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Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g(x)+5$ where $g(x)=(x+1)(x+2)(x+3)(x+4)$
Let's find the maximum value of g(x)
It can be clearly seen that from $x=-6$ to $x=6$, maximum value of $g(x)$ is at $x=6$.
$$g(6)=5040$$
Let's find the minimum value of $g(x)$
From the sign scheme one can see that negative value of $g(x)$ occurs in the interval $(-4,-3)$ and $(-2,-1)$
Hence intuitively it feels that the minimum value of g(x) would be at $x=-\dfrac{3}{2}$ or at $x=-\dfrac{7}{2}$ as in case of parabola also, minimum value is at average of both the roots.
So $g\left(-\dfrac{3}{2}\right)=-\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{5}{2}=-\dfrac{15}{8}$
At $x=-\dfrac{7}{2}$, $g\left(-\dfrac{7}{2}\right)=-\dfrac{5}{2}\cdot\dfrac{3}{2}\cdot\dfrac{1}{2}\cdot\dfrac{1}{2}=-\dfrac{15}{8}$
So range of $g(x)$ would be $[\dfrac{-15}{8},5040]$
Hence range of $f(x)$=$\left[\dfrac{25}{8},5045\right]$
But it is given that the range is $[a,b]$ where $a,b\in N$
I am stuck here. I am also not able to prove mathematically that at $x=-\dfrac{3}{2}$ or $x=-\dfrac{7}{2}$, minimum value of $g(x)$ will occur.
Please help me in this.
|
Continuing discussion in comments, use first derivative test.
$$
f’(x)=2x^3+15x^2+35x+25=(2x+5)(x^2+5x+5)
$$
Now the roots for the quadratic are
$x = \frac{-5+\sqrt{5}}{2}, \frac{-5-\sqrt{5}}{2}$. So to check for maximum and minimum value, check $f(-6)=120$, $f(\frac{-5+\sqrt{5}}{2})=-1$, $f(\frac{5}{2})>0$ cannot be max/min by your discussion, $f(\frac{-5-\sqrt{5}}{2})=-1$ and $f(6)=5040$. So we just take the max/min now.
|
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|
Prove convergence of $n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ I am working on some old analysis exams and i got stuck on this exercise :
Using the epsilon definition show that $a_{n} = n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ converges and determine its limit.
Knowing that the limit is 1/2, I know need to find an $ N \in \mathbb{N}$ so that : $ \forall \epsilon > 0 n > N \implies \left| a_n - \frac{1}{2} \right| < \epsilon $
Next step I simplify $a_n$ : $ a_n = \frac{n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right) \cdot \sqrt{1+ \frac{1}{n}} + 1 } {\sqrt{1+ \frac{1}{n}} + 1} = \frac{1}{\sqrt{1+ \frac{1}{n}} + 1}$
And then I got stuck,what am I supposed to do with : $ \left|\frac{1}{\sqrt{1+ \frac{1}{n}} + 1} - \frac{1}{2} \right|$
|
$$n\left(\sqrt{1+\frac1n}-1\right)<n\left(\sqrt{1+\frac1n+\frac1{4n^2}}-1\right)=\frac12$$
and
$$n\left(\sqrt{1+\frac1n}-1\right)=\frac1{\sqrt{1+\dfrac1n}+1}>\frac1{\sqrt{1+\dfrac1n+\dfrac1{4n^2}}+1}=\frac12-\frac1{4n+1}.$$
|
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|
Showing that solutions to $x^3y^{\prime\prime\prime}+2x^2y^{\prime\prime}-4xy^\prime+4y=0$ are linearly independent
Find solutions for $x^3y^{\prime\prime\prime}+2x^2y^{\prime\prime}-4xy^\prime+4y=0$ which have the form $y(x)=x^r$ and then show that they are linearly independent.
So my method for solving this was to let $y(x)=x^r$ and then substituting that into the original equation:
$x^3\cdot r(r-1)(r-2)x^{r-3}+2x^2r(r-1)x^{r-2}-4xrx^{r-1}+4x^r=0$
Then solving $r(r-1)(r-2)+2\cdot r(r-1)-4r+4=0$ I guessed the solution $r=1$
Then used polynomial long division to get $r(r-1)(r-2)=0$
And so have the general solution $y(x)=c_1+c_2x+c_3x^2$
So I believe my $3$ linearly independent solutions should be $x^0, x^1, x^2$ however when I try these solutions, the only one that seems to work is $y(x)=x$.
So my first question is, is my general solution correct?
And then are solutions supposed to be any linear combination of $x^0,x^1,x^2$ or do only some linear combination actually solve the equation?
As for showing they are linearly independent I was going to use a theorem that polynomials of different degrees are linearly independent.
|
Then solving $ r(r−1)(r−2)+2⋅r(r−1)−4r+4=0$ I guessed the solution $r=1$
You need to solve completely your equation :
$r(r−1)(r−2)+2⋅r(r−1)−4r+4=0$
$r(r−1)(r−2)+2⋅r(r−1)−4(r-1)=0$
$(r-1)(r(r-2)+2r-4)$
$(r-1)(r^2-4)=0 \implies (r-1)=0 \text{ or } (r^2-4)=0$
$r^2-4 =0 \implies r=2,r=-2$
$r-1 =0 \implies r=1$
$$S_r=\{1,2,-2\}$$
So the general solution is :
$$y=c_1x+c_2x^2+\frac {c_3}{x^2}$$
|
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|
Proving the Sum to Product Identities (without using the product to sum identities) I got this project for my students from another teacher, and it did not come with a key. There are four versions, each one is proving a different Sum to Product identity, but they can't use the product to sum identity. The instructions say "Using reciprocal, Pythagorean, sum/difference, double and half-angle identities verify the equation" then they are given one of the identities.
We have figured out the sine ones, but we haven't figured out the cosine ones. This is what I have so far on one of them. Can anyone help?
|
Let's correct your calculation.
\begin{align*}
-2&\sin\left(\frac{\alpha + \beta}{2}\right)\sin\left(\frac{\alpha - \beta}{2}\right)\\
& = -2\sin\left(\frac{\alpha}{2} + \frac{\beta}{2}\right)\sin\left(\frac{\alpha}{2} - \frac{\beta}{2}\right)\\
& = -2\left[\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right) + \cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\right]\left[\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\right) - \cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\right]\\
& = -2\left[\sin^2\left(\frac{\alpha}{2}\right)\cos^2\left(\frac{\beta}{2}\right) - \sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\cos\left(\frac{\beta}{2}\right)\right.\\
& \qquad \left. + \sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)\sin\left(\frac{\beta}{2}\right)\cos\left(\frac{\beta}{2}\right)\right.
\left. - \cos^2\left(\frac{\alpha}{2}\right)\sin^2\left(\frac{\beta}{2}\right)\right]\\
& = -2\left[\sin^2\left(\frac{\alpha}{2}\right)\cos^2\left(\frac{\beta}{2}\right) - \cos^2\left(\frac{\alpha}{2}\right)\sin^2\left(\frac{\beta}{2}\right)\right]\\
& = -2\left[\left(\frac{1 - \cos\alpha}{2}\right)\left(\frac{1 + \cos\beta}{2}\right) - \left(\frac{1 + \cos\alpha}{2}\right)\left(\frac{1 - \cos\beta}{2}\right)\right]\\
& = -2\left[\frac{1 + \cos\beta - \cos\alpha - \cos\alpha\cos\beta}{4} - \frac{1 + \cos\alpha - \cos\beta - \cos\alpha\cos\beta}{4}\right]\\
& = -2\left[\frac{2\cos\beta - 2\cos\alpha}{4}\right]\\
& = \cos\alpha - \cos\beta
\end{align*}
|
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Convergence of $\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\cdots$ Does the following series converge? If yes, what is its value in simplest form?
$$\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6} \right)^2+\left( \frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10} \right)^2+\dots$$
I have no idea how to start. Any hint would be really appreciated. THANKS!
|
Suggestion.
This is merely a suggestion that is too large to fit in a comment. Euler's integral formula for the harmonic numbers, $H_n=\int_0^1\frac{1-x^n}{1-x}\ \mathrm{d}x$, gives us a formula for the series as $a\to1^-$:
$$S=4\sum_{n=1}^{\infty}\left[f_a(n(n+1))-f_a(n(n-1))\right]^2$$
where $f_a(x)=\int_{0}^{a}\frac{t^{x+1}}{1-t^{2}}\ \mathrm{d}t$. The function, $f_a$, can also be given in terms of the hypergeometric function, as $\frac{a^{x+2}}{x+2}\cdot{_2F}_1\left(1,\frac{x}2+1;\frac{x}2+2;a^{2}\right)$.
Thus, it could potentially aid in finding an analytic solution to $S$ (or showing whether that would be impossible) by finding a closed form for the square of the difference of hypergeometric functions, $\left[f_a(n(n+1))-f_a(n(n-1))\right]^2$. However, no such solution has leapt out to me so far.
|
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|
Why does this process map every fraction to the golden ratio? Start with any positive fraction $\frac{a}{b}$.
First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$
Then add the (new) numerator to the denominator:
$$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$
So $\frac{2}{5} \rightarrow \frac{7}{5} \rightarrow \frac{7}{12}$.
Repeating this process appears to map every fraction to
$\phi$ and $\frac{1}{\phi}$:
$$
\begin{array}{ccccccccccc}
\frac{2}{5} & \frac{7}{5} & \frac{7}{12} & \frac{19}{12} & \frac{19}{31} & \frac{50}{31} & \frac{50}{81} & \frac{131}{81} & \frac{131}{212} & \frac{343}{212} & \frac{343}{555} \\
0.4 & 1.40 & 0.583 & 1.58 & 0.613 & 1.61 & 0.617 & 1.62 & 0.618 & 1.62 & 0.618 \\
\end{array}
$$
Another example:
$$
\begin{array}{ccccccccccc}
\frac{11}{7} & \frac{18}{7} & \frac{18}{25} & \frac{43}{25} & \frac{43}{68} & \frac{111}{68} & \frac{111}{179} & \frac{290}{179} & \frac{290}{469} & \frac{759}{469} & \frac{759}{1228} \\
1.57143 & 2.57 & 0.720 & 1.72 & 0.632 & 1.63 & 0.620 & 1.62 & 0.618 & 1.62 & 0.618 \\
\end{array}
$$
Q. Why?
|
First consider the sequence of every second fraction:
$$
\frac{a_{2n+2}}{b_{2n+2}} = \frac{a_{2n}+b_{2n}}{a_{2n}+2b_{2n}} = \frac{\frac{a_{2n}}{b_{2n}} +1}{\frac{a_{2n}}{b_{2n}} + 2} = f(\frac{a_{2n}}{b_{2n}})
$$
where $f(x)$ is defined as
$$
f(x) = \frac{x+1}{x+2} = 1 - \frac{1}{x+2}
$$
for $x \ge 0$.
Use the monotony of $f$ to show that $\left(\frac{a_{2n}}{b_{2n}}\right)_n$ is a monotonic and bounded sequence, and determine its limit $L$ as the (unique positive) fixed point of $f$.
Then consider the fractions with odd indices: $\frac{a_{2n}}{b_{2n}} \to L$ implies
$$
\frac{a_{2n+1}}{b_{2n+1}} = \frac{a_{2n} + b_{2n}}{b_{2n}} \to L + 1 \, .
$$
|
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|
Solve the equation $\cos^{-1}\frac{x^2-1}{x^2+1}+\tan^{-1}\frac{2x}{x^2-1}=\frac{2\pi}{3}$ $\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$
Let's first find the domain
$$-1<=\dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1}{x^2+1}>=-1 \text { and } \dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1+x^2+1}{x^2+1}>=0 \text { and } \dfrac{x^2-1-x^2-1}{x^2+1}<=0$$
$$\dfrac{2x^2}{x^2+1}>=0 \text { and } \dfrac{-2}{1+x^2}<=0$$
$$x\in R$$
$$x^2-1\ne0$$
$$x\ne\pm1$$
$$\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$$
$$\pi-\cos^{-1}\dfrac{1-x^2}{1+x^2}-\tan^{-1}\dfrac{2x}{1-x^2}=\dfrac{2\pi}{3}$$
$$\dfrac{\pi}{3}=\cos^{-1}\dfrac{1-x^2}{1+x^2}+\tan^{-1}\dfrac{2x}{1-x^2}$$
Substituting $x$ by $\tan\theta$
$$x=\tan\theta$$
$$\tan^{-1}x=\theta$$
$$\theta\in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)-\{-\dfrac{\pi}{4},\dfrac{\pi}{4}\}$$
$$\dfrac{\pi}{3}=\cos^{-1}\dfrac{1-\tan^2\theta}{1+\tan^2\theta}+\tan^{-1}\dfrac{2\tan\theta}{1-\tan^2\theta}$$
$$\dfrac{\pi}{3}=\cos^{-1}(\cos2\theta)+\tan^{-1}(\tan2\theta)$$
$$2\theta\in(-\pi,\pi)-\{-\dfrac{\pi}{2},\dfrac{\pi}{2}\}$$
Now we break the range of $2\theta$ into various parts:-
Case $1$: $2\theta\in\left(-\pi,-\dfrac{\pi}{2}\right)$,$\theta\in\left(-\dfrac{\pi}{2},-\dfrac{\pi}{4}\right)$
$$\dfrac{\pi}{3}=2\pi+2\theta+\pi+2\theta$$
$$-\dfrac{8\pi}{3}=4\theta$$
$$-\dfrac{2\pi}{3}=\theta$$
But it is not the range of $\theta$ we assumed
Case $2$: $2\theta\in\left(-\dfrac{\pi}{2},0\right]$,$\theta\in\left(-\dfrac{\pi}{4},0\right]$
$$\dfrac{\pi}{3}=-2\theta+2\theta$$
$$\dfrac{\pi}{3}=0 \text { not possible }$$
Case $3$: $2\theta\in\left(0,\dfrac{\pi}{2}\right)$,$\theta\in\left(0,\dfrac{\pi}{4}\right)$
$$\dfrac{\pi}{3}=2\theta+2\theta$$
$$\dfrac{\pi}{12}=\theta$$
It is coming in the range of $\theta$, so its a valid solution.
$$\tan^{-1}x=\dfrac{\pi}{12}$$
$$x=\tan\dfrac{\pi}{12}$$
$$x=2-\sqrt{3}$$
Case $4$: $2\theta\in\left(\dfrac{\pi}{2},\pi\right)$, $\theta\in\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)$
$$\dfrac{\pi}{3}=2\pi-2\theta+2\theta-\pi$$
$$\dfrac{\pi}{3}=\pi \text { not possible }$$
So only solution is $2-\sqrt{3}$, but actual answer is $2-\sqrt{3}, \sqrt{3}$
|
Following your approach, you are letting $x=\tan(\theta)$ for some $\theta \in (-\pi/2,\pi/2)$, and arriving at $\frac{\pi}{3}=\arccos(\cos(2\theta))+\arctan(\tan(2\theta))$. The important thing is that $\arccos$ maps to $[0,\pi]$ while $\arctan$ maps to $(-\pi/2,\pi/2)$. Thus:
*
*if $2 \theta \not \in [0,\pi]$, then $\arccos(\cos(2\theta)) \neq 2\theta$. In this situation that occurs when $2\theta \in (-\pi,0)$, in which case the arccosine is $-2\theta$ (which geometrically is the corresponding point in the upper half plane).
*if $2\theta \not \in (-\pi/2,\pi/2)$ then $\arctan(\tan(2\theta)) \neq 2\theta$. This occurs here when $2\theta \in (-\pi,-\pi/2)$ and when $2\theta \in (\pi/2,\pi)$. In the first case you have a positive value of tangent in the third quadrant, so you need the corresponding point in the first quadrant, which is $2\theta+\pi$ (represented in $(-\pi/2,\pi/2)$). In the second case, you have a negative value of tangent in the second quadrant, so you need the corresponding point in the fourth quadrant (represented in $(-\pi/2,\pi/2)$). This is $2\theta-\pi$.
Thus you have two cases for the arccosine: $2\theta \in [0,\pi)$ in which case $\arccos(\cos(2\theta))=2\theta$ and $2\theta \in (-\pi,0]$ in which case $\arccos(\cos(2\theta))=-2\theta$. You have three cases for the arctangent: $2 \theta \in (-\pi/2,\pi/2)$ gives $\arctan(\tan(2\theta))=2\theta$, $2 \theta \in (-\pi,-\pi/2)$ gives $\arctan(\tan(2\theta))=2\theta+\pi$, and $2\theta \in (\pi/2,\pi)$ which gives $\arctan(\tan(2\theta))=2\theta-\pi$. Overall this results in four cases:
*
*$2\theta \in (-\pi,-\pi/2] \Rightarrow \arccos(\cos(2\theta))+\arctan(\tan(2\theta))=\pi$
*$2\theta \in (-\pi/2,0] \Rightarrow \arccos(\cos(2\theta))+\arctan(\tan(2\theta))=0$
*$2\theta \in [0,\pi/2) \Rightarrow \arccos(\cos(2\theta))+\arctan(\tan(2\theta))=4\theta$
*$2\theta \in [\pi/2,\pi) \Rightarrow \arccos(\cos(2\theta))+\arctan(\tan(2\theta))=4\theta-\pi$.
Case 3 gives $\theta=\pi/12$; case 4 gives $\theta=\pi/3$.
|
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|
If $x+y+xy=3$ , $y+z+yz=8$ and $z+x+xz=15$ then Find $6xyz$. If $x+y+xy=3$ , $y+z+yz=8$ and $z+x+xz=15$ then
Find $6xyz$.
How can I approach this problem?
I need some hints. Thanks.
|
For convenience, we just need to look at one equation, $$x+y+xy=3 \implies (x+1)(y+1)=4$$
hence we define
$a=x+1,b=y+1,c=z+1$
and we obtain
$ab=4,bc=9,ac=16$
You solve easily by that hint then.
|
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Stuck on a probability law problem I'm currently trying to solve a problem, I completed the first question but I am stuck at the second one, here is the problem:
One person roll a die until the result is a $1$, a second person toss a coin until he gets $3$ tails.
*
*How many tries are they going to make on average.
$X$ = number of die rolls
$Y$ = number of time we toss a coin
$X$ follows a geometric law with parameter $1/6$, $E(X) = 6$ and $V(X) = 30$. $Y$ follows a negative binomial law with $n = 3$ and $p = 1/2,$ $E(Y) = 3/(1/2) = 6$ and $V(Y) = n(1-p)/p^2 = 6.$
*What is the probability that they both stop at the same time $(p(X = Y)).$
I found $p(X=Y) = \sum_{k=3}^\infty p(X=k, Y=k) $ so the same sum with $p(X=k)p(Y=k)$ since they are independent. Then $p(X=k)p(Y=k) = (5/6)^{k-1} \cdot\frac16 \begin{pmatrix} k-1 \\ 2 \\ \end{pmatrix}
\left(\frac12\right)^{k-3}\left(\frac12\right)^3$
But now i don't know how to compute it so i can get the value of that probability? Any help would be very appreciated.
Thanks
|
To show that $$\sum_{k=3}^\infty\left[\left(\frac56\right)^{k-1}\cdot\frac16\binom{k-1}2\left(\frac12\right)^{k-3}\left(\frac12\right)^3\right]= \frac{25}{343},\tag1$$ it suffices to show that $$\sum_{k=3}^\infty\left(\frac5{12}\right)^k\binom{k-1}2= \frac{125}{343}\tag2$$ after combining constants. The binomial term can be rewritten as $(k-1)(k-2)/2$ so that $(2)$ is equivalent to $$\sum_{k=3}^\infty\left(\frac5{12}\right)^k(k-1)(k-2)= \frac{250}{343}\tag3.$$ Now the LHS can be split to give \begin{align}\small\sum_{k=3}^\infty\left(\frac5{12}\right)^k(k-1)(k-2)&=\small\sum_{k=3}^\infty k(k-1)\cdot\left(\frac5{12}\right)^k-2\sum_{k=3}^\infty k\cdot\left(\frac5{12}\right)^k+2\sum_{k=3}^\infty\left(\frac5{12}\right)^k\\&=\small\left(\frac5{12}\right)^2\sum_{k=3}^\infty k(k-1)\cdot\left(\frac5{12}\right)^{k-2}-2\cdot\frac5{12}\sum_{k=3}^\infty k\cdot\left(\frac5{12}\right)^{k-1}+\frac{125}{504}\tag4\end{align} using standard geometric series. Notice that for $|a|<1$, we have \begin{align}\sum_{k=3}^\infty k(k-1)\cdot a^{k-2}&=\sum_{k=3}^\infty\frac{d^2}{da^2} a^{k}=\frac{d^2}{da^2}\sum_{k=3}^\infty a^{k}=\frac{d^2}{da^2}\left[-\frac{a^3}{a-1}\right]=-2-\frac2{(a-1)^3}\end{align} and \begin{align}\sum_{k=3}^\infty k\cdot a^{k-1}&=\sum_{k=3}^\infty\frac{d}{da} a^{k}=\frac{d}{da}\sum_{k=3}^\infty a^{k}=\frac{d}{da}\left[-\frac{a^3}{a-1}\right]=-\frac{a^2(2a-3)}{(a-1)^2}.\end{align} Letting $a=5/12$, equation $(4)$ becomes $$\small\sum_{k=3}^\infty\left(\frac5{12}\right)^k(k-1)(k-2)=\frac{25}{144}\cdot\left(-2-\frac2{\left(\frac5{12}-1\right)^3}\right)-\frac56\cdot\left(-\frac{\left(\frac5{12}\right)^2\left(2\cdot\frac5{12}-3\right)}{\left(\frac5{12}-1\right)^2}\right)+\frac{125}{504}=\frac{250}{343}$$ which proves $(1)$.
|
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Solve $2 \cos^2 x+ \sin x=1$ for all possible $x$ $2\cos^2 x+\sin x=1$
$\Rightarrow 2(1-\sin^2 x)+\sin x=1$
$\Rightarrow 2-2 \sin^2 x+\sin x=1$
$\Rightarrow 0=2 \sin^2 x- \sin x-1$
And so:
$0 = (2 \sin x+1)(\sin x-1)$
So we have to find the solutions of each of these factors separately:
$2 \sin x+1=0$
$\Rightarrow \sin x=\frac{-1}{2}$
and so $x=\frac{7\pi}{6},\frac{11\pi}{6}$
Solving for the other factor,
$\sin x-1=0 \Rightarrow \sin x=1$
And so $x=\frac{\pi}{2}$
Now we have found all our base solutions, and so ALL the solutions can be written as so:
$x= \frac{7\pi}{6} + 2\pi k,\frac{11\pi}{6} + 2\pi k, \frac{\pi}{2} + 2\pi k$
|
Your method's fine, the answer's right. The only improvement I can suggest is to make the definition of "base solution" clear upfront. Each "and so" acts as if a specific value of $\sin x$ has finitely many solutions rather than finitely many per period, so before you obtain them you should mention a restriction to $[0,\,2\pi)$ and then extend to $\Bbb R$ at the end.
|
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|
Evaluate $\lim\limits_{n \to \infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}}$
$$\lim_{n \to
\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}}$$
Maybe we can believe with assurance that
$$\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}}\approx\sum_{k=1}^{n}\frac{1}{\sqrt{n-k}\cdot\sqrt{k}}=\frac{1}{n}\sum_{k=1}^n\frac{1}{\sqrt{\frac{k}{n}-\left(\frac{k}{n}\right)^2}}.$$
Thus, we can obtain
$$\lim_{n \to
\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}}=\int_0^1 \frac{{\rm d}x}{\sqrt{x-x^2}}=\pi$$
This is true? If so, how to prove it rigorously?
|
You can turn the sum into two Riemann sums each of which converges to $\frac \pi 2$:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}} = \sum_{k=1}^{n}\frac{n+1-k + k}{\sqrt{n+1-k}\cdot\sqrt{k}}\cdot \frac 1{n+1}$$
$$=\frac 1{n+1}\sum_{k=1}^{n}\sqrt{\frac{n+1-k}{k}} + \frac 1{n+1}\sum_{k=1}^{n}\sqrt{\frac{k}{n+1-k}}$$
$$= \underbrace{\frac 1{n+1}\sum_{k=1}^{n}\sqrt{\frac{1}{\frac{k}{n+1}}-1}}_{\stackrel{n\to \infty}{\rightarrow}\int_0^1\sqrt{\frac 1x-1}dx=\frac \pi2} + \underbrace{\frac 1{n+1}\sum_{k=1}^{n}\sqrt{\frac{1}{\frac{1}{\frac{k}{n+1}}-1}}}_{\stackrel{n\to \infty}{\rightarrow}\int_0^1\sqrt{\frac{1}{\frac 1x -1}}dx=\frac \pi2}$$
|
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|
Show that $\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$ Let $ m_a, m_b $ and $ m_c $ be the medians relative to the $ a, b, c $ sides of a triangle, show that:
$$\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$$
What i tried:
ust use Stewart’s theorem. We have $m=n=a/2,d=m_a$ so
$$\frac{a^3}{4}+m_a^2a=\frac{ab^2}{2}+\frac{ac^2}{2}\implies m_a^2=\frac{2b^2+2c^2-a^2}{4},$$and likewise for the other two medians.
