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How to find $f(x)$ in the composite function $f(f(x)) = 6x-8$ where f(x) is a linear function Hi just want to know how to solve this.
find $f(x)$ in the composite function $f\bigl(f(x)\bigr) = 6x-8$
|
If you consider a linear function $f(x)=mx+b$, we have $f(f(x))=m(mx+b)+b$.
This is equal to $m^2x+mb+b$.
We have that $m^2=6$, so $m=\sqrt{6}$.
Therefore, $\sqrt{6}b+b=8$,
We have $b(\sqrt{6}+1)=8$, so $b=\displaystyle \frac{8}{\sqrt{6}-1}=\frac{8\sqrt{6}+8}{5}$
Therefore, $\displaystyle f(x)=x\sqrt{6}+\frac{8\sqrt{6}+8}{5}$, is a solution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $\int_0^7 g(r)~dr$ from $ g(r)=\begin{cases} \sqrt{9-(r-3)^2}, & 0\le r<3 \\ r+2, & 3\le r<5 \\ \sqrt{4-(r-5)^2}, & 5 \le r \le 7\end{cases} $ I have some piece-wise integration that I for some reason am not getting.
$$
g(r) =
\begin{cases}
\sqrt{9-(r-3)^2} && 0 \le r \lt 3 \\
r+2 && 3 \le r \lt 5 \\
\sqrt{4-(r-5)^2} && 5 \le r \le 7
\end{cases}
$$
I must find the area of $\int_0^7g(r) \ dr$.
I know that this must be equivalent to:
$$
\int_0^2\sqrt{9-(r-3)^2} \ dr \ + \ \int_3^4(r+2) \ dr \ + \ \int_5^7\sqrt{4-(r-5)^2} \ dr
$$
But my solution of $4.125 + 5.5 + \pi$ yields an incorrect answer?
|
Note: I just add this as a complement to @projectilemotion's answer.
We can still the integral into three parts as
$$\int_0^7 g(r)\; dr=\int_0^3 g(r)\; dr+\int_3^5 g(r)\; dr+\int_5^7 g(r)\; dr$$
*
*For the first one, we have $y=g(r)=\sqrt{9-(r-3)^2}$, so $y^2+(r-3)^2=9$. So the region bounded by the graph of the function, the $x$-axis and the vertical lines $x=0$ and $x=3$ is a quarter of circle of center $(3,0)$ and radius. Therefore,
$$\int_0^3 g(r)\; dr=\frac{1}{4}\pi \times 3^2=\frac{9\pi}{4}$$
*Similarly, the third integral is the area of a quarter of circle of radius $2$, so
$$\int_5^7 g(r)\; dr=\frac{1}{4}\pi \times 2^2=\pi$$
*On $[3,5]$, the function is linear, so the region is a right trapezoid of bases $g(3)=5$ and $g(5)=7$ and height $5-3=2$, so
$$\int_3^5 g(r)\; dr=\frac{5+7}{2}\times 3=12$$
As a consequence,
$$\int_0^7 g(r)\; dr=\frac{9\pi}{4} + 12 + \pi = \frac{13\pi}{4} + 12$$
|
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|
Help to understand polynomial factoring I'm following some proof, but got stuck at how the factoring works. I can follow this part:
$$\begin{align*}
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\
&= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\
\end{align*}$$
The next two steps are not so clear to me anymore:
$$\begin{align*}
&= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\
&= \frac{(k+1)^2(k+2)^2}{4}.\\
\end{align*}$$
I understand that first $(k+1)^3$ was changed to have the same denominator as the main term (which is $4$). Can someone help me break down the steps how the polynomials are added then after that, the powers confuse me a bit.
|
\begin{align}
k^2(k+1)^2 + 4(k+1)^3&=k^2(k+1)^2 + 4(k+1)(k+1)^2\\
&=(k+1)^2[k^2 + 4(k+1)]\\
&=(k+1)^2(k^2 + 4k+4)\\
&=(k+1)^2(k+2)^2\\
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$ then find $B^{16}$.
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$, then find $B^{16}$.
My Method:
Given $$AB^2=BA \tag{1}$$ Post multiplying with $B^2$ we get
$$AB^4=BAB^2=B^2A$$ Hence
$$AB^4=B^2A$$ Pre Multiplying with $A$ and using $(1)$ we get
$$A^2B^4=(AB^2)A=BA^2$$ hence
$$A^2B^4=BA^2 \tag{2}$$ Now post multiplying with $B^4$ and using $(2)$we get
$$A^2B^8=B(A^2B^4)=B^2A^2$$ hence
$$A^2B^8=B^2A^2 \tag{3}$$
Now Pre Multiply with $B^2$ and use $(3)$ we get
$$B^2A^2B^8=B^4A^2$$ $\implies$
$$A^2B^8B^8=B^4A^2$$
$$A^2B^{16}=B^4A^2$$
Now pre multiply with $A^2$ and use $(2)$we get
$$A^4B^{16}=A^2B^4A^2$$ $\implies$
$$B^{16}=BA^4=B$$
is there any other approach to solve this?
|
My Method:
Given $$AB^2=BA $$ Pre multiplying with $A^3$ we get
$$A^4B^2=A^3BA$$
$$B^2=A^3BA \tag{1}$$ Using $(1)$ we get
$$B^{16}=A^3B^8A \tag{2}$$
$$B^{8}=A^3B^4A \tag{3}$$
$$B^{4}=A^3B^2A \tag{4}$$
$$B^{2}=A^3BA \tag{5}$$
Now combining $(5)$ in $(4)$ , $(4)$ in $(3)$ , $(3)$ in $(2)$ and
we get
$$B^{16}=A^{12}BA^4$$
$$B^{16}=IBI$$
$$B^{16}=B$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the following quadratic equation: $4\tan^2(\theta)x^4-4x^2+1=0$ $$4\tan^2(\theta)x^4-4x^2+1=0$$
It can be seen that as $\theta$ goes to $0$, the leftmost term disappears and $x=\pm \frac{1}{2}$.
But when I try to solve this using the quadratic formula I get:
$$x=\pm \sqrt{\frac{1\pm \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)}}$$
In this expression as $\theta$ goes to $0$ the denominator goes to zero.
I guess there may be a way of manipulating this expression such that as $\theta$ goes to zero $x$ goes to $\pm 0.5$.
How could this expression be manipulated?
|
Basically, your question can be reformulated to
What is $$\lim_{\theta\to 0}\sqrt{\frac{1\pm \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)}}$$
To calculate that, first of all, since $\sqrt{}$ is a continuous function, all you need to do is to caluclate
$$\lim_{\theta\to 0}\frac{1\pm \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)}$$
and take the square root.
This limit has two distinct possibilities:
The first:
Taking the plus sign makes $$\lim_{\theta\to 0}\frac{1+ \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)} = \infty$$
which makes sense because, as $\theta$ becomes small but not zero, two of the roots of your polynomial become larger and larger.
The second:
We can introduce a new variable $y=\tan(\theta)$ and get
$$\lim_{\theta\to 0}\frac{1- \sqrt{1-\tan^2(\theta)}}{2\tan^2(\theta)} = \lim_{y\to 0} \frac{1-\sqrt{1-y^2}}{2y^2}$$
which can easily be calculated either by using L'Hospital or by seeing that
$$\frac{1-\sqrt{1-y^2}}{2y^2} = \frac{1-\sqrt{1-y^2}}{2y^2}\cdot\frac{1+\sqrt{1-y^2}}{1+\sqrt{1-y^2}} = \frac{1-(1-y^2)}{2y^2(1+\sqrt{1-y^2})} =\\=\frac{y^2}{2y^2(1+\sqrt{1-y^2})} = \frac{1}{2(1+\sqrt{1-y^2})}$$
|
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|
Finding minima and maxima to $f(x,y) = x^2 + x(y^2 - 1)$ in the area $x^2 + y^2 \leq 1$ I'm asked to find minima and maxima on the function
$$f(x,y) = x^2 + x(y^2 - 1)$$
in the area
$$x^2 + y^2 \leq 1.$$
My solution:
$$\nabla (f) = (2x + y^2 - 1, 2xy)$$
Finding stationary points
$$2xy = 0$$
$$2x + y^2 - 1 = 0$$
gets me $(0,1),(0,-1),(\frac{1}{2},0)$.
Stuyding the boundary:
$$x^2 + y^2 = 1$$
$$y^2 = 1 - x^2 $$
$$f(x,y) = f(x, 1-x^2) = x^2 -x^3$$
Finding stationary points on the boundary:
$$f'(x,y) = 2x - 3x^2 = 0$$
gives $(0, 1), (0,-1), (\frac{2}{3}, \sqrt{\frac{5}{9}}),(\frac{2}{3}, -\sqrt{\frac{5}{9}})$.
So in total i've got the stationary points
$$(\frac{1}{2} , 0) | (0,1)| (0,-1)| (\frac{2}{3},\sqrt{\frac{5}{9}})|(\frac{2}{3}, -\sqrt{\frac{5}{9}}).$$
which give the function values of:
$$-\frac{1}{4}, 0,0,0.15,0.15$$.
which gives the minima: $-\frac{1}{4}$ and maxima $0.15$.
The minima is correct but the maxima should be 2. Why? What stationary point am I missing?
|
HINT: the searched maximum is given by $2$ and will be attained for $$x=-1,y=0$$
|
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|
In an equilateral triangle, prove that $|BQ| + |PQ| + |CP| > 2l$ I am trying to solve the following problem:
Let $ABC$ be an equilateral triangle with side $l$.
If $P$ and $Q$ are points respectively in sides $AB$ and $AC$, different from the triangle vertices, prove that $$|BQ| + |PQ| + |CP| > 2l$$
I can see that, as point $P$ tends to $A$, $|CP|+|PQ|$ tends to $|AC|+|AQ|$. If I could prove this, the problem would be solved (the rest follows from the triangle inequality).
However I have no clue on how to do this. I tried to play with triangle inequality and relations between sides and angles but nothing worked.
How can I proceed?
|
Let $AQ=x$, $AP=y$ and $l=1$.
Thus, $$PQ=\sqrt{x^2-xy+y^2},$$
$$PC=\sqrt{y^2-y+1}$$ and
$$BQ=\sqrt{x^2-x+1}$$ and we need to prove that
$$\sqrt{x^2-xy+y^2}+\sqrt{x^2-x+1}+\sqrt{y^2-y+1}\geq2.$$
Now, by Minkowwski
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}+\sqrt{\left(y-\frac{1}{2}\right)^2+\frac{3}{4}}\geq$$
$$\geq\sqrt{\left(x-\frac{1}{2}+y-\frac{1}{2}\right)^2+\left(\frac{\sqrt3}{2}+\frac{\sqrt3}{2}\right)^2}=\sqrt{(x+y-1)^2+3}$$ and
$$\sqrt{x^2-xy+y^2}\geq\frac{x+y}{2}.$$
Let $x+y=2a$.
Hence, $a\leq1$ and we need to prove that
$$a+\sqrt{(2a-1)^2+3}\geq2$$ or
$$\sqrt{4a^2-4a+4}\geq2-a$$ or
$$4a^2-4a+4\geq a^2-4a+4,$$
which is obvious.
Done!
By my solution easy to make a geometric proof.
|
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|
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x\cos x}{2\sin x+ 2\sin x\cos x} & \text{since $\sin(2x) = 2\sin x\cos x$}&\\
& = \dfrac{2\sin x(1 - \cos x)}{2\sin x(1 + \cos x)} &&\\
& = \dfrac{(1- \cos x)}{(1+ \cos x)} &&\\
& = \dfrac{(1- \cos x)(1+ \cos x)}{(1+ \cos x)(1+ \cos x)}& \text{since $\dfrac{(1+ \cos x)}{(1+ \cos x)}=1$}&\\
& = \dfrac{(1)^2-(\cos x)^2}{(1+ \cos x)^2} & \text{since $a^2-b^2 = (a+b)(a-b)$}&\\
& = \dfrac{(\sin x)^2}{(1+ \cos x)^2} & \text{since $(\sin x)^2 + (\cos x)^2 =1$, so $(\sin x)^2 = 1- (\cos x)^2$.}&
\end{array}$$
Now, I understand that I have the $\sin x$ part on the numerator. What I have to do is get the denominator to be $\cos x$ somehow and also make the angles $\frac{x}{2}$ instead of $x$. How do I do that?
Please be through, and you can't use half-angle or triple angle or any of those formulas.
Also, we have to show left hand side is equal to right hand side, we can't do it the other way around. So please do not take $(\tan\frac{x}{2})^2$ and solve the equation.
Thank you for understanding and have a nice day :)
|
All you've done is fine.
Now: since we want $\frac{x}{2}$ angles to appear, use $\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1+\cos x=\cos 0+\cos x=2(\cos\frac{x}{2})^2$ (a formula for the sum of cosines), and you're done.
you cant use half-angle or triple angle or any of those formulas.
I don't understand - how to obtain $\tan\frac{x}{2}$ without using half-angles? ... since $\frac{x}{2}$ is in the answer, it has to appear somehow..?
|
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|
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
|
Hint. Use the identity
$$z^{m+n}+\frac{1}{z^{m+n}}=\left( z^m+\frac{1}{z^{m}}\right)\left(z^{n}+\dfrac{1}{z^{n}}\right)-\left(z^{m-n}+\frac{1}{z^{m-n}}\right).$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof $13 \mid (k\cdot 2^n+1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$?
Proof $13 \mid (k \cdot 2^n + 1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$
Hint: for $k$ odd: $2^n \equiv-k' \pmod p$ and $kk' \equiv1 \pmod p$
My thoughts:
$13\mid(k-3) \Rightarrow k=13a+3$
and
$12|(n-2) \Rightarrow n=12b+2$
so
$\begin{align}k\cdot 2^n+1 &=(13a+3)2^{12b+2}+1 \\
&=4(13a+3)(2^{b})^{12}+1\\
\textrm{ or }
&=(k-3+3)\cdot 2^{n-2+2}+1\\
&=4\cdot 2^{n-2}(k-3+3)+1\\
&=4\cdot 2^{n-2}(k-3)+3\cdot2^n+1\end{align}$
I don't kow how to use the hint :(
|
By Fermat we have $2^{12} \equiv 1 \pmod{13}$ so $2^{12b}=1+13c$.
So
\begin{eqnarray*}
k2^n+1=(3+13a)(4+4 \times 13 c)+1=13(1+4a+12c+52ac)
\end{eqnarray*}
|
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|
If the roots of an equation are $a,b,c$ then find the equation having roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$. Actually I have come to know a technique of solving this kind of problem but it's not exactly producing when used in a certain problem.
Say we have an equation
$$2x^3+3x^2-x+1=0$$
the roots of this equation are $a,b,c$. If I were to find the equation having the roots $\frac{1}{2a}, \frac{1}{2b}$ and $\frac{1}{2c}$ then I can use the root co-efficient relations and use the rule of creating equations from roots.
This gives me the result : $$4x^3 + 2x^2 -3x -1 =0$$
A much much simpler way to solve this math is: if we pick that, $f(x)= 2x^3+3x^2-x+1$ and the values of $x$ are $a,b$ and $c$ then $f\left(\frac{1}{2x}\right)$ will denote an equation (if we just write $f\left(\frac{1}{2x}\right)=0$ ) which has the roots $\frac{1}{2a}, \frac{1}{2b}$ and $\frac{1}{2c}$. And this way the result also matches the result of my previous work.
But now I face an equation
$$x^3+3x+1=0$$ and if the roots are $a,b,c$ then I have to find the equation with the roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$.
If I use the root-coefficient relations then I have the result
$$x^3+6x^2+9x+5=0$$ (which is correct I think).
But here $f\left(\frac{1-x}{x}\right)= 3x^3-6x^2+3x-1$. So writing $f\left(\frac{1-x}{x}\right)=0$ won't give me the actual equation.
Where am I mistaking actually?
|
A much much simpler way to solve this math is; If we pick that, $f(x)= 2x^3+3x^2-x+1$ and the values of x are a,b and c then f(1/2x) will denote an equation (if we just write f(1/2x)=0 ) which has the roots 1/2a, 1/2b, 1/2c.
This is not true. Suppose we have $f(x) = (x - 1)^2$. This has the root 1. And $f(1/2x) = (1/2x - 1)^2$. But we solve for the root we get $1/2x - 1 = 0$, so $x = 2 \neq 1/2$.
|
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|
Simplify $5(3 \sin(x) + \sqrt3 \cos(x))$ So Wolfram tells me I can reach $10 \sqrt3 \sin(x + π/6)$ from this expression, but I cant grasp how to do it. Any help is appreciated.
|
Given these identities:
$$\frac{1}{2}\sqrt{3} = \cos(\frac{\pi}{6})$$
$$\frac{1}{2} = \sin(\frac{\pi}{6})$$
$$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)$$
We get:
$$5(3\sin(x) + \sqrt{3}\cos(x)) = $$
$$5\sqrt{3}(\sqrt{3}\sin(x) + \cos(x)) =$$
$$10\sqrt{3}(\cos(\frac{\pi}{6})\sin(x) + \frac{1}{2}\cos(x)) =$$
$$10\sqrt{3}(\cos(\frac{\pi}{6})\sin(x) + \sin(\frac{\pi}{6})\cos(x)) =$$
$$10\sqrt{3}\sin(x + \frac{\pi}{6})$$
|
{
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"url": "https://math.stackexchange.com/questions/2340143",
"timestamp": "2023-03-29T00:00:00",
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|
parametric spiral arc length I am attempting to calculate the arc length of a spiral. Given an Archamedean spiral of the parametric form:
$x(t) = \sqrt{t}\cos\left(\omega\sqrt{t} \right)$ and $y(t) = \sqrt{t}\sin\left(\omega\sqrt{t} \right)$, the arc length $L$ is calculated as
\begin{align}\label{parametricArcLength}
L & = \int_a^b \sqrt{1 + \left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \right)^2}\cdot \frac{dx}{dt}dt \\
& = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ \ dt
\end{align}
with
\begin{align}
\frac{dx}{dt} & = \frac{1}{2\sqrt{t{}}} \cdot \cos (\omega \sqrt{t}) - \sqrt{t} \cdot \frac{\omega}{2\sqrt{t}} \sin (\omega \sqrt{t}) \\
& = \frac{\cos (\omega \sqrt{t})}{2\sqrt{t}} - \frac{\omega \sin (\omega \sqrt{t})}{2} \nonumber
\end{align}
\begin{align} \label{dy_dt}
\frac{dy}{dt} & = \frac{1}{2\sqrt{t}} \cdot \sin (\omega \sqrt{t}) + \sqrt{t} \cdot \frac{\omega}{2\sqrt{t}} \cos (\omega \sqrt{t}) \\
& = \frac{\sin (\omega \sqrt{t})}{2 \sqrt{t}} + \frac{\omega \cos (\omega \sqrt{t})}{2} \nonumber
\end{align}
yielding
\begin{split}
L = &\int_{t_a}^{t_b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ \ dt \\
= & \int_{t_a}^{t_b} \left[\frac{1}{4}\left(\frac{\cos^2 \left(\omega \sqrt{t}\right)}{t} - \frac{2\omega\cos\left(\omega\sqrt{t} \right) \sin \left(\omega \sqrt{t}\right)}{\sqrt{t}} + \omega^2 \sin^2\left(\omega \sqrt{t}\right)\right) \right.\\
& \left. + \frac{1}{4}\left( \frac{\sin^2\left( \omega\sqrt{t} \right)}{t} + \frac{2\omega \sin \left(\omega \sqrt{t}\right) \cos\left(\omega \sqrt{t} \right)}{\sqrt{t}} + \omega^2 \cos\left(\omega \sqrt{t} \right) \right) \right]^{1/2}dt\\
= & \int_{t_a}^{t_b} \left[\frac{1}{4}\left( \frac{\sin^2\left(\omega\sqrt{t}\right) + \cos^2\left(\omega\sqrt{t}\right)}{t} + \omega^2\sin^2\left(\omega\sqrt{t}\right) + \omega^2\cos^2\left(\omega\sqrt{t}\right)\right)\right]^{1/2}dt\\
= & \int_{t_a}^{t_b} \left[ \frac{1}{4} \left( \frac{1}{t} + \omega^2 \right)\right]^{1/2}dt\\
= & \frac{1}{2}\int_{t_a}^{t_b}\left[\frac{1}{t} + \omega^2 \right]^{1/2}dt
\end{split}
Is the above integral the correct solution? If so, what method should be used to solve it?
|
Parametric curves are just screaming out to be solved in the complex plane. Consider that
$$z=x+iy=\sqrt{t}~e^{i\omega\sqrt{t}}$$
The arc length is given by
$$s=\int |\dot z|dt$$
Thus,
$$
\dot z=\left(\frac{1}{2\sqrt{t}}+\frac{i\omega}{2} \right)e^{i\omega\sqrt{t}}\\
|\dot z|=\frac{1}{2}\sqrt{\frac{1}{t}+\omega^2}
$$
Then
$$
\begin{align}
s=\int |\dot z|dt
&=\frac{1}{2}\int \sqrt{\frac{1}{t}+\omega^2}~dt\quad \text{(same as in the OP)}\\
&=\frac{1}{\omega}\int \sqrt{1+u^2}~du,\quad u=\omega\sqrt{t}\\
&=\frac{1}{2\omega}\left[u\sqrt{1+u^2}+\sinh^{-1}u \right]\\
&=\frac{1}{2\omega}\left[\omega\sqrt{t}\sqrt{1+\omega^2t}+\sinh^{-1}(\omega\sqrt{t}) \right]\\
\end{align}
$$
Notice, in particular, how this method circumvents all of the trigonometry in order to arrive at the same expression for the arc length integral as in the OP.
