Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find an angle not on the circumference of a circle problem. I know this has to be extremely easy but I'm not going to solve this problem.
The task is to find the angle at point $A$.
Thanks!
| Notice that $\angle EOF+\angle BOC+49^\circ+175^\circ=360^\circ$. It follows that $\angle EOF+\angle BOC=136^\circ$.
$\angle OCA=\dfrac{180^\circ-\angle BOC}{2}$ and $\angle OEA=\dfrac{180^\circ-\angle EOF}{2}$ because $\triangle OBC$ and $\triangle OEF$ are isosceles respectively.
Now $\angle OAC+\angle OCA+\angle BOC+\angle BOA=180^\circ$ and $\angle OAE+\angle OEA+\angle EOF+\angle FOA=180^\circ$. Summing these two gives $$\begin{align}
(\angle OAC+\angle OAE)+\angle OCA+\angle OEA+(\angle BOC+\angle EOF)+(\angle BOA+\angle FOA)&=360^\circ\\
\angle EAC+\angle OCA+\angle OEA+136^\circ+49^\circ&=\\
\angle EAC+\frac{180^\circ-\angle BOC}{2}+\frac{180^\circ-\angle EOF}{2}+136^\circ+49^\circ&=\\
2\angle EAC+(180^\circ-\angle BOC)+(180^\circ-\angle EOF)+370&=720^\circ\\
2\angle EAC+360^\circ-(\angle BOC+\angle EOF)+370&=\\
2\angle EAC+360^\circ-136^\circ+370&=\\
2\angle EAC&=126^\circ\\
\angle EAC&=63^\circ
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707749",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove that the sequence $b_n = 1$, $b_{n+1} = \frac{1}{1+b_n}$ converges. I need to prove that the sequence $b_1 = 1$, $b_{n+1} = \frac{1}{1+b_n}$ converges.
If $\lim_{n\to\infty} b_n = L$ then $\lim_{n\to\infty} b_{n+1} = L$, so $\lim_{n\to\infty} b_{n+1} = \frac{1}{1+\lim_{n\to\infty} b_n} \implies L = \frac{1}{1+L} \implies L^2 + L - 1 = 0 \implies L = \frac{-1+\sqrt{5}}{2}$.
So I know that if the sequence converges to $L$ then $L = \frac{-1+\sqrt{5}}{2}$. How can I prove that the sequence converges?
EDIT:
With the help of the answers I got my resolution:
Let's prove that $b_n$ converges by proving that the subsequences $b_{2n}$ and $b_{2n-1}$ are monotonic and bounded.
I want to prove that $\forall n \in \mathbb N$, $b_{2n} < b_{2(n+1)} < L$ and $b_{2n-1} > b_{2(n+1)-1} > L$.
I'm going to prove using induction.
First Step:
$\frac{1}{2} < \frac{3}{5} < \frac{-1 + \sqrt{5}}{2} \implies b_2 < b_4 < L $
and
$1> \frac{2}{3} > \frac{-1 + \sqrt{5}}{2} \implies b_1 > b_3 > L $
Second Step:
Let's prove that $b_{2n-1} > b_{2n+1} > L \implies b_{2n+1} > b_{2n+3} > L$
and $b_{2n} < b_{2n+2} < L \implies b_{2n+2} < b_{2n+4} < L$
$b_{2n-1}>b_{2n+1}>L \implies 1+b_{2n-1} > 1+b_{2n+1}>1+L \implies$
$\frac{1}{1+b_{2n-1}} = b_{2n} < \frac{1}{1+b_{2n+1}} = b_{2n+2} < \frac{1}{1+L} = L \implies $
$1+b_{2n} < 1+b_{2n+2}<1+L \implies \frac{1}{1+b_{2n}} = b_{2n +1} > \frac{1}{1+b_{2n+2}} = b_{2n+3} > \frac{1}{1+L} = L \implies$
$1+b_{2n+1} > 1+b_{2n+3} > 1+L \implies \frac{1}{1+b_{2n+1}} = b_{2n +2} < \frac{1}{1+b_{2n+3}} = b_{2n+4} < \frac{1}{1+L} = L$.
Thus $b_n$ converges.
| Try showing that the subsequences of even and odd terms are monotonic and bounded. If this is the case, then both subsequences must converge. From there it shouldn't be too hard to show that they converge to the same thing, and therefore the overall sequence must converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708114",
"timestamp": "2023-03-29T00:00:00",
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Triangle inequality with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$ Given a triangle with the exradii $r_{a}$, $r_{b}$, $r_{c}$, the medians $m_{a}$, $m_{b}$, $m_{c}$. Show that
$$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq 3\sqrt{3}. S+ \left ( m_{a}- m_{b} \right )^{2}+ \left ( m_{b}- m_{c} \right )^{2}+ \left ( m_{c}- m_{a} \right )^{2}$$
I can only show the weaker inequality
$$r_{a}^{2}+ r_{b}^{2}+ r_{c}^{2}\geq 3\sqrt{3}. S$$
I need to the help. Thanks!
| There is another proof bases on $R, r, s$ relationship as follows:
Let $R, r, s$ be the circumcircle radius, incircle radius, and semi-perimeter of the triangle. Then $a, b, c, r_a, r_b, r_c$ satisfy the following relationship:
$$ R\ge 2r$$
$$ r_a+r_b+r_c=4R+r $$
$$r_ar_b+r_br_c+r_cr_a=s^2 $$
$$m_a^2+m_b^2+m_c^2=\dfrac{3}{4}(a^2+b^2+c^2) $$
$$ a^2+b^2+c^2=2(s^2-4Rr-r^2)$$
$$ s^2\le 4R^2+4Rr+3r^2\ \text{(Gerretsen inequality)}$$
We have: $R\ge 2r$ then: $4R^2\ge 16r^2$. Thus:
$$32R^2+28Rr+5r^2\ge 28R^2+28Rr+21r^2$$
By Gerretsen inequality we have: $32R^2+28Rr+5r^2\ge 7s^2$.
Hence: $32R^2+16Rr+2r^2+12Rr+3r^2\ge7s^2$ or equivalently:
$$2(4R+r)^2+12Rr+3r^2\ge 7s^2$$
$$\Leftrightarrow 2(4R+r)^2-4s^2\ge 3s^2-12Rr-3r^2$$
$$\Leftrightarrow 2\Big((4R+r)^2-2s^2\Big)\ge 3(s^2-4Rr-r^2)$$
$$\Leftrightarrow (4R+r)^2-2s^2\ge \dfrac{3}{2}(s^2-4Rr-r^2)$$
$$ \Leftrightarrow r_a^2+r_b^2+r_c^2\ge 2\cdot\dfrac{3}{4}(a^2+b^2+c^2)=m_a^2+m_b^2+m_c^2$$
Now, note that $m_a, m_b, m_c$ are also the sidelength of a triangle, with area equals to $\dfrac{3}{4}S$, then by the well-known Finsler-Hadwiger inequality we have:
$$ m_a^2+m_b^2+m_c^2 \ge 3\sqrt{3}S+(m_a-m_b)^2+(m_b-m_c)^2+(m_c-m_a)^2 $$
Combine the two results above we obtain: $$ r_a^2+r_b^2+r_c^2 \ge 3\sqrt{3}S+(m_a-m_b)^2+(m_b-m_c)^2+(m_c-m_a)^2 $$
| {
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"url": "https://math.stackexchange.com/questions/2711511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $ $ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \ $ then show that $ \ a=1, \ b=-1 \ $
Answer:
$ \lim_{x \to \infty} [\frac{x^2+1}{x+1}-ax-b]=0 \\ \Rightarrow \lim_{x \to \infty} [\frac{x^2+1-ax^2-ax-bx-b}{x+1}]=0 \\ \Rightarrow \lim_{x \to \infty} \frac{2x-2ax-a-b}{1}=0 \\ \Rightarrow 2x-2ax-a-b=0 \ \ (?) $
Comparing both sides , we get
$ 2-2a=0 \\ a+b=0 \ $
Solving , we get
$ a=1 , \ b=-1 \ $
But I am not sure about the above line where question mark is there.
Can you help me?
| Let $f(x)=\dfrac{x^2+1}{x+1}$. If
$$
\lim_{x\to\infty}(f(x)-ax-b)=0
$$
then also
$$
\lim_{x\to\infty}\frac{f(x)-ax-b}{x}=0
$$
Thus we must have
$$
\lim_{x\to\infty}\left(\frac{x^2+1}{x(x+1)}-a\right)=0
$$
and therefore $a=1$. Now
$$
\frac{x^2+1}{x+1}-x=\frac{x^2+1-x^2-x}{x+1}=\frac{-x}{x+1}
$$
so
$$
\lim_{x\to\infty}(f(x)-x-b)=-1-b
$$
and therefore $b=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Last three digits of $6^{2002}$ Find the last three digits of $6^{2002}$. I did some work and figured out that the last two digits is 36. Can anyone help me with the hundredth digit? By the way, I used modular arithmetic and the recursion method for the tens digit, but it fell short when I attempted to do the hundreds digit. Thank you in advance!
How I figured out the last two digits:
I used the formula $\frac {1}{10^k} [n-a_k (n)]$. The last digit is obviously 6. I obtained the tens digit this way: $\frac {1}{10}(6^{2002}-6)=\frac {6}{10}(6^{2001}-1)=\frac {3}{5} (6^{2001}-1)=\frac {3}{5}(6-1)(6^{2000}+6^{1999}+...+6^2+6+1)=3(6^{2000}+6^{1999}+...+6^2+6+1)\equiv3(6+6+...+6+6+1)=3(2000\cdot6+1)\equiv3 (mod 10)$
Therefore, the last two digits of $6^{2002}$ are $36$.
| $3^{400}\equiv 1\mod 1000$ so $3^{2002}\equiv 9\mod 1000$.
$2^{100}\equiv 1\mod 125$ so $2^{2002}\equiv 4\mod 125$.
And $2^{2002}\equiv 0\mod 8$. So by Chinese Remainder theorem $2^{2002}\equiv 504 \mod 1000$.
So $6^{2002}\equiv 9*504\equiv 536\mod 1000$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to find a real $d \ne 0$ such that the equation $(2-3i) z^2 -(d-1)z + 4+ 3i = 0$ has a real root? I need to find a real value for d so that the equation $(2-3i)z^2 -(d-1)z + 4+3i= 0$ has real roots.
I've tried using the discriminant formulae and then I got
$$\frac {(d-1) \pm \sqrt{d^2 -2 d-67 +24i}}{4-6i} $$
Here I got stuck because I get complex numbers.
I appreciate your help.
| Even if you don't have gimusi's insight into considering the conjugate equation, it's possible to "plod" through to a solution.
Let $z = x + yi$. Ultimately, we want to set $y=0$ for the real root(s), but for now, let's do the algebra. $z^2 = x^2 - y^2 + 2xyi$, so the equation becomes:
$$(2-3i)(x^2-y^2+2xyi) - (d-1)(x+yi) + 4+3i = 0$$
$$2(x^2 - y^2) + 4xyi - 3(x^2-y^2)i + 6xy - (d-1)x - (d-1)yi + 4 + 3i = 0$$
And here it should be intuitive that we need to gather terms and set the real and imaginary parts separately to zero.
$$2(x^2 - y^2) + 6xy - (d-1)x + 4 = 0$$
$$4xy - 3(x^2 - y^2) - (d-1)y + 3 = 0$$
Set $y = 0$ in both equations. The second immediately yields: $x = \pm 1$. Substituting these two possible values for $x$ into the first equation yields the two possible values $d = -5, 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714425",
"timestamp": "2023-03-29T00:00:00",
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Finding unit digit of $f(10)$
If we define $$f(x)=\left\lfloor \frac {x^{2x^4}}{x^{x^2}+3}\right\rfloor$$ and we have to find unit digit of $f(10)$
I had tried approximation, factorization and substitutions like $x^2=u$ but it proved of no use. Moreover the sequential powers are feeling the hell out of me. Can someone please provide me with some hints
| When $x=10$, $\displaystyle \frac {x^{2x^4}}{x^{x^2}+3}=\frac {10^{20000}}{10^{100}+3}$.
Let $y=10^{100}$. $\displaystyle \frac {10^{20000}}{10^{100}+3}=\frac{y^{200}}{y+3}$.
When $y^{200}$ (as a polynomial) is divided by $y+3$, the remainder is $(-3)^{200}=3^{200}$.
So, $\displaystyle \frac{y^{200}}{y+3}=Q(y)+\frac{3^{200}}{y+3}$ for some polynomial $Q(y)$.
Put $y=0$, $Q(0)=-3^{199}=(-1)(81)^{49}(27)\equiv 3$ (mod $10$).
Note that $\displaystyle \frac{3^{200}}{10^{100}+3}=\frac{9^{100}}{10^{100}+3}<1$.
$$ \frac {10^{20000}}{10^{100}+3}=Q(10^{100})=10^{100}\times \textrm{some positive integer}+3+\frac{9^{100}}{10^{100}+3}$$
The unit digit of $\displaystyle \left\lfloor\frac {10^{20000}}{10^{100}+3}\right\rfloor$ is $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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In any triangle, if $\frac {\cos A+2\cos C}{\cos A+2\cos B}=\frac {\sin B}{\sin C}$, then the triangle is either isosceles or right-angled In any triangle, if $\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$, prove that the triangle is either isosceles or right angled.
My Attempt:
Given:
$$\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$$
$$\dfrac {\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+b^2-c^2}{2ab}}{\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+c^2-b^2}{2ac}}=\dfrac {b}{c}$$
On simplification,
$$\dfrac {ab^2+ac^2-a^3+2a^2c+2b^2c-2c^3}{ab^2+ac^2-a^3+2a^2b+2bc^2-2b^3}=\dfrac {b}{c}$$
| You are very close to answer.
*
*cross multiply.
*subtract 1 from both side.
*take b-c common from Numerator.
*It's done!! :)
| {
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Integrate $\int\frac{1}{x^3+1}dx$ The problem is, as stated:
$$\int\frac{1}{x^3+1}dx$$
I tried using substitution: $t^3 = x^3 + 1$ but didn't get far with that. I also tried setting: $t = x^3 + 1$, with no luck again.
I tried partial decomposition but I didn't know how to integrate $$\int\frac{1}{x^2-x+1}$$ and I kept getting that term when expanding $x^3 + 1$
Any help would be much appreciated.
| $$
\begin{aligned}
I &=\int \frac{4}{4 x^2-4 x+1} d x \\
&=4 \int \frac{d x}{(2 x-1)^2+ (\sqrt 3)^2} \\
&=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)+C
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$ Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$.
Want to make sure that my proof is correct.
Suppose there are rational solutions.
Note $2x^2 + 3y^2 = 1 \iff 2x^2 + 3y^2 = z^2\ |\ x,y.z \in \mathbb{Z}$
Since taking $(\frac{x}{z}, \frac{y}{z})$ as our rational solution, $2(\frac{x}{z})^2 + 3(\frac{y}{z})^2 = 1 \implies 2x^2 + 3y^2 = z^2$.
Now reducing $mod\ 3$, we have $2x^2 + 3y^2 \equiv 1 \ mod\ 3 \implies 2x^2 \equiv 1 \ mod\ 3 \implies -1x^2 \equiv 1\ mod\ 3 \implies x^2 \equiv -1\ mod\ 3 $, which is impossible.
| As José Carlos Santos was pointing out in the comments, we have
$$2x^2 \equiv z^2 \mod 3$$
But this does not imply your statement that
$$2x^2 \equiv 1 \mod 3$$
If $z$ were divisible by $3$ and we would have
$$2x^2 \equiv 0 \mod 3$$
| {
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In any $\triangle ABC$, prove that: $\frac {\cos B-\cos C}{\cos A +1}=\frac {c-b}{a}$ In any $\triangle ABC$, prove that: $\dfrac {\cos B-\cos C}{\cos A +1}=\dfrac {c-b}{a}$
My Attempt:
$$\begin{align}
\text{R.H.S.}&=\dfrac {c-b}{a} \\[4pt]
&=\frac {a\cos B+b\cos A-a\cos C-c\cos A}{b\cos C+c\cos B} \\[4pt]
&=\dfrac {a(\cos B-\cos C)+(b-c)\cos A}{b\cos C+c\cos B}
\end{align}$$
| For $\triangle ABC$,
$$\frac{\sin\left(A\right)}{a}=\frac{\sin\left(B\right)}{b}=\frac{\sin\left(C\right)}{c}=\frac{1}{k}$$
therefore,
$$k\sin\left(A\right)=a,\:k \sin\left(B\right)=b,\:k\sin\left(C\right)=c$$
\begin{align*}
RHS= &\frac{k\sin\left(C\right)-k\sin\left(B\right)}{k\sin\left(A\right)}\\&=\frac{\sin\left(C\right)-\sin\left(B\right)}{\sin\left(A\right)} \\&=\frac{2\cos\left(\frac{B+C}{2}\right)\cdot \sin\left(\frac{C-B}{2}\right)\cdot \cos\left(\frac{A}{2}\right)}{2\sin\left(\frac{A}{2}\right)\cdot \cos\left(\frac{A}{2}\right)\cdot \cos\left(\frac{A}{2}\right)}\\&=\frac{2\sin\left(\frac{A}{2}\right)\cdot \sin\left(\frac{C-B}{2}\right)\cdot \cos\left(\frac{A}{2}\right)}{2\sin\left(\frac{A}{2}\right)\cdot \cos^2\left(\frac{A}{2}\right)+1-1}\\&=\frac{2\sin\left(\frac{B+C}{2}\right)\cdot \sin\left(\frac{C-B}{2}\right)}{\cos\left(A\right)+1}\\&=\frac{\cos\left(B\right)-\cos\left(C\right)}{\cos\left(A\right)+1}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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proof verification $\frac{3+2\sqrt{6}}{1-\sqrt{6}}$ is an algebraic integer
Is $$\frac{3+2\sqrt{6}}{1-\sqrt{6}}$$
an algebraic integer?
An algebraic integer means an algebraic number in some algebraic number field $K\supset \Bbb Q$ that is the root of a monic polynomial $f\in \Bbb Z[x]$. Here I guess we are in $\Bbb Q({\sqrt{6}})=\Bbb{Q}(\alpha)/(\alpha^2-6)$, and with that:
$$\beta =\frac{3+2\sqrt{6}}{1-\sqrt{6}}\implies \beta= \frac{(3+2\sqrt{6})(1+\sqrt6)}{-5}\implies -5\beta =15+5\sqrt{6}\implies (-5\beta-15)^2=25(6)\implies 5^2\beta^2+2(15)(5)\beta+15^2-5^2(6)=0,$$
and thus $5^2x^2+6(5)^2x+5^2(3)\in\Bbb{Z}[x]$ is the desired polynomial.
Is this correct?
| $$ \beta =\frac{3+2\sqrt{6}}{1-\sqrt{6}}\implies $$
$$\beta= \frac{(3+2\sqrt{6})(1+\sqrt 6)}{-5}\implies $$
$$-5\beta =15+5\sqrt{6}\implies $$
$$ \beta =-3-\sqrt 6$$
$$ \beta ^2 + 6\beta +3=0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728010",
"timestamp": "2023-03-29T00:00:00",
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Proving determinant with variables I have a problem that asks:
Prove that det$\begin{pmatrix}
1 && 1 && 1 \\
a && b && c \\
a^2 && b^2 && c^2 \end{pmatrix} = (c-a)(c-b)(b-a)$
I was thinking of solving it like normal and see if I can end up with $(c-a)(c-b)(b-a)$
What I did:
det$\begin{pmatrix}
b && c \\
b^2 && c^2 \end{pmatrix}$ -
det$\begin{pmatrix}
a && c \\
a^2 && c^2 \end{pmatrix}$ +
det$\begin{pmatrix}
a && b \\
a^2 && b^2 \end{pmatrix}$
which gives
$(bc^2 - cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$
which simplifies to
$(b-a)c^2 + (a-c)b^2 + (c-b)a^2$
and this is where I got stuck. Am I even approaching this problem correctly? Any help would be appreciated.
| It might be useful to look for helpful column/row operations before expanding.
Here, subtraction of the first column from the second and the third one simplify the determinant considerably:
$$\det\begin{pmatrix}
1 && 1 && 1 \\
a && b && c \\
a^2 && b^2 && c^2 \end{pmatrix} =
\det \begin{pmatrix}
1 && 0 && 0 \\
a && b-a && c-a \\
a^2 && b^2-a^2 && c^2-a^2 \end{pmatrix} = \ldots $$
$$ \ldots =(b-a)(c-a)\det\begin{pmatrix}
1 && 1 \\
b+a && c+a \end{pmatrix} = (b-a)(c-a)(c-b)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the value of $ax^5 + by^5$ given that $a$, $b$, $x$, and $y$ satisfy the following system: Given
$\begin{cases} ax + by = 4 \\ ax^2 + by^2 = 8 \\ ax^3 + by^3 = 17 \\ ax^4 + by^4 = 42 \end{cases} $
Determine the value of $ax^5 + by^5$
Do you have any brilliant idea to solve this problem? By looking at the right-hand side, it is actually a sequence of the sum of quadratic numbers. But, of course it is so irrational to say that $ax^5 + by^5 = 42 + 36 = 78$ just by looking at that point.
| You have
$$ (x+y)(ax^2+by^2) = ax^3+by^3+xy(ax+by) \implies 8(x+y) = 17+4xy $$
$$ (x+y)(ax^3+by^3) = ax^4+by^4 + xy(ax^2+by^2) \implies 17(x+y) = 42 + 8xy $$
It follows that
$$ x+y = 8 $$
$$ xy = \frac{47}{4} $$
Also, from the same relations
$$ ax^5 + by^5 = (x+y)(ax^4+by^4) - xy(ax^3+by^3) = 42(x+y)-17xy = \frac{545}{4} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
} |
Find Jordan canonical form and basis of a linear operator. Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear operator such that: $T(x,y,z)=(-y-2z,x+3y+z,x+3z)$, I need to find a Jordan canonical form and a basis.
This is what i did:
In the first place, I found the associated matrix to this linear operator in the canonical basis which is this one:
$$A=\begin{pmatrix}
0 & -1 & -2 \\
1 & 3 & 1 \\
1 & 0 & 3 \\
\end{pmatrix}$$
After that i found the characteristic polynomial which is: $(\lambda-2)^3=0$ so we have this polynomial that has only one root with multiplicity 3.
After finding the eigenvalue I found the eigenvector associated to 2, which is $V_3=(-1,0,1)$.
Now the Jordan canonical form should be this one(If I have done it correctly):
$$J=\begin{pmatrix}
2 & 0 & 0 \\
1 & 2 & 0 \\
0 & 1 & 2 \\
\end{pmatrix}$$
We know the a Jordan basis is formed with these vectors: $B=v_1,v_2,v_3$
First I found $v_2$ :
$A.v_2=2.v_2+1.v_3$, which is equal to :
$$
\begin{align*}
\begin{cases}
2x+y+2z &=1\\
x+y+z&=0\\
x+z&=1
\end{cases}
\end{align*}
$$So $v_2=(-1,-1,1)$
The problem is here with the last vector: After doing the same operation I get to this point:
$A.v_1=2.v_1+1.v_2$, which is equal to:
$$
\begin{align*}
\begin{cases}
2x+y+2z &=1\\
x+y+z&=0\\
x+z&=1
\end{cases}
\end{align*}
$$ and clearly this system does not have solution... what am I doing wrong?
| Well, $(A - 2 I )^3 = 0,$ but $(A - 2 I )^2 \neq 0.$ So the minimal polynomial and the characteristic polynomial agree, meaning each eigenvalue occurs in just a single block. Indeed
$$
(A - 2 I )^2 =
\left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 0 & 0 \\
-1 & -1 & -1
\end{array}
\right)
$$
Now find some $u$ such that $(A - 2 I )^2 u \neq 0.$ Since you want the extra $1$s below the main diagonal, we will put $u$ as the left hand column of $P,$ where we are solving $P^{-1} A P = J$ is the Jordan form you want.
