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Verifying the trigonometric identity $\cos{x} - \frac{\cos{x}}{1 - \tan{x}} = \frac{\sin{x} \cos{x}}{\sin{x} - \cos{x}}$ I have the following trigonometric identity
$$\cos{x} - \frac{\cos{x}}{1 - \tan{x}} = \frac{\sin{x} \cos{x}}{\sin{x} - \cos{x}}$$
I've been trying to verify it for almost 20 minutes but coming up with nothing
Thank you
| $$
\frac{\cos}{1-\tan x}\cdot\frac{\cos x}{\cos x} = \frac{\cos^2 x}{\cos x-\sin x}
$$
$$
\cos x\cdot\frac{\cos x-\sin x}{\cos x-\sin x} - \frac{\cos^2 x}{\cos x-\sin x} = \frac{-\sin x\cos x}{\cos x-\sin x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/418476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Distinguishable telephone poles being painted Each of n (distinguishable) telephone poles is painted red, white, blue or yellow. An odd number are painted blue and an even number yellow. In how many ways can this be done?
Can some give me a hint how to approach this problem?
| The exponential generating function (EGF) for the number of red or white poles is
$$1 + z + \frac{1}{2!}z^2 + \frac{1}{3!}z^3 + \dots = e^z$$
The EGF for the number of blue poles is
$$z + \frac{1}{3!}z^3 + \frac{1}{5!}z^5 + \dots = \frac{1}{2}(e^z-e^{-z})$$
The EGF for the number of yellow poles is
$$1 + \frac{1}{2!}z^2 + \frac{1}{4!}z^4 + \dots = \frac{1}{2}(e^z + e^{-z})$$
So the EGF for the number of acceptable permutations of red, white, blue and yellow poles is
$$f(z) = (e^z)^2 \cdot \frac{1}{2}(e^z-e^{-z}) \cdot \frac{1}{2}(e^z + e^{-z}) = \frac{1}{4}(e^{4z}-1)$$
The number of acceptable permutations of $n$ poles is the coefficient of
$\frac{1}{n!}z^n$
in $f(z)$, which by inspection is $4^{n-1}$ for $n > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/418520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Upper bound for expression involving logarithms Let $N = 2^p$ for some $p \in \mathbb{N}$. Find the smallest upper bound for $\frac{N}{2}\log\left(\frac{N}{2}\right) + \frac{N}{4}\log\left(\frac{N}{4}\right) + \ldots + 1$
I guess I could first rewrite this to $\frac{2^p}{2}\log\left(\frac{2^p}{2}\right) + \frac{2^p}{4}\log\left(\frac{2^p}{4}\right) +\ldots+ 1$ and then to
$2^{p-1}\log(2^{p-1}) + 2^{p-2}\log(2^{p-2}) +\ldots+1$ but I still don't know how I should proceed.
All help appreciated.
Edit: Now I though also writing it to the form $(p-1)2^{p-1} + (p-2)2^{p-2} + ... + 1 \Leftrightarrow \sum_{i=1}^{log N} (p-i)2^{p-i}$ but I still feel like ...... ;-( (yes, the log is base 2, sorry, forgot to mention that)
| You want
$\begin{align}
\sum_{k=1}^p \frac{n}{2^k}\ln \frac{n}{2^k}
&=\sum_{k=1}^p \frac{n}{2^k}(\ln n- \ln {2^k})\\
&=\sum_{k=1}^p \frac{n}{2^k}(\ln n- k\ln {2})\\
&=n \ln n\sum_{k=1}^p \frac{1}{2^k}
-n\ln 2\sum_{k=1}^p \frac{k}{2^k} \\
\end{align}
$
For not small $p$,
$\sum_{k=1}^p \frac{1}{2^k} \approx 1$.
Since $\sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$,
$\sum_{k=1}^p \frac{k}{2^k}
\approx \frac{1/2}{(1-1/2)^2}
=2
$,
so your sum $\approx n \ln n- 2 n \ln 2$.
You can get it more exactly by
getting the exact form for
$\sum_{k=1}^p k x^k$,
but the difference is of order
$\frac{p}{2^p}=\frac{\ln n}{n}$,
so I will leave that to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/418579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Elementary lower bounds for $n^{1/n}$ I can show that
$n^{1/n} > 1+1/n$
for integer $n \ge 3$
by completely elementary means -
no logs, exponentials,
or calculus.
Are there better bounds that
can be proved in an elementary way?
Here is my proof:
The bound is equivalent to
$n > \frac{(n+1)^n}{n^n}$
or
$\frac{(n+1)^n}{n^{n+1}} < 1$.
For $n=3$,
this is $\frac{4^3}{3^4}
=\frac{64}{81}
< 1$.
The ratio of consecutive terms is
$\begin{align}
\frac{\frac{(n+2)^{n+1}}{(n+1)^{n+2}}}{\frac{(n+1)^n}{n^{n+1}}}
&=\frac{(n(n+2))^{n+1}}{(n+1)^{2n+2}}\\
&=\frac{(n^2+2n)^{n+1}}{(n^2+2n+1)^{n+1}}\\
&< 1
\end{align}
$
so the terms are decreasing
and thus less than $1$.
Of course, since $e^x > 1+x$,
$n^{1/n} = e^{\ln n/n} > 1+\ln n/n$,
but this is non-elementary.
| $$\left(1+\frac1n\right)^n=\sum_{k=0}^n\frac{n!}{k!(n-k)!\cdot n^k}$$
After cancelling $n!$ against $(n-k)!$ we are left with $k$ factors $\le n$ in the numerator. Since we have $k$ factors $=n$ in the denominator, we have $\frac{n!}{k!(n-k)!\cdot n^k}\le \frac1{k!}$. This brings us almost to the series $\sum \frac 1{k!}$ for $e$, but remains elementary enough, I guess. All but the $0$th summand are $\le \frac12$, so we get the very weak $\left(1+\frac1n\right)^n\le 1+\frac n2$, hence for $n> 2$
$$\sqrt[n]n>\sqrt[n]{1+\frac n2} \ge 1+\frac1n.$$
Of course, we can easily improve this by noting $\frac1{k!}<\frac1{2^{k-1}}$ for $k\ge1$ and hence $\left(1+\frac1n\right)^n<3$.
This makes
$$ \sqrt[n]n\ge\sqrt[n]9> \left(1+\frac1n\right)^2=1+\frac2n+\frac1{n^2}\qquad \text{for }n\ge 9$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/418644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find the values of the constants in the following identity $2x^3+3x^2-14x-5=(ax+b)(x+3)(x+1)+C$ I'm working through identities but I can't figure out how to get further than multiplying out the above to get :
$$2x^3+3x^2-14x-5=2ax^3+3ax^2+3ax+bx^2+3bx+bx+3b+C$$
can someone give me a hint on what to do next?
| The now deleted answer by amWhy used polynomial long division. Since $(x+3)(x+1)= x^2+4x+3$, if we perform the following polynomial long division
we get rightaway the quotient $2x-5=ax+b$, thus $a=2,b=-5$, and the remainder $10=C$.
have you any tips on spotting when to use this ?
Given two polynomials $A(x)$ and $B(x)$, with degree of $B(x)$ greater than $0$, we can find two other polynomials $Q(x)$ and $R(x)$ such that $$A(x)=B(x)Q(x)+R(x),$$ and the degree of $R(x)$ is lower than the degree of $B(x)$.
In your case $A(x)=2x^3+3x^2-14x-5$, $B(x)=x^2+4x+3$, $Q(x)=2x-5$ and $R(x)=10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/418815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
How can $\left({1\over1}-{1\over2}\right)+\left({1\over3}-{1\over4}\right)+\cdots+\left({1\over2n-1}-{1\over2n}\right)+\cdots$ equal $0$?
Let
$$\begin{align*}x &= \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + \cdots\\
y &= \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots\\
z &= \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots \end{align*}$$
so we have
$$x = y + z.$$
However, $x = 2\cdot z$, so $y$ = $z$ or
$$\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-1} + \cdots = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots + \frac{1}{2n} + \cdots$$
This looks ok if I interpret it as
$$\frac{1}{1} = \left (\frac{1}{2} - \frac{1}{3} \right ) + \left (\frac{1}{4} - \frac{1}{5} \right ) + \left (\frac{1}{6} - \frac{1}{7} \right ) + \cdots + \left (\frac{1}{2n} - \frac{1}{2n+1} \right ) + \cdots$$
However, it's a bit weird if I write it as
$$\left (\frac{1}{1} - \frac{1}{2} \right ) + \left (\frac{1}{3} - \frac{1}{4} \right ) + \left (\frac{1}{5} - \frac{1}{6} \right ) + \cdots + \left (\frac{1}{2n-1} - \frac{1}{2n} \right ) + \cdots = 0.$$
How can a sum of positive numbers equal $0$?
| Almost everything in your proof works fine until you write "this looks ok if I interprete it as..."
Until then, you are manipulating infinite sums of series with positive terms, these are extended nonnegative real numbers (numbers $x$ such that $0\leqslant x\leqslant+\infty$, if you like) hence adding them and equating them is perfectly legal.
The trouble begins when you substract them, since there is no substraction on the set of extended nonnegative real numbers. Unsurprisingly, you soon must deal with $(+\infty)-(+\infty)$ differences, and chaos ensues.
A less sophisticated example, flawed quite similarly, is to start with the correct identity
$$
1+1+1+\cdots=\underline{\mathbf 1}+(\color{red}{1}+\color{blue}{1}+\color{green}{1}+\cdots),
$$
and to deduce from it that
$$
0=(1-\color{red}{1})+(1-\color{blue}{1})+(1-\color{green}{1})+\cdots=\underline{\mathbf 1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/420047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Show that $\frac{x}{1-x}\cdot\frac{y}{1-y}\cdot\frac{z}{1-z} \ge 8$.
If $x,y,z$ are positive proper fractions satisfying $x+y+z=2$, prove that $$\dfrac{x}{1-x}\cdot\dfrac{y}{1-y}\cdot\dfrac{z}{1-z}\ge 8$$
Applying $GM \ge HM$, I get $$\left[\dfrac{x}{1-x}\cdot\dfrac{y}{1-y}\cdot\dfrac{z}{1-z}\right]^{1/3}\ge \dfrac{3}{\frac 1x-1+\frac 1y-1+\frac 1z-1}\\=\dfrac{3}{\frac 1x+\frac 1y+\frac 1z-3}$$
Then how to proceed. Please help.
| We begin by setting $a=1-x, b=1-y, c=1-z$, and noting that $a+b+c=1$. The conditions of the problem imply that $a,b,c\in (0,1)$.
We now need Maclaurin's inequality: $$\frac{a+b+c}{3}\ge \sqrt{\frac{ab+bc+ac}{3}}\ge \sqrt[3]{abc}$$
Ignoring the middle part temporarily, we have $\frac{1}{3}\ge \sqrt[3]{abc}$ or $3\sqrt[3]{abc}\le 1$. Ignoring the first part and squaring, we have $\frac{ab+bc+ac}{3}\ge (\sqrt[3]{abc})^2\ge 3abc$, where we multiplied by $3\sqrt[3]{abc}\le 1$ in the last step. We rearrange this to get $$ab+bc+ac\ge 9abc$$
then rearrange that to get $$1-a-b-c+ab+bc+ac-abc\ge 8abc$$
The left side factors as $(1-a)(1-b)(1-c)$; divide by $abc$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Explanation of this step in a modular arithmetic problem
The multiplicative inverse of $5$ is $7$, when using mod $34$.
$$\begin{align*}
5\cdot x&=3\\[0.1in]
7\cdot 5\cdot x &=7\cdot 3\\[0.1in]
1\cdot x &=7\cdot 3\\[0.1in]
x&=21
\end{align*}$$
I don't understand this part:
$$\begin{align*}
7\cdot 5\cdot x &=7\cdot 3\\[0.1in]
1\cdot x &=7\cdot 3
\end{align*}$$
How is 7*5*x the same as 1*x?
| Recall that $a\equiv b\bmod n$ precisely when $n$ divides the difference $a-b$. Therefore, we have
$$35\equiv 1\bmod 34.$$
It is also true that if $a\equiv b \bmod n$, then $ac\equiv bc\bmod n$ for any $c$. Therefore, whatever $x$ is,
$$35x\equiv x\bmod 34,$$
so that we can go from
$$35x\equiv 21\bmod 34$$
to
$$x\equiv 21\bmod 34.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving non homogeneous recurrence relation I am having a hard time understanding these questions. I know I need to find the associated homogeneous recurrence relation first, then its characteristic equation. I cant figure out how to find the particular solution to the non homo recurrence relation though.
Ex: $$a_{n}= 4a_{n-1} + 4a_{n-2} + (n+1)2^n$$
My characteristic equation is $r^2-4r-4=0$ and $r=2(1+ \sqrt{2}), r=2(1- \sqrt{2})$. Next I need to guess some equation for my $f(n)=(n+1)2^n$ and plug it into the original to find some constants,, I am having the trouble here,, I dont understand how to come up with these guess equations.
I know the theorem that says the general solution (of the non homo recurrence relation) is the general solution of the associated recurrence relation + the particular solution:
$a_{n}=a_{n}^{(h)} + a_{n}^{(p)}$
So far I have $a_{n}=A2(1+ \sqrt{2})^n + B2(1- \sqrt{2})^n + a_{n}^{(p)}$
| Forget all this, use generating functions directly. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write:
$$
a_{n + 2} = 4 a_{n + 1} + 4 a_n + 8 (n + 3) \cdot 2^n
$$
Multiply by $z^n$, sum over $n \ge 0$, recognize:
\begin{align}
\sum_{n \ge 0} a_{n + r} z^n
&= \frac{A(z) - a_0 - a_1 z - \ldots - a_{r - 1} z^{r - 1}}{z^r} \\
\sum_{n \ge 0} 2^n z^n
&= \frac{1}{1 - 2 z} \\
\sum_{n \ge 0} n 2^n
&= z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - 2 z} \\
&= \frac{2 z}{(1 - 2 z)^2}
\end{align}
and get:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 4 \frac{A(z) - a_0}{z} + 4 A(z)
+ \frac{16 z}{(1 - 2 z)^2} + \frac{3}{1 - 2 z}
$$
This gives, written as partial fractions (partially):
$$
A(z)
= \frac{1 - 2 a_0 - 2 (a_1 - 1) z}{2 (1 - 2 z^2)} +\frac{1}{2 (1 - 2 z)}
$$
The $(1 - 2 z)^{-1}$ gives rise to a $2^n$ in the solution, while you can write:
\begin{align}
\frac{1}{1 - 2 z^2}
&= \sum_{n \ge 0} 2^n z^{2 n} \\
\frac{z}{1 - 2 z^2}
&= \sum_{n \ge 0} 2^n z^{2 n + 1}
\end{align}
Thus you get expressions for even/odd indices. Or you could split that into partial fractions too, and mess with the resulting irrationals.
If you are simply interested in a particular solution, pick any easy values, like $a_0 = 0$ and $a_1 = 1$, and expand the above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Recursion problem help The following are the teachers example problems. The issue is that I don't understand the exact steps they took to go from $f(0)$ to $f(1)$ to $f(2)$ to $f(3)$. What I'm asking here is if someone could be so kind to show me how the answers for $f(0)\ldots f(3)$ are derived for each problem.
Let $f$ be defined as follows: $f(0) = 3$
$f(n) = (−1)^nf(n−1) + 4n$ for $n\ge1$.
Find $f(7)$
The correct answer is $3$
$f(0) = 3$
$f(1) = 1$
$f(2) = 9$
$f(3) = 3$
$f(4) = 19$
$f(5) = 1$
$f(6) = 25$
$f(7) = 3$
Let $f$ be defined as follows:
$f(0) = 1$
$f(n+1) = f(n) + n$ for $n≥0$
Find $f(6)$
The correct answer is $22$
$f(0) = 1$
$f(1) = 2$
$f(2) = 4$
$f(3) = 7$
$f(4) = 11$
$f(5) = 16$
$f(6) = 22$
Let $f$ be defined as follows:
$f(0) = 4$
$f(1) = 3$
$f(n) = f(n−1) + 3f(n−2)$ for $n≥2$.
Find $f(6)$
The correct answer is $348$
$f(0) = 4$
$f(1) = 3$
$f(2) = 15$
$f(3) = 24$
$f(4) = 69$
$f(5) = 141$
$f(6) = 348$
| The first recurrence is $f(n)=(-1)^nf(n-1)+4n$ for $n\ge 1$, with initial value $f(0)=3$. Just plug in successive values of $n$, starting with $n=1$:
$$\begin{align*}
f(1)&=(-1)^1f(0)+4\cdot1=(-1)(3)+4=-3+4=1\\
f(2)&=(-1)^2f(1)+4\cdot2=1\cdot1+8=9\\
f(3)&=(-1)^3f(2)+4\cdot3=(-1)(9)+12=3\\
f(4)&=(-1)^4f(3)+4\cdot4=1\cdot3+16=19\;,
\end{align*}$$
and so on.
The second recurrence is $f(n+1)=f(n)+n$ for $n\ge 0$, with initial value $f(0)=1$. To find $f(1)=f(0+1)$, you need to take $n=0$:
$$f(1)=f(0+1)=f(0)+0=1+0=1\;.$$
Continue in similar fashion, taking $n$ to be in succession $1,2,3$, and so on:
$$\begin{align*}
f(2)&=f(1+1)=f(1)+1=1+1=2\\
f(3)&=f(2+1)=f(2)+2=2+2=4\\
f(4)&=f(3+1)=f(3)+3=4+3=7\;,
\end{align*}$$
and so on.
The third recurrence is $f(n)=f(n-1)+3f(n-2)$ for $n\ge 2$, with initial values $f(0)=4$ and $f(1)=3$. Like the first one, this one gives you $f(n)$ instead of $f(n)$, so you can proceed very straightforwardly, just as with the first one:
$$\begin{align*}
f(2)&=f(1)+3f(0)=3+3\cdot4=15\\
f(3)&=f(2)+3f(1)=15+3\cdot3=24\\
f(4)&=f(3)+3f(2)=24+3\cdot15=24+45=69\;,
\end{align*}$$
and so on.
In every case it’s just a matter of substituting the appropriate value of $n$ into the recurrence, and if you’ve already computed the values of $f$ that appear on the righthand side of the recurrence, you can use them to compute the desired value.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A problem on Cauchy sequences Let $\langle x_n\rangle $ be a sequence defined recursively by $ 0<a \le x_1 \le x_2\le b $ and $ x_{n+2} =\sqrt{x_nx_{n+1}} $ for each n $ \in \Bbb N $ show that $|x_{n+2} -x_{n+1}| \le \frac {b}{b+a} |x_n -x_{n+1}|$ Deduce that $\langle x_n \rangle$ is Cauchy and find its limit.
Can some one help me on this especially in deducing its Cauchy! I think the 1st part can be proven by induction though I had some problem there too.
| We can estimate
\begin{eqnarray}
|x_{n+2}-x_{n+1}| & = & |\sqrt{x_{n+1} x_{n+1}}-\sqrt{x_n x_{n+1}}| = \frac{|x_{n+1}(x_{n+1}-x_n)|}{\sqrt{x_{n+1} x_{n+1}} + \sqrt{x_n x_{n+1}}} \\
& = & \frac{1}{1+\sqrt{x_n/x_{n+1}}}|x_{n+1}-x_n| \leq \frac{1}{1+\sqrt{a/b}}|x_{n+1}-x_n| \\
& \leq & \frac{1}{1+a/b/\sqrt{a/b}} |x_{n+1}-x_n| \leq \frac{1}{1+a/b} |x_{n+1}-x_n| \\
& = & \frac{b}{a+b} |x_{n+1}-x_n| \ .
\end{eqnarray}
Assume $m>n$. Now
\begin{eqnarray}
|x_m-x_n| & = & \bigg| \sum_{k=n}^{m-1} x_{k+1}-x_k \bigg| \leq \sum_{k=n}^{m-1} |x_{k+1}-x_k| \leq \sum_{k=n}^\infty \bigg(\frac{b}{a+b}\bigg)^{k-1} |x_2-x_1| \\
& = & \frac{1}{1-\frac{b}{a+b}} \bigg(\frac{b}{a+b}\bigg)^{n-1} |x_2-x_1| \rightarrow 0 \ ,
\end{eqnarray}
as $m,n \rightarrow \infty$, where the second inequality is estalished by induction to the index $m-1$. This shows that the sequence is a Cauchy-sequence and has a limit $x$.
The sequence can be mapped using the logarithm function, $y_i = \ln(x_i)$, and we obtain $y_{i+2} = \frac{1}{2}(y_{i+1}+y_i)$ and hence $y_{i+2}-y_{i+1} = -\frac{1}{2}(y_{i+1}-y_i)$. Now
\begin{eqnarray}
y & = & y_1 + \sum_{k=1}^\infty y_{k+1}-y_k = y_1 + \sum_{k=1}^\infty \bigg(-\frac{1}{2}\bigg)^{k-1} (y_2-y_1) = y_1 + \frac{1}{1+\frac{1}{2}} (y_2-y_1) \\
& = & \frac{y_1+2y_2}{3} \ ,
\end{eqnarray}
where the second equation is established termwise by induction. Hence
\begin{eqnarray}
x = e^y = e^\frac{y_1+2y_2}{3} = (e^{y_1} e^{y_2^2})^\frac{1}{3}= (x_1 x_2^2)^\frac{1}{3} \ .
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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$\sqrt{x + \sqrt{2x -1}} + \sqrt{x- \sqrt{2x-1}} = A $ I am puzzling over the following problem, which involves an equation of the form:
$$\sqrt{x + \sqrt{2x -1}} + \sqrt{x- \sqrt{2x-1}} = A $$
The problem involves finding real values of x corresponding to
A = $ \sqrt{2}$, A = 1, and A = 2, where the roots must be of non-negative real numbers.
So far, I have found that A = $\sqrt{2}$ when x = 1:
$$\sqrt{1 + \sqrt{2(1) -1}} + \sqrt{x- \sqrt{2(1)-1}} = \sqrt{1 + 1} + \sqrt{1 - 1} = \sqrt{2} $$
But it is not so obvious to me how to go about obtaining values of x when A = 1, or A = 2, for example. And how can I be certain that I have found all values of x, if there is more than one? And how can I be absolutely sure that there is only one solution?
And if we think about it the other way around- in general, for which values of A are there solutions?
Some thoughts on any of the above would be much appreciated!
| We look for real solutions. Let $y=2x-1$. Note that $x\ge \frac{1}{2}$. We have $x=\frac{y+1}{2}$. Then
$$x+\sqrt{2x-1}=\frac{y+1}{2}+\sqrt{y}=\frac{1}{2}(1+\sqrt{y})^2.$$
Taking the square root, we find that
$$\sqrt{x+\sqrt{2x-1}}=\frac{1}{\sqrt{2}}(1+\sqrt{y}).$$
Almost similarly, we find that
$$\sqrt{x-\sqrt{2x-1}}=\frac{1}{\sqrt{2}}|1-\sqrt{y}|.$$
Adding, we end up with the equation
$$(1+\sqrt{y})+|1-\sqrt{y}|=\sqrt{2}A.$$
If $\sqrt{y}\le 1$, we get $A=\sqrt{2}$, with no other conditions on $y$. That gives us an interval of solutions when $A=\sqrt{2}$.
