Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find $a\in \mathbb{R}$ such that a root of $ax^3-13x^2+(15a)x-25$ is $2+i$ Find $a\in \mathbb{R}$ such that a root of the polynomial $$p(x)=ax^3-13x^2+(15a)x-25$$ is $2+i$
Solution:
$q(x)=\frac{p(x)}{a}$
$=x^3-\frac{13}{a}x^2+15x-\frac{25}{a}$
$p(x)$ and $q(x)$ have the same roots, call them $r_1, r_2, r_3$
$r_1=2+i$. Since the coefficients are all real we know $r_2=2-i$
$(x-r_1)(x-r_2)(x-r_3)= x^3-(r_1+r_2+r_3)x^2+(r_1r_2 + r_1r_3 +r_2r_3)x-r_1r_2r_3$
$r_1+r_2+r_3=\frac{13}{a}$
$(2+i)(2-i)+r_3=\frac{13}{a}$
$5+r_3=\frac{13}{a}$
We also have:
$r_1r_2r_3=\frac{25}{a}$
$(2+i)(2-i)r_3=\frac{25}{a}$
$5r_3=\frac{25}{a}$
$r_3=\frac{5}{a}$
Going back to $5+r_3=\frac{13}{a}$
We have $5+\frac{5}{a}=\frac{13}{a}$
Thus $a=\frac{8}{5}$
The statement I am having trouble with is
"$r_1=2+i$. Since the coefficients are all real we know $r_2=2-i$"
Why do we know what $r_2$ is? I understand it's the conjugate and that $r_1r_2=5$ is a real number. But I don't understand why the coefficients being real implies that $r_2=2-i$
| The coefficients of the polynomial is real, hence complex roots comes in conjugate pairs.
If $$\sum_{i=0}^n a_i z^i = 0$$
We can take conjugate and see that
$$\sum_{i=0}^n a_i \bar{z}^i = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Exponential diophantine equation $2^x+7^y=9^z$. The challenge is to solve this equation $2^{x}+7^{y}=9^{z}$ in positive integers. The obvious solution is $x=y=z=1$. Using brute force, I found $3$ possible solutions:
\begin{eqnarray*}
(x_1,y_1,z_1)&=&(3,0,1),\\
(x_2,y_2,z_2)&=&(1,1,1)\\
(x_3,y_3,z_3)&=&(5,2,2).\\
\end{eqnarray*}
There are no other natural solutions for $z≤10000$.
It seems that the equation $2^{x}+7^{y}=9^{z}$ has no other solutions in natural numbers. How can this be proven?
| This is a PARTIAL answer. What remains to show is that there are not an infinite number of solutions for $x=1$. Assume $x\ge 3$. Then as in Conner's answer,
$$2^x = (3^z-7^v)(3^z+7^v).$$
This gives for some positive integer $a$ the equation
$$2^a(3^z-7^v)=(3^z+7^v).$$
Rearranging terms gives
$$(2^a-1)3^z = (2^a+1)7^v,$$
Note that this implies that $a$ must also be at least 3. So this gives $2^a+1=3^z$. As $a$ is at least $3$ [because one can check that there are no solutions to the above equation for $a \in \{1,2\}$], it follows that $2^a+1=3^z$ and thus in particular $3^z$ must be $1$ mod $8$, and so it follows that $z$ must be even, which gives
$$(3^{z/2}-1)(3^{z/2}+1) = 2^a,$$
which implies both $3^{z/2}-1$, $3^{z/2}+1$, must be powers of $2$. As they differ by only $2$, this is possible only if $z=2$ and $a=3$. So if $x$ is at least $3$, then $v$ must be $1$ and thus $y$ must be $2$. Thus $x$ must be $5$ and $z=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$ Q: (a) By comparing the expressions for $(\cos(\theta) + \sin(\theta)^5$ given by De Moivre's theorem and by the binomial theorem prove that $\cos(5\theta) = 16\cos^5(\theta)-20\cos^3(\theta) + 5\cos(\theta)$
(b) By considering the equation $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$
I have completed part (a). I am stuck on part (b) however.
My Workings:
By Factor formulae, $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{1}{2} (\cos(\frac{2\pi}{5}) + \cos(\frac{\pi}{5}))$
Considering roots of $z^5 = 1$. The sum of the roots equals 0.
So $\cos(\frac{2\pi}{5}) + \cos(\frac{4\pi}{5}) + \cos(\frac{6\pi}{5}) + \cos(\frac{8\pi}{5}) + \cos(\frac{10\pi}{5})=0$
$\cos(\frac{2\pi}{5}) = \cos(\frac{8\pi}{5})$, $\cos(\frac{4\pi}{5}) = \cos(\frac{6\pi}{5}) = -\cos(\frac{\pi}{5})$
So $2\cos(\frac{2\pi}{5})-2\cos(\frac{\pi}{5}) = -1$
$\cos(\frac{2\pi}{5})-\cos(\frac{\pi}{5}) = -\frac{1}{2}$
Been on stuck on this problem for an hour. Please help me.
| The equation $ \ \cos(5 \theta) \ = \ 0 \ $ has the set of angle solutions $ \ 5 \theta \ = \ \frac{\pi}{2} \ + \ k·\pi \ \Rightarrow \ \theta \ = \ \frac{(2k + 1) \ · \ \pi}{10} \ \ , \ $ arranged on the unit circle as presented in the graph below. Upon solving the polyomial equation in cosine which is equivalent to $ \ \cos(5 \theta) \ = \ 0 \ \ , \ $ we obtain
$$ 16·\cos^5 \theta \ - \ 20·\cos^3 \theta \ + \ 5·\cos \theta \ \ = \ \ \cos \theta \ · \ ( \ 16·\cos^4 \theta \ - \ 20·\cos^2 \theta \ + \ 5 \ ) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ \cos \theta \ \ = \ \ 0 \ \ \ , $$
$$ 16·\cos^4 \theta \ - \ 20·\cos^2 \theta \ + \ 5 \ \ = \ \ 0 \ \ \Rightarrow \ \ \cos^2 \theta \ \ = \ \ \frac{20 \ \pm \ \sqrt{80}}{32} \ \ = \ \ \frac{5 \ \pm \ \sqrt5}{8} \ \ . $$
[Thus far, these results have also been shown by earlier posters.]
Because the angle solutions are symmetrical about the $ \ y-$axis (as they are odd multiples of $ \ \frac{\pi}{10} \ ) \ , \ $ they are divided into three "families":
$$ \mathbf{\cos \theta \ \ = \ \ 0 \ \ : } \quad \quad \theta \ = \ \frac{5 \pi}{10} \ = \ \frac{\pi}{2} \ \ , \ \ \frac{15 \pi}{10} \ = \ \frac{3 \pi}{2} \ \ \ \text{[points in blue]} \ \ ; $$
$$ \mathbf{\cos^2 \theta \ \ = \ \ \frac{5 \ - \ \sqrt5}{8} \ \ : } \quad \quad \theta \ = \ \frac{3 \pi}{10} \ \ , \ \ \frac{7 \pi}{10} \ \ , \ \ \frac{13 \pi}{10} \ \ , \ \ \frac{17 \pi}{10} \ \ \ \text{[points in green]} \ \ ; $$
$$ \mathbf{\cos^2 \theta \ \ = \ \ \frac{5 \ + \ \sqrt5}{8} \ \ : } \quad \quad \theta \ = \ \frac{ \pi}{10} \ \ , \ \ \frac{9 \pi}{10} \ \ , \ \ \frac{11 \pi}{10} \ \ , \ \ \frac{19 \pi}{10} \ \ \ \text{[points in red]} \ \ . $$
The cosine function has the property $ \ \cos \theta \ = \ \cos (2 \pi - \theta) \ = \ -\cos ( \pi - \theta ) \ = \ -\cos ( \pi + \theta ) \ \ , \ $ so four angles will have a common value of cosine-squared (the exception, of course, being $ \ \cos^2 \theta \ = \ 0 \ $ for which $ \ \theta \ = \ \frac{ \pi}{2} \ = \ \left( \pi - \frac{ \pi}{2} \right) \ \ $ and $ \ \theta \ = \ \frac{ 3 \pi}{2} \ = \ \left( \pi + \frac{ \pi}{2} \right) \ \ . \ ) $
Applying the "triple-angle formula" for cosine (as Eric Towers suggests), $ \ \cos (3 \theta) \ = \ 4 · \cos^3 \theta \ - \ 3 · \cos \theta \ \ , \ $ we can write
$$ \cos \theta \ · \ \cos (3 \theta) \ \ = \ \ 4 · \cos^4 \theta \ - \ 3 · \cos^2 \theta \ \ . \ $$
The foregoing discussion makes it clear which value of $ \ \cos^2 \theta \ $ should be used among the solutions to $ \ \cos (5 \theta) \ = \ 0 \ \ , \ $ so we may calculate
$$ \cos \left( \frac{\pi}{10} \right) \ · \ \cos \left( \frac{3 \pi}{10} \right) \ \ = \ \ 4 · \cos^4 \left( \frac{\pi}{10} \right) \ - \ 3 · \cos^2 \left( \frac{\pi}{10} \right) $$
$$ = \ \ 4 · \left( \ \frac{5 \ + \ \sqrt5}{8} \ \right)^2 \ - \ 3 · \left( \ \frac{5 \ + \ \sqrt5}{8} \ \right) $$ $$ = \ \ 4 · \left( \ \frac{25 \ + \ 5 \ + \ 10·\sqrt5}{64} \ \right) \ - \ \left( \ \frac{15 \ + \ 3·\sqrt5}{8} \ \right) $$
$$ = \ \ \frac{30 \ + \ 10·\sqrt5 \ - \ 30 \ - \ 6·\sqrt5}{16} = \ \ \frac{\sqrt5 }{4} \ \ . $$
| {
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"url": "https://math.stackexchange.com/questions/4299942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how to express the $n$th power of the cosine as a series of cosines? Which is the correct way for expressing the $n$th power of a cosine as a series of cosines without any exponent?
By using the Euler's formula
$\cos^n{(\theta)}=\left( \frac{e^{j\theta}+e^{-j\theta}}{2} \right)^n=
\frac{1}{2^n}\left( e^{j\theta}+e^{-j\theta} \right)^n=\frac{1}{2^n}\left( z+z^{-1} \right)^n.$
with $z=e^{j\theta}$.
Since the term $\left( z+z^{-1} \right)^n$ is the $n$th power of a binomial, I could express it using the binomial identity, thus
$\cos^n{(\theta)} = \frac{1}{2^n} \displaystyle\sum_{k=0}^n \binom{n}{k} z^k(z^{-1})^{n-k}
= \frac{1}{2^n} \displaystyle\sum_{k=0}^n \binom{n}{k} z^{2k-n}.$
If we expand the expression above, we obtain
$\cos^n{(\theta)}=\frac{1}{2^n}\Bigg ( z^{-n} + \binom{n}{1}z^{-(n-2)} + \binom{n}{2}z^{-(n-4)} + \dots + \binom{n}{2}z^{n-4} + \binom{n}{1}z^{(n-2)} + z^n \Bigg )$
which can be rewritten as
$\cos^n{(\theta)}=\frac{1}{2^n} \Bigg ( (z^{-n} + z^n) + \binom{n}{1} \left(z^{-(n-2)} + z^{(n-2)}\right) + \binom{n}{2}\left( z^{-(n-4)} + z^{(n-4)}\right) + \dots \Bigg ) $
Finally, since $z=e^{j\theta}$
$\cos^n{(\theta)}=\frac{1}{2^n} \Bigg ( (e^{-jn\theta} + e^{jn\theta}) + \binom{n}{1} \left(e^{j(n-2)\theta} + e^{-j(n-2)\theta}\right) + \binom{n}{2}\left( e^{j(n-4)\theta} + e^{-j(n-4)\theta}\right) + \dots \Bigg )$
By applying the Euler's formula once again, we obtain
$\cos^n{(\theta)}=\frac{2}{2^{n}} \sum_{k=0}^n \binom{n}{k} \cos{((n-2k)\theta)}.$
Unfortunately, if I plug n=2, I obtain
$\cos^2{(\theta)}= \cos{(2\theta)} + 1.$
instead of the well-known result
$\cos^2{(\theta)}= \frac{1}{2}(\cos{(2\theta)} + 1).$
a) Why my result is scaled by a factor of 2 ?
b) Is the correct general formula
$\cos^n{(\theta)}=\frac{1}{2^{n}} \sum_{k=0}^n \binom{n}{k} \cos{((n-2k)\theta)}.$
c)If so, why ?
| Here is a way to fix the factor of $2$ while still summing from $0$ to $n$
(setting aside the question of whether that is the best way to do the sum).
You have
$$
\cos^n(\theta) =
\frac{1}{2^n}\left(
z^{-n} + \binom n1 z^{-(n-2)} + \binom n2 z^{-(n-4)} + \cdots
+ \binom n2 z^{n-4} + \binom n1 z^{n-2} + z^n \right).
$$
Reversing the order of the sum,
$$
\cos^n(\theta) =
\frac{1}{2^n}\left(
z^n + \binom n1 z^{n-2} + \binom n2 z^{n-4} + \cdots
+ \binom n2 z^{-(n-4)} + \binom n1 z^{-(n-2)} + z^{-n} \right).
$$
Adding termwise,
\begin{align}
2 \cos^n(\theta) &=
\frac{1}{2^n}\bigg(
\left(z^{-n} + z^n\right) + \binom n1 \left(z^{-(n-2)} + z^{n-2}\right)
+ \binom n2 \left(z^{-(n-4)} + z^{n-4}\right) + \cdots \\
&\qquad\qquad + \binom n2 \left(z^{n-4} + z^{-(n-4)}\right)
+ \binom n1 \left(z^{n-2} + z^{-(n-2)}\right) + \left(z^n + z^{-n}\right) \bigg) \\
&=\frac{2}{2^n}\bigg(\cos(n\theta) + \binom n1 \cos((n-2)\theta)
+ \binom n2 \cos((n-4)\theta) + \cdots \\
&\qquad\qquad + \binom n2 \cos(-(n-4)\theta)
+ \binom n1 \cos(-(n-2)\theta) + \cos(-n\theta) \bigg) .
\end{align}
Canceling one factor of $2$ on each side, this simplifies to
$$
\cos^n(\theta) = \frac{1}{2^n} \sum_{k=0}^n \binom nk \cos{((n-2k)\theta)}.
$$
For $n = 2,$ this gives
\begin{align}
\cos^2(\theta)
&= \frac 14\left(\binom 20 \cos(2\theta)
+ \binom 21 \cos(0) + \binom 22 \cos(-2\theta) \right) \\
&= \frac 12\left(\cos(2\theta) + 1\right)
\end{align}
as expected.
Your mistake was you did not write the end of the sum after the three dots, so you did not notice that you had moved terms from the right end of the sum to the left end without replacing them, so about half the terms in your summation were not matched by terms in the $+ \cdots +$ notation. Adding two copies of the sum in reverse order, you don't have to move any terms so you won't make this mistake.
Note that it's conventional to combine the $\cos(m\theta)$ and $\cos(-m\theta)$ terms in the sum and have about half as many terms (exactly half as many for odd powers) at the cost of (usually) having separate summations for even and odd powers of the cosine.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int_a^b \sqrt{(1+ m^2\cosh^2x)}dx$ $$\int_a^b \sqrt{(1+ m^2\cosh^2x)}dx$$ where $a = 0$ and $b= \ln 2$
I just need a little hint to start. Please don't answer it completely. Just a hint.
Edit : Originally the question was to find the arc length of the curve $f(x) = m \sinh x$ on the interval $[0, \ln 2]$. But I cannot evaluate this integral at all.
| If you do not know yet about elliptic integrals, you do not have much choice.
Start using
$$\cosh^2(x)=\frac 1 2(1+\cosh(2x))$$ which makes
$$1+m^2\cosh^2(x)=\frac{1}{2} m^2 \cosh (2 x)+\frac{m^2}{2}+1=\frac{1}{2} m^2\Big[1+\frac{2}{m^2}+\cosh (2 x) \Big]$$
Now, use the series expansion of $\cosh(2x)$. This will give
$$1+m^2\cosh^2(x)=\left(m^2+1\right)+m^2 x^2+\frac{m^2}{3} x^4+\frac{2 m^2 }{45}x^6+\frac{m^2 }{315} x^8+O\left(x^{10}\right)$$
Use the binomial theorem
$$\sqrt{1+m^2\cosh^2(x)}=\sqrt{m^2+1}+\frac{m^2 x^2}{2 \sqrt{m^2+1}}+\frac{m^2 \left(m^2+4\right) x^4}{24
\left(m^2+1\right)^{3/2}}+\frac{m^2 \left(m^4-28 m^2+16\right) x^6}{720
\left(m^2+1\right)^{5/2}}+\frac{m^2 \left(m^6+696 m^4-816 m^2+64\right) x^8}{40320
\left(m^2+1\right)^{7/2}}+O\left(x^{10}\right)$$ Integrate termwise to have an approximate antiderivative. Use the bounds.
Suppose that, with your given bounds, we use this truncated series for $m=3$. This would give as a numerical result $2.3547717$ while using the elliptic integral, the result would be $2.3547702$.
| {
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"timestamp": "2023-03-29T00:00:00",
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About an inequality wich have a link with $\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2$ Hi it's a follow up of show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$:
Problem :
Let $a,x>0$ then (dis)prove :
$$\left(\frac{1}{2}a^{x}+\frac{1}{2}x^{-1}\right)^{\frac{1}{5}}\cdot\left(\frac{2}{a^{-x}+x}\right)^{\frac{4}{5}}+\left(\frac{1}{2}x^{a}+\frac{1}{2}a^{-1}\right)^{\frac{1}{5}}\cdot\left(\frac{2}{x^{-a}+a}\right)^{\frac{4}{5}}\geq 2$$
If true I have tried to show it using the well-know formula :
Let $x$ be a real number then we have :
$$e^x\geq x+1$$
Applying to $a^x$ and $x^a$ .Unfortunately the inequality becomes false .
Also I have tried Tchebytchev's inequality but the application is similar to the use of Am-Gm and the inequality becomes wrong .
I found this refinement as a refinement of Am-Gm-Hm wich is a mix .
Edit :
It seems we have for $a\geq 1$ and $x\in(\frac{1}{a},a)$ :
$$g(x)=\left(\frac{1}{2}a^{x}+\frac{1}{2}x^{-1}\right)^{-\frac{1}{5}}\cdot\left(\frac{2}{a^{-x}+x}\right)^{-\frac{4}{5}}\cdot\left(\frac{1}{2}x^{a}+\frac{1}{2}a^{-1}\right)^{\frac{1}{5}}\cdot\left(\frac{2}{x^{-a}+a}\right)^{\frac{4}{5}}$$
And :
$$h(x)=\left(\frac{1}{2}a+\frac{1}{2}x^{-1}\right)^{-\frac{1}{5}}\cdot\left(\frac{2}{a^{-1}+x}\right)^{-\frac{4}{5}}\cdot\left(\frac{1}{2}x+\frac{1}{2}a^{-1}\right)^{\frac{1}{5}}\cdot\left(\frac{2}{x^{-1}+a}\right)^{\frac{4}{5}}$$
Then it seems we have :
$$h(x)< g(x)$$
Last edit 25/11/2021 correction 26/11/2021:
We need to show that for $a\leq 1$ and $x\ge 1$ :
$$2^{\frac{3}{5}}\cdot\frac{\left(\frac{a^{x}}{x}\right)^{\frac{4}{5}}}{\left(a^{x}+\frac{1}{x}\right)^{\frac{3}{5}}}+2^{\frac{3}{5}}\cdot\frac{\left(\frac{x^{a}}{a}\right)^{\frac{4}{5}}}{\left(x^{a}+\frac{1}{a}\right)^{\frac{3}{5}}}\geq 2 $$
Or :
$$x^{-\frac{5}{5}}r\left(a^{x}x\right)+a^{-\frac{5}{5}}r\left(x^{a}a\right)\geq 2$$
Where the function :
$$r\left(x\right)=2^{\frac{3}{5}}\cdot\frac{x^{\frac{4}{5}}}{\left(x+1\right)^{\frac{3}{5}}}$$
Is reciprocally convex on $(0,\infty)$ .
See this paper for further informations : http://artem.sobolev.name/posts/2021-05-02-reciprocal-convexity-to-reverse-the-jensen-inequality.html at ("defined by Merkle")
We get a lower bound :
$$\left(\frac{1}{x}+\frac{1}{a}\right)\cdot r\left(\frac{\left(\frac{1}{x}+\frac{1}{a}\right)}{\frac{1}{x}\cdot\frac{1}{\left(a^{x}x\right)}+\frac{1}{a}\left(\frac{1}{x^{a}\cdot a}\right)}\right)$$
It seems to be superior to two when $0.75\leq a\leq 1$ and $0<x\leq 6$
Question :
How to (dis)prove it ?
| Conjectures :
Using the comment above and the fact that $r(x)$ increases along the positive real axis we have the first conjecture :
Let $x>0$ then we have :
$$x\cdot r\left(\left(\frac{4x^{2}}{\left(x+2\right)^{2}}\right)^{\frac{-1}{x}}\left(x-\sqrt{\frac{4x^{2}}{\left(x+2\right)^{2}}}\right)^{-1}\right)\geq 2$$
After that we have a second conjecture :
Let $0<x<1$ then $\exists a>1$,$\exists b\in R^{*}$ and $|b|<x$ and $x+a\geq 2$ such that :
$=\left(\left(a-b\right)\left(x+b\right)\right)^{\frac{-1}{x+a}}\left(x+a-\sqrt{\left(x+b\right)\left(a-b\right)}\right)^{-1}\leq \left(ax\right)^{-\frac{1}{x+a}}x^{-\frac{x}{x+a}}a^{-\frac{a}{x+a}}$
This two conjecture are sufficient to show the proposed problem .
Edit :
For the first conjecture and using Bernoulli's inequality we need to show for $x\geq 1$ :
$$x\cdot r\left(\left(1+\frac{1}{x}\left(\frac{4x^{2}}{\left(x+2\right)^{2}}-1\right)\right)^{-1}\left(x-\sqrt{\frac{4x^{2}}{\left(x+2\right)^{2}}}\right)^{-1}\right)\geq 2$$
This last inequality could also be seen as a long polynomial with a root at $x=2$
Edit 2 :
For the second conjecture we can use derivative with respect to $b$ see Wolfram alpha
Edit 3 :
Due to the homogeneity in the second conjecture we can choose :
$$b=a-\frac{a^2}{x}$$
We obtain a one variable inequality with the contraint of the second conjecture .
Last edit :
Unfortunetaly the second conjecture is not sufficient but I propose another conjecture wich seems highly true :
Let $0<a\leq 1$ and $x\geq 1$ such that $2\leq a+x\leq 5$ then we have :
$$g\left(x+a\right)-f\left(x\right)\leq 0$$
Where :
$$f(x)=\left(x+a\right)\cdot r\left(\left(ax\right)^{-\frac{1}{x+a}}x^{-\frac{x}{x+a}}a^{-\frac{a}{x+a}}\right)$$
And :
$$g\left(x\right)=x\cdot r\left(\left(1+\frac{1}{x}\left(\frac{4x^{2}}{\left(x+2\right)^{2}}-1\right)\right)^{-1}\left(x-\sqrt{\frac{4x^{2}}{\left(x+2\right)^{2}}}\right)^{-1}\right)$$
In other word we ensure the inequality including the minimum .
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining all $(a,b)$ on the unit circle such that $2x+3y+1\le a(x+2)+b(y+3)$ for all $(x,y)$ in the unit disk In the middle of another problem, I came up with the following inequality which needed to be solve for $(a, b)$ : $$2x+3y+1\le a(x+2)+b(y+3)$$ for all $(x, y)\in\mathbb{R}^2$ with $x^2+y^2\le1.$
Here the solution for $(a, b)$ must be a subset of the unit circle, and I believe that it is a singleton.
Since I have no clue to solve it in this form, I tried to substitute some points of the closed unit disk and make a system of inequalities. But it seems like there should be a general way of solving these type of problems.
| Rewrite $2x+3y+1\le a(x+2)+b(y+3)$ as $(2-a)x+(3-b)y\leq 2a+3b-1$. Solve for the intersection of the line $(2-a)x+(3-b)y=2a+3b-1$ with $x^2+y^2=1$ to get that the $x$-coordinates of the intersection are $$\frac{-2a^2-3ab+5a+6b-2\pm\sqrt{-(b-3)^2(3a^2+12ab+8b^2-12)}}{a^2-4a+b^2-6b+13}.$$ If this has two real solutions, then the inequality cannot be satisfied for all $(x,y)$ in the unit circle, so we need to find the intersection of $-(b-3)^2(3a^2+12ab+8b^2-12)\leq 0$ and $a^2+b^2\leq 1$. Simultaneously solving these equations, we find that the real solutions for $(a,b)$ are $(\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}})$ and $(-\frac{2}{\sqrt{13}},-\frac{3}{\sqrt{13}})$. It's easy to see that the first solution is valid and the second is invalid - plug in $(x,y)=(0,0)$, for instance.
| {
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$g=\rho^2 (d x^2 + dy^2) \Rightarrow g = ds^2 +\tanh^2 s d\theta^2$ The question is from the 10th page of Topping's Lectures on Ricci flow, the calculation about Hamilton's cigar soliton.
Consider $R^2$ with metric
$$
g=\rho^2 (dx^2 + dy^2),~~~\rho^2= \frac{1}{1+x^2 + y^2}
$$
where $dx^2 = dx \otimes dx $. Topping state that
$$
g=ds^2 + \tanh^2 s d\theta^2
$$
in terms of the geodesic distance from the origin $s$, and the polar angle $\theta$. But I can't get it.
What I try: In polar coordinates
$$
x=r\cos\theta,~~~ y = r\sin\theta
$$
I get
$$
g=\rho^2 dr^2 + r^2 \rho^2 d\theta^2
$$
The process:
$$
\begin{pmatrix}
dx \\
dy
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta & -r\sin\theta \\
\sin\theta & r\cos\theta
\end{pmatrix}
\begin{pmatrix}
dr \\
d\theta
\end{pmatrix}
$$
and
\begin{align}
g
&=
\begin{pmatrix}
dx &dy
\end{pmatrix}
\begin{pmatrix}
\rho^2 & 0 \\
0 & \rho^2
\end{pmatrix}
\begin{pmatrix}
dx \\
dy
\end{pmatrix}
\\
&=
\begin{pmatrix}
dr &d\theta
\end{pmatrix}
\begin{pmatrix}
\cos\theta & \sin\theta \\
-r\sin\theta & r\cos\theta
\end{pmatrix}
\begin{pmatrix}
\rho^2 & 0 \\
0 & \rho^2
\end{pmatrix}
\begin{pmatrix}
\cos\theta & -r\sin\theta \\
\sin\theta & r\cos\theta
\end{pmatrix}
\begin{pmatrix}
dr \\
d\theta
\end{pmatrix}
\\
&=
\rho^2 dr^2 + r^2 \rho^2 d\theta^2
\end{align}
but I don't know how to deal the geodesic distance $s$.
| As you noted, in polar coordinates, the metric reads
$$
g = \frac{1}{1+r^2}\mathrm{d}r^2 + \frac{r^2}{1+r^2}\mathrm{d}\theta^2
$$
Heuristically, you are looking for $s$ such that $\mathrm{d}s^2 = \frac{1}{1+r^2}\mathrm{d}r^2$, which can be stated as ${\mathrm{d}s} = \frac{1}{\sqrt{1+r^2}}{\mathrm{d}r}$. So let's define
$$
s(r) = \int_0^r \frac{1}{\sqrt{1+t^2}}\mathrm{d}t = \sinh^{-1}(r)
$$
It now follows that
$$
\frac{r^2}{1+r^2} = \frac{\sinh(s)^2}{1+\sinh(s)^2} = \frac{\sinh(s)^2}{\cosh(s)^2} = \tanh(s)^2
$$
and finally,
$$
g = \mathrm{d}s^2 + \tanh(s)^2\mathrm{d}\theta^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3} + \sqrt[3]{9}}$ Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}}$
I keep making a mess of this. I tried vewing the denominator as
$a +\sqrt[3]{9}$, where $a=9-3\sqrt[3]{3}$ and secondly as
$b -3\sqrt[3]{3}+\sqrt[3]{9}$, where $b=9$.
Then using the sum and differences in cubes fratorization but this keeps adding radicals to the denominator.
How should I approach this/where could I be going wrong?
| $9-3\sqrt[3]{3}+\sqrt[3]{9} = a^2 -ab+b^2$
$\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}} \cdot \frac{3+\sqrt[3]{9}}{3+\sqrt[3]{9}}=\frac{12+4\sqrt[3]{9}}{30}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Seeking for help to find a formula for $\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}$, where $a>1.$ When tackling the question, I found that for any $a>1$,
$$
I_1(a)=\int_{0}^{\pi} \frac{d x}{a-\cos x}=\frac{\pi}{\sqrt{a^{2}-1}}.
$$
Then I started to think whether there is a formula for the integral
$$
I_n(a)=\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}, $$
where $n\in N.$
After trying some substitution and integration by parts, I still failed and got no idea for reducing the power n. After two days, the Leibniz Rule for high derivatives come to my mind.
Differentiating $I_1(a)$ w.r.t. $a$ by $(n-1)$ times yields
$$
\displaystyle \begin{array}{l}
\displaystyle \int_{0}^{\pi} \frac{(-1)^{n-1}(n-1) !}{(a-\cos x)^{n}} d x=\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\pi}{\sqrt{a^{2}-1}}\right) \\ \displaystyle
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{(-1)^{n-1} \pi}{(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{1}{\sqrt{a^{2}-1}}\right) \tag{*}\label{star}
\end{array}
$$
I am glad to see that the integration problem turn to be merely a differentiation problem.
Now I am going to find the $(n-1)^{th} $ derivative by Leibniz Rule.
First of all, differentiating $I_1(a)$ w.r.t. $a$ yields $$
\left(a^{2}-1\right) \frac{d y}{d a}+a y=0 \tag{1}\label{diffeq}
$$
Differentiating \eqref{diffeq} w.r.t. $a$ by $(n-1)$ times gets $$
\begin{array}{l}
\displaystyle \left(a^{2}-1\right) \frac{d^{n} y}{d a^{n}}+\left(\begin{array}{c}
n-1 \\
1
\end{array}\right)(2 a) \frac{d^{n-1} y}{d a^{n-1}}+2\left(\begin{array}{c}
n-1 \\
2
\end{array}\right) \frac{d^{n-2} y}{d a^{n-2}}+x \frac{d^{n-1} y}{d a^{n-1}}+(n-1) \frac{d^{n-2} y}{d a^{n-2}}=0
\end{array}
$$
Simplifying, $$
\left(a^{2}-1\right) y^{(n)}+(2 n-1) ay^{(n-1)}+(n-1)^{2} y^{(n-2)}=0 \tag{2}\label{diffrec}
$$
Initially, we have $ \displaystyle y^{(0)}=\frac{1}{\sqrt{a^{2}-1}}$ and $ \displaystyle y^{(1)}=-\frac{a}{\left(a^{2}-1\right)^{\frac{3}{2}}}.$
By \eqref{diffrec}, we get $$
y^{(2)}=\frac{2 a^{2}+1}{\left(a^{2}-1\right)^{\frac{5}{2}}}
$$ and $$
\displaystyle y^{(3)}=-\frac{3 a\left(2 a^{2}+3\right)}{\left(a^{2}-1\right)^{\frac{7}{2}}}
$$
Plugging into \eqref{star} yields $$
\begin{aligned}
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{3}} &=\frac{\pi}{2} y^{(2)}=\frac{\pi\left(2 a^{2}+1\right)}{2\left(a^{2}-1\right)^{\frac{5}{2}}} \\
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{4}} &=-\frac{\pi}{6} \cdot \frac{3 a\left(2 a^{2}+3\right)}{\left(a^{2}-1\right)^{\frac{7}{2}}} =-\frac{\pi a\left(2 a^{2}+3\right)}{2\left(a^{2}-1\right)^{\frac{7}{2}}}
\end{aligned}
$$
Theoretically, we can proceed to find $I_n(a)$ for any $n\in N$ by the recurrence relation in $(2)$ .
By Mathematical Induction, we can further prove that the formula is $$
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{\pi P(a)}{\left(a^{2}-1\right)^{\frac{2 n-1}{2}}}
$$
for some polynomial $P(a)$ of degree $n-1$.
