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Proving $\int_{0}^{\pi/2} \ln \sin x ~ \ln \cos x~ dx=-\frac{\pi^3}{48}+\frac{\pi}{2}\ln^22$ Interestingly the integral $$I=\int_{0}^{\pi/2} \ln \sin x ~ \ln \cos x~ dx~~~~~~~~~(1)$$
is doable by hand by using Fourier series: $$ \ln \sin x=-\sum_{j=1}^{\infty} \frac{\cos 2j x}{j}-\ln 2,\quad \ln \cos x=\sum_{k=1}^{\infty} (-1)^{k+1} \cfrac{\cos 2kx}{k}-\ln 2, ~x\in [0,\pi/2].~~~~~~~~~~(2)$$
$$I=\sum_{j=1}^{\infty} \sum_{k=1}^{\infty} (-1)^{k}\int_{0}^{\pi/2} \frac{\cos 2jx \cos 2kx}{jk} dx-\ln 2 \int_{0}^{\pi/2}\left (\sum_{j=1}^{\infty} \frac{\cos 2jx}{j}+\sum_{k=1}^{\infty} (-1)^k \frac{\cos 2kx}{k}\right) dx+\frac{\pi}{2}\ln^22~~~~~~~~~~~(3)$$
The second and third integrals vanish, then
$$\implies I=\frac{1}{2} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} (-1)^k \int_{0}^{\pi/2}\frac{\cos2(j+k)x+\cos2(j-k)x}{jk} dx+\frac{\pi}{2}\ln^22~~~~~~~~~~~~~~~~(4)$$
$$\implies I=\sum_{j=1}^{\infty} \sum_{k=1}^{\infty}(-1)^k \frac{\sin2(j-k)\pi}{4jk(j-k)}+\frac{\pi}{2}\ln^22~~~~~~~~~~~~~~~~(5)$$
Taking limit $(j-k)\to 0$, we get
$$I=\frac{\pi}{4} \sum_{k=1}^{\infty} \frac{(-1)^k }{k^2}+\frac{\pi}{2}\ln^22=
-\frac{\pi^3}{48}+\frac{\pi}{2}\ln^22.~~~~~~~~~~~~~~~(6)$$
The question is: How else we can get this interesting integral (1)?
|
Solution without using beta function:
Let $a=\ln(\sin x)$ and $b=\ln(\cos x)$ in the algebraic identity
$$ab=\frac12a^2+\frac12b^2-\frac12(a-b)^2,$$
we have
$$\ln(\sin x)\ln(\cos x)=\frac12\ln^2(\sin x)+\frac12\ln^2(\cos x)-\frac12\ln^2(\tan x).$$
Integrate both sides from $x=0$ to $\pi/2$,
\begin{gather*}
\int_0^{\frac{\pi}{2}}\ln(\sin x)\ln(\cos x)\mathrm{d}x\\
=\frac12\int_0^{\frac{\pi}{2}}\ln^2(\sin x)\mathrm{d}x+\frac12\int_0^{\frac{\pi}{2}}\ln^2(\cos x)\mathrm{d}x-\frac12\int_0^{\frac{\pi}{2}}\ln^2(\tan x)\mathrm{d}x.
\end{gather*}
The second integral is equivalent to the first one by using the rule $\int_a^b f(x)\mathrm{d}x=\int_a^b f(a+b-x)\mathrm{d}x$. For the third integral, let $\tan x=y$, we have
$$\int_0^{\frac{\pi}{2}}\ln(\sin x)\ln(\cos x)\mathrm{d}x=\int_0^{\frac{\pi}{2}}\ln^2(\sin x)\mathrm{d}x-\frac12\int_0^\infty\frac{\ln^2(y)}{1+y^2}\mathrm{d}y.$$
First integral:
We have
$$\ln^2(2\sin x)=\ln^2(2)+2\ln(2)\ln(\sin x)+\ln^2(\sin x)$$
or
$$\ln^2(\sin x)=\ln^2(2\sin x)-2\ln(2)\ln(\sin x)-\ln^2(2).$$
Integrate both sides from $x=0$ to $\pi/2$,
$$\int_0^{\frac{\pi}{2}} \ln^2(\sin x)\mathrm{d}x=\int_0^{\frac{\pi}{2}}\ln^2(2\sin x)\mathrm{d}x-2\ln(2)\int_0^{\frac{\pi}{2}}\ln(\sin x)\mathrm{d}x-\int_0^{\frac{\pi}{2}}\ln^2(2)\mathrm{d}x.$$
The third integral is $\frac{\pi}{2}\ln^2(2)$ and the second integral is $-\frac{\pi}{2}\ln(2)$. For the first one, integrate both sides of the identity:
$$\ln^2(2\sin x)=\left(\frac{\pi}{2}-x\right)^2+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\cos(2nx)$$
from $x=0$ to $\pi/2$ then change the order of integration and summation,
\begin{gather*}
\int_0^{\frac{\pi}{2}}\ln^2(2\sin x)\mathrm{d}x=\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-x\right)^2\mathrm{d}x+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\int_0^{\frac{\pi}{2}}\cos(2nx)\mathrm{d}x\\
=-\frac13\left(\frac{\pi}{2}-x\right)^3\bigg|_0^{\frac{\pi}{2}}+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\cdot\frac{\sin(2nx)}{2n}\bigg|_0^{\frac{\pi}{2}}\\
=\frac{\pi^3}{24}+2\sum_{n=1}^\infty\frac{H_{n-1}}{n}\cdot\frac{\sin(n\pi)}{2n}\\
\{\text{the sum evaluates to $0$, since $\sin(n\pi)=0$ for integer $n$}\}\\
=\frac{\pi^3}{24}.
\end{gather*}
Therefore,
$$\int_0^{\frac{\pi}{2}} \ln^2(\sin x)\mathrm{d}x=\frac{\pi}{2}\ln^2(2)-\frac{\pi^3}{16}.$$
Second integral:
By the generalization:
$$ \int_0^\infty\frac{\ln^{2a}(x)}{1+x^2}\mathrm{d}x=2^{-2a-1}\pi\lim_{s\to \frac12}\frac{\mathrm{d}^{2a}}{\mathrm{d} s^{2a}}\csc(\pi s),$$
which follows from differentiating Euler's reflection formula, we have
$$ \int_0^\infty\frac{\ln^2(x)}{1+x^2}\mathrm{d}x=\frac{\pi^3}{8}.$$
Combining the two integrals, we reach
$$\int_0^{\frac{\pi}{2}} \ln(\sin x)\ln(\cos x)\mathrm{d}x=\frac{\pi}{2}\ln^2(2)-\frac{\pi^3}{48}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Alternative approaches to maximize $y=x\sqrt{100-x^2}$ I could find three good approaches to find maximum of the function $y=x\sqrt{100-x^2}$. I will explain them briefly :
First: Finding $x$ satisfies $y'=0$ then plugging it in the function.
Second: Using the substitution $x=10\sin\theta$ (or $x=10\cos\theta)$ for
$\theta\in(0,\frac{\pi}2)$ to get $y=100\sin\theta\cos\theta=50\sin(2\theta)$ hence the maximum is $50$.
Third: Using AM-GM inequality: It is obvious that maximum occurs for $x>0$ So we can rewrite $y$ as $y=\sqrt{x^2(100-x^2)}$ . Now the sum of $x^2$ and $100-x^2$ is $100$ so the maximum of product happens when $x^2=100-x^2$ or $x^2=50$ Hence $y_{\text{max}}=50$.
Just for fun, can you maximize $y=x\sqrt{100-x^2}$ with other approaches?
|
Square both sides to get $y^2 = x^2 (100-x^2) \implies -x^4+100x^2-y^2=0$. This is a quadratic in $x^2$: when $\Delta = 0$, $100^2 - 4(-1)(-y^2) = 0 \implies y = ±50$.
The maximum value is $y = 50$ as considering the negative branch, $f(-x) = -f(x)$, hence the maximum of the positive branch is the same as the negative branch.
|
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|
What insight is there on negative number bases? I don't really have much to say, other than that I was curious about how negative number bases work, and I was wondering if I could get some insight on how to understand them.
I know how to convert negative base numbers to base 10, but have no clue as to how to write a base 10 number to an arbitrary negative base, at least not easily.
I have also noticed that there seems to be 2 ways to write every number in a negative base, for example $37_{10}$ can be written as $177_{-10}$ or $-43_{-10}$ and think there could be some way to use that to your advantage, although I still haven't gotten anywhere myself.
|
I know how to convert negative base numbers to base 10, but have no clue as to how to write a base 10 number to an arbitrary negative base, at least not easily.
Write it to the positive absolute value base first. subtract the odd terms from $|b|$ and carry to even terms.
Example: To write $1234$ in base $-6$.
$1234 = 205\times 6 + 4 =$
$34\times 6^2 + 1\times 6 + 4=$
$5\times 6^3 + 4\times 6^2 + 1\times 6 + 4 =5414_6$.
$=5\times 6^3 + 4\times 6^2 + 1\times 6 + 4$
$=(6-1)\times 6^3 + 4\times 6^2 + (6-5)\times 6 + 4=$
$=6^4 - 6^3 + (4+1)\times 6^2 -5\times 6 + 4 = $
$(-6)^4 + (-6)^3 + 5\times (-6)^2 + 5\times(-6) + 4 = 11554_{-6}$
With practice we can do it directly.
$1234= -205(-6) + 4=$
$(35\cdot (-6)+5)(-6) + 4=$
$35(-6)^2 + 5(-6) + 4=$
$(-5\times -6 + 5)(-6)^2 + 5(-6) + 4=$
$-5(-6)^3 + 5(-6)^2 + 5(-6) + 4=$
$(-6 + 1)(-6)^3 + 5(-6)^2 + 5(-6) + 4=$
$(-6)^4 + (-6)^3 + 5(-6)^2 + 5(-6) + 4=11554_{-6}$.
I have also noticed that there seems to be 2 ways to write every number in a negative base, for example 3710 can be written as 177−10 or −43−10 and think there could be some way to use that to your advantage, although I still haven't gotten anywhere myself.
Yes that's true. But I find the negative sign in front of a number that is ultimately positive confusing.
Note a negative sign in front of an even number of digits or a lack of negative sign in front of an odd number of digits means the number is positive (and vice versa).
Actually, that could be your rule. All numbers in negative base are expressed with positive digits. If the number has an even number of digits the number is positive (as $neg^{highest\ even} > 0$ will overpower the number). But if the number has an odd number of digits the number is negative (as $neg^{highest\ odd} < 0$ will overpower the numbers).
|
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|
Solving $(x^3-4)^3=(\sqrt[3]{(x^2+4)^2}+4)^2$ I'm trying to solve the equation below for $x$: $$(x^3-4)^3=(\sqrt[3]{(x^2+4)^2}+4)^2$$
I tried to solve it by doing some algebraic operations such as expanding both sides and then try to remove the third roots by rising both sides of the equation to power of 3 but it gets somehow complicated.
Someone told me that this equation can be solved by using derivation but I have no idea how to use derivation for solving this equation.
Any help would be appreciated.
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As mentioned by @Player0, the equation can be solved using monotonicity noticing that $x=2$ is a solution. But in this answer, I will be solving the equation algebraically (and for real numbers).
We first remove the bizarre cube roots from our equation. To do that, we consider $y=\sqrt{x^3-4}$ and $z=\sqrt[3]{x^2+4}$ where $y,z>0$. So, we can rewrite our equation as $$(y^2)^3=(z^2+4)^2$$ $$\Rightarrow y^3=z^2+4$$ which looks good.
We also have the following systems of equations from our definition of $y$ and $z$ and from the above equation $$\left\{\begin{matrix}x^3-y^2=4\\ y^3-z^2=4\\ z^3-x^2=4\end{matrix}\right.$$ which are really beautiful.
Now in our original equation, since RHS is a square, we have $x^3-4 \geq 0 \implies x\geq\sqrt[3]{4}$. Now, we assume $x \geq z$. From the above systems of equations we have $$\left\{\begin{matrix}x^3-z^3=y^2-x^2\\ y^3-x^3=z^2-y^2 \end{matrix}\right.$$
Now since we had $x \geq z$, we have $y\geq x$ and $z \geq y$. Notice that we have $z\geq y\geq x$ which is not possible unless we have $x=y=z$. (Similarly contradiction happens if $x<z$).
Now, plugging in the values we have,
$x=\sqrt{x^3-4} \Rightarrow x^3-x^2-4=0 \Rightarrow (x-2)(x^2+x+2)=0$
This implies $\boxed{x=2}$ which is the only solution to the problem.
Remark: I think it is quite amazing that this irrational equation can be written as a beautiful system of equations.
|
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|
Does $ a_{\lfloor n + \sqrt{n/2}\rfloor} \le a_n\left(1 + \frac{b}{n^{\frac{1}{2}+\varepsilon}}\right) $ imply that $a_n$ is bounded? Let $(a_n)_{n\ge 1}$ be an increasing sequence of positive real numbers and suppose there is a constant $b > 0$ and $\varepsilon > 0$ such that
$$
a_{\lfloor n + \sqrt{n/2}\rfloor} \le a_n\left(1 + \frac{b}{n^{\frac{1}{2}+\varepsilon}}\right)
$$
for all $n$ sufficiently large. Does it follow that $a_n$ is bounded? i.e., is $\limsup a_n < \infty$?
If the left hand side of the inequality was $a_{n+1}$, then we could use induction to derive
$$
a_{n + 1} \le a_{n_0}\prod_{k=n_0}^{n}\left(1 + \frac{b}{k^{\frac{1}{2}+\varepsilon}}\right)
$$
and even then, it is not clear whether $a_n$ is bounded. Moreover, the form of the index on the left hand side of the inequality makes it difficult to apply this method. Is there a way to do this?
Any help or comments are welcome.
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As the sequence $(a_n)_n$ increases, it tends either to a finite limit (the sequence converges), either to $+\infty$ (the sequence diverges).
We will prove that the second case can't happen. Suppose by contradiction that the sequence tends to $+\infty$, then, there exists $n_0$ such that $a_{n_0} >2$.
Let us define the two sequence $g(k)$ and $f(k)$ with $k\in\Bbb N^*$ and $x>0$ as follows
$$
\begin{cases}
f(k) = \left[f(k-1)+\sqrt{\frac{f(k-1)}{2}}\right] \\
f(0) = a_{n_0}
\end{cases}
$$
$$
\begin{cases}
g(k) = g(k-1)+\sqrt{\frac{g(k-1)}{2}}-1 \\
g(0) = a_{n_0}
\end{cases} \tag{1}
$$
It's easy to prove that the two sequences $f$ and $g$ are increasing and $g(k) \le f(k)$. So,
$$a_{f(n)} \le a_0\prod_{k=1}^n\left( 1+ \frac{b}{(f(k))^{\frac{1}{2}+\epsilon}}\right) \le a_{n_0} \prod_{k=1}^n\left( 1+ \frac{b}{(g(k))^{\frac{1}{2}+\epsilon}}\right) $$
$$\implies \ln a_{f(n)} \le \ln a_{n_0} \sum_{k=1}^n\ln\left( 1+ \frac{b}{(g(k))^{\frac{1}{2}+\epsilon}}\right) \tag{2}$$
Let us study the sequence $g(k)$, it's easy to notice that $g(k) \to +\infty$ with $a_{n_0}>2$.
$$
\begin{align}
\sqrt{g(k)}-\sqrt{g(k-1)} &= \sqrt{g(k-1)} \left(\left(1+ \frac{1}{\sqrt{2g(k-1)}}-\frac{1}{g(k-1)} \right)^{\frac{1}{2}} -1\right) \\
& \approx \sqrt{g(k-1)} \left(1+ \frac{1}{2} \frac{1}{\sqrt{2g(k-1)}}+\mathcal{O}\left(\frac{1}{g(k)}\right) -1\right)\\
& \approx \frac{1}{2\sqrt{2}}+\mathcal{O}\left(\frac{1}{\sqrt{g(k)}}\right) \\
&\to \frac{1}{2\sqrt{2}}
\end{align}
$$
Applying the Stolz–Cesàro theorem, we deduce that $g(k) \simeq \frac{k^2}{8} $ for $k \to +\infty$.
Return back to $(2)$, we have le term
$$
\begin{align}
\ln \left( 1+ \frac{b}{(g(k))^{\frac{1}{2}+\epsilon}}\right) \simeq \frac{b. 8 ^{\frac{1}{2}+\epsilon}}{k^{1+2\epsilon}}
\end{align}
$$
As the sum $\sum_{k=1}^{+\infty}\frac{1}{k^{1+2\epsilon}}$ converges to a limit $l$, we have the sequence $(a_{f(k)})_{k\ge 1}$ bounded. And so the sequence $(a_{n})_{n\ge 1}$ is also bounded (for each $n$, there exists a $k$ such that $f(k)>n$).
This result contradicts to what we supposed that the sequence tends to $+\infty$.
Conclusion:
we can conclude that the sequence $(a_n)_{n\ge 1}$ converges. Q.E.D
|
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|
Using generating functions to count number of ways to plan a semester This is the question:
The semester of a college consists of n days. In how many ways can we separate the semester into sessions if each session has to consist of at least five days?
My work:
If $A(x)$ is the generating function for making subintervals of length $\geq 5$, then
$A(x)=\sum_{k \geq 5}x^k$. Now, If B(x) is the generating function for splitting the n days into disjoint nonempty subintervals, then, by composition, $B(x)=\frac{1}{1-A(x)}$. However, this does not give the answer. (Image from the book "Introduction to Enumerative And Analytic Combinatorics", Miklos Bona)
I want to know where I am going wrong and a solution to this question.
|
You want to count ordered partitions of $n$ into parts of size at least $5$. You have $$A(x)=x^5+x^6+\dots= \frac{x^5}{1-x}$$
and
\begin{align}
B(x) &= \frac{1}{1-A(x)}
= \frac{1}{1-\frac{x^5}{1-x}}
= \frac{1-x}{1-x-x^5} \\
&= 1x^0 + 1x^5 + 1x^6 + 1x^7 + 1x^8 + 1x^9 + 2 x^{10} + 3 x^{11} + 4 x^{12} + 5 x^{13} + 6 x^{14} + 8 x^{15} + \dots
\end{align}
These coefficients look correct. For example, $2x^{10}$ corresponds to $10$ and $5+5$. The $3x^{11}$ corresponds to $11$, $6+5$, and $5+6$. Why do you doubt it?
Alternatively, let $b_n$ be the desired count of partitions. Clearly, $$b_n = \begin{cases}
1 &\text{if $n=0$}\\
0 &\text{if $n\in\{1,2,3,4\}$}\\
\end{cases} \tag1$$
For $n \ge 5$, condition on the size of the first part to obtain recurrence
$$b_n = \sum_{k=5}^n b_{n-k} \tag2$$
Now $(1)$ and $(2)$ imply that
\begin{align}
B(x) &= 1x^0 + \sum_{n=5}^\infty \left(\sum_{k=5}^n b_{n-k}\right)x^n\\
&= 1 +\sum_{k=5}^\infty x^k \sum_{n=k}^\infty b_{n-k} x^{n-k}\\
&= 1 +\sum_{k=5}^\infty x^k B(x) \\
&= 1 +B(x) \frac{x^5}{1-x}.
\end{align}
Solving for $B(x)$ yields
$$B(x)=\frac{1}{1-\frac{x^5}{1-x}}
=\frac{1-x}{1-x-x^5}.$$
|
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Vector quadruple product How did the author arrive at step $(3)$ from step $(1)$ and step $(2)$ in the following definition of vector quadruple product?
Many combinations of vector and scalar products are possible, but we consider only one more, namely the vector quadruple product $(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})$. By regarding $\mathbf{a} \times \mathbf{b}$ as a single vector, we see that this vector must be representable as a linear combination of $\mathbf{c}$ and $\mathbf{d}$. On the other hand, regarding $\mathbf{c} \times \mathbf{d}$ as a single vector, we see that it must also be a linear combination of $\mathbf{a}$ and $\mathbf{b}$. This provides a means of expressing one of the vectors, say $\mathbf{d}$, as linear combination of the other three, as follows:
\begin{align}
(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d}) &=[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}] \mathbf{c}-[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}] \mathbf{d} \tag{1}\\
&=[(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{a}] \mathbf{b}-[(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{b}] \mathbf{a} \tag{2}
\end{align}
Hence
$$
[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}] \mathbf{d}=[(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{d}] \mathbf{a}+[(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{d}] \mathbf{b}+[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}] \mathbf{c}\tag{3}
$$
or
$$
\mathbf{d}=\frac{[(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{d}] \mathbf{a}+[(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{d}] \mathbf{b}+[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}] \mathbf{c}}{[(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}]}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}
$$
(transcribed from this screenshot)
|
We have: $[(a\times b)\cdot d]c - [(a\times b)\cdot c]d = [(c\times d)\cdot a]b - [(c\times d)\cdot b]a$
isolating the intended left side we get $[(a\times b)\cdot c]d = [(c\times d)\cdot b]a - [(c\times d)\cdot a]b + [(a\times b)\cdot d]c$
Recognizing the things in the brackets as the scalar triple product, we can rearrange the first term by cycling $(c\times d) \cdot b = (b\times c) \cdot d$, and rearrange the second term and its sign with a single order swap $-(c\times d)\cdot a = (c \times a)\cdot d$. Plug these manipulations back in and we get the intended result, $$[(a\times b)\cdot c]d = [(b\times c)\cdot d]a + [(c\times a)\cdot d]b + [(a\times b)\cdot d]c$$
|
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if $a+b=2$, finding minimum value of $\frac{a^2+b^2}{\sqrt{a^2+1} \sqrt{b^2+4}}$ if $a+b=2$, Finding minimum value of $\frac{a^2+b^2}{\sqrt{a^2+1} \sqrt{b^2+4}}$.
I only thought of the Lagrangian multiplier method, it’s a lot of calculation, I can’t do it.
$\frac{a^2+b^2}{\sqrt{a^2+1} \sqrt{b^2+4}}+\lambda (a+b-2)$
(1) $-\frac{a \left(a^2+b^2\right)}{\left(a^2+1\right)^{3/2} \sqrt{b^2+4}}+\frac{2 a}{\sqrt{a^2+1} \sqrt{b^2+4}}+\lambda =0$
(2) $-\frac{b \left(a^2+b^2\right)}{\sqrt{a^2+1} \left(b^2+4\right)^{3/2}}+\frac{2 b}{\sqrt{a^2+1} \sqrt{b^2+4}}+\lambda =0$
Is there a simpler way?
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$$ b=2-a$$
$$ f(a) = \frac{a^2+(2-a)^2}{\sqrt{a^2+1} \sqrt{(2-a)^2+4}} $$
$$ f'(a) = \dfrac{2(a^3+6a^2+2a-12)}{\left(\left(2-a\right)^2+4\right)^\frac{3}{2}\left(a^2+1\right)^\frac{3}{2}}$$
The denominator of the derivative is always positive. It is sufficient to analyse the numerator. Observe that $a=-2$ is a root of the cubic in the numerator. It can be factorised as
$$ a^3+6a^2+2a-12 = (a+2)(a^2+4a-6)$$
The critical points of this function are thus $-2, -2 \pm \sqrt{10} $
It's easy to see that $a=-2$ is the local maxima and $a= -2 \pm \sqrt{10} $ are the local minima. Now compare the value of $f(a)$ at these two points to see which one is smaller. The minimum of $f(a)$ comes out to be at $a=-2+\sqrt{10}$
|
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Min and max of $f(x,y)=e^{-xy}$ where $x^2+4y^2 \leq 5$ I am trying to use Lagrange multipliers to find the maximum and minimum values of the function $$f(x,y)=e^{-xy}$$
constrained as $$x^2+4y^2=5$$
I began this problem by setting up the Lagrangian:
$$f(x,y) = e^{-xy}$$
$$g(x,y) = x^2+4y^2-5$$
$$L(x,y) = f(x,y) - \lambda g(x,y) = e^{-xy} - \lambda (x^2+4y^2-5)$$
So our equations are:
$$-ye^{-xy} - 2\lambda x = 0$$
$$-xe^{xy} -8\lambda y = 0$$
$$x^2+4y^2-5 = 0$$
Now from the first equation, $e^{-xy} = 2\lambda x/y$. Substituting into the second equation yields: $2x^2 \lambda / y = 8 \lambda y$ or $x^2+4y^2 = 0$. This clearly violates the third equation, meaning this system has no solution. Doe this mean there are no local maxima or minima?
Any guidance is greatly appreciated!
|
$\min f$ is at $\max (xy)$
If $g = x y$ is defined on $x^2 + 4 y^2 <= 5$
Then maximum of $g$ occurs at the boundary of the elliptical region,
namely the ellipse $x^2 + 4 y^2 = 5$
The semi-major and semi-minor axes are:
$a = \sqrt{5}$
$b = \sqrt{5}/2$
The parameteric equation of the boundary is
$p(t) = ( \sqrt{5} \cos t, \sqrt{5}/2 \sin t )$
$g = x y = 5/4 \sin (2 t)$ , so it is maximum at $t = \dfrac{\pi}{4}, \dfrac{5 \pi}{4}$
and minimum of $g$ occurs at $t = \dfrac{3\pi}{4}, \dfrac{7 \pi}{4}$
Thus $\min f = e^{-5/4} $ and $\max f = e^{5/4}$
The minimum occurs at $( \pm \sqrt{\dfrac{5}{2} } , \pm \sqrt{\dfrac{5}{8}} )$
And the maximum occurs at $( \mp \sqrt{ \dfrac{5}{2} } , \pm \sqrt{ \dfrac{5}{8} } )$
|
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|
Find the smallest odd value of $k$ such that $\displaystyle\int_{10}^{19}{\sin x\over1+x^k}dx<\frac 19$
Find the smallest odd value of $k$ such that $\displaystyle\int_{10}^{19}{\sin x\over1+x^k}dx<\frac 19$
$$\displaystyle\begin{align*}\int_{10}^{19}{\sin x\over1+x^k}dx&<\int_{10}^{19}{\mid\sin x\mid\over1+x^k}dx\\&<\int_{10}^{19}{1\over1+x^k}dx\\&<\int_{10}^{19}{1\over1+10^k}dx\\&<{9\over1+10^k}\end{align*}$$
If $k=3$,
$$\displaystyle\begin{align*}\int_{10}^{19}{\sin x\over1+x^3}dx&<{9\over1+10^3}<\frac 19\end{align*}$$
If $k=1$,
$$\require{cancel}\displaystyle\begin{align*}\int_{10}^{19}{\sin x\over1+x}dx&<{9\over1+10}\bcancel{<}\frac 19\end{align*}$$
However this is not sufficient to prove that the inequality does not hold for $k=1$. Infact the inequality is actually true for $k=1$ (wolfram alpha). Can someone suggest a method to prove that the inequality is true for $k=1$. Thanks in advance.
EDIT-1
I have also tried,
$$\displaystyle\begin{align*}\int_{10}^{19}{\sin x\over1+x}dx&=\int_{10}^{4\pi}{\sin x\over1+x}dx+\int_{4\pi}^{5\pi}{\sin x\over1+x}dx+\int_{5\pi}^{6\pi}{\sin x\over1+x}dx+\int_{6\pi}^{19}{\sin x\over1+x}dx\\&<\int_{4\pi}^{5\pi}{\sin x\over1+x}dx+\int_{6\pi}^{19}{\sin x\over1+x}dx\\&<{5\pi-4\pi\over 1+10}+{19-6\pi\over 1+6\pi}\bcancel{<}\frac 19\end{align*}$$
Again I got stuck.
|
You want to show that
$$\int_{10}^{19}\frac{\sin x dx}{1+x}\leq\frac19.$$
If this is true, it's because the "oscillations" of $\sin x$ cancel out, since the integral of the absolute value is a good bit larger than $1/9$. Since $1/(1+x)$ doesn't vary much in the interval $[10,19]$, you can subtract something close to its average, and see what happens if you then separate out the integrals. Use
$$\frac{1}{1+x}=\frac{1}{15}+\left(\frac{14-x}{15(1+x)}\right)$$
to get
$$\int_{10}^{19}\frac{\sin x dx}{1+x}=\frac1{15}\int_{10}^{19}\sin xdx+\int_{10}^{19}\frac{14-x}{15(1+x)}\sin xdx.$$
The first integral is $1/15$ of
$$-\int_{3\pi}^{10}\sin xdx+\int_{3\pi}^{6\pi}\sin xdx+\int_{6\pi}^{19}\sin xdx,$$
which is
$$-2+\int_0^{10-3\pi}\sin xdx+\int_0^{19-6\pi}\sin xdx.$$
Since $|\sin x|\leq x$, this is at most
$$-2+\frac{(10-3\pi)^2+(19-6\pi)^2}{2}<-2+\frac{0.58^2+0.16^2}{2}=-1.819.$$
The second integral is bounded by
$$\frac9{15}\max_{10\leq x\leq 19}\frac{|14-x|}{1+x}=\frac9{15}\frac4{11}=\frac{12}{55}.$$
So, the total integral is at most
$$-\frac{1.819}{15}+\frac{12}{55}<0.1<\frac19.$$
|
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|
Find the minimum of $P(X=0)$ when $E[X]=1,E[X^2]=2,E[X^3]=5$ using the probability generating function I was given the following exercise.
Let $X$ be a random variable that takes non-negative natural number
values such that $E[X]=1,E[X^2]=2,E[X^3]=5$. Find the minimum value of
$P(X=0)$ using the taylor expansion of the probability generating function
at $z=1$.
I know the method not using the generating function stated in this question.
My attempt:
Let $G(z)=E[z^X]$ be the probability generating function of $X$. Then by definition,
$G(z) = P(X=0)+P(X=1)z+P(X=2)z^2+\cdots $
Also, since $E[\frac{X(X-1)\cdots (X-n+1)}{n!}]=\frac{G^{(n)}(1)}{n!}$ according to Wikipedia, we have
$\displaystyle G(z) = \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](z-1)^n$
Substituting $z=0$ yields
$\displaystyle P(X=0) = \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n$
Note that for $n=0,1,2,3$, the coefficient can be calculated as follows:
$\begin{align} E[1] &= 1 \\ E[X]&= 1 \\ E[X(X-1)/2] &= E[X^2]/2 -E[X]/2 = 1-1/2 =1/2 \\ E[X(X-1)(X-2)/3!] &= E[X^3]/6 -E[X^2]/2 +E[X]/3 = 5/6-2/2+1/3 \\ &= 1/6 \end{align}$
Therefore,
$\displaystyle \begin{align} P(X=0) &= \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n \\ &=1-1+\frac{1}{2} -\frac{1}{6} +\sum_{n=4}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n \\ &= \frac{1}{3} + \sum_{n=4}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n\end{align}$
According to the question linked above, $1/3$ is the minimum, so I think we need to prove that the sum is non-negative to complete the proof. However, I was unable to do so.
Am I on the right path? If not, what is the correct one? If yes, how can I finish it?
|
We have absolute convergence on the (complex) unit ball around $0$.
Using the Taylor expansion of order three with an explicit form of the rest
we have for some $c=c(z)\in(z,1)$:
$$
\begin{aligned}
G(z) &=
\frac 1{0!}G(1)
+ \frac 1{1!}G'(1)(z-1)
\\
&\qquad\qquad
+ \frac 1{2!}G''(1)(z-1)^2
+ \frac 1{3!}G'''(1)(z-1)^3
\\
&\qquad\qquad\qquad\qquad
+ \frac 1{4!}G^{(IV)}(c)(z-1)^4 \ .
\\
&\qquad\text{Taking $z=0$...}
\\
G(0) &=
\frac 1{0!}E[1]
+ \frac 1{1!}E[X](0-1)
\\
&\qquad\qquad
+ \frac 1{2!}E[X(X-1)](0-1)^2
+ \frac 1{3!}E[X(X-1)(X-2)](0-1)^3
\\
&\qquad\qquad\qquad\qquad
+ \frac 1{4!}E[\underbrace{X(X-1)(X-2)(X-3)\cdot c^{X-4}}_{\ge 0}](0-1)^4
\\
&\ge 1 -\frac 1{1!}\cdot 1+\frac 1{2!}(2-1)-\frac 1{3!}(5-3\cdot 2+2\cdot 1)
\\
&=1-1+\frac 12-\frac 16=\frac 13
\ .
\end{aligned}
$$
$\square$
Note: The equality is satisfied if and only if for some (but then also for any) $c\in(0,1)$ we have $X(X-1)(X-2)(X-3)c^{X-4}=0$. This means that $X$ takes only the values $0,1,2,3$.