If $\sqrt{y}\gt 1$, we arrive after a little manipulation at $y=\frac{A^2}{2}$. So there is no solution if $A=1$, and indeed if $A\lt \sqrt{2}$. There is a unique solution if $A\gt \sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Greatest integer $n$ where $n \lt (\sqrt5 +\sqrt7)^6$ I'm really not sure how to do this. I factored out a power of $3$ and squared so that I have $2^3 (6+\sqrt{35})^3 \gt n$ , and I know that if I can prove that $12^3-1 \le (6+\sqrt{35})^3 \lt 12^3$ I am basically done, but I don't know how to do that. Any help is appreciated. Thanks!
| We want a lower bound on
$\sqrt{1-x}$.
From $\sqrt{1-x} = (1-x)^{1/2}
=1-x/2-x^2/8-x^3/16 -5x^4/128 ...
$
I will try
$1-x/2-x^2/4$.
$\begin{align}
(1-x/2-x^2/4)^2
&=1-x-x^2(1/4+1/2)+x^3/4+x^4/16\\
&=1-x-3x^2/4+x^3/4+x^4/16\\
&= 1-x-x^2(3/4-x/4-x^2/16)\\
&< 1-x\\
\end{align}
$
for $0 < x < 1/2$,
so $\sqrt{1-x} > 1-x/2-x^2/4$
for $0 < x < 1/2$.
$\sqrt{35} = \sqrt{36-1}
=
6\sqrt{1-1/36}
> 6(1-1/12-1/(4*36))
=6-1/2-1/24
$
so
$6+\sqrt{35} > 12-1/2-1/24
> 12-1$
as you wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/426631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Parametrization of $x^2+y^2=z^2$ How can we show that any point on $x^2+y^2=z^2$ can be written in the form (z $\cos(\theta)$, z $\sin(\theta)$, z) for some $\theta$? Here is how I tried to approach it:
$$(z \cos(\theta))^2+(z \sin(\theta))^2)^2=(z)^2$$
$$z^2 \cos^2(\theta)+z^2 \sin^2(\theta)=z^2$$
$$z^2 (\cos^2(\theta)+\sin^2(\theta))=z^2$$
$$z^2 (\cos^2(\theta)+\sin^2(\theta))=z^2$$
$$z^2 (1)=z^2 = t^2$$
Is my approach correct or am I way off? If not, any hints would be wonderful with some detailed explanation. The way I attempted definitely does not look right in someway.
| You have shown that if $x=z\cos\theta$ and $y=z\sin\theta$, then $x^2+y^2=z^2$.
The proof could be shortened a bit. Suppose that $x=z\cos\theta$ and $y=z\sin\theta$. Then
$$x^2+y^2=z^2(\cos^2\theta+\sin^2\theta)=(z^2)(1)=z^2.$$
However, that is not what you are being asked to show. You are asked to show that if $x^2+y^2=z^2$, then there exists a $\theta$ such that $x=z\cos\theta$ and $y=z\sin\theta$.
First deal with the special case $z=0$. Then if $x^2+y^2=z^2$, we must have $x=y=0$. Then taking $\theta$ to be any angle (number) we get that $x=z\cos \theta$ and $y=z\sin\theta$. The case $z=0$ is exceptional, in that any $\theta$ will work.
Suppose now that $z\ne 0$. Then if $x^2+y^2=z^2$, we must have $|x|\le |z|$ and $|y|\le |z|$. It follows that $\frac{x}{z}$ and $\frac{y}{z}$ are both between $-1$ and $1$.
Since $\frac{x}{z}$ is between $-1$ and $1$, it is the cosine of something. There is a $\theta$ in the interval $[0,2\pi)$ such that $\frac{x}{z}=\cos\theta$. Actually, this $\theta$ is almost determined by $\frac{x}{z}$.
Typically there are two such $\theta$. For example, if $\frac{x}{z}$ is positive and less than $1$, there is one such $\theta$ in the first quadrant and one in the fourth quadrant. So we almost know $\sin\theta$.
The sign of $\frac{y}{z}$ determines, in this case, whether $\theta$ is in the first or fourth quadrant.
The geometry: We took a relentlessly algebraic approach. However, the geometry provides much more insight. Let $z$ be a fixed non-zero number. Then $x^2+y^2=z^2$ is the equation of a circle of radius $|z|$. For simplicity take $z$ positive. Take a point $P$ on that circle. Then $\theta$ be the angle we must rotate starting at the positive $x$-axis to get to $P$. And now drawing an appropriate triangle, and paying attention to sign, we see that $P=(z\cos\theta,z\sin\theta)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate a $A_{x\times y}B_{y\times z}$ where $ A$ and $B$ are matrices, $x\neq z$ I know how to evaluate a Cx,y * Dy,x (C rows are equals to D columns), but how do I evaluate a matrix multiplication in which the involved matrices (A and B) have respectively different number of rows and columns?
Here is the "mask": $A_{x\times y} B_{y\times z}$ (x,y in A and y,z in B are the matrices order).
EDIT:
I wanna know how to evaluate a multiplication of matrices that have A number of columns = B number of rows, but A number of rows != B number of columns.
You usually multiply the elements of the A rows with the elements of B columns, but if the number of rows are different from the number of columns some elements would remain, what should I do?
| Your getting what the indices denote mixed up: $C_{x\times y}\times D_{y \times x} = E_{x\times x}$ denotes the product of a matrix $C$ with $x$ rows and $y$ columns, times a matrix $D$ having $y$ rows and $x$ columns. This results in a matrix $E_{x\times x}$ which has $x$ rows and $x$ columns: i.e. $E$ is then a square matrix.
Multiplication of matrices $A\cdot B$ is defined if and only if the number of columns of $A$ is equal to the number of rows of $B$.
So in your case $A_{\large {\bf x}\times\color{blue}{\bf y}} \cdot B_{\large \color{blue}{\bf y}\times {\bf z}}$ is defined, and results in a matrix $C_{\large \bf x\times z}$.
Here the first index denotes the number of rows of a matrix, and the second index denotes the number of columns of a matrix. We see that multiplication is defined because there are $\color{blue}{\bf y}$ columns in $A$, and $\color{blue}{\bf y}$ rows in $B$.
So, e.g., $A_{x\times y}$ denotes a matrix with $x$-rows, and $y$-columns, and $x\times y$ is called the dimension of the matrix $A$.
Example:
Lets multiply: $A_{2\times 3}\cdot B_{3\times 3} = \begin{bmatrix}1 & 2 & 3 \\ 1 & 2 & 0\end{bmatrix} \cdot \begin{bmatrix} 0 & 1 & 0\\ 2 & 0 & 1\\ 1 & 0 & 2\end{bmatrix}$
We will obtain a product $$ \begin{align}A\cdot B =C_{2\times 3} & = \begin{bmatrix} (1\cdot 0 + 2\cdot2 + 3\cdot 1) & (1\cdot 1 + 2\cdot 0 + 3\cdot 0) & (1\cdot 1 + 2\cdot 0 + 3 \cdot 2)\\ (1\cdot 0 + 2\cdot 2 + 0\cdot 1) & (1\cdot 1 + 2\cdot 0 + 0\cdot 0) & (1\cdot 0 + 2\cdot 1 + 0\cdot 2)\end{bmatrix} \\ \\ &= \begin{bmatrix} 7 & 1 & 7 \\ 4 & 1 & 2 \end{bmatrix}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/427626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomials Question: Proving $a=b=c$. Question:
Let $P_1(x)=ax^2-bx-c, P_2(x)=bx^2-cx-a \text{ and } P_3=cx^2-ax-b$
, where $a,b,c$ are non zero reals. There exists a real $\alpha$ such
that $P_1(\alpha)=P_2(\alpha)=P_3(\alpha)$. Prove that $a=b=c$.
The questions seems pretty easy for people who know some kind of calculus. Since, the question is from a contest, no calculus can be used. I have got a solution which is bit long(No calculus involved), I'm looking for a simplest way to solve this, which uses more direct things like $a-b|P(a)-P(b)$ .
| You can do this pretty systematically. The given information is equivalent to the statements that $P_1(\alpha) - P_2(\alpha) = 0$ and $P_2(\alpha) - P_3(\alpha) = 0$; in other words that
$$(a-b)\alpha^2 -(b - c)\alpha -(c - a) = 0$$
$$(b-c)\alpha^2 -(c - a)\alpha -(a - b) = 0$$
Next, it is natural to eliminate $\alpha^2$ and then solve for $\alpha$. So, we take $b-c$ times the first equation minus $a - b$ times the second equation to obtain
$$((a-b)(c-a) -(b-c)^2)\alpha + (a-b)^2 - (c-a)(b-c) = 0$$
We rewrite this as
$$((a-b)(c-a) -(b-c)^2)\alpha = (c-a)(b-c) - (a-b)^2$$
By symmetry we have also that
$$((b-c)(a-b) -(c-a)^2)\alpha = (a-b)(c-a) - (b-c)^2$$
$$((c-a)(b-c) - (a-b)^2)\alpha = (b-c)(a-b) -(c-a)^2$$
Adding the three equations gives
$$((a-b)(b-c) + (b-c)(a-b) + (c-a)(b-c) - ((a-b)^2 + (b-c)^2 + (c-a)^2))\alpha = ((a-b)(b-c) + (b-c)(a-b) + (c-a)(b-c) - ((a-b)^2 + (b-c)^2 + (c-a)^2))$$
For this to hold, either $\alpha = 1$ or
$$(a-b)(b-c) + (b-c)(a-b) + (c-a)(b-c) - ((a-b)^2 + (b-c)^2 + (c-a)^2) = 0$$
Plugging in $\alpha = 1$ gives $a = b + c,$ $b = c+a,$ $c = a+ b$, leading immediately to $a = b = c$. One can use the AM-GM inequality on the cross terms to show that
$$(a-b)(b-c) + (b-c)(a-b) + (c-a)(b-c) - ((a-b)^2 + (b-c)^2 + (c-a)^2) \leq 0$$
Equality holds iff $a-b = b-c = c-a$, which once again leads to $a = b = c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
estimate the population numbers $100$ cookies labeled as $1$-$100$ numbers, and a cookie jar. The interviewer randomly selects a number between $1$-$100$ (say he selected $N$), and put cookies number $1$-$N$ to the jar. You randomly pick a cookie from the jar, it is labeled $5$, estimate how many cookies are in the jar.
| Let $X \equiv $ cookie you picked. Note that:
$$
Pr(X=5 \mid N=n)=
\begin{cases}
0 & \text{if } n \in \{1,2,3,4\} \\
1/n & \text{if } n \in \{5,6,...,100\}
\end{cases}
$$
Hence, the expected number of cookies given that you picked one that was labelled $5$ is:
$$ \begin{align*}
E[N \mid X=5] &= \sum_{n=1}^{100} n \cdot Pr(N=n \mid X=5) \\
&= \sum_{n=1}^{100} n \cdot \dfrac{Pr(N=n)Pr(X=5 \mid N=n)}{Pr(X=5)} \\
&= \sum_{n=1}^{100} n \cdot \dfrac{Pr(N=n)Pr(X=5 \mid N=n)}{\sum_{k=1}^{100}Pr(N=k)Pr(X=5 \mid N=k)} \\
&= \sum_{n=5}^{100} n \cdot \dfrac{\dfrac{1}{100} \cdot \dfrac{1}{n}}{\sum_{k=5}^{100}\dfrac{1}{100} \cdot \dfrac{1}{k}} \\
&= \sum_{n=5}^{100} \dfrac{\dfrac{1}{100}}{\dfrac{1}{100}\sum_{k=5}^{100} \dfrac{1}{k}} \\
&= \sum_{n=5}^{100} \dfrac{1}{\sum_{k=5}^{100} \dfrac{1}{k}} \\
&= \dfrac{96}{\sum_{k=5}^{100} \dfrac{1}{k}} \\
&= 30.9274...
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate the derivative$\frac{d}{dx}\left(\ln\sqrt{\frac{4+x²}{4-x²}}\right)$? How can I evaluate this derivative?:
$$\frac{d}{dx}\left(\ln\sqrt{\frac{4+x^2}{4-x^2}}\right)$$
Thank you.
| We'll exploit the properties of logarithms, recalling that $$\ln\left(\frac{a}{b}\right)^b = b \ln\left(\frac ab\right) = b (\ln a - \ln b)$$
$$\begin{align} {\bf f(x)} & = \ln\sqrt{\frac{4+x^2}{4-x^2}} \\ \\
&= \ln\left(\frac{4+x^2}{4-x^2}\right)^{1/2}\\ \\
& = \frac 12\ln\left(\frac{4 + x^2}{4 - x^2}\right)\tag{$\ln\left(\frac ab\right)^c = c\ln\left(\frac ab\right)$}\\ \\
& = \frac12 \left(\ln(4+x^2) - \ln(4-x^2)\right)\tag{$\ln\left(\frac ab\right) = \ln a - \ln b$}\\ \\
{\bf f'(x)} & = \frac 12\left( \frac{2x}{4 + x^2} + \frac{2x}{4 - x^2}\right)\tag{chain rule} \\ \\ & = \frac 12\left(\frac{2x[4-x^2 + 4 + x^2]}{16-x^4}\right) \\ \\ & =
\frac{8x}{16 - x^4}\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating 7^7^7^7^7^7^7 mod 100 What is
$$\large 7^{7^{7^{7^{7^{7^7}}}}} \pmod{100}$$
I'm not much of a number theorist and I saw this mentioned on the internet somewhere. Should be doable by hand.
| A quick hand calculation gives
$$\begin{align}
7^1 &\equiv 7 \pmod{100} \\
7^2 &\equiv 49 \pmod{100} \\
7^3 &\equiv 43 \pmod{100} \\
7^4 &\equiv 1 \pmod{100}
\end{align}$$
So it reduces to the problem of calculating the value of $7^{7^{7^{7^{7^7}}}} \pmod 4$. And $7^2 \equiv 1 \pmod 4$, so it reduces to the problem of calculating $7^{7^{7^{7^7}}} \pmod 2$... and this is easy, it's odd, so it's congruent to $1$ modulo $2$.
Working backwards:
$$7^{7^{7^{7^{7^7}}}} \equiv 7^1 \equiv 3 \pmod{4}\quad \Rightarrow\quad 7^{7^{7^{7^{7^{7^7}}}}} \equiv 7^3 \equiv 43 \pmod{100}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/430633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Induction and convergence of an inequality: $\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\leq \frac{1}{\sqrt{2n+1}}$ Problem statement:
Prove that $\frac{1*3*5*...*(2n-1)}{2*4*6*...(2n)}\leq \frac{1}{\sqrt{2n+1}}$ and that there exists a limit when $n \to \infty $.
, $n\in \mathbb{N}$
My progress
LHS is equivalent to $\frac{(2n-1)!}{(2n)!}=\frac{(2n-1)(2n-2)(2n-3)\cdot ....}{(2n)(2n-1)(2n-2)\cdot ....}=\frac{1}{2n}$ So we can rewrite our inequality as:
$\frac{1}{2n}\leq \frac{1}{\sqrt{2n+1}}$ Let's use induction:
For $n=1$ it is obviously true. Assume $n=k$ is correct and show that $n=k+1$ holds.
$\frac{1}{2k+2}\leq \frac{1}{\sqrt{2k+3}}\Leftrightarrow 2k+2\geq\sqrt{2k+3}\Leftrightarrow 4(k+\frac{3}{4})^2-\frac{5}{4}$ after squaring and completing the square. And this does not hold for all $n$
About convergence: Is it not enough to check that $\lim_{n \to \infty}\frac{1}{2n}=\infty$ and conclude that it does not converge?
| First, note that $2\cdot 4 \cdot 6\cdots (2n)=2^n(n!)$. Next, note that if we multiplied $1\cdot 3\cdot 5\cdots (2n-1)$ by $2\cdot 4\cdot 6\cdots (2n)$, that would exactly fill the gaps and produce $(2n)!$. Hence, the denominator of the LHS is $2^nn!$, while the numerator of the LHS is $\frac{(2n)!}{2^nn!}$ Combining, the LHS equals $$\frac{1}{2^{2n}}\frac{(2n)!}{n!n!}=2^{-2n}{2n\choose n}$$
This is a central binomial coefficient, which are well-studied. For example, one bound given is that ${2n \choose n}\le \frac{4^n}{\sqrt{3n+1}}$; applying it in this case gives $$LHS=4^{-n}{2n\choose n}\le \frac{1}{\sqrt{3n+1}}\le \frac{1}{\sqrt{2n+1}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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divergence of $\int_{2}^{\infty}\frac{dx}{x^{2}-x-2}$ i ran into this question and im sitting on it for a long time.
why does this integral diverge:
$$\int_{2}^{\infty}\frac{dx}{x^{2}-x-2}$$
thank you very much in advance.
yaron.
| Notice the following:
$(x^2 - x - 2) = (x - 2)(x + 1)$
Therefore notice that:
$\frac{2}{x^2 - x - 2} = \frac{2}{(x-2)(x+1)} $
From here we can do a partial fraction decomposition which basically means we want to find:
$A, B$ such that $\frac{A}{x-2} + \frac{B}{x+1} = \frac{2}{x^2 - x - 2}$
This alternatively means that:
$A(x+1) + B(x-2) = 2$
And therefore:
$Ax + Bx = 0 ==> A + B = 0$
$A - 2B = 2$
We solve this system of 2 linear equations in 2 unknowns to find that:
$A = \frac{2}{3}, B = -\frac{2}{3}$
So finally we have:
$\frac{2}{x^2 - x - 2} = \frac{\frac{2}{3}}{x-2} - \frac{\frac{2}{3}}{x+1} $
From here it is obvious that by integrating this expression you are integrating over the asymptote of x - 2 and therefore the sum diverges
Hope that made sense!
| {
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"timestamp": "2023-03-29T00:00:00",
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Trig substitution $\int x^3 \sqrt{1-x^2} dx$ $$\int x^3 \sqrt{1-x^2} dx$$
$x = \sin \theta $
$dx = \cos \theta d \theta$
$$\int \sin^3 \theta d \theta$$
$$\int (1 - \cos^2 \theta) \sin \theta d \theta$$
$u = \cos \theta$
$du = -\sin\theta d \theta$
$$-\int u^2 du$$
$$\frac{-u^3}{3} $$
$$\frac{\cos^3 \theta}{3}$$
With the triangle trick I get:
$$\frac{-\sqrt{1-x^2}^3}{3}$$
This is wrong but I am not sure where I went wrong.
| Let $x=\sin{\theta}$, then $dx = \cos{\theta} \, d\theta$; the integral becomes
$$\int d\theta \, \sin^3{\theta} \, \cos^2{\theta} = \int d\theta \, \sin^3{\theta} -\int d\theta \, \sin^5{\theta} $$
$$\int d\theta \, \sin^3{\theta} = \int d\theta \, \sin{\theta} (1-\cos^2{\theta}) = -\int d(\cos{\theta}) (1-\cos^2{\theta})= -\cos{\theta} + \frac13 \cos^3{\theta}+C$$
Similarly
$$\int d\theta \, \sin^5{\theta} = -\int d(\cos{\theta}) (1-\cos^2{\theta})^2 = -\cos{\theta} + \frac{2}{3} \cos^3{\theta}-\frac15 \cos^5{\theta}+C'$$
Subtracting the two, I get
$$\int d\theta \, \sin^3{\theta} \, \cos^2{\theta} = -\frac13 \cos^3{\theta}+\frac15 \cos^5{\theta}+C$$
Then use $x=\sin{\theta}$ and get
$$\int dx \, x^3 \, \sqrt{1-x^2} = \frac{1}{15} (3 x^4-x^2-2) \sqrt{1-x^2}+C$$
EDIT
I see that the answer can be simplified further to
$$-\frac{1}{15} (1-x^2)^{3/2} (3 x^2+2) + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The golden ratio and a right triangle Assume the square of the hypotenuse of a right triangle is equal to its perimeter and one of its legs is $1$ plus its inradius(the radius inside the circle inscribed inside the triangle.) Find an expression for the hypotenuse $c$ in terms of the golden ratio.
| For any triangle,
$$
2\times\text{area} = \text{inradius}\times\text{perimeter}
$$
For a right triangle,
$$
2\times\text{area} = ab
$$
We are given
$$
\text{inradius}=a-1\quad\text{and}\quad\text{perimeter}=c^2
$$
Therefore,
$$
\begin{align}
\overbrace{(a-1)}^{\text{inradius}}\overbrace{(a+b+c)}^{\text{perimeter}}&=ab\tag{1}\\
(a-1)(a+c)&=b\tag{2}
\end{align}
$$
and
$$
\overbrace{a+b+c}^{\text{perimeter}}=c^2\tag{3}
$$
Combining $(2)$ and $(3)$ to eliminate $b$ yields
$$
\begin{align}
(a-1)(a+c)&=c^2-c-a\tag{4}\\
a(a+c)&=c^2\tag{5}\\
\left(\frac ac\right)^2+\frac ac&=1\tag{6}\\
\frac ac&=\frac1\phi\tag{7}
\end{align}
$$
Equation $(5)$ and $a^2+b^2=c^2$ gives
$$
\begin{align}
b^2&=ac\tag{8}\\
\left(\frac bc\right)^2&=\frac ac\tag{9}\\
\frac bc&=\frac1{\sqrt\phi}\tag{10}
\end{align}
$$
Dividing equation $(3)$ by $c$ yields
$$
\begin{align}
c
&=\frac ac+\frac bc+1\tag{11}\\
&=\frac1\phi+\frac1{\sqrt\phi}+1\tag{12}
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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solution of difference equation I am trying to solve the following difference equation:
$$-\frac{\epsilon}{h^2}U_{n+1}+\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)U_{n}-\left(\frac{\epsilon}{h^2}+\frac{1}{h}\right)U_{n-1}=0,\mbox{ }\mbox{ }\mbox{ }\mbox{ }U_0=1,\mbox{ }U_1=0.$$
I try $U_{n}=Aw^n$ then I get
$$w_{1,2}=\frac{\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)\pm\sqrt{\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)^2-4\frac{\epsilon}{h^2}\left(\frac{\epsilon}{h^2}+\frac{1}{h}\right)}}{2\frac{\epsilon}{h^2}}.$$
This seems a bit far from what I want to get. I am trying to verify that the solution is
$$U_n=\dfrac{1-(1+\rho)^{n-N}}{1-(1+\rho)^{-N}},$$
where $0\leq n\leq N$ and $\rho=h/\epsilon$.
| First of all note that, inside the square root, we have
$$
\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)^2-4\frac{\epsilon}{h^2}\left(\frac{\epsilon}{h^2}+\frac{1}{h}\right) = \frac{1}{h^2}
$$
So the answers read
$$w_{1,2}=\frac{\left(\frac{2\epsilon}{h^2}+\frac{1}{h}\right)\pm \frac{1}{h}}{2\frac{\epsilon}{h^2}}.
\Longrightarrow w_1=1, w_2=1+\frac{h}{\epsilon}
$$
This means that the solution reads $U_n=A+B(1+\rho)^n$. Now let us impose boundary conditions . Since $U_0=1$, we get that $A+B=1$. Also, since $U_1=0$, we get that $A+B(1+\rho)=0$ or equivalently, $A=-(1+\rho)B$. Plugging this into the first equation, we get that $B=\frac{-1}{\rho}$, and that $A=\frac{1+\rho}{\rho}$, leading to a solution of the form
$$U_n=\frac{1+\rho}{\rho} \bigg[ 1-(1+\rho)^{n-1} \bigg].
$$
We can readily check that for $n=1$, we get $U_1=0$, and $n=0$, we get $U_0=1$.
bBut this is not what you are trying to recover. Your desired answer has $N$ in it, which is defined nowhere in the statement of the problem. My conjecture is that, $n$ is limited to the range $0,N$.