Last but not least, how to find the formula for $P(a)$? Would you help me?
| From what you have deduced, we can apply here the Faà di Bruno's formula
$$
\begin{align}
I_n(a) &= \frac{(-1)^{n-1}\pi}{(n-1)!} \frac{d^{n-1}}{da^{n-1}} \left(\frac{1}{\sqrt{a^2 - 1}}\right) \\
&= \frac{(-1)^{n-1}\pi}{(n-1)!} \frac{d^{n-1} \sqrt{b(a)}}{da^{n-1}} \\
&= \sum_{}\frac{(n-1)!}{\prod_{k=1}^{n-1}m_k! k!^{m_k}} \frac{d^{m_1 + \cdots + m_{n-1}}\sqrt{b}}{db^{m_1 + \cdots + m_{n-1}}} \cdot \prod_{j=1}^{n-1} \left(\frac{d^j b(a)}{da^j}\right)^{m_j}
\end{align}
$$
where $b = \frac{1}{a^2 - 1}$ and the summation is over all $n-1$ tuples of non-negative integers $m_i$ such that
$$
\sum_{k=1}^{n-1} k m_k = n-1.
$$
Indeed, we have that $\frac{db}{da} = \frac{-2a}{(a^2 - 1)^2}$ (finding a general formula for this should be not too difficult) and that for any $k\in\mathbb{N}$
$$
\frac{d^k}{db^k}\sqrt{b} = (-1)^{k-1}\frac{(2(k-1))!}{(k-1)!}(4b)^{\frac{1 - 2k}{2}},
$$
where the last equality was obtained via this question.
Following down this path and executing the necessary derivatives should yield a sufficient answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What is the easier way to find the circle given three points? Given three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, if $$\frac{y_2-y_1}{x_2-x_1} \neq \frac{y_3-y_2}{x_3-x_2} \neq \frac{y_1-y_3}{x_1-x_3},$$ then there will be a circle passing through them. The general form of the circle is $$x^2 + y^2 + dx + ey + f = 0.$$ By substituting $x = x_i$ and $y = y_i$, there will be a system of equation in three variables, that is:
\begin{align*}
\begin{pmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{pmatrix}
\begin{pmatrix}
d \\
e \\
f
\end{pmatrix} &=
\begin{pmatrix}
-\left(x_1^2+y_1^2\right) \\
-\left(x_2^2+y_2^2\right) \\
-\left(x_3^2+y_3^2\right)
\end{pmatrix}.
\end{align*}
Solving this system gives the solution
\begin{align*}
d &= \frac{(x_3^2 + y_3^2 -x_1^2+y_1^2)}{x_1 - x_3} - e\left(\frac{y_1 - y_3}{x_1 - x_3}\right) \\
e &= \frac{(x_3^2 + y_3^2 -x_2^2+y_2^2)(x_1-x_3) - (x_3^2 + y_3^2 -x_1^2+y_1^2)(x_2-x_3)}{(y_2-y_3)(x_1-x_3) - (y_1-y_3)(x_2-x_3)} \\
f &= \frac{-(x_3^2 + y_3^2)(x_1-x_3) - (y_1-y_3)x_3}{x_1 - x_3} - e\left(\frac{y_3(x_1 - x_3) - x_3(y_1 - y_3)}{x_1 - x_3}\right)
\end{align*}
As there are a lot of things going around, the solution is prone to errors. Maybe this solution also has an error.
Is there a better way to solve for the equation of the circle?
| Move the circle so that it passes through the origin (subtract a point from all three). The equation loses a coefficient,
$$x^2+y^2+dx+ey=0$$
which is easier to solve. Then reverse-translate.
$$-\frac d2=\dfrac{\begin{vmatrix}x_1^2+y_1^2&y_1\\x_2^2+y_2^2&y_2\end{vmatrix}}{2\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}}$$
$$-\frac e2=
\dfrac{\begin{vmatrix}x_1&x_1^2+y_1^2\\x_2&x_2^2+y_2^2\end{vmatrix}}{2\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}}$$
are the coordinates of the center in the moved position.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Proving that equation has at least one solution in the interval (-1,1) We are given two numbers $a>0$ and $b>0$ and the equation given is
$$ \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} = 0 $$
Prove that this equation has at least one solution in the interval $(-1,1)$. Now I can let
$$ f(x) = \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} $$
And then I can think of using intermediate value theorem (IVT) on the interval $[-1,1]$. But the problem is that $ x^3 + 2x^2 - 1 $ has zeroes at $-1$ and $0.618$ and $x^3 + x - 2$ has a real zero at $1$. So, that means that $f(x)$ is not continuous on $[-1,1]$. So, how can IVT be used here ?
Thanks
| The two cubics do not have a common root.That can be exploited as follows:
Let $g(x)=a(x^{3}+x-2)+b(x^{3}+2x^{2}-1)$. Then $g(1)=2b>0$ and $g(-1)=-4a<0$. Hence, there exists $x \in (-1,1)$ such that $g(x)=0$. At this point $x$ neither $x^{3}+x-2$ nor $x^{3}+2x^{2}-1$ can be $0$: If $x^{3}+x-2=0$ we get $g(x)=b(x^{3}+2x^{2}-1)\neq 0$ and if $x^{3}+2x^{2}-1=0$ then $g(x)=a(x^{3}+x-2) \neq 0$. Hence, we can divide the equation $g(x)=0$ by $(x^{3}+x-2) (x^{3}+2x^{2}-1)$.
EDIT: A simple proof of the fact that the two cubics do not have a common root is posted below by user2661923
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Choosing at least 2 women from 7 men and 4 women In how many different ways can we choose six people, including at least two women, from a group made up of seven men and four women?
Attempt:
As we have to have at least two women in the choices, then $\displaystyle\binom{4}{2}$, leaving a total of $4$ out of $9$ people to be chosen. So $$\binom{9}{4} \cdot \binom{4}{2}=756$$
The answer is $371$. Where is my error?
| The number of ways is
$$\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2} =6\times35+4\times35+21 = 371$$
"At least two women" means either $2$, $3$ or $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Spivak Calculus, Ch. 5 Limits: Finding the smallest $b$ in $|x-3|Consider the function $f(x)=x^2$.
In the text of chapter 5 of Spivak's Calculus, he goes through the following argument to show that $f(x)$ approaches the limit $9$ near $x=3$.
We want that
$$|x^2-9|=|x-3||x+3|<\epsilon$$
Assume $|x-3|<1$.
$$\implies 2<x<4$$
$$\implies 5<x+3<7$$
$$\implies |x+3|<7$$
Therefore
$$|x^2-9|=|x-3||x+3|<7|x-3|<\epsilon$$
$$\implies |x-3|<\frac{\epsilon}{7}$$
provided $|x-3|<1$ as we assumed initially, ie $|x-3|<min(1,\frac{\epsilon}{7})$
I have a few observations about this argument:
*
*$|x+3|=7$ only happens if $|x-3|=1$, the largest distance from $3$ that we allow, by assumption.
*Assuming this "worst case scenario", $|x-3|$ has to be relatively smaller so that the product $|x-3||x+3|$ is smaller than $\epsilon$.
*In fact, depending on $\epsilon$, $|x-3|$ may have to be smaller than $1$. However, $\epsilon$ may also be so large that $|x-3|$ has to be smaller than a number larger than 1.
*All we know is that if $|x-3|<1$ then $|x+3|$ is definitely smaller than 7, and so to achieve a product smaller than $\epsilon$, we need $|x-3|<\frac{\epsilon}{7}$.
*Assume that $\epsilon<7$. Then, $|x-3|<\frac{\epsilon}{7}$<1. But then we can be more precise about how large $|x+3|$ is:
$$-\frac{\epsilon}{7}<x-3<\frac{\epsilon}{7}$$
$$6-\frac{\epsilon}{7}<x+3<6+\frac{\epsilon}{7}$$
$$|x+3|<6+\frac{\epsilon}{7}<7$$
$$|x^2-9|<(6+\frac{\epsilon}{7})|x-3|<\epsilon$$
$$|x-3|<\frac{\epsilon}{6+\frac{\epsilon}{7}}$$
The interval allowed for $x$ near $3$ is now larger. But that means we can go back and calculate what this means for $|x+3|$ again:
$$-\frac{\epsilon}{6+\frac{\epsilon}{7}}<x-3<\frac{\epsilon}{6+\frac{\epsilon}{7}}$$
$$6-\frac{\epsilon}{6+\frac{\epsilon}{7}}<x+3<6+\frac{\epsilon}{6+\frac{\epsilon}{7}}$$
Note that $\frac{\epsilon}{6+\frac{\epsilon}{7}}>\frac{\epsilon}{7}$, so the interval for $|x+3|$ is now larger than before. And if we go back and recalculate the interval for $|x-3|$ it will now be smaller again.
My questions are
*
*Are my calculations correct? Ie, is it ok to do this thing where I narrow down the intervals one at a time?
*How do I figure out what the smallest interval is for $x-3$ such that we have $|x^2-9|=|x-3||x+3|<\epsilon$?
| Let $x = 3+h$ then the problem is to estimate $h$, now you need $$-\varepsilon< (3+h)^2 - 9<\varepsilon$$
which is same as saying $ 9-\varepsilon < (3+h)^2 < 9+\varepsilon$, Now we need to check cases if $\varepsilon < 9$ and othercases, using that you will be able to get the interval for $h$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve this nonhomegenous ode $y''+4y = \cos(2x)$ Solve the given nonhomogenous linear ODE by variation of parameters or undetermined coefficients $y'' + 4y = \cos(2x)$.
A general solution is $y_1 = \cos(2x), y_2 = \sin(2x)$, and the Wronksin determinant is equal to 2.
Plugging this into the equation for the method of variation of parameters, I get
$$-\cos(2x)\int\frac{\sin(2x)\cos(2x)}{2}dx + \sin(2x)\int\frac{\cos^2(2x)}{2}dx$$
The integrals cancel out to $0$. How can I approach this correctly with the method proposed?
| Null space solution
$m^2 +4 =0$
$\implies m=\pm 2i$
Hence, $ y_N = C_1 \sin 2x + C_2 \cos 2x$.
Since, $\cos2x$ already in null space solution, take particular integral as $$y_p =x(A \sin 2x +B \cos 2x) $$
Then, \begin{align} y'_p &{=A \sin 2x +B \cos 2x +2x(A \cos 2x -B \sin 2x )}\\ \end{align}
$y''_p =4A \cos 2x -4B \sin 2x -4x(A\sin 2x + B \cos 2x )$
$$y''_p +4y'_p = \cos 2x$$
\begin{align} {4A\cos 2x -4B\sin 2x -4x(A\sin 2x + B\cos 2x )+4x(A \sin 2x +B \cos 2x) =\cos 2x } \end{align}
$4A\cos 2x -4B\sin 2x= \cos 2x$
$4A=1$ , $4B=0$
$A=1/4$ , $B=0$
Hence, $y_p = \frac {1}{4}{x\sin 2x }$
So, $$y_\text{complete} =y_N + y_p$$
$$y_\text{complete}=C_1\sin 2x + C_2\cos 2x + \frac {1}{4}{x \sin 2x }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Number of line of curvature meets at one point I am now consider the surface given by $$f=(x,y,x^3-3xy^2)$$
And I have been asked to prove that there are three line of curvature meet at origan point.
I can compute that
\begin{align*}
f_x&=(1,0,3x^2-3y^2)\\
f_y&=(0,1,-6xy)\\
f_x\times f_y&=(-3(x^2-y^2),6xy,1)\\
|f_x\times f_y|&=\sqrt{9(x^2-y^2)^2+36x^2y^2+1}=\sqrt{9(x^2+y^2)^2+1}\\
n&=\frac{f_x\times f_y}{|f_x\times f_y|}=\frac{(-3(x^2-y^2),6xy,1)}{\sqrt{9(x^2+y^2)^2+1}}\\
f_{xx}&=(0,0,6x)\\
f_{xy}&=(0,0,-6y)\\
f_{yy}&=(0,0,-6x)\\
E&=<f_x,f_x>=1+9(x^2-y^2)^2\\
F&=<f_x,f_y>=-18(x^2-y^2)xy\\
G&=<f_y,f_y>=1+36x^2y^2\\
L&=<n,f_{xx}>=\frac{6x}{\sqrt{9(x^2+y^2)^2+1}}\\
M&=<n,f_{xy}>=\frac{-6y}{\sqrt{9(x^2+y^2)^2+1}}\\
N&=<n,f_{yy}>=\frac{-6x}{\sqrt{9(x^2+y^2)^2+1}}.\\
\end{align*}
When $x=y=0$, we have that $$(g)=\begin{pmatrix}
E & F\\
F & G
\end{pmatrix}=\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}\mbox{ and }(b)=\begin{pmatrix}
L & M\\
M & N
\end{pmatrix}=\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix}.$$
Hence the shape operator $$(s)=(g^{-1})(b)=\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix}.$$
So the principal curvatures at point $x=y=0$ is that $k_1=k_2=0$.
I do not know how to continuous and I think I can not even calculate the principal direction at the origan and how can I know how the number of lines of curvature.
But I have observed that this surface is mirror symmetrical to $(x,z)-$plane. Does it have any relations?
I have also calculated that a line of curvature $f\circ \gamma$ with $\gamma=\begin{pmatrix}
\gamma_1\\
\gamma_2
\end{pmatrix}$ should fullfill the following equation $$\begin{vmatrix}
\gamma_2^{\prime 2} & -\gamma_1^\prime \gamma_2^\prime & \gamma_2^{\prime 2}\\
E & F & G\\
L & M & N \end{vmatrix}=0$$
but how can I know a line of curvature cross the origan point?
| Principal directions at $(x,0)$ (for $x\ne 0$) are $(1,0)$ and $(0,1)$.
The computations are very difficult unless you happen to recognize that $x^3-3xy$ is the real part of $(x+iy)^3$. Thus, this surface has rotational symmetry about the origin, with rotations of $\pm 2\pi/3$. If one does the computation with a polar coordinate parametrization, these things become more evident. So the $x$-axis and its $\pm 2\pi/3$ rotated images give the three lines of curvature coming into the origin.
P.S. Without any hints, I think this problem is rather unfair.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $A$, $B$ be square matrices of order $2$ such that $|I_2 + AB| = 0$. Prove that $|I_2 + BA| = 0$. In this question, I denote
$$A=\begin{bmatrix}a&b\\c&d\end{bmatrix},$$
$$B=\begin{bmatrix}e&f\\g&h\end{bmatrix}.$$
So, from $|I_2 + AB| = 0$, I'll have:
$$(ae+fc+1)(gb+hd+1) = (ga+hc)(eb+fd).$$
If $|I_2 + BA| = 0$ (which we have to prove here), then
$$(ae+bg+1)(cf+hd+1) = (af+bh)(ce+dg).$$
From both equations, I can see that what we're going to prove here is $cf=bg$, more precisely, $g=f$ and $b=c$.
Is my thinking path right or wrong?
I still have not figured out the solution yet, I appreciate any help.
| It is just a corollary (taking $n = m = 2$) of the result below:
For $A \in F^{m \times n}, B \in F^{n \times m}$, it follows that
\begin{equation}
\det(I_{(m)} + AB) = \det(I_{(n)} + BA). \tag{$*$}
\end{equation}
To prove identity $(*)$, consider the block matrix:
\begin{align}
\Delta = \begin{pmatrix} I_{(m)} & A \\
-B & I_{(n)}
\end{pmatrix},
\end{align}
and verify decompositions below:
\begin{align*}
& \begin{pmatrix}
I_{(m)} & A \\
-B & I_{(n)}
\end{pmatrix}
\begin{pmatrix} I_{(m)} & 0 \\
B & I_{(n)}
\end{pmatrix} =
\begin{pmatrix}
I_{(m)} + AB & A \\
0 & I_{(n)}
\end{pmatrix}, \tag{1} \\
& \begin{pmatrix}
I_{(m)} & A \\
-B & I_{(n)}
\end{pmatrix}
\begin{pmatrix} I_{(m)} & -A \\
0 & I_{(n)}
\end{pmatrix} =
\begin{pmatrix}
I_{(m)} & 0 \\
-B & I_{(n)} + BA
\end{pmatrix} \tag{2}
\end{align*}
Taking determinants on both sides of $(1)$ and $(2)$ then yields
$$\det(\Delta) = \det(I_{(m)} + AB) = \det(I_{(n)} + BA).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4333104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Calculate Indefinite Integral $\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx$ The following integration is given by Wolfram Alpha
$$\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx=\frac{x^6(28x^4+16x^3+39x^2+16x+28)}{168(1+x)^{16}}.$$
My question is: what is the best (meaning least work), method to achieve this result by hand ? There are two approaches I see, partial fractions, maybe setting $y=1+x$, but this is still alot of work. Or perform successive integration by parts. Both still involve a lot of calculation. The compact form of the answer makes me hope there is a nice way to achieve it. Guessing the ultimate form and then calculating the derivative is, to me, a less desirable method.
| We have the following : \begin{aligned}\frac{x^{5}\left(1-x^{6}\right)}{\left(1+x\right)^{18}}&=\frac{\left(1+x-1\right)^{5}}{\left(1+x\right)^{18}}-\frac{\left(1+x-1\right)^{11}}{\left(1+x\right)^{18}}\\ &=\frac{\sum\limits_{k=0}^{5}{\left(-1\right)^{k}\binom{5}{k}\left(1+x\right)^{5-k}}}{\left(1+x\right)^{18}}-\frac{\sum\limits_{k=0}^{11}{\left(-1\right)^{k}\binom{11}{k}\left(1+x\right)^{11-k}}}{\left(1+x\right)^{18}}\\ \frac{x^{5}\left(1-x^{6}\right)}{\left(1+x\right)^{18}}&=\sum_{k=0}^{5}{\binom{5}{k}\frac{\left(-1\right)^{k}}{\left(1+x\right)^{13+k}}}-\sum_{k=0}^{11}{\binom{11}{k}\frac{\left(-1\right)^{k}}{\left(1+x\right)^{7+k}}}\end{aligned}
Thus : $$ \int{\frac{x^{5}\left(1-x^{6}\right)}{\left(1+x\right)^{18}}\,\mathrm{d}x}=\sum_{k=0}^{5}{\binom{5}{k}\frac{\left(-1\right)^{k+1}}{\left(12+k\right)\left(1+x\right)^{12+k}}}-\sum_{k=0}^{11}{\binom{11}{k}\frac{\left(-1\right)^{k+1}}{\left(6+k\right)\left(1+x\right)^{6+k}}}+C $$
The common denominator is clearly $ \left(1+x\right)^{17} $, so I guess, we could explicit the sums above, simplify and factor, then we'll get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
minimum polynomial of $\sqrt[3]{3}+\sqrt[3]{5}$ over $\mathbb {Q}$ I have tried putting $x=\sqrt[3]{3}+\sqrt[3]{5}$ but i got stuck because after elevating to the third power I obtain again cube roots and I am stuck . How can I solve?
| \begin{align}
x &= \sqrt[3]{3} + \sqrt[3]{5} \\
x^3 &= 3 + 3\sqrt[3]{3\cdot 3 \cdot 5} + 3\sqrt[3]{3 \cdot 5 \cdot 5} + 5 \\
x^3 - 8 &= 3\sqrt[3]{3 \cdot 5}(\sqrt[3]{3} + \sqrt[3]{5}) \\
x^3 - 8 &= 3\sqrt[3]{15}x\\
(x^3-8)^3 &= 405 x^3 \\
x^9 - 24x^6 - 213 x^3 - 512 &= 0.
\end{align}
Therefore the minimum polynomial of $\sqrt[3]{3} + \sqrt[3]{5}$ is a factor of $x^9 - 24x^6 - 213 x^3 - 512$.
If shouldn't be difficult to prove that the defining field is of degree 9, and therefore this is the minimal polynomial.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove that $(p,mp]$ will always contain at least $y-1$ cubes
If $n, m\in \mathbb{Z}_{>0}$ and $(n, mn]$ contains $y$ cubes, prove that $(p,mp]$ will always contain at least $y-1$ cubes (integers $a$ so that $a=k^3$ for some integer k) for all $p\in [n,\infty)$.
This works for small examples, such as for $n=7, m=4$ when the cubes are $8$ and $27$ as $(p, 4p]$ always contains at least one cube for $p\ge n$, but I'm not sure how to generalize the results. I know that between any two real numbers that are at least one apart there is an integer. Also, clearly, the ratio between $k^3$ and $(k+1)^3$ decreases as $k$ increases. For $n\ge 8$, there exists an integer $k$ so that $k \in (n, mn]$ is a perfect cube because $\sqrt[3]{mn} - \sqrt[3]{n} \ge \sqrt[3]{n} > 1$.
| Lemma (without proof): Show that $ ( n, mn ] $ contains at least $y$ cubes iff
$$ \lfloor \sqrt[3]{mn} \rfloor - \lfloor \sqrt[3] {n} \rfloor \geq y.$$
Corollary: The problem can be restated to:
For $ p \geq n \geq 1$, and $ m \geq 1$, show that
$$ \lfloor \sqrt[3]{mp} \rfloor - \lfloor \sqrt[3] {p} \rfloor \geq \lfloor \sqrt[3]{mn} \rfloor - \lfloor \sqrt[3] {n} \rfloor - 1.$$
Proof of inequality: This is true because with $ \sqrt[3]{m} \geq 1$,
$$ \lfloor \sqrt[3]{mp} \rfloor - \lfloor \sqrt[3]{mn} \rfloor \geq \lfloor \sqrt[3]{mp} - \sqrt[3]{mn} \rfloor \geq \lfloor \sqrt[3]{m} \times \lfloor \sqrt[3]{p} - \sqrt[3]{n} \rfloor \rfloor \\ \geq \lfloor \sqrt[3]{p} - \sqrt[3]{n} \rfloor \geq \lfloor \sqrt[3]{p} \rfloor - \lfloor\sqrt[3]{n} \rfloor - 1.$$
The absolute-function inequalities used at each stage of the inequality are derived from the well-known inequalities (for suitable values of $x, y$, and rearranged as needed):
*
*$ \lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x+y \rfloor$
*$ \lfloor xy \rfloor \geq \lfloor x \lfloor y \rfloor \rfloor$.
*For $ x \geq 1$, $ \lfloor x \lfloor y \rfloor \rfloor \geq \lfloor y \rfloor$
*$ \lfloor x \rfloor + \lfloor y \rfloor \geq \lfloor x+y \rfloor - 1$
| {
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Integration of Bounded Region I'm trying to solve this problem:
The region of integration is the triangle $D$ with vertexes $A(0,0),B(1,1),C(10,1)$.
Find the solution of $\iint_D \sqrt{x^2-y^2}\,dx\,dy$.
MY SOLUTION (as @ryang suggested):
We can write the triangle as $D=\{(x,y)\in\mathbb R^2|0\le y\le 1,y\le x \le 10y\}$.
After that the integral becomes:
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy$$
We can solve first this indefinite integral $$\int\sqrt{x^2-y^2}\,dx$$
where $y\in[0,1]$ is a constant.
So$$\begin{align}\int\sqrt{x^2-y^2}\,dx=x\sqrt{x^2-y^2}-\int\frac{x^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\frac{x^2-y^2+y^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{1}{\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{(1+\frac{x}{\sqrt{x^2-y^2}})}{x+\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\log (x+\sqrt{x^2-y^2})\end{align}$$
The result is:
$$\int\sqrt{x^2-y^2}\,dx=\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})+C$$
We can now solve the definite integral
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy=\int_0^1\Bigl(\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})\Bigr)\Big|_y^{10y}\,dy\\=\int_0^1 \Bigl(5y^2\sqrt{99}-\frac{1}{2}y^2\log (10+\sqrt{99})\Bigr)\,dy\\=\frac{5\sqrt{99}}{3}-\frac{\log (10+\sqrt{99})}{6}
$$
| Using your parametrization we have:
$$\begin{align}\iint_{D_1}\sqrt{x^2-y^2}dxdy&=\int_0^1dx\int_{y=\frac{1}{10}x}^{y=x}\sqrt{x^2-y^2}dy\\&=\int_0^1xdx\int_{y=\frac{1}{10}x}^{y=x}\sqrt{1-\left(\frac{y}{x}\right)^2}dy \:\: \blacktriangle\end{align}$$
Now, suppose we know that (there's pleny of proofs of this fact on the net): $$\int\sqrt{1-u^2}du=\frac{\arcsin(u)+u\sqrt{1-u^2}}{2}+C$$
So: $$\int\sqrt{1-\left(\frac{y}{x}\right)^2}dy=\frac{x\arcsin\left(\frac{y}{x}\right)+y\sqrt{1-\left(\frac{y}{x}\right)^2}}{2}+C \:\: \star$$
Combining $\blacktriangle$ and $\star$ together we obtain:
$$\begin{align}\iint_{D_1}\sqrt{x^2-y^2}&=\int_0^1x\left[\frac{x\arcsin\left(\frac{y}{x}\right)+y\sqrt{1-\left(\frac{y}{x}\right)^2}}{2}\right]_{y=\frac{1}{10}x}^{y=x}dx\\&=\frac{1}{6}\left[\frac{\pi}{2}-\left(\arcsin\left(\frac{1}{10}\right)+\frac{\sqrt{99}}{100}\right)\right]\end{align}$$
In a similar way you can calculate $\iint_{D_2}\sqrt{x^2-y^2}dxdy$.
As suggested, there are better ways to parametrize your domain of integration
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve system of linear equations that came up in looking at a probability problem, any tricks? The following system of $k$ equations came up in a probability problem I was looking at:
$$(n+1)x_1 - x_2 = 1, \quad x_1 + nx_2 - x_3 = 1 , \quad x_1 + nx_3 - x_4 = 1, \quad x_1 + nx_4 - x_5 = 1, \quad \ldots \quad x_1 + nx_{k-1} - x_{k} = 1, \quad x_1 + nx_{k} = 2,$$which involves solving$$\begin{pmatrix}
n+1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0\\
1 & n & -1 & 0 & \cdots & 0 & 0 & 0 \\
1 & 0 & n & - 1 & \cdots & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
1 & 0 & 0 & 0 & \cdots & 0 & n & -1\\
1 & 0 & 0 & 0 & \cdots & 0 & 0 & n\\
\end{pmatrix} \begin{pmatrix}
x_1\\
x_2 \\
x_3 \\
\vdots \\
x_{k - 1}\\
x_k\\
\end{pmatrix} = \begin{pmatrix}
1\\
1 \\
1 \\
\vdots \\
1\\
2\\
\end{pmatrix}$$Now, given that this matrix looks pretty "sparse", I'm wondering if there's any shortcuts that can get us the answer quickly without me having to painstakingly row reduce this augmented matrix by brute force. I've started in the direction of brute force but couldn't see any immediate pattern/simplification.
For the record, the answer in the back of my book is$$x_r = {{n^r - n^{r - 1} + n^{k} - 1}\over{n^{k+1} - 1}}.$$
| Another approach, by solving a linear 2-term recurrence relation.
Let $ y_i = x_i - \frac{ 1-x_1}{ n-1}$.
Then, the equations become
$ny_i - y_{i+1} = 0$ for $ i = 1$ to $k-1$ and $y_1 + ny_k = 2 - \frac{n+1}{n-1} ( 1 - x_1)$,
which then become
$ y_{i+1} = n^{i} y_1 $ for $ i = 1 $ to $k-1$ and $y_1 (1+n^k) = 2 - \frac{n+1}{n-1} ( 1 - x_1)$.
Since $y_1 = x_1 - \frac{1-x_1} {n-1}$, the previous equation is only in terms of $x_1$, from which we get $x_1 = \frac{n^k + n - 2 } { n^{k+1} - 1 } $. So
$$x_i = y_i + \frac{1-x_1}{n-1} = n^{i-1} y_1 + \frac{1-x_1}{n-1} = \frac{n^i - n^{i-1} + n^k - 1 } { n^{k+1} - 1 }.$$
Notes
*
*Once $y_{i+1} = n^i y_1$, we can conclude that $x_i = n^{i-1} (x_1 - \frac{ 1-x_1}{ n-1})+ \frac{1-x_1}{n-1}$ which was Alan's $x_r = n^ra + b $form.
*If helpful to verify the calculations, we have $ \frac{1-x_1}{n-1} = \frac{n^k-1}{n^{k+1} - 1 } $, $y_1 = \frac{n-1}{n^{k+1} - 1 } $,
| {
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"source": "stackexchange",
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Find the minimum value of this
Let $a,b,c,d$ be nonnegative integers, such that $a+b+c+d=4$. Find the minimum value of $$\sum_{cyc}\frac{a}{b^3+4}.$$
My textbook says that the answer is $2/3$ achieved at $(2,2,0,0)$, but my method show that it is $1/17$ here it is
Since $a+b+c+d=4$, we have $a,b,c,d\le 4$ so $b^3+4\le 68$ hence $$\frac{a}{b^3+4} \ge\frac{a}{68}$$
Cycling we get the min is $1/17$, achieved at $(4,0,0,0)$, so where is the mistake?
| You have a mistake in replacement. Actually, the point $(a,b,c,d)=(4,0,0,0)$ leads to
$$
\sum_{cyc}\frac{a}{b^3+4}=\frac{4}{0^3+4}+\frac{0}{0^3+4}+\frac{0}{0^3+4}+\frac{0}{4^3+4}=1.
$$
What you found, is the value of $\sum_{cyc}\frac{a}{a^3+4}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the ratio of area in increasing order for the given figure?
Consider the attached image for the diagram. I'm interested in finding the area of the 4 parts so formed as the ratio of increasing order. Here line segment BD is diagonal for square and AE is so-called median for the square as it divides the line segment DC such that DE = EC.
I tried a lot but unfortunately couldn't find it.
I was able to find the length of AE and BD to be a√5 and 2a√2 assuming each side as 2a. Can we apply heron's formula to find the area of triangles?
Can you help me? Anyone?
| The sketch given below might help you.
$\underline{\text{Added at the request of @Utkarsh}}$
Let us denote the point of intersection of $AE$ and $BD$ as $V$. We draw a line segment perpendicular to the side $AB$ through $V$ to cut the side $CD$ at $U$. Another line segment is drawn through $V$ perpendicular to the side $AD$ to meet it at $T$. For brevity, we let $AB = BC= a$, and $VT = x$.
Area of $\triangle DAV$ can be expressed as,
$$\text{Area of }DAV = \frac{1}{2}ax. \tag{1}$$
The area of $\triangle ABD$ is equal to the half of that of $\square ABCD$. Therefore,
$$\text{Area of }ABD = \frac{1}{2}a^2. \tag{2}$$
Using (1) and (2), we can determine the area of $\triangle ABV$ as,
$$\text{Area of }ABV = \text{Area of }ABD - \text{Area of }AVD = = \frac{1}{2}a^2 - \frac{1}{2}ax = \frac{1}{2}a \left(a-x\right). \tag{3}$$
Since $V$ is located on the diagonal of $\square ABCD$, $VU = VT = x$. This let us express the area of $\triangle EDV$ in terms of $a$ and $x$ as shown below.
$$\text{Area of }EDV = \frac{1}{4}ax \tag{4}$$
An expression for the area of the quadrilateral $BCEV$ can be derived using (4).
$$\text{Area of } BCEV = \frac{1}{2}a^2 - \frac{1}{4}ax = \frac{1}{4}a \left(2a-x\right) \tag{5}$$
Now, we need to find $x$ in terms of $a$. To do that consider the two similar triangles $EDV$ and $ABV$. According to the theorem Euclid VI. 19., the ratio of the areas of two similar triangles is equal to the ratio of the squares on any of the three pairs of corresponding sides. Therefore, we shall write,
$$\text{Area of }EDV : \text{Area of }ABV = DE^2 : AB^2 = \frac{1}{4}a^2 : a^2 = 1: 4.$$
We can express the same ratio using (3) and (4) as,
$$\text{Area of }EDV : \text{Area of }ABV = \frac{1}{4}ax : \frac{1}{2}a \left(a-x\right) =x : 2 \left(a-x\right). $$
Therefore,
$$\dfrac{x}{2 \left(a-x\right)}=\dfrac{1}{4} \qquad\longrightarrow\qquad x=\dfrac{1}{3}a.$$
We use this to eliminate $x$ in formulae (1), (3), (4), and (5) to get,
$$\text{Area of }DAV = \dfrac{1}{6}a^2, \quad \text{Area of }ABV = \dfrac{1}{3}a^2, \quad \text{Area of }EDV =\dfrac{1}{12}a^2, \quad\text{and}\quad$$ $$\text{Area of } BCEV = \dfrac{5}{12}a^2.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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if $\cos^{10}x + \sin^{10}x=11/36$, find $\cos^{12}x+\sin^{12}x$ If $\cos^{10}x+ \sin^{10}x=11/36$, find $\cos^{12}x+\sin^{12}x$. I've tried solving using index manipulation but no way. I think we're to use $\sin^2 x+\cos^2 x=1$. But I don't know how. Need help please.
| Here is one solution. Let $A=\sin^2 x\cos^2 x$
$$\sin^4+\cos^4=\sin^2+\cos^2-2\sin^2 x\cos^2 x=1-2A$$
$$\sin^6+\cos^6=\sin^4+\cos^4-A(\sin^2 +\cos^2 x)=1-3A$$
$$\sin^8+\cos^8=\sin^6+\cos^6-A(\sin^4 +\cos^4 x)=1-4A+2A^2$$
$$\sin^{10}+\cos^{10}=\sin^8+\cos^8-A(\sin^6 +\cos^6 x)$$
$$=1-5A+5A^2$$
Thus we have $$5A^2-5A+1=\frac{11}{36}$$
Which gives $$A=\frac{1}{6}$$ using the fact that $A\leq \frac{1}{4}$
So finally
$$\sin^{12}+\cos^{12}=\sin^{10}+\cos^{10}-A(\sin^8 +\cos^8 x)$$
$$=1-6A+9A^2-2A^3=\frac{13}{54}$$
Hope I havent made a mistake.
| {
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Find the area $ S_ {OGBH}$ in the figure below For reference:
In the figure, calculate area $ S_ {OGBH} $ if the triangle area $ABC=9\ \mathrm{m^2}$ and $AB=3\ \mathrm m$ and $AO = 2\ \mathrm m$. (Answer: $5\ \mathrm{m^2}$)
My progress:
Draw $BO.$
Let $AG = x$ and $HO=GO=R.$
$BG = BH.$
$S_{BGO} = S_{BOH}$
$\displaystyle \frac{9}{S_{CHO}} = \frac{BC\cdot CH}{AC\cdot CO}$
$\displaystyle \frac{S_{BGO}}{S_{CHO}}=\frac{R\cdot BG}{R\cdot CH}=\frac{BG}{CH}$
$\displaystyle \frac{S_{AGO}}{S_{BGO}}=\frac{Rx}{R\cdot BG}=\frac{x}{BG}$
$\displaystyle S_{OAG} = \frac{Rx}{2}$
$\displaystyle \frac{S_{ABC}}{S_{ABO}} = \frac{3AC}{3\cdot2}\implies \frac{9}{S_{ABO}} = \frac{3AC}{6}\implies S_{ABO}=\frac{18}{AC}$
....???
| Say radius of the semicircle is $r$, Then $OG = OH = r$. Now drop a perp from $B$ to $AC$. Say the foot of perp is $E$.
