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Roll a fair die and denote the outcome by $Y$ . Then flip $Y$ fair coins and let $X$ denote the number of tails observed.
Roll a fair die and denote the outcome by $Y$. Then flip $Y$ fair coins and let $X$ denote the number of tails observed.
*
*Find the probability mass function of $X$
*Compute the expectation of $X$
My Attempt
The number of possible combinations of $y$ coins is $2^y$ and the number of combinations with $x$ tails is $\binom{y}{x}$. Thus, the probabilities are $p_{X,Y}(x,y)=\frac{\binom{y}{x}}{2^y}$. Then the joint p.m.f. is
\begin{array}{c|c|ccc}
& & & & Y & & & \\ \hline
& & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
& 1 & 1/2 &\ 1/2 & 3/8 & 1/4 & 5/32 & 3/32 \\
& 2 & 0 & 1/4 & 3/8 & 3/8 & 5/16 & 15/64 \\
X& 3 & 0 & 0 & 1/8 & 1/4 & 5/16 & 5/16 \\
& 4 & 0 & 0 & 0 & 1/16 & 5/32 & 15/64\\
& 5 & 0 & 0 & 0 & 0 & 1/32 & 3/32\\
& 6 & 0 & 0 & 0 & 0 & 0 & 1/64\\
\end{array}
This is my thinking but I know that it is wrong because the sum of all the probabilities exceeds $1$. Can I get some pointers on where I went wrong?
After I find the joint p.m.f., I can sum the values in the rows to find the p.m.f. for $X$.
|
If Y represents the outcome of a fair die and X represents the number of tails observed then,
$ \ 0 \leq x \leq y$ and $1 \leq y \leq 6 \ $
$ \displaystyle P(X = x|Y=y) = {y \choose x} \cdot \frac{1}{2^y}$
So the table that you have put together is correct but you missed $x = 0$, probability of which is same as for $x = y$.
Edit:
To find pmf of X,
$ \displaystyle P(X = x) = \frac{1}{6} \sum \limits_{y=1}^6 P(X = x|Y=y) \ \ (0 \leq x \leq 6)$
To find expectation of X,
$\mathbb{E}(X) = \sum \limits_{x=0}^6 x \cdot P(X = x)$
|
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|
Paradox of $-i$ seemingly equal to $1$ via the Wallis product for $\pi$ and the Euler sine product Assuming $x$ is a real variable throughout,$$\frac{\sinh(ix)}{i}=\sin(x)$$
$$\frac{\sinh(\pi ix)}{\pi ix} = \frac{\sin(\pi x)}{\pi x}$$
$$\frac{e^{\pi ix}-e^{-\pi ix}}{2\pi ix}=\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2}\right)$$
Letting $x=\frac{1}{2}$ yields:
$$\frac{2}{\pi i}=\prod_{n=1}^{\infty}\left(1-\frac{1}{4n^2}\right)$$
$$\frac{1}{i}=-i=\frac{\pi}{2}\prod_{n=1}^{\infty}\left(\frac{4n^2-1}{4n^2}\right)=\frac{\pi}{2}\prod_{n=1}^{\infty}\left(\frac{(2n-1)(2n+1)}{4n^2}\right)$$
The famous Wallis product for $\pi$ tells us:
$$\frac{\pi}{2}=\frac{2\cdot 2}{1\cdot 3}\cdot\frac{4\cdot 4}{3\cdot 5}\cdot\frac{6\cdot 6}{5\cdot 7}\dots=\prod_{n=1}^{\infty}\left(\frac{4n^2}{(2n-1)(2n+1)}\right)$$
Therefore, in a multiplication which I hope to be legitimate, we seem to get:
$$-i=\prod_{n=1}^{\infty}\left(\frac{4n^2}{(2n-1)(2n+1)}\right)\cdot\prod_{n=1}^{\infty}\left(\frac{(2n-1)(2n+1)}{4n^2}\right)=\prod_{n=1}^{\infty}\left(\frac{(2n-1)(2n+1)}{(2n-1)(2n+1)}\right)=1$$
Clearly, $1\neq-i$, so what have I done wrong here? I randomly came up with this while half asleep, so there may be some trivial error here - perhaps the products cannot be combined like this - but I'm at a loss!
Thanks for any help.
|
when you let $x=\frac{1}{2}$ in $\frac{e^{\pi ix}-e^{-\pi i x}}{2\pi i x}$ your numerator should be $2i$ not 2
|
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|
Calculating the following integral: $\int_0^{\frac{z}{1+z}}\frac{-X+z-Xz}{z}dX$ I'm trying to calculate the following integral: $\int_0^{\frac{z}{1+z}}\frac{-X+z-Xz}{z}dX$ however, I seem to be getting the wrong answer and would like some support on where I went wrong. The answer should be: $\frac{z}{2(z+1)}$
What I have tried:
$$\int_0^{\frac{z}{1+z}}\frac{-X+z-Xz}{z}dX \implies \frac{-(1+z)}{z}\int_0^{\frac{z}{1+z}}X\space dX+\int_0^{\frac{z}{1+z}}+1 \space dX$$
Because:
$-X-Xz = -X(1+z)$ & $\frac{-X(1+z)+z}{z} = \frac{-X(1+z)}{z}+1$
After simplification I get:
$$\left(\frac{-(1+z)}{z}\cdot \frac{(\frac{z}{1+z})^2}{2}\right) \space + \frac{z}{1+z} = \frac{-z^2+z}{2\left(z+1\right)}$$
Where might I have gone wrong?
|
Everything is correct besides your last equality:
\begin{align}
\left(\frac{-(1+z)}{z}\cdot \frac{(\frac{z}{1+z})^2}{2}\right)+ \frac{z}{1+z} &= \left(\frac{-(1+z)}{z}\cdot \frac{z^2}{2(1+z)^2}\right)+ \frac{z}{1+z} \\
&= \frac{-z}{2(1+z)}+ \frac{z}{1+z}\\
&= \frac{-z}{2(1+z)} + \frac{2z}{2(1+z)} \\
&= \frac{z}{2(z+1)}
\end{align}
|
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|
About the inequality $x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12$
Problem: Let $x > 0$. Prove that
$$x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12.$$
Remark 1: The problem was posted on MSE (now closed).
Remark 2: I have a proof (see below). My proof is not nice.
For example, we need to prove that
$\frac{3x^2 - 3}{x^2 + 4x + 1} + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7} \le 0$ for all $0 < x < 1$ for which my proof is not nice.
I want to know if there are some nice proofs. Also, I want my proof reviewed for its correctness.
Any comments and solutions are welcome and appreciated.
My proof (sketch):
We split into cases:
i) $x \ge 1$:
Clearly, $x^{x^{x^{x^{x^x}}}}\ge x^x$.
By Bernoulli's inequality, we have
$x^x = (1 + (x - 1))^x \ge 1 + (x - 1)x = x^2 - x + 1 \ge \frac12 x^2 + \frac12$. The inequality is true.
ii) $0 < x < 1$:
It suffices to prove that
$$x^{x^{x^{x^x}}}\ln x \ge \ln \frac{x^2 + 1}{2}$$
or
$$x^{x^{x^{x^x}}} \le \frac{\ln \frac{x^2 + 1}{2}}{\ln x}$$
or
$$x^{x^{x^x}}\ln x \le \ln \frac{\ln \frac{x^2 + 1}{2}}{\ln x}$$
or
$$x^{x^{x^x}}\ge \frac{1}{\ln x}\ln \frac{\ln \frac{x^2 + 1}{2}}{\ln x}.$$
It suffices to prove that
$$x^{x^{x^x}}\ge \frac{7}{12} \ge \frac{1}{\ln x}\ln \frac{\ln \frac{x^2 + 1}{2}}{\ln x}. \tag{1}$$
First, it is easy to prove that $$x^x \ge \mathrm{e}^{-1/\mathrm{e}}
\ge \frac{1}{\ln x}\ln\frac{\ln\frac{7}{12}}{\ln x}.$$
Thus, the left inequality in (1) is true.
Second, let $f(x) = x^{7/12}\ln x - \ln \frac{x^2 + 1}{2}$. We have
\begin{align*}
f'(x) &= \frac{7}{12x^{5/12}}
\left(\ln x + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7}\right)\\
&\le \frac{7}{12x^{5/12}}
\left(\frac{3x^2 - 3}{x^2 + 4x + 1} + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7}\right)\\
&\le 0 \tag{2}
\end{align*}
where we have used
$\ln x \le \frac{3x^2 - 3}{x^2 + 4x + 1}$ for all $x$ in $(0, 1]$.
Also, $f(1) = 0$. Thus, $f(x) \ge 0$ for all $x$ in $(0, 1)$.
Thus, the right inequality in (1) is true.
Note: For the inequality $\frac{3x^2 - 3}{x^2 + 4x + 1} + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7} \le 0$ for all $0 < x < 1$,
we let $x = y^{12}$ and it suffices to prove that
$11y^{47} + \cdots + 3 \ge 0$ (a polynomial of degree $47$, a long expression) for all $0 < y < 1$.
We are done.
|
A partial answer
First Fact :
For $x\in(0,1)$ we have :
$$x^{x^{x^{x^{x^{x}}}}}\geq x^{x^{x}}$$
Proof: see the Reference in my other answer .
Second Fact
For $x\in(0,1)$ we have :
$$ x^{x^{x}}\geq x^{\left(1+\left(x-1\right)x\right)}$$
Hint :use Bernoulli's inequality.
Third Fact
For $x\in[0.65,1]$ we have :
$$x^{\left(1+\left(x-1\right)x\right)}\geq b(x)=\left(x\left(1+\left(x-1\right)\cdot\left(\left(x-1\right)x\right)+0.5\left(x-1\right)^{2}\cdot\left(\left(x-1\right)x\right)\cdot\left(\left(\left(x-1\right)x\right)-1\right)\right)\right)$$
Rewrite $x^{\left(1+\left(x-1\right)x\right)}=xx^{\left(\left(x-1\right)x\right)}$ and use the binomial theorem for $p(x)=x^a$ at $x=1$ .We stop the power series at the second order .
Remains to show :
$$0.5 (x - 1)^2 (x^5 - 2 x^4 + 3 x^2 - 1)=b(x)-0.5x^2-0.5\geq0$$
Or
$$(x^5 - 2 x^4 + 3 x^2 - 1)\geq 0$$
Wich is left to the reader and easy using derivatives .
A lemma :
We have for $a,x\in(0,1)$:
$$x^{a^{a^{1.86a\left(1+a\left(a-1\right)\right)}}}\leq x^{a^{a^{a^{a^{a}}}}}\quad\quad(I)$$
Wich is a refinement if $x=a$
The inequality $(I)$ is equivalent to :
$$a^{a^{a}}\leq 1.86a\left(1+a\left(a-1\right)\right)$$
It seems that we have for $a\in(0.03,1)$
$$a^{a^{a}}\leq a^{0.86\left(1+\left(a-1\right)a\right)} \leq 1.87a\left(1+a\left(a-1\right)\right)$$
We start from :
$$a^{a^{a}}\leq a^{0.86\left(1+\left(a-1\right)a\right)}$$
Wich is equivalent to :
$$a^{a}\geq 0.86\left(1+\left(a-1\right)a\right)$$
The function $f(a)=a^{a}$ is convex so we have :
$$f(x)\geq f'(b)(x-b)+f(b)$$
Remains to choose judicious points wich is not hard using a graphic so I let it to the reader .
Also see the reference .
Now we start from :
$$a^{0.86\left(1+\left(a-1\right)a\right)} \leq 1.87a\left(1+a\left(a-1\right)\right)$$
A trick is : put $a$ in exponent on both side and the inequality have the form :
$$(1.87u)^v\geq v^{0.86u}$$
The inequality in $u,v$ reminds me the inequality :
Let $a,b>0$ and $k\in(0,1)$ then we have :
$$a(1-k)+bk\geq a^{1-k}b^{k}$$
Using this we have :
$$\left(1.87u\left(x\right)\right)^{-1}\left(v\left(x\right)^{\left(\frac{x}{\left(0.86u\left(x\right)\right)}\right)^{-1}}\left(x\right)+u\left(x\right)\cdot1.87\cdot\left(1-x\right)\right)\geq \left(\left(1.87u\left(x\right)\right)^{-x}\right)x^{\left(0.86u\left(x\right)\right)}$$
Where :
$$u(x)=x\left(1+\left(x-1\right)x\right)$$ and : $$v(x)=x$$ and $x\in(0.03,1)$
The rest is smooth using the lemma 7.1 (p.136 see the first reference for that) .
End lemma
Second lemma
Let $a,x\in(0,1)$ then we have :
$$x^{a^{\frac{7}{12}}}\leq x^{a^{a^{1.87a\left(1+\left(a-1\right)a\right)}}}$$
Proof :
It's equivalent to :
$$a^{1.87a\left(1+\left(a-1\right)a\right)}\geq \frac{7}{12}$$
The function :
$$n(a)=a^{1.87a\left(1+\left(a-1\right)a\right)}$$
Is convex on $(0,1)$ so admits a global minimum on $(0,1)$. The rest is smooth again !
End Second lemma
Remains to show for $x,a\in(0,1)$ and $x\geq a$:
$$0.5a^{2}+0.5\leq x^{a^{\frac{7}{12}}}$$
I pursue it later thanks for advices or comments !
Reference :
Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions",
The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137.
https://eudml.org/doc/223938
https://www.planetmath.org/convexfunctionslieabovetheirsupportinglines
|
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Integral $\int x^2\sqrt{1-x^2} dx$ I tried to solve the integral $$\int x^2\sqrt{1-x^2} dx.$$
(I know there are already solutions here on MSE but I need some help to find my error)
Substituting $x(y):=\cos(y)$ and using trigonometirc rules yields:
$$\int \cos^2(y)\sqrt{1-\cos(y)^2}(-1)\sin(y) dy= (-1)\int \cos^2(y)\sin^2(y) dy\\= \left(-\frac{1}{4}\right)\int \sin^2(2y)dy=\left(-\frac{1}{4\cdot 2}\right)\int 2\sin^2(2y)dy.$$
Then substituting again, $z(y):=2y$:
$$
\left(-\frac{1}{8}\right)\int \sin^2(z)dz= \left(-\frac{1}{8}\right)\left(\frac{1}{2}z-\frac{1}{2}\sin(z)\cos(z)\right).$$
Resubstitution yields:
$$
\left(-\frac{1}{8}\right)\left(\frac{1}{2}z-\frac{1}{2}\sin(z)\cos(z)\right)= \left(-\frac{1}{8}\right)\left(y-\frac{1}{2}\sin(2y)\cos(2y)\right) = \left(-\frac{1}{8}\right)\left(\arccos(x)-\frac{1}{2}\sin(2\arccos(x))\cos(2\arccos(x))\right)=\left(-\frac{1}{8}\right)\left(\arccos(x)+\sin(\arccos(x))x\left(1-2x^2\right)\right)=
\left(-\frac{1}{8}\right)\left(\arccos(x)+\sqrt{1-x^2}x\left(1-2x^2\right)\right).
$$
The result resembles the offical solution $\left(\text{which is }\left(\frac{1}{8}\right)\left(\arcsin(x)-\sqrt{1-x^2}x\left(1-2x^2\right)\right)\right)$ except that mine has $\arccos(x)$ instead of $\arcsin(x)$ and the "$-$" sign is flipped. If I perform the substitution $x(y):=\sin(y)$. Then I arrive at the official soultion.
After revising my solution several times I have no idea where my mistake is? Maybe someone else sees it?
|
$$\left(-\frac{1}{8}\right)\left(\arccos(x)+\sqrt{1-x^2}x\left(1-2x^2\right)\right)+C\\=\left(-\frac{1}{8}\right)\left(\frac{\pi}2-\arcsin(x)+\sqrt{1-x^2}x\left(1-2x^2\right)\right)+C\\=-\frac{\pi}{16}+\frac{\arcsin(x)}{8}-\frac{1}{8}\sqrt{1-x^2}x(1-2x^2)+C\\=\frac18\left(\arcsin(x)-\sqrt{1-x^2}x(1-2x^2)\right)+C_1$$
|
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|
Find the number of 5-digit number divisible by 6 which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
Find the number of $5$-digit number divisible by $6$ which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
I started by considering the following cases:
*
*Unit digit = $0$
I can fill four places with digits ${1,2,4,5}$ , so number of 5-digit numbers $= P(4,4) = 24$
*Unit digit = $2$
I can fill four places with digits ${0,1,3,4,5}$ such that (a) $0$ does not come in the first place (b) either of $1,4$ is used (c) $0,3,5$ are always used. I have to fill 4 places with $0,(1,4),3,5$. So, total number of arrangements $=P(4,4)$. Number of arrangements when $0$ comes as first digit $=P(3,3)$. Number of arrangements of $(1,4) = 2$. Therefore , number of 5-digit numbers $=2(P(4,4)-P(3,3)) = 36$
*Unit digit = $4$
Similarly, number of 5-digit numbers $= 36$
Therefore, the total number of 5-digit numbers $= 24+36+36 = 96$.
But the correct answer is $108$. Where did I make a mistake?
|
The sum of the digits must be divisible by $3$, so either $0$ or $3$ must be absent (since $0+1+2+3+4+5=15$ is divisible by $3$).
If $0$ is absent, then we have a permutation of $12345$ where the last digit is $2$ or $4$, giving $(2)(4!)=(2)(24)=48$ possibilities.
If $3$ is absent, then we have a permutation of $01245$ where the first digit is not $0$ and the last digit is $0, 2,$ or $4$. Such permutations could be counted as follows:
First, consider the case where the last digit is $0$. In this case, the remaining digits must form a permutation of $1245$, giving $4!=24$ possibilities.
Second, consider the case where the last digit is $2$ or $4$. In this case, the remaining digits must form a permutation of $0145$ (if the last digit is $2$) or $0125$ (if the last digit is $4$) where the first digit is not $0$. This gives $(2)(4!-3!)=(2)(24-6)=(2)(18)=36$ possibilities.
Thus, there are $48+24+36=108$ possible numbers.
|
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|
Proving $\sum_{k=1}^{n}\cos\frac{2\pi k}{n}=0$ I want to prove that the below equation can be held.
$$\sum_{ k=1 }^{ n } \cos\left(\frac{ 2 \pi k }{ n } \right) =0, \qquad n>1 $$
Firstly I tried to check the equation with small values of $n$
$$ \text{As } n=2 $$
$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 2 } \right) + \cos\left(\frac{ 2 \pi \cdot 2 }{ 2 } \right) $$
$$ = \cos\left(\pi\right) + \cos\left(2 \pi\right) $$
$$ = -1+ 1 =0 ~~ \leftarrow~~ \text{Obvious} $$
But
$$ \text{As}~~ n=3 $$
$$ \cos\left(\frac{ 2 \pi \cdot 1 }{ 3 } \right) +\cos\left(\frac{ 2 \pi \cdot 2 }{ 3 } \right) + \cos\left(\frac{ 2 \pi \cdot 3 }{ 3 } \right) $$
$$ = \cos\left(\frac{ 2 \pi }{ 3 } \right) + \cos\left(\frac{ 4 \pi }{ 3 } \right) + \cos\left( 2\pi \right) $$
$$ = \cos\left(\frac{ 2 \pi }{ 3 } \right) + \cos\left(\frac{ 4 \pi }{ 3 } \right) + 1 =?$$
What formula(s) or property(s) can be used to prove the equation?
|
Observe that
$$\sum_{k=1}^n\cos(\frac{2\pi k}{n})=\operatorname{Re}\left(\sum_{k=1}^ne^{\frac{2\pi ki}{n}}\right).$$
Consider the sum
$$\sum_{k=1}^ne^{\frac{2\pi ki}{n}}=\sum_{k=1}^n(e^{\frac{2\pi i}{n}})^k$$
and use the fact that
$$1+\sum_{k=1}^n r^k =\frac{1-r^{n+1}}{1-r}$$
to show it's zero.
|
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|
If $3\sin x +5\cos x=5$, then prove that $5\sin x-3\cos x=3$
If $3\sin x +5\cos x=5$ then prove that $5\sin x-3\cos x=3$
What my teacher did in solution was as follows
$$3\sin x +5\cos x=5 \tag1$$
$$3\sin x =5(1-\cos x) \tag2$$
$$3=\frac{5(1-\cos x)}{\sin x} \tag3$$
$$3=\frac{5\sin x}{(1+\cos x)} \tag4$$
$$5\sin x-3\cos x=3 \tag5$$
However this should not be true when $\sin x=0$, as division by zero is not defined; and also, if $\sin x=0$, then the expression we have to prove evaluates to $-3$. In other words question is incomplete but my teacher denied it.
Am I correct in my reasoning?
|
Yes you are right,he/she should have broken the problem into two cases.
Case $1.$ When $\sin x\ne 0$, then according to your teacher $5\sin x-3\cos x=3$.
Case $2.$ If $\sin x=0$, then from the given equation we get $\cos x=1$. Plugging these in to $5\sin x-3\cos x$ we get the required value as $-3$.
We can do the same problem in an alternative way:
$$3\sin x+5\cos x=5$$
Let $$5\sin x-3\cos x =k$$
Squaring and adding both the above equations we get
$$k^2+25=34$$
$\implies$
$$k=\pm 3$$
|
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|
Solve $y' = (1+\frac{y-1}{2x})^2$ Solve $y' = (1+\frac{y-1}{2x})^2$
My first thought was to expand to see if I can get a linear form:
$$y' = 1 + \frac{y-1}{x} + (\frac{y-1}{2x})^2 = 1 + \frac{y-1}{x} + \frac{y^2-2y+1}{4x^2}$$
$$y' = 1 + \frac{y}{x} -\frac{1}{x} + \frac{y^2}{4x^2} + \frac{-2y}{4x^2} + \frac{1}{4x^2} \rightarrow y'+y(\frac{2}{4x^2}-\frac{1}{x}) = \frac{(2x-1)^2+y^2}{4x^2}$$
However, this is not even a bernoulli equation. How should I approach this problem?
|
$$y' = \left(1+\frac{y-1}{2x}\right)^2$$
Substitute :
$$z=1+\frac{y-1}{2x}$$
$$y'=2(z-1+xz')$$
Then the DE is separable.
$$2xz'=(z-1)^2+1$$
|
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Looking for other approaches to evaluate $\lim _{x\to 0^+}\frac{\sin 3x}{\sqrt{1-\cos ^3x}}$ Here is my approach: $$\lim _{x\to 0^+}\frac{\sin 3x}{\sqrt{1-\cos ^3x}}=\lim _{x\to 0^+}\frac{1}{\sqrt{\cos^2x+\cos x+1}}\times\frac{\sin 3x}{\sqrt{1-\cos x}}$$
$$=\lim _{x\to 0^+}\frac1{\sqrt3}\times\frac{\sin 3x}{\sqrt{1-\cos x}}\times\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}=\frac{\sqrt2}{\sqrt3}\times\lim _{x\to 0^+}\frac{\sin 3x}{\sin x}=\sqrt6$$
Can you evaluate $\lim _{x\to 0^+}\dfrac{\sin 3x}{\sqrt{1-\cos ^3x}}$ with other approaches?
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I would write something like that,
$$\sin(3x) \sim 3x$$
$$\cos^3(x) = \left(1-\dfrac{x^2}{2}+o(x^2)\right)^3 = 1-3\dfrac{x^2}{2}+o(x^2)$$
Hence,
$$1-\cos^3(x) \sim 3\dfrac{x^2}{2} $$
Hence,
$$\sqrt{1-\cos^3(x)} \sim \sqrt{3/2}x $$
Finally,
$$\dfrac{\sin(3x)}{\sqrt{1-\cos^3(x)}} \sim \dfrac{3}{\sqrt{3/2}} \sim \sqrt{6}$$
|
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|
Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of $x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP then $p + q$ is?
Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of
$x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP
then $p + q$ is ?
My solution approach :-
I have assumed $α,β,γ,δ$ to be as follows ;
$α = \frac{a}{r^3}$
$β = \frac{a}{r}$
$γ = ar$
$δ = ar^3$
Here the common ratio is $r^2$.
Now $α+β = 1$ and $γ+δ = 4$ and solving these two equations in terms of $a$ and $r$ gives the value of $r= \sqrt{2} \text{ and } \sqrt{-2}$ which leads to the values of $a$ to be $\frac{2\sqrt{2}}{3}$ and $\frac{-2\sqrt{2}}{3}$ respectively.
In both the above cases, $α,β,γ,δ$ turns out to be the same i.e. $\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}$ which gives the value of $p+q = \frac{34}{9}$.
Now if I assume $α,β,γ,δ$ to be as follows ;
$α = a$
$β = ar$
$γ = ar^2$
$δ = ar^3$
Here the common ratio is just $r$.
But in this case $r$ values turn out to be $\pm2$ and for $r=2$ we get $α,β,γ,δ$ to be $\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}$ whereas for $r=-2$ we get $α,β,γ,δ$ to be $-1,2,-4,8$ and hence $p+q = \frac{34}{9} \text{ or }-34 $.
Now I am getting confused why there is this difference in both the approaches. What am I doing wrong? Please help me on this !!!
Thanks in advance !!!
|
In the first approach: when you solve for $r$ you would have got an equation $r^4=4$. This gives you $r^2=2$ and $r^2=-2$. You have only considered the first case (i.e. when $r^2=2$). If you consider the second case $r^2=-2$, then you get $r=\pm i\sqrt{2}$. In which case you will get $p+q=-34$. So both approaches will yield the same answers.
Some details:
$$\alpha+\beta=1 \implies a\frac{(1+r^2)}{r^3}=1.$$
Likewise
$$\gamma+\delta=4 \implies ar(1+r^2)=4.$$
Thus
$$r^4=4.$$
This means $r^2=\pm 2$. Consequently $a=\frac{4}{3r}$ or $a=\frac{-4}{r}$ (the second case occurs when $r^2=-2$). Now we can use,
$$\alpha \cdot \beta=p \implies p=\frac{a^2}{r^4}=\frac{a^2}{4}.$$
Likewise we have
$$q=a^2r^4=4a^2.$$
So,
$$p+q=\frac{17}{4}a^2$$
Now plug in the two possible values of $a^2=\frac{16}{9r^2}=\frac{8}{9}$ OR $a^2=\frac{16}{r^2}=-8$ to get both the possibilities.
Additional remark:
In fact, from the conditions of the problem that $p$ and $q$ are integers, we can see that $r^2=2$ is NOT a valid solution because it gives $a^2=\frac{8}{9}$, in which case $p=\frac{a^2}{4}=\frac{2}{9} \not\in \Bbb{Z}$. Thus $\color{red}{r^2=-2}$ is the ONLY valid possibility. In which case
$$\color{magenta}{p+q=-34}$$
is the ONLY valid answer.
|
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|
Solve : $(2x - 4y + 5) \frac{dy}{dx} + x - 2y + 3 = 0 , y(2) = 2.5$ I have to find the solution of following differential equation:
$$(2x - 4y + 5) \frac{dy}{dx} + x - 2y + 3 = 0 , y(2) = 2.5$$
It can be re-written as
$$(2(x - 2y) + 5) \frac{dy}{dx} + (x - 2y) + 3 = 0$$
let $u = (x-2y)$
so, on differentiating both sides with respect to $x$ we get
$$\frac{du}{dx} = 1 - 2\cdot\frac{dy}{dx}$$
$$\therefore \frac{dy}{dx} = \frac{1 - \frac{du}{dx}}{2}$$
using this value in initial equation we get,
$$(2u + 5) (\frac{1-\frac{du}{dx}}{2}) + u + 3 = 0$$
On further solving we get
$$\frac{2u + 5}{4u + 11}.\frac{du}{dx} = 1$$
Integrating both sides with respect to $x$, we get
$$\int \frac{2u + 5}{4u + 11}.du = \int dx$$
$$\implies \int (\frac{1}{2} - \frac{1}{8u + 22})\cdot du = x + C$$
$$\implies \frac{1}{2}\cdot u - \frac{1}{8} \ln(8u + 22) = x + C$$
putting the value of $u$ in the above equation we get,
$$\implies \frac{1}{2} (x-2y) - \frac{1}{8} ln(8(x - 2y)+ 22) = 2 + C$$
we have, when $x = 2 , y = 2.5.$
When I put these value of $x$ and $y$ in the above equation to obtain the value of $C$
$ln(-negative Value)$ is obtained, what is wrong here?
The answer given in the book is :
$$4x + 8y + ln(4x - 8y + 11) = 28$$
|
The answer was given by Lalit Tolani. This is an additional information.
From the implicit equation
$$4x+8y+\ln(4x-8y+11)=28$$
the explicit solution $y(x)$ cannot be expressed with a finite number of elementary functions. It can be analytically expressed on closed form thanks to a special function namely the Lambert's W function.
Let $X=8y-4x-11\quad\implies y=\frac18(X+4x+11)$
$$4x+(X+4x+11)+\ln(X)=28$$
$$X+\ln(X)=-8x+17$$
$$Xe^X=e^{-8x+17}$$
From the definition of the Lambert's W function :
$$X=W(e^{-8x+17})$$
$$8y-4x-11=W(e^{-8x+17})$$
$$\boxed{y(x)=\frac{4x+11+W(e^{-8x+17})}{8}}$$
|
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|
Let $K/F$ be a finite extension of finite fields and suppose $K=F[\alpha]$ with $\alpha^{15}=1$ then $\mid K:F \mid \leq 4$.
Let $K/F$ be a finite extension of finite fields and suppose $K=F[\alpha]$. If $\alpha^{15}=1$ then show that $\mid K:F \mid \leq 4$.
Let $F=\mathbb F_q$, where $q$ is power of some prime $p$, say, and let $\mid K:F\mid=d$, so that $K=\mathbb F_{q^d}$. If $\min_{\alpha,F}(x)=f(x)$, then we have $\deg(f)=d$ and $f(x)\mid(x^{15}-1)$, hence $f(x)\mid (x^{16}-x)$. I wanted to use the fact that $g(x)$ monic irreducible in $\mathbb F_q[x]$. Then $g(x)\mid (x^{q^{n}}-x) \iff \deg(g)\mid n$. But how do I show that characteristic of the field $F$ is $2$. I need help. Thanks.
|
Claim:$\;$If $K$ is a field with an element $\alpha$ and a subfield $F$ such that
*
*$F$ is a finite field.$\\[4pt]$
*$K=F(\alpha)$.$\\[4pt]$
*$\alpha^{15}=1$.
then $[K:F]\le 4$.
Proof:
For the subfield $F(\alpha^3,\alpha^5)$ of $K$, we have
\begin{align*}
&
\frac{\alpha^5}{\alpha^3}\in F(\alpha^3,\alpha^5)
\\[4pt]
\implies\;&
\alpha^2\in F(\alpha^3,\alpha^5)
\\[4pt]
\implies\;&
\frac{\alpha^3}{\alpha^2}\in F(\alpha^3,\alpha^5)
\\[4pt]
\implies\;&
\alpha\in F(\alpha^3,\alpha^5)
\\[4pt]
\end{align*}
hence $F(\alpha^3,\alpha^5)=F(\alpha)=K$.
Next consider $4$ cases . . .
Case $(1)$:$\;\alpha^3,\alpha^5\in F$.
Then $K=F$, hence $[K:F]=1$.
Case $(2)$:$\;\alpha^3\in F,\alpha^5\not\in F$.
\begin{align*}
\text{Then}\;\;&
\alpha^{15}-1=0
\\[4pt]
\implies\;&
(\alpha^5-1)(\alpha^{10}+\alpha^5+1)=0
\\[4pt]
\implies\;&
\alpha^{10}+\alpha^5+1=0
\\[4pt]
\end{align*}
so $\alpha^5$ is a root of $x^2+x+1$, hence $[K:F]=[F(\alpha^5):F]\le 2$.
