Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$
$$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$
then what I will do??
| Let $t=2-\sqrt 3$. Note that $2+\sqrt 3=\frac 1t$. Then, we have
$$\begin{align}x=t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}&\Rightarrow x^3=t+\frac 1t+3\left(t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}\right)\\&\Rightarrow x^3=4+3x\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
} |
Substituting the value $x=2+\sqrt{3}$ into $x^2 + 1/x^2$ My teacher gave me a question which I am not able to solve:
If $x=2+\sqrt{3}$ then find the value of $x^2 + 1/x^2$
I tried to substitute the value of x in the expression, but that comes out to be very big.
| Hint:
$$(2-\sqrt{3})(2+\sqrt{3})=1$$
$$(2-\sqrt{3})(x)=1$$
so
$$\frac{1}{x}=2-\sqrt{3}$$
$$(x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}$$
$$x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2$$
$$x^2+\frac{1}{x^2}=(2+\sqrt{3}+2-\sqrt{3})^2-2=14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 3
} |
How to find sum of the infinite series $\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)}$ $$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$
How to find sum of this series?
I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that I am not able to solve this.
| The sum can be found using our favourite alternative method of converting the sum into a double integral.
Noting that
$$\int_0^1 x^{n - 1} \, dx = \frac{1}{n} \quad \text{and} \quad \int_0^1 y^{2n} \, dy = \frac{1}{2n + 1},$$
the sum can be rewritten as
\begin{align*}
\sum_{n = 1}^\infty \frac{1}{n (2n + 1)} &= \sum_{n = 1}^\infty \int_0^1 \int_0^1 x^{n - 1} y^{2n} \, dx dy\\
&= \int_0^1 \int_0^1 \sum_{n = 1}^\infty x^{n - 1} y^{2n} \, dx dy \tag1 \\
&= \int_0^1 \int_0^1 \frac{1}{x} \sum_{n = 1}^\infty (xy^2)^n \, dx dy\\
&= \int_0^1 \int_0^1 \frac{1}{x} \cdot \frac{xy^2}{1 - xy^2} \, dx dy \tag2\\
&= \int_0^1 \int_0^1 \frac{y^2}{1 - xy^2} \, dx dy\\
&= -\int_0^1 \ln (1 - xy^2) \Big{|}_0^1 \, dy\\
&= - \int_0^1 \ln (1 - y^2) \, dy\\
&= 2 \int_0^1 \frac{y(1 - y)}{1 - y^2} \, dy \tag3\\
&= 2 \int_0^1 \frac{y}{1 + y} \, dy\\
&= 2 \int_0^1 \frac{(1 + y) - 1}{1 + y} \, dy\\
&= 2 \int_0^1 \left (1 - \frac{1}{1 + y} \right ) \, dy\\
&= 2 \left [y - \ln (1 + y) \right ]_0^1\\
&= 2 (1 - \ln 2).
\end{align*}
Explanation
(1) Changing the summation with the double integration.
(2) Summing the series which is geometric.
(3) Integrating by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
} |
$x^2 + 3x + 7 \equiv 0 \pmod {37}$ I'm trying to solve the following
$x^2 + 3x + 7 \equiv 0 \pmod {37}$
What I've tried -
I've tried making the left side as a square and then I know how to solve
but couldn't make it as a square root..
We also learned in class that you can multiply the left side and the modulo by $4a$
(that is $4\cdot 1 = 1$) and continue somehow - which I can't figure out how.
any help will be appreciated.
| In the real numbers, a method of finding a solution to a quadratic equation is to complete the square. This would involve adding and subtracting $(b/2)^2$. $b=3$ in your case, and remember that $1/2 = 19 \mod 37$.
Specifically notice: $$(x+3 \cdot 19)^2 \equiv x^2 + 2\cdot 3 \cdot 19 x + (3 \cdot 19)^2$$ $$\equiv x^2 + 3x + (20)^2 \mod 37$$
Note that $3 \cdot 19 \equiv 20 \mod 37$. Also $20^2 = 400 \equiv 30 \mod 37$.
Thus the method of completing the square is as follows $$x^2 + 3x + 7 \equiv x^2 + 3x + 20^2 - 20^2 + 7 \equiv (x+20)^2 - 23 \mod 37$$
Finally this means you need to solve $$(x+20)^2 \equiv 23 \mod 37$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Definite solution for the mean distance from an external point to the surface of a sphere.
Sphere, radius $E$, is centred at point $O$ $[0,0,0]$.
External point $Q$ is at $[D,0,0]$.
I can slice the sphere by making multiple planar cuts parallel to the $YZ$ plane to produce circular zones (quasi-discs) of equal infinitessimal width $dx$ and position $x_i$. The curved surface of each zone has the same surface area. The radius $R_i$ of any zone $i$ lies in the $YZ$ plane and has magnitude: $R_i = \sqrt{E^2 - x_i^2} .$
The distance $L_i$ from point $Q$ to any point $P_i$ in the zone $i$ is given by:-
$$L_i = \sqrt{ (D-x_i)^2 + R_i^2}
= \sqrt{ D^2 -2Dx_i + x_i^2 + E^2 - x_i^2}
= \sqrt{ D^2 -2Dx_i + E^2 } $$
$$L_i = D \sqrt{ 1 -\frac{2x_i}{D} + \frac{E^2}{D^2} }. $$
Now I wish to integrate this expression over the range $-E\cdots+E$ and then divide by $2E$ to obtain the average value $\bar{L}$ thus
$$\bar{L} = \frac{D}{2E} \int_{-E}^{+E} \sqrt{ 1 -\frac{2x_i}{D} + \frac{E^2}{D^2} } dx.$$
I can make a Taylor Series approximation of the square root term using the standard formula $(1+x)^{0.5} = 1 + x/2 - x^2/8 + x^3/16 -\cdots$
Applying this to the square root term I obtain
$$
\sqrt{ 1 -\frac{2x_i}{D} + \frac{E^2}{D^2} } = 1-x_i/D + (1/2)E^2/D^2 -(1/2)x_i^2/D^2 + (1/2)x_iE^2/D^3 + \cdots$$ where the subsequent terms on the RHS diminish in magnitude. By dropping terms with odd powers of $x_i$ (because
they will go to zero when integrating over the range $-E \le x_i \le +E$ ) and consolidating and integrating I come up with the following approximate result:-
$$\bar{L} = \frac{D}{2E} \int_{-E}^{+E} \sqrt{ 1 -\frac{2x_i}{D} + \frac{E^2}{D^2} } dx
\approx
D\left(1 + \frac{E^2}{3D^2} + A\right) $$
where A is a small term whose expression depends on the number of terms evaluated in the Taylor series approximation. From numerical modelling it appears plausible that the term $A$ vanishes if the Taylor Series is extended to infinite terms and thus the definite solution would be given by:-
$$\bar{L} =
D\left(1 + \frac{E^2}{3D^2} \right)
= D + \frac{E^2}{3D}
$$
MY QUESTION
Is there a way to derive a definite solution to this problem?
| Following suggestions from coffeemath & user2566092 I found the following standard rule:-
$$
\int (ax + b) ^{p/n} = \frac{n}{(n+p)a} (ax + b) ^{1+p/n} +C
$$
for $ p = \pm1, \pm2, \cdots p \ne -n .$
(Source: The Universal Encyclopedia of Mathematics, page 590; a similar equation is here).
Applying that to the expression for $\bar {L}$
$$\bar{L} = \frac{D}{2E} \int_{-E}^{+E} \sqrt{ 1 -\frac{2x_i}{D} + \frac{E^2}{D^2} } dx
=
\frac{D}{2E} \int_{-E}^{+E} \left[ \left(\frac{-2}{D}\right)x_i + \left(1 + \frac{E^2}{D^2}\right) \right]^{1/2} dx$$
whence $a=-2/D,b = 1 +(E^2/D^2), p = 1, n = 2$ and so
$$\bar{L} =
\frac{D}{2E}
\frac{2}{3(-2/D)}
\left[
\left(
\left(\frac{-2}{D}\right)x_i + \left(1 + \frac{E^2}{D^2}\right) \right)^{3/2}+constant\right]_{-E}^{+E} $$
then cancelling and dropping the constant leads to
$$\bar{L} =
\frac{-D^2}{6E}
\left[ \left(
\frac{-2Dx_i+D^2+E^2}{D^2} \right)^{3/2}\right]_{-E}^{+E}
=
\frac{-D^2}{6E} \left( \frac{1}{D^2}\right)
^{3/2}
\left[ \left(
-2Dx_i+D^2+E^2 \right)^{3/2}\right]_{-E}^{+E} $$
expanding and shuffling minus signs
$$\bar{L} =
\frac{1}{6DE}
\left[
\left(2DE+D^2+E^2 \right)^{3/2}
-
\left(-2DE+D^2+E^2 \right)^{3/2}
\right]
$$
this simplifies to
$$\bar{L} =
\frac{1}{6DE}
\left[
\left( (D+E)^2 \right)^{3/2}
-
\left((D-E)^2 \right)^{3/2}
\right]
=
\frac{1}{6DE}
\left[
(D+E)^3 -(D-E)^3
\right]
$$
and then
$$\bar{L} =
\frac{1}{6DE}
\left[
6D^2E +2E^3
\right]
=
D + \frac{E^2}{3D}
$$
and finally
$$\bar{L} =
D \left( 1+ \frac{E^2}{3D^2} \right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Calculate in closed form $\sum_{n=1}^{\infty} \frac{\arctan(1/n) H_n}{n}$ Playing with Taylor series is not helpful enough. What else would you try out?
$$\sum_{n=1}^{\infty} \frac{\arctan(1/n) H_n}{n}$$
$$\approx 2.1496160413898356727147400526167103602143301206321$$
It's easy to see the series converges since $\arctan(1/n) \approx 1/n$ when $n$ large.
Maybe its integral representation makes us feel more comfortable
$$1/4\int_0^1 \frac{ 2(\gamma \pi x \coth (\pi x)+\gamma) +i x \left(\psi ^{(0)}(-i x)^2-\psi ^{(0)}(i x)^2-\psi ^{(1)}(-i x)+\psi ^{(1)}(i x)\right)}{ x^2} \, dx$$
| Let $f(z)=\dfrac{(\gamma+\psi_0(-z))^2}{z^2+x^2}$. On $z=Re^{i[0,2\pi]}$, $f(z)\sim\mathcal{O}\left(\dfrac{\ln^2{R}}{R^2}\right)$, so the residue theorem gives
\begin{align}
\sum^\infty_{n=1}\operatorname{Res}\left(f(z),n\right)+\sum_{\pm}\operatorname{Res}\left(f(z),\pm ix\right)+\operatorname{Res}\left(f(z),0\right)=0\tag1
\end{align}
At the positive integers,
\begin{align}
\sum^\infty_{n=1}\operatorname{Res}\left(f(z),n\right)
&=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\frac{1}{(z^2+x^2)(z-n)^2}+\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\frac{2H_n}{(z^2+x^2)(z-n)}\\
&=2\sum^\infty_{n=1}\frac{H_n}{n^2+x^2}-2\sum^\infty_{n=1}\frac{n}{(n^2+x^2)^2}\tag2
\end{align}
At $z=\pm ix$,
\begin{align}
\sum_{\pm}\operatorname{Res}\left(f(z),\pm ix\right)
&=\frac{(\gamma+\psi_0(-ix))^2}{2ix}-\frac{(\gamma+\psi_0(ix))^2}{2ix}\\
&=\operatorname{Im}\frac{(\gamma+\psi_0(-ix))^2}{x}\tag3
\end{align}
At $z=0$,
\begin{align}
\operatorname{Res}(f(z),0)
&=\operatorname*{Res}_{z=0}\frac{1}{z^2(z^2+x^2)}=0\tag4
\end{align}
Substituting $(2)$, $(3)$, $(4)$ into $(1)$,
\begin{align}
\sum^\infty_{n=1}\frac{H_n}{n^2+x^2}=\sum^\infty_{n=1}\frac{n}{(n^2+x^2)^2}+\operatorname{Im}\frac{(\gamma+\psi_0(ix))^2}{2x}\tag5
\end{align}
and integrating $(5)$ from $0$ to $1$ gives
\begin{align}
\sum^\infty_{n=1}\frac{H_n\operatorname{arccot}{n}}{n}
&=\color{darkblue}{\operatorname{Im}\int^1_0\frac{(\gamma+\psi_0(ix))^2}{2x}\ {\rm d}x}+\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n(n^2+1)}+\frac{1}{2}\sum^\infty_{n=1}\frac{\operatorname{arccot}{n}}{n^2}\\
&=\color{darkblue}{\mathcal{I}}+\frac{1}{2}\sum^\infty_{n=1}\left(\frac{1}{n}-\frac{1}{2(n-i)}-\frac{1}{2(n+i)}\right)+\frac{1}{2}\int^1_0\sum^\infty_{n=1}\frac{1}{n(n^2+x^2)}\ {\rm d}x\\
&=\color{darkblue}{\mathcal{I}}+\frac{\gamma}{2}+\frac{1}{2}\operatorname{Re}\psi_0(1+i)+\int^1_0\frac{\gamma+\operatorname{Re}\psi_0(1+ix)}{2x^2}{\rm d}x\\
&=\frac{\gamma}{2}+\frac{1}{2}\operatorname{Re}\psi_0(1+i)+\operatorname{Re}\int^1_0\frac{\gamma+\psi_0(1+ix)-ix(\gamma+\psi_0(ix))^2}{2x^2}\ {\rm d}x\\
&\tag6
\end{align}
The remaining integral is quite hard to approach as of now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Concluding three statements regarding $3$ real numbers.
$\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$
Conclusion $I.)\ 1<b<3$
Conclusion $II.)\ 2<a<3$
Conclusion $III.)\ 0<c<1$
Options
By the given statements
$\color{green}{a.)\ \text{Only conclusion $I$ can be derived}}$.
$b.)\ $ Only conclusion $II$ can be derived.
$c.)\ $ Only conclusion $III$ can be derived.
$d.)\ $ Conclusions $I,\ II,\ III$ can be derived.
$e.)\ $ None of the three conclusions can be derived.
$\quad\\~\\$
I tried $(a+b+c)^2=36 \implies a^2+b^2+c^2=18$
and found that $(a,b,c)\rightarrow \{(-1,1,4),(-3,0,3)\}$ satisfies the two conditions
$a^2+b^2+c^2,\ a<b<c $ but not this one $a+b+c=6$
I thought a lot but can't find any suitable pairs.
I look for a simple and short way.
I have studied maths upto $12$th grade.
|
$\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$
Conclusion I.) $1<b<3$
Conclusion II.) $2<a<3$
Conclusion III.) $0<c<1$
Let $\{p,q,r\}\in \mathbb{R}: a=\min(p,q,r),\ c=\max(p,q,r),
b=\max(\min(p,q),\min(q,r),\min(r,p))$.
Let's solve a system
\begin{align}
p+q+r&=6 ,\\
pq+qr+rp&=9
\end{align}
for $q$ and $r$ in terms of $p$:
\begin{align}
q(p)&=3-\frac{1}{2} p-\frac{1}{2} \sqrt{3p (4-p)},
\\
r(p)&=3-\frac{1}{2} p+\frac{1}{2} \sqrt{3p (4-p)}.
\end{align}
The solution suggests that
$0\le p\le 4$ (to ensure real values).
This diagram illustrates relationship between $p,q$ and $r$:
Analysis of $q(p)$ and $r(p)$
(boundary end extreme points)
shows that
\begin{align}
a&=\begin{cases}
p,& 0<p<1\\
q(p),& 1<p<4
\end{cases}
\\
b&=\begin{cases}
q(p),& 0<p<1\\
p,& 1<p<3\\
r(p),& 3<p<4
\end{cases}
\\
c&=\begin{cases}
r(p),& 0<p<3\\
p,& 3<p<4\\
\end{cases}
\end{align}
And
$0<a<1$
$1<b<3$
$3<c<4$:
so the answer is
a.) Only conclusion I can be derived.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to get the results of this logarithmic equation? How to solve this for $x$:
$$\log_x(x^3+1)\cdot\log_{x+1}(x)>2$$
I have tried to get the same exponent by getting the second
multiplier to reciprocal and tried to simplify $(x^3+1)$.
| We have that
$$\begin{align}
\log_{x+1}x&=\frac{\log_xx}{\log_x(x+1)}\\\\
&=\frac{1}{\log_x(x+1)} \tag 1
\end{align}$$
and
$$\log_x(x^3+1)=\log_x (x+1)+\log_x(x^2-x+1) \tag 2$$
Using $(1)$ and $(2)$ reveals that
$$\begin{align}
\log_x(x^3+1)\log_{x+1}x&=\left(\log_x (x+1)+\log_x(x^2-x+1)\right)\frac{1}{\log_x(x+1)}\\\\
&=1+\frac{\log_x(x^2-x+1)}{\log_x(x+1)} \tag 3
\end{align}$$
We note that if the right-hand side of $(3)$ is to be greater than $2$, we must have $\frac{\log_x(x^2-x+1)}{\log_x(x+1)}>1\implies \frac{x^2-x+1}{x+1}>1\implies x(x-2)>0\implies x>2 \,\,\text{for real-valued solutions}$
Thus, we have that
$$\bbox[5px,border:2px solid #C0A000]{\log_x(x^3+1)\log_{x+1}x>2\,\,\text{for}\,\,x>2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Counting points on the Klein quartic In Moreno's book "Algebraic Curves over Finite Fields", he mentions the following in passing with no further comments ($K$ denotes the Klein quartic defined by $X^3 Y + Y^3 Z + Z^3 X = 0$):
The Jacobian of $K$ is a product of three elliptic curves all isogenous to the elliptic curve with complex multiplication over the field $Q(\sqrt{-7})$. This last fact implies the existence of a formula for counting the number of rational points on $K$ over the field $\mathbb{F}_p$ which depends on how the prime $p$ splits in the ring $\mathbb{Z}[\sqrt{-7}]$.
Does anyone know what he is referring to? The Klein quartic is the same as the modular curve $X(7)$ which has genus 3, so I'd be very interested in such a formula!
EDIT: After some computation, it appears that the number of points on $X(7)$ over $\mathbf{F}_q$ is precisely $q+1$ as long as $q \not\equiv 1 \bmod 7$ (*). Otherwise, the behaviour of the error term $a_q = q+1-\#X(7)(\mathbf{F}_q)$ is kind of complicated, but it appears to be constant if we restrict attention to primes $q$ lying in certain quadratic progressions, for example when $q$ is of the form $28n^2 - 28n + 43$ my data suggests that we always have $a_q = -12$. Does anyone have any idea what's going on here? Note that when $q \equiv 1 \bmod 7$ then $a_q/3 = b_q$ where $b_n$ are the Fourier coefficients of the unique cusp form of weight 2 for the congruence subgroup $\Gamma_0(49)$.
Furthermore, what would the moduli interpretation be of statement (*) above? We know $X(7)$ has 24 cusps, but the only time there are elliptic curves defined over $\mathbf{F}_q$ with complete rational 7-torsion is when $q \equiv 1 \bmod 7$ and the trace of Frobenius is 2 mod 7... so why should there be precisely $q-23$ noncuspidal points on $X(7)$ when $q \not\equiv 1 \bmod 7$?
| $\def\FF{\mathbb{F}}$I'm switching to another answer to discuss relations with $X_0(49)$ and complex multiplication. Again, Elkies notes are a good reference and $\zeta$ is a primitive $7$-th root of unity.
Let $\Delta \subset PSL_2(\FF_7)$ be the group of diagonal matrices. Let $Y = \Delta \backslash K$. Then $Y$ is the modular curve corresponding to the arithmetic subgroup $\left( \begin{smallmatrix} a & 7b \\ 7c & d \end{smallmatrix} \right)$. Conjugating by $\left( \begin{smallmatrix} 7 & 0 \\ 0 & 1 \end{smallmatrix} \right)$ turns this group into $\left( \begin{smallmatrix} a & b \\ 49c & d \end{smallmatrix} \right)$ which is $\Gamma_0(49)$, so $Y \cong X_0(49)$. On the level of moduli, when our base field contains a $7$-th root of unity, $K$ parametrizes elliptic curves with a basis for $E[7]$ (up to a condition on the determinant) and $Y$ parametrizes curves with a chosen splitting of $E[7]$ into two cyclic subgroups. See the penultimate paragraph of Elkies Section 4.2 for why $Y \cong X_0(49)$ from a moduli perspective.
We can also use this perspective to explain why $Y$ has CM by $(1+\sqrt{-7})/2$.
The curve $Y$ has genus $1$; make it into an elliptic curve by choosing the cusp at $\infty$ as the origin. Let $y \in Y$ and let $z \in K$ be a preimage of $y$. Write $\pi$ for the map $K \to Y$. Define
$$\alpha(y) = \pi \left( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} z \right)+ \pi \left( \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} z \right)+ \pi \left( \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} z \right)$$
and
$$\beta(y) = \pi \left( \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} z \right)+ \pi \left( \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} z \right)+ \pi \left( \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix} z \right)$$
where the sums are in the group law on $Y$.
Note that replacing $z$ by a different preimage $\left( \begin{smallmatrix} a & 0 \\ 0 & a^{-1} \end{smallmatrix} \right) z$ would just turn $\alpha(y)$ into
$$\pi \left( \begin{pmatrix} a & a^{-1} \\ 0 & a^{-1} \end{pmatrix} z \right)+ \pi \left( \begin{pmatrix} a & 2 a^{-1} \\ 0 & a^{-1} \end{pmatrix} z \right)+ \pi \left( \begin{pmatrix} a & 4 a^{-1} \\ 0 & a^{-1} \end{pmatrix} z \right)=$$
$$\pi \left( \begin{pmatrix} 1 & a^{-2} \\ 0 & 1 \end{pmatrix} z \right)+ \pi \left( \begin{pmatrix} 1 & 2 a^{-2} \\ 0 & 1 \end{pmatrix} z \right)+ \pi \left( \begin{pmatrix} 1 & 4 a^{-2} \\ 0 & 1 \end{pmatrix} z \right) = \alpha(y).$$
To get to the second line, we used that $\pi$ is the quotient by $\Delta$, so $\pi\left( \left( \begin{smallmatrix} a & 0 \\ 0 & a^{-1} \end{smallmatrix} \right) w \right) = \pi(w)$. So $\alpha$ and $\beta$ are well defined endomorphisms of $Y$.
We claim that $\alpha+\beta+1=0$ and $\alpha \beta = \beta \alpha = 2$. So $\alpha$ and $\beta$ generate a quadratic ring, with $\alpha$ and $\beta$ corresponding to $\frac{-1 \pm \sqrt{-7}}{2}$.
We verify that $\alpha+\beta+1=0$: Writing this out, we must show that
$$\sum_{b=0}^6 \pi\left( \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} z \right) =0.$$
Write $\psi$ for the map $Y \to X_0(7)$ which forgets one of the two summands into which $Y$ decomposes $E[7]$. Then the above sum is $\sum_{\psi(y')= \psi(y) } y'$. But $X_0(7)$ has genus $0$, so the class of $\sum_{\psi(y') = w} y'$ is independent of the choice of point $w \in X_0(7)$, and taking $y$ to be the cusp shows that this constant value is $0$.