$\triangle ABE \sim \triangle AOG$ with hypotenuse $3$ and $2$ respectively.
So, $S_{AOG} = \frac 49 S_{ABE}$ and $BE = \dfrac{3r}{2}$
Next, using similarity of $\triangle COH$ and $\triangle CBE$,
$S_{COH} = \frac 49 \cdot S_{BCE}$
But also, $S_{ABC} = S_{ABE} + S_{CBE} = 9$. That leads to $S_{BGOH} = 5$.
| {
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Show that $(b_1+b_2)^2\leq (a_1+a_2)(c_1+c_2)$ How to show that if $a_k,b_k,c_k\in \mathbb{R}^+$ for which $a_kc_k-b_k^2\geq 0$ for $k=1,2$, then
$$
(b_1+b_2)^2\leq (a_1+a_2)(c_1+c_2)?
$$
I tried to check some well-known identities such as Young's inequality, but I failed to make it work.
| HINT
You can start with the RHS as follows
\begin{align*}
(a_{1} + a_{2})(c_{1} + c_{2}) & = a_{1}c_{1} + a_{1}c_{2} + a_{2}c_{1} + a_{2}c_{2}\\\\
& \geq b^{2}_{1} + a_{1}c_{2} + a_{2}c_{2} + b^{2}_{2}
\end{align*}
Now you can apply the AM-GM inequality in order to obtain
\begin{align*}
a_{1}c_{2} + a_{2}c_{1} & \geq 2\sqrt{a_{1}c_{2}a_{2}c_{1}}\\\\
& = 2\sqrt{a_{1}c_{1}a_{2}c_{2}}\\\\
& \geq 2\sqrt{b^{2}_{1}b^{2}_{2}}\\\\
& = 2b_{1}b_{2}
\end{align*}
Can you take it from here?
| {
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How does $ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$ reduce to $ 1 - \frac{1}{\sqrt{n+2}}\;$? $$ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$$
Reduces to:
$$ 1 - \frac{1}{\sqrt{n+2}} $$
I have no clue how. What is the exact trick here and how can I practice this? I can't properly google the problem.
| $\require{cancel}$I'd us a combination putting over a common denominator and factoring.
Using that $foo = \sqrt{foo}^2$ we can factor $\sqrt{n+1}$ out of $(n+2)\sqrt{n+1} + (n+1)\sqrt{(n+2)}= \color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]$
So
$1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}=$
$1 -\frac 1{\color{green}{\sqrt{n+1}}} + \frac 1{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1- \frac {(n+2) + \sqrt{n+1}\sqrt{n+2}}{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}+\frac 1{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1-\frac {(n+2) + \sqrt{n+1}\sqrt{n+2}-1}{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1-\frac {(n+1) + \sqrt{n+1}\sqrt{n+2}}{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1-\frac {\color{green}{\sqrt{n+1}}(\sqrt{n+1} + \sqrt{n+2})}{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1-\frac {\cancel{\color{green}{\sqrt{n+1}}}(\sqrt{n+1} + \sqrt{n+2})}{\color{green}{\cancel{\color{green}{\sqrt{n+1}}}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1-\frac {\sqrt{n+1}+\sqrt{n+2}}{(n+2)+\sqrt{n+1}\sqrt{n+2}}=$
$1 -\frac {\sqrt{n+1}+\sqrt{n+2}}{\color{red}{\sqrt{n+2}}(\sqrt{n+2} + \sqrt{n+1})}=$
$1 -\frac {\cancel{\sqrt{n+1}+\sqrt{n+2}}}{\color{red}{\sqrt{n+2}}\cancel{(\sqrt{n+2} + \sqrt{n+1})}}=$
$1 - \frac 1{\color{red}{\sqrt{n+2}}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a$ and $b$ be positive integers such that $an + 1$ s a cube if and only if $ bn + 1$ is a cube. Prove that $a = b.$ Let $a$ and $b$ be positive integers such that $an + 1$ is a cube if and only if $bn + 1$ is a cube. Prove that $a = b.$
By choosing $p_n^3 \equiv 1 \mod b$ , we find that there are infinitely many numbers $n$ such that $:bn+1=p_n^3\Rightarrow an+1=q_n^3$
So we have : $(a-b)n = q_n^3-p_n^3 \Rightarrow (a-b)$ is a divisor of an infinite number of numbers $T$ of the form$:T=p^3-q^3$
This is a quite hard problem for me. Any assistance would be appreciated.
| Here's a fairly short proof.
There are infinitely many positive integers $n$ for which $abn+1$ is a cube $x^3$ (any $x\equiv 1\pmod{ab}$ works). Then, $a^2n+1$ and $b^2n+1$ must both be cubes; say $a^2n+1=y^3$ and $b^2n+1=z^3$. We have
$$(y^3-1)(z^3-1)=(a^2n)(b^2n)=(abn)^2=(x^3-1)^2;$$
rearranging gives
$$y^3+z^3-2x^3=(yz)^3-x^6.$$
If $x^2=yz$, then $y^3+z^3=2x^3$, and so $(a^2+b^2)n=2abn$ and $a=b$. So, if $a\neq b$, $x^2\neq yz$, which means
$$\left|y^3+z^3-2x^3\right|=\left|(yz)^3-x^6\right|\geq\min_{t\in\mathbb Z_{\neq 0}}\left|(x^2-t)^3-x^6\right|=3x^4-3x^2+1.$$
If $y^3+z^3\leq 2x^3$, then $|y^3+z^3-2x^3|<2x^3$, which contradicts the above inequality for $x>1$. So $y^3+z^3>2x^3$, and
$$\max(y^3,z^3)\geq \frac{y^3+z^3}2\geq x^3+\frac{3x^4-3x^2+1}2\geq x^4+1.$$
Let $z\geq y$, so that $z^3-1\geq x^4$. Then $b^2n\geq (abn+1)^{4/3}$, which can't hold for large $n$.
| {
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"url": "https://math.stackexchange.com/questions/4361692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\sin 2x$ from transforming $\sin^4 x+ \cos^4 x = \frac{7}{9}$ using trigonometric identities While studying for the first exams of the year, the following question found on one of Kognity's questionbank was extremely challenging for myself, a pre calculus student. The current topic is that of trigonometric equations, mostly making use of basic trigonometric identities (see image below for a screenshot of my formula booklet). Even after seeing the resolution, I just did not understand what magic was used to transform the sine and cosine.
The question below stated: Given that X is in quadrant III and $\sin^4 x + \cos^4 x = \frac{7}{9}$, work out the value of $\sin 2x$.
The question in its original format.
At first, one may notice that: $$\sin^4 x + \cos ^4 x=\left(\sin^2 x +\cos^2x\right)^2=1$$
because of the fundamental trigonometrical identity ($\sin^2x+\cos^2x=1)$. However to my frustration, this was as far as I got. Substituting by 1 did not yield any result, and I was forced to admit defeat and to submit the question without a resolution. Fortunately, I was at least pleased that I would have access to the resolution. Yet upon seeing the resolution, my confusion only doubled.
Resolution of the question.
By making use of black magic, the resolution transformed $$(\sin^4 x + \cos^4x)$$ into $$(\sin^4 x + \cos^4x)=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$$ Analysing and transforming the terms with the identities, one reaches that these two terms could be rewritten as: $$1-\sin^22x$$
The rest of the question was very straightforward, simple algebra to find an expression for X and noticing that $\sin2x=2\sin x\cos x$ were both in the third quadrant, therefore a negative multiplying a negative resulting in a positive result. My question, therefore, is how does one transform $$(\sin^4 x + \cos^4x)$$ into $$(\sin^4 x + \cos^4x)=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$$ It seems illogical since the second term appeared out of nowhere.
|
$\displaystyle \sin^4(x) + \cos^4(x) = \frac{7}{9}.$
$\sin(2x) = ~$ ?
$\underline{\text{Preliminary Results}}$
$$\cos(2x) = 2\cos^2(x) - 1 \implies 2\cos^2(x) = \cos(2x) + 1.\tag{R-1}$$
$$\cos^4(x) = \frac{1}{8} \times [\cos(4x) + 4\cos(2x) + 3]. \tag{R-2}$$
$$\sin^4(x) = \frac{1}{8} \times [\cos(4x) - 4\cos(2x) + 3]. \tag{R-3}$$
See the Addendum for a proof of (R-2) and (R-3).
Using (R-2) and (R-3),
$\displaystyle \frac{7}{9} = \frac{1}{4} \times
[\cos(4x) + 3] \implies $
$\displaystyle \frac{28}{9} = \cos(4x) + 3 \implies $
$\displaystyle \frac{1}{9} = \cos(4x) =
\cos^2(2x) - \sin^2(2x) = 1 - 2\sin^2(2x) \implies $
$\displaystyle \frac{8}{9} = 1 - \frac{1}{9} = 2\sin^2(2x) \implies $
$\displaystyle \sin^2(2x) = \frac{4}{9} \implies $
$\displaystyle \sin(2x) = \pm \frac{2}{3}.$
Given that $x$ is in quadrant $3$, you have that
$180^\circ < x < 270^\circ$.
This implies that $360^\circ < 2x < 540^\circ.$
This implies that $\displaystyle \sin(2x) = \frac{2}{3}.$
Addendum
Proof of (R-2) and (R-3).
Proof of (R-2)
Using (R-1),
$\cos(4x) = 2\cos^2(2x) - 1 $
$\displaystyle = 2 \left[2\cos^2(x) - 1\right]^2 - 1$
$ = 2[4\cos^4(x) - 4\cos^2(x) + 1] - 1$
$= 8\cos^4(x) - 8\cos^2(x) + 1$
$= 8\cos^4(x) - 4[\cos(2x) + 1] + 1$
$= 8\cos^4(x) - 4\cos(2x) - 3.$
This implies that
$\cos(4x) + 4\cos(2x) + 3 = 8\cos^4(x).$
Proof of (R-3)
$\sin^4(x) = [\sin^2(x)]^2 = [1 - \cos^2(x)]^2$
$\displaystyle = 1 - 2\cos^2(x) + \cos^4(x)$
$\displaystyle = 1 + \cos^4(x) - [\cos(2x) + 1]$
[using R-2]
$\displaystyle = \left\{ ~\frac{1}{8} \times [\cos(4x) + 4\cos(2x) + 3] ~\right\} - \cos(2x).$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A sum that's possibly equal to the Euler-Mascheroni Constant $\sum_{n=1}^\infty \frac{\ln n!}{n^3}$ The following interesting sum seems to approach the Euler-Mascheroni constant $\gamma$.
$$\sum_{n=1}^\infty \frac{\ln n!}{n^3} \overset{?}{=} \gamma$$
I've looked at the different ways to express the Euler-Mascheroni constant and tried to apply those methodes to this sum.
I also tried using the series representation of $\displaystyle \frac{\ln n!}{n^3}$ for $\displaystyle n=0,1,2$
and using the log gamma function
$$\displaystyle\ln \Gamma(n+1)=-\gamma-\gamma n-\ln (n+1)+\sum\limits_{k=1}^\infty\frac{n+1}{k}-\ln\left(1+\frac{n+1}{k}\right)$$
This problem might be quite well out of my mathmatical reach but I still would love to know the answer. It would be awesome if anyone could prove or disprove that the sum equals the Euler-Mascheroni constant and show their method.
| It is known that for $n\geq 1$,
$$
\log n! > \left( {n + \frac{1}{2}} \right)\log n - n + \log \sqrt {2\pi } + \frac{1}{{12n}} - \frac{1}{{360n^3 }}.
$$
Thus
\begin{align*}
\sum\limits_{n = 1}^\infty {\frac{{\log n!}}{{n^3 }}} & > \sum\limits_{n = 1}^\infty {\frac{{\log n}}{{n^2 }}} + \frac{1}{2}\sum\limits_{n = 1}^\infty {\frac{{\log n}}{{n^3 }}} - \sum\limits_{n = 1}^\infty {\frac{1}{{n^2 }}} \\ &\quad + \log \sqrt {2\pi } \sum\limits_{n = 1}^\infty {\frac{1}{{n^3 }}} + \frac{1}{{12}}\sum\limits_{n = 1}^\infty {\frac{1}{{n^4 }}} - \frac{1}{{360}}\sum\limits_{n = 1}^\infty {\frac{1}{{n^6 }}}
\\ & = - \zeta '(2) - \frac{1}{2}\zeta '(3) - \zeta (2) + \zeta (3)\log \sqrt {2\pi } + \frac{1}{{12}}\zeta (4) - \frac{1}{{360}}\zeta (6) \\ & = 0.583661 \ldots>\gamma.
\end{align*}
For the values of the derivative of $\zeta$ I used the OEIS A073002 and A244115. The other values can be expressed in terms of $\pi$.
| {
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About integration of $\frac{1}{(1+a^2 x^2)(1+x^2)}$ in differentiation under integral sign Consider the function $F(a)=\int_0^{+\infty} \frac{\arctan(ax)}{x(1+x^2)} dx$. I assumed $a \ge 0$ because $F$ is odd, I applied the theorem of differentiation under the integral sign and arrive to $F'(a)=\int_0^{+\infty} \frac{1}{(1+a^2 x^2)(1+x^2)}dx$; while doing the partial fraction decomposition of $\frac{1}{(1+a^2 x^2)(1+x^2)}$, I get
$$\frac{1}{(1+a^2 x^2)(1+x^2)}=\frac{A}{1+a^2 x^2}+\frac{B}{1+x^2}=\frac{A+Ax^2+B+Ba^2 x^2}{(1+a^2 x^2)(1+x^2)}$$
$$=\frac{A+B+(A+Ba^2)x^2}{(1+a^2 x^2)(1+x^2)}$$
So
$$\begin{cases} A+B=1 \\ A+Ba^2=0 \end{cases} \iff \begin{cases} A=1-B \\ 1-B+Ba^2=0 \end{cases} \iff \begin{cases} A=1-B \\ B(1-a^2)=1 \end{cases} \iff \begin{cases} A=-\frac{a^2}{1-a^2} \\ B=\frac{1}{1-a^2} \\ a \ne 1\end{cases}$$
I assumed $a \ne 1$ because I divided by $1-a^2$ (and not $a \ne -1$ too, because I'm working with $a \ge 0$). So I obtain the same result of the question I linked above, that is $F'(a)=\frac{\pi}{2(1+a)}$ and so $F(a)=\frac{\pi}{2}\log(1+a)$ for $a \ge 0$, by oddity then for $a<0$ it is $F(a)=-\frac{\pi}{2} \log (1-a)$.
The question is: what about $a=1$ and $a=-1$? To obtain the antiderivative with the partial fraction decomposition I divided by $1-a^2$, so I believe that I can't use the same expression when $a=1$ (or $a=-1$ by oddity), however the value of the integral for $a=1$ or $a=-1$ is respectively $\frac{\pi}{2} \log 2$ and $-\frac{\pi}{2}\log 2$, so it is coherent with the result $F(a)=\text{sgn}(a)\frac{\pi}{2} \log(1+|a|)$ even if $a=1$ or $a=-1$. I don't understand why this works even if, at some step, I divided by something that could be $0$ and still get the right result. Moreover, I can't fix $a=1$ or $a=-1$ to do the partial fraction decomposition of $\frac{1}{(1+x^2)^2}$ because all the point of differentiation under the integral sign is having an parameter to fix appropriately at the end and get the desired value of the integral; so fixing it makes all this senseless (moreover, $1/(1+x^2)^2$ is already decomposed).
I searched other answers and no one seems to consider this possibility of dividing by $0$ when using partial fraction decomposition, so I assume that I am missing something about partial fractions decomposition theory or somewhere else. What am I missing?
Edit. I tried, failing, to see if $F(1)$ was integrable with some tricks typical of definite integrals, like trying to use the substitution $t=\frac{1}{x}$ and using the identity $\arctan\frac{1}{t}=\pi/2-\arctan t$ valid for $t>0$, but it doesn't seem to help. So I'm still stuck.
| If $a=\pm1$ your fraction becomes $\frac 1{(1+x^2)^2}$ The partial fraction decomposition is modified in the case of a repeated root in the denominator. Over the reals this one is already decomposed. Alpha gives the integral of this as
$$\int \frac 1{(1+x^2)^2} dx= \frac 12 \left(\frac x{(x^2 + 1)} + \arctan(x)\right) + c$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum \frac{a^3}{a^2+b^2}\le \frac12 \sum \frac{b^2}{a}$
Let $a,b,c>0$. Prove that
$$ \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\le \frac12 \left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).\tag{1}$$
A idea is to cancel the denominators, but in this case Muirhead don't work because the inequality is only cyclic, not symmetric.
Another idea would be to apply Cauchy reverse technique:
$$(1)\iff\sum \left(a-\frac{a^3}{a^2+b^2}\right)\ge \frac12 \sum (2a-b^2/a)\iff \sum\frac{ab^2}{a^2+b^2}\ge \frac12\sum\frac{2a^2-b^2}{a}$$
$$\iff \sum \frac{(ab)^2}{a^3+ab^2}\ge \frac12\sum \frac{2a^2-b^2}a.$$
Now we can apply Cauchy-Schwarz, and the problem reduces to
$$\frac{(\sum ab)^2}{\sum a^3+\sum ab^2}\ge \frac12\sum \frac{2a^2-b^2}{a},$$
and at this point I am stuck. Here the only idea is to cancel the denominators, but as I say above it can't work.
| Using Cauchy-Bunyakovsky-Schwarz,
\begin{align*}
\frac{ab^2}{a^2 + b^2} + \frac{bc^2}{b^2 + c^2} + \frac{ca^2}{c^2 + a^2}
\ge \frac{(b + c + a)^2}{\frac{a^2 + b^2}{a} + \frac{b^2 + c^2}{b} + \frac{c^2 + a^2}{c}}
= \frac{(a + b + c)^2}{a + b + c + \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}}.
\end{align*}
It suffices to prove that
$$\frac{(a + b + c)^2}{a + b + c + \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}}
\ge a + b + c - \frac12\left(\frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}\right).$$
Denote $p = a + b + c$ and $Q = \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}$.
It suffices to prove that
$$\frac{p^2}{p + Q} \ge p - \frac12 Q$$
or
$$p^2 \ge (p + Q)(p - Q/2) = p^2 + pQ/2 - Q^2/2$$
or
$$Q \ge p$$
i.e.,
$$\frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c} \ge a + b + c$$
which is true using Cauchy-Bunyakovsky-Schwarz (easy).
| {
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"timestamp": "2023-03-29T00:00:00",
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Six people (half are female, half are male) for seven chairs. Problem:
Suppose there are $7$ chairs in a row. There are $6$ people that are going to randomly
sit in the chairs. There are $3$ females and $3$ males. What is the probability that
the first and last chairs have females sitting in them?
Answer:
Let $p$ be the probability we seek. Out of $3$ females, only $2$ can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
\begin{align*}
p &= \dfrac{ {3 \choose 2 } 3(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(3)(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(4)(3)(2) } { 7(5)(4)(3)(2) } = \dfrac{ 3(3)(2) } { 7(5)(3)(2) } \\
p &= \dfrac{ 18 } { 35(3)(2) } \\
p &= \dfrac{ 3 } { 35 }
\end{align*}
Am I right?
Here is an updated solution.
Let $p$ be the probability we seek. Out of $3$ females, only $2$ can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
\begin{align*}
p &= \dfrac{ 3(2) (5)(4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ (5)(4)(3)(2) } { 7(5)(4)(3)(2) } \\
p &= \dfrac{1}{7}
\end{align*}
Now is my answer right?
| We can think of the seating assignment as a random permutation of $7$ items (the three males, the three females, and the empty seat). This random permutation puts a female in the first chair with probability $\frac37$. Conditional on having done that, there are $2$ females left, so one of them ends up in the last chair with probability $\frac26$.
Overall, $p = \frac37 \cdot \frac26 = \frac17$.
There's also the brute force approach: $p = \frac{20}{140}$ by counting.
With an approach whose denominator is $7!$, the numerator should be $\binom32 \cdot 2 \cdot 5!$: we pick the $2$ females at the ends, pick the order they sit in, and then pick the permutation of the middle. This also gives $\frac{\binom 32 \cdot 2 \cdot 5!}{7!} = \frac{3\cdot 2}{7 \cdot 6} = \frac17$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is wrong with this projection onto two basis vectors? Given
\begin{align}
b_1 &= \begin{bmatrix} 2 \\ -1 \end{bmatrix} \\
b_2 &= \begin{bmatrix} 1 \\ 3 \end{bmatrix} \\
v &= \begin{bmatrix} 3 \\ 2 \end{bmatrix}
\end{align}
it is clear that
\begin{align}
v &= b_1 + b_2 \\
&= \begin{bmatrix} b_1 & b_2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix}
\end{align}
However, I want to compute the vector $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ manually using projections. I know that
$$
v = \left(v \cdot \frac{b_1}{||b_1||}\right)\frac{b_1}{||b_1||} + \left(v \cdot \frac{b_2}{||b_2||}\right)\frac{b_2}{||b_2||}
$$
and so I expect that
\begin{align}
v \cdot \frac{b_1}{||b_1||^2} &= 1 \\
v \cdot \frac{b_2}{||b_2||^2} &= 1
\end{align}
However, what I get instead is
\begin{align}
v \cdot \frac{b_1}{||b_1||^2} &= \begin{bmatrix} 3 & 2 \end{bmatrix} \begin{bmatrix} \frac{2}{5} \\ \frac{-1}{5} \end{bmatrix} \\
&= \frac{4}{5} \\
v \cdot \frac{b_2}{||b_2||^2} &= \begin{bmatrix} 3 & 2 \end{bmatrix} \begin{bmatrix} \frac{1}{10} \\ \frac{3}{10} \end{bmatrix} \\
&= \frac{9}{10}
\end{align}
Where am I going wrong?
| The flaw lies in the fact that equation $v = \left(v \cdot \frac{b_1}{||b_1||}\right)\frac{b_1}{||b_1||} + \left(v \cdot \frac{b_2}{||b_2||}\right)\frac{b_2}{||b_2||}$ is valid only when $b_1$ and $b_2$ are orthogonal. (It also requires $v$ to be in the span of $\{b_1,b_2\}$, although that is satisfied here.)
Side note: I recommend against writing things like $b_1 + b_2
= \begin{bmatrix} b_1 & b_2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$, because it will lead to confusion about when matrix multiplication is and isn't valid. Multiplying by $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ happens to be a proxy for adding two real numbers, but that doesn't mean it should be used as a proxy for adding any two mathematical objects.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the maximum value of $x^2y^3$ subject to the condition that $3x+2y=1$ I am trying to use G.M.$\leq $ A.M.as follows
$(\frac{2x^2y^3}{4})^\frac{1}{5} \leq \frac{x+2x+y+\frac{y}{2} +\frac{y}{2}}{5}$
$\implies (\frac{x^2y^3}{2})^\frac{1}{5} \leq \frac{1}{5}$
$\implies$ $x^2y^3\leq \frac{2}{5^5}$.
Is it correct ?
| For positive $x$ and $y$, you may use AM-GM:
$$\frac{1}{5}=\frac{3x+2y}{5}=\frac{\frac{3x}{2}+\frac{3x}{2}+\frac{2y}{3}+\frac{2y}{3}+\frac{2y}{3}}{5}\ge \sqrt[5]{\frac{2}{3}x^2y^3}$$
From where one gets $x^2y^3\le\frac{3}{2\cdot 5^5}=\frac{3}{6250}$
The equality takes place when $\frac{3x}{2}=\frac{2y}{3}$, which coupled with $3x+2y=1$ yields $x=\frac{2}{15}$ and $y=\frac{3}{10}$
But of course, the general method (not precalculus level as tagged) to solve this optimization problem is by using the method of Lagrange Multipliers.
As others noted, if we omit $x,y>0$ there is no global maximum.
| {
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Suppose X$\sim$ Cauchy(0,1). Then what will be the distribution of $\frac{1-X}{1+X}$? In order to find distribution of $\frac{1-X}{1+X}$ below approach I followed,
Let,
\begin{align}
Y = \frac{1-X}{1+X}
\end{align}
Then, cdf of Y is
\begin{align}
F_{Y}(y) = P(Y \leq y)
\end{align}
\begin{align}
= P\left(\frac{1-X}{1+X} \leq y\right)
\end{align}
\begin{align}
= 1 - P\left(X < \frac{1-y}{1+y}\right)
\end{align}
\begin{align}
= 1 - \int_{-\infty}^{\frac{1-y}{1+y}} f(x) \,dx
\end{align}
\begin{align}
= 1 - \int_{-\infty}^{\frac{1-y}{1+y}} \frac{1}{\pi}\cdot \frac{1}{1+x^2} \,dx
\end{align}
\begin{align}
= 1 - \frac{1}{\pi}\cdot \left[tan^{-1}x\right]_{-\infty}^{\frac{1-y}{1+y}}
\end{align}
\begin{align}
F_{Y}(y) = \frac{1}{\pi}\cdot \left[-\frac{\pi}{2}+tan^{-1}\left({\frac{1-y}{1+y}}\right)\right]
= \frac{1}{2} -\frac{1}{\pi}.tan^{-1}\left({\frac{1-y}{1+y}}\right)
\end{align}
and then
\begin{align}
f_{Y}(y) = \frac{d F_{Y}(y)}{dy} = \frac{1}{\pi}\cdot \frac{1}{1+y^2}
\end{align}
But I have a little confusion here how to find range of Y from X? And CDF of Y is doesn't looks like cdf of a cauchy distribution.
| Remark (Theorem of transformation for one random variable): Let be $X$ a continuous r.v. with pdf given by $f_X$ and let be Y=g(X) with g a diffeomorphism, then the pdf $f_Y$ is given by:
$f_Y(y)=f_X(g^{-1}(y))\cdot|{{d\over{dy}}g^{-1}(y)}|$.
If $Y=g(X)={{1+X}\over{1-X}}$ then $X=g^{-1}(Y)={{1-Y}\over{1+Y}}$. We find that ${dx\over{dy}}=-{{2\over{1+y^2}}}$. Now:
$f_Y(y)=f_X({{1-y}\over{1+y}})\cdot{2\over{1+y^2}}$.
I'll let you do the math, then you get:
$f_Y(y)={1\over{\pi{(1+y^2)}}}$.
So $Y\sim{Cauchy(0,1)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375593",
"timestamp": "2023-03-29T00:00:00",
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"answer_count": 4,
"answer_id": 3
} |
Representing the cube of any natural number as a sum of odd numbers I'm expanding my notes on exercises from Donald Knuth's The Art of Computer Programming, and found something rarely mentioned in the Internet, but still useful to prove Nicomachus' Theorem about the sum of cubes.
Knuth phrases this in the following way in exercise 8(a) to chapter 1.2.1:
Prove the following theorem of Nicomachus (A.D. c. 100) by induction: $1^3=1$, $2^3=3+5$, $3^3=7+9+11$, $4^3=13+15+17+19$, etc.
In the answers to exercises, the author gives the following formula: $(n^2-n+1)+(n^2-n+3)+...+(n^2+n-1)=n^3$
My question is, how do I get to that formula from the sample sums given in the problem? It looks kind of odd, especially because the last summand doesn't give me any idea on how it is connected with first ones. Usually one is able to see this clearly, but not here.
I tried to get to that formula by the following set of thoughts:
To get $n^3$ one must sum up $n$ odd numbers, starting from the $(n-1)$th central polygonal number, meaning that a cube of number $n$ is made by summing up odd numbers, starting from $(n-1)$th central polygonal number to $n$th triangular number.
Since odd numbers form arithmetic progression with $a_1=1, d=2$, it's possible to use the following formula of summing up $p$th to $q$th member of this progression:
$$S_{p,q}=\dfrac{a_p+a_q}2\cdot(q-p+1)$$
We can simplify $\dfrac{a_p+a_q}2$, putting $a_p=2p-1$ and $a_q=2q-1$:
$$\dfrac{a_p+a_q}2=\dfrac{(2p-1)+(2q-1)}2=\dfrac{2p+2q-2}2=p+q-1$$
Thus, the formula of summing up $p$th to $q$th member of this progression is:
$$(p+q-1)(q-p+1)=(q+(p-1))(q-(p-1))=q^2-(p-1)^2$$
Substituting in $p=\dfrac{n(n-1)}{2}+1=\dfrac{n^2-n+2}{2}$, and $q=\dfrac{n(n+1)}{2}$, we get the formula:
$$\left(\dfrac{n(n+1)}{2}\right)^2-\left(\dfrac{n(n-1)}{2}+1-1\right)^2=\left(\dfrac{n(n+1)}{2}\right)^2-\left(\dfrac{n(n-1)}{2}\right)^2=\\
\dfrac{(n(n+1))^2-(n(n-1))^2}{4}=\dfrac{(n(n+1)-n(n-1))(n(n+1)+n(n-1))}{4}=\\
\dfrac{n^2(n+1-n+1)(n+1+n-1)}{4}=\dfrac{4n^3}{4}=n^3$$
This... kind of... proves the sum formula for any $n$, really. But it doesn't give out the formula in question, i.e. $(n^2-n+1)+(n^2-n+3)+...+(n^2+n-1)$.
What is the correct way to get this formula? Any hints are greatly appreciated.
| I've approached the problem wrongly, as it turned out. Below is the concise solution I came up with and haven't seen elsewhere.
Let's start out from analyzing initial data:
*
*To get $1^3$, add $1$ odd number, starting from the $1$st odd number.
*To get $2^3$, add $2$ odd numbers, starting from the $2$nd odd number.
*To get $3^3$, add $3$ odd numbers, starting from the $4$th odd number.
*To get $4^3$, add $4$ odd numbers, starting from the $7$th odd number.
That leads to the following conclusions:
*
*To get $1^3$, it's needed to sum $1$st to $1$st odd numbers.
*To get $2^3$, it's needed to sum $2$nd to $3$rd odd numbers.
*To get $3^3$, it's needed to sum $4$th to $6$th odd numbers.
*To get $4^3$, it's needed to sum $7$th to $10$th odd numbers.
Hence it's needed to demonstrate that a cube of any natural number $n$ is equal to sum of $n$ odd numbers, starting from the next odd number after the last one used to get a cube of $(n-1)$.
To avoid 'notice that <...>' kind of wording with the goal of avoiding the risk of misunderstanding the way things are, it's necessary to base further logic directly on facts stated above.
If for getting, for example, the value of $2^3$ it's necessary to add $2$nd to $3$rd odd number, then it's necessary to subtract the sum of the first one odd number (technically, it's just the first odd number) from the sum of the first three odd numbers. In mathematical terms, $2^3=(1+3+5)-(1)$.
This same principle can be applied further, which can be seen below:
*
*$3^3=(1+3+5+7+9+11)-(1+3+5)$
*$4^3=(1+3+5+7+9+11+13+15+17+19)-(1+3+5+7+9+11)$
For purposes of clarity and concision, the first such sum be denoted as minuend sum, and the second such sum as the subtrahend sum. For example, $3^3$ has the minuend sum $1+3+5+7+9+11$ and the subtrahend sum $1+3+5$.
