Author Topic: Spec '10 - Help forum (Read 11237 times) Tweet Share

Chavi · « **on:** December 15, 2009, 02:44:16 pm »

Could i get some help with:
Factorize over R: z^4 + 64
and Factorize over C z^4 + 64

Thanks

TrueTears · « **Reply #1 on:** December 15, 2009, 03:15:05 pm »

Over $\mathbb{R} \implies \left(z^2\right)^2 + \left(\left(2\sqrt{2}\right)^2\right)^2$

Let $u = z^2$ and $x = \left(2\sqrt{2}\right)^2$ and let $y = \left(z^2\right)^2 + \left(\left(2\sqrt{2}\right)^2\right)^2$

$y=u^2+x^2$

$y = u^2+2ux+x^2-2ux$

$y = (u+x)^2-2ux$

$y = (u+x)^2 - \left(\sqrt{2ux}\right)^2$

$y = \left(u+x-\sqrt{2ux}\right)\left(u+x+\sqrt{2ux}\right)$

$y = \left(z^2+\left(2\sqrt{2}\right)^2-4z\right)\left(z^2+\left(2\sqrt{2}\right)^2+4z\right)$

$y = \left(z^2-4z+8\right)\left(z^2+4z+8\right)$

$\therefore z^4+64 = \left(z^2-4z+8\right)\left(z^2+4z+8\right)$

Ahmad · « **Reply #2 on:** December 15, 2009, 03:19:54 pm »

This is a special case of Sophie Germain's Identity.

TrueTears · « **Reply #3 on:** December 15, 2009, 03:24:24 pm »

Quote from: Ahmad on December 15, 2009, 03:19:54 pm

This is a special case of Sophie Germain's Identity.

lol I did that AIME question when I saw it in the algebra section in AnC

QuantumJG · « **Reply #4 on:** December 15, 2009, 03:38:09 pm »

Quote from: TrueTears on December 15, 2009, 03:15:05 pm

Over $\mathbb{R} \implies \left(z^2\right)^2 + \left(\left(2\sqrt{2}\right)^2\right)^2$

Let $u = z^2$ and $x = \left(2\sqrt{2}\right)^2$ and let $y = \left(z^2\right)^2 + \left(\left(2\sqrt{2}\right)^2\right)^2$

$y=u^2+x^2$

$y = u^2+2ux+x^2-2ux$

$y = (u+x)^2-2ux$

$y = (u+x)^2 - \left(\sqrt{2ux}\right)^2$

$y = \left(u+x-\sqrt{2ux}\right)\left(u+x+\sqrt{2ux}\right)$

$y = \left(z^2+\left(2\sqrt{2}\right)^2-4z\right)\left(z^2+\left(2\sqrt{2}\right)^2+4z\right)$

$y = \left(z^2-4z+8\right)\left(z^2+4z+8\right)$

$\therefore z^4+64 = \left(z^2-4z+8\right)\left(z^2+4z+8\right)$

That is pretty cool!

Chavi · « **Reply #5 on:** December 15, 2009, 09:13:08 pm »

Quote from: TrueTears on December 15, 2009, 03:15:05 pm

$y = u^2+2ux+x^2-2ux$

Is that the only conventional way of factorising u^2 + v^2 over C?

Thanks

Mao · « **Reply #6 on:** December 15, 2009, 09:23:15 pm »

I would like to clarify:

Factorization over R:

$u^2 + v^2 = u^2 + 2uv + v^2 - 2uv = (u+v)^2 - 2uv = (u + v - \sqrt{2uv})(u + v + \sqrt{2uv})$

Factorization over C:

$u^2 + v^2 = u^2 - (vi)^2 = (u + vi)(u - vi)$

TT's method was purely over R. To factorize over C:

$\begin{align*} z^4 + 64 & = z^2 - (8i)^2 \\ & = (z^2 + 8i)(z^2 - 8i) \\ & = (z^2 - (i\sqrt{8i})^2)(z^2 - (\sqrt{8i})^2) \\ & = (z + i\sqrt{8i})(z - i\sqrt{8i})(z + \sqrt{8i})(z - \sqrt{8i}) \\ & = (z + 2\sqrt{2}i^{3/2})(z - 2\sqrt{2}i^{3/2})(z + 2\sqrt{2}i^{1/2})(z - 2\sqrt{2}i^{1/2}) \\ \end{align*} $

This can be further simplified with $i^{1/2} = \frac{\sqrt{2}}{2}(1+i)$ and $i^{3/2} = \frac{\sqrt{2}}{2}(-1+i)$

$\begin{align*} & = (z -2 + 2i)(z + 2 - 2i)(z + 2 + 2i)(z - 2 - 2i) \\ & = ((z - 2)+2i)((z-2)-2i)((z+2)+2i)((z+2)-2i)) \\ & = ((z-2)^2 -4i^2)((z+2)^2 - 4i^2) \\ & = (z^2 - 4z + 4 + 4)(z^2 + 4z + 4 + 4) \\ z^4 + 64 & = (z^2 - 4z + 8)(z^2 + 4z + 8)\\ \end{align*}$

Takes much more effort.

