+2}+....+\frac{A_n}{x+n}
$$
$$
A_1=(x+1)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-1}=\frac{n!}{(n-1)!}=n
$$
$$
A_2=(x+2)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-2}=\frac{n!}{(-1)(n-2)!}=-n(n-1)
$$
$$
A_3=(x+3)\prod_{k=1}^n \frac{k}{x+k} \bigg|_{x=-3}=\frac{n!}{(-1)(-2)(n-3)!}=+\frac{n(n-1)(n-2)}{2!}
$$
$$
A_k=(x+k)\prod_{m=1}^n \frac{m}{x+m} \bigg|_{x=-k}=\frac{n!}{(-1)(-2)(-3)..(-(k-1))(n-k)!}=(-1)^{k+1}\frac{n!k}{1.2.3..(k-1).k(n-k)!}=(-1)^{k+1}\frac{n!k}{k!(n-k)!}=(-1)^{k+1}k\dbinom{n}k
$$
$$
\prod_{k=1}^n \frac{k}{x+k}= \frac{A_1}{x+1}+\frac{A_2}{x+2}+....+\frac{A_n}{x+n}=\sum_{k=1}^n \frac{A_k}{x+k}=\sum_{k=1}^n \binom{n}{k}(-1)^{k+1} \frac{k}{x+k}
$$
If we use equation 1
$$
\prod_{k=1}^n \frac{k}{x+k}= \sum_{k=0}^n \binom{n}{k}(-1)^{k} \frac{x}{x+k}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int\!\sqrt{\cot x}\,dx $ Find the integral
$$\int\!\sqrt{\cot x}\,dx $$
How can one solve this using substitution?
Can this be solved by complex methods?
| Consider $I_1=\int (\sqrt{\tan x}+\sqrt{\cot x})dx$
Put $ x=\arctan t^2\implies dx= \frac{2t}{1+t^4}dt$
Then, $I_1=2\int \frac{t(t+\frac{1}{t})}{1+t^4}dt=2\int \frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2}dt$
Put $t-\frac{1}{t}=z\implies (1+\frac{1}{t^2})dt=dz$
Then $I_1=2\int \frac{1}{z^2+2}dz =\sqrt{2}\tan^{-1}\left(\frac{z}{\sqrt{2}}\right)+c=\sqrt{2}\tan^{-1}\left(\frac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt{2}}\right)+c$
Similarly, $I_2=\int (\sqrt{\tan x}-\sqrt{\cot x})dx$
First substitute is same, then, in denominator make term $(t+1/t)^2-1$ and substitute $t+1/t=z$ which gives $I_2$
Then $I_3=\int \sqrt{\cot x}dx=\frac{I_1-I_2}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Given $d(x) = \sqrt x + 3$ and $g(x) = 3\sqrt x - 2$, find $k(x) = 4\left(\frac{g}{d}\right)(x)$.
Given: $d(x) = \sqrt x + 3$ and $g(x) = 3\sqrt x - 2$, find $k(x) = 4\left(\frac{g}{d}\right)(x)$.
The answer I got was:
$$\frac{12x-44\sqrt{x}+24}{x-9}$$
Any input would be much appreciated thanks.
| You're correct.$$4\left(\frac{g}{d}\right)(x)=4\cdot\frac{g(x)}{d(x)}=4\cdot\frac{3\sqrt{x}-2}{\sqrt{x}+3}=\frac{12\sqrt{x}-8}{\sqrt{x}+3}\cdot\frac{(\sqrt{x}-3)}{(\sqrt{x}-3)}=\frac{12x-44\sqrt{x}+24}{x-9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$ \frac{\pi}{4} = \frac{3}{4} \times \frac{5}{4} \times \frac{7}{8} \times \frac{11}{12} \times \cdots$ (just primes in the numerator) How to prove the following equality?
$$ \frac{\pi}{4} = \frac{3}{4} \times \frac{5}{4} \times \frac{7}{8} \times \frac{11}{12} \times \frac{13}{12} \times \frac{17}{16} \times \frac{19}{20} \times \frac{23}{24} \times \frac{29}{28} \times \frac{31}{32} \times \cdots$$
I've seen this formula on Wikipedia.
The numerators are the odd primes; each denominator is the multiple of four nearest to the numerator.
Thanks.
| As I mentioned in my comment yesterday, this question is very similar to this one, except that here we want to prove the identity
\begin{align*}
\dfrac{\pi}{4}=\prod_{k=2}^{\infty}\left(1+\dfrac{(-1)^{\frac{p_{{k}}+1}{2}}}{p_{k}} \right )^{-1},
\end{align*}
which turns out to be even simpler.
So I almost literally reproduce the relevant half of my previous solution:
*
*Decompose the product on the right as
$$B=\prod_{\text{primes}\; p\\ \text{ of the form }4k+1}\left(1-\frac{1}{p}\right)^{-1}\prod_{\text{primes}\; p\\ \text{ of the form }4k+3}\left(1+\frac{1}{p}\right)^{-1}.\tag{1}$$
*Consider an odd integer $n=2m+1$. It is easy to understand that if primes of the form $4k+3$ appear in its prime number decomposition an even number of times, then $n$ is of the form $4K+1$ [since $(4k_1+3)(4k_2+3)=1\; \mathrm{mod}\;4$]. If the number of such appearances is odd, then $n$ is of the form $4K+3$.
*Rewrite (1) (expanding its factors into geometric series) as
$$B=\sum_{m=0}^{\infty}\frac{(-1)^{r(m)}}{2m+1}$$
where $r(m)$ counts the number of appearances of primes of the form $4k+3$ in the decomposition of $2m+1$. But then Step 2 allows to write $(-1)^{r(m)}=(-1)^m$, so that
$$ B=\sum_{m=0}^{\infty}\frac{(-1)^{m}}{2m+1}=\frac{\pi}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/428656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 0
} |
How find $dy/dx$ $y= \frac{x^2+\sin2x}{2x+\cos^2 x}$ $$dy/dx\space\space y= \frac{x^2+\sin2x}{2x+\cos^2x}$$
| We need the quotient rule and the chain rule to find $\frac {dy}{dx} = y'$ given $$y= \frac{x^2+\sin2x}{2x+\cos^2x}$$
Given a quotient of functions: $f(x) = \frac{g(x)}{h(x)}$
$$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$$
In your case, we have $$f(x) = \frac{x^2+\sin2x}{2x+\cos^2x}$$
So put $g(x) = x^2 + \sin 2x,\;$ and $\;h(x) = 2x + \cos^2x$.
Now, $g'(x) = 2x + 2\cos 2x,\;$ and $\;h'(x) = 2 - 2\cos x \sin x = 2 - \sin(2x)$.
So $$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$$ $$f'(x) = \frac{dy}{dx} = \frac{(2x + 2\cos 2x)(2x + \cos^2 x) + (x^2 + \sin 2x)(2 - \sin 2x)}{\left(2x + \cos^2 x\right)^2}$$
The rest is of the work is merely algebraic simplification.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Manipulating Algebraic Expression
$a + b + c = 7$ and $\dfrac{1}{a+b} + \dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{7}{10}$. Find the value of $\dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b}$.
I algebraically manipulated the second equation to get:
$\dfrac{(b+c)(c+a) + (a+b)(c+a) + (a+b)(b+c)}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$
$\dfrac{bc+ab+c^2+ac+a^2+bc+ba+ab+ac+b^2+bc}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$
$\dfrac{(a+b+c)^2}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$
$\dfrac{7^2}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$
$(a+b)(b+c)(c+a) = 70$
I'm stuck after this.
| HINT:
$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}$$
$$=\frac a{b+c}+1-1+\frac b{c+a}+1-1+\frac c{a+b}+1-1$$
$$=-3+(a+b+c)\left(\frac 1{b+c}+\frac 1{c+a}+\frac 1{a+b}\right)$$
Using summation notation, $$\sum_{a,b,c} \frac a{b+c}=-3+\sum_{a,b,c} \left(\frac a{b+c}+1\right)=-3+(a+b+c)\sum_{a,b,c}\frac1{b+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluating the $\int \frac{x+2}{x^2-4x+8}$ - a doubt I have to find the antiderivative of $f(x) = \dfrac{x+2}{x^2-4x+8}$
I rewrote it to the form $$ \dfrac{x-2}{x^2 -4x + 8} + \dfrac{1}{\frac{1}{4} (x-2)^2 +1}$$
The next step supposedly is $$F(x) = \dfrac{1}{2}\ln|x^2-4x+8| + 2 \arctan\left(\dfrac{1}{2}(x-2)\right) + c$$
But now there is a problem which I've always had: I don't know why the $\dfrac{1}{2}$ in front of the $\ln(x)$ is there, same for the $2$ and the $\dfrac{1}{2}$ for the $\arctan(x)$. For me simplifying the function to standard integrals is easy, however finding the right factors and such is hard.
| When we use substitution, we need to determine $u$ as a function of $x$, and so, as a sort of "reverse" chain rule, we need also two substitute $du$ as a function of $dx$:
$$\int \dfrac{x-2}{x^2 -4x + 8}\,dx$$
Here, $u = x^2 - 4x + 8 \implies \,du = 2x - 4 = 2(x-2)\,dx \iff \frac 12 du = x - 2 \,dx$
This gives our substituted integrand:
$$\begin{align} \int \dfrac{x-2}{x^2 -4x + 8}\,dx & = \int \frac{\frac 12 \,du}{u} \\ \\ & = \frac 12 \int \frac {du}{u} \\ \\ & = \frac 12 \ln|u| + C =\frac 12 \ln|x^2 - 4x + 8| + C\tag{1}\end{align}$$
For the second term:
$$\int\dfrac{1}{\frac{1}{4} (x-2)^2 +1}\,dx = \int \dfrac{1}{\left(\frac 12(x-2)\right)^2 + 1} \,dx$$
We let $$u = \frac 12(x - 2) \implies du = \frac 12 dx \implies 2 du = dx$$
Substituting we have $$\begin{align} \int\dfrac{1}{\left(\frac 12(x-2)\right)^2 + 1} \,dx & = \int \frac 1{u^2 + 1}\cdot(2\,du) \\ \\ & = 2 \int \cdot \dfrac{1}{u^2 + 1} \,du \\ \\& = 2\arctan\left(u\right) + C \\ \\ & = 2 \arctan\left(\frac 12(x - 2)\right) + C \tag{2} \end{align}$$
Putting $(1)$ and $(2)$ together:
$$F(x) = \dfrac{1}{2}\ln|x^2-4x+8| + 2 \arctan\left(\dfrac{1}{2}(x-2)\right) + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding Eigenvalues and Eigenvectors weird equations I have matrix:
$A = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$
And I have to find eigenvalues and coresponding eigenvectors.
I get $det(A-xI_3) = -x(x-1)^2$, so eigenvalues are $x_1 = 0$ and $x_2 = 1$
Now I want to find coresponding eigenvectors. I get:
$A - 0I_3 = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right] $
and
$A - I_3= \left[\begin{array}{ccc}2&-1&-2\\2&-1&-2\\2&-1&-2\end{array}\right]$
To first equation I have solution $y = x, z = x$, but I dont know what eigenvectors are and to the second I dont even know the solution to equation $(A-I_3)[x,y,z]^T$. Could you help me?
| Since you have the eigenvalues, you can consider finding a basis of the eigenvectors associated to each eigenvalues, which means resolving (with Gauss method) this equation (for the 0 eigenvalue) :
$X=
\begin{pmatrix}x\\y\\z\end{pmatrix} \in E_0 \Leftrightarrow
\begin{cases}
3x-y-2z&=0 \\
2x-2z&=0 \\
2x-y-z&=0
\end{cases}\Leftrightarrow AX=0.X$
$\begin{cases}
3x-y-2z&=0 \\
2x-2z&=0 \\
2x-y-z&=0
\end{cases}
\Leftrightarrow
\begin{cases}
2x-y-z&=0 \\
-y+z&=0 \\
0&=0
\end{cases}
\Leftrightarrow
\begin{pmatrix}x\\y\\z\end{pmatrix}\in \{\begin{pmatrix}s\\s\\s\end{pmatrix},s\in\mathbb{K}\}$
Finally :
$E_0 = Vect(\begin{pmatrix}1\\1\\1\end{pmatrix})$
The case of the eigenvalue 1 starts exactly the same way :
$X=
\begin{pmatrix}x\\y\\z\end{pmatrix} \in E_1 \Leftrightarrow
\begin{cases}
3x-y-2z&=x \\
2x-2z&=y \\
2x-y-z&=z
\end{cases}\Leftrightarrow AX=1.X$ and you find the above eigenvectors ;-).
So, you have to find a basis of the plane : $P: 2x-y-2z=0$.
One way is to parameter the problem :
$\begin{pmatrix}x\\y\\z\end{pmatrix}\in P \Leftrightarrow \begin{pmatrix}x\\y\\z\end{pmatrix}\in \lbrace\begin{pmatrix}\frac{1}{2}s+t\\s\\t\end{pmatrix},(s,t)\in\mathbb{K}\rbrace$
So you find immediately that $E_1 = Vect(\begin{pmatrix}\frac{1}{2}\\1\\0\end{pmatrix},\begin{pmatrix}1\\0\\1\end{pmatrix})$
This is the classic method of resolution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the integers satisfying $a^3+b^3-a^2b^2-(a+b)^2c=0$ Suppose $a,b,c\in \mathbb{Z}_{\ge0}$. How do I find all such $a$, $b$ and $c$ which satisfy
$$a^3+b^3-a^2b^2-(a+b)^2 c=0$$
| If $a>0,b>0,$ $a^3+b^3-a^2b^2-(a+b)^2 c=0 $
$\Rightarrow a^3+b^3\equiv a^2b^2 \pmod {(a+b)^2}$
$\Rightarrow ab\equiv 0 \pmod {a+b}$
Denote $\dfrac{ab}{a+b}=k\in \mathbb Z,(a,b)=d,a=a_1 d,b=b_1 d,(a_1,b_1)=1,\dfrac{a_1b_1d}{a_1+b_1}=k.$
Since $(a_1b_1,a_1+b_1)=1,a_1+b_1\mid d.$
Let $d=t(a_1+b_1),$ then $k=a_1b_1t,a=a_1(a_1+b_1)t,b=b_1(a_1+b_1)t.$
$c=\dfrac{a^3+b^3-a^2b^2}{(a+b)^2}=(a_1^2-a_1b_1+b_1^2-ta_1^2b_1^2)t<(a_1^2+b_1^2-a_1^2b_1^2)t=(1-(a_1^2-1)(b_1^2-1))t$
If $a_1>1$ and $b_1>1,$ then $c<0.$ So $a_1=1$ or $b_1=1,$ but no matter in any case, since $\dfrac{c}{t}\in \mathbb Z,$ $\dfrac{c}{t}<1$ means $\dfrac{c}{t}=0.$
Hence if $a>0$ and $b>0$ then $c=0.$
*
*If $a>0,b>0,c=0,$ then $a^3+b^3-a^2b^2=0,(a_1^3+b_1^3)=da_1^2b_1^2,$ since $(a_1^3+b_1^3,a_1^2b_1^2)=1,$ we have $a_1=b_1=1,d=2.$ We get $a=b=2,c=0.$
*If $a=0,$ then $b^3-b^2c=0, b=0$ or $c=b.$
Hence the only solutions are $(a,b,c)=(0,0,0)(2,2,0)(0,0,n)(0,n,n)(n,0,n).(n>0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/431593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
relation between arithmetic series and `square` arithmetic series For example:
$$1+2+\text{...}+n=\frac{n(n+1)}{2}~~~(1)$$
$$1^2+2^2+\text{...}+n^2=\frac{n(n+1)(2n+1)}{6}~~~(2)$$
In this equality, I sometimes recall by heart
$\frac{n(2n+1)(2n+3)}{6}$ or others.
Why I cannot memorize some formulas exactly over these years?
Question1
How to derive $\frac{n(n+1)(2n+1)}{6}$ from $1^2+2^2+\text{...}+n^2$?
Question2
Now I happen to notice that $n(n+1)$ are the same to the summation in
$1+2+\text{...}+n=\frac{n(n+1)}{2}$
What's the relation between (1) and (2).
I think there must be large materials about more general cases
$$\sum _{i=0}^n i^k\text{ },k=1,2,\text{...},n~~~(3)$$
What's the formal terminology about that?
| First, we look at your observation that
$$1+2+3+\cdots+n=\frac{n(n+1)}{2}.\tag{1}$$
If you are somewhat familiar with combinatorics, there is a nice way of seeing it via the binomial coefficient $\binom{n+1}{2}$, sometimes called $C(n+1,2)$ or $C_2^{n+1}$ (there are other names).
The right-hand side of (1) is the number of ways to choose $2$ numbers from the $n+1$ numbers $1$ to $n+1$. Let's count the number of ways to choose $2$ numbers in another way.
Maybe the second largest number chosen is $n$. There is then $1$ way of choosing the largest number.
Maybe the second largest number chosen is $n-1$. There are then $2$ ways to choose the largest number.
Maybe the second largest number chosen is $n-2$. There are then $3$ ways to choose the largest number.
Continue. Finally, if the second largest number chosen is $1$, there are $n$ ways to choose the largest number.
It follows that $\binom{n+1}{2}=1+2+3+\cdots +n$.
The nice thing about this viewpoint is that it generalizes. The same reasoning shows, for example, that
$$\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\cdots+\binom{n}{2} =\binom{n+2}{3}.\tag{2}$$
The binomial coefficient on the right of (2) is equal to $\frac{(n+2)(n+1)(n)}{3!}.$
The sum on the left of (2) is
$$\frac{(1)(2)}{2}+\frac{(2)(3)}{2}+\frac{(3)(4)}{2}+\cdots +\frac{n(n+1)}{2}.\tag{3}.$$
The sum (3) is a close relative of the sum of the first $n$ squares. For it is equal to
$$\frac{1}{2}\left(1^2+1+2^2+2+3^2+3+\cdots+n^2+n\right).$$
Since we already know the sum $1+2+3+\cdots +n$, we can now find a formula for the sum of the first $n$ squares. Putting things together, it is
$$2\frac{(n+2)(n+1)(n)}{3!} -\frac{n(n+1)}{2}.$$
Algebraic simplification (bring to common denominator $6$) then yields the hard to remember formula for the sum of the first $n$ squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/432091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Show that there are infinitely many solutions of distinct natural numbers $m,n$ such that $n^3+m^2\mid m^3+n^2$
Show that there are infinitely many solutions of distinct natural numbers $m,n$ such that $n^3+m^2\mid m^3+n^2$.
This question appeared in Round $2$ of British Math Olympiad $2007-08$. I have been trying this problem since two days. And got no sequence of numbers satisfying it.
| Choose $a\in\mathbb{N}$, and set $m=an$. Then the condition becomes $n^3+a^2n^2|a^3n^3+n^2$, which is equivalent to $$n+a^2|a^3n+1$$
We now set $n=a^5-a^2-1$ and see that $$(a^3-1)(n+a^2)=(a^3-1)(a^5-1)=a^8-a^5-a^3+1=a^3n+1$$
Update, pulling back the veil: I looked for $k$ such that $k(n+a^2)=a^3n+1$ and I could solve for $n$. After one or two false attempts I tried $k=a^3-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/433711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
the limit $\lim_{a \to \infty}\int_{a}^{a+1}\frac{x+\sqrt{x}}{x-\sqrt{x}}\,\mathrm dx=1$ I proved this using mean value theorem
$$\lim_{a \to \infty}\int_{a}^{a+1}\frac{x+\sqrt{x}}{x-\sqrt{x}}\,\mathrm dx=1$$
because $$\frac{x+\sqrt{x}}{x-\sqrt{x}}-1=\frac{2\sqrt{x}}{x-\sqrt{x}}$$
it suffices prove that $$\int_{a}^{a+1}\frac{2\sqrt{x}}{x-\sqrt{x}}\,\mathrm dx\rightarrow0 \text{ as} \ a\to \infty $$
the mean value theorem applies in the interval $[a,a+1]$.
