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Looking for a cleaner/quicker way to evaluate the integral of a quadratic Looking for a clean/quick way to evaluate
$$\int_{2-\sqrt{3}}^{2+\sqrt{3}}{ \left(x^2-4x+1\right)\textrm{d}x}=\left.\frac{x^3}{3}-2x^2+x \right|_{2-\sqrt{3}}^{2+\sqrt{3}}$$
So I evaluated all of this out leaving this expression as it was and there was some pretty nice cancelation that led me to the answer $-\sqrt{3}$. However, this was a question from a contest so I'm expecting that if we factor out the $x$ and then evaluate at the bounds there is something we can do to make the algebra nicer.
$$=\left.x(\frac{x^2}{3}-2x+1) \right|_{2-\sqrt{3}}^{2+\sqrt{3}}$$
Anyone see any quick way to do this? Thanks!
|
Notice that $x^2 - 4x + 1 = (x - 2)^2 - 3$, so a substitution $y = x - 2$ yields
\begin{align*}
\int_{2-\sqrt{3}}^{2+\sqrt{3}} (x^2-4x+1)dx = \int_{-\sqrt{3}}^{\sqrt{3}}(y^2-3)dy = \left[\frac{y^3}{3} - 3y\right]_{-\sqrt{3}}^{\sqrt{3}} = 2\sqrt{3}-6\sqrt{3} = -4\sqrt{3}
\end{align*}
|
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"url": "https://math.stackexchange.com/questions/4421818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
If $-1 \le x \le 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method) If $-1 \leq x \leq 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method)
As stated in the problem, I can't use calculus. Therefore, I'm using things I've learnt so far instead:
One of the things I have tried most successfully is using trig substitutions...
For example, if I substitute $x = \sin \phi$, this yields $\sin \phi + \cos \phi$
But what should I do next? Or there are any methods else to solve the problem?
|
Substitute $$x = \sin \theta$$ The expression is $$x + \sqrt{-x^2 + 1} = \sin \theta + |\cos \theta|$$
We want to find the max. value of $\sin \theta + |\cos \theta|$. Now, two cases: $\cos \theta < 0$ and $\cos \theta \geq 0$.
$$\textbf{Case 1:} \cos \theta < 0$$
We need to find the max. value for $\sin \theta - \cos \theta$ for $\theta \in (\frac{\pi}{2}, \frac{3 \pi}{2})$
Note that $\sin \theta - \cos \theta \leq \sqrt{2}$ always.
In this range, this value is achievable at $\theta = \frac{3 \pi}{4} + 2\pi n, n \in \mathbb{Z}$
$$\textbf{Case 2:} \cos \theta \geq 0$$
We need to find the max. value for $\sin \theta + \cos \theta$ for $\theta \in [0, \frac{\pi}{2}] \cup [\frac{3 \pi}{2}, 2 \pi)$
Note that $\sin \theta + \cos \theta \leq \sqrt{2}$ always.
In this range, this value is achievable at $\theta = \frac{\pi}{4} + 2\pi n, n \in \mathbb{Z}$
$$\textbf{Thus, the max. value for} \sin \theta + |\cos \theta| \textbf{ is } \sqrt{2} \textbf{ achieved at } x = \frac{1}{\sqrt{2}}$$
NOTE $1$: Note that mentioning the values for $\theta$ for which the max. value occurs is important. Min-max problems are two step problems:
$1.$ Show that some expression is bounded.
$2.$ Show that the bound is achievable for the values we are concerned with.
As an exercise, try to find the min. value of the original expression. By following the same case-work and not following step-$2$, one would arrive at the incorrect conclusion that $-\sqrt{2}$ is the min. value for $\sin \theta + |\cos \theta|$.
NOTE $2$: We use the fact that $$-\sqrt{a^2 + b^2} \leq a \sin \theta + b \cos \theta \leq \sqrt{a^2 + b^2}$$
Why is it true?
Consider the polar co-ordinates of $(a, b)$. Let it be $(r, \phi)$ where $r = \sqrt{a^2 + b^2}$. $$a = r \cos \phi$$ $$b = r \sin \phi$$ Substitute, and we get $$a \sin \theta + b \cos \theta = r \sin(\theta + \phi) = \sqrt{a^2 + b^2} \sin(\theta + \phi)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is this interval notation for the solution of an inequality problem correct? Is this notationally correct?
$$
\begin{align*}
5 - x^2 &< -2, \\[1ex]
5 - x^2 + (-5) &< -2 + (-5), \\[1ex]
-x^2 &< -7, \\[1ex]
(-x^2)(-1) &> (-7)(-1), \\[1ex]
x^2 &> 7, \\[1ex]
x^2 - 7 &> 0, \\[1ex]
(x + \sqrt{7})(x - \sqrt{7}) &> 0, \\[1ex]
x < -\sqrt{7} \,\lor\, x &> \sqrt{7}.
\; \llap{\mathrel{\boxed{\phantom{\;x < -\sqrt{7} \,\lor\, x > \sqrt{7}.}}}}\end{align*}
$$
or would the following be better notation?
$$
\begin{align*}
(x < -\sqrt{7}) \,\lor\, (x &> \sqrt{7}).
\; \llap{\mathrel{\boxed{\phantom{\;(x < -\sqrt{7}) \,\lor\, (x > \sqrt{7}).}}}}
\end{align*}
$$
Thank you.
|
I will write the final part in question as $x\in (-\infty,-\sqrt7)\cup (\sqrt 7,\infty)$ because the inequality is in $x$.
|
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|
Relation between roots and coefficients of equation If the roots of the equation $x^4 - x^3 +2x^2+x+1 = 0 $ are given by $a,b,c,d$ then find the value of $(1+a^3)(1+b^3)(1+c^3)(1+d^3)$
I found out that:
$$ (1+a^3)(1+b^3)(1+c^3)(1+d^3) = (abcd)^3+\sum(abc)^3 +\sum(ab)^3+\sum(a)^3 +1$$
But how do I calculate $\sum(abc)^3$ and $\sum(ab)^3$?
|
With a lot of these problems involving symmetric sums of the roots of a polynomial, there are two ways to do it. There is a long and tedious way using Vieta’s formulas and Newton sums, and a “clever” way that just involves evaluating the polynomial a couple times. I will go over the clever way for this problem, and I’ll encourage you to look into vieta’s formulas and Newton sums yourself.
For the clever way, we start by factoring the polynomial:
$$P(x)=x^4-x^3+2x^2+x+1=(x-a)(x-b)(x-c)(x-d)$$
Now, observe that if we let $x=-1$, something interesting happens:
$$P(-1)=(-1-a)(-1-b)(-1-c)(-1-d)=(1+a)(1+b)(1+c)(1+d)$$
This is related to the product we wish to evaluate, but it doesn’t have the exponents. We can fix this by noticing that our product can be factored in the complex numbers. If we let $\omega=e^{\frac{2\pi i}{3}}$, then we have:
$$(1+a^3)(1+b^3)(1+c^3)(1+d^3)=(1+a)(1+\omega a)(1+\omega^2 a)(1+b)(1+\omega b)\ldots$$
$$=(-1-a)(-1-b)\ldots (-\omega^2-a)(-\omega^2-b)\ldots(-\omega -a)(-\omega -b)\ldots$$
$$=P(-1)P(-\omega^2)P(-\omega)$$
You’d then need to evaluate the product $P(-1)P(-\omega^2)P(-\omega) $. While it is a little tedious, it is doable in a reasonable amount of time, and it’s significantly faster than other methods.
|
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|
Calculate $\lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{1}{n^2}-\frac{\pi^2}{6}\right)N$ How to calculate the limit
$$\lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{1}{n^2}-\frac{\pi^2}{6}\right)N?$$
By using the numerical method with Python, I guess the right answer is $-1$ but how to prove? I have no idea.
|
Let $N\geq 1$. Note that
$$
\frac{{\pi ^2 }}{6} - \sum\limits_{n = 1}^N {\frac{1}{{n^2 }}} = \sum\limits_{n = N + 1}^\infty {\frac{1}{{n^2 }}} = \sum\limits_{n = N + 1}^\infty {\frac{1}{{n(n - 1)}}} - \sum\limits_{n = N + 1}^\infty {\frac{1}{{n^2 (n - 1)}}} = \frac{1}{N} - \sum\limits_{n = N + 1}^\infty {\frac{1}{{n^2 (n - 1)}}} .
$$
Then
$$
\sum\limits_{n = N + 1}^\infty {\frac{1}{{n^2 (n - 1)}}} \le \sum\limits_{n = N}^\infty {\frac{1}{{n^3 }}} \le \frac{1}{{N^{3/2} }}\sum\limits_{n = N}^\infty {\frac{1}{{n^{3/2} }}} \le \frac{{\zeta (3/2)}}{{N^{3/2} }}.
$$
Accordingly,
$$
\left( {\sum\limits_{n = 1}^N {\frac{1}{{n^2 }}} - \frac{{\pi ^2 }}{6}} \right)N = - 1 + \mathcal{O}\!\left( {\frac{1}{{N^{1/2} }}} \right),
$$
showing that the limit is indeed $-1$.
|
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|
Does the pattern of identities for $(x+y+z)^{2n-1}$ continue past $2 n - 1 = 5$? We have the following identities.
\begin{align}
(x+y+z)^3&=x^3+y^3+z^3+3(x+y)(y+z)(z+x)\\
(x+y+z)^5&=x^5+y^5+z^5+5(x+y)(y+z)(z+x)(x^2+y^2+z^2+xy+yz+zx)
\end{align}
I'm wondering is there any pattern like above in the expansion of $(x+y+z)^{2n-1}$ where $n\in\mathbb{Z}^+$? The pattern I see so far is appearance of the coefficient $2n-1$ and $(x+y)(y+z)(z+x)$.
The above expansions can be found by factoring $(x+y+z)^{t}-x^{t}-y^{t}-z^{t}$ by using Cyclic Polynomials. For example for $t=5$ the factorization is provided on Brilliant.
I tried factoring $(x+y+z)^{7}-x^{7}-y^{7}-z^{7}$ using the method I got:
$$(x+y)(y+z)(z+x) [A(x^4+y^4+z^4)+B(x^3y+y^3z+z^3x)+C(x^2y^2+y^2z^2+z^2x^2)]$$
Here I'm not sure whether I should include $D(xy^3+yz^3+zx^3)$ too or not.
|
Denote $$\Delta_m(x, y, z) := (x + y + z)^m - x^m - y^m - z^m .$$
For odd $m$, $\Delta_m(x, y, y) = 0$ so $y + z$ divides $\Delta_m(x, y, z)$; by symmetry so do $z + x$ and $x + y$. Hence, $$(x + y + z)^m = x^m + y^m + z^m + (y + z)(z + x)(x + y) q_m(x, y, z)$$ for some symmetric, homogeneous polynomial $q_m$ of (even) degree $m - 3$. One can always write a symmetric polynomial as a polynomial in elementary symmetric polynomials, in this case,
$$1, \quad x + y + z, \quad y z + z x + x y, \quad x y z .$$
In general, the coefficient of $x^i y^j z^k$, $i + j + k = m$, in the expansion of $(x + y + z)^m$ is the multinomial coefficient ${m \choose{i, j, k}} = \frac{m!}{i! j! k!} .$ If $m = p$ is prime, then except when one of $i, j, k$ is equal to $p$, corresponding to the monomials $x^p, y^p, z^p$, respectively, we have $i, j, k < p$ and so $p \mid {p\choose{i, j, k}}$. Thus $p \mid \Delta_p(x, y, z)$ (over $\Bbb Z$), and hence also $p \mid q_p(x, y, z)$.
Example ($m = 7$) By our previous observation, $7 \mid q_7(x, y, z)$ because $7$ is prime, and because $\deg q_7 = 4$ we get one potential summand for each (unordered) partition of $4$ into sums of positive integers no larger than $3$ (namely, $[1,1,1,1]$, $[2,1,1]$, $[2,2]$, $[3,1]$). More precisely,
\begin{multline}
q_7(x, y, z) = 7[A (x + y + z)^4 + B (y z + z x + x y) (x + y + z)^2 \\ + C (y z + z x + x y)^2 + D x y z (x + y + z)]
\end{multline}
for some integer coefficients $A, B, C, D$. Specializing the above formulae to $z = -y$ and comparing like coefficients immediately gives $A = C = 1$ and $B + D = -1$. Then instead specializing to $y = x, z = 0$ gives $B = -2$ and thus $D = 1$.
Alternatively, we can decompose
\begin{multline}
q_7(x, y, z)
= 7 [
E (x^2 y z + y^2 z x + z^2 x y)
+ F (y^2 z^2 + z^2 x^2 + x^2 y^2) \\
+ G (y^3 z + y z^3 + z^3 y + z y^3 + x^3 y + x y^3)
+ H (x^4 + y^4 + z^4) ] ,
\end{multline}
and again solving for the coefficients gives $E = 5, F = 3, G = 2, H = 1$.
For general odd $m$, for both decompositions of $q_m$ there are $\left[\frac{m^2}{12}\right]$ terms, where $[\,\cdot\,]$ denotes rounding to the nearest integer; see A001399.
Counterexample ($m = 9$) As alex.jordan pointed out in the comments, if $m$ is composite it need not divide $q_m(x, y, z)$, and Oscar Lanzi observed in a comment replying to the other answer that's already the case for $m = 9$, the smallest odd composite number: Indeed the coefficients of the monomials $x^i y^j z^k$ of $q_9(x, y, z)$ turn out to be $9, 27, 57, 69, 105, 189, 273$, some of which are not divisible by $9$, so $9 \nmid q_9(x, y, z)$
Probably one could use Newton's Identities to derive an explicit formula for $q_m$ in terms of $m$. In the standard notation $e_i(x, y, z)$ for elementary symmetric polynomials and $p_i(x, y, z)$ for power sums, $(y + z)(z + x)(x + y) = e_1 e_2 - e_3 = \frac{1}{3} (p_1^3 - p_3)$, so your observation is that for odd $m$ the difference $e_1^m - p_m$ factors as $$e_1^m - p_m = (e_1 e_2 - e_3) q_m = \frac{1}{3} (p_1^3 - p_3) q_m$$ for some polynomial $q_m$. I don't know formulae offhand, but Bell numbers might play a role.
|
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Proving by induction that $\frac{1}{n} \ge \frac{n!}{n^n}$ I have been trying to prove by induction that for all $n \ge 0, \frac{1}{n} \ge \frac{n!}{n^n} $
Here is my proof so far:
Prove that for $ n = 1, \frac{1}{n} \ge \frac{n!}{n^n} $
$ \frac{1}{1} = 1 $
$ \frac{1!}{1^1} = 1 \le 1 $
Next, assume there exists $ n \ge 0 $ such that $ \frac{n!}{n^n} \le \frac{1}{n}. $.
We'll prove that $ \frac{1}{n+1} \ge $ $\frac{(n+1)!}{(n+1)^{n+1}} $.
$\frac{(n+1)!}{(n+1)^{n+1}} = \frac{(n!)(n+1)}{(n+1)^{n+1}} = \frac{n!}{(n+1)^n} \lt \frac{n!}{n^n} \le \frac{1}{n}$
However, I don't know how to show now that $\frac{n!}{(n+1)^n} \le \frac{1}{n +1}$
|
Just note that $$\frac{1}{n}\ge \frac{n!}{n^n} \iff n^n\ge n! \cdot n$$
$$\iff n^{n-1}\ge n!$$
Now assume that for some $n\ge 1$ the above proposition holds
$$(n+1)^n\ge (n+1)!$$
$$\iff (n+1)^{n-1}\ge n!$$
But this is trivial since $(n+1)^{n-1}\ge n^{n-1}\ge n!$
|
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|
Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3$ Prove that if $a,b,c,d$ are positive reals we have:
$$\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{b^2+c^2}}+\frac{c}{\sqrt{c^2+d^2}}+\frac{d}{\sqrt{d^2+a^2}}\leq3.$$
I think that I have found a equality case, for example, when $a=x^4,b=x^3,c=x^2,d=x$, and as $x$ tends to $\infty$, the LHS tends to $3$, but this means that the inequality is very unlikely to be solved with traditional methods, such as Cauchy-Schwartz (my starting idea), so I got stuck.
|
Partial Hint :
For $x,y>0$ and $x,y\leq 1$ such that $\left(y\right)^{2}+\left(x-\frac{1}{5}\right)^{2}\geq\frac{2}{50}$ we have the inequality :
$$\sqrt{\frac{x^{2}}{y^{2}+x^{2}}}-\left(f\left(x\right)-f\left(y\right)\right)-\frac{3}{4}\le0$$
Where :
$$f\left(x\right)=x\sqrt{\frac{1}{x^{2}+1}}$$
Summing gives the inequality
Edit : we supposed that for :
$$h\left(x,y\right)=\frac{3x^{2}}{2\left(x^{2}+y^{2}\right)} $$
We have :
$$h\left(a,b\right)+h\left(b,c\right)+h\left(c,d\right)+h\left(d,a\right)\leq3$$
Wich can be done since the inequality is homogeneous .
Now we have the inequality for $x,y>6$ and $x/5\leq y$ :
$$\sqrt{\frac{x^{2}}{x^{2}+y^{2}}}-f\left(x\right)+f\left(y\right)-\frac{3x^{2}}{2\left(x^{2}+y^{2}\right)}\le0$$
Where :
$$f\left(x\right)=0.5\ln\left(\sqrt{\frac{x}{x^{2}+1}}\right)$$
I haven't a proof yet but I come back again on it .
Hope it helps .
|
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|
Prove that the equation has solutions in two different intervals [-1,1] and [1,2] $a,b,c \in \mathbb{R}$ and $a,b,c$ positive numbers
$\frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=1$
Show that the equation has a solution in the interval $[-1,1]$ and a solution in the interval $[1,2]$.
$\frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=1$
$\iff [(a+b)\cdot x+a-b] \cdot (x-2) + c \cdot (x^2-1) = (x-2)\cdot (x^2-1)$
$\iff [(a+b)\cdot x+a-b] \cdot (x-2) + c \cdot (x^2-1) - (x-2)\cdot (x^2-1) = 0$
$f(x) := [(a+b)\cdot x+a-b] \cdot (x-2) + c \cdot (x^2-1) - (x-2)\cdot (x^2-1)$
$f(-1) = [-a+b+a-b]\cdot (-3)+c \cdot 0-(-3) \cdot 0 = 0$
$f(1) = [a-b+a-b] \cdot (-1)+0-(-1)\cdot 0 = 2b-2a \lt 0$, for $a \gt b$
$f(2) = [(a-b)\cdot 2 +a-b] \cdot 0 +c \cdot 3 -0\cdot 3= 3c \gt0$
How do I go on from here?
Or is this even the right way?
|
Hint: $$\lim_{x \to -1+} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=\infty$$
$$\lim_{x \to 1-} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=-\infty$$
$$\lim_{x \to 1+} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=\infty$$
and
$$\lim_{x \to 2-} \frac{(a+b)\cdot x+a-b}{x^2-1}+\frac{c}{x-2}=-\infty$$.
By continuity, there are points in $(-1,1)$ and $(1,2)$ where any real value is attained, in particular the value $1$.
|
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|
A number theory question that is probably wrong Prove that $a^3-b^3 = 2011$ has no integer solutions.
I think the question is wrong as
$a^3-b^3 = 2011$
$(a-b)(a^2+ab+b^2) = 2011$
As $2011$ is prime so the only factors $2011$ has are $1$ and $2011$ itself.
So if $a-b = 1$ and $a^2+ab+b^2 = 2011$ , then we can say that
$a^2-2ab+b^2 = 1$
So , $a^2+b^2 = 2ab+1$
Putting the value in the second equation
$3ab +1 = 2011$
$3ab = 2010$
And as $2010$ is divisible by $3$ it has integer solutions , which is contradictory to the original question. Can anybody help me with this?
|
Well, you seem to be doing a very good job up until when you realize that
one must have $(a-b)(a^2+ab+b^2)= 2011$.
You then come to the key observation $a-b$ must be a divisor of $2011$.
At this point I would consider the problem "almost solved", it seems like now we must only do some case checking and everything will be fine.
Case $1$: $a-b = 1$. We get $b=a-1$ and so we have $a^2 + (a-1)a + a^2 = 2011$.
Case $2$: $a-b= 2011$. We get $b= a-2011$ and so we must have $(2011)(a^2 + (a-2011)a + (a-2011)^2) = 2011$.
Case $3$: $a-b= -1$. We get $b= a+1$ and so we must have $-(a^2 + (a+1)a + (a+1)^2) = 2011$.
Case $4$: $a-b= -2011$. We get $b= a+2011$ and so we must have $-2011(a^2 + (a+2011)a + (a+2011)^2) = 2011$.
Now, do these cases have solutions? These are the sort of equations that can be "solved" with the quadratic formula ! So you can get the answers needed.
|
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|
Is $\tan^{-1}(0)=\pi$ or is $\tan^{-1}(0)=0$? This question came in the Dhaka University admission exam 2006-7
Q) The value of $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$ is -
(a) $0$
(b) $\frac{\pi}{2}$
(c) $\pi$
(d) $2\pi$
My attempt:
$$\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$$
$$=\tan^{-1}\frac{1+2+3-1\cdot2\cdot3}{1-1\cdot2-2\cdot3-1\cdot3}$$
$$=\tan^{-1}\frac{0}{-10}$$
$$=\tan^{-1}0$$
Now, as no range is mentioned, I should pick the angle that is within the principal range of $\tan^{-1}$: (a). (c) and (d) are also acceptable solutions, but they don't fall within the principal range. So, I'll go with (a).
However, my question bank says that $\tan^{-1}0=\pi$, so the answer is (b). I don't understand their reasoning.
Which is the correct option?
|
There's a bit of a discrepancy between the page title and your actual question.
Conventionally, $\tan^{-1}(0)$ is defined as 0.
However, the individual terms $\tan^{-1} 1$, $\tan^{-1} 2$, and $\tan^{-1} 3$ are all positive numbers, so their sum can't be zero.
*
*$\tan^{-1} 1 = \frac{\pi}{4}$
*$\tan^{-1} 2$ and $\tan^{-1} 3$ are hard to express exactly, but each must be between $\tan \sqrt{3} = \frac{\pi}{3}$ and $\tan^{-1} \infty = \frac{\pi}{2}$
So, you can determine that $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3$ must have a value somewhere in the interval $[\frac{\pi}{4} + \frac{\pi}{3} + \frac{\pi}{3}, \frac{\pi}{4} + \frac{\pi}{2} + \frac{\pi}{2}] = [\frac{11}{12} \pi, \frac{5}{4}\pi]$. And it happens that $\pi$ is the only answer choice that falls within that range.
|
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|
Evaluation of $\int\sqrt{1+2\sin(\theta)^2\cos(\theta)^2}\mathrm{d}\theta~~\text{where}~~\theta\in\left[0,{\pi\over4}\right]$ $$\begin{align}
I&:=\int \sqrt{1+2\sin(\theta)^2\cos(\theta)^2} \mathrm{d} \theta ~~~\text{where}~~~~\theta\in\left[0,{\pi\over4}\right]\\
\cos(\theta)^2&=1-\sin(\theta)^2\\
I&=\int\sqrt{1+2\sin(\theta)^2 \left(1-\sin(\theta)^2 \right) } \mathrm{d} \theta\\
&= \underbrace{\int\sqrt{1+2\sin(\theta)^2-2\sin(\theta)^4} \mathrm{d} \theta}_{\text{Seems nothing can be done from this form} } \\\sin(2\theta)&=2\sin(\theta)\cos(\theta)\\
&\iff {\sin(2\theta) \over2 } = \sin(\theta)\cos(\theta)
\\ {\sin(2\theta)^2 \over 4 }&= \sin(\theta)^2\cos(\theta)^2
\\I&=\int \sqrt{1+2 \left({\sin(2\theta)^2 \over 4 } \right) } \mathrm{d} \theta\\&=\int \sqrt{1+{\sin(2\theta)^2 \over 2 }} \mathrm{d} \theta\\&=\int \sqrt{ {2+\sin(2\theta)^2 \over 2 }} \mathrm{d} \theta\\&= {1 \over \sqrt{2}}\underbrace{\int\sqrt{2+\sin(2\theta)^2} \mathrm{d} \theta}_{\text{It is confusing for me} }
\end{align}$$
How can I integrate the originally given integral?
|
$$I=\int_0^{\frac \pi 4}\sqrt{1+\frac{1}{2} \sin ^2(2 \theta )}\,d\theta=\frac{1}{2}\int_0^{\frac \pi 2} \sqrt{1+\frac{1}{2}\sin ^2(x)}\,dx=\frac{1}{2}E\left(-\frac{1}{2}\right)$$ where appears the complete elliptic integral of the second kind.
If it had been
$$J=\int_0^{t}\sqrt{1+\frac{1}{2} \sin ^2(2 \theta )}\,d\theta=\frac{1}{2}\int_0^{2t} \sqrt{1+\frac{1}{2}\sin ^2(x)}\,dx=\frac{1}{2} E\left(2 t\left|-\frac{1}{2}\right.\right)$$where appears the incomplete elliptic integral of the second kind.
Have a look here for definitions and properties.
In the worst case, write
$$\sqrt{1+\frac{1}{2}\sin ^2(x)}=\frac 1 {\sqrt \pi}\sum_{n=0}^\infty (-1)^{n+1}\frac{ \Gamma \left(n-\frac{1}{2}\right) }{ 2^{n+1}\,\Gamma(n+1)}\sin ^{2 n}(x)$$ and using
$$\int_0^{\frac \pi 2}\sin ^{2 n}(x)\,dx=\frac{\sqrt{\pi } \Gamma \left(n+\frac{1}{2}\right)}{2 \Gamma (n+1)}$$
$$\int_0^{\frac \pi 2} \sqrt{1+\frac{1}{2}\sin ^2(x)}\,dx=\sum_{n=0}^\infty (-1)^{n+1} \frac{ \Gamma \left(n-\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{2^{n+2}\,\Gamma(n+1)^2}$$
If
$$a_n=\frac{ \Gamma \left(n-\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{2^{n+2}\,\Gamma(n+1)^2}\implies \frac{a_{n+1}}{a_n}=\frac{4 n^2-1}{8 (n+1)^2}= \frac{1}{2}-\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ and since it is alternating it will converge quite fast.
Summing from $n=0$ to $n=5$ would give $1.75185$ while $E\left(-\frac{1}{2}\right)=1.75177$.
If this has to be computed, write
$$\sum_{n=0}^\infty (-1)^{n+1} \frac{ \Gamma \left(n-\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{2^{n+2}\,\Gamma(n+1)^2}=$$ $$\sum_{n=0}^p (-1)^{n+1} \frac{ \Gamma \left(n-\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{2^{n+2}\,\Gamma(n+1)^2}+\sum_{n=p+1}^\infty (-1)^{n+1} \frac{ \Gamma \left(n-\frac{1}{2}\right) \Gamma \left(n+\frac{1}{2}\right)}{2^{n+2}\,\Gamma(n+1)^2}$$ and tou want to know $p$ such that
$$R_p=\frac{ \Gamma \left(p+\frac{1}{2}\right) \Gamma \left(p+\frac{3}{2}\right)}{2^{p+3}\,\Gamma (p+2)^2}\leq \epsilon$$ Taking logarithms and using Stirling approximation, this is almost
$$2^{p+3} p^2 \geq \frac 1 \epsilon \implies p \geq \frac{2}{\log (2)}\,W\left(\frac{\log (2)}{4 \sqrt{2\epsilon } }\right)$$ where $W(.)$ is Lambert function.
Using $\epsilon=10^{-16}$, $p \geq 39.5403$ while the exact solution is $p=39.4815$. So, coding, no more comparison tests.
|
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|
Evaluate $\int \frac {x+3}{\sqrt{5-4x-2x^2}} dx$ Question:
The question is about evaluating the following indefinite integral:-
$$\int \frac {x+3}{\sqrt{5-4x-2x^2}}dx$$
My Work:
We know the numerator can be expressed in the form of :
$$ x+3 = A \frac {d}{dx}(5-4x-2x^2) + B$$
So we get ,
$$ x+3 = A(-4-4x) +B $$
Equating $x$ and $3$ we get,
$$ A = \frac {1}{4} , B={4} $$
So integrating,
$$ \int \frac {1}{4} ({{-4-4x}})\frac{1}{\sqrt{5-4x-2x^2}}dx + \int \frac {4}{\sqrt{5-4x-2x^2}}dx$$
Taking the constants out,
Assume $5-4x-2x^2 = t$
Differentiating it we get,
$$(-4-4x)dx= dt$$
By completing the square method we get,
$$5-4x-2x^2 = (2)(\frac {5}{2}-2x-x^2)$$
So doing the calculation we get ,
$$(2)[{(x+1)^2-(\sqrt \frac{7}{2})^2}$$
Now again integrating all the values obtained,
$$\frac {1}{4} \int \frac {dt}{\sqrt {t}} - \frac {4}{\sqrt 2} \int \frac {dx}{(x+1)^2-(\sqrt \frac{7}{2})^2}$$
Taking the minus sign out,
$$\frac{1}{2}\sqrt{t}dt+\frac{4}{\sqrt{2}}\int_{ }^{ }\frac{dx}{\left(\sqrt{\frac{7}{2}}\right)^{2}-(x+1)^{2}}$$
$$\frac{1}{2}\sqrt{5-4x-2x^{2}}dt+\frac{4}{\sqrt{2}}\sin^{-1}\left(\frac{x+1}{\left(\sqrt{\frac{7}{2}}\right)^{2}}\right)$$
I cannot integrate it further any help or advice will be very much appreciated
|
Another method using Completing the square and Integration by substitution.
We have,
$$\begin{align}I &= \int \dfrac{x + 3}{\sqrt{5 - 4x - 2x^2}} dx\\& = \int \frac{x+3}{\sqrt{5 - 2(x^2 +2x)}}dx\\& = \int \frac{x+3}{\sqrt{5 - 2[(x + 1)^2 - 1]}}dx\\& = \int \frac{x+3}{\sqrt{7 - 2(x + 1)^2}}dx\end{align}$$
Let $(x+1) = \sqrt{\frac{7}{2}} \sin\theta \implies dx = \sqrt{\frac72}\cos\theta\ d\theta.$
So we have,
$$\begin{align}
I& = \int \frac{\left(\sqrt{\frac{7}{2}} \sin \theta + 2\right)}{\sqrt{7 - 2\cdot \frac72\sin^2\theta}}\cdot \sqrt{\frac72}\cos\theta\ d\theta
\\& = \frac1{\sqrt2}\int \frac{\left(\sqrt{\frac{7}{2}} \sin \theta + 2\right)}{\sqrt{7 - 7\sin^2\theta}}\cdot \sqrt{7}\cos\theta\ d\theta
\\& = \frac1{\sqrt2}\int \frac{\left(\sqrt{\frac{7}{2}} \sin \theta + 2\right)}{\sqrt{7}\cos\theta}\cdot \sqrt{7}\cos\theta\ d\theta
\\& = \frac1{\sqrt2}\int \sqrt{\frac{7}{2}} \sin \theta + 2\ d\theta
\\& = -\frac{\sqrt7}{2} \cos \theta +\sqrt{2}\theta + C
\end{align}$$
Consider our substitution,
$$ \sqrt{\frac{2}{7}} (x+1) =\sin\theta \implies \cos\theta = \sqrt{1 -\sin^2\theta} = \sqrt{1 - \frac27 (x+1)^2}.$$
Thus, our integral becomes,
$$ I = -\frac{\sqrt7}{2} \cos \theta +\sqrt{2}\theta + C$$
$$ -\frac{\sqrt7}{2} \cdot\sqrt{1 - \frac27 (x+1)^2} +\sqrt{2}\cdot \sin^{-1}\left(\sqrt{\frac{2}{7}} (x+1)\right)+ C$$
$$ \color{blue}{\boxed{-\frac{1}{2}\sqrt{-2x^{2}-4x+5}+\sqrt{2}\cdot \sin^{-1}\left(\sqrt{\frac{2}{7}} (x+1)\right)+ C}}$$
This is the final answer.
|
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|
If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
If $A=\begin{bmatrix}2&1&1\\0&1&0\\1&1&2\end{bmatrix}$, find $A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$
Attempt $1:$ $A=I+B$, where $B=\begin{bmatrix}1&1&1\\0&0&0\\1&1&1\end{bmatrix}$
$B^n$ comes out to be $2^{n-1}B$
After that I tried doing $A^8=(I+B)^8$ and then opening RHS with binomial. But it didn't help.