Note: The above tries to follow as close as possible the path from the OP. But there are some details related to the application of Taylor's Formula, if we insist to go this way. (Usually we want $G$ to be differentiable in the suitable class on an interval around the point we develop. This can be arranged as follows. Use some $a\in (0,1)$ instead of one first, then use the continuity of the $G$-derivatives to replace $a$ by $1^-$, and use their continuity in the compact ball of absolute convergence of the series for $G(z)$.)
|
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|
Solving a first order homogenous differential equation Problem:
Solve the following differential equation.
$$ x^2 \, dy + (y^2 -xy) \, dx = 0 $$
Answer:
$$ \left( \dfrac{x^2}{y^2} \right) \, dy
+ \left( \dfrac{y^2}{x^2} - \dfrac{y}{x} \right) \, dx = 0 $$
Hence we have a homogeneous differential equation. Let $y = vx$.
\begin{align*}
\dfrac{dy}{dx} &= v + x\dfrac{dv}{dx} \\
v^2 \left( v + x\dfrac{dv}{dx} \right) &= v^2 - v \\
v^3 + v^2 x \dfrac{dv}{dx} &= v^2 - v \\
v^2 x \dfrac{dv}{dx} &= -v^3 + v^2 - v \\
x \dfrac{dv}{dx} &= -v + 1 - v^{-1} \\
\dfrac{dv}{-v + 1 - v^{-1}} &= \dfrac{dx}{x}
\end{align*}
Using an online integral calculator, we find:
$$ \int \dfrac{1}{-v + 1 - v^{-1}} \, dv =
-\dfrac{\ln\left(v^2-v+1\right)}{2}
- \dfrac{\arctan\left(\frac{2v-1}{\sqrt{3}}\right)}{\sqrt{3}} $$
\begin{align*}
-\dfrac{\ln\left(v^2-v+1\right)}{2}
- \dfrac{\arctan\left(\frac{2v-1}{\sqrt{3}}\right)}{\sqrt{3}}
&= \ln{|x|} + C
\end{align*}
However, the book gets:
$$ y = \dfrac{x}{\ln{|x|}+C} $$
My answer is not going to match the book's answer. Where did I go wrong?
|
Let $y = vx$
$\cfrac{dy}{dx} = v + x\dfrac{dv}{dx}$
$x^2 \, dy + (y^2 -xy) \, dx = 0$
$ \cfrac{dy}{dx} + \left(\cfrac{y^2}{x^2} - \cfrac{y}{x}\right) = 0 \ $ (this is where you have a mistake)
$v + x\dfrac{dv}{dx} = v - v^2$
$\cfrac{dx}{x} = - \cfrac{dv}{v^2}$
$ \ln |x| + C = \cfrac{1}{v} = \cfrac{x}{y}$
$y = \cfrac{x}{\ln |x| + C}$
|
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|
How to derive this series for $\gamma$ that is only involving odd integer values of $\zeta(s)$? With $\gamma$ being the Euler Mascheroni constant, this series is well known:
$$1- \sum_{n=2}^{\infty} \frac{\zeta(n)-1}{n} = \gamma \tag{1}$$
The following series involving $\zeta(2n+1)$ also seems to converge to the same value, albeit slower:
$$1- \sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{(n+1)\,(2n+1)} = \gamma \tag{2}$$
Is there a way to derive (2) from (1) ?
|
Note that
$$\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}=2\frac{\zeta(2n+1)-1}{2n+1}-\frac{\zeta(2n+1)-1}{n+1}+\frac{2}{2n+1}-\frac{2}{2n+2}.$$
Hence
$$\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}=
2\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{2n+1}-\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{n+1}+2\sum_{n=3}^{\infty}\frac{(-1)^{n-1}}{n}.$$
Finally we apply Question about series involving zeta function? and A Tough Series $\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$ :
$$\begin{align}\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}&=
2\left(1-\gamma-\frac{\log(2)}{2}\right)-\left(-\gamma+\log(2)\right)+2\left(\log(2)-1+\frac{1}{2}\right)\\
&=\left(2-2\log(2)-\gamma\right)+2\log(2)-1\\
&= 1-\gamma.\end{align}$$
For a selfcontained proof, just note that
$$\begin{align}2\sum_{n=1}^{\infty}&\frac{\zeta(2n+1)-1}{2n+1}-\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{n+1}\\
&=\sum_{k=2}^\infty \sum_{n=1}^\infty \frac{2}{(2n+1)k^{2n+1}}-\sum_{k=2}^\infty\sum_{n=1}^\infty\frac{1}{(n+1)k^{2n+1}}\\
&=\sum_{k=2}^\infty \left(-\log \left(1-\frac{1}{k}\right)+\log \left( 1+\frac{1}{k}\right)-\frac{2}{k}\right)\\
&\qquad +\sum_{k=2}^\infty
\left(\frac{1}{k}+ k\log \left( 1+\frac{1}{k}\right)+ k\log \left( 1-\frac{1}{k}\right)\right)\\
&=\sum_{k=2}^\infty \left((k+1)\log \left( 1+\frac{1}{k}\right)
+(k-1)\log \left( 1-\frac{1}{k}\right)-\frac{1}{k}\right)\\
&=\sum_{k=2}^\infty \left((k+1)\log \left(k+1\right)-2k\log(k)
+(k-1)\log \left( k-1\right)-\frac{1}{k}\right)\\
&=\lim_{N\to \infty}\left((N+1)\log \left(N+1\right)
-2\log(2)-N\log(N)-H_N+1\right)\\
&=\lim_{N\to \infty}\left((N+1)\log \left(N+1\right)
-2\log(2)-N\log(N)-\log(N)-\gamma+1\right)\\
&=\lim_{N\to \infty}\left((N+1)\log \left(1+\frac{1}{N}\right)
-2\log(2)-\gamma+1\right)\\
&=2-2\log(2)-\gamma.
\end{align}$$
|
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|
Find the number of solutions of the equation $e^z = 2z+1$ in the open unit disc $\{z \in \Bbb C : |z| < 1\}$ Find the number of solutions of the equation $e^z = 2z+1$ in the open unit disc $\{z \in \Bbb C : |z| < 1\}$.
My Attempt:
Let $f(z) = e^z-2z-1$. Let $A \subseteq \{z \in \Bbb C : |z| < 1\}$ be the solution set. For solution,we will put $f(z) = 0$. Since $f(0) = 0$ so $0 \in A$. Therefore $A \neq \varnothing$. Also let $z = x+iy$ then $f(z) = 0$ gives $$e^{x+iy}-2(x+iy)-1 = 0 \implies e^xe^{iy}-(2x+1)-i(2y) =0\\
\implies e^x(\cos y + i\sin y) = (2x+1) + i(2y)$$
On comparing real and imaginary parts on both sides, we get
$$e^x\cos y = 2x + 1,\ e^x\sin y = 2y.$$
On dividing, we get $\tan y = \dfrac{2y}{2x+1}$.
I have no general formula. Please help me.
|
You already figured out that $z=0$ is one solution of the equation $e^z=2z+1$. Using the Taylor series of the exponential function and simple estimates one can show that there are no other solutions in the unit disk:
For $0 < |z| < 1$ is
$$
e^z - (1+2z) = -z + \frac{z^2}{2!}++ \frac{z^3}{3!} + \cdots
$$
and therefore
$$
\left| \frac{e^z-(1+2z)}{z}\right| = \left|-1 + \frac{z}{2!} + \frac{z^2}{3!} + \cdots\right| \ge \ 1- \frac{|z|}{2!} - \frac{|z|^2}{3!} - \cdots \\
\ge 1 - \frac{1}{2!} - \frac{1}{3!} - \cdots = 3 - e > 0 \, .
$$
|
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|
Find the value of $\int_0^1{4dx\over 4x^2-8x+3}$
Find the value of $\displaystyle\int_0^1{4dx\over 4x^2-8x+3}$
$$\begin{align*}\int_0^1{4dx\over 4x^2-8x+3}&=\int_0^1{dx\over (x-1)^2-(\frac 12)^2}
\\&=\int_0^1{dx\over (x)^2-(\frac 12)^2}
\\&=\int_{0}^{1/2}{dx\over (x)^2-(\frac 12)^2}+\int_{1/2}^{1}{dx\over (x)^2-(\frac 12)^2}
\\&=\bigg[\ln\bigg|\frac{x-\frac12}{x+\frac12}\bigg|\bigg]_0^{1/2}+\bigg[\ln\bigg|\frac{x-\frac12}{x+\frac12}\bigg|\bigg]_{1/2}^{1}
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|\frac{x-\frac12}{x+\frac12}\bigg|\bigg)-\ln(1)+\ln(\frac13)-\lim_{x\to{\frac{1^-}{2}}}\bigg(\ln\bigg|\frac{x-\frac12}{x+\frac12}\bigg|\bigg)
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)-\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x+\frac12}\bigg|\bigg)-\lim_{x\to{\frac{1^-}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)+\lim_{x\to{\frac{1^-}{2}}}\bigg(\ln\bigg|{x+\frac12}\bigg|\bigg)+\ln(\frac13)
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)-\ln(1)-\lim_{x\to{\frac{1^-}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)+\ln(1)+\ln(\frac13)
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)-\lim_{x\to{\frac{1^-}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)+\ln(\frac13)
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)-\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{(1-x)-\frac12}\bigg|\bigg)+\ln(\frac13)
\\&=\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)-\lim_{x\to{\frac{1^+}{2}}}\bigg(\ln\bigg|{x-\frac12}\bigg|\bigg)+\ln(\frac13)
\\&=\ln(\frac13)
\end{align*}$$
Can someone please tell me where I made the mistake (wolframalpha says that the integral does not converge). Also how do we know the integral does not converge? Any help will be appreciated.
|
Find the value of $\int_{0}^{1} \frac{4}{4x^2-8x+3} dx$
$$\int_{0}^{1} \frac{4}{4x^2-8x+3} dx = 4 \int_{0}^{1} \frac{1}{4x^2-8x+3} dx $$
Splitting the denominator we get,
$$4 \int_{0}^{1} \frac{1}{4x^2-8x+3} dx = 4 \int_{0}^{1} \frac{1}{(2x-3)(2x-1)} dx$$
Now using the method f partial fractions we get,
$$ 4 \int_{0}^{1} \frac{1}{(2x-3)(2x-1)} dx = 4 \int_{0}^{1} \frac{1}{2(2x-3)} -\frac{1}{2(2x-1)} dx = 4 \left[ \frac{1}{2} \int_{0}^{1} \frac{1}{2x-3} dx - \frac{1}{2} \int_{0}^{1} \frac{1}{2x-1} dx \right] = 2 \left[ \int_{0}^{1} \frac{1}{2x-3} dx - \int_{0}^{1} \frac{1}{2x-1} dx \right]$$
And,
$$\int_{0}^{1} \frac{1}{2x-3} dx = \left[\frac{1}{2} \ln|2x-3|\right|_{0}^{1} = \frac{1}{2} \left( \ln|-1|- \ln|-3| \right)~~~~~~~~~ \textbf{(1)} $$
$$\int_{0}^{1} \frac{1}{2x-1} dx = \left[ \frac{1}{2} \ln|2x-1| |\right|_{0}^{1} = \frac{1}{2} \left( \ln|1|- \ln|-1| \right)= \frac{1}{2} \ln|-1|~~~~~~~~~ \textbf{(2)}$$
Thus we have, $$2 \left[ \int_{0}^{1} \frac{1}{2x-3} dx - \int_{0}^{1} \frac{1}{2x-1} dx \right] = 2 \left[ \frac{1}{2} \left( \ln|-1|-\ln|-3| \right) -\frac{1}{2} \ln|-1| \right] $$ $$ = -\ln|-3| = \ln \left( \frac{1}{3} \right)$$
You can use Cauchy principal value to verify that the integral diverges.
https://en.wikipedia.org/wiki/Cauchy_principal_value
|
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|
Converting $\intop_{0}^{a}\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}dx$ to elliptic integral Tried using $x=a\sin\left(\theta\right)$ $$\rightarrow \intop_{0}^{\pi/2}\sqrt{\frac{a^{2}-\left(a\sin\left(\theta\right)\right)^{2}}{1-\left(a\sin\left(\theta\right)\right)^{2}}}a\cos{\left(\theta\right)d\theta}$$
$$\iff \intop_{0}^{\pi/2}\frac{a^{2}\cos^{2}\left(\theta\right)}{\sqrt{1-\left(a\sin\left(\theta\right)\right)^{2}}}{d\theta}$$
Which looks similar to the complete elliptical of first or second kind, is there a way to make the conversion? Thanks
|
Approximating the elliptic integral seems to be difficult.
What you could do is to expand the integrand around $a=1$
$$\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}=1+\frac{a-1}{1-x^2}-\frac{(a-1)^2 x^2}{2 \left(x^2-1\right)^2}-\frac{(a-1)^3 x^2}{2
\left(x^2-1\right)^3}-\frac{(a-1)^4 \left(x^2 \left(x^2+4\right)\right)}{8
\left(x^2-1\right)^4}+O\left((a-1)^5\right)$$ and integrate termwise to have for the integral (without simplifications)
$$a+(a-1) \tanh ^{-1}(a)+\frac{(a-1) \left(\left(a^2-1\right) \tanh ^{-1}(a)+a\right)}{4 (a+1)}-$$ $$\frac{1}{16} (a-1)^3 \left(\tanh ^{-1}(a)-\frac{a
\left(a^2+1\right)}{\left(a^2-1\right)^2}\right)+\frac{(a-1) \left(-9 a^5+40 a^3+9 \left(a^2-1\right)^3 \tanh ^{-1}(a)+9
a\right)}{384 (a+1)^3}$$
|
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|
Conditional Distribution Brownian process $X(t) = W^{2}(t)$ This is a homework problem. Please tell me if my solution is correct, if I am not correct please point out my error. If there is a better way to solve the problem, please let me know.
Problem
W(t) is a standard Brownian process $X(t) = W^{2}(t)$
a) find $f_{X}(x;t)$
b) find $f_{X}(x_{2}|x_{1};t_{2}. t_{1})$
Solution:
a)
$$ f_{W(t)} = \frac{1}{\sqrt{2 \pi t}}e^{-\frac{W(t)^{2}}{2t}} \\
g(w) = w^{2} \\
w_{1} = \sqrt{x} \hspace{5mm} w_{2} = -\sqrt{x} \\
f_{X(t)}(x) = \frac{f_{W(t)}(\sqrt{x})}{|2 \sqrt{x}|} + \frac{f_{W(t)}(-\sqrt{x})}{|-2 \sqrt{x}|}
= \frac{1}{\sqrt{2 \pi t X(t)}}e^{- \frac{X(t)}{2t}} \hspace{3mm}, X(t) \in (0, \infty)
$$
b)
Assume $ t_{1} < t_{2}$
let $W_{1} = X(t_{1}) \hspace{3mm}W_{2} = X(t_{2}) $
W(t) is Brownian so $W_{1}$ and $W_{2}$ are jointly normal $\therefore$
$$
E[W_{2}|W_{1} = w] = \mu_{W_{2}} + \rho \sigma_{W_{2}}\frac{w - \mu_{W_{1}}}{\sigma_{W_{1}}} \\
Var(W_{2} | W_{1} = w) = (1- \rho^{2})\sigma^{2}_{W_{2}} \\
\rho = \frac{Cov(W_{1}, W_{2})}{\sigma_{W_{1}} \sigma_{W_{2}}} \\
W_{1} \sim N(0, t_{1}) \hspace{3mm} W_{2} \sim N(0, t_{2}) \\
\therefore
\rho = \sqrt{\frac{t_{1}}{t_{2}}} \\
$$
$Cov(W_{1}, W_{2}) = t_{1}$ because $t_{1} < t_{2}$
$ X_{1}(t) = W_{1}^{2}(t) $ , $ X_{2}(t) = W_{2}^{2}(t)$
$\pm \sqrt{X_{1}(t)} = W_{1}(t) $ , $\pm \sqrt{X_{2}(t)} = W_{2}(t) $
for a given $(w_{1}, w_{2}) $ we have four points
$( \sqrt{x_{1}}, \sqrt{x_{2}}), ( -\sqrt{x_{1}}, \sqrt{x_{2}}), ( \sqrt{x_{1}}, -\sqrt{x_{2}}), ( -\sqrt{x_{1}}, -\sqrt{x_{2}})$
$$
\frac{\partial x_{1}}{\partial w_{1}} = 2w_{1} \hspace{3mm}\frac{\partial x_{1}}{\partial w_{2}} = 0 \\
\frac{\partial x_{2}}{\partial w_{1}} = 0 \hspace{3mm}\frac{\partial x_{2}}{\partial w_{2}} = 2w_{2} \\
J(w_{1}, w_{2}) = det \begin{bmatrix}2w_{1} & 0 \\0 & 2w_{2} \end{bmatrix} = 4w_{1}w_{2}
$$
$$
f_{X(t_{1}),X(t_{2})}(x_{1}, x_{2}) = \frac{1}{\pi \sqrt{1- \rho^{2}}}\frac{1}{|4 \sqrt{x_{1}} \sqrt{x_{2}}|}[exp[- \frac{1}{2(1- \rho^{2})}(x_{1} - 2 \rho \sqrt{x_{1}}\sqrt{x_{2}}+x_{2}) + exp[- \frac{1}{2(1- \rho^{2})}(x_{1} + 2 \rho \sqrt{x_{1}}\sqrt{x_{2}}+x_{2})]]
$$
$$
f_{X(t_{2})|X(t_{2})}(x_{2}| x_{1}) = \frac{\sqrt{2 \pi t_{1}x_{1}}}{\pi \sqrt{1- \rho^{2}}}\frac{e^{\frac{x_{1}}{2t_{1}}}}{|4 \sqrt{x_{1}} \sqrt{x_{2}}|}[exp[- \frac{1}{2(1- \rho^{2})}(x_{1} - 2 \rho \sqrt{x_{1}}\sqrt{x_{2}}+x_{2}) + exp[- \frac{1}{2(1- \rho^{2})}(x_{1} + 2 \rho \sqrt{x_{1}}\sqrt{x_{2}}+x_{2})]]
$$
|
(a)
$$P(W_t^2\leq x)=P(-\sqrt{x}\leq W_t\leq \sqrt{x})=\int_{[-\sqrt{x},\sqrt{x}]}f_{W_t}(y,t)dy=\Phi\bigg(\sqrt{\frac{x}{t}}\bigg)-\Phi \bigg(-\sqrt{\frac{x}{t}}\bigg) $$
where $\Phi$ is the standard normal cdf. Thus
$$f_{X_t}(x,t)=\frac{1}{2\sqrt{xt}}\phi \bigg(\sqrt{\frac{x}{t}}\bigg)+ \frac{1}{2\sqrt{xt}}\phi \bigg(-\sqrt{\frac{x}{t}}\bigg)= \frac{1}{\sqrt{xt}}\phi \bigg(\sqrt{\frac{x}{t}}\bigg) $$
where $\phi$ is the standard normal pdf.
(b)
Consider for $s < t$
$$P(X_t\leq x_1,X_s\leq x_2)=P(W_t\in [-\sqrt{x_1},\sqrt{x_1}],W_s\in [-\sqrt{x_2},\sqrt{x_2}])$$
we have
$$f_{W_t|W_s}(y_1,y_2)=\frac{1}{\sqrt{2\pi (t-s)}}\exp\bigg\{-\frac{(y_1-y_2)^2}{2(t-s)}\bigg\}$$
thus
$$P(X_t\leq x_1,X_s\leq x_2)=\int_{[-\sqrt{x_2},\sqrt{x_2}]}\bigg(\Phi\bigg(\frac{\sqrt{x_1}-y_2}{\sqrt{t-s}}\bigg)-\Phi\bigg(\frac{-\sqrt{x_1}-y_2}{\sqrt{t-s}}\bigg)\bigg)f_{W_s}(y_2)dy_2$$
so
$$\partial_{x_1}P(X_t\leq x_1,X_s\leq x_2)=\int_{[-\sqrt{x_2},\sqrt{x_2}]}\frac{1}{2\sqrt{x_1(t-s)}}\bigg(\phi\bigg(\frac{\sqrt{x_1}-y_2}{\sqrt{t-s}}\bigg)+\phi\bigg(\frac{-\sqrt{x_1}-y_2}{\sqrt{t-s}}\bigg)\bigg)f_{W_s}(y_2)dy_2$$
and
$$\partial_{x_1,x_2}P(X_t\leq x_1,X_s\leq x_2)=\frac{1}{4\sqrt{x_1x_2(t-s)}}\bigg(\phi\bigg(\frac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{t-s}}\bigg)+\phi\bigg(\frac{-\sqrt{x_1}-\sqrt{x_2}}{\sqrt{t-s}}\bigg)\bigg)f_{W_s}(\sqrt{x_2})+ \\
+\frac{1}{4\sqrt{x_1x_2(t-s)}}\bigg(\phi\bigg(\frac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{t-s}}\bigg)+\phi\bigg(\frac{-\sqrt{x_1}+\sqrt{x_2}}{\sqrt{t-s}}\bigg)\bigg)f_{W_s}(-\sqrt{x_2})= \\
=\frac{1}{2\sqrt{x_1x_2(t-s)}}\bigg(\phi\bigg(\frac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{t-s}}\bigg)+\phi\bigg(\frac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{t-s}}\bigg)\bigg)\frac{1}{\sqrt{s}}\phi\bigg(\frac{\sqrt{x_2}}{\sqrt{s}}\bigg)$$
So ultimately
$$f_{X_t|X_s}(x_1,x_2)=\frac{1}{2\sqrt{x_1(t-s)}}\bigg(\phi\bigg(\frac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{t-s}}\bigg)+\phi\bigg(\frac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{t-s}}\bigg)\bigg)$$
|
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|
Solving by Lagrange's auxiliary equations $ \frac {dx}{z} =\frac {dy}{z} =\frac {(3z+1)(dz)}{x+y} $ By taking 1st and 2nd equations 1st solution is $$ x=y+c_{0}$$
but i am getting two different solutions slightly differing :
Solution 1 : using lagrange multiplier 1,1,0 we get
$$ \frac{dx+dy}{2z}=\frac{(3z+1)dz}{x+y}$$
$$ \frac {(x+y)d(x+y)}{2z}= {(3z+1)(dz)}$$
$$ \frac {d(x+y)^{2}}{2} = (2z)(3z+1)(dz) =(6z^{2}+2z)dz$$
$$ \frac {(x+y)^2}{2} = {2z^3}+(z^2)+c_{1} $$
$$ (x+y)^{2}= 4z^{3}+2z^{2}+c_{1'} \to(1)$$
Solution 2 : taking x,y,-z as multipliers
$$ \frac {xdx+ydy-(zdz)(3z+1)}{xz+yz-z(x+y)} =\frac {xdx+ydy-(zdz)(3z+1)}{0}$$
$$ xdx+ydy= (3z+1)(zdz)=(3z^{2}+z)dz$$
$$ \frac{x^2}{2} +\frac{y^2}{2}= z^{3}+ \frac{z^2}{2}+c_{2} $$
$$ x^{2}+y^{2}=2z^{3}+z^{2}+c_{2'}$$
$$ 2(x^{2}+y^{2})=4z^{3}+2z^{2}+c_{2''} \to(2)$$
here the first solution $(1)$ is different and the term of $$xy$$ is extra .. i believe i have done mistake in my first solution but i am unable to find out where?
solution $(1)$ is my solution and solution $(2)$ is the solution from the the institute
|
There is no mistake in your calculus.
Both :
$$ (x+y)^{2}= 4z^{3}+2z^{2}+c_{1'} \to(1)$$
$$ 2(x^{2}+y^{2})=4z^{3}+2z^{2}+c_{2''} \to(2)$$
are correct, each one on different characteristic curves (because different constants of integrations).
This is not contradictory (on the characteristic curves, but not elsewhere) because
$$4z^3+2z^2=(x+y)^2-c_{1''}=2(x^2+y^2)-c_{2''}$$
$$2(x^2+y^2)-(x+y)^2=c_{2''}-c_{1''}$$
$$x^2+y^2-2xy=c_{2''}-c_{1''}$$
$$(x-y)^2=c_{2''}-c_{1''}$$
Since $x-y=c_0$
$$c_0^2=c_{2''}-c_{1''}$$
This denotes the relationship between the constants of integrations as expected.
|
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|
Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of inequality and using the equality leaves me to prove: $ab + bc + ca \le 3$.
Final edit: I found a easy way to prove above.
$18 = 2(a+b+c)^2 = (a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) + 4ab + 4bc + 4ca \ge 6(ab + bc + ca) \implies ab + bc + ca \le 3$
(please let me know if there is a mistake in above).
Attempt 2: multiplying both sides of inequality by $2$, we get:
$(a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$. By substituting $x = a+b, y = b+c, z = c+a$ and using $x+y+z = 6$ we will need to show:
$x^2 + y^2 + z^2 \ge 12$. This doesnt seem trivial either based on am-gm.
Edit: This becomes trivial by C-S.
$(a+b).1 + (b+c).1 + (c+a).1 = 6 \Rightarrow \sqrt{((a+b)^2 + (b+c)^2 + (c+a)^2)(1 + 1 + 1)} \ge 6 \implies (a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$
Attempt 3:
$x = 1-t-u$, $y = 1+t-u$, $z = 1 + 2u$
$(1-u-t)^2 + (1-u+t)^2 + (1+2u)^2 + (1-u-t)(1-u+t) + (1+t-u)(1+2u) + (1-t-u)(1+2u)$
$ = 2(1-u)^2 + 2t^2 + (1 + 2u)^2 + (1-u)^2 - t^2 + 2(1+2u)$
expanding we get:
$ = 3(1 + u^2 -2u) + t^2 + 1 + 4u^2 + 4u + 2 + 4u = 6 + 7u^2 + t^2 + 2u\ge 6$.
Yes, this works.. (not using am-gm or any such thing).
|
By the Cauchy-Schwarz, we have
$$a^2+b^2+c^2+ab+bc+ca=\frac{(a+b)^2+(b+c)^2+(c+a)^2}{2} \geqslant \frac{[(a+b)+(b+c)+(c+a)]^2}{2 \cdot 3} $$
$$=\frac{2}{3}(a+b+c)^2 = 6.$$
Done.
|
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|
Conditional inequality $2a^3+b^3≥3$
Non-negative $a$ and $b$ such that $a^5+a^5b^5=2$. How then do I prove
the following inequality $2a^3+b^3≥3$?
So, we can try using the Lagrange multiplier method:
Let $f(a, b)=2 a^{3}+b^{3}+\lambda(a^{5}+a^{5} b^{5}-2), \quad a, b \geq 0$
$$\tag1 \frac{\partial f}{\partial a}=6 a^{2}+\lambda(5 a^{4}+5 a^{4} b^{5})=0 \ldots$$
$$\tag2 \frac{\partial f}{\partial b}=3 b^{2}+\lambda(5 a^{5} b^{4})=0 \ldots$$
$$\left\{\begin{array}{c} 6+\lambda(5 a^{2}+5 a^{2} b^{5})=0 \\ 3+\lambda(5 a^{3} b^{4})=0 \end{array}\right. $$
$$ \lambda=-\frac{6}{5 a^{2}+5 a^{2} b^{5}}=-\frac{3}{5 a^{3} b^{4}} $$
Multiply by $a^3$, $\,2 a^{6} b^{4}=a^{5}+a^{5} b^{5}=2$.
$$ a^{6} b^{4}=1,\, a^{3} b^{2}=1 \Rightarrow b=\frac{1}{a^{\frac{3}{2}}}\quad a, b \geq 0 $$
$$ \begin{gathered} a^{5}+a^{5} b^{5}=2 \Rightarrow a^{5}+\frac{a^{5}}{a^{\frac{15}{2}}}=2 \Rightarrow a^{5}+\frac{1}{a^{\frac{5}{2}}}=2 \Rightarrow a^{5}-2 a^{\frac{5}{2}}+1=0 \\ \Rightarrow\left(a^{\frac{5}{2}}-1\right)^{2}=0 \Rightarrow a=1 \end{gathered} $$
And the minimum is at $(a,b)=(1,1)$.
I'm not sure I solved this inequality correctly, I would like to see a more beautiful way.
|
Proof by contradiction. Suppose $2a^3 + b^3 < 3$.
Then $ 0 \leq a \leq \sqrt[3]{3/2}$ and $a^5 + a^5b^5 < a^5 + a^5 (3-2a^3)^{5/3} $.
Let $ f(a) = a^5 ( 1 + (3-2a^3)^{5/3})$.
Verify that $f'(a) = 5a^4 ( 1 + (3-4a^3) (3-2a^3)^{2/3}) $.
Verify that $f'(a)$ is $0$ on $\{ 0, 1 \}$, positive on $(0, 1)$, and negative on $(1, \sqrt[3]{3/2}]$.
So $f(a)$ achieves its maximum at $a = 1$.
Thus, $f(a) \leq 2$.
So $a^5 + a^5b^5 < f(a) \leq 2 $, which is a contradiction.
I wanted to do this without calculus, but wasn't successful.
|
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|
Prove that $(a+b)^c\cdot(b+c)^a\cdot (c+a)^b\leq \left[\frac{2}{3}\cdot (a+b+c)\right]^{a+b+c}$ where $a,b,c\in \mathbb{Q}^{+}$ unless $a=b=c$.
If $a,b,c$ be positive rational numbers, prove that
$$(a+b)^c\cdot (b+c)^a\cdot (c+a)^b\leq \left[\frac{2}{3}\cdot
(a+b+c)\right]^{a+b+c}$$ unless $a=b=c$
My try :
Consider $(a+b),(b+c),(c+a)$ be three numbers with associated weights $c,a,b$
By weighted AM-GM inequality,
$$\frac{c\cdot (a+b)+a\cdot (b+c)+b\cdot (c+a)}{a+b+c}\ge \biggl[(a+b)^c\cdot (b+c)^a\cdot (c+a)^b\biggl]^\frac{1}{a+b+c}$$
the equality occurs when $(a+b)=(b+c)=(c+a)$ that is $a=b=c$
Therefore
$$\left[\frac{2(ab+bc+ca)}{a+b+c}\right]^{a+b+c}\gt (a+b)^c\cdot (b+c)^a\cdot (c+a)^b-----(1)$$
unless $a=b=c$
Now $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$$
Therefore $$(a+b+c)^2\gt 2(ab+bc+ca)-------(2)$$
using equation $(1)$ and $(2)$, we get,
$$\left(a+b+c\right)^{a+b+c}\gt (a+b)^c\cdot (b+c)^a\cdot (c+a)^b$$
I am stuck. please give me some hint. Thanks in advance.
|
$\displaystyle\left[\frac{2(ab+bc+ca)}{a+b+c}\right]^{a+b+c}\gt (a+b)^c\cdot (b+c)^a\cdot (c+a)^b-----(1)$
unless $a=b=c$
You can continue from here as follows
Note that $a^2+b^2+c^2> ab+bc+ca$
It follows that $\displaystyle\frac{(a+b+c)^2}{3}>ab+bc+ca$ and equivalently
$\implies\displaystyle\frac{2(a+b+c)^2}{3}>2(ab+bc+ca)$
$\implies\displaystyle\frac{2(a+b+c)}{3}>\frac{2(ab+bc+ca)}{a+b+c}$
Thus you have proved
$\displaystyle{\frac{2(a+b+c)}{3}}^{a+b+c}>{\frac{2(ab+bc+ca)}{a+b+c}}^{a+b+c}>(a+b)^c\cdot (b+c)^a\cdot (c+a)^b$
|
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|
Calculate the angle measure $TBA$ for reference:
On the internal bisector of $\measuredangle C$ of an isosceles triangle $ABC (AB = BC)$ if
externally marks point $T$, in such a way that $\measuredangle CTA = \measuredangle TAB = 30^o$.
Calculate $\measuredangle TBA$ (Answer $20^o$)
My progress:
$\triangle ABC(Isósceles) \rightarrow \measuredangle BAC = \measuredangle C\\
2\alpha+\alpha =120^o \therefore \alpha =40^0\rightarrow \measuredangle ABC = 180^o - 160^o = 20^0\\
\theta +x = 120^o$ $\text{but I can't finish...there's some equation missing}$
|
Longer method with the use of calculator!
Drop altitude $TD$ to the side $AB$. Then we have:
$$\sin 30^\circ = \frac{DT}{AT} \Rightarrow DT=\frac12AT\\
\begin{cases}\tan x=\frac{DT}{BD}\\ \tan 30^\circ =\frac{DT}{AD}\end{cases}\Rightarrow AD+BD=\frac{DT}{\tan x}+\sqrt3\cdot DT\Rightarrow AB=\frac12AT\left(\frac1{\tan x}+\sqrt3\right) \quad (1)\\
\frac{AT}{\sin 40^\circ}=\frac{AC}{\sin 30^\circ} \quad (2)\\
\frac{AB}{\sin 80^\circ}=\frac{AC}{\sin 20^\circ} \quad (3)$$
From $(2)$ and $(3)$:
$$AB=\frac{AT\cdot \sin80^\circ}{2\sin40^\circ \sin20^\circ} \quad (4)$$
From $(1)$ and $(4)$:
$$\frac1{\tan x}+\sqrt3=\frac{\sin80^\circ}{\sin40^\circ \sin20^\circ}=\frac{1}{\sin 40^\circ \cdot \sin 10^\circ}\Rightarrow \\
x=\arctan\left(\frac{1}{\frac1{\sin 10^\circ\cdot \sin 40^\circ}-\sqrt3}\right)=20^\circ.$$
Wolframalpha answer.
|
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|
Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$. I want to find limit of $a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}$ as $n\to \infty$.
$\displaystyle a_{n} =\frac{\left( 2n^{\frac{1}{n}} -1\right)^{n}}{n^{2}} =\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right)^{n}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\log a_{n} =n\log\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right) =\frac{\log\left( 1+\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)\right)}{\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)} .\left(\frac{2n}{n^{\frac{1}{n}}} -\frac{n}{n^{\frac{2}{n}}} -n\right)\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\log\left( 1+\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)\right)}{\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)} .\left(\frac{n^{\frac{1}{n}} -1}{n^{\frac{1}{n}}}\right)^{2} .( -n)
\end{array}$
The first term on RHS has limit equal to $\displaystyle 1\ $but the second term is giving me problem.
Please help. Thanks.
|
Note that $b_n=n^{1/n}-1\to 0$ and by Taylor series we have $$nb_n= \log n+\frac{(\log n) ^2}{2n}+ o\left(\frac{(\log n) ^2}{n}\right)=\log n+o(1)$$ and hence we have $$\log a_n=n\log (1+ 2b_n) -2\log n$$ Next we have via Taylor series $$\log a_n=2nb_n-2\log n-2nb_n^2+o(nb_n^2)=o(1)$$ (as $nb_n-\log n=o(1)$ and $nb_n^2=o(1)$) and hence $a_n\to 1$.
