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Random walk (Exercise 3.11.39 from Grimmett and Stirzaker) A particle performs a random walk on the non-negative integers as follows. When at the point $n\ (> 0)$ its next position is uniformly distributed on the set $\{0, 1, 2, \ldots, n + 1\}$. When it hits $0$ for the first time, it is absorbed. Suppose it starts at the point $a$.
(a) Find the probability that its position never exceeds $a$, and prove that, with probability $1$, it is absorbed ultimately.
(b) Find the probability that the final step of the walk is from $1$ to $0$ when $a = 1$.
(c) Find the expected number of steps taken before absorption when $a = 1$.
I completed part (a) and obtained the probability of not exceeding $a$ to be
$$
p = \frac{(a + 1)!}{1 + \sum_{n=1}^{a + 1} n!}.
$$
Now I am stuck on part (b). Here is what I have so far:
Let $p_n$ be the probability that the final step of the walk is from $1$ to $0$. Then
$$
p_n = \frac{1}{n + 2}(p_{n + 1} + p_{n - 1} + \cdots + p_1),
$$
and,
$$
p_{n - 1} = \frac{1}{n + 1}(p_n + p_{n - 1} + \cdots + p_1).
$$
From these two equations we obtain
\begin{align}
p_{n + 1} - p_n &= (n + 1)(p_n - p_{n - 1}), \\
\implies p_{n + 1} - p_n &= \frac{1}{2}(n + 1)!(p_2 - p_1).
\end{align}
Summing the above expression for $p_{n + 1} - p_n$ from $2$ to $n$ gives
$$
p_{n + 1} - p_2 = \frac{1}{2}(p_2 - p_1) \sum_{i = 3}^{n + 1} i!
$$
But we also know that
$$
p_1 = \frac{1}{3}(p_2 + p_1 + 1).
$$
Therefore $p_2 = 2 p_1 - 1$. Substituting for $p_2$ in the expression for $p_{n + 1} - p_2$ we obtain
$$
p_{n + 1} = p_1 + \frac{1}{2}(p_1 - 1) \sum_{i = 2}^{n + 1} i!
$$
Now I am stuck because the last expression does not make sense. This is because for all $p_1 \neq 1$, $p_n \rightarrow -\infty$ as $n \rightarrow \infty$. Where have I gone wrong? Please can you provide a hint on how to approach this problem in the correct way?
Edit (2016/02/28)
With help from the comments of @zhoraster and @mick-a below, the expression for $p_{n + 1}$ actually turns out to be
$$
p_{n + 1} = 5p_1 - 3 + \frac{3 p_1 - 2}{6}\sum_{i = 3}^n (i + 1)!
$$
For $p_{n + 1}$ to be a valid probability the coefficient of the factorial sum on the right hand side must be equal to $0$ as $n \rightarrow \infty$. This implies that $p_1 = 2/3$ and hence that $p_n = 5 \times 2/3 - 3 = 1/3$ for $n \geq 2$. I would never have expected that!
|
For completeness, here is the solution to part(c):
Assuming that a transition from $i$ to $j$ is always a step of length 1 (regardless of the values of $i$ and $j$) we can write the expected number of steps, $\lambda_n$, given that the particle starts at $n \geq 1$ as,
\begin{align}
\lambda_n &= \frac{1}{n + 2}((\lambda_{n + 1} + 1) + (\lambda_n + 1) + \cdots + (\lambda_1 + 1) + 1), \\
&= \frac{1}{n + 2}(\lambda_1 + \lambda_2 + \cdots + \lambda_{n + 1}) + 1.
\end{align}
And for $n \geq 2$,
$$
\lambda_{n - 1} = \frac{1}{n + 1}(\lambda_1 + \lambda_2 + \cdots + \lambda_n) + 1.
$$
Subtracting these two equations gives for $n \geq 2$,
$$
(n + 1)(\lambda_n - \lambda_{n + 1}) = \lambda_{n + 1} - \lambda_n + 1.
$$
The next step is the crucial step which allows the recurrence relation to be solved (and I have to admit that I needed to peek into the solutions manual for this hint). Let $v_{n + 1} = (\lambda_{n + 1} - \lambda_n) / (n + 1)!$. Then,
\begin{align}
v_{n + 1} - v_n &= \frac{\lambda_{n + 1} - \lambda_n}{(n + 1)!} - \frac{\lambda_n - \lambda_{n - 1}}{n!}, \\
&= -\frac{1}{(n + 1)!}.
\end{align}
Adding $v_{n + 1} - v_n$ from $n = 2$ to $\infty$ gives,
\begin{align}
-v_2 &= -\sum_{i = 3}^{\infty} \frac{1}{i!} \\
\implies v_2 &= -\frac{1}{0!} - \frac{1}{1!} - \frac{1}{2!} + \sum_{i = 0}^{\infty} \frac{1}{i!} \\
\implies v_2 &= -\frac{5}{2} + e.
\end{align}
But $v_2 = \frac{1}{2!}(\lambda_2 - \lambda_1)$ and $\lambda_1 = \frac{1}{3}(\lambda_1 + \lambda_2 + 3)$. Solving for $\lambda_2$ gives $\lambda_2 = 2 \lambda_1 - 3$. Therefore,
\begin{align}
v_2 = \frac{\lambda_2 - \lambda_1}{2} &= -\frac{5}{2} + e \\
\implies \frac{2 \lambda_1 - 3 - \lambda_1}{2} &= -\frac{5}{2} + e \\
\implies \lambda_1 &= 2 e - 2.
\end{align}
|
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|
Prove by induction that $n^2 < 2^n$ for all $n \geq 5$? So far I have this:
First consider $n = 5$. In this case $(5)^2 < 2^5$, or $25 < 32$. So the inequality holds for $n = 5$.
Next, suppose that $n^2 < 2^n$ and $n \geq 5$. Now I have to prove that $(n+1)^2 < 2^{(n+1)}$.
So I started with $(n+1)^2 = n^2 + 2n + 1$. Because $n^2 < 2^n$ by the hypothesis, $n^2 + 2n + 1$ < $2^n + 2n + 1$. As far as I know, the only way I can get $2^{n+1}$ on the right side is to multiply it by $2$, but then I get $2^{n+1} + 4n + 2$ on the right side and don't know how to get rid of the $4n + 2$. Am I on the right track, or should I have gone a different route?
|
First, we prove a simple lemma.
$$
\begin{align}
n & \gt 5\\
\implies n-1 &\gt 4\\
\implies (n-1)^2 &\gt 16 \gt 2 \\
\implies n^2 - 2n + 1 &\gt 2\\
\implies n^2 &\gt 2n +1\\
\end{align}
$$
Now, we start with our induction step.
$$
\begin{align}
n^2 &\lt 2^n \text{ From hypothesis}\\
\implies 2n^2 & \lt 2^{n+1}\\
\implies n^2 + n^2 &\lt 2^{n+1}\\
\implies n^2 + 2n + 1 &\lt 2^{n+1} (n^2 \gt 2n+1)\\
\implies (n+1)^2 &\lt 2^{n+1} \\
\end{align}$$
|
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|
Proving correctness of a recursive function of multiplication by induction. Define multiplication of two numbers y and z as:
$$m_c(y,z)=\begin{cases}0&z=0\\m_c\left(cy,\left\lfloor \frac zc\right\rfloor\right)+y(z\mod c)&z\neq 0\end{cases}\tag{$\forall c\geq 2$}$$
Now I need to show that this is correct using induction.
It is obviously true for the case $z=0$. Otherwise:
$$\begin{align}\forall z\neq 0,\;m_c(y,z)=&m_c\left(cy,\left\lfloor \frac zc\right\rfloor\right)+y(z\mod c)\\=&m_c\left(c^2y,\left\lfloor\frac 1c\left\lfloor \frac zc\right\rfloor\right\rfloor\right)+y(z\mod c)+cy\left(\left\lfloor\frac zc\right\rfloor\mod c\right)\\=&m_c\left(c^3y,\left\lfloor\frac 1c\left\lfloor\frac 1c\left\lfloor \frac zc\right\rfloor\right\rfloor\right\rfloor\right)+y(z\mod c)+cy\left(\left\lfloor\frac zc\right\rfloor\mod c\right)+\\&c^2y\left(\left\lfloor\frac 1c\left\lfloor\frac zc\right\rfloor\right\rfloor\mod c\right)\\=&\ldots\\=&m_c\left(c^ny,\underbrace{\left\lfloor\frac 1c\left\lfloor\frac 1c\left\lfloor\frac 1c\ldots\left\lfloor\frac zc\right\rfloor\ldots\right\rfloor\right\rfloor\right\rfloor}_{\text{n times c}}\right)\\&+\sum_{r=0}^{n-1} c^ry\left(\underbrace{\left\lfloor\frac 1c\left\lfloor\frac 1c\left\lfloor\frac 1c\ldots\left\lfloor\frac zc\right\rfloor\ldots\right\rfloor\right\rfloor\right\rfloor}_{\text{n-1 times c}}\mod c\right)\end{align}$$
Now for induction, where do I start? From n=1 , then k then k+1. Or something else? A hint is enough.
|
You should do induction on $z$, so that you can assume in the inductive step that $m_c\left(cy,\left\lfloor \frac zc\right\rfloor\right)=cy\cdot\left\lfloor \frac zc\right\rfloor$.
|
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|
Strengthen inequality It is known that for positive integers $ a, b, c, d $ and $n $, the inequalities $ a + c <n $ and $ \frac ab + \frac cd <1 $ hold. Prove that
$$ \frac ab + \frac cd <1 - \frac {1} {n ^ 3} $$
My work so far:
$ a, b, c, d, n -$ positive integers $\Rightarrow a+c \le n-1$
and $ad+bc \le bd-1.$
|
From $\frac ab+\frac cd \lt 1$ and $a+c<n$ it is clear that $a<b;\space c<d$, $n\ge 3$ and $2\le a+c\le n-1$. Furthermore at least one of the two fractions should be less than $\frac 12$ therefore we can put $\frac {a}{2a+k}+\frac cd\lt 1$ where $k\ge 1$ so $$\frac ab+\frac cd\le \frac {a}{2a+1}+\frac cd\lt 1\iff \frac cd\lt \frac{a+1}{2a+1}\qquad (*)$$
Now, fixing $a$, we have for all the possible values of $\frac cd$
$$\frac cd\in \left\{\frac{1}{1+k}\right\}_{k>0}\cup \left\{\frac{2}{2+k}\right\}_{k>0}\cup…..\cup \left\{\frac{n-1-a}{n-1-2+k}\right\}_{k>0}$$
and the maximum of $\frac cd$ is in the set
$$\left\{\frac 12\right\}\cup \left\{\frac 23\right\}\cup…..\cup \left\{\frac{n-1-a}{n-a}\right\}\qquad (**)$$
Without restrictions this maximum is clearly $\left\{\frac{n-1-a}{n-a}\right\}$ but we have to take into account the restriction $(*)$,
Consider the functions $f(x)=\frac{x+1}{2x+1}$ and $g(x)=\frac{x}{x+1}$; $f$ is decreasing and $g$ increasing in its domains of definition. Furthermore the positive solution $x_0$ of the equation $f(x)=g(x)$ gives exactly the value in which $\frac cd=\frac{a+1}{2a+1}$. This solution (unexpectedly it is the Golden ratio) is $x_0 \approx 1.6180$ and to the right of $x_0$ one has $g(x)>f(x)$ (so for example
$\frac 23\gt \frac {3}{5}$)
Hence the value of $\frac cd$ we must take is $\frac 12$.
Hence $$\frac ab+\frac cd +\frac {1}{n^3}\le \frac {a}{2a+1}+\frac 12 +\frac{1}{n^3}\le \frac{n-2}{2n-3}+\frac 12+\frac{1}{n^3}=\frac{4n^4-7n^3+4n-6}{4n^4-6n^3}$$
Now, by the absurd, suppose that $$\frac{4n^4-7n^3+4n-6}{4n^4-6n^3}\ge 1\iff 4n\ge n^3+6$$ This is not true for $n\ge 3$.
Thus $$\frac ab+\frac cd +\frac {1}{n^3}\lt 1$$
|
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|
What is the maximum of $\ 2(a+b)-ab$ if we have: $\ a^2+b^2=8\ (a,b\ real)$ What is the maximum of $\quad 2(a+b)-ab\ $ if we have: $a^2+b^2=8 \ (a,b\ real) $
My work is as follows:
According to AM-GM inequality:
$$\frac{a^2+b^2}2\ge\sqrt{a^2b^2} \Rightarrow $$
$$ab\le4 $$
On the other hand , from $\ a^2+b^2=8\ ,\ $we have: $\ 2(a+b)-ab=2\sqrt{8+2ab}-ab$
We can get this: $$2\sqrt{8+2ab}\le8$$
Is it possible to somehow combine these two inequalities to find the maximum value of $\ 2(a+b)-ab$ ?
I think in this solution technique we must find the minimum value of $ab\ $ , but how???
|
let $u=\sqrt{8+2ab} \implies ab=\dfrac{u^2-8}{2} \implies 2u-\dfrac{u^2-8}{2}=-\dfrac{(u-2)^2}{2}+6,|ab| \le \dfrac{a^2+b^2}{2}=4 \implies 0\le u \le 4$
|
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|
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides a, b and c. This is an A-level trigonometric problem.
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides $a$, $b$ and $c$.
Answer: $$\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{4b^2c^2}$$
The last step of the solution requires going from:
Expression 1: $$-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2$$
to:
Expression 2: $$(a+b-c)(a-b+c)(a+b+c)(-a+b+c)$$
I can see that these two expressions are equal by working in reverse and multiplying out the second expression to get the first but how does one go from Exp.1 to Exp.2?
|
$a+b-c=2(s-c)$ etc.
$$\implies\dfrac{(a+b-c)(b+c-a)(c+a-b)(a+b+c)}{b^2c^2}=\dfrac{16s(s-a)(s-b)(s-c)}{(bc)^2}$$
Now use $\triangle=\sqrt{s(s-a)(s-b)(s-c)}=\dfrac12bc\sin A$
|
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|
What's wrong with this transformation? I have the equation $\tan(x) = 2\sin(x)$ and I'd like to transform it in this way:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)$$
But I'm getting a wrong result so I suppose that I can't do it in this way. Why?
EDIT:
This is my solution:
$$\tan(x) = 2\sin(x)
\Longleftrightarrow
\frac{\sin(x)}{\cos(x)} = 2\sin(x)
\Longleftrightarrow
\sin(x) = 2\sin(x)\cos(x)
\Longleftrightarrow
\sin(x) = \sin(2x)
\Longleftrightarrow
\sin(2x) - \sin(x) = 0
\Longleftrightarrow
2\cos(\frac{3x}{2})\sin(\frac{x}{2}) = 0
\Longleftrightarrow
\cos(\frac{3x}{2}) = 0 \vee \sin(\frac{x}{2}) = 0$$
Proper solution is $\cos(x) = \frac{1}{2} \vee \sin(x) = 0$
|
There is no problem. Your solution is correct, and the other solution is correct. But you need to do a little work to see that they give the same answers:
*
*$\cos \frac {3x}2 = 0$ gives $x = \frac{(2k+1)\pi}3:\ x \in \{\pm \frac{\pi}3, \pm\pi,\pm \frac{5\pi}3, \pm \frac{7\pi}3, \pm 3\pi, \ldots\}$
*$\sin \frac {x}2 = 0$ gives $x = 2k\pi:\ x \in \{0, \pm 2\pi, \pm 4\pi, \ldots\}$
while
*
*$\cos x = \frac 12$ gives $x =\pm \frac{\pi}3 + 2k\pi:\ x \in \{\pm \frac{\pi}3, \pm \frac{5\pi}3, \pm \frac{7\pi}3, \ldots\}$
*$\sin x = 0$ gives $x = k\pi:\ x \in \{0, \pm\pi, \pm 2\pi, \pm 3\pi,\pm 4\pi, \ldots\}$
Comparison of the values shows that the same ones are in both lists.
|
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|
Integral $I=\int \frac{dx}{(x^2+1)\sqrt{x^2-4}} $ Frankly, i don't have a solution to this, not even incorrect one, but, this integral looks a lot like that standard type of integral $I=\int\frac{Mx+N}{(x-\alpha)^n\sqrt{ax^2+bx+c}}$ which can be solved using substitution $x-\alpha=\frac{1}{t}$ so i tried to find such subtitution that will make this integral completely the same as this standard integral so i could use substitution i mentioned, so i tried two following substitutions
$x^2-4=t^2 \Rightarrow x^2=t^2+4 \Rightarrow x=\sqrt{t^2+4}$ then i had to determine $dx$
$2xdx=2tdt \Rightarrow dx=\frac{tdt}{\sqrt{t^2+4}}$
from here i got:
$\int\frac{dt}{(t^2+5)\sqrt{(t^2+4)}}$ but i have no idea what could i do with this, so i tried different substitution
$x^2+1=t^2$ and then, by implementing the same pattern i used with the previous substitution i got this integral
$\int\frac{dt}{t\sqrt{(t^2-1)(t^2-5)}}$ but again, i don't know what to do with this, so i could use some help.
|
HINT:
Your integral:
$$\text{I}=\int\frac{1}{\left(x^2+1\right)\sqrt{x^2-4}}\space\text{d}x=$$
Subsitute $x=2\sec(u)$ and $\text{d}x=2\tan(u)\sec(u)\space\text{d}u$.
Then $\sqrt{x^2-4}=\sqrt{4\sec^2(u)-4}=2\tan(u)$ and $u=\text{arcsec}\left(\frac{x}{2}\right)$:
$$\int\frac{\sec(u)}{4\sec^2(u)+1}\space\text{d}u=\int\frac{\cos(u)}{5-\sin^2(u)}\space\text{d}u=$$
Substitute $s=\sin(u)$ and $\text{d}s=\cos(u)\space\text{d}u$:
$$\int\frac{1}{5-s^2}\space\text{d}s=\frac{1}{5}\int\frac{1}{1-\frac{s^2}{5}}\space\text{d}s=$$
Substitute $p=\frac{s}{\sqrt{5}}$ and $\text{d}p=\frac{1}{\sqrt{5}}\space\text{d}s$:
$$\frac{1}{\sqrt{5}}\int\frac{1}{1-p^2}\space\text{d}p=\frac{\text{arctanh}\left(p\right)}{\sqrt{5}}+\text{C}=$$
$$\frac{\text{arctanh}\left(\frac{s}{\sqrt{5}}\right)}{\sqrt{5}}+\text{C}=\frac{\text{arctanh}\left(\frac{\sin\left(u\right)}{\sqrt{5}}\right)}{\sqrt{5}}+\text{C}=$$
$$\frac{\text{arctanh}\left(\frac{\sin\left(\text{arcsec}\left(\frac{x}{2}\right)\right)}{\sqrt{5}}\right)}{\sqrt{5}}+\text{C}=\frac{\text{arctanh}\left(\frac{\sqrt{x^2-4}}{x\sqrt{5}}\right)}{\sqrt{5}}+\text{C}$$
|
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|
Complicated Indefinite Integration $\displaystyle\int{\dfrac{\left(\sqrt{x}+2\right)^2}{3\sqrt{x}}\;\mathrm{d}x} =\frac{2}{3}\big(\sqrt{x}+2\big) $
I do not know how to get $\dfrac{2}{3}$. I assume $\sqrt{x}+2 = u$, then $\mathrm{d}u= \dfrac{1}{2}\,x^{1/2}$
But it is wrong answer....
So help me
Thank you
|
Assuming you mean $\displaystyle \int \frac{\left(\sqrt{x}+2\right)^2}{3\sqrt{x}}\,dx$:
Let $\,u=\sqrt{x}+2\,$ and $\,du=\dfrac{dx}{2\sqrt{x}}$
Thus, we now have $$\dfrac{2}{3} \int u^2 \,du=\dfrac{2u^3}{9}+C=\dfrac{2\left(\sqrt{x}+2\right)^3}{9}+C$$
|
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|
Find the value of $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$
Find the value of $$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$=\sin\frac{2\pi}{14}\sin\frac{4\pi}{14}\sin\frac{6\pi}{14}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$=\sin\frac{\pi}{14}\sin\frac{2\pi}{14}\sin\frac{3\pi}{14}\sin\frac{4\pi}{14}\sin\frac{5\pi}{14}\sin\frac{6\pi}{14}$$
How can I compute $$\prod\limits_{r=1}^{n}\sin\frac{r\pi}{c}$$
|
By the basic properties of Chebyshev polynomials the numbers $x_k=\sin(k\pi/14)$, $k=-6,-5,\ldots,5,6, k\neq0$ are the twelve zeros of the polynomial
$$
\frac{U_{13}(x)}x=8192 x^{12}-24576 x^{10}+28160 x^8-15360 x^6+4032 x^4-448 x^2+14.
$$
By the usual relations (aka Vieta relations) of the zeros of a polynomial and its coefficients it follows that
$$
A=\prod_{k=-6..6, k\neq0}x_k=\frac{14}{8192}=\frac{7}{2^{12}}.
$$
Because sine is known to be an odd function, the products $N=\prod_{k=-6}^{-1}x_k$ and $P=\prod_{k=1}^6x_k$ are equal. You are asking about the product
$P$. It is clearly a positive number, so from the equation
$$A=NP=P^2=\frac{7}{2^{12}}
$$
we can solve
$$
P=\frac{\sqrt7}{2^6}.
$$
The method generalizes for other products of sines as long as you take the product of ALL the numbers $\sin (k\pi/(2N))$, $k=1,2,\ldots,N-1$.
|
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|
How many $4$ digit integer elements $X$ having no digit $0$ are in the set $C$ such that $X$ has exactly one $1$ or $X$ has exactly one $5$? Let $A = \{4~\text{digit integers}~X~\text{having no}~0~\text{such that}~X~\text{has exactly one}~1\}$.
Let $B = \{4~\text{digit integers}~X~\text{having no}~0~\text{such that}~X~\text{has exactly one}~5\}$.
Let $A+B = \{4~\text{digit integers}~X~\text{having no}~0\}$.
Then $|C| = |A| + |B| - |A + B|$.
Set A has $4C1 \cdot 8 \cdot 8 \cdot 8 = 2048$ elements.
Set B has $4C1 \cdot 8 \cdot 8 \cdot 8 = 2048$ elements.
Set $A+B$ has $8 \cdot 8 \cdot 8 \cdot 8 = 4096$ elements.
Therefore $|A| + |B| - |A + B| = 2048 + 2048 - 4096 = 0$.
I think my steps are correct, but doubt that I got the correct answer.
|
You correctly calculated the number of four digit integers with non-zero digits that have exactly one $5$ and the number of four digit integers with non-zero digits having exactly one $1$. However, their intersection is the set of four digit integers with non-zero digits that have exactly one $1$ and exactly one $5$. The number of four digit integers with non-zero digits that have exactly one $1$ and exactly one $5$ is
$$4 \cdot 3 \cdot 7^2 = 588$$
since there are four ways to place the $1$, three ways to place the $5$ in one of the remaining places, and $7$ choices (other than $0$, $1$, or $5$) for each of the two remaining digits.