Attempt $2:$ Finding characteristic equation and putting $A$ in it. It didn't help either.
The equation I got here is $A^3-5A^2+7A-3I=0$
Note: Concept of minimal polynomial is not in syllabus. Also, if we can do this without characteristic polynomial, that would be great.
|
Hint:
$$A^5(A^3-5A^2+7A-3I)+A(A^3-5A^2+7A-3I)+A^2+A+I=\cdots$$
|
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|
Volume of $y=x^2+1; y=-x^2+2x+5; x=0; x=3$ about $x$ axis (Shell Method). I was working on this exercise for an assignment. However, I get stuck in the following part.
$
y=-x^2+2x+5
$
Complete the square
$y=-(x^2-2x)+5$
$(b/2)^2=(-2/2)^2=1$
$y=-(x^2-2x+1-1)+5$
$y=-(x^2-2x+1)+5+1$
$y=-(x-1)^2+6$
$y-6=-(x-1)^2$
$\sqrt{6-y}+1=x$
For $x=0$, $\sqrt{6-y}+1=x$ has no real answer and I was wondering, is it possible to solve this with the Shell Method about the $x$-axis? I tried with Washer Method, and the expected volume is $277π/3$.
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Note that in $[0,3]$, $-x^2+2x+5\geq x^2+1$ for $x\in [0,2]$. Therefore, by the Washer Method, the volume is given by
$$V=\pi\int_0^2((-x^2+2x+5)^2-(x^2+1)^2)\,dx+\pi\int_2^3((x^2+1)^2-(-x^2+2x+5)^2)\,dx=\frac{277\pi}{3}$$
Using the Shell Method is a bit more complicated because we have to split the evaluation into 4 pieces:
$$\begin{align}
V=&2\pi\int_1^5\sqrt{y-1}y\,dy+2\pi\int_2^5(3-(1+\sqrt{6-y}))y\,dy
\\&+2\pi\int_5^{10}(3-\sqrt{y-1})y\,dy+
2\pi\int_5^6((1+\sqrt{6-y})-(1-\sqrt{6-y}))y\,dy=\frac{277\pi}{3}
\end{align}$$
|
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|
Why are the two calculations of $ \int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x $ give two distinct answers? One of the calculations:
$$
\begin{aligned}
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x &=\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{1}{\sqrt{x}} \mathrm{~d} x \\
&=2 \int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \mathrm{~d}(\sqrt{x})=2 \arcsin \sqrt{x}+C .
\end{aligned}
$$
The other:
$$
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x = \int \frac{d\left(x-\frac{1}{2}\right)}{\sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}} = \arcsin \frac{x-1 / 2}{1 / 2}+C=\arcsin(2x-1)+C
$$
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Because the integral is indefinite.
If you differentiate the first form,
$${d2\arcsin{(\sqrt{x})} \over dx}=2{1 \over \sqrt{1-(\sqrt{x})^2}}\cdot {1 \over 2\sqrt{x}}={1 \over \sqrt{x(1-x)}}$$
As for the second one,
$${d\arcsin{(2x-1)} \over dx}={1 \over \sqrt{1-(2x-1)^2}}\cdot 2={1 \over \sqrt{x(1-x)}}$$
Therefore, the both forms are correct.
|
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|
Find the integer solution: $a+b+c=3d$, $\: a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$ Find the integer solutions:
$$a+b+c=3d$$
$$a^{2} + b^{2} + c^{2}= 4d^{2}-2d+1$$
Attempt:
Notice that $a=b=c=d=1$ is a solution.
Other facts:
Notice that $a^{2} + b^{2} + c^{2} > 0$, so $4d^{2}-2d+1>0$. Notice that $4d^{2}-2d+1$ has negative discriminant: $D = -12$, and so it is always one sign, which is positive in this case. So we cannot narrow $d$-solution this way.
Next, since $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab+ac+bc)$, we get
$$ 9d^{2} - 2(ab+ac+bc) = 4d^{2}-2d+1 $$
$$ 5d^{2} + 2d -1 = 2(ab+ac+bc) $$
$5d^{2} + 2d - 1 = 0$ has solutions:
$$ d_{1,2} = \frac{-2 \pm \sqrt{24}}{10} = \frac{-1 \pm \sqrt{6}}{5}$$
and when $d > (-1 + \sqrt{6})/5$, OR, $d < (-1 - \sqrt{6})/5$ we have
$5d^{2}+2d-1 > 0$. And when $d$ is between the 2 roots we have $5d^{2}+2d-1<0$.
Then also nocitce that $(a+b+c)d= 3d^{2}$, and so
$$a^{2}+b^{2}+c^{2}-(a+b+c)d = (d-1)^{2}$$
Now if $d \ne 1$ we must have
$$ a^{2}+b^{2}+c^{2}-(a+b+c)d > 0$$
Another fact:
$$d = \frac{a+b+c}{3} \ge (abc)^{1/3} $$
by AM-GM.
Also notice $d < \max(a,b,c)$, because
$$a^{2}+b^{2}+c^{2} =d^{2} + d^{2} + d^{2} + (d-1)^{2}$$
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This isn't a full answer, but noting that $$2a+2b+2c=6d$$ first subtract this from the second equation to obtain $$(a-1)^2+(b-1)^2+(c-1)^2=4(d-1)^2$$
Then set $A=a-1, B=b-1, C=c-1, D=d-1$ to get the system$$A^2+B^2+C^2=4D^2$$
and $$A+B+C=3D$$
So for given $D$ the solutions lie on a sphere of radius $2D$ centred at the origin.
@Will Jagy has completed this to a full solution in the comments below - noting that if we don't have $A=B=C=D=0$ then modulo $4$ considerations give a contradiction.
One motivation for looking for something like this is the existence of the solution $a=b=c=d=1$: how would one encode that in an equation or (in different circumstances) produce a factor which could be divided out to simplify the search for other solutions.
Will should get credit for the solution here.
In response to request to complete this rather than leaving the finish in comments:
Now suppose we have a non-zero solution. If $A,B,C,D$ are all even we can divide through by $2$ to get another solution (because the two equations we now have are homogeneous). So if there is a non-zero solution, there is a non-zero solution where at least one of $A,B,C,D$ is odd.
However the sum of three squares can only be divisible by $4$ if it is the sum of three even squares (any odd square is $\equiv 1 \bmod 4$ - indeed $\equiv 1 \bmod 8$), so $A,B,C$ must all be even to satisfy the first equation. Then the second equation tells us that $D$ is even. And this is a contradiction to the fact that we can find a solution with at least one odd number. Hence there is no non-zero solution.
|
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|
To show $\int_{0}^{\infty} \frac{x}{1+x^6\sin^2x}\,dx$ is convergent
Show that $\displaystyle \int\limits_{0}^{\infty} \frac{x}{1+x^6\sin^2x}\,dx$ is convergent
Since we need to check if the given improper integral is convergent, I considered Abels/Dirichlet test for convergence, but for that, although I got monotone bounded or monotone functions converging to zero, the integral of the other function was not bounded.
Certainly though we can break the domain into intervals, $[n\pi, (n+1)\pi]$, thereby bounding the integral over each interval, and the integrand being continuous over each interval is thereby integrable. Then performing summation of the integrals over each interval we check whether its bounded and if so the convergence follows.
BUT, I was in search for a function of form $\frac{1}{x^p}$ with $p>1$ such that it is always larger than the given integrand over the domain $[a,\infty]$, $a>0$, but with $\sin^2x$ in the denominator, I couldn't. I also tried to think of some function larger than the integrand over the domain $[a, \infty]$, upon which I could apply the Abels/Dirichilet tests, but that didn't worked also.
So any solutions to the given problem considering comparison tests or any other approach that would make the solution brief would be highly appreciated.
|
You can proceed as in this answer:
Note that $\left| \sin x \right| \geq \frac{2}{\pi}|x|$ for $|x| \leq \frac{\pi}{2} $. Using this, we get:
$$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+a \sin^2 x}
\leq \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\mathrm{d}x}{1+a\bigl(\frac{2}{\pi}x\bigr)^2}
\leq \int_{-\infty}^{\infty} \frac{\mathrm{d}x}{1+a\bigl(\frac{2}{\pi}x\bigr)^2}
= \frac{C}{\sqrt{a}} $$
for some constant $C$. From this, it follows that
\begin{align*}
&\int_{0}^{\infty} \frac{x}{1+x^6\sin^2 x} \, \mathrm{d}x \\
&= \int_{0}^{\frac{\pi}{2}} \frac{x}{1+x^6\sin^2 x} \, \mathrm{d}x + \sum_{n=1}^{\infty} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x+n\pi}{1+(x+n\pi)^6\sin^2 x} \, \mathrm{d}x \\
&\leq \int_{0}^{\frac{\pi}{2}} \frac{x}{1+x^6\sin^2 x} \, \mathrm{d}x + \sum_{n=1}^{\infty} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(n+\frac{1}{2})\pi}{1+(n-\frac{1}{2})^6\pi^6\sin^2 x} \, \mathrm{d}x \\
&\leq \int_{0}^{\frac{\pi}{2}} \frac{x}{1+x^6\sin^2 x} \, \mathrm{d}x + \sum_{n=1}^{\infty} \frac{C(n+\frac{1}{2})\pi}{(n-\frac{1}{2})^3\pi^3} \\
&< \infty.
\end{align*}
|
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|
Evaluating integral $\int_{-\infty}^\infty \frac{x}{\sqrt{x^2 + c^2}} e^{-(x-m)^2} dx$ and one more similar I'm trying to evaluate two integrals:
$$f(m,c) = \int_{-\infty}^\infty \frac{x}{\sqrt{x^2 + c^2}} e^{-(x-m)^2} dx$$
and
$$g(m,c) = \int_{-\infty}^\infty \frac{1}{\sqrt{x^2 + c^2}} e^{-(x-m)^2} dx$$
where $c,m\in\mathbb{R}$
For $f(m,c)$ I was trying to substitute $x$ with $t = \sqrt{x^2 + c^2}$, which eliminates the fraction, but complicates the expression under the exponent. Also, I'm not sure what the new limits of integration should be:
$$f(m,c) = \int_{?}^{?} e^{-(\sqrt{t^2-c^2}-m)^2} dt$$
Maybe this integral is related, but I couldn't bring $f(m,c)$ to it.
For $g(m,c)$ I've tried to substitute $x$ with $x=c\sinh{t}$, which leads to integral:
$$g(m,c) = \int_{?}^{?} e^{-(c\sinh{t}-m)^2} dt$$
with which I don't know what to do next.
Maybe this integral is related to $g(m,c)$.
|
If $m$ is "small" you could expand the exponential term as
$$e^{-(x-m)^2}=e^{-x^2}\,\sum_{n=0}^\infty P_n(x)\, m^n$$ where the first polynomials are
$$\left(
\begin{array}{cc}
n & P_n(x) \\
0 & 1 \\
1 & 2 x \\
2 & 2 x^2-1 \\
3 & \frac{4 x^3}{3}-2 x \\
4 & \frac{2 x^4}{3}-2 x^2+\frac{1}{2} \\
5 & \frac{4 x^5}{15}-\frac{4 x^3}{3}+x \\
6 & \frac{4 x^6}{45}-\frac{2 x^4}{3}+x^2-\frac{1}{6} \\
7 & \frac{8 x^7}{315}-\frac{4 x^5}{15}+\frac{2 x^3}{3}-\frac{x}{3} \\
8 & \frac{2 x^8}{315}-\frac{4 x^6}{45}+\frac{x^4}{3}-\frac{x^2}{3}+\frac{1}{24} \\
9 & \frac{4 x^9}{2835}-\frac{8 x^7}{315}+\frac{2 x^5}{15}-\frac{2
x^3}{9}+\frac{x}{12} \\
10 & \frac{4 x^{10}}{14175}-\frac{2 x^8}{315}+\frac{2
x^6}{45}-\frac{x^4}{9}+\frac{x^2}{12}-\frac{1}{120} \\
11 & \frac{8 x^{11}}{155925}-\frac{4 x^9}{2835}+\frac{4 x^7}{315}-\frac{2
x^5}{45}+\frac{x^3}{18}-\frac{x}{60} \\
12 & \frac{4 x^{12}}{467775}-\frac{4 x^{10}}{14175}+\frac{x^8}{315}-\frac{2
x^6}{135}+\frac{x^4}{36}-\frac{x^2}{60}+\frac{1}{720}
\end{array}
\right)$$
and use
$$I_{2n}= \int_{-\infty}^{+\infty} \frac{x^{2n}}{\sqrt{x^2 + c^2}} e^{-x^2}\, dx$$ and face linear combinations of Bessel functions
|
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|
Is it possible to generalize this equation more? This is one of my discoveries, and my question is: can it be more general?
Let $x, y, z$ be three arbitrary complex numbers and set
$x_1=3 x^2+y^2$,
$x_2=1-z$,
$x_3=1+z$,
$x_4=3 x^2+y^2+4 x y z$,
$x_5=3 x^2+y^2-4 x y z$,
$x_6=3 x^2+y^2+z (3 x^2+2 x y-y^2)$,
$x_7=3 x^2+y^2-z (3 x^2+2 x y-y^2)$,
$x_8=3 x^2+y^2+z (3 x^2-2 x y-y^2)$,
$x_9=3 x^2+y^2-z (3 x^2-2 x y-y^2)$.
Then for $k=0,1,2,3,4,5$,
$2(x_1^k+x_1^kx_2^k+x_1^kx_3^k)=x_4^k+x_5^k+x_6^k+x_7^k+x_8^k+x_9^k$.
In case $x, y, z$ are integers, we get a Diophantine equation.
For example,
$2(7^1+21^1+(-7)^1)=(-9)^1+23^1+1^1+13^1+17^1+(-3)^1$,
$2(7^2+21^2+(-7)^2)=(-9)^2+23^2+1^2+13^2+17^2+(-3)^2$,
$2(7^3+21^3+(-7)^3)=(-9)^3+23^3+1^3+13^3+17^3+(-3)^3$,
$2(7^4+21^4+(-7)^4)=(-9)^4+23^4+1^4+13^4+17^4+(-3)^4$,
$2(7^5+21^5+(-7)^5)=(-9)^5+23^5+1^5+13^5+17^5+(-3)^5$.
My discovery published at: https://demonstrations.wolfram.com/author.html?author=Minh%20Trinh%20Xuan
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This is very nice. It relies to the Tarry-Escott problem, on which there is a vast literature going back more than 100 years. I do not know how much you know of this. I suggest you consult Dickson, the History of the Theory of Numbers.
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|
Finding a closed form of $f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$. I'm trying to find the closed form of
$f(n) = \frac{n-4}{n}f(n-1)+1\ ,\ n \geq 5$ where $f(4) = 1$.
Empirically, it turns out the answer is simply $f(n)=1+\frac{n-4}{5}\ ,\ n \geq 4$, but I'm having a hard time getting there. I've tried two ways but neither is very successful. Could someone please suggest a way through?
Attempt 1: I try to set up a generating function
$$G(x)=f(4)x^4+f(5)x^5+\dots=\sum_{k=4}^\infty f(k)x^k$$
Substituting in the recurrence (edit: noticed I forgot to add the +1)
$$\begin{align}G(x)&=x^4+\sum_{k=5}^\infty(1-\frac{4}{k})f(k-1)x^k \\
&=x^4+x\sum_{k=5}^\infty (1-\frac{4}{k}) f(k-1)x^{k-1} \\
&= x^4+x\sum_{k=4}^\infty (1-\frac{4}{k+1})f(k)x^k \\
&= x^4+xG(x)-x\sum_{k=4}^\infty \frac{4}{k+1}f(k)x^k
\end{align} $$
But I don't know how to express the last part in terms of G(x).
Attempt 2: Just brute force it:
With arbitrary $k$,
$$\begin{align}f(n) &= \frac{(n-4)(n-5) \dots (n-(k+3))}{n(n-1)\dots(n-(k-1))}f(n-k)\\
&\ \ + \left(1+\frac{n-4}{n}+\frac{(n-4)(n-5)}{n(n-1)}+\dots+\frac{(n-4)\dots(n-(k+2))}{n\dots(n-(k-2))}\right)\end{align}$$
Setting $n-k=4 \implies k = n -4$ and using $f(4)=1$,
$$\begin{align} f(n) &= \frac{(n-4)!}{n!/4!}+\left(1+\frac{n-4}{n}+\frac{(n-4)(n-5)}{n(n-1)}+\dots+\frac{(n-4)\dots2}{n \dots 6}\right) \\
&= {n \choose 4}^{-1}+\left(1+\frac{(n-4)!}{n!}\left(\frac{(n-1)!}{(n-5)!} + \frac{(n-2)!}{(n-6)!}+\dots +\frac{5!}{1!}\right)\right)
\\
&= {n \choose 4}^{-1}+\frac{(n-4)!}{n!}\sum_{k=1}^{n-5}\frac{(k+4)!}{k!} \\
&= {n \choose 4}^{-1}+\frac{1}{n(n-1)(n-2)(n-3)}\sum_{k=1}^{n-5}(k+4)(k+3)(k+2)(k+1) \\
&= {n \choose 4}^{-1}+\frac{1}{n(n-1)(n-2)(n-3)}\sum_{k=5}^{n-1}k(k-1)(k-2)(k-3)\end{align}$$
Empirically this seems to be correct, but even Mathematica refuses to simplify it all the way down. Surely there is a better way?
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From the recursive definition, we have the identity
$$\sum_{n=5}^\infty f(n) x^n = \sum_{n=5}^\infty \frac{n-4}n f(n-1) x^n + \sum_{n=5}^\infty x^n$$
With $G(x) = \sum\limits_{n=4}^\infty f(n)x^n$, for $|x|<1$ we have
$$\begin{align*}
G(x) - x^4 &= \sum_{n=5}^\infty f(n-1) x^n - 4 \sum_{n=5}^\infty \frac{f(n-1)}n x^n + \left(\frac1{1-x} - 1 - x - x^2 - x^3 - x^4\right) \\[1ex]
G(x) &= x \sum_{n=5}^\infty f(n-1) x^{n-1} - 4 \sum_{n=5}^\infty \frac{f(n-1)}n x^n + \frac{x^4}{1-x} \\[1ex]
G(x) &= x G(x) - 4 \sum_{n=5}^\infty \frac{f(n-1)}n x^n + \frac{x^4}{1-x} \\[1ex]
(1-x) G(x) &= -4 \phi(x) + \frac{x^4}{1-x}
\end{align*}$$
where $\phi'(x) = G(x)$. Differentiating both sides and isolating $G'(x)$ yields the linear differential equation
$$G'(x) + \frac3{1-x}G(x) = \frac{4x^3-3x^4}{(1-x)^3}$$
which is solved below using the integrating factor method.
$$\begin{align*}
\frac1{(1-x)^3} G'(x) + \frac3{(1-x)^4}G(x) &= \frac{4x^3-3x^4}{(1-x)^6} \\[1ex]
\left(\frac1{(1-x)^3} G(x)\right)' &= \frac{4x^3-3x^4}{(1-x)^6} \\[1ex]
G(x) &= (1-x)^3 \int_0^x \frac{4\xi^3-3\xi^4}{(1-\xi)^6} \, d\xi \\[1ex]
G(x) &= \frac{x^4 (5-4x)}{5 (1-x)^2}
\end{align*}$$
Next, get the power series expansion of $G(x)$ to determine $f(n)$. Polynomial division yields
$$G(x) = -\frac{4x^3}5 - \frac{3x^2}5 - \frac{2x}5 - \frac15 + \frac15 \frac1{(1-x)^2}$$
Use the series for the derivative of $\frac1{1-x}$ to wrap up.
$$\frac1{1-x} = \sum_{n=0}^\infty x^n \implies \frac1{(1-x)^2} = \sum_{n=0}^\infty nx^{n-1}$$
Hence
$$G(x) = -\frac{4x^3}5 - \frac{3x^2}5 - \frac{2x}5 - \frac15 + \frac15 \left(1 + 2x + 3x^2 + 4x^3 + \sum_{n=5}^\infty nx^{n-1}\right) \\ G(x) = \frac15 \sum_{n=4}^\infty (n+1)x^n = \sum_{n=4}^\infty f(n)x^n$$
$$\implies f(n) = \frac{n+1}5$$
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|
Find the maximum value of $M$ with given constraints.
Let $a, b \in \mathbb{R}$ such that $n^{4a-\log_5{n^2}} \leq 25^{40-b^2}$ for all positive real number $n$. Find the maximum value of $M=a^2+b^2+a-3b$.
What I have tried:
Taking the logarithm base 5 of both sides from the inequality, we get $(4a-2\log_5{n})\log_5n\leq2(40-b^2)$
Now, let $x=\log_5 n$, we get $x(4a-2x)\leq2(40-b^2)$, and this is equivalent to $x^2+2ax-b^2+40
\geq0$
The new inequality is true if and only if the discriminant $4a^2-4(40-b^2)\leq0$, this is equivalent to $a^2+b^2\leq40$, so we have $\max(a^2+b^2)=40$
This gave me hope of finding the maximum value of $M$ by finding the maximum value of $a-3b$ but I have no idea how to do it.
I'd like to call for some help. Thank you!
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I want to contribute my solution using geometry :D
As Cesareo told above, our problem now is to find the maximum value of $f(a,b)=a^2+b^2+a-3b$ knowing that $a^2+b^2 \leq 40$.
Geometrically, $\sqrt{a^2+b^2}$ is the distance between the point $M(a,b)$ and the origin $O(0,0)$, so the condition $a^2+b^2 \leq 40$ means that $M(a,b)$ lies inside the circle $(C)$ whose center and radius are respectively $O(0,0)$ and $\sqrt{40}$. By some calculations, we have
$$f(a,b)=\left(a+\dfrac{1}{2}\right)^2+\left(b-\dfrac{3}{2}\right)^2-\dfrac{5}{2}=MI^2-\dfrac{5}{2},$$
where $I\left(-\dfrac{1}{2},\dfrac{3}{2}\right)$. Hence, to find the maximum of $f(a,b)$ means to find the maximum of $MI$.
At this point, note that $I$ lies inside $(C)$ since $OI^2=\dfrac{5}{2}<40$. Therefore the position of $M$ maximizing $f(a,b)$ is one of two nodes of the diameter of $(C)$ which passes through $M$; that is,
$$\max f(a,b)=\max \left\{IA^2-\dfrac{5}{2},IB^2-\dfrac{5}{2}\right\},$$
where $AB$ is a diameter of $(C)$ containing $M$.
The following part is my calculations. The equation of the line $OI$ is $3x+y=0$. The coordinates of $A$ and $B$ satisfy the system
$$\begin{cases}
3x+y=0 \\
x^2+y^2=40
\end{cases} \Leftrightarrow \begin{cases}
y=-3x \\
x^2+(-3x)^2=40
\end{cases} \\ \begin{cases}
y=-3x \\
10x^2=40
\end{cases} \Leftrightarrow \begin{cases}
x=2 \\
y=-6
\end{cases} \vee \begin{cases}
x=-2 \\
y=6
\end{cases}$$
Thus $\max f(a,b)=\max \left\{IA^2-\dfrac{5}{2},IB^2-\dfrac{5}{2}\right\}=\max \left\{60,20\right\}=60.$
Hope this helps. ^^
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why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1) \cdot (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$
Why does $1/((x-1)(x^2 - 1)) = (x^4-1)/(x-1) \cdot (x^8 - 1)/(x^2 - 1) \cdot (x^{16} - 1)/(x^4 - 1) \cdots$?
I think the equality holds as formal power series. Expanding the LHS, one should get $(1+x+x^2 +\cdots )(1+x^2 + x^4+\cdots).$ But doing cancellation for the RHS gives that it equals $(x^2+1)/(x-1) \cdot (x^4 + 1)(x^8 + 1)\cdots,$ which clearly contradicts the equality. If however, one multiplies both sides by $x^2 - 1$, one gets an obvious equality after cancellation.
I was wondering if someone could explain what I'm doing wrong here (i.e. why I get two different "answers")?
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It is a telescoping product. Everything on the RHS cancels in pairs, except the $1-x$ and $1-x^2$, which appear in the denominator but not the numerator.
|
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|
Conditional Variance of PDF A random vector $(X,Y)$ has a continuous distribution with a density function $$f(x,y)=\begin{cases}c⋅x & \text{when }0 ≤ x ≤ 2, \max\{0,1−x\} ≤ y ≤2−x\\ 0& \text{otherwise}\end{cases}$$ where $c > 0$ is a constant. Find variance of a $Y$ conditioned on $X = 1.5$, $Var(Y |X = 1.5)$.
I found $c = \frac 67$ with the given integral. Now I want to ask how can I find variance?
I found variance as -87/3136. Is it possible ?
Here is my attempt
Thank you
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You have the region being the area trapped between the triangles $B$ and $A$ . Let it be denoted by $S$ .
You have $$\iint_{S}f(x,y)\,dxdy=1$$
$$\iint_{S}f(x,y)\,dxdy=\iint_{B}f(x,y)\,dxdy-\iint_{A}f(x,y)\,dxdy=\frac{8c}{6}-\frac{c}{6}=\frac{7c}{6}$$.
Hence $c=\frac{6}{7}$ as you correctly calculated.
Now the marginal density for $X$ is found by integrating the joint pdf over $y$ .
Hence $f_{X}(x)=\begin{cases}\int_{1-x}^{2-x}cx\,dy\,,0\leq x< 1\\\int_{0}^{2-x}cx\,dy\,,1\leq x\leq 2\end{cases}$
Hence $f_{X}(x)=\begin{cases}cx\,,0\leq x<1\\cx(2-x)\,,1\leq x\leq 2\end{cases}$
Now $$f_{Y|X=x}(y)=\frac{f(x,y)}{f_{X}(x)}$$
Hence $f_{Y|X=x}(y)=\begin{cases}\frac{cx}{cx}\,,0\leq x <1,\,,\max\{0,1-x\}\leq y\leq 2-x\\ \frac{cx}{cx(2-x)}\,,1\leq x<2 ,\,\max\{0,1-x\}\leq y\leq 2-x\end{cases}$
That is : $$f_{Y|X}=\begin{cases}1\,,0\leq x<1 \,,\max\{0,1-x\}\leq y\leq 2-x \\ \frac{1}{2-x}\,,1\leq x \leq 2 ,\max\{0,1-x\}\leq y\leq 2-x\end{cases}$$
Hence as $x=\frac{3}{2}$ we have
$f_{Y|X=x}(y)=2\,,0\leq y\leq 2-\frac{3}{2}$
That is $$f_{Y|X=\frac{3}{2}}(y)=2\cdot\mathbf{1}_{\{0\leq y\leq \frac{1}{2}\}}$$ .
Then $$\Bbb{E}(Y|X=\frac{3}{2})=\int_{0}^{\frac{1}{2}}2y\,dy=\frac{1}{4}$$
And $$\Bbb{E}(Y^{2}|X=\frac{3}{2})=\int_{0}^{\frac{1}{2}}2y^{2}\,dy=\frac{1}{12}$$
Hence $$\text{Var}(Y|X=\frac{3}{2})=\frac{1}{12}-\frac{1}{4^{2}}=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}$$ .
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For a triangle, prove that $\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\frac{1}{r^2} $
Context: I made this question and am contemplating about submitting it as a contest question. The contest is not a large one; its scope does not comprise even our whole class. Its just a friendly, small contest within a small group. This originally occurred to me when we were asked to prove $\tan \frac A2\tan\frac B2 +\tan \frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1$ in class. I tried a geometrical proof but failed, but I got the following for my efforts.
Derivation: Given $\triangle ABC$ and its incircle $\Im$ with inradius $r$, we have the known (and easily provable) formula $$\tan \frac A2\tan\frac B2 +\tan \frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=1.\tag{1}\label{1}$$ We know, $IE=IF=IG=r$. Now, $$\tan \frac A2=\frac{r}{AE}=\frac{r}{AF}$$$$\tan \frac B2=\frac{r}{BE}=\frac{r}{BG}$$$$\tan \frac C2=\frac{r}{CG}=\frac{r}{CF}$$ Putting appropriate values in the formula $(1)$, we get $$\frac{r^2}{AE\cdot BE}+ \frac{r^2}{CG\cdot BG}+ \frac{r^2}{AF\cdot CF}=1$$
or $$\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\frac{1}{r^2}$$ Finally, using the identities $\Delta=rs, s=\dfrac{a+b+c}{2}, \Delta=\dfrac{abc}{4R}$ (where R is the circumradius) and the sine law we get $$\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}=\dfrac{s^2}{\Delta^2}=\dfrac{(a+b+c)^2}{4\Delta^2}=\dfrac{(a+b+c)^2\cdot 16R^2}{4a^2b^2c^2} $$ Thus, $$\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=2R\frac{a+b+c}{abc}$$ $$=2R\frac{2R(\sin A+\sin B+\sin C)}{8R^3\sin A\sin B\sin C}$$$$=\frac{1}{2R}\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$ so that $$2R\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$
Question: Prove that for a triangle ABC, $$2R\sqrt{\frac{1}{AE\cdot BE}+ \frac{1}{CG\cdot BG}+ \frac{1}{AF\cdot CF}}=\left(\frac{1}{\sin B\sin C}+ \frac{1}{\sin A\sin C}+ \frac{1}{\sin B\sin A}\right)$$
The result looks pretty symmetrical and nice, and also reminds one of Ceva’s theorem because of the segments taken consecutively.
If you were given this question and did not know the above derivation, how would you approach it? Basically, I am looking for alternative solutions to this problem just for reference.
If the question has any flaws, please tell me. Also, I would really like to know how you would rate this question for a high school level informal contest. If any of you can come up with variants of this, especially as inequalities, I would be grateful. Thanks in advance.
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Note that $AE=s-a, BE=s-b, BG=s-b, GC=s-c, CF=s-c, FA=s-a$. The expression under the square root becomes
$$\sqrt{\frac{1}{(s-a)(s-b)}+\frac{1}{(s-b)(s-c)}+\frac{1}{(s-a)(s-c)}}$$
By Heron's formula, $K=\sqrt{s(s-a)(s-b)(s-c)}\implies \frac{1}{(s-b)(s-c)}=\frac{s(s-a)}{K^2}$, etc. and so the expression under square root becomes
$$\frac{1}{K}\sqrt{s(s-a)+s(s-b)+s(s-c)}=\frac{s}{K}$$
So we want to show that
$$\frac{2Rs}{K}=\sum_{cyc}\frac{1}{\sin A\sin B}$$
Now, by the extended law of sines, $\frac{a}{\sin A}=2R\implies \frac{1}{\sin A}=\frac{2R}{a}$, so the RHS is
$$(2R)^2\sum_{cyc}\frac{1}{ab}=\frac{(2R)^2(a+b+c)}{abc}$$
Thus we want to show
$$\frac{s}{K}=\frac{2R(a+b+c)}{abc}$$
Now this is easy because it rearranges to $\frac{1}{K}=\frac{4R}{abc}$, which is $K=\frac{abc}{4R}$ which is true.
|
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|
show that $0\le \frac a{1+b}+\frac b{1+a} \le 1$ Let $0\le a < b \le 1$
prove that $0\le \frac a{1+b}+\frac b{1+a} \le 1$
This is the answer from book :
Obviously $0\le \frac a{1+b}+\frac b{1+a}$ so :
\begin{align}
&1\ge \frac a{1+b}+\frac b{1+a} \\
\iff & (1+a)(1+b) \ge a(1+a)+b(1+b)\\
\iff & 1-a^2\ge b^2-ab \\
\iff &(1-a)(1+a)\ge b(b-a) \tag{*}
\\ \iff & 1+a\ge b-a
\end{align}
which is obvious. Q.E.D.