How to proceed?
|
Note,
$$m_a^4+m_b^4+m_c^4 = (m_a^2+m_b^2+m_c^2 )^2 - 2(m_a^2m_b^2+m_b^2m_c^2+m_c^2m_a^2) $$
Then, evaluate,
$$m_a^2+m_b^2+m_c^2 = \frac{2b^2+2c^2-a^2}{4} +\frac{2c^2+2a^2-b^2}{4} +\frac{2a^2+2b^2-c^2}{4} = \frac34 (a^2+b^2+c^2)$$
$$m_a^2m_b^2+m_b^2m_c^2+m_c^2m_a^2 = \frac9{16}(a^2b^2+b^2c^2+c^2a^2)$$
Thus,
$$\frac{m_a^4+m_b^4+m_c^4}{a^4+b^4+c^4} = \frac9{16}\frac{(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)}{a^4+b^4+c^4} =\frac{9}{16}$$
|
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|
Mathematical Induction Involving The Floor Function I need to prove the identity $\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$ for all natural numbers $n$.
I wanted to use mathematical induction. The identity is true for $n=1$. Then, I assume $$\lfloor \sqrt{k}+\sqrt{k+1}\rfloor=\lfloor\sqrt{4k+2}\rfloor$$
and need to show that it is also true for $n=k+1$
$$\lfloor \sqrt{k+1}+\sqrt{k+2}\rfloor=\lfloor\sqrt{4k+6}\rfloor.$$
I thought about the inequality $m\leq\sqrt{4k+6}<m+1$ for some integer $m$, and the same for the LHS, but this doesn't seem to help and I don't have other ideas.
|
Here is an (maybe) easier approach.
To prove $\lfloor\sqrt n+\sqrt {n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$, and it's easy to notice that $\lfloor\sqrt{4n+2}\rfloor>\lfloor\sqrt n+\sqrt {n+1}\rfloor$ and that $\lfloor\sqrt{4n+2}\rfloor-\lfloor\sqrt n+\sqrt {n+1}\rfloor<1$, it suffices to prove that there's no integer between $\sqrt n+\sqrt {n+1}$ and $\sqrt{4n+2}$.
We suppose that there is an integer $k$ between $\sqrt n+\sqrt {n+1}$ and $\sqrt{4n+2}$. Then
\begin{equation*}
\sqrt n+\sqrt {n+1}\leq k\leq\sqrt{4n+2}\\
n+n+1+2\sqrt{n(n+1)}\leq k^2\leq 4n+2
\end{equation*}
Since $\sqrt{n(n+1)}>n$, we have
\begin{equation*}
4n+1<k^2\leq4n+2
\end{equation*}
So the only possibility is $k^2=4n+2$. So $k=\sqrt{4n+2}=2\sqrt{n+1/2}=2\sqrt{(2n+1)/2}$. So
\begin{equation}
\sqrt{\frac{2n+1}{2}}=x\quad (1),\quad or\\
\sqrt{\frac{2n+1}{2}}=y+\frac{1}{2}\quad(2)
\end{equation}
for integers $x$ and $y$. Since $2n+1$ is an odd number, $(2n+1)/2$ cannot be an integer, so $(1)$ cannot be the case. For $(2)$, we notice that $(y+1/2)^2=(2y+1)^2/4$, which is an odd number over $4$. But $(2n+1)/2$ is an odd number over $2$, namely an even number over $4$, thus they can not be equal. We have the desired contradiction.
|
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|
How to remove square roots from denominator in $\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}$? I have this math problem: Remove all square roots in the denominator of
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = ?$
The obvious solution is to multiply the fraction with $\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$.
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} \cdot \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{\sqrt{x}\sqrt{x} - \sqrt{x}\sqrt{y} + \sqrt{y}\sqrt{x}-\sqrt{y}\sqrt{y}}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - \sqrt{x}\sqrt{y} + \sqrt{y}\sqrt{x}-y}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{(\sqrt{x}-\sqrt{y})(\sqrt{x}-\sqrt{y})}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{\sqrt{x}\sqrt{x}-\sqrt{x}\sqrt{y} - \sqrt{y}\sqrt{x}+\sqrt{y}\sqrt{y}}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{x-\sqrt{x}\sqrt{y} - \sqrt{y}\sqrt{x}+y}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{x-2\sqrt{x}\sqrt{y}+y}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{(\sqrt{x} - \sqrt{y})^2}$
Then I tried to multiply the fraction by $\frac{(\sqrt{x} - \sqrt{y})^2}{(\sqrt{x} - \sqrt{y})^2}$.
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{(\sqrt{x} - \sqrt{y})^2} \frac{(\sqrt{x} - \sqrt{y})^2}{(\sqrt{x} - \sqrt{y})^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{x - y}{x - 2\sqrt{x}\sqrt{y} + y} \frac{(\sqrt{x} - \sqrt{y})^2}{x - 2\sqrt{x}\sqrt{y} + y}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-x2\sqrt{x}\sqrt{y}+xy -2\sqrt{x}\sqrt{y}x + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y + yx - y2\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-x2\sqrt{x}\sqrt{y} -2\sqrt{x}\sqrt{y}x + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y - y2\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-2x\sqrt{x}\sqrt{y} -2x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y - y2\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2\sqrt{x}\sqrt{y}y - y2\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 2y\sqrt{x}\sqrt{y} - 2y\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + (2\sqrt{x}\sqrt{y})(2\sqrt{x}\sqrt{y}) - 4y\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + 2\cdot2\sqrt{x}\sqrt{x}\sqrt{y}\sqrt{y} - 4y\sqrt{x}\sqrt{y}+y^2}$
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = \frac{(x - y)(\sqrt{x} - \sqrt{y})^2}{x^2-4x\sqrt{x}\sqrt{y} + 4xy - 4y\sqrt{x}\sqrt{y}+y^2}$
This seems like a dead end. Where exactly did I make a mistake?
|
Instead multiply by $\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}$.
|
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|
Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^2+n}}{n} \rightarrow 1$, so $\lim d_n=1 $
So, $\lim\sqrt[n]{n^2+n} = \lim (1+d_n)^3 = \lim(1+3d_n^2+3d_n+d_n^3) =8$
However, $\sqrt[n]{n^2+n}$ tends to $1$. Where is the problem of my solution ? Can you give me a hint of how can I solve it with Bernoulli's rule ?
|
First, we will show that $\lim_{n\to\infty}n^{1/n}=1$. We define the sequence $x_n$ as
$$x_n=n^{1/n}-1\tag1$$
ASIDE:
Note that $n^{1/n}\ge 1$for $n\ge1$. This is true since if $0\le y\le 1$, then $0\le y^n\le 1$ for all $n\in \mathbb{N}$.
From $(1)$, it is easy to see that $(1+x_n)^n=n$. Then, using the binomial theorem, we see that
$$\begin{align}
n&=(1+x_n)^n\\\\
&=\sum_{k=0}^n \binom{n}{k}x_n^k\\\\
&\ge \binom{n}{2}\,x_n^2\\\\
&=\frac{n(n-1)}{2}\,x_n^2 \tag2
\end{align}$$
from which we conclude that
$$\begin{align}
0\le x_n \le \sqrt{\frac{2}{n-1}}\tag3
\end{align}$$
Applying the squeeze theorem to $(3)$ reveals
$$\lim_{n\to\infty}n^{1/n}=1\tag4$$
Finally, we write
$$\left(n^2+n\right)^{1/n}=n^{1/n}\left(1+\frac1n\right)^{1/n}\tag5$$
Inasmuch as the limit of the second term on the right-hand side of $(5)$ is not of indeterminate form, rather is of the form $1^1=1$, we conclude from using $(4)$ that
$$\lim_{n\to\infty}\left(n^2+n\right)^{1/n}=1$$
|
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|
Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Tried dividing by $y$ to get $$y^\prime+\frac{x}{y}=\frac{\sqrt{x^2+y^2}}{y}$$
$$(y^\prime)^2+2\frac{x}{y}+\frac{x^2}{y^2}=\frac{x^2+y^2}{y^2}$$
$$(y^\prime)^2+2\frac{x}{y}=1$$
$$y^\prime=\sqrt{1-2\frac{x}{y}}$$
Tried using $v=\frac{y}{x}$
$$y^\prime=v^\prime x+v$$
$$v^\prime x + v=\sqrt{1-2v^{-1}}$$
Not sure what to do now or if I've even done something right until now.
|
Let $r=\sqrt{x^2+y^2}$, then the equation becomes
$$
rr'=r
$$
which means that $r'=1$; that is, $r=x+c$ which is the horizontal parabola
$$
y^2=2cx+c^2
$$
|
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|
Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$
Let's see what happens:-
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$
In the same way for further terms, we will get $0$
Let's also confirm for general term
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$
So the whole expression $E$ will be zero
But actual answer is $1$
Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$
Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?
I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.
But I got the following counter thought:-
$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$
As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$
$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$
$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$
Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.
But surprisingly $\dfrac{1}{3}$ is the correct answer.
I am feeling very confused in these two things. Please help me.
|
Observe that
$$\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots+\frac{n}{n^2+n}$$ lies between
$$\frac{n}{n^2+n}+\frac{n}{n^2+n}+\cdots+\frac{n}{n^2+n}=\frac{n}{n+1}
$$ and
$$\frac{n}{n^2}+\frac{n}{n^2}+\cdots+\frac{n}{n^2}=1.$$
|
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|
Interval for which $x^{12}-x^9+x^4-x+1>0$
The largest interval for which $x^{12}-x^9+x^4-x+1>0$ is :
$$
x^{12}-x^9+x^4-x+1>0\\
x(x^3-1)(x^8+1)+1>0\\
x(x-1)(x^2+x+1)(x^8+1)+1>0
$$
I can see that this is satisfied for $x\in(-\infty,0]\cup[1,\infty)$. But how can I verify the above function is greater than zero when $x\in(0,1)$ ?
|
$$x^{12}+x^4(1-x^5)+(1-x)>0$$ for $0<x<1$.
|
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|
Evaluate in closed form: $ \sum_{m=0}^\infty \sum_{n=0}^\infty \sum_{p=0}^\infty\frac{m!n!p!}{(m+n+p+2)!}$
Evaluate in closed form:
$$ \sum_{m=0}^\infty \sum_{n=0}^\infty \sum_{p=0}^\infty\frac{m!n!p!}{(m+n+p+2)!}$$
I tried to use same method for similar question and two variables.Was not able to get the final answer. Question proposed by Jalil Hajimir --
Infinite Series $\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m!\:n!}{(m+n+2)!}$
|
Playing with Mathematica a bit and you will see that the sum is
$$
S(3)=\sum_{p=0}^\infty \sum_{m=0}^\infty \sum_{n=0}^\infty\frac{m!n!p!}{(m+n+p+3)!}
=
\frac{1}{12} \left(3 \zeta (3)+2 \pi ^2 \log (2)\right)+
\sum _{n=1}^{\infty } \left(\frac{\psi ^{(1)}(n)}{4 n (2 n-1)}-\frac{\psi ^{(1)}\left(n+\frac{1}{2}\right)}{4 n (2 n-1)}\right)
$$
where $\psi$ is the digamma function. It does not look like it can be further simplified.
Curiously though, a slightly different sum do has a closed form
$$
S(2)=\sum_{p=0}^\infty \sum_{m=0}^\infty \sum_{n=0}^\infty\frac{m!n!p!}{(m+n+p+2)!}
=\pi^2/4
$$
To see this, one can check
$$
S(2)
=\sum_{p=0}^\infty \sum_{m=0}^\infty\frac{m! p!}{(m+p+1)^2 (m+p)!}
=\sum_{p=0}^\infty\frac{\, _3F_2(1,1,p+1;p+2,p+2;1)}{(p+1)^2}
=:\sum_{p=0}^\infty a_p
$$
Mathematica cannot simplify this anymore, but look at the first few terms
$$
\frac{\pi ^2}{6},\frac{1}{6} \left(12-\pi ^2\right),\frac{1}{6} \left(\pi ^2-9\right),\frac{1}{18} \left(31-3 \pi ^2\right),\frac{1}{72} \left(12 \pi ^2-115\right),\frac{3019-300 \pi ^2}{1800},\frac{1}{600} \left(100 \pi ^2-973\right),\frac{48877-4900 \pi ^2}{29400}
$$
The pattern is quite obvious, we should have
$$
a_{2p}+a_{2p+1}= \frac{2}{(2 p+1)^2}, \quad p \ge 0.
$$
So
$$
S(2)=\sum_{p \ge 0} \frac{2}{(2 p+1)^2} = \pi^2/4.
$$
|
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|
Prove $x^2+\frac{1}{x^2}=2\cos(2\theta)$ Prove $x^2+\frac{1}{x^2}=2\cos(2\theta)$ and $x^3+\frac{1}{x^3}=2\cos(3\theta)$
knowing that there exist a number $x$ given angle $\theta$ such that $x+\frac{1}{x}=2\cos(\theta)$
Doesn't really know how to start this problem, thought that I would some how need to use the double angle identities
|
$$\left(x+\frac 1x\right)^2 = x^2+2+\frac 1{x^2} = 4\cos^2\theta \\\implies x^2+\frac{1}{x^2} = 2(2\cos^2\theta-1) = 2\cos2\theta$$
Similarly
$$\left(x+\frac 1x\right)^3 = x^3 + 3\left(x+\frac 1x\right) + \frac{1}{x^3} = x^3+\frac{1}{x^3} + 6\cos\theta = 8\cos^3\theta$$
$$ \implies x^3+\frac{1}{x^3} = 2(4\cos^3\theta-3\cos\theta) = 2\cos3\theta$$
|
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|
Solve equation $(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$ Find the value of $x$ given the equation,
$$(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$$
I think they are powers of $ \frac {1} {\Phi} $ or $ \Phi $. In case I think the first would be $ \Phi ^ 3 $ and the second $ (\frac {1} {\Phi}) ^ 4 $. Is it true?
How to solve the problem?
|
There are two solutions (see below).
Here is how we can prove this fact in a rigorous way. Let :
$$f(x):=(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}\tag{1}$$
which has the form
$$f(x)=a^x+2b^x \ \text{with} \ a>1, b<1.\tag{2}$$
Its derivative being $f'(x)=\ln(a) a^x +2\ln(b) b^x,$
we have $$f'x)<0 \ \iff \ \ln(a) a^x +2\ln(b) b^x<0$$
$$\iff \left(\frac{a}{b}\right)^x<\underbrace{-2\dfrac{\ln b}{\ln a}}_C$$
$$\iff x<\dfrac{\ln C}{\ln\left(\frac{a}{b}\right)}$$
(the RHS, named $C$, being a positive number because $\ln a>0$ whereas $\ln b<0$).
Therefore, $f$ is first decreasing, till a certain value $x_0$ then increasing.
Besides, we have, taking (2) into account :
$$\lim_{x \to -\infty}f(x)=\lim_{x \to +\infty}f(x)=+\infty$$
Therefore, as $f(0)=3$, the initial equation $f(x)=5$ has two solutions. One of them is clearly
$$x=2$$
(indeed $f(2)=(2+ \sqrt{5})+2\sqrt{\frac{7 - 3 \sqrt{5}}{2}}=2+\sqrt{5}+3-\sqrt{5}=5$.)
The other one is negative :
$$x\approx-1.7863876...$$
(see figure below)
|
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|
Generating Function & Sequence Find the generating functions of the sequences
2, 1, 2, 1, 2, 1, . . .
I get $\frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2}$
But the solution ends up with $\frac{2}{1-x^2} + \frac{x}{1-x^2} = \frac{2+x}{1-x^2}$.
The solutions starts with $\sum_{n\ge 0} (2)x^{2n}+\sum_{n\ge 0} (1) x^{2n+1}$
I couldn't come up with anything like that. I feel like I'm confused with something.
|
As an alternative to the other answers, you can simply use a variant of one function, that being
$$\frac{1}{1-x^2}=\sum_{n=0}^{\infty} x^{2n}$$
If you consider that the $2$'s are even indexed (assuming in the sequence $\{a_n\}$ the first term is $a_0$) and the $1$'s are odd indexed, you have that
$$\frac{2}{1-x^2}$$
will generate the $2$'s and
$$\frac{x}{1-x^2}$$
will generate the $1$'s. This is because
$$\frac{1}{1-x^2}=1+x^2+x^4+...$$
and so
$$\frac{x}{1-x^2}=x(1+x^2+x^4+...)=x+x^3+x^5+...$$
Thus, you sequence will be generated by
$$f(x)=\frac{2+x}{1-x^2}$$
|
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|
Verify $JJ' = 1$, given $x = e^u \cos v$, $y = e^u \sin v$. I was trying to solve this jacobian problem but I'm not getting the required solution i.e $JJ' = 1$
I'm getting $J = e^{2u}$ and $J' = \frac{x^2 - y^2}{x^2 + y^2}$ So multiplying both
$J*J'$ will not return 1 in my case.
What am I doing wrong? Here's my solution,
|
You're almost there. First, a miscalculation: your $\frac{\partial v}{\partial x}$ has the wrong sign. In particular: we indeed have
$$
u = \frac 12 \log(x^2 + y^2), \quad v = \tan^{-1}(y/x).
$$
It follows that
$$
\frac{\partial u}{\partial x} = \frac{x}{x^2 + y^2}, \quad
\frac{\partial u}{\partial y} = \frac{y}{x^2 + y^2},
$$
which you calculated correctly. However,
$$
\frac{\partial v}{\partial x} = \frac{-y}{x^2} \cdot \frac{1}{1 + (y/x)^2} =
\frac{-y}{x^2 + y^2}, \qquad
\frac{\partial v}{\partial y} = \frac{1}{x} \cdot \frac{1}{1 + (y/x)^2} = \frac{x}{x^2 + y^2}.
$$
Now, in order to see that $JJ' = 1$, write both Jacobians as a function of the same set of variables. That is, either as a function of $x$ and $y$ or as a function of $u$ and $v$.
If we write $J'$ as a function of $u$ and $v$, then we find
$$
J' = \frac{x^2 + y^2}{(x^2 + y^2)^2} = \frac{1}{x^2 + y^2} = \frac{1}{[e^u \cos v]^2 + [e^u \sin v]^2} = \frac{1}{e^{2u}}.
$$
In greater detail: we have $v(x,y) = \tan^{-1}(y/x)$. We find the derivative of this using the chain rule. Because $\frac{d}{dt}\tan^{-1}(t) = \frac{1}{1 + t^2}$, we have
$$
\frac{\partial v}{\partial x} = \frac{1}{1 + (y/x)^2} \cdot \frac{\partial}{\partial x}[(y/x)].
$$
To calculate the partial derivative of $y/x$ with respect to $x$, we use the power rule. In particular:
$$
f(x,y) = y/x = y \cdot x^{-1} \implies
\frac{\partial f}{\partial x} = -y x^{-2} = -y \cdot \frac{1}{x^2} = \frac{-y}{x^2}.
$$
|
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|
$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$ Any suggestions how to solve: $$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$$
I can simplify the system and get a homogeneous polynomial of degree $2$, but I think it must have an easier way.
|
Hint Let
$$a=x+y \\ b=x-y$$
Then, the equations become
$$ab=7 \\
3a^2+b^2=4 \cdot 37$$
Thus
$$3a^2+\frac{49}{a^2}=148$$
This is a quadratic in $a^2$.
|
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|
$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$ I've tried to solve the limit
$$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$$
but I'm not sure.
$ (1+ \frac{3}{n^2+n^4})^n = \sqrt [n^3]{(1+ \frac{3}{n^2+n^4})^{n^4}} \sim \sqrt [n^3]{(1+ \frac{3}{n^4})^{n^4}} \sim \sqrt [n^3]{e^3 } \rightarrow 1$
Is it right?
|
You may also just use the binomial formula and squeeze by applying
*
*$(\star)$: $\frac 1{n^k}\binom nk \leq \frac 1{k!}$ for $0\leq k\leq n$
Hence,
$$1\leq \left(1 +\frac{3}{n^2+n^4}\right)^n\leq \left(1 +\frac{3}{n^4}\right)^n$$
$$\leq 1+\sum_{k=1}^n\left( \frac{3}{n^4}\right)^k\binom nk \stackrel{(\star)}{\leq} \sum_{k=1}^n\left( \frac{3}{n^3}\right)^k\frac 1{k!}$$
$$\leq 1 + \frac 1{n^3}\sum_{k=1}^n\frac{3^k}{k!} \leq 1+\frac{e^3-1}{n^3}\stackrel{n\to\infty}{\longrightarrow}1$$
|
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|
Find the inverse element of an unknown complex root. I got asked if I could solve the following task: Let $f:=X^3-3X+4 \in \mathbb{Q}[X]$ be the polynomial with $\alpha \in \mathbb{C}$, $f(\alpha)=0$ and $K=\mathbb{Q}(\alpha)$.
Furthermore, let $\beta := \alpha^2+\alpha+1$.
Find the inverse of $\beta$.
I tried solving it in a pretty naive way: 1=$\beta*\beta^{-1}=\beta^{-1}(\alpha^2+\alpha+1).$
Therefore if $\beta^{-1}*\alpha=0$ I should have $\beta^{-1}$.
If $\beta^{-1}:=\alpha^2-3+4*\alpha^{-1}$ I should be done.
However I don't think this is the correct answer, since the task further said, that $\beta^{-1}$ is of the form $\beta^{-1}=a_0+a_1\alpha+a_2\alpha^2$ for some $a_0,a_1,a_2 \in \mathbb{Q}$.
Am i wrong?
And if so, where did I go wrong?
Thank you in advance!
|
Since $\alpha$ is a root of $x^3-3x+4$, we have that $\alpha^3-3\alpha+4=0$, so
$$\alpha^3=3\alpha-4.$$
Since, $\beta=\alpha^2+\alpha+1$ we have that $\beta\alpha=\alpha^3+\alpha^2+\alpha$. Hence
$$\beta\alpha=\alpha^2+4\alpha-4.$$
Similarly, $\beta\alpha^2=\alpha^3+4\alpha^2-4\alpha.$ Hence
$$\beta\alpha^2=4\alpha^2-\alpha-4.$$
Now we can find a system of linear equations to solve to find $\beta^{-1}$. We have that $1=\beta\left(a_0+a_1\alpha+a_2\alpha^2\right)$. So
$$1=a_0\beta+a_1\beta\alpha+a_2\beta\alpha^2=a_0\left(\alpha^2+\alpha+1\right)+a_1\left(\alpha^2+4\alpha-4\right)+a_2\left(4\alpha^2-\alpha-4\right).$$
Hence
$$1=\left(a_0+a_1+4a_2\right)\alpha^2+\left(a_0+4a_1-a_2\right)\alpha+\left(a_0-4a_1-4a_2\right).$$
Which gives us the following system:
$$a_0+a_1+4a_2=0$$
$$a_0+4a_1-a_2=0$$
$$a_0-4a_1-4a_2=1.$$
It follows that $a_0=\dfrac{17}{49}$, $a_1=\dfrac{-5}{49}$, and $a_2=\dfrac{-3}{49}$. So
$$\dfrac{1}{\alpha^2+\alpha+1}=\dfrac{17-5\alpha-3\alpha^2}{49}.$$
|
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|
Number of ordered pairs of $A,B$ in Probability
Let $|X|$ denote the number of elements in a set $X,$
Let $S = \{1,2,3,4,5,6\}$ be a sample space,
where each element is equally likely to occur.
If $A$ and $B$ are independent events associated with
$S,$
Then the number of ordered pairs $(A,B)$
such that $1 \leq |B| < |A|,$ equals
what i try
Let number of elements in $A,B,A \cap B$ is $x,y,z$ respectively
conditions given as $1\leq y\leq x.$
For $2$ Independent events $A,B$
is $\displaystyle P(A \cap B)=P(A)\cdot P(B)\Rightarrow \frac{z}{6}=\frac{x}{6}\cdot \frac{y}{6}\Rightarrow z=xy/6$
|
The condition should be $1\le y\lt x$, not $\le$.
Since $z=xy/6$, $x$ or $y$ must be divisible by $2$ and by $3$, and $xy\le6$. With $1\le y\lt x$, that allows only two solutions: $x=6$, $y=1$ and $x=3$, $y=2$.
In the first case, we have $6$ choices for the element in $y$.
In the second case we have $z=3\cdot2/6=1$, so there must be exactly one common element. That gives us $6$ choices for the common element, then $5$ choices for the other element in $y$ and then $\binom42=6$ choices for the other two elements in $x$.
Thus the total number of ordered pairs is $6+6\cdot5\cdot6=186$.
|
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|
Find the area bounded by the curve $\left(\frac{x}a+\frac{y}b\right)^5=\frac{x^2y^2}{c^4}$ Find the area bounded by the curve $\left(\dfrac{x}a+\dfrac{y}b\right)^5=\dfrac{x^2y^2}{c^4}$.