Thus, by the Inclusion-Exclusion Principle, the number of four digit integers having no zero and exactly one $1$ or exactly one $5$ is
$$|A \cup B| = |A| + |B| - |A \cap B| = 2048 + 2048 - 588 = 3508$$
|
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|
Extracting the divergent part of an integral I want to evaluate the integral
$$ \int_0^1 \frac{2x(x-2)(1-x)}{(1-x)^2 + ax} \, \mathrm{d}x$$
in the limit of small $a$. For $a = 0$ this integral is divergent due to the $1/(1-x)$ pole. The exact expression for this integral is, according to Mathematica, a horrible combination of logs and inverse hyperbolic functions. However, I don't really care about all of the terms that go to zero as $a$ goes to zero. In, other words, I want to deduce that this integral is
$$1 + \ln a + \mathcal{O}(a)$$
(I believe this is the correct answer) by making some approximation at an early stage that saves me from having to compute the entire integral and then take the small $a$ limit.
|
Notice, $$\int_0^1\frac{2x(x-2)(1-x)}{(1-x)^2+ax}\ dx$$
using property of definite integral: $\int_a^bf(x)\ dx=\int_a^bf(1+b-x)\ dx$,
$$=\int_0^1\frac{2(1-x)(1-x-2)(1-1+x)}{(1-1+x)^2+a(1-x)}\ dx$$
$$=\int_0^1\frac{2x^3-2x}{x^2-ax+a}\ dx$$
$$=\int_0^1\left(2(x+a)+\frac{(a^2-a-1)x-a^2}{x^2-ax+a}\right)\ dx$$
$$=2\int_0^1(x+a)\ dx+\int_0^1\frac{(a^2-a-1)(2x-a)+(a^3-3a^2-a)}{x^2-ax+a}\ dx$$
$$= 2\left(\frac{x^2}{2}+ax\right)_0^1+(a^2-a-1)\int_0^1\frac{(2x-a)dx}{x^2-ax+a}+(a^3-3a^2-a)\int_0^1\frac{1}{x^2-ax+a}\ dx$$
$$=\small 2\left(\frac{1}{2}+a\right)+(a^2-a-1)\int_0^1\frac{d(x^2-ax+a)}{x^2-ax+a}+(a^3-3a^2-a)\int_0^1\frac{1}{\left(x-\frac a2\right)^2+a-\frac{a^2}{4}}\ dx$$
$$=1+2a+(a^2-a-1)\left(\ln|x^2-ax+a|\right)_0^1+(a^3-3a^2-a)\int_0^1\frac{1}{\left(x-\frac a2\right)^2+a-\frac{a^2}{4}}\ dx$$
$$=1+2a+(a+1-a^2)\ln |a|+(a^3-3a^2-a)\int_0^1\frac{1}{\left(x-\frac a2\right)^2+a-\frac{a^2}{4}}\ dx$$
I hope you can prove your approximation now.
|
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|
How can I find the maximum and minimum of $f(x,y)$ in $D=\{ (x,y) \in \Bbb{R}^2 \mid x^2+y^2 \le 4 \}$? How can I find the maximum and minimum of $f(x,y)$ in $D=\{ (x,y) \in \Bbb{R}^2 \mid x^2+y^2 \le 4 \}$
where
$$f(x,y)=x^4+y^4-x^2-2xy-y^2$$
Answer:
$$f_x=4x^3-2x-2y$$
$$f_y=4y^3-2y-2x$$
Critical points are $(0,0), (1,1), (-1,-1)$.
All of these critical points are in D.
$$f(0,0)=0$$
$$f(1,1)=-2$$
$$f(-1,-1)=-2$$
Now we can find the extreme values of $f$ on the boundary of the region $x^2+y^2=4$
$$x^2=4-y^2$$
Therefore we have,
$$g(y)=2y^4-8y^2-2y \sqrt{4-y^2}+16$$
$$g'(y)=8y^3-16y-2 \sqrt{4-y^2}+2y^2(4-y^2)^{-1/2}=0$$
$\color{red}{problem:}$
I could do only up to here. I don't know how to solve the above equation to get the values for $y$.
After finding extreme values of this, we can compare those values with the above and we can find the maximum and minimum values.
|
Probably easier to write the boundary as
$$x=2\cos\theta\ ,\quad y=2\sin\theta\ .$$
Then
$$\eqalign{f(x,y)
&=16\cos^4\theta+16\sin^4\theta-4\cos^2\theta-8\cos\theta\sin\theta-4\sin^2\theta\cr
&=4(1+\cos2\theta)^2+4(1-\cos2\theta)^2-4\sin2\theta-4\cr
&=4+8\cos^22\theta-4\sin2\theta\cr
&=12-4\sin2\theta-8\sin^22\theta\cr}$$
and it should not be hard to find extreme values of this.
|
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|
How to solve the non-homogeneous linear recurrence $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$? The problem: $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$
First I solved the associated homogeneous recurrence and got $a_n = A(1)^n = A$, where A is a constant, but I got stuck solving the rest. My final answer was $a_n=2n^2+n+1$ while my textbook has $a_n=n^2+2n+1$
Here is my work:
$ a_{n+1} - a_n = 2n+3 $
$pn=nd_1+d_2$ for all n
plug in and solve for $d_1$:
$(n+1)d_1+d_2-nd_1-d_2=2n+3$
$d_1=2n+3$
then solve for $d_2$:
$a_0=1=a_{n+1}-2n-3=(n+1)d_1+d_2-2n-3$
$d_2=1-n(2n+3)$
therefore
$p_n=nd_1+d_2 = 2n^2+n+4$ and $a_n=A+2n^2+n+4$
then solve for A:
$a_0=1=A +0+0+4$ so $A=-3$
therefore
$a_n=2n^2+n+1$
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Let $f(z) = \sum_{n=0}^\infty a_n z^n$. Multiplying both sides of the recurrence by $z^n$ and summing over $n$ yields
$$\sum_{n=0}^\infty a_{n+1}z^n - \sum_{n=0}^\infty a_nz^n = \sum_{n=0}^\infty(2n+3)z^n $$
and hence
$$z^{-1}(f(z)-a_0) - f(z) = 2\sum_{n=0}^\infty (n+1)z^n + \sum_{n=0}^\infty z^n, $$ so that
$$f(z)(1-z) = 1 + \frac{2z}{(1-z)^2}+\frac z{1-z} $$
and
\begin{align}
f(z) &= \frac1{1-z}+\frac z{(1-z)^2} + \frac{2z}{(1-z)^3}\\
&= \frac{(1-z)^2+z(1-z)+2z}{(1-z)^3}\\
&= \frac{1 - 2z + z^2 + z -z^2 + 2z}{(1-z)^3}\\
&= \frac{1+z}{(1-z)^3}.
\end{align}
Expanding $f$ in a power series about $z=0$ yields
$$f(z) = \sum_{n=0}^\infty(n+1)^2 z^n, $$
so that $a_n=(n+1)^2, n\geqslant0$.
|
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|
How to evaluate $\int_{0}^{\infty }\frac{\ln( 1+x^{4} )}{1+x^{2}}{d}x~,~\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}{d}x$ How to evaluate these two integrals below
$$\int_{0}^{\infty }\frac{\ln\left ( 1+x^{4} \right )}{1+x^{2}}\mathrm{d}x$$
$$\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}\mathrm{d}x$$
For the first one, I tried to use
$$\mathcal{I}'(s)=\int_{0}^{\infty }\frac{x^4}{(1+sx^4)(1+x^{2})}\mathrm{d}x$$
but it seems hard to solve.
|
For the first integral, we introduce a new parameter $a$. Let $I(a)$ be the integral
$$I(a)=\int_0^\infty \frac{\log(a+x^4)}{1+x^2}\,dx \tag 1$$
We note that it can be shown that $I(0)=0$. Our integral of interest is $I(1)$.
Proceeding, we differentiate $(1)$ to obtain
$$\begin{align}
I'(a)&=\int_0^\infty\frac{1}{(1+x^2)(a+x^4)}\,dx\\\\
&=\frac{1}{2(a+1)}\int_{-\infty}^\infty\left(\frac{1}{x^2+1}+\frac{1-x^2}{x^4+a}\right)\,dx\\\\
&= \frac{2\pi i}{2(a+1)}\left(\frac{1}{2i}+\frac{1}{i\,2^{3/2}a^{3/4}}-\frac{1}{i\,2^{3/2}a^{1/4}}\right)\\\\
&=\frac{\pi}{2(1+a)}+\frac{\pi}{2^{3/2}(1+a)a^{3/4}}-\frac{\pi}{2^{3/2}(1+a)a^{1/4}}\tag 2
\end{align}$$
Integrating the right-hand side of $(2)$ reveals
$$\begin{align}
I(1)&=\frac{\pi}{2}\log(2)+\frac{\pi}{2^{3/2}}\int_0^1\left(\frac{1}{(1+x)x^{3/4}}-\frac{1}{(1+x)x^{1/4}}\right)\,dx\\\\
&=\frac{\pi}{2}\log(2)+\pi\sqrt 2\int_0^1\frac{1-x^2}{1+x^4}\,dx\\\\
&=\frac{\pi}{2}\log(2)+\pi\sqrt 2\left.\left(\frac{\log(x^2+\sqrt2 x+1)-\log(-x^2+\sqrt 2x-1)}{2^{3/2}}\right)\right|_{0}^{1}\\\\
&=\frac{\pi}{2}\log(2)+\frac{\pi}{2}\log\left(\frac{2+\sqrt 2}{2-\sqrt 2}\right)\\\\
&=\pi\log(2+\sqrt 2)
\end{align}$$
|
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|
Induction Proof $k^2 \times 2^k$ I need help on this proof. I am not able to do it after setting m=m+1.
Prove by induction on n that sum of $k^2 \times 2^k$ from $k=1$ to $n$ is equal to $(n^2-2n+3) \times 2^{n+1}-6$
Base case:
Let $k=1$ so L.H.S side is $2$
Let $n=1$ so R.H.S side is $2$
Inductive hypothesis:
Let $n=m$ so $(m^2-2m+3) \times 2^{m+1}-6$
Proof:
Let $n=m+1$ so prove that $((m+1)^2-2(m+1)+3) \times 2^{m+2}-6=(m^2-2m+3) \times 2^{m+1}-6$
But I am not able to prove that they are equal.
|
The induction hypothesis is that
$$\tag1\sum_{k=1}^m k^22^k=(m^2-2m+3)2^{m+1}-6$$
Now,
$$\sum_{k=1}^{m+1}k^22^k=\sum_{k=1}^mk^22^k+(m+1)^22^{m+1}$$
Inductive Step:
From $(1)$,
$$\sum_{k=1}^{m+1}k^22^k=\sum_{k=1}^mk^22^k+(m+1)^22^{m+1}=(m^2-2m+3)2^{m+1}-6+(m+1)^22^{m+1}=2^{m+1}(m^2-2m+3+m^2+2m+1)-6=2^{m+1}(2m^2+4)-6=2^{m+2}(m^2+2)-6$$
Moreover,
$$(m+1)^2-2(m+1)+3=m^2+2m+1-2m-2+3=m^2+2$$
Hence,
$$\sum_{k=1}^{m+1}k^22^k=2^{m+2}(m^2+2)-6=2^{m+2}((m+1)^2-2(m+1)+3)-6$$
Thus, given the statement is true for $m$, it is true for $m+1$.
By the principle of mathematical induction, it is true for all $m\in\mathbb N$.
|
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|
Derive a polynomials formula by setting $n = 1$, $n = 2$, $n = 3$ to determine the coefficients. Derive a formula for the sum of squares $1^2 + 2^2 + 3^2 + … + n^2$.
Hint: assume the formula is a polynomial of degree 3, i.e. $an^3 + bn^2 + cn + d$, and use the cases of $n=0, n=1, n=2$, and $n=3$ to determine its coefficients.
|
Another derivation is using proof by induction; these are the $\color{green}{\mathrm{three}}$ steps to carry out:
Step 1: Basis Case: For $i=1$: $$\sum^{i=k}_{i=1} i^2=\frac{1(1+1)(2\times 1+1)}{6}= \frac{2\times 3}{6}=1$$ So statement holds for $i=1$.
Step 2: Inductive Assumption: Assume statement is true for $i=k$:
$$\sum^{i=k}_{i=1} i^2=\frac{k(k+1)(2k+1)}{6} $$
Step 3: Prove Statement holds for $i=k+1$. You need to prove that for $i=k+1$: $$\sum^{i=k+1}_{i=1} i^2=\color{blue}{\frac{(k+1)(k+2)(2k+3)}{6}}$$
To do this you cannot use: $$\sum^{i=k}_{i=n} i^2=\color{red}{\frac{n(n+1)(2n+1)}{6}} $$ as this is what you are trying to prove.
So what you do instead is notice that:
$$\sum^{i=k+1}_{i=1} i^2= \underbrace{\frac{k(k+1)(2k+1)}{6}}_{\text{sum of k terms}} + \underbrace{(k+1)^2}_{\text{(k+1)th term}}$$
$$\sum^{i=k+1}_{i=1} i^2= \frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$$
$$\sum^{i=k+1}_{i=1} i^2= \frac{(k+1)\left(k(2k+1)+6(k+1)\right)}{6}$$
$$\sum^{i=k+1}_{i=1} i^2= \frac{(k+1)(2k^2+\color{green}{7k}+6)}{6}=\frac{(k+1)(2k^2+\color{green}{4k+3k}+6)}{6}=\frac{(k+1)\left(2k(k+2)+3(k+2)\right)}{6}=\color{blue}{\frac{(k+1)(k+2)(2k+3)}{6}}\quad \forall \space k \in \mathbb{N}$$
Which is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\color{blue}{\mathrm{blue}}$ equation at the end.
|
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|
Geometry - Pentagon This is a tough program I have hard time to find the answer. Can anyone help me? Thank you very much in advance.
Problem - In regular pentagon $ABCDE$, point $M$ is the midpoint of side $AE$, and segments $AC$ and $BM$ intersect at point $Z$. If $ZA = 3$, what is the value of $AB$? Express your answer in simplest radical form.
This is a drawing for the problem:
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Let $F$ be the intersection of $AC$ and $BE$. By angle chasing, we obtain the following picture:
Now $\triangle BAF$ and $\triangle BEA$ are similar, so
$$BF:BA = BA:BE,$$
or
$$\frac yx = \frac{x}{x+y}=\frac{1}{1+y/x}.\tag{1}$$
Solving that for $y$ in terms of $x$ we have
$$\frac yx = \frac{\sqrt{5}-1}2.$$
Now apply Menelaus' Theorem to $\triangle AEF$ with $M$, $Z$, $B$ colinear we get
$$\frac{EM}{MA}\cdot \frac{ZA}{ZF}\cdot \frac{BF}{BE} = 1.\tag{2}$$
Note that $EM = MA$,
$$\frac{BF}{BE} = \frac{y}{x+y}=1-\frac{x}{x+y}=\frac{3-\sqrt{5}}2,$$
and
$$\frac{ZA}{ZF} = \frac{3}{y-3}.$$
Thus, (2) becomes
$$\frac{y-3}{3} = \frac{3-\sqrt{5}}2.$$
So
$$ y = \frac{9-3\sqrt{5}}{2}+3 = \frac{15-3\sqrt{5}}2.$$
And we get
$$ x = \frac xy \cdot y = \frac{\sqrt{5}+1}2 \cdot \frac{15-3\sqrt{5}}2$$
Note:
*
*I'm sure there are more elegant solutions.
*One can avoid Menelaus' Theorem by connecting $M$ with the midpoint $N$ of $AF$.
|
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|
find $\sum_{k=2}^{\infty}\frac{1}{k^2-1}$
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}$$
I found that:
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$$
and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{1}{2(n+1)}=\frac{1}{2}$ but the answer is $\frac{3}{4}$
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Notice that the first few terms are:
$(1/2-1/6)+(1/4-1/8)+(1/6-1/10)+(1/8-1/12)\cdots$
which gives: $1/2+1/4=3/4$.
|
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|
An approximation for $e^{ix}$ By the Taylor expansion of $\log(1+z)$ one can show that
$$e^{ix} = (1+ix)e^{-x^2/2+r(x)}$$
where $\lvert r(x)\rvert \leq \lvert x\rvert^3$ for all $x$ such that $\lvert x\rvert \leq 1/2$.
In a paper I saw the same claim except for the part where the author claims the the bound on $r(x)$ is true for all $x$. I have no clue how they arrived at this. Could this be an error? Or does someone have a proof for this?
|
If we define \begin{align}
r(x)=ix+\frac{x^2}{2}-\log (1+ix)\tag{1}\end{align}
then $e^{ix} = (1+ix)e^{-x^2/2+r(x)}$.
Using the Taylor expansion of $\log(1+z)$ in $(1)$ we have \begin{align}
r(x)&= ix+\frac{x^2}{2}-\left(ix+\frac{x^2}{2}-i\frac{x^3}{3}-\frac{x^4}{4}-i\frac{x^5}{5}+\cdots\right)\\
&=i\frac{x^3}{3}+\frac{x^4}{4}+i\frac{x^5}{5}+\cdots.
\end{align}
Therefore we have
\begin{align}
|r(x)|&\le \frac{|x|^3}{3}+\frac{|x|^4}{4}+\frac{|x|^5}{5}+\cdots\le\frac{|x|^3}{3}+\frac{|x|^4}{3}+\frac{|x|^5}{3}+\cdots\\
&=\frac{|x|^3}{3}\left( 1+|x|+|x|^2+\cdots\right)=\frac{|x|^3}{3}\cdot\frac{1}{1-|x|}\\
&\le \frac{2\,|x|^3}{3}\le |x|^3
\end{align}
for $|x|\le \frac{1}{2}$.
EDIT:
If we want to argue it for all values of $x$, use inequalities \begin{align}
&|x^2-\log (1+x^2)|\le |x|^3,\tag{1}\\
&|x-\tan^{-1}x|\le \frac{|x|^3}{3}\tag{2}
\end{align}
instead of the Taylor expansion of $\log (1+z)$.
Then \begin{align}
|r(x)|&=\left|ix+\frac{x^2}{2}-\log (1+ix)\right|\\
&=\left|\frac{\,1\,}{2}\left(x^2-\log (1+x^2)\right)+i(x-\tan^{-1} x)\right|\\
&\le \frac{\,1\,}{2}|x|^3+\frac{\,1\,}{3}|x|^3\le |x|^3.
\end{align}
|
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|
Compute all possible values of $\cot{\theta} - \frac{6}{z}$. Note that $ 0 < \theta < \frac{\pi}{2}$.
Let $\theta = \arg{z}$ and suppose $z$ satisfies $|z - 3i| = 3$.
Compute all possible values of $\displaystyle \cot{\theta} - \frac{6}{z}$. Note that $\displaystyle 0 < \theta < \frac{\pi}{2}$.
$\bf{My\; Try::}$ Let $z-3i=3e^{i\theta}\Rightarrow z=3i+3\cos \theta+3i\sin \theta = 3\cos \theta+3i(1+\sin \theta)$
So we get $$z=3\sin \left(\frac{\pi}{2}-\theta\right)+3i\left[1+\cos\left(\frac{\pi}{2}-\theta\right)\right]$$
So we get $$z=3i\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)e^{-i\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}$$
Now how can i solve after that, Thanks
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Since $\theta=\arg z$, note that $z=3\cos\theta+3i(1+\sin\theta)$ is not correct.
Let $z=x+yi$ where $x,y\in\mathbb R$. Then, since we have
$$x^2+(y-3)^2=3^2$$we can write
$$x=3\cos\alpha,\qquad y=3+3\sin\alpha$$
where $-\pi/2\lt\alpha\lt \pi/2$.
So, we can have
$$\cos\theta=\frac{x}{\sqrt{x^2+y^2}}=\frac{3\cos\alpha}{\sqrt{6(3+3\sin\alpha)}}$$
$$\sin\theta=\frac{y}{\sqrt{x^2+y^2}}=\frac{3+3\sin\alpha}{\sqrt{6(3+3\sin\alpha)}}$$
and so
$$\begin{align}\cot\theta-\frac{6}{z}&=\frac{3\cos\alpha}{3+3\sin\alpha}-\frac{6}{3\cos\alpha+(3+3\sin\alpha)i}\\&=\frac{\cos\alpha}{1+\sin\alpha}-\frac{2(\cos\alpha-(1+\sin\alpha)i)}{\cos^2\alpha+(1+\sin\alpha)^2}\\&=\frac{\cos\alpha}{1+\sin\alpha}-\frac{\cos\alpha-(1+\sin\alpha)i}{1+\sin\alpha}\\&=\frac{1+\sin\alpha}{1+\sin\alpha}i\\&=\color{red}{i}\end{align}$$
|
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|
Solving Recurrence Relation problem I am trying to solve the recurrence relation:
$$G_n = \frac{1-G_{n-1}}{4}$$
$$G_0 = 0$$
$$G_1 = \frac{1}{4}$$
I am told that the answer is
$$G_n = \frac{1}{5}\left(1+\left(\frac{-1}{4}\right)^{n+1}\right)$$
I have found the characteristic equation to be
$$G_n = B\left(\frac{-1}{4}\right)^{n}$$
But then I am not sure how to continue.
|
You have $G_n=\frac{1}{4}-\frac{1}{4}G_{n-1}$. With the ansatz $G_n=r^n$ we have for the homogeneous
$$r^n=-\frac{1}{4}r^{n-1}$$
$$r=-\frac{1}{4}$$
and for the particular
$$K=\frac{1}{4}-\frac{1}{4}K$$
$$K=\frac{1}{5}$$
so the general solution is $G_n=\frac{1}{5}+B\left(-\frac{1}{4}\right)^n$ and we plug in the initial conditions to get $0=\frac{1}{5}+B$, so $B=-\frac{1}{5}$.
Thus $G_n=\frac{1}{5}\left( 1-\left(-\frac{1}{4}\right)^n\right)$.
By the way, the answer you were given is not consistent with the initial values. If you plug in $n=0$ you get $G_0=\frac{3}{20}$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
determinant of 3x3 matrix algebra not matching CAS Find
$$\begin{vmatrix}
i & 2 & -1
\\ 3 & 1+i & 2
\\ -2i & 1 & 4-i \end{vmatrix}$$
we will expand along first row
$$ \begin{aligned}
\begin{vmatrix}
i & 2 & -1
\\ 3 & 1+i & 2
\\ -2i & 1 & 4-i \end{vmatrix}
&=i \begin{vmatrix} 1+i & 2 \\ 1 & 4-i \end{vmatrix}
-2 \begin{vmatrix} 3 & 2 \\ -2i& 4-i \end{vmatrix}
-1 \begin{vmatrix} 3 & 1+i \\ -2i & 1 \end{vmatrix}
\\&=i[(1+i)(4-i)-2]
-2[3(4-i)-2(-2i)]
-1[3-(1+i)(-2i)]
\\&=i[1(4-i)+i(4-i)-2]
-2[12-3i+4i]
-1[3+(2i)(1+i)]
\\ &= i[4-i+4i-i^2-2]
-2[12+i]
-1[3+2i+2i^2]
\\ &=i[4-3i+1-2]-2[12+i]-1[3+2i-2]
\\ &= i[3-3i]-2[12+i]-1[1+2i]
\\ &=i(3-3i)-2(12+i)-1(1+2i)
\\ &=3i-3i^2-24-2i-1-2i
\\ &=3i+3-24-2i-1-2i
\\ &=3-24-1+3i-2i-2i
\\& =22-i
\end{aligned}
$$
Did something wrong. Some silly algebra mistake, did not use the cofactor expansion formula right?? Cannot catch and is time to ask for help.