My question is: Can I prove this with another argument?
My idea was this : $$\int_{a}^{a+1}\frac{2\sqrt{x}}{x-\sqrt{x}}dx=\int_{a}^{a+1}\frac{2}{\sqrt{x}-1}dx=\frac{2}{\sqrt{c}-1}$$ for some $c\in [a,a+1]$
| Similar idea: Suppose $x>0$ and let $f(x) = \frac{x+\sqrt{x}}{x-\sqrt{x}}= \frac{1+\frac{1}{\sqrt{x}}}{1-\frac{1}{\sqrt{x}}}$. It is clear that $\lim_{x \to \infty} f(x) = 1$.
Let $\epsilon>0$ and choose $N$ large enough so that if $x \ge N$, then $|f(x)-1| < \epsilon$. Hence if $a > N$, we have
$ |\int_a^{a+1} ( f(x) - 1)dx| \le \int_a^{a+1} | f(x) - 1|dx < \epsilon $.
Alternatively:
Note that if $x>1$, we have $f(x) = (1+\frac{1}{\sqrt{x}})(1+\frac{1}{(\sqrt{x})^1}+\frac{1}{(\sqrt{x})^2}+\cdots)$, or,
$f(x) = 1 + 2(\frac{1}{(\sqrt{x})^1}+\frac{1}{(\sqrt{x})^2}+\cdots)$, and furthermore, for any $N>1$, the convergence is uniform for $x \ge N$. Hence, for $a >1$, we can form the estimate
\begin{eqnarray}
|\int_a^{a+1} f(x) dx -1| &=& 2\int_a^{a+1} (\frac{1}{(\sqrt{x})^1}+\frac{1}{(\sqrt{x})^2}+\cdots) dx \\
&\le& 2 \sum_{n=1}^\infty \frac{1}{(\sqrt{a})^n} \\
&=& 2 \frac{1}{1-\frac{1}{\sqrt{a}}}
\end{eqnarray}
The desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/434212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Prove an inequality concerning $\sqrt[3]{4a^3+4b^3}+\sqrt[3]{4b^3+4c^3}+\sqrt[3]{4c^3+4a^3}$ Let $a,b,c$ be positive. I need to prove
$\sqrt[3]{4a^3+4b^3}+\sqrt[3]{4b^3+4c^3}+\sqrt[3]{4c^3+4a^3}\leq \dfrac{4a^2}{a+b}+\dfrac{4b^2}{b+c}+\dfrac{4c^2}{c+a}$
Thanks!
| Since $\sqrt[3]{\frac{a^3+b^3}{2}}\leq\frac{a^2+b^2}{a+b}\Leftrightarrow(a-b)^4(a^2+ab+b^2)$, we obtain
$$\sum_{cyc}\sqrt[3]{4(a^3+b^3)}\leq\sum_{cyc}\frac{2(a^2+b^2)}{a+b}=\sum_{cyc}\left(\frac{2(a^2+b^2)}{a+b}+2(a-b)\right)=\sum_{cyc}\frac{4a^2}{a+b}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that: $a^2+b^2+(1-a-b)^2\ge \frac {1}{3}$ Where $a$ and $b$ are any given real number. I have tried solving it using partial derivative.
$$
s=a^2+b^2+(1-a-b)^2$$
$$\frac{\partial s}{\partial a}=2a-2(1-a-b) \tag{1}$$
$$\frac{\partial s}{\partial b}=2b-2(1-a-b) \tag{2}$$
for maxima both (1) and (2) are 0..from here we get two equations from where we get values of $a$ and $b$
$$2a-2(1-a-b)=0 \tag{3}$$
$$2b-2(1-a-b)=0 \tag{4}$$
putting the values of $a$ and $b$ we find from (3) and (4) in the function the maximum value of the function should be $\frac {1}{3}$.which in this case is not
| You want to estimate $R=a^2+b^2+c^2$, where $a+b+c=1$. Note that $R$ is the squared length of the radius vector to $(a,b,c)$. The equation $a+b+c=1$ gives you some plane, and the shortest radius vector to the plane is the perpendicular on it. It's easy to calculate that its length(height of a pyramid) is equal to $\frac{1}{ \sqrt{3}}$, so $R\ge\frac 1 3$.
Anyway, I like the way with Cauchy-Schwarz inequality more - it is the most obvious and easy way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Partial fractions $\int \frac{(3x^2 - 4x + 5)\,dx}{(x-1)(x^2+2)}$ $$\int \frac{(3x^2 - 4x + 5 )\, dx}{(x-1)(x^2+2)}$$
I am going to use undetermined coefficients since it seems straightforward, no wacky matrices or tables to memorize.
$$\int \frac{(3x^2 - 4x + 5 )\, dx}{(x-1)(x^2+2)} \quad = \quad\int \left(\frac {A}{x} + \frac{B}{(x-1)^2}+\frac{C}{(x-1)}+D\right)\,dx$$
I get $$A(x^3 - 3x^2 + 3x - 1) +B(x^3 - 2x^2 + x) + C(x^2 - x) + Dx$$
This gives me the sets of
$$\begin{align}
A + B & = 0\\ \\
-3A + -2B + C & = 0 \\ \\
3A + B - C + D & = 0 \\ \\
-A &= 0
\end{align}$$
This is obviously wrong because A and B are now 0. What do I do?
| We need to use the factors of the denominator of the following integral, to decompose into partial fractions. We need for the numerator of the second degree factor (second fraction below) to be of the form of a degree one polynomial: $Bx + C$, since $x^2 + 2$ cannot be factored into the product of first-degree factors:
$$\int \frac{3x^2 - 4x + 5}{(x-1)(x^2+2)}\,dx = \int \frac{A}{x- 1} + \frac{Bx + C}{x^2 + 2}\,dx\tag{1}$$
Then we need for $$A(x^2 + 2) + (Bx + C)(x-1) = 3x^2 - 4x + 5\tag{2}$$
We expand the terms on the left-hand side of equation $(2)$, group coefficients for $\,x^2,\, x,\; \text{and constants},\;$ and equate with the right-hand side of $(2)$:
$$\begin{align} A(x^2 + 2) + (Bx + C)(x-1) & = (A x^2 + 2A) + (Bx^2 + Cx - Bx - C)\\ \\ & = \color{blue}{\bf (A + B)}x^2 + \color{red}{\bf (C - B)} x + \color{green}{\bf (2A - C)} \\ \\ & = \color{blue}{\bf 3}x^2 + \color{red}{\bf - 4}x + \color{green}{\bf 5}\end{align}$$
Now we match up coefficients of the left hand side with those on the right (I'll use color to distinguish between the three matchings we'll need to make), to obtain three equations in three unknowns $A,\, B, \,C :$:
This gives us the following system of linear equations to solve:
$$\begin{align} \color{blue}{\bf A + B} & = \color{blue}{\bf 3}\\
\color{red}{\bf C - B} & = \color{red}{\bf -4} \\
\color{green}{\bf 2A - C} & = \color{green}{\bf 5} \end{align}$$
suppose we add the first two of these equations. Then we get: $$\begin{align} A + B - B + C & = 3 - 4 \\ A + C & = -1\end{align}$$. Now adding this "new" equation with the third equation above, we get $$\begin{align} 2A - C + A + C & = 5 - 1\\ 3A & = 4 \\ A & = \dfrac 43\end{align}$$
Now, we know that $2A -C = 5,$ so $$\begin{align} 2A - C & = 5 \\ C & = 2A - 5 \\
C & = 2\cdot \dfrac 43 -\dfrac{15}{3} \\
C &= \left(-\dfrac 73\right)
\end{align}.$$
And we know that $$\begin{align} A + B & = 3 \\ B & = 3 - A = 3 - \dfrac 43 \\ B& = \dfrac 53\end{align}$$
That gives us, in all: $$A = \dfrac 43,\;B = \frac 53,\;C = -\dfrac 73$$
Now, we have the integral:
$$\begin{align} \int \frac{A}{x- 1} + \frac{Bx + C}{x^2 + 2}\,dx & = \int \frac 13 \left(\frac{4}{x - 1} + \frac{5x - 7}{x^2 + 2}\right)\,dx \\
& = \frac 43 \int \dfrac {dx}{x - 1} + \frac {1\cdot 5}{3\cdot 2} \int \frac{2x \,dx}{x^2 + 2} - \frac 73 \int \frac {dx}{x^2 + (\sqrt 2)^2} \\
& = \frac 43 \ln|x - 1| + \frac 56 \ln(x^2 + 2) - \frac 7{3\sqrt 2} \tan^{-1}\left(\frac{x}{\sqrt 2}\right) + C
\end{align}$$
We can get really clever and write the sum (first two terms) $ \dfrac 43\ln|x - 1| + \dfrac 56\ln(x^2 + 2)$ as $\ln(x - 1)^{4/3} + \ln(x^2 + 2)^{5/6} = \ln\left((x-1)^{4/3}(x^2 + 2)^{5/6}\right) = \ln\left((x - 1)^8(x^2 + 2)^5\right)^{1/6}$, giving us an answer:
$$\ln\left((x - 1)^8(x^2 + 2)^5\right)^{1/6} - \frac 7{3\sqrt 2} \tan^{-1}\left(\frac{x}{\sqrt 2}\right) + C = $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/439272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Double integral over a region Given $f(x,y)=\displaystyle\frac{x^2}{x^2+y^2}$ and $D=\{(x,y) : 0 \leq x \leq 1, x^2 \leq y \leq 2-x^2\}$ i have to solve $\displaystyle\int\displaystyle\int_Df(x,y)dA$.
Here's my try:
(1) Changing variables
$x = \sqrt{v-u}$, $y= v+u$.
(1.1) Since $0 \leq x \leq 1$, then $0 \leq v-u \leq 1 \rightarrow u \leq v \leq 1+u$
(1.2) Since $x^2 \leq y \leq 2-x^2$, then $v-u \leq v+u \leq 2-v+u \rightarrow -u \leq u \rightarrow 0\leq u$ and $v \leq 2-v \rightarrow v \leq 1$
(1.3) It seems that now i should integrate over $S = \{(u,v) : 0\leq u \leq v \leq 1 \}$ (the upper triangle in $[0,1]\times[0,1]$ ?), so i may as well put $S = \{(u,v) : 0 \leq v \leq 1, 0 \leq u \leq v \}$.
(2) Alright, what do i need to calculate the integral?
(2.1) First, i should calculate the Jacobian
$ \displaystyle\frac{\partial(x,y)}{\partial(u,v)} = \left| \begin{array}{cc}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\
\end{array} \right| = \left| \begin{array}{cc}
-\frac{1}{2\sqrt{v-u}} & \frac{1}{2\sqrt{v-u}} \\
1 & 1 \\
\end{array} \right| = -\frac{1}{\sqrt{v-u}}$
(2.2) Then i have to solve $\displaystyle\int_0^1\displaystyle\int_0^v \frac{v-u}{(v-u)+(v^2+2uv+u^2)}\bigg(-\frac{1}{\sqrt{v-u}} \bigg)dvdu$
$=-\displaystyle\int_0^1\displaystyle\int_0^v \frac{\sqrt{v-u}}{v^2+v(1+2u) + (u^2-u) }dvdu$
(2.3) Well, here i'm stuck.I've been thinking about taking $z = \sqrt{v-u}$ and then $dz = \displaystyle\frac{1}{2\sqrt{v-u}}dv$ wich means $dv = 2zdz$, this would lead to an integral of the form
$2\displaystyle\int\displaystyle\int \frac{z^2}{z^2+z(1+4u)+ 4u^2 }dvdu = 2\displaystyle\int\displaystyle\int \frac{z^2}{(z+(\frac{1}{2}+2u))^2-2u }dvdu$ and if i put $w = z+(\frac{1}{2}+2u)$ i'll have $2\displaystyle\int\displaystyle\int \frac{(w-\frac{1}{2}-2u)^2}{w^2-2u }dwdu$ but it seems that the last one will lead to some ugly shaped solution and i would have a hard time getting the final answer.
What would be the best way to solve this?
| $$
\begin{align}
&\int_0^1\int_{x^2}^{2-x^2}\frac{x^2}{x^2+y^2}\,\mathrm{d}y\,\mathrm{d}x\tag{1}\\
&=\int_0^1\int_{x}^{\frac2x-x}\frac{x}{1+y^2}\,\mathrm{d}y\,\mathrm{d}x\tag{2}\\
&=\int_0^1x\left(\arctan\left(\frac2x-x\right)-\arctan(x)\right)\,\mathrm{d}x\tag{3}\\
&=\int_0^1x\arctan\left(\frac{2(1-x^2)}{x(3-x^2)}\right)\,\mathrm{d}x\tag{4}\\
&=\int_0^1\frac{x^2}{2}\frac{\frac{2(3+x^4)}{x^2(3-x^2)^2}}{1+\left(\frac{2(1-x^2)}{x(3-x^2)}\right)^2}\,\mathrm{d}x\tag{5}\\
&=\int_0^1\frac{x^2(3+x^4)}{x^2(3-x^2)^2+4(1-x^2)^2}\,\mathrm{d}x\tag{6}\\
&=\int_0^1\left(1-\frac1{2(1+x^2)}+\frac{5x^2-4}{2(x^4-3x^2+4)}\right)\,\mathrm{d}x\tag{7}\\
&=1-\frac\pi8+\int_0^1\frac14\left(\frac{\sqrt7x-2}{x^2-\sqrt7x+2}-\frac{\sqrt7x+2}{x^2+\sqrt7x+2}\right)\,\mathrm{d}x\tag{8}\\
&=1-\frac\pi8+\int_0^1\frac{\sqrt7}{8}\left(\frac{2x-\sqrt7}{x^2-\sqrt7x+2}-\frac{2x+\sqrt7}{x^2+\sqrt7x+2}\right)\,\mathrm{d}x\\
&+\int_0^1\frac32\left(\frac1{(2x-\sqrt7)^2+1}+\frac1{(2x+\sqrt7)^2+1}\right)\,\mathrm{d}x\tag{9}\\
&=1-\frac\pi8+\frac{\sqrt7}{8}\log\left(\frac{3-\sqrt7}{3+\sqrt7}\right)\\
&+\frac34(\arctan(2-\sqrt7)+\arctan(2+\sqrt7))\tag{10}\\
&=1+\frac\pi{16}+\frac{\sqrt7}{8}\log(8-3\sqrt7)\tag{11}
\end{align}
$$
Justification:
$\ \;(1)$: get limits from the problem
$\ \;(2)$: change variables $y\mapsto xy$
$\ \;(3)$: integrate in $y$
$\ \;(4)$: combine arctans
$\ \;(5)$: integrate by parts
$\ \;(6)$: algebra
$\ \;(7)$: partial fractions
$\ \;(8)$: integrate and partial fractions
$\ \;(9)$: separate into easily integrable pieces
$(10)$: integrate
$(11)$: simplify: $\quad\frac{3-\sqrt7}{3+\sqrt7}=8-3\sqrt7\quad$ and $\quad\arctan(2-\sqrt7)+\arctan(2+\sqrt7)=\frac\pi4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/440183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Maxima/Minima Problem I am working on this Maxima and Minima Problem :
Determine the Max and Min distance of origin from the curve $3x^2+4xy+6y^2=140$
I tried it solving using the lagrange's method of multipliers. I get the following equations
$x+3x\theta+2y\theta=0$
$y+4y\theta+2x\theta=0$
$2z=0$
$3x^2+4xy+6y^2=140$
So i get $z=0$
Please suggest how to go about solving for $x$ and $y$ ?
| \begin{align}
\min_{x,y}&x^2+y^2\\s.t.\\
3x^2+6y^2+4xy&=140\\
L(x,\lambda)&=x^2+y^2-\lambda(3x^2+6y^2+4xy-140)\\
\nabla L(x,\lambda)&=\binom{2x}{2y}-\lambda\binom{6x+4y}{12y+4x}=0\\
\text{Necessary Conditions:}\\
2x-\lambda(6x+4y)&=0\\
2y-\lambda(12y+4x)&=0\\
3x^2+6y^2+4xy&=140\\
\end{align}
Solve these 3 equations and you have your answer.
\begin{align}
2x&=\lambda(6x+4y)\\
\lambda&=\frac{2x}{6x+4y}\\
2y-\lambda(12y+4x)&=0\\
2y-\frac{2x}{6x+4y}(12y+4x)&=0\\
y&=\frac{x}{6x+4y}(12y+4x)\\
(6x+4y)y&=x(12y+4x)\\
6xy+4y^2&=12xy+4x^2\\
2y^2&=3xy+2x^2\\
3x^2+6y^2+4xy&=140\\
\text{Solving you get,}\\
x=\pm2,y=\pm4\\
d=20
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/441137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Where does $\sin 3° =3\sin 1° -4 \sin^3 1°$ come from? Wikipedia makes the claim:
"Though a complex task, the analytical expression of $\sin 1°$ can be obtained by analytically solving the cubic equation $\sin 3° =3\sin 1° -4 \sin^3 1°$ from whose solution one can analytically derive trigonometric functions of all angles of integer degrees."
Where did this equation come from? I did a quick google search and I didn't find much.
P.S. If possible do not answer this using series.
| You can derive the identity $$\sin(3x) = 3\sin(x)\cos^2(x) - \sin^3(x)$$ by applying the angle sum rules $$\begin{align*} \cos(a + b) &= \cos(a)\cos(b) - \sin(a)\sin(b) \\ \sin(a+b) &= \sin(a)\cos(b) + \sin(b)\cos(a)\end{align*}$$
as follows:
$$\begin{align*} \sin(3x) &= \sin(2x)\cos(x) + \cos(2x)\sin(x) \\
&= \sin(x)\cos^2(x) + \sin(x)\cos^2(x) + \cos^2(x)\sin(x) - \sin^3(x) \\
&= 3\sin(x)\cos^2(x) - \sin^3(x)\end{align*}$$
The angle sum formulas are also how one can obtain $\sin(n), \cos(n)$ for all other integers once $\sin(1)$ is known, as $$\begin{align*} \cos(n+ 1) &= \cos(n)\cos(1) - \sin(n)\sin(1) \\ \sin(n+1) &= \sin(n)\cos(1) + \sin(1)\cos(n)\end{align*}.$$
(If you have some experience with complex numbers, you should figure out why the angle sum formulas are just a consequence of $$e^{i\theta} = \cos(\theta) + i\sin(\theta),$$ so all this comes from just one identity!)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/442537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Summation of weighted squares of binomial coefficients Show that
$$\sum_{k=0}^n \left[ \frac{n-2k}{n} {n\choose k}\right]^2=\frac{2}{n}{2n-2 \choose n-1}.$$
| $$
\begin{align}
\sum_{k=0}^n\left[\frac{n-2k}{n}\binom{n}{k}\right]^2
&=\sum_{k=0}^n\left[\binom{n}{k}-2\binom{n-1}{k-1}\right]^2\\
&=\sum_{k=0}^n\left[\binom{n-1}{k}-\binom{n-1}{k-1}\right]^2\\
&=\sum_{k=0}^n\binom{n-1}{k}^2+\binom{n-1}{k-1}^2-2\binom{n-1}{k}\binom{n-1}{k-1}\\
&=2\binom{2n-2}{n-1}-2\sum_{k=0}^n\binom{n-1}{k}\binom{n-1}{n-k}\\
&=2\binom{2n-2}{n-1}-2\binom{2n-2}{n}\\
&=2\binom{2n-2}{n-1}-2\frac{n-1}{n}\binom{2n-2}{n-1}\\
&=\frac2n\binom{2n-2}{n-1}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/443026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Proof by induction that the sum of terms is integer I'm having some trouble in order to solve this induction proof.
Proof that $\forall{n} \in \mathbb{N}$ the number $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n$ is an integer.
I've tried proving this by induction, but I've not succeed so far. What I did was:
Let's assume that $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n = k$, with $k \in \mathbb{Z}$. So, by induction,
$n=1$
$\frac{1}{5} + \frac{1}{3} + \frac{7}{15} = 1$ and $1 \in \mathbb{Z}$.