But I can't understand how we get last inequality from $(*)$.
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An easier proof for me is the following:
Firstly, $\frac a{1+b}+\frac b{1+a}\geq \frac{b}{1+a}>0,\ $ as numerator and denominator of $\frac{b}{1+a}$ are both positive.
Secondly,
$$\frac a{1+b}+\frac b{1+a} = \frac{a(1+a)+b(1+b)}{(1+b)(1+a)} = \frac{a+b+a^2+b^2}{a+b+ab+1}. $$
So if we show that $ab+1 \geq a^2 + b^2, $ and then we are done. To this end:
$$\text{Since } b>a,\ \text{ and } a\geq 0,\ ab \geq a^2.\quad \text{Also, } 1\geq b, \text{ and so } 1 \geq b^2.$$
$$\text{Therefore, } ab+1 \geq a^2 + b^2. $$
The strict inequality holds on the left side only:
$$0 < \frac a{1+b}+\frac b{1+a} \leq 1.$$
Equality on the RHS occurs when $b=1$ and $a=0.$
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Find the general solutions of the PDEs below...
a)$3u_x-4u_y = x^2$
b)$u_x - 4u_y + u = x + y + 1$
I'm starting in PDEs, I saw two resolution methods, but I think I didn't understand them right.
a) $\frac{dx}{3}= \frac{dy}{-4}=\frac{dz}{x^2}$
$\frac{dx}{3}= \frac{dy}{-4} \to -4x + c = 3y \to c_1= 4x+3y$
$\frac{dy}{-4}=\frac{dz}{x^2} \to x^2y = -4z + c_2 \to c_2 = x^2y + 4z $
$u= \frac{x^2y - f(4x+3)}{-4} $
$\dfrac{dx}{dt}=3 \to x(0)=0 \to x=3t$
$\dfrac{dy}{dt}=-4 \to y(0)=y_0 \to y= -4t +y_0 \to y= \frac{-4x}{3} + y_0$
$\dfrac{du}{dt}=x^2 = (3t)^2 \to u(0)=f(y_0) \to u(x,y)=3t^3 + f(y_0) = x^3+f(\frac{4x} {3}+y)$
b) $\frac{dx}{1}= \frac{dy}{-4}=\frac{dz}{x + y + 1}$
$\frac{dx}{1}= \frac{dy}{-4} \to -4x + c = y \to c_1= 4x+y$
$\frac{dy}{-4}=\frac{dz}{x + y + 1} \to x^2 + y + x = -4z + c_2 \to c_2 = x^2 + y + x + 4z $
$u= \frac{x^2 + y + x - f(4x+y)}{-4} $
$\dfrac{dx}{dt}=1 \to x(0)=0 \to x=t$
$\dfrac{dy}{dt}=-4 \to y(0)=y_0 \to y= -4t +y_0 \to y= -4x + y_0$
$\dfrac{du}{dt}=x + y + 1 = -3t + 1 \to u(0)=f(y_0) \to u(x,y)=\frac{-3t^2}{2} + t f(y_0) = \frac{-3t^2}{2} + t + f(4x + y)$
Shouldn't the answers be the same? Are any of them correct?
Thanks.
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a)
$$3u_x-4u_y=x^2$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{3}=\frac{dy}{-4}=\frac{du}{x^2}$$
A first characteristic equation comes from solving $\frac{dx}{3}=\frac{dy}{-4}$ :
$$4x+3y=c_1$$
A second characteristic equation comes from solving $\frac{dx}{3}=\frac{du}{x^2}$ :
$$u-x^3=c_2$$
The solution of the PDE expressed on implicit form $c_2=F(c_1)$ is :
$$u-x^3=F(4x+3y)$$
$F$ if an arbitrary function.
$$\boxed{u(x,y)=x^3+F(4x+3y)}$$
This is equivalent to
$$u(x,y)=x^3+f\left(\frac{4x}{3}+y\right)$$
$f$ is an arbitrary function. Both arbitrary functions $F$ and $f$ are related : $F(3X)=f(X)$.
Thus you correctly found the solution of problem (a).
b)
$$u_x-4u_y+u=x+y+1$$
$$u_x-4u_y=-u+x+y+1$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{1}=\frac{dy}{-4}=\frac{du}{-u+x+y+1}$$
You forgot $u$ in the denominator. That is a cause of failure of your calculus.
A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{-4}$ :
$$4x+y=c_1$$
A second characteristic equation comes from solving $\frac{dx}{1}=\frac{du}{-u+x+y+1}$. This cannot be donne rightaway because $x$ and $y$ are related on the caracteristic curves. In the present case $y=c_1-4x$.
$\frac{dx}{1}=\frac{du}{-u+x+(c_1-4x)+1}\quad\implies\quad \frac{dx}{1}=\frac{du}{-u-3x+c_1+1}\quad\implies\quad \frac{du}{dx}=-u-3x+c_1+1.\quad$ Solving it leads to :
$u=c_2e^{-x}-3x+c_1+4$
The solution of the PDE expressed on implicit form $c_2=F(c_1)$ is :
$u=-3x+c_1+4+e^{-x}F(c_1)$
$F$ is an arbitrary function.
$u=-3x+(4x+y)+4+e^{-x}F(4x+y)$
$$\boxed{u(x,y)=x+y+4+e^{-x}F(4x+y)}$$
|
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|
pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$8x^4+27y^4+64z^4$$
where $$x+y+z=\frac{13}{4}$$
I tried to apply C-S inequality but it has little difference,
The form what I know is :
$$(a^4+b^4+c^4)(x^4+y^4+z^4)\ge (ax+by+cz)^4$$
But in this problem, coefficient is form of $()^3$, not a $()^4$.
I tried to rewrite $8x^4=(2x)^4\times\frac{1}{2}$, but It wasn't helpful :
$$\left(\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}\right)((2x)^4+(3y)^4+(4z)^4)\ge (x+y+z)^4$$
So, is there any nice transform to apply C-S inequality?
or should I apply Lagrange Multiplier?
|
Define:
$$u^4=8x^4,~v^4=27y^4,~w=64z^4$$
$$f=u^4+v^4+w^4$$
Constraint becomes: $$8^{-1/4}u+27^{-1/4}v+64^{-1/4}w=13/4$$
So we have:
$$a=8^{-1/4},~~b=27^{-1/4},~~c=64^{-1/4}$$
Now you can use C-S inequality.
|
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|
$\binom{1/2}{n}=\frac{2(-1)^{n-1}}{n4^n}\binom{2n-2}{n-1}$ There is a "direct" way to prove this equality.
$\displaystyle \binom{1/2}{n}=\frac{2(-1)^{n-1}}{n4^n}\binom{2n-2}{n-1}$
I am trying to skip the induction. Maybe there is a rule or formula that will help me.
Thank you
|
another way
Let's rewrite part of the RHS in terms of gamma
$$
\frac{1}{n}\left( \begin{array}{c}
2n - 2 \\ n - 1 \\ \end{array} \right)
= \frac{{\Gamma \left( {2n - 1} \right)}}
{{n\Gamma \left( n \right)\Gamma \left( n \right)}}
= \frac{{\Gamma \left( {2n - 1} \right)}}{{\Gamma \left( {n + 1} \right)\Gamma \left( n \right)}}
$$
To the above we can apply the Gamma Duplication Formula](https://en.wikipedia.org/wiki/Multiplication_theorem#Gamma_function%E2%80%93Legendre_formula)
$$
\Gamma \left( {2\,z} \right)
= \frac{{2^{\,2\,z - 1} }}{{\sqrt \pi }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)
= 2^{\,2\,z - 1} \frac{{\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)}}{{\Gamma \left( {1/2} \right)}}
$$
wherefrom
$$
\Gamma \left( {2n - 1} \right) = \Gamma \left( {2\left( {n - 1/2} \right)} \right)
= 2^{\,2n - 2} \frac{{\Gamma \left( {n - 1/2} \right)\Gamma \left( n \right)}}{{\Gamma \left( {1/2} \right)}}
$$
so getting
$$
\begin{array}{l}
\frac{1}{n}\left( \begin{array}{c}
2n - 2 \\ n - 1 \\
\end{array} \right) = \frac{{\Gamma \left( {2n - 1} \right)}}{{\Gamma \left( {n + 1} \right)\Gamma \left( n \right)}}
= 2^{\,2n - 2} \frac{{\Gamma \left( {n - 1/2} \right)}}{{\Gamma \left( {1/2} \right)\Gamma \left( {n + 1} \right)}} = \\
= 2^{\,2n - 2} \frac{{\Gamma \left( {n - 1/2} \right)}}
{{\left( { - \frac{1}{2}} \right)\Gamma \left( { - 1/2} \right)\Gamma \left( {n + 1} \right)}}
= - 2^{\,2n - 1} \left( \begin{array}{c}
n - 3/2 \\ n \\
\end{array} \right) = \\
= - 2^{\,2n - 1} \left( { - 1} \right)^n \left( \begin{array}{c}
1/2 \\ n \\
\end{array} \right) \\
\end{array}
$$
where:
*
*1st line : application of the duplication formula,
*2nd line: passing from $\Gamma(1/2)$ to $\Gamma(-1/2)$ to write the fraction as a binomial
*3rd line : applying the "upper negation".
|
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|
Finding the definite integral of $\frac{\log x}{1+ x^3}$ The problem is apparently simple:
Find $\displaystyle{\int_0^\infty \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x }$.
However, I have not been able to get around it. The usual "big circle" strategy is useless since we don't have parity. With substitution I was able to prove that, choosing $t = 1/x$, we have
$$
\int_0^1 \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x =- \int_1^\infty \!\!\frac{t \log t}{1+ t^3} \,\mathrm{d} t
$$
And, analogously
$$
\int_1^\infty \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x =- \int_0^1 \!\!\frac{t \log t}{1+ t^3} \,\mathrm{d} t
$$
However, this doesn't seem to be helpful. I could now write my integral as \begin{align}
\int_0^\infty \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x= &
\int_0^1 \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x + \int_1^\infty \!\! \frac{\log x}{1+ x^3} \,\mathrm{d} x \\
= &\int_0^1 \!\!\frac{\log x}{1+ x^3} \,\mathrm{d} x - \int_0^1 \!\!\frac{x \log x}{1+ x^3} \,\mathrm{d} x
\\
= &\int_0^1 \!\!\frac{1-x}{1+ x^3} \log x \,\mathrm{d} x\\
\end{align}
This is a proper definite integral and I hoped it could be useful, e.g. to use Integration by series, but I didn't manage to do it.
|
Continue with
\begin{align}
\int_0^\infty&\frac{\log x}{1+x^3}{d}x
=
\int_0^1\frac{(1-x)\log x}{1+x^3}{d}x \\
&=\int_0^1\frac{\log x}{1+x}
- \underset{x^3\to x}{\frac{x^2\log x}{1+x^3}}\ {dx}
= \frac89\int_0^1\frac{\log x}{1+x}{d}x
=-\frac{2\pi^2}{27}
\end{align}
$\int_0^1\frac{\log x}{1+x}{d}x=-\frac{\pi^2}{12}$
|
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|
Evaluating $\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$. Why doesn't $x^2$ overpower $2x$ when $x\to\infty$, so that the answer is $0$ (instead of $1$)?
Evaluate $$\lim_{x \to \infty} \sqrt{x^2 + 2x} - x$$
Hello!
I was solving this problem, and here is my approach:
Inside the square root $x^2$ overpowers $2x$ as $x \to \infty$, so $$\lim_{x \to \infty} \sqrt{x^2 + 2x} - x = \lim_{x \to \infty} \sqrt{x^2} - x = \lim_{x \to \infty} |x| - x = 0$$ which is wrong answer. Actual answer is $1$.
This is the way I have solved so many problems when $x \to \infty$, especially when the terms are in denominator.
For example, consider $$\lim_{x \to \infty} \frac{x^2 + x}{2x^2 + 4x + 10}$$
Now, we ignore the smaller power terms, and answer is $\frac{1}{2}$. Why can we ignore smaller powers here but not in the original question on top?
Sorry if this is a stupid question.
|
Let us try to eliminate or cancel the square term & the root term :
$\lim_{x \to \infty} {\sqrt{x^2 + 2x} - x} = \lim_{x \to \infty} {\sqrt{x^2 + 2x + 1 -1} - x} = \lim_{x \to \infty} {\sqrt{(x + 1)^2 -1} - x}$
We can now see that the square term will Over-Power the Constant.
We then get a way to eliminate or cancel the square term & the root term :
$\lim_{x \to \infty} {\sqrt{x^2 + 2x} - x} = \lim_{x \to \infty} {\sqrt{(x + 1)^2} - x} = \lim_{x \to \infty} {(x + 1) - x} = 1$
|
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|
Define the term $a_n$ in Laurent's series expansion $\sum_{n=-\infty}^{\infty} a_n(z-a)^n$ of $f$ centered on $z=a.$...
Define the term $a_n$ in Laurent's series expansion $\sum_{n=-\infty}^{\infty} a_n(z-a)^n$ of $f$ centered on $z=a.$ classify the singularity $ z=a.$
a)$f(z)=\frac{z^2+1}{z(z+1)}, a=0.$
b)$f(z)=\frac{1}{z(z-1)}$ for $a=0.1.$
we have the following formula $a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz$
So I need to do
a) $a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz = \frac{1}{2 \pi i} \int_{\gamma} \frac{\frac{z^2+1}{z(z+1)}}{z}dz$ ?
b)$a=0 \to a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz = \frac{1}{2 \pi i} \int_{\gamma} \frac{\frac{1}{z(z-1)}}{z}dz$?
$a=1 \to a_n = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz = \frac{1}{2 \pi i} \int_{\gamma} \frac{\frac{1}{z(z-1)}}{z-1}dz$?
If it is that, how do I solve this integral?
Thanks.
|
You don't need that formula.
a) For each $z\in\Bbb C$ with $|z|<1$, you have\begin{align}\frac{z^2+1}{z+1}&=z-1+\frac2{z+1}\\&=z-1+2\sum_{n=0}^\infty(-1)^nz^n\\&=1-z+2\sum_{n=2}^\infty(-1)^nz^n.\end{align}Therefore, if $z\ne0$,$$\frac{z^2+1}{z(z+1)}=\frac1z-1+2\sum_{n=1}^\infty(-1)^{n+1}z^n.$$So, $0$ is a simple pole.
b) Near $0.1$, $\frac1{z(z-1)}$ is the quotient of two analytic functions, and therefore it is an analytic function. It follows that $0.1$ is a removable singularity of $f$.
|
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|
Finding the minimum value of the expression $xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$ From R. D. Sharma's Objective Mathematics,
Given that $x + y + z = 1$, ($x,y,z$ are positive real numbers) find the minimum value of $$A = xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2$$
My attempt:
By A.M.-G.M. inequality,
$$\begin{align}
(x+y)^2 &\geq 4(xy) \\
(y+z)^2 &\geq 4(yz) \\
(z+x)^2 &\geq 4(zx)
\end{align}$$
By multiplying by $xy$, $yz$, $zx$ in these equations respectively, gives
$$\begin{align}
xy(x+y)^2 &\geq 4(xy)^2 \\
yz(y+z)^2 &\geq 4(yz)^2 \\
zx(z+x)^2 &\geq 4(zx)^2 \\
\end{align}$$
Adding these equations, we get
$$ A \geq 4[(xy)^2 + (yz)^2 + (zx)^2] $$
Using AM GM inequality once again on RHS, we get
$$ A \geq 12(xyz)^{\tfrac43}$$
However, provided answer key says that minimum value is $4xyz$. I do not want a solution but only wants to know that is something wrong in my procedure? Why?
EDIT : This question has been identified as a possible duplicate of another question . However, readers will believe that my problem is different if they re-read the paragraph just above .
|
Note that,
If $x,y\to 0$ with $z\to 1$, we have
$$\inf \left\{xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2\mid x+y+z=1 \wedge x>0\wedge y>0\wedge z>0\right\}=0$$
As for obtaining the inequality $A≥4xyz$, you can easily obtain this result using the Cauchy-Schwarz inequality:
$$(xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2)\left(\frac 1{xy}+\frac 1{yz}+\frac 1{xz}\right)≥4(x+y+z)^2=4$$
$$\frac A{xyz}≥4\implies A≥4xyz.$$
But, note that $xyz$ is not a constant. Therefore, $4xyz$ is not a "minimum". We only proved that,
$$xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2≥4xyz$$
The inequality you want to prove.
|
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|
Spivak: How do we divide $1$ by $1+t^2$, to obtain $\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}$? In Chapter 20, "Approximation by Polynomial Functions" in Spivak's Calculus, there is the following snippet on page $420$
The equation
$$\arctan{x}=\int_0^x \frac{1}{1+t^2}dt$$
suggests a promising method of finding a polynomial close to $\arctan$
*
*divide $1$ by $1+t^2$, to obtain a polynomial plus a remainder:
$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+...+(-1)^nt^{2n}+\frac{(-1)^{n+1}t^{2n+2}}{1+t^2}\tag{1}$$
How exactly do we perform this division of $1$ by $1+t^2$ to obtain $(1)$?
|
$$\frac{1}{1+t^2}= 1 - \frac{t^2}{1+t^2}$$ implies
$$\frac{1}{1+t^2}=1-t^2\left(1 - \frac{t^2}{1+t^2}\right)= 1 -t^2 + \frac{t^4}{1+t^2}$$ and continue in this manner.
|
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|
If $\frac{x}{y}+\frac{y}{x}\ge2$ and $\sum_{cyc}\frac{x}{y+z}\ge\frac32$, can we say $\sum_{cyc}\frac{w}{x+y+z}\ge\frac43$? We know that $$\frac{x}{y}+\frac{y}{x}\ge2$$ and $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\ge\frac32$$ Can we say that
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}\ge\frac43$$ Or in general, where there are $n$ variables can we say that their sum (in the same format as above) is no less than $\frac{n}{n-1}$$?$
All the variables used here are positive real numbers.
My thoughts:
I think that we might need induction here, and then we can try completing the squares to find the minimum value. Any help is greatly appreciated.
|
Problem: Let $x_i \ge 0, i= 1, 2, \cdots, n$. Prove that
$$\sum_{i=1}^n \frac{x_i}{S - x_i} \ge \frac{n}{n - 1}$$
where $S = \sum_{j=1}^n x_j$.
Proof.
Since the desired inequality is homogeneous, assume that $\sum_{j=1}^n x_j = 1$. We need to prove that
$$\sum_{i=1}^n \frac{x_i}{1 - x_i} \ge \frac{n}{n - 1}.$$
We have
$$\frac{x_i}{1 - x_i}
- \frac{1}{n - 1} - \frac{n^2}{(n-1)^2}\left(x_i - \frac{1}{n}\right) = \frac{(1-nx_i)^2}{(n-1)^2(1 - x_i)} \ge 0$$
which results in
$$\sum_{i=1}^n \frac{x_i}{1 - x_i} - \frac{n}{n-1} \ge \frac{n^2}{(n-1)^2}\sum_{i=1}^n\left(x_i - \frac{1}{n}\right) = 0.$$
We are done.
|
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|
Prove that: $\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$
$\color{red}{\textbf{Problem:}}$
Let, $x_i>0,1\le i\le n$, then
Prove that:
$$\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$$
$\color{red}{\textbf{Proof:}}$
Using AM-GM inequality we have,
\begin{align}&\sum_{1\le i<j\le n} (x_i^2+x_j^2)=(n-1)\sum_{1\le i\le n} x_i^2\ge 2\left(\sum_{1\le i < j \le n}x_ix_j\right)\\
\implies &\sum_{1\le i\le n} x_i^2\ge \frac 2{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\\
\implies &\left(\sum_{1\le i\le n} x_i\right)^2\ge\left( 2+\frac 2{n-1}\right)\left(\sum_{1\le i < j \le n}x_ix_j\right)\\
&\qquad\qquad\quad =\frac{2n}{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\end{align}
Finally, applying Cauchy-Schwars, we obtain
\begin{align}\left(\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\right)\left(\sum_{1\le j\le n} \left({x_j}{\sum_{1\le i\le n} x_i-x_j}\right)\right)&\ge \frac{ \left(\sum_{1\le j\le n} x_i\right)^2}{2\left(\sum_{1\le i < j \le n}x_ix_j\right)}\\
&\ge \frac {\frac{2n}{n-1}\sum_{1\le i < j \le n}x_ix_j}{2\sum_{1\le i < j \le n}x_ix_j}\\
&=\boxed {\frac{n}{n-1}.}\end{align}
I need to know if there is something wrong with my solution.
|
OP's work is correct however $x_j>0$ needs to be mentioned. Here is another approach:
Let $s=\sum_{j=1}^{n} x_j,$ then
$$f(x_j)=\frac{x_j}{s-x_j} \implies f''(x)=\frac{2s}{(s-x_j)^3}>0,\quad 0<x_j<s.$$
So by Jensen's inequality
$$\frac{1}{n}\sum_{j=1}^{n} f(x_j) \ge f(\frac{s}{n}) \implies \sum_{j=1}^{n} \frac{x_j}{s-x_j}\ge n \frac{s/n}{s-s/n}=\frac{n}{n-1}. $$
|
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|
Calculating eigenvalues of a matrix I have this quadratic matrix:
\begin{align*}
A = \begin{bmatrix}
0 & -1 & -2\\
-1 & 0 & -2\\
-2 & -2 & -3
\end{bmatrix} \implies A - \lambda I = \begin{bmatrix}
0 - \lambda & -1 & -2\\
-1 & 0 - \lambda & -2\\
-2 & -2 & -3 - \lambda
\end{bmatrix}
\end{align*}
After forming the characteristic polynomial and using the Rule of Sarrus, I get this equation:
$$-\lambda^3 - 3\lambda^2 + 9\lambda - 5$$
However the correct form of this term would be:
$$-(5 + \lambda)(1 - \lambda)^2$$
So the eigenvalues are $1$ and $-5$.
The thing I do not understand, is how to form this equation to get the values. Which approach and rules can solve this?
Thank you very much.
|
$$p(\lambda)=-\lambda^3 - 3\lambda^2 + 9\lambda - 5$$
Proceed from your work,
$$\begin{align}
p(\lambda)&=-\lambda^3-5\lambda^2+2\lambda^2+9\lambda-5\\
\\
&=-\lambda^2(\lambda+5)+(2\lambda-1)(\lambda+5)\\
\\
&=(\lambda+5)(-\lambda^2+2\lambda-1)\\
\\
&=-(\lambda+5)(\lambda^2-2\lambda+1)\\
\\
&=-(\lambda+5)(\lambda-1)^2\end{align}
$$
|
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|
How do I make a change of variable for $\;\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$? I can't use l'hopital, so change of variable is the only way.
$$\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$$
|
You can write:
$$\sqrt{x^2+p^2}-p=p\cdot\left(\sqrt{1+\frac{x^2}{p^2}}-1\right)$$
And:
$$\sqrt{x^2+q^2}-q=q\cdot\left(\sqrt{1+\frac{x^2}{q^2}}-1\right)$$
We know that $(1+t)^\alpha-1 \,\, \sim\,\, \alpha \cdot t$ when $x\to 0$; notice that in our cases $t=\frac{x^2}{p^2}$ for the numerator and $t =\frac{x^2}{q^2}$ for the denominator.
So, we have:
$$\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}=\lim_{x\to 0}\frac{p\cdot\left(\sqrt{1+\frac{x^2}{p^2}}-1\right)}{q\cdot\left(\sqrt{1+\frac{x^2}{q^2}}-1\right)}\,\,\sim\,\,\lim_{x\to 0}\frac{p\cdot \frac{1}{2}\frac{x^2}{p^2}}{q\cdot \frac{1}{2}\frac{x^2}{q^2}}=\frac{q}{p}$$
|
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|
Evaluate ${\cos ^{ - 1}}\sqrt {1 - {x^2}} = $ for $x<0$ Question:
Let $x<0$ then $\cos^{-1}\sqrt{1-x^2}=$
(1) $\pi-\cos^{-1}x$
(2) $-\sin^{-1}x$
(3) $\pi-\sin^{-1}x$
(4) $\sin^{-1}x$
My approach is as follow
Given $x < 0,{\cos ^{ - 1}}\sqrt {1 - {x^2}} $
Let $x = \sin \theta $, therefore ${\cos ^{ - 1}}\sqrt {1 - {{\sin }^2}\theta } = {\cos ^{ - 1}}\sqrt {{{\cos }^2}\theta } = {\cos ^{ - 1}}\left| {\cos \theta } \right| = \theta $
Hence ${\cos ^{ - 1}}\sqrt {1 - {{\sin }^2}\theta } = {\cos ^{ - 1}}\sqrt {{{\cos }^2}\theta } = {\cos ^{ - 1}}\left| {\cos \theta } \right| = {\sin ^{ - 1}}x$
But as per the graph answer is $-\sin^{-1}x$ for $x<0$.
|
Because $x=\sin \theta <0$ and $\cos \theta = \sqrt{1-x^2} >0$, angle $\theta$ is in the fourth quadrant. Thus, $\cos^{-1} |\cos \theta|=-\theta$
|
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|
Prove $\int_{- \infty}^{\infty} \frac{\mathrm{dx}}{({x}^{2} + {b}^{2}) {({x}^{2} + {a}^{2})}^{2}} = \frac{\pi (2 a + b)}{2 {a}^{3} b {(a + b)}^{2}}$ I am in chapter 6 of Whittaker and Watson, and I am currently confused attempting to find the mistake I made in computing the following integral. The problem asks me to prove the following relationship
\begin{align}\int_{- \infty}^{\infty} \frac{\mathrm{dx}}{\left({x}^{2} + {b}^{2}\right) {\left({x}^{2} + {a}^{2}\right)}^{2}} = \frac{\pi \left(2 a + b\right)}{2 {a}^{3} b {\left(a + b\right)}^{2}}\end{align}
Attempted solution: I thought it best to try an evaluate this integral using Jordan's lemma. I already feel comfortable with my proof for demonstrating that the integral along a semicircle in the upper half plane will collapse to $0$ as its radius trends towards infinity. For simplicity, I am examining the case $a > 0$ and $b > 0$. (I believe that the case when one of them is zero could be solved by producing another semicircular domain around the real axis.)
I assert
\begin{align}
I & = \int_{- \infty}^{\infty} \frac{\mathrm{dx}}{\left({x}^{2} + {b}^{2}\right) {\left({x}^{2} + {a}^{2}\right)}^{2}} \\
& = \int_{- \infty}^{\infty} \frac{\mathrm{dx}}{\left(x - i b\right) \left(x + i b\right) {\left(x - i a\right)}^{2} {\left(x + i a\right)}^{2}} \\
& = \int_{C} \frac{\mathrm{dz}}{\left(z - i b\right) \left(z + i b\right) {\left(z - i a\right)}^{2} {\left(z + i a\right)}^{2}} \\
\end{align}
where $C$ is a semicircular contour in the upper half plane formed by the line from $- R$ to $R$ and a semicircle $\Gamma$.
For concision, I define
\begin{align}
f \left(z\right) = \frac{1}{\left(z - i b\right) \left(z + i b\right) {\left(z - i a\right)}^{2} {\left(z + i a\right)}^{2}} \\
\end{align}
such that
\begin{align}
\int_{C} f \left(z\right) \mathrm{dz} = 2 \pi i \left[{\text{Res}}_{z = i b} f \left(z\right) + {\text{Res}}_{z = i a} f \left(z\right)\right] \\
\end{align}
The point $i b$ is an order $1$ pole, so I use the formula
\begin{align}
{\text{Res}}_{z = i b} f \left(z\right) & = \lim_{z \to i b} \left(z - i b\right) f \left(z\right) \\
& = \lim_{z \to i b} \frac{1}{\left(z + i b\right) {\left({z}^{2} + {a}^{2}\right)}^{2}} \\
& = \frac{1}{2 i b \left({a}^{2} - {b}^{2}\right)} \\
\end{align}
Likewise, $i a$ is an order $2$ pole, so
\begin{align}
{\text{Res}}_{z = i a} f \left(z\right) & = \lim_{z \to i a} \frac{d}{\mathrm{dz}} \left({\left(z - i a\right)}^{2} f \left(z\right)\right) \\
& = \lim_{z \to i a} \frac{d}{\mathrm{dz}} \left(\frac{1}{\left({z}^{2} + {b}^{2}\right) {\left(z + i a\right)}^{2}}\right) \\
& = \lim_{z \to i a} \frac{- 2 z {\left(z + i a\right)}^{2} - 2 \left({z}^{2} + {b}^{2}\right) \left(z + i a\right)}{{\left({z}^{2} + {b}^{2}\right)}^{2} {\left(z + i a\right)}^{2}} \\
& = \frac{8 i {a}^{3} - 4 i a {b}^{2} + 4 i {a}^{3}}{\left({b}^{2} - {a}^{2}\right) \left(- 4 {a}^{2}\right)} \\
\end{align}
Then,
\begin{align}
I & = \frac{2 \pi i}{{\left({a}^{2} - {b}^{2}\right)}^{2}} \left(\frac{1}{2 i b} + \frac{8 {a}^{3} i - 4 i a {b}^{2} + 4 i {a}^{3}}{\left(- 4 {a}^{2}\right)}\right) \\
& = \pi \left(\frac{{a}^{2} + 4 {a}^{3} b - 2 a {b}^{3} + 2 b {a}^{3}}{{a}^{2} b {\left({a}^{2} - {b}^{2}\right)}^{2}}\right) \\
\end{align}
However, I believe that I have made a mistake as using the identity, $\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$ and multiplying by $\frac{a}{a}$, I find
\begin{align}
I = \pi \left(\frac{{a}^{4} + 4 {a}^{4} b - 2 {a}^{2} {b}^{3} + 2 b {a}^{4}}{{a}^{2} b {\left({a}^{2} - {b}^{2}\right)}^{2}}\right) \\
\end{align}
with the contradiction that
\begin{align}
\left(2 a + b\right) {\left(a - b\right)}^{2} & = \left(2 a + b\right) \left({a}^{2} + {b}^{2} - 2 a b\right) \\
& = 2 {a}^{3} + 2 a {b}^{2} - 4 {a}^{2} b + b {a}^{2} + {b}^{3} - 2 a {b}^{2} \\
& = 2 {a}^{3} - 3 {a}^{2} b + {b}^{3} \\
& \ne {a}^{4} + 4 {a}^{4} b - 2 {a}^{2} {b}^{3} + 2 b {a}^{4} \\
\end{align}
Question: Where is the mistake in my above reasoning? Is my computation of the residues incorrect? (I hope that it was not a simple algebraic oversight given how long it took to type my question.)
|
Actually,\begin{align}\lim_{z\to ai}\frac{\mathrm d}{\mathrm dz}\frac1{(z^2+b^2)(z+ai)^2}&=\lim_{z\to ai}\frac{-2z(z+ai)^2-2(z^2+b^2)(z+ai)}{(z^2+b^2)^2(z+ai)^{\color{red}4}}\\&=\frac{i \left(3 a^2-b^2\right)}{4 a^3 (a-b)^2 (a+b)^2}.\end{align}
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|
A deceptively difficult linear recurrence problem Consider the sequence $b_0 = 0$, $b_1 = 1$, $b_{n+2} = 2b_{n+1} - 3b_n$ for all $n\geq 0$. Prove that the only positive integers $m$ with $b_m = \pm 1$ are $m = 1$ and $m = 3$. (In fact, I expect that $b_m = -1$ never occurs but not proving this is not an issue for the purpose described below.)
