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Combinatorial proofs that $\binom{2^n}{k}$ is even for every $n\geq 1$ and $1\leq k \leq 2^n-1$. I am trying to prove that $\binom{2^n}{k}$ is even for every $n\geq 1$ and $1\leq k \leq 2^n-1$.
The one I could come up with is that the equivalence relation where $A\sim B$ if $B$ is of the form $A+k$ for some $k$ with the addition taken $\bmod 2^n$ splits the subsets into classes, no class has size $1$ and the size is a divisor of $2^n$ so they are all even. I am wondering if there are other combinatorial proofs.
|
I am trying to prove hat $\binom{2^n}{k}$ is even for every $n\geq 1$ and $1\leq k \leq 2^n-1$.
Assumed that it is intended that $n,k \in \Bbb{Z}.$
Alternative (direct) approach that uses
Legendre's Formula.
While less elegant than other approaches, this approach provides a nice fail-safe, if elegance eludes you.
For $p$ prime, and $n \in \Bbb{Z_{\geq 2}},~$ let $~V_p(n)~$ denote the largest non-negative integer $~\alpha~$ such that $~p^\alpha ~| ~(n!)$.
For $r \in \Bbb{R}$, let $\lfloor r\rfloor$ denote the largest integer $k$ such that $k \leq r.$
That is, $\lfloor r\rfloor$ is the floor of $r$.
Preliminary result
(1)
Given: $~\displaystyle a,b,c \in \Bbb{Z^+},
~\left\lfloor \frac{a}{c} \right\rfloor ~+
~\left\lfloor \frac{b}{c} \right\rfloor ~\leq
~\left\lfloor \frac{a + b}{c} \right\rfloor.$
Proof:
Denote $\displaystyle ~\left(\frac{a}{c}\right) ~\text{as}
~P_1 + r_1, ~~~\text{and} ~~~
~\left(\frac{b}{c}\right) ~\text{as}
~P_2 + r_2$
where $~P_1, P_2 \in \Bbb{Z}, ~0 \leq r_1, r_2 < 1.$
Then, $~\displaystyle
~\left\lfloor \frac{a}{c} \right\rfloor ~+
~\left\lfloor \frac{b}{c} \right\rfloor = P_1 + P_2.$
$\underline{\text{Case 1:} ~~0 \leq (r_1 + r_2) < 1}.$
Then $\displaystyle ~\left\lfloor \frac{a + b}{c} \right\rfloor = P_1 + P_2.$
$\underline{\text{Case 2:} ~~1 \leq (r_1 + r_2) < 2}.$
Then $\displaystyle ~\left\lfloor \frac{a + b}{c} \right\rfloor = P_1 + P_2 + 1.$
Per Legendre's formula,
$\displaystyle V_p(n) = \sum_{i=1}^\infty \left\lfloor \frac{n}{p^i}\right\rfloor. $
It is understood that the start of the summation consists of a finite number of non-zero terms, and that all of the remaining terms in the summation are $= (0)$.
For $\displaystyle 1 \leq k \leq 2^n - 1, ~\binom{2^n}{k} = \frac{\left(2^n\right)!}{k!\left[\left(2^n\right) - k\right]!}.$
$$\displaystyle V_2\left(2^n\right) = \sum_{i=1}^{n-1} \left\lfloor \frac{2^n}{2^i}\right\rfloor
~~+ ~~\left\lfloor \frac{2^n}{2^n}\right\rfloor. \tag{A}$$
$$\displaystyle V_2\left(k\right) = \sum_{i=1}^{n-1} \left\lfloor \frac{k}{2^i}\right\rfloor
~~+ ~~\left\lfloor \frac{k}{2^n}\right\rfloor. \tag{B}$$
$$\displaystyle V_2\left(2^n - k\right) = \sum_{i=1}^{n-1} \left\lfloor \frac{2^n - k}{2^i}\right\rfloor
~~+ ~~\left\lfloor \frac{2^n - k}{2^n}\right\rfloor. \tag{C}$$
Since $k < 2^n~~~$ and $~~~\left(2^n - k\right) < 2^n$
the right hand terms in (B) and (C) above are both $= 0$,
while the right hand term in (A) $= 1.$
For each of the other $(n-1)$ terms in the summations of (A), (B), and (C)
you can invoke Preliminary Result (1) above.
That is:
$\displaystyle \left\lfloor \frac{2^n}{2^1} \right\rfloor
~\geq ~\left\lfloor \frac{k}{2^1} \right\rfloor
~+ ~\left\lfloor \frac{2^n - k}{2^1} \right\rfloor.$
$\displaystyle \left\lfloor \frac{2^n}{2^2} \right\rfloor
~\geq ~\left\lfloor \frac{k}{2^2} \right\rfloor
~+ ~\left\lfloor \frac{2^n - k}{2^2} \right\rfloor.$
$\cdots$
$\displaystyle \left\lfloor \frac{2^n}{2^{(n-1)}} \right\rfloor
~\geq ~\left\lfloor \frac{k}{2^{(n-1)}} \right\rfloor
~+ ~\left\lfloor \frac{2^n - k}{2^{(n-1)}} \right\rfloor.$
Therefore, the summation of $(n-1)$ terms in (A) is $\geq$ the corresponding combined summations in (B) and (C) above.
Therefore,
$$ V_2\left(2^n\right) > V_2\left(k\right) +
V_2\left(2^n - k\right).$$
Addendum
As a corrollary to this answer's analysis, you can immediately conclude that
$$2^1 ~| ~\binom{2^n}{2^{[n-1]}}~~~\text{and}
~~~2^2 ~\not| ~\binom{2^n}{2^{[n-1]}}.$$
This follows, per examination of (A), (B), and (C) in the answer by noting that when $~c|a~$ and $~c|b~$ then
$$\left\lfloor \frac{a}{c} \right\rfloor +
\left\lfloor \frac{b}{c} \right\rfloor =
\left(\frac{a}{c} + \frac{b}{c}\right) =
\left(\frac{a + b}{c}\right) =
\left\lfloor \frac{a + b}{c} \right\rfloor.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Differential equation. $ \sqrt{3+y^2}dx-x\cdot dy=x^2\cdot dy $ Check my steps I have equation $ \sqrt{3+y^2}dx-x\cdot dy=x^2\cdot dy $ and I need to solve it with separation method.
My try:
$-x^2\cdot dy-x\cdot dy=-\sqrt{3+y^2}\cdot dx$
$-x(x+1)\cdot dy=-\sqrt{3+y^2}\cdot dx $
$\frac{-x(x+1)}{dx}=-\frac{\sqrt{3+y^2}}{dy}$
And another step should be integrals for both sides. Are my steps correct or not?
|
Just rewrite the equation as
$$
\dfrac{1}{x+x^2} dx = \frac{1}{\sqrt{3+y^2}}dy.
$$
By integration, you get
$$
\log \frac{|x|}{|1+x|} = \textrm{arcsinh} y + C
$$
So, the solution can be given in an explicit form:
$$
y = \sinh\left(\log \dfrac{|x|}{|x+1|}-C\right).
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all complex numbers which make the following equations true: $ |z+1| =1 $ and $ |z^2+1| =1 $ Find all complex numbers which make the following equations true:
$$ |z+1| =1 $$
$$ |z^2+1| =1 $$
Solution:
If $ |z+1| =1 $ holds true, then
$$z+1 = 1.e^{i2n\pi}$$
$$z = 1.e^{i2n\pi}-1$$
$$z = 1.e^{i2n\pi}-1e^{i2m\pi}$$
If $ |z^2+1| =1 $ holds true, then
$$z^2+1 = 1.e^{i2p\pi}$$
$$z^2 = 1.e^{i2p\pi}-1$$
$$z^2 = 1.e^{i2p\pi}-1e^{i2q\pi}$$
*
*How to proceed after this?
*Am I supposed to do in this manner or break complex number z into real part x and imaginary part y and get two equations and thus solve for x and y?
|
If you write $z$ as $a+bi$, then you get the system$$\left\{\begin{array}{l}(a+1)^2+b^2=1\\(a^2-b^2+1)^2+(-2ab)^2=1,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}a^2+b^2+2a=0\\2 a^2 b^2+a^4+2 a^2+b^4-2 b^2=0.\end{array}\right.$$From the first equation, you get that $b^2=-2a-a^2$, and if you replace $b^2$ by $-2a-a^2$ in the second equation, you get $8a^2+4a=0$. So, $a=0$ or $a=-\frac12$, and therefore the solutions of your system are $0$ and $-\frac12\pm\frac{\sqrt3}2i$.
|
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|
Solve for $x$ in $\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$
Solve for $x$:
$$\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$$
I tried to substitute $y=x+2$ and then I try to solve the equation by again and again squaring.
Then I got equation, $$(y-2)(3y^{14}-(y-2)^{15})=0$$
One solution is $y = 2$ and another is $y = 5.$ (I found $5$ as a solution of the equation by hit and trial method).
Therefore, $x = 0$ or $3.$
I'm wondering if there's any another method to solve it as the repeated squaring step seems to be somewhat absurd.
|
We have,
$$\begin{align}&\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x≥0\\
\iff &2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=2x\\
\iff &x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}=3x\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=\sqrt{3x}\\
\iff &2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=2\sqrt{3x}\\
\iff &x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}=x+2\sqrt{3x}\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}}}} = \sqrt{x+2\sqrt{3x}}\end{align}$$
Let, $\sqrt{x+2\sqrt{3x}}=u,\thinspace u≥0$ then we get,
$$\begin{align}\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}}=u\end{align}$$
Then, we see that,
$$\begin{align}&\sqrt{x+2u} = u\\
\iff &\sqrt{x+2\sqrt{x+2u}} = u\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}=u\\
\iff &\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{x+2u}}}}= u\end{align}$$
Therefore, we have
$$\begin{align}&x+2u= x+2\sqrt{3x}\\
\iff &u^2=3x\\
\iff &x+2\sqrt{3x}=3x\\
\iff &\sqrt{3x}=x\\
\iff &x\in\left\{0,3\right\}.\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\log_{a}{(\frac{b^2}{ac}-b+ac})\cdot\log_{b}{(\frac{c^2}{ab}-c+ab})\cdot\log_{c}{(\frac{a^2}{bc}-a+bc})\geq1 $ Prove $$\log_{a}{(\frac{b^2}{ac}-b+ac})\cdot\log_{b}{(\frac{c^2}{ab}-c+ab})\cdot\log_{c}{(\frac{a^2}{bc}-a+bc})\geq1,
$$
where $a,b,c \in (0,1)$.
I tried to solve it in this way:
$$\log_{a}{(\frac{b^2}{ac}-b+ac})\geq1$$
$$\frac{b^2}{ac}-b+ac\geq a$$
$$\frac{(b-ac)^2+abc}{ac}\geq a$$
$$\frac{(b-ac)^2}{ac}+b\geq a$$
We ca affirm that
$$\frac{(b-ac)^2}{ac}\geq 1,b \geq1, a\geq 1$$ then the product
$\frac{(b-ac)^2}{ac}+b\geq a$ is true.
Analogous for $\log_{b}{(\frac{c^2}{ab}-c+ab})$ and $\log_{c}{(\frac{a^2}{bc}-a+bc})$.
The question is whether this solution is properly addressed?
|
Your idea is good but I think it would be more rigorous in the following way:
Before proving
$$F(a,b,c) = \log_{a}\left(\frac{b^2}{ac}-b+ac\right)\cdot\log_{b}\left(\frac{c^2}{ab}-c+ab\right)\cdot\log_{c}\left(\frac{a^2}{bc}-a+bc\right)\geq1$$
one can prove that
*
*$\frac{b^2}{ac}-b+ac \ge b$
*$\frac{c^2}{ab}-c+ab \ge c$
*$\frac{a^2}{bc}-a+bc \ge a$
For example, the first one can be proven as follows (others are the same):
$$\begin{align}
&\frac{b^2}{ac}-b+ac \ge b \\
&\iff\frac{b^2 - 2bac + (ac)^2}{ac} \ge 0 \\
&\iff\frac{(b-ac)^2}{ac}\ge 0 \qquad\square\end{align}$$
Hence,
$$F(a,b,c) \ge \log_ab\cdot\log_bc\cdot\log_ca = \frac{\ln b}{\ln a}\cdot\frac{\ln c}{\ln b}\cdot\frac{\ln a}{\ln c} = 1$$
|
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|
Integral$ \int_0^{\pi\over3} \frac{\cos^2x}{\sqrt{1+\cos^2x}} dx$ $$\int_0^{\pi\over3} \frac{\cos^2x}{\sqrt{1+\cos^2x}} dx$$
I tried to substitute $1+\cos^2x$ , tried to change cos in sin by complementary formula etc . But nothing seems to work out. I think it's not as simple as I thought ?
|
$$ I = \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1+\cos^2 x}} dx = \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{2-\sin^2 x}} dx = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1-\frac{1}{2}\sin^2 x}} dx $$
Recall the elliptic integral of the first and second kind, respectively:
$$F(\phi,k) = \int_{0}^{\phi} \frac{1}{\sqrt{1-k^2\sin^2 \theta}}d\theta$$
$$E(\phi,k) = \int_{0}^{\phi} \sqrt{1-k^2\sin^2 \theta}d\theta$$
Is easy to show that
$$\int_{0}^{\phi} \frac{\sin^2 \theta}{ \sqrt{1-k^2\sin^2 \theta}} d\theta = \frac{F(\phi,k)-E(\phi,k)}{k^2}$$
$$\frac{F(\phi,k)-E(\phi,k)}{k^2} = \frac{1}{k^2}\int_{0}^{\phi} \frac{1}{\sqrt{1-k^2\sin^2\phi}} - \sqrt{1-k^2\sin^2\phi} d\phi= \frac{1}{k^2}\int_{0}^{\phi} \left[\frac{1}{\sqrt{1-k^2\sin^2\phi}} -\frac{1-k^2\sin^2 \phi}{\sqrt{1-k^2\sin^2\phi}}\right]d\phi = \int_{0}^{\phi} \frac{\sin^2 \theta}{ \sqrt{1-k^2\sin^2 \theta}} d\theta$$
Hence:
$$ \int_{0}^{\phi} \frac{\cos^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = \int_{0}^{\phi} \frac{1-\sin^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = \int_{0}^{\phi} \frac{1}{\sqrt{1-k^2\sin^2 \theta}} d\theta - \int_{0}^{\phi} \frac{\sin^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = F(\phi,k) -\frac{F(\phi,k)-E(\phi,k)}{k^2} = \frac{E(\phi,k)-(1-k^2)F(\phi,k)}{k^2} = \frac{E(\phi,k)-k'^2F(\phi,k)}{k^2}$$
where $k$ is the modulus and $k'$ is the complementary modulus and $|k|<1, \quad k=\sqrt{1-k'^2}$.
Hence
$$\boxed{\int_{0}^{\phi} \frac{\cos^2 \phi}{\sqrt{1-k^2\sin^2 \theta}} d\theta = \frac{E(\phi,k)-k'^2F(\phi,k)}{k^2}}$$
Therefore
$$\boxed{I = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{3}} \frac{\cos^2 x}{\sqrt{1-\frac{1}{2}\sin^2 x}} dx = \frac{2}{\sqrt{2}}\left[E\left(\frac{\pi}{3},\frac{1}{\sqrt{2}}\right)-\frac{1}{2}F\left(\frac{\pi}{3},\frac{1}{\sqrt{2}}\right)\right]}$$
We have to take care evaluating elliptic functions since the software sometimes may use a slight different notation
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sqrt{6}+\sqrt{3}$ is not rational proof I want to prove $\sqrt{6}+\sqrt{3}$ is not rational; here is my attempt:
Assume for the sake of contradiction that $\sqrt{6}+\sqrt{3}$ is rational. Then $(\sqrt{6}+\sqrt{3})^2$ must also be rational. Since $$(\sqrt{6}+\sqrt{3})^2=9+2\sqrt{6}\sqrt{3}=9+2\sqrt{2}\sqrt{3}\sqrt{3}=9+6\sqrt2,$$
we see $9+6\sqrt2$ must be rational. But, since $\sqrt{2}$ is not rational we have a contradiction and hence $\sqrt{6}+\sqrt{3}$ is not rational.
Is it correct?
|
You can also appeal to the rational root theorem,
$$ \begin{align*} x &= \sqrt{3} + \sqrt{6}\\
x^2 &= 9 + 6\sqrt{2}\\
x^4-18x^2 &+9=0 \end{align*} $$
So if $x = p/q$ with $p,q$ coprime then $p |9, q|1$ or $p = 1,3,9$ and $q=1,-1$, but none of the options qualify as $x$ is not an integer
|
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|
Probability of randomly assigning elements to groups
10 students are tested in an exam with 4 different versions. Each
student is randomly assigned to one of the versions. What is the
probability that there are exactly $i$ versions in which exactly > $3$ students were assigned. Answer separately for $i=2, i = 3$.
I've just started my probability course and it's been a while since I've touched combinatorics (and admittedly I was never good at it), and I'm having trouble getting it right..
My attempt:
Note that as we choose the cards randomly we have a symmetric probability space.
We have $10$ students, each student "chooses" one of $4$ versions. Order matters and repetition allowed, therefore
$|\Omega| = 4^{10}$
For $i = 2$ we first choose two versions which will have exactly $3$ students
${4 \choose 2}$
Then we choose $6$ students to be assigned to the versions we've chosen
${10 \choose 6}$
Then we count the number of different combinations for $6$ students in $2$ versions, where there are exactly $3$ students in each version. We choose $3$ students out of $6$ to place in one version and the remaining students will be in the second version.
${6 \choose 3}$
Now we are left with $4$ students to assign between the remaining two versions. We either have two versions with $2$ students each, or one version with $4$ students. We calculate total options to allocate students and subtract the options in which there are $3$ students in the same version.
$3$ Students in $1$ version - We choose the version ${2 \choose 1}$, allocate $3$ out of $4$ students to it with the remaining student in the last version $\frac{4!}{\left(4-3\right)!}$, so in total
${2 \choose 1}\frac{4!}{\left(4-3\right)!} = 8$
Total options to allocate $4$ students into $2$ versions without restrictions $2^{4} = 16$
Therefore in total $|A| = {4 \choose 2}{10 \choose 6}{6 \choose 3}(16-8)={4 \choose 2}{10 \choose 6}{6 \choose 3}8$ and we have
$P(A) = \frac{\left|A\right|}{\left|Ω\right|} = \frac{{4 \choose 2}{10 \choose 6}{6 \choose 3}8}{4^{10}} = 0.384$
For $i = 3$ we choose $3$ versions to have $3$ students.
${4 \choose 3}$
We choose $9$ students to allocate to these versions, with the remaining one on the other version.
${10 \choose 9}$
Total options to organize $9$ students in $3$ versions.
$3^{9}$
Therefore $|A| = {4 \choose 3}{10 \choose 9}3^{9}$ and we have
$P(A) = \frac{\left|A\right|}{\left|Ω\right|} = \frac{{4 \choose 3}{10 \choose 9}3^{9}}{4^{10}} = 0.750$
I can't seem to find any mistakes, but I would also be really grateful if you can comment whether my way of thinking is correct and maybe give a few tips on how to approach these types of questions!
|
$10$ students are tested in an exam with $4$ different versions. Each student is randomly assigned to one of the versions. What is the probability that there are exactly $i$ versions in which exactly $3$ students were assigned. Answer separately for $i=2,i=3$.
$i = 2$
As you observed, the sample space has size $|\Omega| = 4^{10}$.
There are indeed $\binom{4}{2}$ ways to select the two versions which exactly three students will receive. We can assign three of the ten students the lower numbered of the two selected versions in $\binom{10}{3}$ ways and three of the remaining seven students the higher numbered of the two selected version in $\binom{7}{3}$ ways. You are also correct that there are eight ways to assign versions to the remaining students so that neither of the remaining versions is taken by exactly three students since there are $\binom{4}{2}$ ways to assign exactly two students to take the lower-numbered of the remaining two versions and one way to assign the remaining two students the remaining version and $2$ ways to assign all four students to take one of the remaining versions.
Therefore, the probability that there are exactly two versions which exactly three students receive is
$$\frac{\dbinom{4}{2}\dbinom{10}{3}\dbinom{7}{3}\left[\dbinom{4}{2} + \dbinom{2}{1}\right]}{4^{10}} \approx 0.1923$$
Where did you make your mistake?
You made two mistakes.
You meant to write that the number of ways of assigning exactly three of the remaining four students one of the two remaining versions is
$$\binom{2}{1}\binom{4}{3} = 8$$
since there are two ways to select the version three of the remaining four students will receive, $\binom{4}{3}$ to select the three students who will receive that version, and one way to assign the remaining student the remaining version.
Note that
$$\frac{4!}{(4 - 3)!} = \frac{4!}{1!} = \frac{4!}{1} = 24$$
Evidently, what you wrote is not what you meant.
Also, you made a computational error.
$$\Pr(A) = \frac{|A|}{|\Omega|} = \frac{\dbinom{4}{2}\dbinom{10}{6}\dbinom{6}{3}\cdot 8}{4^{10}} \approx 0.1923$$
$i = 3$
There are $\binom{4}{3}$ ways to select the three versions which exactly three students each will receive, $\binom{10}{3}$ ways to assign three of the ten students the lowest numbered of those versions, $\binom{7}{3}$ to assign three of the remaining seven students the next lowest numbered of the selected versions, and $\binom{4}{3}$ ways to assign three of the remaining four students the highest numbered of the selected versions. The other student must receive the fourth version. Hence, the probability that there are three versions which exactly three students each will receive is
$$\frac{\dbinom{4}{3}\dbinom{10}{3}\dbinom{7}{3}\dbinom{4}{3}}{4^{10}} \approx 0.0641$$
Where did you make your mistake?
After you selected which nine students would receive the three versions which exactly three students each would receive, the number of ways of distributing versions to students so that exactly three students each would receive them is
$$\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
Therefore, you should have obtained
$$\Pr(A) = \frac{|A|}{|\Omega|} = \frac{\dbinom{4}{3}\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{4^{10}} = 0.0641$$
Your term $3^9$ is the number of ways of distributing nine students to take those versions without restriction, which is why you obtained a much larger number.
|
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|
Factoring a third degree polynomial with a given root My professor gave us the following polynomial:
$f(x) = 3x^3-4x^2-x+2$
Given is that $x = 1$ is a root of this function. We are asked to find the other ones.
He then told us that, given $x=1$ is a root, we now know that we can factorize this polynomial into $(x-1)$, and a second factor starting with $3x^2...$.
I'm familiar with the process of factoring polynomials, and of finding the roots by setting each factor to zero. But what are you actually doing when you factor a polynomial? In my mind it's just a formulaic way to find zeros of such functions. That's why I was a little confused when we are asked to do the reverse; finding the factor of a given root. More generally, if you know a root of a polynomial, for example $x=7$, can you then always conclude that one factor is $(x-7)$?
How should I proceed finding the other factor(s)?
|
Factoring a polynomial is basically the same as factoring an integer into primes. $34= 2(17), 36= 2^2(3^2$). Any polynomial can be factored, over the real numbers, into either linear terms, such as "$(x- a)$" where $a$ is a real number, or a irreducible quadratic term. (quadratics that cannot be factored further with real numbers)
Any polynomial can be factored, over the Complex numbers, into all linear factors.
The fact that $x = 1$ is a root of this equation means that $x - 1$ is a factor of $f(x)= 3x^3- 4x^2- x+ 2$.
So we can write $f(x) = (x- 1)(ax^2+ bx+ c) = ax^3+ (b- a)x^2+ (c- b)x- c$. So we must have $a= 3, b- a= b- 3= -4$.
So $b= -1$, and $c- b= c+ 1= -1$ so $c= -2$.
That is $3x^3- 4x^2- x+ 2= (x- 1)(3x^2- x- 2)$.
Now, can $3x^3- x- 2$ be factors (with integer coefficients)? If it did one factor would have to have $3x$ and the other $x$ (so that they multiply to $3x^2$) and one factor would have to have $-1$ and the other $2$ or one would have to have $1$ and the other $-2$.
Putting those together gives four possibilities:-
To decide which, just multiply them.
$(3x- 1)(x+ 2)= 3x^2+ 5x- 2$
$(3x+ 2)(x- 1)= 3x^2- x- 2$
$(3x+ 1)(x- 2)= 3x^2- 5x- 2$
$(3x- 2)(X+ 1)= 3x^2+ x- 2.$
But, $3x^2- x- 2= (3x+ 2)(x- 1).$
So $3x^3- 4x^2- x+ 2= (x- 1)(3x+ 2)(x- 1)$
$\rightarrow (x- 1)^2(3x+ 2)$
|
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|
Calculating the joint probability density $f(x,y)=\frac{1}{y}$ If the joint probability density of $X$ and $Y$ is given by:
$$f(x,y) = \frac{1}{y}$$
for $0<x<y, 0<y<1$
Find the probability that the sum of the values of $X$ and $Y$ will exceed 1/2.
What I have tried:
I have tried sketching out the region and got this:
However, I was not sure on which areas to shade. The corrected areas have been provided in the image as I looked at the solution. I got the framework on the graph correct, just not which areas to shade.
What should I look out for when trying to shade regions of my graph when given a pdf as described? I sort of understand why the area above 1/2 is shaded, but I do not understand why the triangle below 1/2 is also shaded.
Secondly here's my working on the integrals in the y direction:
$$\int_0^{\frac{1}{2}}dx\int^{\frac{1}{2}-x}_0 \frac{1}{y}dy + \int_{\frac{1}{2}}^1\int_0^{1-x}\frac{1}{y}dy$$
|
According to nejimban's comment you have to integrate on the set
$$
\begin{align}
A&=\{(x,y):0<x<y<1,\ \frac{1}{2}< x+y\}\\
\end{align}
$$
Given this information you find
$$y>\frac{1}{2}-x>\frac{1}{2}-y$$
such that
$$1>y>\frac{1}{4}$$
Furthermore you find, that
$$y>x>\frac{1}{2}-y$$
However, note that for $y>\frac{1}{2}$
$$0 > \frac{1}{2}-y$$
Because of this, you have to add another boundary at $y=\frac{1}{2}$. Integrating your density function with these information in mind, gives the asked probability
$$
\begin{align}
\int_\frac{1}{4}^\frac{1}{2}\int_{\frac{1}{2}-y}^y\frac{1}{y}dx\ dy + \int_\frac{1}{2}^1\int_{0}^y\frac{1}{y}dx\ dy&=\int_\frac{1}{4}^\frac{1}{2}\frac{1}{y}\Big(\int_{\frac{1}{2}-y}^ydx\Big)\ dy + \int_\frac{1}{2}^1\frac{1}{y}\Big(\int_{0}^ydx\Big)\ dy\\
&=\int_\frac{1}{4}^\frac{1}{2}\frac{1}{y}(2y-\frac{1}{2})\ dy + \int_\frac{1}{2}^1\ dy\\
&=\int_\frac{1}{4}^\frac{1}{2}(2-\frac{1}{2y})\ dy + \int_\frac{1}{2}^1\ dy\\
&=\frac{1}{2}-\frac{1}{2}\ln(2x)\Big|_{x=\frac{1}{4}}^{x=\frac{1}{2}}+\frac{1}{2}\\
&=1-\frac{1}{2}\ln(2)
\end{align}
$$
|
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|
What is the value of the inradius of triangle $PHQ$? For refrence: The right triangle $ABC$, right at $B$, the height $BH$ is drawn. Let $P$ and $Q$ be the intersections of triangles $AHB$ and $BHC$,
$PQ$ intersects at $E$ the $BH$, where $\frac{BE}{EH} = 5\sqrt2$ and the inradius of the triangle $ABC$ measures $10$.
Calculate the inradius of triangle $PHQ$.(Answer: $1$)
My progress..It is a complicated drawing ..not to scale...
I found it a difficult question and didn't get much out of it.... Maybe there is some relationship between the inradius but I don't know
I I know this(by property): $ r_1+r_2+R = BH\\
r_1+r_2+10 = BH$
Also: $\triangle MPH \sim \triangle NQH \implies\\
\frac{r_1}{r_2} = \frac{PH}{QH}$
By Geogebra $\triangle PQH$ is right?? $\implies$ T.Poncelet: $PH+HQ = PQ+2r$
...
|
$S$ the point of touching incircle ABH with BH. $T$ is the point of intersection of straight lines $PS$ and $QN$. Triangles $PES$ and $PQT$ are right and similar. $PS=r_1$, $PT=r_1+r_2$, $QT=r_2-r_1$. From similarity:
$$ES=PS\cdot QT/PT=r_1\frac{r_2-r_1}{r_1+r_2}$$
$$EH=ES+SH=r_1\frac{r_2-r_1}{r_1+r_2}+r_1=\frac{2r_1r_2}{r_1+r_2}$$
From similarity of triangles $AHB$, $CHB$ and $ABC$ and Pythagoras theorem follows $$R^2=r_1^2+r_2^2\Rightarrow 2r_1r_2=(r_1+r_2)^2-R^2\Rightarrow EH=r_1+r_2-\frac{R^2}{r_1+r_2}$$
$$BH=BE+EH=5\sqrt{2}\cdot EH+EH=EH(5\sqrt{2}+1)$$
Also $BH=r_1+r_2+R$, then
$$r_1+r_2+R=\left(r_1+r_2-\frac{R^2}{r_1+r_2}\right)(5\sqrt{2}+1)$$
Let $r_1+r_2=xR$, then
$$x+1=\left(x-\frac{1}{x}\right)(5\sqrt{2}+1)\Rightarrow x=(x-1)(5\sqrt{2}+1)\Rightarrow x=1+\frac{\sqrt{2}}{10}$$
The incircle radius of tectangular triangle $PHQ$ is $$\frac{PH+HQ-PQ}{2}=\frac{r_1\sqrt{2}+r_2\sqrt{2}-\sqrt{2r_1^2+2r_2^2}}{2}=R\sqrt{2}\ \frac{x-1}{2}=\frac{R}{10}=1$$
The answer will transform to 2 if one changes $5\sqrt{2}$ to $\frac{5}{\sqrt{2}}$.
|
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|
Where is the mistake? Finding an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
Find an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
We consider the foci in the coordinate system $XY$ such that $X=x-2$ and $Y=y-1-x$, the coordinates of the foci in this system are $(-1,0)$ and $(1,0)$, furthermore $2a=5$, the equation of the ellipse in $XY$ is
\begin{equation}
\left( \frac{X}{2.5}\right)^2 + \left( \frac{Y}{\sqrt{5.25}} \right)^2 = 1
\end{equation}
and this can be expressed in $xy$ as
$$\left( \frac{x-2}{2.5} \right) + \left( \frac{y-1-x}{\sqrt{5.25}}\right) = 1$$
I have made the graph of the last equation and it is not the case that foci are $(1,2)$ and $(3,4)$, so, can anyone help me to see the mistake please?
|
As shown in other answer, you can just form an equation for sum of distances to foci. Going by your method, the goal would be to write
$$\left( \frac{\text{Distance from Line perp to Axis, thro' center }}{\text{Semi-major axis}} \right)^2 + \left( \frac{\text{Distance from Axis}}{\text{Semi-minor axis}} \right)^2 = 1$$
which are proper substitutes for $X,Y$ in the standard form.
*
*Axis is the line passing through foci : $y=x+1$
*Center of ellipse is midpoint of foci : $(2,3)$
*Line perpendicular to axis and passing through center $(2,3)$ is $x+y=5$
*You can calculate semi-major axis $=a$ and semi-minor axis $=b$.
Then required equation of ellipse is
$$\frac{1}{a^2}\left( \frac{x+y-5}{\sqrt{2}} \right)^2 + \frac{1}{b^2}\left( \frac{x-y+1}{\sqrt{2}} \right)^2 = 1$$
This simplifies to $$84x^2 -32xy + 84y^2 + \cdots = 0$$
|
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|
Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now I need to factor the quadratic $y^2-12y+12$.
So, I calculated discriminant
$$D=12^2-4\times 12=96\implies \sqrt D=4\sqrt 6.$$
This means that the multipliers of quadratic are not rational. So I don't know how to proceed anymore.
|
Or
$$
\begin{align}
4x^2-2xy-4x+3y-3 &= \\
4x^2-4x-3-y(2x-3) &= \\
(2x-3)(2x+1)-y(2x-3) &= \\
(2x-3)(2x+1-y)
\end{align}
$$
As far as factoring quadratics is concerned (your question about factoring $y^2-12y+12$) if the roots are $\alpha$ and $\beta$ the quadratic factors as
$$
(y-\alpha)(y-\beta)
$$
|
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|
Minimize the ratio involving the ellipse
Let $P$ be any point on the curve $\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$,
and $A,B$ be two fixed points $\left(\dfrac{1}{2},0\right)$ and
$(1,1)$ respectively. Find the minimum value of
$\dfrac{|PA|^2}{|PB|}$.
Assume $x=2\cos\theta,y=\sqrt{3}\sin\theta$, then
$$\dfrac{|PA|^2}{|PB|}=\frac{\left(2\cos\theta-\frac{1}{2}\right)^2+(\sqrt{3}\sin\theta-0)^2}{\sqrt{\left(2\cos\theta-1\right)^2+(\sqrt{3}\sin\theta-1)^2}},$$
but which is not so easy to tackle.
WolframeAlpha gives the result $1$.
|
Doing what @Paul Sinclair already commented and using multiple angle formulae
$$\dfrac{|PA|^4}{|PB|^2}=\frac{-256 \cos (\theta )+92 \cos (2 \theta )-16 \cos (3 \theta )+2 \cos
(4 \theta )+259}{8 \left(-4 \sqrt{3} \sin (\theta )-8 \cos (\theta
)+\cos (2 \theta )+11\right)}$$ Computing the derivative, its numerator is "just"
$$(15-8 \cos (\theta )+2 \cos (2 \theta )) \times$$
$$\left(24 \sqrt{3}-40 \sin (\theta )+45 \sin (2 \theta )-12 \sin (3 \theta )+\sin
(4 \theta )-40 \sqrt{3} \cos (\theta )-8 \sqrt{3} \cos (2 \theta )+6
\sqrt{3} \cos (3 \theta )\right)$$ The first factor does not cancel and with all these $\sqrt{3} $ in the remaining term, if the solution is simple, it is a multiple of $\frac \pi 3$ and, just trying, one zero of the derivative is $\theta=\frac {5\pi} 3$ and then the result.
|
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|
How should I evaluate $\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$? $$\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$$
$$\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}\frac{1}{\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}x$$
$$\frac1{\pi}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}\frac{x^2}{x^4+x^2+1}\ \mathrm{d}x$$
then substituting $t^3=x$
$$\frac3{\pi}\int_0^{\infty}\frac{t^9}{(1+t)(t^{12}+t^6+1)}\ \mathrm{d}t$$
what should i do after this should i write $\frac{1}{1+t}$ as $\sum(-1)^kt^k$?
|
After some truly horrific partial fraction decomposition (I'm trusting WolframAlpha (check near the bottom of the page) for now, until I can verify it), we get
\begin{align}
\frac{ t^3 }{3 (t^6 - t^3 + 1)} + \frac{t}{6 (t^6 - t^3 + 1)} - \frac{t}{2 (t^6 + t^3 + 1)} - \frac{1}{6 (t^6 - t^3 + 1)} + \frac{1}{2 (t^6 + t^3 + 1)} + \frac{t^5}{3 (t^6 - t^3 + 1)} - \frac{t^4}{3 (t^6 - t^3 + 1)} - \frac{t^2}{6 (t^6 - t^3 + 1)} + \frac{t^2}{2 (t^6 + t^3 + 1)} - \frac{1}{3 (t + 1)}.
\end{align}
If I plug this guy into FriCAS, it simplifies to
\begin{align}
\frac{t^5 - t^4 + t^3 + t^2 - t}{3}.