This is what maxima spits out it say the det is -28-i
|
Why not first add twice the first row to the third one to make things slightly simpler?
$$\begin{vmatrix}i&2&\!\!-1\\3&1+i&2\\0&5&2-i\end{vmatrix}=-5\begin{vmatrix}i&\!\!-1\\3&2\end{vmatrix}+(2-i)\begin{vmatrix}i&2\\3&1+i\end{vmatrix}\stackrel{\text{develop by third row}}=$$
$$=-5(2i+3)+(2-i)(-1+i-6)=-10i-15+(2-i)(-7+i)=$$
$$=-10i-15-13+9i=-28-i$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the ratio of $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$ $$I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx$$ and $$I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx$$ Then find $\frac{I_1}{I_2}$
I tried by subtracting $I_1$ and $I_2$
$$I_1-I_2=\int_{0}^{1}\left(1-x^{50}\right)^{100}\left(1-(1-x^{50}\right))dx$$ so
$$I_1-I_2=\int_{0}^{1} \left(1-x^{50}\right)^{100} x^{50} dx$$ Now using Integration by Parts we get
$$I_1-I_2= \left(1-x^{50}\right)^{100} \times \frac{x^{51}}{51}\bigg|_{0}^{1} -\int_{0}^{1} 100 \left(1-x^{50}\right)^{99} \times -50 x^{49} \times \frac{x^{51}}{51} dx$$
So
$$I_1-I_2=\frac{5050}{51} \times \int_{0}^{1}\left(1-x^{50}\right)^{99} x^{100} dx$$
Now $x^{100}=\left(1-x^{50}\right)^2-(1-x^{50}-x^{50})$ so
$$\frac{51}{5050}(I_1-I_2)=\int_{0}^{1} \left(1-x^{50}\right)^{99} \times \left(\left(1-x^{50}\right)^2-(1-x^{50})+x^{50}\right)dx=I_2-I_1+\int_{0}^{1} \left(1-x^{50}\right)^{99} x^{50} dx$$
Need a hint to proceed further.
|
Another approach is to note that $$\int_0^1\left(1-x^a\right)^bdx=\int_0^1\left(1-y\right)^b\tfrac{1}{a}y^{1/a-1}dy=\frac{1}{a}\text{B}\left(\frac{1}{a},\,b+1\right)=\frac{b!\Gamma\left(\tfrac{1}{a}+1\right)}{\Gamma\left(b+\tfrac{1}{a}+1\right)},$$ so $$\frac{\int_0^1\left(1-x^{50}\right)^{100}dx}{\int_0^1\left(1-x^{50}\right)^{101}dx}=\frac{\Gamma\left(102.02\right)}{101\Gamma\left(101.02\right)}=\frac{5051}{5050}.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Using Lagrangian multipliers to find maximum and minimum values of $f(x)=\frac{1}{\sqrt{(x+1/2)^2+y^2+z^2}}$ for $x^2+y^2+z^2=1$ We define $$f(x)=\frac{1}{\sqrt{(x+1/2)^2+y^2+z^2}}$$ and find the maximum and minimum values of $f$ such that $$x^2+y^2+z^2=1$$
Here is what I have so far:
$f(x)$ approaches a maximum as $1/f(x)$ approaches a minimum and vice versa also $1/(f(x))^2$ does the same. So we define a new function $h(x)=1/(f(x))^2=(x+1/2)^2+y^2+z^2$ (side question: is it possible to shift $h(x)$ so it is simply $x^2+y^2+z^2$? or is this invalid?)
So we define the Lagrangian as $$L=((x+1/2)^2+y^2+z^2)+\lambda(x^2+y^2+z^2-1)$$
Then we get the following after taking partial derivatives:
$$L_x=2x+1+2\lambda x=0\\L_y=2y(1+\lambda)=0 \\ L_z=2z(1+\lambda)=0\\L_\lambda=x^2+y^2+z^2-1=0$$
But from here I am stuck, do we want to solve for $x,y,z,\lambda$? And how do we proceed from there if that is the case?
|
You have to solve the four equations simultaneously for $\lambda, x, y, z$. The second equation gives either $\lambda=-1$ or $y=0$. But if $\lambda=-1$ then the first equation becomes $1=0$. Therefore $y=0$, and similarly $z=0$ from the third equation. The fourth equation then gives $x^2=1$, so either $x=1$ or $x=-1$.
You can verify that $x=\pm1$ and $y=z=0$ are the values where $h(x,y,z)$ attains its max and min by using @lab bhattacharjee's hint: Given the constraint $x^2+y^2+z^2=1$, the function $h$ reduces to $x+\frac54$, which attains its max and min under that constraint when $x=\pm1$.
|
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|
How do I solve the equation $x^5 - 5x^4 - 5x^3 +25x^2 +4x -20 = 0$ given that its roots are of the form $+a, -a, +b, -b, c$? I understand that it is an easy problem, but I am not able to solve it at all! Any clue on how to approach this problem will be amazing! Thank you. (PS: I am just beginning to learn math).
|
Let $P(x)=x^5 - 5x^4 - 5x^3 +25x^2 +4x -20$.
Note a equation that would have $-a,a,-b,b,c$ as it's roots would be of the form $(x-a)(x+a)(x-b)(x+b)(x-c)=(x^2-a^2)(x^2-b^2)(x-c)$.
So we get
$$P(x)=x^5 - 5x^4 - 5x^3 +25x^2 +4x -20 = (x^2-a^2)(x^2-b^2)(x-c)=(x^4-(a^2+b^2)x^2+a^2b^2)(x-c)=x^5-cx^4+\dots$$
This tells us that $c=5$.
By polynomial division, if we divide $(x^5 - 5x^4 - 5x^3 +25x^2 +4x -20)$ by $(x-5)$, note that we get $x^4-5x^2+4$.
However, note $$x^4-5x^2+4=(x^2)^2-5x^2+4=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2)$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Evaluation of $\int_{0}^{3a} \frac{x\sqrt{3a-x}}{\sqrt{a+x}} dx$ How can we evaluate following integral:
$$I=\int_{0}^{3a} \frac{x\sqrt{3a-x}}{\sqrt{a+x}} dx$$
I tried using the property $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$ but it not helping because of the denominator. How should I proceed?
|
Assume $a>0$. One may write
$$
\begin{align}
\int_{0}^{3a} \frac{x\sqrt{3a-x}}{\sqrt{a+x}} dx &=a^2\int_{0}^{3} \frac{t\sqrt{3-t}}{\sqrt{1+t}} dt \qquad (x=at)
\\\\&=2a^2\int_{1}^{2} (u^2-1)\sqrt{4-u^2}\: du \qquad \left(u=\sqrt{1+t}\:\right)
\\\\&=8a^2\int_{\pi/6}^{\pi/2} (3-4\cos^2 v)\cos^2 v\: dv \qquad \left(u=2\sin v\:\right)
\\\\&=\frac{3\sqrt{3}}2 a^2,
\end{align}
$$ where we have used
$$
\cos^2 v=\frac{1+\cos(2v)}2
$$ twice.
|
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"timestamp": "2023-03-29T00:00:00",
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|
(Easier) way to determine determinant of a almost-circulant matrix Consider the matrix:
$$\begin{pmatrix} \color{red}{-x+y} &-1&0&-1 \\ -1&\color{blue}{x+y}&-1&0
\\ 0&-1&\color{red}{-x+y}&-1 \\ -1&0&-1&\color{blue}{x+y} \end{pmatrix}$$
Is there any easy way to calculate the determinant of this matrix?
I have read up on circulant matrices and there is some nice theory on them, however this is not quite circulant, the only difference is that the main diagonal contains 2 different values.
So I was wondering if there is a 'short cut' to find the determinant of these kind of almost-circulant matrices.
Thanks.
|
Substract line two from line four and develop by first column (factoring out $\;x+y\;$ from last row):
$$\begin{vmatrix}-x+y&-1&0&-1\\-1&x+y&-1&0\\0&-1&-x+y&-1\\0&-x-y&0&x+y\end{vmatrix}=(-x+y)(x+y)\begin{vmatrix}x+y&-1&0\\-1&-x+y&-1\\-1&0&1\end{vmatrix}+$$
$$+(x+y)\begin{vmatrix}-1&0&-1\\-1&-x+y&-1\\-1&0&1\end{vmatrix}=$$$${}$$
$$=(y^2-x^2)\left(y^2-x^2-2\right)+(x+y)\left(2x-2y\right)=(y^2-x^2)(y^2-x^2-2)-2(y^2-x^2)=$$
$$=(y^2-x^2)(y^2-x^2-4)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that a positive integer is composite Let $ n $ be a positive integer that can be written as a sum of two relatively prime squares in two distinct ways, that is $ n = a^2 + b^2 = c^2 + d^2 $ so that $ gcd(a, b) = gcd(c, d) = 1, $ then $ n $ is composite.
So I have successfully factored $ n $ into $ \displaystyle \frac{ac + bd}{a - d}.\frac{ac - bd}{a + d} $ and have proved that $ \displaystyle (a - d) | (ac + bd) $ and $ \displaystyle (a + d) | (ac - bd). $ So each of the $ \displaystyle \frac{ac + bd}{a - d} $ and $ \displaystyle \frac{ac - bd}{a + d} $ is an integer, but I am currently stuck on proving that $ n $ is composite. My intention is to prove that both factors of $ n $ are greater than $ 1, $ but no success at this point. Any hint or suggestion? I have tried to use the fact that $ gcd(a, b) = gcd(c, d) = 1, $ but still no progress.
Thanks.
|
I carefully read Brillhart's first paper and realized I was unsure about the final step. So, naive or not, here is a lemma.
LEMMA Suppose, with positive integers $n,H,E,F,$ such that $$ nH = EF, $$
while
$$ 1 < E < F < n. $$
Define
$$ g = \gcd(n,E). $$
THEN both $g > 1$ and $g < n.$
PROOF First, $g \leq E < n.$ Next, we know there are integers $x,y$ such that $$ nx - E y = g. $$
$$ nFx - EF y = gF, $$
$$ nFx - nHy = gF, $$
$$ n(Fx - Hy) = gF > 0, $$
$$ n \leq n(Fx - Hy), $$
$$ n \leq gF. $$
Recall $$F < n.$$
$$ n \leq gF < gn. $$
$$ n < gn. $$
$$ 1 < g.$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
Using the letters in the original question, $n,a,b,c,d \geq 1,$ demand $a > d.$ Note that $a^2 - d^2 = c^2 - b^2 > 0.$ Next
$$ a^2 c^2 - b^2 d^2 = a^2 c^2 - a^2 b^2 + a^2 b^2 - b^2 d^2 $$
$$ a^2 c^2 - b^2 d^2 = a^2 (c^2 - b^2) + b^2 (a^2 - d^2) $$
$$ a^2 c^2 - b^2 d^2 = a^2 (a^2 - d^2) + b^2 (a^2 - d^2) $$
$$ a^2 c^2 - b^2 d^2 = (a^2 + b^2) (a^2 - d^2) $$
$$ a^2 c^2 - b^2 d^2 = n (a^2 - d^2), $$
$$ n (a^2 - d^2) = a^2 c^2 - b^2 d^2 = (ac-bd)(ac+bd) > 0.$$
It is not difficult to check the identity
$$ n^2 = (ac-bd)^2 + (ad+bc)^2, $$ that is one of those Brahmagupta things.
If we had $ac = bd$ we would also have $a/b = d/c,$ leading to $n = b^2 (1 + (a^2 / b^2)) = c^2 (1 + (d^2 / c^2)),$ finally $b=c$ and $a=d.$ This is ruled out, so $ac-bd \geq 1.$
It is also not difficult to check the identity
$$ n^2 = (ad-bc)^2 + (ac+bd)^2, $$ that is another of those Brahmagupta things, derivable from the first by negating $b.$
If we had $ad = bc$ we would also have $a/b = c/d,$ leading to $n = b^2 (1 + (a^2 / b^2)) = d^2 (1 + (c^2 / d^2)),$ finally $b=d$ and $a=c.$ This is ruled out, so $|ad-bc| \geq 1,$ AND $ac+bd < n.$
SO FAR,
$$ 1 \leq ac-bd < ac + bd < n. $$ However, recall
$$ n (a^2 - d^2) = (ac-bd)(ac+bd).$$ Since
$$ (ac-bd)(ac+bd) \geq n$$ but
$$ n > ac+bd, $$ we find
$$ (ac-bd)(ac+bd) > ac+bd $$ and
$$ ac-bd > 1. $$
We have reached
$$ 1 < ac-bd < ac + bd < n. $$
Using $ n (a^2 - d^2) = (ac-bd)(ac+bd),$ apply my LEMMA far above with $H = a^2 - d^2,$ $E = ac-bd,$ $F = ac+bd.$ The conclusion is that $g = \gcd(n, ac-bd)$ results in $1 < g < n.$ Therefore we have the nontrivial factorization $$ n = g \cdot \frac{n}{g}. $$
|
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|
recurrence relation unable to solve I am trying to solve recurrence relation : $$z_n = 2z_{n-1} + z_{n-2}
\;\;\;\;\;z_0=1\;\;\;z_1=3$$
Could you please help to provide a solution. I got stuck with Lamdas..
Are there some simple methods to solve any kind of these problems perhaps ? Like a good cookbook ?
Thanks
|
Here is an answer based upon generating functions.
With
\begin{align*}
A(t)=\sum_{n=0}^\infty{z_nt^n}
\end{align*}
we obtain
\begin{align*}
\sum_{n=2}^\infty z_nt^n&=2\sum_{n=2}^\infty z_{n-1}t^n+\sum_{n=2}^\infty z_{n-2}t^n\tag{1}\\
&=2t\sum_{n=1}^\infty z_{n}t^{n}+t^2\sum_{n=0}^\infty z_{n}t^{n}\tag{2}\\
\end{align*}
Comment:
*
*In (1) we start with the index $n\geq 2$ which is convenient since we have a second order recurrence relation.
*In (2) we shift the index in the right hand series by $1$ resp. $2$ in order to represent the series by $A(t)$.
From (2) we obtain an equation in terms of $A(t)$:
\begin{align*}
A(t)-1-3t&=2t\left(A(t)-1\right)+t^2A(t)\\
A(t)&=\frac{1+t}{1-2t-t^2}
\end{align*}
Partial fraction decomposition and expansion of the series $A(t)$ results in
\begin{align*}
A(t)&=\frac{1+t}{1-2t-t^2}\\
&=\frac{1-\sqrt{2}}{2}\cdot\frac{1}{1-\left(1-\sqrt{2}\right)t}+\frac{1+\sqrt{2}}{2}\cdot\frac{1}{1-\left(1+\sqrt{2}\right)t}\\
&=\frac{1-\sqrt{2}}{2}\sum_{n=0}^{\infty}\left(1-\sqrt{2}\right)^nt^n
+\frac{1+\sqrt{2}}{2}\sum_{n=0}^{\infty}\left(1+\sqrt{2}\right)^nt^n
\end{align*}
We conclude
The numbers $z_n$ fulfilling the recurrence relation
\begin{align*}
z_n=2z_{n-1}+z_{n-2}\qquad \qquad z_0=1, z_1=3
\end{align*}
are given by
\begin{align*}
z_n=\frac{1}{2}\left[\left(1-\sqrt{2}\right)^{n+1}
+\left(1+\sqrt{2}\right)^{n+1}\right]\qquad\qquad n\geq 0
\end{align*}
Hint: This approach and a lot of other nice material can be found in Generatingfunctionology by H.S. Wilf which is a great starter for problems of this kind.
|
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|
Why do we not include $c$ in the computation of the definite integral? Why is it when evaluating the definite integral we commonly opt to omit the constant $c$
$$\int_1^2x^2 \, dx= \left.\frac{x^3}{3} \right|_1^2 =\frac{2^3}{3}-\frac{1^3}{3}=\frac{7}{3}$$
But when evaluating the indefinite integral we include it?
$$\int x^2dx = \frac{x^3}{3} + c$$
Would it not be more accurate to instead do the following?
$$\int_1^2 x^2\,dx=\left.\frac{x^3}{3}+c\right|_1^2 = \left(\frac{2^3}{3}+c\right) - \left(\frac{1^3}{3}+c\right) = \left(\frac{7}{3}+c\right)$$
Is there a historical reason to all of this?
|
$$
\text{Becuase } \left(\frac{2^3}{3}+c\right) - \left(\frac{1^3}{3}+c\right) = \frac{7}{3} \ne \left(\frac{7}{3}+c\right).
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Integrate $\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$ Integrate $$\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$$
I have tried to do Tangent substitution, but there is huge power.
What method is better to use here?
|
I suspect the OP did not intend for this, but let's go with the residue theorem. Even if we make no simplification to the denominator, things end up pretty simple. Note that, when $z=e^{i x}$,
$$\cos^6{x}+\sin^6{x} = \left (\frac{z+z^{-1}}{2} \right )^6 - \left (\frac{z-z^{-1}}{2} \right )^6 = \frac1{16} \left (3 z^4+10+3 z^{-4} \right )$$
As usual, sub $z=e^{i x}$ in the integral; then the integral is equal to
$$-i 16 \oint_{|z|=1} \frac{dz}{z} \frac2{3 z^4+10+3 z^{-4}} = -i 32 \oint_{|z|=1} dz \frac{z^3}{3 z^8+10 z^4+3}$$
The poles of the denominator occur inside the unit circle where $z_k^4=-1/3$. By the residue theorem, the integral is equal to $i 2 \pi$ times the sum of the residues at these poles inside the unit circle, or
$$64 \pi \sum_{k=1}^4 \frac{z_k^3}{24 z_k^7+40 z_k^3} = 8 \pi \sum_{k=1}^4 \frac1{3 z_k^4+5} = 8 \pi (4) \left (\frac14 \right ) $$
Thus,
$$\int_0^{2 \pi} dx \frac{2}{\cos^6{x}+\sin^6{x}} = 8 \pi$$
|
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|
Grouping students into groups of 3 I have 6 students.
I wanted to divide the students into discussion groups of size 3.
I wished to do this such that all possible groups of 3 students get to discuss together for exactly 1 round.
I could do this trivially in 10 rounds so that the assignment of students into groups (A and B) at different rounds were:
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Round& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\ \hline
Student & & &\\ \hline
1 & A & A & A & A & A & A & A & A & A & A\\ \hline
2 & A & A & A & A & B & B & B & B & B & B\\ \hline
3 & A & B & B & B & A & A & A & B & B & B\\ \hline
4 & B & A & B & B & A & B & B & A & A & B\\ \hline
5 & B & B & A & B & B & A & B & A & B & A\\ \hline
6 & B & B & B & A & B & B & A & B & A & A\\ \hline
\end{array}
What if I would have had 9 students?
How to then find the optimal partition of students over the time, so that all possible groups of size 3 would be covered in the minimum number of rounds?
|
$$\frac1{3!}\binom9{3,3,3}=\frac1{3!}\frac{9!}{3!3!3!}=280$$
The second factor of LHS equals the number of arrangments when $9$ students are divided in $3$ distinghuishable groups of $3$. Since the groups are not distinghuishable we must repair the multiple counting. For that the first factor of LHS.
Applying it on $6$ (mentioned in your question) gives $\frac1{2!}\binom6{3}=10$ as it should.
|
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|
Why is $a_{214}$ nonintegral where $a_0 = 1, a_{n+1}=\frac{1 + a_0^5 + \cdots + a_n^5}{n + 1}?$ Define the sequence $a_n$ by the following.
$$a_0 = 1,$$
$$a_{n+1}=\frac{1 + a_0^5 + \cdots + a_n^5}{n + 1}$$
$a_{214}$ is nonintegral.
（https://oeis.org/A108394）
Please tell me the proof.
(By computing the sequence modulo 251 it is easy to show that $a_{251}$ is not integral.)
|
EDIT: This isn't fully correct and contains some imprecise calculations. The core is using modulo 215. To misquote the OP: "By computing the sequence modulo 215 it is easy to show that $a_{215}$ is not integral."
ORIGINAL:
Doing the calculations modulo 215 gives the following:
$$a_0=1$$
$$a_1=\frac{1+1^5}{1}=2$$
$$a_2=\frac{1+1^5+2^5}{2}=17$$
$$a_3=\frac{1+1^5+2^5+17^5}{3}=473297\equiv82\pmod{215}$$
$$a_4\equiv\frac{1+1^5+2^5+17^5+82^5}{4}\equiv42\pmod{215}$$
$$a_5\equiv\frac{1+1^5+2^5+17^5+82^5+42^5}{5}\equiv171\pmod{215}$$
$$a_6\equiv\frac{1+1^5+2^5+17^5+82^5+42^5+171^5}{6}\equiv171\pmod{215}$$
All consecutive terms are 171 module 215 which leads to your conclusion.
|
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|
Solve: $\int x^3\sqrt{4-x^2}dx$ $$\int x^3\sqrt{4-x^2}dx$$
I tried changing the expression like this:
$$\int x^3\sqrt{2^2-x^2}dx=\int x^3\sqrt{(2-x)(2+x)}dx$$
And that's where I got stuck. I tried integration by parts but it doesn't simplify the expression.
|
We can write the integral as follows
\begin{align*}
\int x^3\sqrt{4-x^2}dx&=\int x(x^2-4)\sqrt{4-x^2}dx+\int 4x\sqrt{4-x^2}dx\\
&=\int(4-x^2)\sqrt{4-x^2}(-xdx)-2\int (4-x^2)^{1/2}(-2xdx)\\
&=\int(4-x^2)^{3/2}(-xdx)-2\int (4-x^2)^{1/2}(-2xdx)\\
&=\frac{1}{2}\frac{(4-x^2)^{5/2}}{5/2}-2\frac{(4-x^2)^{3/2}}{3/2}+C\\
&=\frac15(4-x^2)^{5/2}-\frac43(4-x^2)^{3/2}+C
\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
for $f(x)=\ln({1-\frac{1}{x^2})}$ find $\lim_{n\to\infty}{f(1)+f(2)+...+f(n)}$
Let $f:(1, \infty)\to \Bbb R$ with $f(x)=\ln\left({1-\frac{1}{x^2}}\right)$
Find $$\lim\limits_{n\to\infty}\left[f(1)+f(2)+...+f(n)\right]$$
What I have done so far is
$$\begin{align}
S_n&=\sum_{k=2}^n{f(k)} \\
&= \ln\left(\prod_{k=2}^n \left(1-\frac{1}{k^2}\right)\right)\\
&= \ln\left(\prod_{k=2}^n\left(\frac{k-1}{k^2}\right)\prod_{k=2}^n\left(\frac{k+1}{k^2}\right)\right)\\
&=\ln\left(\prod_{k=2}^n\left(\frac{1}{k}-\frac{1}{k^2}\right)\prod_{k=2}^n\left(\frac{1}{k}+\frac{1}{k^2}\right)\right)
\end{align}$$
|
Since $f(n)=\ln \frac{(n-1)(n+1)}{n^2}$ for all $n\ge 2$,
$$
\sum_{k=2}^n f(n)=\ln \frac{1\cdot 3}{2\cdot 2}\cdot \frac{2\cdot 4}{3\cdot 3}\cdot \frac{3\cdot 5}{4\cdot 4}\cdot \cdots \cdot \frac{(n-1)(n+1)}{n\cdot n} =\ln \frac{n+1}{2n}.
$$
Therefore,
$$
\lim_{n\to\infty}\sum_{k=2}^n f(n)= -\ln 2.
$$
(I excluded $f(1)$, because it is not defined.)
|
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|
Understanding Leibniz formula: $D^n (f g) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}f D^kg$ I would like some clarification for better understanding of Leibniz formula: $$D^n (f g) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}f D^kg$$
If I use the formula with the following expression: $f(x)= x^3e^x$
$$D^n (f g) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}x^3 D^k e^x = \binom{n}{0} 6e^x + \binom{n}{1} 6x e^x + \binom{n}{2} 3x^2e^x + \binom{n}{3} x^3 e^x$$
But when using the commutative law I get a different result...
$$D^n (gf) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}e^x D^k x^3 = \binom{n}{0} e^x x^3 + \binom{n}{1} e^x 3x^2 + \binom{n}{2} e^x 6x + \binom{n}{3} e^x6$$
Since my question was to calculate $D^nf$ of $f(x) = x^3e^x$ I get two different expressions depending on which "order" I use which seems strange...
|
$$D^n (gf) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}e^x D^k x^3 = \binom{n}{0} e^x x^3 + \binom{n}{1} e^x 3x^2 + \binom{n}{2} e^x 6x + \binom{n}{3} e^x6$$
Second expression you wrote is correct but the first formula you wrote you made mistake.