Calculating the first $50$ terms does not give much of a hint since the sequence is quite oscillating above and below $0$ and even $|b_n|$ is not monotonic all the time. An explicit formula is $b_n = (\sqrt{3})^n\frac{\sin(n \cdot \arctan \sqrt{2})}{\sqrt{2}}$ (derived by solving the characteristic equation etc.) but it does not really help me either.
(Interesting context: if one proves this, then it is not hard to prove, using $\mathbb{Z}[\sqrt{-2}]$, that the only positive integer solutions to $x^2 + 2 = 3^n$ are $(1,1)$ and $(5,3)$.)
Any help appreciated!
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We will work in the ring
$$ R=\Bbb Z[a]\ ,\qquad\text{ where } a=\sqrt{-2}\ .
$$
Then the characteristic equation of the given linear recursion is
$\lambda^2 -2\lambda +3=0$, it has the roots $(1+a)$ and $(1-a)$, both in $R$.
The general formula for $b_n$ is then:
$$
b_n =\frac 1{2a}\Big[\ (1+a)^n -(1-a)^n \ \Big]\ ,
$$
since it is true for $n=0$ and $n=1$, and since it is one and the same linear combination of the $n$-powers of the roots $(1\pm a)$. Let us assume there is some $n$ such that $b_n=\pm 1$. Then we have
$$
\pm 2a = (1+a)^n - (1-a)^n\ ,
$$
and thus $(1+a)^n$ is an element from $R$ of the shape
$$
y\pm 1\cdot a\ ,\qquad\text{ for some suitable } y\in \Bbb Z\ .
$$
Then we also know the conjugate, so the product is:
$$
3^n =(\ (1+a)(1-a)\ )^n=(1+a)^n(1-a)^n=(y+a)(y-a)=y^2-a^2 =y^2+2\ .
$$
This is a relation that lives in $\Bbb Z$, so let us see if it can be satisfied. To reduce some cases, although not needed, let us consider the above relation modulo $13$. Then $3$ has the multiplicative order three, since $3^3 =27=26+1\equiv 1$ modulo $13$. Its powers are $1,3,9$. Which powers are of the shape $x^2+2$ in $\Bbb F_{13}$?
$$
\begin{array}{|r|r|r|r|r|r|r|r|}
\hline
y & 0 & \pm 1 & \pm2 & \pm3 &\pm4 &\pm5 &\pm6
\\\hline
y^2 + 2 & 2 & 3 & 6 & 11 & 5 & 1 & 12
\\\hline
\end{array}
$$
So there is no match for the nine. Then $n$ is either $n=3m$ or $n=3m+1$.
In the first case, we have $y^2 = 3^{3m}-2$, in the second case after multiplication with nine $(3y)^2=3^{3(m+1)}-18$. So we obtain an integral point on the elliptic curve
$y^2=x^3-2$ or $Y^2=X^3 -18$. The integral points of these elliptic curves can be computed algorithmically, i am using sage:
sage: EllipticCurve(QQ, [0, -2]).integral_points()
....: EllipticCurve(QQ, [0, -18]).integral_points()
....:
[(3 : 5 : 1)]
[(3 : 3 : 1)]
so let us see if the power-of-three condition for $x$, respectively $X$ is matched, oh, yes, indeed, (note that sage gives only the integral point with positive second component, except we explicitly claim we also want the other one... well, we know what we are doing...)
*
*the first point $(x,y)=(3,\pm5)$ on $y^2 = x^3-2$ corresponds to the solution $(\pm5)^2=3^3-2$, matching $3^n=y^2+2$ for $n=3$,
*the second point $(X,Y)=(3,\pm3)$ on $Y^2 = X^3-18$ corresponds to the solution $(\pm3)^2=3^3-18$, we take back the multiplication with nine, matching $3^n=y^2-2$ for $n=1$.
And indeed, we compute for $n=1$ and $n=3$ the values $b_1=1$, given,
$b_3=\frac 1{2a}(\ (1+a)^3-(1-a)^3\ )=\frac 1a(3a+a^3)=3+a^2=1$. (Or use the linear recursion to get $b_2=2b_1-3b_0=2-0=2$, $b_3=2b_2-3b_1=4-3=1$.)
$\square$
|
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|
Convergence of $1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$ I was asked to prove whether
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} - ...$$
converges absolutely, conditionally or diverges, and was wondering whether my proof is correct or not.
I began by noticing that
$$1-\frac{1 \times 3}{3!} + \frac{1 \times 3 \times 5}{5!}- \frac{1 \times 3 \times 5 \ \times 7}{7!} +....+ \frac{1 \times 3 \times 5 \times ... \times(2n-1)}{(2n-1)!} -...$$
$$= 1- \frac{1}{2} + \frac{1}{2\times4} - \frac{1}{2 \times 4 \times 6}+...+\frac{1}{(2n)\times(2n-2)... \times2} -...$$
In other words, each $n$th term is the inverse of the product of all even natural numbers, and thus the initial sum is simply $ 1 + \sum a_n$ with
$$a_n := (-1)^n \frac{1}{\prod_{j=1}^n2j}$$
We can show that
$$\lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|} = \lim_{n\to\infty} \frac{\prod_{j=1}^{n}2j}{\prod_{j=1}^{n+1}2j}=\lim_{n\to\infty} \frac{1}{2(n+1)} = 0 < 1$$
and therefore the series converges absolutely according to the ratio test.
Is this proof correct?
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Yes, this is correct; there's nothing of note to mention otherwise that I can think of.
I'm choosing to answer like this and mark it as Community Wiki since I have nothing further to add, but don't want the question to end up in the unanswered queue.
|
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|
Find the intervals to which $p$ belong for these $4$ conditions
Suppose there is a quadratic equation $$f(y)=y^2-(p+1)y+p^2+p-8=0$$ then find the intervals in which $p$ should belong to fulfill the following conditions
$1.$ both the roots are less than $2$
$2.$ one root is greater than $2$ and the other one is greater than $2$
$3.$ at least one root is less than $2$
$4.$ at least one root is greater than $2$
For the $1$st one $($actually for all conditions to happen$)$, $\Delta\ge0$ $\implies$
$$-3p^2-2p+33\ge0$$ or $$p\in\left[\frac{-11}{3},3\right]$$
Now as both roots are less than $2$, $f(2)>2$ $\implies$ $p>3$ or $p<-2$. By intersecting it with the bound of $p$ obtained from discriminant, we see that $p\in\left[\frac{-11}{3},-2\right)$ but the answer is given $(-3,-2)$ I don't know why$?$
For the $2$nd one, $f(2)<0$ $\implies$ $$p\in(-2,3)$$
but the answer given to this sub part is $p\in(2,3)$ I can't understand why$?$
For the rest two parts, I don't know the approach. Any help is greatly appreicated.
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Part 3.
$$y = \frac{1}{2} \left[~(p+1) \pm \sqrt{(p+1)^2 - 4(p^2 + p - 8)} ~\right] \implies $$
$$y = \frac{1}{2} \left[~(p+1) \pm \sqrt{(p^2 + 2p + 1) - 4(p^2 + p - 8)} ~\right] \implies $$
$$y = \frac{1}{2} \left[~(p+1) \pm \sqrt{-3p^2 -2p + 33} ~\right].$$
As the OP (i.e. original poster) has already determined,
$$-3p^2 - 2p + 33 \geq 0 \iff -\frac{11}{3} \leq p \leq 3. \tag1 $$
For part 3, the smaller root must be less than $(2)$.
$$\frac{1}{2} \left[~(p+1) - \sqrt{-3p^2 -2p + 33} ~\right] < 2 \iff $$
$$(p+1) - \sqrt{-3p^2 -2p + 33} < 4 \iff $$
$$ - \sqrt{-3p^2 -2p + 33} < 4 - (p+1) \iff $$
$$ - \sqrt{-3p^2 -2p + 33} < (3 - p). \tag2 $$
Compare (1) and (2) above. If (p = 3), then the LHS of (2) above and the RHS of (2) above both equal $(0)$. Given the bounds on $(p)$, set by (1) above, this is the only way that the inequality in (2) can fail. That is, if $p < 3$, and $p$ is still in the boundary set by (1) above, then the LHS will be non-positive, and the RHS will be positive.
Therefore, the answer to part (3) is
$$ -\frac{11}{3} \leq p < 3.$$
Part 4.
For part 4, the larger root must be greater than $(2)$.
$$\frac{1}{2} \left[~(p+1) + \sqrt{-3p^2 -2p + 33} ~\right] > 2 \iff $$
$$(p+1) + \sqrt{-3p^2 -2p + 33} > 4 \iff $$
$$ + \sqrt{-3p^2 -2p + 33} > 4 - (p+1) \iff $$
$$ + \sqrt{-3p^2 -2p + 33} > (3 - p). \tag3 $$
Here, the analysis is not so simple. For one thing, whatever constraint is implied by (3) above must be melded with the constraint by (1) above. For another, to attack the constraint on (3) above, I have to square both sides. This risks extraneous values. This implies that whatever implied constraint is determined, after squaring both sides, must be manually verified against (3) above.
$$-3p^2 -2p + 33 > 9 - 6p + p^2 \iff $$
$$0 > 4p^2 - 4p - 24 \iff $$
$$0 > p^2 - p - 6 = (p-3)(p+2) \iff -2 < p < 3. \tag4 $$
Since the interval shown in (4) above is a subset of the interval shown in (1) above, the problem reduces to determining which portion of the interval in (4) above actually satisfies the constraint in (3) above.
However, in (3) above, both the LHS and the RHS must be positive, when $-2 < p < 3.$
In Real Analysis, if $0 < r,s$ then
$\displaystyle \sqrt{r} < \sqrt{s} \iff r < s.$
Therefore, the final answer to part (4) is $-2 < p < 3.$
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|
Integration $\int_{0}^{\infty} \frac{2x-1}{x^{2/3}\sqrt{(x+1)((x+1)^3-x)}}\text{d}x=0.$ Here is the integral:
$$\int_{0}^{\infty} \frac{2x-1}{x^{2/3}\sqrt{(x+1)((x+1)^3-x)}}\text{d}x=0.$$
This is an elliptic integral, with such an easy result, maybe some clever substitutions or integrating methods can solve it, but so far I don't know exactly how to attack it.
Thought 1:
To separate the integration interval into two parts:$(0,\frac12]$ and $[\frac12,+\infty)$, for the second one, doing a substitution $x\mapsto \frac1x$ is my first thought(possibly not valid).
Thought 2:
It looks 100%, even 90%, like pseudo elliptic integral(its anti-derivative is elementary). I think this one is elementary as well.
THOUGHT 3:
Maybe this is a very STUPID question. Perhaps some mathematical softwares can done it(especially for Mathematica). For some good reasons, I can't use them.
Waiting for your replies.
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I would like to solve it directly.
For $|t|$ large enough, let
$$
\tag{1}J\left(t\right)=\int_{0}^{\infty}\frac{2x-1}{x^{2/3}\sqrt{(x+1)((x+1)^3-x/t^2)} }\text{d}x.
$$
Firstly, apply the substitution $y=xt$, and
$$
(1)={t^{2/3}}\int_{0}^{\infty}\frac{2y-t}{y^{2/3}\sqrt{(y+t)((y+t)^3-y)} }\text{d}y.
$$
Secondly, let $z=y^{1/3}$, and
$$
(1)=-3t^{2/3}\int_{0}^{\infty}\frac{t-2z^3}{\sqrt{(z^3+t)((z^3+t)^3-z^3)} }\text{d}z.
$$
Note that
$$
\frac{t-2z^3}{\sqrt{(z^3+t)((z^3+t)^3-z^3)} }
=\frac{1}{\sqrt{1+\left ( \frac{2z^3+z+2t}{z^3-z+t} \right )^3}}\cdot\frac{\mathrm{d}}{\mathrm{d}z} \frac{2z^3+z+2t}{z^3-z+t}.
$$
Eventually, we apply the substitution
$$
s= \frac{2z^3+z+2t}{z^3-z+t}.
$$
Since $z\in[0,\infty)$, $s\in(2,2]$. Therefore $J(t)$ must be $0$. For complex $t$, these substitutions need to relate contour integration. But it's also good to show that $J(t)=0$.
One checks,
$$
\int_{\sqrt{5}-2}^{1} \frac{2x-1}{x^{2/3}\sqrt{(1-x^2)
(x^2+4x-1)} }\text{d}x=0.
$$
Actually, by performing substitution $ s=\frac{2z^a+z+2t}{z^a-z+t},a>1$ ($a$ no need to be an integer), we obtain
$$
\int_{0}^{\infty}\frac{(a-1)x-1}{x^{1-1/a}\sqrt{(x+1)\left((x+1)^3-tx^\frac3a\right)} }\text{d}x\equiv 0.
$$
*
*Setting $a=5,6,12$, then the following three integrals are derived.
$$
\int_{0}^{\infty}\frac{4x-1}{x^{4/5}\sqrt{(x+1)\left (
(x+1)^3+x^\frac35\right ) } }\text{d}x=0,\\
\int_{0}^{\infty}\frac{5x-1}{x^{5/6}\sqrt{(x+1)\left (
(x+1)^3+\sqrt{x}\right ) } }\text{d}x=0,\\
\int_{0}^{\infty}\frac{11x-1}{x^{11/12}\sqrt{(x+1)\left (
(x+1)^3+\sqrt[4]{x}\right ) } }\text{d}x=0.
$$
*For non-integer $a$: set $a=\frac{9}{2}$, and
$$
\int_{0}^{\infty}
\frac{7x^3-2}{x^{1/3}\sqrt{(x^3+1)\left ( (x^3+1)^3-t x^2 \right ) } }
\text{d}x\equiv0.
$$
And we deduce the final one
$$
\int_{0}^{\infty}
\frac{(7x^3-2)f\left(\frac{tx^2}{(1+x^3)^3}\right)}{x^{4/3}\sqrt{1+x^3} }
\text{d}x\equiv0.
$$
Where $f(x)$ could be almost any function that makes the integral convergent.
|
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|
How to derive the polar equation of a circle.
I have learnt that the standard equation of a circle whose centre is at (a,b) is
$$(x-a)^2 + (y-b)^2 = c^2
$$
I am trying to derive the polar equation of this circle but I am unfortunately stuck.
$$\begin{align}(x-a)^2+(y-b)^2&=c^2\\
(r\cos\theta-a)^2+(r\sin\theta-b)^2&=c^2\\
r^2\cos^2\theta-2ra\cos\theta+a^2+r^2\sin^2\theta-2rb\sin\theta+b^2&=c^2\\
r(r-2a\cos\theta-2b\sin\theta)&=c^2-a^2-b^2\end{align}$$
The following is stated in James Stewart's book on Precalculus but there is no proof stated in this book. So I am trying to reach a similar result by beginning at the standard equation of a circle.
|
You got off to a good start: Just plug in $x = r \cos\theta$ and $y = r \sin\theta$ in the standard rectangular-coordinate equation $(x-a)^2 + (y-b)^2 = c^2$
$$(r \cos\theta-a)^2 + (r \sin\theta-b)^2 = c^2$$
Or, if you need the equation expressed as a polynomial of $r$,
$$r^2 \cos^2\theta - 2ar\cos\theta + a^2 + r^2\sin^2\theta - 2br\sin\theta + b^2 = c^2$$
$$r^2(\cos^2\theta + \sin^2\theta) - 2r(a\cos\theta + b\sin\theta) + a^2 + b^2 - c^2 = 0$$
$$r^2 - 2(a\cos\theta + b\sin\theta)r + (a^2 + b^2 - c^2) = 0$$
Or, if you need to have an explicit formula for $r$ in terms of $\theta$, just use the Quadratic Formula,
$$r = \frac{2(a\cos\theta + b\sin\theta) \pm \sqrt{4(a\cos\theta + b\sin\theta)^2 - 4(a^2 + b^2 - c^2)}}{2}$$
$$r = \frac{2a\cos\theta + 2b\sin\theta \pm \sqrt{4a^2\cos^2\theta + 8ab\cos\theta\sin\theta+4b^2\sin^2\theta - 4a^2 - 4b^2 + 4c^2}}{2}$$
$$r = \frac{2a\cos\theta + 2b\sin\theta \pm 2 \sqrt{a^2(\cos^2\theta - 1) + 2ab\cos\theta\sin\theta+b^2(\sin^2\theta - 1) + c^2}}{2}$$
$$r = a\cos\theta + b\sin\theta \pm \sqrt{a^2(-\sin^2\theta) + 2ab\cos\theta\sin\theta+b^2(-\cos^2\theta) + c^2}$$
$$r = a\cos\theta + b\sin\theta \pm \sqrt{c^2 - (a\sin\theta - b\cos\theta)^2}$$
|
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Determinant of block matrices. $$
X = \begin{pmatrix}
1+b_1 & 1 & 0 & 0 & 0 & \frac{1}{a_{6}} \\
1+b_2 & 1 & 1 & 0 & 0 & -\frac{a_1}{a_6} \\
b_3 & 1 & 1 & 1 & 0 & -\frac{a_2}{a_6} \\
b_4 & 0 & 1 & 1 & 1 & -\frac{a_3}{a_6} \\
b_5 & 0 & 0 & 1 & 1 & 1-\frac{a_4}{a_6} \\
b_6 & 0 & 0 & 0 & 1 & 1-\frac{a_5}{a_6}
\end{pmatrix}$$
The Schur complement w.r.t. the first and last row/column gives
$$S = \begin{pmatrix}
1+b_1 & \frac{1}{a_6}\\
b_6 & 1 - \frac{a_5}{a_6}
\end{pmatrix}
-\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 &1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 &1 &1&1\\0&0&1&1 \end{pmatrix}^{-1}
\begin{pmatrix}1+b_2 & -\frac{a_1}{a_6} \\ b_3 & -\frac{a_2}{a_6}\\ b_4 & -\frac{a_3}{a_6}\\ b_5 & 1-\frac{a_4}{a_6}\end{pmatrix}.$$
$$S = \begin{pmatrix} b_1 - b_2 + b_4 - b_5 & - \frac{-1-a_1+a_3-a_4+a_6}{a_6}\\-1-b_2+b_3-b_5+b_6 & \frac{a_1-a_2 + a_4 - a_5}{a_6}\end{pmatrix}$$
Then $\det(X) = \det\Biggr(\begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 &1 &1&1\\0&0&1&1 \end{pmatrix}\Biggl). \det(S)$.
How matrix $S$ is obtained? I am not sure why and how to take the blocks here. How $S$ is derived? I can see that in matrix $M$ but how to approach it in matrix $X$?
Suppose $M = \begin{pmatrix} A & B \\ C & D\end{pmatrix}$. The Schur complement of $D
$ w.r.t $M$ is given by $M/D = A - B D^{-1} C$. It is easy to see when there are contiguous blocks. But when they are not contiguous, how we apply the formula? Like how we get matrix $S$. But I am not sure how to connect this with matrix $X$.
How matrix $X$ will be different from $Y$ below, by clubbing the blocks together the blocks used to construct matrix $S$.
$$
Y = \begin{pmatrix}
1+b_1 & \frac{1}{a_6} & 1 & 0 & 0 & 0 \\
b_6 & 1-\frac{a_5}{a_6} & 0 & 0 & 0 & 1 \\
1+b_2 & -\frac{a_1}{a_6} & 1 & 1 & 0 & 0 \\
b_3 & -\frac{a_2}{a_6} & 1 & 1 & 1 & 0 \\
b_4 & -\frac{a_3}{a_6} & 0 & 1 & 1 & 1\\
b_6 & 1-\frac{a_4}{a_6} & 0 & 0 & 1 & 1
\end{pmatrix}$$
|
The blocks in the Schur complement need not be contiguous. In your case, the diagonal blocks corresponds to the submatrices
$$\begin{aligned}
A &=\begin{bmatrix}X_{11} & X_{1n} \\ X_{n1} & X_{nn}\end{bmatrix}
=\begin{bmatrix}1+b_1 & \frac{1}{a_6} \\ b_6& 1 - \frac{a_5}{a_6}\end{bmatrix}
\\
D&=\begin{bmatrix}X_{2,2} & ⋯ & X_{2,n-1} \\ ⋮&&⋮\\ X_{n-1,2} & ⋯& X_{n-1,n-1}\end{bmatrix}
= \begin{bmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 1 & 0\\ 0 &1 &1&1\\0&0&1&1 \end{bmatrix}
\end{aligned}$$
This is justified as follows: Consider the matrix $M' = P^⊤MP$ obtained by permuting both the rows and columns of a $n×n$ matrix $M$ by some permutation matrix $P∈ℙ_n$. We can compute the Schur complement of a contiguous block partition of $M'$:
$$\begin{aligned} M'
&=\begin{bmatrix}A&B\\C&D\end{bmatrix}
=\begin{bmatrix}I_{p}&BD^{-1}\\0&I_{q}\end{bmatrix}
\begin{bmatrix}A-BD^{-1}C&0\\0&D\end{bmatrix}
\begin{bmatrix}I_{p}&0\\D^{-1}C&I_{q}\end{bmatrix}.
\end{aligned}$$
But since $\det(M') = \det(P^⊤ M P) = \det(M)$ we have
$$ \det(M) = \det(D) ⋅ \det(\underbrace{A-BD^{-1}C}_{=M'/D =S})$$
In your case, $P$ is the permutation matrix corresponding to the cycle $π=(2…n)$. And the size of the diagonal blocks is $2×2$ and $(n-2)×(n-2)$ respectively.
|
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|
Limit of $(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$ Find $a,b$ of:
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-ax-b) = -\frac{1}{2}$$
I can't use L'hopital, I tried multiplying by the conjugate, and solving it,
$$\lim_{x \to \infty}(\sqrt{4x^2+2x+1}-(ax+b))\cdot \frac{\sqrt{4x^2+2x+1}+(ax+b)}{\sqrt{4x^2+2x+1}+(ax+b)}$$
$$=\lim_{x \to \infty}\frac{4x^2+2x+1-(ax+b)^2}{\sqrt{4x^2+2x+1}+(ax+b)} = \lim_{x \to \infty}\frac{x^2\left(4+\frac{2}{x}+\frac{1}{x^2}-a^2-\frac{2ab}{x}-\frac{b^2}{x^2}\right)}{x^2\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
Applying limit on the numerator and denominator
$$\frac{\lim_{x \to \infty}\left(4-a^2+\frac{2-2ab}{x}+\frac{1-b^2}{x^2}\right)}{\lim_{x \to \infty}\left(\sqrt{\frac{4x^2+2x+1}{x^4}}+\frac{a}{x}+\frac{b}{x^2}\right)}$$
It can be seen that the denominator tends to $0$
|
HINT
I would start with noticing that:
\begin{align*}
\lim_{x\to\infty}(\sqrt{4x^{2} + 2x + 1} - (ax + b)) & = \lim_{x\to\infty}\frac{(4x^{2} + 2x + 1) - (ax + b)^{2}}{\sqrt{4x^{2} + 2x + 1} + (ax + b)}\\\\
& = \lim_{x\to\infty}\frac{(4 - a^{2})x^{2} + (2 - 2ab)x + 1 - b^{2}}{\sqrt{4x^{2} + 2x + 1} + ax + b}\\\\
& = \lim_{x\to\infty}\frac{(4 - a^{2})x + (2 - 2ab) + \frac{(1 - b^{2})}{x}}{\sqrt{4 + \frac{2}{x} + \frac{1}{x^{2}}} + a + \frac{b}{x}}
\end{align*}
Based on such expression, can you set the corresponding system of equations in terms of $a$ and $b$?
|
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|
What happens with indexes in this proof? I am trying to understand a proof we did in our discrete math class.
Theorem: For $n \geq 0,$ $$x^{\overline{n}} = \sum_k s(n,k) x^k,$$
where $x^\overline{n} = x (x+1)\cdots (x+n-1)$ and $s(n,k)$ is the Stirling number of the first kind.
Proof:
We prove the theorem by induction. For $n=0$ we get $1$ on both sides of the equation. So it remains to show $n-1 \to n$.
$$
\begin{align}
x^{\overline{n}} &= x (x+1) \cdots (x+n-1) \\
&= (x+n-1)\cdot x^{\overline{n-1}} \\
&= (x+n-1)\sum_ks(n-1,k)x^k \\
&= x \sum_ks(n-1,k )x^k + (n-1)\sum_ks(n-1,k)x^k \\
&= \sum_k s(n-1,k)x^{k+1} + (n-1) \sum_k s(n-1,k)x^k \\
&\stackrel{?}{=} \sum_k s(n-1,k-1)x^k + \sum_k (n-1) s(n-1,k)x^k \\
&= \sum_k (s(n-1, k-1) + (n-1)s(n-1,k))x^k \\
&= \sum_k s(n,k)x^k
\end{align}$$
The part which confuses me is marked with "$?$". It seems like there was some shift in indexes, which is further made unclear because the notation on $\sum$ simply states $k$ instead of where the $k$ is running from/to.
This is how I understand it: $k$ runs from $1$ to $n$ (or simply $0$ for $n=0$). So we have:
$$
\begin{align}
&\sum_{k=1}^{n} s(n-1,k)x^{k+1} + (n-1) \sum_{k=1}^n s(n-1,k)x^k \\
=& \sum_{k=2}^{n+1} s(n-1,k-1)x^k + \sum_{k=1}^n (n-1) s(n-1,k)x^k
\end{align}$$
So how can we add these two sums together? What happens with the indexes and what am I doing wrong?
|
We obtain using the recurrence relation
\begin{align*}
\begin{bmatrix}n\\k\end{bmatrix}&=\begin{bmatrix}n-1\\k-1\end{bmatrix}+(n-1)\begin{bmatrix}n-1\\k\end{bmatrix}\qquad\qquad n\geq k>0\tag{1}\\
\begin{bmatrix}0\\0\end{bmatrix}&=1\qquad\begin{bmatrix}n\\0\end{bmatrix}=\begin{bmatrix}0\\n\end{bmatrix}=0\qquad\qquad\qquad\ n>0
\end{align*}
of the unsigned Stirling numbers of the first kind:
\begin{align*}
\color{blue}{x^{\overline{n}}}&=x(x+1)\cdots(x+n-1)\\
&=(x+n-1)x^{\overline{n-1}}\\
&=(x+n-1)\sum_{k=0}^{n-1}\begin{bmatrix}n-1\\k\end{bmatrix}x^k\\
&=\sum_{k=0}^{n-1}\begin{bmatrix}n-1\\k\end{bmatrix}x^{k+1}
+(n-1)\sum_{k=0}^{n-1}\begin{bmatrix}n-1\\k\end{bmatrix}x^k\\
&=\sum_{k=1}^{n}\begin{bmatrix}n-1\\k-1\end{bmatrix}x^{k}
+(n-1)\sum_{k=0}^{n-1}\begin{bmatrix}n-1\\k\end{bmatrix}x^k\\
&=\begin{bmatrix}n-1\\n-1\end{bmatrix}
+\sum_{k=1}^{n-1}\left(\begin{bmatrix}n-1\\k-1\end{bmatrix}+(n-1)\begin{bmatrix}n-1\\k\end{bmatrix}\right)x^k\\
&\qquad
+(n-1)\begin{bmatrix}n-1\\0\end{bmatrix}\\
&=1+\sum_{k=1}^{n-1}\begin{bmatrix}n\\k\end{bmatrix}x^k+0\tag{$\to \mathrm{(1)}$}\\
&\,\,\color{blue}{=\sum_{k=0}^{n}\begin{bmatrix}n\\k\end{bmatrix}x^k}
\end{align*}
and the induction step follows.
|
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|
Proving that $x^n=a$, for $n>0$ an odd natural number, has exactly one real root In my school book, I read this theorem
Let $n>0$ is an odd natural number (or an odd positive integer), then the equation $$x^n=a$$ has exactly one real root.
But, the book doesn't provide a proof, only tells $x=\sqrt [n]a$.
How can I prove this theorem?
I tried to prove some special cases
$$x^3=8$$
$$(x-2)(x^2+2x+4)=0$$
$$x=2 \vee x^2+2x+4=0$$
But the Discriminant of $x^2+2x+4=0$ equals to $2^2-4×4=-12<0$. So $x=2$ is an only root. But for $x^5=32$, I got $x=2$ and $x^4+2x^3+4x^2+8x+16=0$.
I don't know how I can proceed.
|
If $n$ is even then $x^n=a$ has two real roots, $x=\pm\sqrt[n]{a}$, since $x^n=\left(x^{(n/2)}\right)^2$ is always positive, but solutions are restricted to $a\geq0$. If $n$ is odd then $x^n=a$ does not have the negative root $x=-\sqrt[n]{a}$ since $(-\sqrt[n]{a})^n=(-1)^n (\sqrt[n]{a})^n=-a\neq a$. It may also have solutions when $a$ is negative.
One can prove uniqueness of the solution to $\sqrt
[n]{a}$ using properties of the real numbers. If there is a number such that $b^n=(\sqrt[n]{a})^n=a$ then $(\sqrt[n]{a})^{-n}b^n=1$ thus $b\sqrt[n]{a}=1$ and by uniqueness of inverse $b=\sqrt[n]{a}$. (If $n$ is even you also need to take into account negative roots.)
|
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|
Solve in exact form: $x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$ ( WolframAlpha failed)
Solve the polynomial in closed form:
$$x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1=0$$
WolframAlpha obviously failed.
I tried several ways:
*
*I tried the Rational Root Thereom, but there is no rational root.
*I tried possible factorisations
$$x^6-x^5+4x^4-4x^3+4x^2-x+1= (x^2+a_1x+a_2)(x^4+a_3x^3+a_4x^2+a_5x+a_6)$$
and
$$x^6-x^5+4x^4-4x^3+4x^2-x+1= (x^3+a_1x^2+a_2x+a_3)(x^3+a_4x+a_5)$$
But, the expansions are terrible!
|
The trick here is that the coefficients are palindromic. Let $$f(x)=x^6 - x^5 + 4 x^4 - 4 x^3 + 4 x^2 - x + 1,$$ and let $$g(x)=f(x)/x^3=(x^3+1/x^3) - (x^2+1/x^2) +4(x+1/x)-4.$$ We can rewrite this in terms of $w=x+1/x$. We note that $w^2=(x+1/x)^2=x^2+1/x^2+2$, and $w^3=x^3+1/x^3 + 3(x+1/x)$, and so $x^2+1/x^2=w^2-2$ and $x^3+1/x^3=w^3-3w$.Therefore
$$\begin{split}g(x)&=(w^3-3w)-(w^2-2)+4w-4 \\
&=w^3-w^2+w-2=0\end{split}$$
Unfortunately, the roots of this cubic aren't particularly nice (as far as I can tell), but one can solve for them using the cubic formula, call them $r_1, r_2, r_3$, and then solve each of the equations of the form $x+1/x=r_i$, which yields a quadratic $x^2-r_i x +1=0$.
We remark that if the middle coefficient had been 5 instead of 4, then the resulting polynomial in w would have factored nicely and the problem would have a pretty solution.
|
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|
A Gaussian elimination problem on Matrices Considering these linear equations :-
$-3x_1 - 3x_2 + 9x_3 = 12$
$2x_1 + 2x_2 - 4x_3 = -2$
$ -2x_2 - 4x_3 = -8$
Now to solve this problem I used the Gaussian elimination method.