Inductive step
I want to prove that if $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n = k$, with $k \in \mathbb{Z}$ then $\frac{1}{5}(n+1)^5+\frac{1}{3}(n+1)^3 + \frac{7}{15}(n+1) = l$, with $l \in \mathbb{Z}$
And then I'm stuck. I've tried developing every binomial but, for example, for $\frac{1}{5}(n+1)^5$ I get a $\frac{1}{5}n^4$ that is not in the inductive hypothesis, so therefore, I cannot get rid of it and it's not an integer, so I cannot conclude that $l \in \mathbb{Z}$.
Any help or ideas from where I can follow? Thanks in advance!
| Hint
$$5|(n+1)^5-n^5-1 \,.$$
$$3|(n+1)^3-n^3-1 \,.$$
Write $a=\frac{(n+1)^5-n^5-1}{5} \,;\, b=\frac{(n+1)^3-n^3-1}{3}$, and express $P(n+1)-P(n)$ in terms of $a,b$, where $P(n)=\frac{1}{5}n^5+\frac{1}{3}n³ + \frac{7}{15}n$.
To prove the Hint, you can use either Fermat Little Theorem or binomial Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/443538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Find the value of $\cos^{12}\theta + 3\cos^{10}\theta + 3\cos^{8}\theta + \cos^6\theta + 2\cos^4\theta + 2\cos^2\theta - 2$ We are given that $\sin\theta + \sin^3\theta + \sin^2\theta = 1$
Find the value of $\cos^{12}\theta + 3\cos^{10}\theta + 3\cos^{8}\theta + \cos^6\theta + 2\cos^4\theta + 2\cos^2\theta - 2$
Now, I was able to establish the following from the first equation:
$\sin\theta + \sin^3\theta + \sin^2\theta = 1 = \sin^2\theta + \cos^2\theta \implies \sin\theta + \sin^3\theta = \cos^2\theta$
The next obvious step was to simplify the second expression. I let $\cos^2\theta = x$:
$f(x) = x^6 + 3x^5 + 3x^4 + x^3 + 2x^2 + 2x - 2$
$f(-1) = 0 \implies f(x) = (x + 1)(x^5 + 2x^4 + x^3 + 2x - 2)$
I was stuck after this.
| Observe that the powers of $\cos \theta$ are all even, suggesting that we should use the conversion $\cos^2 \theta = 1 - \sin ^2 \theta$. For simplicity, let $x = \sin \theta$. We are given that
$$(x + x^2 + x^3) = 1$$
and want to find
$$(1-x^2)^6 + 3(1-x^2)^5 + 3(1-x^2)^4 + (1-x^2)^3 + 2(1-x^2)^2 + 2(1-x^2) - 2 $$
By long division, we could factor out $x^3 + x^2 + x -1 $ to get $214x^2 + 20x - 68$, but then it's not clear what to do after that.
You could use the fact that $x^3 + x^2 + x - 1 = 0 $ has 1 real root; but I doubt that is how they want you to proceed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/444219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove a limit using the epsilon-delta definition $\lim_{x\to-2}\frac{4x-1}{x+1} = 9$
Given $\epsilon>0$, $$(\exists \delta(\epsilon)>0) \left( |x+2|<\delta \implies \left|{\frac{4x-1}{x+1} - 9}\right| < \epsilon \right)$$
So, if $|x+2|<\delta$, then:
$$\left|{\frac{4x-1}{x+1} - 9}\right| = \left|{\frac{4x-1-9(x+1)}{x+1}}\right| = \left|{\frac{-5x-10}{x+1}}\right| = \left|{\frac{-1 (5x+10)}{x+1}}\right| = {\frac{|-1||5x+10|}{|x+1|}} = {\frac{5|x+2|}{|x+1|}}$$
I've found the expression $5|x+2|$ in the numerator:
$|x+2| < \delta = 1/2$
$-1/2<x+2<1/2$
$-5/2<x<-3/2$
$-3/2<x+1<-1/2 \implies$ ?
A colleague told me to leave both members of this inequality as positives. Why do we need both sides to be positives?
| You're right to want to bound values of $x$ away from $-1$, and to do so by $\frac{1}{2}$ is fine. The inequality you obtain by requiring $x$ to be within $\frac{1}{2}$ of $2$ is
$$-\frac{5}{2}<x<-\frac{3}{2}$$
which you have written. Then, as you've written, we have
$$-\frac{3}{2}<x+1<-\frac{1}{2}$$
Notice that the above inequality implies that $|x+1|>\frac{1}{2}.$ It follows that
$$\frac{5|x+2|}{|x+1|}<10|x+2|$$
So let $\delta(\epsilon)<\min\{\frac{1}{2},\frac{\epsilon}{10}\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/446014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How to make this difference equation continuous? $$ F_n=F_{n-1}-F_{n-2} $$
How can I convert this oscillating sequence into a continuous function? IE get it in terms of n.
| This sequence has exactly 6 values:
($a$,$\ $ $b$, $\ $ $b-a$, $\ $ $-a$, $\ $ $-b$, $\ $ $a-b$) in cycle.
So, there is a reason to build trigonometric function with period of $6$:
$$
F(n) = A \sin \frac{\pi n}{3} + B \cos\frac{ \pi n}{3},
$$
If $F_0 = a$, $F_1 = b$, then
$$
\left\{
\begin{array}{c}
A\cdot 0 + B\cdot 1 = a;\\
A\cdot \dfrac{\sqrt{3}}{2} + B \cdot \dfrac{1}{2} = b; \\
A\cdot \dfrac{\sqrt{3}}{2} - B \cdot \dfrac{1}{2} = b-a; \\
\end{array}
\right.
$$
$B=a$, $A=\dfrac{2b-a}{\sqrt{3}}$.
So,
$$
F(n) = \dfrac{2b-a}{\sqrt{3}} \cdot \sin \frac{\pi n}{3} \;+ \; a \cdot \cos\frac{\pi n}{3},
$$
where $a=F_0$, $b=F_1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/446333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$y=\{x\}$ represents greatest integer less than $x$ then solve $x^2-7\{x\}+5=0$. $y=\{x\}$ represents greatest integer less than $x$,
e.g.
*
*$\{1.1\}=1$
*$\{2\}=1$
*$\{5\}=4$
*$\{3.7\}=3$
Then what are the solutions of
$$x^2-7\{x\}+5=0?$$
Note that this is not same as greatest integer function. So its more difficult to solve.
| Your problem is equivalent to solving
$$ x^2-7\lceil x\rceil +12=0. $$
If $x\le 1$, then $$x^2-7\lceil x\rceil+12\ge -7\lceil x\rceil +12\ge5$$
hence we obtain $x>1$.
If $x>7$, then using $\lceil x\rceil < x+1 $ we get $$x^2-7\lceil x\rceil +12>x^2-7x+5=(x-7)x+5>0.$$
We conclude $\lceil x\rceil \in\{2,3,4,5,6,7\}$ and accordingly $x^2=7\lceil x\rceil -12\in\{2,9,16,23,30,37\}$. As $x>0$, we conlcude $$x\in\{\sqrt 2, 3, 4, \sqrt{23},\sqrt{30}\sqrt{37}\}.$$ Checking against the original equation shows that all of these are in fact valid solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/448792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Dividing polynomial by binomial using remainder theorem An A level text book claims that one can find the quotient by first:
1.) Setting up an identity, $f(x)≡ Q(x)(divisor) + remainder$
2.) Finding the coefficients
However, another A level text book says, "Note. This theorem gives a (simple) method for evaluating the remainder only. If the quotient is required, long division must be used."
The question is: Divide $x^3 + x^2 - 7$ by $x-3$ using the remainder theorem.
In this example,
1.) They set up the identity: $x^3 + x^2 - 7 ≡ (Ax^2 + Bx + C)(x-3) + D$
2.) They let $x=3$ to find coefficient $D$
3.) They let $x=0$ to find coefficient $C$
4.) To find coefficients $A$ and $B$, the text book then goes on to "comparing the coefficients". No more detail is given as to how "comparing the coefficients" to find $A$ and $B$ is achieved.
Can you find coefficients $A$ and $B$ using this method ONLY? If so, how?
| Assuming the equation was meant to be:
$$x^3+x^2-7 = (Ax^2+Bx+C)(x-3)+D$$
Setting $x=3$, we sse that $3^3+3^2-7 = 29 = D$.
Setting $x=0$, we see that $-7 = D-3C=29-3C$, so $C=12$.
Setting $x=1$, we see that $-5=(A+B+12)(-2)+29$, or $A+B+12 = 17$.
Setting $x=2$, we see that $5 = (4A+2B+12)(-1)+29 = 4A+2B+12 = 24$.
So now you have two linear equations for $A,B$, and you can solve those.
(Definitely redo my arithemetic, it could be in error.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/448910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Minimize a nonlinear sum subject to a quadratic constraint Currently I am solving an optimization problem that could be written as follows:
$$\min J= \sum_{i=1}^N {(q_i^H\Lambda q_i)}^{\frac{1}{3}} $$
subject to $\{q_i\}_{i\in [1..N]}$ forming an orthonormal basis.
$\Lambda$ is a symmetric matrix with size $N\times N$ with $u_i$ eigenvectors and $\lambda_i$ eigenvalues.
The $q_i$ are column vectors with size $N\times 1$ that collectively form an orthonormal basis.
Do you know any method to solve this problem? What we want to find out are expressions of the $q_i$.
| We are asked to minimize objective function $J$ by selecting an orthonormal basis $\{q_1,\ldots ,q_n\}$, where:
$$ J = \sum_{i=1}^n (q_i^H \Lambda q_i)^{\frac{1}{3}} $$
Since $\Lambda$ is (real?) symmetric (Hermitian?), $\Lambda$ can be assumed diagonal without loss of generality, by conflating the required orthogonal (unitary?) similarity transformation with our choice of basis $\{q_i\}$.
The odd-root exponent $\frac{1}{3}$ makes it possible to treat (real) eigenvalues $\lambda_i$ of either sign as well as zero.
Outside of perhaps $n=2$, I don't see much prospect of "expressions for the $q_i$" that explicitly minimize $J$. The case $n=2$ is illuminating for suggesting methods for numerical optimization.
First note that for $\Lambda = \left( \begin{array} {cc} A & 0 \\ 0 & B \end{array} \right)$ the function $J$ is:
$$ \left[ \left( \begin{array} {cc} \cos \theta & \sin \theta \end{array} \right)
\Lambda \left( \begin{array} {c} \cos \theta \\ \sin \theta \end{array} \right)
\right]^{\frac{1}{3}} +
\left[ \left( \begin{array} {cc} -\sin \theta & \cos \theta \end{array} \right)
\Lambda \left( \begin{array} {c} -\sin \theta \\ \cos \theta \end{array} \right)
\right]^{\frac{1}{3}} = (A\cos^2 \theta + B\sin^2 \theta)^{\frac{1}{3}} + (A\sin^2 \theta
+ B\cos^2 \theta)^{\frac{1}{3}} $$
Trivially, if $A=B$ then identically $J = 2A^{\frac{1}{3}}$, and if $A=-B$ then identically $J = 0$. More generally, assuming (say) $A \gt B$:
$$ J = ((A-B)\cos^2 \theta + B)^{\frac{1}{3}} + (A + (B-A)\cos^2 \theta)^{\frac{1}{3}} $$
and we can replace $x = (A-B)\cos^2 \theta \in [0,A-B]$:
$$ J = (x + B)^{\frac{1}{3}} + (A - x)^{\frac{1}{3}} $$
Applying the standard optimization technique of (univariate) calculus, the value of $J$ at the endpoints is the same, $A^{\frac{1}{3}} + B^{\frac{1}{3}}$. Taking the derivative:
$$ \frac{dJ}{dx} = \frac{1}{3} (x+B)^{\frac{-2}{3}} - \frac{1}{3} (A-x)^{\frac{-2}{3}} $$
we find a critical point where $|x+B|=|A-x|$, i.e. midpoint $x = \frac{A-B}{2}$. There:
$$ J\bigg\rvert_{x=\frac{A-B}{2}} = 2\left(\frac{A+B}{2}\right)^{\frac{1}{3}} $$
That said, we can determine by inspection whether this critical point is a relative minimum or a relative maximum. Consider the derivative of $J$ at $x=0$. If negative, then the critical point is our minimum; otherwise the endpoints' value will do. Moreover:
$$ \frac{dJ}{dx}\bigg\rvert_{x=0} = \frac{1}{3}\left(B^{\frac{-2}{3}}
- A^{\frac{-2}{3}}\right) $$
so the function $J$ is decreasing at $x=0$ if and only if $|A| \lt |B|$. corrected
For $n \gt 2$ this seems to be motivation for a greedy numerical method. Consider all pairs of orthonormal basis vectors and find the pair $q_i$ and $q_j$ whose mutual adjustment gives the greatest decrease in $J$. Stop when no pair adjustment gives a decrease.
Concretely, suppose that a current selection of orthonormal basis vectors $\{q_i\}$ is represented by the columns of an orthogonal (or unitary) matrix $Q$. Adjustment can be made to only two columns at one time by supplying a Givens rotation $R$ to multiply by on the right, $QR$.
Note that excluding reflections among orthogonal matrices, if we wish, is not really an issue because replacing basis vector $q_i$ by $-q_i$ does not affect the value of $J$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/449147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of real positive roots of a polynomial? Consider the polynomial
$$f(x)=x((1+x^n)^n+a^n)-a(1+x^n)^n,$$
where $n\geq 2$ is a positive integer and $a$ is a positive real number. I'm interesting in deducing the number of positive real roots of $f$ as a function of $a$ and $n$. Checking numerically, it seems that if $a$ is sufficiently small in comparison to $n$ there is a single positive root, otherwise there are three.
Initially, I thought of simply using Descartes' rule of signs. Applying the binomial theorem and re-arranging we have that
$$f(x)=\begin{pmatrix}n\\0\end{pmatrix}x^{n^2+1}-a\begin{pmatrix}n\\0\end{pmatrix}x^{n^2}+\begin{pmatrix}n\\1\end{pmatrix}x^{n^2-n+1}-a\begin{pmatrix}n\\1\end{pmatrix}x^{n^2-n}+\dots+\left[\begin{pmatrix}n\\n\end{pmatrix}+a^n\right]x-a\begin{pmatrix}n\\n\end{pmatrix}$$
from which it is straightforward to see that $f$ has $2n+1$ sign changes, that is, the number of positive real roots of $f$ is $2n+1$ or less than $2n+1$ by an even number.
Are there any other relevant results that could be used to narrow down the number of positive roots any further (with the aim of proving or disproving that there are either $1$ or $3$)?
| I've stumbled onto the value of $a$ at which $f(x)$ transitions from one positive real zero to three positive real zeros. I have no proof though.
When
$$
a = \frac{n}{(n-1)^{1+1/n}},
$$
$f$ has a triple zero at
$$
x = \frac{1}{(n-1)^{1/n}}.
$$
So suppose that $a$ has this value. We will show that
$$
f\left(\frac{y}{(n-1)^{1/n}}\right) = (n-1)^{-\frac{1+n+n^2}{n}} \left[n^n y+\Bigl((n-1)y-n\Bigr) \left(n-1+y^n\right)^n\right]
$$
has a triple zero at $y=1$. For convenience let
$$
g(y) = n^n y+\Bigl((n-1)y-n\Bigr) \left(n-1+y^n\right)^n.
$$
We can check that $g$ has a zero at $y=1$ by simply substituting it in. Differentiating yields
$$
g'(y) = n^n + (n-1)\left(n-1+y^n\right)^n + n^2 y^{n-1} \Bigl((n-1)y-n\Bigr) \left(n-1+y^n\right)^{n-1},
$$
and we can again check that this has a zero at $y=1$. Differentiating once more yields
$$
\begin{align}
g''(y) &= n^2 (n-1) y^{n-2} (n-1+y^n)^{n-1} \\
&\qquad \times \Bigl(n-n^2+(n^2-1)y-(n^2+n)y^n+(n^2+1)y^{n+1}\Bigr),
\end{align}
$$
which also has a zero at $y=1$.
These values for $a$ and $x$ were found numerically with the help of the Inerse Symbolic Calculator. I don't see how to use this information to answer the problem in its entirety. Perhaps someone else will have an idea.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
Find the point of intersection of the straight line $\frac{X+1}{4}=\frac{Y-2}{-2}=\frac{Z+6}{7}$ and plane $3X+8Y-9Z=0$ Find the point of intersection of the straight line
$$\frac{X+1}{4}=\frac{Y-2}{-2}=\frac{Z+6}{7}$$ and plane $3X+8Y-9Z=0$
the point of the line is $M(-1,2,-6)$ and direction vector of the line is $A(4,-2,7)$
I would like to get some advice how to do that.
Thanks!
| Write all the parameters as one of them, say:
$$y=\frac{-x-1}2+2=\frac{-x+3}2\;,\;\;z=\frac{7x+7}4-6=\frac{7x-17}{4}$$
Substitute in the plane's formula:
$$3x+8\left(\frac{-x+3}2\right)-9\left(\frac{7x-17}4\right)=0\iff -\frac{67}4x+\frac{201}4=0\implies$$
$$ x=3\;,\;y=0\;,\;z=1$$
From here it follows that I've no idea what you call $\,M\,$ to as the intersection point between the given line and given plane is $\;(3,0,1)\;\ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/451690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The partial fraction expansion of $\frac{1}{\cos z}$ According to the Mittag-Leffler partial fraction expansion theorem for meromorphic functons, $$
\frac{1}{\cos z}= 1+\sum_{n=-\infty}^{\infty}(-1)^{n+1}\Big (\frac{1}{z-\frac{\pi}{2}-\pi n}+\frac{1}{\frac{\pi}{2}+\pi n}\Big). $$
But how does one rearrange terms (and justify the rearrangement) to show that $$\frac{1}{\cos z} = \sum_{n=0}^{\infty}(-1)^{n+1}\frac{(2n+1)\pi}{z^{2}-(\frac{2n+1}{2})^{2}\pi^2} \ ? $$
EDIT:
An alternative version of the theorem (which requires showing that $\int \frac{\sec w}{w-z} \ dw$ vanishes around an appropriate contour) allows one to conclude that
$$ \begin{align} \frac{1}{\cos z} &= \sum_{n=-\infty}^{\infty} \frac{(-1)^{n+1}}{z - \frac{\pi}{2}- \pi n} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{(-n-1)+1}}{z - \frac{\pi}{2} - \pi (-n-1)} + \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{z - \frac{\pi}{2} - \pi n} \\ &= \sum_{n=0}^{\infty} (-1)^{n+1} \left( \frac{-1}{z+ \frac{\pi}{2} + \pi n} + \frac{1}{z- \frac{\pi}{2} - \pi n} \right) \\ &= \sum_{n=0}^{\infty} (-1)^{n+1} \frac{(2n+1) \pi}{z^{2}- (\frac{2n+1}{2})^{2}\pi^{2}}. \end{align}$$
| The terms in
$\Big (\frac{1}{z-\frac{\pi}{2}-\pi n}+\frac{1}{\frac{\pi}{2}+\pi n}\Big)$
are not absolutely convergent,
since
$\sum_{n=1}^{\infty} \frac1{z-\frac{\pi}{2}-\pi n}$
diverges.
So you may have problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to integrate : $\sqrt{\frac{a-x}{x-b}}$ Problem :
How to integrate : $\sqrt{\frac{a-x}{x-b}}$
Unable to find the substitution for this :
$\sqrt{\frac{a-x}{x-b}}$
Please help how to proceed ...........thanks..
| Why so much regress maths in simple Solutions.
