# Summation question within complex numbers

1. Jan 4, 2013

### jj364

1. The problem statement, all variables and given/known data

Find the sum of the series

$\displaystyle S_1=1 + \frac{x^3}{3!}+\frac{x^6}{6!}+\,\dots$

Can't seem to get the bit above to show up nicely, should be 1+x^3/3! +x^6/6! +... Sorry!!

2. Relevant equations

In a prior part of the question I had to find the complex roots of z3-1=0 which I did and got:

1, e$\frac{2\pi}{3}$, e$\frac{4\pi}{3}$

and then by writing z3-1=0 as being equal to the product of its roots I was able to show that $\omega$2+$\omega$+1=0 where $\omega$ is a complex root

3. The attempt at a solution

Using all of this I then have to solve the summation so I'm sure that the prior questions have something to do with solving the summations I'm just not really sure how to relate them. I've only really been able to look at maclaurin series' for various functions but have had no joy as to further inspiration.

Any help would be greatly appreciated, thank you in advance!

Last edited: Jan 4, 2013
2. Jan 4, 2013

### SammyS

Staff Emeritus
You can't put LaTeX in a [ SUP] superscript.

Do the whole expression in LaTeX.

[ itex]\displaystyle S_1=1 + \frac{x^3}{3!}+\frac{x^6}{6!}+\,\dots[/itex]

gives

$\displaystyle S_1=1 + \frac{x^3}{3!}+\frac{x^6}{6!}+\,\dots$

3. Jan 4, 2013

### jj364

Ah thanks! Makes it a lot easier to read. Not so good at LaTeX!

4. Jan 4, 2013

### Millennial

The above looks awfully like the exponential series in the sense that it only has multiples of 3 as its summation values as a difference. In fact, if you look at the sum you want to find and express it as a function of x, you'll see that it satisfies the differential equation $f^{(3)}(x)=f(x)$ with initial conditions f(0)=1, f'(0)=0 and f''(0)=0. Can you solve this?

(Note: You actually have to solve the equation $x^3-1=0$ during the solution to this differential equation, so it reveals the purpose of the previous exercise.)

Last edited: Jan 4, 2013
5. Jan 4, 2013

### Ray Vickson

The sum above is not easy. You can find it by entering sum(x^(3*k)/(3*k)!,k=0..infinity) into a computer algebra system such as Maple or Wolfram Alpha. Give it a try.

6. Jan 5, 2013

### Millennial

I just solved the question using this method and compared it with Mathematica's and WolframAlpha's answers, and it checks out. Another tip if you need it: Take the Laplace transform of both sides.

7. Jan 5, 2013

### jj364

I'm still a little confused I think. I managed to get the the differential equation whose auxiliary equation is m^3-1=0 which is what I solved in the previous section but for complex numbers not real, or does it not matter?

Because if I use my solutions for z^3-1=0 then I end up with very funny answers. Or do I have to take real components? Just quite confused as to where I actually go from here to get an answer?

8. Jan 5, 2013

### Millennial

Well, the method I suggested was using the Laplace transform. I wouldn't know about other methods, but your one seems like it'd work to me. The funny answers you end up with, try finding trigonometric patterns inside them.

9. Jan 7, 2013

### jj364

Ok I think I've worked it out now, I wrote out the differential equation and the auxiliary equation gave me answers of

m=Aex + Bex($\frac{-1}{2}$+$\frac{\sqrt{3}}{2}$) + Cex($\frac{-1}{2}$+$\frac{-\sqrt{3}}{2}$)

Which I can solve using initial conditions to get A=B=C=1/3 so the final sum is

S(x)=$\frac{1}{3}$ex+$\frac{2}{3}$e$\frac{-x}{2}$cos($\frac{\sqrt{3}x}{2}$)

Is this right?

10. Jan 7, 2013

### SammyS

Staff Emeritus
(I feel like I'm playing the role of Mr. LaTeX here!) LOL!

Using \frac{}{} for fractions in superscripts in LaTeX inline mode is bad enough. Putting a LaTeX \frac{}{} in a SUPerscript tag [] make reading even harder.

I like to use the \displaystyle modifier & do the whole expression in LaTeX:
$\displaystyle \large m=Ae^x+Be^{x\left(\frac{-1}{2}+\frac{\sqrt{3}}{2}\right)}+Ce^{x\left(\frac{-1}{2}+\frac{-\sqrt{3}}{2}\right)}$​
(I also bumped up the size one notch to make the exponents somewhat readable.)

For fractional exponents with simple denominators, using / rather than ]frac{}{} works well in normal size.
$\displaystyle S(x)=\frac{1}{3}e^x+\frac{2}{3}e^{-x/2}\cos\left( \frac{\sqrt{3\,}\,x}{2}\right)$​

11. Jan 7, 2013

### awkward

Hint:

Notice that $$e^z + e^{-z} = 2 \left(1 + \frac{1}{3!} z^2 + \frac{1}{4!} z^4 + \dots \right)$$
because the odd-numbered terms in the power series cancel.

This is a standard trick with power series; to isolate the even powers of z in the series for f(z), just consider f(z) + f(-z).

Now, what would happen if we looked at the series $$e^z + e^{\omega z} + e^{\omega^2 z}$$ where $\omega= \exp(2 \pi i / 3)$?
Some of the terms in the power series will cancel. Which terms will cancel, and which terms will remain?

12. Jan 8, 2013

### Millennial

Indeed it is.