Let $x=ar\cos\varphi$, $y=br\sin\varphi$, then
$$r=\dfrac{a^2b^2\cos^2\varphi\sin^2\varphi}{c^4(\cos\varphi+\sin\varphi)^5}.$$
But how to find the polar angle?
|
It's probably better to use the substitution
$$\begin{cases} x = ar\cos^2\theta \\ y = br\sin^2\theta \\ \end{cases} $$
This has a Jacobian of $abr\sin(2\theta)$. From looking at the equation, it only encloses a loop in the first quadrant. Then plugging in we get that
$$ r^5 = \frac{a^2b^2}{c^4}r^4\sin^4\theta\cos^4\theta \implies r = \frac{a^2b^2}{16c^4}\sin^42\theta$$
Now we can set up our integral
$$\int_0^{\frac{\pi}{2}} \int_0^{\frac{a^2b^2}{16c^4}\sin^42\theta} abr\sin 2\theta \:dr \: d\theta = \frac{a^5b^5}{512c^8} \int_0^{\frac{\pi}{2}} \sin^92\theta \: d\theta = \frac{a^5b^5}{1024c^8} \int_{-1}^1 (1-x^2)^4dx$$
The last integral evaluates to $\frac{256}{315}$, making the value of the area
$$A = \iint_{\text{Loop}} 1 \: dA = \frac{a^5b^5}{1260c^8}$$
|
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Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$
My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos^23x+\cos x\cos3x\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\dfrac{1+\cos6x}{2}+\dfrac{1}{2}\left(\cos4x+\cos2x\right)\right)}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{4}\left(1+\cos6x+\cos4x+\cos2x\right)}{x^2}$$
Applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-6\sin6x-4\sin4x-2\sin2x\right)}{2x}$$
Again applying L'Hospital as we have $\dfrac{0}{0}$ form
$$\lim_{x\to0}\dfrac{-\dfrac{1}{4}\left(-36\cos6x-16\cos4x-4\cos2x\right)}{2}=\dfrac{36+16+4}{8}=\dfrac{56}{8}=7$$
But actual answer is $3$, what am I messing up here?
|
Hint:
Method$\#1:$
For $x\to0,$
$$1-\cos2ax\cos2bx\cos2cx\cdots=1-(1-2\sin^2ax)(1-2\sin^2bx)(1-2\sin^2cx)\cdots$$
$$\approx2(ax)^2+2(bx)^2++2(cx)^2+\cdots+O(x^4)$$ as $\lim_{x\to0}\dfrac{\sin px}{px}=1$
Method$\#2:$
$$1-\cos2ax\cos2bx\cos2cx\cdots=\dfrac{1-(1-\sin^22ax)(1-\sin^22bx)(1-\sin^22cx)}{1+\cos2ax\cos2bx\cos2cx\cdots}$$
For $x\to0,$ the numerator $\approx(2ax)^2+(2bx)^2+(2cx)^2+\cdots+O(x^4)$
|
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|
Functional equation $ f(x+yf(x^2))=f(x)+xf(xy) $ I have trouble solving the following functional equation: $f:\mathbb{R}\to\mathbb{R}$ such that $f(x+yf(x^2))=f(x)+xf(xy)$.
I think $f(x)=x$ is the unique solution, especially, $f(x+y)=f(x)+f(y)$. But how do we show that?
|
The beginning steps are the same as in the answer of @MohsenShahriari, which I repeat for sake of completeness.
Our equation is $$f(x + yf(x^2)) = f(x) + xf(xy).\tag 0$$
Setting $y = 0$ in $(0)$, we get $xf(0) = 0$. Setting again $x = 1$ gives us $f(0) = 0$.
Setting $x = 1$ in $(0)$, we get $$f(1 + yf(1)) = f(1) + f(y).\tag 1$$
If $f(1) \neq 1$, then we may set $y = \frac{1}{1 - f(1)}$ in $(1)$ and get $f(1) = 0$. Then $(1)$ implies that $f(y) = 0$ for all $y\in \Bbb R$, which is one solution to the equation.
Now assume that $f(1) = 1$. Then $(1)$ becomes $$f(1 + y) = 1 + f(y).\tag 2$$
Using induction with $(2)$ gives $$f(n + y) = n + f(y), \forall n\in \Bbb Z_{> 0}.\tag 3$$
Setting $y = y - n$ in $(3)$, we get $f(-n + y) = -n + f(y), \forall n \in \Bbb Z_{> 0}$. Combined with $(3)$, we have $$f(n + y) = n + f(y), \forall n\in \Bbb Z.\tag 4$$
Setting $y = 0$ in $(4)$, we get $f(n) = n, \forall n \in \Bbb Z$.
For integer $n \neq 0$, setting $x = n$ and $y = \frac y n$ in $(0)$ gives $n + f(ny) = n + nf(y)$, hence we get $f(ny) = nf(y)$. In particular, for every rational number $q = \frac u v$ with $u, v \in \Bbb Z$, setting $y = q$ and $n = v$ gives $f(q) = q$.
Setting $x = -x$ in $(0)$, we get $$f(-x + y f(x^2)) = f(-x) - xf(-xy) = -f(x) + xf(xy).\tag 5$$
Taking difference of $(0)$ and $(5)$, we get $$f(x + yf(x^2)) - f(-x + yf(x^2)) = 2f(x).\tag 6$$
Now for any $t\neq 0$, we have $f(t^2)\neq 0$, otherwise setting $x = t$ and $y = \frac 1 t$ in $(0)$ would give $f(1) = 0$, contradicting our hypothesis.
Therefore, if $x\neq 0$, then we may set $y = \frac{x + y}{f(x^2)}$ in $(6)$ and get $f(2x + y) - f(y) = 2f(x) = f(2x)$. Setting again $x = \frac x 2$, we get $$f(x + y) = f(x) + f(y).\tag 7$$
Using $(7)$, we may rewrite $(0)$ and get $$f(yf(x^2)) = xf(xy).\tag 8$$
Setting $y = x$ in $(8)$, we get $f(xf(x^2)) = xf(x^2)$.
Setting $y = f(x^2)$ in $(8)$, we get $f(f(x^2)^2) = xf(xf(x^2)) = x^2f(x^2)$. Since $x^2$ can be any non-negative real number, we have $$f(f(x)^2) = xf(x), \forall x \geq 0. \tag 9$$
Setting $x = -x$ in $(9)$, we get $f(f(x)^2) = xf(x), \forall x \leq 0$. Combined with $(9)$, we get $f(f(x)^2) = xf(x), \forall x$.
Setting $x = f(x)$ in $(8)$, we get $f(yf(f(x)^2)) = f(x)f(f(x)y)$, hence $f(xyf(x)) = f(x)f(yf(x))$. When $f(x) \neq 0$, setting again $y = \frac y {f(x)}$ gives $f(xy) = f(x)f(y)$.
When $f(x) = 0$, we note that $f(x + 1) = f(x) + f(1) = 1 \neq 0$, hence we have $f((x + 1)y) = f(x + 1) f(y)$, which again leads to $f(xy) = f(x)f(y)$. Therefore we have $f(xy) = f(x)f(y)$ for all $x, y$.
It follows that $f$ is an increasing function. In fact, for any $x \geq 0$, we have $f(x) = f(\sqrt x)^2 \geq 0$. Hence whenever $x \geq y$, we have $f(x) - f(y) = f(x - y) \geq 0$.
Since we already know $f(q) = q$ for any rational number $q$, this implies $f(x) = x$ for any real number $x$.
Otherwise, assume there is $x$ such that $f(x) \neq x$. Replacing $x$ with $-x$ if necessary, we may assume that $f(x) > x$. Then there is a rational number $q$ such that $f(x) > q > x$. But since $f$ is increasing, $q > x$ implies $q = f(q) > f(x)$, contradiction.
|
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Sum of coefficients of $x^i$ (Multinomial theorem application)
A polynomial in $x$ is defined by
$$a_0+a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}=(x+2x^2+ \cdots +nx^n)^2.$$
Show that the sum of all $a_i$, for $i\in\{n+1,n+2, \ldots , 2n\}$, is
$$ \frac {n(n+1)(5n^2+5n+2)} {24}.$$
I don't know how to proceed. I know the Multinomial theorem, however, I have problems in applying it. Any help will be appreciated as it will help me understand the theorem well.
Thanks!
|
Let $S_n$ be the required sum. By expanding the right hand side of $$a_0 + a_1 x + \ldots + a_{2n} x^{2n} = (x + 2x^2 + \ldots + nx^n)^2$$ we have
$$a_{n+i} = n \cdot i + (n-1) \cdot (i+1) + (n-2) \cdot (i+2) + \cdots \ + i \cdot n$$ for $i=1, 2, \ldots, n.$
Summing over $i=1,2,\ldots, n$ gives
$$ S_n = n(1 + 2 + 3+ \ldots + n) + (n-1)(2+ 3 + \ldots + n) + \ \ldots \ + 2( (n-1) + n) + 1(n) $$
The terms in the brackets are the sum of the first $n$ integers (which equals $\binom{n+1}{2}$) minus the sum of the first $k$ integers (which equals $\binom{k+1}{2}$) so we get
$$ S_n = \sum_{k=0}^{n-1} (n-k) \left( \binom{n+1}{2} - \binom{k+1}{2} \right)$$
$$ = \binom{n+1}{2} \sum_{k=0}^{n-1} (n-k) - \sum_{k=1}^{n-1} \binom{n-k}{1} \binom{k+1}{2}$$
Again, we have $\sum_{k=0}^{n-1} (n-k) = \binom{n+1}{2}$ so the first term above simplifies to $\binom{n+1}{2}^2.$
To pick a $4$-element subset from $n+2$ elements we follow this scheme - Pick element $k+2$ ($ \ k$ from $1$ to $n-1$) to be $3$-rd element of the subset, then pick $2$ from the $k+1$ elements on its left and $1$ element from the $n-k$ on its right. Summing over all possibilities of what the $3$rd element could be yields the second term above.
Therefore, $$S_n = \binom{n+1}{2}^2 - \binom{n+2}{4}$$
$$ \ \ \ \ \ \ \ \ \ = \frac {n(n+1)(5n^2+5n+2)} {24}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3484240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
$\frac{x}{y}-\frac{y}{x}=\frac56$ and $x^2-y^2=5$
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \\ x^2-y^2=5 \end{array}$$
First, we have $x,y \ne 0$. Let's write the first equation as:
$$\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \Leftrightarrow \dfrac{x^2-y^2}{xy}=\dfrac{5}{6}$$
We have $x^2-y^2=5$, therefore $xy=6$. What to do next?
|
The given problem is equivalent to finding the intersections between a rectangular hyperbola and two lines through the origin, since the first equation gives $\frac{y}{x}\in\left\{\frac{2}{3},-\frac{3}{2}\right\}$. These lines are orthogonal, so it is pretty simple to locate the solutions $(3,2)$ and $(-3,-2)$. The line with slope $-\frac{3}{2}$ does not intersect the hyperbola, whose asymptotes are $y=\pm x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3484688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
|
How to evaluate $\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$ I am trying to calculate this integral, but I find it is very challenging
$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$$
but somehow I have managed to local its closed form to be $$(-1)^n\left(\frac{1}{2}-n\ln 2+\sum_{j=0}^{n-1}H^{*}_j\right)$$
where $n\ge 0$, $H^{*}_0=0$ and $H^{*}_k=\sum_{j=1}^{k}\frac{(-1)^{j+1}}{j}$
I have try
$$\frac{1-x}{1+x}\cdot \frac{x^n}{(x^2-1)^{1/2}}$$
$$-x^n\sqrt{\frac{x-1}{(x+1)^2}}$$
$$-\int_{0}^{1}x^n\sqrt{\frac{x-1}{(x+1)^3}}\mathrm dx$$
from this point I tried to use the binomial to expand $$\sqrt{\frac{x-1}{(x+1)^3}}$$ but it seem not possible
|
In $0\le x<1$
$$\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}=\dfrac{x^n}{(1+x)^2}$$
If $\displaystyle I_n=\int_{0}^{1}\dfrac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$
$$I_n+2I_{n-1}+I_{n-2}=\int_0^1x^{n-2} dx$$
Now $I_0=?,I_1=?$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3486268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Determinant of the matrix $(\omega^i \theta^j)_{i,j = 0,1,2}$ Given that $\omega=\frac{-1+\sqrt{-3}}{2}$ is a complex cube root of 1, $\theta$ is a real cube root of 2 and $\omega^{2}+\omega+1=0$, I am trying to maipulate the matrix
\begin{pmatrix}
1 & \theta &\theta^{2} \\
\omega & \omega \theta & \omega \theta^{2} \\
\omega^{2} & \omega^{2}\theta & \omega^{2}\theta^{2}
\end{pmatrix}
to end up at
\begin{pmatrix}
1 & \theta &\theta^{2} \\
1 & \omega \theta & \omega^{2}\theta^{2} \\
1 & \omega^{2}\theta & \omega \theta^{2}
\end{pmatrix}
However I am struggling to come up with a way of manipulating the matrix in order to end up here.
|
From properties of determinants we get, by adding row 2 and row 3 to the first row
$$\det\begin{pmatrix}
1 & \theta &\theta^{2} \\
\omega & \omega \theta & \omega \theta^{2} \\
\omega^{2} & \omega^{2}\theta & \omega^{2}\theta^{2}
\end{pmatrix}=\det\begin{pmatrix}
1+\omega+\omega^2 & \theta+\omega\theta+\omega^2\theta &\theta^{2}+\omega\theta^2+\omega^2\theta^2 \\
\omega & \omega \theta & \omega \theta^{2} \\
\omega^{2} & \omega^{2}\theta & \omega^{2}\theta^{2}
\end{pmatrix}$$
The last matrix has a row of zeros, since $1+\omega+\omega^2=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3488010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Find polynomials $P(x)P(x-3) = P(x^2)$
Find all polynomials $P \in \Bbb R[x]$ such that $P(x)P(x-3) = P(x^2) \quad \forall x \in \Bbb R$
I have a solution but I'm not sure about that. Please check it for me.
It is easy to see that $P(x) = 0, P(x) = 1$ satisfied.
Consider $P(x) \neq c$ :
We have $P(x+3)P(x) = P((x+3)^2)$
If $a$ is a root of $P(x) $ then $a^2$ and $(a+3)^2$ are roots of $P(x)$.
If $|a| > 1$, then $ |a^2| = |a||a| > |a| \Rightarrow P(x)$ has infinitely many roots.
If $0 < |a| < 1$, then $|a^2| = |a||a| < |a| \Rightarrow P(x)$ has infinitely many roots.
If $a = 0 \Rightarrow (a+3)^2 = 9$ is a root of $P(x)$, so $P(x)$ has infinitely many roots.
Hence, every root $a$ of $P(x)$ has $|a| = 1$. So $1 = |(a+3)^2| = (|a+3|^2) \Rightarrow |a+3|=1$.
We have $a + 3 = \cos\alpha + i\sin\alpha + 3 \Rightarrow (\cos\alpha+3)^2 + \sin^2\alpha = 1 \Rightarrow \cos\alpha = -\frac{3}{2}\ (\text{impossible}).$
$P(x) = 0, P(x) = 1$ are all the results.
|
Considering
$$
P(x)P(x-b)=P(x^2),\ \ \ b \ne 0\ \ \ \ (1)
$$
As long as $P(x^2)$ is even then $P(x)P(x-b)$ is also even then
$$
P(b)P(0) = P(b^2)
$$
because making $x = b$ in $(1)$ we have $P(b)P(0) = P(b^2)$, As this is for a generic $b$ we have $P(x)P(0) = P(x^2)$
or
$$
P(x)P(0) = P(x^2) = P(-x)P(-(x+b))
$$
which implies that $P(x)$ is even as well as $P(x+b)$
This should be true for all $b \ne 0$ so the solution is $P(x) = C_0$
Finally, according to $C_0^2=C_0$ we conclude that $C_0 = 0$ or $C_0 = 1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3488111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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|
How to prove this formula for the determinant of a $4 \times 4$ tridiagonal matrix? This following is a problem from B. S. Grewal's Higher Engineering Mathematics.
Show
$$\begin{vmatrix} 2\cos(\theta) & 1 & 0 & 0 \\ 1 & 2 \cos(\theta) & 1
& 0 \\ 0 & 1 & 2 \cos(\theta) & 1 \\ 0 & 0 & 1 & 2 \cos(\theta)
\end{vmatrix} = \frac{\sin(5\theta)}{\sin(\theta)}.$$
If I take the $3 \times 3$ matrix after deleting the first row and first column, the value is $\frac{\sin(4\theta)}{\sin(\theta)}$, but I am unable to solve this $4\times 4$ matrix. I tried solving the RHS but I am still unable to solve.
|
https://en.wikipedia.org/wiki/Tridiagonal_matrix
$f_n = \left|
\begin{array}{llll}
a_1 & b_1 &0 &0 &0 \\
c_1 & a_2 & b_2 &0 & 0\\
0 & c_2 & \ddots & \ddots & 0\\
0 &0 & \ddots & \ddots & b_{n-1}\\
0 &0 &0 & c_{n-1} & a_n
\end{array}
\right|
$
with $f_0 = 1, f_{-1} = 0$.
$f_n
= a_n f_{n-1}-c_{n-1}b_{n-1}f_{n-2}
$.
If all
$a_i = a, b_i = b, c_i = c$
then
$f_n
= a f_{n-1}-cbf_{n-2}
$.
In this case,
$b = c = 1,
a = 2\cos(t)
$
so
$f_n
= 2\cos(t)f_{n-1}-f_{n-2}
$
so
$f_n = \left|
\begin{array}{ccccc}
2\cos(t) & 1 &0& 0&0\\
1 & 2\cos(t) & 1 & 0& 0\\
0 & 1 & \ddots & \ddots &0 \\
0 &0 & \ddots & \ddots & 1\\
0&0 &0 & 1 & 2\cos(t)
\end{array}
\right|
$
Therefore
$f_n\sin(t)
= 2f_{n-1}\sin(t)\cos(t)-\sin(t)f_{n-2}
$.
Let
$g_n
=f_n\sin(t)
$,
so
$g_n
= 2\cos(t)g_{n-1}-g_{n-2}
$.
Since
$f_0 = 1$
and
$f_1 = 2\cos(t)
$,
$g_0 = \sin(t)$
and
$g_1
=2\sin(t)\cos(t)
=\sin(2t)
$.
We have
$\begin{array}\\
\sin((n+1)t)+\sin((n-1)t)
&=\sin(nt)\cos(t)+\cos(nt)\sin(t)+\sin(nt)\cos(t)-\cos(nt)\sin(t)\\
&=2\sin(nt)\cos(t)\\
\end{array}
$
or
$\sin((n+1)t)=2\sin(nt)\cos(t)-\sin((n-1)t)
$.
Therefore
$g_n(t)
=\sin((n+1)t)$,
so
$f_n(t)
=\dfrac{\sin((n+1)t)}{\sin(t)}
$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3489325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$ but got stucked, and made it more complex, any help?
Sorry if I made any silly mistake, it's been while since I practiced complex equation and finding roots. Was helping my brother with his doubts :)
|
If you are going the hard way of finding explicitly the roots, then the roots and their exponential forms are
$$\alpha=-\frac{1}{2}-\frac{i\sqrt{3}}{2}=-e^{\frac{i\pi}{3}} \Rightarrow\alpha^6=e^{i2\pi}=1$$
$$\beta=\frac{1}{2}+\frac{i\sqrt{3}}{2}=e^{\frac{i\pi}{3}}\Rightarrow\beta^6=e^{i2\pi}=1$$
$$\gamma=\frac{1}{2}-\frac{i\sqrt{3}}{2}=-e^{\frac{i2\pi}{3}}\Rightarrow\gamma^6=e^{i4\pi}=1$$
$$\delta=-\frac{1}{2}+\frac{i\sqrt{3}}{2}=e^{\frac{i2\pi}{3}}\Rightarrow\delta^6=e^{i4\pi}=1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
}
|
How many pairs of integers have a fixed GCD? Let $k \ge 0, a \ge 0$ be a fixed integer and $f(x,k, a)$ be the number of positive integers pairs $(m,n), m < n \le x$ such that $\gcd(km-n, m + kn) = a$.
Question: Is there a closed form of expression $\lim_{x \to \infty}\dfrac{f(x,k,a)}{x^2}$ in terms of $a$ and $k$?
My experimental data suggest that a closed form exists.
$$
f(x,0,1) \sim 3 \pi^2 x^2
\\
f(x,1,1) \sim 2 \pi^2 x^2
\\
f(x,2,1) \sim \frac{5}{2} \pi^2 x^2
\\
f(x,3,1) \sim \frac{5}{3} \pi^2 x^2
\\
f(x,4,1) \sim \frac{17}{6} \pi^2 x^2
\\
f(x,5,1) \sim \frac{13}{7} \pi^2 x^2
\\
f(x,2,3) \sim \frac{5}{18} \pi^2 x^2
\\
f(x,1,5) \sim \frac{2}{25} \pi^2 x^2
$$
Code
s = 2
j = 0
a = 3
test = 2
target = set = 10^2
while True:
r = 1
while r < s:
if gcd(a*s - r,s + a*r) == test:
j = j + 1
r = r + 1
if s > target:
target = target + set
print s, j, (pi^2*j/s^2).n()
s = s + 1
|
$f(n,1,1) = \sum_{m < n, \gcd(m-n,m+n) = 1} 1 = \sum_{m < n} \sum_{d \mid m-n, d \mid m+n} \mu(d) = \sum_{d \mid 2n} \mu(d)\sum_{m < n, m \equiv n \mod d} 1 = \sum_{d \mid 2n} \mu(d)\left[\frac{n}{d}+O(1)\right] = n\sum_{d \mid 2n} \frac{\mu(d)}{d}+O(\tau(2n))$.
So, $\sum_{n \le x} f(n,1,1) = \sum_{n \le x} \left[n\sum_{d \mid 2n} \frac{\mu(d)}{d}+O(n^{1/2})\right] = \sum_{d \le 2x} \frac{\mu(d)}{d}\sum_{n \le x, d \mid 2n} n + O(x^{3/2})$. Ignore the $O(x^{3/2})$. We get $\sum_{d \le 2x, d \equiv 1 \pmod{2}} \frac{\mu(d)}{d}\left[\frac{1}{2}\frac{x^2}{d}+O(x)\right]+\sum_{d \le 2x, d \equiv 0 \pmod{2}} \frac{\mu(d)}{d}\left[\frac{x^2}{d}+O(x)\right]$. Similarly, we can ignore the $O(x)$ terms, so we end up with $\frac{1}{2}x^2\sum_{d \le 2x} \frac{\mu(d)}{d^2}+\frac{1}{2}x^2\sum_{d \le 2x, d \equiv 0 \pmod{2}} \frac{\mu(d)}{d^2}$. The first term is basically $\frac{1}{2}x^2\sum_{d=1}^\infty \frac{\mu(d)}{d^2} = \frac{3}{\pi^2} x^2$. The second term can be figured out also. So that's how the $\pi^2$'s come about.
One can do $f(n,a,k)$ similarly.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3494730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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|
A complex Banach space satisfying the parallelogram law is Hilbert Let $H$ be a Banach space with associated norm $\|-\|.$ Suppose that for any $x,y\in H,$ we have:
$$\|x+y\|^2+\|x-y\|^2=2\left(\|x\|^2+\|y\|^2\right),$$
which we call the parallelogram law.
Then it is a well-known standard fact that $H$ becomes a Hilbert space. This is true both for real and complex coefficients.
I managed to prove this fact for an Hilbert space over $\mathbb{R},$
defining the inner product
$$(x,y) \mapsto \langle x,y\rangle=\frac{\|x+y\|^2-\|x-y\|^2}{4}= \frac{\|x+y\|^2-\|x\|^2-\|y\|^2}{2}.$$
Question
How to prove this for the complex case?
The inner product should be in this case
$$(x,y)\mapsto\ \alpha(x,y)= \frac{1}{4} \left(\|x + y\|^2 - \|x-y\|^2 + i\|x + iy\|^2 -i\|x-iy\|^2 \right)= \langle x,y \rangle + i \langle x,iy \rangle$$
but I am not able to replicate the proof of the real case.