I will take
$$
u =
\left(
\begin{array}{c}
1 \\
0 \\
0
\end{array}
\right)
$$
The middle column will be $v = (A - 2 I) u,$ or
$$
v =
\left(
\begin{array}{c}
-2 \\
1 \\
1
\end{array}
\right)
$$
and finally $w = (A - 2 I) v =(A - 2 I)^2 u ,$ so that $ (A - 2 I) w =(A - 2 I)^3 u = 0 u = 0,$ so that $w$ is a genuine eigenvector
$$
w =
\left(
\begin{array}{c}
1 \\
0 \\
-1
\end{array}
\right)
$$
The columns of $P$ will be $u,v,w$ so
$$
P =
\left(
\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right)
$$
next
$$
P^{-1} =
\left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right)
$$
With
$$
A =
\left(
\begin{array}{ccc}
0 & -1 & -2 \\
1 & 3 & 1 \\
1 & 0 & 3
\end{array}
\right)
$$
we get to
$$
\left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right)
\left(
\begin{array}{ccc}
0 & -1 & -2 \\
1 & 3 & 1 \\
1 & 0 & 3
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right) =
\left(
\begin{array}{ccc}
2 & 0 & 0 \\
1 & 2 & 0 \\
0 & 1 & 2
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right)
\left(
\begin{array}{ccc}
2 & 0 & 0 \\
1 & 2 & 0 \\
0 & 1 & 2
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 0 \\
0 & 1 & -1
\end{array}
\right) =
\left(
\begin{array}{ccc}
0 & -1 & -2 \\
1 & 3 & 1 \\
1 & 0 & 3
\end{array}
\right)
$$
If you would like to get the $1$s above the main diagonal instead, start over with the columns of $P$ in order $w,v,u,$ then calculate the new $P^{-1}$ Indeed, the fact that a single Jordan block is similar to its transpose gives a cheap proof that any matrix is similar to its transpose.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the limit of sequence $\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}$ without using of derivatives and etc. I need to find limit of sequence
$$
\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)
$$
I tried to solve it and stopped here
$$
f(n+1) = \frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}+\frac{2n+1}{2^{n+1}}
$$
$$
2f(n+1) = 1+\frac{3}{2}+\frac{5}{2^{2}}+\cdots+\frac{2n-1}{2^{n-1}}+\frac{2n+1}{2^{n}}
$$
$$
2f(n+1) -f(n) = 1+ \left(1 + \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\right) = 1 + g(n)
$$
I can find the limit of $g$, but what to do with the other parts?
| Let's assume that the limit exists. If we call the limit $s$ then
$s = \sum_1^{\infty}\frac{2n-1}{2^n}$
$\Rightarrow 2s = 1 + \sum_1^{\infty}\frac{2n+1}{2^n} = 1 + \sum_1^{\infty}\frac{2n-1}{2^n} + \sum_1^{\infty}\frac{2}{2^n}$
$\Rightarrow 2s = 1 + s + \sum_0^{\infty}\frac{1}{2^n}$
$\Rightarrow 2s = 1 + s + 2$
$\Rightarrow s=3$
So if the limit exists then it must be 3.
Now you just have to prove that the limit exists i.e. the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Solve system:$yz(y+z-x)= a(x+y+z)\\ zx(z+x-y)= b(x+y+z)\\ xy(x+y-z)= c(x+y+z)$ Solve system:
$$yz(y+z-x)= a(x+y+z)\\
zx(z+x-y)= b(x+y+z)\\
xy(x+y-z)= c(x+y+z)$$
$$a, b, c>0$$
I can only solve the system at $a= 2, b= 5, c= 10$. I have tried all things but it's hard with me. Somebody help me?
| From the first equation
$$y z \left(( x + y + z ) - 2 x \right) = a ( x + y + z ) \quad\to\quad
-2x y z = ( a - y z )( x + y + z ) \tag{$\star$}$$
Likewise,
$$-2xyz=(b-zx)(x+y+z) \qquad -2 x y z = ( c - x y )( x + y + z )$$
If, say, $x=0$, then $(a-yz)(y+z) = b(y+z) = c(y+z)=0$; thus, either $y=-z$, or $b=c=0$ and $a=yz$ (or both). Similarly for $y=0$ and $z=0$.
Going forward, we take $x$, $y$, $z$ non-zero, noting that $x+y+z$ is then also non-zero. Consequently, we can write
$$a - y z = b - z x = c - x y = k$$
for some (non-zero) $k$. Therefore,
$$x^2 = \frac{xy\cdot zx}{yz}= \frac{(b-k)(c-k)}{a-k} \qquad y^2 = \frac{(c-k)(a-k)}{b-k} \qquad z^2=\frac{(a-k)(b-k)}{c-k}$$
Since $x^2$, $y^2$, $z^2$ are positive (we've handled the zero case), we see that an even number of $a-k$, $b-k$, $c-k$ can be negative. Let's take $a-k$ to be positive, and define $s = \pm1$ to have the common sign of $b-k$ and $c-k$. Then,
$$ x = \frac{r}{a-k} \qquad y = \frac{sr}{b-k} \qquad z = \frac{sr}{c-k} \qquad r := \sqrt{(a-k)(b-k)(c-k)} \tag{$\star\star$}$$
Since
$$b = k + z x = k + \frac{s(a-k)(b-k)(c-k)}{(c-k)(a-k)} = k + s(b-k)$$
we have that $s$ is unambiguously $1$. Substituting $x$, $y$, $z$ into $(\star)$, and reducing, ultimately gives
$$k^3 - k( b c + c a + a b ) + 2 a b c = 0 \tag{$\star\star\star$}$$
While there is a Cubic Formula for solving such equations explicitly, we won't bother. (Actually, see below.)
However, we might as well solve the specific case of $a=2$, $b=5$, $c=10$ as a sanity check. We can factor $(\star\star\star)$ to get
$$(k+10)(k^2-10k+20) = 0 \quad\to\quad k = -10\;\text{or}\; 5 \pm \sqrt{5}$$
Since all of $a-k$, $b-k$, $c-k$ must be positive, $a=2$ eliminates $k=5\pm\sqrt{5}$ from consideration; therefore,
$$x = \frac{\sqrt{12\cdot 15\cdot 20}}{12} = \frac{60}{12} = 5 \qquad y = 4 \qquad z = 3$$
Edit. The question has been updated to ask for proof that a positive solution exists for positive $a$, $b$, $c$.
First of all, the parameters being strictly positive allow us to say specifically that, if one of $x$, $y$, $z$ are zero, then the other two are mutual negatives. So, the trivial solutions are slightly streamlined.
For non-zero $x$, $y$, $z$, the previous argument holds as-is. Considering $(\star\star\star)$ with $k\to-k$, we can invoke the Descartes Rule of Signs, counting just one sign change in the coefficient sequence $\{-1, bc + ca + ab, 2abc\}$. Thus, the polynomial has one strictly-negative root $k$, so that $a-k$, $b-k$, $c-k$ are positive, as are the corresponding $x$, $y$, $z$. $\square$
In the interest of completeness, I'll give the explicit roots of $(\star\star\star)$. Since the equation involves a "depressed" cubic (that is, its second-degree term is zero), we can jump to the trigonometric form.
$$k = 2 \sqrt{\frac13(a b + b c + c a )}
\;\cos\left(\frac{\pi}{3} ( 2 n-1 ) + \frac13 \operatorname{arccos}\frac{3^{3/2} a b c}{( a b + b c + c a )^{3/2}} \right)
$$
with $n = 0, 1, 2$. The Arithmetic-Geometric Mean Inequality guarantees that the argument of the inverse cosine doesn't exceed $1$, so that all roots are real. The case $n=2$ reduces to
$$k = -2 \sqrt{\frac13(a b + b c + c a )}
\;\cos\left(\frac13 \operatorname{arccos}\frac{3^{3/2} a b c}{( a b + b c + c a )^{3/2}}\right)
$$
giving the negative $k$ that guarantees $a-k$, $b-k$, $c-k$, and $x$, $y$, $z$, are positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Mathematical physics problem. Eletric and Magnetic field. Velocity of propagation of the electromagnetic energy.
If we define the velocity of propagation of the electromagnetic energy, for an arbitrary field in the vacuum, for $\vec{S} = U\vec{v}$ , where $U$ is the density of electromagnetic energy, then
$$\left(1-\frac{v^{2}}{c^{2}}\right)U^{2} = (U_{E} - U_{M})^{2} + \frac{\epsilon_{0}}{\mu_{0}}(\vec{E}.\vec{B})^{2}$$
I'm trying to use the equations:
$$\vec{S}=\frac{1}{\mu_{0}}(\vec{E}\times\vec{B})$$
and
$$(\vec{a}\times\vec{b})^{2} + (\vec{a}.\vec{b}) = a^{2}b^{2}$$
But I'm not getting that equality. Can someone help me?
| This is straightforward. Use $c^2=1/\sqrt{\epsilon_0\mu_0}$.
\begin{align}
\left(1-\frac{v^2}{c^2}\right)U^2&=U^2-\frac{v^2}{c^2}U^2=U^2-\frac{1}{c^2}\left\|\mathbf{S}\right\|^2\\
&=\left[\frac{1}{2}\left(\epsilon_0\left\|\mathbf{E}\right\|^2+\frac{1}{\mu_0}\left\|\mathbf{B}\right\|^2\right)\right]^2-\frac{1}{c^2}\left\|\frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}\right\|^2\\
&=\frac{1}{4}\left(\epsilon_0^2\left\|\mathbf{E}\right\|^4+2\frac{\epsilon_0}{\mu_0}\left\|\mathbf{E}\right\|^2\left\|\mathbf{B}\right\|^2+\frac{1}{\mu_0^2}\left\|\mathbf{B}\right\|^4\right)-\frac{1}{c^2\mu_0^2}\left(\left\|\mathbf{E}\right\|^2\left\|\mathbf{B}\right\|^2-\left(\mathbf{E}\cdot\mathbf{B}\right)^2\right)\\
&=\frac{1}{4}\left(\epsilon_0^2\left\|\mathbf{E}\right\|^4+2\frac{\epsilon_0}{\mu_0}\left\|\mathbf{E}\right\|^2\left\|\mathbf{B}\right\|^2+\frac{1}{\mu_0^2}\left\|\mathbf{B}\right\|^4\right)-\frac{\epsilon_0}{\mu_0}\left(\left\|\mathbf{E}\right\|^2\left\|\mathbf{B}\right\|^2-\left(\mathbf{E}\cdot\mathbf{B}\right)^2\right)\\
&=\frac{1}{4}\left(\epsilon_0^2\left\|\mathbf{E}\right\|^4-2\frac{\epsilon_0}{\mu_0}\left\|\mathbf{E}\right\|^2\left\|\mathbf{B}\right\|^2+\frac{1}{\mu_0^2}\left\|\mathbf{B}\right\|^4\right)+\frac{\epsilon_0}{\mu_0}\left(\mathbf{E}\cdot\mathbf{B}\right)^2\\
&=\left[\frac{1}{2}\left(\epsilon_0\left\|\mathbf{E}\right\|^2-\frac{1}{\mu_0}\left\|\mathbf{B}\right\|^2\right)\right]^2+\frac{\epsilon_0}{\mu_0}\left(\mathbf{E}\cdot\mathbf{B}\right)^2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof Verification: Prove $\sqrt{x}$ is uniformly continuous on [0, $\infty$) Proof:
Fix $\epsilon \gt 0$. We want to find $\delta \gt 0$ such that:
$$|x-a| \lt \delta \Rightarrow |\sqrt{x} - \sqrt{a}| \lt \epsilon$$
$$\phantom{2000i11111}\Rightarrow |\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \epsilon$$
If $|x-a|\lt \delta$ then $|\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \frac{\delta}{\sqrt{a+1}+\sqrt{a}}$ for $\delta \lt 1$
Thus, choose $\delta = min\{1, (\sqrt{a+1}+\sqrt{a})\epsilon\}$. Then we have:
$$|x-a| \lt \delta \Rightarrow |\sqrt{x} - \sqrt{a}|=|\frac{x-a}{\sqrt{x}+\sqrt{a}}| \lt \frac{\delta}{\sqrt{a+1}+\sqrt{a}}=\frac{(\sqrt{a+1}+\sqrt{a})\epsilon}{\sqrt{a+1}+\sqrt{a}}=\epsilon$$
| Simply, if $|x - a| < \delta(\epsilon) = \epsilon^2$, then, since $|\sqrt{x} - \sqrt{a}| \leqslant |\sqrt{x} + \sqrt{a}|$, we have
$$|\sqrt{x} - \sqrt{a}|^2 = |\sqrt{x}-\sqrt{a}|\, |\sqrt{x}-\sqrt{a}|\leqslant |\sqrt{x}-\sqrt{a}|\, |\sqrt{x}+\sqrt{a}| = |x - a|< \epsilon^2\\ \implies|\sqrt{x} - \sqrt{a}|< \epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Let A be a matrix then $A^{50}$ is?
If $$A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\\ \end{bmatrix}$$ then ${A^{50}}$ is?
How to calculate easily? What is the trick behind it? Please tell me.
| Use the fact that:
$$A^2={\begin{pmatrix}1&0&0\\1&1&0\\1&0&1\\ \end{pmatrix}}=I+\begin{pmatrix}0&0&0\\1&0&0\\1&0&0\\ \end{pmatrix}\equiv I+ B$$
Also note that $B^n=0, \forall n>1$, thus using the binomial theorem, and noting that all higher powers of $B$ are dropped:
$$A^{2n}=(I+B)^n=I+nB+ {n\choose 2}B^2+\cdots+ B^n=I+nB$$
Thus,
$$A^{50}=I+25B=\begin{pmatrix}1&0&0\\25&1&0\\25&0&1\\ \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
$y^2+5xy+6x^2-9x-4y=0$ where $x$ and $y$ are integers Need some help solving this for integers $x$ and $y$:
$$
y^2+5xy+6x^2-9x-4y=0
$$
I managed to make something like this:
$$
(y+3x-4)(y+2x)=x\\
(y+3x)(y+2x-3)=y
$$
Find integers for $x$ and $y$ that satisfy the equations above.
But, what do I do next, or is this a bad approach?
| We have $$(y+3x)(y+2x)-(9x+4y)=0$$
so if we put $a=y+3x$ and $b=y+2x$ we get $$ab-a-3b=0\implies b={a\over a-3}$$
so $a-3\mid a$ and thus $a-3\mid 3$ so $a-3\in\{-3,-1,1,3\}$ so $a\in\{0,2,4,6\}$ and corresponding $b\in\{0,-2,4,2 \}$ Now for each pair $(a,b)$ solve for $x$ and $y$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to solve this equation for c? I want to solve the equation
$$
-\frac{a}{2}\left(c+\sqrt{c^2+4}\right)=-\frac{a-1}{2}\left(c-\sqrt{c^2+4}\right)
$$
for $c$, where $a$ is just a constant.
What I get is
$$
\frac{c-\sqrt{c^2+4}}{c+\sqrt{c^2+4}}=\frac{a}{a-1}.
$$
I think there now is some "trick" to solve this for $c$.
| Try to isolate the square root, it is much simpler:
$$-a \sqrt{c^2+4}+a c+\sqrt{c^2+4}-c=a \sqrt{c^2+4}+a c\\
-2 a \sqrt{c^2+4}+\sqrt{c^2+4}-c=0\\
(1-2 a) \sqrt{c^2+4}-c=0\\
\sqrt{c^2+4}=\frac{c}{1-2a}$$
Square both sides and you get:
$$c^2+4=\frac{c^2}{(1-2a)^2}$$
Rearrange that and you have a quadratic equation:
$$((1-2a)^2-1)c^2+4(1-2a)^2=0$$
Using the quadratic formula gives you:
$$c=\begin{cases}
-\frac{\sqrt{-4 a^2+4 a-1}}{\sqrt{a^2-a}}\Rightarrow -\frac{i (1-2 a)}{\sqrt{a-1} \sqrt{a}}\\
\frac{\sqrt{-4 a^2+4 a-1}}{\sqrt{a^2-a}}\Rightarrow \frac{i (1-2 a)}{\sqrt{a-1} \sqrt{a}}
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Using the $N - \varepsilon$ definition to find the limit of a sequence Yesterday I had a post about this, and it cleared a lot up; however, even though I feel like I understand how to go about solving problems like this, I don't seem to get the right answers. For example:
$u_n = \frac{2n+3}{2n+1}$. We know that this sequence converges to 1, but I want to prove it using the definition.
So I want to show that , $\forall \varepsilon > 0$, $\exists N$ where,$ n > N \implies u_n - 1 < \varepsilon$.
We can manipulate the expression such that $u_n = \frac{2n+3}{2n+1} =\frac{n}n \frac{2+\frac{3}n}{2+\frac{1}n} = \frac{2+\frac{3}n}{2+\frac{1}n}$
Thus, we can show that $\forall \varepsilon > 0$, $\exists N$ where,$ n > N \implies \frac{2+\frac{3}n}{2+\frac{1}n} - 1 < \varepsilon$
Now, we can say that 0 < $\frac{2+\frac{3}n}{2+\frac{1}n}$ < $\frac{2+\frac{3}n}{2} = 1 + \frac{3}{2n}$.
We can use this logic to say that
$\frac{2+\frac{3}n}{2+\frac{1}n} - 1 < 1 + \frac{3}{2n} - 1 < \varepsilon$ (since $1 + \frac{3}{2n}$ converges to 1).
So, finding an $N$ for $\frac{3}{2n} - 1$ would be the same as finding an $N$ for $u_n$, so we do:
$1 + \frac{3}{2n} - 1 < \varepsilon$ is the same as $\frac{3}{2n} < \varepsilon$.
Thus, if $ N > \frac{3}{2\varepsilon}$, then the initial implication follows. However, I'm not sure how to complete things from here; am I close, or did I fall off the rails at some point? Thanks.
| I think you are giving yourself too much work here.
Straightforwardly $u_n-1=\frac {2}{2n+1}$ and if $\frac 2{2n+1} \lt \epsilon$ we have $n\gt \frac 1{\epsilon}-\frac 12$
This is reversible, so this will find you a candidate for $N$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2741249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$, then $\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$
If $$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$$
show that
$$\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$$
where $k$ is an integer such that $2 < k \neq 4$, and where $a$, $b$, $c$ are sides of $\triangle ABC$.
Actually, I don't have any idea. Please someone help. It would be great if someone who answers include how he/she found where to begin.
| \begin{align}
\text{If }\quad
\frac{b+c}{2k-1}&=\frac{c+a}{2k}=\frac{a+b}{2k+1}
\tag{1}\label{1}
,\\
\text{show that }\quad
\frac{\sin A}{k+1}&=\frac{\sin B}{k}=\frac{\sin C}{k-1}
\tag{2}\label{2}
.
\end{align}
From \eqref{1} we have
by
the rules based on componendo and dividendo,
\begin{align}
\frac{b+c}{2k-1}&=\frac{c+a}{2k}=\frac{a+b}{2k+1}
\\
&=
\frac{-(b+c)+(c+a)+(a+b)}{-(2k-1)+(2k)+(2k+1)}
=
\frac{2a}{2k+2}
=
\frac{a}{k+1}
\\
&=
\frac{(b+c)-(c+a)+(a+b)}{(2k-1)-(2k)+(2k+1)}
=
\frac{2b}{2k}
=
\frac{b}{k}
\\
&=
\frac{(b+c)+(c+a)-(a+b)}{(2k-1)+(2k)-(2k+1)}
=
\frac{2c}{2k-2}
=
\frac{c}{k-1}
.
\end{align}
Now we have
\begin{align}
\frac{a}{k+1}&=
\frac{b}{k}=
\frac{c}{k-1}
\tag{3}\label{3}
.
\end{align}
Recall that by the sine rule
for $\triangle ABC$
with the radius of circumscribed circle $R$
\begin{align}
a&=2R\sin A,\quad
b=2R\sin B,\quad
c=2R\sin C,
\end{align}
so \eqref{3} becomes
\begin{align}
\frac{2R\sin A}{k+1}&=
\frac{2R\sin B}{k}=
\frac{2R\sin C}{k-1}
\end{align}
and \eqref{2} follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Maximum Triangle Inscribed So here is an interesting but frustrating problem I have been working on: Inside a square with side length 10, two congruent equilateral triangles are drawn such that they share one side and each has one vertex on a vertex of the square. What is the side length of the largest square that can be inscribed in the space inside the square and outside of the triangles?
T I had no idea how to solve it. Any help would be GREATLY appreciated. Thank you]1
| Start by drawing it out and labeling it like so (sorry, it's not perfect, but good enough):
First find the side length of the equilateral triangle. $G$ is the midpoint of $\overline{FE}$. Let $\overline{GE} = x$. $\overline{DG} = \overline{GB}$, so using basic trigonometry, $\overline{DG} \land \overline{GB}=x\sqrt{3}$. Thus, $\overline{DB} = 2x\sqrt{3}$. Since $\overline{DB}$ is the diagonal of $ABCD$, it has length $10\sqrt{2}$. Then we can set up the equation $2x\sqrt{3}=10\sqrt{2}$. So the side length of the triangle is $2x = \frac{10\sqrt{2}}{\sqrt{3}}$.
Now look at the diagonal $\overline{AC}$, it is made up of twice the diagonal of the small square plus the side length of the triangle. Let the side length of the small square be $y$:
$$\overline{AC} = \overline{AF}+\overline{FE}+\overline{EC}=y\sqrt{2}+\frac{10\sqrt{2}}{\sqrt{3}}+y\sqrt{2}=10\sqrt{2}.$$
Solve for y:
$$y\sqrt{2}+y\sqrt{2}+\frac{10\sqrt{2}\sqrt{3}}{3} = 10\sqrt{2} \\ y\sqrt{2}+y\sqrt{2}+\frac{10}{3}\left(\sqrt{6}\right) = 10\sqrt{2} \\ \frac{3y\sqrt{2}+3y\sqrt{2}+10\sqrt{6}}{3} = 10\sqrt{2} \\ \frac{6y\sqrt{2}+10\sqrt{6}}{3} = 10\sqrt{2} \\ 6y\sqrt{2}+10\sqrt{6} = 30\sqrt{2} \\ 6y\sqrt{2} = 30\sqrt{2}-10\sqrt{6} \\ y = \frac{30\sqrt{2}-10\sqrt{6}}{6\sqrt{2}} \\ y = \frac{\left(30\sqrt{2}-10\sqrt{6}\right)*\sqrt{2}}{12} \\ y = \frac{60-20\sqrt{3}}{12}.$$
Which simplifies to $\frac{5\left(3-\sqrt{3}\right)}{3}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all solutions of the equation $13[x]+25\{x\}=271$
For a real number $x$, let $[x]$ denote the largest integer $\le x$, and let $\{x \}$ denote $x-[x]$. Find all solutions of the equation:
$$13[x]+25\{x\}=271.$$
I tried to simplify the equation by:
$$13[x]+25(x-[x])=271$$
$$\implies 13[x]+25x-25[x]=271$$
$$\implies 25x-12[x]=271$$
$$\implies x= \dfrac{271+12[x]}{25}.$$
Now I saw that for $[x]=17$, we get $x=19$, but this is not a solution.
Do I have to go on manually checking for such solution, if it exists? Is there any generic/empirical way to solve this?
Some more facts I have observed (I don't know whether I am hitting at the right point or not).
If $x=[x]$,
$$13x=271$$
$$\implies x=\dfrac{271}{13}\approx 2.84.$$
Thus $[x]=2.$
But if $x>[x]$, then $$13[x]+25\{x\}=271,$$
where $\{x\}$ is the fractional part of $x$.
Thus $25\{x\}=271-[x]$.
But as $[x]$ is always an integer, $271-[x] \ne 25\{x\}.$
So we have only solution for $x=[x]$.
Possibly I am wrong at this. Please show me the correct way to solve it.
| $\displaystyle \frac{25x-271}{12}=\lfloor x \rfloor$.
\begin{align*}
x-1<\frac{25x-271}{12}&\le x\\
19+\frac{12}{13}< x&\le20+\frac{11}{13}\\
\lfloor x \rfloor&=19 \textrm{ or }20
\end{align*}
If $\lfloor x \rfloor=19$, $\displaystyle x=\frac{12(19)+271}{25}=19.96$.
If $\lfloor x \rfloor=20$, $\displaystyle x=\frac{12(20)+271}{25}=20.44$
The only answers are $19.96$ and $20.44$.
The other way to solve this is to write $\displaystyle x=\frac{12\lfloor x\rfloor+271}{25}$. Then
\begin{align*}
\lfloor x\rfloor\le \frac{12\lfloor x\rfloor+271}{25}&<\lfloor x\rfloor+1\\
18+\frac{12}{13}<\lfloor x\rfloor&\le 20+\frac{11}{13}
\end{align*}
As $\lfloor x\rfloor$ is an integer, $\lfloor x\rfloor=19$ or $20$.
When $\lfloor x\rfloor=19$, $\displaystyle x=\frac{12(19)+271}{25}=19.96$.
When $\lfloor x\rfloor=20$, $\displaystyle x=\frac{12(20)+271}{25}=20.44$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$
Find $$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$$
without L'Hôpital's rule.