I think it's a good solution.
$$\begin{align}&\int\sqrt{\frac{a-x}{x-b}}dx\\ & =\int \frac{a-x}{\sqrt{(a-x)(x-b)}} dx \\ &= \int \frac{\frac12(a+b-2x)}{\sqrt{(a-x)(x-b)}} dx+\int \frac{(\frac a2-\frac b2)}{\sqrt{(a-x)(x-b)} } dx \\ &=\sqrt{(a-x)(x-b)}+\frac{(a-b)}{2}\int\frac{dx}{\sqrt{(\frac{a+b}2)^2-(x-\frac{a+b}2)^2-ab}}\\ &=\sqrt{(a-x)(x-b)}+\frac{(a-b)}{2}\int\frac{dx}{\sqrt{{{(\frac{a-b}2)^2}}-(x-\frac{a+b}2)^2}}\\ &=\sqrt{(a-x)(x-b)}+\left(\frac{a-b}2\right)\arcsin\left(\frac{x-\frac{a+b}2}{\frac{a-b}{2}}\right)+C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
$2^a +1$ is not divisible by $2^b-1$. Let $a,b>2$ be positive integers. We need to show that $2^a +1$ is not divisible by $2^b-1$.
Could any one give me hint?
| Let $b$ be a fixed positive integer. If there is a $k$ such that $2^b-1$ divides $2^k+1$, then there is a smallest such $k$. Call this smallest $k$ by the name $a$.
We first show that $a\lt b$. Suppose to the contrary that $a\ge b$. Note that
$$2^a+1=2^{a-b}(2^b-1) +2^{a-b}+1.$$
Thus if $2^b-1$ divides $2^a+1$, then $2^{b}-1$ divides $2^{a-b}+1$, contradicting the minimality of $a$.
It follows that $2^b-1$ divides $2^a+1$ for some $a\lt b$. In particular, $2^b-1\le 2^a+1\le 2^{b-1}+1$.
From $2^b-1\le 2^{b-1}+1$, we conclude that $2^{b}-2^{b-1}=2^{b-1}\le 2$. Thus $b-1\le 1$ and therefore $b\le 2$.
Remark: If $b=1$, then $2^b-1$ divides $2^a+1$ for all $a$. If $b=2$, then $2^b-1$ (that is, $3$) divides $2^a+1$ for all odd values of $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find $C$ for which improper integral is convergent Find $C$ for which $\displaystyle I=\int\limits_{1}^{\infty} \left(\dfrac{1}{\sqrt{x^2+4}}-\dfrac{C}{x+2}\right)dx$ is convergent.
| The only problem is at $+\infty$.
We have
$$\dfrac{1}{\sqrt{x^2+4}}-\dfrac{C}{x+2}=\frac{1}{x}\left(\left(1+\frac{4}{x^2}\right)^{-1/2}-C\left(1+\frac{2}{x}\right)^{-1}\right)=\frac{1}{x}\left(1-C+\frac{2C}{x}+O\left(\frac{1}{x^2}\right)\right)$$
so if $C=1$ we have
$$\dfrac{1}{\sqrt{x^2+4}}-\dfrac{C}{x+2}\sim_\infty\frac{2}{x^2}$$
and the integral is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\det(I+A) = 1 + tr(A) + \det(A)$ for $n=2$ and for $n>2$? Let $I$ the identity matrix and $A$ another general square matrix. In the case $n=2$ one can easily verifies that
\begin{equation}
\det(I+A) = 1 + tr(A) + \det(A)
\end{equation}
or
\begin{equation}
\det(I+tA) = 1 + t\ tr(A) + t^2\det(A)
\end{equation}
for some scalar $t \in \mathbb{R}$.
I have tried to see if there exists a similar formula for $n>3$. This is a natural question. But the calculations are very big and difficulty to see. Then I do the answer. Is there a similar formula for $n>2$?
| One can formally expand $\det(I+tA)$ as a power series of $t$ and get:
$$\begin{align}
& \det(I + tA)\\
= & \exp\left(\text{Tr}\log(I+tA)\right)\\
= & \exp\left(t\,\text{Tr}A - \frac{t^2}{2}\text{Tr}A^2 + \frac{t^3}{3}\text{Tr}A^3 + \cdots
\right)\\
= & 1 + t\,\text{Tr}A + \frac{t^2}{2!}\left( (\text{Tr}A)^2 - \text{Tr}A^2 \right)
+ \frac{t^3}{3!}\left((\text{Tr}A)^3 - 3 (\text{Tr}A)(\text{Tr}A^2) + 2 \text{Tr}A^3 \right)
+ \cdots
\end{align}$$
When $A$ is a $n \times n$ matrix, the above expansion terminate at the $t^n$ term with coefficient equal to $\det A$. With this, you can obtain formula similar to what you have for $n = 2$:
$$
\det(I+tA) = \begin{cases}
1 + t\,\text{Tr}A + t^2 \det(A) & n = 2\\
\\
1 + t\,\text{Tr}A + \frac{t^2}{2!}\left( (\text{Tr}A)^2 - \text{Tr}A^2 \right) + t^3 \det(A) & n = 3\\
\\
1 + t\,\text{Tr}A + \frac{t^2}{2!}\left( (\text{Tr}A)^2 - \text{Tr}A^2 \right)
\\\;\;\;+ \frac{t^3}{3!}\left((\text{Tr}A)^3 - 3 (\text{Tr}A)(\text{Tr}A^2) + 2 \text{Tr}A^3 \right)
+ t^4 \det(A) & n = 4\\
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 2
} |
Infinite solutions of Pell's equation $x^{2} - dy^{2} = 1$
Let $d > 1$ be a squarefree integer. Prove that the equation $x^{2} - dy^{2} = 1$ has infinitely many solutions in $\mathbb{Z} \times \mathbb{Z}$.
What I have done: let $ \ \mathbb{K} = \mathbb{Q}[\sqrt{d}]$. If $ d \not\equiv 1$ mod $4$, then $O_{\mathbb{K}} = \mathbb{Z}[\sqrt{d}]$ and the statement follows by the Dirichlet Unit Theorem. What about the case $ d \equiv 1 $ mod $4$ ?
| If x solves $x^2-4 = dy^2$ then so does $x^2-2$
Put $(x^2-2)^2 = x^2 - 4x + 4$ , which is $d(xy)^2 + 4$
Dividing $x$, $y$ by $2$ will give this equation in units.
Now, if $x'$ and $x''$ are consecutive solutions, then $x x'' - x'$ is also a solution. Put
Since we have $2$ and $x$ gives $x^2-2$, it then follows that $x_{n+1} = x x_n - x_{n-1}$ os also a solution, by
$x_{n+1} + x_{n-1} = x x_n$
One shows that by this rather exotic arrangement, that if two values solve this equation, then the third one does also, and thus that this series is recursive over powers of $x$.
Suppose $a, b, c, d$ form a series such that $xb = a+c$. Then the measure $b^2 - ac$ is a constant, since $d = xc-b$.
Now, $c^2 - bd = c^2 - xcb + b^2 = (c-xb)c + b^2 = b^2-ac$
Thus, if eg $2, x, x-2, x^3-3x$ form such a series based on this solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Minimal polynomial of $\zeta+\zeta^{-1}$
Find the minimal polynomial of $\zeta+\zeta^{-1}\in \mathbb{Q}(\zeta)$ over $\mathbb{Q}$, where $\zeta$ is primitive $13^{th}$ root of unity.
All I know is that the minimal polynomial should be of degree 6.
My thougths
Usually, given an element, say $\sqrt{2}+\sqrt{3}$, to find the minimal polynomial, we take $\alpha=\sqrt{2}+\sqrt{3}$ and the square it and do further simplifications to get a linear combination of powers of $\alpha$ (polynomial in $\alpha$) equal to zero, If the resulting polynomial is irreducible, we say it is a minimal polynomial for the given element over the given field. However, for the element $\zeta+\zeta^{-1}$ this way is too complicated.
Could you suggest some other procedure (if any) or a hint to simplify the calculation?
| We have the extensions
$$\mathbb{Q}\subset \mathbb{Q}( \zeta + \zeta^{-1}) \subset \mathbb{Q}(\zeta)$$
We have $[\mathbb{Q}(\zeta)\colon \mathbb{Q}( \zeta + \zeta^{-1})]\le 2$, and also $\ne 1$, since one is real, and the other is not. Therefore, the degree is $2$, and therefore
$$ [\mathbb{Q}( \zeta + \zeta^{-1})\colon \mathbb{Q}] = \frac{1}{2} [\mathbb{Q}(\zeta)\colon \mathbb{Q}]= \frac{\phi(n)}{2}$$
The minimal polynomial for $\zeta = \zeta_n$ is the cyclotomic polynomial of order $n$
$$\Phi_n(X)= \prod_{d \mid n} (X^{\frac{n}{d}}-1)^{\mu(d)} = \prod_{d \mid n} (1 + X + \cdots X^{\frac{n}{d} - 1})^{\mu(d)}$$
We see from the above that $\Phi_n(X)$ is palyndromic. We have therefore
$$\Phi_n(X) = X^{\frac{\phi(n)}{2}} Q( X+ \frac{1}{X})$$ where $\deg Q = \frac{\phi(n)}{2}$
In our case $n=13$,
$$\Phi_{13}(X) = \frac{X^{13}-1}{X-1}= (1+X + \cdots +X^{12}) = X^6( X^6 + \frac{1}{X^6} + X^{5}+ \frac{1}{X^5} + \cdots + X + \frac{1}{X} + 1)$$
We can use a CAS (WA) to express $\frac{x^{13}-1}{x^6(x-1)}$ as a polynomial in $x+ \frac{1}{x}$
$$\frac{x^{13}-1}{x^6(x-1)} =z^6 + z^5 - 5 z^4 - 4 z^3 + 6 z^2 + 3 z - 1$$
where $z= x+ \frac{1}{x}$. Therefore
$Q(z) = z^6 + z^5 - 5 z^4 - 4 z^3 + 6 z^2 + 3 z - 1$ is the minimal polynomial of $\zeta_{13} + \zeta_{13}^{-1}$, as one can check.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/460930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Arc length $y = \frac{1}{4} x^2 - \frac{1}{2} \ln x$ $$y = \frac{1}{4} x^2 - \frac{1}{2} \ln x$$
$$\int_1^{2e} \sqrt{1 + (y')^2}$$
$$y' = \frac{x}{2} - \frac{1}{2x}$$
$$y' = \frac{2x^2-1}{2x}$$
$$\left(\frac{2x^2-1}{2x}\right)^2$$
$$\frac{4x^4-4x^2+1}{4x^2}$$
$$\int_1^{2e} \sqrt{1 + \frac{4x^4-4x^2+1}{4x^2} }$$
The 1 cancels out the negative term in the numerator
$$\int_1^{2e} \sqrt{ \frac{4x^4+1}{4x^2} }$$
So now if i have done this right I have now idea how to integrate this, subsitution doesn't seem to help. What is the trick here?
| $$y' = \frac{x}{2} - \frac{1}{2x}$$
The only problem I can see in your work is the fact that you slipped when finding a common denominator for the fractions above. Doing so correctly gives us:
$$y' = \frac{x^2-1}{2x};\quad (y')^2 = \dfrac{x^4 - 2x^2 + 1}{4x^2}$$
Now, back to the integral:
$$\begin{align} \int_1^{2e} \sqrt{1 + (y')^2} \,dx & = \int_1^{2e} \sqrt{1 + \frac{x^4 - 2x^2 + 1}{4x^2} }\,dx \\ \\
& = \int_1^{2e} \sqrt{\frac{x^4 + 2x^2 + 1}{4x^2} }\,dx \\ \\
& = \int_1^{2e} \sqrt{\frac{x^2 + 2 +\frac 1{x^2}}{4} }\,dx\\ \\
& = \int_1^{2e} \sqrt{\dfrac{1}{4}\left(x^2 + 2 +\frac 1{x^2}\right) }\,dx\\ \\
& = \int_1^{2e} \frac 12\sqrt{(x + \dfrac 1x)^2}\,dx\tag{factored}\\ \\
& = \dfrac 12\int_1^{2e} \left(x + \dfrac 1x\right)\,dx \\ \\
& = \left.\dfrac 12\left(\frac{x^2}{2} + \ln x\right)\right|_1^{2e}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$ How would you solve the following
$$\int^1_0 \frac{\operatorname{Li}_3(x)}{1-x} \log(x)\, \mathrm dx$$
I might be able to relate the integral to Euler sums .
| \begin{align}
\int^1_0\frac{\log{x} \ {\rm Li}_3(x)}{1-x}{\rm d}x
&=\sum^\infty_{n=1}H_n^{(3)}\int^1_0x^n\log{x} \ {\rm d}x\\
&=-\sum^\infty_{n=1}\frac{H_n^{(3)}}{(n+1)^2}\\
&=\sum^\infty_{n=1}\frac{1}{(n+1)^5}-\sum^\infty_{n=1}\frac{H_{n+1}^{(3)}}{(n+1)^2}\\
&=\zeta(5)-\underbrace{\sum^\infty_{n=1}\frac{H_{n}^{(3)}}{n^2}}_{S}
\end{align}
Consider $\displaystyle f(z)=\frac{\pi\cot{\pi z} \ \Psi^{(2)}(-z)}{z^2}$. We know that
\begin{align}\pi\cot{\pi z}&=\frac{1}{z-n}-2\sum^\infty_{k=1}\zeta(2k)(z-n)^{2k-1}\\&\approx\frac{1}{z-n}-2\zeta(2)(z-n)\end{align}
(see here for a proof) and
\begin{align}\Psi^{(2)}(-z)&=\frac{2}{(z-n)^3}+\sum^\infty_{k=2}(-1)^{k}k(k-1)\left(H_n^{(k+1)}+(-1)^{k+1}\zeta(k+1)\right)(z-n)^{k-2}\\&\approx\frac{2}{(z-n)^3}+2\left(H_n^{(3)}-\zeta(3)\right)\end{align}
At the positive integers,
\begin{align}
{\rm Res}(f,n)
&=\operatorname*{Res}_{z=n}\left[\frac{2}{z^2(z-n)^4}-\frac{4\zeta(2)}{z^2(z-n)^2}+\frac{2\left(H_n^{(3)}-\zeta(3)\right)}{z^2(z-n)}\right]\\
&=-\frac{8}{n^5}+\frac{8\zeta(2)}{n^3}+\frac{2H_n^{(3)}}{n^2}-\frac{2\zeta(3)}{n^2}
\end{align}
At the negative integers,
\begin{align}
{\rm Res}(f,-n)
&=\frac{\Psi^{(2)}(n)}{n^2}\\
&=\frac{2H_{n}^{(3)}}{n^2}-\frac{2\zeta(3)}{n^2}-\frac{2}{n^5}
\end{align}
At $z=0$,
\begin{align}
{\rm Res}(f,0)
&=[z^1]\left(\frac{1}{z}-2\zeta(2)z\right)\left(-2\zeta(3)-12\zeta(5)z^2\right)\\
&=-12\zeta(5)+4\zeta(2)\zeta(3)
\end{align}
Hence
\begin{align}4S&=8\zeta(5)-8\zeta(2)\zeta(3)+2\zeta(2)\zeta(3)+2\zeta(2)\zeta(3)+2\zeta(5)+12\zeta(5)-4\zeta(2)\zeta(3)\\&=22\zeta(5)-8\zeta(2)\zeta(3)\end{align}
which implies
$$\color{blue}{\int^1_0\frac{\log{x} \ {\rm Li}_3(x)}{1-x}}=\zeta(5)-\frac{22\zeta(5)-8\zeta(2)\zeta(3)}{4}=\color{blue}{2\zeta(2)\zeta(3)-\frac{9}{2}\zeta(5)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/465490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Finding the length of $AE$ In the figure there're $28$ points .The distance between each point is $1$ unit .The segment $AB$ intersect with the segment $CD$ in the point $E$. How to find the length of $AE$
|
$$\bigtriangleup AEF \sim\bigtriangleup BED $$
This implies
$$ \frac{EB}{AE} = \frac{BD}{AF} $$
$$ \frac{EB}{AE} = \frac{4}{5} $$
Add $1$ to both sides
$$ \frac{AB}{AE} = \frac{9}{5} $$
$$ AE = \frac{5}{9} \times AB $$
$$ AE = \frac{5}{9} \times \sqrt{3^2 + 6^2 } = \frac{5}{9} \times \sqrt{45} $$
$$ AE = \frac{5}{9} \times 3\sqrt{5} $$
$$ \boxed{AE = \displaystyle\frac{5\sqrt{5}}{3}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/465955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving a function for square numbers Essentially I'm curious; could a perfect square($x$ squared) be less than the sum of all lesser perfect squares by a perfect square, and if so, what would the smallest solution be. Take $36$ for example, $36 < 25+16+9+4+1$ by $19$, $19$ is not a perfect square.
$\dfrac{(x(x+1)(2x+1))}{6}$ sums the squares so I substitute $x-1$ and take away the highest square and I get $\dfrac{(x(x-1)(2x-1))}{6}-x^2=$ "Perfect Square" how would you find if there exists any integers $x$ that satisfy such an equation?
Thank you!
| Picking up where Prometheus left off:
The forgotten case $x = 3a^2,\, y = 2b^2$ produces
$$\begin{align}
2b^2 &= 18a^4 - 27a^2 + 1\\
\iff 16b^2 &= 144a^4 - 216a^2 + 8 = (12a^2 - 9)^2 - 73\\
\iff 73 &= (12a^2 - 9)^2 - (4b)^2 = (12a^2-9-4b)(12a^2-9+4b).
\end{align}$$
$73$ is prime, so that forces $12a^2 - 9 - 4b = 1$ and $12a^2-9+4b = 73$, whence $b = 9$ and $12a^2 = 1 + 4\cdot 9 + 9 = 46$ which obviously is impossible.
Then, looking further at the case $x = 6a^2,\,y = b^2$ and the equation
$$\begin{align}
b^2 &= 72a^4 - 54a^2 + 1\\
\iff 8b^2 &= (24a^2)^2 - 2\cdot 9(24a^2) + 8 = (24a^2 - 9)^2 - 73\\
\iff 73 &= (24a^2 - 9)^2 - 2(2b)^2.
\end{align}$$
The equation $u^2 - 2v^2 = 73$ has infinitely many solutions, but none of them has the required form.
The ring $\mathbb{Z}[\sqrt{2}]$ is Euclidean, hence factorial. The rational prime $73$ is reducible in $\mathbb{Z}[\sqrt{2}]$, $73= (19 + 12\sqrt{2})(19-12\sqrt{2})$, and all solutions of $u^2 - 2v^2 = 73$ arise from the solution $u = 19,\, v = 12$ by multiplication with a unit of norm $+1$. The smallest solution of $x^2 - 2y^2 = 1$ is $x = 3,\, y = 2$, so all solutions are generated by $(3+2\sqrt{2})^k(19+12\sqrt{2})$. Looking at the remainders modulo $24$ of the solutions, we find a short period,
$$(19,12),\, (9,2),\, (11,0),\, (9,22),\, (19,12)$$
and the cycle closes. The first component never has the remainder $-9 \equiv 15 \pmod{24}$.
Thus:
$$\frac{x(x-1)(2x-1)}{6} - x^2$$
is never a perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Calculating $\lim_{x\to+\infty}(\sqrt{x^2-3x}-x)$ Let $f(x) = \sqrt{x^2-3x}$ and $g(x) = x$. Calculate the following limits, showing all working.