We sketch the verification of $\alpha \beta = \beta \alpha = 2$. Multiplying out both sides, we must show
$$3 y + \sum_{b=1}^6 \pi\left( \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} z \right) = 2y.$$
Cancelling $2y$ from both sides, this turns into our previous verification.
Of course, all of this is just using that $\mathbb{Z}[\zeta]$ acts on the Jacobian of $K$ and that $\zeta+\zeta^2+\zeta^4 = \frac{-1+\sqrt{-7}}{2}$, but I think it is fun to see it in terms of matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
probability and expected value Hey I am not sure if I thinking correctly on this question? In a carnival, there is game which charges you $3$ dollars to play a game. You win $1$ dollar for every consecutive head you get and you you can play till you get tail. if you get head head tail you get back two dollars. What is expected profit?
my theory is your probability of t,ht,hht,hhht and so on will be $1/2,1/4,1/8,1/16$ and so on, and income will be $0,1,2,3$ and so on.
so average income will be
$$ 0 * \frac{1}{2} + 1 * \frac{1}{4} + 2 * \frac{1}{8} + 3 *\frac{1}{16} + ....$$
which will come to a dollar?
so average profit will be, $1-3 = -2$??
| Taking "for every consecutive head" to mean "number of heads before a tail", your formulation is correct, and here is another way to get its sum.
A = $0\times\frac{1}{2} + 1\times\frac{1}{4} + 2\times\frac{1}{8} + 3\times\frac{1}{16} + ...$
$\frac{A}{2}$ = ............ $0\times\frac{1}{4} + 1\times\frac{1}{8} + 2\times\frac{1}{16} + ...$
Subtracting, $\frac{A}{2} = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + ...$ , a well known series with sum = 1/2
thus expected net gain = 1 - 3 = -2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$ Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$
The answer should be: $y = \frac{1}{12} x^2 -3$
But how to arrive at the answer?
I tried replacing r with $\sqrt{x^2 + y^2}$, then $\sinθ$ as $\frac{y}{r}$ but to no avail.
I also ended up with
$r = \frac{6}{\frac{r-y}{r}} $ ->
$ 1 = \frac{6}{\sqrt{x^2+y^2}-y}$ -> x = +- 6
but that's not the answer...
| Just multiply by denominator:
$$
r = \frac{6}{1-\sin\theta} \Longrightarrow r(1 - \sin\theta) = 6\Longrightarrow
r - y = 6
$$
Now
$$
\sqrt{x^2+y^2} = 6+y\Longrightarrow x^2 + y^2 = (6+y)^2,
$$
or
$$
x^2 = 36 + 12y \Longrightarrow y = \frac{x^2}{12} - 3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
finding $a_1$ in an arithmetic progression Given an arithmetic progression such that: $$a_{n+1}=\frac{9n^2-21n+10}{a_n}$$
How can I find the value of $a_1$?
I tried using $a_{n+1}=a_1+nd$ but I think it's a loop..
Thanks.
| We are given the condition $a_{n+1}=\dfrac{9n^2-21n+10}{a_n}$ or more compactly
$a_{n} a_{n+1}=9n^2-21n+10 \tag{1}$
Solving for d
By evaluating (1) at $n-1$ we are able to get another term with an $a_n$ factor:
$a_{n-1} a_{n}=9(n-1)^2-21(n-1)+10 = 9n^2-39n+40 \tag{2}$
so by subtracting (1) - (2) : $(a_{n+1} - a_{n-1}) a_{n}=18n-30$ but $a_{n+1} - a_{n-1} = 2d$ so
$2d \cdot a_n = 18n - 30$ so at $n+1$ we have $2d \cdot a_{n+1} = 18(n+1) - 30 = 18n - 12$
$2d \cdot (a_{n+1}-a_n) = 18$ or $2d^2 = 18$ so $\boxed{d = \pm 3}$
For d=3
Now by (1), with $a_n = a_1 + 3n - 3$ and $a_{n+1} = a_1 + 3n$ we have
$(a_1+3n-3)(a_1+3n) = 9n^2-21n+10$
which (eventually) reduces to
${a_1}^2 + 6na_1 - 3a_1 = -12n+10$
Since this must be true for all $n$, we equate coefficients of $n$ (and check the rest) to see the only solution is $\boxed{a_1=-2}$
For d=-3
Now by (1), with $a_n = a_1 - 3n + 3$ and $a_{n+1} = a_1 - 3n$ we have
$(a_1-3n+3)(a_1-3n) = 9n^2-21n+10$
which reduces to
${a_1}^2 - 6na_1 + 3a_1 = -12n+10$
Since this must be true for all $n$, we equate coefficients of $n$ (and check the rest) to see the only solution is $\boxed{a_1=2}$
Remarks
If when solving for $a_1$ (for a given value of $d$) you ony apply (1) to a single $n$, say $n=1$, you will get two possible values for each value of $d$, and hence four solutions for $a_1$. Two of these are incorrect and will give $a_3 = a_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Differentiate the Function: $f(x)=\ln (\sin^2x)$ $$\begin{align}f(x)&=\ln (\sin^2x)\\
f'(x)&=\frac{1}{\sin^2x}\cdot 2(\sin x)(\cos x)\\
&=\frac{2(\sin x)(\cos x)}{\sin^2x}\\
&=\frac{2\ \ (\cos x)\ }{\sin x}\\
&=2\cot x
\end{align}$$
= $2\ cot\ x$
Is this answer right?
| Not quite; you have misapplied the Chain Rule. If $f(x) = g(h(x))$, where here $g(x) = \ln x$ and $h(x) = \sin^2 x$, then $f'(x) = g'(h(x))h'(x)$, so we get
$$
\frac{1}{\sin^2 x}\cdot\frac{d}{dx}(\sin^2 x) = \frac{2\sin x\cos x}{\sin^2 x}
= \frac{2\cos x}{\sin x} = 2\cot x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
solve $|x-6|>|x^2-5x+9|$
solve $|x-6|>|x^2-5x+9|,\ \ x\in \mathbb{R}$
I have done $4$ cases.
$1.)\ x-6>x^2-5x+9\ \ ,\implies x\in \emptyset \\
2.)\ x-6<x^2-5x+9\ \ ,\implies x\in \mathbb{R} \\
3.)\ -(x-6)>x^2-5x+9\ ,\implies 1<x<3\\
4.)\ (x-6)>-(x^2-5x+9),\ \implies x>3\cup x<1 $
I am confused on how I proceed.
Or if their is any other short way than making $4$ cases than I would like to know.
I have studied maths up to $12$th grade. Thanks.
| Another way is to note the inequality is equivalent to
$$(x-6)^2>(x^2-5x+9)^2\iff -(x-1)(x-3)(x^2-6x+15)>0$$
The quadratic is always positive, so this is the same as $(x-1)(x-3)<0$ which means $x\in (1,3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving the closed form for an infinite sum (related to Chebyshev polynomials) How do I prove the following identity? For $y\not= 0$, we have
$$
\sum_{n=0}^{\infty} \dfrac{1}{2y}\left( (x+y)^{n+1}-(x-y)^{n+1}\right)
= \dfrac{1}{(x+y-1)(x-y-1)}.
$$
I am trying to find the closed form for the left hand side. Thanks in advance!
| $$
\begin{align}
\sum_{n=0}^{\infty} \dfrac{1}{2y}\left( (x+y)^{n+1}-(x-y)^{n+1}\right)
&= \sum_{n=0}^{\infty} \sum_{k=0}^n (x+y)^{k}(x-y)^{n-k} \\
&= \sum_{k=0}^\infty \sum_{n=k}^{\infty} (x+y)^{k}(x-y)^{n-k}\\
&= \sum_{k=0}^\infty (x+y)^{k} \sum_{n=k}^{\infty} (x-y)^{n-k}\\
&= \sum_{k=0}^\infty (x+y)^{k} \sum_{m=0}^{\infty} (x-y)^{m}\\
&= \frac{1}{1-(x-y)}\sum_{k=0}^\infty (x+y)^{k} \\
&= \frac{1}{1-(x-y)}\cdot\frac{1}{1-(x+y)}\\
&= \frac{1}{(x+y-1)(x-y-1)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a system of non-linear equations Let
$$(\star)\begin{cases}
\begin{vmatrix}
x&y\\
z&x\\
\end{vmatrix}=1, \\
\begin{vmatrix}
y&z\\
x&y\\
\end{vmatrix}=2, \\
\begin{vmatrix}
z&x\\
y&z\\
\end{vmatrix}=3.
\end{cases}$$
Solving the above system of three non-linear equations with three unknowns.
I have a try.
Let$$A=\begin{bmatrix}
1& 1/2& -1/2\\
1/2& 1& -1/2\\
-1/2& -1/2& -1
\end{bmatrix}$$
We have $$(x,y,z)A\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=0.$$
There must be a orthogonal matrix $T$,such that $T^{-1}A T=diag \begin{Bmatrix}
\frac{1}{2},\frac{\sqrt{33}+1}{4},-\frac{\sqrt{33}-1}{4}
\end{Bmatrix}.$
$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=T\begin{pmatrix}
x^{'}\\
y^{'}\\
z^{'}
\end{pmatrix}\Longrightarrow\frac{1}{2} {x'}^{2}+\frac{\sqrt{33}+1}{4} {y'}^{2}-\frac{\sqrt{33}-1}{4}{z'}^{2}=0.$$
But even if we find a $\begin{pmatrix}
x_0^{'}\\
y_0^{'}\\
z_0^{'}
\end{pmatrix} $ satisfying $\frac{1}{2} {x_0'}^{2}+\frac{\sqrt{33}+1}{4} {y_0'}^{2}-\frac{\sqrt{33}-1}{4}{z_0'}^{2}=0,\begin{pmatrix}
x_0\\
y_0\\
z_0
\end{pmatrix}=T\begin{pmatrix}
x_0^{'}\\
y_0^{'}\\
z_0^{'}
\end{pmatrix}$ may not be the solution of $(\star)$
If you have some good ideas,please give me some hints. Any help would be appreciated!
| Note: This is a slightly clumsy but systematic approach. On the plus side, this allow you solving similar equations of the form
$$\begin{cases}
x^2 - Ayz &= D\\
y^2 - Bxz &= E\\
z^2 - Cxy &= F
\end{cases}$$
without knowing how to complete the squares. On the minus side, you need to factor a quartic polynomial in the middle of the process.
Notice the LHS of given set of equations are all homogenous with degree $2$,
$$\begin{cases}
x^2 - yz &= 1\\
y^2 - xz &= 2\\
z^2 - xy &= 3
\end{cases}\tag{*1}$$
we can simplify it by looking at the ratios first. i.e. let $u = \frac{y}{x}$ and $v = \frac{z}{x}$, we have
$$
\begin{cases}
\frac{1-uv}{1} &= \frac{u^2 - v}{2}\\
\frac{1-uv}{1} &= \frac{v^2 - u}{3}
\end{cases}
\iff
\begin{cases}
2(1-uv) &= u^2 - v\\
3(1-uv) &= v^2 - u
\end{cases}
\implies
\begin{cases}
v &= \frac{2-u^2}{2u-1} &(*2a)\\
u &= \frac{3-v^2}{3v-1} &(*2b)
\end{cases}
$$
Substitute $(*2a)$ into $(*2b)$, we get
$$u = \frac{3 - \left(\frac{2-u^2}{2u-1}\right)^2}{3\left(\frac{2-u^2}{2u-1}\right) - 1}
\iff
\frac{5u^4+u^3-5u-1}{6u^3+u^2-16u+7} =
\frac{(u-1)(5u+1)(u^2+u+1)}{(2u-1)(3u^2+2u-7)}
= 0\\
$$
Since $u^2 + u + 1 = (u+\frac12)^2 + \frac34 > 0$ for all real $u$, there are
only two choices for $u$:
*
*Case 1 : $u = 1 \implies v = \frac{2 - 1^2}{2-1} = 1 \implies 1 - uv = 0$.
However, this contradicts with the requirement $x^2 - yz = x^2(1-uv) = 1$,
this doesn't lead to any solution for $(*1)$.
*Case 2 : $u = -\frac15 \implies v = \frac{2 - \left(-\frac15\right)^2}{2\left(-\frac15\right)-1} = -\frac75 \implies 1 - uv = \frac{18}{25}$.
We have $x = \frac{1}{\sqrt{1-uv}} = \pm\frac{5}{3\sqrt{2}}$ now. This
leads to two real solutions for $(*1)$.
$$(x,y,z) = (x,xu,xv) =
\left(\pm\frac{5}{3\sqrt{2}},
\mp\frac{1}{3\sqrt{2}},
\mp\frac{7}{3\sqrt{2}}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
How to determine congruence manually How is it possible to determine if the the following congruence is true manually?
$$
2015^{53} \equiv 8 \pmod{11}
$$
| Since $2015 = 11 \cdot 183 + 2$, then $2015 \equiv 2 \pmod{11}$.
Our problem has now been reduced to showing that $2^{53} \equiv 8 \pmod{11}$.
Since $2^{53} = \left(2^5\right)^{10} \cdot 2^3 \pmod{11}$ and $2^5 = 32\equiv -1 \pmod{11}$ then $$2^{53} \equiv (-1)^{10} \cdot 2^3 \pmod{11} = 8 \pmod{11}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Differentiate the Function $ h(z)=\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$
Differentiate the function $$h(z)=\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$$
My try:
$$h(z) = \frac{1}{2}\ln\left(a^2-z^2\right)-\frac{1}{2}\ln\left(a^2+z^2\right)$$
so
$$h'(z) = \frac{1}{2}\cdot\frac{2a-2z}{a^2-z^2}-\frac{1}{2}\cdot\frac{2a+2z}{a^2+z^2}$$
My answer is therefore $$h'(z) = \frac{-2a^2z+2z^2a}{(a^2-z^2)(a^2+z^2)}$$
Is it correct?
| No, this is not correct, though you are on the good path. $a^2$ is a constant, so when differentiated, it will be gone. The correct derivative is:
$$\frac{1}{2}\cdot\frac{-2z}{a^2-z^2}-\frac{1}{2}\cdot\frac{2z}{a^2+z^2}=\frac{1}{2} \frac{-2z(a^2+z^2)-2z(a^2-z^2)}{(a^2-z^2)(a^2+z^2)}=\\=\frac{1}{2} \frac{-4a^2z}{(a^2-z^2)(a^2+z^2)}=\frac{-2a^2z}{(a^2-z^2)(a^2+z^2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do i evaluate this integral $ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $? Is there some one show me how do i evaluate this integral :$$ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $$
Note :By mathematica,the result is :
$\frac{Gamma\left(\frac1 4\right)Gamma\left(\frac5 4\right)}{\sqrt{\pi}}-\sqrt{2} Hypergeometric2F1\left(\frac1 4,\frac3 4,\frac5 4,\frac1 4\right).$
and i think it elliptic integral .
Thank you for any kind of help
| This is not an answer but it is too long for a comment.
As I wrote in comment, there is something wrong somewhere since $$\int\frac{\sqrt{\tan x}}{\sin x}dx=-2 \sqrt{\cos (x)} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\cos
^2(x)\right)$$ and $$I=\int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx=\frac{\Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{5}{4}\right)}{\sqrt{\pi
}}-\sqrt{2} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\frac{1}{4}\right)\approx 0.379133$$ which does not match with Mercy's result $$\ln\left(\frac{1+\sqrt2}{\sqrt3}\right) \approx 0.332067$$
Reusing all steps done for Mercy, I thing that in fact, we should arrive to $$I=\int_1^{\sqrt{3}}\frac{1}{\sqrt t \,\sqrt{1+t^2}}\,dt=\frac{4 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}-\frac{2 \,
_2F_1\left(\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{1}{3}\right)}{\sqrt[4]{3}}\approx 0.379133$$ which is still a nightmare ( the Inverse Symbolic Calculator did not find any solution for this number).
Totally unable to find any closed form derivative, I came back to the original problem and expanded the integrand as a Taylor series built at $x=\frac \pi 4$. This gave $$\frac{\sqrt{\tan x}}{\sin x}=\sqrt{2}+\sqrt{2} \left(x-\frac{\pi }{4}\right)^2+\frac{7 \left(x-\frac{\pi
}{4}\right)^4}{3 \sqrt{2}}+\frac{139 \left(x-\frac{\pi }{4}\right)^6}{45
\sqrt{2}}+\frac{5473 \left(x-\frac{\pi }{4}\right)^8}{1260
\sqrt{2}}+\cdots$$ Integrating between the given bounds leads to $$I\approx \frac{\pi }{6 \sqrt{2}}+\frac{\pi ^3}{2592 \sqrt{2}}+\frac{7 \pi ^5}{3732480
\sqrt{2}}+\frac{139 \pi ^7}{11287019520 \sqrt{2}}+\frac{5473 \pi
^9}{58511909191680 \sqrt{2}}+\cdots$$ which gives the correct value for six significant figures.
Edit
For sure, the exact solution exits but it involves elliptic integrals. Using the tangent half-angle substitution $t=\tan(\frac x2)$ and the fiven bounds, the result is $$I=-2 i \sqrt{2} \left(F\left(\left.\sin
^{-1}\left(\sqrt[4]{3}\right)\right|-1\right)-F\left(\left.\sin
^{-1}\left(\sqrt{1+\sqrt{2}}\right)\right|-1\right)\right)$$ where appear elliptic integrals of the first kind.
Edit
If we change variable $x=y+\frac \pi 4$, expanding the trigonometric functions and simplifying, we can arrive at $$I=\int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx=\sqrt{2}\int_{0}^{\pi /12}\frac{dy}{\sqrt{\cos (2 y)}}=\sqrt{2}\, F\left(\left.\frac{\pi }{12}\right|2\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How do I prove this nice inequality $x+3\sqrt[3]{xy^2}\geq4\sqrt{xy} $?
Let $x,y\geq0$. Prove that:
$$
x+3\sqrt[3]{xy^2}\geq4\sqrt{xy}
$$
Note: It's seems easy but when I tried to show it I went to complicated formula.
| Hint :$x+3\sqrt[3]{xy^2}=x+\sqrt[3]{xy^2}+\sqrt[3]{xy^2}+\sqrt[3]{xy^2}\ge 4\sqrt[4]{x\cdot\sqrt[3]{xy^2}\cdot\sqrt[3]{xy^2}\cdot\sqrt[3]{xy^2}}=4\sqrt{xy}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $\int \frac{\sin x}{\tan x+\cos x} \, dx$ How to calculate $$\int \frac{\sin x}{\tan x+\cos x} \, dx\text{ ?}$$ I got to $$\int \frac{-u}{u^2-u-1} \, du$$ while $u=\sin x$ but can I continue from here?
| First complete the square:
$$
u^2-u-1 = \left(u^2-u+\frac 1 4\right)-\frac 5 4 = \left( u - \frac 1 2 \right)^2 -\frac 5 4 = \left( u - \frac 1 2 - \frac{\sqrt 5} 2 \right)\left( u - \frac 1 2 + \frac{\sqrt 5} 2 \right)
$$
Then use partial fractions.
$$
\frac{-u}{u^2-u-1} = \frac A {u - \frac 1 2 - \frac{\sqrt 5} 2} + \frac B {u - \frac 1 2 + \frac{\sqrt 5} 2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $
Evaluate
$$ \prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right) $$
I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$
but I couldn't find the product.
Any help will be appreciated.
Thanks.
| Let $\displaystyle\text{C}=\prod_{r=1}^{7}\cos{\left(\dfrac{r\pi}{15}\right)}$
and
$\displaystyle\text{S}=\prod_{r=1}^{7}\sin{\left(\dfrac{r\pi}{15}\right)}$
Now,
$\text{C}\cdot\text{S}=\left(\sin{\dfrac{\pi}{15}} \cdot \cos{\dfrac{\pi}{15}}\right) \cdot \left(\sin{\dfrac{2\pi}{15}}\cdot\cos{\dfrac{2\pi}{15}}\right)\cdot \ldots \cdot\left(\sin{\dfrac{7\pi}{15}} \cdot \cos{\dfrac{7\pi}{15}}\right) $
$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \left(2\sin{\dfrac{\pi}{15}} \cdot \cos{\dfrac{\pi}{15}}\right) \cdot \left(2\sin{\dfrac{2\pi}{15}}\cdot\cos{\dfrac{2\pi}{15}}\right)\cdot \ldots \cdot\left(2\sin{\dfrac{7\pi}{15}} \cdot \cos{\dfrac{7\pi}{15}}\right) $
$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \ \sin{\dfrac{2\pi}{15}}\cdot\sin{\dfrac{4\pi}{15}}\cdot \ldots \cdot\sin{\dfrac{14\pi}{15}} $
$\{\because \sin(2x) = 2\sin (x) \cos (x) \}$
$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \ \sin{\dfrac{\pi}{15}}\cdot\sin{\dfrac{2\pi}{15}} \cdot \ldots \cdot \sin{\dfrac{7\pi}{15}} \\\\ \{\because \sin(\pi-x)=\sin(x)\} $
$\implies \text{C}\cdot\text{S}= \dfrac{1}{2^7} \cdot \text{S}$
since $\text{S} \neq 0$,
$\therefore \boxed{\text{C}=\dfrac{1}{2^7}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 5,
"answer_id": 0
} |
Solving for y' in a fraction Given the equation $x+xy^2 = \tan^{-1}(x^2y)$ find $y'$.
I have tried doing this but solving for $y'$ I need some help and would like your advice.
Work so far...
$$1+y^2+2xy\left(\frac{dy}{dx}\right)= \frac{2xy+x^2\left(\frac{dy}{dx}\right)}{1+x^4y^2}$$
What can be done now to solve for $y'$?
| Multiply the both sides by $1+x^4y^2$ to get
$$(1+y^2+2xyy')(1+x^4y^2)=2xy+x^2y'.$$
Now you'll get
$$(2xy(1+x^4y^2)-x^2)y'=2xy-(1+y^2)(1+x^4y^2).$$
So, simplify the following :
$$y'=\frac{2xy-(1+y^2)(1+x^4y^2)}{2xy(1+x^4y^2)-x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$ Any ideas fot evaluating:
$$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$$
thanks.
| HINT:
Multiply by
$$\frac{\sqrt{2x^2+1} + x\sqrt{2}}{\sqrt{2x^2+1} + x\sqrt{2}}$$
to get
$$x(\sqrt{2x^2+1} - x\sqrt{2})\cdot \frac{\sqrt{2x^2+1} + x\sqrt{2}}{\sqrt{2x^2+1} + x\sqrt{2}}= x\cdot \frac{x}{\sqrt{2x^2+1} + x\sqrt{2}}$$
At this point, you should see that the numerator has degree 1 and (in some sense) the denominator has degree 1 too (square root of a square), as David suggests on the comments, dividing by $x$ should gives you a light (L'Hopital should work too).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving set of 2 equations with 3 variables I'm working through an example and my answer is not coming out right.
Two equations are given and then the solution is shown.