My work:
1) I know that $$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$$
2) Let $x=t-\frac{\pi}{6}$. Then
$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=\lim_{t\rightarrow 0}\frac{3t}{\sin3t}\cdot\frac{1-2\sin \left(t-\frac{\pi}6\right)}{3t}$$
| Let $x=t+\frac{\pi}{6}$ with $t\to 0$
$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=\lim_{t\rightarrow 0}\frac{1-2\sin \left(t+\frac{\pi}{6}\right)}{\cos\left(3t+\frac{\pi}{2}\right)}=\lim_{t\rightarrow 0}\frac{1-\sqrt 3\sin t-\cos t}{-\sin 3t}=\frac{\sqrt 3}{3}$$
indeed
$$\frac{1-\sqrt 3\sin t-\cos t}{-\sin 3t}=\frac{3t}{-\sin 3t}\frac{1-\sqrt 3\sin t-\cos t}{3t}=\\
=-\frac{3t}{\sin 3t}\left(-\frac{\sqrt 3}{3}\frac{\sin t}{t}+\frac{1-\cos t}{3t}\right)
\to-1\cdot\left(-\frac{\sqrt 3}{3}+0 \right)
=\frac{\sqrt 3}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding value of $\prod_{k=1}^{n-1}\cos\frac{k\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$
Finding value of $$\prod_{k=1}^{n-1}\cos\frac{k\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$$
Try: let $$z^{2n}=1\Rightarrow z=(1)^{\frac{1}{2n}}=e^{\frac{k\pi}{n}}$$
So $$z^{2n}-1= \prod^{2n}_{k=1}(z-e^{\frac{k\pi}{n}})=(z^2-1)\prod^{n-1}_{k=1}(z-e^{\frac{k\pi}{n}})\cdot (z-e^{\frac{-k\pi}{n}})$$
So $$z^{2n}-1=(z^2-1)\prod^{n-1}_{k=1}(z^2-2z\cos(k\pi/n)+1$$
Could some help me how to prove my original formula, Thanks
| You are pretty close to derive the answer yourself.
Let $S$ be the product at hand, we have
$$\begin{align}S^2 &= \prod_{k=1}^{n-1} \cos^2\frac{k\pi}{2n} =
\prod_{k=1}^{n-1}\frac12\left(1+\cos\frac{k\pi}{n}\right) =
4^{1-n}\prod_{k=1}^{n-1}\left(2+2\cos\frac{k\pi}{n}\right)\\
&= 4^{1-n}\lim_{z\to-1}\prod_{k=1}^{n-1}\left(1+z^2-2z\cos\frac{k\pi}{n}\right)
= 4^{1-n}\lim_{z\to-1}\frac{z^{2n}-1}{z^2-1}
= \frac{n}{4^{n-1}}
\end{align}
$$
Since $S$ is clearly positive, taking square root give us $S = \frac{\sqrt{n}}{2^{n-1}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Largest cone that can be inscribed in a sphere
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is $\frac{8}{27}$ of the volume of the sphere.
My Attempt
r : radius of the base
h : height of the cone
$$
x=\sqrt{R^2-r^2}\implies h=R+x=R+\sqrt{R^2-r^2}\\
V(r)=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi r^2(R+\sqrt{R^2-r^2})\\
V'(r)=\frac{\pi}{3}\big( 2r.(R+\sqrt{R^2-r^2})+r^2\frac{-r}{\sqrt{R^2-r^2}} \big)\\=\frac{\pi}{3}r\big( 2R+2\sqrt{R^2-r^2})-\frac{r^2}{\sqrt{R^2-r^2}} \big)\\=\frac{\pi}{3}r\big( \frac{2R\sqrt{R^2-r^2}+2R^2-2r^2-r^2}{\sqrt{R^2-r^2}} \big)\\=\frac{-\pi}{3}r\big( \frac{2r^2-2R\sqrt{R^2-r^2}-2R^2}{\sqrt{R^2-r^2}} \big)
$$
How do I proceed further and find the points where $V'(r)=0$ and prove the above statement ?
Note: I would like to stick with $r$ as the variable.
| $2R\sqrt {R^2 - r^2} +2R^2 - 3r^2 = 0\\
4R^2 (R^2 - r^2) =(2R^2 - 3r^2)^2\\
4R^4 - 4R^2r^2 = 4R^4 - 12R^2r^2 + 9r^2\\
8R^2 = 9r^2\\
r = \sqrt {\frac 89} R$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\sin(\tfrac{\pi}3) + \tfrac12\sin(\tfrac{2\pi}3) + \tfrac13\sin(\tfrac{3\pi}3)+\dots$ up to infinity Initially, I thought of doing this by first evaluating the function
f(x) = cos(x) + cos(2x) + cos(3x) +.....
and then integrating it. However, I cant seem to find a proper integratable function (if that's a word) to replace f(x) with. Am I on the right track, or is the solution on a completely different path? Anyways, I would appreciate solutions of 10+2 level, if it exists.
| $$\sum^{\infty}_{j=1}\frac{1}{j}\sin \bigg(\frac{j\pi}{3}\bigg)$$
$$=\frac{\sqrt{3}}{2}\bigg[1+\frac{1}{2}-\frac{1}{4}-\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots\bigg]$$
$$=\frac{\sqrt{3}}{2}\int^{1}_{0}\bigg(1+x-x^3-x^4+x^6+x^7-x^9-x^{10}\cdots\bigg)dx=\frac{\sqrt{3}}{2}\int^{1}_{0}\frac{1+x}{1+x^3}dx$$
So Sum is $$\frac{\sqrt{3}}{2}\int^{1}_{0}\frac{1}{1-x+x^2}dx=\frac{\pi}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Taking common $2$ from $R_1$
$$2\left|
\begin{matrix}
(b+c)&(a+c)&(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
How do I proceed further?
| The determinant of the given matrix is a symmetric polynomial with degree $3$ in the variables $a,b,c$. By the Laplace expansion the determinant equals zero if $a,b$ or $c$ equal zero, hence the determinant is a constant multiple of $abc$. In order to find which multiple, it is enough to evaluate the determinant at $(a,b,c)=(1,2,3)$, for instance:
$$\det\begin{pmatrix}5 & 1 & 1\\ 2 & 4 & 2 \\ 3 & 3 & 3 \end{pmatrix}=6\det\begin{pmatrix}5 & 1 & 1\\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{pmatrix}=6\det\begin{pmatrix}4 & 0 & 0\\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}=24.$$
This leads to $\det M(a,b,c) = 4abc$ as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Sum to $n$ terms of the given series Find the sum to $n$ terms of the given series:
$$0.3+0.33+0.333+0.3333+\cdots$$
My Attempt:
Let
$$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$
$$=\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots$$
$$=\frac {3}{10} \left[1+\frac {11}{10}+\frac {111}{100}+\frac {1111}{1000}+\cdots \text{ to $n$ terms}\right]$$
How do I continue from here?
| Let
$$S_n=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$
$$=\underbrace{\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots}_n$$
$$=3\left(\frac {1}{10}+\frac {11}{100}+\frac {111}{1000} + \frac {1111}{10000}+\cdots\right)$$
$$=3\left(\frac {1}{10^1}+\frac {11}{10^2}+\frac {111}{10^3} + \frac {1111}{10^4}+\cdots\right)$$
$$=3\left(\frac {\sum_{k=0}^010^k}{10^1}+\frac {\sum_{k=0}^110^k}{10^2}+\frac {\sum_{k=0}^210^k}{10^3} + \frac {\sum_{k=0}^310^k}{10^4}+\cdots+\frac {\sum_{k=0}^{(n-1)}10^k}{10^n}\right)$$
And $$\frac{1}{10^{n}}\left(\sum_{k=0}^{n-1}10^k\right)= \frac{1}{10^{n}}\left(\frac{10^{n}-1}{9}\right) = \frac{1}{9}\left(1 - \frac{1}{10^n}\right) = \frac{1}{9}\left(1 - 10^{-n}\right)$$
Therefore
\begin{align}
S_n &= 3\left(\sum_{k=1}^n\left[\frac{1}{9}\left(1 - 10^{-k}\right)\right]\right)\\
&= \frac{3}{9}\sum_{k=1}^n\left(1 - 10^{-k}\right)\\
&= \frac{1}{3} \left( \sum_{k=1}^n 1 - \sum_{k=1}^n10^{-k}\right)\\
&= \frac{1}{3} \left( n- \sum_{k=1}^n10^{-k}\right)\\
&= \frac{1}{3} \left( n- \frac{1}{9}\left(1 - 10^{-n}\right)\right)
\end{align}
$$S_n = \frac{1}{27}\left(10^{-n}+ 9n -1\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\cos(x_n)$ we get
$$\cos(x_{n+1})=\cos(x_n-\theta)$$ where $\cos(\theta)=\frac{4}{5}$
Now Since $f_0=0$ we have $b_0=0$ and $x_0=\frac{\pi}{2}$
Now we have
$$x_{n+1}=x_n-\theta$$
Putting $n=0,1,2,3 \cdots 10$ and adding all we get
$$x_{10}=\frac{\pi}{2}-10\theta$$
Hence
$$b_{10}=\cos\left(\frac{\pi}{2}-10\theta\right)=\sin(10\theta)=\sin\left(10\arcsin\left(\frac{3}{5}\right)\right)$$
How to proceed further?
| Let $g_n = \sqrt{4^n - f_n^2}$
You have noticed that the sequence seem well-behaved at first, and in particular, $f_{n+1}$ and $g_{n+1}$ should look like linear combinations of $f_n$ and $g_n$ except for the issue of the sign of the square root.
$f_{n+1} = \frac 15 (8 f_n + 6 g_n)$ is easy to get, and
$g_{n+1}^2 = 4^{n+1}-f_{n+1}^2 = 4(g_n^2+f_n^2)-\frac 1{25}(64f_n^2+96f_ng_n+36g_n^2) = \frac 1{25}(36f_n^2-96f_ng_n+64g_n^2) = (\frac 15(6f_n-8g_n))^2$
And since $g_n$ is positive, we have $g_{n+1} = \frac 15|6f_n - 8g_n|$ (this absolute value is a giant trap and is also why your general trigonometric formula is wrong)
Now I only did that so that it was easier to compute the sequence (no need to take any square root !)
You start with $(f_0,g_0) = (0,1)$ and you apply those recurrence relations 10 times and you will get
$f_{10} = 983.04 = \frac{24576}{25}$
Despite the square root sign apparently screwing things up, it's still possible to say some more :
Let $z_n = f_n + ig_n$. The recurrence relation say that $z_{n+1} = \frac 15(8-6i)z_n$ or its conjugate (whichever has positive imaginary part)
So letting $\theta_n$ be an argument of $z_n$, we can pick $\theta_0 = \pi/2$, then $\theta_{n+1} = \theta_n - \arcsin(3/5)$ if it's positive, and $\theta_{n+1} = - \theta_n + \arcsin(3/5)$ otherwise.
If the second case happens, $\theta_{n+1} - \arcsin(3/5) = - \theta_n < 0$, and so $\theta_{n+2} = -\theta_{n+1}+\arcsin(3/5) = \theta_n$ : we enter an infinite $2$-cycle.
Which means that $z_{n+2} = 4z_n$ forall $n$ from that point on.
It turns out that $\theta_2 = \pi/2 - 2\arcsin(3/5) < \arcsin(3/5)$ and so $z_{10} = 4^4 z_2$
Thus $f_{10} = 256 f_2$ (which is why the denominator of $f_{10}$ isn't $5^{10}$ but $5^2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Let $x_1=a>0$ and $x_{n+1}=x_n+\frac{1}{x_n} \forall n\in \mathbb N$. Check whether the following sequence converges or diverges.
Let $x_1=a>0$ and $x_{n+1}=x_n+\frac{1}{x_n} \forall n\in \mathbb N$.
Check whether the following sequence converges or diverges.
When I was in UG my teacher used derivative test for monotonicity.
$f(x)=x+\frac{1}{x}, f'(x)=1-\frac{1}{x^2}>0(x>1).$ So, $f(x)$ is increasing. How to prove the sequence is monotonic? Differentiation is coming after the sequences and series. By AM-GM inequality sequence is bounded below. $x_{n+1}=x_n+\frac{1}{x_n}\ge 2\sqrt{x_n.\frac{1}{x_n}}=2 \forall n\in \mathbb N$. How can I judge whether the sequence bounded above or not? Please help me.
| Extending Clark's answer.
$x_{n+1}^2
=x_n^2+2+\dfrac1{x_n^2}
$
so
$x_{n+1}^2-x_n^2 \ge 2$.
Summing,
$x_{n+1}^2-x_1^2 \ge 2n$
so
$x_{n+1}^2
\ge x_1^2 + 2n
= 2n+a^2
$
so
$x_{n+1}
\ge \sqrt{2n+a^2}
\gt \sqrt{2n}
$.
Therefore
$x_{n+1}
=x_n+\frac{1}{x_n}
\le x_n+\frac{1}{\sqrt{2(n-1)}}
$
or
$x_{n+1}- x_n
\le \frac{1}{\sqrt{2(n-1+a^2)}}
\lt \frac{1}{\sqrt{2(n-1)}}
$.
Summing,
$\begin{array}\\
x_{m+1}- x_2
&=\sum_{n=2}^m (x_{n+1}- x_n)\\
&\lt \sum_{n=2}^m\frac{1}{\sqrt{2(n-1)}}\\
&= \dfrac1{\sqrt{2}}\sum_{n=1}^{m-1}\frac{1}{\sqrt{n}}\\
&= \dfrac1{\sqrt{2}}(1+\sum_{n=2}^{m-1}\frac{1}{\sqrt{n}})\\
&< \dfrac1{\sqrt{2}}(1+\int_1^m \dfrac{dt}{t^{1/2}})\\
&= \dfrac1{\sqrt{2}}(1+\dfrac{t^{1/2}}{1/2}|_1^m)\\
&= \dfrac1{\sqrt{2}}(1+2(\sqrt{m}-1))\\
&= \dfrac1{\sqrt{2}}(2\sqrt{m}-1)\\
&= \sqrt{2m}-\dfrac1{\sqrt{2}}\\
\end{array}
$
so that
$\sqrt{2m}
\lt x_{m+1}
\lt \sqrt{2m}+x_2-\dfrac1{\sqrt{2}}
$.
Extending this,
I am sure that we can prove that
$\lim_{m \to \infty} (x_m-\sqrt{2m})
$
exists
(and I have a feeling that
I already have),
but I'll leave it at this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
What is the angle between the median and the bisector? In a triangle ABC, what is the angle $\theta$ between the median AM and the bisector AD?
I want a way to know the measure of that angle, given the lenghts of the sides and angles of the triangle. I tried solving the triangle, with the sine and cosine laws, and I arrived at
$$
\theta = \frac{A}{2}
- \arcsin\left(\frac{2\sqrt{(a/2)^2+b^2
-ab\cos C}}
{a\sin C}\right).
$$
I'm not even totally sure if it's correct and it's a bit unwieldy... I am wondering if there is a smarter aproach that gives a nicer formula. Thank you!
| By law of sines:
$$
\frac{\sin CAM}{\sin BAM}=\frac{\sin C}{\sin B}=:\lambda.
$$
Setting $\sin BAM=x$, and solving the resulting equation for cosine of the sum:
$$
\sqrt{1-x^2}\cdot\sqrt{1-\lambda^2x^2}-x\cdot\lambda x=\cos A\\
$$
one obtains:
$$
\sin BAM=\frac{\sin A}{\sqrt{1+\lambda^2+2\lambda\cos A}}
=\frac{\sin A\sin B}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}}
$$
and similarly
$$
\sin CAM=\frac{\sin A\sin C}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}}.
$$
Correspondingly:
$$
\cos BAM=\frac{\sin B\cos A+\sin C}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}},\\
\cos CAM=\frac{\sin B+\sin C\cos A}{\sqrt{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}}.
$$
Writing:
$$\small{
\cos(BAM-CAM)=\frac{(\sin B\cos A+\sin C)(\sin B+\sin C\cos A)+\sin B\sin C\sin^2A}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}\\
=\frac{(\sin^2 B+\sin^2 C)\cos A+2\sin B\sin C}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}
}
$$
and noticing that $|\widehat{BAM}-\widehat{CAM}|$ is the doubled value of the angle in question one finally obtains:
$$
\theta=\frac{1}{2}\arccos\frac{(\sin^2 B+\sin^2 C)\cos A+2\sin B\sin C}{\sin^2 B+\sin^2 C+2\sin B\sin C\cos A}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Intuitive method of finding probability of getting an onto function from all possible functions from a set to another. The question I was dealing with is as follows:
Let sets $A$ and $B$ have $7$ and $5$ elements, respectively. If one function is selected from all possible defined functions from $A$ to $B$, what is the probability that it is onto?
Now, this is how my thought process was:
For a relation to classify as a function only an element of $A$ should possess only one image in the co-domain $B$. Thus, each element of $A$ has $5$ elements to choose as its image. Since it is a function from $A$ to $B$, two elements from $A$ mapping to the same image is allowed.
And so the total number of possible functions would be $= 5^7$
Next, to choose the number of onto functions, what I thought was: each element of $B$ must be mapped, but not to the same element. Therefore, the first element of $B$ would have $7$ options to choose from, the next one would have $6$ and so on... After every element of $B$ is mapped, there would be $2$ remaining elements in $A$ and they have to be mapped to any element of $B$ so as to become a function. Therefore, those two elements would have to choose any $2$ out of $5$ elements of $B$.
Thus the total number of possible onto functions would be =
$$(7 \cdot 6 \cdot 5 \cdot 4 \cdot 3) \cdot (5 \cdot 5) = \frac{7! \cdot 5^2}{ 2!}$$
Thus the probability required = $$\frac{\frac{7! \cdot 5^2}{2!}}{5^7}$$
But, the answer was $$\frac{7! \cdot 2}{3 \cdot 5^6}$$
Where did I go wrong?
| Both your answer and the stated answer of $$\frac{7 \cdot 2!}{3 \cdot 5^6}$$ are incorrect.
Let sets $A$ and $B$ have $7$ and $5$ elements, respectively. If one function is selected from all possible defined functions from $A$ to $B$, what is the probability that it is onto?
You counted the number of functions correctly.
Here are two approaches to counting the number of surjective (onto) functions from a set of seven elements to a set of five elements.
Method 1: We use the Inclusion-Exclusion Principle.
There are $5^7$ functions from a set of seven elements to a set of five elements since there are $7$ elements in the codomain to which each of the $5$ elements in the domain could be mapped.
From these, we must subtract those in which fewer than five elements in the codomain are in the range. There are $\binom{5}{k}$ ways to exclude $k$ of the five elements in the codomain from the range and $(5 - k)^7$ ways to map the seven elements in the domain to the remaining $5 - k$ elements in the codomain. Thus, by the Inclusion-Exclusion Principle, the number of surjective functions from a set with seven elements to a set with five elements is
\begin{align*}
\sum_{k = 0}^{7} (-1)^k\binom{5}{k}(5 - k)^7 & = \binom{5}{0}5^7 - \binom{5}{1}4^7 + \binom{5}{2}3^7 - \binom{5}{3}2^7 + \binom{5}{4}1^7 - \binom{5}{5}0^7\\
& = 78625 - 81920 + 21870 - 1280 + 5 - 0\\
& = 16800
\end{align*}
Method 2: If a function from a set of seven elements to a set of elements is surjective, either one element of the codomain is the image of three elements in the domain and the others are each the image of one element of the domain or two elements in the codomain are each the image of two elements of the domain and the others are each the image of one element of the domain.
One element of the codomain is the image of three elements in the domain and each of the other four elements in the codomain is the image of one element in the domain: There are five ways to select the element in the codomain that is the image of three elements in the domain, $\binom{7}{3}$ ways to select which three elements in the domain are mapped to that element, and $4!$ ways to map the remaining four elements in the domain to the remaining four elements in the codomain in such a way that those elements of the codomain are each the image of one element of the domain. There are
$$\binom{5}{1}\binom{7}{3}4!$$
such cases.
Two elements of the codomain are each the image of two elements of the domain and each of the remaining three elements in the codomain are each the image of one element of the domain: There are $\binom{7}{2}$ ways to select two elements in the codomain to each be the images of two elements of the domain. There are $\binom{7}{2}$ ways to select which two of the seven elements in the domain map to the smaller of those elements and $\binom{5}{2}$ to select which two of the remaining five elements in the domain map to the larger of those elements. The remaining three elements in the domain can be mapped to the remaining three elements in the codomain in such a way that each such element in the codomain is the image of one of those elements in $3!$ ways. Hence, there are
$$\binom{5}{2}\binom{7}{2}\binom{5}{2}3!$$
such surjective functions.
Therefore, there are
$$\binom{5}{1}\binom{7}{3}4! + \binom{5}{2}\binom{7}{2}\binom{5}{2}3! = 4200 + 12600 = 16800$$
surjective functions from a set of seven elements to a set of five elements.
Thus, the probability that a randomly selected function from a set of seven elements to a set of five elements is surjective is
$$\frac{16800}{5^7}$$
Where did I go wrong?
Look at the cases in method 2.
You counted each surjective function in which one element of the codomain is the image of three elements in the domain three times, once for each way of designating one of those three elements as the one that is mapped to that element.
You counted each surjective function in which two elements of the codomain is the image of two elements in the domain four times, once for each of the two ways you could designated one of the two elements in the domain that maps to each such element in the codomain as the one that maps to that element of the codomain.
Notice that
$$3 \cdot \binom{5}{1}\binom{7}{3}4! + 2^2 \cdot \binom{5}{2}\binom{7}{2}\binom{5}{2}3! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 5^2$$
| {
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Partial fractions integral. Compute $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$
Evaluate $\int_{0}^{1} \frac {x^4+1}{x^6+1}dx$.
I directly approached this with partial fractions and rewritten $x^6+1=(x^2)^3+1=(x^2+1)(x^4+x^2+1)$.
Therefore the integral is:
$$\int_{0}^{1} \frac {x^4+1}{(x^2+1)(x^4+x^2+1)}dx=-2\int_{0}^{1} \frac {x}{1+x^2}dx+2\int_{0}^{1}\frac 1{1+x^2}dx-\int_{0}^1\frac {(x-1)^2}{(x^2+1)^2+(\frac {\sqrt{3}}2)^2}dx$$
But I don't know how to solve the last term.
I answered the question... I realized my silly mistake, sorry for bothering...
| Note that $x^6+1=(x^2)^3+1=(x^2+1)(x^4-x^2+1)$ and
use the fact that you are integrating over $[0,1]$,
$$\int_0^1 \frac{x^4+1\pm x^2}{x^6+1}dx =\int_0^1 \frac{x^4-x^2+1}{x^6+1}dx+ \frac{1}{3}\int_0^1 \frac{3x^2}{x^6+1}dx\\=\int_0^1 \frac{1}{x^2+1}dx+ \frac{1}{3}\int_0^1 \frac{1}{t^2+1}dt=\left(1+\frac{1}{3}\right)\frac{\pi}{4}=\frac{\pi}{3}$$
where, in the second integral, $t=x^3$.
| {
"language": "en",
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Verify my proof of divisibility of a rational fraction I tried to prove that the following rational fraction can be divided only when n=1,2,5 without using mathematical induction or function, just by using basic method. $$\frac{n^2+2}{2n-1}$$ this expression can be divided only when the remainder of $$\frac{n^2}{2n-1}$$ is $$\frac{2n-3}{2n-1}$$ and the quotient of the following expression $$\frac{n^2}{2n-1}$$ has the same quotient with the following expression, except in the case of n=1$$\frac{n^2}{2n}$$ because when some dividend is divided by these two different divisor, in order to $$2n-1$$ has much more quotient than $$2n$$ It's necessary to meet particular condition that when dividend is greater than or equal to $$2n$$ also when the dividend is greater than or equal to $$(2n)(2n-1)$$ then $$2n-1$$ can has much more quotient than $$2n$$ and when the dividend is less than $$2n$$ also when the value of the dividend is equal to the value of $$2n-1$$ then $$2n-1$$ can has much more quotient than $$2n$$ So $$n^2$$ is greater than or equal to $$2n$$ except when n=1, and when n=1, the value of $$n^2$$ is less than $$2n$$ and is equal to the value of $$2n-1$$ So when n=1, $$2n-1$$ has much more quotient than $$2n$$ and when n>1, $$n^2$$ is greater than or equal to $$2n$$ and is less than $$(2n)(2n-1)$$ So $$\frac{n^2}{2n-1}$$ has the same quotient with $$\frac{n^2}{2n}$$ except in the case of n=1. and $$\frac{n^2}{2n}=\frac{n}{2}$$ so the quotient of $$\frac{n}{2}$$ can be expressed by a formula $$\frac{n}{2}-\frac{1+(-1)^{n+1}}{4}$$ and in the following expression$$\frac{n}{2}$$ when n is even number there is no remainder and when n is odd number the remainder is always $$\frac{1}{2}$$ It also can be expressed as $$\frac{n}{2n}$$ and in the following expression $$\frac{n^2}{2n-1}$$ when n is even number the remainder of the expression is as much as Its quotient. and when n is odd number the remainder of the division is $$n+quotient$$ now I can answer the question about when the remainder become $$2n-3$$ is, by using the formula $$\frac{n}{2}-\frac{1+(-1)^{n+1}}{4}$$ when n is even number It can be expressed as $$\frac{n}{2}-\frac{1+(-1)^{n+1}}{4}=2n-3$$ this equation can be changed into $$12-(1+(-1)^{n+1})=6n$$ by multiplying both sides by 4 and by transposition. and when n is odd number, It can be expressed as $$n+\frac{n}{2}-\frac{1+(-1)^{n+1}}{4}=2n-3$$ this equation can be changed into $$12-(1+(-1)^{n+1})=2n$$ so, the following expression $$\frac{n^2+2}{2n-1}$$ is can be divided only when n=2,5 but this equation works only when n>1. and when=1 the expression can be divided. so this is what I tried to prove the question. I wanna know this proof is correct or incorrect, Could you please check my proof? roughly If you want. and I wanna know this kind of approach is proper or not when I'm gonna study advanced mathematics.
| Yet another way to get the answer:
If $2n - 1$ divides $n^2 + 2$ then there is an integer $k$ such that
$n^2 + 2 = k(2n - 1),$
that is,
$$ n^2 - 2kn + (k + 2) = 0.$$
Solving this as a quadratic equation in $n,$
\begin{align}
n &= \frac{2k \pm \sqrt{4k^2 - 4(k + 2)}}{2} \\
&= k \pm \sqrt{k^2 - k - 2)},
\end{align}
which is possible only if $k^2 - k - 2 = m^2$
for some non-negative integer $m.$
Observing that $k^2 - k - 2 = (k - 2)(k + 1),$
the product can be a square only if it is non-negative, that is,
$k \geq 2$ or $k \leq = -1.$
If $k \geq 2$ then $m \leq k - 1$; it follows that
$m^2 \leq k^2 - 2k + 1,$ and therefore
$$k^2 - k - 2 \leq k^2 - 2k + 1.$$
Canceling the $k^2$ terms and collecting the other terms appropriately, this is equivalent to $k \leq 3.$ We can check that $k = 2$ and $k = 3$ both are valid solutions.
