Find function is analytic on R and has Maclaurin expansion

In summary, the conversation is about proving the function e^{x}cosx is analytic and finding its Maclaurin expansion. The person has already shown that e^{x} and cosx are analytic and have Maclaurin expressions. However, they are struggling with finding the Maclaurin expansion and proving that it is analytic for e^{x}cosx. Another person suggests using the Cauchy product to multiply the two series and obtain a power series representation. After some work, the person is able to find the Maclaurin series and asks for help in proving its analyticity.
  • #1
HF08
39
0
Prove the function is analytic on R and find its Maclaurin expansion.

(a) e[tex]^{x}[/tex]cosx


Well, I did some work. I can show that

e[tex]^{x}[/tex]=[tex]\sum[/tex]x[tex]^{k}[/tex]/k!

cosx=[tex]\sum[/tex]((-1)[tex]^{k}[/tex]x[tex]^{2k}[/tex])/(2k)!

are analytic and have the above Maclaurin expression.

My problem is multiplying these two expressions together, finding the Maclaurin
expression and proving it is analytic.

I would be grateful for all your help. I think the hardest part will be showing how
it is analytic.

Thanks,
HF08
 
Physics news on Phys.org
  • #2
If you were to multiply the two series like you would two polynomials, wouldn't that give you a power series representation of exp(x)cos(x)?

Here is a link that may be useful: http://en.wikipedia.org/wiki/Cauchy_product
 
Last edited:
  • #3
I know the Maclaurin Series now :-)

e[tex]^{x}[/tex]cos x = [tex]\sum[/tex][tex]\sum[/tex][(-1)[tex]^{j}[/tex]]/((2j)!(k-2j)1))

Whew! That was a lot of work for me, but I got it. Thanks. The result is correct. So how do I show this is analytic? I'm stuck, please, please help me.

Thanks,
HF08
 

Related to Find function is analytic on R and has Maclaurin expansion

What does it mean for a function to be analytic on R?

A function is analytic on R if it can be represented by a convergent power series on the entire real line. This means that the function has a well-defined derivative at every point of the real line.

What is a Maclaurin expansion?

A Maclaurin expansion is a special case of a Taylor series, where the series is centered at x=0. It is a way to represent a function as an infinite sum of polynomial terms, and is useful for approximating the value of the function at specific points.

Why is it important for a function to have a Maclaurin expansion?

Having a Maclaurin expansion allows us to approximate the value of the function at any point, as long as the series converges. This is useful in many applications, such as in physics and engineering, where we often need to approximate values of functions.

Can any function have a Maclaurin expansion?

No, not all functions have a Maclaurin expansion. A function must be analytic on R for a Maclaurin expansion to exist. This means that the function must have a well-defined derivative at every point on the real line.

What is the significance of having a Maclaurin expansion for a function?

A Maclaurin expansion represents a function in terms of simpler polynomial terms, making it easier to understand and manipulate. It also allows us to approximate the value of the function at any point, which is useful in many applications. Additionally, having a Maclaurin expansion is a necessary condition for a function to be considered analytic on R.

Similar threads

  • Calculus and Beyond Homework Help
2
Replies
38
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
5K
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
202
  • Calculus and Beyond Homework Help
2
Replies
36
Views
3K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
Back
Top