Case $(3)$:$\;\alpha^5\in F,\alpha^3\not\in F$.
\begin{align*}
\text{Then}\;\;&
\alpha^{15}-1=0
\\[4pt]
\implies\;&
(\alpha^3-1)(\alpha^{12}+\alpha^9+\alpha^6+\alpha^3+1)=0
\\[4pt]
\implies\;&
\alpha^{12}+\alpha^9+\alpha^6+\alpha^3+1=0
\\[4pt]
\end{align*}
so $\alpha^3$ is a root of $x^4+x^3+x^2+1$, hence $[K:F]=[F(\alpha^3):F]\le 4$.
Case $(4)$:$\;\alpha^3,\alpha^5\not\in F$.
Then the order, $o(\alpha)$, of $\alpha$ in $K^*$ is $15$.
Hence $1,\alpha,\alpha^2,\alpha^3,...\alpha^{14}$ are all distinct.
As was shown in case $(2)$, $\alpha^5$ is a root of $x^2+x+1$, hence $[F(\alpha^5):F]\le 2$.
As was shown in case $(3)$, $\alpha^3$ is a root of $x^4+x^3+x^2+x+1$.
Moreover, $\alpha^6,\alpha^9,\alpha^{12}$ are also roots of $x^4+x^3+x^2+x+1$.
Since $\alpha^3\not\in F$, it follows that
*
*$\alpha^{6},\alpha^{9}\not\in F$, else $\alpha^{18}\in F$, contradiction, since $\alpha^{18}=\alpha^{3}$.$\\[4pt]$
*$\alpha^{12}\not\in F$ since $\alpha^{12}=\alpha^{-3}$.
Thus $\alpha^3,\alpha^6,\alpha^9,\alpha^{12}$ are distinct roots of $x^4+x^3+x^2+x+1$, and none of them are in $F$.
It follows that $x^4+x^3+x^2+x+1$ is either irreducible in $F[x]$, or else factors in $F[x]$ as a product of two irreducible quadratics.
Hence $[F(\alpha^3):F]$ equals $2$ or $4$.
Either way, since $[F(\alpha^5):F]\le 2$, it follows that $F(\alpha^5)\subseteq F(\alpha^3)$.
The above line is the only place where we use the hypothesis that $F$ is a finite field.
Hence $K=F(\alpha^3,\alpha^5)=F(\alpha^3)$, so $[K:F]=[F(\alpha^3):F]\le 4$.
Thus in all $4$ cases, we have $[K:F]\le 4$, as was to be shown.
Note:
With some minor adjustments to the above argument (a future edit), we can obtain the slightly stronger result: $[K:F]\in\{1,2,4\}$.
|
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|
Mathematicals inequalities For $$x,y,z>0 $$
Prove that
$$(2xyz)^2 \ge (x^3+y^3+z^3+xyz)(x+y-z)(y+z-x)(z+x-y)$$
I have tried a famous inequalities: $$(x+y-z)(y+z-x)(z+x-y) \le xyz$$
So the problem is:
$$3xyz \ge x^3+y^3+z^3$$
But in fact, this isn’t true
Help me plz
|
If $\prod\limits_{cyc}(x+y-z)<0$ so the inequality is obvious.
Thus, it's enough to assume that $\prod\limits_{cyc}(x+y-z)\geq0.$
Now, since $x+y-z<0$ and $x+z-y<0$ gives $x<0$, which is a contradiction, we can assume that
$x+y-z=c\geq0,$ $x+z-y=b\geq0$ and $y+z-x=a\geq0$ and we need to prove that
$$(a+b)^2(a+c)^2(b+c)^2\geq4abc(abc+\sum_{cyc}(a^3+2a^2b+2a^2c))$$ or
$$\sum_{sym}(a^4b^2-a^4bc+a^3b^3-2a^3bc+a^2b^2c^2)\geq0$$ or
$$\prod_{cyc}(a-b)^2+4\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0,$$ which is true by Schur.
|
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|
Prove that $(n+1)a\leq a^{n+1}+n, \forall a,n\in\mathbb{N}$. We can start from the fact that:
\begin{align*}
0\leq a^n + a^{n-1} +\ldots + a^2 + a-n,\forall a,n\in\mathbb{N}.
\end{align*}
The above is true, since if $a>1$, then $a^n>n, \forall n \in\mathbb{N}$. Also if $a = 1$, then we will have $1 + 1 + \ldots + 1 + 1-n = n-n = 0, \forall n \in \mathbb{N}$. Thus we have:
\begin{align*}
0 &\leq (a^{n} + a^{n-1} + \ldots + a^2 + a-n)(a-1)\\
0 &\leq (a^{n} + a^{n-1} + \ldots + a^2 + a)(a-1)-n(a-1)\\
0 &\leq a(a^{n-1} + a^{n-2} + \ldots + a + 1)(a-1) -na + n\\
0 &\leq a(a^{n}-1) -na + n\\
0 &\leq a^{n + 1} -a-na + n\\
0 &\leq a^{n + 1} -(n + 1) a + n\\
(n + 1)a &\leq a^{n + 1} + n\\
\end{align*}
Another form of reason is using the inequality of the arithmetic mean with the geometric mean, as follows:
\begin{align*}
\frac{a^{n + 1} + n}{n + 1} &= \frac{a^{n + 1} + 1 + 1 + \ldots + 1 + 1}{n + 1}\\
&\geq \sqrt[n + 1]{a^{n + 1}(1)(1)\ldots(1)(1)}\\
&= \sqrt[n + 1]{a^{n + 1}}\\
& = a.
\end{align*}
Finally it is solved and we have $a^{n + 1} + n \geq (n + 1)a$.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
|
Alternatively, the result follows from the Bernoulli’s inequality:
$$[1+(a-1)]^{n+1}\geq 1+(n+1)(a-1),$$ which is true for all real numbers $a\geq 0$, and all integers $n\geq -1$. (Exclude the case when $a=0,n=-1$ if necessary to avoid the ambiguity $0^0$.)
|
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|
Cartesian product of finitely generated integral monoids I would like to understand the fact that the Cartesian product of finitely generated integral monoids is also a finitely generated monoid.
$\textbf{Definition:}$ A monoid is a set of vectors in Q and is defined under addition. When it only contains integer vectors, then it is called an integral monoid.
Let $M_1, M_2, \cdots, M_k$ be finitely generated integral monoids. Then, my goal is to investigate $M_1 \times M_2 \times \cdots \times M_k$.
The proof goes as follows. For the sake of simplicity, let us assume $k=2$. Let ${r}^1_1, {r}^1_2, \cdots {r}^1_n$ and ${r}^2_1, {r}^2_2, \cdots {r}^2_n$ correspond the finite set of vectors used to generate $M_1$ and $M_2$, respectively. Then, the following vectors can be used to generate $M_1 \times M_2$.
$
\begin{pmatrix}
{r}^1_1 \\
0
\end{pmatrix}
,
\cdots,
\begin{pmatrix}
{r}^1_n \\
0
\end{pmatrix},
\begin{pmatrix}
0 \\
{r}^2_1
\end{pmatrix}
,
\cdots,
\begin{pmatrix}
0 \\
{r}^2_n
\end{pmatrix} $
Could someone explain to me how these vectors are used to generate $M_1 \times M_2$ and provide a small example if possible?
|
Take any $\begin{pmatrix} r^1 \\ r^2\end{pmatrix} \in M_1\times M_2$. Since $M_1$ is generated by $r_1^1,\dots,r_{n_1}^1$ and $M_2$ is generated by $r_1^2,\dots,r_{n_2}^2$, we may write
\begin{align}
r^1 &= k_1^1 r_1^1 + \cdots + k_{n_1}^1 r_{n_1}^1 \\
r^2 &= k_1^2 r_1^2 + \cdots + k_{n_2}^2 r_{n_2}^2
\end{align}
for some integers $k_i^1,k_j^2$ for $i=1,\dots,n_1$ and $j=1,\dots,n_2$.
Hence we have
\begin{align}
\begin{pmatrix} r^1 \\ r^2\end{pmatrix} &=
\begin{pmatrix} k_1^1 r_1^1 + \cdots + k_{n_1}^1 r_{n_1}^1+0+0+\dots+0+0 \\
0+0+\dots+0+0+k_1^2 r_1^2 + \cdots + k_{n_2}^2 r_{n_2}^2 \end{pmatrix} \\
&=
k_1^1 \begin{pmatrix} r_1^1 \\ 0 \end{pmatrix}
+\cdots+
k_{n_1}^1 \begin{pmatrix} r_{n_1}^1 \\ 0 \end{pmatrix}
+
k_1^2 \begin{pmatrix} 0 \\ r_1^2 \end{pmatrix}
+\cdots+
k_{n_2}^2 \begin{pmatrix} 0 \\ r_{n_2}^2 \end{pmatrix}.
\end{align}
This shows that $\begin{pmatrix} r_1^1 \\ 0 \end{pmatrix},\dots,\begin{pmatrix} r_{n_1}^1 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ r_1^2 \end{pmatrix},\dots,\begin{pmatrix} 0 \\ r_{n_2}^2 \end{pmatrix}$ generate $M_1\times M_2$.
|
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|
The $x^2+ax+b = 0$ has only one solution and this is $x = -1+1/2$. What are the values of $a$ and $b$? This is a two part question. I was able to solve the first part. I need help with the second part.
a) The equation $x^2 + ax + b = 0$ has solutions $x = 2$ and $x = -5$. Find $a$ and $b$.
I was able to solve this one. we just have to set $f(2)$ equal to $f(-5)$. We will get $a=3$ and $b=-10$
b) The $x^2+ax+b = 0$ has only one solution and this is $x = -1+1/2$.
What are the values of $a$ and $b$?
I don't understand how a quadratic equation can have only one solution. Isn't a quadratic equation always supposed to have two solutions. How do I solve this when only one solution is given? What do I set the expression equal to?
|
*
*
$$\begin{align}&x^2+ax+b \\
\iff &(2x+a)^2=a^2-4b\end{align}$$
If the quadratic equation has one real root, then $\Delta=a^2-4b=0$.
This implies,
$$\begin{cases}2x+a=0 \\a^2=4b \end{cases}\implies \begin{cases} a=-2x \\ b=x^2\end{cases}$$
*
If the quadratic has one real root, then, $x_1=x_2$.
By the Vieta's formulas, we have
$$\begin{cases} x_1+x_2=-a \\x_1x_2=b \end{cases} \implies \begin{cases} a=-2x \\ b=x^2\end{cases}$$
*
If the quadratic has one real root, then
$$x^2+ax+b=(x-x_1)^2$$
$$\begin{align}&\implies x^2+ax+b=x^2-2xx_1+x_1^2 \\
&\implies a=-2x_1,~ b=x_1^2.\end{align}$$
|
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|
$\lim_{x\to 0^+}\dfrac{\tan^2(\frac1{\sqrt{1-x^2}}-1)}{\left(1-\cos(\sqrt{2x})\right)^n}=a$. What is the value of $a+n$?
Suppose $\lim_{x\to
0^+}\dfrac{\tan^2(\frac1{\sqrt{1-x^2}}-1)}{\left(1-\cos(\sqrt{2x})\right)^n}=a$.
What is the value of $a+n$ ?
$1)\frac74\qquad\qquad2)\frac94\qquad\qquad3)\frac{15}4\qquad\qquad4)\frac{17}4$
The limit is $\frac00$. For $x\to0^+$ both numerator and denominator tends to $0^+$, so $a>0$. I can use equivalences for $\tan u\sim u$ For $u\to0$. Hence in the numerator we can write $\dfrac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}}$. But I'm not sure if this helps.
I think I should find $n$ first then evaluate the limit but I don't know how to find $n$ at first place.
|
We have, from Newton's series:
$$(1-x^2)^{-\frac 12}=1+\frac {x^2}{2}+O(x^4)$$
Also, Taylor series for $\tan x$ yields:
$$\tan x=x+\frac {x^3}{3}+O(x^5)$$
Hence, $$\tan\left(\frac {1}{\sqrt {1-x^2}}-1\right)=\frac {x^2}{2}+O(x^4)$$
Squaring results in:
$$\tan^2\left(\frac {1}{\sqrt{1-x^2}}-1\right)=\frac {x^4}{4}+O(x^6)$$
Now, Taylor Series for $$\cos(\sqrt {2x})=1-\frac {(\sqrt {2x})^2}{2!}+\frac {(\sqrt {2x})^4}{4!}+O(x^3)$$
Thus, denominator evaluates to:
$$(1-\cos(\sqrt{2x}))^n=(x-\frac {x^2}{6}+O(x^3))^n$$
Thus, it is clear that non-zero limiting value
$$L=\frac {\frac {x^4}{4}+O(x^6)}{x^n(1-\frac {x}{6}+O(x^2))^n}$$
exists at $n=4$, and at that value $a=\frac 14$. Hence, we get option D.
|
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|
Showing $\sum_{n=1}^\infty\frac1{n^2}\left(1+\frac1{2^2}+\cdots+\frac1{n^2}\right)=\frac{7\pi^4}{360}$ Help me calculate the amount in a simpler way:
\begin{aligned}
\mathcal{S}&=\sum\limits_{n=1}^{\infty }\frac{1}{n^2}\left ( 1+\frac{1}{2^2}+\ldots+\frac{1}{n^2} \right )=\sum\limits_{n=1}^{\infty }\frac{H^{\left ( 2 \right )}_n}{n^2}=-\sum\limits_{n=1}^{\infty }H^{\left ( 2 \right )}_n\int\limits_{0}^{1}x^{n-1}\ln xdx=&&\\
&=-\int\limits_{0}^{1}\frac{\ln x}{x}\sum\limits_{n=1}^{\infty }H^{\left ( 2 \right )}_nx^ndx=\int\limits_{0}^{1}\frac{\ln x}{x}\frac{\operatorname{Li}_2x}{1-x}dx=-\int\limits_{0}^{1}\left ( \frac{1}{x}+\frac{1}{1-x} \right )\ln x\operatorname{Li}_2xdx=&&\\
&=-\int\limits_{0}^{1}\frac{\ln x\operatorname{Li}_2x}{x}dx-\int\limits_{0}^{1}\frac{\ln x\operatorname{Li}_2x}{1-x}dx=&&\\
&=-\frac{\ln^2 \operatorname{Li}_2x}{2}\Bigg|_0^1-\frac{1}{2}\int\limits_{0}^{1}\ln^2 x\cdot \frac{\ln \left ( 1-x \right )}{x}dx+\ln x\operatorname{Li}_2x\ln \left ( 1-x \right )\Bigg|_0^1-&&\\
&-\int\limits_{0}^{1}\ln \left ( 1-x \right )\left ( \frac{\operatorname{Li}_2x}{x}-\frac{\ln x\ln \left ( 1-x \right )}{x} \right )dx=&&\\
&=-\frac{1}{2}\int\limits_{0}^{1}\frac{\ln^2 x\ln \left ( 1-x \right )}{x}dx+\int\limits_{0}^{1}\frac{\ln^2 \left ( 1-x \right )\ln x}{x}dx+\int\limits_{0}^{1}\operatorname{Li}_2xd\left ( \operatorname{Li}_2x \right )=&&\\
&=\frac{1}{2}\sum\limits_{n=1}^{\infty }\frac{1}{n}\int\limits_{0}^{1}x^{n-1}\ln^2 xdx+2\sum\limits_{n=1}^{\infty }\left ( \frac{H_n}{n}-\frac{1}{n^2} \right )\int\limits_{0}^{1}x^{n-1}\ln x+\frac{\operatorname{Li}^2_2x}{2}\Bigg|_0^1=&&\\
&=\frac{1}{2}\sum\limits_{n=1}^{\infty }\frac{1}{n}\cdot \frac{2}{n^3}+2\sum\limits_{n=1}^{\infty }\left ( \frac{H_n}{n}-\frac{1}{n^2} \right )\left ( -\frac{1}{n^2} \right )+\frac{\pi^4}{72}=&&\\
&=\zeta \left ( 4 \right )-2\sum\limits_{n=1}^{\infty }\left ( \frac{H_n}{n^3}-\frac{1}{n^4} \right )+\frac{\pi^4}{72}=\frac{\pi^4}{90}-2\left ( \frac{\pi^4}{72}-\frac{\pi^4}{90} \right )+\frac{\pi^4}{72}=\frac{\pi^4}{30}-\frac{\pi^4}{72}=\frac{7\pi^4}{360}
\end{aligned}
|
You don’t need any integrals.
$$S=\sum_i \sum_{j<=i} 1/(i^2 j^2)= \sum_i \sum_j 1/(2i^2 j^2) + \sum_i 1/(2i^4) =(\zeta(2)^2 +\zeta(4))/2= (\pi^4/36+\pi^4/90)/2=7\pi^4/360$$
|
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Proving $\frac{2(1+y)\sqrt{1+x}+y\sqrt{1+y}}{2(1+x)\sqrt{1+y}+x\sqrt{1+x}} = \frac{1}{(1+x)^2}$ with $x\sqrt{1+y}+y\sqrt{1+x}=0$ I want to find $\frac{dy}{dx}$ in $x\sqrt{1+y}+y\sqrt{1+x}=0$ and I proceed through 2 different ways expecting the same answer.
Method:-1
$x\sqrt{1+y}=-y\sqrt{1+x}$
$\implies x^2(1+y)=y^2(1+x)$
$\vdots$
$\implies y=-1+\frac{1}{1+x}$
$$\frac{dy}{dx}=\frac{-1}{(1+x)^2}$$
Method:-2
$\phi(x,y)=x\sqrt{1+y}+y\sqrt{1+x}=0$
$$\frac{dy}{dx}=-\frac{\frac{\partial {\phi}}{\partial x}}{ \frac{\partial {\phi}}{\partial y} }$$
$$\frac{dy}{dx}=-\frac{2(1+y)\sqrt{1+x}+y\sqrt{1+y}}{2(1+x)\sqrt{1+y}+x\sqrt{1+x}}$$
But I could not find any way to convert $\frac{2(1+y)\sqrt{1+x}+y\sqrt{1+y}}{2(1+x)\sqrt{1+y}+x\sqrt{1+x}}$ to
$\frac{1}{(1+x)^2}$.
Thank you for your help!
|
Your expression is $$\frac{2(1+y)\sqrt{1+x}+y\sqrt{1+y}}{2(1+x)\sqrt{1+y}+x\sqrt{1+x}}$$
Taking $\sqrt{1+y}$ common from numerator and $\sqrt{1+x}$ common from denominator, we get $$\frac{\sqrt{1+y}}{\sqrt{1+x}}\cdot\frac{2\sqrt{1+y}\sqrt{1+x}+y}{2\sqrt{1+x}\sqrt{1+y}+x}$$
Given that $x\sqrt{1+y}+y\sqrt{1+x}=0\implies\frac{\sqrt{1+y}}{\sqrt{1+x}}=-\frac yx$. Also, $\sqrt{1+x}\sqrt{1+y}=-\frac yx(1+x)$. Putting these in the working expression, we get $$-\frac yx\cdot\frac{-2\frac yx(1+x)+y}{-2\frac yx(1+x)+x}\\=-\frac yx\cdot\frac{-2y-2xy+xy}{-2y-2x+x^2}\\=\frac {y^2}x\cdot\frac{x+2}{-2y-2x+x^2}$$
Since our objective is to obtain the final answer only in $x$, so, we have to take out the value of $y$ solely in terms of $x$ the way you have done in your first method. So, putting $y=-1+\frac{1}{1+x}$ here, our expression does reduce to the following$$\frac1{(1+x)^2}$$
|
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Simple proof that $\sum_{k = 1}^\infty \sum_{n = 1}^\infty a_k b_n / (k + n) \lesssim (\sum_k a_k^2)^{1/2} (\sum_n b_n^2)^{1/2}$ I have a complicated proof involving real interpolation + restricted bounds on operators, but I can only imagine there is a simpler proof of this statement involving some combinations of basic inequalities, such as Cauchy-Schwartz.
|
To avoid the ambiguity about the convergence, assume $a_n,b_n \ge 0$ and $\sum a_n^2 =1$.
We have:
$$
\begin{align}
\left(\sum_{k\ge 1}\sum_{n \ge 1} \frac{a_kb_n}{n+k}\right)^2 &\le \left( \sum_{k \ge 1} a_k^2 \right)\left( \sum_{k \ge 1} \left( \sum_{n \ge 1} \frac{b_n}{n+k} \right)^2 \right)
\\
& \le \sum_{k \ge 1} \left( \sum_{n \ge 1} \frac{b_n^2\sqrt{n}}{n+k} \right) \underbrace{ \left( \sum_{n \ge 1} \frac{1}{\sqrt{n}(n+k)}\right)}_{=\sum_{n=1}^{k} \frac{1}{\sqrt{n}(n+k)}+ \sum_{n \ge k+1} \frac{1}{\sqrt{n}(n+k)}}
\\
& \le \sum_{k \ge 1} \left( \sum_{n \ge 1} \frac{b_n^2\sqrt{n}}{n+k} \right)\left( 2\frac{1}{\sqrt{k}}+2\frac{1}{\sqrt{k}} \right)
\\
&= 4\sum_{n \ge 1} b_n^2 \sqrt{n}\left( \sum_{k \ge 1} \frac{1}{\sqrt{k}(k+n)} \right)
\\
& \le 4\sum_{n \ge 1} b_n^2 \sqrt{n} \frac{4}{\sqrt{n}} (\text{ like above})
\\
& \le 16 \sum_{n \ge 1} b_n^2
\end{align}
$$
Hence the conclusion.
|
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|
Prove $\sum_{n=2}^{\infty}\frac{1}{n^2+e}<\frac{1}{2}$ Of course, you can use the following formula
$$\sum_{n = 1}^\infty \frac{1}{n^2 + a^2} = \frac{\pi\coth(\pi a)}{2a} - \frac{1}{2a^2},$$
but which is too "advanced". We want to find a solution only depending on inequality estimation only.
Maybe, we can obtain
\begin{align*}
\sum_{n=2}^{\infty} \frac{1}{n^2+e}&\le \sum_{n=2}^{100}\frac{1}{n^2+e}+\sum_{101}^{\infty}\frac{1}{n^2}=\sum_{n=2}^{100}\left(\frac{1}{n^2+e}-\frac{1}{n^2}\right)+\sum_{n=2}^{\infty}\frac{1}{n^2}\\
&=\frac{\pi^2}{6}-1+\sum_{n=2}^{100}\left(\frac{1}{n^2+e}-\frac{1}{n^2}\right)<\frac{1}{2},
\end{align*}
which is true by checking on machine, but too hard to compute by hand.
|
I only use $\sum_{n=1}^\infty \frac{1}{n^k}$ for $k=2,4,6$.
Using $\frac{1}{n^2}-\frac{1}{n^2+e} = \frac{e}{n^2(n^2+e)}$, it suffices to show that $\sum_{n=2}^\infty \frac{1}{n^2(n^2+e)} > \frac{\pi^2-9}{9e}$. Then using $\frac{1}{n^4}-\frac{1}{n^2(n^2+e)} = \frac{e}{n^4(n^2+e)}$, it suffices to show $\sum_{n=2}^\infty \frac{1}{n^4(n^2+e)} < \frac{\pi^4-90}{90e}-\frac{\pi^2-9}{9e^2}$. Equivalently, we wish to show $\sum_{n=3}^\infty \frac{1}{n^4(n^2+e)} < \frac{\pi^4-90}{90e}-\frac{\pi^2-9}{9e^2}-\frac{1}{16(4+e)}$. So it suffices to show $\sum_{n=3}^\infty \frac{1}{n^6} < \frac{\pi^4-90}{90e}-\frac{\pi^2-9}{9e^2}-\frac{1}{16(4+e)}$, which is equivalent to $\frac{\pi^6}{945}-\frac{65}{64} < \frac{\pi^4-90}{90e}-\frac{\pi^2-9}{9e^2}-\frac{1}{16(4+e)}$, which is easy to do by hand/calculator (there's actually some room to spare).
|
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If the roots of the equation $x^4 + ax^3 + bx^2 + cx + d = 0$ are in geometric progression then, If the roots of the equation $x^4 + ax^3 + bx^2 + cx + d = 0$ are in geometric progression then,
$a) b^2 = ac$
$b) a^2 = b$
$c) c^2 = a^2d$
Using Vieta's relations, finding values of coefficients in terms of the assumed roots is what first comes to mind. That process works out well to give the answer ($c)$) But, as it can be seen, this method is quite lengthy. Is there a shorter method to solve this question or a trick?
|
Here is the OP's original method. Assume the roots are $k, kr, kr^2, kr^3$. Then:
$$k(1+r+r^2+r^3) = -a$$
$$k^2r+k^2r^3+k^2r^4+k^2r^3+k^2r^4+k^3r^5 = b$$
$$k^3r^3 + k^3r^4+k^3r^5+k^3r^6 = k^3 r^3(1+r+r^2+r^3) = -c$$
$$k^4r^6 = d$$
By comparing the maximum powers of $r$, a) is not correct ($r^{10}$ vs $r^9$) and so is b). But $c/a = k^2r^3$ or $c^2/a^2 = k^4r^6 = d$, hence c) is correct.
This relationship between $a$ and $c$ can be extended to any power $n$ in general. $-c$ can also be written as $k^4 r^6 (1/k + 1/(kr) + 1/(kr^2) + 1/(kr^3))$ as you miss out one term in the product of three (or $n$) roots, and in general this is $k^n r^{n(n-1)/2} (1/k + \cdots + \frac{1}{kr^{n-1}}) = k^{n-1} r^{(n-1)(n-2)/2} (1 + \cdots + r^{n-1})$.
|
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For each value for $a$ and $b$, being real numbers, $a^2 + b^2 \ge ab$ For each value for $a$ and $b$, being real numbers, $a^2 + b^2 \ge ab$
Should I solve this by replacing all possible real number formulas in these to be able to prove this? As an example for odd numbers, by replacing $2x + 1$, etc...?
|
Try to write it as $a^2+b^2-ab=\frac{2a^2+2b^2-2ab}{2}=\frac{a^2+b^2+(a-b)^2}{2}$. Now all three terms are $\geq 0$, so you get $a^2+b^2\geq ab$.
|
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|
Help with $\int _0^{\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x$ I want to know how to prove that
$$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=\frac{4G}{\pi }$$
Here $G$ denotes Catalan's constant, I obtained such result with the help of mathematica.
I also found that the integral equals a certain infinite series
$$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=\sum _{n=0}^{+\infty }\frac{\binom{2n}{n}^2}{16^n\left(2n+1\right)}=\frac{4G}{\pi }$$
which can also be found in this link.
So I've $2$ questions
$1)$$¿$How can we transform the integral into the mentioned series?
$2)$$¿$Is there a simple way to evaluate the main integral without resorting to series expansion?
What I did for question $\#2$ is to employ the substitution $x=\ln\left(t\right)$
$$\int _0^{+\infty }\frac{\sinh \left(x\right)}{x\cosh ^2\left(x\right)}\:\mathrm{d}x=-2\int _1^{\infty }\frac{1-t^2}{\ln \left(t\right)\left(1+t^2\right)^2}\:\mathrm{d}t$$
But I'm not sure how to proceed.
|
I present another approach, utilising a well-known property of the Laplace Transform i.e.
$$\int_{0}^{+\infty} f\left(x\right) g\left(x\right) \, dx = \int_{0}^{+\infty} \left(\mathcal{L} f\right)\left(y\right)\left(\mathcal{L}^{-1} g\right)\left( y\right) \, dy$$
Letting $f\left(x\right) = \tanh \left(x\right) \text{sech} \left(x\right)$ and $g(x) = \frac{1}{x}$:
$$\left( \mathcal{L} f \right) (y) = 1 + \frac{1}{2} y \left( \psi^{(0)} \left(\frac{1+y}{4}\right)- \psi^{(0)} \left(\frac{3+y}{4}\right)\right)$$
$$\left( \mathcal{L}^{-1} g \right) \left(y \right) = 1$$
Where $\psi$ represents the polygamma function.
$$\implies I = \int_{0}^{\infty} \left(1 + \frac{1}{2} y \left( \psi^{(0)} \left(\frac{1+y}{4}\right)- \psi^{(0)} \left(\frac{3+y}{4}\right)\right)\right) \, dy$$
I will now proceed to integrate indefinitely, then take limits at the end.
$$\int 1 \, dy + \frac{1}{2} \int y \, \psi^{(0)} \left(\frac{1+y}{4}\right) \, dy - \frac{1}{2} \int y \, \psi^{(0)} \left(\frac{3+y}{4}\right) \, dy$$
We proceed with integrating by parts.
$$=y + \frac{1}{2} \left( 4y \ln \Gamma \left( \frac{1+y}{4} \right) - 4 \int \ln \Gamma \left( \frac{1+y}{4} \right) \, dy \right)-\frac{1}{2} \left( 4y \ln \Gamma \left( \frac{3+y}{4} \right) - 4 \int \ln \Gamma \left( \frac{3+y}{4} \right) \, dy \right)$$
$$=y + 2 y \ln \Gamma \left( \frac{1+y}{4}\right) - 8 \psi^{(-2)} \left(\frac{1+y}{4}\right) - 2y \ln \Gamma \left( \frac{3+y}{4} \right) + 8 \psi^{(-2)} \left( \frac{3+y}{4} \right)$$
Taking limits to $\infty$ and $0$, and then subtracting gives us:
$$2\ln (2\pi) + 8 \psi^{(-2)} \left( \frac{1}{4} \right)-8 \psi^{(-2)} \left( \frac{3}{4} \right)$$
$$=\boxed{\frac{4G}{\pi}}$$ (the equality was checked with WolframAlpha) as required.
|
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|
Convergence of $\sum_{n=0}^{\infty} \frac{4^n}{3^n+7^n}$ I need help with this.
$\sum_{n=0}^{\infty} \frac{4^n}{3^n+7^n}$
I know that it converges but i can not proove why.
I tried to rewrite it, it seems to be a geometric serie. I tried to do a common factor between $3^n+7^n \rightarrow 3^n(1+\frac{7^n}{3^n})$
So I have $\sum_{n=0}^{\infty} (\frac{4}{3})^n \frac{1}{1+(\frac{7}{3})^n}$ And I do not know if that helps.
I can also make different the common factor and I would have $\sum_{n=0}^{\infty} (\frac{4}{7})^n \frac{1}{1+(\frac{3}{7})^n}$
|
Comparison test: Observe that
$$\frac{4^n}{3^n+7^n} \le \frac{4^n}{7^n} = \left(\frac47\right)^n$$
which is a convergent geometric series
Root test:
Observe that
$$\frac{4}{\sqrt[n]{7^n+7^n}}\le \sqrt[n]{\frac{4^n}{3^n + 7^n}} \le \frac{4}{\sqrt[n]{7^n}} \\
\frac{4}{7\sqrt[n]{2}}\le \sqrt[n]{\frac{4^n}{3^n + 7^n}} \le \frac{4}{7}$$
Now, by Squeeze theorem
$$\lim_{n\to\infty} \sqrt[n]{\frac{4^n}{3^n + 7^n}} = \frac47 < 1$$
Ratio test: Calculate the limit of the ratio of two consecutive terms:
$$\frac{4^{n+1}}{3^{n+1}+7^{n+1}}\cdot \frac{3^n + 7^n}{4^n} = 4\frac{(3/7)^n + 1}{3(3/7)^n + 7}\to \frac47 < 1$$
|
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Order on three real numbers given they are roots of a cubic, their sum and sum of pairwise product
Let $a,b,c$ be three real numbers which are roots of a cubic polynomial and satisfy $a+b+c=6$ and $ab+bc+ca=9.$ Suppose $a<b<c.$ Show that $$0<a<1 <b<3<c<4$$
Source: ISI Bmath Entrance 18-Jul-2021
Let the polynomial be $p(x) = (x-a)(x-b)(x-c)= x^3 - x^2 (6) + 9x - abc$, now clearly $p(a)=p(b)=p(c)$, adding the three:
$$p(a)+p(b)+p(c)=0$$
$$ (a^3 + b^3 + c^3)- 6(a^2 + b^2 +c^2) + 9 (a+b+c) - abc=0$$
Using identites we find the expression above becomes:
$$a^3 + b^3+c^3 - 6(6^2- 2 \cdot 9) + 9 \cdot 6 -abc=0$$
or
$$ a^3 + b^3 + c^3 -54 -abc=0$$
To be frank, I didn't do the manipulatons with any plan in mind, under desperation in the exam hall, I took the above to be a cubic in $a$ and applied Vieta assuming roots to be $\{a_1,a_2,a_3 \}$, where I found:
$$a_1 + a_2 + a_3 = 0$$
$$a_1 a_2 + a_2 a_3 + a_1 a_3 =0$$
$$a_1 a_2 a_3 = b^3 + c^3 - 54 - abc$$
We can write similar relation by considering it as a cubic in $b$ and $c$... but what after this..?
|
I approached the problem differently. It was a long write-up, here are the key points:
▪︎Define a cubic $f(x)=x^3-6x^2+9x+t$. It is clear that this is the cubic with roots $a,b,c$.