Equations:
$$\begin{aligned}20q_{1}+15q_{2}+7.5q_{3}&=10\\
q_{1}+q_{2}+q_{3}&=1\end{aligned}$$
Solution Given:
$(X, (1/3)(1-5X), (1/3)(2+2X))$ for arbitrary $X$.
I set up a matrix and solved and my solution is
$((3/2)X+1,(-5/2)X-2,X)$
What's going on here?
| The matrix representing the system is
$$(A|B) = \left(\begin{array}{ccc|c}
4 & 3 & 1.5 & 2\\
1 & 1 & 1 & 1
\end{array}\right) \longrightarrow \left(\begin{array}{ccc|c}
4 & 3 & 1.5 & 2\\
-1.5 & -1.5 & -1.5 & -1.5
\end{array}\right) \longrightarrow \left(\begin{array}{ccc|c}
4 & 3 & 1.5 & 2\\
2.5 & 1.5 & 0 & 0.5
\end{array}\right)$$
where we see that
$$2.5q_1 + 1.5q_2 = 0.5 \implies q_2 = \frac13 - \frac53q_1.$$
Substituting into the first equation we get
$$4q_1 + 1 - 5q_1 + 1.5q_3 = 2 \implies q_3 = \frac23 + \frac23q_1.$$
Now let $q_1 = X$ and the set of solutions is
$$\left\{\left(X, \frac13 - \frac53X, \frac23 + \frac23X\right)\colon X \in \mathbb R\right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I prove that there is no other :$k=9,12,18$ for which this fails :$\sigma^k(114) \equiv 0\mod 6 $? let $\sigma(n)$ be the sum of divisors for a positive integer for example :
$$\sigma(6)=1+2+3+6=12$$ .
I have performed some calculations in wolfram alpha about the sum divisors of this number:
$q=114$ such that I got this :$\sigma(114)=240\equiv 0\mod 6 $ and $\sigma(\sigma(114))=744\equiv 0\mod 6 $ and $\sigma(\sigma(\sigma(114)))=1920\equiv 0\mod 6 ,\cdots$ .
My question here is :How do I prove that there is no other :$k=9,12,18$ for which this fails :$$\sigma^k(114) \equiv 0\mod 6 $$ then what is the place of this number in number thoery ?
Note (01):* I think that is the only integer fails only for $k=9,12,18$
Note (02) :$\sigma^k(114)=\sigma(\sigma(\sigma(\sigma(114\cdots))))))),k-th $
Thank you for any help .
| Hint for fast calculation of sigma in a chain...
Let a number be written as
$$
\prod p^{n_p},
$$
where $p$ are prime numbers. Then we have
$$
\sigma\Big(\prod p^{n_p}\Big) = \prod \frac{p^{n_p+1}-1}{p-1}.
$$
Example:
$$
\sigma(114) = \sigma(2 \times 3 \times 19)
= \frac{2^2-1}{2-1} \frac{3^2-1}{3-1} \frac{19^2-1}{19-1}
= 3 \times 4 \times 20
= 2^4 \times 3 \times 5
$$
$$
\sigma^2(114) = \sigma(2^4 \times 3 \times 5)
= 31 \times 4 \times 6
= 2^3 \times 3 \times 31
$$
$$
\sigma^3(114) = \sigma(2^3 \times 3 \times 31)
= 15 \times 4 \times 32
= 2^7 \times 3 \times 5
$$
$$
\sigma^4(114) = 2^3 \times 3^2 \times 5 \times 17\\
\sigma^5(114) = 2^2 \times 3^4 \times 5 \times 13\\
\sigma^6(114) = 2^2 \times 3 \times 7^2 \times 11\\
\sigma^7(114) = 2^2 \times 3 \times 7^2 \times 19^2\\
\sigma^8(114) = 2^2 \times 3^2 \times 7 \times 19 \times 127\\
\sigma^9(114) = 2^{13} \times 5 \times 7 \times 13\\
$$
and $\sigma^9(114)$ is not dividable by $3$...
So-far, I see no pattern...
You could write a computer program to run the sequence and check it...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to prove that the Fibonacci sequence is periodic mod 5 without using induction? The sequence $(F_{n})$ of Fibonacci numbers is defined by the recurrence relation
$$F_{n}=F_{n-1}+F_{n-2}$$
for all $n \geq 2$
with $F_{0} := 0$
and
$F_{1} :=1$.
Without mathematical induction,
how can I show that
$$F_{n}\equiv F_{n+20}\pmod 5$$
for all $n \geq 2$?
| $$\begin{bmatrix}
F_{n+1} & F_{n} \\
F_{n} & F_{n - 1}
\end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$$
So
$$\begin{align}
\begin{bmatrix}
F_{n+21} & F_{n+20} \\
F_{n+20} & F_{n + 19}
\end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{20 + n} &\pmod 5 \\
%
&= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{20}
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n} &\pmod 5 \\
%
&= \begin{bmatrix} 10946 & 6765 \\ 6765 & 4181 \end{bmatrix}
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n} &\pmod 5 \\
%
&= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n} &\pmod 5 \\
%
&= \begin{bmatrix}
F_{n+1} & F_{n} \\
F_{n} & F_{n - 1}
\end{bmatrix} &\pmod 5 \\
\end{align}$$
Other cycles can be found similarly, for example
$$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{24} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \pmod 6$$
Giving a cycle of length $24$ modulo $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 13,
"answer_id": 9
} |
Probability that team $A$ has more points than team $B$
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
I got that since, Team $B$ already has one loss, it doesnt matter how many games team $B$ wins.
We must find the probability that team $A$ wins the rest $5$ games.
Since it says: "outcome of games is independent, I am confused."
My first approach was:
$$P(\text{A Wins 5}) = \frac{1}{32} \implies m + n = 33$$
This was wrong.
Second approach.
Suppose $A$ has a match with Team $C$.
$$P(\text{Team A wins, Team C loses}) = \frac{1}{2}\frac{1}{2} = \frac{1}{4}$$
But then overall: $\frac{1}{4^5} > 1000$ too big of an answer ($m + n < 1000$ requirement).
HINTS ONLY PLEASE!!
EDIT:
I did some casework and the work is very messy and I don't think I got the right answer anyway. I have to find:
$$\binom{5}{1} = 5, \binom{5}{2} = 10, \binom{5}{3} = 10, \binom{5}{4} = 5, \binom{5}{5} = 1.$$
Let $A = x$
$$P(B=1, x \ge 1) + P(B = 2, x\ge 2) + P(B=3, x\ge 3) + ... + P(B=5, x = 5)$$
$$P(B=1, x \ge 1) = \binom{5}{1}(0.5)^{5} \cdot \bigg(\binom{5}{1} (0.5)^5 + \binom{5}{2} (0.5)^5 + ... + (0.5)^5\binom{5}{1} \bigg) = \frac{5}{1024} \cdot \bigg(31\bigg) = \frac{155}{1024} $$
$$P_2 = \frac{10}{1024} \bigg(\binom{5}{2} + ... + \binom{5}{5}\bigg) = \frac{260}{1024}$$
$$P_3 = \frac{10}{1024} \bigg(\binom{5}{3} + ... + \binom{5}{5} \bigg) = \frac{160}{1024}$$
$$P_4 = \frac{5}{1024} \bigg( \binom{5}{4} + \binom{5}{5}\bigg) = \frac{60}{1024}$$
$$P_5 = \frac{1}{1024} \bigg( 1\bigg) = \frac{1}{1024}$$
$$P(\text{Total}) = \frac{636}{1024} = \frac{318}{512} = \text{wrong}$$
What is wrong with this method?
| Let us compute some figures so that you can confirm answer !
Firstly, notice that when p = q = 1/2, the binomial distribution formula simplifies to P(X) = $\dfrac{n\choose X}{2^n}$
To simply computations, we can leave the division by $2^n$ till the end.
n(A wins) = $n[A = X]\cdot n[B\le X] =1\cdot1 + 5\cdot6 + 10\cdot16 + 10\cdot26 + 5\cdot31 + 1\cdot32$= 638
and P(A wins) = 638/1024 = 319/512
Finally, m+n = 831
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Differentiate the Function: $y=\frac{ae^x+b}{ce^x+d}$ $y=\frac{ae^x+b}{ce^x+d}$
$\frac{(ce^x+d)\cdot [ae^x+b]'-[(ae^x+b)\cdot[ce^x+d]'}{(ce^x+d)^2}$
numerator only shown (') indicates find the derivative
$(ce^x+d)\cdot(a[e^x]'+(e^x)[a]')+1)-[(ae^x+b)\cdot (c[e^x]'+e^x[c]')+1]$
$(ce^x+d)\cdot(ae^x+(e^x))+1)-[(ae^x+b)\cdot (ce^x+e^x)+1]$
$\frac{(ace^x)^2+de^x+dae^x+de^x+1-[(ace^x)^2(ae^x)^2(bce^x)(be^x)+1]}{(ce^x+d)^2}$
Am I doing this problem correctly? I got to this point and was confused as to how to simplify. Can someone help me?
| Here are the steps
$$\frac{d}{dx}\left[\frac{ae^x+b}{ce^x+d}\right]$$
$$=\frac{\left(ce^x+d\right)\frac{d}{dx}\left[ae^x+b\right]-\left(ae^x+b\right)\frac{d}{dx}\left[ce^x+d\right]}{\left(ce^x+d\right)^2}$$
$$=\frac{\left(ce^x+d\right)\left(ae^x\right)-\left(ae^x+b\right)\left(ce^x\right)}{\left(ce^x+d\right)^2}$$
$$=\frac{e^x\left(ace^x+ad-ace^x-bc\right)}{\left(ce^x+d\right)^2}$$
$$=\frac{e^x\left(ad-bc\right)}{\left(ce^x+d\right)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find the roots of this 6th degree polynomial Hey guys I'm reviewing for a test and I'm getting stuck on one part, I can't remember what to do next.
$x^6+16x^3+64$
$(x^3)^2+16x^3+64$
let $x^3=w$
$w^2+16w+64$
$(w+8)^2$ now substitute again
$(x^3+8)^2$
Now what do I do?
| We can actually factor anything in the form $x^3+a^3$ into $(x+a)(x^2-ax+a^2)$.
$x^3+8$ can be factored into the following:
$$x^3+8 = x^3+2^3 =(x+2)(x^2-2x+4)$$
We actually have $(x^3+8)^2$, however we are setting this value to equal $0$ to find the roots so we can just take the square root of both sides and obtain:
$$x^3+8 = 0$$
$$(x+2)(x^2-2x+4) = 0$$
We know now that $x+2 = 0$ so $-2$ is a solution. We also know that $x^2-2x+4=0$ gives us a solution. So if we use the quadratic formula, we get:
\begin{align*}
\frac{-(-2)\pm\sqrt{(-2)^2-4\cdot1\cdot4}}{2\cdot1}&= \frac{2\pm\sqrt{-12}}{2} \\
&= \frac{2\pm2\sqrt{3}i}{2} \\
&= {1\pm\sqrt{3}i}
\end{align*}
So our solutions are:
$$\boxed{-2, {1+\sqrt{3}i}, {1-\sqrt{3}i}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$
let $0<a\le b\le c\le d$, and such $abcd=1$,show that
$$a+b^2+c^3+d^4\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}+\dfrac{1}{d^4}$$
it seems harder than This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
| Necessarily $\color{green}{d\ge 1}$ and at least one between $a,b,c$ is less or equal than $1$. We have the equivalent inequality
$$a+b^2+c^3+d^4 - \frac{1}{a}-\frac{1}{b^2}-\frac{1}{c^3}-\frac{1}{d^4}\ge 0\qquad (*)$$
Arranging otherwise to compare the larger numbers involved with smaller ones
$$(d^4-\frac 1a)+(a-\frac {1}{d^4}) +c^3+b^2-\frac{1}{c^3}-\frac{1}{b^2}\ge 0\qquad (**)$$
$$d^4-\frac 1a=d^4-bcd=d(d^3-bc)$$ Since $d^2\ge bc$ and $d\ge 1$ one has $d^3-bc$ is positive; it follows $$d^4-\frac 1a\ge 0\iff a-\frac{1}{d^4}\ge 0$$ therefore, if $b$ and $c$ are greater or equal than $1$ then the proof of $(**)$ is finished. Because of this we consider the restriction $\color{green}{0\lt a\le b\le c\lt 1\le d}$.
$$*********************$$
Now, by the known $A-M\ge H-M$ and the statement’s inequality, one has the system $$\begin{cases}LHS\ge \frac {16}{RHS}\ge 0\\LHS\ge RHS\ge 0\end{cases} \Rightarrow (LHS)^2\ge 16\iff LHS\ge 4$$
Hence we have to show $$a+b^2+c^3+d^4\ge 4\qquad (***)$$
One has $$\frac{a^4+b^4+c^4+d^4}{4}\ge \sqrt[4]{a^4b^4c^4d^4}=1$$
But (since $\color{green}{0\lt a\le b\le c\lt 1}$) $$a\ge a^4\\b^2\ge b^4\\c^3\ge c^4\\d^4=d^4$$
Consequently $$a+b^2+c^3+d^4\ge a^4+b^4+c^4+d^4\ge 4$$ as we wanted to prove.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital's: $$\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{2}{x^3}- \frac{2\cos x}{\sin ^3x} = \frac{0}{0}\,\ldots$$ The terms are getting endless here. Any help? Thanks.
| $\lim_{x \to 0}\frac{\sin(x)-x\cos(x)}{x\sin(x)}=\lim_{x \to 0}\frac{\cos(x)-\cos(x)+x\sin(x)}{\sin(x)+x\cos(x)}=\lim_{x \to 0}\frac{\sin(x)}{\frac{\sin(x)}{x}+\cos(x)}=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Find the integral: $ \int{\frac{\sin x+2\cos x}{\sin^2x+3\sin x+2}\,dx}$ I would like some help with the following integral
$$ \int{\frac{\sin x+2\cos x}{\sin^2x+3\sin x+2}\,dx}$$
I tried do split the fraction to $\displaystyle\int\frac{\sin x}{\sin^2x+3\sin x+2}\,dx + \displaystyle\int\frac{2\cos x}{\sin^2x+3\sin x+2}\,dx$
and for $\displaystyle\int\frac{2\cos x}{\sin^2x+3\sin x+2}\,dx$, letting $[ \sin x=t,\frac{dt}{dx}=\cos x ]$ and we get $\displaystyle \int\frac{2dt}{t^2+3t+2}$
but what can I do with $\displaystyle \int\frac{\sin x}{\sin^2x+3\sin x+2}\,dx$?
thx
| hint:$\dfrac{\sin x}{\sin^2x+3\sin x + 2} =\dfrac{2}{\sin x+2}-\dfrac{1}{\sin x+1}$, and for each of them use $t = \tan\left(\frac{x}{2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\circ}))$$ However I don't seem to get anywhere with this.
Unfortunately I don't know how to solve this question. I would be really grateful for any help or suggestions. Many thanks in advance!
| Here's a solution:
$$\cos(\theta + 120^\circ) = \cos\theta \cos 120^\circ-\sin\theta \sin 120^\circ = \frac{\cos \theta}{2} - \frac{\sqrt{3}}{2}\sin \theta$$
$$ \cos(\theta + 120^\circ) = \cos\theta \cos 120^\circ+\sin\theta \sin 120^\circ = \frac{\cos \theta}{2} + \frac{\sqrt{3}}{2}\sin \theta$$
$$ \left(\frac{\cos \theta}{2} - \frac{\sqrt{3}}{2}\sin \theta\right)^2 = \frac{\cos^2\theta}{4} - \frac{\sqrt3}{2} \sin \theta \cos \theta+ \frac{3}{4} \sin^2\theta$$
$$ \left(\frac{\cos \theta}{2} + \frac{\sqrt{3}}{2}\sin \theta\right)^2 = \frac{\cos^2\theta}{4} + \frac{\sqrt3}{2}\sin \theta \cos \theta+ \frac{3}{4} \sin^2\theta$$
$$\cos^2(\theta + 120^\circ) + \cos^2(\theta - 120^\circ) = \frac{1}{2}\cos^2\theta + \frac{3}{2} \sin^2\theta$$
$$\cos^2\theta +\cos^2(\theta + 120^\circ) + \cos^2(\theta - 120^\circ) = \frac{3}{2}(\cos^2\theta + \sin^2\theta) = \frac{3}{2} $$
Note: $\cos^2\theta + \sin^2\theta = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$
| Here's the thing about this question: It is a trick.For most numbers of the form $a+b\sqrt 5$ the cube root is a mess. But some such numbers have nice cube roots and when they do we can find them.
The one thing you have to know is that if $x$ is any number of the form $a+b\sqrt5$ then $x^2, x^3$, etc. also have that form. For example: $$\begin{align}
(a+b\sqrt 5)^2 & = a^2 + 2ab\sqrt 5 + 5b^2 \\
& = (a^2 + 5b^2) + 2ab\sqrt 5
\end{align}$$
Or similarly $$(a+b\sqrt 5)^3 = (a^3 + 15ab^2) + (3a^2b+5b^3)\sqrt 5.\tag{$\star$}$$
If $17\sqrt5 + 38$ has a simple cube root, it will be something of the form $a+b\sqrt 5$. From $(\star)$ we would want $$\begin{align}
a^3+ 15ab^2 & = 38 \\
3a^2b+5b^3 &= 17 \\
\end{align}$$
Looking at the first equation we see that maybe $15ab^2 = 30$ is almost 38, and then $a=2, b=1$ would work. This also works in the second equation. (Or we could look at the second equation first and see right away that the only way to get $3x+5y=17$ is to have $x=4, y=1$, so $a=2, b=1$.) This is called solving equations "by inspection". It is not quite a lucky guess; it is a thoughtful lucky guess based on the hope that the solution will be simple. But it also wouldn't be hard to try every possibility for $a$ and $b$: they can't be too big because then $a^3+15ab^2$ would be much bigger than 38.
Now we have $a=2, b=1$ we know that $\sqrt[3]{17\sqrt 5+ 38} = 2+\sqrt 5$. We can use the same method to find $\sqrt[3]{17\sqrt 5- 38}$ exactly, and from there the solution is easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Maximize the area of a triangle inscribed in a semicircle. http://i.imgur.com/Q5gjaSG.png
Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.
My attempt so far, I'm new at these problems and have only done a few so far. Thanks for any hints. I feel I'm not setting up the correct equation.
$x^{2}$+$y^2$=$r^2$
$y=\sqrt{r^2 - x^2}=\sqrt{1-x^2}\;(\text{since}\:r=1).$
The area is A=$\dfrac{1}{2}$($x+1$)($y$)
A'=$\frac{2x^2-x-1}{2\sqrt{1-x^2}}$
A'=$0$ When $x$ = - $\dfrac{1}{2}$
$y$=$\dfrac{\sqrt{3}}{2}$
Plugging everything back into the original area equation I get $\dfrac{ \sqrt{3}}{8}$
| Almost: Since the area is
$$
A = \frac{(1+x)\sqrt{1-x^2}}{2}
$$
we want to find the zero of the derivative
$$
\frac{dA}{dx} = \frac{1-x-2x^2}{2\sqrt{1-x^2}}
$$
Since $-1 < x < 1$ (do you see why?), we can just focus on the numerator, and then
$$
2x^2+x-1 = 0
$$
or $x = -1$ or $x = 1/2$. The first solution is discarded, and we use the second solution: $x = 1/2$, so $y = \sqrt{3}/2$, and $A = (1+x)y/2 = 3\sqrt{3}/8 \doteq 0.64952$.
You just lost a minus sign, is all, I think. But it is useful to evaluate your answer to see if it makes sense. $\sqrt{3}/8 \doteq 0.21651$ is too small for something that must be larger than half the unit square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question:
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\left(\frac{6^\circ+66^\circ}{2}\right)\sin\left(\frac{6^\circ-66^\circ}{2}\right) $$
$$ 2\cos(60^\circ)\sin(18^\circ) + 2\cos(36^\circ)\sin(-30^\circ) $$
$$ 2\frac{1}{2}\sin(18^\circ) - 2\cos(36^\circ)\cdot\frac{1}{2} $$
$$ \sin(18^\circ) - \cos(36^\circ) $$
At this point I had to use a calculator. Does anyone know a way to solve it without a calculator.Thanks in advance.
| There is not fast but strightforward way. $18^\circ = \pi/10$; let $s=\sin{(\pi/10)}$ so, we should evaluate
$$
a=\sin\frac\pi{10}-\cos\frac{\pi}{5}=2s^2 + s - 1.
$$
We may express $\sin5x$ in terms of $\sin x$:
$$
\sin5x=16\sin^5 x - 20\sin^3 x + 5\sin x
$$
(i.e. from $\sin5x=\sin(4x + x), \sin4x=2\sin2x\cos2x$, or by de Moivre). For $x=\frac{\pi}{10}$ we have $\sin5x=1$:
$$
16s^5 - 20s^3 + 5s -1 =0,
$$
or
$$
(s-1)(-1+2s+4s^2)^2 = 0;
$$
So,
$$
4s^2 + 2s - 1= 0=2a+1 \Longrightarrow a = -\frac12
$$
is answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Contradiction between integration by partial fractions and substitution Integration by substitution:
$$\int \frac {dx}{x^2-1}$$
Let $x=\sec\theta$ and $dx=\sec\theta\tan\theta \,d\theta$
$$\int \frac {dx}{x^2-1} = \int \frac{\sec\theta\tan\theta \,d\theta}{\sec^2\theta-1} = \int \frac {\sec\theta\tan\theta\, d\theta}{\tan^2\theta} = \int \frac {\sec\theta \,d\theta}{\tan\theta} $$
$$\int \frac {\sec\theta \,d\theta}{\tan\theta} = \int \frac {\cos \theta \,d\theta}{\cos\theta \sin \theta} = \int \csc\theta\, d\theta$$
$$=\ln|\csc\theta-\cot\theta|+C$$
$$=\ln| \frac{x}{\sqrt{x^2-1}}-\frac{1}{\sqrt{x^2-1}}|+C=\ln| \frac{x-1}{\sqrt{x^2-1}}|+C$$
Which is $Undefined$ for $|x|<1$
Integration by partial fractions:
$$\int \frac {dx}{x^2-1}$$
$$\int \frac {dx}{x^2-1}= \frac 12\int\frac{dx}{x-1}- \frac12\int \frac{dx}{x+1} = \frac12 \ln | x-1| - \frac12 \ln|x+1| +C$$
$$ = \frac12 \ln | \frac{x-1}{x+1}|+C$$
Which is $Defined$ for $|x|<1$ and this is right because the integrand is defined for $|x|<1$
What is the problem in the substitution method ?
| On simplifying
\begin{align}
& \ln \frac{x-1}{\sqrt{x^2-1}}=\ln\frac{\sqrt{x-1}^2}{\sqrt{(x-1)(x+1)}} \\[6pt]
= {} & \ln\frac{\sqrt{x-1}}{\sqrt{x+1}} \\[6pt]
= {} & \frac 1 2 \ln \frac{x-1}{x+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Non-linear system of equations Solve following system of equations over real numbers:
$$
x-y+z-u=2\\
x^2-y^2+z^2-u^2=6\\
x^3-y^3+z^3-u^3=20\\
x^4-y^4+z^4-u^4=66
$$
This does not seem as hard problem. I have tried what is obvious here, to write $x^2-y^2$ as $(x-y)(x+y)$, $x^3-y^3$ as $(x-y)(x^2+xy+y^2)$ etc. Problem is because if I substitute $x-y=u-z+2$ in second equation, I cannot eliminate $x+y$ or if I substitute $x^2-y^2=u^2-z^2+6$ in fourth equation, I cannot eliminate $x^2+y^2$ etc. When I realized that this is not the best way, I made a second attempt.