I have verified this result numerically for randomly selected values of $\omega$.
|
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|
A limit involving square roots Transcribed from photo
$$\require{cancel}
\lim_{n\to\infty}\sqrt{n+\tfrac12}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)\\[12pt]
$$
$$
\lim_{n\to\infty}\sqrt{n+\tfrac12}\lim_{n\to\infty}\left(\sqrt{2n+1}-\sqrt{2n+3}\right)\frac{\sqrt{2n+1}+\sqrt{2n+3}}{\sqrt{2n+1}+\sqrt{2n+3}}\\[12pt]
$$
$$
\lim_{n\to\infty}\underbrace{\sqrt{\frac{\cancel{n}1}{\cancel{n}}+\cancelto{0}{\frac1{2n}}}}_1\lim_{n\to\infty}\frac{(2n+1)-(2n+3)}{\sqrt{2n+1}+\sqrt{2n+3}}\\[24pt]
$$
$$
\lim_{n\to\infty}\frac{(\cancel{2n}+1)-(\cancel{2n}+3)}{\sqrt{2n+1}+\sqrt{2n+3}}=\lim_{n\to\infty}\frac{-2}{\sqrt{2n+1}+\sqrt{2n+3}}\cdot\frac1{\sqrt{n}}\\[12pt]
$$
$$
\lim_{n\to\infty}\frac{\cancelto{0}{\frac2{\sqrt{n}}}}{\sqrt{2+\cancelto{0}{\frac1n}}+\sqrt{2+\cancelto{0}{\frac3n}}}=\frac0{2\sqrt2}=0\\[24pt]
$$
$$
0\cdot1=0
$$
I solved it, but i dont know if it is right.
I got $0$
|
$$\sqrt{n+\frac{1}{2}}\cdot (\sqrt{2n+1}-\sqrt{2n+3})
=\sqrt{n+\frac{1}{2}}\cdot\frac{-2}{\sqrt{2n+1}+\sqrt{2n+3}}\\
=\sqrt{n}\cdot \sqrt{1+\frac{1}{2n}}\cdot\frac{-2}{\sqrt{2n}\cdot (\sqrt{2n+1}+\sqrt{2n+3})}\\
=\sqrt{1+\frac{1}{2n}}\cdot\frac{-\sqrt{2}}{\sqrt{1+\frac{1}{2n}}+\sqrt{1+\frac{3}{2n}}}$$
Thus, the required limit is $1\cdot \frac{-\sqrt 2}{1+1}=\frac{-1}{\sqrt2}$
|
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|
Solve an indefinite integral Find the value of
$$ \int\frac{\sqrt{1+x^2}}{1-x^2}dx $$
My Attempt: I tried to arrange the numerator as follows,
$$ \sqrt{1+x^2} = \sqrt{1-x^2 + 2x^2} $$
but that didn't help me.
Any help will be appreciated.
|
HINT: Make the substitution $1+x^2 \to u$ or $x \to \sqrt{u-1}$. Then you can transform the integral into
$$\int -\frac{\sqrt u}{u}\cdot\frac{1}{2\sqrt{u-1}}du$$
$$\int -\frac{1}{\sqrt u}\cdot\frac{1}{2\sqrt{u-1}}du$$
$$-\frac{1}{2}\int \frac{1}{\sqrt{u^2-u}}du$$
Then complete the square in the denominator:
$$-\frac{1}{2}\int \frac{1}{\sqrt{(u-\frac{1}{2})^2-\frac{1}{4}}}du$$
Now see if you can get it in the form
$$\int \frac{1}{\sqrt{a^2-1}}$$
so that you can use the formula
$$\int \frac{1}{\sqrt{a^2-1}}=\ln\bigg(\sqrt{x^2-1}+x\bigg)+C$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find $\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)$ I did it as follows: $$\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)=\tan\Bigg(\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)\Bigg)=\frac{\sqrt{2}-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}\sqrt{2}}=\frac{\sqrt{2}}{4}.$$ But there is no such an answer. What is wrong with it?
|
The angle you are looking for is the red angle:
$$\arctan\sqrt{2}-\arctan\frac{1}{\sqrt{2}} = \arctan\left(\frac{\sqrt{2}-\frac{1}{\sqrt{2}}}{1+1}\right)=\color{red}{\arctan\frac{1}{\sqrt{8}}} $$
|
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|
If $f(x)=\frac{1}{x^2+x+1}$, how to find $f^{(36)} (0)$?
If $f(x)=\frac{1}{x^2+x+1}$, find $f^{(36)} (0)$.
So far I have tried letting $a=x^2+x+1$ and then finding the first several derivatives to see if some terms would disappear because the third derivative of $a$ is $0$, but the derivatives keep getting longer and longer. Am I on the right track? Thanks!
|
Use $$x^2+x+1=\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)$$
By this hint we obtain:
$$\left(\frac{1}{x^2+x+1}\right)^{(36)}_{x=0}=\left(\frac{1}{\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)}\right)^{(36)}_{x=0}=$$
$$=\left(\frac{1}{\sqrt3i}\left(\frac{1}{x+\frac{1}{2}-\frac{\sqrt3}{2}i}-\frac{1}{x+\frac{1}{2}+\frac{\sqrt3}{2}i}\right)\right)^{(36)}_{x=0}=$$
$$=\frac{1}{\sqrt3i}\left(\frac{36!}{\left(x+\frac{1}{2}-\frac{\sqrt3}{2}i\right)^{37}}-\frac{36!}{\left(x+\frac{1}{2}+\frac{\sqrt3}{2}i\right)^{37}}\right)_{x=0}=$$
$$=\frac{1}{\sqrt3i}\left(\frac{36!}{\left(\frac{1}{2}-\frac{\sqrt3}{2}i\right)^{37}}-\frac{36!}{\left(\frac{1}{2}+\frac{\sqrt3}{2}i\right)^{37}}\right)=\frac{1}{\sqrt3i}\left(\frac{36!}{\frac{1}{2}-\frac{\sqrt3}{2}i}-\frac{36!}{\frac{1}{2}+\frac{\sqrt3}{2}i}\right)=36!$$
|
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|
Prove that it is impossible to find any positive integers a, b, c such that $(2a+b)(2b+a) = 2^c$ Prove that it is impossible to find any positive integers a, b, c such that $(2a+b)(2b+a) = 2^c$.
This problem has been driving me crazy. Thanks for helping.
|
We have $2a+b=2^k$ and $2b+a=2^m$, where $k$ amd $m$ are non-negative integer numbers.
Thus, $a=\frac{2^{k+1}-2^m}{3}$ and $b=\frac{2^{m+1}-2^k}{3}$, which gives
$k+1>m$ and $m+1>k$ and from here $k+1\geq m+1$ and $m+1\geq k+1$, which gives $k=m$, $a=\frac{2^k}{3}$, which is impossible.
|
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|
In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$? In $\triangle ABC$, we have $AB=3$ and $AC=4$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?
Alright, so if d is the midpoint then my two triangles are ADB and ADC.
Using the law of cosines I get:
$$3^2=a^2 + 2a^2 - 2\cdot a\cdot 2a\cdot\cos \angle ADB$$
for the first triangle,
$$4^2=a^2 + 2a^2 - 2\cdot a\cdot 2a\cdot\cos\angle ADC$$
for the second one. Then I'm stuck. Any help?
|
Using $BD=DC=a $, we can write
$$\begin{align} 3^2 &=a^2+(2a)^2-2 \cdot a \cdot 2a \cos(\alpha) \\
&=5a^2-4a^2\cos (\alpha) \end{align}$$
with $\alpha= \angle ADB$.
Also,
$$\begin{align} 4^2&=a^2+(2a)^2-2 \cdot a \cdot 2a\cos (\pi-\alpha) \\
&=5a^2+4a^2\cos (\alpha). \end{align}$$
By addition, we get
$$9+16=25=10a^2. $$
Hence
$$BC = 2a = \sqrt {10}.$$
|
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|
Simplifying the Solution to the Cubic I am trying to solve the cubic. I currently have that, for $ax^3+bx^2+cx+d=0$, a substitution to make this monic. Dividing by $a$ gives
$$x^3+Bx^2+Cx+D=0$$
where $B=\frac{b}{a}, C=\frac{c}{a}, D=\frac{d}{a}$. Then, with the substitution $x=y-\frac{B}{3}$, I got
$$y^3+\left(C-\frac{B^2}{3}\right)y+\left(D-\frac{BC}{3}+\frac{2B^3}{27}\right)=0$$
Thus, to make things simpler, i made the substitution $p=C-\frac{B^2}{3}$ and $q=D-\frac{BC}{3}+\frac{2B^3}{27}$ we have the "depressed cubic"
$$y^3+py+q=0$$
Now, using the identity,
$$(m+n)^3=3mn(m+n)+(m^3+n^3)$$
we let $y=m+n$. This then translates to $p=-3mn,$ and $q=-(m^3+n^3)$ and gives us a system of equations in $m$ and $n$. Solving for $n$ gives $n=-\frac{p}{3m}$ and back substituting yields
$$q=-m^3+\frac{p}{3m}\qquad \Rightarrow \qquad m^6+qm^3-\frac{p^3}{27}=0$$ and now we can solve the quadratic for $m$;
$$m=\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}$$
and then that means, by back substitution
$$n=-\frac{p}{3\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}}$$
So, I think I am almost here, because now,
$$y=m+n=\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}-\frac{p}{3\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}}$$
But how can I simplify this expression? I know I can back substitute for the original $a,b,c,d$ and solve for $x$. But this sum looks complicated and my attepts to simplify the sum have not worked.
|
You can't. This is as simple as you can get, unless you want to re-obtain Cardano's formula, which is basically what you got.
|
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|
Convert vector differential equation's order
Here is a 2nd order vector differential equation: $$\overrightarrow{Y}''= \begin{pmatrix}a & b \\c & d \end{pmatrix} \overrightarrow{Y}$$ Don't work it out, but write it as a vector differential eqn in $1$st order in higher dimensions.
I am not sure where to begin.
How does one convert from 2nd to 1st order? Hints are appreciated.
|
Let
$$
\vec{x} =
\begin{pmatrix}
x_1 \\
x_2 \\
\end{pmatrix} =
\begin{pmatrix}
y_1' \\
y_2' \\
\end{pmatrix} = \vec{y}'.
$$
Then
$$
\vec{x}' =
\begin{pmatrix}
x_1' \\
x_2' \\
\end{pmatrix}
=
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}
\begin{pmatrix}
y_1\\
y_2 \\
\end{pmatrix} =
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} \vec{y}.
$$
In coordinates,
$$
\begin{pmatrix}
\vec{x}' \\
\vec{y}' \\
\end{pmatrix}
=
\begin{pmatrix}
x_1' \\
x_2' \\
y_1' \\
y_2' \\
\end{pmatrix}
= \begin{pmatrix}
0 & 0 & a & b \\
0 & 0 & c & d \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
y_1 \\
y_2 \\
\end{pmatrix}
= \begin{pmatrix}
\vec{x} \\
\vec{y} \\
\end{pmatrix},
$$
or
$$
\begin{pmatrix}
\vec{x}' \\
\vec{y}' \\
\end{pmatrix}
=
\begin{pmatrix}
0 & A \\
I & 0 \\
\end{pmatrix}
\begin{pmatrix}
\vec{x} \\
\vec{y} \\
\end{pmatrix},
$$
where
$$
A=
\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}
\mbox{ and }
I =
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}.
$$
|
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|
What is the value of the expression $\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$? This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.
What is the value of
$\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$
Do I just use
$\cos(a-b)-\cos(a+b) = 2\sin(a)\sin(b)$
|
$\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}$
$\sin \frac{8\pi}{7} = \sin \frac{-6 \pi}{7}$
-> $\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{-6\pi}{7}$
$\sin a \sin (b-c) + \sin b \sin (c-a)+ \sin c \sin (a-b)$ is always equal to "$0$".
Can you see it?
|
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|
Calculate $ \lim_{x\to 0}\frac{ \ln(x+\cos(x))-xe^{-x} }{x^{3}}$ Calculate $\displaystyle \lim\limits_{x\to 0}\dfrac{ \ln\left(x+\cos(x)\right)-xe^{-x} }{x^{3}}$
My attempts
by l’hôpital
*
*$\left( \ln\left(x+\cos(x)\right)-xe^{-x} \right)'=\dfrac{(x+\cos(x))' }{(x+\cos(x))} -(e^{-x}-xe^{-x} )=\dfrac{(1-\sin(x)) }{(x+\cos(x))} -(e^{-x}-xe^{-x} ) $
*$(x^{3})'=3x^{2}$
|
L’Hôpital's rule seems to be hard to use here.
This is an alternative method by using Taylor expansion at $0$:
\begin{align*}
\ln\left(x+\cos(x)\right)-xe^{-x}&=
\ln\left(1+x-\frac{x^2}{2}+o(x^3)\right)-x\left(1-x+\frac{x^2}{2}+o(x^2)\right)\\
&=\left(x-\frac{x^2}{2}\right)-\frac{1}{2}\left(x-\frac{x^2}{2}\right)^2+\frac{1}{3}\left(x-\frac{x^2}{2}\right)^3-x+x^2-\frac{x^3}{2}+o(x^3)\\
&=x-\frac{x^2}{2}-\frac{x^2}{2}+\frac{x^3}{2}+\frac{x^3}{3}-x+x^2-\frac{x^3}{2}+o(x^3)\\
&=\frac{x^3}{3}+o(x^3).
\end{align*}
Hence
$$\lim_{x\to 0}\frac{ \ln\left(x+\cos(x)\right)-xe^{-x} }{x^{3}}=\lim\limits_{x\to 0}\frac{\frac{x^3}{3}+o(x^3) }{x^{3}}=\frac{1}{3}.$$
|
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|
If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this question, but not able to proceed. How do I solve this?
|
This is the same as
$$a^8+b^8+c^8> a^2b^3c^3+a^3b^2c^3+a^3b^3c^2.$$
You can get this by using AM/GM to get an upper bound for each term
on the right.
|
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|
Find the maximum and minimum value of $S= \log_{2}a+\log_{2}b+\log_{2}c$. Let $a, b, c \geq 1$ and $a+b+c=4$. Find the maximum and minimum value of $S= \log_{2}a+\log_{2}b+\log_{2}c$.
I found the maximum, it is easy to prove $S_\max = 3\log_{2}\frac{4}{3}$. I think the minimum is $1$ when there are two number are $1$ and the remain is $2$. But I do not know, how to prove it.
|
The max/min of $S$ will occur on the same instance as the max/min of $2^S = abc$. By $AM-GM$ the max occurs at $a= b=c = 4/3$ as you figured.
To figure out the min, fix $c$ and and set $b = (4-c) -a = K -a$.
So $abc = aK - a^2$ the derivative is $K - 2a$ which has a maximum at $a = \frac K2$ (which means $b = a = \frac K 2$) and will be increasing for $a < \frac K 2$ and decreasing for $a > \frac K2$ so the minimum will occur at either the minimum possible value of $a$ which is $1$ (and therefore $a = 1; b= K-1$) or at the maximum possible value of $a$ which is $K - 1$ and (therefore $a= K-1; b = 1$). Wolog we may assume $a \le b$ and $a = 1$ and $b = K-1$.
So lets "unfix" $c$ but fix $a$ so as to find min of $abc$ which means fixing $a = 1$. We have $a = 1; b = 4- c - 1= 3-c$ and so $abc = 3c - c^2$. By a similar argument above, This achieves a maximum at $c = \frac 32$ and minimum at either $c = 1, b=2$ or at $c= 2, b= 1$.
So if, wolog, $a \le b \le c$ then max occurs at $a= b= c = \frac 43$ and min occurs at $a=b=1; c= 2$.
|
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|
Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$
Evaluate the following limit.
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
My Attempt:
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-a} + \sqrt {bx}}$$
$$=\lim_{x\to \infty} \dfrac {x-a-bx}{\sqrt {x-a} + \sqrt {bx}}.$$
How do I proceed?
|
An often good trick is to make the substitution $t=1/x$. However, in this case $t=1/\sqrt{x}$ seems better, because we get
$$
\lim_{t\to0^+}\frac{\sqrt{1-at^2}-\sqrt{b}}{t}
$$
The numerator has limit $1-\sqrt{b}$, therefore we see that, if $0\le b<1$ the limit is $\infty$, whereas for $b>1$ the limit is $-\infty$.
If $b=1$, this is the derivative at $0$ of the function $f(t)=\sqrt{1-at^2}$. Since
$$
f'(t)=-\frac{at}{\sqrt{1-at^2}}
$$
we have $f'(0)=0$.
In conclusion
$$
\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})=
\begin{cases}
\infty & 0\le b<1 \\[4px]
0 & b=1 \\[4px]
-\infty & b>1
\end{cases}
$$
|
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|
Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$? Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$, if $x, y$ are reals greater than $1$, and $a, b, c$ are positive reals?
A proof with all the math to go from one to the other would be nice.
|
Multiply the inequality by $4a^2$ then add & subtract $(a^2+b^2-c^2)y^2$
\begin{eqnarray*}
\underbrace{4a^4 x^2 +4a^2(a^2+b^2-c^2)xy +\color{red}{(a^2+b^2-c^2)^2y^2}}_{} -\color{red}{(a^2+b^2-c^2)^2y^2}+4a^2b^2 y^2>0
\end{eqnarray*}
Now complete the square
\begin{eqnarray*}
\left(2a^2x+(a^2+b^2-c^2)y\right)^2+(4a^2b^2-(a^2+b^2-c^2)^2)y^2>0
\end{eqnarray*}
The first term is zero when $2a^2x+(a^2+b^2-c^2)y=0$ . So
\begin{eqnarray*}
(4a^2b^2-(a^2+b^2-c^2)^2)y^2>0
\end{eqnarray*}
or
\begin{eqnarray*}
(a^2+b^2-c^2)^2-4a^2b^2<0
\end{eqnarray*}
|
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|
In how many ways the sum of 5 thrown dice is 25? What I thought about is looking for the number of solutions to
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=25$$ such that $1\leq x_{i}\leq6$
for every $i$.
Now I know that the number of solutions to this such that
$0\leq x_{i}$ for every $i$ is $${5+25-1 \choose 5-1}={29 \choose 4}$$
How can I continue from here?
Thanks
|
In addition to my other answer using generating functions, I want to offer a more elementary solution. First note that
$$ 5 + 5 + 5 + 5 + 5 = 25 $$
and this is the only way to make $25$ without using a $6$.
Now make one of the dice a $6$ (5 possible ways) to get the equation
$$ a + b + c + d = 19 $$
where $1 \le a, b, c, d \le 19$. Again, the best you can do without a $6$ is
$$4 + 5 + 5 + 5 = 19$$
and there are $4$ of these (one for each position of the $4$.
Now remove a second $6$ to get the equation
$$ a + b + c = 13. $$
Without $6$'s this has two solutions:
$$ 3 + 5 + 5 = 13, $$
which occurs three times, and
$$ 4 + 4 + 5, $$
which also occurs three times.
Remove three $6$'s to get
$$ a + b = 7 $$
which has $4$ solutions not involving a $6$.
Finally remove four $6$'s to get
$a = 1$
with exactly one solution.
Hence in total we have
$$ \binom{5}{0}1 + \binom{5}{1} 4 + \binom{5}{2} 6 + \binom{5}{3} 4 + \binom{5}{4} = 126. $$
The binomial coefficients indicate how many $6$'s we have.
|
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|
Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understand what the author tries to say here. Can't this problem be done in another manner?
|
Clearly $n^4$ is a square, since $(n^2)^2=n^4$. The next smaller square is $(n^2-1)^2 = n^4-2n^2+1$, which is clearly less than the given expression. So if $n^4-n^2+64$ is a square, it needs to be not less than $n^4$, that is, we need $64\ge n^2 \implies |n|\le8$ (meaning the number of solutions finite and small)
We can see by inspection that $n=\pm8$ provides a solution where $n^4-n^2+64=n^4$, which is square without further checking. Any other solutions will involve a square greater than $n^4$, that is for some $k>0,$ $(n^2+k)^2 = n^4+2kn^2+k^2$. Then we would need $2kn^2+k^2 = -n^2+64$ giving $64=(2k+1)n^2+k^2$. Checking values of $k$ up to $8$, we can find intermediate solutions $(k,n^2)=\{(0,64), (1,21),(2,12), (7,1),(8,0)\}$ of which only $k=\{8,7,0\}$ give integer values for $n$ of $\{0, \pm1,\pm8\}$ (the last of which we already noted).
|
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|
perfect powers of the form $\frac{x^2-1}{y^2-1}$
Is there a natural number $k>1$ such that there are infinitely many
pairs $(x,y)$ of natural numbers such that $\frac{x^2-1}{y^2-1}$ is a
power of $k$?
In particular, is $k=2$ or $k=10$ good in this sense? (are there finitely or infinitely many solutions (in natural numbers) to equations $\frac{x^2-1}{y^2-1}=2^n$? or $\frac{x^2-1}{y^2-1}=10^n$?)
In case we put $x=y^2$, we would obtain equation $y^2+1=k^n$, which has been considered during investigations on Catalan conjecture.
|
Plugging in the fraction $\dfrac{x^2-1}{y^2-1}$ values $x=2^{2 n + 1} - 1;\;y=2^n$ we get
$$\dfrac{\left(2^{2 n+1}-1\right)^2-1}{2^{2 n}-1}=2^{2n+2}$$
so there are infinite pairs $(x,\;y)$ which give $2^k$
I think there are also infinite solution to $10^k$ but I still can't find a closed form
Hope this helps
|
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|
Solving trigonometric equations
If we have $$\sin (A) +\cos (A) + \csc (A) + \sec (A) +\tan (A) +\cot (A)= 7$$ and
$$\sin(2A) =a-b\sqrt{7}= 2\sin(A)\cos(A).$$
What values can $a$ and $b$ take?
|
Let $A=x$ and $\sin{x}+\cos{x}=t$.