Let us now consider the following set of boundary conditions:
$$U_0=1,U_N=0$$
Imposing the first condition on $U_n=A+B(1+\rho)^n$, we get $A+B=1$. Imposing the second condition, we get $A+B(1+\rho)^N=0$, which can be equivalently expressed as $A=-B(1+\rho)^N$. Plugging this into the first equation, we obtain $B=\frac{1}{1-(1+\rho)^N}$
. Also for $A$ we obtain $A=\frac{ -(1+\rho)^N}{1-(1+\rho)^N}$.
For the solution, we use the values of $A,B$ obtained above, and arrive at
$$
U_n=\frac{(1+\rho)^n-(1+\rho)^N}{1-(1+\rho)^N}$$
Dividing both the numerator and the denominator by $-(1+\rho)^N$, we arrive at
$$
U_n=\frac{1-(1+\rho)^{n-N}}{1-(1+\rho)^{-N}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a+\sqrt{b}=c+\sqrt{d}$, is it true that $a=c$ and $b=d$?
Assume that $a,b,c,d$ are all positive integers. If $a+\sqrt{b}=c+\sqrt{d}$, is it true that $a=c$ and $b=d$?
I am grading some problems and I don't think this true, but all of a sudden I am doubting myself.
| If $a+\sqrt{b}=c+\sqrt{d}$, then $(a-c)+\sqrt{b}=\sqrt{d}$, so it clearly suffices to study when is it possible to have
$$a+\sqrt{b}=\sqrt{d}.$$
If $b,d$ are arbitrary positive real numbers, then given any $a$ and $b$ such that $a+\sqrt{b}\geq 0$, then there is a $d\geq 0$, namely $d=(a+\sqrt{b})^2$, such that $\sqrt{d}=a+\sqrt{b}$. So I will assume from now on that the OP meant $a\in\mathbb{Q}$ and $b,d\in \mathbb{Q}^{\geq 0}$.
Suppose $a+\sqrt{b}=\sqrt{d}$ holds. Then, by squaring both sides we obtain
$$a^2+2a\sqrt{b} + b = d,$$
If $a=0$, then $b=d$. Otherwise, if $a\neq 0$, then
$$\sqrt{b}=\frac{d-a^2-b}{2a}\in \mathbb{Q}.$$
In particular, $b$ must be a perfect square, and $a+\sqrt{b}\geq 0$.
Conversely, if $b$ is a perfect square, say $b=B^2$ for some $B\geq 0$, so that $B=\sqrt{b}\in \mathbb{Q}$, and $a+\sqrt{b}=a+B\geq 0$, then one can take $d=(a+B)^2$, so that
$$a+\sqrt{b}=a+B=\sqrt{d}.$$
Hence, $a+\sqrt{b}=\sqrt{d}$ has a solution if and only if $b$ is a perfect square, $b=B^2$ for some $B\geq 0$, and $a+B\geq 0$, or $a=0$ and $b=d$.
Thus, $a+\sqrt{b}=c+\sqrt{d}$ has a solution if and only if $b$ is a perfect square, $b=B^2$ for some $B\geq 0$, and $(a-c)+B\geq 0$ (so that $d=((a-c)+B)^2$ is also a perfect square), or $a=c$ and $b=d$.
Example: In the example by David Mitra in the comments,
$$1+\sqrt{4}=2+\sqrt{1}$$
we have $b=4=2^2$, so $B=2$, and $(1-2)+2=1\geq 0$.
| {
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Partial fraction integration $\int \frac{dx}{(x-1)^2 (x-2)^2}$ $$\int \frac{dx}{(x-1)^2 (x-2)^2} = \int \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}\,dx$$
I use the cover up method to find that B = 1 and so is C. From here I know that the cover up method won't really work and I have to plug in values for x but that won't really work either because I have two unknowns. How do I use the coverup method?
| $$
\frac1{(x-1)^2(x-2)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}\tag{1}
$$
Multiply both sides by $(x-1)^2$ and evaluate at $x=1$: $B=1$
Multiply both sides by $(x-2)^2$ and evaluate at $x=2$: $D=1$
$$
\frac{1-(x-1)^2-(x-2)^2}{(x-1)^2(x-2)^2}=\frac{A}{x-1}+\frac{C}{x-2}\tag{2}
$$
Multiply both sides by $x-1$ and use L'Hospital at $x=1$: $A=2$
Multiply both sides by $x-2$ and use L'Hospital at $x=2$: $C=-2$
$$
\frac1{(x-1)^2(x-2)^2}=\frac{2}{x-1}+\frac{1}{(x-1)^2}+\frac{-2}{x-2}+\frac{1}{(x-2)^2}\tag{3}
$$
Details of $(2)$
It is pretty simple to subtract off the parts we got from $(1)$:
$$
\frac1{(x-1)^2(x-2)^2}-\frac1{(x-1)^2}-\frac1{(x-2)^2}
=\frac{1-(x-1)^2-(x-2)^2}{(x-1)^2(x-2)^2}\tag{4}
$$
$(4)$ gives us $(2)$.
In the next part, we can use a small bit of simplification. Suppose we have $(x-a)^nP(x)$ in the denominator, and we want to look at $x\to a$. We are going to need to apply L'Hospital $n$ times. This will give $n!P(a)$ in the denominator.
Multiply $(2)$ by $(x-1)$ and let $x\to1$. The denominator is $(x-1)(x-2)^2$, so we apply L'Hospital once.
Take one derivative of the denominator and evaluate at $\color{#C00000}{x=1}$: $1!(\color{#C00000}{1}-2)^2=1$.
Take one derivative of the numerator and evaluate at $\color{#C00000}{x=1}$: $-2(\color{#C00000}{1}-1)-2(\color{#C00000}{1}-2)=2$
Thus, $A=\frac21=2$
Multiply $(2)$ by $(x-2)$ and let $x\to2$. The denominator is $(x-1)^2(x-2)$, so we apply L'Hospital once.
Take one derivative of the denominator and evaluate at $\color{#C00000}{x=2}$: $1!(\color{#C00000}{2}-1)^2=1$.
Take one derivative of the numerator and evaluate at $\color{#C00000}{x=2}$: $-2(\color{#C00000}{2}-1)-2(\color{#C00000}{2}-2)=-2$
Thus, $C=\frac{-2}1=-2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Differentiate the following w.r.t. $\tan^{-1} \left(\frac{2x}{1-x^2}\right)$ Differentiate : $$ \tan^{-1} \left(\frac {\sqrt {1+x^2}-1}x\right) \quad w.r.t.\quad \tan^{-1} \left(\frac{2x}{1-x^2}\right) $$
| Let $y=\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-1}x\right)$
and let $u= \tan^{-1} \left(\dfrac{2x}{1-x^2}\right)$.
We want to find $dy/du$. Note that:
$$
\dfrac{dy}{dx} = \dfrac{1}{2(1+x^2)}
$$
similarly for $u$, we obtain:
$$
\dfrac{du}{dx} = \dfrac{2}{1+x^2} \iff \dfrac{dx}{du} = \dfrac{1+x^2}{2}
$$
Hence, by Chain Rule, we obtain:
$$
\dfrac{dy}{du} = \dfrac{dy}{dx} \cdot \dfrac{dx}{du} = \dfrac{1}{2(1+x^2)} \cdot \dfrac{1+x^2}{2} = \dfrac{1}{4}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve a ratio question Studying for the GRE. In the GRE guide, it says that
If the ratio is $2x:5y$, and this equals the ratio $3:4$, what is the ratio of $x:y$?
I tried cross multiplying but I don't get the answer. It says the answer is $15:8$. I get $8:15$. Which step am I missing?
| We are given: $$\dfrac {2x}{5y} = \frac 34$$
$$2x\cdot (4) = 5y \cdot (3)\tag{1}$$ $$ \iff 8x = 15 y\tag{2: cross-multiplied}$$ $$\iff \frac {8x}{y} = 15\tag{divide by y}$$ $$ \iff \frac xy = \frac{15}{8}\tag{divide by 8}$$
It seems as though you went from $(2)$ to $\dfrac {8x}{15y} = 1$, concluding the ratio is $8:15$. But we want $x: y$ which is the value of $\dfrac xy$, so $$\frac {8x}{15y} = 1 \iff \dfrac{8x}{15y}\cdot \dfrac{15}{8} = 1\cdot \dfrac{15}{8} \iff \dfrac xy = \dfrac{15}{8}$$
Rather than cross-multiplying, it makes more sense in this problem to start from the given $$\frac {2x}{5y} = \frac 34 \iff \frac {2x}{5y}\cdot \frac 52 = \frac 34 \cdot \frac 52 \iff \frac xy = \frac{15}{8}$$
| {
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solving equations by the method of substitution $\dfrac{a}{x}+\dfrac{b}{y}=\dfrac{a}{2}+\dfrac{b}{3},$
$x+1=y$
We have to solve for $x$ and $y$.I have tried to solve for them by finding value of $x$ or $y$ from the second equation and place them in the second.It is obvious that the answers would be $2$ and $3,$but we need something else. I tried to find the relation between $a$ and $b$ and them place them again in the first equation along with the value of $x$ or $y$ , but it yields something bizarre. So how do we solve it? A tiny hint will be appreciated.
| $\dfrac{a}{x}+\dfrac{b}{y}=\dfrac{a}{2}+\dfrac{b}{3}$...........$(1)$
$x+1=y$............$(2)$
From (2),$x=y-1$. Substituting it into $(1)$,
$\dfrac{a}{y-1}+\dfrac{b}{y}=\dfrac{a}{2}+\dfrac{b}{3}$
$\implies \dfrac{a}{y-1}-\dfrac{a}{2}=\dfrac{b}{3}-\dfrac{b}{y}$
$\implies \dfrac{2a-ay+a}{2(y-1)}=\dfrac{b(y-3)}{3y}$
$\implies \dfrac{3a-ay}{2(y-1)}=\dfrac{b(y-3)}{3y}$
$\implies \dfrac{a(3-y)}{2(y-1)}=\dfrac{b(y-3)}{3y}$
$\implies \dfrac{a(3-y)}{2(y-1)}=-\dfrac{b(3-y)}{3y}$ [Multiplying the fraction on the right side by $\dfrac{-1}{-1}$]
$\implies \dfrac{a(3-y)}{2(y-1)}+=0$
$\implies (3-y)[\dfrac{a}{2(y-1)}+\dfrac{b}{3y}]=0$
Since the product of two numbers is 0,at least one of them must be 0.........
No quadratic formula needed!
| {
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What is wrong with this Jordan normal form computation? The question I am working on is to compute the Jordan normal form of $$A := \begin{pmatrix} 2 & 1 & 5 \\ 0 & 1 & 3\\ 1 & 0 & 1\end{pmatrix}.$$ The characteristic polynomial and minimal polynomial of $A$ is $x^{2}(x - 4)$. Then the Jordan normal form of $A$ is given by $$J := \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 4 \end{pmatrix}.$$ Then there exist a matrix $P$ such that $P^{-1}AP = J$.
I am having an issue finding $P$. From the theory of Jordan normal forms, $P = [w_{1}\, w_{2}\, w_{3}]$ where $w_{1}, w_{2}$ is the basis of the nullspace of $(A - 0\cdot I)^{2}$ (where $I$ is the identity matrix) and $w_{3}$ is the basis of the nullspace of $(A - 4\cdot I)$. We first consider $(A - 0 \cdot I)^{2}$. As $$A^{2} = \begin{pmatrix} 9 & 3 & 18\\ 3 & 1 & 6\\ 3& 1 & 6\end{pmatrix} \sim \begin{pmatrix} 1 & 1/3 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ (where the $\sim$ denotes row equivalence), the nullspace of $A^{2}$ is spanned the column vectors $\{(-2, 0, 1)^{t}, (-1, 3, 0)^{t}\}$. Next we consider $(A - 4 \cdot I)$. We have $$A - 4I = \begin{pmatrix} -2 & 1 & 5\\ 0 & -3 & 3\\ 1 & 0 & -3 \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & -3\\0 & 1 & -1\\ 0 & 0 & 0\end{pmatrix}.$$ Then the nullspace is spanned by the column vector $(3, 1, 1)^{t}$.
Therefore we should have $$P = \begin{pmatrix} -2 & -1 & 3\\0 & 3 & 1\\ 1 & 0 & 1 \end{pmatrix}.$$ However, when I compute $P^{-1}AP$, I get $$P^{-1}AP = \begin{pmatrix} -1 & -1 & 0\\ 1 & 1 & 0\\0 & 0 & 4\end{pmatrix} \neq J.$$ Where did I go wrong? Is there something wrong on how I computed $P$?
| To calculate the eigenvectors for $\lambda=0$ you want to find a chain. You need to find $\vec{u}_1$ and $\vec{u}_2$ such that $A\vec{u}_2=\vec{u}_1$ and $A\vec{u}_1=0$. Clearly there exists such a solution as zero is an e-value of $A$ hence $A$ is singular. For example,
$\vec{u}_1 = ( -2, -6, 2)$ and $\vec{u}_2 = (-1,-6,0)$. I believe these will put the matrix in the form you desire.
| {
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Prove that $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$? $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin 10^\circ \sin 10^\circ \sin 100^\circ$
This is a competition problem which I got from the book "Art of Problem Solving Volume 2". I'm not sure how to solve it because there's just so many possibilities of using most of the trig identities that I don't know which path to take. One way I tried is canceling out the $\sin 10$ on both sides:
$\sin 20^\circ \sin 30^\circ= \sin 10^\circ \sin 100^\circ$
$2 \sin 10^\circ \cos 10^\circ \sin 30^\circ=\sin 10^\circ \sin 100^\circ$
$2 \sin 80^\circ \sin 30^\circ= \sin 100^\circ$
$2 \sin 80^\circ \sin 30^\circ= \sin 30^\circ \cos 70^\circ+\sin 70^\circ \cos 30^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ+\sin 70^\circ \sin 60^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ + 2\sin 70^\circ \sin 30^\circ \cos 30^\circ$
$2 \sin 80^\circ \sin 30^\circ=\sin 20^\circ \sin 30^\circ + 2\sin 70^\circ \sin 30^\circ \sin 60^\circ$
$1=\large \frac {\sin 20^\circ}{2 \sin 80^\circ}+ \frac {\sin 70^\circ \sin 60^\circ}{\sin 80^\circ}$
Now I set $\sin^2 \theta=\large \frac {\sin 20^\circ}{2 \sin 80^\circ}$ and $\cos^2 \theta =\large \frac {\sin 70^\circ \sin 60^\circ}{\sin 80^\circ}$
I solves for $\sin \theta$ and $\cos \theta$ and set up a right triangle, which then led me to the equation
$\sin 70^\circ \sin 60^\circ=\sin 80^\circ-\frac 12 \sin 20^\circ$
Another approach I tries is from the beginning having $\sin 10^\circ \sin 20^\circ \sin 30^\circ=\sin^2 10^\circ \sin 100^\circ$ and the using the power-reducing formula, but that also got me nowhere. Any help is appreciated. Thanks.
| Write $\displaystyle\sin10\sin100=\sin10\sin(90+10)=\sin10\cos10=\frac{1}{2}\sin20=\sin20\sin30$.
| {
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$x \sin x=2$ why is my proof that there no solutions wrong? $\frac 12 x \sin x=1$ . Let's look at a right triangle with base $x$ and altitude $\sin x$ . Then our equation is for the area of this triangle. Let the sides of the triangle be $a=x$ , $b=\sqrt {x^2+sin^2 x}$ , and $c= \sin x$ . According to wikipedia, Heron's formula can be written as $$A=\large \frac { \sqrt {4a^2c^2-(a^2+b^2-c^2)^2}}{4}$$
Plugging in:
$4=\large \sqrt{4x^2 \sin^2 x-(x^2+x^2+\sin^2 x-\sin ^2 x)^2}$
$4=x^2 \sin^2 x -x^4$
$x^2(x^2- \sin^2 x)=-4$
$x^2$ will always be positive, and $\sin^2 x$ is never greater than $x^2$ , so this equation can have no real solutions. The original has solutions, so why is this wrong?
| There are solutions on each interval $\left[2k\pi,2k\pi+\frac\pi2\right]$ for positive integer $k$ by the intermediate value theorem because $x\sin(x)$ is $0$ on the left end and $2k\pi+\frac\pi2$ on the right.
Heron's Formula should be
$$
A=\frac{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}{4}
$$
Does that cause the same problem?
| {
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Indefinite integral $\int{\frac{dx}{x^2+2}}$ I cannot manage to solve this integral:
$$\int{\frac{dx}{x^2+2}}$$
The problem is the $2$ at denominator, I am trying to decompose it in something like $\int{\frac{dt}{t^2+1}}$:
$$t^2+1 = x^2 +2$$
$$\int{\frac{dt}{2 \cdot \sqrt{t^2-1} \cdot (t^2+1)}}$$
But it's even harder than the original one. I also cannot try partial fraction decomposition because the polynomial has no roots. Ho to go on?
| I find it much more versatile when encountering a denominator of the form $x^2 + a^2$, rather than only having learned what to do when $a = 1$, I use the fact that : $$\int \dfrac{dx}{x^2 + a^2} = \dfrac 1a\arctan\left(\frac x{a}\right) + C$$
Why? $$\frac{dx}{x^2+a^2} = \frac{dx}{a^2 \left(\frac{x^2}{a^2} + 1\right)} =\frac{dx}{a^2\left(\left(\frac{x}{a}\right)^2+1\right)} = \dfrac 1a\cdot\frac{(1/a) \,dx}{\left(\left(\frac{x}{a}\right)^2+1\right)} = \frac{1}{a}\cdot\frac{du}{u^2+1}, \;\;u = \frac xa$$
Applying this fact to your integral is rather straightforward then:
$$\int{\frac{dx}{x^2+2}} = \int\frac{dx}{x^2 + \left(\sqrt 2\right)^2} = \frac 1{\sqrt 2} \arctan\left(\frac x{\sqrt 2}\right) + C$$
| {
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Volume of a wine barrel This is a famous calculus problem and is stated like this
Given a barrel with height $h$, and a small radius of $a$ and
large radius of $b$. Calculate the volume of the barrel
given that the sides are parabolic.
Now I seem to have solved the problem incorrectly because here it seems 2
that the volume should be
$ \displaystyle \hspace{1cm}
V(a,b,h) = \frac{h\pi}{3}\left(2b^2 + a^2\right)\,.
$
Below is my attempt. As in the picture I view the barrel from the side, and try to find a formula for the parabola. So i solve
$ \displaystyle \hspace{1cm}
f(x) := A x^2 + B x + C
$
given $f(0) = f(h) = a/2$ and $f(h/2) = b/2$. This yields
$ \displaystyle \hspace{1cm}
f(x) = \frac{2(a-b)}{h^2} \cdot x^2 -
\frac{2(a-b)}{h} \cdot x +
\frac{a}{2}
$
Using the shell method integrating now gives the volume as
$ \displaystyle \hspace{1cm}
V(a,b,h) := \pi \int_0^h \bigl[f(x)\bigr]^2\,\mathrm{d}x
= \frac{\pi}{60} \cdot h (a+2b)^2 + \frac{\pi}{30} \cdot h(a^2+b^2)
$
Alas according to the formula above this seems incorrect! Where is my mistake?
| Note that in their case $a$, $b$ are radiuses, not diameters. Another important thing to say is that the formula from that external site is not exact!
For simplicity, take $h=1$. Let's define our function on $[-1/2,1/2]$ as $f(x) = 4(a-b)x^2+b$. We write
$$\frac{V}{\pi}= 2\int _{0}^{1/2}(4(a-b)x^2+b)^2 \mathrm dx= 16 (a-b)^2 \cdot \frac{1}{5\cdot 16}+ 8 b(a-b)\cdot \frac{1}{3\cdot 4} + b^2$$
$$= (a-b)^2/5 + 2 b(a-b)/3 + b^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Double integral $\iint_D |x^3 y^3|\, \mathrm{d}x \mathrm{d}y$ Solve the following double integral
\begin{equation}
\iint_D |x^3 y^3|\, \mathrm{d}x \mathrm{d}y
\end{equation}
where $D: \{(x,y)\mid x^2+y^2\leq y \}$.
Some help please? Thank you very much.
| Since
$$
D=\{(x,y)|\ x^2+y^2\le y\}=\left\{(x,y)|\ x^2+(y-\frac12)^2\le \frac14\right\},
$$
we have $y\in [0,1]$ for all $(x,y) \in D$.
Setting
$$
x=r\cos\theta,\ y=r\sin\theta+\frac12,\quad \theta \in [-\pi/2,3\pi/2],\ r \in [0,\frac12],
$$
we get
\begin{eqnarray}
\int_D|x^3y^3|\,dxdy&=&\int_Dy^3|x|^3\,dxdy=\int_{-\pi/2}^{3\pi/2}\int_0^{\frac12}r^4|\cos^3\theta|(r\sin\theta+\frac12)^3\,drd\theta\\
&=&\int_0^{\frac12}\int_{-\pi/2}^{\pi/2}r^4\cos^3\theta(r\sin\theta+\frac12)^3\,drd\theta\\
&&-\int_0^{\frac12}\int_{\pi/2}^{3\pi/2}r^4\cos^3\theta(r\sin\theta+\frac12)^3\,drd\theta\\
&=&\int_0^{\frac12}\int_{-\pi/2}^{\pi/2}r^4\cos^3\theta(r\sin\theta+\frac12)^3\,drd\theta\\
&&-\int_0^{\frac12}\int_{-\pi/2}^{\pi/2}r^4\cos^3\theta(r\sin\theta-\frac12)^3\,drd\theta\\
&=&2\int_0^{\frac12}\int_0^{\pi/2}r^4\cos^3\theta\left[(r\sin\theta+\frac12)^3-(r\sin\theta-\frac12)^3\right]\,drd\theta\\
&=&2\int_0^{\frac12}\int_0^{\pi/2}r^4\cos^3\theta(3r^2\sin^2\theta+\frac14)\,drd\theta\\
&=&\int_0^1(\frac{3t^2}{7\cdot2^6}+\frac{1}{320})(1-t^2)\,dt\\
&=&\int_0^1\left[\frac{3}{7\cdot2^6}(t^2-t^4)+\frac{1}{320}(1-t^2)\right]\,dt\\
&=&\frac{3}{7\cdot2^6}(\frac13-\frac15)+\frac{1}{320}(1-\frac13)=\frac{1}{336}.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/445688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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convert ceil to floor Mathematically, why is this true?
$$\left\lceil\frac{a}{b}\right\rceil= \left\lfloor\frac{a+b-1}{b}\right\rfloor$$
Assume $a$ and $b$ are positive integers.
Is this also true if $a$ and $b$ are real numbers?
| If $\frac{a}{b}$ is not an integer and $\frac{a}{b} > n + \frac{1}{b}$, where $n$ is an integer, then it is easy to see that $$\left\lceil\frac{a}{b}\right\rceil= \left\lfloor\frac{a}{b}+1-\frac{1}{b}\right\rfloor.$$
If $\frac{a}{b}$ is an integer then the relationship
$$\left\lceil\frac{a}{b}\right\rceil= \left\lfloor\frac{a}{b} + 1 - \frac{1}{b}\right\rfloor$$ holds.