Alternatively let $a_n=c_n^n$ so that $$c_n=\frac{2n^{1/n}-1}{n^{2/n}}$$ and $$n(c_n-1)=-n\left(1-n^{-1/n}\right)^2=-\left(\frac{1-\exp(-(\log n) /n) }{-(\log n) /n}\right)^2\cdot\frac{(\log n) ^2}{n}$$ so that $n(c_n-1)\to 0$ and $a_n=c_n^n\to 1$ via lemma of Thomas Andrews.
|
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|
calculate cubic equation discriminant Let $p(x)=x^3+px+q$ a real polynomial.
Let $a,b,c$ the complex root of $p(x)$.
What is the easiest way to calculate $\Delta=(a-b)^2(b-c)^2(a-c)^2$ in function of $p,q$?
The result is $\Delta=-4p^3-27q^2$
Ps. $f=(x-a)(x-b)(x-c)$ and so $a+b+c=0$; $ab+bc+ac=p$ and $abc=-q$; how to proceed?
|
Let me offer you a different method which can be useful when calculating other symmetric functions of the roots.
$\Delta$ is symmetric in the roots, of total degree $6$.
In general then, where the cubic is $X^3-e_1X^2+e_2 X-e_3$, $\Delta$ must be a sum of monomials $e_1^{k_1}e_2^{k_2}e_3^{k_3}$ with $k_1+2k_2+3k_3=6$. When $e_1=0$ we must have $\Delta= A e_2^3 +B e_3^2$.
We can find $A,B$ by evaluating $\Delta$ for the two easy cubics $X^3-X$ and $X^3-1$ whose discriminants are $4$ and $-27$.
|
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|
Find the largest volume of a cuboid inside a sphere with radius $3m$ I've solved this one by using geometry, but is there a way of finding the max by using derivatives?
My work:
So, because of geometry it has to be a cube.
The diagonal of a cube is equal to $a\sqrt{3}$ and also to $2r$
So: $a\sqrt{3}=2r$
$a=\dfrac{2r}{\sqrt{3}}$
The volume of a cube is $a^3$, so by replacing $a$ with the above result, we get:
$V=\dfrac{8r^3}{3\sqrt{3}}$, replace $r$ with $3m$ and we get $V=24\sqrt{3} \;m^3$
Any help would be appreciated!
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Consider a sphere with the equation $x^2 + y^2 + z^2 = 9m^2$ with volume generated by $V = 8|x||y||z|$.
Consider $0 \leq A \leq 3m$ such that $x^2 + y^2 = A^2$
To maximize $xy$, it is clear that since $xy =\frac{(x+y)^2 - A^2}{2}$ that a maximum occurs when $x + y$ is maximized.
Since $\frac{d}{dx} (x+y) = \frac{d}{dx}(x + \sqrt{A^2-x^2}) = 1 - \frac{x}{y}$, a maximum occurs when $x = y$
Then $2x^2 + z^2 = 9m^2$ and $V^2 = x^4(9m^2 - 2x^2)$
$\frac{dV^2}{dx} = x^3(36m^2 - 12x^2)$
$\frac{dV^2}{dx} = 0$ when $x = 0$ or $x^2 = 3m^2$, and clearly since $V = 0$ when $x = 0$, a maximum occurs at $x = \pm \sqrt{3}m$. From the equations found previously, $y = \pm\sqrt{3}m$ and $z = \pm\sqrt{3}m$
Thus $V = 8 \cdot 3\sqrt{3} \cdot m^3 = 24\sqrt{3} \cdot m^3$
(Note: You can also differentiate $xy$ directly to obtain $\sqrt{A^2-x^2} - \frac{x^2}{\sqrt{A^2-x^2}} = y - \frac{x^2}{y}$ which also results in $x = y$)
|
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|
Relation of coefficient of Cubic polynomial If $a.b $ & $c$ are rational number satisfying the equation $x^3+ax^2+bx+c=0$, then which of the following can also be true .
A) $a+b^2+c^3=0$
B) $a+b^2+c^3=5$
C) $a+b^2+c^3=1$
D) None of these
If it is a cubic function the equation has at least one real root. So for any value of a, b & c we have one real root, hence i am confused. The question i presume is cotrect
|
Since no relation between $a, b$ and $c$ need to hold in order to satisfy $$x^3+ax^2+bx+c=0 \tag{i}$$ as there always exist some real $x$ which is one of the roots of $\text{(i)}$, I'm assuming that $a, b$ and $c$ satisfy $\text{(i)}$ for all $x \in \mathbb R$
Thus, for $x=0$, equation $\text{(i)}$ becomes
$0+0+0+c=0 \implies c=0$
Also,
$1+a+b+c=0$ and $ -1+a-b+c=0$ for $x=1$ and $x=-1$ in $\text{(i)}$ respectively.
Solving them will give you $a=0,b=-1,c=0$
But for $x=2$,we get $8+4a+2b+c=0 \implies 8-2=0$ which is a contradiction, so there do not exists any rational numbers $a,b,c$ that can satisfy $\text{(i)}$ for all $x \in \mathbb R$
|
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|
Divisibility rules with prime number Let $n \in \mathbb{Z}$ with the property that
$$
7 \mid\left(n^{3}+1\right)
$$
but $7$ does not divide $\left(n^{2}-2 n-3\right) .$ Prove that $7 \mid(4 n+1)$.
So far I got:
$n^3+1 = (n+1)\cdot(n^2-n+1)$, since $7$ is a prime, so
$7\mid(n+1)$ or $7\mid(n^2-n+1)$.
And since $7$ does not divide $\left(n^{2}-2 n-3\right) = (n-3)\cdot(n+1)$, I know that $7\mid(n^2-n+1)$.
And $7$ does not divide the difference of $(n^2-n+1)$ and $(n^2-2n-3)$ which is $n+4$.
Same since $7$ is prime, $7$ does not divide $4n+16\Rightarrow$$7$ does not divide $4n+2$.
Then I am stuck here, how could I use the information I got so far to prove $7 \mid(4 n+1)$?
Really hope someone could help/hint me with it! Thank you!
|
Further factor $n^2 - n + 1 \equiv (n - 3)(n - 5) \mod 7$. Since $n - 3 \neq 0 \mod 7$, we see that $n - 5 \equiv 0 \mod 7$. So $n \equiv 5 \mod 7$. So $4n + 1 \equiv 0 \mod 7$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $m^3-n^3=2mn+8$ We have to find integer solutions to the given equation, this is what i tried :-
For ease, let us denote $x=-m, \enspace n=-y$, and then we are basically considering $x,m,y,n$ as nonnegative wherever they occur below :-
We have four cases :-
CASE $1$ : $mn<0$ and $x^3+n^3=-(8+2xn)$
This case is clearly rejected as $x,n$ are both positive.
CASE $2$ : $mn<0$ and $m^3+y^3=8+2my$
For this we apply AM - GM to get,
$$9 + 2my = m^3 + y^3 + 1 \geq 3my \implies my \le 9 $$
Case bash (sadly) gives that the solutions to this case are
$$(m,y)=(0,2), \enspace (2,0), \enspace (2,2)$$
Now we are left with two cases
CASE $3$ : $mn>0$ and $m^3-n^3=8-2mn$
CASE $4$ : $mn>0$ and $x^3-y^3=2xy-8$
Now how to finish off these last $2$ cases?
Thanks in advance!
|
Another way to solve this is to introduce $x=m-n$. Then you have quadratic equation on $n$:
$$(3x-2)n^2+(3x^2-2x)n +x^3-8=0$$
This equation has integer solution if it discriminant is perfect square:
$$(3x^2-2)^2-4(3x-2)(x^3-8) = d^2$$
In particular $$3x^4-8x^3+12x^2-96x+60\leq 0$$ which can not occurre many times...
|
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|
The number of solutions of $a+b+c+d=n,\ a\geq b,\ d,\ {\rm and}\ c\geq b+1,\ d$ Define $P(n)$ to be the number of solutions : $a,\ b,\ c,\ d$ are nonnegative integers s.t.
$$a+b+c+d=n $$ and $$ a\geq b,\ d,\ {\rm and}\ c\geq b+1,\ d$$
Define $Q(n)$ to be the number of solutions : $x,\ y,\ z,\ w $ are nonnegative integers s.t.
$$ x+y+2z+3w= n-1 $$
Prove that $P(n)= Q(n)$ for all $n$
Proof : First we consider the case where $b=z,\ d=w$. I think that this is the easiest case. (Surely we can use the other case. For instance, $b={\rm max}\ \{w,z\} - {\rm min}\ \{ w,z\}$.)
If we set
$$a = x+{\rm max}\ \{w,z\},\ b=z,\ d=w,\ c=y+{\rm max}\ \{w,z+1\} $$
then we have the condition $$a\geq b,\ d\ {\rm and}\ c\geq b+1,\ d$$
If $w\geq z+1$, then $$ a+b+c+d = x+w+z+w+y+w =n-1-z$$
How can we prove this ?
|
Let
\begin{align*}
P_n &= \{(a, b, c, d) \in \mathbb{N}_0^4 : a + b + c + d = n, a \geq b, d, c \geq b + 1, d\}\\
Q_n &=\{(x, y, z, w) \in \mathbb{N}_0^4 : x + y + 2z + 3w = n - 1\}
\end{align*}
so $|P_n| = P(n)$ and $|Q_n| = Q(n)$.
We will proceed by induction to show $P(n) = Q(n)$ for all $n \geq 0$.
Note that the base cases are true since $P(0) = Q(0) = 0$ and $P(1) = Q(1) = 1$.
Now suppose true for $n$. Then for $n + 1$, we will show that $P(n+1) - P(n) = Q(n+1) - Q(n)$:
Note that if $(x, y, z, w) \in Q_n$, then $(x + 1, y, z, w) \in Q_{n+1}$ and for all $(x, y, z, w) \in Q_{n+1}$ such that $x \neq 0$, $(x - 1, y, z, w) \in Q_n$. Hence, $Q(n+1) - Q(n)$ represents the number of non-negative integer solutions to $y + 2z + 3w = n$.
Also note that if $(a, b, c, d) \in P_n$, then $(a + 1, b, c, d) \in P_{n+1}$ and for all $(a, b, c, d) \in P_{n+1}$ such that $a > \max(b, d)$, $(a - 1, b, c, d) \in P_n$. Hence, $P(n+1) - P(n)$ represents the number of non-negative integer solutions to $\max(b, d) + b + c + d = n + 1$ with $c \geq b + 1, d$.
In the case where $b \geq d$, we have $b = d + k$, $c = d + k + l + 1$ with $k, l \geq 0$, so $l + 3k + 4d = n$. Hence, this case counts the number of solutions to $y + 2z + 3w = n$ where $z$ is even ($z = 2d$).
In the case where $d > b$, we have $d = b + k + 1$, $c = b + k + l + 1$ with $k, l \geq 0$, so $l + 3k + 4b = n - 2$. Hence, this case counts the number of solutions to $y + 2z + 3w = n$ where $z$ is odd ($z = 2b + 1$).
But these two cases count the total number of solutions so we get that $Q(n+1) - Q(n) = P(n+1) - P(n)$.
Since $P(n) = Q(n)$, we get that $P(n+1) = Q(n+1)$.
Therefore, by induction we're done.
Note that this also yields the following bijection from $Q_n \to P_n$:
\begin{align*}
(x, y, z, w) \mapsto \begin{cases}
\left ( x + \frac{z}{2} + w, \frac{z}{2} + w, y + \frac{z}{2} + w + 1, \frac{z}{2}\right ) & \text{if $z$ is even}\\
\left ( x + \frac{z + 1}{2} + w, \frac{z - 1}{2}, y + \frac{z + 1}{2} + w, \frac{z + 1}{2} + w\right ) & \text{if $z$ is odd}
\end{cases}
\end{align*}
|
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|
Trigonometric elimination between two variables Eliminate $\theta$ and $\phi$ between the following equations: $$\begin{cases}\sin \theta + \sin \phi = x \\ \cos \theta + \cos \phi = y \\ \tan \frac {\theta}{2} \tan \frac {\phi}{2} = z\end{cases}$$
What I've done so far
I've established that $$\tan \left(\frac {\theta+\phi}{2}\right) = \frac {\sin \theta + \sin \phi}{\cos \theta + \cos \phi}$$ so that $$\tan \frac {\theta+\phi}{2} = \frac {x}{y}.$$
I then used the trigonometric identity $$\tan \left(\frac {\theta+\phi}{2}\right) = \frac {\tan \frac {\theta}{2} + \tan \frac {\phi} {2}}{1-\tan \frac {\theta}{2}\tan \frac {\phi} {2}}$$ and with a little manipulation got to $$\tan \frac {\theta}{2} + \tan \frac {\phi} {2} = \frac {x(1-z)}{y}$$
I'm stumped on the next steps...would I have to find the difference of the roots also (i.e. $\tan \frac {\theta}{2} - \tan \frac {\phi} {2}$)? Or is there a simpler way? (Side note: I also tried $\tan \frac {\theta}{2}$ substitution but that went nowhere.)
|
Write $u = \tan \theta/2$, $v = \tan \phi/2$. Then we get
$$
\begin{cases}
2(u+v)(1 + uv) = x(1+u^2)(1+v^2) \\
2 -2u^2v^2 = y(1+u^2)(1+v^2) \\
uv = z.
\end{cases}
$$
Now setting $s = u + v$ and using the third relation, we obtain
$$
\begin{cases}
2s(1+z) = x(1 - 2z + z^2 + s^2) \\
2(1-z^2) = y(1 - 2z + z^2 + s^2).
\end{cases}
$$
Dividing, we obtain the relation $s = x(1-z)/y$, which you've already mentioned. Substituting this relation into either of the equations, we find
$$2y(1 + z) = (x^2 + y^2)(1-z).$$
|
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|
Calculating the Image of $A.$ Where did I err? $$A=\begin{pmatrix}
4 &-1& 1\\
8&-2&2\\
-6&1&-2\\
\end{pmatrix}$$
I have to show $p=\begin{pmatrix}
1 \\
2\\
-2 \\
\end{pmatrix} \in \mathrm{Im}A=\left\{Ax \mid x\in \mathbb{R^3} \right\}$
If I let $x=\begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix},$ then $Ax=\begin{pmatrix}
1 \\
2\\
-2 \\
\end{pmatrix}=p$ thus $p\in \mathrm{Im}A$.
But when I try to find what $\mathrm{Im}A$ is, I'm pazzled.
In order to find Im$A,$ I did elementary transformations for $A$, and I got $$A\to \begin{pmatrix}
1 & 0 & \frac{1}{2} \\
0&1&1\\
0&0&0\\
\end{pmatrix}.$$
And $$\begin{pmatrix}
1 & 0 & \frac{1}{2} \\
0&1&1\\
0&0&0\\
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2\\
x_3\
\end{pmatrix}=(x_1+\frac{1}{2}x_3)\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}
+
(x_2+x_3)
\begin{pmatrix}
0 \\
1 \\
0\\
\end{pmatrix}$$
so Im$A=$span$\left\{
\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix},
\begin{pmatrix}
0 \\
1 \\
0\\
\end{pmatrix}
\right\}$
But I cannot write $p$ as a linear combination of $\begin{pmatrix}
1\\
0\\
0\\
\end{pmatrix}$ and $
\begin{pmatrix}
0 \\
1 \\
0\\
\end{pmatrix}$
Where did I err ?
|
The image is the span of linearly independent vectors in the column of the matrix. It suffices to show that $p$ is redundant and can be written as a linear combination of the other columns. After your row reduction, notice that only the first two columns have free variables whereas the third doesn't. This automatically implies that the third is redundant and the linear combination should be obvious.
Alternatively, you could have set up the augmented matrix system below:
$$\begin{pmatrix}
4 &-1& | &1 &\\
8&-2& | &2\\
-6&1& | &-2\\
\end{pmatrix}$$
After row reducing, we have,
$$\begin{pmatrix}
1 &0& | &\frac{1}{2} &\\
0&1& | &1\\
0&0& | &0\\
\end{pmatrix}$$
, which is analogous to what you found. In other words, we have that,
$$
\begin{pmatrix}
1 \\
2 \\
-2\\
\end{pmatrix} = \frac{1}{2}\begin{pmatrix}
4 \\
8 \\
-6\\
\end{pmatrix} + \begin{pmatrix}
-1 \\
-2 \\
1\\
\end{pmatrix}$$
, which is easily verified.
|
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|
Wrong answer in Thomas Calculus 14th Edition textbook There is this question on derivatives to which the answer is given as $\frac{43}{75}$rad/sec in the answers section. This answer appears to be wrong.
My Solution
$$
\theta+\tan^{-1}\left(\frac{6}{4-x}\right)+\tan^{-1}\left(\frac{3}{x}\right)=\pi
$$
which gets reduced to
$$
\theta=\pi-\tan^{-1}\left(\frac{3x+12}{4x-x^2-18}\right)
$$
Therefore,
$$
\frac{d\theta}{dt}=\left(\frac{3x^2+24x-102}{(4x-x^2+18)^2+(3x+12)^2}\right)\frac{dx}{dt}
$$
Given $x=4$ and $\frac{dx}{dt}=2\text{ cm/sec}$,
$$
\frac{d\theta}{dt}=-\frac{7}{75}\text{rad/sec}
$$
I've got the graph of $\theta$ as a function of $x$ here, which also indicates my answer is correct. Or am I? Kindly help.
|
Here's an alternative explicit calculation.
Let $D$ be the end of the $6$-cm segment opposite $C.$
Let $E$ be the end of the $3$-cm segment opposite $A.$
The angle $\theta$ is the sum of the angles $\angle BDC$ and $\angle BEA.$
We can see that when $x=4,$ then $B$ has just reached $C$ and is moving perpendicular to the $6$ cm segment $BC$ at the rate $2$ cm per second,
so the angular velocity of $B$ around the point $D$ is
$2/6 = \frac13$ cm per second clockwise
and therefore $\angle BDC$ is changing at the rate
$-\frac13$ radian per second.
Also when $x = 4,$ the distance from $B$ to $E$ is $5$ cm.
Consider the radial and tangential components of the velocity vector of $B$
relative to the circle of radius $5$ around $E.$
The velocity vector and its components make a $3,4,5$ right triangle
with the magnitude of the tangential component equal to $\frac35$ the magnitude of the velocity; that is, the tangential component is
$\frac35\cdot 2 = \frac65$ cm per second clockwise.
This means that the angle $\angle BEA$ is increasing at the rate
$\frac{6/5}{5} = \frac{6}{25}$ cm per second.
So the answer is
$$ -\frac13 + \frac{6}{25} = - \frac{25}{75} + \frac{18}{25} = - \frac{7}{75}, $$
exactly as you found.
Note that
$$ \frac{25}{75} + \frac{18}{25} = \frac{43}{75}, $$
so the supposed answer in the book appears to be the result of a sign error.
|
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|
For an acute angled triangle $ABC,$ if $p=\frac{\sqrt3+\sin A+\sin B+\sin C}{2\sin A\sin B\sin C}$, find the range of $p$ $$ p=\frac{\sqrt3+\sin A+\sin B+\sin C}{2\sin A\sin B\sin C}$$
$\displaystyle \sin A+\sin B+\sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$ and
$\displaystyle \sin A\sin B\sin C=8\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$ in a triangle
$A<90^\circ, \text{so} \space A/2<45^\circ$ then $ \sin\frac{A}{2}<1/\sqrt2$. However, $$ \cos\frac{A}{2}>1/\sqrt2$$
So this probably doesn't lead us anywhere :( .
Can anyone please help, thanks.
|
The lower bound.
For $\alpha=\beta=\gamma=60^{\circ}$ we obtain a value $\frac{10}{3}$.
We'll prove that it's a minimal value.
Indeed, in the standard notation we need to prove that $$\sqrt3+\sum_{cyc}\frac{2S}{bc}\geq\frac{20}{3}\cdot\frac{8S^3}{a^2b^2c^2}$$ or
$$\sqrt3+\frac{2S(a+b+c)}{abc}\geq\frac{160S^3}{3a^2b^2c^2}.$$
Here $S$ is the area of the triangle
Now, let $a=\frac{y+z}{2}$, $b=\frac{x+z}{2}$ and $c=\frac{x+y}{2}$.
Thus, $x$, $y$ and $z$ are positives and we need to prove that:
$$\sqrt3+\frac{4\sqrt{xyz(x+y+z)^3}}{\prod\limits_{cyc}(x+y)}\geq\frac{160\sqrt{x^3y^3z^3(x+y+z)^3}}{3\prod\limits_{cyc}(x+y)^2}$$ or
$$\prod_{cyc}(x+y)+12\sqrt{xyz\left(\frac{x+y+z}{3}\right)^3}\geq\frac{160\sqrt{x^3y^3z^3(x+y+z)^3}}{3\sqrt3\prod\limits_{cyc}(x+y)}.$$
Now, by AM-GM we obtain:
$$\prod_{cyc}(x+y)+12\sqrt{xyz\left(\frac{x+y+z}{3}\right)^3}=2\cdot\frac{\prod\limits_{cyc}(x+y)}{2}+3\cdot4\sqrt{xyz\left(\frac{x+y+z}{3}\right)^3}\geq$$
$$\geq5\sqrt[5]{\left(\frac{\prod\limits_{cyc}(x+y)}{2}\right)^2\left(4\sqrt{xyz\left(\frac{x+y+z}{3}\right)^3}\right)^3}$$ and it's enough to prove that:
$$5\sqrt[5]{\left(\frac{\prod\limits_{cyc}(x+y)}{2}\right)^2\left(4\sqrt{xyz\left(\frac{x+y+z}{3}\right)^3}\right)^3}\geq\frac{160\sqrt{x^3y^3z^3(x+y+z)^3}}{3\sqrt3\prod\limits_{cyc}(x+y)}$$ or
$$3^6\prod_{cyc}(x+y)^{14}\geq2^{42}(xyz)^{12}(x+y+z)^6.$$
Now, use $$\prod_{cyc}(x+y)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)\geq\frac{8}{9}(x+y+z)\sqrt{3xyz(x+y+z)}$$ and $$x+y+z\geq3\sqrt[3]{xyz}.$$
The upper bound is $+\infty$.
Try $\alpha=\beta\rightarrow\left(90^{\circ}\right)^-$ and $\gamma\rightarrow\left(0^{\circ}\right)^+$.
|
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|
Why is $\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)=\frac n4$? I found the relation for $n\geq3$ $$\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)=\frac n4$$But despite my best efforts, I still have no idea as to how to prove it.
Things I've tried:
*
*Adding $\cos^2$ terms. Of course, $\sin^2x+\cos^2x$ and $\cos^2x-\sin^2x$ are both simplifiable, so I thought about adding $\cos^2$ terms to make the sum into $\frac n2$ and hope that the $\sin^2$ and $\cos^2$ terms sum to equal amounts. They don't in the $n$ odd case, so I didn't know what to do with this approach.
*Adding more $\sin^2$ terms: Since $\sin^2 x=\sin^2(\pi-x)$, we can add terms to this sum, but honestly, it didn't make the sum any easier to evaluate.
Any help here?
Edit:
Based on the comments, here are my attempts.
$$\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)=\frac12\cdot\left\lfloor\frac n2\right\rfloor-\frac12\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\cos\left((2k-1)\frac{2\pi}n\right)$$$$=\frac12\cdot\left\lfloor\frac n2\right\rfloor-\frac12\mathfrak{Re}\left(\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\exp\left((2k-1)\frac{2i\pi}n\right)\right)$$How do I finish?
|
Let$$T=\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\cos\left((2k-1)\frac{2\pi} n\right)$$$$S=\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)$$
We can get the relation $T+2S=\lfloor\frac n2\rfloor$
If $n$ is even then using summation of cosines whose angles are in AP we get,
$$T=\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\cos\left((2k-1)\frac{2\pi} n\right)=\frac {\sin\lfloor\frac n2\rfloor\frac {2\pi}n}{\sin\frac {2\pi}n}\cos\bigg[\frac{2\pi}n+(\lfloor\frac n2\rfloor-1)\frac{2\pi}n\bigg]=0$$
$$\implies S=\frac n4$$
If $n$ is odd then let $n=2m+1$. We get,
$$\begin{align*}T=\sum\limits_{k=1}^{m}\cos\left((2k-1)\frac{2\pi} {2m+1}\right)&=\frac {\sin \frac {2\pi m}{2m+1}}{\sin\frac {2\pi}{2m+1}}\cos\bigg[\frac{2\pi}{2m+1}+(m-1)\frac{2\pi}{2m+1}\bigg]
\\&
\\&=\frac {\sin \frac {2\pi m}{2m+1}\cos \frac {2\pi m}{2m+1}}{\sin\frac {2\pi}{2m+1}}
\\&
\\&=\frac {\sin \frac {4\pi m}{2m+1}}{2\sin\frac {2\pi}{2m+1}}
\\&
\\&=\frac {-\sin \frac {2\pi }{2m+1}}{2\sin\frac {2\pi}{2m+1}}=\frac{-1}{2}\end{align*}$$
$$\implies S=\frac{2m+1}2=\frac n4$$
|
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|
Is it possible to prove equality between complicated algebraic expression and integer without mathematical programs or websites? Prove:
$\sqrt[3]{3+10i\sqrt{\frac{1}{27}}}+\sqrt[3]{3-10i\sqrt{\frac{1}{27}}}=3$
From WolframAlpha:
First:
by
https://www.wolframalpha.com/input/?i=cubrt%283%2B10isqrt%281%2F27%29%29
$\sqrt[3]{3+10i\sqrt{\frac{1}{27}}}=(1.499...9...)+(0.288675134...)i$
Second:
by
https://www.wolframalpha.com/input/?i=cubrt%283-10isqrt%281%2F27%29%29
$\sqrt[3]{3-10i\sqrt{\frac{1}{27}}}=(1.499...9...)-(0.288675134...)i$
By first and second:
$\sqrt[3]{3+10i\sqrt{\frac{1}{27}}}+\sqrt[3]{3-10i\sqrt{\frac{1}{27}}}=3$
So is there way to prove this without mathematical programs or websites?
|
Recall the algebraic identity
$$a^3+b^3 + c^3 - 3abc = \frac12(a+b+c)((a-b)^2 + (b-c)^2+(c-a)^2)$$
Whenever $a + b + c = 0$, we have $a^3 + b^3 + c^3 = 3abc$.
Let $\rho = 3 \pm \frac{10}{\sqrt{27}}i$ and $u = \sqrt[3]{\rho} + \sqrt[3]{\bar{\rho}} = 2\Re(\sqrt[3]{\rho})$. Substitute $a,b,c$ by $\rho$, $\bar{\rho}$ and $-u$, we obtain
$$\rho_{+} + \rho_{-} - u^3 = 3\sqrt[3]{\rho_{+}\rho_{+}} u$$
Since $\rho_{+} + \rho_{-} = 6$ and $\rho_{+}\rho_{-} = 3^2 + \frac{100}{27} = \left(\frac{7}{3}\right)^3$, this reduces to
$$6-u^3 = 7u \iff u^3-7u-6 = (u-3)(u+1)(u+2) = 0$$
Since $\frac{10}{\sqrt{27}} < 3$,
$$\arg(\rho_{+}) \in (0,\frac{\pi}{4})
\implies \arg(\sqrt[3]{\rho_{+}}) \in (0,\frac{\pi}{12}) \implies \Re( \sqrt[3]{\rho_{+}}) > 0$$
This means $u > 0$ and hence $u$ can only be $3$.
|
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|
Is the answer of a problem about the equation of a straight line given by my book wrong? This is the problem and the answer given by my book:
My solution:
$3x+\sqrt{3}y+2=0...(i)$
$x\cos\alpha+y\sin\alpha=p...(ii)$
Since (i) & (ii) are equations of the same straight line,
$$\frac{3}{\cos\alpha}=\frac{\sqrt{3}}{\sin\alpha}=\frac{2}{-p}$$
$$\implies -3p=2\cos\alpha...(i)$$
$$\implies -\sqrt{3}p=2\sin\alpha...(ii)$$
$(i)^2+(ii)^2:-$
$$9p^2+3p^2=4\cos^2\alpha+4\sin^2\alpha$$
$$\implies 12p^2=4$$
$$\implies p^2=\frac{1}{3}$$
$$\implies \sqrt{p^2}=\sqrt{\frac{1}{3}}$$
$$\implies |p|=\frac{1}{\sqrt{3}}$$
$$\implies p=\pm \frac{1}{\sqrt{3}}$$
Why did the book pick only the negative value of $p$?
Related
|
You are correct that $p$ needn't be negative: $$3x+\sqrt{3}y+2=0,$$ $$x\cos\left(\frac{\pi}{6}\right)+y\sin\left(\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}},$$ and $$x\cos\left(\frac{7\pi}{6}\right)+y\sin\left(\frac{7\pi}{6}\right)=\frac{1}{\sqrt{3}}$$ are all represented by the same line.
|
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|
Norm of the trimming map Let $M_n$ denote the linear space of $n\times n$ matrices over $\mathbb{C}$. Consider $M_n$ as a normed space with operator norm. Is it true that the trimming map
$$
T:M_n\to M_n:
\begin{pmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\
a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\
\ldots & \ldots & \ldots & \ldots \\
a_{n,1} & a_{n,2} & \ldots & a_{n,n} \\
\end{pmatrix}
\mapsto
\begin{pmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\
0 & a_{2,2} & \ldots & a_{2,n} \\
\ldots & \ldots & \ldots & \ldots \\
0 & 0 & \ldots & a_{n,n} \\
\end{pmatrix}
$$
has norm 1? I did a few numerical tests and got that $T$ is indeed contractive, but I don't know how to prove this.
|
*
*$T$ is a contraction for the operator $\left\|\cdot\right\|_p$ norms for $p=1$ and $p = \infty$ since those \begin{align}
\left\|A\right\|_1 &= \max_{j}\sum_{i=1}^n \left|a_{i,j}\right| &\text{and}\quad \left\|A\right\|_\infty &= \max_{i}\sum_{j=1}^n \left|a_{i,j}\right|.
\end{align}
*However $T$ is not a contraction for the $\left\|\cdot\right\|_2$. For example take $A = \begin{pmatrix}1 & 1\\ -1 & 1\end{pmatrix}$, \begin{align}
\left\|A\right\|_2 &= \sqrt{\lambda_{\max}\left(A^*A\right)}\\
&=\sqrt{\lambda_{\max}\left(\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}\right)}\\
&= \sqrt{2}\end{align} and $T(A) = \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}$ so
\begin{align}
\left\|T(A)\right\|_2 &= \sqrt{\lambda_{\max}\left(T(A)^*T(A)\right)} \\
&= \sqrt{\lambda_{\max}\left(\begin{pmatrix}1 & 1\\ 1 & 2\end{pmatrix}\right)} = \sqrt{\frac{3+\sqrt5}{2}} > \sqrt2\end{align}
|
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|
Find an ellipse from two points and the tangent at a third point I am searching an ellipse in the form $\frac{\left(x - x_\circ\right)^2}{a^2} + \tfrac{\left(y - y_\circ\right)^2}{b^2} = 1$ (its axis parallel to the coordinate axes). What I know are three points, and for one of them I additionally know the direction ϑ of the tangent. I only need to consider solvable situations where the three points are distinct, not in a line, and the tangent will not pass between the two other points.
How can I get $x_0$, $y_0$, a and b for this ellipse? From visual guesswork, I am pretty sure there is only one solution, but I have no idea how to get there.
|
You have $4$ unknowns, which are, $a, b, x_0, y_0$, and you can write down $4$ equations, so from dimensionality the problem should be solvable. Let the three points be $P_1(x_1, y_1), P_2(x_2, y_2), P_3 (x_3, y_3)$. The equation of the ellipse to be identified is
$ A x^2 + B xy + C y^2 + D x + E y + F = 0 $
Since, the axes of the ellipse are parallel to the coordinate axes, $B = 0$.
We can also take $A = 1$. Hence,
$ x^2 + C y^2 + D x + E y + F = 0 \hspace{24pt} (1)$
which is an equation with $4 $ unknowns $C, D, E, F$
Differentiating $(1)$ implictly, we obtain,
$ 2 x + 2 C y y' + D + E y' = 0 \hspace{24pt} (2)$
Now plug in the three points that you have into $(1)$ , and also the slope value $y'_1 = \tan \theta$ into (2), you obtain the following system of equations:
$ x_1^2 + C y_1^2 + D x_1 + E y_1 + F = 0 \hspace{24pt} (3)$
$ x_2^2 + C y_2^2 + D x_2 + E y_2 + F = 0 \hspace{24pt} (4)$
$ x_3^2 + C y_3^2 + D x_3 + E y_3 + F = 0 \hspace{24pt} (5)$
and assuming the point with the known slope is the first one, then
$ 2 x_1 + 2 C y_1 \tan \theta + D + E \tan \theta = 0 \hspace{24pt} $
To account for any value of $\theta$, multiply the last equation by $\cos \theta$:
$ 2 x_1 \cos \theta + 2 C y_1 \sin \theta + D \cos \theta + E \sin \theta = 0\hspace{24pt}(6)$
Equations $(3), (4), (5), (6)$ is a linear system of four equations in the four unknowns $C, D, E, F$, and can be readily solved.
Once we have the values of $C, D, E, F$, then it should be straight forward to find $a, b, x_0, y_0$.