To be able to generalize these cases, it's necessary to have the formula of the quantity of terms in each of these two sums for any natural $n$.
According to the pattern discovered directly from the initial data, incrementing $n$ adds $n$ new terms to the minuend sum. For example, the minuend sum of $2^3$ has $2$ more terms than the minuend sum of $1^3$. Hence the minuend sum of $n^3$ will have $1+2+...+n$ terms, and the subtrahend sum will have as much terms as the minuend sum of $(n-1)^3$, i.e. $1+2+...+(n-1)$ terms.
The sequence A005408 from OEIS gives the formula for $n$th odd number: $(2n-1)$. The sum of first $n$ natural numbers has a special name in mathematics: triangular numbers. The sequence A000217 provides a formula $1+2+...+n=\dfrac{n(n+1)}2$, provable by numerous ways, one being mentioned in this question.
From here, we get:
$1+2+...+n=\dfrac{n(n+1)}2$,
$1+2+...+(n-1)=\dfrac{(n-1)((n-1)+1)}2=\dfrac{n(n-1)}2$.
Plugging these into the formula of $n$th odd number, we get the last term of the minuend sum:
$2\cdot\dfrac{n(n+1)}2-1=n(n+1)-1=n^2+n-1$
Same approach is used to get the last term of the subtrahend sum:
$2\cdot\dfrac{n(n-1)}2-1=n(n-1)-1=n^2-n-1$
Thus, this is how we generalize the initial case:
$n^3=(1+3+...+(n^2+n-1))-(1+3+...+(n^2-n-1))$
After subtracting the subtrahend sum from the minuend sum, all terms up to $(n^2-n-1)$ inclusive will cancel out, so we are left with the sum of the odd numbers from the one next to $(n^2-n-1)$, which is $(n^2-n-1+2)=(n^2-n+1)$ to $(n^2+n-1)$. Thus, we have just discovered the formula in question!
$n^3=(n^2+n+1)+(n^2+n-2)+...+(n^2+n-1)$.
Hope this will be helpful to those struggling to understand this task.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Is there an elegant method to find the minimum value of $\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$ for positive $x$?
Find the minimum value of $\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$ for $x\gt0$.
In some textbook, the problem is usually tackled by calculus. So I started to investigate the problem using triangle inequality only. The answer is so interesting and simple that the minimum value of $S(x)$ is $\sqrt{10}$ when $x=\frac{1+\sqrt{13}}{6}$.
My Question:
How many elegant methods are there to find the minimum point?
| We are going to find the minimum value of}
$$ S(x)=\dfrac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$$
geometrically using Triangle Inequality.
$$\displaystyle \begin{aligned} S(x)&=\sqrt{x^{2}+1+\frac{2}{x}+\frac{1}{x^{2}}} +\sqrt{x^{2}-2 x+5-\frac{4}{x}+\frac{1}{x^{2}}} \\&=\sqrt{x^{2}+\left(\frac{1}{x}+1\right)^{2}} +\sqrt{(x-1)^{2}+\left(\frac{1}{x}-2\right)^{2}}\end{aligned}$$
$\textrm{which is the sum of distances of any point }P\text{ on a rectangular hyperbola from }(0,-1)$
$\textrm{and }(1,2) \textrm{ as shown below:}$
Since $S(x) =A P+P B \geqslant A B =\sqrt{(1-0)^{2}+(2-(-1))^{2}}=\sqrt{10},$
therefore the minimum value of S(x) is $\sqrt{10},\\ $ when and only when A, B and P are collinear$ \quad \textrm{ i.e. } \dfrac{2-\frac{1}{x}}{1-x}=\dfrac{2-(-1)}{1-0 }
\Leftrightarrow \dfrac{2 x-1}{1-x}=3 x \Leftrightarrow 3 x^{2}-x-1=0 \Leftrightarrow x=\dfrac{1+\sqrt{13}}{6}. $
We can now conclude that the minimum value of S(x) is $\sqrt{10}$ when $x=\dfrac{1+\sqrt{13}}{6}.$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve for theta , $0<\theta<90^\circ$
Solve $$\cos60=\frac{\cos(2θ)+\frac32\sin(2θ))}{\sqrt{4-\sin^2 (2θ)}},$$ where $0<θ<90^\circ $
Question: Is the following solution correct?
Solution attempt:
$$\sqrt{4-\sin^2 (2θ)}=2\cos(2θ)+3\sin(2θ)$$
$$4-\sin^2 (2θ)=(2\cos(2θ)+3\sin(2θ))^2$$
$$4-\sin^2 (2θ)=4\cos^2 (2θ)+12 \sin(2θ) \cos(2θ)+9\sin^2 (2θ)$$
$$4=4\cos^2 (2θ)+12\sin(2θ) \cos(2θ)+10\sin^2 (2θ)$$
$$4=4(1-\sin^2 (2θ) +12 \sin(2θ) \cos(2θ)+10\sin^2 (2θ)$$
$$6\sin^2 (2θ)+12 \sin(2θ) \cos(2θ)=0$$
$$\sin(2θ) (\sin(2θ)+2 \cos(2θ) )=0$$
$$\sin(2θ)=0 \text{ or } \sin(2θ)+2 \cos(2θ)=0$$
$$θ=0 \ \text{or} \ \tan(2θ)+2=0$$
$θ=0$ or $\tan^{-1} (-2)=-63.4$
$θ=0$ or $2θ=-63.4, \theta=-31.7$
| When you arrive to $$\sin(2θ)=0 \text{ or } \sin(2θ)+2 \cos(2θ)=0$$ you should conclude as follows:
Since $\theta \in (0,90^\circ)$, then $2\theta \in (0,180^\circ)$, so $\sin(2\theta)\ne 0$.
From the other equation, $\tan(2\theta) = -2$, so $2\theta = \arctan(-2) + 180^\circ k$ for some integer $k$. Then $\theta = \frac{1}{2}(\arctan(-2) + 180^\circ k)$, and the only value of $k$ that makes $\theta \in (0,90^\circ)$ is $k=1$, so $\theta = \frac{1}{2}(180^\circ-\arctan(2)) = 58.2825\ldots^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating probability of values for multiple dice rolls where sum is equal or greater than 4 Assuming we have 6 sided dice and play game of rolling dice each turn until we get sum value of at least 4.
For example we know that game will always finish with value less or equal than 9 because on 1. turn max value we can get without finishing the game is 3, then we can get highest value of 6, all values on 2. turn will result in finishing game.
My question is, what is the probability that game will finish with value 5, 6 or 7?
What's the correct approach to solve this probabilities? My difficulty is that each next turn is conditional on probability of previous turns.
| You know that the final sum will be in the range of $4$ through $9$. Determine all possible rolls that yield the final result. Add up the probabilities as each event will be disjoint.
Example: to get to 4, you have the following roll patterns:
$$4 \\ 3, 1 \\ 2, 2 \\ 2, 1, 1 \\ 1, 3\\ 1,2,1 \\1, 1, 2 \\ 1, 1, 1, 1$$
This gives:
$$P(4) = P(5) = P(6) = \dfrac{1}{6}+3\cdot \dfrac{1}{6^2}+3\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4} = \dfrac{344}{1296} \\ P(7) = 3\cdot \dfrac{1}{6^2}+3\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4} = \dfrac{127}{1296} \\ P(8) = 2\cdot \dfrac{1}{6^2}+3\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4}=\dfrac{91}{1296} \\ P(9) = \dfrac{1}{6^2}+2\cdot \dfrac{1}{6^3}+\dfrac{1}{6^4}=\dfrac{49}{1296}$$
As a semi-verification, let's add it all up to make sure it adds to $1$:
$$\dfrac{3\cdot 343+127+91+49}{1296} = 1$$
This appears to be the correct probability for each result.
Another way to think about this: Let's consider the game in rounds. It can end on round $1$ with a roll of $4,5,6$. So that is a $\dfrac{1}{6}$ probability for each.
It can end on round $2$ with a first roll of $1,2,3$ for round $1$ to reach any of $4,5,6,7$ (so $3\dfrac{1}{6^2}$ probability of each), a first round roll of $2,3$ can reach $8$ for a probability of $2\cdot \dfrac{1}{6^2}$, or $3$ can reach $9$ for a probability of $\dfrac{1}{6^2}$.
It can end on round $3$ with the first two rolls as $(1,1),(1,2),(2,1)$ and reach $4,5,6,7,8$ for a probability of $3\cdot \dfrac{1}{6^3}$. And the first two rolls being $(1,2),(2,1)$ can reach $9$ with probability $2\cdot \dfrac{1}{6^3}$.
And it can reach any result on the fourth round by rolling three consecutive $1$'s followed by any roll of the $d6$ for a probability of $\dfrac{1}{6^4}$ each.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral replacement I have the following integral
$φ=\frac{M}{\sqrt{2m}}\int_{ }^{ }\frac{dr}{r^{2}\sqrt{E-\frac{M^{2}}{2mr^{2}}-\frac{kr^{2}}{2}}}$
As a result, I want to get the following result: $r=\frac{p}{1-ε\cosφ}$ This satisfies the ellipse equation in polar coordinates
I know I have to get an integral that reduces to an arccosine or an arcsine, but I can't think of a replacement for $p$, $ε$
| $E=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\theta}^2+\frac{1}{2}kr^2=\frac{1}{2}m\dot{r}^2+\frac{1}{2} \frac{ L^2}{mr^2} +\frac{1}{2}kr^2$
$k/m=\omega^2$
$\frac{2E}{m}=\dot{r}^2+\frac{L^2}{m^2r^2}+\omega^2r^2$
$\frac{2E}{m}r^2=r^2\dot{r}^2+\frac{L^2}{m^2}+\omega^2r^4$
$\frac{2E}{m}r^2-\omega^2r^4-\frac{L^2}{m^2}=r^2\dot{r}^2$
$\sqrt{ \frac{2E}{m}r^2-\omega^2r^4-\frac{L^2}{m^2}}=r\dot{r}$
$dt=\frac{rdr}{\sqrt{\frac{2E}{m}r^2-\omega^2r^4-\frac{L^2}{m^2}}}, u=r^2$
$dt=\frac{du}{2\sqrt{\frac{2E}{m}u-\omega^2u^2-\frac{L^2}{m^2}}}$
Complete the square.
$dt=\frac{du}{2\omega\sqrt{\frac{-E^2}{m^2\omega^2}+ \frac{2E}{m\omega^2}u-u^2+\frac{E^2}{m^2\omega^2}-\frac{L^2}{m^2\omega^2}}}$
$dt=\frac{du}{2\omega\sqrt{\alpha^2-(u-\frac{E}{m\omega^2})^2}}$
Now with some variable substitution, you should be able to get it to something of the form $\int \frac{du}{\sqrt{1-u^2}}$
Then you have $\omega t=\arcsin(Au+B)+C$ for appropriate constants.
From there: $(\sin(\omega t-C)-B)/A=u$
Once you have $u$, you have $r$. From $r$ you can get $\dot{\theta}=\frac{L}{mu}$
By the chain rule:
$\frac{dr}{d\theta}=\frac{\dot{r}}{\dot{\theta}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve simultaneous equations of the form $\frac{a}{x}+\frac{b}{y}=1$ really fast? Context:
*
*Taking the major and minor axes of an ellipse as the $x$ and $y$ axes respectively find the equation of the ellipse passing through the points $(1,\sqrt{6})$ and $(3,0)$.
*Taking the major and minor axes of an ellipse as the $x$ and $y$ axes respectively find the equation of the ellipse passing through the points $(2,4)$ and $(5,\sqrt{2})$.
In problems like these, I frequently find myself solving simultaneous equations of the form $\frac{a}{x}+\frac{b}{y}=1$. For example, when doing math 2, I had to solve the following system of equations (I assumed $p=a^2$, $q=b^2$).
$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$
$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$
I'll now show how I solved them.
$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$
$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$
Now in $(1)$,
$$\frac{4q+16p}{pq}=1$$
$$4q+16p=pq$$
Again in $(2)$,
$$\frac{25}{p}+\frac{2}{q}$$
$$\frac{25q+2p}{pq}=1$$
$$25q+2p=pq$$
Now,
$$4q+16p=25q+2p$$
$$21q-14p=0$$
$$21q=14p$$
$$q=\frac{14p}{21}$$
Now,
$$\frac{4}{p}+\frac{16}{q}=1$$
$$\frac{4}{p}+\frac{16}{\frac{14p}{21}}=1$$
$$\frac{4}{p}+\frac{16}{\frac{14p}{21}}=1$$
$$\frac{4}{p}+\frac{336}{14p}=1$$
$$\frac{56+336}{14p}=1$$
$$14p=392$$
$$p=28$$
Now,
$$\frac{4}{28}+\frac{16}{q}=1$$
$$\frac{1}{7}+\frac{16}{q}=1$$
$$\frac{16}{q}=\frac{6}{7}$$
$$6q=112$$
$$q=\frac{56}{3}$$
So, the equation of the ellipse is $$\frac{x^2}{28}+\frac{y^2}{\frac{56}{3}}=1\ \text{(Ans.)}$$
As you can see, the process is pretty long and time-taking. Is there any way to find the values of $p,q$ in 2-3 lines?
| I've found a quicker way:
$$\frac{4}{p}+\frac{16}{q}=1\tag{1}$$
$$\frac{25}{p}+\frac{2}{q}=1\tag{2}$$
Now, $(2)\times8-(1)$:
$$\frac{200}{p}+\frac{16}{q}-\frac{4}{p}-\frac{16}{q}=8-1$$
$$\frac{196}{p}=7$$
$$p=\frac{196}{7}$$
$$p=28$$
Now, inputting $p=28$ in $(1)$,
$$\frac{4}{28}+\frac{16}{q}=1$$
$$\frac{16}{q}=\frac{6}{7}$$
$$q=7\cdot\frac{16}{6}$$
$$q=\frac{56}{3}$$
$$(p,q)=(28,\frac{56}{3})$$
| {
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"answer_count": 2,
"answer_id": 0
} |
Compute $\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$ with the residue theorem $$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$$
I calculated the above integral several times using the residue theorem. The answer I get is $\pi / 2$ but this answer is not in the options please help
*
*$\frac{\pi i}{2}$
*$\frac{-\pi i}{2}$
*$\frac{\pi i}{1}$
*0
My answer
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$
$1+z+z^{2}+z^{3}=0 \rightarrow z=\pm i, z=-1$
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}=2 \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=i}\right)$
$+\pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=-1}\right)$
$=2 \pi i(-0.25-0.25 i)+0.5 \pi i=\frac{\pi}{2}$
| You have:
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}$
$1+z+z^{2}+z^{3}=0 \rightarrow z=\pm i, z=-1$
After this you didn't apply the Residue Theorem correctly. Instead of
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}=2 \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=i}\right)+\pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=-1}\right)$
It should be:
$\oint_{|z|=2} \frac{d z}{1+z+z^{2}+z^{3}}=2 \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=i}\right)\color{red}{+ 2 \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=-i}\right)}+\color{red}{2} \pi i\left(\left.\operatorname{Res}\left(\frac{1}{1+z+z^{2}+z^{3}}\right)\right|_{z=-1}\right)$
I don't know why you ignored the pole at $-i$, and you missed a $2$ in the term of the pole $-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
matrix rearrangement that preserves row/column sum I have the following symmetric matrix
$$
\begin{pmatrix}
1 & 5 & 3 \\
5 & 25 & 15 \\
3 & 15 & 9 \\
\end{pmatrix}
$$
whose rows and columns sum to 9, 45 and 27. I would like to transform this matrix such that the diagonal is zero, but the row and columns still sum to the same values as the original and the matrix is still symmetric. Is there an algorithm that can find such matrices?
For instance,
$$
\begin{pmatrix}
0 & 13.5 & -4.5 \\
13.5 & 0 & 63/2 \\
-4.5 & 63/2 & 0 \\
\end{pmatrix}
$$
However, all elements must be strictly non-negative. Can this be achieved?
| As you want to obtain the column/row sum, the symmetry and positivity you can start by combinatorics. Fortunately you don't have many possibilities. Let's start with the first row/column. By having zero on the diagonal, the numbers $5$ and $3$ have to be replace by two which sum up to $9$
$$
\begin{pmatrix}
0 & 0 & 9 \\
0 & 0 & . \\
9 & . & 0 \\
\end{pmatrix},
\begin{pmatrix}
0 & 1 & 8 \\
1 & 0 & . \\
8 & . & 0 \\
\end{pmatrix},
\begin{pmatrix}
0 & 2 & 7 \\
2 & 0 & . \\
7 & . & 0 \\
\end{pmatrix},
\begin{pmatrix}
0 & 3 & 6 \\
3 & 0 & . \\
6 & . & 0 \\
\end{pmatrix},
\begin{pmatrix}
0 & 4 & 5 \\
4 & 0 & . \\
5 & . & 0 \\
\end{pmatrix},
\begin{pmatrix}
0 & 5 & 4 \\
5 & 0 & . \\
4 & . & 0 \\
\end{pmatrix},
\begin{pmatrix}
0 & 6 & 3 \\
6 & 0 & . \\
3 & . & 0 \\
\end{pmatrix},
\begin{pmatrix}
0 & 7 & 2 \\
7 & 0 & . \\
2 & . & 0 \\
\end{pmatrix},
\begin{pmatrix}
0 & 8 & 1 \\
8 & 0 & . \\
1 & . & 0 \\
\end{pmatrix} \text{ or }
\begin{pmatrix}
0 & 9 & 0 \\
9 & 0 & . \\
0 & . & 0 \\
\end{pmatrix}.
$$
In the next step you want to get $45$ in the second row, which impossible under the constraint that the last row has to sum up to $27$. As there is no such matrix, there can't be any transformation.
EDIT: As you don't allow only natrual number, consider the following case. By this you also see that it is impossible for real nonnegative numbers:
$$\begin{pmatrix}
0 & 9-a & a \\
9-a & 0 & 36+a \\
a & 36+a & 0 \\
\end{pmatrix} \text{ for } a\in[0,9].
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
radius of circle tangent to the parabola y=x^2 In the following diagram, what's the amount of r=?
I supposed the center of the small circle to be $(0,y)$. So I have
its distance from $(x, x^2)$ is $(x-0)^2+(y-x^2)^2=1400^2$ this question must have just one answer so the discriminant of the equation must be zero. But then I don't have any idea. how can I relate this to the bigger circle :(
| You have the lower circle of radius $1400$. Assuming its center is $(0, a)$ then its equation is
$ x^2 + (y - a)^2 = 1400^2 $
At the tangency points $y = x^2 $, therefore
$ x^2 + (x^2 - a)^2 = 1400^2 $
Expanding, this becomes,
$ x^4 + x^2 (1 - 2 a) + a^2 - 1400^2 = 0 $
Since we want only one intersection point (actually two, but a single value of $x^2$), the discriminant must be zero, hence
$ (1 - 2 a)^2 - 4 (a^2 - 1400^2 ) = 0 $
Simplifying this becomes
$ 1 - 4 a + 4 (1400^2 ) = 0 $
From which,
$ a = \dfrac{ 1 + 4 (1400^2) }{4} = 1400^2 + \dfrac{1}{4} $
Now, we can move to the upper circle, its center is at $( 0, a + r )$ and radius $r$, thus its equation is
$ x^2 + ( y - (a + r) )^2 = r^2$
Again at the points of tangency, $y = x^2 $, so
$ x^2 + (x^2 - (a + r) )^2 = r^2 $
which simplifies to
$ x^4 + x^2 (1 - 2 (a + r) ) + (a + r)^2 - r^2 = 0 $
And we want the value of $x^2$ that solves this quadratic to be a single value, so again, the discriminant must be zero,
$ ( 1 - 2(a + r) )^2 - 4 ( a^2 + 2 a r ) = 0 $
Expanding,
$ 1 - 4 (a + r) + 4 (a^2 + 2 a r + r^2 ) - 4 (a^2 + 2 a r ) = 0 $
and this simplifies to
$ 1 - 4 (a + r) + 4 r^2 = 0 $
Knowing $a$ from the first circle, we can proceed to solve for $r$, there will be two values, and we should take the positive one. The other negative value corresponds to the circle that can drawn tangent to the '$1400$' circle but from below.
Can you take it from here ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find area of $x^2+axy+y^2=1$, $|a|\leq1$ I was wondering how to find the area of
$$x^2+axy+y^2=1,\>\>\>\>\>|a|\leq1$$
I have solved
$$\rho^{2}(\theta)=\frac{1}{1+\frac{a}{2}\sin 2\theta}$$
However, integrating this equation using trigonometric substitution is very cumbersome.
| Solving
$$x^2+axy+y^2=1,\>\>\>\>\>|a|\leq1$$
for $y$, one has
$$ y_{1,2}=\frac12(-ax\pm\sqrt{a^2x^2-4(x^2-1}))=\frac12(-ax\pm\sqrt{4-(4-a^2)x^2})$$
where $4-(4-a^2)x^2\ge0$ or $|x|\le\frac{2}{\sqrt{4-a^2}}$. So the are is
\begin{eqnarray}
A&=&\int_{-\frac{2}{\sqrt{4-a^2}}}^{\frac{2}{\sqrt{4-a^2}}}|y_1-y_2|dx\\
&=&\int_{-\frac{2}{\sqrt{4-a^2}}}^{\frac{2}{\sqrt{4-a^2}}}\sqrt{4-(4-a^2)x^2}dx.
\end{eqnarray}
Letting $x=\frac{2}{\sqrt{4-a^2}}\sin t$, then
$$ A=\frac4{\sqrt{4-a^2}}\int_0^{\pi/2}\cos^2 td= \frac{2 \pi}{\sqrt{4-a^2}}\tag{1} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solving a probability inequality Given three probability density functions $p$, $q$ and $h$ of Normal distributions. $p$ and $h$ are fixed and I want to solve for $q$, is it possible to identify a condition on $q$ apart from the trivial cases q=h or q=p such that:
$$\int_{z} \frac{p(z)}{q(z)} \: h(z) \: dz \leqslant 1$$
| Another approach: WLOG assume $p \sim N(a, \sigma_1^2)$ and $h \sim N(b, \sigma_2^2)$ where $0<\sigma_1^2 \leq \sigma_2^2$. Define $q\sim N(r, \sigma_2^2)$ for some $r \in \mathbb{R}$ that we will use to achieve the result.
Then
\begin{align}
\frac{h(z)}{q(z)} &= \exp\left(-\frac{(z-b)^2}{2\sigma_2^2}\right)\exp\left(\frac{(z-r)^2}{2\sigma_2^2}\right) \\
&= \exp\left(-\frac{1}{2\sigma_2^2}\left[(z-b)^2 - (z-r)^2 \right]\right)\\
&=\exp\left(\frac{1}{2\sigma_2^2}\left[-2z(b-r) + b^2 -r^2 \right]\right)
\end{align}
So then if we define $\lambda = \sigma_1^2/\sigma_2^2$, we have $0<\lambda \leq 1$ and
\begin{align}
\frac{p(z)h(z)}{q(z)}&=\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(z-a)^2}{2\sigma_1^2}\right)\exp\left(\frac{1}{2\sigma_2^2}\left[-2z(b-r) + b^2 -r^2 \right]\right)\\
&=\frac{1}{\sqrt{2\pi \sigma_1^2}}\exp\left(-\frac{1}{2\sigma_1^2}\left[ (z-a)^2 + 2\lambda z(b-r) -\lambda b^2 + \lambda r^2\right]\right)\\
&=\frac{1}{\sqrt{2\pi \sigma_1^2}}\exp\left(-\frac{1}{2\sigma_1^2}\left[ (z-a+\lambda (b-r))^2 - (-a+\lambda(b-r))^2 - \lambda b^2 + \lambda r^2\right]\right)\\
\end{align}
So
\begin{align}
\int_{-\infty}^{\infty} \frac{p(z)h(z)}{q(z)}dz &= \exp\left(\frac{1}{2\sigma_1^2}\left[(-a+\lambda(b-r))^2 +\lambda b^2- \lambda r^2\right]\right)\\
&=\exp\left(\frac{1}{2\sigma_1^2}\left[ (\lambda^2-\lambda)r^2 + 2ar\lambda + d\right]\right)\\
\end{align}
for some $d \in \mathbb{R}$. We know $0\leq \lambda \leq 1$. If $0<\lambda<1$ then $\lambda^2-\lambda < 0$ and we can choose $r$ suitably large to make the result strictly less than 1. If $\lambda=1$ then, if $a\neq 0$, we can again choose $r$ to make the result strictly less than 1.
The only remaining case is if $\lambda=1$ and $a=0$, which gives
\begin{align}
\int_{-\infty}^{\infty} \frac{p(z)h(z)}{q(z)}dz &= \exp\left(\frac{1}{2\sigma_1^2}\left[(-a+\lambda(b-r))^2 +\lambda b^2- \lambda r^2\right]\right)\\
&=\exp\left(\frac{1}{2\sigma_1^2}\left[(b-r)^2+b^2-r^2\right]\right)\\
&=\exp\left(\frac{1}{2\sigma_1^2}\left[2b(b-r)\right]\right)\\
\end{align}
and if $b\neq 0$ we can choose $r$ to make the result strictly less than 1.
The only remaining remaining case is when $\lambda=1$ and $a=b=0$, meaning that we have $p =h\sim N(0,\sigma^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Volume between paraboloid $x^2 +y^2 -4a(z+a)=0$ and sphere $x^2 + y^2 +z^2 =R^2$ I'm trying to obtein the volume via triple integral but think I'm setting the wrong radius. The solid in particular is bounded by the sphere $x^2 + y^2 +z^2 =R^2$ and above the parabolloid $x^2 +y^2 -4a(z+a)=0$ (consedering $R>a>0$). I'm setting cylindrical coordinates and I do eventually get $\theta \in [0,2\pi[$ and $(\rho^2 / 4a) -a\leq z \leq R^2-\rho^2$. I deduce that the radius must be limited in between $0$ and $4Ra-4a^2$ (Intersection is at height $z=R-2a$), so the volume should be:
$$\int_{0} ^{2\pi} \int_0 ^{4Ra-4a^2} \int_{(\rho^2 /4a)-a} ^{R^2 -\rho^2} \mathrm{d}V$$
, but this integral results in a different expression from the original solution. My teacher told us the solution is $2\pi \left ( \frac{a^3}{3} - aR^2 + \frac{2R^3}{3} \right )$.
| First we find the intersection of the sphere with the paraboloid.
Since $x^2 + y^2 = 4a(z + a)$ then $4a(z + a) + z^2 = R^2$ so $z^2 + 4az + 4a^2 -R^2 = 0$. Then $$z = \frac{-4a \pm \sqrt{16a^2 + 4R^2 - 16a^2}}{2} = -2a \pm R$$ so the intersection is at height $R-2a$, as you said. The maximum radius would occur at the intersection so $r= \sqrt{x^2 + y^2} \sqrt{4a(R-2a +a)} = \sqrt{4a(R-a)}$. And $z$ is bounded below by the paraboloid so $r^2/4a - a$ and above by the sphere so $\sqrt{R^2 - r^2}$. Then the volume integral is given by $$\int_0^{2\pi} \int_0^{\sqrt{4a(R-a)}} \int_{r^2/4a - a}^{\sqrt{R^2 - r^2}} r dz dr d\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
calculate one determinant to show 【National College Entrance Exam, old China】
$ \omega =\frac{1 \pm \sqrt{3} i } {2}$
Please calculate and show:
$ \begin{vmatrix} 1 &\omega& \omega^2 & 1\\ \omega& \omega^2 & 1 &1\\ \omega^2& 1& 1 & \omega \\ 1& 1& \omega& \omega^2\\ \end{vmatrix} =3\sqrt{-3} $
Here's what I have done until now
$$
D=\begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1 \\
\omega & {{\omega }^{2}} & 1 & 1 \\
{{\omega }^{2}} & 1 & 1 & \omega \\
1 & 1 & \omega & {{\omega }^{2}} \\
\end{vmatrix} \\= \begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1+1+\omega +{{\omega }^{2}} \\
\omega & {{\omega }^{2}} & 1 & 1+1+\omega +{{\omega }^{2}} \\
{{\omega }^{2}} & 1 & 1 & 1+1+\omega +{{\omega }^{2}} \\
1 & 1 & \omega & 1+1+\omega +{{\omega }^{2}} \\
\end{vmatrix} \\= \begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1 \\
\omega & {{\omega }^{2}} & 1 & 1 \\
{{\omega }^{2}} & 1 & 1 & 1 \\
1 & 1 & \omega & 1 \\
\end{vmatrix} \\= \begin{vmatrix}
1 & \omega & {{\omega }^{2}} & 1 \\
\omega -1 & {{\omega }^{2}}-\omega & 1-{{\omega }^{2}} & 0 \\
{{\omega }^{2}}-1 & 1-\omega & 1-{{\omega }^{2}} & 0 \\
0 & 1-\omega & \omega -{{\omega }^{2}} & 0 \\
\end{vmatrix}
$$
| One approach is to note that this matrix (which I will refer to as $A$) is a circulant matrix. As is explained on the page, we can thereby deduce that the eigenvalues of $A$ are of the form
$$
\lambda_j = 1 + i^j + \omega^2 i^{2j} + \omega i^{3j}, \quad i = \sqrt{-1}, \quad j = 0,1,2,3.
$$
So, we find that the eigenvalues are
$$
\lambda_0 = 1 + (1 + \omega + \omega^2) = 1\\
\lambda_1 = 1 + i - \omega^2 - i\omega = (1 + 1/2 + \sqrt{3}/2) +i(1 + \sqrt{3}/2 + 1/2)
\\
\lambda_2 = 1 -1 + \omega^2 - \omega = -i\sqrt{3}\\
\lambda_3 = 1 - i - \omega^2 + i\omega = (1 + 1/2 - \sqrt{3}/2) - i(1 - \sqrt{3}/2 + 1/2).
$$
It now suffices to compute the product of these eigenvalues.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $|ax^2+bx+c|\leq 2\ \ \forall x\in[-1,1]$ then find the maximum value of $|cx^2+2bx+4a|\ \ \forall x\in [-2,2]$.
If
$$\left|ax^2+bx+c\right|\leq 2\quad \forall x\in[-1,1]$$
then find the maximum value of
$$\left|cx^2+2bx+4a\right|\quad \forall x\in [-2,2].$$
My Attempt
Let $f(x)=ax^2+bx+c$, then
$$|cx^2+2bx+4a|=x^2|4\frac{a}{x^2}+2\frac{b}{x}+c|=x^2\left|f\left(\frac{2}{x}\right)\right|$$
But then I couldn't make any headway.
I also tried to express $a,b,c$ in terms of $f(-1),f(0),f(1)$ i.e $$c=f(0);b=\frac{f(1)-f(-1)}{2};a=\frac{f(-1)-2f(0)+f(1)}{2}$$ but again couldn't go further
| A solution using elementary calculus.
First observe by changing variables $x\mapsto\frac{x}{2}$ that the question becomes finding the maximum value of $4|cx^2+bx+a|$ on $[-1,1]$. Let's set $f(x)=ax^2+bx+c$ and $g(x)=cx^2+bx+a$. Also note that by swapping between $f$ and $-f$ (and $g$ and $-g$ respectively), we can assume wlog that $c\geq 0$. Also wlog, since the domains are symmetric around the origin, we can assume $b\geq 0$.
Now there are four main ingredients to the elementary proof.
Step 1: As Maverick pointed out, one can get bounds on the coefficients. Indeed, $a=\frac{f(-1)-2f(0)+f(1)}{2}$, $b=\frac{f(1)-f(-1)}{2}$, and $c=f(0)$ implying with the above that $b,c\in[0,2]$ and $|a|\leq4$.
Step 2: The extremal values a parabola attains on a closed interval are either at the endpoints of the interval or at the global extremum of the parabola.
Therefore
\begin{equation*}
\max_{[-1,1]}|g(x)|\in\{|g(1)|,|g(-1)|,|g(-\frac{b}{2c})|\}=\{|c+b+a|,|c-b+a|,|a-\frac{b^2}{4c}|\}=\{|f(1)|,|f(-1)|,|a-\frac{b^2}{4c}|\}.
\end{equation*}
So $\max_{[-1,1]}|g(x)|\leq \max(2,|a-\frac{b^2}{4c}|)$.
As a technicality, if $c=0$, $g$ is linear, so it will attain extremal values only at the endpoints of the interval, so we don't need to worry about $g(-\frac{b}{2c})$.
Step 3: Note that by taking $a=-4,b=0,c=2$, we have that $|g(0)|=4$, so we are only further interested in looking at the case when the extremal value of $|g|$ is attained at the global extremal value of $g$ and when this value is attained inside the interval $[-1,1]$.