The only situation which causes problem is when $\frac{a}{b} \in [n, n+\frac{1}{b})$, where $n$ is an integer.This is not going to happen since $a$ is a positive integer. Clearly, the above statement doesn't hold true if $a$ is real.
| {
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"url": "https://math.stackexchange.com/questions/448300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Integration $\int \frac{\sqrt{x^2-4}}{x^4}$ Problem :
Integrate $\int \frac{\sqrt{x^2-4}}{x^4}$
I tried : Let $x^2-4 =t^2 \Rightarrow 2xdx = 2tdt$
$\int \frac{\sqrt{x^2-4}}{x^4} \Rightarrow \frac{t^3 dt}{\sqrt{t^2+4}(t^4-8t+16)}$
But I think this made the integral too complicated... please suggest how to proceed.. Thanks..
| $\displaystyle\int\frac{\sqrt{x^2-4}}{x^4}dx=\int\frac{\sqrt{1-\frac{4}{x^2}}}{x^3}dx$
Put $\frac1x=z\implies \frac{-1}{x^2}dx=dz$
So it boils to, $-\displaystyle\int z\sqrt{1-4z^2}dz$
Again put $1-4z^2=t\implies -8zdz=dt$
Hence we get, $\frac18\displaystyle\int \sqrt{t}dt=\frac{1}{12}t^{3/2}+C=\frac{1}{12}(1-4z^2)^{3/2}+C=\frac{(1-4x^2)^{3/2}}{12x^3}+C$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplifying compound fraction: $\frac{3}{\sqrt{5}/5}$ I'm trying to simplify the following:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ }.$$
I know it is a very simple question but I am stuck. I followed through some instructions on Wolfram which suggests that I multiply the numerator by the reciprocal of the denominator.
The problem is I interpreted that as:
$$\frac{3}{\ \frac{\sqrt{5}}{5} \ } \times \frac{5}{\sqrt{5}},$$
Which I believe is:
$$\frac{15}{\ \frac{5}{5} \ } = \frac{15}{1}.$$
What am I doing wrong?
| You have the following fraction to simplify:
$$\begin{align} \frac{3}{\sqrt{5}/5} &=\frac{5\times 3}{\sqrt{5}} \\ &=\frac{15}{\sqrt{5}} \\ &=\sqrt{\bigg(\frac{15}{\sqrt{5}}\bigg)^2} \\ &=\sqrt{\frac{15^2}{\sqrt{5}^2}} \\ &=\sqrt{\frac{225}{5}} \\ &=\sqrt{45} \\ &= \sqrt{9\times 5} \\ &= \sqrt{9}\sqrt{5} \\ &= 3\sqrt{5} \\ \therefore \frac{15}{\sqrt{5}}\times \frac{5}{\sqrt{5}} &=\frac{15}{\sqrt{5}}\times \frac{\big(\frac{15}{\sqrt{5}}\big)}{3} \\ &=\frac{15}{\sqrt{5}}\times \frac{15}{\sqrt{5}}\times \frac{1}{3} \\ &=\bigg(\frac{15}{\sqrt{5}}\bigg)^2 \times \frac{1}{3} \\ &=45\times \frac{1}{3} \\ &=\frac{45}{3} \\ &=15 \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Differentiating $\tan\left(\frac{1}{ x^2 +1}\right)$ Differentiate: $\displaystyle \tan \left(\frac{1}{x^2 +1}\right)$
Do I use the quotient rule for this question? If so how do I start it of?
| Answering this one too now, I'll write the steps straight away
$$\frac{d}{dx}\left(\tan \frac{1}{1+x^2} \right)=
\frac{d}{dy}\left.\left(\tan y \right)\right|_{y=\frac{1}{1+x^2}}
\cdot\frac{d}{dx}\left( \frac{1}{1+x^2} \right)$$
$$=\sec^2y|_{y=\frac{1}{1+x^2}} \cdot \left( - \frac{2x}{1+x^2} \right)$$
$$=\sec^2 \left( \frac{1}{1+x^2} \right) \cdot \left( - \frac{2x}{1+x^2} \right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Power series for $(1+x^3)^{-4}$ I am trying to find the power series for the sum $(1+x^3)^{-4}$ but I am not sure how to find it. Here is some work:
$$(1+x^3)^{-4} = \frac{1}{(1+x^3)^{4}} = \left(\frac{1}{1+x^3}\right)^4 = \left(\left(\frac{1}{1+x}\right)\left(\frac{1}{x^2-x+1}\right)\right)^4$$
I can now use
$$\frac{1}{(1-ax)^{k+1}} = \left(\begin{array}{c} k \\ 0 \end{array}\right)+\left(\begin{array}{c} k+1 \\ 1 \end{array}\right)ax+\left(\begin{array}{c} k+2 \\ 2 \end{array}\right)a^2x^2+\dots$$
on the $\frac{1}{1+x}$ part but I am not sure how to cope with the rest of formula.
| Just use the generalized binomial series. For natural $m$:
$$
(1 + u)^{-m}
= \sum_{k \ge 0} \binom{-m}{k} u^k
= \sum_{k \ge 0} (-1)^k \binom{k + m - 1}{m - 1} u^k
$$
and plug in $u = x^3$, $m = 4$:
\begin{align}
(1 + x^3)^{-4}
&= \sum_{k \ge 0} (-1)^k \binom{k + 3}{3} x^{3 k} \\
&= \sum_{k \ge 0} (-1)^k \frac{(k + 3) (k + 2) (k + 1)}{3!} x^{3 k} \\
&= \frac{1}{6} \sum_{k \ge 0} (-1)^k (k^3+ 6 k^2 + 11 k + 6) x^{3 k}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/450900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Simplifying a radical to solve a problem $$
L = \sqrt{(x+8)^2 + \left(\dfrac{10(x+8)}{x}\right)^2}
$$
$$
L = \sqrt{(x+8)^2 + \dfrac{100(x+8)^2}{x^2}}
$$
$$
L = \sqrt{(x+8)^2\left(1 + \dfrac{100}{x^2}\right)}
$$
$$
L = (x+8)\sqrt{1 + \dfrac{100}{x^2}}
$$
Here am stuck, the answer is
$$
L = \frac{(x+8)}{x}\sqrt{x^2 + 100}
$$
| Find the common denominator in the radicand:
$$\begin{align} L = (x+8)\sqrt{1 + \dfrac{100}{x^2}} & = (x+8)\sqrt{\dfrac{x^2 + 100}{x^2}} \\ \\ & = \dfrac{(x+8)}{x}\sqrt{x^2 + 100} \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Functional Equation - Am I right? Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $$x^2+y^2+2f(xy)=f(x+y)(f(x)+f(y))$$
So here's my solution,
If $x=y=0$,
$2f(0)=2f(0)^2$
$\implies f(0)=0$ or
$f(0)=1$.
Case $1$: $f(0)=0$
If $y=0$,
$x^2+2f(0)=f(x)(f(x)+f(0))$
$$f(x)^2=x^2$$
Now suppose for the sake of contradiction that $f(x)=-x$ for some $x \in \mathbb{R/\{0\}}$.
Then, $x^2+y^2-2xy=(-x-y)(-x-y)$
$$\implies (x-y)^2=(x+y)^2$$ which is a contradiction as $x+y=\pm(x-y)$ forces $x$ or $y$ to equal $0$ which isn't permitted.
Hence $f(x)=x$, conversely one readily checks that this satisfies the given equation.
Case $2$: $f(0)=1$
Again let $y=0$ and then $$f(x)^2+f(x)-x^2-2=0$$ (remember we're assuming $f(0)=1$)
Solving for $f(x)$ we get $$f(x)=\frac{-1 \pm \sqrt{4x^2+9}}{2}$$
If you allow the plus or minus to be a minus $f(0) \not= 1$. Therefore it must be a plus.
But when you sub it back in to the original equation and then putting $x=y=1$ it does not work. Hence it does not work for all $x$ contradiction hence $f(x)=x$ as in case one is the only solution. QED
My concern here is the last paragraph - Can I do that? Is it tight? If so, is there a better and more mathematically "professional" (for the lack of a better word) way to phrase it?
Any help would be greatly appreciated.
| In Case 1, we can substitute $y = -x$ into the original equation. We obtain $$x^2+x^2+2f(-x^2)=0(f(x)+f(-x))$$ which means that $$f(-x^2)=-x^2$$ holds for all $x$. This means that $$f(y)=y \text{ holds for all }y\leq 0.\tag{*}$$
Now, let $x>0$ and $y < -x$. Substitute $x$ and $y$ into the original equation. By $(*)$ this gives us $$x^2+y^2 + 2xy=(x+y)(f(x)+y),$$ where we used the facts that $y,xy$ and $x+y$ are negative. Divide by $x+y$ (which is nonzero by our assumptions) and simplify to get $f(x) = x$. This proves that $f(x) = x$ holds also for positive $x$.
In Case 2, as you noticed, we obtain $$f(x)=\frac{-1 \pm \sqrt{4x^2+9}}{2},$$ i.e. there are only two possibilities for each $x\in\mathbb R$. Therefore there are four possible choices of $f(1)$ and $f(2)$. But each of them leads to a contradiction if we substitute $x=y=1$ into the original equation. Therefore Case 2 does not lead to a solution.
Edit: the thing to notice here is that in $$f(x)=\frac{-1 \pm \sqrt{4x^2+9}}{2},$$ we can choose a plus or a minus sign for each $x$ independently, so it is not enough to consider only the cases $$f(x)=\frac{-1 + \sqrt{4x^2+9}}{2}$$ and $$f(x)=\frac{-1 - \sqrt{4x^2+9}}{2}.$$ We also have to prove that functions like $$f(x)=\begin{cases}\frac{-1 + \sqrt{4x^2+9}}{2};&\text{if }x>0,\\
\frac{-1 - \sqrt{4x^2+9}}{2};&\text{if }x\leq 0.
\end{cases}$$
cannot occur. By showing that there is no value for $f(1)$ that could possibly work, we have achieved exactly this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$ without using trigonometry? $$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$
Is there any way to find the answer without using trigonometry, like this?
Hint by Parth Thakkat:
$$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$
$$ = \int \dfrac {\sqrt{x+1}} {x^{1/2}} \cdot \dfrac{dx} {x^{3}}$$
$$ = \int \sqrt{1+\dfrac 1 x} \cdot \dfrac{dx} {x^{3}}$$
Take $t^2 = 1 + \dfrac 1 x$
and note that $ \dfrac 1 x = t^2 - 1$
More Hint:
$t^2 = 1 + \dfrac 1 x$
$\implies 2tdt = -\dfrac 1 {x^2}dx$
Next
$$ = \int t dx/x^3$$
Since $dx = -2tx^2 dt$
$$= \int t(-2t)(t^2-1) dt$$
$$ = -2\int t^4-t^2 dt$$
$$ = -2(\dfrac{t^5} {5} - \dfrac{t^3}{3} + C$$
C = constant
$$ = -\dfrac{2}{5}(1+\dfrac{1}{x})^\dfrac{5}{2}+\dfrac{2}{3}(1+\dfrac{1}{x})^\dfrac{3}{2} + C$$
Thanks alot :)
| Hint:
$$\int \dfrac {\sqrt{x+1}} {x^{7/2}} dx$$
$$ = \int \dfrac {\sqrt{x+1}} {x^{1/2}} \cdot \dfrac{dx} {x^{3}}$$
$$ = \int \sqrt{1+\dfrac 1 x} \cdot \dfrac{dx} {x^{3}}$$
Take $t^2 = 1 + \dfrac 1 x$
and note that $ \dfrac 1 x = t^2 - 1$
More Hint:
$t^2 = 1 + \dfrac 1 x$
$\implies 2tdt = -\dfrac 1 {x^2}dx$
$\implies 2t(1-t^2)dt = \dfrac 1 {x^3}dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/454206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Partial Fractions I here would like to clear my doubt on the question below:
$$\frac{1}{x(x-1)(x-2)}\;,$$
that is, we want to bring it into the form:
$$\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}\;,$$
in which the unknown parameters are $A,B$, and $C$. Multiplying these formulas by $x(x − 1)(x − 2)$ turns both into polynomials, which we equate:
$$A(x-1)(x-2) + Bx(x-2) + Cx(x-1) = 1\;,$$
or, after expansion and collecting terms with equal powers of $x$:
$$(A+B+C)x^2 - (3A+2B+C)x + 2A = 1\;.$$
At this point it is essential to realize that the polynomial $1$ is in fact equal to the polynomial $0x^2 + 0x + 1$, having zero coefficients for the positive powers of $x$. Equating the corresponding coefficients now results in this system of linear equations:
$$\left\{\begin{align*}
&A+B+C = 0\\
&3A+2B+C = 0\\
&2A = 1\;.
\end{align*}\right.$$
Solving it results in:
$$A = \frac{1}{2},\, B = -1,\, C = \frac{1}{2}\;.$$
So from my solving I had different values of $A,B$, and $C$ which gave me:
$$\left\{\begin{align*}
&A=\frac12\\
&B= 2\\
&C= -\frac52\;.
\end{align*}\right.$$
Can someone please tell me if these answers are correct because when I substitute these values into equation $A+B+C= 0$, it still gave me a zero.
But this time for the $2$nd equation, instead of $3A+2B+C= 0$, I used $-3A+2B+C= 0$, which then by substituting the values of $A, B$, and $C$ I had, also gave me a zero. Only $A= \frac12$ was the same as obtained from $2A= 1$.
Does this mean that the values that I have obtained for $A, B$, and $C$ are also correct? Kindly can someone please give a clear explanation to this?
Many thanks.
| $$A=\frac 12,B-1=C=\frac 12$$
these valuse are correct
from the step:
$$A(x-1)(x-2)+Bx(x-2)+Cx(x-1)=1$$
put $x=1,x=2,x=0$ you will get right values
even from this equations you also get same values:
$$\left\{\begin{align*}
&A+B+C = 0\\
&3A+2B+C = 0\\
&2A = 1\;.
\end{align*}\right.$$
from 3rd equation $A=\dfrac 12$
after perform (2)-(1)
$$2A+B=0\implies B=-1$$
and in eqn (1)
$$A+B+C=0\implies \frac 12-1+C=0\implies C=\frac 12$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$
Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$
I have tried $\sin^3\alpha(3\sin\alpha - 4 \sin^3\alpha) = 3\sin^4\alpha - 4\sin^6\alpha$ and $\cos^3\alpha(4\cos^3\alpha - 3\cos\alpha) = 4\cos^6\alpha - 3\cos^4\alpha$ to give
$$\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = 3\sin^4\alpha - 4\sin^6\alpha + 4\cos^6\alpha - 3\cos^4\alpha$$
I can't work out how to simplify this to $\cos^32\alpha$.
I also noticed that the LHS of the question resembles $\cos(A-B)$, but I can't figure a way of making that useful.
| Using $$\cos3A=4\cos^3A-3\cos A,\sin3A=3\sin A-4\sin^3A,$$
$$4(\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha)$$
$$=\sin3\alpha (3\sin\alpha-\sin3\alpha) + \cos3\alpha (\cos3\alpha+3\cos\alpha)$$
$$= \cos^23\alpha-\sin^23\alpha +3(\cos3\alpha\cos\alpha+\sin3\alpha \sin\alpha)$$
$$=\cos6\alpha+3\cos(3\alpha-\alpha)$$
(using $\cos2A=\cos^2A-\sin^2A,\cos(A-B)=\cos A\cos B+\sin A\sin B$)
$$=\cos(3\cdot2\alpha)+3\cos2\alpha$$
$$ = 4\cos^32\alpha-3\cos2\alpha+3\cos2\alpha (\text{ again applying }\cos3A)$$
$$ = 4\cos^32\alpha $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum and minimum function on area Find maximum and minimum value of function $f(x,y) = 3x+14y$ on $ \left\{ (x,y): 3x^4 + xy + y^4 =6\right\} $.
I will grateful for hints and yours help.
| It is clear that the tangents of the smooth curve $$3x^4+y^4+xy-6=0\,\,\,\, (1)$$ at the extremal points have to be parallel to the level lines of the target function, i.e. $3x+14y=C$. Using the implicit differentiation and equating the slopes, we obtain the equation $$-\frac {12x^3+y} {4y^3+x}= - \frac 3 {14} \Leftrightarrow 168x^3+14y-12y^3-3x=0.\,\,\,\, (2) $$ In order to solve the system (1)-(2) in $x$ an $y$, we find its resultant in $y$ with help of Maple:
$$797154048x^{12}+56899584x^{10}-1327104x^8+113799168x^6+10434789x^4+8232x^2-2677344. $$
It is well known that its rational roots must have the form $\pm \frac {\mathop{\rm divisor\,\, of} 2677344}{ \mathop{\rm divisor\,\, of} 797154048}.$ Factoring $2677344=3\cdot 2^5\cdot 167^2$ and $797154048=37\cdot 2^8\cdot 3^4\cdot 1039$ and considering $6480=6\cdot 2 \cdot 3 \cdot 9 \cdot 5 \cdot 2 \cdot 2$ cases, we find $\pm \frac 1 2$ to be the rational roots of the discriminant. The numerical solution shows the discriminant has only the two real roots (The others are complex.). Therefore, we obtain the maximum of the target function ar $x= \frac 1 2 ,\,y= \frac 3 2$. The value of $y$ is found as the only real root of $-24y^3+28y+39=0$. The last equation is formed by the substitution $ x= \frac 1 2$ in (2). The minimum of the target function is achieved at $x= - \frac 1 2 ,\,y= - \frac 3 2 .$
| {
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"timestamp": "2023-03-29T00:00:00",
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A geometric property of the graph of $y = x^2$ Consider $n\geq 3$ and two points $A, B$ on the graph of $y = x^2$. Now choose points $P_1,...,P_{n-2}$ on this graph such that the area of the convex $n$-gon $AP_1...P_{n-2}B$ is maximum (they do exist). Let $S_n$ be the area of this $n$-gon and let $S$ be the area between the chord $AB$ and the graph. I want to show that $\frac {S_n}{S}$ is only a function of $n$ and is independent from (the position of) $A$ and $B$.
| The area between $AB$ and the graph is just
$$\int_a^b a^2(1-\frac{x-a}{b-a})+b^2\frac{x-a}{b-a}-x^2=a^2x-\frac{a^2}{b-a}(\frac12 x^2-ax)+\frac{b^2}{b-a}(\frac 12x^2-ax)-\frac13x^3\bigg|_a^b$$
$$=a^2b-a^3+\frac12(b+a)(a-b)^2-\frac13 b^3+\frac13 a^3$$
$$=(a-b)(-a^2+\frac12(a^2-b^2)+\frac 13(a^2+ab+b^2))$$
$$=\frac 16(b-a)(a^2-2ab+b^2)=\frac16(b-a)^3$$
Now we note: maximizing the convex polygon is the same as minimizing the collection of smaller subarcs underneath them. If my points are $a_i$, with $a_0=a$ and $a_n=b$, we want to find the $a_i, 0<i<n$ that minimize
$$A=\frac 16\sum_{i=0}^{n-1}(a_{i+1}-a_i)^3$$
now we just use some simple calculus:
$$\frac{\partial A}{\partial a_i}=\frac 16\left(3(a_i-a_{i-1})^2-3(a_{i+1}-a_i)^2\right)$$
Solving $$\frac{\partial A}{\partial a_i}=0$$ gives
$$(a_i-a_{i-1})^2=(a_{i+1}-a_i)^2$$
Since we assume that $\forall i a_{i+1}>a_i$:
$$a_i-a_{i-1}=a_{i+1}-a_i$$
$$a_i=\frac12(a_{i+1}+a_{i-1})$$
In other words, the $x$-coordinate of each point is the average of its neighbors. This happens when the points are equally spaced on the $x$-axis. In this case, $a_i=a+(b-a)\frac in$, $a_{i+1}-a_i=\frac{b-a}n$, and the total area of these smaller arcs is
$$A=\frac 16\sum_{i=0}^{n-1}(\frac{b-a}n)^3=\frac 16\frac{(b-a)^3}{n^2}$$
The area of the polygon $S_n$ is then $S-A$:
$$\frac16(b-a)^3(1-\frac 1{n^2})$$
and we see that $\frac{S_n}S$ is independent of $a$ and $b$, and only dependent on $n$.
| {
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Prove that if $a$ and $b$ are relatively prime, then $\gcd(a+b, a-b) = 1$ or $2$ Prove that if $a$ and $b$ are relatively prime, then $\gcd(a+b, a-b) = 1$ or $2$.
I started off by putting $\gcd(a+b, a-b) = d$.
This implies that there are two relatively prime integers $x_1, x_2$, such that
$dx_1 = a+b$
$dx_2 = a -b$
Adding the first equation to the second gives us: $d(x_1 + x_2) = 2a$, and subtracting the second from the first gives us $d(x_1 - x_2) = 2b$. This implies that $d\mid2a, d\mid2b \implies \gcd(2a, 2b) = d \implies d = 2\cdot \gcd(a,b) \implies d = 2$.
