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Prove for any positive real numbers $a,b,c$ $\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} \geq \frac{a+b+c}{3}$ Since the problem sheets says I should use Cauchy-Schwarz inequality, I used
$\frac{{a_1}^2}{x_1}+\frac{{a_2}^2}{x_2}+\frac{{a_3}^2}{x_3}$
$\geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}$
I first multiplied each term by $a,b,c$ to get a perfect square on top like
$\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}$
$=\frac{a^4}{a(a^2+ab+b^2)}+\frac{b^4}{b(b^2+bc+c^2)}+\frac{c^4}{c(c^2+ca+a^2)}$
$\hspace{120pt}\geq \frac{(a^2+b^2+c^2)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a)} $
But I am still stuck for few hours. This is not a homework, but is a set of problems to prepare for AMC/USAMO.
(Note: I started off with a general question and got closed as off topic. I have another genuine Math Question now, and I am just trying to see if math.se is going to be useful)
|
$
\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}
=\frac{a^4}{a(a^2+ab+b^2)}+\frac{b^4}{b(b^2+bc+c^2)}+\frac{c^4}{c(c^2+ca+a^2)}
\geq \frac{(a^2+b^2+c^2)^2}{a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a)} \tag{1}
$
Remember
$\frac{{a_1}^2}{x_1}+\frac{{a_2}^2}{x_2}+\frac{{a_3}^2}{x_3} \geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3} \tag{2}$
$
a^3+b^3+c^3+ab(a+b)+bc(b+c)+ca(c+a) $
$= a^3+b^3+c^3+a^{2}b+ab^2+b^{2}c+bc^2+c^{2}a+ca^2$
$= (a+b+c)(a^2+b^2+c^2) \tag{3}$
From $(1) $ and $(3)$
$
\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2} $
$\geq \frac{a^2+b^2+c^2}{a+b+c}$
$\geq\frac{a^2}{a+b+c}+\frac{b^2}{a+b+c}+\frac{c^2}{a+b+c}$
Now applying $(2)$ again to the expression on the right
$
\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}$
$\geq \frac{a^2}{a+b+c}+\frac{b^2}{a+b+c}+\frac{c^2}{a+b+c}$
$\geq \frac{(a+b+c)^2}{3(a+b+c)} = \frac{a+b+c}{3}$
|
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|
Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$
*
*$\displaystyle
0\cdot \binom{n}{0} + 1\cdot \binom{n}{1} + 2\binom{n}{2}+\cdots+(n-1)\cdot \binom{n}{n-1}+n\cdot \binom{n}{n}$
*$\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}$
How do you find the sum of these and prove it by induction? Can someone help me get through this?
|
1) Take $\displaystyle f(x)= \sum_{i=0}^n\binom{n}{i} x^i=(x+1)^n$. Now consider $f'(1)$.
2) Use that $\frac{1}{(k-1)\cdot k} = \frac{1}{k-1} -\frac{1}{k}$.
|
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|
How to compute $\int{\frac{5x^3+8x^2+x+2}{x^2(2x^2+1)}} dx$? $$\int{\frac{5x^3+8x^2+x+2}{x^2(2x^2+1)}} dx$$
So ... how do I start? Numerator cant be factorized it seems, and this looks like a complicated expression ...
I tried expanding the denominator to see if integration by substitution will work, but it didn't give any ideas
$$\int{\frac{5x^3+8x^2+x+2}{2x^4+x^2}} dx$$
I don't see a clear way to integrate by parts either. Can anyone enlighten me?
|
Hint :
Use partial fraction decomposition :
$$\frac{5x^3+8x^2+x+2}{x^2(2x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{2x^2+1}$$
|
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|
No Integer Solutions and Congruences Using congruences, I seek to prove two things:
1) $x^2 - 4y^2 = 3$ has no solutions in integers $x,y,z$.
I think this can be done using modulo 4? How so?
2) $3x^3 - 7y^3 + 21 z^3 = 2$ has no solutions in integers $x,y,z$.
Not sure how to attack this one...
|
(1) If $x$ is even, is $x^2-4y^2$ odd or even? If $x$ is odd, can $x^2-3$ be divisible by $4$ if $x$ is an integer?
(2) $3x^3-7y^3+21z^3=7(3z^3-y^3)+3x^3$; can $3x^3-2$ be divisible by $7$ if $x$ is an integer?
|
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|
Differentiation of $y = \tan^{-1} \Bigl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\Bigr\}$ How do i differentiate the following: $$y = \tan^{-1} \biggl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\biggr\}$$
I know that $\text{derivative}$ of $\tan^{-1}{x}$ is $\frac{1}{1+x^{2}}$ but not sure as to how to do this.
|
Here are all the steps in excruciating detail.
Using the tangent of a sum formula gives
$$
\frac{1-\tan{\phi}}{1+\tan{\phi}}=\tan\left(\frac\pi4-\phi\right)\tag{1}
$$
and letting $\tan(\phi)=\sqrt{\frac{1-x^2}{1+x^2}}$ yields
$$
\begin{align}
\tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)
&=\tan^{-1}\left(\frac{1-\sqrt{\frac{1-x^2}{1+x^2}}}{1+\sqrt{\frac{1-x^2}{1+x^2}}}\right)\\
&=\frac\pi4-\tan^{-1}\left(\sqrt{\frac{1-x^2}{1+x^2}}\right)\tag{2}
\end{align}
$$
Furthermore, we have
$$
\begin{align}
\sqrt{\frac{1-\cos(\psi)}{1+\cos(\psi)}}
&=\frac{\sin(\psi/2)}{\cos(\psi/2)}\\
&=\tan(\psi/2)\tag{3}
\end{align}
$$
so letting $x^2=\cos(\psi)$ yields
$$
\begin{align}
\tan^{-1}\left(\sqrt{\frac{1-x^2}{1-x^2}}\right)
&=\psi/2\\
&=\frac12\cos^{-1}(x^2)\tag{4}
\end{align}
$$
Combining $(2)$ and $(4)$ shows
$$
\begin{align}
\tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)
&=\frac\pi4-\frac12\cos^{-1}(x^2)\\
&=\frac12\sin^{-1}(x^2)\tag{5}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)
&=\frac{\mathrm{d}}{\mathrm{d}x}\frac12\sin^{-1}(x^2)\\
&=\frac{x}{\sqrt{1-x^4}}\tag{6}
\end{align}
$$
|
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|
How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
|
\begin{align*}
\dfrac{n(n^4-1)}{30}&=\dfrac{n(n^2-1)\color{#c00}{(n^2+1)}}{30}\\&=\dfrac{(n-1)n(n+1)\color{#c00}{(5+(n+2)(n-2))}}{30}\\&=\dfrac{(n-1)n(n+1)}{6}+\dfrac{\color{#c00}{(n-2)}(n-1)n(n+1)\color{#c00}{(n+2)}}{30}\\&=\binom{n+1}{3}+4\binom{n+2}{5}\in\Bbb Z
\end{align*}
|
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|
How do I solve $(x-1)(x-2)(x-3)(x-4)=3$ How to solve $$(x-1) \cdot (x-2) \cdot (x-3) \cdot (x-4) = 3$$
Any hints?
|
I assume the hints would already have given you the answer. If not, here is the full answer:
Let y = x-2.5
(y+1.5)(y-1.5)(y+0.5)(y-0.5) = 3.
so $(y^2-2.25)(y^2-0.25) = 3$.
Let $z = y^2-1.25$.
(z-1)(z+1) = 3.
So $z^2-1 = 3$.
Hence $z^2 = 4$.
z = -2 gives $y^2 = z + 1.25 = -0.75$.
So $y = \pm \sqrt{0.75}i$. Clearly, this should be ignored if you only want real roots.
$x = 2.5 + y = 2.5 \pm \sqrt{0.75}i$.
z = 2 gives $y^2 = z + 1.25 = 3.25$.
So $y = \pm \sqrt{3.25}$. Clearly, this should be ignored if you only want real roots.
$x = 2.5 + y = 2.5 \pm \sqrt{3.25}$.
If you want all the terms in the product to be positive, then obvious ly $x = 2.5 + \sqrt{3.25}$ is the only one that works. This is roughly 4.30277564.
|
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|
Prime number in a polynomial expression Will be glad for a little hint: let x and n be positive integer such that $1+x+x^2+\dots+x^{n-1}$ is a prime number then show that n is prime
|
Assuming that you meant that $1+x+x^2+\ldots+x^{n-1}$ is prime, note that this sum is $\frac{x^n-1}{x-1}$. If $n=ab$, we have $$\frac{x^n-1}{x-1}=\frac{(x^a)^b-1}{x-1}=\frac{(x^a-1)(1+x^a+x^{2a}+\ldots+x^{(b-1)a})}{x-1}\;,$$ and $\dfrac{x^a-1}{x-1}=\;$?
|
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|
Taking the derivative of $y = \dfrac{x}{2} + \dfrac {1}{4} \sin(2x)$ Again a simple problem that I can't seem to get the derivative of
I have $\frac{x}{2} + \frac{1}{4}\sin(2x)$
I am getting $\frac{x^2}{4} + \frac{4\sin(2x)}{16}$
This is all very wrong, and I do not know why.
|
$y = \dfrac{1}{2} x + \dfrac {1}{4} \sin(2x)$
$y' = \dfrac{1}{2} + \dfrac{1}{4}\cos(2x)*2$
$y' = \dfrac{1}{2} + \dfrac{2}{4}\cos(2x)$
$y' = \dfrac{1}{2} + \dfrac{1}{2}\cos(2x)$
$y' = \dfrac{1}{2} \big(1 + \cos (2x)\big) $
Which is equivalent to $\cos^2x$.
|
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|
What is awry with this proof? Let $x=5$, $y=7$, $z=6$
$x+y = 2z$
Rearranging, $x-2z = -y$
and $x = -y+2z$
Multiply both sides respectively. $x^2-2xz = y^2-2yz$
$$x^2-2xz+z^2 = y^2-2yz+z^2$$
$$(x-z)^2 = (y-z)^2$$
$$x-z = y-z$$
Hence $x=y$, or $5 = 7$
Well, the conclusion is clearly false, but what went wrong? I think it may be the step in which one square roots both sides because it's taking out one solution?
|
If $(x-z)^2 = (y-z)^2$ then $x-y = \pm(y-z)$.
|
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|
What is this series called? $\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \dots \pm \frac{x^n}{n!}$ I remember learning about this series in Precalculus the other day but I neglected to get the name of it. It looks something like this:
$
\begin{align*}
\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \dots \pm \frac{x^n}{n!}
\end{align*}
$
All I remember is that it helps in the modeling of $\sin{x}$.
|
This
$$
\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} \dots \pm \frac{x^n}{n!}
$$
Is called a Maclaurin polynomial for the sine function.
|
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|
More highschool math $\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3} = \frac{n^3+3n^2+3n+1}{3n^3} \to \frac{1}{3}$ So the question I am trying to work through is:
Test the series
$$\frac{1}{3}+\frac{2^3}{3^2}+\frac{3^3}{3^3}+\frac{4^3}{3^4}+\frac{5^3}{3^5}+\cdot\cdot\cdot$$
for convergence.
The solution (using D'Alembert's ratio test) is:
$$u_n=\frac{n^3}{3^n}\;,$$
so
$$\begin{align*}
\frac{|u_{n+1}|}{|u_n|} &=\frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3}\\
&=\frac{3^n(n+1)^3}{3^{n+1}\cdot n^3}\\
&=\frac{3^n(n+1)^3}{3^{n+1}\cdot n^3}\\
&=\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3}\;.
\end{align*}$$
How do we get from there to...
$$=\frac{n^3+3n^2+3n+1}{3n^3}$$
What happens with $3^n$ in the numerator and power of $n+1$ in the denominator? How do they cancel out?
Also, in the very next step that all goes to being equal to
$$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=\frac{1}{3}<1\;,$$
which means the series is convergent.
But how do we get to $\dfrac{1}{3}$?
|
Here is one way to simplify the limit and arrive at the answer. Hopefully, it will let you see how terms cancel out.
\begin{align*}
\left|\frac{u_{n+1}}{u_{n}}\right| &= \frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3} \\
&= \frac{(n+1)^3}{n^3}\cdot \frac{3^n}{3^{n+1}} \\
&= \left(\frac{n+1}{n}\right)^3 \cdot \frac{1}{3} \\
&= \left(1 + \frac{1}{n}\right)^3 \cdot \frac{1}{3} \\
\Rightarrow \lim_{n \rightarrow \infty} \left|\frac{u_{n+1}}{u_{n}}\right| &= 1^3 \cdot \frac{1}{3} = \frac{1}{3}
\end{align*}
|
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|
Determine convergence of $\sum_{n=1}^{\infty} (\cos{\frac{2}{n}}-\cos{\frac{4}{n}})$ Determine convergence of $$\sum_{n=1}^{\infty} \left(\cos{\frac{2}{n}}-\cos{\frac{4}{n}}\right)$$
In the answer, it says
$$\cos{\frac{2}{n}}-\cos{\frac{4}{n}} = 2\sin{\frac{3}{n}}\sin{\frac{1}{n}} \le 2\cdot \frac{3}{n} \cdot \frac{1}{n} = \frac{6}{n^2}$$
But how do I get the above trig substitution? I guess removing the fractions, I will get $\cos{x}-\cos{2x}=2\sin{(2x-1)}\sin{(x-1)}$ ... probably this is wrong, but how do I get that?
|
The result follows from the Addition Law for Cosines.
We have $\cos(x+y)=\cos x\cos y-\sin x\sin y$ and $\cos(x-y)=\cos x\cos y+\sin x\sin y$. Subtract. We get
$$\cos(x-y)-\cos(x+y)=2\sin x\sin y.$$
Let $x-y=\frac{2}{n}$ and $x+y=\frac{4}{n}$. Solve for $x$ and $y$. We get $x=\frac{3}{n}$ and $y=\frac{1}{n}$.
|
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|
Evaluating $\int y^4(1-y)^3 dy$ using integration by parts Here is the function which could easily be solved using expansion method but how could I solve it using integration by parts
$$\int y^4(1-y)^3 dy$$
The problem is, when I apply integration by parts to solve it, it is never ending solution and I am not able to get the answer.
For example,
I let $u = (1-y)^3$ and $dv = (y^4)$, so
$du = 3(1-y)^2$ and $v = \dfrac{y^5}{5}$
When I apply the Integration by Parts formula,
$$uv - \int v du$$
I got the kind of same equation as I started with, so I need to apply integration by parts once again, and then again. How many times is it required to apply before I get the answer ?
|
$$I(4,3) = \int y^4 (1-y)^3 \mathrm{d}y$$
You were heading in the right direction, i.e. $\mathrm{d}v=y^4 \Rightarrow v = \frac{y^5}{5}$
$$
\begin{align*}
I(4,3) &= \frac{y^5(1-y)^3}{5} + \frac{3}{5} \int y^5 (1-y)^2 \mathrm{d}y\\
&= \frac{y^5(1-y)^3}{5} + \frac{3}{5} I(5,2)\\
\end{align*}
$$
Similarly use $\mathrm{d}v=y^5$, $u=(1-y)^2$ to evaluate $I(5,2)$
$$
\begin{align*}
I(5,2) &= \frac{y^6(1-y)^2}{6} +\frac{1}{3} I(6,1)\\
&= \frac{y^6(1-y)^2}{6} + \frac{1}{3}\left(\frac{y^7}{7} - \frac{y^8}{8}\right)\\
I(4,3) &= \frac{y^5(1-y)^3}{5} + \frac{1}{10}y^6(1-y)^2+\frac{1}{5}\left(\frac{y^7}{7} - \frac{y^8}{8}\right)+ Constant
\end{align*}
$$
|
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|
Evaluating $\lim\limits_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$ Evaluate $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$
|
If we draw the graph of $x^p$ from $x=1$ to $x=n,$ divide it into unit length intervals and approximate each segment of area by a trapezium (this is known as the trapezoidal rule) then we see that $$\int^n_1 x^p dx \approx \sum_{k=1}^n k^p - \frac{n^p+1}{2}.$$ The integral on the left is precisely $\displaystyle \frac{n^{p+1} -1}{p+1},$ so for large $n$ (where the major contribution is from the dominant terms) we have $$\sum_{k=1}^n k^p \approx \frac{n^{p+1}}{p+1} + \frac{n^p}{2}$$ so your limit is $1/2.$
For a precise solution, we need the error term along with the trapezoidal rule, which is derived here. It gives : $$\int^b_a f(x) dx = \frac{b-a}{2} ( f(a) + f(b) ) - \frac{(b-a)^3 }{12} f''(\zeta) $$ for some $\zeta \in [a,b].$ For $f(x)=x^p$ we have $f''(x) = p (p-1)x^{p-2}$ which is largest at $x=b$, the right end point. So the sum of the error terms in our application of the trapezoidal rule is at largest $$\frac{p(p-1)}{12} (2^{p-2} + 3^{p-2} + \cdots + n^{p-2}).$$ The sum in the brackets is overestimated by $\int^{n+1}_1 x^{p-2} dx= \frac{(n+1)^{p-1}-1}{p-1},$ so we get that $$\sum_{k=1}^n k^p = \frac{n^{p+1}}{p+1} + \frac{n^p}{2} + E_n$$ where $E_n$ is an error term that satisfies $\displaystyle \lim_{n\to\infty} \frac{E_n}{n^p} = 0$ which proves your limit.
|
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|
Complex number: calculate $(1 + i)^n$. I have to solve the following complex number exercise: calculate $(1 + i)^n\forall n\in\mathbb{N}$ giving the result in $a + ib$ notation.
Basically what I have done is calculate $(1 + i)^n$ for some $n$ values.
$$(1 + i)^1 = 1 + i$$
$$(1 + i)^2 = 2i$$
$$(1 + i)^3 = - 2 + 2i$$
$$\boxed{(1 + i)^4 = - 4}$$
$$(1 + i)^5 = (1 + i)^4\cdot(1 + i)^1 = (-4)\cdot(1 + i) = - 4 - 4i$$
$$(1 + i)^6 = (1 + i)^4\cdot(1 + i)^2 = (-4)\cdot2i = - 8i$$
$$(1 + i)^7 = (1 + i)^4\cdot(1 + i)^3 = (-4)\cdot(- 2 + 2i) = 8 - 8i$$
$$(1 + i)^8 = (1 + i)^4\cdot(1 + i)^4 = (-4)\cdot(-4) = (-4)^2 = 16$$
We can write $n = 4\cdot q + r$ (Euclidean division), so we have:
$$(1 + i)^n = (1 + i)^{(4\cdot q + r)} = ((1 + i)^4)^q\cdot(1 + i)^r = (-4)^q\cdot(1 + i)^r$$
Finally if you want to calculate say... $(1 + i)^n$ for $n = 625$ you have:
$$625 = 4\cdot156 + 1\Rightarrow q = 156, r = 1$$
$$(1 + i)^{625} = (-4)^{156}\cdot(1 + i)^1 = (-4)^{156} + (-4)^{156}i$$
What other approach would you suggest? Mine works, but you have to find $q$ and $r$ in order to do the calculation, and I think it is not "calculate" technically speaking which was what the exercise is asking (although I am not sure what they mean by calculate).
|
Perhaps, proceed via the trigonometric form using De Moivre's formula:
$$\left(1+i\right)^{n}=2^{\frac{n}{2}}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{n}=2^{\frac{n}{2}}\cos\frac{\pi n}{4}+i2^{\frac{n}{2}}\sin\frac{\pi n}{4}$$
|
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|
Integral of $\int \frac{du}{u \sqrt{5-u^2}}$ I am trying to find this integral and I can get the answer on wolfram of course but I do not know what is wrong with my method, having gone through it twice.
$$\int \frac{du}{u \sqrt{5-u^2}}$$
$u = \sqrt{5} \sin\theta$ and $du= \sqrt{5} \cos \theta$
$$\int \frac{\sqrt{5} \cos \theta}{\sqrt{5} \cos \theta \sqrt{5-(\sqrt{5} \sin \theta)^2}}$$
$$\int \frac{1}{\sqrt{5-(5 \cos^2 \theta)}}$$
$$\int \frac{1}{\sqrt{5(1- \cos^2 \theta)}}$$
$$\int \frac{1}{\sqrt{5(\sin^2 \theta)}}$$
$$\frac{1}{\sqrt5}\int \frac{1}{(\sin \theta)}$$
$$\frac{1}{\sqrt5}\int \csc\theta$$
$$\frac{\ln|\csc \theta - \tan \theta|}{\sqrt5} + c$$
|
You made several mistakes throughout your answer. Once corrected the answer should be
$$\int \frac{du}{u \sqrt{5-u^2}}$$
$u = \sqrt{5} \sin\theta$ and $du= \sqrt{5} \cos \theta d\theta$
\begin{eqnarray}
\int \frac{du}{u \sqrt{5-u^2}} &=& \int\dfrac{ \sqrt{5} \cos \theta d\theta}{\sqrt{5} \sin\theta \sqrt{5 - (\sqrt{5} \sin\theta)^2}} \\
&=& \int\dfrac{ \cos \theta d\theta}{ \sin\theta \sqrt{5 - 5 \sin^2\theta}} \\
&=& \int\dfrac{ \cos \theta d\theta}{ \sin\theta \sqrt{5cos^2\theta}} \\
&=& \dfrac{1}{\sqrt{5}}\int\dfrac{ \cos \theta d\theta}{ \sin\theta \cos\theta} \\
&=& \dfrac{1}{\sqrt{5}}\int\dfrac{ d\theta}{ \sin\theta } \\
&=& \dfrac{1}{\sqrt{5}}\int \sec\theta d\theta \\
&=& \dfrac{1}{\sqrt{5}}\int \csc\theta \dfrac{ \csc \theta + \cot \theta }{ \csc \theta + \cot \theta } d\theta\\
&=& -\dfrac{1}{\sqrt{5}} \ln|\csc\theta+\cot\theta| +C
\end{eqnarray}
Then since $\sin \theta = \dfrac{u}{\sqrt{5}}$, drawing the triangle we find out that the remaining side is $\sqrt{\sqrt{5}^2 -u^2} = \sqrt{5 -u^2}$. Therefore $\csc \theta = \dfrac{1}{\sin \theta}= \dfrac{\sqrt{5}}{u}$ and $\cot\theta = \dfrac{1}{\tan \theta} = \dfrac{\sqrt{5 -u^2}}{u}$. So
\begin{eqnarray}
-\dfrac{1}{\sqrt{5}} \ln|\csc\theta+\tan\theta| +C &=& -\dfrac{1}{\sqrt{5}} \ln\left|\dfrac{\sqrt{5}}{u}+\dfrac{\sqrt{5 -u^2}}{u}\right| + C \\
&=& -\dfrac{1}{\sqrt{5}}\left(-\ln|u| + \ln\left|\sqrt{5}+\sqrt{5 -u^2}\right|\right) + C
\end{eqnarray}
|
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|
Factorize the polynomial $P(z) = z^4 - 2z^3-z^2+2z+10$, into linear and/or quadratic factors with real coefficients $2+i$ is given to be one of the roots of the polynomial.
I am doing this as a practice for exam prep.
Since $2+i$, is a root, then $(z-2-i)$ is a factor?
So I have:
$(z-2-i)(z^3-Az^2-Bz+C) = z^4-2z^3-z^2+2z+10$
Then, I group the $z$ terms of the same degrees after I multiply out?