\end{align}
This still looks like it'll diverge, so something else funky is going on here.
|
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|
How to find the exact value of the integral $ \int_{0}^{\infty} \frac{d x}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}}$? $\textrm{I first reduce the power two to one by Integration by Parts.}$
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6}}{\left(x^{6}+1\right)^{2}} d x\\&=\displaystyle -\frac{1}{6} \int_{0}^{\infty} x d\left(\frac{1}{x^{6}+1}\right)\\&
=\displaystyle -\left[\frac{x}{6\left(x^{6}+1\right)}\right]_{0}^{\infty}+\frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x \quad \textrm{ (Via Integration by Parts})\\&=\displaystyle \frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x\end{aligned}$
$\textrm{Then I am planning to evaluate }\displaystyle I= \int_{0}^{\infty} \frac{1}{x^{6}+1}\text{ by resolving }\frac{1}{x^{6}+1} \text{ into partial fractions.}$
But after noticing that $$I=\int_{0}^{\infty} \frac{d x}{x^{6}+1}\stackrel{x\mapsto\frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{4}}{x^{6}+1} d x,$$
I changed my mind and started with $3I$ instead of $I$ as below:
$$
\begin{aligned}
3 I &=\int_{0}^{\infty} \frac{x^{4}+2}{\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)} d x \\
&=\int_{0}^{\infty}\left(\frac{1}{x^{2}+1}+\frac{1}{x^{4}-x^{2}+1}\right) d x \\
&=\left[\tan ^{-1} x\right]_{0}^{\infty}+\int_{0}^{\infty} \frac{\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&=\frac{\pi}{2}+\frac{1}{2} \int_{0}^{\infty} \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+\frac{1}{x^{2}}-1} d x\\
&=\frac{\pi}{2}+\frac{1}{2}\left[\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1}-\int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-3}\right] \\
&=\frac{\pi}{2}+\frac{1}{2}\left[\tan ^{-1}\left(x-\frac{1}{x}\right)\right]_{0}^{\infty}-0 \\
&=\frac{\pi}{2}+\frac{1}{2}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\
&=\pi \\ \therefore I &=\frac{\pi}{3}
\end{aligned}
$$
Now I can conclude that
$$\boxed{\displaystyle \quad \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x=\frac{\pi}{18} }.$$
:|D Wish you enjoy the solution! Opinions and alternative methods are welcome.
|
An alternative method using some known identities .
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6}}{\left(x^{6}+1\right)^{2}} d x.\end{aligned}$
Let $\displaystyle x=t^{\frac{1}{6}}$.
Then we get $$\int_{0}^{\infty}\frac{1}{6}\frac{t\cdot t^{\frac{-5}{6}}}{(1+t)^{2}}dt=\int_{0}^{\infty}\frac{1}{6}\frac{t^{\frac{1}{6}}}{(1+t)^{2}}dt.$$
We know that $\text{B}(m,n)=\int_{0}^{\infty}\frac{x^{m-1}}{(1+x)^{m+n}}dx.$
So we see that our integral is nothing but:
$$\frac{1}{6}\text{B}(\frac{7}{6},\frac{5}{6}).$$
Now expanding we get
$$\frac{1}{6}\frac{\Gamma(\frac{7}{6})\Gamma(\frac{5}{6})}{\Gamma(2)}=\frac{1}{6}\frac{\frac{1}{6}\Gamma(\frac{1}{6})\Gamma(\frac{5}{6})}{1}.$$
Now using Euler's reflection formula. We get: $\Gamma(n)\Gamma(1-n)=\pi\csc(n\pi)$
$$\frac{1}{36}\csc(\frac{\pi}{6})=\frac{\pi}{18}.$$
This is just an alternate method. It's no way even close to being as elegant as your solution. But it is perhaps a little easier for people who know these identities.
|
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|
Simplifying $\cosh x + \sinh x$, $\cosh^2 x + \sinh^2 x$, $\cosh^2 x - \sinh^2 x$ using only the Taylor Series of $\cosh,\sinh$ I was trying to solve the following question:
\begin{align*}
\sinh x &= x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \\
\cosh x &= 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots
\end{align*}
Using only this information, calculate
$\cosh x + \sinh x$, $\cosh^2 x + \sinh^2 x$, and $\cosh^2 x - \sinh^2 x$.
Calculating $\cosh x + \sinh x$ was easy because it's just the Taylor Series of $e^x$, but dealing with squaring is where it gets difficult because the Taylor Series are infinite. How can I circumvent the infinite portion to get $\cosh^2 x$ and $\sinh^2 x$?
|
In order to multiply two power series, say
\begin{align*}
\def\bl#1{\color{blue}{#1}}
\def\gr#1{\color{green}{#1}}
\bl{A}(x) &= \bl{a_0} + \bl{a_1}x + \bl{a_2}x^2 + \cdots \\
\gr{B}(x) &= \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots,
\end{align*}
we have to imagine opening parentheses:
\begin{align*}
\bl{A}(x) \, \gr{B}(x)
&= \bigl( \bl{a_0} + \bl{a_1}x + \bl{a_2}x^2 + \cdots \bigr) \,
\bigl( \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots \bigr) \\
&= \bl{a_0} \,
\bigl( \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots \bigr) \\
&\quad {}+ \bl{a_1}x \,
\bigl( \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots \bigr) \\
&\qquad {}+ \bl{a_2}x^2 \,
\bigl( \gr{b_0} + \gr{b_1}x + \gr{b_2}x^2 + \cdots \bigr) \\
&\qquad\quad {}+\cdots \\
&= \bigl( \bl{a_0}\gr{b_0} + \bl{a_0}\gr{b_1}x
+ \bl{a_0}\gr{b_2}x^2 + \cdots \bigr) \\
&\quad {}+ \bigl( \bl{a_1}\gr{b_0}x
+ \bl{a_1}\gr{b_1}x^2 + \bl{a_1}\gr{b_2}x^3 + \cdots \bigr) \\
&\qquad {}+ \bigl( \bl{a_2}\gr{b_0}x^2 + \bl{a_2}\gr{b_1}x^3
+ \bl{a_2}\gr{b_2}x^4 + \cdots \bigr) \\
&\qquad\quad {}+\cdots
\end{align*}
Now, we collect like terms:
\begin{align*}
\bl{A}(x) \, \gr{B}(x)
&= \bigl( \bl{a_0}\gr{b_0} \bigr) \\
&\quad {}+ \bigl( \bl{a_0}\gr{b_1} + \bl{a_1}\gr{b_0} \bigr) x \\
&\qquad {}+ \bigl( \bl{a_0}\gr{b_2} + \bl{a_1}\gr{b_1}
+ \bl{a_2}\gr{b_0} \bigr) x^2 \\
&\qquad\quad {}+\cdots
\end{align*}
In general, the $x^n$ coefficient in $\bl{A}(x) \gr{B}(x)$ is
$$
\bl{a_0}\gr{b_n} + \bl{a_1}\gr{b_{n-1}}
+ \cdots + \bl{a_{n-1}}\gr{b_1} + \bl{a_n}\gr{b_0},
$$
i.e. a sum of terms $\bl{a_i}\gr{b_j}$,
where $\bl{i} + \gr{j} = n$.
Can you see how to use these observations to square the series for the hyperbolic trig. functions? There are some rather straightforward patterns that you should recognize before having to compute too many coefficients. Try it!
|
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|
What's the measure of the segment $EF$ in the triangle below? For reference:
The triangle $ABC$ where $AB = 7$, $BC = 8$ and $AC = 9$ is inscribed in a circle. Calculate the measure of the arrow ($EF$) of the side $AC$. (Answer: $\frac{3\sqrt5}{2}$)
My progress:
Here are the relations I found:
$EF = R - OE \\
\triangle AEF: EF^2 + OE^2 = AF^2\implies\\
(R-OE)^2 + (\frac{9}{2})^2 = AF^2 \implies:
(R-OE)^2 + \frac{81}{4} = AF^2\\
\triangle AOE: OE^2+AE^2 = AO^2 \implies OE^2 +(\frac{9}{2})^2 =R^2 \implies OE^2+ \frac{81}{4} = R^2\\
\triangle AOF: AF^2 = R^2 +OF^2-2OF.OE \implies\\
AF^2 = R^2 +R^2-2R.OE \therefore AF^2 = 2R^2-2R.OE = 2R\underbrace{(R-OE)}_{=EF}$
...??
|
Here is another approach -
Using Heron's formula, $\triangle_{ABC} = 12 \sqrt5$
$\triangle_{ABC} = \frac12 \cdot BH \cdot AC = 12 \sqrt5 \implies BH = \frac{8 \sqrt5}{3}$
Using Pythagoras, $ \displaystyle AH = \frac{11}{3} \implies HE = \frac92 - \frac{11}3 = \frac56$
As $BG$ is angle bisector, $ \displaystyle AG = \frac{7}{7+8} \cdot 9 = \frac{21}5 \implies GE = \frac3{10}$
That leads to $ \displaystyle HG = \frac8{15}$
As $ \triangle BHG \sim \triangle FEG, ~ \displaystyle \frac{EF}{BH} = \frac{GE}{HG} \implies EF = \frac{3 \sqrt5}{2}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What's the the value of $x$ in the circumference below? For reference:
In the figure; calculate $x$, if $r =\sqrt2$.
(Answer: $x = \sqrt2$)
My progress:
Draw $PO_1\perp HG\:(O_1 \in HG).$
Let $O$ be the center of the largest circle.
Using Euclid's Theorem:
$\triangle OPF:OP^2 = OQ^2+PQ^2-2\cdot OQ\cdot FQ$
$\implies ((R-x)^2 =(R-r_2)^2+(r_2+x)^2-2(R-r_2)((r_2-x)$
$\implies R^2 -2Rx+x^2 = R^2-2Rr_2+r_2^2 +r_2^2+2r_2x+x^2 -2Rr_2+2Rx+2r_2^2-2r_2x$
$\therefore\boxed{
r_2^2-r_2R-Rx = 0}$
$\triangle MJR: ((r_1+r)^2 = IH^2 +(r_1-x)^2$
$\implies r_1^2+2r_1r+r^2=IH^2+r_1^2-2r_1x+x^2$
$\therefore \boxed{2r_1(r+x)-x^2 = IH^2}$
$\triangle PFQ: PQ^2=PF^2+FQ^2 $
$\implies (r_2+x)^2=PF^2 + (r_2-x)^2 $
$\implies r_2^2+2r_2x+x^2=PF^2+r_2^2-2r_2x+x^2$
$\therefore \boxed{4r_2x = PF^2}$
...?
|
$AB = 2R\\
r_1+r_2= R\\
AO = R$
Euclid's Th.: $\triangle OO_1O_3$ and $OO_2O_4$
$(R-r)^2=(r_1+r)^2+r_2^2-2(r_1-r)r_2\\R^2-2Rr+r^2 = r_1^2+2r_1r+{r^2}+r_2^2-2r_1r_2+2rr_2\implies\\\boxed{R^2-2Rr = (r_1-r_2)^2+2r(r_1+r_2)}(I)$
$(R-x)^2=(r_2+x)^2+r_1^2-2(r_2-x)r_1=\\
R^2-2Rx+{x^2} =r_2^2+2r_2x+{x^2}+r_1^2-2r_2r_1+2xr_1=\\
\boxed{R^2-2Rx=(r_1-r_2)^2+2x((r_1+r_2)}(II)$
$(I)-(II):2Rx-2Rr=2r(r_1+r_2)+2x(r_1+r_2)=$
$2R(x-r)=(\underbrace{r_1+r_2}_{=R})(2(r+x))$
$\therefore\boxed{\color{red}x-r = r+x \implies x = r}$
|
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|
Linear transformation of a line Question:
Find the image of the line $y=5x+9$ by rotation with center $O(0,0)$ with rotation angle of $270^{\circ}$ followed by dilatation at the center $O(0,0)$ with a scale factor of $3$.
My Attempt:
Given the line $y=5x+9$, it passes through $(-1.8,0)$ and $(0,9)$. I used these two points to perform the rotation of $270^{\circ}$ about the origin. When rotated anti-clockwise, the image of these two points becomes $(0,1.8)$ and $(9,0)$ respectively. With a dilatation at the center with a scale factor of $3$, these points becomes $(0, 5.4)$ and $(27,0)$, hence the image of the line $y=5x+9$ becomes $y=-0.2x+5.4$ or $10y=-2x+54$.
When rotated clockwise, the image of the line $y=5x+9$ becomes $y=-0.2x-5.4$ or $10y=-2x-54$.
I would like to check if the answer/ approach is correct for this question? If not correct, how should I approach it.
Thank you.
|
This is correct. You can verify the answer by using linear algebra as well. The original line is given by
$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0\\9 \end{pmatrix}+ t \begin{pmatrix} 1 \\ 5 \end{pmatrix}; (t\in\mathbb{R})$$
The rotation by $270$° represented by the matrix R, and the dilation by 3 represented by the matrix D are given by:
$$R=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}; D=\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$$
Finding the transformed line is equivalent to performing the following operation:
$$\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\bigg(\begin{pmatrix} 0\\9 \end{pmatrix}+ t \begin{pmatrix} 1 \\ 5 \end{pmatrix}\bigg)$$
This results in the following equation:
$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 27\\0 \end{pmatrix}+ t \begin{pmatrix} 15 \\ -3 \end{pmatrix}$$
Which can be written as this system
$$x=27+15t$$
$$y=-3t$$
solving for $y(x)$ yields the same answer you obtained.
|
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|
Integral of $1 / \sqrt x$ using Limits Actually the problem here is to find out the INTEGRAL of $\frac{1}{\sqrt x}$ using the limit definition. I am very well able to solve the question using POWER rule but that is not allowed in the question.
$$b-a=nh$$
where $h$ is very small.
Then
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}{\frac{h}{\sqrt x} + \frac{h}{\sqrt{x+h}}+...+\frac{h}{\sqrt{x+(n-1)h}}}$$
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}({\frac{1}{\sqrt x} + \frac{1}{\sqrt{x+h}}+...+\frac{1}{\sqrt{x+(n-1)h}}})h$$
Every problem starts from this very point as from here I am unable to cancel out the $h$
1
Tried to remove the sqrt from the denominator
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}({\frac{\sqrt x}{x} + \frac{\sqrt{x+h}}{x+h}+...+\frac{\sqrt{x+(n-1)h}}{x+(n-1)h}})h$$
After which it seems much more complicated
2
Taking the $h$ in the numerator to the denominator
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}{\frac{1}{\sqrt{\frac{x}{h^2}}} + \frac{1}{\sqrt{\frac{x+h}{h^2}}}+...+\frac{1}{\sqrt{\frac{x+(n-1)h}{h^2}}}}$$
After some cancelling
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}{\frac{1}{\sqrt{\frac{x}{h^2}}} + \frac{1}{\sqrt{\frac{x}{h^2}+\frac{1}{h}}}+...+\frac{1}{\sqrt{\frac{x}{h^2}+\frac{(n-1)}{h}}}}$$
Which left me in a frenzy
3
Lastly i tried to rake common first
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}{\frac{h}{\sqrt x} + \frac{h}{\sqrt{x+h}}+...+\frac{h}{\sqrt{x+(n-1)h}}}$$
Taking $\frac{h}{\sqrt x}$ common
$$\int_a^b{\frac{1}{\sqrt x}} dx= \lim_{h \to 0}\frac{h}{\sqrt x}({1 + \frac{1}{\sqrt{1+\frac{h}{x}}}+...+\frac{1}{\sqrt{1+\frac{(n-1)h}{x}}}})$$
I was not able to remove $h$ from this method even
|
I'm posting another answer based on fixed-width partitions and a "harmonic mean approximation" for fun. I think this answer can hardly be generalized to other rational powers of $x$. The image and the arguments were copied from my former messages on Discord.
We would make a little trick of approximating $1/\sqrt{x}$ by the harmonic mean of the integrand evaluated at midpoints of neighbouring partitions
$$\frac{1}{\sqrt{x}} \approx \frac{2}{\sqrt{x - h/2} + \sqrt{x + h/2}}.\label{hma}\tag{$\Large\star$}$$
Using the difference of squares identity, it's easy to see that
$$\frac{2}{\sqrt{x - h/2} + \sqrt{x + h/2}} = \frac{2 (\sqrt{x + h/2} - \sqrt{x - h/2})}{h},$$
so the right Riemann sum can be approximated by
$$\begin{aligned} S &= \sum_{k = 1}^n \frac{1}{\sqrt{x_k}} \cdot h \\
&\approx \sum_{k = 1}^n \frac2h \cdot \left(\sqrt{x_k + h/2} - \sqrt{x_k - h/2} \right) \cdot h \\
&= 2 \left(\sqrt{b+h/2} - \sqrt{a+h/2} \right)
\end{aligned}$$
Here we have $n = (b-a)/h$ partitions with partition points $x_k = a + kh$ for $k \in \{1, \dots, n\}$. Using the following elementary identity, I'm going to find the order of the error in terms of mesh $h$.
Exercise: Show that for all $a,b\ge0$, $\sqrt{\mathstrut a+b} \le \sqrt{\mathstrut a} + \sqrt{\mathstrut b}.$
Hence, show that for all $a>0$ and $b \in [0,a]$, $\sqrt{\mathstrut a-b} \le \sqrt{\mathstrut a} - \sqrt{\mathstrut b}.$
These two basic inequalities enable us to establish upper and lower bounds for the denominator in our "harmonic mean approximation" \eqref{hma}
$$2\sqrt{\mathstrut x} - \sqrt{\mathstrut h/2} < \sqrt{\mathstrut x - h/2} + \sqrt{\mathstrut x + h/2} < 2 \sqrt{\mathstrut x} + \sqrt{\mathstrut h/2}.$$
Take reciprocal, multiply by 2, then minus the integrand $1/\sqrt{x}$ to observe that the actual error in our "harmonic mean approximation" \eqref{hma} can be bounded by other another fraction.
Calculate
$$\frac{2}{2\sqrt{x} \pm \sqrt{h/2}} - \frac{1}{\sqrt{x}} = \frac{\mp\sqrt{h/2}}{(2\sqrt{x} \pm \sqrt{h/2}) \sqrt{x}}$$
to get the bound
$$\left|\frac{2}{2\sqrt{x} \pm \sqrt{h/2}} - \frac{1}{\sqrt{x}}\right| = \frac{\sqrt{h/2}}{\sqrt{a} \cdot \sqrt{a}} < \frac{1}{\sqrt2 a} \, h^{1/2}$$
for $x\in[a,b]$.
Use the triangle inequality $|a+b|\le|a|+|b|$ to take summation out of the absolute sign.
$$
\begin{aligned}
& \left|\underbrace{\sum_{k=1}^n\frac{2}{\sqrt{x_k - h/2} + \sqrt{x_k + h/2}} \cdot h }_{2(\sqrt{b+h/2} - \sqrt{a+h/2})} - \underbrace{\sum_{k=1}^n\frac{1}{\sqrt{x_k}} \cdot h}_{S} \right| \\
&\le \sum_{k=1}^n \underbrace{\left| \frac{2}{\sqrt{x_k - h/2} + \sqrt{x_k + h/2}} - \frac{1}{\sqrt{x_k}} \right|}_{< h^{1/2}/(\sqrt2 a)} \cdot h \\
&< nh \, \frac{h^{1/2}}{\sqrt{2} a} \\
&= \frac{b-a}{\sqrt2 a} \, h^{1/2}
\end{aligned}
\label{err1}\tag{main error}
$$
It remains to take away the $h/2$ in $\sqrt{a+h/2}$ and $\sqrt{b+h/2}$.
\begin{gather*}
\sqrt{\mathstrut b} < \sqrt{\mathstrut b + h/2} < \sqrt{\mathstrut b} + \sqrt{\mathstrut h/2} \\
-\sqrt{\mathstrut a} - \sqrt{\mathstrut h/2} < -\sqrt{\mathstrut a + h/2} < -\sqrt{\mathstrut a}
\end{gather*}
Add these two inequalites to see that
$$\left|\left(\sqrt{\mathstrut b + h/2}-\sqrt{\mathstrut a + h/2}\right)-\left(\sqrt{\mathstrut b}-\sqrt{\mathstrut a}\right)\right|< \sqrt{\mathstrut h/2}.\label{err2}\tag{minor error}$$
Use \ref{err1}, \ref{err2} and the triangle inequality to see that
$$\left|S - 2\left(\sqrt{\mathstrut b} - \sqrt{\mathstrut a}\right)\right| < \left(\frac{b-a}{\sqrt2 a} + 2 \right) h^{1/2}.$$
|
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|
Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x+2}=0\\(x+1)(x+2)-2x\sqrt{3x+2}=0$$ What next? Thank you!
|
First multiply by $\sqrt{3x+2}$ to get
$$3x+2 + x^2 = 2x \sqrt{3x+2}$$
Then square both sides
\begin{align*}
(x^2+3x+2)^2 &= 4x^2(3x+2)\\
x^4+6x^3+13x^2+12x+4 &= 12x^3+8x^2 \\
x^4-6x^3+5x^2+12x+4 &=0
\end{align*}
Try to solve this polynomial equation, giving you the true solution and one false solution that you have to exclude. It is helpful to observe that
$$x^4-6x^3+5x^2+12x+4 = x^4-2x^2(3x+2) + (3x+2)^2.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What's the rate of the stream's current? At his usual rate a man rows 15 miles downstream in five hours less than it takes him to return. If he doubles his usual rate, the time dowstream is only hour less than the time uqstream. What's the rate of the stream's current?
Let the man's rate in still water be $r$ and the rate of the current be $c$.
\begin{align*}
\frac{15}{r-c}-\frac{15}{r+c} &= 5 \tag{1}\\
\frac{15}{2r-c}-\frac{15}{2r+c} &= 1 \tag{2}
\end{align*}
Considering equation 1.
\begin{align*}
\frac{15(r+c)}{(r-c)(r+c)}-\frac{15(r-c)}{(r+c)(r-c)} &= 5 \\
\frac{15r+15c}{(r-c)(r+c)}-\frac{15r-15c}{(r+c)(r-c)} &= 5 \\
\frac{15r+15c}{r^2-c^2}-\frac{15r-15c}{r^2-c^2} &= 5 \\
\frac{30c}{r^2-c^2} &= 5 \\
5r^2-5c^2 &= 30c \tag{3}
\end{align*}
Considering equation 2.
\begin{align*}
\frac{15(2r+c)}{(2r-c)(2r+c)}-\frac{15(2r-c)}{(2r+c)(2r-c)} &= 1 \\
\frac{30r+15c}{(2r-c)(2r+c)}-\frac{30r-15c}{(2r+c)(2r-c)} &= 1 \\
\frac{30r+15c}{4r^2-c^2}-\frac{30r-15c}{4r^2-c^2} &= 1 \\
\frac{30c}{4r^2-c^2} &= 1 \\
4r^2-c^2 &= 30c \tag{4}
\end{align*}
We create a system of equations using 3 and 4.
\begin{align*}
5r^2-5c^2 &= 30c \tag{3}\\
4r^2-c^2 &= 30c \tag{4}
\end{align*}
\begin{align*}
5r^2 &= 30c +5c^2\\
\frac{5r^2}{5} &= \frac{30c +5c^2}{5}\\
\frac{5r^2}{5} &= \frac{30c +5c^2}{5}\\
r^2 &= 6c +c^2
\end{align*}
We subsitute into equation 4.
\begin{align*}
4r^2-c^2 &= 30c \\
4(6c +c^2)-c^2 &= 30c \\
24c +3c^2 &= 30c \\
\frac{3c^2}{3c} &= \frac{6c}{3c} \\
c &= 2
\end{align*}
I'm not feeling very confident of my answer.
|
This is correct. I would have made things a little simpler, noting that I have $5r^2-5c^2=30c=4r^2-c^2$, and so $r^2=4c^2$. Substituting $r=2c$ in either of these equations then gives the result a little more easily.
|
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|
Real function $f$ such that $f(f(x)) = x^3 - 3x^2 + 3x$ Does there exist a function $f:\mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = x^3 - 3x^2 + 3x$?
Since $f(x) = f(y)$ implies $f(f(x)) = f(f(y))$ and so $(x-1)^3 - 1 = (y-1)^3 - 1$, i.e. $x=y$, we have that $f$ must be injective. In particular, $f(f(0)) = 0$ and $f(z) \neq 0$ for $z\neq f(0)$.
I was thinking to try to mimic a known nice construction for $f(f(x)) = ax$ for $a > 0$ - namely $f(x) = -x$ if $x\geq 0$ and $f(x) = -ax$ if $x<0$. But if we try to use it here e.g. like $f(x) = -x$ for $x\geq 0$ and $f(x) = -x(x^2+3x+3)$ since $g(x) = x^2 + 3x + 3$, even though positive, does not satisfy $g(x) = g(-x)$.
Any idea how to fix this or for perhaps some different construction? (If it is "easy to describe" it would be awesome, but I am open to any suggestions.) Any help appreciated!
|
Take the function
$$
f(x)=
\begin{cases}
(x-1)^{\sqrt 3}+1,\qquad x\geq 1,\\
1-(1-x)^{\sqrt 3},\qquad x< 1.
\end{cases}
$$
Then for $x\geq 1$ we have
$$
f(f(x))=\left((x-1)^{\sqrt 3}+1-1\right)^{\sqrt 3}+1=\left((x-1)^{\sqrt 3}\right)^{\sqrt 3}+1=(x-1)^3+1=x^3-3x^2+3x.
$$
If $x<1$ we get
$$
f(f(x))=1-\left(1-\left(1-(1-x)^{\sqrt 3}\right)\right)^{\sqrt 3}=1-\left((1-x)^{\sqrt 3}\right)^{\sqrt 3}=1-(1-x)^3=x^3-3x^2+3x.
$$
|
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|
Proving that $x^2-2xy+6y^2-12x+2y+41\ge 0$ where $x,y \in\Bbb{R}$ We use the result that $AX^2+BX+C \ge 0, \forall X ~ \in \Bbb{R} $ if $A>0$ and $B^2-4AC \le 0$
for proving that $f(x,y)=x^2-2xy+6y^2-12x+2y+41\ge 0$, where $x,y \in\Bbb{ R}.$
Let us re-write the quadratic of $x$ and $y$ as a quadratic of $x$ as
$$f(x,y)=x^2+x(-2y-12)+6y^2+2y+41\ge 0, \forall x \in \Bbb{R}.$$
$$\implies B^2-4AC=(2y+12)^2-4(6y^2+2y+41)=-20(y-1)^2 \le 0,$$
which is true and the equality holds if $y=1$ this further means that $f(x)=(x-7)^2.$
So $f(x,y) \ge 0$, the equality holds if $x=7$ and $y=1$.
The question is: What could be other ways of proving this.
Edit: it will be interesting to note that this quadratic of $x$ and $y$ would represent just a point that is $(7,1)$. It is more clear by the solution of @Aqua given below.
|
Let $z=x-y$ then $$f(x,y) = z^2+5y^2-12z-10y+41 $$ $$ = (z-6)^2+5(y-1)^2$$
|
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|
Integral representation of Bessel function $J_1(x)$ In "The Handbook of Mathematical Functions" by Abramovitz and Stegun, according to Eq. 9.1.24,
\begin{align}
J_0(x)=&\frac{2}{\pi}\int_{1}^\infty \frac{\sin(xt)}{\sqrt{t^2-1}}dt,\quad x>0.
\end{align}
Naively using $J_1(x)=-dJ_0(x)/dx$,
\begin{align}
J_1(x)=&-\frac{2}{\pi}\int_{1}^\infty \frac{t\cos(xt)}{\sqrt{t^2-1}}dt,\quad x>0.
\end{align}
Neither one of these two integrals seem absolutely convergent, with the latter being particularly bad as the non-oscillatory part of the integrand becomes a constant for large $t$.
However, on feeding the left-hand-side of the following expression to WolfraAlpha, I obtain $J_1(x)$:
\begin{align}
-\frac{2}{\pi}\int_{0}^\infty \bigg[\frac{t\Theta(t-1)}{\sqrt{t^2-1}}-1\bigg]\cos(xt)dt= J_1(x),\quad x>0.
\end{align}
($\Theta(x)$ is the Heaviside step function.) I am wondering how to analytically obtain this result.
|
If $x>0$, then
$$ \begin{align}J_{0}(x) &= \frac{2}{\pi} \int_{1}^{\infty} \frac{\sin (xt)}{\sqrt{t^{2}-1}} \, \mathrm dt \\ &= \frac{2}{\pi} \int_{1}^{\infty} \left(\frac{1}{\sqrt{t^{2}-1}} -\frac{1}{t} \right) \sin(xt) \, \mathrm dt + \frac{2}{\pi} \int_{1}^{\infty}\frac{\sin (xt)}{t} \, \mathrm dt \\ &= \frac{2}{\pi} \int_{1}^{\infty} \left(\frac{1}{\sqrt{t^{2}-1}} -\frac{1}{t} \right) \sin(xt) \, \mathrm dt+ \frac{2}{\pi} \int_{x}^{\infty} \frac{\sin (u)}{u} \, \mathrm du \\ &= \frac{2}{\pi} \int_{1}^{\infty} \left(\frac{1}{\sqrt{t^{2}-1}} -\frac{1}{t} \right) \sin(xt) \, \mathrm dt + \frac{2}{\pi} \left(\frac{\pi}{2}- \operatorname{Si}(x) \right). \end{align}$$
Differentiating both sides of the equation, we get $$\begin{align} - J_{1}(x) &= \frac{2}{\pi} \int_{1}^{\infty} \left(\frac{t}{\sqrt{t^{2}-1}} -1 \right) \cos(xt) \, \mathrm dt- \frac{2}{\pi} \frac{\sin (x)}{x} \\ &= \frac{2}{\pi} \int_{\color{red}{0}}^{\infty} \left(\frac{t \theta(t-1)}{\sqrt{t^{2}-1}} -1 \right) \cos(xt) \, \mathrm dt + \frac{2}{\pi} \int_{0}^{1} \cos(xt) \, \mathrm dt- \frac{2}{\pi} \frac{\sin (x)}{x}\\ &= \frac{2}{\pi} \int_{0}^{\infty} \left(\frac{t \theta(t-1)}{\sqrt{t^{2}-1}} -1 \right) \cos(xt) \, \mathrm dt. \end{align}$$
|
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|
Coefficient extraction I want to show:
\begin{equation*}
[z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha}).
\end{equation*}
where $[z^n]$ means the $n$-th coefficient of the power series and
\begin{equation}
H_{n+\alpha} - H_{\alpha} = \sum^{n}_{k=1}{\frac{1}{\alpha + k}}
\end{equation}
So far I got
\begin{equation*}
\frac{1}{(1 - z)^{\alpha + 1}} = (1-z)^{-(\alpha + 1)} = \sum_{n \geq 0}{\binom{-\alpha - 1}{k}(-1)^k z^k} = \sum_{n \geq 0}{\binom{\alpha + n }{n} z^n}
\end{equation*}
where I used $\binom{-\alpha}{k} = (-1)^k \binom{\alpha + k - 1}{k}$and
\begin{equation*}
\log \frac{1}{1 - z} = - \log 1- z = \sum_{n \geq 1}\frac{z^n}{n} = z\sum_{n \geq 0}\frac{z^{n}}{n+1}
\end{equation*}
therefore
\begin{align*}
\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} &= z\left( \sum_{n \geq 0}{\binom{\alpha + n }{n} z^n} \right) \left( \sum_{n \geq 0}\frac{z^{n}}{n+1} \right) \\
&= z \sum_{n \geq 0}\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} z^{n} \\
&= \sum_{n \geq 0}\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} z^{n + 1}.
\end{align*}
Now the $n$-th coefficient is
\begin{align*}
\sum^{n}_{k=0}\binom{\alpha + k}{k} \frac{1}{n - k + 1} &= \binom{\alpha + n}{n} \sum^n_{k = 1}{\frac{n!}{(n-k)! k!} \frac{1}{\binom{\alpha + n}{n - k}} \frac{1}{n - k + 1}} + \frac{1}{n+1}\\
&= \binom{\alpha + n}{n} \sum^n_{k = 1}{\binom{n}{k} \frac{1}{\binom{\alpha + n}{n - k}} \frac{1}{n - k + 1}} + \frac{1}{n+1}\\
&= \binom{\alpha + n}{n} \sum^n_{k = 1}{\binom{n}{k - 1} \frac{1}{\binom{\alpha + n}{n - k - 1}} \frac{k-1}{n - k -1} \frac{1}{\alpha + k + 1} } + \frac{1}{n+1}\\
&= \binom{\alpha + n}{n} \sum^{n+1}_{k = 2}{\binom{n}{k - 2} \frac{1}{\binom{\alpha + n}{n - k}} \frac{k- 2 }{n - k} \frac{1}{\alpha + k} } + \frac{1}{n+1}
\end{align*}
but from now on I can't see how to proceed.
I also know that
\begin{equation}
[z^n] \frac{1}{1-z} \log \frac{1}{1-z} = H_n
\end{equation}
somehow I also tried to use some sort of transformation law for
coefficient extraction, but I am not aware of any kind of transformation law for coefficient extraction. Using $1-u = (1-z)^{\alpha + 1}$ or $z = 1 - (1-u)^{1/(\alpha + 1)}$gives
\begin{equation}
\frac{1}{\alpha + 1}\frac{1}{1-u} \log \frac{1}{1 - u}
\end{equation}
and for this I get the coefficient $\frac{1}{\alpha + 1} H_n$.
Another approach I did was the following:
Since
\begin{equation*}
(1-z)^{-m} = \exp(-m \ln(1-z))
\end{equation*}
I get
\begin{equation*}
\frac{\partial }{\partial m} \exp(-m \ln(1-z)) = -\ln(1-z) \exp(- m \ln(1-z)) = - \ln(1-z) \frac{1}{(1-z)^m} = \frac{1}{(1-z)^m} \ln \frac{1}{1-z}
\end{equation*}
therefore
\begin{align*}
\frac{\partial }{\partial m} (1-z)^{-m} &= \sum_{n \geq 0}{\frac{\partial }{\partial m} \binom{m + n - 1}{n} z^n} \\
&= \sum_{n \geq 0}{\frac{\partial }{\partial m} \frac{\Gamma(m + n)}{n!\Gamma(m)} z^n}
\end{align*}
But I don't know any identities for the derivative of the gamma function.
|
Here is another variation. We show the identity
\begin{align*}
\color{blue}{[z^n]\frac{1}{(1-z)^{\alpha+1}}\log\frac{1}{1-z}=\binom{n+\alpha}{n}\sum_{k=1}^n\frac{1}{\alpha+k}}\tag{1}
\end{align*}
is valid for integral $n>0$ and $\alpha \in\mathbb{C}\setminus\{-1,\ldots,-n\}$.
We start with the LHS of (1) and obtain
\begin{align*}
\color{blue}{[z^n]}\color{blue}{\frac{1}{(1-z)^{\alpha+1}}\log\frac{1}{1-z}}
&=[z^n]\frac{1}{(1-z)^{\alpha+1}}\sum_{k=1}^\infty\frac{z^k}{k}\tag{2.1}\\
&=\sum_{k=1}^n\frac{1}{k}[z^{n-k}]\frac{1}{(1-z)^{\alpha+1}}\tag{2.2}\\
&=\sum_{k=1}^n\frac{1}{k}\binom{-\alpha-1}{n-k}(-1)^{n-k}\tag{2.3}\\
&\,\,\color{blue}{=\sum_{k=1}^n\frac{1}{k}\binom{n+\alpha-k}{n-k}}\tag{2.4}
\end{align*}
Comment:
*
*In (2.1) we use the logarithmic series expansion.
*In (2.2) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$ and restrict the upper limit of the sum to $n$, since other terms do not contribute.
*In (2.3) we select the coefficient of $[z^{n-k}]$ of the binomial series expansion.
*In (2.4) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
It remains to show the binomial identity
\begin{align*}
\color{blue}{\sum_{k=1}^n\frac{1}{k}\binom{n+\alpha-k}{n-k}=\binom{n+\alpha}{n}\sum_{k=1}^n\frac{1}{\alpha+k}}\tag{3}
\end{align*}
A nice aspect is, that when considering the expressions in (3) as polynomial in $\alpha$ we can show, that they both are the derivative $\frac{\partial}{\partial\alpha}\binom{n+\alpha}{n}$.
We start with the RHS of (3) which is the easy part. We obtain
\begin{align*}
\color{blue}{\frac{\partial}{\partial \alpha}}\color{blue}{\binom{n+\alpha}{n}}
&=\frac{1}{n!}\frac{\partial}{\partial \alpha}\prod_{j=0}^{n-1}(n+\alpha-j)\tag{4.1}\\
&=\frac{1}{n!}\frac{\partial}{\partial \alpha}\prod_{j=0}^{n-1}(\alpha+j+1)\tag{4.2}\\
&=\frac{1}{n!}\frac{\partial}{\partial \alpha}\prod_{j=1}^{n}(\alpha+j)\tag{4.3}\\
&=\frac{1}{n!}\sum_{k=1}^n\left(\frac{\partial}{\partial \alpha}(\alpha+k)\right)\prod_{{j=1}\atop {j\ne k}}^n(\alpha+j)\tag{4.4}\\
&=\frac{1}{n!}\sum_{k=1}^n\frac{1}{\alpha+k}\prod_{j=1}^n(\alpha+j)\tag{4.5}\\
&\,\,\color{blue}{=\binom{n+\alpha}{n}\sum_{k=1}^n\frac{1}{\alpha+k}}
\end{align*}
and we get the RHS of (3).