This is based on the fact that $\gcd(x_1 + x_2, x_1 - x_2) = 1$, which is in fact the statement of the problem. How do I prove this? Can I somehow use the Euclidean Algorithm for this.
| If $ax+by=1$ then $$\begin{align}2&=2ax+2by \\&=(a-b)x + (a+b)x + (b-a)y+(b+a)y \\&= (a-b)(x-y)+(a+b)(x+y)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Trigonometric functions I wonder how does a WolframAlpha get this relation where input is a LHS and output is RHS:
$$\cos^2(x)\cos(2x) = \tfrac{1}{4}\cos(4x) + \tfrac{1}{2}\cos(2x) + \tfrac{1}{4}$$
| \begin{align*}
\cos^2(x)\cos(2x) &= \frac{1}{2}(1+\cos(2x))\cos(2x)\\
&= \frac{1}{2}\cos(2x) +\frac{1}{2}\cos^2(2x) \\
&= \frac{1}{2}\cos(2x) + \frac{1}{4} + \frac{1}{4}\cos(4x)
\end{align*}
by two applications of the double angle formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all the values of $x \in \mathbb R$ from an equation Find all the values of $x \in \mathbb{R}$ from this inequality:
$$\left|\frac3{x^3-8}\right|=\left|\frac1{x-2}\right|$$
This is my work:
$$\frac{\left|\frac3{x^3-8}\right|}{\left|\frac1{x-2}\right|}=1$$
$$\left|\frac{3(x-2)}{x^3-8}\right|=1$$
$$\left|\frac {x-2}{x^3-2^3}\right|=\frac13$$
| HINT:
$x^3-8=x^3-2^3=(x-2)\{x^2+2x+2^2\}=(x-2)\{(x+1)^2+3\}$
So assuming $x-2\ne0,$ we can safely cancel $x-2$ and utilize $|x^2+2x+4|=x^2+2x+4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/459552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is this the correct solution? Determine the coordinates of the vector $U=(4,5,-3)\;\text{of}\; R^3$ with respect to base ${(1,0,0), (0,1,0), (0,0, 1)}$
$$x(1,0,0) + y (0,1,0) + z (0,0,1) = (4,5, -3)$$
$$(x, 0,0) + (0, y, 0) + (0,0,z) = (4,5, -3)$$
$$x +0 +0 = 4 \Longrightarrow x = 4$$
$$0 +0 + y = 5 \Longrightarrow y = 5$$
$$0 +0 + z = -3 \Longrightarrow z = -3$$
Thus, the coordinates are $$(4,5, -3).$$
$$$$ $$$$And the base $\{(1,1,1),(1,2,0),(3,1,0)\}$, just find the values of $x$, $y$ and $z$ such that $$x(1,1,1)+y(1,2,0)+z(3,1,0)=(4,5,-3)??$$
$$(x,x,x)+(y,2y,0)+(3z,z,0)=(4,5,-3)$$$\begin{cases}x+y+3z=4&\\ x+2y+z=5&\\ x+0+0=-3&\end{cases}$$$x=-3$$$$y=\frac{53}{5}$$and$$z=-\frac{6}{5}$$
Is this correct?
| Yes, note that definition of representation of a vector $\vec v$ with respect to a basis $B=\left\langle {\vec{\beta}_{1},...,\vec{\beta}_{n}} \right\rangle $ of a $n$-dimensional space $V$ is
$$Rep_{B}(\vec v)=\left( {\begin{array}{*{20}{c}}
{c_{1}} \\
{ \vdots } \\
{ c_{n} }
\end{array}} \right)$$
where $c_{1}, ..., c_{n}$ are coefficients such that $\vec v=c_{1}\vec{\beta}_{1}+...+c_{n}\vec{\beta}_{n}$. With your basis
$$B=\left\langle \left( {\begin{array}{*{20}{c}}
{ 1 } \\
{ 1 } \\
{ 1 }
\end{array}} \right),
\left( {\begin{array}{*{20}{c}}
{ 1 } \\
{ 2 } \\
{ 0 }
\end{array}} \right),
\left( {\begin{array}{*{20}{c}}
{ 3 } \\
{ 1 } \\
{ 0 }
\end{array}} \right)
\right\rangle$$
and
$$\vec v = \left( {\begin{array}{*{20}{c}}
{ 4 } \\
{ 5 } \\
{ -3 }
\end{array}} \right)$$
then you have to find $c_1,c_2,c_3$ such that
$$
c_1 \left( {\begin{array}{*{20}{c}}
{ 1 } \\
{ 1 } \\
{ 1 }
\end{array}} \right) +
c_2 \left( {\begin{array}{*{20}{c}}
{ 1 } \\
{ 2 } \\
{ 0 }
\end{array}} \right) +
c_3 \left( {\begin{array}{*{20}{c}}
{ 3 } \\
{ 1 } \\
{ 0 }
\end{array}} \right) =
\left( {\begin{array}{*{20}{c}}
{ 4 } \\
{ 5 } \\
{ -3 }
\end{array}} \right)
$$
so solve this system and then your answer will be
$$Rep_{B} (
\left({\begin{array}{*{20}{c}}
{ 4 } \\
{ 5 } \\
{ -3 }
\end{array}} \right))=
\left({\begin{array}{*{20}{c}}
{ c_1 } \\
{ c_2 } \\
{ c_3 }
\end{array}} \right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expected number of people sitting in the right seats. There was a popular interview question from a while back: there are $n$ people getting seated an airplane, and the first person comes in and sits at a random seat. Everyone else who comes in either sits in his seat, or if his seat has been taken, sits in a random unoccupied seat. What is the probability that the last person sits in his correct seat?
The answer to this question is $1/2$ because everyone looking to sit on a random seat has an equal probability of sitting in the first person's seat as the last person's.
My question is: what is the expected number of people sitting in their correct seat?
My take: this would be $\sum_{i=1}^n p_i$ where $p_i$ is the probability that person $i$ sits in the right seat..
$X_1 = 1/n$
$X_2 = 1 - 1/n$
$X_3 = 1 - (1/n + 1/n(n-1))$
$X_4 = 1 - (1/n + 2/n(n-1) + 1/n(n-1)(n-2))$
Is this correct? And does it generalize to $X_i$ having an $\max(0, i-1)$ term of $1/n(n-1)$, a $\max(0, i-2)$ term of $1/n(n-1)(n-2)$ etc?
Thanks.
| Now that I have cleaned my glasses, I'll try again. Thank you, Lord Farin.
Some observations.
The answer must be greater than 1. The probability that the first person, regardless of the number of people, sits in the proper seat is $\frac{1}{n}$, and there are $n$ people, so that expectation is 1. Even if the first person sits in the wrong seat, there is non-zero probability of other people sitting in the correct seat, so the answer must be greater than or equal to 1, and I'm pretty sure equality only exists in the case $n=2$.
There is some form of recurrence going on here. Enumerating the possibilities, I get (if I haven't erred):
$$
E_2 = \frac{1}{2}\cdot 2 + \frac{1}{2}\cdot 0 = 1\\
E_3 = \frac{1}{3}\cdot 3 + \frac{2}{3}\left[\frac{1}{2}\cdot 1 + \frac{1}{2}\cdot 0\right] = 1+\frac{1}{3}=\frac{4}{3}\\
E_4 = \frac{1}{4}\cdot 4 + \frac{3}{4}\left[\frac{1}{3}\cdot2 + \frac{2}{3}\left[\frac{1}{2}\cdot 1 + \frac{1}{2}\cdot 0\right]\right] = 1+\frac{3}{4}\cdot\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{7}{4}\\
E_5 = \frac{1}{5}\cdot 5 + \frac{4}{5}\left[\frac{1}{4}\cdot3 + \frac{3}{4}\left[\frac{1}{3}\cdot2 + \frac{2}{3}\left[\frac{1}{2}\cdot 1 + \frac{1}{2}\cdot 0\right]\right]\right] = 1+\frac{4}{5}\left(\frac{3}{4} + \frac{3}{4}\cdot\left(\frac{2}{3}+\frac{1}{3}\right)\right)=\frac{11}{5}\\
$$
Let $E_k$ be the expected number of people in the correct seats when the starting population is k. The relationship seems to be:
$$
E_k = 1 + \left(E_{k-1} - 1 + \frac{k-2}{k-1}\right)\frac{k-1}{k}
$$
The first 1 is the expected value of person 1 taking the correct seat. The next term is broken into two parts. If the first person did not take the correct seat, then the second person can "fix" the error by swapping and taking the previous person's seat, leaving the remaining $n-2$ people their proper seats. If the second person also takes the wrong seat, the problem restarts on the $n-1$ scale. In both the latter two cases, there is "one less" correct seat than the initial, since the previous term used up a seat with an incorrect choice. If the above supposition is correct, the expected value would just be the sum of the recurrence relation applied to $n$ or $\sum_{k=1}^n E_k$.
Generating the first few terms using this relationship gives:
$$
1, 1, \frac{4}{3}, \frac{7}{4}, \frac{11}{5}, \frac{16}{6}, \frac{22}{7}, \frac{29}{8}, \frac{37}{9}, \frac{46}{10}, \ldots
$$
The numerator is a quadratic and the denominator is just $n$ so the expected value should be:
$$
E_n = \frac{n^2-n+2}{2n}
$$
Edit: The long-term proportion of people sitting in the proper seats intuitively would be $$
\lim_{n \to \infty} \frac{E_n}{n}\\
=\lim_{n \to \infty} \frac{n^2-n+2}{2n^2}\\
=\frac{1}{2}
$$
Which dovetails nicely with the long-term probability of the last person sitting in his or her proper seat.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Simplify : $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7) $ The question is to simplify $( \sqrt 5 + \sqrt6 + \sqrt7)(− \sqrt5 + \sqrt6 + \sqrt7)(\sqrt5 − \sqrt6 + \sqrt7)(\sqrt5 + \sqrt6 − \sqrt7)$ without using a calculator .
My friend has given me this challenge . I solved it by expanding $$(a+b+c)(-a+b+c)(a-b+c)(a+b-c) = -a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4$$ and then substituting $a,b,c=\sqrt 5 , \sqrt6 , \sqrt7$ respectively to get the answer $104$ .
But I suppose there is a more elegant and easy way to solve this problem .
Can anyone find it ?
| Consider Heron's formula: the area of a triangle with sides $a, b, \text{and } c$ is
$$
\sqrt{s(s-a)(s-b)(s-c)}
$$
where $s$ is the semi-perimeter $\frac12 (a + b + c)$.
Let $a, b, \text{and } c$ be $\sqrt{5}, \sqrt{6}, \text{and } \sqrt{7}$. Then the area is the square root of your expression divided by $4$. So, what is the area of this triangle? Use the law of cosines to find the cosine of the angle $C$ opposite $c$:
$$
\begin{align}
7 &= 5 + 6 - 2 \sqrt{5}\sqrt{6}\cos{C}\\
2\sqrt{30}\cos{C} &= 4\\
\cos{C} &= \frac{2}{\sqrt{30}}
\end{align}
$$
But the area of the triangle is $\frac12 ab\sin{C}$.
$$
\frac12 ab\sin{C} = \frac12 \sqrt{30} \frac{\sqrt{26}}{\sqrt{30}} = \frac12\sqrt{26}.
$$
Your expression is therefore the square of $2\sqrt{26}$, which is $104$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/465103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
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} |
Factorise: $2a^4 + a^2b^2 + ab^3 + b^4$ Factorize : $$2a^4 + a^2b^2 + ab^3 + b^4$$
Here is what I did:
$$a^4+b^4+2a^2b^2+a^4-a^2b^2+ab^3+b^4$$
$$(a^2+b^2)^2+a^2(a^2-b^2)+b^3(a+b)$$
$$(a^2+b^2)^2+a^2(a+b)(a-b)+b^3(a+b)$$
$$(a^2+b^2)^2+(a+b)((a^2(a-b)) +b^3)$$
$$(a^2+b^2)^2+(a+b)(a^3-a^2b+b^3)$$
At this point I don't know what to do and am feeling that my direction is wrong. Please help me.
( Wolfram alpha says that the answer is $(a^2-a b+b^2) (2 a^2+2 a b+b^2)$ but how? )
| Different Hint
Since your polynomial is homogeneous, this is equivalent to factoring the degree 4 univariate polynomial $2x^4+x^2+x+1.$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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How prove this $|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\frac{1}{20n^3}$ Prove that
$$|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\dfrac{1}{20n^3}$$
let $t=\{n\sqrt{2}\}-\{n\sqrt{3}\}$ and $k=[n\sqrt{3}]-[n\sqrt{2}]$
then we have $$t=k-(\sqrt{3}-\sqrt{2})n=k-\sqrt{5-2\sqrt{6}}n\neq 0$$
so
\begin{align*}
t&=\dfrac{(k-(\sqrt{3}-\sqrt{2})n)(k-(\sqrt{3}+\sqrt{2})n)(k+(\sqrt{3}-\sqrt{2})n)(k+(\sqrt{3}+\sqrt{2})n)}{(k-(\sqrt{3}+\sqrt{2})n)(k+(\sqrt{3}-\sqrt{2})n)(k+(\sqrt{3}+\sqrt{2})n)}\\
&=\dfrac{k^4-10k^2n^2+n^4}{(t-2\sqrt{2}n)(t+2(\sqrt{3}-\sqrt{2})n)(t+2\sqrt{3}n)}
\end{align*}
notice that
$$|t-2\sqrt{2}n|\le 2\sqrt{2}n+\dfrac{1}{20}\le(2\sqrt{2}+\dfrac{1}{20})n$$
$$|t+2(\sqrt{3}-\sqrt{2})n|\le2(\sqrt{3}-\sqrt{2})n+\dfrac{1}{20}\le(2\sqrt{3}-2\sqrt{2}+\dfrac{1}{20})n$$
$$|t+2\sqrt{3}n|\le2\sqrt{3}n+\dfrac{1}{20}\le(2\sqrt{3}+\dfrac{1}{20})n$$
so
$$|t|\ge\dfrac{1}{(2\sqrt{2}+\dfrac{1}{20})(2\sqrt{3}+2\sqrt{2}-\dfrac{1}{20})(2\sqrt{3}+\dfrac{1}{20})n^3}>\dfrac{1}{7n^3}$$
so
$$|t|\ge\min{\left(\dfrac{1}{20},\dfrac{1}{7n^3}\right)}\ge\dfrac{1}{20n^3}$$
\
This post https://math.stackexchange.com/questions/465419/how-prove-this-t-2-sqrt2n-le2-sqrt2n-frac120 is not true? so This methods is wrong, so How prove it? Thank you
| What I have done is not a full solution, but it may even lead to the solution of a more general problem. Also, using this method I can get only a lower bound. Here it is.
For any positive integer $n$, let $$\sqrt {3} = \frac {a_n} {n} + \varepsilon_n \qquad \text {and} \qquad \sqrt {2} = \frac {b_n} {n} + \delta_n,$$ where $a_n$ and $b_n$ are positive integers, $\varepsilon_n$ and $\delta_n$ are positive real numbers, and $\max (\varepsilon_n, \delta_n) < \frac {1} {n}$. Such a configuration exists, for, let $a_n = [\sqrt {3} n]$ and $b_n = [\sqrt {2} n]$. Since $$a_n < \sqrt {3} n < a_n + 1 \qquad \text {and} \qquad b_n < \sqrt {2} n < b_n + 1,$$ we have $0 < \varepsilon_n, \delta_n < \frac {1} {n}$.
Then, we have
$$\begin {eqnarray}
|\{n\sqrt{3}\}-\{n\sqrt{2}\}| & = & |\{a_n + n \varepsilon_n\}|
- |\{b_n + n \delta_n\}| \nonumber \\ & = & |n \varepsilon_n - n \delta_n| \nonumber \\ & = & n |\varepsilon_n - \delta_n|.
\end{eqnarray}$$
Since we know very crudely that $\varepsilon_n = O \left (\frac {1} {n^2} \right)$ and $\delta_n = O \left (\frac {1} {n^2} \right)$, we have $$|\{n\sqrt{3}\}-\{n\sqrt{2}\}| = O \left (\frac {1} {n} \right),$$ that is, there exists an absolute constant $c$ such that $$|\{n\sqrt{3}\}-\{n\sqrt{2}\}| < \frac {c} {n}.$$
Note: This does not answer the question, it is rather related to it. The reason I put it here (rather than not) is that I feel it may be useful. If you guys feel otherwise, delete it.
| {
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"timestamp": "2023-03-29T00:00:00",
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System of quadratic equations $x^2 + y = 4$ and $x + y^2 = 10$ How would you solve the following system of equations:
$$
x^2 + y = 4 \\
x + y^2 = 10
$$
Thanks very much!
I tried defining y in terms of x and then inserting in to the second equation:
$$
y = 4 - x^2 \\
x + (4 - x^2)^2 = 10
$$
Expand the second equation:
$$
x + 16 - 8x^2 + x^4 = 10
$$
Rearrange the terms:
$$
x^4 - 8x^2 +x + 6 = 0
$$
I tried factoring out this polynomial to simplify it for solving, but didn't succeed :(
| To solve general equations of the form $ax^4+bx^3+cx^2+dx+e=0$ requires quartic formula.
| {
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Triangle with integral side lengths and $\angle A=3\angle B$
$ABC$ is a triangle with integral side lengths. Given that $\angle A=3\angle B$, find the minimum possible perimeter of $ABC$.
I got this problem from an old book (which did not provide even a hint). I can think of some approaches, but all of them result in complicated Diophantine equations that would not be solvable without the help of a computer. Any suggestions?
| Giving it a try using elementary techniques.
In $\triangle ABC$, let $B=\theta$, $A=3\theta$, $C=\pi-4\theta$. By sine-rule
$$\frac{a}{\sin 3\theta}=\frac{b}{\sin \theta}=\frac{c}{\sin 4\theta}$$
so that $$a=b(3-4\sin^2\theta) \quad , \quad c=b(4\cos \theta \cos 2\theta)$$
On further simplification,
$$a=b(4\cos^2\theta-1) \quad , \quad c=b \cdot 4\cos \theta (2\cos^2 \theta-1)$$
$c>0$ places bound on $\cos \theta$ : $\cos \theta > 1/\sqrt{2} \Rightarrow \theta < 45^{\circ}$ which makes sense since $45^{\circ}+3\cdot 45^{\circ}=180^{\circ}$
We may choose any suitable value of $\cos \theta > 1/\sqrt{2} \approx 0.7071$.
Taking $\cos \theta = 3/4=0.75$, yields $a/b=5/4$, $c/b=3/8$. So one class of triangles would be multiples of
$$(a,b,c)=(10,8,3)$$
Taking $\cos \theta = 5/6$, yields $a/b=16/9$, $c/b=35/27$. So another class of triangles would be multiples of
$$(a,b,c)=(48,27,35)$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Derivative of $\frac {x\cdot\left(1 - 3x\right)}{\sqrt{x-1}}$ Problem. Find the first derivative of $$ \dfrac {x \left( 1 - 3x \right)}{\sqrt{x-1}} $$
Work. Let $u = x-1$ and $y = \dfrac {(u+1)(-3u-2)}{\sqrt{u}} $
Using the chain rule, I got$$\dfrac{(-9x^2-5x+2)}{(2(x-1)^\frac{3}{2})}$$
But the answer is $$\dfrac{(-9x^2+13x+2)}{\left(2(x-1)^\frac{3}{2}\right)}$$
I'm not sure what I did wrong, maybe something related to the $(-3u-2)$?
| If you have some patience to do directly, you will get it.
$$
\frac{d}{dx}\left(\frac{x(1-3x)}{\sqrt{x-1}}\right) =\frac{\sqrt{x-1}(1-6x) - \frac{1}{2\sqrt{x-1}}(x-3x^2)}{x-1} \to \frac{-9x^2 +13x-2}{2(x-1)^\frac{3}{2}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of a sequence I need guidance for the following question.
Using the fact that $\sum_1^{\infty}\frac{(-1)^{n+1}}{n}=\log2$, $\sum_1^{\infty}\frac{(-1)^{n}}{n(n+1)}$ equals
$1.$ $1-2\log2$
$2.$ $1+2\log2$
$3.$ $(\log2)^2$
$4.$ $-(\log2)^2$
The given sequence gives us $1-\frac12+\frac13-\frac14+\cdots=log2$, but I am unable to think how this would help me to solve $-\frac12+\frac16-\frac1{12}+\frac1{20}-\cdots$
I wish somebody could help. Thanks in advance!
| $$\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)} = \sum_{n=1}^\infty(-1)^n\left(\frac{1}{n} - \frac{1}{n+1}\right)$$
This is true because:
$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
$$\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)} = \sum_{n=1}^\infty-\frac{(-1)^{n+1}}{n} - \frac{(-1)^n}{n+1} $$
First term converges to $- \log{2}$. Now we work on the second term.
If we set $n=k-1$, we can write:
$$\sum_{n=1}^\infty\frac{(-1)^n}{n+1} = \sum_{k=2}^\infty \frac{(-1)^{k+1}}{k}$$
Now we have:
$$\sum_{k=2}^\infty \frac{(-1)^{k+1}}{k} = \sum_{k=2}^\infty \frac{(-1)^{k+1}}{k} + \frac{1}{1} - \frac{1}{1} = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} - \frac{1}{1} = \log 2 - 1$$
Now we substitute:
$$\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)} = - \log{2} - \log{2} + 1 = 1 - 2\log{2} = 1-\log{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can someone explain this trigonometric limit without L'Hopital? I can not solve this limit:
$$\lim \limits_{x\to 0}\frac{x^2}{1-\sec(x)}$$
$$\lim \limits_{x\to 0} \frac{x^2}{1-\sec(x)}=\lim \limits_{x\to 0}\frac {x^2}{1-\sec(x)}\cdot{\frac{1+\sec(x)}{1+\sec(x)}}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{1-\sec^2(x)}=\lim \limits_{x\to 0}\frac{x^2(1+\sec(x))}{-\tan^2(x)}$$
| Note:
I fixed an error noted by triple_sec .
$\dfrac{x^2}{1-\sec x}
=\dfrac{x^2}{1-1/\cos x}
=\dfrac{ x^2 \cos x}{\cos x-1}
$.
Using
$\cos(2x)
=\cos^2(x)-\sin^2(x)
=1-2\sin^2(x)
$,
$\cos(x)-1
=-2\sin^2(x/2)
$
(I originally had $+$ here instead of $-$)
so
$\dfrac{x^2}{1-\sec x}
=\dfrac{ x^2 \cos x}{\cos x-1}
=\dfrac{ x^2 \cos x}{-2\sin^2(x/2)}
=- \cos x\dfrac{ x^2 }{-2\sin^2(x/2)}
$.
Since,
as $x \to 0$,
$\cos x \to 1$
and
$\dfrac{\sin x}{x}
\to 1$,
$-\cos x\dfrac{ x^2}{2\sin^2(x/2)}
=-2\cos x\left(\dfrac{ x/2 }{\sin(x/2)}\right)^2
\to -2
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/468674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Is $(-3)^n + 5^n$ monotone? How can I establish if this sequence is monotone?
If it isn't, is it permanently monotone from a certain n* to infinite?
| To show that a sequence is stictly increasing, we must show that $a_{n+1} > a_n$ for all $n$.
\begin{array}
1(-3)^{n+1}+5^{n+1} &>& (-3)^n + 5^n \\
5^{n+1} - 5^n &>& (-3)^n - (-3)^{n+1} \\
5^n (5-1) &>& (-3)^n (1+3) \\
5^n &>& (-3)^n
\end{array}
When $n$ is even then we have $5^n > 3^n$ which is clearly true. When $n$ is odd we have $5^n > -3^n$ which is clearly true since $5^n$ is positive and $-3^n$ is negative. Hence your sequence increases for all $n\ge 1.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
The Laurent series of the digamma function at the negative integers To find the Laurent series of $\psi(z)$ at $z= 0$, I would first find the Taylor series of $\psi(z+1)$ at $z=0$ and then use the functional equation of the digamma function.
Specifically,
$$\begin{align} \psi(z + 1) = \frac{1}{z} + \psi(z) &= \psi(1) + \psi'(1)z + \mathcal{O}(z^{2}) \\ &= -\gamma + \zeta(2) + \mathcal{O}(z^{2}) \\ &= -\gamma + \frac{\pi^{2}}{6}z + \mathcal{O}(z^{2}) \end{align}$$
which implies
$$ \psi(z) = -\frac{1}{z} - \gamma + \frac{\pi^{2}}{6} z + \mathcal{O}(z^{2}).$$
But I'm having trouble finding the Laurent series of $\psi(z)$ at the negative integers.
Since $\psi(z)$ has simple poles at the negative integers with residue $-1$, I know that the first term of the series must be $ \displaystyle\frac{-1}{z+n}$.
But I would like to determine more terms in the series.
EDIT:
The series appears to be $$\begin{align} \psi(z) &= - \frac{1}{z+n} + (H_{n} - \gamma)+ \Big( H_{n}^{(2)} + \zeta(2) \Big) (z+n) + \Big( H_{n}^{(3)} - \zeta(3) \Big) (z+n)^{2} \\ &+ \Big( H_{n}^{(4)} + \zeta(4) \Big) (z+n)^{3} + \Big( H_{n}^{(5)} - \zeta(5) \Big) (z+n)^{4} + \ldots \end{align}$$
SECOND EDIT:
We can find the constant term by evaluating $\lim_{z \to -n} \Big( \psi(z) + \frac{1}{z+n} \Big)$.