If $k \leq -1,$ let $p = -k$; then $p \geq 1$ and $m^2 = p^2 + p - 2,$
which has solution $m = 0$ if $p = 1,$ solution $m = 2$ if $p = 2,$
and otherwise $m \geq p + 1,$ so $m^2 \geq p^2 + 2p + 1,$
so $p^2 + p - 2 \geq p^2 + 2p + 1,$ so $-p \geq 3,$ but that contradicts the case assumption that $k \leq -1.$
Hence there are no other solutions.
So the solutions are $k = 2,$ $k = 3,$ $k = -1,$ and $k = -2$.
The possible values of $n$ therefore are given in this table:
\begin{array}{ccccl}
k &\qquad k^2 - k - 2 \qquad& &n& \\ \hline
\phantom{-}2 & 4 - 2 - 2 = 0 & 2 \pm \sqrt0 &=& 2, \\
\phantom{-}3 & 9 - 3 - 2 = 4 & 3 \pm \sqrt4 &=& 1 \text{ or } 5, \\
-1 & 1 - (-1) - 2 = 0 & -1 \pm \sqrt0 &=& -1, \\
-2 & 4 - (-2) - 2 = 4 & -2 \pm \sqrt4 &=& 0 \text{ or } -4.
\end{array}
It's easier to solve $2n - 1 \mid 9,$ so I would go with one of the other solutions.
| {
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"source": "stackexchange",
"question_score": "6",
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How to calculate probability of $5$ in a row with at least one rare vs $5$ in a row without a single rare? There are $36$ possible outcomes. Event $A$ (later - "rare group") happens in $11$ outcomes. Event $B$ (normal group) in $25$ outcomes ($36-11$). Events $A$ and $B$ cannot occur at the same time - only one of them is possible.
We make $5$ Bernoulli trials and observe the composite result of all $5$ trials. After any event happens, number of "positive" outcomes for that event decreases by one (if event $A$ happened, next trial there are $35$ (not $36$) outcomes in general for both $A$ and $B$, but now only $10$ for $A$ (that is minus one), still $25$ for $B$ (nothing changed for B positive outcomes)). I need to calculate and compare probability that in $5$ trials
*
*Event $A$ never appears,
*Event $A$ appears at least once.
I have an idea how to calculate, but I get a result more than $1$, which looks bad. So I am wrong somewhere.
The question is about $5$ Bernoulli trials (tossing a single "die" with $36$ sides). Each next trial (appearance of next number) is a conditional probability, because trial $2$ is equivalent of tossing a single "die" with $35$ sides ($36$ - what appeared in trial 1). So in trial 1 probability to get a number within rare group is $11/36$, probability to get a number not in rare group is $1-11/36$. In trial 2: probability to get a number within rare group is $11/35$ (UNLESS THAT NUMBER APPEARED IN TRIAL 1! Then it is $10/35$), probability to get a number not in rare group is $1 - 11/36$. Finally, probability that all $5$ numbers (results of all $5$ trials) are not from rare group is just a product of respective probabilities of trials 1-5:
$$\left(1 - \frac{11}{36}\right)\left(1 - \frac{11}{35}\right)\left(1 - \frac{11}{34}\right)\left(1 - \frac{11}{33}\right)\left(1 - \frac{11}{32}\right) = \boldsymbol{14\%}$$
Same as $$\frac{\dbinom{25}{5}}{\dbinom{36}{5}}$$
$C = \binom{25}{5}$ ways to choose $5$ without replacement out of $25$, $\binom{36}{5}$ - total combinations.
So, the opposite event (probability that a ball from rare group would appear at least once) is $100\% - 14\% = 86\%$. Right? But why...
On the other hand, probability that a ball from rare group would appear at least once in $5$ trials is a sum of respective probabilities: $$\frac{11}{36} + \frac{11}{35} + \frac{11}{34} + \frac{11}{33} + \frac{11}{32} = 1.62 = 162\%$$
(no adjustments in case of appearing of rare ball in trials 1-4, but even all five rare balls is huge (rough estimation):
$$\frac{11}{36} + \frac{10}{35} + \frac{9}{34} + \frac{8}{33} + \frac{7}{32} = 1.31 = 131\%$$
| Since the selections are made without replacement, this is actually a hypergeometric distribution.
You correctly calculated that the number of ways no members of the rare group are selected in five trials is
$$\frac{\dbinom{25}{5}}{\dbinom{36}{5}}$$
Therefore, the probability that at least one member of the rare group is selected in five trials is
$$1 - \frac{\dbinom{25}{5}}{\dbinom{36}{5}}$$
The number of ways of selecting exactly $k$ members of the rare group and $5 - k$ members of the normal group in five trials is
$$\binom{11}{k}\binom{25}{5 - k}$$
Hence, another way of calculating the probability that at least one member of the rare group is selected is to add the probabilities that from $1$ to $5$ members of the rare group are selected, which is
$$\frac{\dbinom{11}{1}\dbinom{25}{4} + \dbinom{11}{2}\dbinom{25}{3} + \dbinom{11}{3}\dbinom{25}{2} + \dbinom{11}{4}\dbinom{25}{1} + \dbinom{11}{5}\dbinom{25}{0}}{\dbinom{36}{5}}$$
| {
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"source": "stackexchange",
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Maximum Value of $a+b+c$
Given that $a+\frac {11}b+\frac c4=20$, what is the maximum value of $a+b+c$? Here, $a,b,c$ are positive integers.
I tried to find the maximum from the AM-GM relationship but failed.
| Notice
$$b > 0 \implies \frac{11}{b} > 0
\implies a + \frac{c}{4} < 20 \iff 4a+c < 80 \implies 4a + c \le 79$$
This leads to
$$b = \frac{11}{20- \left(a + \frac{c}{4}\right)} = \frac{44}{80 - (4a + c)} \le \frac{44}{80-79} = 44$$
Furthermore, $a \ge 1$ implies
$$a + c \le 79 - 3a \le 79 - 3 = 76$$
Combine these, we can conclude $a + b + c \le 76 + 44 = 120$.
Since this value $120$ is achieved by $(a,b,c) = (1,44,75)$ and
$1 + \frac{11}{44} + \frac{75}{4} = 20$. The maximum value of $a + b + c$ equals to $120$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A limit involving $\cos x$ and $x^2$ The question is finding the value of $L = \lim_{x \to 0}\frac{1 - \cos x \cos 2x \cos 3x}{x^2}$ if it exists . I've found the answer using taylor's formula but I'm looking for other solutions maybe using trigonometric identities .
| An answer not using trig identities: rewrite the function as
$$
\begin{align}
&\frac{1}{x^2}\left[1-\left(1-x^2\frac{1-\cos x}{x^2}\right)\left(1-4x^2\frac{1-\cos 2x}{4x^2}\right)\left(1-9x^2\frac{1-\cos 3x}{9x^2}\right)\right]=\\
&\quad=\frac{1}{x^2}[1-(1-x^2f(x))(1-4x^2f(2x))(1-9x^2f(3x))]=\\
&\quad=f(x)+4f(2x)+9f(3x)-x^2[4f(x)f(2x)+9f(x)f(3x)+36f(2x)f(3x)]+36x^4f(x)f(2x)f(3x)
\end{align}
$$
with
$$
f(x)=\frac{1-\cos x}{x^2}
$$
In the limit $x\to0$, only the first three terms survive and they give
$$
\frac{1}{2}(1+4+9)=7
$$
| {
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"question_score": "2",
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"answer_id": 6
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Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$ Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$
I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$
$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$
$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $$
From here, I got stuck. I have to remove y from here to solve it.
The answer in the textbook gave $ y= (k(1+x^2) - 1)/{2}$ I believe $k$ is the integration constant. How do I remove the $\ln$ from both sides?
| First double both sides to delete the halves, then exponentiate. Thus $k=\pm\exp 2C$, the sign being due to removing the modulus signs.
| {
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"url": "https://math.stackexchange.com/questions/2768904",
"timestamp": "2023-03-29T00:00:00",
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Solve the differential equation: $\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} } $ $$\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} }$$
Simplified it to: $$ \int \frac{1}{\sqrt{1-y^2}} dy = \int \frac{1}{\sqrt{1-x^2}} dx$$
$$\implies\sin^{-1} y = ( \sin^{-1} x + C)$$
$$y = (\sin (\sin^{-1} x) \cos C_1) + \cos(\sin^{-1} x) (\sin C_1)$$
How to I simplify $\cos(\sin^{-1} x)$ to get the final answer ?
| Put,
$y$ = $\sin \alpha$ and $x$ = $\sin \beta$
calculate $dy$ and $dx$,
$dy$ = $\cos \alpha$$.d \alpha$ and $dx$ = $\cos \beta$$.d\beta$.
The equation will be something like,
=> $\int \cot\alpha $ $d\alpha$= $\int \cot\beta$ $d\beta$
=> $\log$ |$sin$ $\alpha$| + $c_1$ = $\log$ |$sin$ $\beta$| + $c_2$
=> $\left(\frac{sin \alpha} {sin\beta}\right)$ = $e^{c_2 - c_1}$ = $c$
=> $sin$ $\alpha$ = $sin$ $\beta$ $.c$
Now, replace $sin$ $\alpha$ and $sin$ $\beta$, with $y$ and $x$.
$y$ = $x$ $.c$
| {
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Limit of a Function involving tangent function and limits at infinity Determine $$\lim_{x \to \infty}\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$.
Attempt
Let $$y=\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$
Put $\frac{1}{x}=p$.
$$\lim_{p \to 0}\left(\tan{\frac{\pi}{2+p}}\right)^p$$.
We have
$$\lim_{x \to \infty} y=\lim_{p \to 0}\left(\tan{\frac{\pi}{2+p}}\right)^p$$.
Now consider the function $y$ in variable $p$
Taking $ln$ both sides
$$ln\left(y\right)=p.ln\left(\tan{\frac{\pi}{2+p}}\right)$$.
$$ln\left(y\right)=p.\frac{ln\left(\tan{\frac{\pi}{2+p}}\right)}{\tan{\frac{\pi}{2+p}}}.\tan{\frac{\pi}{2+p}}$$.
Putting $\tan{\frac{\pi}{2+p}}=m$
We have
$$ln\left(y\right)=p.\frac{ln\left(m\right)}{m}.\tan{\frac{\pi}{2+p}}$$.
As $x \to \infty$ we have $p \to 0$ and hence $m \to \infty$
Hence the limit of $\frac{ln\left(m\right)}{m}$ is $0$.
But I am unable to show the limit of other to part of the product.
Please help me out.
| $$L =\lim_{x \to \infty}\left(\tan{\frac{\pi x}{2x+1}}\right)^\frac{1}{x}$$
$$\lim_{x \to \infty}g(x) = \lim_{x \to \infty}\left(\tan{\frac{\pi x}{2x+1}}\right) = \lim_{x \to \infty}\left(\frac{2 + 4 x}{\pi}\right)\tag{1}$$
So
$$L = e^{\left(\lim_{x \to \infty}\frac{1}{x}\ln(g(x))\right)}$$
$$\lim_{x \to \infty}\left(\frac{\ln\left(\frac{2 + 4 x}{\pi}\right)}{x}\right) \stackrel{I´H}{\implies} \lim_{x \to \infty}\left(\frac{2}{2x + 1}\right) = 0$$
Therefore
$$L = e^0 = 1$$
(1) Proof :
For $x \to\infty$
$$g(x) = \tan\left(\frac{\pi x}{2x+1}\right) =\tan\left(\frac{\pi}{2}\frac{1 }{\left(1+\frac{1}{2x}\right)}\right) =\tan\left(\frac{\pi}{2}\frac{1 }{\left(1+\frac{1}{2x}\right) } - \frac{\pi}{2} + \frac{\pi}{2}\right)$$
$$\tan\left(\frac{\pi}{2}\left(\frac{1 }{\left(1+\frac{1}{2x}\right) } - 1\right) + \frac{\pi}{2}\right) = \tan\left(\frac{\pi}{2}- \frac{\pi}{2}\left(\frac{1}{2x+1 }\right)\right)$$
Set $\frac{1}{\left(2x+1\right) } = t$, observe that, as $t \to 0$
$$
\begin{align}
\tan\left(\frac{\pi}{2} - \frac{\pi}{2}t\right)
= \tan\left(\frac{\pi}{2}(1-t)\right) = \cot\left(\frac{\pi}{2}t\right) = \frac{\cos(\pi t) +1 }{\sin(\pi t)}=\\
\end{align}$$
$$\frac{\pi t}{\pi t}\cdot\frac{\cos(\pi t) +1 }{\sin(\pi t)}=$$
$$\frac{1}{t\pi}\cdot\underbrace{\left(\frac{\pi t }{\sin(\pi t)}\left(\cos(\pi t) +1\right)\right)}_{t \to 0 \implies 2}=
\frac{1}{t\pi}\cdot 2 = $$
$$\frac{1}{\left(\frac{1}{\left(2x+1\right)}\right)\pi}\cdot 2 =
\frac{(2 + 4 x)}{\pi}\tag*{$\Box$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Shortest distance between ellipsoid and plane Find the shortest distance between the points $A = (x_1,y_1,z_1)$ and $B = (x_2,y_2,z_2)$ if $A$ lies on the plane $x+y+z=2a$ and $B$ lies on the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
| The plane containing $A$:
$$x+y+z=2a \tag{1}$$
Direction ratio of the normal at $A$ is $1:1:1$
Equation of tangent plane at $B(x_2,y_2,z_2)$:
$$\frac{x_2 x}{a^2}+\frac{y_2 y}{b^2}+\frac{z_2 z}{c^2}=1 \tag{2a}$$
Equating the direction ratios of their normals
$$(x_2,y_2,z_2)=t(a^2,b^2,c^2)$$
As $B$ lies on the ellipsoid,
$$\frac{(a^2t)^2}{a^2}+\frac{(b^2t)^2}{b^2}+\frac{(c^2t)^2}{c^2}=1$$
$$t=\frac{1}{\sqrt{a^2+b^2+c^2}}$$
So the tangent plane is
$$x+y+z=\sqrt{a^2+b^2+c^2} \tag{2b}$$
The distance between the two planes
$$\fbox{$\frac{2a}{\sqrt{3}}-\sqrt{\frac{a^2+b^2+c^2}{3}}$}$$
provided $3a^2>b^2+c^2$ for no intersections between the plane and the ellipsoid.
Useful facts
Distance between two parallel planes $Ax+By+Cz=D$ and $Ax+By+Cz=D'$:
$$d=\frac{|D-D'|}{\sqrt{A^2+B^2+C^2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum splitting field of the polynomial over GF(2) I need to find the minimum splitting field of the polynomial over GF(2): $$x^5+x^4+1.$$
I find that $x^5+x^4+1 = (x + \alpha)^2(x + \alpha + 1)(x + \alpha^2)(x + \alpha^2 + \alpha)$ over GF(8). But I don't know if it's right, because $x =\alpha $ is a multiple root.
Am I right? And what is right? Help me please!
| $x^5+x^4+1=(x^2 + x + 1) (x^3 + x + 1)$ in $GF(2)$.
These factors are irreducible because they have no root in $GF(2)$.
The splitting field of $x^2 + x + 1$ has degree $2$ over $GF(2)$.
The splitting field of $x^3 + x + 1$ has degree $3$ over $GF(2)$.
Therefore, the splitting field of $x^5+x^4+1$ has degree $6$ over $GF(2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Matrices satisfying $(A-B)C=BA^{-1}$ where $A$ is Non singular If $A$,$B$,$C$ are Square matrices satisfying $$(A-B)C=BA^{-1},$$ where $A$ is nonsingular.
Then which is true among these?
*
*$C(A-B)=BA^{-1}$
*$(A-B)C=A^{-1}B$
*$C(A-B)=A^{-1}B$
*$(A-B)^{-1}=C+A^{-1}$
My try:
Since $A$ is invertible,
$$(A-B)CA=B$$ $\implies$
$$ACA=B(CA+I)$$
$$ACA-ACB=BCA-ACB+B$$ $\implies$
$$AC(A-B)=BCA-ACB+B.$$
Now pre multiplying with $A^{-1}$ we get
$$A^{-1}AC(A-B)=A^{-1}BCA-CB+A^{-1}B$$ $\implies$
$$C(A-B)=A^{-1}BCA-CB+A^{-1}B.$$
Any help here?
| (1) and (2) are incorrect. Take $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $C = \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix}$. Verify that the given equation is satisfied.
$$(A-B)C = \begin{pmatrix} ? & -1 \\ ? & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\
BA^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$
but
$$C(A-B) = \begin{pmatrix} 0 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix} \ne BA^{-1}$$
$$A^{-1}B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ -1 & 0 \end{pmatrix} \ne (A-B)C.$$
(4) is correct. To show this, consider
$$(A-B)(C+A^{-1}) = (A-B)C + I - BA^{-1} = [(A-B)C - BA^{-1}] + I = I.$$
Use (4) to prove (3). From (4), we have $C = (A-B)^{-1} - A^{-1}$.
$$C(A-B) = [(A-B)^{-1} - A^{-1}] (A-B) = I - A^{-1} (A-B) = A^{-1}B$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the sum of numbers between 250 and 350 which are divisible by 7?
What is the sum of numbers between $250$ and $350$ which are divisible by $7$?
My attempted solution:
7|250|35
|21 |
-----
40
35
----
5
The 1st number divisible by $7$ is: $(250-5)+7 = 252$.
7|350|50
|35 |
-----
0
The last number divisible by $7$ is: $350$.
Total numbers between 250 and 350 divisible by $7$ is:
$((350-252)/7) + 1$
$ = (98/7)+1$
$ = 14+1$
$ = 15$
$ Sum = \frac{15}{2} [2 \cdot 252 + (15 - 1) \cdot 7 ]$
$ = \frac{15}{2} [2 \cdot 252 + 14 \cdot 7 ]$
$ = \frac{15}{2} [504 + 98 ]$
$ = \frac{15}{2} \cdot 602$
$ = 15 \cdot 301$
$ = 4515 $
(Ans.) $4515$
Is there a more efficient way to do this?
| $$
252 + 259 + 266 + 273+\cdots\cdots+336+343+350
$$
Since this is an arithmetic sequence, i.e. the difference between each number and the next is the same in all cases, the average of all the numbers in the list must be the same as the average of the first one and the last one:
$$
\frac{252+350} 2 = 301
$$
Therefore the sum is $(301\times\text{the number of terms}).$
Since you have to add $7$ fourteen times, there are $15$ numbers, so the answer will be $301\times15.$
| {
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Is this Evaluation of definite Integral correct? To Evaluate $$I=\int_{-1}^{1} \frac{dx}{x^2+x+1+\sqrt{x^4+3x^2+1}}$$
I rationalized the denominator getting as
$$I=\int_{-1}^{1}\frac{x^2+x+1-\sqrt{x^4+3x^2+1}}{2x(x^2+1)}$$ $\implies$
$$I=\int_{-1}^{1}\frac{dx}{2x}+\int_{-1}^{1}\frac{dx}{2(x^2+1)}+\int_{-1}^{1}\frac{\sqrt{x^4+3x^2+1}\:dx}{2x(x^2+1)}$$
First and Third integrals are zero as the Integrands are odd functions and hence
$$I=\int_{-1}^{1}\frac{dx}{2(x^2+1)}=\frac{\pi}{4}$$
|
Is this Evaluation of definite Integral correct?
No.
Your final result is right, but we have
$$
\int_{-1}^{1}\frac{dx}{x}\ne0,\qquad \int_{-1}^{1}\frac{\sqrt{x^4+3x^2+1}\:dx}{x(x^2+1)}\ne0, \tag1
$$ each integral being divergent, it is the difference of the preceding integrals which vanishes, that is
$$
\int_{-1}^{1}\left(\frac{1}{x}-\frac{\sqrt{x^4+3x^2+1}\:dx}{x(x^2+1)}\right)dx=0. \tag2
$$To prove $(2)$, one may observe the integrand is an odd function over $[-1,1]$ and one may observe the integral is convergent, since the singularity of the integrand as $x \to 0$ is removable,
$$
\left(\frac{1}{x}-\frac{\sqrt{x^4+3x^2+1}\:dx}{x(x^2+1)}\right) \sim -\frac x2,
$$ the function being continuous elsewhere.
| {
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Let $P(x)=x^2 -1$. Find out number of distinct real roots of $P(P(\cdots P(x))\cdots)=0$ where there are $2018$ $P$. Let $P(x)=x^2 -1$. Find out number of distinct real roots of $P(P(\cdots P(x))\cdots)=0$ where there are $2018$ $P$.
My Attempt
Let $Q(x)=P(P(\cdots P(x))\cdots)=0$ where there are $2017$ $P$ is a root of $P(x)=0$. Thus $Q(x)=\pm1$. But from this point I am lost how to proceed. Can anyone help. Thanks in advance.
| Let $a_n$ be the number of roots of the equation with $n P$'s.
Now we have $a_1= 2$. and $a_2 =3$.
For simplicity, denote $P_n$ with the equation of $n P$'s.
For $n \ge 2$, we have : $$ P_n^2 -1 = ((P_{n-2}^2-1)^2-1)^2 -1 = (P_{n-2}^2 -1)^2((P_{n-2}^2-1) -\sqrt{2})((P_{n-2}^2-1) + \sqrt{2}). $$ (Here $P_0(x) =x$).
Now the first factor has roots $a_{n-2}$. The third factor has no roots. We show that the second factor has 2 roots.
Let $g_n(x)= P_{n-2}^2 - (\sqrt{2}+1)$.We claim that the roots are $$\pm \sqrt{\sqrt{\sqrt{\cdots \sqrt{\sqrt{2}+1~ (\text{n -1 times})}\cdots +1}+1}+1} .$$
This can be verified by factorizing and seeing that we have a expression of form $(g^2-1)^2 - b$ where $b>1$, then only the factor $g^2 - (\sqrt{b}+1)$ can only real root in the factorization of the expression.
Thus we have $a_n = a_{n-2}+2$. Thus the sequence is natural numbers. So the answer shall be then $2017$.
| {
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"url": "https://math.stackexchange.com/questions/2781913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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degree of $\sqrt{2} + \sqrt[3]{3}$ over $\mathbb Q$ How can i find the degree of the minimal polynomial $P \in \mathbb Q[x]$ such that $P(\sqrt{2} + \sqrt[3]{3}) = 0$ ?
Recently i proved that $\mathbb Q[\sqrt{2} + \sqrt{3}] = \mathbb Q[\sqrt{2}, \sqrt{3}]$ using $(\sqrt{2} + \sqrt{3})^{-1} = \sqrt{3} - \sqrt{2}$, so $2\sqrt{3} = (\sqrt{3} - \sqrt{2}) + (\sqrt{2} + \sqrt{3})$ etc.
But how can i express $\sqrt{2}$ or $\sqrt[3]{3}$ with $\sqrt{2} + \sqrt[3]{3}$?
Is $\mathbb Q[\sqrt{2} + \sqrt[3]{3}]$ equal to $\mathbb Q[\sqrt{2}, \sqrt[3]{3}]$?