I've done the first two -
$$\lim_{x\to0}f(x)\\
=\lim_{x\to0}\sqrt{x^2-3x}=0$$
$$\lim_{x\to-\infty}\frac{f(x)}{g(x)}\\
=\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{x}=\sqrt{1-\frac{3}{x}}=1$$
How do I calculate this one?
$$\lim_{x\to+\infty}(f(x)-g(x))$$
| To find $\lim \limits_{x \to \infty} \sqrt{x^2-3x}-x:$
Rewrite $\lim \limits_{x \to \infty} \sqrt{x^2-3x}-x=\lim \limits_{x \to \infty} x \left(\sqrt{1-\frac{3}{x}}-1\right)=\lim \limits_{x \to \infty} \frac{\sqrt{1-\frac{3}{x}}-1}{1/x}.$ This gives an indeterminate $\frac{0}{0}$ form. Applying L'Hopital gives us:
$\lim \limits_{x \to \infty} \frac{\left(\frac{3/x^2}{2 \sqrt{1-3/x}}\right)}{(-1/x^2)}=\lim \limits_{x \to \infty} \frac{-3}{2\sqrt{1-\frac{3}{x}}}=\frac{-3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Describe $R=\mathbb{Z}[X]/(X^2-3,2X+4)$
I need to describe a ring $R=\mathbb{Z}[X]/(X^2-3,2X+4)$
I know that its element would be of the form $\{f(x)+(X^2-3,2X+4)|f(x)\in \mathbb{Z}[X]\}$
After that will i divide $f(x)$ by $X^2-3$ first and then by $2X+4$ to reduce the elements in $R$?
| it looks to me like it is $\mathbb Z_2[X]/\left<X^2+1\right>$.
$2$ is in your ideal because:
$$2 = 2(X^2-3)-(2X+4)(X-2)$$
Also $X^2+1$ is in your ideal because:
$$\begin{align}X^2+1 &= (X^2-3) + 4 = (X^2-3) + 4(X^2-3) - (2X+4)(2X-4) \\&= 5(X^2-3) - (2X+4)(2X-4) \end{align}$$
So we know that $\left<2,X^2+1\right>$ is contained in your ideal.
But it is obvious that the generators for your ideal are in $\left<2,X^2+1\right>$.
So we are done.
Note: Given that $(X+1)^2=X^2+1$ in $\mathbb Z_2$, this can be rewritten as isomorphic to $\mathbb Z_2[Y]/\left<Y^2\right>$ with $Y$ corresponding to $X+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/472235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{n\to\infty}$ of this quotient. Find, with proof, the value of this limit
$$\lim_{n\to\infty}\frac{\sum^n_{r=0}\binom{2n}{2r}\cdot2^r}{\sum^{n-1}_{r=0}\binom{2n}{2r+1}\cdot2^r}$$
I have tried using binomial identities but two problems occur:
*
*Only even binomial coefficients in numerator and only odd in denominator.
*The binomial coefficient occurs with $2^r$ and not with $2^{2r}$ which would be the binomial identity.
| Generalization :
Using Binomial Theorem,
$$(1+x)^{2n}=1+\binom{2n}1x+\binom{2n}2x^2+\cdots+\binom{2n}{2n-1}x^{2n-1}+x^{2n}$$
$$\implies (1-x)^{2n}=1-\binom{2n}1x+\binom{2n}2x^2+\cdots-\binom{2n}{2n-1}x^{2n-1}+x^{2n}$$
$$\implies (1+x)^{2n}+(1-x)^{2n}=2\{1+\binom{2n}2x^2+\binom{2n}4x^4+\cdots+\binom{2n}{2n-2}x^{2n-2}+x^{2n}\}$$
$$\implies (1+x)^{2n}-(1-x)^{2n}=2\{\binom{2n}1x+\binom{2n}3x^3+\cdots+\binom{2n}{2n-3}x^{2n-3}+\binom{2n}{2n-1}x^{2n-1}\}$$
On division, $$\frac{\sum_{0\le r\le n}\binom{2n}{2r}x^{2r}}{x\sum_{0\le r\le n-1}\binom{2n}{2r+1}x^{2r}}=\frac{(1+x)^{2n}+(1-x)^{2n}}{(1+x)^{2n}-(1-x)^{2n}}=\frac{1+\left(\frac{1-x}{1+x}\right)^{2n}}{1-\left(\frac{1-x}{1+x}\right)^{2n}}$$
This will $\to 1$ when $n\to\infty$ if $-1<\frac{1-x}{1+x}<1$
Now, $-1<\frac{1-x}{1+x}\iff 0<\frac{1-x}{1+x}+1=\frac2{1+x}\iff 1+x>0\iff x>-1\ \ \ \ (1)$
$\frac{1-x}{1+x}<1\iff \frac{1-x}{1+x}-1<0\iff \frac{-2x}{1+x}<0$
$\iff x(x+1)>0$ multiplying by $-\frac{(x+1)^2}2$ which is $<0$
$\implies $ either $(x>0$ and $x+1>0\iff x>-1)\implies x>0\ \ \ \ (2)$
or $(x<0$ and $x+1<0\iff x<-1)\implies x<-1$ which contradicts $(1)$
Combining $(1),(2)$ we need $x>0$
Here $x=\sqrt2>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/476790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}\end{equation*} and that \begin{equation*}\text{and that}~A+B = 180^\circ-C.\end{equation*}
Therefore $\tan(A+B) = -\tan C.$
From here, I got stuck.
| HINT:
Using $\displaystyle \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B},$
we can prove $$\tan(A+B+C)=\frac{\sum_\text{cyc}\tan A-\prod \tan A}{1-\sum_\text{cyc}\tan A\tan B}$$
Now, if $A+B+C=n180^\circ,$ where $n$ is any integer we know $\tan(n180^\circ)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 0
} |
Let $X$ be the smaller of the two numbers drawn and $Y$ the larger. Find the joint discrete density function of $X$ and $Y$ Consider a sample of size 2 drawn without replacement from an urn containing three balls, numbered 1,2, and 3. Let $X$ be the smaller of the two numbers drawn and $Y$ the larger.
(a) Find the joint discrete density function of $X$ and $Y$. $$\begin{array}{c|cc|c} x/y & 2 & 3 & f_x(x)\\ \hline \\ 1 & 2/6 & 2/6 & 4/6 \\ 2 & 0 & 2/6 & 2/6 \\ \hline \\ f_y(y) & 2/6 & 4/6 & \end{array} $$
(b) Find the conditional distribution of $Y$ given $X=1$
$$P(Y=2|X=1)=\frac{P(X=1,Y=2)}{P(X=1)}=\frac{1}{2}$$
$$P(Y=3|X=1)=\frac{P(X=1,Y=3)}{P(X=1)}=\frac{1}{2}$$
(c) Find $cov[X,Y]$$
$E[XY]=(1)(2)(\frac{2}{6})+(1)(3)(\frac{2}{6})+(2)(3)(\frac{2}{6})=\frac{22}{6}$
$E[X]=(1)(\frac{4}{6})+(2)(\frac{2}{6})=\frac{8}{6}$
$E[Y]=(2)(\frac{2}{6})+(3)(\frac{4}{6})=\frac{16}{6}$
$$cov[X,Y]=\frac{22}{6}-(\frac{8}{6})(\frac{16}{6})=\frac{1}{9}$$
Could someone tell me if my answers are corrects? Thanks for your help :)
| Almost everything is right. There is only an error of arithmetic in calculating the covariance.
I do not know why you consistently used $\frac{2}{6}$ instead of $\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/478794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Don't know how to find all the roots So i got this problem :
Find all the roots of $r^{3}=(-1)$
i can only think to use :
$\sqrt[n]{z} =\sqrt[n]{r}\left[\cos \left(\dfrac{\theta + 2\pi{k}}{n}\right) + i \sin\left(\dfrac{\theta + 2\pi k}{n}\right)\right] $
i know that $\theta = \tan^{-1}(\dfrac{b}{a}) $as in from $ z=a + bi$
$r=\sqrt[]{a^2 + b^2}$ , $k=0,1,2,3,4,.. n-1$
so anyone could explain this to me ?
lets say that $r^{3}=(-1) == z^{3}=(-1)$ a since i dont like it and it messes me up using the formula
so far :
$ z^{3} = (-1) $ so $z = \sqrt[3]{(-1 + 0 \cdot i)}$
$ r= \sqrt[]{(-1)^{2} + 0^{2}}=1$
$\tan^{-1}(\dfrac{b}{a})=\tan^{-1}(\dfrac{0}{-1})=\tan^{-1}(0)=0$ <- not sure about his part
k=0;
$ z^{3}=\sqrt[3]{1}(\cos(\dfrac{\pi + 2\pi \cdot 0}{3})+i \cdot \sin(\dfrac{\pi + 2\pi \cdot 0}{3})$
$ =\cos(\dfrac{\pi}{3}) +i\cdot \sin(\dfrac{\pi}{3})= \dfrac{\sqrt[]{3}}{2} +\dfrac{i}{2}$
k=1;
$ \sqrt[3]{1}(\cos(\dfrac{\pi+2\pi \cdot 1}{3})+i \cdot \sin(\dfrac{\pi+2\pi \cdot 1}{3})$
$=\cos(\pi) + i \cdot \sin(\pi)=(-1) + 0 = -1$
k=2;
$\sqrt[3]{1}(\cos(\dfrac{\pi + 2\pi \cdot 2}{3})+i \cdot \sin(\dfrac{\pi + 2\pi \cdot 2}{3})$
$=\cos(\dfrac{5\pi}{3})+i \cdot \sin(\dfrac{5\pi}{3})=-\dfrac{\sqrt[]{2}}{2}-i\cdot\dfrac{\sqrt[]{2}}{2}$
so is this correctet or im missing something here?
Roots $(\dfrac{\sqrt[]{3}}{2} +\dfrac{i}{2};-1;-\dfrac{\sqrt[]{2}}{2}-i\cdot\dfrac{\sqrt[]{2}}{2})$
| Hints:
Using $\,cis(\theta):=\cos\theta+i\sin\theta=cis(\theta+2k\pi)\;,\;\;k\in\Bbb Z\;$ , we have:
$$z=r\,cis(\theta)\implies -1=cis(\pi+2k\pi)=z^3=r^3\,cis(3\theta)\;\;\text{(de Moivre's Theorem)}\implies$$
$$z_k:=r\,cis\left(\frac\pi3(1+2k)\right)\;,\;\;k=0,1,2\;\;\text{(why is it enough to take these values of}\;\;k)?$$
Now calculate the roots, taking into account that $\,r=1\;$ (why?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/479009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find the determinant, inverse of a matrix and under what condition it is positive The matrix is
$B=[(1-\rho)I_n+\rho\textbf{1}\textbf{1}']$ where $\textbf{1}=[1\;\cdots\;1]'$, an $n\times 1$ vector with every entry $1$.
So what's the determinant, inverse of this matrix and under what condition it is positive?
(I just know the matrix is equal to $\begin{bmatrix}
1 &\rho & \cdots & \rho\\
\rho& 1 &\cdots & \rho\\
\cdots&\cdots &\cdots &\cdots \\
\rho & \rho & \cdots & 1
\end{bmatrix}$, but what's then?)
| If we subtract the first row from all of the remaining rows, we obtain the matrix
$$\begin{bmatrix}
1 &\rho & \rho & \cdots & \rho\\
\rho-1& 1-\rho & 0 & \cdots & 0\\
\rho-1 & 0 & 1-\rho & \cdots & 0\\
\cdots&\cdots &\cdots &\cdots \\
\rho-1 & 0 & 0 & \cdots & 1-\rho
\end{bmatrix}$$
which has the same determinant (link).
If we compute the determinant, we obtain: $$(1-\rho)^{n-1}-(n-1) \times \rho(\rho-1)(1-\rho)^{n-2}$$ accounting for the diagonals marked with $(*)$ below:
$$\begin{bmatrix}
1 (*) & \rho & \rho & \cdots & \rho\\
\rho-1 & 1-\rho (*) & 0 & \cdots & 0\\
\rho-1 & 0 & 1-\rho (*) & \cdots & 0\\
\cdots&\cdots &\cdots &\cdots \\
\rho-1 & 0 & 0 & \cdots & 1-\rho (*)
\end{bmatrix}$$
for the first term
$$\begin{bmatrix}
1 & \rho (*)& \rho & \cdots & \rho\\
\rho-1 (*) & 1-\rho & 0 & \cdots & 0\\
\rho-1 & 0 & 1-\rho (*) & \cdots & 0\\
\cdots&\cdots &\cdots &\cdots \\
\rho-1 & 0 & 0 & \cdots & 1-\rho (*)
\end{bmatrix}, \qquad \begin{bmatrix}
1 & \rho & \rho (*)& \cdots & \rho\\
\rho-1 & 1-\rho (*) & 0 & \cdots & 0\\
\rho-1 (*) & 0 & 1-\rho & \cdots & 0\\
\cdots&\cdots &\cdots &\cdots \\
\rho-1 & 0 & 0 & \cdots & 1-\rho (*)
\end{bmatrix}, \qquad \text{etc.}$$
for the second term (noting that there are $n-1$ ways this type of diagonal can be found).
The above equation simplifies to $$\det=(1-(n-1)\rho)(1-\rho)^{n-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/482628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $1^3 + 2^3 + 3^3 +\cdots+ n^3 = \frac14n^4 + \frac12n^3 + \frac14n^2$ I have to prove that this is true using mathematical induction.
I have this:
for every $n \in \mathbb N$: $1^3 + 2^3 + 3^3 + ... + n^3 = \frac 14n^4 + \frac 12n^3 + \frac 14n^2$
for $n = 1: 1^3 = 1/4 + 1/2 + 1/4$, hence $P(1)$ is true.
Let $N \in \mathbb N$ be given and assume that $P(N)$ is true, that is $$1^3 + 2^3 + 3^3 + ... + N^3 = \frac 14N^4 + \frac 12N^3 + \frac 14N^2$$
For n = $N$ + 1:
And now what? I just couldn't solve it.
| We have \begin{equation*}
1^3 + 2^3 + 3^3 + ... + N^3 +(N+1)^3= \frac 14N^4 + \frac 12N^3 + \frac 14N^2+(N+1)^3
\end{equation*}
by the induction hypothesis. Also
\begin{equation*}
\frac{1}{4}(N+1)^4+\frac{1}{2}(N+1)^3+\frac{1}{4}(N+1)^2=\frac 14N^4 + \frac 12N^3 + \frac 14N^2+(N+1)^3
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Calculation of $x$ in $x \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 88$ How can I calculate real values of $x$ in $x \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 88$, where $\lfloor x\rfloor$ is the floor function?
My attempt:
Let $\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = k\in \mathbb{Z}$. Then
$$
k \leq x\lfloor x\lfloor x\rfloor\rfloor<k+1
$$
and our equation becomes
$$
x\cdot k = 88 \implies x=\frac{88}{k}
$$
For $x>0$, simple guessing shows that $3.1<x<3.2$. But how can we account for $x<0$?
| Through guess and check
$x=\dfrac{22}{7}$
We know apriori, $x\approx 3.16 \Longrightarrow \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 28$, so setting $k=28$ yields $x=22/7$
You can verify this yourself:
$$
\tfrac{22}{7}\lfloor\tfrac{22}{7}\lfloor \tfrac{22}{7}\lfloor \tfrac{22}{7}\rfloor\rfloor\rfloor =88
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/488120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 0
} |
Math inequality proof If $a, b$ are positive real numbers and $a + b = 1$, prove that
$$
\left(a +\frac{1}{a}\right)^2 + \left(b +\frac{1}{b}\right)^2 \geq \frac{25}{2}
$$
Thank you.
| It's easily seen that, for $a+b=1$ with $a,b\gt0$, the expression
$$ab+{1\over ab}$$
is minimized when $a=b=1/2$. (In general, $x+{1\over x}$ is smallest when $x$ is as close to $1$ as possible. If $a+b=1$, the closest $ab$ gets to $1$ is when $a=b=1/2$.) It follows that
$$\begin{align}
\left(a+{1\over a} \right)^2+\left(b+{1\over b} \right)^2&=a^2+b^2+{1\over a^2}+{1\over b^2}+4\cr
&=(a-b)^2+\left({1\over a}-{1\over b} \right)^2+2\left(ab+{1\over ab}\right)+4\cr
&\ge 2\left(ab+{1\over ab}\right)+4\cr
&\ge 2\left({1\over4}+{1\over1/4}\right)+4\cr
&={25\over2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/488186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Proving that $\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=\pi$ How do I prove that
$$\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=\pi?$$
I'm just wondering if LHS even equal to the RHS in the first place?
Thanks for the help!
| Here is another method using square completion:
$$
\begin{align}
\int_a^b \frac{dx}{\sqrt{(x-a)(b-x)}}&=\int_a^b \frac{dx}{\sqrt{-\left(x^2-(a+b)x+ab\right)}}\\
&=\int_a^b \frac{dx}{\sqrt{\frac{1}{4}(a-b)^2-\left(x-\frac{b}{2}-\frac{a}{2}\right)^2}} \, \text{square completion}\\
&=\int_{\frac{a-b}{2}}^{\frac{b-a}{2}}\frac{2du}{\sqrt{(a-b)^2-4u^2}}, \,\text{substituing } u=x-b/2-a/2
\end{align}
$$
Let $u=\frac{(a-b)}{2}\sin(v)$ to complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/488558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Entropy of geometric random variable? I am wondering how to derive the entropy of a geometric random variable? Or where I can find some proof/derivation? I tried to search online, but seems not much resources is available.
Here is the probability density function of geometric distribution:
$(1 - p)^{k-1}\,p$
Here is the entropy of a geometric distribution: $\frac{-(1-p)\log_2 (1-p) - p\log_2 p}{p}$
Where $p$ is the probability for the event to occur during each single experiment.
Thanks a lot.
| Let $ P(X=k) = (1-p)^{k-1}p $, where $ k \in Z^{+} $, then the entropy is given as
$$ \begin{aligned}
H(X) & = \sum_{k=1}^{+\infty} -(1-p)^{k-1}p \cdot \log_{2}{((1-p)^{k-1}p)} \\
& = -p \cdot \log_{2}{(p)} \sum_{k=1}^{+\infty} (1-p)^{k-1} - p \cdot \log_{2}{(1-p)} \sum_{k=1}^{+\infty} (k-1)(1-p)^{k-1} \\
\end{aligned} $$
Now, the first series is a straightforward infinite geometric progression sum, the second summation can be evaluated as follows
$$ \begin{aligned}
S = \sum_{k=1}^{+\infty} -(1-p)^{k-1}p \cdot \log_{2}{((1-p)^{k-1}p)}\\
S = -p \cdot \log_{2} (1-p) \cdot (0 +(1-p)+2(1-p)^2+3(1-p)^3 ....)\\
(1-p) \cdot S = -p \cdot \log_{2} (1-p) \cdot (0 +(1-p)^2+2(1-p)^3+3(1-p)^4 ....)\\
S -(1-p)S = -p \cdot \log_{2} (1-p) \cdot (0 +(1-p)+(1-p)^2+(1-p)^3 ....)\\
S \cdot p = -p \cdot \log_{2} (1-p) \cdot ( \frac{1-p}{1 -(1-p)}) \\
S = -\log_{2} (1-p) \cdot ( \frac{1-p}{p})
\end{aligned} $$
Now, combining both the summations gives us,
$$\begin{aligned}
H(X) = -\log_{2}(p) - (\frac{1-p}{p})\log_{2}(1-p)
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$n^2 + (n-1)^2 + \dots +2^2 + 1^2$ equals what? $$n^2 + (n-1)^2 + \dots +2^2 + 1^2 = \text{??}$$
I am reading that this equals
$$\frac{1}{3} n \left( n + \frac{1}{2} \right) (n+1) $$
But have no clue how.. The thing that strikes me most is the fact that the latter has a cubed factor when expanded (i.e. in big O notation it's O$(n^3)$).
| Hint: Use induction on $n$
$$n=1:\quad1^2=\frac{1}{3}(1)\left(\frac{3}{2}\right)(2)=1$$
Assume statement is true for $n$, then
\begin{align*}
(n+1)^2+n^2+\dots+2^2+1^2&=n^2+2n+1+\frac{1}{3}(n)\left(n+\frac{1}{2}\right)(n+1)\\
\end{align*}
Then expand and factorise and you'll get your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Locating possible complex solutions to an equation involving a square root Note after reviewing answers. This question illustrates in a non-trivial way how the choice of how to compute the square root in the complex plane can make real difference to the answer you get.
In this question finding all the solutions came down to solving:
$$ab-1=\sqrt{a^2+b^2+1}$$ subject to $$(a+1)(b+1)=1$$
The condition implies $a^2+b^2+1=(ab-1)^2$, but this leaves open the possibility of $\sqrt {a^2+b^2+1}=1-ab$ rather than $ab-1$.