▪︎Try to put a bound on $t$ for which three distinct roots are obtained. The graphical argument can be used. The defined $f(x)$ is merely a shifting of $g(x)=x^3-6x^2+9x$ along the $+y$-axis by $t$ units. If $t\leq-4$ or $t\geq0$, three distinct roots are not obtained. Hence $t\in(-4,0)$.
▪︎We have to find solutions for $x^3-6x^2+9x=k$. Here $k=-t$.We have shown that $k\in(0,4)$.
▪︎To finally solve the problem, we will apply intermediate value theorem upon $g(x)$. Note that $g(0)=0$ and $g(1)=4$. Hence, $g(x)=k$ has a root in $x\in(0,1)$. This is the smallest solution, so $0<a<1$. Also, this interval can have ONLY one solution, since $g$ is monotonic here.
▪︎ Apply the same argument in intervals $(1,3)$ and $(3,4)$. We have finished our proof.
|
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Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modular arithmetic or induction? Below is my proof:
$\text{Case 1, }n\bmod3=0,\text{then $n=3k$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k}+5^{3k}+56
\\ & = 8^k+125^k+56
\\ & \equiv (-1)^k+(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 2, }n\bmod3=1,\text{then $n=3k+1$ for some even integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+1}+5^{3k+1}+56
\\ & = 2\cdot8^k+5\cdot125^k+56
\\ & \equiv 2\cdot(-1)^k+5\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 9\equiv0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 3, }n\bmod3=2,\text{then $n=3k+2$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+2}+5^{3k+2}+56
\\ & = 4\cdot8^k+25\cdot125^k+56
\\ & \equiv 4\cdot(-1)^k+25\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv -27\equiv0\quad&\left(\bmod9\right)
\end{align}$$
|
Here's one way (not sure if it is the easiest, though).
Let us start with a lemma which you may easily prove by induction.
Lemma: For each natural number $k$, $2^{2k+1}+1$ and $2^{2k}-1$ are divisible by $3$.
Next, we have the following
Claim:
For $n=2k+1$, one has
$$2^n+5^n+56\equiv-2(2^{2k+1}+1)(2^{2k}-1)\mod 9.$$
In fact,
$$2^n+5^n+56\equiv 2^{2k+1}+(-2^2)^{2k+1}+2\equiv 2(2^{2k}-2\cdot 2^{4k}+1) \mod 9.$$
Observe that the RHS is equal to the RHS of the claim.
By the lemma, the RHS of the claim is divisible by $3\cdot 3=9$, hence the LHS as well.
|
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Evaluate $\iiint_{V} (x^2+y^2+z^2)\,dx\,dy\,dz$ in the common part of $2az > x^2+y^2$ and $x^2+y^2+z^2 < 3a^2$ Evaluate $$\iiint_{V} x^2+y^2+z^2 \,dx\,dy\,dz$$
Where $V$ (the integration region) is the common part of the paraboloid $x^2 + y^2 \leq 2az$ and the sphere $x^2+y^2+z^2 \leq a^2$.
I first found the intercept of the paraboloid and the sphere: $x^2+y^2 = 2a^2$. In the $z$ direction, the area is bounded by the sphere on top and the paraboloid on the bottom. Hence I setup the integration:
$$4\int_{0}^{\sqrt{2a^2-x^2}} dy \int_{0}^{\sqrt{2}a} dx\int_{(x^2+y^2)/2}^{\sqrt{3a^2-x^2-y^2}} (x+y+z)^2\,dz$$
However, I have been told that the $z$ direction integral should be from $0$ to $\sqrt{3a^2-x^2-y^2}$. I do not understand why this is the case if we are interested in the common part of the circle and paraboloid. Could someone please explain this?
|
You are correct, and the person who said the greatest lower limit is $z=0$ is wrong:
|
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"source": "stackexchange",
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Extrema of $f(x,y)=(1-x^2-y^2)\cdot xy$ subject to $x^2+y^2\leq 1$ I want to determine the extrema of $f(x,y)=(1-x^2-y^2)\cdot xy=xy-x^3y-xy^3$ subject to $x^2+y^2\leq 1$.
We use Lagrange Multipliers to check the critical points on the circle $x^2+y^2=1$.
To check the critical points inside the circle, $x^2+y^2<1$, we calculate the critical points of $f(x,y)$.
Let's start with the Lagrange-Multipliers.
We have $f(x,y)=(1-x^2-y^2)\cdot xy=xy-x^3y-xy^3$ und $g(x,y)=x^2+y^2-1=0$.
We consider
\begin{equation*}L(x,y,\lambda )=f(x,y)+\lambda g(x,y)=xy-x^3y-xy^3+\lambda \left (x^2+y^2-1\right )\end{equation*}
We calculate the partial derivatives of $L$.
\begin{align*}&\frac{\partial{L}}{\partial{x}}=y-3x^2y-y^3+2\lambda x \\ &\frac{\partial{L}}{\partial{y}}=x-x^3-3xy^2+2\lambda y \\ &\frac{\partial{L}}{\partial{\lambda }}= x^2+y^2-1 \end{align*}
To calculate the extrema we set each equation equal to zero and solve the system:
\begin{align*}&y-3x^2y-y^3+2\lambda x=0 \ \ \ \ \ \ \ \ (1) \\ & x-x^3-3xy^2+2\lambda y =0 \ \ \ \ \ \ \ \ (2) \\ & x^2+y^2-1=0 \ \ \ \ \ \ \ \ (3)\end{align*}
We get the critical points \begin{equation*}P_1\left (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right ) \ \ , \ \ \ P_2\left (-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right )\ \ , \ \ \ P_3\left (-1, 0\right )\ \ , \ \ \ P_4\left (1, 0\right ) \ \ , \ \ \ P_5\left (0, -1\right ) \ \ , \ \ \ \ P_6\left (0, 1\right )\end{equation*}
Now we check inside the circle.
We set the partial derivatives of $f$ equal to zero and solve the system:
\begin{align*}&f_x=0 \Rightarrow y-3x^2y-y^3=0 \ \ \ \ \ \ \ \ (1)\\ &f_y=0 \Rightarrow x-x^3-3xy^2=0\ \ \ \ \ \ \ \ (2)\end{align*}
Then we get the critical points \begin{align*}&Q_1\left (0, 0\right ) \ \ , \ \ \ Q_2\left (\frac{1}{2}, \frac{1}{2}\right )\ \ , \ \ \ Q_3\left (-\frac{1}{2}, -\frac{1}{2}\right )\ \ , \ \ \ Q_4\left (0, -1\right )\ \ , \ \ \ Q_5\left (0, 1\right )\ \ , \ \ \ Q_6\left (-2, -1\right ) \ \ , \ \ \ Q_7\left (0, -1\right )\ \ , \ \\ & \ Q_8\left (-\frac{3}{2}, -\frac{1}{2}\right ) \ \ , \ \ \ Q_9\left (\frac{1}{2}, -\frac{1}{2}\right )\ \ , \ \ \ Q_{10}\left (2, 1\right )\ \ , \ \ \ Q_{11}\left (\frac{3}{2}, \frac{1}{2}\right ) \ \ , \ \ \ Q_{12}\left (-\frac{1}{2}, \frac{1}{2}\right )\end{align*}
Then we have to find the maximal and minimum value of $f$ at all these critical points.
Is everything correct? I am not really sure about the part inside the circle, if all critical points are correct.
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The poser of this problem seems to have been having a little fun, as the behavior of this function $ \ f(x,y) \ = \ xy·(1 - x^2 - y^2) \ $ is a bit peculiar. The factor $ \ 1 - x^2 - y^2 \ $ has "four-fold" symmetry about the origin, but the factor $ \ xy \ $ "breaks" this down to just "diagonal" symmetry. So we may expect extremal points to be found at $ \ (\pm x \ , \ \pm x) \ $ or at $ \ (\pm x \ , \ \mp x) \ \ . $ The level-curves for $ \ f(x,y) \ = \ c \ > \ 0 \ $ lie in the second and fourth quadrants, while those for $ \ c \ < \ 0 \ $ are in the first and third; this holds with one exception we will come to. (If this seems counter-intuitive, keep in mind that outside the unit circle, $ \ 1 - x^2 - y^2 < 0 \ \ . ) $ For $ \ c \ = \ 0 \ \ , $ the level curve "degenerates" to the union of the two coordinate axes and the unit circle; this marks a "boundary" at which the level-curves change quadrants.
This makes any "critical points" found on the unit circle or the axes degenerate, since all points there correspond to $ \ c \ = \ 0 \ \ . $
What will prove to be of greater interest are the level-curves for $ \ 0 < c < c^{*} \ $ and $ \ -c^{*} < c < 0 \ \ , $ where "loops" appear within the unit disk: in the first and third quadrants for positive $ \ c \ $ and in the other two quadrants for negative $ \ c \ \ . $ We can find the value of $ \ c^{*} \ $ by considering what happens with $ \ f(x,y) \ $ for $ \ y = \pm \ x \ \ . $ We find that
$$ \ f(x,x) \ = \ x^2·(1 - 2x^2) \ = \ c^{*} \ \ \Rightarrow \ \ 2x^4 - x^2 + c^{*} \ = \ 0 \ \ \Rightarrow \ \ x^2 \ = \ \frac{1 \ \pm \ \sqrt{1 \ - \ 4·2·c^{*}}}{4} \ \ , $$
so $ \ x^2 \ $ has just a single value for $ \ c^{*} = \frac18 \ \ ; $ a similar calculation for $ \ f(x,-x) \ $ gives a symmetrical result. Hence, the "loops" in the interior of the unit disk are only present for $ \ 0 < c < \frac18 \ \ $ and $ \ -\frac18 < c < 0 \ \ ; $ otherwise, the level-curves are always outside the unit disk for $ \ |c| > \ \frac18 \ \ . $
As for the Lagrange equations $ \ y - y^3 - 3x^2y \ = \ \lambda · 2x \ $ and $ \ x - x^3 - 3xy^2 \ = \ \lambda · 2y \ \ , $ we can solve for $ \ \lambda \ $ to produce
$$ \lambda \ \ = \ \ \frac{y \ - \ y^3 \ - \ 3x^2y}{2x} \ \ = \ \ \frac{x \ - \ x^3 \ - \ 3xy^2}{2y} $$ $$ \Rightarrow \ \ 2x^2 \ - \ 2x^4 \ - \ 6x^2y^2 \ = \ 2x^2 \ - \ 2x^4 \ - \ 6x^2y^2 \ \ \Rightarrow \ \ x^2 · (1 - x^2) \ = \ y^2 · (1 - y^2) \ \ , $$
with this "cross-multiplication" being "safe" since we have already dispensed with points for which $ \ x = 0 \ $ or $ \ y = 0 \ \ . $ The possible pairing of factors in this equation are
$$ x^2 \ = \ y^2 \ \ , \ \ 1 - x^2 \ = \ 1 - y^2 \ \ \Rightarrow \ \ y \ = \ \pm \ x \ \ , $$
$$ x^2 \ = \ 1 - y^2 \ \ , \ \ 1 - x^2 \ = \ y^2 \ \ \Rightarrow \ \ x^2 + y^2 \ = \ 1 \ \ . $$
The result we are interested in comes from the critical points you found from $$ \ y·(1 - y^2 - 3x^2) \ = \ 0 \ , \ x·(1 - x^2 - 3y^2) \ = \ 0 \ \ \Rightarrow \ \ 3x^2 + y^2 = 1 \ , \ x^2 + 3y^2 = 1 \ \ , $$
which intersect at the four points $ \ \left( \pm \frac12 \ , \ \pm \frac12 \right) \ . $ [I don't know where your points $ \ Q_6 , Q_8 , Q_{10} \ $ and $ \ Q_{11} \ $ are coming from.]
The points $ \ \left( \ \pm \frac{1}{2} \ , \ \pm \frac{1}{2} \ \right) \ \ [y \ = \ x] \ \ $ correspond to $$ \ f\left( \ \pm \frac{1}{2} \ , \ \pm \frac{1}{2} \ \right) \ = \ \left(\pm \frac{1}{2} \right) · \left(\pm \frac{1}{2} \right) · \left(1 - \frac{1}{4} - \frac{1}{4} \right) \ \ = \ \frac18 \ \ , $$
making this the absolute maximum of the function in the interior of the unit disk. (This is the value $ \ c^{*} \ $ at which the loops in the first and third quadrants just "disappear".) The other set of points $ \ \left( \ \pm \frac{1}{2} \ , \ \mp \frac{1}{2} \ \right) \ \ [y \ = \ -x] \ \ $ correspond to $$ \ f\left( \ \pm \frac{1}{2} \ , \ \mp \frac{1}{2} \ \right) \ = \ \left(\pm \frac{1}{2} \right) · \left(\mp \frac{1}{2} \right) · \left(1 - \frac{1}{4} - \frac{1}{4} \right) \ \ = \ -\frac18 \ \ , $$
the absolute minmum in the region (or $ \ -c^{*} \ $ at which the loops in the second and fourth quadrants vanish.)
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"url": "https://math.stackexchange.com/questions/4211762",
"timestamp": "2023-03-29T00:00:00",
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|
$\tan\frac{\pi}{9} +4\sin\frac{\pi}{9} = \sqrt 3 $
Prove that $$ \tan\frac{\pi}{9} +4\sin\frac{\pi}{9} = \sqrt 3 $$
There seem to be a lot of similar identities that are provable, for example, by using roots of unity. However, here I cannot get things to work out nicely.
If $u=e^{\frac{2\pi i}{9}} $, then $$i\left(\tan\frac{\pi}{9} +4\sin\frac{\pi}{9}\right) =\frac{u-1}{2(u+1)} +2(u^4-u^5)=\frac{-4u^6 +4u^4+u-1}{2(u+1)} $$ and so $$\left(\tan\frac{\pi}{9} +4\sin\frac{\pi}{9}\right)^2 = 3 \\ \iff (-4u^6+4u^4+u-1)^2+12(u+1)^2 =0 \\ \iff 16u^8-8u^7+8u^6+8u^5-8u^4+16u^3+13u^2-10u+13 =0$$ Unfortunately, the LHS is not of the form $k(u^8+u^7 +\dots+1)$ making the equality unobvious. How to proceed?
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Here's a purely trigonometric solution, in case you're interested:
$$\tan 20°+4\sin 20°=\frac {\sin 20°+2(2\sin 20° \cos 20°)}{\cos 20°}=\frac {(\sin 20°+\sin 40°)+\sin 40°}{\cos 20°}=\frac {2\frac 12 \cos 10°+\cos 50°}{\cos 20°}=\frac {2\cos 30°\cos 20°}{\cos 20°}=2 \frac {\sqrt 3}{2}=\sqrt 3$$
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{
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"timestamp": "2023-03-29T00:00:00",
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$a^a\cdot{b^b}\ge \bigl(\frac{a+b}{2}\bigl)^{a+b}\ge{a^b}\cdot{b^a}$ If $a$ and $b$ are positive rational numbers, prove that
$$a^a\cdot b^b\ge \left(\frac{a+b}{2}\right)^{a+b} \ge a^b \cdot{b^a}$$
My try:
consider $\frac{a}{b}$ and $\frac{b}{a}$ be two positive numbers with associated weights $b$ and $a$.
Then $\displaystyle\frac{b\cdot\frac{a}{b}+a\cdot\frac{b}{a}}{a+b}\ge \biggl[\left(\frac{a}{b}\right)^b\cdot \left(\frac{b}{a} \right)^a\biggl]^\frac{1}{a+b}$ implies $a^a\cdot{b^b}\ge a^b\cdot{b^a}$
Please help me to solve this problem. Thanks
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From weighted AM-GM, we have $$\left(\frac{\frac 1a\cdot a+\frac 1 b\cdot b}{a+b}\right )^{a+b}\geq \frac{1}{a^ab^b}\\ \implies \left(\frac{2}{a+b}\right)^{a+b}\geq \frac{1}{a^ab^b}$$which is equivalent to the first inequality.
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{
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"url": "https://math.stackexchange.com/questions/4214747",
"timestamp": "2023-03-29T00:00:00",
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|
Solving $2x^3-5ix^2+3x+4i=0$, need $(21x_1-1)^{-2}+(21x_2-1)^{-2}+(21x_3-1)^{-2}$ I have this polynomial from Problemas selectos (Lumbreras editors):
$$2x^3-5ix^2+3x+4i=0$$
I make $x=it$. then:
$$2(it)^3-5i(it)^2+3it+4i=0$$
$$i(-2it^3+5it^2+3it+4)=0$$
$$2t^3-5t^2-3t-4=0$$
$$8t^3-20t^2-12t-16=0$$
$$(2t)^3-3(2t)^2(5/3)+(50/3)t-125/27-(14/3)t-307/27=0$$
$$(2t-5/3)^3-(86/3)t-307/27=0$$
If $2t-5/3=z$, then:
$$z^3-(43/3)z-952/27=0$$
Where $p=-\frac{43}{3}$ and $q=-\frac{952}{27}$
When $A=-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$ and $B=-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$
We have $z_1=\sqrt[3]{A}+\sqrt[3]{B}$, $z_2=-\frac{\sqrt[3]{A}+\sqrt[3]{B}}{2}+i\sqrt{3}\frac{\sqrt[3]{A}-\sqrt[3]{B}}{2}$, $z_3=-\frac{\sqrt[3]{A}+\sqrt[3]{B}}{2}-i\sqrt{3}\frac{\sqrt[3]{A}-\sqrt[3]{B}}{2}$
There for $x_n=\frac{i}{2}(z_n+\frac{5}{3})$,
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Assume $x$ is pure imaginary.
$x=iy$, with $y$ real.
We have:
$$- 2i y^3 +5 i y^2 +3iy +4 i =0$$
or
$$2 y^3 -5 y^2 -3y -4 =0$$
This leads to the solution
$$y \approx 3.17174.$$
The other solutions are
$$y \approx -0.335868 \mp 0.719557 i$$
and the solutions to the original problem are $x=iy.$
There are techniques to solve cubic equations exactly. I admit that I am not so familiar with them (we could look them up!), so you've got here a numerical solution.
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Computing $I=\int \frac{dx}{(x+1)(3x^2+3x+1)^{1/3}}$ I am trying to solve this integral and find its primitive, unfortunately I have not been successful. I made some variable changes, but I think it has become even more complicated.
My solution is as follows:
$$I=\int \frac{dx}{(x+1)(3x^2+3x+1)^{1/3}}$$
$$u=x+\frac{1}{2}\iff u+\frac{1}{2}=x+1$$
$$du=dx$$
$$I=\int \frac{du}{(u+\frac{1}{2})(u^2+\frac{1}{12})^{1/3}}$$
$$\alpha=\frac{1}{2}$$
$$I(\alpha)=\int \frac{(u-\alpha)}{(u^2-\alpha^2)(u^2+\frac{\alpha^2}{3})^{1/3}}du$$
$$I(\alpha)=\frac{1}{3^{1/3}}\int \frac{(u-\alpha)}{(u^2-\alpha^2)(3u^2+\alpha^2)^{1/3}}du$$
I hope you can help me. Regards
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\begin{gather*}
\int \frac{dx}{( x+1)\left( 3x^{2} +3x+1\right)^{\frac{1}{3}}}\\
=\int \frac{dx}{( x+1)\left( x^{3} +3x^{2} +3x+1-x^{3}\right)^{\frac{1}{3}}}\\
=\int \frac{dx}{( x+1)\left(( x+1)^{3} -x^{3}\right)^{\frac{1}{3}}}\\
=\int \frac{dx}{( x+1)^{2}\left( 1-\left(\frac{x}{x+1}\right)^{3}\right)^{\frac{1}{3}}}\\
Let\ \frac{x}{x+1} =t\\
\frac{dx}{( x+1)^{2}} =dt\\
The\ integral\ becomes\\
\int \left( 1-t^{3}\right)^{\frac{-1}{3}} dt
\end{gather*}
Does this help in some way?
Edit: For solving the last integral, just change it to
\begin{equation*}
-\int \frac{dt}{\left( t^{3} -1\right)^{\frac{1}{3}}}
\end{equation*}
and make a substitution
\begin{equation*}
u=\frac{\left( t^{3} -1\right)^{\frac{1}{3}}}{t}
\end{equation*}
After some manipulation, you will get an integrand that can be solved by partial fractions.
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The Diophantine equation $x^5-2y^2=1$ I'm trying to solve the Diophantine equation $x^5-2y^2=1$.
Here's my progress so far. We can write the Diophantine equation as
$$\frac{x-1}{2}\cdot(x^4+x^3+x^2+x+1)=y^2.$$
If $x\not\equiv1\pmod{5}$, then $\gcd(\frac{x-1}{2},x^4+x^3+x^2+x+1)=1$, so both $\frac{x-1}{2}$ and $x^4+x^3+x^2+x+1$ must be perfect squares (note: $x^4+x^3+x^2+x+1>0$).
In particular, $4(x^4+x^3+x^2+x+1)$ is a perfect square.
Comparison with $(2x^2+x)^2$ and $(2x^2+x+1)^2$ forces $-1\leq x\leq3$.
This results in the solutions $(3,\pm11)$.
If $x\equiv1\pmod{5}$, then we can write the Diophantine equation as
$$\frac{x-1}{10}\cdot\frac{x^4+x^3+x^2+x+1}{5}=\left(\frac{y}{5}\right)^2,$$
where $\gcd(\frac{x-1}{10},\frac{x^4+x^3+x^2+x+1}{5})=1$, so both $\frac{x-1}{10}$ and $\frac{x^4+x^3+x^2+x+1}{5}$ must be perfect squares.
Thus,
$$x=10a^2+1,$$
$$x^4+x^3+x^2+x+1=5b^2.$$
Unfortunately, this is where I get stuck.
I can substitute the first equation into the second, giving
$$10000a^8+5000a^6+1000a^4+100a^2+5=5b^2,$$
$$2000a^8+1000a^6+200a^4+20a^2+1=b^2,$$
$$2000a^8+1000a^6+200a^4+20a^2=(b-1)(b+1),$$
$$5a^2(100a^6+50a^4+10a^2+1)=\frac{b-1}{2}\cdot\frac{b+1}{2},$$
but this doesn't seem to be making progress, even with modular arithmetic considerations.
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You don't really say if you just want to know the solutions, or if you want a nice elementary argument for why the solutions are only $(3, \pm 11)$, if you just want a proven answer and not an elementary argument the following works, its a bit overkill but its easier than thinking if you know these methods already:
The equation $x^5 - 2y^2 = 1$ considered over the rationals defines a hyperelliptic curve, of genus 2.
So there is a big hammer called Chabauty's method that often determines all rational points on such curves.
Our curve is isomorphic via change of variables ($y\mapsto 4y,x\mapsto 2x$) to the curve
$$y^2 = x^5 - \frac{1}{2^5}$$
or even to an integral model
$$y^2 = -2x^6 + 2x.$$
The computer algebra system Magma can determine the rank of the Mordell-Weil group of the Jacobian of this curve (using the integral model above) to be 1 (and hence Chabauty's method applies), using a generator Magma can also run Chabauty's method automatically in this case, and provably find all rational points:
> R<x>:=PolynomialRing(Rationals());
> H:=HyperellipticCurve(x^5-1/(2^5));
> HH,ma:=MinimalWeierstrassModel(H);
> a,b,P:=RankBounds(Jacobian(HH):ReturnGenerators);
> a,b,P;
1 1 [ (x^2 - 1/3*x, 22/9*x, 2) ]
> pts:=Chabauty(P[1]);
> pts;
{ (1 : 22 : 3), (1 : -22 : 3), (0 : 0 : 1), (1 : 0 : 1) }
> [P : P in pts][1]@(ma^(-1));
(3/2 : -11/4 : 1)
> [P : P in pts][3]@(ma^(-1));
(1 : 0 : 0)
From this list we see that translating back to the original equation/curve the only interesting rational solutions are those you found already.
If you only wanted integral solutions to begin with there should be less high-tech methods to do this!
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Finding foot of perpendicular of centre of ellipse on its variable tangent
Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2\over a^2} +{y^2\over b^2} = 1$$
on the chord joining the points whose eccentric angles differ by $π/2.$
Ref:Locus of foot of perpendicular of origin from tangent of ellipse, related
The points may be given as $(a\cos \alpha, b\sin \alpha)$ and $(-a\sin \alpha , b \cos \alpha )$, then their midpoint is given as: $\left( \frac{a}{2} (\cos \alpha - \sin \alpha) , \frac{b}{2} (\cos \alpha + \sin \alpha) \right)=(\kappa, \beta)$, the equation of chord given midpoint of ellipse is:
$$ \frac{\kappa X}{a^2} + \frac{\beta Y}{b^2} = \frac{\kappa^2}{a^2} + \frac{\beta^2}{b^2} $$
Let $u= \frac{\cos \alpha - \sin \alpha}{2}$ and $ v=\frac{\cos \alpha + \sin \alpha}{2}$, then our equation becomes:
$$\frac{ uX}{a} + \frac{ vY}{b} = u^2 + v^2 $$
Foot of perpendicular from origin for above line is given as:
$$ \frac{ax}{u}= \frac{by}{v}= \frac{(ab)^2 (u^2 +v^2)}{b^2u^2 +a^2v^2}$$
$ax= u(\frac{(ab)^2 (u^2 +v^2)}{b^2u^2 +a^2v^2})$
and $by =v (\frac{(ab)^2 (u^2 +v^2)}{b^2u^2 +a^2v^2})$, following the last line of this answer, I squared and added:
$$(ax)^2 + (by)^2 = (ab)^4 \left[ \frac{u^2 +v^2}{b^2 u^2 +a^2v^2} \right]^2= \frac{(ab)^4}{4} \left[ \frac{1}{b^2 u^2 + a^2 v^2}\right]^2$$
How do I write bracketed term in purely $(x,y)$?
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You are missing a $(u^2+v^2)$ when you add $(ax)^2$ and $(by)^2$.
Just in case, you want to finish continuing from the point you got to,
$(ax)^2 + (by)^2 = (ab)^4 (u^2+v^2) \left[ \frac{u^2 +v^2}{b^2 u^2 +a^2v^2} \right]^2$
$ = \cfrac{(ab)^4}{2} \left[ \frac{(u/v)^2 +1}{b^2 (u/v)^2 + a^2} \right]^2$
Now note that $\cfrac{u}{v} = \cfrac{ax}{by}$
That leads to $2 (x^2+y^2)^2 = (ax)^2 + (by)^2$
or we could have substituted $u = k a x, v = k by$ in RHS.
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|
To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$ To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$
My Attempt:
First Method: we know that $\cot^{-1}x = \tan^{-1}\frac{1}{x}$ for $x>0$ and $\tan^{-1}x+\tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}, xy<1$
Now $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18$ = $\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{18}$ = $(\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{8})+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\frac{1}{8}}+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{15}{56}}{\frac{55}{56}}+\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{3}{11} +\tan^{-1}\frac{1}{18}$ = $\tan^{-1}\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\frac{1}{18}}$ = $\tan^{-1}\frac{\frac{65}{198}}{\frac{195}{198}}$ = $\tan^{-1}\frac{1}{3}$ = $\cot^{-1}3$
Second Method: we know that $\tan^{-1}x+\cot^{-1}x = \frac{π}{2}$ for $x \in \Bbb R$. So $\cot^{-1}x = \frac{π}{2} - \tan^{-1}x$.
Now $$\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \frac{π}{2} - \tan^{-1}7+ \frac{π}{2} - \tan^{-1}8+ \frac{π}{2} - \tan^{-1}18 \implies \\ \frac{3π}{2} - (\tan^{-1}7+\tan^{-1}8+\tan^{-1}18) = \frac{3π}{2} - \tan^{-1}\frac{7+8+18-7×8×18}{1-7×8-8×18-7×18} \\ = \frac{3π}{2} - \tan^{-1}\frac{-975}{-325} = \frac{3π}{2} - \tan^{-1}3 = π + (\frac{π}{2} - \tan^{-1}3) = π + \cot^{-1}3 .$$ Also we know that $\tan x$ and $\cot x$ are periodic function with period $π$. Please help me in Second Method. Is $π + \cot^{-1}3 = \cot^{-1}3$ ?. If yes, then elaborate it.
|
I can't resist, even though your request for a critique makes this response somewhat off-topic. Let the 3 LHS angles be denoted as $a,b,c$, and use the formula
$$\cot(a + b) = \frac{[\cot(a)\cot(b)] - 1}{\cot(a) + \cot(b)}.$$
This gives $\cot(a + b) = \frac{55}{15} = \frac{11}{3}.$
Therefore,
$$\cot[(a+b) + c] = \frac{\left[\left(\frac{11}{3}\right)\left(18\right)\right] - 1}{\frac{11}{3} + 18} = 3.$$
Further, since each of $a,b,c$ are in the first quadrant, as is $(a + b + c)$, you have that $(a + b + c) = \text{Arccot}(3).$
|
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|
Find all primes $p$ and $r$ such that $ pr+1+r^3=p^2 $ I tried some small values, and I found that $p=7$ and $r=3$ was a solution.
I ve also find a way to factorise the equation:
$$ p(p-r)=(r+1)(r^2-r+1)$$
Since we know that $ p > r+1 $ then $$p| (r^2\color{blue}{-}r+1) $$
That's all what i've found , thank you in advance fo your precious help !
|
As you've already noted,
$$p(p - r) = (r + 1)(r^2 - r + 1) \tag{1}\label{eq1A}$$
Also, there's a positive integer $k$ such that
$$p \mid (r^2 - r + 1) \; \; \to \; \; r^2 - r + 1 = kp \tag{2}\label{eq2A}$$
Similar to how Find all pairs of prime numbers $p$ and $q$ such that $\,p^2-p-1=q^3.$ that Dietrich Burde's question comment suggested is solved, substitute \eqref{eq2A} into \eqref{eq1A} and divide both sides by $p$ to get
$$p - r = (r + 1)k \; \; \to \; \; p = (r + 1)k + r \; \; \to \; \; p = (k + 1)r + k \tag{3}\label{eq3A}$$
Substituting this into \eqref{eq2A} gives
$$\begin{equation}\begin{aligned}
& r^2 - r + 1 = k((k + 1)r + k) \\
& r^2 - r + 1 = (k^2 + k)r + k^2 \\
& r^2 + (-k^2 - k - 1)r + (1 - k^2) = 0
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This is a quadratic equation in $r$, so the quadratic formula gives
$$r = \frac{k^2 + k + 1 \pm \sqrt{(-k^2 - k - 1)^2 - 4(1 - k^2)}}{2} \tag{5}\label{eq5A}$$
For $r$ to be integral requires the discriminant (note I use $(-k^2 - k - 1)^2 = (-(k^2 + k + 1))^2 = (k^2 + k + 1)^2$ for convenience), i.e.,
$$\begin{equation}\begin{aligned}
d & = (k^2 + (k + 1))^2 - 4(1 - k^2) \\
& = k^4 + 2k^2(k + 1) + (k + 1)^2 - 4 + 4k^2 \\
& = k^4 + 2k^3 + 2k^2 + k^2 + 2k + 1 - 4 + 4k^2 \\
& = k^4 + 2k^3 + 7k^2 + 2k - 3
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
must be the square of an integer. Consider the values of $k$ for which
$$(k^2 + k + 2)^2 = k^4 + 2k^3 + 5k^2 + 4k + 4 \lt d \lt (k^2 + k + 3)^2 = k^4 + 2k^3 + 7k^2 + 6k + 9 \tag{7}\label{eq7A}$$
is true. With the first inequality, we get
$$\begin{equation}\begin{aligned}
k^4 + 2k^3 + 5k^2 + 4k + 4 & \lt k^4 + 2k^3 + 7k^2 + 2k - 3 \iff \\
5k^2 + 4k + 4 & \lt 7k^2 + 2k - 3 \iff \\
-2k^2 + 2k & \lt -4 - 3 \iff \\
-2k(k - 1) & \lt -7
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
This is true for all $k \ge 3$. Note the second inequality in \eqref{eq7A} is true for all positive $k$. Thus, \eqref{eq7A} holds for all $k \ge 3$. This means $d$ is between two consecutive perfect squares and, thus, can't be a perfect square itself, so we must have $k \lt 3$.