Thanks
| Hint: It is enough to prove that $\mathbb Q[\sqrt{2} + \sqrt[3]{3}] =\mathbb Q[\sqrt{2},\sqrt[3]{3}]$ because then the facts below imply that the degree of $\sqrt{2} + \sqrt[3]{3}$ is exactly $6$:
*
*$\mathbb Q[\sqrt{2},\sqrt[3]{3}]$ has degree at least $6$
*$\sqrt{2} + \sqrt[3]{3}$ is a root of a polynomial of degree $6$
*$\mathbb Q[\sqrt{2} + \sqrt[3]{3}]$ has degree at most $6$
Alternatively,
$\sqrt{2} + \sqrt[3]{3}$ is a root of $x^6 - 6 x^4 - 6 x^3 + 12 x^2 - 36 x + 1$, which is irreducible because it is irreducible mod $13$. But that's a lot of work to do by hand.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the polynomial $x^2+2x+1$ divides the polynomial $x^{2(2n+1)}+2x^{2n+1}+1$ for every $n\in \mathbb{N}$ I've been trying to prove that the polynomial $x^2+2x+1$ divides the polynomial $x^{2(2n+1)}+2x^{2n+1}+1$ for all $n\in \mathbb{N}$ as follows and reached a "roadblock":
I wrote $x^2+2x+1$ as $(x+1)^2$ and showed that $x=-1$ is a root of the 2nd polynomial, hence by Bezout's Little Theorem the polynomial $x+1$ divides $x^{2(2n+1)}+2x^{2n+1}+1$, but how do I continue from here?
| Take $x^{2m+1}=a$
Now, $x^{2(2m+1)}+2x^{2m+1}+1=a^2+2a+1$
$\Rightarrow$$(a+1)^2$ $\Rightarrow$$(x^{2m+1}+1)^2$
So, $x+1|x^{2m+1}+1$ $\Rightarrow$$(x+1)^2|x^{2(2m+1)}+2x^{2m+1}+1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding value of $\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$
Finding value of $$\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$$
Solution I tried:
$$(1+x)^{10}=1+\binom{10}{1}x+\binom{10}{2}x^2+\cdots +\binom{10}{10}x^{10}$$
$$(x+1)^{10}=x^{10}+\binom{10}{1}x^9+\binom{10}{2}x^7+\cdots +\binom{10}{10}$$
I did not find how do multiply terms such that i get my result. Help me
| To complete the method sketched in the comments.
The answer is $\binom {13}5$.
To see this, suppose we are choosing $5$ values from $\{1,\cdots, 13\}$. Order the choices as $n_1<\cdots <n_5$. Let $n_3=k+1$. Clearly $2≤k≤10$. Given a choice of such an $n_3$ we see that we are asked to choose $2$ values from $\{1,\cdots, k\}$ and $2$ from $\{k+2,\cdots, 13\}$. There are $\binom k2$ ways to do the former and $\binom {13-(k+2)+1}2=\binom {12-k}2$ ways to do the latter. Hence the number of choices with a given $n_3$ is the product $\binom {k}2\times \binom {12-k}2$. Summing over $k$ we see that $$\binom {13}5=\sum_{k=2}^{10}\binom {k}2\times \binom {12-k}2$$ as desired.
| {
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Calculate the flux of $F=(x+3y^5, y+10xz, z-xy),$ $ S$ the closed hemisphere bounded by $x^2+y^2+z^2=1, z\ge0$. Calculate the flux across the surface $S$ of $F=(x+3y^5, y+10xz, z-xy),$ $ S$ the hemisphere bounded by $x^2+y^2+z^2=1, z\ge0$.
I have done $n=(x,y,z)$
thus $\iint F\cdot n dS= \iint(x+3y^5, y+10xz, z-xy)\cdot(x,y,z)dS = \iint(1+3xy^5+9xyz)dS$
since $z=(1-x^2+y^2)^{1/2}$ it follows $\int_{-1}^1 \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}}(1+3xy^5+9xy(1-x^2+y^2)^{1/2})dS$
then
$\int_{-1}^1 \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}}1dydx=\pi,$
$\int_{-1}^1 \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}}3xy^5dydx=0$ and $\int_{-1}^1 \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}}9xy(1-x^2+y^2)^{1/2}dydx=0,$
so $\iint F\cdot n dS=\pi$
I want to know if this result is correct by using the Divergence Theorem, Does anybody can help me?
| $$div F=(F_1)'_x+(F_2)'_y+(F_3)'_z=3$$
$$\int\int\int_V3dxdydz=3.\frac {2}{3}\pi 1^3=2\pi $$
| {
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If three sides of an acute triangle is $3$ cm, $4$ cm, and $x$ cm. What are the possible values of $x$? Problem: If three sides of an acute triangle is 3 cm, 4 cm, and $x$ cm. What are the possible values of $x$?
2 of the choices: (A) $1<x<5$, and (B) $0<x<7$
Solution: According to triangle theorems, 2 sides of a triangle is greater than the 3rd side and their difference is less than the 3rd side.
From one of the choices, (ranges $0<x<7$) when $x = 6$:
*
*$3 + 4 > 6$
*$6 + 4 > 3$
*$3 + 6 > 4$
From the above, it means $6$ is a possible value for $x$, and since choice A doesn't have $6$ in it, I chose B as the correct answer. However that isn't the case, the correct answer according to the book is A, which I don't know why. So how was it $1<x<5$? Any help would be appreciated.
| Let the angles opposite the sides of lengths $x,3,4$ be $\theta,\omega, \phi$ respectively. Then by the Cosine Rule we obtain the relations: $$x^2=3^2+4^2-2\cdot3\cdot4\cos \theta,$$ $$3^2=x^2+4^2-2\cdot x\cdot4\cos \omega$$ and $$4^2=x^2+3^2-2\cdot x\cdot3\cos \phi$$ for $0<\cos \theta, \cos \omega,\cos \phi<1$.
From the first relation we have that $x^2=25-24\cos\theta$, which tells us that the extremes of $x^2$ are $25-24\times 1=1$ and $25-24\times 0=25$, so that $1<x^2<25$, or $1<x<5$.
From the second relation we deduce that $x=4\cos\omega\pm\sqrt{16\cos^2\omega-7}$, from which we learn that $x=\sqrt{-7}$ when $\cos\omega=0$ and $x=1,7$ when $\cos\omega=1$. Any way we look at these, they tell us nothing interesting.
From the third relation we have that $x=3\cos\phi\pm\sqrt{9\cos^2\phi+7}$, which tells us that $x=\sqrt7$ when $\cos\phi=0$ and $x=-1,7$ when $\cos\phi=1$. We have the possible interesting relations $\sqrt7<x<7$.
Finally, combining the inequalities $1<x<5$ and $\sqrt 7<x<7$, we have the tighter bounds $\sqrt 7<x<5$.
PS. I have tacitly assumed throughout that the sinusoidal functions $x(\theta),x(\omega),x(\phi)$ which we analysed are either strictly increasing or strictly decreasing (indeed they're all strictly decreasing) wherever they are defined in the intervals $0<\theta,\omega,\phi<π/2$ respectively. This can however be easily justified if needed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int \frac {dx}{1+\sin x}$ Integrate $\int \dfrac {dx}{1+\sin x}$
My attempt:
$$\begin{align}&=\int \dfrac {dx}{1+\sin x}\\
&=\int{\dfrac{dx}{1+\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1+\tan ^2\left( \dfrac{x}{2} \right)}}}\\
&=\int{\dfrac{1+\tan ^2\left( \dfrac{x}{2} \right)}{\left( 1+\tan \left( \dfrac{x}{2} \right) \right) ^2}}\,dx\end{align}$$
| Hint:$$\int\frac1{1+\sin x}\,\mathrm dx=\int\frac{1-\sin x}{\cos^2x}\,\mathrm dx=\int\frac1{\cos^2x}\,\mathrm dx-\int\frac{\sin x}{\cos^2x}\,\mathrm dx.$$
| {
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Find $\int\arcsin(\sqrt{x})dx$
Find $\displaystyle\int\arcsin(\sqrt{x})dx$
My Attempt
Put $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$
$$
\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\
=y^2\arcsin(y)-\int\frac{y^2}{\sqrt{1-y^2}}dy.
$$
How do I proceed further and find the solution or is there any easier way ?
| Hint. Note that
$$-\int\frac{y^2}{\sqrt{1-y^2}}\,dy=\int\frac{1-y^2}{\sqrt{1-y^2}}\,dy-\int\frac{dy}{\sqrt{1-y^2}}=\int\sqrt{1-y^2}\,dy-\arcsin(y).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.$ I tried to use the series for $\sin \pi x$ and maybe find something related to $\zeta(3)$, but didn't work. I'm guessing this integral needs more than the little calculus that I know.
\begin{equation}
\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}.
\end{equation}
| $\displaystyle f(a):=\int\limits_0^1 x e^{ax}dx = \frac{1+e^a(a-1)}{a^2}$
$\displaystyle g(a):=\int\limits_0^1 x^2 e^{ax}dx = \frac{-2+e^a(a^2-2a+2)}{a^3}$
$\displaystyle \int\limits_0^1 \frac{x^2-x}{\sin(\pi x)}dx = i2\int\limits_0^1\frac{x^2-x}{e^{i\pi x}-e^{-i\pi x}}dx = i2\sum\limits_{k=0}^\infty \int\limits_0^1 (x^2-x)e^{-i\pi x(2k+1)}dx $
$\displaystyle = i2\sum\limits_{k=0}^\infty (g(-i\pi(2k+1))-f(-i\pi(2k+1)))$
$\displaystyle = 2\sum\limits_{k=0}^\infty i\frac{-2+i\pi(2k+1) + e^{-i\pi(2k+1)}(2+i\pi(2k+1))}{(-i\pi(2k+1))^3} \enspace$ with $\enspace e^{-i\pi(2k+1)}=-1$
$\displaystyle = -8\sum\limits_{k=0}^\infty\frac{1}{(\pi(2k+1))^3}=-\frac{8}{\pi^3}(1-\frac{1}{2^3})\zeta(3)=-\frac{7\zeta(3)}{\pi^3}$
| {
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} |
Derivative of a trigonometric function $\arctan$ Will someone help me explain this example please?
$$y=\arctan \dfrac{\sqrt{1-\cos (x)}}{\sqrt{1+\cos (x)} }$$
Solution:
$$\sin^2 (x/2)=\dfrac{1-\cos (x)}{2}\Rightarrow \sqrt{2}\sin (x/2)=\sqrt{1-\cos (x)}$$
$$\cos^2 (x/2)=\dfrac{1+\cos (x)}{2}\Rightarrow \sqrt{2} \cos (x/2)=\sqrt{1+\cos (x)}$$
$$y=\arctan\dfrac{\sqrt{1-\cos (x)}}{\sqrt{1+\cos (x)}}=\arctan\dfrac{\sin (x/2)}{\cos (x/2)}=\arctan (\tan (x/2))=\dfrac{x}{2}$$
$$y=\dfrac{x}{2}\rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}$$
| Without the preliminary simplifications.
Let $$y=\tan^{-1}(u) \qquad \text{with} \qquad u=\dfrac{\sqrt{1-\cos (x)}}{\sqrt{1+\cos (x)} } $$
$$\frac{dy}{dx}=\frac 1 {1+u^2}\frac{du}{dx}=\cos ^2\left(\frac{x}{2}\right)\,\frac{du}{dx}$$ Now $$\log(u)=\frac 12 \log(1-\cos(x))-\frac 12 \log(1+\cos(x))$$ Differentiate both sides
$$\frac{du}u=\frac{1}{2} \left(\frac{\sin (x)}{1-\cos (x)}+\frac{\sin (x)}{1+\cos (x)}\right)\,dx=\csc (x)\,dx\implies \frac{du}{dx}=\csc (x)\dfrac{\sqrt{1-\cos (x)}}{\sqrt{1+\cos (x)} }$$ Now, we make the simplifcation to get
$$\frac{du}{dx}=\sqrt{\tan ^2\left(\frac{x}{2}\right)} \csc (x)\implies \frac{dy}{dx}=\frac{1}{2} \frac{\sqrt{\tan ^2\left(\frac{x}{2}\right)}}{ \tan \left(\frac{x}{2}\right)}$$ and you see the problem related to the sign of $\tan \left(\frac{x}{2}\right)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I know when to take ordering into account?
I'm quite confused as to why in (c) the ordering matters but not in (b)
I kind of understand in (a) ordering doesn't really matter since all the numbers are the same therefore switching places doesn't really make any difference.
However, for (c) the solution is showing that $3* \frac{1}{6} * \frac{5}{6}$ I assume it means that we have to take ordering into account. That means $5,1,5$ is different from $5,5,1$.
Following the same logic, for (b) why don't we also take ordering into account? Because clearly $6,2,3$ is different from $2,3,6$, right?
What am I missing here?
| It helps to distinguish between the dice. Suppose the dice are, respectively, blue, green, and red. Then an outcome can be represented by the ordered triple $(b, g, r)$, where $b, g, r \in \{1, 2, 3, 4, 5, 6\}$.
I roll three six-sided fair dice. What is the probability that all three dice show the same number?
Since there are six possible outcomes for each die, there are $6^3$ possible outcomes in the sample space.
Of these, just six show the same number on each die.
Hence,
$$\Pr(\text{same number on each die}) = \frac{6}{6^3} = \frac{1}{6^2} = \frac{1}{36}$$
I roll three six-sided fair dice. What is the probability that each die shows a different number?
The sample space is the same as above.
For each of the six possible numbers on the blue die, there are five possible numbers on the green die that are different from the one on the blue die, and four possible numbers on the red die that differ from both the number on the blue die and the number on the green die. Hence, there are $6 \cdot 5 \cdot 4$ favorable outcomes.
Hence,
$$\Pr(\text{different number on each die}) = \frac{6 \cdot 5 \cdot 4}{6^3} = \frac{5 \cdot 4}{6^2} = \frac{5}{9}$$
I roll three six-sided fair dice. What is the probability that all two dice show one number and the other die shows a different number?
Again, the sample space is the same as above.
There are $\binom{3}{2}$ ways to select which two of the three dice show the same outcome. There are six possible numbers for both of those dice to show. For each of these possibilities, there are five possible numbers the third die could show. Hence, the number of favorable outcomes is
$$\binom{3}{2} \cdot 6 \cdot 5 = 3 \cdot 6 \cdot 5$$
Hence,
$$\Pr(\text{two dice show the same number and the other die shows a different number})\\
= \frac{3 \cdot 6 \cdot 5}{6^3} = \frac{3 \cdot 5}{6^2} = \frac{5}{12}$$
Since these three cases are mutually exclusive and exhaustive, their probabilities must have sum $1$, which you can verify by direct calculation.
Note: The reason for distinguishing between the dice is to ensure that all outcomes are equally likely. For instance, if all three dice were white, the outcome $(1, 2, 3)$ would occur $3! = 6$ times as often as $(1, 1, 1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\int e^{-3x} \cos^3x\,dx$
Evaluate $\int e^{-3x}\cos^3x\,dx$
My Attempt
\begin{align}
& \int e^{-3x} \cos^3x\,dx=\cos^3x \cdot \frac{e^{-3x}}{-3}-\int3\cos^2x\sin x \cdot \frac{e^{-3x}}{-3} \, dx \\
= {} &\frac{-1}{3}\cos^3x \cdot e^{-3x}+\int\cos^2x\sin x \cdot e^{-3x} \, dx\\
= {} & \frac{-1}{3}\cos^3x\cdot e^{-3x}+\cos^3x\cdot e^{-3x} - \int\bigg[-2\cos x\sin x\cdot e^{-3x}-3\cos^2x\cdot e^{-3x}\bigg](-\cos x)\,dx\\
= {} & \frac{-1}{3}\cos^3x\cdot e^{-3x}+\cos^3x\cdot e^{-3x}+2\int \cos^2x\sin x \cdot e^{-3x} \, dx-3\int\cos^3x\cdot e^{-3x}\,dx
\end{align}
How do I solve this integral ?
| Hint: try to use the trigonometric identity
$$\cos^3(x)= \frac{1}{4}(3\cos(x)+\cos(3x))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Solving recurrent equation $a_1=3; a_2=1; a_n = a_{n-1} + 2a_{n-2}+2$ I have the following recurrent equation
$
a_1=3\\a_2=1\\a_n=a_{n-1}+2a_{n-2}+2
$
and I cannot seem to be able to solve it correctly. According to wolframalpha and my math textbook (that contains results but not the actual process), the result should be $a_n=-2(-1)^n+2^n-1$, but I cannot seem to actually get that result and I cannot find where I am making a mistake.
My current approach (based on the textbook) is:
$a_{n=1} = 2$, needs to add 1.
$a_{n=2} = 5$, needs to substract 4.
The universal equation therefore is $a_n=a_{n-1}+2a_{n-2}+2+[n=1]-4[n=2]$.
$A(x)=xA(x) + 2x^2A(x)+2+x-4x^2$, divide both sides by $A(x)$:
$1= x + 2x^2 + {\frac {2+x-4x^2} {A(x)}}$, move $A(x)$ to one side:
$A(x) = {\frac {-4x^2+x+2} {-2x^2 -x +1}}$, divide the polynomials:
$A(x) = 2+{\frac {3x} {-2x^2-x+1}}$, split into multiple fractions:
$A(x) = 2+3x * \frac {1} {1-2x} * \frac {1} {1+x}$.
This is the point where I get lost, how exactly am I supposed to expand something like this into a power series? If I made a mistake in the process, could someone explain, why? Does anyone have other solutions for this problem?
Thanks a lot.
| Well I would do like this. Since for all $n$ we have:
$$
a_n-a_{n-1}-2a_{n-2}=2 $$
we have also (change $n$ to $n+1$) $$a_{n+1}-a_{n}-2a_{n-1}=2$$
so if we substract these two we get: $$a_{n+1}-2a_n-a_{n-1}+2a_{n-2}=0$$
so zeroes of characteristic equation $x^3-2x^2-x+2=0$ are $1,2,-1$ so $$a_n = a\cdot (-1)^n+b\cdot 2^n+c\cdot 1^n$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the marginal probability density function The random vector $[\,X \,\,\, Y \,]'$ has probability density function
$f_{X,Y} (x,y) = ke^{-2x^2-3xy-\frac{9}{2}y^2}$, where $k$ is some constant
Find $k.$
Find the marginal probability density functions of $X$ and $Y.$
I know for it to be a valid pdf its integral from negative to positive infinity must be equal to one, and that it must be greater than $0$ for all $x.$ But for starters I'm not sure on the integration.
| A standard form of the bivariate normal density with expected value $(0,0)$ is this:
$$
f(x,y) = \frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}}
\exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{x^2}{\sigma_X^2} + \frac{y^2}{\sigma_Y^2} - \frac{2\rho xy}{\sigma_X \sigma_Y} \right] \right)
$$
The exponent is this:
$$
-\frac 1 {2(1-\rho^2)}\left[ \frac{x^2}{\sigma_X^2} + \frac{y^2}{\sigma_Y^2} - \frac{2\rho xy}{\sigma_X \sigma_Y} \right] = -2x^2-3xy-\frac{9}{2}y^2
$$
So
$$
\frac 1 {1-\rho^2} \left[ \left( \frac x {\sigma_X}\right)^2 + \left( \frac y {\sigma_Y} \right)^2 - 2\rho\left( \frac x {\sigma_X} \right)\left( \frac y {\sigma_Y} \right) \right] = 4x^2 + 6xy +9y^2 \tag 1
$$
We can see that this is positive-definite by completing the square and writing it as
$$
\left(2x + \frac 3 2 y \right)^2 + \frac {27} 4 y^2,
$$
so it's always positive.
Equating coefficients, we get
\begin{align}
& \frac 1 {(1-\rho^2)\sigma_X^2} = 4 \\[10pt]
& \frac 1 {(1-\rho^2)\sigma_Y^2} = 9 \\[10pt]
& \frac{-2\rho}{(1-\rho^2)\sigma_X\sigma_Y} = 6
\end{align}
From this it follows that
\begin{align}
\sigma_X^2 & = 1/3, \\[8pt]
\sigma_Y^2 & = 4/27, \\[8pt]
\sigma_X \sigma_Y & = 2/9, \\[8pt]
\rho & = -1/2.
\end{align}
The marginal density for $X$ is
$$
\frac 1 {\sigma_X\sqrt{2\pi}} \exp\left( \frac {-1} 2 \left( \frac x {\sigma_X} \right)^2 \right)
$$
and similarly for $Y.$ (If we had $\mu_X\ne0$ then instead of $\dfrac x {\sigma_X}$ we'd have $\dfrac {x-\mu_X}{\sigma_X}.$)
The constant $k$ should be $\dfrac{1}{2\pi\sigma_X\sigma_Y(1-\rho^2)}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove that $\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$ Recently I have been reading physics book and saw interesting equation, like this:
$$\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
But I still don't understand how to get the right part of the equation from left part, and I ask for the explaining of this. Thanks a lot!
| We can prove this in a reverse way
Lets start with $$\frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
$$=\frac{1/a}{\sqrt{\frac{a^2-b^2}{a^2}}}$$
$$=\frac{1/a}{\frac{\sqrt{a^2-b^2}}{a}}$$
$$=\frac{\frac{1}{a}(a)}{\sqrt{a^2-b^2}}$$
$$=\frac{1}{\sqrt {a^2 - b^2}}$$
Hence,$$\frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}=\frac{1}{\sqrt {a^2 - b^2}}$$
which is nothing but,
$$\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integral $I=\int_0^1 \frac{\ln x\arctan(ax)}{1+b^2x^2}dx$ Greetings I am trying to evaluate $$I=\int_0^1 \frac{\ln x\arctan(ax)}{1+b^2x^2}dx$$ Where $a$ and $b$ are positive numbers. My try was to derivate the integral with respect to $a$ in order to get: $$I'(a)=\int_0^1 \frac{\ln x}{(1+a^2x^2)(1+b^2x^2)}dx=\frac{1}{a^2-b^2}\left(\int_0^1\frac{a^2\ln x}{1+a^2x^2}dx-\int_0^1\frac{b^2\ln x}{1+b^2x^2}dx\right)$$ expanding into geometric series using: $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n$$ $$I'(a)=\frac{1}{a^2-b^2}\left(a^2\sum_{n=0}^{\infty}(-1)^na^{2n}\int_0^1x^{2n}\ln xdx-b^2\sum_{n=0}^{\infty}(-1)^nb^{2n}\int_0^1x^{2n}\ln xdx\right)$$ Using the following relation:$$I(k)=\int_0^1x^{2k}dx=\frac{1}{2k+1}\rightarrow I'(k)=\int_0^1 x^{2k}\ln xdx=\frac{-2}{(2k+1)^2}$$ gives: $$I'(a)=\frac{1}{a^2-b^2}\left(-2a\sum_{n=0}^{\infty}(-1)^n\frac{a^{2n+1}}{(2n+1)^2}+2b\sum_{n=0}^{\infty}(-1)^n\frac{b^{2n+1}}{(2n+1)^2}\right)$$ $$I'(a)=\frac{2}{a^2-b^2}\left(b\sin b-a\sin a\right)$$ And finally $$I=I(a)=2\int \frac{b\sin b- a\sin a}{a^2-b^2}da$$ Now I am stuck, could you help me finish this? And do I have any mistakes?
Edit: With the mistake pointed in the comments it yields to: $$I'(a)=\frac{2}{a^2-b^2}\left(\sum_{n=1}^{\infty}(-1)^n\frac{a^{2n}}{(2n)^2}-\sum_{n=1}^{\infty}(-1)^n\frac{b^{2n}}{(2n)^2}\right)=\frac{1}{2}\frac{1}{a^2-b^2}(Li_2(-a^2)-Li_2(-b^2))$$
| For one, you're missing an $x$ in the numerator when you differentiating $\arctan ax$.
Let $F(a)$ denote our integral and differentiating with respect to the parameter gives
$$\begin{align*}F'(a) & =\int\limits_0^1 dx\,\frac {x\log x}{(1+a^2x^2)(1+b^2x^2)}\\ & =\frac {a^2}{a^2-b^2}\int\limits_0^1dx\,\frac {x\log x}{1+a^2x^2}-\frac {b^2}{a^2-b^2}\int\limits_0^1dx\,\frac {x\log x}{1+b^2x^2}\end{align*}$$
Both integrals can be evaluated, like you've done with your work, using the infinite geometric sequence. The first integral, with $a$, is equal to
$$\begin{align*}\int\limits_0^1dx\,\frac {x\log x}{1+a^2x^2} & =-\sum\limits_{n\geq1}\frac {(-a^2)^n}{(2n+2)^2}\\ & =-\frac 14\left[\sum\limits_{n\geq1}\frac {(-a^2)^{n-1}}{n^2}-1\right]\\ & =-\frac 14\left[-1-\frac {\operatorname{Li}_2(-a^2)}{a^2}\right]\\ & =\frac {1}{4}+\frac 1{4a^2}\operatorname{Li}_2(-a^2)\end{align*}$$
Similarly, the other integral is the exact same; but with $a$ replaced with $b$. Putting everything together
$$F'(a)=\frac 1{a^2-b^2}\left[\frac {a^2}4+\frac 14\operatorname{Li}_2(-a^2)-\frac {b^2}4-\frac 14\operatorname{Li}_2(-b^2)\right]$$
Now integrate that monster to retrieve $F(a)$. I will update this when I finish figuring out how to efficiently integrate $F'(a)$.