Hence, $|t|\leq\sqrt2$, $\sin{x}\cos{x}=\frac{t^2-1}{2}$ and we need to solve
$$t+\frac{t}{\frac{t^2-1}{2}}+\frac{1}{\frac{t^2-1}{2}}=7$$ or
$$t+\frac{2}{t-1}=7$$ or
$$t^2-8t+9=0$$ or
$$(t-4)^2=7,$$ which gives
$t=4+\sqrt7$, which is impossible, or $t=4-\sqrt7$,
which gives $\sin2x=t^2-1=(4-\sqrt7)^2-1=22-8\sqrt7$.
Id est, $a=22$ and $b=8$ if you mean that $a$ and $b$ are naturals.
Done!
|
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|
generating function formula Suppose $\lambda$ is a young tableau, with one color boxes.
Prove the generating function,
$F(x) = \Sigma_{\lambda} x^{i} = \prod_{n \geq 1} \frac{1}{1-x^n} $, where $i$ is the number of boxes in the young tableau.
attempt:
$\Sigma_{\lambda} x^{i} = \prod_{n \geq 1} \frac{1}{1-x^n} = \frac{1}{1-x} \cdot \frac{1}{1-x^2} \cdot \frac{1}{1-x^3} \cdot \frac{1}{1-x^4}\cdots = (1 + x + x^2 + x^3 + \cdots)(1 + x^2 + x^4+ \cdots )(1 + x^3 + x^6 + x^9 + \cdots)( 1+ x^4 + x^8 + x^{12} \cdots ) = 1 + x + 2x^2 + 3x^3 + \cdots$ .
Here this one comes from when we have zero boxes so we would have $x^0$, and when there is only one box, then we have $x^1 $, similarly, there are $2x^2$ since there are two ways of having a young tableau with two boxes (horizontally, and vertically), and so on.
Can someone please help me?I am stuck after this. I would really appreciate it.
Thank you.
|
Suppose we have a Young tableaux with shape
\begin{eqnarray*}
(\underbrace{k,\cdots,k}_{a_k \text{times}}, \underbrace{k-1,\cdots,k-1}_{a_{k-1} \text{times}},\cdots,\underbrace{2,\cdots,2}_{a_2 \text{times}}\underbrace{1,\cdots,1}_{a_1 \text{times}})
\end{eqnarray*}
This will give a contribution of $x^n$ to the generating function, where $n=a_1+2a_2+\cdots+(k-1)a_{k-1}+ka_k$.
To get all possible Young tableauxs, $k$ can take any value $1,2,\cdots$ and the values $a_k$ can take any value $0,1,\cdots$. So the generating function is given by
\begin{eqnarray*}
\sum_{a_1=0}^{\infty} \sum_{a_2=0}^{\infty} \cdots \sum_{a_{k-1}=0}^{\infty} \sum_{a_k=0}^{\infty} \cdots x^n
\end{eqnarray*}
All of the sums will "decouple" and this can be rewritten as
\begin{eqnarray*}
\sum_{a_1=0}^{\infty} x^{a_1} \sum_{a_2=0}^{\infty} x^{2 a_2}\cdots \sum_{a_{k-1}=0}^{\infty} x^{(k-1)a_{k-1}} \sum_{a_k=0}^{\infty} x^{ka_k}\cdots
\end{eqnarray*}
Now sum each of these geometric plums & we get
\begin{eqnarray*}
\frac{1}{(1-x)} \frac{1}{(1-x^2)} \cdots \frac{1}{(1-x^{k-1})} \frac{1}{(1-x^k)} \cdots = \prod_{k=0}^{\infty} \frac{1}{(1-x^k)}
\end{eqnarray*}
|
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Series Solution to the ODE $(x-1)y'' - xy' + y = 0$ with I.C. $y(0) = -3$ and $y'(0)=4$ \begin{align*}
(x-1)y'' - xy' + y = 0 &\iff (x-1)\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2} - x\sum_{n=1}^{\infty} nc_n x^{n-1} + \sum_{n=0}^{\infty} c_n x^n = 0 \\
&\iff \sum_{n=2}^{\infty} n(n-1)c_n x^{n-1} - \sum_{n=2}^{\infty} n(n-1)c_n x^{n-2} - \sum_{n=1}^{\infty} nc_n x^n + \sum_{n=0}^{\infty} c_n x^n = 0 \\
&\iff \sum_{k=1}^{\infty} (k+1)kc_{k+1} x^k - \sum_{k=0}^{\infty} (k+2)(k+1)c_{k+2} x^k - \sum_{k=1}^{\infty} kc_k x^k + \sum_{k=0}^{\infty} c_k x^k = 0 \\
&\iff -2c_2 + c_0 + \sum_{k=1}^{\infty} x^k \left[ k(k+1)c_{k+1} - (k+2)(k+1)c_{k+2} - kc_k + c_k \right] = 0 \\
\end{align*}
Therefore, $-2c_2 + c_0 = 0 \Rightarrow c_2 = \frac{c_0}{2!}$. And,
$$c_{k+2} = \dfrac{c_k (1-k) + k(k+1)c_{k+1}}{(k+2)(k+1)}$$
To find the solution, first we let $c_0 = 1$ and $c_1 = 0$, then $c_2 = \frac{1}{2!}$. Then, for $k=1$ we have:
$$c_3 = \dfrac{2c_2}{3\cdot 2} = \dfrac{1}{3!}$$
For $k=2$ we have:
$$c_4 = \dfrac{-2c_2 + 6c_3}{4 \cdot 3} = \dfrac{1}{4!}$$
And the pattern continues so for this solution say $y_1$ we have:
$$y_1 = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \dots = 1 + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots = e^x - x$$
Now for $y_2$ we let $c_0 = 0$ and $c_1 = 1$, so that $c_2 = 0$ too. Then, we find that $c_3 = c_4 = \dots = 0$, so that
$$y_2 = c_0 + c_1 x + c_2 x^2 + \dots = x$$
Then, the final solution should be $a_0 y_1 + a_1 y_2 = a_0 e^x + (a_1 - a_0)x$. However, from wolfram it says the solution is $y = a_0 e^x + a_1 e^{-x}$. Where did I go wrong?
|
This ODE belongs to the class which is dealt with in here Yet another example of linear second order ODEs being reduced to hypergeometric functions. .
Using that approach we get the solution:
\begin{eqnarray}
y(x) &=& \exp(x/2) \sqrt{x-1} \left( C_1 W_{1/2,1}(x-1) + C_2 M_{1/2,1}(x-1) \right)\\
&=& (x-1)^2 \left( C_1 U(1,3,x-1) + C_2 F_{1,1}[1,3,x-1]\right)\\
&=& (x-1)^2 \left( C_1 \frac{e^{x-1} \Gamma(2,x-1)}{(x-1)^2} + C_2 \frac{2(e^{x-1}-x)}{(x-1)^2}\right)\\
&=& C_1 x + 2 C_2 (e^{x-1}-x)
\end{eqnarray}
Her $W_{\kappa,\mu}()$ and $M_{\kappa,\mu}()$ are the Whittaker functions and $U$ is the confluent hypergeometric function.
|
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$x^2 + y^2 = 10 , \,\, \frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ I couldn't solve the above equation. Below, I describe my attempt at solving it.
$$x^2 + y^2 = 10 \tag{1}$$
$$\frac{1}{x} + \frac{1}{y} = \frac{4}{3} \tag{2}$$
Make the denominator common in the RHS of $(2)$.
$$\frac{y + x}{xy} = \frac{4}{3} \tag{2.1}$$
Multiply $(2.1)$ through by $3$:
$$\frac{3(y + x)}{xy} = 4 \tag{2.2}$$
Let $y = mx$
Substitute $y = mx$ into $(1)$ and $(2.2)$:
$$x^2 + m^2x^2 = 10 \tag{1.1}$$
$$\frac{3(mx + x)}{mx^2} = 4$$
Factorise:
$$\frac{3x(m + 1)}{x(mx)} = 4$$
$$\frac{3(m + 1)}{(mx)} = 4$$
Cross multiply:
$$3(m+1) = 4mx$$
Collect like terms:
$$4mx - 3m = 3$$
Factorise:
$$m(4x - 3) = 3$$
Divide through by $(4x - 3)$:
$$m = \frac{3}{4x - 3} \tag{3}$$
Substitute $(3)$ into $(1.1)$
$$x^2 + \left(\frac{3}{4x - 3}\right)^2x^2 = 10$$
$$x^2 + \frac{9x^2}{16x^2 - 24x + 9} = 10$$
Multiply through by ${16x^2 - 24x + 9}$:
$$16x^2 - 24x^3 + 9x^2 + 9x^2 = 10(16x^2 - 24x + 9)$$
Divide through by $2$:
$$8x^4 - 12x^3 + 9x^2 = 80x^2 - 120x + 45$$
$$8x^4 - 12x^3 - 71x^2 +120x - 45 = 0$$.
$x = 0$ is not a solution of the above equation, so divide through by $x^2$:
$$8x^2 - 12x - 71 + \frac{120}{x} - \frac{45}{x^2} = 0 \tag{1.2}$$
Let $v = x - \frac{k}{x}$.
$$v^2 = x^2 - 2k + \frac{k^2}{x^2}$$
Rewriting $(1.2)$:
$$\left(8x^2 - 71 - \frac{45}{x^2}\right) + \left(-12x + \frac{120}{x}\right)$$
$$\left(8x^2 - 71 - \frac{45}{x^2}\right) + \left(-12(x - \frac{10}{x}\right) \tag{1.3}$$
Putting $k = 10$, works for $v$, but not for $v^2$.
I don't know any other approach to solving the equation from $(1.3)$.
|
The right approach is via symmetric functions: set $s=x+y$, $p=xy$. Then
*
*$x^2+y^2=s^2-2p,\:$ so we have the relation: $\;s^2-2p=10 \tag{1}$
*$\dfrac 1x+\dfrac 1y=\dfrac sp=\dfrac43,\:$ whence a second relation: $\;3s=4p\tag{2}$.
Relation $(1)$, taking relation $(2)$ into account, yields a quadratic equation in $s$:
$$2s^2-3s-10.$$
Solve for $s$, then $p$, and it comes down to the high school classical problem of finding two numbers, given their sum and their product.
You should find $4$ pairs of solutions corresponding to the fact that two conics intersect in $4$ points.
|
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|
What is the probability of rolling at least two 6's with 3 Dice and 2 Rolls? Question: What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls? (with your first roll you keep dice only if they are 6's and roll the remainder for your second roll).
(6*6*6 = 216 = outcomes when rolling 3 dice)
(6*6 = 36 = outcomes when rolling 2 dice)
1st roll: (3 dice)
A = 125/216 = 0 sixes
B = 75/216 = 1 six
C = 15/216 = 2 sixes
D = 1/216 = 3 sixes
Outcomes C and D fulfill requirement.
For outcome A: (first roll = no 6's, pick up all dice throw 3 dice again)
2nd roll:
a = 125/216 = 0 sixes
b = 75/216 = 1 six
c = 15/216 = 2 sixes
d = 1/216 = 3 sixes
Outcomes c and d fulfill requirements.
For outcome B: (first roll = one 6, pick up two non 6's and roll again)
2nd roll:
z = 25/36 = 0 sixes
y = 10/36 = 1 six
x = 1/36 = 2 sixes
Outcomes y and x fulfill requirements.
So would the formula below give me my answer?
C + D + Ac + Ad + By + Bx = X
What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls?
If I substituted correctly and did the math correct the answer I got was 22.3%
Is this correct?
|
Your approach and your calculations are correct. Here is a variation based upon generating functions. We encode the roll of three dice with
\begin{align*}
(5+t)^3
\end{align*}
marking an occurrence of $6$ with $t$ and collecting all other five possibilities with $5$. The probability to get $j$ sixes $0\leq j \leq 3$ in the first roll can be written as
\begin{align*}
[t^j](5+t)^3\cdot\frac{1}{6^3}
\end{align*}
Encoding the second roll with $(5+u)^{3-j}$ we calculate
\begin{align*}
\sum_{{0\leq j,k\leq 3}\atop{j+k\geq 2}}&[t^j](5+t)^3[u^k](5+u)^{3-j}\cdot\frac{1}{6^{6-j}}\\
&=[t^0](5+t)^3\left([u^2]+[u^3]\right)(5+u)^3\cdot\frac{1}{6^6}\\
&\qquad+[t^1](5+t)^3\left([u^1]+[u^2]\right)(5+u)^2\cdot\frac{1}{6^5}\\
&\qquad+[t^2](5+t)^3\left([u^0]+[u^1]\right)(5+u)^1\cdot\frac{1}{6^4}\\
&\qquad+[t^3](5+t)^3\\
&=\binom{3}{0}5^3\left(\binom{3}{2}5^1+\binom{3}{3}5^0\right)\cdot\frac{1}{6^6}\\
&\qquad+\binom{3}{1}5^2\left(\binom{2}{1}5^1+\binom{2}{2}5^0\right)\cdot\frac{1}{6^5}\\
&\qquad+\binom{3}{2}5^1\left(\binom{1}{0}5^1+\binom{1}{1}5^0\right)\cdot\frac{1}{6^4}\\
&\qquad+\binom{3}{3}5^0\cdot\frac{1}{6^3}\\
&=\frac{125\cdot16}{6^6}+\frac{75\cdot11}{6^5}+\frac{15\cdot6}{6^4}+\frac{1}{6^3}\\
&=\frac{5\,203}{23\,328}\\
&\doteq 0.223
\end{align*}
|
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|
Is there an error in this problem? $\sin^4(\frac{23\pi}{12})-\cos^4(\frac{13\pi}{12})$
$\sin^4(\frac{23\pi}{12})-\cos^4(\frac{13\pi}{12})$
If there were written $\frac{23\pi}{12}$ instead of $\frac{13\pi}{12}$ it could be solved as $$\begin{align}[\sin^2(\frac{23\pi}{12})+\cos^2(\frac{23\pi}{12})][\sin^2(\frac{23\pi}{12})-\cos^2(\frac{23\pi}{12})]=&\\ -[\cos^2(\frac{23\pi}{12})-\sin^2(\frac{23\pi}{12})]=& \\ -\cos(\frac{23\pi}{6})=&\\ -\cos(\frac{11\pi}{6})=&\\ -\frac{\sqrt{3}}{2}\end{align}$$ Otherwise, there seems to be no way towards solving it.
|
$\frac{23}{12} \pi = 2\pi - \frac{\pi}{12}$, and $\frac{13}{12} \pi = \pi + \frac{\pi}{12}$. So you have:
$$\sin^4 \frac{\pi}{12} - \cos^4 \frac{\pi}{12}$$
which surprisingly gives the same answer you reached.
|
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|
Prove that $\sqrt{m-1}$ is an integer
Let $k$ be a fixed odd positive integer. Let $m,n$ be positive integers such that $(2+\sqrt{3})^k=m+n\sqrt{3}$. Prove that $\sqrt{m-1}$ is an integer.
Let $(2+\sqrt{3})^n = a_n+b_n\sqrt{3}$. From $$a_{n+1}+b_{n+1}\sqrt{3} = (a_n+b_n\sqrt{3})(2+\sqrt{3}) = (2a_n+3b_n)+(a_n+2b_n)\sqrt{3},$$ we get $a_{n+1} = 2a_n+3b_n$ and $b_{n+1} = a_n+2b_n$ with $a_1 = 2$ and $b_1 = 1$.
How can we show that $\sqrt{a_n-1}$ is an integer for odd $n$?
|
Since $k$ is odd $m-1$ is not multiple of $3$. This is because it is a sum of multiples of $3$ and a $2^{k}=2\cdot 4^{(k-1)/2}=2\mod{3}$.
Now, $m^2-3n^2=1$. So,
$$(m-1)(m+1)=3n^2$$
Since $m-1$ is not divisible by $3$, the $3$ divides the factor $m+1$.
The only common divisor of $m-1$ and $m+1$ can be $2=(m+1)-(m-1)$. But $m$ is even. Therefore, $m-1$ and $m+1$ are relatively prime.
So, $m-1$ is a square.
|
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|
Infinite square root muliplication $(x=3\sqrt{y\sqrt{3\sqrt{y}....}})$ I have this problem which says
$$
x=3\sqrt{y\sqrt{3\sqrt{y\cdots}}}\\
y=3\sqrt{x\sqrt{3\sqrt{x\cdots}}}
$$
What is $x+y$
I tried
$$
x^2=9y\sqrt{x}\\
x^3=81y^2\\
y^3=81x^2\\
x^3+y^3=81(x^2+y^2)\\
(x+y)(x^2-xy+y^2)=81(x^2+y^2)
$$
I don't know what to do or if I am do that right
|
From second equality we obtain
$$x^2=9x\sqrt{x},$$ which gives $x=0$ or $x=81$. The rest is smooth.
|
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|
If I select 9 random numbers from 1 to 10, what is the probability that their sum is less than 20?
If I select 9 random numbers from 1 to 10, what is the probability that their sum is less than 20?
Is there someway to figure this out without manually figuring out each possible scenario in which the 9 digits don't sum to 20?
|
A generating function approach yields the result without too much computation.
If we choose 9 integers from 1 to 10, there are $10^9$ possible outcomes, all of which we assume are equally likely. We want to count the outcomes in which the total is less than or equal to 19, i.e. the number of solutions in integers to
$$x_1 + x_2 + x_3 + \dots +x_9 \le 19$$
subject to $1 \le x_i \le 10$ for $i = 1,2,3, \dots ,9$.
An equivalent problem is to count the number of solutions to
$$x_1 + x_2 + x_3 + \dots +x_9 + y = 19$$
subject to $1 \le x_i \le 10$ for $i = 1,2,3, \dots ,9$ and $0 \le y$.
The generating function for the number of solutions to
$$x_1 + x_2 + x_3 + \dots +x_9 + y = r$$ subject to the above constraints is
$$\begin{align}
f(z) &= (z + z^2 + z^3 + \dots + z^{10})^9 \; (1 + z + z^2 + \dots) \\
&= z^9 \; \left( \frac{1-z^{10}}{1-z} \right)^9 \frac{1}{1-z}\\
&= z^9 \; (1-z^{10})^9 \; (1-z)^{-10} \\
&= z^9 \cdot \sum_{i=0}^9 (-1)^i \binom{9}{i}z^{10 i} \cdot \sum_{j=0}^{\infty} \binom{10+j-1}{j} z^j
\end{align}$$
From the last equation we can easily find the coefficient of $z^{19}$, which is
$$[z^{19}] f(z) = 1 \cdot \binom{19}{10} - \binom{9}{1} \cdot 1 = 92,369$$
So the probability of getting a sum less than or equal to 19 is
$$\frac{92,369}{10^9} = 9.2369 \times 10^{-5}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$ Find $\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$
$$I=\int e^{x^4}(x+x^3+2x^5)e^{x^2}dx$$
$$I=\int e^{x^4}(1+x^2+2x^4)e^{x^2}xdx$$Let $x^2=t$
$I=\int e^{t^2}(1+t+2t^2)e^t\frac{dt}{2}=\frac{1}{2}\int e^{t^2+t}(1+t+2t^2)dt$
I am stuck here.
|
Hint:
As $\dfrac{d(x^4+x^2)}{dx}=4x^3+2x$
$$\int e^{x^4+x^2}(2x+2x^3+4x^5)dx=\int[e^{x^4+x^2}x^2(2x+4x^3)+e^{x^4+x^2}(2x)dx$$
which is clearly of the from $$\int e^{f(x)}[g'(x)+g(x)f'(x)]dx=e^{f(x)}g(x)+K$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Calculate $\frac{1}{{1 - 100 + 500}} + \frac{{{2^2}}}{{{2^2} - 200 + 500}} + \ldots + \frac{{{{99}^2}}}{{{{99}^2} - 9900 + 500}} $ I need a hint to solve this problem : Find the sum of
$$ \frac{1}{{1 - 100 + 500}} + \frac{{{2^2}}}{{{2^2} - 200 + 500}} + \frac{{{3^2}}}{{{3^2} - 300 + 500}} + + \ldots + \frac{{{{99}^2}}}{{{{99}^2} - 9900 + 500}}
$$
|
Hint:
$\frac{N^2}{N^2-100N+500} = \frac{N^2}{N^2-100N+2500-2500+500} = \frac{N^2}{(N-50)^2-(10\sqrt{20})^2)} $
$ a = (50-10\sqrt{20})$
$S = \frac{AN+B}{N-a} + \frac{C}{N+a}$
Simplifying this
$S = \sum_{N=1}^{99} 1+ a^2\frac{1}{N^2-a^2} = \sum_{N=1}^{99}1+ a^2\frac{1}{(\frac{N}{a})^2-1}$
$S= 99+(50-10\sqrt{20})^2*(-6.67) $
I evaluated the last sum by Wolfram Alpha, if anyone has an idea of how to evaluate that will be great
|
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|
Matrix to power 30 using eigen values issue
$$ \text{If matrix } A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} \text{ then find } A^{30}.$$
I tried to approach through diagonalization using eigen values method.
I got eigen values as $-1, 1, 1$
As per diagonalization $ A = P*D*P^{-1}.$ So $ A^{30} = P*D^{30}*P^{-1}. $
But $ D^30 = I.$
So, $ A^{30} = P*I*P^{-1} = I $
But $ A^{30} $ is not equal to I. If we do general multiplication without all these.