With $k = 1$, \eqref{eq6A} gives $d = 9$, so \eqref{eq5A} gives $r = 0, 3$, with only $r = 3$ being valid. In \eqref{eq2A}, this gives $p = 7$, as you've already determined. Next, using in $k = 2$ in \eqref{eq6A} gives $d = 61$, which is not a perfect square.
This means there's only the one solution you've already found, i.e., $p = 7$ and $r = 3$.
|
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|
Finding the anti-derivative of $ \frac{e^{-c y^2 }}{y\sqrt{y^2-1}}$ I am trying to evaluate the integral
\begin{align}
\frac{1}{2\sqrt{2}\pi}\int_{0^{-}}^{t} ds \ \frac{e^{-x^2/2S^2(t,s) }}{\Sigma(s) S(t,s)}
\end{align}
where $S(t, s) = 2D(t-s)+\frac{\Sigma(s)}{2}$ and $\Sigma(s)= \sigma^2+2Ds$. I found a rather neat change of variable by taking $\xi=S^{-1}(t,s)$ so that
\begin{align}
\Sigma(s)&=\sqrt{2} \ \xi^{-1}\sqrt{\xi^2\Sigma^2(t)-1}\\
2D(t-s)&=2 \ \xi^{-2}-\Sigma^2(t)\\
ds&=\frac{2}{D\xi^3}\ d\xi
\end{align}
Applying this to the integral gives the result
\begin{align}
\frac{1}{2 \pi D}\int_{\sqrt{2}/\Sigma (2 t)}^{\sqrt{2}/\Sigma (t)} d{\xi} \ \frac{e^{-x^2 \ \xi^2/2 }}{\xi\sqrt{\Sigma^2(t)\xi^2-1}}=\frac{1}{2 \pi D}{\int_{\xi_L}^{\xi_H} d{\xi} \frac{e^{-x^2 \ \xi^2/2 }}{\xi\sqrt{\Sigma^2(t)\xi^2-1}}}.
\end{align}
and at last applying the change of variable $y=\Sigma(t)\xi$ gives
\begin{align}
\frac{1}{2 \pi D}{\int_{\xi_L}^{\xi_H} d{\xi} \frac{e^{-x^2 \ \xi^2/2 }}{\xi\sqrt{\Sigma^2(t)\xi^2-1}}} = \frac{1}{2 \pi D}{\int_{\sqrt{2}\Sigma(t)/\Sigma(2t)}^{\sqrt{2}} d{y} \frac{e^{-(x^2/2\Sigma^2(t)) y^2 }}{y\sqrt{y^2-1}}} =\frac{1}{2 \pi D}\color{blue}{\int_{a}^{b} d{y} \frac{e^{-c y^2 }}{y\sqrt{y^2-1}}}
\end{align}
Above calculation indeed check out numerically.
It seems the integral in blue could have an anti-derivative. However, no luck just yet! Someone who knows a way forward? Many thanks in advance!
|
As others have pointed out, no closed-form solution in terms of elementary functions can be found. However, if you can live with the error function and the Owen T function then a closed-form solution can be found in terms of these special functions.
I will assume the error function $\operatorname{erf} (x)$ is well known to you. An integral representation for the Owen T function we intent to use can be found here. It is
$$\operatorname{T}(h,a) = \frac{1}{2\sqrt{2\pi}} \int_h^\infty e^{-x^2/2} \operatorname{erf} \left (\frac{ax}{\sqrt{2}} \right )\, dx.$$
Enforcing a substitution of $x \mapsto x \sqrt{2}$ gives the following form that will be found most convenient for the integral considered here:
$$\int_z^\infty e^{-x^2} \operatorname{erf}(ax) \, dx = 2 \sqrt{\pi} \operatorname{T}(z \sqrt{2}, a). \tag1$$
Now let us look at the integral in question. Let
$$I(t) = \int_a^b \frac{e^{-tx^2}}{x\sqrt{x^2 - 1}} \, dx, \quad t > 0.$$
I will assume $1 < a < b = \sqrt{2}$. Note that $I(\infty) = 0$ and we are required to find $I(c)$ where $c > 0$. Using Feynman's trick of differentiating under the integral sign with respect to $t$ gives
$$I'(t) = - \int_a^b \frac{x e^{-tx^2}}{\sqrt{x^2 - 1}} \, dx.$$
Enforcing a substitution of $u^2 = x^2 - 1$ yields
\begin{align*}
I'(t) &= e^{-t} \int_{\alpha_1}^{\alpha_2} e^{-tu^2} \, du = -\frac{\sqrt{\pi} e^{-t}}{2\sqrt{t}} \operatorname{erf}(u\sqrt{t}) \Big{|}_{\alpha_1}^{\alpha_2}\\
&= -\frac{\sqrt{\pi} e^{-t}}{2\sqrt{t}} \left [\operatorname{erf}(\alpha_2 \sqrt{t}) - \operatorname{erf}(\alpha_1 \sqrt{t}) \right ].
\end{align*}
Here $\alpha_1 = \sqrt{a^2 - 1}$ and $\alpha_2 = \sqrt{b^2 - 1}$.
Now, as
$$\int_c^\infty I'(t) \, dt = I(\infty) - I(c) = -I(c),$$
since $I(\infty) = 0$, we have
$$I(c) = -\frac{\sqrt{\pi}}{2} \int_c^\infty \frac{e^{-t}}{\sqrt{t}} \left [\operatorname{erf} \left (\alpha_2 \sqrt{t} \right ) - \operatorname{erf} \left (\alpha_1 \sqrt{t} \right ) \right ] \, dt.$$
On enforcing a substitution of $t \mapsto t^2$ one obtains
$$I(c) = \sqrt{\pi} \int_{\sqrt{c}}^\infty e^{-t^2} \left [\operatorname{erf}(\alpha_2 t) - \operatorname{erf}(\alpha_1 t) \right ] \, dt.$$
Evaluating this integral in terms of the Owen T function, from (1) one immediately sees that
$$I(c) = 2\pi \left (\operatorname{T} \left (\sqrt{2c}, \alpha_2 \right ) - \operatorname{T} \left (\sqrt{2c}, \alpha_1 \right ) \right ),$$
or
$$\int_a^b \frac{e^{-cy^2}}{y\sqrt{y^2 - 1}} \, dy = 2\pi \left (\operatorname{T} \left (\sqrt{2c}, \sqrt{b^2 - 1} \right ) - \operatorname{T} \left (\sqrt{2c}, \sqrt{a^2 - 1} \right ) \right ),$$
the required closed-form expression for our integral.
Note, if $b = \sqrt{2}$, as it appears to be from the question, further simplification of one of the Owen T functions is possible. As can be seen here when the right argument of the Owen T function is unity
$$\operatorname{T} \left (\sqrt{2c}, \sqrt{b^2 - 1} \right ) = \operatorname{T} \left (\sqrt{2c}, 1 \right ) = \frac{1}{8} \left [1 - \operatorname{erf}^2 \left (\sqrt{c} \right ) \right ],$$
allowing one to write
$$\int_a^{\sqrt{2}} \frac{e^{-cy^2}}{y\sqrt{y^2 - 1}} \, dy = -\frac{\pi}{4} \left [8 \operatorname{T} \left (\sqrt{2c}, \sqrt{a^2 - 1} \right ) + \operatorname{erf}^2 \left (\sqrt{c} \right ) - 1 \right ].$$
|
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|
Integrate $ \int \frac{2\tan x+1}{\sqrt{\tan^2 x+2\tan x+2}} \, \mathrm dx $ I tried writing $\tan^2 x+2\tan x+2$ as $(\tan x+1)^2+1$ and letting $\tan\theta = \tan x+1$ yields
$$ \int \frac{2\tan\theta-1}{\tan^2\theta-2\tan\theta+2}\,\sec\theta\,\mathrm d\theta $$
Then letting $ t = \tan(\frac{\theta}{2})$ yields
$$ \int\frac{t^2+4t-1}{t^4+2t^3-2t+1}\,\mathrm dt $$
Obviously what's next is factoring out the denominator and doing partial fraction decomposition, but this seems very hard.
I wanted to know if there's an easier method.
|
$$\int\dfrac{2\tan x+1}{\sqrt{\tan^2x+2\tan x+2}}~dx=\int\dfrac{2(\tan x+1)-1}{\sqrt{(\tan x+1)^2+1}}~dx$$
Let $u=\tan x+1$ ,
Then $x=\tan^{-1}(u-1)$
$dx=\dfrac{1}{(u-1)^2+1}~du$
$$\therefore\int\dfrac{2(\tan x+1)-1}{\sqrt{(\tan x+1)^2+1}}~dx=\int\dfrac{2u-1}{((u-1)^2+1)\sqrt{u^2+1}}~du$$
Introduce the Euler substitution:
Let $v=u+\sqrt{u^2+1}$ ,
Then $u=\dfrac{v^2+1}{2v}$
$du=\dfrac{v^2-1}{2v^2}~dv$
$$\begin{align}\therefore\int\dfrac{2u-1}{((u-1)^2+1)\sqrt{u^2+1}}~du&=\int\dfrac{\dfrac{2(v^2+1)}{2v}-1}{\left(\left(\dfrac{v^2+1}{2v}-1\right)^2+1\right)\left(v-\dfrac{v^2+1}{2v}\right)}\dfrac{v^2-1}{2v^2}~dv\\&=\int\dfrac{\dfrac{2(v^2-v+1)}{2v}}{\dfrac{(v^2-2v+1)^2+4v^2}{4v^2}\dfrac{v^2-1}{2v}}\dfrac{v^2-1}{2v^2}~dv\\&=\int\dfrac{4(v^2-v+1)}{(v^2-2v+1)^2+4v^2}~dv\end{align}$$
|
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|
Inductively prove $1 + \frac12 + \frac14 + \cdots + \frac{1}{2^n} = 2 - \frac{1}{2^n}$. Inductively prove that the formula holds for all $n\in\Bbb{N}$:
$$1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}=2-\frac{1}{2^n}.$$
What I have so far:
base: n = 1: $$1+\frac{1}{2}=2-\frac{1}{2}=1.5$$
inductionstep: n = k: $$1+\frac{1}{2}+\cdots+\frac{1}{2^k}=2-\frac{1}{2^k}$$
inductionhypothesis: n=k+1: $$1+\frac{1}{2}+\cdots+\frac{1}{2^k+1}=1+\frac{1}{2}+\cdots+\frac{1}{2^k}+\frac{1}{2^(k+1)}=(2-\frac{1}{2})+\frac{1}{2^k+1}$$
This is where I am stuck and not sure what to do next.
|
Use the following structure:
Step 1. $n = 0$, means $1 = 2 - \frac{1}{2^0} = 2 - 1 = 1$, which is true. We can proceed to induction step.
Step 2. Use simple induction: $p(n) \to p(n + 1)$. So, if:
$$1 + \dots + \frac{1}{2^n} = 2 - \frac{1}{2^n}$$
We have:
$$1 + \dots + \frac{1}{2^n} + \frac{1}{2^{n + 1}} = 2 - \frac{1}{2^n} + \frac{1}{2^{n+1}} = 2 - \frac{2}{2^{n + 1}} + \frac{1}{2^{n + 1}} = 2 - \frac{1}{2^{n + 1}}$$
This ends the proof. $\blacksquare$
|
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|
How to find the maximum value of $\frac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$? Let $\theta\in\left(0,\frac{\pi}{2}\right)$, then find the maximum value of $\dfrac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$
I found the derivative, which is equal to $\dfrac{(\cos\theta−\sin\theta)(\sin\theta+\cos\theta+1)^2}{(1+\sin\theta+\cos\theta+\sin\theta\cos\theta)^2}$
But there were complicated calculations involved. Is there a simpler way to find the maximum of this function?
|
The first derivative is:
$\frac{2\sqrt{2}cos(\theta+\frac{\pi}{4})}{(1+\cos\theta)(1+\sin\theta)}$,
which becomes null in:
$\theta=\frac{5\pi}{4}$, $\theta=\frac{-3\pi}{4}$, $\theta=\frac{\pi}{4}$.
The maximum is for
$\theta=\frac{\pi}{4}$ and,
and is equal to $-4\sqrt{2}+6$.
|
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|
Proof by induction: induction hypothesis question In this question I found online:
*
*Show that $$ S(n):0^2 + 1^2 + 2^2 + · · · + n^2 =
\frac{n(n + 1)(2n + 1)}{6}$$
I don't understand why for S(k+1) they wrote:
$$S(k+1):1^2+2^2+3^2+⋯+k^2+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
instead of:
$$S(k+1):1^2+2^2+3^2+⋯+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
Why is the $k^2$ included in the $S(k+1)$ step I don't get it surely you just substitute $k+1$ for $n$ so I don't know why $k^2$ is needed there because in other proof by induction questions I've done for example
for this proof: $$n<2^n$$
for the $k+1$ step the answer was not $$k + k + 1 < 2^k+1$$
it was:
$$k+1 < 2^k+1$$
EDIT
People have mentioned that my version is correct and they just wrote it in a different way but why could I not prove both sides were equal? This is what I did:
$$S(k+1):1^2+2^2+3^2+⋯+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
I expanded out $$(k+1)^2$$ which gave me $$k^2+2k+1$$
so going back to the k+1 statement we have:
$$S(k+1):1^2+2^2+3^2+⋯+k^2+2k+1=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
I subbed in $$\frac{k(k+1)(2k+1)}{6}$$ for $$1^2+2^2+3^2+⋯+k^2$$ (the induction hypothesis) in the k+1 statement and got:
$$\frac{k(k+1)(2k+1)}{6} + 2k+1=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
long story short my proof failed and I couldn't prove both sides were equal but why was this if my version of S(k+1) was not incorrect?
|
The sum you show is simply going up to $k+1$, which means the preceding term is $k^2$, and the final term is $(k+1)^2$. In the other proof you reference there is no sum.
|
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|
Troubleshooting a trigonometry/geometry question - spot the (silly) mistake! I'm back again!
Again, another error - not sure if it's them or me this time...
Here is the question
In order to calculate the shaded region, I added some lines to the diagram. I realise that there are other ways to solve this (and indeed the solution give uses a different breakdown and there's seems correct).
I then calculated the area of the sector ADB - the area of the triangle ADB and added this to the area of the sector BOD. I then multiplied that value by 2 to get the area of the shaded region.
Here is my working (Let a = alpha and t = theta)
$Area = 2[\frac12(3r)^2(\frac{a}2)-\frac12(3r)^2\sin(\frac{a}2)+\frac12(2r)^2(\frac{t}2)]$
Giving me a final answer of $r^2[(\frac{9a}2)-9\sin(\frac{a}2)+2t]$
Can you see where I went wrong...?
EDIT:
Here is the solution they provided:
|
Another way to compute the area is to add the part on the right of chord $CD$
and the part on the left of chord $CD$.
The portion of the shaded region on one side of the chord is called a circular segment.
You can compute the area of the circular segment on the right by finding the area of the circular sector between $AC$ and $AC$ and subtracting the area of the triangle
$\triangle ACD$:
$$ \frac92 r^2 \alpha - \frac92 r^2 \sin\alpha. $$
Similarly the area of the circular segment on the left is the area of the sector between $BC$ and $BD$ minus the area of $\triangle BCD$:
$$ 2r^2 \theta - 2r^2 \sin\theta. $$
Add them together and you get:
$$ \frac92 r^2 \alpha - \frac92 r^2 \sin\alpha + 2r^2 \theta - 2r^2 \sin\theta. $$
Factor out $r^2$:
$$
r^2\left(\frac92 \alpha - \frac92 \sin\alpha + 2 \theta - 2 \sin\theta\right).\tag1
$$
Excluding the final term $- 2 \sin \theta,$ the part in the brackets looks a lot like your result, but it has a term $-\frac92 \sin\alpha$ where you have
$-9 \sin\frac\alpha2.$
Those are not the same terms.
Let $M$ be the point where chord $CD$ intersects the segment $AB$
and consider the two right triangles $\triangle AMD$ and $\triangle BMD.$
You can find that $DM = 3 \sin\frac\alpha2 = 2 \sin\frac\theta2.$
Moreover, $AM = 3 \cos\frac\alpha2$ and $BM = 2 \cos\frac\theta2,$
so $3 \cos\frac\alpha2 + 2 \cos\frac\theta2 = AB = 3.$
You can also verify via the double-angle formula for the sine that
$\frac92 \sin\alpha = 9 \sin\frac\alpha2 \cos\frac\alpha2$
and $2 \sin\theta = 4 \sin\frac\theta2 \cos\frac\theta2.$
Then
\begin{align}
\frac92 \sin\alpha + 2 \sin \theta
&= 9 \sin\frac\alpha2 \cos\frac\alpha2 + 4 \sin\frac\theta2 \cos\frac\theta2 \\
&= \left(3 \sin\frac\alpha2\right)\left(3 \cos\frac\alpha2\right)
+ \left(2 \sin\frac\theta2\right)\left(2 \cos\frac\theta2\right) \\
&= \left(3 \sin\frac\alpha2\right)\left(3 \cos\frac\alpha2\right)
+ \left(3 \sin\frac\alpha2\right)\left(2 \cos\frac\theta2\right) \\
&= \left(3 \sin\frac\alpha2\right)
\left(3 \cos\frac\alpha2 + 2 \cos\frac\theta2\right) \\
&= \left(3 \sin\frac\alpha2\right) \times 3\\
&= 9 \sin\frac\alpha2.
\end{align}
So if we now write Equation $(1)$ in the form
$$
r^2\left(\frac92 \alpha - \left(\frac92 \sin\alpha + 2 \sin\theta\right)
+ 2 \theta\right),
$$
by the substitution $\frac92 \sin\alpha + 2 \sin\theta = 9 \sin\frac\alpha2$
we can see this is equal to your result.
If you generalize the problem so that the circles can have any radius and $B$ is not necessarily on the circumference of the circle about $A,$
with the restriction only that the two circles intersect at two distinct points, then the provided solution can be adapted to a formula to solve the more general problem. That is one reason it is likely to have been chosen.
Your solution, which effectively simplifies
$\frac92 \sin\alpha + 2 \sin\theta$ to $9 \sin\frac\alpha2,$
is possible because $B$ is on the circumference of the other circle.
|
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|
The solution set of the equation $|\lfloor x|x|-7\rfloor|<5$ is? My solution approach :-
$|\lfloor x|x|-7\rfloor|<5$
When $x\geq 0$, then;
$|\lfloor x^2-7\rfloor|<5$
$\Rightarrow -5 \lt \lfloor x^2-7\rfloor \lt5$
After this I do know that I'll have to take another case in which $x\lt0$ but I am not able to get how to solve the floor function.
|
So far, when $x\geq 0$ you got here
$$
-5 \lt \lfloor x^2-7\rfloor \lt5
$$
This means $\lfloor x^2-7\rfloor \in \{-4,-3,-2,-1,0,1,2,3,4\}$ or in other words $x^2 - 7 \in [-4,5)$.
Therefore we have
$$
x^2 \in [3,12),
$$
or
$$
x\in[\sqrt{3},2\sqrt{3}).
$$
Note that $x\in(-2\sqrt{3}, -\sqrt{3}]$ is excluded because we only take $x\geq 0$.
For the case $x\leq 0$, we have
$$
-5 < \lfloor -x^2 - 7 \rfloor < 5.
$$
But $\lfloor -x^2 - 7 \rfloor \leq -7 < -5$, therefore there are no solutions for the case $x\leq 0$.
To sum it all up, the solution set of the equation $|\lfloor x|x|-7\rfloor|<5$ is $x\in[\sqrt{3},2\sqrt{3})$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Generalization of Remainder Theorem Technique Generalization of this question
$$\frac{x^{2021}}{x^3 +x^2+x+1}$$
We wish to determine the remainder for the expression above. As in the linked question, write
$$x^{2021}=({x^3 +x^2+x+1})P(x)+ R(x)$$
To eliminate the term with P(x) let
$$x^3 +x^2+x+1=0\Rightarrow x^4 =1$$
We must then have:
$$R(x)= x^{2021}=x^{2020}\times x=1 \times x =x$$
Second Example
The remainder of $x^{2023}$ when dividing by $x^3+x^2+x+1$ will be after reduction:
$$x^{2023}=x^3=-(x^2+x+1)$$
(You can write this more formally using the Remainder Theorem, but this is a shorter version).
Further generalization
We can generalize this technique. To find the remainder when dividing by
$$1+x+x^2+...+x^n$$
substitute
$$x^{n+1}=1,x\neq1$$
Is this correct?
|
As opposed to blindly using "eliminate the term", a simple explanation of why that works is just algebraic manipulation:
$$\begin{align} & x^{2021} \\
= & (x^4 -1 ) A(x) + x \\
= & (x^3+x^2+x+1) B(x) + x.\\
\ \\
& x^{2023} \\
= & (x^4 -1 ) C(x) + x^3 \\
= & (x^3+x^2+x+1) D(x) + x^3 \\
= & (x^3+x^2+x+1)E(x) + (-x^2-x-1).\end{align}$$
Note: Of course, I'm using that $ x^{2020} - 1 = (x^4 - 1) \times F(x)$, which could be demonstrated directly via algebraic manipulation.
General idea: If we want to divide by the polynomial $P(x)$, which has a "simpler" form of $P(x) Q(x)$, then we can write it out step by step.
This could be applied even if $P(x)$ isn't a factor of $ x^n - 1 $. The hard part is guessing what $Q(x)$ could be to "simplify" the polynomial.
Here's a worked example, with the solution hidden.
What is the remainder when $x^{2021} $ is divided by $x^2 + 2x + 2$?
Using Remainder-Factor theorem might be a slight pain (esp if someone isn't comfortable with complex numbers) as the roots are $-1 \pm i$, which we have to take to the 2021 power and equate to $Ax+B$.
However, if you recognize the Sophie-Germain factorization $ x^4 + 4 = (x^2 - 2x + 2 ) (x^2 + 2x + 2)$, we see that
$$x^{2021} = (x^4 + 4 ) A(x) + 4^{403} x^5 = (x^4 + 4)B(x) - 4^{404} x.$$
Here's an example of you to practice on:
What is the remainder when $x^{2021} $ is divided by $x^4 - 2x^2 + 2$?
|
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|
USAMO 1973 (Simultaneous Equations)
Determine all the roots, real or complex, of the system of simultaneous equations (USAMO 1973/4)
$$x+y+z=3$$
$$x^2+y^2+z^2=3$$
$$x^3+y^3+z^3=3$$
Multiply equation I by 2 and subtract it from Equation II:
$$x^2-2x+y^2-2y+z^2-2z=-3$$
Complete the square for all three variables:
$$(x-1)^2+(y-1)^2+(z-1)^2=0$$
Since the RHS is zero, and the LHS has only perfect squares, by the trivial inequality, we must have:
$$(x-1)^2=0\Rightarrow x=1$$
$$(y-1)^2\Rightarrow y=1$$
$$(z-1)^2=0\Rightarrow z=1$$
Hence, the only solution for Equations I and II is
$$x=y=z=1$$
This satisfies Equation III as well. And hence, this is also the only solution to the overall system of equations.
1.
Would this be enough to get full marks?
Anything missing, or needed to be added?
2.
The complex roots is neither needed nor used anywhere. This was just used to (artificially?) increase the "complexity" of the problem. Is this correct.
|
The nicer solutions have been posted already, so here is a different one, just for fun.
Moving the constant terms $\,3\,$ to the LHS, distributing and factoring, the system can be written as:
$$
\begin{cases}
(x-1) & +\; (y-1) & +\; (z-1) & = 0 \\
(x-1) \cdot (x+1) & +\; (y-1) \cdot (y+1) & +\; (z-1) \cdot (z+1) & = 0 \\
(x-1) \cdot (x^2+x+1) & +\; (y-1) \cdot (y^2+y+1) & +\; (z-1) \cdot (z^2+z+1) & = 0
\end{cases}
$$
Considering this as a linear system in $\,x-1,y-1,z-1\,$ its determinant is:
$$
\left|
\begin{matrix}
\;1 \;&\; 1 \;&\; 1 \; \\
\;x+1 \;&\; y+1 \;&\; z+1 \; \\
\;x^2+x+1 \;&\; y^2+y+1 \;&\; z^2+z+1 \; \\
\end{matrix}
\right|
$$
After the obvious manipulations, this reduces to a Vandermonde determinant in variables $\,x,y,z\,$, which is non-zero iff all variables are different. But that case leads to the contradiction that the trivial solution $\,x-1=y-1=z-1=0\,$ has all variables equal. So the determinant must be $\,0\,$ i.e. two of the variables be equal. Assuming $\,x=y\,$ for example, it is straightforward to verify that the only solution is, again, $\,x=y=z=1\,$.
|
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|
Finding the inverse of a block $2\times2$ square matrix $ \begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} $ where $A$ is a square invertible matrix. The hint I got was to rewrite the original matrix as a product of 3 matrices and use the property for inverse of product of matrices $(XYZ)^{-1} = Z^{-1} Y^{-1} X^{-1}$
$\begin{align} \begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} &= \begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix} \begin{bmatrix} I & A \\ 0 & I \end{bmatrix}
\end{align}\\ $
Then
$\begin{align}
\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix}^{-1}= \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} ^{-1} \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix}^{-1} \begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix}^{-1}
\end{align}$
The first 2 matrices on the RHS are invertible but the third is not, so I might have gone down the wrong path with the product of 3 matrices I think.
The actual inverse for $\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} $ is $\begin{bmatrix} 0 & A^{-T} \\ A^{-1} & -A^{-1}A^{-T} \end{bmatrix}$ which can be found by using the formula at the end of section 3 in Lu and Shiou. I would like to know how to solve the problem without resorting to using a formula. Thanks for any tips.
|
Your product formula is incorrect. Computing the product you have written yields
$$
\begin{bmatrix} I & 0 \\ A^T & 0 \end{bmatrix} \begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix} \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} =
\begin{bmatrix}
I & A\\A^T & A^TA
\end{bmatrix}.
$$
A correct formula is
\begin{align}
\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix} &=
\begin{bmatrix} I & 0 \\ A^T & I \end{bmatrix}
\begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix} \begin{bmatrix} I & A \\ 0 & I \end{bmatrix},
\end{align}
which coincides with the formula referenced here. Taking the inverse yields
$$
\begin{bmatrix} I & A \\ 0 & I \end{bmatrix}^{-1}
\begin{bmatrix} I & 0 \\ 0 & -A^T A \end{bmatrix}^{-1}
\begin{bmatrix} I & 0 \\ A^T & I \end{bmatrix}^{-1} = \\
\begin{bmatrix} I & -A \\ 0 & I \end{bmatrix}
\begin{bmatrix} I & 0 \\ 0 & -A^{-1}A^{-T} \end{bmatrix}
\begin{bmatrix} I & 0 \\ -A^T & I \end{bmatrix}.
$$
Taking the product of these leads to the correct answer of
$$
\begin{bmatrix} I & A \\ A^T & 0 \end{bmatrix}^{-1} = \begin{bmatrix} 0 & A^{-T} \\ A^{-1} & -A^{-1}A^{-T} \end{bmatrix}.
$$
|
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|
What will be the maximum value of expression: 7sin²x + 5cos²x +√2[sin(x/2) + cos(x/2)]? This is what I've managed to do:
$=5 + 2\sin²x +√2[\sin(x/2) + \sin(π/2 - x/2)]$
$=5 + 2\sin²x + √2[2.\sin((x/2 + π/2 -x/2)/2)·\cos((x/2 - π/2 + x/2)/2)]$
$=5 + 2\sin²x + √2[2\sin(π/4)\cos(x/2 - π/4)]$
$=5 + 2\sin²x + 2\cos(x/2 - π/4)$
From here I've tried to reduce everything to $\sin(x/2)$ and $\cos(x/2)$ but unfortunately I can't do anything further.
Please help and provide a solution.
|
HINT
I would start with noticing that
\begin{align*}
\sqrt{2}\left[\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}
{2}\right)\right] & = 2\times \left[\frac{1}{\sqrt{2}}\times\sin\left(\frac{x}{2}\right) + \frac{1}{\sqrt{2}}\times\cos\left(\frac{x}{2}\right)\right]\\\\
& = 2\sin\left(\frac{x}{2} + \frac{\pi}{4}\right)
\end{align*}
Moreover, we do also have that
\begin{align*}
7\sin^{2}(x) + 5\cos^{2}(x) & = 7\times [\sin^{2}(x) + \cos^{2}(x)] - 2\cos^{2}(x)\\\\
& = 7 - 2\cos^{2}(x)\\\\
& = 6 - \cos(2x)
\end{align*}
Hence the proposed function can be described as
\begin{align*}
f(x) & := 7\sin^{2}(x) + 5\cos^{2}(x) + \sqrt{2}\times\left[\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right]\\\\
& = 6 - 2\cos(2x) + 2\sin\left(\frac{x}{2} + \frac{\pi}{4}\right)
\end{align*}
whose first derivative is given by:
\begin{align*}
f'(x) = -4\sin(2x) + \cos\left(\frac{x}{2} + \frac{\pi}{4}\right)
\end{align*}
Can you take it from here?
|
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|
Prove that $\int_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$ Prove that $$\int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$$
On simplifying by parts we get:
$$\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$$
Thus if we prove that$$\max\left(\displaystyle\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx\right)<\frac{1}{2(n+1)(n+2)}$$
We will be able to prove the above inequality
|
Inspired by the other answers: $f(x) = \frac{x}{1+x}$ is concave on $[0, \infty)$, and therefore
$$
f(x) \le f(1) + (x - 1) f'(1) = \frac 1 4(1+x)
$$
for $x \ge 0$. It follows that
$$
\int_0^1 \frac{x^n}{1+x}\, dx \le \frac 1 4 \int_0^1 \left(x^{n+1} + x^{n+2} \right) \, dx \\
= \frac 1 4 \left( \frac 1{n+2} + \frac 1{n+3}\right) < \frac{1}{2(n+2)} \, ,
$$
which is slightly sharper than the desired estimate.
|
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|
Stuck solving limit without L'Hospital's rule We need to solve limit:
$$\lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{4\cos^2{x}-1}$$ without using derivation (L'Hospital's rule). With the substitution $t = 4\cos^2{x}-1$, I got it to $$\lim_{t \to 0} \frac{\ln{(\sqrt{9-3t}-2)}}{t}$$ but can't progress any further without going in circles.
|
Here's another way. First use $\cos^2{x}=1-\sin^2{x}$ and
let $y=\ln{(2\sqrt{3}\sin{x}-2})$, $\frac{e^{y}+2}{2\sqrt{3}}=\sin{x}$, $y\to0$ as $x\to\frac{\pi}{3}$
$$\begin{align}
\lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{4\cos^2{x}-1}
&=\lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{4(1-\sin^2{x})-1}\\
&=\lim_{x \to \frac{\pi}{3}} \frac{\ln{(2\sqrt{3}\sin{x}-2})}{3-4\sin^2{x}}\\
&=\lim_{y \to 0} \frac{y}{3-4 \left(\frac{e^{y}+2}{2\sqrt{3}}\right)^2}\\
%&=\lim_{y \to 0} \frac{y}{3-4 \left(\frac{e^{2y}+4e^{y}+4}{4\times3}\right)}\\
&=\lim_{y \to 0} \frac{3y}{5-\left(e^{2y}+4e^{y}\right)}\\
&=\lim_{y \to 0} \frac{3y}{5-\left(\displaystyle\sum_{n=0}^{\infty}\frac{(2y)^n}{n!}+4\displaystyle\sum_{n=0}^{\infty}\frac{y^n}{n!}\right)}\\
&=\lim_{y \to 0} \frac{3y}{5-\displaystyle\sum_{n=0}^{\infty}\left(\frac{(2y)^n+4y^n}{n!}\right)}\\
&=\lim_{y \to 0} \frac{3y}{5-\left(\frac{(2y)^0+4y^0}{0!}+\frac{(2y)^1+4y^1}{1!}+\displaystyle\sum_{n=2}^{\infty}\frac{(2y)^n+4y^n}{n!}\right)}\\
&=\lim_{y \to 0} \frac{3y}{5-\left(\frac{1+4}{1}+\frac{2y+4y}{1}+\displaystyle\sum_{m=0}^{\infty}\frac{(2y)^{m+2}+4y^{m+2}}{(m+2)!}\right)}\\
&=\lim_{y \to 0} \frac{3y}{5-\left(5+6y+\displaystyle\sum_{m=0}^{\infty}\frac{(2y)^{m+2}+4y^{m+2}}{(m+2)!}\right)}\\
&=\lim_{y \to 0} \frac{3y}{-6y-\displaystyle\sum_{m=0}^{\infty}\left(\frac{(2y)^{m+2}+4y^{m+2}}{(m+2)!}\right)}\\
&=\lim_{y \to 0} \frac{1}{-2-\frac{1}{3y}\displaystyle\sum_{m=0}^{\infty}\frac{(2y)^{m+2}+4y^{m+2}}{(m+2)!}}\\
&=\lim_{y \to 0} \frac{1}{-2-\displaystyle\sum_{m=0}^{\infty}\frac{(2y)^{m+1}+4y^{m+1}}{3(m+2)!}}\\
&=\frac{1}{-2}\\
\end{align}$$
|
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|
Integrate $\int_{0}^{\infty}\frac{\tan^{-1}x}{x\left(1+x^{2}\right)}dx$ Q:
Integrate $I=\int_{0}^{\infty}\frac{\tan^{-1}x}{x\left(1+x^{2}\right)}dx$
My Approach:
Put $$\tan^{-1}x=t\to x=\tan t$$
Also we have, $$\frac{dx}{1+x^{2}}=dt$$
We get, $$I=\int_{0}^{\frac{\pi}{2}}\frac{t}{\tan t}dt$$
I'm stuck here, how do I proceed further? I tried integration by parts but it doesn't seem to work out for me.