We have
$ x^2 + C y^2 + D x + E y + F = 0 \hspace{24pt} (1)$
Completing the squares in $x$ and $y$:
$ (x + \dfrac{D}{2} )^2 - \dfrac{D^2}{4} + C (y + \dfrac{E}{2 C})^2 - \dfrac{E^2}{4C} + F = 0 $
Hence, $x_0 = - \dfrac{D}{2}$, $y_0 = - \dfrac{E}{2 C} $
So that the equation now is
$ (x - x_0)^2 + C (y - y_0)^2 = K $
where $ K = - F + \dfrac{D^2}{4} + \dfrac{E^2}{4C} $
Dividing by $K$ ,
$ \dfrac{(x - x_0)^2}{K} + \dfrac{ (y - y_0)^2 }{ K / C } = 1 $
So $a = \sqrt{K} $, and $b = \sqrt{\dfrac{K}{C} } $
|
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|
If $ \frac{X_n}{1024^n} $ is an odd integer, find the smallest possible value of $n$, where $n\ge2$ is an integer.
For each integer $k\ge2$, the decimal expansions of the numbers $1024, 1024^2, \dots, 1024^k$ are concatenated, in that order, to obtain a number $X_k$. (For example, $X_2 = 10241048576$.) If $ \frac{X_n}{1024^n}$ is an odd integer, find the smallest possible value of $n$, where $n\ge2$ is an integer..
*
*The answer is $\boxed{5}.$
*I tried for $X_1,X_2,X_3$ and noticed they weren't coming out as integers.
After this, the numbers were too big.
*
*I think that we can also try expanding $X_i$ terms. But then we would like t know the number of digits in $X_i.$
For example,
we have
*$$X_2 = 1024 \cdot 10^7 + 1024^2 = 2^{17} \cdot 5^7 + 2^{20} = 2^{17} (5^7 + 2^3).$$
So $$X_2/{1024^2}$$which is not a integer.
*$$X_3= 1024 \cdot 10^{17} + 1024^2 \cdot 10^{10} + 1024^3 = 2^{27} \cdot 5^{17} + 2^{30} \cdot 5^{10} + 2^{30} = 2^{27} (5^{17} + 2^3 \cdot 5^{10} + 2^3). $$
*We have $$V_2(X_3)<V_2(1024^3).$$
Any hints?
|
As $1024$ is just a little larger than $10^3, 1024^k$ will be a little larger than $10^{3k}$, so will have $3k+1$ decimal digits. According to Alpha, the number of digits does not increase above this until $k=98$. I think you can rely on $n$ being smaller than that.
|
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|
Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$
With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
3 y z^2 + 6 x y z$$
taking advantage of the fact that there is an $xyz=-1$ in the equation, but I'm not getting anywhere, someone less myopic than me.how to solve it?
Thanks
Edit : Will there be any university way to solve this problem , they posed it to a high school friend and told him it was just manipulations of remarkable products.
His answers I understand to some extent but I don't think my friend understands all of it.
|
This is an expansion of my comment. Recall Vieta's formulas (in this case, for cubics):
If $p, q, r$ are the roots of $t^3 + bt^2 + ct + d$, where $a \neq 0$, then
\begin{align*}
p + q + r &= -b \\
pq + qr + pr &= c \\
pqr &= -d.
\end{align*}
This follows easily by just expanding $(t - p)(t - q)(t - r)$ and equating coefficients with $t^3 + bt^2 + ct + d$.
Using these formulas, we can see, as others have commented, that the numbers $x, y, z$ in your question are just roots of the polynomial $t^3 - t + 1$. We can also use Vieta's formulas, and a little computation along the lines of what you've attempted already, to obtain a polynomial whose roots are precisely the squares of the roots of the original polynomial.
Let's say that $t^3 + bt^2 + ct + d$ has roots $p, q, r$ as above. Let $t^3 + b't^2 + c't + d'$ have roots $p^2, q^2, r^2$. We wish to compute $b', c' d'$ vie Vieta's formulas, i.e. to compute $p^2 + q^2 + r^2$, $p^2 q^2 + q^2 r^2 + p^2 r^2$, and $p^2 q^2 r^2$ in terms of $a, b, c$.
There's a quick and easy one: $p^2 q^2 r^2 = (pqr)^2 = d^2$. Thus $d' = -d^2$.
To compute $p^2 + q^2 + r^2$, consider
$$b^2 = (p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) = -b' + 2c,$$
thus $b' = 2c - b^2$.
Lastly, we compute
\begin{align*}
c^2 &= (pq + qr + pr)^2 \\
&= p^2 q^2 + q^2 r^2 + p^2r^2 + 2p^2qr + 2pq^2r + 2pqr^2 \\
&= c' + 2pqr(p + q + r) \\
&= c' + 2bd,
\end{align*}
so $c' = c^2 - 2bd$.
Thus, our transformation which squares roots takes a monic polynomial $t^3 + bt^2 + ct + d$ and transforms it into the monic polynomial
$$t^3 + (2c - b^2)t^2 + (c^2 - 2bd)t - d^2.$$
Let's apply it to $t^3 - t + 1$ to get a polynomial whose roots are $x^2, y^2, z^2$. We obtain the polynomial:
$$t^3 + (2(-1) - 0^2)t^2 + ((-1)^2 - 2(0)(1))t - 1^2 = t^3 - 2t^2 + t - 1.$$
Apply again to get a polynomial whose roots are $x^4, y^4, z^4$:
$$t^3 + (2(1) - (-2)^2)t^2 + (1^2 - 2(-2)(-1))t - (-1)^2 = t^3 - 2t^2 - 3t - 1.$$
We could apply this transformation one final time to get the polynomial whose roots are $x^8, y^8, z^8$, but really we just need the coefficient of $t^2$. So, partially applying the above formula, we get:
$$t^3 + (2(-3) - (-2)^2)t^2 + \underline{\hspace{12pt}} t + \underline{\hspace{12pt}} = t^3 - 10t^2 + \underline{\hspace{12pt}} t + \underline{\hspace{12pt}}.$$
So, by Vieta,
$$x^8 + y^8 + z^8 = -(-10) = 10.$$
|
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|
Showing that $\int_{0}^{\infty}\frac{x^{m-1}-x^{n-1}}{1-x}dx$ is convergent if $ 0
Showing that $\int_{0}^{\infty}\frac{x^{m-1}-x^{n-1}}{1-x}dx$ is convergent if $ 0<m<1 , 0<n <1$
Let $I = \int_{0}^{\infty}\frac{x^{m-1}-x^{n-1}}{1-x}dx$ then $I = \int_{0}^{\infty}\frac{x^{m-1}}{1-x}dx - \int_{0}^{\infty}\frac{x^{n-1}}{1-x}dx$
We consider $I_1 = \int_{0}^{\infty}\frac{x^{m-1}}{1-x}dx$
Let $f_1(x) = \frac{x^{m-1}}{1-x}$ then $f_1(x)$ has three points of discontinuity at $x = 0,1 ,\infty$ if $0<m<1$
Let us consider $g_1(x) = \frac{1}{x^{1-m}}$ then $\lim_{x \to 0+}\frac{f_1(x)}{g_1(x)} = 1$
Now the integral of the function $g_1(x)$ from $0$ to $a(<1)$ is convergent so the $\int_{0}^{a}f_1(x)dx$ is convergent.
Now we consider the function $f_1(x)$ at the point $x = 1$ then we consider the function $g_1(x) = 1-x$ then $\lim_{x \to 1+}\frac{f_1(x)}{g_1(x)} = \frac{1-x}{x^{m-1}(1-x)} = 1$
Now we see that the function $\int_{a}^{1}g(x) = \int_{a}^{1}\frac{1}{1-x}dx $ is not convergent so the integral $\int_{a}^{1}f_1(x)dx$ is not convergent.
Where am I going wrong in this step?
|
A simple way to show that the integral is convergent is by cutting the integral into three parts
\begin{align}
\int^{1/2}_0 + \int^{3/2}_{1/2} + \int^\infty_{3/2}\frac{x^{m-1}-x^{n-1}}{1-x}d x = I_1+I_2+I_3.
\end{align}
It is clear that $I_3$ is convergent since
\begin{align}
I_3 \le C\int^\infty_{3/2}\left(\frac{1}{x^{2-m}}+\frac{1}{x^{2-n}}\right)dx<\infty
\end{align}
for some constant $C>0$.
It is also clear that $I_1$ is convergent since
\begin{align}
I_1 \le C\int^{1/2}_0 \left(\frac{1}{x^{1-m}}+\frac{1}{x^{1-n}}\right)dx.
\end{align}
Lastly, let us deal with $I_2$. Assume $n\ge m$. Notice by the mean value theorem, we have that
\begin{align}
I_2 = \int^{3/2}_{1/2} x^{m-1}\frac{1-x^{n-m}}{1-x} \le C\int^{3/2}_{1/2} x^{m-1} d x
\end{align}
|
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|
How to solve $\sin(2\theta)$ questions Given that: $\sin\theta=\displaystyle{}\frac{12}{13}$ and $0<\theta<\displaystyle{}\frac{\pi}{2}$ the value of $\sin(2\theta)$ is:
I figured out a way to solve it, though I'm not sure if it is the best solution.
Here we will combine two different trigonemtric identities. First:
$\begin{align}
\sin(2\theta) & = 2\sin\theta\cos\theta \\
& = 2\cdot\frac{12}{13}\cdot\cos\theta \\
& = \frac{24}{13}\cdot\cos\theta
\end{align}$
Also:
$\begin{align}
1 & = \cos^2\theta+\sin^2\theta \\
1 & = \cos^2\theta+\Bigg(\frac{12}{13}\Bigg)^2 \\
1 & = \cos^2\theta+\frac{144}{169} \\
1-\frac{144}{169} & = \cos^2\theta \\
\frac{25}{169} & = \cos^2\theta \\
\sqrt\frac{25}{169} & = \sqrt{\cos^2\theta} \\
\frac{5}{13} & = \cos\theta \\
\end{align}$
Then we insert this into the previous equation:
$\begin{align}
\sin(2\theta) & = \frac{24}{13}\cdot\cos\theta \\
& = \frac{24}{13}\cdot\frac{5}{13} \\
& = \frac{120}{169}
\end{align}$
And I believe this is the correct answer. I'm just not sure if this was a super round about way of solving it or if there is something better.
|
Another way to do this is to first find $\cos\theta$. This is easier if you recognise small Pythagorean triads. ;)
Let
$y/r=\sin\theta$ and $x/r=\cos\theta$, where $r^2=x^2+y^2$.
We have $\sin\theta=5/13$, so
$$x^2 = 13^2-5^2 = (13+5)(13-5) = 18\cdot8=12^2$$
thus $\cos\theta=12/13$.
Now, we know that $\sin(2\theta)=2\sin\theta\cos\theta$. But that's the middle term of
$$(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$$
And of course $\sin^2\theta + \cos^2\theta = 1$
So
$$(\sin\theta + \cos\theta)^2 = 1 + \sin(2\theta)$$
Therefore,
$$\begin{align}\\
\sin(2\theta) & = (5/13 + 12/13)^2 - 1\\
& = (17/13)^2 - 1\\
& = (17^2 - 13^2)/13^2\\
& = 120/169
\end{align}$$
Of course, we need to check the signs of our trig ratios to make sure they're all in the correct quadrant.
|
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|
Prove that: $\sum\limits_{cyc}\frac{1}{\sqrt{2a^2+5ab+2b^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$ \dfrac{1}{\sqrt{2a^2+5ab+2b^2}}+\dfrac{1}{\sqrt{2b^2+5bc+2c^2}}+\dfrac{1}{\sqrt{2c^2+5ca+2a^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}.$$
I solved this problem by Hölder:
$$\left(\sum_{cyc}\dfrac{1}{\sqrt{2a^2+5ab+2b^2}}\right)^2\sum_{cyc}\frac{(a+b)^3}{(2a^2+5ab+2b^2)^2}\geq\left(\sum_{cyc}\frac{a+b}{2a^2+5ab+2b^2}\right)^3$$
and it remains to prove that
$$(ab+ac+bc)\left(\sum_{cyc}\frac{a+b}{2a^2+5ab+2b^2}\right)^3\geq3\sum_{cyc}\frac{(a+b)^3}{(2a^2+5ab+2b^2)^2},$$ which is true by BW and by using computer.
In this topic https://artofproblemsolving.com/community/c6h542992 there is a proof (from gxggs), but it's very very complicated.
I found another way, a smooth enough, but it's still a very hard solution.
I am looking for a nice proof by hand, for which there is a possibility to find this proof during a competition.
Thank you!
|
Remark: I give a proof using the so-called isolated fudging.
It suffices to prove that
$$\frac{\sqrt{\frac{ab + bc + ca}{3}}}{\sqrt{2a^2 + 5ab + 2b^2}}\ge \frac{8c^2 + 9(a + b)c + 8ab}{8(a^2 + b^2 + c^2) + 26(ab + bc + ca)}. \tag{1}$$
Note: Taking cyclic sum on (1), we get the desired inequality.
If $c = 0$, it is easy.
If $c > 0$, WLOG, assume that $c = 1$. Let $p = a + b, ~ q = ab$. Then $0 \le q \le p^2/4$. It suffices to prove that
$$\frac{\sqrt{(q + p)/3}}{\sqrt{2(p^2 - 2q) + 5q}}
\ge \frac{8 + 9p + 8q}{8(p^2 - 2q + 1) + 26(q + p)}.$$
Squaring both sides, it suffices to prove that
\begin{align*}
&-92\,{q}^{3}+ \left( -224\,{p}^{2}+188\,p-224 \right) {q}^{2}+ \left(
64\,{p}^{4}-288\,{p}^{3}+313\,{p}^{2}+144\,p-128 \right) q \\
&\quad +64\,{p}^{5}
-70\,{p}^{4}-60\,{p}^{3}+32\,{p}^{2}+64\,p \ge 0.
\end{align*}
Denote LHS by $f(q)$.
We have $f''(q) = - 448p^2 + 376p - 448 - 552q$.
So, we have $f''(q) < 0$ on $q \ge 0$.
So, $f(q)$ is concave on $q \ge 0$.
Also, we have $f(0) = 64p^5 - 70p^4 - 60p^3 + 32p^2 + 64p \ge 0$
and $f(p^2/4) = \frac{1}{16}\,p \left( 9\,{p}^{3}+96\,{p}^{2}+256\,p+256 \right) \left( p-2
\right) ^{2} \ge 0$.
Thus, $f(q) \ge 0$ for all $q\in [0, p^2/4]$.
We are done.
|
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|
How to prove $\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}$ Prove that
$$
\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}
$$
My attempt :I tried to use the beta function, but I couldn't.
|
We have
$$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos(\theta)}}{1+\cos^{2}(\theta)}d\theta=\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}(-1)^{n}\cos^{2n+\frac{1}{2}}(\theta)d\theta$$
$$=\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{\frac{\pi}{2}}\cos^{2n+\frac{1}{2}}(\theta)d\theta$$
by Fubini/Tonelli theorems. Then using the trigonometric representation of the Beta function
$$\frac{1}{2}\beta\left(\frac{1+n}{2},\frac{1+m}{2}\right)=\int_{0}^{\frac{\pi}{2}}\cos^{n}(x)\sin^{m}(x)dx,\space \text{for}\space n,m>-1$$
we obtain $$\int_{0}^{\frac{\pi}{2}}\cos^{2n+\frac{1}{2}}(\theta)d\theta=\frac{1}{2}\beta\left(n+\frac{3}{4},\frac{1}{2}\right)=\frac{1}{2}\frac{\Gamma\left(n+\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(n+\frac{5}{4}\right)}$$
and the integral becomes
$$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos(\theta)}}{1+\cos^{2}(\theta)}d\theta=\frac{\sqrt{\pi}}{2}\sum_{n=0}^{\infty}(-1)^{n}\frac{\Gamma\left(n+\frac{3}{4}\right)}{\Gamma\left(n+\frac{5}{4}\right)}$$
$$=\frac{\sqrt{\pi}}{2}\left(\frac{5\Gamma\left(\frac{7}{4}\right)\, _2F_1\left(\frac{3}{4},1;\frac{9}{4}; -1\right)\, }{3\, \Gamma \left(\frac{9}{4}\right)}-\frac{4\Gamma\left(\frac{11}{4}\right)\, _2F_1\left(\frac{7}{4},2;\frac{13}{4}; -1\right)\, }{7\, \Gamma \left(\frac{13}{4}\right)}\right)=\frac{\pi}{4}$$
where $_2F_1(a,b;c;x)$ is the hypergeometric function.
|
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"url": "https://math.stackexchange.com/questions/4250267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find the minimum value of: $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$ Let $a,b,c>0$, find the minimum value of: $$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$$
I have tried:
$\bullet$ The minimum value is $\dfrac{3}{4}$ occur when $a=b=c$
$\bullet \dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}= \dfrac{a^2}{3a^2+ab}+\dfrac{b^2}{3b^2+bc}+\dfrac{c^2}{3c^2+ca}\ge\dfrac{(a+b+c)^2}{3(a^2+b^2+c^2)+ab+bc+ca}=\dfrac{(a+b+c)^2}{3(a+b+c)^2-5(ab+bc+ca)}$
$\bullet$ $ab+bc+ca\le\dfrac{(a+b+c)^2}{3}$ so we have $\dfrac{(a+b+c)^2}{3(a+b+c)^2-5(ab+bc+ca)}\le \dfrac{3}{4}$ but we need to prove $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c} \ge \dfrac{3}{4}$
Can you give me some hint ?
|
Another way:
We know that our expression is closed to $\frac{1}{3}$ but
$$\sum_{cyc}\frac{a}{b+3a}\geq \sum_{cyc} \frac{a}{3b+3a+3c}=\frac{1}{3}$$ and we got the same result again.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4251200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solving Inequality $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$ If someone could help with solving the inequality above, that would be awsome!
Here is my thinking of using AGM
*
*$\sqrt{\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)}\le \frac{\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}}{2}$
*$\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)\le \frac{\left(x\sqrt{x}+y\sqrt{y}\right)^2}{4xy}$ (Square both sides)
*$\sqrt{x}\sqrt{y}\le \:\frac{\left(x\sqrt{x}+y\sqrt{y}\right)^2}{4xy}$ (Simplify Left Side)
*$4xy\sqrt{x}\sqrt{y}\le x^3+2xy\sqrt{x}\sqrt{y}+y^3$ (Move 4xy to the other side and Square remaining)
*But now I see that this will not end up in $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$
Please help????
|
Note that we must have $x > 0$ and $y > 0$ for the inequality to even be defined at all.
Start by squaring both sides to get the equivalent inequalty
$$\frac{x^2}{y} + \frac{y^2}{x} + 2 \sqrt{\frac{x^2}{y} \frac{y^2}{x}} = \frac{x^2}{y} + \frac{y^2}{x} + 2 \sqrt{xy} \geq x + y + 2 \sqrt{xy}$$
Equivalently,
$$\frac{x^2}{y} + \frac{y^2}{x} \geq x + y$$
Multiply both sides by $xy$ to get the equivalent inequality
$$x^3 + y^3 = (x + y)(x^2 - xy + y^2)\geq x^2 y + y^2 x = xy(x + y)$$
Divide both sides by $x + y$ to get the equivalent inequality $x^2 - xy + y^2 \geq xy$. Subtract $xy$ from both sides to get the equivalent inequality
$$(x - y)^2 = (x^2 - 2x + y^2) \geq 0$$
Which is clearly true.
|
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"url": "https://math.stackexchange.com/questions/4252299",
"timestamp": "2023-03-29T00:00:00",
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|
How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:
$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$$
and I wanted to try and prove it.
Inspired by this answer by Jack D'Aurizio, I took the Weierstrass product for $\cosh(x)$ to obtain
$$
\cosh\left(\frac{\pi x}{2} \right) = \prod_{n \ge 1}\left(1 + \frac{x^2}{(2n-1)^2} \right)
$$
And by logarithmically differentiating twice we get
$$
\frac{\pi^2}{4}\text{sech}^2\left(\frac{\pi x}{2} \right) = \sum_{n \ge 1} \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}
$$
Which means we get
\begin{align*}
\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx & =\frac{4}{\pi^2}\sum_{n\ge 1} \int_{0}^{\infty} \frac{(1-x^2)}{(1+x^2)^2}\left( \frac{4(2n-1)^2}{\left(x^2 + (2n-1)^2\right)^2} - \frac{2}{x^2 + (2n-1)^2}\right)\, dx
\end{align*}
However, after this, I couldn't figure out how to evaluate the resulting integral.
Does anyone know how I could continue this method? Or alternatively, does anyone know another way in which the result can be proven? Thank you very much!!
Edit:
Per jimjim's request, I'll add that I found this integral on the Wikipedia article for $\zeta(3)$. I believe the reference is to this text where the following formula is given
$$
(s-1) \zeta(s) = 2\pi \int_{\mathbb{R}}\frac{\left(\frac{1}{2} + xi \right)^{1-s}}{\left(e^{\pi x} +e^{-\pi x} \right)^2}\, dx
$$
which for the case of $s=3$ reduces to the surprisingly concise
$$
\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx = \frac{\zeta(3)}{\pi}
$$
And I presume that one can modify the previous equation to get to the original integral from the question, but it is not apparent to me how this may be done.
Edit 2:
Random Variable has kindly posted in the comments how to go from $\int_{\mathbb{R}}\frac{\text{sech}^2(\pi x)}{(1+2xi)^2} \, dx$ to $ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx$. Thank you very much!
|
Following K.defaoite's suggestion, I've managed to get a partial solution using the Residue Theorem. Rewriting the integrand with the Weierstrass product for $\cosh$ squared in the denominator we get
$$
f(x) =\frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2} = \frac{(1-x^2) }{(x+i)^4(x-i)^4 \prod_{n\ge2}\left(\frac{1}{(2n-1)^4}\left(x+i(2n-1)\right)^2\left(x-i(2n-1)\right)^2 \right)}
$$
from where we see that we have poles of order $2$ at $z = \pm i(2n-1), \ k \ge 2$ and we have poles of order $4$ at $z = \pm i$.
To calculate the residues we'll use that
$$
\lim_{z \to (2n-1)i} \frac{(z -(2n-1)i)^2}{\cosh^2\left(\frac{\pi z}{2}\right)} \overset{\text{L.H.}}{=}\frac{4}{\pi}\lim_{z \to (2n-1)i} \frac{z -(2n-1)i}{\sinh\left(\pi z\right)}\overset{\text{L.H.}}{=}\frac{4}{\pi^2}\lim_{z \to (2n-1)i} \frac{1}{\cosh\left(\pi z\right)} =-\frac{4}{\pi^2}\tag{1}
$$
as $\cosh(iz) = \cos(z)$ and $\cos(-\pi) = -1$. And also
\begin{align*}
\lim_{z \to (2n-1)i} \frac{d}{dz} \left(\frac{z-(2n-1)i}{\cosh\left( \frac{z \pi}{2}\right)}\right)&\overset{u = z/i}{=}\frac{i}{i}\lim_{u \to 2n-1} \frac{d}{du} \left( \frac{u-(2n-1)}{\cos\left( \frac{u \pi}{2}\right)}\right)\\
& = \lim_{u \to 2n-1}\frac{\cos\left( \frac{u \pi}{2}\right) +\frac{\pi}{2}u\sin\left( \frac{u \pi}{2}\right) - \frac{\pi}{2}(2n-1)\sin\left( \frac{u \pi}{2}\right)}{\cos^2\left( \frac{u \pi}{2}\right)}\\
& \overset{\text{L.H.}}{=}\lim_{u \to 2n-1}\frac{\frac{\pi^2}{4}(u-(2n-1))\cos\left( \frac{u \pi}{2}\right)}{-\pi \sin\left( \frac{u \pi}{2}\right)\cos\left( \frac{u \pi}{2}\right)}\\
& = 0\tag{2}
\end{align*}
So we get
\begin{align*}
\text{Res}\left(f, i(2n-1)\right) & =\lim_{z \to (2n-1)i}\frac{d}{dz}\left(\frac{1-z^2}{(1+z^2)^2}\cdot \frac{(z - (2n-1)i)^2}{\cosh^2\left(\frac{\pi z}{2} \right)}\right)\\
& = \lim_{z \to (2n-1)i}\frac{2z(z^2-3)}{(1+z^2)^3}\cdot\underbrace{ \lim_{z \to (2n-1)i} \frac{(z - (2n-1)i)^2}{\cosh^2\left(\frac{\pi z}{2} \right)}}_{-\frac{4}{\pi^2}} \\
&\qquad + \lim_{z \to (2n-1)i}\left(\frac{1-z^2}{(1+z^2)^2}\right)\cdot 2 \lim_{z \to (2n-1)i} \left(\frac{z-(2n-1)i}{\cosh\left( \frac{z \pi}{2}\right)}\right)\underbrace{\lim_{z \to (2n-1)i} \frac{d}{dz} \left(\frac{z-(2n-1)i}{\cosh\left( \frac{z \pi}{2}\right)}\right) }_{0}\\
& = -\frac{4}{\pi^2}\frac{2((2n-1)i)(((2n-1)i)^2-3)}{(1+((2n-1)i)^2)^3}\\
& =-\frac{i}{2\pi^2}\left(\frac{1}{n^3} + \frac{1}{(n-1)^3} \right)
\end{align*}
By a similar (yet longer) procedure we can also get that
$$
\text{Res}\left(f, i\right) = -\frac{i}{2\pi^2}
$$
as can be verified here.
So we get that the residues are:
\begin{align*}
\text{Res}\left(f, i(2n-1)\right) &= -\frac{i}{2\pi^2}\left(\frac{1}{n^3} + \frac{1}{(n-1)^3} \right), \qquad n \ge 2\\
\text{Res}\left(f, i\right) &= -\frac{i}{2\pi^2}
\end{align*}
Now we take a semicircular contour with diameter from $-a$ to $a$, where the semicircle is on the upper part of the complex plane over which to integrate. As we tend to an infinite radius said contour will engulf the singularities $z = i(2n-1), \ k \ge 1$, which means that
\begin{align}
\lim_{a \to \infty}\oint_{C}\frac{(1-z^2) \, \text{sech}^2\left(\frac{\pi z}{2} \right)}{(1+z^2)^2} \, dz & = 2 \pi i \left(-\frac{i}{2\pi^2}+\sum_{n=2}^{\infty} -\frac{i}{2\pi^2}\left(\frac{1}{n^3} + \frac{1}{(n-1)^3} \right)\right)\\
& =\frac{2}{\pi} \zeta(3)\\
\end{align}
But remembering that
$$\lim_{a \to \infty}\oint_{C} f = \int_{\mathbb{R}} f + \lim_{a \to \infty}\int_{\text{Arc}} f =2\int_{0}^{\infty} f + \lim_{a \to \infty}\int_{\text{Arc}} f $$
it sufficies to show that the limit of the integral over the arc goes to $0$ to finish the solution, but I haven't found a way to justify this.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\frac{7}{x+4}\\
-1(x+4)<7\\-x < 11\\
x>-11.$$
This results in the answer $x>-11, x>3,$ which doesn't make sense.
|
$$\frac{7}{x+4}>-1 \implies \frac{x+11}{x+4} >0$$
This means
(1): $(x+11)>0 \& x+4>0 \implies x>-4 \& x >-11 \implies x>-4$
(2): $(x+11) <0 \& x+4 <0 \implies x<-11 \& x <-4 \implies x<-11$
So the total answer is union of the two: $x\in (-\infty, -11) \cup (-4, \infty)$ or
$x<-11$ or $x >-4$.
|
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"url": "https://math.stackexchange.com/questions/4253491",
"timestamp": "2023-03-29T00:00:00",
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|
Error in my approach to prove that $(m-1)^{m-1} \equiv m - 1 \pmod m$ using a concrete example I read somewhere that if $m$ is composite then $(m - 1)^{m - 1} \equiv m - 1 \pmod m$ and I was curious to try to prove it myself.
So I took as $m = 6$.
Now I can see that $5^5 \equiv 5 \pmod 6$
I was thinking along the following lines:
Since $\gcd(6,5) = 1$ this means that if we multiply $5$ with all the numbers from $1$ to $5$ we have to get all the numbers in different order.
I.e.
The numbers less than $6$: $1,2,3,4,5$.
Multiply the above by $5$: $5,4,3,2,1$ i.e. we get the same set of numbers but in different order.
Since: $1\cdot 2\cdot 3\cdot 4\cdot 5 = 5\cdot 4\cdot 3\cdot 2\cdot 1$ we also have:
$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow 1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 1 \equiv 5^5$
Which is actually wrong.
What is the problem in my approach and thought process?
|
The fallacy in your argument is in the last step, where you "cancel" $1\cdot2\cdot3 \cdot 4\cdot 5$ from each side of the congruence.
As in ordinary arithmetic, you can only cancel factors that aren't $0$. (In fact, you can only reliably cancel factors that are relatively prime to the modulus.)
So in this case the assertion you are trying to prove is correct but your argument is not. The argument does work for arithmetic with a prime modulus.
|
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"timestamp": "2023-03-29T00:00:00",
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|
problem with power series $\sum \frac{x^n}{n+3}$ I need to evaluate the following power series, and I don’t really know how to do it, this is the series $$\sum \frac{x^n}{n+3}$$
This is how I tackled this problem, but the final solution looks very dumb.
So we know that $$\frac{1}{1-x}=\sum x^n$$
But we also know that we can get that $n+3$if we integrate $x^n+2$
So what we want is this series $$\sum x^{n+2}$$
Now, this series is equal to the original times $x^2$, so we can do this$$\sum x^{n+2} = x^2 \frac{1}{1-x}$$ Now we take the integral of both sides and we get something like this $$\sum \frac{x^{n+3}}{n+3} = (-\frac{x^2}{2}-1-\ln|x-1|)$$ But know we see that the left-hand side is $$x^3 \sum \frac{x^n}{n+3}$$ so we can multiply both sides by $\frac{1}{x^3}$ and we get the following $$\sum \frac{x^n}{n+3}=-\frac{1}{2x}-\frac{1}{x^3}-\frac{\ln|x-1|}{x^3}$$Can someone tell me if this answer is acceptable or if I have completely messed up?
|
Denote
\begin{equation}
f(x)=\sum_{n=1}^\infty\frac{x^n}{n+k}, \quad k\in\{0,1,2,\dotsc\}.
\end{equation}
Then
\begin{equation}
x^kf(x)=\sum_{n=1}^\infty\frac{x^{n+k}}{n+k}
\end{equation}
and
\begin{equation}
\bigl[x^kf(x)\bigr]'=\sum_{n=1}^{\infty}x^{n+k-1}=x^k\sum_{n=0}^{\infty}x^n=\frac{x^k}{1-x}, \quad -1\le x<1.
\end{equation}
Integrating with respect to $x$ over the interval between $0$ and $t\in[-1,1)$ on both sides of the above euqation gives
\begin{equation}
t^kf(t)=\int_0^t\frac{x^k}{1-x}\textrm{d}\,x, \quad t\in[-1,1).
\end{equation}
This means that
\begin{equation}
f(t)=\begin{cases}\displaystyle
\frac{1}{t^k}\int_0^t\frac{x^k}{1-x}\textrm{d}\,x, & t\in[-1,1)\setminus\{0\};\\
0, & t=0.
\end{cases}
\end{equation}
Taking $k=3$ in the above equation, directly computing the definite integral, and simplifying yiled
\begin{equation}
f(t)=\begin{cases}
-\dfrac{1}{3}-\dfrac1{2t}-\dfrac1{t^2}-\dfrac{\ln(1-t)}{t^3}, & t\in[-1,1)\setminus\{0\};\\
0, & t=0.
\end{cases}
\end{equation}
|
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|
Solve integral $\int_0^{2\pi}\frac{\sin^2\frac{(N+1)x}{2}}{2\pi(N+1)\sin^2\frac x2} dx $ My question is to solve
$$\int_0^{2\pi}\frac{\sin^2\left(\frac{N+1}{2}x\right)}{2\pi(N+1)\sin^2(x/2)} dx $$
Up to $1000$ or more, Wolfram-Alpha solves this integral and gives the value $1$ for every integer. But, I couldn't prove that this is the case for any $N\in\mathbb{N}$.
I know if we didn't have the $N+1$ part and squares, this integral should be $1$. We also have
$$ \frac{\sin^2\left(\frac{N+1}{2}x\right)}{\sin^2(x/2)}=\frac{\cos((N+1)x)-1}{\cos(x)-1}=\sum_{n=0}^N\left(\sum_{k=-n}^n e^{inx} \right)$$
and
$$ \frac{\sin\left((N+\frac{1}{2})x\right)}{\sin(x/2)}=\sum_{n=-N}^Ne^{inx}= \sum_{k=1}^N2\cos(kx)+1$$
|
Apply
$$\frac{\sin \frac{(N+1)x}2- \sin\frac{(N-1)x}2}{\sin \frac x2}=2\cos\frac{Nx}2
$$
to the integral
\begin{align}
I_{N+1}&
= \int_{0}^{\pi}{\frac{\sin^2\frac{(N+1)x}2}{\sin^2\frac x2}}dx
=\int_{0}^{\pi}\left({\frac{\sin\frac{(N-1)x}2}{\sin\frac x2}}+2\cos\frac{Nx}2 \right)^2dx\tag1
\end{align}
Note that the integration over the cross terms vanishes as shown below
\begin{align}
&\int_{0}^{\pi} {\frac{2\sin\frac{(N-1)x}2 \cos\frac{Nx}2 }{\sin\frac x2}}dx\\
=&\int_{0}^{\pi} \left( {\frac{\sin{(N-\frac12)x}}{\sin\frac x2}}-1\right)dx
=\int_{0}^{\pi}\sum_{k=1}^{N-1}2\cos kx dx=0
\end{align}
and (1) establishes a recursive relationship
\begin{align}
I_{N+1}= I_{N-1}+2\pi
\end{align}
which, with $I_1=\pi$ and $I_2 =2 \pi$, leads to $I_{N}=N\pi$. As a result
$$\int_0^{2\pi}\frac{\sin^2\frac{(N+1)x}{2}}{2\pi(N+1)\sin^2\frac x2} dx =\frac{2I_{N+1}}{2\pi(N+1)}=1 $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The limit of $n^2 f \left(a-\frac{b}{n}, b+\frac{a}{n} \right)$. Let $f : \mathbb{R}^2\to \mathbb R$ be $C^2$ class function s.t. $$f(x,y)=0 \ \mathrm{ and } \ (f_x(x,y))^2+(f_y(x,y))^2=1$$ for $(x.y)\in C:=\{ (x,y) \mid x^2+y^2=1 \},$ and suppose $f$ is monotonically increasing as getting away from $(0,0)$ on the half line from $(0,0)$ to arbitrary direction.