Where is the mistake in my approach?
|
Since $A$ is not diagonalizable we can use Jordan form
\begin{align}
A^{30} &=
\begin{pmatrix}
0 & 2 & 0 \\
-1 & 1 & 1 \\
1 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
-1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 1 & 1
\end{pmatrix}^{30}
\begin{pmatrix}
1/4 & -1/2 & 1/2 \\
1/2 & 0 & 0 \\
-1/4 & 1/2 & 1/2
\end{pmatrix} \\
&=
\begin{pmatrix}
0 & 2 & 0 \\
-1 & 1 & 1 \\
1 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 30 & 1
\end{pmatrix}
\begin{pmatrix}
1/4 & -1/2 & 1/2 \\
1/2 & 0 & 0 \\
-1/4 & 1/2 & 1/2
\end{pmatrix} \\
&= \begin{pmatrix}
1 & 0 & 0 \\
15 & 1 & 0 \\
15 & 0 & 1
\end{pmatrix}
\end{align}
|
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|
Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$
Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$
Also it is a question of S.L. Loney's Plane Trignonometry
What I've tried by now:
\begin{align}
& =\frac{1+\sin\theta-\sin(90-\theta)}{1+\cos(90-\theta)+\cos\theta} \\[10pt]
& =\frac{1+2\cos45^\circ \sin(\theta-45^\circ)}{1+2\cos45^\circ \cos(45-\theta)} \end{align}
Cause I do know
\begin{align} & \sin c + \sin d = 2\sin\left(\frac{c+d}{2}\right)\cos\left(\frac{c-d}{2}\right) \\[10pt] \text{and } & \cos c + \cos d = 2\cos\left(\frac{c+d}{2}\right)\sin\left(\frac{c-d}{2}\right)
\end{align}
I can't think of what to do next..
|
Let $\theta = 2 \phi$, then the thing to be proven is:
Prove that $$\frac{1 + \sin(2\phi) - \cos(2\phi)}{1 + \sin(2\phi) + \cos(2\phi)} = \tan(\phi)$$
Then use:
$$\sin(2\phi) = 2 \sin \phi \cos \phi$$
$$\cos(2\phi) = \cos^2 \phi - \sin^2 \phi$$
and:
$$\sin^2 \phi + \cos^2 \phi = 1$$
|
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|
The derivative of the logistic function The logistic function is $\frac{1}{1+e^{-x}}$, and its derivative is $f(x)*(1-f(x))$. In the following page on Wikipedia, it shows the following equation:
$$f(x) = \frac{1}{1+e^{-x}} = \frac{e^x}{1+e^x}$$
which means $$f'(x) = e^x (1+e^x) - e^x \frac{e^x}{(1+e^x)^2} = \frac{e^x}{(1+e^x)^2}$$
I understand it so far, which uses the quotient rule $$\left(\frac{g(x)}{h(x)}\right)' = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}.$$
However, why is it then transformed into $f(x) * (1-f(x))$?
|
Just start from the answer and work backwards.
$$\begin{aligned}
f(x)(1-f(x)) &= \frac{e^x}{1+e^x} \left( 1 - \frac{e^x}{1+e^x} \right)\\
&= \frac{e^x}{1+e^x} \left( \frac{1+e^x}{1+e^x} - \frac{e^x}{1+e^x} \right)\\
&= \frac{e^x}{1+e^x} \left( \frac{1}{1+e^x} \right) = \frac{e^x}{(1+e^x)^2} = f'(x)
\end{aligned}$$
|
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|
Condition of existence of a triangle It's easy to prove that the triangle inequality holds for any triangle with the lengths of sides $a$, $b$ and $c$.
But how can one prove that if the triangle inequality holds for any given positives $a$, $b$ and $c$ then a triangle (geometric figure) with the lengths of the sides equal to $a,b$ and $c$ can necessarily be formed?
|
Let $B(0,0)$ and $C(a,0)$.
Hence, $BC=a$ and we need to prove that there exists $A(x,y)$ such that $AB=c$ and $AC=b$.
You can write equations of two circles and prove that there are intersection points.
For example $x^2+y^2=c^2$ and $(x-a)^2+y^2=b^2$.
Thus, $-2ax+a^2+c^2=b^2$ or $x=\frac{a^2+c^2-b^2}{2a}$ and
$$y^2=c^2-\left(\frac{a^2+c^2-b^2}{2a}\right)^2$$ or
$$y^2=\frac{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}{4a^2},$$
which says that there are two intersection points.
|
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|
Integrating factor $(3y^2-x) + 2y(y^2-3)y'=0$ is a function
Show that the differential equation $(3y^2-x) + 2y(y^2-3)y' = 0$ admits an integrating factor which is a function of $(x+y^2)$. Hence solve the equation.
I know how to solve this by using Integrating Factor of an exact equation
But question specifically asking solving using a function. How to approach this question.
|
$(3y^2-x)+2y(y^2-3)y'=0$
$2y\dfrac{dy}{dx}=\dfrac{x-3y^2}{y^2-3}$
$\dfrac{d(y^2)}{dx}=\dfrac{x-3y^2}{y^2-3}$
$\dfrac{du}{dx}=\dfrac{x-3u}{u-3}$ $(\text{Let}~u=y^2)$
Let $\begin{cases}x=t+a\\u=v+b\end{cases}$ ,
Then $\dfrac{dv}{dt}=\dfrac{t+a-3v-3b}{v+b-3}$
Take $a=9$ and $b=3$ , the ODE becomes
$\dfrac{dv}{dt}=\dfrac{t-3v}{v}$
$\dfrac{dv}{dt}=\dfrac{t}{v}-3$
Let $w=\dfrac{v}{t}$ ,
Then $v=tw$
$\dfrac{dv}{dt}=t\dfrac{dw}{dt}+w$
$\therefore t\dfrac{dw}{dt}+w=\dfrac{1}{w}-3$
$t\dfrac{dw}{dt}=\dfrac{1}{w}-3-w$
$t\dfrac{dw}{dt}=-\dfrac{w^2+3w-1}{w}$
$\dfrac{w}{w^2+3w-1}~dw=-\dfrac{dt}{t}$
$\int\dfrac{w}{w^2+3w-1}~dw=-\int\dfrac{dt}{t}$
$\dfrac{(13+3\sqrt{13})\ln(2w+3+\sqrt{13})}{26}+\dfrac{(13-3\sqrt{13})\ln(2w+3-\sqrt{13})}{26}=-\ln t+c$
$(2w+3+\sqrt{13})^{13+3\sqrt{13}}(2w+3-\sqrt{13})^{13-3\sqrt{13}}t^{26}=C$
$\left(\dfrac{2v}{t}+3+\sqrt{13}\right)^{13+3\sqrt{13}}\left(\dfrac{2v}{t}+3-\sqrt{13}\right)^{13-3\sqrt{13}}t^{26}=C$
$(2v+(3+\sqrt{13})t)^{13+3\sqrt{13}}(2v+(3-\sqrt{13})t)^{13-3\sqrt{13}}=C$
$(2(u-3)+(3+\sqrt{13})(x-9))^{13+3\sqrt{13}}(2(u-3)+(3-\sqrt{13})(x-9))^{13-3\sqrt{13}}=C$
$(2(y^2-3)+(3+\sqrt{13})(x-9))^{13+3\sqrt{13}}(2(y^2-3)+(3-\sqrt{13})(x-9))^{13-3\sqrt{13}}=C$
Which is impossible to solve by considering the integrating factor of the function type $x+y^2$
|
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|
Inequality : $\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq 1+\sqrt[3]{3}$
Let $x>0$, $y>0$ and $z > 0$. Prove that $$\sqrt[3]{\frac{x^3+y^3+z^3}{xyz}} + \sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}} \geq
1+\sqrt[3]{3}.$$
From Micheal Rozenberg's answer :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq\sqrt[3]x\left(\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2}\right)$,
Prove that $(x+2)^2(x+2\sqrt{x}+3)\geq9\sqrt[3]{x(x^3+2)^2}$,
LHS :
$(x+2)\sqrt{x^2+2}\left(\sqrt{x^2+2}+\sqrt{2x+1}\right)\geq (x+2) \frac{x+2}
{\sqrt{3}}\left(\frac{x+2}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)= \frac{(x+2)^2}{3}(x+2\sqrt{x}+3)$
RHS :
$\sqrt[3]{3x(x^3+2)}\leq \sqrt[3]{(x^3+2)^2}$
$\sqrt[3]{9x^2}\leq \sqrt[3]{(x^3+2)^2}$
so $\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2}\leq 3\sqrt[3]{(x^3+2)^2}$
$\sqrt[3]{x}(\sqrt[3]{(x^3+2)^2}+\sqrt[3]{3x(x^3+2)}+\sqrt[3]{9x^2})\leq 3\sqrt[3]{x(x^3+2)^2}$
Thus,
$\frac{(x+2)^2}{3}(x+2\sqrt{x}+3) \geq 3\sqrt[3]{x(x^3+2)^2}$
$(x+2)^2(x+2\sqrt{x}+3) \geq 9\sqrt[3]{x(x^3+2)^2}$
|
$$x+y+z=p,\ xy+yz+zx=q,\ xyz=r$$
we can assume : $x+y+z=1$
$x^3+y^3+z^3=1-3q+3r,\ x^2+y^2+z^2=1-2q\ \ \ \left( 0 < q \le \dfrac{1}{3}\right)$
$$\left( \dfrac{1-3q}{r}+3\right)^{1/3}+\left( \dfrac{q}{1-2q}\right)^{1/2}\ge 1+\sqrt[3]3$$
We have
$$q^2=(xy+yz+zx)^2\ge 3xyz(x+y+z) =3r.$$
So: $LHS \ge \sqrt[3]3\cdot\left( \dfrac{1-3q}{q^2}+1\right)^{1/3}+\left( \dfrac{q}{1-2q}\right)^{1/2}=f(q)$
$$\dfrac{q}{1-2q}=t \le 1$$
$g(t)= \sqrt[3]3\cdot\left( \dfrac{1+t}{t^2}-1\right)^{1/3}+ \sqrt{t}\ge g\left(1\right )=1+\sqrt[3]3$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the inequality $\sum_{cyc}\frac{a}{1+\left(b+c\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$
Let $a>0$, $b>0$ and $c>0$ such that $a+b+c=3$. Prove the inequality
$$\frac{a}{1+\left(b+c\right)^2}+\frac{b}{1+\left(c+a\right)^2}+\frac{c}{1+\left(a+b\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}.$$
$$LHS=\frac{a}{1+\left(3-a\right)^2}+\frac{b}{1+\left(3-b\right)^2}+\frac{c}{1+\left(3-c\right)^2}$$
We have inequality: $$\frac{a}{1+\left(3-a\right)^2}\le \frac{9}{25}a-\frac{4}{25}\Leftrightarrow -\frac{\left(a-1\right)^2\left(9a-40\right)}{25\left(\left(a-3\right)^2+1\right)}\le 0\forall 0<a\le 1$$
$$\Rightarrow LHS\le \frac{9}{25}\left(a+b+c\right)-\frac{4}{25}\cdot 3=\frac{3}{5}$$
Need prove: $$\frac{3}{5}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$$
$$\Leftrightarrow a^2+b^2+c^2+12abc\le 5\left(a^2+b^2+c^2\right)$$
$$\Leftrightarrow 12abc\le 4\left(a^2+b^2+c^2\right)\Leftrightarrow 3abc\le a^2+b^2+c^2$$
By AM-GM: $$3abc\le \frac{\left(a+b+c\right)^3}{9}=3=\frac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2$$
Right or Wrong. I think it's wrong. Help me
|
We need to prove that
$$\sum_{cyc}\left(\frac{a}{1+(b+c)^2}-a\right)\leq\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}-3$$ or
$$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}\geq\frac{36abc}{a^2+b^2+c^2+12abc},$$
which seems better, but I did not find something nice here.
By the way, our inequality is obviously true after homogenization and full expanding.
Indeed, we need to prove that
$$\sum_{cyc}\frac{a}{(a+b+c)^2+9(b+c)^2}\leq\frac{a^2+b^2+c^2}{(a^2+b^2+c^2)(a+b+c)+36abc}$$ or
$$\sum_{sym}(10a^7b+130a^6b^2+250a^5b^3+130a^4b^4-68a^6bc-159a^5b^2c+255a^4b^3c-225a^4b^2c^2-323a^3b^3c^2)\geq0,$$
which is true by Muirhead:
$$\sum_{sym}10a^7b\geq\sum_{sym}10a^6b^2;$$
$$\sum_{sym}68a^6b^2\geq\sum_{sym}68a^6bc;$$
$$\sum_{sym}72a^6b^2\geq\sum_{sym}72a^5b^3;$$
$$\sum_{sym}159a^5b^3\geq\sum_{sym}159a^5b^2c;$$
$$\sum_{sym}163a^5b^3\geq\sum_{sym}163a^4b^2c^2;$$
$$\sum_{sym}62a^4b^4\geq\sum_{sym}62a^4b^2c^2;$$
$$\sum_{sym}68a^4b^4\geq\sum_{sym}68a^3b^3c^2$$ and
$$\sum_{sym}255a^4b^3c\geq\sum_{sym}255a^3b^3c^2.$$
After summing of these inequalities we'll get the needed inequality.
Done!
|
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|
Argument for why $a^2 + 1$ is never divisible by a $3 \mod 4$ integer How do you show that $$a^2 + 1$$ is never divisible by a $3 \mod 4$ integer (which is equivalent to showing that it has no $3 \mod 4$ prime factor) for any non-negative integer $a$ by analysing the arithmetic series representation of $a^2$, $1 + 3 + 5 + ... + (2a-1)$?
|
Not using the series:
*
*Case 1: $a$ is even, say, $a=2k$. Then $a^2+1 = 4k^2 + 1$.
*Case 2: $a$ is odd, say, $a=2k+1$. Then $a^2+1 = 4k^2+4k+2$.
Using the series $a^2=1+3+5+...+(2a-5)+(2a-3)+(2a-1)$.
*
*Note that consecutive terms add to multiples of $4$.
*Therefore, $a^2$ is either $(1+3)+(5+7)+...+((2a-3)+(2a-1))$, which is a sum of multiples of 4, or $1 + (3+5)+(7+9)+...+((2a-3)+(2a-1))$, which is one more than a multiple of $4$.
*in any case, the result follows for $a^2+1$
Explicitly in $\mathbb{Z}_4$:
*
*$a$ is either 0, 1, 2 or 3. Therefore $a^2$ is either 0 or 1, so $a^2+1$ is 1 or 2.
|
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|
Evaluate $\lim_{x \to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}}$ Evaluate $$L=\lim_{x \to 0}\frac{x}{\sqrt{1-\sqrt{1-x^2}}} \tag{1}$$
I have used L Hopital's Rule we get
$$L=\lim_{x \to 0} 2\frac{ \sqrt{1-\sqrt{1-x^2}} \sqrt{1-x^2}}{x}$$
$\implies$
$$L=2 \lim_{x \to 0}\frac{ \sqrt{1-\sqrt{1-x^2}}}{x}$$ and from $(1)$ we get
$$L=\frac{2}{L}$$
$$L=\sqrt{2}$$
is this correct appraoach?
|
Alternatively:
$$L=\lim_\limits{x\to 0} \frac{sgn(x) \sqrt{x^2}}{\sqrt{1-\sqrt{1-x^2}}}=$$
$$\lim_\limits{x\to 0} sgn(x) \sqrt{\lim_\limits{x\to 0} \frac{x^2}{\sqrt{1-\sqrt{1-x^2}}}} =(LR)= $$
$$\lim_\limits{x\to 0} sgn(x) \sqrt{\lim_\limits{x\to 0} 2\sqrt{1-x^2}}=$$
$$\lim_\limits{x\to 0} sgn(x) \sqrt{2}.$$
Hence, the limit exists if $x$ approaches $0$ from left or right, otherwise it does not exist.
|
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|
How to find the volume of a part of sphere from $z=0.5r$ in spherical coordinates?
Given sphere $x^2+y^2+z^2=a^2$ find the volume above $z=0.5a, a>0$. The solution must use spherical coordinates.
It looks like the radius is not constant as it depends on the angle $\phi$ so $0.5a\le r\le\frac{0.5a}{\cos\phi}$
I think regarding the ranges:
$$
0\le\theta\le2\pi\\
0\le\phi\le\pi/3\quad\text{because }\cos\phi=\frac{0.5a}{a}=1/2
$$
Because we're using spherical coordinates we also need to multiply by the Jacobian:
$$
\iiint r^2\sin\phi
$$
Finally the computation:
$$
\iint\sin\phi\bigg[\frac{r^3}{3}\bigg]_{0.5a}^{\frac{0.5a}{\cos\phi}}=\iint\frac{a^3\sin\phi}{24\cos^3\phi}-\iint \frac{a^3\sin\phi}{24}
$$
We can solve the integrals separately:
$$
\iint\frac{a^3\sin\phi}{24\cos^3\phi}\quad\text{using substitution}\quad u=\sin\phi\\
\iint\frac{a^3\sin\phi}{24\cos^3\phi}=\iint-\frac{a^3}{24u^3}=\frac{a^3}{96}\int\bigg[\frac{1}{\cos^4\phi}\bigg]_0^{\pi/3}=\frac{a^3}{96}\int 15=\frac{15a^3}{96}\cdot 2\pi
$$
I'm really not sure about the ranges I chose and the result seems not correct to me.
|
Using spherical coordinates, the radius $r$ should be $\frac{a}{2 \cos \phi}<r<a$ and indeed the angle $0<\phi<\pi/3$. So you get
$$
2 \pi \int_0^{\pi/3}\sin\phi\bigg[\frac{r^3}{3}\bigg]_{\frac{a}{2 \cos \phi}}^{a} \rm d \phi
= 2 \pi \int_0^{\pi/3}\frac{a^3\sin\phi}{3} \rm d \phi -2 \pi \int_0^{\pi/3} \frac{a^3\sin\phi}{24 \cos^3 \phi} \rm d \phi \\
= 2 \pi \frac{a^3}{3} \bigg[ 1 - \cos\pi/3 -\frac{1}{16} ( \cos^{-2} \phi)_0^{\pi/3} \bigg] \\
= 2 \pi \frac{a^3}{3} \bigg[ 8/16 -\frac{1}{16} (4-1 )\bigg]\\
= 2 \pi \frac{a^3}{3} \frac{5}{16} = \pi {a^3} \frac{5}{24}
$$
Compare this to the standard result (which uses cylindrical coordinates) which is (see my comment above):
$$
V = \dfrac{\pi (a/2)^2}{3} (3a -a/2 ) = \dfrac{5 \pi a^3}{24}
$$
Here you go!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Give a sequence such that root test works while ratio test fails
Question: Give a sequence $(a_n)_{n=1}^\infty$ with $a_n>0$ such that root test works while ratio test does not work, that is,
$$\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}} \text{ exists}$$
while
$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} \text{ does not exist}.$$
My attempt:
Define a sequence $(a_n)_{n=1}^\infty$ such that
$$ a_n = \begin{cases}
2^{n-1} & \text{if }n \text{ is odd,} \\
2^{n+1} & \text{if }n \text{ is even}.
\end{cases}
$$
Odd subsequence $(a_{n_j})_{j=1}^\infty$ implies that $\lim_{j\rightarrow\infty}(a_{n_j})^{\frac{1}{n_j}} = \lim_{j\rightarrow\infty} 2^{1-\frac{1}{n_j}} = 2 $ while even subsequence $(a_{n_k})_{k=1}^\infty$ implies that $\lim_{k\rightarrow\infty}(a_{n_k})^{\frac{1}{n_k}} = \lim_{k\rightarrow\infty} 2^{1+\frac{1}{n_k}} = 2.$
It follows that $\lim_{n\rightarrow\infty}(a_n)^{\frac{1}{n}} = 2,$ that is, root test works.
However, for odd subsequence $(a_{n_j})_{j=1}^\infty,$
$$\lim_{j\rightarrow\infty}\frac{a_{n_j+1}}{a_{n_j}} = \lim_{n\rightarrow\infty} \frac{2^{n_j+1}}{2^{n_j-1}} = \lim_{n\rightarrow\infty} 4 = 4.$$
For even subsequence $(a_{n_k})_{k=1}^\infty,$
$$\lim_{k\rightarrow\infty}\frac{a_{n_k+1}}{a_{n_k}} = \lim_{k\rightarrow\infty} \frac{2^{n_k-1}}{2^{n_k+1}} = \lim_{k\rightarrow\infty} \frac{1}{4} = \frac{1}{4}.$$
Therefore, $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}$ does not exist, that is, ratio test does not work.
Does my example work?
|
Your example is right. You can construct a similar exemple such that the series $\sum\limits_{n=1}^{\infty} a_n$ converges as follows:
$$ a_n = \begin{cases}
\frac{1}{2^{n+1}} & \text{if $n$ is odd,} \\
\frac{1}{2^{n}} & \text{if $n$ is even}.
\end{cases}
$$
Then for the odd subsequence $(a_{n_j})_{j=1}^\infty$ we have $\lim\limits_{j\rightarrow\infty}(a_{n_j})^{\frac{1}{n_j}} = \lim\limits_{j\rightarrow\infty} \frac{1}{2^{1+\frac{1}{n_j}}}= \frac{1}{2}$ and for the even subsequence $(a_{n_k})_{k=1}^\infty$ we have $\lim\limits_{k\rightarrow\infty}(a_{n_k})^{\frac{1}{n_k}} = \lim\limits_{k\rightarrow\infty} \frac{1}{2}=\frac{1}{2}$, this implies that $(a_n)^{\frac{1}{n}} \to \frac{1}{2}$, therefore the root test works and the series converges absolutely.