My second attempt is to find come constraints, because all of these numbers are real. I have proven that if $m+n=a$ for some $m,n\in\mathbb{R}^+$, then $\frac{a^2}2\le m^2+n^2\le a^2$. I have applied this formula to second and fourth equation (because in these equations variables have even exponents and that means they are positive), but this didn't give me anything.
What should be the easiest way to solve it?
| \begin{align}
x-y+z-u&=2 \quad (1)\\
x^2-y^2+z^2-u^2&=6 \quad (2)\\
x^3-y^3+z^3-u^3&=20 \quad (3)\\
x^4-y^4+z^4-u^4&=66 \quad (4)
\end{align}
The easiest way to solve it
is to start of testing the simplest solutions first.
The first eqn suggests to check if $x-y=1$, $z-u=1$ fits.
In this case (2) becomes
\begin{align}
x+y+z+u&=6, \quad(3)
\\
(1)+(3)\Rightarrow&
\\
x+y&=4,
\\
z+u&=2
\end{align}
and we have a solution $u=0,z=1,x=3,y=2$,
which is indeed a valid solution of the system (1)--(4).
And due to the symmetry, there are three more
valid combinations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trig Equations Using Identities How would you solve:
$2\csc^2x=3\cot^2x-1$
I said:
*
*Turn the cosecant to $1+\cot^2~x$.
*Distribute to get $3=\cot^2~x$.
*Turn it into tan. To get $\tan x=\frac{1}{\pm \sqrt3}$.
Is this correct?
| Notice, we can also solve this as follows $$2\csc^2 x=3\cot^2 x-1$$ $$\implies \frac{2}{\sin^2 x}=\frac{3\cos^2 x}{\sin^2x}-1$$ $$\implies 2=3\cos^2 x-\sin^2x$$ $$\implies 3\cos^2 x-(1-\cos^2x)=2$$ $$\implies 4\cos^2 x=3$$ $$\implies \cos^2 x=\frac{3}{4}=\left(\frac{\sqrt{3}}{2}\right)^2$$ $$\implies \cos^2 x=\left(\cos\frac{\pi}{6}\right)^2$$ Now, writing the general solution as follows
$$\implies \color{blue}{x=n\pi\pm\frac{\pi}{6}}$$ Where, $\color{blue}{\text{n is any integer}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.
| Let $a=\frac{x^2}{yz},\, b=\frac{y^2}{zx},\,c=\frac{z^2}{xy}.$ Thus$,$ we need to prove$:$
$$\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8zx}} +\frac{z}{\sqrt{z^2+8xy}} \geqq 1$$
By AM-GM$:$ \begin{align*} \text{LHS} &=\sum\limits_{cyc} \frac{x}{\sqrt{x^2+8yz}} =\sum\limits_{cyc} \frac{x(x+y+z)}{\sqrt{(x^2+8yz)(x+y+z)^2}}\\&\geqq 2\sum\limits_{cyc} \frac{x(x+y+z)}{(x^2+8yz)+(x+y+z)^2} \geqq 1 \end{align*}
Where the last inequality equivalent to $$\frac{1}{2} \sum\limits_{cyc} \left( 8\,{x}^{3}y+31\,{x}^{2}{y}^{2}+8\,x{y}^{3}+202\,x{y}^{2}z+262
\,xy{z}^{2}+202\,x{z}^{3}+79\,{z}^{4} \right) \left( x-y \right) ^{2}
\geqq 0$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 5
} |
Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find the answer of the following expression, without solving the original equation?
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a}
$$
| We have
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a} = \frac{a^2c+b^2a+c^2b}{abc}
$$
Since from Vieta's relations we know
$$
a+b+c =0,\quad ab+bc+ca =-3,\quad abc =-1,
$$
our goal is to calculate
$$
s = a^2c+b^2a+c^2b.
$$
Let's introduce
$$
p = ac^2+ba^2+cb^2.
$$
Than we have
$$
0 = (ab+bc+ac)(a+b+c) = p+s+3abc
$$
and $p+s = 3$.
Now let's multiply
$$
s\cdot p = a^3b^3 + a^3c^3 +b^3c^3 + 3(abc)^2+ abc(a^3+b^3+c^3)
$$
Since $a,b,c$ are the roots of polynomial the last equation can be rewritten as
$$
sp = (3a-1)(3b-1)+(3a-1)(3c-1)+(3b-1)(3c-1) + 3(abc)^2
+ abc(3(a+b+c)-3)=
$$
$$
=9(ab+ac+bc)-6(a+b+c)+3 + 3(abc)^2 + abc(3(a+b+c)-3) =-27+3+3+3=-18.
$$
So, $s+p=3$ and $sp=-18$. From here one can deduce that $s= 6$. (see @mathlove answer)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Find the sum: $\sum_{i=1}^{n}\dfrac{1}{4^i\cdot\cos^2\dfrac{a}{2^i}}$ Find the sum of the following :
$S=\dfrac{1}{4\cos^2\dfrac{a}{2}}+\dfrac{1}{4^2\cos^2\dfrac{a}{2^2}}+...+\dfrac{1}{4^n\cos^2\dfrac{a}{2^n}}$
| A very different approach to calculate the sum. Note that
$$
\frac{1}{4^k\cos^2\frac{x}{2^k}}=\left(-\ln\cos\frac{x}{2^k}\right)''.
$$
Let's calculate (using in the $3^\text{d}$ equality the general form of Morrie's law for $\alpha=\frac{x}{2^n}$)
$$
F(x)=\sum_{k=1}^n-\ln\cos\frac{x}{2^k}=-\ln\prod_{k=1}^n\cos\frac{x}{2^k}=-\ln\frac{\sin x}{2^n\sin\frac{x}{2^n}}=\ln\sin\frac{x}{2^n}-\ln\sin x+n\ln 2.
$$
Finally, our sum is
$$
F''(a)=\frac{1}{\sin^2a}-\frac{1}{4^n\sin^2\frac{a}{2^n}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
$U_n=\int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ . $U_n= \int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ where
Find $\lim_{n\to \infty} U_n$ without finding the integration
I don't know how to start
| $$\int_{n^2+n+1}^{n^2+1}\frac{\arctan x}{\sqrt{x}}\,dx = \frac{\pi}{2}\int_{n^2+n+1}^{n^2+1}\frac{dx}{\sqrt{x}}+\int_{n^2+1}^{n^2+n+1}\frac{\arctan\left(\frac{1}{x}\right)dx}{\sqrt{x}}$$
Now:
$$\frac{n}{\sqrt{n^2+n+1}}\leq\int_{n^2+1}^{n^2+n+1}\frac{dx}{\sqrt{x}}\leq\frac{n}{\sqrt{n^2+1}}$$
and:
$$ 0\leq \int_{n^2+1}^{n^2+n+1}\frac{\arctan\left(\frac{1}{x}\right)dx}{\sqrt{x}}\leq \int_{n^2+1}^{n^2+n+1}\frac{dx}{x\sqrt{x}} \leq \frac{n}{(n^2+1)^{3/2}}$$
so the limit is $\displaystyle\color{red}{-\frac{\pi}{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Evaluating the limit: $\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1)$ I need to solve this question: $$\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1)$$
I tried the graphical method and observed that the graph was approaching $0$ as $x$ approached $-1$ but I need to know if there's a way to calculate this.
| $$\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1) = \lim_{x\rightarrow 0^+}\sqrt[3]{x}\ln(x) = \lim_{x\rightarrow 0^+} e^{ln (x)\frac{1}{3}}\ln(x) = \lim_{x\rightarrow -\infty} e^{\frac{1}{3}x} x$$
$$= -\lim_{x\rightarrow +\infty} \frac{x}{e^{\frac{1}{3}x}} = -\lim_{x\rightarrow +\infty} \frac{x}{1 + \frac{x}{3} + \frac{x^2}{18} +O(x^3)} = -\lim_{x\rightarrow +\infty} \frac{1}{\frac{1}{x} + \frac{1}{3} + \frac{x}{18} +O(x^3)} = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solution of Differential equation Question:
Find solution of differential equation
$$ 3e^{4x} \frac{dy}{dx} = -16\frac{x}{y^2} $$
which satisfies the initial condition y(0)=1
Solution:
I know that I have to bring it in the general form of :
$$ \frac{dy}{dx} + P(x) y = Q(x)$$
However in the equation there is no P(x)y component so how do i do it ?
| An interesting aspect:
The differential equation has the form
\begin{align}
3 \, e^{a x} \, y' = - 16 \cdot \frac{x}{y^{2}}
\end{align}
for which
\begin{align}
3 \, y^{2} \, y' = - 16 \, x \, e^{-a x}
\end{align}
where $a$ is a constant, here $a = 4$. This last equation can be seen to have the form
\begin{align}
\frac{d (y^{3})}{dx} &= \frac{d}{da} \left(16 \, e^{-a x} \right) = \frac{d}{dx} \left[ \frac{d}{da} \left( - \frac{16}{a} \, e^{-ax} \right) \right].
\end{align}
Integration with respect to $x$ provides:
\begin{align}
y^{3} = \frac{d}{da} \left( - \frac{16}{a} \, e^{-ax} \right) + c_{1}
\end{align}
or
\begin{align}
y(x) &= \left.\left[ \frac{d}{da} \left( - \frac{16}{a} \, e^{-ax} \right) + c_{1} \right]^{1/3}\right|_{a=4} \\
&= \left.\left[ \frac{16}{a} \, \left(1 + \frac{1}{a}\right) \, e^{-ax} + c_{1} \right]^{1/3} \right|_{a=4} \\
&= \left[ (4 x + 1) \, e^{-4x} + c_{1} \right]^{1/3}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Cauchy-Schwarz inequality problem The problems:
*Prove that $$\frac{\sin^3 a}{\sin b} + \frac{\cos^3 a}{\cos b} \geqslant \sec (a-b),$$ for all $a,b \in \bigl(0,\frac{\pi}{2}\bigr)$.
*Prove that $$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{2\sqrt[3]{abc}} \geqslant \frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)},$$ for all $a,b,c > 0$.
Can someone give me a hint for the two problems. They are all based on the Cauchy-Schwarz inequality.
Just a hint.
| One form of the Cauchy-Schwarz Inequality is
$$\sum_i^n \frac{x_i^2}{y_i}\ge \frac{\left(\sum_i^n x_i\right)^2}{\sum_i^n y_i}\tag 1$$
Let $n=2$ and $x_1=\sin^2a$, $x_2=\cos^2a$, $y_1=\sin a\sin b$, and $y_2=\cos a\cos b$
The result follows immediately after expanding $\sec (a-b)=\frac{1}{\cos a\cos b+\sin a\sin b}$
To see this explicitly, we use $(1)$ and write
$$\begin{align}
\sum_i^n \frac{x_i^2}{y_i}&=\frac{\sin^4 a}{\sin a \sin b}+\frac{\cos^4 a}{\cos a\cos b}\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{\sin^3a}{\sin b}+\frac{\cos^3 a}{\cos b}}\\\\
&\ge \frac{\left(\sum_i^n x_i\right)^2}{\sum_i^n y_i}\\\\
&=\frac{\sin^2 a +\cos^2 a}{\sin a\sin b+\cos a\cos b}\\\\
&=\frac{1}{\cos(a-b)}\\\\
&=\bbox[5px,border:2px solid #C0A000]{\sec(a-b)}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Power serie of $f'/f$ It seems that I'm [censored] blind in searching the power series expansion of $$f(x):=\frac{2x-2}{x^2-2x+4}$$ in $x=0$.
I've tried a lot, e.g., partiell fraction decomposition, or regarding $f(x)=\left(\log((x+1)^2+3)\right)'$ -- without success.
I' sure that I'm overseeing a tiny little missing link; dear colleagues, please give me a hint.
| Given
$$
f(x) = \frac{ 2 x - 2 }{ x^2 - 2 x + 4 }. \tag 1
$$
Assuming you want to expand $f(x)$.
Let
$$
\phi_\pm = 1 \pm \mathtt{i} \sqrt{3}. \tag 2
$$
We can write (1) as
$$
f(x) =
\frac{ 1 }{ x - \phi_+ } +
\frac{ 1 }{ x - \phi_- }. \tag 3
$$
Whence
$$
f(x) =
- \sum_{k=0}^\infty \left(
\frac{1}{{\phi_+}^{k+1}} +
\frac{1}{{\phi_-}^{k+1}}
\right) x^k. \tag 4
$$
Note that
$$
\phi_\pm = 1 \pm \mathtt{i} \sqrt{3} =
2 {\exp}\big( \pm \mathtt{i} \pi / 3 \big). \tag 5
$$
Putting (5) in (4) yields
$$
\bbox[16px,border:2px solid #800000] {
\frac{ 2 x - 2 }{ x^2 - 2 x + 4 } =
- \sum_{k=0}^\infty {\cos}\big( [k+1] \pi / 3 \big)
\left( \frac{x}{2} \right)^k.
} \tag {I}
$$
Simple expansion yields
$$
\bbox[16px,border:2px solid #800000] { \frac{ 2 x - 2 }{ x^2 - 2 x + 4 } =
- \frac{1}{2}
+ \frac{x}{4} + \frac{x^2}{4} + \frac{x^3}{16}
- \frac{x^4}{32} - \frac{x^5}{32} - \frac{x^6}{128}
+ \cdots
} \tag{II}
$$
Note that
$$
f(x) = \frac{g'(x)}{g(x)}, \tag 6
$$
where
$$
g(x) = x^2 - 2x + 4. \tag 7
$$
If you want to expand
$$
\frac{f'(x)}{f(x)}, \tag 8
$$
then use the simple relation
$$
\frac{f'(x)}{f(x)} = \frac{g''(x)}{g'(x)} - \frac{g'(x)}{g(x)}. \tag 9
$$
Note that
$$
\frac{g''(x)}{g'(x)} = \frac{2x}{2x-2} = - \sum_{k=0}^\infty x^k \tag {10}.
$$
Combination with (II) yields
$$
\bbox[16px,border:2px solid #800000] {
\frac{f'(x)}{f(x)} =
+ \sum_{k=0}^\infty \left(
2^{-k} {\cos}\big( [k+1] \pi / 3 \big) - 1
\right)
x^k.
} \tag {III}
$$
Simple expansion yields
$$
\bbox[16px,border:2px solid #800000] { \frac{f'(x)}{f(x)} =
- \frac{1}{2}
- \frac{5x}{4} - \frac{5x^2}{4} - \frac{17x^3}{16}
- \frac{31x^4}{32} - \frac{31x^5}{32} - \frac{127x^6}{128}
\cdots
} \tag{IV}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find a Basis $B$ of $R^2$ so that $B$ matrix of $T$ is diagonal $T([1,1]^t) = [3,7]^t$
$T([1,-1]^t) = [1,1]^t$
Here's what I get:
$T=
\left(\begin{array}{cc}3 & 1 \\7 & 1\end{array}\right)
$
The eigenvectors of $T$ is $E = \left(\begin{array}{cc} .4798 & -.2527 \\.8774 & .9675\end{array}\right)$.
$E^{-1}TE$ gives us a diagonal matrix. Am I doing this right? So my answer is that the columns of $E$ are the bases.
The answer I was given was $v_1 = (\frac{\sqrt(33)-3}{6},1)^t$ and $v_2 = (\frac{-(\sqrt(33) +3)}{6},1)^t$. Is this right? If so, how do I arrive at this solution?
| It is convenient to work with the standard basis $\left( \begin{matrix} 1 \\ 0 \end{matrix} \right), \left( \begin{matrix} 0 \\ 1 \end{matrix} \right)$ to write the answers directly as vectors in the standard basis.
So we must solve, with $T= \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)$
$$ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) = \left( \begin{matrix} a+b \\ c+d \end{matrix} \right) = \left( \begin{matrix} 3 \\ 7 \end{matrix} \right) \textrm{ and } \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 \\ - 1 \end{matrix} \right) = \left( \begin{matrix} a-b \\ c-d \end{matrix} \right) = \left( \begin{matrix} 1 \\ 1 \end{matrix} \right)$$
So we have two sets equations, namely $a+b =3$ and $ a -b =1$ which yields $a=2$ and $b=1$, and $c+d =7$ and $c-d=1$ which yields $c=4$ and $d=3$.
So we find $T= \left( \begin{matrix} 2 & 1 \\ 4 & 3 \end{matrix} \right)$.
Then we can indeed find $E$ such that $E^{-1} T E = \left( \begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix} \right)$ is diagonal, which gives us
$$ E^{-1} T E \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \lambda_1\left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \textrm{ which gives } TE \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \lambda_1 E \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) $$
So if we take our basis the column vectors of $E$ we have that $E \left( \begin{matrix} 1 \\ 0 \end{matrix} \right)$ is the first column of $E$ and $E \left( \begin{matrix} 0 \\ 1 \end{matrix} \right)$ is the second column of $E$.
Then we indeed have that this basis makes $T$ a diagonal matrix as we saw with $E \left( \begin{matrix} 1 \\ 0 \end{matrix} \right)$, and $E \left( \begin{matrix} 0 \\ 1 \end{matrix} \right)$ can be shown at the same way.
This yields in the standard basis the vectors $v_1 =\left( \begin{matrix} \frac{1}{8} \left( \sqrt{17} -1 \right) \\ 1 \end{matrix} \right), v_2 =\left( \begin{matrix} \frac{-1}{8} \left( \sqrt{17} +1 \right) \\ 1 \end{matrix} \right)$. And if we write it in the basis $\left( \begin{matrix} 1 \\ 1 \end{matrix} \right), \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)$ we find the vectors $v_1 =\left( \begin{matrix} \frac{-1}{4} \left( \sqrt{17} +5 \right) \\ 1 \end{matrix} \right),v_2 =\left( \begin{matrix} \frac{1}{4} \left( \sqrt{17} -5 \right) \\ 1 \end{matrix} \right)$. So the answer given to you is not correct, whereas your approach is the right approach. I would only recommend to write $T$ in the standard basis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that a polynomial does not have real roots? For example, let's take the polynomial
$$x^8-x^7+x^2-x+15$$
Here, the power ($n=8$) is even so it can have real roots or it might not have real roots.
Something which I thought was to find the minima and show that if the minima of $p(x)$ is greater than $0$ and $a_1$ that is the coefficient of $x^8$ are both greater than $0$ then we cannot have real roots . But in this case the derivative is $8x^7-7x^6+2x-1$ and I cannot find minima for it . So what should I do in this example? Well it is already given this polynomial does not have real roots, but I have to prove it.
Also even if I get that this does not have any real roots then is this a general method for all kinds of polynomials?
Edit: I know Strum's theorem is one general way to solve such questions but this question is from an undergrad entrance paper and I guess a method under the reach of calculus or something similar will suffice better.
| To show that a polynomial has no real roots, we will try to write it as an equation where the sum of some positive numbers equals a strictly negative number. As the sum of positive numbers cannot be strictly negative, there is a contradiction, which means there's no real root.
$x^8-x^7+x^2-x+15 = 0$
Subtract 15: $x^8-x^7+x^2-x = -15$
Multiply by 2: $2x^8-2x^7+2x^2-2x = -30$
Rearrange the terms: $x^8-2x^7+x^6 + x^8-2x^6+x^4 + x^6-2x^4+x^2 + x^4 + x^2-2x = -30$
Add 1: $x^8-2x^7+x^6 + x^8-2x^6+x^4 + x^6-2x^4+x^2 + x^4 + x^2-2x+1 = -29$
$(x^4-x^3)^2 + (x^4-x^2)^2 + (x^3-x)^2 + x^4 + (x-1)^2 = -29$
All terms on the left side are always positive. The sum of positive numbers cannot equal -29 so there is no real root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 4
} |
find total integer solutions for $(x-2)(x-10)=3^y$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
How many integer solutions ($x$, $y$) are there of the equation $(x-2)(x-10)=3^y$?
(A)1 (B)2 (C)3 (D)4 (E)5
If let $y=0$, we had
$$x^2 - 12x + 20 = 1$$
$$x^2 - 12x + 19 = 0$$
no integer solution for x
let $y=1$, we had $x^2 - 12x + 17 = 0$, no integer solution too.
let $y=2$, we had $x^2 - 12x + 11 = 0$, we had $x = 1, 11$.
let $y=3$, we had $x^2 - 12x - 7 = 0$, no integer solution.
let $y=4$, we had $x^2 - 12x - 61 = 0$, no integer solution.
and going on....
is there any other efficient way to find it? "brute-forcing" it will wasting a lot of time.
| Since their product is a power of $3$, both $x-2$ and $x-10$ must be powers of $3$, perhaps with minus signs (as Andre Nicolas pointed out). Note, however, that $x-2$ and $x-10$ cannot simultaneously be divisible by $3$. The only power of $3$ which is not divisible by $3$ is $1$, hence we must have that $x-2 = \pm 1$ or $x-10 = \pm 1$.
With this in mind, there are four values of $x$ to try: $1, 3, 9,$ and $11$. Two of these give you integer solutions to the equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Where did I go wrong with the Gram-Schmidt orthogonalisation process? Problem: Let $\alpha = \left\{(1,2,0), (1,0,1), (2,3,1)\right\}$ be a basis vor $\mathbb{R}^3$. Apply the Gram-Schmidt orthogonalisation process to turn $\alpha$ into an orthonormal basis for $\mathbb{R}^3$ with respect to the standard innerproduct.
Attempt at solution: I change this set to an orthogonal set first, and normalize afterwards.