Am I on the right track? Because it seems like this will take quite a while in exam conditions? Thanks for any direction!
|
As Thomas has said, if $2+i$ is a root then $2-i$ is also root. Hence you have two factors: $z-(2-i)$ and $z-(2+i)$ whose product is $z^2-4z+5$. So we can can assume that there is another quadratic factor $z^2+bz+c$ where $b$ and $c$ are constants to be determined. Obviously their product should yield the given polynomial. Thus,
$$(z^2-4z+5)(z^2+bz+c)= z^4-2z^3-z^2+2z+10.$$
Equating the constant terms you obtain $5c=10$ which gives $c=2$.
Now equating the coefficients of $z$, we obtain the equation;
$$-4c+5b=2 $$ from which we obtain $b=2$. Replacing $b$ and $c$ we now solve for $z$ in the equation $$z^2+2z+2=0$$ as follows:
$$(z+1)^2+1=0.$$ Thus, $z+1=i$ and $z+1=-i$, which gives $z=-1+i,z=-1-i$.
Hence we have obtained all the factors.
|
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|
Numbers arranged in square arrays - observe patterns, make a conjecture, and prove or disprove it Please can anyone tell me what the below question means and how to solve it:
Observe patterns. Make a conjecture. Prove or disprove.
$$\begin{matrix}1\end{matrix}\qquad\begin{matrix}1 & 1 \\ 1 & 2\end{matrix}\qquad\begin{matrix}1 & 1 & 1\\ 1 & 2 & 2 \\ 1 & 2 & 3\end{matrix}\qquad\begin{matrix}1 & 1 & 1 & 1\\ 1 & 2 & 2 & 2\\ 1 & 2 & 3 & 3 \\ 1 & 2 & 3 & 4\end{matrix}$$
Compute the sum of all numbers in each square array.
|
Assuming the pattern is $A_{ij}=\min(i,j)+1$
Each number $m$ in the $n \times n$ matrix is written $2(n-m)+1$ times. So we have
$$\sum_{m=1}^n m(2n-2m+1)$$
$$(2n+1)\sum_{m=1}^n m - 2\sum_{m=1}^n m^2$$
If you know the formula for sums of values and sums of squares, then you're golden.
$$=(2n+1)\cdot n(n+1)/2-n(n+1)(2n+1)/3$$
$$=\frac {n(n+1)(2n+1)} 6$$
It's a sum of squares in disguise. Look: each matrix can be expanded as
$$\begin{matrix}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}+\begin{matrix}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1\end{matrix}+\begin{matrix}0 & 0 & 0 & 0\\ 0 & 1 & 1 & 1\\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1\end{matrix}+\begin{matrix}1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1\end{matrix}$$
which should make that a little more obvious. Each matrix of size $n \times n$ has a sum of squares up to $n$.
|
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|
Minimal polynomial of the root of algebraic number I have obtained the minimal polynomial of $9-4\sqrt{2}$ over $\mathbb{Q}$ by algebraic operations:
$$ (x-9)^2-32 = x^2-18x+49.$$
I wonder how to calculate the minimal polynomial of $\sqrt{9-4\sqrt{2}}$ with the help of this sub-result? Or is there a smarter way to do this (not necessarily algorithmic)?
|
Since $x = 9 - 4\sqrt{2}$ satisfies $x^2 - 18x + 49 = 0$, your number $y = \sqrt{x} = \sqrt{9 - 4\sqrt{2}}$ satisfies $y^4 - 18y^2 + 49 = 0$. This could be your minimal polynomial, but the polynomial factorizes as $$y^4 - 18y^2 + 49 = (y^2 + 2y - 7)(y^2 - 2y - 7).$$ Since the product is zero if and only if at least one of them is zero, we get that either $y^2 + 2y - 7 = 0$ or $y^2 - 2y - 7 = 0$. Since the minimal polynomial must have degree at least $2$, one of those must be your minimal polynomial. (In this case it is the latter, $y^2 - 2y - 7$.)
|
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|
Finding $\int \frac {dx}{\sqrt {x^2 + 16}}$ I can not get the correct answer.
$$\int \frac {dx}{\sqrt {x^2 + 16}}$$
$x = 4 \tan \theta$, $dx = 4\sec^2 \theta$
$$\int \frac {dx}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{\sqrt {16 \sec^2 \theta}}$$
$$\int \frac {4 \sec^ 2 \theta}{4 \sec \theta}$$
$$\int \sec \theta$$
$$\ln| \sec \theta + \tan \theta|$$
Then I solve for $\theta$:
$x = 4 \tan \theta$
$x/4 = \tan \theta$
$\arctan (\frac{x}{4}) = \theta$
$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tan (\arctan (\tfrac{x}{4}))|$$
$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tfrac{x}{4}))| + c$$
This is wrong and I do not know why.
|
$$\int \frac {dx}{\sqrt {x^2 + 16}}$$
Let $$t=\sqrt {x^2+16}$$ this will change the integral into $$ \int \frac {dt}{\sqrt {t^2 - 16}}$$
$$t=4\sec (\theta)$$ changes the integral into $$\int \sec \theta d \theta =ln| \sec \theta + \tan \theta|+c$$
$$=\ln (t/4 + \frac {\sqrt {t^2-16}}{4} )+c$$
$$= \ln ( \sqrt \frac {x^2+16}{4}+x/4 )+c$$
|
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|
Integral of $\int \sqrt{1-4x^2}$ I know I am messing up something with the substitutions but I am not sure what.
$$\int \sqrt{1-4x^2}$$
$$u = 4x, du = 4 \,dx$$
$$\frac{1}{4}\int \sqrt{1-u^2}$$
$u = \sin \theta$
$$\frac{1}{4}\int \sqrt{1-\sin^2 \theta} = \frac{1}{4}\int \sqrt{ \cos^2 \theta} = \frac{1}{4}\int \cos \theta = \frac{\sin \theta}{4}$$
Replace $\theta$ with $u$
$$\frac{\sin (\arcsin u)}{4} = \frac{u}{4} = \frac{4x}{4} = x$$
This is wrong and I have no idea why.
|
$$\int \sqrt{1-4x^2}\ dx$$
Substitute
$$\begin{align*}x &= \frac{\sin (\theta)}{2}\\ dx &= \frac{\cos(\theta)}{2}d \theta\end{align*}$$
So the intregral become
$$\begin{align*}
\int \sqrt{1-4x^2}\ dx &=\frac{1}{2}\int \cos^2(\theta)d\theta\\
&=\frac{1}{2}\int \frac{\cos(2\theta) + 1}{2}d\theta\\
&=\frac{\sin(2\theta)}{8}+\frac{\theta}{4}
\end{align*}$$
Again
$$\theta = \arcsin(2x)$$
And
$$\begin{align*}
\sin 2\theta &= 2\sin \theta \cos \theta\\
&=4x\sqrt{1-4x^2}
\end{align*}$$
So the answer is
$$\frac{x\sqrt{1-4x^2}}{2}+\frac{\arcsin(2x)}{4}+C$$
|
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|
Integral of $\int \frac{5x+1}{(2x+1)(x-1)}$ I am suppose to use partial fractions
$$\int \frac{5x+1}{(2x+1)(x-1)}$$
So I think I am suppose to split the top and the bottom. (x-1)
$$\int \frac{A}{(2x+1)}+ \frac{B}{x-1}$$
Now I am not sure what to do.
|
Bring the expression $\frac{A}{2x+1}+\frac{B}{x-1}$ to the common denominator $(2x+1)(x-1)$. We get
$$\frac{A(x-1)+B(2x+1)}{(2x+1)(x-1)}.$$
The top is $(A+2B)x -A+B$. We want this to be identically equal to $5x+1$.
The coefficients of $x$ must match, and the constant terms must match.
So we want $A+2B=5$, and $-A+B=1$. Solve for $A$ and $B$. We get $B=2$ and $A=1$.
Check by cross-multiplying that we did not make a mistake.
Now find
$$\int\left(\frac{1}{2x+1}+\frac{2}{x-1}\right)\,dx.$$
Remark: Another popular technique is to stop at $A(x-1)+B(2x+1)=5x+1$. Plug in $x=1$. We get $3B=6$, so $B=2$. Plug in $x=-1/2$ (to kill the $2x+1$ term). We get $(-3/2)A=-3/2$, so $A=1$.
|
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|
Need help with the integral $\int \frac{2\tan(x)+3}{5\sin^2(x)+4}\,dx$ I'm having a problem resolving the following integral, spent almost all day trying.
Any help would be appreciated.
$$\int \frac{2\tan(x)+3}{5\sin^2(x)+4}\,dx$$
|
$$
\begin{aligned}
& \int \frac{2 \tan x+3}{5 \sin ^{2} x+4} d x \\
=& \int \frac{2 \sec ^{2} x \tan x+3 \sec ^{2} x}{5 \tan ^{2} x+4 \sec ^{2} x} d x \\
=& 2 \int \frac{\sec ^{2} x \tan x}{5 \tan ^{2} x+4 \sec ^{2} x} d x+3 \int \frac{\sec ^{2} x}{5 \tan ^{2} x+4 \sec ^{2} x} d x \\
=& \int \frac{2sd s}{9 s^{2}-5}+\int \frac{3d t}{9 t^{2}+4} \quad \textrm{ where }s=\sec x \textrm{ and }t=\tan x.\\
=& \ln \left|9 s^{2}-5\right|+\frac{1}{2} \arctan\left(\frac{3 t}{2}\right)+C \\
=& \ln \left|9 \sec ^{2} x-5\right|+\frac{1}{2} \arctan \left(\frac{3 \tan x}{2}\right)+C
\end{aligned}
$$
|
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|
What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 x}{ \cos x}dx\\
&=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\
&=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\
&=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\
&=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\
&=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\
&=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\
&=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\
&=\ln|\sec x| + \frac{\cos 2x}{4} + C
\end{align}$$
This is the wrong answer, I have went through and back and it all seems correct to me.
|
You have been simplifying things up until line 6 and then kind of turned back into complications. It would be natural to notice that
$$\sin x dx = -d\left(\cos x\right)$$
Then the integral looks as follows:
$$I=-\int\frac{1-\cos^2x}{\cos x}d\left(\cos x\right)$$
So it appears that $\cos x$ plays the role of a variable in its own right, so why not let for example $t=\cos x$. Now
$$I=-\int\frac{1-t^2}{t}dt=-\int\frac{dt}{t}+\int tdt=-\ln t+\frac{t^2}{2}+C$$
Now plug $\cos x$ back into place.
Recognising distinct "reusable" blocks within the expression is the most natural way to arrive at useful substitutions.
|
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|
Formula to estimate sum to nearly correct : $\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$ Estimate the sum correct to three decimal places :
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$$
This problem is in my homework. I find that n = 22 when use Maple to solve this. (with some programming) But, in my homework, teacher said find the formula for this problem.
Thanks :)
|
Averaging the 9th and 10th partial sums will do it, as in robjohn's answer and my comment there. Just for fun, as an alternative to alternating series methods, you could group consecutive terms to give a positive series and then use integral approximation.
$$\begin{align}
S:=\sum_{n=1}^\infty\frac{(-1)^n}{n^3}
&=-1+\sum_{n=1}^\infty\frac{1}{8n^3}-\frac{1}{(2n+1)^3}=-1+\sum_{n=1}^\infty f(n)\\
\end{align}
$$
And then
$$\begin{align}
-1+\sum_{n=1}^{N}\frac{1}{8n^3}-\frac{1}{(2n+1)^3}+\int_{N}^\infty\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx<S<-1+\sum_{n=1}^N\frac{1}{8n^3}-\frac{1}{(2n+1)^3}+\int_{N-1}^\infty\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx\\
\end{align}
$$
The difference between the outer bounds is
$$\begin{align}
\int_{N-1}^{N}\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx
&=\left[{-{\frac1{16x^2}}}+\frac{1}{4(2x+1)^2}\right]_{N-1}^{N}\\
&={-{\frac1{16N^2}}}+\frac{1}{4(2N+1)^2}+{{\frac1{16(N-1)^2}}}-\frac{1}{4(2N-1)^2}\\
\end{align}$$
We'd like this difference to be less than $0.001$. This way we could average the two outer bounds as an estimate for the sum and know that our estimate was within $0.0005$. This happens for $N=5$. So an estimate that will work out to within $0.0005$ is
$$\sum_{n=1}^{11}\frac{(-1)^n}{n^3}+\frac12\int_4^{5}\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx$$
which works out to
$$\sum_{n=1}^{11}\frac{(-1)^n}{n^3}+\frac{24209}{125452800}\approx-0.9016748\ldots$$
|
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|
Spivak Calculus Prologue I'm completely blown away by the difficulty of Spivak. I've managed to work through the first 3 problems, but I feel I'm missing something important to solve these basic inequalities in his 4th problem:
$$x^2 + x + 1 > 2$$
&
$$x^2 + x + 1 > 0$$
any suggestions?
|
Complete the square:
$$
x^2 + x + 1 = (x^2 + x + \tfrac 1 4) + \frac 3 4 = \left(x + \frac 1 2 \right)^2 + \frac 3 4.
$$
That gets you the second one.
For the first one, put everything on one side of the inequality and $0$ on the other side, and procede similarly.
Later addendum in response to vitno's question in the comments below:
In general, the process of completing the square looks like this:
$$
\begin{align}
ax^2 + bx + c & = a\left(x^2 + \frac b a x\right) + c \\[12pt]
& = a\left(x^2 + \frac b a x + \frac{b^2}{4a^2}\right) + c - a\left(\frac{b^2}{4a^2}\right) \tag{$\begin{array}{c} \text{completing} \\ \text{the square}\end{array}$} \\[12pt]
& = a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a}.
\end{align}
$$
Say you have a particular case:
$$
3x^2 + 20 x + 7.
$$
Proceed as follows:
$$
3\left(x^2 + \frac{20}{3} x \right) + 7.
$$
Take half the coefficient of the first-degree term and square it, getting $(10/3)^2$. Add this in the appropriate place, and substract it out later:
$$
3\underbrace{\left(x^2 + \frac{20}{3} x + \left(\frac{10}{3}\right)^2 \right)}_{\text{a perfect square}} + 7 - 3\left(\frac{10}{3}\right)^2
$$
$$
= 3\left(x + \frac{10}{3}\right)^2 - \frac{79}{3}.
$$
Knowing how and when to complete the square is useful.
Remember this: The purpose of completing the square is always to reduce a quadratic polynomial with a first-degree term to a quadratic polynomial with no first-degree term.
|
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|
Another Congruence Proof I've been asked to attempt a proof of the following congruence. It is found in a section of my textbook with Wilson's theorem and Fermat's Little theorem. I've pondered the problem for a while and nothing interesting has occurred to me.
$1^23^2\cdot\cdot\cdot(p-4)^2(p-2)^2\equiv (-1)^{(p+1)/2} (\text{mod} p)$
|
To Prove : $1^2.2^2.3^2....(p-1)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$
We know that
$$k \equiv -(p-k) \pmod p$$
Applying this repeatedly we will get
$$2.4.6.......(p-1) \equiv (-1)^{\frac{p-1}{2}} 1.3.5.......(p-2) \pmod p$$
$$\implies 1.3.5.......(p-2) \equiv (-1)^{\frac{p-1}{2}} 2.4.6.......(p-1) \pmod p$$
$$\implies 1^2.3^2.5^2.......(p-2)^2 \equiv (-1)^{\frac{p-1}{2}} 1.2.3.4.5.6.......(p-1) \pmod p$$
$$\implies 1^2.3^2.5^2.......(p-2)^2 \equiv (-1)^{\frac{p-1}{2}} (p-1)! \pmod p$$
By Wilson's Theorem
$$\implies 1^2.3^2.5^2.......(p-2)^2 \equiv (-1)^{\frac{p-1}{2}} (-1) \pmod p$$
$$\implies 1^2.3^2.5^2.......(p-2)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$$
|
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|
Limit involving $(\sin x) /x -\cos x $ and $(e^{2x}-1)/(2x)$, without l'Hôpital Find:
$$\lim_{x\to 0}\ \frac{\dfrac{\sin x}{x} - \cos x}{2x \left(\dfrac{e^{2x} - 1}{2x} - 1 \right)}$$
I have factorized it in this manner in an attempt to use the formulae.
I have tried to use that for $x$ tending to $0$, $\dfrac{\sin x}{x} = 1$ and that $\dfrac{e^x - 1}x$ is also $1$.
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$$\dfrac{\sin(x)}{x} = 1 - \dfrac{x^2}{3!} + \mathcal{O}(x^4)$$
$$\cos(x) = 1 - \dfrac{x^2}{2!} + \mathcal{O}(x^4)$$
$$\exp(2x) = 1 + 2x + \dfrac{(2x)^2}{2!} + \mathcal{O}(x^3)$$
Hence, $$\dfrac{\dfrac{\sin(x)}{x} - \cos(x)}{\exp(2x) - 1 -2x} = \dfrac{-\dfrac{x^2}{3!} + \dfrac{x^2}{2!} + \mathcal{O}(x^4)}{2x^2 + \mathcal{O}(x^3)} = \dfrac{\dfrac{x^2}{3} + \mathcal{O}(x^4)}{2x^2 + \mathcal{O}(x^3)} = \dfrac{\dfrac13 + \mathcal{O}(x^2)}{2 + \mathcal{O}(x)}$$
Hence, $$\lim_{x \rightarrow 0} \dfrac{\dfrac{\sin(x)}{x} - \cos(x)}{\exp(2x) - 1 -2x} = \dfrac16$$
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"timestamp": "2023-03-29T00:00:00",
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|
Complex numbers lead a trigonometric equality Times ago, I used to think about some trigonometric equalities. Now, I have faced a new one with different one:
Show that if $z^7+1=0$ then cos$(\frac{\pi}{7})$+cos$(\frac{3\pi}{7})$+cos$(\frac{5\pi}{7})=\frac{1}{2}$ wherein $z\in\mathbb C$.
Thanks
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if $z^7+1=0$ then $$(z+1)(z^6-z^5+z^4-z^3+z^2-z+1)=0$$
Putting $z=e^{\frac{i \pi}{7}}$ we have $z + 1 \neq 0$ and $z^7 +1 =0$, then :
$$z^6-z^5+z^4-z^3+z^2-z+1=0$$
This gives :
$$\mathcal Re(z^6-z^5+z^4-z^3+z^2-z+1)=0$$
Since : $\cos \frac{\pi}{7} = - \cos \frac{6\pi}{7}$ and $\cos \frac{2\pi}{7} = - \cos \frac{5\pi}{7}$ abd $\cos \frac{3\pi}{7} = - \cos \frac{4\pi}{7}$ we have :
$$2\left(- \cos \frac{\pi}{7}- \cos \frac{3\pi}{7}- \cos \frac{5 \pi}{7} \right) +1 = 0$$ which gives the desired relation.
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|
Simplifying $\frac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}$ I'm stuck in the follow equation:
$$\dfrac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}$$
As all the bases are equal, I got $\dfrac{3n + 5}{2n - 3}$
Where I've to go now ?
Thanks
EDIT:
Then, my initial idea was totally wrong, starting again, in the right way I got:
$$\dfrac{2^n(2^4 + 2^2 + 2^{-1})}{2^n(2^{-2} + 2^{-1})}$$
but it's still wrong, I didn't get the right idea on the divisions you have shown to me in the answers.
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You’ve nothing to solve. My guess is that you’re supposed to simplify the fraction:
$$\begin{align*}
\dfrac{2^{n + 4} + 2^{n + 2} + 2^{n - 1}}{2^{n - 2} + 2^{n - 1}}&=\frac{2^{n-1}(2^5+2^3+1)}{2^{n-2}(1+2)}\\
&=\frac{2^{n-1}}{2^{n-2}}\cdot\frac{32+8+1}{3}\;,
\end{align*}$$
and you should have no trouble finishing it.
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|
Smallest number in a set
A is the set of seven consequtive two digit numbers, none of these being multiple of 10. Reversing the numbers in set A forms numbers in set B. The difference between the sum of elements in set A and those in set B is 63. The smallest number in set A can be :
I tried to write some sets and reverse them and calculate their value, but I am not able to arrive at the answer.
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Let $A = \{10a+b,10a+b+1,10a+b+2,10a+b+3,10a+b+4,10a+b+5,10a+b+6\}$, where $a \in \{1,2,\ldots,9\}$ and since none of them is divisible by $10$, we have that $b\in\{1,2,3\}$. Then $$B =\{10b+a,10b+a+10,10b+a+20,10b+a+30,10b+a+40,10b+a+50,10b+a+60\}$$
Sum of elements in $A$ is $70a+7b+21$ and sum of elements in $B$ is $70b+7a+210$. We are given $$\vert 63(a-b) - 189 \vert = 63 \implies \vert (a-b) - 3 \vert = 1$$
If $b=1$, we get that $\vert a-4\vert = 1 \implies a=3,5$.
If $b=2$, we get that $\vert a-5\vert = 1 \implies a=4,6$.
If $b=3$, we get that $\vert a-6\vert = 1 \implies a=5,7$.
The smallest number in $A$ is obtained by choosing $a=3$ and $b=1$.
Hence, the smallest number in the set $A$ is $31$.
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$|2^x-3^y|=1$ has only three natural pairs as solutions Consider the equation $$|2^x-3^y|=1$$ in the unknowns $x \in \mathbb{N}$ and $y \in \mathbb{N}$. Is it possible to prove that the only solutions are $(1,1)$, $(2,1)$ and $(3,2)$?
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Assume that $x>3$ and $y>2$.
$3^2=1\pmod{8}$. Since $3\not=-1\pmod{8}$ we know that $3^y\not=-1\pmod{8}$. Thus, if $|2^x-3^y|=1$, we must have $2^x-3^y=-1$ and $y$ must be even. Thus, we get that
$$
2^x=(3^{y/2}-1)(3^{y/2}+1)\tag{1}
$$
The only factors of a power of $2$ are other powers of $2$, and the only powers of two that differ by $2$ are $2$ and $4$, but since $y>2$, we have $3^{y/2}-1>2$ and $3^{y/2}+1>4$.
Therefore, there are no $x>3$ and $y>2$ so that $|2^x-3^y|=1$.
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Two proofs I'm having difficulty with I've been given an assignment. Almost done except the last two are tripping me up. They are as follows:
1) if $2x^2-x=2y^2-y$ then $x=y$
2) if $x^3+x=y^3+y$ then $x=y$
I imagine they use a similar tactic as they both involve powers, but I've tried factoring,completing the square, difference of squares and difference of cubes and nothing seems to help.
Any hints would be appreciated.
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Question $1$: The equation is equivalent to $2x^2-2y^2=x-y$. The left-hand side factors as $2(x+y)(x-y)$. So our equation can be rewritten as
$$2(x+y)(x-y)=x-y.$$
Thus any pair $(x,y)$ such that $x\ne y$ and $2(x+y)=1$ is a counterexample to the assertion that $x$ must be equal to $y$. This was pointed out by ncmathsadist.
Edit: It turns out that one is supposed to show that the only integer solutions have $x=y$. This follows from the above calculation, since $2(x+y)=1$ has no integer solutions.
Question $2$: We look at the question as amended. We can factor and rewrite the equation as $(x-y)(x^2+xy+y^2)=-(x-y)$. If $x\ne y$, we can cancel $x-y$, and obtain $x^2+xy+y^2=-1$.