But for odd $n$, $$a_n = \frac{1}{2^{n+1}}= a_{n+1}.$$ So, if the limit $\lim\limits_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}$ exists, it must be equal to $1$, which makes the ratio test inconclusive. Actually, the limit does not exist, because for even $n$, $\frac{a_{n+1}}{a_n} = \frac{1}{4}$.
|
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|
Evaluating the limit $\lim_{x\to1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)$ In trying to evaluate the following limit:
$$\lim_{x\to1}\left(\frac{1}{1-x}-\frac{3}{1-x^3}\right)$$
I am getting the indefinite form of:
$$\frac{1}{\mbox{undefined}}-\frac{3}{\mbox{undefined}}$$
What would be the best solution to evaluating this limit?
|
Hint. Note that for $x\not=1$,
$$\frac{1}{1-x}-\frac{3}{1-x^3}=\frac{x^2+x+1-3}{(1-x)(x^2+x+1)}
=\frac{(x+2)(x-1)}{(1-x)(x^2+x+1)}=-\frac{x+2}{x^2+x+1}.
$$
|
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|
How to evaluate $\binom{2}{2}\binom{10}{3} + \binom{3}{2}\binom{9}{3} + \binom{4}{2}\binom{8}{3} + \ldots + \binom{9}{2}\binom{3}{3}$ I really can't understand how to approach exercise 41.
I tried by comparing the 5th term of binomial expansion and something like that, but getting 12C5, I think the answer should be 13C6. How to approach?
|
Consider the number of non-negative integral solutions to $X_1 + X_2 + X_3 .. + X_7 = 7$. This should equal $\binom{7+ 7 - 1}{7 - 1} = \binom{13}{6}$.
These solutions can also be counted as follows:
Case 1: Let $X_1 + X_2 + X_3 = 0$ and $X_4 + X_5 + X_6 + X_7 = 7$. The combined number of solutions for these are $\binom{2}{2} \cdot \binom{10}{3}$
Case 2: Let Let $X_1 + X_2 + X_3 = 1$ and $X_4 + X_5 + X_6 + X_7 = 6$. The combined number of solutions for these are $\binom{3}{2} \cdot \binom{9}{3}$.
.
.
.
Case 8: Let Let $X_1 + X_2 + X_3 = 7$ and $X_4 + X_5 + X_6 + X_7 = 0$. The combined number of solutions for these are $\binom{9}{2} \cdot \binom{3}{3}$.
Thus, we get that $$\binom{13}{6} = \binom{2}{2} \cdot \binom{10}{3} + \binom{3}{2} \cdot \binom{9}{3} + \ .. \ + \binom{9}{2} \cdot \binom{3}{3}$$
|
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|
Let $\frac13\sin a-\frac17\cos a=\frac{1}{2017}.$ ... Let $a$ in $[0, \pi]$ such that $\frac13\sin a-\frac17\cos a=\frac{1}{2017}.$ If $\lvert\tan{a}\rvert=\frac mn$, where m and n are relatively prime, find m+n.
My work:
$\frac13\sin a-\frac17\cos a=\frac{1}{2017}$From the trig identity $a \cos{t} + b \sin{t} = \sqrt{a^2 + b^2} \; \sin(t + \tan^{-1} \frac{a}{b})$ it follows that:$\sqrt{\frac{1}{9}+\frac{1}{49}}\sin (a-\arctan{\frac37})=\frac{1}{2017}$$a=\arcsin{\frac{1}{2017\sqrt{\frac{58}{441}}}}+\arctan{\frac37}$...
From here I bashed out long equations using more trig identities/cancelling by taking the sine of both sides, but I did not get a fraction and the work is not worth putting here. I'm assuming the approach of using $a \cos{t} + b \sin{t} = \sqrt{a^2 + b^2} \; \sin(t + \tan^{-1} \frac{a}{b})$ was somehow incorrect or I made a simple error somewhere later on.
When I originally tried solving this problem I multiplied through by 3*7*2017 but this leads to the same issue and gigantic numbers everywhere.
Is there a better approach to solving this?
This question was from the AMSP 2017 Test C entrance exam.
|
As $\tan=\frac \sin\cos$, we may assume $\sin a= mu$, $\cos a=\pm nu$. Thus
$$\frac m3u\mp\frac n7u =\frac1{2017}$$
or
$$ u=\frac{21}{2017\cdot (7m\mp 3n)}.$$
Additionally,
$$ 1=\sin^2a+\cos^2a=(m^2+n^2)u^2$$
so that elimination gives
$$ 2017^2\cdot(7m\mp 3n)^2=21^2\cdot (m^2+n^2).$$
The left hand side is either odd or a multiple of $4$;
the right hand side is either odd (if exactly one of $n,m$ is odd) or $\equiv 2\pmod 8$ (if both are odd). We conclude that both sides are odd and exactly one of $n,m$ is even and one is odd.
Now let $p$ be any prime $\ne 2017$ and assume $p^k\mid m^2+n^2$ with $k\ge 1$ (so $p\ne2$). It follows that $k$ is even and $p^{ k/2}\mid 7m\mp 3n$, as well as $p^{k/2}\mid (7m-3n)(7m+3n)=49m^2-9n^2$, as well as $p^{k/2}\mid 49m^2-9n^2+9(m^2+n^2)=58m^2$, and similarly $p^{k/2}\mid-58n^2$. From $\gcd(m,n)=1$, it follows that $p=29$ and $k=2$.
It cannot happen that $29^3\mid m^2+n^2$
Therefore,
$$ 7m\mp 3n=21\cdot 29^a\cdot2017^b,\qquad m^2+n^2=29^{2a}\cdot2017^{2b+2}$$
with $a\in\{0,1\}$ and $b\ge0$.
The simplest integer solution to this are obtained for $a=0$, $b=0$ from the observation that $44^2+9^2=2017$ and $(44+9i)^2=1855+792i$. Luckily, $3\cdot 1855-7\cdot 792=21$, which suggests the solution $m=792$, $n=1855$.
As of now, I do not see a fast way to show that there is no other solution (i.e., no solution with $a=1$ or $b>0$),
|
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|
Show that if $a+b+c=0$, $2(a^4 + b^4+ c^4)$ is a perfect square
Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square.
This question is from a math olympiad contest.
I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful direction after that.
Note: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question.
|
Denote:
$$a+b+c=0; ab+ac+bc=k; abc=t$$
Then $a,b,c$ are the roots of:
$$x^3+kx+t=0$$
Note:
$$a^3+ka+t=0 \Rightarrow a^4+ka^2+ta=0,$$
$$b^3+kb+t=0 \Rightarrow b^4+kb^2+tb=0,$$
$$c^3+kc+t=0 \Rightarrow c^4+kc^2+tc=0.$$
Add and multiply by $2$:
$$2(a^4+b^4+c^4)=-2k(a^2+b^2+c^2)-t(a+b+c)=-2k((a+b+c)^2-2k)-0=(2k)^2.$$
|
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|
find a and b such that the limit will exist and find the limit I am given this question $$\lim_{x\rightarrow 0} \frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}$$
And asked to find the $a$ and $b$ such that the limit exists and also to compute the limit
My solution:
Let $t$=$\sqrt{1+x}$.
Then the Maclaurin polynomial is : $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\mathcal{O}(x^3)$$
Now plugging in $x^2$ for x we get $$\sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)$$
and from the common Maclaurin polynomial we have that $e^t=1+t+\frac{t^2}{2}+ \mathcal{O}(t^3)$. Plugging in $\sqrt{x^2+1}$ for $t$ we get:
$$e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}$$
which in turn is:
$$e^{1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)
}=1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{O}(x^6)$$
hence we have:
$$\lim_{x\rightarrow 0} \frac{1+\left(1+\frac{x^2}{2}-\frac{x^4}{8}+\mathcal{O}(x^6)\right)+\mathcal{}O(x^6)-a-bx^2}{x^4}$$
And i argue that $a=2, b=1/2$ and the $\lim=-1/8 $
However the book disagrees with me and argues that the they should be $a=e, b=e/2$ and limit $=0$
i mean i can see how theyd done it, by not expanding $e^t$ but by only expanding $\sqrt{x^2+1}$ however what i dont understand how come we didnt get the same values or at least the same value for the limit?
|
Simplified computation:
Let $\sqrt{1+x^2}=t+1$, so that
$$\lim_{x\to0}\frac{e^{\sqrt{1+x^2}}-a-bx^2}{x^4}=\lim_{t\to0}\frac{e\cdot e^t-a-bt(2+t)}{t^2(2+t)^2}\\
=\lim_{t\to0}\frac{e+et+\dfrac{et^2}2+\dfrac{et^3}{3!}+\cdots-a-2bt-bt^2}{4t^2}.$$
Then it is obvious that we need
$$a=e,b=\frac e2$$ and the third coefficient is
$$\frac e2-\frac e2=0.$$
|
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|
Complex number $i^{{i^{i^{.^{.^.}}}}}$ If $A+iB=i^{{i^{i^{.^{.^.}}}}}$
Principal values only being considered,
Prove that
(a)tan $ \frac {\pi}{2} $A= $\frac{B}{A}$
(b) $A^2 + B^2 = e^{-\pi B}$
I tried the concept A+iB= $y=i^y$
$i= e^{ \frac{i\pi}{2}}$
$\ln(A+iB)=i \frac{\pi}{2}$(A+iB)
After this step not able to proceed
|
Whatever the value of $x = i^{i^{i^\cdots}}$ (if it exists), it should satisfy the equation
$$
x = i^x.
$$
But what is the definition of $i^x$ ? Since it is ambiguous, we must loosen the constraint even further: Whatever the value of $x$, it should satisfy
$$
x = e^{x \log i} \text{ for } \textit{some} \text{ value of } \log i
$$
Now, the set of logarithms of $i$ (the set of $y$ such that $e^y = i$) is $\{2 \pi i k + \tfrac{\pi i}{2} \}_{k \in \mathbb{Z}}$.
So our condition is now
$$
x = \exp\left(2 \pi i k x + \frac{\pi i}{2} x\right) \tag{1}
$$
Letting $x = A + Bi$, we get
\begin{align*}
A + Bi
&= \exp\left[ (A + Bi)\left(2 \pi i k x + \frac{\pi i}{2} x\right)\right] \\
&= \exp\left[ (-B + Ai)\left(2 \pi k + \frac{\pi}{2} \right)\right] \\
&= \underbrace{\exp\left[ (-B)\left(2 \pi k + \frac{\pi}{2} \right)\right]}_{\text{magnitude}}
\underbrace{\exp\left[ A \left(2 \pi k + \frac{\pi}{2} \right) i \right]}_{\text{direction}} \\
\end{align*}
Equate the magnitude squared of both sides and the slope (tangent of the angle) of both sides. For magnitude squared we get:
$$
A^2 + B^2 = e^{- \pi B + 4 \pi k B}. \tag{2}
$$
For tangent of the angle we get
$$
\frac{B}{A} = \tan \left((2 \pi k + \pi / 2) A\right). \tag{3}
$$
Your desired results follow from (2) and (3) if we assume $k = 0$.
This corresponds to taking the principal logarithm in defining $i^x$.
For $k = 0$:
\begin{align*}
A^2 + B^2 &= e^{- \pi B} \\
\frac{B}{A} &= \tan \left(\pi A / 2\right)
\end{align*}
|
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|
$\sin(ωt) + \sin(ωt+\Delta\theta) + \sin(ωt+2\,\Delta\theta) + \cdots + \sin(ωt+\overline{n-1}\,\Delta\theta) ={}$? How will we add this series?
Any simple process?
This is one of the equation derived from the equation to find intesity of ray at any particular point in Fraunhofer diffraction.
I think this can be solved using simple series but my teacher told any process involving Imaginary part iota, which I couldn't understand.
|
The hint.
Thy to multiply it by $2\sin\frac{\Delta\theta}{2}$ and use $$2\sin\alpha\sin\beta=\cos(\alpha-\beta)-\cos(\alpha+\beta).$$
\begin{align}
& \sin\omega t + \sin(\omega t+∆θ) + \sin(ωt+2∆θ) + \cdots + \sin(ωt+(n-1)∆θ) \\[10pt]
= {} & \frac{2\sin\frac{\Delta\theta}{2}\sin\omega t +2\sin\frac{\Delta\theta}{2} \sin(\omega t+∆θ)+ \cdots + 2\sin\frac{\Delta\theta}{2}\sin(ωt+(n-1)∆θ)}{2\sin\frac{\Delta\theta}{2}} \\[10pt]
= {} & \tfrac{\cos\left(\omega t-\frac{\Delta\theta}{2}\right)-\cos\left(\omega t + \frac{\Delta\theta}{2}\right) +\cos\left(\omega t+\frac{\Delta\theta}{2} \right) - \cos\left(\omega t+\frac{3\Delta\theta}{2} \right) + \cdots+\cos\left(\omega t+\left(n-\frac{3}{2}\right)\Delta\theta\right)-\cos\left(\omega t+\left(n-\frac{1}{2}\Delta\right)\theta\right)}{2\sin\frac{\Delta\theta}{2}} \\[10pt]
= {} & \frac{\cos\left(\omega t-\frac{\Delta\theta}{2}\right)-\cos\left(\omega t + \left(n-\frac{1}{2}\right)\Delta\theta\right)}{2\sin\frac{\Delta\theta}{2}}
\end{align}
|
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|
Solve $\lfloor \frac{2x-1}{3} \rfloor + \lfloor \frac{4x+1}{6} \rfloor=5x-4$ I came across this problem and it doesn't seem so tricky, but I didn't do it.
Solve the equation
$$\lfloor {\frac{2x-1}{3}}\rfloor + \lfloor {\frac{4x+1}{6}}\rfloor=5x-4.$$
My thoughts so far:
Trying to use the inequality $k-1 < \lfloor k\rfloor \leq k$
$5x-4$ is integer, so $x=\frac{a}{5},$ where $a$ is from $\mathbb Z,$ then replace in the original equation, to get
$$\lfloor \frac{4a-10}{30}\rfloor + \lfloor \frac{4a+5}{30}\rfloor =a-4$$
Let $k= \frac{4a-10}{30}$
so $\lfloor k\rfloor +\lfloor k+\frac{1}{2}\rfloor =a-4$
but $\lfloor k\rfloor =\lfloor k+\frac{1}{2}\rfloor $ or $\lfloor k\rfloor +1=\lfloor k+\frac{1}{2}\rfloor $, so we have 2 cases
I strongly think that a shorter solution exists, I would appreciate a hint!
|
$$\lfloor{\frac{2x-1}{3}}\rfloor + \lfloor{\frac{4x+1}{6}}\rfloor=5x-4.$$
As you have been mentioned;
$5x-4$ must be an integer,
so $x=\frac{a}{5},$
where $a$ is from $\mathbb Z,$
then replace in the original equation, to get
$$\lfloor\frac{4a-10}{30}\rfloor + \lfloor\frac{4a+5}{30}\rfloor=a-4.$$
Notice that by Euclid's algorithm
there are integers $n,t \in \mathbb{Z}$;
such that $a=15n+t$, with $0 \leq t \leq 14$,
by replacing we get that:
$$
\lfloor\frac{60n+4t-10}{30}] + \lfloor\frac{60n+4t+5}{30}]=15n+t-4
\Longrightarrow
\\
2n+\lfloor\frac{4t-10}{30}] + 2n+\lfloor\frac{4t+5}{30}]=15n+t-4
\Longrightarrow
\\
\lfloor\frac{4t-10}{30}] + \lfloor\frac{4t+5}{30}] -t +4 =11n
\
;
\ \ \ \ \
\color{Blue}{\star}
$$
but notice that $0 \leq t \leq 14$; implies the following inequalities:
$$
-10 \leq 4t-10 \leq 46
\ \
%%\text{and}
;
\ \ \
5 \leq 4t+5 \leq 61
\ \
%%\text{and}
;
\ \ \
-10 \leq -t+4 \leq 4
\ \
\Longrightarrow
\\
\color{Red}{-1} \leq \lfloor\frac{4t-10}{30}\rfloor \leq \color{Red}{1}
\ \
%%\text{and}
;
\ \ \
\color{Red}{0} \leq \lfloor\frac{4t+5}{30}\rfloor \leq \color{Red}{2}
\ \
%%\text{and}
;
\ \ \
-10 \leq -t+4 \leq 4
\ \
\Longrightarrow
\\
-1+0-10
\leq
\lfloor\frac{4t-10}{30}\rfloor +
\lfloor\frac{4t+ 5}{30}\rfloor
-t +4
\leq
1 + 2 + 4
\ \
\color{Green}{\star}
;
\overset{\color{Blue}{\star}}{\Longrightarrow}
\\
-11
\leq
11n
\leq
7;
$$
so we have only two choices for $n$ ;
*
*$n=-1$, gives:
$$ -1+0-10 = -11 =
\lfloor\frac{4t-10}{30}\rfloor +
\lfloor\frac{4t+ 5}{30}\rfloor
-t +4
;
$$
which implies that:
$$
4t-10 < 0
\ \
;
\ \ \
4t +5 < 30
\ \
;
\ \ \
-t+4 = -10
;
$$
so there is no possibilities for $t$ in this case.
*$n=0$, gives:
$$ 0 =
\lfloor\frac{4t-10}{30}\rfloor +
\lfloor\frac{4t+ 5}{30}\rfloor
-t +4
;
$$
which implies that:
$$
-3=-(\color{Red}{1+2})
\leq
-t+4
\leq
-(\color{Red}{-1+0})=1
;
$$
so the only possibilities for $t$ is:
$t=3, 4, 5, 6, 7$.
Now you can check by-hand that only $\color{Green}{t=4}$ gives an answer.
|
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|
Find polynomial : $P(x)=x^3+ax^2+bx+c$ Find all polynomials in $\mathbb{Q}[x]$ is of the form $P(x)=x^3+ax^2+bx+c$ which has $a, b, c$ as its roots.
Is my answer below correct?
|
By Vieta's formulas,
$$a+b+c=-a,\\ab+bc+ca=b,\\abc=-c.$$
Then if $c\ne 0$,
$$ab=-1$$
$$c=-2a-b$$
$$-1-(a+b)(2a+b)=-2a^2+2-b^2=b$$ and
$$2a^2b^2=(2-b-b^2)b^2=1.$$
The last equation has no rational roots and the only remaining option is $c=0$, giving
$$2a+b=0,ab=b$$
with the two solutions $a=b=0$, and $a=1,b=-2$.
|
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If $(x^2-5x+4)(y^2+y+1)<2y$ for all real values of $y$ then what interval does $x$ belong to? Okay so I'm stuck on this one for a long time :
If $(x^2-5x+4)(y^2+y+1)<2y$ for all real values of $y$ then what interval does $x$ belong to?
My Try: Since $y$ can take any real value, I tried multiplying the first two terms and formed a quadratic equation in $y$ with its coefficients and the constant term in terms of $x$. Then I applied the two conditions as:
*
*Coefficient of $y^2$ should be negative ; and
*Discrimnant should also be negative.
Am I correct? What am I doing wrong? Is there a better approach to solve similar problems like this one? Please let me know.
|
First, notice that $y^2+y+1$ is positive for all real values of $y$.
This means that we can divide by $y^2+y+1$ on both sides of the inequality, without having to worry about negative values that would change the direction of the inequality.
Divide by $y^2+y+1$ on both sides:
$$x^2 -5x + 4 < \frac{2y}{y^2+y+1}$$
We can find the minima and maxima of the rational function of $y$ on the right hand side, by taking its derivative and setting it equal to zero:
$$\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{2y}{y^2+y+1}\right) = 0$$
$$\frac{2-2y^2}{\left(y^2+y+1\right)^2}=0$$
Since the denominator is always positive, we can simply eliminate it and divide by $2$:
$$1-y^2 = 0$$
$$y=\pm 1$$
This means that $2y/\left(y^2+y+1\right)$ achieves a minimum value of $-2$ at $y=-1$, and achieves a maximum value of $2/3$ at $y=1$. Therefore,
$$-2 \leq \frac{2y}{y^2+y+1}\leq \frac{2}{3}$$
At the boundary values for the right hand side, the inequality becomes:
$$x^2 - 5x + 4 < -2\,\, \implies\,\, x^2 - 5x + 6 < 0$$
$$x^2 - 5x + 4 < \frac{2}{3}\,\, \implies\,\, x^2 -5x + \frac{10}{3} <0 $$
Here, the first inequality is a tighter bound for $x$ than the second inequality, so we may only consider the first:
$$x^2 - 5x + 6 < 0$$
Factoring gives:
$$(x-2)(x-3)<0$$
This is a concave-up parabola with zeroes at $x=2$ and $x=3$. This means that the range of values for which the quadratic is negative is exactly between the two roots:
$$\boxed{2 < x < 3\,}$$
|
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|
Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$ or $a^3-b^3$ will be divisible by $7$. This is a very interesting word problem that I came across in an old textbook of mine:
Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$ or $a^3-b^3$ will be divisible by $7$.
So I know it's got something to do with Euler's totient function and the Euler-Fermat theorem, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance.
|
There's always brute force:
If $m \equiv 0 \mod 7$ then $m^3 \equiv 0 \mod 7$.
If $m \equiv \pm 1,\pm 2, \pm 3$ then $m^3 \equiv \pm 1, \pm 8, \pm 27 \equiv \pm 1, \pm 1, \mp 1 \mod 7$.
So either i) $a^3 \equiv 0 \mod 7$ and $7|a^3$, ii) $b^3 \equiv 0 \mod 7$, and $7|b^3$, iii) $a^3 \equiv b^3 \mod 7$ and $7|a^3 - b^3$, or iv) $a^3 \equiv -b^3 \mod 7$ and $7|a^3 + b^3$.