This implies $c>0$ and $-1\leq -\frac{b}{2c}\leq 0$. Observe that if $g(-\frac{b}{2c})\geq 0$, since we assumed $c\geq0$, $g$ increases away from $-\frac{b}{2c}$ and therefore $g(1)$ and $g(-1)$ will both be bigger than $g(-\frac{b}{2c})$. Therefore we can restrict our analysis to the case when $g(-\frac{b}{2c})<0$.
Step 4: Arguing that $g$ is bounded below on $[-1,1]$ by $-4$.
Assume this is not the case. Then necessarily $g(-\frac{b}{2c})<-4$. We now apply the mean value theorem on the interval $[-1,-\frac{b}{2c}]\subset [-1,0]$ to find some $t\in[-1,0]$ such that
\begin{equation*}
g'(t)=\frac{g(-\frac{b}{2c})-g(-1)}{-\frac{b}{2c}-(-1)}\leq g(-\frac{b}{2c})-g(-1)<- 2,
\end{equation*}
because $0\geq-\frac{b}{2c}\geq -1, g(-1)=f(-1)\geq -2$ and because we assumed $g(-\frac{b}{2c})<-4$.
But this implies that $2ct+b<-2$, or, since $t\in[-1,0]$, that $c>\frac{b}{2|t|}+\frac{1}{|t|}\geq\frac{1}{|t|}\geq 1$.
Now since we assumed that $g(-\frac{b}{2c})<-4$, we have
\begin{equation*}
-4>a-\frac{b^2}{4c}=(a-b+c)+b-c-\frac{b^2}{4c}\geq -2 +b-c-\frac{b^2}{4c}.
\end{equation*}
Rearranging, this means $\frac{1}{4c}b^2-b+c-2>0$. Viewing this as a degree $2$ inequality in $b$, we can compute the discriminant $\Delta=\sqrt{\frac{2}{c}}$ and roots $b_{1,2}=\frac{1\pm\sqrt{\frac{2}{c}}}{\frac{1}{2c}}=2c\pm 2\sqrt{2c}$. For the inequality to hold, we must therefore have that either $b<2c-2\sqrt{2c}=2\sqrt{c}(\sqrt{c}-\sqrt{2})<0$ (since $c\in(1,2]$), impossible since $b\geq 0 $, or that $b>2c+2\sqrt{2c}>2c>2$ (since $c>1$), impossible as $b\leq 2$. So we have a contradiction, concluding Step 4.
This means that the required maximum is $16$ and it can be achieved by taking $a=-4,b=0,c=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Maclaurin series of $\ln \left( 1+\frac{\ln(1+x)}{1+x} \right)$ So the first thing I done was
$$\begin{align}\ln(1+x)&=x-\frac{1}{2}x^2+\frac{1}{3}x^3+o(x^3)\\&=(1+x)\left(x-\frac{3}{2}x^2+\frac{11}{6}x^3\right)+o(x^3)\end{align}$$
I've never seen this done but I'm pretty sure I can do this.
Now I want to divide:
$$(1+x)\left(x-\frac{3}{2}x^2+\frac{11}{6}x^3\right)+o(x^3)$$By $(1+x)$ but I don't know what
$\frac{o(x^3)}{1+x}$ will be.
Also is there another way to do this?
| Following @jjagmaths approach we show
\begin{align*}
\color{blue}{\ln\left(1+\frac{\ln(1+x)}{1+x}\right)}
&\color{blue}{=\sum_{n=1}^\infty(-1)^{n-1}\sum_{m=1}^{n}\frac{1}{m}
\sum_{{k_1+\cdots +k_m=n}\atop{k_1,\ldots,k_m\geq 1}}H_{k_1}\cdots H_{k_m}x^n}\tag{1}\\
&=x-2x^2+\frac{11}{3}x^3\color{blue}{-\frac{163}{24}}x^4+\cdots
\end{align*}
Denoting with $[x^n]$ the coefficient of $x^n$ of a series we obtain for $n\geq 1$:
\begin{align*}
[x^n]&\ln\left(1+\frac{\ln(1+x)}{1+x}\right)\\
&=\sum_{m=1}^{n}\frac{(-1)^{m-1}}{m}[x^n]\frac{\ln^m(1+x)}{(1+x)^m}\tag{2}
\end{align*}
In (2) we note that since $\ln(1+x)$ starts with $x$, the $m$-th power of $\frac{\ln(1+x)}{1+x}$ starts with $x^m$ and we can restrict the sum with the upper limit $n$. We have
\begin{align*}
A(x)&=\frac{\ln(1+x)}{1+x}\\
&=\left(\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}x^k\right)\left(\sum_{l=0}^\infty(-1)^lx^l\right)\\
&=\sum_{q=1}^\infty\left(\sum_{{k+l=q}\atop{k\geq 1, l\geq 0}}\frac{(-1)^{k-1+l}}{k}\right)x^q\\
&=\sum_{q=1}^\infty(-1)^{q-1}\left(\sum_{k=1}^q\frac{1}{k}\right)x^q\\
&=\sum_{q=1}^\infty(-1)^{q-1}H_qx^q\\
\end{align*}
It follows for $m\geq 1$:
\begin{align*}
A^m(x)&=\sum_{q=m}^{\infty}\left(\sum_{{k_1+\cdots+k_m=q}\atop{k_1,\ldots,k_m\geq 1}}
(-1)^{k_1-1}H_{k_1}\cdots(-1)^{k_m-1}H_{k_m}\right)x^q\\
&=\sum_{q=m}^{\infty}(-1)^{q-m}\sum_{{k_1+\cdots+k_m=q}\atop{k_1,\ldots,k_m\geq 1}}
H_{k_1}\cdots H_{k_m}x^q\\
\end{align*}
and for $n\geq m$:
\begin{align*}
[x^n]A^m(x)&=(-1)^{n-m}\sum_{{k_1+\cdots+k_m=n}\atop{k_1,\ldots,k_m\geq 1}}
H_{k_1}\cdots H_{k_m}\tag{3}
\end{align*}
Putting (3) in (2) we obtain
\begin{align*}
\color{blue}{[x^n]\ln\left(1+\frac{\ln(1+x)}{1+x}\right)
=\sum_{m=1}^{n}\frac{(-1)^{n-1}}{m}\sum_{{k_1+\cdots+k_m=n}\atop{k_1,\ldots,k_m\geq 1}}
H_{k_1}\cdots H_{k_m}}\tag{4}
\end{align*}
and the claim (1) follows.
Calculating for instance the coefficient of $x^4$ we obtain from (4)
\begin{align*}
\color{blue}{[x^4]}&\color{blue}{\ln\left(1+\frac{\ln(1+x)}{1+x}\right)}\\
&=\sum_{m=1}^{4}\frac{(-1)^{n-1}}{m}\sum_{{k_1+\cdots+k_m=4}\atop{k_1,\ldots,k_m\geq 1}}
H_{k_1}\cdots H_{k_m}\\
&=-\left(H_4+\frac{1}{2}\left(2H_1H_3+H_2^2\right)+\frac{1}{3}\left(3H_1^2H_2\right)+\frac{1}{4}H_1^4\right)\\
&=-\left(\frac{25}{12}+\frac{1}{2}\left(2\cdot 1\cdot \frac{11}{6}+\left(\frac{3}{2}\right)^2\right)+\frac{1}{3}\left(3\cdot 1\cdot\frac{3}{2}\right)+\frac{1}{4}\cdot 1\right)\\
&\,\,\color{blue}{=-\frac{163}{24}}
\end{align*}
in accordance with the claim (1) and Wolfram Alpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What could be these polynomials? Trying to build an approximation, I encountered the following polynomials
$$\left(
\begin{array}{cc}
1 & 1 \\
2 & x-3 \\
3 & x^2-30 x+45 \\
4 & x^3-273 x^2+1575 x-1575 \\
5 & x^4-2460 x^3+43470 x^2-132300 x+99225 \\
6 & x^5-22143 x^4+1123650 x^3-8004150 x^2+16372125 x-9823275 \\
7 & x^6-199290 x^5+28423395 x^4-431531100 x^3+1830403575 x^2-2809456650 x+1404728325
\end{array}
\right)$$
What I noticed is that
*
*the constant term is $(-1)^{n}\,\frac{2^{2 n+1} \Gamma \left(n+\frac{1}{2}\right) \Gamma
\left(n+\frac{3}{2}\right)}{\pi }$
*the coefficient of the second highest power is $\frac{3}{8} \left(1-9^n\right)$
*the coefficient of the third highest power is $\frac{15}{128} \left(2-3^{2 n+5}+5^{2 n+4}\right)$ but this does not help me much.
Any idea or suggestion would be very welcomed
| $\{-1575, -132300, -8004150, -431531100,...\}$:
$\dfrac{35}{1024}\bigg(5 - 3^{2 n + 8} + 5^{2 n + 7} - 7^{2 n + 6}\bigg)$
$\{1, -30, 1575, -132300, 16372125, -2809456650,...\}$:
$-(-1)^{n + 1} 2^{2 n} (n + 1) Pochhammer\bigg(\dfrac{3}{2}, n\bigg) Pochhammer\bigg(\dfrac{5}{2}, n\bigg)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4426856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Orthogonal projection of $D=\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix}$ on $W=\{A\in M_{2\times 2}(\mathbb{R})~|~Trace(A)=0\}$ in product space $V$. This was an assigned homework problem (Turned in 4/14):
$V=M_{2\times 2}(\mathbb{R})$ with the inner product $\langle A,B\rangle = Trace(B^t,A),~D=\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix}$, and $W=\{A\in M_{2\times 2}(\mathbb{R})~|~Trace(A)=0\}$.
Using definition of $A$ having trace of zero, I used $\begin{pmatrix}a & b\\c & -a\end{pmatrix}$ to establish a basis for $W, ~ \beta_{W}=\left\{ \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix},\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}\right\}$
I tried applying Gram-Schmidt orthogonalization process and I got the same basis that I had for W. I think that I figured out that is because the basis $\beta_W$ is already orthogonal. I tested this by taking inner products of all the basis matrices and getting 0.
I'm struggling with where to go next.
Is it the correct approach to use:
$\frac{\langle D,w_1\rangle}{\langle w_1,w_1\rangle}w_1+\frac{\langle D,w_2\rangle}{\langle w_2,w_2\rangle}w_2+\frac{\langle D,w_3\rangle}{\langle w_3,w_3\rangle}w_3=w^{\prime}$
$\frac{tr\left(\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix} \right)}{tr\left(\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix} \right)}\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}+\frac{tr\left(\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix} \right)}{tr\left(\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix} \right)}\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}+\frac{tr\left(\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix} \right)}{tr\left(\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix} \right)}\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}=w^{\prime}$
$=\frac{tr\begin{pmatrix}1 & 1\\3 & -1\end{pmatrix}}{tr\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}}\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}+\frac{tr\begin{pmatrix}0 & 0\\1 & 1\end{pmatrix}}
{tr\begin{pmatrix}0 & 0\\0 & 1\end{pmatrix}}\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}+\frac{tr\begin{pmatrix}-3 & 1\\0 & 0\end{pmatrix}}{tr\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}}\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}=w^{\prime}$
and $w^{\prime}=0\cdot\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}+1\cdot\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}-3\cdot\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}=\begin{pmatrix}0 & 1\\-3 & 0\end{pmatrix}$
I'm not sure how to verify the solution and I'm not sure what the solution means in relation to the given "vector" $D$.
| In vector form
$D = [1, 1, -3, 1]^T $
And a basis for $W$ is
$\{ [1, 0, 0, -1]^T , [0, 1, 0, 0]^T , [0, 0, 1, 0]^T \} $
The orthogonal space is spanned by
$N = [1, 0, 0, 1]^T$ (which corresponds to the identity matrix)
Therefore,
$D - Proj(D) = c_1 N $
$ D \cdot N - 0 = c_1 (N \cdot N) $
Hence $c_1 = \dfrac{D \cdot N}{N \cdot N} = \dfrac{1 + 1}{1 + 1} = 1 $
Therefore,
$Proj(D) = D - c_1 N = [1, 1, -3, 1]^T - (1) [1, 0, 0, 1]^T = [0, 1, -3, 0]^T $
which is the matrix
$\begin{bmatrix} 0 && 1 \\ -3 && 0 \end{bmatrix} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4427969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help with a surface integral Let G(x, y, z) = $\ (1-x^{2}-y^{2})^{\frac{3}{2}}$
Evaluate the surface integral:
$\int\int_{S}^{}G(x, y,z)dS$
where S is the hemisphere:
z =$\ (1-x^{2}-y^{2})^{\frac{1}{2}}$
This is my work thus far:
$\int_{}^{}\int_{S}^{}G(x,y,z)dS=\int_{}^{}\int_{R}^{}(1-x^{2}-y^{2})^{\frac{3}{2}}\sqrt[]{1+\frac{x^{2}}{z^{2}}+\frac{y^{2}}{z^{2}}} \ dxdy$
Factoring 1/z^2 out of the square root we obtain:
= $\int_{}^{}\int_{R}^{}(1-x^{2}-y^{2})^{2}\ dxdy$
Converting to polar coordinates (x = cos$\theta$; y = sin$\theta$, dxdy = rdrd$\theta$):
$\int_{0}^{2\pi}\int_{0}^{1}(1-r^{2})^{2}r \ drd\theta =\frac{\pi}{3}$
The answer provided in the textbook is pi/2.
| Since,
$$\boxed{\iint_{S}f(x,y,z)\,{\rm d}S=\iint_{D}f(x,y,g(x,y))\sqrt{z_{x}^{2}+z_{y}^{2}+1}\,{\rm d}A}$$
So, we have
\begin{align*}
\iint_{S}(1-x^{2}-y^{2})^{3/2}\, {\rm d}S&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{z_{x}^{2}+z_{y}^{2}+1}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\left(-\frac{x}{\sqrt{1-x^{2}-y^{2}}} \right)^{2}+\left(-\frac{y}{\sqrt{1-x^{2}-y^{2}}}\right)^{2}+1}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\frac{x^{2}}{1-x^{2}-y^{2}}+\frac{y^{2}}{1-x^{2}-y^{2}}+1}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\frac{x^{2}+y^{2}}{1-x^{2}-y^{2}}+1}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\frac{x^{2}+y^{2}+1-x^{2}-y^{2}}{1-x^{2}-y^{2}}}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})^{3/2}\sqrt{\frac{1}{1-x^{2}-y^{2}}}\, {\rm d}A,\\
&=\iint_{D}(1-x^{2}-y^{2})\, {\rm d}A,\\
&=\int_{0}^{2\pi}\int_{0}^{1}(1-r^{2})r\, {\rm d}{\rm d}\theta,\\
&=\int_{0}^{2\pi} \frac{1}{4}\, {\rm d}\theta,\\
&=\frac{2\pi}{4},\\
&=\boxed{\frac{\pi}{2}}
\end{align*}
The change of variables $x=r\cos \theta$ and $y=r\sin \theta$ with $r\in \mathbb{R}^{*}_{+}$ and $\theta\in \left[0,2\pi\right[$ has Jacobian $r$ and the limits of integration is given because $S$ is given by $x^{2}+y^{2}+z^{2}=1$ in the top upper.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trouble with a triple integral on a region bounded by a sphere and two planes
I would like to compute the integral $\int_A zdzdydx,$ where $A$ is the region bounded by the sphere $x^2+y^2+z^2=R^2,$ plane $\frac{x}a+\frac{y}b=1$ and coordinate planes (which doesn't contain the origin on its boundary and is in the first quadrant).
I considered switching to either spherical or cylindrical coordinates, but I don't see any symmetry or pattern.
In spherical coordinates, I tried expressing the lower bound for radius $r$ in terms of $\theta\in[0,\pi/2]$ $$r(\theta, \varphi)=\frac{ab}{a\sin\varphi\sin\theta+b\cos\varphi\sin\theta}$$ and the integral becomes
\begin{aligned}&\color{white}=\int_0^{\pi/2}\int_0^{\pi/2}\int_{\frac{ab}{a\sin\theta\sin\varphi+b\sin\theta\cos\varphi}}^Rr^2\sin\theta r\cos\theta drd\theta d\varphi\\&=\int_0^{\pi/2}\int_0^{\pi/2}\sin\theta\int_{\frac{ab}{a\sin\theta\sin\varphi+b\sin\theta\cos\varphi}}^Rr^3drd\theta d\varphi\end{aligned}
In cylindrical coordinates
$$r(\varphi)=\frac{ab}{a\sin\varphi+b\cos\varphi}$$ and the integral is
$$\int_0^{\pi/2}\int_{\frac{ab}{a\sin\varphi+b\cos\varphi}}^R\int_0^{\sqrt{R^2-r^2}}zrdzdrd\varphi$$
However, I have to deal with a powers of $a\sin\theta\sin\varphi+b\sin\theta\cos\varphi$ and $a\sin\varphi+b\cos\varphi$ too early, and if I swap the order of integration, I need to express the angles in terms of $r$ which seems worse.
I saw the substitution in this answer, but it isn't so smooth here and I've seen the reduction formula.
How should one attack this task?
| Proceed in cylindrical coordinates as follows
\begin{align}
&\int_0^{\pi/2}\int_{r(\varphi)}^R\int_0^{\sqrt{R^2-r^2}}z \>rdzdrd\varphi\\
= & \> \frac12 \int_0^{\pi/2}\int_{\frac{ab}{a\sin\varphi+b\cos\varphi}}^R (R^2-r^2)r dr d\varphi\\
= & \> \frac12 \int_0^{\pi/2}\left( \frac14R^4–\frac12\frac{a^2b^2R^2}{(a\sin\varphi+b\cos\varphi)^2}
+ \frac14\frac{a^4b^4}{(a\sin\varphi+b\cos\varphi)^4 } \right)d\varphi\\
= & \> \frac\pi{16}R^4 -\frac 14abR^2 +\frac1{24}ab(a^2+b^2)\\
\end{align}
where
$$\int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{(a\sin\varphi+b\cos\varphi)^2} =\frac1{ab}, \>\>\>
\int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{(a\sin\varphi+b\cos\varphi)^4} =\frac{a^2+b^2}{3a^3b^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
minimizing the distance between values
Find all values of $a$ that minimize the expression $|a- x_1| + |a-x_2|+\cdots + |a-x_n|$, where $x_1\leq x_2\leq \cdots \leq x_n$.
I know the minimum is achieved when $a=x_{\lfloor n/2\rfloor + 1}$ if $n$ is odd and when $a\in [x_{\lfloor n/2\rfloor}, x_{\lfloor n/2\rfloor + 1}]$ when $n$ is even.
I think one approach is mostly "brute force" and is relatively complicated. If I let $f(a) = |a-x_1|+\cdots + |a-x_n|$. Or maybe I could use derivatives? For $a$ so that $x_i < a<x_{i+1}$, $f'(a)$ equals $i - (n-i)$, which is less than $0$ for $i < n/2$ and $>0$ for $i > n/2$.
I think the following approach might work, but even if it does, it's way too tedious.
I could show directly that $f(x_i) < f(a) $ or $f(x_{i+1}) < f(a)$, depending on whether $i < \frac{n}2$ or $i > \frac{n}2$. I could then evaluate $f(a)$ directly by setting $a = x_k$ for some $k < \lfloor n/2\rfloor, x_k < x_{\lfloor n/2\rfloor}$ and for $k > \lfloor n/2\rfloor + 1, x_k > x_{\lfloor n/2\rfloor + 1}$. Then I could show the distance would be decreased by choosing $a=x_{\lfloor n/2\rfloor }$ or $x_{\lfloor n/2\rfloor + 1}.$ Finally, I could show that when $n$ is even, the distance will be the same if we choose any $a\in [x_{\lfloor b/2\rfloor}, x_{\lfloor n/2\rfloor + 1}$.
Here are some more details regarding my approach:
First observe that there cannot exist an index i so that $x_i < a < x_{i+1}$ unless $i = \frac{n}2$ (in which case of course n has to be even).
To see why, observe that for such an a, we have
\begin{align}
f(x_i) - f(a) &= \sum_{j=1}^i (x_i - a) + \sum_{j=i+1}^n (a- x_i)\\
&= (a- x_i)(n- 2i)\tag{1}
\end{align}
and
\begin{align}
f(x_{i+1}) - f(a) &= \sum_{j=1}^i (x_{i+1} - a) + \sum_{j=i+1}^n (a - x_{i+1})\\
&= (a-x_{i+1})(n-2i) \tag{2}
\end{align}
Clearly, $(1)$ shows $f(x_i) - f(a) < 0$ for $i > \frac{n}2$ and $(2)$ shows that $f(x_{i+1} ) - f(a) < 0$ for $i < \frac{n}2$. Thus we know that if $a$ lies strictly between two indices $x_i$ and $x_{i+1}$ and $f(a)$ is minimum, then $i = \frac{n}2$. Now we find the values of $i$ so that $f(x_i)$ is minimum.
Now we claim that if $n$ is odd, and $x_k < x_{\lfloor \frac{n}2\rfloor + 1}$ then $f(x_{\lfloor \frac{n}2\rfloor + 1}) < f(x_k)$.
Observe that $f(x_k) = \sum_{i=1}^k (x_k - x_i) + \sum_{i=k+1}^n (x_i - x_k)$. Then if $x_k < x_{\lfloor \frac{n}2\rfloor}$, we have
\begin{align}f(x_{\lfloor \frac{n}2\rfloor}) - f(x_k) &= \sum_{j=1}^k (x_{\lfloor \frac{n}2\rfloor} - x_k) + \sum_{j=k+1}^{\lfloor \frac{n}2\rfloor - 1} (x_k + x_{\lfloor \frac{n}2\rfloor } - 2x_j) + \sum_{j= \lfloor \frac{n}2\rfloor}^n (x_k - x_{\lfloor \frac{n}2\rfloor })\\
&\leq (k- (n-\lfloor \frac{n}2\rfloor + 1)) (x_{\lfloor \frac{n}2\rfloor} - x_k) + (x_{\lfloor \frac{n}2\rfloor} - x_k) (\lfloor \frac{n}2\rfloor - k - 1) \\
&= (2 \lfloor \frac{n}2\rfloor - n - 2)(x_{\lfloor \frac{n}2\rfloor }- x_k)
< 0,\end{align}
Now if $x_k > x_{\lfloor \frac{n}2\rfloor + 1}$, then we have
\begin{align}
f(x_{\lfloor \frac{n}2\rfloor + 1}) - f(x_k) &= \sum_{j=1}^{\lfloor \frac{n}2\rfloor +1} (x_{\lfloor \frac{n}2\rfloor + 1} - x_k) + \sum_{j=\lfloor \frac{n}2\rfloor + 2}^{k} (2x_j - (x_k + x_{\lfloor \frac{n}2\rfloor + 1} )) + \sum_{j= k + 1}^n (x_k - x_{\lfloor \frac{n}2\rfloor + 1})\\
&\leq (\lfloor\frac{n}2 \rfloor + 1- (n-k )) (x_{\lfloor \frac{n}2\rfloor + 1} - x_k) - (x_{\lfloor \frac{n}2\rfloor + 1} - x_k) (k - \lfloor\frac{n}2 \rfloor - 1) \\
&= (2 \lfloor \frac{n}2\rfloor - n + 2)(x_{\lfloor \frac{n}2\rfloor + 1}- x_k)
< 0\end{align}
So to minimize $f(a)$, we must have $x_{\lfloor \frac{n}2 \rfloor }\leq a \leq x_{\lfloor \frac{n}2 \rfloor + 1}$. Then we have for $k=\lfloor \frac{n}2\rfloor$ that (3)
\begin{align}f(x_{\lfloor \frac{n}2\rfloor + 1}) - f(a) &= \sum_{j=1}^k (x_{\lfloor \frac{n}2\rfloor + 1} - a) + \sum_{j=k+1}^{\lfloor \frac{n}2\rfloor} (a + x_{\lfloor \frac{n}2\rfloor + 1} - 2x_j) + \sum_{j= \lfloor \frac{n}2\rfloor + 1}^n (a - x_{\lfloor \frac{n}2\rfloor + 1})\\
&\leq (k- (n-\lfloor \frac{n}2\rfloor)) (x_{\lfloor \frac{n}2\rfloor + 1} - a) + (x_{\lfloor \frac{n}2\rfloor + 1} - a) (\lfloor \frac{n}2\rfloor - k) \\
&= (2 \lfloor \frac{n}2\rfloor - n)(x_{\lfloor \frac{n}2\rfloor + 1}- a)
\leq 0,\end{align}
with equality when $n$ is even. Since the minimum cannot be achieved for $x < \lfloor n/2\rfloor$ by a proof above, this shows that if $n$ is odd we cannot have $x_{\lfloor n/2\rfloor + 1} > a$ if $f(a)$ is minimum (as either $a < x_{\lfloor n/2\rfloor}$, in which case $f(a)$ is not minimum, or $x_{\lfloor n/2\rfloor} \leq a < x_{\lfloor n/2\rfloor + 1}$, in which case by the proof above ((3)) $f(a)$ is not minimum. If n is even, we thus have the the minimum is achieved for $a \in [x_{n/2}, x_{n/2+1}]$.
| This is a long-winded comment.
It is not an answer.
The approach that I would suggest is a minor variation of the approach that you described in your posting. Because (as you indicated) it is tedious, I will simply explore the individual cases of $n = 9$ and $n = 10$, for illustrative purposes.
Further, I will map out partial illustrations that should readily expand into full-blown proofs.
For $n = 9$, you have the candidate value of $a = x_5$.
Instead, consider $k > 0$, where $x_6 - x_5 > k$ and $x_5 - x_4 > k$. Within any satisfying (positive) value for $k$, consider the alternative value of $A = x_5 + k$.
When comparing the computations, using $A$ versus $a$, you will end up increasing $5$ computations by $k$ each. Namely $|A - x_1|, \cdots, |A - x_5|.$
Then, you will end up decreasing $4$ computations by $k$ each. Namely $|A - x_6|, \cdots, |A - x_9|$.
So, the net effect is $+5k - 4k = +k$, which has increased the computation. For the specific value of $n = 9$, you should be able to expand this type of analysis to cover any $k > 0$, including (for example) $k \geq |x_6 - x_5|.$ Then, symmetric reasoning should lead to the same result for any $k < 0$.
So, the result will have been proven specifically for $n = 9.$ Then, this result should be reasonable extensible, to cover $n = $ any odd positive integer.
A possible shortcut here would be induction. You could show that if the analysis applies to the middle value for any odd positive integer $n$, it must also apply to the middle value for the integer $(n+2)$.
For $n = 10$ the approach is very similar. The first thing to do would be to demonstrate that if $a$ is any value in $[x_5,x_6]$, then the computation yields the same value. Then, you could examine (for example) what happens if you consider $A = x_6 + k, k > 0$, first examining what happens if (for example) $k < x_7 - x_6.$
The idea, when focusing on the specific value for $n=10$ would be to examine how many terms are increased by $+k$ and how many terms are decreased by $-k$. The analysis should be very similar to that of $n=9.$
Again, a possible shortcut is induction, where you show that if the optimal value is any value between the $2$ middle values for the even positive integer $n$, you get the same result for the integer $(n+2)$.
The induction approach, if workable, has the advantage that you only have to examine (for example), the base cases of $n=2$ and $n=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding closed form expression for $n^{th}$ term of sequence with generating function $F(x)$? I have been asked the above question regarding the generating function
$$F(x) = \frac{x^2(1-x)}{(1-x)^3}$$
I have no idea what procedure this type of question follows. The solution gives that $F(x)$ can be written as
$$\frac{x^2}{(1-x)^3}-\frac{x^3}{(1-x)^3}$$
This I understand. But then it then says for $F(x) = \dfrac{1}{(1-x)^3}$ we get
$$\frac{F^n(0)}{n!} = \frac{(n+1)(n+2)}{2} \tag1$$
From which we can get the coefficients.
Can someone help me understand what is computed here? Why do we use $F(x) = \dfrac{1}{(1-x)^3}$, and how is $(1)$ computed? All help appreciated!
| Simplify and then use partial fraction decomposition to obtain
\begin{align}
\sum_{n=0}^\infty a_n x^n
&= \frac{x^2(1-x)}{(1-x)^3} \\
&= \frac{x^2}{(1-x)^2} \\
&= 1 - \frac{2}{1-x} + \frac{1}{(1-x)^2} \\
&= 1 - 2\sum_{n=0}^\infty x^n + \sum_{n=0}^\infty (n+1) x^n \\
&= 1 + \sum_{n=0}^\infty (n-1) x^n,
\end{align}
which immediately implies that
$$a_n =
\begin{cases}
0 &\text{if $n=0$} \\
n-1 &\text{if $n>0$}.
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving $\frac{x^2+y^2+z^2}{x+y+z}+\frac 32\sqrt[3]\frac{xy+yz+xz}{x+y+z}\geq \frac 52$, for positive values with $xyz=1$, without expansion?
Let, $x,y,z>0$ and $xyz=1$, then prove that
$$\frac{x^2+y^2+z^2}{x+y+z}+\frac 32\sqrt[3]\frac{xy+yz+xz}{x+y+z}\geq \frac 52$$
I know that $$x^2+y^2+z^2\geq x+y+z$$ by Cauchy-Schwarz. So, $$\frac{x^2+y^2+z^2}{x+y+z}\geq 1$$
But, $xy+yz+xz\geq x+y+z$ is not always true. My question is, is there a way to prove this inequality without expansion? I mean, I want to avoid getting the cube of each side. Is it possible?
| Here is a solution using only various mean inequalities:
First use AM-GM:
$$2\frac{x^2+y^2+z^2}{x+y+z}+3\sqrt[3]{\frac{xy+yz+zx}{x+y+z}}\ge5\sqrt[5]{\frac{(x^2+y^2+z^2)^2(xy+yz+zx)}{(x+y+z)^3}}$$
Then using $xyz=1$:
$$\frac{(x^2+y^2+z^2)^2(xy+yz+zx)}{(x+y+z)^3}=\frac{(x^2+y^2+z^2)^2}{(x+y+z)^4}(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$$
Using AM-HM:
$$(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge 9$$
Using AM-QM:
$$\frac{(x^2+y^2+z^2)^2}{(x+y+z)^4}\ge\frac{1}{9}$$
Equality takes place for $x=y=z=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$p$ ways to write $p$ as sum of primes I hope this question is valid as I'm just curious.
in a tweet from AlgebraFact I read the following:
"There are 17 ways to write 17 as a sum of primes": $17, 2+2+13, 3+3+11, 3+7+7, 5+5+7, 2+2+2+11, 2+3+5+7, 2+5+5+5, 2+2+3+3+7, 2+2+3+5+5, 3+3+3+3+5, 2+2+2+2+2+7, 2+2+2+3+3+5, 2+3+3+3+3+3, 2+2+2+2+2+2+5,2+2+2+2+3+3+3, 2+2+2+2+2+2+2+3$
It seemed to me a curious property and I would like to know if there are more primes that satisfy this property or if there is any result in this regard (another thing that seems strange to me is that 17 is included)
| Here is an approach to prove that there are only finitely many numbers with this property. We only pay attention to expressions of the form $m \cdot 2 + n \cdot 3 +p\cdot 5 +7=N$. The point is that the number of expressions of this type grows faster than $N$, so there will eventually be far too many expressions of this type and we have ignored others. There are a lot of asymptotic approximations made, so you need to fill in a lot of blanks.
There are about $N/6$ ways to express $N$ as an unordered sum of $2$s and $3$s. If you only use $2$s there are $N/2$ of them, then you can replace $3\ 2s$ by $2\ 3s$ up to $N/6$ times. For each of these $N/6$ expressions, we can replace $2+3$ by $5$, or $5\ 2$s by $2\ 5s$, or $5\ 3$s by $3\ 5$s. For each sum of $2$s and $3$s there are at least $N/15$ replacements we can do of this sort. We therefore have at least $N^2/90$ ways to express $N$ as a sum of $2,3,5$s For a given $q$ we have at least $\frac {(q-7)^2}{90}$ ways to express $q$ as a sum of primes, which exceeds $q$ when $q \gt 100$. There are not many numbers to check. In fact the number of expressions will grow much faster than this, so once it is greater than $N$ you will be done. $17$ should be the only one.
| {
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"url": "https://math.stackexchange.com/questions/4438778",
"timestamp": "2023-03-29T00:00:00",
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Constructing a right triangle given $a+b-c$ and $\alpha$ The exercise is to construct a right triangle given $a+b-c$ and $\alpha$.
I know we then have $\beta=90^\circ -\alpha$. I tried to draw the right triangle $\triangle ABC$ and find where I can use $a+b-c$. By adding $a$ to $b$ to get $\overline{DA}, |DA|=a+b$, I get an isosceles triangle $\triangle DCB$ so $|\angle CDB|=45^\circ$. I then subtract the length $c$ from $\overline{DA}$ and have $|DN|=a+b-c$. I thought I could construct the triangle $\triangle DNB$ so I could find $C$ as the intersection of the line $DN$ and the perpendicular bisector of $\overline{DB}$. However, I'm missing an element in order to be able to even construct $\triangle DNB$. Additionally, I tried adding lengths to the hypotenuse, but also without much success.
| Alternative approach.