Define: \begin{align*} v_1 &= (1,2,0) \\ v_2 &= (1,0,1) - \frac{ \langle (1,0,1), (1,2,0) \rangle}{\langle (1,2,0), (1,2,0) \rangle} (1,2,0) \\ v_3 &= (2,3,1) - \frac{ \langle (2,3,1), (1,0,1) \rangle}{ \langle (1,0,1), (1,0,1) \rangle} (1,0,1) - \frac{ \langle (2,3,1), (1,2,0) \rangle}{\langle (1,2,0), (1,2,0) \rangle} (1,2,0) \end{align*}
Then this gives \begin{align*} v_2 = (1,0,1) - \frac{1}{1+4} (1,2,0) = (1,0,1) - (\frac{1}{5}, \frac{2}{5}, 0) = (\frac{4}{5}, - \frac{2}{5}, 1)
\end{align*} and \begin{align*} v_3 &= (2,3,1) - \frac{(2+1)}{(1+1)} (1,0,1) - \frac{(2+6)}{(1+4)}(1,2,0) \\ &= (2,3,1) - (\frac{3}{2}, 0, \frac{3}{2}) - (\frac{8}{5}, \frac{16}{5}, 0) \\ &= (- \frac{11}{10}, - \frac{1}{5}, - \frac{1}{2}). \end{align*}
Now I wanted to check my work. So I compute the innerproduct of $v_2$ with $v_3$, i.e. \begin{align*} \langle (\frac{4}{5}, - \frac{2}{5}, 1), (- \frac{11}{10}, - \frac{1}{5}, - \frac{1}{2}) \rangle \end{align*} but I don't get zero? When I compute $\langle v_1, v_2 \rangle$, however, I do get zero. So something must be wrong with my $v_3$. I triple checked my work and I can't find the mistake. Please help!
| The problem is that you projected two non-orthogonal vectors, $v_1$ and $v_2$, out of $v_3$. Each of those orthogonalizations kills the other. In orthogonalizing $v_3$, you need to use the result of orthogonalizing $v_2$ against $v_1$, not $v_2$ itself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a polynomial from an equality Find all polynomials for which
What I have done so far:
for $x=8$ we get $p(8)=0$
for $x=1$ we get $p(2)=0$
So there exists a polynomial $p(x) = (x-2)(x-8)q(x)$
This is where I get stuck. How do I continue?
UPDATE
After substituting and simplifying I get
$(x-4)(2ax+b)=4(x-2)(ax+b)$
For $x = 2,8$ I get
$x= 2 \to -8a+b=0$
$x= 8 \to 32a+5b=0$
which gives $a$ and $b$ equal to zero.
| The following is essentially @drhab's solution, but uses only one idea repeatedly.
From $$ (x-8)p(2x) = 8(x-1)p(x) $$ we see $x-8$ divides $p(x)$. Let $p(x) = (x-8)p_1(x)$ and substitute, yielding $$ 2(x-8)(x-4)p_1(2x) = 8(x-1)(x-8)p_1(x) $$
From this we see $x-4$ divides $p_1(x)$. Let $p_1(x) = (x-4)p_2(x)$ and substitute, yielding $$ 4(x-8)(x-4)(x-2)p_2(2x) = 8(x-1)(x-4)(x-8)p_2(x) $$
From this we see $x-2$ divides $p_2(x)$. Let $p_2(x) = (x-2)p_3(x)$ and substitute, yielding $$ 8(x-8)(x-4)(x-2)(x-1)p_3(2x) = 8(x-1)(x-2)(x-4)(x-8)p_3(x) $$
( ... and our recursive process stops because the new $x-1$ factor divides the $x-1$ that's been lingering on the right all along.)
But now we simplify to $p_3(2x) = p_3(x)$ and the rest of @drhab's argument finishes the argument.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Exponential function negative: $\left(\frac{81}{4}\right)^{1/4}\left(\frac{1}4\right)^{-3/4}$ This is another example. $\left(\dfrac{81}{4}\right)^{1/4}\left(\dfrac{1}4\right)^{-3/4}$
Multiply on both sides equals $\dfrac{81^{1/4}}{4^{1/4}}\cdot \dfrac{1^{-3/4}}{4^{-3/4}}$
This should be $\dfrac{3}{4^{1/4}}\cdot \dfrac{1^{-3/4}}{4^{-3/4}}$ which I get to -3/$\sqrt 4^{-1/2}$
Is that correct? I should get the answer 6.
| An alternative way (bring everything under a single exponent):
$$(\frac{81}{4})^\frac 14 \cdot (\frac{1}{4})^\frac {-3}{4} = (\frac{81}{4})^\frac 14 \cdot [(\frac 14)^{-3}]^{\frac 14} = (\frac{81}{4})^\frac 14 \cdot(4^{3})^{\frac 14} = (\frac{81}{4}\cdot 64)^{\frac 14} = (81 \cdot 16)^{\frac 14} = 3 \cdot 2 = 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Volume of Solid Enclosed by an Equation I'm having problems finding the triple integrals of equations. I guess it has to do with the geometry. Can someone solve the two questions below elaborately such that I can comprehend this triple integral thing once and for all:
Compute the volume of the solid enclosed by
*
*$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1, x=0,y=o, z=0$
*$x^2+y^2-2ax=0, z=0, x^2+y^2=z^2$
| for the #1 if you want to use a geometrical solution you can say that it's the volume of a pyramid with apex at $(0,0,0)$ intersecting the axes at $(a,0,0),(0,b,0),(0,0,c)$ now because of the orthogonality of axes we can say the base is a right triangle having the area $\frac{1}{2}ab$ and the length of the height is $c$ then:
$$volume\; of\; pyramid=\frac{1}{3}(area\; of\; base)\times height=\frac{1}{6}abc$$
For a calculus solution we have an enclosed space limited by the planes $z=0$ and $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ from bottom and top so the range for $z$ will be $0\le z\le c-\frac{c}{a}x-\frac{c}{b}y$
consider the side of the pyramid that is lying on the $xy$-plane, it is a triangle enclosed by the lines $x=0$ and $y=0$ and $\frac{x}{a}+\frac{y}{b}=1$ on $xy$-plane, so the range for y will be $0\le y\le b-\frac{b}{a}x$
and finally the range for $x$ is $0\le x\le a$ so we have:
$$\begin{align}
\int^{a}_{0}\int^{b-\frac{b}{a}x}_{0}\int^{c-\frac{c}{a}x-\frac{c}{b}y}_{0}dz\,dy\,dx&=\int^{a}_{0}\int^{b-\frac{b}{a}x}_{0}(c-\frac{c}{a}x-\frac{c}{b}y)dy\,dx\\
&=c\int^{a}_{0}[y-\frac{1}{a}xy-\frac{1}{2b}y^2]^{b-\frac{b}{a}x}_{0}dx\\
&=c\int^{a}_{0}(\frac{b}{2}-\frac{b}{a}x+\frac{b}{2a^2}x^2)dx\\
&=bc[\frac{1}{2}x-\frac{x^2}{2a}+\frac{x^3}{6a^2}]^{a}_{0}\\
&=bc(\frac{a}{2}-\frac{a}{2}+\frac{a}{6})\\
&=\frac{1}{6}abc
\end{align}$$
And about your question for #2 that you have commented in the other answer, I've created the following figure for you in matlab:
you should find the volume of the space enclosed by $z=0$ at bottom and by the cone $z^2=x^2+y^2$ at top within a cylinder along the z-axis which its base is a circle centered at $(a,0)$ and has radius $a$. The height of the cylinder will be $2a$
Samir Khan has used the fact that this volume is the volume below the function $z=\sqrt{x^2+y^2}$ in the area of a circle centered at $(a,0)$ with radius $a$ then has used polar coordinates to solve the integral
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Eliminate the parameter of a
Eliminate the parameter to find a description of the following circles
or circular arcs in terms of $x$ and $y$. Give the center and radius,
and indicate the positive orientation.
$x=4\cos{(t)} ,\ y=3\sin{(t)} ;\ 0 \leq t \leq 2\pi$
So,
$\displaystyle x^2=4^2\cos^2{(t)} ,\ y^2=3^2\sin^2{(t)} \implies \frac{x^2}{4^2}=\cos^2{(t)} ,\ \frac{y^2}{3^2}=\sin^2{(t)}$
But I detect no radius. I'm rather confused, on this whole question. It doesn't even explicitly define "the parameter". Insight?
| Notice, $$x=4\cos t \implies \cos t=\frac{x}{4}\tag 1$$ & $$y=3\sin t \implies \sin t=\frac{y}{3}\tag 2$$ Now, for eliminating $t$, squaring & adding (1) & (2), we get $$\cos^2t+\sin^2t=\left(\frac{x}{4}\right)^2+\left(\frac{y}{3}\right)^2$$ $$\color{blue}{\frac{x^2}{16}+\frac{y^2}{9}=1}$$ The above equation is in the form of the standard form of equation of an ellipse: $\color{blue}{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$ where, $a=4$ & $b=3$.
Hence, the curve represents an ellipse not a circle, hence we have $$\text{center of ellipse, origin}\equiv(0, 0)$$ $$\text{major axis}, 2a=2\times 4=8$$ $$\text{minor axis}, 2b=2\times 3=6$$
O.P.'s detection is right that there is no radius as that is not a circle but the curve represents an ellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $x$ in the triangle
the triangle without point F is drawn on scale, while I made the point F is explained below
So, I have used $\sin, \cos, \tan$ to calculate it
Let $\angle ACB = \theta$, $\angle DFC = \angle BAC = 90^\circ$, and $DF$ is perpendicular to $BC$ (the reason for it is to have same $\sin, \cos, \tan$ answer)
$$\sin \angle ACB = \frac {DF}{CD} = \frac{AB}{BC}$$
$$\cos \angle ACB = \frac {CF}{CD} = \frac{AC}{BC}$$
$$\tan \angle ACB = \frac {DF}{CF} = \frac{AB}{AC}$$
putting known data into it
\begin{align}
\frac {DF}{CD} &= \frac {AB}{12} \quad(1) \\
\frac {EF+3}{CD} &= \frac {2CD}{12} \\
\frac {EF+3}{CD} &= \frac {CD}{6} \quad(2)\\
\frac {DF}{EF+3} &= \frac {AB}{2CD} \quad(3)
\end{align}
I've stuck at here, how do I find their length?
|
$K$ will be like that so - $DK\perp CD$
we know that $DK\perp CD$ than $DK||AB$ (because $AC\perp AB$), Also we know that $CD=AD$, because of that we can understand $DK$ is median of triangle $\Delta ABC$ so $CK=KB=6$.
We know that $CE=3$, than $CE=EK=3$. $\Delta CDK$ is a right triangle, than we can understand that - $DE=CE=EK=3$ (Because $DE$ is median)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Maximal $n$ such the the additive partition with a given product is unique. Given $n$, there are many tuples with $a + b + c = n,0 < a < b < c$. For large $n$, different tuples may give the same products. E.g. $2+8+9=19=3+4+12,2\times8\times9=144=3\times4\times12$.
What is the largest value of $n$, such that there is no tuples with the same product? The computer tells me it's $22$. Maybe directly proving from $22$ is hard. We can just prove that for sufficiently large $n$, there are always tuples with the same product.
Any ideas?
|
For sufficiently large $n$, there are always unique tuples $a,b,c, \ 0<a<b<c$ and $a',b',c', \ 0<a'<b'<c'$ such that $a+b+c=n=a'+b'+c'$ and $a \times b \times c = a' \times b' \times c'$.
Proof. Say we have $b$ divisible by $3$ and $c$ divisible by $2$. This assumption will be justified later. Relate $a,b,c$ with $a',b',c'$ as follows:
$$
a' = 2a \qquad b' = \frac{b}{3} \qquad c' = \frac{3c}{2}
$$
Clearly these are all integers, and clearly we have $a' \times b' \times c' = a \times b \times c$. The difference between the new sum and the original sum, which we need to be $0$, is
$$a-\frac{2b}{3}+\frac{c}{2} = 0$$
which has solutions $0<a<b<c$ of the form
$$b = 3x, \quad c=4x-2a \quad | \quad a,x \in \mathbb{Z}^+ \text{ and } x>2a$$
which also justifies our divisibility assumptions about $b$ and $c$.
So if we can write $n=a+b+c=a + (3x) + (4x - 2a) = 7x - a$ where $x$ is a positive integer greater than $2a$, then we can certainly find a working solution.
It is easy to see that any sufficiently large $n$ can be written in this form. Given some $n$, we choose $x$ so that $7x$ is the smallest multiple of $7$ greater than $n$, and we pick and $a=7x-n$ to correct for the error. Then $a$ will always be between $1$ and $7$. So if $x \geq 15$, we can ensure that $x>2a$ holds.
The last step is to ensure that the inequality $0<a'<b'<c'$ holds. We have positive $a$, so certainly $0<2a=a'$. Since $x>2a$, we have that $a'=2a<x=\frac{3x}{3}=\frac{b}{3}=b'$. Finally, $b'=\frac{b}{3}<b<c<\frac{3c}{2}=c'$.
Hence, for all sufficiently large $n$ (specifically $n \geq 98$), we can find a solution pair $a,b,c$ and $a',b',c'$ as described above. Q.E.D.
Summary. Given $n \geq 98$, calculate $x=\lceil \frac{n+1}{7} \rceil$. Then the following is a working pair of tuples:
$$\boxed{(a,b,c)=(7x-n, \ 3x, \ 2n-10x)} \\
\boxed {(a',b',c')=(14x-2n, \ x, \ 3n-15x)}$$
This follows from the proof and the fact that $\lceil \frac{n+1}{7} \rceil$ is the $x$ such that $7x$ is the smallest multiple of $7$ greater than $n$. The rest is substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding a triangle ABC if $2\prod (\cos \angle A+1)=\sum \cos(\angle A-\angle B)+\sum \cos \angle A+2$ Find $\triangle ABC$ if $\angle B=2\angle C$ and $$2(\cos\angle A+1)(\cos\angle B+1)(\cos\angle C+1)=\cos(\angle A-\angle B)+\cos(\angle B-\angle C)+\cos(\angle C-\angle A)+\cos\angle A+\cos\angle B+\cos\angle C+2$$
| Re-writing the equation as
$$0 = 2\prod (\cos A + 1) - \sum \cos(B-C) - \sum \cos A - 2$$
we begin by multiplying-out the product, and carrying-on from there:
$$\begin{align}
0 &= 2\cos A \cos B \cos C + 2\sum \cos B \cos C \color{blue}{+ 2\sum \cos A} \color{red}{+ 2} \\[4pt]
&\quad-\sum\cos(B-C) \color{blue}{- \sum\cos A} \color{red}{- 2} \\[8pt]
&= 2\cos A \cos B \cos C + \sum\left(\; 2 \cos B \cos C + \cos A -\cos(B-C)\;\right) \\[8pt]
&= 2\cos A \cos B \cos C + \sum\left(\; 2 \cos B \cos C - \cos(B+C) -\cos(B-C)\;\right) \\[8pt]
&= 2\cos A \cos B \cos C + \sum\left(\; 2 \cos B \cos C - 2 \cos B \cos C \;\right)\\[8pt]
&= 2 \cos A \cos B \cos C
\end{align}$$
Thus, the exercise reduces to
$$B = 2 C \qquad\text{and}\qquad \cos A \cos B \cos C = 0$$
which is readily solved.
Edit. Since the solutions were posted as comments, I'll provide arguments for them.
Note that the equation $\cos A \cos B \cos C = 0$ implies that one of $A$, $B$, $C$ is a right angle. Taking these case by case ...
$$\begin{align}
A = 90^\circ &\quad\implies\quad 90^\circ = B+C = 2 C + C = 3C \implies \color{blue}{( A, B, C ) = (90^\circ, 60^\circ, 30^\circ)} \\[4pt]
B = 90^\circ &\quad\implies\quad C = B/2 = 45^\circ \implies \color{blue}{(A, B, C) = (45^\circ, 90^\circ, 45^\circ)} \\[4pt]
C = 90^\circ &\quad\implies\quad B = 2 C = 180^\circ \implies \color{red}{\text{invalid triangle}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to show that this difference of products is $O \left( \frac{1}{n^2} \right) $ Let $k \leq n$. Consider the following difference of products:
$$ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right)$$
For $n=1,2,3$, this is clearly $O \left( \frac{1}{n^2} \right) $ for $k=1,2,3$ respectively. How to show it for arbitrary $n$? And is it possible to give an estimate $C$ such that the difference is less than $\frac{C}{n^2}$?
| We have
$$\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n+1}\right)-\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n}\right)=\left(\prod_{i=1}^{k-2}\,\left(1-\frac{i}{n}\right)\right)\left(\left(1+\frac{1}{n}\right)^{-(k-1)}-\left(1-\frac{k-1}{n}\right)\right)\,.$$
For large $n\in\mathbb{N}$, $\prod_{i=1}^{k-2}\,\left(1-\frac{i}{n}\right)=1+O\left(\frac{1}{n}\right)$ and $$\left(1+\frac{1}{n}\right)^{-(k-1)}=1-\frac{k-1}{n}+\frac{k(k-1)}{2n^2}+O\left(\frac{1}{n^3}\right)\,.$$
Consequently,
$$\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n+1}\right)-\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n}\right)=\frac{k(k-1)}{2n^2}+O\left(\frac{1}{n^3}\right)=O\left(\frac{1}{n^2}\right)\,.$$
Indeed, you can easily show that, for $k>1$,
$$\frac{k(k-1)}{2n^2}\left(1-\frac{(k-2)(3k-1)}{6n}\right)\leq \prod_{i=1}^{k-1}\,\left(1-\frac{i}{n+1}\right)-\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n}\right)\leq \frac{k(k-1)}{2n^2}\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Trying to solve a pair of trigonometric simultaneous equations I have a machine that has two shafts which are the inputs and their position is set by 2 servo motors.
Depending on the angle of these two shafts (shaft 1 has an angle designated $Ta$ degrees, shaft 2 has an angle designated $Ba$ degrees) a set of gimbals rotates and tilts an attached plate, which is the output of the machine. ( A mirror for reflecting a light beam.)
I have worked out that the plate position designated by the two angles $Pa$ and $Pb$ are related by the following two equations:
$Cos(Pa) = Cos(Ta) * Cos(Ba)$
$Tan(Pb) = {sin(Ta) \over Tan(Ba)}$
$Tb$ has a range of $0$ to $180$ degrees
$Ta$ has a range of $0.001$ to $90$ degrees
Using these 2 equations I can work out the position of the plate given $Ta$ and $TB$
What I would like to do is solve the 2 equations so I can specify $Pa$ and $Pb$ (the required plate location) and calculate $Ta$ and $Ba$ so I can then set the angle of my 2 input shafts accordingly.
Any help would be greatly appreciated As I am no mathematician and struggling with this.
| I'll simplify your notation a bit, using shaft angles $T$ and $B$ (rather than $T_a$ and $B_a$), and plate angles $P$ and $Q$ (rather than $P_a$ and $P_b$). The given equations are
$$\cos P = \cos T \cos B \qquad\qquad
\tan Q = \frac{\sin T}{\tan B} \tag{$\star$}$$
Solving these for $\cos T$ and $\sin T$, we eliminate $T$ via the relation $\cos^2 T + \sin^2 T = 1$:
$$\begin{align}
\left(\frac{\cos P}{\cos B}\right)^2 + \left(\tan Q \tan B\right)^2 = 1 &\quad\to\quad \frac{\cos^2 P}{\cos^2 B} + \frac{\tan^2 Q \sin^2 B}{\cos^2 B} = 1 \\[6pt]
&\quad\to\quad \cos^2 P + \tan^2 Q \sin^2 B = \cos^2 B \\
&\quad\to\quad \cos^2 P + \tan^2 Q (1-\cos^2 B) = \cos^2 B \\
&\quad\to\quad \cos^2 P + \tan^2 Q = \cos^2 B ( 1 + \tan^2 Q) = \frac{\cos^2 B}{ \cos^2 Q }
\end{align}$$
Therefore,
$$\begin{align}
\cos^2 B &= \cos^2 Q \left( \cos^2 P + \tan^2 Q \right) = \cos^2 P \cos^2 Q + \sin^2 Q \\
&= ( 1 - \sin^2 P)\cos^2 Q + \sin^2 Q = \cos^2 Q + \sin^2 Q - \sin^2 P \cos^2 Q \\
&= 1 - \sin^2 P \cos^2 Q \\[6pt]
\sin^2 B &= \sin^2 P \cos^2 Q
\end{align}$$
and we can write
$$ \cos B = \pm \sqrt{ 1 - \sin^2 P \cos^2 Q } \qquad\qquad \sin B = |\sin P \cos Q|$$
where the "$\pm$" depends $B$'s quadrant, and the "$|\cdot|$" ensures that that quadrant is I or II. Substituting these into $(\star)$, we can solve for functions of $T$:
$$\begin{align}
\cos T &= \frac{\cos P}{\cos B} = \frac{|\cos P|}{\sqrt{1-\sin^2 P \cos^2 Q}} \\[6pt]
\sin T &= \tan B \tan Q =\frac{|\sin P \cos Q \tan Q|}{\sqrt{1-\sin^2 P \cos^2 Q}} = \frac{|\sin P \sin Q|}{\sqrt{1-\sin^2 P \cos^2 Q}} \\[6pt] \tan T &= \left|\frac{\sin P \sin Q}{ \cos P}\right| = \left| \tan P \sin Q \right|\end{align}$$
where the "$\pm$"s vanish but "$|\cdot|$"s remain, to ensure that $T$ is a first-quadrant angle. In fact, when $\cos T \neq 0$, the relation $\cos P = \cos T \cos B$ tells us that $\cos B$ and $\cos P$ share a sign, which resolves the "$\pm$" ambiguity for $\cos B$. (However, when $\cos T = 0$, we also have $\cos P = 0$. In this case, the "$\pm$" remains for $\cos B$, although the expression simplifies to $\cos B = \pm \sin Q$; likewise, $\sin B = |\cos Q|$.)
This agrees with @Daniel's answer, although the final expressions are perhaps a little cleaner here (and maybe also the derivation).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate a quotient with fractional power of a quadratic polynomial I need help finding the indefinite integral of
$$\int\,\frac{x}{(7x - 10 - {x^2})^{3/2}}\,\text{d}x\,.$$
| $$\int { \frac { x }{ \sqrt { { \left( 7x-10-{ x }^{ 2 } \right) }^{ 3 } } } dx } =\int { \frac { x }{ \sqrt { { -\left( { x }^{ 2 }-7x+10 \right) }^{ 3 } } } dx= } \int { \frac { x }{ \sqrt { { -\left( { x }^{ 2 }-7x+\frac { 49 }{ 4 } -\frac { 49 }{ 4 } +10 \right) }^{ 3 } } } dx= } \\ =\int { \frac { x }{ \sqrt { { { \left( \frac { 9 }{ 4 } -{ \left( x-\frac { 7 }{ 2 } \right) }^{ 2 } \right) }^{ 3 } } } } dx } \\ x-\frac { 7 }{ 2 } =\frac { 3 }{ 2 } \sin { t } \\ x=\frac { 1 }{ 2 } \left( 7+3\sin { t } \right) \Rightarrow dx=\frac { 3 }{ 2 } \cos { t } dt\\ \int { \frac { \frac { 3 }{ 2 } \cos { t } }{ \sqrt { { { \left( \frac { 9 }{ 4 } -{ \frac { 9 }{ 4 } }\sin ^{ 2 }{ t } \right) }^{ 3 } } } } dt } =\int { \frac { \frac { 3 }{ 2 } \cos { t } }{ { \left( \frac { 3 }{ 2 } \right) }^{ 3 }\sqrt { { { \left( 1-\sin ^{ 2 }{ t } \right) }^{ 3 } } } } dt } =\frac { 4 }{ 9 } \int { \frac { \cos { t } }{ \cos ^{ 3 }{ t } } dt=\frac { 4 }{ 9 } \int { \frac { dt }{ \cos ^{ 2 }{ t } } =\frac { 4 }{ 9 } } } \tan { t } +C\\ \sin { t } =\left( \frac { 2x-7 }{ 3 } \right) \Rightarrow t=\arcsin { \left( \frac { 2x-7 }{ 3 } \right) } \\ \frac { \\ 4 }{ 9 } \tan { \arcsin { \left( \frac { 2x-7 }{ 3 } \right) } } +C=\frac { 4 }{ 9 } \frac { \left( \frac { 2x-7 }{ 3 } \right) }{ \sqrt { 1-{ \left( \frac { 2x-7 }{ 3 } \right) }^{ 2 } } } +C=\frac { 8x-28 }{ 9\sqrt { -4x^{ 2 }+28x-40 } } +C$$
so
$$\int { \frac { x }{ \sqrt { { \left( 7x-10-{ x }^{ 2 } \right) }^{ 3 } } } dx } =\frac { 8x-28 }{ 9\sqrt { -4x^{ 2 }+28x-40 } } +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
find the complex number $z^4$
Let $z = a + bi$ be the complex number with $|z| = 5$ and $b > 0$ such that the distance between $(1 + 2i)z^3$ and $z^5$ is maximized, and let $z^4 = c + di$.