But the equation $x^2+xy+y^2=-1$ has no real solutions, since $x^2+xy+y^2\ge 0$ for all real $x$ and $y$. One way to see this is to complete the square, getting
$$x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2.$$
The right-hand side is clearly never negative for real $x$ and $y$. So it is true that the only solutions of the original equation have $x=y$.
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Some method to solve $\int \frac{1}{\left(1+x^2\right)^{2}} dx$ and some doubts. First approach.
$\int \frac{1}{1+x^2} dx=\frac{x}{1+x^2}+2\int \frac{x^2}{\left(1+x^2\right)^2} dx=\frac{x}{1+x^2}+2\int \frac{1}{1+x^2}dx-2\int \frac{1}{\left(1+x^2\right)^2}dx$
From this relationship, I get:
$2\int \frac{1}{\left(1+x^2\right)^2}dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}dx$ Then:
$\int \frac{1}{\left(1+x^2\right)^2}dx=\frac{1}{2}\left[\frac{x}{1+x^2}+\arctan x\right]+C$ This is a recursive solution.
Second approach.
$x=\tan t$ in $t\in (- \pi/2, \pi/2)$, i.e. $t=\arctan x$, then $dx=(1+x^2) dt$.
$\int \frac{1}{\left(1+x^2\right)^2}dx=\int \frac{1}{1+x^2}dt=\int \frac{\cos^2t}{\sin^2t+\cos^2t}dt=\int \cos^2t dt=\frac{1}{2}\int \left(1+\cos 2t \right) dt=\frac{t}{2}+\frac{1}{4}\sin 2t$
This result can be rewritten (using trigonometric formulas):
$\frac{t}{2}+\frac{1}{4}\sin 2t=\frac{t}{2}+\frac{1}{2}\sin t \cos t$
From $\cos^2 t=\frac{1}{1+x^2}$, I have:
$|\cos t|=\sqrt{\frac{1}{1+x^2}}$ but in $t\in (- \pi/2, \pi/2)$, $|\cos t|=\cos t$. So:
$\cos t=\sqrt{\frac{1}{1+x^2}}$. Now I have a problem: $|\sin t|=\sqrt{\frac{1}{1+x^2}}$, but $|\sin t|\neq \sin t$ for $t\in (- \pi/2, \pi/2)$. Any suggestions, please? This integral can be solved in other ways?
Thanks.
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One way to make the trigonometric solution substitution solution end nicely, without worries about signs, is to note that
$$\sin t \cos t =\frac{\sin t}{\cos t} \cos^2 t.$$
Since $\cos^2 t=\frac{1}{\sec^2 t}=\frac{1}{1+\tan^2 t}$, we get that
$$\sin t \cos t =\tan t\frac{1}{1+\tan^2 t}=\frac{x}{1+x^2}.$$
Another way: The following is closer to your calculation. Use the fact that $\sin t\cos t$ has the same sign as $\tan t$ to resolve ambiguities of sign. We know from your calculation that $\sin t \cos t =\pm \frac{x}{1+x^2}$. But this has the same sign as $\tan t$, that is, the same sign as $x$. That resolves the $\pm$ problem in favour of $\frac{x}{1+x^2}$, which has the same sign as $x$.
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Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives
$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$
and on $abc$ which gives
$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$
Since both inequalities have the same righthand side, I have tried to deduce something about the lefthand sides, but to no avail. Can somebody help me, please? I am sure it is something simple I have missed.
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Norbert's answer explains the case when $a,b,c$ are not the sides of a triangle.
Let $x,y,z$ be as in this picture, where $AB=c, BC=a, CA=b$.
Then $(a,b,c)=(y+z,x+z,x+y)$.$\,$ Inequality becomes $$8xyz\le (x+y)(y+z)(z+x),$$
which as Norbert says is true by $2\sqrt{xy}\le x+y$ (proof: $\Leftrightarrow (\sqrt{x}-\sqrt{y})^2\ge 0$) and $2\sqrt{yz}\le y+z$ and $2\sqrt{zx}\le z+x$.
Some overkill approaches: by Hölder's inequality: $$(x+y)(y+z)(z+x)\ge (\sqrt[3]{xyz}+\sqrt[3]{yzx})^3=8xyz$$
By AM-GM: $$(x+y)(y+z)(z+x)-xyz=(x+y+z)(xy+yz+zx)$$ $$\ge (3\sqrt[3]{xyz})(3\sqrt[3]{xy\cdot yz\cdot zx})=9xyz$$
By Cauchy-Schwarz: $$(z+x+y)(xy+yz+zx)\ge (\sqrt{z}\cdot \sqrt{xy} + \sqrt{x}\cdot \sqrt{yz}+\sqrt{y}\cdot \sqrt{zx})^2=9xyz$$
By Muirhead's inequality:
$$(x+y)(y+z)(z+x)-2xyz=\sum_{\text{sym}}x^2y^1z^0\ge \sum_{\text{sym}}x^1y^1z^1=6xyz,$$
because $(2,1,0)\succ (1,1,1)$. Last one is also true by rearrangement inequality because hint: we can let WLOG $x\ge y\ge z$ and $xy \ge zx\ge yz$.
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Trigonometric Identities To Prove
*
*$\tan\theta+\cot\theta=\dfrac{2}{\sin2\theta}$
Left Side:
$$\begin{align*}
\tan\theta+\cot\theta={\sin\theta\over\cos\theta}+{\cos\theta\over\sin\theta}={\sin^2\theta+\cos^2\theta\over\cos\theta\sin\theta}
= \dfrac{1}{1\sin\theta\cos\theta}
\end{align*}$$
Right Side:
$$\begin{align*}
\dfrac{2}{\sin2\theta}=\dfrac{2}{2\sin\theta\cos\theta}=\dfrac{1}{1\cos\theta\sin\theta}
\end{align*}$$
I got it now. Thanks!
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I'll just put together what you wrote...
$$\begin{align*}
\tan\theta+\cot\theta=\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\cos\theta \cdot\sin\theta} = \dfrac{1}{\cos \theta \cdot \sin \theta}
\end{align*}$$
Where the penultimate inequality is what you should have written.
Can you take it from here?
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How to evaluate $\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx$ by hand How can I evaluate$$\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx\text{ and }\int_0^\infty\frac{\sin x}{e^x-1}\,\mathrm dx.$$
Thanks in advance.
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For the second one,
$$ \begin{align*}
\int_{0}^{\infty} \frac{\sin x}{e^x - 1} \; dx
&= \int_{0}^{\infty} \frac{\sin x \, e^{-x}}{1 - e^{-x}} \; dx \\
&= \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} \sin x \, e^{-nx} \right) \; dx \\
&\stackrel{\ast}{=} \sum_{n=1}^{\infty} \int_{0}^{\infty} \sin x \, e^{-nx} \; dx \\
&= \sum_{n=1}^{\infty} \frac{1}{1+n^2},
\end{align*}$$
where the starred identity is justified by the following formula
$$ \begin{align*}
\int_{0}^{\infty} \frac{\sin x}{e^x - 1} \; dx
&= \int_{0}^{\infty} \left( \frac{1 - e^{-Nx}}{1 - e^{-x}} e^{-x} + \frac{e^{-Nx}}{e^x - 1} \right) \sin x \; dx \\
&= \sum_{n=1}^{N} \int_{0}^{\infty} \sin x \, e^{-nx} \; dx + \int_{0}^{\infty} \frac{\sin x \, e^{-Nx}}{e^x - 1} \; dx,
\end{align*}$$
together with the dominated convergence theorem. Now the resulting infinite summation can be evaluated in numerous ways. For example, exploiting identities involving the digamma function,
$$ \sum_{n=1}^{\infty} \frac{1}{1+n^2} = \frac{1}{2i} \sum_{n=1}^{\infty} \left( \frac{1}{n-i} - \frac{1}{n+i} \right) = \frac{\psi_0(1+i) - \psi_0(1-i)}{2i} = -\frac{1}{2} + \frac{\pi}{2} \coth \pi. $$
Similar techniques apply to the first integral.
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How to prove that $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$ Help me prove $a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)\geq6abc$
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LHS=$\sum a^2\ +\ \sum(ab)^2$
Applying A.M. ≥ G.M.,
$\sum a^2 ≥ 3(abc)^{\frac{2}{3}} $
$\sum (ab)^2 ≥ 3(abc)^{\frac{4}{3}} $
Taking summation,$ LHS ≥ 3((abc)^{\frac{2}{3}} + (abc)^{\frac{4}{3}}) $
But $(abc)^{\frac{2}{3}} + (abc)^{\frac{4}{3}}$ ≥ 2 abc (applying A.M. ≥ G.M.,)
LHS ≥ 3(2abc)
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Sum of the series : $1 + 2+ 4 + 7 + 11 +\cdots$ I got a question which says
$$ 1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$$
I got the solution by dividing by $7$ and subtracting it from original sum. Repeated for two times.(Suggest me if any other better way of doing this).
However now i am interested in understanding the series 1,2,4,7,11,..... In which the difference of the numbers are consecutive natural numbers.
How to find the sum of
$1+2+4+7+11+\cdots nterms$
This is my first question in MSE. If there are some guidelines i need to follow, which i am not, please let me know.
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Note that starting with your first element, $a_0=1$; to get to $a_1=2$ we sum $1$, to get to $a_2=4$, we sum $2$, to get to $a_3=7$, we sum $3$. So in general, we can say that
$$a_{n}=a_{n-1}+n$$
This is
$$
\begin{align}
2&=1+1 \\[8pt]
4& =2+2\\[8pt]
7&=4+3\\[8pt]
11& =7+4\\[8pt]
\cdots&=\cdots\\[8pt]
a_n&=a_{n-1}+n
\end{align}
$$
We can get several solutions to you problem, I will share $2$:
Solution 1 Use the recursion to obtain a closed formula:
Since we know that $a_{n}=a_{n-1}+n$ we can write
$a_{n-1}=a_{n-2}+n-1$
so we get $$a_{n}=a_{n-2}+(n-1)+n$$
Repeating this process, we get that
$$a_{n}=a_{n-3}+(n-2)+(n-1)+n$$
$$a_{n}=a_{n-4}+(n-3)+(n-2)+(n-1)+n$$
$$a_{n}=a_{n-5}+(n-4)+(n-3)+(n-2)+(n-1)+n$$
...so in general we can say that
$$a_n=a_{n-k}+(n-k+1)+\cdots+n$$ for any $k$ a natural number.
(actually we should be proving the above by induction, but let it be)
So we can choose $k=n$, which means...
$$a_n=a_{n-n}+(n-n+1)+\cdots+n$$
$$a_n=a_0+(1+\cdots+n)$$
$$a_n=1+\frac{n(n+1)}{2}$$
Solution 2
Starting from $$a_{n}=a_{n-1}+n$$ we use generating functions:
$$\eqalign{
& \sum\limits_{n = 1}^\infty {{a_n}} {x^n} = \sum\limits_{n = 1}^\infty {{a_{n - 1}}} {x^n} + \sum\limits_{n = 1}^\infty n {x^n} \cr
& \sum\limits_{n = 1}^\infty {{a_n}} {x^n} = x\sum\limits_{n = 1}^\infty {{a_{n - 1}}} {x^{n - 1}} + x\sum\limits_{n = 1}^\infty {n{x^{n - 1}}} \cr
& \sum\limits_{n = 0}^\infty {{a_n}} {x^n} - {a_0} = x\sum\limits_{n = 0}^\infty {{a_n}} {x^n} + x\frac{d}{{dx}}\sum\limits_{n = 0}^\infty {{x^n}} \cr
& A\left( x \right) - 1 = xA\left( x \right) + x\frac{d}{{dx}}\frac{1}{{1 - x}} \cr
& A\left( x \right) - xA\left( x \right) = 1 + \frac{x}{{{{\left( {1 - x} \right)}^2}}} \cr
& \left( {1 - x} \right)A\left( x \right) = 1 + \frac{x}{{{{\left( {1 - x} \right)}^2}}} \cr
& A\left( x \right) = \frac{1}{{1 - x}} + \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr} $$
The generating sequence of $f(x)=\frac{1}{{1 - x}}$ is $a_n=1$, while the generating sequence of $g(x)=\frac{x}{{{{\left( {1 - x} \right)}^3}}}$ can be obtained by diferentiation of the first one:
$$\eqalign{
& \sum\limits_{n = 0}^\infty {{x^n}} = \frac{1}{{1 - x}} \cr
& \sum\limits_{n = 1}^\infty {n{x^{n - 1}}} = \frac{1}{{{{\left( {1 - x} \right)}^2}}} \cr
& \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right){x^{n - 2}}} = \frac{2}{{{{\left( {1 - x} \right)}^3}}} \cr
& \sum\limits_{n = 2}^\infty {\frac{{n\left( {n - 1} \right)}}{2}{x^{n - 1}}} = \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr
& \sum\limits_{n = 1}^\infty {\frac{{n\left( {n + 1} \right)}}{2}{x^n}} = \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr} $$
so we finally get that
$$a_n=1+{\frac{{n\left( {n + 1} \right)}}{2}}$$
ADD I forgot to add the sum of the $a_n$s!
You need to evaluate
$$\sum_{k=0}^{n-1} a_k=\sum_{k=0}^{n-1} 1+\sum_{k=0}^{n-1}\frac{k(k+1)}{2}$$
$$\sum_{k=0}^{n-1} a_k=n+\sum_{k=0}^{n}\frac{k(k-1)}{2}$$
$$\sum_{k=0}^{n-1} a_k=n+\sum_{k=0}^{n}{k\choose 2}$$
Using the binomial identity
$$\sum_{k=0}^n {k\choose l}={{n+1}\choose {l+1}}$$ we get
(you can find how to obtain it here
$$\sum_{k=0}^{n-1} a_k=n+{n+1\choose 3}$$
$$\sum_{k=0}^{n-1} a_k=n+\frac{(n+1)n(n-1)}{6}$$
which is what you wanted.
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|
Evaluating $\int (x^6+x^3)\sqrt[3]{x^3+2}dx$ I am trying to evaluate:
$$\int (x^6+x^3)\sqrt[3]{x^3+2} \ \ dx$$
My solution:
$$\int (x^5+x^2)\sqrt[3]{x^6+2x^3} \ \ dx$$
Let $$(x^6+2x^3) = t^3 \ \ \text{and} \ \ (x^5+x^2) \ \ dx = \frac{1}{2}t^2 \ \ dt$$
$$\frac{1}{2}\int t^2\cdot t \ \ dt = \frac{1}{2}.\frac{t^4}{4}+C $$
So $$\int (x^5+x^2)\sqrt[3]{x^6+2x^3} \ \ dx = \frac{1}{8}(x^6+2x^3)^{{4}/{3}}+C$$
Is that right? And is there a different way ?
|
The $t$ stuff is not necessary. You can directly let $u=x^6+2x^3$. Then $(x^5+x^2)\,dx=\frac{1}{6}\,du$. But the initial step was the key one.
|
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|
A question about arithmetic progressions, a.k.a. arithmetic sequences What is the relation between the general formula of the sum of $n$ terms of an arithmetic progression, $a n^2+b n+c$, and the first term $a+b+c$ and the common difference?
|
If the sum of $n$ terms of an arithmetic sequence is
$$
s(n)=\color{#C00000}{an^2+bn+c}\tag{1}
$$
then $c$ is the sum of $0$ terms; that is,
$$
c=s(0)=0\tag{2}
$$
The general term of the arithmetic sequence would be
$$
\begin{align}
a(n)
&=s(n)-s(n-1)\\
&=(b-a)+(2a)n\tag{3}
\end{align}
$$
The first term of the arithmetic sequence would be
$$
a(1)=\color{#C00000}{a+b}\tag{4}
$$
Thus, the common difference of the arithmetic sequence is
$$
\begin{align}
a(n)-a(n-1)
&=((b-a)+(2a)n)-((b-a)+(2a)(n-1))\\
&=\color{#C00000}{2a}\tag{5}
\end{align}
$$
As for a relation among $(1)$, $(4)$, and $(5)$, we get
$$
\color{#C00000}{an^2+bn+c}=n(\color{#C00000}{a+b})+\frac{n^2-n}{2}\color{#C00000}{2a}\tag{6}
$$
|
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|
General solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$ Airy differential equation.
$y''(x)=xy(x)$
$y'''(x)=y(x)+x y'(x)$
$y'^v(x)=x^2y(x)+2 y'(x)$
$y^v(x)=4xy(x)+x^2 y'(x)$
$y^{(6)}(x)=(x^3+4)y(x)+6x y'(x)$
.
.
$y^{(n)}(x)=A_n(x)y(x)+B_n(x) y'(x)$
Where $A_n(x)$ and $B_n(x)$ are polynomials
($y^{(n)}(x)$ means $n$-th order derivative of $y(x)$ )
I have found that
$C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$
If $C_2(x)=x$ ;$C_3(x)=1$ ; $C_4(x)=x^2$ are initial condition
$C_n(x)=A_n(x)$
If $C_2(x)=0$;$C_3(x)=x$;$C_4(x)=2$ are initial condition
$C_n(x)=B_n(x)$
How can we find the general solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$ ?
Can we express $C_n(x)$ as Airy function?
Thanks for answers.
$EDIT:$
I tried generating function method as Sam recommended.
$$g(z,x)=\sum_{n=3}^\infty z^n C_n(x)$$
$$g(z,x)=z^3C_3(x)+z^4C_4(x)+\sum_{n=5}^\infty z^n C_n(x)$$
$$g(z,x)=z^3C_3(x)+z^4C_4(x)+\sum_{n=3}^\infty z^{n+2} C_{n+2}(x)$$
$$g(z,x)=z^3C_3(x)+z^4C_4(x)+\sum_{n=3}^\infty z^{n+2}(xC_n(x)+nC_{n-1}(x))$$
$$g(z,x)=z^3C_3(x)+z^4C_4(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+z^4\sum_{n=3}^\infty nz^{n-2}C_{n-1}(x)$$
$$g(z,x)=z^3C_3(x)+z^4C_4(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+3z^5C_2(x)+z^4\sum_{n=4}^\infty nz^{n-2}C_{n-1}(x)$$
$$g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+z^4\sum_{n=3}^\infty (n+1)z^{n-1}C_{n}(x)$$
$$g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+z^4\sum_{n=3}^\infty nz^{n-1}C_{n}(x)+z^3\sum_{n=3}^\infty z^{n}C_{n}(x)$$
We can get that
$$\frac{\partial g(z,x)}{\partial z}=\sum_{n=3}^\infty n z^{n-1} C_n(x)$$
And finally,
$$g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)+xz^2g(z,x)+z^4\frac{\partial g(z,x)}{\partial z}+z^3g(z,x)$$
$$-z^4\frac{\partial g(z,x)}{\partial z}- (z^3+xz^2-1)g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)$$
$$\frac{\partial g(z,x)}{\partial z}+\frac{ z^3+xz^2-1}{z^4}g(z,x)=-\frac{1}{z}C_3(x)-C_4(x)-3zC_2(x)$$
I will let you know after solving the first order linear differential equation.
|
Let $G(z)$ be the exponential generating function defining $G(z) = \sum_{n=0}^\infty C_n \frac{z^n}{n!}$. The recurrence equation $C_{n+2} = x C_n + n C_{n-1}$ translates into a differential equation:
$$
G^{\prime\prime}(z)- (x+z) G(z)= C_2 - x C_0
$$
This equation admits a closed form solution:
$$
G(z) = \kappa_1 Ai(x+z) + \kappa_2 Bi(x+z) + (x+z)^2(C_2 - x C_0) h(x+z)
$$
where
$$
h(u) = \frac{1}{2} \cdot {}_{0}F_{1}\left(\frac{2}{3}, \frac{u^3}{9}\right) \cdot {}_1 F_{2}\left(\frac{2}{3}; \frac{4}{3}, \frac{5}{3}; \frac{u^3}{9}\right) -
{}_{0}F_{1}\left(\frac{4}{3}, \frac{u^3}{9}\right) \cdot {}_1 F_{2}\left(\frac{1}{3}; \frac{2}{3}, \frac{4}{3}; \frac{u^3}{9}\right)
$$
The initial condition $G(0) = C_0$ and $G^\prime(0) = C_1$ determines $\kappa_1$ and $\kappa_2$.
Added: After substitution of $C_0 = 1$, $C_1=0$, $C_2=x$ for $A_n(x)$ we get:
$$
\sum_{n=0}^\infty \frac{z^n}{n!} A_n(x) = \pi \left( Bi^\prime(x) Ai(x+z) - Ai^\prime(x) Bi(x+z) \right)
$$
Similarly, for $B_n(x)$ with $C_0=0$, $C_1=1$, $C_2=0$ we obtain:
$$
\sum_{n=0}^\infty \frac{z^n}{n!} B_n(x) = \pi \left( Ai(x) Bi(x+z) - Bi(x) Ai(x+z) \right)
$$
The simplification is made possible by simple expression of ${}_0F_1$ in terms of Airy functions:
$$\begin{eqnarray}
{}_{0}F_1\left(\frac{2}{3}; \frac{u^3}{9}\right) &=& -\frac{\Gamma\left(-\frac{1}{3}\right)}{ 3^{5/6}} \frac{1}{2} \left( \sqrt{3} Ai(u) + Bi(u) \right) \\
{}_{0}F_1\left(\frac{4}{3}; \frac{u^3}{9}\right) &=& -\frac{\Gamma\left(-\frac{2}{3}\right)}{ 3^{5/6}} \frac{1}{u} \left( \sqrt{3} Ai(u) - Bi(u) \right)
\end{eqnarray} $$
Of course, appearance of $y(x+z)$ should be anticipated, knowing the origin of the problem:
$$
\left(\sum_{n=0}^\infty A_n(x) \frac{z^n}{n!} \right) y(x) + \left(\sum_{n=0}^\infty B_n(x) \frac{z^n}{n!} \right) y^\prime(x) = \sum_{n=0}^\infty \frac{z^n}{n!} y^{(n)}(x) = y(x+z)
$$
|
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|
Solving $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$ $$E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$$
I got no idea how to find the solution to this. Can someone put me on the right track?
Thank you very much.
|
We have
\begin{eqnarray*}
E&=&\frac{\cos 10^\circ-\sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\\
&=&4\frac{(1/2)\cos 10^\circ-(\sqrt{3}/2)\sin 10^\circ}{2\sin 10^\circ \cos 10^\circ}\\
&=&4\frac{\cos 60^\circ\cos 10^\circ-\sin 60^\circ\sin 10^\circ}{\sin 20^\circ}\\
&=&4\frac{\cos(60^\circ+10^\circ)}{\sin 20^\circ}\\
&=&4\frac{\cos 70^\circ}{\sin 20^\circ}\\
&=&4\frac{\sin 20^\circ}{\sin 20^\circ}\\
&=&4.
\end{eqnarray*}
|
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|
What can be the possible value of $a+b+c$ in the following case? What can be the possible value of $a+b+c$ in the following case?
$$a^{2}-bc=3$$
$$b^{2}-ca=4$$
$$c^{2}-ab=5$$
$0, 1, -1$ or $1/2$?