It would be awesome if there was a slick proof that used the real case to deduce the complex case, but I would still be happy with any kind of direct proof too. Since this is a standard result, if you can provide a reference where a detailed proof is given, that would be excellent too.
|
Let $\|\cdot\|: H\to \mathbb{R}$ be a norm in a complex Banach space $H$ which fulfills for all $x,y\in H$ the parallelogram law
\begin{align*}
\|x+y\|^2+\|x-y\|^2=2\left(\|x\|^2+\|y\|^2\right)\tag{1}
\end{align*}
We show the map $\alpha:H\times H\to\mathbb{C}$ defined by
\begin{align*}
\alpha(x,y)= \frac{1}{4} \left\{\|x + y\|^2 - \|x-y\|^2 + i\|x + iy\|^2 -i\|x-iy\|^2 \right\}\tag{2}
\end{align*}
fulfills for all $x,y,z \in H$ and $c\in\mathbb{C}$
\begin{align*}
c\alpha(x,y)=\alpha(cx,y)\tag{3}
\end{align*}
Let $x,y,z\in H$. We have
\begin{align*}
&\color{blue}{\alpha(x,y)+\alpha(z,y)}\\
&\qquad=\frac{1}{4}\left\{\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2\right.\\
&\qquad\qquad\quad\left.+\|z+y\|^2-\|z-y\|^2+i\|z+iy\|^2-i\|z-iy\|^2\right\}\tag{4}\\
&\qquad=\frac{1}{4}\left\{
\left\|\left(\frac{x+z}{2}+y\right)+\frac{x-z}{2}\right\|^2+\left\|\left(\frac{x+z}{2}+y\right)-\frac{x-z}{2}\right\|^2\right.\\
&\qquad\qquad\quad\left.-\left\|\left(\frac{x+z}{2}-y\right)+\frac{x-z}{2}\right\|^2+\left\|\left(\frac{x+z}{2}-y\right)-\frac{x-z}{2}\right\|^2\right.\\
&\qquad\qquad\quad+i\left\|\left(\frac{x+z}{2}+iy\right)+\frac{x-z}{2}\right\|^2+i\left\|\left(\frac{x+z}{2}+iy\right)-\frac{x-z}{2}\right\|^2\\
&\qquad\qquad\quad\left.-i\left\|\left(\frac{x+z}{2}-iy\right)+\frac{x-z}{2}\right\|^2+i\left\|\left(\frac{x+z}{2}-iy\right)-\frac{x-z}{2}\right\|^2\right\}\tag{5}\\
&\qquad=\frac{1}{2}\left\{\left\|\frac{x+z}{2}+y\right\|^2+\left\|\frac{x-z}{2}\right\|^2
-\left\|\frac{x+z}{2}-y\right\|^2-\left\|\frac{x-z}{2}\right\|^2\right.\\
&\qquad\qquad\quad\left.+i\left\|\frac{x+z}{2}+iy\right\|^2+i\left\|\frac{x-z}{2}\right\|^2
-i\left\|\frac{x+z}{2}-iy\right\|^2-i\left\|\frac{x-z}{2}\right\|^2\right\}\tag{6}\\
&\qquad\,\,\color{blue}{=2\alpha\left(\frac{x+z}{2},y\right)}\tag{7}
\end{align*}
Comment:
*
*In (4) we use the definition of $\alpha$ from (2).
*In (5) we do some preparatory work in order to apply the parallelogram law.
*In (6) we apply the parallelogram law (1).
Since we have from (2)
\begin{align*}
\alpha(x,0)&=\frac{1}{4} \left\{\|x\|^2 - \|x\|^2 + i\|x\|^2 -i\|x\|^2 \right\}=0
\end{align*}
we obtain by substituting $z=0$ in (7)
\begin{align*}
\alpha(x,y)=2\alpha\left(\frac{x}{2},y\right)\tag{8}
\end{align*}
We obtain from (7) and (8) by induction that for all $n,m\in\mathbb{N}_0$:
\begin{align*}
2^{-n}m\alpha(x,y)=\alpha\left(2^{-n}m x,y\right)
\end{align*}
Case $c\geq 0$:
If $c \geq 0$, then we have numbers $c_k=2^{-n(k)}m(k)$ such that $c_k\to c$ as $k\to \infty$. Since the norm $\|\cdot\|$ fulfills for all $x,y\in H$:
\begin{align*}
\left|\|x\|-\|y\|\right|\leq \|x\pm y\|
\end{align*}
it follows
\begin{align*}
&\left|\|c_kx\pm y\|-\|cx\pm y\|\right|\leq \left|c_k-c\right|\|x\|\\
&\left|\|ic_kx\pm y\|-\|icx\pm y\|\right|\leq \left|c_k-c\right|\|x\|\\
\end{align*}
and we obtain
\begin{align*}
\alpha\left(c_kx,y\right)\to\alpha(cx,y)\qquad \text{as }k\to\infty
\end{align*}
We conclude
\begin{align*}
\color{blue}{c\alpha(x,y)}=\lim_{k\to\infty}c_k\alpha(x,y)=\lim_{k\to\infty}\alpha\left(c_kx,y\right)\color{blue}{=\alpha(cx,y)}
\end{align*}
Case $c\in\mathbb{R}$:
Since
\begin{align*}
\color{blue}{\alpha(-x,y)}&=\frac{1}{4} \left\{\|-x + y\|^2 - \|-x-y\|^2 + i\|-x + iy\|^2 -i\|-x-iy\|^2 \right\}\\
&=-\frac{1}{4} \left\{-\|x - y\|^2 + \|x+y\|^2 - i\|x - iy\|^2 +i\|x+iy\|^2 \right\}\\
&\,\,\color{blue}{=-\alpha(x,y)}
\end{align*}
we have $\alpha(cx,y)=c\alpha(x,y)$ for all $c \in \mathbb{R}$.
Case $c\in\mathbb{C}$:
Since
\begin{align*}
\color{blue}{\alpha(ix,y)}&=\frac{1}{4} \left\{\|ix + y\|^2 - \|ix-y\|^2 + i\|ix + iy\|^2 -i\|ix-iy\|^2 \right\}\\
&=\frac{i}{4} \left\{-i\|x - iy\|^2 + i\|x+iy\|^2 +\|x +y\|^2 -\|x-y\|^2 \right\}\\
&\,\,\color{blue}{=i\alpha(x,y)}
\end{align*}
we have $\alpha(cx,y)=c\alpha(x,y)$ for all $c \in \mathbb{C}$ and the claim (3) follows.
Note: This post follows closely the proof provided in section 1.2 of Linear Operators in Hilbert Spaces by J. Weidmann.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3496778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
}
|
Evaluating $\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{2^k}\tan\frac{x}{2^k}$ So I was solving a previous year question paper and stumbled upon the following question-
$${\lim_{n\to {\infty}}}\biggl(\tan x+\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\frac{1}{2^3}\tan \frac{x}{2^3}+{\cdots}+\frac{1}{2^n}\tan \frac{x}{2^n}\biggl)$$
The only limit with $\tan$ that I have learnt is ${\lim_{x\to 0}}\frac{\tan x}{x}=1$. I have tried the following with no success-
1)Representing $\tan x$ in terms of $\tan \frac x2$ 2)Simplifying into $\sin x$ and $\cos x$
I do not want a complete solution but would greatly appreciate a hint on how to simplify the series.
|
Hints:
*
*$\ln\prod\cos\frac{x}{2^n}=\sum\ln\cos\frac{x}{2^n}$
*$\left(\sum\ln\cos\frac{x}{2^n}\right)'=-\sum\frac{1}{2^n}\tan\frac{x}{2^n}$
*$\prod\cos\frac{x}{2^n}$ is something well known.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3496972",
"timestamp": "2023-03-29T00:00:00",
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|
Area between 2 curves and $X$-axis I have trouble doing this. I have to find the area of the region bounded by the parabola ($4X-X^2$), $y=X$ and the $X$-axis. I found the interception points but don't understand what the common area is between all of them. The answer is $\frac{37}{6}$
Thanks guys.
|
It is helpful to visualize the intersection between the two curves and $x$-axis by a graphing program such as Wolfram Alpha or Desmos.
Labeling $f(x)=4x-x^2, g(x)=x, h(x)=0$, we see that $g(x) < f(x)$ when $0< x < 3$. So, for $0\le x \le 3$ the area of the region bounded by the curves and the $x$-axis is the triangle between the points $(0,0), (3,3),$ and $(3,0)$. The area of the triangle is
$$\frac{1}{2}{(\text{base})}{(\text{height})}=\frac{1}{2}(3)(3)=\frac{9}{2}$$
Next, when $3<x\le 4$ we have that $f(x) < g(x)$. Therefore, the area of the region bounded by the curves and the $x$-axis is the integral between $f(x)$ and $h(x)$
$$\int_3^4 \big(f(x)-h(x)\big)dx=\int_3^4 \big(4x-x^2\big)dx=\left(2x^2-\frac{x^3}{3}\right)\bigg|_3^4=\frac{5}{3}$$
the total area is the sum of these two regions
$$\frac{9}{2}+\frac{5}{3}=\frac{27}{6}+\frac{10}{6}=\frac{37}{6}$$
|
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|
$\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$
I should solve the following system: $$\begin{cases} x^4+x^2y^2+y^4=21 \\
x^2+xy+y^2=3 \end{cases}$$
by reducing the system to a system of second degree.
What can I look for in such situations? What is the way to solve this kind of systems? The only thing I see here is that we can factor:
$$\begin{cases} x^2(x^2+y^2)+y^4=21 \\
x(x+y)+y^2=3 \end{cases}$$
|
Hint:
$$x^4+y^4+x^2y^2=(x^2+y^2)^2-(xy)^2=?$$
So, we know $x^2+y^2,xy=9>0$
So, $x,y$ will have the same sign
Hope you can take it from here using $y=\dfrac9x$
|
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|
Help with inequality problem
Given $a$ , $b$ , $c \ge 0$ show that
$$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} \ge \frac{3}{4}.$$
I tried using Titu's lemma on it, resulting in
$$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2 + 3(ab + bc + ca)} $$
And I am stuck here.
|
Another way.
We need to prove that $$4\sum_{cyc}(a^2b+a^2c)\geq3\prod_{cyc}(a+b)$$ or
$$\sum_{cyc}c(a-b)^2\geq0.$$
|
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|
$\sum_{n=1}^\infty \frac{n^k}{1+2^n+(-2)^{nk}}$ $S = \sum_{n=1}^\infty \frac{n^k}{1+2^n+(-2)^{nk}}$
Find $k \in N: S$ coverages.
I started with checking the Cauchy's theorem for $k = 2m, m \in N$ and at this point everything is fine. But for $k = 2m - 1, m \in N$ things start to get complicated and I don't really have an idea how to proceed.
|
Let be
$$
a_n \left( k \right) = \frac{{n^k }}
{{1 + 2^n + \left( { - {\text{2}}} \right)^{nk} }}
$$
If $k=1$ then the series is not convergent. Indeed, you have that
$$
a_n(1) = \frac{n}
{{1 + 2^n + \left( { - {\text{1}}} \right)^n 2^n }}
$$
Thus, when $n$ is odd you have that $a_n=n$. And the necessary condition for the convergence is not satisfied.
If $k$ is odd and greater than $1$ then your series is convergent. Indeed,
$$
\begin{gathered}
a_n \left( k \right) = \frac{{n^k }}
{{1 + 2^n + \left[ {\left( { - {\text{1}}} \right)^k } \right]^n 2^{nk} }} = \hfill \\
\hfill \\
= \frac{{n^k }}
{{1 + 2^n + \left( { - {\text{1}}} \right)^n 2^{nk} }} = \hfill \\
\hfill \\
= \frac{{n^k }}
{{\left( { - {\text{1}}} \right)^n 2^{nk} \left[ {1 + \frac{{1 + 2^n }}
{{\left( { - {\text{1}}} \right)^n 2^{nk} }}} \right]}} = \hfill \\
\hfill \\
= \frac{{n^k \left( { - {\text{1}}} \right)^n }}
{{2^{nk} \left[ {1 + \frac{{\left( {1 + 2^n } \right)\left( { - {\text{1}}} \right)^n }}
{{2^{nk} }}} \right]}} \hfill \\
\end{gathered}
$$
so that
$$
\begin{gathered}
\left| {a_n \left( k \right)} \right| \leqslant \frac{{n^k }}
{{2^{nk} \left[ {1 - \frac{{\left( {1 + 2^n } \right)}}
{{2^{nk} }}} \right]}} \leqslant \hfill \\
\hfill \\
\leqslant \frac{{n^k }}
{{2^{nk} \left[ {1 - \frac{1}
{2}} \right]}} = \frac{{2n^k }}
{{2^{nk} }} \hfill \\
\end{gathered}
$$
which is true if $n$ is large enough. Now it is easy to prove that the series
$$
\sum\limits_{n = 1}^{ + \infty } {\frac{{2n^k }}
{{2^{nk} }}} = 2\sum\limits_{n = 1}^{ + \infty } {\frac{{n^k }}
{{2^{nk} }}}
$$
is convergent and so the given series is absolutely convergent.
|
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|
Solve $y' + y^2 = \frac{1}{x^2}$ by introducing $z = xy$ as a new function. Question:
Solve the equation:
$$y' + y^2 = \frac{1}{x^2},~~~~~~~~~x > 0$$
by introducing $z = xy$
Attempted answer:
$z = xy \Rightarrow y = \frac{z}{x}$
Taking the derivative of $y$ with respect to $x$ using he product rule:
$$y' = \frac{z'}{x} - \frac{z}{x}$$
Adding this into the original equation:
$$\frac{z'}{x} - \frac{z}{x^2} + \frac{z^2}{x^2} = \frac{1}{x^2}$$
Putting $z$ and $x$ on each side:
$$\frac{z'}{1-z^2 + z} = \frac{1}{x}$$
Integrating on each side:
$$\int \frac{1}{1-z^2 + z}dz = \int \frac{1}{x} dx$$
This produces
$$z(x) = \frac{(1+\sqrt{5})c_1x^{\sqrt{5}}+1-\sqrt{5}}{(2c_1 x^{\sqrt{5}}+2)}$$
Substituting back to $y$:
$$y(x) = \frac{(1+\sqrt{5})c_1x^{\sqrt{5}}+1-\sqrt{5}}{x(2c_1 x^{\sqrt{5}}+2)}$$
This seems all and well, but the expected answer contains yet another solution:
$$y = \frac{1\pm\sqrt{5}}{2x}$$
How does this second solution arise? Since there is an $x^2$ in the question, I think that a solution might have been dropped at some point.
|
When you divide by $1-z^2+z$, you're actually excluding the possibility of this term to be $0$. But note that if it is $0$, you get the desired solution.
|
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|
Solve $2^m=7n^2+1$ Solve $2^m=7n^2+1$ with $(m,n)\in \mathbb{N}^2$
Here is what I did:
First try, I have seen first that the obvious solutions are $n=1$ and $m=3$ , and $n=3$ and $m=6$, then I proved by simple congruences that $m$ must be divisible by $3$ so $m=3k$, If we add $27$ to the equation we will have $2^{3k}+3^3=7(n^2+2^2)$, but unfortunately I tried to do something with Legendre symbol or the multiplicative order but I found nothing interesting.
Second try,I let $n=2k+1$ then I worked in $\mathbb{Z}\left[ \frac{-1+\sqrt{-7}}{2} \right] $ and the equation becomes $7\times 2^{m-2}=\left( 7k+4+\frac{-1+\sqrt{-7}}{2} \right) \left( 7k+3-\frac{-1+\sqrt{-7}}{2} \right) $ but I didn't find something interesting because the two factors are not coprime.
|
If $2^m=7n^2+1$ with $m=3k$, as the OP found must be the case (since $2^m\equiv1$ mod $7$), we have
$$7n^2=2^{3k}-1=(2^k-1)(2^{2k}+2^k+1)$$
Now for $k\ge1$ we have
$$\begin{align}
\gcd(2^k-1,2^{2k}+2^k+1)
&=\gcd(2^k-1,2^{2k}+2^{k+1})\\
&=\gcd(2^k-1,2^{k+1}(2^{k-1}+1))\\
&=\gcd(2^k-1,2^{k-1}+1)\\
&=\gcd(2^k+2^{k-1},2^{k-1}+1)\\
&=\gcd(3\cdot2^{k-1},2^{k-1}+1)\\
&=\gcd(3,2^{k-1}+1)\\
&=\begin{cases}
3\quad\text{if $k$ is even}\\
1\quad\text{if $k$ is odd}
\end{cases}
\end{align}$$
If $k$ is even, we proceed as in Mastrem's answer: $k=2h$ implies $7n^2=2^{6h}-1=(2^{3h}-1)(2^{3h}+1)$ with $\gcd(2^{3h}-1,2^{3h}+1)=1$ and $7\mid(2^{3h}-1)$, so $2^{3h}+1$ must be a square, but $2^{3h}+1=N^2$ implies $2^{3h}=(N-1)(N+1)$, which holds only for $N=3$, corresponding to the known solution with $m=6$.
If $k$ is odd, then we must have $7$ divide one of the factors in $(2^k-1)(2^{2k}+2^k+1)$ and the other factor be a square. If $k\ge3$, $2^k-1\equiv-1$ mod $8$, which is not a square, so we must have $7\mid2^k-1$, which implies $k=3h$ (with $h$ odd, but that's no longer important), from which it follows that $2^{6h}+2^{3h}+1$ is a square. But $2^{6h}+2^{3h}+1\equiv1+1+1\equiv3$ mod $7$, which is not a square. So the case of odd $k$ leaves only $k=1$, corresponding to the other known solution, $m=3$.
|
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|
$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx$ I was attempting to solve an MIT integration bee problem (1) when I misread the integral and wrote (2) instead.
$$\int\sqrt{x\cdot \sqrt[3]{x\cdot \sqrt[4]{x\cdot\sqrt[5]{x\ldots } }}}\,dx\tag{1}$$
$$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx\tag{2}$$
I was able to solve (1), as the integrand simplifies to $x^{e-2}$, however, I'm struggling with solving (2).
If we rewrite the roots as powers, we get:
$$\int x^\frac{2}{2}\cdot x^\frac{3}{4}\cdot x^\frac{4}{8}\cdot x^\frac{5}{16}\ldots\,dx$$
combining the powers we get:
$$\int x^{\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\ldots}$$
the exponent is the infinite sum
$$\sum^{\infty}_{n=1}\frac{n+1}{2^n}\tag{3} $$
we can split this into:
$$\sum^{\infty}_{n=1}\frac{n}{2^n}+\sum^{\infty}_{n=1}\frac{1}{2^n} $$
The right sum is well known except here the sum begins at $n=1$, meaning that the right sum evaluates to 1. Messing around with desmos, the integrand appears to be $x^3,x>0$ implying that (3) converges to 3 and the $\sum^{\infty}_{n=1}\frac{n}{2^n}$ converges to 2.
Which is part I'm struggling with. Any ideas?
$$\sum^{\infty}_{n=1}\frac{n}{2^n}$$
|
Consider
$$\sum_{n=1}^\infty n x^n=x\sum_{n=1}^\infty n x^{n-1}=x\left(\sum_{n=1}^\infty x^{n}\right)'$$
When finished, make $x=\frac 12$
|
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|
Natural numbers equal to the sum of the squares of their four smallest divisors I was in the process of answering this question when I fell asleep; when I woke up, I found that the question has been closed for being too vague:
Find all positive integers
Anyhow, the mathematical problem is as follows:
Given a natural number $n > 0$, let $1$, $a$, $b$, and $c$ be the smallest four divisors of $n$, such that $1 < a < b < c$. Find all possible natural numbers $n$ such that $1^2 + a^2 + b^2 + c^2 = n$.
|
$n$ must be even, because if it were odd the four divisors would be odd and the sum of squares would be even.
The two smallest divisors are $1$ and $2$. $n$ cannot be a multiple of $4$ because if the fourth divisor is $p$ we have $n=1^2+2^2+4^2+p^2=21+p^2$ and squares $\bmod 4$ are $0,1$ so the right cannot be a multiple of $4$.
If the four smallest divisors were $1,2,p,q$ for $p,q$ prime the sum of squares would be odd, so the four smallest divisors are $1,2,p,2p$ for $p$ an odd prime.
If $n$ has a factor $3$, the four smallest divisors would be $1,2,3,6$ but then $n=50$ and it is not divisible by $3$.
$n=1^2+2^2+p^2+(2p)^2=5(p^2+1)$. As $n$ has a factor $5$, $p$ must be $5$. This gives $n=130$ as the only solution.
|
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|
Infinite product $\prod_{n=1}^{\infty}(1+z^n)(1-z^{2n-1}) = 1$ I have to show that if $|z| < 1$, $z \in \mathbb{C}$, $$\prod_{n=1}^{\infty}(1+z^n)(1-z^{2n-1}) = 1$$
This exercise if taken from Remmert, Classical Topics in Complex Function Theory.
I don't know where to start. Could anyone give a hint ?
|
The finite product is
$$
\prod_{k=1}^n\left(1+z^k\right)\left(1-z^{2k-1}\right)
=\prod_{k=n+1}^{2n}\left(1-z^k\right)\tag1
$$
This can be proven by induction. Suppose $(1)$ is true for $n$
$$
\begin{align}
\prod_{k=1}^{n+1}\left(1+z^k\right)\left(1-z^{2k-1}\right)
&=\left(1+z^{n+1}\right)\left(1-z^{2n+1}\right)\prod_{k=1}^n\left(1+z^k\right)\left(1-z^{2k-1}\right)\tag2\\
&=\left(1+z^{n+1}\right)\left(1-z^{2n+1}\right)\prod_{k=n+1}^{2n}\left(1-z^k\right)\tag3\\
&=\left(1-z^{2n+2}\right)\left(1-z^{2n+1}\right)\prod_{k=n+2}^{2n}\left(1-z^k\right)\tag4\\
&=\prod_{k=n+2}^{2n+2}\left(1-z^k\right)\tag5
\end{align}
$$
Explanation:
$(2)$: pull the $k=n+1$ term out front
$(3)$: apply $(1)$ for $n$
$(4)$: pull the $k=n+1$ term out front
$(5)$: bring the $k=2n+1$ and $k=2n+2$ terms inside
Thus, $(1)$ is true for $n+1$. $\large\square$
Since
$$
\begin{align}
\left|\sum_{k=n+1}^{2n}\log\left(1-z^k\right)\right|
&\le\sum_{k=n+1}^{2n}\left|\log\left(1-z^k\right)\right|\tag7\\
&\le\sum_{k=n+1}^\infty\left|\log\left(1-z^k\right)\right|\tag8\\
&\le\sum_{k=n+1}^\infty\frac{|z|^k}{1-|z|^k}\tag9\\
&\le\frac1{1-|z|^{n+1}}\sum_{k=n+1}^\infty|z|^k\tag{10}\\
&=\frac{|z|^{n+1}}{\left(1-|z|\right)\left(1-|z|^{n+1}\right)}\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(7)$: generalized triangle inequality
$\phantom{1}(8)$: absolute value is non-negative
$\phantom{1}(9)$: $|\log(1+z)|\le\frac{|z|}{1-|z|}$
$(10)$: $\frac1{1-|z|^k}\le\frac1{1-|z|^{n+1}}$
$(11)$: sum of a geometric series
Equation $(1)$ and inequality $(11)$ say that
$$
\left|\log\left(\prod_{k=1}^n\left(1+z^k\right)\left(1-z^{2k-1}\right)\right)\right|\le\frac{|z|^{n+1}}{\left(1-|z|\right)\left(1-|z|^{n+1}\right)}\tag{12}
$$
Thus,
$$
\log\left(\prod_{k=1}^\infty\left(1+z^k\right)\left(1-z^{2k-1}\right)\right)=0\tag{13}
$$
and therefore,
$$
\prod_{k=1}^\infty\left(1+z^k\right)\left(1-z^{2k-1}\right)=1\tag{14}
$$
|
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Convex cyclic hexagon $ABCDEF$. Prove $AC \cdot BD \cdot CE \cdot DF \cdot AE \cdot BF \geq 27 AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FA$
Convex hexagon $ABCDEF$ inscribed within a circle. Prove that
$$AC \cdot BD \cdot CE \cdot DF \cdot AE \cdot BF \geq 27 AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FA\,.$$
I was thinking of represending the inequalities in trigonometry then use Language multiplier. For example let $\angle AOB = \theta_1$, $\angle BOC = \theta_2$, represent the inequality in trigonometry, subject to constraint $\theta_1 + \theta_2 + ... + \theta_6 = 2\pi$. But it's still quite a bit of work and I didn't manage to get to the end. It also seems a bit overkill -- might be better solution? Would like to see any approach.
|
This problem had been posted here, but was deleted by the owner for an unspecified reason. I flagged the moderators about this, but they didn't do anything. Here is the same solution I gave in that link.
Ptolemy's theorem with $\square ABCD$ yields
$$AB\cdot CD+AD\cdot BC=AC\cdot BD.$$
Ptolemy's theorem with $\square ACDE$ yields
$$AC\cdot DE+EA\cdot CD=AD\cdot CE.$$
Hence,
$$AD=\frac{AC\cdot DE+EA\cdot CD}{CE}$$
so that
$$AC\cdot BD=AB\cdot CD+AD\cdot BC=AB\cdot CD+\left(\frac{AC\cdot DE+EA\cdot CD}{CE}\right)\cdot BC.$$
Hence
$$AC\cdot BD=AB\cdot CD+\frac{AC}{CE}(BC\cdot DE)+\frac{EA}{CE}(BC\cdot CD).$$
By AM-GM,
$$AC\cdot BD\geq 3\sqrt[3]{(AB\cdot CD)\left(\frac{AC}{CE}(BC\cdot DE)\right)\left(\frac{EA}{CE}(BC\cdot CD)\right)}=3\sqrt[3]{AB\cdot BC^2\cdot CD^2\cdot DE\cdot \frac{AC\cdot EA}{CE^2}}.$$
This shows that
$$\sqrt[3]{\frac{AC^2\cdot BD^3\cdot CE^2}{EA}}\geq 3\sqrt[3]{AB\cdot BC^2\cdot CD^2\cdot DE}.$$
Similarly,
$$\sqrt[3]{\frac{BD^2\cdot CE^3\cdot DF^2}{FB}}\geq 3\sqrt[3]{BC\cdot CD^2\cdot DE^2\cdot EF},$$
$$\sqrt[3]{\frac{CE^2\cdot DF^3\cdot EA^2}{AC}}\geq 3\sqrt[3]{CD\cdot DE^2\cdot EF^2\cdot FA},$$
$$\sqrt[3]{\frac{DF^2\cdot EA^3\cdot FB^2}{BD}}\geq 3\sqrt[3]{DE\cdot EF^2\cdot FA^2\cdot AB},$$
$$\sqrt[3]{\frac{EA^2\cdot FB^3\cdot AC^2}{CE}}\geq 3\sqrt[3]{EF\cdot FA^2\cdot AB^2\cdot BC},$$
and
$$\sqrt[3]{\frac{FB^2\cdot AC^3\cdot BD^2}{DF}}\geq 3\sqrt[3]{FA\cdot AB^2\cdot BC^2\cdot CD}.$$
Multiplying all six inequalities above gives
$$(AC\cdot BD\cdot CE\cdot DF\cdot EA\cdot FB)^2\geq (27\cdot AB\cdot BC\cdot DE\cdot EF\cdot FA)^2\,,$$
which is equivalent to the required inequality. The equality holds if and only if $ABCDEF$ is a regular hexagon.
|
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|
Prove $\int^{n+1}_1\frac{1}{x} dx \leq 1 + 1/2 + \cdots + 1/n$.