But it's probably more practical to note:
For $p$ an odd prime, either $a \equiv 0 \mod p$ and $p|a^{\frac {p-1}2}$. Or $b \equiv 0 \mod p$ and $p|b^{\frac {p-1}2}$ or neither $a$ not $b$ is congruent modulo $p$ so $a^{p-1}\equiv b^{p-1} \equiv 1 \mod p$ in which case $a^{p-1}- b^{p-1} = (a^{\frac {p-1}2} + b^{\frac {p-1}2})(a^{\frac {p-1}2} - b^{\frac {p-1}2}) \equiv 0 \mod p$.
So if neither $p|a^{\frac {p-1}2}$ or $p|b^{\frac {p-1}2}$ then $p|a^{\frac {p-1}2} + b^{\frac {p-1}2}$ or $p|a^{\frac {p-1}2} - b^{\frac {p-1}2}$
|
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|
Let $h(4x-1)=2x+7$. For what value of $x$ is $h(x)=x$? Let $h(4x-1)=2x+7$. For what value of $x$ is $h(x)=x$?
If $h(a)=a$, then $4x-1=2x+7$ which implies $x=4$. So $a=15$ when I substitute $x=4$ into both linear equations. Is the value of $x$ $15$?
|
First, we find an expression for $h(x)$. From our definition of $h$, we have $h(4y-1) = 2y+7$. So, if we let $x=4y-1$, so that $y = (x+1)/4$, we have[h(x) = 2\cdot\frac{x+1}{4} + 7 = \frac{x+1}{2} + 7.]Setting this equal to $x$ gives[x =\frac{x+1}{2} + 7.]Multiplying both sides by 2 gives $2x = x+1 + 14$, so $x = \boxed{15}$.
|
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Solve $\frac{1}{x}<\frac{x}{2}<\frac{2}{x}$. First I consider the case $\frac{1}{x}<\frac{x}{2}$, manipulating to $\frac{(x-\sqrt{2})(x+\sqrt{2})}{2x}<0\Leftrightarrow x < -\sqrt{2}.$ Did this by subtracting $\frac{x}{2}$ from both sides and writing everything with a common denominator.
Second case to consider is $\frac{x}{2}<\frac{2}{x}\Leftrightarrow\frac{(x-2)(x+2)}{2x}<0\Leftrightarrow x<-2.$
Since both of these inequalities have to be satisfied simultaneously, one can combine them to get $x<-2.$ Correct answer is $x\in(\sqrt{2},2)$
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$\dfrac1x<\dfrac2x$ is only possible for positive $x$.
Then we may multiply by $2x$ and take the square roots,
$$\sqrt2<x<2.$$
|
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If $a$ is a root of $x^2+ x + 1$, simplify $1 + a + a^2 +\dots+ a^{2017}.$
If $a$ is a root of $x^2 + x + 1$ simplify $$1 + a + a^2 + a^3 + \cdots + a^{2017}.$$
my solution initially starts with the idea that $1 + a + a^{2} = 0$ since $a$ is one of the root then using the idea i grouped $$1 + a + a^2 + a^3 + \cdots + a^{2017}$$ just like this $$(1 + a + a^{2}) + a^3(1 + a + a^2) + a^6(1 + a + a^2) + \cdots + a^{2013} + (1 + a + a^2) + a^{2016} + a^{2017}$$
which is equivalent to $$(0) + a^3(0) + a^6(0) + \cdots + a^{2013}(0) + a^{2016} + a^{2017}$$ and finally simplified into $$a^{2016} + a^{2017}$$. That's my first solution, but eventually I noticed that since $1 + a + a^2 = 0$ then it can be $a + a^2 = -1 $ using this idea I further simplified $$a^{2016} + a^{2017}$$ into $$a^{2015}(a + a^{2})$$ ==> $$a^{2015}(-1)$$ ==> $$-a^{2015}$$ But answers vary if I start the grouping at the end of the expression resulting to $1 + a$ another solution yields to $-a^{2017}$ if I set $1 + a = -a^2$ several solutions rise when the grouping is being made anywhere at the body of the expression... my question is, Is $$a^{2016} + a^{2017} = 1 + a = -a^{2015} = -a^{2017} \text{ etc.?}$$ How does it happen? I already forgot how to manipulate complex solutions maybe that's why I'm boggled with this question...any help?
|
I haven't looked through your working, but here's a much easier method:
If $a^2+a+1=0$, then $(a-1)(a^2+a+1)=0$, that is, $a^3-1=0$, ie, $a^3=1$.
On the other hand, $1+a+\dots+a^{2017}=\frac{a^{2018}-1}{a-1}$. Since $a^3=1$, it follows that $a^{2018}=a^2$. So, your sum simplifies to $\frac{a^2}{a-1}$. <-- wrong! Read on....
Edit:
I had a typo (thanks @Math lover for pointing that out). The sum simplifies to $\frac{a^2-1}{a-1}$, which simplifies easily to $a+1$.
|
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|
If $z$ is a complex number and $Re(z)>1$ then prove that $|\frac{1}{z}-\frac{1}{2}|<\frac{1}{2}$. If $z$ is a complex number and $Re(z)>1$ then prove that $|\frac{1}{z}-\frac{1}{2}|<\frac{1}{2}$.
I tried replacing $z=a+bi$ but it makes problem too long.
|
You can use the theory of Moebius transformations/inversive geometry to see the behavior of the map $z\mapsto\frac{1}{z}$. This will tell you that the line $x=1$ will map to a circle. However, we can do this more explicitly as follows:
The line $x=1$ has points $1+yi$ on it. Now, the transformation $z\mapsto\frac{1}{z}$ takes this line to
$$
1+yi\mapsto\frac{1}{1+y^2}-\frac{y}{1+y^2}i.
$$
The square of the distance between these points and $\frac{1}{2}$ is given by
$$
\left(\frac{1}{1+y^2}-\frac{1}{2}\right)^2+\left(\frac{y}{1+y^2}\right)^2=\frac{1}{(1+y^2)^2}-\frac{1}{1+y^2}+\frac{1}{4}+\frac{y^2}{(1+y^2)^2}=\frac{1}{4}.
$$
Therefore, the line $x=1$ becomes the circle centered at $\frac{1}{2}$ of radius $\frac{1}{2}$. Now, all you need to do is to figure out the inside/outside using a test point.
|
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|
Interesting way to evaluate $ \int \cos^3 x\ dx$ I have read these days a nice way for integrating $\cos^3x$:
First differentiate:
$$f=\cos^3(x)$$$$f'=-3\cos^2(x)\sin(x)$$$$f''=6\cos(x)\sin^2(x)-3\cos^3(x)$$$$f''=6\cos(x)(1-\cos^2(x))-3f$$$$f''=6\cos(x)-6\cos^3(x)-3f$$$$f''=6\cos(x)-9f$$
Then integrate:
$$f'= 6\sin(x) -9\int f(x)dx$$
Therefore,
$$\int f(x)dx=\frac 23 \sin(x) - \frac {f'} 9$$
Do you know any other way to calculate easily this integral?
|
Use that $\cos 3x = 4\cos^3 x - 3\cos x$ to get:
$$\int \cos^3 x \,dx = \frac{1}{4}\int\left(\cos 3x +3\cos x\right)\,dx$$
and the right side is easy to compute as $\frac{1}{12}\sin 3x +\frac{3}{4}\sin x$.
My original answer included a second approach, but I had the wrong formula for $dx$ in the substitution.
Use the tangent-half-angle substitution, $t=\tan(x/2)$ so $dx = \frac{2}{1+t^2}\,dt$ (corrected) and $\cos(x)=\frac{1-t^2}{1+t^2}$.
This reduces to:
$$\int \frac{2(1-t^2)^3}{(1+t^2)^4}\,dt$$
But this is a bit harder to integrate than the nice formula I had before. You might still solve this if you rewrite $1-t^2=2-(1+t^2)$ and expand, and then compute:
$$\int \frac{dt}{\left(1+t^2\right)^k}$$
for $k=1,2,3,4.$
|
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Prove that inequality $\sum _{cyc}\frac{a^2+b^2}{a+b}\le \frac{3\left(a^2+b^2+c^2\right)}{a+b+c}$
Let $a>0$,$b>0$ and $c>0$. Prove that:
$$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$
$$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)$$
$\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)$
$\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2$
$\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2$
$\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)$
I can't continue. Help me
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Now, by C-S $$\sum_{cyc}\frac{1}{a+b}\geq\frac{9}{\sum\limits_{cyc}(a+b)}=\frac{9}{2(a+b+c)}.$$
Thus, it remains to prove that
$$a^2+b^2+c^2+\frac{9abc}{a+b+c}\geq2(ab+ac+bc)$$ or
$$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$
which is Schur.
Done!
|
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For what value of $k$ is one root of the equation
For which value of $k$ is one root of the equation $x^2+3x-6=k(x-1)^2$ double the other?
My Attempt:
$$x^2+3x-6=k(x-1)^2$$
$$x^2+3x-6=k(x^2-2x+1)$$
$$x^2+3x-6=kx^2-2kx+k$$
$$(1-k)x^2+(3+2k)x-(6+k)=0$$
|
Now, $k\neq1$, $\Delta\geq0$ and since $x_1=2x_2$, we obtain:
$$3x_2=\frac{2k+3}{k-1}$$ and
$$2x_2^2=\frac{k+6}{k-1},$$
which gives
$$\frac{2(2k+3)^2}{9(k-1)^2}=\frac{k+6}{k-1}$$ or
$$k^2+21k-72=0,$$
which gives $k=3$ or $k=-24.$
Done!
|
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The flux crossing a unit ball I am doing sample test provided by my professor. One of the question is:
Find the flux integral of
$$ \overrightarrow{F}=\left(x^3, y^3, z^3\right) $$
across the surface $\Sigma:x^2+y^2+z^2=1$, and oriented by normal vectors pointing away from the origin.
I do it by this way:
Let $\overrightarrow{r}(x,y,z)=\overrightarrow{r}\big(x(\theta,\varphi),y(\theta,\varphi),z(\theta,\varphi)\big)$
\begin{cases}
x=\sin\theta \cos\varphi \\
y=\sin\theta\sin\varphi \\
z=\cos\theta\\
\theta\in[0,\pi], \varphi\in[0,2\pi]
\end{cases}
Thus:
\begin{cases}
r_\theta=(\cos\theta \cos\varphi,\cos\theta\sin\varphi, -\sin\theta) \\
r_\varphi=(-\sin\theta \sin\varphi,\sin\theta\cos\varphi, 0) \\
E= \|r_\theta\|^2=1, F=r_\theta \cdot r_\varphi=0, G=\|r_\varphi\|^2=\sin^2\theta\\
\sqrt{EG-F^2}=\sin\theta
\end{cases}
The unit normal vector at point $(x,y,z)$ is just $(x,y,z)$ since the radius of the surface is $1$.
Then:
\begin{align*}
\int_\Sigma \overrightarrow{F}\cdot \overrightarrow{n}\ \mathrm{d}\sigma&=
\int_\Sigma \left(x^4+y^4+z^4\right)\ \mathrm{d}\sigma\\
&=8\int_{ \left\{\Sigma:\ x>0,y>0,z>0 \right\} } \left[\sin^4\theta(\cos^4\varphi+\sin^4\varphi)+\cos^4\theta\right]\ \mathrm{d}\sigma\\
&=8\iint_{[0,\frac{\pi}{2}]^2}\left[\sin^4\theta(\cos^4\varphi+\sin^4\varphi)+\cos^4\theta\right]\sqrt{EG-F^2}\ \mathrm{d}\theta\mathrm{d}\varphi\\
&=8\left(\int_{0}^{\frac{\pi}{2}}\sin^5{\theta}\ \mathrm{d}\theta\int_{0}^{\frac{\pi}{2}}(\cos^4\varphi+\sin^4\varphi)\ \mathrm{d}\varphi+\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\sin\theta\cos^4\theta\ \mathrm{d}\theta \right)\\
&=8\left( \frac{\sqrt{\pi}}{2} \frac{\Gamma\left(3\right)}{\Gamma\left(\frac{7}{2}\right)}\cdot 2\cdot\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{5}{2}\right)}{\Gamma\left(3\right)}+\frac{\pi}{2}\cdot \frac{1}{2}B\left(1,\frac{5}{2}\right) \right)\\
&=8\left( \frac{\pi}{2} \cdot \frac{1}{\frac{5}{2}}+\frac{\pi}{4}\cdot\frac{1}{\frac{5}{2} } \right)\\
&=8\left(\frac{\pi}{5} + \frac{\pi}{10} \right)\\&=\frac{12\pi}{5}
\end{align*}
I spent more than 15 minutes on this problem so I'm thinking of whether there is an easier way to do this problem so that I can save more time. Is there a simpler way? Thank you for your help!
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HINT: Divergence (Gauss') Theorem would come pretty handy and reduce the amount of calculations.
Namely using it we have:
$$\iint_\Sigma \vec{F} \cdot \vec{n} \ d\sigma = \iiint_V \nabla \cdot F \ dV = 3\iiint_V x^2 + y^2 + z^2 \ dV$$
Now using the spherical coordinates we have:
$$\iint_\Sigma \vec{F} \cdot \vec{n} \ d\sigma = 3\int_0^{2\pi}\int_0^{\pi}\int_0^{1} r^4\sin \phi \ drd\phi d\theta = \frac 35 \int_0^{2\pi}\int_0^{\pi} \sin \phi \ d\phi d\theta = \frac 65 \int_0^{2\pi} \ d\theta = \frac{12}{5}\pi$$
|
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A.P.: Sum of numbers in a particular group Question: A series of odd positive integers are divided into the following groups $(1), (3,5), (7,9,11),\dotsm$ . Prove that the sum of numbers in the $n^{\text{th}}$ group is $n^3$.
My attempt:
It is observed that the number of terms in the $n^{\text{th}}$ group is equal to $n$. The common difference is same in all groups and is equal to 2. If we do not consider the grouping, then the resulting sequence $1,3,5,\dotsm$ is in A.P. with first term $t_1$ equal to 1 and common difference equal to 2. Now the first term of the first group is equal to $t_1=t_{1+0}$, that of the second group is equal to $t_2=t_{1+1}$, that of the third group is equal to $t_4=t_{1+1+2}$, that of the fourth group is equal to $t_7=t_{1+1+2+3}$ and so on. Thus the first term of the $n^{\text{th}}$ group is equal to $t_{1+S_{n-1}}=t_{1+\frac{(n-1)n}{2}}=t_{\frac{n^2-n+2}{2}}=t_1+(\frac{n^2-n+2}{2}-1)2=t_1+\frac{n^2-n}{2}\times2=t_1+n^2-n=1+n^2-n$
Here $S_{n-1}$ denotes the sum of first $(n-1)$ natural numbers.
Therefore the first term of the $n^{\text{th}}$ group is equal to $n^2-n+1$.
Sum of all numbers in the $n^{\text{th}}$ group $=S^{'}_n=\frac{n}{2}[2(n^2-n+1)+(n-1)2]=\frac{n}{2}[2n^2-2n+2+2n-2]=n^3$.
My problem: I am looking for other methods to prove this result.
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To form $n$ groups you need $1+2+3+\cdots+n=\frac{n(n+1)}{2}$ terms. To form $n-1$ groups you need $1+2+3+\cdots+n-1=\frac{n(n-1)}{2}$ terms. To get the sum of the terms in the $n$-th group you sum all the terms from the $1$st group to the $n$th and subtract the sum from the $1$st group to the $n-1$th which is
$$S_{n(n+1)/2}-S_{n(n-1)/2}=\frac{n(n+1)}{4}\left(2+2\left(\frac{n(n+1)}{2}-1\right)\right)-\frac{n(n-1)}{4}\left(2+2\left(\frac{n(n-1)}{2}-1\right)\right)=\frac{n(n+1)}{2}(\frac{n(n+1)}{2})-\frac{n(n-1)}{2}(\frac{n(n-1)}{2})=\frac{n^2}{4}((n+1)^2-(n-1)^2)=n^3$$
|
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Factorization of $18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc$ I solved a following problem in my mathematics homework today:
Factorize $18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc$.
That's easy. I solved by the following way and it was same as sample answer of the book:
$\ \ \ \ 18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc$
$=(-12b+18c)a^2 + (18b^2 - 19bc - 12c^2) + (18bc^2 - 12b^2c)$
$=-6a^2 (2b-3c) + a(2b-3c)(9b+4c) - 6bc(2b-3c)$
$=(2b-3c)(-6a^2 + 9ab + 4ac - 6bc)$
$=(2b-3c)(2a-3b)(2c-3a)$.
(if there is a typo, please inform this)
Though the process took much times (highly implementing one), and I think it's too straight forward way.
The question and answer looks simple, so I think there must be more elegant and way, which has less calculation time.
So, I want to ask the simplier way to solve this.
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Without loss of generality, assume $a, b=ax, c=axy$. Then:
$$18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc=$$
$$18a^3x(x + x^2y^2 + y) - 12a^3x(1 + x^2y + xy^2) - 19a^3x^2y=$$
$$a^3x[18x+18x^2y^2+18y-12-12x^2y-12xy^2-19xy]=$$
$$a^3x[6x(3-2xy)+6y(3-2xy)-(3-2xy)(4+9xy)]=$$
$$a^3x(3-2xy)(6x+6y-4-9xy)=$$
$$a^3x(3-2xy)(2-3y)(3x-2)=$$
$$(3a-2axy)(2ax-3axy)(3ax-2a)=$$
$$(3a-2c)(2b-3c)(3b-2a).$$
|
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|
Induction: Bounds on sum of inverses of first $n$ square roots I am trying to show by induction that $2(\sqrt{n}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$.
For the upper bound, I have that if $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}<2\sqrt{n}$, then $1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}<2\sqrt{n}+\dfrac{1}{\sqrt{n+1}}=$
$2\sqrt{n}+\dfrac{2}{2\sqrt{n+1}}<2\sqrt{n}+\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\sqrt{n}+2(\sqrt{n+1}-\sqrt{n})=2\sqrt{n+1}$.
However, despite hours of trying, I cannot figure out how to show that $2(\sqrt{n+1}-1)<1+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}$.
|
Note: I can see that there are already fine answers using induction!
An ad in favor of graph paper. The outcome of the standard sum/integral comparison is
$$ 2 \sqrt {n+1} - 2 \; < \; 1 + \frac{1}{\sqrt 2 } + \frac{1}{\sqrt 3 } + \cdots + \frac{1}{\sqrt n } \; < \; 2 \sqrt n - 1 $$
|
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|
Find $\int \frac{\sqrt{2-3x}\,dx}{x^2+1}$ Find $$\int \frac{\sqrt{2-3x}\,dx}{x^2+1}$$
I used substitution $x=\frac{2}{3} \sin^2 y$ So we get $dx=\frac{4}{3} \sin y \cos y dy$
hence
$$I= \frac{36\sqrt{2}}{3} \int \frac{\sin y \cos^2 y\, dy}{4 \sin^4 y+9}$$
Now if again i use substitution $\cos y=t$ we get
$$I=-12\sqrt{2} \int \frac{t^2 \,dt}{4t^4-8t^2+13}$$
$$I=-12 \sqrt{2} \int \frac{dt}{4t^2+\frac{13}{t^2}-8}$$
Any other approach?
|
Let $u=\sqrt{2-3x}$, then $1+x^2 = \frac{1}{9}\left((2-u^2)^2+9\right)$ and $x=\frac{1}{3}\left(2-u^2\right)$ so $dx=-\frac{2}{3}u\,du$ so you want:
$$-6\int\frac{u^2\,du}{(2-u^2)^2+9}$$
Now attempting partial fractions, we get a factoring:
$$\begin{align}(2-u^2)^2+9 &= u^4-4u^2+13\\&=\left(u^2+\left(\sqrt{4+2\sqrt{13}}\right)u+\sqrt{13}\right)\left(u^2-\left(\sqrt{4+2\sqrt{13}}\right)u+\sqrt{13}\right)
\end{align}$$
That's gonna involve some very ugly formula to get a final result. I'd be surprised if there is a clean formula.
If $\alpha = \sqrt{1+\sqrt{\frac{13}{4}}}$ and $\beta = \sqrt{13}$, then you need to solve:
$$f(u)=\frac{u^2}{u^4-4u^2+13}=\frac{au+b}{u^2+2\alpha u +\beta}+\frac{cu+d}{u^2-2\alpha u +\beta}$$
Since $f(u)=f(-u)$, you have that $-c=a,b=d$. Since $b+d=0$, you get $b,d=0$.
You get $$f(u)=\frac{au}{u^2+2\alpha u+\beta}+\frac{-au}{u^2-2\alpha u+\beta}$$
And you get $-4a\alpha =1$ of $a=-\frac{1}{4\alpha}$.
You are going to get some horrible mix of $\arctan, \log$ and lots of square roots of $13$.
