Rewriting Taylor Series in Sigma Notation

[Stratosphere]

Homework Statement

I understand the whole concept of Taylor Series and Maclaurin series but I don't know how to rewrite them in sigma notation.

I'll use this generic example. Find the Maclaurin series of the function \ f(x)=e^{x}

Homework Equations

The Attempt at a Solution

\ e^{x}+\frac{1}{1!}(x-0)+\frac{1}{2!}(x-0)^{2}+\frac{1}{3!}(x-0)^{3}\cdot\cdot\cdot

I know the answer is \sum^{\infty}_{n=0} \frac{x^{n}}{n!} but I don't know how to get it from the expanded form.

 

[Mark44]

Homework Statement

I understand the whole concept of Taylor Series and Maclaurin series but I don't know how to rewrite them in sigma notation.

I'll use this generic example. Find the Maclaurin series of the function \ f(x)=e^{x}

Homework Equations

The Attempt at a Solution

\ e^{x}+\frac{1}{1!}(x-0)+\frac{1}{2!}(x-0)^{2}+\frac{1}{3!}(x-0)^{3}\cdot\cdot\cdot

You're missing the first term. What you have above should be
\ e^{x} = \frac{1}{0!}(x - 0)^0 + \frac{1}{1!}(x-0)+\frac{1}{2!}(x-0)^{2}+\frac{1}{3!}(x-0)^{3} + \cdot\cdot\cdot

When simplified, the expression on the right is
\ e^{x} = \frac{1}{0!}x^0 + \frac{1}{1!}x+\frac{1}{2!}x^{2}+\frac{1}{3!}x^{3}+ \cdot\cdot\cdot

What will be the general term in the sum above? Do you notice anything that's common to each of the terms above?

I know the answer is \sum^{\infty}_{n=0} \frac{x^{n}}{n!} but I don't know how to get it from the expanded form.

 

[Stratosphere]

You're missing the first term. What you have above should be
\ e^{x} = \frac{1}{0!}(x - 0)^0 + \frac{1}{1!}(x-0)+\frac{1}{2!}(x-0)^{2}+\frac{1}{3!}(x-0)^{3} + \cdot\cdot\cdot

When simplified, the expression on the right is
\ e^{x} = \frac{1}{0!}x^0 + \frac{1}{1!}x+\frac{1}{2!}x^{2}+\frac{1}{3!}x^{3}+ \cdot\cdot\cdot

What will be the general term in the sum above? Do you notice anything that's common to each of the terms above?

I noticed that there should be an x^{n} in there and there should be an n!. I guess I need to use a slightly more complicated example. How about sin(x)
The expanded form is (after the 0s are dropped out) sin(x)+\frac{1}{1!}x-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5}-\frac{1}{7!}x^{7}+\cdot\cdot\cdot.

I'll try and get an answer for this one now since I don't know the answer.

sin(x)+\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!} I know that that can't possibly be correct but that's my attempt.

 

[Mark44]

You're making the same mistake you made earlier. The Maclaurin series for sin(x) should not start off with sin(x) + ... What you should be saying is that sin(x) = <the series>.

Other than that, your summation is exactly right. Here's what you have with the correction.
sin(x) = \sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}

 

[Stratosphere]

You're making the same mistake you made earlier. The Maclaurin series for sin(x) should not start off with sin(x) + ... What you should be saying is that sin(x) = <the series>.

Other than that, your summation is exactly right. Here's what you have with the correction.
sin(x) = \sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}[/QUOT
Oops. I'll make note of that.

Thanks for the help though, I think I understand it now.

 

[Mark44]

What you're doing with Taylor's series (and in this case Maclaurin series) is writing a function f(x) as a power series. For a Maclaurin series,

f(x) = f(0) + \frac{f&#039;(0)x}{1!} + \frac{f&#039;&#039;(0)x^2}{2!} + \frac{f&#039;&#039;&#039;(0)x^3}{3!} + ... + \frac{f^{(n)}(0)x^n}{n!} + ...

 

[HallsofIvy]

I noticed that there should be an x^{n} in there and there should be an n!. I guess I need to use a slightly more complicated example. How about sin(x)
The expanded form is (after the 0s are dropped out) sin(x)+\frac{1}{1!}x-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5}-\frac{1}{7!}x^{7}+\cdot\cdot\cdot.

As Mark44 told you, this should be
sin(x)= \frac{1}{1!}x^2- \frac{1}{3!}x^3+ \frac{1}{5!}x^5- \frac{1}{7!}x^7+ \cdot\cdot\cdot

The first thing I would not is that the powers of x are all odd. You can write any even number as "2n" and any odd number as "2n+1", so I would try "2n+1" as as the powers and not that if n=0, 2n+1= 1, if n= 1, 2n+1= 2+ 1= 3, etc.

I would then notice that the signs alternate and think of (-1) to some power. That will alternate + and - for even and odd powers. In some cases, you might need to experiment with (-1)^n or (-1)^{n+1} to get the right parity but here I see that the first term corresponds to n= 0 and (-1)⁰= 1 is the right sign: I want (-1)ⁿ.

Finally, I see that the denominator is the factorial of the power- since I am writing the powers as "2n+1" I want to do the same in the denominators- each term is of the form
\frac{(-1)^n}{(2n+1)!}x^{2n+1}
and the sum is
\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}

I'll try and get an answer for this one now since I don't know the answer.

sin(x)+\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!} I know that that can't possibly be correct but that's my attempt.[/QUOTE]