Since $$ \begin{align} &\psi(z) + \frac{1}{z+n} \\ &= \psi(z+1) - \frac{1}{z} + \frac{1}{z+n} \\ &= \psi(z+2) - \frac{1}{z+1} - \frac{1}{z} + \frac{1}{z+n} \\ &= \ ... \ = \psi(z+n+1) - \frac{1}{z+n} - \frac{1}{z+n-1} - \ldots - \frac{1}{z+1} - \frac{1}{z} + \frac{1}{z+n}, \end{align}$$
we have $$ \lim_{z \to -n} \Big( \psi(z) + \frac{1}{z+n} \Big) = \psi(1) + 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} = \psi(1) + H_{n} = H_{n} - \gamma.$$
Similarly, we can find the coefficient of the $(z+n)$ term by evaluating $ \lim_{z \to -n} \frac{\psi(z) + \frac{1}{z+n} - H_{n} + \gamma}{z+n}.$
Since $$ \psi_{1}(z) = \psi_{1}(z+n+1) + \frac{1}{(z+n)^{2}} + \frac{1}{(z+n-1)^{2}} + \ldots + \frac{1}{z^2}, $$
we have $$ \lim_{z \to -n} \frac{\psi(z) + \frac{1}{z+n} - H_{n} + \gamma}{z+n} = \lim_{z \to -n} \Big(\psi_{1}(z) - \frac{1}{(z+n)^{2}} \Big) = \psi_{1}(1) + H_{n}^{(2)} = H_{n}^{(2)} + \zeta(2) .$$
And we can find the coefficient of $(z+n)^{2}$ by evaluating $ \lim_{z \to -n} \frac{\psi(z) + \frac{1}{z+n} - H_{n} + \gamma -\big( H_{n}^{(2)} + \zeta(2) \big) (z+n)}{(z+n)^{2}} . $
Since $$\psi_{2}(z) = \psi_{2} (z+n+1) - \frac{2}{(z+n)^{3}} - \frac{2}{(z+n-1)^{3}} - \ldots - \frac{2}{z^{3}},$$
we have $\begin{align} \lim_{z \to -n} \frac{\psi(z) + \frac{1}{z+n} - H_{n} + \gamma -\big( H_{n}^{(2)} + \zeta(2) \big) (z+n)}{(z+n)^{2}} &= \lim_{z \to -n} \frac{\psi_{1}(x) - \frac{1}{(z+n)^{2}} -H_{n}^{(2)} - \zeta(2)}{2(z+n)} \\ &= \lim_{z \to -n} \frac{\psi_{2}(x) + \frac{2}{(z+n)^{3}}}{2} \\ &= \frac{1}{2} \Big( \psi_{2}(1) + 2 H_{n}^{(3)} \Big) \\ &= H_{n}^{(3)} - \zeta(3). \end{align}$
$ $
And so on.
| A few terms for starters....
$$
\Psi \left( x \right) = -{x}^{-1}-\gamma+1/6\,{\pi }^{2}x-\zeta
\left( 3 \right) {x}^{2}+{\frac {1}{90}}\,{\pi }^{4}{x}^{3}-\zeta
\left( 5 \right) {x}^{4}+{\frac {1}{945}}\,{\pi }^{6}{x}^{5}+O
\left( {x}^{6} \right)
$$
$$
\Psi \left( x \right) =- \left( x+1 \right) ^{-1}+1-\gamma+ \left( 1+
1/6\,{\pi }^{2} \right) \left( x+1 \right) + \left( 1-\zeta \left( 3
\right) \right) \left( x+1 \right) ^{2}+ \left( {\frac {1}{90}}\,{
\pi }^{4}+1 \right) \left( x+1 \right) ^{3}+ \left( 1-\zeta \left( 5
\right) \right) \left( x+1 \right) ^{4}+ \left( {\frac {1}{945}}\,{
\pi }^{6}+1 \right) \left( x+1 \right) ^{5}+O \left( \left( x+1
\right) ^{6} \right)
$$
$$
\Psi \left( x \right) =- \left( x+2 \right) ^{-1}+{\frac {3}{2}}-
\gamma+ \left( {\frac {5}{4}}+1/6\,{\pi }^{2} \right) \left( x+2
\right) + \left( {\frac {9}{8}}-\zeta \left( 3 \right) \right)
\left( x+2 \right) ^{2}+ \left( {\frac {1}{90}}\,{\pi }^{4}+{\frac {
17}{16}} \right) \left( x+2 \right) ^{3}+ \left( {\frac {33}{32}}-
\zeta \left( 5 \right) \right) \left( x+2 \right) ^{4}+ \left( {
\frac {1}{945}}\,{\pi }^{6}+{\frac {65}{64}} \right) \left( x+2
\right) ^{5}+O \left( \left( x+2 \right) ^{6} \right)
$$
$$
\Psi \left( x \right) =- \left( x+3 \right) ^{-1}+{\frac {11}{6}}-
\gamma+ \left( {\frac {49}{36}}+1/6\,{\pi }^{2} \right) \left( x+3
\right) + \left( {\frac {251}{216}}-\zeta \left( 3 \right) \right)
\left( x+3 \right) ^{2}+ \left( {\frac {1}{90}}\,{\pi }^{4}+{\frac {
1393}{1296}} \right) \left( x+3 \right) ^{3}+ \left( {\frac {8051}{
7776}}-\zeta \left( 5 \right) \right) \left( x+3 \right) ^{4}+
\left( {\frac {1}{945}}\,{\pi }^{6}+{\frac {47449}{46656}} \right)
\left( x+3 \right) ^{5}+O \left( \left( x+3 \right) ^{6} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Complex Numbers: Finding solutions to $ (z^2-3z+1)^4 = 1 $ I have to find all solutions to
$$ (z^2-3z+1)^4 = 1 $$
What I thought could work was
$$z^2-3z+1= 1^{1/4} $$
Given that the 4 4th-roots of 1 are $1, i, -i, -1$ my idea was to look at each case separately. Starting with $1$ and with $z=a+bi \quad a,b \in R$:
$$z^2-3z+1= 1\\z^2-3z=0\\a^2+2abi-b^2=3a+3bi$$
From where you get
$$2ab=3b\ \to \ a=\frac 32\\a^2-b^2=3a\ \to \ \frac94-b^2=\frac92 \ \to \ b\notin R$$
So no possible solutions in this case.
I think the idea is okay but when I try to do the same with $i$:
$$z^2-3z+1= i\to (a^2-b^2-3a+1)+i(2ab-3b)=i $$
And I get:
$$a^2-b^2-3a+1 = 0\\2ab-3b=1 $$
Which can be solved but seems overly complicated...
Is what I've done correct? Any simpler ideas?
| Ron Gordon's comment on the first case is exactly the way to go for the real roots of unity: simply factor the expression. For the two complex roots, the quadratic formula should suffice. You may find it worthwhile to convert the complex radicands to polar form by
$$x+iy=re^{i\theta},\,\,\,r=\sqrt{x^2+y^2},\,\,\,\theta=\tan^{-1}\left(\frac{y}{x}\right).$$
This will make taking that square root a whole lot easier. For an example, let's take the second case: z^2-3z+(1+i)=0. The quadratic formula gives us
$$z=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(1-i)}}{2(1)}=\frac{3\pm\sqrt{9-(4-4i)}}{2}=\frac{3\pm\sqrt{5+4i}}{2}.$$
The radicand here is $5+4i$ and has magnitude $r=\sqrt{5^2+4^2}=\sqrt{41}$ and argument $\theta=\tan^{-1}\left(\frac{4}{5}\right)$. So
$$5+4i=\sqrt{41}e^{i\tan^{-1}\left(\frac{4}{5}\right)}.$$
Then the square root is simply
$$\sqrt{5+4i}=(5+4i)^{\frac{1}{2}}=\sqrt[4]{41}e^{\frac{1}{2}i\tan^{-1}\left(\frac{4}{5}\right)}.$$
Substitute back into the quadratic formula to obtain those two roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If n be a positive integer, then prove by binomial theorem that the integral part of $(7+4\sqrt3)^n$ is an odd number. I wanted to know, how can i prove this?
If $n$ be a positive integer, then prove by binomial theorem that the integral part of $(7+4\sqrt3)^n$ is an odd number.
| I'm working with the hypothesis that "integral part" means "floor".
$$\begin{align}
(7+4\sqrt{3})^n + (7 - 4\sqrt{3})^n &= \sum_{k=0}^n \binom{n}{k}7^{n-k}4^k\sqrt{3}^k + \sum_{k=0}^n \binom{n}{k}7^{n-k}(-1)^k4^k\sqrt{3}^k\\
&= 2\sum_{m=0}^{\lfloor n/2\rfloor} \binom{n}{2m}7^{n-2m}4^{2m}3^m
\end{align}$$
is even. $0 < 7 - 4\sqrt{3} < 1$. Thus the integral part of $(7+4\sqrt{3})^n$ is
$(7+4\sqrt{3})^n + (7-4\sqrt{3})^n - 1$, an odd number.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove that: $ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$ How to prove the following trignometric identity?
$$ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$$
Using half angle formulas, I am getting a number for $\cot7\frac12 ^\circ $, but I don't know how to show it to equal the number $\sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$.
I would however like to learn the technique of dealing with surds such as these, especially in trignometric problems as I have a lot of similar problems and I don't have a clue as to how to deal with those.
Hints please!
EDIT:
What I have done using half angles is this: (and please note, for convenience, I am dropping the degree symbols. The angles here are in degrees however).
I know that
$$ \cos 15 = \dfrac{\sqrt3+1}{2\sqrt2}$$
So,
$$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$
$$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$
$$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} $$
| using half angle identities,
$$\cot(x)=\frac{1}{\tan(x)}=\frac{\sin(2x)}{1-\cos(2x)}=\frac{\left(\frac{1-\cos(4x)}{2}\right)^{\frac{1}{2}}}{1-\left(\frac{1+\cos(4x)}{2}\right)^\frac{1}{2}}$$
with $x=7.5^o=\frac{\pi}{24}$ and $\cos(\frac{4\pi}{24})=\frac{\sqrt{3}}{2}$ we can substitute and simplify by multiplying through by $\sqrt{2}$ twice,
$$
\cot(\frac{\pi}{24})
=
\frac
{
\frac
{ \left( 1-\frac{\sqrt{3}}{2} \right) ^{\frac{1}{2}} }
{ ( {2} )^{\frac{1}{2}} }
}
{
1- \frac
{ \left( 1+\frac{\sqrt{3}}{2} \right) ^{\frac{1}{2}} }
{ ( {2} )^{\frac{1}{2}}
}
}
=
\frac
{
\frac
{ ( 2-\sqrt{3} ) ^{\frac{1}{2}} }
{ ( {2} ) ^{\frac{1}{2}} }
}
{
\sqrt{2} - \frac
{(2+\sqrt{3}) ^{\frac{1}{2}}}
{{(2)}^{\frac{1}{2}}}
}
=
\frac
{
( 2-\sqrt{3} ) ^{\frac{1}{2}}
}
{
2 - (2+\sqrt{3}) ^{\frac{1}{2}}
}
$$
to simplify from here, we could look to 'complete the square' under the radicals (leaving NO remainder) in order for the the powers $2$ and $1/2$ to multiply and cancel
look for $a, b$ such that $(a + b)^2 = 2+\sqrt{3}$ ideally (and similarly for the $2-\sqrt{3}$ radical), but note that any scaler multiple of this RHS would do as this multiple can be either factored out, i.e. completing for $k(2-\sqrt{3}) = (a-b)^2$ works if we replace $(2-\sqrt{3})$ with $\left(\frac{1}{\sqrt{k}}\right)^2(a-b)^2$ instead, or alternatively pulled in from outside of the radical, i.e. by multiplying the main equation through by the factor $\sqrt{k}$
this is the weakest part of the solution, as only trial and error led me to $k=2, a=1, b=\pm\sqrt{3}$
$$
\cot(\frac{\pi}{24})
=
\frac
{\left(\left(\frac{1}{\sqrt{2}}\right)^2(1-\sqrt{3})^2\right)^\frac{1}{2}}
{2-\left(\left(\frac{1}{\sqrt{2}}\right)^2(1+\sqrt{3})^2\right)^\frac{1}{2}}
=
\frac
{\frac{1}{\sqrt{2}}|1-\sqrt{3}|}
{2-\frac{1}{\sqrt{2}}(1+\sqrt{3})}
=
\frac
{|1-\sqrt{3}|}
{2\sqrt{2}-(1+\sqrt{3})}
$$
to simplify the (disgusting) modulus (which exists to maintain the exact radical solution), multiply through by its conjugate. being very careful with signs (positive $|1-\sqrt{3}|$ scaled by positive $(1+\sqrt{3})$ is positive) we have,
$$
\cot(\frac{\pi}{24})
=
\frac
{
|1-\sqrt{3}|(1+\sqrt{3})
}
{
2\sqrt{2}(1+\sqrt{3})-(1+\sqrt{3})^2
}
=\frac{2}{2\sqrt{2}+2\sqrt{6}-(4+2\sqrt{3})}
$$
finally, multiplying through by $\frac{1}{2}(\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6})$, the denominator simplifies to $1$, and we're left with your required solution
| {
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"source": "stackexchange",
"question_score": "23",
"answer_count": 8,
"answer_id": 7
} |
Solve $x^2+y^2=2$ for $x,y\in\mathbb Q$.
Solve $x^2+y^2=2$ for $x,y\in\mathbb Q$.
I think the answer should be in terms of 1 integer variable $\in\mathbb Z$ only. I rewrite the equation to $(x+y)^2+(x-y)^2=2^2$, then by the formula of pythagorean triples, $x+y=u^2-v^2,x-y=2uv,2=u^2+v^2$. How can I proceed? Thanks.
| Solutions to the equation $x^2+y^2=2$ with $x,y\in \mathbb{Q}$ can be parametrized by $$\left(x,y\right)=\left(\frac{1+2t-t^{2}}{1+t^{2}},\ \frac{1-2t-t^{2}}{1+t^{2}}\right),$$ where $t\in \mathbb{Q}$.
This follows by rewriting the equation as $$\left(\frac{x+y}{2}\right)^{2}+\left(\frac{x-y}{2}\right)^{2}=1,$$ and using the parametrization for $u^2+v^2=1$. Consequently, we see that $$\left(\frac{x+y}{2},\ \frac{x-y}{2}\right)=\left(\frac{1-t^{2}}{1+t^{2}},\ \frac{2t}{1+t^{2}}\right),$$ and so we arrive at $$\left(x,y\right)=\left(\frac{1+2t-t^{2}}{1+t^{2}},\ \frac{1-2t-t^{2}}{1+t^{2}}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/473280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
If $ \cos x +2 \cos y+3 \cos z=0 , \sin x+2 \sin y+3 \sin z=0$ and $x+y+z=\pi$. Find $\sin 3x+8 \sin 3y+27 \sin 3z$
Problem : If $ \cos x +2 \cos y+3 \cos z=0 , \sin x+2 \sin y+3 \sin z=0$ and $x+y+z=\pi$. Find $\sin 3x+8 \sin 3y+27 \sin 3z$
Solution: Adding $ \cos x +2 \cos y+3 \cos z=0$ and $\sin x+2 \sin y+3 \sin z=0$,we get
$ (\cos x+\sin x) +2(\cos y+\sin y)+3(\cos z+\sin z) =0$
Am I doing right ?
| Hint: use complex exponentials.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Factoring $a^3-b^3$ I recently got interested in mathematics after having slacked through it in highschool. Therefore I picked up the book "Algebra" by I.M. Gelfand and A.Shen
At problem 113, the reader is asked to factor $a^3-b^3.$
The given solution is:
$$a^3-b^3 = a^3 - a^2b + a^2b -ab^2 + ab^2 -b^3 = a^2(a-b) + ab(a-b) + b^2(a-b) = (a-b)(a^2+ab+b^2)$$
I was wondering how the second equality is derived. From what is it derived, from $a^2-b^2$? I know that the result is the difference of cubes formula, however searching for it on the internet i only get exercises where the formula is already given. Can someone please point me in the right direction?
| Being able to quickly derive the factorization for a difference of cubes (and most other things) prevented me from having to memorize many things in school. I cannot help but share this with you as it was something I must have done a hundred times in a pinch. I will answer your question by delivering a proper full derivation of the standard factorization for a difference of cubes. Close inspection of the result will answer your question I am sure. Begin by expanding $(a-b)^3$ in any manner you choose.
$$(a-b)^3 = a^3-3a^2b+3ab^2-b^3.$$
We will now solve this equality for $a^3-b^3$ and manipulate the other side of the equality into the standard formula.
\begin{align*}
(a-b)^3 &= a^3-3a^2b+3ab^2-b^3 \\
\Rightarrow a^3-b^3 &=(a-b)^3+3a^2b-3ab^2 \\
&=(a-b)^3+3ab(a-b) \\
&=(a-b)\left((a-b)^2+3ab\right) \\
&=(a-b)(a^2-2ab+b^2+3ab) \\
&=(a-b)(a^2+ab+b^2).
\end{align*}
This is one handy derivation. The sum of cubes can be derived in a similar manner.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/484281",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find all functions $f(x+y)=f(x^{2}+y^{2})$ for positive $x,y$ Find all functions $f:\mathbb{R}^{+}\to \mathbb{R}$ such that for any $x,y\in \mathbb{R}^{+}$ the following holds:
$$f(x+y)=f(x^{2}+y^{2}).$$
| For any positives $a,b$ we have:
\begin{aligned} f(a) &=f\left(\frac{a+b}{2}+\frac{a-b}{2}\right) \\
&=f\left(\big(\frac{a+b}{2}\big)^2+\big(\frac{a-b}{2}\big)^2\right) \\
&=f\left(\big(\frac{a+b}{2}\big)^2+\big(\frac{b-a}{2}\big)^2\right) \\
&=f\left(\frac{a+b}{2}+\frac{b-a}{2}\right) \\
&=f(b)\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 3
} |
Solving a radical equation $\sqrt{x+1} - \sqrt{x-1} = \sqrt{4x-1}$ $$
\sqrt{x+1} - \sqrt{x-1} = \sqrt{4x-1}
$$
How many solutions does it have for $x \in \mathbb{R}$?
I squared it once, then rearranges terms to isolate the radical, then squared again.
I got a linear equation, which yielded $x = \frac54$, but when I put that back in the equation, it did not satisfy.
So I think there is no solution, but my book says there is 1.
Can anyone confirm if there is a solution or not?
| As $(x+1)-(x-1)=2$ and given that $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}\ \ \ \ (1)$
$$\sqrt{x+1}+\sqrt{x-1}=\frac2{\sqrt{4x-1}}\ \ \ \ (2)$$
On addition, $$2\sqrt{x+1}=\sqrt{4x-1}+\frac2{\sqrt{4x-1}}=\frac{4x+1}{\sqrt{4x-1}}$$
$$\implies 2\sqrt{x+1} \sqrt{4x-1}=4x+1 \ \ \ \ (3)$$
Squaring we get $4x=5\iff x=\frac54$ which does not satisfy $(1)$ (the given equation) and $(2)$ but satisfies $(3)$
In fact, $x=\frac54$ is a root of $\sqrt{x+1}+\sqrt{x-1}=\sqrt{4x-1}$ and is an Extraneous root of $(1),(2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/486484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Does the derivative of $x^{-1}$ and of $x^3-x$ equal $-\frac{1}{x^{2}}$ and $3x^2-1$? I want to check my answers concerning the derivative of the following functions: $\displaystyle f(x)= \frac{1}{x}$ and of $\displaystyle j(x)= x^3-x$
$$\displaystyle f(x)= \frac{1}{x}$$
$$\begin{align}
f'(x) & = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\
& = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} \\
& = \lim_{h \to 0} \frac{\frac{x}{x(x+h)} - \frac{(x+h)}{x(x+h)}}{h} \\
& = \lim_{h \to 0} \frac{\frac{x - (x+h)}{x^2+hx}}{h} \\
& = \lim_{h \to 0} \frac{\frac{-h}{x^2+hx}}{h} \\
& = \lim_{h \to 0} \frac{-h}{x^2+hx} \times \frac{1}{h} \\
& = \lim_{h \to 0} \frac{-1}{x^2+hx} \\
& = -\frac{1}{x^2}
\end{align}$$
$$j(x) = x^3 - x$$
$$\begin{align}
j'(x) & = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\
& = \lim_{h \to 0} \frac{(x+h)^3 - x^3 - x}{h} \\
& = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x - h - x^3 + x}{h} \\
& = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3 - h}{h} \\
& = \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2 - 1)}{h} \\
& = \frac{3x^2 - 1}{1} \\
& = 3x^2 - 1
\end{align}$$
Please, be rude, if you see any error correct me.
| The title of your post says the derivative of $\frac 1x$ is $\frac{1}{x^2}$ but in your post, you found that the derivative is $-\frac{1}{x^2}$, which is correct.
Also, in the title you wrote that the derivative of $x^3-x$ is $3x^2$, but you found that it is $3x^2-1$
Your calculations are correct, keep up the good work!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/488787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How prove this is an equilateral triangle in $\Delta ABC$,$AB=c,AC=b,BC=a$and such
$$ab^2\cos{A}=bc^2\cos{B}=ca^2\cos{C}$$
show that
$\Delta ABC$ is an equilateral triangle
this problem I have solution,But not nice, and I think this problem have more nice methods,Thank you everyone.
my solution:
$$ab^2\cdot\dfrac{b^2+c^2-a^2}{2bc}=bc^2\cdot\dfrac{a^2+c^2-b^2}{2ac}=ca^2\cdot\dfrac{a^2+b^2-c^2}{2ab}$$
$$\Longrightarrow a=b=c$$
My other idea:
$$\Longleftrightarrow\dfrac{\sin{B}}{\sin{C}}\cos{A}=\dfrac{\sin{C}}{\sin{A}}\cos{B}=\dfrac{\sin{A}}{\sin{B}}\cos{C}$$
$$\Longleftrightarrow \sin{(2A)}\sin{B}=\sin{(2B)}\sin{C}=\sin{(2C)}\sin{A}$$
then How prove
$$A=B=C$$
and have other nice methods? Thank you
| Dividing throughout by $abc$, we get
$$ \frac{b^2+c^2 - a^2}{c^2} = \frac{a^2 + c^2 - b^2}{a^2} = \frac{a^2+b^2-c^2}{b^2}.$$
Subtracting 1 from each term,
$$ \frac{b^2- a^2}{c^2} = \frac{ c^2 - b^2}{a^2} = \frac{a^2-c^2}{b^2}.$$
Applying Componendo et dividendo, these fractions are equal to
$$ \frac{ b^2-a^2+c^2-b^2+a^2-c^2} { c^2 + a^2 + b^2}$$
Hence, all of them are 0, which means $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find min $P$: $P=\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b)$ Let $a,b,c\geq 0$ and $a+b+c=1$. Know that never have two numbers both zero. Find min $P$:
$$P=\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b)$$
| \begin{align}
P=&\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b)
\\
=&\frac{1}{1-c}\left(\frac{1}{b+c}+\frac{1}{c+a}\right)+(c+1)(4-c)
\\
\ge&\frac{1}{1-c}\frac{4}{(b+c)+(c+a)}+(c+1)(4-c)
\\
=&\frac{4}{1-c^2}+(c+1)(4-c)=:f(c),
\end{align}
the equality holds iff $a=b$.
$$f'(c)=-2c+3+\frac{8c}{(c^2-1)^2}>0,\quad\textrm{ when }0\le c\le1.$$
So $\min f(c)=f(0)=8$. Hence, $\min P=\min f(c)=8$, when $a=b=1/2$ and $c=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$, we can write the polar equation of the ellipse as $r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}$. And we can find the area enclosed by a curve $r(\theta)$ by integrating
$$\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.$$
So we should be able to find the area of the ellipse by
$$\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta$$
$$= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta$$
$$= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta$$
$$= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}$$
$$=\pi\frac{a^2+b^2}{2}.$$
First of all, this is not the area of an ellipse. Second of all, when I plug in $a=1$, $b=2$, this is not even the right value of the integral, as Wolfram Alpha tells me.