It is assumed that $c$ is the hypotenuse, that $a$ is the side opposite the angle $\alpha$, and that the value $(a+b-c)$ equals a given fixed value $T$.
Then, you have the following two equations in the two unknowns, $a,b$.
*
*$c^2 = (a + b - T)^2 = a^2 + b^2.$
*$\displaystyle \frac{a}{b} = \tan(\alpha)$.
Here, I am assuming that since $\alpha$ is known, that $\tan(\alpha)$ is known.
Once $a,b$ are solved, then $c$ is immediately solved by $a + b - c = T.$ Therefore, the problem reduces to using the above two equations to solve for $a,b$.
Using the first equation above, you have that
$$T^2 + 2ab = 2T(a+b). \tag1 $$
Let $R = \tan(\alpha).$
This implies that
$$a = bR \tag2 .$$
Using (1) and (2) above, you have that
$$T^2 + 2Rb^2 = 2Tb(R+1) \implies$$
$$b^2(2R) - b(2T[R+1]) + T^2 = 0. \tag3 $$
(3) above represents a quadratic in $b$. Therefore,
$$b = \frac{1}{4R}\left[2T(R+1) \pm \sqrt{[2T(R+1)]^2 - 8RT^2}\right]. \tag4 $$
(4) above simplifies to
$$b = \frac{T}{2R}\left[(R+1) \pm \sqrt{R^2 + 1}\right]. \tag5 $$
Temporarily, both roots in (5) will be carried, until $c$ is evaluated. Since $c$ must be $> 0$, one of the roots will then be eliminated.
$$a = \frac{T}{2R}\left[(R+1) \pm \sqrt{R^2 + 1}\right] \times R. \tag6 $$
Using that $c = a + b - T$, this implies that
$$c = \frac{T}{2R}\left[(R^2+1) \pm (R+1)\sqrt{R^2 + 1}\right]. \tag7 $$
At this point, since $R > 0 \implies \sqrt{R^2 + 1} < (R+1)$
you can infer that the smaller roots in (5), (6), and (7) are inappropriate.
You can then manually verify that the resulting values for $a,b,c$ satisfy $a^2 + b^2 = c^2.$
Edit
After reading the answer of Grab a Coffee, I was intrigued enough to sanity-check my answer.
$\displaystyle R = \frac{\sin(\alpha)}{\cos(\alpha)} \implies $
*
*$\displaystyle R^2 + 1 = \frac{1}{\cos^2(\alpha)}.$
*$\displaystyle (R + 1) = \frac{\sin(\alpha) + \cos(\alpha)}{\cos(\alpha)}.$
Therefore,
$$c = \frac{T}{2} \times \frac{\cos(\alpha)}{\sin(\alpha)} \times
\left[\frac{1}{\cos^2(\alpha)} + \frac{\sin(\alpha) + \cos(\alpha)}{\cos(\alpha)} \frac{1}{\cos(\alpha)}\right]. $$
This simplifies to
$$\frac{T}{2} \times \frac{1}{\sin(\alpha) \cos(\alpha)} \times [\sin(\alpha) + \cos(\alpha) + 1]
\times \frac{\sin(\alpha) + \cos(\alpha) - 1}{\sin(\alpha) + \cos(\alpha) - 1}$$
$$ = \frac{T}{2} \times \frac{\sin^2(\alpha) + \cos^2(\alpha) + 2\sin(\alpha)\cos(\alpha) - 1}{\sin(\alpha)\cos(\alpha) \times [\sin(\alpha) + \cos(\alpha) - 1]}$$
$$ = \frac{T}{2} \times \frac {2}{\sin(\alpha) + \cos(\alpha) - 1}.$$
| {
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If $abc=1$ ,and $n$ is a natural number,prove $ \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2} $ If $a, b, c$ are distinct positive real numbers such that $abc=1$,and $n$ is a natural number,prove
$$
\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} \geqslant \frac{n(n-1)}{2}
$$
I know for $n=3$,the answer is here,and how to go further?
| As can be found here and here , we have that
$$
P_n = \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} =
\displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}
$$
Now by AM-GM, it follows that
$$
P_n =
\displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}
\geq Q [\displaystyle\prod_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3} ]^{1/Q}
$$
where $Q$ is the number of terms in the sum $\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}$.
Now note that by symmetry,
$$
P_n
\geq Q [\displaystyle\prod_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3} ]^{1/Q} = Q [a b c ]^{\frac{Q(n-2)}{3 Q}} = Q
$$
where the last one is true by the condition $abc =1 $. So it remains to find $Q$. This derives from the sum's conditions: we can select $n_1$ from $0$ to $n-2$, i.e. $n-1$ possibilites. Then we have that $n_2$ can be taken from $0$ to $n-2-n_1$, i.e. $n-1-n_1$ possibilites. So $Q = \sum_{n_1 =0}^{n-2} (n-1-n_1) = \frac12 n (n-1)$ which proves the claim. $\qquad \Box$
Some examples, using AM-GM directly:
$$P_2 = 1 \geq 1 = \frac12 \cdot 2 \cdot (1)\\
P_3 = a + b +c \geq 3 [abc]^{\frac{1}{3}} = 3 = \frac12 \cdot 3 \cdot(2)\\
P_4 = a^2 + b^2 + c^2 + ab + bc + ac \geq 6 [abc]^{\frac{2}{3}} = 6 = \frac12 \cdot 4 \cdot(3)
$$
| {
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solve the recurrence relation $h_n - 0h_{n - 1} - 3h_{n - 2} + 2h_{n - 3}= 0$ $h_n - 0h_{n - 1} - 3h_{n - 2} + 2h_{n - 3}= 0$
$h_0=2, h_1=0, h_2=7$, and n≥3
is given
here is what I did
$x^3-3x+2=0$
$h_n=a1^n+b(-2)^n$
roots of the polynomial are 1 and -2 but there are 3 equations($h_0,h_1,h_2$) and 2 variables so we need 3 three roots. what is the third root? how do you solve this type of recurrences?
| What you did is correct.
The characteristic equation is
$ x^3 - 3 x + 2 = 0 $
By inspection, one of the roots is $x = 1 $. By synthetic division,
$ (x^3 - 3 x + 2) = (x - 1) (x^2 + x - 2) = (x - 1)^2 (x + 2) $
So the roots are $ 1, 1, -2 $
Therefore, the solution to this homogeneous difference equation is
$ h_n = A + B n + C (-2)^n $
From the initial conditions $h_0, h_1, h_2$ the constants $A,B,C$ can be computed.
We have
$h_0 = 2 = A + C $
$ h_1 = 0 = A + B - 2 C $
$ h_2 = 7 = A + 2 B + 4 C $
And these equations solve to $ A = B = C = 1 $
Hence, the sequence is
$ h_n = 1 + n + (-2)^n $
| {
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Solve $a_n=n(a_{n-1}+a_{n-2})$ where $a_0=1,a_1=2$ using generating functions I am trying to solve a recurrence relation,$a_n=n(a_{n-1}+a_{n-2})$ where $a_0=1,a_1=2$, using generating functions. So, I did: let $$A(x)=\sum_{n\geq 0}a_n\frac{x^{n}}{n!}$$
$$\sum_{n\geq 0}a_{n+2}\frac{x^{n+2}}{(n+2)!}=\sum_{n\geq 0}a_{n+1}\frac{x^{n+2}}{(n+1)!}+\sum_{n\geq 0}a_n\frac{x^{n+2}}{(n+1)!}$$
$$[A(x)-a_1x-a_0]=\frac{[A(x)-a_0]}{x}+\frac{A(x)x^2}{(n+1)}$$
However, $$\sum_{n\geq 0}a_n\frac{x^{n+2}}{(n+1)!}=\frac{A(x)x^2}{(n+1)}$$ does not allow me make process further, so I need help to find $A(x)$.
Note that the answer should be $a_n=(n+1)!$. However, I want to arrive at it using only generating functions not differential equations or linear algebra techniques.
Edit: I saw that I made a silly mistake in $$\sum_{n\geq 0}a_n\frac{x^{n+2}}{(n+1)!}=\frac{A(x)x^2}{(n+1)}$$ Do you have any suggestion to write $\sum_{n\geq 0}a_n\frac{x^{n+2}}{(n+1)!}$ in terms of $A(x)=\sum_{n\geq 0}a_n\frac{x^{n}}{n!}$ ?
Original question: Formula for $a_n$ where $a_n$ = n*($a_{n-1}$+$a_{n-2}$)
| A standard way to do this does end up with a differential equation.
Define
$$
A(x) := \sum_{n=0}^\infty a_n \frac{x^n}{n!}
$$
We need similar sums with $na_{n-1}$ and $na_{n-2}$.
Compute
$$
\sum_{n=1}^\infty n a_{n-1} \frac{x^n}{n!}
= \sum_{n=1}^\infty a_{n-1}\frac{x^n}{(n-1)!}
= \sum_{n=0}^\infty a_n\frac{x^{n+1}}{n!}
= x\sum_{n=0}^\infty a_n\frac{x^{n}}{n!}
=xA(x)
$$
Next
\begin{align}
\sum_{n=2}^\infty n a_{n-2}\frac{x^n}{n!}
&=\sum_{n=2}^\infty a_{n-2}\frac{x^n}{(n-1)!}
=x\sum_{n=2}^\infty \frac{a_{n-2}}{n-1}\frac{x^{n-1}}{(n-2)!}
\\&
=x\sum_{n=0}^\infty \frac{a_{n}}{n+1}\frac{x^{n+1}}{n!}
=x \int_0^x \sum_{n=0}^\infty a_n\frac{t^n}{n!}\;dt = x\int_0^x A(t)\;dt
\end{align}
Now the recurrence equation $a_n = n(a_{n-1}+a_{n-2})$ yields
\begin{align}
0 &= \sum_{n=2}^\infty \big(a_n - na_{n-1}-na_{n-2}\big)\frac{x^n}{n!}
\\
&=\sum_{n=2}^\infty a_n \frac{x^n}{n!}
-\sum_{n=2}^\infty na_{n-1}\frac{x^n}{n!}
-\sum_{n=2}^\infty na_{n-2}\frac{x^n}{n!}
\\ &=
\left(-a_0-a_1x+\sum_{n=0}^\infty a_n \frac{x^n}{n!}\right)
-\left(-a_0 x+\sum_{n=1}^\infty na_{n-1}\frac{x^n}{n!}\right)
-\left(\sum_{n=2}^\infty na_{n-2}\frac{x^n}{n!}\right)
\\ &=
\big(-1-2x+A(x)\big) - \big(-x+xA(x)\big) - \left(x\int_0^x A(t)\;dt\right)
\end{align}
We need to solve this integral equation. Since I know much more about differential equations, let's consider this a differential equation for $F(x) := \int_0^x A(t)\,dt$.
\begin{align}
0 &= \big(-1-2x+F'(x)\big) - \big(-x+xF'(x)\big) - \big(xF(x)\big)
\\\implies
(1-x)F'(x) - xF(x) &= x+1
\end{align}
with initial condition $F(0) = 0$. We get solution $F(x) = \frac{x}{x-1}$,
so $A(x) = F'(x) = \frac{1}{(1-x)^2}$.
Finally, compute the Taylor series for this:
$$
\frac{1}{(1-x)^2} = 1+2x+3x^2+4x^3+\cdots = \sum_{n=0}^\infty (n+1)x^n
$$
To find the formula for $a_n$, reason like this:
$$
\sum_{n=0}^\infty a_n \frac{x^n}{n!} = \sum_{n=0}^\infty (n+1)x^n
\\
a_n \frac{1}{n!} = (n+1)
\\
a_n = (n+1)n! = (n+1)!
$$
| {
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What is the value of $a_1a_2\cdots a_{2019}$? Let $a_1=\frac 34$ and for any $n\geq2$ $4a_n=4a_{n-1}+\frac {2n+1}{1^3+2^3+\cdots n^3 }$. What is the value of $a_1a_2\cdots a_{2019}$?
I tried $1^3+2^3+\cdots +n^3=\frac {n^2(n+1)^2}{4}$ and
I wroted
$4(a_n-a_{n-1})=\frac {2n+1}{1^3+2^3+\cdots n^3 }$.
Then
$$4(a_2-a_1)=\frac {2×2+1}{1^3+2^3 }\\ 4(a_3-a_2)=\frac {2×3+1}{1^3+2^3+3^3 }\\ \cdots$$
When we add the equations side by side, the left side becomes simpler. But the right side gets complicated. I tried to find closed form for $a_n$ but it seems very complicated.
| Observe $$a_n - a_{n-1} = \frac{2n+1}{n^2(n+1)^2} = \frac{(n+1)^2 - n^2}{n^2 (n+1)^2} = \frac{1}{n^2} - \frac{1}{(n+1)^2}. \tag{1}$$ So $$a_m - a_1 = \sum_{n=2}^m a_n - a_{n-1} = \frac{1}{2^2} - \frac{1}{(m+1)^2},$$ hence $$a_m = 1 - \frac{1}{(m+1)^2} = \frac{m(m+2)}{(m+1)^2}. \tag{2}$$ Therefore, $$\prod_{m=1}^{2019} a_m = \prod_{m=1}^{2019} \frac{m}{m+1} \frac{m+2}{m+1} = \frac{2021}{4040}.$$
| {
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Proving $\frac{\sqrt 2\ \Gamma\left(\frac m 2\right)}{\sqrt{m - 1}\ \Gamma\left(\frac{m - 1}{2}\right)} < 1$ I would like to prove, that for $m \ge 2$ we have the following inequality:
$$\frac{\sqrt 2\ \Gamma\left(\frac m 2\right)}{\sqrt{m - 1}\ \Gamma\left(\frac{m - 1}{2}\right)} < 1$$
My thoughts
This inequality is equivalent to that one:
$$\sqrt 2\ \Gamma\left(\frac m 2\right) < \sqrt{m - 1}\ \Gamma\left(\frac{m - 1}{2}\right)$$
And now, when $m \ge 2$, then of course $\sqrt{m - 1} \ge 2$, but I don't know how can I prove, that
$$\Gamma\left(\frac m 2\right) < \Gamma\left(\frac{m - 1}{2}\right)$$
For example, when $m \ge 2$ and $m = 2k$, where $k \in \mathbb N$. We have that
$$(k - 1)! < \Gamma\left(k - \frac 1 2\right)$$
But how this inequality can be justified?
| Gautschi's inequality states that
$$
x^{1 - s} < \frac{{\Gamma (x + 1)}}{{\Gamma (x + s)}}
$$
for $x>0$ and $0<s<1$. Let $m>1$. Taking $x=\frac{m-1}{2}$ and $s=\frac{1}{2}$ gives
$$
\sqrt {\frac{{m - 1}}{2}} < \frac{{\Gamma \left( {\frac{{m - 1}}{2} + 1} \right)}}{{\Gamma \left( {\frac{m}{2}} \right)}},
$$
i.e.,
$$
\frac{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{m}{2}} \right)}}{{\Gamma \left( {\frac{{m - 1}}{2} + 1} \right)}} < 1.
$$
But note that
$$
\frac{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{m}{2}} \right)}}{{\Gamma \left( {\frac{{m - 1}}{2} + 1} \right)}} = \frac{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{m}{2}} \right)}}{{\frac{{m - 1}}{2}\Gamma \left( {\frac{{m - 1}}{2}} \right)}} = \frac{{\Gamma \left( {\frac{m}{2}} \right)}}{{\sqrt {\frac{{m - 1}}{2}} \Gamma \left( {\frac{{m - 1}}{2}} \right)}} = \frac{{\sqrt 2 \Gamma \left( {\frac{m}{2}} \right)}}{{\sqrt {m - 1} \Gamma \left( {\frac{{m - 1}}{2}} \right)}}.
$$
| {
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Simpler proof that $y^3[d^2y/dx^2]$ is a constant if $y^2=ax^2+bx+c$? here's my question
If $y^2=ax^2+bx+c$ then prove that $y^3[d^2y/dx^2]$ is a constant .
I have solved this using the conventional method, taking square root, differentiating w.r.t to x and using chain and quotient rule
But can't think of some alternative smaller and more efficient method ??
Can you ?
Also can someone suggest me any alternative to quotient rule
I need something more non conventional, time saving and vastly
Applicable method .
| Using implicit derivation:
$$
(y^2)^\prime = 2yy^\prime = 2ax+b \Rightarrow y^\prime=\frac{2ax+b}{2y}
$$
$$
(y^2)^{\prime\prime} = 2(y^\prime)^2+2yy^{\prime\prime} = 2a \Rightarrow y^{\prime\prime}=\frac{a-(y^\prime)^2}{y}
$$
Thus
\begin{align}
y^3y^{\prime\prime}=y^2(a-(y^\prime)^2) &= y^2(a-\left(\frac{2ax+b}{2y}\right)^2) \\
&= ay^2-\frac{1}{4}(2ax+b)^2 \\
&= (a^2x^2+abx+ac)-\frac{1}{4}(4a^2x^2+4abx+b^2) \\
&= (a^2x^2+abx+ac)-(a^2x^2+abx+\frac{b^2}{4}) \\
&= ac - \frac{b^2}{4} \ \ \ \text{(which is constant)}
\end{align}
| {
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What did I do wrong in solving $\int\sec^{-1} x\,{dx}$? I used integration by parts:
let u=$\sec^{-1}\,x$, dv=dx,
then du=$\frac{1}{|x|\sqrt{x^2-1}}$, v=x.
I = $x\sec^{-1}\,x\;-\;\int\frac{x}{|x|\sqrt{x^2-1}}dx\\$
Integration of $\int\frac{x}{|x|\sqrt{x^2-1}}dx$:
Let x=$\sec\theta$, then dx = $\sec\theta\tan\theta\,d\theta$. $\theta\in(0, \frac{\pi}2)\cup(\frac{\pi}2, \pi)\\$.
$\int\frac{x}{|x|\sqrt{x^2-1}}dx\,=\,\int\frac{\sec\theta}{|\sec\theta|\sqrt{\sec^2\theta-1}}\sec\theta\tan\theta\,d\theta\\\qquad\qquad\qquad=\,\int\frac{\sec\theta}{|\sec\theta|}\frac{\tan\theta}{|\tan\theta|}\sec\theta\,d\theta$
When $\theta\in(0, \frac{\pi}2),\;\sec\theta=x\text{, which is}\gt0$. $\tan\theta\gt0\text{, and }\tan\theta=\sqrt{x^2-1}$,
$\int\frac{x}{|x|\sqrt{x^2-1}}dx\,=\,\int\frac{\sec\theta}{\sec\theta}\frac{\tan\theta}{\tan\theta}\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\int\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\ln|\sec\theta+\tan\theta|+c\\\qquad\qquad\qquad=\,\ln|x+\sqrt{x^2-1}|+c$.
Then, I = $x\sec^{-1}\,x\;-\;\ln|x+\sqrt{x^2-1}|+c$.
I know that this is the right answer, but as I continue, I get a different answer for $\theta\in(\frac{\pi}2, \pi)$.
When $\theta\in(\frac{\pi}2, \pi)$, $\sec\theta=x\text{, which is }\lt0$, $\tan\theta\lt0\text{, and }\tan\theta=-\sqrt{x^2-1}$,
$\int\frac{x}{|x|\sqrt{x^2-1}}dx\,=\,\int\frac{\sec\theta}{-\sec\theta}\frac{\tan\theta}{-\tan\theta}\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\int\sec\theta\,d\theta\\\qquad\qquad\qquad=\,\ln|\sec\theta+\tan\theta|+c\\\qquad\qquad\qquad=\,\ln|x-\sqrt{x^2-1}|+c\\\qquad\qquad\qquad=\,-\ln|x+\sqrt{x^2-1}|+c$
Then, I = $x\sec^{-1}\,x\;\boldsymbol{+}\;\ln|x+\sqrt{x^2-1}|+c$
The answer is $x\sec^{-1}\,x\;\boldsymbol{-}\;\ln|x+\sqrt{x^2-1}|+c\\$, what did I do wrong in the last part?
| Note that the hyperbolic relation $y^2-2xy+1=1$ is equivalent to $y=x\pm\sqrt{x^2-1}$. And one of those branches, $y=x-\sqrt{x^2-1}$ is equivalent to $y=\frac{1}{x+\sqrt{x^2-1}}$ (if you rationalize its denominator). And once you take the logarithm of $\frac{1}{x+\sqrt{x^2-1}}$, you have the negative of the logarithm of $y=x+\sqrt{x^2-1}$.
So the two answers are related as they capture a certain portion of the relationship $y^2-2xy+1=1$.
However as another answer explains, given the context with the domain of $\sec^{-1}$, working with $y=x+\sqrt{x^2-1}$ makes sense for positive $x$, and working with $y=\left\lvert\frac{1}{x+\sqrt{x^2-1}}\right\rvert$ makes sense for negative $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to show that $f(x)=\frac{\arcsin x}x$ is increasing when $x\ge0$? How to show that $f(x)=\frac{\arcsin x}x$ is increasing when $x\ge0$?
My Attempt:
$f'(x)=\frac{\frac x{\sqrt{1-x^2}}-\arcsin x}{x^2}$
Since $0\le x\le1\implies0\le\sqrt{1-x^2}\le1$
And $0\le\arcsin x\le1.57$
But not able to show that $f'(x)\ge0$
| You can also find a series expansion of $\arcsin(x)$ as follows:
$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\dots$$
Put $x=\frac{t^2}{1-t^2}$, and we have
$$\frac{1}{\sqrt{1-t^2}}=1+\frac{t^2}{2}+\frac{3t^4}{8}+\frac{5t^6}{16}+\dots$$
Now integrating we have
$$\arcsin(x)=\int_0^x\frac{1}{\sqrt{1-t^2}}dt=x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+\dots$$
and so $$\frac{\arcsin(x)}{x}=1+\frac{x^2}{6}+\frac{3x^4}{40}+\frac{5x^6}{117}+\dots$$
which is clearly increasing for $x\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Let $a$,$b$ be the roots of $x^2-6cx-7d=0$, and let $c$, $d$ be the roots of $x^2-6ax-7b=0$. Find $b+d$.
Let $a$,$b$ be the roots of the equation $x^{2}-6 c x-7 d=0$ and let $c$,$d$ be the roots of the equation $x^{2}-6ax-7 b=0$. $a,b,c,d$ are distinct. Find value of $b+d$.
Using the relation between roots and coefficients of an equation, I got $a+b+c+d=6(a+c)$, and hence getting $b+d= 5(a+c)$, but I don't know what to do next.
Also, $abcd= 49bd$, but I don't know how to use it here.
| Alternative approach:
From the original equations you have that:
*
*$[E_1:] ~ a + b = 6c \implies b = 6c - a.$
*$[E_2:] ~ c + d = 6a \implies d = 6a - c.$
Using the two results above, you also have, from the original equations that:
*
*$[E_3:] ~ 6ac - a^2 = ab = -7d = -7(6a - c) = 7c - 42a.$
*$[E_4:] ~ 6ac - c^2 = cd = -7b = -7(6c - a) = 7a - 42c.$
Subtracting $E_3 - E_4$ above gives
$(c-a)(c+a) = c^2 - a^2 = 7(c-a) + 42(c-a) = 49(c-a).$
This implies that $c+a = 49.$
Then, adding $E_1 + E_2$ gives
$(a + c) + (b + d) = 6(a + c) \implies (b + d) = 5(a + c) = 5 \times 49.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral of this weird function $ \int \frac{1}{(x^2+x+1)^2}dx $ I put this equation into Symbolab and it produced me a very complex result
Basically this is the result but I believe there is a simpler way to solve this question
$$
\frac{2}{3\sqrt{3}}\left(2\arctan \left(\frac{2x+1}{\sqrt{3}}\right)+\sin \left(2\arctan \left(\frac{2x+1}{\sqrt{3}}\right)\right)\right)+C
$$
This was my question
$$
\int \frac{1}{\left(x^2+x+1\right)^2}dx
$$
| Under $x+\frac12=\frac{\sqrt3}{2}\tan\theta$ or $\theta=\arctan\bigg(\frac{2x+1}{\sqrt3}\bigg)$, one has
\begin{eqnarray}
&&\int \frac{1}{(x^2+x+1)^2}dx\\
&=& \int \frac{1}{((x+\frac12)^2+\frac34)^2}dy\\
&=& \frac{8\sqrt3}{9}\int\frac{1}{\sec^4\theta}\sec^2\theta d\theta\\
&=& \frac{8\sqrt3}{9}\int\cos^2\theta d\theta\\
&=& \frac{4\sqrt3}{9}\int(1+\cos(2\theta)) d\theta\\
&=& \frac{2\sqrt3}{9}\bigg[2\theta+\sin(2\theta)\bigg]+C.
\end{eqnarray}
Using
$$ \sin(2x)=\frac{2\tan x}{1+\tan^2x} $$
one has
$$ \sin(2\theta) = \frac{2x+1}{2\sqrt3}\frac{1}{x^2+x+1}$$
and hence
$$ \int \frac{1}{(x^2+x+1)^2}dx=\frac{2\sqrt3}{9}\bigg[2\arctan\bigg(\frac{2x+1}{\sqrt3}\bigg)+\frac{2x+1}{2\sqrt3}\frac{1}{x^2+x+1}\bigg]+C. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all odd primes $p$ such that $3p-8$ is equal to the cube of a positive integer I recently tackled this problem from one of my country's regional competitions. My solution is as follows:
Let $$3p - 8 = n^3$$ From this we know that $$p = \frac{n^3 + 8}{3}$$ It follows that $$p = \frac{(n+2)(n^2 - 2n + 4)}{3}$$
Now, since $p$ is prime, we require $\frac{n + 2}{3} = 1$ or $n^2 - 2n + 4 = 1$. (not both, since then it would be $p = \frac{1}{3}$, which is a contradiction). From both of our conditions we obtain $n = 1$, and so $p = 3$.
The "accepted" solution to this problem looked at 4 total different cases and is way longer, but comes to the exact same result as mine. Is my solution inadequate?
| I'd say that in getting the factorization of $p$ that you did is a crux of the problem, but you still had more math to do. You did not exhaust all possibilities. However, you could note the following:
Claim 1. If $a$ and $b$ are integers that are at least $4$ each, then either $\frac{ab}{3}$ is nonintegral or composite.
If $\frac{ab}{3}$ is integral, then at least one of $a,b$, say WLOG $b$, has $3$ as a factor. So write $b=3c$ for some integer $c$. Then $\frac{ab}{3} = ac$. However, that $a,b$ satisfy $a,b \ge 4$ gives $a \ge 4$ and $c \ge 2$. And so if $\frac{ab}{3}$ is integral then $\frac{ab}{3} = ac$, for some integer $c \ge 2$ and $a \ge 4$, and so then $\frac{ab}{3}$ must indeed be composite. ■
Then setting $a=n+2$, $b=n^2-2n+4$, it remains to find the set of positive integers $n$ such that $p =\frac{ab}{3}$ is both integral and prime. Fortunately both $a=n+2$, $b=n^2-2n+4$ is at least $4$ for all $n\ge 2$, so by Claim 1 all you need to check is $n=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4460734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}=\cot\left(\frac{3\pi}{2}-\alpha\right)$ Prove that $$\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}=\cot\left(\dfrac{3\pi}{2}-\alpha\right)$$
The RHS is equal to $\tan\alpha,$ so we are to show $$\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}=\tan\alpha$$
My try for simplifying the LHS: $$\dfrac{2\sin2\alpha\cos2\alpha}{1+\cos^22\alpha-\sin^22\alpha}\cdot\dfrac{\cos^2\alpha-\sin^2\alpha}{1+\cos^2\alpha-\sin^2\alpha}=\\=\dfrac{4\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)}{1+(\cos^2\alpha-\sin^2\alpha)^2-(2\sin\alpha\cos\alpha)^2}\cdot\dfrac{\cos^2\alpha-\sin^2\alpha}{1+\cos^2\alpha-\sin^2\alpha}\\=\dfrac{4\sin\alpha\cos\alpha(\cos\alpha-\sin\alpha)(\cos\alpha+\sin\alpha)}{1+\sin^4\alpha-2\cos^2\alpha\sin^2\alpha+\cos^4\alpha-4\sin^2\alpha\cos^2\alpha}\cdot\dfrac{\cos^2\alpha-\sin^2\alpha}{2\cos^2\alpha}$$
| In fact, using
$$ 1-\cos 2x=2\sin^2x, 1+\cos2x=2\cos^2x $$
one has
\begin{eqnarray}
&&\dfrac{\sin4\alpha}{1+\cos4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}\\
&=&\dfrac{\sin4\alpha(1-\cos4\alpha)}{1-\cos^24\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}\\
&=&\dfrac{1-\cos4\alpha}{\sin4\alpha}\cdot\dfrac{\cos2\alpha}{1+\cos2\alpha}\\
&=&\dfrac{1-\cos4\alpha}{2\sin2\alpha}\cdot\dfrac{1}{1+\cos2\alpha}\\
&=&\dfrac{2\sin^22\alpha}{2\sin2\alpha}\cdot\dfrac{1}{2\cos^2\alpha}\\
&=&\dfrac{\sin2\alpha}{2\cos^2\alpha}=\tan\alpha\\
&=&\cot\left(\dfrac{3\pi}{2}-\alpha\right).
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to simplify the calculations of reflecting a ray about an ellipse
I wrote the script that made these images several days ago, the segments each depict a ray of light, as the light hits the boundary of the ellipse, it is reflected by the ellipse according to the laws of reflection, and the reflected ray of light is again reflected by the ellipse, and the reflection of the reflection is again reflected by the ellipse...
The light keeps bouncing back and forth, over and over again, until the light has been reflected certain number of times.
Step by step on how I made these images.
First, you make an ellipse.
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
The above equation describes the points on the ellipse, it is also the boundary of the ellipse. (Assume the ellipse is horizontal).
Construct a right triangle from one semi-minor axis and one semi-major axis, let the acute angle adjacent to the semi-major axis be $\alpha$, then the relationship between a, b and $\alpha$ can be written as:
$b = a \cdot tan(\alpha)$
Rewrite the equation of the ellipse:
$\frac{x^2}{a^2} + \frac{y^2}{a^2 \cdot tan(\alpha)^2} = 1$
Pick a random point inside the ellipse, for all points inside by the ellipse, simply use this equation:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} <= 1$
Then the random point is given as:
$(a \cdot cos(\alpha) \cdot m, b \cdot sin(\alpha) \cdot n)$
Where $\alpha$ is in range $[0, 2\pi]$, and m, n are in range $[0, 1]$.
Then pick a random angle $\beta$, to construct a line passes the chosen point at angle $\beta$ with the x-axis.
I use the slope-intercept form
of line equation ($y = k \cdot x + c$):
Let the chosen point be $(x_0, y_0)$, then the equation of the incident ray is:
$y = tan(\beta) \cdot x + y_0 - x_0 \cdot tan(\beta)$
But if $\beta$ is a multiple of $\frac{\pi} {2}$ things get complicated, because tan(0) = 0 and $tan(\frac{\pi} {2})$ is undefined.
So I use $y = y_0$ if the line is parallel to x-axis and $x = x_0$ if the line is perpendicular to x-axis.
Now to calculate the intersections between the incident ray and ellipse.
If:
$c^2 < a^2 \cdot k^2 + b^2$
Then there can be two intersections $(x_0, y_0)$ and $(x_1, y_1)$.
$$\begin{align}
n_1 &= a^2 \cdot k^2 + b^2 \\
n_2 &= 2 \cdot a^2 \cdot k \cdot c \\
n_3 &= a^2 \cdot (c^2 - b^2) \\
n_4 &= \sqrt{(n_2^2 - 4 \cdot n_1 \cdot n_3)} \\
x_0 &= \frac{(-n_2 + n_4)} {(2 \cdot n_1)} \\
x_1 &= \frac{(-n_2 - n_4)} {(2 \cdot n_1)} \\
y_0 &= k \cdot x_0 + c \\
y_1 &= k \cdot x_1 + c
\end{align}$$
For lines like $x = x_0$ and $y = y_0$ however:
First set:
If $abs(x_0) <= a$
$y_0 = \sqrt{b^2 - x_0^2 \cdot \frac{b^2} {a^2}}$
Intersections are $(x_0, +y_0)$ and $(x_0, -y_0)$.
Second set:
If $abs(y_0) <= b$
$x_0 = \sqrt{a^2 - y_0^2 \cdot \frac{a^2} {b^2}}$
Intersections are $(+x_0, y_0)$ and $(-x_0, y_0)$
Then choose the intersection to do further calculations, for $\beta$ in range $[0, \frac{\pi} {2}]$ and $[\frac{3 \pi} {2}, 2 \pi]$ I choose the right intersection, else I choose the left one.
Then I calculate the tangent of the ellipse at the intersection.