Finally, $$\begin{align}\int\frac{u\,du}{u^2+2\alpha u+\beta} &= \frac{1}{2}\int \frac{(2u+2\alpha)\,du}{u^2+2\alpha u+\beta}-\alpha\int\frac{du}{u^2+2\alpha u+\beta}
\\&=\frac{1}{2}\log(u^2+2\alpha u+\beta)+\frac{\sqrt{\beta-\alpha^2}}\arctan\left(\frac{u+\alpha}{\sqrt{\beta-\alpha^2}}\right)
\end{align}$$
And likewise:
$$\int\frac{u\,du}{u^2-2\alpha u+\beta}=\frac{1}{2}\log(u^2-2\alpha u+\beta)+\frac 1{\sqrt{\beta-\alpha^2}}\arctan\left(\frac{u-\alpha}{\sqrt{\beta-\alpha^2}}\right)$$
Finally, you get, with $\gamma = \sqrt{\beta-\alpha^2}=\sqrt{-1+\sqrt{\frac{13}4}}$:
$$\frac{3}{2\alpha}\left(\frac{1}{2}\log\frac{u^2+2\alpha u+\beta}{u^2-2\alpha u+\beta}+\frac1\gamma\left(\arctan\left(\frac{u+\alpha}{\gamma}\right)-\arctan\left(\frac{u-\alpha}{\gamma}\right)\right)\right)$$
As it turns out, $\alpha\gamma = \frac{3}{2}$, so you get some simplifications. You can also use that $\arctan x-\arctan y = \arctan\frac{x-y}{1+xy}.$
Then replace $u=\sqrt{3-2x}$.
|
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|
Euler squared sum of order 2 : $\sum \limits_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2}$ Let $\mathcal{H}_n$ denote the $n$ - th harmonic number. What techniques would one use to prove that
$$\sum \limits_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2} = \zeta^2(3) + \frac{19 \zeta(6)}{24}$$
where $\zeta$ denotes the Riemann zeta function.
|
Since
$$\left(\mathcal{H}_n^{(2)}\right)^2=\mathcal{H}_n^{(4)}+2\sum_{k=1}^{n}\frac{\mathcal{H}_{k-1}^{(2)}}{k^2}$$
we can rewrite the sum in terms of Multiple zeta functions:
\begin{align*}\sum_{n=1}^{\infty} \frac{\left( \mathcal{H}_n^{(2)} \right)^2}{n^2}
&=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n^{(4)}}{n^2}+
2\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=1}^{n}\frac{\mathcal{H}_{k-1}^{(2)}}{k^2}\\
&=\left(\zeta(6)+\zeta(2,4)\right)+2(\zeta(2,2,2)+\zeta(4,2))\\
&=\zeta(6)+\left(\frac{25}{12}\zeta(6)-\zeta(3)^2\right)+2\left(\frac{\zeta(2)^3}{6}+\frac{\zeta(6)}{3}-\frac{\zeta(2)\zeta(4)}{2}\right)\\
&\qquad+2\left(\zeta(3)^2-\frac{4\zeta(6)}{3}\right)\\
&=\zeta(3)^2+\frac{19\pi^6}{22680}=\zeta(3)^2 + \frac{19 \zeta(6)}{24}
\end{align*}
where we also used the explicit values of
$$\zeta(2)=\frac{\pi^2}{6},\quad \zeta(4)=\frac{\pi^4}{90}, \quad \zeta(6)=\frac{\pi^6}{945}.$$
|
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|
Find maximum of $f(x)=3\cos{2x}+4\sin{2x}.$ Find maximum of $f(x)=3\cos{2x}+4\sin{2x}.$ The shortest way to do this is to only consider $2x\in[0,2\pi)$, set $t=2x$ and note that
*
*Max$(\cos{t})=1$ for $t=0.$
*Max$(\sin{t})=1$ for $t=\frac{\pi}{2}.$
Max of these two functions added is when $t$ equals the angle exactly in the middle of $[0,\pi/2]$, which is $t=\pi/4.$ For this $t$ we have that $\cos{\pi/4}=\sin{\pi/4}=\sqrt{2}/2.$ So; $$f\left(\frac{\pi}{4}\right)=3\frac{\sqrt{2}}{2}+4\frac{\sqrt{2}}{2}=\frac{7\sqrt{2}}{2}.$$
Correct answer: $5$. My answer is slightly less than $5$. Why?
|
We can express $3\cos 2x + 4\sin 2x$ in the form $R\cos(2x-\alpha)$, where $R>0$ and $\alpha$ is acute.
Use the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$ to find the values of $R$ and $\alpha$.
\begin{eqnarray*}
R\cos(2x-\alpha) &=& R(\cos 2x\cos\alpha + \sin 2x\sin \alpha) \\ \\
&=& (R\cos\alpha)\cos 2x + (R\sin\alpha)\sin 2x
\end{eqnarray*}
We need $R\cos\alpha = 3$ and $R \sin \alpha = 4$, i.e. $\cos\alpha = \frac{3}{R}$ and $\sin\alpha = \frac{4}{R}$. So we can draw a right-angled triangle with angle $\alpha$, hypotenuse $R$, opposite $4$ and adjacent $3$.
Pythagoras tells us that $R = \sqrt{3^2+4^2} = 5$ and $\alpha = \arctan\frac{4}{3}$. Meaning that
$$3\cos 2x + 4\sin 2x \ \ \equiv \ \ 5\cos\left(2x-\arctan\tfrac{4}{3}\right)$$
Hence $-5$ and $+5$ are the minimum and maximum respectively.
|
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|
How can one find $\lim_{x \to \pm\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$ without graphing? I'm trying to find both of these:
$$\lim_{x \to +\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$$
$$\lim_{x \to -\infty}\frac{2x+1}{\sqrt{x^2+x+1}}$$
I know that they end up being $2$ and $-2$ by graphing, but what process can I use to find them without graphing (in the simplest way possible)?
|
First recall that $\sqrt{x^2} = x$ if $x\ge0$ and $\sqrt{x^2} = -x$ (which is positive) if $x<0.$
Then treat the two cases, $x>0$ and $x<0,$ separately:
\begin{align}
\frac{2x+1}{\sqrt{x^2+x+1}} & = \frac{2 + \frac 1 x}{\frac 1 x \sqrt{x^2 + x + 1}} \\[10pt]
& = \frac{2+\frac 1 x}{\frac 1 {\sqrt{x^2}}\sqrt{x^2 + x+1}} \quad \text{if } x>0 \\[10pt]
& = \frac{2+ \frac 1 x}{\sqrt{1 + \frac 1 x + \frac 1 {x^2}}} \\[10pt]
& \to 2 \text{ as } x\to\infty. \\[10pt]
\frac{2x+1}{\sqrt{x^2+x+1}} & = \frac{2 + \frac 1 x}{\frac 1 x \sqrt{x^2 + x + 1}} \\[10pt]
& = \frac{2+\frac 1 x}{\frac {-1} {\sqrt{x^2}}\sqrt{x^2 + x+1}} \quad \text{if } x<0 \\[10pt]
& = \frac{2+ \frac 1 x}{-\sqrt{1 + \frac 1 x + \frac 1 {x^2}}} \\[10pt]
& \to -2 \text{ as } x\to\infty.
\end{align}
|
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|
Another way of expanding Taylor series
I have to show that if $$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\ldots, $$ then prove that $$ f(x)=f(a)+2\left[ \frac{(x-a)}{2}f'\left(\frac{x+a}{2}\right)+ \frac{(x-a)^3}{3!\times 8}f'''\left(\frac{x+a}{2}\right)+ \frac{(x-a)^5}{32\times 5!}f^{(v)}\left(\frac{x+a}{2}\right)+\ldots\right] $$
I am unable to proceed. Any hint will be appriciated.
|
We write the Taylor expansion stated in OPs first line as
\begin{align*}
f(x)=\sum_{j=0}^\infty\frac{(x-a)^j}{j!}f^{(j)}(a)\tag{1}
\end{align*}
and the other expansion as
\begin{align*}
f(x)=f(a)+2\sum_{j=0}^\infty \frac{1}{(2j+1)!}\left(\frac{x-a}{2}\right)^{2j+1}f^{(2j+1)}\left(\frac{x+a}{2}\right)\tag{2}
\end{align*}
Expansion at $\frac{x+a}{2}$ instead of $a$ in (1) results in
\begin{align*}
f(x)&=\sum_{j=0}^\infty\frac{1}{j!}\left(x-\frac{x+a}{2}\right)^j f^{(j)}\left(\frac{x+a}{2}\right)\\
&=\sum_{j=0}^\infty\frac{1}{j!}\left(\frac{x-a}{2}\right)^j f^{(j)}\left(\frac{x+a}{2}\right)\tag{3}
\end{align*}
Exchanging the role of $x$ and $a$ in (3) gives
\begin{align*}
f(a)&=\sum_{j=0}^\infty\frac{1}{j!}\left(\frac{a-x}{2}\right)^j f^{(j)}\left(\frac{x+a}{2}\right)\\
&=\sum_{j=0}^\infty\frac{(-1)^j}{j!}\left(\frac{x-a}{2}\right)^j f^{(j)}\left(\frac{x+a}{2}\right)\tag{4}
\end{align*}
subtracting (4) from (3) gives
\begin{align*}
\color{blue}{f(x)-f(a)}&=\sum_{j=0}^\infty\frac{1-(-1)^j}{j!}\left(\frac{x-a}{2}\right)^j f^{(j)}\left(\frac{x+a}{2}\right)\\
&\color{blue}{=2\sum_{j=0}^\infty \frac{1}{(2j+1)!}\left(\frac{x-a}{2}\right)^{2j+1}f^{(2j+1)}\left(\frac{x+a}{2}\right)}
\end{align*}
and the claim follows.
|
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|
Finding $\int\frac{x}{\sqrt{3-2x-x^2}} dx$. I was looking for the integral of
$$\int\frac{x}{\sqrt{3-2x-x^2}} dx$$
My work:
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(3)+(-2x-x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(2x+x^2)}} dx $$
$$ = \int \frac{x}{\sqrt{(3)-(1+2x+x^2) +1}} dx $$
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x}{\sqrt{(4)-(x+1)^2}} dx $$
Then I often remember this integral $\frac{u}{\sqrt{a^2 - x^2}} du$. So I modified the above integral to
look like the integral $\frac{u}{\sqrt{a^2 - x^2}}$.
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x+1-1}{\sqrt{(4)-(x+1)^2}} dx $$
$$\int \frac{x}{\sqrt{3-2x-x^2}} dx = \int \frac{x+1}{\sqrt{(4)-(x+1)^2}} dx + \int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$$
I recognized the the last integral $\int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$ has the form $\int \frac{1}{\sqrt{a^2-u^2}} du$, where $a =2$ and $u = x+1$.
It's corresponding integral would be $\arcsin \left( \frac{u}{a}\right) + c$.
Evaluating $\int \frac{-1}{\sqrt{(4)-(x+1)^2}} dx$, it would be $-\arcsin \left( \frac{x+1}{2}\right)$
Here's the problem: I couldn't find the integral of $\int \frac{x+1}{\sqrt{(4)-(x+1)^2}} dx, $ because my Table of Integral doesn't show what is the
integral of $\frac{u}{\sqrt{a^2 - x^2}} du$.
How to evaluate the integral of $\frac{x}{\sqrt{3-2x-x^2}} dx$ properly?
|
$$\int \frac{x+1}{\sqrt{3-2x-x^2}}dx=-\frac{1}{2}\int\frac{-2-2x}{\sqrt{3-2x-x^2}}dx=-\frac{1}{2}\int\frac{g'(x)}{\sqrt{g(x)}}dx$$
$$=-\frac{1}{2}\int g'(x)[g(x)]^{-1/2}dx=-(3-2x-x^2)^{1/2}=-\sqrt{3-2x-x^2}+C$$
This is the chain rule:
$$\int g'(x)f(g(x))dx=F(g(x))+C$$
where $F'(x)=f(x)$. In this case $f=x^{-1/2}$ and $g=3-2x-x^2$.
|
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|
Prove that for any integer $n, n^2+4$ is not divisible by $7$. The question tells you to use the Division Theorem, here is my attempt:
Every integer can be expressed in the form $7q+r$ where $r$ is one of $0,1,2,3,4,5$ or $6$ and $q$ is an integer $\geq0$.
$n=7q+r$
$n^2=(7q+r)^2=49q^2+14rq+r^2$
$n^2=7(7q^2+2rq)+r^2$
$n^2+4=7(7q^2+2rq)+r^2+4$
$7(7q^2+2rq)$ is either divisible by $7$, or it is $0$ (when $q=0$), so it is $r^2+4$ we are concerned with.
Assume that $r^2+4$ is divisible by 7. Then $r^2+4=7k$ for some integer $k$.
This is the original problem we were faced with, except whereas $n$ could be any integer, $r$ is constrained to be one of $0,1,2,3,4,5$ or $6$.
Through trial and error, we see that no valid value of $r$ satisfies $r^2+4=7k$ so we have proved our theorem by contradiction.
I'm pretty sure this is either wrong somewhere or at the very least not the proof that the question intended. Any help would be appreciated.
|
since we have $$n\equiv 0,1,2,3,4,5,6\mod 7$$ we get $$n^2\equiv 0,1,2,4\mod 7$$
therefore $$n^2+4\equiv 1,4,5,6\mod 7$$
|
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|
for $abc = 1$ and $a \le b \le c$ prove that $(a+1)(c+1)>3$ This inequality has been given to me by my teacher to keep me occupied and after hours of fumbling around with it, and later trying to google it. I found nothing at all.
For any three positive real numbers $a$, $b$ and $c$, where $abc = 1$ and $a\le b \le c$, prove that:
$$(a+1)(c+1)>3.$$
|
First simplify the expression :
$ac + a + c + 1 \gt 3 \iff ac + a+c \gt 2 \iff 1/b + a+ c \gt 2 \iff 1 + ab+bc \gt 2b $
We should prove $1+ab+bc \gt 2b$ . Use the condition $a\le b\le c $ :
$a\le b\le c \iff 2a \le a+b \le a+c \to a+b\le a+c$
From these we have :
$a+b\le a+c \iff ab + b^2 \le ab + bc \iff ab + b^2 +1 \le ab + bc+ 1$
If we prove $ab + b^2 +1 \gt 2b $ then problem is solved :
$ab + b^2 +1 \gt 2b \iff ab + (b-1)^2 \gt 0$
which is obvious because $ab \gt 0 $ and $(b-1)^2 \gt 0$
Done !
|
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|
How to split $\frac{x^3}{(x^2+5)^2}$ into partial fractions?
Split $\frac{x^3}{(x^2+5)^2}$ into partial fractions.
I have tried letting $$\frac{x^3}{(x^2+5)^2}=\frac{ax+b}{x^2+5}+\frac{cx^3+dx^2+ex+f}{(x^2+5)^2},$$ but this yields a system of only four simultaneous equations.
I have also tried looking at $$\frac{x^3}{(x^2+5)^2} = x \cdot \left( \frac{x}{x^2+5} \right) ^2,$$ but sadly in vain.
|
Note that, by adding and subtracting $5x$ at the numerator, we obtain
$$\frac{x^3}{(x^2+5)^2}=\frac{x^3+5x-5x}{(x^2+5)^2}=\frac{x(x^2+5)-5x}{(x^2+5)^2}=\frac{x}{x^2+5}-\frac{5x}{(x^2+5)^2}.$$
|
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|
Prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ The title is quite clear. I am required to prove $\lim_{x\to 1} \frac{x+2}{x^2+1}=\frac{3}{2}$ using the epsilon-delta definition of the limit.
Given any $\varepsilon \gt 0$, there exists a $\delta =$
Such that $0 \lt \lvert x-1 \rvert \lt \delta \Rightarrow \lvert\frac{x+2}{x^2+1} -\frac{3}{2} \rvert \lt\varepsilon$
$\frac{x+2}{x^2+1} -\frac{3}{2}= \frac{-3x^2+2x+1}{2(x^2+1)}$
I have managed to express both the numerator and the denominator as such
$-3x^2+2x+1 = -(x-1)(3x+1)\\
2(x^2+1) = 2x^2+2= 2(x-1)^2 + 4(x-1) +4$
Returning back to the fraction $\frac{-(x-1)(3x+1)}{2(x-1)^2 + 4(x-1) +4}$
I am unsure on how to continue and would really appreciate some guidance.
|
So you simplified the expression
\begin{align}
\left | \frac{x+2}{x^2+1} -\frac{3}{2} \right | = \left | \frac{(x-1)(3x+1)}{2x^2+2} \right |.
\end{align}
The good news is we have an $|x-1|$ appearing, but there is also a $|3x+1|$ and $|2x^2+2|,$ which are expressions we don't necessarily want. The thing to do here is to bound these unwanted expressions by numbers, which I demonstrate in this problem.
Suppose $\delta < 1.$ What happens if $|x-1|< \delta$ ?
First, we see $0 < x < 2.$ Multiplying all sides of the inequality by $3$ and then adding $1$ to all sides, we see $1<3x+1<7,$ so $|3x+1|<7.$ Second, we see upon squaring all sides of $0<x<2$ and then adding $2$, $2<2x^2+2<10,$ so $\frac{1}{|2x^2+2|} < \frac{1}{2}.$
Now, for any $\epsilon>0,$ let us set $\delta= \min \lbrace 1, \frac{2 \epsilon}{7} \rbrace.$ Why does this work ?
If $\delta=1,$ then we see $|x-1| < \delta$ implies $|x-1|<\frac{2 \epsilon}{7}$ and also
\begin{align}
\left|f(x)-\frac{3}{2} \right| = \left | \frac{x+2}{x^2+1} -\frac{3}{2} \right | = \left | \frac{(x-1)(3x+1)}{2x^2+2} \right |< \frac{7|x-1|}{2}< \frac{7}{2} \frac{2 \epsilon}{7}= \epsilon
\end{align}
If $\delta=\frac{2 \epsilon}{7}$ then already we have $|x-1|< \delta$ implies
\begin{align}
\left|f(x)-\frac{3}{2} \right|=\left | \frac{x+2}{x^2+1} -\frac{3}{2} \right | = \left | \frac{(x-1)(3x+1)}{2x^2+2} \right |< \frac{7|x-1|}{2}< \frac{7}{2} \frac{2 \epsilon}{7}= \epsilon.
\end{align}
|
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|
Prove the sequence $\frac{n^2}{n^2-n-5}$ converges I know the basic procedure for proving convergence of sequences, but I'm having difficulty reducing this sequence. I believe the sequence converges to 1, so I've set up the following two approaches:
1:
$\left|\frac{n^2}{n^2 - n - 5} - 1\right| = \left|\frac{n^2}{n^2 - n - 5} -
\frac{n^2 - n - 5}{n^2 - n - 5}\right| = \frac{n + 5}{n^2 - n - 5}$
2:
$\left|\frac{n^2}{n^2-n-5} - 1\right| = \left|\frac{1}{1-\frac{1}{n}-\frac{5}{n^2}} - \frac{1-\frac{1}{n}-\frac{5}{n^2}}{1-\frac{1}{n}-\frac{5}{n^2}}\right| = \frac{\frac{1}{n}+\frac{5}{n^2}}{1-\frac{1}{n}-\frac{5}{n^2}}$
Both assume $n \geq 3$. Neither of these approaches appear to be leading me in the right direction. I can technically solve these for $f(n) < f(\epsilon)$, but both result in something very nasty, and I feel like I'm missing something simple.
|
For $n\to \infty$ we have $\frac{n^2}{n^2-n-5}=\frac{1}{1-\frac{1}{n}-\frac{5}{n^2}}$ (dividing numerator and denominator by $n^2$). All terms on the denominator go to 0 except 1 so the limit is 1.
|
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|
If $a,b$ are positive integers and $x^2+y^2\leq 1$ then find the maximum of $ax+by$ without differentiation. If $x^2+y^2\leq 1$ then maximum of $ax+by$
Here what I have done so far.
Let $ax+by=k$ . Thus $by=k-ax$.
So we can have that $$b^2x^2+(k-ax)^2 \leq b^2$$
$$b^2x^2+k^2-2akx +a^2x^2-b^2\leq 0 $$
By re-writing as a quadratic of $x$ ,
$$(a^2+b^2)x^2-2akx +k^2-b^2\leq 0 $$
Since $a^2+b^2$ is positive , above quadratic has a minimum. Thus it to be negeative it must have roots.
So $$(-2ak)^2-4(a^2+b^2)(k^2-b^2) \geq 0$$
$$a^2k^2-(a^2+b^2)k^2+(a^2+b^2)b^2 \geq 0$$
$$(a^2+b^2) \geq k^2$$
So maximum of $k$ is $\sqrt{a^2+b^2}$
Is this correct ?
If it is correct any shorter method ? Thanks in advance.
|
Let $x=r \cos \theta$ and $y=r \sin \theta$ with $r \leq 1$ so that $x^2+y^2 \leq 1$. Then,
$$ax+by=r(a\cos \theta+b\sin \theta)$$
But one may show that we may write,
$$a \cos \theta+b \sin \theta=\sqrt{a^2+b^2} \cos (\theta+\phi)$$
For some $\phi$.
Hence $ax+by \leq (1)(\sqrt{a^2+b^2})$.
|
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|
How to integrate $\int \frac {e^y}{y} dy$? The question is to evaluate $$\iint_R \frac {x}{y} e^y dx dy$$ where R is the region bounded by $0 \leq x \leq 1$ and $x^2 \leq y \leq x$.
So i write it as $$\int_0^1 \int_{x^2}^{x} \frac{x}{y} e^y dy dx$$.
The thing is, how do i evaluate $I=\int_{x^2}^{x} \frac{1}{y} e^y dy$? I tried integration by parts but failed. Then i tried to use infinite series.
Use $$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$
we get \begin{align}
I & =\int_{x^2}^{x} \frac{1}{y}+1+\frac{y}{2!}+\frac{y^2}{3!}+\cdots dy \\
&=ln(\frac{1}{x})+(x-x^2)+\frac{1}{4}(x^2-x^4)+\frac{1}{18}(x^3-x^6)+ \cdots
\end{align}
Now get back to $\int_0^1 I dx$. We need to find $P=\int_0^1 ln(\frac{1}{x})$ and $Q=\int_0^1 (x-x^2)+\frac{1}{4}(x^2-x^4)+\frac{1}{18}(x^3-x^6)+ \cdots$
integrate by parts, we find $P=0$.