Find $c+d$.
I got that the distance is:
$$|z^3|\cdot|z^2 - (1 + 2i)| = 125|z^2 - (1 + 2i)|$$
So I need to maximize the distance between those two points.
$|z^2| = 25$ means since: $z^2 = a^2 - b^2 + 2abi$ that:
$625 = (a^2 - b^2)^2 + 4a^2b^2 = a^4 + b^4 + 2a^2b^2$
But that doesnt help much.
| You are trying to maximize the value of $|z^2-(1+2i)|$ where $z^2$ is a point on the circle, centre $0$ radius $25$
As has been pointed out by Terra Hyde, $z^2$ must be the point on the circle on the other side of the origin from $1+2i$ which would be collinear with $1+2i$ and the origin. Therefore
$$z^2=25(-\cos\theta-i\sin\theta)$$, where $\tan\theta=2$
Therefore $$z^4=625(\cos2\theta+i\sin2\theta)=375+500i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Coefficient of binomial expansion The coefficient of $x^3$ is $4$ times the coefficient of $x^2$ in the new expansion of $(1+x)^n$. Find the value of $n$.
| coefficient of $x^3$ is four times the coefficient of $x^2$
$$(1+x)^n=\\\binom{n}{0}1^{n}x^{0}+\binom{n}{1}1^{n-1}x^{1}+{\color{DarkBlue} {\binom{n}{2}1^{n-2}x^{2} }}+{\color{Red}{\binom{n}{3}1^{n-3}x^{3}} }+...+\binom{n}{n}1^{n-n}x^{n}\\$$so
$$\binom{n}{3}1^{n-3}=4*\binom{n}{2}1^{n-2}\\\binom{n}{3}=4\binom{n}{2}\\\frac{n(n-1)(n-2)}{3!}=4 \frac{n(n-1)}{2!} \\\frac{n-2}{6}=\frac{4}{2}\\n-2=12\\n=14$$ note that $$n(n-1) \neq 0 \rightarrow n \neq 0 ,n \neq 1$$ because if n=1 $(1+x)^1=1+x$
and if n=0 then $(1+x)^0=1$ these are not acceptable
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Math Subject GRE 1268 Question 55 If $a$ and $b$ are positive numbers, what is the value of $\displaystyle \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$. | A slight variation of the accepted solution begins on line $3$:
\begin{align}
I &= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\
&= \int_{0}^{\infty} \frac{dx}{1 + e^{bx}} - \int_{0}^{\infty} \frac{dx}{1 + e^{ax}} \\
&= \int_{0}^{\infty} \frac{e^{-bx} dx}{(1 + e^{bx})e^{-bx}} - \int_{0}^{\infty} \frac{e^{-ax}dx}{(1 + e^{ax})e^{-ax}} \\
&=\int_{0}^{\infty} \frac{e^{-bx} dx}{(1 + e^{-bx})} - \int_{0}^{\infty} \frac{e^{-ax}dx}{(1 + e^{-ax})} \\
\end{align}
For the integral involving $b$, let $u=1+e^{-bx}$ so that $\frac{-1}{b}du=e^{-bx}dx$. Then $x=0 \implies u=2$, and $x \to \infty \implies u \to 1^+$. With a similar result for the integral involving $a$, and changing the order of integration, we end up with
\begin{align}
I &= \left( \frac{1}{b} - \frac{1}{a} \right) \lim _{\epsilon \to 0^+}\int_{1+\epsilon}^{2}\frac{1}{u}du = \left( \frac{1}{b} - \frac{1}{a} \right) \ln 2 \ . \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
} |
find the interval of convergence of the power series like the title said i have to find the interval of convergence of this power series :
$$\sum_{n=1}^\infty{ ((-1)^n *(x-1)^{2n-1})\over 3^n}$$
I applied the ratio test and i got something like this:
$$\left|\frac{(-1)*(x-1)^2}3\right|$$
I know that I have to find the interval and then study the end points but I dont know how to do it. I'm stuck at this step
thanks in advance
| The absolute value you got was correct.
\begin{align*}
\left|\frac{(-1)(x-1)^2}{3}\right| & = \left|\frac{(x-1)^2}{3}\right|\\
&= \frac13\left|(x-1)^2\right| <1
\end{align*}
Now we have
$$\left|(x-1)^2\right| < 3$$
$$1-\sqrt3 < x < 1+\sqrt3$$
Writing out some terms: $$\sum_{n=1}^\infty{(-1)^n \cdot (x-1)^{2n-1}\over 3^n} = -\frac{x-1}{3} + \frac{(x-1)^3}{9}-\frac{(x-1)^5}{27} + \cdots$$
Plugging in $1+\sqrt3$ for $x$:
$$-\frac{\sqrt3}{3} + \frac{\sqrt{3}}{3}- \frac{\sqrt3}{3} + \cdots$$
Which is a geometric series with common ratio $-1$ so it diverges.
If we plug in $1-\sqrt3$ for $x$ we get the series $$\frac{\sqrt3}{3} - \frac{\sqrt3}{3}+\frac{\sqrt{3}}{3} - \cdots$$
Which diverges by the same logic as above. There is no endpoint convergence, so our interval of convergence is:
$$\boxed{1 - \sqrt3 < x < 1+\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\sum_{j=1}^n \frac{1}{\sqrt{j}} > \sqrt{n}$ with induction Problem: Prove with induction that \begin{align*} \sum_{j=1}^n \frac{1}{\sqrt{j}} > \sqrt{n} \end{align*} for every natural number $n \geq 2$.
Attempt at proof: Basic step: For $n = 2$ we have $1 + \frac{1}{\sqrt{2}} > \sqrt{2}$ which is correct.
Induction step: Suppose the assertion holds for some natural number $n$, with $n > 2$. Then we now want to prove that it also holds for $n +1$, i.e. that \begin{align*} \sum_{j=1}^{n+1} \frac{1}{\sqrt{j}} > \sqrt{n+1} \end{align*}
Now we have that \begin{align*} \sum_{j=1}^{n+1} \frac{1}{\sqrt{j}} = \sum_{j=1}^n \frac{1}{\sqrt{j}} + \frac{1}{\sqrt{n+1}} > \sqrt{n} + \frac{1}{\sqrt{n+1}} \end{align*} or \begin{align*} \sum_{j=1}^n \frac{1}{\sqrt{j}} + \frac{1}{\sqrt{n+1}} > \frac{\sqrt{n} \sqrt{n+1} + 1}{\sqrt{n+1}} \end{align*}
Now I'm stuck, and I don't know how to get $\sqrt{n+1}$ on the right hand side. Help would be appreciated.
| As Daniel Fischer points out in the comments, since you have
$$
\sum_{j=1}^{n+1} \frac{1}{\sqrt{j}} > \sqrt{n} + \frac{1}{\sqrt{n+1}}
$$
it is enough to show
$$
\sqrt{n} + \frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}
$$
or equivalently $
\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1} - \sqrt{n}
$. A way to show this final inequality is to recall the identity $(a-b)(a+b) = a^2-b^2$ and multiply both sides by $\sqrt{n+1} + \sqrt{n}$, i.e. to use the identity with $a=\sqrt{n+1}$ and $b=\sqrt{n}$: what happens to the LHS? To the RHS?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the length of latus rectum of an ellipse If the lines $2x+3y=10$ and $2x-3y=10$ are tangents at the extremities of its same latus rectum to an ellipse whose center is origin,then the length of the latus rectum is
$(A)\frac{110}{27}\hspace{1cm}(B)\frac{98}{27}\hspace{1cm}(C)\frac{100}{27}\hspace{1cm}(D)\frac{120}{27}$
I found point of intersection of two tangents at $(5,0)$ but dont know how to solve further.Any guidance will be helpful.
| Using standard ellipse notation and relations for
$ a, b, c, p $.
Tangent equation of ellipse
$$ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} =1 \tag{1}$$
Given tangent equation
$$ \frac{x}{5} + \frac{y}{10/3} =1 \tag{2}$$
Comparing $ x, y $ coefficients,
$$ \frac{x_1}{a^2}= \frac{1}{5} \tag{3}$$
$$ \frac{y_1}{b^2}= \frac{3}{10} = \frac{p}{b^2} = \frac{1}{a} \tag{4}$$
$$ a =\frac{10}{3}\tag{5} $$
$$ x_1= \frac{a^2}{5} = \frac{100}{45} =c \tag{6}$$
$$b^2 = a^2 -c^2 = \frac{400}{81}, b =\frac{20}{9} \tag{7}$$
Semi-latus rectum or latus rectum?
$$ p =\frac{b^2}{a}= \frac{40}{27} \tag{8} $$
I checked it graphically also, none of the given options tally. But there ought to be a more elegant way than this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Radical under Radical expression how to find the sum of $\sqrt{\frac54 + \sqrt{\frac32}} + \sqrt{\frac54 - \sqrt{\frac32}} $ ? Is there a method to solve these kind of equations ?
| $$
x = \sqrt{\frac54 + \sqrt{\frac32}} + \sqrt{\frac54 - \sqrt{\frac32}}\\
x^2 = \frac54 + \sqrt{\frac32} + 2\sqrt{\left(\frac54\right)^2 - \left(\sqrt{\frac32}\right)^2} + \frac54 - \sqrt{\frac32} = \frac52 + 2\sqrt{\frac{25}{16} - \frac32} = \frac52 + 2\cdot\frac14 = 3\\
x = \sqrt3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove the series has positive integer coefficients How can I show that the Maclaurin series for
$$
\mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4}
\\
= 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\,
{x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots
$$
has positive integer coefficients? (I have others to do, too, but this one will be a start.)
possibilities
(a) The coefficients $Q(n)$ satisfy the recurrence
$$
(n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0
$$
(b) $\mu(x)$ satisfies the differential equation
$$
(x^3+9x^2+7x-3)\mu(x)+(x^4+12x^3+14x^2-12x+1)\mu'(x)=0
$$
(c) factorization of $x^4+12x^3+14x^2-12x+1$ is
$$
\left( x-\sqrt {5}+3+\sqrt {15-6\,\sqrt {5}} \right) \left( x-\sqrt
{5}+3-\sqrt {15-6\,\sqrt {5}} \right) \left( x+\sqrt {5}+3-\sqrt {15+
6\,\sqrt {5}} \right) \left( x+\sqrt {5}+3+\sqrt {15+6\,\sqrt {5}}
\right)
$$
(d) $(1-8X)^{-1/4}$ has positive integer coefficients, But if $X$ is defined by $1-8X=x^4+12x^3+14x^2-12x+1$, then $X$ does not have integer coefficients.
(e) Can we compute the series for $\log\mu(x)$
$$3\,x+{\frac {29}{2}}{x}^{2}+99\,{x}^{3}+{\frac {3121}{4}}{x}^{4}+{
\frac {32943}{5}}{x}^{5}+{\frac {348029}{6}}{x}^{6}+\dots
$$
and then recognize that its exponential has integer coefficients?
| Here is an "almost" solution for integer coefficients.
The zeros of $x^4+12x^3+14x^2-12x+1$ are algebraic integers. So are their reciprocals. Now in the series
$$
(1-xa)^{-1/4} = 1+\frac{a}{4} x+\frac{5 a^2}{32} x^2 +\frac{15a^3}{128} x^3 + \cdots
$$
where $a$ is an algebraic integer, the coefficients are algebraic integers (except for power-of-2 denominators). Multiply four of these
$$
(x^4+12x^3+14x^2-12x+1)^{-1/4} = \prod_a (1-xa)^{-1/4}
$$
where the product is over all $a$ such that $1/a$ is a zero of $x^4+12x^3+14x^2-12x+1$. Its coefficients are also algebraic integers, except possibly for power-of-2 denominators. But this is invariant under permutation of the roots, so the coefficients are rational numbers. (We can also see this more directly.) Therefore the coefficients are rational integers, except possibly for power-of-2 denominators.
After doing this, I thought I would be able to finish by using 2-adic numbers, to analyze these denominators. But it seems this polynomial has no 2-adic zeros. So I would have to go to an algebraic extension of the 2-adics...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
How to evaluate infinite series $\sum\limits_{n=0}^\infty\sqrt{B^2+n^2} e^{-an}$ I'm trying to evaluate an infinite series:
$$
\sum\limits_{n=0}^\infty\sqrt{B^2+n^2} e^{-an}
$$
where $a$ and $B$ are real parameters, or equivalently:
$$\sum\limits_{n=0}^\infty\sqrt{B^2+n^2} x^n$$
where $0<x<1$.
When $n$ becomes much larger than $B$, the terms will begin to look like:
$$nx^n$$
If the entire sum were comprised of these, there's a closed analytic form, which I think is:
$$\sum\limits_{n=0}^\infty n x^n = \frac{x}{(1-x)^2}$$
It would be great to know the sum in the first line in closed form though. If anyone could assist either with a closed analytic expression, or even just ideas or techniques as to how to go about evaluating this, It'd be much appreciated.
| This series can not be evaluated directly. One approximation method is as follows. Let the Lerch transcendent be defined by
$$ \phi(z;s,\alpha) = \sum_{n=0}^{\infty} \frac{z^{n}}{(n+\alpha)^{s}}.$$
The series expansion of $\sqrt{1+x}$ is, for the first few terms,
$$\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^{2}}{8}+ \frac{3 \, x^{3}}{16} - \cdots. $$
and leads to
\begin{align}
S &= \sum_{n=0}^{\infty} \sqrt{b^{2}+n^{2}} \, e^{-an} = \sqrt{b^{2}} + \sum_{n=1}^{\infty} n \, \sqrt{1 + \frac{b^{2}}{n^{2}}} \, e^{-an} \\
&= \sqrt{b^{2}} + \sum_{n=1}^{\infty} n \, e^{-an} \, \left( 1 + \frac{b^{2}}{2 \, n^{2}} - \frac{b^{4}}{8 \, n^{4}} + \frac{3 \, b^{6}}{16 \,n^{6}} - \cdots \right) \\
&= \sqrt{b^{2}} - \frac{d}{da} \, \left( \frac{e^{-a}}{1 - e^{-a}} \right) + \frac{b^{2}}{2} \, \sum_{n=1}^{\infty} \frac{e^{-an}}{n} - \frac{b^{4}}{8} \, \sum_{n=0}^{\infty} \frac{e^{-a(n+1)}}{(n+1)^{3}} + \frac{3 \, b^{6}}{16} \, \sum_{n=0}^{\infty} \frac{e^{-a(n+1)}}{(n+1)^{5}} - \cdots \\
&= \sqrt{b^{2}} - \frac{d}{da} \, \left( \frac{1}{e^{a} - 1} \right) - \frac{b^{2}}{2} \, \ln(1-e^{-a}) - \frac{b^{4} \, e^{-a}}{8} \, \phi(e^{-a}; 3,1) + \frac{3 \, b^{6} \, e^{-a}}{16} \, \phi(e^{-a}; 5,1) - \cdots \\
&= \sqrt{b^{2}} + \frac{e^{a}}{(e^{a}-1)^{2}} - \frac{b^{2}}{2} \, \ln(1 - e^{-a}) - \frac{b^{4} \, e^{-a}}{8} \, \phi(e^{-a}; 3,1) + \frac{3 \, b^{6} \, e^{-a}}{16} \, \phi(e^{-a}; 5, 1) - \cdots
\end{align}
By using
$$\sqrt{1+x} = \sum_{r=0}^{\infty} \frac{(-1)^{r} \, \left(- \frac{1}{2}\right)_{r}}{r!} \, x^{r},$$
where $(x)_{n}$ is the Pochhammer symbol, then
\begin{align}
S &= \sum_{n=0}^{\infty} \sqrt{b^{2} + n^{2}} \, e^{-an} = \sqrt{b^{2}} + \sum_{n=1}^{\infty} n \, \sqrt{1 + \frac{b^{2}}{n^{2}}} \, e^{-an} \\
&= \sqrt{b^{2}} + \sum_{n=1}^{\infty} n \, e^{-an} + \frac{b^{2}}{2} \, \sum_{n=1}^{\infty} \frac{e^{-an}}{n} + \sum_{r=2}^{\infty} \frac{(-1)^{r} \, \left(- \frac{1}{2}\right)_{r} \, b^{2r}}{r!} \cdot \sum_{n=1}^{\infty} \frac{e^{-an}}{n^{2r-1}} \\
&= \sqrt{b^{2}} - \partial_{a}\left(\frac{1}{e^{a}-1}\right) - \frac{b^{2}}{2} \, \ln(1 - e^{-a}) + \sum_{r=2}^{\infty} \frac{(-1)^{r} \, \left(- \frac{1}{2}\right)_{r} \, b^{2r}}{r!} \, Li_{2r-1}(e^{-a}) \\
&= \sqrt{b^{2}} + \frac{e^{a}}{(e^{a}-1)^{2}} - \frac{b^{2}}{2} \, \ln(1 - e^{-a}) + \sum_{r=2}^{\infty} \frac{(-1)^{r} \, \left(- \frac{1}{2}\right)_{r} \, b^{2r}}{r!} \, Li_{2r-1}(e^{-a}),
\end{align}
where $Li_{n}(z)$ is the polylogarithm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How do I find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17? Recently I came across a question,
Find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17?
At first I applied sum of G.P. formula but ended up with the expression $1\cdot \dfrac{4^{41}-1}{4-1}$. I couldn't figure out how to proceed further. Secondly I thought of using the fact $(a+b+\cdots) \pmod {17} = (r_a+r_b\dots) \pmod {17}$ but it is getting more messier.
Please explain in detail. And also mention the formula being used.
| $$\frac{4^{41}-1}{4-1}\equiv \frac{2^{82}-1}{4-1}\equiv \frac{2^2-1}{4-1}\equiv 1\pmod{17}.$$
Fermat's little theorem was used: $2^{16}\equiv 1\pmod{17}$ implies:
$$ 2^{82} = 2^{5\cdot 16+2} = 4\cdot \left(2^{16}\right)^5 \equiv 4\pmod{17}.$$
Another approach. Let $a_n = 4^0+4^1+\ldots+4^n$. Then obviously $a_{n+1}=4a_n+1$, so given $a_n\pmod{17}$, to compute $a_{n+1}\pmod{17}$ is straightforward. $a_0=1$, so our sequence $\pmod{17}$ goes this way:
$$ 1 \to 5 \to 4 \to 0 \to 1 \to 5 \to 4 \to \ldots $$
hence it is $4$-periodic. That implies $a_{40}\equiv a_{0}\equiv 1\pmod{17}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?
| If you know a bit about complex numbers and roots of unity, here is another approach:
The $n^\text{th}$ roots of unity are vertices of a regular $n$-gon. Let $z=1$ denote $A_1$ and $w = e^{\frac{2\pi\iota}{n}}$ denote $A_2$ in the complex plane. Then $w^2$ denotes $A_3$ and $w^3$ denotes $A_4$. Now, the given equation can be written as
$$\frac{1}{|1-w|} = \frac{1}{|1-w^2|} + \frac{1}{|1-w^3|}$$
Multiplying with $|1-w|$ throughout, we get
$$1 = \frac{1}{|1+w|} + \frac{1}{|1+w+w^2|}$$
Now since $|w|=1$, pulling out $|w|$ from the denominator of the second term in the R.H.S,
$$1 = \frac{1}{|1+w|} + \frac{1}{|1+w+\bar{w}|}$$
Notice that $w+\bar{w}$ is purely real and can be written $w+\frac{1}{w}$, hence
$$1 - \frac{w}{1+w+w^2} = \frac{1+w^2}{1+w+w^2} = \frac{1}{|1+w|}$$
Squaring both sides and using $|z|^2 = z\bar{z}$,
$$\Bigg(\frac{1+w^2}{1+w+w^2}\Bigg)^2 = \frac{1}{(1+w)} \cdot \frac{1}{(1+\frac{1}{w})}= \frac{w}{(1+w)^2}$$
Cross multiply,
$$(1+w+w^2+w^3)^2 = w (1+w+w^2)^2$$
and simplify to get
$$1+w+w^2+w^3+w^4+w^5+w^6 = 0$$
Therefore,
$$1-w^7 =0$$
and $w \neq 1$, so $w$ must be a $7$th root of unity. Argue that $n =7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
Find the rank of matrix. "Find rank of matrix $\begin{bmatrix}
0&0&-3\\
9&3&5\\
3&1&1\\
\end{bmatrix}$
using echelon form?"
"I am getting an answer equal to 2."
"When I reduce this I get $\begin{bmatrix}
1&1/3&0\\
0&0&1\\
0&0&0\\
\end{bmatrix}$ so two pivots so rank must me 2 right ? However the book I am using says rank =1."
| It is indeed a rank 2 matrix.
$\begin{bmatrix}
0&0&-3\\
9&3&5\\
3&1&1\\
\end{bmatrix}\overset{r_1\leftrightarrow r_3}{\rightarrow}
\begin{bmatrix}
3&1&1\\
9&3&5\\
0&0&-3\\
\end{bmatrix}
\overset{r_2=r_2-3\times r_1}{\rightarrow}
\begin{bmatrix}
3&1&1\\
0&0&2\\
0&0&-3\\
\end{bmatrix}
\overset{r_3=r_3+\frac{3}{2}\times r_2}{\rightarrow}
\begin{bmatrix}
3&1&1\\
0&0&2\\
0&0&0\\
\end{bmatrix}
\overset{r_1=r_1/3}{\rightarrow}
\begin{bmatrix}
1&\frac{1}{3}&\frac{1}{3}\\
0&0&2\\
0&0&0\\
\end{bmatrix}
\overset{r_2=r_2/2}{\rightarrow}
\begin{bmatrix}
1&\frac{1}{3}&\frac{1}{3}\\
0&0&1\\
0&0&0\\
\end{bmatrix}
\overset{r_1=r_1-r_2/3}{\rightarrow}
\begin{bmatrix}
1&\frac{1}{3}&0\\
0&0&1\\
0&0&0\\
\end{bmatrix}
$
Your book has made a mistake.