Let me show where you made mistake
$$D^n (f g) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}x^3 D^k e^x \neq \binom{n}{0} 6e^x + \binom{n}{1} 6x e^x + \binom{n}{2} 3x^2e^x + \binom{n}{3} x^3 e^x$$
The first expression you wrote must be:
$$D^n (f g) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}x^3 D^k e^x =\binom{n}{0} \frac{d^n(x^3)}{dx^n} e^x + \binom{n}{1} \frac{d^{n-1}(x^3)}{dx^{n-1}} e^x+\binom{n}{2} \frac{d^{n-2}(x^3)}{dx^{n-2}} e^x +....+\binom{n}{n-3} \frac{d^3(x^3)}{dx^3}e^x + \binom{n}{n-2} \frac{d^2(x^3)}{dx^2}e^x+\binom{n}{n-1} \frac{d(x^3)}{dx}e^x+\binom{n}{n} x^3e^x $$
To use the fact: $$\frac{d^m(x^3)}{dx^m}=0$$where $m>3$
You can get
$$D^n (f g) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}x^3 D^k e^x =\binom{n}{n-3} \frac{d^3(x^3)}{dx^3}e^x + \binom{n}{n-2} \frac{d^2(x^3)}{dx^2}e^x+\binom{n}{n-1} \frac{d(x^3)}{dx}e^x+\binom{n}{n} x^3e^x $$
To use the fact: $$\binom{n}{n-k}=\binom{n}{k}=\frac{n!}{k!(n-k)!}$$
$$D^n (f g) = \sum\limits_{k=0}^{n} \binom{n}{k} D^{n-k}x^3 D^k e^x =\binom{n}{3} \frac{d^3(x^3)}{dx^3}e^x + \binom{n}{2} \frac{d^2(x^3)}{dx^2}e^x+\binom{n}{1} \frac{d(x^3)}{dx}e^x+\binom{n}{0} x^3e^x $$
|
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|
Find all integer solutions to $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$
Find all integer solutions $(x, y)$ of the equation
$$\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$$
What have done is that:
$$\frac{1}{x}= \frac{2y-3}{3y}$$
so,
$$x=\frac{3y}{2y-3}$$
If $2y-3 = +1 \text{ or } {-1}$, $x$ will be an integer, so we choose integer $y$
to make $2y-3=1 \text{ or } {-1}$. $y = 2$ or $y = 1$ is such a solution.
Also, $2y - 3$ can be deleted by numerator $3$, so $2y - 3$ can be $3$ or $-3$ too. This gives $y = 3$ or $y = 0$, but $y$ can not be $0$. So far, we have $y=1,2,3$. Finally, $2y-3$ can be deleted by numerator $y$, but how can we find such a $y$?
|
If $0<x\le y$ then $\frac23 = \frac1x+\frac1y \le \frac1x+\frac1x$ and so $x\le3$.
If $x<0$ then $\frac23 < \frac1{-x}+\frac23=\frac1y$ and so $y=1$.
Therefore $(x,y) \in \{ (3,3) , (-3,1) , (1,-3), (2,6),(6,2)\}$.
|
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|
Find a polynomial f(x) of degree 5 such that 2 properties hold. I have been trying to find a polynomial $f(x)$ such that these $2$ properties hold:
*
*$f(x)-1$ is divisible by $(x-1)^3$
*$f(x)$ is divisible by $x^3$
To start, I set $f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f$.
This is divisible by $x^3$, so $d, e, f $ must be $0$. So polynomial $f(x) = ax^5 + bx^4 + cx^3$
This minus $1$ is divisible by $(x-1)^3$. So I used synthetic division and got that the remainder of $ax^5 + bx^4 + cx^3$ divided by $(x-1)$ is $a+b+c-1$. But it should divide out evenly, so $a+b+c-1 = 0$.
From here, I can't find any other equations involving the three variables. Could someone have any suggestions for how to continue? Thanks in advance. :)
|
When a repeated factor $(x-a)^3$ is present in polynomial $P(x)$, then $P(a) = P'(a) = P''(a) = 0$
So set $f(1) - 1 = 0, f'(1) = 0, f''(1) = 0$
$f(x) - 1 = ax^5 + bx^4 + cx^3 - 1$
$f'(x) = 5ax^4 + 4bx^3 + 3cx^2$
$f''(x) = 20ax^3 + 12bx^2 + 6cx$
Hence your three equations are:
$a + b + c - 1 = 0 $
$5a + 4b + 3c = 0$
$20a + 12b + 6c = 0$
and you should be able to solve them to get $(a,b,c) = (6,-15, 10)$ so $f(x) = 6x^5 - 15x^4 + 10x^3$ = $x^3(6x^2-15x + 10)$.
|
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|
Find all continuous functions $f$ over real numbers such that $f(x)+x = 2f(2x)$
Find all continuous functions $f$ over real numbers such that $f(x)+x = 2f(2x)$.
We have $f(0) = 0$ and $f(x) = 2f(2x) - x$, but I am not sure how to convert this functional equation into something that is easier to solve. Maybe using induction may work, but I don't see an easy way to induct since we only have one variable.
|
We prove that there is only one function $f:\mathbb R\to\mathbb R$ that satisfies $2f(2x)=f(x)+x$ and
\begin{align*}
\lim_{t\to0}tf(xt)=0
\end{align*}
for all $x\in\mathbb R$, namely $f(x)=x/3$. Note that if $f$ is continuous, it satisfies this limit condition (why?)!
By iterating $f(x)=\frac{1}{2}f(x/2)+x/4$ one finds the formula
\begin{align*}
f(x)=\frac{1}{2^n}f\left(\frac{x}{2^n}\right)+\frac{x}{4^n}\sum_{k=0}^{n-1}4^k,\quad n\geq 1,x\in\mathbb R.\qquad (\star)
\end{align*}
This can also be shown by induction: For $n=1$, this is just the definition of $f$. Assume that $(\star)$ has been proven for some $n\geq 1$. Then, again by definition,
\begin{align*}
f\left(\frac{x}{2^n}\right)=\frac{1}{2}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{2^{n+2}},
\end{align*}
and if we put this into our formula (induction!), we obtain
\begin{align*}
f(x)=
\frac{1}{2^n}\left(\frac{1}{2}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{2^{n+2}}\right)+\frac{x}{4^n}\sum_{k=0}^{n-1}4^k=
\frac{1}{2^{n+1}}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{4^{n+1}}+\frac{x}{4^n}\sum_{k=0}^{n-1}4^k=
\frac{1}{2^{n+1}}f\left(\frac{x}{2^{n+1}}\right)+\frac{x}{4^{n+1}}\sum_{k=0}^{n}4^k,
\end{align*}
which proves $(\star)$.
Now, if we fix $x\in\mathbb R$ and take the limit $n\to\infty$ in $(\star)$ we obtain due to the limit condition that $f(x)=x/3$.
|
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|
Cubic modular equation
If I found the sol of (a), I could find sol of (b) and (c) by using Hensel's lemma.
I want to know the way of solving (a) without testing all the cases (ex testing 1,2,3,4,...10)
I think that there will be other way not testing all residues. It takes such a long time...
Or is there short cut for testing..?
|
Note the polynomial in a) can be written as
$$x^3-3x^2-x-1=(x-1)^3-3x+1-x-1=(x-1)^3-4(x-1)-4.$$
Hence, setting $x-1=y$, it is enough to solve $y^3=4y+4 $.
We'll use Cardano's method: set $y=u+v$. The last equation becomes
$$u^3+v^3+3uv(u+v)=4(u+v)+4,$$
and adding the condition $3uv=4$, it comes down to
$$\begin{cases}
uv=5\\u^3+v^3=4
\end{cases} \iff \begin{cases}
u^3v^3=4\\u^3+v^3=4
\end{cases} $$
Thus $u^3, v^3$ are solutions to the quadratic equation $\;t^2-4t+4=(t-2)^2=0$: $\;u^3=v^3=2$. If a draw a table of cubes in $\mathbf Z/11\mathbf Z$:
$$\begin{matrix}
x&0&\pm1&\pm2&\pm3&\pm4&\pm5\\\hline
x^3&0&\pm1&\mp3&\pm5&\mp2&\pm4\end{matrix}$$
we see that $u=v=-4$, whence $y=3$ and finally $\color{red}{x=4.}$
To obtain the other roots, divide $y^3-4y-4$ by $y-3$:
$$y^3-4y-4=(y-3)(y^2+3y+5)=(y-3)(y^2-8y+5)=(y-3)(y-4)^2,$$
so the equation in $x$ also has a double root: $\color{red}{x=5.}$
|
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|
Series $1^3-1.5^3+2^3-2.5^3+...$ I came across a question today...
Sum to $n$ term of the series $$1^3-1.5^3+2^3-2.5^3+...$$ is?
I divided this series in two different series $$\left(1^3+2^3+3^3+...\right)-\left(1.5^3+2.5^3+3.5^3+...\right)$$
$$\Rightarrow\sum^{\frac{n+1}{2}}_{r=1}r^3-\sum^{\frac{n-1}{2}}_{r=1}\left(r+\frac{1}{2}\right)^3$$
$$\Rightarrow\sum^{\frac{n+1}{2}}_{r=1}r^3-\sum^{\frac{n-1}{2}}_{r=1}\left(r^3+\frac{1}{8}+1.5r^2+0.75r\right)$$
I solved this and couldn't get to the answer. Am I doing it in the right way?
|
Hints: You can represent the series as $$(\frac{-n}2)^3$$
Assume even number of terms
$$n^3 - (n+1)^3 = -(3n^2+3n+1)$$
|
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|
How to compute $\int \frac{1}{(x^2+1)^2}dx$? Suppose we know $\int \frac{1-x^2}{(x^2+1)^2}=\frac{x}{x^2+1}+C$
How to compute $\int \frac{1}{(x^2+1)^2}dx$?
I tried writing it as $\frac{1+x^2-x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}+\frac{x^2}{(x^2+1)^2}$. But then how do you deal with $\frac{x^2}{(x^2+1)^2}$?
Some ideas?
|
The approach shown by Elliot G is perfectly fine, but if you like a viable alternative, you may consider that:
$$ \frac{1}{x^2+1} = \frac{1}{(x+i)(x-i)} = \frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right) $$
hence by squaring both sides:
$$ \frac{1}{(x^2+1)^2} = -\frac{1}{4}\left(\frac{1}{(x-i)^2}+\frac{1}{(x+i)^2}-\frac{2}{x^2+1}\right)$$
so:
$$ \int\frac{dx}{(x^2+1)^2} = C+\frac{1}{4}\left(\frac{1}{x-i}+\frac{1}{x+i}\right)+\frac{1}{2}\arctan x $$
or:
$$ \int\frac{dx}{(x^2+1)^2} = C+\frac{1}{2}\left(\frac{x}{x^2+1}+\arctan x\right).$$
|
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|
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$.
$4\cos x - 3 \sin x = k\sin(x - \alpha)$
=> $k(\sin x \cos \alpha - \cos x \sin \alpha)$
=> $k\cos \alpha \sin x - \sin \alpha \cos x$
Equating coefficients:
$k\cos \alpha = - 3$
$k\sin \alpha = 4$
$k = \sqrt{(-3)^2 + 4^2} = 5$
$\alpha$ is in the fourth quarter because $ \cos \alpha$ is positive and $\sin \alpha$ is negative.
$\alpha = \arctan\frac{4}{-3} = -53.1$
$\alpha = 360 - 53.1 = 306.9$
$4\cos x - 3 \sin x = 5\sin(x - 306.9)$
But the answer is $4\cos x - 3 \sin x = 5\sin(x - 233.1)$
|
We have
$$
\begin{align}
&4 \cos x − 3 \sin x \\
&= 5 \left( \frac{4}{5} \cos x - \frac{3}{5} \sin x \right) \\
&= 5 \left( \sin (\tan^{-1} (\tfrac{4}{3})) \cos x
- \cos (\tan^{-1} (\tfrac{4}{3})) \sin x \right) \\
&= 5 \sin\left(\tan^{-1} (\tfrac{4}{3}) - x \right) \\
&= 5 \sin\left(x - (\tan^{-1} (\tfrac{4}{3}) - 180^\circ{}) \right)
\end{align}
$$
|
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|
Given $a+b+..=a^7+b^7+..=0$ show that $a(a+b)..=0$ Question:
Suppose $a,b,c,d$ are real numbers such that
$a+b+c+d=a^7+b^7+c^7+d^7=0$
Show that
$a(a+b)(a+c)(a+d)=0$
My attempt: Using $a+b+c+d=0$, I get
$a(a+b)(a+c)(a+d)= 0
\implies a=0, \text{or}$ $ a(bc+cd+db)+bcd=0$
How can I use $a^7+b^7+c^7+d^7=0$ to prove $a(bc+cd+db)+bcd=0$ ?
Edit:(courtesy @mathguy)
Replacing $d$ by $-(a+b+c)$ we see that the hypothesis is equivalent to
$(a+b+c)^7=a^7+b^7+c^7$, and the conclusion equivalent to $a(a+b)(a+c)(b+c)=0$. The polynomial $(a+b+c)^7-a^7-b^7-c^7$ is divisible by $(a+b)(a+c)(b+c)$, and another irreducible fourth degree symmetric polynomial $P(a,b,c)$ .
Also, $P(0,b,c)= b^4+2b^3c+3b^2c^2+2bc^3+c^4$. It remains to be shown that $P(a,b,c)=0$ when $a=0$
|
Since $d=-(a+b+c)$ and $d^7=-(a^7+b^7+c^7)$, we obtain
$$(a+b+c)^7-a^7-b^7-c^7=0$$ or
$$7(a+b)(a+c)(b+c)\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$$ or
$$(a+b)(a+c)(a+d)\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0.$$
Thus, it remains to prove that if $\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$ so $a=0$.
We'll prove that $\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)\geq0$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$,where $v^2$ can be negative, and $abc=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where $f$ is a linear function.
But the linear function gets a minimal value for en extremal value of $w^3$.
Since $a$, $b$ and $c$ are real roots of the equation $(x-a)(x-b)(x-c)=0$ or
$x^3-3ux^2+3v^2x=w^3$,
we see that the graph of $g(x)=x^3-3ux^2+3v^2x$ and a line $y=w^3$ have three common points
and $w^3$ gets an extremal value, when a line $y=w^3$ is a tangent line to the graph of $g$,
which happens for equality case of two variables.
Since our inequality is symmetric, we can assume $c=b$,
which gives $a^4+4a^3b+11a^2b^2+14ab^3+9b^4\geq0$ or
$$(a+b)^4+ 5(a+b)^2b^2+3b^4\geq0,$$
which is obvious.
The equality occurs for $a=b=0$ and we proved that
if
$\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+5a^2bc)=0$ so $a=b=c=0$
and we are done!
|
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|
Easy partial fraction decomposition with complex numbers There is an easy method to perform a partial fraction decomposition - described here, under the "Repeated Real Roots" title, for the coefficient A2.
The problem is - this method doesn't work in some cases, for example:
$$
\frac{1}{(2 + j \omega)^2 \, (4 + j \omega)}
= \frac{A}{2 + j \omega}
+ \frac{B}{(2 + j \omega)^2}
+ \frac{C}{4 + j \omega}
$$
The coefficients B,C get their proper value:
$$
B = \frac{1}{4 + j \omega} \Big|_{\omega=2j} = \frac 12
$$
$$
C = \frac{1}{(2 + j \omega)^2} \Big|_{\omega=4j} = \frac 14
$$
But the coefficient A get a false value:
$$
A = \frac{d}{d \omega} \left( \frac{1}{4 + j \omega} \right)\Big|_{\omega=2j}
= \frac{-j}{4}
$$
The true/proper value for A is $-\frac 14$.
I wonder what is the general formula for this kind of elegant/quick/easy trick to find the coefficients during partial fraction decomposition?
|
The partial fraction expansion is
$$\frac{1}{(2+j\omega)^2(4+j\omega)}=\frac{A}{2+j\omega}+\frac{B}{(2+j\omega)^2}+\frac{C}{4+j\omega}\tag 1$$
To find $C$, we multiply both sides of $(1)$ by $4+j\omega$ and take the limit as $j\omega \to -4$. Proceeding, we find
$$\begin{align}
\lim_{j\omega \to -4}\frac{(4+j\omega)}{(2+j\omega)^2(4+j\omega)}&=\frac14\\\\
&=\lim_{j\omega \to -4}\left(\frac{A(4+j\omega)}{2+j\omega}+\frac{B(4+j\omega)}{(2+j\omega)^2}+\frac{C(4+j\omega)}{4+j\omega}\right)\\\\
&=C
\end{align}$$
To find $B$ , we multiply both sides of $(1)$ by $(2+j\omega)^2$ and take the limit as $j\omega \to -2$. Proceeding, we find
$$\begin{align}
\lim_{j\omega \to -2}\frac{(2+j\omega)^2}{(2+j\omega)^2(4+j\omega)}&=\frac12\\\\
&=\lim_{j\omega \to -2}\left(\frac{A(2+j\omega)^2}{2+j\omega}+\frac{B(2+j\omega)^2}{(2+j\omega)^2}+\frac{C(2+j\omega)^2}{4+j\omega}\right)\\\\
&=B
\end{align}$$
To find $A$ , we multiply both sides of $(1)$ by $(2+j\omega)^2$, take a derivative with respect to $j\omega$, and take the limit as $j\omega \to -2$. Proceeding, we find
$$\begin{align}
\lim_{j\omega \to -2}\frac{d}{d(j\omega)}\left(\frac{(2+j\omega)^2}{(2+j\omega)^2(4+j\omega)}\right)&=-\frac14\\\\
&=\lim_{j\omega \to -2}\frac{d}{d(j\omega)}\left(\frac{A(2+j\omega)^2}{2+j\omega}+\frac{B(2+j\omega)^2}{(2+j\omega)^2}+\frac{C(2+j\omega)^2}{4+j\omega}\right)\\\\
&=A
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Considering the power series $\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n+1)(2n-1)}$. Consider the power series $\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n+1)(2n-1)}$. Find a closed form expression for all x which converge and hence evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)(2n-1)}$.
Attempt at the solution: The radius of convergence is 1. We can rewrite the summands by:
\begin{eqnarray}
\sum_{n=1}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)(2n-1)} &=& \frac{1}{2}\Big[(x^3-\frac{x^3}{3}) - (\frac{x^5}{3} - \frac{x^5}{5}) + (\frac{x^7}{5} - \frac{x^7}{7}) + \dots \Big]\\
&=& \frac{1}{2}\Big[(x^3 -\frac{x^5}{3}+ \frac{x^7}{5} + \dots ) + (\frac{-x^3}{3} + \frac{x^5}{5}-\frac{x^7}{7} +...)\Big]\\
&=& \frac{1}{2}\Big[x^2\int\frac{1}{1+x^2}dx + \int\frac{1}{1+x^2}dx-x\Big]\\
&=& \frac{x^2}{2}\arctan(x)+\frac{1}{2}\arctan(x) -\frac{x}{2}
\end{eqnarray}
Substituting $x=1$ then gives $\frac{\pi}{4}-\frac{1}{2}$ .
The issue I have is two fold. Firstly, when dealing with evaluations at the boundary, term by term differentiation may not be valid. In particular, we used the fact that $\arctan(x) = x-\frac{x^3}{3} + ...$ by integrating power series for $\frac{1}{1+x^2}$, valid for |x|<1. This means that the arctan formula can only be guaranteed to hold within the interior (-1,1). What are the conditions needed to talk about power series validity at boundary points?
(Abelian/Tauberian theorems came to mind at first, but the conditions in this problem weren't strong enough. Alternatively, I noted that uniform convergence of the terms meant that the limit function of $x-\frac{x^3}{3} + ...$ had to be continuous. So $\arctan(1) = \frac{\pi}{4}$ by continuous extension. Do correct me if I'm wrong.
The other issue that I have not been able to justify is that of conditional convergence. Clearly, the arctan series is conditionally convergent at $x=1$. How do we justify the rearrangments carried out above then?
|
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\color{#f00}{\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{x^{2n + 1} \over \pars{2n + 1}\pars{2n - 1}}} =
\half\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{x^{2n + 1} \over 2n - 1} - \half\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{x^{2n + 1} \over 2n + 1}
\\[4mm] = &\
\half\,x^{3} + \half\,x^{2}\sum_{n = 1}^{\infty}
\pars{-1}^{n}\,{x^{2n + 1} \over 2n + 1} - \half\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{x^{2n + 1} \over 2n + 1}
\\[4mm] = &\
\half\,x^{3} + \half\,\pars{x^{2} + 1}x\
\underbrace{\sum_{n = 1}^{\infty}
\pars{-1}^{n}\,{x^{2n} \over 2n + 1}}_{\ds{\equiv\ \,\mathcal{J}}}\tag{1}
\end{align}
\begin{align}
\,\mathcal{J} & =
\sum_{n = 1}^{\infty}
\pars{-1}^{n}\,{x^{2n} \over 2n + 1} =
\sum_{n = 1}^{\infty}\pars{-1}^{n}\,x^{2n}\int_{0}^{1}y^{2n}\,\dd y =
\int_{0}^{1}\sum_{n = 1}^{\infty}\pars{-x^{2}y^{2}}^{n}\,\dd y
\\[4mm] &\ =\
\overbrace{\int_{0}^{1}{-x^{2}y^{2} \over 1 + x^{2}y^{2}}\,\dd y}
^{\ds{-1 + {\arctan\pars{x} \over x}}}
\end{align}
$$
\color{#f00}{\sum_{n = 1}^{\infty}
\pars{-1}^{n - 1}\,{x^{2n + 1} \over \pars{2n + 1}\pars{2n - 1}}} =
\color{#f00}{\half\bracks{\pars{x^{2} + 1}\arctan\pars{x} - x}}
$$
When $\ds{x \to 1^{-}}$, $\ds{\sum\cdots \to {\pi \over 4} - \half}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is there a difference of $\frac{-1}{6}$ when integrating $(1-2x)^2$ in two different ways? I integrated $(1-2x)^2$ by expanding the expression first, that is $1 - 4 x + 4 x^2$ and I get $$x - 2 x^2 + \frac{4 x^3}{3} +C$$
When I integrate by substitution I get $$\frac{-(1-2x)^3}{6} + C$$
Expanding $(-(1-2x)^3)/6 +C$ get $$-\frac{1}{6} + x - 2 x^2 + \frac{4 x^3}{3} +C$$
Why the difference in the two answers?
|
Indefinite integral can differ by a constant (you denote it by $C$). In this case you just have $C=-1/6$.
This is simply given by the fact that indefinite integral (or for this purpose better name would be antiderivative) of $f(x)$ is a function (not the function) $F(x)$ such that $F'(x) = f(x)$. And since derivative of the constant is $0$ (zero), this works for both of your results, namely you have:
$$\left(x - 2 x^2 + \frac{4 x^3}{3}\right)' = 1-4x+4x^2$$
$$\left(x - 2 x^2 + \frac{4 x^3}{3}-\frac{1}{6}\right)' = 1-4x+4x^2$$
Also it is not hard to end up with another constant. For example by choosing substitution $t=2-2x$ you can easily verify that after expanding and integrating you end up with constant $C=-1/3$.
