4 General power series

A general power series has the form

b 0 + b 1 ( x x 0 ) + b 2 ( x x 0 ) 2 + = p = 0 b p ( x x 0 ) p

Exactly the same considerations apply to this general power series as apply to the ‘special’ series p = 0 b p x p except that the variable x is replaced by ( x x 0 ) . The radius of convergence of the general series is obtained in the same way:

R = lim p b p b p + 1

and the interval of convergence is now shifted to have centre at x = x 0 (see Figure 4 below). The series is absolutely convergent if x x 0 < R , diverges if | x x 0 | > R and may or may not converge if | x x 0 | = R .

Figure 4

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Task!

Find the radius of convergence of the general power series

1 ( x 1 ) + ( x 1 ) 2 ( x 1 ) 3 +

First find an expression for the general term:

p = 0 ( x 1 ) p ( 1 ) p so b p = ( 1 ) p

Now obtain the radius of convergence:

lim p b p b p + 1 = lim p ( 1 ) p ( 1 ) p + 1 = 1.

Hence R = 1 , so the series is absolutely convergent if x 1 < 1 .

Finally, decide on the convergence at x 1 = 1 (i.e. at x 1 = 1 and x 1 = 1 i.e. x = 0 and x = 2 ):

At x = 0 the series is 1 + 1 + 1 + which diverges and at x = 2 the series is 1 1 + 1 1 which also diverges. Thus the given series only converges if x 1 < 1 i.e. 0 < x < 2 . No alt text was set. Please request alt text from the person who provided you with this resource.

Exercises
  1. From the result 1 1 x = 1 + x + x 2 + x 3 + , x < 1
    1. Find an expression for ln ( 1 x )
    2. Use this expression to obtain an approximation to ln ( 0.9 ) to 4 d.p.
  2. Find the radius of convergence of the general power series  1 ( x + 2 ) + ( x + 2 ) 2 ( x + 2 ) 3 +
  3. Find the range of values of x for which the power series 1 + x 4 + x 2 4 2 + x 3 4 3 + converges.
  4. By differentiating the series for ( 1 + x ) 1 3 find the power series for ( 1 + x ) 2 3 and state its radius of convergence.
    1. Find the radius of convergence of the series 1 + x 3 + x 2 4 + x 3 5 +
    2. Investigate what happens at the points x = 1 and x = + 1
  1. ln ( 1 x ) = x x 2 2 x 3 3 x 4 4 ln ( 0.9 ) 0.1054 (4 d.p.)
  2. R = 1 . Series converges if 3 < x < 1 . If x = 1 series diverges. If x = 3 series diverges.
  3. Series converges if 4 < x < 4 .
  4. ( 1 + x ) 2 3 = 1 2 3 x + 5 3 x 2 + valid for x < 1 .
    1. R = 1 .
    2. At x = + 1 series diverges. At x = 1 series converges.