Comment:
*
*In (4.1) we use the definition of binomial coefficients in the form $\binom{p}{q}=\frac{1}{q!}\prod_{j=0}^{q-1}(p-j)$.
*In (4.2) we change the order of multiplication $j\to n-1-j$.
*In (4.3) we shift the index and start with $j=1$.
*In (4.4) we use the product rule for derivation of products with $n$ terms.
*In (4.5) we do the derivation and expand numerator and denominator with $\alpha+k$.
Now the somewhat more challenging part. We do the derivation by using first principles. We obtain
\begin{align*}
\color{blue}{\frac{\partial}{\partial \alpha}}&\color{blue}{\binom{n+\alpha}{n}}
=(-1)^n\frac{\partial}{\partial \alpha}\binom{-\alpha-1}{n}\tag{5.1}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\left(\binom{-\alpha-1-q}{n}-\binom{-\alpha-1}{n}\right)\tag{5.2}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\left(\sum_{k=0}^n\binom{-q}{k}\binom{-\alpha-1}{n-k}-\binom{-\alpha-1}{n}\right)\tag{5.3}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\sum_{k=1}^n\binom{-q}{k}\binom{-\alpha-1}{n-k}\\
&=(-1)^n\lim_{q\to 0}\frac{1}{q}\sum_{k=1}^n\frac{-q}{k}\binom{-q-1}{k-1}\binom{-\alpha-1}{n-k}\tag{5.4}\\
&=(-1)^{n+1}\sum_{k=1}^n\frac{1}{k}\binom{-1}{k-1}\binom{-\alpha-1}{n-k}\\
&=\sum_{k=1}^n\frac{(-1)^{n-k}}{k}\binom{-\alpha-1}{n-k}\tag{5.5}\\
&\,\,\color{blue}{=\sum_{k=1}^n\frac{1}{k}\binom{n+\alpha-k}{n-k}}\tag{5.6}
\end{align*}
which is the LHS of (3) and the claim (3) follows.
Comment:
*
*In (5.1) we use again the binomial identity as in (2.4).
*In (5.2) we do the derivation according to first principles.
*In (5.3) we apply the Chu-Vandermonde identity.
*In (5.4) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.
*In (5.5) we use the binomial identity $\binom{-1}{q}=\frac{1}{q!}(-1)(-2)\cdots(-q)=(-1)^q$.
*In (5.6) we use again the binomial identity as in (2.4).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4316307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Show that if $|z|=2$, $\text{Im}(1-\bar{z}+z^2)\le 7$. Show that if $|z|=2$, $|\text{Im}(1-\bar{z}+z^2)|\le 7$.
My attempt:
Let $z=x+yi$, $|z|=\sqrt{x^2+y^2}$, $|z|^2=(\sqrt{x^2+y^2})^2=x^2+y^2$.
Since $|z|=2$, then $|z|^2=2^2=4$. So $x^2+y^2=4$. Also, $(x-y)^2\ge0$ $\implies x^2+y^2\ge2xy$;
$$xy\le2 $$
So $|y|\le2$ and $|x|\le 2$.
Now,
$$\begin{aligned}
1-\bar{z}+z^2&=1-(x-yi)+(x+yi)^2\\
&=1-x+yi+x^2+2xyi-y^2\\
&=1-x+x^2-y^2+(2xy+y)i
\end{aligned}$$
Thus,
$$\begin{aligned}
\ |\text{Im}(1-\bar{z}+z^2)|&=|y+2xy|\\
&\le|y|+|2xy|\\
&\le |y|+|x^2+y^2|\\
&\le2+4=6
\end{aligned}$$
How about the case when it is equal to $7$?
|
By the triangle inequality:
$$|1 + (-\bar{z}) + z^2 | ≤ |1| + |-\bar{z}| + |z^2| ≤ 1+2+2^2 = 7$$
(think about what happens to $z = 2e^{i \theta}$).
|
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|
Amount of solutions to an equation Briliant.org asks for how many solutions there are to this equation:
$x^3 - 8x^2 + 3x^2 + 3x^3 - 7x^2 = x + 2x^3 - 2x - 2x^3 + x$
Now if I put them in order:
$ 3x^3 + x^3 -8x^2 -7x^2 + 3x^2 = 2x^3 - 2x^3 - 2x + x + x$
and combine
$ 4x^3 -12x^2 = 0$
I could add $12x^2$ to both sides and divide by $4x^2$ so it will become
$x = 3$
Now I would say there is one solution, also khan academy told me there is either one, infinite or no solution to an equation. But according to Brilliant.org, there are 2 solutions: either 3 or 0. They use the distributive property after combining so that
$ 4x^3 -12x^2 = 0$ becomes $ 4x^2(x-3) = 0$. They conclude that either x can be 3 or 0 so one of the fractions become 0 and so will the product, using the zero product property.
It does make perfect sense since I can just put the answer in the original equations and they work... But now I am also confused, isn't this true for most equations and can equations have multiple solutions, maybe even more than 2?
|
A cubic equation has either $3$ real roots or $1$ real root and $2$ complex roots. Here, we have three real roots, two of which are identical.
\begin{align*}
x^3 - 8x^2 + 3x^2 + 3x^3 - 7x^2 &= x + 2x^3 - 2x - 2x^3 + x\\
4 x^3 - 12 x^2 &= 0\\
(x-0)(x-0)(x-3) & =\frac{0}{4}
\implies x\in\big\{0,0,3\big\}
\end{align*}
|
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|
Solve the equation $12x^5+16x^4-17x^3-19x^2+5x+3=0$ Solve the equation $$12x^5+16x^4-17x^3-19x^2+5x+3=0$$ The divisors of $3$ are $\pm1;\pm3$ and the divisors of $12$ are $\pm1;\pm2;\pm3;\pm4;\pm6;\pm12$, so the possible rational roots are $$\pm1;\pm\dfrac12;\pm\dfrac13;\pm\dfrac14;\pm\dfrac16;\pm\dfrac{1}{12};\pm3;\pm\dfrac32;\pm\dfrac34$$ which makes $22$ possible roots. We can use Horner, but how can I reduce the number of possible roots?
|
Step 1:
Try $\pm1$, they work so we can reduce the problem to a cubic by dividing by $(x-1)(x+1)=x^2-1$:
$$
P(x)=(12x^5+16x^4-17x^3-19x^2+5x+3)/(x^2-1)\\=12 x^3 + 16 x^2 - 5 x - 3
$$
Step 2: To find a 3rd solution we find the extrema of $P(x)$ we solve
$$
P'(x)=36x^2-32x-5=0
$$
Finding that it has 2 solutions at $x=-1.02..$ and $x=0.135..$. The first one is a maximum and the $P$ is positive there so there must be a zero of $P(x)$ to the left of $-1.02...$ and it must be $-3/2$.
Step 3: Now we can divide by $x+3/2$, to get a quadratic polynomial:
$$
P(x)/(x+3/2)=12x^2-12x-2
$$
whose zeros are the remaining 2 zeros: $-1/3$ and $1/2$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate this limit $(\infty/\infty)$ I want to prove that $1=0.999...$ in this way, we have
$$0.9...9=0.9 \left(10^{-n-1}\right)\left(\sum_{i=0}^n 10^i\right)$$
Hence $$0.999...=\lim_{n \to \infty} 0.9 \left(10^{-n-1}\right)\left(\sum_{i=1}^n 10^i\right) \\ =0.9 \lim_{n \to \infty} \frac{\sum_{i=0}^n 10^i}{10^{n+1}}$$
Both the numerator and the denominator goes to $\infty$ so we xan use hôpital but even if we did we’d get $\infty/\infty$ again, how can I evaluate this limit?
|
When summing a geometric sequence, we want the common ratio to be less than $1$ in magnitude:
$$
\sum_{n=0}^\infty r^n = 1 + r + r^2 + \cdots = \frac{1}{1-r}
$$
as long as $\lvert r \rvert < 1$.
Thus, to evaluate $0.\overline{9} = 0.999\dots$, we write it in terms of a geometric series with common ratio $r=\tfrac{1}{10} < 1$. Try it yourself, then click to reveal the calculation.
\begin{align} 0.\overline{9} &= \tfrac{1}{10} \bigl( 9 + \tfrac{9}{10} + \tfrac{9}{10^2} + \cdots \bigr) \\ &= \tfrac{9}{10} \bigl( 1 + \tfrac{1}{10} + \tfrac{1}{10^2} + \cdots \bigr) \\ &= \frac{9}{10} \sum_{n=0}^\infty \Bigl(\frac{1}{10} \Bigr)^n \\ &= \frac{9}{10} \cdot \frac{1}{1 - \tfrac{1}{10}} \\ &= \frac{9}{10 - 1} \\ &= 1 \end{align}
|
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|
The constraints are not all considered
The solution says that domain of integration is delimited above by the sphere of equation $x ^ 2 + y ^ 2 + z ^ 2 = 2$ and below by the cone $z = \sqrt{x ^ 2 + y ^ 2}$. I have the impression that we do not consider the fact that $x$ is between $0$ and $1$ and that $y$ is between $0$ and $\sqrt{1-x ^ 2}$. Am I wrong? I had the impression we could find an $x$ great than $1$ or less than $0$ respecting the fact that the $z$ could be selected between the cone and the sphere.
EDIT:
As we have to select a $y$ such that $0 \leq y \leq \sqrt{1-x^2}$, then $x$ is necessarily between $0$ and $1$, because otherwise $\sqrt{1-x^2}$ would be a complex number. However, we work in the real numbers. So it seems obvious that the condition $0 \leq y \leq \sqrt{1-x^2}$ is not necessary to delimit the domain of integration. How about the condition that $0 \leq y \leq \sqrt{1-x^2}$?
|
You are right and the solution is incorrect. If we wished to consider the region bounded by the sphere $x^2 + y^2 + z^2 = 2$ and the cone $z = \sqrt{x^2 + y^2}$, then the correct bounds would be
$$\int_{-1}^1 \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}}\int_{\sqrt{x^2 + y^2}}^{\sqrt{2 - x^2 - y^2}} \frac{1}{\sqrt{x^2 + y^2}} \, \mathrm{d}z \, \mathrm{d}y \, \mathrm{d}x.$$
Plotting the cone and sphere, you'll find that the intersection is a circle lying on the $z = 1$ plane, and the region is a conical segment of the sphere. If $(x, y, z)$ lies on the intersection, then
$$\sqrt{x^2 + y^2} = z = \sqrt{2 - x^2 - y^2} \implies 2(x^2 + y^2) = 2 \implies x^2 + y^2 = 1.$$
The entire intersection of the two regions lies in the cylinder defined by $x^2 + y^2 \le 1$, so the bounds of our integrals with respect to $x$ and $y$ must precisely capture the disk $x^2 + y^2 \le 1$ in the $x$-$y$ plane$^1$. That's what the integral bounds above do for us: the $y$ variable must range between the bottom semicircle $y = -\sqrt{1 - x^2}$ and the top semicircle $y = \sqrt{1 - x^2}$. The $x$ variable, on the disk $x^2 + y^2 \le 1$, ranges between $-1$ and $1$.
The proposed solution only considers the region in the positive orthant; the region is further bounded by $x \ge 0$ and $y \ge 0$. Given the symmetry of the region and the function, the proposed solution will be one quarter of the true value.
$^1$ Suppose that $(x, y, z) \in \Bbb{R}^3$ lay in our intersection. That is, $x^2 + y^2 + z^2 \le 2$ and $z \ge \sqrt{x^2 + y^2}$. Then $$x^2 + y^2 \le z^2 \le 2 - x^2 + y^2 \implies 2(x^2 + y^2) \le 2 \implies x^2 + y^2 \le 1,$$
i.e. basically the same argument as above, with inequalities instead.
|
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|
Upper bound: Given $L$-smooth convex $f$; $( y- x)^T \left( \nabla f(z)-\nabla f(x)\right)\leq(L/2) ( \| x-z\|^2+\| x-y\|^2+\| z-y\|^2)$? Given $L$-smooth convex $f$, I would highly appreciate if you can confirm whether the following bound is correct or not.
\begin{align}
\left( y- x\right)^T \left( \nabla f(z)-\nabla f(x)\right)
\leq \frac{L}{2} \left( \| x-z\|^2+\| x-y\|^2+\| z-y\|^2 \right) \tag{$\clubsuit$}.
\end{align}
Attempt:
Since function $f$ is both $L$-smooth and convex, I particularly make use of the following two inequalities (e.g., can be found here or many other books such as in Yurii Nesterov's book).
*
*Three points descent lemma:
\begin{align}
f(x) \leq f(y) + \left( x - y \right)^T \nabla f(z) + \frac{L}{2} \| x - z \|^2
\end{align}
*$$0 \leq f(y) - f(x) - \left( y - x\right)^T \nabla f(x) \leq \frac{L}{2} \| x - y \|^2 $$
To this end, we rewrite the above two respective inequalities
1.
\begin{align}
f(x) &\leq f(y) + \left( x - y \right)^T \nabla f(z) + \frac{L}{2} \| x - z \|^2 \\
\Longleftrightarrow \left( y - x \right)^T \nabla f(z) &\leq f(y) - f(x) + \frac{L}{2} \| x - z \|^2 \tag{1}
\end{align}
2.
\begin{align}
f(y) - f(x) - \left( y - x\right)^T \nabla f(x) &\leq \frac{L}{2} \| x - y \|^2 \\
\Longleftrightarrow \left( y - x\right)^T \nabla f(x) &\geq f(y) - f(x) - \frac{L}{2} \| x - y \|^2 \tag{2}
\end{align}
Now,
\begin{align}
\left( y- x\right)^T \left( \nabla f(z)-\nabla f(x)\right)
=& \underbrace{\left( y- x\right)^T \nabla f(z)}_{ \text{upper bound using} \ (1) } - \underbrace{\left( y- x\right)^T \nabla f(x)}_{ \text{lower bound using} \ (2) } \\
=& \underbrace{\left( y- x\right)^T \nabla f(z)}_{ \leq f(y) - f(x) + \frac{L}{2} \| x - z \|^2 }
- \underbrace{\left( y- x\right)^T \nabla f(x) }_{ \geq f(y) - f(x) - \frac{L}{2} \| x - y \|^2 } \\
\leq & f(y) - f(x) + \frac{L}{2} \| x - z \|^2 - \left[ f(y) - f(x) - \frac{L}{2} \| x - y \|^2 \right] \\
=& \frac{L}{2} \left( \| x - z \|^2 + \| x - y \|^2 \right) \\
\leq& \frac{L}{2} \left( \| x - z \|^2 + \| x - y \|^2 + \underbrace{\| z - y \|^2}_{\geq 0} \right)
\end{align}
This completes the proof of $(\clubsuit)$.
|
What about this simple estimate:
$$
\left( y- x\right)^T \left( \nabla f(z)-\nabla f(x)\right)
\le \|y-x\| \cdot \|\nabla f(z)-\nabla f(x)\|
\le L \|y-x\| \cdot \|z-x\|
\le \frac L2(\|y-x\|^2 + \|z-x\|^2)
$$
??
|
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|
number of real roots of the equation $f(f(x)) = c$ The question is stated as follows:
Let $f(x) = x^4 − 4x + 1$ for all real number $x$.
$\ $Determine, for each real number $c$, the
number of real roots of the equation $f(f(x)) = c.$
By calculating $f(f(x))$ explicitly it is easy to see that $f(f(x))=u^4+4u^3+6u^2-2$ with $u=x^4-4x$. The range of the outer function is $[-2, +\infty)$ so the number of real roots of the equation $f(f(x)) = c$ is zero if $c<-2$. How should I deal with case when $c \geq 2$?
|
drew some graphs, fiddled with that. I don't see how you would know what to do without some sketches.
$$f(x) = x^4 - 4x + 1$$
$$g(x) = f(f(x)) = (x^4-4x+1)^4-(x^4-4x+1) + 1 \; , \; \; $$
then
$$ 2+g(x) = x^2 \; (x^3-4)^2 \; \left( \; 2 + (x^4 -4x+2)^2 \; \right) $$
where
$$ 2 + (x^4 -4x+2)^2 \; > \; \; 0 $$
We see that $2+g(x) $ has double roots (local minima) at $x=0$ and $x= 4^{1/3}$
You also need to know that there is a local max, the whole graph is a W shape.
$$ -25+g(x) = (x-1)^2 \; (x^2 + 2x+3) \; \cdot \left( x^{12} - 12 x^9 + x^8 + 48 x^6 -8x^5 +3x^4 -64 x^3 + 16 x^2 -12x - 9 \right) $$
The factor of degree 12 has two real roots, as had to happen because of the W shape. Indeed, taking
$$ q(x) = x^{12} - 12 x^9 + x^8 + 48 x^6 -8x^5 +3x^4 -64 x^3 + 16 x^2 -12x - 9 , $$
we find
$$ 9+q(x) = x (x^3-4) \left(x^8 - 8x^5 + x^4 + 16x^2 - 4x + 3 \right) , $$
where $$ x^8 - 8x^5 + x^4 + 16x^2 - 4x + 3 > 0$$
for some reason.
YES,
$$ x^8 - 8x^5 + x^4 + 16x^2 - 4x + 3 = \frac{11 + (2x^4 -8x+1)^2}{4}$$
I had it draw $y = \frac{g(x)}{10}$
Just to check, the derivative
$$g'(x) = 16x^{15} - 208x^{12} + 48x^{11} + 960x^9 - 432x^8 + 48x^7 - 1792x^6 + 1152x^5 - 240x^4 + 1024x^3 - 768x^2 + 192x $$
factorws as
$$g'(x) = x (x-1) (x^3-4) (x^2 + x+1) \left(x^8 - 8x^5 + 3x^4 + 16x^2 - 12x + 3 \right) $$
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Evaluate $\int_{\pi}^{\infty}\left(x^{2}-\sin\left(x\right)-1\right)^{-1}dx=?$ It's an integral which seems simple but I confess I cannot evaluate this :
$$\int_{\pi}^{\infty}\left(x^{2}-\sin\left(x\right)-1\right)^{-1}dx=?$$
I can evaluate another integral where I start from :
$$\int_{\pi}^{\infty}\left(x^{2}-x-1\right)^{-1}dx=\frac{2\coth^{-1}\left(\frac{2\pi-1}{\sqrt{5}}\right)}{\sqrt{5}}$$
Show the convergence is not hard using bound for $\sin(x)$.
Question :
Can we hope to find a closed form ?
Thanks .
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Here is an attempt at an antiderivative. You can consider it a comment.
Attempt 1:
Here is a series expansion for the antiderivative using geometric series which includes the $[\pi,\infty)$ interval of convergence:
$$\int \frac{dx}{x^2-\sin(x)-1}=-\int \sum_{n=0}^\infty \left(i\frac{e^{-ix}-e^{ix}}{2}\right)^n(x^2-1)^{-n-1}dx$$
Which cannot be integrated in closed form. Let’s also use a binomial theorem expansions which have an infinite radius of convergence since they are truncated.
$$-\int \sum_{n=0}^\infty \left(i\frac{e^{-ix}-e^{ix}}{2}\right)^n(x^2-1)^{-n-1}dx = -\int \sum_{n=0}^\infty i^n2^{-n} \left(e^{-ix}-e^{ix}\right)^n\sum_{k_1=0}^n\frac{n!}{(n-k_1)!k_1!}e^{-ix(n-k_1)}e^{ixk_1}\sum_{k_2=0}^n(x^2-1)^{-n-1}dx $$
Then use a Binomial Series which would constrict the series expansion.
Please let me know if there is a simpler series expansion.
Attempt 2:
$$\int \frac{dx}{x^2-\sin(x)-1} =\sum_{n=1}^\infty \frac{\frac{d^{n-1}}{dx^{n-1}}\frac1{x^2-\sin(x)-1}\big|_{x=a}}{n!}(x-a)^n$$
With the nth derivative and Gauss Hypergeometric function for a convergence interval:
$$\int \frac{dx}{x^2-\sin(x)-1} =\sum_{n=1}^\infty\left((-2)^{n-1} a^{n-1} (n-1)!(a^2-\sin(a)-1)^{-(n-1)-1}\,_2\text F_1\left(\frac{1-(n-1)}2,-\frac {n-1}2;-(n-1);1-\frac{\sin(a)+1}{a^2}\right)\right) \frac{(x-a)^n}{n!}= \sum_{n=0}^\infty\left((-2)^n a^n n!(a^2-\sin(a)-1)^{-n-1}\,_2\text F_1\left(\frac{1-n}2,-\frac n2;-n;1-\frac{\sin(a)+1}{a^2}\right)\right) \frac{(x-a)^{n+1}}{(n+1)!}= \sum_{n=0}^\infty\,_2\text F_1\left(\frac{1-n}2,-\frac n2;-n;1-\frac{\sin(a)+1}{a^2}\right) \frac{(-2a)^n(x-a)^{n+1}}{(n+1) (a^2-\sin(a)-1)^{n+1}} $$
Even with the $-n$ in the hypergeometric function, the sum terms exist.
It can be shown that:
$$_2\text F_1\left(\frac{1-n}2,-\frac n2;-n;1-\frac{\sin(a)+1}{a^2}\right) =2^{-n-1 } \left(\sqrt{\frac{\sin(x) + 1}{x^2}} + 1\right)^n + 2^{-n-1} \frac{\left(\sqrt{\frac{\sin(x) + 1}{x^2}} + 1\right)^n}{\sqrt{\frac{\sin(x) + 1}{x^2}}}= 2^{-n-1 } \left(\sqrt{\frac{\sin(x) + 1}{x^2}} + 1\right)^n\left(1+ \frac1{\sqrt{\frac{\sin(x) + 1}{x^2}}}\right)$$
Therefore:
$$\int \frac{dx}{x^2-\sin(x)-1} =\sum_{n=0}^\infty 2^{-n-1 } \left(\sqrt{\frac{\sin(a) + 1}{a^2}} + 1\right)^n\left(1+ \frac1{\sqrt{\frac{\sin(a) + 1}{a^2}}}\right) \frac{(-2a)^n(x-a)^{n+1}}{(n+1) (a^2-\sin(a)-1)^{n+1}} =C+ \frac12 \left( \frac1{\sqrt{\frac{\sin(a) + 1}{a^2}}}+1\right)\sum_{n=0}^\infty \left(\sqrt{\frac{\sin(a) + 1}{a^2}} + 1\right)^n\frac{(-a)^n(x-a)^{n+1}}{(n+1) (a^2-\sin(a)-1)^{n+1}} $$
This result is based on this result and this computation. Please correct me and give me feedback!
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Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$ Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$.
So far,
\begin{align*}
x^2 + y^2 + 121 + 2xy + 22x + 22y &= x^2 + y^2 + 121\\
2xy + 22x + 22y &= 0\\
(2x+22)y &= -22x\\
(x+11)y &= -11x
\end{align*}
At least 1 of $x,y$ must be a multiple of 11?
Dont know where to progress after this. All help appreciated.
|
Since $y$ is an integer,$\frac{-11x}{x+11}$ will be an integer. When $\frac{x}{x+11}$ is an integer, it refers that $\frac{11}{x+11}$ is an interger, or $x=0$. Then you can find all such $x$ in this situation. Otherwise $\frac{x}{x+11}$ is not an integer but $\frac{-11x}{x+11}$ is. Thus $x$ is a multiple of $11$, write $x=11k$ where $k$ is an integer and solve $k$ in the same way.
|
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Finding the Maclaurin polynomial of order 6 of: $f(x)=x\ln(1+x^{3})\ln(1-x^{2})$ Find the Maclaurin polynomial of order 6 of:
$$f(x)=x\ln(1+x^{3})\ln(1-x^{2})$$
The result I get doesn't make sense.
my try:
because: $\ln(1+x)=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n+1}}{n+1}$
$$\ln(1+x^{3})=x^{3}-\frac{x^{6}}{2}+\frac{x^{9}}{3}+x^{9}\varepsilon_{1}(x^{3}) = x^{3}+x^{5}\varepsilon_{2}(x)$$
$$\ln(1-x^{2})=-x^{2}-\frac{x^{4}}{2}-\frac{x^{6}}{6}-x^{6}\varepsilon_{1}(-x^{2}) = -x^{2}-\frac{x^{4}}{2}+x^{5}\varepsilon_{3}(x)$$
$f(x)=x\ln(1+x^{3})(1-x^{2})=x(x^{3}+x^{5}\varepsilon_{2}(x))(-x^{2}-\frac{x^{4}}{2}+x^{5}\varepsilon_{3}(x))$
$=(x^{4}+x^{6}\varepsilon_{2}(x))(-x^{2}-\frac{x^{4}}{2}+x^{5}\varepsilon_{3}(x))=-x^{6}+x^{6}\varepsilon_{4}(x)\Longrightarrow P_{6,0}(x)=-x^{6}$
But this doesn't make sense at all, I guess I wrote nonsense :(
Would appreciate any help :)
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after writing out the first terms of the maclaurin expansion of $\ln(1+x^3)$ and $\ln(1-x^2)$ , observe that the only way to get an $x^6$ term is to multiply $x$ and the 2 first terms $x^3$ and $-x^2$. thus the order 6 maclaurin polynomial has only one term $-x^6$.
the maclaurin expansion does not have an $x^7$ term as there are no positive integer solutions $(a,b)$ for the eqn $1+2a+3b=7$.
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|
Find the area of the CEN triangular region For reference: If ABCD is a square, AB = DE.
Calculate the area of the CEN triangular region.(Answer:$6m^2$)
My progress:
$S_{\triangle MCE}=X=\frac{MC⋅DC}{2}=\frac{MC⋅ℓ}{2}(I)\\
I+S_{\triangle MBA}=\frac{l^2}{2}\\
\therefore X+5 = \frac{l^2}{2}\implies 9 + S_{\triangle CEN} = \frac{l^2}{2}\\
\frac{4}{MN}=\frac{S_{\triangleÇNE}}{EN}=\frac{CN.DE}{2EN}\\
\frac{5}{BM}=\frac{4}{MC}\\
S_{ABCE} = \frac{3AB^2}{2}$
...?
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You can use a system of $4$ equations :
$\begin{cases} \frac{l \cdot BM}{2} = 5 \\ \frac{(l -BM)\cdot CN}{2} = 4 \\ \frac{(l -BM)\cdot l}{2} = A_{CNE}+4 \\ \frac{l\cdot CN}{2} = A_{CNE} \end{cases} $
and solve for $A_{CNE}$.
|
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A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain a $5$ and $Y$ be the number of rolls needed to obtain as $6$. A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain a $5$ and $Y$ be the number of rolls needed to obtain as $6$.
Calculate $E(X|Y=2)$.
My attempt
As $X$ is the first time a 5 is rolled among independent trials, $X$ follows a geometric distribution with $p=\frac{1}{6}$.
$Y=2 \Rightarrow$ the first roll was not a $6$ and the second roll was a $6$.
This means a $5$ was obtained for the first time either on the first roll with $p=\frac{1}{5}$ or on at least the third roll with $p=1-\frac{1}{5}=\frac{4}{5}$.
$P(X=1|Y=2)=\frac{1}{5}, P(X=2|Y=2)=0, P(X=3|Y=2)=\frac{4}{5} \cdot 1 \cdot \frac{1}{6}, P(X=4|Y=2)=\frac{4}{5} \cdot 1 \cdot \frac{5}{6} \cdot \frac{1}{6}$
$E(X|Y=2)=(1/5)+(3)(4/5)(1/6)+4(4/5)(5/6)(1/6)+...= (1/5)+(4/5)\sum_{x=3}^{\infty}x \cdot (\frac{5}{6})^{x-3} \cdot (1/6)$
By letting $k=x-3$, I see that this sum equals $0.2(1)+0.8(8)=6.6$
One solution to this problem has the following: $E(X|X \ge 3)= \frac{1}{p}+2=6+2=8$. Thus $E(X|Y=2)=0.2(1)+0.8(8)=6.6$. But why is $E(X|X \ge 3)=8$?
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$\mathbb E(X\mid X\geq3)$ is the expectation of the number of rolls needed to obtain a $5$ under the condition that the rolls $1$ and $2$ do not provide a $5$.
We might say that the sequence of relevant rolls actually starts after $2$ rolls that are in vain, so that:$$\mathbb E(X\mid X\geq3)=\mathbb E(2+X)=2+\mathbb EX$$
Here $\mathbb EX=6$ since $X$ has geometric distribution (number of trials needed) with parameter $p=\frac16$.
This together explains that: $$\mathbb E(X\mid X\geq3)=2+6=8$$
|
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The Frobenius norm is submultiplicative
Let $M_{d \times d} (\Bbb R)$ be the space of real $d \times d$ matrices. For $A = (a_{ij} ) \in M^{d×d}$, define $$ \| A \|_2 := \left( {\sum_{i,j=1}^{d}|a_{ij}|^2} \right)^\frac{1}{2} $$ Show that $$\|AB\|_2 \leq \|A\|_2 \|B\|_2$$
I don't know where to start this step. How do I show that $I+A+A^2+\cdots$ converges to a limit if $\|A\|_2 \leq 1$? If anyone could please help me, I would appreciate it.
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In my humble opinion brute-force calculation should suffice.
Denote $a_{ij}$ to be the element on the $i$-th row and $j$-th column of matrix $\mathbf{A}$; similar for $b_{ij}$. Then
$$\mathbf{A B} =
\begin{bmatrix}
\sum a_{1j} b_{j1} & \sum a_{1j} b_{j2} & \cdots & \sum a_{1j} b_{jd} \\
\sum a_{2j} b_{j1} & \sum a_{2j} b_{j2} & \cdots & \sum a_{2j} b_{jd} \\
\vdots & \vdots & \ddots & \vdots \\
\sum a_{dj} b_{j1} & \sum a_{dj} b_{j2} & \cdots & \sum a_{dj} b_{jd}
\end{bmatrix} \, .
$$
Then
$$
\begin{split}
\left\Vert \mathbf{A B} \right\Vert_2^2 & = \sum_i \sum_k \left\vert \sum_j a_{ij} b_{jk} \right\vert ^ 2 \\
& \leq \sum_i \sum_k \left[ \left( \sum_j \left\vert a_{ij} \right\vert ^ 2 \right) \left( \sum_j \left\vert b_{jk} \right\vert ^ 2 \right) \right] \ (*)\\
& = \sum_i \sum_k \left[ \left( \sum_j \left\vert a_{ij} \right\vert ^ 2 \right) \left( \sum_{\color{red}{l}} \left\vert b_{\color{red}{l} k} \right\vert ^ 2 \right) \right] \\
& = \sum_i \sum_k \sum_j \sum_l \left\vert a_{ij} \right\vert ^ 2 \left\vert b_{lk} \right\vert ^ 2 \\
& = \left( \sum_i \sum_j \left\vert a_{ij} \right\vert ^ 2 \right) \left( \sum_k \sum_l \left\vert b_{lk} \right\vert ^ 2 \right) \\
& = \left\Vert \mathbf{A} \right\Vert_2^2 \left\Vert \mathbf{B} \right\Vert_2^2 \, .
\end{split}
$$
$\left(*\right)$ follows from Cauchy-Schawarz inequality. So there is $$ \left\Vert \mathbf{A B} \right\Vert_2 \leq \left\Vert \mathbf{A} \right\Vert_2 \left\Vert \mathbf{B} \right\Vert_2 \, . $$
To show that the equality sign can hold, just take this example:
\begin{align}
\mathbf{A} & = \begin{bmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{bmatrix} \, ,
\quad \mathbf{B} = \begin{bmatrix}
2 & 2 & 2 \\
2 & 2 & 2 \\
2 & 2 & 2
\end{bmatrix} \\
\mathbf{A B} & = \begin{bmatrix}
6 & 6 & 6 \\
6 & 6 & 6 \\
6 & 6 & 6
\end{bmatrix} \, .
\end{align}
Apparently $\left\Vert \mathbf{A} \right\Vert_2 = 3$, $\left\Vert \mathbf{B} \right\Vert_2 = 6$, $\left\Vert \mathbf{A B} \right\Vert_2 = 18 = \left\Vert \mathbf{A} \right\Vert_2 \left\Vert \mathbf{B} \right\Vert_2$.
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How to determine the basis with given vectors? So I tried to solve this problem, but I can't seem to find an answer.
So the question is: "Using determinants, indicate the sets of values of $x$, $y$, and $z$ for which the vector sequence is a basis.
$$
S = ((0, z, -y), (-z, 0, x), (y, -x, 0))
$$
I tried to put these three vectors in columns to calculate the determinant and it gives me $0$. But I did something wrong because the determinant can't be $0$, otherwise, this couldn't be a basis. Can someone help me, please?
|
Have you considered the possibility that you have not made a mistake?
It seems like you already know that if the determinant of the matrix $\begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix}$ is $0,$ then the column vectors $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$, $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix},$ and $\begin{pmatrix} y \\ -x \\ 0 \end{pmatrix}$ are not linearly independent, hence cannot be the elements of a basis.
You have also calculated this determinant, and found that it was $0$. So, you should conclude that for any $x, y, z$, the vectors $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$, $\begin{pmatrix} -z \\ 0 \\ x \end{pmatrix},$ and $\begin{pmatrix} y \\ -x \\ 0 \end{pmatrix}$ are always linearly dependent, hence they never form a basis.
Indeed, we can show this directly. First, suppose that $z = 0$. Then, $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -y \end{pmatrix}$ and $\begin{pmatrix} -z \\ 0 \\ x \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ x \end{pmatrix}$ are clearly linearly dependent, hence the vectors $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$, $\begin{pmatrix} -z \\ 0 \\ x \end{pmatrix},$ and $\begin{pmatrix} y \\ -x \\ 0 \end{pmatrix}$ do not form a basis.
Now, suppose $z \neq 0.$ Then, $-xz^{-1}$ and $-yz^{-1}$ are well-defined scalars. Note that we have
$$ (-xz^{-1})\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix} + (-yz^{-1})\begin{pmatrix} -z \\ 0 \\ x \end{pmatrix} = \begin{pmatrix} 0 \\ -x \\ xyz^{-1} \end{pmatrix} + \begin{pmatrix} y \\ 0 \\ -xyz^{-1} \end{pmatrix} = \begin{pmatrix} y \\ -x \\ 0 \end{pmatrix}.$$ So, the vectors $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$, $\begin{pmatrix} -z \\ 0 \\ x \end{pmatrix},$ and $\begin{pmatrix} y \\ -x \\ 0 \end{pmatrix}$ are linearly dependent, hence do not form a basis.
Having considered both possibilities for $z$, we've shown that regardless of what $x, y, z$ are, the three vectors do not form a basis. So, your calculation was correct.
|
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Find the area of the quadrangular region $EBCD$ For reference:
Calculate the area of the quadrangular region $EBCD$, if $BC = 5$, and $AD = AC$. $P$ is a tangent point. (Answer: $64$)
My progress:
$\triangle OPD \sim \triangle ADC$
$AD=2OD \implies k=1:2$
Therefore $P$ is the midpoint $CD$ and $AP$ is perpendicular bisector of $CD$.
Quadrilateral $ABCP$ is cyclic $(\because\angle ABC=\angle APC = 90^\circ).$
$\implies \angle PAC=\angle CBP= 26.5^\circ$
$AP$ is angle bisector.
$\triangle ACD \implies DAC = 53^\circ$
$\therefore \angle ADC= 63.5^\circ$
$AC \parallel OP$
$BP$ is tangent to the circumference at $P$.
$\implies AC \perp BP$
$\therefore CAB = 26.5^\circ$
$BC \parallel ED (\perp AB)$
$\triangle ABC(\text{right}): (26.5^\circ, 63.5^\circ, 90^\circ) \implies(k, 2k, k\sqrt5)$
$\therefore k = 10\sqrt5 \implies AB = 2k = 10\\ BF = 2\sqrt5, BC = 5=CP=DP$
$AD = 5\sqrt5$
Any hint to finish???.....
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From the relations you've discovered, we find $AC=5\sqrt5$ and so is $AD$.
Considering the angle $26.5^\circ$ is approximately equal to the angle we get when bisecting the second largest angle in a $3:4:5$ right triangle, we have $\tan26.5^\circ\approx\frac12$. (See also)
Therefore we can easily get $$\sin(3\cdot26.5^\circ)\approx\frac{11}{5\sqrt5}.$$ Also $$\cos(3\cdot 26.5^\circ)\approx\frac2{5\sqrt5}.$$
Now we see $\triangle AED$'s perpendicular sides are in the ratio $2:11$.
Therefore $AE=2$ and $ED=11$.