After doing $II-I$, $III-I$ and $III-II$, I got,
$$(a+b+c)(b-a)=1$$
$$(a+b+c)(c-a)=2$$
$$(a+b+c)(c-b)=1$$
I'm unable to solve further, please help.
|
$(a-b)^2+(b-c)^2+(c-a)^2=2(3+4+5)=24$
Now $a-b=\dfrac1{a+b+c}$ etc.
Putting the values of $a-b, b-c, c-a;$
$$\frac1{(a+b+c)^2}(1^2+2^2+1^2)=24$$
$$\implies(a+b+c)^2=\frac14$$
$$\implies a+b+c=\pm\frac12$$
|
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|
How to solve this System of Polynomial Equations? I have to complete a summer packet of 90 Algebra 2 questions. I have completed 89 of them, the only one I could not get was this. I know the answer is $y = \frac {47}2$, $\frac 17$ according to WolframAlpha, but I have no idea how to reach that answer, my Algebra 2 Honors teacher couldn't figure it out.
Directions: Use substitution or linear combination to solve each system.
$$\dfrac 3{(x-1)} + \dfrac 4{(y+2)} = 2$$
$$\dfrac 6{(x-1)} - \dfrac 7{(y+2)} = -3$$
|
How about this approach: It's not hard to see that $2 \cdot \left(\frac{3}{x-1}\right) = \frac{6}{x-1}$. Therefore, we can do a linear combination approach:
Add $-2$ times the first equation to the second equation. We get:
$$ \left(\frac{6}{x-1} - \frac{7}{y+2} \right) - 2 \left( \frac{3}{x-1} + \frac{4}{y+2} \right) = -3 - 2\cdot 2.$$
Simplifying gives $$\frac{-15}{y+2} =\frac{-7-8}{y+2} + \frac{6-6}{x-1} = -7.$$
Can you take it from here?
|
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|
Trigonometric eigenvalue equation In solving an eigenvalue problem, I've come to following equation ($\lambda=1$):
$$\begin{pmatrix}
\cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta)
\end{pmatrix}\begin{pmatrix}
a \\ b
\end{pmatrix}=\begin{pmatrix}
a \\ b
\end{pmatrix}$$
Now, the solution says, "This matrix equation can be reduced to a single equation":
$$a \sin(\frac{1}{2}\theta)=b\cos(\frac{1}{2}\theta)$$
I've been rotating trigonometric formulas to get to this, but I simply can't find the way. Could you help me with this, or at least give me a hint?
|
$$\begin{pmatrix}
\cos(\theta) & \sin(\theta) \\ \sin\theta & -\cos\theta
\end{pmatrix}\begin{pmatrix}
a \\ b
\end{pmatrix}\\=\begin{pmatrix}
a\cos\theta+b\sin\theta \\ a\sin\theta-b\cos\theta
\end{pmatrix}=\begin{pmatrix}
a \\ b
\end{pmatrix}$$
Which gives us the following system of equations:
$$a\cos\theta+b\sin\theta=a\tag{1}$$
$$a\sin\theta-b\cos\theta=b\tag{2}$$
So from $(1)$ you have:
$$b\sin\theta=a(1-\cos\theta)$$
$$2b\sin(\frac \theta{2})\cos(\frac \theta{2})=a(2\sin^2(\frac \theta{2}))$$
$$b\cos(\frac \theta{2})=a\sin(\frac \theta{2})$$
|
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|
For which angles we know the $\sin$ value algebraically (exact)? For example:
*
*$\sin(15^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$
*$\sin(18^\circ) = \frac{\sqrt{5}}{4} - \frac{1}{4}$
*$\sin(30^\circ) = \frac{1}{2}$
*$\sin(45^\circ) = \frac{1}{\sqrt{2}}$
*$\sin(67 \frac{1}{2}^\circ) = \sqrt{ \frac{\sqrt{2}}{4} + \frac{1}{2} }$
*$\sin(72^\circ) = \sqrt{ \frac{\sqrt{5}}{8} + \frac{5}{8} }$
*$\sin(75^\circ) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$
*?
Is there is a list of known exact values of $\boldsymbol \sin$ somewhere?
Found a related post here.
|
We use radian notation. Every rational multiple of $\pi$ has trigonometric functions that can be expressed using the ordinary arithmetic operations, plus $n$-th roots for suitable $n$.
This is almost immediate if we use complex numbers, since $(\cos(2\pi/n)+i\sin(2\pi/n)^n=1$.
But it is known, for example, that there is no expression for $\sin(\pi/9)$ that starts from the integers, and uses only the ordinary operations of arithmetic and roots in which every component is real.
The following more restricted problem has a long history because of its close connection with the problem of which angles are constructible by straightedge and compass.
Let $\theta=\frac{m}{n}\pi$, where $m$ and $n$ are relatively prime. Restrict our algebraic operations to the ordinary operations of arithmetic, plus square roots only, The trigonometric functions of $\theta$ are so expressible iff $n$ has the form
$$n=2^k p_1p_2\cdots p_s,$$
where the $p_i$ are distinct Fermat primes.
A Fermat prime is a prime of the form $2^{\left(2^t\right)}+1$. There are only five Fermat primes known: $3$, $5$, $17$, $257$, and $65537$. It is not known whether or not there are more than five.
|
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|
Prove that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers Show that $(n+\sqrt{n^2 -1})^k$ will always be of the form$ (t+\sqrt{t^2 -1})$ where $n$, $k$, $t$ are natural numbers
|
I enjoyed working this out-cool question man. Here's a straightforward inductive proof:
I'm writing $c$ instead of $n$.
Inductively assume $(c + \sqrt{c^2 - 1})^k = a + b \sqrt{c^2 - 1}$, where $b^2(c^2 - 1) = a^2 - 1$ (the base case is clear). That is, not only can you write it as $t + \sqrt{t^2 - 1}$, but also $t^2 - 1$ is a square times $c^2 - 1$ (I guessed this by working out the first $k$'s for $c =2,3$).
Now to see $k \Rightarrow k+1$, we have $$(c + \sqrt{c^2 - 1})^{k+1} = (a + b \sqrt{c^2 - 1})(c + \sqrt{c^2 - 1}) $$$$= (ac + b(c^2 - 1)) + (a + cb)\sqrt{c^2 - 1}$$It suffices to check that $$(ac + b(c^2 - 1))^2 - 1 = (a + cb)^2(c^2 - 1)$$Expanding both sides gives $$c^2 a^2 + 2c(c^2 - 1)ab + (c^2 - 1)^2 b^2 - 1 = (c^2 - 1)a^2 + 2c(c^2 - 1)ab + c^2(c^2 - 1)b^2$$Cancelling the $ab$ terms, pulling the $a^2$ term to the LHS, and the $b^2$ terms to the RHS, we get $$a^2 - 1=b^2(c^2 - 1) $$our inductive hypothesis.
|
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|
How to show x and y are equal? I'm working on a proof to show that f: $\mathbb{R} \to \mathbb{R}$ for an $f$ defined as $f(x) = x^3 - 6x^2 + 12x - 7$ is injective. Here is the general outline of the proof as I have it right now:
Proof: For a function to be injective, whenever $x,y \in A$ and $x\neq y$, then $f(x) \neq f(y)$, i.e., where $A$, $B$ are finite sets, every two elements of $A$ must have distinct images in $B$, which also implies that there must be at least as many elements in $B$ as in $A$ such that the cardinality of $A$ is less than or equals the cardinality of $B$.
We shall prove the contra-positive: If $\exists$ $f(x) = f(y)$, then $x=y.$
Let $x^3 - 6x^2 + 12x - 7 = y^3 - 6y^2 + 12y - 7$.
Then by addition and some algebra, we get $x(x^2 - 6x + 12) = y(y^2 - 6y + 12)$
This feels dumb to ask but how do I continue to finally get the result that $x = y$?
|
Note that $w^3-6w^2+12w-8=(w-2)^3$. So $w^3-6w^2+12w-7=(w-2)^3+1$.
So we want to show that if $(x-2)^3+1=(y-2)^3+1$ then $x=y$.
Equivalently, we want to show that if $(x-2)^3=(y-2)^3$ then $x=y$. This is easy, the cube function is increasing.
Remark: We can use the basic algebra of ordered fields to show that if $s^3=t^3$ then $s=t$. For $s^3-t^3=(s-t)(s^2+st+t^2)$. But
$$2(s^2+st+t^2)=(s+t)^2+s^2+t^2,$$
so $s^2+st+t^2$ can only be $0$ when $s=t=s+t=0$.
|
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|
How do I proceed with these quadratic equations? The question is
$$ax^2 + bx + c=0 $$ and $$cx^2+bx+a=0$$ have a common root, if $b≠ a+c$, then what is $$a^3+b^3+c^3$$
|
The value of $a^3+b^3+c^3$ is not determined. Just choose $a=c$. But leaving out the condition $a\ne c$ is probably an oversight, so assume from now on that $a\ne c$.
If $q$ is a common root, then $aq^2+bq+c=cq^2+bq+a=0$. Subtracting, we find that $(a-c)q^2-(a-c)=0$. Since $a\ne c$, we get $q^2=1$.
We cannot have $q=-1$, for that implies that $b=a+c$. We are left with $q=1$, which gives $a+b+c=0$. Conversely, if $a+b+c=0$, then $1$ is a common root of the two equations.
That still leaves many possibilities for the value of $a^3+b^3+c^3$. For example, let $a=1$, $b=-3$, $c=2$. Then $a^3+b^3+c^3=-18$. Let $a=1$, $b=-4$, $c=2$. Then $a^3+b^3+c^3=-36$.
However, we can say something interesting about $a^3+b^3+c^3$ in the case $a\ne c$.
Use the general identity
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-bc-ca-ab),\tag{$1$}$$
which can be verified by multiplying out. Since in our case $a+b+c=0$, we conclude that
$$a^3+b^3+c^3=3abc.$$
Perhaps this is the intended answer.
|
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|
Proving $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
Possible Duplicate:
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove trigonometry identity?
If $A$, $B$, and $C$ are to be taken as the angles of a triangle, then I beg someone to help me the proof of
$$\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}.$$
Thanks!
|
How about a proof with a geometric flavor?
Let $a$, $b$, $c$ be the sides that oppose angles $A$, $B$, $C$, respectively. By the Law of Sines,
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
where $d$ is the circumdiameter of the triangle. If we conveniently scale the triangle so that $d=1$, then we can say simply that
$$a = \sin A \qquad b = \sin B \qquad c = \sin C$$
This is a common simplification technique, as it nicely blurs the distinction between edges and angles, giving us things like this wonderfully-symmetric area formula:
$$\mathrm{Area} \; = \frac{1}{2}a b c \qquad \left(\;=\frac{1}{2}ab\sin C =\frac{1}{2}ac\sin B = \frac{1}{2}bc\sin A\right)$$
(When you have this mindset, you can't look at the expression "$\sin A + \sin B + \sin C$" and not think, "That's perimeter!" ... and then you find yourself pursuing proof approaches like this one.) In what follows, I'll continue to use "$a$", "$b$", "$c$", because they're more compact than "$\sin A$", etc, but you should read them as "$\sin A$", etc.
By the Law of Cosines,
$$\cos{C} = \frac{a^2+b^2-c^2}{2ab}$$
By the half-angle formula for cosines,
$$\cos^2\frac{C}{2} = \frac{1+\cos{C}}{2}=\frac{a^2+2ab+b^2-c^2}{4ab}=\frac{(a+b)^2-c^2}{4ab}=\frac{(a+b+c)(a+b-c)}{4ab}$$
Likewise for $\cos(A/2)$ and $\cos(B/2)$, so that
$$\cos^2\frac{A}{2} \cos^2\frac{B}{2} \cos^2\frac{C}{2}=\frac{(a+b+c)^2}{4a^2b^2c^2}\cdot \frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16}$$
The conveniently-separated second factor just happens to be Heron's formula for the square of the area of the triangle; re-writing the area in wonderfully-symmetric form gives ...
$$\cos^2\frac{A}{2} \cos^2\frac{B}{2} \cos^2\frac{C}{2}=\frac{(a+b+c)^2}{4a^2b^2c^2}\cdot \left(\frac{1}{2}abc\right)^2 = \frac{1}{16}\left(a+b+c\right)^2$$
We can now clear the fraction, expand the symbols "$a$", "$b$", "$c$" as the sines they represent, and take square roots (secure in the knowledge that none of the trig values is negative), so that
$$4 \cos \frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2} = \sin A + \sin B + \sin C$$
as desired.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the conjugate hyperbola Assume we are given a general proper hyperbola $ a_{11}x^2 + 2 a_{12} x y + a_{22} y^2 + 2 a_{13}x +2 a_{23}y + a_{33} = 0$ with
$D =\det \begin{pmatrix}
a_{11} & a_{12} \\
a_{12} & a_{22}
\end{pmatrix} < 0$
and
$
A = \det \begin{pmatrix}
a_{11} & a_{12} & a_{13} \\
a_{12} & a_{22} & a_{23} \\
a_{13} & a_{23} & a_{33}
\end{pmatrix} \neq 0$.
Is there an easy way to find the implicit description of the conjugate hyperbola?
I could translate and rotate the hyperbola and obtain a canonical representation of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = \pm 1$, find the conjugate of the canonical one and the transform back. However, I'm looking for a more straightforward way to achieve this.
|
Using the right translation and rotation one can obtain for a general hyperbola, as given in the OP, a canonical congruent hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = \pm 1$. The canonical conjugate is then given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = \mp 1$, and can be transformed back (using inverse transformations) to obtain a general form of the conjugate hyperbola of the one we started with. Doing some algebra its the implicit representation of the conjugate hyperbola is given by $a_{11}x^2 +2a_{12}xy+a_{22}y^2 +2a_{13}x+2a_{23}y+a_{33} −2\frac{A}{D}$ with $D,A$ are the determinants given in the OP.
|
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|
Find the value of $x$ in the displayed figure Find $x$ in the following figure.
$AB,AC,AD,BC,BE,CD$ are straight lines.
$AE=x$, $BE=CD=x-3$, $BC=10$, $AD=x+4$
$\angle BEC=90^{\circ}$, $\angle ADC=90^{\circ}$
NOTE: figure not to scale.
|
By the Pythagorean theorem we have
$$\begin{equation*}
CE=\sqrt{10^{2}-\left( x-3\right) ^{2}}=\sqrt{91-x^{2}+6x}
\end{equation*}$$
and
$$\begin{equation*}
CD^{2}+AD^{2}=AC^{2}=\left( CE+AE\right) ^{2}
\end{equation*}.$$
So we have to solve the following irrational equation
$$\begin{equation*}
\left( x-3\right) ^{2}+\left( x+4\right) ^{2}=\left( \sqrt{91-x^{2}+6x}
+x\right) ^{2},\tag{1}
\end{equation*}$$
which can be simplified to
$$\begin{equation*}
x^{2}-2x-33=\sqrt{-x^{4}+6x^{3}+91x^{2}}.
\end{equation*}$$
After squaring both sides and grouping the terms of the same degree we get the quartic equation
$$\begin{equation*}
2x^{4}-10x^{3}-153x^{2}+132x+1089=0.\tag{2}
\end{equation*}$$
The coefficient of $x^{4}$ is $2=1\times 2$ and the constant term is $1089=1\times 3^{2}11^{2}$. To find possible rational roots of this equation, we apply the rational root theorem and test the numbers of the form
$$\begin{equation*}
x=\pm \frac{p}{q},
\end{equation*}$$
where $p\in \left\{ 1,3,9,11,33,99,121,363,1089\right\} $ is a divisor of $1089$ and $q\in \left\{ 1,2\right\} $ is a divisor of $2$. It turns out that $x=3$ and $x=11$ are roots. Now we divide the LHS by $x-3$
$$
\begin{equation*}
\frac{2x^{4}-10x^{3}-153x^{2}+132x+1089}{x-3}=2x^{3}-4x^{2}-165x-363
\end{equation*}$$
and this quotient by $x-11$
$$\begin{equation*}
\frac{2x^{3}-4x^{2}-165x-363}{x-11}=2x^{2}+18x+33.
\end{equation*}$$
So we have the equivalent equation
$$\begin{equation*}
\left( x-3\right) (x-11)\left( 2x^{2}+18x+33\right) =0\tag{3}
\end{equation*}$$
Since the solutions of $2x^{2}+18x+33$ are both negative and $x=3$ is not a solution of the original irrational equation, the solution is therefore $$x=11.
$$
|
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|
How to find $\lim\limits_{x\to0}\frac{e^x-1-x}{x^2}$ without using l'Hopital's rule nor any series expansion? Is it possible to determine the limit
$$\lim_{x\to0}\frac{e^x-1-x}{x^2}$$
without using l'Hopital's rule nor any series expansion?
For example, suppose you are a student that has not studied derivative yet (and so not even Taylor formula and Taylor series).
|
Consider fundamental limit: $e = \lim\limits_{n\to \infty}(1+\frac{1}{n})^n$ and $e^x = \lim\limits_{n\to\infty}(1+\frac{x}{n})^n$
Proof
$e^x = [\lim\limits_{k\to\infty}(1+1/k)^k]^x = \lim\limits_{k\to \infty}((1+1/k)^{kx})\Rightarrow kx = n
\Rightarrow e^x = \lim\limits_{n\to\infty}(1+\frac{x}{n})^n$.
Understand the first expression:
$P = \large\frac{e^x-1}{x}$
Note that $e^x - 1 - x = x.[\large\frac{(e^x-1)}{x} - 1]\,\,\therefore\,\,$ $\boxed{\lim\limits_{x\to 0}\frac{e^x-1-x}{x^2}=\lim\limits_{x\to 0}\frac{P-1}{x}}$
Lets go to understand the expression $\,\,P-1$.
$P - 1= \frac{e^x - 1}{x} - 1 = \lim\limits_{n\to\infty}\left(\large\frac{[(1+\frac{x}{n})^n - 1]}{x} - 1\right)=$
Using that tool:
$\boxed{b^n - 1 = (b-1).(b^{n-1}+b^{n-2}+...+1)}$
$=\lim\limits_{n\to\infty}\left((1+\frac{x}{n}-1).\large\frac{[(1+x/n)^{n-1} + (1+x/n)^{n-2} + ... + {1+x/n}]}{x}-1 \right) =\\
\\
= \lim\limits_{n\to\infty}\left(\frac{1}{n}.[(1+x/n)^{n-1} + (1+x/n)^{n-2} + ... + (1+x/n)]-1\right) = \\
\\
=\lim\limits_{n\to\infty}\frac{1}{n}.\left((1+x/n)^{n-1} + (1+x/n)^{n-2} + ... + (1+x/n)-n\right)$
Writing the last "$n$" as $\underbrace{1+1+1...+1}_{n\,\, times}$ and inputing these $1`s$ into it:
$P-1 = \lim\limits_{n\to\infty} (1/n).[((1+x/n)^{n-1} - 1)+ ((1+x/n)^{n-2} - 1) + ... + ((1+x/n) - 1)]$
Using again that tool in each expression:
$=\lim\limits_{n\to\infty}(\frac{1}{n}).(\frac{x}{n}) [((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... +1)+((1+x/n)^{n-3}+...+1)+...+1]$
Finally,
$L = \lim\limits_{x\to 0}\frac{P-1}{x} =\lim\limits_{x\to 0}\lim\limits_{n\to\infty}(\frac{1}{n}).(\frac{x}{n})[\large\frac{((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1)}{x}]=$
$=\lim\limits_{n\to\infty}\lim\limits_{x\to0}(\frac{1}{n}).(\frac{x}{n})[\large\frac{((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1)}{x}] =\\$
$=\lim\limits_{n\to\infty}\lim\limits_{x\to 0}\left(\frac{1}{n^2}\right).((1+x/n)^{n-2} + (1+x/n)^{n-3} + ... + 1 ) + ( (1+x/n)^{n-3} + ... + 1 ) + ... + 1) =$
$=\lim\limits_{n\to\infty}\left(\frac{1}{n^2}\right)(n-1 + n-2 + n-3 + ... + 1) = \lim\limits_{n\to\infty}\left(\frac{1}{n^2}\right)(n-1)(\frac{n}{2}) = \lim\limits_{n\to\infty}\frac{n-1}{2n} = \boxed{\large\frac{1}{2}}$.
|
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|
$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$ Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$
prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$
|
Let $a=b$. Then $c=\frac{1}{a^2}$ and the formula is :
$$f(a)=\sqrt{\frac{a}{2}}+\sqrt{a+\frac{1}{a^2}}$$
$$f'(a)=\frac{1}{2\sqrt{2a}}+(\frac{1}{2}-\frac{1}{a^3}).\frac{1}{\sqrt{a+\frac{1}{a^2}}}$$
The only root in $[0,\infty)$ of f'(a) is 1, hence $f(1)=\frac{3}{\sqrt{2}}$ is a minimum.
Now, what happens if $a\neq b$ ?
Use the same method, but say $b=k.a+(1-k)$ (so let $k=\frac{b-1}{a-1}$ be a constant, and if $a=1$ exchange $a$ and $c$). Then $c=\frac{1}{ka^2+a(1-k)}$ and :
$$f_k(a)=\frac{a}{\sqrt{a.(1+k)+(1-k)}}+\frac{ka+(1-k)}{\sqrt{ka+(1-k)+\frac{1}{ka^2+a(1-k)}}}+\frac{\frac{1}{ka^2+a(1-k)}}{\sqrt{a+\frac{1}{ka^2+a(1-k)}}}$$
Once again, obtain $f'_k$ and show that its only root is $1$ (This is quite technical, use mathematica ?). I agree there should have something more simple.
|
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|
Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \frac{x(2x - 13) - 15(2x+3)}{x(2x + 3)} \lt 0$$
$$\iff \frac{2x^2 - 13x - 30x - 45}{2x(x + \frac{3}{2})} \lt 0 \iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 $$
$$\iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 \iff \frac{(x - \frac{43}{4})^2 - \frac{1939}{4}}{x(x+\frac{3}{2})} \lt 0$$
I don't know how to get any further, and I'm starting to get too high values to handle.
The next step as I can see would be to find an $x$ that makes $(x - \frac{43}{4})^2 = \frac{1939}{4}$. This, alongside the obvious ones for $x$ and $x+\frac{3}{2}$ (creating division by $0$), would help me find the possible values for $x$.
But how do I get the last step? Or am I already dead wrong?
|
One approach is to find the (potential) "switchover" points. Note that one rational function can only become bigger than another after either intersecting or after a hole or vertical asymptote (of one or both). (Why?) There our no holes, and our asymptotes occur at $x=0$ and $x=-\frac32$. Intersection occurs when $x\neq 0$, $x\neq-\frac32$ and $$\frac{2x-13}{2x+3}=\frac{15}x$$ $$2x^2-13x=30x+45$$ $$2x^2-43x-45=0$$ $$2x^2-45x+2x-45=0$$ $$(2x-45)(x+1)=0.$$ Thus, our other possible switchover points are $x=-1$ and $x=\frac{45}2$.