Then, I have to calculate
$$\lim_{n\to \infty} n^2 f\left(u-\dfrac{v}{n}, v+\dfrac{u}{n}\right)$$
where $(u,v)\in C.$
At the pre-step of this problem, I had to prove
(i) $vf_x(u,v)=uf_y(u,v) $
(ii) $f_x(u,v)=u, f_y(u,v)=v $
(iii)
$\begin{pmatrix}
f_{xx}(u,v) & f_{xy}(u,v) \\
f_{yx}(u,v) & f_{yy}(u,v) \\
\end{pmatrix}
\begin{pmatrix}
-v \\
u \\
\end{pmatrix}
=\begin{pmatrix}
-v \\
u \\
\end{pmatrix}$
Thus I can use these (i)(ii)(iii) to calculate $\displaystyle\lim_{n\to \infty} n^2 f\left(u-\dfrac{v}{n}, v+\dfrac{u}{n}\right)$, but I don't know how.
Letting $\dfrac{1}{n}=m,$ I get $$\lim_{n\to \infty} n^2 f\left(u-\dfrac{v}{n}, v+\dfrac{u}{n}\right)=\lim_{m\to 0}\dfrac{f(u-vm, v+um)}{m^2}.$$
This appears to be what is relative to derivative, but I couldn't proceed.
Thanks for your help.
|
Hint: If you choose $f(x, y) = \frac{1}{2}x^2+\frac{1}{2}y^2-\frac{1}{2}$, then you see it satisfies all the required conditions. Observe we have that
\begin{align}
\lim_{n\rightarrow \infty} n^2f(a-\frac{b}{n}, b+\frac{a}{n})=&\ \lim_{n\rightarrow \infty} \frac{1}{2}n^2\left((a-\frac{b}{n})^2+(b+\frac{a}{n})^2 -1\right)\\
=&\ \lim_{n\rightarrow \infty} \frac{1}{2}n^2\left( a^2+b^2 + \frac{1}{n^2}(a^2+b^2)-1 \right)\\
=&\ \frac{1}{2}(a^2+b^2).
\end{align}
In general, by Taylor's theorem, we have that
\begin{align}
f(x, y) = f(a, b) + f_x(a, b)(x-a)+f_y(a, b)(y-b) + \frac{1}{2}f_{xx}(a, b)(x-a)^2+ f_{xy}(a, b)(x-a)(y-b) + \frac{1}{2}f_{yy}(a, b)(y-b) + R_2(a, b, x, y).
\end{align}
where $|R_2| \le C((x-a)^2+(y-b)^2)^{3/2}$.
Finally, we have that
\begin{align}
f(a-\frac{b}{n}, b+\frac{a}{n}) = -\frac{b}{n}f_x(a, b)+\frac{a}{n}f_y(a, b)+ \frac{1}{2}f_{xx}(a, b)\frac{b^2}{n^2}-f_{xy}(a, b)\frac{ab}{n^2}+\frac{1}{2}f_{yy}(a, b)\frac{a^2}{n^2}+R_2(a, b, a-\frac{b}{n},b+\frac{a}{n})
\end{align}
which means
\begin{align}
n^2f(a-\frac{b}{n}, b+\frac{a}{n})=&\ -nbf_x(a, b)+naf_y(a, b)+ \frac{1}{2}f_{xx}(a, b)b^2-f_{xy}(a, b)ab+\frac{1}{2}f_{yy}(a, b)a^2+n^2R_2(a, b, a-\frac{b}{n},b+\frac{a}{n})\\
=&\ n (-b, a)\cdot (f_x(a, b), f_y(a, b))+ \frac{1}{2}f_{xx}(a, b)b^2-f_{xy}(a, b)ab+\frac{1}{2}f_{yy}(a, b)a^2+n^2R_2(a, b, a-\frac{b}{n},b+\frac{a}{n}).
\end{align}
Notice that
\begin{align}
(-b, a)\cdot (f_x(a, b), f_y(a, b)) = 0
\end{align}
(Why?).
Next, also notice that
\begin{align}
n^2|R_2|\le Cn^2\frac{1}{n^3}\rightarrow 0
\end{align}
as $n\rightarrow \infty$.
Hence, we have that
\begin{align}
\lim_{n\rightarrow\infty} n^2f(a-\frac{b}{n}, b+\frac{a}{n}) = \frac{1}{2}[-b, a]
\begin{bmatrix}
f_{xx}(a, b) & f_{xy}(a, b)\\
f_{yx}(a, b) & f_{yy}(a, b)
\end{bmatrix}
\begin{bmatrix}
-b\\
a
\end{bmatrix}.
\end{align}
Combine with your (i), (ii) and (iii) yields the desired result of
\begin{align}
\lim_{n\rightarrow\infty} n^2f(a-\frac{b}{n}, b+\frac{a}{n}) = \frac{1}{2}(a^2+b^2)
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum $\sum_{r=1}^n \cos(2.(\frac {3^rx}{3})).\csc (3^rx)$ Prove that $$\sum_{r=1}^n \cos(2.(\frac {3^rx}{3})).\csc (3^rx) = \frac{1}{2\sin x}-\frac{1}{2\sin (3^nx)}$$
My attempt:
$$\Sigma \frac{\cos(\frac {2.3^rx}{3})}{\sin(3^rx)} = \Sigma \frac{1-2\sin^2(\frac {3^rx}{3})}{\sin(3^rx)}$$
I am trying to convert it to a form $f(r+1)-f(r)$ where $f(r)$ represents a general term in the series. But I am unable to do so.
|
\begin{align}
\frac{1-2\sin^2(3^{r-1}x)}{\sin(3^rx)} &= \frac{1-2\sin^2(3^{r-1}x)}{\sin(3\cdot3^{r-1}x)} \\
&= \frac{1-2\sin^2(3^{r-1}x)}{3\sin(3^{r-1}x)-4\sin^3(3^{r-1}x)} \\
&= \frac{(3-4\sin^2(3^{r-1}x))-1}{2\sin(3^{r-1}x)\left(3-4\sin^2(3^{r-1}x)\right)} \\
&= \frac{1}{2\sin(3^{r-1}x)}-\frac{1}{2\sin(3\cdot3^{r-1}x)} \\
&= f(r-1)-f(r)
\end{align}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4259884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the minimum value of $a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ Let $a,b,c$ be the lengths of the three sides of the triangle, $a+b+c=3$. Find the minimum value of $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)$$
My attempts:
$\bullet$ The minimum value is $3$, equality holds iff $a=b=c=1$ so by AM-GM, we have: $$a^8+b^8+c^8\ge 8(a+b+c)-21=3$$
$\bullet$ We need to prove $$3+2(a-1)(b-1)(c-1)\ge3$$ or $$abc-(ab+bc+ca)+a+b+c-1\ge0$$
$\bullet$ Note that $ab+bc+ca\le \dfrac{(a+b+c)^2}{3}=3 $ so we need to prove: $$abc-3+3-1\ge0$$ or $$abc\ge1$$
But I have no idea from here, please help me
|
Another way.
We'll prove that the inequality
$$a^8+b^8+c^8+2(a-1)(b-1)(c-1)\geq3$$ is true for any positives $a$, $b$ and $c$ such that $a+b+c=3$.
Indeed, since $$\prod_{cyc}((a-1)(b-1))=\prod_{cyc}(a-1)^2\geq0,$$ we can assume $a(b-1)(c-1)\geq0$ and by AM-GM and C-S we obtain:
$$\sum_{cyc}a^8+2\prod_{cyc}(a-1)-3\geq\sum_{cyc}(4a^2-3)+2\prod_{cyc}(a-1)-3=$$
$$=\sum_{cyc}(4a^2-3)+2(abc-ab-ac-bc+3-1)-3=$$
$$=3(a^2+b^2+c^2-3)+a^2+b^2+c^2+2abc+1-2(ab+ac+bc)\geq$$
$$\geq a^2+b^2+c^2+2(ab+ac-a)+1-2(ab+ac+bc)=(b-c)^2+(a-1)^2\geq0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Euler substitution inconsistency in evaluating $\int_{-1}^1 \frac{\mathrm dx}{\sqrt{1-x}+2+\sqrt{1+x}}$
$$I = \int_{-1}^1 \frac{\mathrm dx}{\sqrt{1-x}+2+\sqrt{1+x}} = 4\sqrt2-2-\pi$$
Euler's substitutions immediately come to mind. But as is, the integrand is free of $\sqrt{ax^2+bx+c}$ (with $a\neq0$). So, I considered rewriting the integrand as
$$\frac{1}{\sqrt{1-x}+2+\sqrt{1+x}} = \dfrac{-\sqrt{1-x}+2-\sqrt{1+x}}{-(1-x)-2\sqrt{1-x^2}+4-(1+x)} = \frac{\sqrt{1-x}-2+\sqrt{1+x}}{2\sqrt{1-x^2}-2}$$
That is, if the denominator is $a+b+c$, I multiplied by $-a+b-c$. Now I can set up the second Euler substitution:
$$\sqrt{1-x^2} = xt-1 \implies 1-x^2 = x^2t^2-2xt+1 \implies x=\frac{2t}{t^2+1}$$
Then $\mathrm dx = \frac{2(1-t^2)}{(1+t^2)^2}\,\mathrm dt$. The integration range remains unchanged.
In terms of $t$, the square roots are
$$\begin{cases}\sqrt{1-x^2} = xt-1 = \frac{2t^2}{1+t^2}-1 = \frac{t^2-1}{1+t^2} & \color{red}{(1)}\\[1ex] \sqrt{1\pm x} = \sqrt{1\pm\frac{2t}{1+t^2}} = \frac{|t\pm1|}{\sqrt{1+t^2}}\end{cases}$$
Since $|t|<1$, we have $|t+1|=t+1$ and $|t-1|=1-t$. In the integral,
$$\begin{align}
I &= \frac12 \int_{-1}^1 \frac{\sqrt{1-x}-2+\sqrt{1+x}}{\sqrt{1-x^2}-1} \,\mathrm dx \\[1ex]
&= \int_{-1}^1 \frac{1-t^2}{1+t^2} - \frac{1-t^2}{(1+t^2)^{3/2}}\,\mathrm dt
\end{align}$$
But this leads to a different value of $-2-2\sqrt2+\pi+2\sinh^{-1}(1)$.
It seems the problem lies within the substitution:
$$\sqrt{1-x^2} = \sqrt{1-\frac{4t^2}{(t^2+1)^2}} = \sqrt{\frac{(t^2-1)^2}{(t^2+1)^2}} = \frac{|t^2-1|}{t^2+1} = \frac{1-t^2}{t^2+1} \quad \color{red}{(2)}$$
since $|t|<1$. But shouldn't $\color{red}{(1)}$ and $\color{red}{(2)}$ be in agreement?
|
Using instead $\sqrt{1-x^2} = xt\color{red}{+}1$ is indeed the correct move, as this ensures $xt+1\ge0$ for $(x,t)\in[-1,1]^2$. Then
$$\begin{cases}x = -\frac{2t}{1+t^2} \implies \mathrm dx = -\frac{2(1-t^2)}{(1+t^2)^2}\\\\ \sqrt{1-x^2} = \frac{1-t^2}{1+t^2} \\\\ \sqrt{1\pm x} = \frac{1\mp t}{\sqrt{1+t^2}}\end{cases}$$
The integral then reduces drastically to
$$\int_{-1}^1 \left(\frac1{t^2(1+t^2)} - \frac1{t^2(1+t^2)^{3/2}} - \frac1{1+t^2} + \frac1{(1+t^2)^{3/2}}\right) \,\mathrm dt$$
and subsequent evaluation yields the correct value, $4\sqrt2-2-\pi$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How could I rewrite $\dfrac{6 + 4 {i}} { -9 - 4 {i}}$ in a+bi form? Peace to all. When I solve the problem I get $\dfrac{70} { 97}$ - $\dfrac{12{i}} {97}$ and it's the wrong answer. How exactly do you go about solving this problem?
This is my work: I received that answer by multiplying both the numerator and denominator by the conjugate partner of "-9 - 4i" which is "-9 + 4i".
$\dfrac{-54 + 24 {i} - 36{i} +16i^2} { 81 - 36 {i} + 36 {i} - 16i^2}$
Combining like terms: $\dfrac{-54 + 24 {i} - 36{i} +16(-1)} { 81 - 36 {i} + 36 {i} - 16(-1)}$ = $\dfrac{-70 - 12i} { 81 + 16}$ = $\dfrac{-70} { 97}$ - $\dfrac{12{i}} {97}$
|
Another approach that can be taken is through the application of the definition of division. If $ \ \frac{6 + 4 i}{ -9 - 4 i} \ = \ a + bi \ \ , $ then
$$ 6 \ + \ 4i \ \ = \ \ (-9 \ - \ 4i)·(a \ + \ bi) \ \ = \ \ -9a \ - \ 4ai \ - \ 9bi \ - \ 4b·i^2 \ \ . $$
Equating the real and imaginary parts produces the pair of equations $ \ -9a + 4b \ = \ 6 \ \ , \ \ -4a - 9b \ = \ 4 \ \ . $ Solving this system by multiplication and addition
[first array: multiply first equation by 9 , second by 4 ; second array: multiply first equation by (-4) , second by 9] yields
$$ \begin{array}{ccc} -81a & + \ 36b & \ = \ 54 \\ -16a & - \ 36b & \ = \ 16 \end{array} \ \ \ , \ \ \ \begin{array}{ccc} 36a & - \ 16b & \ = \ -24 \\ -36a & \ - \ 81b & \ = \ 36 \end{array} $$
[then add equations together]
$$ \Rightarrow \ \ -97a \ \ = \ \ 70 \ \ \ , \ \ \ -97b \ \ = \ \ 12 \ \ . $$
Hence, the ratio is $ \ \large{\frac{6 \ + \ 4 i}{ -9 \ - \ 4 i}} \ = \ \normalsize{-\frac{70}{97} - \frac{12}{97}i} \ \ , $
in agreement with the result from other posters.
$$ \ \ $$
The multiplication of the numerator and denominator of a ratio of complex numbers by the complex conjugate in order to "rationalize" the denominator is, of course, the conventional method of simplification. It can be "reduced" to a formula (although the calculations are usually simple enough that the formula isn't generally taught):
$$ \frac{a \ + \ bi}{c \ + \ di} \ · \ \frac{c \ - \ di}{c \ - \ di} \ \ = \ \ \frac{ac \ + \ bd}{c^2 \ + \ d^2} \ + \ i·\left(\frac{bc \ - \ ad}{c^2 \ + \ d^2} \right) \ \ ; $$
for this problem, we then obtain
$$ \frac{6·(-9) \ + \ 4·(-4)}{(-9)^2 \ + \ (-4)^2} \ + \ i·\left(\frac{4·(-9) \ - \ 6·(-4)}{(-9)^2 \ + \ (-4)^2} \right) $$ $$ = \ \ \frac{(-54) \ + \ (-16)}{81 \ + \ 16} \ + \ i·\left(\frac{(-36) \ - \ (-24)}{81 \ + \ 16} \right) \ \ . $$
Because the method of solving a system of two linear equations is equivalent, there is a way of representing complex numbers as $ \ 2 \times 2 \ $ matrices, $ \ a + bi \ \rightarrow \ \left[\begin{array}{cc} a & b \\ -b & a \end{array} \right] \ \ . $ [I mention this approach since many "algebra/pre-calculus" course at least introduce matrices.] Multiplying two complex numbers then gives the same result as multiplying two such matrices and division involves multiplying the matrix representing the numerator by the inverse of the matrix representing the denominator. For this problem,
$$ 6 + 4i \ \rightarrow \ \left[\begin{array}{cc} 6 & 4 \\ -4 & 6 \end{array} \right] \ \ , \ \ -9 - 4i \ \rightarrow \ \left[\begin{array}{cc} -9 & -4 \\ 4 & -9 \end{array} \right] $$ $$ \Rightarrow \ \ \frac{1}{-9 - 4i} \ \ = \ \ \left[\begin{array}{cc} -9 & -4 \\ 4 & -9 \end{array} \right]^{-1} \ \ = \ \ \frac{1}{(-9)·(-9) \ - \ 4·(-4)} \left[\begin{array}{cc} -9 & 4 \\ -4 & -9 \end{array} \right] \ \ = \ \ \frac{1}{97} \left[\begin{array}{cc} -9 & 4 \\ -4 & -9 \end{array} \right] \ \ ; $$
$$ \frac{6 \ + \ 4 i}{ -9 \ - \ 4 i} \ \ \rightarrow \ \ \frac{1}{97} \left[\begin{array}{cc} -9 & 4 \\ -4 & -9 \end{array} \right] \ \left[\begin{array}{cc} 6 & 4 \\ -4 & 6 \end{array} \right] \ \ = \ \ \frac{1}{97} \left[\begin{array}{cc} -54 - 16 & -36+24 \\ -24 + 36 & -16-54 \end{array} \right] $$
$$ = \ \ \frac{1}{97} \left[\begin{array}{cc} -70 & -12 \\ 12 & -70 \end{array} \right] \ \ , $$
which represents the complex number $ \ -\frac{70}{97} - \frac{12}{97}i \ \ . $
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrating $\int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx$ I found the following integral and wanted to know if there is a nice closed form solution in terms of elementary or some special functions (Polylogarithm, Clausen, etc).
$$\displaystyle \int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx$$
I know that the integral converges numerically to $\approx 0.403926$
Here is my try using integration by parts:
Let
$$ du = \frac{\arctan(x)}{x^2} \Longrightarrow u = -\frac{1}{2}\ln(1+x^2) + \ln(x) - \frac{\arctan(x)}{x} $$
$$ v = \arctan(x^2) \Longrightarrow dv = \frac{2x}{x^4+1}$$
Hence
$$\displaystyle \int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx \stackrel{IBP}{=} -\frac{\pi^2}{16} - \frac{1}{8}\pi \ln(2) -2\underbrace{\int_{0}^{1} \frac{x\ln(x)}{x^4+1} dx}_{I_{1}} + 2\underbrace{\int_{0}^{1}\frac{\arctan(x)}{x^4+1}dx}_{I_{2}} + \underbrace{\int_{0}^{1} \frac{x\ln(1+x^2)}{x^4+1}dx}_{I_{3}} $$
Can be proven that
$$I_{1} = -\frac{C}{4}$$
where $C$ is the Catalan constant and
$$I_{3} = \frac{1}{16} \pi \ln(2) $$
but I'm stuck with $I_{2}$
Another way could be:
Define
$$\Psi(a) = \int_{0}^{1} \frac{\arctan(ax)\arctan(x^2)}{x^2} dx$$
Hence
$$\Psi'(a) = \int_{0}^{1} \frac{\arctan(x^2)}{x(a^2x^2+1)}dx = \frac{1}{2}\int_{0}^{1} \frac{\arctan(w)}{w(a^2w+1)}dx$$
Using integration by parts, we have:
$$du = \frac{\arctan(w)}{1+a^2w} \Longrightarrow u= \frac{1}{1+a^4}\ln\left( \frac{1+a^2w}{\sqrt{1+w^2}} \right) - \frac{a^2-w}{(1+a^4)(1+a^2w)} \arctan(w) $$
$$ v = \frac{1}{w} \Longrightarrow dv = \ln(w) $$
However, this path seems even more rugged that the other.
One last hint could be the following integral:
$$\int_{0}^{1} \frac{\arctan(x) \arctan(x^3)}{x} dx = \frac{7}{72}\zeta(3) + \frac{\pi}{3}C - \frac{5\pi}{12}\operatorname{Cl}_{2}\left(\frac{2\pi}{3} \right)$$
where $\operatorname{Cl}_{2}$ is the Clausen function of order 2. However, I do not know the proof of this result either.
|
By continuing the OP's work and using Cornel's closed-form of $\displaystyle \int_0^1\frac{\arctan(x)}{1+x^4}\textrm{d}x$, we get that
$$\int_0^1\frac{\arctan(x)\arctan(x^2)}{x^2}\textrm{d}x$$
$$=G+\frac{11}{192}\sqrt{2}\pi^2-\frac{1}{16}\log(2)\pi-\frac{3}{16}\sqrt{2}\log(\sqrt{2}-1)\pi-\frac{1}{4}\sqrt{2}\operatorname{Li}_2(1-\sqrt{2})-\frac{1}{32}\psi^{(1)}\left(\frac{1}{8}\right).$$
|
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What is the proof for variance of triangular distribution? In Wikipedia, the formula for the variance of the triangular distribution is given here. However, I don't know how to find it. I have tried a brute force method but the formula is quite complicated (polynomial of degree 5 in a, b, c) and I can't simplify it (I tried manually and with Xcas).
The following part is edited thanks to @Imaosome remark:
I came to this question with the following problem:
Say $X$ and $Y$ are independent random variables with uniform distribution between $0$ and $1$.
I want to study $Z = X-Y$. With convolution, I find that the distribution is triangular, centered in $0$ with extremities $-1$ and $1$ (the proof is also available in this pdf here). With Wikipedia notations, it gives $a=-1, b=1, c=0$. And in this case, we sum $2$ independent variables therefore the variance shoud be
$Var(X)+Var(-Y) = Var(X)+Var(Y)=2 Var(X)$. The variance of $X$ is $1/12$ (see for instance formula here). Therefore the variance of $Z$ is $Var(Z) = 2 * 1/12 = 1/6$.
If I come back to Wikipedia formula, I find:
$$
\frac{a^2+b^2+c^2-ab-ac-bc}{18}=\frac{1+1+0+1-0-0}{18}=1/6.
$$
This shows, at least in that particular case that the formula is correct. But I would like to try and prove Wikipedia result. Once again, I know that it should be possible to prove it by integration but I did not succeed and I hope somebody has a simple way to get this formula.
Here are the details. By definition, we want to compute:
\begin{align*}
\sigma^2 &= \int_a^c \frac{2(x-a)}{(b-a)(c-a)} \left( x - \frac{a+b+c}{3} \right)^2 dx + \int_c^b \frac{2(b-x)}{(b-a)(b-c)} \left( x - \frac{a+b+c}{3} \right)^2 dx \\
&= \frac{2}{(b-a)(c-a)} \left[ \frac{1}{3} (x-a) \left( x - \frac{a+b+c}{3} \right)^3 - \frac{1}{12} \left( x - \frac{a+b+c}{3} \right)^4 \right]_a^c \\
& ~~~~~~ \frac{2}{(b-a)(b-c)} \left[ \frac{1}{3} (b-x) \left( x - \frac{a+b+c}{3} \right)^3 - \frac{1}{12} \left( x - \frac{a+b+c}{3} \right)^4 \right]_c^b
\end{align*}
From there, one can see that terms in $\left( x - \frac{a+b+c}{3}\right)$ cancel out leaving us with:
\begin{align*}
\sigma^2 &= \frac{2}{12(b-a)(c-a)} \left( - \left( \frac{2c-a-b}{3} \right)^4 + \left( \frac{2a-b-c}{3} \right)^4 \right) \\
& ~~~~~~ \frac{2}{12(b-a)(b-c)} \left( - \left( \frac{2b-a-c}{3} \right)^4 + \left( \frac{2c-a-b}{3} \right)^4 \right) \\
&= \frac{((c-a)-(b-c))^5 +(b-c)((b-a)+(c-a))^4 - (c-a)((b-a)+(b-c))^4}{ 2 \times 3^5 (b-a)(b-c)(c-a) }
\end{align*}
And from this point, I am stuck. After all, maybe the last line is not helping much. I don't know.
|
The computation is best performed on the location-transformed variable such that the mode is at $0$. In particular, suppose $$X \sim \operatorname{Triangular}(a,b,c), \\ f_X(x) = \begin{cases} 0, & x < a \\ \frac{2(x-a)}{(b-a)(c-a)}, & a \le x \le c \\ \frac{2(b-x)}{(b-a)(b-c)}, & c < x \le b, \\ 0, & x > b. \end{cases}$$ Then consider $$Y = X - c$$ which then has density
$$f_Y(y) = \begin{cases}
0, & y < a - c \\
\frac{2(y-(a-c))}{(b-a)(c-a)}, & a-c \le y \le 0 \\
\frac{2((b-c)-y)}{(b-a)(b-c)}, & 0 < y \le b-c \\
0, & x > b-c.
\end{cases}$$
All we did was shift the density, which lets us now use the auxiliary parameters $\alpha = a-c$, $\beta = b-c$, to write
$$f_Y(y) = \begin{cases}
0, & y < \alpha \\
\frac{2(y-\alpha)}{(\beta - \alpha)(-\alpha)}, & \alpha \le y \le 0 \\
\frac{2(\beta-y)}{(\beta - \alpha)\beta}, & 0 < y \le \beta \\
0, & y > \beta.
\end{cases}$$
Now the variance calculation performed on $Y$ proceeds as follows:
$$\begin{align}
\operatorname{E}[Y^k] &= \int_{y=\alpha}^0 y^k \frac{2(y-\alpha)}{(\beta-\alpha)(-\alpha)} \, dy + \int_{y=0}^\beta y^k \frac{2(\beta-y)}{(\beta-\alpha)\beta} \, dy \\
&= \frac{2}{(\beta - \alpha)(-\alpha)} \left[\frac{y^{k+2}}{k+2} - \alpha \frac{y^{k+1}}{k+1}\right]_{y=\alpha}^0 + \frac{2}{(\beta - \alpha)\beta} \left[\beta\frac{y^{k+1}}{k+1} - \frac{y^{k+2}}{k+2} \right]_{y=0}^\beta \\
&= \frac{2}{(\beta-\alpha)} (\beta^{k+1} - \alpha^{k+1}) \left(\frac{1}{k+1} - \frac{1}{k+2} \right) \\
&= \frac{2(\beta^{k+1} - \alpha^{k+1})}{(k+1)(k+2)(\beta-\alpha)}.
\end{align}$$
For $k \in \{1, 2\}$ we easily get
$$\operatorname{E}[Y] = \frac{\alpha + \beta}{3} \\
\operatorname{E}[Y^2] = \frac{\alpha^2 + \alpha\beta + \beta^2}{6}.$$
Consequently, $$\operatorname{Var}[X] = \operatorname{Var}[Y] = \frac{\alpha^2 - \alpha \beta + \beta^2}{18} = \frac{a^2 + b^2 + c^2 - (ab + bc + ca)}{18}.$$
Note that we have used the fact that a location transformation of a random variable does not change its variance. This calculation minimizes the number of integrations we need to perform and reduces the number of parameters we need to account for. We also get the additional result for the $k^{\rm th}$ raw moment of $Y$, which has a particularly convenient form.
Also note that we could have written the density of the transformed variable $Y$ simply by choosing $c = 0$, $a = \alpha$, $b = \beta$ from the density of $X$; however, I wanted to emphasize how the density of $Y$ is obtained by shifting $X$ by $c$.
|
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|
How to calculate a permutation to the $-100$th power? I have a this permutation element $(1,2)(3,4,6,5,8)(7,9,10)$ and I need find the $-100$th power of this element.
I think firstly we need find $-1$ power, and I have this element: $(1,2)(3,8,5,6,4)(7,10,9)$.
Here $(1,2)$ have order $2$, so after $100$ permutation we have $(1 \to 1)$, $(2 \to 2)$ and we don't need write this. The same goes with $(3,8,5,6,4)$. But $(7,10,9)$ changes to $(7,10,9)$ in power $1$, that is $(7,10,9)$ doesn't change.
The answer is $(7,10,9)$.
Are my calculations correct?
|
Your calculations are fine.
An alternative approach would be to notice that the order of the element is the least common multiple of the orders of its disjoint cyclic components, so the order is ${\rm lcm}(2,5,3)=30$. Now note that
$$\begin{align}
((1,2)(3,4,6,5,8)(7,9,10))^{-100}&=((1,2)(3,4,6,5,8)(7,9,10))^{-90}\\
&\times((1,2)(3,4,6,5,8)(7,9,10))^{-10}\\
&={\rm id}\times((1,2)(3,4,6,5,8)(7,9,10))^{-10}\\
&=(1,2)^{-10}(3,4,6,5,8)^{-10}(7,9,10)^{-10}\tag{1}\\
&={\rm id}\times(7,9,10)^{-9}(7,9,10)^{-1}\\
&={\rm id}\times(7,9,10)^{-1}\\
&=(7,10,9),
\end{align}$$
where $(1)$ holds because disjoint cycles commute.
|
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|
Probability of having an all-heterogeneous genotype
Find the probability that the genotype of an offspring will be $\mathrm{AaBbCcDdEeFfGgHhIiJj}$ from the cross $\mathrm{AaBbCcDdEeFfGgHhIiJj \times AaBbCcDdEeFfGgHhIiJj}$.
Making a Punnett square for this problem will be too tedious. However, I noticed that this problem can be treated as a combinatorics problem. For the simplest case, the possible outcomes of the cross $\mathrm{Aa \times Aa} = \{\mathrm{AA, Aa, Aa, aa}\}$ where two of them are desired. As there are 10 traits, we have $2^{10}$ of this genotype from the cross. Now, considering the number of genotypes that are not of this form, there are $2^1 + 2^2 + \cdots + 2^{10} = 2^{11} - 2$. Getting the number of all possible genotypes, we have
\begin{align*}2^{11} + 2^{10}-2 &= 2^{10}(2 + 1) - 2 \\ &= 3(2^{10}) - 2\end{align*} Therefore, the probability of getting the genotype is $$\frac{2^{10}}{3(2^{10}) - 2} = \frac{2^9}{3(2^9) - 1}$$
Can someone check if I overcounted or undercounted something? If not, is there a better way to solve these types of problems?
|
There are four possible outcomes for each of the ten traits. For instance, for the first trait, they are $AA, Aa, aA, aa$, where we first list the allele from the father, then the allele from the mother. Since we are not told the traits are linked in some way, it is reasonable to assume independence. Thus, there are $4^{10}$ possible genotypes.
As you observed, there are two possible heterozygous outcomes for each trait. For instance, for the first trait, they are $Aa, aA$. Again, assuming independence, there are $2^{10}$ possible favorable outcomes.
Thus, the probability that the genotype of an offspring of two parents who are heterozygous for each of the ten traits is heterozygous for each of the ten traits is
$$\frac{2^{10}}{4^{10}} = \left(\frac{2}{4}\right)^{10} = \left(\frac{1}{2}\right)^{10} = \frac{1}{2^{10}} = \frac{1}{1024}$$
Where did you make your mistake?
