Interactive Real Analysis - part of MathCS.org

• Real Analysis
• 1. Sets and Relations
• 2. Infinity and Induction
• 3. Sequences of Numbers
• 4. Series of Numbers
• 5. Topology
• 6. Limits, Continuity, and Differentiation
• 7. The Integral
• 8. Sequences of Functions
• 8.1. Pointwise Convergence
• 8.2. Uniform Convergence
• 8.3. Series and Power Series
• 8.4. Taylor Series
• 8.5. Approximation Theory
• 9. Historical Tidbits
• Java Tools

Example 8.3.12 (b): Differentiating and Integrating Power Series

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Integrate the series f(x) = twice and simplify your answer. Can you use your result to figure out what function this power series represents?

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The series converges for all x (confirm!) and can therefore be integrated (and differentiated) term-by-term. We have:

and integrating again:

Now, the above integration is technically correct, but with all the summation signs, integrals, and indeces it is difficult to see what is really going on. Things get a little clearer if we write out the series and integrate 'manually':

f(x) = =
     = 1 - 1/2! x² + 1/4! x⁴ - 1/6! x⁶ + ...

Integrating gives us a new function g(x):

g(x) = f(x) dx = x - 1/3! x³ + 1/5! x⁵ - 1/7! x⁷ + ... + C

Integrating again gives us:

g(x) dx = 1/2 x² - 1/4! x⁴ + 1/6! x⁶ - 1/8! x⁸ + ... + Cx + D =
     = -f(x) + 1 + Cx + D' =
     = -f(x) + Cx + D''

where we subsume the +1 into the new constant D'.

In other words, if we ignore the constants we have two functions f and g such that

f(x) dx = g(x) and g(x) dx = -f(x)

Hmmm ... what well-known functions could f and g be? To help your memory, here is the graph of these functions:

f(x) =	g(x) =