I suggest to check other answers related to this where constant is even more tricky, for example:
Different results from the same integral with two different methods, Two different solutions to integral or Is $\int \frac{1}{2x} \, dx $ equal to $\frac{\ln|2x|}{2}+ C$ or $\frac{\ln|x|}{2}+ C$.
|
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|
The function $\lim_{n\to\infty}({4^n+x^{2n}+\frac{1}{x^{2n}})}^{1/n}$ is non-derivable at
The function $$f(x)=\lim_{n\to\infty}\left(4^n+x^{2n}+\frac{1}{x^{2n}}\right)^{\frac{1}{n}}$$ is non-derivable at how many points?
The limit is of $\infty^0$ form. Is it an indeterminate form or is it simply equal to $1$?
$$\lim_{n\to\infty}\left(4^n+x^{2n}+\frac{1}{x^{2n}}\right)^{\frac{1}{n}}=\lim_{n\to\infty}\frac{(4^nx^{2n}+x^{4n}+1)^{\frac{1}{n}}}{x^2}=\frac{1}{x^2}$$
|
If $\frac14 < x^2 < 4$, then we can write
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)^{1/n}&=\lim_{n\to \infty}4\left(1+\left(\frac{x^2}{4}\right)^n+\left(\frac{x^{-2}}{4}\right)^n\right)^{1/n}\\\\
&=4
\end{align}$$
If $x^2>4$, then
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)^{1/n}&=\lim_{n\to \infty}x^2\left(1+\left(\frac{4}{x^2}\right)^n+\left(\frac{1}{x^4}\right)^n\right)^{1/n}\\\\
&=x^2
\end{align}$$
If If $x^2<1/4$, then
$$\begin{align}
\lim_{n\to \infty}\left(4^n+x^{2n}+x^{-2n}\right)^{1/n}&=\lim_{n\to \infty}x^{-2}\left(1+\left(4x^2\right)^n+\left(x^4\right)^n\right)^{1/n}\\\\
&=x^{-2}
\end{align}$$
Therefore, we can write
$$f(x)=\begin{cases}x^2&,|x|>2\\\\4&\frac12\le |x|\le 2\\\\\frac{1}{x^2}&,0<|x|<\frac12\end{cases}$$
Obviously, $f$ is differentiable everywhere it is defined except when $x\ne \pm \frac12$ and $x=\pm 2$. Note that $f$ is undefined at $x=0$.
|
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|
How do I go about solving this? I have tried substitution, but it is not working for me.
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+1)}+\sin(x)+n\cos(x)}=\int_0^\pi \frac{n dx}{\sqrt{(n^2+1)}+n\sin(x)+\cos(x)}=2
$$
General form of this integral is
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+m^2)}+m\sin(x)+n\cos(x)}=\frac{2}{m}
$$
|
Let
$$
I_1=\int_0^\pi \frac{dx}{\sqrt{n^2+1}+\sin(x)+n\cos(x)}, I_2=\int_0^\pi \frac{n dx}{\sqrt{n^2+1}+n\sin(x)+\cos(x)}.
$$
Let $u=\tan\frac{x}{2}$. Then $x=2\arctan u$, $\sin x=\frac{2u}{1+u^2},\cos x=\frac{1-u^2}{1+u^2}$. Thus
\begin{eqnarray}
I_1&=&\int_0^\infty \frac{1}{\sqrt{n^2+1}+\frac{2u}{1+u^2}+n\frac{1-u^2}{1+u^2}}\frac2{1+u^2}du\\
&=&2\int_0^\infty \frac{1}{\sqrt{n^2+1}(1+u^2)+2u+n(1-u^2)}du\\
&=&2\int_0^\infty \frac{1}{(\sqrt{n^2+1}-n)u^2+2u+(\sqrt{n^2+1}+n)}dx\\
&=&2\int_0^\infty \frac{\sqrt{n^2+1}+n}{u^2+2(\sqrt{n^2+1}+n)u+(\sqrt{n^2+1}+n)^2}du\\
&=&2\int_0^\infty \frac{\sqrt{n^2+1}+n}{(u+\sqrt{n^2+1}+n)^2}du\\
&=&2.
\end{eqnarray}
Similarly
$$ I_2=2. $$
Done.
|
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|
How is the last "=" true? How can the last equality be true?
$$
G(u)=\frac{g}{(1+u)^g-1}-\frac1u=\frac{g}{gu + \cdots + u^g} - \frac{1}{u}=\sum_{k=0}^\infty c_ku^k
$$
|
If you look at the left-hadn side of the equals sign, you have
$$
\begin{align}
\frac{g}{gu + \cdots + u^g} - \frac{1}{u}
&= \frac{1}{u}\frac{g}{g + {g \choose 2} u + \cdots + u^{g-1}} - \frac{1}{u} \\
&= \frac{1}{u}\frac{1}{1 + \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}} - \frac{1}{u}
\end{align}
$$
If you then expand the first term as a geometric series, you will see that the terms with $\frac{1}{u}$ cancel out, leaving a series of the form $\sum_{k=0}^\infty c_ku^k$ for some undetermined coefficents $c_k$.
Edit: Let's expand on this a little bit. Note that we have (formally!) the expansion
$$
\frac{1}{1 + A} = 1 - A + A^2 - A^3 + \cdots
$$
for any $A$. In particular, if we let $A = \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}$, then we have
$$
\begin{align}
\frac{1}{u}\frac{1}{1 + \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}} - \frac{1}{u}
&=
\frac{1}{u}\frac{1}{1 + A} - \frac{1}{u} \\
&=
\frac{1}{u}(1 - A + A^2 - A^3 + \cdots) - \frac{1}{u} \\
&=
\frac{1}{u} - \frac{1}{u}A + \frac{1}{u}A^2 - \frac{1}{u}A^3 + \cdots - \frac{1}{u} \\
&=
- \frac{1}{u}A + \frac{1}{u}A^2 - \frac{1}{u}A^3 + \cdots
\end{align}
$$
Since $A = \frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}$ this now becomes
$$
- \frac{1}{u}\Big(\frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}\Big) + \frac{1}{u}\Big(\frac{1}{g}{g \choose 2} u + \cdots + \frac{1}{g}u^{g-1}\Big)^2 - \cdots
$$
which is a series that only contains positive exponent terms in $u$.
|
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|
Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$ If we have
$$\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$$
Then, what will be the set of $x$ for which this equation is true?
I tried to solve it by putting $x = \sin a$ or $\cos a$ but got no result.
I am totally stuck on how to do it.
|
Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
$$
2\arcsin x =\begin{cases}
\arcsin(2x\sqrt{1-x^2}) \;\;;2x^2\le 1 \\
\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2 > 1, 0< x\le 1\\
-\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2> 1, -1< x \le 0
\end{cases}
$$
So the given equation clearly holds true for $2x^2\le1\iff-\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$
For the other two cases, we need $$\arcsin(2x\sqrt{1-x^2})=\pm\dfrac\pi2$$
$$\iff2x\sqrt{1-x^2}=\sin\left(\pm\dfrac\pi2\right)=\pm\sin\dfrac\pi2=\pm1$$
$$\iff(2x^2-1)^2=0$$ which has already been covered in the first case
|
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|
Prove when $abc=1$: $ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$
Question: Prove the following inequality which holds for all positive reals $a$, $b$ and $c$ such that $abc=1$:
$$ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$$
My thoughts were turning the right hand side to $abc$ as $abc=1$ however I think this will make the proving even more harder. I also attempted applying the Cauchy-Schwarz inequality and Hölder's inequality but to no avail.
Could someone please show me how to prove it using the inequalities above or another method.
|
If $a,b,c> 0$ and $abc=1$, then by Cauchy-Schwarz: $$\sum_{\text{cyc}}\frac{a}{2+bc}=\sum_{\text{cyc}}\frac{a^2}{2a+abc}=\sum_{\text{cyc}}\frac{a^2}{2a+1}\ge \frac{(a+b+c)^2}{2(a+b+c)+3}$$
And by AM-GM $a+b+c\ge 3\sqrt[3]{abc}=3$, so: $$(a+b+c)^2-2(a+b+c)-3=((a+b+c)-3)((a+b+c)+1)\ge 0$$
Equality holds if and only if $a=b=c=1$.
|
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|
How many values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
How many different values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
I was wondering about this problem and didn't think it was immediately obvious. The answer can't be $2^{n-1}$ since the combinations all might not be unique. I was thinking of looking at partitions of $\{1,2,\ldots,n, k\}$ where $1 \pm 2 \pm 3 \pm \cdots \pm n = k$ and thus the sum must be even. Therefore, $\dfrac{n(n+1)}{2}+k = \dfrac{n(n+1)+2k}{2}$ must be even which means $n(n+1) + 2k$ must be a multiple of $4$ so $n(n+1)+2k = 4m$. Thus if $n \equiv 1,2 \bmod 4$, then $k \equiv 1,3 \bmod 4$ and if $n \equiv 3,0 \bmod 4$ then $k \equiv 0,2 \bmod 4$.
|
Hint:
$$ 1\pm 2\pm 3\pm 4\ldots \pm n = \frac{n(n+1)}{2}-a_2-a_3-\ldots-a_n, \qquad a_k\in\{0,2k\}.$$
Consider $(1+x^4)(1+x^6)\cdot\ldots\cdot(1+x^{2n})$. How many monomials appear in the expansion?
|
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|
Is this an acceptable trig-sub and reversion from trig at the answer? Today I had to take the indefinite integral $\int x^3 \sqrt{x^2-1} \, dx$
My steps:
*
*$x=\sec\theta$, $dx=\sec\theta\tan\theta\, d\theta$
*$\displaystyle \int \sec^3\theta \sqrt{\sec^2\theta - 1} \, \sec\theta\tan\theta\, d\theta$
*${\displaystyle \int \sec^3\theta \sqrt{\tan^2\theta}\sec\theta\tan\theta d\theta}$
*${\displaystyle \int \sec^3\theta \tan\theta \cdot \sec\theta \tan\theta d\theta}$
*${\displaystyle \int \sec^4\theta \cdot \tan^2\theta d\theta}$
*${\displaystyle \int \sec^2\theta \sec^2\theta\tan^2\theta d\theta}$
*$u=\tan\theta$, $du= \sec^2\theta d\theta$
*${\displaystyle \int u^2(u^2 + 1)}$
*${\displaystyle \int u^4 + u^2}$
*${\displaystyle = \frac{1}{5}u^5 + \frac{1}{3}u^3 + C}$
here is where I'm even more shaky:
*${\displaystyle \frac{1}{5}\tan^5 + \frac{1}{3}\tan^3 + C}$
*since $\tan= \sin/\cos$, $\tan$ also $= \sec/\csc$, and if $\sec\theta=x$, per the substitution, couldn't $\tan$ also be written $x/(1/x)=x^2$, right?
*I replaced all $\tan$s in the final answer with $x^2$, leaving a final answer of ${\displaystyle \frac{1}{5}x^{10}+\frac{1}{3}x^6+C}$
|
You are good except for your last step; $\tan(\theta)$ is not $x^2$, and in particular $\csc$ is not $1/\sec$.
One way to identify what $\tan(\theta)$ is in terms of $x$ is to consider the triangle that justifies the substitution $x=\sec(\theta)$ in the first place. This has a leg of $\sqrt{x^2-1}$, which means the hypotenuse is $x$ and the other leg is $1$. By choosing $u=\sec(\theta)$ you make $\theta$ adjacent to the leg of length $1$ and opposite the leg of length $\sqrt{x^2-1}$. Thus $\tan(\theta)=\sqrt{x^2-1}$. Alternately you could have figured that out by simply using the identity $\sec^2(\theta)=\tan^2(\theta)+1$.
|
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|
Solve by factoring, then perform synthetic division. $${x^4 - 3x^3 + 3x^2 - x = 0}$$
$${x(x^3 - 3x^2 + 3x - 1)}$$
Now, we have to perform synthetic division, if I'm correct we get the 1 to divide by from the X outside of the parentheses from the second equation, which is above this paragraph, and is also the factored form of the first equation.
$$\frac{x - 3 + 3 - 1}{1}$$
The answer is ${x = 0, 1}$. Does the 0 come from the remainder, and do the 1 come from the divisor?
|
$$x^3-3x^2+3x-1=x^3-x^2-2x^2+2x+x-1=x^2(x-1)-2x(x-1)+(x-1)=(x-1)(x^2-2x+1)=(x-1)^3$$
|
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|
Does $\sum_{n=2}^\infty \frac{n^2-4}{(n-1)^2(n+3)^2} $ converge? Does $\sum_{n=2}^\infty \frac{n^2-4}{(n-1)^2(n+3)^2} $ converge?
I used the integral test and found that it does, but it was a bit cumbersome. Is there an easier way?
Thanks
|
By partial fraction decomposition,
$$\begin{eqnarray*}\sum_{n\geq 3}\frac{n^2-2n-3}{(n-2)^2 (n+2)^2}&=&\frac{1}{32}\sum_{n\geq 3}\left(\color{blue}{\frac{7}{x-2}-\frac{7}{x+2}}+\color{red}{\frac{10}{(x+2)^2}-\frac{6}{(x-2)^2}}\right)\\&=&\frac{1}{32}\left[\color{blue}{\frac{175}{12}}\color{red}{-6\,\zeta(2)+10\left(\zeta(2)-\frac{1}{1^2}-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}\right)}\right]\\&=&\frac{1}{32}\left[\frac{25}{72}+4\,\zeta(2)\right]=\color{purple}{\frac{25}{2304}+\frac{\pi^2}{48}}.\end{eqnarray*}$$
The blue sum is a telescopic sum, the red sum is a well-known sum.
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Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.
Update 1: In this forum, somebody said that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant.
Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by Peter Scholze in here may help.
Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.
Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality".
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Let's reform this inequality in a way such that we can comprehend it better. Define $a=\dfrac{y}{x}$ and $b=\dfrac{z}{y}$, therefore $\dfrac{x}{z}={1\over ab}$. We can suume without lose of generality that $a,b\le1$ We need to prove that $$\dfrac{x}{8+5\left(\dfrac{y}{x}\right)^3}+\dfrac{y}{8+5\left(\dfrac{z}{y}\right)^3}+\dfrac{z}{8+5\left(\dfrac{x}{z}\right)^3}\ge\dfrac{x+y+z}{13}$$by dividing the two sides of the inequality by $x$ and substituting $a,b,c$ we have that$$\dfrac{1}{8+5\left(\dfrac{y}{x}\right)^3}+\dfrac{\dfrac{y}{x}}{8+5\left(\dfrac{z}{y}\right)^3}+\dfrac{\dfrac{z}{y}}{8+5\left(\dfrac{x}{z}\right)^3}\ge\dfrac{1+\dfrac{y}{x}+\dfrac{z}{x}}{13}$$and $$\dfrac{1}{8+5a^3}+\dfrac{a}{8+5b^3}+\dfrac{a^4b^4}{5+8a^3b^3}\ge \dfrac{1}{13}+\dfrac{a}{13}+\dfrac{ab}{13}$$which is equivalent to $$\left(\dfrac{1}{8+5a^3}-\dfrac{1}{13}\right)+\left(\dfrac{a}{8+5b^3}-\dfrac{a}{13}\right)+\left(\dfrac{a^4b^4}{5+8a^3b^3}-\dfrac{ab}{13}\right)\ge 0$$by simplifying each of the components and multiplying both sides in $\dfrac{13}{5}$ we obtain$$\dfrac{1-a^3}{8+5a^3}+\dfrac{a(1-b^3)}{8+5b^3}+\dfrac{a^4b^4-ab}{5+8a^3b^3}\ge0$$below is a depiction of $f(a,b)=\dfrac{1-a^3}{8+5a^3}+\dfrac{a(1-b^3)}{8+5b^3}+\dfrac{a^4b^4-ab}{5+8a^3b^3}$ for $0\le a,b\le 1$
which proves the inequality graphically (I believe that Lagrange multipliers or any other method based on 1st order derivations may help but i hadn't much time to think on it hope you find an analytic way) but neither such a time i spent on the problem nor a computer is given us in the exam :) also i appreciate if any one updates his/her comment with such an analytical method. I'm really curious about that.....
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|
Show that, $2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$ Show that,
$$2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$$
There is a mixed of sin and tan, how can I simplify this to $\frac{\pi}{4}$
We know the identity of $\arctan\left(\frac{1}{a}\right)+\arctan\left(\frac{1}{b}\right)=\arctan\left(\frac{a+b}{ab-1}\right)$
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From the Article $240,$ Ex$-5$ of Plane Trigonometry(by Loney),
$$\arctan x+\arctan y=\begin{cases} \arctan\dfrac{x+y}{1-xy} &\mbox{if } xy<1\\ \pi+\arctan\dfrac{x+y}{1-xy} & \mbox{if } xy>1\end{cases} $$
$$\implies 2\arctan x=\arctan \frac{2x}{1-x^2}\text{ if }x^2<1$$
Finally use $$\arcsin x=\arctan\dfrac x{\sqrt{1-x^2}}$$
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Find the area inside the circle $r = 10 \sin \theta$ and above the line $r = 2 \csc \theta$. Function Plotter graph:
I think the formula is
$$A = \frac 1 2 \int_{\alpha}^{\beta} (\text{outer})^2 - (\text{inner})^2 d\theta$$
where $\alpha, \beta$ are where they intersect in $[0, 2\pi]$.
This is what I got based on that
$$A = \frac 1 2 \int_{x}^{\pi-x} (10 \sin \theta)^2 - (2 \csc \theta)^2 d\theta$$
where $x= \sin^{-1}(\frac {1}{\sqrt{5}} )$
Is that right?
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We can do this all with classical geometry.
The region in question is a circle or radius 5 less a section of a circle with angle $2\cos^{-1} \frac 35$ plus 2 $3-4-5$ right triangles.
$25\pi - 25\cos^{-1} \frac 35 + 12$
Lets see if calculus gives us the same answer.
limits of integration.
$2\csc \theta = 10\sin \theta\\
\frac 15 = \sin^2 \theta\\
\sin\theta = \pm \frac 1{\sqrt 5}$
$\frac 12\int_{\sin^{-1} \frac 1{\sqrt 5}}^{\pi - \sin^{-1} \frac 1{\sqrt 5}} (10\sin\theta)^2 - (2\csc\theta )^2\ d\theta\\
\frac 12\int_{\sin^{-1} \frac 1{\sqrt 5}}^{\pi - \sin^{-1} \frac 1{\sqrt 5}} 50(1-\cos2\theta) - 4\csc^2\theta\ d\theta\\
25 (\theta -\sin\theta\cos\theta) + 2\cot\theta |_{\sin^{-1} \frac 1{\sqrt 5}}^{\pi - \sin^{-1} \frac 1{\sqrt 5}}\\
25 \pi -50\sin^{-1}\frac {1}{\sqrt 5} +20 - 8
$
Does $25 \cos^{-1} \frac 35 = 50 \sin^{-1} \frac 1{\sqrt5}?$
Indeed it does!
Set up in Cartesian
inside:
$x^2 + y^2 = 10 y$
above $y = 2$
$\int_2^{10} \sqrt {10y - y^2} dy$
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|
Calculating cosine of dihedral angle
Let $O,A,B,C$ be points in space such that $\angle AOB=60^{\circ},\angle BOC=90^{\circ},\angle COA=120^{\circ}$ Let $\theta$ be the acute angle between the planes $AOB$ and $AOC$. Find $\cos\theta.$
Here is what I did:
I let $AO=OB=BA=1, OC=1$. Then we can compute $BC=\sqrt{2},AC=\sqrt{3}$. From here, I let $A=(0,0,0),B=(1,0,0),C=(\frac{1}{2},\frac{\sqrt{3}}{2},0),C=(1,0,\sqrt{2})$ (I found the coordinates of $C$ by noticing $\angle ABC=90^{\circ}$ since $AB^2+BC^2=AC^2$). Then I found the equation of $AOC$ to be $-\sqrt{6}x+\sqrt{2}y+\sqrt{3}z=0$ and the equation of $AOB$ to be $z=0$, so using the formula here, I got $\cos\theta=\frac{\sqrt{3}}{11}$, but the answer is $\frac{1}{3}$. Where did I mess up?
Also a solution without using coordinates would be very nice to see!
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You can also use the Spherical Law of Cosines.
We imagine a sphere centered at $O$, and we take $A$, $B$, $C$ the points where the rays from $O$ meet that sphere as vertices of a curvilinear triangle on the surface of that sphere. The "sides" of that triangle are arcs of great circles, and the lengths of those arcs are directly proportional to the measures of the angles between the rays. (With an appropriate choice of radius, we can say that the lengths are equal to the measures.) The "angles" of the triangle match the dihedral angles between the planes formed by those rays.
Now, since angle $\theta$ is opposite side $\overline{BC}$, we calculate as follows ...
$$\cos\theta = \frac{\cos \angle BOC-\cos \angle AOB \cos \angle AOC}{\sin \angle AOB \sin \angle AOC} = \frac{\cos 90^\circ - \cos 60^\circ \cos 120^\circ}{\sin 60^\circ \sin 120^\circ} \\[20pt]
=\frac{0 - \frac{1}{2}\frac{-1}{2}}{\frac{\sqrt{3}}{2}\frac{\sqrt{3}}{2}} = \frac{1/4}{3/4} = \frac{1}{3}$$
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Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square.
Since $n^2+n+43$ is odd,if it's a perfect square it can be written as: $8k+1$,then:
$$n^2+n+43=8k+1\Rightarrow\ n^2+n+42=8k\ \Rightarrow\ n(n+1)\equiv6\pmod8$$
So $n$ becomes $2,5,...$ but $5$ doesn't work!!
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$n^2+n+43=m^2$ iff $4n^2+4n+172=4m^2$ iff $(2n+1)^2+171=(2m)^2$.
Let $u=2n+1$ and $v=2m$. Then $171 = v^2-u^2 =(v+u)(v-u)$.
Since $171 = 3 \cdot 3 \cdot 19$, we can try all possibilities for $v+u$ and $v-u$ and then find $n$ and $m$.
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Trying to compute a Riemann-sum as an integral where the $\lim_{n\to\infty} a\cdot n +\sum_{n^2}^{4n^2} \dfrac{1}{k+n}$
I have the following problem: I want to figure out , for what value of a does this limit converge and what value does it then converge to.
I'm assuming that I'm supposed to rewrite the sum as an integral where I can evaluate the integral in the form of "n's". Given the integral $\int_{1}^{4}\dfrac{1}{\sqrt{x}}dx = [2\sqrt{x}]_{1}^{4}$, where $x= k+n$. I think I can finish the problem.
However, I am unsure of how to get the integral $\int_{1}^{4}\dfrac{1}{\sqrt{x}}dx$
Any help will be grateful.
Thanks for your time.
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Unless there is some miraculous cancellation -- and there will not be -- the term $a \cdot n$ with $a \neq 0$ will cause divergence. So let's take $a = 0$ and focus on the behavior of the sum. This would be a typical Riemann sum for $f(x) = (1 + x)^{-1}$ if not for the limits that are given.
Nevertheless, we can find upper and lower bounds and apply the squeeze theorem to find the limit.