Edit: I tried again and I got the answer
|
$$\begin{align*}
\int_0^\infty \frac{\tan^{-1}(x)}{x(1+x^2)} \, dx &= \int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)} \, dx + \int_1^\infty \frac{\tan^{-1}(x)}{x(1+x^2)} \, dx \\[1ex]
&= \int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)} \, dx - \int_1^0 \frac{\tan^{-1}\left(\frac1x\right)}{\frac1x\left(1+\frac1{x^2}\right)} \frac{dx}{x^2} & (1) \\[1ex]
&= \int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)} \, dx + \int_0^1 \frac{\frac\pi2 x-x\tan^{-1}(x)}{1+x^2} \, dx & (2)\\[1ex]
&= \frac\pi2 \int_0^1 \frac x{1+x^2} \, dx + \int_0^1 \frac{\tan^{-1}(x)}{1+x^2} \left(\frac1x-x\right) \, dx \\[1ex]
&= \frac{\pi\ln(2)}4 + \int_0^1 \frac{\tan^{-1}(x)}x \, dx - \int_0^1 \frac{2x\tan^{-1}(x)}{1+x^2} \, dx & (3) \\[1ex]
&= \frac{\pi\ln(2)}4 + G - \left(\frac{\pi\ln(2)}4 - \int_0^1 \frac{\ln(1+x^2)}{1+x^2} \, dx\right) & (4) \\[1ex]
&= \boxed{\frac{\pi\ln(2)}2} & (5)
\end{align*}$$
*
*$(1)$ substitute $x\mapsto\frac1x$
*$(2)$ if $x>0$, then $\tan^{-1}(x)+\tan^{-1}\left(\frac1x\right)=\frac\pi2$
*$(3)$ partial fractions
*$(4)$ $G$ is Catalan's constant; integration by parts
*$(5)$ this integral with $a=1$
|
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|
Find the sum of series $ \sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$ Let it be known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}
{6}.$$ Given such—find $$\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$$
Attempt:
I have tried using the fact that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and then expanding or using known sum types as $\displaystyle \sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ or $\displaystyle \sum_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4}$
but nothing seems to lead to anything!
|
HINT:
Use the partial fraction decomposition
$$\frac{1}{n^3(n+1)^3} = \left(\frac{1}{n^3} - \frac{1}{(n + 1)^3}+ \frac{6}{n} - \frac{6}{n+1}\right) - \left(\frac{3}{n^2} + \frac{3}{(n + 1)^2}\right) $$
The sum equals $(1+6) -( 3 \zeta(2) +3 \zeta(2)- 3)= 10 - \pi^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $. Simply bringing it to a common denominator does not lead me to success How can I solve this equation?
$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $$
Simply bringing it to a common denominator does not lead me to success
What I tried
|
If you insist on not dealing with a cubic equation, then maybe this can work. Notice:
$$\dfrac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)} = \dfrac{(1+x)(-1-2x)(1+3x)}{(4+x)(-4-2x)(4+3x)} = \dfrac{a(b-c)(c-4)}{b(a-c)(c-1)} = 4$$
where $c = 3x+5$, $a = x+1$ and $b = x+4.$ If you expand this quadratic in $c:$
$$c^2(a-4b)+c(3ab+4b-4a) = 0$$
and this one is easy to solve.
But I must say that learning how to factor cubic if they have nice solutions by the Rational Root Theorem etc is much more useful than trying to find clever substitutions. The other answer already covers that.
|
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|
Making $6$ digits numbers by the digits $2,3,3,5,5,5,5$
How many $6$ digits number we can generate by the digits $2,3,3,5,5,5,5$ ?
To solve this problem I considered three cases because we have $7$ digits and looking for six digit number.
*
*Numbers without $2\Rightarrow\quad\dfrac{6!}{4!2!}=15$
*Numbers without one $3\Rightarrow\quad\dfrac{6!}{4!}=30$
*Numbers without one $5\Rightarrow\quad\dfrac{6!}{3!2!}=60$
So the answer is $15+30+60=105$.
But The book I'm reading from used another method:
Number of $6$ digits permutation from these $7$digits=Number of $7$ digits permutation from these $7$digits= $\dfrac{7!}{2!4!}=105$.
But I don't completely why permutation of $6$ digits is equal to permutation of all the digits.
|
\begin{array}{|c|c|c|c|}
\hline\hline
2s&3s&5s&6-\text{digit numbers}\\
\hline
1&1&4&\frac{6!}{4!}=30\\
\hline
1&2&3&\frac{6!}{2!3!}=60\\
\hline
0&2&4&\frac{6!}{2!4!}=15
\end{array}
\begin{array}{|c|c|}
\hline
\hline
\text{Total 6-digit numbers}&30+60+15=105
\end{array}
Your calculations are correct, as shown in the table. Regarding your question as to why the seven-digit permutations work, observe that every six-digit permutation can be changed into a seven-digit permutation by appending the unused digit. There is a one-to-one correspondence between the two.
$$233555\color{magenta}{5}$$
$$253553\color{magenta}{5}$$
$$255553\color{magenta}{3}$$
|
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|
Convergence of $1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } $ as $n \to \infty$ I stumbled upon this problem while reading about the bias of the sample standard deviation.
How to show that:
$$\bigg(1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \bigg) \sim \frac{1}{4 n}$$
as $n \to \infty$ and n is positive integer.
I thought I should use Stirling's formula, or Gautschi's inequality. The following equality might also lead in the right direction:
$$\Gamma(1/2+n)=\frac{(2n)!}{4^nn!}\sqrt{\pi}$$
Can you help?
|
By Stirling's formula we have
$$1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \sim 1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \sqrt{2\pi \left(\frac n 2 -1\right)}\left(\frac {\frac n 2 -1} e\right)^{\frac n 2 -1} }{\sqrt{2\pi \left(\frac n 2 -\frac 32\right)}\left(\frac {\frac n 2 -\frac 32} e\right)^{\frac n 2 -\frac32} } $$
$$=1 - \sqrt{ \frac{2}{n-1} } \cdot \frac1{\sqrt{2e}}\frac{n-3}{\sqrt{n-2} }\left(\frac{n-2}{n-3}\right)^\frac n 2 = \ldots$$
and since
$$\sqrt{ \frac{2}{n-1} } \cdot \frac{n-3}{\sqrt{n-2} }=\sqrt 2 \sqrt{1-\frac{3n-6}{n^2-3n+3}}=\sqrt 2 \left(1-\frac 3{2n}+O\left(\frac1{n^2}\right)\right)$$
$$\left(\frac{n-2}{n-3}\right)^\frac n 2= \left(1+\frac{1}{n-3}\right)^\frac n 2= \sqrt e\left(1+\frac 5{4n}+O\left(\frac1{n^2}\right)\right)$$
we have
$$\ldots = 1-\left(1-\frac 3{2n}+O\left(\frac1{n^2}\right)\right)\left(1+\frac 5{4n}+O\left(\frac1{n^2}\right)\right)=1-\left(1-\frac 3{2n}+\frac 5{4n}+O\left(\frac1{n^2}\right)\right)=\frac1{4n}+O\left(\frac1{n^2}\right)$$
|
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|
The series representation of $\sum _{i=0}^{n-1} (-1)^{i+1} \left(\frac{i}{n}\right)^{r}$ Lets consider the:
$$\left(\frac{1}{n}\right)^r-\left(\frac{2}{n}\right)^r+\left(\frac{3}{n}\right)^r-...+\left(\frac{n-1}{n}\right)^r$$
Trying to find any formal serise representation for variable $n$. Have tried the Euler–Maclaurin summation, seems there is no simple way to go to series from there.
I have separated the odd and even parts:
$$\sum _{u=0}^{n-1} (-1)^{u+1}\left(\frac{u}{n}\right)^r=\sum _{u=0}^{\frac{n-2}{2}} \left(\frac{2 u+1}{n}\right)^r-\sum _{u=0}^{\frac{n-1}{2}} \left(\frac{2 u}{n}\right)^r$$
Then for each sum:
$$\sum _{u=0}^{\frac{n-2}{2}} \left(\frac{2 u+1}{n}\right)^r=\frac{1}{2} \left(\left(\frac{2 \left\lfloor \frac{n-2}{2}\right\rfloor +1}{n}\right)^r+\left(\frac{1}{n}\right)^r\right)+\int_0^{\frac{n-2}{2}} \left(\frac{2 u+1}{n}\right)^r \, du + \sum _{k=1}^{\infty } \frac{B_{2 k} \frac{\partial ^{2 k-1}}{\partial u^{2 k-1}}\left(\frac{2 u+1}{n}\right)^r}{(2 k)!} \biggr|_0^{\lfloor \frac{n-2}{2} \rfloor}$$
$$\sum _{u=0}^{\frac{n-1}{2}} \left(\frac{2 u}{n}\right)^r=\frac{1}{2} \left(\frac{2 \left\lfloor \frac{n-1}{2}\right\rfloor }{n}\right)^r+\int_0^{\frac{n-1}{2}} \left(\frac{2 u}{n}\right)^r \, du+\sum _{k=1}^{\infty } \frac{B_{2 k} \frac{\partial ^{2 k-1}}{\partial u^{2 k-1}}\left(\frac{2 u}{n}\right)^r}{(2 k)!} \biggr|_0^{\lfloor \frac{n-1}{2} \rfloor}$$
Then after extracting for first 2 terms for example for odd $n$ we get nothing simple in terms of series representation:
$$\frac{(n-2)^r-(n-1)^r+\frac{r}{r+1}}{2 n^r}+... +\sum _{k=1}^{\infty }... - \sum _{k=1}^{\infty }...$$
Any hint is appreciated.
|
Hint: We can write the expression as
\begin{align*}
\color{blue}{\sum_{i=0}^{n-1}(-1)^{i+1}\left(\frac{i}{n}\right)^r}
&=\frac{1}{n^r}\sum_{i=0}^{n-1}(-1)^{i+1}i^r\\
&=\frac{1}{n^r}\left(\sum_{i=0}^{n-1}i^r-\sum_{i=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(2i)^r\right)\tag{1}\\
&\,\,\color{blue}{=\frac{1}{n^r}\left(\sum_{i=1}^{n-1}i^r-2^r\sum_{i=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}i^r\right)}\tag{2}\\
\end{align*}
Comment:
*
*In (1) we split the sum by adding even terms to the left sum and subtracting them twice in the right sum.
*In (2) we note the terms with index $i=0$ are zero and we factor out $2^r$.
You can then apply the Euler-Maclaurin formula to these somewhat easier expressions or you may recall Faulhaber's formula
which is in fact the result of application of the Euler-Maclaurin formula.
|
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|
Struggling to solve an equation containing a floor function I am trying to solve the following equation ($x\in \mathbb R)$
$$
2\lfloor{(x+1)^2+8}\rfloor=(x+1)(2x+3) \tag{1}
$$
I have tried substituting $x$ with its split form (integer part ($n$) + decimal part ($\alpha$)), i.e., $x=n+\alpha$ with $n=\lfloor x \rfloor.$
The substitution yields:
$$
\lfloor 2\alpha^2 + 4\alpha n + 4\alpha\rfloor=2\alpha^2+4\alpha n+5\alpha + 3n -15\tag{2}
$$
or equivalently
$$
\lfloor 2\alpha(\alpha + 2n +2)\rfloor= \alpha(2\alpha + 4n + 5) + 3n -15\tag{3}
$$
and here I am stuck and I don't know how to proceed, or whether I've already complicated matters more than needed. Any hints would be much appreciated.
|
Let $x$ be an integer.
Set $y = x+1$, then we have $2\lfloor y^{2}+8 \rfloor=y(2y+1)=2y^{2}+y. $
Since $\lfloor a+c \rfloor = \lfloor a \rfloor +c$ for all integers $a$, $2\lfloor y^{2}+8 \rfloor=2(\lfloor y^{2} \rfloor + 8)=2\lfloor y^{2} \rfloor +16=2y^{2}+y$ from above.
If $y > 16$, then $2\lfloor y^{2} \rfloor = 2y^{2} + c$ for some $c > 0 \implies 2y^{2}<2 \lfloor y^{2} \rfloor \implies y^{2}<\lfloor y^2 \rfloor$ which is a contradiction as $\lfloor a \rfloor \leq a$ for all $a$ by definition.
Hence, we must check the finite number of cases of the form $y \leq 16 \implies x \leq 15$.
Since $x$ is an integer ($\implies y$ is an integer), then $2y^{2}+16=2\lfloor y^{2} \rfloor +16=2y^{2}+y \implies$
$y = 16 \implies x = 15.$
Therefore $15$ is the only integer solution.
|
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|
Prove that $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge ab + bc + ca$ For all positive $a,b,c $ satisfying $a+b+c = 3$,Prove:
$$
\sum_{cyc} \sqrt[3]{a} \ge \sum_{cyc} ab
$$
This is a hard problem and I tried it myself, but it's really hard without using advanced techniques(e.g. EV theorem or HCF theorem).
Is there any easy way without computer and HCF to this problem?
(Updated) Sorry for that I posted this in a hurry. I will show you more in this updated part.
First of all, HCF kills it.(If you don't know it, click here) Consider
$$
2\sum_{cyc} ab = (\sum_{cyc} a)^2 - \sum_{cyc} a^2 = 9 - \sum_{cyc} a^2
$$
Thus, we only need to prove
$$
\sum_{cyc} (a^2 + 2 \sqrt[3]{a}) \ge 9
$$
or
$$
\sum_{cyc} f(a) \ge 9
$$
where $f(x)=x^2 + 2\sqrt[3]{x}$. And because $f''(x)=2-\frac{4}{9\sqrt[3]{x^5}}$, Thus we know $f(x)$ is convex on $\left [ \frac{\sqrt[5]{8}}{3\sqrt[5]{3}}, 3\right)$. So by HCF theorem we only need to prove when $b=a$,$c=3-2a$ or
$$
6a^2 - 12a + 9 + 4\sqrt[3]{a} +2 \sqrt[3]{3-2a} \ge 9
$$
The rest is easy with derivative.
Second, a friend of mine told me that using tangent line can solve it. But I didn't see how that's relevant.
Third, I found this problem on the Internet, but I'm sure I've solved it or saw the solution to it before.
|
pqr method:
Using AM-GM, we have
$$\sqrt[3]{a} = \frac{a}{\sqrt[3]{a \cdot a \cdot 1}} \ge \frac{a}{(a + a + 1)/3} = \frac{3a}{2a + 1}.$$
It suffices to prove that
$$\frac{3a}{2a + 1} + \frac{3b}{2b + 1} + \frac{3c}{2c + 1} \ge ab + bc + ca.$$
Let $p = a + b + c = 3, q = ab + bc + ca, r = abc$.
It suffices to prove that
$$\frac{36r + 12q + 3p}{8r + 4q + 2p + 1} \ge q$$
or
$$(q + 1)(9 - 4q) + (36 - 8q)r \ge 0.$$
If $9 - 4q \ge 0$, since $q \le p^2/3 = 3$,
the inequality is true.
If $9 - 4q < 0$, using $r \ge \frac{4pq - p^3}{9} = \frac{4}{3}q - 3$ (Schur's inequality), it suffices to prove that
$$(q + 1)(9 - 4q) + (36 - 8q)\cdot \left( \frac{4}{3}q - 3\right) \ge 0$$
or
$$\frac{11}{3}(4q - 9)(3 - q) \ge 0$$
which is true.
We are done.
|
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|
Eliminating $\theta$ from $\cos^3\theta +a\cos\theta =b$ and $\sin^3\theta +a\sin\theta =c$
Eliminate $\theta$ from the equations.
$$\cos^3\theta +a\cos\theta =b$$
$$\sin^3\theta +a\sin\theta =c$$
Can anyone solve this question?
|
$\def\¿{\mathcal}$
Maybe I am complicating this, yet another method.
I will use these identities:
$\cos^6x+\sin^6x=1-\frac34\sin^22x$
$\cos^6x-\sin^6x=\cos 2x(1-\frac14\sin^22x)$
$\cos^4x+\sin^4x=1-\frac12\sin^22x$
$\cos^4x-\sin^4x=\cos2x$
Let
$c=\cos x$,
$s=\sin x$,
$\¿ C=\cos 2x$,
$\¿S=\sin2x$
I use $d$ instead of $c$ in the original problem to avoid confusion.
Now we have $$c^3+ac=b\tag{1}$$ $$s^3+as=d\tag{2}$$
$\small\mathit{(1)^2+(2)^2}$, $$c^6+s^6+2a(c^4+s^4)+a^2(c^2+s^2)=b^2+d^2$$ $$\left(1-\frac34\¿S^2\right)+2a\left(1-\frac12\¿S^2\right)+a^2=b^2+d^2$$ $$\¿S^2=\frac{4\left[\left(a+1\right)^2-(b^2+d^2)\right]}{(3+4a)}$$
$\small\mathit{(1)^2-(2)^2}$, $$c^6-s^6+2a(c^4-s^4)+a^2(c^2-s^2)=b^2-d^2$$ $$\¿C\left(1-\frac14\¿S^2\right)+2a\¿C+a^2\¿C=b^2-d^2$$ $$\¿C\left[(a+1)^2-\frac14\¿S^2\right]=b^2-d^2$$ $$\¿C=\frac{(3+4a)(b^2-d^2)}{2(a+1)^2(2a+1)+(b^2+d^2)}$$ Now we can use $$\¿C^2+\¿S^2=1$$
|
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|
solve the equation $1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$ im trying to solve the equation $$(E)\quad 1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$$
attempt : because $1$ isnt a solution we have $$\begin{aligned}
1+2 z+2 z^{2}+\cdots+2 z^{n-1}+z^{n} &=2\left(z^{0}+z+z^{2}+\cdots+z^{n-1}\right)-1+z^{n} \\
&=2 \frac{1-z^{n}}{1-z}-1+z^{n} \\
&=\frac{2-2 z^{n}+z^{n}-z^{n+1}}{1-z}-1 \\
&=\frac{2-z^{n}-z^{n+1}}{1-z}-1
\end{aligned}$$ then (E) becomes $$z^{n+1}+z^{n}-z-1=0$$ but i dont know how to get any further.
|
Hint: \begin{align}
z^{n+1}+z^n-z-1&=z^n(z+1)-(z+1)\\
&=(z^n-1)(z+1)
\end{align}
Also, I think you should justify why the second equality holds, i.e. explain why you can assume $1-z\neq 0$.
|
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|
Find the largest value of $P=\frac{3x+2y+1}{x+y+6}$ I know there has been a similar question here, but my question is a little bit different
The Problem:
Given that $3(x+y)=x^2+y^2+xy+2$,
then find the maximum value of
\begin{align}
P=\frac{3x+2y+1}{x+y+6}
\end{align}
What I was thinking about is that I'm trying to transform P into a first-degree equation on $x$, which is totally possible using the condition in the problem, and then apply the basic Cauchy inequalities.
However, I haven't been able to figure it out. The other possibility is to apply differentiation like in the problem I mentioned at the beginning of the question. The problem is that I took this problem from Olympiad training, which often does not use differentiation (so there should be a way that doesn't use the method)
So any help is appreciated
|
$3(x+y)=x^{2}+y^{2}+xy+2$ must be a conic section because it can be written in the form $Ax^2+Bxy+Cy^2+Dx+Ey+F = 0$. It also is symmetric across the line $y=x$ as it is invariant after swapping $x$ and $y$.
Now if the maximum value is $P$, then $3x+2y+1 = P(x+y+6)$, equation $1$. This can be seen below:
If we can rotate this ellipse by $45º$, we can write it in parametric form in terms of an angle $\theta$. Using the rotation matrix and a scaling factor, we need the transformation $x \to x - y, y \to x + y$ as $\cos \pi/4 = \sin \pi/4$:
$$3(x-y + x+y) = (x-y)^2 + (x+y)^2 + (x-y)(x+y) + 2$$
$$6x = 2(x^2 + y^2) + (x^2-y^2) + 2$$
$$6x = 3x^2 + y^2 + 2 \tag{2}$$
where from $(1)$, we now have:
$$3(x-y) + 2(x+y) + 1 = P \left( (x-y) + (x+y) + 6 \right)$$
$$5x-y+1 = P(2x+6)$$
$$y = (5-2P)x-6P+1 \tag{3}$$
Just to check, the lines are still tangent to the ellipse after transformation:
and now we can substitute $(3)$ into $(2)$:
$$6x = 3x^2 + ((5-2P)x-6P+1)^2 + 2$$
$$6x = 3x^2 + (25-20P+4P^2)x^2 + 2(5-2P)(-6P+1)x + 36P^2-12P+1+2$$
$$(28-20P+4P^2)x^2 + (24P^2-64P+4)x + 36P^2-12P+3 = 0$$
$$\Delta = 0: (24P^2-64P+4)^2 - 4(28-20P+4P^2)(36P^2-12P+3) = 0$$
$$4(6P^2-16P+1)^2 - 48(7-5P+P^2)(P^2-4P+1) = 0$$
and miraculously, this is a quadratic $-752P^2 + 1072P - 320 = 0 \implies (47-20P)(P-1) = 0$, so the maximum value is $P = 1$.
|
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|
Isn't my book doing this math about differentiation wrongly? Problem:
Differentiate with respect to $x$: $\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$
My attempt:
Let,
$$y=\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$$
Both $e^x$ and $\frac{x-1}{x+1}$ are positive: $e^x$ can never be negative, and you can take the square root of only positive numbers, so $\frac{x-1}{x+1}$ is positive as well. So, we can apply logarithm properties:
$$=\ln e^x+\ln(\frac{x-1}{x+1})^{\frac{3}{2}}$$
$$=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$
Now,
$$\frac{dy}{dx}=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{d}{dx}(\frac{x-1}{x+1})$$
$$=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{x+1-x+1}{(x+1)^2}$$
$$=1+\frac{3}{2}.\frac{x+1}{x-1}.\frac{2}{(x+1)^2}$$
We can cancel $(x+1)$ and $(x+1)$ in the numerator and denominator because the graph doesn't change.
$$=1+\frac{3}{x^2-1}$$
$$=\frac{x^2+2}{x^2-1}$$
My book's attempt:
Let,
$$y=\ln(e^x(\frac{x-1}{x+1})^{\frac{3}{2}})$$
$$=\ln e^x+\ln(\frac{x-1}{x+1})^{\frac{3}{2}}$$
$$=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$
$$=x+\frac{3}{2}(\ln(x-1)-\ln(x+1))\tag{1}$$
$$...$$
$$\frac{dy}{dx}=\frac{x^2+2}{x^2-1}$$
Question:
*
*In my book's attempt, is line $(1)$ valid? I think my book's assumption in $(1)$ that $(x-1)$ and $(x+1)$ must be positive is unfounded. I deduced that $\frac{x-1}{x+1}$ is positive; $\frac{x-1}{x+1}$ can be positive even when both $(x-1)$ and $(x+1)$ is negative. So, is line $(1)$ of my book valid?
|
My book's attempt:
$$y=x+\frac{3}{2}\ln(\frac{x-1}{x+1})$$
$$=x+\frac{3}{2}(\ln(x-1)-\ln(x+1))\tag{1}\\\ldots$$
$$\frac{dy}{dx}=\frac{x^2+2}{x^2-1}$$
is line $(1)$ valid? I think my book's assumption that $(x-1)$ & $(x+1)$ must be positive is unfounded. $\frac{x-1}{x+1}$ can be positive even when both $(x-1)$ & $(x+1)$ are negative.
The author is hand-waving; pretending that $x-1$ and $x+1$ are nonnegative doesn't affect the result.
To be rigorous: on our domain of interest, $$\frac{\mathrm d}{\mathrm dx}\left(x+\frac32\ln \left(\frac{x-1}{x+1}\right)\right)\\ =\frac{\mathrm d}{\mathrm dx}\left(x+\frac32\ln \left|\frac{x-1}{x+1}\right|\right)\\ =\frac{\mathrm d}{\mathrm dx}\left(x+\frac32\Big(\ln \left|{x-1}\right|-\ln\left|{x+1}\right| \Big)\right),$$ noting that $$\frac{\mathrm d}{\mathrm dx}\ln|f(x)|=\frac{f’(x)}{f(x)}.$$
This hand-wavy issue of implicit modulus signs when there are logarithmic functions, is also common when solving differential equations: see here and here.
And here is a related answer: How is taking $\ln$ on both sides justified?
Addendum
For negative $f(x),$ $$\frac{\mathrm d}{\mathrm dx}\ln\left|f(x)\right|\\=\frac{\mathrm d}{\mathrm dx}\ln\left(-f(x)\right)\\=\frac{-f'(x)}{-f(x)}\\=\frac{f'(x)}{f(x)};$$ for positive $f(x),$ again $$\frac{\mathrm d}{\mathrm dx}\ln\left|f(x)\right|=\frac{f'(x)}{f(x)}.$$
|
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|
$3 \sin x + 4 \cos y = 5$, $4 \sin y + 3 \cos x = 2$ How to find $\sin x$, $\sin y$, $\cos x$, $\cos y$, 2020 contest question https://www.cemc.uwaterloo.ca/contests/past_contests/2020/2020CSMC.pdf
https://www.cemc.uwaterloo.ca/contests/past_contests/2020/2020CSMCSolution.pdf
Question 5 from 2020 CSMC math contest:
$$3 \sin x + 4 \cos y = 5,$$ $$4 \sin y + 3 \cos x = 2.$$
Find $\sin(x+y)$.
The solution page added that it is also possible to find $\sin x$, $\sin y$, $\cos x$, $\cos y$ with a different approach but I could not figure it out.
|
We have $4\cos y=5-3\sin x$. Similarly we have $4\sin y=2-3\cos x$. Squaring and add:
$$
16=(5-3\sin x)^2+(2-3\cos x)^2=38-30\sin x-12\cos x
$$
so $15\sin x+6\cos x=11$
which gives you possible $\sin x,\cos x$ (i.e., solve the system $15\sin x+6\cos x=11, \sin^2 x+\cos^2 x=1$ to get $\sin x=\frac{55\pm 4\sqrt{35}}{87}$, and similar for $\cos x$, or solve for $\tan\frac12x=\frac{15\pm 2\sqrt{35}}{17}$) and hence $\sin y,\cos y$.
|
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|
How to use the epsilon delta definition to prove that $\lim\limits_{x\to 1} \frac{x^3-1}{x-1} = 3$
Not sure if I am doing this right, however, this is what I have:
Let $\epsilon > 0$. We need to find a $\delta > 0$ such that $0<|x-1|<\delta$ leads to the conclusion $|f(x)-3|<\epsilon$.
$$|\frac{x^3-1}{x-1} - 3| < \epsilon$$
We know that $x^3-1 = (x-1)(x^2+x+1)$, so
$$|x^2+x+1-3|<\epsilon$$
$$|(x+2)(x-1)|<\epsilon$$
Do I need to use the triangle inequality? Not really sure where to go from here. Any help is appreciated.
|
Following @Ben's comment...
We want to find $\delta$ such that $|x-1|<\delta \Rightarrow |(x-2)(x-1)|<\epsilon$.
As $|(x-2)(x-1)|=|x-2||x-1|\leq(|x-1|+3)|x-1|$, because of the triangle inequality..then it is enough to find $\delta$ such that
$|x-1|<\delta \Rightarrow (|x-1|+3)|x-1|<\epsilon$
Notice that $(|x-1|+3)|x-1|<\epsilon \iff z^2+3z-\epsilon<0$ when $z=|x-1|\geq0$. From here one can conclude $|x-1|=z \in \Big[0,\frac{-3+\sqrt{9+4\epsilon}}{2}\Big)$ by solving the inequation.
This is equivalent to affirm that $|x-1|<\frac{-3+\sqrt{9+4\epsilon}}{2}$ and so for any $\epsilon>0$ one can take $\delta=\frac{-3+\sqrt{9+4\epsilon}}{2}$. Note that $\frac{-3+\sqrt{9+4\epsilon}}{2}>0$ for any $\epsilon>0$.
|
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|
If $\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}$, prove that $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
Problem: Let $a, b, c, x, y, z$ be real numbers
such that $abc \ne 0$, $ ~ a + b + c \ne 0$, $~ (a^2 - bc)(b^2 - ca)(c^2 - ab)\ne 0$, $xyz\ne0$, $x+y+z\ne0$, $(x-y)^2+(y-z)^2+(z-x)^2\ne0$
and $$\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab} \ne 0.$$
Prove that $$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}.$$
The converse can be proved easily.
My Attempt:
$
\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}\\
\Rightarrow\frac{x^2-yz-y^2+zx}{a^2-bc-b^2+ca}=\frac{y^2-zx-z^2+xy}{b^2-ca-c^2+ab}=\frac{z^2-xy-x^2+yz}{c^2-ab-a^2+bc}\ [By\ Addendo]\\\Rightarrow\frac{x-y}{a-b}\cdot\frac{x+y+z}{a+b+c}=\frac{y-z}{b-c}\cdot\frac{x+y+z}{a+b+c}=\frac{z-x}{c-a}\cdot\frac{x+y+z}{a+b+c}\\\Rightarrow\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}\ \left[Considering\ (x+y+z)\ and\ (a+b+c)\ \neq0\right]
$
Known Exceptional Counter-examples:-
Macavity: $a=1, b=2, c=3, x=y=z$
River Li: $a = 1, b=2, c = -3, x = 7, y = 7, z = -14$
The conditions in the question have been edited to exclude exceptional counter-examples.
