• Finding the radius of convergence using ratio test
• Finding the interval of convergence
• Term-by-term differentiation
• Term-by-term integration
• Multiplication of power series
1. Radius of convergence using ratio test
   • u_n = xⁿ/n then apply ratio test to å | u_n | :
     

     | un+1 un | = n/(n+1) | x | ® | x |

     converges if | x | < 1. Diverges if > 1. converges at x = 1 since alternating harmoic series converges while diverges at x = -1. So converges for -1 < x £ 1.
   • u_n = [(xⁿ)/n!] goes to x/(n+1) which converges for all x.

2. Finding the interval of convergence
   • Use ratio test to find interval where the series converges absolutely. Ordinarily this is an open interval. If the interval is finite, test for convergence or divergence at the each endpoint. If interval of convergence if a-R < x < a+R the series diverges for | x-a | > R (it does not even converge conditionally) because the nth term does not approach zero for those values.

3. The convergence of a power series if absolute at every point in the interior of the interval. If the series converges absolutely for all x we say its radius of convergence is infinite. If only have convergence for x = a the radius of convergence is zero.
4. Term-by-term differentiaton:differentiate term by term, this series converges at every interior point of the interval of convergence of the original series!
5. Term-by-term integration:Same for integration.
6. Multiplication of power series:
   • Absolutely convergent series can be multiplied as we multiply polynomials, has the same radius of convergence.

• Know that within interval of convergence sum of power series is a continuous function with derivatives of all orders but if a function has derivates of all orders on an interval I can it be expressed as a power series....
• If function can be represented as a power series we would have
  

  f(x) = a0+a1(x-a)+...an(x-a)n+...

  then since we can perform term by term differentiation:
  

  f¢(x) = a1+2a2(x-a)+...+(n+1)an+1(x-a)n+....

  and
  

  f(k) = k!ak+ (k+1)! 1! ak+1(x-a)+...+ (n+k)! n! an+k(x-a)n+....

  Since this holds at x = a we have f'(a) = a_1,
  f''(a) = 2!a_2,
  f^(n)(a) =n!a_n. So since f has a series representation
  

  f(x) = a0+a1(x-a)+...an(x-a)n+...

  it must be
  

  f(x) = f(a)+f¢(a)(x-a)+f¢¢(a)/2! (x-a)2+....+f(n)/n! (x-a)n+....

  since we know that
  

  an = f(n)(a) n! .

• 
  

  f(x) = ¥ å k = 0  f(k) k! (x-a)k.

Maclaurin series when a = 0.

• More Taylor series
• Application of power series
• Fourier series
• Sample paper

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