Power series: x^3/3 + x^9/9 + x^15/15 .......

In summary, the conversation discusses finding a function whose power series is represented by x^3/3! + x^9/9! + x^15/15! ..., with a connection to the power series expansion of e^x. The suggestion is to try differentiating the series multiple times to determine a differential equation it satisfies. It is noted that the 6th, 12th, 18th, etc. derivative of the function would be itself, and the hint is given to use complex roots of unity to solve the differential equation.
  • #1
Shubhan672
2
0
Moved from technical forum section, so missing the homework template

Let x be a real number. Find the function whose power series is represented as follows: x^3/3! + x^9/9! + x^15/15! ...

I see that there is a connection to the power series expansion of e^x but am having difficulty finding the function.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Shubhan672 said:
Moved from technical forum section, so missing the homework template
Let x be a real number. Find the function whose power series is represented as follows: x^3/3! + x^9/9! + x^15/15! ...

I see that there is a connection to the power series expansion of e^x but am having difficulty finding the function.
Notice that ##x^3 = (x^3)^1## and ##x^9 = (x^3)^3##. Does that give you any ideas?
 
  • #3
Try differentiating the series multiple times and see if you can determine a differential equation it satisfies.
 
  • Like
Likes etotheipi
  • #4
vela said:
Try differentiating the series multiple times and see if you can determine a differential equation it satisfies.
f(x) =
1581222250175.png


f'(x) =
1581222301684.png


f''(x) =
1581234957297.png


I can see that the 6 th, 12th, 18th ... derivative of f(x) would be f(x) itself but am having difficulty understanding its relevance and am having difficulty proceeding from here,
 

Attachments

  • 1581222032075.png
    1581222032075.png
    669 bytes · Views: 89
  • #5
Shubhan672 said:
f(x) = View attachment 256852

f'(x) = View attachment 256853

f''(x) = View attachment 256862

I can see that the 6 th, 12th, 18th ... derivative of f(x) would be f(x) itself but am having difficulty understanding its relevance and am having difficulty proceeding from here,

##f^{(6)}(x) = f(x)## is a differential equation. Can you solve it?

Hint: complex roots of unity.
 
  • Like
Likes etotheipi

What is a power series?

A power series is an infinite series of the form ∑n=0 an(x-c)n, where an represents the coefficient and c is the center point. It is a type of mathematical series that can be used to represent a function as an infinite sum of powers of x.

How is a power series different from a Taylor series?

A power series is a type of Taylor series that is centered at 0. This means that c=0, and the series becomes ∑n=0 anxn. In contrast, a Taylor series can be centered at any point, not just 0.

What is the purpose of using a power series?

A power series can be used to approximate a function, especially if the function is difficult to integrate or differentiate. It can also be used to represent functions in a simpler form, making calculations and analyses easier.

How is the convergence of a power series determined?

The convergence of a power series can be determined by using the ratio test or the root test, which check the limit of the absolute value of the ratio or root of the terms in the series. If the limit is less than 1, the series will converge.

Can a power series represent any function?

No, not all functions can be represented by a power series. The function must be analytic, meaning it can be represented as an infinite sum of powers of x. Functions with discontinuities or singularities cannot be represented by a power series.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
265
  • Calculus and Beyond Homework Help
Replies
7
Views
713
  • Calculus and Beyond Homework Help
Replies
3
Views
292
  • Calculus and Beyond Homework Help
Replies
17
Views
888
  • Calculus and Beyond Homework Help
Replies
1
Views
917
  • Calculus and Beyond Homework Help
Replies
22
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
2
Replies
38
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
Back
Top