I converted into an augmented matrix of the same
$$ \begin{bmatrix} -3 & -3 & 9 & 12 \\ 2 & 2 & -4 & -2 \\ 0 & -2 & -4 & -8 \end{bmatrix} $$
Then I performed these operations :-
*
*$[ \frac {-1} 3 R_1 \to R_1]$
$$ \begin{bmatrix} 1 & 1 & -3 & -4 \\ 2 & 2 & -4 & -2 \\ 0 & -2 & -4 & -8 \end{bmatrix} $$
*$[ R_2 - R_1 \to R_2]$
$$ \begin{bmatrix} 1 & 1 & -3 & -4 \\ 0 & 0 & -2 & 6 \\ 0 & -2 & -4 & -8 \end{bmatrix} $$
*$[ \frac {-1} {2} R_3 \to R_3]$
$$ \begin{bmatrix} 1 & 1 & -3 & -4 \\ 0 & 0 & -2 & 6 \\ 0 & 1 & 2 & 4 \end{bmatrix} $$
*$[ R_3 + R_2 \to R_2]$
$$ \begin{bmatrix} 1 & 1 & -3 & -4 \\ 0 & 1 & 0 & 10 \\ 0 & 1 & 2 & 4 \end{bmatrix} $$
*$[ R_1 - R_2 \to R_1]$ $[ R_3 - R_2 \to R_3]$
$$ \begin{bmatrix} 1 & 0 & -3 & -14 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 2 & -6 \end{bmatrix} $$
*$[ \frac {1} {2} R_3 \to R_3]$
$$ \begin{bmatrix} 1 & 0 & -3 & -14 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 1 & -3 \end{bmatrix} $$
*And finally $[ R_1 + 3R_3 \to R_1]$
$$ \begin{bmatrix} 1 & 0 & 0 & -17 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 1 & -3 \end{bmatrix} $$
$\therefore$ $x_1 = -17$, $x_2 = 10$ and $x_3 = -3$.
But the answer is given as $x_1 = 7$ , $x_2 = -2$ and $x_3 = 3$.
Where did I go wrong?
|
You went wrong in the second step (where you should also write $[R_{2}-2R_{1}\to R_{2}]$) and again in the last step.
With problems like this, you may want to consider leaving it be for a moment and coming back some time later. That makes it easier to spot your own mistakes.
|
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|
Olympiad Math Problem About Simultaneous Equations Given the question:
*
*$x + y = 1$
*$x^2 + y^2 = 2$
*$x^5 + y^5 = ?$
After a bunch of manipulation of the above equations we find the solution to be 19/4. Yet could the above be solved by simultaneous equations?
From $1)$ we can conclude that $x = 1 - y$.
Substituting this into $2)$ we get a quadratic $2y^2 - 2y - 1 = 0$.
However, when we solve for $y$ we get two possible solutions. Hence how do we proceed from here? Why doesn't simultaneous equations work to solve the above problem?
|
We might also think of the first two equations as describing the intersections of a line with intercepts $ \ (0 \ , \ 1) \ \ $ and $ \ (1 \ , \ 0 ) \ $ and the circle of radius $ \ \sqrt2 \ $ centered on the origin. These intersections are symmetric about the line $ \ y \ = \ x \ \ . $
Since the axis intercepts are within the circle, the intersection points occur in the second and fourth quadrants, so their coordinates are $ \ (-X \ , \ Y) \ \ $ and $ \ (Y \ , \ -X) \ \ . \ $ The numerical result for $ \ x^5 + y^5 \ \ $ will thus be the same for both points.
We can solve the two equations simultaneously then as
$$ Y \ \ = \ \ 1 \ - \ X \ \ \Rightarrow \ \ Y^2 \ \ = \ \ (1 - X)^2 \ \ = \ \ 2 \ - \ X^2 \ \ \Rightarrow \ \ 2X^2 - 2X - 1 \ \ = \ \ 0 \ \ , $$
for which the solutions are $ \ X \ = \ \frac12 \pm \frac{\sqrt3}{2} \ \ , \ $ corresponding to the fourth- and second-quadrant solutions.
From this, we obtain
$$ X^2 \ \ = \ \ X \ + \ \frac12 \ \ = \ \ 1 \pm \frac{\sqrt3}{2} \ \ , \ \ Y^2 \ \ = \ \ 2 \ - \ X^2 \ \ = \ \ 1 \ \mp \frac{\sqrt3}{2} $$
$ \Rightarrow \ \ \mathbf{X^5 \ + \ Y^5} \ \ = \ \ X·(X^2)^2 \ + \ Y·(Y^2)^2 $
$$ = \ \ \left( \frac12 \pm \frac{\sqrt3}{2} \right)·\left( 1 \pm \frac{\sqrt3}{2} \right)^2 \ + \ \left( \frac12 \mp \frac{\sqrt3}{2} \right)·\left( 1 \mp \frac{\sqrt3}{2} \right)^2 $$ $$ = \ \ \left( \frac12 \pm \frac{\sqrt3}{2} \right)·\left( \frac74 \pm \sqrt3 \right) \ + \ \left( \frac12 \mp \frac{\sqrt3}{2} \right)·\left( \frac74 \mp \sqrt3 \right) $$
[all of the "middle terms" involving $ \ \sqrt3 \ $ cancel!]
$$ = \ \ \frac12·\frac74 \ + \ \frac32 \ + \ \frac12·\frac74 \ + \ \frac32 \ \mathbf{= \ \frac{19}{4} } \ \ . $$
The green curve in the graph represents this last equation.
|
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|
prove $\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\ge \frac{2}{n}-\frac{1}{n^2}$
Let $x_1,x_2,\cdots,x_n$ be $n$ non-negative numbers such that
$x_1+x_2+\cdots+x_n=1$. Show $$\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j}\ge \frac{2}{n}-\frac{1}{n^2}.$$
We may consider using Cauchy. That is
$$\sum_{i=1}^n\sum_{j=1}^i x_j\cdot \sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i
x_j}\ge \left(\sum_{i=1}^nx_i\right)^2=1.$$
How to go on?
|
We'll assume $x_1>0,$ since the left hand side is undefined if $x_1=0.$ But if ww treat the first term as zero when $x_1=0,$ we can just use that the right side is decreasing on $n.$
Try induction on $n.$ When $n=1,$ you have $x_1=1,$ the inequality becomes $1\geq 1.$
If true for $n,$ take $x_1,\dots,x_{n+1}.$
Since $x_{1}>0,$ then the $x_{n+1}<1.$
Let $y_i=\frac{x_i}{1-x_{n+1}}$ for $i=1,\dots,n.$
Then $\sum_1^n y_i=1,$ so we have:
$$\frac2n-\frac1{n^2}\leq \sum_{i=1}^n \frac{y_i^2}{\sum_{j=1}^i y_j}=\frac1{1-x_{n+1}}\sum_{i=1}^n\frac{x_i^2}{\sum_{j=1}^i x_j}$$
So $$\begin{align}\sum_{i=1}^{n+1}\frac{x_i^2}{\sum_{j=1}^i x_j}&\geq (1-x_{n+1})\left(\frac2n-\frac1{n^2}\right)+x_{n+1}^2\\\end{align}$$
In general, the minimum value for $f(x)=(1-x)a+x^2$ when $0<a\leq 1$ and $x\in [0,1]$ is at $x=a/2$ and the value is $a-\frac{a^2}{4}=1-(1-a/2)^2.$
So this is the recursion for the absolute minimum value of the left hand side:
$$m_1=1, m_{n+1}=\inf_{m\in[m_n,1]}1-(1-m/2)^2=1-(1-m_n/2)^2.$$
Because $1-(1-u/2)^2=u-\frac{u^2}4$ is increasing on $[0,1].$
We have, if $m_n\geq \frac2n-\frac1{n^2}$ then:
$$\begin{align}m_{n+1}&\geq 1-\left(1-\frac 1n+\frac1{2n^2}\right)^2\end{align}$$
Now, $$\frac2{n+1}-\frac1{(n+1)^2}=1-\left(1-\frac1{n+1}\right)^2$$
So you need $$\frac1{n+1}\leq \frac1{n}-\frac1{2n^2}=\frac{2n-1}{2n^2}$$
or $$2n^2\leq (n+1)(2n-1)=2n^2+n-1$$ which is obviously true for $n\geq1.$
|
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|
How to prove you found ALL limit points: $x_n = (-1)^n + 1/n + 2\sin(\frac{n\pi}{2})$ How does one find all limit points? How do you prove you're not missing a limit point? For example, if $x_1,x_2,\cdots$ converges to $x$ and $y_1,y_2,\cdots$ converges to $y$, and we consider the shuffled sequence, obvious limit points are $x$ and $y$ but how can we prove those are the only two limit points?
For my problem, we have $$x_n = (-1)^n + 1/n + 2\sin\left(\frac{n\pi}{2}\right)$$ An observation is that for $n$ even, the third term is 0, so if we remove all the odd $n$, the resulting subsequence is defined such that $x_j = 1 + 1/2j$ which converges to 1. So one limiting point is 1. Something to note about this subsequence, moreover, is that its $\sup$ is $3/2$. If we don't find a larger $\sup$ in the subsequence we removed from the original sequence, then $3/2$ is the $\sup$ for the entire sequence. But we realize that $x_1 = -1 + 1 + 2 = 2$, which is the $\sup$ for the whole sequence. Now, let's look at the subsequence we removed. First, observe the first term is always -1, the second term converges to 0, and the third term bounces back and forth between 2 and -2. Hence, we have two limit points: $-1 + 0 + 2 = 1$ and $-1 + 0 - 2 = -3$. The $\inf$, then, is $-3$, as well as the $\liminf$, and the $\limsup = 1$.
How can I prove that I pointed out all limit points?
|
We have that, starting from $n=4k$
*
*$x_{4k} = (-1)^{4k} + \frac1{4k} + 2\sin\left(\frac{4k\pi}{2}\right)=1+\frac1{4k}+0=1+\frac1{4k}$
*$x_{4k+1} = (-1)^{4k+1} + \frac1{4k+1} + 2\sin\left(\frac{4k\pi}{2}+\frac \pi 2\right)=-1+\frac1{4k+1}+2=1+\frac1{4k+1}$
*$x_{4k+2} = (-1)^{4k+2} + \frac1{4k+2} + 2\sin\left(\frac{4k\pi}{2}+ \pi \right)=1+\frac1{4k+2}+0=1+\frac1{4k+2}$
*$x_{4k+3} = (-1)^{4k+3} + \frac1{4k+3} + 2\sin\left(\frac{4k\pi}{2}+\frac {3\pi} 2\right)=-1+\frac1{4k+3}-2=-3+\frac1{4k+3}$
and the same pattern repeats, indeed $4k+4=4(k+1)$, $4k+5=4(k+1)+1$, $4k+6=4(k+1)+2$, $4k+7=4(k+1)+3$ and so on, therefore we can take $k\to \infty$ and determine the limit points.
|
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|
Find all real values of x and y such that $\left(x+yi\right)^3$ is real and greater than $8$, represent these values in the xy-plane My solution
$\left(x+yi\right)^3>8$
$\left(x^3-3xy^2-8\right)+\left(3x^2y-y^3\right)i>0$
and now I'm stuck with these equations
$\left(x^3-3xy^2-8\right)>0,\left(3x^2y-y^3\right)>0$
|
We need the cube is real then
$$3x^2y-y^3 =y(3x^2-y^2)=0$$
(and not $3x^2y-y^3>0$) then we have three cases
*
*$y=0 \land x\neq 0 \implies x^3-8>0 \implies \begin{cases}x>2\\y=0 \end{cases}$
*$y\neq 0 \land 3x^2=y^2\neq 0 \implies x^3-9x^3-8>0 \implies \begin{cases}x<-1\\y=\pm \sqrt 3 x \end{cases}$
*$y=0 \land x= 0 \implies -8>0$
As an alternative by polar coordinates
$$z=re^{i\theta} \implies z^3=r^3e^{i3\theta}$$
that is
$$r^3\cos (3\theta)>8 \land \sin (3\theta)=0$$
therefore we have
*
*$3\theta = 2k\pi \implies \theta = \frac 2 3 k\pi$, with $k=0,1,2$
*$r>2$
which indicates that solutions are three oriented lines out of the circle $x^2+y^2=4$.
|
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|
Pairs $(x,y)$ such that $3x-1$ is divisible by $y$ and $3y-1$ is divisible by $x$. How many possible pairs of integers $x$ and $y$ exist such that $3x-1$ is divisible by $y$ and $3y-1$ is divisible by $x$?
Since $y| 3x-1$, there exists some $k$ such that $3x-1 = ky$.
On the other hand since $x| 3y-1$ we have $3y-1 = lx$ for some $l$.
i.e. $9y-3 = l(3x) = l(ky+1)$ which implies $(9-kl)y = l+3$.
I am not able to procced further.
Any hints will be helpful. Thanks.
|
Multiplying the two divisibility relations together gives
$$xy\mid(3x-1)(3y-1)=9xy-3(x+y)+1\implies xy\mid3(x+y)-1$$
Now neither $x$ nor $y$ can be zero, since the original relations would then force $3x-1=3y-1=0$ which is impossible for integral $x,y$, so the quotient $\frac{3(x+y)-1}{xy}$ exists and is an integer with absolute value at least $1$. If we then assume without loss of generality that $|x|\ge|y|$, $-6\le y\le5$:
This means we can get all solutions by checking each possibility for $y$ in turn, which is easy since $x$ must then be a factor of the fixed integer $3y-1$. The solutions with $|x|\ge|y|$ are
$$(x,y)\in\{(-8, -5), (-13, -4), (-7, -2), (7, -2), (-4, -1), (-2, -1), (-1, -1), (1, -1), (2, -1), (4, -1), (-2, 1), (-1, 1), (1, 1), (2, 1), (-5, 2), (5, 2), (11, 4), (7, 5)\}$$
|
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|
How to differentiate $y=\sqrt{\frac{1-x}{1+x}}$? It is an example question from "Calculus Made Easy" by Silvanus Thompson (page 68-69). He gets to the answer $$\frac{dy}{dx}=-\frac{1}{(1+x)\sqrt{1-x^2}}$$ The differentiation bit of the question is straightforward, but I'm having trouble simplifying it to get the exact answer. My working is the following:
$$y=\sqrt{\frac{1-x}{1+x}}$$
This can be written as $$y=\frac{(1-x)^\frac{1}{2}}{(1+x)^\frac{1}{2}}$$
Using the quotient rule we get $$\frac{dy}{dx}=\frac{(1+x)^\frac12\frac{d(1-x)^\frac12}{dx}-(1-x)^\frac12\frac{d(1+x)^\frac12}{dx}}{1+x}$$
Hence $$\frac{dy}{dx}=-\frac{(1+x)^\frac12}{2(1+x)\sqrt{1-x}}-\frac{(1-x)^\frac12}{2(1+x)\sqrt{1+x}}$$
$$=-\frac{1}{2\sqrt{1+x}\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{(1+x)^3}}$$
What is the next step?
|
Alternative approach:
Start out differently: Let
$$A = \frac{1-x}{1+x}, ~B = A^{(1/2)}.$$
Then
$$\frac{d [B]}{dx} = \frac{1}{2} \times {A}^{-(1/2)} \times \frac{dA}{dx}. \tag1 $$
and
$$\frac{1}{2} \times {A}^{-(1/2)} = \frac{1}{2} \times \sqrt{\frac{1+x}{1-x}}.\tag2 $$
So, what remains is to compute $~\displaystyle \frac{dA}{dx},~$ combine that with (2) above, and then try to simplify it.
$$\frac{dA}{dx} = \frac{d\left[\frac{1-x}{1+x}\right]}{dx} $$
$$= \frac{[(1 + x)(-1)] - [(1-x)(1)]}{(1 + x)^2} = \frac{-2}{(1 + x)^2} \tag3$$
Combining (2) and (3) results in
$$(-1) \times (1+x)^{(1/2)} \times (1-x)^{(-1/2)} \times (1+x)^{(-2)}$$
$$ = (-1) \times (1-x)^{(-1/2)} \times (1 + x)^{(-3/2)}$$
$$= (-1) (1+x)^{(-1)} \times [(1+x)(1 - x)]^{(-1/2)}$$
$$= \frac{-1}{1+x} \times \frac{1}{\sqrt{1 - x^2}}.$$
|
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|
For all $x \in \Bbb R$ there exists a $y \in (-\infty, 1)$ such that $3yx^2+y-x=0$ I have this question:
Prove that for all real numbers $x$ there exist a number $y$ in the interval $(-\infty, 1)$ such that $$3x^2y+y-x=0$$
I proved that the range of the function
$$y = \frac{x}{3x^2 + 1}$$
is $(-1/2\sqrt{3},1/2\sqrt{3})$.
But the teacher said I should use another way.
I didn't find any.
|
The universe is the set of real numbers.
$\forall x\exists y<1:3x^2y+y-x=0$
$\iff\forall x\exists y<1:3x^2y+y=x$
$\iff\forall x\exists y<1:y(3x^2+1)=x$
$\iff\forall x\exists y<1:y=\frac{x}{3x^2+1}$, because we know that $3x^2+1\not=0$
$\iff\forall x\exists y(y<1\land y=\frac{x}{3x^2+1})$
$\iff\forall x\exists y(y=\frac{x}{3x^2+1}\land y<1)$
$\iff\forall x\exists y(y=\frac{x}{3x^2+1}\land \frac{x}{3x^2+1}<1)$
$\iff\forall x(\exists y(y=\frac{x}{3x^2+1})\land \frac{x}{3x^2+1}<1)$
$\iff\forall x\frac{x}{3x^2+1}<1$
$\iff\forall x:x<3x^2+1$, because we know that $3x^2+1>0$
$\iff\forall x:3x^2-x+1>0$
$\iff\forall x:x^2-\frac{1}{3}x+\frac{1}{3}>0$
$\iff\forall x:(x-\frac{1}{6})^2-(\frac{1}{6})^2+\frac{1}{3}>0$
$\iff\forall x:(x-\frac{1}{6})^2-\frac{1}{36}+\frac{1}{3}>0$
$\iff\forall x:(x-\frac{1}{6})^2>\frac{-11}{36}$, which is true, because the square of every real number is positive.
|
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|
Simplify $\frac{\sin(2022) + \sin(2022+\alpha)}{\cos(2022)-\cos(2022+\alpha)}$ as much as possible
Let $\alpha\in\mathbb{R}$ be so that $\cos\alpha\neq 1$. Simplify as much as possible $\dfrac{\sin(2022) + \sin(2022+\alpha)}{\cos(2022)-\cos(2022+\alpha)}$. To clarify, if for a finite number of values of $\alpha,$ the expression takes a numerical value, find that numeric value for each of those values of $\alpha$. If not, find a function of $\alpha$ that is as simplified as possible.
Clearly $\alpha$ is not an integer multiple of $2\pi.$ Let $f(\alpha)$ denote the expression that needs to be simplified. We have that $f(\alpha + 2k\pi) = f(\alpha)$ for any integer $k$, so we may assume WLOG that $0 < \alpha < 2\pi$. It might be easier to generalize and replace $2022$ with $m$. Perhaps we only need the fact that $2022$ is an integer? Using the addition law, we get that $f(\alpha) = \dfrac{\sin(m)(1+ \cos\alpha) + \sin(\alpha)\cos m}{\cos(m)(1 - \cos(\alpha)) + \sin(m)\sin(\alpha)}$. I'm not sure if Euler's formula will help simplify $f(\alpha)$. Multiply the denominator and numerator of $f(\alpha)$ by $\cos m(1+\cos \alpha) - \sin m \sin \alpha$. The denominator becomes $\cos^2 m \sin^2 \alpha + 2\sin m \cos m \sin \alpha \cos \alpha - \sin^2m\sin^2\alpha = \sin\alpha( \cos(2m) \sin \alpha + \sin(2m)\cos \alpha) = \sin \alpha \sin(2m+\alpha)$. The numerator becomes $\sin m \cos m (1+\cos \alpha)^2 -\sin^2 m \sin \alpha(1+\cos \alpha) + \cos^2 m \sin \alpha(1+\cos \alpha) - \sin m \cos m \sin^2 \alpha = \cos(2m) \sin (\alpha) ( 1+\cos \alpha) + \sin m \cos m(2\cos \alpha + 2\cos^2 \alpha) = (1+\cos \alpha)(\cos (2m) \sin (\alpha) + \sin(2m)\cos (\alpha)) = (1+\cos\alpha)\sin(2m+\alpha).$ Hence $f(\alpha) = \dfrac{1+\cos \alpha}{\sin \alpha}.$ Note that $m$ could've actually been any real number by the same proof as above.
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Let $x = 2022$ and $y = 2022 + \alpha$. Observe that using sum-to-product identities
\begin{align*}
\frac{\sin(x) + \sin(y)}{\cos(x) - \cos(y)} &= \frac{2\sin\left(\frac{x + y}{2}\right) \cos\left(\frac{x - y}{2}\right)}{-2\sin\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right)}\\
&=\cot\left (\frac{y - x}{2} \right )\\
&= \cot\left ( \frac{\alpha}{2}\right )
\end{align*}
|
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|
Check if this series is convergent or not I've been going crazy for a long time in determining if this series converges or diverges. Most likely it converges.
$\displaystyle \sum_{n=1}^\infty \left(\frac{n}{2}\, \sin\frac{1}{n}\right)^\frac{n^2+1}{n+2}$
I am stuck in the necessary condition.
I tried to solve in this way :
$\displaystyle \frac{1}{2}\lim_{n \to \infty} \left(\frac{\sin\frac{1}{n}}{\frac{1}{n}}\right)^\frac{n^2+1}{n+2} = \frac{1}{2}\lim_{n \to \infty} (1)^\infty$
To solve this indeterminate form, I tried to use the exponential in this way
$\displaystyle \frac{1}{2}\lim_{n \to \infty} \left(e^{\frac{n^2+1}{n+2}log \left(\frac{sin \frac{1}{n}}{\frac{1}{n}}\right)}\right)$
In this way I always get an indeterminate form
$\displaystyle \frac{1}{2}\lim_{n \to \infty} (e^{\infty *0})$
could you kindly give me support for the resolution of the character of this series?
|
Using the ratio test
$$a_n
=
\left(
\frac{n}{2}
\,
\sin\Bigl(\frac{1}{n}\Bigr)
\right)
^{\large{\frac{n^2+1}{n+2}}}\implies
\log(a_n)=\frac{n^2+1}{n+2}\log\left(
\frac{n}{2}
\,
\sin\Bigl(\frac{1}{n}\Bigr)
\right)$$
For large $n$, compose Taylor expansion
$$\log(a_n)=\frac{n^2+1}{n+2}\bigg[-\log (2)-\frac{1}{6 n^2}+O\left(\frac{1}{n^4}\right) \bigg]$$
$$\log(a_n)=n \log (2)+2 \log (2)+\frac{-\frac{1}{6}-5 \log (2)}{n}+\frac{\frac{1}{3}+10 \log (2)}{n^2}+O\left(\frac{1}{n^3}\right)$$ Replace $n$ by $n+1$ and continue with long division
$$\log(a_{n+1})-\log(a_n)=-\log (2)+\frac{\frac{1}{6}+5 \log (2)}{n^2}+O\left(\frac{1}{n^3}\right)$$
Now
$$\frac {a_{n+1}}{a_n}=e^{\log(a_{n+1})-\log(a_n)}=\frac{1}{2}+\frac{\frac{1}{6}+5 \log (2)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
|
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|
Find numbers divisible by 6 Find the number of all $n$, $1 \leq n \leq 25$ such that $n^2+15n+122$ is divisible by 6.
My attempt. We know that:
\begin{align*}
n^2+15n+122 & \equiv n^2+3n+2 \pmod{6}
\end{align*}
But $n^2+3n+2=(n+1)(n+2)$, then $n^2+15n+122 \equiv (n+1)(n+2) \pmod{6}$, now we have
\begin{align*}
n(n^2+15n+122) & \equiv n(n+1)(n+2)\pmod{6} \\
n^3+15n^2+122n & \equiv 0 \pmod{6}
\end{align*}
I have done this and I think I have complicated the problem even more.
|
If $n$ must be an integer: HINT: First, $n^2+15n+122$ is even for all integers $n$. Then $n^2+15n+122$ will be divisble by $6$ iff $n^2+15n+122$ is a muliple of $3$. So what integers $n \pmod 3$ is $n^2+15n+122$ a multiple of $3$?
You can tell working $\pmod 3$, that $n^2+15n+122$ is divisible by $3$ for every integer $n$ that is not a multiple of 3. [Indeed, $$n^2+15n+122 \pmod 3 = n^2+2 $$ $$= 0 \ \text{ if $n \pmod 3 \not = 0$}.] $$ Thus, $n^2+15n+122$ is divisible by $3$ iff $n \pmod 3$ is in $\{1,2\}$.
Thus, from this and the top paragraph, $n^2+15n+122$ is divisible by $6$ iff $n \pmod 3$ is in $\{1,2\}$.
If $n$ is allowed to be any real number and not just integral: Note that, as $n^2+15n+122$ is continuous on $[0,25]$, the function $n^2+15n+122$ in $[0,25]$ takes all values in $[122,1122]$. There are precisely $166$ multiples of $6$ in $[122,1122]$. As $n^2+15n+122$ is strictly increasing on $[0,25]$, it follows that there is at most one real number $n \in [0,25]$ such that $n^2+15n+122 = y$.
So for each of these $166$ multiples $y$ of $6$ in $[122,1122]$ there is exactly one real number $n \in [0,25]$ such that $n^2+15n+122 = y$.
|
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|
System of first order differential equation. Consider the linear system $y’=Ay+h$ where
$$
A=\begin{bmatrix}
1 & 1\\
4 & -2
\end{bmatrix}
$$and $$h=\begin{bmatrix}
3t+1\\
2t+5
\end{bmatrix}$$ Suppose $y(t)$ is a solution such that $$\lim_{t\to\infty}\frac{y(t)}{t}=k\in\mathbb R^2$$ What is the value of $k?$
$1$. $\begin{bmatrix}
\frac{-4}{3}\\
\frac{-5}{3}
\end{bmatrix}$.
$2$. $\begin{bmatrix}
\frac{4}{3}\\
\frac{-5}{3}
\end{bmatrix}$.
$3$. $\begin{bmatrix}
\frac{2}{3}\\
\frac{-5}{3}
\end{bmatrix}$.
$4$. $\begin{bmatrix}
\frac{-2}{3}\\
\frac{-5}{3}
\end{bmatrix}$.
I find eigen value of corresponding homogeneous system as $2$ and $-3$ and corresponding eigen vectors as $\begin{bmatrix}
1\\
1
\end{bmatrix}$ and $\begin{bmatrix}
1\\
-4
\end{bmatrix}$. Therefore general solution of corresponding homogeneous differential equation is given by $y_c=\Phi(x)c=\begin{bmatrix}
e^{2t} & e^{-3t}\\
e^{2t} & -4e^{-3t}
\end{bmatrix}c$, for arbitrary constant $c$.
Now as I know that by using variation of parameter general solution is given by $$y=\Phi(t)c+\Phi(t)\int {\Phi(x)}^{-1}h(x)dx= \begin{bmatrix}
e^{2t} & e^{-3t}\\
e^{2t} & -4e^{-3t}
\end{bmatrix}c+ \begin{bmatrix}
e^{2t} & e^{-3t}\\
e^{2t} & -4e^{-3t}
\end{bmatrix}\int {\begin{bmatrix}
e^{2x} & e^{-3x}\\
e^{2x} & -4e^{-3x}
\end{bmatrix}}^{-1} \begin{bmatrix}
3x+1\\
2x+5
\end{bmatrix}dx$$ Now I am unable to reach at final answer . Please help. Thank you.
|
We are given the linear system
$$y'(t) = Ay(t) + h(t) =\begin{bmatrix}1 & 1\\4 & -2\end{bmatrix}y(t) + \begin{bmatrix}3t+1\\2t+5\end{bmatrix}$$
I will use Example $2$ of these notes as a guide.
We find the eigenvalues and eigenvectors as
$$\lambda_1 = -3, v_1 = \begin{bmatrix} -1\\4 \end{bmatrix}\\ \lambda_2 = 2, v_2 = \begin{bmatrix} 1\\1 \end{bmatrix}$$
We can write the complimentary solution as
$$y_c(t) = c_1e^{-3t} \begin{bmatrix} -1\\4 \end{bmatrix} + c_2 e^{2t}\begin{bmatrix} 1\\1 \end{bmatrix}$$
From this we have $$X = \begin{bmatrix}
-e^{-3 t} & e^{2 t} \\
4 e^{-3 t} & e^{2 t} \\
\end{bmatrix}$$
The inverse is
$$X^{-1} = \begin{bmatrix}
-\dfrac{e^{3 t}}{5} & \dfrac{e^{3 t}}{5} \\
\dfrac{4 e^{-2 t}}{5} & \dfrac{e^{-2 t}}{5} \\
\end{bmatrix}$$
The multiplication inside the integral is
$$X^{-1} h =\begin{bmatrix}
\dfrac{4 e^{3 t}}{5}-\dfrac{1}{5} e^{3 t} t \\
\dfrac{14}{5} e^{-2 t} t+\dfrac{9 e^{-2 t}}{5} \\
\end{bmatrix}$$
Now do the integral
$$\int X^{-1}h dt = \int \begin{bmatrix}
\dfrac{4 e^{3 t}}{5}-\dfrac{1}{5} e^{3 t} t \\
\dfrac{14}{5} e^{-2 t} t+\dfrac{9 e^{-2 t}}{5} \\
\end{bmatrix}~dt = \begin{bmatrix}
\dfrac{14 e^{3 t}}{135}-\dfrac{1}{45} e^{3 t} t \\
\dfrac{7}{10} e^{-2 t} t+\dfrac{23 e^{-2 t}}{20} \\
\end{bmatrix}$$
Now we can get the particular solution
$$y_p(t) = X \int X^{-1}h~dt = \begin{bmatrix}
-\dfrac{4 t}{3}-\dfrac{17}{9} \\
-\dfrac{5 t}{3}-\dfrac{4}{9} \\
\end{bmatrix}$$
Next, use $y_p(t)$ (a solution of the DEQ) and find
$$\lim_{t\to\infty}\frac{y_p(t)}{t}=\lim_{t\to\infty}\begin{bmatrix}
-\dfrac{4 }{3}-\dfrac{17}{9t} \\
-\dfrac{5}{3}-\dfrac{4}{9t} \\
\end{bmatrix}$$
You arrive at choice $1.$
|
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|
Complex contour integral using Cauchy integral formula I am trying to get integral in contour $C=\left \{ 3\cos(t)+2i\sin(t) : 0\leq t\leq 2\pi \right \}$
$$\oint_{C} \frac{z}{(z+1)(z-1)^2}dz$$
What I tried was using partial fractions and integrate them separately
$$\oint_{C} \frac{z}{(z+1)(z-1)^2}dz =\oint_{C} \frac{1}{4(z-1)}dz +\oint_{C} \frac{1}{-4(z+1)}dz+\oint_{C} \frac{1}{2(z-1)^2}dz$$
For $\int_{C}^{} \frac{1}{4(z+1)}dz$ +$\int_{C}^{} \frac{1}{-4(z-1)}dz$, using Cauchy's integral fomula they are zero.
For $\int_{C}^{} \frac{1}{2(z-1)^2}dz$, I tried to use Cauchy's integral formula like:
$$\frac{1}{2}\int_{C}^{} \frac{\frac{1}{(z-1)}}{(z-1)}dz$$
but can't solve it... What am I missing??
|
Using the Residue at Infinity
You need not evaluate the residues at $-1$ and $1$. Rather, note that the Residue at $\infty$ is given by
$$\begin{align}
\text{Res}\left(\frac{z}{(z+1)(z-1)^2},z=\infty\right)&=\text{Res}\left(-\frac1{z^2}\frac{1/z}{(1/z+1)(1/z-1)^2},z=0\right)\\\\
&=-\text{Res}\left(\frac{1}{(z+1)(z-1)^2},z=0\right)\\\\
&=0\tag1
\end{align}$$
Therefore, we find using $(1)$ that
$$\begin{align}
\oint_{C}\frac{z}{(z+1)(z-1)^2}\,dz&=-2\pi i \text{Res}\left(\frac{z}{(z+1)(z-1)^2},z=\infty\right)\\\\
&=0
\end{align}$$
And we are done!