EDIT: There's no way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show that $(9-5\sqrt 3)(2-\sqrt 2) $ is not a square in $\mathbb Q (\sqrt 2, \sqrt 3) $
Show that $(9-5\sqrt 3)(2-\sqrt 2) $ is not a square in $\mathbb Q (\sqrt 2, \sqrt 3) $; equivalently, prove that $\mathbb Q (\sqrt 2, \sqrt 3, u)$ is of degree $2$ over $\mathbb Q (\sqrt 2, \sqrt 3) $ where $u $ is a number such that $u^2 = (9-5\sqrt 3)(2-\sqrt 2) $.
This question is a small section of what is required for this question: Showing that $\mathbb{Q}(\sqrt{2},\sqrt{3}, \sqrt{(9 - 5\sqrt{3})(2-\sqrt{2})})$ is normal over $\mathbb{Q}$, and finding its Galois group.
The answer of the above question reads as follows:
"Note that $Gal(\mathbb Q(\sqrt 2, \sqrt 3)/ \mathbb Q(\sqrt 3))$ has order $2$, generated by $σ_{1,0} ∈ Gal(\mathbb Q(\sqrt 2, \sqrt 3)/\mathbb Q)$, where $σ_{1,0}$ fixes $\sqrt 3$ and sends $\sqrt 2$ to its negative. Let $N$ denote the norm of the extension $\mathbb Q(\sqrt 2, \sqrt 3)/\mathbb Q(\sqrt 2)$; if $α^2 = x^2$ for some $x ∈ \mathbb Q(\sqrt 2, \sqrt 3)$, then $N(α^2) = N(x^2) = N(x)^2 ∈ \mathbb Q(\sqrt 3)$, i.e. $N(α^2)$ is a square in $\mathbb Q(\sqrt 3)$. We note $N(α^2) = (9 − 5√3)^2(2)$so if $N(α^2)$ is a square in $\mathbb Q(\sqrt 3)$, then so is $2$, a contradiction.
However, as a person totally unfamiliar with field norms, I was wondering if there is any other way to solve this, or at least explaining the similar idea of the proof in simpler notions. Brute-forcingly writing $u^2 = (a+b\sqrt2+c\sqrt3+d\sqrt6)^2$ with all letters as rationals and expanding it seems a lot tedius - may take hours to solve everything, and there's no gaurantee that this would lead to a contradiction. Standard result in Galois theory&Group theory are welcome.
| Here is a more "hands-on" (and systematic) approach. Suppose that
$$u^2=(2-\sqrt{2})(9-\sqrt{3})\tag{1}$$
for some $u\in{\mathbb Q}(\sqrt{2},\sqrt{3})$. You can write $u=v+w\sqrt{3}$
with $v,w\in{\mathbb Q}(\sqrt{2})$. Expanding in (1) yields
$$
(v^2+3w^2)+(2vw)\sqrt{3}=9(2-\sqrt{2})-(2-\sqrt{2})\sqrt{3} \tag{2}
$$
Since $1$ and $\sqrt{3}$ are linearly independent over ${\mathbb Q}(\sqrt{2})$, it follows that
$$
v^2+3w^2=9(2-\sqrt{2}), \ 2vw = -(2-\sqrt{2}) \tag{3}
$$
The second equality yields $w=-\frac{2-\sqrt{2}}{2v}$, and reinjecting this into
the first equality we obtain
$$
v^2+3\bigg(\frac{2-\sqrt{2}}{2v}\bigg)^2=9(2-\sqrt{2}) \tag{4}
$$
which can be rewritten as
$$
v^4-9(2-\sqrt{2})v^2+3\bigg(\frac{2-\sqrt{2}}{2}\bigg)^2=0 \tag{5}
$$
or
$$
v^4-9(2-\sqrt{2})v^2+3(1-\sqrt{2})=0 \tag{6}
$$
Complating the square, we find that this is equivalent to
$$
\bigg(v^2-\frac{9(2-\sqrt{2})}{2}\bigg)^2=78(\sqrt{2}-1) \tag{7}
$$
Writing $v^2-\frac{9(2-\sqrt{2})}{2}$ as $a+b\sqrt{2}$ with $a,b\in\mathbb{Q}$, as in the deduction
of (3) we obtain
$$
a^2+2b^2=-78, 2ab=78 \tag{8}
$$
and the first equality is impossible for rational $a,b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding roots using $\displaystyle\int_0^1(1+\cos^8x)(ax^2+bx+c)dx=\int_0^2(1+\cos^8x)(ax^2+bx+c)dx $ I have the following relation-
$$\displaystyle\int_0^1(1+\cos^8x)(ax^2+bx+c)dx=\int_0^2(1+\cos^8x)(ax^2+bx+c)dx $$
$$\text{then what is the interval in which the root of the equation}\space ax^2+bx+c=0\space\text{lie?}$$
Where $a,b,c$ are non-zero numbers.
I think one should apply mean value theorem here. But I can't proceed. What to do?
| $$\int_0^1(1+\cos^8x)(ax^2+bx+c)dx=\int_0^2(1+\cos^8x)(ax^2+bx+c)dx$$
$$\Leftrightarrow$$
$$\int_0^1(1+\cos^8x)(ax^2+bx+c)dx - \bigg(\int_0^1 (1+\cos^8x)(ax^2+bx+c)dx $$
$$+ \int_1^2 (1+\cos^8x)(ax^2+bx+c)dx \bigg)= 0$$
$$=$$
$$\int_1^2(1+\cos^8x)(ax^2+bx+c)dx=0$$
Observe that :
$$1+\cos^8(x) >0 \; \forall \; x\in \mathbb R \; \text{since} \; -1 \leq \cos x \leq1 $$
But, since $1+\cos^8(x) > 0$ and the integral equals to zero, what can you conclude about $ax^2 + bx + c$ in the interval $[1,2]$ which the integral is defined to be zero ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Obliques asymptotes of a logarithmic and rational Function with absolute value. I'm trying to find the obliques asymptotes of the following function:
$$f(x)=|x-1| \log\Big(\sqrt{x^2 +3x+3}-x-1\Big)$$
Its domain should be this: $D(f)=(-\infty, +\infty)$;
I discussed the absolute values: $$|x-1|=\left\{
\begin{array}{c}
x-1 <=>x\geq1 \\
1-x <=>x\lt 1
\end{array}
\right. \ \text{and}\ \ |x|=\left\{
\begin{array}{c}
x <=>x\geq0 \\
-x <=>x\lt0
\end{array}
\right.; $$
When $f(x)$ approaches to the left border:
$$\lim_{x\to\ - \ \infty} (1-x) \log\Bigg((-x)\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg)=+\infty \ ;$$
When it goes to the right border:
$$\lim_{x\to\ + \ \infty} (x-1) \log\Big(\sqrt{x^2 +3x+3}-x-1\Big)=-\infty \ ;$$
Now I want to find obliques asymptotes.
$$m = \lim_{x\to\ - \ \infty} \frac{f(x)}{x}=\lim_{x\to\ - \ \infty}\frac1x(1-x) \log\Bigg((-x)\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg)= - \infty \ ;$$
For $x\to-\infty$ I do not have oblique asymptote. So, I check for the right border:
$$m = \lim_{x\to\ + \ \infty} \frac{f(x)}{x}=\lim_{x\to\ + \ \infty}\frac1x(x-1) \log\Bigg((x)\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg)= \frac1x x \Bigg(1-\frac1x\Bigg) \log\Bigg(\frac12\Bigg)= \log\Bigg(\frac12\Bigg) \ ;$$
Here comes my problem, I'm not able to find $q$, the "known term".
$$q = \lim_{x\to\ + \ \infty} f(x) - mx=\lim_{x\to\ + \ \infty}(x-1) \log\Bigg((x)\sqrt{1 +\frac3x+\frac3{x^2}}-x-1\Bigg) -\log\Bigg(\frac12\Bigg)x = + \infty\cdot 0 \ ;$$
But this is undetermined form and I'm not able to solve it.
Could someone point me out where I'm doing wrong or just give me a hint to find the correct solution? Thank you.
| We have
\begin{align}
\log\bigl(\sqrt{x^2 +3\,x+3}-x-1\bigr)-\log\Bigl(\frac{1}{2}\Bigr)&=
\log\Bigl(\frac{x+2}{\sqrt{x^2 +3\,x+3}+x+1}\Bigr)-\log\Bigl(\frac{1}{2}\Bigr)\\
&=\log\Bigl(\frac{2\,x+4}{\sqrt{x^2 +3\,x+3}+x+1}\Bigr)\\
&=\log\Bigl(1+\frac{x+3-\sqrt{x^2 +3\,x+3}}{\sqrt{x^2 +3\,x+3}+x+1}\Bigr)\\
&=\log\Bigl(1+\frac{3\,x+6}{\bigl(\sqrt{x^2 +3\,x+3}+x+1\bigr)\bigl(\sqrt{x^2 +3\,x+3}+x+3\bigr)}\Bigr)\\
&\sim\frac{3}{4\,x}\quad\text{as}\quad x\to+\infty.
\end{align}
Then
\begin{align}
f(x)-\log\Bigl(\frac{1}{2}\Bigr)\,x&=
x\Bigl(\log\bigl(\sqrt{x^2 +3\,x+3}-x-1\bigr)-\log\Bigl(\frac{1}{2}\Bigr)\Bigr)-\log\bigl(\sqrt{x^2 +3\,x+3}-x-1\bigr)\\&\sim\frac34+\log2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Laurent series of $f(z) = \frac{5}{z^2 + (2i - 1)z - 2i}$ Let $f: \mathbb{C} \setminus \left\{ 1, -2i \right\} \rightarrow \mathbb{C}$ with $$f(z) = \frac{5}{z^2 + (2i - 1)z - 2i}$$
Determine the Laurent series of $f$ in the annuli $\left\{ z \in \mathbb{C} : 0 < |z + 2i| < \sqrt{5} \right\}$.
I think I didn't understand properly how the Laurent series works. Can someone explain me how to approach problems like this?
| By a standard fraction decomposition, one has
$$
f(z) = \frac{5}{z^2 + (2i - 1)z - 2i}=\frac{1-2i}{z+2i}+\frac{2i-1}{1-z}
$$wich rewrites
$$
\frac{1-2i}{z+2i}+\frac{2i-1}{1-z}=\frac{1-2i}{z+2i}+\frac{2i-1}{1+2i}\cdot\frac{1}{1-\frac{z+2i}{1+2i}}=\frac{1-2i}{1+2i}\cdot \frac1Z+\frac{2i-1}{1+2i}\cdot\frac{1}{1-Z}
$$with
$$
Z=\frac{z+2i}{1+2i},\qquad |Z|<1,
$$ giving
$$
f(z)=\tilde{f}(Z)= \frac{1-2i}{1+2i}\cdot \frac1Z+\frac{2i-1}{1+2i}\cdot\sum_{n=0}^\infty Z^n.
$$ Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $x+y$, if integers $x$ and $y$ satisfy the equation $y+1/x=25/3$ Find $x+y$, if $x$ and $y$ are integers and satisfy the equation $y+1/x=25/3$
so I got to the answer by placing $3$ in the $x$ cause it looked like a fraction and $8$ was left for $y$,
My question is: Is there any other/better way to solve this algebraically?
| We can solve it algebraicly by considering divisiblity by 3 (see below).
But we can also justify why your intuition yields the only possible answer.
No $\frac {25}3 = 8\frac 13$ and $8< 8\frac 13 < 8+1$.
If $x > 1$ then $y < y+ \frac 1x< y + 1$.
So $y = 8; \frac 1x = \frac 13; x=3$.
If $x < -1$ then $y-1 < (y-1) + (1-\frac 1{|x|}) < y$.
So $y-1 = 8$ and $1 -\frac 1{|x|} =\frac 13$ so $\frac 1x = -\frac 23; x = -\frac 32$ which is not an integer.
If $x =\pm 1$ then $y+\frac 1x= y\pm 1$ is an integer and can not equal $8\frac 13$.
And obviously $x \ne 0$.
==== number theory divisibility by $3$ argument below =====
$y+1/x=25/3$
$xy +1 = \frac {25x}3 $
$3$ is prime. $3\not \mid 25$ so $3|x$.
Let $x = 3k$
$3ky + 1 = 25k$.
RHS is a multiple of $K$. LHS is one more than a multiple of $k$. That's only possible if $k =\pm 1$.
$\pm 3y +1 = \pm 25$ or
$3y \pm 1 = 25$
$3y = 24, 26$. Only $24$ is divisible by $3$.
So $k = 1; x = 3; y =8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Prove that $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n-1}I tried using mathematical induction to prove this, but the problem I faced was that there are a lot of numbers between $\frac{1}{2^k-1}$ and $\frac{1}{2^{k+1}-1}$. Is it possible to prove this with induction or is there a better method?
| $$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n-1}=1+\left(\frac{1}{2^1}+\frac{1}{2^2-1}\right)+\left(\frac{1}{2^2}+\cdots+\frac{1}{2^3-1}\right)+\cdots+\left(\frac{1}{2^{n-1}}+\cdots+\frac{1}{2^n-1}\right)<n,
$$
since
$$
\frac{1}{2^{k-1}}+\cdots+\frac{1}{2^k-1}<2^{k-1}\cdot\frac{1}{2^{k-1}}=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Cool way of finding $\cos\left(\frac{\pi}{5}\right)$ while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?
Consider
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right).$$
Using the difference of cosines identity, we have
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = -2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right).$$
Now we change the RHS using the identity $\sin(x) = \cos\left(\frac{\pi}{2}-x\right)$ and the fact that $\sin(x)$ is odd.
$$-2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
So,
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
Now we make use of the identity $\cos(2x)=2\cos^2(x)-1$.
$$\cos\left(\frac{\pi}{5}\right) - 2\cos^2\left(\frac{\pi}{5}\right)+1 = 2\cos\left(\frac{\pi}{5}\right) \left(2\cos^2\left(\frac{\pi}{5}\right)-1\right)$$
Let $y=\cos\left(\frac{\pi}{5}\right)$ and we have
$$y-y^2+1=2y(2y^2-1)$$
$$4y^3+2y^2-3y-1=0$$
which has the correct solution
$$ y=\frac{\sqrt{5}+1}{4} =\cos\left(\dfrac{\pi}{5}\right) $$
One of the roots is also $\sin\left(\dfrac{\pi}{10}\right)$ which I'm guessing is because you end up with the same cubic if you apply the above to sin too.
| Not so cool. The following may be cooler.
Let $\Delta ABC$ be a triangle with $\angle BAC=36^o$, $\angle ABC=\angle ACB=72^o.$ $BD$ bisects $\angle ABC$. $BE\perp AC$. You would find that $AD=BD=BC$ and $\Delta ABC \sim \Delta BDC.$ Thus, as the figure shows, we may obtain $$\frac{x}{2y}=\frac{x+2y}{x}.$$ Thus $$\left(\frac{x}{y}\right)^2-2\left(\frac{x}{y}\right)-4=0.$$
Solve it. We have $$\frac{x}{y}=1+\sqrt{5}.$$
Another negative root is not what we want.Thus, $$\cos 36^o=\frac{AE}{AB}=\frac{x+y}{x+2y}=\dfrac{\dfrac{x}{y}+1}{\dfrac{x}{y}+2}=\frac{1+\sqrt{5}}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Common tangent(s) to two parabolas without calculus We have the parabolas: $y = 2x^{2} + 2x + 1$ and $y = 2x^{2} - 2x - 1$.
Finding the common tangents with calculus is as straight forward as just solving a system of equations with the derivatives of both parabolas and $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$; but how about in analytical geometry?
| A line $y=kx+m$ is a tangent to the parabola $y=ax^2+bx+c$ iff the quadratic equation
$$
kx+m=ax^2+bx+c
$$
has exactly one solution, which makes the discriminant to be zero. In our case
$$
kx+m=2x^2+2x+1\iff 2x^2+(2-k)x+1-m=0,\\
kx+m=2x^2-2x-1\iff 2x^2-(2+k)x-1-m=0
$$
should each have a unique solution $x$, that is the discriminants are zeros for both equations
$$
\begin{cases}
(2-k)^2-8(1-m)&=&0,\\
(2+k)^2+8(1+m)&=&0.
\end{cases}
$$
Solve the system: $k=-2$, $m=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Splitting up an infinite sum I am playing with the following example trying to determine if $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is in $\ell^2$.
Here is what I have,
\begin{align}
\sum\limits_{n=1}^\infty \left( \frac{1}{n}-\frac{1}{\sqrt{n}}\right)^2 &= \sum\limits_{n=1}^\infty \left(\frac{\sqrt{n}-n}{n\sqrt{n}}\right)^2\\
&=\sum\limits_{n=1}^\infty\frac{n-2n\sqrt{n}+n^2}{n^3}\\
&=\sum\limits_{n=1}^\infty\frac{1}{n^2}-2\sum\limits_{n=1}^\infty\frac{1}{n\sqrt{n}}+\sum\limits_{n=1}^\infty\frac{1}{n}
\end{align}
The first summand is in $\ell^2$, the second summand is in $\ell^2$, but the third summand is not. Thus the sequence $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is not in $\ell^2$. I think I may be making an error when I split the sums. Allowing such splitting would result in things like,
\begin{align}
\sum\limits_{n=1}^\infty\frac{1}{n^2} &= \sum\limits_{n=1}^\infty\frac{1-n^2}{n^2}+\sum\limits_{n=1}^\infty\frac{n^2}{n^2}
\end{align}
What am I missing here?
| It is not in $\ell^2$:
\begin{align}
\sum_{n=1}^\infty\frac{n-2n\sqrt n+n^2}{n^3}&=\sum_{n=1}^\infty\frac{1-2\sqrt n+n}{n^2} \\
&= \sum_{n=1}^\infty\frac{(\sqrt{n}-1)^2}{n^2} \\
&\ge \sum_{n=1}^3\frac{(\sqrt{n}-1)^2}{n^2} + \sum_{n=4}^\infty\frac{\left(\frac12\sqrt{n}\right)^2}{n^2} \\
&= \sum_{n=1}^3\frac{(\sqrt{n}-1)^2}{n^2} + \frac14 \sum_{n=4}^\infty\frac1n \\
&= +\infty
\end{align}
because $\sqrt{n}-1 \ge \frac12 \sqrt{n}$ for all $n \ge 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Finding $f(x)$ from the functional equation $(e^x-1)f(2x)= \left(e^{2x}-1\right)f(x)$ and $f'(0) =1 $ Let $f: \mathbb R \to \mathbb R$ be a non constant continuous function such that $(e^x-1)f(2x)= \left(e^{2x}-1\right)f(x)$. If $f'(0) =1 $, then $\lim_{x\to 0}\left(\dfrac {f(x)}{x}\right)^{\frac 1x}= ? $
I am trying hard to find $f(x)$ but unable to.
Attempt:
Differentiating both sides of the equation and putting $x=0$, we obtain $f(0)= 0$.
After that can we directly substitute $f(x)=$ exact $0$ and say that limit doesn't exist (which is not the answer)?
How do I go about solving it?
The answer is:
$e^{\frac12}$
| This solution only assumes $f$ is differentiable at $0$.
Fix $x\ne 0$ and notice that
$$f(x)\left(e^{\frac{x}2} - 1\right) = f\left(2\cdot \frac{x}2\right)\left(e^{\frac{x}2} - 1\right) = f\left(\frac{x}2\right)(e^x-1)$$
so inductively we get
$$f\left(\frac{x}{2^n}\right)(e^x-1) = f(x)\left(e^{\frac{x}{2^n}} - 1\right)$$
Letting $n\to\infty$ and using continuity of $f$ at $0$, we get
$$f(0)(e^x-1) = 0 \implies f(0) = 0$$
Now for $x \ne 0$ we have
$$\frac{f(x)}{e^x-1} = \frac{f\left(\frac{x}{2^n}\right)}{e^{\frac{x}{2^n}}-1} = \frac{f\left(\frac{x}{2^n}\right)}{\frac{x}{2^n}}\frac{\frac{x}{2^n}}{e^{\frac{x}{2^n}}-1} = \frac{f\left(\frac{x}{2^n}\right) - f(0)}{\frac{x}{2^n}}\frac{\frac{x}{2^n}}{e^{\frac{x}{2^n}}-1} \xrightarrow{n\to\infty} f'(0)\cdot1 = 1$$
Therefore $f(x) = e^x-1$ for all $x \ne 0$, and by inspection we also have $f(0) = 1 = e^0-1$.
Hence $f(x) = e^x-1, \forall x \in \mathbb{R}$.
Now the desired limit is
$$\lim_{x\to 0}\left(\frac{f(x)}x\right)^{1/x} = \lim_{x\to 0}\left(\frac{e^x-1}x\right)^{1/x}$$
which can be computed by taking the logarithm and applying l'Hopital thrice:
$$\lim_{x\to 0}\frac{\ln (e^x-1) - \ln x}{x} \stackrel{0/0}= \lim_{x\to 0}\frac{\frac{e^x}{e^x-1} - \frac1x}{1} = \lim_{x\to 0} \frac{xe^x-e^x+1}{x(e^x-1)} \stackrel{0/0}= \lim_{x\to 0}\frac{xe^x+e^x-e^x}{xe^x+e^x-1} = \lim_{x\to 0}\frac{xe^x}{e^x(x+1)} \stackrel{0/0}= \lim_{x\to 0}\frac{xe^x+e^x}{e^x(x+1) + e^x} = \frac{0+1}{1\cdot1 + 1} = \frac12$$
Hence
$$\lim_{x\to 0}\left(\frac{e^x-1}x\right)^{1/x} = e^{\lim_{x\to 0}\frac{\ln (e^x-1) - \ln x}{x}} = e^{1/2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify: $\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$ Simplify:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$$
So what I've tried was:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=\frac{1+\cos^2x-\sin^2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=$$
$$=\frac{2\cos^2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=\frac{2\cos^2x\,\tan^2x-4\cos^2(\frac{x}{2})(1-\cos x)}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{2\sin^2x-4\cos^2(\frac{x}{2})+4\cos^2(\frac{x}{2})\cos x}{(1-\cos x)(\tan^2x)}=\frac{2\cdot2\sin^2(\frac{x}{2})\cos^2(\frac{x}{2})-4\cos^2(\frac{x}{2})+4\cos^2(\frac{x}{2})\cos x}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{4\cos^2(\frac{x}{2})(\sin^2(\frac{x}{2})-1+\cos x)}{(1-\cos x)(\tan^2x)}=\frac{4\cos^2(\frac{x}{2})(\sin^2(\frac{x}{2})-1+\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{4\cos^2(\frac{x}{2})(-1+\cos^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=\frac{4\cos^2(\frac{x}{2})(-\sin^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=\frac{-2\sin^2(x)}{(1-\cos x)(\tan^2x)}$$
This is farthest I got but the answer is supposed be: $0$,
according to the solutions, I might've missed something on the process even though I checked it a couple of time..
I'm preparing for the Tel-Aviv univeristy math entry test and I took this exercise from math-exercises.com to practise (that is where the solution is from).
| The computation can be much shorter if one uses the relevant duplication/linearisation formulæ:
*
*$\displaystyle \frac{1+\cos 2x}{1-\cos x}=\frac{2\cos^2x}{1-\cos x}=\frac{\cos^2x}{\,\smash[b]{\cfrac{1-\cos x}2}\,}=\frac{\cos^2x}{\,\sin^2\dfrac x2\,}$.
$\phantom{fg}$
*
*$\displaystyle \frac{4\cos^2\dfrac x2}{\tan^2x}=\frac{4\cos^2\dfrac x2\,\cos^2x}{\sin^2x}=\frac{4\cos^2\dfrac x2\,\cos^2x}{\Bigl(2\sin\dfrac x2\cos\dfrac x2\Bigr)^2}=\frac{\cos^2x}{\,\sin^2\dfrac x2\,}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability that all three shirts were worn Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.
What is the probability that he wears all three shirts after 6 days?
My try:
$$\binom{6}{1}\binom{5}{1}\binom{4}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}\cdot[\binom{3}{2}(\frac{1}{2})^2\cdot[\frac{1}{3}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{3})^2\cdot[\frac{1}{2}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{6})^2\cdot[\frac{1}{2}+\frac{1}{3}]+\binom{3}{1}\binom{2}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}]$$
First, we pick possessions for red, green and white. Then, we multiply it by the other combinations possible of the other three possessions left, and summing it up (I didn't count all possibilities). The problem is the answer is bigger than 1, so it must be wrong. Also, I guess there is more elegant way.
Thanks in advance.
| I will try to replicate @Mustafa's answer in a different way.