Now, check the inequality on the intervals $\left(-\infty,-\frac32\right)$, $\left(-\frac32,-1\right)$, $(-1,0)$, $\left(0,\frac{45}2\right)$, and $\left(\frac{45}2,\infty\right)$ by using test points in each interval. Note that for the first interval a sign chart would readily tell us that the desired inequality fails (without need for a test point), since $\cfrac{2x-13}{2x+3}$ is positive and $\cfrac{15}x$ is negative.
|
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|
Numerically evaluating the limit of $\frac{x^4-1}{x^3-1}$ as $x\rightarrow 1$ What is the limit as $x \to 1$ of the function
$$ f(x) = \frac{x^4-1}{x^3-1} . $$
|
Applying L'Hôpital's rule makes it quite straightforward.
$$ \lim_{x \rightarrow 1}\; \frac{x^4-1}{x^3-1} = \lim_{x \rightarrow 1} \;\frac{4x^3}{3x^2} = \lim_{x \rightarrow 1} \;\frac{4}{3}x = \frac{4}{3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why is $a^n - b^n$ divisible by $a-b$? I did some mathematical induction problems on divisibility
*
*$9^n$ $-$ $2^n$ is divisible by 7.
*$4^n$ $-$ $1$ is divisible by 3.
*$9^n$ $-$ $4^n$ is divisible by 5.
Can these be generalized as
$a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer?
But why is $a^n$ $-$ $b^n$$ = (a-b)N$ ?
I also see that $6^n$ $- 5n + 4$ is divisible by $5$ which is $6-5+4$ and $7^n$$+3n + 8$ is divisible by $9$ which is $7+3+8=18=9\cdot2$.
Are they just a coincidence or is there a theory behind?
Is it about modular arithmetic?
|
It’s a standard identity:
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1})\;.$$
It’s most neatly verified using summation notation, but you can also see what’s going on when you write everything out in extended form, as I did above. First,
$$\begin{align*}
a(a^{n-1}&+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1})\\
&=a^n+a^{n-1}b+a^{n-2}b^2+\ldots+a^3b^{n-3}+a^2b^{n-2}+ab^{n-1}\;.\tag{1}
\end{align*}$$
Next,
$$\begin{align*}
b(a^{n-1}&+a^{n-2}b+a^{n-3}b^2+\ldots+a^2b^{n-3}+ab^{n-2}+b^{n-1})\\
&=a^{n-1}b+a^{n-2}b^2+a^{n-3}b^3+\ldots+a^2b^{n-2}+ab^{n-1}+b^n\;.\tag{2}
\end{align*}$$
Now subtract $(2)$ from $(1)$:
$$\begin{align*}
a^n&+\color{red}{a^{n-1}b+a^{n-2}b^2+\ldots+a^3b^{n-3}+a^2b^{n-2}+ab^{n-1}}\\
&\color{red}{-a^{n-1}b-a^{n-2}b^2-\ldots-a^3b^{n-3}-a^2b^{n-2}-ab^{n-1}}-b^n\;;
\end{align*}$$
the red terms cancel out leaving $a^n-b^n$.
|
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|
Inequality $(a+\frac{1}{b})^2+(b+\frac{1}{c})^2+(c+\frac{1}{a})^2\ge 16$ For every real positive number $a,b,c$ such that $ab+bc+ca=1$, how to prove that:
$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2\ge 16$$
|
Expanding the left hand side, we find that because for all non-zero real numbers $x$, $x^2 + x^{-2} \geq 2$ that
$$\left(a + \frac{1}{a}\right)^2 + \ldots \left(c + \frac{1}{c}\right)^2 \geq \frac{2a}{b} + \frac{2b}{c} + \frac{2c}{a} + 6.$$
Can you complete the argument?
|
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|
Easy polynomials question? Please do this without using the quadratic formula.
If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2 -6x + a$ then find the value of "$a$" if $3\times \alpha + 2\times \beta = 20$
Thank you for the help
There is also a second question of this sort, i dont get that either. Would help if both were answered.
if $\alpha$ and $\beta$ are zeroes of the polynomial $x^2 -5x + k$ such that alpha- beta = 1. Find K
|
First of all don't confuse $a$ with $\alpha$ !(It is better to substitute A or m for $a$)
We know that the sum of the roots of a quadratic equation $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$
So: $\alpha +\beta =6$
On the other hand:$3\times \alpha + 2\times \beta =20$
Solving this two equations two variables yields:$\beta = -2$ and $\alpha=8$
We also know that the product of the roots of a quadratic equation $ax^2+bx+c$ is : $x_1 \times x_2=\frac{c}{a}$
so $$\alpha \times \beta = -16 = a $$
Second question:
Again:
We know that the sum of the roots of a quadratic equation $ax^2+bx+c$ is $x_1+x_2=-\frac{b}{a}$
So: $\alpha +\beta =5$
On the other hand:$\alpha - \beta =1$
Solving this two equations two variables yields:$\beta = 2$ and $\alpha=3$
We also know that the product of the roots of a quadratic equation $ax^2+bx+c$ is : $x_1 \times x_2=\frac{c}{a}$
so $$\alpha \times \beta = 6 = k $$
|
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|
Perimeter of an ellipse How can I calculate the perimeter of an ellipse? What is the general method of finding out the perimeter of any closed curve?
|
If the semi-major axis has length $a$ and the eccentricity is $e$, then the perimeter is
$$
2πa \left(\frac{1}{1} \left(\frac{(-1)!!}{0!!} e^0\right)^2 + \frac{1}{-1} \left(\frac{1!!}{2!!} e^1\right)^2 + \frac{1}{-3} \left(\frac{3!!}{4!!} e^2\right)^2 + \frac{1}{-5} \left(\frac{5!!}{6!!} e^3\right)^2 + ⋯\right)
$$
where $(-1)!! = 1 = 0!!$, and for all $n ∈ \{⋯, -7, -5, -3, -1, 0, 1, 2, 3, 4, 5, ⋯ \}$: $(n+2)!! = (n+2) n!!$. The sum extends to negative powers of $e$, if you take $n!! = ∞$ for $n ∈ \{⋯, -8, -6, -4, -2\}$; and if you work out the double-factorials, using:
*
*$(2n)!! = 2^n n!$, for $n ≥ 0$;
*$(2n-1)!! = \frac{(2n)!}{2^n n!}$, for $n ≥ 0$;
*$(2n-1)!! = \frac{(-1)^n}{(-2n - 1)!!} = (-2)^{-n} \frac{(-n)!}{(-2n)!}$, for $n ≤ 0$;
then it comes out to the following:
$$
2πa \left(1 - \left(\frac{1}{2} e\right)^2 - \frac{1}{3} \left(\frac{1}{2} \frac{3}{4} e^2\right)^2 - \frac{1}{5} \left(\frac{1}{2} \frac{3}{4} \frac{5}{6} e^3\right)^2 + ⋯\right).
$$
Im my previous life oops I said that out loud I offered the following estimate:
$$\mbox{Ramanujan's Formula}:
π(3(a+b) - \sqrt{(3a+b)(a+3b)}) = π(a+b) (3 - \sqrt{4 - h}),$$
where $b = a \sqrt{1 - e^2}$ is the length of the semi-minor axis and $h = ((a - b)/(a + b))^2$, while secretly holding onto the other, much better, estimate:
$$π(a+b) \left(\frac{12 + h}{8} - \sqrt{\frac{2 - h}{8}}\right).$$
|
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|
Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities Please help to solve the following inequality using rearrangement inequalities.
Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that
\begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\end{equation}
Thanks.
|
We can assume $a \leq 1 \leq b$. Applying rearrangement inequalities to
$$
\begin{align}
a &\leq 1 \\
1 &\leq b
\end{align}
$$
we get
$$
a + b \geq 1 + ab = 2
$$
and
$$
b + 3a \geq 2a + 2 \\
a + 3b \geq 2b + 2
$$
Therefore
$$
\begin{align}
\frac{a}{a^2+3}+\frac{b}{b^2+3} &= \frac{1}{a + 3b} + \frac{1}{b + 3a} \leq\\ &\frac{1}{2b + 2} + \frac{1}{2a + 2} = \frac{a}{2a + 2} + \frac{1}{2a + 2} = \frac 1 2
\end{align}
$$
|
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|
Evaluation of $ \sum_{k=0}^n \cos k\theta $ I just wanted to evaluate
$$ \sum_{k=0}^n \cos k\theta $$
and I know that it should give
$$ \cos\left(\frac{n\theta}{2}\right)\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin(\theta / 2)} $$
I tried to start by writing the sum as
$$ 1 + \cos\theta + \cos 2\theta + \cdots + \cos n\theta $$
and expand each cosine by its series representation. But this soon looked not very helpful so I need some clue about how this partial sum is calculated more efficiently ...
|
Use:
$$
\sin \frac{\theta}{2} \cdot \cos(k \theta) = \underbrace{\frac{1}{2} \sin\left( \left(k+\frac{1}{2}\right)\theta\right)}_{f_{k+1}} - \underbrace{\frac{1}{2} \sin\left( \left(k-\frac{1}{2}\right)\theta\right)}_{f_{k}}
$$
Thus
$$ \begin{eqnarray}
\sin \frac{\theta}{2} \cdot \sum_{k=1}^n \cos(k \theta) &=& \sum_{k=1}^n \left(f_{k+1} - f_k\right) = f_{n+1}-f_1 \\ &=& \frac{1}{2} \underbrace{\sin\left(\left(n+\frac{1}{2}\right)\theta\right)}_{\sin(\alpha+\beta)}-\frac{1}{2} \underbrace{\sin \frac{\theta}{2}}_{\sin(\alpha-\beta)} = \cos(\alpha) \sin(\beta) \\
&=& \cos\left(\frac{n+1}{2}\theta\right) \sin\left(\frac{n}{2} \theta\right)
\end{eqnarray}
$$
where $\alpha = \frac{n+1}{2} \theta$ and $\beta = \frac{n}{2} \theta$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
An intriguing definite integral: $\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$ I need some hints, suggestions for the following integral
$$\int_{\pi/4}^{\pi/2} \frac{x^2+8}{(x^2-16)\sin (x) + 8 x \cos(x)} \ dx$$
Since it's a high school problem, I thought of some variable change, integration by parts, but I can't see yet how to make them work. I don't know where I should start from. Thanks!
|
Following up on Blue's comment:
In general
$$
a\sin x+b\cos x = \sqrt{a^2+b^2}\Big( \frac{a}{\sqrt{a^2+b^2}}\sin x + \frac{b}{\sqrt{a^2+b^2}}\cos x \Big) = \sqrt{a^2+b^2} \left(\cos\varphi\sin x+\sin\varphi\cos x\right) = \sqrt{a^2+b^2}\sin(x+\varphi),
$$
so $\tan\varphi = \dfrac b a$.
Now putting in $a=x^2-16$ and $b=8x$, we have
$$
\begin{align}
\Big( x^2 - 16 \Big)^2 + \Big(8x\Big)^2 & = \Big(x^4 - 32x^2 + 256\Big) + \Big(64x^2\Big) \\[10pt]
& = x^4 + 32x^2 + 256 = \Big( x^2 + 16 \Big)^2
\end{align}
$$
Therefore
$$
(x^2-16)\sin x + 8x\cos x = (x^2+16)\sin(x+\varphi)
$$
where
$$
\varphi = \arctan\frac{8x}{x^2-16}. \tag{1}
$$
The answer from sos440 doesn't explain how s/he thought of that susbstitution in the first place, but conjoining that with $(1)$, I'm thinking: let's see if that last fraction in $(1)$ is the tangent of a double angle. Remember that
$$
\tan(2\alpha) = \frac{2\tan\alpha}{1-\tan^2\alpha}.
$$
Therefore
$$
\frac{8x}{x^2-16} = \frac{2(-x/4)}{1-(-x/4)^2} = \tan(2\alpha)\text{ where } \tan\alpha = \frac{-x}{4}.
$$
After that, proceed as in the answer from sos440.
|
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|
how to calculate the exact value of $\tan \frac{\pi}{10}$ I have an extra homework: to calculate the exact value of $ \tan \frac{\pi}{10}$.
From WolframAlpha calculator I know that it's $\sqrt{1-\frac{2}{\sqrt{5}}} $, but i have no idea how to calculate that.
Thank you in advance,
Greg
|
$$\tan\frac{3\pi}{10}=\tan(\frac{\pi}{2}-\frac{2\pi}{10})=\cot\frac{2\pi}{10}$$
$$\frac{3\tan\frac{\pi}{10}-\tan^3\frac{\pi}{10}}{1-3\tan^2\frac{\pi}{10}}=\frac{\cot^2\frac{\pi}{10}-1}{2\cot\frac{\pi}{10}}$$
$$(3\tan\frac{\pi}{10}-\tan^3\frac{\pi}{10})(2\cot\frac{\pi}{10})=(\cot^2\frac{\pi}{10}-1)(1-3\tan^2\frac{\pi}{10})$$
$$6-2\tan^2\frac{\pi}{10}=\cot^2\frac{\pi}{10}-4+3\tan^2\frac{\pi}{10}$$
$$5\tan^2\frac{\pi}{10}-10+\cot^2\frac{\pi}{10}=0$$
$$5\tan^4\frac{\pi}{10}-10\tan^2\frac{\pi}{10}+1=0$$
$$\tan^2\frac{\pi}{10}=\frac{10\pm\sqrt{100-20}}{10}=\frac{10\pm4\sqrt{5}}{10}=1+\frac{2}{\sqrt{5}}\;\textrm{or}\;1-\frac{2}{\sqrt{5}}(\textrm{rej.})$$
$$\tan\frac{\pi}{10}=\sqrt{1+\frac{2}{\sqrt{5}}}\;\textrm{or}\;-\sqrt{1+\frac{2}{\sqrt{5}}}(\textrm{rej.})$$
$$∴\tan\frac{\pi}{10}=\sqrt{1+\frac{2}{\sqrt{5}}}$$
|
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|
Could there be a formula of this? Say you have a function: f (m,n)
f (m,n) = m if n = 1
f (m,n) = n if m = 1
otherwise f (m,n) = f (m - 1, n) + f (m, n - 1)
Pre-calculated value:
1 2 3 4 5 6
2 4 7 11 16
3 7 14 25
4 11 25
5 16
6
Just wondering if there could be formula of f(m,n), instead of doing a dynamic progrmming or recursive calculation to get the value.
|
There is indeed. Let $g(m,n)=f(m-n+1,n)$, so that for instance $g(5,3)=f(3,3)=14$. The corresponding table for $g$ is:
$$\begin{array}{}
1\\
2&2\\
3&4&3\\
4&7&7&4\\
5&11&14&11&5\\
6&16&25&25&16&6
\end{array}$$
Now
$$\begin{align*}
g(m,n)&=f(m-n,n)\\
&=f(m-n-1,n)+f(m-n,n-1)\\
&=f\big((m-1)-n,n\big)+f\big((m-1)-(n-1),n-1\big)\\
&=g(m-1,n)+g(m-1,n-1)\;,
\end{align*}$$
which is the recurrence that generates Pascal’s triangle of binomial coeffients. Moreover, $g$’s table looks a lot like Pascal’s triangle in overall form:
$$\begin{array}{}
1\\
1&1\\
1&2&1\\
1&3&3&1\\
1&4&6&4&1\\
1&5&10&10&5&1\\
1&6&15&20&15&6&1
\end{array}$$
Ignore that first column of Pascal’s triangle:
$$\begin{array}{}
1\\
2&1\\
3&3&1\\
4&6&4&1\\
5&10&10&5&1\\
6&15&20&15&6&1
\end{array}\tag{1}$$
Subtract this from $g$’s triangle:
$$\begin{array}{}
0\\
0&1\\
0&1&2\\
0&1&3&3\\
0&1&4&6&4\\
0&1&5&10&10&5
\end{array}\tag{2}$$
Ignore the first column of this, and put $1$’s along the diagonal, and Pascal’s triangle shows up again. Now the $(m,n)$-entry in $(1)$ is $\binom{m}n$, and if $(2)$ really is Pascal’s triangle, its $(m,n)$-entry is $\binom{m-1}{n-2}$, so we conjecture that $g(m,n)=\binom{m}n+\binom{m-1}{n-2}$ and hence that
$$f(m,n)=g(m+n-1,n)=\binom{m+n-1}n+\binom{m+n-2}{n-2}\;.$$
We first verify the recurrence:
$$\begin{align*}
\binom{m+n-1}n+\binom{m+n-2}{n-2}&=\binom{m+n-2}{n-1}+\color{red}{\binom{m+n-2}n}+\binom{m+n-3}{n-3}+\color{blue}{\binom{m+n-3}{n-2}}\\
&=\binom{m+(n-1)-1}{n-1}+\color{red}{\binom{m+(n-1)-2}{n-3}}\\
&\qquad\qquad+\binom{(m-1)+n-1}n+\color{blue}{\binom{(m-1)+n-2}{n-2}}\\
&=f(m,n-1)+f(m-1,n)\;,
\end{align*}$$
as desired. Finally,
$$f(1,n)=\binom{1+n-1}n+\binom{1+n-2}{n-2}=1+(n-1)=n$$
and
$$f(m,1)=\binom{m+1-1}1+\binom{m+1-2}{-1}=m+0=m\;,$$
so the initial conditions are also satisfied. To repeat,
$$f(m,n)=\binom{m+n-1}n+\binom{m+n-2}{n-2}\;,$$
which can easily be manipulated into a variety of other forms involving binomial coefficients, e.g.,
$$f(m,n)=g(m+n-1,n)=\binom{m+n-2}n+\binom{m+n-2}{n-1}+\binom{m+n-2}{n-2}\;.$$
|
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|
finding $\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}$ If:
$$\frac{\cos x}{\cos y}=\frac{1}{2}$$ and $$\frac{\sin x}{\sin y}=3$$
How to find
$$\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}$$
|
$$\frac{\sin{2x}}{\sin{2y}}=\frac{2\sin{x}\cos{x}}{2\sin{y}\cos{y}}=\frac{\sin{x}}{\sin{y}}\cdot \frac{\cos{x}}{\cos{y}}=3\cdot \frac{1}{2}=\frac{3}{2}. \tag{1}$$
$$\frac{\cos{2x}}{\cos{2y}}=\frac{\cos^{2}{x}-\sin^{2}{x}}{\cos^{2}{y}-\sin^{2}{y}}. \tag{2}$$
$$\frac{\cos{x}}{\cos{y}}=\frac{1}{2} \Leftrightarrow 4 \cdot\cos^{2}{x}=\cos^{2}{y}.\tag{3}$$
$$\frac{\sin{x}}{\sin{y}}=3 \Leftrightarrow \frac{1}{9}\cdot \sin^{2}{x}=\sin^{2}{y}.\tag{4}$$
We know that (using $(3)$ and $(4)$):
\begin{cases}
\sin^{2}{x}+\cos^{2}{x}=1\\
4\cdot \cos^{2}{x}+\frac{1}{9}\cdot\sin^{2}{x}=1
\end{cases}
So: $\displaystyle 35\cdot\cos^{2}{x}=8 \Rightarrow \cos^{2}{x}=\frac{8}{35}\tag{5}$ and $\displaystyle \sin^{2}{x}=1-\frac{8}{35}=\frac{27}{35}. \tag{6}$
But using $(3)$ and $(4)$ we obtain that:
$$\sin^{2}{y}=\frac{3}{35}\tag{7}$$ and $$\cos^{2}{y}=\frac{32}{35}\tag{8}$$
So $(3)$ is equivalent with :
$$\large\frac{\frac{8}{35}-\frac{27}{35}}{\frac{32}{35}-\frac{3}{35}}=\frac{-\frac{19}{35}}{\frac{29}{35}}=-\frac{19}{29}.\tag{9}$$
The final answer is obtained using $(1)$ and $(9)$:
$$\frac{3}{2}-\frac{19}{29}=\frac{49}{58}.$$
I hope it is all right, I hope not to mistake to calculations.
|
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|
Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
|
I'm pretty new to proofs, but it seems like you could just evaluate $\sqrt{1+i\sqrt{3}}+\sqrt{1-i\sqrt{3}}$ and if it equals $\sqrt{6}$, then it's proof enough (though this is admittedly a convoluted step-by-step that I derived from WolframAlpha):
\begin{aligned}
\sqrt{1+i\sqrt{3}}+\sqrt{1-i\sqrt{3}} &= \sqrt{\frac{3}{2} + i\sqrt{3} -\frac{1}{2}} + \sqrt{\frac{3}{2} - i\sqrt{3} - \frac{1}{2}} \\
&= \sqrt{\frac{9}{6} + \frac{6i\sqrt{3}}{6} -\frac{3}{6}} + \sqrt{\frac{9}{6}-\frac{6i\sqrt{3}}{6}-\frac{3}{6}} \\
&= \sqrt{\frac{9+6i\sqrt{3}-3}{6}} + \sqrt{\frac{9-6i\sqrt{3}-3}{6}} \\
&= \sqrt{\frac{9+6i\sqrt{3}+(i\sqrt{3})^2}{6}} + \sqrt{\frac{9-6i\sqrt{3}+(i\sqrt{3})^2}{6}} \\
&= \sqrt{\frac{(3+i\sqrt{3})^2}{6}}+\sqrt{\frac{(3-i\sqrt{3})^2}{6}} \\
&= \frac{\sqrt{(3+i\sqrt{3})^2}}{\sqrt{6}}+\frac{\sqrt{(3-i\sqrt{3})^2}}{\sqrt{6}} \\
&= \frac{3+i\sqrt{3}}{\sqrt{6}}+\frac{3-i\sqrt{3}}{\sqrt{6}} \\
&= \frac{\sqrt{6}(3+i\sqrt{3})}{6}+\frac{\sqrt{6}(3-i\sqrt{3})}{6} \\
&= \frac{3\sqrt{6}+3i\sqrt{2}}{6}+\frac{3\sqrt{6}-3i\sqrt{2}}{6} \\
&= \frac{3\sqrt{6}+3i\sqrt{2}+3\sqrt{6}-3i\sqrt{2}}{6} \\
&= \frac{6\sqrt{6}}{6} = \sqrt{6}
\end{aligned}
|
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|
Problem using Jensen's Inequality For some convex function $f$, and elements $x_{1},x_{2},x_{3}$ of its domain, show that:
$$f(x_{1})+f(x_{2})+f(x_{3})+f\left( \frac{1}{3}(x_{1}+x_{2}+x_{3}\right)\ge \frac{4}{3}\left( f\left(\frac{x_{1}+x_{2}}{2} \right)+f\left(\frac{x_{2}+x_{3}}{2} \right)+f\left(\frac{x_{3}+x_{1}}{2} \right) \right)$$
So far I have that the left hand side is greater than or equal to:
$$4f\left(\frac{x_{1}+x_{2}+x_{3}}{3}\right)=4f\left(\frac{x_{1}+x_{2}}{6}+\frac{x_{2}+x_{3}}{6}+\frac{x_{1}+x_{3}}{6}\right)$$
by Jensen, which we can say is less than or equal to the right hand side (again, by Jensen). My problem is that this doesn't imply the conclusion above, since $a\ge b\le c$ doesn't tell us anything about the relationship between $a$ and $c$. This makes me think I've made a mistake, which I would be very grateful if someone could spot.
|
Use these:
$f\left(\displaystyle\frac{x_i+x_j}2\right) \le \displaystyle\frac{f(x_i)+f(x_j)}{2}$, and
$f\left(\displaystyle\frac{y_1+y_2+y_3}3\right) \le \displaystyle\frac{f(y_1)+f(y_2)+f(y_3)}{3}$.