P.S. Welcome to Math.SE! Next time please learn to use some basic MathJaX to prevent people from downvoting you and misunderstanding you. You may start from this tutorial on Math.SE.Meta.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determining the minimal polynomial over $\Bbb{Q}$ I was working on a homework assignment from Hungerford:
Find the minimal polynomial of the element $\sqrt{1+\sqrt{5}}$ over $\Bbb{Q}$.
Naturally the solution would be the polynomial with roots
$$ \pm \sqrt{1 \pm \sqrt{5}} $$
Which is found as
$$ x = \pm \sqrt{1 \pm \sqrt{5}} \rightarrow x^2 -1 = \pm \sqrt{5} \rightarrow (x^2-1)^2 = 5 \rightarrow $$
$$\text{Minimal Polynomial = } (x^2-1)^2-5$$
Problem is, I quite frankly don't know how to prove this. Hypothetically what if, in the same vein that
$$ \sqrt{5 + 2\sqrt{6}} = \sqrt{2} + \sqrt{3}$$
There exists some subtle factorization for $\sqrt{1 + \sqrt{5}}$ with the additionaly property that it is the root of a cubic or quadratic. How do I definitely rule out any of those cases?
| It is enough to show that $(x^2-1)^2-5$, that is, $x^4-2x^2-4$ is irreducible over the rationals. There are no rational roots, so we only need to rule out factorization as a product of quadratics with integer coefficients.
Since there is no $x^3$ term. we can confine attention to factorizations $(x^2+ax+b)(x^2-ax+c)$. If $a=0$, we get $b+c=-2$, $bc=-4$, impossible in integers. If $a\ne 0$, then because there is no $x$ term we need $b=c$, but then $bc$ cannot be $-4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solve the complex equation The equation is $$z^2 -4z +4+ 2i = 0$$
I know that i am supposed to use $$(a+bi)^2 = a^2 + 2abi + bi^2$$ to solve the equation but i am stuck on how to expand the equation.
Can you help out with which term to expand?
| Notice, we have $$z^2-4z+4+2i=0$$
$$(z-2)^2+2i=0$$
$$(z-2)^2=-2i\iff (z-2)^2=2i^3$$
$$z-2=\sqrt{2}i^{3/2}$$ Since, $i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}$, hence we get $$z-2=\sqrt{2}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{3/2}=\sqrt{2}\left(\cos\left(2k\pi+\frac{\pi}{2}\right)+i\sin\left(2k\pi+\frac{\pi}{2}\right)\right)^{3/2}$$
$$=\sqrt{2}\left(\cos\left(\frac{4k\pi+\pi}{2}\right)+i\sin\left(\frac{4k\pi+\pi}{2}\right)\right)^{3/2}$$
$$z-2=\sqrt{2}\left(\cos\frac{3(4k\pi+\pi)}{4}+i\sin\frac{3(4k\pi+\pi)}{4}\right)$$
Now, setting $k=0$, we get first root as follows
$$z=\sqrt{2}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)+2$$
$$=\sqrt{2}\left(-\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}\right)+2$$
$$=-1+i+2=1+i$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{z=1+i}}$$
Now, setting $k=1$, we get second root as follows
$$z=\sqrt{2}\left(\cos\frac{15\pi}{4}+i\sin\frac{15\pi}{4}\right)+2$$
$$=\sqrt{2}\left(\frac{1}{\sqrt2}-i\frac{1}{\sqrt2}\right)+2$$
$$=1-i+2=3-i$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{z=3-i}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Give the equation of a line that passes through the point (5,1) that is perpendicular and parallel to line A. The equation of line $A$ is $3x + 6y - 1 = 0$. Give the equation of a line that passes through the point $(5,1)$ that is
*
*Perpendicular to line $A$.
*Parallel to line $A$.
Attempting to find the parallel,
I tried $$y = -\frac{1}{2}x + \frac{1}{6}$$
$$y - (1) = -\frac{1}{2}(x-5)$$
$$Y = -\frac{1}{2}x - \frac{1}{10} - \frac{1}{10}$$
$$y = -\frac{1}{2}x$$
| Line perpendicular to A that passes through the point $(5,1)$: it is directed by the normal vector to A: $(3,6)$ \ or $(1,2)$. Hence its equation is
$$\frac{x-5}1=\frac{y-1}2\iff 2x-y-9=0. $$
Line parallel to A:it has the smame normal vector as A:
$$3x+6y=3\cdot 5+6\cdot 1\iff 3x+6y-21=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Perpendicular Bisector of Made from Two Points For a National Board Exam Review:
Find the equation of the perpendicular bisector of the line joining
(4,0) and (-6, -3)
Answer is 20x + 6y + 29 = 0
I dont know where I went wrong. This is supposed to be very easy:
Find slope between two points:
$${ m=\frac{y^2 - y^1}{ x^2 - x^1 } = \frac{-3-0}{-6-4} = \frac{3}{10}}$$
Obtain Negative Reciprocal:
$${ m'=\frac{-10}{3}}$$
Get Midpoint fox X
$${ \frac{-6-4}{2} = -5 }$$
Get Midpoint for Y
$${ \frac{-0--3}{2} = \frac{3}{2} }$$
Make Point Slope Form:
$${ y = m'x +b = \frac{-10}{3}x + b}$$
Plugin Midpoints in Point Slope Form
$${ \frac{3}{2} = \frac{-10}{3}(-5) + b}$$
Evaluate b
$${ b = \frac{109}{6}}$$
Get Equation and Simplify
$${ y = \frac{-10}{3}x + \frac{109}{6}}$$
$${ 6y + 20x - 109 = 0 }$$
Is the problem set wrong? What am I doing wrong?
| To find the midpoint, you don't need to negate the coordinates. So the midpoint is $\left(\frac{-6+4}{2}, \frac{0+(-3)}{2}\right)=\left(-1, \frac{-3}{2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
The asymptotic of the number of integers that are sums of three nonnegative cubes Let $c(n) $ be the number of distinct integers between $0 $ and $n $ of the form $ a^3 + b^3 + c^3$, meaning the sum of $3$ nonnegative cubes.
$C(n) = O( n \space \ln(n)^x ) $
Find and prove the optimal real value of $x$.
| Maybe this could help:
The amount of cubes lower than or equal to $n$ is $\lfloor\sqrt[3]{n}\rfloor$.
If we just have $A^3+B^3$ instead of $A^3+B^3+C^3$ we get:
$$C(n)=\dfrac{1}{2}\sum_{i=1}^{\lfloor\sqrt[3]{n-1}\rfloor}\lfloor\sqrt[3]{n-i^3}\rfloor$$
And for $A^3+B^3+C^3$ (which is the result you want):
$$C(n)=\dfrac{1}{6}\sum_{i=1}^{\lfloor\sqrt[3]{n-2}\rfloor}\sum_{j=1}^{\lfloor\sqrt[3]{n-i^3-1}\rfloor}\lfloor\sqrt[3]{n-i^3-j^3}\rfloor$$
This formula basically just checks whether there exists a $C$ for every combination of $A$ and $B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Alternative area of a triangle formula The problem is as follows:
There is a triangle $ABC$ and I need to show that it's area is: $$\frac{1}{2} c^2 \frac{\sin A \sin B}{\sin (A+B)}$$
Since there is a half in front I decided that base*height is equivalent to $c^2 \frac{\sin A \sin B}{\sin (A+B)}$. So I made an assumption that base is $c$ and went on to prove that height is $c \frac{\sin A \sin B}{\sin (A+B)}$. But I end up expressing height in terms of itself.. i.e. $h \equiv \frac{ch}{a\cos B + b \cos A}$. How do I prove this alternative area of triangle formula?
| Let's denote by $[ABC]$ the area of $\triangle ABC$, its known that $$[ABC]=\frac{1}{2}ab\sin C$$
From Sine Law we have $a\sin C=c\sin A$ and $b\sin C=c\sin B$, also $\sin (A+B)=\sin(\pi-C)=\sin C$, then
\begin{align*}
[ABC]&=\frac{1}{2}\frac{(a\sin C)(b\sin C)}{\sin C}\\
&=\frac{1}{2}\frac{(c\sin A)(c\sin B)}{\sin (A+B)}\\
&=\frac{1}{2}c^2\frac{\sin A\sin B}{\sin (A+B)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find $\int\frac{x^2+x}{(e^x+x+1)^2}dx$ $\int\frac{x^2+x}{(e^x+x+1)^2}dx$
I tried solving it but could not finish.I tried putting $e^x=t$ but not getting integrable.Please help me in solving it.
| \begin{align}
\int\frac{x(x+1)}{(1+e^x+x)^2}dx&=\int\frac{(1+e^x+x)^2-(1+e^x)(1+e^x+x)-xe^x}{(1+e^x+x)^2}dx\\
&=\int1dx-\int\frac{1+e^x}{(1+e^x+x)}dx-\int\frac{xe^x}{(1+e^x+x)^2}dx\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\int \frac{4x^5-7x^4+8x^3-2x^2+4x-7}{x^2(x^2+1)^2}dx$ $\displaystyle\int \frac{4x^5-7x^4+8x^3-2x^2+4x-7}{x^2(x^2+1)^2}dx$
I attempted but in vain.
$\displaystyle\int \frac{4x^5-7x^4+8x^3-2x^2+4x-7}{x^2(x^2+1)^2}dx=\int \frac{4x(x^4+1)}{x^2(x^2+1)^2}-7\frac{(x^4+1)}{x^2(x^2+1)^2}+2\frac{x^2(4x-1)}{x^2(x^2+1)^2}dx$
I got stuck. Please help....
| By partial fractions decomposition we have $$\int\left(\frac{12}{\left(1+x^{2}\right)^{2}}-\frac{7}{x^{2}}+\frac{4}{x}\right)dx
$$ now for the first integral we can use the substitution $x=\tan\left(u\right)
$ to get $$12\int\frac{dx}{\left(1+x^{2}\right)^{2}}=12\int\cos^{2}\left(u\right)du=6\int\cos\left(2u\right)du+6u=3\sin\left(2u\right)+6u+C_{1}
$$ for the second $$-7\int\frac{1}{x^{2}}dx=\frac{7}{x}+C_{2}
$$ and $$4\int\frac{1}{x}dx=4\log\left(x\right)+C_{3}
$$ so if we substitute back we have $$3\sin\left(2\tan^{-1}\left(x\right)\right)+6\tan^{-1}\left(x\right)+\frac{7}{x}+4\log\left(x\right)+C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving a Complicated Trig Problem I am preparing for AIME and I came across this problem which I need help solving:
$$\begin{eqnarray} 10^{10^{10}} \sin\left( \frac{109}{10^{10^{10}}} \right) - 9^{9^{9}} \sin\left( \frac{101}{9^{9^{9}}} \right) - 8^{8^{8}} \sin\left( \frac{17}{8^{8^{8}}} \right) + 7^{7^{7}} \sin\left( \frac{76}{7^{7^{7}}} \right) + 6^{6^{6}} \sin\left( \frac{113}{6^{6^{6}}} \right) \end{eqnarray} $$
There must be some simple way to do this for sure, but I don't know how to.
| Since $\sin x\approx x$ for small $x$, this should be pretty darn close to $$109−101−17+76+113=180.$$
How close? By far the largest error comes from the $6^{6^6}\sin\frac{113}{6^{6^6}}$ term, because that argument to the sine is hugely larger than the other arguments (power towers grow fast). And we can estimate the relative difference between $\sin\frac{113}{6^{6^6}}$ and $\frac{113}{6^{6^6}}$ by the mean value theorem: the relative error is between $0$ and $1-\cos\frac{113}{6^{6^6}}$. And we can again estimate the cosine as $1-\cos x\approx \frac12x^2$ for small $x$, so the absolute difference from 180 should be close to
$$ 113 \cdot \frac12 \cdot \frac{113^2}{6^{2\cdot 6^6}} \approx \frac{113^3}{2\cdot 6^{2\cdot 6^6}} $$
where the $6^{2\cdot 6^6}$ in the denominator dominates absolutely everything.
Since $\log_{10}(6^{2\cdot 6^6}) = \log_{10}(6)\cdot 2\cdot 6^6 > 6^6 = 46656$, the answer is
$$ 179.\underbrace{999999\ldots999999}_{\text{at least 46,650 nines}}\ldots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$\int_{0}^{\infty}\frac{\ln x dx}{x^2+2x+2}$ $$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}$$
$$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}=\int_{0}^{\infty}\frac{\ln x .dx}{(x+1)^2+1}\\
=\ln x\int_{0}^{\infty}\frac{1}{(x+1)^2+1}-\int_{0}^{\infty}\frac{1}{x}\frac{1}{(x+1)^2+1}dx$$ and then lost track,answer is $\frac{\pi \ln 2}{8}$.
Any hint will solve my problem.
| Let $$\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2+2x+2}dx = \int_{0}^{\infty}\frac{\ln x}{(x+1)^2+1^2}dx$$
Put $(x+1) = \tan \theta \;,$ Then $dx = \sec^2 \theta d\theta$ and changing Limits
We get $$\displaystyle I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\ln\left(\tan \theta - 1\right)}{1+\tan^2 \theta }\cdot \sec^2 \theta d\theta = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln \left(\sin \theta -\cos \theta \right)d\theta -\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln \cos \theta d\theta$$
So $$\displaystyle I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln\left(\sqrt{2}\cdot \sin \left(\theta-\frac{\pi}{4}\right)\right)d\theta-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln\cos \theta d\theta$$
$$\displaystyle I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln\sqrt{2}d\theta+\int_{0}^{\frac{\pi}{4}}\ln\sin \left(\theta\right)d\theta -\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln \cos \theta d\theta $$
Now In Second Integral $\displaystyle \theta = \left(\frac{\pi}{2}-\phi\right)\;,$ then $d\theta = -d\phi$ and Changing Limit
We Get $$\displaystyle I = \frac{\pi}{8}\ln 2-\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\ln \cos \phi d\phi-\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\ln \cos \theta d\theta$$
So $$\displaystyle I = \frac{\pi}{8}\ln 2+\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}\ln \cos \theta d\theta-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln \cos \theta d\theta = \frac{\pi}{8}\ln 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 - 2\sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int -2 e^x \sin^2x \mathrm{d}x + 2 e^x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x -4 \int e^x \sin^2x \mathrm{d}x \end{align*}
First two steps use integration by parts. In the first step we differentiate $\sin^2 x$. In the second step we differentiate $\sin x \cos x$.
Using this, we reach $$5\int e^x \sin^2 x \mathrm{d}x = e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x$$
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$
I can't reach the form that most integral calculators give, which has terms $\cos(2x)$ and $\sin(2x)$ by just using trig identities, so I wonder whether the result is correct. I would also be interested in a method that immediately gives the form $$-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$
| You can use the Reduction formula:
$$I_n=\int e^{ax}\sin^n bx\mathrm. dx\\
=\frac{e^{ax}\sin^{n-1} bx (a\sin bx-nb\cos bx)}{a^2+n^2b^2}+\frac{n(n-1)b^2}{a^2+n^2b^2}I_{n-2}$$
Use $n=2,a=1,b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
limit involving $e$, ending up without $e$. Compute the limit
$$ \lim_{n \rightarrow \infty} \sqrt n \cdot \left[\left(1+\dfrac 1 {n+1}\right)^{n+1}-\left(1+\dfrac 1 {n}\right)^{n}\right]$$
we have a bit complicated solution using Mean value theorem. Looking for others
| Clearly if $a_{n} = \left(1 + \dfrac{1}{n}\right)^{n}$ then we have
\begin{align}
L &= \lim_{n \to \infty}\sqrt{n}\{a_{n + 1} - a_{n}\}\notag\\
&= \lim_{n \to \infty}\sqrt{n}\left[\exp\left\{(n + 1)\log\left(1 + \frac{1}{n + 1}\right)\right\} - \exp\left\{n\log\left(1 + \frac{1}{n}\right)\right\}\right]\notag\\
&= \lim_{n \to \infty}\sqrt{n}\left(1 + \frac{1}{n}\right)^{n}\left[\exp\left\{(n + 1)\log\left(1 + \frac{1}{n + 1}\right) - n\log\left(1 + \frac{1}{n}\right)\right\} - 1\right]\notag\\
&= e\lim_{n \to \infty}\sqrt{n}\cdot t\cdot\frac{e^{t} - 1}{t}\text{ (where }t = (n + 1)\log\left(1 + \frac{1}{n + 1}\right) - n\log\left(1 + \frac{1}{n}\right) \to 0)\notag\\
&= e\lim_{n \to \infty}\sqrt{n}\left\{(n + 1)\log\left(1 + \frac{1}{n + 1}\right) - n\log\left(1 + \frac{1}{n}\right)\right\}\notag\\
&= e\lim_{n \to \infty}\sqrt{n}\left\{\log\left(1 + \frac{1}{n + 1}\right) + n\log\left(\frac{n(n + 2)}{(n + 1)^{2}}\right)\right\}\notag\\
&= e\lim_{n \to \infty}\sqrt{n}\left\{\log\left(1 + \frac{1}{n + 1}\right) + n\log\left(1 - \frac{1}{(n + 1)^{2}}\right)\right\}\notag\\
&= e\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left\{n\log\left(1 + \frac{1}{n + 1}\right) + n^{2}\log\left(1 - \frac{1}{(n + 1)^{2}}\right)\right\}\notag\\
&= e\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left\{\frac{n}{n + 1}\cdot(n + 1)\log\left(1 + \frac{1}{n + 1}\right) + \frac{n^{2}}{(n + 1)^{2}}\cdot(n + 1)^{2}\log\left(1 - \frac{1}{(n + 1)^{2}}\right)\right\}\notag\\
&= e\cdot 0\{1\cdot 1 - 1\cdot 1\}\notag\\
&= 0\notag
\end{align}
This is done without Taylor/Mean Value Theorem (essentially without using derivatives). If $\sqrt{n}$ is replaced by $n^{2}$ (as mentioned in Lucian's comment to the original question) then we will need these higher level tools.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Linear equations (solving by substitution) I'm having trouble solving this equation with fractions.
$$\frac 23 x+\frac 16 y=\frac 23 \\
-y=12-2x$$
| Notice, we have $$\frac{2}{3}x+\frac{1}{6}y=\frac{2}{3}\tag 1$$ $$-y=12-2x\tag 2$$
$$\implies y=2x-12$$ setting this value of $y$ in the eq(1), we get $$\frac{2}{3}x+\frac{1}{6}(2x-12)=\frac{2}{3}$$
$$2x+x-6=2$$ $$3x=2+6=8$$ $$x=\frac{8}{3}$$ Setting the value of $x$ in (2), we get $$-y=12-2\frac{8}{3}=\frac{36-16}{3}=\frac{20}{3}$$ $$y=-\frac{20}{3}$$
Hence, we get
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x=\frac{8}{3}, \ y=-\frac{20}{3}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the distance from the origin to the surface $xy^2 z^4 = 32$ using the method of Lagrange Multipliers Problem: Find the distance from the origin to the surface $xy^2z^4 = 32$.
Attempt: The Lagrange equation for this problem is $L(x,y,z, \lambda) = x^2 + y^2 + z^2 + \lambda (xy^2 z^4 - 32)$. Setting the first partials to zero we have \begin{align*} \frac{\partial L}{\partial x} &= 2x + \lambda y^2 z^4 = 0 \qquad (1) \\ \frac{\partial L}{\partial y} &= 2y + 2 \lambda x y z^4 = 0 \qquad (2) \\ \frac{\partial L}{\partial z} &= 2z + 4 \lambda x y^2 z^3 = 0 \qquad (3) \\ \frac{\partial L}{\partial \lambda} &= xy^2 z^4 - 32 = 0 \qquad (4)
\end{align*}
Now I'm having a hard time solving this system for $x,y$ and $z$. Here is what I did so far. From $(1)$ and $(2)$ we get \begin{align*} \frac{2x}{y^2 z^4} = - \lambda \qquad \text{and} \qquad \frac{1}{xz^4} = - \lambda \end{align*} Thus $\frac{2x}{y^2 z^4} = \frac{1}{xz^4} $ or $y^2 = 2x^2$ after simplification. Also, from $(2)$ and $(3)$ we can deduce that \begin{align*} \frac{1}{xz^4} = - \lambda = \frac{2z}{4xy^2 z^3} \end{align*} so that $2y^2 = z^2$ after simplification. Now I used all this and substituted it into $(4)$. This gave me \begin{align*} x(2x^2) (4y^4) - 32 = 0 \end{align*} or (since $y^4 = 4x^4)$ \begin{align*} 8x^3 (4x^4) - 32 = 0 \end{align*}
This means that $32x^7 - 32 = 0$, so that $x = 1$. Then $y^2 = 2$, so that $y = \pm \sqrt{2}$. Then $z^2 = 4$, so that $z = \pm 2$. So I found the points $(x,y,z) = (1, \sqrt{2}, 2)$ and $(1, - \sqrt{2}, -2)$. They both give me the distance $\sqrt{x^2 + y^2 + z^2} = \sqrt{7}$, so I'm guessing they are equal? Is my reasoning correct?
| Your solution is correct, except the last part. You should get four minimizing points. (You mistakenly assumed that $y$ and $z$ must be both positive or both negative.)
However, there is a solution without using Lagrange multipliers. Note by AM-GM that
$$x^2+y^2+z^2=x^2+2\left(\frac{y^2}{2}\right)+4\left(\frac{z^2}{4}\right)\geq 7\sqrt[7]{x^2\left(\frac{y^2}{2}\right)^2\left(\frac{z^2}{4}\right)^4}=7\sqrt[7]{\left(\frac{xy^2z^4}{32}\right)^2}=7\,.$$
The equality holds iff $x^2=\frac{y^2}{2}=\frac{z^2}{4}$, which means $(x,y,z)=\big(1,\pm\sqrt{2},\pm 2\big)$. Hence, the required distance is $\sqrt{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $(a+b)(b+c)(c+a) \ge8$
Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$,
Prove that $(a+b)(b+c)(c+a)\ge8$.
My attempt: By AM-GM inequality, we have
$$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$
and similarly
$$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$
$$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$
Multiplying $(1)$, $(2)$ and $(3)$ together we reach
$$(a+b)(b+c)(c+a) \ge8abc.$$
Now, I need to show that $abc = 1$.
Again by AM-GM inequality we have
$$\frac{a+b+c}{3}\ge\sqrt[3]{abc}
\implies \frac{\frac{3}{abc}}{3}\ge\sqrt[3]{abc}
\implies abc\le1$$
Now if we show somehow that $abc\ge1$, we are done and that's where I am stuck. Can someone please explain how to show that?
Other solutions to the above question are also welcomed.
| If the numbers are positive here is a solution
$$3=abc(a+b+c)\ge 3abc(abc)^{\dfrac 1 3} \Rightarrow abc\le 1 \\$$
$$3=abc(a+b+c)\le \dfrac {(a+b+c)^3} {27} \cdot (a+b+c) \Rightarrow a+b+c \ge 3 \\ $$
$$ (ab+bc+ca)^2 \ge 3abc(a+b+c)=9 \Rightarrow ab+bc+ca \ge 3$$
$$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc \ge 9-1=8$$
equality holds if and only if $a=b=c=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
An arctan integral $\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$ According to Mathematica, we have that
$$\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx=\pi \left(\frac{\log (2)}{2}-\frac{1321}{6144}\right)$$
that frankly speaking looks pretty nice.