Source: https://math.mit.edu/~choiks/Pset%202%20solutions.pdf
I don't understand the third line $\int^{n+1}_1\frac{1}{x}dx \leq 1 + 1/2 + \cdots + 1/n$.
I think it should rather be $\int^{n+1}_1\frac{1}{x}dx \geq 1 + 1/2 + \cdots + 1/n$ because
$\int^{n}_1\frac{1}{x} < \int^{n+1}_1\frac{1}{x}$ and $1 + 1/2 + \cdots + 1/n$ is just addition of $f(x) = 1/x$ at certain values (integers). Therefore:
$1 + 1/2 + \cdots + 1/n < \int^{n}_1\frac{1}{x} < \int^{n+1}_1\frac{1}{x}$.
Thank you!
|
There are simple diagrams that compare sums and integrals.... Here what I call $f(x)$ is real valued and continuous, also positive. Things change (order of the inequalities) depending on $f$ increasing or decreasing. I guess $a < b$ are fixed integers.
if we have $f(x) > 0$ and $f'(x) > 0,$ then
$$ \int_{a-1}^{b} \; f(x) \; dx \; < \; \sum_{j=a}^b \; f(j) \; < \; \int_{a}^{b+1} \; f(x) \; dx $$
if we have $f(x) > 0$ but $f'(x) < 0,$ then
$$ \int_a^{b+1} \; f(x) \; dx \; < \; \sum_{j=a}^b \; f(j) \; < \; \int_{a-1}^b \; f(x) \; dx $$
We use this second one, as we are going to integrate $1/x$ we take $a=2$ along with $b=n$
$$ \int_2^{n+1} \; \frac{1}{x} \; dx \; < \; \sum_{j=2}^n \; \frac{1}{j} \; < \; \int_{1}^n \; \frac{1}{x} \; dx $$
$$ 1+ \int_2^{n+1} \; \frac{1}{x} \; dx \; < \; 1+ \sum_{j=2}^n \; \frac{1}{j} \; < \; 1+ \int_{1}^n \; \frac{1}{x} \; dx $$
$$ 1+ \log (n+1) - \log 2 \; < \; \sum_{j=1}^n \; \frac{1}{j} \; < \; 1+ \log n $$
|
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|
Solution verification:$\lim_{x\to 2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$
Evaluate without L'Hospital:$$\lim_{x\to
2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$$
My attempt:
I used: $$\lim_{f(x)\to 0}\frac{\ln(1+f(x))}{f(x)}=1\;\&\;\lim_{f(x)\to 0}\frac{a^{f(x)}-1}{f(x)}=\ln a$$
$$
\begin{split}
L &= \lim_{x\to 2} \frac{\ln(x-1)}{3^{x-2}-5^{-x+2}} \\
&= \lim_{x\to 2} \frac{\dfrac{\ln(1+(x-2))}{x-2}\cdot(x-2)}
{(x-2)\cdot\dfrac{3^{x-2}-1+1-5^{-x+2}}{x-2}} \\
&= \lim_{x\to 2} \frac{\dfrac{\ln(1+(x-2))}{x-2}}
{\dfrac{3^{x-2}-1}{x-2}+\dfrac{5^{2-x}-1}{2-x}} \\
&=\frac{1}{\ln3+\ln5} \\
&=\frac{1}{\ln(15)}
\end{split}
$$
Is this correct?
|
Your answer a bit rephrased.
$y=x-2;$
$\dfrac{\log (1+y)}{3^y-5^{-y}}$;
Numerator :
$f(y):=y\dfrac{\log (1+y)-\log 1}{y}$
Denominator:
$g(x)=\dfrac{15^y-1}{5^y}=$
$5^{-y}(15^y-1)=$
$5^{-y}(y\log 15)\dfrac{e^{y\log 15}-1}{y\log 15}$;
$\dfrac{f(x)}{g(x)}=$
$[\dfrac{\log (y+1)-\log 1}{y}]\cdot$
$[\dfrac{5^y}{\log 15}]$
$ \big [\dfrac{1}{\dfrac{e^{y\log15}-1}{y\log 15}}\big ].$
Take the limit $y \rightarrow 0$.
|
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|
A generalization of the (in)famous IMO 1988 problem 6: If $\frac{a^2 + b^2 - abc}{ab + 1}$ is a positive integer then it is a square. This question is motivated by the famous IMO $1988$ problem $6$. Is the following true?
Let $a,b$ be positive integers and $c \ge 0$ be a non-negative integer. If $\dfrac{a^2 + b^2 - abc}{ab + 1}$ is a positive integer then it is a square.
|
Yes, this fact is true. We have:
$$\frac{a^2+b^2-abc}{ab+1}=k \implies a^2-b(c+k)a+(b^2-k)=0$$
Following the standard procedure for Vieta jumping, assume that $(a,b)$ is the smallest solution with respect to the sum of solutions $a+b$, such that $k$ is not a perfect square. If one root of $$x^2-b(c+k)x+(b^2-k)$$ is $a$, then:
$$x=b(c+k)-a$$
$$x=\frac{b^2-k}{a}$$
WLOG let $a>b$. We can note that the first equation gives $x$ is an integer, while the second gives:
$$x<\frac{b^2}{a}<\frac{a^2}{a}=a \implies x<a$$
By the minimality of the solution $(a,b)$, we cannot have $(b,x)$ as a solution. Thus, we must have $x$ to be non-negative. We cannot have $x=0$ as this would give $b^2=k$ contradicting the assumption that $k$ is not a square.
If $x<0$, we must have $b^2<k$. This gives:
$$(k-b^2)(ab+1)=a^2+b^2-abc-ab^3-b^2=a(a-bc-b^3) \implies \frac{k-b^2}{a}=\frac{a-bc-b^3}{ab+1}$$
$$-x=\frac{a-bc-b^3}{ab+1}<\frac{a}{ab+1}<1 \implies x>-1$$
This is a contradiction since there is no integer $0>x>-1$. Thus, there were no solutions to begin with, where $k$ is not a square.
|
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|
How to multiply $\sqrt{5x^2+2x+1}$ with $\frac{1}{x}$? I believe the answer says, $\sqrt{5+\frac{2}{x}+\frac{1}{x^2}}$, but I don't see how they obtained it. I'm quite rusty with radicals.
|
$f(x) = \sqrt{5 x^2 + 2x +1} $ is the positive function whose square $f^2$ is $f^2=5 x^2 + 2x +1$.
Hence the square $(\frac f x)^2$ is $(\frac f x)^2 = 5 +\frac 2 x + \frac 1 {x^2}$, which implies that
$$\Big | \frac{\sqrt{ 5x^2 +2x+1}}{x}\Big|= \sqrt { 5 +\frac 2 x + \frac 1 {x^2}}.$$
|
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|
Why does't quadratic formula work to factor polynomial when $a \ne 1$? $$2x^2 + 3x + 1$$
applying quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$a=2, b=3, c=1$$
$$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot1}}{2\cdot2}$$
$$x = \frac{-3 \pm \sqrt{9-8}}{4}$$
$$x = \frac{1}{4}[-3 + 1],~~~x=\frac{1}{4}[-3-1]$$
$$x_1 = -1/2,~~~x_2 = -1$$
therefore:
$$2x^2 + 3x + 1 = (x + 1/2)(x+1)$$
Now I check it:
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x \cdot x + x \cdot 1 + \frac{1}{2}\cdot x + \frac{1}{2}\cdot 1$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + x + \frac{1}{2}x + \frac{1}{2}$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + x + \frac{1}{2}x + \frac{1}{2}$$
$$2x^2 + 3x + 1 = (x + \frac{1}{2})(x+1) = x^2 + \frac{3}{2}x + \frac{1}{2}$$
but:
$$2x^2 + 3x + 1 \ne x^2 + \frac{3}{2}x + \frac{1}{2}$$
Why does't quadratic formula work when $a \ne 1$?
however, I can pull out the 1/2.
$$2x^2 + 3x + 1 \ne \frac{1}{2}[2x^2 + 3x + 1]$$
I feel that this is something I must have missed in grade school...
Does this mean you need to multiple by "a" if $a \ne 1$?**
|
Your error is in forgetting about the leading coefficient. If a monic polynomial $p(x) $ has roots $r_1,\ldots, r_n$, with multiplicity, then
$$p(x) =(x-r_1) \cdots(x-r_n) $$
If it is not monic but instead has a leading coefficient of $a$, then
$$p(x) =a(x-r_1) \cdots(x-r_n)$$
Note that for this to work you need to include all of the roots. For polynomials with real coefficients, some of these roots may be complex.
|
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|
Prove that if $x$ and $y$ are both not $0$ Prove that if $x$ and $y$ are both not $0,$ then
$$x^4+x^3y+x^2y^2+xy^3+y^4>0$$
I know this seems fairly easy but I'm fairly new to calculus and need some help proving that this is true. Appreciate the help!
|
\begin{align}
& x^4 + x^3y + x^2y^2 + xy^3 + y^4 \\[8pt]
= {} & \left( \left( \tfrac x y \right)^4 + \left( \tfrac x y \right)^3 + \left( \tfrac x y \right)^2 + \left( \tfrac x y \right) + 1 \right) y^4 \\[8pt]
= {} & \Big(u^4 + u^3 + u^2 + u + 1\Big) y^4.
\end{align}
Since $y^4>0$ except when $y=0,$ it is enough to look at this $4$th-degree polynomial in $u.$
Recall from algebra that
$$
(u-1)(u^4+u^3+u^2+u+1) = u^5-1.
$$
If you solve $u^5-1=0$ for $u$ you get the $5$th roots of $1$ as solutions. De Moivre's formula says these are
$$
\cos \tfrac {2\pi k} 5 + i\sin \tfrac{2\pi k} 5 \text{ for } k = 0,1,2,3,4.
$$
When $k=0$ this number is $1$ and that is a root of $u-1=0,$ so the other four are the roots of $u^4+u^3+u^2+1=0.$ Since the coefficients of this polynomial are real, the roots must come in complex-conjugate pairs. And here we note that the cases $k=1$ and $k=4$ are complex conjugates of each other, as are the cases $k=2$ and $k=3,$ so we have these roots:
$$
\cos\tfrac{2\pi} 5 \pm i \sin\tfrac {2\pi}5 \text{ and } \cos\tfrac{4\pi}5 \pm i \sin\tfrac{4\pi}5.
$$
Therefore
\begin{align}
& u^4+u^3+u^2+u+1 \\[8pt]
= {} & \left( u - \left( \cos\tfrac{2\pi} 5 + i \sin\tfrac {2\pi}5 \right) \right) \left( u - \left(\cos\tfrac{2\pi} 5 - i \sin\tfrac {2\pi}5\right) \right) \\
& {} \times \left( u - \left( \cos\tfrac{4\pi} 5 + i \sin\tfrac {2\pi}5 \right) \right) \left( u - \left(\cos\tfrac{2\pi} 5 - i \sin\tfrac {4\pi}5\right) \right) \\[8pt]
= {} & \left( u^2 - u\cos\tfrac{2\pi} 5 + 1 \right) \left( u^2 - u \cos\tfrac{4\pi} 5 + 1 \right)
\end{align}
Each of these is a quadratic polynomial whose roots are not real. Each has a graph that is a parabola that opens upward, not downward. Therefore each factor is always positive.
I would call this algebra rather than calculus. I would also guess that most students who have taken the prerequisite courses for calculus would not handle this well at all.
|
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|
Prove that $A^2=a_1^2+a_2^2-a_3^2-a_4^2$ for all integers $A$.
Prove that:
For every $A\in\mathbb Z$, there exist infinitely many $\{a_1,a_2,a_3,a_4\}\subset\mathbb Z$ given $a_m\neq a_n $ such that $$A^2=a_1^2+a_2^2-a_3^2-a_4^2$$
After many hours, I found a general formula satisfying the statement for all $A$ and $B$. $$A^2=(3A+B)^2+(9A+2B)^2-(5A+B)^2-(8A+2B)^2$$ This was derived by noticing the following pattern, to which I used the first equation below as a seed and multiplied through $A^2$. I was then able to find a second paramater $B$ due to the arithmetic progression of the pattern, in order to prove the infinitude of $\{a_n\}_{n=1}^{4}$. $$\begin{align}1^2+5^2+8^2 &= 3^2+9^2 \\ 1^2+6^2+10^2 &= 4^2+11^2 \\ 1^2+7^2+12^2 &= 5^2+13^2 \\ &\vdots\end{align}$$
My question is, is there a way one could prove the statement without the involvement of such curious patterns surrounding square numbers? Apologies if this question is somewhat vague.
Edit:
Fun fact, it appears there is also a general formula for the equation $$A=a_1^2+a_2^2+a_3^2-a_4^2+a_5^2$$ That is, $$A^2=(A+B)^2+(A+3B)^2+(A+8B)^2-(A+5B)^2-(A+7B)^2$$
Edit 2: It appears that the first general equation at which I arrived in this question is actually part of an even more general equation $$(pq +s)^2=\big\{p(3q+r)+3s\big\}^2+\big\{p(9q+r)+3s\big\}^2-\big\{p(5q+r)+s\big\}^2-\big\{2p(4q+r)+4s\big\}^2+4pqs$$ where $(p,q,r,s)=(1,A,B,0)$. Note that an interesting fact implies when $p$, $q$ and $s$ are square numbers.
|
$a^2=a1^2+a2^2-a3^2-a4^2\tag{1}$
Let assume $p^2+q^2-r^2-s^2 = 1\tag{2}$
Equation $(2)$ has many parametric solutions.
We use one of the solutions, $(p,q,r,s)=(2n+1, n-1, n+1, 2n).$
( By consider two identies, $(n+1)^2 - (n-1)^2 = 4n, (2n+1)^2 - (2n)^2 = 4n+1$ )
$n$ is arbitrary.
Substitute $a1=pt+c, a2=qt+d, a3=rt+c, a4=st+d, a=t$ to equation $(1)$, then we get $$c=s-q$$ $$d=p-r.$$
Thus, we get a parametric solution below.
\begin{eqnarray}
&a& = t \\
&a1& = (2n+1)t+n+1 \\
&a2& = (n-1)t+n \\
&a3& = (n+1)t+n+1 \\
&a4& = 2nt+n \\
\end{eqnarray}
$t$ is arbitrary.
Example:
\begin{eqnarray}
&(t)^2& = (3t+2)^2 + (1)^2 - (2t+2)^2 - (2t+1)^2 \\
&(t)^2& = (5t+3)^2 + (t+2)^2 - (3t+3)^2 - (4t+2)^2 \\
&(t)^2& = (7t+4)^2 + (2t+3)^2 - (4t+4)^2 - (6t+3)^2 \\
&(t)^2& = (9t+5)^2 + (3t+4)^2 - (5t+5)^2 - (8t+4)^2 \\
&(t)^2& = (11t+6)^2 + (4t+5)^2 - (6t+6)^2 - (10t+5)^2 \\
\end{eqnarray}
|
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|
The equation $a^{4n}+b^{4n}+c^{4n}=2d^2$ Recently, I found that if $a+b=c$, then $a^4+b^4+c^4=2d^2$ for some positive integer $d$. The parametric equation is:
$$m^4+n^4+(m+n)^4=2(m^2+mn+n^2)^2$$
The condition $a+b=c$ (assuming $c \geqslant a,b$) isn't necessary. For example:
$$7^4+7^4+12^4=2 \cdot 113^2$$
We can note that when we make the equation in the form $a^{4n}+b^{4n}+c^{4n}=2d^2$, and we impose the condition $a^n+b^n=c^n$ for the parametric solution:
(i) When $n=1$, we can have any positive integers $a+b=c$
(ii) When $n=2$, we can have any Pythagorean Triple $(a,b,c)$.
(iii) When $n>2$, there are no solutions by Fermat's Last Theorem.
Checking when $n=2$, I saw that there are no solutions for $a \leqslant b \leqslant c \leqslant 3000$ where $a^2+b^2 \neq c^2$. I have not run a program for any value $n>2$ though.
For positive integers $a \leqslant b \leqslant c$ where
$\gcd(a,b,c)=1$ :
$1$. Are there any solutions for $a^8+b^8+c^8=2d^2$ where $a^2+b^2
\neq c^2$ ?
$2$. Are there any solutions for $a^{4n}+b^{4n}+c^{4n}=2d^2$ where
$n>2$?
$3$. For the solutions of $a^4+b^4+c^4=2d^2$ which do not follow
$a+b=c$, is there any way of generating more solutions from primitive
solutions? From primitive solution $(a,b,c,d)$, can we get more
solutions $(A,B,C,D)$?
EDIT : First off, it suffices to focus on solutions for $a^{4n}+b^{4n}+c^{4n}=2d^2$ for prime $n$ alone, since if we have a solution for some $n$, then we have a solution for the divisors of $n$ as well. An accepted answer would be one of:
$(i)$ Verifying problem $1$ for $a \leqslant b \leqslant c \leqslant
1000000$.
$(ii)$ Verifying problem $2$ for $a \leqslant b \leqslant c \leqslant
100000$ (for odd primes $n<100$).
$(iii)$ Verifying problem $1$ for $a \leqslant b \leqslant c \leqslant
100000$ and problem $2$ for $a \leqslant b \leqslant c \leqslant
10000$ (for odd primes $n<100$).
$(iv)$ Proof or Counterexample for either problems $1$ or $2$.
$(v)$ Relations, generation or parametric characterization of the
non-trivial solutions of $$a^4+b^4+c^4=2d^2$$
|
Problem 3
This is a scheme to generate the solutions which, like your example of $(7,7,12,113)$, have two of $a,b,c$ equal.
Consider the following system of three closely related equations.
E: $2x^4-y^4=z^2$
F: $x^4+8y^4=z^2$
G: $x^4-2y^4=z^2$
A 'base solution' $(x,y,z)$ of E can be used to generate a solution $(z,xy,2x^4+y^4)$ of F.
Each solution $(x,y,z)$ of F can be used to generate a solution $(z,2xy,|x^4-8y^4|)$ of G.
Each solution $(x,y,z)$ of G can be used to generate a further solution $(z,xy,x^4+2y^4)$ of F.
Each solution $(x,y,z)$ of F can be used to generate the solution $(x,x,2y,z)$ of the required equation.
Example starting with the solution $(1,1,1)$ of E.
The scheme generates F$(1,1,3)$, G$(3,2,7)$,F$(7,6,113)$, G$(113,84,7967)$, F$(7967,9492,262621633)$, .....
The required solutions are then $$(1,1,2,3),(7,7,12,113),(7967,7967,18984,262621633),...$$
|
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|
First order PDE problem $yu_{x} - xu_{y} = x^2$ I am a beginner to PDE, trying to solve $$yu_{x} - xu_{y} = x^2$$ using characteristic line, first I have \begin{align*}
\frac{dx}{y} &= -\frac{dy}{x} = \frac{du}{x^2}
\end{align*}
Then I get the characteristic line is given by $$C = \frac{1}{2}y^2 - \frac{1}{2}x^2$$
Next, I solve the first term and third term, I have \begin{align*}
\frac{du}{x^2} &= \frac{dx}{y} \\
du &= \frac{x^2}{y}dx
\end{align*}
Here is my problem, that is we can not integrate right-hand side without eliminating the variable $y$, but if we try to replace $y$ in terms of $x$ and $C$, the result does not look integrable and really messy.
Can someone help me, please? Thanks!
|
\begin{align*}
\frac{dx}{y} &= -\frac{dy}{x} = \frac{du}{x^2}
\end{align*}
From the first two ratios,
$$\frac{dx}{y} = -\frac{dy}{x} \implies x~dx~+~y~dy~=0 $$
Integrating, $~x^2+y^2=c^2~\tag1$where $~c~$ is integrating constant.
From the last two ratios, $$ -\frac{dy}{x} = \frac{du}{x^2} \implies du=-x~dy\implies du=-\sqrt{c^2-y^2~}~dy \qquad\text{[using equation $(1)$]}$$
Integrating, $$u=-\dfrac 12~y\sqrt{c^2-y^2}~-~\dfrac {c^2}{2} \sin^{-1}\dfrac yc~+~d\qquad\text{(using direct formula)
}$$
$$u=-\dfrac 12~xy~-~\dfrac {1}{2}(x^2+y^2) \sin^{-1}\left(\dfrac y{\sqrt{x^2+y^2}}\right)~+~d\qquad\text{[using equation $(1)$]}$$where $d$ is integrating constant.
Hence the general solution is of the form$$u=-\dfrac 12~xy~-~\dfrac {1}{2}(x^2+y^2) \sin^{-1}\left(\dfrac y{\sqrt{x^2+y^2}}\right)~+~\phi\left(\sqrt{x^2+y^2~}\right)$$where $~\phi~$ is arbitrary function of $~x,~y~$.
or, in the form of$$f\left(\sqrt{x^2+y^2}~,~u~+~\dfrac 12~xy~+~\dfrac {1}{2}(x^2+y^2) \sin^{-1}\left(\dfrac y{\sqrt{x^2+y^2}}\right)\right)=0$$where $~f~$ is arbitrary function of $~x,~y~$.
Integration formula :
$$\int\sqrt{a^2-x^2~}~dx=\dfrac 12~x\sqrt{a^2-x^2~}+\dfrac{a^2}{2}~\sin^{-1}\left(\dfrac{x}{a}\right)+c$$ where $~c~$ is a constant of integration.
|
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|
Identity involving falling factorial Let $(x)_n$ denote the falling factorial,
$$(x)_n = x(x-1) \dots (x - n + 1).$$
I came across the following identity in trying to solve another problem.
$$\sum_{k=1}^m \frac{(a)_k}{(x)_k} = \frac{a}{x - a + 1}\left( 1 - \frac{(a-1)_m}{(x)_m}\right).$$
This identity allows you to find closed forms for sums such as
$$\sum_{k=0}^m \frac{\Gamma(x + k)}{\Gamma(y + k)}$$
or
$$\sum_{k = 0}^c \frac{a \choose k}{b \choose k}.$$
It is similar to a series mentioned here (Infinite sum of falling factorial quotients. ), except that my sum is finite.
I proved the identity by induction, which isn't too hard, as long as you know what you're looking for. I have a couple of questions, though.
*
*I tried to find this identity mentioned somewhere, but couldn't find it. Is it well-known, or a special case of a better known identity?
*I found the expression on the right-hand side through trial and error. Is there a more natural derivation of the identity, particularly if you know the left-hand side and are looking for the right-hand side?
For example, after multiplying through by $(x - a + 1)(x)_m$, the identity becomes one of polynomials in $a$ and $x$, hence it would be sufficient to prove it for sufficiently large integers $a$ and $x$. Perhaps there is a combinatorial explanation.
|
We prove
\begin{align*}
\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}=\frac{b+1}{b-a+1}\left(\frac{\binom{a}{m}}{\binom{b+1}{m}}-\frac{\binom{a}{n+1}}{\binom{b+1}{n+1}}\right)\tag{1}
\end{align*}
We start with the right-hand side of (1) and obtain
\begin{align*}
\color{blue}{\frac{b+1}{b-a+1}}&\color{blue}{\left(\frac{\binom{a}{m}}{\binom{b+1}{m}}-\frac{\binom{a}{n+1}}{\binom{b+1}{n+1}}\right)}\\
&=\frac{b+1}{b-a+1}\sum_{k=m}^n\left(\frac{\binom{a}{k}}{\binom{b+1}{k}}-\frac{\binom{a}{k+1}}{\binom{b+1}{k+1}}\right)\tag{2}\\
&=\frac{b+1}{b-a+1}\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}\left(\frac{1}{\frac{b+1}{b-k+1}}-\frac{\frac{a-k}{k+1}}{\frac{b+1}{k+1}}\right)\tag{3}\\
&=\frac{b+1}{b-a+1}\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}\left(\frac{b-k+1}{b+1}-\frac{a-k}{b+1}\right)\\
&=\frac{1}{b-a+1}\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}\left(b-k+1-(a-k)\right)\\
&\,\,\color{blue}{=\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}}
\end{align*}
and the claim (1) follows.
