# The Point of Series/Convergence?

1. Jul 31, 2009

### Jammin_James

Could someone please explain to me what the point of all of this material is? What is it used for? I know it's used for something, but I'm having trouble seeing the "bigger picture" I guess. I'm doing well on quizzes or w/e... but it feels like I'm missing something here.

I liked everything before this point and liked it's applications, but I can't seem to appreciate this material and it's bothering me a bit.

2. Aug 1, 2009

### Office_Shredder

Staff Emeritus
Infinite series are used to represent functions often. One immediate point is whether the series actually exists at a given point.

Taylor series and Fourier series are two examples

3. Aug 1, 2009

### HallsofIvy

Series, in general, whether Taylor's series, or Fourier series, give us a way of representing complicated functions a sums of simpler functions, then working with just the simpler individual terms of the series.

Here's a simple example of their use.

Solve y'= y with y(0)= 1.

Okay, we know that y(0)= 1 and, since y'= y, y'(0)= y(0)= 1. Differentiating both sides of the equation, y"= y' so y"(0)= y'(0)= 1. Differentiating again, y"'= y" so y"'= y"(-0)= 1. It should be easy to see that that leads to the nth derivative of y at x=0 equal to 0 for all n. That's only at x= 0, not other x, but now we can say, from the Taylor's series, that $y(x)= y(0)+ y'(0)x+ (1/2)y"(0)x^2+ (1/6)y"'(0)x^3...= 1+ x+ (1/2)x^2+ (1/6)x^3+ ...$ and we now know y as a series expanion. If we happened to have found the Taylor's series earlier for $e^x$ and recognize that this is the same thing, then we no that $y(x)= e^x$ satisfies that but even if it is not, we have a solution none the less.

Another important application is to extend functions to other "number systems" like the complex numbers or even matrices.

I know how to add and multiply complex numbers and matrices but what would $e^x$ be for x a complex number or a matrix?

Well, I know that the McLaurin series for $e^x$ is $1+ x+ (1/2)x^3+ (1/6)x^3+ ...+ (1/n!) x^n+ ...$ and that only involves multiplication and addition!

$$e^{bi}= 1+ bi+ (1/2)(bi)^2+ (1/6)(bi)^3+ ...+ (1/n!)(bi)^n+ ...$$

Now, it is not too hard to see that $i^2= -1$, $i^3= -i$, $i^4= 1$ and then it starts over again: the powers of i are: 1, i, -1, -i, 1, i, -1, -i, 1, ...

So
$$e^{bi}= 1+ bi- (1/2)b^2- (1/6)ib^3+ ...$$

Or, separating real and imaginary parts,
$$e^{bi}= 1- (1/2)b^2+ (1/4!)b^4- (1/6!)b^6+ ...+ i(b- (1/6)b^3+ (1/5!)b^6- (1/7!)b^7+ ...)[/itex] And, having studied series, we recognize those two series as being the power series for cosine and sine at x= 0. Thus: [tex]e^{bi}= cos(b)+ i sin(b)$$.

Looks pretty useful to me!

4. Aug 2, 2009

### Jammin_James

Tyvm, I guess I'm not really missing anything XD.