For $k \leqslant x \leqslant k+1,$ we have
$$\frac{1}{k+1+n} \leqslant \int_k^{k+1} \frac{1}{n+x} \, dx = \int_{k/n}^{(k+1)/n} \frac{1}{1+u} \, du \leqslant \frac{1}{k+n}.$$
Summing from $p = n^2$ to $q = 4n^2$ we have
$$ \int_{p/n}^{(q+1)/n} \frac{1}{1+u} \, du = \sum_{k=p}^q\int_{k/n}^{(k+1)/n} \frac{1}{1+u} \, du \leqslant \sum_{k=p}^q \frac{1}{k+n}.$$
Summing from $p-1$ to $q-1$ we have
$$ \sum_{k=p}^q \frac{1}{k+n} = \sum_{k=p-1}^{q-1} \frac{1}{k+1+n} \leqslant \sum_{k=p-1}^{q-1}\int_{k/n}^{(k+1)/n} \frac{1}{1+u} \, du = \int_{(p-1)/n}^{q/n} \frac{1}{1+u} \, du .$$
Hence,
$$\int_{p/n}^{(q+1)/n} \frac{1}{1+u} \, du \leqslant \sum_{k=p}^q \frac{1}{k+n} \leqslant \int_{(p-1)/n}^{q/n} \frac{1}{1+u} \, du. $$
Integrating we find,
$$\log\left( \frac{n + q +1 }{n +p} \right) \leqslant \sum_{k=p}^q \frac{1}{k+n} \leqslant \log\left( \frac{n + q}{n + p -1}\right). $$
Substituting $p = n^2$ and $q = 4n^2$ we get,
$$\log\left( \frac{n + 4n^2 +1 }{n + n^2} \right) \leqslant \sum_{k=n^2}^{4n^2} \frac{1}{k+n} \leqslant \log\left( \frac{n + 4n^2}{n + n^2 -1}\right). $$
Applying the squeeze theorem, we find the limit
$$\lim_{n \to \infty} \sum_{k=n^2}^{4n^2} \frac{1}{k+n} = \log(4).$$
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Find $\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms Find $S=\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms
I first multiplied and divided $S$ with $1\cdot3\cdot5$
$$\frac{S}{15}=\frac{1}{1\cdot3\cdot5\cdot7}+\frac{1\cdot3}{1\cdot3\cdot5\cdot7\cdot9}+\frac{1\cdot3\cdot5}{1\cdot3\cdot5\cdot7\cdot9\cdot11}+\cdots$$
Using the expansion of $(2n)!$
$$1\cdot3\cdot5\cdots(2n-1)=\frac{(2n)!}{2^nn!}$$
$$S=15\left[\sum_{r=1}^{20}\frac{\frac{(2r)!}{2^rr!}}{\frac{(2(r+3))!}{2^{r+3}(r+3)!}}\right]$$
$$S=15\cdot8\cdot\left[\sum_{r=1}^{20}\frac{(2r)!}{r!}\cdot\frac{(r+3)!}{(2r+6)!}\right]$$
$$S=15\sum_{r=1}^{20}\frac{1}{(2r+5)(2r+3)(2r+1)}$$
How can I solve the above expression? Or is there an simpler/faster method?
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Here's another approach, using recurrence relations.
Note that
$$\begin{align}
a_1&=\frac 17\qquad\text{and}\\
a_n&=a_{n-1}\left(\frac{2n-1}{2n+5}\right)\\
(2n+1)a_n+4a_n&=(2n-1)a_{n-1}\\
a_n&=\frac 14\big(b_{n-1}-b_n\big)\\
\end{align}$$
where $b_m=(2m+1)a_m$.
Summing from $2$ to $n$ by telescoping and adding $a_1$:
$$\begin{align}
\sum_{i=1}^{n}a_i
&=\frac 14(b_1-b_n)+a_1\\
&=\frac 14\big(1-(2n+1)a_n\big)
\qquad\qquad\qquad\text{as $b_1=3a_1$ and $a_1=\frac 17$}\\
&=\frac 14\bigg(1-\color{green}{(2n+1)}\cdot\frac{1\cdot 3\cdot 5\cdot \color{blue}{7\cdot 9\cdots(2n-1)}}{\color{blue}{7\cdot 9\cdot 11\cdots (2n-1)}\color{green}{(2n+1)}(2n+3)(2n+5)}\bigg)\\
&=\frac 14\bigg(1-\frac{15}{(2n+3)(2n+5)}\bigg)\\
&=\frac{n(n+4)}{(2n+3)(2n+5)}
\end{align}$$
Putting $n=20$:
$$\begin{align}
\sum_{i=1}^{20}a_i
&=\frac{20\cdot 24}{43\cdot 45}=\frac{32}{129}\qquad\blacksquare
\end{align}$$
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Prove that $\frac{1}{7\sqrt{2}} \leq \int_{0}^1 \frac{x^6}{\sqrt{1+x^2}}dx \leq \frac{1}{7}$
Prove that $\displaystyle \dfrac{1}{7\sqrt{2}} \leq \int_{0}^1 \dfrac{x^6}{\sqrt{1+x^2}}dx \leq \dfrac{1}{7}$.
My book says let $f(x) = \dfrac{1}{\sqrt{1+x^2}}$ and $g(x) = x^6$. Then $\displaystyle \int_0^1 \dfrac{x^6}{\sqrt{1+x^2}}dx = \dfrac{1}{\sqrt{1+\xi^2}} \int_{0}^1 x^6 dx$ where $0 \leq \xi \leq 1$. Thus $$\dfrac{1}{7\sqrt{2}} \leq \dfrac{1}{\sqrt{1+\xi^2}} \leq \dfrac{1}{7}.$$
I don't understand how they are getting all these results. The problem did say to use other results in the book, but I didn't see any that related. For reference it is in chapter 13 of Michael Spivak's calculus book.
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Achtung: overkill. $g(x)=\frac{1}{\sqrt{1+x}}$ is a convex function on $[0,1]$ (since its logarithm is convex) bounded between $1$ and $\frac{1}{\sqrt{2}}$. Since $g'(1)=-\frac{1}{4\sqrt{2}}$,
$$\forall x\in(0,1),\qquad \frac{(1-x)}{4\sqrt{2}}+\frac{1}{\sqrt{2}}\leq g(x)\leq (1-x)+\frac{x}{\sqrt{2}}\tag{1}$$
holds by convexity. Now:
$$ I=\int_{0}^{1}\frac{x^6}{\sqrt{1+x^2}}\,dx =\frac{1}{2}\int_{0}^{1} x^{5/2}g(x)\,dx \tag{2}$$
can be bounded by using $(1)$:
$$ \color{red}{\frac{1}{10}}<\frac{19}{126\sqrt{2}} < \color{red}{I} < \frac{4+7\sqrt{2}}{126}<\color{red}{\frac{1}{9}}. \tag{3}$$
If we use integration by parts and the Cauchy-Schwarz inequality we also get the interesting lower bound:
$$ I = \sqrt{2}-\int_{0}^{1}5x^4\sqrt{1+x^2}\,dx \geq \sqrt{2}-\sqrt{\frac{12}{7}}.\tag{4} $$
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Approximation of $\sqrt{2}$ I got the following problem in a chapter of approximations:
If $\frac{m}{n}$ is an approximation to $\sqrt{2}$ then prove that $\frac{m}{2n}+\frac{n}{m}$ is a better approximation to $\sqrt{2}.$(where $\frac{m}{n}$ is a rational number)
I am stuck in this problem. Any help will be appreciated.
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$\frac{m}{n}$ approximately equals $\sqrt 2$
Suppose $\epsilon$ is our the error in our estimate. i.e. $\frac{m}{n} + \epsilon = \sqrt 2$ or
$(\frac{m}{n} + \epsilon)^2 = 2$
$\frac{m^2}{n^2} + 2\frac{m}{n}\epsilon + \epsilon^2 = 2$
Now we need to solve for $\epsilon.$
If $\epsilon$ is small then then $\epsilon^2$ is very small. Which is a little fuzzy. Certainly we can say that $\epsilon^2 < 2\frac{m}{n}\epsilon$ and if we solved for $\epsilon^*$ in this equation:
$2\frac{m}{n}\epsilon^* = 2 - \frac{m^2}{n^2}$
then:
$|(\frac{m}{n} + \epsilon^*) - \sqrt 2| < |\frac{m}{n} - \sqrt2|$
$\epsilon^* = \frac{n}{m} - \frac{m}{2n}\\
\frac{m}{n} + \epsilon^* = \frac{n}{m} + \frac{m}{2n}$
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|
Prove $\frac{2\cos x}{\cos 2x + 1 }= \sec x$ Prove that $\dfrac{2\cos x}{\cos 2x + 1 }= \sec x$.
So far I have:
$\dfrac{2\cos x}{\cos 2x + 1 }= \dfrac 1 {\cos x}$
Where do I go from here?
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$$\frac{2\cos(x)}{1+\cos(2x)}=\sec(x)\Longleftrightarrow$$
Use $\sec(x):=\frac{1}{\cos(x)}$
$$\frac{2\cos(x)}{1+\cos(2x)}=\frac{1}{\cos(x)}\Longleftrightarrow$$
$$2\cos^2(x)=1+\cos(2x)\Longleftrightarrow$$
Use $\cos(2x)=2\cos^2(x)-1$
$$2\cos^2(x)=1+2\cos^2(x)-1\Longleftrightarrow$$
Use $1-1+2\cos^2(x)=0+2\cos^2(x)=2\cos^2(x)$
$$2\cos^2(x)=2\cos^2(x)$$
|
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|
Sum of the series $\sum \frac{n}{2^{n}}$ I know that the series converges by d'Alembert ratio test, where $\lim\left ( \frac{A_{n+1}}{A_{n}} \right )= \frac{1}{2}$, but I don't know how to calculate the sum of the serie. Thanks for the help.
|
\begin{align*}
S&= \frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+.....\\
\frac{1}{2}S&=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}\\
S-\frac{1}{2}S&=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......\\
S-\frac{1}{2}S&=1
\end{align*}
Thus $S=2$
|
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|
Integrate $ \int\frac{1+x^2}{(1-x^2)(\sqrt{1+x^4})}dx $ $$ \int\frac{1+x^2}{(1-x^2)(\sqrt{1+x^4})}dx $$
I thought of substituting $ x-\frac{1}{x} $ as $t$ but it gets stuck midway. I am close but I think I need to sustitute something else here.
|
Let $$I = \int\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^4}}dx = -\int\frac{(1+x^2)}{x^2(x-\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}}dx$$
So $$I = -\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})\sqrt{(x-\frac{1}{x})^2+2}}dx$$
Now Put $\displaystyle x-\frac{1}{x} = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)dx = dt$
So $$I = -\int\frac{1}{t\sqrt{t^2+2}}dt$$
Now Put $t = \sqrt{2}\tan \theta\;,$ Then $dt=\sqrt{2}\sec^2 \theta d\theta$
So $$I = -\int\frac{\sqrt{2}\sec^2 \theta}{2\tan \theta \sec \theta}d\theta = -\frac{1}{\sqrt{2}}\int\csc \theta d \theta$$
So $$I = \ln \left|\csc \theta +\cot \theta\right|+\mathcal{C}=\frac{1}{\sqrt{2}}\ln \left|\frac{\sqrt{2}+\sqrt{2+t^2}}{t+\sqrt{2+t^2}}\right|+\mathcal{C}$$
So $$I = \frac{1}{\sqrt{2}}\ln \left|\frac{\sqrt{2}x^2+x\sqrt{1+x^4}}{x^2-1+\sqrt{1+x^4}}\right|+\mathcal{C}$$
|
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Error in solving $\int \sqrt{1 + e^x} dx$ . I want to solve this integral for $1 + e^x \ge 0$
$$\int \sqrt{1 + e^x} dx$$
I start by parts $$\int \sqrt{1 + e^x} dx = x\sqrt{1 + e^x} - \int x \frac{e^x}{2\sqrt{1 + e^x}} dx $$
Substitute $\sqrt{1 + e^x} = t \implies dt = \frac{e^x}{2\sqrt{1 + e^x}} dx$
So I remain with $$x\sqrt{1 + e^x} - \int \ln((t-1)(t+1)) dt = x\sqrt{1 + e^x} + 2t +(-t+1)\ln(t-1)+ (-t+1)\ln(t+1) $$
that is
$$(x+2)\sqrt{1 + e^x} + (-\sqrt{1 + e^x} + 1) \ln(+\sqrt{1 + e^x} - 1) + (-\sqrt{1 + e^x} + 1) \ln(+\sqrt{1 + e^x} + 1) + C $$
But the solution should be $$2\sqrt{1 + e^x} + \ln(+\sqrt{1 + e^x} - 1) - \ln(+\sqrt{1 + e^x} + 1) $$
Where is the mistake hidden?
|
Let $x = \ln t$. Then $dx = dt/t$ and we have
$$\int \sqrt{1 + e^x} \, dx = \int \frac{\sqrt{1 + t}}t \, dt.$$
Now let $u = t + 1$. Then $du = dt$ and we have
$$\int \frac{\sqrt{1+t}}t \, dt = \int\frac{\sqrt{u}}{u-1} \, du.$$
Now let $v = \sqrt{u}$. Then $u = v^2$ and so $du = 2v \, dv$, and we have
$$\int\frac{\sqrt{u}}{u-1} \, du = 2\int\frac{v^2}{v^2 - 1} \, dv.$$
Consider then:
\begin{align*}
2\int\frac{v^2}{v^2 - 1} \, dv
&= 2\int\frac{v^2 \color{red}{-1+1}}{v^2 - 1} \, dv \\[0.3cm]
&= 2\int\frac{v^2-1}{v^2-1} \, dv + 2\int\frac{dv}{v^2-1}\\[0.3cm]
&= 2v + \ln(v-1) - \ln(v+1) + C
\end{align*}
Back substitute:
$$
2v + \ln(v-1) - \ln(v+1) + C =
2\sqrt{u} + \ln(\sqrt{u}-1) - \ln(\sqrt{u}+1) + C
$$
Back substitute:
$$
2\sqrt{u} + \ln(\sqrt{u}-1) - \ln(\sqrt{u}+1) + C =
2\sqrt{t+1} + \ln(\sqrt{t+1}-1) - \ln(\sqrt{t+1}+1) + C
$$
Back substitute $t = e^x$ to finally get:
$$
\int \sqrt{1+e^x} \, dx =
2\sqrt{e^x+1} + \ln(\sqrt{e^x+1}-1) - \ln(\sqrt{e^x+1}+1) + C
$$
|
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|
How to solve the equation $ x^{13} = 1$ by radicals? (And $x^{19}=1$) Is there any elementary way to solve the equation $ x^{13}= 1 $ by means of radicals? If not, how to get all the solutions?
Remark: The transcendental form of the solution by means of sines and cosines is not allowed, but only radicals, since this equation is solvable.
The solutions in terms of sines and cosines are $ \cos (k\pi/13)+i\sin(k\pi/13) $ for $ k = 0.1... 12$. So, what I say is that I am looking for an explicit solution in terms of radicals. The trivial answer $\sqrt[13]{1} $ is also excluded.
|
First, we establish some ground rules. All cubics and quartics can be solved by radicals, agreed? Thus, given,
$$\frac{x^{13}-1}{x-1}=x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1=0\tag1$$
to solve for $x$ in radicals all we have to do is to "break-up" the $12$-deg into quadratics, cubics, or quartics.
I. Method 1
Equivalently, $(1)$ is the palindromic quartic,
$$F(x)=x^4 + u x^3 + (u^2 - u - 1)x^2 + u x + 1 =0\tag2$$
with its coefficients determined by,
$$u^3 - u^2 - 4u - 1 = 0\tag3$$
$F(x)=0\,$ has $4$ roots, but as there are $3$ choices of $u$, using each one will yield the $4\times3=12$ roots of $(1)$. As proof, eliminating $u$ between $(2),(3)$ will recover the $12$-deg. The WA command is,
Resultant[x^4 + u x^3 + (u^2 - u - 1)x^2 + u x + 1, u^3 - u^2 - 4u - 1, u]
II. Method 2
For primes $p=6n+1$, we can "factor" the cyclotomic polynomial into $n$ palindromic sextics with coefficients determined by an equation of degree $n$. But a palindromic sextic is just a cubic in disguise since given,
$$x^6 + a x^5 + b x^4 + c x^3 + b x^2 + a x + 1=0$$
then one can solve it as $x+1/x=y$ where $y$ are the roots of the cubic,
$$y^3+a y^2 + (b-3) y +(-2a+c)= 0$$
$p=13:$
Resultant[1 - u x + 2 x^2 - (1 + u) x^3 + 2 x^4 - u x^5 + x^6, -3 + u + u^2, u]
$p=19:$
Resultant[1 - u x - (2 - u^2) x^2 - (6 + 2 u - u^2) x^3 - (2 - u^2) x^4 - u x^5 + x^6, -7 - 6 u + u^2 + u^3, u]
and so on. Thus, you only need to solve $-3 + u + u^2=0$ and $-7 - 6 u + u^2 + u^3=0$, respectively. (Click link for more on $p=19$). For $p=31$, you'll have to tackle the quintic $-5 + u + 21 u^2 - 12 u^3 - u^4 + u^5=0$. But that's another story.
|
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|
Positive integers $a,b$ satisfying $a^3+a+1=3^b$ How to prove that $a=b=1$ is the only positive integer solution to the following Diophantine equation?$$a^3+a+1=3^b$$
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Partial solution, Check the equation mod 9 $\big(\forall b>1,3^b\equiv 0 \pmod 9\big)$. Nine cases to check:
$a=0 \pmod 9, a^3+a+1\equiv 1 \pmod 9\\
a=1 \pmod 9, a^3+a+1\equiv 3 \pmod 9 \\
a=2 \pmod 9, a^3+a+1\equiv 2 \pmod 9\\
a=3 \pmod 9, a^3+a+1\equiv 4 \pmod 9\\
a=4 \pmod 9, a^3+a+1\equiv 6 \pmod 9\\
a=5 \pmod 9, a^3+a+1\equiv 5 \pmod 9\\
a=6 \pmod 9, a^3+a+1\equiv 7 \pmod 9\\
a=7 \pmod 9, a^3+a+1\equiv 0 \pmod 9\\
a=8 \pmod 9, a^3+a+1\equiv 8 \pmod 9 $
so the only cases possible greater than a=b=1, is if a, is 7 mod 9, and b>1 . plugging that in we get:
$(9j+7)^3+(9j+7)+1 = 729j^3 + 1701j^2 + 1332j + 351$ ( done with PARI/GP) which reduces to 26j when done mod 27. Without j a multiple of 27, this won't be 0 mod 27 which $3^b$ is for b>2.
|
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$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find $(x+y)$.
We know that $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find the expression $(x+y)$.
My work so far:
$$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$$
$$\color{red}{\left(x-\sqrt{x^2+1}\right)\cdot}\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\color{red}{\left(x-\sqrt{x^2+1}\right)\cdot}1$$
$$-\left(y+\sqrt{y^2+1}\right)=\left(x-\sqrt{x^2+1}\right)$$
$$-\color{red}{\left(y-\sqrt{y^2+1}\right)\cdot}\left(y+\sqrt{y^2+1}\right)=\color{red}{\left(y-\sqrt{y^2+1}\right)\cdot}\left(x-\sqrt{x^2+1}\right)$$
$$1=\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)$$
I need help here.
|
Notice that we have
$$-\left(y+\sqrt{y^2+1}\right)=\left(x-\sqrt{x^2+1}\right)$$ From your third line. Now $$x+y=\sqrt{x^2+1}-\sqrt{y^2+1}\tag{1}$$ In the same fashion, we can get$$x+y=\sqrt{y^2+1}-\sqrt{x^2+1}\tag{2}$$ Now adding $\text{(1)}$ and $\text{(2)}$ together gives $x+y=0$. We are done.
|
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|
problem with Combination and Permutation Four married couples have bought 8 seats in a row for a concert.
In how many ways can they be seated:
a)if each couple is to sit together?
(8)(1)(6)(1)(4)(1)(2)(1)
b)if all men sit together?
n-k+1=8-4+1=5
so, (5)(4! 4!)=5! 4!
c)If no man sits next to his wife ?(This is a non-trivial questions)
8*6*6*4*4*2*2*1=18432 ways
could you please check it for me ?
|
a) and b) are correct.
For c), we will use Inclusion/Exclusion. There are $8!$ arrangements without restriction. From this we need to subtract the number of bad arrangements, where at least one couple are next to each other.
First we count the number of arrangements where Couple A are together. Tie them together with rope. There are then $7$ objects to be arranged. This can be done in $7!$ ways. Now untie Couple A. They can occupy $2$ different positions, for a total of $2\cdot 7!$. Multiply by $\binom{4}{1}$ for the number of ways to choose a couple. Our first estimate of the number of bad arrangements is $\binom{4}{1}\cdot 2\cdot 7!$.
However, this grossly overcounts the number of bad arrangements, for it double-counts, among others, the arrangements where Couple A and B are both next to each other. An analysis similar to the previous one shows that there are $2^2\cdot 6!$ arrangements in which Couple A and B are together. Thus our adjusted count for the number of bad arrangements is $\binom{4}{1}\cdot 2\cdot 7!-\binom{4}{2}\cdot 2^2\cdot 6!$.
However, we have subtracted too much, for we have subtracted one too many times, for example, the arrangements where couples A, B, C are together. Adjusting for this gives the adjusted count $\binom{4}{1}\cdot 2\cdot 7!-\binom{4}{2}\cdot 2^2\cdot 6!+\binom{4}{3}\cdot 2^3\cdot 5!$.
A final adjustment needs to be made, we must subtract the $\binom{4}{4}\cdot 2^4\cdot 4!$ arrangements in which all couples are together. This will give us the total number of bad arrangements, and now we are nearly finished.
|
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|
$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52$ for $a,b,c >0$ and $a+b+c=3$
Let $a,b,c >0$ with $a+b+c=3$. Prove that
$$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52.$$
What I did:
It is cyclic inequality so I assume $c= \min\{ a,b,c \}$.
I consider the first case where $a\ge b\ge c$ then
$$a^{|b-a|}+b^{|c-b|}+c^{|a-c|} > \frac52$$
$$\Leftrightarrow \frac{a^a}{a^b} +\frac{b^b}{b^c}+\frac{c^a}{c^c}> \frac52$$
I check function $f(x) =x^x$ to see if it is a strictly monotonic function or not. It turns out that it is a concave up function so I get stuck here.
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Perhaps it would help to take $a=1+x$, $b=1+y$, $c=1+z$, so that $x+y+z=0$
Now the inequality looks like:
$$(1+x)^{|y-x|} + (1+y)^{|z-y|} + (1+z)^{|x-z|} \geq \frac{5}{2}$$
Using bernoulli's inequality, the expression is more or less
$$1 + x|y-x| + 1 + y|z-y| + 1 + z|x-z| \geq \frac{5}{2}$$
$$x|y-x| + y|z-y| + z|x-z| \geq \frac{-1}{2}$$
And we can split it in two cases depending on how x, y and z are sorted.
For example, if $x<y<z$, we must prove that:
$$x(y-x) + y(z-y) + z(z - x) = -x^2 - y^2 + z^2 + xy + yz - xz \geq \frac{-1}{2}$$
We recall that $x+y+z=0$, and thus $z = -x-y$:
$$-x^2 - y^2 + z^2 + xy + yz - xz = -x^2 - y^2 + x^2 + y^2 - 2xy + xy - (y+x)(y-x) = x^2 - y^2 - xy$$
It should not be hard to check if this is more that $\frac{-1}{2}$ given our restraints.
|
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Discriminant formula issue Could you please explain to me, how to get the formula of discriminant ? How can I visualize it, any articles, lectures?
I can memorize it $b^2 - 4ac$ But, want to understand it.
Thanks.
|
You must have seen a method called "completing the square" before you saw the quadratic formula in school right? This is where it comes from, in fact, if you know how to complete the square, you can derive the whole quadratic formula on your own.