The number of genotypes that are not heterozygous for all traits is $4^{10}−2^{10}$. It looks like you are saying that there are $2^k$ genotypes that different from those that are heterozygous for all traits in $k$ traits. However, that is not true. There are $$10⋅2^k⋅2^{10 - k} = \binom{10}{k} \cdot 2^{10}$$
ways for exactly one of the traits to be homozygous since there are $\binom{10}{k}$ ways to choose exactly $k$ of the traits to be homozygous, $2$ ways to obtain a homozygous version of that trait, and $2$ possible ways for each of the remaining $10 - k$ traits to be heterozygous. Notice that
$$\sum_{k = 1}^{10} \binom{10}{k} 2^{10} = 4^{10} - 2^{10}$$
|
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|
Does this integral have an analytical solution? (The residue theorem method seems to fail here) The integral is the following:
\begin{equation}
\int_{-\infty}^{+\infty}dx \frac{e^{-{\sigma}^2(x+b)^2}}{x^2+a^2}
\end{equation}
I know there is an analytical solution when $b=0$. But what about $b\neq0$ ?
|
May be, we could try to expand the exponential as
$$e^{-\sigma ^2 (x+b)^2}=e^{-\sigma ^2 x^2}\sum_{n=0}^\infty (-1)^n P_n(x)\, b^n$$ and face integrals
$$I_n=\int_{-\infty}^\infty \frac {x^n} {x^2+a^2}e^{-\sigma ^2 x^2}\, dx=a^{n-1}\int_{-\infty}^\infty \frac{t^n}{t^2+1}e^{-k t^2}\,dt\qquad \text{where}\qquad k=a^2\sigma ^2$$ For sure, for odd $n$, the result is zero
$$I_{2n}= e^k \,\Gamma \left(\frac{2n+1}{2}\right)\, \Gamma \left(\frac{1-2n}{2},k\right)$$
Let $k=c^2$ and the $I_{2n}$ write
$$I_{2n}=(-1)^{n+1} \Bigg[\frac{\sqrt \pi}{2^{n-1} c^{2 n-1} } Q_{n}(c)-\pi e^{c^2} \text{erfc}(c) \Bigg]$$
$$\left(
\begin{array}{cc}
n & Q_n(c) \\
1 & 1 \\
2 & 2 c^2-1 \\
3 & 4 c^4-2 c^2+3 \\
4 & 8 c^6-4 c^4+6 c^2-15 \\
5 & 16 c^8-8 c^6+12 c^4-30 c^2+105 \\
6 & 32 c^{10}-16 c^8+24 c^6-60 c^4+210 c^2-945 \\
7 & 64 c^{12}-32 c^{10}+48 c^8-120 c^6+420 c^4-1890 c^2+10395 \\
8 & 128 c^{14}-64 c^{12}+96 c^{10}-240 c^8+840 c^6-3780 c^4+20790 c^2-135135
\end{array}
\right)$$
where clear and simple patterns appear.
|
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|
Proving $\lim_{x \to +\infty}\frac{\ln(x)}{2\cdot \ln(x+1)} = \frac{1}{2}$ with $\epsilon- \delta$ Today, I spent some hours trying to solve this problem but I can't reach the conclusion:
Show with $\epsilon-\delta $ definition the following limit:
$$\lim_{x \to +\infty}\frac{\ln(x)}{2\cdot \ln(x+1)} = \frac{1}{2}$$
So, first of all, I set up the inequality:
$$\left|\frac{\ln(x)}{2\cdot \ln(x+1)}-\frac{1}{2}\right|<\epsilon$$
Then:
$$\frac{1}{2}-\epsilon<\frac{\ln(x)}{2\cdot \ln(x+1)}<\frac{1}{2}+\epsilon \implies (1-2\epsilon)\cdot \ln(x+1) < \ln(x) < (1+2\epsilon)\cdot \ln(x+1)$$
And elevating with base $e$ to cancel the $\ln(x)$:
$$(x+1)^{1-2\epsilon}<x<(x+1)^{1+2\epsilon}$$
I want to apply the limit $\lim_{x \to +\infty}\frac{(1+f(x))^\alpha-1}{\alpha\cdot f(x)}=1$ with $f(x) \to_{x \to 0} 0$. So, I have:
$$x^{1-2\epsilon}\cdot\left(1+\frac{1}{x}\right)^{1-2\epsilon}<x<x^{1+2\epsilon}\cdot\left(1+\frac{1}{x}\right)^{1+2\epsilon}$$
Then:
$$
\left\{\begin{matrix}
x>x^{1-2\epsilon}\cdot\left(1+\frac{1}{x}\right)^{1-2\epsilon}
\\ x<x^{1+2\epsilon}\cdot\left(1+\frac{1}{x} \right )^{1+2\epsilon}
\end{matrix}\right.
$$
So:
$$
\left\{\begin{matrix}
x^{2\epsilon} - 1>\left(1+\frac{1}{x}\right)^{1-2\epsilon} - 1
\\ \frac{1}{x^{2\epsilon}}-1<\left(1+\frac{1}{x} \right )^{1+2\epsilon}-1
\end{matrix}\right.
$$
When $x$ approaches $+\infty$:
$$
\sim
\left\{\begin{matrix}
x^{2\epsilon}-1>\frac{1-2\epsilon}{x}
\\\frac{1}{x^{2\epsilon}}-1<\frac{1+2\epsilon}{x}
\end{matrix}\right.
$$
Notice that the second inequality is always true because $\frac{1}{x^{2\epsilon}}\to 0^+$ when $x \to +\infty$, so that left side is negative while right side $\frac{1-2\epsilon}{x}$ is positive.
Now, in some way I have to find $\delta = N$. I noticed that: $$x^{2\epsilon}-1\sim x^{2\epsilon} \;\;\; x \to +\infty$$
and then for the first inequality:
$$x^{2\epsilon}>\frac{1-2\epsilon}{x} \implies x> \delta(\epsilon)= N =(1-2\epsilon)^{1+2\epsilon}$$
Here, tere is a problem because when $\epsilon \to 0^+$, $\delta = N \to 1$ and not to $+\infty$. Is this anyway correct? Or, I can't apply asympothics reduction with $\epsilon - \delta$?
Edit:
I all three solutions proposed, $\delta = N$ goes to infinity when $\epsilon \to 0^+$. In my answer, $N$ tends to $1$, so is my solution not corret? If it's yes, why?
Thanks.
|
Fix $x>e^2-1$. Using
$$ \ln(1+\frac1x)<\frac1x $$
we have
$$ \left|\frac{\ln(x)}{2\cdot \ln(x+1)}-\frac{1}{2}\right|=\frac12\frac{\ln(1+\frac1x)}{\ln(x+1)}<\frac{1}{4x}. $$
Then for $\forall \varepsilon>0$, denote $N$ by $N=\max\{e^2-1,\frac1{4\varepsilon}\}$. Then when $x>N$,
$$ \left|\frac{\ln(x)}{2\cdot \ln(x+1)}-\frac{1}{2}\right|<\varepsilon. $$
|
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|
Given $a+b+c =1$ prove $\displaystyle \sum_{\text{cyclic}}\sqrt{abc+4ab+4ac}\ge 8(ab+bc+ca)+\sqrt{abc}$
If $a,b,c$ are non-negative real numbers such that: $a+b+c=1.$ Prove that: $$\sqrt{abc+4ab+4ac}+\sqrt{abc+4bc+4ba}+\sqrt{abc+4ca+4cb}\ge8(ab+bc+ca)+\sqrt{abc}$$
Anyone can help me give a hint to get proof?
I guess equality holds iff: two of them are equal to $0.$ Since $ab+bc+ca\le\frac{1}{3}$ and $abc\le\frac{1}{27}$.
But the last inequality is not true by a countable example. Also, it is hard for me with this kind of equality. I tried squaring both side, the rest is very complicated.
Thank you for your interest!
|
Hint:
Observe,
$$\begin{align*}
8(ab+bc+ca)+\sqrt{abc}&=4a(b+c)+4b(c+a)+4c(a+b)+\sqrt{abc} \\
(b+c-a)^2&=(1-2a)^2=1-4a(1-a)=1-4a(b+c)
\end{align*}$$
Using the Cauchy-Schwarz inequality,
$$\sum_{\text{cyc}}\sqrt{\left((b+c-a)^{2}+4a(b+c)\right)\left(abc+4a(b+c)\right)}\geq \sum_{\text{cyc}} \left(\sqrt{abc}(b+c-a)+4a(b+c)\right)\\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=8(bc+ca+ab)+\sqrt{abc}.$$
|
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|
How do you prove by induction that $\frac{1}{2} + \frac{2}{2^2} + \ldots + \frac{n}{2^n} = 2 - \frac{n+2}{2^n}$? For $n=1$ this is true because $\frac{1}{2^{1}}=2-\frac{1+2}{2^{1}}=\frac{1}{2}$. Further, it is a little more complicated, can we now assume that this is true up to the number $n-1$? Then do the induction step from $n-1$ to $n$.
So what I've tried:
$$\frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{n-1}{2^{n-1}}+\frac{n}{2^{n}}=$$
$$=2-\frac{n+1}{2^{n-1}}+\frac{n}{2^{n}}=$$
|
Another solution : for sequences $\{a_{n}\}, \{b_{n}\}$
*
*Prove $\sum_{k=1}^{n} a_{k} b_{k} = \sum_{k=1}^{n-1} S_{k} (b_{k} - b_{k+1}) + S_{n} b_{n}$ where $S_n = \sum_{k=1}^{n} a_{k}.$ ($n \in \mathbb{N}$)
$pf)$ $(RHS) = b_{n} (S_{n} - S_{n-1}) + b_{n-1} (S_{n-1} - S_{n-2}) + \cdots + b_{1} S_{1} = \sum_{k=1}^{n} a_{k} b_{k}$
since $S_{n} - S_{n-1} = a_{n}$ for $n \ge 1.$
*Evaluate $\sum_{k=1}^{n} \frac{k}{2^{k}}$, $n \in \mathbb{N}$.
let $a_n = 2^{-n}, b_{n} = n, n \in \mathbb{N}$. By the formula shown in 1,
$$\sum_{k=1}^{n} a_{k} b_{k} = \sum_{k=1}^{n-1} S_{k} (b_{k} - b_{k+1}) + S_{n} b_{n}$$
$$= \sum_{k=1}^{n-1} \frac{2^{-1} \cdot (1 - 2^{-k})}{1 - 2^{-1}} \cdot (-1) + \frac{2^{-1} \cdot (1 - 2^{-n})}{1 - 2^{-1}} \cdot n$$
$$= \sum_{k=1}^{n-1} (2^{-k} - 1) + n(1 - \frac{1}{2^{n}})$$
$$= \frac{2^{-1} \cdot (1 - 2^{1-n})}{1 - 2^{-1}} - n + 1 + n - \frac{n}{2^{n}}$$
$$= 1 - \frac{1}{2^{n-1}} + 1 - \frac{n}{2^{n}}$$
$$= 2 - \frac{n+2}{2^{n}}.$$
|
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|
Can we approximate $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$? It seems like from the graph $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$ is somehow alike to the graph $e^{-x^2}$, the main problem is that the limitations of the software makes it hard to graph for large numbers. Is it possible to do so or it diverges?
|
Calling $\Pi$ the partial product up to $N$, for $0<|x|<\pi/\sqrt{3}$ we have
$$ \begin{array}{ll}
-\ln\Pi & =\displaystyle \sum_{n=1}^N \ln\Bigg(\frac{1}{1-\frac{(2n+1)x^2}{n^2\pi^2}}\Bigg) \\[5pt] & > \displaystyle \sum_{n=1}^N \ln\Big(1+\frac{(2n+1)x^2}{n^2\pi^2}\Big) \\[5pt]
& >\displaystyle \sum_{n=1}^N \left[\frac{(2n+1)x^2}{n^2\pi^2}-\frac{1}{2}\left(\frac{(2n+1)x^2}{n^2\pi^2}\right)^2 \right] \\[5pt]
& >\displaystyle \left[\frac{2x^2}{\pi^2}\sum_{n=1}^N \frac{1}{n}\right] - \mathcal{O}(1) ~~~\to\infty
\end{array} $$
Thus, $\ln \Pi\to-\infty$ so $\Pi\to0^+$ as $N\to\infty$.
|
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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.
|
Probably not the fastest, but I would like to suggest you use the general factorization method:
We have,
$$
\begin{align}
& 4 x^{2}-2 x y-4 x+3 y-3=4 x^{2}-x(2 y+4)+(3 y-3) \\
\implies & \Delta=(y+2)^{2}-4(3 y-3)=(y-4)^{2} \\
\implies & x_{1}=\frac{y+2+(y-4)}{4}=\frac{y-1}{2} \\
\implies & x_{2}=\frac{y+2-(y-4)}{4}=\frac{3}{2} \\
\implies & 4 x^{2}-2 x y-4 x+3 y-3=4\left(x-\frac{3}{2}\right)\left(x-\frac{y-1}{2}\right) \\
\implies & 4 x^{2}-2 x y-4 x+3 y-3=(2 x-3)(2 x-y+1).
\end{align}
$$
|
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|
Prove $\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}=1$ Use $f(n) = \frac{1}{n}$ and $f(n) = \frac{1}{n^2}$ when:
*
*$\sum_{n=1}^{\infty}[f(n+1) - f(n)] = \lim_{n \to \infty} f(n) - f(1)$
to prove $\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}=1$
What I have tried:
$$\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2} =\sum_{n=1}^{\infty}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2} \right)$$
Therefore;
$$\implies \lim_{n \to \infty}f(n) - f(1) = \lim_{n \to \infty}\frac{\left(\frac{1}{n^2}\right)}{n}-\frac{1}{(1+1)^2}$$
Though it does not look to me that this equals to 1, only hints on where do I go from here?
|
$f(n)-f(1)=\frac 1 {n^{2}}-1 \to -1$ so $\sum_{n=1}^{\infty}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2} \right)=-\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2} \right) =-\lim [f(n)-1]=1$
|
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|
$\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....?$ $$\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....?$$
I tried to solve it by using this product formula,
$$\frac 1{\Gamma (x)}=xe^{\gamma x} \prod_{n=1}^{\infty} \left(1+\frac x{n}\right)e^{-\frac x{n}}$$
I tried by differentiating after taking $\log$
I'm interested in finding the answer for the above summation $\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....$
|
Moreover,
\begin{align*}
\sum_{n=2}^{\infty} \frac{\zeta(n)}{k^{n}}&= \frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{n!k^n} \Gamma(n+1)\zeta(n+1) \\&= \frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{n!k^n} \int_{0}^{\infty} \frac{x^n}{e^{x}-1} \, dx \\&= \frac{1}{k}\int_{0}^{\infty} \frac{e^{x/k}-1}{e^{x}-1} \, dx \\&= - \frac{1}{k}\left(\gamma + \psi\left(1 - \frac{1}{k}\right) \right)
\end{align*}
where $\psi$ is the digamma function.
|
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|
Mistake computing $\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}} \right)\frac{dx}{x}$ I am looking to evaluate the integral
$$\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}} \right)\frac{dx}{x}=\ln\left(\frac{\pi}{\Gamma^2\left(\frac{1+a}{2b}\right)\cos\left(\frac{a\pi}{2}\right)}\right)$$
To this end I considered
$$I(w)=\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(bx)}-\frac{ad}{e^{cx}} \right)\frac{e^{-wx}}{x}\,dx \tag{1}$$
Note that as $w \to \infty$ the integrand vanishes. And as $w =0$ we recover the desired integral.
Differentiating $(1)$ w.r. to $w$ we obtain
$$
\begin{aligned}
I^\prime(w)&=-\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(bx)}-\frac{ad}{e^{cx}} \right)e^{-wx}\,dx\\
&=ad\int_0^\infty e^{-(c+w)x}\,dx- \int_0^\infty \frac{\sinh(ax)}{\sinh(bx)}e^{-wx}\,dx\\
&=\frac{ad}{c+w}-\int_0^\infty \frac{e^{ax}-e^{-ax}}{e^{bx}-e^{-bx}}e^{-wx}\,dx\\
&=\frac{ad}{c+w}-\int_0^\infty \frac{e^{-(w-a)x}-e^{-(w+a)x}}{e^{bx}-e^{-bx}}\,dx\\
&=\frac{ad}{c+w}-\int_0^\infty \frac{e^{-bx}}{e^{-bx}}\cdot\frac{e^{-(w-a)x}-e^{-(w+a)x}}{e^{bx}-e^{-bx}}\,dx\\
&=\frac{ad}{c+w}-\int_0^\infty \frac{e^{-(w-a+b)x}-e^{-(w+a+b)x}}{1-e^{-2bx}}\,dx\\
&=\frac{ad}{c+w}-\frac{1}{2b}\int_0^\infty \frac{e^{-\frac{(w-a+b)}{2b}x}-e^{-\frac{(w+a+b)}{2b}x}}{1-e^{-x}}\,dx \qquad (2bx \to x)\\
&=\frac{ad}{c+w}-\frac{1}{2b}\int_0^1 \frac{x^{\frac{(w-a+b)}{2b}-1}-x^{\frac{(w+a+b)}{2b}-1}}{1-x}\,dx \qquad (e^{-x} \to x)\\
&=\frac{ad}{c+w}-\frac{1}{2b}\left(\psi\left(\frac{w+a+b}{2b}\right)-\psi\left(\frac{w-a+b}{2b}\right)\right)\\
I(w)&=ad\int\frac{1}{c+w}\,dw-\frac{1}{2b}\left(\int\psi\left(\frac{w+a+b}{2b}\right)\,dw-\int\psi\left(\frac{w-a+b}{2b}\right)\,dw\right)\\
&=ad\ln(c+w)-\left(\ln\left(\Gamma\left(\frac{w+a+b}{2b}\right)\right)\,-\ln\left(\Gamma\left(\frac{w-a+b}{2b}\right)\right)\right)\\
&=ad\ln(c+w)+\ln\left(\frac{\Gamma\left(\frac{w-a+b}{2b}\right)}{\Gamma\left(\frac{w+a+b}{2b}\right)}\right)\\
\end{aligned}
$$
Now,our integral is equal to
$$I=-\int_0^\infty I^\prime(w)\,dw=I(0)$$
Letting $w=0$
$$\begin{aligned}
\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(bx)}-\frac{ad}{e^{cx}} \right)\frac{dx}{x}&=\ln\left(\frac{c^{ad}\Gamma\left(\frac12-\frac{a}{2b}\right)}{\Gamma\left(\frac12+\frac{a}{2b}\right)}\right)\\
&=\ln\left(\frac{c^{ad}\Gamma\left(\frac12-\frac{a}{2b}\right)\Gamma\left(\frac12+\frac{a}{2b}\right)}{\Gamma\left(\frac12+\frac{a}{2b}\right)\Gamma\left(\frac12+\frac{a}{2b}\right)}\right)\\
&=\ln\left(\frac{c^{ad}\pi}{\Gamma^2\left(\frac12+\frac{a}{2b}\right)\cos\left(\frac{a\pi}{2b}\right)}\right) \qquad \blacksquare\\
\end{aligned}$$
setting $b=1$, $c=2$ and $d=1$ I obtained
$$\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}} \right)\frac{dx}{x}=\ln\left(\frac{2^{a}\pi}{\Gamma^2\left(\frac{1+a}{2b}\right)\cos\left(\frac{a\pi}{2}\right)}\right)$$
Which has an extra term leading to an incorrect answer. Can someone please point out where I am mistaking?
|
For this specific integral We can evaluate as follows:
$$
\begin{aligned}
I(a)&=\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}}\right)\frac{dx}{x}\\
& \\
I^\prime(a)&=\int_0^\infty \left(\frac{x\cosh(ax)}{\sinh(x)}-\frac{1}{e^{2x}}\right)\frac{dx}{x}\\
&=\int_0^\infty \left(\frac{\cosh(ax)}{\sinh(x)}-\frac{e^{-2x}}{x}\right)dx\\
&=\int_0^\infty \left(\frac{e^{ax}+e^{-ax}}{e^x-e^{-x}}-\frac{e^{-2x}}{x}\right)dx\\
&=\int_0^1 \left(\frac{x^{a}+x^{-a}}{x^{-1}-x}+\frac{x^2}{\ln(x)}\right)\frac{dx}{x} & (e^{-x} \to x)\\
&=\int_0^1 \left(\frac{x}{x}\cdot\frac{x^{a}+x^{-a}}{x^{-1}-x}+\frac{x^2}{\ln(x)}\right)\frac{dx}{x}\\
&=\int_0^1 \left(\frac{x^{a+1}+x^{1-a}}{1-x^2}+\frac{x^2}{\ln(x)}\right)\frac{dx}{x}\\
&=\frac12\int_0^1 \left(\frac{x^{\frac{a+1}{2}}+x^{\frac{1-a}{2}}}{1-x}+\frac{2x}{\ln(x)}\right)\frac{dx}{\sqrt{x}\sqrt{x}} & (x^2 \to x)\\
&=\frac12\int_0^1 \left(\frac{1}{\ln(x)}+\frac{x^{\frac{a+1}{2}-1}}{1-x}\right)dx+\frac12\int_0^1 \left(\frac{1}{\ln(x)}+\frac{x^{\frac{1-a}{2}-1}}{1-x}\right)dx\\
&=-\frac12\psi\left(\frac{1+a}{2} \right)-\frac12\psi\left(\frac{1-a}{2} \right)
\end{aligned}
$$
Then
$$
\begin{aligned}
I(a)&=-\frac12\int_0^a\psi\left(\frac{1+u}{2} \right)\,du-\frac12\int_0^a\psi\left(\frac{1-u}{2} \right)\,du\\
&=-\int_{1/2}^{\frac{1+a}{2}}\psi\left(u \right)\,du+\int_{1/2}^{\frac{1-a}{2}}\psi\left(u \right)\,du\\
&=\ln\left(\Gamma\left(\frac{1-a}{2} \right) \right)-\ln\left(\Gamma\left(\frac{1}{2} \right) \right)-\ln\left(\Gamma\left(\frac{1+a}{2} \right) \right)+\ln\left(\Gamma\left(\frac{1}{2} \right) \right)\\
&=\ln\left(\frac{\Gamma\left(\frac{1-a}{2} \right)}{\Gamma\left(\frac{1+a}{2} \right) } \right)\\
&=\ln\left(\frac{\Gamma\left(\frac{1-a}{2} \right)\Gamma\left(\frac{1+a}{2} \right)}{\Gamma\left(\frac{1+a}{2} \right)\Gamma\left(\frac{1+a}{2} \right) } \right)\\
&=\ln\left(\frac{\pi}{\Gamma^2\left(\frac{1+a}{2} \right)\cos\left(\frac{a \pi }{2} \right) } \right) \qquad \blacksquare\\
\end{aligned}
$$
We used that:
$\int_0^1 \left(\frac{x^{z-1}}{\ln(x)}+\frac{x^{w-1}}{1-x}\right)dx=\ln(z)-\psi(w)$
and
$
\Gamma\left(\frac{1}{2}-x\right) \Gamma\left(\frac{1}{2}+x\right)=\frac{\pi}{\cos \pi x}
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4296809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
}
|
determining the cubic function from a graph This question may be a bit below the regular level for this forum, but I really struggle finding the solution. All of my attempts seem to end up describing $y = 1$, probably because all the points I can read from the graph are on that line. While this was given as a homework assignment, it wasn't to me, and I have been struggling with it for days now, since I was asked to help. The deadline has long passed. I have, in the meantime, studied several of my old text books, to no avail.
The question, as stated, was: given the graph below, determine the cubic function expression on the form $f(x) = ax^3 + bx^2 + cx + d$.
Note that the graph crosses $(0,1)$, has a local maximum at $x=1$ and minimum at $x=3$, and satisfies $f(3)=1$.
This corresponds to $f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$. I know that the $a$ value must be positive, given the graph.
I have determined visually that $f(0) = 1$, $f(3) = 1$, $f'(1) = 0$, $f'(3) = 0$, $f''(2) = 0$.
The exact values of the root or the value at $f(1)$ are impossible to tell.
This gives me this set of equations:
$$d = f(0) = 1 \tag{1}$$
$$f(3) = 1 \Rightarrow 27a + 9b + 3c + 1 = 1 \Rightarrow 27a + 9b + 3c = 0 \tag{2}$$
$$f'(1) = 0 \Rightarrow 3a + 2b + c = 0 \Rightarrow c=-3a-2b\tag{3}$$
$$f'(3) = 0 \Rightarrow 27a + 6b + c = 0 \tag{4}$$
$$f''(2) = 0 \Rightarrow 12a + 2b = 0 \Rightarrow b = -6a \tag{5}$$
Substituting $(3)$ and $(5)$ into $(2)$ gives me:
$$27a + 9(-6a) + 3(-3a-2(-6a)) = 0 \Rightarrow 27a - 54a + 27a = 0\Rightarrow 0=0\tag{6}$$
I get a similar result if I substitute into $(4)$:
$$27a + 6(-6a) + (-3a -2(-6a)) = 0 \Rightarrow 27a-36a-3a+12a = 0\Rightarrow0=0\tag{7}$$
But $(6)$ and $(7)$ are useless. I realize that all of the points I'm using are on $y = 1$ and the derivatives hold true for that line, too. Given that the graph isn't precise enough to actually read the values from either the root or the value of the local maximum - how do I proceed here to find the value of $a$? It seems to me I've pretty much found the rest once I lock down that one.
|
The derivative of the cubic function has two zeroes at $x = 1 $ and $x = 3$. Therefore,
$f'(x) = A (x - 1)(x - 3) = A (x^2 - 4 x + 3) $
Integrate that from $0$ to $x$,
$f(x) = f(0) + A (\dfrac{x^3}{3} - 2 x^2 + 3 x ) $
Now from the graph, $f(0) = 1$ and $f(3) = 1$ , hence,
$f(x) = 1 + A (\dfrac{x^3}{3} - 2 x^2 + 3 x ) $
Substitute $f(3) = 1$
$ 1 = 1 + A ( 9 - 18 + 9 ) $
which is $0 = 0$ , so $A$ cannot be determined from the information used so far.
So let's consider $x = 2$ and let's assume that the value of $y$ at that $x$ is 1.75
Then $1.75 = 1 + A (\dfrac{8}{3} - 8 + 6 ) $
From which, $ 0.75 = A (\dfrac{2}{3} )$, which means that $A = \dfrac{9}{8} $
Hence the cubic function equation is
$ y = f(x) = 1 + \dfrac{9}{8} (\dfrac{x^3}{3} - 2 x^2 + 3 x ) $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4299684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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|
FUN with f̶̶l̶̶a̶̶g̶̶s̶ Newton Cotes Quadrature formula and Bernoulli polynomials of the second kind I was told to phrase my question in a more exciting way when I asked it last time. The following is a preliminary consideration. If you don't need it, just scroll down to START HERE. Here we go then:
Imagine you have a really cool integral,
\begin{align}
I = \int_a^b f(x) \ dx ,
\end{align}
which you wanted to solve numerically. And what do you take for it: A quadrature formula. And one of the most popular quadrature formulas are the Newton-Cotes formulas. Don't be put off, this just means that the supports for my quadrature formula are at a fixed distance from each other. It applies to an interval $[a,b]$:
\begin{align*}
x_k = a + k \cdot h, \hspace{15pt} h = \frac{b-a}{n}, \hspace{15pt} k = 0,1,\ldots,n
\end{align*}
The quadrature error $E_n[f] := I_n - I$ of a Newton-Cotes formula $n$-th degree to $n$ support points,
\begin{align*}
I = \int_a^b f(x) \, d x, \hspace{30pt} I_n = I_n\left(f,[a,b]\right) = \sum_{k=0}^n \, w_kf(x_k)\,,
\end{align*}
for a $(n+1)$ times continuously differentiable function $f(x)$ has the representation
\begin{align*}
E_n[f] &= \frac{f^{(n+1)}(\xi)}{(n+1)!} \int_a^b \omega(x) \,d x, \hspace{30pt} \omega(x) = \prod_{k=0}^n\,(x-x_k) \\
&\leq \frac{1}{(n+1)!} \,\max_{\xi\in[a,b]}\left|f^{(n+1)}(\xi)\right|\,\int_a^b \omega(x) \,d x
\end{align*}
BUT ALL OF THIS DOESN'T GIVE A SHIT NOW. Because I want to specify the error exactly. It follows:
\begin{align*}
\int_a^b \omega(x) \,d x &= \int_a^b (x-x_0)(x-x_1)\cdots(x-x_n)\,d x \\
&= \int_a^b (x-x_0)(x-x_0 - h)(x- x_0 - 2h)\cdots(x-x_0-nh)\,d x \\
&= \int_a^b h^{n+1}\,\left(\frac{x-x_0}{h}\right)\left(\frac{x-x_0}{h} - 1\right)\cdots\left(\frac{x-x_0}{h}-n\right)\,d x, \hspace{30pt} \color{red}{u = \frac{x-x_0}{h}} \\
&= h^{n+2} \int_{\frac{a-x_0}{h}}^{\frac{b-x_0}{h}} u(u-1)\cdots(u-n)\,d u \\
&= h^{n+2}\,(n+1)!\,\psi_{n+2}(u)\,\Big|_{u_1=\frac{a-x_0}{h}}^{u_2=\frac{b-x_0}{h}} \\
&= h^{n+2}\,(n+1)!\,\left[\psi_{n+2}(u_2)-\psi_{n+2}(u_1)\right]
\end{align*}
And please do not run away in panic now. These $\psi_n(u)$ here are Bernoulli polynomials of the second kind, certain weird polynomials like Legendre polynomials, which have great properties that I want to use, see Wikipedia.
\begin{align}
(n+1)! \cdot \psi_{n+2}(u) = \int u(u-1)\cdots(u-n)\ d u
\end{align}
For the limits $u_1 = 0$, $u_2 = n$ we find the following expression:
\begin{align*}
\int_a^b \omega(x) \, d x &= h^{n+2}\,(n+1)!\,\left[\psi_{n+2}(n) - \psi_{n+2}(0)\right] \\
&= h^{n+2}\,(n+1)!\,\left[-|G_{n+2}| - G_{n+2}\right] \color{white}{\frac{1}{2}} \\
&= h^{n+2}\,(n+1)!\,\left[(-1)^n-1\right]\,|G_{n+2}| \color{white}{\frac{1}{2}}
\end{align*}
where $G_n$ are the Gregory coefficients defined as follows:
\begin{align*}
\frac{x}{\ln(1+x)} = 1 + \frac{1}{2}x - \frac{1}{12}x^2 + \frac{1}{24}x^3 - \ldots = 1 + \sum_{n=1}^\infty\,G_n\,x^n
\end{align*}
This means that for even $n$ we cannot find an expression. Thus, for even $n$, we must use a different approximation. For this we double the term $x_{n/2} = \frac{a+b}{2}$. So we calculate in such a way that we consider an additional grid point and thus a double zero for the interpolation polynomial.
\begin{align*}
\int_a^b \omega(x)\,d x &= h^{n+3} \int_{0}^{n} u(u-1)\cdots(u-\tfrac{n}{2})^2\cdots(u-n)\,d u \\
&= h^{n+3} \bigg[\underbrace{\int_0^{n} u\cdot u(u-1)\cdots(u-n)\,d u}_{\color{red}{=\,I}} - \frac{n}{2}\underbrace{\int_0^{n} u(u-1)\cdots(u-n)\,d u}_{\color{red}{=\,0}}\bigg]
\end{align*}
START HERE:
And now we have finally reached the point where I need your help. Your superhuman knowledge of mathematics is now required. Unfortunately, there was probably still no mathematician who has set up a suitable relation for me, which I can use to solve this integral:
\begin{align}
I = \int_0^{n} u\cdot u(u-1)\cdots(u-n)\,d u
\end{align}
I want to use this:
\begin{align}
(n+1)! \cdot \psi_{n+2}(u) = \int u(u-1)\cdots(u-n)\ d u
\end{align}
I had the following ideas:
*
*Integration by parts:
\begin{align}
I &= u\cdot (n+1)!\,\psi_{n+2}(u)\,\Big|_0^{n} - (n+1)! \int_0^{n} \psi_{n+2}(u)\,d u \\
&= (n+1)!\left[n\cdot\psi_{n+2}(n) - \int_0^n \psi_{n+2}(u)\,d u\right] \\
&= -(n+1)!\left[n\cdot|G_{n+2}| + \int_0^n \psi_{n+2}(u)\,d u\right]
\end{align}
But what is now
\begin{align}
\int_0^n \psi_{n+2}(u)\,d u \ ?
\end{align}
*Consider the case $n+1$
\begin{align}
(n+2)!\cdot\psi_{n+3}(u) &= \int u(x-1)(u-2)\cdots(u-n)(u-(n+1))\ du \\
&=\underbrace{\int u \cdot u(u-1)(u-2)\cdots(u-n) \ du}_{\color{red}{=\,I(u)}} - (n+1) \underbrace{\int u(u-1)(u-2)\cdots(u-n) \ du}_{\color{red}{=\,(n+1)!\,\cdot\,\psi_{n+2}(u)}} \\
&= I(u) - (n+1)\cdot(n+1)!\,\cdot\,\psi_{n+2}(u)
\end{align}
Thus applies:
\begin{align}
I(u) &= (n+1)\cdot(n+1)!\,\cdot\,\psi_{n+2}(u) + (n+2)! \cdot \psi_{n+3}(u) \color{white}{\frac{1}{2}} \\ \\
I &= I(u) \ |_0^n = I(n) - I(0) \\
&= (n+1)! \cdot (n+1) \,\underbrace{\left[\psi_{n+2}(n) - \psi_{n+2}(0)\right]}_{\color{red}{=\,0; \text{ $n$ is even}}} \color{white}{\frac{1}{2}} \\
&\hspace{15pt} + (n+2)! \left[\psi_{n+3}(n) - \psi_{n+3}(0)\right] \color{white}{\frac{1}{2}} \\
&= (n+2)! \left[\psi_{n+3}(n) - \psi_{n+3}(0)\right]
\end{align}
But what is now
\begin{align}
\psi_{n+3}(n) \ ?
\end{align}
|
According to Wikipedia:
\begin{align}
\psi_{2k}(k-1+y) = \psi_{2k}(k-1-y)
\end{align}
If we now choose $y = k - 2$, it follows:
\begin{align}
&\psi_{2k}(k-1+(k-2)) = \psi_{2k}(k-1-(k-2)) \color{white}{\frac{1}{2}}\\
& \Longrightarrow \hspace{10pt} \psi_{2k}(2k-3) = \psi_{2k}(1) \color{white}{\frac{1}{2}} \\
\end{align}
If we consider an even $n$ as described in my problem, i.e. $n = 2k$, the following holds:
\begin{align}
&\psi_{n}(n-3) = \psi_{n}(1) = G_{n-1} + G_n \\
&\psi_{n+3}(n) = \psi_{n+3}(1) = G_{n+2} + G_{n+3}
\end{align}
We get for the integral:
\begin{align}
I &= (n+2)! \cdot \left[\psi_{n+3}(n) - \psi_{n+3}(0)\right] \color{white}{\frac{1}{2}}\\
&= (n+2)! \cdot \left[G_{n+2} + G_{n+3} - G_{n+3}\right] \color{white}{\frac{1}{2}} \\
&= (n+2)! \cdot G_{n+2}
\end{align}
Thus, for Newton-Cotes formulas:
\begin{align*}
E_n[f] = \left\{\begin{array}{ll} \displaystyle -f^{(n+1)}(\xi)\,h^{n+2}\cdot 2|G_{n+2}|, & n \text{ ungerade} \\ \displaystyle -f^{(n+2)}(\xi)\,h^{n+3}\cdot G_{n+2}, & n \text{ ungerade}
\end{array}\right.