The solution and discussion on the original question was getting rather long, and dealt mainly with real roots, so I've extracted this sub-question on complex roots.
Could someone please explain what constraints can be put on the complex solutions of this equation - the values of $a$ and $b$ - which would be consistent with the square root operation. It would help if answers were elementary and didn't assume a prior knowledge of complex square roots.
| (A solution using too much algebra.)
I show that
the solutions are
$a=0, b=0$
and
$a=1, b=-1/2$
and,
in both of these,
the minus sign has to be taken
in the $\sqrt{}$.
If
$(a+1)(b+1)=1$,
then
$b = \dfrac1{a+1}-1$
so
$\begin{align}
ab-1
&=a(\frac1{a+1}-1)-1\\
&=a\frac1{a+1}-(a+1)\\
&=\frac{a-(a+1)^2}{a+1}\\
&=\frac{-a^2-a-1}{a+1}\\
&=-\frac{a^2+a+1}{a+1}\\
\end{align}
$
and
$\begin{align}
a^2+b^2+1
&=a^2+(\frac1{a+1}-1)^2+1\\
&=a^2+(\frac1{a+1})^2-2\frac1{a+1}+2\\
&=\frac{a^2(a+1)^2+1-2(a+1)+2(a+1)^2}{(a+1)^2}\\
&=\frac{(a^2+2)(a+1)^2+1-2(a+1)}{(a+1)^2}\\
&=\frac{(a^2+2)(a^2+2a+1)-2a-1}{(a+1)^2}\\
&=\frac{a^4+2a^3+a+2a^2+4a+2-2a-1}{(a+1)^2}\\
&=\frac{a^4+2a^3+2a^2+3a+1}{(a+1)^2}\\
\end{align}
$
Putting these in
$ab-1=\sqrt{a^2+b^2+1}
$,
$-\dfrac{a^2+a+1}{a+1}
=\sqrt{\dfrac{a^4+2a^3+2a^2+3a+1}{(a+1)^2}}
$.
Multiplying both sides by
$a+1$,
$-(a^2+a+1)
=\sqrt{a^4+2a^3+2a^2+3a+1}
$.
Since
$(a^2+a+1)^2
=a^4+2a^3+3a^2+2a+1
$,
squaring and subtracting gives
$3a^2+2a = 2a^2+3a$
or
$a^2 = a$,
so
$a=0$
or $a=1$.
If $a=0$,
then $b=0$
and we have to take
the negative sign in the $\sqrt{}$.
If $a=1$,
then $b=-1/2$,
$ab-1 = -3/2$,
and
$a^2+b^2+1
=1+1/4+1
=9/4
= (3/2)^2
$,
and,
again,
we have to take
the negative sign in the $\sqrt{}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Finding two rationals that satisfy an equation So the homework question goes as follows:
Find rational numbers such that $ \sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$
The homework only asks for one pair of $\alpha$ and $\beta$ and it is almost trivial to find the solution $\alpha=\beta = 1$
Solution:
$\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$
$7+5\sqrt{2}=(\alpha+\beta\sqrt{2})^3$
$7+5\sqrt{2}=\alpha^3+6\beta^2\alpha +\sqrt{2}(3 \alpha^2 \beta +2\beta^3)$
By identification of coefficients:
$7=\alpha^3+6\beta^2\alpha \space and \space 5=3 \alpha^2 \beta +2\beta^3$
So I just "saw" that $\alpha=\beta = 1$ is a solution
My question is, how do we find other solutions?
Thank you!
| The real solutions of a cube root of a positive number are positive numbers, and if you have two positive real-valued solutions then they must be equal. So if $\alpha_1 + \beta_1 \sqrt{2} = \alpha_2 + \beta_2 \sqrt{2}$ are two different solutions where all $\alpha_i, \beta_i$ are rational then you will be able to solve for $\sqrt{2}$ as a rational number, which is impossible. So there is only at most one solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Binomial coefficient question? I'm unsure how to do these types of questions, so any help would be great:
Find the coefficient of $x^2$ in the expansion of $(x+1/x)^3(x-1/x)^5$
Thanks
| Here is how you advance. Try to simplify the expression as
$$(x+1/x)^3(x-1/x)^5= -\frac{1}{x^8}(1+x^2)^3(1-x^2)^5= -\frac{1}{x^8}(1-x^4)^3(1-x^2)^2$$
$$ = -\frac{1}{x^8}(1-x^2)^2(1-x^4)^3$$
$$ = -\frac{1}{x^8}(1-2x^2+x^4)(1-3x^{4}+3x^8-x^{12})=\dots\,. $$
Can you finish it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How find this maximum and minimum of$|x+1|+|x-1|+\sqrt{4-x^2}$
show that
$$2+\sqrt{3}\le|x+1|+|x-1|+\sqrt{4-x^2}\le2\sqrt{5}$$
This problem have nice methods? Thank you
my ugly methods,
since $-2\le x\le 2$,and $f(x)=|x-1|+|x+1|+\sqrt{4-x^2}\Longrightarrow f(x)=f(-x)$
so we only find $x\in [0,2]$ $f(x)_{\max},f(x)_{\min}$
so when
(1):
$$0\le x\le 1\Longrightarrow f(x)=2+\sqrt{4-x^2}\le 4$$
(2): when $1\le x\le 2$, then
$$f(x)=2x+\sqrt{4-x^2}\Longrightarrow f'(x)=0\Longrightarrow x=\dfrac{4}{\sqrt{5}}$$
so $\cdots\cdots$
My question This problem have nice methods?Thank you
| If $x=a+ib,$ where $a,b$ are real
$\implies 4-x^2=4-(a+ib)^2=4+b^2-a^2-2ab i$
This needs to be real and $\ge0$
For real value of $4-x^2, ab=0$
Case $1:$ If $b=0,$
$4-a^2\ge0\iff a^2\le 4\iff -2\le a\le 2$
We know for real $x,$ $|x|=\begin{cases} x &\mbox{if } x\ge0 \\
-x & \mbox{if } x<0 \end{cases}$
$$|x+1|=|a+1|=\begin{cases} a+1 &\mbox{if } a+1\ge0\iff a\ge-1 \\
-(a+1) & \mbox{if } a<-1 \end{cases}$$
and
$$|x-1|=|a-1|=\begin{cases} a-1 &\mbox{if } a\ge1 \\
-(a-1)=1-a & \mbox{if } a<1 \end{cases}$$
Check region by region $[-2,-1)[-1,1],(1,2]$
Case $2:$ If $a=0,$
$4+b^2\ge0$ which is true for real $b$
$4-x^2=4-(ib)^2=4+b^2$
$|x+1|=|1+ib|=\sqrt{1+b^2}$
$|x-1|=|-1+ib|=\sqrt{1+b^2}$
$|x+1|+|x-1|+\sqrt{4-x^2}=2\sqrt{1+b^2}+\sqrt{4+b^2}\ge 1+2=3$
Observe that it has no finite maximum value
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove complex identity of $\frac{1}{z}$ $\log(\cdot)$ and $\sqrt{\cdot} $ are any branches of the logarithm and square root, respectively. Show that if $z \notin \{0, \pm i\}$, then the following identity holds:
$$\frac 1z =\cot\left [\frac{1}{i}\log\left (\sqrt{\frac{1+iz}{1-iz}}\right ) \right ]$$
| Whenever applicable
\begin{align}
\cot \left [\frac 1i \ln\left ( \sqrt{\frac {1+iz}{1-iz}} \right )\right ] &= i \cdot\frac {\exp \left [\ln\left ( \sqrt{\frac {1+iz}{1-iz}} \right )\right ] + \exp \left [-\ln\left ( \sqrt{\frac {1+iz}{1-iz}} \right )\right ]}{\exp \left [\ln\left ( \sqrt{\frac {1+iz}{1-iz}} \right )\right ] - \exp \left [-\ln\left ( \sqrt{\frac {1+iz}{1-iz}} \right )\right ]} = \\
&= i \cdot \frac {\sqrt{\frac {1+iz}{1-iz}} + \sqrt{\frac {1-iz}{1+iz}}}{\sqrt{\frac {1+iz}{1-iz}} - \sqrt{\frac {1-iz}{1+iz}}} = i \cdot \frac {1+iz + 1 - iz}{1+iz - 1 + iz} = \\
&= i \cdot \frac 2{2iz} = \frac 1z
\end{align}
Some info
\begin{align}
\cot z &= \frac {\cos z}{\sin z} = \frac {\frac {e^{iz} + e^{-iz}}2}{\frac {e^{iz} - e^{-iz}}{2i}} = i \cdot \frac {e^{iz} + e^{-iz}}{e^{iz} - e^{-iz}} \tag 1 \\
-\ln z &= \ln \frac 1z \tag 2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
show that $a^3+b^3+c^3-3abc\ge2(\frac{b+c}{2}-a)^3$ let $a,b,c\ge 0$,show that: $$a^3+b^3+c^3-3abc\ge2 \left(\dfrac{b+c}{2}-a\right)^3$$
my try:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
then let $b-a=x,c-a=y$
But following I don't can't prove it,Thank you
| First of all the inequality is trivial for $b+c-2a<0$.
Secondly, you have:
$$
2(a^2+b^2+c^2-ab-bc-ac)=(a-b)^2+(c-b)^2+(a-c)^2
$$
Also you have:
$$
(a-b)^2+(a-c)^2\geq 2(b-a)(c-a)
$$
and therefore:
$$
2(a-b)^2+2(a-c)^2+2(c-b)^2\geq (a-b)^2+(a-c)^2+2(b-a)(c-a) =(b+c-2a)^2
$$
hence:
$$
(a^2+b^2+c^2-ab-bc-ac)\geq (\frac{b+c-2a}{2})^2
$$
But you also have $a+b+c\geq b+c-2a$. Now if you multiply this by the previous inequality you get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Complex numbers and their matrix form. I have a line starting at the origin, and i extend it to a point $(a,b)$ in the plane. This thing can be called a vector and be represented as $(a,b), [a\text{ }b]^T$ (column vector) or by $a\mathbf{i}+b\mathbf{j}$, where $(\mathbf{i},\mathbf{j})$ is the stardard basis in $\mathbb{R}^2$ Or it could be seen as a visual representation of a complex number where $(a,b)=a+bi,$ where $i=\sqrt{-1}$.
So I want to rotate this vector $(a,b)$ $90$ degrees counter clockwise, so i know I can use my trusty matrix for rotations
$\begin{bmatrix} \cos(90) & -\sin(90) \\ \sin(90) & \cos(90)\\ \end{bmatrix}$=$\begin{bmatrix} 0 & -1 \\ 1 & 0\\ \end{bmatrix}$ and we find that
$$\begin{bmatrix} 0 & -1 \\ 1 & 0\\ \end{bmatrix}\begin{bmatrix} a \\ b\\ \end{bmatrix}=\begin{bmatrix} -b \\ a\\ \end{bmatrix}$$
Or, I could choose the complex multiplication way and say,
$i(a+bi)=ai+bi^2=ai-b=-b+ai$
So we all know that, but what are some of the advantages and disadvantages to having two things that are completely identical operation in different systems?
| There is a homeomorphism between the the complex numbers
$$
\color{blue}{a} + \color{red}{b}i
$$
and the rotation matrices
$$
\left(
\begin{array}{rc}
\color{blue}{a} & \color{red}{b} \\
-\color{red}{b} & \color{blue}{a} \\
\end{array}
\right)
$$
where $\color{blue}{a}^{2} + \color{red}{b}^{2} = 1.$ We see the familiar rotation matrix
$$
R(\theta) =
\left(
\begin{array}{rc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta \\
\end{array}
\right)
$$
which in the form
$$
x'=R(\theta)x
$$
rotates the $2-$vector $x$ about the origin by $\theta$, producing the $2-$vector $x'$.
Verify homeomorphism
Start with two complex numbers $z_{1}$ and $z_{1}$. The Cartesian forms are
$$
z_{1} = \color{blue}{a} + \color{red}{b}i, \quad
z_{2} = \color{blue}{c} + \color{red}{d}i
$$
where the numbers $a$, $b$, $c$, and $d$, are all real. Blue numbers signify the real component of $z$, and red the imaginary component.
Equivalent matrix forms are defined as
$$
z_{1} = \left(
\begin{array}{rc}
\color{blue}{a} & \color{red}{b} \\
-\color{red}{b} & \color{blue}{c} \\
\end{array}
\right),
\quad
z_{2} = \left(
\begin{array}{rc}
\color{blue}{c}& \color{red}{d} \\
- \color{red}{d} & \color{blue}{c}\\
\end{array}
\right)
$$
Verify basic properties of of the homeomorphism.
Addition
$$
%
z_{1} +
z_{2} =
(\color{blue}{a} + \color{red}{b}i) +
(\color{blue}{c} + \color{red}{d}i)
=
(\color{blue}{a}+\color{blue}{c}) + (\color{red}{b} + \color{red}{d} )i
$$
$$
%
z_{1} +
z_{2} =
\left(
\begin{array}{rc}
\color{blue}{a} & \color{red}{b} \\
-\color{red}{b} & \color{blue}{c} \\
\end{array}
\right) +
\left(
\begin{array}{rc}
\color{blue}{c} & \color{red}{d} \\
-\color{red}{d} & \color{blue}{c} \\
\end{array}
\right)
= \left(
\begin{array}{rc}
\color{blue}{a}+\color{blue}{c}& \color{red}{b}+ \color{red}{d} \\
-\color{red}{b}- \color{red}{d} & \color{blue}{a}+\color{blue}{c}\\
\end{array}
\right)
$$
Multiplication
$$
z_{1} z_{2} =
(\color{blue}{a} + \color{red}{b}i)
(\color{blue}{c} + \color{red}{d}i)
=
(\color{blue}{ac}- \color{red}{bd}) +
(\color{red}{b}\color{blue}{c}+\color{blue}{a}\color{red}{d})i
%
$$
$$
%
z_{1} z_{2} =
\left(
\begin{array}{rc}
\color{blue}{a} & \color{red}{b} \\
-\color{red}{b} & \color{blue}{c} \\
\end{array}
\right)
\left(
\begin{array}{rc}
\color{blue}{c} & \color{red}{d} \\
-\color{red}{d} & \color{blue}{c} \\
\end{array}
\right)
=
%
\left(
\begin{array}{rc}
\color{blue}{a}\color{blue}{c}- \color{red}{b} \color{red}{d} & \color{red}{b}\color{blue}{c}+\color{blue}{a} \color{red}{d} \\
- \color{red}{b} \color{blue}{c}-\color{blue}{a} \color{red}{d} & \color{blue}{a}\color{blue}{c}- \color{red}{b} \color{red}{d} \\
\end{array}
\right)
$$
Inversion
$$
%
\frac{1}{z}
= \frac{1}{\color{blue}{a} + \color{red}{b} i}
= \left( \frac{\color{blue}{a} - \color{red}{b} i}{\color{blue}{a} - \color{red}{b} i} \right)
\frac{1}{\color{blue}{a} + \color{red}{b} i}
= \left( \color{blue}{a}^{2} + \color{red}{b}^{2} \right)^{-1} \left( \color{blue}{a} - \color{red}{b} i \right) %
$$
$$
z^{-1}
= \left(
\begin{array}{cr}
\color{blue}{a} & - \color{red}{b} \\
\color{red}{b} & \color{blue}{a} \\
\end{array}
\right)^{-1}
= \frac{\text{adj }z}{\det z}
= \left( \color{blue}{a}^{2} + \color{red}{b}^{2} \right)^{-1}
\left(
\begin{array}{cr}
\color{blue}{a} & - \color{red}{b} \\
\color{red}{b} & \color{blue}{a} \\
\end{array}
\right)
$$
where adj $z$ is the adjugate matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Factor $(x+y)^7-(x^7+y^7)$ I encountered the following problem while preparing for upcoming math contests.
Factor $(x+y)^7-(x^7+y^7)$.
I got zero for $(x+y)^7-(x^7+y^7)$, however, the solutions said it's
$$7xy(x+y)(x^2+xy+y^2)(x^2+xy+y^2)$$
Can someone explain how this is possible?
| A more compact display : If $σ_n = x^n + y^n$ and $P = xy$, then from 7-th power binomial expansion we get $$σ_1^7 = σ_7 + 7P(σ_5 + 3σ_3P + 5σ_1P^2)$$ and since $σ_1 | σ_n$for odd $n$, viz., $σ_3 = σ_1(σ_2 - P)$;$σ_5 = σ_3σ_2 - σ_1P^2 = σ_1\left(σ_2^2 - P(σ_2 + P)\right),$ $$σ_1^7 = σ_7 + 7σ_1P\left[σ_2^2 \overbrace{-\ P(σ_2 + P)} + \ 3P(σ_2 - P) + 5P^2\right]$$ $$= σ_7 + 7σ_1P\left[(σ_2 + P)(σ_2 + 2P) \overbrace{-\ P(σ_2 + P)}\right]$$ $$= σ_7 + 7σ_1P(σ_2 + P)^2,$$i.e., $$σ_1^7 - σ_7 = 7σ_1P(σ_2 + P)^2,$$ or more explicitly, $$(x + y)^7 - (x^7 + y^7) = 7xy(x + y)(x^2 + xy + y^2)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 5
} |
Common tangent lines of two quadratic functions Find all such lines that are tangent to the following curves:
$$y=x^2$$ and $$y=-x^2+2x-2$$
I have been pounding my head against the wall on this. I used the derivatives and assumed that their derivatives must be equal at those tangent point but could not figure out the equations. An explanation will be appreciated.
|
Let $f(x) = x^2$ and $g(x) = -x^2 + 2x - 2 \implies \dfrac{d}{dx}(f(x))|_{x = a} = 2a$ and $\dfrac{d}{dx}(g(x))|_{x = b} = -2b + 2$
$2a = -2b + 2 \implies a = 1 - b \implies$
$A: (1 - b, (1 - b)^2), B: (b, -b^2 + 2b - 2)
\implies m_{AB} = \dfrac{2b^2 - 4b + 3}{1 - 2b} = -2b + 2 \implies $
$2b^2 - 2b - 1 = 0 \implies b = \dfrac{1 \pm \sqrt{3}}{2}$
Using $b = \dfrac{1 + \sqrt{3}}{2} \implies $
$A: (\dfrac{1 - \sqrt{3}}{2},\dfrac{2 - \sqrt{3}}{2}), B: (\dfrac{1 + \sqrt{3}}{2},\dfrac{\sqrt{3} - 4}{2}) \implies m_{AB} = 1 - \sqrt{3} \implies $
$ \boxed{y = (1 - \sqrt{3})x + \dfrac{\sqrt{3} - 2}{2}} $
Using $b = \dfrac{1 - \sqrt{3}}{2} \implies A^{'}: (\dfrac{1 + \sqrt{3}}{2}, \dfrac{2 + \sqrt{3}}{2}), B^{'}: (\dfrac{1 - \sqrt{3}}{2}, \dfrac{-\sqrt{3} - 4}{2}) \implies$
$m_{A^{'}B^{'}} = 1 + \sqrt{3} \implies \boxed{y = (1 + \sqrt{3})x - \dfrac{\sqrt{3} + 2 }{2}}. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Existence of Quadratic Non-Residues less than square root of a prime I had a homework question that I was unable to solve even after the solutions were released!
If $p$ is a prime $\ne 3, 7, 23$ then there is a quadratic nonresidue $\le \sqrt{p}$
I noticed that with very few quadratic residues $\lt \sqrt{p}$ I would be able to construct many more by multiplying them together. I was hoping to use this to get a contradiction by creating $\gt p/2$ quadratic residues but I stalled at a conjecture that turned out to be false. Given the first N integers, how many large prime factors can I disallow and still have half the set remaining?
I cannot follow most of the solution given at http://www.math.nyu.edu/~tschinke/teaching/Fall13/hw2.pdf (#4)
I would like a better exposition of the answer given in the link, or another elementary solution.
| This is essentially a rewrite of the linked solution, just with more details.