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An idea would be to show that if
$$[p\colon q\colon r] = [x^2 - y z \ \colon y^2 - x z\ \colon z^2 - x y]$$
then
$$[x\ \colon y \colon z] = [p^2 - q r\ \colon q^2 - p r\ \colon r^2 - p q]$$
In other words, the map from $\mathbb{P}^2$ to itself given by
$$[x\ \colon y \colon z]\mapsto [x^2 - y z \ \colon y^2 - x z\ \colon z^2 - x y]$$
is an involution.
This can be checked by a direct calculation. Alternatively, we can see that for the matrix
$$A = \left (\begin{matrix} x & y & z \\ y & z & x \\ z & x & y\end{matrix} \right)$$
the adjugate matrix is of similar form
$$\operatorname{adj} A = \left (\begin{matrix} x^2 - y z & y^2 - x z & z^2 - x y \\ y^2 - x z & z^2 - x y & x^2 - y z \\ z^2 - x y & x^2 - y z & y^2 - x z \end{matrix} \right)$$
and use the formula
$$\operatorname{adj}\operatorname{adj}A = \det A \cdot A $$
Notice that
$$\det A = \frac{1}{2}(x+y+z) ((x-y)^2 + (y-z)^2 + (z-x)^2)$$
so according to the conditions in OP, it is $\ne 0$.
$\bf{Added:}$
Some calculations related to Hirst transform. Given a quadric $Q(p)=0$ and a point $p_0$ not on the quadric, to consider the "inversion" wr to $Q$ and center $p_0$. It is like the inversion wr to a sphere, but with a projective description. I am following @Jean Marie:
For a point $p$ in projective space, consider the intersection of the line $p_0p$ with the hyperplane polar to $p$. Denote by $\langle \cdot,\cdot\rangle$ the bilinear form giving $Q$. We have $\phi(p) = \alpha p_0 + \beta p$, such that
$$\alpha p_0 + \beta p \perp p$$
So we can take
$$\phi(p) = \langle p,p \rangle p_0 - \langle p, p_0 \rangle p $$
Check that
$$\langle \phi(p), p \rangle = 0 \\
\langle \phi(p), p_0\rangle = G(p, p_0)\\
\langle \phi(p), \phi(p)\rangle = G(p, p_0) \cdot \langle p, p\rangle $$
where $G(p, p_0)$ is the Gram determinant of $p$, $p_0$, which gives
$$\phi\circ \phi(p) = \langle p, p_0 \rangle\cdot G(p, p_0) \cdot p$$
so projectively $\phi$ is an involution.
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|
What's the measure of the segment $ON$ in the circle below? For reference:On the circle with center $O$ and perpendicular diameters
perpendiculars $AC$ and $BD$ a string $AE (E~ in~ \overset{\LARGE{\frown}}{BC})$ is drawn
such that AE intersects $BD$ at $M$; if $BM = MO$. Calculate $ON$, if $OA = 12$. ($N =ED\cap AC$).
My progress:
$\triangle ABO(right): AB = 12\sqrt2 = AD\\\triangle ABM \sim \triangle DME (A.A)\\
\frac{12\sqrt2}{DE} = \frac{BM}{ME} = \frac{AM}{DM}\\
\angle ABD = \angle BEA = \angle AED =45^o\\
\angle DBE = \angle OND \implies\\
\frac{ON}{BE}=\frac{DN}{24}=\frac{6}{DE}$
Extend $DC$ and $AE (F=DC \cap AE)$
$\angle MAD = \angle AND\\
\triangle NOD \sim \triangle ADF \implies: \frac{OD}{ON} = \frac{FD}{AD}\implies ON = \frac{144\sqrt2}{FD}...$
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Denote $\angle OAM =\alpha, \angle EAB =\beta.$
We have $\tan\alpha =\frac{|MO|}{|OA|} =\frac{1}{2}$ thus $\cos\alpha =2\sin\alpha$ and hence $$\sin\alpha =\frac{\sqrt{5}}{5} , \cos\alpha =\frac{2\sqrt{5}}{5}.$$
Now $$\sin\beta =\sin (\frac{\pi }{4} -\alpha ) =\frac{\sqrt{2}}{2}\cdot \frac{2\sqrt{5}}{5} -\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{5}}{5}=\frac{\sqrt{10}}{10},\cos\beta = \frac{3\sqrt{10}}{10}$$
Now it is easy to observe that $\angle BDE =\beta$ and hence
$$\frac{|ON|}{|OD|} =\tan\beta =\frac{1}{3}$$
therefore $$|ON| =4.$$
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"url": "https://math.stackexchange.com/questions/4286822",
"timestamp": "2023-03-29T00:00:00",
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|
Find $f: \mathbb R \to \mathbb R$ which satisfies $f\left(xf(y)-y^2\right)=(y+1)f(x-y)$
Find $f: \mathbb R \to \mathbb R$ which satisfies $f\left(xf(y)-y^2\right)=(y+1)f(x-y)$.
My attempt:
\begin{align}
&P(x, -1): f\bigl(xf(-1)-1\bigr)=0. \\
&\text{If } f(-1) \ne 0 \implies x \leftarrow \frac {x}{f(-1)}: f(x-1)=0 \Rightarrow f \equiv 0, \text{ which is contradiction.} \\
& \implies f(-1)=0. \\
\\
&P(0, y): f\left(-y^2\right) = (y+1)f(-y). \\
\ \\
&P(x, 0): f\bigl(xf(0)\bigr)=f(x). \\
&\text{If } f(0)=1 \implies P\left(\frac {y^2}{f(y)}, y\right): 1=f(0)=(y+1)f\left(\frac{y\bigl(y-f(y)\bigr)}{f(y)}\right) \\
&f\left(\frac{y\bigl(y-f(y)\bigr)}{f(y)}\right)=\frac {1}{y+1}.
\end{align}
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For any $y\ne -1$, choose $x$ such that $x f(y) -y^2=-1$. This is possible because, when $f$ is not constant, its only zero is $-1$. Then,
$$
(y+1) f\left(\frac{y^2-1}{f(y)}-y\right) = 0\Leftrightarrow \frac{y^2-1}{f(y)}-y = -1,
$$
This means that any non-constant solution satisfies $f(y)=y+1$ for any $y\ne -1$. Finally, since this candidate solution also satisfies $f(-1)=0$, it is in fact the only non-constant solution to this equation.
To conclude, this equation has two solutions: $f(x) \equiv 0$ and $f(x) = x+1$.
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|
Dealing more efficiently with fractional forms in system of equations As an example, suppose we have to solve the following system of two equations and two unknowns:
$$
\begin{cases}
-\frac{10}{x}-\frac{8}{y} &= \frac{8}{3} \\
-\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}
\end{cases}
$$
My approach and solution
I opted to solve it by combination, referring to the first equation as (1) and the second equation as (2), I started by eliminating the y's:
\begin{align}
\frac{6}{8}(1)+(2): -\frac{60}{8x}-\frac{6}{x} &= \frac{6}{3}-\frac{1}{3} \\
\frac{-60-48}{8x} &= \frac{5}{3} \tag{*}\\
40x &=3(-108) \tag{**} \\
x &=-\frac{81}{10} \tag{1'}
\end{align}
Substituting (1') into (2):
\begin{align}
\frac{100}{81}-\frac{8}{y}&=\frac{8}{3} \\
y &= -\frac{162}{29}
\end{align}
So I find the tuple of $(-\frac{81}{10};-\frac{162}{29})$ as solution.
Questions
*
*For such systems, should we not be concerned about the conditions of existence of the system? Namely, in this case both $x\neq 0$ and $y\neq 0,$ in the same way that we would do when solving an equation. If yes, how do we formally write the domain of existence for a system?
*Is my transition from step $(*)$ to $(**)$ allowed? My understanding is that, yes for all $x\neq 0.$
*I am really eager to learn whether there are simple ideas that simplify the system (and other similar systems) before we start solving it. As shown in my approach, it was slightly awkward dealing with the fractions throughout and the "large" numbers, which inherently may render the approach more prone to mistakes. Any alternative, quicker approach (I only know by combination and substitution) would be much appreciated.
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As requested in a comment, here are the details of what I suggested in a comment. I've arranged the work to minimize the arithmetic details. In fact, I not only didn't have to resort to a calculator, but I did not have to resort to any "paper and pencil" multiplication. For example, $81 \cdot 2$ comes up at one point, but this can be done by sight: twice $80 + 1$ is $160 + 2.$
$$\begin{array}{r} -\frac{10}{x} \; - \; \frac{8}{y} & = & \frac{8}{3}
\\ -\frac{6}{x} \; + \; \frac{6}{y} & = & -\frac{1}{3} \end{array}$$
Changing variables by letting $a = \frac{1}{x}$ and $b = \frac{1}{y}$ converts this into two linear equations in two unknowns.
$$\begin{array}{r} -10a \; - \; 8b & = & \frac{8}{3}
\\ -6a \; + \; 6b & = & -\frac{1}{3} \end{array}$$
Now clear fractions by multiplying both sides of each equation by $3.$
$$\begin{array}{r} -30a \; - \; 24b & = & 8
\\ -18a \; + \; 18b & = & -1 \end{array}$$
Divide both sides of the first equation by $2.$
$$\begin{array}{r} -15a \; - \; 12b & = & 4
\\ -18a \; + \; 18b & = & -1 \end{array}$$
Ignoring signs, the coefficients of $a$ are $15 = 3 \cdot 5$ and $18 = 3 \cdot 6.$
Ignoring signs, the coefficients of $b$ are $12 = 6 \cdot 2$ and $18 = 6 \cdot 3.$
From this we can see that it will be simpler, when using the addition and subtraction method, to eliminate $b$ (multiply equations by $2$ and $3)$ than to eliminate $a$ (multiply equations by $5$ and $6).$
In the most recent pair of displayed equations, multiply both sides of the upper equation by $3$ and multiply both sides of the lower equation by $2.$
$$\begin{array}{r} -45a \; - \; 36b & = & 12
\\ -36a \; + \; 36b & = & -2 \end{array}$$
Adding the two equations above gives
$$ -81a \; = \; 10 \;\;\; \implies \;\;\; a = -\frac{10}{81} $$
Plugging this value of $a$ into what appears to be the simplest to-work-with equation above gives
$$ -18\left(-\frac{10}{81}\right) \; + \; 18b \; = \; -1 $$
$$ -2\left(-\frac{10}{9}\right) \; + \; 18b \; = \; -1 $$
Now clear fractions by multiplying both sides by $9,$ and continue in some standard way until $b$ is isolated.
$$ -2(-10) \; + \; 9(18b) \; = \; -9 $$
$$ 20 \; + \; (9 \cdot 9)(2b) \; = \; -9 $$
$$ 20 \; + \; 81(2b) \; = \; -9 $$
$$ 20 \; + \; 162b \; = \; -9 $$
$$ 162b \; = \; -9 - 20 \; = \; -29 $$
$$ b \; = \; -\frac{29}{162} $$
Therefore, we have $a = -\frac{10}{81} = \frac{1}{x}$ and $b = -\frac{29}{162} = \frac{1}{y},$ which implies $x = -\frac{81}{10}$ and $y = -\frac{162}{29}.$
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|
Prove that $f(x)^2+f(x+1)^2 = f(2x+1)$ for the function $f$ satisfying $f(x)=f(x-1)+f(x-2), f(1)=f(2)=1.$
Prove that $f(x)^2+f(x+1)^2 = f(2x+1)$ for the function $f$ satisfying $f(x)=f(x-1)+f(x-2), f(1)=f(2)=1.$
I know how to prove it, but it is interesting so I am posting it.
The first hint is:
\begin{align} &\text{Try to prove this first: } \\ &f(x)=f(x-1)+f(x-2) \\ &=2f(x-2)+f(x-3) \\ &=3f(x-3)+2f(x-4) \\ &= 5f(x-4)+3f(x-3) \\ &= \cdot \cdot \cdot \end{align}
The second hint is:
$\text{According to the first hint: } f(x)=f(k+1)f(x-k)+f(k)f(x-k-1). $
Check this to see that your proof is the same as mine.
\begin{align} &f(x)=f(x-1)+f(x-2) = f(2)f(x-1)+f(1)f(x-2). \ \\ \ \\ &f(x-1)=f(x-2)+f(x-3) \\ &\Rightarrow f(x)=f(2)(f(x-2)+f(x-3))+f(1)f(x-2)=(f(1)+f(2))f(x-2) \\ &+f(2)f(x-3)=f(3)f(x-2)+f(2)f(x-3). \ \\ \ \\ &f(x-2)=f(x-3)+f(x-4). \\ &\Rightarrow f(x)=f(3)(f(x-3)+f(x-4))+f(2)f(x-3)=(f(2)+f(3))f(x-3) \\ &+f(3)f(x-4)=f(4)f(x-3)+f(3)f(x-4). \\ &\cdot \\ &\cdot \\ &\cdot \\ &\therefore f(x)=f(k+1)f(x-k)+f(k)f(x-k-1). \\ &x=2k+1; \ f(2k+1)=f(k)^2+f(k+1)^2. \\ \ \\ &\therefore f(x)^2+f(x+1)^2=f(2x+1).\end{align}
p.s. if you have another solution, please post it as an answer with a spoiler. (>! another answer with no enter)
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My answer above:
\begin{align} &f(x)=f(x-1)+f(x-2) = f(2)f(x-1)+f(1)f(x-2). \ \\ \ \\ &f(x-1)=f(x-2)+f(x-3) \\ &\Rightarrow f(x)=f(2)(f(x-2)+f(x-3))+f(1)f(x-2)=(f(1)+f(2))f(x-2) \\ &+f(2)f(x-3)=f(3)f(x-2)+f(2)f(x-3). \ \\ \ \\ &f(x-2)=f(x-3)+f(x-4). \\ &\Rightarrow f(x)=f(3)(f(x-3)+f(x-4))+f(2)f(x-3)=(f(2)+f(3))f(x-3) \\ &+f(3)f(x-4)=f(4)f(x-3)+f(3)f(x-4). \\ &\cdot \\ &\cdot \\ &\cdot \\ &\therefore f(x)=f(k+1)f(x-k)+f(k)f(x-k-1). \\ &x=2k+1; \ f(2k+1)=f(k)^2+f(k+1)^2. \\ \ \\ &\therefore f(x)^2+f(x+1)^2=f(2x+1).\end{align}
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General form for $\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$ I'm wondering if there is a general form for the following sum:
$$\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$$ for $m \in \mathbb{N}$
I have obtained the following closed-forms for these special cases:
Where $G$ is Catalan's constant and $\text{Cl}_2$ is the Clausen function of order 2.
$$\sum_{n=1}^{\infty}(-1)^n \left(2n \, \text{arccoth} \, (2n) - 1\right) = \frac{1}{2} - \frac{2G}{\pi}$$
$$\sum_{n=1}^{\infty} (-1)^{n}\left( 3n \, \text{arccoth} \, (3n)-1\right) = \frac{1}{2} - \frac{5}{2\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{3}\right) + \frac{1}{4} \ln (3)$$
$$\sum_{n=1}^{\infty} (-1)^n \left(4n \, \text{arccoth} \, (4n) - 1\right) = \frac{1}{2}+ \frac{G}{\pi} - \frac{4}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln \left(3- 2\sqrt{2}\right)$$
etc.
Given that $G = \text{Cl}_2 \left(\frac{\pi}{2}\right)$, I am curious to know if the general sum is expressible in terms of the Clausen function. These above sums were determined by using the Mittag-Leffler expansion of $\csc (z)$, i.e $\csc(z) = \frac{1}{z} + 2z \sum_{n=1}^{\infty} (-1)^n \frac{1}{z^2 - \left(\pi n\right)^2}$ and substituting it into the integral $\int_{0}^{\pi/m} x \csc (x) \, dx$
If one uses the following other method, we can determine the odd and even terms of the sums
$$\sum_{n=1}^{\infty} \left( 4n \, \text{arccoth} \, (4n)-1\right) = \frac{1}{2} - \frac{G}{\pi}- \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left( (4n-2) \, \text{arccoth} \, (4n-2) - 1\right) = \frac{G}{\pi} - \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left(6n \, \text{arccoth} \, (6n) - 1\right) = \frac{1}{2} - \frac{3}{2\pi} \, \text{Cl}_2 \left( \frac{\pi}{3}\right)$$
$$\sum_{n=1}^{\infty} \left( (6n-3) \, \text{arccoth} \, (6n-3) - 1\right) = \frac{1}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{3}\right) - \frac{1}{4} \ln(3)$$
$$\sum_{n=1}^{\infty} \left( 8n \, \text{arccoth} \, (8n) - 1\right) = \frac{1}{2} - \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln (2-\sqrt{2})$$
$$\sum_{n=1}^{\infty} \left( (8n-4) \, \text{arccoth} \, (8n-4) - 1\right) = \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{G}{\pi} - \frac{1}{4} \ln(2+\sqrt{2})$$
EDIT
I am now interested in
$$\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$$
for $|m|>1$ as asked here.
Here is how I originally proved it for the odd terms:
For the first odd term sum, begin with the known result $$2G = \int_{0}^{\frac{\pi}{2}} x \csc (x) \, dx$$
then integrate by parts to get:
$$2G = \int_{0}^{\frac{\pi}{2}} \ln \left(\cot (x) + \csc (x) \right) \, dx = \int_{0}^{\frac{\pi}{2}} \ln (\cos (x) + 1) \, dx - \int_{0}^{\frac{\pi}{2}} \ln (\sin (x)) \,dx$$
Then use the well-known result $\int_{0}^{\frac{\pi}{2}} \ln (\sin(x)) \,dx = - \frac{\pi}{2} \ln (2)$ and use the identity $\cos (x) + 1 = 2 \cos^2\left( \frac{x}{2}\right)$ and make the substitution $\frac{x}{2} = u$.
$$\implies 2G - \pi \ln (2) = 4 \int_{0}^{\frac{\pi}{4}} \ln (\cos (u)) \, du$$
Now use the Weierstrass product for $\cos (z)$, namely $\cos(z) = \prod_{n=1}^{\infty} \left(1-\frac{4z^2}{\pi^2 (2n-1)^2}\right)$ to obtain:
$$2G - \pi \ln (2) = 4 \sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{4}} \ln \left( 1-\frac{4u^2}{\pi^2 (2n-1)^2}\right) \, du$$
After integrating, obtain $\pi \sum_{n=1}^{\infty} \ln \left(1-\frac{1}{4(1-2n)^2}\right) = -\frac{\pi}{2} \ln (2)$ and the result quickly follows. The other odd sums are the same idea. The even sums just comes from combining the two results from the alternating sum and the odd term sum.
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Assume that $m >1$.
Similar to my answer for the non-alternating version of the series, we can exploit the fact that $\csc(\pi z)$ has the Laurent series expansion $$\csc(\pi z) = \frac{1}{\pi z} + \frac{2}{\pi z} \sum_{n=1}^{\infty} \eta(2n)z^{2n}, \quad 0 <|z| < 1, $$ where $\eta(z)$ is the Dirichlet eta function.
Then again using the the Laurent series expansion of $\operatorname{arcoth}(z)$ at $\infty$, we get $$ \begin{align} S&= \sum_{n=1}^{\infty} (-1)^{n} \left(mn \operatorname{arcoth}(mn)-1 \right) \\ &= \sum_{n=1}^{\infty} (-1)^{n} \left(mn \sum_{k=0}^{\infty} \frac{1}{(2k+1)(mn)^{2k+1}} -1\right) \\ &= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{ (-1)^{n}}{(2k+1)(mn)^{2k}} \\ &= \sum_{k=1}^{\infty} \frac{1}{(2k+1)m^{2k}}\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2k}} \\ &= -\sum_{k=1}^{\infty} \frac{\eta(2k)}{(2k+1)m^{2k}} \\ &= \frac{m}{2} \int_{0}^{1/m} \mathrm dx - \frac{m \pi}{2} \int_{0}^{1/m} x\csc(\pi x) \, \mathrm dx \\ &= \frac{1}{2} + \frac{mx}{2} \log \left(\cot \frac{\pi x}{2} \right)\Bigg|_{0}^{1/m} - \frac{m}{2} \int_{0}^{1/m} \log \left(\cot \frac{\pi x}{2} \right) \, \mathrm dx \\ &= \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right)- \frac{m}{2} \int_{0}^{1/m} \log \left(\cos \frac{\pi x}{2} \right) \, \mathrm dx + \frac{m}{2} \int_{0}^{1/m} \log \left(\sin \frac{\pi x}{2} \right) \, \mathrm dx \\ &= \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right) - \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(\cos \frac{u}{2} \right) \, \mathrm du + \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(\sin \frac{u}{2} \right) \, \mathrm du \\ &= \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right)- \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(2 \cos \frac{u}{2} \right) \, \mathrm du + \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(2 \sin \frac{u}{2} \right) \, \mathrm du \\ &=\frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right) - \frac{m}{2 \pi} \int_{0}^{\pi/m} \log \left(2 \cos \frac{u}{2} \right) \, \mathrm du - \frac{m}{2 \pi} \operatorname{Cl}_{2} \left(\frac{\pi}{m} \right). \end{align}$$
The substitution $w= \pi -u$ shows that $$-\int_{0}^{\pi/m} \log \left(2 \cos \frac{u}{2} \right) \, \mathrm du = - \operatorname{Cl}_{2} \left(\pi - \frac{\pi}{m} \right)=\operatorname{Cl}_{2} \left(\frac{\pi}{m}- \pi \right). $$
Therefore, $$\sum_{n=1}^{\infty} (-1)^{n} \left(mn \operatorname{arcoth}(mn)-1 \right) = \frac{1}{2} + \frac{1}{2} \log \left(\cot \frac{\pi}{2m} \right) + \frac{m}{2 \pi} \left(\operatorname{Cl}_{2} \left(\frac{\pi}{m}- \pi \right) - \operatorname{Cl}_{2} \left(\frac{\pi}{m} \right) \right), $$ which is the same result that skbmoore got.
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Proving $\sum_{n=1}^\infty \frac{1}{(n^2+a^2)^2}=\frac{\pi}{4a^3}\left[\coth(\pi a)+\frac{\pi a}{\sinh^2 \pi a}-\frac{2}{\pi a}\right]$ Use the function $$f(z)=\frac{\pi\text{cosec}(\pi z)}{z^2+a^2}$$
to show that
$$\sum_1^\infty \frac{1}{(n^2+a^2)^2}=\frac{\pi}{4a^3}\left[\coth(\pi a)+\frac{\pi a}{\sinh^2 \pi a}-\frac{2}{\pi a}\right]$$
I'm trying to use Mittag-Leffler decomposing
$$f(z)=f(0)+\sum_{j=1}^\infty \text{Res}_{z=z_j}\left(\frac{1}{z-z_j}+\frac{1}{z_j}\right)$$
Using this on our function, We have
$$\frac{\pi\text{cosec}(\pi z)}{z^2+a^2}=\frac{\pi \csc(\pi ai)}{2ai}\left(\frac{1}{z-ai}+\frac{1}{ai}\right)+\frac{\pi\text{cosec}(\pi z)}{z^2+a^2}+\frac{\pi \csc(\pi ai)}{2ai}\left(\frac{1}{z+ai}-\frac{1}{ai}\right)+\sum_{n=-\infty}^\infty \frac{\pi}{n^2+a^2}\left(\frac{1}{z-n}+\frac{1}{n}\right)$$
which can be simplified to
$$=\frac{\pi \csc(\pi ai)}{ai}\left(\frac{z}{z^2+a^2}\right)+\sum_{-\infty}^\infty \frac{\pi}{n^2+a^2}\left(\frac{z}{n(z-n)}\right)$$
I don't know how to proceed. Apart from this $f(z)$ has a singularity at origin so how to compute $f(0)$. Please help me with this.
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Without the hint.
$$\frac{1}{(n^2+a^2)^2}=\frac{i}{4 a^3 (n+i a)}-\frac{i}{4 a^3 (n-i a)}-\frac{1}{4 a^2 (n+i
a)^2}-\frac{1}{4 a^2 (n-i a)^2}$$ Computing the partial sums
$$\sum_{n=1}^p\frac{i}{4 a^3 (n+i a)}=-\frac{i (\psi ^{(0)}(i a+1)-\psi ^{(0)}(i a+p+1))}{4 a^3}$$
$$\sum_{n=1}^p\frac{i}{4 a^3 (n-i a)}=-\frac{i (\psi ^{(0)}(1-i a)-\psi ^{(0)}(-i a+p+1))}{4 a^3}$$
$$\sum_{n=1}^p \frac{1}{4 a^2 (n+i a)^2}=\frac{\psi ^{(1)}(i a+1)-\psi ^{(1)}(i a+p+1)}{4 a^2}$$
$$\sum_{n=1}^p \frac{1}{4 a^2 (n-i a)^2}=\frac{\psi ^{(1)}(1-i a)-\psi ^{(1)}(-i a+p+1)}{4 a^2}$$
Now, using asymptotics,
$$\sum_{n=1}^p \frac{1}{(n^2+a^2)^2}=\frac{\pi a \left(\coth (\pi a)+\pi a \text{csch}^2(\pi a)\right)-2}{4
a^4}-\frac{1}{3 p^3}+\frac{1}{2 p^4}+O\left(\frac{1}{p^5}\right)$$
|
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What is my mistake in this counting problem?
Question: How many 6-digit numbers, written in decimal notation, have at least one 1, one 2, and one 3 among their digits?
Define $A, B, C$ to be the set of $6$-digit numbers with at least one $1$, at least one $2$, and at least one $3$, respectively.
We may count the cardinality of $A$ with two cases: either $A$ has first digit $1$ or it is not a first digit. If the first digit is $1$, then there are $1$ choices for the first digit ($1$) and $10^5$ choices for the remaining digits. If the first digit of $A$ is not $1$, then there are $8$ choices ($\{2,3,\dots,9 \}$) choices for the first digit, $5$ choices for where $1$ is, and $10^4$ choices for the remaining digits. In total, there are $8\cdot5\cdot10^4 + 10^5$ digits with at least one $1$. By symmetry, we see that $|A| = |B| = |C|$.
We shall now count $A\cap B$ in a similar manner. Either the first digit is a $1$ or a $2$, or it isn't. There are $2$ choices for the first digit ($1$ or $2$), and $5$ choices for the other digit which wasn't first. Then there are $10^4$ choices for the remaining digits. If the first digit is not a $1$ or a $2$, then there are $7$ choices for the first digit ($\{3,4,\dots,9\}$) and $5\cdot4$ choices for the $1$ and $2$, and $10^3$ choices for the remaining digits. By symmetry, we have that $|A \cap B| = |A \cap C| = |B \cap C| = 2\cdot 5 \cdot 10^4 + 7\cdot 5 \cdot 4 \cdot 10^3$.
Finally, we count $A \cap B \cap C$ using the same method. Either the first digit is a $1, 2$ or $3$ or it is not. If it is, then we have $3$ choices for which of the three it is, then $5\cdot4$ choices for where to place the other two. Then we have $10^3$ ways to fill the remaining digits. If the first digit is not a $1,2$ or $3$, then we have $6$ choices for the first digit, and $5\cdot4\cdot3$ ways to pick places for the $1,2,3$, and $10^2$ ways to fill the remaining digits.
By the Principle of Inclusion-Exclusion, we have that $$ |A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C| + |B \cap C|) + |A \cap B \cap C|, $$ so we have that $$ 3(8\cdot5 \cdot 10^4 + 10^5) - 3(2 \cdot 5 \cdot 10^4 + 7\cdot 5 \cdot 4 \cdot 10^3) + 3 \cdot 5 \cdot 4 \cdot 10^3 + 6 \cdot 5 \cdot 4 \cdot 3 \cdot 10^2, $$ which is much larger than the supposed answer that you would get using complementary counting.
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We have to find $|A \cap B \cap C|$.
Note that the complement of $A$ is the set of $6$-digit numbers which have no digit $1$, therefore $|A^c|=8\cdot 9^5$. Hence in your attempt you should have $|A|=9\cdot 10^5-8\cdot 9^5$ which less than $8\cdot5\cdot10^4 + 10^5$ because you are overcounting! For instance the number $211111$ is counted several times in $8\cdot5\cdot10^4$.
Similarly, $|B^c|=|C^c|=8\cdot 9^5$, $|A^c \cap B^c|=|B^c \cap C^c|=|A^c \cap C^c|=7\cdot 8^5$ and $|A^c \cap B^c \cap C^c|=6\cdot 7^5$.
Finally, by the Principle of Inclusion-Exclusion, we have
$$\begin{align*}|A \cap B \cap C|&=9\cdot 10^5-|A^c \cup B^c \cup C^c|\\
&=9\cdot 10^5-|A^c| - |B^c| - |C^c|+(|A^c \cap B^c| + |A^c \cap C^c| + |B^c \cap C^c|)\\
&\qquad - |A^c \cap B^c \cap C^c|\\
&=9\cdot 10^5-3\cdot 8\cdot 9^5+3\cdot 7\cdot 8^5-6\cdot 7^5=70110.
\end{align*}$$
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Number of solutions of $\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right) =0$
Solve in the interval $[0,2\pi]$ : $$\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right)=0 $$
I tried expanding the L.H.S by applying the formula of $\cos(A+B)$ but it results in a quintic polynomial in terms of $\cos x $. Wolfram alpha has simplified the left hand side all the way to $\frac{15}{16}\cos3x$ but I'm unable to think of any method to reach there.
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Clearly the roots of $$4c^3-3c-\cos3x=0$$ are $c_r\cos(x+2r\pi/3), r=0,1,2$
Let $p=c_0,c_1=q,c_2=r$
$p+q+r=0\implies p^3+q^3+r^3=3pqr$
$pq+qr+rp=-3/4\implies p^2+q^2+r^2=0^2-2(-3/4)$
$pqr=\dfrac{\cos3x}4$
As $4c^5=3c^3+c^2\cos3x,$
$4(p^5+q^5+r^5)$
$=3(p^3+q^3+r^3)+(p^2+q^2+r^2)\cos3x$
Replace the values of $p^3+q^3+r^3, p^2+q^2+r^2$
|
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Find $a\in \mathbb{R}$ such that a root of $ax^3-13x^2+(15a)x-25$ is $2+i$ Find $a\in \mathbb{R}$ such that a root of the polynomial $$p(x)=ax^3-13x^2+(15a)x-25$$ is $2+i$
Solution:
$q(x)=\frac{p(x)}{a}$
$=x^3-\frac{13}{a}x^2+15x-\frac{25}{a}$
$p(x)$ and $q(x)$ have the same roots, call them $r_1, r_2, r_3$
$r_1=2+i$. Since the coefficients are all real we know $r_2=2-i$
$(x-r_1)(x-r_2)(x-r_3)= x^3-(r_1+r_2+r_3)x^2+(r_1r_2 + r_1r_3 +r_2r_3)x-r_1r_2r_3$
$r_1+r_2+r_3=\frac{13}{a}$
$(2+i)(2-i)+r_3=\frac{13}{a}$
$5+r_3=\frac{13}{a}$
We also have:
$r_1r_2r_3=\frac{25}{a}$
$(2+i)(2-i)r_3=\frac{25}{a}$
$5r_3=\frac{25}{a}$
$r_3=\frac{5}{a}$
Going back to $5+r_3=\frac{13}{a}$
We have $5+\frac{5}{a}=\frac{13}{a}$
Thus $a=\frac{8}{5}$
The statement I am having trouble with is
"$r_1=2+i$. Since the coefficients are all real we know $r_2=2-i$"
Why do we know what $r_2$ is? I understand it's the conjugate and that $r_1r_2=5$ is a real number. But I don't understand why the coefficients being real implies that $r_2=2-i$
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$(x-2)^2 = -1 $ or $x^2 - 4x + 5 = 0$ represents the pair of complex roots.
$$\begin {aligned}
p(x) &=ax^3-13x^2+(15a)x-25 \\
& = (ax^3 - 4ax^2 + 5ax) - ((13-4a) x^2 - 10ax + 25) \\
& = ax (x^2 - 4x + 5) - 5 \left(\frac{13-4a}{5} x^2 - 2ax + 5 \right)
\end{aligned}$$
That gives $a = 2$ as the solution.
|
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Inequalities when matrix is negative definite
Let $$A = \begin{pmatrix} x&1&y \\ 1&-1&0 \\ y&0&-2 \end{pmatrix} $$ $A$ is negative definite $\iff x<a \text{ and } bx^2 + cy^2 + dx + 2 < 0$. Find $a,b,c,d$
I know that for a matrix to be negative definite all of its eigenvalues must be negative (I thought this may link to the $<0$ of the equation from the question)
I have attempted to form an equation for the eigenvalues of A, by forming the characteristic polynomial as follows:
$$|A-\lambda*I_3|=
\begin{vmatrix} x-\lambda&1&y \\ 1&-1-\lambda&0 \\ y&0&-2-\lambda\end{vmatrix}$$
$$ =x-\lambda \begin{vmatrix} -1-\lambda&0 \\ 0&-2-\lambda \end{vmatrix} -1 \begin{vmatrix} 1&0 \\ y&-2-\lambda \end{vmatrix}+y\begin{vmatrix} 1&-1-\lambda \\ y&0 \end{vmatrix}$$
$$=\lambda^2x+3\lambda x+2x-\lambda^3-3\lambda^2-2\lambda+2+\lambda-y^2-\lambda y^2$$
I am then unsure what to do with this to find the answer, I tried factoring out $x$ and $y$ but that did not seem to help?