Chavi · « **Reply #7 on:** December 15, 2009, 09:30:52 pm »

Quote from: Mao on December 15, 2009, 09:23:15 pm

$\begin{align*} (z + 2\sqrt{2}i^{3/2})(z - 2\sqrt{2}i^{3/2})(z + 2\sqrt{2}i^{1/2})(z - 2\sqrt{2}i^{1/2}) \\ \end{align*}$

How do you take the sqrt of i, or other indicy variants such as i^(1/4)?

Thanks

Mao · « **Reply #8 on:** December 15, 2009, 09:38:54 pm »

Quote from: Chavi on December 15, 2009, 09:30:52 pm

Quote from: Mao on December 15, 2009, 09:23:15 pm

$\begin{align*} (z + 2\sqrt{2}i^{3/2})(z - 2\sqrt{2}i^{3/2})(z + 2\sqrt{2}i^{1/2})(z - 2\sqrt{2}i^{1/2}) \\ \end{align*}$

How do you take the sqrt of i, or other indicy variants such as i^(1/4)?

Thanks

De Moivre's Theorem (polar form) [Note, $\mbox{cis}(\theta)$ is equivalent to $e^{i\theta}$ ]

$\begin{align*} i & = e^{ i \pi / 2} \\ i^{1/2} & = e^{i \pi / 4} \\ & = \cos \left( \frac{\pi}{4} \right) + i \cdot \sin \left(\frac{\pi}{4} \right) \\ & = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \\ \end{align*}$

And so on.

Chavi · « **Reply #9 on:** December 15, 2009, 09:42:27 pm »

Quote from: Mao on December 15, 2009, 09:38:54 pm

De Moivre's Theorem (polar form) [Note, $\mbox{cis}(\theta)$ is equivalent to $e^{i\theta}$ ]

$\begin{align*} i & = e^{ i \pi / 2} \\ i^{1/2} & = e^{i \pi / 4} \\ & = \cos \left( \frac{\pi}{4} \right) + i \cdot \sin \left(\frac{\pi}{4} \right) \\ & = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \\ \end{align*}$

And so on.

Thanks - it makes sense now. but what is the purpose of the "e" notation?

Mao · « **Reply #10 on:** December 16, 2009, 09:03:44 am »

Exponential form is a lot more useful, since you can use rules that already apply to exponentials, as opposed to using the cis notation where you have to create a new set of rules, e.g.

$w = r \mbox{cis} (\theta)$

$w^n = (r \mbox{cis} (\theta))^n = r^n \mbox{cis}(n\theta)$

compared to

$w = r\cdot e^{i \theta}$

$w^n = \left( r \cdot e^{i \theta} \right)^n = r^n \cdot e^{i (n\theta)}$

The exponential form is used extensively in complex analysis.

Chavi · « **Reply #11 on:** December 20, 2009, 12:57:24 pm »

Could I please get some help with: $\int sin^3(x/2) \ dx$

Thanks

TrueTears · « **Reply #12 on:** December 20, 2009, 01:02:37 pm »

Quote from: Chavi on December 20, 2009, 12:57:24 pm

Could I please get some help with: $\int sin^3(x/2) \ dx$

Thanks

$\sin^3\left(\frac{x}{2}\right) = \sin^2\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) = \left(1-\cos^2\left(\frac{x}{2}\right)\right)\sin\left(\frac{x}{2}\right)$

Let $u = \cos\left(\frac{x}{2}\right)$

$\implies \frac{du}{dx} = -\frac{1}{2}\left(\sin\left(\frac{x}{2}\right)\right)$

$\implies du = -\frac{1}{2}\left(\sin\left(\frac{x}{2}\right)\right) dx$

$\int \sin^3\left(\frac{x}{2}\right) dx = \int \left(1-\cos^2\left(\frac{x}{2}\right)\right)\sin\left(\frac{x}{2}\right) dx = -2\int \left(1-u^2\right)du$

Rest is trivial.

Chavi · « **Reply #13 on:** January 03, 2010, 04:29:49 pm »

How would I integrate the following expression?
$\int x(x^2+1){\sqrt{x^2-1}} dx$

And also out of interest, is there a fast way to insert math expressions into your messages?

Thanks

TrueTears · « **Reply #14 on:** January 03, 2010, 04:33:05 pm »

Try let $u = x^2-1$

So $u+1=x^2$

$\frac{du}{dx} = 2x$

Rest is trivial.

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