But for Q,we have $\left((\frac{1}{2}x^2-\frac{1}{3}x^3)+\frac{1}{4}(\frac{1}{3}x^3-\frac{1}{5}x^5)+ \cdots \right)_0^1$
Can we simplify the infinite sum in the bracket as a simple result?
|
$$\int_0^1 x \left(\int_{x^2}^{x} \frac{1}{y} e^y dy \right) dx$$
You can integrate by parts. Let $u(x)=\int_{x^2}^{x} \frac{1}{y} e^y dy$ and $dv=xdx$. Then we have,
$$du=\left(\frac{e^x}{x}-(2x)\frac{e^{x^2}}{x^2}\right) dx$$
$$=\left(\frac{e^x}{x}-2\frac{e^{x^2}}{x} \right) dx$$
We may let $v=\frac{x^2}{2}$. Then because $\int u dv=uv-\int v du$ we get that the value we want is,
$$\int_{0}^{1} \left(xe^{x^2}-\frac{1}{2}xe^{x} \right) dx$$
This is a simple exercise in further integration by parts and substitution. One may show this leads to,
$$=\frac{1}{2}(e-2)$$
|
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|
If $B$ is added to $A$, under what condition does the resultant vector have a magnitude equal to $A+B$? If $B$ is added to $A$, under what condition does the resultant vector have a magnitude equal to $A+B$? Under what conditions is the resultant vector equal to zero?
My Attempt:
Let $\theta $ be the angle between $\vec {A}$ and $\vec {B}$. Then,
$$R=A+B$$
$$\sqrt {A^2+B^2+2A.B\cos \theta}=A+B$$
$$A^2+B^2+2A.B\cos \theta=A^2+B^2+2A.B$$
$$\cos \theta =1$$
$$\cos \theta = \cos 0$$
$$\theta =0°$$
The resultant have a magnitude $A+B$ when the angle between the vectors is $0°$. How do I solve the second part of the question.?
|
Though the obvious fact that $\vec B=-\vec A$, we can infer this from your previous expression:
$$A^2+B^2+2AB\cos\theta=0$$
If $A=0$, then trivially $B^2=0 \to B=0$, and viceversa, and the angle can be any value.
If $A,B\gt0$, this leads to:
$$
\cos\theta=-{A^2+B^2 \over 2AB}
$$
Knowing that $\cos(\cdot)$ must be bounded in [-1,1]:
$$
-1\le {A^2+B^2 \over 2AB}\le1\\
-2AB\le A^2+B^2\le2AB\\
-4AB\le (A-B)^2\le0\\
$$
which can only be true for $A=B$, and the angle in this case is $
\cos\theta=-{2\over2}=-1 \to \theta=\pi$
|
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|
A floor function equation I want to solve below equation analytically $$\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor=1$$ but I have no idea to start .
Implicit: I solve it by graphing and the solution was $x\in[\frac {1}{4},1)$
I am thankful for any idea in advance .
I try to use $x=n+p \\0\leq p<1 $ but can't go further $$\lfloor \frac{2n+2p+1}{3}\rfloor +\lfloor \frac{\lfloor4n+4p\rfloor+2}{3}\rfloor=1\\0\leq 2p<2 \\0\leq 4p<4 $$
|
If $x<0$ then $a=\lfloor \frac{2x+1}3\rfloor$ and $b=\lfloor \frac{\lfloor 4x\rfloor+2}3\rfloor$ are both inferior or equal to $0$, so the sum cannot be $1$.
Now since $a$ is smaller than $b\quad$ [ $(\lfloor 4x\rfloor+2)-(2x+1)\ge 4x-1+2+2x-1\ge 2x\ge 0$ ]
The only way to get $a+b=1$ with $0\le a\le b\quad$ integers is $a=0$ and $b=1$.
$\begin{cases} a=\lfloor \frac{2x+1}3\rfloor=0\\ b=\lfloor \frac{\lfloor 4x\rfloor+2}3\rfloor=1\end{cases}\iff
\begin{cases} 0\le2x+1<3\\ 3\le\lfloor 4x\rfloor+2<6 \end{cases}\iff
\begin{cases} -\frac 12\le x<1\\ 1\le\lfloor 4x\rfloor<4 \end{cases}\iff
\begin{cases} -\frac 12\le x<1\\ \frac 14\le x<1 \end{cases}$
Finally we get $\bbox[5px,border:2px solid]{x\in[\frac 14,1[}$
|
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|
Eavaluating the roots of quadratic equation If $b>a , c>0$
Determine the intervals that the roots of the equation
$(x-a)(x-b) -c =0$ belong to
My work is to get the values of the roots in terms of a , b and c using the general form but i couldn't determine those intervals
|
First solve for $x$ the quadratic equation $(x-a)(x-b)-c=0$ by the general form we get that $x_1 =\frac{1}{2} \left(\sqrt{a^2-2 a b+b^2+4 c}+a+b\right)$ and $x_2=\frac{1}{2} \left(-\sqrt{a^2-2 a b+b^2+4 c}+a+b\right)$
substituting instead of $b$ the value $a$ we get that $x_1 >\frac{1}{2} \left(2 a+2 \sqrt{c}\right) $ and $x_2 > \frac{1}{2} \left(2 a-2 \sqrt{c}\right)$ because $b>a$.
substituting instead of $a$ the value $b$ we get that $x_1< \frac{1}{2} \left(2 b+2 \sqrt{c}\right)$ and $x_2 <\frac{1}{2} \left(2 b-2 \sqrt{c}\right) $
the smallest value of $\frac{1}{2} \left(2 a+2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 a-2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 b+2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 b-2 \sqrt{c}\right) $
is $\frac{1}{2} \left(2 a-2 \sqrt{c}\right)$ since $a<b$ and $c>0$.
the biggest value of $\frac{1}{2} \left(2 a+2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 a-2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 b+2 \sqrt{c}\right)$ and $\frac{1}{2} \left(2 b-2 \sqrt{c}\right) $
is $\frac{1}{2} \left(2 b+2 \sqrt{c}\right)$ since $b>a$ and $c>0$.
thus the roots are in the interval $[\frac{1}{2} \left(2 a-2 \sqrt{c}\right),\frac{1}{2} \left(2 b+2 \sqrt{c}\right)]$
|
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|
Integration changing limits - does the question have an error? I am asking about changing the limits of integration.
I have the following integral to evaluate -
$$\int_2^{3}\frac{1}{(x^2-1)^{\frac{3}{2}}}dx$$ using the substitution $x = sec \theta$.
The problem states
Use the substitution to change the limits into the form $\int_a^b$ where $a$ and $b$ are multiples of $\pi$.
Now, this is what I did.
$$ x= \sec \theta$$
$$\frac{dx}{d\theta} = \sec\theta \tan\theta$$
$$dx = sec\theta tan\theta \ d\theta$$
$$\begin{align}\int_2^{3}\frac{1}{(x^2-1)^{\frac{3}{2}}}\,dx \\
&= \int\frac{1}{(\sec^2\theta-1)^{\frac{3}{2}}}\sec\theta \tan\theta \,d\theta \\
&= \int\frac{\sec\theta \tan\theta}{(\tan^2\theta)^{\frac{3}{2}}} \,d\theta \\
&= \int\frac{\sec\theta \tan\theta}{\tan^3\theta} \, d\theta \\
&= \int\frac{\sec\theta}{\tan^2\theta} \, d\theta \\
&= \int\frac{\cos\theta}{\sin^2\theta} \, d\theta \\
&= \int \csc\theta \cot\theta \, d\theta
\end{align}$$
But here is my problem.
I know that when $x = 2$,
$$2 = \sec \theta$$
$$\frac{1}{2} = \cos \theta$$
$$\frac{\pi}{3} = \theta$$
but when $x = 3$
$$3 = \sec \theta$$
$$\frac{1}{3} = \cos \theta$$
$$\arccos\left(\frac{1}{3}\right) = \theta = $$
but this does not give me a definite result in $\pi$.
The book says the following -
where $\arccos\left(\frac{1}{3}\right) = \frac{\pi}{3}$
Am I the only one or is the book wrong in this instance?
|
The book is indeed incorrect.
$$arccos(\frac{1}{3}) \neq \frac{\pi}{3}$$
Note if it was the case that $arccos(\frac{1}{3}) = \frac{\pi}{3}$ as stated in the solutions,
then the integral would total 0.
|
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|
Binomial series $\sum^{\infty}_{n=0} {{n+k}\choose{k}} a^{n}$ Does someone know how to derive this sum?
$$\sum^{\infty}_{n=0} {{n+k}\choose{k}} a^{n}$$ where $|a|<1$ and $k$ is given integer.
|
$\binom{n+k}{k} = [x^k]:(1+x)^{n+k}$
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{n+k}{k} a^k &=& [x^k]: \sum_{n=0}^{\infty} a^k (1+x)^{n+k} \\
&=& [x^k]: (1+x)^{k} \frac{1}{1-a(1+x)} \\
&=& [x^k]: (1+x)^{k} \frac{1}{1-a} \frac{1}{1-\frac{ax}{1-a}}\\
&=& \frac{1}{1-a}[x^k]: \left( \sum_{j=0}^{k} \binom{k}{j} x^{k-j} \right)\left( \sum_{l=0}^{\infty} \left(\frac{ax}{1-a} \right)^l \right) \\
&=&\frac{1}{1-a}\sum_{j=0}^{k} \binom{k}{j} \left(\frac{a}{1-a} \right)^j \\
&=&\frac{1}{1-a}\left(1+ \frac{a}{1-a}\right)^k = \color{blue}{\frac{1}{(1-a)^{k+1}}}
\end{eqnarray*}
|
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|
Prove $ \forall n \ge 4$, $n^{3} + n < 3^{n}$
Prove that $\forall n \ge 4$, $n^{3} + n < 3^{n}$
My attempt: Base case is trivial.
Suppose $ \ n \ge 4$, $n^{3} + n < 3^{n}$. Then,
$$ (n+1)^{3} + (n+1) = n^3 + n + 3n^2 + 3n + 1 +1 < 3^{n} + 3n^2 + 3n + 2 \\< 3\cdot 3^{n} + 3n^2 + 3n +2.$$
Not sure how to get rid of $3n^2 + 3n +2$.
|
You proved correctly that $(n+1)^3+(n+1)<3^n+3n^2+3n+2$. So, prove now that$$(\forall n\geqslant4):3n^2+3n+2\leqslant2\times3^n.$$This is easy to do using induction again.
|
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|
If $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$, show that $x+y=0$
For $\{x,y\}\subset \Bbb R$, $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1.$
Prove that $x+y=0.$
Problem presented in a book, as being from Norway Math Olympiad 1985. No answer was presented. My developments are not leading to a productive direction. Sorry if this is a duplicate. Hints and answers are welcomed.
|
Since
$$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1 / \cdot (x-\sqrt{x^2+1})$$
we get
$$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x$$ so
$$\sqrt{x^2+1}-\sqrt{y^2+1}=x+y\;/^2$$
and thus $$-\sqrt{y^2+1}\cdot \sqrt{x^2+1}+1= xy$$
so $$x^2y^2 +x^2+y^2+1=x^2y^2-2xy+1$$
and finaly $(x+y)^2=0$ so $x+y=0$
|
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|
Calculating the limit $\lim\limits_{x \to 0^+} \frac{\sqrt{\sin x}-\sin\sqrt{ x}}{x\sqrt{x}}$
Calculate $$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}$$
without use Taylor serie and L'Hôpital.
$$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}\cdot\dfrac{\sqrt{\sin x}+\sin \sqrt{x}}{\sqrt{\sin x}+\sin \sqrt{x}}=\lim\limits_{x \to 0^+} \dfrac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin \sqrt{x})}$$
now what ?
|
note that
$$
\begin{split}
\lim_{x\to0^{+}}
\frac{\sin x-\sin^2\sqrt{x}}{x^2}&\overset{t=\sqrt{x}}{=}
\lim_{t\to0^{+}}
\frac{\sin(t^2)-\sin^2t}{t^4}\\
&=\lim_{t\to0^{+}}
\frac{(t^2-\frac{t^6}{6}+\frac{t^{10}}{120}-\ldots)-(t^2-\frac{t^4}{3}+\frac{2t^{6}}{45}+\frac{t^8}{315}-\ldots}{t^4}\\
&=\lim_{t\to0^{+}}
\frac{\frac{t^4}{3}-\frac{19t^6}{90}+\frac{t^8}{315}+\ldots)}{t^4}\\
&=\frac{1}{3}
\end{split}
$$
Hence
$$
\lim_{x\to0^{+}}
\frac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin\sqrt{x})}=
\lim_{x\to0^{+}}
\frac{\frac{\sin x-\sin^2\sqrt{x}}{x^2}}{\sqrt{\frac{\sin x}{x}}+\frac{\sin\sqrt{x}}{\sqrt{x}}}
=\frac{\frac{1}{3}}{\sqrt{1}+1}=\frac{1}{6}
$$
|
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|
Contractive Principle to Prove Convergence How do I use the contraction principle to show ${a_{n}}$ converges?
$a_{1}=1 \:$ and$\: a_{n+1}= 1+ \frac{1}{1+a_{n}}.$
I know I need to show that $\left | a_{n+2}-a_{n+1} \right |=k\left | a_{n+1}-a_{n} \right |$ with $k \in (0,1)$
So far I have:
$\left | a_{n+2}-a_{n+1} \right |= \left | \left ( 1+ \frac{1}{1+a_{n+1}} \right )-\left ( 1+ \frac{1}{1+a_{n}} \right ) \right | =\left | \frac{1}{1+a_{n+1}} -\frac{1}{1+a_{n}}\right |=\left | \frac{a_{n}-a_{n+1}}{(1+a_{n+1})(1+a_{n})} \right |$
I am stuck from here...
I see the $\lim_{n\to\infty} a_{n}= \sqrt 2$ and I believe $\frac{1}{2}\leq {a_{n}} \leq \frac{3}{2}.$
Can I get some nudges to push me along?
|
Show that $f(x) = 1 + {1\over 1 +x}$ maps $[1,\infty) \to [1, \infty)$.
Show that $|f'(x)| \le {1 \over 4} < 1$ for $x \ge 1$.
Show that $|f(x)-f(y)| \le {1 \over 4} |x-y|$ for $x,y \ge 1$.
Note that $a_{n+1} = f(a_n)$ and the contraction mapping theorem shows that there is a unique $a$ such that $a_n \to a$. Since $a=f(a)$ we can solve for $a$ to get $a^2=2$ and so $a= \sqrt{2}$ (we are living in $[1,\infty)$.
|
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|
Prove that for $n \ge 2$ the follow inequality holds $\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}$. As already shown above I need to prove that $$\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}\qquad \forall n \ge 2.$$
What I've come up with is the following:
$\underline{n=2:}\qquad$ $$\frac {16}{3} \lt 6$$
$\underline{n=k:}\qquad $$$\frac {4^k}{k+1} \lt \frac{(2k)!}{(k!)^2} $$
$\underline{n+1}:$ $$\frac {4^{k+1}}{k+2} \lt \frac{(2(k+1))!}{((k+1)!)^2} $$
$$\frac{4^k \cdot 4}{k+2} \lt \frac {(2k+2)!}{(k!)^2\cdot(k+1)\cdot(k+1)}$$
$$\frac{4^k \cdot 4}{k+2} \lt \frac{(k+1)\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)\cdot(2k+2)}{(k!)^\require{enclose}\enclose{updiagonalstrike}2\cdot(k+1)\cdot(k+1)}$$
$$\frac{4^k \cdot 4}{k+2} \lt 2 \cdot\frac{\require{enclose}\enclose{updiagonalstrike}{(k+1)}\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}}{(k!)\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}}$$
$$\frac{4^k \cdot 4}{k+2} \lt 2 \cdot\frac{{(k+2)}\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)}{(k!)}$$
From here on I'm stuck. Can somebody help me please?
|
Note that $$\frac{\frac{4^{n+1}}{n+2}}{\frac{4^n}{n+1}}=4\frac{n+1}{n+2}$$ and that $$\frac{\frac{(2(n+1))!}{(n+1)!^2}}{\frac{(2n)!}{n!^2}}=\frac{(2n+2)!}{(2n)!}\cdot\frac{n!^2}{(n+1)!^2}=\frac{(2n+2)(2n+1)}{(n+1)^2}=2\frac{2n+1}{n+1}. $$So, when is it true that$$4\frac{n+1}{n+2}\leqslant2\frac{2n+1}{n+1}?$$ Well, the previous inequaliy is equivalent to $2(n+1)^2\leqslant(2n+1)(n+2)=2n^2+5n+2$, which clearly holds always. So, since you proved that $\frac{4^2}{2+1}<\frac{(2\times2)!}{2!^2}$, the inequality that you want to prove is true whenever $n\geqslant2$.
|
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Prove that two of the straight lines represented by $x^3+bx^2y+cxy^2+y^3 = 0$ will be at right angles if $b+c$=-2.
Prove that two of the straight lines represented by $x^3+bx^2y+cxy^2+y^3 = 0$ will be at right angles if $b+c=-2$.
I divided whole equation by $y^3$ to get a cubic in $\frac{x}{y}$. I guess product of two roots of this equation should be $-1$ but I am unable to derive a condition for that.
|
Let $y=mx$.
Thus, me need
$$m^3+bm^2+cm+1=0$$ with $m_1m_2=-1$ and since $m_1m_2m_3=-1$,
we obtain $m_3=1$ and $$1+b+c+1=0$$ or
$$b+c=-2.$$
Id est, we proved a bit of more:
Two of the straight lines represented by $$x^3+bx^2y+cxy^2+y^3=0$$ will be at right angles iff $b+c=-2$.
Because if $b+c=-2$ we see that one of roots of the following equation
$$m^3+bm^2+cm+1=0$$ is $m_1=1$, which gives $m_2m_3=-1$ and we are done!
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2437543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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|
Show that $\left(\sum_{r=1}^n r\right)^2=\sum_{r=1}^n r^3$ without expanding to closed form It is well know that $$\left(\sum_{r=1}^n r\right)^2=\sum_{r=1}^n r^3$$
i.e.
$$(1+2+3+\cdots+n)(1+2+3+\cdots+n)=1^3+2^3+3^3+\cdots+n^3$$
and this is usually proven by showing that the closed form for the sum of cubes is $\frac 14 n^2(n+1)^2$ which can be written as $\left(\frac 12 n(n+1)\right)^2$, and then noticing that $\frac 12 n(n+1)$ is the sum of integers.
Is it possible to prove the result, preferably from LHS to RHS, without first expanding the summation to closed form, as described above?
Note: The intention is to arrive at RHS from LHS by manipulation of limits and summands without full expansion to closed form, rather than using induction, or graphical methods, like Nicomachus' method.
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$$
\underbrace{(1+2+3+\cdots+n)(1+2+3+\cdots+n) = 1^3+2^3+3^3+\cdots+n^3}_{\Large\text{This will be our induction hypothesis.}}
$$
\begin{align}
& \Big( \underbrace{1+2+3+\cdots+n}_{\Large A} +\underbrace{\Big(n+1\Big)}_{\Large B}~~\Big)^2 \\[10pt]
= {} & (A+B)^2 = A^2+2AB+B^2 \\[10pt]
= {} & \overbrace{(1+2+3+\cdots+n)^2}^{\Large A^2} {}+{} \overbrace{2(1+2+3+\cdots+n)(n+1)}^{\Large 2AB} + \overbrace{(n+1)^2}^{\Large B^2} \\[10pt]
= {} & \Big(1^3+2^3+3^3+\cdots+n^3\Big) + 2(1+2+3+\cdots+n)(n+1) + (n+1)^2 \\
& \quad \text{by the induction hypothesis} \\[10pt]
= {} & \Big(1^3+2^3+3^3+\cdots+n^3\Big) + \Big(n(n+1)\Big)(n+1) + (n+1)^2 \\
& \quad \text{(since $1+2+3+\cdots+n = \dfrac {n(n+1)} 2$)} \\[10pt]
= {} & \Big(1^3+2^3+3^3+\cdots+n^3\Big) + (n+1)^3.
\end{align}
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2437635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Find all the points that lie $ 3$ units from each of the points $ (2,0,0), (0,2,0), \text{ and }(0,0,2)$
I calculated the result $$ \{(x,y,z) \in\mathbb{R^3}: x=y=z\}.$$
I'm wondering whether I did this problem correctly and if I did how to draw the set of solutions.
I used the euclidean distance formula with the square root (the euclidean norm) and set up three equations for each point $(2,0,0)$, $(0,2,0)$ and $(0,0,2)$ that made sure the distance from arbitrary $(x,y,z)$ to these three points equal to $3$ and arrived at the solution set. –
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The set of all points equidistqnt from $A=(2,0,0)$ and $B=(0,2,0)$:
\begin{align}
\sqrt{(x-2)^2+y^2+z^2}&=\sqrt{x^2+(y-2)^2+z^2} \\
-4x+4 &= -4y+4 \\
x&=y
\end{align}
Similarly, the set of all points equidistant from $A=(2,0,0)$ and $C=(0,0,2)$ is described by $x=z$.
Hence the set of all points equidistant from $A, B$, and $C$ is described by $x=y=z$, that is, the line P(t) = (t,t,t). If that distance is $3$, then $d(A,P(t))=3$ implies
\begin{align}
\sqrt{(t-2)^2+t^2+t^2}&=3 \\
3t^2-4t+4 &= 9 \\
3t^2-4t-5 &= 0 \\
t &= \frac{2 \pm \sqrt{19}}{3}
\end{align}
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2439824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.