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$A$ is negative definite iff $(-A)$ is positive definite. To test the positive definiteness of $(-A)$, we have to calculate all three principal determinants of $(-A)$
$-A = \begin{bmatrix} -x && - 1 && - y \\ -1 && 1 && 0 \\ - y && 0 && 2 \end{bmatrix}$
$\Delta_1 = - x \gt 0 \Longleftrightarrow x \lt 0$
$\Delta_2 = -x - 1 \gt 0 \Longleftrightarrow x \lt -1 $
$\Delta_3 = 2 (-x - 1) - y (y) = -2 x - y^2 - 2 \gt 0 \Longleftrightarrow y^2 + 2 x + 2 \lt 0 $
Hence, $a = -1, b = 0, c = 1, d = 2 $
|
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|
$\int_c 3y dx +5x dy + \frac{2x+3}{z^2} dz$ for the intersection between two surfaces For the closed curve, $c = \{(x,y,z) \vert x^2 + y^2 - z^2 =0\} \cap \{(x,y,z) \vert (x-1)^2 + y^2 =4\}$
Find the $\int_c 3y dx +5x dy + \frac{2x+3}{x^2 + y^2} dz$
First I focused the surface $\{(x,y,z) \vert x^2 + y^2 - z^2 =0\}$ and tried to simplify like the $\int_c 3y dx +5x dy + \frac{2x+3}{z^2} dz$.
Next step is parameterizing the boundary surface $X(r,\theta)= (1+rcos\theta, rsin\theta, \pm\sqrt{1+r^2 + 2rcos\theta})$ for $0\leq r \leq 2, 0\leq \theta \leq 2\pi$
But the problem happened there are two curves. To applying the Stoke's thm, Which curve do I choose? Plus Is there are another methods without stokes thm? It is too complicated to solve by the Stokes thm. Help me.
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$x^2 + y^2 - z^2 = 0$ represents a cone with axis along the $z$ axis, and $(x-1)^2 + y^2 = 4 $ is a right circular cylinder with axis parallel to the $z$ axis and passing through $(1, 0)$ and a radius of $2$.
Let $x = 1 + 2 \cos \theta , y = 2 \sin \theta $, then this point lies on the cylinder, the intersection with the cone results in
$z^2 = x^2 + y^2 = 5 + 4 \cos \theta $
Take $z$ to be positive, then $z = \sqrt{ 5 + 4 \cos \theta } $
Differentials:
$dx = -2 \sin \theta d\theta $
$dy = 2 \cos \theta d\theta $
$dz = \dfrac{- 2 \sin \theta d \theta }{ \sqrt{5 + 4 \cos \theta }} $
$\displaystyle \int_C 3 y dx + 5 x dy + \dfrac{2x+3}{x^2 + y^2} dz $
Becomes
$\displaystyle \int_0^{2 \pi} (-6 \sin^2 \theta + 10(1 + 2 \cos \theta) \cos \theta + \dfrac{ 2(1 + 2 \cos \theta) + 3 }{5 + 4 \cos \theta } \dfrac{- 2 \sin \theta}{ \sqrt{5 + 4 \cos \theta }}) d\theta $
which simplifies to
$\displaystyle \int_0^{2 \pi} (-12 \sin^2 \theta + 20 \cos^2 \theta + 10 \cos \theta + \dfrac{- 2 \sin \theta}{ \sqrt{5 + 4 \cos \theta }}) d\theta = \dfrac{1}{2} (-12 + 20)(2 \pi) = \boxed{8 \pi}$
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Determining all $(a,b)$ on the unit circle such that $2x+3y+1\le a(x+2)+b(y+3)$ for all $(x,y)$ in the unit disk In the middle of another problem, I came up with the following inequality which needed to be solve for $(a, b)$ : $$2x+3y+1\le a(x+2)+b(y+3)$$ for all $(x, y)\in\mathbb{R}^2$ with $x^2+y^2\le1.$
Here the solution for $(a, b)$ must be a subset of the unit circle, and I believe that it is a singleton.
Since I have no clue to solve it in this form, I tried to substitute some points of the closed unit disk and make a system of inequalities. But it seems like there should be a general way of solving these type of problems.
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Since $(a,b)$ lies on the unit circle and $(a,b)$ satisfies the linear inequality, it must be tangent to the circle. So first find the tangent line to the unit circle at $(\cos t, \sin t)$. This is:
$$y - \sin t = (-\cot t)(x - \cos t) \implies y = -(\cot t)x + \csc t.$$
Now the equality case of the condition $2x+3y+1 = a(x+2)+b(y+3)$ rearranges to $(3-b)y = (a-2)x$ $ + (2a + 3b - 1)$. Using the identity $\csc^2 t - \cot^2 t = 1$ and comparing coefficients:
$$\frac{1}{(3-b)^2} \left((2a + 3b - 1)^2 - (a-2)^2 \right) = 1$$
$$(4a^2 + 12ab + 9b^2 - 4a - 6b + 1) - (a^2 - 4a + 4) = 9 - 6b + b^2$$
$$3a^2 + 12ab + 8b^2 - 12 = 0$$
and now substitute $12 = 12(a^2+b^2)$:
$$-9a^2 + 12ab - 4b^2 = 0$$
$$-(3a - 2b)^2 = 0 \implies b = 3a/2$$
thus $(3b/2)^2 + b^2 = 1 \implies a = \frac{2}{\sqrt{13}}, b = \frac{3}{\sqrt{13}}$. We discard the negative solutions as substituting $(x,y) = (0,0)$ into the condition we need $1 ≤ 2a + 3b$.
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Integrate: $\frac{x^2+x-1}{x^2-1}$ with respect to $x$ Ok so here is where I am up to:
I can find some similarity with the numerator and denominator and managed to reduce to the following:
$\int\frac{x^2+x-1}{x^2-1}dx = \int\frac{x^2-1}{x^2-1}+\frac{x}{x^2-1}dx$
$=\int1+\frac{x}{x^2-1}dx.$
I tried to reduce it further by factoring out an $(x-1)$ but did not get anywhere after that.
Is there any other way to transpose this into a simpler format so I can integrate it?
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Yeah!
My first mathSE integration problem answer, in I don't remember how long.
$\displaystyle \frac{1}{x^2-1} = \frac{1}{2}
\times \left[\frac{1}{x-1} - \frac{1}{x+1}\right].$
$\displaystyle \frac{x}{x^2-1} = \frac{1}{2}
\times \left[\frac{x}{x-1} - \frac{x}{x+1}\right].$
This equals $\displaystyle
\frac{1}{2} \left\{ \left[1 + \frac{1}{x-1}\right]
- \left[1 - \frac{1}{x+1}\right] \right\}.$
This equals $\displaystyle
\frac{1}{2} \left[\frac{1}{x-1} + \frac{1}{x+1}\right].$
Therefore $$\int \frac{x}{x^2 - 1}~dx = \frac{1}{2}
\left[ \log(x-1) + \log(x+1) \right].$$
Edit
Sad to say...
The shortcut is that
$\displaystyle \frac{d}{dx}\log(x^2 - 1) = \frac{2x}{x^2 - 1}.$
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convert hyperbola in rectangular form to polar form i am trying to convert the rectangular equation of a conic (hyperbola) to a polar form.
the rectangular equation is:
$$3y^2 - 16y -x^2 + 16 = 0.$$
i substituted $r\sin\theta$ for y and $r\cos\theta$ for x, and tried to simplify, but I am stuck.
i tried substituting $1-\sin^2\theta$ for the resulting $\cos^2\theta$ term, and ended up with the following expression after some manipulation:
$$r^2(2\sin\theta-1)(2\sin\theta+1)-16r\sin\theta + 16 = 0.$$
the solution to the question is supposed to be:
$$r = \frac{4}{1 + 2\sin\theta}.$$
I just can't figure out how to get from where I am to the expected solution. Perhaps I made the wrong substitution? I just need to figure out the intermediate steps to get from the original rectangular equation to the final equation in polar form given above.
Hope someone can help me -- thanks!
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Staring with $3 y^2 - 16 y - x^2 + 16 = 0 $
Put it first in the standard format. Complete the square in $y$
$ 3 \left(y - \dfrac{8}{3} \right)^2 - \dfrac{64}{3} - x^2 + 16 = 0 $
$ 3 \left(y - \dfrac{8}{3} \right)^2 - x^2 = \dfrac{16}{3} $
$ \dfrac{9}{16} \left( y - \dfrac{8}{3} \right)^2 - \dfrac{3}{16} x^2 = 1 $
Thus the center is at $\left(0, \dfrac{8}{3} \right) $ , $ a = \dfrac{4}{3} $ and $b = \dfrac{4}{\sqrt{3}} $
The focal distance $ c = \sqrt{a^2 + b^2} = 4 \sqrt{ \dfrac{1}{9} + \dfrac{1}{3} } = \dfrac{8}{3} $
Hence, the foci are at $ (0, 0) $ and $ \left(0, \dfrac{16}{3} \right) $
Taking the focus that is at the origin, then $ x = r \cos \theta, y = r \sin \theta $
Back to the original equation
$3 y^2 - 16 y - x^2 + 16 = 0 $
Substituting the polar expressions,
$ 3 (r \sin \theta)^2 - 16 r \sin \theta - r^2 \cos^2 \theta + 16 = 0 $
Using $\cos^2 \theta = 1 - \sin^2 \theta$ and collecting terms
$ r^2 ( 4 \sin^2 \theta - 1 ) - 16 r \sin \theta + 16 = 0$
From the quadratic formula, and taking the positive root
$ r = \dfrac{1}{ 2(4 \sin^2 \theta - 1 ) } \left ( 16 \sin \theta - \sqrt{ 256 \sin^2 \theta - (256 \sin^2 \theta - 64 ) } \right) $
And this simplifies to
$ r = \dfrac{1}{2 (4 \sin^2 \theta - 1) } \left( 16 \sin \theta - 8 \right) $
Factoring the denominator and cancelling equal terms
$ r = \dfrac{ 4 }{ 2 \sin \theta + 1 } $
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Find the number of ways in which the number 30 can be partitioned into three unequal parts.(Please use multinomial theorem) OfLet $a,b,c$ be the parts such that $a<b<c$
Now, let $a−b=x, b−c=y,$ implies $x,y>0$
$⇒a+b+c=30\\
⇒(b+x)+b+c=30\\
⇒x+2(c+y)+c=30\\
⇒x+2y+3c=30$ ,
$c≤27$
Sum =30,Co-eff =$1,2,3$
$(x^1+x^2+.......)×(x^2+x^4+.......)×(x^3+x^6+.......+x^{27})$
Co-efficient of $x^{30}$ in the above product will give us the required answer.=$x×x^2×x^3×(1−x)^{−1}×(1−x^2)^{−1}×(1−x^3)^{−1}\\
=x^6\times (1−x)^{−1}\times(1−x^2)^{−1}\times(1−x^3)^{−1}$
Please help me to understand the last ,how to find the coefficient of p^24 in those 3 brackes
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You are pretty much on the right track. The last part is pretty tedious, so it might be better that you just use Wolfram Alpha or possibly some other program. I think you defined your variables a little wrong. For example, if $a<b$, then $a-b<0$. However, you defined $x=a-b$, but you are saying that $x>0$. It looks like you're fine if you instead force $c<b<a$.
As you found before, we want to find the coefficient of $x^{24}$ in
$$\frac{1}{(1-x)(1-x^2)(1-x^3)}$$
Our goal will be to convert this into something of the form
$$\frac{P(x)}{(1-x^a)^b}$$
We can then use the fact that
$$\frac{1}{(1-x^a)^b}=\sum_{n=0}^\infty \binom{n+b-1}{b-1}x^{na}$$
to find the coefficient of $x^{24}$ in the resulting expansion. The most tedious part of this process is determining $P(x)$. We will start with some manipulations of
$$\frac{1}{(1-x)(1-x^2)(1-x^3)}$$
Note that $1-x, 1-x^2,$ and $1-x^3$ are all factors of $1-x^6$ (I think this would be called the LCM). Using this, we get
$$=\frac{(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^4)(1+x^3)}{(1-x)(1+x+x^2+x^3+x^4+x^5)(1-x^2)(1+x^2+x^4)(1-x^3)(1+x^3)}$$
$$=\frac{(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^4)(1+x^3)}{(1-x^6)^3}$$
$$=(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^4)(1+x^3)\sum_{n=0}^\infty \binom{n+2}{2}x^{6n}$$
Well that was simple. However, the hard part is yet to come. We need to find the coefficient of $x^{24}$ in that expansion. Denote the product of the leftmost $3$ polynomials (the ones not in the summation) as $P(x)$. We will be finding the coefficient $x^{24}$ in the expansion of
$$P(x)\sum_{n=0}^\infty \binom{n+2}{2}x^{6n}$$
Since $\sum_{n=0}^\infty \binom{n+2}{2}x^{6n}$ only has terms of powers $0\mod 6$, we only care about the terms with power $0\mod 6$ in the expansion of $P(x)$. Although this doesn't help us too much, it does reduce some of the calculations. Expanding $P(x)$ while only caring about the final terms whose powers are $0\mod 6$ gives
$$P(x)$$
$$=(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^4)(1+x^3)$$
$$=(1+x+x^2+x^3+x^4+x^5)(1+x^2+x^3+x^4+x^5+x^7)$$
$$\sim 1+x^6+x^6+x^6+x^6+x^{12}$$
$$=1+4x^6+x^{12}$$
So we only need to find the coefficient of $x^{24}$ in
$$(1+4x^6+x^{12})\sum_{n=0}^\infty \binom{n+2}{2}x^{6n}$$
$$\binom{4+2}{2}+4\cdot \binom{3+2}{2}+\binom{2+2}{2}$$
$$=15+40+6$$
$$=61$$
Indeed this matches what Wolfram Alpha gives (just click more terms on the taylor series expansion until you get to $x^{24}$)
|
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|
Number of real solutions of $\begin{array}{r} {\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} =\frac{3 x-1}{2} \end{array}$ Solve for $x \in \mathbb{R}$ $$\begin{array}{r}
{\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]}
=\frac{3 x-1}{2}
\end{array}$$ where $[x]$ denotes greatest integer less than or equal to $x$.
My try:
Letting $a=\frac{2x+1}{3}$ we get
$$\left[a\right]+\left[a+\frac{1}{2}\right]=\frac{9 a-5}{4} \tag{1}$$
Now knowing that:
$$a-1<[a]\leq a$$ and $$a-\frac{1}{2}<\left[a+\frac{1}{2}\right]\leq a+\frac{1}{2}$$
Adding both the above inequalities and using $(1)$ we get
$$2a-\frac{3}{2}<\frac{9a-5}{4}\leq 2a+\frac{1}{2}$$
So we get $$a \in (-1, 7]$$
Any help here?
|
Hint :
$$ \lfloor a \rfloor + \lfloor a + \frac{1}{2} \rfloor = \lfloor 2a \rfloor $$
This follows from Hermite's identity.
|
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|
Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3} + \sqrt[3]{9}}$ Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}}$
I keep making a mess of this. I tried vewing the denominator as
$a +\sqrt[3]{9}$, where $a=9-3\sqrt[3]{3}$ and secondly as
$b -3\sqrt[3]{3}+\sqrt[3]{9}$, where $b=9$.
Then using the sum and differences in cubes fratorization but this keeps adding radicals to the denominator.
How should I approach this/where could I be going wrong?
|
in general we may factor the norm form
$ x^3 + d y^3 + d^2 z^3 - 3dxyz$
as
$$ x^3 + d y^3 + d^2 z^3 - 3dxyz = $$ $$ \left( x+ d^{1/3}y + d^{2/3} z \right) \left( x^2 + d^{2/3} y^2 + d^{4/3} z^2 - dyz -d^{2/3}zx - d^{1/3} x y \right) $$
so that
$$ \frac{1}{ x+ d^{1/3}y + d^{2/3} z} = $$ $$ \frac{x^2 + d^{2/3} y^2 + d^{4/3} z^2 - dyz -d^{2/3}zx - d^{1/3} x y}{ x^3 + d y^3 + d^2 z^3 - 3dxyz} $$
I guess it is desirable to write it according to the exponent of $d,$
$$ \frac{1}{ x+ d^{1/3}y + d^{2/3} z} = $$ $$ \frac{(x^2-dyz)+(dz^2 -xy) d^{1/3} + (y^2 - zx) d^{2/3} }{ x^3 + d y^3 + d^2 z^3 - 3dxyz} $$
For you $d=3$
|
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|
Seeking for help to find a formula for $\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}$, where $a>1.$ When tackling the question, I found that for any $a>1$,
$$
I_1(a)=\int_{0}^{\pi} \frac{d x}{a-\cos x}=\frac{\pi}{\sqrt{a^{2}-1}}.
$$
Then I started to think whether there is a formula for the integral
$$
I_n(a)=\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}, $$
where $n\in N.$
After trying some substitution and integration by parts, I still failed and got no idea for reducing the power n. After two days, the Leibniz Rule for high derivatives come to my mind.
Differentiating $I_1(a)$ w.r.t. $a$ by $(n-1)$ times yields
$$
\displaystyle \begin{array}{l}
\displaystyle \int_{0}^{\pi} \frac{(-1)^{n-1}(n-1) !}{(a-\cos x)^{n}} d x=\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\pi}{\sqrt{a^{2}-1}}\right) \\ \displaystyle
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{(-1)^{n-1} \pi}{(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{1}{\sqrt{a^{2}-1}}\right) \tag{*}\label{star}
\end{array}
$$
I am glad to see that the integration problem turn to be merely a differentiation problem.
Now I am going to find the $(n-1)^{th} $ derivative by Leibniz Rule.
First of all, differentiating $I_1(a)$ w.r.t. $a$ yields $$
\left(a^{2}-1\right) \frac{d y}{d a}+a y=0 \tag{1}\label{diffeq}
$$
Differentiating \eqref{diffeq} w.r.t. $a$ by $(n-1)$ times gets $$
\begin{array}{l}
\displaystyle \left(a^{2}-1\right) \frac{d^{n} y}{d a^{n}}+\left(\begin{array}{c}
n-1 \\
1
\end{array}\right)(2 a) \frac{d^{n-1} y}{d a^{n-1}}+2\left(\begin{array}{c}
n-1 \\
2
\end{array}\right) \frac{d^{n-2} y}{d a^{n-2}}+x \frac{d^{n-1} y}{d a^{n-1}}+(n-1) \frac{d^{n-2} y}{d a^{n-2}}=0
\end{array}
$$
Simplifying, $$
\left(a^{2}-1\right) y^{(n)}+(2 n-1) ay^{(n-1)}+(n-1)^{2} y^{(n-2)}=0 \tag{2}\label{diffrec}
$$
Initially, we have $ \displaystyle y^{(0)}=\frac{1}{\sqrt{a^{2}-1}}$ and $ \displaystyle y^{(1)}=-\frac{a}{\left(a^{2}-1\right)^{\frac{3}{2}}}.$
By \eqref{diffrec}, we get $$
y^{(2)}=\frac{2 a^{2}+1}{\left(a^{2}-1\right)^{\frac{5}{2}}}
$$ and $$
\displaystyle y^{(3)}=-\frac{3 a\left(2 a^{2}+3\right)}{\left(a^{2}-1\right)^{\frac{7}{2}}}
$$
Plugging into \eqref{star} yields $$
\begin{aligned}
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{3}} &=\frac{\pi}{2} y^{(2)}=\frac{\pi\left(2 a^{2}+1\right)}{2\left(a^{2}-1\right)^{\frac{5}{2}}} \\
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{4}} &=-\frac{\pi}{6} \cdot \frac{3 a\left(2 a^{2}+3\right)}{\left(a^{2}-1\right)^{\frac{7}{2}}} =-\frac{\pi a\left(2 a^{2}+3\right)}{2\left(a^{2}-1\right)^{\frac{7}{2}}}
\end{aligned}
$$
Theoretically, we can proceed to find $I_n(a)$ for any $n\in N$ by the recurrence relation in $(2)$ .
By Mathematical Induction, we can further prove that the formula is $$
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{\pi P(a)}{\left(a^{2}-1\right)^{\frac{2 n-1}{2}}}
$$
for some polynomial $P(a)$ of degree $n-1$.
Last but not least, how to find the formula for $P(a)$? Would you help me?
|
Express the integral as
\begin{align}
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}
=\sum_{k=0}^{[\frac{n-1}2]} \frac{\binom{n-1}{2k}a^{n-2k-1}}{(a^2-1)^{n-1/2}}\int_0^{\pi}\cos^{2k}x \ dx
= \frac{\pi P(a)}{\left(a^{2}-1\right)^{n-{1}/{2}}}
\end{align}
which leads to
$$P(a) = \sum_{k=0}^{[\frac{n-1}2]}
\binom{n-1}{2k} \frac{(2k-1)!!}{(2k)!!} a^{n-2k-1}
$$
|
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|
Choosing at least 2 women from 7 men and 4 women In how many different ways can we choose six people, including at least two women, from a group made up of seven men and four women?
Attempt:
As we have to have at least two women in the choices, then $\displaystyle\binom{4}{2}$, leaving a total of $4$ out of $9$ people to be chosen. So $$\binom{9}{4} \cdot \binom{4}{2}=756$$
The answer is $371$. Where is my error?
|
Consider three cases:
*
*choosing 2 women and 4 men
*choosing 3 women and 3 men
*choosing 4 women and 2 men
So the total number of arrangements is:
$\binom{4}{2} \times \binom{7}{4}+ \binom{4}{3} \times \binom{7}{3}+ \binom{4}{4} \times \binom{7}{2}=371$
|
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|
Solve indefinite integral $\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}dx$ $$\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}dx$$
I multiply the integral so that I can get $-x^2$ in the numerator. I then expand the fraction so I can split the integral into easier integrals.
$$-\int\frac{-x^2}{1-x^2+\sqrt{1-x^2}}dx$$
$$-\int\frac{1-x^2+\sqrt{1-x^2}-1-\sqrt{1-x^2}}{1-x^2+\sqrt{1-x^2}}dx$$
$$-\int 1 dx
-\int\frac{-1-\sqrt{1-x^2}}{1-x^2+\sqrt{1-x^2}}dx$$
$$-\int 1 dx
+\int\frac{1}{1-x^2+\sqrt{1-x^2}}dx
+\int\frac{\sqrt{1-x^2}}{1-x^2+\sqrt{1-x^2}}dx$$
I then tried to do the same thing with one of the other integrals I got out of this (the fatal flaw).
$$-\int 1 dx
+\int\frac{1-x^2+\sqrt{1-x^2}+x^2-\sqrt{1-x^2}}{1-x^2+\sqrt{1-x^2}}dx
+\int\frac{\sqrt{1-x^2}}{1-x^2+\sqrt{1-x^2}}dx$$
$$-\int 1 dx
+\int 1 dx
+\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}dx
-\int\frac{\sqrt{1-x^2}}{1-x^2+\sqrt{1-x^2}}
+\int\frac{\sqrt{1-x^2}}{1-x^2+\sqrt{1-x^2}}dx$$
Then some integrals cancel out and all that remains is:
$$\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}dx$$
The problem is that this the original integral and I can't think of another way to approach this.
Could someone point out the thing I'm missing? This problem can't be solved with per partes or substitution as I can see.
|
The terms $\sqrt{1-x^2}$ motivates us to use change of variable by trigonometric functions. The term $\sqrt{1-x^2}$ naturally implies $x\in[-1,1]$ and so we can assume $x=\sin(t)$, $t\in[-\frac{\pi}{2},\frac{\pi}{2}]$. Then
\begin{align*}
\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}\,dx&=\int\frac{\sin^2(t)}{\cos^2(t)+\cos(t)}\,\cos(t)\,dt\\
&=\int\frac{1-\cos^2(t)}{1+\cos(t)}\,dt\\
&=\int\left(1-\cos(t)\right)\,dt\\
&=t-\sin(t)+C\\
&=\arcsin(x)-x+C.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
How does $\tan^2(x) \sec(x) + \sec^3(x)$ turn in to $2\sec^3(x) - \sec(x)$ Can someone explain how $\tan^2 $ disappeared and $\sec^3$ turn into $2\sec^3$ ???
The derivatives of the function $2\sec(x)\tan(x)$ is apparently $2(-\sec(x) + 2\sec^3(x))$
|
Need these
$$
\tan=\frac{\sin}{\cos} \tag{$\color{green}{\blacksquare}$}
$$
$$
\sec=\frac{1}{\cos} \tag{$\color{blue}{\blacksquare}$}
$$
$$
\sin^2+\cos^2=1\longrightarrow \sin^2=1-\cos^2 \tag{$\color{red}{\blacksquare}$}
$$
Just use properties and algebra
\begin{align*}
\tan^2\sec+\sec^3
&=
\left(\frac{\sin}{\cos}\right)^2\frac{1}{\cos}+\frac{1}{\cos^3}
\tag*{($\color{blue}{\blacksquare}$),($\color{green}{\blacksquare}$)}\\
&=
\frac{\sin^2}{\cos^2}\frac{1}{\cos}+\frac{1}{\cos^3}
\\&
=
\frac{1}{\cos^3}(\color{red}{\sin^2}+1)
\\&
=
\frac{1}{\cos^3}(\color{red}{1-\cos^2}+1) \tag{$\color{red}{\blacksquare}$}
\\&
=
\frac{1}{\cos^3}({2-\cos^2})
\\&=
2\frac{1}{\cos^3}-\cos^2\frac{1}{\cos^3}\\
&=
2\sec^3-\frac{1}{\cos}
\\&=
\boxed{2\sec^3-\sec} \tag*{$\blacksquare$}
\end{align*}
$\bf IF$ there are any issues ore clarifications needed, $\bf THEN$ comment or send an email to $\tt TheGreatJRB@Berkeley.edu$
Thanks,
Jason
Applied Mathematics Undergraduate
University of California, Berkeley
|
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|
Limit without hospital's rule $\lim_{x\rightarrow 0} (\frac{1}{x^2} - \frac{x}{\sin^3(x)})$
Calcule $\lim_{x\rightarrow 0} (\frac{1}{x^2} - \frac{x}{\sin^3(x)})$ without using L'Hospital
I first was trying to replace $x = \frac{\pi}{2} + u$ but I can't find a rule or eliminate anything.
After that I try using cubes:
$\frac{\sin^3(x) - x^3}{x^2\sin^3(x)} = \frac{(\sin(x)-x)(\sin^2(x)+x^2+x\sin(x))}{x^2\sin^3(x)}$ but again i couldn't find any rule or formula.
How can I solve this problem without using L'Hospital? Thank you very much in advance
|
$$
\begin{align}
\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{x}{\sin^3x}\right) &= \lim_{x\to 0}\left(\frac{\sin^3x-x^3}{x^2\sin^2x}\right) \\
&= \lim_{x\to 0}\left(\frac{x^3}{\sin^3x}\right)\left(\frac 1{x^2}\right)\left(\frac{\sin^3x}{x^3}-1\right) \\
&=1\cdot\lim_{x\to 0}\left(\frac{1}{x^2}\right)\left(\frac{\sin x}{x}-1\right)\left(\frac{\sin^2x}{x^2}+\frac{\sin x}{x}+1\right) \\
&=1\cdot\lim_{x\to 0}\left(\frac{\sin x-x}{x^3}\right)\cdot3 \\
&=1\cdot\left(\frac {-1}6\right)\cdot3 \\
&=\frac{-1}{2}
\end{align}
$$
|
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About an inequality wich is an upper bound for Am-Gm. Hi it's a upper bound for New bound for Am-Gm of 2 variables :
Problem :
Let $0<a\leq 1$ and $x\geq 1$ such that $2\leq a+x\leq 5$ then (dis)prove :
$$\left(1+\frac{1}{x+a}\left(\frac{4\left(x+a\right)^{2}}{\left(x+a+2\right)^{2}}-1\right)\right)\left(x+a-\frac{2\left(x+a\right)}{\left(x+a+2\right)}\right)-\left(ax\right)^{\frac{1}{x+a}}x^{\frac{x}{x+a}}a^{\frac{a}{a+x}}\geq0$$
In the link above you can find a proof for the lower bound .
If true i have tried to use Bernoulli's inequality wich is more than unsufficient .
Edit :
Playing with Taylor series and improving it manually we have :
$$x+\ln\left(ax\right)-a\ln\left(x\right)+a\ln\left(a\right)+\frac{1-a}{x}+\frac{1}{92.95}\left(\frac{1-a}{x}\right)^{2}\simeq \left(ax\right)^{\frac{1}{a+x}}x^{\frac{x}{a+x}}a^{\frac{a}{a+x}}$$
For $0.75\leq a \leq 1$ and $x\geq 1$ wich is a good approximation but in reality unsufficient .
Edit 2:
Let $1\leq x\leq 2$ and $0.5\leq a \leq 1$ then it seems we have :
$$\frac{d}{dx}\left(\frac{d}{dx}g\left(x\right)\right)\geq 0$$
Where :
$$g\left(x\right)=\left(ax\right)^{\frac{1}{x+a}}x^{\frac{x}{x+a}}a^{\frac{a}{a+x}}$$
We can try to show the convexity using :
for $a,x>0$:
$$2\left(a+x\right)\ln\frac{a+x}{2}-\left(a+1\right)\ln a-\left(x+1\right)\ln x\geq 0$$
And :
$$a\ln a+x\ln x-\left(a+x\right)\ln\left(a+x-\sqrt{ax}\right)\geq 0$$
Using this supposition and using a chord on $g(x)$ we have for $x\in[1,2-a]$ and $0.5\leq a\leq 1$ :
$$g(x)\leq x+a-1$$
The rest seems easy .
Last edit :
Using directly differentiation and logarithm and multiplying by $x+a$:
We have using WA :
$$\frac{\left(12-7\left(a+x\right)^{2}\right)}{\left(a+x\right)^{2}\left(a+x+7\right)-4}+\frac{6}{a+x+2}+\ln\left(\frac{\left(a+x\right)\left(\left(a+x\right)^{2}\left(a+x+7\right)-4\right)}{\left(a+x+2\right)^{3}}\right)-\frac{1}{x}-\ln\left(x\right)$$
We can replace $-\frac{1}{x}$ by $\frac{-2}{a+x}$ and $-\ln\left(x\right)$ by $-\ln\left(0.5(x+a)\right)$ around the zero of the derivative
Now it's a single variable inequality in $x+a$
Question :
How to (dis)prove it ?
|
Partial answer :
We prove a first step :
we have for $x\in[1,2-a]$ and $0.5\leq a\leq 1$ :
$$g(x)\leq x+a-1$$
or we need to show :
$$f(x)=\left(a+1\right)\ln\left(a\right)+\left(x+1\right)\ln\left(x\right)-\left(a+x\right)\ln\left(a+x-1\right)$$
It's easy using a computer or WA to see that the function is convex on the interval above because :
$$f''(x)=\frac{(a-1)(a\left(x-1\right)+x^{2}-3x+1)}{something positive}\geq 0$$
So using a chord we have :
$$f(1)=0\quad, f(2-a)\leq 0$$
Again it's easy to show it
So using the convexity we have :
$$f(x)\leq \frac{\left(f\left(1\right)-f\left(2-a\right)\right)}{1-\left(2-a\right)}\left(x-1\right)+f\left(1\right)\leq 0$$
Now we need to show for $x\in[1,2-a]$ and $0.5\leq a\leq 1$:
$$\left(1+\frac{1}{x+a}\left(\frac{4\left(x+a\right)^{2}}{\left(x+a+2\right)^{2}}-1\right)\right)\left(x+a-\frac{2\left(x+a\right)}{\left(x+a+2\right)}\right)-(x+a-1)\geq 0$$
Wich is a single variable inequality or a polynomial with an obvious root .