For a prime $p$, let $a$ be the smallest QNR. Our goal is to show that $a \leq \sqrt{p}$.
By the division algorithm, $p = qa + r$ where $ 0 \leq r < a $. Since $p$ is prime and $a \neq p$, hence $ r \geq 1$.
Since $ a (q+1) = qa+a \equiv a - r \pmod{p}$. Since $0 < a-r < a $ by construction, hence $a-r$ must be a quadratic residue. By the Law of quadratic reciprocity $q+1$ must be a QNR. By the definition of $a$, we know that $ a \leq q+1$. Hence, $ a^2 \leq a(q+1) = aq+a = p-r+a$.
This is almost close to $a^2 \leq p$, which is what we want. How close are we? We actually have $$a^2 -a + \frac{1}{4} \leq p - r + \frac{1}{4} \leq p - 1 + \frac{1}{4} \Rightarrow a - \frac{1}{2} \leq \sqrt{p - \frac{3}{4} } \Rightarrow a \leq \sqrt{p - \frac{3}{4} } + \frac{1}{2}.$$
This is not yet good enough, so we shall consider the cases where $ \sqrt{p} < a \leq \sqrt{ p - \frac{3}{4} } - \frac{1}{2}$.
If $p\equiv 1 \pmod{4}$, then we know that $-1$ is a QR. Since $a-1$ is a quadratic residue by construction, hence $ a (a-1)$ is a QNR. Evaluating it, we see that $p-a <a^2 -a \leq p-r$. Since $-1$ is a QR, hence $-a(a-1)$ is a QNR. But, we are given that $ r \leq -a(a-1) < a$, which contradicts the minimality of $a$.
If $p \equiv 3 \pmod{4}$, and $p > 49$. Recall that $-1$ is a QNR. Consider the term $(a-1)(a+1)$. We have $ p-1 < a^2 -1 \leq p - r + a - 1 $, hence when reduced modulo $p$, $ 0 \leq a^2 - 1 \leq a-r-1 < a$. This tells us that $a^2-1$ is a QR. Since $a-1$ is a QR, thus $a+1$ is also a QR.
Now consider $(a-2)(a+1)$. Since $p-a-2 < (a-2)(a+1) \leq p-r-2$ is the product of 2 QR, it is a QR. Remember that $-1, -2, \ldots -(a-1)$ are all QNR and $p-a-1$ is also a QNR, thus the only possibility for a QR in the range is $p-a$. Hence, we must have $p-a = (a-2)(a+1)$, which gives us $p = a^2 - 2$. Since $(a-2)(a+2) = a^2 - 4 \equiv -2 \pmod{p}$ is a QNR, and $a-2$ is a QR, hence $a+2$ is a QNR. Similarly, since $(a-2)(a+3) = a^2 +a - 6 \equiv a-4 \pmod{p}$ is a QR, hence $a+3$ is a QR. Finally, since $(a-3)(a+3) = a^2 - 9 \equiv -7 pmod{p}$ is a QNR (Recall that $7 \leq \sqrt{p} < a$ is a QR and $-1$ is a QNR), hence $(a-3)$ is a QNR. But that contradicts the assumption that $a$ is the smallest QR.
Now, it remains to check for $p \equiv 3 \pmod{4} $ and $ p < 49$. The primes are 3, 7, 11, 19, 23, 31, 43, 47. I leave it to you to verify that the only ones with a QNR that is not smaller than $\sqrt{p}$ are 3, 7, 23. Hence we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Problem with Newton's Method I am trying to solve this problem:
Use Newton's method to find the intersection points of the two circles defined by
$$x^2 + y^2 = 2, \quad (x-1)^2 + (y+1.5)^2 = 1.$$
I used this code in Maple:
Newton(4+(-y^2+2)*(2*y+3)^2-4*(-y^2+2), y = .5, output = sequence);
and this gave me:
.5, -1.062500000, -.7532718336, -1.001899321, -.5138663374,
-.7942990696, -1.069866630, -.7693943666, -1.025772773,
-.6418803790, -.8854387837
From this I do not understand what the solution is, and also how can I choose my initial value more accurately?
| First I couldn't understand why you used that particular formula. the formula should be:
$$(x-1)^2 + (y + \frac 32)^2 = 1$$
$$x^2 - 2x + 1 + y^2 + 3y + \frac 94 = 1$$
$$(x^2 + y^2) - 2x + 3y + \frac 94 = 0$$
$$2 - 2x + 3y + \frac94 = 0$$
$$3y - 2x + \frac{17}{4} = 0$$
$$y = \frac{2x - \frac{17}{4}}{3}$$
Now substituting we have:
$$x^2 + \left(\frac{2x - \frac{17}{4}}{3}\right)^2 - 2 = 0$$
$$x^2 + \frac{4x^2 - 17x + \frac{289}{16}}{9} - 2 = 0$$
$$9x^2 + 4x^2 - 17x + \frac{289}{16} - 18 = 0$$
$$144x^2 + 64x^2 - 272x + 289 - 288 = 0$$
$$208x^2 - 272x + 1 = 0$$
Now it should be easy to find the solutions using Newton's method.
Also Newton's method may fail for numbers of reasons. First is "bad" starting point. For some starting points the value of x doesn't converges to the right value, but diverges
For more info you can read here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove through induction that $3^n > n^3$ for $n \geq 4$ I'm new to induction and have not done induction with inequalities before, so I get stuck at proving after the 3rd step.
The question is:
Use induction to show that $3^n > n^3$ for $n \geq 4$.
I have so far:
Step 1:
Prove for $n=4$ (since question states this)
$3^4 > 4^3$
$81 > 64 $
which is true
Step 2:
Assume true for $n=k$
$3^k > k^3$
Step 3:
Prove for $n = k+1$
$3^{k+1} > (k+1)^3$
Here I expand to:
$3^k \cdot 3 > k^3 + 3k^2 + 3k + 1$
However I have no idea how to prove this.
Thanks for any help given
| $3^n > n^3$ for $n\geq 4$
Base case:
If $n=4$, we get
$\text{LHS}=3^4=81$
$\text{RHS}=4^3=64$.
We can see that $\text{LHS} > \text{RHS}$.
Let us assume the above hypothesis holds true for some $n=k$ such that $k\geq 4$.
Therefore,
$$3^k > k^3$$
$$3(3^k) > 3k^3$$
$$3^{k+1} > k^3 + k^3 + k^3 > k^3 + 3k^2 + 3k^2 > k^3 + 3k^2 + k^2 + k^2 + k^2 > k^3 + 3k^2 + 3k + 3k + 3k > k^3 + 3k^2 + 3k + 1 = (k+1)^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/503842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Integrating $\int \frac{x^2+1}{x(x^2-1)}$ How would I integrate the following.
$$\int \frac{x^2+1}{x(x^2-1)}$$
I have done the following.
$$\frac{x^2}{(x)(x+1)(x-1)}$$
$$\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}$$
I then did $\quad \displaystyle A(x^2-1)+B(x-1)+C(x+1)=x^2+1$
Then $\quad Ax^2-1A+Bx-1B+Cx+C=x^2+1$
Grouping I get $$Ax^2=x^2,\quad A=1, \quad C=1,\quad Bx+Cx=0, \quad B=-1$$
Then plug back in
$$\int\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x-1}$$
But I am not sure if I did it correctly.
| $$\frac{x^2}{(x)(x+1)(x-1)} = \frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}$$
$$A(x^ - 1) + Bx(x-1) + Cx(x+ 1) = (A + B + C)x^2 + (-B+ C)x + -A = x^2 + 1$$
$$A + B + C = 1$$
$$C - B = 0\iff B = C$$
$$-A = 1\iff A = -1$$
Now, we have $$A = -1,\;B = C, \implies A + B + C = -1 + 2B = 1 \iff 2B = 2 \iff B = C = 1$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
A strange "pattern" in the continued fraction convergents of pi? From the simple continued fraction of $\pi$, one gets the convergents,
$$p_n = \frac{3}{1}, \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \frac{103993}{33102}, \frac{104348}{33215}, \frac{208341}{66317}, \frac{312689}{99532}, \frac{833719}{265381}, \frac{1146408}{364913}, \dots,$$
starting with $n=1$, where the numerators and denominators are A002485 and A002486, respectively. If you stare at it hard enough, a pattern will emerge between three consecutive convergents. Define,
$$\left(a_n,\,b_n,\,c_n\right) = \left(p_{n}-3,\;\; p_{n+1}-3,\;\; p_{n+2}-3\right),$$
$$v_n=\text{Numerator}\,(a_n)\,\text{Numerator}\,(b_n).$$
Then, for even $n \ge 2$,
$$F(n) = \sqrt{\frac{a_n c_n}{a_n-c_n}-v_n}\in\mathbb{Z}\text{ (often)}.$$
For example, for $n = 2$,
$$\left(a_2,\,b_2,\,c_2\right) = \left(\frac{22}{7}-3,\; \frac{333}{106}-3,\; \frac{355}{113}-3\right),$$
$$F(2) = 1.$$
More generally,
$$\begin{array}{cc}
n&F(n) \\
2&1 \\
4&16\\
6&4703\\
8&14093\\
10&51669\\
12&122126\sqrt{2}\\
14&7468474\\
16&\frac{18549059}{\sqrt{2}}\\
\end{array}$$
and so on. For even $n<100$, I found half of the $F(n)$ were either integer or half-integer. (And all the non-integers were of form $N\sqrt{d}$ for some very small d.)
Some questions:
*
*For $n<500$, $n<1000$, etc, how many $F(n)$ are integers or half-integers?
*More importantly, why is $F(n)$ often an integer?
| The $q_n = p_n-3$ are the convergent fractions of $\pi-3$ (it really doesn't matter to do this change by the way, you could have started straight from $\pi$, only by picking $n \ge 3$ odd instead of $n \ge 2$ even)
3 consecutive convergent fractions are of the form $\frac ab, \frac cd, \frac{a+kc}{b+kd}$ for some integers $a,b,c,d,k$ and $ad-bc=1$ (because we picked $n$ even).
$F(n) = \sqrt{\frac {a(a+kc)}{a(b+kd)-b(a+kc)}-ac} = \sqrt{\frac{a^2+kac}k-ac} = a/\sqrt k$
From the wikipedia page of $\pi$ I can only see the first $3$ relevant $k$, and they are all $1$, so $F(n) = numerator(a_n)$ for $n=2,4,6$ at least.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/508190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
How to prove the identity $3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$? I'm trying to prove a trigonometric identity but I can't. I've been trying a lot but I can't prove it. The identity says like this:
$$3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$$
The identity would be easy if $1-\cos^4x=\sin^4x$ and $1-\cos^6x=\sin^6x$ but we know that $\sin^4x+\cos^4x$ isn't equal with $1$ and $\sin^6x+\cos^6x$ isn't equal with $1$.
Can anybody help me?!
Thank you!
| L.H.S=$3\sin^4x-2\sin^6x$
=$3(\sin^2x)^2-2(\sin^2x)^3$
=$3(1-\cos^2x)^2 -2 (1-\cos^2x)^3$
=$3(1+\cos^4x -2\cos^2x) -2 (1-\cos^6x -3 \cos^2x+3\cos^4x)$
=$ 3+3\cos^4x -6\cos^2x -2+2\cos^6x +6\cos^2x-6\cos^4x)$
=$1-3\cos^4x+2\cos^6x$
=R.H.S
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/508763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How find this function $f(1+xy)=f(x)f(y)+f(x+y)$ let $f:\mathbb R\longrightarrow \mathbb R$,and for any real numbers $x,y$ have
$$f(1+xy)-f(x+y)=f(x)f(y)$$
and $f(-1)\neq 0$.
Find the $f(x)$
My try:let $x=y=0$,then we have
$$f(1)-f(0)=[f(0)]^2$$
let
$$x=1,y=-1,\Longrightarrow f(0)-f(0)=f(1)f(-1)$$
since $f(-1)\neq 0$,
so
$$f(1)=0$$
so
$f(0)=0 $ or $f(0)=-1$
Then I dont know what to do.
Edit: This is from IMO2012 Shortlist, Problem A5.
| try to find solution in this form :
f(x)=x²+ax+b
if we put y=-x in equation we get
f(1-x²)-f(x-x)=f(x)f(-x)
so f(1-x²)=-1+f(x)f(-x)
so (1-2x²+x^4)+a(1-x²)+b=-1+(x²+ax+b)(x²-ax+b)
this need to be true for all x so
-a-2=2b-a², a+b+1=b²-1
so a=b²-b-2=(b+1)(b-2)
also a(a-1)=2(b+1)
so (b+1)(b-2)(b²-b-3)=2(b+1)
so one solution is b=-1,a=0 so f(x)=x²-1 is solution
other is maybe when b³-3b²-b+4=0
i.e. if we put b-1=t we have t³-4t+1=0
solutions are t1~-2.1149 ,t2~0.25410 , t3~1.8608
then a=(b+1)(b-2)=(t+2)(t-1)
but a must be 0,so these 3 solutions aren't our solutions
just f(x)=x²+1
if we assume f(x)=x+c
then 1-x²+c=-1+(x+c)(c-x)
so 1+c=-1+c²
i.e. c²-c-2=0 i.e. (c+1)(c-2)=0
so c=-1 or c=2
but when we check we see f(x)=x-1 is solution,
so f(x)=x-1 or f(x)=x²-1
also we can prove that f(x)=x^n-1 isn't solution for n>2,n natural number
that means that these two are the only two polynomial with positive integer degrees that satisfy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
How to prove that if $2x^2-x=2y^2-y$, then $x=y$, for $x,y\in\mathbb{Z}.$ How to prove that if
$2x^2-x=2y^2-y$, then $x=y$, for $x,y\in\mathbb{Z}.$
| Let $y=x+k$ for some $k \in \mathbb{Z}^{\geq 0}$ (if $k<0$ then swap $x$ and $y$). Then $$2x^2-x=2y^2-y$$ implies that
\begin{align*}
2x^2-x &= 2(x+k)^2-x+k \\
&= (2x^2-x)+4kx+2k^2+k
\end{align*}
or equivalently that $k(4x+2k+1)=0$. If $k \neq 0$, this implies $4x+2k+1=0$, and this gives a contradiction, since $4x+2k+1$ is odd and $0$ is even (like Andres Caicedo's answer).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/514087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Verify this identity: $\sin x/(1 - \cos x) = \csc x + \cot x$ Verify this identity $\frac{\sin {x}} {1 - \cos {x}}\ = \csc x + \cot x$ I got the left side to $\frac{1-\cos x} {\csc x}$, but I can't get any farther. Am I on the right path? Can I get some help?
| To prove sin x/1-cos x = cosec x + cot x
We know that,
Sin^2 x = 1- cos^2
So from RHS we get,
Cosec x+cot x
=1/sin x + cos x/sin x
So taking LCM,
We get,
1+cos x/sin x
Now multiply both numerator and denominator by sin x to get denominator as sin^2 x
We get,
(1+cos x)(sin x)/sin^2 x
Now we can write sin^2 x as 1- cos^2 x
So, we get
(1+cos x)(sin x)/1-cos^2 x
(1+cos x)(sin x)/(1+cos x)(1-cos x)
applying a^2-b^2=(a+b)(a-b)
so by cancellation
We get sin x/1-cosx=cosec x+cot x
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/514229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
if $f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$ If $\displaystyle f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$
$\underline{\bf{My\;\; Try}}::$ Given $\displaystyle f(x) = x-\frac{1}{x} = \frac{x^2-1}{x}.$ Now Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get.
$\displaystyle f(f(x)) = x-\frac{1}{x}-\frac{x}{x^2-1} = \frac{x^2-1}{x}-\frac{x}{x^2-1} = \frac{(x^2-1)^2-x^2}{x.(x^2-1)} = \frac{x^4-3x^2+1}{x.(x^2-1)}$
again Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get
$\displaystyle f(f(f(x))) = \frac{\left(\frac{x^2-1}{x}\right)^4-3\left(\frac{x^2-1}{x}\right)^2+1}{\left(\frac{x^2-1}{x}\right).\left\{\left(\frac{x^2-1}{x}\right)^2-1\right\}}$
Now I did not understand how can I solve This $8^{th}$ Degree equation,
So Help Required,
Thanks
| Hint: $f^{(-1)}(x)=\frac{x\pm\sqrt{x^2+4}}{2}$, thus $f^{(-1)}(1)=(1\pm\sqrt{5})/2$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/518683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How can I prove that one of $n$, $n+2$, and $n+4$ must be divisible by three, for any $n\in\mathbb{N}$ Intuitively it's true, but I just can't think of how to say it "properly".
Take for example, my answer to the following question:
Let $p$ denote an odd prime. It is conjectured that there are infinitely many twin primes $p$, $p+2$. Prove that the only prime triple $p$, $p+2$, $p+4$ is the triple $3,\ 5,\ 7$.
And my solution:
Given an odd integer $n$, between the three integers $n$, $n+2$ and $n+4$, one of them must be divisible by $3$... Three possible cases are $n=3k$, $n+2=3k$, and $n+4=3k$. The only such possible $k$ that makes $n$ prime is $k=1$. In this case, given an odd prime $p$, either $p=3$, $p+2=3$, or $p+4=3$. This would imply that $p=3$, $p=1$, or $p=-1$. The only of these three that is prime is $p=3$, therefore the only three evenly distributed primes are $3$, $5$, and $7$.
Is there a "better" way that I can assert that one of the integers is divisible by 3? This feels too weak.
| Suppose that n is not divisible by 3.
It means :
n % 3 = {1,2} let call it x
and
n + 2 % 3 = (x + 2) % 3
and n + 4 % 3 = (x + 4) % 3
So if x = 1
n + 2 % 3 = 3 % 3 = 0
n + 4 % 3 = 5 % 3 = 2
n + 2 is divisible by 3
n + 4 is not divisible by 3
else if x = 2
n + 2 % 3 = 4 % 3 = 1
n + 4 % 3 = 6 % 3 = 0
n + 2 is not divisible by 3
n + 4 is divisible by 3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/522585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 14,
"answer_id": 11
} |
Modulo operation of large powers I came through this property in a cryptography book.
$(ab)\bmod n=\bigl((a \bmod n)(b \bmod n)\bigr)\bmod n$.
There is an example in the book,
$10^n\bmod 3= (10\bmod n)^n$.
Now if I have to calculate $8^15 \bmod 17$ can I calculate $8 \bmod 17$ and multiply the answer (which is $9$) $15$ times, so the answer becomes $16$?
| Yes, except that $8 \bmod 17$ is just $8$. You can also continually square to save some operations. So you get:
$$8 \equiv 8 \pmod {17},$$
$$8^2 = 64 \equiv 13 \pmod {17},$$
$$8^4 \equiv 13^2 = 169 \equiv 16 \pmod {17},$$
$$8^8 \equiv 16^2 = 256 \equiv 1 \pmod {17},$$
$$8^{15} = 8^8 \cdot 8^4 \cdot 8^2 \cdot 8 \equiv 1 \cdot 16 \cdot 13 \cdot 8 = 1664 \equiv 15 \pmod {17}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many Integer values of $n$ are possible for $n^2+25n+19$ to be a perfect square. [1] How many Integer values of $n$ are possible for $n^2+25n+19$ to be a perfect square.
[2] How many Integer values of $n$ are possible for $n^2-19n+99$ to be a perfect square.