This applied to $y_1=\displaystyle\frac{x_2+x_3}2$,.. we also have
$f\left(\displaystyle\frac{x_1+x_2+x_3}3\right) \le \displaystyle\frac{f\left(\frac{x_1+x_2}2\right) + f\left(\frac{x_2+x_3}2\right) + f\left(\frac{x_3+x_1}2\right)}{3} \le\displaystyle\frac{f(x_1)+f(x_2)+f(x_3)}{3}$
|
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|
Prove $\sqrt[3]{60}>2+\sqrt[3]{7}$ Prove $$\sqrt[3]{60}>2+\sqrt[3]{7}$$
I try to both sides of the cubic equation, but it is quite complicated
|
Well, I don't see any alternative way..
$$60\overset{?}> 8+6\sqrt[3]{7^2}+12\sqrt[3]7+7 $$
$$45\overset{?}> 6\sqrt[3]{7^2}+12\sqrt[3]7 $$
$$15\overset{?}> 2\sqrt[3]{7^2}+4\sqrt[3]7 $$
Well, we could raise it to cubic, but that's really not nice.
What about considering the roots of $2x^2+4x-15 =2(x+1)^2-17$, and finally comparing if $\sqrt[3]7$ is between its roots..
$$\sqrt{\frac{17}2} -1 \overset{?}> \sqrt[3]7 $$
A bit nicer perhaps.. taking cubes:
$$\frac{17}{2}\sqrt{\frac{17}{2}}-3\cdot\frac{17}2+3\cdot\sqrt{\frac{17}2}-1 \overset{?}>7
$$
$$23\sqrt{\frac{17}2} \overset{?}> 16+3\cdot 17 = 67$$
and finally this leads to
$$8993 = 23^2\cdot 17 > 67^2\cdot 2 = 8978 $$
|
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|
Evaluate $x+4/x$ when $x$ is a solution of $x-6/\sqrt{x}=11$ I have a question that I see in a math text.
I asked to my teacher but she couldn't find the solution, too.
If
$$x - \frac{6}{\sqrt{x}}=11,$$
then to what is equal
$$x + \frac{4}{x}\text{ ?}$$
|
AS $x-\frac{6}{\sqrt{x}}-11=(\sqrt{x}+3)(\sqrt{x}-\frac{2}{\sqrt{x}}-3)$, so from $x-\frac{6}{\sqrt{x}}=11$ and $\sqrt{x}\geq0$, we can obtain that $\sqrt{x}-\frac{2}{\sqrt{x}}=3$, squaring at both sides, i.e. $(\sqrt{x}-\frac{2}{\sqrt{x}})^{2}=x+\frac{4}{x}-4=9$, so $x+\frac{4}{x}=13$.
|
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|
How to prove an Inequality I'm beginner with proofs and I got the follow exercise:
Prove the inequality $$(a + b)\Bigl(\frac{1}{a} + \frac{4}{b}\Bigr) \ge 9$$ when $a > 0$ and $b > 0$. Determine when the equality occurs.
I'm lost, could you guys give me a tip from where to start, or maybe show a good resource for beginners in proofs ?
Thanks in advance.
|
The first thing to do is simplify the expression on the lefthand side of the inequality.
$$(a + b)\left(\frac{1}{a} + \frac{4}{b}\right)=\frac{(a+b)(4a+b)}{ab}=\frac{4a^2+5ab+b^2}{ab}=\frac{4a}b+5+\frac{b}a\;.$$
Now notice that the resulting expression contains both $\frac{a}b$ and $\frac{b}a$; this is an indication that it might simplify matters to introduce a new quantity, $x=\frac{a}b$, and rewrite the inequality as $$4x+5+\frac1x\ge 9\;,$$ or $$4x+\frac1x\ge 4\;.\tag{0}$$
The natural thing to do now is to multiply through by $x$ to get rid of the fraction, but be careful: since this is an inequality, the sign of $x$ matters. If $x\ge 0$ we get $4x^2+1\ge 4x$, or $4x^2-4x+1\ge 0$, but if $x<0$ we get $4x^2+1\le 4x$, or $4x^2-4x+1\le 0$. In either case, though, we recognize that $4x^2-4x+1=(2x-1)^2$, so either $$x\ge 0\quad\text{and}\quad(2x-1)^2\ge 0\tag{1}$$ or $$x<0\quad\text{and}\quad(2x-1)^2\le 0\;.\tag{2}$$
Now $(2)$ is impossible: $(2x-1)^2\le 0$ if and only if $(2x-1)^2=0$, in which case $2x=1$, $x=\frac12$, and $x\not<0$. Thus, any solution must come from $(1)$: $x>0$, and $(2x-1)^2\ge 0$. If $2x\ne 1$, then $2x-1\ne0$, so $(2x-1)^2>0$, and we have a solution. If $2x=1$, then $x=\frac12>0$ and $(2x-1)^2=0\ge0$, and again we have a solution. In short, every positive $x$ is a solution, no negative $x$ is a solution, and $x$ can’t be $0$. (Why not?) We could actually have discovered this just by looking more closely at $(0)$, but one won’t always have so nice an inequality as that.
What does this mean in terms of $a$ and $b$? Recall that $x=\dfrac{a}b$; thus, $x>0$ if and only if $\dfrac{a}b>0$, which is true if and only if $a$ and $b$ have the same algebraic sign: both are positive, or both are negative. Since we were told that both are positive, we know that the inequality holds for all $a$ and $b$ in the given domain.
Finally, we have equality in $(0)$ if and only if $4x^4-4x+1=0$, or $(2x-1)^2=0$, i.e., if and only if $x=\frac12$. Since $x=\frac{a}b$, that’s equivalent to $\frac{a}b=\frac12$, or $b=2a$.
|
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|
Two Diophantine equations What is known about solutions in integers of the equations; $x^4 - y^2 = z^6$
I got $x=4st(s^4 - t^4)$ , $z=4st(s^2 - t^2)$ , $y=(4st(s^2 - t^2))^2 (s^4 + t^4 - 6 (st)^2) $
and, the equation $x^2 - y^4 = z^6 $
I got $y=4st(s^4 - t^4) , z=4st(s^2 + t^2) , x=(4st(s^2 + t^2))^2 (s^4 + t^4 + 6 (st)^2)$
|
Here is another solution of $x^4 - y^2 = z^6$. Suppose we have a Pythagorean triple $a^2 + b^2 = c^2$, and c is itself a square, say $d^2$. There are many such solutions, a sufficient condition being that c can be factorised into primes of the form $4n+1$, with an even exponent of each prime. Where this is the case each factor can be expressed as a sum of two squares (Fermat's 4n+1 Theorem), then we can use Brahmagupta's identity to express c as a sum of two squares, say $m^2 + n^2$. Then using the standard formula for Pythagorean triples we put a and b equal to $m^2 - n^2$ and $2mn$ (in either order).
Given $a^2 + b^2 = d^4$ we have:
$d^4 - a^2 = b^2$
$(d^4)(b^4) - (a^2)(b^4) = (b^2)(b^4)$
$(bd)^4 - (ab^2)^2 = b^6$
The smallest non-trivial solution of this type has a = 24, b = 7, c = 25, d = 5, yielding $35^4 - 1176^2 = 7^6$. Each such Pythagorean triple yields two solutions by reversing a and b, the second solution in this case with a = 7, b = 24 being $120^4 - 4032^2 = 24^6$. The second solution is an instance of the solution in the question (with s = 2, t = 1), but the first is not (since 4 does not divide 35 or 7). More generally, any solution of the above form with b and d both odd will result in x odd and therefore not be an instance of the solution in the question.
|
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|
How to solve the cubic equation $x^3-12x+16=0$ Please help me for solving this equation $x^3-12x+16=0$
|
This is a cubic equation. The forme of cubic equation is $x^3+px+q=0$ where
$1)$ $p$, $q$ $\in$ R
$2)$ $D=(\frac{q}{2})^2+(\frac{p}{3})^3$
$3)$ $x=u+v=\sqrt[3]{-\frac{q}{2}+\sqrt{(\frac{q}{2})^2 +(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(\frac{q}{2})^2 +(\frac{p}{3})^3}}$
For the given equation we have:
$p=-12$, $q=16$, $D=(\frac{16}{2})^2+(\frac{-12}{3})^3=8^2+(-4)^3=64-64=0$.
Because $D=0$, $u=v=\sqrt[3]{\frac{-q}{2}}$ $\Rightarrow$ $u=v=\sqrt[3]{\frac{-16}{2}}=\sqrt[3]{-8}$ $\Rightarrow$ $u_1=v_1=-2$.
Definitly
$x_1=2u_1$ $\Rightarrow$ $x_1=-4$
$x_2=x_3=-u_1$ $\Rightarrow$ $x_2=x_3=2$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
"answer_count": 5,
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|
Check my answer for the derivative of $y=\tan^{-1}(40/h) -\tan^{-1}(32/h)$
Differentiate
$$y=\tan^{-1}{\frac {40}{h}}-\tan^{-1}{\frac {32}{h}}$$
My answer
Using the identity $\frac{d}{dy}\tan^{-1}(x)=\frac{1}{1+x^2}$, can I conclude that
$$\frac{dy}{dh}=\frac{1}{1+(\frac{40}{h})^2}(-\frac {40}{h^2})-\frac{1}{1+(\frac{32}{h})^2}(-\frac {32}{h^2})$$
|
Why don't we use
$$\tan^{-1}\frac a h= \frac \pi 2-\cot^{-1}\frac a h=\frac \pi 2-\tan^{-1}\frac h a.$$
So, $$\frac{d(\tan^{-1}\frac a h)}{dh}=\frac{d(\frac \pi 2-\tan^{-1}\frac h a)}{dh}=-\frac{d(\tan^{-1}\frac h a)}{dh}=-\frac{1}{1+(\frac h a)^2}\frac 1 a=-\frac{a}{h^2+a^2}$$
So, $$\frac{d(\tan^{-1}\frac {40} h)}{dh}=-\frac{40}{h^2+40^2}$$ and
$$\frac{d(\tan^{-1}\frac {32} h)}{dh}=-\frac{32}{h^2+32^2}$$
So, $$\frac{dy}{dh}=-\frac{40}{h^2+40^2}-\left(-\frac{32}{h^2+32^2}\right)=\frac{32}{h^2+32^2}-\frac{40}{h^2+40^2}=\frac{32\cdot 40\cdot 8-8h^2}{(h^2+32^2)(h^2+40^2)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/209175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the distance between the line and plane Find the distance between the line $x-z=3, x+2y+4z=6$ and the plane $3x+2y+2z=5$.
What I have done so far,
Found the vector by crossing $1,0-1$ and $1,2,4$ from the first line
The line is parallel to plane $\langle 2,-5,2\rangle \cdot \langle 3,2,2 \rangle = 0$
I think my next step is to find a point on Line 1 which satisfies both equations and then insert those values into the plane $3(x)+2(y)+2(z)=5$ and use the formula, $$
\frac{(3(x)+2(y)+2(z)-5)}{(3^2+2^2+2^2)}$$ to find the distance.
The values that I calculuate do not match the posted answer of $7/\sqrt{17}$
|
Put $(P_1): x - z -3 = 0$, $(P_2): x+2y+4z=6$, $(P_3): 3x + 2y +2z -5 = 0$ and $\Delta$ is intersection of two plane $(P_1)$ and $(P_2)$. We have, a normal vector of the plane $(P_1)$ is $a = (1,0,-1)$, a normal vector of the plane $(P_2)$ is $b = (1,2,4)$. A direction vector $v$ of $\Delta$ is cross product of $a$ and $b$, therefore $v = (3, 2, 2)$. Another way, $M=(3, \dfrac32, 0)$ lies on $\Delta$ and not belongs to $(P_3)$, thus $\Delta$ is parallel to the plane $(P_3)$. We have, the distance between the line $\Delta$ to the plane $(P_3)$ is the distance from the point $M$ to the plane $(P_3)$. From here, we have answer is $$\dfrac{|3\cdot 3 + 2\cdot\dfrac{3}{2} + 2\cdot 0 - 5|}{\sqrt{3^2 + 2^2 + 2^2}} = \dfrac{7\sqrt{17}}{17}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/209910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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|
Simplification of Sum I can't see how to simplify from step 1 to step 2 in the following example:
*
*$$
\frac{1}{3}n(n+1)(n+2)+(n+2)(n+1)
$$
*$$
(\frac{1}{3}n+1)(n+1)(n+2)
$$
Thanks to the answers this is how I got from 1 to 2:
1.1
$$
\frac{1}{3}n(n+1)(n+2)+1(n+2)(n+1)
$$
1.2
$$
(n+2)\left(\frac{1}{3}n(n+1)+1(n+1)\right)
$$
1.3
$$
\left((n+1)(\frac{1}{3}n+1)\right)(n+2)
$$
Then you get to step 2.
Or factor out both (n+1) and (n+2) from the whole sum at once:
$$
(n+1)(n+2)\left((\frac{1}{3}n+1)\right)
$$
In case you wonder why all this - now I can show that
$$
\sum_{i=1}^{n+1} (i + 1)i = \left(\sum_{i=1}^n (i + 1)i\right) + (n+2)(n+1)
$$
$$
= \frac{1}{3}(n+1)(n+2)(n+3)
$$
which should proof (by using mathematical induction) that
$$
\forall n \in N : \sum_{i=1}^n (i + 1)i = \frac{1}{3}n(n+1)(n+2).
$$
|
Factor out $(n+1)(n+2)$. What's left in each term? What's the sum of those two expressions?
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/210102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solve $5a^2 - 4ab - b^2 + 9 = 0$, $ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0$
Solve $\left\{\begin{matrix} 5a^2 - 4ab - b^2 + 9 = 0\\ - 21a^2 - 10ab + 40a - b^2 + 8b - 12 = 0. \end{matrix}\right.$
I know that we can use quadratic equation twice, but then we'll get some very complicated steps. Are there any elegant way to solve this? Thank you.
|
By arrange the terms, you can note that
$$\begin{cases}
5a^{2}-4ab-b^2+9=0\\
-21a^{2}-10ab+40a-b^{2}+8b-12=0
\end{cases}
\Leftrightarrow
\begin{cases}
9(a^{2}+1)=(2a+b)^{2}\\
4(a^{2}+1)=(5a+b-4)^{2}
\end{cases}$$
So we get
$$\begin{cases}
9(a^{2}+1)=(2a+b)^{2}\\
4(2a+b)^{2}=9(5a+b-4)^{2}
\end{cases}$$
According to the second equation, we can get two cases of first relation between $a$ and $b$, then substitute them into the first equation respectively, we can get all the cases of values of pairs $(a,b)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/210454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
number of integral solutions for $x^2+y^2=5^k$ Prove that the equation $x^2+y^2=5^k$ has $4k+4$ integral solution.
Any ideas would be appreciated.
Thanks
|
Lets use Strong form of mathematical induction.
Base Case: $k=0$ and $k=1$ are obvious.
Assume $x^2+y^2=5^k$ has exactly $8$ solutions $(x,y)$ such that $x$ and $y$ are not divisible by $5$ along with $4k-4$ solutions of the form $(5a,5b)$ where $(a,b)$ are solution of $a^2+b^2=5^{k-2}$. We need to show similar for $k+1$.
Note that $(x,y)$ are all obtained from each other by permutations of $x$ and $y$ and changes of signs,lets call them new solutions.
If $x^2 + y^2$ be divisible by $5$. Then $(x + 2y)(x − 2y) = x^2 + y^2 − 5y^2$ is also divisible by $5$. Hence, one of the numbers $x + 2y$ and $x − 2y$ is divisible by $5$. Note that if both are divisible by $5$, then both $x$ and $y$ are divisible by $5$.
If $(x, y)$ is a new solution of equation $x^2 + y^2 = 5^k$ , then $(x + 2y, 2x − y)$ and $(x − 2y, 2x + y)$ are solutions of equation $m^2+n^2=5^{k+1}$ and precisely one of them is new. Hence each new solution pair $(x,y)$ of $x^2+y^2=5^k$ yields $1$ new solution and $1$ not new solution for $x^2+y^2=5^{k+1}$. Hence we are done as there are exactly $8$ new solutions for $x^2+y^2=5^k$ yielding $8$ new and $8$ not new solutions for $x^2+y^2=5^{k+1}$ in the form of $$(\pm (2x \pm y),\pm(x \pm 2y)) \text{ and }(\pm (x \pm 2y), \pm(2x \pm y)).$$
Note that the $8$ not new solutions will be contained in case of $x^2+y^2=5^{k-1}$.
Thanks
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/211270",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
How to construct a bijection from $(0, 1)$ to $[0, 1]$?
Possible Duplicate:
Bijection between an open and a closed interval
How do I define a bijection between $(0,1)$ and $(0,1]$?
I wonder if I can cut the interval $(0,1)$ into three pieces:
$(0, \frac{1}{3})\cup(\frac{1}{3},\frac{2}{3})\cup(\frac{2}{3},1)$, in which I'm able to map point $\frac{1}{3}$ and $\frac{2}{3}$ to $0$ and $1$ respectively.
Now the question remained is how to build a bijection mapping from those three intervels to $(0,1)$.
Or, my method just goes in a wrong direction. Any correct approaches?
|
The idea you mention, with mild modification, will work. Please note that what is below is a minor variant of the solution given by Patrick da Silva.
Your decomposition of $(0,1)$ is not quite complete. We want to write
$$(0,1)=\left(0,\frac{1}{3}\right) \cup\left\{\frac{1}{3}\right\} \cup \left(\frac{1}{3},\frac{2}{3}\right)\cup \left\{\frac{2}{3}\right\}\cup \left(\frac{2}{3},1\right),$$
so $5$ "intervals," two of them kind of boring. From now on, we will call $\frac{2}{3}$ by the more cumbersome name $1-\frac{1}{3}$.
Use the identity function on the two endintervals $\left(0,\frac{1}{3}\right)$ and $\left(1-\frac{1}{3},1\right)$, and map $\frac{1}{3}$ to $0$, and $1-\frac{1}{3}$ to $1$.
This leaves $\left(\frac{1}{3},1-\frac{1}{3}\right)$, which needs to be bijectively mapped to $\left[\frac{1}{3},1-\frac{1}{3}\right]$.
Use the same trick on the interval $\left(\frac{1}{3},1-\frac{1}{3}\right)$ that we used on $(0,1)$. So this time the two special points "inside" that will be mapped to $\frac{1}{3}$ and $1-\frac{1}{3}$ respectively are $\frac{1}{3}+\frac{1}{9}$ and $1-\frac{1}{3}-\frac{1}{9}$. That leaves $\left(\frac{1}{3}+\frac{1}{9},1-\frac{1}{3}-\frac{1}{9}\right)$ to be mapped bijectively to $\left[\frac{1}{3}+\frac{1}{9},1-\frac{1}{3}-\frac{1}{9}\right]$. Continue, forever. As pointed out by Patrick da Silva, the point $\frac{1}{2}$ is not dealt with in this process: simply map it to itself.
It would be notationally a little simpler to use the same idea to map $(-1,1)$ bijectively to $[-1,1]$, and use linear functions to take $(0,1)$ to $(-1,10$, and $[-1,1]$ to $[0,1]$ toadjust to our situation.
The advantage is that first "middle" interval is $\left(-\frac{1}{3},\frac{1}{3}\right)$, the second middle interval is $\left(-\frac{1}{9},\frac{1}{9}\right)$, and so on.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/213391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Get the equation of a circle when given 3 points Get the equation of a circle through the points $(1,1), (2,4), (5,3) $.
I can solve this by simply drawing it, but is there a way of solving it (easily) without having to draw?
|
Based on what I've learnt from this page by Stephen R. Schmitt (thanks to @Dan_Uznanski for pointing the concept out), there's a faster way to do it using matrices:
$$\text{let }〈x_1, y_1〉, 〈x_2, y_2〉, 〈x_3, y_3〉\text{ be your 3 points, and let }〈x_0, y_0〉\text{ represent the center of the circle.}\\\
\\
\text{let }A =
\left[\begin{array}{cccc}
x^2+y^2 & x & y & 1\\
x_1^2+y_1^2 & x_1 & y_1 & 1\\
x_2^2+y_2^2 & x_2 & y_2 & 1\\
x_3^2+y_3^2 & x_3 & y_3 & 1\\
\end{array}\right] =
\left[\begin{array}{cccc}
x^2+y^2 & x & y & 1\\
1^2+1^2 & 1 & 1 & 1\\
2^2+4^2 & 2 & 4 & 1\\
5^2+3^2 & 5 & 3 & 1\\
\end{array}\right] =
\left[\begin{array}{cccc}
x^2+y^2 & x & y & 1\\
2 & 1 & 1 & 1\\
20 & 2 & 4 & 1\\
34 & 5 & 3 & 1\\
\end{array}\right]\\\ \\\
\\\
\text{Note: }M_{11} = 0 ⟹ 〈x_0, y_0〉\text{ is undefined}∧\text{the points lie on a line rather than a circle.}\\\
\\\
x_0 =
\frac{1}{2} ⋅ \frac{M_{12}}{M_{11}}
= \frac{1}{2} ⋅
\left|\begin{array}{cccc}
2 & 1 & 1\\
20 & 4 & 1\\
34 & 3 & 1\\
\end{array}\right|
\Bigg{/}
\left|\begin{array}{cccc}
1 & 1 & 1\\
2 & 4 & 1\\
5 & 3 & 1\\
\end{array}\right|
=
\frac{1}{2} ⋅ \frac{-60}{-10} = 3
$$
$$
y_0 =
-\frac{1}{2} ⋅ \frac{M_{13}}{M_{11}}
= -\frac{1}{2} ⋅
\left|\begin{array}{cccc}
2 & 1 & 1\\
20 & 2 & 1\\
34 & 5 & 1\\
\end{array}\right|
\Bigg{/}
\left|\begin{array}{cccc}
1 & 1 & 1\\
2 & 4 & 1\\
5 & 3 & 1\\
\end{array}\right|
=
-\frac{1}{2} ⋅ \frac{40}{-10} = 2
$$
The radius, $r$, can then be calculated with Pythogoras' theorem, or using matrices again:
$$r^2 = x_0^2 + y_0^2 + \frac{M_{14}}{M_{11}}$$
*
*$M_{12}(A)$ is a minor of A — the determinant of $A$ without row $1$ or column $2$.
*Lines on either side of a matrix indicate the determinant (e.g. $|A|$) (sometimes written $\det{(A)}$)
If you have the time, it's really worth learning about matrices — they make a lot of things much faster, and are just generally awesome! KhanAcademy has a pretty easy introduction to matrices and linear algebra.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving trigonometric Identity: $\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ I would like to try and prove
$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$$
using $LHS=RHS$ methods, i.e. pick a side and rewrite it to make it identical to the other side.
I found a quick way by doing this:
$$LHS = \frac{1+\sin x}{\cos x} = \frac{1+\sin x}{\cos x} \cdot \frac{1 - \tan x + \sec x}{1 - \tan x + \sec x}= \frac{1+\sin x+\cos x}{1-\sin x+\cos x} = RHS$$
but I feel that this is not a good way because I am manipulating the denominator of the LHS somewhat artificially, because I know it must be, in the end, $1-\sin x+\cos x$.
Does anyone have a better way of doing this?
|
We'll work on the RHS. We'll multiply $\frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ with the conjugate of the denominator. We'll do this from right to left, i.e. we pick the RHS and walk through to LHS.