Hence the area of trapezium, $$[BCDE]=\frac{(5+11)}2\cdot8=64.$$
|
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If $n$ is a positive integer, $a$, $b$, and $\sqrt[n]{a}-\sqrt[n]{b}$ are rational, then each of $\sqrt[n]{a}$ and $\sqrt[n]{b}$ must be rational. Assume that $a$ and $b$ are rational numbers with $a\neq b$ such that $\sqrt{a}-\sqrt{b}$ is rational. The each of $\sqrt{a}$ and $\sqrt{b}$ must be rational. Because if $\sqrt{a}-\sqrt{b}$ is rational so it is its reciprocal which is $$\frac{\sqrt{a}+\sqrt{b}}{a-b}.$$ This implies that $\sqrt{a}+\sqrt{b}$ must be rational as must $2{\sqrt a}$ and $2{\sqrt b}$, and hence each of $\sqrt{a}$ and $\sqrt{b}$ must be rational.
I am trying to prove the following assertion:
Assume that $a$ and $b$ are rational numbers such that $\sqrt[3]{a}-\sqrt[3]{b}$ is rational. The each of $\sqrt[3]{a}$ and $\sqrt[3]{b}$ must be rational.
More generally, if $a$ and $b$ are rational numbers such that $\sqrt[n]{a}-\sqrt[n]{b}$ is rational for any positive integer $n$, then each of $\sqrt[n]{a}$ and $\sqrt[n]{b}$ must be rational.
For the case $n=3$, I was able to conclude that since $$a-b =(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a^2}+\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b^2})$$ is rational, it must be the case that $$\sqrt[3]{a^2}+\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b^2}$$ must be rational. Not sure where to go from here!
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Easiest is to say the following. Assume $x^k-a$ and $x^l-b$ are irreducible over $\mathbb{Q}$ then $\sqrt[k]{a}-\sqrt[l]{b}$ is irrational, for otherwise if $k\leq l$,
$$a=(\sqrt[l]{b}+r)^k$$ is an $k$th degree equation for $\sqrt[l]{b}$, or $k-1$ degree if they are equal.
Note that if $$x^n-a=f(x)g(x)$$ is reducible with the degree of $f$ being $k$ then the roots of $f$ are of the form $\epsilon \sqrt[n]{a}$ and their product, the constant coefficient of the form $\eta(\sqrt[n]{a})^k\in \mathbb{Q}$ which implies that $(\sqrt[n]{a})^k\in \mathbb{Q}$ and so $a=q^{n/k}$. This is one case of a much more general theorem proved by other means.
|
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Prove $\int_0^1 \frac{\ln x\ +\ \ln(\sqrt x\ +\sqrt {1+x})}{\sqrt {1-x^2}} dx=0$ If a simple way exists, I am looking to show that
$$\boxed{K=\int_0^1 \frac{\ln(x)+\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}} dx=0}$$
say, with symmetry, a clever change of variables, or integrations by parts, without evaluating integrals separately. It is similar to @Zacky question, of proving $$\boxed{\int_0^\frac{\pi}{2}\left(\frac{\pi}{3}-x\right)\frac{\ln(1-\sin x)}{\sin x}dx=0}$$
without calculating separately integrals.
If we take separately
$$I=\int_0^1 \frac{\ln(x)}{\sqrt {1-x^2}}dx,\>\>\>\>J=\int_0^1 \frac{\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}}dx$$the integrals $I$ and $J$ define the same series to a sign (two series of opposite sums)
$$ I=-\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}},\>\>\>\>\>J=\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}}$$
Explanation By Fourier series $\ln\left(\sqrt{1+\sin x}+\sqrt{\sin x}\right)=\sum_{k=0}^\infty\frac{(2k)!}{4^k(2k+1)(k!)^2}\sin((2k+1)x)$ then $J=\int_0^{\frac{\pi}{2}} \ln\left(\sqrt{\sin t}+\sqrt{1+\sin t}\right)dt=\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}}$
$ I=\int_{0}^{1}\frac{\log(t)}{\sqrt{1-t^{2}}}dt$ We know that $\frac{1}{\sqrt{1-t^{2}}}=\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}}t^{2n} $ and $ \int_{0}^{1}\log(t)t^{2n}dt=-\frac{1}{(2n+1)^{2}}$ then $I=-\sum_{n=0}^{\infty}\frac{{2n \choose n}}{4^{n}(2n+1)^{2}} $
We can write K as
$$K=\int_0^{\dfrac{\pi}{2}} \Big(\ln(\sin t )+\ln\left(\sqrt{\sin t}+\sqrt{1+\sin t}\right)\Big)dt=0 $$
Remark : Wolframalpha can calculate K, but do not know how to calculate $$\int_0^1 \frac{\ln(\sqrt x+\sqrt {1+x})}{\sqrt {1-x^2}} dx$$
I believe that Wolframe uses a simple way to see that $K$ is zero . same observation for the Zacky’s integral
|
I know this is not exactly what it is asked for.
On the path of Zacky,
\begin{align*}
J&=\int_0^1\frac{\text{arcsinh}\left(\sqrt{x}\right)}{\sqrt{1-x^2}}dx\\
&\overset{\text{IBP}}=\Big[\arcsin (x)\text{arcsinh}\left(\sqrt{x}\right)\Big]_0^1-\frac{1}{2}\underbrace{\int_0^1 \frac{\arcsin x}{\sqrt{x}\sqrt{1+x}}dx}_{_{z=\sqrt{\frac{1-x}{1+x}}}}\\
&=\frac{\pi}{2}\ln\left(1+\sqrt{2}\right)-\sqrt{2}\int_0^1 \frac{z\arcsin\left(\frac{1-z^2}{1+z^2}\right)}{(1+z^2)\sqrt{1-z^2}}dz\\
&=\frac{\pi}{2}\ln\left(1+\sqrt{2}\right)-\sqrt{2}\int_0^1 \frac{z\left(\frac{\pi}{2}-2\arctan z\right)}{(1+z^2)\sqrt{1-z^2}}dz\\
&=\frac{\pi}{2}\left(\ln\left(1+\sqrt{2}\right)-\sqrt{2}\underbrace{\int_0^1 \frac{z}{(1+z^2)\sqrt{1-z^2}}dz}_{=\text{K}}\right)+2\sqrt{2}\int_0^1 \frac{z\arctan z}{(1+z^2)\sqrt{1-z^2}}dz\\
K&=\frac{1}{2\sqrt{2}}\left[\ln\left(\frac{\sqrt{2}-\sqrt{1-z^2}}{\sqrt{2}+\sqrt{1-z^2}}\right)\right]_0^1=\frac{1}{\sqrt{2}}\ln\left(1+\sqrt{2}\right)\\\
J&=2\sqrt{2}\int_0^1 \frac{z\arctan z}{(1+z^2)\sqrt{1-z^2}}dz
\end{align*}
Define on $[0,1]$,
\begin{align*}F(a)&=\int_0^1 \frac{z\arctan (az)}{(1+z^2)\sqrt{1-z^2}}dz\\
F^\prime(a)&=\int_0^1 \frac{z^2}{(1+z^2)(1+a^2z^2)\sqrt{1-z^2}}dz\\
&=-\frac{1}{2}\left[\frac{\sqrt{2}\arctan\left(\frac{z\sqrt{2}}{\sqrt{1-z^2}}\right)-\frac{2}{\sqrt{1+a^2}}\arctan\left(\frac{z\sqrt{1+a^2}}{\sqrt{1-z^2}}\right)}{1-a^2}\right]_{z=0}^{z=1}\\
&=\frac{\pi\left(\sqrt{\frac{2}{1+a^2}}-1\right)}{2(1-a^2)\sqrt{2}}
\end{align*}
Since $F(0)=0$ then,
\begin{align*}\int_0^1 \frac{z\arctan z}{(1+z^2)\sqrt{1-z^2}}dz&=F(1)-F(0)\\
&=\frac{\pi}{2\sqrt{2}}\int_0^1\frac{\left(\sqrt{\frac{2}{1+a^2}}-1\right)}{1-a^2}da\\
&=\frac{\pi}{2\sqrt{2}}\left[\text{arctanh}\left(\frac{\sqrt{2}a}{\sqrt{1+a^2}}\right)-\text{arctanh}(a)\right]_0^1\\
&=\frac{\pi}{2\sqrt{2}}\lim_{a\rightarrow 1}\ln\left(\frac{\sqrt{1+a^2}+a\sqrt{2}}{1+a}\right)\\
&=\frac{\pi\ln 2}{4\sqrt{2}}
\end{align*}
Therefore,
\begin{align*}\boxed{J=\frac{\pi\ln 2}{2}}\end{align*}
|
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"url": "https://math.stackexchange.com/questions/4357422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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"answer_id": 0
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|
Algebraic proof that $\sum\limits_{k=0}^{n} {n \choose k}\cdot \frac{(-1)^k}{(k+1)(k+2)} = \frac{1}{n+2}$ I tried evaluating the integral $\int\limits_{[0,1]^n} \min(x^1,x^2,\ldots,x^n) \lvert d^nx\rvert$.
In my first attempt, I used a recursive approach and managed to defined the integral as being $I_n^k = \int\limits_{[0,1]^n} \min(x^1,x^2,\ldots,x^n)^k \lvert d^nx\rvert$, where I could evaluate
$I_n^k = I_{n-1}^ k - \frac{k}{k+1} \cdot I_{n-1}^{k+1}$, and
$I_1^k = \frac{1}{k+1}$.
Using this recursive approach, I managed to extract the
$I_n^1 = \sum\limits_{k=0}^{n-1} {n-1 \choose k}\cdot \frac{(-1)^k}{(k+1)(k+2)}$
I plugged numbers into this sum for multiple values of $n$, and saw that I constantly get $\frac{1}{n+1}$.
During my second attempt, I managed to eventually find the integral by splitting the area into $n!$ areas in which there exists some order for each element $x_0$ such that
$x_0^{i_1} \leq x_0^{i_2} \leq \ldots \leq x_0^{i_n}$, and proved that the integral over each of these area is $\frac{1}{(n+1)!}$ which gave me showed me more definitively that the integral is equal to $\frac{1}{n+1}$.
That said, after trying for a while, I couldn't come up with any combinatorial/algebraic proof that the sum I found is indeed $\frac{1}{n+1}$. I tried evaluating it as a telescoping sum, giving me the expression
$\sum_{k=0}^{\frac{n}{2}}{n \choose 2k}\cdot\frac{\left(4k+3-n\right)}{\left(2k+3\right)\left(2k+2\right)\left(2k+1\right)}$, but couldn't expand this sum to anything useful either. I haven't worked much with sums of this form and was wondering whether I'm missing something that can help me show this without the integral.
|
This answer comes a bit late but I thought it is worth mentioning it because it shows a simple algebraic proof of the formula in question.
First note the following two facts:
*
*(1): $\frac 1{(k+1)(k+2)}\binom nk = \frac 1{(n+1)(n+2)}\binom{n+2}{k+2}$, Indeed
\begin{eqnarray*} \frac 1{(k+1)(k+2)}\binom nk
& = & \frac{n!}{k!(k+1)(k+2)(n-k)!} \\
& = & \frac 1{(n+1)(n+2)}\frac{(n+2)!}{(k+2)!((n+2)-(k+2))!} \\
& = & \frac 1{(n+1)(n+2)}\binom{n+2}{k+2}
\end{eqnarray*}
*(2): According to the binomial formula, we have
$$0 = (1-1)^{n+2} = \sum_{k=0}^{n+2}(-1)^k\binom{n+2}{k}$$
Now, we have
\begin{eqnarray*} \sum\limits_{k=0}^{n} {n \choose k}\cdot \frac{(-1)^k}{(k+1)(k+2)}
& \stackrel{(1)}{=} & \frac 1{(n+1)(n+2)} \sum\limits_{\color{blue}{k=0}}^{n} (-1)^k \cdot \binom{n+2}{k+2}\\
& \stackrel{(-1)^{k+2}=(-1)^k}{=} & \frac 1{(n+1)(n+2)} \sum\limits_{\color{blue}{k=2}}^{n+2} (-1)^k \cdot \binom{n+2}{k} \\
& \stackrel{(2)}{=} & \frac 1{(n+1)(n+2)}\left(0 - \binom{n+2}{0} + \binom{n+2}{1}\right) \\
& = & \frac{n+1}{(n+1)(n+2)} = \frac 1{n+2}
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4360564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
}
|
Integral $\int_0^{\pi/3}\frac{x}{\cos(x)}\,dx$ I am trying to compute the integral
$$\int_0^{\pi/3}\frac{x}{\cos(x)}\,dx \tag{1}$$
Context: Originally I was trying to prove the following result:
$$\sum_{n=0}^\infty\frac{1}{(2n+1)^2\binom{2n}{n}}=\frac83\beta(2)-\frac{\pi}3\ln(2+\sqrt{3})\tag{2}$$
Where $\beta(2)$ is the Catalan´s constant
To this end I started with the well known result
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^\infty\frac{4^n x^{2n+1}}{(2n+1)\binom{2n}{n}} \tag{3}$$
Dividing both sides of $(3)$ by $x$ and integrating from $0$ to $1/2$ we obtain
$$\int_0^{1/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx=\frac12\sum_{n=0}^\infty\frac{ 1}{(2n+1)^2\binom{2n}{n}} \tag{4}$$
So the task reduces to compute the integral in $(4)$. Therefore
$$
\begin{aligned}
\sum_{n=0}^\infty\frac{ 1}{(2n+1)^2\binom{2n}{n}}&=2\int_0^{1/2}\frac{\arcsin(x)}{x\sqrt{1-x^2}}\,dx\\
&=2\int_0^{\pi/6}\frac{x}{\sin(x)}\,dx &(x \to \sin(x))\\
&=2\int_0^{\pi/2}\frac{x}{\sin(x)}\,dx-2\int_{\pi/6}^{\pi/2}\frac{x}{\sin(x)}\,dx\\
&=4\beta(2)-2\int_{\pi/6}^{\pi/2}\frac{x}{\sin(x)}\,dx\\
&=4\beta(2)-2\int_{0}^{\pi/3}\frac{\left(\frac{\pi}{2}-x\right)}{\cos(x)}\,dx & (x \to \frac{\pi}{2}-x)\\
&=4\beta(2)-\pi\int_{0}^{\pi/3}\sec(x)\,dx+2\int_{0}^{\pi/3}\frac{x}{\cos(x)}\,dx\\
&=4\beta(2)-\pi\ln\left(\sec(x)+\tan(x) \right)\Big|_0^{\pi/3}+2\int_{0}^{\pi/3}\frac{x}{\cos(x)}\,dx\\
&=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+2\int_{0}^{\pi/3}\frac{x}{\cos(x)}\,dx\\
&=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\int_{0}^{\pi/3}\frac{x}{e^{ix}+e^{-ix}}\,dx\\
&=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\int_{0}^{\pi/3}\frac{xe^{-ix}}{1+e^{-2ix}}\,dx\\
&=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\int_{0}^{\pi/3}xe^{-ix}\sum_{k=0}^\infty(-1)^ke^{-2ikx}\,dx\\
&=4\beta(2)-\pi\ln\left(2+\sqrt{3}\right)+4\sum_{k=0}^\infty(-1)^k\int_{0}^{\pi/3}xe^{-ix(2k+1)}\,dx\\
\end{aligned}
$$
The integral in the last line is $(1)$. I integrated by parts, but ended up with some nasty series not very promising.
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Using integration by parts:
\begin{align*}
\int{\frac{x}{\cos(x)}dx} &= \int{x\sec(x)\ dx} \\
&= x\sec(x)\tan(x) - \int{\sec(x)\tan(x)dx} \\
&= x\sec(x)\tan(x) - \int{\cos(x)\sec^2(x)\tan(x)dx} && ...multiply \ by \cos(x)\sec(x) = 1 \\
&= x\sec(x)\tan(x) - [\tan(x)\cos(x) + \int{\sin(x)\tan(x)dx}]\\
&= x\sec(x)\tan(x) - [\tan(x)\cos(x) + \int{\frac{\sin^2(x)}{\cos(x)}dx}] \\
&= x\sec(x)\tan(x) - [\tan(x)\cos(x) + \int{\frac{1-\cos^2(x)}{\cos(x)}dx}] \\
&= x\sec(x)\tan(x) - [\tan(x)\cos(x) + \int{[\sec(x)-\cos(x)]dx }] \\
&= x\sec(x)\tan(x) - [\tan(x)\cos(x) + [\sec(x)\tan(x)-\sin(x)]] \\
\end{align*}
After puttting limits, we get answer = $\frac{2\pi}{\sqrt{3}} - [\frac{\sqrt{3}}{2}+[2\sqrt{3} - \frac{\sqrt{3}}{2}]]$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4360729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ improve $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.
Here is my initial question:
Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ lead to an improvement to the upper bound $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$?
MOTIVATION
Here is a proof that $I(n^2) \leq 2 - \frac{5}{3q}$ holds in general.
Suppose to the contrary that $I(n^2) > 2 - \frac{5}{3q}$ is true.
Note that
$$I(n^2) = \frac{2}{I(q^k)} = \frac{2q^k (q - 1)}{q^{k+1} - 1} = 2 - 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg),$$
so that we have
$$I(n^2) > 2 - \frac{5}{3q} \iff 2 - 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg) > 2 - \frac{5}{3q} \iff \frac{5}{3q} > 2\cdot\Bigg(\frac{q^k - 1}{q^{k+1}-1}\Bigg) \iff 5q^{k+1} - 5 > 6q^{k+1} - 6q \iff 0 > q^{k+1} - 6q + 5,$$
which then implies that $k=1$. (Otherwise, if $k > 1$, we have
$$0 > q^{k+1} - 6q + 5 \geq q^6 - 6q + 5,$$
since $k \equiv 1 \pmod 4$, contradicting $q \geq 5$.) Now, since $k=1$, we get
$$0 > q^2 - 6q + 5 = (q - 1)(q - 5),$$
which implies that $1 < q < 5$. This contradicts $q \geq 5$. This concludes the proof.
Now, let $Q = 2 - \frac{5}{3q}$.
Since $I(q^k) < I(n^2)$, then we obtain
$$I(q^k) < I(n^2) \leq Q \iff (I(q^k) - Q)(I(n^2) - Q) \geq 0$$
$$\iff 2 + Q^2 = I(q^k)I(n^2) + Q^2 \geq Q(I(q^k) + I(n^2) \iff I(q^k) + I(n^2) \leq \frac{2}{Q} + Q.$$
But $\frac{2}{Q} + Q$ can be rewritten as
$$\dfrac{2}{Q} + Q = \dfrac{2}{2 - \dfrac{5}{3q}} + \Bigg(2 - \dfrac{5}{3q}\Bigg) = 3 - \Bigg(\dfrac{5}{3q} - \dfrac{5}{6q-5}\Bigg) = 3 - \dfrac{5(3q - 5)}{3q(6q - 5)} = \dfrac{54q^2 - 60q + 25}{3q(6q - 5)}.$$
Let
$$f(q) = \dfrac{54q^2 - 60q + 25}{3q(6q - 5)}.$$
Then the derivative
$$f'(q) = \dfrac{5}{3q^2} - \dfrac{30}{(6q - 5)^2}$$
is positive for $q \geq 5$. This means that $f$ is an increasing function of $q$, which implies that
$$I(q^k) + I(n^2) \leq \dfrac{2}{Q} + Q < \lim_{q \rightarrow \infty}{f(q)} = 3.$$
FINAL QUESTION
Can we do better? If that is not possible, can you explain why?
|
In this post, I will attempt to improve on the upper bound
$$I(n^2) \leq \dfrac{2q}{q+1}.$$
(Although this attempt is unsuccessful, I am retaining this answer here, mainly for my own benefit.)
Following mathlove's lead in the comments, I tried to find $a, b, c, d, e, f \in \mathbb{R}$ such that
$$2 - 2\cdot\Bigg(\dfrac{q^k - 1}{q^{k+1} - 1}\Bigg) = \frac{2q^k (q - 1)}{q^{k+1} - 1} \leq 2 - \Bigg(\dfrac{dq^2 + eq + f}{q^3 + aq^2 + bq + c}\Bigg) \leq 2 - \dfrac{2}{q+1} = \dfrac{2q}{q+1}.$$
This was very messy. But these inequalities are equivalent
to
$$\dfrac{2}{q+1} \leq \dfrac{dq^2 + eq + f}{q^3 + aq^2 + bq + c} \leq 2\cdot\Bigg(\dfrac{q^k - 1}{q^{k+1} - 1}\Bigg),$$
and I set
$$a = b = e = f = 0,$$
and then noticed that
$$(c, d) = (25, 2)$$
seems to work.
PROOF
Let
$$S = 2 - \dfrac{2q^2}{q^3 + 25}.$$
We compute
$$S - I(n^2) = 2\cdot\Bigg(\dfrac{q^k - 1}{q^{k+1} - 1}\Bigg) - \dfrac{2q^2}{q^3 + 25} = \dfrac{2(25q^k - q^3 + q^2 - 25)}{(q^3 + 25)(q^{k+1} - 1)}.$$
Set
$$f(k) := \dfrac{2(25q^k - q^3 + q^2 - 25)}{(q^3 + 25)(q^{k+1} - 1)}.$$
Differentiating once with respect to $k$, we obtain
$$f'(k) = \dfrac{2(q - 1){q^k}\log(q)}{(q^{k+1} - 1)^2},$$
which is positive. This means that $f(k)$ is an increasing function of $k$, which implies that
$$f(k) \geq f(1) = -\dfrac{2(q^2 - 25)}{(q + 1)(q^3 + 25)}.$$
This implies that
$$I(n^2) \leq S + \dfrac{2(q^2 - 25)}{(q + 1)(q^3 + 25)} = \Bigg(2 - \dfrac{2q^2}{q^3 + 25}\Bigg) + \dfrac{2(q^2 - 25)}{(q + 1)(q^3 + 25)} = \dfrac{2q}{q+1},$$
which seems to show, at least, for these particular values for
$$(a, b, c, d, e, f) = (0, 0, 25, 2, 0, 0)$$
that
$$I(n^2) \leq \dfrac{2q}{q+1}$$
is best-possible.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods. Then I learnt a very useful trick to evaluate the integral.
Noting that
$$I(1):= \int_{0}^{\infty} \frac{d x}{x^{4}-x^{2}+1} \stackrel{x \mapsto \frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{2}}{x^{4}-x^{2}+1}
$$
Combining them yields
\begin{aligned}
I(1)&=\frac{1}{2} \int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x\\&= \frac{1}{2}\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&= \frac{1}{2}\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1} \\
&= \frac{1}{2}\tan ^{-1}\left(x-\frac{1}{x}\right)_{0}^{\infty} \\
&= \frac{\pi}{2}
\end{aligned}
Later, I started to investigate the integrands with higher powers.
Similarly,
$$
I(2):= \int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{2}} \stackrel{x \mapsto \frac{1}{x}}{=}\int_{0}^{\infty} \frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}} d x
$$
By division, we decomposed $x^6$ and obtain $$
\frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}}=\frac{x^{2}+1}{x^{4}-x^{2}+1}-\frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}
$$$$
I(2)=\int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x-\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}dx
$$
We can now conclude that $$I(2)=I(1)=\frac{\pi}{2} $$
My Question:
How about the integral $$\displaystyle I_{n}=\int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{n}}$$ for any integer $n\geq 3$?
|
Glad to see there are several nice answers in various perspectives. I am now going to give one more by inverse substitution followed by integration by parts.
Using $x \mapsto \frac{1}{x}$ yields
$$
I(m, n, r):=\int_{0}^{\infty} \frac{x^{r} d x}{\left(x^{m}+1\right)^{n}} \stackrel{x \mapsto \frac{1}{x}}{=} \int_{0}^{\infty} \frac{\frac{1}{x^{r}}}{\left(\frac{1}{x^{m}}+1\right)^{n}}\left(\frac{d x}{x^{2}}\right)
$$
Simplifying and then performing integration by parts,
$$
\begin{aligned}
I(m, n, r) &=\int_{0}^{\infty} \frac{x^{m n-r-2}}{\left(1+x^{m}\right)^{n}} d x \\
&=-\frac{1}{m(n-1)} \int_{0}^{\infty} x^{(n-1)m-r-1} d\left(\frac{1}{\left(1+x^{m}\right)^{n-1}}\right) \\
&=\frac{m(n-1)-r-1}{m(n-1)} \int_{0}^{\infty} \frac{x^{m(n-1)-r-2}}{\left(1+x^{m}\right)^{n-1}} d x \\
&=\left(1-\frac{r+1}{m(n-1)}\right) I(m, n-1, r) \\
&\qquad\qquad \vdots\\ &=\left(1-\frac{r+1}{m(n-1)}\right)\left(1-\frac{r+1}{m(n-2)}\right) \cdots\left(1-\frac{r+1}{m}\right) I\left(m,1,r\right) \\
&=\prod_{j=1}^{n-1}\left(1-\frac{r+1}{j m}\right)\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x
\end{aligned}
$$
Using the formula proved in my post
$$
\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x=\frac{\pi}{m} \csc \frac{(r+1) \pi}{m},
$$
by putting $r=2k$ and $m=6$, we can conclude
$$
\begin{aligned}
\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}} d x
=& \int_{0}^{\infty}\left(\frac{x^{2}+1}{x^{6}+1}\right)^{n} d x \\
=& \sum_{k=0}^{n}\left(\begin{array}{c}
n \\
k
\end{array}\right) \int_{0}^{\infty} \frac{x^{2 k}}{\left(1+x^{6}\right)^{n}} d x \\=& \boxed{\sum_{k=0}^{n}\left[\left(\begin{array}{c}
n \\
k
\end{array}\right) \frac{\pi}{6} \csc \frac{(2 k+1) \pi}{6} \prod_{j=1}^{n-1}\left(1-\frac{2 k+1}{6 j}\right)\right]}
\end{aligned}
$$
|
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"url": "https://math.stackexchange.com/questions/4367988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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|
Approximating factorial using identity $\frac{1}{x}!\frac{2}{x}!\cdots\cdot\frac{x}{x}!=\frac{ {x}!\cdot(2\pi)^{\frac{x-1}{2}} }{ x^x\cdot\sqrt{x} }$ I created a function that describes the product of the inverse multiples of a factorial
$$ m(x) = \frac{1}{x}!\cdot\frac{2}{x}!\cdot\frac{3}{x}!\cdots\frac{x-1}{x}!\cdot\frac{x}{x}!$$
for some reasons i thought this function might be useful, thats why i'm posting an incomplete explanation to the best of my knowledge but after some rigorous calculation, i was able to express $m(x)$ as a formula
$$ m(x) = \frac{ {x}!\cdot(2\pi)^{\frac{x-1}{2}} }{ x^x\cdot\sqrt{x} }$$
it turns out that the graph of $m(x)$ is a very simple one, makes me suggest it would be easy to interpolate and create a super factorial approximation from it
$$ {x}! = x^x\cdot\sqrt{x}\cdot\sqrt{2\pi}\cdot m(x)\cdot\sqrt{{2\pi}^{-x}}$$
and it resembles striling's approximation
$${x}! \approx x^x\cdot\sqrt{x}\cdot\sqrt{2\pi}\cdot e^{-x} $$
$$
m(1)=1, m(2)=\frac{1}{2}!, m(3)=\frac{4\pi}{3^{5/2}}, m(4)=\frac{3\pi^{3/2}}{2^{9/2}}, m(5)=\frac{96\pi^2}{5^{9/2}}, m(6)=\frac{45\pi^{5/2}}{3^{13/2}} , m(7)=\cdots
$$
now my question, can you help with the approximation, does the function gives any more information
|
Here is the proof of how I got $m(x)$
and since we're till speaking on approximation i also want to share this to you (degrees) @claude leibovici
$$ \sin{5} \approx \frac{1}{6+\sqrt{3}+\sqrt{14}}$$
The factorial function above is actually extended to the gamma function, but i am using factorial notation for simplicity $ \Gamma(x) = (x-1)!$ and i was able to write out the $m(x)$ function using help from euler reflextion formula check here
$$ {x}!(-x)! = \frac{\pi x }{\sin{\pi x }}$$
I reduced each term in the factorial of $m(x)$ so that i would always get it's negative face by using ${x}!=x(x-1)!$ and ignore product of unity there
$$
\begin{array}
\\
m(1) = 1\\
m(2) = \frac{1}{2}!\\
m(3) = \frac{1}{3}!\cdot\frac{2}{3}!\\
m(3) = \frac{1}{3}!\cdot\frac{-1}{3}!\cdot\frac{2}{3}\\
m(4) = \frac{1}{4}!\cdot\frac{2}{4}!\cdot\frac{3}{4}!\\
m(4) = \frac{1}{4}!\cdot\frac{-1}{4}!\cdot\frac{2}{4}!\cdot\frac{3}{4}\\
m(5) = \frac{1}{5}!\cdot\frac{2}{5}!\cdot\frac{3}{5}!\cdot\frac{4}{5}!\\
m(5) = \frac{1}{5}!\cdot\frac{-1}{5}!\cdot\frac{2}{5}!\cdot\frac{-2}{5}!\cdot\frac{3}{5}\cdot\frac{4}{5}\\
m(6) = \frac{1}{6}!\cdot\frac{2}{6}!\cdot\frac{3}{6}!\cdot\frac{4}{6}!\cdot\frac{5}{6}!\\
m(6) = \frac{1}{6}!\cdot\frac{-1}{6}!\cdot\frac{2}{6}!\cdot\frac{-2}{6}!\cdot\frac{3}{6}!\cdot\frac{4}{6}\cdot\frac{5}{6}\\
m(x) = \frac{1}{x}!\cdot\frac{2}{x}!\cdot\frac{3}{x}!\cdot\frac{4}{x}\cdots\frac{x-1}{x}!\\
m(x) = \frac{1}{x}!\cdot\frac{-1}{x}!\cdot\frac{2}{x}!\cdot\frac{-2}{x}!\cdot\frac{3}{x}!\cdot\frac{-3}{x}!\cdots\frac{n}{x}!\cdot\frac{-n}{x}!\cdots\cdots\frac{x-n}{x}\cdots\frac{x-3}{x}\cdot\frac{x-2}{x}\cdot\frac{x-1}{x}\cdot f(n)\\
\end{array}
$$
above, $n$ represents a function of the variable $x$ and $f(n)$ is a function of $n$
$$n=
\begin{cases}
\frac{x}{2}-1, &\text{if $x$ is even}\\
\frac{x+1}{2}-1, &\text{if $x$ is odd}
\end{cases}
$$
$$f(n)=
\begin{cases}
1, &\text{if $x$ is odd}\\
\frac{1}{2}!, &\text{if $x$ is even}
\end{cases}
$$
$\frac{1}{2}! = m(2)$ and from the $m(x)$ formula it is directly equal to $\frac{\sqrt{\pi}}{2}$
$m(x)$ is variant, so i'll solve each part of it separately and apply some trigonometry identity to simplify
$$\frac{1}{x}!\cdot\frac{-1}{x}!\cdot\frac{2}{x}!\cdot\frac{-2}{x}!\cdot\frac{3}{x}!\cdot\frac{-3}{x}!\cdots\frac{n}{x}!\cdot\frac{-n}{x}! =
\frac{\frac{\pi}{x} }{\sin{\frac{\pi}{x} }}\cdot\frac{ \frac{2\pi}{x} }{\sin{\frac{2\pi}{x} }}\cdot\frac{\frac{3\pi}{x} }{\sin{\frac{3\pi}{x} }}\cdots\frac{\frac{n\pi}{x} }{\sin{\frac{n\pi}{x} }}$$
$$\frac{1}{x}!\cdot\frac{-1}{x}!\cdot\frac{2}{x}!\cdot\frac{-2}{x}!\cdot\frac{3}{x}!\cdot\frac{-3}{x}!\cdots\frac{n}{x}!\cdot\frac{-n}{x}! = \frac{ {n}!\cdot \pi^{n}}{ x^n \cdot (\sin{\frac{\pi}{x}}\cdot\sin{\frac{2\pi}{x}}\cdot\sin{\frac{3\pi}{x}}\cdots\sin{\frac{n\pi}{x}}) }$$
$$\frac{1}{x}!\cdot\frac{-1}{x}!\cdot\frac{2}{x}!\cdot\frac{-2}{x}!\cdot\frac{3}{x}!\cdot\frac{-3}{x}!\cdots\frac{n}{x}!\cdot\frac{-n}{x}! = \frac{{n}!\cdot \pi^n\cdot \sqrt{2^{x-1}}}{x^n\cdot \sqrt{x}}$$
$$\frac{x-n}{x}\cdots\frac{x-3}{x}\cdot\frac{x-2}{x}\cdot\frac{x-1}{x}\cdot f(n) = { (x-n)\cdots(x-3)(x-2)(x-1) }\cdot \frac{f(n)}{x^n}$$
$${x}! = x(x-1)(x-2)(x-3)\cdots(x-n)\cdot(x-n-1)!$$
$$(x-1)(x-2)(x-3)\cdots(x-n) = \frac{{x}!}{x(x-n-1)!}$$
$$\frac{x-n}{x}\cdots\frac{x-3}{x}\cdot\frac{x-2}{x}\cdot\frac{x-1}{x}\cdot f(n) = \frac{ {x}!f(n)}{x(x-n-1)!x^n}$$
$$m(x) = \frac{{n}!\cdot\pi^n\cdot\sqrt{2^{x-1}}\cdot{x}!\cdot f(n)}{x^{2n}\cdot\sqrt{x}\cdot x\cdot (x-n-1)!}$$
if we simplify further to remove the $n$ and $f(n)$, we would arrive at the formula i wrote in the question
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the vertical asymptotes of this equation, $f(x)=2\tan(4x-32)$? Could someone show how to use the vertical asymptote formula? I am having a hard time getting it into the right form. I thought maybe I had to put $(4x-32)$ equal to the vertical asymptote equation. I can't get the math to work out.
Asymptote formula $$x= \frac CB + \frac{\pi}{ 2 \lvert B \rvert} k $$
where $k$ is an integer. Here is my attempt but clearly it is wrong. $$ 4x-32 = \frac{32}{4} + \frac {\pi}{2 \lvert 4 \rvert}k$$ When I solve the equation, I get: $$ x= 2+\frac{\pi}{32}k +8$$ When I put a $0$ in for $k$, the equation $= 10$. It should $= 8$.
What am I doing wrong?
|
The tangent function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \tan x$ has vertical asymptotes at $x = \dfrac{\pi}{2} + k\pi, k \in \mathbb{Z}$.
We can solve for the vertical asymptotes of $g(x) = A\tan(Bx - C)$, where $B \neq 0$, by setting
$$Bx - C = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$
then solving for $x$. Doing so yields
\begin{align*}
Bx & = C + \frac{\pi}{2} + k\pi, k \in \mathbb{Z}\\[2 mm]
x & = \frac{C}{B} + \frac{\pi}{2B} + \frac{k\pi}{B}, k \in \mathbb{Z}
\end{align*}
which does not agree with the formula you stated.
If we substitute $4$ for $B$ and $32$ for $C$ in the above formula, we obtain the vertical asymptotes
\begin{align*}
x & = \frac{32}{4} + \frac{\pi}{2 \cdot 4} + \frac{k\pi}{4}, k \in \mathbb{Z}\\
& = 8 + \frac{\pi}{8} + \frac{k\pi}{4}, k \in \mathbb{Z}
\end{align*}
If $k = 0$, then we obtain
$$x = 8 + \frac{\pi}{8} \approx 8.392699082$$
which is a vertical asymptote of the graph of $f(x) = 2\tan(4x - 32)$, as you can check with a graphing calculator.
|
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|
Integral involving product of dilogarithm and an exponential I am interested in the integral
\begin{equation}
\int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u , ~~~~ (\ast)
\end{equation}
where $\mathrm{Li}_2$ is the dilogarithm. This integral arose in my attempt to evaluate the double integral
\begin{equation}
I (a,b) = \int_{\mathbb{R}^2} \arctan^2{ \left( \frac{y+b}{x+a} \right) } e^{- (x^2 + y^2) } d x dy ,
\end{equation}
where $a,b \in \mathbb{R}$. See, for example, this question: Interesting $\arctan$ integral. In fact, I am able to show that
\begin{equation}
I (a,0) = \frac{\pi}{2} \left( \frac{\pi^2}{6} e^{-a^2} + a^2 \int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u \right) .
\end{equation}
I would very much like to see the trailing term (i.e., ($\ast$)) evaluated in terms of commonly used special functions. I am have not worked with the dilogarithm very much, so I am not sure what might be the best approach to rewrite this. Does anyone here have any suggestions?
|
This is a partial answer that is too long for a comment.
Let
\begin{align*}
J(a) = \int_0^1 \mathrm{Li}_2 (u) e^{-a^2 u} d u .