Exploiting Cauchy's Integral Theorem
Alternatively, we note that the the integrand is analytic in the region $|z|>1$. Therefore, any integral over a contour that contains both poles at $\pm1$ will have the same value as any other such integral. Therefore, we assert that
$$\oint_{C}\frac{z}{(z+1)(z-1)^2}\,dz=\oint_{|z|=R>1}\frac{z}{(z+1)(z-1)^2}\,dz \tag2$$
Now, letting $R\to \infty$, the right-hand side approaches $0$. And we conclude that the value of the integral of interest is also $0$, as expected!
|
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|
Finding the closed formula of a generating function for flower selection Find the generating function for the number of ways to select $n$ flowers from a boundless supply of dandelion, tulip, rose and blossoms if we have at least one flower of each type, the number of dandelions is less than 4, the number of tulips is a multiple of 5, and the number of roses are a multiple of 2? Find generating function in CLOSED form.
So we have:
$$
F_1(x) = x + x^2 + \dots = \frac{x}{1-x}
$$
$$
F_2(x) = x + x^2 + x^3 = \frac{x(1-x^3)}{1-x}
$$
$$
F_3(x) = x^5 + x^{10} + \dots = \frac{x^5}{1-x^5}
$$
$$
F_4(x) = x^2 + x^4 + \dots = \frac{x^2}{1-x^2}
$$
I'm not sure what it means to take it closed form from here? Do we just multiply them together and we are done?
|
Yes, the desired generating function is the product
$$\frac{x^9 (1-x^3)}{(1 - x)^2(1-x^2) (1-x^5)}=\frac{x^9 (1+x+x^2)}{(1+x) (1 - x)^3 (1+x+x^2+x^3+x^4)}.$$
|
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|
Find an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $A=QDQ^T$
Let $$A=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$$
Find an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $A=QDQ^T$.
I already got these Eigenvalues $D=\begin{bmatrix}0&0&0\\0&0&0\\0&0&3\end{bmatrix}$and Eigenvectors $P=\begin{bmatrix}-1&-1&1\\1&0&1\\0&1&1\end{bmatrix}$
\begin{align}
\langle v_1,v_2\rangle &= \quad \!1+0+0=1 \\
\langle v_1,v_3\rangle &=-1+1+0=0 \\
\langle v_2,v_3\rangle &=-1+0+1=0
\end{align}
$\because \langle v_1,v_3\rangle= \langle v_2,v_3\rangle=0 \space \therefore A=A^T$
$\because \langle v_1,v_2\rangle=1+0+0=1 \space \therefore v_1\perp v_2$
But I have no idea what is the next step.
|
I try to give my input from a methodical approach. From your above computations, we can see that:
For $\lambda = 0$, these eigenvectors $\{v_1,v_2\}$ span the eigenspace $E_{\lambda=0}$:
\begin{Bmatrix}
\begin{bmatrix}
-1\\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
-1\\
0 \\
1
\end{bmatrix}
\end{Bmatrix}
and for $\lambda = 3$, this eigenvector $\{v_3\}$ span the eigenspace $E_{\lambda=3}$:
\begin{Bmatrix}
\begin{bmatrix}
1\\
1 \\
1
\end{bmatrix}
\end{Bmatrix}
These eigenvectors are linearly independent. However, there are repeated eigenvalues for $\lambda = 0$. Next, we check $\langle v_1, v_2 \rangle$. We observe that these eigenvectors are not orthogonal, as what you have shown $\langle v_1, v_2 \rangle =1\neq0$.
Hence, we can use Gram-Schmidt orthogonalization to obtain our orthonormal set.
For $E_{\lambda=0}$, we find our orthonormal vectors $\{e_1, e_2\}$:
\begin{align}
u_1 & = v_1 \\
e_1 & = \frac{u_1}{||u_1||} = \frac{1}{\sqrt2}
\begin{bmatrix}
-1\\
1 \\
0
\end{bmatrix} \\ \DeclareMathOperator{\proj}{proj}
& \\
u_2 & = v_2 - \proj_{u_1} (v_2) \\
& = v_2 - \langle v_2, u_1\rangle u_1\\
& =
\begin{bmatrix}
-1\\
0 \\
1
\end{bmatrix} -
\left\langle
\begin{bmatrix}
-1\\
0 \\
1
\end{bmatrix},
\frac{1}{\sqrt2}
\begin{bmatrix}
-1\\
1 \\
0
\end{bmatrix}
\right\rangle
\frac{1}{\sqrt2}
\begin{bmatrix}
-1\\
1 \\
0
\end{bmatrix} \\
& =
\begin{bmatrix}
-1/2\\
-1/2 \\
1
\end{bmatrix} = \frac{1}{2}
\begin{bmatrix}
-1\\
-1 \\
2
\end{bmatrix} \\
e_2 & = \frac{u_2}{||u_2||} = \frac{1}{\sqrt6}
\begin{bmatrix}
-1 \\
-1 \\
2
\end{bmatrix} \\
\end{align}
For $E_{\lambda=3}$:
\begin{align}
u_3 & = v_3 \\
e_3 & = \frac{u_3}{||u_3||} =
\frac{1}{\sqrt 3}
\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}
\\
\end{align}
Hence,
$$
\begin{align}
Q & = [e_1 \space e_2 \space e_3]
=
\begin{bmatrix}
-1/ \sqrt2 & -1/ \sqrt6 & 1/ \sqrt3\\
1/ \sqrt2 & -1/ \sqrt6 & 1/ \sqrt3\\
0/ \sqrt2 & 2/ \sqrt6 & 1/ \sqrt3\\
\end{bmatrix}: \\
D &
=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 3 \\
\end{bmatrix}
\end{align}
$$
|
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|
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$ by partial fractions
Integration of $\int \frac{x^2}{(x-3)(x+2)^2}$
I know that we can use substitution to make the expression simplier before solving it, but I am trying to solve this by using partial fractions only.
$\frac{x^2}{(x-3)(x+2)^2} = \frac{A}{x-3} + \frac{B}{x+2} + \frac{C}{(x+2)^2} = \frac{A(x+2)^2+B(x-3)(x+2) + C(x-3)}{(x-3)(x+2)} = \frac{Ax^2 + 4Ax + 4A + Bx^2 - Bx - 6B +Cx - 3C}{(x-3)(x+2)}$
Looking at the numerator: $x^2(A+B) + x(4A-B+C) + (4A-6B-3C)$
So, comparing coefficients:
$A+B=1$,
$4A-B+C=0$
$4A-6B-3C=0$
I am struggling to solve these 3 equations to find A,B,C
|
You have\begin{align}-4&=0-4\\&=4A-B+C-4(A+B)\\&=-5B+C\end{align}and\begin{align}-4&=0-4\\&=4A-6B-3C-4(A+B)\\&=-10B-3C.\end{align}And now, you have\begin{align}4&=(-2)\times(-4)-4\\&=(-2)(-5B+C)-10B-3C\\&=-5C,\end{align}and therefore $C=-\frac45$. Now, since $-5B+C=-4$, you get that $B=\frac{16}{25}$. And, since $A+B=1$, $A=\frac9{25}$.
|
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|
how do I evaluate this definite integral? $I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$ $$I=\int_3^5{\left( \frac{x^2+x}{\sqrt[3]{(x-3)(5-x)}} \right)dx}$$
I tried using substitution and some common definite integration results like
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx \\
\int_a^bf(x)dx = (b-a)\int_0^1f((b-a)x+a)dx$$
and done substitutions like
$$x=3\cos^2(t)+5\sin^2(t)$$
but it just becomes messy.
I have tried adding the different results hoping that stuff would cancel out or simplify,
but I'm not able to figure it out.
Please help. Thanks!
|
Noticing that $(x-3)(5-x)=1 -(x-4)^2$, we use the substitution $x-4=\sin \theta$ to transform the integral into
$$
\begin{aligned}
I &=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(\sin \theta+4)(\sin \theta+5)}{\cos^{\frac 2 3} \theta} \cos \theta d \theta \\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\sin ^2 \theta+9 \sin \theta+20\right) \cos ^{\frac{1}{3}} \theta d \theta \\
&=2 \int_0^{\frac{\pi}{2}} \sin ^2 \theta \cos ^{\frac{1}{3} }\theta+40 \int_0^{\frac{\pi}{2}} \cos ^{\frac{1}{3} }\theta d \theta \\
&=B\left(\frac{3}{2}, \frac{2}{3}\right)+20 B\left(\frac{1}{2}, \frac{2}{3}\right) \\
&=\frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{13}{6}\right)}+\frac{20\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{7}{6}\right)}\\&= \frac{143 \sqrt{\pi}}{7} \cdot \frac{\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{7}{6}\right)},
\end{aligned}
$$
where the last result using the facts $\Gamma(z+1)=z\Gamma(z)$ and $\Gamma(\frac 12)=\sqrt{\pi}$.
|
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|
Proving that $\sum_{i=0}^{n/2} (-1)^i \frac{n}{n-i} {n-i \choose i} = 2\cos(\frac{\pi n}{3})$ I was investigating the Girard-Waring identity, specifically for two variables:
$$x_1^n + x_2^n = \sum_{i=0}^{\frac{n}{2}} (-1)^i \frac{n}{n-i} {n-i \choose i}(x_1+x_2)^{n-2i}(x_1x_2)^i $$
This lead to me considering the following simpler expression:
$$\sum_{i=0}^{\frac{n}{2}} (-1)^i \frac{n}{n-i} {n-i \choose i} $$
Plugging this into WolframAlpha tells us that the expression is actually equal to $2\cos\left(\frac{\pi n}{3}\right)$, but I don't know how to prove this.
Using Chebyshev polynomials or De Moivre's Theorem I made some progress:
$$2\cos\left(\frac{\pi n}{3}\right) = 2\sum_{i=0}^{\frac{n}{2}} (-1)^i {n \choose 2i} \cos^{n-2i}\left(\frac{\pi}{3}\right) \sin^{2i}\left(\frac{\pi}{3}\right)$$
$$= 2\sum_{i=0}^{\frac{n}{2}} (-1)^i {n \choose 2i} \frac{3^i}{2^n}$$
But I'm not sure how to show this is equal to the original expression. Any help in proving this identity would be much appreciated!
|
I saw that one binomial coefficient, so that got me thinking about Fibonacci polynomials, and what do you know I find that this is an evaulation of a Lucas polynomial
$$L_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor}\frac{n}{n-k}\binom{n-k}{k}x^{n-2k}$$
By the Wikipedia page, $L_n(x) = \alpha^n(x) + \beta^n(x)$ where $\alpha(x) = \frac{x+\sqrt{x^2+4}}{2}$ and $\beta(x) = \frac{x-\sqrt{x^2+4}}{2}$
Now note that $(-1)^k = i^{-n} i^{n-2k}$
Using the identities $e^{ix} = \cos(x) + i\sin(x)$ and $\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$, we obtain
\begin{align*}
&\sum_{k=0}^{n/2} (-1)^k \frac{n}{n-k}\binom{n-k}{k} \\
&= i^{-n} L_n(i) \\
&= i^{-n} (\alpha^n(i)+\beta^n(i))\\
&= i^{-n} \left(\left(\frac{i+\sqrt{3}}{2}\right)^n+\left(\frac{i-\sqrt{3}}{2}\right)^n\right) \\
&= \left(\frac{1+i\sqrt{3}}{2}\right)^n+\left(\frac{1-i\sqrt{3}}{2}\right)^n \\
&= e^{\frac{n\pi i}{3}} + e^{-\frac{n\pi i}{3}} \\
&= 2\cos\left(\frac{n\pi}{3}\right)
\end{align*}
|
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|
Proving $\frac1{2^1}+\frac2{2^2}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$ by induction I need to prove that
$$\frac1{2^1}+\frac2{2^2}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$$ by induction.
The base case was fine, and after that we have the induction hypothesis:
$$\frac1{2^1}+\frac2{2^2}+\cdots+\frac{k}{2^k}=2-\frac{k+2}{2^k} \tag1$$
holds for any given $n=k$.
The induction step is where I get stuck. We have
$$\frac1{2^1}+\frac2{2^2}+\cdots+\frac{k+1}{2^{k+1}} \tag2$$ and by the induction hypothesis we get
$$\begin{align}
2-\frac{k+2}{2^{k+1}} + \frac{k + 1}{2^{k+1}} &=2-\frac{2(k+2)}{2^{k+1}} +\frac{k+1}{2^{k+1}} \tag3 \\[6pt]
&= 2- \frac{3k+5}{2^{k+1}} \tag4
\end{align}$$ I certainly have made a mistake somewhere, but I can’t for the life of me figure out where.
|
Some simple mistake in your algebra. $-2(k+2)+k+1=-((k+1)+2)$.
|
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|
Change of variables for an ODE Question: Use the substitution $u = 2 \sqrt{x}$ to show that the equation:
$$x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-(1-x)y=0$$
becomes:
$$u^2\frac{d^2y}{du^2}+u\frac{dy}{du}-(u^2-u)y = 0$$
My attempt:
$$x^2\left[\frac{d^2y}{du^2}\left(\frac{du}{dx} \right)^2+\frac{dy}{du}\left(\frac{d^2u}{dx^2}\right)\right]+x\frac{dy}{du}\frac{du}{dx}-(1-x)y=0$$
simplifying with
$$\frac{du}{dx}=\frac{1}{\sqrt{x}},\ \frac{d^2u}{dx^2}=-\frac{1}{2\sqrt{x}^3}$$
yielding:
$$x\frac{d^2y}{du^2}+\frac{1}{2}\sqrt{x}\frac{dy}{du}-(1-x)y = 0$$
now using $\sqrt{x}=\frac{1}{2}u$, $x = \frac{1}{4}u^2$ and multiplying across by $4$:
$$\implies u^2\frac{d^2y}{du^2}+u\frac{dy}{du}-(4-u^2)y= 0$$
Which disagrees with the proposed solution. Have I made an error somewhere here so far or is there a further manipulation possible?
Attempting the substitution in reverse I find:
$$ x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}-(x-\frac{1}{2}\sqrt{x})y=0$$
in disagreement with the starting equation.
|
You must write everything in terms of $u$.
First derivative is easy, noting that $x=\frac{u^2}{4}$:
$$\frac{dy}{dx}=\frac{\frac{dy}{du}}{\frac{dx}{du}}=\frac{\frac{dy}{du}}{\frac{u}{2}}=\frac{2}{u}\frac{dy}{du}$$
Second derivative is difficult and key step:
$$\frac{d^2y}{dx^2}=\frac{\frac{d^2y}{du^2}-\frac{d^2x}{du^2}\frac{dy}{dx}}{(\frac{dx}{du})^2}=\frac{\frac{d^2y}{du^2}-\frac{1}{2}\frac{2}{u}\frac{dy}{du}}{(\frac{u}{2})^2}=\frac{4}{u^2}\frac{d^2y}{du^2}-\frac{4}{u^3}\frac{dy}{du}.$$
Then we substitute in the equation $x^2y''+xy'-(1-x)y=0$ and it becomes:
$$(\frac{u^2}{4})^2(\frac{4}{u^2}\frac{d^2y}{du^2}-\frac{4}{u^3}\frac{dy}{du})+(\frac{u^2}{4})(\frac{2}{u}\frac{dy}{du})-(1-\frac{u^2}{4})y=0$$
$$(\frac{u^4}{16})(\frac{4}{u^2}\frac{d^2y}{du^2}-\frac{4}{u^3}\frac{dy}{du})+(\frac{u}{2})(\frac{dy}{du})+(\frac{u^2}{4}-1)y=0$$
$$\frac{u^2}{4}\frac{d^2y}{du^2}-\frac{u}{4}\frac{dy}{du}+\frac{u}{2}\frac{dy}{du}+(\frac{u^2}{4}-1)y=0$$
$$\frac{u^2}{4}\frac{d^2y}{du^2}+\frac{u}{4}\frac{dy}{du}+(\frac{u^2}{4}-1)y=0$$
$$u^2\frac{d^2y}{du^2}+u\frac{dy}{du}+(u^2-4)y=0.$$
You wrongly wrote $u$ for $4$... $u^2-u$???
|
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|
Find the all positive integer solutions to: $x^3-x^2+x=3y^3$ Number theory problem:
Find the all positive integer solutions to:
$$x^3-x^2+x=3y^3$$
Here are my attempts:
$$x(x^2-x+1)=3y^3$$
$$x(x+1)(x^2-x+1)=3y^3(x+1)$$
$$x(x^3+1)=3(x+1)y^3$$
$$x^4+x^2=3(x+1)y^3$$
I can not see how can I proceed.
|
As hinted in the LMFDB database (see my comments under the question), this elliptic curve is birationally equivalent to the Fermat cubic. In addition to the transformation described in the comments, I needed to bring it into the short Weierstrass form $r^2=s^3-432$, and then use the transformation given here. A bit of cleaning up gave me the following.
By expanding we see that
$$
(x+1)^3+(2x-1)^3=9x^3-9x^2+9x.
$$
This means that multiplying the given equation by nine we can rewrite it in the form
$$
(x+1)^3+(2x-1)^3=(3y)^3.
$$
By the well-studied case $n=3$ of Fermat's last this equation is possible for rational $x,y$ only if one of the three numbers $x+1$, $2x-1$ or $3y$ vanishes. This quickly leads to the following:
All the rational solutions are $(0,0)$, $(-1,-1)$ and $(1/2,1/2)$. Hence there are no solutions with positive integers.
|
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|
Evaluate $\int_0^1 \cos^{-1} x\ dx$ by first finding $\frac{d}{dx}(x\cos^{-1} x)$ Question
Evaluate $$\int_0^1 \cos^{-1} x\ dx$$ by first finding the value of $$\frac{d}{dx}(x\cos^{-1} x).$$
My Working
As the question said to evaluate $$\frac{d}{dx}(x\cos^{-1} x),$$ I used the product rule to differentiate. We first have to let $u=x$ and $v=\cos^{-1} x$, so
\begin{align}
& \quad\frac{d}{dx}(x\cos^{-1} x)\\
&=u^\prime v+v^\prime u\\
&=1\cdot \cos^{-1}x + x\cdot\frac{-1}{\sqrt{1-x^2}}\\
&=\cos^{-1}x-\frac{x}{\sqrt{1-x^2}}
\end{align}
Unfortunately, after that, I have no idea about how to proceed, but I think that we should go somewhere from
$$\int\left[\frac{d}{dx}(x\cos^{-1}x)\right]\ dx+\int\frac{x}{\sqrt{1-x^2}}\ dx=\int \cos^{-1}x???$$
Thank you for your help!
|
Integration by parts gives
$$
\begin{aligned}
\int_0^1 \cos ^{-1} x d x &=\left[x \cos ^{-1} x\right]_0^1+\int_0^1 \frac{x}{\sqrt{1-x^2}} d x \\
&=-\frac{1}{2} \int_0^1u^{-\frac{1}{2}} d u, \quad (\textrm{ where }u=1-x^2)\\
&=-\frac{1}{2} \cdot 2\left[u^{\frac{1}{2}}\right]_1^0 \\
&=1
\end{aligned}
$$
|
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|
Verification of solution of contest problem related to algebra Given that $m, n$ are positive integers such that $mn<2023$ and $|n^2-mn-m^2|=1$, find the maximum value of $mn$.
This was problem 3 on the preliminary IMO exam where I live. My answer was $6$. However, there is space for $4$ numbers on the answer sheet, indicating that I might have got it wrong. Please help me find where I made a mistake (or if I even made one). I'd appreciate correction as well.
Below is my solution:
\begin{align*}
|n^2-mn-m^2|=1 &\Longrightarrow n^2-mn-m^2=\pm 1\\
&\Longrightarrow 1 = n^2-mn-m^2 \quad \textbf{or} \quad m^2+mn-n^2\\
&\Longrightarrow \text{WLOG, }\quad n^2 \pm mn - m^2 = 1
\end{align*}
Where $m>n>0$.
$$\pm mn = 1+m^2-n^2$$
Because we want the maximum value of $mn$, we'll take the positive value, that is, we seek to find the maximum $mn$ that satisfies
\begin{align*}
&mn = 1+m^2-n^2\\
&\Longrightarrow n^2+mn-(m^2+1) = 0\\
&\Longrightarrow n = \frac{-m\pm \sqrt{5m^2+4}}{2}
\end{align*}
Because $n$ is natural, the determinant $D = 5m^2+4$ is a perfect square, let $$5m^2+4 = k^2$$, where $k$ is natural.
Notice
$$5m^2 = k^2-2^2 = (k+2)(k-2)$$
Consider the "gap" between two squares. For example, $2^2-1^2=3$, $3^2-2^2=5$ and so on. It is trivial that no two squares have the difference of $4$, that is, the criteria $$m^2 = k^2-4$$ cannot be satisfied. So it must be that $5$ divides (or, equals), $(k+2) \quad \textbf{or} \quad (k-2)$. From there we have two values of $k$.
\begin{align*}
k = 3 \quad &\textbf{or} \quad 7\\
m = 1 \quad &\textbf{or} \quad 3\\
n = 1 \quad &\textbf{or} \quad 2
\end{align*}
The two sets of possible $(k, m, n)$ are listed respectively.
Therefore, there are only two values of $mn$, $6$ and $1$. We choose the larger. Hence, our answer is $6$.
|
If anything, here's a counter-example and its derivation:
$$|21^2-21\cdot 13-13^2|=1$$
$$21\cdot 13=273\gt 6\quad$$
derivation: one of the pell equations in $\quad |m^2-mn-n^2|=1\quad$is:
$$(m+2n)^2-5m^2=4,\quad subs\quad N=m+2n,\quad\text{consider}\quad N^2-5m^2=1\quad (9-4\sqrt5)(9+4\sqrt 5)=1\implies (9N+20m)^2-5(4N+9m)^2=1\\
N_2=m_2+2n_2=9N_1+20m_1=29m_1+18n_1,\quad m_2=4N_1+9m_1=13m_1+8n_1$$
$2n_2=29m_1+18n_1-(13m_1+8n_1)=16m_1+10n_1\to\quad $
$$m_2=13m_1+8n_1, \quad n_2=8m_1+5n_1$$
$$\begin{bmatrix}m\\n\end{bmatrix}_{k+1}=\begin{bmatrix}13&8\\8&5\end{bmatrix}\cdot \begin{bmatrix}m\\n\end{bmatrix}_k,\qquad \begin{bmatrix}1\\1\end{bmatrix}\to\begin{bmatrix}21\\13\end{bmatrix}\to\cdots$$
|
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|
Symmetric point of a point wrt a line I am trying to solve the following question.
Let $d$ be the
$$
\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-3}{5},
$$
and let $P$ be the point of coordinates $(4,3,10)$. Find the coordinates of the symmetric point $P^{\prime}$ of $P$ with respect to $d$.
Book has the following solution.
Let $M\left(x_0, y_0, z_0\right)$ be the midpoint of $P P^{\prime}$. Then $M \in d, P M \perp$ d, i.e.,
$$
\begin{gathered}
\frac{x_0-1}{2}=\frac{y_0-2}{4}=\frac{z_0-3}{5} \\
2\left(x_0-4\right)+3\left(y_0-3\right)+5\left(z_0-10\right)=0 .
\end{gathered}
$$
This yields $x_0=129 / 41, y_0=258 / 41, z_0=343 / 41$. Now, the coordinates of $P^{\prime}$ are given by $\left(P, M ; P^{\prime}\right)=-2$ (which is the simple ratio), and we have $P^{\prime}(-94 / 41$, $-393 / 41,-276 / 41)$
However, I don't understand why he takes $2\left(x_0-4\right)+3\left(y_0-3\right)+5\left(z_0-10\right)=0$ instead of $2\left(x_0-4\right)+4\left(y_0-3\right)+5\left(z_0-10\right)=0$. Since $PM$ is orthogonal to $d$, I am in $PM\cdot (2,4,5)$ must be equal to zero since $(2,4,5)$ is direction vector of the line. If we use the dot product that I wrote, I calculated $x_0=135/13, y_0=-270/13$.
Is my solution wrong?
I appreciate any help in advance.
|
The equations are
$ (x_0, y_0, z_0) = (1,2,3) + t (2, 4, 5) $
$PM = M - P = (1, 2, 3) - (4, 3, 10) + t (2, 4, 5) = (-3, -1, -7) + t (2, 4, 5)$
$ PM \perp d $ , so
$ \bigg( (-3, -1, -7) + t(2, 4, 5) \bigg) \cdot (2, 4, 5) = 0 $
So that
$ -45 + 45 t = 0 $
$\Longrightarrow t = 1 $
Thus $M = (1,2,3) + (2,4,5) = (3, 6, 8) $
$M = \dfrac{1}{2} (P + P') $
So
$P' = 2 M - P = 2 (3, 6, 8) - (4,3,10) = (2, 9, 6) $
|
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|
Use of differentiation to find the power series and determine the sum of the series We were asked to differentiate $f(x) = \frac{1}{4-{x^2}}$ to determine the sum of the series $\sum_{n=1}^\infty \frac{n}{16^{n}}$
so far I got the power series representation which is
$$\sum_{n=0}^\infty \frac{x^{2n}}{4^{n+1}}$$
But when I differentiated the power series, my prof said it was wrong. The differentiated $f(x)$ and power series I got was:
$$f'(x) = \frac{2x}{(4-{x^2})^2}$$
$$\sum_{n=1}^\infty \frac{nx^{2n-1}}{2^{2n+1}}$$
How can I differentiate the power series and determine the sum of the series $\sum_{n=1}^\infty \frac{n}{16^{n}}$?
|
From $$f(x)=\frac{1}{4-x^2}=\frac{1}{4}\frac{1}{1-\left(\frac{x}{2}\right)^2}=\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{2n}$$
Now
$$f'(x)=\frac{1}{4}\sum_{n=0}^{\infty}n\left(\frac{x}{2}\right)^{2n-1}=\frac{1}{4}\frac{2}{x}\sum_{n=0}^{\infty}n\left(\frac{x}{2}\right)^{2n}$$
Put $x=\frac{1}{2}$ and from the derivative you got you should get $\sum_{n=1}^{\infty}\frac{n}{16^n}=\frac{16}{225}$
|
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|
How can you show the results from substituting $x=a\cos{\theta}$ and $x=a\sin{\theta}$ are the same when finding $\int{\frac{16}{\sqrt{9-x^2}}}$? How can you show the results from substituting $x=a\cos{\theta}$ and $x=a\sin{\theta}$ are the same when finding $\int{\frac{16}{\sqrt{9-x^2}}}$?
I am learning trig substitution right now. I know from seeing this problem I should substitute x for $a\sin{\theta}$. After working this out:
$$\int{\frac{16}{\sqrt{9-x^2}}}$$
$$x=3\sin{\theta}$$
$$dx=3\cos{\theta}d\theta$$
$$=\int{\frac{16}{\sqrt{9-x^2}}}=\int{\frac{48\cos{\theta}d\theta}{\sqrt{9-9\sin^2}{\theta}}}$$
$$=\int{\frac{48\cos{\theta}d\theta}{3\sqrt{1-\sin^2}{\theta}}}$$$$=16\int{\frac{\cos{\theta}d\theta}{\sqrt{\cos^2}{\theta}}}$$
$$=16\int{\frac{\cos{\theta}d\theta}{\cos{\theta}}}=16\int{d\theta}$$
$$=16\theta=16\sin^{-1}\left({\frac{x}{3}}\right)+C_1$$
However, substituting $x=a\cos{\theta}$ yields a different result.
$$\int{\frac{16}{\sqrt{9-x^2}}}$$
$$x=3\cos{\theta}$$
$$dx=-3\sin{\theta}d\theta$$
$$=\int{\frac{16}{\sqrt{9-x^2}}}=\int{\frac{-48\sin{\theta}d\theta}{\sqrt{9-9\cos^2}{\theta}}}$$
$$=\int{\frac{-48\sin{\theta}d\theta}{3\sqrt{1-\cos^2}{\theta}}}$$$$=-16\int{\frac{\sin{\theta}d\theta}{\sqrt{\sin^2}{\theta}}}$$
$$=-16\int{\frac{\sin{\theta}d\theta}{\sin{\theta}}}=-16\int{d\theta}$$
$$=-16\theta=-16\cos^{-1}\left({\frac{x}{3}}\right)+C_2$$
$$\implies\quad{-16\cos^{-1}\left({\frac{x}{3}}\right)}+C_2={16\sin^{-1}\left({\frac{x}{3}}\right)}+C_1$$
Unless I did something incorrectly or there is something that prohibits different types of substitutions, how can this be true? I have tried to think it over how $-\cos^{-1}{x}$ can equal $\sin^{-1}{x}$.
I thought about it being a negative angle, perhaps making the opposite leg of the triangle negative. This however leads to a contradiction that says $-\cos^{-1}{x}=\sin^{-1}{x}$ but $\cos^{-1}{x}\neq-\sin^{-1}{x}$.
Any thoughts, answers, or corrections are appreciated.
|
If you do a simpler example of $\int \frac{1}{\sqrt{1 - x^2}} dx$, you will get $\sin^{-1}(x) + C$ and $-\cos^{-1}(x) + C$ respectively. If you draw the diagrams, you will see that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$, since it's the two angles that's not a right angle. Therefore, you see that $\sin^{-1}(x)$ and $-\cos^{-1}(x)$ differ by a constant $\frac{\pi}{2}$.
|
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|
Arrangements of $a,a,b,b,b,c,c,c,c$ in which no two consecutive letters are the same I have the following question:
We are going to generate permutations from a,a,b,b,b,c,c,c,c. Please compute the number of permutations such that:
(a) for any consecutive 4 elements, they are not all the same;
(b) for any consecutive 3 elements, they are not all the same;
(c) for any consecutive 2 elements, they are not all the same.
See Arrangements of $a,a,b,b,b,c,c,c,c$ in which no four/three/two consecutive letters are the same for the first two parts.
I got an answer for part c but I am not sure if it is correct or not.
My step is the following.
I found that there are six pairs containing adjacent identical letters. I will define them as$S_1, S_2,..., S_6$ the small number is the number of pairs of adjacent identical letters.
$$S_0 =\binom{9}{2,3,4}$$
$$S_1 =\binom{8}{3,4} + \binom{8}{2,4} + \binom{8}{2,3,2}$$
$$S_2 = \frac{7!}{4!}+\binom{7}{3,2}+\binom{7}{4,2} + \binom{7}{3,2} + \binom{6}{3,2}+\binom{7}{2,2}$$
$$S_3 = \frac{6!}{4!}+\frac{6!}{3!}+\frac{5!}{3!} + \binom{6}{2,2} + \frac{6!}{2!} + \frac{5!}{2!}$$
$$S_4 = \frac{5!}{2!}+\frac{5!}{3!}+\frac{5!}{2!} + \frac{5!}{2!} + {5!}+{4!}$$
$$S_5 = {4!}+{3!}+\frac{4!}{2!} $$
$$S_6 = 3!$$
I sum these up by the inclusion-exclusion principle, and I got the answer of the following:
$$N = S_0 - S_1+ S_2- S_3 +S_4 -S_5+ S_6 = 473$$
I want to know if there is any miscalculation or if the wrong steps exist. Could anyone please help?
|
This answer is rather a supplement which could be used as crosscheck for manual calculations. We consider a $3$-ary alphabet built from letters $\mathcal{V}=\{a,b,c\}$. Words which do not have any consecutive equal letters are called Smirnov words. A generating function for Smirnov words is given as
\begin{align*}
\left(1-\frac{az}{1+az}-\frac{bz}{1+bz}-\frac{cz}{1+cz}\right)^{-1}\tag{1}
\end{align*}
The coefficient $[z^n]$ of $z^n$ in the series (1) gives the number of $3$-ary words of length $n$ which do not have any consecutive letters.