We are done in this case .
Edit :
Following the same way we can improve the proof for $x\in[1,2-a+\frac{1}{2}\left(a-1\right)^{2}]$ and $0.5\leq a\leq 1$
Curious fact :
Let $a=0.5$ and $x=e$ then define :
$$k(x)=\left(1+\frac{1}{x+a}\left(\frac{4\left(x+a\right)^{2}}{\left(x+a+2\right)^{2}}-1\right)\right)\left(x+a-\frac{2\left(x+a\right)}{\left(x+a+2\right)}\right)-\left(ax\right)^{\frac{1}{x+a}}x^{\frac{x}{x+a}}a^{\frac{a}{a+x}}$$
And :
$$j(x)=\frac{d}{dx}\left(\frac{d}{dx}\left(g\left(x\right)\right)\right)$$
Then :
$$j(e)-k(e)>0$$
It seems that the function :
$$k(x)+g(x)$$
Is convex for $x\in(1,3)$
We can repeat the same reasoning as above on differents interval include in $(1,3)$ until getting the minimum .
edit 2:
We have :
$$r(x+a)=\left(1+\frac{1}{x+a}\left(\frac{4\left(x+a\right)^{2}}{\left(x+a+2\right)^{2}}-1\right)\right)\left(x+a-\frac{2\left(x+a\right)}{\left(x+a+2\right)}\right)=\frac{(x+a)((x+a)^3+7(x+a)^2-4)}{(x+a+2)^3}$$
The second derivative is :
$$r''(x)=\frac{-36x^2+144x+48}{somethingpositive}$$
So solving the quadratics we see that the function $r(x)$ is convex on $x\in[1,4]$
Using the fact above (convexity) we can do the same thing at first taking :
$$k(x)=r'\left(2+b\right)\left(x+a-\left(2+b\right)\right)+r\left(2+b\right)$$
And $b=\left(2-a\right)\left(2.25-a\right)\left(a-1\right)^{2}$ where $0.5\leq a \leq 0.9$
on the interval $[2-a+1.5\left(a-1\right)^{2},c]$ where $c$ is :
$$k(c)-\left(ac\right)^{\frac{1}{c+a}}c^{\frac{c}{c+a}}a^{\frac{a}{a+c}}=0$$
We can choose the interval $x\in[2-a+0.5(a-1)^2,c]$ for $0.95\leq a\leq 0.999$ .With my previous work it remains to find the interval for $0.9\leq a\leq 0.95$
Now it includes the minimum on $x\in|1,2.5)$
...To be continued
|
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Using limits properties when proving using epsilon-delta definition. I want to prove $\lim\limits_{x \to 5} \sqrt{2x+6} = 4$
using the epsilon-delta definition.
My intuition to proving this is to use the root rule and square both ends of the equation to end up with:
$\lim\limits_{x \to 5} 2x+6 = 16$
I would then continue to prove the later limit as I would any other.
Is this a valid way to approach this proof, or am I missing something?
|
Note: This is extra detailed so you can use it as a template for similar problems.
When trying to use the $\epsilon, \delta$ method to show that
$$\lim_{x\to a}f(x)=L$$
Suppose that for some punctuated interval $I=(a-t)\cup(a+t)$ we can find an upper bound $B$ such
$$ \left|\frac{f(x)-L}{x-a}\right|\le B $$
for $x\in I$. Then for $\epsilon >0$ if we let $\delta =\min\left(t,\frac{\epsilon}{B}\right)$. It will follow that if $\left|\frac{f(x)-L}{x-a}\right|\le B$
and $|x-a|<\delta$, then we can multiply the two inequalities and get
$$ |f(x)-L|=\left|\frac{f(x)-L}{x-a}\right|\cdot|x-a|\le B\cdot\frac{\epsilon}{B}=\epsilon$$
To apply this to $\lim\limits_{x \to 5} \sqrt{2x+6} = 4$ we pick a punctured $t$ interval such as $t=1$ (this value of $t$ is frequently sufficient, but you can pick a smaller $t$ if necessary). So we will try to find an upper bound $B$ for $\left|\dfrac{\sqrt{2x+6}-4}{x-5}\right|$ for $x\in(4,5)\cup(5,6)$.
We can deal with the square root by rationalizing the numerator.
For $x\ne5$
$$ \left|\dfrac{\sqrt{2x+6}-4}{x-5}\right|=\left|\dfrac{\sqrt{2x+6}-4}{x-5}\cdot\frac{\sqrt{2x+6}+4}{\sqrt{2x+6}+4}\right|=\left|\frac{2}{\sqrt{2x+6}+4}\right|$$
Now we want to find an upper bound $B$ on the value of this expression on the interval $(4,6)$ excluding $x=5$.
We will work our way up to the correct inequality in stages.
\begin{eqnarray}
4&<x&<6\\
8&<2x&<12\\
14&<2x+6&<18\\
\sqrt{14}&<\sqrt{2x+6}&<\sqrt{18}\\
\sqrt{14}+4&<\sqrt{2x+6}+4&<\sqrt{18}+4
\end{eqnarray}
The reciprocals inequalities will have the opposite sense, so we have for $x\in(4,6)$, and $x\ne5$ it is the case that
$$ \left|\frac{2}{\sqrt{2x+6}+4}\right|<\frac{2}{\sqrt{18}+4} $$
So this will do for an upper bound. We can choose a larger upper bound if it makes the problem simpler. For example, it will be a larger upper bound if we use $\sqrt{16}$ instead of $\sqrt{18}$, giving $B=\frac{1}{4}$, then $\frac{\epsilon}{B}=4\epsilon$
Now we are ready to make our $\epsilon, \delta$ proof. Everything up to now is preparation.
Let $\epsilon>0$. Then if $\delta =\min\left(1,4\epsilon\right)$ it follows that if $|x-5|<4\epsilon$
$$ |\sqrt{2x+6}-4|=\left|\frac{\sqrt{2x+6}-4}{x-5}\right|\cdot|x-5|<\frac{1}{4}\cdot4\epsilon=\epsilon $$
|
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|
$10$ distinct color balls and we want to divide them into $3$ identical boxes with hard restrictions and probability
We have $10$ distinct color balls and we want to divide them into $3$ identical boxes .We decide that two of the groups can contain at most $3$ balls and the rest one contain at most $4$ balls. Moreover, there are two balls called $A$ and $B$ in $10$ balls.
$1-)$ How many ways are there to distribute these $10$ balls to $3$ groups obeying size restriction of groups and $A$ and $B$ are in the same box?
$2-)$ What is the probability that $A$ and $B$ are in the same boxes obeying size restriction of boxes?
$3-)$ How many ways are there to distribute these $10$ balls to $3$ groups obeying size restriction of groups and $A$ and $B$ are not in the same box?
$4-)$ What is the probability that $A$ and $B$ are not in the same boxes obeying size restriction of boxes?
$5-)$ Solve the foregoing $4$ questions if the number of balls is $7$ instead of $10$.
Solution
$1-)$ $$\bigg (\frac{\binom{8}{1}\binom{7}{3}\binom{4}{4}}{2!}=140 \bigg)+ \bigg (\frac{\binom{8}{3}\binom{5}{3}\binom{2}{2}}{2!} =280\bigg )=420$$
$2-)$ $$\bigg (\frac{\binom{8}{1}\binom{7}{3}\binom{4}{4}}{2!}+ \frac{\binom{8}{3}\binom{5}{3}\binom{2}{2}}{2!} \bigg )/\frac{\binom{10}{3}\binom{7}{3}\binom{4}{4}}{2!}=42/168$$
$3-)$ $$\frac{\binom{10}{3}\binom{7}{3}\binom{4}{4}}{2!}-420 =2100-420=1680$$
$4-)$ By using $3$ and $4$ , $$1-42/168=126/168$$
$5-)$ I could not solve this part because of low number of balls, and I do not know how to distribute.
Are my solutions correct? If not, can you correct me and help for $5$?
|
First question:
The first term is incorrect. It should be,
$ \displaystyle \binom{8}{1}\binom{7}{3}\binom{4}{4}=280$
Unlike second term, it should not be divided by $2!$ as the boxes having $3$ balls each are different groups - one of them has $A$ and $B$ and the other is without them.
So the answer should be $560$
Second question:
The denominator, as you mentioned, $ \displaystyle \binom{10}{3}\binom{7}{3}\binom{4}{4} / 2! = 2100$
So the probability is $\displaystyle \frac{560}{2100} = \frac{4}{15}$
Now you can use above results to solve $(3)$ and $(4)$.
Coming to the question if there are $7$ balls instead of $10$, you can break it into cases: $i$) three groups with $1, 2, 4$ balls $ii$) three groups with $2, 2, 3$ balls $iii$) $1, 3, 3$ balls
|
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Diophantine problem: Find all pairs of $(x, y)$ such that $x^3-4xy+y^3=-1$
Find all pairs of integers $x, y$ such that:
$$x^3-4xy+y^3=-1$$
My analysis:
*
*$x$ and $y$ can't be both negative because all terms would be negative and $(x,y)=(-1,-1) \implies (-1)^3-4(-1)(-1)+(-1)^3 =-6 $ would already be too small.
*Reducing modulo $4$ gives: $x^3+y^3\equiv 3 \pmod 4
\implies (x,y)\equiv(0,3)$ or $(3,0) \pmod 4$
*WLOG let $x$ be the larger (+ve) one in a solution pair$\implies x\ge y+1$:
$-1 = x^3-4xy+y^3 \ge (y+1)^3 - 4(y+1)y +y^3$
$-2 \ge 2y^3 - y^2 -y$
$-2 \ge (y-1)(2y^2 + y)$
Since $(2y^2+y)\ge 0$ and $y\le-1$
Indeed $(x,y)=(0,-1)$ is a solution fitting all the above (inconclusive) results. However, I don't know how to find a lower bound for $y$ to lock down all possible pairs of $(x,y).$ What's the missing step? Am I completely off?
|
Hint: A messy quadratic is better than a simple cubic while dealing with integers.
$$x+y=s\implies x^3-4x(s-x)+(s-x)^3=-1 $$
Simplifying, $$ (3s+4)x^2-(3s^2+4s)x+s^3+1=0$$
The discriminant must be non-negative, $$\Delta=-(3s+4)(s^3-4s^2+4)\ge 0$$
If $s\in (-\infty,-2]$ or $s\in [4,\infty)$, $\Delta<0$. Therefore, $s\in\{-1,2,3\}$. Only $s=-1$ results in an integer value for $x$.
$$(x,y)\in \{(-1,0),(0,-1)\}$$
|
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Taylor expansion of $\sin \pi z$ at $z = -1$. Taylor expansion of $\sin \pi z$ at $z = -1$ is
$$\sin\pi z = -\sin(\pi(z+1)) = -\sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}$$ so that
$$\sin\pi z = \sum_{n=0}^\infty \frac{(-1)^{n+1}\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}. \tag{$\dagger$}$$
But if I try this
\begin{align}
\sin\pi z & = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}z^{2n+1} \\& = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}(z+1-1)^{2n+1} \\ & = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}\left(\sum_{k=0}^{2n+1}\binom{2n+1}{k}(z+1)^k(-1)^{2n-k+1}\right).\tag{$\dagger^*$}
\end{align}
In this case, how can I reduce $(\dagger^*)$ as the above form $(\dagger)$? Just direct calculation?
|
You can continue by changing the order of summation and then splitting the sum into two parts:
\begin{align*}
& \sum\limits_{n = 0}^\infty {\frac{{( - 1)^{n + 1} \pi ^{2n + 1} }}{{(2n + 1)!}}\left( {\sum\limits_{k = 0}^{2n + 1} {( - 1)^k \binom{2n + 1}{k}(z + 1)^k } } \right)}
\\ & = \sum\limits_{k = 0}^\infty {\left( {\sum\limits_{n = \left\lceil {\frac{{k - 1}}{2}} \right\rceil }^\infty {\frac{{( - 1)^{n + k + 1} \pi ^{2n - k + 1} }}{{(2n - k + 1)!}}} } \right)\frac{{\pi ^k }}{{k!}}(z + 1)^k }
\\ & = \sum\limits_{k = 0}^\infty {\left( {\sum\limits_{n = k}^\infty {\frac{{( - 1)^{n + 2k + 1} \pi ^{2n - 2k + 1} }}{{(2n - 2k + 1)!}}} } \right)\frac{{\pi ^{2k} }}{{(2k)!}}(z + 1)^{2k} } \\ & \quad + \sum\limits_{k = 0}^\infty {\left( {\sum\limits_{n = k}^\infty {\frac{{( - 1)^{n + 2k + 2} \pi ^{2n - 2k} }}{{(2n - 2k)!}}} } \right)\frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} }
\\ & = \sum\limits_{k = 0}^\infty {\left( {( - 1)^{k + 1} \sum\limits_{j = 0}^\infty {\frac{{( - 1)^j \pi ^{2j + 1} }}{{(2j + 1)!}}} } \right)\frac{{\pi ^{2k} }}{{(2k)!}}(z + 1)^{2k} } \\ & \quad + \sum\limits_{k = 0}^\infty {\left( {( - 1)^k \sum\limits_{j = 0}^\infty {\frac{{( - 1)^j \pi ^{2j} }}{{(2j)!}}} } \right)\frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} }
\\ & = \sum\limits_{k = 0}^\infty {\left( {( - 1)^{k + 1} \sin \pi } \right)\frac{{\pi ^{2k} }}{{(2k)!}}(z + 1)^{2k} } + \sum\limits_{k = 0}^\infty {\left( {( - 1)^k \cos \pi } \right)\frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} } \\ & = \sum\limits_{k = 0}^\infty {( - 1)^{k + 1} \frac{{\pi ^{2k + 1} }}{{(2k + 1)!}}(z + 1)^{2k + 1} } .
\end{align*}
|
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$1^2 - 2^2 + 3^2 - 4^2 + \dots + 1999^2$ using closed form formula for sum of squares of first $n$ natural numbers The question is simple, the solution, not so much
Q.Find the sum of the given expression $1^2- 2^2 + 3^2 - 4^2 + \dots + 1999^2$
My idea is we know $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$
So for $n=1999$ I get the sum as $2,66,46,67,000$
From this I need to subtract the squares of the even terms twice because subtracting once leaves with only the sum of the squares of the odd nos.
I observed something :
$2^2 + 4^2 + 6^2 + \dots + 1998^2 = (2 \cdot 1)^2 + (2 \cdot 2)^2 + \dots + (2 \cdot 999)^2$
Therefore to obtain the sum of the square of the even terms, I can take $4$ as common and use the aforementioned formula for $n=999$ and multiply it by $4$.
therefore sum of square of even terms = $1,33,13,34,000$
I need to subtract this sum twice to get the answer, because subtracting once simply leaves me with the sum of the squares of the odd numbers.
The answer is now $1999000$, which still doesn't match the answer key. Can someone explain where I am going wrong ?
|
If you first add all the squares:
$$
1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + \cdots
$$
and then subtract the even squares:
$$
\phantom{1^2} - 2^2\phantom{ + 3^2} - 4^2 \phantom{+ 5^2} - 6^2 \phantom{-}\cdots
$$
then you are left with only the odd squares:
$$
1^2 \phantom{+ 2^2} + 3^2 \phantom{+ 4^2} + 5^2 \phantom{+ 6^2} + \cdots
$$
You need to subtract the even squares one more time to get
$$
1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + \cdots
$$
|
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$a,b>0$ then: $\frac{1}{a^2}+b^2\ge\sqrt{2(\frac{1}{a^2}+a^2)}(b-a+1)$
Let $a,b>0$. Prove that: $$\frac{1}{a^2}+b^2\ge\sqrt{2\left(\frac{1}{a^2}+a^2\right)}(b-a+1)$$
Anyone can help me get a nice solution for this tough question?
My approach works for 2 cases:
Case 1: $b-a+1>0$ then squaring both side, we get equivalent inequality: $$\frac{1}{a^4}+b^4+2\frac{b^2}{a^2}\ge2\left(\frac{1}{a^2}+a^2\right)(b^2+a^2+1-2ab-2a+2b)$$ Or: $$\frac{1}{a^4}+b^4\ge\frac{2}{a^2}(a^2+1-2ab-2a+2b)+2a^2(b^2+a^2+1-2ab-2a+2b)$$
The rest is so complicated. Is there nice idea etc: AM-GM, C-S to prove this inequality.
Case 2: $b-a+1<0$ which is obviously true.
I hope we can find a better approach for the inequality. Thank you very much!
|
Due to @Calvin Lin, I will post solution based on his hint later.
Now I get solution by AM-GM. Notice that: $\frac{1}{a^2}+a^2=\left(\frac{1}{a}+a\right)^2-2=\left(\frac{1}{a}+a+\sqrt{2}\right)\left(\frac{1}{a}+a-\sqrt{2}\right)$
Then using AM-GM: $$2\frac{\sqrt{2}\frac{1}{a^2}}{\sqrt{\frac{1}{a^2}+a^2}}+\frac{\frac{1}{a}+a+\sqrt{2}}{2+\sqrt{2}}+\frac{\frac{1}{a}+a-\sqrt{2}}{2-\sqrt{2}}\ge\frac{4}{a}$$ Or:$$\frac{\sqrt{2}\frac{1}{a^2}}{\sqrt{\frac{1}{a^2}+a^2}}+a\ge\frac{1}{a}+1$$
Similarly, $$\frac{\sqrt{2}b^2}{\sqrt{\frac{1}{a^2}+a^2}}+a+\frac{1}{a}\ge2b+1$$
The rest is obvious: $$\frac{\frac{1}{a^2}+b^2}{\sqrt{\frac{1}{a^2}+a^2}}\ge\sqrt{2}(b-a+1)$$
|
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Calculate Indefinite Integral $\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx$ The following integration is given by Wolfram Alpha
$$\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx=\frac{x^6(28x^4+16x^3+39x^2+16x+28)}{168(1+x)^{16}}.$$
My question is: what is the best (meaning least work), method to achieve this result by hand ? There are two approaches I see, partial fractions, maybe setting $y=1+x$, but this is still alot of work. Or perform successive integration by parts. Both still involve a lot of calculation. The compact form of the answer makes me hope there is a nice way to achieve it. Guessing the ultimate form and then calculating the derivative is, to me, a less desirable method.
|
$$\int \frac{x^5 \left(1-x^6\right)}{(x+1)^{18}}\,dx=\frac{P(x)}{(x+1)^{17}}$$
$$\frac{x^5 \left(1-x^6\right)}{(x+1)^{18}}=\frac{(x+1) P'(x)-17 P(x)}{(x+1)^{18}}$$ So, $P(x)$ is a polynomial of degree $11$.
Let
$$P(x)=\sum_{n=0}^{11} a_n\,x^n$$ Expand and group powers to get
$$0=(a_1-17 a_0)+(2 a_2-16 a_1) x+(3 a_3-15 a_2) x^2+(4 a_4-14 a_3) x^3+$$ $$(5
a_5-13 a_4) x^4+(-12 a_5+6 a_6-1) x^5+(7 a_7-11 a_6) x^6+(8 a_8-10 a_7)
x^7+$$ $$(9 a_9-9 a_8) x^8+(10 a_{10}-8 a_9) x^9+(11 a_{11}-7 a_{10}) x^{10}+(1-6
a_{11}) x^{11}$$ which is simple.
If I not wrong the coefficients are
$$\left\{0,0,0,0,0,0,\frac{1}{6},\frac{11}{42},\frac{55}{168},\frac{55}{168},\frac{11}{42},\frac{1}{6}\right\}$$ and factoring
$$P(x)=\frac{1}{168} x^6 (x+1) \left(28 x^4+16 x^3+39 x^2+16 x+28\right)$$
|
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|
If $\frac{n^2}{3} < xy \leq n^2 - 5$, then does it follow that $n > \max(x,y)$? My question here is pretty basic:
If $\frac{n^2}{3} < xy \leq n^2 - 5$, then does it follow that $n > \max(x,y)$?
Here, $n, x, y$ are positive integers.
MY ATTEMPT
Let $n, x, y$ be positive integers such that
$$\frac{n^2}{3} < xy \leq n^2 - 5.$$
Suppose to the contrary that we have
$$n \leq \max(x,y) = \frac{x+y+|x-y|}{2}.$$
We get the bounds
$$\frac{n^2}{3x} < y \leq \frac{n^2 - 5}{x}$$
which implies that
$$\frac{5 - n^2}{x} \leq -y < -\frac{n^2}{3x}.$$
It follows that
$$n \leq \frac{\bigg(x + \dfrac{n^2 - 5}{x}\bigg) + |x - \dfrac{n^2}{3x}|}{2}.$$
I then asked WolframAlpha to solve the last inequality for $x$ and then for $n=a$, but I am not sure I can interpret the outputs by myself, as they are a bit messy.
Can anybody help? Thanks!
|
For every $n\ge 3$, a counter example is given by $x=n-2$ and $y=n+1$
indeed in that case we have
\begin{equation}
\frac{n^2}{3}< n^2 - n - 2 = x y \le n^2 - 5
\end{equation}
|
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prove that $(p,mp]$ will always contain at least $y-1$ cubes
If $n, m\in \mathbb{Z}_{>0}$ and $(n, mn]$ contains $y$ cubes, prove that $(p,mp]$ will always contain at least $y-1$ cubes (integers $a$ so that $a=k^3$ for some integer k) for all $p\in [n,\infty)$.
This works for small examples, such as for $n=7, m=4$ when the cubes are $8$ and $27$ as $(p, 4p]$ always contains at least one cube for $p\ge n$, but I'm not sure how to generalize the results. I know that between any two real numbers that are at least one apart there is an integer. Also, clearly, the ratio between $k^3$ and $(k+1)^3$ decreases as $k$ increases. For $n\ge 8$, there exists an integer $k$ so that $k \in (n, mn]$ is a perfect cube because $\sqrt[3]{mn} - \sqrt[3]{n} \ge \sqrt[3]{n} > 1$.
|
The statement to prove is obviously true for $y = 0$ and $y = 1$, so consider just $y \gt 1 \; \to \; y - 1 \gt 0$. Since $(n, mn]$ contains $y$ cubes, this means there's an $a \in \mathbb{Z}_{>0}$ such that
$$n \lt a^3 \lt (a + (y - 1))^3 \le mn \tag{1}\label{eq1A}$$
Since the ratio of the upper & lower limits, i.e., $mn$ and $n$, must be greater than the ratio of the two intermediate values, this leads to
$$\frac{mn}{n} = m \gt \left(\frac{a + y - 1}{a}\right)^3 = \left(1 + \frac{y-1}{a}\right)^3 \tag{2}\label{eq2A}$$
For $n \le p \lt a^3$, no cubes will become out of range of $(p, mp]$ at the bottom or top, so there will be at least $y$ cubes in $(p, mp]$. Next, for $p \in [a^3, \infty)$, let $b$ be the number of cubes in $(p, mp]$. Thus, for some integer $c \ge 1$, we have $p \lt (a + c)^3$ and $(a + c + (b - 1))^3 \le mp$, plus also
$$(a + c - 1)^3 \le p \lt mp \lt (a + c + b)^3 \tag{3}\label{eq3A}$$
Handling the ratios as discussed for \eqref{eq1A} gives that
$$\left(\frac{a + c + b}{a + c - 1}\right)^3 = \left(1 + \frac{b + 1}{a + c - 1}\right)^3 \gt \frac{mp}{p} = m \tag{4}\label{eq4A}$$
Since $f(x) = x^3$ is a strictly increasing function for $x \gt 0$, from \eqref{eq2A} and \eqref{eq4A} we have
$$\left(1 + \frac{b + 1}{a + c - 1}\right)^3 \gt m \gt \left(1 + \frac{y-1}{a}\right)^3 \; \to \; \frac{b + 1}{a + c - 1} \gt \frac{y - 1}{a} \tag{5}\label{eq5A}$$
Since $c \ge 1$, then $a + c - 1 \ge a$, so
$$b + 1 \gt y - 1 \; \to \; b \ge y - 1 \tag{6}\label{eq6A}$$
This concludes proving that for all $p \in [n, \infty)$, there are at least $y - 1$ cubes in $(p, mp]$.
Note the above proof techniques apply not only to cubes, they also work for any positive power (e.g., "cubes" may be replaced by "squares").
|
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Integration of Bounded Region I'm trying to solve this problem:
The region of integration is the triangle $D$ with vertexes $A(0,0),B(1,1),C(10,1)$.
Find the solution of $\iint_D \sqrt{x^2-y^2}\,dx\,dy$.
MY SOLUTION (as @ryang suggested):
We can write the triangle as $D=\{(x,y)\in\mathbb R^2|0\le y\le 1,y\le x \le 10y\}$.
After that the integral becomes:
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy$$
We can solve first this indefinite integral $$\int\sqrt{x^2-y^2}\,dx$$
where $y\in[0,1]$ is a constant.
So$$\begin{align}\int\sqrt{x^2-y^2}\,dx=x\sqrt{x^2-y^2}-\int\frac{x^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\frac{x^2-y^2+y^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{1}{\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{(1+\frac{x}{\sqrt{x^2-y^2}})}{x+\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\log (x+\sqrt{x^2-y^2})\end{align}$$
The result is:
$$\int\sqrt{x^2-y^2}\,dx=\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})+C$$
We can now solve the definite integral
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy=\int_0^1\Bigl(\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})\Bigr)\Big|_y^{10y}\,dy\\=\int_0^1 \Bigl(5y^2\sqrt{99}-\frac{1}{2}y^2\log (10+\sqrt{99})\Bigr)\,dy\\=\frac{5\sqrt{99}}{3}-\frac{\log (10+\sqrt{99})}{6}
$$
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MY SOLUTION (as @ryang suggested):
We can write the triangle as $D=\{(x,y)\in\mathbb R^2|0\le y\le 1,y\le x \le 10y\}$.
After that the integral becomes:
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy$$
We can solve first this indefinite integral $$\int\sqrt{x^2-y^2}\,dx$$
where $y\in[0,1]$ is a constant.
So$$\begin{align}\int\sqrt{x^2-y^2}\,dx=x\sqrt{x^2-y^2}-\int\frac{x^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\frac{x^2-y^2+y^2}{\sqrt{x^2-y^2}}\,dx\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{1}{\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\Bigl(\int\frac{(1+\frac{x}{\sqrt{x^2-y^2}})}{x+\sqrt{x^2-y^2}}\,dx\Bigr)\\=x\sqrt{x^2-y^2}-\int\sqrt{x^2-y^2}\,dx-y^2\log (x+\sqrt{x^2-y^2})\end{align}$$
The result is:
$$\int\sqrt{x^2-y^2}\,dx=\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})+C$$
We can now solve the definite integral
$$\int_0^1\int_y^{10y} \sqrt{x^2-y^2}\,dx\,dy=\int_0^1\Bigl(\frac{1}{2}x\sqrt{x^2-y^2}-\frac{1}{2}y^2\log (x+\sqrt{x^2-y^2})\Bigr)\Big|_y^{10y}\,dy\\=\int_0^1 \Bigl(5y^2\sqrt{99}-\frac{1}{2}y^2\log (10+\sqrt{99})\Bigr)\,dy\\=\frac{5\sqrt{99}}{3}-\frac{\log (10+\sqrt{99})}{6}
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4343599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
How to solve system of linear equations that came up in looking at a probability problem, any tricks? The following system of $k$ equations came up in a probability problem I was looking at:
$$(n+1)x_1 - x_2 = 1, \quad x_1 + nx_2 - x_3 = 1 , \quad x_1 + nx_3 - x_4 = 1, \quad x_1 + nx_4 - x_5 = 1, \quad \ldots \quad x_1 + nx_{k-1} - x_{k} = 1, \quad x_1 + nx_{k} = 2,$$which involves solving$$\begin{pmatrix}
n+1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0\\
1 & n & -1 & 0 & \cdots & 0 & 0 & 0 \\
1 & 0 & n & - 1 & \cdots & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
1 & 0 & 0 & 0 & \cdots & 0 & n & -1\\
1 & 0 & 0 & 0 & \cdots & 0 & 0 & n\\
\end{pmatrix} \begin{pmatrix}
x_1\\
x_2 \\
x_3 \\
\vdots \\
x_{k - 1}\\
x_k\\
\end{pmatrix} = \begin{pmatrix}
1\\
1 \\
1 \\
\vdots \\
1\\
2\\
\end{pmatrix}$$Now, given that this matrix looks pretty "sparse", I'm wondering if there's any shortcuts that can get us the answer quickly without me having to painstakingly row reduce this augmented matrix by brute force. I've started in the direction of brute force but couldn't see any immediate pattern/simplification.
For the record, the answer in the back of my book is$$x_r = {{n^r - n^{r - 1} + n^{k} - 1}\over{n^{k+1} - 1}}.$$
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Note that we have $\forall r\in [2,k]$
$$x_r=nx_{r-1}+x_1-1$$
With the condition that $x_k=\frac{2-x_1}{n}$. Note that we can actually solve this recursion for a general formula. Shift the indices up by $1$ to get that
$$x_{r+1}=nx_r+x_1-1$$
and subtract the two equations to get
$$x_{r+1}-(n+1)x_r+nx_{r-1}=0$$
Note that we dropped the constant $x_1-1$ term(s) and increased the degree of our recursion. We will need to "reimplement" that term(s) later.
This has characteristic polynomial $x^2-(n+1)x+n=(x-n)(x-1)$. Hence, $x_r=an^r+b$ for some constants $a,b$. We know that $x_1=an+b$, $x_2=an^2+b$ and $x_k=an^k+b$. From our given problem, we know
$$\begin{cases}(an^2+b)=(n+1)(an+b)-1\\n(an^k+b)=2-an-b\end{cases}$$
This first equation is where we "reimplement" the term(s) we dropped (we account for the dropping of the terms by increasing the degree and therefore requiring another initial condition, which can be created by using any specific value of $r$ in the initially found recurrence).
$$\begin{cases}an^2+b=an^2+bn+an+b-1\\an^{k+1}+bn=2-an-b\end{cases}$$
$$\begin{cases}1=bn+an\\an^{k+1}+bn=2-an-b\end{cases}$$
$$\begin{cases}a+b=\frac{1}{n}\\an^{k+1}+b=2-(an+bn)\end{cases}$$
$$\begin{cases}a+b=\frac{1}{n}\\an^{k+1}+b=1\end{cases}$$
$$\begin{cases}a+b=\frac{1}{n}\\a(n^{k+1}-1)=1-\frac{1}{n}\end{cases}$$
$$\begin{cases}a+b=\frac{1}{n}\\a=\frac{n-1}{n(n^{k+1}-1)}\end{cases}$$
$$\begin{cases}b=\frac{1}{n}-\frac{n-1}{n(n^{k+1}-1)}\\a=\frac{n-1}{n(n^{k+1}-1)}\end{cases}$$
$$\begin{cases}b=\frac{n^k-1}{n^{k+1}-1}\\a=\frac{n-1}{n(n^{k+1}-1)}\end{cases}$$
So we have that $x_r=an^r+b=\frac{(n-1)n^{r-1}}{n^{k+1}-1}+\frac{n^k-1}{n^{k+1}-1}$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4344226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
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