Comment:
*
*In (2) we write the expression as telescoping sum.
*In (3) we factor out $\frac{\binom{a}{k}}{\binom{b}{k}}$ and simplify in the following steps.
|
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What is the sum of the coefficients of the terms containing $x$ (Problem 105, Algebra, Gelfand) In problem 105 described below, the solution provided is:
The sum of the terms containing $x$ is the sum of the terms not
containing y less the sum of the constant terms. There can be only one
constant term which is the result of cubing the 1 in $ (1+x-y)^3 $. Then
the sum of the coefficients of the terms containing $x$ (based on the
previous problem) must be $8-1=7.$
I think his solution is wrong as it's only taking into account terms that only contain $ x $ and not also the terms that include $xy$
Is my logic correct or am I missing something?
|
You are right, the Problem 105 must read: "What is the sum of the coefficients of the terms containing $\color{red}{\text{only}}$ $x$?"
Note that $(1+x-y)^3$ is a trinomial and its expansion is:
$$f(x,y)=(1+x-y)^3=\sum_{i,j,k\\ i+j+k=3} {3\choose i,j,k}1^ix^j(-y)^k=\\
{3\choose 3,0,0}+{3\choose 2,1,0}x+{3\choose 2,0,1}(-y)+{3\choose 1,2,0}x^2+{3\choose 1,1,1}x(-y)+{3\choose 1,0,2}(-y)^2+{3\choose 0,3,0}x^3+{3\choose 0,2,1}x^2(-y)+{3\choose 0,1,2}x(-y)^2+{3\choose 0,0,3}(-y)^3$$
Problem 103. The sum of coefficients is:
$$f(1,1)=(1+1-1)^3=1$$
Problem 104. The sum of coefficients of the terms not containing $y$ is:
$$f(1,0)=(1+1-0)^3=8$$
Problem 105. The sum of coefficients of the terms containing $\color{red}{\text{only}}$ $x$ is:
$$f(1,0)-f(0,0)=(1+1-0)^3-(1+0-0)^3=7$$
Problem 105*. The sum of coefficients of the terms containing $x$ is:
$$f(1,1)-f(0,1)=(1+1-1)^3-(1+0-1)^3=1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can diagonalization of a positive definite matrix be used to compute its determinant and inverse? I have a positive definite matrix $A= E^T DE$. $A\in \mathcal{R}^{n \times n} $.
$E$ is a orthonormal matrix $E \in \mathcal{R}^{m \times n}$ consisting of n orthonormal columns where $m >n $. And, $E^T E=I_{n \times n}$ (this follows directly from the orthonormality).
Also, $D \in \mathcal{R}^{m \times m}$ is a diagonal matrix.
Notice that this diagonalization is somewhat similar to the singular value decomposition(svd)but not quite as svd requires $A=udv^t$ where $u$ has orthonormal columns not rows as is the case here.
Given, $E$ and $D$, is it possible to compute the inverse and determinant of A?
|
I don't see any simple method for computing the determinant. What I can confirm is that there is no function for $\det A$ purely in terms of $D$; the entries in $E$ must also be considered in any formula. To prove this, all we need is a diagonal matrix, and two orthogonal $m \times n$ matrices $E_1, E_2$ such that $\det A_1 \neq \det A_2$, where $A_i = E_i^T D E_i$.
Let
\begin{align}
D &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \\
E_1 &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} \\
E_2 &= \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1 \end{pmatrix}.
\end{align}
Then,
\begin{align}
A_1 &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \\
A_2 &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1 \end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix},
\end{align}
which have respective determinants $2$ and $3$.
|
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|
Summation and factorial. If $$2^{n}a_{n}=\sum_{k=0}^{n}{a_{k}a_{n-k}},\phantom{a}\forall n\in\mathbb{N},$$ with $a_{0}=1,$ then $$a_{n}=\dfrac{a_{1}^{n}}{n!},\phantom{a}\forall n\in\mathbb{N}.$$
Attempt: Notice that
(1) For $ n=2t + 1, $ we have $$2^{2t+1}a_{2t+1}=2(a_{0}a_{2t+1}+a_{1}a_{2t}+a_{2}a_{2t-1}+\cdots+a_{t-1}a_{t+2}+a_{t}a_{t+1}).$$
Then
\begin{align}
(2^{2t}-1)a_{2t+1}&=a_{1}a_{2t}+a_{2}a_{2t-1}+\cdots+a_{t-1}a_{t+2}+a_{t}a_{t+1}\\
&=a_{1}^{2t+1}\left(\dfrac{1}{(2t)!}+\dfrac{1}{2!(2t-1)!}+\cdots+\dfrac{1}{(t-1)!(t+2)!}+\dfrac{1}{t!(t+1)!}\right)\\
&=\dfrac{a_{1}^{2t+1}}{(2t)!}\left(1+\dfrac{2t}{2!}+\cdots+\dfrac{(t+3)(t+4)\cdots(2t-1)(2t)}{(t-1)!}+\dfrac{(t+2)(t+3)(t+4)\cdots(2t-1)(2t)}{t!}\right).
\end{align}
(2) For $n=2t,$ we have $$2^{2t}a_{2t}=2a_{0}a_{2t}+2a_{1}a_{2t-1}+2a_{2}a_{2t-2}+\cdots+2a_{t-1}a_{t+1}+a_{t}^{2}.$$
Then
\begin{align}
2(2^{2t-1}-1)a_{2t}&=2a_{1}a_{2t-1}+2a_{2}a_{2t-2}+\cdots+2a_{t-1}a_{t+1}+a_{t}^{2}\\
&=a_{1}^{2t}\left(\dfrac{2}{(2t-1)!}+\dfrac{2}{2!(2t-2)!}+\cdots+\dfrac{2}{(t-1)!(t+1)!}+\dfrac{1}{t!t!}\right)\\
&=\dfrac{a_{1}^{2t}}{(2t-1)!}\left(2+\dfrac{2(2t-1)}{2!}+\cdots+\dfrac{2(t+2)(t+3)\cdots(2t-1)}{(t-1)!}+\dfrac{(t+1)(t+2)(t+3)\cdots(2t-1)}{t!}\right).
\end{align}
Is there a formula to reduce the above expressions?
Any hints would be appreciated.
|
The sum is a convolution of $a_n$ with itself, so the generating function of the RHS is $A(z)^2$, where $A(z)=\sum_n a_n z^n$ is the generating function of $a_n$. The generating function of $2^n a_n$ is $A(2z)$, so we have $A(2z)=A(z)^2$. Hence $A(z)=\exp(c z)$ for some constant $c$, which implies that $a_n=c^n/n!$. Now evaluate at $n=1$ to obtain $a_1=c^1/1!=c$, so $a_n=a_1^n/n!$.
|
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|
Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x(3+x)}{\sqrt{3-x^2}(3+x)^2}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
At this point, I want to transform this derivative into the form of $\dfrac{3(x+1)}{(3+x)^2\sqrt{3-x^2}}$
How do I do this? This form is given by Wolfram:
https://www.wolframalpha.com/input/?i=derivative+%283-x%5E2%29%5E%281%2F2%29%2F%283%2Bx%29
|
You just need to take the LCM of the denominators.
$$\begin{align}\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2} &= \frac{-x(3 + x) - \sqrt{3 - x^2}\sqrt{3 - x^2}}{(3 + x)^2\sqrt{3 - x^2}} \\ &= \dfrac{-3x - x^2 - 3 + x^2}{(3 + x)^2\sqrt{3 - x^2}} \\ &= -\dfrac{3(x + 1)}{(3 + x)^2\sqrt{3 - x^2}}\end{align}$$
|
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|
Given the points $A(2a, a)$ and $B(2b, b)$ find the coordinates of the point $M$ such that $\vec{AM} = 3 \vec{MB}$. Consider the points:
$$A(2a, a) \hspace{2cm} B(2b, b)$$
with $a \ne b$ and $a, b \in \mathbb{R}$. Find the point $M(x, y)$ such that $\vec{AM} = 3 \vec{MB}$.
First thing I tried was to plot everything we know so far. Here's what the graph looks like with $a = 2$ and $b = 4$:
So it's clear that what I have to do is find the point $M(x, y)$ on the segment $[AB]$ such that we have $\vec{AM} = 3 \vec{MB}$. I thought that an interactive graph would be helpful to see how thing move around, so here it is.
What I did first was create the vectors
$$\vec{OA} = \begin{pmatrix}
2a \\
a
\end{pmatrix}
\hspace{2cm}
\vec{OB} = \begin{pmatrix}
2b \\
b
\end{pmatrix}$$
Then I got the vector $\vec{AB}$:
$$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix}
2(b - a) \\
b - a
\end{pmatrix}$$
Now since we need $\vec{AM} = 3\vec{MB}$ that means that the point $\vec{AM}$ is situated at three quarters length of the whole vector $\vec{AB}$. So since we now have the vector $\vec{AB}$ we just need to shorten this vector, while keeping its direction. So I created the matrix:
$$X = \begin{pmatrix}
\dfrac{3}{4} & 0 \\
0 & \dfrac{3}{4}
\end{pmatrix}$$
The matrix $X$ squishes each vector to three quarters of its length.
So now we have:
$$\vec{AM} = X \cdot \vec{AB} =
\begin{pmatrix}
\dfrac{3}{4} & 0 \\
0 & \dfrac{3}{4}
\end{pmatrix}
\begin{pmatrix}
2(b - a) \\
b - a
\end{pmatrix}
=
\begin{pmatrix}
\dfrac{3(b - a)}{2} \\
\dfrac{3(b - a)}{4}
\end{pmatrix}$$
So we can now find the coordinates of $M$.
$$\vec{OM} = \vec{OA} + \vec{AM} = \begin{pmatrix}
2a \\
a
\end{pmatrix}
+
\begin{pmatrix}
\dfrac{3(b - a)}{2} \\
\dfrac{3(b - a)}{4}
\end{pmatrix}
=
\begin{pmatrix}
\dfrac{a+3b}{2} \\
\dfrac{a+3b}{4}
\end{pmatrix}$$
So we can conclude that the point $M$ is at:
$$M \bigg ( \dfrac{a+3b}{2}, \dfrac{a+3b}{4} \bigg )$$
My question is: Is this correct? We've never used matrices like this before in class, it's just an idea that came to me, so I have no idea if my work is correct, even though it seems right. And if it is correct, what is another way of solving this problem? I couldn't come up with any other solution, so that's why I had to do this "slight of hand".
|
What you're doing looks correct. Using vectors and matrices is a valid way to proceed. However, as you requested, here is an alternate method to consider.
The line going through $A$ and $B$ has a slope of
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{b - a}{2b - 2a} = \frac{1}{2} \tag{1}\label{eq1A}$$
Thus, the general equation for it would be
$$y = \left(\frac{1}{2}\right)x + c \tag{2}\label{eq2A}$$
Using $y = a$ and $x = 2a$ gives that $c = 0$, so $y = \left(\frac{1}{2}\right)x \implies x = 2y$. Note this can be parametrized by having $y = a + t$ so $x = 2a + 2t$, giving
$$f(t) = (2a + 2t, a + t) \tag{3}\label{eq3A}$$
with $f(0) = A(2a, a)$ and $f(b - a) = B(2b, b)$. Since $M$ is three-quarters of the way between $A$ and $B$, then the point is
$$\begin{equation}\begin{aligned}
M(x,y) & = f\left(\left(\frac{3}{4}\right)\left(b-a\right)\right) \\
& = \left(2a + 2\left(\frac{3}{4}\right)\left(b-a\right), a + \left(\frac{3}{4}\right)\left(b-a\right)\right) \\
& = \left(\frac{4a}{2} + \frac{3b-3a}{2}, \frac{4a}{4} + \frac{3b-3a}{4}\right) \\
& = \left(\frac{a + 3b}{2}, \frac{a + 3b}{4}\right)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This matches your answer of $M\left(\frac{a + 3b}{2}, \frac{a + 3b}{4}\right)$.
|
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|
Is the function $f$ continuous at $(0,0)$
Is the function $f$ continuous at $(0,0)$?
$f(x, y)$ := $\frac{xy}{|x|+|y|}$, if $(x, y)$ $\not= (0, 0)$
and $f(0, 0) := 0$
My attempt:
$f(x, y)$ = $\frac{xy}{x+y}$ if $(x,y) > (0,0)$ and
$f(x, y)$ = $\frac{xy}{-(x+y)}$ if $(x,y) < (0,0)$
So using polar coordinates:
$f(x, y)$ = $\frac{xy}{x+y}$ = $\frac{rcos\theta sin\theta}{cos\theta + sin\theta}$, thus $|f(x,y)| <= r = {(x^2 + y^2)}^{0.5} $ and thus $f$ is continuous.
$f(x, y)$ = $\frac{xy}{-(x+y)}$ = $\frac{-rcos\theta sin\theta}{cos\theta + sin\theta}$, thus $|f(x,y)| <= - r = {-(x^2 + y^2)}^{0.5} $ and thus $f$ is not continuous.
Is my answer correct?
|
$(x,y)\not =(0,0)$;
1) $x\not =0$;
$|\dfrac{xy}{|x|+|y|}| \le |y| \lt \sqrt{x^2+y^2};$
2) $y \not =0$;
$|\dfrac{xy}{|x|+|y|}| \le |x| \lt \sqrt{x^2+y^2};$
$\epsilon >0$ given.
Choose $\delta =\epsilon$.
|
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|
the $\max$ of $\min$ $a,b,c>0$,
$$\begin{cases}a+b+c=8\\ab+bc+ac=16\end{cases}$$
$m=\min\{ab,bc,ca\}$ , how to get $m_{\max}$
With Mathematica
Maximize[{Min[a b, b c, c a], a + b + c == 8, a b + b c + c a == 16, a > 0, b
> 0, c > 0}, {a, b, c}]
$$\left\{\frac{16}{9},\left\{a\to \frac{4}{3},b\to \frac{4}{3},c\to \frac{16}{3}\right\}\right\}$$
but how to do it by hand?
|
Let $a\geqslant b\geqslant c$, $b=c+u$ and $a=c+u+v$.
Thus, $u$ and $v$ are non negatives, $$3c+2u+v=8$$ and
$$(c+u)(c+u+v)+c(c+u)+c(c+u+v)=16$$ or
$$3c^2+(4u+2v)c+u(u+v)=16$$ or
$$9c^2+(4u+2v)3c+3u(u+v)=48,$$ which gives
$$(8-2u-v)^2+(4u+2v)(8-2u-v)+3u^2+3uv=48$$ or
$$u^2+v^2+uv=16.$$
Now, since $$ab\geqslant ac\geqslant bc,$$ we'll prove that
$$c(c+u)\leqslant\frac{16}{9}$$ or
$$\frac{8-2u-v}{3}\left(\frac{8-2u-v}{3}+u\right)\leqslant\frac{16}{9}$$ or
$$2u^2-uv-v^2+8(u+2v)\geqslant48$$ or
$$2u^2-uv-v^2+2(u+2v)\sqrt{u^2+uv+v^2}\geqslant3(u^2+uv+v^2)$$ or
$$2(u+2v)\sqrt{u^2+uv+v^2}\geqslant(u+2v)^2$$ or
$$4u^2+4uv+4v^2\geqslant u^2+4uv+4v^2,$$ which is obvious.
The equality occurs for $u=0$ and $v^2=16,$ id est for $(u,v)=(0,4)$, $c=b=\frac{4}{3}$ and $a=\frac{16}{3},$
which says that $$m_{\max}=\frac{16}{9}.$$
|
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|
Find $\int_0^{2\pi} \frac{1}{3 + \cos x} dx$. I have to find the integral
$$\int_0^{2\pi} \dfrac{1}{3 + \cos x} dx$$
I tried using the Weierstrass subtitution, but replacing the bounds, I get:
$$t_1 = \tan \dfrac{0}{2} = \tan 0 = 0$$
$$t_2 = \tan \dfrac{2 \pi}{2} = \tan \pi = 0$$
Resulting in the integral:
$$\int_0^0 \dfrac{1}{3 + \dfrac{1 - t^2}{1 + t^2}} \cdot \dfrac{2}{1 + t^2} dt$$
which obviously equals $0$ since the bounds are the same. Is this correct? It feels wrong.
|
Using the identity
$$\frac{1}{a+b\cos t}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\ a>b$$
we have
$$\frac{1}{3+\cos x}=\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{2}}\sum_{n=1}^{\infty}\left(\sqrt{8}-3\right)^n\cos{(nx)}$$
giving us
$$\int_0^{2\pi}\frac{1}{3+\cos x}\ dx=\frac{1}{\sqrt{8}}\int_0^{2\pi}\ dx+\frac{1}{\sqrt{2}}\sum_{n=1}^{\infty}\left(\sqrt{8}-3\right)^n\underbrace{\int_0^{2\pi}\cos{(nx)}\ dx}_{0}=\frac{\pi}{\sqrt{2}}$$
|
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|
I don't understand how -12 is the numerator in this difference quotient. Find the difference quotient of $$\frac{\left(\frac{1}{12(x+h)+3}\right)-\left(\frac{1}{12x+3}\right)}{h}$$
When I solve it I end up canceling h's at the end and only having -12 as the answer. However, the correct answer is -12 over the denominator of the fractions in the numerator.
I solved it like this:
$$\frac{\left(\frac{1}{12x+12h+3}\right)-\left(\frac{1}{12x+3}\right)}{h}$$
Then multiplied each fraction by $(12x+12h+3)(12x+3)$ to get:
$$\frac{12x+3-12x-12h-3}{h}$$
Then simplify the numerator to this:
$$\frac{-12h}{h}$$
Then cancel h which gave me -12.
|
$$
\frac{\frac{1}{12(x+h)+3}-\frac{1}{12x+3}}{h}=\frac{1}{h}\cdot \left(\frac{1}{12(x+h)+3}-\frac{1}{12x+3}\right)=\\
\frac{1}{h}\cdot \frac{12x+3-12(x+h)-3}{(12(x+h)+3)(12x+3)}=\\
\frac{1}{h}\cdot \frac{-12h}{(12(x+h)+3)(12x+3)}=\frac{-12}{(12(x+h)+3)(12x+3)}
$$
|
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|
Solving the sequence $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$: proving that $2+2a_n$ is a perfect square Question:
Let $a_1=a_2=97$ and $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$ for $n>1$. Prove that
(a) $2+2a_n$ is a perfect square, and (b) $2+\sqrt{2+2a_n}$ is a perfect square.
I changed the given recursive formula by squaring, and the result was as follows:
$$(a_{n+1}^2+a_n^2+a_{n-1}^2)-2a_{n+1}a_na_{n-1}-1=0$$
$\Rightarrow$ $$(a_{n+1}+a_n+a_{n-1})^2-2(a_{n+1}a_n+a_na_{n-1}+a_{n-1}a_{n+1}+a_{n+1}a_na_{n-1})-1=0$$
$\Rightarrow$ $$(1+a_{n+1})^2+(1+a_n)^2+(1+a_{n-1})^2-2(a_{n+1}+a_n+a_{n-1}+a_{n+1}a_n+a_na_{n-1}+a_{n-1}a_{n+1}+a_{n+1}a_na_{n-1}+1)-2=0$$
$\Rightarrow$ $$(1+a_{n+1})^2+(1+a_n)^2+(1+a_{n-1})^2-2(1+a_{n+1})(1+a_n)(1+a_{n-1})-2=0$$
And consequently, I lost the way :(
I thought of proving in a inductive way, that is :
$$2+2a_1=196=14^2$$
When we set $2+2a_n=k^2$, $2+2a_{n-1}=l^2$,
$$ 2+2_{n+1}=\frac{(k^2-2)(m^2-2)}{4}+\sqrt{\left(\left(\frac{k^2-2}{2}\right)^2-1\right)\left(\left(\frac{m^2-2}{2}\right)^2-1\right)}$$
As a result, I again lost the way :/
I think I sill have not got the main points. Could you give me some clues about this problem? Thanks.
|
$ $(Essentially taken from Sequence problem on AoPS.)
The key point is to recognize the connection between the recurrence formula
$$
a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}
$$
and the addition formula for the inverse hyperbolic cosine:
$$
\DeclareMathOperator{\arcosh}{arcosh}
\arcosh u + \arcosh v=\arcosh \left(uv + {\sqrt {(u^{2}-1)(v^{2}-1)}}\right) \, .
$$
First note that $a_n > 1$ for all $n$, so that the sequence is well-defined, and we can set $x_n = \arcosh a_n$. Then
$$
x_{n+1} = x_n + x_{n-1} \, ,
$$
which together with $x_1 = x_2$ implies that $(x_n)$ is a multiple of the Fibonacci sequence: $x_n = F_n x_1$. (In the following we need only that $x_n$ is an integer multiple of $x_1$.)
So we have
$$
a_n = \cosh x_n = \cosh \left( F_n x_1 \right) \, ,
$$
and using the half-angle formula for $\cosh$ we conclude that
$$
2 + 2a_n = \left( 2 \cosh \left( \frac{F_n x_1}{2} \right)\right)^2
$$
and then
$$
2+\sqrt{2+2a_n} = \left( 2 \cosh \left( \frac{F_n x_1}{4} \right)\right)^2
$$
It remains to show that
$$
y_n = 2 \cosh \left( \frac{F_n x_1}{4} \right) = e^{F_n x_1/4} + e^{-F_n x_1/4} = \alpha^{F_n} + \left(\frac{1}{\alpha}\right)^{F_n}
$$
with $\alpha = e^{x_1/4}$ is an integer for all $n$. Setting $n=1$ allows to determine $\alpha$:
$$
y_1 = \sqrt{2+\sqrt{2+2 \cdot 97}} = 4 = \alpha + \frac 1\alpha
$$
has the solution
$$
\{ \alpha, \frac 1\alpha \} = \{ 2 + \sqrt 3, 2 - \sqrt 3 \} \, .
$$
So finally we get
$$
y_n = \left( 2 + \sqrt 3\right)^{F_n} + \left( 2 - \sqrt 3\right)^{F_n}
$$
and that is indeed an integer for all $n$ (see for example The number $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is an integer).
|
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"timestamp": "2023-03-29T00:00:00",
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|
No triple of primes Prove that $p^5+q^5+r^5$ is not divisible by $10pqr$ where $p,q,r$ are prime numbers.
I proved using orders that if $p>q>r$ then $r=2$ and $q|p+2$.
I proved (it is easy) using orders that there are no primes $p,q,r$ for which $6pqr|p^3+q^3+r^3$ .
And I tried to solve this one, but it couldn't solve it.
|
We use Fermat little theorem:
$P^5+q^5+r^5 ≡ (p+q+r) \ mod (5)$
⇒ $(p^5-p)+(q^5-q)+(r^5-r)≡ 0 \ mod(5)$
It can be shown that $30 \big | n^5-n$; $n∈ \mathbb N$: If n is prime, then:
$n^5-n≡ 0 \ mod(5)$
$n^5-5=n(n^2-1)(n^2+1)$
$n^2-1≡ 0 \ mod(3)$
Also $n(n+1)$ is even, that is $n^5-n$ is divisible by $30$. (Notice that this fact is true for all natural numbers).
Now we can write:
$(p^5-p)+(q^5-q)+(r^5-r)=30 k$
⇒ $p^5+q^5+r^5≡ (p+q+r) \ mod (30)= 30 m +(p+q+r)$; $m∈ \mathbb N$.
For $pqr$ we do not see a common factor like $pqr$ on RHS of following identity(the fourth term does not have factor $pqr$):
$p^5+q^5+r^5=\frac{(p+q+r)^5}{6}-\frac{5(p+q+r)^3(p^2+q^2+r^2)}{6}+\frac{5(p+q+r)^2(p^3+q^3+r^3)}{6}+\frac{5(p^2+q^2+r^2)(p^3+q^3+r^3)}{6}$
Hence $10 pqr $ does not divide $p^5+q^5+r^5$.
Moreover you may make your question stronger:
Prove that $p^5+q^5+r^5$ is not divisible by $30pqr$.
|
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|
integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
Find all possible integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
what i try $y^2-(x+2)y+x^2-2x=0$
$$y=\frac{x+2\pm\sqrt{(x+2)^2-4(x^2-2x)}}{2}$$
$$y=\frac{x+2\pm \sqrt{-3x^2+12x+4}}{2}$$
How do i solve it Help me please
|
Look at the discriminant which should be positive definite so the $y$ is real for real $x$. This demands
$$-3x^2+12x+4 \ge 0 \implies 3x^2-12x-4 \le 0 \implies (x-4.3)(x+.3) \le 0 \implies -.3 \le x \le 4.3 ~~(roughly)$$
So the possible integral values of $x$ are $x=0,1,2,3,4$
out of these we accept the ones which give $y$ as integrals.
