Rewriting Taylor Series in Sigma Notation

In summary: 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)!}
  • #1
Stratosphere
373
0

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 [tex]\ f(x)=e^{x}[/tex]

Homework Equations


The Attempt at a Solution


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

I know the answer is [tex]\sum^{\infty}_{n=0} \frac{x^{n}}{n!}[/tex] but I don't know how to get it from the expanded form.
 
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  • #2
Stratosphere said:

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 [tex]\ f(x)=e^{x}[/tex]

Homework Equations





The Attempt at a Solution


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

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

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

Stratosphere said:
I know the answer is [tex]\sum^{\infty}_{n=0} \frac{x^{n}}{n!}[/tex] but I don't know how to get it from the expanded form.
 
  • #3
Mark44 said:
You're missing the first term. What you have above should be
[tex]\ 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[/tex]

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

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 [tex]x^{n}[/tex] 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) [tex]sin(x)+\frac{1}{1!}x-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5}-\frac{1}{7!}x^{7}+\cdot\cdot\cdot[/tex].

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

[tex]sin(x)+\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}[/tex] I know that that can't possibly be correct but that's my attempt.
 
Last edited:
  • #4
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.
[tex]sin(x) = \sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}[/tex]
 
  • #5
Mark44 said:
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.
[tex]sin(x) = \sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}[/tex][/QUOT
Oops. I'll make note of that.

Thanks for the help though, I think I understand it now.
 
  • #6
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,

[tex]f(x) = f(0) + \frac{f'(0)x}{1!} + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + ... + \frac{f^{(n)}(0)x^n}{n!} + ...[/tex]
 
  • #7
Stratosphere said:
I noticed that there should be an [tex]x^{n}[/tex] 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) [tex]sin(x)+\frac{1}{1!}x-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5}-\frac{1}{7!}x^{7}+\cdot\cdot\cdot[/tex].
As Mark44 told you, this should be
[tex]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[/tex]

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 [itex](-1)^n[/itex] or [itex](-1)^{n+1}[/itex] 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
[tex]\frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex]
and the sum is
[tex]\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}[/tex]

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

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

FAQ: Rewriting Taylor Series in Sigma Notation

What is a Taylor Series Rewriting?

A Taylor Series Rewriting is a mathematical technique used to approximate a function with an infinite polynomial series. It allows for the evaluation of a function at a specific point without having to use the original function.

Why is Taylor Series Rewriting important?

Taylor Series Rewriting is important because it allows for the simplification and approximation of complex functions, making them easier to work with. It is also used in many real-world applications, such as in engineering and physics, to model and solve problems.

How is Taylor Series Rewriting calculated?

Taylor Series Rewriting is calculated using the derivatives of a function at a specific point. The derivatives are then plugged into the Taylor Series formula, which involves an infinite sum of terms that are raised to different powers.

What are the limitations of Taylor Series Rewriting?

The limitations of Taylor Series Rewriting include the fact that it only provides an approximation of a function, not an exact solution. It also may not converge for some functions, making the approximation unreliable. Additionally, it can be computationally expensive to calculate for functions with many terms.

How is Taylor Series Rewriting used in real life?

Taylor Series Rewriting is used in many real-life applications, such as in engineering, physics, and economics. It is used to approximate physical phenomena, model complex systems, and solve differential equations. It is also used in computer graphics to create smooth curves and surfaces.

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