Solution 1:
$$\require{cancel}\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1+\sin x-\cos x}{1+\sin x-\cos x}\\&=\frac{1+\sin x\cancel{-\cos x}+\sin x+\sin^{2}x\cancel{-\sin{x}\cos{x}}\cancel{+\cos x}\cancel{+\sin{x}\cos{x}}-\cos^{2}x}{1\cancel{+\sin x}\cancel{-\cos x}\cancel{-\sin x}-\sin^{2}x+\sin{x}\cos{x}\cancel{+\cos x}+\sin{x}\cos{x}-\cos^{2}x}\\&=\frac{1+2\sin x+\sin^{2}x-\cos^{2}x}{1-\sin^{2}x+2\sin{x}\cos{x}-\cos^{2}x}\\&=\frac{1+2\sin x+\sin^{2}x-\cos^{2}x}{\cancel{\cos^{2}x}+2\sin{x}\cos{x}\cancel{-\cos^{2}x}}\\&=\frac{1+2\sin x+\sin^{2}x-\cos^{2}x}{2\sin{x}\cos{x}}\\&=\frac{\sin^{2}x\cancel{+\cos^{2}x}+2\sin x+\sin^{2}x\cancel{-\cos^{2}x}}{2\sin{x}\cos{x}}\\&=\frac{\cancel2\sin^{2}x+\cancel2\sin x}{\cancel2\sin{x}\cos{x}}\\&=\frac{\sin^{2}x+\sin x}{\sin{x}\cos{x}}\\&=\frac{\cancel{\sin{x}}\left(\sin x+1\right)}{\cancel{\sin{x}}\cos{x}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
Solution 2:
$$\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1-\sin x-\cos x}{1-\sin x-\cos x}\\&=\frac{1\cancel{-\sin x}\cancel{-\cos x}\cancel{+\sin x}-\sin^{2}x-\sin{x}\cos{x}\cancel{+\cos x}-\sin{x}\cos{x}-\cos^{2}x}{1-\sin x\cancel{-\cos x}-\sin x+\sin^{2}x\cancel{+\sin{x}\cos{x}}\cancel{+\cos x}\cancel{-\sin{x}\cos{x}}-\cos^{2}x}\\&=\frac{1-\sin^{2}x-2\sin{x}\cos{x}-\cos^{2}x}{1-2\sin x+\sin^{2}x-\cos^{2}x}\\&=\frac{\cancel{\sin^{2}}x+\cancel{\cos^{2}x}\cancel{-\sin^{2}x}-2\sin{x}\cos{x}\cancel{-\cos^{2}x}}{\sin^{2}x-2\sin x+\sin^{2}x}\\&=\frac{-\cancel2\sin x\cos x}{\cancel2\sin^{2}x-\cancel2\sin x}\\&=\frac{-\sin{x}\cos{x}}{\sin^{2}x-\sin x}\\&=\frac{\sin{x}\cos{x}}{\sin x-\sin^{2}x}\\&=\frac{\cancel{\sin{x}}\cos{x}}{\cancel{\sin x}(1-\sin x)}\\&=\frac{\cos{x}}{1-\sin{x}}\cdot\frac{1+\sin{x}}{1+\sin{x}}\\&=\frac{\cos{x}(1+\sin{x})}{1-\sin^2{x}}\\&=\frac{\cancelto{1}{\cos{x}}(1+\sin{x})}{\cancelto{\cos{x}}{\cos^2{x}}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
Solution 3:
$$\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1-\sin x+\cos x}{1-\sin x+\cos x}\\&=\frac{1+\sin x\cancel{-\cos x}+\sin x+\sin^{2}x\cancel{-\sin{x}\cos{x}}\cancel{+\cos x}\cancel{+\sin{x}\cos{x}}-\cos^{2}x}{1+\cancel{\sin x}\cancel{-\cos x}\cancel{-\sin x}-\sin^{2}x+\sin{x}\cos{x}\cancel{+\cos x}+\sin{x}\cos{x}-\cos^{2}x}\\&=\frac{1+2\cos x-\sin^{2}x+\cos^{2}x}{1-2\sin x+2\cos x+\sin^{2}x-2\sin{x}\cos{x}+\cos^{2}x}\\&=\frac{\cancel{\sin^{2}x}+\cos^{2}x+2\cos x\cancel{-\sin^{2}x}+\cos^{2}x}{\sin^{2}x+\cos^{2}x-2\sin x+2\cos x+\sin^{2}x-2\sin{x}\cos{x}+\cos^{2}x}\\&=\frac{\cancel2\cos^{2}x+\cancel2\cos x}{\cancel2\sin^{2}x+\cancel2\cos^{2}x-\cancel2\sin x+\cancel2\cos x-\cancel2\sin{x}\cos{x}}\\&=\frac{\cos^{2}x+\cos x}{\sin^{2}x+\cos^{2}x-\sin x+\cos x-\sin{x}\cos{x}}\\&=\frac{\cos{x}\left(1+\cos x\right)}{1-\sin x-\sin{x}\cos{x}+\cos x}\\&=\frac{\cos{x}\left(1+\cos x\right)}{1-\sin x+\cos{x}\left(1-\sin x\right)}\\&=\frac{\cos{x}\cancel{\left(1+\cos x\right)}}{\cancel{\left(1+\cos x\right)}\left(1-\sin x\right)}\\&=\frac{\cos{x}}{1-\sin{x}}\cdot\frac{1+\sin{x}}{1+\sin{x}}\\&=\frac{\cos{x}(1+\sin{x})}{1-\sin^2{x}}\\&=\frac{\cancelto{1}{\cos{x}}(1+\sin{x})}{\cancelto{\cos{x}}{\cos^2{x}}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
Solution 4:
$$\begin{aligned}\frac{1+\sin x+\cos x}{1-\sin x+\cos x}&=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}\cdot\frac{1+\sin x+\cos x}{1+\sin x+\cos x}\\&=\frac{1+\sin x+\cos x+\sin x+\sin^{2}x+\sin{x}\cos{x}+\cos x+\sin{x}\cos{x}+\cos^{2}x}{1\cancel{+\sin x}+\cos x\cancel{-\sin x}-\sin^{2}x\cancel{-\sin{x}\cos{x}}+\cos x\cancel{+\sin{x}\cos{x}}+\cos^{2}x}\\&=\frac{1+2\sin x+2\cos x+\sin^{2}x+2\sin{x}\cos{x}+\cos^{2}x}{1+2\cos x-\sin^{2}x+\cos^{2}x}\\&=\frac{\cancelto{1}{2}+\cancel2\sin x+\cancel2\cos x+\cancel2\sin{x}\cos{x}}{\cancel2\cos x+\cancel2\cos^{2}x}\\&=\frac{1+\sin x+\cos x+\sin{x}\cos{x}}{\cos x+\cos^{2}x}\\&=\frac{1+\sin x+\cos{x}\left(1+\sin x\right)}{\cos{x}\left(1+\cos x\right)}\\&=\frac{\cancel{\left(1+\cos x\right)}\left(1+\sin x\right)}{\cos{x}\cancel{\left(1+\cos x\right)}}\\&=\frac{1+\sin x}{\cos x}\end{aligned}$$
I hope this helps.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Algebraic manipulation help I'm self-teaching geometric progressions and was asked to find the sum to $n$ terms of $$x + 1 + \frac{1}{x} + \cdots$$ so I used the formula $$\frac{a(r^n - 1)}{r - 1}$$ with $a = x$ and $r = \frac{1}{x}$, and I arrived at $$\frac{x^{1-n} - x}{x^{-1} - 1}$$ The answer in the book is $$\frac{x^n-1}{x^{n-2}(x - 1)}$$
I cannot see how to manipulate my answer to get to the one in the book (though I know they're equivalent as Wolfram Alpha says so). Hope someone can show me how to approach this?
|
Here’s a computational verification.
$$\begin{align*}
\frac{x^{1-n}-x}{x^{-1}-1}&=\frac{x^{1-n}-x}{x^{-1}-1}\cdot\frac{x^{n-1}}{x^{n-1}}\\
&=\frac{1-x^n}{x^{n-2}-x^{n-1}}\\
&=\frac{1-x^n}{x^{n-2}\left(1-x\right)}\\
&=\frac{1-x^n}{x^{n-2}\left(1-x\right)}\cdot\frac{-1}{-1}\\
&=\frac{x^n-1}{x^{n-2}(x-1)}
\end{align*}$$
Had I been confronted with your result, however, I’d automatically have turned negative exponents into positive ones and then simplified the four-storey fraction:
$$\begin{align*}
\frac{x^{1-n}-x}{x^{-1}-1}&=\frac{\frac1{x^{n-1}}-x}{\frac1x-1}\\
&=\frac{\frac1{x^{n-1}}-x}{\frac1x-1}\cdot\frac{x^{n-1}}{x^{n-1}}\\
&=\frac{1-x^n}{x^{n-2}-x^{n-1}}\\
&=\frac{1-x^n}{x^{n-2}(1-x)}\;.
\end{align*}$$
Whether I leave it like that or flip the differences around depends on whether I expect to be dealing with $x<1$ (leave it) or $x>1$ (flip it).
|
{
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"url": "https://math.stackexchange.com/questions/214150",
"timestamp": "2023-03-29T00:00:00",
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|
Mechanism Behind Dot Product and Least Square Sorry for my ignorance, but I want to know how the mechanism of finding the least square solutions or the closest points in Euclidean space works.
For example:
Find the closest point or points to $b =(−1,2)^T$ that lie on the line $x + y = 0$.
I know the answer is
$$\frac{\left( \begin{matrix} -1 \\ 2 \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)}{\left( \begin{matrix} 1 \\ -1 \end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)} \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right) = \left( \begin{matrix} -\frac{3}{2} \\ \frac{3}{2} \end{matrix} \right)$$
But what does the dot product between $\left( \begin{matrix} -1 \\ 2\end{matrix} \right) \cdot \left( \begin{matrix} 1 \\ -1 \end{matrix} \right)$ tell you? If I commute it the answer is $-3$, but what exactly is $-3$? Also, the denominator is $2$, again, what exactly is $2$ telling us here?
|
Clearly your first equation must be incorrect -- the dot product returns a scalar, so there is no way to obtain a vector on the right hand side, unless you are overloading your nomenclature.
The dot product returns the sum of the element-wise products of vectors. When you take the dot product of a vector with itself, then that dot product returns the sum of the squares of the component of that vector.
Thus, when you perform a transform such as $\frac{y^Tx}{x^Tx} = \frac{y\cdot x}{x\cdot x}$ you are dividing by the sum of the squares of $x$. The sum of the squares of a vector is, of course, the square of the distance of that vector from the origin.
|
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"url": "https://math.stackexchange.com/questions/214577",
"timestamp": "2023-03-29T00:00:00",
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|
If $\dfrac{4x^2-1}{4x^2-y^2}$ is an integer, then it is $1$ The problem is the following:
If $x$ and $y$ are integers such that $\dfrac{4x^2-1}{4x^2-y^2}=k$ is
also an integer, does it implies that $k=1$?
This equation is equivalent to $ky^2+(1-k)4x^2=1$ or to $(k-1)4x^2-ky^2=-1$. The first equation is a pell equation (if $k$ is a perfect square) and the second is a pell type equation (if $k-1$ is a perfect square). I've tried setting several values of $k$ to get some solutions but i got nothing. I'm starting to think that $k$ must be $1$.
|
This is a fun problem! Where did you find it?
There are no solutions except for $k=1$. Assume from now on that $k \neq 1$. Since $k$ is clearly odd, it is also not $0$ and we deduce that $k(k-1)>0$. It is convenient to set $M = k(k-1)$. Also, we may assume WLOG that $x$ and $y\geq0$.
As you did, rewrite the equation to
$$ky^2 - 4 (k-1) x^2 = 1$$
or
$$(ky)^2 - k(k-1) (2x)^2 = k.$$
Set $Y=ky$ and $X=2x$ so the equation is
$$Y^2 - M X^2 = k. \quad (\ast)$$
We will study the equation $(\ast)$. In the end, we will see that there are no solutions with $X$ even and $Y$ divisible by $k$.
The rest of this proof works inside the ring $R:=\mathbb{Z}[\sqrt{M}]$. Note that $M=k(k-1)$ is not square, so this is an integral domain. For an element $\alpha = a+b \sqrt{M} \in R$, set $\bar{\alpha} = a - b \sqrt{M}$.
Set $\epsilon = (2k-1) + 2 \sqrt{M}$. Note that $\epsilon \bar{\epsilon} = (2k-1)^2 - 4 k(k-1) = 1$, so $\epsilon$ is a unit of $R$.
Set $\delta = Y+X \sqrt{M}$. Since $\delta$ is a positive real, and $\epsilon>1$, there is some integer $n$ such that $\epsilon^n \leq \delta < \epsilon^{n+1}$. Write $\delta = \gamma \epsilon^n$. Since $\epsilon$ is a unit, $\gamma$ is in the ring $R$ and, by construction,
$$1 \leq \gamma < \epsilon.$$
We have
$$\epsilon = 2k-1 + 2 \sqrt{k(k-1)} < 2k+ 2k = 4k.$$
So
$$1 \leq \gamma < 4k.$$
But, also $\gamma \bar{\gamma} = \delta \bar{\delta} =k$. So
$$\frac{1}{4} \leq \bar{\gamma} \leq k.$$
Write $\gamma = U + V \sqrt{M}$. So
$$\begin{matrix}
1 & \leq & U+V\sqrt{M} & < & 4k \\
\frac{1}{4} & \leq & U-V\sqrt{M} & < & k \\
\end{matrix}.$$
Solving for $V$, we have
$$\frac{-k}{2 \sqrt{M}} < V < \frac{2k}{\sqrt{M}}.$$
The LHS is $\approx -1/2$, and the right hand side is slightly larger than $2$. So $V$ is $0$, $1$ or $2$. We break into cases:
$\bullet$ If $V=0$, then the equation $\gamma \bar{\gamma} =k$ gives $U^2 =k$. So $k$ is a square, say $k=m^2$ and $M=m^2(m^2-1)$. We have
$$Y+X \sqrt{M} = m ((2k-1)+2 \sqrt{M})^n = m (2m^2-1 + 2 m \sqrt{m^2-1})^n.$$
An easy induction on $n$ shows that $Y \equiv (-1)^n m \mod m^2$ so, except when $m=1$, we do not have $Y$ divisible by $k=m^2$. Of course, the case $m=1$ corresponds to $k=1$.
$\bullet$ If $V=1$ then $U^2 - k(k-1) = k$ so $U=k$. We have
$$Y+ X \sqrt{M} = (k+\sqrt{M}) \cdot ((2k-1)+2 \sqrt{M})^n.$$
An easy induction on $n$ shows that $X$ is odd.
$\bullet$ If $V=2$ then $U^2 - 4k(k-1) = k$ so $4k^2-3k = U^2$. This gives $64k^2 - 48 k + 9= 64 U^2+9$ or $(8k-3)^2 - (8U)^2 = 9$. The only ways to write $9$ as a difference of squares are $3^2-0^2$ and $5^2-4^2$. The former gives $k=0$, but we saw that $k$ must be odd; the latter gives no solution since $8$ does not divide $4$.
|
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|
Find the value of : $\lim_{n\to\infty}\prod_{k=1}^n\cos\left(\frac{ka}{n\sqrt{n}}\right)$ Find the limit (where a is a constant)
$\lim_{n\to\infty}\prod_{k=1}^n\cos\left(\frac{ka}{n\sqrt{n}}\right)$
I think the answer is $1-a^2/6$
|
Let $$f(n) = \prod_{k=1}^n \cos \left(\dfrac{ka}{n \sqrt{n}}\right)$$
$$g(n) = \log (f(n)) = \sum_{k=1}^{n} \log \left(\cos \left(\dfrac{ka}{n \sqrt{n}}\right) \right) = \sum_{k=1}^{n} \log \left(1 - \dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right)$$
$$\log \left(1 - \dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right) = -\left(\dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right) + \mathcal{O} \left(\dfrac{k^4}{n^6} \right)$$
$$\sum_{k=1}^{n} \log \left(1 - \dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right) = \sum_{k=1}^{n} \left( -\dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right)\\ = -\dfrac{a^2}{2n^3} \dfrac{n(n+1)(2n+1)}{6} + \mathcal{O}(1/n)$$
$$\lim_{n \to \infty }\sum_{k=1}^{n} \log \left(1 - \dfrac{\left(\dfrac{ka}{n \sqrt{n}}\right)^2}2 + \mathcal{O} \left( \dfrac{k^4}{n^6}\right)\right) = -\dfrac{a^2}{6}$$
Hence, $$\prod_{k=1}^{\infty} \cos \left(\dfrac{ka}{n \sqrt{n}}\right) = \exp(-a^2/6)$$
The solution you have $1-a^2/6$ is a first order approximation to $\exp(-a^2/6)$ since $$\exp(x) = 1 + x + \mathcal{O}(x^2)$$
|
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|
$x^2 \equiv 2x \pmod m$ Toward counting the solutions for the congruence $x^2 \equiv 2x \pmod n$, if we write $m$
as $m = p_1^{a_1}p_2^{a_2}...p_r^{a_r}$
we have the following equivalent system of congruence equations:
$p_1^{a_1}\mid x(x-2)$
$p_2^{a_2}\mid x(x-2)$
.
.
.
$p_r^{a_r}\mid x(x-2)$
I am thinking of the two cases of the relationships of $x$ and $x-2$ where
*
*$gcd(x,x-2) = 1$
*$gcd(x,x-2) = 2$
and consider the number of solutions for each case. Am I on the right track?
|
Using your approach, let for prime $p,$ $p\mid x$ and $p\mid (x-2)$
$\implies p$ divides $x-(x-2)=2\implies p=2$ for $p^n\mid x(x-2)$
So, if prime $p>3,$ either $p^n\mid x$ or $p^n\mid (x-2)$
So, there are $2$ in-congruent solutions, namely, $0,2\pmod {p^n}$ of $x^2\equiv2x \pmod {p^n}$.
If case $p=2,x$ must be even $\implies (x,x-2)=2\implies(\frac x2,\frac{x-2}2)=1$
So, $2^n\mid x(x-2)\implies 2^{n-2}\mid \frac x2\cdot\frac{(x-2)}2$ if $n\ge 3$
So, either $2^{n-2}\mid \frac x2$ or $2^{n-2}\mid \frac{(x-2)}2$
So, $x\equiv0,2 \pmod{2^{n-1}}\implies x\equiv0,0+2^{n-1},2,2+2^{n-1}\pmod{2^n}$ i.e., there are $4$ in-congruent solutions.
If $n=2, x(x-2)\equiv 0\pmod4\implies x\equiv0,2\pmod4$ i.e., $2$ in-congruent solutions.
If $n=1, x(x-2)\equiv 1\pmod2\implies x\equiv 1\pmod2$ i.e., exactly $1$ in-congruent solution.
Now, if $ax^2+bx+c\equiv 0\pmod {m_1}$ and $ax^2+bx+c\equiv 0\pmod {m_2}$ have $t_1,t_2$ solutions respectively(where $(m_1,m_2)=1$),
$ax^2+bx+c\equiv 0\pmod {m_1\cdot m_2}$ will have $t_1\cdot t_2$ solutions.
Alternatively,
$x^2\equiv2x\pmod n\implies (x-1)^2\equiv 1\pmod n\implies y^2\equiv1$ if $y=x-1$
If $y^2\equiv1 \pmod {q^n}$ where prime $q>2$
So, $q^n \mid (y-1)(y+1)$ but $(y-1,y+1)\mid 2$
either $q^n\mid (y-1)$ or $q^n\mid (y+1)$
So, there are $2$ in-congruent solutions, namely, $y\equiv \pm 1\pmod {q^n}$ of $y^2\equiv1 \pmod {q^n}$.
If $y^2\equiv1 \pmod {2^n}$,
so, $2^n\mid (y-1)(y+1)$ ,here $(y-1,y+1)=2$ as $y$ is odd.
so, $2^{n-2} \mid\frac{(y-1)}2\cdot\frac{(y+1)}2$ where
$(\frac{y-1}2,\frac{y+1}2)=1$ if $n\ge 3$
so, either $2^{n-1}\mid (y-1)$ or $2^{n-1}\mid (y+1)$
So, there are $2$ in-congruent solutions, namely, $y\equiv \pm 1\pmod {2^{n-1}}$
So, there will be $4$ in-congruent solutions, namely, $y\equiv \pm 1,2^{n-1}\pm1\pmod {2^n}$
If $n=2, y^2\equiv 1\pmod4\implies y\equiv\pm 1\pmod4$ i.e., $2$ in-congruent solutions.
If $n=1, y^2\equiv 1\pmod2\implies y\equiv 1\pmod2$ i.e., exactly $1$ in-congruent solution.
|
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|
Solving the recurrence relation $T(n)=2T(n/4)+\sqrt{n}$ I've solved $T(n)=2T(n/4)+\sqrt{n}$ to equal $2^{\log_{4}n}(\log_{4}n+1)$, but I'm not sure how to solve it directly.
I have:
$2(2T(\frac{n}{16})+\sqrt{\frac{n}{4}})+\sqrt{n} = 4T(\frac{n}{16})+2\sqrt{n}$
$2(4T(\frac{n}{64})+2\sqrt{\frac{n}{16}})+\sqrt{n} = 8T(\frac{n}{64})+2\sqrt{n}$
$2(8T(\frac{n}{256})+2\sqrt{\frac{n}{64}})+\sqrt{n}=16T(\frac{n}{256})+\frac{5}{4}\sqrt{n}$
I'm not seeing a pattern here and I'm not sure I'm modifying $n$ as necessary. What's the problem?
|
Let's turn the equation $T(n) = 2 T(n/4) + \sqrt{n}$ into a recurrence equation. To this end, let $f(m) = T(4^m p)$ for some $p>0$. Then
$$
f(m) = 2 f(m-1) + \sqrt{p} 2^m
$$
which can be systematically solved. First rewrite it as
$$
2^{-m} f(m) - 2^{-(m-1)} f(m-1) = \sqrt{p}
$$
Then sum equations from $m=1$ to some upper bound $k$:
$$
\sum_{m=1}^k \left(2^{-m} f(m) - 2^{-(m-1)} f(m-1) \right) = \sum_{k=1}^m \sqrt{p}
$$
The sum on the left-hand-side telescopes:
$$\begin{eqnarray}
\sum_{m=1}^k \left(2^{-m} f(m) - 2^{-(m-1)} f(m-1) \right) &=& \left(2^{-m} f(m) - \color\green{2^{-(m-1)} f(m-1)}\right) + \\
&\phantom{=}& \left(\color\green{2^{-(m-1)} f(m-1)} - 2^{-(m-2)} f(m-2)\right) + \\
&\phantom{=}& \vdots \\
&\phantom{=}& \left( 2^{-2} f(2) - \color\green{2^{-1} f(1)} \right) +\\
&\phantom{=}& \left( \color\green{2^{-1} f(1)} - 2^{-0} f(0) \right) \\
&=& 2^{-m} f(m) - f(0)
\end{eqnarray}
$$
Hence we arrive at the solution
$$
f(m) = 2^m \left( m \sqrt{p} + f(0) \right)
$$
since $m = \log_4 \left(\frac{n}{p}\right)$ we get:
$$
T(n) = \sqrt{\frac{n}{p}} \left( \sqrt{p} \cdot \log_4 \frac{n}{p} + f(0)\right) = \sqrt{n} \log_4(n) + d \sqrt{n}
$$
where $d$ is a free constant to be determined by the initial condition.
|
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|
General term of the series Given the series:
$$\sqrt c + \sqrt{c\sqrt c} + \sqrt{c\sqrt{c\sqrt c}} + \ldots$$
where $0 < c < 1$
What is the general term of this series?
|
$a_1 = \sqrt{c} = c^{1/2}$, $a_2 = \sqrt{c\sqrt{c}} = \sqrt{c^{3/2}} = c^{3/4}$, $a_3 = \sqrt{c\sqrt{c\sqrt{c}}} = \sqrt{c \times c^{3/4}} = c^{7/4}$. In general, $$a_{n+1} = \sqrt{c a_{n-1}}$$ with $a_0 = 1$.