\end{align*}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4302421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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:
$EF = R - OE \\
\triangle AEF: EF^2 + OE^2 = AF^2\implies\\
(R-OE)^2 + (\frac{9}{2})^2 = AF^2 \implies:
(R-OE)^2 + \frac{81}{4} = AF^2\\
\triangle AOE: OE^2+AE^2 = AO^2 \implies OE^2 +(\frac{9}{2})^2 =R^2 \implies OE^2+ \frac{81}{4} = R^2\\
\triangle AOF: AF^2 = R^2 +OF^2-2OF.OE \implies\\
AF^2 = R^2 +R^2-2R.OE \therefore AF^2 = 2R^2-2R.OE = 2R\underbrace{(R-OE)}_{=EF}$
...??
|
*
*$\angle ABC=\frac12\angle AOC=\angle AOE$
*$\sin B=\sin\angle AOE=\frac{AC/2}{R}=\frac{b}{2R}$
*Similarly
$$2R=\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}$$
*Area of $\triangle ABC$ is $$K=\frac12ab\sin C=\frac{abc}{4R}$$
*Use Heron's formula to calculate area $K$
*Get $R$
*Use Pythagoras to get $OE$
*Get $EF$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4302567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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|
Find the value of $ \frac{1}{{2(2^2 - 1)}} + \frac{1}{{3(3^2 - 1)}} + \frac{1}{{4(4^2 - 1)}} + \cdots \;. $ To find the value of $$
\frac{1}{{2(2^2 - 1)}} + \frac{1}{{3(3^2 - 1)}} + \frac{1}{{4(4^2 - 1)}} + \cdots \;.
$$
I presented it as $$
\sum\limits_{n = 2}^\infty {\frac{1}{{n(n^2 - 1)}}} .
$$
Then using partial fractions method
I wrote the sum as
$$
\sum\limits_{n = 2}^\infty {\left( { - \frac{1}{n} + \frac{1}{{2(n + 1)}} + \frac{1}{{2(n - 1)}}} \right)} .
$$
Then I wrote down the individual terms to see if most of the terms cancel out like they do sometimes.
$$
- \frac{1}{2} + \frac{1}{6} + \frac{1}{2} - \frac{1}{3} + \frac{1}{8} + \frac{1}{4} - \frac{1}{4} + \frac{1}{{10}} + \frac{1}{6} - \frac{1}{5} + \frac{1}{{12}} + \frac{1}{8} - \frac{1}{6} + \frac{1}{{14}} + \frac{1}{{10}} + \cdots \; .
$$
Unfortunately they don't seem to cancel out. Is there a way to find out the sum in this method or any other method?
|
You can continue with the partial sums as
\begin{align*}
\sum\limits_{n = 2}^N {\left( { - \frac{1}{n} + \frac{1}{{2(n + 1)}} + \frac{1}{{2(n - 1)}}} \right)} & = \sum\limits_{n = 2}^N {\left( {\left( {\frac{1}{{2(n - 1)}} - \frac{1}{{2n}}} \right) + \left( {\frac{1}{{2(n + 1)}} - \frac{1}{{2n}}} \right)} \right)} \\ & =
\sum\limits_{n = 2}^N {\left( {\frac{1}{{2(n - 1)}} - \frac{1}{{2n}}} \right)} + \sum\limits_{n = 2}^N {\left( {\frac{1}{{2(n + 1)}} - \frac{1}{{2n}}} \right)} \\
& = \frac{1}{4}- \frac{1}{{2N}} + \frac{1}{{2(N + 1)}} = \frac{1}{4} - \frac{1}{{2N(N + 1)}}.
\end{align*}
Now take the limit $N\to +\infty$ and conclude.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4303104",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
find derivative of $\sqrt{x}+\sqrt{y}=\sqrt{a}$ I have an Implicit Function $\sqrt{x}+\sqrt{y}=\sqrt{a}$
the graph of the function is
I need to prove that $p+q=a$
and I need to find $\frac{d}{dx}$ to find the the slop to prove that.
result:
after rearranged the function I got
$a=x+2\sqrt{xy}+y$
and the derivative of that function is
$\frac{d}{dx}=\frac{-\sqrt{y}}{\sqrt{x}}$
how can I prove $p+q=a$ with $\frac{d}{dx}$?
|
Let $y'$ denote $\displaystyle\frac{dy}{dx}.$
Given: $\sqrt{x} + \sqrt{y} = \sqrt{a}.$
Using implicit differentiation, you have that
$\displaystyle \frac{1}{2\sqrt{x}} + \frac{y'}{2\sqrt{y}} = 0 \implies y' = \frac{-\sqrt{y}}{x} =
\frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}.$
The distance $Q$ may be represented by the point $(q,0)$.
Then, the slope going from $(x,y)$ to $(q,0)$ is $y'$.
Therefore,
$\displaystyle \frac{-y}{q - x} = \frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}
\implies
\frac{-\left(\sqrt{a} - \sqrt{x}\right)^2}{q - x} = \frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}.$
Dividing both sides by
$\displaystyle -\left(\sqrt{a} - \sqrt{x}\right)$
gives
$\displaystyle \frac{\sqrt{a} - \sqrt{x}}{q - x} = \frac{1}{\sqrt{x}} \implies
q - x = \sqrt{ax} - x \implies q = \sqrt{ax}.$
The distance $P$ may be represented by the point $(0,p)$.
Then, the slope going from $(0,p)$ to $(x,y)$ is $y'$.
Therefore,
$\displaystyle \frac{y - p}{x - 0} = \frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}
\implies
\frac{\left(\sqrt{a} - \sqrt{x}\right)^2 - p}{x} = \frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}.$
This implies that
$\displaystyle \frac{\left(\sqrt{a} - \sqrt{x}\right)^2 - p}{\sqrt{x}}
= \frac{-\left(\sqrt{a} - \sqrt{x}\right)}{1}.$
This implies that
$\displaystyle a + x - 2\sqrt{ax} - p = -\sqrt{ax} + x \implies
a - \sqrt{ax} = p.$
Thus, $\displaystyle (p + q) = \left(a - \sqrt{ax}\right) + \sqrt{ax} = a.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4304423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $
$\implies r_2^2+2r_2x+x^2=PF^2+r_2^2-2r_2x+x^2$
$\therefore \boxed{4r_2x = PF^2}$
...?
|
$PG = r_1 - r, PR = r_1 + r$ and so by Pythagoras, it is easy to see that $RG = 2 \sqrt{r r_1}$
Now if $O$ is the center of the circle with radius $R$,
$AG = 2r_1 - r ~$ and
$ |OG| = |AG - AO| = |2r_1 - r - R| ~ $
By Pythagoras, $OG^2 = OR^2 - RG^2$
$\implies (2r_1 - r - R)^2 = (R-r)^2 - 4 r r_1$
Solving, $R r = R r_1 - r_1^2 \implies R r = r_1 (R - r_1)$. But as $R = r_1 + r_2$,
$R r = r_1 r_2 \tag1$
Now $KQ = r_2 - x, QS = r_2 + x$ and we obtain $SK = 2 \sqrt{xr_2}$
$OK = OB - KB = R - (2r_2 - x)$
$OK^2 = (R - 2r_2 + x)^2 = OS^2 - SK^2 = (R-x)^2 - 4 xr_2$
Solving, $Rx = r_2 (R - r_2) = r_1 r_2 \tag2$
From $(1)$ and $(2)$, we conclude that $r = x$.
|
{
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"url": "https://math.stackexchange.com/questions/4304972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove, for every nonnegative integer $n$, that $5^{2n} + 2^{2n} ≡ 2^{2n+1}\pmod{21}$. Is my proof correct? Thanks in advance.
Base step: n = 0
$$5^{2 \times 0} + 2^{2 \times 0} ≡ 2^{2 \times 0+1} \pmod{21}$$
$$= 1 + 1 ≡ 2 \pmod{21} \checkmark$$
Inductive hypothesis: assume $5^{2n} + 2^{2n} ≡ 2^{2n+1} \pmod{21}$ is true when $n=k$:
$$ 5^{2k} + 2^{2k} ≡ 2^{2k+1} \pmod{21}$$
Since the statement above is true we can write the following:
$$5^{2k} + 2^{2k} - 2^{2k+1} = 21x$$
Simplify and rewrite:
$$2^{2k} = 5^{2k}-21x$$
Now we proceed with the inductive step: $n = k + 1$:
$$5^{2(k+1)} + 2^{2(k+1)} ≡ 2^{2(k+1)+1} \pmod{21}$$
$$5^{2k+2} + 2^{2k+2} ≡ 2^{2k+3} \pmod{21}$$
Rephrase the statement above:
$$21 \mid (5^{2k+2} + 2^{2k+2} - 2^{2k+3})$$
$$21 \mid (5^{2k+2} - 4 \times 2^{2k})$$
Replace $2^{2k}$ with $5^{2k}-21x$:
$$21 \mid (5^{2k+2} - 4 \times (5^{2k}-21x))$$
Multiply out:
$$21 \mid (5^{2k+2} - 4\times 5^{2k} + 4\times 21x))$$
$$ = 21 \mid (25 \times 5^{2k} - 4\times 5^{2k} + 4\times 21x))$$
$$ = 21 \mid (21 \times 5^{2k} + 4\times 21x))$$
$$ = 21 \mid 21(5^{2k} + 4x))\checkmark$$
Hence
$$5^{2(k+1)} + 2^{2(k+1)} ≡ 2^{2(k+1)+1} \pmod{21}$$
$\blacksquare$
|
The proof you gave is correct.
However, it can actually be done with a lot less work and without induction:
$$5^{2n}=\left(5^2\right)^n=25^n\equiv 4^n=2^{2n} (\mathrm{mod}\ 21)$$
Hence $5^{2n}+2^{2n}\equiv 2\cdot 2^{2n}=2^{2n+1} (\mathrm{mod}\ 21)$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Laurent expansion on annulus Let $F(z)= \frac{1}{(z-1)^2(z+2)}$.
I need to find a Laurent expansion for $F$ on the annulus $A = \{z: \sqrt{2}<|z-i|<\sqrt{5}$}.
So far I have managed to write $F$ as $F(z)=\frac{-1}{9(z-1)}+\frac{1}{9(z+2)}+\frac{1}{3(z-1)^2}$. Then writing $z-i = w$, I got the following expansions:
$\frac{-1}{9(z-1)}$ = $\frac{-1}{9w} \sum_{n=0}^{\infty} (-1)^{n}(\frac{i-1}{w})^n $
$\frac{1}{3(z-1)^2}$ = $\frac{1}{3w^2} \sum_{n=0}^{\infty} (-1)^{n}(1+n)(\frac{i-1}{w})^n$.
But how do we get the expansion for $\frac{1}{9(z+2)}$?
|
You are interested in the Laurent series centered at $i$, rather than $0$. So, use the fact that\begin{align}F(z)&=F\bigl((z-i)+i\bigr)\\&=\frac1{\bigl((z-i)+i-1\bigr)^2\bigl((z-i)+i+2\bigr)}\\&=\frac1{9\bigl((2+i)+(z-i)\bigr)}+\frac1{9\bigl((1-i)-(z-i)\bigr)}+\frac1{3\bigl((1-i)-(z-i)\bigr)^2}\\&=\frac1{9(2+i)}\frac1{1-\frac{z-i}{2+i}}+\frac1{9(1-i)}\frac1{1-\frac{z-i}{1-i}}+\frac1{3(1-i)^2}\frac1{\left(1-\frac{z-i}{1-i}\right)^2}\\&=\frac1{9(2+i)}\sum_{n=0}^\infty\left(\frac{z-i}{2+i}\right)^n-\frac1{9(1-i)}\sum_{n=-\infty}^{-1}\left(\frac{z-i}{1-i}\right)^n+\\&\phantom{=}\quad-\frac1{3(1-i)^2}\sum_{n=-\infty}^{-2}(n+1)\left(\frac{z-i}{1-i}\right)^n.\end{align}It's only at the last step that I've used the fact that $\sqrt2<|z-i|<\sqrt5$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4307562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find the sum of the squares of the medians of triangle APC below For reference: The diameter of a circle measures 8m. On this diameter are located the points A and B equidistant 1 m from the center; through B draw any chord PC , determine the sum of the squares of the medians of triangle APC.(Answer: $73,5$)
My progress:
sum of the squares of the median
$4(PE^2+CD^2+AB^2) = 3(AC^2+AP^2+CP^2)\therefore\\
PE^2+CD^2+AB^2=\frac{3}{4}(AC^2+AP^2+CP^2)\\
AC = AP \implies PE^2+CD^2+AB^2=\frac{3}{4}(2AC^2+CP^2)\\
CB=\frac{CP}{2} \\
\triangle ABC: 4+(\frac{CP}{2})^2=AC^2\implies \frac{16+CP^2}{4} =AC^2\\
\therefore PE^2+CD^2+AB^2 = \frac{3}{8}(16+3CP^2)
$
...??
|
If $O$ is the center of the circle, $OP$ is median of $\triangle APB$. Radius of the circle is $4$.
So, $4 OP^2 = 64 = 2 AP^2 + 2 PB^2 - AB^2$
$ \implies AP^2 + PB^2 = 34$
Similarly, $AC^2 + BC^2 = 34$
Now by intersecting chords theorem, $PB \cdot BC = 5 \cdot 3 = 15$
So, $AP^2 + AC^2 + PB^2 + BC^2 + 2 PB \cdot BC = 98$
$\implies AP^2 + AC^2 + CP^2 = 98$
Given you know the sum of squares of the sides of the triangle, you can find the sum of squares of the medians.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4307773",
"timestamp": "2023-03-29T00:00:00",
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|
Can a ratio of sinusoidal functions with the same frequency always be written as a tangent function? In general, two sinusoidal functions $y_1$ and $y_2$ with angular frequency $\omega$ can be written as
*
*$y_1 = A\sin(\omega x) + B \cos(\omega x)$,
*$y_2 = C\sin(\omega x) + D \cos(\omega x)$.
Where $A, B, C, D$ are real numbers and it is not the case that $A=B=0$ or $C=D=0$.
Can the ratio $\frac{y_1}{y_2}$ always be written in the form
$$ \frac{y_1}{y_2} = E \tan(\omega x + h) + k$$
for some constants $E, h$, and $k$?
Edit:
For example
$$\frac{12 \sin (x) + 3 \cos(x)}{ 4 \sin(x) + 2 \cos(x)}$$
has a graph like this:
Which looks like the tangent function transformed by scaling and shifting.
|
Yes: for $C,D$ not both zero, $$\frac{A\sin \omega x+B\cos \omega x}{C\sin \omega x+D\cos \omega x}=\frac{AD-BC}{C^2+D^2}\;\tan\left(\omega x-\arctan\frac CD\right)+\frac{AC+BD}{C^2+D^2}.$$ This is true even when $D=0,$ if we let $\displaystyle\arctan\frac CD=\begin{cases}\displaystyle&\frac\pi2&\text{if }C>0;
\\ \displaystyle-&\frac\pi2 &\text{if }C<0.\end{cases}$
*
*The above
expressions
for the parameters $E,h,k$ were obtained by equating
$$\frac{A\sin \omega x+B\cos \omega x}{C\sin \omega x+D\cos \omega x}$$ and the given tangent function $$E \tan(\omega x + h) + k,$$ expanding the
latter into sines and cosines, observing from the denominators that
$$\frac CD=-\tan h,$$ and setting $$h=-\arctan\frac CD.$$
Now, $h$ is in either quadrant $4$ or $1$ depending on the relative
signs of $C$ & $D,$ so $$\sin h=\frac{\mp C}{C^2+D^2},\\\cos
h=\frac{D}{C^2+D^2}.$$
Substituting these two equalities into the equation where we left off
gives $$\sqrt{C^2+D^2}(E\cos h-k\sin h)=A,\\\sqrt{C^2+D^2}(E\sin
h+k\cos h)=B.$$ Solving this system finally gives the above
expressions for $E$ and $k.$
*Here's a rigorous proof: $$\frac{AD-BC}{C^2+D^2}\;\tan\left(\omega
x-\arctan\frac CD\right)+\frac{AC+BD}{C^2+D^2}\\
=\frac{AD-BC}{C^2+D^2}\left(\frac{\tan \omega x-\frac CD}{1+(\tan \omega x)\frac CD}\right)+\frac{AC+BD}{C^2+D^2}\\
=\frac{AD-BC}{C^2+D^2}\left(\frac{D\sin \omega x-C\cos \omega x}{D\cos \omega x+C\sin \omega x}\right)+\frac{AC+BD}{C^2+D^2}\\
=\frac{(AD-BC)(D\sin \omega x-C\cos \omega x)+(AC+BD)(D\cos \omega x+C\sin \omega x)}{(C^2+D^2)(C\sin \omega x+D\cos \omega x)}\\
=\frac{(AD^2-BCD+AC^2+BCD)\sin \omega x+(-ACD+BC^2+ACD+BD^2)\cos \omega x}{(C^2+D^2)(C\sin \omega x+D\cos \omega x)}\\
=\frac{A\sin \omega x+B\cos \omega x}{C\sin \omega x+D\cos \omega x}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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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!
|
There is a clever solution here: it requires using AM-GM.
Assume that $\sqrt{3x+2} \geq 0$, which clearly must be true. Then, By AM-GM, we have
$$
\sqrt{3x+2} + \frac{x^2}{\sqrt{3x+2}} \geq 2\sqrt{x^2} = 2x
$$
with equality if and only if $\sqrt{3x+2} = \frac{x^2}{\sqrt{3x+2}}$. This is true when $x^2 = 3x+2$, or $x = \frac{3 \pm \sqrt{17}}{2}$. Clearly, we must have $x \geq \frac{-2}{3}$, so the only real solution is $\frac{3 + \sqrt{17}}{2}$.
|
{
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"url": "https://math.stackexchange.com/questions/4309941",
"timestamp": "2023-03-29T00:00:00",
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|
For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that...
For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that $$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$$
Here's what I've done so far:
$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}+3\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
Not sure where to go from here, any help's appreciated. I think the $AM-GM$ inequality should be used here in some way.
|
$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)=$ (1)
We know that $a+b+c=1$, so
$b+c=1-a$,
$a+c=1-b$,
$a+b=1-c$.
$(1)=\frac{1-a}{a}+2+\frac{1-b}{b}+2+\frac{1-c}{c}+2\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
We know from the $A.M-G.M$ inequality that
$\frac{1-x}{x}+2\ge2\sqrt{2*\frac{1-x}{x}}$
So we have
$\frac{1-a}{a}+2\ge2\sqrt{2*\frac{1-a}{a}}$
$\frac{1-b}{b}+2\ge2\sqrt{2*\frac{1-b}{b}}$
$\frac{1-c}{c}+2\ge2\sqrt{2*\frac{1-c}{c}}$
Summing them we get
$\frac{1-a}{a}+2+\frac{1-b}{b}+2+\frac{1-c}{c}+2\ge2\sqrt{2*\frac{1-a}{a}}+2\sqrt{2*\frac{1-b}{b}}+2\sqrt{2*\frac{1-c}{c}}$
$=\frac{1-a}{a}+2+\frac{1-b}{b}+2+\frac{1-c}{c}+2\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
So we've proved the original inequality.
|
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|
A special inequality (related to 2021) Let $x,y,z\in[0,1]$ such that: $$x+y+z\leq 1+2xyz$$
I want to prove the following inequality (for a student in high school):
$$x^{2021}+y^{2021}+z^{2021}\leq 1+2\sqrt{xyz}^{2021}$$
Edit:
This question was sent to me by a student from high school, his teacher ask them to look for proof. I think it's a kind of question given in "International Mathematical Olympiad", that's why the "2021" number is used here. I tried by recurrence..
|
Alternative proof:
Let us prove that, for any integer $n \ge 2$,
$$x^n + y^n + z^n \le 1 + 2 \sqrt{xyz}^n.$$
Let
$$f(n) = 1 + 2\sqrt{xyz}^n - x^n - y^n - z^n.$$
We have
$$f(n + 1) - f(n) = 2\sqrt{xyz}^{n + 1} - x^{n + 1} - y^{n + 1} - z^{n + 1} - 2\sqrt{xyz}^n + x^n + y^n + z^n,$$
and
\begin{align*}
&[f(n + 2) - f(n + 1)] - [f(n + 1) - f(n)]\\
=\,\, & 2\sqrt{xyz}^n (1 - \sqrt{xyz})^2
- x^n(1 - x)^2 - y^n(1 - y)^2 - z^n(1 - z)^2\\
\le\,\,& 2\sqrt{xyz}^n (1 - \sqrt{xyz})^2 -
3\sqrt[3]{x^n(1 - x)^2 \cdot y^n(1 - y)^2 \cdot z^n(1 - z)^2}\\
=\,\,& (xyz)^{n/3}
\left[2(xyz)^{n/6} (1 - \sqrt{xyz})^2 - 3\sqrt[3]{(1 - x)^2(1 - y)^2(1 - z)^2}\right]\\
\le\,\,& (xyz)^{n/3}
\left[2(xyz)^{2/6} (1 - \sqrt{xyz})^2 - 3\sqrt[3]{(1 - x)^2(1 - y)^2(1 - z)^2}\right]\\
\le\,\,& (xyz)^{n/3}
\left[2(xyz)^{2/6} (1 - \sqrt{xyz})^2 - 2\sqrt[3]{(1 - x)^2(1 - y)^2(1 - z)^2}\right]\\
\le\,\,& 0. \tag{1}
\end{align*}
(The proof of (1) is given at the end.)
Thus, we have
$$f(n + 2) - f(n + 1) \le f(n + 1) - f(n), \quad \forall n \ge 2.$$
Also, clearly $\lim_{n\to \infty} [f(n + 1) - f(n)] = 0$.
Thus, we have
$$f(n + 1) - f(n) \ge 0, \quad \forall n\ge 2.$$
Also, we have
$f(2) = 1 + 2xyz - x^2 - y^2 - z^2 \ge 1 + 2xyz - x - y - z \ge 0$.
Thus,
$$f(n) \ge f(2) \ge 0, \quad \forall n \ge 2.$$
We are done.
Proof of (1):
It suffices to prove that
$$\sqrt{xyz} (1 - \sqrt{xyz})^3 \le (1 - x)(1 - y)(1 - z).$$
Let $p = x + y + z, q = xy + yz + zx, r = xyz$.
We have $1 + 2r \ge p$.
Using $q^2 \ge 3pr$, we have
\begin{align*}
&(1 - x)(1 - y)(1 - z)\\
=\,\,& 1 - xyz + (xy + yz + zx) - (x + y + z)\\
=\,\,& 1 - r + q - p\\
\ge\,\,& 1 - r + \sqrt{3pr} - p\\
=\,\,& 1 - r - \left(\sqrt{p} - \frac12\sqrt{3r}\right)^2 + \frac34 r\\
\ge\,\,& 1 - r - \left(\sqrt{1 + 2r} - \frac12\sqrt{3r}\right)^2 + \frac34 r
\end{align*}
where we have used $1 + 2r \ge p$ and $\sqrt{p} \ge \frac{1}{2}\sqrt{3r}$.
It suffices to prove that
$$\sqrt{r} (1 - \sqrt{r})^3 \le 1 - r - \left(\sqrt{1 + 2r} - \frac12\sqrt{3r}\right)^2 + \frac34 r$$
or
$$\sqrt{r}\left(\sqrt{3 + 6r} - 1 - 3r + r\sqrt{r}\right)\ge 0$$
which is true (using $r\in [0, 1]$).
We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int_{-\infty}^\infty f(x)dx$ vs. $\lim_{b\rightarrow\infty}\int_{-b}^b f(x)dx$ for odd $f(x)$. I think the below one is a problem which demonstrates a beautiful subtlety (at least for someone like me who's a beginner).
Show that
$$
\int_{-\infty}^\infty \frac{2x}{1+x^2}dx
$$
diverges and
$$
\lim_{b\rightarrow\infty}\int_{-b}^b \frac{2x}{1+x^2}dx=0
$$
My Solution/Proof:
Let's start with
$$
\begin{aligned}
\lim_{b\rightarrow\infty}\int_{-b}^b \frac{2x}{1+x^2}dx &= \lim_{b\rightarrow\infty} \left[\ln(1+x^2)\right]_{-b}^b \\
&= \lim_{b\rightarrow\infty}(0) \\
&= 0
\end{aligned}
$$
That's one part. Next we have
$$
\begin{aligned}
\int_0^\infty \frac{2x}{1+x^2}dx &= \lim_{a\rightarrow\infty}\int_0^a\frac{2x}{1+x^2}dx \\
&= \lim_{a\rightarrow\infty}\ln(1+a^2) \\
&= \infty
\end{aligned}
$$
and also
$$
\begin{aligned}
\int_{-\infty}^0\frac{2x}{1+x^2}dx &= \lim_{b\rightarrow-\infty}\int_b^0\frac{2x}{1+x^2}dx \\
&= \lim_{b\rightarrow-\infty}\left[-\ln(1+b^2)\right] \\
&= -\infty
\end{aligned}
$$
Therefore
$$
\begin{aligned}
\int_{-\infty}^\infty \frac{2x}{1+x^2}dx &= \int_{-\infty}^0\frac{2x}{1+x^2}dx+\int_0^\infty \frac{2x}{1+x^2}dx \\
&= \lim_{b\rightarrow-\infty}\int_b^0\frac{2x}{1+x^2}dx+\lim_{a\rightarrow\infty}\int_0^a\frac{2x}{1+x^2}dx \\
&= -\infty+\infty \\
&= \text{undefined}
\end{aligned}
$$
and hence the assertion is proven. But...but...but...
But consider
$$
\begin{aligned}
\lim_{b\rightarrow-\infty}\int_b^0\frac{2x}{1+x^2}dx+\lim_{a\rightarrow\infty}\int_0^a\frac{2x}{1+x^2}dx &= \lim_{b\rightarrow-\infty}\left[-\ln(1+b^2)\right]+\lim_{a\rightarrow\infty}\ln(1+a^2) \\
& = \lim_{\color{red}{b\rightarrow\infty}}\left[-\ln(1+b^2)\right]+\lim_{a\rightarrow\infty}\ln(1+a^2)
\end{aligned}
$$
because $\ln(1+x^2)$ is even. Now, if $a=b$, this limit would become zero. Right?
If so, what exactly is stopping me from considering $a=b$ and hence the second limit as zero? I mean what's stopping the second integral from converging?
|
No. The symbols $a$ and $b$ are just variables. You could also write
$$
\int_{-\infty}^{\infty} \frac{2 x}{1 + x^2} \ \text{d}x= \left(\lim_{c \to \infty} - \ln(1 + c^2)\right)
+ \left(\lim_{c \to \infty} \ln(1 + c^2)\right).
$$
But this is not equal to
$$
\lim_{c \to \infty} \left(- \ln(1 + c^2) + \ln(1 + c^2)\right)
= \lim_{c \to \infty} 0
= 0.
$$
The sum of the limits is only the limit of the sum, if both limits exist and are finite, which is not the case here.
A more simple example for this statement is the following:
$$
0
= \lim_{n \to \infty} 0
= \lim_{n \to \infty} \left(n + \left(- n\right)\right)
\ne \left(\lim_{n \to \infty} n\right) + \left(\lim_{n \to \infty} - n\right)
= \infty + (- \infty)
= \text{undefined.}
$$
|
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|
Number of solution of ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ Number of solution of the equation ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ in the interval $[0,2\pi]$ is equal to_____
My approach is as follow
$a = \sin x - 1;b = \cos x - 1;c = \sin x$
${a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3}$
$ \Rightarrow {a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3}$
${\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 6abc + 3ab\left( {a + b} \right) + 3bc\left( {b + c} \right) + 3ac\left( {a + c} \right)$
${a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3} \Rightarrow 6abc + 3ab\left( {a + b} \right) + 3bc\left( {b + c} \right) + 3ac\left( {a + c} \right) = 0$
$ \Rightarrow 2abc + ab\left( {a + b} \right) + bc\left( {b + c} \right) + ac\left( {a + c} \right) = 0$
How do we approach from here
|
Hint
$$(a+b+c)^3-c^3=(a+b+c-c)((a+b+c)^2+(a+b+c)c+c^2)$$
$$a^3+b^3=?$$
Clearly $a+b$ is a common factor.
By symmetry, we can claim that the other two factors of $$(a+b+c)^3-a^3-b^3-c^3$$ will be $b+c,c+a$
Alternatively simplify
$$((a+b+c)^2+(a+b+c)c+c^2)-(a^2-ab+b^2)$$
|
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|
How do we calculate the expected value of $Y$? Let $\Omega = \{−1, 0, 1, 3\}$ with the probability function $p$ given by
I want to calculate the expected value and the variance of $X(\omega)=\omega$, $Y(\omega)=5\omega-3$, $Z(\omega)=(\omega-1)^2$.
For $X$ I have done the following :
We have that $X\in \{-1,0,1,3\}$.
The expected value is \begin{equation*}E[X]=\sum_{x\in X(\Omega)}xP[X=x]=(-1)\cdot \frac{1}{10}+0\cdot \frac{3}{10}+1\cdot \frac{2}{10}+3\cdot \frac{4}{10}= \frac{13}{10}\end{equation*}
The variance is \begin{align*}Var[X]&=\sum_{x\in X(\Omega)}(x-E[X])P[X=x]\\ & =\left (-1-\frac{13}{10}\right )\cdot \frac{1}{10}+\left (0-\frac{13}{10}\right )\cdot \frac{3}{10}+\left (1-\frac{13}{10}\right )\cdot \frac{2}{10}+\left (3-\frac{13}{10}\right )\cdot \frac{4}{10}\\ & =0\end{align*}
Doe $Y$ we have $Y=5\omega-3\in \{-8,-3,2,12\}$, or not? But then we don't knowthe probabilities.
Or do weuse the property $E[aX+b]=aE[X]+b$ ?
|
You can use the formula $$\mathbb{E}[aX+b] = a\mathbb{E}[X]+b$$ to see that
$$\mathbb{E}[Y] = 5\cdot \frac{13}{10} -3 = \frac{7}{2}\ .$$
However, you can also calculate it directly because you do have the probabilities for $Y$. For example
$$\Pr(Y=-8) = \Pr(5\omega-3 = -8) = \Pr(\omega=-1)\ .$$
Since $Y$ is defined in terms of $\omega$, its distribution is also determined by the distribution for $\omega$. So once you get the distributions for $Y$ and $Z$ you can calculate their expectations and variances just like you did for $X$.
|
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|
Show that $\ln(\sin(-20)+21)>3$ by hand Hi I hope this problem is new :
Show that :
$$\ln(\sin(-20)+21)>3$$
I have tried the power series of $\ln(x)$ and $\sin(x)$ without success because it's becomes hard by hand .
You can also find the inequality due to Michael Rozenberg wich states for $x\geq 1$ :
$$\ln(x)\leq (x-1)\left(\frac{2}{x^2+x}\right)^{\frac{1}{3}}$$
But unfortunately it's an upper bound not a lower bound so maybe we can play with it .
Edit :
We have :
$$k=\sin(-20)+21$$
Then it seems we have for $5\leq x\leq k$ :
$$\left(\frac{2x}{2+x}\left(1-b\right)+b\frac{\left(x-1\right)}{\sqrt{x}}\right)<\ln(x)$$
Where $b=0.484092$
I used the lower bound
Use Jensen's inequality to show $\frac{2x}{2+x} < \log(1+x) < \frac{2x+x^2}{2+2x}$ for $x>0$
And another well-know upper bound for the logarithm .
Question :
How to show it by hand (without a calculator) ?
|
Let's approximate $\sin(-20)$ using a Taylor series centered near $-20$. Since $\pi \approx 3.14$, we have $-6.5\pi \approx -20.41$, so this is what we'll use as the center of the approximation. Note that $\sin(-6.5\pi) = -1$ and $\cos(-6.5\pi) = 0$. So we have
$$\sin(x) = \sum_{k=0}^{\infty} (-1)^k \frac{\sin(-6.5\pi)}{(2k)!} \cdot (x + 6.5\pi)^{2k} = \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(2k)!} (x + 6.5\pi)^{2k}.$$
We'll use the quadratic approximation:
$$\sin(x) = -1 + \frac{1}{2} \cdot (x + 6.5\pi)^2 + O((x + 6.5\pi)^4).$$
Since $3.14159 < \pi < 3.14160$, we have $0.42 < -20 + 6.5\pi < 0.421$. We compute
$$\sin(-6.5\pi + 0.42) \approx -1 + \frac{1}{2} \cdot 0.42^2 = -0.9118$$
and
$$\sin(-6.5\pi + 0.421) \approx -1 + \frac{1}{2} \cdot 0.421^2 = -0.9113795.$$
By Taylor's theorem, these approximations have error bounded above by $0.421^4/4! < 0.002$. So
$$-0.914 < \sin(-6.5\pi + 0.42) < \sin(-20) < \sin(-6.5\pi + 0.421) < -0.909.$$
Thus,
$$20.086 < \sin(-20) + 21 < 20.091.$$
At this point, we could try to approximate the natural logarithm, but its Taylor series converges rather slowly, so it's more efficient to bound $e^3$. We have $e < 2.7183$, so
$$e^3 < 2.7183^3 = 20.08593... < 20.086 < \sin(-20) + 21.$$
Thus $3 < \ln(\sin(-20) + 21)$.
|
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|
Find the area of the shaded region in the triangle below? For reference: In figure $G$ is the centroid of the triangle $ABC$; if the area of the $FGC$ triangle is $9m^2$, the area of the FGB triangle is $16m^2$
Calculate the area of the shaded region. (Answer:$7m^2$)
If possible by geometry
My progress:
$S_{FGC} = \frac{b.h_1}{2} = \frac{FG.h_1}{2}\implies FG = \frac{18}{h_1}\\
S_{FGB}=\frac{b.h_2}{2} = \frac{FG.h_2}{2} \implies FG = \frac{32}{h_2}\\
\therefore \frac{18}{h_1} = \frac{32}{h_2}\implies \frac{h_1}{h_2} = \frac{32}{18}=\frac{16}{9}\\
S_{ABG} = S_{BCG} = S_{ACG}$
...??? I'm not able to develop this
|
$\begin{array}{} A=(p,q) & B=(a,0) & C=(0,0) & F=(x,y) \end{array}$
$S_{FGC}=|\frac{1}{2}\left| \begin{array}{} x & y & 1 \\ x_{G} & y_{G} & 1 \\ 0 & 0 & i \\ \end{array} \right| |=9 \\ S_{FGB}=|\frac{1}{2}\left| \begin{array}{} x & y & 1 \\ x_{G} & y_{G} & 1 \\ a & 0 & i \\ \end{array} \right| |=16$
$\begin{array}{} |x·y_{G}-x_{G}·y|=18 & & |x·y_{G}-x_{G}·y+a·y-a·y_{G}|=32 \end{array}$
Solving the modular equations we get four solutions. Replace $x_{G}=\frac{a+p}{3}$ and $y_{G}=\frac{q}{3}$
$\begin{array}{} \left( x=\frac{aq(a+p)±6(2a-25p)}{3a},y=\frac{aq∓150}{3a} \right) & ⇒ & S_{AFG}=7 \\ \left( x=\frac{aq(a+p)±6(34a+7p)}{3a},y=\frac{aq±42}{3a} \right) & ⇒ & S_{AFG}=25 \\ \end{array}$
|
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|
Prove instability using Lyapunov function Consider the system:
\begin{align}
x' &= x^3 + xy \\
y' &= -y + y^2 + xy - x^3 \\
\end{align}
I want to prove the origin is an unstable point by using the Lyapunov function $V(x,y) = \tfrac{x^4}{4} - \tfrac{y^2}{2}$ (this is a hint provided in the exercise).