With some help of Wolfram Alpha we calculate the answer from (1) as
\begin{align*}
\color{blue}{[z^{9}a^2b^3c^4]\left(1-\frac{az}{1+az}-\frac{bz}{1+bz}-\frac{cz}{1+cz}\right)^{-1}=79}
\end{align*}
Note: Smirnov words can be found for instance in example III.24 in Analytic Combinatorics by P. Flajolet and R. Sedgewick.
|
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|
write an explicit formula for $f(x)$ when $x$ is in $[\frac{1}{n+1},\frac{1}{n}]$
this is from spivak calculus ,i tried to find an explicit formula when $x$ is in $[\frac{1}{n+1},\frac{1}{n}]$
my attempt:
in $[\frac{1}{n+1},\frac{1}{n}]$ the function $f$ is a linear function so we can write it as $f(x)=\alpha x+ \beta$
-finding $\alpha$
we can see that the function equal $0$ when $x$ exists in the midle of $\frac{1}{n}$ and $\frac{1}{n+1}$,because we can imagine the function graph in $[\frac{1}{n+1},\frac{1}{n}]$ as a diameter of a rectangle with length $2$ and width $\frac{1}{n}-\frac{1}{n+1}$,so here the both diameters of our rectangle have to intersect in the middle ($\frac{\frac{1}{n}-\frac{1}{n+1}}{2}=\frac{1}{2n(n+1)}$) so that's means each of the diameters have to pass from the middle of the rectangle which is in our case the middle point in $[\frac{1}{n+1},\frac{1}{n}]$ in the horizontal axis ,and that's means $f(x)=0$,when $x=\frac{1}{n+1}+\frac{1}{2n(n+1)}=\frac{2n+1}{2n(n+1)}$(for understand my idea , you can draw a rectangle in the graph from $\frac{1}{3}$ to $\frac{1}{2}$)
so $f(\frac{2n+1}{2n(n+1)})=\alpha \frac{2n+1}{2n(n+1)}+ \beta=0$
$\Rightarrow$$\alpha=-\beta .\frac{2n(n+1)}{2n+1}$
so $f(x)=-\beta .\frac{2n(n+1)}{2n+1}x+ \beta$
-finding $\beta $
we know that $f(x)=-\beta .\frac{2n(n+1)}{2n+1}x+ \beta$ and $ f(\frac{1}{n})=(-1)^{n+1}$
so $f(\frac{1}{n})=-\beta .\frac{2(n+1)}{2n+1} +\beta =(-1)^{n+1}$
$\Rightarrow $$\beta(\frac{1+2n -2n -2}{1+2n})=(-1)^{n+1}$$\Rightarrow $$\beta=(-1)^{n+2}.(1+2n)$
so $f_n(x)=-(-1)^{n+2}.2n(n+1)x+(-1)^{n+2}.(1+2n)$ when $x$ is in $[\frac{1}{n+1},\frac{1}{n}]$
finally we can add this :
$f_n(x) =
\begin{cases}
1 & |x| \geq 1 \\
-(-1)^{n}.2n(n+1)x+(-1)^{n}.(1+2n) & x \in [\frac{1}{n+1},\frac{1}{n}] \\
&
\end{cases}$
-so does my attempt is right ?.
-can you find a better formula for $f$?.
|
Your attempt is right except $f$ instead of $f_{\color{red}n}$ and $x>0$(just say the function is even). I will make a few remarks in addition to the remarks made in comments :
Let $a,b\in\Bbb R$: the middle of $[a,b]$ is $\frac12(a+b)$
The middle of $[\frac{1}{n+1},\frac1n]$ is $\frac{\frac{1}{n}+\frac{1}{n+1}}{2}=\frac{2n+1}{2n(n+1)}$(you could have gone faster that way)
Anyway, for $a\neq b, \color{red}{\text{the}}$ equation $y=\alpha x+\beta$ ("linear function") of the line going through $(a,A)$ and $(b,B)$ is $y-A=(x-a)\frac{B-A}{b-a}$
Here, with $(a,A)=(\frac{1}{n+1},(-1)^{n})$ and $(b,B)=(\frac{1}{n},(-1)^{n-1})$
$$\forall n>0, \forall x\in [\frac{1}{n+1},\frac1n], f(x)$$$$=(-1)^n+(x-\frac{1}{n+1})\frac{(-1)^{n-1}-(-1)^n}{\frac{1}{n}-\frac{1}{n+1}}$$$$=(-1)^n+2(-1)^{n-1}n(n+1)(x-\frac{1}{n+1})$$
$$=2(-1)^{n-1}n(n+1)x-2(-1)^{n-1}n+(-1)^n$$
$$=2(-1)^{n-1}n(n+1)x+(-1)^n(1+2n)$$
As Spivak said, "good exercise".
|
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Struggles solving : $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
Solve $\frac23xyy'=\sqrt{x^6-y^4}+y^2.$
(Hint: the equation can be boiled down to a homogeneous one.)
My attempt:
I didn't know how to obtain a homogeneous equation, so I'm going to show what I did instead.
We can write $\frac23xyy'=|x|^3\sqrt{1-\frac{y^4}{x^6}}+y^2.$ Let $z=\frac{y^2}{x^3}=y^2x^{-3}.$ Then
$$\begin{aligned}z'&=2yy'x^{-3}-3y^2x^{-4}\\ \implies x^4z'&=2xyy'-3y^2\\ \implies\frac23xyy'&=\frac{x^4z'+3y^2}3=\frac{x^4z'}3+y^2\\ \implies\frac{x^4z'}3+y^2&=|x|^3\sqrt{1-z^2}+y^2\\ \implies\frac{|x|z'}3&=\sqrt{1-z^2}\\ \implies\frac{dz}{\sqrt{1-z^2}}&=3\frac{dx}{|x|}\\ \implies\arcsin z&=3\ln|x|+C, C\in\Bbb R\\ \implies z&=\sin\ln(C|x|^3), C>0\\ \implies y^2&=x^3\sin\ln(C|x|^3)C>0.\end{aligned}$$
I'm not sure my answer is correct. I didn't get far plugging the result into the equation.
How do we solve the given ODE? Or should I better ask, how do we obtain a homogeneous ODE from it?
|
$$\frac23xyy'=\sqrt{x^6-y^4}+y^2.$$
$$x^3(y^2)'=3x^2\sqrt{x^6-y^4}+3x^2y^2$$
$$\dfrac {x^3(y^2)'-3x^2y^2}{x^6}=\dfrac 3 {x^4}\sqrt{x^6-y^4}$$
$$\left (\dfrac {y^2}{x^3}\right)'=\dfrac 3 {x^4}\sqrt{x^6-y^4}$$
$$\left (\dfrac {y^2}{x^3}\right)'=\dfrac 3 {x}\sqrt{1-\left (\dfrac {y^2}{x^3}\right)^2}$$
The DE is separable.
Substitute $u=\left (\dfrac {y^2}{x^3}\right)$:
$$\dfrac {du}{ \sqrt{1- u^2}}=\dfrac 3x dx$$
|
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|
Are there any other decent methods to evaluate $\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x?$ We first split the integrand into 3 parts as
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &= \underbrace{\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x}_J+\underbrace{\int_0^1 \frac{\ln (1+x)}{1+x^2} d x}_K+ \underbrace{\int_0^1 \frac{\ln (1-x)}{1+x^2} d x}_L
\end{aligned}
Denotes the Catalan’s constant by $G$.
By my post, $$J=\frac{\pi}{2}\ln2-G$$
Dealing with the last $2$ integrals, we use a powerful substitution $x=\frac{1-t}{1+t} ,$ then $dx=-\frac{2dt}{(1+t)^2}.$
$$
\begin{aligned}
K&=\int_1^0 \frac{\ln 2-\ln (1+t)}{\frac{2+2 t^2}{(1+t)^2}} \frac{-2 d t}{(1+t)^2} \\
&=\ln 2 \int_0^1 \frac{d t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2}
\end{aligned}
$$
Hence $$K=\frac{\pi}{8} \ln 2 $$
$$
\begin{aligned}
L=& \int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\
&=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t . \\
&=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\
&=\frac{\pi}{8} \ln 2-G
\end{aligned}
$$
Combining them to get
$$
\begin{aligned}
I &=\left(\frac{\pi}{2} \ln 2-G\right) +\frac{\pi}{8} \ln 2 +\left(\frac{\pi}{8} \ln 2-G\right)\\
&=\frac{3 \pi}{4} \ln 2-2 G
\end{aligned}
$$
I do want to know if it can be solved by any other elegant methods. Your comments and methods are highly appreciated.
|
Letting $x=\tan \theta$ yields
$$
\begin{aligned}
\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &=\int_0^{\frac{\pi}{4} } \ln \left[\left(1+\tan ^2 \theta\right)\left(1-\tan ^2 \theta\right)\right] d \theta \\
&=\int_0^{\frac{\pi}{4}} \ln \left(\sec ^2 \theta\right) d \theta+\ln \left(\frac{\cos^2 \theta-\sin ^2 \theta}{\cos ^2 \theta}\right) d \theta \\
&=-4 \int_0^{\frac{\pi}{4}} \ln (\cos \theta) d \theta+\int_0^{\frac{\pi}{4}} \ln (\cos 2 \theta) d \theta \\
&=-4\left(-\frac{\pi}{4} \ln 2+\frac{G}{2}\right)-\frac{\pi}{4} \ln 2 \\
&=\frac{3 \pi}{4} \ln 2-2G
\end{aligned}
$$
where the first integral refers to the post.
|
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|
Solve the equation $\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$ Solve the equation $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=|5-x|$$
$x=5$ is the only real solution
We can note that $x^2-9x+24>0$ and $6x^2-59x+149>0$ for all $x$. The first thing I decided to try: as $$|5-x|=\begin{cases}5-x,x\le5\\x-5,x>5\end{cases},$$ we can look at two different cases based on the sign of $5-x$: let's $x>5$. Then the equation becomes $$\sqrt{x^2-9x+24}-\sqrt{6x^2-59x+149}=x-5\\\sqrt{x^2-9x+24}=x-5+\sqrt{6x^2-59x+149}\\x^2-9x+24=(x-5)^2+6x^2-59x+149+2(x-5)\sqrt{6x^2-59x+149}\\30x-3x^2-7=(x-5)\sqrt{6x^2-59x+149}$$ We have to raise both sides to the power of 2 again, so I decided to stop here.
Something else we can try is to raise both sides of the initial equation to the power of 2, as $|5-x|^2=(5-x)^2,$ so we have $$7x^2-68x+173-2\sqrt{(x^2-9x+24)(6x^2-59x+149)}=x^2-10x+25\\3x^2-29x+72=\sqrt{(x^2-9x+24)(6x^2-59x+149)}$$ I think it's obvious I am missing something.
|
Observe that
\begin{align*}
x^2-9x+24&=(x-5)^2+(x-5)+4\\
6x^2-59x+149&=6(x-5)^2+(x-5)+4
\end{align*}
Say $(x-5)^2+(x-5)+4=\alpha$ (note $\alpha \geq 0$), then we have
$$\sqrt{\alpha}-\sqrt{5(x-5)^2+\alpha}=|5-x|$$
The expression on the left is $\leq 0$ and that on the right is $\geq 0$. For the equality to hold, both have to be $0$.
Thus $x=5$ is the only solution.
|
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|
Why $\lim\limits_{n \to \infty} \frac{\frac1{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim\limits_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$? I just need clarification about answer to this question:
$$\lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n}} = \lim_{n \to \infty} \frac{\sum\limits_{i = 1}^{n+1} \frac{1}{\sqrt{i}} - \sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}}{\sqrt{n+1} - \sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = 2$$
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n}} \sum\limits_{i = 1}^{n} \frac{1}{\sqrt{i}}\right)^k = 2^k$$
In this case I don't understand why
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}$$
My reasoning. $$\frac{ \frac{1}{a} }{b - c} = \frac{1}{a(b-c)}$$Hence, we get:
$$\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}}=\lim_{n \to \infty} \frac{1}{n+1 - \sqrt{n}\sqrt{n+1}} = \lim_{n \to \infty} \frac{ \frac{1}{n} }{ \frac{n}{n} + \frac{1}{n} + \sqrt{\frac{n^2}{n^2}+\frac{n}{n^2} } } = \lim_{n \to \infty}\frac{0}{1+0+\sqrt{1+0}}=0$$
While $\lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} = \lim_{n \to \infty}\frac{ \sqrt{\frac{n}{n} + \frac{1}{n}} + \sqrt{\frac{n}{n}} }{ \sqrt{\frac{n}{n} + \frac{1}{n} } } = \lim_{n \to \infty}\frac{\sqrt{1+0} + \sqrt{1} }{\sqrt{1+0}} = 2$
Any explanation, what's going on? Thank you
|
Let $s = \sqrt{n+1}$ and $t = \sqrt{n}.$ Then:
$$\mathcal{L} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{s}}{s-t} = \\
= \lim_{n \to \infty}\displaystyle\frac{1}{(s-t)}\frac{1}{s} = \lim_{n \to \infty}\displaystyle\frac{s+t}{(s-t)(s+t)}\frac{1}{s}.$$
But:
$$(s-t)(s+t) = s^2 -t^2 = (n+1) - n = 1,$$
and hence:
$$\mathcal{L} = \lim_{n \to \infty} \frac{s+t}{1}\frac{1}{s} = \lim_{n \to \infty} \frac{s+t}{s}= \lim_{n \to \infty}\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Verify that $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots As the title states, the goal is to verify that the quadratic equation: $abc^2x^2+c(3a^2+b^2)x+3a^2-ab+b^2=0$ has real roots. This problem comes from an interschool mathematics contest for High-schoolers, here's my (brute force) attempt:
Since we need to verify that the roots are rational, we need to prove that the discriminant of this equation (which I'll denote with $D$) is greater than $0$. In other words:
$D=B^2-4AC>0$ where $B=c(3a^2+b^2)$, $A=abc^2$ and $C=3a^2-ab+b^2$ Therefore:
$D=[c(3a^2+b^2)]^2-4(abc^2)(3a^2-ab+b^2)$
$D=[c^2(9a^4+6a^2b^2+b^4)]+[c^2(-12a^3b+4a^2b^2-4ab^3)]$
$D=c^2(9a^4+10a^2b^2+b^4-12a^3b-4ab^3)$
I'm not quite sure where to proceed from here. And this certainly does not prove that the roots are real as this expression can be negative. Is there a way to proceed from here? Are there any better or alternative ways to answer this? Please share your approaches!
|
$$
p(x) = ab{c}^{2}{x}^{2}+c \left( 3\,{a}^{2}+{b}^{2} \right) x+3\,{a}^{2}-ab+{b}^{2} .
$$
Let's try to factor this with the idea of finding a factor $Ax+B$ where $A$ is a divisor of $abc^2$ and $B$ is a divisor of $3a^2-ab+b^2$. Since $3a^2-ab+b^2$ cannot be factored over the rationals, we try without factoring it. (If this doesn't work, we have to do something harder.) The possible factors of $p(x)$ are $Ax+B$ where
$$
A \in \big\{1, a, b, ab, c, ac, bc, abc, c^2, ac^2, bc^2, abc^2\big\} ,
\\
B \in \big\{1, -1, (3a^2-ab+b^2), -(3a^2-ab+b^2)\big\} .
$$
Trying these four candidates for $B$ until one works, we arrive at:
$$
p(x)= \left( cx+1 \right) \left( abcx+(3\,{a}^{2}-ab+{b}^{2})\right)
$$
|
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|
Trig and de Moivre's theorem
A) Use de Moivre's theorem to prove that $\cos^4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$
B) Therefore deduce that $\cos(\pi/8) = \left(\frac{2 + \sqrt{2}}{4}\right)^{1/2}$
C) and write down an expression for $\cos(3\pi/8)$.
I have proved the first part of the question but I am not sure where to go from there.
My attempt
de Moivre's theorem states that $(\cos x +i\sin x)^n = \cos nx + i\sin nx$
using this and $\cos ^2θ + \sin ^2θ =1$, I did the following:
\begin{align}
\cos 4θ + i\sin 4θ &= (\cos θ +i\sin θ)^4 + \cos 4θ + 4i\cos ^3θ\sin θ - 6\cos ^2θ\sin ^2θ-4i\cos θ\sin 3θ+\sin ^4θ\\
\cos 4θ &= \cos ^4θ -6\cos ^2θ\sin ^2θ +\sin ^4θ\\
\cos 4θ &= \cos ^4θ - 6\cos ^2θ(1-\cos ^2θ) + (1-\cos ^2θ)^2\\
\cos 4θ &= \cos ^4θ - 6\cos ^2θ + 6\cos ^4θ +1 - 2\cos ^2θ + \cos ^θ\\
\cos 4θ &= 8\cos ^4θ -8\cos ^2θ +1
\end{align}
|
With $\cos(4\theta)$ on the lhs, you can substitute $\theta = \frac{\pi}{8}$ in your first equation, which gives a polynomial equation for $\cos(\frac{\pi}{8})$.
$$0 = 8\cos^4\frac{\pi}{8}- 8\cos^2\frac{\pi}{8} + 1$$
This equation is biquadratic and can be solved with the quadratic formula to show that,
$$\cos\frac{\pi}{8} = \frac{1}{2}\sqrt{2+\sqrt{2}}$$
You can use related methods to show that,
\begin{align}
\cos\frac{\pi}{4} &= \frac{1}{2}\sqrt{2} \\
\cos\frac{\pi}{8} &= \frac{1}{2}\sqrt{2+\sqrt{2}} \\
\cos\frac{\pi}{16} &= \frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2}}} \\
\cos\frac{\pi}{32} &= \frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}} \\
\cos\frac{\pi}{64} &= \frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}
\end{align}
and the pattern goes on
|
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|
Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. What is its general solution formula? Diophantine equation ${x ^ 2}+{y ^ 2}+{z ^ 2}={m ^ 2}+{n ^ 2} $ has infinite solutions. For examples,
$\begin{array}{l}
{1^2}+ {3^2} + {4^2} = {1^2} + {5^2}\\
{1^2} + {2^2} + {6^2} = {4^2} + {5^2}\\
{322^2} + {562^2} + {597^2} = {601^2} + {644^2}\\
{938^2} + {1063^2} + {4722^2} = {536^2} + {4901^2}\\
\vdots
\end{array}$
What is its general solution formula? How to get the general solution formula?
|
It is clear that there are infinitely many solutions (see my comment above). We know that the identity $$(a^2+b^2+c^2+d^2)^2=(a^2-b^2-c^2-d^2)^2+(2ab)^2+(2ac)^2+(2ad)^2$$ gives all the solutions of $x^2+y^2+z^2+w^2=u^2$.
Now we have $$x^2+y^2+z^2=m^2+n^2\iff x^2+y^2+z^2+(im)^2=n^2$$ because the identity is valid in the ring of the Gaussian integers, so for the parameters $a,b,c,id$ we get $$(a^2+b^2+c^2-d^2)^2=(a^2-b^2-c^2+d^2)^2+(2ab)^2+(2ac)^2-(2ad)^2$$ so we have
$$\boxed{(a^2+b^2+c^2-d^2)^2+(2ad)^2=(a^2-b^2-c^2+d^2)^2+(2ab)^2+(2ac)^2}$$ This new identity giving consequently all the solutions of the proposed equation.
|
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|
Show that scalene triangle $\triangle ABC$ is a right-triangle if $\sin(A)\cos(A)=\sin(B)\cos(B)$ As the title suggests, this is a college entrance exam practice problem from Japan, it is as follows:
Given a scalene triangle $\triangle ABC$, prove that it is a right triangle if $\sin(A)\cos(A)=\sin(B)\cos(B)$
I found this problem pretty interesting, and after some thinking, I found a way to solve it, and I'll show my attempt here. I want to know, are there any other/better ways to solve this? Or is there anything about my solution that can be improved? Please let me know!
Here's my attempt:
Some have that: $$\sin(A)\cos(A)=\sin(B)\cos(B)$$
From Law of Sines we know that:
$$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}=2R$$
And from Law of Cosines:
$$\cos(B)=\frac{a^2+c^2-b^2}{2ac}$$ $$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$
Therefore:
$$(\frac{a}{2R})(\frac{b^2+c^2-a^2}{2bc})=(\frac{b}{2R})(\frac{a^2+c^2-b^2}{2ac})$$
Dividing and multiplying by $a$ on the left side and by $b$ on the right side gives us:
$$\frac{(a^2)(b^2+c^2-a^2)}{4Rabc}=\frac{(b^2)(a^2+c^2-b^2)}{4Rabc}$$
$$a^2b^2+a^2c^2-a^4=a^2b^2+b^2c^2-b^4$$
$$(a^2-b^2)c^2-(a^2-b^2)(a^2+b^2)=0$$
$$(a+b)(a-b)(-a^2-b^2+c^2)=0$$
Now, obviously the case $a=-b$ cannot be true. We also know that $a=b$ does not work in this particular case since this triangle is scalene, therefore we're left with:
$$-a^2-b^2+c^2=0$$
$$a^2+b^2=c^2$$
And this proves that the triangle is right angled
|
Here's an identity-free approach:
Writing $D$ and $E$ for the feet of the altitudes from $A$ and $B$, we have
$\require{cancel}$
$$\color{blue}{|\triangle ABE|}+\color{red}{|\triangle CBE|} \;=\; |\triangle ABC| \;=\; \color{blue}{|\triangle BAD|}+\color{red}{|\triangle CAD|} \tag1$$
$$\cancel{\color{blue}{\tfrac12c^2\sin A\cos A}}+\color{red}{\tfrac12 a^2\sin C\cos C} \;=\; \cancel{\color{blue}{\tfrac12c^2\sin B\cos B}}+\color{red}{\tfrac12 b^2\sin C\cos C} \tag2 $$
$$\underbrace{(a^2-b^2)}_{\neq0\;\text{(scalene)}}\cdot\underbrace{\sin C}_{\neq0}\cdot\cos C \;=\; 0 \quad\to\quad \cos C = 0 \quad\to\quad C = 90^\circ\tag3$$
Done! $\square$
|
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|
The sum $\frac{1}{2\sqrt1+1\sqrt 2}+\frac{1}{3\sqrt2+2\sqrt 3}+\cdots +\frac{1}{100\sqrt{99}+99\sqrt{100}}$ is equal to: From an exercise of a textbook of an high school of 15 years old, I have this partial sum
$$S_n=\frac{1}{2\sqrt1+1\sqrt 2}+\frac{1}{3\sqrt2+2\sqrt 3}+\cdots +\frac{1}{100\sqrt{99}+99\sqrt{100}}$$
Considering that every fraction is $<1$ I have seen that the general term is $$\frac{1}{100\sqrt{99}+99\sqrt{100}}=\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}$$$$=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}=\dfrac{1}{\sqrt{n}} - \dfrac{1}{\sqrt{n+1}}$$
$$A) \frac{999}{1000}, \quad B) \frac{99}{100}, \quad C) \frac{9}{10}, \quad D) 9, \quad E) 1$$
How should I arrive at the result without to use the calculator? My students not have studies the $\sum$.
|
The general term
$$\begin{align}
\frac{1}{(n+1)\sqrt{n}+ n\sqrt{n+1}} &= \frac{1}{\sqrt{n(n+1)}}\cdot\frac{1}{\sqrt{n+1}+\sqrt{n}} \\
&=\frac{1}{\sqrt{n(n+1)}}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\\
&=\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}
\end{align}$$
Make the sum, the result must be equal to
$$\sum_{n=1}^{99}\frac{1}{(n+1)\sqrt{n}+ n\sqrt{n+1}} =\sum_{n=1}^{99}\left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \right)= \frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}} = \frac{9}{10}$$
$C$ is the final answer ;)
|
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|
Why does cancelling by $\sin x$ when solving $4\tan x = 5\sin x$ for $0\leq x < 2\pi $ miss solutions? So the solution to the problem is:
$$4 \tan x = 5 \sin x$$
$$4 \frac{\sin x}{\cos x} = 5 \sin x$$
$$4 \sin x = 5 \sin x \cos x$$
So, either $\sin x = 0$ or $\cos x = 4/5$. I'm fine with the logic of this. But is there some deeper reason why cancelling the $\sin x$ straight away loses solutions?
|
For the same reason why solving $x = x^2$ by dividing by $x$ would lose solutions.
If you divide both sides by something, you implicitly give rise to two cases.
For $x=x^2$, if you divide by $x$, then you get $1=x$, but only if $x \ne 0$. If $x$ (what you divided by) is equal to zero, then you just divided by zero, and hence have to treat that case separately.
That is,
$$x = x^2 \implies \underbrace{\frac{x}{x} = \frac{x^2}{x}}_{1 = x, \text{ if } x \ne 0} \text{ or } x = 0$$
In your scenario,
$$4 \sin x = 5 \cos x \sin x
\implies \underbrace{\frac{4 \sin x}{\sin x} = \frac{5 \cos x \sin x}{\sin x}}_{4 = 5 \cos x\text{ if } \sin x \ne 0} \text{ or } \sin x = 0$$
Sometimes that equals-to-zero case gives rise to other solutions, sometimes not, it just depends. Here, it does, because if $\sin x = 0$, then both sides of the equation are equal to zero, so what remains is to find the $x$ that work.
|
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|
Solve $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ I would like to solve the equation $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$ analytically, without using Wolfram Alpha. I have tried several substitutions, including $x=\cos(t)$, $\sqrt{1-x^2}=t$, but they all fail and eventually result in a quartic polynomial which is not easily solved. Any suggestions would be helpful.
|
$$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1-x^2}$$
substitute $x=\sqrt{y}$
$$\sqrt y\left(1+\sqrt{1-y}\right)=\sqrt{1-y}$$
divide both sides by $\sqrt{y}$
$$1+\sqrt{1-y}=\sqrt{y^{-1}-1}$$
square both sides
$$2-y+2\sqrt{1-y}=y^{-1}-1$$
isolate the square root
$$2\sqrt{1-y}=y+y^{-1}-3$$
square both sides
$$4-4y=y^2+y^{-2}-6y-6y^{-1}+11$$
multiply both sides by $y^2$
$$4y^2-4y^3=y^4-6y^3+11y^2-6y+1$$
subtract the left hand side from the right
$$y^4-2y^3+7y^2-6y+1=0$$
Factoring $y^4-2y^3+7y^2-6y$ we get
$$y^4-2y^3+7y^2-6y=y(y^3-2y^2+7y-6)=y(y-1)(y^2-y+6)=(y^2-y)(y^2-y+6)$$
Substitute $y^2-y=z$
$$y^4-2y^3+7y^2-6y=z^2+6z$$
Plugging it back in we get
$$z^2+6z+1=0$$
$$(z+3)^2-8=0$$
$$z=-3±\sqrt{8}$$
Undoing the substitution we get
$$y^2-y+3±\sqrt{8}=0$$
$$y=\frac{-(-1)±\sqrt{(-1)^2-4(1)(3±\sqrt{8})}}{2(1)}=\frac{1±\sqrt{-11±4\sqrt{8}}}{2}$$
Knowing $x=\sqrt{y}$ we get the following
$$x=\sqrt\frac{1±\sqrt{-11±4\sqrt{8}}}{2}$$
Three of these solutions are extraneous
|
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|
In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead anywhere. My actual approach, which I will post as an answer below, uses the law of Cosines. Please share your own approaches especially if they use a different method!
|
From $\cos(A-B)=\frac{7}{8}$, we have $\sin(\frac{A-B}{2})=\sqrt{\frac{1-\cos(A-B)}{2}}=\frac{1}{4}. {\tag 1}$
This is a nice property about interior angles: $$\sin(\frac{A-B}{2})=\frac{a-b}{c}\cos(\frac{C}{2}).$$ From this property and $(1)$ with $a=5$, $b=4$, we have $\cos(\frac C2)=\frac c4.\tag 2$
By using the half-angle formula $\cos C=2\cos^2(\frac C2)-1$, we have $\cos(C)=\frac{c^2-8}{8}.\tag3$
By using the law of cosines $c^2=a^2+b^2-2ab\cos C$ for the side $c$ and $(3)$ we can easily find $c=\frac{3\sqrt3}{\sqrt2}.$ And finally, by using $(3)$ again
$$\cos C=\frac{\frac{27}{2}-8}{8}=\frac{11}{16}.$$
|
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|
Find $a_n$ such that the formula is equal to $\zeta(2)$ My question start with the observation :
$$\sqrt{e}\simeq \frac{\pi^2}{6}$$
At first glance it's not really convincing but after some work I found :
$$\sqrt{e-5\left(\frac{1}{\pi}-\frac{1}{e}\right)^{2}+2\left(\frac{1}{\pi}-\frac{1}{e}\right)^{3}+9\left(\frac{1}{\pi}-\frac{1}{e}\right)^{4}-4\left(\frac{1}{\pi}-\frac{1}{e}\right)^{5}}\simeq \frac{\pi^2}{6}$$
Wich is really better .
So now my question is clear :
If we have :
$$\sqrt{e-5\left(\frac{1}{\pi}-\frac{1}{e}\right)^{2}+2\left(\frac{1}{\pi}-\frac{1}{e}\right)^{3}+9\left(\frac{1}{\pi}-\frac{1}{e}\right)^{4}-4\left(\frac{1}{\pi}-\frac{1}{e}\right)^{5}+\cdots+a_n\left(\frac{1}{\pi}-\frac{1}{e}\right)^{n}+\cdots}=\frac{\pi^2}{6}$$
Where $a_n\in Z$
How to find $a_n$ ?
I don't find it in the librairy of integer .On the other hand it seems a bit artificial anyway perhaps there is some interest but I ignore it currently .
Thanks for your help .
Side notes :
$$\sqrt{e}=1+\sum_{k=1}^{\infty}\frac{1}{2^{k}k!}$$
We have also :
$\lim_{a\to \infty}\exp\left(\frac{a}{a+\frac{a}{\exp\left(a-\frac{a}{\exp\left(a\frac{a}{\exp\left(a+\frac{a}{\exp\left(a-\frac{a}{\exp\left(a-a\cdot\frac{a}{\exp\left(a...\right)}\right)}\right)}\right)}\right)}\right)}}\right)=e^{\frac{1}{2}}$
And see the wiki page about the Vieta's formula for $\pi$
|
As @Gevorg Hmayakyan says, the problem can be rephased as finding a sequence $a_n\in \mathbb{Z}$ such that
\begin{equation}
\sum_{i=1}^\infty a_i \left(\frac{1}{\pi}-\frac{1}{e}\right)^{i}=\frac{\pi^4}{36}-e
\end{equation}
To to this, I use a greedy algorithm, i.e., in each step the error is maximally reduced. Therefore, $a_{n+1}=\mathrm{nint}(\tilde{a}_{n+1})$, where $\mathrm{nint}(\cdot)$ denotes the nearest integer,
\begin{equation}
\tilde{a}_{n+1}:= \frac{\frac{\pi^4}{36}-e-\sum_{i=1}^n a_i \left(\frac{1}{\pi}-\frac{1}{e}\right)^{i}}{\left(\frac{1}{\pi}-\frac{1}{e}\right)^{n+1}}
\end{equation}
and $a_0=0$.