This gives us
$a_n = c^{1-1/2^n}$
|
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|
integral of exponential divided by polynomial I would like to solve the integral $$A\int_{-\infty}^\infty\frac{e^{-ipx/h}}{x^2+a^2}dx$$
where h and a are positive constants.
Mathematica gives the solution as $\frac\pi{a}e^{-|p|a/h}$, but I have been trying to reduce my reliance on mathematica. I have no idea what methods I would use to solve it.
Is there a good (preferably online) resource where I could look up methods for integrals like this fairly easily?
|
$\mathbf{Method\;1: }$ Integral Fourier Transform
Consider the function $f(t)=e^{-a|t|}$, then the Fourier transform of $f(t)$ is given by
$$
\begin{align}
F(\omega)=\mathcal{F}[f(t)]&=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\\
&=\int_{-\infty}^{\infty}e^{-a|t|}e^{-i\omega t}\,dt\\
&=\int_{-\infty}^{0}e^{at}e^{-i\omega t}\,dt+\int_{0}^{\infty}e^{-at}e^{-i\omega t}\,dt\\
&=\lim_{u\to-\infty}\left. \frac{e^{(a-i\omega)t}}{a-i\omega} \right|_{t=u}^0-\lim_{v\to\infty}\left. \frac{e^{-(a+i\omega)t}}{a+i\omega} \right|_{t=0}^v\\
&=\frac{1}{a-i\omega}+\frac{1}{a+i\omega}\\
&=\frac{2a}{\omega^2+a^2}.
\end{align}
$$
Next, the inverse Fourier transform of $F(\omega)$ is
$$
\begin{align}
f(t)=\mathcal{F}^{-1}[F(\omega)]&=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\,d\omega\\
e^{-a|t|}&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2a}{\omega^2+a^2}e^{i\omega t}\,d\omega\\
\frac{\pi e^{-a|t|}}{a}&=\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{\omega^2+a^2}\,d\omega.
\end{align}
$$
Comparing the last integral to the problem yields $t=-\frac{p}{h}$. Thus,
$$
\int_{-\infty}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx=\frac{\pi e^{-a\left|\frac{p}{h}\right|}}{a}.
$$
$\mathbf{Method\;2: }$
Note that:
$$
\int_{y=0}^\infty e^{-(x^2+a^2)y}\,dy=\frac{1}{x^2+a^2},
$$
therefore
$$
\int_{x=0}^\infty\int_{y=0}^\infty e^{-(x^2+a^2)y}\;e^{-\frac{ipx}{h}}\,dy\,dx=\int_{x=0}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx
$$
Rewrite
$$
\begin{align}
\int_{x=0}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx&=\int_{y=0}^\infty\int_{x=0}^\infty e^{-(yx^2+\frac{ip}{h}x+a^2y)}\,dx\,dy\\
&=\int_{y=0}^\infty e^{-a^2y} \int_{x=0}^\infty e^{-\left(yx^2+\frac{ip}{h}x\right)}\,dx\,dy.
\end{align}
$$
In general
$$
\begin{align}
\int_{x=0}^\infty e^{-(ax^2+bx)}\,dx&=\int_{x=0}^\infty \exp\left(-a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right)\right)\,dx\\
&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\
\end{align}
$$
Let $u=x+\frac{b}{2a}\;\rightarrow\;du=dx$, then
$$
\begin{align}
\int_{x=0}^\infty e^{-(ax^2+bx)}\,dx&=\exp\left(\frac{b^2}{4a}\right)\int_{x=0}^\infty \exp\left(-a\left(x+\frac{b}{2a}\right)^2\right)\,dx\\
&=\exp\left(\frac{b^2}{4a}\right)\int_{u=0}^\infty e^{-au^2}\,du.\\
\end{align}
$$
The last form integral is Gaussian integral that equals to $\frac{1}{2}\sqrt{\frac{\pi}{a}}$. Hence
$$
\int_{x=0}^\infty e^{-(ax^2+bx)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\exp\left(\frac{b^2}{4a}\right).
$$
Thus
$$
\int_{x=0}^\infty e^{-(yx^2+\frac{ip}{h}x)}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(\frac{\left(\frac{ip}{h}\right)^2}{4y}\right)=\frac{1}{2}\sqrt{\frac{\pi}{y}}\exp\left(-\frac{p^2}{4h^2y}\right).
$$
Next
$$
\int_{x=0}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx=\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-a^2y-\frac{p^2}{4h^2y}\right)}{\sqrt{y}}\,dy.
$$
In general
$$
\begin{align}
\int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv\\
&=2\int_{v=0}^\infty \exp\left(-a\left(v^2+\frac{b}{av^2}\right)\right)\,dv\\
&=2\int_{v=0}^\infty \exp\left(-a\left(v^2-2\sqrt{\frac{b}{a}}+\frac{b}{av^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dv\\
&=2\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dv\\
&=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\
\end{align}
$$
The trick to solve the last integral is by setting
$$
I=\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv.
$$
Let $t=-\frac{1}{v}\sqrt{\frac{b}{a}}\;\rightarrow\;v=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dv=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then
$$
I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt.
$$
Let $t=v\;\rightarrow\;dt=dv$, then
$$
I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Adding the two $I_t$s yields
$$
2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then
$$
I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}.
$$
Thus
$$
\begin{align}
\int_{y=0}^\infty \frac{\exp\left(-ay-\frac{b}{y}\right)}{\sqrt{y}}\,dy&=2\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\
&=\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}\\
\end{align}
$$
and
$$
\begin{align}
\int_{-\infty}^{\infty}\frac{e^{-\frac{ipx}{h}}}{x^2+a^2}\,dx&=2\cdot\frac{\sqrt{\pi}}{2}\int_{y=0}^\infty \frac{\exp\left(-a^2y-\frac{p^2}{4h^2y}\right)}{\sqrt{y}}\,dy\\
&=\sqrt{\pi}\cdot\sqrt{\frac{\pi}{a^2}}\;e^{-2\sqrt{a^2\cdot\frac{p^2}{4h^2}}}\\
&=\frac{\pi}{a}\;e^{-\frac{pa}{h}}.
\end{align}
$$
$$
\text{# }\mathbb{Q.E.D.}\text{ #}
$$
|
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Inequality $\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{z^4+x+y}}\geq 2\sqrt{3}$ If we Let $x,y,z>0$ such that $x+y+z=3$. how to prove that
$$\frac{x^3+1}{\sqrt{x^4+y+z}}+\frac{y^3+1}{\sqrt{y^4+z+x}}+\frac{z^3+1}{\sqrt{z^4+x+y}}\geq 2\sqrt{3}$$
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We shall prove the inequality with the weaker condition $x, y, z \geq 0$.
$$\sum_{cyc}{\frac{x^3+1}{\sqrt{x^4+y+z}}}=\sum_{cyc}{\frac{x^3+1}{\sqrt{x^4-x+3}}}$$
Let $f(x)={\frac{x^3+1}{\sqrt{x^4-x+3}}}$. Differentiate twice, and note that $f''(x)$ is positive for $0 \leq x \leq 1$, and that $f''(x)=0$ for exactly 1 value of $x$ between $0$ and $2$.
We shall show that the minimum must be achieved when 2 of the variables are equal. Consider 2 cases.
Case 1: 2 of the 3 variables are $\leq 1$. Then $f(x)$ is convex over $[0, 1]$, so by Jensen's inequality, the minimum must be achieved when the 2 variables are equal.
Case 2: At most 1 of the variables are $\leq 1$. Then all the variables are $\leq 2$. Since $f(x)$ is differentiable and has 1 inflection point in $[0, 2]$, by (n-1)-equal value principle, we have extrema only when 2 of the variables are equal.
(Refer to page 15 of http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf for a proof of (n-1)-equal value principle)
As such, we have shown that the minimum is achieved when 2 of the variables are equal. Set $y=x, z=3-2x$ to get the following 1 variable inequality:
$$2\left(\frac{x^3+1}{\sqrt{x^4-x+3}}\right)+\frac{(3-2x)^3+1}{\sqrt{(3-2x)^4+2x}} \geq 2 \sqrt{3}$$
Differentiating to find extrema over $[0, 1.5]$, we get $x=1$ as the only extrema. It thus suffices to check $x=0, x=1, x=1.5$.
$x=0$ gives $\frac{2}{\sqrt{3}}+\frac{28}{9}>2\sqrt{3}$. $x=1$ gives $2\sqrt{3}$. $x=1.5$ gives $2(\frac{\frac{35}{8}}{\sqrt{\frac{105}{16}}})+\frac{1}{\sqrt{3}}>2 \sqrt{3}$.
Thus $$\sum_{cyc}{\frac{x^3+1}{\sqrt{x^4+y+z}}} \geq 2\sqrt{3}$$
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|
How to take the gradient of the quadratic form? It's stated that the gradient of:
$$\frac{1}{2}x^TAx - b^Tx +c$$
is
$$\frac{1}{2}A^Tx + \frac{1}{2}Ax - b$$
How do you grind out this equation? Or specifically, how do you get from $x^TAx$ to $A^Tx + Ax$?
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Yet another approach.
We will utilize the following the identities
*
*Trace and Frobenius product relation $$\left\langle A, B \right\rangle={\rm tr}(A^TB) = A:B$$ or $$\left\langle A^T, B \right\rangle ={\rm tr}(AB) = A^T:B$$
*Cyclic property of Trace/Frobenius product
\begin{align}
\left\langle A, B C \right\rangle \equiv A:BC
&= AC^T:B\\
&= B^TA:C\\
&= BC : A\\
&= {\text{etc.}} \cr
\end{align}
Let $f(x) := \left( \frac{1}{2} x^T A x - b^T x + c \right) $.
We obtain the differential first, and then the gradient subsequently.
\begin{align}
d\,f(x)
&= d\left( \frac{1}{2} x^T A x - b^T x + c \right) \\
&= d\left( \frac{1}{2} \left( x: A x \right) - \left( b : x \right) + c \right) \\
&= \frac{1}{2} \left[ \left( dx: A x \right) + \left( x: A dx \right) \right] - \left( b : dx \right) \\
&= \frac{1}{2} \left[ \left( A x : dx \right) + \left( A^Tx: dx \right) \right] - \left( b : dx \right) \\
&= \frac{1}{2} \left[ \left(A + A^T \right)x: dx \right] - \left( b : dx \right) \\
&= \left( \frac{1}{2} \left[ \left(A + A^T \right)x\right] - b \right): dx \\
\end{align}
Thus the gradient is
$$ \eqalign {
\frac { \partial} {\partial x}f(x) &= \frac{1}{2} \left[ \left(A + A^T \right)x\right] - b \\ &= \frac{1}{2} A^T x + \frac{1}{2} A x - b . \cr
} $$
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|
Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
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$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} =\frac{a^4}{a^2+ab}+\frac{b^4}{b^2+bc}+\frac{c^4}{c^2+ac}\geq \frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2+bc+ca+ab}.\tag{1}$$ But $$a^2+b^2+c^2 \geq ab+bc+ca $$ so
$$a^2+b^2+c^2+ab+bc+ca \leq 2(a^2+b^2+c^2)$$
$$(1) \geq \frac{(a^2+b^2+c^2)^2}{2(a^2+b^2+c^2)} \geq \frac{ab+bc+ca}{2}.$$
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Solving 4 Quadratic Simultaneous Equations
Consider the equations:
$$\begin{cases}
ab+c+d=3\\
bc+d+a=5\\
cd+a+b=2\\
da+c+b=6
\end{cases}$$
Where $a,b,c,d \in \mathbb{R}$. How can we find $a,b,c,d$?
The furthest I've got is by adding the first two equations and the last two equations together to get:
$ab+bc+c+d+d+a=(b+1)(a+c)+2d=8$ and $(d+1)(a+c)+2b=8$
We can rearrange these (assuming $b,d \neq -1$) to get $$\frac{8-2d}{b+1}=\frac{8-2b}{d+1} \implies -d^{2}+3d+4=-b^{2}+3b+4$$
Which leads us further to
$$b^{2}-d^{2}-3(b-d)=0 \implies (b-d)(b+d-3)=0$$
Now since $b=d$ gives us an absurdity (equations 1 and 4 reduce to 3=6), we must have $b+d=3$. This is as far as I can get unfortunately, since trying the same approach with the other pairs doesn't work in the same way.
Help would be much appreciated; the problem is from the BMO1 2004 paper.
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Take $\#1 - \#2 + \#3 - \#4$ to get $(b-d)(a-c)=-6$
$\#1 + \#2 - \#3 - \#4$ gives $(b-d)(a+c-2) = 0$
$\#1 - \#2 - \#3 + \#4$ gives $(b+d-2)(a-c)=2$.
So now $a+c-2 = 0$, and $$\frac{1}{a-c} = \frac{b-d}{-6} = \frac{b+d-2}{2}$$
That gives us $d=3-2b$ and $a-c = \dfrac{-2}{b-1}$ (in particular, $b \ne 1$).
Together with $a+c-2=0$ we get
$a = \dfrac{b-2}{b-1}$, $c = \dfrac{b}{b-1}$. Finally, substitute into $\#1$ to get
$-b+3=3$, or $b=0$, and thus $a=2$, $c=0$, $d=3$.
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|
Triple integral for Volume of a region $Q$ is the solid bounded by the plane $x+2y+2z=2$ and above paraboloid $x=z^2+y^2$. Setup the triple integral for the volume of $Q$.
I tried to find vertices and go from there bu t got confused. I need help with this problem.
Thanks in advance.
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When I approach a problem like this I consider a point $(x,y,z)$ and work towards discovering a set of 3 inequalities which describe an arbitrary point in the given region. The paraboloid $x=z^2+y^2$ opens into the positive $x$-half volume. On the other hand $x = 2-2y-2z$ intersects the paraboloid when $z^2+y^2=2-2y-2z$ which is also written as $(z+1)^2+(y+1)^2=4$. Observe $z^2+y^2 \leq x \leq 2-2y-2z$ where $y,z$ are bounded by $0 \leq (z+1)^2+(y+1)^2 \leq 4$. We can unwrap the $y,z$ inequality as $ (z+1)^2 \leq 4-(y+1)^2$ or $-\sqrt{4-(y+1)^2} -1 \leq z \leq \sqrt{4-(y+1)^2} -1$ where $-3 \leq y \leq 1$. To summarize:
*
*$-3 \leq y \leq 1$ puts the point $(x,y,z)$ between the $y=-3$ and $y=1$ plane.
*$-\sqrt{4-(y+1)^2} -1 \leq z \leq \sqrt{4-(y+1)^2} -1$ places $(x,y,z)$ inside the cylinder $(z+1)^2+(y+1)^2=4$.
*$z^2+y^2 \leq x \leq 2-2y-2z$
When setting up the volume integral you want to put the numerical bounds on the outside and the double-variable inequality on the inside. Remember the integral ,if it exists, is a number so the bounds must be ordered so that the result is a number.
Also, pragmatically speaking, if you were to calculate this then I would urge you to use polar coordinates $y+1=r\cos \theta$ and $z+1 = r\sin \theta$ so the $y,z$ bounds are replaced with $0 \leq r \leq 2$ and $0 \leq \theta \leq 2\pi$. Then only the $x$ bounds would involve variables.
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|
How do I find a value for $ 2^{156221} - 1\pmod 9$? Again the problem is: Calculate the value of:
$$\left(2^{156221} - 1\right) \bmod 9$$
I have no idea how to find a solution to this and need help urgently!!
Thank you in advance.
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The problem has already been solved several times. Perhaps a marginally different point of view will be helpful. The hard part of the problem is to calculate
$$2^{156221}\pmod{9}.$$
Let us calculate $2^n\pmod{9}$, starting at $n=0$.
We have $2^0\pmod{9}=1$, $2^1\pmod{9}=2$, $2^2\pmod{9}=4$, $2^3\pmod{9}=8$, and $2^{4}\pmod{9}=8$. This last one is because $2^4=16$, and $16\pmod{9}=7$.
Continue. Let us calculate $2^5\pmod{9}$. There are two ways to do this. We can calculate $2^5$, getting $32$, and then reduce modulo $9$, getting $5$. Or else we can just multiply the previous answer, which is $7$, by $2$, and reduce modulo $9$.
Now calculate $2^6\pmod{9}$. Again, we could calculate in two ways. Either find $2^6$, and reduce modulo $9$. Or else take the previous answer of $5$, multiply by $2$, and reduce modulo $9$. We get $1$.
Continue, or else imagine continuing. What is $2^7\pmod{9}$? Take the previous answer, which is $1$, multiply by $2$, and reduce modulo $9$. We get $2$. What is $2^8\pmod{9}$? Take the previous answer, multiply by $2$, and reduce modulo $9$. We get $4$. You should continue this process a few more times.
So the pattern of remainders that we get goes like this:
$$1,2,4,8,7,5,1,2,4,8,7,5,1,2,4,8,7,5,1,2,4,\dots.$$
Note that the remainder is $1$ when the exponent is any multiple of $6$.
Now divide $156221$ on your calculator. We get a quotient of $q=26036$ (not important) and a remainder of $5$. So
$$156221=6q+5=156216+5.$$
Because $156216$ is a multiple of $6$, when we reach $n=156216$, our remainder is $1$, and we are starting the pattern of remainders all over again, and have to step forward until $216221$. That advances us forward by $5$ in our pattern, and there the remainder (by coincidence) is $5$.
More algebraically,
$$2^{156221}=2^{6q+5}=(2^6)^q 2^5.$$
But $2^{6}$ gives remainder $1$ when you divide by $9$. Therefore so does $(2^6)^q$. So our remainder is the same as the remainder when $2^5$ is divided by $9$, and that is $5$.
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A functional equation: $f\left(x^2-1\right)+2f\left(\frac{2x-1}{(x-1)^2}\right)=2-\frac{4}{x}+\frac{3}{x^2}$
If for all $ x > 1 $
$$
f\left(x^2-1\right)+2f\left(\frac{2x-1}{(x-1)^2}\right)=2-\frac{4}{x}+\frac{3}{x^2} \text,
$$
then $f(x)=?$
I don't know how to solve such equations. Help me please. Thank you.
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The solution is based on finding a change of variable $x\to\phi(x)$ such that $\phi(\phi(x))=x$ leading to a system of 2 equations in 2 variables.
Let's rewrite
$$\frac{2x-1}{\left(x-1\right)^{2}}=\frac{-x^{2}+2x-1+x^{2}}{\left(x-1\right)^{2}}=\frac{x^{2}-\left(x-1\right)^{2}}{\left(x-1\right)^{2}}=\frac{x^{2}}{\left(x-1\right)^{2}}-1$$
Which suggests that
$$x\to\frac{x}{x-1}$$
might be a good candidate. Indeed
$$\frac{2\left(\frac{x}{x-1}\right)-1}{\left(\frac{x}{x-1}-1\right)^{2}}=x^{2}-1$$
Making the change and not remembering to do this in the RHS as well we obtain:
$$f\left(\frac{2x-1}{\left(x-1\right)^{2}}\right)+2f(x^2-1)=2-\frac{4}{x}\left(x-1\right)+\frac{3}{x^{2}}\left(x-1\right)^{2}$$
Now multiply this equation by 2 and subtract the original one
$$3f(x^{2}-1)=2-\frac{4}{x}\left(2\left(x-1\right)-1\right)+\frac{3}{x^{2}}\left(2\left(x-1\right)^{2}-1\right)=\\2-\frac{4}{x}\left(2x-3\right)+\frac{3}{x^{2}}\left(2x^{2}-4x+1\right)=\\2-8+\frac{12}{x}+6-\frac{12}{x}+\frac{3}{x^{2}}=\frac{3}{x^{2}}$$
$$f(x^{2}-1)=\frac{1}{(x^{2}-1)+1}$$
$$f(x)=\frac{1}{x+1}$$
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|
Riemann-integrable function The following multiple choice question was asked in my exam but I don't know how to proceed:
Define $f:[0,1]\to [0,1]$ by $\displaystyle f(x)=\frac{2^{k}-1}{2^{k}}$ for $\displaystyle x\in [\frac{2^{k-1}-1}{2^{k-1}},\frac{2^{k}-1}{2^{k}}],k\geq 1$. Then $f$ is a Riemann-integrable function such that
1.$\displaystyle \int_{0}^{1}f(x)dx=\frac{2}{3}$
2.$\displaystyle \frac{1}{2}<\int_{0}^{1}f(x)dx<\frac{2}{3}$
3.$\displaystyle \int_{0}^{1}f(x) dx=1$
4.$\displaystyle \frac{2}{3}<\int_{0}^{1}f(x)dx<1$.
I know that function is said to be Riemann-integrable if its upper Riemann-integral and lower Riemann-integral exists and are same.
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$$\int_0 ^1f=\int_0^{\frac{1}{2}} \frac{1}{2}+\int_{\frac{1}{2}}^{\frac{3}{4}}\frac{3}{4}+\int_{\frac{3}{4}} ^{\frac{7}{8}}\frac{7}{8}+... =\sum_{k=1}^\infty \int_{\frac{2^{k-1}-1}{2^{k-1}}} ^{\frac{2^k-1}{2^k}} \frac{2^k-1}{2^k}=\sum_{k=1}^\infty (\frac{2^k-1}{2^k})(\frac{2^k-1}{2^k}-\frac{2^{k-1}-1}{2^{k-1}})=\sum_{k=1} ^\infty (\frac{2^k-1}{2^k})(\frac{1}{2^k})=\sum_{k=1} ^\infty \frac{2^k-1}{2^{2k}}=1-\frac{1}{3}=\frac{2}{3}$$
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If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple. Pythagoras stated that there exist positive natural numbers, $a$, $b$ and $c$ such that $a^2+b^2=c^2$. These three numbers, $a$, $b$ and $c$ are collectively known as a
Pythagorean triple. For example, $(8, 15, 17)$ is one of these triples as $8^2 + 15^2 = 64 + 225= 289 = 17^2$. Other examples of this triple are $(3, 4, 5)$ and $(5, 12, 13)$.
Using Proof by Contradiction, show that: If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple.
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We must suppose that $(a+1,b+1,c+1)$ is in fact a Pythagorean triple, and that $(a,b,c)$ is, too. Then we have $$(a+1)^2+(b+1)^2=(c+1)^2\tag{1}$$ and $$a^2+b^2=c^2.\tag{2}$$
Expand $(1)$--using for example that $(a+1)^2=a^2+2a+1$--and then use $(2)$ to eliminate all the squared terms from the resulting equation. You should be able to conclude (after gathering the $a,b,c$ terms on one side and other terms on the other) that $1$ is an even number.
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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.