Guess there is $6$ white tshirts, $3$ green, and $2$ red. For this person to have picked one atleast from each of all of them during 6 days it sums to $\sum\binom{1}{6}^i\binom{1}{3}^j\binom{1}{2}^k$ for $i$ $j$ $k$ strictly positive and $i+j+k<=6$.
Taking count of ordering implicitely between distinct colors gives: $6!(\frac{2*3*6^4}{4!}+\frac{2*3^2*6^3}{2!3!}+\frac{2^2*3*6^3}{2!3!}+\frac{2*3^3*6^2}{2!3!}...=84180$
All possible combinations restrictlessly of any choice sums up the previous value to $6!(\frac{6^6+3^6+2^6}{6!}+\frac{3^5(2+6)+2^5(3+6)+6^5(2+3)}{5!}+\frac{3^4(2^2+6^2)+2^4(3^2+6^2)+6^4(2^2+3^2)}{2!4!}+\frac{3^3(2^3+6^3)+6^32^3}{3!3!})$ which equals $761761$
Probability is $\frac{84180}{761761+84180} \approx 0.1$ I'm surprised it's the same amount !!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$3^n-1$ is divisible by $4 \implies n $ is even What is the easiest ways to prove this: $3^n-1$ is divisible by $4 \implies n $ is even? Moreover, how would I figure out that $n$ must be even if I didn't know the result?
My approach is this: suppose $n=2k+1$. Consider $3^{2k+1}-3$. If we show that it is divisible by $4$, it will follow that $3^n-1$ has remainder $2$ in the division by $4$ if $n$ is odd. We have $3^{2k+1}-3=3(3^k-1)(3^k+1)$. Each of the two brackets is divisible by $2$, so the whole thing is divisible by $4$. Is this correct?
| We can use the identity
$a^n - b^n = (a - b) \displaystyle \sum_0^{n - 1} a^{n - 1 - k}b^k: \tag 0$
$3^n - 1 = (3^1 - 1^1) \displaystyle \sum_0^{n - 1} 3^{n - 1 - k} 1^k = 2 \sum_0^{n - 1} 3^{n - 1 - k}; \tag 1$
now if
$4 = 2^2 \mid 3^n - 1 = 2 \displaystyle \sum_0^{n - 1} 3^{n - 1 - k}, \tag 2$
then
$2 \mid \displaystyle \sum_0^{n - 1} 3^{n - 1 - k}; \tag 3$
now if $n$ were odd, the sum would consist of an odd number of odd integers $3^{n - 1 - k}$, hence would itself be odd. If, however, $n$ is even, the sum consists of an even number of odd integers, hence is itself even. Thus (3) binds precisely when $n$ is even, and thus
$4 \mid 3^n -1 \Longleftrightarrow n \; \text{is even.} \tag 4$
| {
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"url": "https://math.stackexchange.com/questions/2821936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve for $y$ in the equation $x = \frac{3y^2 + 14y + 16}{6y^2 + 24y + 24}$ As part of a larger calculation, I came across $x = \frac{3y^2 + 14y + 16}{6y^2 + 24y + 24}$, which I now have to solve for $y$. My initial idea, besides a failed attempt to use the general quadratic formula, was, incorrectly:
\begin{align}
\dots \Leftrightarrow 6xy^2 + 24xy + 24x &= 3y^2 + 14y + 16 \\
\Leftrightarrow \frac{d}{dy} 6xy^2 + 24xy + 24x &= 3y^2 + 14y + 16 \frac{d}{dy} \\
\Leftrightarrow 12xy + 24x &= 6y + 14 \\
\Leftrightarrow (12x-6)y &= 14 - 24x \\
\Leftrightarrow y &= \frac{14 - 24x}{12x-6} \\
\Leftrightarrow y &= \frac{7 - 12x}{6x-3}
\end{align}
This is very "close" to the correct solution as given by WolframAlpha: $y = \frac{8 - 12x}{6x - 3}$. However, the derivation step is incorrect, because $x$ is not independent of $y$ and thus can't be regarded as a constant.
Obviously, the result from WA doesn't help much without a derivation, so that's what I'm looking for here (and maybe some useful tips for dealing with such expressions in general).
Thanks in advance!
| $$ (3y+8)(y+2) = 3 y^2 + 14 y + 16 $$
$$ 6 (y+2)^2 = $$
$$ x = \frac{3y+8}{6y+12} $$
This is a Mobius transformation
$$ y = \frac{12x - 8}{-6x+3} $$
Audience Request:
If we have constants $a,b,c,d$ with $ad-bc \neq 0,$ and
$$ y = \frac{ax+b}{cx+d} $$
when $cx+d \neq 0,$ then
$$ x = \frac{dy - b}{-cy + a} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $a_{n+1}^2=a_n·a_{n+2}+(-1)^n$ for Fibonacci sequence $\{a_n\}$ Let $a_1,a_2,a_3,...,a_n$ be the sequence of Fibonacci numbers.
Then we are required to prove by induction that $$a_{n+1}^2=a_n \cdot a_{n+2} + (-1)^n$$
My Attempt:
Initially putting a few values like $n=1,2 \: \text{or}\ 3$ makes it certain that we can move ahead.
Following the conventional way we can assume that the proposition holds true for some $k$ s.t.$$a_{k+1}^2=a_k \cdot a_{k+2} + (-1)^k \tag1$$
To prove that the proposition holds for all values of $n$, we need to prove for the same for $n=k+1$ too.
In short we need to prove that:
$$a_{k+2}^2=a_{k+1} \cdot a_{k+3} + (-1)^{k+1} \tag2$$
We know that
$$ a_{k+2}^2=(a_{k+3}-a_{k+1})^2$$
$$\Rightarrow a_{k+2}^2=a_{k+3}^2+a_{k+1}^2-2\cdot a_{k+3}\cdot a_{k+1}$$
$$\Rightarrow a_{k+2}^2=a_{k+3}^2+a_k \cdot a_{k+2} + (-1)^k-2\cdot a_{k+3}\cdot a_{k+1}$$
$$\Rightarrow a_{k+2}^2=a_{k+3}[a_{k+3}- 2\cdot a_{k+1}]+a_k \cdot a_{k+2} + (-1)^k \tag{*}$$
It is easy to see that $$a_{k+3}-a_{k+1}=a_{k+2} \tag3$$
And $$a_{k+2}-a_{k+1}=a_{k} \tag4$$
Adding equation $3$ and $4$, we get$$a_{k+3}-2a_{k+1}=a_{k} \tag5$$
Putting eq$(5)$ in eq$(*)$
$$a_{k+2}^2=a_{k+3}[a_k]+a_k \cdot a_{k+2} + (-1)^k $$
$$\Rightarrow a_{k+2}^2=a_k[a_{k+3}+a_{k+2}]+(-1)^k $$
$$\Rightarrow a_{k+2}^2=a_k(a_{k+4})+(-1)^k$$
This I indeed completely different from something that I wanted to prove. Is there any way to get ahead of this or is there something wrong with what I have done?
| Induction step $n\to n+1$
$$\begin{align}a_{n+1}^2-a_n \cdot a_{n+2} &= a_{n+1}^2-a_n \cdot (a_{n+1}+a_n) \\
& =a_{n+1}(a_{n+1}-a_n)-a_n^2
\\&= a_{n+1}a_{n-1}-a_n^2
\\&= (a_n^2-a_{n+1}a_{n-1})
\\&= -(-1)^{n-1}\\
&= (-1)^n\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Where is the mistake in my reasoning? I have a statement that says:
If $a^2 + b^2 + c^2 = 2$ and $(a + b + c)(1 + ab + bc + ac) = 3^2$
What is the value of $( a + b + c )$ ?
My reasoning was:
$a^2 + b^2 + c^2 = 2$, rewritten as:
*
*$(a + b + c)^2 = 2 + 2ab + 2ac + 2bc$
Since, $(a + b + c)(1 + ab + bc + ac) = 3^2$
*$(1 + ab + bc + ac) = \frac{9}{(a + b + c)}$
Now, replacing in the 1.
*Factorize $(a + b + c)^2 = 2(1 + ab + ac + bc)$
*Replacing $(a + b + c)^2 = \frac{18}{a + b + c}$
*Multiplying by $(a + b + c)$
*$(a + b + c)^3 = 18$. Thus $a + b + c = \sqrt[3]{18}$. That is my result.
But, the correct result should be $4$, then where is my mistake ?
| Suppose $a^2+b^2+c^2=2$ and $(a+b+c)(1+ab+bc+ac)=k$. Then, setting $s=a+b+c$ and $q=ab+bc+ac$, we get
$$
s^2=2+2q=2(1+q),\qquad s(1+q)=k
$$
Therefore $1+q=s^2/2$ and
$$
s^3=2k
$$
so that $s=\sqrt[3]{2k}$. If $s$ has to be $4$, then $2k=64$, that is, $k=32$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does one construct general forms that certain variables in an equation must take? Srinivasa Ramanujan was one of the greatest mathematicians of all time $-$ the greatest in the $20^\text{th}$ century. One day, he stumbled across the equation $$\rm3^3+4^3+5^3=6^3\tag1$$ and only days later, he was able to discover the general form that this equation takes place in: $$\big(3a^2+5ab-5b^2\big)^3+\big(4a^2-4ab+6b^2\big)^3+\big(5a^2-5ab-3b^2\big)^3=\big(6a^2-4ab+4b^2\big)^3\tag2$$ for all integers $a$ and $b$, with Eq. $(1)$ being created when $a=1$ and $b=0$. Now I know that barely anyone knows how Ramanujan obtained his results $-$ but he did, and they are brilliant.
My question is, are there any useful techniques or methods that I can learn so I may be able to pull something like finding Eq. $(2)$? What algorithms are there to discover some general equations? I was not convinced when I looked at it, so I evaluated, and the left hand side weighs the same as the right hand side.
But are there any such algorithms at all? For example, $a^2-b^2$ looks like some regular expression, but just add in the clever substitution $ab - ab$ and then this happens: $$\begin{align}a^2-b^2&=a^2+ab-ab-b^2\\ &= a(a+b)-b(a+b) \\ &= (a+b)(a-b).\end{align}$$ So now when anybody says what $915^2-914^2$ is, I can say it right away: it is $915+914=1829$.
How does anybody know to put such clever substitutions? When I was first given the problem to factorise $a^2-b^2$, I just did this: $$\begin{align}(a-b)^2&=a^2-2ab+b^2 \\ \Leftrightarrow a^2-b^2&=(a-b)^2 + 2ab - 2b^2 \\ &= (a-b)^2 + 2b(a-b) \\ &= (a-b)(a-b+2b) \\ &= (a+b)(a-b).\end{align}$$ I got the same result, but it was not as efficient as the first method.
I want to become a great mathematician one day; I don't want to just tell my friends that I do math for a living $-$ I want to prove something, or make a theorem, especially if it involves prime numbers. What are some helpful algebraic formulae that I should know when it comes to constructing generalised equations and such? I know the quadratic and cubic formula... but that is it.
For example: $$x^2 + 3y^2 = 7z^2.$$ These are the forms that $x$, $y$ and $z$ must take for integers $p$ and $s$: $$\begin{align}x&=2\big(3p^2+3ps-s^2\big) \\ y&=-3p^2+4ps+s^2 \\ z&=3p^2+s^2.\end{align}$$
Go here to find some similar general equations (and let's not forget about pythagorean triples!).
How does one find such equations for $x$, $y$ and $z$? Is it just trial and error?
Thank you in advance, and I apologise if the post is too long and/or too broad.
| (OP) enquiry about derivation of the Ramanujan Identity. Derivation is given below:
Assume that below equation is true:
$\big(3p^2+mp-5\big)^3+\big(4p^2-np+6\big)^3+\big(5p^2-mp-3\big)^3=\big(6p^2-np+4\big)^3\tag1$
Let:
$u=3p^2+mp-5$
$v=4p^2-np+6$
$w=5p^2-5p-3$
$z=6p^2-np+4$
We have identity:
$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$
Hence $(u^3+v^3+w^3-z^3)$=
$(3p+m)(mp-5)(3p^2-5)-(5p-m)(mp+3)(5p^2-3)-2(3p^2+2)(4-np)(6p-n)+2(2p^2+3)(6-np)(4p-n)=0$
or
$4p^4(5n-4m)+2p^3(4m^2-n^2-84)-2p(4m^2-n^2-84)-4(5n-4m)=0$ -----(2)
Equation (2) has solution at $(m,n)=(5,4)$
Hence after substituting in equation (1) $(m,n,p)=(5,4,(a/b))$ we get:
$\big(3a^2+5ab-5b^2\big)^3+\big(4a^2-4ab+6b^2\big)^3+\big(5a^2-5ab-3b^2\big)^3=\big(6a^2-4ab+4b^2\big)^3$
| {
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"question_score": "5",
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What is $x$ if $\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$? I need to find $x$, given that
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
I simplified this to $x^4+14x^3+105x^2+68x-188=0$. According to Symbolab, that is not correct. That's why I'm goint to write out my attempt so you can point out where my mistake is.
My attempt
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
$$\left(\sqrt{x+4+2\sqrt{x+3}}\right)^2=\left(\frac{x+8}{3}\right)^2$$
$$x+4+2\sqrt{x+3}=\frac{x^2+16x+64}{9}$$
$$9x+36+18\sqrt{x+3}=x^2+16x+64$$
$$-x^2-7x-28=-18\sqrt{x+3}$$
$$x^2+7x+28=18\sqrt{x+3}$$
$$(x^2+7x+28)^2=(18\sqrt{x+3})^2$$
$$x^4+7x^3+28x^2+7x^3+49x^2+196x+28x^2+196x+784=324x+972$$
$$x^4+14x^3+105x^2+68x-188=0$$
Where is my mistake? Even if this were true, I still wouldn't be able to solve it without a calculator (I can't use Rational Root Theorem on such a big numbers!).
By the way, the solution should be (again, according to Symbolab) $x \in \{1,-2\}$.
| What you did is fine. Now, you can use the rational root theorem in order to find the roots $1$ and $-2$. Since your polynomial is $(x-1)(x+2)(x^2+13x+94)$, there are no more real roots. Note however that you still must check whether or not $-2$ and $1$ are solutions of the original equation.
| {
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Ellipse: Most direct algebraic demonstration that $\left|x\right|$ is maximal when $y=0$ Yes, I know this is the equation of an ellipse.
This is the starting point of the problem:
$$
r_1+r_2=2a,
$$
where
$$r_1\equiv\sqrt{(x+c)^2+y^2}; r_2\equiv\sqrt{(x-c)^2+y^2}.$$
I want to establish certain properties prior to deriving a parametric expression for the ellipse, or even the "standard form",
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
$$
I can do all of this with a couple tacks, a string and a pencil, but
I want to show things algebraically. Preferably in a way that is directly applicable to all second-degree plane curves.
The first property I wish to establish is the range of $x$ values.
I know from experience that the answer is going to be that the extreme
values of $x$ will occur when $y=0$. The question that I am asking
in this post is: what is the most direct (algebraic) way to show that $\left|x\right|$
is greatest when $y=0$?
I had hoped to establish that fact, and then use another (absolute value)
method of determining the actual values of $x$ when $y=0$. But, the best
I cam up with is this very long-winded demonstration.
First some definitions: Let $a,c,x,y\in\mathbb{R}$ where $a>c>0$
are constants, and $x$ and $y$ are variables such that, for $r_1\equiv\sqrt{(x+c)^2+y^2}$
and $r_2\equiv\sqrt{(x-c)^2+y^2}$ we have the equation
$$
r_1+r_2=2a.
$$
The degenerate case has been omitted.
I will presume the greatest value to be $\left|x\right|=a$, and take
the case of $x=a$. We square our equation
$$
(r_1+r_2)^2=4a^2
$$
$$
=r_1^2+2r_1 r_2+r_2^2
$$
$$
=(a+c)^2+y^2+2r_1 r_2+(a-c)^2+y^2
$$
$$
=2(a^2+c^2+y^2)+2r_1 r_2.
$$
So
$$
r_1(2a-r_1)=a^2-c^2-y^2
$$
$$
=2a\sqrt{(a+c)^2+y^2}-((a+c)^2+y^2)=a^2-c^2-y^2.
$$
Transpose, cancel and simplify
$$
2a\sqrt{(a+c)^2+y^2}=2a^2+2ac,
$$
$$
r_1=\sqrt{(a+c)^2+y^2}=a+c.
$$
And, therefore
$$
r_2=2a-r_1=a-c.
$$
Now, by definition we have
$$
\sqrt{\left(a+c\right)^2}\equiv\left|a+c\right|=r_1
$$
and
$$
\sqrt{(a-c)^2}\equiv\left|a-c\right|=r_2.
$$
If $y\ne0$ while $x=a$, then
$$
r_1=\sqrt{(a+c)^2+y^2}=\left|a+c\right|+\delta_1,
$$
and
$$
r_{2}=\sqrt{(a-c)^2+y^2}=\left|a-c\right|+\delta_2,
$$
where both $\delta_1>0$ and $\delta_2>0.$
Since $a>c>0,$ we have $r_1+r_2=2a+\delta_1+\delta_2$, in
contradiction to the equation we seek to satisfy. So $x=a\implies y=0.$
If $x>a$, let $\delta_3=x-a>0.$
So
$$
r_1+r_2=a+c+\delta_3+a-c+\delta_3>2a;
$$
another contradiction.
Thus far we have dertermined that $x=a\implies y=0,$ and for any
$y$ that $x\le a$.
If we set $x=-a,$ then
$$
(r_1+r_2)^2=4a^2
$$
$$
=r_1^2+2r_1 r_2+r_2^2
$$
$$
=(-a+c)^2+y^2+2r_1 r_2+(-a-c)^2+y^2
$$
$$
=2(a^2+c^2+y^2)+2r_1 r_2
$$
$$
r_{1}\left(2a-r_{1}\right)=a^{2}-c^{2}-y^{2}
$$
$$
2a\sqrt{\left(-a+c\right)^{2}+y^{2}}-\left(\left(-a+c\right)^{2}+y^{2}\right)=a^{2}-c^{2}-y^{2}
$$
$$
2a\sqrt{\left(-a+c\right)^{2}+y^{2}}=2a^{2}-2ac
$$
$$
r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c>0.
$$
$$
r_{2}=2a-r_{1}=a+c.
$$
Since we now have
$$
r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c,
$$
and
$$
r_{2}=\sqrt{\left(a+c\right)^{2}+y^{2}}=a+c,
$$
the argument showing that $x=-a\implies y=0,$ is essentially the
same. Likewise for the argument that $x\ge-a.$
And that's just the $x$ range.
NB: I don't "need" any of this to fully characterize the ellipse. My goal is to find multiple ways of examining the same construct.
| This is a constrained optimization problem that can be attacked directly with the method of Lagrange multipliers. Form the Lagrangian $x-\lambda(r_1+r_2-2a)$ and differentiate, producing the equations $$1-\lambda\left({x+c \over r_1}+{x-c \over r_2}\right) = 0 \\
-\lambda\left({y\over r_1}+{y\over r_2}\right) = 0.$$ The multiplier $\lambda$ cannot be zero and the two distances are always positive, so we must have $y=0$. Substituting this back into the equation $r_1+r_2=2a$ yields $x=\pm a$.
Strictly speaking, you should also show that $|x|$ can’t be greater than $a$, but that’s almost trivial.
The above is one of the most straightforward solutions that I can think of, but if you insist on avoiding calculus, here’s another approach. The perimeter of the triangle formed by the foci and an arbitrary point on the ellipse is fixed at $r_1+r_2+2c=2(a+c)$. Using Heron's formula and the defining equation $r_1+r_2=2a$, the square of this triangle’s area can be expressed as a function of $r_1$: $$A^2 = (c-a+r_1)(c+a-r_1)(a+c)(a-c).\tag1$$ Since $a\gt c$, for this to be nonnegative we must have $(c-a+r_1)(c+a-r_1)\ge0$, or $$(r_1-a)^2\le c^2.\tag2$$ On the other hand, by construction every point on the ellipse is also the intersection of the circles $(x+c)^2+y^2=r_1^2$ and $(x-c)^2+y^2=r_2^2=(2a-r_1)^2$. From these equations we get $r_1=\frac ca x+a$. Substituting into (2) and simplifying results in $|x|\le a$. Plugging $\pm a$ into the defining equation of the ellipse and solving for $y$ yields $y=0$, as desired.
The bounds for $y$ can be found in a similar way. The $y$-coordinate of a point on the ellipse is the altitude of the triangle, so the maximum $y$ value will maximize the triangle’s area. Equation (1) is quadratic in $r_1$ and the $r_1^2$ term has a negative coefficient, so the value of $r_1$ that maximizes the area can be computed using standard algebraic techniques, and from that the corresponding values of $y$.
That wasn’t too bad, but I don’t really see much advantage to working directly with the radicals from the equation $r_1+r_2=2a$. It’s only a couple of more steps from $ar_1=cx+a^2$ to the standard equation of an ellipse. Once you have that equation, the ranges for $x$ and $y$ can be found by inspection, as Michael Hardy wrote in his now-deleted answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating function of a recurrent relation Problem: Find the generating function for the recurrent relation:
$$f_n=2f_{n-1}+\frac 12 f_{n-2},$$
where
$$f_0=f_1=1.$$
My idea was to first find a few of the beginning values and then try to make into some sum of an infinite series, but I had no luck. Any ideas?
|
We denote with $F(x)=\sum_{k=0}^\infty f_n x^n$ a generating function for the recurrence relation
\begin{align*}
f_n&=2f_{n-1}+\frac{1}{2}f_{n-2}\qquad\qquad n\geq 2\tag{1}\\
f_0&=1\\
f_1&=1
\end{align*}
It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain according to (1) for $n\geq 2$
\begin{align*}
[x^n]F(x)&=2[x^{n-1}]F(x)+\frac{1}{2}[x^{n-2}]F(x)\\
&=2[x^n]xF(x)+\frac{1}{2}[x^{n}]x^2F(x)\tag{2}\\
\end{align*}
From which
\begin{align*}
[x^n]F(x)\left(1-2x-\frac{1}{2}x^2\right)&=0\qquad\qquad n\geq 2
\end{align*}
follows.
In (2) we use the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
We therefore get for $n\geq 0$
\begin{align*}
F(x)\left(1-2x-\frac{1}{2}x^2\right)=a+bx\tag{3}\\
\end{align*}
with $a,b \in \mathbb{R}$.
We can calculate $a$ and $b$ by using $f_0=f_1=1$ and noting that only coefficients of $x^0$ and $x^1$ from the left-hand side of (3) contribute anything to $a+bx$ at the right-hand side of (3).
We obtain from (3)
\begin{align*}
a&=[x^0]F(x)\left(1-2x-\frac{1}{2}x^2\right)\\
&=[x^0](1+\sum_{k=1}^\infty f_nx^n)\left(1-2x-\frac{1}{2}x^2\right)\\
&=1\\
b&=[x^1]F(x)\left(1-2x-\frac{1}{2}x^2\right)\\
&=[x^1](1+x+\sum_{k=2}^\infty f_nx^n)\left(1-2x-\frac{1}{2}x^2\right)\\
&=[x^1](1+x)(1-2x)\\
&=-1
\end{align*}
We conclude
\begin{align*}
\color{blue}{F(x)}&\color{blue}{=\frac{1-x}{1-2x-\frac{1}{2}x^2}}\\
&\color{blue}{=1+x+\frac{5}{2}x^2+\frac{11}{2}x^3+\frac{49}{4}x^4+\cdots}
\end{align*}
| {
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Limit with Epsilon - Delta method Prove using the $\epsilon - \delta$ definition of limits that $\lim_{x\to3} \frac{5}{4x-11} = 5$.
I know how the setup should be given $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that $|x-3| \lt \delta$ and $|\frac{5}{4x-11} - 5| \lt \epsilon$ but I can't do the computation to help me find $\delta$ can someone guide me in the right direction?
| Let $\varepsilon > 0$ and define $\delta = \min\left\{\frac{\varepsilon}{4(5+\varepsilon)}, \frac14\right\}$. For $|x-3| < \delta$ we have
$$\left|5 - \frac{5}{4x-11}\right| = \left|\frac{20x-60}{4x-11}\right| \le \frac{20\left|x-3\right|}{\left|4x-11\right|} \le \frac{20\left|x-3\right|}{1 - \left|4x-12\right|} = \frac{20\left|x-3\right|}{1 - 4\left|x-3\right|} < \varepsilon$$
because
\begin{align}
\frac{20\left|x-3\right|}{1 - 4\left|x-3\right|} < \varepsilon &\iff 20\left|x-3\right| < \varepsilon(1 - 4\left|x-3\right|) \\
&\iff (20 + 4\varepsilon)\left|x-3\right| < \varepsilon\\
&\iff |x-3| < \frac{\varepsilon}{4(5+\varepsilon)}
\end{align}
which is true.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
| If the numbers are $x, x+1, x+2, x+3$.
