Why Does Differentiating a Geometric Series Lead to an Alternating Series?

Click For Summary
SUMMARY

The discussion centers on differentiating the geometric series to derive a power series representation for the function f(x) = 1/(1+x)^2. The key steps involve recognizing that the function is the derivative of 1/(1+x) and applying differentiation to the summation of the geometric series. The confusion arises when transitioning to the alternating series, where the differentiation of the summation leads to the expression d/dx [summation (-1)^n (x^n)]. This highlights the relationship between the geometric series and its alternating counterpart.

PREREQUISITES
  • Understanding of geometric series and their summation.
  • Knowledge of power series representation.
  • Proficiency in differentiation techniques in calculus.
  • Familiarity with the concept of alternating series.
NEXT STEPS
  • Study the derivation of power series for different functions.
  • Learn about the properties and convergence of alternating series.
  • Explore the relationship between geometric series and their derivatives.
  • Investigate the rigorous treatment of infinite sums in calculus.
USEFUL FOR

Students studying calculus, particularly those focusing on series and sequences, as well as educators looking to clarify the concepts of power series and differentiation.

fk378
Messages
366
Reaction score
0

Homework Statement


Use differentiation to find a power series representation for f(x) = 1/ (1+x)^2

Homework Equations


geometric series sum = 1/(1+x)

The Attempt at a Solution


(1) I see that the function they gave is the derivative of 1/(1+x).
(2) Therefore, (-1)*(d/dx)summation(x^n) = -1/(1+x)^2
(3) Differentiating the summation gives:
(-1)*[summation (n)x^(n-1)]

However, the book is telling me that for my second step (2) I should be getting
d/dx [summation (-1)^n (x^n)].

Why is it becoming an alternating series here?
 
Physics news on Phys.org
Remember:

\frac{1}{1-x} = 1+x+x^2+...

so that, after substituting -x for x:

\frac{1}{1+x} = 1 - x + x^2 + ...

You can remember the denominator in the first equation is 1-x by multiplying both sides by (1-x), giving:

1 = (1-x)(1+x+x^2+...) = 1 + x + x^2 + ... - x - x^2 -x^3 - ... = 1

which is consistent, as opposed to what you'd get if you assume 1+x+... was 1/(1+x). (By the way, these manipulations of infinite sums aren't strictly valid, but they can be made more rigorous by restricting to finite sums and taking a limit at the end).
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
14
Views
2K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K