What am I doing wrong?
| Here you go - this person even made your mistake, then someone else corrected it.
Link
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Show that $x(x+1) = y^4+y^3+ay^2+by+c$ has a finite number of positive integral solutions. More precisely,
If $a$, $b$, and $c$
are integers,
show that
there are only a finite number
of positive integers $x$ and $y$
such that
$x(x+1) = y^4+y^3+ay^2+by+c$.
I have a solution,
which I will show in two days
if no better one is found.
| Here is my answer.
Interestingly, it uses no divisibility properties.
We are looking at
$x(x+1) = y^4+y^3+ay^2+by+c$.
I will show that,
if $a$, $b$, and $c$ are integers,
there are at finite number of solutions in
positive integral $x$ and $y$.
This will be done
by finding bounds for $y$
in terms of $a$, $b$, and $c$.
Multiplying by 4,
$(2x+1)^2-1
=4y^4+4y^3+4ay^2+4by+4c
$
or
$(2x+1)^2
=4y^4+4y^3+4ay^2+4by+d
$,
where
$d = 4c+1$.
My goal is to show algebraically
that this polynomial in $y$
is between two consecutive
integer squares
for large enough $y$.
First,
$(2y^2+y)^2
=4y^4+4y^3+y^2
$.
Next,
$\begin{align}
(2y^2+y+a)^2
&=4y^4+4y^3+y^2 +2a(2y^2+y)+a^2 \\
&=4y^4+4y^3+(4a+1)y^2+2ay+a^2 \\
\end{align}
$.
Finally,
$\begin{align}
(2y^2+y+a-1)^2
&=4y^4+4y^3+y^2
+2(a-1)(2y^2+y)+(a-1)^2 \\
&=4y^4+4y^3+(4a-3)y^2+(2a-2)y+(a-1)^2 \\
\end{align}
$.
Since
$(2x+1)^2 =4y^4+4y^3+4ay^2+4by+d
$,
for
$(2x+1)^2$
to be between these consecutive squares,
we need
$(4a-3)y^2+(2a-2)y+(a-1)^2
<4ay^2+4by+d
<(4a+1)y^2+2ay+a^2
$.
I will now show
that both inequalities are true
for large enough $y$.
The first inequality is the same as
$0
<3y^2+(4b-2a+2)y+d-(a-1)^2
$
or
$0
<9y^2+6(2b-a+1)y+3(d-(a-1)^2)
$
or
$0 <
(3y-(2b-a+1))^2-(2b-a+1)^2 +3(d-(a-1)^2)
$
or
$ (3y-(2b-a+1))^2
>(2b-a+1)^2 -3(d-(a-1)^2)
$
and this is certainly true for
$y$ large enough.
For the second inequality to be true,
we need
$4ay^2+4by+d
<(4a+1)y^2+2ay+a^2
$
or
$y^2+(2a-4b)y+a^2-d
> 0
$
or
$(y+(a-2b))^2-(a-2b)^2+a^2-d
> 0
$
or
$(y+(a-2b))^2
>(a-2b)^2-a^2+d
=a^2-4ab+4b^2-a^2+d
=4b^2-4ab+d
$.
This is true for
$y$ large enough,
so the equation has no solutions for large enough $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that for each positive integer m, the number $9\cdot 2^m$ can be written as a sum of three squares of positive integers. Prove that for each positive integer $m$, the number $9\cdot2^m$ can be written as a sum of three squares of positive integers.
I think that induction might be used here.!! I could write the first step, i.e true for $m=1$ but couldn't proceed further. This might be solved in a different way.
| We have $$9\cdot 2^1=18=4^2+1^2+1^2$$ and $$9\cdot 2^2=36=4^2+4^2+2^2.$$ Now, if $$9\cdot 2^m=x^2+y^2+z^2,$$ then $$9\cdot 2^{m+2}=4(x^2+y^2+z^2)=(2x)^2+(2y)^2+(2z)^2,$$ and we are done.
(Note the result also holds with $m=0$, as $9=2^2+2^2+1$. And, yes, if needed, the argument can be formalized as an induction that has two base cases, and deals differently with even and odd values of $m$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Expressing a $3\times 3$ determinant as the product of four factors I am attempting to express the determinant below as a product of four linear factors
$$\begin{vmatrix}
a & bc & b+c\\
b & ca & c+a\\
c & ab & a+b\\
\end{vmatrix}
=
a\begin{vmatrix}
ca & c+a\\
ab & a+b\\
\end{vmatrix}
-
bc\begin{vmatrix}
b & c+a\\
c & a+b\\
\end{vmatrix}
+(b+c)\begin{vmatrix}
b & ca\\
c & ab\\
\end{vmatrix}$$
This is as far as I get before it gets too messy
$$
=a^3(c-b)-bc\{(b-c)(b+c)+a(b-c)\}+a(b-c)(b+c)^2
$$
But I cant seem to arrive at the answer in the book, which is given as
$$
(a-b)(b-c)(c-a)(a+b+c)
$$
Am I doing something wrong as I have been stuck on this question for three days.
Thanks in advance!
$$
$$
| $$\begin{vmatrix}
a & bc & b+c\\
b & ca & c+a\\
c & ab & a+b
\end{vmatrix}
=
\begin{vmatrix}
a & bc & b+c\\
b -a & c(a-b)& a-b\\
c -a& b(a-c)& a-c
\end{vmatrix}=(a-b)(c-a)
\begin{vmatrix}
a & bc & b+c\\
-1&c&1\\
1&-b&-1
\end{vmatrix}$$
$$=(a-b)(c-a)\begin{vmatrix}
a& bc& b+c\\
-1&c&-1\\
0&b-c&0
\end{vmatrix}
=(a-b)(c-a)(b-c)(a+b+c)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $3|n(n+1)(n+2)$ by induction I tried proving inductively but I didn't really go anywhere. So I tried:
Let $3|n(n+1)(n+2)$.
Then $3|n^3 + 3n^2 + 2n \Longrightarrow 3|(n(n(n+3)) + 2)$
But then?
| Here's a somewhat tedious (but fairly elementary) way. By the division algorithm, we can write $n=3k+r$ for some integers $k,r$ with $0\le r\le 2,$ whence $$\begin{align}n(n+1)(n+2) &= (3k+r)(3k+r+1)(3k+r+2)\\ &= (3k+r)(9k^2+6kr+9k+r^2+3r+2)\\ &= 27k^3+27k^2r+27k^2+9kr^2+18kr+6k+r^3+3r^2+2r\\ &= 3(9k^3+9k^2r+9k^2+3kr^2+6kr+2k+r^2)+r^3+2r.\end{align}$$ All that's left is to show that $r^3+2r$ is a multiple of $3$ for $r=0,1,2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
a - b > 0 algebra correction Simple Algebra
$\frac{1}{2}-\frac{1}{5} > 0$
$\frac{1}{2} > \frac{1}{5}$
$5 >2$
Looks correct, but where am I wrong in this,
$\frac{1}{2}-\frac{1}{5} > 0$
$-\frac{1}{5} > -\frac{1}{2}$
$\frac{1}{5} > \frac{1}{2}$
$ 2 > 5$
| $-1>-2$, but does that imply that $1>2$? In general, if $a<b$ then $-a>-b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Complex Integral with exponential I've been struggling with this:
$$\int_{0}^{\infty }\frac{e^{-px}}{x^{2}+1}\mathrm{d}x, \; \; p\ge 0.$$
| \begin{align*}
\int_0^\infty \frac{e^{-px}}{x^2 + 1}dx &\overset{(1)}{=} \int_0^\infty \int_0^\infty e^{-px} e^{-sx} \sin(s)ds dx \\
&\overset{(2)}{=} \int_0^\infty \int_0^\infty e^{-(p+s)x} \sin (s)dx ds\\
&\overset{(3)}{=} \int_0^\infty \frac{\sin(s)}{(p+s)} ds \\
&\overset{(4)}{=} \text{Ci}(p) \sin (p)+\frac{1}{2} (\pi -2 \text{Si}(p)) \cos (p)\\
\end{align*}
$\displaystyle(1): \int_0^{\infty} e^{-sx} \sin(s)dx = \frac{1}{1+x^2}$
$(2):$ change of order of integration.
$\displaystyle(3): \int_{0}^\infty e^{-(p+s)x} dx = \frac{1}{(p+s)}$
$(4):$
\begin{align*}
\int_0^\infty \frac{\sin(s)}{(p+s)} ds &= \int_p^\infty \frac{\sin(y - p)}{y }dy \\
&= \int_p^\infty \frac{\cos(p)\sin(y) - \cos(y) \sin(p)}{y }dy \\
&= - \sin(p) \int_p^\infty \frac{\cos(y)}{y}dy + \cos(p)\int_p^\infty \frac{\sin(y)}{y}dy\\
&= \sin(p) \text{Ci}(p) + \cos(p) \left( \int_0^{\infty } \frac{\sin(y)}{y}dy - \int_0^{p } \frac{\sin(y)}{y}dy\right )\\
&= \text{Si}(p)\cos(p) + \frac \pi 2 \cos(p) - \sin(p) \text{Ci}(p)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is the probability that the student knew the answer to at least one of the two questions? A student takes a true-false examination containing 20
questions. On looking at the examination the student and that he
knows the answer to 10 of the questions which he proceeds to answer
correctly. He then randomly answers the remaining 10 questions. The
instructor selects 2 of the questions at random and that the students answered both questions correctly.
What is the probability that the student knew the answer to at least one of the two questions?
| The probability of the teacher selecting 2 question, for which the student don't know the answer is: $\frac{10}{20} \times \frac{9}{19}$. There are $\frac{1}{2}$ chance that he'll know the answer so for the both question the probability is:
$$\frac{10}{20} \times \frac{9}{19} \times \frac{1}{4} = \frac{9}{152}$$
The probability of the teacher selecting 1 question that he don't know and one that the student know is $\frac{10}{20} \times \frac{10}{19}$ and we know that there are $\frac 12$ chance to guess the answer right. So the probability is:
$$\frac{10}{20} \times \frac{10}{19} \times \frac{1}{2} = \frac{5}{38}$$
The probability of the teacher selecting 2 questions that the student knows the answers of is $\frac{10}{20} \times \frac{9}{19}$
$$\frac{10}{20} \times \frac{9}{19} = \frac{9}{38}$$
So the total probability is:
$$\frac{9}{152} + \frac{5}{38} + \frac {9}{38} = \frac{65}{152}$$
The probability of selecting at least one question that the student knows is:
$$\frac{5}{38} + \frac {9}{38} = \frac{7}{19}$$
Now divide them and you'll get:
$$\frac{\frac{7}{19}}{\frac{65}{152}} = \frac{56}{65} \approx 86.15\% $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Help with the Integral of $x \arcsin x\,dx$ So I've started the integration using integration by parts, but I'm stuck on the second part of it.
$$ \int_0^\frac{1}{2} x \sin^{-1}x dx $$
*
*Step 1. Set up my variables
$$ u = \arcsin x $$
$$du = \frac{1}{\sqrt{1-x^{2}}} dx $$
$$ dv = x dx $$
$$ v = \frac{x^{2}}{2} $$
*Step 2. $uv - \int v\,du$
$$ \frac{x^2}{2} \arcsin x - \frac{1}{2} \int x^{2}\frac{1}{\sqrt{1-x^{2}}}$$
*Step 3. Now I need to integrate $ \frac{x^2}{\sqrt{1-x^2}} $
This is where I'm stuck. Do I do a trig sub? Let $x = \sin\theta$ which $1-\sin^2\theta$ would then equal $\sqrt{\cos^2\theta}$ turning the second part into $\int \sec \theta$?
Edit. Attempting to continue the problem with the suggestions provided.
*
*Step 3 (continued). $$ \frac{-1}{2} \int \frac{x^2}{\sqrt{1-x^2}}dx $$
$$ \text{Let }x = \sin \theta $$
$$ -\frac{1}{2} \int \frac{\sin^2\theta \cos\theta}{\sqrt{1-\sin^2\theta}} $$
$$ -\frac{1}{2} \int \frac{\sin^2\theta \cos\theta}{\sqrt{\cos^2\theta}} $$
$$ -\frac{1}{2} \int \frac{\sin^2\theta \cos\theta}{\cos\theta} $$
$$ -\frac{1}{2} \int \sin^2\theta $$
$$ -\frac{1}{2} \int \frac{1 - \cos2\theta}{2} $$
$$ -\frac{1}{4} [\theta-\sin\theta \cos\theta] $$
*Step 4. Find what $\theta$ is equal to in terms of x.
Since we let $x = \sin\theta$, that would mean we have $$x^2 + b^2 = 1^2$$
So, does that mean I substitute $$ 1 - x$$ in for $\theta$?
| If you want to continue with step (3), then you can use integration by parts again with $u=x$
$$ \int \frac{x^2}{\sqrt{1-x^2}} dx = -x\sqrt {-{x}^{2}+1}+\int \!\sqrt {1-{x}^{2}}{dx}.$$
For the last integral, you can use the trig. subs $x=\sin(\theta)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplest or nicest proof that $1+x \le e^x$ The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is accessible to students with limited calculus experience, and does not involve too much analysis of different cases.
| Repeatedly using $1 + x \le \left(1 + \frac{x}{2} \right)^2$, we have
\begin{align}
1 + x
\le
\left(1 + \frac x 2\right)^2
\le
\left(1 + \frac x 4\right)^4
\le
\left(1 + \frac x 8\right)^8
\le
\dots
\le
\left(1 + \frac x {2^k}\right)^{2^k}.
\end{align}
Taking the limit of $k \rightarrow \infty$ yields
$$
1 + x \le e^x. \qquad\qquad(1)
$$
Another proof using the technique in this post. By the AM-GM inequality,
$$
\sqrt[n]{1 \times \cdots \times 1 \times (1 + x)}
\le
\frac{1 + \dots + 1 + (1 + x)}{n}
=1 + \frac{x}{n}.
$$
So,
$$
1+x \le \left(1 + \frac{x}{n} \right)^n.
$$
Taking the limit of $n \rightarrow \infty$ yields (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/504663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "113",
"answer_count": 27,
"answer_id": 11
} |
Limit $ \sqrt{2\sqrt{2\sqrt{2 \cdots}}}$
Find the limit of the sequence $$\left\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots\right\}$$
Another way to write this sequence is $$\left\{2^{\frac{1}{2}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}}2^{\frac{1}{8}},\hspace{5 pt} \dots,\hspace{5 pt}2^{\frac{1}{2^{n+1} - 2}}\right\}$$
So basically we have to find $\lim_{n\to\infty}2^{\frac{1}{2^{n+1} - 2}}$. This equates to $$\lim_{n\to\infty}2^{\frac{1}{2^{\infty+1} - 2}} \Longrightarrow 2^{\frac{1}{2^{\infty}}} \Longrightarrow 1$$
Is this correct? P.S. Finding that $S(n)$ was a pain!
| $$\frac 1 2 + \frac 1 4 + \frac 1 8 + \dots = 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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calculate $x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^{6}+1$ given $(x+x^{-1})^2 = 3$ If $\left(x+\dfrac 1 x\right)^2=3$ then the value of $$x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^{6}+1.$$
I'm trying to solve it like this $$x^2+\dfrac {1}{x^2}=1\text{ and }; x^6+\dfrac {1}{x^6}=-2$$
then solve the expression like this
$$(x^6+1)(x^{200}+x^{84}+x^{12}+1).$$
from here I am stuck.
Thanks in advance.
| Since, we have
$x^2=x^4+1 \implies x^4=x^2-1 \implies x^8 = -x^2$.
Then, we get
$x^6 = x^2 .x^4= -1 $
$x^{12} = x^6 . x^6 = 1$
$x^{18} = x^{12}.x^6= (1)(-1)=-1$
$x^{84}= x^{(18).(4)}.x^{12}=(-1)^4 .(1)=1$
$x^{90}=x^{(18)(5)}=(-1)^5=-1$
$x^{200}=x^{(12)(16)}.x^6.x^2=(1)^{16}.(-1).x^2=-x^2$
$x^{206}=x^{200}.x^6=(-x^2).(-1)=x^2$
Therefore,
$x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1=x^2-x^2 -1+1-1+1-1+1=0.$
| {
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Find a closed form of $\sum_{i=1}^n i^3$ I'm trying to compute the general formula for $\sum_{i=1}^ni^3$. My math instructor said that we should do this by starting with a grid of $n^2$ squares like so:
$$
\begin{matrix}
1^2 & 2^2 & 3^2 & ... & (n-2)^2 & (n-1)^2 & n^2 \\
2^2 & 3^2 & 4^2 & ... & (n-1)^2 & n^2 & 1^2 \\
3^2 & 4^2 & 5^2 & ... & n^2 & 1^2 & 2^2 \\
\vdots &\vdots & \vdots&\ddots & \vdots & \vdots & \vdots \\
(n-2)^2 & (n-1)^2 & n^2 & ... & (n-5)^2 & (n-4)^2 & (n-3)^2 \\
(n-1)^2 & n^2 & 1^2 & ... & (n-4)^2 & (n-3)^2 & (n-2)^2 \\
n^2 & 1^2 & 2^2 & ... & (n-3)^2 & (n-2)^2 & (n-1)^2 \\
\end{matrix}
$$
And then sum the rows, and then sum all of the sums of rows. How would I then compute the general formula for $\sum_{i=1}^ni^3$?
| Notice
$$k^3 = k(k+1)(k+2) - k(3k+2) = k(k+1)(k+2) - 3k(k+1) + k$$
and following identities for positive integer $m$:
$$\sum_{k=1}^n k(k+1)(k+2)\cdots(k+m-1) = \frac{1}{m+1} n(n+1)(n+2)(n+3)\cdots(n+m)$$
One get
$$\begin{align}
\sum_{k=1}^n k^3 = & \frac{1}{4}n(n+1)(n+2)(n+3) - n(n+1)(n+2) + \frac12 n(n+1)\\
= & \frac{1}{4}n(n+1)((n+2)(n+3) - 4(n+2) + 2)\\
= & \frac{1}{4}\left(n(n+1)\right)^2\\
= & \left(\sum_{k=1}^n k\right)^2
\end{align}$$
| {
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Help checking proof of reverse triangle inequality $|x| - |y| \le |x + y|$? Let $x, y \in \mathbb{R}$. Prove $|x| - |y| \le |x + y|$.
By the the triangle inequality $|x| + |y| \ge |x + y|$, hence
$$
\begin{align}
&|y| \ge |x+y| - |x| \\
&|x+y| \ge |x+y| - |y| \\
\end{align}
$$
Subtracting the first inequality from the second, we have
$$
\begin{align}
&|x+y|-|y| \ge |x+y| - |y| - (|x+y| - |x|) = |x| - |y| \\
&|x+y| \ge |x+y| - |y| \ge |x| - |y| \\
&|x+y| \ge |x|- |y| \\
\end{align}
$$
The mistake in this proof is that subtracting the inequalities isn't generally valid. For example, $101 \ge 100$ and $100 \ge 1$ but $101 - 100 = 1 \not\ge 99$.
Taking someone's hint, a shorter proof is
$$|x| = |x + y - y| = |x + y + (-y)| \le |x + y| + |-y| = |x + y| + |y| $$
by the triangle inequality. Then
$$
\begin{align}
&|x| \le |x+y| +|y| \\
&|x| - |y| \le |x + y|
\end{align}
$$
| Sorry, I can't do comments yet.
Your first line after subtracting inequalities is incorrect.
$$|x+y|−|y|≥|x|−|y|$$
if $x=1/2$, $y=-1/2$ we get $-1/2 \ge 0$
We can't subtract inequalities like that, the inequality we are subtracting will be reversed.
e.g. $3>2$ and $3>1$ so $0=3-3 \not> 2-1=1$
but $2=3-1>2-3=-1$
| {
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Prove that $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq 27$ How can I prove $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq 27$, given that $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 9$ and $x+y+z=1$.
I've already tried using that: $\frac{1}{x} +\frac{1}{y} +\frac{1}{z}\geq 9$ But I can't seem to manipulate that to prove the above.
| Cauchy-Schwarz inequality tells us that
\begin{equation}
\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2 \leq 3\times \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\qquad (\star)
\end{equation}
But the left-hand term is $\geq 9^2$, so
$$
\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq \frac{9^2}{3} = 27.
$$
Note that in this particular case, it very is easy to prove $(\star)$. Let
$$
A = 3 \left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right),\qquad B = \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2.
$$
We need to show that $A \geq B$. Taking their difference,
\begin{align}
A - B & = \frac{1}{x^2}+\frac{1}{y^2} - 2\frac{1}{xy} + \frac{1}{y^2} + \frac{1}{z^2} - 2\frac{1}{yz} + \frac{1}{z^2}+\frac{1}{x^2} - 2\frac{1}{zx}\\
& = \left(\frac{1}{x}-\frac{1}{y}\right)^2 + \left(\frac{1}{y}-\frac{1}{z}\right)^2 + \left(\frac{1}{z}-\frac{1}{x}\right)^2\\
& \geq 0.
\end{align}
| {
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How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$ let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}
{5}\cdots (1)$$
I find sometimes,and I find this same problem:
let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$ \frac{1}{5a^2-4a+11}+\frac{1}{5b^2-4b+11}+\frac{1}{5c^2-4c+11}\leq\frac{1}{4} $$
and this problem have some methods,you can see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=223910&start=20
and I like this can_hang2007 methods and Honey_S methods,But for $(1)$ I can't prove it.Thank you
| Let $f(x) = \dfrac{1}{x^2+(3-x)^2}$. WLOG let $a \le b \le c$.
We note: $f(x) = f(3-x)$ and
if $x < 0$, then $f(x) < f(-x)$ as is obvious from signs or from $f(x)-f(-x) = \dfrac{12x}{4x^4+81}$. Using these, if $a < 0$, we also note
$$f(a)+f(b)+f(c) < f(-a) + f(3-b)+f(3-c)$$
where the new arguments also fulfil the constraint.
Thus if $a < 0$, to prove the inequality holds for $(a, b, c)$, it is sufficient to show it holds for $(-a, 3-b, 3-c)$. If this also has negative variables, then one more application turns $(-a, 3-b, 3-c) \to (3+a, b, c-3)$, where the lower most value has increased by $3$ and the largest has decreased by $3$.