Consider the quadratic equation that you're used to $$ax^2 + bx + c = 0.$$
To complete the square we need the coefficient of $x^2$ (the term in front of $x^2$, i.e. the $a$) to be $1$, so we need to divide both sides by $a$. (If $a = 0$, then the equation isn't a quadratic and we actually have $bx + c = 0$, which you can solve easily). This gives:
$$x^2 + \frac{b}{a}x + \frac{c}{a} = 0.$$
We now complete the square as usual:
$$\Bigg(x + \frac{b}{2a}\Bigg)^2 + \frac{c}{a} - \frac{b^2}{4a^2} = 0.$$
We now need to make $x$ the subject:
\begin{align}\Bigg(x + \frac{b}{2a}\Bigg)^2 &= -\Bigg(\frac{c}{a} - \frac{b^2} {4a^2}\Bigg)\\
\Bigg(x + \frac{b}{2a}\Bigg)^2 &= -\Bigg(\frac{4ac}{4a^2} - \frac{b^2} {4a^2}\Bigg)\\
\Bigg(x + \frac{b}{2a}\Bigg)^2 &= -\Bigg(\frac{4ac - b^2} {4a^2}\Bigg)\\
\Bigg(x + \frac{b}{2a}\Bigg)^2 &= \Bigg(\frac{b^2-4ac} {4a^2}\Bigg)\\
x + \frac{b}{2a} &= \pm\sqrt{ \Bigg(\frac{b^2-4ac} {4a^2} \Bigg) }\\
x + \frac{b}{2a} &= \pm\frac{\sqrt{b^2-4ac}}{2a}\\
x &= -\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\\
x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a}.
\end{align}
Finally, the part in the square root is the discriminant.
|
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If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $. I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.
|
$(1+a)(1+b)(1+c)=1+(a+b+c)+(ab+bc+ca)+(abc)$
Applying AM≥GM on $(a+b+c)+(ab+bc+ca)+(abc)$
$$\frac{(a+b+c)+(ab+bc+ca)+(abc)}{7}≥(a^{4}b^{4}c^{4})^\frac{1}{7}$$
So $(a+b+c)+(ab+bc+ca)+(abc)≥7(a^{4}b^{4}c^{4})^\frac{1}{7}$
If we add 1 on LHS so now LHS>RHS
So $(1+a)^{7}(1+b)^{7}(1+c)^{7}>7^{7}(a^{4}b^{4}c^{4})$
|
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In how many ways can $2m$ objects be paired and assigned to $m$ boxes? Been refreshing my statistics knowledge and came across this problem while going through a book.
How many ways can $2m$ objects be paired and assigned to $m$ boxes?
The book claims the answer is
$$ \frac{(2m)!}{2^m} $$
Which makes sense, however, it further claims that if we want to arrange the pairs without assigning them to boxes, since there are $m!$ ways to assign the $m$ pairs to $m$ boxes, the answer is:
$$ \frac{(2m)!}{m!2^m} $$
Can someone explain why we have to divide by $m!$ if we are no longer assigning the pairs to boxes?
|
If $m = 1$, we have a set $\{1,2\}$ and there's only $1$ unordered pair we can form.
If $m = 2$, we have a set $\{1,2,3,4\}$ and there are $3$ pairings we can form, namely,
$$\{\{1,2\}, \{3,4\}\}$$
$$\{\{1,3\}, \{2,4\}\}$$
$$\{\{1,4\}, \{2,3\}\}$$
For a general $m$, we have a set $\{1,2,\dots,2m\}$ and the number of pairings we can form is the number of partitions of size $2$ of the set, which is given by the following double factorial
$$\begin{array}{rl} (2 m - 1)!! &= (2m-1) \cdot (2m-3) \cdot (2m-5) \cdots 3 \cdot 1\\\\ &= \dfrac{(2m-1) \cdot (2m-2) \cdot (2m-3) \cdots 2 \cdot 1}{(2m-2) \cdot (2m-4) \cdot (2m-6) \cdots 4 \cdot 2}\\\\ &= \dfrac{(2m) \cdot (2m-1) \cdot (2m-2) \cdot (2m-3) \cdots 2 \cdot 1}{(2m) \cdot(2m-2) \cdot (2m-4) \cdot (2m-6) \cdots 4 \cdot 2}\\\\ &= \dfrac{(2m) \cdot (2m-1) \cdot (2m-2) \cdot (2m-3) \cdots 2 \cdot 1}{2^m \cdot m \cdot(m-1) \cdot (m-2) \cdot (m-3) \cdots 2 \cdot 1}\\\\ &= \dfrac{(2m)!}{2^m \cdot m!}\end{array}$$
The total number of assignments of $m$ partitions of size $2$ to $m$ boxes is
$$m! \cdot (2m-1)!! = m! \cdot \dfrac{(2m)!}{2^m \cdot m!} = \dfrac{(2m)!}{2^m}$$
which is also given by
$$\binom{2m}{2} \binom{2m-2}{2} \binom{2m-4}{2} \cdots \binom{2}{2} = \dfrac{(2m)!}{2^m}$$
|
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Integrating trig function I'm stuck at this problem:
$$ \int{\sqrt{(\sin^2 x)^2 + (2\sin x \cos x)^2}dx} = \int{\sqrt{\sin^2 x \sin^2 x + 4\sin^2 x \cos^2 x} dx}$$
I tried a few trig identities: $\sin^2 x = \frac{1-\cos 2x}{2} $ and $ \cos^2 x = \frac{1+\cos 2x}{2} $ and $\sin^2 x + \cos^2 x = 1$. Keep hitting dead end. Any tips?
|
Try a substitution with the last expression that you have
$$\int\sqrt{\sin^4(x)+4\sin^2(x)\cos^2(x)}dx\quad u=\cos(x)$$.
This leads us to the following integral where we do another substitution:
$$\int\sqrt{3u^2+1}du \quad\!\! \text{substitute}\quad\!\! u=\sqrt{\dfrac{a}{b}}\tan (v) \quad\!\!\text{where}\quad\!\! a=1 \quad\!\!\text{and} \quad\!\!b=3$$
Then, the integral becomes the following:
$$\int\dfrac{\sqrt{\tan^2(v)+1}+\sec^2(v)}{\sqrt{3}}$$
and the rest is more or less is using trig identities.
|
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Solutions to equation
Let $\alpha(n)$ be the number of pairs $(x, y)$ of integers such that $x+y = n, 0 \le y \le x$, and let $\beta(n)$ be the number of triples $(x, y, z)$ such that $x + y + z = n$ and $0 \le z \le y \le x.$ Find a simple relation between $\alpha(n)$ and the integer part of the number $\frac{n+2}{2}$ and the relation among $\beta(n), \beta(n -3)$ and $\alpha(n).$
Attempt:
We see that $\alpha(n) = \dfrac{\binom{n+1}{1}}{2} = \dfrac{n+1}{2}$ if $n$ is odd and $\alpha(n) = \dfrac{\binom{n+1}{1}-1}{2}+1 = \dfrac{n+2}{2}$ if $n$ is even. Thus, $\alpha(n) = \left[\dfrac{n+2}{2}\right]$.
How do I relate $\beta(n), \beta(n -3)$ and $\alpha(n)$?
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The triples that satisfy your condition are of two types. Type 1 are the triples with smallest element $0$, and Type 2 are the ones with smallest element $\ge 1$.
There are just as many triples of Type 1 with sum $n$ as there are pairs $(y,x)$ with $0\le y\le x$ such that $y+x=n$.
There are just as many triples of Type 2 with sum $n$ as there are triples $0\le z'\le y'\le x'$ with sum $n-3$. For let $z'=z-1$, $y'=y-1$, and $x'=x-1$. That gives a bijection between the Type 2 triples with sum $n$ and all good triples with sum $n-3$.
Thus we obtain the recurrence
$$\beta(n)=\alpha(n)+\beta(n-3).$$
|
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Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
I think we'll have to use number theory to do it. Simply solving the equations won't do.
If we divide the second equation by the first, we get:
$$x^2 - xy + y^2 = 1 + z.$$
Also, since they are integers $z^2 \ge z \implies -z^2 \le -z$. This implies
$$x + y = 1 - z \ge 1 - z^2 = x^3 + y^3.$$
This shows that atleast one of $x$ and $y$ is negative with the additive inverse of the negative being larger than that of the positive.
I have tried but am not able to proceed further. Can you help me with this?
Thanks.
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$$y=-x, z=1$$
$${y = \frac{-1 + x - \sqrt{3} \sqrt{3 - 2 x - x^2}}{2}, z = \frac{3 - 3 x + \sqrt{3} \sqrt{3 - 2 x - x^2}}2}$$
$${y = \frac{-1 + x + \sqrt{3} \sqrt{3 - 2 x - x^2}}{2}, z = \frac{3 - 3 x - \sqrt{3} \sqrt{3 - 2 x - x^2}}2}$$
All integer solutions:
$x=-3, y=-2, z=6$
$x=-2, y=-3, z=6$
$x=-2, y=0, z=3$
$x=0, y=-2, z=3$
$x=0, y=1, z=0$
|
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|
Extreme values of $\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}$ Let $a,b,c$ be side lengths of a triangle. What are the minimum and maximum of $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}?$$
When $a=b=c$, the value is $9$. In addition, we can write $a=x+y,b=y+z,c=z+x$ since they are side lengths of a triangle. The expression becomes
$$\frac{8(x+y+z)(xy+yz+xz)}{(x+y)(y+z)(z+x)}.$$
|
For the maximum values, we could also show it using derivatives.
Consider the function $$F=\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}$$ Compute derivatives; after simplifications, we get $$F'_x=\frac{y z \left(y z-x^2\right)}{(x+y)^2 (x+z)^2 (y+z)}$$ $$F'_y=\frac{x z \left(x z-y^2\right)}{(x+y)^2 (x+z) (y+z)^2}$$ $$F'_z=\frac{x y \left(x y-z^2\right)}{(x+y) (x+z)^2 (y+z)^2}$$ If none of $x,y,z$ is zero, then, for cancelling each derivative, we have $$yz-x^2=xz-y^2=xy-z^2=0$$ which admits as only real solutions $x=y=z=1$ for which $F=\frac 98$
|
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|
Solve the equation $|2x-1| -|x+5| = 3$
Problem : Solve the equation $|2x-1| - |x+5| = 3$
In my attempt to solve the problem, I only manage to get one of the solutions.
Attempted Solution
$$\begin{equation}
\begin{split}
|x|-|y| & \leq |x-y| \\
\implies |2x-1| - |x+5| & \leq |(2x-1)-(x+5)| \\
\implies 3 &\leq |x-6| \\
\implies 3 &= |x-6| \\
\implies (x = 9) \ & \lor \ (x= 3 \ (\text{Reject})) \\
\implies x &= 9
\end{split}
\end{equation}
$$
However plugging the correct solution set to this equation is $x \in \{-\frac{7}{3}, 9\}$. Why does my attempted solution, not reach $-\frac{7}{3}$, as one of the possible solutions? Furthermore, how would you go about solving a problem of this nature?
|
Because $|2x-1|-|x+5|$ is also $\leq |(x+5) - (2x-1)|$. (Multiplying through by $-1$ doesn't change the absolute value...)
The generic method is this:
*
*$x+5$ is nonnegative for $x \geq -5$ and negative for $x < -5$.
*$2x-1$ is nonnegative for $x \geq 1/2$ and negative for $x < 1/2$.
*On the interval $(-\infty,-5)$, both are negative, so we solve $$ |2x-1|-|x+5| = -(2x-1)--(x+5) = -x + 6 = 3 \text{,} $$ so $x = 3$, but this is not in $(-\infty, -5)$, so we reject it.
*On the interval $[-5,1/2)$, we solve $$ |2x-1|-|x+5| = -(2x-1)-(x+5) = -3x - 4 = 3 \text{,} $$ so $x = -7/3$, which is in $[-5,1/2)$.
*On the interval $[1/2,\infty)$, we solve $$ |2x-1|-|x+5| = (2x-1)-(x+5) = x - 6 = 3 \text{,} $$ so $x = 9$, which is in $[1/2,\infty)$.
So the solution set is $\{-7/3,9\}$.
|
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|
Decomposition of a rational function in partial fractions I am trying to decompose the following rational function: $\dfrac{1}{(x^2-1)^2}$ in partial fractions (in order to untegrate it later).
I have notices that $(x^2-1)^2 = (x+1)^2(x-1)^2$
Therefore $\exists A, B, C, D$ s.t: $\dfrac{1}{(x^2-1)^2} = \dfrac{1}{(x-1)^2(x+1)^2} = \dfrac{Ax+B}{(x+1)^2}+\dfrac{Cx+D}{(x-1)^2}$
We then have $Ax+B =\left. \dfrac{1}{(x-1)^2} \right\vert _{x=-1} \implies B-A=1/4$
Same thing for Cx+D: $Cx+D =\left. \dfrac{1}{(x+1)^2} \right\vert _{x=1} \implies C+D=1/4 $
How do I find A, B, C, D from here?
|
Consider:
$$ \frac{1}{(x^2-1)^2} = \frac{1}{(x+1)^2(x-1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1} + \frac{D}{(x-1)^2}$$
Try to decompose the above
|
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|
Finding $\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$ without L'hôpital I have this $\lim_{}$.
$$\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$$
Indetermation:
$$\lim_{x\to 0} \frac{\ln(0+4)-\ln(4)}{0}$$
$$\lim_{x\to 0} \frac{0}{0}$$
Then i started solving it:
$$\lim_{x\to 0} \frac{\ln\frac{(x+4)}{4}}{x}$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} \frac{(x+4)}{4}$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} (\frac{x}{4}+1)$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} (1 + \frac{x}{4})$$
$$\ln\lim_{x\to 0} (1 + \frac{x}{4})^\frac{1}{x}$$
Then multiply the power, by 4.
$$\ln\lim_{x\to 0} (1 + \frac{x}{4})^\frac{1.4}{x.4}$$
$$\ln \begin{bmatrix}\lim_{x\to 0} (1 + \frac{x}{4})^\frac{4}{x} \end{bmatrix}^\frac{1}{4} $$
$$ \ln \phantom{2}\mathcal e^\frac{1}{4} = \frac{1}{4}$$
Until here is fine, but someone's telling me that it can be made like this:
$$\lim_{x\to 0} \frac{\ln(x) + \ln(4) -\ln(4)}{x}$$
$$\lim_{x\to 0} \frac{\ln(x)}{x} = 1$$
Alright, it uses logarithmic property to separate the expression, but i said it can't be, because of the indetermination, and $\frac{1}{4} $ is closer of $0$ than $1$. There's another reasonable explanation for this ? Why it can't be or when can it ?
And, there's a trick to way out quickly with this limit, keeping L'Hôspital aside ?
|
Yet another method —Taylor expansion (to first order):
We use that $\ln(1+u) = u + o(u)$ when $u\to 0$.
Then $\ln(x+4) = \ln\left(1+\frac{x}{4}\right) + \ln 4 = \frac{x}{4} + o(x) + \ln 4$, and
$$
\frac{\ln(x+4)- \ln 4}{x} = \frac{\frac{x}{4} + o(x)}{x} = \frac{1}{4} + o(1) \xrightarrow[x\to0]{} \frac{1}{4}.
$$
|
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|
Computing an infinite product $\prod_{n=1}^{\infty} \frac{1}{2}(1+\cos\frac{x}{2^n})$ I would like to compute the infinite product
$\displaystyle f(x)=\prod_{n=1}^{N\rightarrow\infty} \frac{1}{2}\left({1+\cos\frac{x}{2^n}}\right)$ for a given real $x$.
Since the terms in the product are positive and smaller than $1$, the product is bounded by $0$ and decreases. Therefore the limit $N\rightarrow\infty$ is well defined.
I know that
$\displaystyle \prod_{n=1}^{\infty} \cos\frac{x}{2^n}=\frac{\sin(x)}{x}$ and I wonder if $f$ also has a simple analytical expression.
Thank you.
EDIT : Using the expression $\cos^2(x)=(1+\cos(2x)/2)$ I can reducethe first sum to the second and get
$f(x)=\left(\frac{\sin(x)}{x \cos(x/2)}\right)^2$
|
Take the well known product and square:
$$\prod_{n=1}^{\infty} \cos\frac{x}{2^n}=\frac{\sin(x)}{x}$$
$$\prod_{n=1}^{\infty} \cos^2\frac{x}{2^n}=\frac{\sin^2(x)}{x^2}$$
Now use the formula:
$$\cos^2 \frac{t}{2}=\frac{1}{2} (1+\cos t)$$
To get:
$$\prod_{n=0}^\infty \frac{1}{2} \left( 1+\cos \frac{x}{2^n} \right)=\frac{\sin^2 x}{x^2}$$
|
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|
Recursive Sequence $a_n = \frac{1}{2} (a_{n-1} + 5) $ I got this question in which they ask me to explain why it is convergent and evaluate its limit.
$$a_1=3\;and\;a_n = \frac{1}{2} (a_{n-1} + 5) \\ n=2,3,4,... $$
To prove it's convergent, I show that it is increasing and bounded above by 5. Also, I find its limit by showing that
Let $L=\lim_{n\to\infty} a_n$
Notice that $a_n = \frac{1}{2} (a_{n-1} + 5)$
Hence $\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{1}{2} (a_{n-1} + 5)$
$\Rightarrow L = \frac{1}{2}(L+5)$
$\Rightarrow 2L = L+5$
$\Rightarrow L = 5$
As the sequence is non-decreasing, $$L =\lim_{n\to\infty} a_n=5 $$
That's what I got. However, the book's answer for this question's limit is $\frac{5}{2}$
Is there anything wrong with my proof?
Thanks in advance.
|
It might help to solve the recurrence relation
Define $b_n\equiv a_n+c$ and find the value of $c$ that makes $b_n$ a geometric sequence
$$a_n = \frac{1}{2} (a_{n-1} + 5)\implies b_n-c = \frac{1}{2} (b_{n-1}-c + 5)$$
$$\implies b_n = \frac{1}{2} b_{n-1}-\frac c2 + \frac 52 +c$$
$$\implies b_n = \frac{1}{2} b_{n-1} + \frac {c+5}2$$
SO if $c=-5$ we have $ b_n = \frac{1}{2} b_{n-1} $ which is a geometric sequence having solution
$$ b_n = b_1 \left( \frac 12\right) ^{n-1}$$
Where $b_1=a_1-5 = 3-5 = -2$
So $$ a_n = b_n + 5 = 5 -2 \left( \frac 12\right) ^{n-1}=5 - \left( \frac 12\right) ^{n-2} $$
|
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|
Integrating the following $\int \sqrt{\tan x+1}\,dx$
Question: Integrate the following, $$\int\sqrt{\tan x+1}\;dx.$$
Wolfram Alpha returns a non-elementary answer. Can someone please spot the mistake I have made here:
First consider this integral:
$$\int \frac{1}{(x+1)\sqrt{x+3}} \, dx = -\sqrt{2}\tanh^{-1}\frac{\sqrt{x+3}}{\sqrt{2}} + c$$
Wolfram Alpha confirms that result.
Then, we have
$$I=\int \sqrt{\tan x+1} \, dx, \quad \tan x=u+2,
\quad dx=\frac{du}{\sec^{2}x}=\frac{du}{(u+2)^{2}-1}=\frac{dx}{(u+3)(u+1)}$$
So this transforms the integral to the first integral on this post, which we can evaluate. Then after evaluation and resubstitution I get:
$$I=-\sqrt{2}\tanh^{-1}\frac{\sqrt{\tan x+1}}{\sqrt{2}}+c$$
However differentiating this with Wolfram Alpha gives me a messy trigonometric expression which doesn't seem to be equal (I tested some values in both expressions and get different answers). I also estimated the area under the integral between some values and also obtained different answers using the closed form. Any ideas why?
EDIT: I used the wrong identity. Nevertheless, we can still use this method to integrate sqrt(tanhx integrals). E.g:
$$I=\int \sqrt{\tanh x+1} \, dx, \tanh x=u+2,\quad
-dx = \frac{du}{\operatorname{sech}^2 x} = \frac{du}{(u+2)^2-1} = \frac{dx}{(u+3)(u+1)}$$
To obtain:
$\int \sqrt{\tanh x+1} \, dx = I=\sqrt{2}\tanh^{-1} \dfrac{\sqrt{\tanh x+1}}{\sqrt{2}}+c$
|
Let $I=\int \sqrt{\tan x+1} d x$ and $u=\sqrt{\tan x+1}$.
\begin{aligned}
I &=\int \frac{2 u^{2} d u}{\left(u^{2}-1\right)^{2}+1} \\
&=2 \int \frac{u^{2}}{u^{4}-2 u^{2}+2} d u \\
&=2 \int \frac{d u}{u^{2}+\frac{2}{u^{2}}-2} \\
&= 2 \int \frac{\left(1+\frac{\sqrt{2}}{u}\right)+\left(1-\frac{\sqrt{2}}{u}\right)}{u^{2}+\frac{2}{u^{2}}-2} d u\\
&=\int \underbrace{\frac{1+\frac{\sqrt{2}}{u}}{u^{2}+\frac{2}{u^{2}}-2} d u}_{I}+\underbrace{\int \frac{1-\frac{\sqrt{2}}{u}}{u^{2}+\frac{2}{u^{2}}-2} d u}_{J}\end{aligned}
$$I=\int \frac{d\left(u-\frac{\sqrt{2}}{u}\right)}{\left(u-\frac{\sqrt{2}}{u}\right)^{2}+2(\sqrt{2}-1)}=\frac{1}{\sqrt{2(\sqrt{2}-1)}} \tan ^{-1}\left(\frac{u-\frac{\sqrt{2}}{u}}{\sqrt{2 (\sqrt{2}-1)}}\right)$$
$$J=\int \frac{d\left(u+\frac{\sqrt{2}}{u}\right)}{\left(u+\frac{\sqrt{2}}{u}\right)^{2}-2(\sqrt{2}+1)}=\frac{1}{2 \sqrt{2(\sqrt{2}+1)}} \ln \left|\frac{u+\frac{\sqrt{2}}{u}-\sqrt{2 (\sqrt{2}+1)}}{\left.u+\frac{\sqrt{2}}{u}+\sqrt{2(\sqrt{2}+1}\right)}\right|$$
Now we can conclude that
$$I=\frac{1}{\sqrt{2(\sqrt{2}-1})} \tan ^{-1}\left(\frac{\tan x+1-\sqrt{2}}{\sqrt{2(\sqrt{2}-1)(\tan x+1})}\right)+ \frac{1}{2\sqrt{2(\sqrt{2}+1)}} \ln \left| \frac{\tan x+1-\sqrt{2(\sqrt{2}+1)(\tan x+1)}+\sqrt2}{\tan x+1+\sqrt{2(\sqrt{2}+1)(\tan x+1)}+\sqrt2}\right|+C$$
|
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|
How to prove $|a_{2n}-a_{n}|<\frac{10}{27}$ Let sequence $$a_{1}=1,a_{n+1}=\dfrac{2}{2a_{n}+1}$$
show that
$$\left|a_{2n}-a_{n}\right|<\dfrac{10}{27}$$
and the constant $\frac{10}{27}$ A smaller number instead
$$a_{n+1}+\dfrac{1+\sqrt{17}}{4}=2(1+\sqrt{17})\cdot\dfrac{a_{n}+\frac{1+\sqrt{17}}{4}}{4(2a_{n}+1)}\tag{1}$$
$$a_{n+1}-\dfrac{-1+\sqrt{17}}{4}=-2(-1+\sqrt{17})\cdot\dfrac{a_{n}-\frac{-1+\sqrt{17}}{4}}{4(2a_{n}+1)}\tag{2}$$
$\dfrac{(1)}{(2)}$,that
$$\dfrac{a_{n+1}+\dfrac{1+\sqrt{17}}{4}}{a_{n+1}-\dfrac{-1+\sqrt{17}}{4}}=\dfrac{1+\sqrt{17}}{1-\sqrt{17}}\cdot\left(\dfrac{a_{n}+\dfrac{1+\sqrt{17}}{4}}{a_{n}-\dfrac{-1+\sqrt{17}}{4}}\right)$$
so we I get ugly even more can't solve this problem.can you some recommended or solve this?
|
For convenience, let $p=\frac{1+\sqrt{17}}{4}$ and $q=\frac{-1+\sqrt{17}}{4}$
Let $b_{n}=\frac{a_{n}+p}{a_{n}-q}$, then your last equation can be rewritten as follows:
$b_{n+1}=\frac{p}{q}b_{n}$
Since the above is a geometric sequence, we have $b_{n}=b_{1}r^{n-1}$, where $r=\frac{p}{q}$.