Check that $x=0,1,2,4$ give integral values of $y$, such pairs of $x$ and $y$ are the solutions.
|
{
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|
How to calculate $\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$ $$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$$
someone please help i’m not sure how to compute this. i’ve tried to do it this way:$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})} = \frac {1-1/7^{4x} }{1+ 1/21^{4x}} $$
$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}
= \dfrac {\lim\limits_{ x \to \infty } 1-1/7^{4x} }{ \lim\limits_{ x \to \infty } 1+ 1/21^{4x} }$$
but that’s defiantly wrong
|
In that limes, $x\to\infty$ is equivalent to $7^{2x}\to\infty$, hence:
$$\lim\limits_{x \to \infty} \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}
= \lim\limits_{z \to \infty} \frac {z +z^{-1}}{ 3(z - z^{-1})}
= \frac13\lim\limits_{z \to \infty} \frac {z^2+1}{z^2 - 1}
= \frac13
$$
|
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|
If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$$
$$h(x)=\dfrac{x^2+x^2-2+2}{x-\dfrac{1}{x}}$$
$$h(x)=x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}$$
Case $1$: $x-\dfrac{1}{x}>0$
$$\dfrac{x^2-1}{x}>0$$
$$x\in(-1,0) \cup (1,\infty)$$
$$AM\ge GM$$
$$\dfrac{x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}}{2}>\sqrt{2}$$
$$h(x)\ge 2\sqrt{2}$$
$$x-\dfrac{1}{x}=\dfrac{2}{x-\dfrac{1}{x}}$$
$$x^2+\dfrac{1}{x^2}-2=2$$
$$x^2+\dfrac{1}{x^2}=4$$
$$x^4-4x^2+1=0$$
$$x^2=2\pm\sqrt{3}$$
Only $x=\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}$ are the valid solutions.
Case $2$: $x-\dfrac{1}{x}<0$
$$x\in(-\infty,-1) \cup (0,1)$$
$$h(x)=-\left(\dfrac{1}{x}-x+\dfrac{2}{\dfrac{1}{x}-x}\right)$$
By $AM\ge GM$, $h(x)\ge-2\sqrt{2}$
We will get this minimum value at $-\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}}$
So answer should have been $-2\sqrt{2}$ but actual answer is $2\sqrt{2}$. What am I missing here.
|
With $t:=x-\dfrac1x$
$$h(x)=\left(x^2+\frac1{x^2}\right)\left(x-\frac1x\right)=(t^2-1)t=t^3-2t.$$
Then by the chain rule
$$h'(x)=0\iff (3t^2-2)\left(1-\frac1{x^2}\right)=0.$$
We have the two solutions $x=\pm1$, and the solutions of
$$x-\frac1x=\pm\sqrt{\frac23},$$
which are
$$\dfrac{\pm\sqrt 6\pm\sqrt{42}}{6}.$$
|
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|
Trouble with trig substitution I have a few questions and a request for an explanation.
I worked this problem for a quite a while last night. I posted it here.
Here is the original question: $$\int\frac{-7 x^2}{\sqrt{4x-x^2}} dx$$
And here is the work that I did on it:
Help with trig sub integral
Sorry that the negative sometimes gets cut off in the photo, and yes I know it's not fully simplified there.
\begin{align}
-7 \int\frac{x^2}{\sqrt{4x-x^2}} dx
&= -7 \int\frac{x^{3/2}}{\sqrt{4-x}} dx \\
&= -7 \int\frac{8\sin^3\theta\ 2\cos\theta}
{\sqrt{4-4\sin^2\theta}} d\theta
&& \begin{array}{c}
2\sin\theta = \sqrt x \\
2\cos\theta = dx \\
(2\sin\theta)^3 = x^{3/2}
\end{array} \\
&= -7 \cdot 8 \int\frac{\sin^3\theta\ 2\cos\theta}
{2\sqrt{1 - \sin^2\theta}} d\theta \\
&= -56 \int \sin^3\theta\, d\theta \\
&= -56 \int (1 - \cos^2\theta) \sin\theta\, d\theta \\
&= -56 \int(\sin\theta - \cos^2\theta\sin\theta)\,d\theta \\
&= 56 \cos\theta - 56 \int u^2\, du &&
\begin{array}{c}
u = \cos\theta \\
du = \sin\theta
\end{array}\\
&= 56 \cos\theta - 56 \frac{\cos^3\theta}{3} + C \\
&= 56 \left(\frac{\sqrt{4-x}}{2}
- \frac13 \left( \frac{\sqrt{4-x}}{2}\right)^3 \right)+ C
&& \cos\theta = \frac{\sqrt{4-x}}{2}\\
&= 56 \frac{\sqrt{4-x}}{2}
\left(1 - \frac{4-x}{12}\right) + C\\
\end{align}
My first question is the more involved one: Is the algebra in my original work sound? If it is, why doesn't it work in this instance?
My second question is: is this a correct solution?
$$14\left(\frac{\sqrt{4x-x^2}(x-2)}{2}-2\sqrt{4x-x^2}-3\arcsin\left(\frac{x-2}{2}\right)\right)+C$$
It is for webwork, and I used two out of three chances. I'd prefer to keep my perfect webwork mark, obviously ;p
Finally, I was kind of impressed with Ans4's square completion and had to run it through to see that it was correct. That's such a useful skill. Do you have some specific advice about how I could improve my math tricks to that point?
|
I'll try to convey general lessons by colouring some coefficients. The crux of this problem is to change the square-rooted expression $4x-x^2=\color{blue}{2}^2-(x-\color{limegreen}{2})^2$ to a squared trigonometric function with $x=\color{limegreen}{2}+\color{blue}{2}\sin t=2(1+\sin t)$, so the integral becomes$$\begin{align}\int\frac{-7x^2dx}{\sqrt{4-(2-x)^2}}&=-28\int(1+\sin t)^2dt\\&=14\int(-3-4\sin t+\cos 2t)dt\\&=14(-3t+4\cos t\color{red}{+}\tfrac12\sin 2t)+C\\&=14(-3\arcsin\tfrac{x-2}{2}\color{red}{+}2\sqrt{4x-x^2}+\tfrac{x-2}{\color{red}{4}}\sqrt{4x-x^2})+C,\end{align}$$where in the last line we use$$\sin t=\frac{x-2}{2},\,\cos t=\sqrt{1-\left(\frac{x-2}{2}\right)^2}=\frac12\sqrt{4x-x^2},\,\sin 2t=2\sin t\cos t.$$Of course, you already knew most of this. But everything I've marked in red corrects a sign error in your result (plus a coefficient error @Jam noted), probably due to forgetting that $\cos^\prime u=\color{red}{-}\sin u$ but $\int\cos u=\color{red}{+}\sin u+C$.
|
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|
Integer solutions of $\frac{(a+b)(a+c)(b+c)}{2} + (a+b+c)^3 = 1 -abc $ (AusPol 1994) Find all integer solutions of
$$\frac{(a+b)(a+c)(b+c)}{2} + (a+b+c)^3 = 1 -abc $$
Attempt:
I noticed that $a,b,c>0$ or $a,b,c<0$ can't happen. Besides, if one of them is zero, we can find some solutions. Suppose $c=0$, we get
$$(a+b)\frac{ab}{2} + (a+b)^3 = 1 $$
$$\implies (a+b)(ab+ 2(a+b)^2) = 2 $$
So, we have $(a+b) \in \{\pm 1,\pm 2\}$. Just $(a+b) = 1$ has solution, with $a= 1$ and $b= 0$ or $a= 0$ and $b=1$.
So, due to the symmetry of the equation, just the case $a,b>0$ and $c<0$ is missing. Besides, $\gcd(a,b,c) =1$.
|
It's $$(a+b+c)(ab+ac+bc)+abc+2(a+b+c)^3=2$$ or
$$(a+b+c)^3+\sum_{cyc}a(a+b+c)^2+\sum_{cyc}ab(a+b+c)+abc=2$$ or
$$\prod_{cyc}(a+b+c+a)=2$$ or $$(2a+b+c)(2b+a+c)(2c+a+b)=2.$$
Can you end it now?
|
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|
Finding the limit of $\sqrt{4x^2+x+7}+2x$ I've been working on this problem for a while now, but I can't solve it
$$\lim\limits_{x\to-\infty}{\sqrt{4x^2+x+7}+2x}$$
I've tried multiplying by
$$\frac{\sqrt{4x^2+x+7}-2x}{\sqrt{4x^2+x+7}-2x}$$
but I didn't get it. Am I missing something really obvious? Can someone help me with this question?
|
Hint: Multiplying by $\ \frac{\sqrt{4x^2+x+7}-2x}{\sqrt{4x^2+x+7}-2x}\ \left(=1\right)\ $ shows that
\begin{align}
\sqrt{4x^2+x+7}+2x&=\frac{x+7}{\sqrt{4x^2+x+7}-2x}\\
&=\frac{-1+\frac{7}{|x|}}{\sqrt{4-\frac{1}{|x|}+ \frac{7}{x^2}}+2}\ \ \text{for }\ x<0\ .
\end{align}
Can you see what the limit of this last expression is?
|
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|
What is the locus of $z^2+\bar{z}^2=2$? I have to prove that it's the equation of an equilateral hyperbola
$$z^2+z^{-2}=2$$
I try this
$z^2+z^{-2} +2 = 2+2$
$(z^2+1/z^2 +2 ) = 4$
$(z+1/z)^2=4$
$ z+1/z= 2 $
$ x+yi + 1/(x+yi) = 2 $
$ ((x+yi)^2+1)/(x+yi)=2$
$ x^2+2xyi-y^2+1=2x+2yi$
$ x^2-2x+1 +2xyi -y^2=2yi $
$(x-1)^2 +2xyi-y^2= 2yi$
I don't know how to end it.
|
Let $z = x+iy$. Then, $x=\frac12(z+\bar z)$, $y=-\frac i2(z-\bar z)$, and
$$x^2-y^2=\frac14(z+\bar z)^2 + \frac14(z-\bar z)^2 =\frac12(z^2+\bar z^2)= 1$$
Thus, the locus is an equal-axis hyperbola.
|
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How to solve $ \frac{d^2 x}{dt^2}=e^x $? I want to solve the following ODE:
$$
\frac{d^2 x}{dt^2}=e^x
$$
I can't think of any immediate available tool to help me with this. Any suggestions?
|
Consider $p=\frac{dx}{dt}$ therefore $\frac{dp}{dt}=\frac{dp}{dx}\frac{dx}{dt}=p\frac{dp}{dx}$ therefore we have
$$p\frac{dp}{dx}=e^x\Rightarrow pdp=e^xdx\Rightarrow \frac{p^2}{2}=e^x+c'_1\Rightarrow p=\sqrt{2e^x+c_1}$$
Now we have
$$\frac{dx}{dt}=\sqrt{2e^x+c_1}\Rightarrow \frac{dx}{\sqrt{2e^x+c_1}}=dt \\ \Rightarrow \frac{1}{2\sqrt{c_1}}\left(\ln \left(\frac{\sqrt{2e^x+c_1}-\sqrt{c_1}}{\sqrt{2e^x+c_1}+\sqrt{c_1}}\right)\right) = t+c_2$$
because ($2e^x+c_1=s^2\Rightarrow 2e^xdx=2sds$)
$$
\begin{align}
I=&\int \frac{dx}{\sqrt{2e^x+c_1}} \\
=&\int \frac{sds}{\frac{s^2-c_1}{2}}\frac{1}{s} \\
=&\frac{1}{2\sqrt{c_1}}\int \left(\frac{1}{s-\sqrt{c_1}}-\frac{1}{s+\sqrt{c_1}} \right)ds \\
I=&\frac{1}{2\sqrt{c_1}}\left(\ln \left(\frac{s-\sqrt{c_1}}{s+\sqrt{c_1}}\right)\right) \\
\end{align}
$$
|
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|
Calculus 2: Integration by Parts Stuck on Integral of Product With ArcTan Inside I'm stuck on the following problem:
$$\int_0^{1/3} y \tan^{-1}(3y)\,dy$$
I think my last line in my work below is correct but I don't know what to do beyond that.
A step through of the problem would be appreciated.
|
Let $x = 3y$ then $ 3dy = dx $,
$$\displaystyle \int_0 ^{\frac {1}{3}} y \ \tan^{-1}(3y) dy = \frac {1}{9} \int_0 ^1 x\ \tan^{-1}(x) dx$$
Use ILATE
$\displaystyle \Rightarrow \frac {1}{18} [x^2 \tan^{-1}(x) |_0 ^1 - \int_0 ^1 \frac {x^2}{1+x^2} dx ] $
$\displaystyle \Rightarrow \frac {1}{18} [x^2 \tan^{-1}(x) |_0 ^1 - \int_0 ^1 \frac {x^2}{1+ x^2} dx ] $
$\displaystyle \Rightarrow \frac {1}{18} [x^2 \tan^{-1}(x) |_0 ^1 - ( \int_0 ^1 dx - \int_0 ^1 \frac {1}{1+ x^2} dx )] $
$\displaystyle \Rightarrow \frac {1}{18} [x^2 \tan^{-1}(x) - x + \tan^{-1}(x) ]_0 ^1 $
$\displaystyle \Rightarrow \frac {1}{18} [(x^2 + 1) \tan^{-1}(x) - x ]_0 ^1
= \frac {\pi}{36} - \frac {1}{18}$
Please tell if there are errors in this, because I haven't studied calculus yet.
|
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|
Revisiting Ahmed Integral in $(0,\infty)$ A recent post in MSE:
Evaluate $\int_0^{\infty } \frac{\tan ^{-1}\left(a^2+x^2\right)}{\left(x^2+1\right)\sqrt{a^2+x^2}} \, dx$
re-emphasizes that when Ahmed Integral is converted to a two-dimensional integral in $x,y$, then the sameness of domain the $(0,1)$ of $x$ and $y$ makes it do-able. Further, it asks for the evaluation of a slightly different Integral, which is challenging.
In this light, now the question here is: How to find $$\int_{0}^{\infty} \frac{\tan^{-1}\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}} dx?$$
Mind the upper limit that is $\infty$, there is square root sign in the argument of $\tan^{-1}$, and yes the integrand is the same as that of Ahmed integral.
|
$$I=\int_{0}^{\infty} \frac{\tan^{-1}\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}~ dx= \int_{0}^{\infty} \frac{\pi/2-\tan^{-1}(1/\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}~dx$$
$$\implies I=\frac{\pi}{2}\int_{0}^{\infty} \frac{1}{(1+x^2)\sqrt{2+x^2}} dx
-\int_{0}^{\infty} \frac{\tan^{-1}(1/\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}} dx =I_1-I_2$$
Let $x=\tan t$, then $ \sin t=\sqrt{2} \sin f$, we get
$$I_1=\frac{\pi}{2} \int_{0}^{\pi/2} \frac{\cos t~dt}{\sqrt{2-\sin^2 t}} =\int_{0}^{\pi/4} df=\frac{\pi^2}{8} $$
For $I_2$, we use the integral representation:
$$\frac{1}{z}\tan^{-1}\frac{1}{z}=\int_{0}^{1} \frac{dy}{y^2+z^2} $$
Then we get $$I_2=\int_{0}^{1} dy \int_{0}^{\infty} \frac{dx}{(1+x^2)(2+y^2+x^2)}$$ $$=\int_{0}^{1}\frac {dy}{1+y^2} \int_{0}^{\infty} \left(\frac{1}{1+x^2}-\frac{1}{2+y^2+x^2} \right) dx=\frac{\pi}{2}\int_{0}^{1} \frac{dy}{[2+y^2+\sqrt{2+y^2}]}$$
Next let $2+y^2=u^2$, then
$$I_2=\frac{\pi}{2} \int_{\sqrt{2}}^{\sqrt{3}} \frac{du}{(u+1)\sqrt{u^2-2}}=-
\frac{\pi}{2}\int_{\sqrt{2}-1}^{(\sqrt{3}-1)/2} \frac{dv}{\sqrt{2-(v+1)^2}}$$
We used $u+1=1/v$ in above. further
$$I_2=-\frac{\pi}{2}\left .\sin^{-1}\frac{v+1}{\sqrt{2}}\right|_{\sqrt{2}-1}^{(\sqrt{3}-1)/2}=- \frac{\pi}{2} \left(\sin^{-1}\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{\pi}{2}\right)=\frac{\pi^2}{24}$$
Finally, $$I=I_1-I_2=\frac{\pi^2}{8}-\frac{\pi^2}{24}=\frac{\pi^2}{12}.$$
|
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|
Find the open intervals where $x(x-6)^3$ is concave upwards and concave downwards. Find the open intervals where $f(x)=x(x-6)^3$ is concave upwards and concave downwards.
Solution: We want to find the concavity of $f(x)=x(x-6)^3$, so we have to take the second derivative, find the critical points (where it equals zero and DNE), then split up the number line by these points and test for the sign of $f''(x)$ in each of them. First lets get to the second derivative by using the product rule and the chain rule where appropriate:
$f'(x)=(x)'(x-6)^3+x((x-6)^3)'$
$f'(x)=(1)(x-6)^3+x(3(x-6)^2)(x-6)'$
$f'(x)=(x-6)^3+x(3(x-6)^2)(1)$
$f'(x)=(x-6)^3+3x(x-6)^2$
Okay, now let's do it again
$f''(x)=((x-6)^3)'+(3x)'(x-6)^2+(3x)((x-6)^2)'$
$f''(x)=3(x-6)^2(x-6)'+3(x-6)^2+(3x)(2(x-6))(x-6)'$
$f''(x)=3(x-6)^2(1)+3(x-6)^2+6x(x-6))(1)$
$f''(x)=3(x-6)^2+3(x-6)^2+6x(x-6))$
$f''(x)=6(x-6)^2+6x(x-6))$
Foiling out and collecting like terms yields:
$f''(x)=12x^2-108x+216$
$f''(x)=12(x^2-9x+18)$
$f''(x)=12(x-6)(x-3)$
We want to know when this equals zero...
$f''(x)=12(x-6)(x-3)=0$
$\rightarrow x=6,x=3$
So let's split up our number line by $x=3,6$ and test each region:
$f''(0)=(12)(-6)(-3)=12*18>0$
$f''(5)=12(-1)(2)<0$
$f''(7)=12(1)(4)>0$
So our function is concave upward on $(-\infty,0)$ and $(6,\infty)$, and concave downwards on $(3,6)$
|
This may help to confirm the basic derivatives and regions:
|
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|
Evaluate $\lim_{n\to\infty}\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k}}$ & $\lim_{n\to\infty}n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$. This is a homework question. I have to find two limits:
i. $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}$$
ii. $$\lim_{n\to \infty} n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$$
I was able to solve the first one:
$\sqrt{n^8+1}\le \sqrt{n^8+k}\le \sqrt{n^8+n}$
$\implies \displaystyle\dfrac{1}{\sqrt{n^8+1}}\ge \dfrac{1}{\sqrt{n^8+k}}\ge \dfrac{1}{\sqrt{n^8+n}}$
$\implies \displaystyle\dfrac{k^3}{\sqrt{n^8+1}}\ge \dfrac{k^3}{\sqrt{n^8+k}}\ge \dfrac{k^3}{\sqrt{n^8+n}}$
$\implies \displaystyle\sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+1}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+k}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+n}}$
$\implies \displaystyle\dfrac{n^2(n+1)^2}{4\sqrt{n^8+1}}\ge \sum_{k=1}^n\dfrac{k^3}{\sqrt{n^8+k}}\ge \dfrac{n^2(n+1)^2}{4\sqrt{n^8+n}}$
The upper fraction goes to $\frac{1}{4}$ and the lower as well. From the Squeeze theorem:
$$\lim\limits_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}} = \frac{1}{4}$$
The same method did not work for the second limit. Can I get a clue or a hint, please?
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There is another way which would allow to answer both questions at the same time using the binomial theorem
$$\frac{k^3}{\sqrt{n^8+k}}=\sum_{n=3}^\infty (n-3)^{20-8 n} \binom{-\frac{1}{2}}{n-3} k ^n=(n-3)^{20-8 n} \binom{-\frac{1}{2}}{n-3}\sum_{n=3}^\infty k^n$$ Now, use Faulhaber's formulae and you should arrive to something like
$$S_n=\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k}}=\frac{1}{4}+\frac{1}{2 n}+\frac{1}{4 n^2}+O\left(\frac{1}{n^7}\right)$$
Trying for $n=10$; the summation evaluates to $0.302499987$ while the above truncates expansion gives $\frac{121}{400}=0.302500000$.
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the indefinite integral $\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$ $$\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$$
My Attempt:
$$I = \int \frac{\sin^2\frac\theta2\cdot\sin\frac\theta2\cos\frac\theta2}{\cos^2\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}}
d\theta \\ = \frac12 \int \frac{(1-\cos\theta)\sin\theta}{(1+\cos\theta)\sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}}
d\theta$$
$$\text{let }\cos\theta = t \implies -\sin\theta d\theta = dt$$
$$I = \frac12 \int \frac{t-1}{(t+1) \sqrt{ t^3+t^2+t } } dt $$
I am not sure how to proceed further from here. Any hints/solutions on how to resolve the cubic expression under the square root?
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Following from your approach
\begin{align}I=\frac{1}{2}\int \frac{(t-1)(t+1)}{(t+1)^2\sqrt{t^3+t^2+t}}\,dx=\frac{1}{2}\int \frac{t^2(1-\frac{1}{t^2})}{t(t+\frac{1}{t}+2)t\sqrt{t+\frac{1}{t}+1}}\,dx\\=\frac{1}{2}\int \frac{(1-\frac{1}{t^2})}{(t+\frac{1}{t}+2)\sqrt{t+\frac{1}{t}+1}}\,dx\end{align}
Let $t+\frac{1}{t}=u; \,du=1-\frac{1}{t^2}$ Then:
\begin{align}I=\frac{1}{2}\int\frac{\,du}{(u+2)\sqrt{u+1}}\end{align}
Let $u+1=t^2, 2\,dt=\frac{\,du}{t}$. Then:
\begin{align}I=\int \frac{\,dt}{t^2+1}=\arctan(\sqrt{u+1})=\arctan(\sqrt{\cos(x)+\sec(x)+1})\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why does $3x^4 + 16x^3 + 20x^2 - 9x - 18$ = $(x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3$? $$3x^4 + 16x^3 + 20x^2 - 9x - 18 $$
When simplified I arrive to:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6}) $$
But the math book wrote:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3 $$
with that extra 3 at the end. The graph calculator seem to agree with that extra 3 as well. what did I do wrong?
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By the rational root test the polynomial has the two roots $x=-2$ and $x=-3$ so that
$$
3x^4 + 16x^3 + 20x^2 - 9x - 18=(x+2)(x+3)(3x^2+x-3)
$$
The extra $3$ is not an exponent but is the leading coefficient in $3x^2+x-3$.
|
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|
solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$ solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$. what is the general solution for $x$.
I wrote $\cos2x$ as $2\cos^2x-1$ and $\cos3x$ as $4\cos^3x-3\cos x$. Then the expression gets reduced to $\cos x(2\cos^2x-1)(4\cos^3x-3\cos x)=\frac{1}{4}$. substituted $\cos x=t$ and got
$t(2t^2-1)(4t^3-3t)=\frac{1}{4}$. how do i proceed further?
|
$$\cos x\cdot \cos 2x\cdot\cos 3x=1/4$$
$$\implies (2\cos x\cdot\cos 3x)(2\cos 2x)=1$$
$$\implies (\cos 4x+\cos 2x)(2\cos 2x)=1$$
$$\implies 2\cos 4x\cdot\cos 2x+(2\cos^22x-1)=0$$
$$\implies 2\cos 4x\cdot\cos 2x+\cos 4x=0$$
$$\implies \cos 4x~(2\cos 2x+1)=0$$
Either, $~\cos 4x=0\implies 4x=(2n+1)\dfrac π2\implies x=(2n+1)\dfrac π8~$, for any integer $~n~$.
Or, $~2\cos 2x+1=0\implies 2\cos 2x=-1 \implies \cos 2x=-\dfrac 12 \implies \cos 2x=\cos\left(π-\dfrac π3\right)~$
$~\implies \cos 2x=\cos \left(\dfrac {2π}3\right) \implies 2x=2nπ\pm \dfrac {2π}3 \implies x=nπ\pm\dfrac π3 ~$, for any integer $~n~$.
|
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|
Modular arithmetic and inverse functions The question is shown below:
Suppose $S=\{0,1,2,3,4,5,6,7,8,9,10\}$ and that the function $f:S\rightarrow S$ is given by:
$f(x)=6x^2+3x+8$ (mod 11)
Let $T=\{0,5\}$.
Find $f^{-1}\left(T\right)$.
Alright,
so my initial approach with this question was to find the inverse function,
$f^{-1}(x)=\dfrac{\pm\sqrt{24x-183}-3}{12}$, but $f^{-1}x\in S\geq0$
thus,
$f^{-1}(x)=\dfrac{\sqrt{24x-183}-3}{12}$
and then just churn out congruent values of $x=0,5$ in modulo 11 until I find a suitable value for $f^{-1}(x)$ within $S$, but it seems a bit tedious. Anyways, I repeated this churning process until I received the values of $\{2,3,6,10\}$.
Can anyone suggest a faster approach and please advise me if my approach is logically sound.
Many thanks :)
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If $\ a\equiv \alpha^2\pmod{11}\ $ is a quadratic residue $\mod{11}\ $, then
$$
a^6 \equiv \alpha^{12}\equiv \alpha^2\equiv a\pmod{11}\ ,
$$
so the square roots of $\ a\pmod{11}\ $ are $\ \pm a^3\pmod{11}\ $.