Now, we prove that the series converges. Note that, $|a_n-\tilde{a}_n|\leq \frac{1}{2}$, which implies that
\begin{equation}
\left| \left(\frac{1}{\pi}-\frac{1}{e} \right)^n \tilde{a}_n - \left(\frac{1}{\pi}-\frac{1}{e}\right)^n a_n \right| \leq \frac{1}{2} \left|\frac{1}{\pi}-\frac{1}{e}\right|^{n}
\end{equation}
Using the definition of $\tilde{a}_n$,
\begin{equation}
\left| \frac{\pi^4}{36} - e - \sum_{i=1}^n \left(\frac{1}{\pi}-\frac{1}{e}\right)^i a_i \right| \leq \frac{1}{2} \left|\frac{1}{\pi}-\frac{1}{e}\right|^{n}
\end{equation}
Since $\left(\frac{1}{e}-\frac{1}{\pi}\right) \approx 0.0496<1$, the series converges to $\frac{\pi^4}{36}- e$. Interestingly, note that $|\tilde{a}_n|\leq\frac{e \cdot \pi}{2(e-\pi)}\approx 10.08$, therefore, $|a_n|\leq 10$. Finally, any sequence $b_n$,
\begin{equation}
b_{n+1}=\mathrm{nint}\left(\frac{h_n-\sum_{i=0}^n b_i g_n^i}{g_n^{n+1}}\right)
\end{equation}
where $b_0=0$, $h_n\rightarrow \frac{\pi^4}{36}-e$, and $g_n\rightarrow \frac{1}{\pi}-\frac{1}{e}$ as $n\rightarrow \infty$, satisfies
\begin{equation}
\sum_{i=1}^\infty b_n \left(\frac{1}{\pi}- \frac{1}{e} \right)^i = \frac{\pi^4}{36}-e
\end{equation}
This follows since, $|h_n-\frac{\pi^4}{36}+e|\leq E_n$, where $E_n \rightarrow 0$, and
\begin{equation}
\left|h_{n-1}-\sum_{i=1}^n g_n^i b_i\right| \leq \frac{g_n^n}{2}
\end{equation}
Therefore, $ \left|\frac{\pi^4}{36}-e-\sum_{i=1}^n g_n^i b_i\right|\leq \frac{g_n^n}{2}+ E_n$. Now, using that $g_n=\frac{1}{\pi}-\frac{1}{e}+\epsilon_n$, where $\epsilon_n\rightarrow 0$,
\begin{align}
\left|\frac{\pi^4}{36}-e-\sum_{i=1}^n \left(\frac{1}{\pi}-\frac{1}{e}\right)^i b_i\right|&\leq \frac{g_n^n}{2}+ E_n+ \sum_{i=1}^n \sum_{j=0}^{i-1} \binom{i}{j} \left(\frac{1}{e}-\frac{1}{\pi}\right)^j \epsilon_n^{i-j}\\ & =\frac{g_n^n}{2}+ E_n+ \epsilon_n \sum_{i=1}^n \sum_{j=0}^{i-1} \binom{i}{j} \left(\frac{1}{e}-\frac{1}{\pi}\right)^j \epsilon_n^{i-j-1} \\& \leq \frac{g_n^n}{2}+ E_n+ \epsilon_n \left(\frac{1}{e}-\frac{1}{\pi}\right)^{-1}\sum_{i=1}^n \left[2\left(\frac{1}{e}-\frac{1}{\pi}\right)\right]^i \\ & \leq \frac{g_n^n}{2}+ E_n+ \epsilon_n \frac{2}{1-2\left(\frac{1}{e}-\frac{1}{\pi}\right)}
\end{align}
For example
\begin{equation}
h_n= \left(\sum_{i=1}^n \frac{5}{2i^4}-\frac{1}{i!}\right) -1
\end{equation}
\begin{equation}
g_n= \sum_{i=0}^n \frac{2\sqrt{2}}{9801} \frac{(4i)!(1103+26390i)}{(i!)^4 396^{4i}}- \frac{(-1)^i}{i!}
\end{equation}
|
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}
|
Find the real solutions (if any) of $x^{12}=3 x^9+4 x^6+4 x^3+2 x+2$
Find the real solutions(if any) of $$x^{12}=3 x^9+4 x^6+4 x^3+2 x+2$$
Note: I need only HINT:
My effort:
I let $v=x^3$, then we get
$$v^4=3v^3+4v^2+4v+2\sqrt[3]{v}+2 \Rightarrow v^4-3v^3-4v^2-4v=2(1+\sqrt[3]{v})$$
$$\begin{aligned}
& \Rightarrow v^4-3 v^3-4 v^2-3 v+1=(v+1)+2(\sqrt[3]{v}+1) \\
& \Rightarrow v^2\left(\left(v+\frac{1}{v}\right)^2-3\left(v+\frac{1}{v}\right)-6\right)=(v+1)+2(\sqrt[3]{v}+1)
\end{aligned}$$
Any way to proceed with an hint?
|
Let $$f(x)=x^{12}-3x^9-4x^6-4x^3-2x-2$$
By Intermediate value theorem, we have $f(1)<0, f(2)>0$, so $\exists$ atleast one root on $(1,2)$. But by Descarte's rule of signs, the number of changes in $f(x)$ is one, so only one positive root. Obviously there should be one more real root. So there exists a quadratic factor $x^2+px+q$ with positive discriminant, so we have:
$$f(x)=\left(x^2+p x+q\right)\left(x^{10}-p x^9+b x^8+c x^7+d x^6+e x^5+f x^4+g x^3+h x^2+m x-\frac{2}{q}\right)$$
Comparing the coefficients:
$$b=p^2-q$$
$$c+pb-pq=-3$$
$$d+pc+qb=0$$
$$e+pd+qc=0$$
$$f+pe+qd=-4$$
$$g+pf+qe=0$$
$$h+pg+qf=0$$
$$m+ph+qg=-4$$
$$\frac{-2}{q}+pm+hq=0$$
$$\frac{-2p}{q}+mq=-2$$
Now let us assume or try whether the factorization is possible over $\mathbb{Z}$. So let us start with $q=2$. So the last equation above becomes $$-p+2m=-2 \Rightarrow p=2(m+1)$$ Also the last but one equation will be $$-1+2m(m+1)=-hq$$ This is impossible since LHS is odd, RHS is even. The same reasoning applies when $q=-2$.
\newline
Now let $$q=1$$
Since $x^2+px+q$ should have real roots, we have
$$p^2 \geq 4$$
Now if $q=1$, we get $$-2p+m=-2 \Rightarrow m=2p-2 \Rightarrow h=2-2p^2+2p$$
$$\Rightarrow g=2p^3-2p^2-4p-2 ,f=-2p^4+2p^3+6p^2-2\Rightarrow e=2p^5-2p^4-8p^3+2p^2+6p+2$$
$$\Rightarrow d=-2p^6+2p^5+10p^4-4p^3-12p^2-2p-2-----(*)$$
Also from first two equations, we have
$$b=p^2-1,c=2p-3-p^3$$
From third equation, we get
$$\Rightarrow d=p^4-3p^2+3p+1------(**)$$
From $(*),(**)$, we get
$$2p^6-2p^5-9p^4+4p^3+9p^2+5p+3=0 $$
So that means $p$ can only be odd integer. Since $p^2 \geq 4$, we have $p=3,5,7,...$ or $p=-3,-5,-7,...$. By growth argument, it is easy to say that we cannot find any root of the above sixth degree polynomial with $p \geq 3, p \leq -3$
Thus we have $$\boxed{q \ne \pm 2, 1}$$
Now the only choice left over for $q$ is $q=-1$.For $q=-1$, the discriminant of $x^2+px+q$ is $D=p^2+4>0$.Now
Repeating the above backward substitutions, we get
$$2p^6+2p^5+9p^4+12p^3+9p^2+11p+5=0$$ Obviously the above sixth degree polynomial cannot have non negative roots. So $p<0$. By Inspection $p=-1$ satisfies. Hence $$p=-1,q=-1 \Rightarrow x^2+px+q=x^2-x-1 | f(x)$$ Also if at all there are any integer roots of the above sixth degree polynomial, the root should be an odd integer, that is $p=-1,-3,..$. We have already figured out $p=-1$. Again by growth argument we cannot have any root $p \leq -3$.Thus finally we have successfully found the quadratic factor of $f(x)$ which is $x^2-x-1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4604762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
Frullani like Trig integral $$\int_{0}^{\infty}\frac{\left|\cos\left ( x-\frac{\pi}{4} \right ) \right|- \left|\cos\left ( x+\frac{\pi}{4} \right ) \right| }{x}dx=\sqrt{2}\ln(1+\sqrt{2})$$ The above integral seems to look like a frullani type integral and has a closed form in terms of natural log. I tried to indefinitely integrate it. But the closed form are in terms $\text{Si}(x)$ and $\text{Ci}(x)$. I would highly appreciate if there's any method or unique substitution to evaluate this Integral.
|
We invoke the Fourier series for $\left|\cos x\right|$:
$$ \left| \cos x \right| = \frac{2}{\pi} - \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{(-1)^n}{4n^2 - 1} \cos(2nx) $$
From this, we obtain the following series representation for the numerator of the integrand:
\begin{align*}
\left|\cos\left(x-\frac{\pi}{4}\right)\right| - \left| \cos\left(x+\frac{\pi}{4}\right) \right|
= \frac{8}{\pi} \sum_{k=0}^{\infty} \frac{(-1)^k}{(4k+2)^2 - 1} \sin((4k+2)x) \tag{1}
\end{align*}
Then it can be proved that, when $\text{(1)}$ is plugged into OP's integral, the order of summation and integral can be interchanged.[1] Hence, if we denote the integral by $I$, then
\begin{align*}
I
&= \int_{0}^{\infty} \frac{ \left|\cos\left(x-\frac{\pi}{4}\right)\right| - \left| \cos\left(x+\frac{\pi}{4}\right) \right| }{x} \, \mathrm{d}x \\
&= \frac{8}{\pi} \sum_{k=0}^{\infty} \frac{(-1)^k}{(4k+2)^2 - 1} \int_{0}^{\infty} \frac{\sin((4k+2)x)}{x} \, \mathrm{d}x \\
&= 4 \sum_{k=0}^{\infty} \frac{(-1)^k}{(4k+2)^2 - 1} \\
&= 2 \left( 1 - \frac{1}{3} - \frac{1}{5} + \frac{1}{7} + \frac{1}{9} - \frac{1}{11} - \frac{1}{13} + \frac{1}{15} + \cdots \right)
\end{align*}
The last sum can be evaluated by invoking the Fourier series for $\log \left|\tan x\right|$:
$$ \log\left|\tan x\right| = -2 \sum_{k=0}^{\infty} \frac{\cos(2(2k+1)x)}{2k+1} \tag{2} $$
Indeed, plugging $x = \frac{\pi}{8}$ into $\text{(2)}$, we get
\begin{align*}
\log\left(\tan \frac{\pi}{8}\right)
&= -2 \sum_{k=0}^{\infty} \frac{\cos((2k+1)\frac{\pi}{4})}{2k+1} \\
&= -\sqrt{2} \left( 1 - \frac{1}{3} - \frac{1}{5} + \frac{1}{7} + \frac{1}{9} - \frac{1}{11} - \frac{1}{13} + \frac{1}{15} + \cdots \right) \\
&= -\frac{I}{\sqrt{2}}.
\end{align*}
Therefore
$$ I
= -\sqrt{2}\log\left(\tan \frac{\pi}{8}\right)
= \sqrt{2}\log(1+\sqrt{2}). $$
Addendum. Let us justify the claim [1] that the order of summation and integral can be interchanged. This will follow from the following more general claim:
Theorem. Suppose $\sum_{n=1}^{\infty} |a_n| < \infty$. Then the following equality holds:
$$ \begin{aligned}
\int_{0}^{\infty} \frac{1}{x} \left( \sum_{n=1}^{\infty} a_n \sin(nx) \right) \, \mathrm{d}x
&= \frac{\pi}{2} \sum_{n=1}^{\infty} a_n \\
&= \sum_{n=1}^{\infty} a_n \int_{0}^{\infty} \frac{\sin(nx)}{x} \, \mathrm{d}x
\end{aligned} \tag{3} $$
The convergence of the improper integral in the left-hand side of $\text{(3)}$ is part of the statement to be established within the proof.
Proof. Let $\operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} \, \mathrm{d}t$ be the sine integral. Then by the Weierstrass M-test, $ \sum_{n=1}^{\infty} a_n \frac{\sin(nx)}{x} $ converges uniformly on any $[a, b] \subset (0, \infty)$. Hence,
\begin{align*}
\int_{a}^{b} \frac{1}{x} \left( \sum_{n=1}^{\infty} a_n \sin(nx) \right) \, \mathrm{d}x
&= \sum_{n=1}^{\infty} a_n \int_{a}^{b} \frac{\sin(nx)}{x} \, \mathrm{d}x \\
&= \sum_{n=1}^{\infty} a_n \int_{na}^{nb} \frac{\sin t}{t} \, \mathrm{d}t \\
&= \sum_{n=1}^{\infty} a_n (\operatorname{Si}(bn) - \operatorname{Si}(an)).
\end{align*}
Since $\operatorname{Si}(\cdot)$ is bounded, by the dominated convergence theorem for series, we get
\begin{align*}
\lim_{\substack{b \to \infty \\ a \to 0}} \sum_{n=1}^{\infty} a_n (\operatorname{Si}(bn) - \operatorname{Si}(an))
&= \sum_{n=1}^{\infty} a_n \lim_{\substack{b \to \infty \\ a \to 0}} (\operatorname{Si}(bn) - \operatorname{Si}(an)) \\
&= \frac{\pi}{2} \sum_{n=1}^{\infty} a_n.
\end{align*}
Therefore the desired claim follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4605246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
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|
Help Solve $\int \sqrt{x^{2}+x-2}\,dx$ I have this integral $\int \sqrt{x^{2}+x-2}\,dx=\int |x-1|\sqrt{\frac{x+2}{x-1}}\,dx$.
My textbook advices to use the substitution $t^{2}=\frac{x+2}{x-1}$.
Observation: This integral should be solvable without using integral of modules as it is placed before moduled function integrals chapter.
I did the substitution and i obtaind $-18\int \frac{{t^{2}}}{|(t^{2}-1)^{3}|}\,dt$ . At this point i thought that since the substitution has to be an invertible function, i may consider a restrinction to cancel the absolute value.
I checked on wolfram alpha both integrals but it gives different results. Wolfram solution:
$$\int \sqrt{(x^2 + x - 2)}\,dx = \frac14 (2 x + 1) \sqrt{(x^2 + x - 2)} - \frac98 \log\left(2 \sqrt{(x^2 + x - 2)} + 2 x + 1\right) + c$$
Any help?
The purpose of the exercise is to trasform the irrational function into a rational one and use the Hermite's equation to solve it
|
Well, we are trying to solve:
$$\mathcal{I}\left(x\right):=\int\sqrt{x^2+x-2}\space\text{d}x\tag1$$
First, complete the square:
$$\mathcal{I}\left(x\right)=\int\sqrt{\left(x+\frac{1}{2}\right)^2-\frac{9}{4}}\space\text{d}x\tag2$$
Substitute $\text{u}=x+\frac{1}{2}$:
$$\mathcal{I}\left(x\right)=\int\sqrt{\text{u}^2-\frac{9}{4}}\space\text{du}\tag3$$
Now, substitute $\text{u}=\frac{3\sec\left(\text{s}\right)}{2}$:
$$\mathcal{I}\left(x\right)=\frac{9}{4}\int\tan^2\left(\text{s}\right)\sec\left(\text{s}\right)\space\text{ds}\tag4$$
Now, write $\tan^2\left(\text{s}\right)=\sec^2\left(\text{s}\right)-1$:
$$\mathcal{I}\left(x\right)=\frac{9}{4}\cdot\left(\int\sec^3\left(\text{s}\right)\space\text{ds}-\int\sec\left(\text{s}\right)\space\text{ds}\right)\tag5$$
Now, you can use:
*
*The reduction formula:
$$\int\sec^\text{n}\left(\text{s}\right)\space\text{ds}=\frac{\sin\left(\text{s}\right)\sec^{\text{n}-1}\left(\text{s}\right)}{\text{n}-1}+\frac{\text{n}-2}{\text{n}-1}\int\sec^{\text{n}-2}\left(\text{s}\right)\space\text{ds}\tag6$$
*And:
$$\int\sec\left(\text{s}\right)\space\text{ds}=\ln\left|\tan\left(\text{s}\right)+\sec\left(\text{s}\right)\right|+\text{C}\tag7$$
So, you will get:
\begin{equation}
\begin{split}
\mathcal{I}\left(x\right)&=\frac{9}{4}\cdot\left(\int\sec^3\left(\text{s}\right)\space\text{ds}-\int\sec\left(\text{s}\right)\space\text{ds}\right)\\
\\
&=\frac{9}{4}\cdot\left(\frac{\sin\left(\text{s}\right)\sec^{3-1}\left(\text{s}\right)}{3-1}+\frac{3-2}{3-1}\int\sec^{3-2}\left(\text{s}\right)\space\text{ds}-\int\sec\left(\text{s}\right)\space\text{ds}\right)\\
\\
&=\frac{9}{4}\cdot\left(\frac{\sin\left(\text{s}\right)\sec^2\left(\text{s}\right)}{2}+\frac{1}{2}\int\sec\left(\text{s}\right)\space\text{ds}-\int\sec\left(\text{s}\right)\space\text{ds}\right)\\
\\
&=\frac{9}{4}\cdot\left(\frac{\sin\left(\text{s}\right)\sec^2\left(\text{s}\right)}{2}-\frac{1}{2}\int\sec\left(\text{s}\right)\space\text{ds}\right)\\
\\
&=\frac{9}{4}\cdot\left(\frac{\sin\left(\text{s}\right)\sec^2\left(\text{s}\right)}{2}-\frac{1}{2}\cdot\ln\left|\tan\left(\text{s}\right)+\sec\left(\text{s}\right)\right|+\text{C}\right)\\
\\
&=\frac{9}{8}\cdot\left(\sin\left(\text{s}\right)\sec^2\left(\text{s}\right)-\ln\left|\tan\left(\text{s}\right)+\sec\left(\text{s}\right)\right|+\text{C}\right)\\
\\
&=\frac{9\left(\sin\left(\text{s}\right)\sec^2\left(\text{s}\right)-\ln\left|\tan\left(\text{s}\right)+\sec\left(\text{s}\right)\right|\right)}{8}+\text{C}
\end{split}\tag8
\end{equation}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4608272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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|
Any better way to find $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
Find the value of $\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)$
My Method:
I used the following Identities:
\begin{aligned}
& \sin A \cos B-\cos A \sin B=\sin (A-B) \\
& 2 \sin A \sin B=\cos (A-B)-\cos (A+B) \\
& \cos (2 A)=2 \cos ^2 A-1 \\
& \cos (3 A)=4 \cos ^3 A-3 \cos A \\
& \sin (2 A)=2 \sin A \cos A \\
& 2 \sin A \cos B=\sin (A+B)+\sin (A-B) \\
&
\end{aligned}
$$\begin{aligned}
S_1 & =\cot \left(10^{\circ}\right)+\cot \left(70^{\circ}\right)-\cot \left(50^{\circ}\right)=\cot(10^{\circ})-\cot \left(50^{\circ}\right)+\tan(20^{\circ}) \\
\\
\Rightarrow S_1 & =\frac{\cos \left(10^{\circ}\right)}{\sin \left(10^{\circ}\right)}-\frac{\cos \left(50^{\circ}\right)}{\sin \left(50^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\
\\
\\
\Rightarrow S_1 & =\frac{2 \sin \left(40^{\circ}\right)}{2 \sin \left(10^{\circ}\right) \sin \left(50^{\circ}\right)}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\
\\
\Rightarrow S_1 & =\frac{2 \sin \left(40^{\circ}\right)}{\cos \left(40^{\circ}\right)-\frac{1}{2}}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)} \\ \\
\Rightarrow S_1 & =\frac{4 \sin 40^{\circ}}{4 \cos ^2\left(20^{\circ}\right)-3}+\frac{\sin \left(20^{\circ}\right)}{\cos \left(20^{\circ}\right)}=\frac{6 \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)-3 \sin \left(20^{\circ}\right)}{0.5}
\end{aligned}$$
$$\begin{aligned}
& \Rightarrow S_1=\frac{3}{0.5}\left(2 \sin \left(40^{\circ}\right) \cos \left(20^{\circ}\right)-\sin \left(20^{\circ}\right)\right) \\ \\
& \Rightarrow S_1=6\left(\sin \left(60^{\circ}\right)\right)=3 \sqrt{3}
\end{aligned}$$
|
Since $$\tan{x}+\tan\left(60^{\circ}+x\right)+\tan\left(120^{\circ}+x\right)=$$
$$=\tan{x}+\frac{\sqrt3+\tan{x}}{1-\sqrt3\tan{x}}+\frac{-\sqrt3+\tan{x}}{1+\sqrt3\tan{x}}=3\tan3x,$$ we obtain:
$$\cot10^{\circ}+\cot70^{\circ}-\cot50^{\circ}=\tan80^{\circ}+\tan20^{\circ}+\tan140^{\circ}=3\tan(3\cdot20^{\circ})=3\sqrt3.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4608452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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|
Prove a binomial identity: $\sum_{i=1}^n i \binom{2n}{n-i}=\frac12(n+1) \binom{2n}{n-1}$ I want to prove the product of even/odd power with a combinatorial number:
\begin{aligned}
\sum_{i=1}^n i \binom{2n}{n-i}&=\frac12(n+1) \binom{2n}{n-1}, \\
\sum_{i=1}^n i^2 \binom{2n}{n-i}&=2^{2n-2} n.
\end{aligned}
I am sure these results are correct since WolframAlpha verifies them. I wonder if the first formula can be proved, and also the general version can be proved: Formula related to combinatorial number.
Below is proof for the second formula:
First, note that $r\binom{n}{r}=n\binom{n-1}{r-1}$, and $\sum_{i=0}^n\binom{n}{i}=2^n$. We thus have
\begin{aligned}
\sum_{i=0}^n i\binom{n}{i}&=2^{n-1}n, \\
\sum_{i=0}^n i^2\binom{n}{i}
&= \sum_{i=0}^n i(i-1)\binom{n}{i}+\sum_{i=0}^n i\binom{n}{i} \\
&= 2^{n-2}(n^2-n) + 2^{n-1} n \\
&= n(n+1)2^{n-2}.
\end{aligned}
Therefore,
\begin{aligned}
\sum_{i=0}^{2n}(n-i)^2\binom{2n}{i}
&= n^2\sum_{i=0}^{2n}\binom{2n}{i}-2n\sum_{i=0}^{2n}i\binom{2n}{i}+\sum_{i=0}^{2n}i^2 \binom{2n}{i} \\
&= 2^{2n}n^2-2^{2n+1}n^2+n(2n+1)2^{2n-1} \\
&= 2^{2n-1} n,
\end{aligned}
and thus
\begin{aligned}
\sum_{i=0}^n i^2 \binom{2n}{n-i}
= \sum_{i=0}^n (n-i)^2 \binom{2n}{i}
= \frac12 \sum_{i=0}^{2n} (n-i)^2 \binom{2n}{i}
= 2^{2n-2} n.
\end{aligned}
|
Let $$S_n=\sum_{k=0}^{n} {2n \choose k}$$
and let
$$S'_n=\sum_{k=0}^{n} k {2n \choose n-k}=\sum_{k=0}^{n} (n-k) {2n \choose k}=nS_n-T_n$$
$$2^{2n}=\sum_{k=0}^{2n} {2n \choose k}=\sum_{k=0}^{n} {2n \choose k}+\sum_{k=n+1}^{2n} {2n \choose k}$$
$$\implies S_n+\sum_{k=n+1}^{2n} {2n \choose k}=2^{2n}$$
Change $k$ to $2n-j$, then
$$S_n+\sum_{j=n-1}^{0} {2n \choose 2n-j}=2^{2n}$$
$$\implies S_n+ \sum_{j=0}^{n} {2n \choose j}-{2n \choose n}=2^{2n}$$
$$\implies S_n=2^{2n-1}+\frac{1}{2}{2n \choose n}.$$
Next let
$$T_n=\sum_{k=0}^{n} k {2n \choose k}=\sum_{k=0}^{n} k\frac{(2n)!}{k!(2n-k)!}=2n\sum_{k=0}^{n}{2n-1 \choose k-1}=2n\sum_{j=0}^{n-1}{2n-1 \choose j} $$
We have $$2^{2n-1}= \sum_{k=0}^{2n-1} {2n-1\choose k}=\sum_{k=0}^{n-1} {2n-1 \choose k}+\sum_{k=n}^{2n-1} {2n-1 \choose k}=\sum_{k=0}^{n-1} {2n-1 \choose k}+\sum_{j=0}^{n-1} {2n-1 \choose n+j}$$ $$2^{2n-1}\implies \sum_{k=0}^{n-1} {2n-1 \choose k}+\sum_{j=0}^{n-1} {2n-1 \choose n+j}=\sum_{k=0}^{n-1} {2n-1 \choose k}+\sum_{j=0}^{n-1} {2n-1 \choose n-1-j}$$ $$\implies 2^{2n-1}= \sum_{k=0}^{n-1} {2n-1 \choose k}+\sum_{m=0}^{n-1} {2n-1 \choose m} \implies \sum_{k=0}^{n-1} {2n-1 \choose k}=2^{2n-2}$$
Finally, we have $$S'_n=n2^{2n-1}+\frac{n}{2} {2n \choose n}-n2^{2n-1}=\frac{n}{2}{2n \choose n}$$
|
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"url": "https://math.stackexchange.com/questions/4611896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
A limit of a sequence that yields different answers This is the problem:
$\lim_{n\rightarrow\infty}\frac{1}{n\sqrt{n}}\sum^n_{k=1}\sqrt{k}$
A friend and I tried to solve it using different methods and our results are very different. Our attempts:
First attempt. Using Riemann sums
$\lim_{n\rightarrow\infty}\frac{1}{n\sqrt{n}}\sum^n_{k=1}\sqrt{k}=\lim_{n\rightarrow\infty}\frac{1}{n\sqrt{n}}(\sqrt{1}+...+\sqrt{n})=\lim_{n\rightarrow\infty}\frac{1}{n}(\sqrt{\frac{1}{n}}+...+\sqrt{\frac{n}{n}})=\lim_{n\rightarrow\infty}\frac{1}{n}\sum^n_{k=1}\sqrt{\frac{k}{n}}$
If we choose the partition $\Pi_n= (0, \frac{1}{n}, \frac{2}{n},..., \frac{n}{n}=1)$ and let $f(\frac{k}{n})=\sqrt{\frac{k}{n}}$:
$\lim_{n\rightarrow\infty}\frac{1}{n}\sum^n_{k=1}\sqrt{\frac{k}{n}}=\int_0^1\sqrt{x}\ dx=\frac{2}{3}$
Second attempt. Using Stolz-Cèsaro theorem
$\frac{\sqrt{1}+...+\sqrt{n+1}-\sqrt{1}-...-\sqrt{n}}{(n+1)\sqrt{n+1}-n\sqrt{n}}=\frac{1}{(\sqrt{(n+1)^3}-\sqrt{n^3})(\sqrt{n+1}+\sqrt{n})}\rightarrow 0$
We can't figure out why one of these (or both) may be wrong, so any hint or help would be really appreciated.
|
The second attempt is incorrect. You have made a mistake in your algebra as the first identity written in that part is false. You can try the following
\begin{align*}
\frac{\sqrt{n+1}}{(n+1)\sqrt{n+1}- n\sqrt n} &= \frac{\sqrt{n+1}}{(n+1)\sqrt{n+1}- n\sqrt n} \times \frac{(n+1)\sqrt{n+1}+n\sqrt n }{(n+1)\sqrt{n+1}+n\sqrt n} \\
&=\frac{\sqrt{n+1}\left( (n+1)\sqrt{n+1}+n\sqrt n \right)}{(n+1)^2(n+1)- n^2 n}\\
&= \frac{(n+1)^2 +n\sqrt{n(n+1)} }{(n+1)^3- n^3} \\
&= \frac{(n+1)^2}{(n+1)^3-n^3} + \frac{n\sqrt{n(n+1)} }{(n+1)^3- n^3}
\end{align*}
These two terms both converge to $1/3$ so the limit is $2/3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4612253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Limits problem (unable to solve further) Struggling to solve this problem,
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +\dots+ \frac{1}{6n}\right)$
My approach:
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +...+ \frac{1}{6n}\right)$
= $\displaystyle\lim\limits_{n \to \infty} \int_{0}^{1}(x^{n-1} + x^{n} + x^{n-2}+...+ x^{6n-1}) dx$
= $\displaystyle\lim\limits_{n \to \infty}\int_{0}^{1}\left(x^{n-1} \cdot \frac{x^{5n+1} - 1}{x-1}\right)dx$
got stuck here and don't know how to solve it further (other approaches which are simpler would also help)
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You also can use Riemann-sums:
$$
\left(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots + \frac{1}{6n}\right)
$$
$$
=
\left(1+\frac{n}{n+1}+\frac{n}{n+2}+\dots + \frac{n}{n+ (5n-1)}\right)\frac{1}{n}+ \frac{1}{6n}
$$
$$
=\left(1+\frac{1}{1+1/n}+\frac{1}{1+2/n}+\dots + \frac{1}{1+(5n-1)/n}\right)\frac{1}{n} + \frac{1}{6n}
$$
$$
\to \int_0^5\frac{1}{1+x}dx + 0 = \ln(6) \quad (n \to \infty).
$$
Edit: On the interval $[0,5]$ we consider the partition
$$
x_k=\frac{k}{n} \quad (k=0, \dots, 5n),
$$
and the function $f(x)=1/(1+x)$. Choosing the left boundary point on each interval $[x_k,x_{k+1}]$ we get the Riemann-sum
$$
\sum_{k=0}^{5n-1} f(x_k) (x_{k+1}-x_k) = \sum_{k=0}^{5n-1} \frac{1}{1+k/n} \cdot \frac{1}{n}.
$$
Now it is known that
$$
\sum_{k=0}^{5n-1} f(x_k) (x_{k+1}-x_k) \to \int_0^5 f(x) dx \quad (n \to \infty);
$$
see https://en.wikipedia.org/wiki/Riemann_sum, for example.
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"url": "https://math.stackexchange.com/questions/4612750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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|
If $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$, prove $abcd \geq 3$. Given four positive real numbers $a,b,c,d$. It is given that $\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1$. Prove $abcd \geq 3$.
I've applied AM-HM inequality and got the result, $a^4+b^4+c^4+d^4 \geq 12$. Then by AM-GM, $a^4+b^4+c^4+d^4\geq 4abcd; 12\geq 4abcd \implies 3\geq abcd$.
But this is contradictory. Then I realized I was doing all wrong. Does anyone have method for this one?
|
Clearing all denominators, we have that
$$
\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{1}{1+c^4}+\frac{1}{1+d^4}=1
$$
is equivalent to
$$
a^4b^4c^4d^4 = (a^4b^4 + a^4c^4 + a^4d^4 + b^4c^4 + b^4d^4 + c^4d^4) + 2(a^4 + b^4 + c^4 + d^4) + 3
$$
Applying arithmetic mean $\ge$ geometric mean to the brackets on the right gives
$$
a^4b^4c^4d^4 \ge 6 a^2b^2c^2d^2 + 8 a bcd + 3
$$
This is again equivalent to
$$
(a bcd - 3) (a bcd +1)^3 \ge 0
$$
Hence $ a bcd \ge 3$.
$\qquad \Box$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4612949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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|
Find $\frac {dy}{dx}$ if $y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$ taking JDs advice i used $(fg)'=f'g+fg'$ rule
$$f=\frac x{\sqrt{(1-x^2)}}$$ $$f'=\frac 1{\sqrt{(1-x)}^3}$$
$$g=sin^{-1}x$$ $$g'=\frac{1}{\sqrt{1-x^2}}$$
so anyway adding together
we get
$$\frac 1{(\sqrt{(1-x)}^3}*sin^-x+\frac x{\sqrt{(1-x^2)}}*\frac{1}{\sqrt{1-x^2}}$$
which can be simplified
$$\frac {sin^{-1}x}{(1-x)^3}+\frac x{\sqrt{(1-x^2)^2}}$$
i then multiplied $$\frac x{\sqrt{(1-x^2)^2}}$$ with $\sqrt{(1-x^2)}$ on both numerator and denominator
getting
$$\frac {x(\sqrt{(1-x^2)})}{\sqrt{(1-x^2)^3}}$$
and combining them both we get
$$\frac{sin^{-1}x+x(\sqrt{(1-x^2)}}{\sqrt{(1-x^2)^3}}$$
ps
i never used arcsin before and am compeletely unfamilliar with it
|
You might see that the function can be rewritten as follows:
$$
y=-\arcsin(x)\Big((1-x^2)^{\tfrac{1}{2}}\Big)'
$$
|
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"url": "https://math.stackexchange.com/questions/4614715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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|
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