Proving x^8+1=P_0^3(x^2-2xcos(2k+1)π/8+1)

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In summary, the equation $x^8+1= P_0^3(x^2-2xcos\frac{(2k+1)\pi}{8}+1)$ can be solved by finding the roots of $x^8+1=0$, which are $x = e^{i/8(\pi +2 \pi n)}$ for $n=0,1, \ldots$, and $7$, and then using DeMoivre's formula to factor the polynomial into 8 factors. This can be further simplified by using the index $k = (n-1)/2$ to get the final form of $x^8+1= \prod_{k = 0}^
  • #1
Suvadip
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I have to prove
\(\displaystyle x^8+1= P_0^3(x^2-2xcos\frac{(2k+1)\pi}{8}+1)\) where \(\displaystyle P_0^3\) means product from \(\displaystyle k=0\) to \(\displaystyle k=3\).I tried it but got \(\displaystyle x^8+1= P_0^3(x^2-a_k^2)\) where \(\displaystyle a_k=cos\frac{(2k+1)\pi}{8}+isin\frac{(2k+1)\pi}{8}\). How to arrive at the correct answer
 
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  • #2
Re: complex number

The roots of $x^8+1=0$ are $x = e^{i/8(\pi +2 \pi n)}$ for $n=0,1, \ldots$, and $7$.

That is, the roots are $ e^{i \pi/8},e^{3 \pi i /8},e^{5 \pi i/8},e^{7 \pi i/8},e^{9 \pi i/8},e^{11 \pi i/8},e^{13 \pi i/8}$, and $e^{15 \pi i/8}$.

Then

$$x^8+1 = (x- e^{i \pi/8})(x-e^{15 \pi i/8})(x-e^{3 \pi i/8})(x-e^{13 \pi i/8})(x- e^{5 \pi i/8})(x-e^{11 \pi i/8})(x-e^{7 \pi i/8})(x-e^{9 \pi i/8})$$

$$ = (x- e^{i \pi/8})(x-e^{- \pi i/8})(x-e^{3 \pi i/8})(x-e^{-3 \pi i/8})(x- e^{5 \pi i/8})(x-e^{-5 \pi i/8})(x-e^{7 \pi i/8})(x-e^{-7 \pi i/8}) $$

$$ = \Big( x^{2} - 2 x \cos(\frac{\pi}{8}) + 1 \Big) \Big( x^{2} - 2 x \cos(\frac{3 \pi}{8}) + 1\Big) \Big(x^{2} - 2 x \cos(\frac{5 \pi}{8}) + 1 \Big)\Big(x^{2} - 2 x \cos(\frac{7\pi}{8}) + 1 \Big)$$
 
  • #3
Re: complex number

This is just RandomVariable's solution, but re-arranged a bit.

We'll start here:

Let $w = \cos(\frac{\pi}{8}) + i\sin(\frac{\pi}{8})$

If you know DeMoivre's formula, you can see that:

$w^8 = -1$, so $w$ is one root of $x^8 + 1$.

With any real polynomial, if a complex number $z$ is a root, so is $\overline{z}$.

This means that $\overline{w}^8 = -1$ as well.

It should also be clear that:

$(w^n)^8 = w^{8n} = (w^8)^n = (-1)^n = -1$ for $n = 1,3,5,7$.

Thus, likewise:

$((\overline{w})^n)^8 = 1$ for $n = 1,3,5,7$.

so this gives us 8 factors.

Let's look at the 4 quadratics:

$(x - w^n)(x - \overline{w}^n)$ for $n = 1,3,5,7$.

Expanding this out, this becomes:

$(x - \cos(\frac{n\pi}{8}) - i\sin(\frac{n\pi}{8}))(x - \cos(\frac{n\pi}{8}) + i\sin(\frac{n\pi}{8}))$

$= x^2 -2\cos(\frac{n\pi}{8})x + \cos^2(\frac{n\pi}{8}) + \sin^2(\frac{n\pi}{8})$

$= x^2 - 2\cos(\frac{n\pi}{8})x + 1$

Since we are only considering odd $n$, we can use the index $k = \frac{n-1}{2}$, in which case we get:

$n = 2k+1$ for $k = 0,1,2,3$ and our quadratics become:

$x^2 - 2\cos(\frac{(2k+1)\pi}{8})x + 1$

so that:

\(\displaystyle x^8 + 1 = \prod_{k = 0}^3 [x^2 - 2\cos\left(\frac{(2k+1)\pi}{8}\right)x + 1]\)

In other words, it's not the product of $(x^2 - a_k^2)$ like you had, but instead the product of:

$(x - a_k)(x - \overline{a_k})$
 

FAQ: Proving x^8+1=P_0^3(x^2-2xcos(2k+1)π/8+1)

1. What is the purpose of proving x^8+1=P_0^3(x^2-2xcos(2k+1)π/8+1)?

The purpose of proving this equation is to demonstrate the mathematical relationship between x^8+1 and P_0^3(x^2-2xcos(2k+1)π/8+1). This proof can also be used to solve for unknown variables or to make predictions in other mathematical equations.

2. How can this equation be proven?

This equation can be proven through various mathematical techniques such as algebraic manipulation, substitution, and proof by induction. It may also involve using known mathematical properties and identities.

3. What are the applications of this equation?

This equation has many applications in mathematics, particularly in the fields of algebra, trigonometry, and calculus. It can also be used in physics and engineering to model and solve complex systems.

4. Are there any real-life examples of this equation?

Yes, there are several real-life examples of this equation. For instance, it can be used to model the motion of a pendulum or the behavior of a spring. It can also be found in the equations used to describe electrical circuits and sound waves.

5. Can this equation be generalized to other forms?

Yes, this equation can be generalized to other forms by replacing the variables with different values or by adding additional terms. It can also be extended to higher powers or different types of functions.

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