However Mathematica shows that
$$\int \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$$
$$=-\frac{1}{2} i \text{Li}_2\left(e^{2 i \tan ^{-1}(x)}\right)-\frac{1}{2} i \tan ^{-1}(x)^2+\tan ^{-1}(x) \log \left(1-e^{2 i \tan ^{-1}(x)}\right)-\frac{65}{256} \sin \left(2 \tan ^{-1}(x)\right)-\frac{23 \sin \left(4 \tan ^{-1}(x)\right)}{1024}-\frac{5 \sin \left(6 \tan ^{-1}(x)\right)}{2304}-\frac{\sin \left(8 \tan ^{-1}(x)\right)}{8192}+\frac{65}{128} \tan ^{-1}(x) \cos \left(2 \tan ^{-1}(x)\right)+\frac{23}{256} \tan ^{-1}(x) \cos \left(4 \tan ^{-1}(x)\right)+\frac{5}{384} \tan ^{-1}(x) \cos \left(6 \tan ^{-1}(x)\right)+\frac{\tan ^{-1}(x) \cos \left(8 \tan ^{-1}(x)\right)}{1024}$$
and this form doesn't look that nice.
Having given the nice form of the closed form I wonder if we can find a very nice and simple way of getting the answer. What do you think?
A supplementary question:
$$\int_0^{\infty } \frac{\arctan^2(x)}{x \left(x^2+1\right)^5} \, dx=\frac{55}{108}-\frac{1321}{12288}\pi^2+\frac{\pi^2}{4} \log (2)-\frac{7 }{8}\zeta (3)$$
| Does it look any nicer?
$$\int\frac{\arctan x}{x\,\left(x^2+1\right)^5}\, dx=\frac12\,\Im\operatorname{Li}_2\left(e^{2\,i\arctan x}\right)\\-\frac x{9216\,\left(x^2+1\right)^4}\left(3963x^6+12995x^4+14525x^2+5637\right)\\+\left[\ln\frac{2x}{\sqrt{x^2+1}}+\frac{12x^6+42x^4+52x^2+25}{24\,\left(x^2+1\right)^4}-\frac{1321}{3072}\right]\cdot\arctan x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
} |
Binomial sum with two parameters Let $m$ and $n$ be two integers. Evaluate
$$S_{m,n}=\sum_{j=0}^{m} (-1)^j \binom{m}{j}\binom{mn-jn}{m+1}$$
At first, for $n=2$ I got $S_{m,2}=2^{m-1}m$, for $n=3$ I obtained $S_{m,3}=3^m m$, then I tried in vain to prove by induction that
$$S_{m,n}=\frac{n^k k(n-1)}{2}$$
| Suppose we seek to evaluate
$$S(m,n) = \sum_{j=0}^m (-1)^j {m\choose j}
{mn-jn\choose m+1}.$$
Introduce
$${mn-jn\choose m+1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+2}} (1+z)^{mn-jn} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+2}} (1+z)^{mn}
\sum_{j=0}^m
(-1)^j {m\choose j} \frac{1}{(1+z)^{jn}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+2}} (1+z)^{mn}
\left(1-\frac{1}{(1+z)^n}\right)^m \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+2}} ((1+z)^n-1)^m \; dz.$$
This is
$$[z^{m+1}] ((1+z)^n-1)^m
= [z^{m+1}] \left(nz+{n\choose 2}z^2+\cdots\right)^m.$$
This evaluates by inspection to
$${m\choose 1} n^{m-1} {n\choose 2}
= \frac{1}{2} n^m (n-1) m$$
because from the $m$ terms being multiplied we must choose one to be
$z^2$ to ge a total of $m+1$ and there are $m-1$ terms in $z$ with
coefficient $n.$
Observe that this formula produces zero when $n=1$ which is correct as well (second binomial coefficient is zero).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$ $\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$
Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$
$\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different way to solve it?
| Let $$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}}{x^2-x+1}dx.............(1)$$
Now Replace $x\rightarrow (1-x)\;,$ We get
$$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{1-x}}{(1-x)^2-(1-x)+1}dx = \int_{0}^{1}\frac{\sin^{-1}\sqrt{1-x}}{x^2-x+1}dx$$
So we get $$\displaystyle I = \int_{0}^{1}\frac{\cos^{-1}\sqrt{x}}{x^2-x+1}dx...............(2)$$
Above we have use the formula $$\bullet\; \displaystyle \sin^{-1}(\sqrt{x})+\sin^{-1}\sqrt{1-x} = \sin^{-1}(\sqrt{x})+\cos^{-1}(\sqrt{x}) = \frac{\pi}{2}$$
Now Add these two equations, We get
$$\displaystyle 2I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}+\cos^{-1}\sqrt{x}}{x^2-x+1} = \frac{\pi}{2}\int_{0}^{1}\frac{1}{x^2-x+1}dx = \frac{\pi}{2}\int_{0}^{1}\frac{1}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx$$
Now Put $\displaystyle \left(x-\frac{1}{2}\right) = t$ and $dx = dt$ and Changing Limt, We get
$$\displaystyle 2I = \int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{t^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx = 2\cdot \frac{\pi}{2}\int_{0}^{\frac{1}{2}} \frac{1}{t^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx = \frac{2\pi}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi^2}{3\sqrt{3}}$$
So we get $$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}}{x^2-x+1}dx = \frac{\pi^2}{6\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$ $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$
My Attempt:
$$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt x\sqrt{1+x(1+x)}}$$
Replacing $x$ by $1-x$,we get
$$\int_{0}^{1}\frac{x}{2-x}.\frac{dx}{\sqrt{1-x}\sqrt{1+(1-x)(2-x)}}$$
Then I got stuck. Please help.
| Hint:
$$I =\int {\frac{{1 - x}}{{1 + x}}\frac{{dx}}{{\sqrt {x + {x^2} + {x^3}} }}} = \int {\frac{{1 - x}}{{1 + x}}\frac{{dx}}{{\sqrt x \sqrt {1 + x + {x^2}} }}} $$
Making $u=\sqrt{x}$ gives
$$I=2\int {\frac{{1 - {u^2}}}{{1 + {u^2}}}\frac{{du}}{{\sqrt {1 + {u^2} + {u^4}} }}} = 2\int {\frac{{\frac{1}{{{u^2}}} - 1}}{{\frac{1}{u} + u}}\frac{{du}}{{\sqrt {\frac{1}{{{u^2}}} + 1 + {u^2}} }}} = - 2\int {\frac{1}{{\frac{1}{u} + u}}\frac{{d(\frac{1}{u} + u)}}{{\sqrt {{{(\frac{1}{u} + u)}^2} - 1} }}}$$
I am sure you can do it now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the rate of change in height of water level. A cone shaped container has a diameter of $0.6m$ and height of $0.5m$. Water is poured into the container with a constant rate of $0.2m^{3}s^{-1}$.
Calculate the rate of change in height of the water level when its height reaches $0.4m$.
Ok, so $\frac {dV}{dt} = 0.2$
V = $\frac 1 3 \pi r^2h$
I want to find $\frac {dh}{dt}$
so can it be found by doing this?
$\frac {dV}{dt} = \frac{dV}{dh} . \frac{dh}{dt}$
if so, then how can $\frac {dV}{dh}$ be calculated?
$\frac {dV}{dh} = \frac1 3 \pi r^2$
This is as far as I can go. I don't know the value of $r$ when the height is $0.4m$.
Update:
Since $\tan \theta = \frac {0.3}{0.5}$
$\theta = \tan^{-1} 0.6 = 30.96$
When, $ h = 0.4$
$r = \tan (30.96) \times 0.4 = 0.24$
$\frac {dV}{dh} = \frac 1 3 \pi (0.24)^2 = 0.0603$
So, $\frac {dh}{dt} = \frac{0.2}{0.0603} = 3.316$
But the real answer is $1.105$. Could it be that $\frac{dV}{dh} = \pi r^2$ and not $\frac 1 3 \pi r^2$ ? Or have I made some seriously wrong calculations?
| Using similar triangles, we have that $\frac{r}{h}=\frac{.3}{.5}$, so $r=\frac{3}{5}h$.
Then $\displaystyle V=\frac{1}{3}\pi r^2h=\frac{\pi}{3}\big(\frac{3}{5}h\big)^2h=\frac{3\pi}{25}h^3,\;\;$ so $\displaystyle\frac{dV}{dt}=\frac{9\pi}{25}h^2\frac{dh}{dt}$.
When $h=.4$, this gives $\displaystyle.2=\frac{9\pi}{25}\big(.4\big)^2\frac{dh}{dt},\;\;$ so $\displaystyle\frac{dh}{dt}=\frac{125}{36\pi}\approx1.105 \;m/s$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
| Using Hermite reduction and the Rothstein-Trager Algorithm for integration of rational functions I get
$$\frac{2}{\sqrt{11}}\left(\arctan\left(\frac{3}{\sqrt{11}}+\frac{2}{\sqrt{11}}x\right)-\arctan\left(\frac{3}{\sqrt{11}}-\frac{8}{\sqrt{11}}x-\frac{6}{\sqrt{11}}x^2-\frac{2}{\sqrt{11}}x^3\right)\right)
$$
which seems correct after differentiating. Its strange that Wolfram Alpha doesn't get this result.
Here some details:
Let $A = x^2+1$ and $D = x^4+3x^3+3x^2-3x+1$, so that we want to find $\int{\frac{A}{D}}{dx}$
Since $D$ is square free (contains no multiple factors) and $\deg(A) < \deg(D)$, we can apply the Rothstein-Trager method described in the paper below to compute this integral:
First we compute the resultant
\begin{align}
\text{res}_x(D, A-t\frac{d}{dx}D) &= \text{res}_x(x^4+3x^3+3x^2-3x+1,-4tx^3+(1-9t)x^2-6tx+3t+1)\\ &= 37(1+11t)^2 =: R(t)
\end{align}
It has two distinct roots $t_1 = \frac{i}{\sqrt{11}}$ and $t_2 = -\frac{i}{\sqrt{11}} = -t_1$
We now can express the integral as a sum of complex logarithms:
\begin{align}
\int{\frac{A}{D}}{dx} &= \sum_{t|R(t) = 0}t\log\left(\gcd\left(D, A-t\frac{d}{dx}D\right)\right)\\ &= \sum_{t|R(t) = 0}t\log\left(-1+\left(\frac{3}{2}+\frac{11}{2}t\right)x+x^2\right) \\ &= \frac{i}{\sqrt{11}}\log\left(-1+\left(\frac{3}{2}+\frac{\sqrt{11}}{2}i\right)x+x^2\right)-\frac{i}{\sqrt{11}}\log\left(-1+\left(\frac{3}{2}-\frac{\sqrt{11}}{2}i\right)x+x^2\right)
\end{align}
Note that the GCD inside the logs was taken over $\mathbb{Q}(t_1)[x]$, rather than just over $\mathbb{Q}[x]$.
Having expressed the integral as a sum of logarithms, the only thing that we have to do, is to convert them to inverse tangents. For this, there also exist algorithms.
See for this "Symbolic Integration I: Transcendental Functions" by Manuel Bronstein.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$
I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I want until the $x^4$ term. So what i found is the following:
$$ (1+e^x)^{-1} = 1 - e^x + e^{2x} + e^{3x} + e^{4x}+\cdots\tag1 $$
Then by expanding each $e^{nx}$ term individually:
$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+\cdots\tag2 $$
$$ e^{2x} = 1 + 2x + 2x^2 + \frac{3x^3}{2} + \frac{2x^4}{3}+\cdots\tag3 $$
$$ e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \frac{27x^4}{8}+\cdots\tag4 $$
$$ e^{4x} = 1 + 4x + 8x^2 + \frac{32x^3}{3} + \frac{32x^4}{3}+\cdots\tag5 $$
So substituting $(2),(3),(4),(5)$ into $(1)$ i get:
$$ (1+e^x)^{-1} = 1 +2x+5x^2+ \frac{22x^3}{3} + \frac{95x^4}{12} +\cdots$$
Which isn't the correct result. Am i not allowed to expand this series binomially? I've seen on this site that this is an asymptotic expansion. However, i don't know about these and i haven't been able to find much information on this matter to solve this. If someone could help me understand how to solve it and why my approach isn't correct, i would be VERY grateful. Thanks in advance.
| Since you're just looking for the first few terms:
$$f(x)=\frac{1}{1+e^x} \implies f(0)=\frac 12$$
$$f'(x)=\frac{-e^x}{(1+e^x)^2} \implies f'(0)=-\frac 14$$
$$f''(x)=\frac{-e^x}{(1+e^x)^2}+2\frac{e^{2x}}{(1+e^x)^3}\implies f''(0)=0$$
$$f'''(x)=f''(x)+4\frac{e^{2x}}{(1+e^x)^3}-3\times 2\frac{e^x}{(1+e^x)^4}\implies f'''(0)=\frac 18$$
You get:
$$\frac{1}{1+e^x}=\frac 1{2\times 0!} -\frac 1{4\times 1!} x+\frac 1{8\times 3!}x^3+\mathcal O (x^5)$$
$$\frac{1}{1+e^x}=\frac 1{2} -\frac 1{4} x+\frac 1{48}x^3+ \mathcal O (x^5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
If $f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$ where $x$ and$y$ are independent variable.Find $f(x).$ If $f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$ where $x$ and$y$ are independent variable.Find $f(x).$
I tried to solve it.
$f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$
$f(x)=x+x\int_{0}^{1}y^2f(y)dy+x^2\int_{0}^{1}yf(y)dy$
I applied integration by parts but the expression did not simplify.Please guide me to reach the answer $f(x)=x+\frac{61}{119}x+\frac{80}{119}x^2.$
| $$
f(x)
= x + \int_{0}^{1}[xy^2+x^2y]\,f(y)\,dy
= x + x \int_{0}^{1}y^2 f(y)\,dy + x^2 \int_{0}^{1}y\,f(y)\,dy
= x + Ax + Bx^2
$$
where
$$
A = \int_{0}^{1}y^2 f(y)\,dy
\qquad\mbox{and}\qquad
B = \int_{0}^{1}y\,f(y)\,dy
$$
So,
$$
A = \int_{0}^{1}y^2 \underbrace{\left(y + Ay + By^2\right)}_{f(y)}\,\,dy
= \frac{A+1}{4} + \frac{B}{5}
\qquad\mbox{and}\qquad
B = \int_{0}^{1}y\, \underbrace{\left(y + Ay + By^2\right)}_{f(y)}\,\,dy
= \frac{A+1}{3} + \frac{B}{4}
$$
which can now be solved for $A$ and $B$, resulting in
$$
A = \frac{61}{119}
\qquad\mbox{and}\qquad
B = \frac{80}{119}
$$
Therefore,
$$
f(x) = x + Ax + Bx^2 = x + \frac{61}{119}\,x + \frac{80}{119}\,x^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to compute the monstrous $ \int_0^{\frac{e-1}{e}}{\frac{x(2-x)}{(1-x)}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{2-2x+x^2}dx} $ A friend told me, that he found a closed form for the following integral:
$$
\int_0^{\frac{e-1}{e}}{\frac{x(2-x)}{(1-x)}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{\left(2-2x+x^2\right)}dx}
$$
I don't know if he's just messing around with me, but I wonder if this integral admits a closed form. I tried to expand the $\log(\log)$ term into a power series, but things got worse. So any help will be appreciated!
| Notice, we have $$\int_{0}^{\frac{e-1}{e}}\frac{x(2-x)}{1-x}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{2-2x+x^2}dx$$
$$=\int_{0}^{\frac{e-1}{e}}\frac{x(2-x)}{1-x}\frac{\log\left(\log\left(\frac{2-2x+x^2}{2-2x}\right)\right)}{2-2x+x^2}dx$$
Let, $$\log\left(\frac{2-2x+x^2}{2-2x}\right)=u$$
$$\implies \frac{d}{dx}\left(\log\left(\frac{2-2x+x^2}{2-2x}\right)\right)=\frac{d}{dx}(u)$$ $$\frac{1}{\left(\frac{2-2x+x^2}{2-2x}\right)}\cdot \left(\frac{(2-2x)(-2+2x)-(2-2x+x^2)(-2)}{(2-2x)^2} \right)=\frac{du}{dx}$$
$$\left(\frac{2-2x}{2-2x+x^2}\right)\cdot \left(\frac{2x(2-x)}{(2-2x)^2} \right)=\frac{du}{dx}$$ $$\frac{x(2-x)}{(1-x)}\frac{1}{(2-2x+x^2)}dx=du$$ Now, we have $$\int_{0}^{\log\left(\frac{e^2+1}{2e}\right)}\log(u)du$$
$$=\left[u\log(u)-u\right]_{0}^{\log\left(\frac{e^2+1}{2e}\right)}$$
$$=\left[u\log\left(\frac{u}{e}\right)\right]_{0}^{\log\left(\frac{e^2+1}{2e}\right)}$$
$$=\log\left(\frac{e^2+1}{2e}\right)\cdot\log\left(\frac{1}{e}\log\left(\frac{e^2+1}{2e}\right)\right)-\lim_{u\to 0}u\log\left(\frac{u}{e}\right)$$
$$=\log\left(\frac{e^2+1}{2e}\right)\cdot\log\left(\frac{1}{e}\log\left(\frac{e^2+1}{2e}\right)\right)-0$$
Hence, we get
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int_{0}^{\frac{e-1}{e}}\frac{x(2-x)}{1-x}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{2-2x+x^2}dx}=\color{blue}{\log\left(\frac{e^2+1}{2e}\right)\cdot \log\left(\frac{1}{e}\log\left(\frac{e^2+1}{2e}\right)\right)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Does there exist a function that is differentiable everywhere with everywhere discontinuous partial derivatives? Does there exist a function that is differentiable everywhere with everywhere discontinuous partial derivatives?
I just had this doubt, talking about first order partials.
| Yes it does. Take $$
f\left( {x,y} \right) = \left\{ \begin{array}{l}
x^2 \sin \left( {\frac{1}{x}} \right) + y^2 \sin \left( {\frac{1}{y}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \ne 0 \\
x^2 \sin \left( {\frac{1}{x}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \ne 0,y = 0 \\
y^2 \sin \left( {\frac{1}{y}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0,y \ne 0 \\
0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0,y = 0 \\
\end{array} \right..
$$ Then, $f$ is differentiable everywere however its partial derivatives are discontinuous
$$f_x=
\left\{ \begin{array}{l}
2x\sin \left( {\frac{1}{x}} \right) - \cos \left( {\frac{1}{x}} \right),\,\,\,\,\,\,\,\,x \ne 0 \\
0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0 \\
\end{array} \right.
$$ and $$f_y=\left\{ \begin{array}{l}
2y\sin \left( {\frac{1}{y}} \right) - \cos \left( {\frac{1}{y}} \right),\,\,\,\,\,\,\,\,y \ne 0 \\
0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 0 \\
\end{array} \right.$$ which are discontinuous at the origin $(0,0)$. This example given in the book of B. Gelbaum and J. Olmsted, "Counterexample in Analysis", 3rd ed., page 119.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Does $(1^a+2^a+3^a+4^a+5^a)^b=1^c+2^c+3^c+4^c+5^c$ imply $(a,b,c)=(1,2,3)$?
Question : Is the following proposition true?
Proposition : For positive integers $a,b,c$ where $b\ge 2$, if
$$(1^a+2^a+3^a+4^a+5^a)^b=1^c+2^c+3^c+4^c+5^c$$then $(a,b,c)=(1,2,3)$.
This is an unsolved case of this question where it has been proven that each of the following propositions is true for positive integers $a,b,c$ where $b\ge 2$ :
*
*$\text{If $(1^a+2^a)^b=1^c+2^c$, then $(a,b,c)=(1,2,3)$.}$
*$\text{If $(1^a+2^a+3^a)^b=1^c+2^c+3^c$, then $(a,b,c)=(1,2,3)$.}$
*$\text{If $(1^a+2^a+3^a+4^a)^b=1^c+2^c+3^c+4^c$, then $(a,b,c)=(1,2,3)$.}$
*$\text{If $(1^a+2^a+\cdots +{11}^a)^b=1^c+2^c+\cdots+11^c$, then $(a,b,c)=(1,2,3)$.}$
*$\text{If $(1^a+2^a+\cdots +{12}^a)^b=1^c+2^c+\cdots+12^c$, then $(a,b,c)=(1,2,3)$.}$
$$\vdots$$
I've been trying to use mod and some inequalities, but every attempt has failed. Can anyone help?
Added : I'm going to add the background of this question.
We know that
$$\left(1^1+2^1+\cdots+n^1\right)^2=1^3+2^3+\cdots+n^3$$
holds for every $n\in\mathbb N$. Also, it is known that, for positive integers $a,b,c$ where $b\ge 2$, if
$$\left(1^a+2^a+\cdots+n^a\right)^b=1^c+2^c+\cdots+n^c$$
holds for every $n\in\mathbb N$, then $(a,b,c)=(1,2,3)$.
Then, I've been interested in the following similar, but completely different question :
For positive integers $a,b,c$ where $b\ge 2$, if
$$\left(1^a+2^a+\cdots+n^a\right)^b=1^c+2^c+\cdots+n^c$$
holds for a specific $n\color{red}{\ge 2}\in\mathbb N$, then can we say that $(a,b,c)=(1,2,3)$?
This question has been asked here. It has been proven that the answer is yes for $n=2,8k-5,8k-4$ where $k\in\mathbb N$.
However, the question has not received any complete answers. For example, it is not known if the answer is yes for $n=5$, which is the smallest unsolved case I'm asking here.
| In general we have
$$
\sum_{k=1}^n k^a = \frac{1}{a+1} \prod_{\jmath=1}^{a+1} \big( n + n_\jmath\big)
\tag 1
$$
Examples
$$
\begin{eqnarray}
\sum_{k=1}^n k &=& \frac{1}{2} n \big( n + 1 \big)\\\
\sum_{k=1}^n k^2 &=& \frac{1}{3} n \big( n + 1 \big) \big( n + 1/2 \big)\\\
\sum_{k=1}^n k^3 &=& \frac{1}{4} n^2 \big( n + 1 \big)^2\\\
\sum_{k=1}^n k^4 &=& \frac{1}{5} n \big( n + 1 \big)
\big( n + 1/2 \big)
\Big( n + 1/2 - \sqrt{7/12} \Big)
\Big( n + 1/2 + \sqrt{7/12} \Big)
\end{eqnarray}
$$
So we can only get
$$
\left( \sum_{k=1}^n k^a \right)^b = \sum_{k=1}^n k^c,\\
\quad \textrm{if at least} \quad
\left( \frac{1}{a+1} \right)^b = \frac{1}{1+c},
\quad \textrm{and} \quad
\big( a + 1 \big) b = c + 1. \tag 2
$$
So the question is, when do we have
$$
\big( a + 1 \big)^{b-1} = b ?
$$
So only $a=1$, $b=2$ and $c=3$ are the solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.