For a point $(x_0, y_0)$ on the ellipse given by a, b, the slope k of the line tangent to the ellipse at that point must satisfy:
$k = \frac{-x_0 \cdot b^2} {a^2 \cdot y_0}$
Then for almost all lines, the equation of the tangential line is:
$y = k \cdot x + \frac{b^2} {y_0}$
But if arctan(k) is a multiple of $\frac{\pi} {2}$ the above relationship breaks down and I instead fall back to constant form.
Then I calculate the normal (line perpendicular to that tangent passing through that intersection):
$y = \frac{-x} {k} + y_0 - \frac{-x_0} {k}$
Where $(x_0, y_0)$ is the intersection, and k is the slope of the normal.
But again the above relationship breaks down if the line is special. I won't show how I deal with exceptions here, I already have shown too many equations, you can see all the calculations in the code.
Then I calculate the signed angle formed by the incident ray and the normal:
Let $k_1$ be the slope of the incident ray, let $k_2$ be the slope of the normal:
$$\begin{align}
\alpha_1 &= atan(k_1) \\
\alpha_2 &= atan(k_2) \\
\alpha_\delta &= \alpha_2 - \alpha_1 \\
\alpha_\delta &= (\alpha_\delta + \pi) \bmod 2 \pi - \pi
\end{align}$$
Again, the above doesn't work if either of these lines are special, other calculations are required.
Then I calculate the reflected ray, simply by rotate the normal line about the intersection by $\alpha_\delta$ (assuming the previous calculations succeeded):
$$\begin{align}
\alpha_3 &= \alpha_2 + \alpha_\delta \\
y &= tan(\alpha_3) \cdot x + y_0 - tan(\alpha_3) \cdot x_0
\end{align}$$
Then I calculated the intersections between the reflected ray and the ellipse, there will be two intersections, this time the intersection needed is the other intersection from the current one.
Then all above calculations are repeated recursively, until a certain number of iteration is reached.
How to simplify all calculations involved, and calculate the reflected ray in as few steps as possible, including all the edge cases?
I want it that way: given the coordinate of a point and an angle, calculate the intersection of the ray with the ellipse, and then calculate the tangent of the ellipse at that intersection, then calculate the reflected ray, all of these in as few steps as possible, using one set of equations without exceptions. Preferably the number of equations involved should be less than or equal to six.
| I assume that the ellipse is given by the equation
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
We are given a starting point $\begin{pmatrix} x_0 \\ y_0 \end{pmatrix}$ inside the ellipse and the ray direction vector $\begin{pmatrix} u \\ v \end{pmatrix}$.
*
*The point of the intersection is
$$\tag{1} \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} + t \cdot \begin{pmatrix} u \\ v \end{pmatrix},$$
where $t \in \mathbb{R}$, satisfying
$$\frac{(x_0+tu)^2}{a^2} + \frac{(y_0+tv)^2}{b^2} = 1.$$
Short calculations yield
$$\tag{2} t = \frac{\sqrt{\frac{u^2}{a^2} + \frac{v^2}{b^2} - \left( \frac{uy_0 - vx_0}{ab} \right)^2} - \left( \frac{ux_0}{a^2} + \frac{vy_0}{b^2} \right)}{\frac{u^2}{a^2} + \frac{v^2}{b^2}},$$
which is the positive solution of the quadratic equation because the ray only moves forward.
*The normal vector $N$ to the ellipse at the point $\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ is the gradient of the function $F \begin{pmatrix} x \\ y \end{pmatrix} = \frac{x^2}{a^2} + \frac{y^2}{b^2}$, i.e.
$$\tag{3} N = \nabla F \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \begin{pmatrix} \frac{2x_1}{a^2} \\ \frac{2y_1}{b^2} \end{pmatrix}.$$
The new ray direction $\begin{pmatrix} u' \\ v' \end{pmatrix}$ is given by the formula
$$\tag{4} \begin{pmatrix} u' \\ v' \end{pmatrix} = \begin{pmatrix} u' \\ v' \end{pmatrix} - 2 P_N \begin{pmatrix} u' \\ v' \end{pmatrix} = \begin{pmatrix} u' \\ v' \end{pmatrix} - 2 \cdot \frac{\left< \begin{pmatrix} u' \\ v' \end{pmatrix}, N \right>}{\left< N, N \right>} \cdot N.$$
So ultimately you need four equations, $(1)$ - $(4)$, to compute the new starting point $\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ and the new ray direction $\begin{pmatrix} u' \\ v' \end{pmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\lim_{x\to2}\frac{\sqrt[3]{x^2-x-1}-\sqrt{x^2-3x+3}}{x^3-8}$ without Hopital rule I want to evaluate this limit without applying Hopital rule,$$\lim_{x\to2}\frac{\sqrt[3]{x^2-x-1}-\sqrt{x^2-3x+3}}{x^3-8}$$
After expanding the denominator I got,
$$\frac{1}{12}\lim_{x\to2}\frac{\sqrt[3]{x^2-x-1}-\sqrt{x^2-3x+3}}{x-2}$$Here I can introduce numerator as $f(x)$ And say the limit is in the form $\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}=f'(2)$ But it is very similar to Hopital rule (if not the same).
I noticed that the fraction is in the form $2.\frac{\sqrt[3]A-\sqrt B}{A-B}$ or $2.\frac{M^2-N^3}{M^6-N^6}$ but not sure if this helps.
| use $(a^6 - b^6) = (a-b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)$
Let $a = \sqrt [3]{x^2 - x - 1}, b = \sqrt{x^2 - 3x + 3}$
that is where we have $\frac {a-b}{x-2}$ we multiply top and bottom by $(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)$ giving
$\frac {a^6-b^6}{(x-2)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)}$
$\frac {(x^2 - x - 1)^2 - (x^2 - 3x+3)^3}{(x-2)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)}$
Multiply out the numerator and combine terms. If you haven't made an aglebra error, what you have will be something that divides by $(x-2).$ Factor out $(x-2)$ and cancel out the $(x-2)$ in the denominator.
Plug $x=2$ to evaluate the numerator.
And as $x$ approaches $2,$ $a,b$ approach $1,$ making the denominator equal $6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4463715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\
&= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)
\end{align}$$
now,
$$\begin{align}
a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\
&= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\
&= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\
&= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\
&= 2017
\end{align}$$
so
$$\begin{align}
&\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\
&= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\
&= 90585
\end{align}$$
correct?
| Holy unsimplified long expressions Batman.
$$
\begin{align}
a^2+b^2
&=(a+b)^2-2ab\\
&=45
\end{align}
$$
and
$$
\begin{align}
a^6+b^6
&= (a^2+b^2)^3-3(ab)^2(a^2+b^2)\\
&= (a^2+b^2)((a^2+b^2)^2-3(ab)^2)\\
&= 45\times(45^2-12)\\
&=90585\\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 6
} |
Getting two different answers on differentiating $\cos^{-1}(\frac{3x+4\sqrt{1-x^2}}{5})$
Question given in my book asks to find $\frac{dy}{dx} $ from the following equation.$$y=\cos^{-1}\left(\frac{3x+4\sqrt{1-x^2}}{5}\right)$$
My Attempt:
Starting with substitutions,
*
*Putting $\frac35=\cos\alpha\implies \frac45 = \sin\alpha$.
*Putting $x = \cos\beta\implies \sqrt{1-x^2} = \sin\beta$.
$$\begin{align}&y=\cos^{-1}\Big(\frac{3x+4\sqrt{1-x^2}}{5}\Big)\\ \implies& y = \cos^{-1}\Big(\frac{3}{5}x + \frac{4}{5}\sqrt{1-x^2}\Big)\\\implies& y = \cos^{-1}(\cos\alpha\cos\beta + \sin\alpha \sin \beta)\\\implies& y = \cos^{-1}[\cos(\alpha-\beta )] \tag{1}\\\implies& y = \alpha-\beta\\\implies& y = \cos^{-1}\left(\frac35\right) - \cos^{-1}(x)\\\implies&\color{blue}{\boxed{ \dfrac{dy}{dx} =\frac{1}{\sqrt{1-x^2}}}}.\end{align}$$
But, if I consider $(1.)$ again,
$$y = \cos^{-1}\left[\cos\color{red}{(\alpha-\beta)}\right]$$
This is also equals to, $$\cos^{-1}\left[\cos\color{red}{(\beta - \alpha)}\right].$$differentiating this, will give the negative of the answer which I got earlier.
My book shows that $\frac{-1}{\sqrt{1-x^2}}$ is correct. But why?
I think the mistakes lie in the very first step of substitution i.e.,$\frac35=\cos\alpha$ doesn't imply $\frac45 = \sin\alpha$. It should be $\sin\alpha = \pm \frac45$.
Similary $x = \cos\beta$ doesn't imply $\sqrt{1-x^2} = \sin\beta$ instead $\sin\beta = \pm \sqrt{1-x^2}$.
But how can I make sure that in the equation $(1.)$, $(\alpha - \beta)$ lies in the principal branch of the inverse cosine function?
|
Starting with substitutions,
*
*Putting $\frac35=\cos\alpha\implies \frac45 = \sin\alpha$.
Instead, just let $$\alpha=\arccos\frac35$$ so that $\frac45$ indeed equals $\sin\alpha\,$ (no need for any $\pm$ sign).
*
*Putting $x = \cos\beta\implies \sqrt{1-x^2} = \sin\beta$.
Instead, let $$\beta=\arccos x$$ so that $\sqrt{1-x^2}$ indeed equals $\sin\beta\,$ (no need for any modulus sign).
\begin{align}
\implies& y = \cos^{-1}\left[\cos(\alpha-\beta )\right] \tag{1}\\
\implies& y = \alpha-\beta\end{align}
This is true only for $$x\in[0.6,1],$$ that is, $\beta\in\left[0,\arccos\frac35\right],$ in which case $(\alpha-\beta)$ is in the first quadrant.
For $$x\in[-1,0.6],$$ that is, $\beta\in\left[\arccos\frac35,\pi\right],$ we have $(\alpha-\beta)$ in the third or fourth quadrant, so $y = \arccos(\cos(\alpha-\beta))= \color{red}{\boldsymbol-}(\alpha-\beta).$
$$y=\cos^{-1}\left[\frac{3x+4\sqrt{1-x^2}}{5}\right]$$
\begin{align}
\implies&\dfrac{dy}{dx} =\frac{1}{\sqrt{1-x^2}}\tag2\end{align}
Note that since $\arccos \theta$ has no derivative at $\pm1$ and $$\frac{3x+4\sqrt{1-x^2}}{5}=\pm1\iff x=0.6,$$ the required derivative is undefined at $x=0.6.$
Furthermore, judging from $(2),$ it is also undefined at $x=\pm1.$
All in all, $$\dfrac{\mathrm dy}{\mathrm dx}=
\begin{cases}\dfrac{-1}{\sqrt{1-x^2}} &&\text{if }\;-1<x<0.6;
\\\dfrac{1}{\sqrt{1-x^2}} &&\text{if }\;0.6<x<1;
\\\text{undefined} &&\text{otherwise}. \end{cases}$$
According to Wolfram Alpha and Desmos, an equivalent rewrite avoiding cases is $$\dfrac{\mathrm dy}{\mathrm dx}=\frac{4x- 3\sqrt{1 - x^2}}{\sqrt{(1 - x^2)(7 x^2 - 24 \sqrt{1 - x^2} x + 9)}};$$ that is, $$\dfrac{\mathrm dy}{\mathrm dx}=\frac{4x- 3\sqrt{1 - x^2}}{\left| 4x- 3\sqrt{1 - x^2} \right|\sqrt{1 - x^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Showing that switching the order of variables gives same sum so as to simplify the required expression for exact sum If $S(x,y)$ = $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)}{(5^x(5^x +5^y))}$ , then if we want to show that $S(x,y) = S(y,x)$ so as to get the simplification by adding both to get the exact sum . I tried to evaluate S(x,y)-S(y,x) , which gives $\sum_{y=0}^{\infty}$$\sum_{x=0}^{\infty}$$\frac{(x+y +xy)(5^y - 5^x)}{(5^{x+y}(5^x +5^y))}$ now how to do we show its value is zero ?
| Your double series has non-negative terms and $\sum_{x\geq 0}\sum_{y\geq 0}=\sum_{y\geq 0}\sum_{x\geq 0}$ is granted by Fubini's theorem. Very crudely
$$ \sum_{x=0}^{M}\sum_{y=0}^{N}\frac{x+y+xy}{5^x(5^x+5^y)}\stackrel{\text{AM-GM}}{\leq}\sum_{x=0}^{M}\sum_{y=0}^{N}\frac{(1+x)(1+y)}{2\cdot 5^x 5^{\frac{x+y}{2}}}=\frac{1}{2}\sum_{x=0}^{M}\frac{x+1}{5^{3x/2}}\sum_{y=0}^{N}\frac{y+1}{5^{y/2}} $$
so
$$S=\sum_{y\geq 0}\sum_{x\geq 0}\frac{x+y+xy}{5^x(5^x+5^y)}=\frac{1}{2}\sum_{x,y\geq 0}\left(\frac{x+y+xy}{5^x(5^x+5^y)}+\frac{x+y+xy}{5^y(5^x+5^y)}\right)=\frac{1}{2}\sum_{x,y\geq 0}\frac{(x+1)(y+1)-1}{5^{x+y}}$$
or
$$ S = \frac{1}{2}\left(\sum_{x\geq 0}\frac{x+1}{5^x}\right)^2 - \frac{1}{2}\sum_{s\geq 0}\frac{s+1}{5^s}=\frac{1}{2}\cdot\frac{25}{16}\cdot\frac{9}{16}=\color{red}{\frac{225}{512}}. $$
| {
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"url": "https://math.stackexchange.com/questions/4476314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\int \sin^4x dx$ In my textbook, I came upon the following problem:
Problem: Find $\int \sin^4x dx$
Since there was only given a solution sketch and I came to a different conclusion in one step, I ask if my following solution is correct.
Solution: In my solution, I use the following identities:
$$\sin^2x = \frac{1-\cos2x}{2} \tag{1} \label{1}$$
$$\cos^2x = \frac{1+\cos2x}{2} \tag{2} \label{2}$$
Now we can write
$$\int \sin^4x dx = \int \left( \frac{1-\cos2x}{2} \right)^2 dx = \int\frac{1}{4}dx - \frac{1}{2}\int \cos2xdx + \frac{1}{4} \int cos^22xdx$$
My textbook gave the last term in this equation as $\int\frac{1}{4}dx - \int \cos2xdx + \frac{1}{4} \int cos^22xdx$. But, since $(1-\cos2x)^2 = 1 - 2\cos2x + \cos^22x$ the middle integral should simplify to $- \frac{1}{2}\int \cos2xdx$. Is this correct? If not, how do we come to the conclusion of $-\int \cos2xdx$?
To complete the integration, we have
$$\int \frac{1}{4} dx = \frac{x}{4}$$
$$ - \frac{1}{2} \int \cos 2xdx = - \frac{1}{4}\int \cos u du = - \frac{1}{4} \sin2x = - \frac{\sin2x}{4}$$
and
$$\frac{1}{4} \int cos^22xdx = \frac{1}{4} \int \frac{1 + cos4x}{2}dx = \frac{4x + \sin 4x}{32}$$
so
$$\int \sin^4x dx = \frac{x}{4} - \frac{\sin2x}{4} + \frac{4x + \sin 4x}{32}$$
Are these conclusions and computations correct? Thanks for any help!
EDIT: According to the correction of @Eevee Trainer I now use the correct identity $\cos^2x = \frac{1+\cos2x}{2}$. The rest of the computation should be correct. Thanks all for your help and comments!.
| To answer the first discrepancy with the textbook, note that
$$\left( \frac{1 - \cos(2x)}{2} \right)^2
= \frac{1- 2 \cos(2x) + \cos^2(2x)}{4}$$
Hence the middle term indeed has a factor of $-1/2$.
Textbooks are just as prone to errors as anything else, being made by humans; it happens. Not much one can do but contact the authors (after doing one's due diligence as you are!).
For the rest of the computations, note that in claiming
$$\int \cos^2(2x) \, \mathrm{d}x
= \int \frac{1 - \cos(4x)}{2} \, \mathrm{d}x$$
you used the identity for $\sin^2(x)$, not $\cos^2(x)$. This should be a $+$, not a $-$.
Your subsequent step then has $\int \cos(x) \, \mathrm{d}x = - \sin(x)+C$, in its essence, when it should be positive.
My guess is you thought about the $\cos^2(x)$ identity the entire time but wrote the middle step with a typo.
Aside from this, as ever, you need a constant of integration in your final answer (a $+C$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Should the vertical "at bar" go before or after the function being differentiated? Suppose I want to calculate the derivative of a long function at a particular point $a$. Is it more common to write
$$
\frac{d}{dx} \left( x^2 \sin(x)^{(3x-1)^2} + \frac{e^x}{2 x^3 -2 x^2 +1} -3 \log(x)\right) \Bigg|_{x=a}
$$
or
$$
\frac{d}{dx}\Bigg|_{x=a} \left( x^2 \sin(x)^{(3x-1)^2} + \frac{e^x}{2 x^3 -2 x^2 +1} -3 \log(x) \right)?
$$
Does either notational convention have any advantages for clarity?
| I agree with MathGeek. Use the first one
$$
\frac{d}{dx} \left( x^2 \sin(x)^{(3x-1)^2} + \frac{e^x}{2 x^3 -2 x^2 +1} -3 \log(x)\right) \Bigg|_{x=a}
$$
(in mathematics).
However, I do not speak for engineering or physics. The second notation
$$
\frac{d}{dx}\Bigg|_{x=a} \left( x^2 \sin(x)^{(3x-1)^2} + \frac{e^x}{2 x^3 -2 x^2 +1} -3 \log(x) \right)
$$
treats $\frac{d}{dx}\big|_{x=a}$ as an operator, and applies it to a function.
This is analogous to the notation for integrals (preferred in some parts of engineering and physics)
$$
\int_a^b\,dx\;\left( x^2 \sin(x)^{(3x-1)^2} + \frac{e^x}{2 x^3 -2 x^2 +1} -3 \log(x) \right)
$$
whereas in mathematics we prefer
$$
\int_a^b\;\left( x^2 \sin(x)^{(3x-1)^2} + \frac{e^x}{2 x^3 -2 x^2 +1} -3 \log(x) \right)\,dx
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $a\ge b \ge c \ge d \ge 1 \implies x^{4} -ax^{3}-bx^{2}-cx-d $ has no integer root. If $a,b,c,d$ are positive integers such that $a\ge b\ge c \ge d\ge 1$, prove that
$$ x^{4} -ax^{3}-bx^{2}-cx-d $$
has no integer root.
Attempt:
If there is an integer solution $x=x_{0}$, then
$$x_{0}^{4} = a x_{0}^{3}+ b x_{0}^{2}+c x_{0}+d$$
Notice that it is clear $x_{0} | d$.
Notice that $x_{0} \ne 0$ because $d>0$. If $x_{0}=1$ then, $1=a+b+c+d$, which is impossible because $a+b+c+d \ge 4$. So $x_{0} \notin \{0,1\}$. If $x_{0} = -1$, then $1 = (b-a) + (d-c) \le 0$, contradiction. So $x_{0} \notin \{-1,0,1\}$. So far we can say that $|x_{0}| \ge 2$.
Now if $x_{0} < 0$, then $x_{0}^{4} \ge 16 > 0$. But $a x_{0}^{3},cx_{0} < -1$ with
$$ |a x_{0}^{3} | > b x_{0}^{2} $$
$$ |c x_{0} | > d $$
so we have $x_{0}^{4} = ax_{0}^{3} + bx_{0}^{2} + cx_{0} + d < 0$ contradiction.
So we must have $x_{0} \ge 2$. Now, since $x_{0}| d$ then $d = e x_{0}$, where $e$ positive integer.
But this means $ a \ge b \ge c \ge d \ge x_{0}$, which means
$$ax_{0}^{3}+bx_{0}^{2}+cx_{0}+d > x_{0}^{4}$$
contradiction.
Some parts of this proof are not necessary I know, I was just working on it while writing it in this post.
Are the better/more elegant solutions?
| The case $x=0$ is trivial.
For $x>0$, you have $x^4=(ax+b)x^2+(cx+d)$, with $ax+b\ge cx+d>1$. Let's say that $A=ax+b$ and $C=cx+d$. That means that $x^4=Ax^2+C$ and $C$ must be divisible by $x^2$ but $A\ge C\ge x^2$ so you can't find a solution.
For $x<0$, we have $A\le C\le0$ so $Ax^2+C$ will never be positive. So once again, no solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the measure of the relationship $\frac{1}{r_1} - \frac{1}{r} $in the figure below In In a right triangle $ABC$ ($A=90°$) with inradio $r$, cevian $AD$ is drawn in such a way that the inradium of $ABD$ and $ADC$ are equal to $r1$.If $AD=2$, calculate $\frac{1}{r1}-\frac{1}{r}$ (Answer:0,5).
My progress:
$\triangle CED \sim \triangle CAB \\
\frac{CE}{AC}=\frac{DE}{AB}=\frac{CD}{BC}\\
\triangle BDL \sim \triangle BCA\\
\frac{DL}{AC}=\frac{BD}{BC}=\frac{LB}{AB}\\
CE = CI\\
BK = BL$
but I still haven't found the necessary relationship to finalize
| In a right triangle, $~b + c = 2r + a$, where $a$ is hypotenuse. If $s$ represents sub-perimeter, then by law of cotangent,
$ \displaystyle \frac{s_{\triangle ABC} - b}{r} = \frac{s_{\triangle ABD} - 2}{r_1}$
$ \displaystyle \frac{s_{\triangle ABC} - c}{r} = \frac{s_{\triangle ACD} - 2}{r_1}$
Adding both,
$~\displaystyle \frac{a}{r} = \frac{s_{\triangle ABC} - 2}{r_1} = \frac{r + a - 2}{r_1}$
$ \implies a (r - r_1) = 2r - r^2 \tag1$
By equating area of $\triangle ABC$ to sum of areas of $\triangle ABD$ and $\triangle ACD$,
$r_1 (a + b + c + 4) = r( a + b + c)$
i.e. $~r_1 (r + a + 2) = r (r + a)$
$\implies a (r - r_1) = r_1 (r + 2) - r^2\tag2$
Equating $(1)$ and $(2)$ and simplifying we get $ \displaystyle \frac{1}{r_1} - \frac{1}{r} = \frac12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\frac{x^2 + 4x}{2x - 1} < \frac{-4}{2x-1}$? I want to solve the following inequality $\frac{x^2 + 4x}{2x - 1} < \frac{-4}{2x-1}$ but I know that it is undefined when $x = 1/2,$ so I can not multiply both sides by $2x -1.$ Is there any suggestion of how to solve it? I also have an idea of transferring one fraction to the other side and then adding, but then how can I complete?
Thanks.
| The basic principle is
$$\frac ab<0 \iff ab<0.$$
From $\frac{x^2+4x}{2x-1}<\frac{-4}{2x-1}$ we have
$$\frac{x^2+4x+4}{2x-1}<0,$$
which is equivalent to
$$(x^2+4x+4)(2x-1)<0.$$
However, $x^2+4x+4=(x+2)^2\geq 0$. Hence, $x\neq -2$ (otherwise $x^2+4x+4=0$) and $2x-1<0$. Therefore, the solution is $x<\frac12$ and $x\neq-2$, i.e.
$$x\in (-\infty,-2)\bigcup\left(-2,\frac12\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\epsilon-N$ for $\lim\limits_{n \to \infty} \sqrt{n^{2} +3n-3} -n = \frac{3}{2}$ First, I tried to use the triangle inequality only once to find an N:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right | = \epsilon
$$
$$
N=\left \lfloor \frac{(\epsilon -\frac{3}{2})^{2}+3}{6-2\epsilon } \right \rfloor +1
$$
I choose epsilon to be 0.01, and N is 1, which is incorrect.
Then I manipulated the inequality again by using the triangle inequality one more time:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right |
$$
$$
\leqslant \left | \sqrt{n^2+3n-3} \right | +n+ \frac{3}{2} =\epsilon
$$
$$
N=\left \lfloor \frac{(\epsilon -\frac{3}{2})^{2}+3}{2\epsilon } \right \rfloor +1
$$
and this time, when epsilon is 0.01, N is 2624, which is correct
I would like to know why the first approach is wrong and the second one is right, Thank you.
| First, notice that both of the methods you followed are incorrect because the sum is larger than 3/2. Hence it can't be less than epsilon (for values small enough). I propose a different approach.$$$$
Notice that for all $n \geq 1$ :
$$|\sqrt{n^2+3n-3}-n-\frac{3}{2}| = -\sqrt{n^2+3n-3}+n+\frac{3}{2}$$
Now we're looking for N such that for all $n \geq N$
$$|\sqrt{n^2+3n-3}-n-\frac{3}{2}| \lt \epsilon$$
Using the formula above we find that:
$$\frac{(\frac{3}{2} - \epsilon)^2+3}{2\epsilon} \lt n$$
Thus, we take:
$$N = \pmb\lfloor{}\frac{(\epsilon - \frac{3}{2})^2+3}{2\epsilon}\pmb\rfloor{} + 1$$
which happens to be the exact same answer you got in the second approach you followed. That's why you got a correct answer for N although the method is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Finding the range of $\frac {2x^2+x-3}{x^2+4x-5}$ I solved for the range as follows:
Setting $f(x)=\frac {2x^2+x-3}{x^2+4x-5} =y$, I rearranged it to get a quadratic in x.
$$(y-2)x^{2}+ (4y-1)x +(3-5y)=0$$
Next, using $\Delta \ge 0$, I got
$$(4y-1)^2-4(y-2)(3-5y) \ge 0$$
Which boiled down to
$$(6y-5)^2 \ge 0$$
And this gave me $$ y \in R$$
Next, the value of $x$ for which $y=2$ is $x=1$, for which the function isn't defined, so that gives me $$y \in R - \{2\}$$
However, the solution is $$y \in R- \{\frac{5}{6}, 2\}$$
I understand that the original function is identical to $$g(x) =\frac{2x+3}{x+5}
\forall x \in R - \{1\}$$
and that $\frac{5}{6} =g(1)$, but in the original function $f(x)$ I found $y=2$ corresponded to $x=1$ and therefore excluded it, but if $x= 1$ also corresponds to $y=\frac{5}{6}$ then wouldn't this $not$ be a function, as $x=1$ would then be associated with two different $y$ values?
I've read the answers in Why D≥0 while finding the range of rational functions and Finding the range of $y =\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\frac{\text{quadratic}}{\text{quadratic}}$ in general) but I'm still unsure of how to apply the information from those to figure out what values of $y$ need to be excluded when dealing with such questions.
While I am familiar with derivatives and limits to a certain degree, we were assumed to $\underline {not}$ know calculus when we were taught this and solved such problems.
I apologise if there are any issues with formatting, this is my first time using Latex.
| Your quadratic equation has $x=1$ as a solution, for any $y$. Indeed, for $y\neq2$ the quadratic formula gives
$$x=\frac{-(4y-1)\pm(6y-5)}{2(y-2)}$$
$$x=\frac{-10y+6}{2y-4}=\frac{-(5y-3)}{y-2}\qquad\text{or}\qquad x=\frac{2y-4}{2y-4}=1$$
and for $y=2$ your equation reduces to $7x-7=0$.
The original function was
$$y=\frac{(x-1)(2x+3)}{(x-1)(x+5)}$$
and you rearranged this to get
$$(x-1)(x+5)y-(x-1)(2x+3)=0$$
which always has $x=1$ as a solution.
Essentially, you added a vertical line to the graph of $g$, so the resulting graph doesn't represent a function.
You should have just worked with $g$ from the beginning, and then excluded any values of $y$ which would have given $x=1$ or $x=-5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Product of matrix and vectors in orthogonal subspaces Let $p < n$, and let $A \in \mathbb{R}^{n \times n}$ be rank $p$. Let $V \in \mathbb{R}^{n \times n}$ be an orthogonal matrix with columns $V = \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix}$.
Suppose that for any linear combination of the first $p$ columns of $V$, that the product $A(c_1 v_1 + \cdots + c_p v_p) \ne 0$. Is it true that $A v_i = 0$ for all $i > p$?
I believe the answer to this is yes. By rank-nullity theorem, $A$ must have nullity $n-p$. Since $Aw \ne 0$ for any $w$ spanned by $v_1, \cdots, v_p$, and since the subspace ${\rm span}(v_1, \cdots ,v_p)$ is orthogonal to the subspace ${\rm span}(v_{p+1}, \cdots ,v_n)$, it must follow that $A v_i = 0$ for all $i > p$.
Any corrections or a more detailed proof would be much appreciated!
| $A=\begin{pmatrix}1&0&1\\0&1&0\\0&0&0\end{pmatrix}$
$\operatorname{Rank} A=p=2$
$V=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$
Then $\begin{align}A(c_1v_1+c_2v_2) &=\begin{pmatrix}1&0&1\\0&1&0\\0&0&0\end{pmatrix}\begin{pmatrix}c_1\\c_2\\0\end{pmatrix}\\ &=\begin{pmatrix}c_1\\c_2\\0\end{pmatrix}\neq 0 [\text{ as $c\neq 0$ }]\end{align}$
$\begin{align}Av_3&=\begin{pmatrix}1&0&1\\0&1&0\\0&0&0\end{pmatrix}\begin{pmatrix}0\\0\\1\end{pmatrix}\\ &=\begin{pmatrix}1\\0\\0\end{pmatrix}\neq 0\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For an arbitrary $n$, evaluate $\left(z^n+\frac{1}{z^{n}}\right)\left(z^n +\frac{1}{z^n}+1\right)$ Let me put you in context, I have the following problem:
Let $z$ be a complex number such that $$\left(z+\frac{1}{z}\right)\left(z+\frac{1}{z}+1\right)=1.$$
For an arbitrary $n$, evaluate
$$\left(z^n +\frac{1}{z^n}\right)\left(z^n+\frac{1}{z^n}+1\right)$$
I've been trying to get the answer, however I am not sure if my proof is completely rigorous. My solution:
Given the hypothesis, we have:
\begin{align*}
\left(z+\frac{1}{z}\right)\left(z+\frac{1}{z}+1\right) & = 1\\
z^2+1+z+1+\frac{1}{z^2} + \frac{1}{z} &= 1 \\
z^2+1+z+\frac{1}{z^2} + \frac{1}{z} &= 0 \\
\frac{z^4+z^2+z^3+1+z}{z^2}&= 0\\
\frac{(z^4+z^2+z^3+1+z)(z-1)}{z^2(z-1)}&= 0\\
\frac{(z^5-1)}{z^2(z-1)}&= 0,z\neq 1
\end{align*}
Also, we have the expression to evaluate:
$$\left(z^{n}+\frac{1}{z^{n}}\right)\left(z^{n}+\frac{1}{z^{n}}+1\right)$$
If $n=0$:
$$\left(z^{0}+\frac{1}{z^{0}}\right)\left(z^{0}+\frac{1}{z^{0}}+1\right)= (1+1)(1+1+1) =6$$
If $n\neq 0$:
\begin{align*}
\left(z^{n}+\frac{1}{z^{n}}\right)\left(z^{n}+\frac{1}{z^{n}}+1\right) & = 1\\
z^{2n}+1+z^n+1+\frac{1}{z^{2n}} + \frac{1}{z^n} &= 1 \\
z^{2n}+1+z^n+\frac{1}{z^{2n}} + \frac{1}{z^n} &= 0 \\
\frac{z^{4n}+z^{2n}+z^{3n}+1+z^n}{z^{2n}}&= 0\\
\frac{(z^{4n}+z^{2n}+z^{3n}+1+z^n)(z^{n}-1)}{z^{2n}(z^{n}-1)}&= 0\\
\frac{(z^{5n}-1)}{z^{2n}(z^{n}-1)}&= 0, z^n\neq 0
\end{align*}
Therefore,
$$\left(z^{n}+\frac{1}{z^{n}}\right)\left(z^{n}+\frac{1}{z^{n}}+1\right)=\begin{cases}
6, & n=0\\
1, & n\neq 0
\end{cases}$$
What do you think?
| Let $\left(z+\frac{1}{z}\right)=x$
Then $\left(z+\frac{1}{z}\right)\left(z+\frac{1}{z}+1\right)=(x)(x+1)=x^2+x$
$x^2+x = 1 \to x^2 + x -1 = 0$
By the fundemental theorem of algebra, we know this has exactly two solutions. By the quadratic formula $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ we know that those solutions are $\frac{\left(-1+\sqrt{5}\right)}{2}$ and $\frac{\left(-1-\sqrt{5}\right)}{2}$.
So $\left(z+\frac{1}{z}\right)=x = \frac{\left(-1\pm\sqrt{5}\right)}{2}$
Multiplying by $z$ gives us $z^2 + 1 = xz \to z^2 -xz+1=0$ which has two solutions for each value of $x$, which we can also find using the quadratic formula, so we should have no more than four solutions for $z$.