Let $\frac m2 = x+1.5$ be the midpoint of the four consecutive integers, so that the integers are $\frac {m-3}2, \frac {m-1}2, \frac {m+1}2, \frac {m+3}2$. (Note: $m$ is odd and $\frac m2$ is not an integer.)
So $x(x+1)(x+2)(x+3) + 1 = $
$\frac {(m-3)(m+3)(m-1)(m+1)}{16} + 1=$
$\frac {(m^2 -9)(m^2 - 1) + 16}{16} =$
$\frac {(m^2-10m^2 + 9) +16}{16} =\frac {m^2-10m^2 + 25}{16}=$
$(\frac {m^2 -5}{4})^2$.
Now $m$ is odd. So let $m = 2n+1$ then
$x(x+1)(x+2)(x+3) + 1 = (\frac {(2n+1)^2 -5}{4})^2=$
$(\frac {4n^2 +4n + 1 -5}{4})^2 = (\frac {4n^2 +4n -4}{4})^2=$
$(n^2 +n - 1)^2$.
====
So ....
If $x$ is the first integer and $\frac {2n + 1}2 = x + \frac 32$ then
$n = x + 1$.
So $x(x+1)(x+2)(x+3) + 1 = ((x+1)^2 + (x+1) - 1)^2 = (x^2 + 3x +1)^2$
....
So as you got $x^4 + 6x^3 + 11x^2 + 6x + 1$ that actually equals $(x^2 + 3x +1)^2$
Indeed $x^4 + 6x^3 + 11x^2 + 6x + 1 = x^2(x^2 + 3x + 1) + 3x^3 +10x^2 + 6x + 1$
$= x^2(x^2 + 3x+ 1) + 3x(x^2 + 3x + 1) +x^2 +3x + 1$
$= (x^2 + 3x + 1)^2$.
.... addendum.....
D'oh.
If $x(x+1)(x+2)(x+3) + 1 = a^2$ then
$x(x+ 1)(x+ 2)(x+3) = a^2 - 1 = (a + 1)(a-1)$
To get factors that close together they'd have to be
$a = x(x+3) \pm 1= (x+ 1)(x+2) \mp 1$
and indeed $a= x(x+3) + 1 = (x+1)(x+2) - 1= x^2 +3x + 1$.
proves the statement! (If we work backwards.)
| {
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"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 10
} |
Find the set of primes $p$ which $6$ is a quadratic residue $\mod p$ Since $6$ is not prime (law of quadratic reciprocity could have been used), how does one find the set of primes $p$ for which $6$ is a quadratic residue $\pmod p$? I noticed that $6$ is a quadratic residue $\pmod p$ for the following $p$ up to $100: 2, 3, 5, 19, 23, 29, 43, 47, 53, 67, 71, 73, 97$. I don't recognize an immediate congruence. I know that every prime $p$ that $2$ is a quadratic residue is congruent to $1, 7 \pmod 8$, and every prime $p$ that $3$ is a quadratic residue is congruent to $1, 11 \pmod {12}$, yet I see primes congruent to $1, 3, 5, 7 \pmod 8$ and also primes congruent to $1, 5, 7, 11 \pmod {12}$. Any help appreciated.
| Use the fact that Legendre symbol is completely multiplicative:
$$1=\left(\frac{6}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{3}{p}\right)$$
which means either
$$\left(\frac{2}{p}\right)=\left(\frac{3}{p}\right)=1$$
or
$$\left(\frac{2}{p}\right)=\left(\frac{3}{p}\right)=-1$$
and (same link)
$${\displaystyle \left({\frac {2}{p}}\right)=(-1)^{\tfrac {p^{2}-1}{8}}={\begin{cases}1&{\mbox{ if }}p\equiv 1{\mbox{ or }}7{\pmod {8}}\\-1&{\mbox{ if }}p\equiv 3{\mbox{ or }}5{\pmod {8}}.\end{cases}}}$$
$${\displaystyle \left({\frac {3}{p}}\right)=(-1)^{{\big \lfloor }{\frac {p+1}{6}}{\big \rfloor }}={\begin{cases}1&{\mbox{ if }}p\equiv 1{\mbox{ or }}11{\pmod {12}}\\-1&{\mbox{ if }}p\equiv 5{\mbox{ or }}7{\pmod {12}}.\end{cases}}}$$
Now you have 8 cases to analyse.
*
*$p=8k+1=12t+1 \Rightarrow p=24m+1$
*$p=8k+1=12t+11$ has no integer solutions
*$p=8k+7=12t+1$ has no integer solutions
*$p=8k+7=12t+11 \Rightarrow p=24m+23$
*$p=8k+3=12t+5$ has no integer solutions
*$p=8k+3=12t+7 \Rightarrow p=24m+19$
*$p=8k+5=12t+5 \Rightarrow p=24m+5$
*$p=8k+5=12t+7$ has no integer solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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how to classify critical points for a 2 variable function For each of the following functions, find and classify all critical points. [That is, use the second-derivative test to deduce whether each critical point is a local max, a local min, or a saddle.]
$f(x,y) = (x+y)(1-xy)$
my solution:
I first equated gradient of $f$, $\nabla f(a) = 0$
so, $(1-2xy-y^2, 1-2xy-x^2) = (0,0)$
solving two equations I found $y = x,-x$.
and then I found the Hessian matrix
$H = \begin{bmatrix}
-2y & -2x-2y \\
-2x-2y & -2x \end{bmatrix}$
Now, I am having trouble in how to put in my critical points and check if they are local max, local min or saddle point.
| With
$f(x, y) = (x + y)(1 - xy) \tag 1$
and
$\nabla f = (f_x, f_y), \tag 2$
we see that
$f_x = 1 - xy + (x + y)(-y) = 1 - xy - xy - y^2 = 1 - 2xy - y^2, \tag 3$
and likewise
$f_y = 1 - xy + (x + y)(-x) = 1 - xy - x^2 - xy = 1 - 2xy - x^2; \tag 4$
at critical points of $f(x, y)$, we have
$\nabla f(x, y) = 0, \tag 5$
whence from (3) and (4), at the critical points,
$1 - 2xy - y^2 = 0 = 1 - 2xy - x^2; \tag 6$
we solve this system by observing that it implies
$y^2 = 1 - 2xy = x^2, \tag 7$
so that
$y = \pm x; \tag 8$
using (8) in (6) we may write an equation for $x$:
$0 = x^2 + 2xy - 1 = x^2 \pm 2x^2 - 1; \tag 9$
$x$ must thus obey
$3x^2 - 1 = 0 \tag{10}$
or
$x^2 + 1 = 0; \tag{11}$
we rule out (11) since $x$ is real; thus
$x = \pm \dfrac{1}{\sqrt 3} = \pm \dfrac{\sqrt 3}{3}; \tag{12}$
again from (6),
$y = \dfrac{1 - x^2}{2x} = \dfrac{\dfrac{2}{3}}{2x} = \dfrac{1}{3x}; \tag{13}$
therefore the critical points are
$(x, y) = \left ( \dfrac{\sqrt 3}{3}, \dfrac{\sqrt 3}{3} \right ), \; (x, y) = \left ( -\dfrac{\sqrt 3}{3}, -\dfrac{\sqrt 3}{3} \right ); \tag{14}$
the Hessian $H_f$ of $f(x, y)$ has been provided for us courtesy of our OP kronos:
$H_f = \begin{bmatrix} -2y & -2x-2y \\ -2x-2y & -2x \end{bmatrix}; \tag{15}$
we thus see that
$\det(H_f) = 4xy - 4(x + y)^2 = 4(xy - (x + y)^2) = -4(x^2 +xy + y^2); \tag{16}$
since at the critical points we have
$x = y = \pm \dfrac{\sqrt 3}{3}, \tag{17}$
it follows that at these points
$\det(H_f) = -4, \tag{18}$
which, as is well-known, implies that each critical point is a saddle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Minimal area of a right triangle with inradius $1$ I got this question, solved it, then forgot how I solved it.
What is the minimal area of a right triangle with inradius $1$?
My attempt:
$r=\frac{a+b-c}2$, so $a+b=c+2$
$a^2+b^2=c^2$
This gives $ab=2(c+1)$
I remember using the AM-GM to prove that equality held, thus giving $a=b=2+\sqrt2$ and getting the area as $3+2\sqrt2$, but I can't do it now. Please help.
| $$1=r=\frac{a+b-c}{2}=\frac{a+b-\sqrt{a^2+b^2}}{2}.$$
We'll prove that
$$S\geq3+2\sqrt2.$$
Indeed, we need to prove that:
$$\frac{ab}{2}\geq(3+2\sqrt2)\left(\frac{a+b-\sqrt{a^2+b^2}}{2}\right)^2$$ or
$$(a+b+\sqrt{a^2+b^2})^2\geq2(3+2\sqrt2)ab,$$
which is true by AM-GM:
$$(a+b+\sqrt{a^2+b^2})^2\geq(2\sqrt{ab}+\sqrt{2ab})^2=2(3+2\sqrt2)ab.$$
The equality occurs for $a=b=2+\sqrt2,$ which says that
$$\min_{r=1}S=3+2\sqrt2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
On the property $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$ for complex numbers Suppose we have complex numbers, $x$ and $y$. My question is: we know, from the triangle inequality, that $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$. It makes sense, geometrically, when the strict inequality would hold. Whenever, however, would we have equality? Would it be necessary that $\lvert x \rvert = \rvert y \rvert$, or is there some other necessary condition?
Thanks.
| \begin{align*}
&|x + y| = |x| + |y| \\
\implies \, &|x + y|^2 = (|x| + |y|)^2 \\
\implies \, &|x + y|^2 = |x|^2 + |y|^2 + 2|x||y| \\
\implies \, &(x + y)\overline{(x + y)} = x\overline{x} + y\overline{y} + 2|x||y| \\
\implies \, &x\overline{x} + y\overline{y} + x\overline{y} + y\overline{x} = x\overline{x} + y\overline{y} + 2|x||y| \\
\implies \, &x\overline{y} + \overline{x\overline{y}} = 2|x||y| \\
\implies \, &2\Re(x\overline{y}) = 2|x\overline{y}| \\
\end{align*}
So, we have a complex number $z$ such that $\Re(z) = |z|$. If $z = a + ib$, then
$$a = \sqrt{a^2 + b^2} \implies a^2 = a^2 + b^2 \implies b^2 = 0 \implies b = 0,$$
hence $z$ is real. Moreover, $z$ must be positive. Therefore, $x\overline{y} \in [0, \infty)$. The argument of $x\overline{y}$ is $0$, hence the difference between the arguments of $x$ and $y$ is $0$. That is, $x$ and $y$ must lie on the same ray from the origin, and therefore must be a positive real multiple of each other.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many integral values of $k$ such that $2x^3+3x^2+6x+k=0$ has exactly three real roots How many integral values of $k$ such that $2x^3+3x^2+6x+k=0$ has exactly three real roots. I am unable to see how I'd start this question. A small hint, or the entire solution, both will be highly appreciated!
| Since this is a cubic polynomial.
By using Vieta's formula we have $a+b+c=-\dfrac{3}{2}$ and $ab+bc+ca=3$. Now squaring $a+b+c$ gives $a^2+b^2+c^2+2(ab+bc+ca)=\dfrac{9}{4}$, so $a^2+b^2+c^2=-\dfrac{15}{4}$ which is always impossible if the roots are real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding the value of $\tan B$
In the triangle $\triangle ABC$
$$\cot A=\dfrac{-3}{4}$$
$$\sin A \cos B - \cos A \cos B = 1 $$
*
*Find the value of $\tan B$
This $\sin A \cos B - \cos A \cos B = 1 $ reminds me of sum and difference formulas in a particular way that made me draw a triangle. However, I couldn't see any way to proceed from there. Your helps will be appreciated.
Regards!
| $$\begin{align*}
1&=\sin A\cos B-\cos A\cos B\\[1ex]
&=\cos A\cos B(\tan A-1)\\[1ex]
&=\cos A\cos B\left(\frac1{\cot A}-1\right)\\[1ex]
&=-\frac73\cos A\cos B
\end{align*}$$
$$\implies\cos B=-\frac37\sec A\quad(*)$$
Then by the Pythagorean identities, we have
$$\begin{align*}
\sin^2B&=1-\cos^2B\\[1ex]
&=1-\left(-\frac37\sec A\right)^2\\[1ex]
&=1-\frac9{49}\sec^2A\\[1ex]
&=1-\frac9{49}(\tan^2A+1)\\[1ex]
&=\frac{40-9\tan^2A}{49}\\[1ex]
&=\frac{40\cot^2A-9}{49\cot^2A}\\[1ex]
&=\frac{24}{49}
\end{align*}$$
$$\implies\cos^2B=1-\frac{24}{49}=\frac{25}{49}$$
Since $A,B$ are angles in a triangle, we have $0<A<\pi$ and $0<B<\pi$. Then we know that $\sin B>0$. Similarly, we know that $\sin A>0$, while $\cot A<0\implies\cos A<0\implies\sec A<0$. This and $(*)$ mean that $\cos B>0$, and taken together we know that $\tan B>0$.
So, we end up with
$$\tan B=\sqrt{\frac{\sin^2B}{\cos^2B}}=\frac{2\sqrt6}5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the given limit: $\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$ Evaluate the given limit
$$\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$
My Attempt :
$$=\lim_{x\to 0} \frac {x.\tan (2x) - 2x.\tan (x)}{(1-\cos (2x))^2}$$
$$=\lim_{x\to 0} \dfrac {x(\tan (2x)-2\tan (x))}{1-2\cos (2x)+\cos^2 (2x)}$$
| With equivalents, there are less calculations:
First rewrite the expression as $\;\dfrac{x(\tan 2x-2\tan x)}{(1-\cos 2x)^2}$.
Now,
*
*it's standard that $\;1-\cos u\sim_0\dfrac{u^2}2$, so $\;(1-\cos 2x)^2\sim_0(2x)^2 $.
*Taylor's formula at order $3$ for the tangent is $\;\tan u=u+\dfrac{u^3}3+o(u^3)$, so
$$\tan 2x-2\tan x=2x+\frac{8x^3}3 -\Bigl(2x+\frac{2x^3}3\Bigr)+o(x^3)=2x^3+o(x^3)\sim_0 2x^3, $$
and eventually
$$\dfrac{x(\tan 2x-2\tan x)}{(1-\cos 2x)^2}\sim_0\frac{2x^4}{4x^4}=\frac12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
Evaluate $\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx$
My attempt:
$I=\int_0^{2\pi}\frac 1{\sin^4x+\cos^4x}dx=\int_0^{2\pi}\frac 1{(\sin^2x+\cos^2x)^2-2\sin^2(2x)}dx=\int_0^{2\pi}\frac {1}{1-2\sin^2(2x)}dx=\frac 12\int_0^{4\pi}\frac 1{1-2\sin^2(x)}dx=\frac 12 \int_0^{4\pi}\frac {1}{\cos(\frac{x}2)}dx=\int_0^{2\pi}\frac 1{\cos x}dx=0$
So it actually is:
$$I=2\int_0^{2\pi}\frac {1}{2-\sin^2(2x)}dx=\int_0^{4\pi}\frac{1}{2-\sin^2(x)}dx=\int_0^{4\pi}\frac 1{1+\cos^2x}dx$$
Now if I try to make the substituion $u=\tan(\frac x2)$ I get integral from $0$ to $0$...Why?
What I am doing wrong?
| Under $x\to\tan x\to x-\frac1x$, one has
\begin{eqnarray}
&&\int_0^{2\pi} \frac 1{\sin^4x+\cos^4x}dx\\
&=&4\int_0^{\pi/2} \frac 1{\sin^4x+\cos^4x}dx\\
&=&4\int_0^{\pi/2} \frac {\sec^2x}{1+\tan^4x}\sec^2xdx\\
&=&4\int_0^{\infty} \frac{1+x^2}{1+x^4}dx\\
&=&4\int_0^{\infty} \frac{1+\frac1{x^2}}{x^2+\frac1{x^2}}dx\\
&=&4\int_0^{\infty} \frac{1}{\left(x-\frac1{x}\right)^2+2}d\left(x-\frac1x\right)\\
&=&4\int_{-\infty}^{\infty} \frac{1}{x^2+2}dx\\
&=&\frac{4}{\sqrt2}\arctan(\frac{x}{\sqrt2})|_{-\infty}^{\infty}\\
&=&2\pi\sqrt2.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 0
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How to evaluate $\lim _{x\to 0}\frac{5-5\cos\left(2x\right)+\sin\left(4x\right)}{x}$ without using L'Hospital's rule? I need to evaluate the following limit without using L'Hopital's rule:
$$\lim _{x\to 0}\left(\frac{5-5\cos\left(2x\right)+\sin\left(4x\right)}{x}\right)$$
I thought the best way was to separate it in two limits:
$$\lim _{x\to 0}\left(\frac{5-5\cos\left(2x\right)}{x}\right)+\lim_{x\to \:0}\left(\frac{\sin\left(4x\right)}{x}\right)$$
Considering that $\lim\,_{x\to \:0}\left(\frac{\sin x}{x}\right)=1$ we can easily know that the second limit is $4$. I also know the result of the original limit is $4$, so the result of the first limit in the second line needs to be $0$ $(0+4=4)$.
The issue is that I can not figure out how to remove the $x$ from the denominator so I can avoid the indeterminate form. I already tried replacing $\cos(2x)$ with $\cos^2 x-\sin^2 x$, but it seems to be useless.
So, to summarize everything, my problem is how to evaluate this limit without using L'Hospital's rule:
$$\lim _{x\to 0}\left(\frac{5-5\cos\left(2x\right)}{x}\right).$$
| Just another way using Taylor series.
$$\cos(x)=1-\frac{x^2}{2}+O\left(x^4\right)\implies \cos(2x)=1-2 x^2+O\left(x^4\right)$$
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)\implies \sin(4x)=4 x-\frac{32 x^3}{3}+O\left(x^4\right)$$
$$5-5\cos\left(2x\right)+\sin\left(4x\right)=4 x+10 x^2-\frac{32 x^3}{3}+O\left(x^4\right)$$
$$\frac{5-5\cos\left(2x\right)+\sin\left(4x\right)}{x}=4+10 x-\frac{32 x^2}{3}+O\left(x^3\right)$$ which shows the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\int \sqrt {3 \tan^2 \theta - 1} d \theta$ Evaluate $I=\int \sqrt {3 \tan^2 \theta - 1} d \theta$
My attempt
$\tan \theta = t, $ then $I = \int \frac{\sqrt {3t^2-1}}{1+t^2} dt $
Now integrating by parts,
I = $\sqrt {3t^2-1} \tan^{-1} t- \int( \frac{6t}{2\sqrt {(3t^2-1)}} \tan^{-1} t) dt$
Now i am struck... How to proceed.
| Hints :
$$=\int \frac {(2\tan \theta\sec^2\theta)(\sqrt {3\tan ^2\theta-1})}{(2\tan \theta\sec^2\theta)} d\theta$$
Let $$\tan^2\theta=\frac {x^2+1}{3}$$ hence $$2\tan \theta\sec^2\theta=2x/3$$
And then change the integral to $$\sqrt 3\int \frac {x^2}{(x^2+4)\sqrt {x^2+1}}dx= \sqrt 3\int \frac {1}{\sqrt {x^2+1}}dx-4\sqrt 3\int \frac {1}{(x^2+4)\sqrt {x^2+1}}dx$$
Evaluate the first integral using the substitution $x=\tan \alpha$ and end the second one with the substitution $x=\frac 1t$ which further can be evaluated by the substitution $$t^2+1=u^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find and classify the singularities of $f(z) = \frac{1}{e^{z^2}-1}$ I am trying to find and classify the singularities of $f(z) = \frac{1}{e^{z^2}-1}$. I'd also like to find the residue for any poles. So far, I have found that the singularities are of the form $z=\sqrt{2\pi k i}$ for $k \in \mathbb{Z}$.
I believe that when $z = 0$, I can write
$$\frac{1}{e^{z^2}-1} = \left(z^2\left(1 + \frac{z^2}{2!} + \frac{z^4}{3!} + ...\right)\right)^{-1}$$ so then there is a double pole at $z=0$ and the residue there is $$\frac{d}{dz}\left(\left(1 + \frac{z^2}{2!} + \frac{z^4}{3!} + ...\right)^{-1}\right)$$ evaluated at 0, which is 0. Does that seem correct?
But, when $z=\sqrt{2 \pi k i}$ for $k \neq 0$, I think I have a removable singularity, but I'm not sure how to show this.
| Here we have three cases: $z=0$, $e^{z^2}=1$ and singularity at infinity as $z=\infty$. The series
\begin{align}
f(z)
&=\dfrac{1}{e^{z^2}-1}\\
&=\dfrac{1}{z^2+\frac{1}{2!}z^4+\frac{1}{3!}z^6+\cdots}\\
&=\dfrac{1}{z^2}\cdot\dfrac{1}{1+\frac{1}{2!}z^2+\frac{1}{3!}z^4+\cdots}\\
&=\dfrac{1}{z^2}\left(1-\frac{1}{2!}z^2+\frac{z^4}{12}+\cdots\right)\\
&=\dfrac{1}{z^2}-\frac{1}{2}+\frac{z^2}{12}+\cdots
\end{align}
shows that $z=0$ is a pole of order $2$ with residue $0$, as you found. Also the series
$$f(\dfrac1z)=z^2-\dfrac12+\dfrac{1}{12z^2}+\cdots$$
shows that the infinity is an essential singularity, and in fact there is no $k$ such that $\displaystyle\lim_{z\to0}z^kf(\dfrac1z)<\infty$.
| {
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"url": "https://math.stackexchange.com/questions/2848228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that $x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$ Show that $ f(x)=x^{20}+x^{15}+x^{10}+x^5$ is divisible by $x^2+1$
I tried to simplify it by putting $x^5=y$
It simplifies the polynomial but I cannot put it in the case of the divisor.
So I assumed that $x^2+1$ is a divisor of $f(x)$
Then examine the assumption is correct then $x^2+1=(x-i)(x+i)$
Now for $x=\pm i$ using synthetic division it leaves remainder 0
Am I at the right direction, please tell
| There is no need for synthetic division. Just evaluate $x^{20} + x^{15} + x^{10} + x^5 $ at $x=\pm i$:
$$
(\pm i)^{20} + (\pm i)^{15} + (\pm i)^{10} + (\pm i)^5
=
1 + \mp i - 1 + \pm i
= 0
$$
Therefore, $x\pm i$ divides $x^{20} + x^{15} + x^{10} + x^5 $ and so does their product, because their are coprime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2848792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
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Is mod of exponent allowed? Lately I was studying about Modular Arithmetic and the way modulus is used to calculate large numbers. What caught my attention was calculating powers with modulus. It's generally that we calculate the mod of the base and then proceed with our further calculations. Now I was thinking, is it possible to take mod of the power and then mod of that answer to produce the same answer as $(a^b)$%m.
I tried on a few examples to see myself but the answers matched sometimes or would differ many times. So is this really possible with some linear relation, maybe, in the answers by the two methods or is it just not possible?
| I'm going to guess what the intent of your question is. Two columns of numbers below are surmounted by arrows pointing downward. In the first of those columns, going from each horizontal row to the next, one multiplies by $3$ and reduces modulo $11$ at each step. In the second of those two columns, one reduces modulo $11$ only after multiplying (unreduced) numbers. My guess is that the intent of the question is this: Will these always be the same? The answer to that is "yes". Further comments appear below this table.
$$
\begin{align}
\downarrow\phantom{\pmod{11}} & & & & \downarrow \\
(3^1 \equiv 3) \pmod {11} & \qquad & & 3^1 = 3 & \equiv 3 \\
(3^2 \equiv 9) \pmod {11} & & & 3^2=9 & \equiv9 \\
(3^3 \equiv 3\times 9= 27 \equiv 5) \pmod {11} & & & 3^3 = 3\times9 = 27 & \equiv 5 \\
(3^4 \equiv 3\times 5 = 15 \equiv 4) \pmod{11} & & & 3^4 = 3\times27 = 81 & \equiv 4 \\
(3^5 \equiv3\times 4 = 12 \equiv1) \pmod{11} & & & 3^5 = 3\times81 = 243 & \equiv 1 \\
(3^6 \equiv 3\times 1 = 3 \equiv 3) \pmod{11} & & & 3^6 = 3\times243 = 729 & \equiv3
\end{align}
$$
That these are always the same follows from this theorem:
\begin{align}
\text{Suppose } & (a \equiv b) & & \pmod n \\
\text{and } & (c \equiv d) & & \pmod n. \\[4pt]
\text{Then } & (ac \equiv bd) & & \pmod n.
\end{align}
Proof:
\begin{align}
& ac-bd \\[8pt]
= {} & (ac-bc) + (bc-bd) \\[8pt]
= {} & c(a-b) + b(c-d) \\[8pt]
= {} & c\cdot\Big(n\times\text{something}\Big) + b\cdot\Big(n\times \text{something}\Big) \\[8pt]
= {} & n\times\text{something.}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.