Unsubstitting, we have $\frac{a_{n}+p}{a_{n}-q}=b_{1}r^{n-1}$.
So $a_{n}=\frac{qb_{1}r^{n-1}+p}{b_{1}r^{n-1}-1}$
I think you can derive your desired inequality using extra conditions such as $a_{1}=1$.
|
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|
Showing that $ \left\lfloor \frac{n-1}{2}\right\rfloor +\left\lfloor \frac{n+2}{4}\right\rfloor + \left\lfloor\frac{n+4}{4} \right\rfloor =n$
Show that :$$ \forall \in \mathbb{Z}, \left\lfloor \dfrac{n-1}{2}\right\rfloor +\left\lfloor \dfrac{n+2}{4}\right\rfloor + \left\lfloor\dfrac{n+4}{4} \right\rfloor =n$$
Solution provide by book:
In separate cases, as the remainder of the Euclidean division of $n$ by $4$ and present the results in a table
\begin{array} {|r|r|r|r|} \hline n & \left\lfloor \dfrac{n-1}{2}\right\rfloor & \left\lfloor \dfrac{n+2}{4}\right\rfloor & \left\lfloor\dfrac{n+4}{4} \right\rfloor & \sum \\ \hline 4k & 2k-1 & k & k+1 & 4k \\ \hline 4k+1 & 2k & k & k+1 & 4k+1\\ \hline 4k+2 & 2k & k+1 & k+1 & 4k+2 \\ \hline 4k+3 & 2k+1 & k+1 & k+1 & 4k+3 \\ \hline \end{array}
This establishes the desired result by examining all cases modulo $4$
*
*I didn't understand that solution which provided by Book could someone elaborate it Please ?
Reference
|
Because the least common multiple of the denominators is $4$, the effect of the floor functions will repeat with period $4$. You can convince yourself of this by trying $n$ from $0$ to $12$ or so. This means there are four cases depending on $n \bmod 4$ and they try all four cases. Each column in the middle of the table corresponds to one of the terms in the expression and they compute the value of that term,then add them up to get the last column. The fact that the first and last columns match establishes the identity. As and example, if $n=11,$ it is of the form $4k+3$ with $k=2$ Then
$$\left\lfloor \dfrac{11-1}{2}\right\rfloor +\left\lfloor \dfrac{11+2}{4}\right\rfloor + \left\lfloor\dfrac{11+4}{4} \right\rfloor =5+3+3=11$$
Added per request: The point is that on adding one to nn exactly one of the floor functions will increase, maintaining the identity. Which one increases depends on $n \bmod 4$ Taking the LCM gets you through one cycle of the pattern.
|
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|
Integrate $\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx$ $\displaystyle\int \dfrac {x^{2}} {\sqrt {x^{2}-16}}dx$
Effort 1:
Let be $x=4\sec u$
$dx=4.\sin u.\sec^2u.du$
Then integral;
$\displaystyle\int \dfrac {\sec^2u \; .4.\sin u.\sec^2u.du} {\sqrt {16\sec^2u-16}}=\displaystyle\int \sec^3.du$
After I didn't nothing.
Effort 2:
$\displaystyle\int \dfrac {x^{2}dx} {\sqrt {x^{2}-16}}$
Let's doing integral by parts;
$du=\dfrac{x}{\sqrt{x^2-16}}$
$v=x$
$\displaystyle\int \dfrac {x^{2}dx} {\sqrt {x^{2}-16}}=x.\sqrt{x^2-16}-\displaystyle\int\sqrt{x^2-16}dx$
We have $\quad \displaystyle\int\sqrt{x^2-16}dx$
let be $\quad x=4\sec j$
$dx=4\dfrac{\sin j}{\cos^2 j}dj$
and;
$\displaystyle\int\sqrt{x^2-16}\;dx=16.\displaystyle\int\dfrac{\sin j}{\cos j}\dfrac{\sin j}{\cos^2 j}dj=16.\displaystyle\int \sec j.tan^2j.dj$
After I didn't nothing.
|
Just another way to compute the antiderivative.
Change variable $$x^2-16=t^2\implies x=\sqrt{t^2+16}\implies dx=\frac{t}{\sqrt{t^2+16}}\,dt$$
So, $$I=\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx=\int \sqrt{t^2+16}\,dt$$ One integration by parts $$u=\sqrt{t^2+16}\implies du=\frac{t}{\sqrt{t^2+16}}\,dt$$ $dv=dt \implies v=t$. So $$I=\int \sqrt{t^2+16}\,dt=t \sqrt{t^2+16}-\int \frac{t^2}{\sqrt{t^2+16}}\,dt$$ $$I=t \sqrt{t^2+16}-\int \frac{t^2+16-16}{\sqrt{t^2+16}}\,dt=\sqrt{t^2+16}-I+\int \frac{16}{\sqrt{t^2+16}}\,dt$$ which makes $$2I=t \sqrt{t^2+16}+\int \frac{16}{\sqrt{t^2+16}}\,dt$$ Now, an obvious change of variable $t=4\sinh(u)$ will give the end result easily.
|
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|
Sum of combinatorics sequence $\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1}$ I need to find sum like $$\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1},\qquad \text{ for even } n$$
Example:
Find the sum of $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$
|
The important binomial theorem states that
\begin{equation}
\sum_{k=0}^n\binom{n}{k} x^ky^{n-k}=(x+y)^n
\end{equation}
Setting $x=1$ and $y=1$, we have the following relation
\begin{equation}
\sum_{k=0}^n\binom{n}{k} =2^n
\end{equation}
and setting $x=-1$ and $y=1$, we have the following relation
\begin{equation}
\sum_{k=0}^n\binom{n}{k} (-1)^k=0
\end{equation}
Hence we have
\begin{equation}
\sum_{k=1}^{n/2}\binom{n}{2k-1} =\frac{1}{2}\left(\sum_{k=0}^n\binom{n}{k}-\sum_{k=0}^{n}\binom{n}{k}(-1)^k\right)=2^{n-1}.
\end{equation}
This may be easier to see with your example
\begin{align}
&\binom{20}{1}+\binom{20}{3}+\cdots+\binom{20}{19}=\\
&\frac{1}{2}\left[\color{red}{\binom{20}{0}}+\binom{20}{1}+\color{red}{\binom{20}{2}}+\cdots+\binom{20}{20}-\left(\color{red}{\binom{20}{0}}-\binom{20}{1}+\color{red}{\binom{20}{2}}+\cdots+\binom{20}{20}\right)\right]
\end{align}
|
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|
Series expansion of $\int x^xdx$ The indefinite integral:
$$J=\int x^xdx$$
has no known closed form solution.
Expanding in series the function $f=x^x$ we get:
$$f\simeq\sum_{k=0}^N \dfrac{x^k\ln(x)^k}{k!}$$
So we can write:
$$J\simeq\sum_{k=0}^N\int\dfrac{x^k\ln(x)^k}{k!}dx$$
and we get:
$$J\simeq x+\sum_{k=1}^N\dfrac{\ln(x)^k\left[-(k+1)\ln(x)\right]^{-k}\Gamma\left(k+1,-(k+1)\ln(x)\right)}{(k+1)!}$$
where $\Gamma(x,a)$ is the incomplete gamma function.
It seems to work well in approximating J, for $N\ge 20$ and $x\in\mathbb{R}$.
Is there a better way to calculate numerically $J$ avoiding the use of the usual methods to evaluate definite integrals? Thanks.
|
$\int x^x~dx=\int e^{x\ln x}~dx=\int\sum\limits_{n=0}^\infty\dfrac{x^n(\ln x)^n}{n!}dx$
For $\int x^n(\ln x)^n~dx$ , where $n$ is any non-negative integers,
$\int x^n(\ln x)^n~dx$
$=\int(\ln x)^n~d\left(\dfrac{x^{n+1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{x^{n+1}}{n+1}d((\ln x)^n)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)x}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^n(\ln x)^{n-1}}{n+1}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{n(\ln x)^{n-1}}{n+1}d\left(\dfrac{x^{n+1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{x^{n+1}}{n+1}d\left(\dfrac{n(\ln x)^{n-1}}{n+1}\right)$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^{n+1}(\ln x)^{n-2}}{(n+1)^2x}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^n(\ln x)^{n-2}}{(n+1)^2}dx$
$=\cdots\cdots$
$\vdots$
$\vdots$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}-\int\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times1)x^n}{(n+1)^n}dx$
$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}+\dfrac{(-1)^n(n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+1}}+C$
$=\dfrac{(n+1)x^{n+1}(\ln x)^n}{(n+1)^2}-\dfrac{(n+1)nx^{n+1}(\ln x)^{n-1}}{(n+1)^3}+\cdots\cdots+\dfrac{(-1)^{n-1}((n+1)n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^{n+1}}+\dfrac{(-1)^n((n+1)n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+2}}+C$
$=\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}(n+1)!x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+2}}+C$
$=\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$
$\therefore\int\sum\limits_{k=0}^n\dfrac{(x\ln x)}{n!}dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$
Hence $\int x^x~dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How would you find the exact roots of $y=x^3+x^2-2x-1$? My friend asked me what the roots of $y=x^3+x^2-2x-1$ was.
I didn't really know and when I graphed it, it had no integer solutions. So I asked him what the answer was, and he said that the $3$ roots were $2\cos\left(\frac {2\pi}{7}\right), 2\cos\left(\frac {4\pi}{7}\right)$ and $2\cos\left(\frac {8\pi}{7}\right)$.
Question: How would you get the roots without using a computer such as Mathematica? Can other equations have roots in Trigonometric forms?
Anything helps!
|
Let $p(x) = x^3+x^2-2x-1$, we have $$p(t + t^{-1}) = t^3 + t^2 + t + 1 + t^{-1} + t^{-2} + t^{-3} = \frac{t^7-1}{t^3(t-1)}$$
The RHS has roots of the form $t = e^{\pm \frac{2k\pi}{7}i}$
( coming from the $t^7 - 1$ factor in numerator )
for $k = 1,2,3$. So $p(x)$ has roots of the form $$e^{\frac{2k\pi}{7} i} + e^{-\frac{2k\pi}{7} i} = 2\cos\left(\frac{2 k\pi}{7}\right)$$ for $k = 1,2,3$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Complex Numbers and Binomial Expansion
I was able to show the above by equating the solution of $cos \ 5\theta=0$ (which gives $\pi/10$) and the solution of ${16cos}^5\theta - {20cos}^3\theta+5cos\theta = 0$ (which is what you get when you use De Moivre's Theorem with binomial expansion). However, I do not understand how the second part of the question has anything to do with the first, and thus am lost as to where to begin.
|
What you have to realize about the equation $\cos5\theta=0$ is that $\frac{\pi}{10}$ is a solution because $\cos\frac{\pi}{2}=0$, but also that $\frac{7\pi}{10}$ is a solution because $\cos\frac{35\pi}{10}=\cos\frac{5\pi}2=\cos\left(2\pi+\frac{\pi}2\right)=\cos\frac\pi 2=0$. Therefore, we can use this equation to find both $\cos\frac\pi {10}$ and $\cos\frac{7\pi}{10}$.
We have the following:
$$16\cos^5\theta-20\cos^3\theta+5\cos\theta=0$$
We want to solve this equation to find $\cos \frac{\pi}{10}$ and $\cos \frac{7\pi}{10}$, which we know is not $0$. Therefore, we can divide by $\cos \theta$:
$$16\cos^4\theta-20\cos^2\theta+5=0$$
Now, we can use the quadratic formula to find $\cos^2\theta$:
$$\cos^2\theta=\frac{20\pm\sqrt{20^2-4\cdot16\cdot5}}{2(16)}$$
$$\cos^2\theta=\frac{20\pm4\sqrt 5}{32}$$
$$\cos^2\theta=\frac{5\pm\sqrt5}{8}$$
Now, since $\frac{\pi}{10}$ is very close to $0$, it will have a bigger $\cos$, so we want to take the bigger solution. Therefore, the $\pm$ becomes a $+$.
$$\cos^2\frac{\pi}{10}=\frac{5+\sqrt5}{8}$$
$\frac\pi{10}$ is in Quadrant I, so we take the positive square root of both sides:
$$\cos\frac\pi{10}=\sqrt{\frac{5+\sqrt5}{8}}$$
Now, for $\frac{7\pi}{10}$, we have:
$$\cos^2\frac{7\pi}{10}=\frac{5\pm\sqrt5}{8}$$
For this, you need to ask yourself two questions:
*
*Does the $\pm$ become a $+$ or a $-$? (Think bigger vs smaller $\cos$.)
*Do you take the positive or negative square root? (Think sign of $\cos$ in this quadrant.)
Good luck finding $\cos\frac{7\pi}{10}$!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Compute covariance matrix random walk
Consider a random walk on the square lattice $\mathbb{Z}^2$ with diagonal jumps of size $2$, i.e. the jump probabilities are
$$P(X_1 = x) =
\begin{cases}
\frac{1}{4} & \quad \text{if } x \in \{(2,2), (-2,2), (2,-2), (-2,-2) \} \\
0 & \quad \text{otherwise } \\
\end{cases}$$
Compute the covariance matrix $(\operatorname{Cov}(X_1^i, X_1^j))_{i,j=1,2}$, where $X_1^{(i)}$ denotes the $i$-th component of $X_1$.
In my study material is written:
$$\operatorname{Cov}(X_1^{(i)}, X_1^{(j)}) = E[X_1^{(i)} X_1^{(j)}] = \sum_{x \in \mathbb{Z}^d} x^{(i)} x^{(j)} P(X_1 = x) =
\begin{cases}
\frac{1}{d} & \quad \text{if } i =j, \\
0 & \quad \text{if } i \neq j, \\
\end{cases}$$
since $E[X_1]$ is the $d$-dimensional null-vector.
For $x \in \{(1,1), (1,2), (2,1)\}$ we have $P(X_1 = x) =0$.
For $x = (2,2)$ we have then in the covariance matrix $A$:
$$a_{22} = 2 \cdot 2 \cdot \frac{1}{4} = 1$$
This will give the covariance matrix
$$A = \begin{pmatrix} 0&0 \\ 0&1 \end{pmatrix}$$
Is my way of computing correct?
|
Suppose that $i=1$ and $j=2$ and $d=2$. Then we have
\begin{align}
& \sum_{x \in \mathbb{Z}^d} x^{(i)} x^{(j)} P(X_1 = x) = \sum_{x \in \mathbb{Z}^2} x^{(1)} x^{(2)} P(X_1 = x) \\[10pt]
= {} & \underbrace{2\cdot 2\cdot \frac 1 4}_{x \,=\, (2,2)} + \underbrace{2 \cdot(-2)\cdot \frac 1 4}_{x \,=\, (2,-2)} + \underbrace{(-2)\cdot2\cdot\frac 1 4}_{x \,=\, (-2,2)} + \underbrace{(-2)\cdot(-2)\cdot\frac 1 4}_{x \,=\, (-2,-2)} + \underbrace{\text{terms equal to 0}}_{x\,\in\,\mathbb Z^2 \,\setminus\, (\{\pm2\}\times\{\pm2\}) } \\[12pt]
= {} & 0.
\end{align}
By symmetry of covariance matrices, we get the same thing if $i=2$ and $j=1$. Thus the matrix is
$$
\begin{bmatrix} \text{?} & 0 \\ 0 & \text{?} \end{bmatrix}.
$$
The two question marks would be filled in with the sum
$$
2^2\cdot \frac 1 4 + (-2)^2\cdot \frac 1 4 + (-2)^2 \cdot \frac 1 4 + 2^2 \cdot \frac 1 4 = 4,
$$
so we get
$$
\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1824784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Recursive to non recursive function $$
f(x) =
\begin{cases}
0 & x=1 \\
f(x-1)+1 & \frac{f(x-1)}{x-1} < p \\
f(x-1) & \text{otherwise} \\
\end{cases}
$$
Where $p$ is a constant less that or equal to 1.
And x is a whole number greater than 0.
How do I get a non recursive solution for $f(x)$?
|
To make Nithin's answer more precise, we have
$$f(x) = \begin{cases}
\left\lceil (x-1) p \right\rceil & \text{ if } p \leq 1 \\
x - 1 & \text{ if } p > 1
\end{cases}$$
You can easily prove that by induction on $x$:
*
*If $x = 1$, then $f(x) = 0 = x-1 = \left\lceil (x-1)p \right\rceil$.
*If $p > 1$ and $f(x-1) = x - 2$, then we have $\frac{f(x-1)}{x-1} < p$, so we have $f(x) = x-2 + 1 = x-1$.
*If $p \leq 1$ and $f(x-1) = \left\lceil(x-2)p\right\rceil$, then there are two cases:
*
*If $\frac{f(x-1)}{x-1} < p$, then we have $\left\lceil(x-2)p\right\rceil < (x-1)p$. Thus we have $f(x) = f(x-1)+1 = \left\lceil(x-1)p\right\rceil$.
*Otherwise, we have $\left\lceil(x-2)p\right\rceil < (x-1)p$. So we have $f(x) = f(x-1) = \left\lceil(x-1)p\right\rceil$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
In $\triangle ABC$, if $\tan A$, $\tan B$, $\tan C$ are in harmonic progression, then what is the minimum value of $\cot \frac{B}{2}$?
In a $\triangle ABC$, if $\tan A$, $\tan B$, $\tan C$ are in harmonic progression, then what is the minimum value of $\cot(B/2)$?
$\bf{My\; Try::}$ Here $A+B+C=\pi\;,$ Then $\tan A+\tan B+\tan C=\tan A\cdot \tan B\cdot \tan C$
and given $\tan A,\tan B\;,\tan C$ are in harmonic progression,
So we get $$\frac{2}{\tan B} = \frac{1}{\tan A}+\frac{1}{\tan C} = \frac{\tan A+\tan C}{\tan A\tan C}=\frac{\tan A\tan B\tan C-\tan B}{\tan A\tan C}$$
So $$\tan B -\frac{\tan B}{\tan A\tan C} = \frac{2}{\tan B}\Rightarrow \tan^2 B = \frac{2\tan A\tan C}{\tan A\tan C-1}$$
Now how can i solve after that, Help required, Thanks
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We have $\cot A+\cot C=2\cot B\ \ \ \ (1)$
and Prove that $\tan A + \tan B + \tan C = \tan A\tan B\tan C,$ $A+B+C = 180^\circ$ says $$\cot A\cot B+\cot B\cot C+\cot C\cot A=1$$
Eliminating $\cot C,$
we get $$\cot^2A-2\cot A\cot B+1-2\cot^2B=0$$ which a Quadratic Equation in $\cot A$
As $\cot A$ is real, the discriminant $$(2\cot B)^2-4(1-2\cot^2B)=4(3\cot ^2B-1)$$ must be $\ge0$
$$\implies\cot^2 B\ge\dfrac13\iff\tan^2B\le3\ \ \ \ (2)$$
If $\cot B<0,$ by $(1),$ at least one of $\cot A,\cot C$ must be $<0$
If $\cot A,\cot B<0, A.B>90^\circ\implies A+B>180^\circ$ which is impossible
$\implies B<90^\circ\implies\tan B\ge 0$
By $(2),\tan B\le\sqrt3\implies B\le60^\circ$
Can you take it from here?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that for all odd integer $n, n^{4}=1\pmod {16}$ My answer: $n=2k+1$
$n^{4}=(2k+1)^{4}$=$16k^{4}+32k^{3}+8k^{2}+24k+1$.
I do not know how to conclude; really needed help here.
|
$$n^4-1=(n^2+1)(n+1)(n-1)$$
If $n$ is odd, then all the factors are even, and one of the factors $(n+1)$ or $(n-1)$ is divisible by $4$. Hence $2^4|n^4-1$, or in other words $n^4\equiv1\pmod{16}$.
This can be generalized to see that $n^{2^k}\equiv1\pmod{2^{k+2}}$ for odd $n$:
$$n^{2^k}-1=\left(\left(n^{2^{k-1}}+1\right)\left(n^{2^{k-2}}+1\right)\cdots\left(n+1\right)\right)\left(n-1\right)$$
Since $x$ is odd, all the $k+1$ factors in the above are even, and either $(n+1)$ or $(n-1$) is divisible by $4$, so $2^{k+2}|n^{2^k}-1$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove using induction that $C(n) = C(n/5) + C(3n/4) + n$ is $O(n)$ I was hoping someone could take a look at my answer to this question and check if it's correct and offer some advice/help on how to correct if it's wrong.
Question:
Consider the function $C: \mathbb N \rightarrow \mathbb N$ defined by:
$C(n) \begin{cases} 1& \text{if } n \leq 5, \\
C(\lfloor \frac{n}{5} \rfloor) + C(\lfloor \frac{3n}{4} \rfloor) + n & \text{if } n > 5 \end{cases}$
Prove, by induction on n, that $C(n) = O(n)$.
My attempt:
For $C(n) = O(n), \exists c > 0, \exists n_o > 1$ such that $ C(n) \leq cn, \text{ } \forall n \geq n_o$
Base case: $n=6, C(6) = C(\lfloor \frac{6}{5} \rfloor) + C(\lfloor \frac{3(6)}{4} \rfloor) + 5 = C(1) + C(4) + 6 = 1 + 1 + 6 = 8 $
So $c = 1, n_0 = 8$ will suffice for $C(5) = O(n)$
Inductive step: $n > 6$, we know from the inductive hypothesis that $\exists c>0, \exists n_0>1 $ such that $C(\lfloor \frac{n}{5} \rfloor) \leq \frac{cn}{5}$ and $C(\lfloor \frac{3n}{4} \rfloor) \leq \frac{3nc}{4}$ $\forall n \geq n_0$
So, $C(\lfloor \frac{n}{5} \rfloor) + C(\lfloor \frac{3n}{4} \rfloor) + n \leq \frac{cn}{5} + \frac{3nc}{4} + cn $ $\forall n \geq n_0$
Since $\frac{cn}{5} + \frac{3nc}{4} + cn = cn(\frac{1}{5} + \frac{3}{4} + 1) $
So, $C(n) = C(\lfloor \frac{n}{5} \rfloor) + C(\lfloor \frac{3n}{4} \rfloor) + n \leq c'n,$
where $c' = c \cdot (\frac{1}{5} + \frac{3}{4} + 1)$, $\forall n \geq n_0 $
Thus C(n) = O(n).
I have a feeling I'm doing something wrong here, so any help would be much appreciated.
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I have a feeling I'm doing something wrong
So, anyway, you have to be partially right and partially wrong. :-) But your feeling is right, because if we want to prove inductively the claim with $n_0$ we have to have also $\lfloor \frac{n}{5} \rfloor\ge n_0$ and $\lfloor \frac{3n}{4} \rfloor\ge n_0$, which makes the induction useless.
A simple way to overcome such an obstacle is to prove inductively that $C(n)\le cn$ for all $n$. For the induction step it suffices to have $c\ge \frac c5+\frac {3c}4+1$, that is $c\ge 20$. For the base we need to have $c\ge 1$, so $c=20$ fits the proof.
|
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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.