\end{align*}
Note that
\begin{align*}
\frac{d }{d u} \left( u \mathrm{Li}_2 (u) - u - (1-u)\ln{(1-u)} \right) = \mathrm{Li}_2 (u) .
\end{align*}
Then, we may integrate-by-parts to find
\begin{align*}
J(a) & = \left( \frac{\pi^2}{6} - 1 \right) e^{- a^2} + a^2 \int_0^1 \left( u \mathrm{Li}_2 (u) - u - (1-u)\ln{(1-u)} \right) e^{-a^2 u} d u \\
& = \left( \frac{\pi^2}{6} - 1 \right) e^{- a^2} - \frac{1 - (1 + a^2) e^{-a^2}}{a^2} + \frac{1}{a^2} \left( 1 + e^{-a^2} \left( \ln{(a^2)} - \mathrm{Ei}{(a^2)} + \gamma - 1 \right) \right) + a^2 \int_0^1 u \mathrm{Li}_2 (u) e^{-a^2 u} d u \\
& = \frac{\pi^2}{6} e^{- a^2} + \frac{e^{-a^2}}{a^2} \left( \ln{(a^2)} - \mathrm{Ei}{(a^2)} + \gamma \right) - \frac{a}{2} J' (a) ,
\end{align*}
where $\gamma$ is the Euler constant and $\mathrm{Ei}$ is the exponential integral. Therefore,
\begin{align*}
J'(a) + \frac{2}{a} J(a) = \frac{2}{a} \left( \frac{\pi^2}{6} e^{- a^2} + \frac{e^{-a^2}}{a^2} \left( \ln{(a^2)} - \mathrm{Ei}{(a^2)} + \gamma \right) \right) ,
\end{align*}
and
\begin{align*}
\frac{d }{d a} ( a^2 J (a) ) = 2 a \left( \frac{\pi^2}{6} e^{- a^2} + \frac{e^{-a^2}}{a^2} \left( \ln{(a^2)} - \mathrm{Ei}{(a^2)} + \gamma \right) \right) .
\end{align*}
Finally, we must integrate
\begin{align*}
J(a) = \frac{2}{a^2} \int_0^a s \left( \frac{\pi^2}{6} e^{- s^2} + \frac{e^{-s^2}}{s^2} \left( \ln{(s^2)} - \mathrm{Ei}{(s^2)} + \gamma \right) \right) d s .
\end{align*}
Simplifying some yields
\begin{align*}
J(a) = \frac{\pi^2}{6} \frac{1}{a^2} \left( 1 - e^{-a^2} \right) + \frac{2}{a^2} \int_0^a \frac{e^{-s^2}}{s} \left( \ln{(s^2)} - \mathrm{Ei}{(s^2)} + \gamma \right) d s .
\end{align*}
As a result,
\begin{align*}
I (a,0) = \pi \left( \frac{\pi^2}{12} + \int_0^a \frac{e^{-s^2}}{s} \left( \ln{(s^2)} - \mathrm{Ei}{(s^2)} + \gamma \right) d s \right) .
\end{align*}
Now I just need to understand this remaining integral. I am hopeful that I can leverage a result from https://ia800303.us.archive.org/1/items/jresv73Bn3p191/jresv73Bn3p191_A1b.pdf. Maybe #25 on page 198?
Update: I was able to push this calculation a little further. If we let $s \mapsto s^2$, then
\begin{align}
I_4 (a,0) = \pi \left( \frac{\pi^2}{12} + \frac{1}{2} \int_0^{a^2} \frac{e^{-s}}{s} \left( \ln{(s)} - \mathrm{Ei}{(s)} + \gamma \right) d s \right) .
\end{align}
Noting that
\begin{align*}
\frac{e^{-s}}{s} = \frac{d }{d s} \mathrm{Ei} (-s) ,
\end{align*}
we may integrate-by-parts to find
\begin{align*}
I_4 (a,0) = \pi \left( \frac{\pi^2}{12} + \frac{1}{2} \left( \mathrm{Ei} (- a^2) \left( \ln{(a^2)} - \mathrm{Ei}{(a^2)} + \gamma \right) - \int_0^{a^2} \mathrm{Ei} (-s) \frac{1 - e^s}{s} d s \right) \right) .
\end{align*}
So, I guess now the question is
\begin{align*}
\int_0^{a^2} \mathrm{Ei} (-s) \frac{1 - e^s}{s} d s = ???
\end{align*}
|
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|
Alternative Solution for Cubic Equation I was solving a problem and got to the following equation $x^3 + x^2 + x - 1 = 0 \; \; (1)$, numerically I found that the solution was:
$$x =\frac{1}{3} \left(-1 - \frac{2}{\left(17 + 3 \sqrt{33} \right)^{-1/3}} + \big(17 + 3 \sqrt{33} \big)^{-1/3} \right)$$
Which gives the same result as the professor. However, I'm suppose to write the solution of (1) as:
$$x = \frac{1}{3} \left(-1 -2\sqrt{2}\sinh\left( \frac{1}{3} \sinh^{-1} \left( \frac{17}{2\sqrt{2}}\right) \right)\right)$$
How I do get this expression?
|
I shall follow the steps given here.
We have $\Delta=-44$ so only one real root.
Using $p=\frac 23$, $q=-\frac{34}{27}$ and the hyperbolic method
$$t_0=\frac{2}{3} \sqrt{2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{17}{2
\sqrt{2}}\right)\right)$$
$$x=t_0-\frac b {3a}=\frac{1}{3} \left(-1+2 \sqrt{2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{17}{2\sqrt{2}}\right)\right)\right)$$
I suppose that you have a sign error in what you wrote.
|
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|
Find integer solution for square equation
Does $a^2 + b^2 = 6 c^2$ have any integer solution?
My thought:
(0,0,0) is obviously a solution but I don't think there are any others.
Because if I take equation of modulo 6, it gives $a^2 + b^2 = 0 \pmod 6$
a and b can only be 3 mod 6 or 0 mod 6. However, I am not sure what is the next step here. Any help is appreciated
|
Consider a $\pmod{3}$ argument.
Since $6c^2$ is a multiple of $3$, you must have that
$a^2 + b^2$ is a multiple of $3$.
However, for any integer $n$, either
$n^2 \equiv 0 \pmod{3}$ or $n^2 \equiv (+1) \pmod{3}$.
Therefore, $a$ and $b$ must each be a multiple of $3$.
Let $r$ denote the largest positive integer exponent such that $3^r$ divides $a$.
Let $s$ denote the largest positive integer exponent such that $3^s$ divides $b$.
Since the constraint $a^2 + b^2 = 6c^2$ is symmetrical around $a$ and $b$, you can assume, without loss of generality, that $r \leq s$.
Since $3^r$ divides $a$ and $3^r$ divides $b$, you must have that $3^{(2r)}$ divides $6c^2.$
This implies that $3^{(2r - 1)}$ divides $c^2$.
This implies that $3^r$ divides $c$.
Therefore, you can let
*
*$d = \frac{a}{3^r}$.
*$e = \frac{b}{3^r}$.
*$f = \frac{c}{3^r}$.
Then, $~\displaystyle d^2 + e^2 = 6f^2$
where $d$ is not a multiple of $3$.
This implies that $d^2 \equiv 1\pmod{3}.$
Then, you will have that $e^2 \equiv 0\pmod{3}$ or
$e^2 \equiv 1\pmod{3}$.
In either case, you will not have that $d^2 + e^2$ is a multiple of $3$.
This yields a contradiction, since $d^2 + e^2 = 6f^2$.
|
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|
rolling $4$ dice probability question I thought I understood the $4$ dice rolling probabilities until considering the sum of all possible results.
Probability of getting different results while rolling 4 standard dices:
*
*all $4$ dice have one value - $1/216$
*having $2$ different values:
$\dfrac{4C2 \cdot 6 \cdot 5}{6^4} = \dfrac{30}{216}$
*having $3$ different values:
$4C2 \cdot 6 \cdot 5 \cdot 4= \dfrac{120}{216}$
*having $4$ different values:
$\dfrac{6 \cdot 5 \cdot 4 \cdot 3}{6^4} = \dfrac{60}{216}$
Those are all the possible results so the probabilities should sum to $1$. But it's missing $5/216$??? Where is it? Which one of the above cases is wrong?
After some thinking - the second case above seems wrong. What is the probability of getting $2$ different values when rolling $4$ dice? $\dfrac{6 \cdot 5}{6^4} = \dfrac{5}{216}$ seems wrong.
|
You made an error when considering the case with two different values. There are two possibilities:
*
*One value appears three times and another value appears once.
*Two values each appear twice.
One value appears three times and another value appears once: There are six ways to select the value that appears three times, $\binom{4}{3}$ ways to select the three dice on which that value appears, and five ways to select the value that appears on the other die, giving
$$\binom{6}{1}\binom{4}{3}\binom{5}{1}$$
such cases.
Two values each appear twice: There are $\binom{6}{2}$ ways to select the two values which each appear twice and $\binom{4}{2}$ ways to select the two dice on which the larger of those values appear, giving
$$\binom{6}{2}\binom{4}{2}$$
such cases.
Hence, the probability that exactly two different values appear on the four dice is
$$\frac{\dbinom{6}{1}\dbinom{4}{3}\dbinom{5}{1} + \dbinom{6}{2}\dbinom{4}{2}}{6^4} = \frac{120 + 90}{1296} = \frac{210}{1296} = \frac{35}{216}$$
|
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|
A typical inequality: $\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}$
For $x, y, z\in (0, \infty)$ prove that: $$\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}.$$
My attempts to apply media inequality or other inequalities have been unsuccessful. In desperation I did the calculations and I have a rather complicated question question that I could not write as the sum of squares:
$$ 0\leq 6x^2y^2z^2 +19(x^3y^3+y^3z^3+z^3x^3)+27(x^4yz+xy^4z+xyz^4)-24(x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3).$$
Thanks for any ideas that might help me clarify the issue.
|
First solution: We have
$$8(xy+yz+zx)^3 = [ \sum (xy+yz) ]^3 \leq 9( (xy+yz)^3 + (yz+zx)^3 + (zx+xy)^3 ), $$
so it remains to show that
$$ (xy+yz)^3 + (yz+zx)^3 + (zx+xy)^3 \leq 3 (x^2+yz)(y^2+zx)(z^2 + xy). $$
Let $ M (a, b, c) = \sum_{sym} x^a y^b z^c$. Then, by expanding everything, we get that
$$3(x^2+yz)(y^2+zx)(z^2 + xy) - (xy+yz)^3 - (yz+zx)^3 - (zx+xy)^3 \\ = M(4, 1, 1) + M(3, 3, 0) + M(2, 2, 2) - M (3, 2, 1). $$
By taking the symmetric summations of $ x^4 yz + x^3y^3 + x^2y^2z^2 \geq 3 x^3 y^2 z$, we get that $ M(4, 1, 1) + M(3, 3, 0) + M(2, 2, 2) \geq 3 M(3, 2, 1)$.
Hence we are done.
Alternative solution: Continuing from OP's work, WTS
$$27/2 M(4,1,1) + 19/2 M(3, 3, 0) + 1 M(2, 2, 2) \geq 24 M(3, 2, 1).$$
Muirhead tells us that $M(4,1,1) \geq M(3, 2, 1)$ and $M(3, 3, 0) ) \geq M(3, 2, 1)$.
We just need to AM-GM the $M(2, 2, 2)$ term, which has a very small coefficient, so it likely will work out.
As before, by taking the symmetric summations of $ x^4 yz + x^3y^3 + x^2y^2z^2 \geq 3 x^3 y^2 z$, we get that $ M(4, 1, 1) + M(3, 3, 0) + M(2, 2, 2) \geq 3 M(3, 2, 1)$.
Hence (by summing the relevant inequalities), the result follows.
|
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|
Solving $\sqrt{2x+1}-\sqrt{2x-1}=2$ I solved $\sqrt{2x+1}-\sqrt{2x-1}=2$
by squaring both sides.
$$\sqrt{2x+1}=\sqrt{2x-1}+2$$
$$2x+1=(2x-1)+4\sqrt{2x-1}+4$$
$$-1=2\sqrt{2x-1}$$
$$1=4({2x-1})$$
$$x=\frac{5}{8}$$
Now when I put $x=\frac{5}{8}$ to the given equation,I get $\sqrt{\frac{5}{4}+1}-\sqrt{\frac{5}{4}-1}=1 \neq 2$
What am I missing ?
|
Your re-arranged equation $ \ \sqrt{2x+1} \ = \ \sqrt{2x-1} \ + \ 2 \ \ $ would give the $ \ x-$coordinates for the intersection points between two square-root curves, if such points exist. However, "squaring" this equation reveals a difficulty. The curve equation $ \ y^2 \ = \ 2x + 1 \ \ $ implied by the left side represents a "horizontal" parabola "opening to the right" from the vertex $ \ \left( \ -\frac12 \ , \ 0 \ \right) \ \ , $ for which $ \ y \ = \ \sqrt{2x + 1} \ $ is the "upper arm" of the parabola [violet curve in the graph below] and $ \ y \ = \ -\sqrt{2x + 1} \ $ is the "lower arm" [red curve]. On the right side of the squared equation, the curve equation would be $ \ (y - 2)^2 \ = \ 2x - 1 \ \ , $ a horizontal parabola which also "opens to the right" and has the vertex $ \ \left( \ +\frac12 \ , \ \mathbf{2} \ \right) \ \ , $ The original square-root curve $ \ y \ = \ 2 + \sqrt{2x - 1} \ $ is the upper arm [blue curve] of this parabola; at $ \ x \ = \ \frac12 \ , $ the square-root curve function $ \ y \ = \ \sqrt{2x + 1} \ $ has only attained the value $ \ \sqrt2 \ \ , $ so it does not meet $ \ y \ = \ 2 + \sqrt{2x - 1} \ $ and always remains "below" this latter curve thereafter.
However, the lower arm of that parabola [orange curve], which is the square-root function $ \ y \ = \ 2 - \sqrt{2x - 1} \ $ does intersect $ \ y \ = \ \sqrt{2x + 1} \ $ at $ \ x \ = \ \frac58 \ \ , $ as you found. $ [ \ \sqrt{2·\frac58 + 1} \ = \ \sqrt{\frac94} \ = \ \frac32 \ = \ 2 - \sqrt{2·\frac58 - 1} \ = \ 2 - \sqrt{\frac14} \ = \ 2 - \frac12 \ \ . \ ] $
The parabolas "generated" by the "squared equation" have an intersection point, but since square-roots of real numbers are always taken to be non-negative, we "reject" the intersection with the "negative-of-square-root arm" of one of those parabolas. This leaves no (real-valued) solutions to the original equation.
|
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Evaluate $\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x$. Latest Edit
By the contributions of the writers, we finally get the closed form for the integral as:
$$\boxed{\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x =-\frac{\pi(2 n-3) ! !}{2^{n}(n-1) !} \sum_{j=1}^{n-1} \frac{1}{2j-1}}$$
I first evaluate $$I_1=\int_{0}^{\infty} \frac{\ln x}{x^{2}+1} d x \stackrel{x\mapsto\frac{1}{x}}{=} -I_1 \Rightarrow I_1= 0.$$
and then start to raise up the power of the denominator
$$I_n=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+1)^n} d x .$$
In order to use differentiation, I introduce a more general integral
$$I_n(a)=\int_{0}^{\infty} \frac{\ln x}{(x^{2}+a)^n} d x. $$
Now we can start with $I_1(a)$. Using $I_1=0$ yields
$$\displaystyle 1_1(a)=\int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x \stackrel{x\mapsto\frac{x}{a}}{=} \frac{\pi \ln a}{4 \sqrt a} \tag*{}$$
Now we are going to deal with $I_n$ by differentiating it by $(n-1)$ times
$$
\frac{d^{n-1}}{d a^{n-1}} \int_{0}^{\infty} \frac{\ln x}{x^{2}+a} d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{a}\right)
$$
$$
\int_{0}^{\infty} \ln x\left[\frac{\partial^{n-1}}{\partial a^{n-1}}\left(\frac{1}{x^{2}+a}\right)\right] d x=\frac{\pi}{4} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt a}\right)
$$
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+a\right)^{n}} d x=\frac{(-1)^{n-1} \pi}{4(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)
$$
In particular, when $a=1$, we get a formula for $$
\boxed{\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{n}} d x=\left.\frac{(-1)^{n-1} \pi}{4(n-1)!} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)\right|_{a=1}}
$$
For example, $$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{5}} d x=\frac{\pi}{4 \cdot 4 !}(-22)=-\frac{11 \pi}{48}
$$
and
$$
\int_{0}^{\infty} \frac{\ln x}{\left(x^{2}+1\right)^{10}} d x=\frac{-\pi}{4(9 !)}\left(\frac{71697105}{256}\right)=-\frac{1593269 \pi}{8257536}
$$
which is check by WA.
MY question
Though a formula for $I_n(a)$ was found, the last derivative is hard and tedious.
Is there any formula for $$\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\ln a}{\sqrt{a}}\right)? $$
|
Let $I\left(\lambda\right)=\int_0^{\infty} \frac{x^{2 \lambda-1}}{\left(a^2+x^2\right)^n} d x,$ then
$$I_n=\int_0^{\infty} \frac{\ln x}{\left(a^2 +x^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) .$$
Now we are going to express $I\left(\lambda\right)$ as a beta function by letting $x=a\tan \theta$, then
$$
\begin{aligned}
I(\lambda) &= a^{2(\lambda-n)}\int_0^{\frac{\pi}{2}} \sin ^{2 \lambda-1} \theta \cos ^{2(n-\lambda)-1} \theta d \theta \\
&=\frac{a^{2(\lambda-n)}}{2} B(\lambda, n-\lambda) \\
&=\frac{a^{2(\lambda-n)}}{2\Gamma(n) } \Gamma(\lambda) \Gamma(n-\lambda)
\end{aligned}
$$
By logarithmic differentiation, we get
$$
\begin{aligned}
&\frac{I^{\prime}(\lambda)}{I(\lambda)}=2\ln a+\psi(\lambda)-\psi(n-\lambda) \\
&I^{\prime}(\lambda)=\frac{a^{2(\lambda-n)}}{2 \Gamma(n)} \Gamma(\lambda) \Gamma(n-\lambda)[ 2\ln a+\psi(\lambda)-\psi(n-\lambda)]
\end{aligned}
$$
Putting $\lambda=\frac{1}{2} $ yields
$$
I^{\prime}\left(\frac{1}{2}\right)=\frac{a^{1-2n}\sqrt{\pi}}{2} \cdot \frac{\Gamma\left(n-\frac{1}{2}\right)}{\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right]
$$
Hence $$\boxed{\int_0^{\infty} \frac{\ln x}{\left(a^2+x^2\right)^n} d x= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) = \frac{a^{1-2n}\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] },$$
Back to our integral,
$$\boxed{\int_0^{\infty} \frac{\ln x}{\left(1+x^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) = \frac{\sqrt{\pi}\Gamma \left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] }$$
By Wolfram-Alpha, $\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right) = -H_{n-\frac{3}{2}}-2 \log (2) $ yields
$$I_n= -\frac{\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left(H_{n-\frac{3}{2}}+2\ln 2\right),$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4382586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
For a non-negative integer $k$, $\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l\implies \lim_{x\to \infty}\frac{f(x)}{x^{k+1}}=\frac l{k+1}$ Suppose that $f$ defined on $(a,\infty)$ is bounded on each finite interval $(a,b),a>b$. For a non-negative integer $k$, it is to be shown that $\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l\implies \lim_{x\to \infty}\frac{f(x)}{x^{k+1}}=\frac l{k+1}$
Given an $\displaystyle \epsilon >0,$there exists an integer $\displaystyle N$ such that
$$\displaystyle
x \geq N \Longrightarrow \ \left| \frac{f( x+1) -f( x)}{x^{k}}
-l \right| < \epsilon $$
Noting that
\begin{align*}
f( x) & =\sum _{i=1}^{[ x]} f( x-[ x] +i) -f( x-[ x] +i-1) +\color{red}{f( x-[ x])}
\end{align*}
it follows that
\begin{align*}
f( x) & =\sum _{i=1}^{N}\overbrace{( f( x-[ x] +i) -f( x-[ x] +i-1))}^{h( i)} +\sum _{i=N+1}^{[ x]}( f( x-[ x] +i) -f( x-[ x] +i-1)) +\color{red}{( \ )}
\end{align*}
It follows that
\begin{align*}
|\frac{f( x)}{x^{k+1}} -\frac{l}{k+1} | & \leq \sum _{i=1}^{N} |\frac{h( i)}{x^{k+1}} -\frac{l}{[ x]( k+1)} |+\sum _{i=N+1}^{[ x]} |\frac{h( i)}{x^{k+1}} -\frac{l}{[x](k+1)} |+\color{red}{( )x^{-k-1}}\\
& \leq {\textstyle \frac{\overbrace{M}^{\sup _{1\leq i\leq N} h( i)}}{x^{k+1}} +\frac{N|l|}{[ x]( k+1)} +\frac{\epsilon ([ x] -N)}{x} +|\frac{l}{x} -\frac{l}{[ x]( k+1)} |([ x] -N) +\color{red}{\overbrace{\color{red}{\sup _{t\in [ 0,1)}f(t)}}^{M'} x^{-k-1}}}\\
& =\frac{M+M'}{x^{k+1}} +{\textstyle \frac{N|l|}{[ x]( k+1)} +\color{purple}{|\frac{l[ x]}{x} -\frac{l}{( k+1)} |} -N|\frac{l}{x} -\frac{l}{[ x]( k+1)} |+\frac{\epsilon ([ x] -N)}{x}} \tag 1
\end{align*}
In $(1)$, except the purple term every other term can be made arbitrarily small. How do I take care of the purple term so that I can conlude the desired result by limit definition.
|
First we prove the statement for the case $l=0$, i.e.
$
\lim_{n \to \infty}\frac{f(x+1)-f(x)}{x^k} = 0
$.
Given $\epsilon > 0$ there is a $y > a$ such that for all $x \ge y$:
$$
\left | \frac{f(x+1)-f(x)}{x^k} \right | < \epsilon \, .
$$
Similarly as in If $\lim_{x\to+\infty}[f(x+1)-f(x)]= \ell,$ then $\lim\limits_{x\to+\infty}\frac{f(x)}x=\ell$. we write
$$
\frac{f(x)}{x^{k+1}} =
\frac{1}{x}\left(\sum_{i=1}^{\lfloor x-y\rfloor}\frac{f(x-i+1)-f(x-i)}{(x-i)^k} \cdot \left( 1-\frac ix\right)^k
\right)+\frac{f(x-\lfloor x-y\rfloor)}{x^{k+1}}
$$
which implies
$$
\left| \frac{f(x)}{x^{k+1}}\right| \le \frac{\lfloor x-y\rfloor}{x} \epsilon + \frac{M}{x^{k+1}}
$$
with $M = \sup \{ |f(z)| : y \le z < y+1\}$. It follows that
$$
\limsup_{n \to \infty }\left| \frac{f(x)}{x^{k+1}}\right| \le \epsilon \, .
$$
This holds for all $\epsilon > 0$, and therefore
$$
\lim_{n \to \infty} \frac{f(x)}{x^{k+1}} = 0 \, .
$$
For the general case with $\lim_{x\to \infty} \frac{f(x+1)-f(x)}{x^k}=l$ we set
$$
g(x) = f(x) - \frac{x^{k+1}}{k+1} l \, .
$$
$g$ is bounded on every finite interval $(a, b)$ and satisfies
$$
\frac{g(x+1)-g(x)}{x^k} = \frac{f(x+1)-f(x)}{x^k} - l \frac{(x+1)^{k+1}-x^{k+1}}{(k+1)x^k} \\
= \frac{f(x+1)-f(x)}{x^k} - l \frac{(1+1/x)^{k+1}-1}{(k+1)/x}
\to l - l = 0
$$
and then
$$
\frac{f(x)}{x^{k+1}} = \frac{g(x)}{x^{k+1}} + \frac{l}{k+1} \to 0 + \frac{l}{k+1} = \frac{l}{k+1} \, .
$$
|
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|
To prove: $\bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr) = (\cos A - \sin B)^2 $ How could one show that
$$ \bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr) = (\cos A - \sin B)^2
$$
I have tried the below approach:
\begin{align*}
LHS & = \bigl(1 + \sin(A+B)\bigr)\bigl(1-\sin(A-B)\bigr)\\
& = 1-\sin(A-B)+\sin(A+B)-\sin(A+B)\sin(A-B)\\
& = 1 - \sin{A}\cos{B} + \sin{B}\cos{A} + \sin{A}\cos{B} + \sin{B}\cos{A}\\
& \qquad - (\sin{A})^2 + (\sin{B})^2\\
& = 1 + 2\sin{B}\cos{A} - 1 + (\cos{A})^2 + (\sin{B})^2\\
& = 2\sin{B}\cos{A} + (\cos{A})^2 + (\sin{B})^2\\
& = (\cos{A} + \sin{B})^2
\end{align*}
|
\begin{aligned}
\text { LHS }=&(1+\sin (A+B))(1-\sin (A-B)) \\
=& 1-\sin (A-B)+\sin (A+B)-\sin (A+B) \sin (A-B) \\
=& 1- \sin A \cos B + \cos A \sin B + \sin A \cos B + \cos A \sin B -\sin (A+B) \sin (A-B)\\
=& 1 + 2\cos A \sin B - \sin (A+B) \sin (A-B) \\
=& 1 + 2\cos A \sin B - \frac{1}{2} (cos(A+B-A+B) - cos(A+B+A-B)) \\
=& 1 + 2\cos A \sin B - \frac{1}{2} (cos(2B) - cos(2A)) \\
=& 1 + 2\cos A \sin B - \frac{1}{2} [(1 - 2 \sin^{2} B )-(2 \cos^{2} A - 1)] \\
=& 1 + 2\cos A \sin B - \frac{1}{2} (1 - 2 \sin^{2} B ) +\frac{1}{2} (2 \cos^{2} A - 1) \\
=& 2\cos A \sin B + \sin^{2} B + \cos^{2} A\\
=& \cos^{2} A - 2\cos A \sin B + \sin^{2} B \\
=&(\cos A + \sin B)^{2}
\end{aligned}
Maybe the exercise is really wrong.
I am attaching a link where you can see the formulas used. (https://doza.pro/art/math/geometry/en/trig-formulas)
I hope I helped you.
|
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|
A fair $6$-sided die was rolled $10$ times. Find the probability you rolled exactly two $2$s and exactly two $5$s.
A fair $6$-sided die was rolled $10$ times. Find the probability you rolled exactly two $2$s and exactly two $5$s.
I know the standard way to calculate this, but wasn't sure why another approach - stars and bars - cannot be applied. If we fix the two $2$s and $5$s to their respective boxes and arrange the rest, we have $nCr(10+6-1-2-2, 6-1)$ ways; If we use stars and bars to calculate the total number of results, we have $nCr(10+6-1, 6-1)$. Setting the former as the numerator and the latter as denominator gives a probability larger than the answer. Why is this the case?
|
Although a somewhat offbeat approach, Stars and Bars can be used to solve the problem.
You have $10$ slots, $4$ of which are going to be taken up by $2$ 2's and $2$ 5's. You can construe these as Stars. You will then have $6$ Bars that are divided into $5$ regions, by the $4$ Stars.
So, you compute the number of solutions to
$$x_1 + x_2 + \cdots + x_5 = 6. \tag1 $$
(1) above has $~\displaystyle \binom{6 + [5-1]}{5-1} = \binom{10}{4}~$ solutions.
Now, for each solution, there are $4$ possibilities for each of the bars. So, this results in a factor of
$\left(4^6\right).$
Further, for each solution, there are $~\displaystyle \binom{4}{2}$ ways of selecting which $2$ of the $4$ Stars will be 2's, instead of 5's.
So, in computing the combinatoric probability fraction of
$$\frac{N\text{(umerator)}}{D\text{(enominator)}},$$
where $D = 6^{(10)},$ you end with
$$ N = \binom{10}{4} \times \binom{4}{2} \times (4)^6 = \frac{10!}{(6!)(4!)} \times \frac{4!}{(2!)(2!)} \times (4)^6 = \frac{10!}{(6!)(2!)(2!)} \times (4)^6.$$
Alternatively, in the standard approach,
*
*You have to select $2$ slots for the 2's : factor of $\displaystyle \binom{10}{2} = \frac{10!}{(2!)(8!)}.$
*You then have to select $2$ slots for the 5's : factor of $\displaystyle \binom{8}{2} = \frac{8!}{(2!)(6!)}.$
*You then have to select the number of ways of filling the remaining $6$ positions : factor of $(4)^6.$
$$N = \frac{10!}{(8!)(2!)} \times \frac{8!}{(6!)(2!)} \times (4)^6 = \frac{10!}{(6!)(2!)(2!)} \times (4)^6.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the greatest common divisor of $(n+1)(n+2)^2(n+3)^3(n+4)^4$ Determine the largest positive integer that divides $(n+1)(n+2)^2(n+3)^3(n+4)^4$ for all positive integers n.
First, I noticed that out of $n+1, n+2, n+3,$ and $n+4,$ there must be one multiple of $4$, at least one multiple of $3$, and a multiple of $2$ that isn't the multiple of $4$. Since the problem is asking for a minimal case that satisfies these conditions, $n+1$ should be the multiple of $4$, $n+2$ should be the multiple of $3$, and $n+3$ should be the multiple of $2$, thus giving $4 \cdot 3^2 \cdot 2^3 = 288.$
However, I am unsure as to whether or not this answer is correct, as testing cases has consistently yielded higher GCDs than this.
|
Let $f(x) \in \Bbb Z[x]$. We have $m \mid f(n) \iff f(n)\equiv 0\pmod m$, but when written as this it's clear that $n$ may be replaced by any integer (not only natural numbers) congruent to $n$ modulo $m$, to get a equivalent congruence. The means that
"The largest positive integer that divide $f(n)$ for all positive integers $n$"
is the same as
"The largest positive integer that divide $f(n)$ for all integers $n$"
Now to this particular case:
We have
$$f(-5) =(-5+1)(-5+2)^2(-5+3)^3(-5+4)^4 = (-4)(-3)^2(-2)^3(-1)^4=2^5 3^2$$
(so the answer can't be larger than that) and you argued that $f(n)$ is always multiple of $2^53^2$ so that's the answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Probability Problem with 10 players, Bob and a friend being on a team Please check my work.
Out of 10 players including Bob and his two best friends a team of 5 players will be formed. What is the likelihood of Bob making the team with at most one of his friends?
Total Possible Teams
$$\binom{10}{5} = 252$$
Bob with at least one other friend.
$$\binom{8}{3} = 56$$
Divide total possibilities with possibilities of teams with Bob and his friend.
$$\frac{\mbox{Combinations}}{\mbox{Total Possibilities}} = \frac{2}{9}$$
|
There are indeed $$\binom{10}{5}$$ ways to form a team of five players from among the ten people.
If Bob is selected to be on the team, there are $$\binom{7}{4}$$ ways to select four teammates for Bob from among the seven people who are not his friends.
If Bob is selected to be on the team, there are $$\binom{2}{1}\binom{7}{3}$$ ways to select exactly one of his two friends and three of the other seven people to be on Bob's team.
Hence, the probability that Bob makes the team with at most one of his friends is
$$\frac{\dbinom{7}{4} + \dbinom{2}{1}\dbinom{7}{3}}{\dbinom{10}{5}}$$
Note that the complementary event is that either Bob is not selected for the team, which can occur in $$\binom{9}{5}$$ ways since five of the other nine people must be selected to be on the team, or Bob and both his friends are selected, which can occur in $$\binom{2}{2}\binom{7}{2}$$ ways since both his friends and two of the other seven people must be selected to be on Bob's team. Hence, the probability that Bob makes the team with at most one of his friends is $$1 - \frac{\dbinom{9}{5} + \dbinom{2}{2}\dbinom{7}{2}}{\dbinom{10}{5}}$$
Addendum: In the comments, Graham Kemp suggested another method. There are
$$\binom{9}{4}$$
teams which could include Bob since we must choose four of the other nine people to be his teammates. We know that
$$\binom{2}{2}\binom{7}{2}$$
include both his friends. Therefore, the number of teams which include at most one of his friends is
$$\binom{9}{4} - \binom{2}{2}\binom{7}{2}$$
Consequently, the probability that Bob makes the team with at most one of his friends is
$$\frac{\dbinom{9}{4} - \dbinom{2}{2}\dbinom{7}{2}}{\dbinom{10}{5}}$$
|
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|
Quadratic inequality with negative roots Assume the following quadratic inequality: $$0\lt x^2+4x-100$$ The solutions are: $$
x\lt -2-\sqrt{104},\qquad x\gt-2+\sqrt{104}
$$
In this case, the positive root keeps the original direction of the inequality ($\gt$), but the negative root inverts it.
However, for the general case: $0\lt ax^2+bx+c$, how do I know which root have which inequality direction? Can I say: "if the root is positive the inequality direction of the solution is the same as the original inequality, or the inverted direction otherwise"? Or does it actually depend on the signs and/or values of $a$, $b$ and $c$? and what about $\geq$? What about quadratic inequations where all of their roots are positive or negative?
|
You can tackle this by completing the square. Notice that
$$
\begin{align}
0&<ax^2+bx+c \\
&=a\left(x^2+\frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right)+c \\
&=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c
\end{align}
$$
which means that
$$a>0 \implies \left(x+\frac{b}{2a}\right)^2>\frac{b^2-4ac}{4a^2} \implies
\left|x+\frac{b}{2a}\right|>\frac{\sqrt{b^2-4ac}}{2a}
$$
$$a<0 \implies \left(x+\frac{b}{2a}\right)^2<\frac{b^2-4ac}{4a^2} \implies
\left|x+\frac{b}{2a}\right|<\frac{\sqrt{b^2-4ac}}{2a}
$$
From there, you simply choose between one of the following inequalities for absolute values:
$$|a|<b \implies -b<a<b$$
$$|a|>b \implies a>b \ \ \text{ or } \ a<-b$$
and you're set to isolate $x$.
|
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|
$ \sqrt a+\sqrt b +\sqrt c = 1$ Prove: $\frac{a^2+bc}{\sqrt{2a^2(b+c)}}+\frac{b^2 + ac}{\sqrt{2b^2(a + c)}}+\frac{c^2 +ab}{\sqrt{2c^2(a+b)}}\geq1$ I'm having some trouble with this problem:
a, b, c are positive real numbers where $ \sqrt a + \sqrt b + \sqrt c = 1 $ .
Prove the following inequality:
$$\frac{a^2 + bc}{\sqrt{2a^2(b+c)}} + \frac{b^2 + ac}{\sqrt{2b^2(a + c)}} + \frac{c^2 +ab}{\sqrt{2c^2(a+b)}} \geq 1$$
I've tried substituting out $\geq 1$ for $\geq \sqrt a + \sqrt b + \sqrt c $ , as well as simplifying and rationalising the denominators and putting over a common denominator but I'm struggling to make it much further than that. Any help as to how to untangle this would be greatly appreciated. Many thanks :)
|
$\sum_{cyc}{\frac{a^2+bc}{\sqrt{2a^2(b+c)}}}\geq\sum_{cyc}{\frac{ab+ac}{\sqrt{2a^2(b+c)}}}$……①
WLOG $a\leq b\leq c$
$LHS-RHS=\sum_{cyc}{\frac{(a-b)(a-c)}{\sqrt{2a^2(b+c)}}}$
$=\frac{(c-a)(c-b)(b-a)}{\sqrt2}(\frac{1}{a(c-b)\sqrt{b+c}}-\frac{1}{b(c-a)\sqrt{a+c}}+\frac{1}{c(b-a)\sqrt{a+b}})$……②
$b(c-a)\geq c(b-a),\sqrt{a+c}\geq\sqrt{a+b}$
$\frac{1}{b(c-a)\sqrt{a+c}}\leq\frac{1}{c(b-a)\sqrt{a+b}}$
so that $②\geq0$,thus ① is established.