In order to use Chetaev instability theorem I would like to prove that there is a domain $U$ in a punctured neighborhood of $0$ such that
$$V'(x,y) = \frac{\partial E}{\partial x} (x^3 + xy) + \frac{\partial E}{\partial y} (-y + y^2 + xy - x^3) = x^6 + x^4y + y^2 - y^3 -xy^2 + x^3y$$
is strictly positive, which Wolfram Alpha confirms (indeed, that function is strictly positive in a disk around $0$).
I have tried to show that $0$ is a local minimum, but the Hessian test is not conclusive as the Hessian is positive semidefinite. I have also tried to bound the expression below by $0$ without any luck.
Could you give me any hint to proceed? Thanks in advance
|
The dominant term is $x^6+ x^3 y+ y^2$. This term is positive (complete the square). All other terms are small when $x,y,$ are small, so they can be controlled. Here are the details.
Write
$$ \begin{aligned} V'(x,y) &= x^6 + x^4y + y^2 - y^3 -xy^2 +
x^3y\\ &= x^6 + (1+x)x^3y +(1 - x-y)y^2 \\ & \ge x^6 +(1+x)x^3y + \frac{1}{2}y^2 \quad \text{if} \; |x| + |y| \le \frac{1}{4}\\ &\ge x^6-\frac{5}{4}|x|^3|y| + \frac{1}{2}y^2 \quad \text{since} \; |x| \le \frac{1}{4}\\ &= (|x|^3 - \frac{5}{8}|y|)^2 + \frac{7}{64}y^2 \\ & \ge 0
\end{aligned}
$$
with equality iff $x = y = 0$.
|
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|
Smith normal form of matrix over $Z$? I was wondering if someone could help me find the Smith normal form of the matrix A over $Z$ defined as follows:
$A =
\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
1 & 2 & 4 & 8 & 16\\
1 & 3 & 9 & 27 & 81\\
1 & 4 & 16 & 64 & 256\\
1 & 5 & 25 & 125 & 625\\
\end{bmatrix}$
Clearly this is the same as:
$\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
1 & 2 & 2^2 & 2^3 & 12^4\\
1 & 3 & 3^2 & 3^3 & 3^4\\
1 & 4 & 4^2 & 4^3 & 4^5\\
1 & 5 & 5^2 & 5^3 & 5^4\\
\end{bmatrix}$
Is there some "trick" I am unaware of? Through row and column operations I just get stuck at a matrix looking like this:
$\begin{bmatrix}
1 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 1 & 1\\
0 & 1 & 1 & 1 & 1\\
0 & 1 & 1 & 1 & 1\\
0 & 1 & 1 & 1 & 1\\
\end{bmatrix}$
|
Define
$$P := \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ 1 & -2 & 1 & 0 & 0 \\ -1 & 3 & -3 & 1 & 0 \\ 1 & -4 & 6 & -4 & 1 \end{bmatrix}.$$
Then
$$P A = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 7 & 15 \\ 0 & 0 & 2 & 12 & 50 \\ 0 & 0 & 0 & 6 & 60 \\ 0 & 0 & 0 & 0 & 24\end{bmatrix} =
\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 6 & 0 \\ 0 & 0 & 0 & 0 & 24\end{bmatrix} \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 7 & 15 \\ 0 & 0 & 1 & 6 & 25 \\ 0 & 0 & 0 & 1 & 10 \\ 0 & 0 & 0 & 0 & 1\end{bmatrix} =: DQ.$$
Now since $P$ and $Q$ are unimodular matrices by inspection, it follows that $D$ is the Smith normal form of $A$.
This should generalize nicely to higher values of $n$: if $P$ is the matrix whose $i,j$ element is $(-1)^{i+j} \binom{i-1}{j-1}$, then the $i,j$ element of $PA$ will be $(\Delta^{i-1} x^{j-1}) |_{x = 1}$, which is equal to 0 if $i > j$; equal to $(i-1)!$ if $i = j$; and divisible by $(i-1)!$ if $i < j$. Therefore, we will again be able to get a similar factorization $PA = DQ$ where $D$ is diagonal with $D_{ii} = (i-1)!$; and $Q$ is upper triangular with 1 on the diagonal and therefore unimodular.
|
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|
Solving the trigonometric equation $\sqrt{2}\sin(2x)=-\sqrt{3\sin(x) + 3\cos(x) + 8\cos^4(x-\pi/4)}$ My problem is to solve the following equation:
$$\sqrt{2}\sin(2x)=-\sqrt{3\sin(x) + 3\cos(x) + 8\cos^4(x-\pi/4)}$$
I've narrowed it down to
$$3\sin(x) + 3\cos(x) + 2 + 8\sin(x)\cos(x) = 0.$$
That's as far as I can get. It seems so simple, but I just can't find the next step(s).
|
Using auxiliary angle gives
$$
\begin{aligned}
\sqrt{2} \sin (2 x) &=-\sqrt{3 \sin x+3 \cos x+8 \cos ^{4}\left(x-\frac{\pi}{4}\right)} \\
&=-\sqrt{3 \sqrt{2} \cos \left(x-\frac{\pi}{4}\right)+8 \cos ^{4}\left(x-\frac{\pi}{4}\right)}
\end{aligned}
$$
Let $ y=x-\dfrac{\pi}{4}$, then
$$\sqrt{2} \cos \left(2 y\right)=-\sqrt{3 \sqrt{2} \cos y+8 \cos ^{4} y}$$
Squaring both sides yields $$
2 \cos ^{2}(2 y)=3 \sqrt{2} \cos y+8 \cos ^{4} y
$$
Using double-angle formula gives
$$
\begin{array}{l}
2\left(2 \cos ^{2} y-1\right)^{2}=3 \sqrt{2} \cos y+8 \cos ^{4} y \\
8 \cos ^{4} y-8 \cos ^{2} y+2=3 \sqrt{2} \cos y+8 \cos ^{4} y \\
8 \cos ^{2} y+3 \sqrt{2} \cos y-2=0 \\
\displaystyle \cos y=\frac{-3 \sqrt{2} \pm \sqrt{82}}{16}
\end{array}
$$
After checking for $\sin(2x)<0$, we can conclude that the solutions are$$
\begin{array}{l} \\ \displaystyle x=\frac{(8 n+1) \pi}{4} \pm \cos ^{-1}\left(\frac{\sqrt{82}-3 \sqrt{2}}{16}\right),
\end{array}
$$
where $ n\in Z$.
|
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|
Proving $1/(a+b) + 1/(b+c) + 1/(c+a) > 3/(a+b+c)$ for positive $a, b, c\,$? I have to prove that:
$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} > \frac{3}{a+b+c},$$
where $a, b , c$ are positive real numbers.
I am thinking about using arithmetical and geometrical averages:
$$A_{3} = \frac{a_{1}+a_{2}+a_{3}}{3},$$
$$G_{3} = \sqrt[3]{a_{1}×a_{2}×a_{3}},$$
$$A_{3}\ge G_{3}.$$
However, I am not sure how to do this.
I have tried to substract one side from the other:
$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} - \frac{3}{a+b+c} > 0.$$
I have also tried to reverse both sides of inequality.
I would appreciate if you could just give me a hint.
|
Other solution is Cauchy-Schwarz, that can be applied in one of its several versions.
*
*This is the standard one:
$[(\frac{1}{\sqrt{a+b}})^2+(\frac{1}{\sqrt{b+c}})^2+(\frac{1}{\sqrt{c+a}})^2][((\sqrt{a+b})^2+(\sqrt{b+c})^2+(\sqrt{c+a})^2]\ge 3^2$
From here the factor $9/2$, as already found, follows.
*
*Titu's inequality https://brilliant.org/wiki/titus-lemma/#
Here we directly have:
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{(1+1+1)^2}{2(a+b+c)}$
*
*Holders form: https://brilliant.org/wiki/holders-inequality/ :
$(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})^{1/2}((a+b)+(b+c)+(c+a))^{1/2}\ge 1^{1/2}+1^{1/2}+1^{1/2}$
I wrote this answer also for a reference of mine to the different forms of C.S.
|
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|
Closed form for the sum of the series $\sum^\infty_{n=1} \left( {(-1)^n}\left( 1+n\ln(\frac{2n-1}{2n+1}) \right) \right)$ If $a>2$ then $\sum\limits^\infty_{n=1} \left( {(-1)^n}\left( 1+n\ln(\frac{an-1}{an+1}) \right) \right)$ diverges by divergent test. Does it converge if $a=2$? Is it possible to find an exact form for the sum of the series in case it converges?
|
One has
\begin{align*} {(-1)^n}\left( 1+n\ln\left(\frac{2n-1}{2n+1}\right) \right) &= {(-1)^n}\left( 1+n\ln\left(1 -\frac{2}{2n+1}\right) \right)\\
&= {(-1)^n}\left( 1+n \left(- \frac{2}{2n+1} - \frac{2}{(2n+1)^2} + O \left( \frac{1}{n^3}\right)\right) \right)\\
&= \frac{(-1)^n}{2n+1}\left( \frac{1}{2n+1} + O \left( \frac{1}{n}\right)\right) \\
&= \frac{(-1)^n}{(2n+1)^2} + O \left( \frac{1}{n^2}\right) \\
\end{align*}
which proves the (absolute) convergence of the series for $a=2$.
|
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|
Show that $\frac{1}{2^{2n+1}}\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j} \right]= \frac{1}{2}$ I have a hard time showing that that
$$ \frac{1}{2^{2n+1}}\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j} \right]= \frac{1}{2}$$
Namely, I try to show hat
$$\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j}\right] = 2^{2n} $$
Any help would be appreciated. Thank you all.
Online demo: https://www.desmos.com/calculator/cjo7zggjlf
|
We will show that $\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j} \right]=4^n$
Note that $\sum_{j=i+1}^{n+1}\binom{n+1}{j}=\sum_{j=0}^{n-i}\binom{n+1}{j}$.
Now consider the power series $(1+x)^{n+1}=\binom{n+1}{0}+\binom{n+1}{1}x+\binom{n+1}{2}x^2+\ldots+\binom{n+1}{n+1}x^{n+1}$. We have that the partial sum $\sum_{j=0}^{n-i}\binom{n+1}{j}$ is the coefficient of $x^{n-i}$ in $\frac{(1+x)^{n+1}}{1-x}$.
Now going back to our original sum, we can extend the index to infinity to get that this is equivalent to
$$\sum_{i=0}^\infty \binom{n}{i}\sum_{j=i+1}^{n+1} \binom{n+1}{j}$$
Since $\binom{n}{i}$ is the coefficient of $x^i$ in $(1+x)^n$ and $\sum_{j=i+1}^{n+1} \binom{n+1}{j}$ is the coefficient of $x^{n-i}$ in $\frac{(1+x)^{n+1}}{1-x}$, this cauchy product is the coefficient of $x^n$ in
$$\frac{(1+x)^{2n+1}}{1-x}$$
which is
$$\sum_{i=0}^n \binom{2n+1}{i}$$
We can easily verify that this is just half of $\sum_{i=0}^{2n+1}\binom{2n+1}{i}=2^{2n+1}$, which is $2^{2n}=\boxed{4^n}$
|
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|
What is the minimum of $BP+\frac{1}{2}CP$?
In $Rt\triangle ABC,\ \angle A=90^{\circ},\ AB=4,\ AC=6.$ The radius of $\odot A$ is $2$. What is the minimum of $BP+\frac{1}{2}CP$?
Actually, the question should be "What is the minimum of $BP+\frac{1}{3}CP$", and then it will be simple by using similarity. But I still wonder how to deal with this question.
One possible way I've got is to put it in Rectangular Coordinates and assume $P(p\ ,\ \sqrt{4-p^2})$ . By using $d=\sqrt{(x_1-x_2)^2+(y_1+y_2)^2\ }$ I can write $BP+\frac{1}{2}CP$ with $p$ . And then it can be worked out by calculating the minimum of the function $f(p)=BP+\frac{1}{2}CP$.
However , it is clear that it is very hard to work out. I tried to work it out with better and easier methods but failed. Are there other possible approaches? Thank you for your ideas in advance!
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Here is a solution using angles. Our aim was to investigate whether we could find an answer to OP’s question without much ado. But, we could not. Our answer is as laborious as that of J.W.L. Jan Eerland.
For brevity, we denote $BP = a$, $CP = b$, and $\measuredangle PAB = \omega$. Furthermore, let $L$ be the length, which we expect to find the minimum value of. We apply cosine rule to the triangles $PAB$ and $CAP$ to find expressions for $a$ and $b$ in terms of $\omega$.
$$a^2 = 16 + 4 – 16\cos\left(\omega\right)\qquad\Longrightarrow\qquad a = 2\sqrt{5-4\cos\left(\omega\right)}\tag{1}$$
$$b^2 = 36 + 4 – 24\sin\left(\omega\right)\qquad\Longrightarrow\qquad b = 2\sqrt{10-6\sin\left(\omega\right)}\tag{2}$$
Using (1) and (2), let us express $L$ in terms of $\omega$.
$$L = a + \dfrac{b}{2} = 2\sqrt{5-4\cos\left(\omega\right)} + \sqrt{10-6\sin\left(\omega\right)}$$
The equation $\dfrac{dL}{d\omega} =0= \dfrac{4\sin\left(\omega\right)}{\sqrt{5-4\cos\left(\omega\right)}} - \dfrac{3\cos\left(\omega\right)}{ \sqrt{10-6\sin\left(\omega\right)}}$ must give us the value of $\omega$, for which the length $L$ is a minimum or maximum.
When simplified, it becomes,
$$f\left(\omega\right) = 160\sin^2\left(\omega\right)-96\sin^3\left(\omega\right)-45\cos^2\left(\omega\right)+36\cos^3\left(\omega\right) = 0. \tag{3}$$
The easiest way to solve equation (3) is to apply the numerical method of Newton-Raphson. You need the following Newton-Raphson formula to do so. By the way, $f’\left(\omega\right)$ is the first derivative of $f\left(\omega\right)$. As an educated guess for $\omega_{\text{0}}$, we chose $\dfrac{\pi}{4}$ after some spot checks revealed that the minimum of $L$ occurs for $\dfrac{\pi}{4} \gt \omega \gt 0.$
$\omega_{\text{n+1}} = \omega_{\text{n}} - \dfrac{ f\left(\omega_{\text{n}}\right) }{ f’\left(\omega_{\text{n}}\right) }= \dfrac{160\sin^2\left(\omega\right)-96\sin^3\left(\omega\right)-45\cos^2\left(\omega\right)+36\cos^3\left(\omega\right) }{2\sin\left(\omega\right) \cos\left(\omega\right)\Bigl\{205-144\sin\left(\omega\right) - 54\cos\left(\omega\right)\Bigr\}},\text{ where }\space\omega_{\text{0}} = \dfrac{\pi}{4}$.
When this numerical procedure is executed in an EXCEL worksheet, it took only five iteration steps for $\omega$ to converge to the following values.
$\omega = 0.2701631649520730\enspace \text{rad}= 15.4792091316505000^0$
$L_{\text{min}} = 5.0382236048810800$
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|
Comparing $\frac {9}{\sqrt{11} - \sqrt{2}}$ and $\frac {6}{3 - \sqrt{3}}$ (without calculator) We want to compare the following two numbers:
$$x = \frac {9}{\sqrt{11} - \sqrt{2}} \quad\text{and}\quad y = \frac {6}{3 - \sqrt{3}}$$
My attempts so far:
I multiply both numerator and denominator of $x$ by $\sqrt{11} + \sqrt{2}$ so I get:
$$x = \frac {9(\sqrt{11} + \sqrt{2})}{(\sqrt{11} - \sqrt{2})\cdot(\sqrt{11} + \sqrt{2})}$$ so
$$x = \frac {9(\sqrt{11} + \sqrt{2})}{(11 - 2)} = \sqrt{11} + \sqrt{2}$$
Similarly, $y = 3 + \sqrt{3}$.
But how do I take it from this point forward?
Of course $y>x$ but I must prove it.
I also tried to compare $\sqrt 11$ with $\sqrt 12$ which equals $2 \sqrt3$ but again I am not getting anywhere.
Thank you.
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Not only you can multiply number by the same positive value, you can also add or subtract the same values. The general idea how to solve similar questions is to remove roots one by one (making sure that values to be squared are positive):
$$
a+\sqrt{p}+\sqrt{q}+\ldots+\sqrt{r} \qquad ?\qquad b+\sqrt{u}+\sqrt{v}+\ldots+\sqrt{w},\\
a+\sqrt{p}+\sqrt{q}+\ldots+\sqrt{r}-b-\sqrt{v}-\ldots-\sqrt{w} \qquad ?\qquad \sqrt{u},\\
(a+\sqrt{p}+\sqrt{q}+\ldots+\sqrt{r}-b-\sqrt{v}-\ldots-\sqrt{w})^2 \qquad ?\qquad u,
$$
rinse and repeat.
However, in your case, since the left side has only 2 roots, we can square first, to diminish the number of roots straight away.
$$
\begin{align}\sqrt{11}+\sqrt{2} \qquad &?\qquad 3+\sqrt{3},\\
11 + 2\sqrt{22} + 2 \qquad &?\qquad 9+6\sqrt{3}+3,\\
1 + 2\sqrt{22} \qquad &?\qquad 6\sqrt{3},\\
1 + 4\sqrt{22} + 88 \qquad &?\qquad 108,\\
4\sqrt{22} \qquad &?\qquad 19,\\
352 \qquad &?\qquad 361.
\end{align}
$$
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Equation in integers $x y^2 + 2 y z^2 + z x^2 = 0$ The equation in integers
$$x y^2 + 2 y z^2 + z x^2 = 0$$
has some integral solution $[1\colon 0\colon 0]$, $[0\colon 1\colon 0]$, $[0\colon 0\colon 1]$, $[1\colon -1 \colon 1]$. I wonder if there are other solutions ( non-proportional to the one listed).
Notes: If we expand $(x + y \sqrt[3]{2} + z \sqrt[3]{4})^3$ the coefficient of $\sqrt[3]{4}$ is $3(x y^2 + 2 y z^2 + z x^2)$. I was looking for an $x + y \sqrt[3]{2}+ z\sqrt[3]{4}$ such that
$$(x + y \sqrt[3]{2}+ z\sqrt[3]{4})^3= a+b \sqrt[3]{2}$$
We have the example
$$(1 - \sqrt[3]{2}+\sqrt[3]{4})^3= 9(-1+\sqrt[3]{2})$$
and was wondering if we could get other examples.
What I've tried: the curve above is a smooth cubic, so an elliptic curve. However, no extra rational points using secants and tangents can be gotten out of the given points.
It is easy to show that if $p\mid y$ then $p$ also divides one of the $x$, $z$, but chasing prime divisors does not seem to get a contradiction that quickly.
Thank you for your attention!
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The case where $xyz=0$ is easy, so I will skip it. Assume then that $xyz\neq 0$.
Your equation is then $(\frac{x}{z})(\frac{y}{z})^2+2(\frac{y}{z})+(\frac{x}{z})^2=0,$ that is $(\frac{xy}{z^2})^2+2(\frac{xy}{z})+(\frac{x}{z})^3=0$.
Set $u=-\frac{x}{z}$ and $v=\frac{xy}{z}+1$. Then we get $v^2-1-u^3=0$.
According to LMFDB database , the elliptic curve $v^2=u^3+1$ has rank $0$, and the torsion points are: $(-1,0),(0,-1),(0,1),(2,-3),(2,3)$. This should be enough to solve your question.
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Find $a$ such that ${x_1}^2+{x_2}^2$ takes the minimal value where $x_1, x_2$ are solutions to $x^2-ax+(a-1)=0$ DO NOT USE CALCULUS My thinking:
Let $x_1 = \frac{a+\sqrt{a^2-4a+4}}{2}$ and $x_2 = \frac{a-\sqrt{a^2-4a+4}}{2}$
By the AGM (Arithmetic-Geometric Mean Inequality):
We have
$x_1\cdot x_2\le \left(\frac{x_1\cdot \:x_2}{2}\right)^2$
$=\:x_1\cdot \:x_2\le \:\frac{\left(x_1\cdot \:\:x_2\right)^2}{4}\:$
$=\:\:x_1\cdot \:\:x_2\le \:\:\frac{{x_1}^2+2x_1x_2+{x_2}^2}{4}$
$=4\left(x_1\cdot x_2\right)\:\le {x_1}^2+2x_1x_2+{x_2}^2$
$=4\left(x_1\cdot \:x_2\right)\:-2x_1x_2\le \:{x_1}^2+{x_2}^2$
Substituting in the values for $x_1$ and $x_2$ we get:
$4\left(\frac{a+\sqrt{a^2-4a+4}}{2}\cdot \:\frac{a-\sqrt{a^2-4a+4}}{2}\right)\:-2\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)\le \:\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)^2+\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)^2$
$= 4\left(a-1\right)\:-\left(2a-2\right)\le \:a^2-2a+2$
$ = 0\le a^2-4a+4$
It seems as if I walked in circles through this process, can anyone help? Thanks in advance!
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It looks like we are asked to notice that $ \ x^2 - ax + (a-1) \ = \ (x - 1)·( \ x - [a - 1] \ ) \ = \ 0 \ \ , $ so that one solution to the quadratic equation is always $ \ x_1 \ = \ 1 \ \ , $ with the other being $ \ x_2 \ = \ a - 1 \ \ . $ The function in question is then $ \ x_1^2 + x_2^2 \ = \ 1 + (a - 1)^2 \ \ , $ which will attain its minimum for $ \ a - 1 \ = \ 0 \ \Rightarrow \ a = 1 \ \ . $
As for sorting this out by the use of inequalities, that might work better with the RMS-GM inequality:
$$ \sqrt{\frac{x_1^2 \ + \ x_2^2}{2}} \ \ \ge \ \ \sqrt{x_1·x_2} \ \ \ge \ \ 0 \ \ \ \ \Rightarrow \ \ x_1^2 \ + \ x_2^2 \ \ \ge \ \ 2·x_1·x_2 \ \ = \ \ 2·(a \ - 1) \ \ \ge \ \ 0 \ \ , $$
where we have used the Viete relation for the product of the zeroes.
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How to solve $x$ in terms of $a$ and $b$ in equation $\frac{1}{2}+\frac{a}{b-a}e^{-bx}-\frac{b}{b-a}e^{-ax}=0$ ($x>0,a>0,b>0, a≠b$)? Suppose $x>0,a>0,b>0,a\neq b$. How to solve $x$ in terms of $a$ and $b$ in equation $\frac{1}{2}+\frac{a}{b-a}e^{-bx}-\frac{b}{b-a}e^{-ax}=0$? If a closed-form solution is not possible, a way to approximately solve for $x$ with an expression for approximation error is also okay. Thanks!
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Consider that you look for the zero of function
$$f(x)=\frac{1}{2}+\frac{a}{b-a}e^{-bx}-\frac{b}{b-a}e^{-ax}$$ for which
$$f'(x)=\frac{a b }{b-a}\left(e^{-a x}-e^{-b x}\right)\qquad \text{and} \qquad f''(x)= \frac{a b }{b-a}\left(b e^{-b x}-a e^{-a x}\right)$$
The key points are
$$f(0)=-\frac{1}{2} \qquad \qquad f'(0)=0 \qquad \qquad f''(0)=ab$$
Moreover, if $x\to \infty$, $f(x)\to \frac{1}{2}$; so, there is a root.
For an explicit approximation of the root, the idea is to perform a single iteration of Newton method
$$\color{red}{x_1=x_0-\frac{f(x_0)}{f'(x_0)}}\tag 1$$ but the problem is to find the "best" $x_0$; we have two possibilities.
The first one is based on the Taylor expansion around $x=0$
$$f(x)=-\frac{1}{2}+\frac{1}{2} a b x^2+O\left(x^3\right)\quad\implies\quad \color{blue}{x^{(1)}_0=\frac{1}{\sqrt{ab}}}$$
The second one considers the inflection point
$$f''(x)=0\quad\implies\quad \color{blue}{x^{(2)}_0=\frac {\log \left(\frac{b}{a}\right) }{b-a}}$$
A series of random test reveals that there is not much difference and that, most of the times, $x^{(1)}_0$ is a better choice than $x^{(2)}_0$. So, using it, the approximate formula
$$ \color{red}{x_1=\frac{1}{\sqrt{ab}}+\frac{a \left(2 e^{\sqrt{\frac{a}{b}}}-e^{\frac{a+b}{\sqrt{a b}}}\right)-b \left(2 e^{\sqrt{\frac{b}{a}}}-e^{\frac{a+b}{\sqrt{a b}}}\right) } {2 a b \left(e^{\sqrt{\frac{a}{b}}}-e^{\sqrt{\frac{b}{a}}}\right) }}$$
Just a single test : $a=2$ and $b=5$ lead to $x_0=0.316228$, $x_1=0.545058$; making one more iteration gives $x_2=0.563470 $ while the solution is $x= 0.563693$.
Edit
If we accept to stay with $(1)$, we can improve $x_0$ expanding $f(x)$ as a Taylor series around $x=0$ and using series reversion.
This would give
$$\color{blue}{x_0=t+\frac{a+b}{6} t^2 +\frac{2 a^2+7 a b+2 b^2}{72} t^3 +\frac{
4 a^3+39 a^2 b+39 a b^2+4 b^3}{1080}t^4+\frac{ 4 a^4+172 a^3
b+417 a^2 b^2+172 a b^3+4 b^4}{17280}t^5+O\left(t^6\right)}$$ where $\color{blue}{t=\frac{1}{\sqrt{ab}}}$.
$$\color{blue}{x_1=x_0+\frac{a e^{a x_0} \left(2-e^{b x_0}\right)-b \left(2-e^{a x_0}\right) e^{b x_0}}{2 a b
\left(e^{a x_0}-e^{b x_0}\right)}}$$
For $a=2$ and $b=5$, this gives $x_0=0.536545$ and $x_1=0.563228$ while the solution is $x= 0.563693$.
Pushing the expansion to $O\left(t^{10}\right)$ would give $x_0=0.559250$.
|
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Orbits of the action of $SL(n,\mathbb R)$ on $\mathbb R^n$. We know that $GL(n,\mathbb R)$ acts on $\mathbb R^n$ via left multiplication. We can easily see that there are two orbits viz $\{0\}$ and $\mathbb R^n-\{0\}$. Now we also know that if $G$ acts on $X$ and $H$ is a subgroup of $G$ then $H$ acts on $X$ and the $H$-orbits are actually suborbits of the $G$-orbits. Now I want to know what are the orbits of the action of $SL(n,\mathbb R)$ on $\mathbb R^n$. Does it subdivide the orbit of non-zero vectors further into smaller suborbits?
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Consider the action of $SL(n,\mathbb{R})$ in $\mathbb{R}^n$ via left multiplication.
If $n=1$, then $SL(1,\mathbb{R})\cong\{1\}$ and the action is trivial.
We will study $n=2$, the case $n>2$ is equal with some dots around. Take a generic vector $\begin{pmatrix} x\\y \end{pmatrix}\ne\begin{pmatrix} 0\\0 \end{pmatrix}$. We see this vector is in the same orbit of $\begin{pmatrix} 1\\0 \end{pmatrix}$.
If $y\ne0$
$$\begin{pmatrix} 1&0\\y&1 \end{pmatrix}\begin{pmatrix} 1\\0 \end{pmatrix}=\begin{pmatrix} 1\\y \end{pmatrix}$$
$$\begin{pmatrix} 1&\frac{x-1}{y}\\0&1 \end{pmatrix}\begin{pmatrix} 1\\y \end{pmatrix}=\begin{pmatrix} x\\y \end{pmatrix}.$$
If $y=0$, then $x\ne 0$ and
$$\begin{pmatrix} 1&0\\1&1 \end{pmatrix}\begin{pmatrix} 1\\0 \end{pmatrix}=\begin{pmatrix} 1\\1 \end{pmatrix}$$
$$\begin{pmatrix} 1&x-1\\0&1 \end{pmatrix}\begin{pmatrix} 1\\1\end{pmatrix}=\begin{pmatrix} x\\1 \end{pmatrix}
$$
$$\begin{pmatrix} 1&0\\-\frac{1}{x}&1 \end{pmatrix}\begin{pmatrix} x\\1 \end{pmatrix}=\begin{pmatrix} x\\0 \end{pmatrix}
.$$
So if $n\ge2$ the orbits are exactly the orbits of $GL(n,\mathbb{R})$.
Those kind of matrices that allow you to do operations with the rows of a vector are called shear matrices or transvections. They differ from the identity matrix only for an element which doesn't lie in the diagonal and if $n\ge 2$ they generate $SL(n,\mathbb{F})$ in any field $\mathbb{F}$.
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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?
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Too long to comment :
There is no such upper bound of the form $2−\dfrac{cq+d}{q^2+aq+b}$ better than $2−\dfrac{2}{q+1}$.
Proof :
Suppose that there is $(a,b,c,d)$ such that
$$2 - 2\cdot\dfrac{q^k - 1}{q^{k+1} - 1} \leqslant 2 - \dfrac{cq+d}{q^2 + aq + b} \leqslant 2 - \dfrac{2}{q+1}$$
Then,
$$\dfrac{2q^k - 2}{q^{k+1} - 1} \geqslant \dfrac{cq+d}{q^2 + aq + b} \geqslant \dfrac{2}{q+1}$$
is equivalent to
$$\underbrace{\bigg((2-c)q^2+(2a-d)q+2b\bigg)}_{A}q^k-2q^2+(c-2a)q+d-2b\geqslant 0\tag1$$
$$(c-2)q^2+(c+d-2a)q+d-2b\geqslant 0\tag2$$
Suppose that $2-c\lt 0$. Then, it follows from $(1)$ that $A$ will be negative for a large $q^k$ and so LHS of $(1)$ will be negative. This is a contradiction. So, $2-c\geqslant 0$.
Suppose that $c-2\lt 0$. Then, it follows from $(2)$ that LHS of $(2)$ will be negative for large $q$. This is a contradiction. So, $c-2\geqslant 0$.
Therefore, we have to have $c=2$ for which we have
$$\underbrace{\bigg((2a-d)q+2b\bigg)}_{A}q^k-2q^2+(2-2a)q+d-2b\geqslant 0\tag3$$
$$(2+d-2a)q+d-2b\geqslant 0\tag4$$
It is necessary that $(3)$ holds for $k=1$, so we have to have
$$(2a-d-2)q^2+(2b+2-2a)q+d-2b\geqslant 0\tag5$$
Suppose that $2+d-2a\lt 0$. Then, LHS of $(4)$ will be negative for a large $q$. This is a contradiction. So, $2+d-2a\geqslant 0$.
Suppose that $2a-d-2\lt 0$. Then, LHS of $(5)$ will be negative for a large $q$. This is a contradcition. So, $2a-d-2\geqslant 0$.
Therefore, we have to have $d=2a-2$ for which we have
$$a-1-b\geqslant 0\tag6$$
$$(b+1-a)q+a-1-b\geqslant 0\tag7$$
Suppose that $b+1-a\lt 0$. Then, LHS of $(7)$ will be negative for a large $q$. This is a contradiction. So, $b+1-a\geqslant 0$.
Therefore, from $(7)$, we have to have $b=a-1$.
Now, since $b=a-1,c=2$ and $d=2a-2$, we finally have
$$2 - \dfrac{cq+d}{q^2 + aq + b}=2 - \dfrac{2q+2a-2}{q^2 + aq + a-1}=2-\frac{2(q+a-1)}{(q+1)(q+a-1)}=2-\dfrac{2}{q+1}$$
This means that there is no such upper bound of the form $2−\dfrac{cq+d}{q^2+aq+b}$ better than $2−\dfrac{2}{q+1}$.$\quad\blacksquare$
Similarly, one can prove that there is no such upper bound of the form $2−\dfrac{dq^2+eq+f}{q^3+aq^2+bq+c}$ better than $2−\dfrac{2}{q+1}$.
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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.