$z = \frac{-1\pm\sqrt{x^{2}-4}}{2}$
$x^{2}=\frac{\left(3\pm\sqrt{5}\right)}{2}$
$z=\frac{1\pm\sqrt{\frac{\left(-5\pm\sqrt{5}\right)}{2}}}{2}$
| {
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"url": "https://math.stackexchange.com/questions/4498412",
"timestamp": "2023-03-29T00:00:00",
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$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive 4 real roots.
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive four real roots.
$$(x-a)(x-b)(x-c)(x-d) \\ +(x-a)(x-b)(x-c)(x-e) \\ +(x-a)(x-b)(x-d)(x-e) \\ +(x-a)(x-c)(x-d)(x-e) \\ + (x-b)(x-c)(x-d)(x-e) =0.$$
My attempt:
\begin{align}
& f(x)=\sum_{cyc} (x-a)(x-b)(x-c)(x-d). \\
& f(a)=(a-b)(a-c)(a-d)(a-e). \\
& f(b)=(b-a)(b-c)(b-d)(b-e). \\
& f(c)=(c-a)(c-b)(c-d)(c-e). \\
& f(d)=(d-a)(d-b)(d-c)(d-e). \\
& f(e)=(e-a)(e-b)(e-c)(e-d). \\
\ \\
& f(0)=\sum_{cyc} abcd.\\
\end{align}
I put some arbitrary integers in $a, b, c, d, e$ and drew a graph of $\displaystyle f(x)=\sum_{cyc} (x-a)(x-b)(x-c)(x-d).$
It sure makes 4 distinct roots... How can we prove this?
Values in the graph above:
$a=-1.34, b=-4.67, c=-2.91, d=0.33, e=-6.09$
| Let $F(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$. Then your polynomial is $f(x)=F'(x)$. Because $a,b,c,d,e$ are all distinct, by Rolle's theorem, between any two consecutive roots of $F(x)$ will be a root of $f(x)$. Thus, $f(x)$ will have 4 distinct real roots because $F(x)$ has 5 distinct real roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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Conjectured closed form of $\int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x$ Consider
$$ I = \int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x $$
where $a > 0$ is a constant. We can evaluate this using the Leibnitz theorem for specific values of $n$. Mathematica can solve it for specific values of $n$ as well but not the generalized form. Here, I conjecture a generalization -
$$ I = (-1)^{n+1} \binom{-1/2}{n-1} \cdot \frac{\pi}{2 \, a^{2n-1}} $$
This can be further simplified by
$$ \binom{-1/2}{n} = \left(\frac{-1}{4}\right)^n \binom{2n}{n} $$
Is this conjecture true?
| Let's first try to find a recursive relation of $I_n$. In fact using integration by part,
\begin{align}
I_{n} &= \int_0^\infty \frac{1}{(x^2 + a^2)^n}\mathrm dx\\
&= \int_0^\infty \frac{x^2 + a^2}{(x^2 + a^2)^{n+1}}\mathrm dx\\
&= \frac{1}{2n} I_{n} + a^2 I_{n+1}.
\end{align}
So $$I_{n+1} = \frac{1}{a^2} \frac{2n-1}{2n} I_n = \frac{1}{a^2}\frac{2n(2n-1)}{4n^2} I_n$$
Use induction to prove that:
$$I_n = \frac{1}{a^{2(n-1)}} \frac{(2(n-1))!}{4^{n-1} ((n-1)!)^2} I_1$$
Another way to compute $I_n$ for general case (even with $n$ is non-integer) is
\begin{align}
I_n &= \int_{0}^{\infty} \frac{1}{(x^2 + a^2)^n}\mathrm dx\\
&=_{t := \frac{x}{a}} \frac1{a^{2n-1}}\int_0^{\infty}\frac{1}{(1+t^2)^n}\mathrm dt\\
&=_{u := \frac1{1+t^2}} \frac1{a^{2n-1}}\int_1^0 u^{n} (\frac{1}{2}(-\frac1{u^2}(\frac{1-u}{u})^{-\frac12})) \mathrm d u\\
&= -\frac1{2a^{2n-1}}\int_0^1 u^{n-\frac32}(1-u)^{-\frac12}\mathrm d u\\
&= -\frac{1}{2a^{2n-1}} \beta (n-\frac12, \frac12)
\end{align}
| {
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Solve $x^2 dy + (xy+y^2) dx = 0$ if $y=1$ when $x=1$. Using the given equation I arrived at the following expression:
$$\frac{dy}{dx} = - \left[\frac{y}{x} + \left(\frac{y}{x}\right)^2\right] \tag{1}$$
Since this is a homogenous equation, I assumed $\frac{y}{x}=v$ giving:
$$\frac{dy}{dx} = v + x \frac{dv}{dx}$$
Substituting this in equation $(1)$ and solving I arrived at the following expression:
$$\ln\left|\frac{v}{v+2}\right| = \ln\left(\frac{C}{x}\right)^2$$
where $C$ is an arbitrary constant. Substituting $v$ for $\frac{y}{x}$ and simplifying further I obtained:
$$\left|\frac{y}{y+2x}\right| = \frac{C^2}{x^2}$$
Using the fact that $y=1$ when $x=1$, I find that $C^2 = \frac{1}{3}$. Substituting this value in equation $(2)$, I arrived at the following expression:
$$\left|\frac{y}{y+2x}\right| = \frac{1}{3x^2}$$
This is how I solved it further:
$$\frac{y}{y+2x} = \pm \frac{1}{3x^2}$$
$$\implies y+2x = \pm 3 x^2 y$$
But my book gives the answer as $y + 2x = 3 x^2 y$. Am I missing something? I am not very familiar with solving equations with absolute values, so I'm not sure about my answer. Any help would be appreciated.
|
$$|\frac{y}{y+2x}| = \frac{C^2}{x^2} \tag{2}$$
In your Eq.$(2)$, if you remove the absolute value, you have
$$\frac{y}{y+2x} =\pm \frac{C^2}{x^2} $$
But your initial condition $x=1,y=1$ exculdes the "$-$" sign solution. So you only need to take the positive solution, hence
$$\frac{y}{y+2x} = \frac{C^2}{x^2} $$
| {
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Interesting integral $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}$ Latest Edit
Inspired by @J.G., I find a formula in general,
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{n}} &=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{n}} \\
&=\left.\frac{2(-1)^{n}}{(n-1) !} \frac{\partial^{n}}{\partial a^{n}}\left(\int_{0}^{\pi} \frac{d x}{a-\cos x} \right)\right|_{a=3} \\
&=\left.\frac{2(-1)^{n} \pi}{(n-1) !} \frac{\partial^{n}}{\partial a^{n}}\left(\frac{1}{\sqrt{a^{2}-1}}\right)\right|_{a=3}
\end{aligned}
$$
Multiplying both the numerator and denominator $\sec^4x$ yields
$\displaystyle I=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{4} x}{\left(\sec ^{2} x+\tan ^{2} x\right)^{2}} d x \tag*{} $
Letting $ t=\tan x$ gives
$\displaystyle \begin{aligned}I&= \int_{0}^{\infty} \frac{1+t^{2}}{\left(1+2 t^{2}\right)^{2}} d t\\&=\int_{0}^{\infty} \frac{1+\frac{1}{t^{2}}}{\left(2 t+\frac{1}{t}\right)^{2}} d t\\&=\int_{0}^{\infty} \frac{\frac{3}{4}\left(2+\frac{1}{t^{2}}\right)-\frac{1}{4}\left(2-\frac{1}{t^{2}}\right)}{\left(2 t+\frac{1}{t}\right)^{2}} d t\\&=\frac{3}{4} \int_{0}^{\infty} \frac{d\left(2 t-\frac{1}{t}\right)}{\left(2 t-\frac{1}{t}\right)^{2}+8}-\frac{1}{4} \int_{0}^{\infty} \frac{d\left(2 t+\frac{1}{t}\right)}{\left(2 t+\frac{1}{t}\right)^{2}}\\&=\left[\frac{3}{8 \sqrt{3}} \tan ^{-1}\left(\frac{2 t-\frac{1}{t}}{2 \sqrt{2}}\right)+\frac{1}{4\left(2 t+\frac{1}{t}\right)}\right]_{0}^{\infty}\\&=\frac{3 \pi}{8 \sqrt{2}}\end{aligned}\tag*{} $
Is there any method other than tangent half-angle substitution?
| Even the antiderivative exists $$I_n=\int \frac{d x}{\left(1+\sin ^{2} (x)\right)^{n}}$$
$$I_n=-\frac{\sqrt{-\sin ^2(x)} \sqrt{\cos ^2(x)} \csc (x) \sec (x) }{2 \sqrt{2} (n-1)\,\left(1+\sin^2(x)\right)^{n-1} }\times $$
$$F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{1}{2} \left(1+\sin
^2(x)\right),1+\sin ^2(x)\right)$$
where appears the Appell hypergeometric function of two variables.
For $x\in\big[ 0,\frac \pi 2\big]$
$$I_n=-i \,\,
\frac{F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{1+\sin
^2(x)}{2} ,1+\sin ^2(x)\right)}{2 \sqrt{2} (n-1)\,\left(1+\sin ^2(x)\right)^{n-1} }$$
Using the given bounds, the result write again in terms of the Gaussian hypergeometric function.
For
$$J_n=\int_0^{\frac \pi 2} \frac{d x}{\left(1+\sin ^{2} (x)\right)^{n}}=\frac{\pi }{2 \sqrt{2}}\,\,\frac{a_n}{b_n}$$
where tha $a_n$ form the (unknown ?) sequence
$$\{1,3,19,63,867,3069,22199,81591,2428451,9119601,68993757,\cdots\}$$ and the $b_n$ correspond to sequence $A123854$ in $OEIS$.
Edit
At $x=3$
$$\frac{\partial^{n}}{\partial x^{n}}\left(\frac{1}{\sqrt{x^{2}-1}}\right)$$ satisfy the recurrence relation
$$(n+1)\, u_{n}+3 (2 n+3)\, u_{n+1}+8 (n+2)\,u_{n+2}=0$$ with
$$u_0=\frac{1}{2 \sqrt{2}}\qquad \text{and} \qquad u_1=-\frac{3}{16 \sqrt{2}}$$
| {
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"url": "https://math.stackexchange.com/questions/4508178",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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} |
How many methods to tackle the integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x ?$ $ \text{We are going to evaluate the integral}$
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \tag*{} \\$
by letting $ y=\frac{\pi}{4}-x. $ Then
$$\begin{aligned} \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \displaystyle &=\int_{\frac{\pi}{4}}^{0} \frac{\sin \left(\frac{\pi}{4}-y\right)+\cos \left(\frac{\pi}{4}-y\right)}{9+16 \sin 2\left[(\frac{\pi}{4}-y)\right]}(-d y) \\
\displaystyle &=\int_{0}^{\frac{\pi}{4}} \frac{\frac{1}{\sqrt{2}}(\cos y-\sin y)+\frac{1}{\sqrt{2}}(\cos y+\sin y)}{9+16 \cos 2 y} d y \\
\displaystyle &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\cos y}{9+16\left(1-2 \sin ^{2} y\right)} d y \\
\displaystyle &=\sqrt{2} \int_{0}^{\frac{1}{\sqrt 2} } \frac{d z}{25-32 z^{2}} \text { , where } z=\sin y\\
\displaystyle &=\frac{\sqrt{2}}{10} \int_{0}^{\frac{1}{\sqrt 2} }\left(\frac{1}{5-4 \sqrt{2}z}+\frac{1}{5+4 \sqrt{2} z}\right) d z \\
\displaystyle &=\frac{\sqrt{2}}{10(4 \sqrt{2})}\left[\ln \left|\frac{5+4 \sqrt{2} z}{5-4 \sqrt{2} z}\right|\right]_{0}^{\frac{1}{\sqrt 2} } \\
\displaystyle &=\frac{1}{40}\ln 9 \end{aligned}$$
| $$
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x =& \int_{0}^{\frac{\pi}{4}} \frac{d(\sin x-\cos x)}{5^{2}-4^{2}(\sin x-\cos x)^{2}} \\
=& \frac{1}{40}\left[\ln \left| \frac{4 (\sin x-\cos x)+5}{4(\sin x-\cos x)-5} \right|\right]_{0}^{\frac{\pi}{4}}\\
=& \frac{\ln 9}{40}
\end{aligned}
$$
| {
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"url": "https://math.stackexchange.com/questions/4508889",
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Finding all possible complex solutions to the system $xy = \sqrt 2(x + y)$, $yz = \sqrt 3(y + z)$, $zx = \sqrt 5(z + x)$
Find all possible complex solutions to:
$$\begin{cases}xy = \sqrt 2(x + y) \\
yz = \sqrt 3(y + z)\\
zx = \sqrt 5(z + x)\end{cases}$$
I have some ideas:
$$xy-\sqrt 2y=\sqrt 2x\implies y=\frac{\sqrt 2x}{x-\sqrt 2} $$
and
$$yz-\sqrt 3y=\sqrt 3z\implies y=\frac{\sqrt 3z}{z-\sqrt 3} $$
and
$$xz-\sqrt 5x=\sqrt 5z\implies x=\frac{\sqrt 5z}{z-\sqrt 5} $$
So I got
$$y=\frac{\sqrt 2x}{x-\sqrt 2}=\frac{\sqrt 3z}{z-\sqrt 3}=\frac{\sqrt 2×\frac{\sqrt 5z}{z-\sqrt 5}}{\frac{\sqrt 5z}{z-\sqrt 5}-\sqrt 2}$$
From here we can go to a solution, but that brings up really scary equations. Is there any clever way to solve this? That is a minor Olympiad problem.
| Observe that,
If $x=0$ or $y=0$ or $z=0$, you will get $x=y=z=0$. So, we can assume $x,y,z\neq 0$.
$$\begin{cases}\frac 1x+\frac 1y=\frac {\sqrt 2}{2}\\
\frac 1y+\frac 1z=\frac {\sqrt 3}{3}\\
\frac 1x+\frac 1z=\frac {\sqrt 5}{5}\end{cases}\implies \frac 1x+\frac 1y+\frac 1z=\frac {\frac {\sqrt 2}{2}+\frac {\sqrt 3}{3}+\frac {\sqrt 5}{5}}{2}=a$$
Finally, we have
$$\frac 1x=a-\frac {\sqrt 3}{3}\implies x=\frac{1}{a-\frac {\sqrt 3}{3}}$$
Note that, $y$ and $z$ can be found in the same way.
| {
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$\int_0^\infty\frac{\ln{(x^8+x^4+2x^2+1)}}{x^2+1}dx$ and $\int_0^\infty\frac{1}{x^2+1}\arctan{\left(\frac{x^2+1}{x^4}\right)}dx$ I derived some integrals a while ago
$$I(t)=\int_0^\infty\frac{\ln{(x^4+t(x^2+1))}}{x^2+1}dx$$
$$I(i)=\int_0^\infty\frac{\ln{(x^4+i(x^2+1)}}{x^2+1}dx=2\pi\ln{(1+\sqrt{i}+\sqrt{i+2\sqrt{i}})}$$
I then took the real and imaginary parts of each side and got interesting results
$$\int_0^\infty\frac{\ln{(x^8+x^4+2x^2+1)}}{x^2+1}dx$$$$=\pi\ln{\left(2+\sqrt2+\sqrt{5+2\sqrt2}+2(46+32\sqrt2)^{1/4}\cos{\left(\frac{\pi}{8}-\frac{1}{2}\arctan{\left(1+\frac{1}{\sqrt2}\right)}\right)}\right)}$$
and
$$\int_0^\infty\frac{1}{x^2+1}\arctan{\left(\frac{x^2+1}{x^4}\right)}dx=\pi\arctan{\left(\frac{1+\sqrt{\sqrt{5+2\sqrt2}-\sqrt2}}{1+\sqrt2+\sqrt{\sqrt{5+2\sqrt2}+\sqrt2}}\right)}$$
I checked the numerical answers on WolframAlpha to ensure that they match my answers.
I am curious as to whether there is any way to directly solve the integrals, and I am most interested in the second integral
| We could even compute the antiderivative
$$I=\int\frac{\log{(x^8+x^4+2x^2+1)}}{x^2+1}dx=\sum_{i=1}^4\int\frac{\log(x^2-r_i)}{x^2+1}dx$$
where
$$r_1=\frac{-i-\sqrt{-1-4 i}}{2}\qquad \qquad r_2=\frac{-i+\sqrt{-1-4 i}}{2} $$
$$r_3=\frac{i-\sqrt{-1-4 i}}{2} \qquad \qquad r_4=\frac{i+\sqrt{-1-4 i}}{2} $$
and $$I_a=\int\frac{\log(x^2-a)}{x^2+1}dx$$ is known (have a look here)
For complex $a$ ( provided that $\Re(a)<0$)
$$J_a=\int_0^\infty \frac{\log(x^2-a)}{x^2+1}dx=\frac{\pi}{2} \left(\log (a+1)+2 i \tan ^{-1}\left(\sqrt{a}\right)\right)$$ and the fact that $r_1+r_4=r_2+r_3=0$ must lead to a lot of simplifications.
$$J_{+a}+J_{-a}=\frac{\pi}{2} \left(\log \left(1-a^2\right)+2 i \tan ^{-1}\left(\frac{(1+i)
\sqrt{a}}{1-i a}\right)\right)$$
I did not have the patience to finish the simplications as you did but the results are exactly the same.
| {
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Writing an integral in terms of special functions I would like to find an expression for the following integrals, using special functions or otherwise; I suspect they can be written in terms of elliptic functions but I've played with it for a while and haven't arrived at anything useful. The integrals are:
$$
\mathcal{I}_1 = \int \frac{1}{\sqrt{1-\alpha^2(\beta^2-\cos^2\xi)}}\frac{\xi \cos\xi}{\sqrt{\beta^2-\cos^2\xi}} d\xi
$$
and
$$
\mathcal{I}_2 = \int \frac{1}{\sqrt{1-\frac{1}{4}\sin^2\xi}}\frac{\xi \cos\xi}{\sqrt{\beta^2-\sin^2\xi}} d\xi
$$
I would be happy with expressions for the perhaps simpler form:
$$
\mathcal{I}_a = \int \frac{\sin^{-1}\left(B\cos\xi\right)}{\sqrt{A^2-\sin^2\xi)}} d\xi
$$
and
$$
\mathcal{I}_b = \int \frac{\sin^{-1}\left(B\sin\xi\right)}{\sqrt{A^2-\sin^2\xi)}} d\xi
$$
| Since $\cos^2(x) = 1 - \sin^2(x)$ both $I_1$ and $I_2$ can be rewritten as some constant multiple of an antiderivative of the form:
\begin{align}
I(a,b) =\int \frac{x\cos(x)}{\sqrt{a^2-\sin^2(x)}\sqrt{b^2 - \sin^2(x)}} \,\mathrm{d}x
\end{align}
for some constants $a,b$. Now, from the definition of the Elliptic integral of the First Kind we see that
\begin{align}
\frac{1}{b}F\left(\arcsin\left(\frac{x}{a}\right), \frac{a}{b}\right) =\frac{1}{b} \int_0^{\sin\left(\arcsin\left(\frac{x}{a}\right)\right)} \frac{\mathrm{d}t}{\sqrt{1-\frac{a^2}{b^2}t^2}\sqrt{1-t^2}} \overset{u = at}{=} \int_0^{x} \frac{\mathrm{d}u}{\sqrt{a^2-u^2}\sqrt{b^2-u^2}}
\end{align}
So by taking the substitution $u = \sin(x)$ we get
$$
\int \frac{\cos(x)}{\sqrt{a^2-\sin^2(x)}\sqrt{b^2 - \sin^2(x)}} \mathrm{d}x=\frac{1}{b}F\left(\arcsin\left(\frac{\sin(x)}{a}\right), \frac{a}{b}\right)
$$
And thus
\begin{align}
I(a,b)\overset{\text{I.B.P.}}{=}& x\int \frac{\cos(x)}{\sqrt{a^2-\sin^2(x)}\sqrt{b^2 - \sin^2(x)}} \,\mathrm{d}x - \int\left(\int \frac{\cos(x)}{\sqrt{a^2-\sin^2(x)}\sqrt{b^2 - \sin^2(x)}} \,\mathrm{d}x\right)\mathrm{d}x\\
=&\frac{x}{b}F\left(\arcsin\left(\frac{\sin(x)}{a}\right), \frac{a}{b}\right) - \frac{1}{b}\int F\left(\arcsin\left(\frac{\sin(x)}{a}\right), \frac{a}{b}\right)\mathrm{d}x
\end{align}
And now, even for the simple case of $a=1$ and $b=\frac{1}{\sqrt{2}}$ the antiderivative in the previous equation already involves hypergeometric functions and generalized hypergeometric functions, so I don't have much hope for a possible closed form for the general case.
Hope this helps!
| {
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Integrating $\int^{\infty}_0\frac{\ln^2(x)}{\left(x^2+1\right)^2}\text{ d}x$ I want to integrate $$I=\int^{\infty}_0\frac{\ln^2(x)}{\left(x^2+1\right)^2}\text{ d}x=\frac{\pi^3}{16}$$ using contour integration. I set up the function $$f(z)=\frac{\ln^3(z)}{\left(z^2+1\right)^2}$$
and used a keyhole contour with a branch cut about the positive real axis
The integrals about $\Gamma$ and $\gamma$ I believe both go to $0$ if I did not make any mistakes in my calculations but afaik I should be right.
The integrals about $T$ and $B$ is
$$\int^{\infty}_{0}\frac{\ln^3(x)\text{ d}x}{\left(x^2+1\right)^2}+\int^{0}_{\infty}\frac{\left(\ln(x)+2\pi i\right)^3\text{ d}x}{\left(x^2+1\right)^2}$$ which expands and simplifies into $$\int^{\infty}_{0}\frac{\ln^3(x)\text{ d}x}{\left(x^2+1\right)^2}+\int^{0}_{\infty}\frac{\ln^3(x)\text{ d}x}{\left(x^2+1\right)^2}+\int^{0}_{\infty}\frac{6i\pi\ln^2(x)\text{ d}x}{\left(x^2+1\right)^2}-\int^{0}_{\infty}\frac{12\pi^2\ln(x)\text{ d}x}{\left(x^2+1\right)^2}-\int^{0}_{\infty}\frac{8i\pi^3\text{ d}x}{\left(x^2+1\right)^2}$$
$$\implies -6i\pi I +\int_{0}^{\infty}\frac{12\pi^2\ln(x)\text{ d}x}{\left(x^2+1\right)^2}+\int_{0}^{\infty}\frac{8i\pi^3\text{ d}x}{\left(x^2+1\right)^2}$$
Since $f(z)$ has poles of order $2$ at $z=\pm i$ I use the higher order residue formula and get
$$\implies\mathop{\mathrm{Res}}_{z = i}\frac{\ln^3(z)}{\left(z^2+1\right)^2}=\lim_{z\to i}\frac{\text{d}}{\text{d}z}\left[\frac{\ln^3(z)}{\left(z+i\right)^2}\right]=\lim_{z\to i}\frac{\ln^{2}(z)\left(3i+3z-2z\ln(z)\right)}{z\left(z+i\right)^{3}}=-\frac{\pi^{3}}{32}-\frac{3i\pi^{2}}{16}$$
$$\implies\mathop{\mathrm{Res}}_{z = -i}\frac{\ln^3(z)}{\left(z^2+1\right)^2}=\lim_{z\to -i}\frac{\text{d}}{\text{d}z}\left[\frac{\ln^3(z)}{\left(z-i\right)^2}\right]=\lim_{z\to i}\frac{\ln^{2}(z)\left(3z-3i-2z\ln(z)\right)}{z\left(z-i\right)^{3}}=-\frac{\pi^{3}}{32}+\frac{3i\pi^{2}}{16}$$
which when added and multiplied with $2\pi i$ give $-\dfrac{i\pi^4}{8}$.
Equating only imaginary parts of both sides gives
$$-6\pi I + 2\pi^4 = -\frac{\pi^4}{8}$$
$$\implies I=\frac{17\pi^3}{48}$$
This is obviously wrong. Plus, the real part is nonzero too so my final equation is nonsense. Where did I make a mistake?
Thanks in advance!
| The residue at $-i$ should be
$$\mathop{\mathrm{Res}}_{z = -i}\frac{\ln^3(z)}{\left(z^2+1\right)^2}=\lim_{z\to -i}\frac{\text{d}}{\text{d}z}\left[\frac{\ln^3(z)}{\left(z-i\right)^2}\right]=\lim_{z\to -i}\frac{\ln^{2}(z)\left(3z-3i-2z\ln(z)\right)}{z\left(z-i\right)^{3}}=\frac{27\pi^3}{32}+\frac{27\pi^2 i}{16}$$
where at the last step, due to the branch cut, $\ln(-i)=|-i|+i\text{arg}(-i)=\frac{3\pi i}{2}$.
Hence, equating the imaginary parts, we find
$$-6\pi I + 2\pi^4 = 2\pi\left(-\frac{\pi^3}{32}+\frac{27\pi^3}{32}\right)$$
that is
$$\int^{\infty}_0\frac{\ln^2(x)}{\left(x^2+1\right)^2}\,dx=I = \frac{(32-26)\pi^3}{3\cdot 32}=\frac{\pi^3}{16}.$$
Note that equating the real parts, we get
$$\int^{\infty}_0\frac{\ln(x)}{\left(x^2+1\right)^2}\,dx=-\frac{\pi}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
So I was wondering about a way to find this sum without squaring? It seems impossible, but I still want to ask.
| A polynomial approach.
Note that $a_1,a_2=\sqrt{3\pm\sqrt 5}$ are two of the roots of $x^4-6x^2+4=0.$ The other roots are $-a_1,-a_2.$
So $$x^4-6x^2+4=(x-a_1)(x-a_2)(x+a_1)(x+a_2).$$
Now, $a_1a_2=\sqrt{4}=2.$ If $S=a_1+a_2,$ then this factorization becomes:
$$x^4-6x^2+4=(x^2-Sx+2)(x^2+Sx+2)=x^4+(4-S^2)x^2+4.$$
So $4-S^2=-6,$ or $S^2=10.$
I guess that's sort of squaring, but we never actually numerically square $S.$
More generally, given the roots of $x^4-bx^2+c^2,$ there are two roots $a_1,a_2$ with $a_1a_2=c,$ and then you get a similar result:
$$x^4-bx^2+c^2=(x^2-Sx+c)(x^2+Sx+c)=x^4+(2c-S^2)x^2+c^2,$$ so $2c+b=S^2.$
So this means, at least if $c\geq 0,$ that $$\sqrt{\frac{b+\sqrt{b^2-4c^2}}2}+\sqrt{\frac{b-\sqrt{b^2-4c^2}}2}=\pm\sqrt{2c+b}$$
Multiplying by $\sqrt{2},$ this gives:
$$\sqrt{b+\sqrt{b^2-4c^2}}+\sqrt{b-\sqrt{b^2-4c^2}}=\pm\sqrt{4c+2b}\tag1$$
You also get:
$$\sqrt{b+\sqrt{b^2-4c^2}}-\sqrt{b-\sqrt{b^2-4c^2}}=\pm\sqrt{2b-4c},\tag 2$$ since in this case $a_1a_2=-c.$
The case $b=3,c=1$ in (1) gives your result, since $3\pm\sqrt 5$ are both real and positive, so we know the result has to be positive.
In both $(1)$ and $(2)$ you get the positive sign if $b,c$ and all the square roots are real. If the square roots are complex, you are stuck figuring out the sign.
But if $b,c$ are real, and $0\leq 2c\leq b,$ we can solve $(1)$ and $(2)$ to get:
$$\sqrt{b+\sqrt{b^2-4c^2}}=\frac{\sqrt{2b+4c}+\sqrt{2b-4c}}{2}=\frac{\sqrt{b+2c}+\sqrt{b-2c}}{\sqrt 2}$$
and similarly:
$$\sqrt{b-\sqrt{b^2-4c^2}}=\frac{\sqrt{b+2c}-\sqrt{b-2c}}{\sqrt 2}$$
If $d=b^2-4c^2,$ this gives us:
$$\sqrt{b+\sqrt{d}}=\frac{\sqrt{b+\sqrt{b^2-d}}+\sqrt{b-\sqrt{b^2-d}}}{\sqrt2}$$ when $0\leq d\leq b^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 0
} |
Proving $30x + 3y^2 + \frac{2z^3}{9} + 36 (\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}) \ge 84$
If $x, y, z$ are positive real numbers, prove that $$30x + 3y^2 + \frac{2z^3}{9} + 36 \left(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}\right) \ge 84.$$
I genuinely have no clue on how to proceed. Is it proved using repeated CS?
| Remark: Once we know the equality case $x = 1, y = 2, z = 3$, we apply AM-GM.
Using AM-GM, we have
\begin{align*}
&30\cdot x + 12 \cdot (y/2)^2
+ 6\cdot (z/3)^3 + 18 \cdot \frac{1}{xy/2} + 6\cdot \frac{1}{yz/6} + 12 \cdot \frac{1}{zx/3}\\
\ge\,& 84\sqrt[84]{x^{30}\cdot (y/2)^{24}\cdot (z/3)^{18} \cdot \left(\frac{1}{xy/2}\right)^{18}\left(\frac{1}{yz/6}\right)^6\left(\frac{1}{zx/3}\right)^{12} }\\
=\,& 84.
\end{align*}
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
numerical calculation of 3d integral $\iiint_{[-1,1]^3} \frac{dx dy dz}{r^2} $ I have to calculate the integral
$$I=\iiint_{[-1,1]^3} \frac{dx dy dz}{r^2} $$
where $r^2 = x^2 + y^2 + z^2 $.
My strategy is to split the integral domain into two parts: the unit sphere and the supplement domain. From the unit sphere, I get analytically $4\pi $; for the supplement, I used Monte Carlo. The problem is that the Monte Carlo part is not very efficient, while I want to get the integral to 8 digits.
Could anyone suggest a better scheme?
| $$\iiint_{[-1,1]^3}\frac{d\mu}{x^2+y^2+z^2} = \iint_{[-1,1]^2}\frac{2}{\sqrt{x^2+y^2}}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\,dx\,dy $$
equals
$$ 8\iint_{[0,1]^2}\frac{1}{\sqrt{x^2+y^2}}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\,dx\,dy $$
or
$$ 16\iint_{0\leq y\leq x\leq 1}\frac{1}{\sqrt{x^2+y^2}}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\,dx\,dy \stackrel{y\mapsto tx}{=} 16\iint_{[0,1]^2}\frac{1}{\sqrt{1+ t^2}}\arctan\left(\frac{1}{x\sqrt{1+ t^2}}\right)\,dx\,dt $$
or
$$ 16\int_{0}^{1}\left(\frac{\pi}{2\sqrt{1+t^2}}-\frac{\arctan\sqrt{1+t^2}}{\sqrt{1+t^2}}+\frac{\log(2+t^2)}{2(1+t^2)}\right)\,dt $$
or
$$ 8\pi\log(1+\sqrt{2})+16\int_{0}^{1}\left(-\frac{\arctan\sqrt{1+t^2}}{\sqrt{1+t^2}}+\frac{\log(2+t^2)}{2(1+t^2)}\right)\,dt. $$
The last integral can be written in terms of polylogarithms, or just numerically evaluated through Newton-Cotes or Gaussian quadrature. The outcome, of course, is larger than $4\pi$ and it equals $\color{blue}{15.348248444887\ldots}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{x^2+y^2+x+y-1}{xy-1}$ is an integer for positive integers $x$ and $y$, then its value is $7$. I saw this on quora
and haven't been able to solve it.
If $\dfrac{x^2+y^2+x+y-1}{xy-1}$
is an integer for positive integers
$x$ and $y$,
then its value is $7$.
If
$y=1$ this is
$\dfrac{x^2+x+1}{x-1}
= x+2+\dfrac{3}{x-1}$
which is an integer only when
$x=2$ or $x=4$ and has value $7$.
Looking at the values from 2 to 20
for $x$ and $y$,
this is an integer only for
$x=2, y=12$ (and $x=12, y=2$).
This value is 7.
So it looks like this might be correct,
and I don't know how to show it.
| there are just two flavors of these; For integer $k$ we are looking for integer points on the arc of
$$ x^2 - kxy + y^2 + x + y = 1 -k $$
between the lines $ky = 1 + 2x$ and $kx = 1+2y$
For $k > 7$ the arc includes a point $(t,t)$ with $1 < t < 2$
Calculation will show that, for $x=2,$ we have $y < 1,$ and the (local) minimum $y$ occurs exactly where the line $ky = 1+2x$ intersects the hyperbola.
Here is $k=10$
for $k < 7$ the local min is above $1$ but the slanted lines are closer together...
I mention a second flavor, those are when $ x^2 + kxy + y^2$ is some small number and the picture is rotated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.