$\sum_{cyc}{\frac{ab+ac}{\sqrt{2a^2(b+c)}}}=\sum_{cyc}{\sqrt{\frac{b+c}{2}}}\geq\sum_{cyc}{\frac{\sqrt{b}+\sqrt{c}}{2}}=1$
|
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|
Galois group of splitting field of $x^4-16x^2+4$ isomorphic to $\mathbb{Z}/(2)\times \mathbb{Z}/(2)$ I've already shown irreducible. I know I need to get intermediate fields since they correspond to subgroups (which I can use for isomorphisms). But how do I get the intermediate fields here?
edit: I'm thinking perhaps there is some way to show the order of the extension is $4$, and the only groups of order $4$ are the Klein-$4$ group and cyclic group and eliminating from there.
edit 2:
Treating $x^4-16x^2+4$ as a quadratic in $x^2$, applying the quadratic formula gives $$\frac{16\pm \sqrt{16^2-4\cdot 4}}{2}=\frac{16\pm 4\sqrt{15}}{2}=8\pm 2\sqrt{15}=(\sqrt{3}\pm\sqrt{5})^2.$$ and this is squared so $\pm(\sqrt{3}\pm\sqrt{5})$ gives the roots and there are four of them. Now what's next..
|
Let $K$ be the splitting field of $x^4 - 16x^2 + 4$
$x^2 - 16x + 4$ has roots $8 \pm \sqrt{60}$. Therefore, the roots of $x^4 - 16x^2 + 4$ are $\pm \sqrt{8 \pm \sqrt{60}}$. This is a degree 4 extension since we have $[K : \mathbb{Q}(\sqrt{60})] = 2$ and $[\mathbb{Q}(\sqrt{60}) : \mathbb{Q}] = 2$.
The Galois group is therefore either $\mathbb{Z} / 4 \mathbb{Z}$ or $\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$. We look at each one : $\mathbb{Z} / 4 \mathbb{Z}$ has one subgroup of order 2, and $\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$ has 3. $\mathbb{Q}(\sqrt{60})$ is a subfield of $K$. If we can find another subfield (not $\mathbb{Q}$ or $K$ itself), then we can eliminate the case that the Galois group is $\mathbb{Z} / 4 \mathbb{Z}$ .
Now, $\sqrt{60} = 2 \sqrt{15} $ and $15$ factors as $3 * 5$, so we take a guess that possibly $8 \pm \sqrt{60}$ might be a perfect square. We guess that $(a \sqrt{3} + b \sqrt{5})^2 = 8 \pm \sqrt{60}$, and try to solve this for $a$ and $b$. Indeed, we have that $(\sqrt{3} + \sqrt{5})^2 = 8 + \sqrt{60}$.
Therefore, one pair of roots are $\pm (\sqrt{3} + \sqrt{5})$. To find the other pair, we guess that $(a \sqrt{3} + b \sqrt{5})^2 = 8 - \sqrt{60}$. Again, solving for $a$ and $b$, we get $(\sqrt{3} - \sqrt{5})^2 = 8 - \sqrt{60}$.
Therefore, $\pm (\sqrt{3} - \sqrt{5})$ is another pair of roots.
$K$ contains $\sqrt{3} + \sqrt{5}$ and $\sqrt{3} - \sqrt{5}$, so must contain their sum which is $2\sqrt{3}$. Therefore $K$ contains $\sqrt{3}$ and $\mathbb{Q}(\sqrt{3})$ is another subfield. To show that $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{60})$ are distinct, it suffices to show that the equation $(a + b \sqrt{3})^2 = 60$ has no solutions. Expanding, we get $a^2 + 3b^2 = 60, 2ab\sqrt{3} = 0$. This requires either $a$ or $b$ = 0$.
If $a = 0$, then the first equation becomes $3b^2 = 60 \rightarrow b^2 = 20$. Since $20$ is not a perfect square, this has no rational solutions.
If $b = 0$, then the first equation becomes $a^2 = 60$. Since $60$ is not a perfect square, this has no solutions either.
Therefore, $K$ has the two subfields $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{60})$.
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$ 3^7\cdot (a^9+b^9+c^9)+1\geq 12\cdot (a^3+b^3+c^3)$ Let $a,b,c>0$ s.t. $a+b+c=1$. Show that $$ 3^7\cdot (a^9+b^9+c^9)+1\geq 12\cdot (a^3+b^3+c^3)$$
I tried to apply the next formula:$$a^3+b^3+ c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc$$
Also I made substitutions $ab+bc+ac=x$ and $abc=y$ but I am stuck.
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Hint: Show that $ 3^7 a^9 + a \geq 12 a^3$ using AM-GM.
Corollary: The result follows by summing up the cyclic inequalities.
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Evaluating improper integral $\int_0^1 \frac{\log(x)}{x+\alpha}\; dx$ for small positive $\alpha$ Let $\alpha$ be a small positive real number.
How do I obtain
$$ I = \int_0^1 \frac{\log(x)}{x+\alpha}\; dx = -\frac{1}{2}(\log\alpha)^2 - \frac{\pi^2}{6} - \operatorname{Li}_2(-\alpha)$$? Maxima told me the result, but I do not understand how to get it.
The appearance of $\pi^2/6$ suggests the necessity of complex analysis. I tried to relate it to some contour integral through the change of variables $x=e^{-t}$:
$$ I = -\int_{0}^{\infty} \frac{t e^{-t}}{e^{-t}+\alpha}\;dt, $$
but I could not find the next step.
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Assuming, for $u\leq 1$,
\begin{align}\text{Li}_2\left(u\right)&=-\int_0^1 \frac{u\ln x}{1-ux}dx\\
\text{Li}_2\left(1\right)&=\zeta(2)=\frac{\pi^2}{6}
\end{align}
Therefore, $0<\alpha<1$
\begin{align}\int_0^1 \frac{\log(x)}{x+\alpha}\; dx+\text{Li}_2\left(-\alpha\right)&=\underbrace{\frac{1}{\alpha}\int_0^1 \frac{\log(x)}{1+\frac{x}{\alpha}}\; dx}_{u=\frac{x}{\alpha}}+\underbrace{\int_0^1 \frac{\alpha\log(x)}{1+\alpha x}\; dx}_{u=\alpha x}\\
&=\int_0^{\frac{1}{\alpha}}\frac{\ln(\alpha u)}{1+u}du+\int_0^{\alpha}\frac{\ln(\frac{u}{\alpha} )}{1+u}du\\
&=\int_0^{\frac{1}{\alpha}}\frac{\ln u}{1+u}du+\int_0^{\alpha}\frac{\ln u}{1+u}du-\ln^2 \alpha\\
&=\left(\int_0^1 \frac{\ln x}{1+x}dx+\underbrace{\int_1^{\frac{1}{\alpha}} \frac{\ln x}{1+x}dx}_{u=\frac{1}{x}}\right)+\left(\int_0^1 \frac{\ln x}{1+x}dx-\int_\alpha^1 \frac{\ln x}{1+x}dx\right)-\\&\ln^2 \alpha\\
&=2\int_0^1 \frac{\ln x}{1+x}dx-\int_\alpha^1\frac{\ln x}{x(1+x)}dx-\int_\alpha^1 \frac{\ln x}{1+x}dx-\ln^2\alpha\\
&=2\int_0^1 \frac{\ln x}{1+x}dx+\int_\alpha^1\frac{\ln x}{1+x}dx-\underbrace{\int_\alpha^1\frac{\ln x}{x}dx}_{=-\frac{1}{2}\ln^2\alpha}-\int_\alpha^1 \frac{\ln x}{1+x}dx-\ln^2\alpha\\
&=2\int_0^1 \frac{\ln x}{1+x}dx-\frac{1}{2}\ln^2\alpha\\
&=2\left(\int_0^1 \frac{\ln x}{1-x}dx-\underbrace{\int_0^1 \frac{2x\ln x}{1-x^2}dx}_{u=x^2}\right)-\frac{1}{2}\ln^2\alpha\\
&=2\left(\int_0^1 \frac{\ln x}{1-x}dx-\frac{1}{2}\int_0^1 \frac{\ln u}{1-u}du\right)-\frac{1}{2}\ln^2\alpha\\
&=\int_0^1 \frac{\ln x}{1-x}dx-\frac{1}{2}\ln^2\alpha\\
&=-\text{Li}_2(1)-\frac{1}{2}\ln^2\alpha\\
&=-\frac{\pi^2}{6}-\frac{1}{2}\ln^2\alpha\\
\end{align}
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Find all solutions to $a^3 + b^3 = p^4,$ where $p$ is prime, and $a,b$ are natural Initially, I tried to break the equation down into $(a+b)(a^2-ab+b^2)$, then break it down into two cases:
I: $(a+b) = p, (a^2-ab+b^2) = p^3$, which yields to a contradiction that $p^2 > p^3.$
II: $(a+b) = p^2, (a^2-ab+b^2) = p^2$, which converts to $p^2(p-1)(p+1) = 3ab.$
However, I can't find any way to use this.
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You are on the right track when you render $a+b=p^2,a^2-ab+b^2=p^2$.
You should then plug in
$a^2-ab+b^2=\frac14(a^2+2
ab+b^2)+\frac34(a^2-2ab+b^2)$
$=\frac14(a+b)^2+\frac34(a-b)^2.$
Then conclude that with $a+b=p^2,a^2-ab+b^2=p^2$, you must have
$p^2\ge\frac14p^4.$
From this $p\le2$, thereby limiting your solution candidates.
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Find a limit as product of cos $$\lim\limits_{n\to\infty}\cos\frac{1}{n\sqrt{n}}\cos\frac{2}{n\sqrt{n}}\cdots\cos\frac{n}{n\sqrt{n}}$$
It is not hard to prove that the limit exists, but is that possible to calculate the limit? Suggestions are welcome!
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The infinite product is a bit difficult to work with, so to make things easier we compute the log of the product instead. Our limit becomes
$$\lim_{n\to\infty}\sum^n_{i=1}\ln\cos\frac{i}{n^{\frac{3}{2}}}$$
We use a Taylor series to compute $\ln\cos x$ around $x=0$. We know $\cos x=1-\frac{x^2}{2}+O\left(x^4\right)$ and $\ln(1+x)=x+O\left(x^2\right)$. Hence,
$$\ln\cos x=\ln\left(1-\frac{x^2}{2}+O\left(x^4\right)\right)=-\frac{x^2}{2}+O\left(x^4\right)+O\left(\left(-\frac{x^2}{2}+O\left(x^4\right)\right)^2\right)$$
The last asymptotic is equal to $O\left(x^4\right)$, so our final expression becomes
$$\ln\cos x=-\frac{x^2}{2}+O\left(x^4\right)$$
Applying this in our limit we get:
$$\lim_{n\to\infty}\sum^n_{i=1}-\frac{i^2}{2n^3}+O\left(\frac{i^4}{n^6}\right)=\lim_{n\to\infty}-\sum^n_{i=1}\frac{i^2}{2n^3}+\sum^n_{i=1}O\left(\frac{i^4}{n^6}\right)$$
Now, $\sum^n_{i=1}O\left(i^k\right)=O\left(n^{k+1}\right)$ and $\sum^n_{i=1}i^2=\frac{2n^3+3n^2+n}{6}$ in particular, so our limit becomes
$$\lim_{n\to\infty}-\frac{2n^3+3n^2+n}{12n^3}+O\left(\frac{n^5}{n^6}\right)=-\frac{1}{6}+O\left(n^{-1}\right)$$
Now $O\left(n^{-1}\right)\sim0$ for large $n$, so we conclude
$$\lim_{n\to\infty}\sum^n_{i=1}\ln\cos\frac{i}{n^{\frac{3}{2}}}=-\frac{1}{6}$$
Or equivalently
$$\lim_{n\to\infty}\prod^n_{i=1}\cos\frac{i}{n^{\frac{3}{2}}}=e^{-\frac{1}{6}}$$
The LHS is your limit.
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How to evaluate $\int _{0}^{\frac{\pi}{2}} \frac{\mathrm dx}{a+\cos x}$ using contour integration? I want to evaluate:
$\displaystyle \int _{0}^{\frac{\pi}{2}} \dfrac{\mathrm dx}{a+\cos x} \tag*{}$
With my basic knowledge of Cauchy Residue theorem:
\begin{align*}
\int_0^{\frac{\pi}{2}} \frac{\mathrm dx}{a + \cos x} &= \int_0^{\frac{\pi}{2}}\frac{\mathrm dx}{a + \frac{e^{ix} + e^{-ix}}{2}} \\
&= 2\int_0^{\frac{\pi}{2}} \frac{e^{ix} \ \mathrm dx}{2ae^{ix} + e^{2ix} + 1} && \text{Let } z=e^{ix}, \text{ so } \mathrm dz = ie^{ix} \ \mathrm dx. \\
&= \frac{2}{i} \int_{|z|=1} \frac{\mathrm dz}{z^2 + 2az + 1} \\
&= \frac{2}{i} \int_{|z|=1} \frac{\mathrm dz}{(z-z_1)(z-z_2)}
\end{align*}
where the circle $|z|=1$
My problem is how to proceed further from here? How do I draw circle? Clockwise or Counterclockwise?
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I doubt you can use contour integration to handle this since you can't obtain a closed curve. Noting
$$ \cos x=\frac{1-\tan^2(\frac x2)}{1+\tan^2(\frac x2)}$$
one has, under $\tan(\frac x2)\to t$,
\begin{eqnarray}
&&\int _{0}^{\frac{\pi}{2}} \dfrac{\mathrm dx}{a+\cos x} \\
&=&\int _{0}^{\frac{\pi}{2}} \dfrac{\mathrm dx}{a+\frac{1-\tan^2(\frac x2)}{1+\tan^2(\frac x2)}} \\
&=&\int _{0}^{\frac{\pi}{2}} \dfrac{(1+\tan^2(\frac x2))\mathrm dx}{a(1+\tan^2(\frac x2))+1-\tan^2(\frac x2)} \\
&=&\int _{0}^{\frac{\pi}{2}} \dfrac{(1+\tan^2(\frac x2))\mathrm dx}{(a+1)+(a-1)\tan^2(\frac x2))} \\
&=&\int_0^1\frac{2}{(a+1)+(a-1)t^2}\mathrm dt\\
&=&\frac2{\sqrt{a^2-1}}\arctan\bigg(\sqrt{\frac{a-1}{a+1}}\bigg).
\end{eqnarray}
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How can I solve the following 2nd Order DE? $$y''-\frac{1}{x}y'=2x\cdot cos(x)$$
For the homogeneous part I multiplied through with $x^2$ and got a second order Cauchy Euler equation with the general solution: $$y_h (x)=A x^2 +B$$
Then for the particular solution I tried using the method of undetermined coefficients but the whole thing became too entangled to solve and I couldn't get anywhere!
Maple tells me the solution is: $$2sin(x) - 2x\cdot cos(x)+ \frac{C_1\cdot x^2 }{2}+C_2$$
but I can't figure it out...
Any help would be appreciated!
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This reduces the problem to a first order Diff Eq.
$u= y'\\
xu' - u = 2x^2\cos x$
Choose a candidate for the particular solution that could work.
$u_p = x\sin x$
"Generalize" this by adding terms that might come up in the derivative that we hope will cancel out. Give every term a coefficient.
$u_p = Ax\sin x + Bx\cos x + C\sin x+ D\cos x$
Differentiate and plug into the original equation.
$u_p' = A\sin x + Ax\cos x + B\cos x - Bx\sin x + C\cos x - D\sin x\\
u_p' = - Bx\sin x + Ax\cos x + (A-D)\sin x + (B+C)\cos x$
$xu_p' - u_p = - Bx^2\sin x + Ax^2\cos x -Dx\sin x + Cx\cos x - C\sin x - D\cos x = 2x^2\cos x$
$A = 2, B,C,D = 0$
$u_p = 2x\sin x\\
y_p' = 2x\sin x\\
y_p = \int x\sin x \ dx = -2x\cos x + 2\sin x + C$
$y = y_h + y_p = Ax^2 + B - 2x\cos x + 2\sin x$
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Show that $\frac{\binom{n}{1}}{1} + \frac{\binom{n}{2}}{2} + \frac{\binom{n}{3}}{3} + \cdots + \frac{\binom{n}{n}}{n} = \sum_{r=1}^{n}\frac{2^r-1}{r}$ Good day,
I was solving this problem:
Show that $$\frac{\binom{n}{1}}{1} + \frac{\binom{n}{2}}{2} + \frac{\binom{n}{3}}{3} + \cdots + \frac{\binom{n}{n}}{n} = \sum_{r=1}^{n}\frac{2^r-1}{r}$$
I had already made some progress here, and we can prove using induction $$\int_{0}^{1}{\frac{(1 + x) ^ {n} - 1}{x}}dx = \sum_{r=1}^{n}\frac{2^r-1}{r}$$
But this seems too lengthy. Is there a shorter, more elementary solution?
Thanks
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You need the two identities:
$\displaystyle {n-1\choose k} + {n-1\choose k-1} = {n\choose k}$ and $\displaystyle \binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$
Let $\displaystyle S_n = \sum_{r=1}^{n} \frac{1}{r}{\binom{n}{r}}$ and note that (by def.) $\displaystyle {n-1\choose n} =0. $
$\begin{aligned} \sum_{r=1}^{n} \frac{1}{r}{\binom{n}{r}} & = \sum_{r=1}^{n} \frac{1}{r}{n-1\choose r} + \sum_{r=1}^{n}\frac{1}{r}{n-1\choose r-1} \\& = \sum_{r=1}^{n-1} \frac{1}{r}{n-1\choose r} + \sum_{r=1}^{n}\frac{r}{r \cdot n}{n\choose r} \\& = \sum_{r=1}^{n-1} \frac{1}{r}{n-1\choose r} + \frac{1}{n}\sum_{r=1}^{n}{n\choose r} \\& = \sum_{r=1}^{n-1} \frac{1}{r}{n-1\choose r} + \frac{2^n-1}{n} \\& = \frac{2^n-1}{n} +S_{n-1}\end{aligned} $
Working with this recurrent relation, with $S_0 = 0$ we have:
$\begin{aligned} S_n & = \frac{2^n-1}{n}+S_{n-1} \\& = \frac{2^{n}-1}{n}+\frac{2^{n-1}-1}{n-1}+S_{n-2} \\& = \frac{2^{n}-1}{n}+\frac{2^{n-1}-1}{n-1} + \cdots + \frac{2^{1}-1}{1} +S_0 \\& = \sum_{r=1}^{n} \frac{2^r-1}{r}\end{aligned} $
Alternatively: you've shown that
$$\displaystyle \int_{0}^{1}{\frac{(1 + x) ^ {n} - 1}{x}}dx = \sum_{r=1}^{n}\frac{2^r-1}{r}$$
Since $\displaystyle (1+x)^n-1 = \sum_{r=1}^{n}x^r \binom{n}{r}$ we can write:
\begin{aligned} \sum_{r=1}^{n}\frac{2^r-1}{r} & = \int_{0}^{1}{\frac{(1 + x) ^ {n} - 1}{x}} \, \mathrm dx \\& = \int_{0}^{1}\frac{1}{x} \sum_{r=1}^{n} x^r \binom{n}r \, \mathrm dx \\& = \int_{0}^{1} \sum_{r=1}^{n} x^{r-1} \binom{n}r \, \mathrm dx \\& = \sum_{r=1}^{n} \int_{0}^{1}x^{r-1} \binom{n}r \, \mathrm dx \\& = \sum_{r=1}^{n} \frac{1}{r} \binom{n}r.\end{aligned}
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Prove that $\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$
Prove that
$$J_n=\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$$
For $n=2$, it is OK. For general $n$, it seems impossible by integration by parts. Any other method?
When I calculate an integral $I_n=\int_0^\frac{\pi}{4}\frac{\cos nx}{\cos^nx}dx$, we find $I_n/2^{n-1}-I_{n-1}/2^{n-1}=-1/2^{n-1}J_n$. So we need to find the $J_n$, as the problem states.
The proof of $I_n/2^{n-1}-I_{n-1}/2^{n-1}=-1/2^{n-1}J_n$ is as follows.
\begin{align}
I_n/2^{n-1}-I_{n-1}/2^{n-1}
&=\frac{1}{2^{n-1}} \int_0^\frac{\pi}{4}\left(\frac{\cos nx}{\cos^nx}-\frac{2\cos(n-1)x}{\cos^{n-1}x}\right)dx\\
&=\frac{1}{2^{n-1}}\int_0^\frac{\pi}{4}\frac{\cos[(n-1)x+x]-2\cos(n-1)x\cos x}{\cos^nx}dx\\
&=-\frac{1}{2^{n-1}}\int_0^\frac{\pi}{4}\frac{\cos(n-2)x}{\cos^nx}dx
=-1/2^{n-1}J_n
\end{align}
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Edit: Much simpler answer: \begin{align}J_n(a)=\int_0^a\frac{\cos(n-2)x}{\cos^nx}\,dx&=\int_0^a\frac{\cos(n-1)x\cos x+\sin(n-1)x\sin x}{\cos^nx}\,dx\\&=\int_0^a\frac{\cos(n-1)x}{\cos^{n-1}x}-\frac{(-\sin x)\sin(n-1)x}{\cos^nx}\,dx\\&=\frac1{n-1}\int_0^a\frac d{dx}\frac{\sin(n-1)x}{\cos^{n-1}x}\,dx\\&=\frac{\sin(n-1)a}{(n-1)\cos^{n-1}a}\end{align} and taking $a=\pi/4$ gives the result.
We can also solve this using the complex exponential form of cosine: \begin{align}J_n&=\int_0^{\pi/4}\frac{\cos(n-2)x}{\cos^nx}\,dx\\&=2^{n-1}\int_0^{\pi/4}\frac{e^{(n-2)ix}+e^{-(n-2)ix}}{(e^{ix}+e^{-ix})^n}\,dx\\&=2^{n-2}\int_0^{\pi/2}\frac{e^{(n-1)it}+e^{it}}{(e^{it}+1)^n}\,dt\tag{$t=2x$}\\&=2^{n-2}\int_\gamma\frac{z^{n-2}+1}{i(z+1)^n}\,dz\tag{$\gamma(t)=e^{it},t\in[0,\pi/2]$}\\&=2^{n-2}\left[\frac{z^{n-1}-1}{i(n-1)(z+1)^{n-1}}\right]_{e^{i0}}^{e^{i\pi/2}}\\&=2^{n-2}\cdot\frac{i^{n-1}-1}{i(n-1)(i+1)^{n-1}}\\&=\frac{2^{n-2}}{i(n-1)}(w^{n-1}-\overline w^{n-1})\tag{$w=\frac i{i+1}=\frac{e^{i\pi/4}}{\sqrt2}$}\\&=\frac{2^{n-2}}{i(n-1)}\cdot\frac{2i\sin(n-1)\pi/4}{2^{(n-1)/2}}\\&=\frac{2^{(n-1)/2}}{n-1}\sin\frac{(n-1)\pi}4\end{align}
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Number theory problem (my solution included). Let n be a natural number greater than 1. We call a natural number $a>2$ ,$n-$decomposable if $a^n-2^n$ is divisible by all numbers of the form $a^d + 2^d$, where $d|n$ and $1\le d<n$. Find all $n$ for which there is an $n$-decomposable number.
I got:
$n$ is decomposable $\Leftrightarrow $ n is exactly the form $2^p$.
for "<=", divisors of $n=2^p$ is of form $d=2^q$ where $p\ge q$. since $a^n-2^n=(a-2)(a^1+2^1)(a^2+2^2)(a^4+2^4)(a^8+2^8)\dots(a^n+2^n)$, every $a^d+2^d$ is thereby a factor of $a^n-2^n$
for "=>", by contrapositive, assume $n$ is not $2^p$, then exist odd e and integer $d$ such that $n=de$. Let $b=a^d$ and hence $a^n=b^e$ , then $\frac{a^n-2^n}{a^d+2^d}=\frac{b^e-2^n}{b+2^d}$. by remainder theorem w.r.t. b, the "remainder" is $(-2^d)^e-2^n=-2^n-2^n=-2^{n+1}$.
Now we prove $-2^{n+1}$ is not divisible by $(b+2^d)$. Consider $a=3, b+2^d$ is odd, but $-2^{n+1}$ has only 1 odd factor, namely 1. So $n$ is not decomposable.
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The latter part of your proof doesn’t work - it only shows that $3$ is not $n$-decomposible (or by extension, odd numbers are not $n$-decomposible.) It doesn’t answer whether there is any $n$-decomposible number.
And, indeed, there are other $n.$
If $n=p$ is an odd prime, then the only condition is $a+2$ being a factor of $a^p-2^p.$
So $a\equiv-2\pmod {a+2},$ in which case, by your argument, $$a^{p}-2^{p}\equiv -2^{p+1}\pmod {a+2}.$$
So we only need $a+2\mid 2^{p+1}.$ So $a=2^k-2$ for some $3\leq k\leq p+1.$
For example, $a=2^3-2=6$ is always $p$-decomposible for all odd primes $p$ because $6+2=8$ and $6^p-2^p$ is divisible by $8.$
$a=2^5-2=30$ is $p$-decomposible for prime $p\geq 5.$
It might be the case that $n$ prime or $n>1$ a power of $2$ are all the cases, I don’t know. Certainly, it seems harder to find $a$ when $n$ has a lot of odd factors.
When $n$ is even and not a power of $2,$ showing there are no $n$-reducible numbers reduces to the case of solving $m^2+1=2^k,$ which only has solutions $m=0,1,$ by looking modulo $4.$
So you are left with $n$ odd and not prime. Your proof again shows that $a=2b$ must be even. Then you need $b^d+1$ is a power of $2$ less than or equal to $2^{n+1-d}$ for all factor $d<n.$ That seems unlikely for even $b+1$ and $b^p+1,$ but I don’t have the proof.
It certainly reduces the possible $a,$ since if $n>1$ is odd and not prime and $p$ is the is the smallest fact or $n,$ you get $b^{n/p}\leq 2^{n+p}$ or $b\leq 2^p\cdot 2^{p^2/n}\leq2^{1+\sqrt{n}}.$
|
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|
For $n \geq 2$, show that $\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} < \sqrt{2 ^ {n - 1} n ^ 3}$ Good day,
Can someone help me with giving hints for this problem:
Show that for $n \geq 2, n \in \mathbb{Z}$, $$\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} < \sqrt{2 ^ {n - 1} n ^ 3}$$
I tried $$\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} = \sum_{r = 1}^{n} \sqrt{r ^ 2\binom{n}{r}}$$
$$= \sqrt{n} \sum_{r = 1}^{n} \sqrt{r\binom{n - 1}{r - 1}}$$
So, we need to prove $$\sum_{r = 1}^{n} \sqrt{r\binom{n - 1}{r - 1}} < n \sqrt{2 ^ {n - 1}}$$
This reminds me of $$\sum_{r = 0}^{n} (r + 1) \binom{n}{r} = n 2 ^ {n - 1} + 2 ^ n = 2 ^ {n - 1}(n + 2)$$
but I can't see how to use it.
Also, the only complex math inequality I know is AM-GM, so it's quite possible that it is my theoretical knowledge that is lacking. So, if a well-known inequality is used, please also mention it.
Thanks
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By Cauchy-Schwarz,
\begin{align*}
\sum_{r = 1}^{n} r \sqrt{\binom{n}{r}} &\le \left(\sum_{r = 1}^{n} r^2\right)^{1/2}\left(\sum_{r = 1}^{n} \binom{n}{r}\right)^{1/2} \\
&= \left((2^n-1)\left(\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right)\right)^{1/2} \\
&< \left(2^{n-1}\cdot2\left(\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right)\right)^{1/2}
\end{align*}
Then, verify $2\left(\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right) < n^3$ for $n \ge 4$ by induction while manually checking the $n=2, n=3$ cases.
|
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|
Finding the limits while changing limit of an infinite sum into integral. I was solving the following question.
Find the following limit.
$$\lim_{n\to \infty}\dfrac1n \left(\dfrac{1}{1 + \sin\left(\dfrac{\pi}{2n}\right)} + \dfrac{1}{1 + \sin\left(\dfrac{2\pi}{2n}\right)} + \dfrac{1}{1 + \sin\left(\dfrac{3\pi}{2n}\right)} + ... +\dfrac{1}{1 + \sin\left(\dfrac{n\pi}{2n}\right)}\right)$$
Here's my method:
$$\lim_{n\to \infty}\dfrac1n \left(\dfrac{1}{1 + \sin\left(\dfrac{\pi}{2n}\right)} + \dfrac{1}{1 + \sin\left(\dfrac{2\pi}{2n}\right)} + \dfrac{1}{1 + \sin\left(\dfrac{3\pi}{2n}\right)} + ...+ \dfrac{1}{1 + \sin\left(\dfrac{n\pi}{2n}\right)}\right)$$
$$ = \lim_{n\to \infty}\dfrac1n \left(\sum_{k=1}^n\dfrac{1}{1 + \sin\left(\dfrac{k\pi}{2n}\right)}\right)$$
Now from the Riemann sum the limit is equal to,
$$\int\dfrac{1}{1 + \sin\left(\dfrac{x\pi}{2n}\right)} dx$$
Is it right? I don't know how to determine the limits of the integral as I'm new to this topic. Can anyone help me in this?
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Remark:$$\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n} f(\frac{k}{n})=\int_0^1 f(x)dx$$
$$= \lim_{n\to \infty}\dfrac1n \left(\sum_{k=1}^n\dfrac{1}{1 + \sin\left(\dfrac{k\pi}{2n}\right)}\right)
=\\ \lim_{n\to \infty}\dfrac1n \left(-1+\sum_{k=0}^n\dfrac{1}{1 + \sin\left(\dfrac{k\pi}{2n}\right)}\right)=\\
\lim_{n\to \infty}\dfrac1n \left(\sum_{k=0}^n\dfrac{1}{1 + \sin\left(\dfrac{k\pi}{2n}\right)}\right)-\lim_{n\to \infty }\frac1n=\\
\lim_{n\to \infty}\dfrac1n \left(\sum_{k=0}^nf(\frac{k}{n})\right)-\lim_{n\to \infty }\frac1n=\\\int_{0}^{1}\dfrac{1}{1 + \sin\left(\dfrac{x\pi}{2}\right)} dx
-\lim_{n\to \infty }\frac1n=\\\int_{0}^{1}\dfrac{1}{1 + \sin\left(\dfrac{x\pi}{2}\right)} dx
-0
$$ then you can use substitution $u=\frac{x\pi}{2}$ to solve the integral
$$\int \frac 1{1+\sin u}du=\int \frac{\sec^2(\frac u2)}{(\tan(\frac u2)+1)^2}du =\int \frac{f'}{f^2}du=\\-2\frac{1}{f}=\\\frac{-2}{\tan(\frac u2)+1}+c$$
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|
Probability of drawing 4 aces when drawing 5 cards from a regular deck of cards. I cannot seem to wrap my head around how to calculate the probability of drawing four ace when drawing five cards from a deck of cards. My intuition tells me the math below but its wrong for some reason...
$
\frac{\frac{4}{52}\frac{3}{51}\frac{2}{50}\frac{1}{49}C^{48}_{1}5!}{C^{52}_{5}}.$
$\frac{4}{52}\frac{3}{51}\frac{2}{50}\frac{1}{49}$ is the probability of drawing the four aces.
$C^{48}_{1}$ is all the different combinations we can draw all other cards
$5!$ are all the combinations that this can happen. For example, we can draw any card but an ace first, then four aces. We can draw an ace then the other three aces and some card, and so on. Thus we get 5 factorial.
$C^{52}_{5}$ is all of the combinations one can draw five cards from a deck of cards.
What is wrong with my reasoning and equation above?
It seems from other examples online that if we draw one type of ace (let's say heart) it does not matter that there are three other types of ace nor that we could draw a card that is not an ace at a certain other index (second, third fourth card, etc). How come?
From other posts and my textbook it seems that we can just assume we drew four ace and then we calculate the remaining cards. That yields the equation: $\frac{1*C^{48}_{1}}{C^{52}_{5}}$ where the $1$ is the fact that we are going to draw four ace so we can assume we already have. I am confused about where the different combinations of aces go and the different combinations in which we draw them together with the "extra" non ace card.
I have specified the apparent solution in the above paragraph. It has been given on multiple occasions throughout stackexchange and other parts of the internet. What these answers do not do is explain WHY.
|
The problem can be solved in two ways. We can either take the order of selection into account or not take it into account.
Taking the order of selection into account: There are $52 - 4 = 48$ ways to select the card which is not an ace. There are $5!$ ways to arrange the four aces and that card. There are $P(52, 5) = \binom{52}{5}5!$ ways to select five cards in order. Hence, the probability of selecting four aces when five cards are drawn is
$$\Pr(\text{all four aces are selected}) = \frac{48 \cdot 5!}{P(52, 5)} = \frac{48 \cdot 5!}{\dbinom{52}{5} \cdot 5!} = \frac{48}{\dbinom{52}{5}}$$
Not taking the order of selection into account: We must select all four aces and one of the other $48$ cards in the deck while selecting five of the $52$ cards in the deck. Hence, the probability of selecting four aces is
$$\Pr(\text{all four aces are selected}) = \frac{\dbinom{4}{4}\dbinom{48}{1}}{\dbinom{52}{5}} = \frac{48}{\dbinom{52}{5}}$$
What is wrong with your calculation?
If you take the order of selection into account in the numerator, you must also take it into account in the denominator. Also, in your numerator, you multiplied the probability of selecting four aces by the number of ways of selecting a non-ace. You need to multiply the number of ways of taking all four aces by the number of ways of selecting a non-ace.
Addendum: Here is another way of handling the ordered selection. There are $48$ ways to select the non-ace. There are four ways to select the first ace, three ways to select the second ace, two ways to select the third ace, and one way to select the fourth ace. We will consider cases, depending on the position of the non-ace.
\begin{align*}
\Pr(\text{all four aces are selected}) & = \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49} \cdot \frac{48}{48} + \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{48}{49} \cdot \frac{1}{48}\\
& \quad + \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{2}{49} \cdot \frac{1}{48} + \frac{4}{52} \cdot \frac{48}{51} \cdot \frac{3}{50} \cdot \frac{2}{49} \cdot \frac{1}{48}\\
& \qquad + \frac{48}{52} \cdot \frac{4}{51} \cdot \frac{3}{50} \cdot \frac{2}{49} \cdot \frac{1}{48}\\
& = \frac{5 \cdot 48 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{52 \cdot 53 \cdot 50 \cdot 49 \cdot 48}\\[2 mm]
& = \frac{48 \cdot 5!}{\frac{52!}{47!}}\\[2 mm]
& = \frac{48 \cdot 5!}{\frac{52}{47!5!} \cdot 5!}\\[2 mm]
& = \frac{48 \cdot 5!}{\binom{52}{5}5!}\\[2 mm]
& = \frac{48}{\binom{52}{5}}
\end{align*}
which agrees with the answer obtained above by first selecting the five cards and then arranging them.
|
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|
Find the number of rearrangements of AABBBCCDDD where there are no two consecutive As or Bs I tried using inclusion exclusion where set $A$ is all the rearrangements where two $A$s are consecutive and set $B$ where two $B$s are consecutive. However, I got $8400$ which is incorrect. I think it has to do with there being $3$ $B$s so I am somehow counting them wrong.
This was my process:
Total rearrangements of $AABBBCCDDD:$ ${10\choose 2,3,2,3} = 25200$
$\left|A\right| = {9\choose 1,3,2,3} = 5040$
$\left|B\right| = {9\choose 2,1,1,2,3} = 15120$
$\left|AB\right| = {8\choose 1,1,1,2,3} = 3360$
$A\cup B = 25200 -(5040+15120)+3360 = 8400$
Any help or guidance would be extremely appreciated.
|
This answer is rather a supplement which could be used as crosscheck for manual calculations. We consider a $4$-ary alphabet built from letters $\mathcal{V}=\{A,B,C,D\}$. Words which do not have any consecutive equal letters are called Smirnov words. A generating function for Smirnov words is given as
\begin{align*}
\left(1-\frac{Az}{1+Az}-\frac{Bz}{1+Bz}-\frac{Cz}{1+Cz}-\frac{Dz}{1+Dz}\right)^{-1}\tag{1}
\end{align*}
The coefficient $[z^n]$ of $z^n$ in the series (1) gives the number of $4$-ary words of length $n$ which do not have any consecutive letters.
In the current example we are looking for words which do not have consecutive letters $A$ and $B$. Since we do not have any restriction on $C$ or $D$ we replace $C$ and $D$ with one or more occurrences giving
\begin{align*}
Cz\quad&\to\quad Cz+\left(Cz\right)^2+\left(Cz\right)^3+\cdots=\frac{Cz}{1+Cz}\\
Dz\quad&\to\quad Dz+\left(Dz\right)^2+\left(Dz\right)^3+\cdots=\frac{Dz}{1+Dz}\tag{2}\\
\end{align*}
With some help of Wolfram Alpha we calculate the answer from (1) and (2) as
\begin{align*}
[z^{10}&A^2B^3C^2D^3]\left(1-\frac{Az}{1+Az}-\frac{Bz}{1+Bz}
-\frac{\frac{Cz}{1+Cz}}{1+\frac{Cz}{1+Cz}}-\frac{\frac{Dz}{1+Dz}}{1+\frac{Dz}{1+Dz}}\right)^{-1}\\
&=\color{blue}{[z^{10}A^2B^3C^2D^3]\left(1-\frac{Az}{1+Az}-\frac{Bz}{1+Bz}
-Cz-Dz\right)^{-1}}\\
&\,\,\color{blue}{=9\,660}
\end{align*}
Note: Smirnow words can be found for instance in example III.24 in Analytic Combinatorics by P. Flajolet and R. Sedgewick.