If $\ y\equiv 6x^2+3x+8\pmod{11}\ $ then $\ 2y+4\equiv$$ x^2+6x+9$$\equiv(x+3)^2 \pmod{11}\ $ must be a quadratic residue, so $\ (2y+4)^3\equiv$$\pm(x+3) \pmod{11}\ $, and $\ x\equiv$$\pm(2y+4)^3-3\pmod{11}\ $.
Substituing $\ y=0\ $ in this expression gives $\ x\equiv$$ \pm4^3-3\equiv$$6,10\pmod{11}\ $. Substituting one of $\ x=6,10\ $ back into the polynomial $\ 6x^2+3x+8\ $ shows that it is indeed congruent to $\ 0\mod11\ $ for both of them (this check is necessary for at least one of these two values, to eliminate the possibility that $\ 2\cdot5+4\ $ might have been a quadratic non-residue $\mod 11\ $).
Similarly, substituting $\ y=5\ $ into the expression for $\ x\ $ gives $\ x\equiv$$\pm3^3-3\equiv$$2,3\pmod{11}\ $, which again can be shown to satisfy the congruence $\ 5\equiv 6x^2+3x+8\pmod{11}\ $. Thus, $\ f^{-1}\left(\{0,5\}\right)=$$\{2,3,6,10\}\ $, as you've already established.
|
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|
Uniform convergence of the ratio of a sequence of function
$f_n(x): (0,1) \to (0,1)$ is a sequence of continuous functions that converges uniformly to $f(x): (0,1) \to (0,1)$. Is it true that
$$
\sup_{x \in (0,1) }\frac{f_n(x)}{f_{n+1}(x)} \to 1
$$
I think pointwise convergence follows easily: since $\{f_n\}$ converges uniformly (but pointwise would suffice), then $|f_n(x)-f(x)| < \epsilon_n$ with $\epsilon_n \to 0$, so that:
$$
\frac{f_n(x)}{f_{n+1}(x)} \leq \frac{f(x)+\epsilon_n}{f(x)-\epsilon_{n+1}} \to \frac{f(x)}{f(x)}=1
$$
$$
\frac{f_n(x)}{f_{n+1}(x)} \geq \frac{f(x)-\epsilon_n}{f(x)+\epsilon_{n+1}} \to \frac{f(x)}{f(x)}=1
$$
I am not really sure how to proceed to prove uniform convergence of such ratio (which may actually not hold). Indeed the previous approach does not work, as $f(x)-\epsilon_{n+1}$ may be equal to 0 for some $x \in (0,1)$ - unless we make the stronger assumption that $f(x) > c$ for all $x \in (0,1)$.
If you could provide any help or counter-examples that would be great!
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Let $f_{n}(x)=x^{\frac{n+1}{n+2}}$ and $f(x)=x$, then clearly $0<f_n(x)<1$ and $0<f(x)<1$ for all $x \in (0,1)$.
Next, let's show $f_n(x)$ converges to $f(x)$ uniformly on $(0,1)$:
$\displaystyle h_n(x):= f_n(x) - f(x) = x^{\frac{n+1}{n+2}} - x = x \left( \frac{1}{\sqrt[n+2]{x}} - 1 \right)$.
$\displaystyle [h_n(x)]' = \frac{n+1}{n+2} \frac{1}{\sqrt[n+2]{x}} - 1 = 0 \implies \frac{n+1}{n+2} \frac{1}{\sqrt[n+2]{x}} = 1 \implies x = \left( \frac{n+1}{n+2}\right)^{n+2}$.
Thinking graphically, it can be easily verified that these points maximize $h_n(x)$ on $(0,1)$. Thus,
$\displaystyle \sup_{x \in (0,1)} |f_n(x) - f(x)| = \left( \frac{n+1}{n+2}\right)^{n+2} \frac{1}{n+1} = \left(1 - \frac{1}{n+2}\right)^{n+2} \frac{1}{n+1} \to \frac{1}{e}0 =0$ as $n \to \infty$.
Hence, $f_n(x)$ converges to $f(x)$ uniformly on $(0,1)$. However,
$\displaystyle \frac{f_{n}(x)}{f_{n+1}(x)} = \frac{x^{\frac{n+1}{n+2}}}{x^{\frac{n+2}{n+3}}} = \frac{1}{x^{1/(n+2)(n+3)}}$ gives us $\displaystyle \sup_{x \in (0,1)} \frac{f_{n}(x)}{f_{n+1}(x)} = \sup_{x \in (0,1)} \frac{1}{x^{1/(n+2)(n+3)}} = \infty$ for all $n$.
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|
Prove $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$ $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$
I need to solve this trigonometric identity for a trianlge.
I'm not allowed to use the formula $a+b+c=s$ where 's' is perimeter.
My Try
Using sine rule I was able to simplify,
\begin{align}\text{L. H. S.}& =\frac{(\sin A+\sin B+ \sin C)^2}{\sin^2A+\sin^2B+\sin^2C} \\
&=\frac{4\cos^2\dfrac C2\Bigl(\dfrac{\cos(A-B)}{2}+\sin\frac C2\Bigr)^2}{\sin^2A+\sin^2B+\sin^2C}
\end{align}
I'm unable to simplify further, Please give me hint. Thanks in advance.
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Replace all $\cot x = \frac{\cos x}{\sin x}$ and expand straightforwardly as follows,
$$RHS = \frac{\frac{\cos \frac A2}{\sin \frac A2}+\frac{\cos \frac B2}{\sin \frac B2}+\frac{\cos \frac c2}{\sin \frac C2}}
{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}}$$
$$= \frac{\cos \frac A2\sin \frac B2\sin \frac C2+\cos \frac B2\sin \frac C2\sin \frac A2+\cos \frac C2\sin \frac A2\sin \frac B2}
{\cos A\sin B\sin C+\cos B\sin C\sin A+\cos C\sin A\sin B}
\cdot 8 \cos \frac A2\cos\frac B2\cos \frac C2$$
$$= \frac{(1+\cos A)\sin B\sin C+(1+\cos B)\sin C\sin A+(1+\cos C)\sin A\sin B}
{\cos A\sin B\sin C+\cos B\sin C\sin A+\cos C\sin A\sin B}
$$
$$= 1+ \frac{\sin B\sin C+\sin C\sin A+\sin A\sin B}
{\cos A\sin B\sin C+\cos B\sin C\sin A+\cos C\sin A\sin B}
$$
$$= 1+ \frac{2\sin B\sin C+2\sin C\sin A+2\sin A\sin B}
{\sin A\sin (B+C)+\sin B\sin (C+A)+\sin C\sin (A+B)}
$$
$$= 1+ \frac{2\sin B\sin C+2\sin C\sin A+2\sin A\sin B}
{\sin^2 A+\sin^2 B+\sin^2 C}
$$
$$= 1+ \frac{2bc+2ca+2ab}
{a^2 +b^2 +c^2 } = \frac{(a+b+c)^2}{a^2+b^2+c^2}=LHS$$
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|
Prove $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N$ Prove $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N$
The book I am using asserts a non-trivial way of proving this inequality, but I cannot see why this cannot be proven by rearranging the statement equally as rigorously.
Let
$$\frac{2k+2}{2k+3}=\frac{2k}{2k+1}, k\in\mathbb N$$
$$(2k+2)(2k+1)=2k(2k+3)$$
$$2k^2+3k+1=2k^2+3k$$
$$1=0$$
Contradiction, therefore no solution to where the two sides are equal
Let k=1
$$\frac{2+2}{2+3}= 0.8$$
$$\frac{2}{3}=0.66666$$
Since there are no points of intersection, and for k=1 LHS>RHS, and both sides are continuous on k>0, inequality must hold.
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$\frac{2x+2}{2x+3}>\frac{2x}{2x+1}, X\in\mathbb R^+$
$ f: x \rightarrow \frac{2x}{2x+1} = f: x \rightarrow 1- \frac{1}{2x+1}$ is easily demonstrable to be strictly increasing
$x > y \rightarrow 2x+1 > 2y+1 \rightarrow \frac{1}{2x+1} <\frac{1}{2y+1} \rightarrow 1- \frac{1}{2x+1} > 1-\frac{1}{2y+1}$ therefore $f(k+1) > f(k) $
i.e :
$\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N \subset \mathbb R^+$
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|
Closed form for $\sum_{k=0}^{l}\binom{k}{n}\binom{k}{m}$ Does there exist any closed form for the following sum?
$$\sum_{k=0}^{l}\binom{k}{n}\binom{k}{m}$$
Where
$l \in \mathbb N$ and $m,n \in \mathbb Z$
My try:
$$ \sum_{k=\max\left(m,n\right)}^{l}\binom{k}{n}\binom{k}{m}=\sum_{k=0}^{l}\binom{k}{k-n}\binom{k}{k-m}$$$$=\left(-1\right)^{\left(-n-m\right)}\sum_{k=0}^{l}\binom{-n-1}{k-n}\binom{-m-1}{k-m}$$$$=\left(-1\right)^{\left(-n-m\right)}\sum_{k=0}^{l}\binom{-n-1}{-1-k}\binom{-m-1}{k-m}$$$$=\left(-1\right)^{\left(-n-m\right)}\binom{-n-m-2}{-m-1}$$
$$=\left(-1\right)^{\left(-n-m\right)}\binom{-n-m-2}{-n-1}=\left(-1\right)^{\left(-m-1\right)}\binom{m}{-n-1}$$$$=\left(-1\right)^{\left(-m-1\right)}\binom{m}{m+n+1}=\left(-1\right)^{n}\binom{n}{m+n+1}$$
I'm not sure whether it's right, so can someone verify the solution, and if it's not right then please provide a closed form (of course if that's exist).
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We present a proof of the identity by @Diger, which should be considered
a starting point for additional simplification. We seek to show that
$$\sum_{k=0}^l {k\choose m} {k\choose n} =
\sum_{k=0}^n (-1)^k {l+1\choose m+k+1} {l-k\choose n-k}.$$
The RHS is
$$[z^n] \sum_{k=0}^n (-1)^k {l+1\choose m+k+1}
z^k (1+z)^{l-k}.$$
The coefficient extractor enforces the range:
$$[z^n] \sum_{k\ge 0} (-1)^k {l+1\choose l-m-k}
z^k (1+z)^{l-k}
\\ = [z^n] (1+z)^l [w^{l-m}] (1+w)^{l+1}
\sum_{k\ge 0} (-1)^k w^k
z^k (1+z)^{-k}
\\ = [z^n] (1+z)^l [w^{l-m}] (1+w)^{l+1}
\frac{1}{1+wz/(1+z)}
\\ = [z^n] (1+z)^{l+1} [w^{l-m}] (1+w)^{l+1}
\frac{1}{1+z+wz}
\\ = [z^n] (1+z)^{l+1} [w^{l-m}] (1+w)^{l+1}
\frac{1}{1+z(1+w)}
\\ = [z^n] (1+z)^{l+1} [w^{l-m}]
\sum_{k\ge 0} (-1)^k z^k (1+w)^{k+l+1}
\\ = [z^n] (1+z)^{l+1}
\sum_{k\ge 0} (-1)^k z^k {k+l+1\choose l-m}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n (-1)^k {l+1\choose n-k} {k+l+1\choose l-m}.}$$
The LHS is
$$\sum_{k\ge 0} [[0\le k\le l]] [z^m] (1+z)^k [w^n] (1+w)^k
\\ = [z^m] [w^n] \sum_{k\ge 0} (1+z)^k (1+w)^k
[v^l] \frac{v^k}{1-v}
\\ = [z^m] [w^n] [v^l] \frac{1}{1-v}
\sum_{k\ge 0} (1+z)^k (1+w)^k v^k
\\ = [z^m] [w^n] [v^l] \frac{1}{1-v}
\frac{1}{1-(1+z)(1+w)v}
\\ = [z^m] [w^n]
[v^l] \frac{1}{v-1}
\frac{1/(1+z)/(1+w)}{v-1/(1+z)/(1+w)}.$$
The inner term is
$$\mathrm{Res}_{v=0} \frac{1}{v^{l+1}}
\frac{1}{v-1} \frac{1/(1+z)/(1+w)}{v-1/(1+z)/(1+w)}.$$
Residues sum to zero and the residue at infinity in $v$ is zero.
The contribution from minus the residue at $v=1/(1+z)/(1+w)$ is
$$- [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1}
\frac{1/(1+z)/(1+w)}{1/(1+z)/(1+w)-1}
\\ = - [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1}
\frac{1/(1+z)}{1/(1+z)-(1+w)}
\\ = [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1}
\frac{1/(1+z)}{w+z/(1+z)}
\\ = [z^m] (1+z)^{l+1} [w^n] (1+w)^{l+1}
\frac{1/z}{w(1+z)/z+1}.$$
Now with $l,m,n$ positive integers we must have $l\ge n,m$ or else
there is no contribution to $k^\underline{m} k^\underline{n}.$
This means we continue with
$$[z^m] (1+z)^{l+1}
\sum_{k=0}^n {l+1\choose k}
\frac{1}{z} (-1)^{n-k} \frac{(1+z)^{n-k}}{z^{n-k}}
\\ = \sum_{k=0}^n (-1)^{n-k} {l+1\choose k}
{l+1+n-k\choose m+1+n-k}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n (-1)^{n-k} {l+1\choose k}
{l+1+n-k\choose l-m}.}$$
We have the same closed form for LHS and RHS, thus proving the claim.
For a full proof we also need to show that the contribution from
$v=1$ is zero. We get
$$[z^m] [w^n] \frac{1/(1+z)/(1+w)}{1-1/(1+z)/(1+w)}
= [z^m] [w^n] \frac{1}{(1+z)(1+w)-1}
\\ = [z^m] [w^n] \frac{1}{z+w+zw}
= [z^{m+1}] [w^n] \frac{1}{1+w(1+z)/z}
\\ = [z^{m+1}] (-1)^n \frac{(1+z)^n}{z^n}
= (-1)^n {n\choose n+m+1} = 0.$$
|
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|
Multinomial expansion sum Given,
$$ \frac{x^2+x+1}{1-x}= a_0+a_1x +a_2x^2+\cdots $$
then, find the sum:
$$ \sum^{50}_{r=1}a_r $$
I knew using multinomial expansion, that
$$ (1-x)^{-1} = 1+ x+x^2+\cdots$$
Hence, $$ \frac{x^2+x+1}{1-x}=1\times(1+x+x^2+\cdots) +x\times(1+x+x^2+\cdots)+x^2\times(1+x+x^2+\cdots) $$
Hence, sum
$ = 3(a_0+ a_1+a_2+\cdots+a_{48}) + 2a_{49}+a_{50} $
So, sum should be $147+3 = 150$. But, I was wrong. Any hints?
The correct answer was
149
Edit: from @John's answer, I got to know that I used symbol $a_r$ for multinomial coeff. and sum coeff. both. The latter $a_r$ was for multinomial actually.
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An easier way to handle this is that since $(1-x)(-x-2) = x^2 + x - 2$, you get
$$\begin{equation}\begin{aligned}
\frac{x^2 + x + 1}{1 - x} & = \frac{x^2 + x - 2 + 3}{1 - x} \\
& = -x - 2 + \frac{3}{1 - x} \\
& = -x - 2 + 3(1 + x + x^2 + x^3 + \ldots) \\
& = 1 + 2x + 3x^2 + 3x^3 + 3x^4 + \ldots \\
& = 1 + 2x + \sum_{i=2}^{\infty}3x^i
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, you have
$$\sum^{50}_{r=1}a_r = 2 + 3(49) = 149 \tag{2}\label{eq2A}$$
As for your error, note you wrote that
Hence, sum
$ = 3(a_0+ a_1+a_2+\cdots+a_{48}) + 2a_{49}+a_{50} $
You seem to have the indices mixed around, since the coefficient of $2$ should be for $a_1$, not $a_{49}$, and the coefficient of $1$ should be for $a_0$ instead of for $a_{50}$, along with including extra coefficients. Also, as stated in Eevee Trainer's answer, you should not include $a_0$ as the sum starts at $a_1$.
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|
How many subsets are there with exactly 8 elements in a set of 16 elements? Below is a problem I did which I believe I did correctly. I would like somebody to confirm that I did, or tell me where I went wrong.
Problem:
Consider a set with $16$ elements in it. How many subsets does it have with exactly $8$ elements?
Answer:
Let $c$ be the number of subsets with exactly $8$ elements.
\begin{align*}
c &= \frac{16(15)(14)(13)(12)(11)(10)(9)}{8!} \\
c &= \frac{2(15)(14)(13)(12)(11)(10)(9)}{7!} \\
c &= \frac{2(15)(2)(13)(12)(11)(10)(9)}{6!} \\
c &= \frac{2(15)(2)(13)(2)(11)(10)(9)}{5!} \\
c &= \frac{2(15)(2)(13)(2)(11)(10)(9)}{5(4)(3)(2)} \\
c &= \frac{2(3)(2)(13)(2)(11)(10)(9)}{4(3)(2)} \\
c &= 13(11)(10)(9) \\
c &= 12870
\end{align*}
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Your work appears correct.
The answer is "$16$ choose $8$":
$$\binom {16}8=\dfrac{16!}{8!8!}$$
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|
Differentiate $x^{x^2}$ with respect to $x^2$ I am doing it by $x^2$ as the and differentiating both of the above separately and then deciding them
I got answer $x^{x^2} (2\log x+1))/2$
I don't know if it is correct
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$$\frac{dx^{x^2}}{dx^2}=\frac{de^{x^2\log x}}{2x\,dx}=\frac{2x\log x+x}{2x}x^{x^2}=\frac{2\log x+1}2x^{x^2}.$$
Alternatively,
$$\frac{d\sqrt{x^2}^{x^2}}{dx^2}=\frac{de^{{x^2\log x^2/2}}}{dx^2}=\frac{\log x^2+1}2e^{{x^2\log x^2/2}}=\frac{2\log x+1}2x^{x^2}.$$
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|
Prove that in acute triangle : $\sin A\cos (B-C)=\frac{\sin (2B)+\sin (2C)}{2}$ Im going to prove this identity in acute triangle :
Let $ABC$ acute triangle , $A,B,C$ are angles then :
$$\sin A\cos (B-C)=\frac{\sin (2B)+\sin (2C)}{2}$$
I know that $A,B,C<\frac{π}{2}$
$$\sin A\cos (B-C)=\sin A(\cos B\cos C+\sin A\sin B\sin C)$$
And also :
$$\sin (2B)+\sin (2C)=\sin B\cos B +\sin B\cos B+\sin C\cos C +\sin C\cos C$$
But then I don't know how ?
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Since $A,B,C$ are angles of a triangle, we have that $A+B+C = \pi$. Recall the following trigonometric identities
\begin{align}
\sin(\pi-\theta) & = \sin(\theta)\\
\sin(2\theta) + \sin(2\phi) & = 2 \sin(\theta + \phi) \cos(\theta-\phi)\\
\end{align}
We have that
\begin{align}
\sin(2B) + \sin(2C) & = 2 \sin(B+C) \cos(B-C)\\
& = 2 \sin(\pi-A) \cos(B-C)\\
& = 2 \sin(A) \cos(B-C)
\end{align}
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|
Comparing $2$ infinite continued fractions
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$
I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{A} \\
B = 2 + \dfrac{1}{B}$
Converting to quadratic equations:
$A^2 - A - 1 = 0 \\
B^2 -2B - 1 = 0$
Resulting to:
$2A = 1 + \sqrt{5} > 1 + \sqrt{2} = B$
My Question is:
Are there any more ways to solve this type of problem?
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Isn’t $\frac12B$ equal to
$$
1+\frac1{4+\frac1{1+\cdots}}\qquad <\qquad A
$$
?
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|
Calculate limit of this sequence $I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$ I need to find the limit of the sequence :
$$I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$$
The only thing I have done is : take substitution $\tan x = z$
this gives
$$dx = \dfrac{1}{1 + z^2}\,dz$$
So, integral becomes :
$$I_n = \int_{0}^{\infty} \frac{z^{1/n}}{1 + z^2}\,dz$$
After this I am stuck how should I proceed ?
Can someone help me here ?
Thank you.
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We first convert the integral into a beta function.
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}}(\tan x)^{\frac{1}{n}} d x &=\int_0^{\frac{\pi}{2}} \sin ^{\frac{1}{n}} x \cos ^{-\frac{1}{n}} x d x \\
&=\frac{1}{2} B\left(\frac{1}{2}+\frac{1}{2 n}, \frac{1}{2}-\frac{1}{2 n}\right) \\
&=\frac{\pi}{2} \csc \left(\frac{1}{2}+\frac{1}{2 n}\right) \pi
\end{aligned}
$$
Then \begin{align}
\lim _{n \rightarrow \infty} I_n=\frac{\pi}{2} \csc \frac{\pi}{2}=\frac{\pi}{2}
\end{align}
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|
how to calculate $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1+\cos^2x}$? $$
\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\frac{1}{1+\cos ^2x}\rm{dx}}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{1}{1+\cos ^2x}\rm{dx}+\int_{\frac{\pi }{2}}^{\frac{3\pi}{4}}{\frac{1}{1+\cos ^2x}\rm{dx}}}
$$
I want to split this integral,$
\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\frac{1}{1+\cos ^2x}\rm{dx}}
$
but I don't know how to calculate the two integrals after splitting?
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HINT.-Use first $\cos^2(x)=\dfrac{1+\cos(2x)}{2}$; second $\cos(2x)=\dfrac{1-\tan(x)}{1+\tan^2(x)}$ so you arrive to
$$\int\dfrac{d(\tan(x))}{\tan^2(x)+2}=\int\dfrac{du}{u^2+2}$$ you get finally
$$\left[\frac{1}{\sqrt2}\arctan\left(\frac{\tan(x)}{\sqrt2}\right)\right]_{\dfrac{\pi}{4}}^{\dfrac{3\pi}{4}}$$
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|
Evaluate: $S=\sum_{n=1}^{\infty}\frac{(2n)!}{(2n+1)!!^2}$ How to evaluate this sum?
$$S=\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-1/2)}{(j+1/2)^2}$$
$$\frac{j(j-1/2)}{(j+1/2)^2}=\frac{2j(2j-1)}{(2j+1)^2}$$
$$S=\prod_{j=1}^{1}\frac{2j(2j-1)}{(2j+1)^2}+\prod_{j=1}^{2}\frac{2j(2j-1)}{(2j+1)^2}+\prod_{j=1}^{3}\frac{2j(2j-1)}{(2j+1)^2}+\cdots$$
$$S=\frac{1\cdot2}{3^2}+\frac{1\cdot2}{3^2}\cdot\frac{3\cdot4}{5^2}+\frac{1\cdot2}{3^2}\cdot\frac{3\cdot4}{5^2}\cdot\frac{5\cdot6}{7^2}+\cdots+\frac{(2n)!}{(2n+1)!!^2}$$
$$S=\sum_{n=1}^{\infty}\frac{(2n)!}{(2n+1)!!^2}$$
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Choosing one of the integral representations of Catalan's constant, $$2G=\int_0^{\pi/2}\log\cot\frac{x}{2}\,dx=\frac{1}{2}\int_0^{\pi/2}\log\frac{1+\cos x}{1-\cos x}\,dx=\int_0^{\pi/2}\sum_{n=0}^\infty\frac{\cos^{2n+1}x}{2n+1}\,dx\\=\frac12\sum_{n=0}^\infty\frac{1}{2n+1}\mathrm{B}\left(n+1,\frac12\right)=\sum_{n=0}^\infty\frac{2^n n!}{(2n+1)!!(2n+1)}=\sum_{n=0}^\infty\frac{(2n)!}{(2n+1)!!^2}.$$
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|
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
My attempt is as follows:
$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\right)<0$$
$$\left(x-\dfrac{k-\sqrt{k^2-5}}{5}\right)\left(x-\dfrac{k+\sqrt{k^2-5}}{5}\right)<0$$
$$x\in\left(\dfrac{k-\sqrt{k^2-5}}{5},\dfrac{k+\sqrt{k^2-5}}{5}\right)$$
As it is given that it has got only one integral solution, so there must be exactly one integer between $\dfrac{k-\sqrt{k^2-5}}{5}$ and $\dfrac{k+\sqrt{k^2-5}}{5}$
Let $x_1=\dfrac{k-\sqrt{k^2-5}}{5}$ and $x_2=\dfrac{k+\sqrt{k^2-5}}{5}$ , then $[x_2]-[x_1]=1$ where [] is a greater integer function.
But from here, how to proceed? Please help me in this.
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Well one way of looking at the solution is for $n, k \in Z$ it will have only one integral solution if
$$n-1 \le \dfrac{k-\sqrt{k^2-5}}{5}<n< \dfrac{k+\sqrt{k^2-5}}{5} \le n+1$$
Now, $D \ge 0 \Rightarrow |k| \ge \sqrt5$ and $|\alpha -\beta| \le 2 \Rightarrow k\le \sqrt{30}$
Combining both the conditions we get $k \in ${3,4,5}.
If k=3, then we get
$$n-1 \le \dfrac{1}{5}<n< 1 \le n+1 \Rightarrow n \notin Z$$
Wolfram alpha provides the following integral solutions of the problem
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.