$\underline{\bf{My\;Try}}::$ for first one , Let $k^2 = n^2+25n+99$, where $k,n\in \mathbb{Z}$
So $4k^2 = 4n^2+100n+76\Rightarrow (2k)^2 = (2n)^2+2\cdot (2n)\cdot 25+625+(76-625)$
$(2k)^2 = (2n+25)^2-549\Rightarrow (2n+25)^2-(2k)^2 = 549 = 3^3\cdot 61$
Now Let $x= 2k$ and $y = (2n+25)$,
we get $(x^2-y^2)=(x+y)\cdot(x-y) = 3^2 \cdot 61$
Is it Right or not ,and is there is any other method to solve these type of questions,
If Yes the please explain here
Thanks
| Let $$n^2+25n+19=(n+a)^2\text{ where }a\text{ is some integer }$$
$$\iff n=\frac{a^2-19}{25-2a}$$
Let integer $d$ divides both $a^2-19,25-2a$
$\implies d$ divides $\{2(a^2-19)+a(25-2a)\}=25a-38$
$\implies d$ divides $\{2(25a-38)+25(25-2a)\}=549=9\cdot61$
So, the divisors$(d)$ of $549$ are $\pm1,\pm3,\pm9\pm61,\pm183,\pm549$
As $25-2a$ must divide $a^2-19,$ check which values of $d(=25-2a)$ make $n$ integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Can't solve following limit: $\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right)$ Need to solve following problem:
$$\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right) $$ I've tried to do something like this: $$\lim_{x \to \infty} x\left(\sqrt[3]{5+8x^3} - 2x\right) =\lim_{x \to \infty}x\left( \sqrt[3]{ \left( \frac {5}{x^3}+8 \right)x^3 } - 2x\right) = \lim_{x \to \infty} x\left( \left ( \sqrt[3] { \frac{5}{x^3}+8} \right)x - 2x\right) $$
It seems to be the right way, but I can't do my next step.
| Note that
$$2x\lt \sqrt[3]{5+8x^3}=2x\left(1+\frac{5}{8x^3}\right)^{1/3}\lt 2x\left(1+\frac{5}{24x^3}\right).$$
Then use Squeezing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/525028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Reflection across a line?
The linear transformation matrix for a reflection across the line $y = mx$ is:
$$\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} $$
My professor gave us the formula above with no explanation why it works. I am completely new to linear algebra so I have absolutely no idea how to go about deriving the formula. Could someone explain to me how the formula is derived? Thanks
| Another way. To reflect along a line that forms an angle $\theta$ with the horizontal axis is equivalent to:
*
*rotate an angle $-\theta$ (to make the line horizontal)
*invert the $y$ coordinate
*rotate $\theta$ back.
Further, $y=mx$ implies $\tan \theta = m$, and $1+m^2 = \frac{1}{\cos^2\theta}$ .
Then, assumming you know about rotation matrices, you can write
$$\begin{align}T&=\begin{pmatrix}\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta\end{pmatrix}
\begin{pmatrix}1&0\\ 0 & -1\end{pmatrix}
\begin{pmatrix}\cos \theta & \sin \theta\\ -\sin \theta & \cos \theta\end{pmatrix} \\
&= \begin{pmatrix}\cos \theta & -\sin \theta\\ \sin \theta & \cos \theta\end{pmatrix}
\begin{pmatrix}\cos \theta & \sin \theta\\ \sin \theta & -\cos \theta\end{pmatrix} \\
&= \cos^2 \theta \begin{pmatrix}1 & -\tan \theta\\ \tan \theta & 1\end{pmatrix}
\begin{pmatrix}1 & \tan \theta\\ \tan\theta & -1\end{pmatrix} \\
&= \frac{1}{1 + m^2} \begin{pmatrix}1 & -m\\ m & 1\end{pmatrix}
\begin{pmatrix}1 & m\\ m & -1\end{pmatrix} \\
&=\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/525082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 6,
"answer_id": 3
} |
Compute a $2$D integral $$I=\iint_{D}\cfrac{dxdy}{\sqrt{1-\cfrac{{x}^{2}}{{a}^{2}}-\cfrac{{y}^{2}}{{b}^{2}}}\times\left( {x}^{2}+{y}^{2}+1-\cfrac{{x}^{2}}{{a}^{2}}-\cfrac{{y}^{2}}{{b}^{2}}\right)^{\cfrac{3}{2}}}$$
where $$D=\left\{(x,y);\cfrac{{x}^{2}}{{a}^{2}}+\cfrac{{y}^{2}}{{b}^{2}}\leq 1,x\geq 0,y\geq 0\right\}.$$
How to compute then? What I can image is just the radial transformation: $x=ar\cos t, y=br\sin t$.
| Let
$$\Delta(\theta) = a^2 \cos^2(\theta) + b^2\sin^2(\theta)$$
In terms of the parametrization $(x,y) = (ar\cos\theta,br\sin\theta)$,
the integral $\mathscr{I}$ we want can be rewritten as
$$\begin{align}
\mathscr{I} = & ab \int_0^{\frac{\pi}{2}} d\theta \int_0^1 \frac{rdr}{\sqrt{1-r^2}\sqrt{\Delta(\theta) r^2 + 1 - r^2}^3}\\
= & \frac{ab}{2}\int_0^{\frac{\pi}{2}} d\theta \int_0^1 \frac{dr^2}{(1-r^2)^2}\frac{1}{\sqrt{\Delta(\theta) \frac{r^2}{1-r^2} + 1}^3}
\end{align}$$
Let $u = \frac{r^2}{1-r^2}$, $v = \Delta(\theta) u + 1$ and $t = \tan\theta$.
Notice $\displaystyle du = \frac{dr^2}{(1-r^2)^2} $, we have
$$\mathscr{I} = \frac{ab}{2}\int_0^{\frac{\pi}{2}} d\theta \int_0^{\infty} \frac{du}{\sqrt{\Delta(\theta) u + 1}^3}
= \frac{ab}{2}\int_0^{\frac{\pi}{2}} \frac{d\theta}{\Delta(\theta)} \int_1^{\infty} \frac{dv}{\sqrt{v}^3}\\
= ab \int_0^{\frac{\pi}{2}} \frac{d\theta}{a^2\cos^2(\theta) + b^2\sin^2(\theta)}
= ab \int_0^{\infty} \frac{dt}{a^2 + b^2 t^2}
= \frac{\pi}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/525244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find a positive integer $x$ less than $105$ satisfying the following simultaneous congruence equations.
$$x=2 mod 3$$
$$x=3 mod 5$$
$$x=4 mod 7$$
I have only learnt modulo for 2 weeks so far... really basic theorems.
My attempt using definitions of modulo
From Equation 1, $3a=x-2 \rightarrow 15a=5x-10$
From Equation 2, $5b=x-3 \rightarrow 15b=3x-9$
Adding them together, $15(a+b)=8x-19$ which implies, $8x=19mod15$ or $8x=4mod15$
and hence i am stuck....
please help, i can't seem to understand this at all.
| First, $x=3a+2=5b+3=7c+4$. From the second equality, you get that $3a+2=5b+3\Rightarrow 3a=5b+1$. If $b=3d$ or $b=3d+2$, you reach a contradiction, so $$b=3d+1\Rightarrow x=5b+3=5(3d+1)+3=15d+8.$$ From the last equation, $15d+8=7c+4\Rightarrow 15d+4=7c$. By contradiction you can check that $d=7e+3$, so $$x=15d+8=15(7e+3)+8=105e+53.$$ So, the number you ask for is $53$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq 25/4$
Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{a+b+c+d}\geq \frac{25}{4}$$
given that $a, b, c, d > 0$ and $abcd = 1$
I reach to a point that says $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq \frac{25}{a+b+c+d}$
| Lagrange multipliers.
Let $f(a,b,c,d,\lambda) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{9}{a+b+c+d} + \lambda(abcd - 1)$. Then for a minimum we have
$$\frac{\partial f}{\partial a} = \frac{\lambda}{a} - \frac{1}{a^2} - \frac{9}{(a+b+c+d)^2} = 0,$$
so $\lambda = \frac{1}{a}+\frac{9a}{(a+b+c+d)^2}$. Since the equation is symmetric, we get
$$\lambda = \frac{1}{a} + \frac{9a}{(a+b+c+d)^2} = \frac{1}{b} + \frac{9b}{(a+b+c+d)^2} = \frac{1}{c} + \frac{9c}{(a+b+c+d)^2} = \frac{1}{d} + \frac{9d}{(a+b+c+d)^2}.$$
From this it follows for each pair of variables that they are the same or their product is $\frac{(a+b+c+d)^2}{9}$. In particular we only have to check cases where $a=b$, $c=d$; $a=b=c$; or $a=b=c=d$. All of these are fairly simple.
It remains to check the boundary. By multiplying by $(a+b+c+d)abcd$ we get an equivalent inequality, which holds also whenever one of the variables is $0$. Notice that if one of the variables approaches $\infty$, some other variable must approach $0$, so we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/527285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Comment upon nature of the roots How many roots are there of the following polynomial? How many are real, and how many are complex?
$$\left(x-3\right)^9+\left(x-3^2\right)^9+\left(x-3^3\right)^9+\left(x-3^4\right)^9+...+\left(x-3^9\right)^9=0$$
| I started by trying a simpler version of the same problem to see if the answer shows a pattern which can be used to solve this, more difficult version:
$$(x-3)^3 + (x-3^2)^3 + (x-3^3)^3 = 0$$
Then expanding the brackets. To be honest, the 3 just started confusing matters while looking for a pattern so I've replaced it with a y (I'll substitute it back later)
$$\begin{align}(x-y)^3 &= (x-y)(x-y)(x-y)\\
&= x(x-y)(x-y) - y(x-y)(x-y)\\
&= x^2(x-y) - xy(x-y) - yx(x-y) + y^2(x-y)\\
&= x^3 - x^2y - x^2y + xy^2 - x^2y + xy^2 + xy^2 - y^3\\
&= x^3 - 3x^2y + 3xy^2 - y^3\end{align}$$
$$\begin{align}(x-y^2)^3 &= (x-y^2)(x-y^2)(x-y^2)\\
&= x(x-y^2)(x-y^2) - y^2(x-y^2)(x-y^2)\\
&= x^2(x-y^2) - xy^2(x-y^2) - xy^2(x-y^2) + y^4(x-y^2)\\
&= x^3 - x^2y^2 - x^2y^2 + xy^4 - x^2y^2 + xy^4 + xy^4 - y^6\\
&= x^3 - 3x^2y^2 + 3xy^4 - y^6\end{align}$$
$$\begin{align}(x-3^2)^3 &= (x-y^3)(x-y^3)(x-y^3)\\
&= x(x-y^3)(x-y^3) - y^3(x-y^3)(x-y^3)\\
&= x^2(x-y^3) - xy^3(x-y^3) - xy^3(x-y^3) + y^6(x-y^3)\\
&= x^3 - x^2y^3 - x^2y^3 + xy^6 - x^2y^3 + xy^6 + xy^6 - y^9\\
&= x^3 - 3x^2y^3 + 3xy^6 - y^9\end{align}$$
And I get...
$$x^3 - 3x^2y + 3xy^2 - y^3 +x^3 - 3x^2y^2 + 3xy^4 - y^6 +x^3 - 3x^2y^3 + 3xy^6 - y^9 = 0$$
Seeing patterns but I'm out of time for this morning, I'll post later if I get there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/529576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2}$ Solve the following question :
\begin{eqnarray}
\\\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2}\\
\end{eqnarray}
The answer should be $\frac{1}{128}$.
I try that:
\begin{eqnarray}
\\\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2} &=& \lim_{x\to 16} \frac{(4-\sqrt{x})(4+\sqrt{x})}{(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})}\\
\\ &=& \lim_{x\to 16} \frac{16-x}{(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})}\\
\end{eqnarray}
What can I do?
Thank you for your attention.
| Hint.
$$(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})=\sqrt{x}(4-\sqrt{x})\cdot \sqrt{x}(4+\sqrt{x})^2=x(16-x)(4+\sqrt{x})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/529931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How prove this inequality $2a^ab^bc^cd^d\ge ac+bd$ Let $a,b,c,d$ be positive numbers such that $a+b+c+d=2$. Show that
$$2a^ab^bc^cd^d\ge ac+bd$$
My try: I think maybe I can use this inequality
$$(1+x)^n\ge 1+nx \hspace{12pt} (n>1)$$
then I can't get it to work.
I think this inequality may be solved by nice methods. Thank you.
| If $a \neq c$, then look at what happens if we replace both $a$ and $c$ with $x = (a+c)/2$ :
$x^2 \ge x^2 - (\frac {a-c}2)^2 = ac$.
Since the function $x \mapsto x \log x$ is convex (its second derivative is $1/x$ which is positive),
$\log(a^ac^c) = a \log a + c \log c \ge 2x \log x = \log (x^{2x})$.
Hence $2a^ab^bc^cd^d \ge 2x^xb^bx^xd^d$ , $x^2+bd \ge ac+bd$ , and we still have $x+b+x+d = a+b+c+d = 2$ so it is enough to prove the inequality when $a=c$ (and when $b=d$, for symmetry reasons).
So we are left with proving that if $x+y = 1$ then $2x^{2x}y^{2y} \ge x^2 + y^2$. Let $x = \frac {1+z}2$ and $y = \frac {1-z}2$. After simplifying, we have to prove that for $|z| < 1$, $(\frac{1+z}{1-z})^z \frac {1 - z^2}{1 + z^2} \ge 1$.
Taking logs, we have to show $z \log(1+z) - z \log(1-z) + \log(1-z^2) - \log(1+z^2) \ge 0$.
Using $\log$'s power series at $1$, this is
$(2z^2 + 2z^4/3 + 2z^6/5 + 2z^8/7 + 2z^{10}/9 +\ldots) - (2z^2 + 2z^6/3 + 2z^{10}/5 + \ldots) \ge
(2z^2 + 2z^6/3 + 2z^{10}/5 + 2z^{14}/7 + 2z^{18}/9 + \dots) - (2z^2 + 2z^6/3 + 2z^{10}/5 + \ldots) = 0$
There probably is something smarter to do instead, but this works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/530427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 0
} |
How do I determine the charasteristic function of $X^2$? I'm wondering how I kind show that the charasteristic function of $X^2$ given that $X\in N(0,1)$ is $\varphi_{X^2}(t)=\frac{1}{\sqrt{1-2it}}$.
I have tried using the change of variables such that $Y=X^2\implies X=\sqrt{Y}$ and then finding the derivative which is $\frac{1}{2\sqrt{y}}$ for using this in the transform formula. Then I apply $y$ instead of $x$ in the standard normal distribution $\phi(x)=\frac{1}{2\pi}e^{\frac{-x^2}{2}}$ and then using that for the charasteristic function formula $\varphi_{X^2}(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ity}e^{\frac{-y}{2}}\frac{1}{2\sqrt{y}}dy$. But this integral gets way too complicated and I think that my method is wrong. Is there anyway I can show that $\varphi_{X^2}(t)=\frac{1}{\sqrt{1-2it}}$?
thanks
| $$Y = {X^2} \Rightarrow f\left( y \right) = \frac{1}{{\sqrt {2\pi } }}{y^{ - \frac{1}{2}}}{e^{ - \frac{y}{2}}},x > 0.$$ (because $Y \sim \chi _1^2$) so
$${\varphi _Y}\left( t \right) = \mathop \int \limits_0^\infty {e^{ity}} \cdot \frac{1}{{\sqrt {2\pi } }}{y^{ - \frac{1}{2}}}{e^{ - \frac{y}{2}}}dy = \frac{1}{{\sqrt {2\pi } }}\mathop \int \limits_0^\infty {y^{\frac{1}{2} - 1}}{e^{ - y\left( {\frac{1}{2} - it} \right)}}dy$$
using
$$\mathop \int \limits_0^\infty {x^{\alpha - 1}}{e^{ - \beta x}}dx = \frac{{\Gamma \left( \alpha \right)}}{{{\beta ^\alpha }}}$$, we get
$$ = \frac{1}{{\sqrt {2\pi } }} \cdot \frac{{\Gamma \left( {\frac{1}{2}} \right)}}{{{{\left( {\frac{1}{2} - it} \right)}^{\frac{1}{2}}}}} = \frac{1}{{\sqrt 2 \cdot \sqrt \pi }} \cdot \frac{{\sqrt \pi }}{{\frac{1}{{\sqrt 2 }} \cdot \sqrt {1 - 2it} }} = \frac{1}{{\sqrt {1 - 2it} }}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/530857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find number of solutions of $2^x$+$3^x$+$4^x$=$5^x$ Find number of solutions of
$$2^x+3^x+4^x=5^x$$
I tried using graphs but don't know how to draw graph of L.H.S.
| You can prove it with the intermediate value theorem.
Suppose $f(x) = 2^x + 3^x + 4^x$ and $g(x) = 5^x$. Both are continuous and differentiable for all $x \in \mathbb{R}$. We see that $f(0) = 1 < 3 = g(0)$ and $f(3) = 99 < 125 = g(3)$. So there is at least one solution to $f(x) = g(x)$. Let us call the smallest solution to the equation $z$. We see that
$$
f^{(k)}(x) = (\ln 2)^k \cdot 2^x + (\ln 3)^k \cdot 3^x + (\ln 4)^k \cdot 4^x
$$
and
$$
g^{(k)}(x) = (\ln 5)^k \cdot 5^x.
$$
Now we see that
\begin{align*}
f^{(k)}(z) &< (\ln 5)^k \cdot ( 2^z + 3^z + 4^z) \\
&= (\ln 5)^k \cdot 5^z \\
&= g^{(k)}(z),
\end{align*}
so we know the curves cross rather than being tangential. For values of $x \geq z$, we see that $f(x) < g(x)$ and $f'(x) < g'(x)$, so there cannot be any other solutions to $f(x) = g(x)$. Therefore, there is only one solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/532142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$\cot\theta + \tan \theta = x$ and $\sec \theta - \cos \theta = y$, evaluate $\left(x^2 y + xy^2\right)^{2/3}$
Given that $\cot\theta + \tan \theta = x$ and $\sec \theta - \cos \theta = y$, evaluate $\left(x^2 y + xy^2\right)^{2/3}$
I tried substituting $\cot \theta$ with $\frac{1}{\tan \theta}$ and similarly with $\sec \theta$, but it was of no help.
Please help me in simplifying this preferably with Pythagorean Identities. I found this question from the section of the book where it explains these identities, so I suppose we could use them here.
| $$\left(x^2 y + xy^2\right)^{2/3}=\left(xy(x + y)\right)^{2/3}$$
$$xy=\frac{\sin \theta}{\cos^2 \theta}$$
$$x+y=\frac{1+\sin^3 \theta}{\sin \theta\cos \theta}$$
$$xy(x+y)=\frac{1+\sin^3 \theta}{\cos^3 \theta}$$
$$\left(xy(x + y)\right)^{2/3}=\frac{(1+\sin^3 \theta)^{2/3}}{\cos^2 \theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/533122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$ This is for homework, and I could use a little help. The question asks
Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$.
Here is what I have done so far. Since $f$ is entire, I wrote
$$ f(z) = \sum_{n=0}^{\infty} a_n z^n = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 + \dotsb $$
for some $z \in \mathbb{C}$. Then
$$ f(2z) = a_0 + 2a_1z + 4a_2z^2 + 8a_3z^3 + 16a_4z^4 + \dotsb $$
and
$$ (1-2z)f(z) = a_0 + (a_1-2a_0)z + (a_2-2a_1)z^2+(a_3-2a_2)z^3 + (a_4-2a_3)z^4 + \dotsb. $$
Comparing coefficients, I find that
\begin{align*}
a_0 &= a_0 \\
a_1 &= -2a_0 \\
a_2 &= \frac{2^2}{3}a_0 \\
a_3 &= -\frac{2^3}{7 \cdot 3}a_0 \\
a_4 &= \frac{2^4}{15 \cdot 7 \cdot 3}a_0 \\
a_5 &= \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3} a_0 \\
&\vdots
\end{align*}
Now $f$ looks like
$$ f(z) = a_0 \left( 1 - 2z + \frac{2^2}{3}z^2 - \frac{2^3}{7 \cdot 3}z^3 + \frac{2^4}{15 \cdot 7 \cdot 3}z^4 - \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3}z^5 + \dotsb \right). $$
Does the series in parenthesis represent any elementary function? Besides the denominators, it looks like the Taylor expansion of $e^{-2z}$.
| Your function has zero at $z=1/2$. I think that your formulas should be easier to work with if you expand at that point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/534517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.