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|
Inverse Derivative: Error/Intuition Problem is to find $[f^{-1}(4)]'$ given $f(x)=\frac{x^3+7}{2}$.
Way One: Switch $x$ and $y$ to find inverse function. So, $x=\frac{y^3+7}{2}$. Therefore, $\sqrt[3]{2x-7}=y$ (i.e. our inverse function). So, $f^{-1}(x)=(2x-7)^{\frac{1}{3}}$. So, $\frac{d}{dx}f^{-1}(x)=\frac{2}{3}(2x-7)^{-\frac{2}{3}}\implies \frac{d}{dx}f^{-1}(4)=\frac{2}{3}.$
Way Two: Switch $x$ and $y$ to find inverse function. So, $x=\frac{y^3+7}{2}\implies 2x=y^3+7\implies 2=3y^2\frac{dy}{dx}\implies \frac{1}{\frac{3}{2}y^2}=\frac{dy}{dx}$.
Now, this to me seems like the inverse function is $\frac{d}{dx}f^{-1}|_{x=3}=\frac{dy}{dx}|_{x=3}=\frac{1}{\frac{3}{2}y^2}$ when $x=3$ we know $y=f(3)$ which isn't our answer. So, what am I doing wrong here? Can someone explain what is incorrect with this reasoning? I feel like I am missing something when doing these problems.
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It was tricky for me too so I started doing these problems the following way. You know that $f(f^{-1}(x)) = x$ so differentiating both sides gives you $$f'(f^{-1}(x)) \cdot \frac{d}{dx} f^{-1}(x) = 1 \implies \frac{d}{dx} f^{-1}(x) = \frac{1}{f'(f^{-1}(x)) }.$$ Now $f^{-1}(4) = 1$ so $$\frac{d}{dx} f^{-1}(4) = \frac{1}{f'(f^{-1}(4)) } = \frac{1}{f'(1)} = \frac{1}{3(1)^2/2} = \frac{2}{3}.$$
|
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Any hint on how to prove that the two given conditions may not be fulfilled simultaneously? I have these two conditions for $0<a<2\pi$ and $b>0$ and real.
$$ \sin \left(\frac{\pi a}{2 (\pi -a)}\right)=\frac{a }{2 \pi -a}\;\sin \left(\frac{\pi (a-2 \pi )}{2 (a-\pi )}\right)+\frac{4 b }{2 \pi -a}\;\sin \left(\frac{\pi ^2}{2 (\pi -a)}\right) $$
and
$$ \cos \left(\frac{\pi a}{2 (\pi -a)}\right)=\frac{a}{2 \pi -a} \;\cos \left(\frac{\pi (a-2 \pi )}{2 (a-\pi )}\right)\qquad\qquad\qquad \qquad\quad\qquad$$
As I check them numerically, I see that these two conditions may not be fulfilled simultaneously; Are there any hopes to prove this analytically?
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By the hint given by @Blue , considering $x:=\frac{\pi ^2}{2 (\pi -a)}$ my equations will be
$$ \sin \left( x-\frac\pi2\right)=\frac{a }{2 \pi -a}\;\sin \left(x+\frac\pi2\right)+\frac{4 b }{2 \pi -a}\;\sin \left(x\right) $$
$$ \cos \left(x-\frac\pi2\right)=\frac{a}{2 \pi -a} \;\cos \left(x+\frac\pi2\right)\qquad\qquad\qquad $$
Further, they can be simplified as
$$ -\cos x=\frac{a }{2 \pi -a}\;\cos x+\frac{4 b }{2 \pi -a}\;\sin \left(x\right) \qquad (1) $$
$$ \sin x=\frac{-a}{2 \pi -a} \;\sin x\qquad\qquad\qquad \qquad (2) $$
The equation $(2)$ may be fulfilled either by $\sin x=0$ or $\frac{-a}{2 \pi -a} =1$ where the latter is not valid. Investigating then $(1)$ for $\sin x=0$ in which $\cos x=\pm1$, we get
$$ \mp1=\frac{a }{2 \pi -a} ,$$
which never holds which completes the proof.
|
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Attempt to prove an inequality by induction I am trying to prove an inequality by induction which is as follows:
$$\frac{(2n)!}{2^{2n}\cdot (n!)^2} \le \frac{1}{\sqrt{3n + 1}}$$
Base Case, i.e, for n =1,
$$\frac{(2(1))!}{2^{2(1)}\cdot ((1)!)^2} \le \frac{1}{\sqrt{3(1) + 1}}$$
$$\implies \frac{2!}{2^{2}\cdot (1)^2} \le \frac{1}{\sqrt{3 + 1}}$$
$$\implies \frac{2}{4} \le \frac{1}{2}$$
$$\implies \frac{1}{2} \le \frac{1}{2}$$
which is true
Assume that the inequality is true for $n=m$ where $m \in \mathbb{N}$ , i.e,
$$\frac{(2m)!}{2^{2m}\cdot (m!)^2} \le \frac{1}{\sqrt{3m + 1}}$$
$$\implies \quad \quad \quad\quad\quad\quad\quad \frac{1\cdot2\cdot3\cdot .....\cdot (2m-1)\cdot (2m)}{2^{2n}\cdot (1\cdot2\cdot3\cdot .....\cdot (m-1) \cdot m)^2} \le \frac{1}{\sqrt{3m + 1}} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$
$$\implies \quad \quad \quad \frac{1\cdot2\cdot3\cdot .....\cdot (2m-1)\cdot (2m) \cdot(2m+1)\cdot(2m +2)}{2^{2m}\cdot (1\cdot2\cdot3\cdot .....\cdot (m-1) \cdot m)^2\cdot(2m+2)\cdot (m+1)\cdot2} \le \frac{1}{\sqrt{3m+ 1}}\cdot \frac{(2m +1) }{(m+1)\cdot 2} \quad\quad\quad\quad\quad\quad\quad\quad$$
$$\implies \frac{(2m+2)!}{2^{2m +2}\cdot (1\cdot2\cdot3\cdot .....\cdot (m-1) \cdot m \cdot (m+1))^2 }\le \frac{1}{\sqrt{3m+ 1}}\cdot \frac{(2m +1) }{(m+1)\cdot 2}$$
$$\implies \frac{(2(m+1))!}{2^{2(m +1)}\cdot ((m+1)!)^2 }\le \frac{1}{\sqrt{3m+ 1}}\cdot \frac{(2m +1) }{(m+1)\cdot 2}$$
$$\implies \frac{(2(m+1))!}{2^{2(m +1)}\cdot ((m+1)!)^2 }\le \frac{1}{\sqrt{12m+ 4}}\cdot \frac{(2m +1) }{(m+1)}\quad \quad ......(i)$$$$
Now , I want $LHS$ to be smaller than or equalto $\frac{1}{\sqrt{3m+4}}$. For this I thought to prove that
$$\frac{1}{\sqrt{12m+ 4}}\cdot \frac{(2m +1) }{(m+1)} \le \frac{1}{\sqrt{3m+4}} \quad \quad ......(ii)$$
Now ,
$$12m>3m$$
$$12m + 4 > 3m +4$$
$$\sqrt{12m +4} > \sqrt{3m +4}$$
$$\frac{1}{\sqrt{12m +4}} < \frac{1}{\sqrt{3m +4}}$$
Then, I thought that if $\frac{(2m +1) }{(m+1)}$ is smaller than $1$ , i would get inequality (i) but this is false. So I am lost.
Can anyone help me to proceed further to get the desired result?
Thanking in advance
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I think you messed something up in the computation.
If you want to prove this using induction, what you need to show is that
\begin{equation*}
\frac{1}{\sqrt{3n+1}}\frac{(2n+1)(2n+2)}{4(n+1)^2}\leq \frac{1}{\sqrt{3n+4}}.
\end{equation*}
After simplifying this means
\begin{equation*}
\frac{(2n+1)^2}{(2n+2)^2}\leq \frac{3n+1}{3n+4},
\end{equation*}
which you can easily check by direct computation.
|
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|
Computing $\int_{-1}^{1} \frac{\ln (1+y)}{y} d y$ with simple method. Using the series
$$
\ln (1+y)=\sum_{n=0}^{\infty} \frac{(-1)^{n} y^{n+1}}{n+1} \text { for }|y|<1
$$
to convert the integral into $$
\begin{aligned}
\int_{-1}^{1} \frac{\ln (1+y)}{y} d y &=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} \int_{-1}^{1} y^{n} d y \\
&=\sum_{n=0}^{\infty} \frac{1-(-1)^{n+1}}{(n+1)^{2}} \\
&=\zeta(2)+\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n+1)^{2}} \\
&=\zeta(2)+\frac{1}{2}\zeta(2)\\&=\frac{\pi^{2}}{4}
\end{aligned}
$$
Question: Is there any other simple solution? Your suggestion and solution is warmly welcome.
|
An elementary solution
\begin{align}
\int_{-1}^{1} \frac{\ln (1+y)}{y} &{d y}
\overset{y\to -y} = \frac12 \int_{-1}^{1} \frac{\ln \frac{1+y}{1-y}}{y} d y
\overset{y\to \frac{1+y}{1-y}}=\int_0^\infty
\frac{\ln y}{y^2-1}dy\\
=& \int_0^\infty \int_0^1 \frac{x}{1+(y^2-1)x^2}dx\>dy
=\frac\pi2 \int_0^1 \frac1{\sqrt{1-x^2}}dx=\frac{\pi^2}4
\end{align}
|
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Integral $ \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx$ For $a\in\mathbb R$, I want to evaluate the integral
$$ I = \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx.$$
I tried to integration by parts by considering $\left( \tan^{-1}(x) \right)' = \frac{1}{x^2+1}$, so that
$$I = - \int_{-\infty}^\infty \tan^{-1}(x) \left(\frac{1}{1+x^2} - \frac{1}{1+(x-a)^2}\right).$$
However, I cannot proceed further. Mathematica cannot solve both integrals.
How to evaluate $I$?
|
We proceed by FShrike's comment. We have
$$\frac{\partial I}{\partial a} = \int \frac{1}{x^2+1} \frac{1}{(a-x)^2+1} dx = \frac{2\pi}{a^2+4}.$$
Since $I(a=0)=0$, we obtain
$$I(a) = \int_0^a \frac{2\pi}{a^2+4}da = \pi \tan^{-1}\frac a2.$$
|
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|
How to deal with odd $m$ in integral $\int_{0}^{\frac{\pi}{4}}(\sin^{6}m x+\cos^{6}m x) \ln (1+\tan x) d x $ Latest edit
*
*Thanks to @Quanto for settling down the question by proving the odd one as:$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)$$
By our results for both odd and even multiples $n$ of $x$, we can conclude that
$$
\lim _{n \rightarrow \infty} \displaystyle \int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(nx)+\cos ^{6}(nx)\right] \ln (1+\tan x) d x =\frac{5 \pi\ln 2}{64}
$$
*As asked by @Claude Leibovici for the powers other than 6, I had generalised my result to even powers below as an answer:
$$
I(m,n):=\int_{0}^{\frac{\pi}{4}}\left[\cos ^{2 m}(2 nx)+\sin ^{2 m}(2 n x)\right] \ln (1+\tan x) d x= \frac{\pi \ln 2}{4} \cdot \frac{(2 m-1) ! !}{(2 m) ! !}
$$
In order to evaluate the even case
$$\int_{0}^{\frac{\pi}{4}}\left[\sin^{6}(2 n x)+\cos^{6}(2 nx)\right] \ln (1+\tan x) d x $$
we first simplify
$\displaystyle \begin{aligned}\sin ^{6}(2 n x)+\cos ^{6}(2 n x) =& {\left[\sin ^{2}(2 n x)+\cos ^{2}(2 n x)\right]\left[\sin ^{4}(2 n x)-\sin ^{2}(2 n x) \cos ^{2}(2 n x)\right) } \\&\left.+\cos ^{4}(2 n x)\right] \\=& 1-3 \sin ^{2}(2 n x) \cos ^{2}(2 n x) \\=& 1-\frac{3}{4} \sin ^{2}(4 n x) \\=& 1-\frac{3}{8}(1-\cos 8 n x) \\=& \frac{1}{8}(5+3 \cos (8nx))\end{aligned} \tag*{} $
To get rid of the natural logarithm, a simple substitution transforms the integral into
$\begin{aligned}I &=\frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln (1+\tan x) d x \\& \stackrel{x\mapsto\frac{\pi}{4}-x}{=} \frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \\&=\frac{1}{8} \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x)) \ln \left(\frac{2}{1+\tan x}\right) d x \\&=\frac{1}{8} \ln 2 \int_{0}^{\frac{\pi}{4}}(5+3 \cos (8 n x) )d x-I \\I &=\frac{\ln 2}{16} \int_{0}^{\frac{\pi}{4}}(5+3 \cos 8 n x) d x\\&=\frac{\ln 2}{16}\left[5 x+\frac{3}{8 n} \sin (8 n x)\right]_0^{\frac{\pi}{4} }\\ &=\frac{5 \pi}{64} \ln 2\end{aligned} \tag*{} $
My Question:
How can we deal with the odd one
$$\displaystyle \int_{0}^{\frac{\pi}{4}}\left[\sin ^{6}(2 n +1)x+\cos ^{6}(2 n +1)x\right] \ln (1+\tan x) d x ?$$
Can you help?
|
Since @Quanto provided a detailed and very good answer, I shall make the story short.
Working directly the problem of the difference between two consecutive terms
$$I_{2n+1}-I_{2n}=-\frac{3 }{32 } \frac{1 }{2 n+1}\Phi \left(-1,1,\frac{4n+3}{2}\right)$$ where appears the Lerch transcendent function.
We can also write it as
$$I_{2n+1}-I_{2n}=-\frac{3 }{64 }\frac{1 }{2 n+1} \left(H_{n+\frac{1}{4}}-H_{n-\frac{1}{4}}\right)$$ These are very small numbers. Expanded as series
$$I_{2n+1}-I_{2n}=-\frac{3}{256 n^2}\Bigg[1-\frac{1}{n}+\frac{11}{16 n^2}-\frac{3}{8 n^3}+O\left(\frac{1}{n^4}\right) \Bigg]$$
|
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"url": "https://math.stackexchange.com/questions/4436587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
If a prime divides an integer of the form $n^2+1$ then ... This question is from my problem set in number theory. I tried it a week ago and again today but both times I couldn't solve it.
Problem: If a prime p divides an integer of the form $n^2+1$ , then $p\not\equiv 3 \pmod{4}$.
I thought of attempting this problem by assuming $p\equiv 3\pmod{4} $. But I am unable to move towards anything concrete.
Can you please give some hints?
Thanks!
|
You're right on assuming that $p\equiv 3\pmod{4}$ and then trying arrive at a contradiction. Here's how it may work out.
Suppose that $p\equiv 3\pmod{4}$ and $p$ divides $n^2+1.$ If $p$ divides $n^2+1,$ then we have;
$$n^2+1\equiv 0\pmod{p}$$
$$n^2\equiv -1\pmod{p}.$$
Now, since we have assumed that $p\equiv 3\pmod{4}$ we know that $p=4k+3$ for some positive $k.$ Thus, we have;
\begin{align*}
n^{p}
&\equiv n^{4k+3}\pmod{p}\\
&\equiv n\left(n^{4k+2}\right)\pmod{p}\\
&\equiv n\left(n^2\right)^{2k+1}\pmod{p}.
\end{align*}
Note that $n^2\equiv -1\pmod{p},$ so we have;
\begin{align*}
n^{p}
&\equiv n\left(n^2\right)^{2k+1}\pmod{p}\\
&\equiv n\left(-1\right)^{2k+1}\pmod{p}\\
&\equiv n(-1)\pmod{p}\\
&\equiv -n\pmod{p}.
\end{align*}
But by the Fermat's Little Theorem, we have $n^p\equiv n\pmod{p}.$ So we have $n\equiv -n\pmod{p},$ which simplifies to $p\mid 2n.$ However, we can see that this is incorrect as $p$ also divides $n^2+1$ which is coprime to $2n$. So we have finally arrived a contradiction and your claim is proven.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Expected value of the distance of sample average from the overall average Let $\mathbf{y}_1, \mathbf{y}_2, \dots, \mathbf{y}_N$ be a sequence of $N$ vectors in $\mathbb{R}^d$ and $\bar{\mathbf{y}}_N$ be the overall average, i.e.,
$$
\bar{\mathbf{y}}_N=\frac{1}{N}\sum_{i=1}^{N}\mathbf{y}_i.
$$
Also, let $\bar{\mathbf{y}}$ be any sample average of size $n$ such that $\bar{\mathbf{y}}=\frac{1}{n}\sum_{i \in A}\mathbf{y}_i$ where $A$ is a random set such that $A \subset \{1, \dots, N\}$ and $|A|=n<N$. Note that elements in $A$ are drawn without replacement.
Can we show
$$
\mathbb{E}_A[\|\bar{\mathbf{y}}-\bar{\mathbf{y}}_N\|^2] = \frac{N-n}{n}\frac{1}{N}e
$$
where $e=\frac{1}{N-1}\sum_{i=1}^N\|\mathbf{y}_i-\bar{\mathbf{y}}_N\|^2$?
My try:
The above problem is solved in Sampling: design and analysis page 45-46. I do not know how to do it efficiently by defining $Z_i$'s as random variables that take only zero or one like the scalar case. I also think maybe it is better to write $\mathbb{E}_A[\|\bar{\mathbf{y}}-\bar{\mathbf{y}}_N\|^2]$ as $\mathbb{E}_A[ \langle \bar{\mathbf{y}}-\bar{\mathbf{y}}_N, \bar{\mathbf{y}}-\bar{\mathbf{y}}_N\rangle]$ and write the following:
$$
\mathbb{E}_A[ \langle \bar{\mathbf{y}}-\bar{\mathbf{y}}_N, \bar{\mathbf{y}}-\bar{\mathbf{y}}_N\rangle]
=
\mathbb{E}_A[ \langle \frac{1}{n}\sum_{i=1}^N Z_i\mathbf{y}_i-\bar{\mathbf{y}}_N, \frac{1}{n}\sum_{i=1}^N Z_i\mathbf{y}_i-\bar{\mathbf{y}}_N\rangle]
$$
|
Let $\mathbf{Y} = \begin{pmatrix} \mathbf{y}_1 & \cdots & \mathbf{y}_N \end{pmatrix} \in \mathbb{R}^{n \times N}$. Then notice that
\begin{align*}
\overline{\mathbf{y}} - \overline{\mathbf{y}}_N = \mathbf{Y}\left(n^{-1} \mathbf{1}_A - N^{-1}\mathbf{1}_N\right)
\end{align*}
where $\mathbf{1}_N, \mathbf{1}_A \in \mathbb{R}^{N \times 1}$ with $\mathbf{1}_N$ full of 1's, and $(\mathbf{1}_A)_i = 1$ if $i \in A$, and 0 otherwise. The only thing random in this expression is $\mathbf{1}_A$. So we have
\begin{align*}
\|\overline{\mathbf{y}} - \overline{\mathbf{y}}_N\|^2 &= \left(N^{-1}\mathbf{1}_N - n^{-1} \mathbf{1}_A\right)^\intercal \mathbf{Y}^\intercal\mathbf{Y}\left(N^{-1}\mathbf{1}_N - n^{-1} \mathbf{1}_A\right) \\
&= N^{-2}\mathbf{1}_N^\intercal \mathbf{Y}^\intercal\mathbf{Y}\mathbf{1}_N - 2(nN)^{-1}\mathbf{1}_N^\intercal\mathbf{Y}^\intercal\mathbf{Y}\mathbf{1}_A + n^{-2}\mathbf{1}_A^\intercal \mathbf{Y}^\intercal\mathbf{Y}\mathbf{1}_A \\
\end{align*}
Using the fact that $\mathbb{E}[\mathbf{1}_A] = \frac{n}{N}\mathbf{1}_N$ and $\text{Cov}(\mathbf{1}_A) = \frac{n}{N}(1 - \frac{n-1}{N-1})\mathbf{I} + \frac{n}{N}(\frac{n-1}{N-1} - \frac{n}{N})\mathbf{1}_N\mathbf{1}_N^\intercal \overset{\text{def}}{=}\Sigma$ and facts about quadratic forms, we have
\begin{align*}
\mathbb{E}\|\overline{\mathbf{y}} - \overline{\mathbf{y}}_N\|^2 &= N^{-2}\mathbf{1}_N^\intercal \mathbf{Y}^\intercal\mathbf{Y}\mathbf{1}_N - 2(nN)^{-1}\mathbf{1}_N^\intercal\mathbf{Y}^\intercal\mathbf{Y}\left(\frac{n}{N}\mathbf{1}_N\right) \\
&\qquad + n^{-2}\left(\left(\frac{n}{N}\mathbf{1}_N\right)^\intercal \mathbf{Y}^\intercal\mathbf{Y}\left(\frac{n}{N}\mathbf{1}_N\right) + \text{tr}(\mathbf{Y}^\intercal \mathbf{Y}\Sigma)\right)\\
&= n^{-2}\text{tr}(\mathbf{Y}^\intercal \mathbf{Y}\Sigma) \\
&=n^{-2}\left(\frac{n}{N}\left(1 - \frac{n-1}{N-1}\right)\text{tr}(\mathbf{Y}^\intercal\mathbf{Y}) - \frac{n}{N}\left(\frac{n}{N} - \frac{n-1}{N-1}\right)\text{tr}(\mathbf{1}_N^\intercal \mathbf{Y}^\intercal \mathbf{Y} \mathbf{1}_N)\right) \\
&= n^{-2}\left(\frac{n}{N}\left(1 - \frac{n-1}{N-1}\right)\sum_{i=1}^{N}\|\mathbf{y}_i\|^2 - \frac{n}{N}\left(\frac{n}{N} - \frac{n-1}{N-1}\right)N^2 \overline{\mathbf{y}}_N^\intercal\overline{\mathbf{y}}_N\right) \\
&= n^{-2}\left(\frac{n}{N}\left(1 - \frac{n-1}{N-1}\right)\sum_{i=1}^{N}\|\mathbf{y}_i\|^2 - \frac{n}{N}\left(1 - \frac{n-1}{N-1}\right)N \overline{\mathbf{y}}_N^\intercal\overline{\mathbf{y}}_N\right) \\
&= n^{-2}\frac{n}{N}\left(1 - \frac{n-1}{N-1}\right)\sum_{i=1}^{N}\|\mathbf{y}_i - \overline{\mathbf{y}}_N\|^2 \\
&= \frac{N-n}{n}\frac{1}{N}e
\end{align*}
where
\begin{align*}
\sum_{i=1}^{N} \|\mathbf{y}_i - \overline{\mathbf{y}}_N\|^2 &= \sum_{i=1}^{N} (\mathbf{y}_i^\intercal \mathbf{y}_i - 2 \mathbf{y}_i^\intercal \overline{\mathbf{y}}_N + \overline{\mathbf{y}}_N^\intercal\overline{\mathbf{y}}_N) \\
&= \left(\sum_{i=1}^{N} \mathbf{y}_i^\intercal \mathbf{y}_i\right) - 2 \left(\sum_{i=1}^{N}\mathbf{y}_i\right)^\intercal \overline{\mathbf{y}}_N + N \overline{\mathbf{y}}_N^\intercal\overline{\mathbf{y}}_N) \\
&= \left(\sum_{i=1}^{N} \mathbf{y}_i^\intercal \mathbf{y}_i\right) - 2 N \overline{\mathbf{y}}_N^\intercal \overline{\mathbf{y}}_N + N \overline{\mathbf{y}}_N^\intercal\overline{\mathbf{y}}_N) \\
&= \left(\sum_{i=1}^{N} \mathbf{y}_i^\intercal \mathbf{y}_i\right) - N \overline{\mathbf{y}}_N^\intercal\overline{\mathbf{y}}_N)
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
The number of integer points on the curve $(7x-1)^2+(7y-1)^2=n$ The number of integral solutions to the equation
$$x^2+y^2=n$$
is defined to be $r_2(n)$ and if $n=2^ap_1^{a_1}\dots p_k^{a_k}q^2$ where $p_i\equiv 1\mod 4$ and $q$ is the product of primes which are $3\mod 4$, then
$$r_2(n)=4(a_1+1)\dots(a_k+1).$$
If we restrict $x\equiv y\equiv 1\mod 2$, that is search for solutions to
$$(2x-1)^2+(2y-1)^2=n$$
we find that if $n\equiv 2$ all solutions to the previous one are also solutions to this and otherwise none are. So the number of solutions is $r_2(n)$ or $0$ according to whether $n\equiv 2$ or not.
And if we restrict $x\equiv y\equiv 1\mod 4$, we find that each quadruple of solutions $(\pm x)^2+(\pm y)^2=n$ generates exactly one solution to the new equation. Therefore the number of solutions to $(4x-1)^2+(4y-1)^2=n$ is $\frac{r_2(n)}4$ if $n\equiv 2\mod 4$ and $0$ otherwise.
In general I want to know how many solutions there are for the equation
\begin{equation}\tag{1}
(mx-a)^2+(my-b)^2=n.
\end{equation}
The idea above solves the case $m=2,3,4,5,6$ for all values of $a,b$ but for $m=7$ it doesn't work anymore. This is because $7$ is the first integer for which something like
$$0^1+1^2\equiv 2^2+2^2\mod 7$$
happens. To be precise, $m=7$ is the least so that there are $a,b,c,d$ so that
$$a^2+b^2\equiv c^2+d^2\mod m$$
$$\{a,b\}\neq \{\pm c,\pm d\}$$
for all choices of signs. This means that the number of solutions when $m=7$ is not what you expect. For example the equation $(7x-1)^2+(7y-1)^2=9$ has no solutions, but I expected there to be$\frac{r_2(9)}4=1$. Strangely, the equation $(7x-1)^2+(7y-2)^2=n$ has exactly half as many solutions as I expected for all $n$ up to ten thousand. Since $1^2+2^2=5$, I reasoned that $5$ would behave differently from the other primes, so I excluded its exponent from the product of $r_2(n)$. Therefore I conjecture that the number of solutions to
$$(7x-1)^2+(7y-2)^2=n$$
where $n\equiv 5\mod 7$ and $n=2^a5^bp_1^{a_1}\dots p_k^{a_k}q^2$ is $\frac 12(a_1+1)\dots (a_k+1)=\frac 18r_2\left(\frac n{5^b}\right)$. This product is only odd if $n$ is a square, which is impossible so that doesn't come up.
What I want to know is how many solutions to (1) there are, and in particular the special cases above.
|
Comment: We may use following equation as a particular case:
$$(a-1)^2+(a+1)^2=b^2+1\space\space\space\space\space\space(1)$$
This equation can have infinitely many solution; if a and b satisfy this equation then considering following identity we can have subsequent solutions:
$$(2b+3a-1)^2+(2b+3a+1)^2=(3b+4a)^2+1\space\space\space\space\space\space(2)$$
Now we want to solve equation:
$(mx-1)^2+(my-1)^2=n\space\space\space\space\space\space(3)$
Comparing this equation with (1) we construct following system of equation:
$\begin {cases}mx-1=a-1\Rightarrow a=mx\\my-1=a+1\Rightarrow a=my-2\\ n=b^2+1\end{cases}$
For example $(a, b)=(2, 3)$ satisfies equation (1) and we have:
$(2-1)^2+(2+1)^2=3^2+1=10=2\times 5; 5\equiv 1\bmod 4$
Now we use $(a, b)=(2, 3)$ in identity (2):
$a_1=2\times 3++3\times 2=12$
$b_1=3\times 3+4\times 2=17$
Which gives:
$(12-1)^2+(12+1)^2=17^2+1=290$
Now we can rewrite :
$12-1=2\times 6-1$
$13=2\times 7-1$
Which gives $(x, y, m, n)=(6, 7, 2, 290)$
OR:
$12-1=3\times 5-4$
$13=3\times 6-5$
Which for equation $(mx-c)^2+(my-d)^2=n$ gives:
$(x, y, m, c, d, n)=(5, 6, 3, 4, 5, 290)$
We can continue this and find more solution, one more example:
$a_2=2\times 17+3\times 12=70$
$b_2=3\times 17+4\times 12=99$
$(70-1)^2+(70+1)^2=99^2+1=9802=2\cdot13^2\cdot29$
$13\equiv 1\bmod 4$ also $29\equiv 1\bmod 4$
$70-1=69=7\times 10-1$
$70+1=72=7\times 11-5$
$(x, y, m, c, d, n)=(10, 11, 7, 1, 5, 9802)$
I do not see restrictions for x, y, m and n and limit for number of solutions.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\cot\left(\frac{\pi}{4}+\beta\right)+\frac{1+\cot\beta}{1-\cot\beta}=-2\tan2\beta$ Show that $$\cot\left(\dfrac{\pi}{4}+\beta\right)+\dfrac{1+\cot\beta}{1-\cot\beta}=-2\tan2\beta$$
I'm supposed to solve this problem only with sum and difference formulas (identities).
So the LHS is $$\dfrac{\cot\dfrac{\pi}{4}\cot\beta-1}{\cot\dfrac{\pi}{4}+\cot\beta}+\dfrac{1+\cot\beta}{1-\cot\beta}=\dfrac{\cot\beta-1}{1+\cot\beta}+\dfrac{1+\cot\beta}{1-\cot\beta}=\dfrac{4\cot\beta}{1-\cot^2\beta}$$ I also tried to work with $\sin\beta$ and $\cos\beta$ and arrived at $$\dfrac{4\sin\beta\cos\beta}{\sin^2\beta-\cos^2\beta}$$ I don't see how to get $-2\tan2\beta$ from here (even with other identities).
|
The alternative approach is to stay with your original approach and complete it.
$\displaystyle \frac{4\cot\beta}{1-\cot^2\beta}
= \frac{\frac{4}{\tan(\beta)}}{1 - \frac{1}{\tan^2(\beta)}}
= \frac{\frac{4}{\tan(\beta)}}{\frac{\tan^2(\beta) - 1}{\tan^2(\beta)}}
$
$\displaystyle = ~\frac{4\tan(\beta)}{\tan^2(\beta) - 1}
= (-2) \times \frac{2\tan(\beta)}{1 - \tan^2(\beta)} = (-2) \times \tan(2\beta).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4447522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Difficulty evaluating $\int_0^\infty\frac{1}{(x^3+2)\sqrt{x^2+8}}\,\mathrm{d}x$ In evaluating
$$\int_0^\infty\frac{1}{(x^3+2)\sqrt{x^2+8}}\,\mathrm{d}x$$ I did not have a situation where the polynomial under the square root has lower degree. Neither trigonometric substitutions nor variable changes help, at least how I apply them.
I would like at least to know how to evaluate a similar integral, because I already spent a lot of time on this one.
|
The limits of integration and the form of the radical in the denominator strongly suggest $x = 2\sqrt{2}\sinh(u)$. Performing this substitution gives and expanding in partial fractions gives
$$
\int_0^\infty \frac{dx}{(x^3 + 2)\sqrt{x^2+8}} =\frac{1}{2}\int_0^\infty\frac{du}{1 + 2^{7/2}\sinh^3(u)}
=\frac{1}{6}\sum_{i=-1}^1\int_0^\infty\frac{du}{1 + 2^{7/6}\omega^i \sinh(u)}
$$
where $\omega = \exp(2\pi i/3)$. This last integral is doable, but it's not all that pretty. You get
$$
\int_0^\infty \frac{du}{1+a\sinh(u)} = \frac{\cosh^{-1}(1 + a + a^{-1})}{\sqrt{a}\sqrt{a + a^{-1}}},
$$
where the principal branch of the square root is used. The result is
$$
\int_0^\infty \frac{dx}{(x^3 + 2)\sqrt{x^2+8}} =\frac{1}{6}\sum_{i=-1}^1\frac{\cosh^{-1}(1 + 2^{7/6}\omega^i + 2^{-7/6}\omega^{-i})}{\sqrt{2^{7/6}\omega^i}\sqrt{2^{7/6}\omega^i + 2^{-7/6}\omega^{-i}}}
$$
And to check numerically, the integral equals the sum.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Question on Lagrange multipliers method
My understanding
Finding the minimum of the function using Lagrange multipliers method
\begin{align}
f(x,y,z) &= (y^2+z^2-x^2)^2+4x^2(x+y)^2 \\
x+y+z &= 1 \\
x &\leq y \\
x, y, z &\geq 0
\end{align}
My approach is to let
$$h(x,y,z)=f-\lambda g$$ where $~g=x+y+z-1=0$.
Geometrically to observe all the constraints we get a triangular shape in 1st octant in which the function $f$ has to minimize.
Now using the Lagrange method
\begin{align}
\frac{\partial h }{\partial x} &= -4x(y^2+z^2-x^2)+8x(x+y)^2+8x^2(x+y)-\lambda &=0 \\
\frac{\partial h }{\partial y} &= 4y(y^2+z^2-x^2)+8x^2(x+y)-\lambda &=0 \\
\frac{\partial h }{\partial z} &= 4z(y^2+z^2-x^2)-\lambda &=0 \\
\frac{\partial h}{\partial \lambda} &= x+y+z-1 &=0
\end{align}
Now my doubt, (1) In order to solve these above equation we first find $x,y,z$ in terms of $\lambda $ and put all those value in the last equation to find value of $\lambda, $ But
how to find $x,y,z$ in terms of $\lambda $??
(2) Can we say the given function attains its minimum in the triangle formed by all the constraints and So we first find value of f(x,y,z) on the three side of that triangle and then in order to find the internal minimum point then we apply Lagrange method?? This technique is new to me and it is given in one research paper. Any comment on this ??
he first calculate f(x,y,z) on x=0,x=y and z=0 and he apply Lagrange method to find internal minimum point, and finally he prove that minimum of f is at x=y=1/4 and z=1/2.
Any help is appreciated.
thanks
|
By setting $\lambda=4\mu$ you get
$$
\left\{ \begin{array}{l}
- x\left( {y^2 + z^2 - x^2 } \right) + 2x\left( {x + y} \right)^2 + 2x^2 \left( {x + y} \right) = \mu \\
y\left( {y^2 + z^2 - x^2 } \right) + 2x^2 \left( {x + y} \right) = \mu \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
Thus, from the third equation, you have
$$
\left\{ \begin{array}{l}
- x\left( {y^2 + z^2 - x^2 } \right) + 2x\left( {x + y} \right)^2 + 2x^2 \left( {x + y} \right) = z\left( {y^2 + z^2 - x^2 } \right) \\
y\left( {y^2 + z^2 - x^2 } \right) + 2x^2 \left( {x + y} \right) = z\left( {y^2 + z^2 - x^2 } \right) \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
Hence
$$
\left\{ \begin{array}{l}
2x\left( {x + y} \right)^2 + 2x^2 \left( {x + y} \right) = \left( {x + z} \right)\left( {y^2 + z^2 - x^2 } \right) \\
2x^2 \left( {x + y} \right) = \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
From the second equation, you get
$$
\left\{ \begin{array}{l}
2x\left( {x + y} \right)^2 + \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) = \left( {x + z} \right)\left( {y^2 + z^2 - x^2 } \right) \\
2x^2 \left( {x + y} \right) = \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
Therefore
$$
\left\{ \begin{array}{l}
2x\left( {x + y} \right)^2 = \left( {x + y} \right)\left( {y^2 + z^2 - x^2 } \right) \\
2x^2 \left( {x + y} \right) = \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
Hence
$$
\left\{ \begin{array}{l}
\left( {x + y} \right)\left[ {2x\left( {x + y} \right) - \left( {y^2 + z^2 - x^2 } \right)} \right] = 0 \\
2x^2 \left( {x + y} \right) = \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
*
*First case: if $x+y=0$ you easily get $z=1$ and then $y=1,x=-1$ and this not acceptable.