Using successive application of this as necessary, it is sufficient to consider cases where $a \ge 0$, i.e. where all the variables are non-negative. Now we look at the following version of the inequality,
$$-\sum_{cyc} \left(\frac{5}{2a^2-6a+9} - 1 \right) = \sum_{cyc} \frac{2(a-1)(a-2)}{2a^2-6a+9}$$
$$\frac{2(a-1)(a-2)}{2a^2-6a+9} = -\frac{2(a-1)}{5} + \frac{2(a-1)^2(2a+1)}{5(2a^2-6a+9)} \ge -\frac{2(a-1)}{5}$$
$$\implies -\sum_{cyc} \left(\frac{5}{2a^2-6a+9} - 1 \right) \ge \sum_{cyc}-\frac{2(a-1)}{5}=0 $$
Hence proved.
| {
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"source": "stackexchange",
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Is $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$ Question is to check if $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$
we have $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\prod \limits_{n=2}^{\infty}(\frac{n^2-1}{n^2})=\prod \limits_{n=2}^{\infty}\frac{n+1}{n}\frac{n-1}{n}=(\frac{3}{2}.\frac{1}{2})(\frac{4}{3}.\frac{2}{3})(\frac{5}{4}.\frac{3}{4})...$
In above product we have for each term $\frac{a}{b}$ a term $\frac{b}{a}$ except for $\frac{1}{2}$.. So, all other terms gets cancelled and we left with $\frac{1}{2}$.
So, $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\frac{1}{2}$.
I would be thankful if some one can assure that this explanation is correct/wrong??
I am solving this kind of problems for the first time so, it would be helpful if some one can tell if there are any other ways to do this..
Thank you
| In fact, Euler discovered that
$$\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty (1-z^2/n^2)$$
which we can rearrange to
$$\frac{\sin \pi z}{\pi z (1-z^2)} = \prod_{n=2}^\infty (1-z^2/n^2).$$
By comparison of both sides at $z=1$, your product is $1/2$.
| {
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"url": "https://math.stackexchange.com/questions/513053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
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} |
Infinite Series $\sum_{k=1}^{\infty}\frac{1}{(mk^2-n)^2}$ How can we prove the following formula?
$$\sum_{k=1}^{\infty}\frac{1}{(mk^2-n)^2}=\frac{-2m+\sqrt{mn}\pi\cot\left(\sqrt{\frac{n}{m}}\pi\right)+n\pi^2\csc^2\left(\sqrt{\frac{n}{m}}\pi\right)}{4mn^2}$$
What is the general method for finding sums of the form $\sum\limits_{k=1}^{\infty}\frac{1}{(mk^2-n)^\ell}, \ell\in\mathbb{N}$?
| This sum may be evaluated by considering the following contour integral in the complex plane:
$$\oint_C dz \frac{\pi \cot{\pi z}}{(m z^2-n)^2}$$
where $C$ is a rectangular contour that encompasses the poles of the integrand in the complex plane, up to $z=\pm \left ( N +\frac12\right)$, where we consider the limit as $N \to\infty$.
We note here that we assume that the ratio $n/m$ is not the square of an integer. Now, the contour integral is zero because the individual integrals along each piece of the contour cancel. On the other hand, the contour integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand. Working this out, we find that
$$\sum_{k=-\infty}^{\infty} \frac{1}{(m k^2-n)^2} = -\sum_{\pm}\operatorname*{Res}_{z=\pm \sqrt{n/m}} \frac{\pi \cot{\pi z}}{(m z^2-n)^2}$$
Since the pole is a double pole, we have
$$\sum_{\pm}\operatorname*{Res}_{z=\pm\sqrt{n/m}} \frac{\pi \cot{\pi z}}{(m z^2-n)^2} =\sum_{\pm} \frac{\pi}{m^2} \left [\frac{d}{dz} \frac{\cot{\pi z}}{(z\pm \sqrt{n/m})^2} \right ]_{z=\pm\sqrt{n/m}} $$
I assume that the reader can take the derivatives and do the subsequent algebra. I get for the sum
$$-\frac{\pi}{m^2} \frac{m}{2 n} \left [\sqrt{\frac{m}{n}}\cot{\left(\pi \sqrt{\frac{n}{m}} \right)}+ \pi \csc^2{\left(\pi \sqrt{\frac{n}{m}} \right)}\right ] $$
So we now have
$$\sum_{k=-\infty}^{\infty} \frac{1}{(m k^2-n)^2} = \frac{\pi}{2 m n} \left [\sqrt{\frac{m}{n}}\cot{\left(\pi \sqrt{\frac{n}{m}} \right)}+ \pi \csc^2{\left(\pi \sqrt{\frac{n}{m}} \right)}\right ] $$
We now exploit the evenness of the summand; the result is
$$\sum_{k=1}^{\infty} \frac{1}{(m k^2-n)^2} = -\frac{1}{2 n^2} + \frac{\pi}{4 m n} \left [\sqrt{\frac{m}{n}}\cot{\left(\pi \sqrt{\frac{n}{m}} \right)}+ \pi \csc^2{\left(\pi \sqrt{\frac{n}{m}} \right)}\right ] $$
The result to be proven follows.
For the general sum
$$\sum_{k=1}^{\infty} \frac{1}{(m k^2-n)^{\ell}}$$
where $\ell \gt 2$ and $\ell \in \mathbb{Z}$, we take the same approach. The residue is an $\ell-1$ derivative of the integrand.
| {
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What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots? What are the possible values of $a$ such that $f(x) = (x + a)(x + 1991) + 1$ has two integer roots?
$(x + a)(x + 1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$
This is of the form $ax^2 + bx + c$. Applying the quadratic formula $\left(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\right)$, we get, that the rooots must equal:
$\frac{-(1991 + a) \pm \sqrt{(1991 + a)^2 - 4(1991a + 1)}}{2} = \frac{-(1991 + a) \pm \sqrt{1991^2 + a^2 + 2\times1991a - 4\times1991a -4}}{2} = \frac{-(1991 + a) \pm \sqrt{(1991 - a)^2 - 2^2}}{2} = \frac{-(1991 + a) \pm \sqrt{(1991 - a - 2)(1991-a+2)}}{2}$
For the formula to yield an integer, the discriminant must be a perfect square or $0$. Clearly, the discriminant will be zero if $a = 1993$ or $a = 1989$. The only thing that remains to be shown is that these are indeed the only possible solutions.
I couldn't think of any way to do so. Are there any alternate solutions, perhaps some that involve more number theory and less algebra?
| You are right, $m^2-4$ cannot be a square for $m>2$. Hint: $m^2 - (m-1)^2 = 2m-1$.
| {
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Determine the value of $a$ and $b$ such that $(ax+3)/(bx+5)$ has $f (-2)= 1$ and $f '(-1) = 1/4$? I tried using the quotient rule but I honestly have no clue how to do this!
Can someone try to help?
$f(x)=\dfrac{ax+3}{bx+5}$.
Determine $a$ and $b$ such that $f(-2)=1$ and $f'(-1)=\frac14$.
| I’m assuming that despite the missing required parentheses, $f(x)$ is supposed to be
$$f(x)=\frac{ax+3}{bx+5}\;.$$
You do indeed want the quotient rule to differentiate $f$:
$$\begin{align*}
f\,'(x)&=\frac{(bx+5)(ax+3)'-(ax+3)(bx+5)'}{(bx+5)^2}\\\\
&=\frac{a(bx+5)-b(ax+3)}{(bx+5)^2}\\\\
&=\frac{abx+5a-abx-3b}{(bx+5)^2}\\\\
&=\frac{5a-3b}{(bx+5)^2}\;.
\end{align*}$$
Now you want $$1=f(-2)=\frac{-2a+3}{-2b+5}=\frac{3-2a}{5-2b}\tag{1}$$ and $$\frac14=f\,'(-1)=\frac{5a-3b}{(-b+5)^2}=\frac{5a-3b}{(5-b)^2}\;.\tag{2}$$
Multiply $(1)$ through by $5-2b$ to get $5-2b=3-2a$, and multiply $(2)$ through by $4(5-b)^2$ to get $(5-b)^2=4(5a-3b)$. Each of these can easily be solved for $a$ in terms of $b$:
$$\begin{align*}
a&=b-1\;,\text{ and}\\
a&=\frac1{20}\left(b^2+2b+25\right)\;.
\end{align*}\tag{3}$$
Now equate the righthand sides of $(3)$ to get a quadratic in $b$; you can solve that, and then use $b$ to find $a$.
| {
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Using Exponential Generating Functions on Counting Problems Is it possible to use exponential generating functions to solve problems where repetition is wanted?
For example, if I wanted to solve the following problem which wants distinct possibilities...
How many different 5-letter words can be formed from the letters from the word ABRACADABRA if duplicated letters are allowed but no letter can be used more times than it occurs in the word ABRACADABRA ?
5 distinct letters, A repeated five times, B twice, C once, D once, R twice.
Then, the function would be
$$ F(x) = (1 + \frac {x}{1!} + \frac {x^2}{2!} + \frac {x^3}{3!} + \frac {x^4}{4!} + \frac {x^5}{5!}) (1 + \frac {x}{1!} + \frac {x^2}{2!})^2 (1 + \frac {x}{1!})^2 $$
$$ F(x) = \frac {x^{11}}{120} + \frac {3 x^{10}}{40} + \frac {2 x^9}{5} + \frac {19 x^8}{12}+ \frac{289 x^7}{60}+\frac{331 x^6}{30}+\frac{2221 x^5}{120}+\frac {545 x^4}{24} + \frac{121 x^3}{6}+\frac{25 x^2}{2}+5 x+1 $$
Therefore, the answer would be $5!(\frac {1271}{120})$
However, how would I be able to apply it to the following problem?
How many 5-card hands (from an ordinary deck) have at least one card of each suit?
4 suits, repeated 13 times
$ 52*51*50*49*48 = 311,875,200 $ total possibilities
$$ G(x) = (\frac {x^{13}}{13!} + \frac{x^{12}}{12!}+ \frac{x^{11}}{11!} + \frac{x^{10}}{10!} + \frac{x^9}{9!} + \frac{x^8}{8!} + \frac{x^7}{7!} + \frac{x^6}{6!} + \frac{x^5}{5!} + \frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + \frac{x}{1!} + 1)^4 $$
Of course, this would just give me unique combinations, so it doesn't work. Is there an actual method to solve using exponential generating functions? Is it possible or even recommended?
| The generating function for the second problem will provide you with all possible 5-card hands. But we are interested in the case when we have at least one cord of each suit. So we need to make some restrictions. We already know the suit of 4 of the cards (because the conditon states they are different) and there are 4 options for the fifth card, so there are 4 different combinations if the cards were undistiguishable. But because they aren't, we need to multilply that number by the multinomial:
$$\binom{5}{2,1,1,1} = \frac{5!}{2!\cdot 1! \cdot 1! \cdot 1!} = 60$$
So the total number of combinations is:
$$60 \cdot 4 = 240 \text{ combinations}$$
This is actually the number of possible combinations with a pair of cards of the same suit. We can get the same exact result using the exponential generating function. Actually the exponential generating function is based on the same principle. You can read more about this here, especially on the $7^{th}$ page.
But before applying the exponential generating function, we need to limit the number of cards of one suit. From the condition we can obtain that the maximum number of cards per suit is 2 and the minimum number of cards per suit is 1. So the GF will be:
$$G(x) = \left(x + \frac {x^2}{2!}\right)^4 = x^4+2 x^5+\frac{3 x^6}{2}+\frac{x^7}{2}+\frac{x^8}{16}$$
The coefficient in fron to $x^5$ is 2. So the total number of combinations is:
$$2 \cdot 5! = 2 \cdot 120 = 240 \text{ combinations}$$
And we can see that we've obtained the same results using both methods. That's because, the exponetnial generating function is a more elegant, hidden and simplier version of the multinomial method.
| {
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Is it true that if $n$ is even then $\sum_{k=1}^{n}(n \bmod k)<\frac{8}{45}n^2$? Let $f(n,k)$ be the least non-negative integer such that $n\equiv f(n,k) \bmod k.$
$f(10,k)(k=1,2,\cdots,10)=0, 0, 1, 2, 0, 4, 3, 2, 1, 0.$
Hence $$\sum_{k=1}^{10}f(10,k)=1+2+4+3+2+1=13.$$
Question: Is it true that if $n$ is even then$$\sum_{k=1}^{n}f(n,k)<\frac{8}{45}n^2\tag?$$
This is true for $n<10^5,$ but not true for many odd integers, such as $11,23,29,35,47,53,59,\cdots$
Eidt: Does $$\lim_{n\to \infty}\frac{1}{n^2}\sum_{k=1}^{n}f(n,k)$$ exist?
| Apply $n\bmod k = n - k \big\lfloor \frac{n}{k} \big\rfloor$ to $$g(n) = \sum_{k=1}^n (n\bmod k)$$
to get $$g(n) = n^2 - \sum_{k=1}^n k \bigg\lfloor \frac{n}{k} \bigg\rfloor.$$
Introduce $$q(n) = \sum_{k=1}^n k \bigg\lfloor \frac{n}{k} \bigg\rfloor$$
and observe that $$q(n+1)-q(n) = (n+1) \bigg\lfloor \frac{n+1}{n+1} \bigg\rfloor
+ \sum_{k=1}^n k
\left(\bigg\lfloor \frac{n+1}{k} \bigg\rfloor
- \bigg\lfloor \frac{n}{k} \bigg\rfloor\right)
\\= n+1 + \sum_{d|n+1\atop d <n+1} d = \sigma(n+1).$$
Therefore $$q(n) = \sum_{k=1}^n \sigma(n).$$
Now recall that
$$\sum_{n\ge 1}\frac{\sigma(n)}{n^s} = \zeta(s)\zeta(s-1)
\quad\text{and}\quad
\mathrm{Res}\left(\zeta(s)\zeta(s-1); s=2\right) = \frac{\pi^2}{6}.$$
Hence by the Wiener-Ikehara theorem
$$ \sum_{k=1}^n \sigma(n) \sim \frac{\pi^2}{6} \frac{n^2}{2} = \frac{\pi^2}{12} n^2.$$
It follows that
$$ g(n) \sim \left(1-\frac{\pi^2}{12} \right) n^2$$
and the conjectured limit exists.
This approximation is quite good, e.g. we have $g(2000) = 708989$ and the approximation gives $710132.$
Even better we may use Mellin-Perron summation and include the pole at one which has residue $-1/2$,
$$\mathrm{Res}\left(\zeta(s)\zeta(s-1); s=1\right) = -\frac{1}{2}$$
plus a correction term to get
$$g(n) \sim \left(1-\frac{\pi^2}{12} \right) n^2 + \frac{1}{2} n - \frac{1}{2}\sigma(n).$$
This last approximation is excellent, it gives $708714$ for $n=2000$ and for $n=8000$ with exact value $g(8000)=11356914$ it gives $11356203.$ For $n=16000$, we have $g(16000) = 45437799$ and the approximation gives $45436549.$
Observe that
$$\frac{8}{45} \approx 0.1777777778
\quad\text{and}\quad 1-\frac{\pi^2}{12} \approx 0.1775329664$$
so the conjectured coefficient was very close to the asymptotic one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/519540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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} |
Use the Chinese remainder theorem to find the general solution of $x \equiv a \pmod {2^3}, \; x \equiv b \pmod {3^2}, \; x \equiv c \pmod {11}$ Help! Midterm exam is coming, but i still unable to solve this simple problem using the Chinese remainder theorem.
$$x \equiv a \pmod {2^3}, \quad x \equiv b \pmod {3^2}, \quad x \equiv c \pmod {11}.$$
| The general method to find such a number, is to solve for $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$, where the numbers in brackets represent a number modulo $8$, $9$, $11$. Since the number $(1,1,1) = 1$, it is necessary to solve only for two of them.
For $(1,0,0)$, we start off with $99 = (3, 0, 0)$. Since the first multiple of $3$ to exceed eight by 1, is 9, we then have $3\times 99 = (1,0,0)$.
For $(0,1,0)$, we start with $8\times 11 = (0,7,0)$. It then is a matter of multiplying this by $4$ to get $352 = (0,1,0)$
We could do the same for the next, but note that $(1,1,1)=1+792$, and thence $(0,0,1) = (1,1,1)-(1,0,0)-(0,1,0)$, gives $793 - 297 - 352 = 144$ This is indeed $(0,0,1)$.
So we find here that $a=297$, $b=352$ and $c=144$, the sum is $1 \pmod{8\cdot 9 \cdot 11}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/519930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove some divisibility results by induction Please hint me, I have two questions:
Prove by induction that:
1)
$$ {13}^n+7^n+19^n=39k,\,\, n\in\mathbb O$$
in which $\mathbb O$ is the set of odd natural numbers.
2)
$$ 5^{2n}+5^n+1=31t,~n\not=3k, $$ $n\in \mathbb N$
| For one, let $\displaystyle f(n): 13^n+7^n+19^n=39k$ holds true for $n=m$ where $m$ is odd positive integer
$\implies\displaystyle13^m+7^m+19^m=39k_1$
$\implies\displaystyle13^m+7^m+19^m=13k_2\implies 7^m+19^m=13k_3\ \ \ \ (1)$
Now for $n=m+2,$
$\displaystyle13^{m+2}+7^{m+2}+19^{m+2}\equiv7^m\cdot49+19^m\cdot361\pmod{13}\equiv-3(7^m+19^m)\pmod{13}\equiv0$ [using $(1)$]
$\displaystyle13^{m+2}+7^{m+2}+19^{m+2}-(13^m+7^m+19^m)=13^m(13^2-1)+7^m(7^2-1)+19^m(19^2-1)\equiv0\pmod3$
$\displaystyle\implies 13^{m+2}+7^{m+2}+19^{m+2}\equiv13^m+7^m+19^m\pmod3\equiv0$
$\displaystyle\implies 13^{m+2}+7^{m+2}+19^{m+2}$ is divisible by lcm$(13,3)=39$
if $f(n)$ holds for $n=m$
Show that $f(n)$ holds if $n=1$
For two, let $g(n):5^{2n}+5^n+1=31t$ holds true for $n=m$
Now, $5^{2m}+5^m+1=(5^2)^m+5^m+1\equiv(-6)^m+5^m+1\pmod{31}$
$\displaystyle\implies (-6)^m+5^m+1$ is divisible by $31$
Now for $n=m+3,$
$\displaystyle (-6)^{m+3}+5^{m+3}+1-\{(-6)^m+5^m+1\}=(-6)^m\{(-6)^3-1\}+5^m(5^3-1)=(-6)^m(-217)+5^m(124)\equiv0\pmod{31}$
$\displaystyle\implies (-6)^{m+3}+5^{m+3}+1\equiv(-6)^m+5^m+1\pmod{31}\equiv0$
Now show that $g(n)$ holds for $n=1,2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/522423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Problem finding in simple algebra It is given,
$$x= \sqrt{3}+\sqrt{2}$$
How to find out the value of $$x^4-\frac{1}{x^4}$$/
The answer is given $40 \sqrt{6}$ but my answer was not in a square-root form
I have done in thsi way:
$$x+ \frac{1}{x}= 2 \sqrt{3}$$
Then,
$$(x^2)^2-\left(\frac{1}{x^2}\right)^2= \left(x^2 + \frac{1}{x^2}\right)^2-2$$
But this way is not working. Where I am wrong?
| Note that $\cfrac 1x=\cfrac 1{\sqrt 3+\sqrt 2}=\cfrac {\sqrt 3-\sqrt 2}{\sqrt 3-\sqrt 2}\cdot\cfrac 1{\sqrt 3+\sqrt 2}=\sqrt 3-\sqrt 2$
So you need to find $(\sqrt 3+\sqrt 2)^4-(\sqrt 3-\sqrt 2)^4$
Now note that the terms which have even powers of $\sqrt 2$ will cancel, and the odd powers will double - so we are left with the two terms in the binomial expansion with coefficient $\binom 41=\binom 43=4$ so we get $$2\cdot\binom 41\left((\sqrt3)^3\sqrt2+\sqrt 3(\sqrt 2)^3\right)=2\cdot4\cdot\sqrt 6\cdot(3+2)=40\sqrt 6$$
I've done this longhand, to show the detail.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/522712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that $n!≤((n+1)/2)^n$ by induction I'm new to inequalities in mathematical induction and don't know how to proceed further. So far I was able to do this:
$V(1): 1≤1 \text{ true}$
$V(n): n!≤((n+1)/2)^n$
$V(n+1): (n+1)!≤((n+2)/2)^{(n+1)}$
and I've got : $(((n+1)/2)^n)\cdot(n+1)≤((n+2)/2)^{(n+1)}$ $((n+1)^n)n(n+1)≤((n+2)^n)((n/2)+1)$
| If you really need induction let it be.
Base is $n = 0$: $0! = 1 \le 1 = \left(\frac12\right)^0$.
By induction hypothesis the inequality holds for $n = k$. Let proove it for $n = k+1$.
$$k! \le \left(\frac{k+1}2\right)^k,\\
k!(k+1) \le \left(\frac{k+1}2\right)^k(k+1),$$
$$(k+1)! \le \frac{(k+1)^{k+1}}{2^k}.\tag{*}$$
Now we need to show that $f(x) = \frac{x^{x}}{(x+1)^x} \le \frac12$. Ok, $f(x) = \left(\frac{x}{x+1}\right)^x = e^{x(\ln x - \ln (x+1))}$, then
$$f'(x) = e^{x(\ln x - \ln (x+1))}\left((\ln x - \ln (x+1)) + x\left(\frac1x - \frac1{x+1}\right)\right) =
e^{x(\ln x - \ln (x+1))}\left(\ln \left(1 - \frac1{x+1}\right) + \frac1{x+1}\right) \le 0$$ for any positive $x$, since $\ln y \le y - 1$ for any positive $y$. So $f(x) \le f(1) = 1/2$ for any $x \ge 1$. Then from (*) we get
$$(k+1)! \le \frac{(k+1)^{k+1}}{2^k} \le \frac{(k+1)^{k+1}}{2^k}\cdot \frac1{2f(k+1)} = \frac{(k+1)^{k+1}}{2^k}\cdot \frac{(k+2)^{k+1}}{2(k+1)^{k+1}} = \frac{(k+2)^{k+1}}{2^{k+1}} = \left(\frac{k+2}2\right)^{k+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/523529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Solving $\frac{5}{t-3}-2=\frac{30}{t^2-9}$ I need help with solving this equation:
$$
\frac{5}{t-3}-2=\frac{30}{t^2-9}
$$
I tried to solve, but I always get false result. The result should be $-\frac{1}{2}$ but I always get $-\frac{1}{2}$ and $3$.
This is how I did it:
$$\begin{align}
\frac{5}{t-3}-2&=\frac{30}{t^2-9}\\
\frac{5-2\cdot(t-3)}{t-3}&=\frac{30}{t^2-9}\\
\frac{5-2t+6}{t-3}&=\frac{30}{t^2-9}\\
\frac{-2t+11}{t-3}&=\frac{30}{t^2-9}\\
\frac{-2t+11}{t-3}\cdot\left(t^2-9\right)&=30\\
\frac{(-2t+11)\cdot\left(t^2-9\right)}{t-3}&=30\\
\frac{(-2t+11)\cdot(t-3)\cdot(t+3)}{t-3}&=30\\
(-2t+11)\cdot(t+3)&=30\\
-2t^2-6t+11t+33&=30\\
-2t^2+5t+33&=30\\
-2t^2+5t+33-30&=0\\
-2t^2+5t+3&=0\\
\end{align}$$
Solving
$$
-2t²+5t+3=0
$$
I get:
$$
-\frac{1}{2}\text{ and }3
$$
Am I doing something wrong?
| You have cancelled out $t-3$ which is only allowed if $t-3\ne0\iff t\ne3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/523618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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