*Second case: $x+y\neq 0$. Then, from the first equation you have
$$
\left\{ \begin{array}{l}
2x\left( {x + y} \right) = \left( {y^2 + z^2 - x^2 } \right) \\
2x^2 \left( {x + y} \right) = \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
Now, by substituting the first equation in the second, you have
$$
\left\{ \begin{array}{l}
2x\left( {x + y} \right) = \left( {y^2 + z^2 - x^2 } \right) \\
x^2 = x\left( {z - y} \right) \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
thus
$$
\left\{ \begin{array}{l}
2x\left( {x + y} \right) = \left( {y^2 + z^2 - x^2 } \right) \\
x\left( {x - z + y} \right) = 0 \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
Again, from the second, with $x=0$ you get a non acceptable solution, while, if $x \neq 0$, you have
$$
\left\{ \begin{array}{l}
2x\left( {x + y} \right) = \left( {y^2 + z^2 - x^2 } \right) \\
x - z + y = 0 \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y + z = 1 \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
From the second and the fourth equation, you have
$$
\left\{ \begin{array}{l}
2x\left( {x + y} \right) = \left( {y^2 + z^2 - x^2 } \right) \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y = \frac{1}{2} \\
z = \frac{1}{2} \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
Now you can rewrite the first equation as
$$
\left\{ \begin{array}{l}
2x\left( {x + y} \right) = \left( {y - x} \right)\left( {y + x} \right) + z^2 \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y = \frac{1}{2} \\
z = \frac{1}{2} \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
from which you get
$$
\left\{ \begin{array}{l}
2x - y = \frac{1}{4} \\
z\left( {y^2 + z^2 - x^2 } \right) = \mu \\
x + y = \frac{1}{2} \\
z = \frac{1}{2} \\
x \le y \\
x,y,z \ge 0 \\
\end{array} \right.
$$
And finally we have, from the first and the third $x=y=1/4$, $z=1/2$. Thus the solution of the system is what you indicate. Of course you have to check the boundary.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4450738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove $|1 - (1+\frac{z-1}{n})^m| \leq 1$ where $|z|\leq 1$ and $mHow to prove
$$\Big|1 - \big(1+\frac{z-1}{n}\big)^m\Big| \leq 1$$
where $z\in\mathbb{C}$, $|z|\leq 1$ and $m,n\in\mathbb{N}$ and $m<n$.
I have run numerical experiments and believe the inequality is correct.
|
Based on @onriv's nice answer and user3750444's comment therein:
(Without using Maximum Modulus Principle)
Let $w = 1 + \frac{z - 1}{n}$
and $r = |w|$. We have
\begin{align*}
\left|1 - w^m\right|
&= |1 - w|\cdot |1 + w + w^2 + \cdots + w^{m-1}|\\
&\le |1 - w|\cdot (1 + r + r^2 + \cdots + r^{m-1})\\
&\le |1 - w|\cdot (1 + r + r^2 + \cdots + r^{n - 2}).
\end{align*}
We have $|w - 1 + 1/n|^2 = |z|^2/n^2 \le 1/n^2$
or
$$(1 - 1/n)^2 - (1 - 1/n)(w + \bar{w}) + |w|^2 \le 1/n^2$$
which results in
$$w + \bar{w} \ge \frac{(1 - 1/n)^2 + |w|^2 - 1/n^2}{1 - 1/n}.$$
Thus, we have
$$|1 - w|^2 = 1 + |w|^2 - (w + \bar{w})
\le 1 + |w|^2 - \frac{(1 - 1/n)^2 + |w|^2 - 1/n^2}{1 - 1/n}
= \frac{1 - |w|^2}{n - 1}.$$
Thus, we have
$$|1 - w^m| \le \frac{\sqrt{1 - r^2}}{\sqrt{n - 1}}(1 + r + r^2 + \cdots + r^{n - 2}).$$
We have $|w| = |1 - 1/n + z/n|
\le |1 - 1/n| + |z/n| \le 1$ and
$|w| = |1 - 1/n + z/n| \ge |1 - 1/n| - |z/n| \ge 1 - 2/n$. Thus, $|w| \in [1 - 2/n, 1]$.
It suffices to prove that, for all $n\ge 2$ and $r\in [1 - 2/n, 1]$,
$$\frac{\sqrt{1 - r^2}}{\sqrt{n - 1}}(1 + r + r^2 + \cdots + r^{n - 2})
\le 1. \tag{1}$$
If $r = 1$ or $n = 2, 3, 4$, it is easy to prove the inequality.
In the following, assume that $r < 1$ and $n \ge 5$.
The inequality is written as
$$\frac{\sqrt{1 - r^2}}{\sqrt{n - 1}}\cdot \frac{1 - r^{n-1}}{1 - r}
\le 1.$$
Using Bernoulli inequality $(1 - v)^s \ge 1 - sv$ for all $0 \le v < 1$ and $s\ge 1$, we have
$$r^{(n-1)/4} = [1 - (1 - r)]^{(n-1)/4}
\ge 1 - (1 - r)(n - 1)/4 \ge 0.$$
Also, we have
$$\frac{\sqrt{1 - r^2}}{\sqrt{n - 1}\, (1 - r)}
= \frac{1}{\sqrt{n - 1}}\sqrt{\frac{1 + r}{1 - r}}
\le \frac{1}{\sqrt{n - 1}}\sqrt{\frac{2}{1 - r}}.$$
It suffices to prove that
$$\sqrt{\frac{2}{(n - 1)(1 - r)}}
\left(1 - \left[1 - (1 - r)(n - 1)/4\right]^4\right)\le 1. $$
Letting $(n - 1)(1 - r)/2 = u^2$, it suffices to prove that, for all $u\in [0, \sqrt{1 - 1/n}]$,
$$\frac{1}{u}(1 - (1 - u^2/2)^4) \le 1$$
or
$$u^7 - 8u^5 + 24u^3 - 32u + 16 \ge 0$$
or
$$(1 - u)^3(-u^4 - 3u^3 + 2u^2 + 14u + 9) + 13(u - 19/26)^2 + 3/52 \ge 0$$
which is clearly true.
We are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4451523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Prove $\sum \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$ is convergent.
Assume $\sum\limits_{n=1}^{\infty} \dfrac{1}{a_n}$ is a convergent positive term series and
$p>0$. Prove $$ \sum_{n=1}^{\infty} \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$$ is
convergent.
Since
$$a_1+2^pa_2+\cdots+k^pa_k\ge \sqrt[k]{a_1\cdot2^pa_2\cdots k^pa_k}=\sqrt[k]{a_1a_2\cdots a_k}\cdot \sqrt[k]{(k!)^p},$$
then
\begin{align*} \frac{k^{p+1}}{a_1+2^pa_2+\cdots+k^pa_k}&\le \frac{k^{p+1}}{\sqrt[k]{a_1a_2\cdots a_k}\cdot \sqrt[k]{(k!)^p}}\sim \frac{k^{p+1}}{\sqrt[k]{a_1a_2\cdots a_k}\cdot \frac{k^p}{e^p}}= e^p\cdot \frac{k}{\sqrt[k]{a_1a_2\cdots a_k}}. \end{align*}
This perhaps can not work.
|
Here is another approach
Define
$$
b_m=\sum_{k=2^m+1}^{2^{m+1}}\frac1{a_k}\tag1
$$
By assumption, we have the convergence of
$$
\sum_{m=0}^\infty b_m=\sum_{k=2}^\infty\frac1{a_k}\tag2
$$
Next,
$$
\begin{align}
\sum_{k=2^m+1}^{2^{m+1}}a_kk^p\overbrace{\sum_{k=2^m+1}^{2^{m+1}}\frac1{a_k}}^{b_m}
&\ge\left(\sum_{k=2^m+1}^{2^{m+1}}k^{p/2}\right)^2\tag{3a}\\
&\ge\frac1{(p/2+1)^2}\left(\left(2^{m+1}\right)^{p/2+1}-\left(2^m\right)^{p/2+1}\right)^2\tag{3b}\\
&=\underbrace{\left(\frac{2^{p/2+1}-1}{p/2+1}\right)^2}_{c_p}\,\,2^{m(p+2)}\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: Cauchy-Schwarz
$\text{(3b)}$: underestimating a sum with an integral
$\text{(3c)}$: factor out $c_p$
Thus,
$$
\begin{align}
\sum_{k=1}^{2^{m+1}}a_kk^p
&\ge\sum_{k=2^m+1}^{2^{m+1}}a_kk^p\tag{4a}\\
&\ge\frac{c_p}{b_m}2^{m(p+2)}\tag{4b}
\end{align}
$$
Explanation:
$\text{(4a)}$: sum over fewer terms is smaller
$\text{(4b)}$: apply $(3)$
Therefore,
$$
\begin{align}
\sum_{n=1}^\infty\frac{n^{p+1}}{\sum\limits_{k=1}^na_kk^p}
&=\frac1{a_1}+\sum_{m=0}^\infty\sum_{n=2^m+1}^{2^{m+1}}\frac{n^{p+1}}{\sum\limits_{k=1}^na_kk^p}\tag{5a}\\
&\le\frac1{a_1}+\frac{2^{p+1}}{a_1+a_22^p}+\sum_{m=1}^\infty2^m\frac{2^{(m+1)(p+1)}}{\sum\limits_{k=1}^{2^m}a_kk^p}\tag{5b}\\
&\le\frac1{a_1}+\frac{2^{p+1}}{a_1+a_22^p}+\frac12\sum_{m=1}^\infty\frac{2^{(m+1)(p+2)}}{\frac{c_p}{b_{m-1}}2^{(m-1)(p+2)}}\tag{5c}\\[6pt]
&=\frac1{a_1}+\frac{2^{p+1}}{a_1+a_22^p}+\frac{2^{2p+3}}{c_p}\sum_{m=1}^\infty b_{m-1}\tag{5d}\\[6pt]
&=\frac1{a_1}+\frac{2^{p+1}}{a_1+a_22^p}+\frac{2^{2p+3}}{c_p}\sum_{k=2}^\infty\frac1{a_k}\tag{5e}
\end{align}
$$
Explanation:
$\text{(5a)}$: break the sum into the intervals $\left[2^m+1,2^{m+1}\right]$
$\text{(5b)}$: break out the $m=0$ ($n=2$) term
$\phantom{\text{(5b):}}$ for each $m$, there are $2^m$ terms in the sum
$\phantom{\text{(5b):}}$ for each term, $2^m\lt n\le2^{m+1}$
$\text{(5c)}$: apply $(4)$
$\text{(5d)}$: simplify
$\text{(5e)}$: apply $(2)$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4453465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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|
Remarquable identities $f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}$ Let $n$ be an integer, and
\begin{equation}
f(n) = \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)}
\end{equation}
\begin{equation}
g(n) = \frac{(bc)^n}{(a-b)(a-c)} + \frac{(ac)^n}{(b-a)(b-c)} + \frac{(ab)^n}{(c-a)(c-b)}
\end{equation}
We have the following impressive identities, for all $a,b,c$,
\begin{align}
f(0) &= 0 \\
f(1) &= 0 \\
f(2) &= 1 \\
f(3) &= a+b+c \\
f(4) &= a^2 + b^2 + c^2 + ab + ac + bc \\
f(5) &= a^3 + b^3 + c^3 + a^2b + a^2c + b^2c + ab^2 + ac^2 + bc^2 \\
f(6) &= a^4 + b^4 + c^4 + a^3b + a^3c + b^3c + ab^3 + ac^3 + bc^3 + a^2bc + ab^2c + abc^2 +a^2b^2 + a^2c^2 + b^2c^2 \\
\\
g(0) &= 0 \\
g(1) &= 1 \\
g(2) &= ab + ac + bc \\
g(3) &= a^2b^2 + a^2c^2 + b^2c^2 + a^2bc + ab^2c + abc^2 \\
g(4) &= a^3b^3 + a^3c^3 + b^3c^3 + a^3b^2c + a^3bc^2 + a^2b^3c + ab^3c^2 + a^2bc^3 + ab^2c^3 + a^2b^2c^2
\end{align}
which I have verified by plugging the expressions into Wolfram Alpha. It seems that the general form should be, for $n > 2$.
\begin{align}
f(n) &= \sum_{i+j+k = n-2}a^ib^jc^k \\
g(n) &= \sum_{\substack{i+j+k = 2(n-1)\\1\leq i,j,k \leq n-1}}a^ib^jc^k
\end{align}
The questions are :
*
*How to demonstrate the statements for $n$ general using induction. Intuitively, we should use induction, however I do not see the induction step.
*Could we demonstrate the general case without using induction ? There is a link, between these formulas and Vandermondt matrices (see below), would there be a nice demonstration using matrices ?
=======================================================================
I arrived at such identities when working with partial fractions decomposition,
and after some related work I realized that the Vandermondt matrices where almost the inverse of the matrices which appear when we do partial fractions decomposition,
Then I realised that the Vandermondt matrices have very nice inverse :
\begin{equation}
\begin{pmatrix}
1&1 \\
a&b
\end{pmatrix}
\begin{pmatrix}
-\frac{b}{(a - b)} & \frac{1}{(a - b)} \\
-\frac{a}{(b - a)} & \frac{1}{(b - a)} \\
\end{pmatrix}
= I_{2}
\end{equation}
\begin{equation}
\begin{pmatrix}
1&1&1 \\
a&b&c \\
a^2&b^2&c^2
\end{pmatrix}
\begin{pmatrix}
\frac{b c}{(a - b) (a - c)} & -\frac{b + c}{(a - b) (a - c)} & \frac{1}{(a - b) (a - c)} \\
\frac{a c}{(b - a) (b - c)} & -\frac{a + c}{(b - a) (b - c)} & \frac{1}{(b - a) (b - c)} \\
\frac{a b}{(c - a) (c - b)} & -\frac{a + b}{(c - a) (c - b)} & \frac{1}{(c - a) (c - b)}
\end{pmatrix}
= I_{3}
\end{equation}
\begin{equation}
\begin{pmatrix}
1&1&1&1 \\
a&b&c&d \\
a^2&b^2&c^2&d^2 \\
a^3&b^3&c^3&d^3
\end{pmatrix}
\begin{pmatrix}
-\frac{bcd}{(a - b) (a - c)(a-d)} & \frac{bc + cd + bd}{(a - b) (a - c)(a-d)} &-\frac{b+c+d}{(a - b) (a - c)(a-d)} & \frac{1}{(a - b) (a - c)(a-d)}\\
-\frac{a cd}{(b - a) (b - c)(b-d)} & \frac{ac + ad + cd}{(b - a) (b - c)(b-d)} & -\frac{a + c + d}{(b - a) (b - c)(b-d)}& \frac{1}{(b - a) (b - c)(b-d)}\\
-\frac{a bd}{(c - a) (c - b)(c-d)} & \frac{ab + ad + bd}{(c - a) (c - b)(c-d)} & -\frac{a + b + d}{(c - a) (c - b)(c-d)}&\frac{1}{(c - a) (c - b)(c-d)}\\
-\frac{a bc}{(d - a) (d - b)(d-c)} & \frac{ab + ac + bc}{(d - a) (d - b)(d-c)} & -\frac{a + b + c}{(d - a) (d - b)(d-c)}&\frac{1}{(d - a) (d - b)(d-c)}
\end{pmatrix}
= I_{4}
\end{equation}
The identities $f(0),f(1), f(2)$ are the last column of the inverse equation for 3-dim matrices. However, it seems that such matrix argument is not sufficient to prove the case for $n$ general, and that many similar identities (the other places in the matrices) should exist.
|
With respect to your first question, I don't think induction would be a method that is very helpful here. The problem would be to relate different functions $f(n)$. This is more something that asks for generating functions.
\begin{eqnarray}
f(n)
& = & \frac{a^n}{(a-b)(a-c)} + \frac{b^n}{(b-a)(b-c)} + \frac{c^n}{(c-a)(c-b)} \\ \\
& = & \frac{(a b^n + b c^n + c a^n) - (a^n b + b^n c + c^n a) }{(a-b)(b-c)(c-a)} \\ \\
& = & \frac{(a^n - b^n) c - a b (a^{n-1} - b^{n-1}) - (a - b) c^n}{(a-b)(b-c)(c-a)}
\end{eqnarray}
For $n=0$ and $n=1$ we see from the second line immediately that the numerator will be zero.
So we only need to consider the case $n \geq 2$. The third line tells us that the numerator is a polynomial in $a,b,c$ with all terms of degree $n+1$. In addition to has a factor $(a-b)$, and like-wise due to symmetry it also has a factors $(b-c)$ and $c-a$. It therefore follows that $f(n)$ is a polynomial in $a,b,c$ with all terms of degree $n-2$.
From this point one could in principle assume that the proposed form is correct, multiply with $(a-b)(b-c)(c-a)$, expand everything, and manipulate the sums until the above form is obtained. It is actually not too difficult. There is, however, an easier approach.
Consider the sum $\sum_{n=0}^\infty \epsilon^n f(n)$ in the limit $\epsilon \rightarrow 0$ ( $|\epsilon| \max(|a|,|b|,|c|) < 1$ for which the sum will converge. We then find
\begin{eqnarray}
\sum_{n=0}^\infty \epsilon^n f(n) & = & \sum_{n=0}^\infty \frac{\epsilon^n a^n}{(a-b)(a-c)} + \sum_{n=0}^\infty \frac{\epsilon^n b^n}{(b-a)(b-c)} + \sum_{n=0}^\infty \frac{\epsilon^n c^n}{(c-a)(c-b)} \\ \\
& = &
\frac{1}{(a-b)(a-c)(1-\epsilon a)} + \frac{1}{(b-a)(b-c)(1-\epsilon b)} + \frac{1}{(c-a)(c-b)(1-\epsilon c)} \\ \\
& = & \frac{\epsilon^2}{(1-\epsilon a)(1-\epsilon b)(1-\epsilon c)} \\ \\
& = & \epsilon^2 \left( \sum_{k=0}^\infty (\epsilon a)^k \right) \left( \sum_{l=0}^\infty (\epsilon b)^l \right) \left( \sum_{m=0}^\infty (\epsilon c)^m \right) \\ \\
& = & \epsilon^2 \sum_{n=0}^\infty \epsilon^n \sum_{k+l+m=n} a^k b^l c^m \\ \\
& = & \sum_{n=2}^\infty \epsilon^n \sum_{k+l+m=n-2} a^k b^l c^m
\end{eqnarray}
From which it follows that $f(n)$ is the complete homogeneous symmetric polynomial of degree $n-2$, and that $f(0)=f(1)=0$.
In view of the comment by Blue, the results for $g$ follows.
I don't know whether there is formulation by using matrices that is easier.
|
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"url": "https://math.stackexchange.com/questions/4453585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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|
Solution of $\theta$ when $\tan(\theta)-\sin(\theta)=\frac{\sqrt3}{2}$ I came across this trigonometry problem. If, $$\tan(\theta)-\sin(\theta)=\frac{\sqrt3}{2}$$
What is the value of $\theta$
I got the solution that $\theta$ will be $\frac{\pi}{3}$ by expanding the equation and turning into,
$$\sin^4(\theta)+\sqrt3\sin^3(\theta)+\frac{3}{4}\sin^2(\theta)-\sqrt3\sin(\theta)-\frac{3}{4}=0$$
And then solving $\sin(\theta)$. But this method seems too complicated. Is there any easier and better solution?
|
$$\sqrt{3}\,t^4 + 8\,t^3 - \sqrt{3} =\left(t-\frac{1}{\sqrt{3}}\right) \left(\sqrt{3} t^3+9 t^2+3 \sqrt{3} t+3\right).$$
Using the hyperbolic method for the cubic gives for the real root
$$t=-\sqrt{3}-2 \sqrt{2} \cosh \left(\frac{1}{3} \cosh
^{-1}\left(\sqrt{\frac{3}{2}}\right)\right)$$ from which
$$\theta=2\pi -2 \tan ^{-1}\left(\sqrt{3}+2 \sqrt{2} \cosh \left(\frac{1}{3} \cosh
^{-1}\left(\sqrt{\frac{3}{2}}\right)\right)\right).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4455146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
For what values of $p>0$ is $\lim_{n\rightarrow\infty}\int_0^n\frac{(1-\frac{x}{n})^ne^x}{n^p}dx=0$? For what values of $p>0$ is $\lim_{n\rightarrow\infty}\int_0^n\frac{(1-\frac{x}{n})^ne^x}{n^p}dx=0$?
My thoughts: We know that $(1-\frac{x}{n})^n\leq e^x$, so the numerator is $\leq e^{2x}$. So, we can play with $\frac{e^{2x}}{n^p}$. From here, I am not quite sure what to do. I would really like to be able to find the supremum over $n$, but I can't really minimize the denominator to be able to replace $n$ with something in terms of $x$, because I only have $n^p$ down there. On the other hand, I feel like I should be splitting up the integral from $0$ to $1$ and then from $1$ to $\infty$ based on the denominator. Then, fix some $x$, and just use $p$ integral properties to get that $p\in(0,1)$, but I am not quite sure. Any help is greatly appreciated!
A quick edit: I realize that I made a big mistake above by overlooking the minus sign, so instead the integrand is bounded above by $\frac{1}{n^p}\leq 1$ as $n\rightarrow \infty$, and so we can use DCT and then treat it like a $p$ integral.
A second edit: For Sangchul Lee, I edited the integral to make the upper bound $n$ so he can expand on how he got his approximation. Thank you!
|
Assume that $n \ge 2$.
Let
$$I_n := \int_0^n \left(1 - \frac{x}{n}\right)^n \mathrm{e}^x \,\mathrm{d} x.$$
Using $\ln(1 + u) \le u$ for all $u > -1$, we have, for all $0 \le x < n$,
\begin{align*}
\ln\left(1 - \frac{x}{n}\right)
&= \ln \left(1 - \frac{\sqrt n}{n}\right) + \ln\left(1 - \left(1 - \frac{\sqrt n}{n}\right)^{-1}\frac{x - \sqrt n}{n}\right)\\[6pt]
&\le - \frac{\sqrt n}{n} - \left(1 - \frac{\sqrt n}{n}\right)^{-1}\frac{x - \sqrt n}{n}\\
&= - \frac{x}{n - \sqrt n} + \frac{1}{n - \sqrt n}.
\end{align*}
Thus, we have
\begin{align*}
I_n &\le \mathrm{e}^{n/(n - \sqrt n)}\int_0^n \mathrm{e}^{- nx/(n - \sqrt n) + x}\, \mathrm{d} x\\
&= \mathrm{e}^{n/(n - \sqrt n)}
(\sqrt n - 1)(1 - \mathrm{e}^{-n/(\sqrt n - 1)})\\
&\le \mathrm{e}^{4}\sqrt n.
\end{align*}
On the other hand, we have, for all $x \in [0, \sqrt n)$,
$$\left(1 - \frac{x}{n}\right) \mathrm{e}^{x/n}
\ge \left(1 - \frac{x}{n}\right)\left(1 + \frac{x}{n}\right)
= 1 - \frac{x^2}{n}\cdot \frac{1}{n}
\ge \left(1 - \frac{x^2}{n}\right)^{1/n}$$
where we have used $\mathrm{e}^u \ge 1 + u$ for all $u\ge 0$,
and Bernoulli inequality $(1 + v)^r \le 1 + vr$ for all $0 < r \le 1$ and $v > -1$.
Thus, we have
\begin{align*}
I_n &\ge \int_0^{\sqrt n} \left(1 - \frac{x}{n}\right)^n \mathrm{e}^x \,\mathrm{d} x\\
&\ge \int_0^{\sqrt n} \left(1 - \frac{x^2}{n}\right) \,\mathrm{d} x\\
&= \frac23 \sqrt n.
\end{align*}
Thus, we have
$$\frac23 \sqrt n \le I_n \le \mathrm{e}^4 \sqrt{n}, \quad \forall n\ge 2.$$
Thus, $\lim_{n\to \infty} \frac{I_n}{n^p} = 0$ if and only if $p > 1/2$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4455261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality case, for example, when $a=x^4,b=x^3,c=x^2,d=x$, and as $x$ tends to $\infty$, the LHS tends to $3$, but this means that the inequality is very unlikely to be solved with traditional methods, such as Cauchy-Schwartz (my starting idea), so I got stuck.
|
We first prove the following lemma.
Lemma. Let $\lambda>0$. Then
$$\sup_{x>0}\left(\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+\frac{\lambda}x}}\right)=\begin{cases}\frac2{\sqrt{1+\sqrt\lambda}}&\text{if }\lambda<4\\\sqrt{\frac{\lambda}{\lambda-1}}&\text{if }\lambda\geq 4.\end{cases}$$
Proof. Fixing $\lambda$ constant, let the left side be $f(x)$. The function $f$ is clearly continuous and infinitely differentiable, and so its global extrema are reached when $x\to 0$, $x\to\infty$, or when $f''(x)=0$. It is not hard to compute that
$$f''(x)=0\Longleftrightarrow \lambda^2(1+x)^3=x(\lambda+x)^3;$$
the polynomial on the right factors as $(x^2-\lambda)(x^2-(\lambda^2-3\lambda)x+\lambda)$. So, the extrema are when $x=\sqrt{\lambda}$, or when $x$ and $\lambda/x$ satisfy $x+\frac\lambda x=\lambda^2-3\lambda$. For $\lambda<4$, $x+\frac\lambda x$ is always larger than $\lambda^2-3\lambda$, so the only $x$ with $f''(x)=0$ is $x=\sqrt\lambda$, giving the desired bound. For $\lambda\geq 4$, such a real $x>0$ does exist, and satisfies, letting $y=\lambda/x$ so that $x+y=\lambda^2-3\lambda$,
\begin{align*}
\left(\frac1{\sqrt{1+x}}\right)\left(\frac1{\sqrt{1+y}}\right)&=\frac{1}{\sqrt{1+(\lambda^2-3\lambda)+\lambda}}=\frac1{\lambda-1}\\
\left(\frac1{\sqrt{1+x}}\right)^2+\left(\frac1{\sqrt{1+y}}\right)^2&=\frac{2+x+y}{(1+x)(1+y)}=\frac{\lambda-2}{\lambda-1},
\end{align*}
so
$$\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+y}}=\sqrt{\frac{\lambda-2}{\lambda-1}+\frac2{\lambda-1}}=\frac{\lambda}{\lambda-1}.$$
For $\lambda\geq 4$, this is always at least the value $\frac2{\sqrt{1+\sqrt\lambda}}$ achieved at $x=\sqrt\lambda$, and so is the maximum of $f$. $\square$
Now, under the substitution $w=\frac{b^2}{a^2}$, $x=\frac{c^2}{b^2}$, $y=\frac{d^2}{c^2}$, $z=\frac{a^2}{d^2}$, the problem becomes to show
$$\frac1{\sqrt{1+w}}+\frac1{\sqrt{1+x}}+\frac1{\sqrt{1+y}}+\frac1{\sqrt{1+z}}\leq 3$$
for $w,x,y,z>0$ and $wxyz=1$. If $wx,yz\leq 4$, then the sum is at most
$$\frac2{\sqrt{1+t}}+\frac2{\sqrt{1+\frac1t}}\leq \frac4{\sqrt{2}}=2\sqrt2<3$$
for $t=\sqrt{wx}$, where we have used our lemma on $\lambda=t^2=wx$, $\lambda=1/t^2=yz$, and again on $\lambda=1$. Otherwise, assume without loss of generality that $wx\geq 4$, and let $s=\sqrt{wx}$. Then, by our lemma, the sum is at most
$$\sqrt{\frac{s^2}{s^2-1}}+\frac{2}{\sqrt{1+\frac1s}},$$
where $s\geq 2$. However, for $s\geq 2$,
\begin{align*}
\sqrt{\frac{s^2}{s^2-1}}+\frac{2}{\sqrt{1+\frac1s}}
&=\frac{\sqrt s(\sqrt s+2\sqrt{s-1})}{\sqrt{s^2-1}}\\
&\leq \frac{\sqrt s\left(\sqrt s+2\left[\sqrt s-\frac1{2\sqrt s}\right]\right)}{\sqrt{s^2-1}}\\
&=\frac{3s-1}{\sqrt{s^2-1}},
\end{align*}
which is at most $3$ whenever $s\geq 5/3$ since
$$(3s-1)^2=9s^2-6s+1\leq 9s^2-9\implies \frac{3s-1}{\sqrt{s^2-1}}\leq 3.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4456286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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|
Showing that $x^3 + y = y^3 + x$ is an equivalence relation I am asked to prove that: $x^3 + y = y^3 + x$ is an equivalence relation. So far I have the following:
*
*Reflexive:
$m^3 +m = m^3 +m$
*Symmetric:
$m^3 + n = n^3 + m \rightarrow n^3 + m = m^3 + n$ Then:
$n^3 + m = m^3 +n$ From hypothesis
*Transitivity (here's where I got stuck):
$m^3 + n = n^3 + m \wedge n^3 + o = o^3 + n \rightarrow m^3 + o = o^3 + m$ Then:
$m^3 + o = n^3 + m - n = n^3 - n + m = o^3 + n - o - n + m = o^3 - o +m$
And I cant figure out a way to go from $o^3 - o + m$ to $o^3 + m$; what could I do? Am I missing something?
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Alternative approach:
$x \sim y \iff x^3 + y = y^3 + x \iff (x^3 - x) = (y^3 - y).$
Then, $\{ ~x \sim y ~~~~\text{and}~~~~ y \sim z ~\} \implies $
$(x^3 - x) = (y^3 - y) ~~~~\text{and}~~~~ (y^3 - y) = (z^3 - z).$
This implies that $(x^3 - x) = (z^3 - z) \iff x \sim z.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4456531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Other ways to factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ To factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ I used the fact that $x=-y$ and $y=z$ and $x=-z$ make the expression zero. Hence it factorize to $\lambda (x+y)(y-z)(x+z)$ and we can check that the number $\lambda$ is equal to $1$.
I'm looking for other approaches/ideas to factorize the expression.
|
I think OP approach is the simplest one because it can be done just in mind without paper writing. One of possible another ways
$$A=xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)=x^2(y-z)+x(y^2-z^2)+yz(y-z)$$
$(y-z)$ is common factor of $(y-z)$ and $(y^2-z^2)=(y-z)(y+z)$, so we can factor it out:
$$A=(y-z)(x^2+x(y+z)+yz)$$
Then we need only factorize $$B=x^2+x(y+z)+yz$$
If we don't see that
$$B=x^2+xy+xz+yz=x(x+y)+z(x+y)=(x+z)(x+y)$$
we can use standard way of factorizing quadratic expression with full square separation
$$B=x^2+x(y+z)+yz=x^2+2x\cdot\frac{y+z}2+\frac{(y+z)^2}4-\frac{(y+z)^2-4yz}4$$
$$B=\left(x+\frac{y+z}2\right)^2-\frac{y^2-2yz+z^2}4=\left(x+\frac{y+z}2\right)^2-\left(\frac{y-z}2\right)^2$$
$$B=\left(x+\frac{y+z}2+\frac{y-z}2\right)\left(x+\frac{y+z}2-\frac{y-z}2\right)=(x+y)(x+z)$$
Then $$A=(y-z)B=(y-z)(x+y)(x+z)$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4458244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Calculate $C=\sin3\alpha\cos\alpha$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$. Calculate $$C=\sin3\alpha\cos\alpha$$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$.
My idea was to find $\sin\alpha$ and $\cos\alpha$. Then we have $\sin3\alpha=3\sin\alpha-4\sin^3\alpha$. So $$\tan2\alpha=2=\dfrac{2\tan\alpha}{1-\tan^2\alpha}\iff\tan^2\alpha+\tan\alpha-1=0$$ This equation has solutions $\left(\tan\alpha\right)_{1,2}=\dfrac{-1\pm\sqrt5}{2}$ but as $\alpha\in(0^\circ;45^\circ)\Rightarrow$ $\tan\alpha=\dfrac{\sqrt5-1}{2}$. Now $\sin\alpha=\dfrac{\sqrt5-1}{2}\cos\alpha$ and plugging into $\sin^2\alpha+\cos^2\alpha=1$ got me at $\cos^2\alpha=\dfrac{2}{5-\sqrt5}$
|
For this problem, I like your work, through the conclusion that
$$\tan(\alpha) = \frac{\sqrt{5} - 1}{2}. \tag1 $$
In my opinion, the simplest approach to complete the problem is to forgo any attempt at elegance, and simply use the following identities (one of which you have already referred to):
*
*$\displaystyle \tan^2(\alpha) + 1 = \frac{1}{\cos^2(\alpha)}.$
*$\displaystyle \sin(3\alpha) = 3\sin(\alpha) - 4\sin^3(\alpha).$
What is being asked for is
$$\sin(3\alpha)\cos(\alpha). \tag2 $$
To me, the simplest approach is to manually calculate both $\sin(\alpha)$ and $\cos(\alpha)$, and then use these calculations to evaluate the expression in (2) above.
As you indicated, because of the stated domain for $(\alpha)$, you know that $\sin(\alpha)$ and $\cos(\alpha)$ are both non-negative. Using (1) above,
$$\tan^2(\alpha) = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \implies $$
$$\sec^2(\alpha) = \tan^2(\alpha) + 1 = \frac{5 - \sqrt{5}}{2} \implies $$
$$\cos^2(\alpha) = \frac{2}{5 - \sqrt{5}} \implies \tag3 $$
$$\cos(\alpha) = \sqrt{\frac{2}{5 - \sqrt{5}}}. \tag4 $$
Using (3), you also have that
$$\sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \frac{2}{5 - \sqrt{5}} = \frac{3 - \sqrt{5}}{5 - \sqrt{5}} \implies $$
$$\sin(\alpha) = \sqrt{\frac{3 - \sqrt{5}}{5 - \sqrt{5}}}. \tag5 $$
Now, (2), (4), (5), and the $\sin(3\alpha)$ identity can be used to complete the problem.
$$\sin(3\alpha)\cos(\alpha) = \left[3\sin(\alpha) - 4\sin^3(\alpha)\right] \cos(\alpha) $$
$$= \sin(\alpha)\cos(\alpha) \times \left[3 - 4\sin^2(\alpha)\right] $$
$$ = \sqrt{\frac{3 - \sqrt{5}}{5 - \sqrt{5}}} \times
\sqrt{\frac{2}{5 - \sqrt{5}}} \times
\left[ ~3 - \left(4 \times \frac{3 - \sqrt{5}}{5 - \sqrt{5}} ~\right) ~\right]. \tag 6 $$
In (6) above, you know that $\left(3 - \sqrt{5}\right) \times 2 = \left(6 - 2\sqrt{5}\right) = \left(\sqrt{5} - 1\right)^2.$
Therefore, the 1st two factors in (6) above simplify to
$\displaystyle \frac{\sqrt{5} - 1}{5 - \sqrt{5}}.$
The 3rd factor in (6) above may be re-expressed as
$\displaystyle \frac{\left(15 - 3\sqrt{5}\right) - \left(12 - 4\sqrt{5}\right)}{5 - \sqrt{5}} = \frac{3 + \sqrt{5}}{5 - \sqrt{5}}.$
Putting this all together, the final computation is
$$\frac{\sqrt{5} - 1}{5 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{5 - \sqrt{5}} = \frac{2 + 2\sqrt{5}}{30 - 10\sqrt{5}} $$
$$= \frac{1 + \sqrt{5}}{15 - 5\sqrt{5}} \times \frac{15 + 5\sqrt{5}}{15 + 5\sqrt{5}} = \frac{40 + 20\sqrt{5}}{100} = \frac{2 + \sqrt{5}}{5}.$$
|
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"url": "https://math.stackexchange.com/questions/4460616",
"timestamp": "2023-03-29T00:00:00",
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|
Remainder of $\frac{3x^{2019}+5x^{1019}-7x+4}{x^2-1}$ I don't understand how I should go about solving the following question:
Find the remainder when polynomial $f(x)=3x^{2019}+5x^{1019}-7x+4$ is divided by $x^2-1$.
I tried to use the factor theorem, but I never encountered a problem with a divisor which, in this case, is $(x+1)(x-1)$, so I simply found $f$ of both roots, so $f(1)$ and $f(-1)$.
Allegedly the remainder is a linear polynomial in the form $ax+b$ but I fail to see how they derived that fact. Apparently they made use of simultaneous equations, but I'm not sure how or why.
Any help would be appreciated!
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Degree of the remainder less then degree of quotient then remainder is linear , $R(x)=ax+b$
$f(x)=3x^{2019}+5x^{1019}-7x+4=Q(x)(x-1)(x+1)+ax+b$
$f(1)=a+b=5$
$f(-1)=-a+b=3$
then $a=1,b=4$
$R(x)=x+4$
OR
To find remainder when
$f(x)=3x^{2019}+5x^{1019}-7x+4$ is divided by $x^2-1$.
By using remainder theorem
$x^2-1=0$
then
$x^2=1$
$f(x)=3x^{2018+1}+5x^{1018+1}-7x+4=3((x^2))^{1009}\times x +5((x^2))^{509}\times x -7x+4=3\times 1 \times x +5\times x\times 1\times x-7x+4=x+4$
Hence
$R(x)=x+4$
|
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"timestamp": "2023-03-29T00:00:00",
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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.