I    Power Series.

(A)   Radius and Interval of Convergence.

Suppose that we wish to find the radius of convergence or the interval of convergence of a power series

Normally, we use the ratio test or the root test.   Consider the following examples:

Let a_n = 1/n!  and let c = 0.

>   t :=  n -> x^n /factorial(n);
>  ratio := abs(t(n+1)/t(n));
>  simplify(%);
>  limit(ratio, n = infinity);

Note that since the limit is always 0, the interval of convergence is (-infinity, infinity).

Next, consider the example a_n = 1/n, c = 3.

>   restart;
>   t :=  n -> (x-3)^n / n;
>  ratio := abs(t(n+1)/t(n));
>  simplify(%);
>  ratioLimit := limit(ratio, n = infinity);
>  simplify();
>  solve (ratioLimit <= 1, x);

Thus the radius of convergence is 1.  To investigate endpoint behavior:  If x = 4, then the series is harmonic and thus diverges.  If x = 2, then the series is alternating and converges by Leibniz's rule.  Thus the interval of convergence is [2, 4).

Next, consider the series

We will let

Again, we apply the ratio test:

>  restart;
>   t :=  n -> (-1)^n*x^(2*n+1) / 7^n;
>  ratio := abs(t(n+1)/t(n));
>  simplify(%);
>  ratioLimit := limit(ratio, n = infinity);
>  simplify();
>  solve (ratioLimit <= 1, x);

Thus the radius of convergence is sqrt(7).  To investigate endpoint behavior, let x = sqrt(7).

>  subs( x = sqrt(7), t(n));
>  simplify(%);

Note that the atoms do not tend toward 0, and thus the series diverges.  Similarly show that the series diverges when x = - sqrt(7).
Hence the interval of convergence is (-sqrt(7), sqrt(7)).

(B)   Computing power series.
 

Recall that the power series expansion of a continuously differentiable function f(x) about x = c is given by the infinite series

>   series (sin(x), x = 0, 7);
>    series (cos(x), x = Pi, 8);
>   series (exp(x), x=1, 15);
>   series(ln(x), x=1, 7);
>   series (sinh(x), x=0, 14);
>   series (arctan(x), x=0, 15);

(C)   Taylor and Maclaurin polynomials.

Recall that the n^th degree Taylor polynomial of a function at x = c, is given by:

If c = 0, then such a polynomial is called the n^th degree Maclaurin polynomial of f.

We can use the convert command to obtain Taylor polynomials from the infinite power series of f.  For example:

>  # Let us obtain the 13th degree Taylor polynomial of sec x about x = p/4.
>   seriesSec :=  series(sec(x), x=Pi/4, 14);  #  This gives us all terms up to and including the 14th.
>    Taylor13 :=  convert(seriesSec, polynom);
>    Digits := 3;
>    evalf( Taylor13);

Let us compare the graph of a function with that of one of its Maclaurin polynomials:

>   s1 := series(sin(x), x=0, 6);
>   Mac5 := convert(s1, polynom);
>   plot ({s1, Mac5}, x = -4..4);
 
 

Exercises:
 
 

1.    Find the 8^th degree Maclaurin polynomial of  tanh(x).

2.    Find the 7^th degree Taylor polynomial of ex about x = 1.

3.    Find the 9^th degree Maclaurin polynomial of ln(1+x).

4.   Plot the graph of y = e^x along with the first three Maclaurin polynomials of e^x on the same set of axes.
 

II   Infinite products.

In mathematics, infinite products play an important role, although perhaps not quite as important as infinite series.  Analogous to infinite series, an infinite product is the limit of a sequence of partial products.  Capital pi is used to indicate a product.  For example,

>  restart;
>  a := n -> 1-exp(-n);
>   p :=  n -> product(a(k), k=1..n);
>   product(a(n), n=1..infinity);
>   evalf(%);
>  plot (p(n), n = 4..50);

>  restart;
>  a := n -> 3^(n/3^n);
>   p :=  n -> product(a(k), k=1..n);
>   product(a(n), n=1..infinity);
>   evalf(%);
>  plot (p(n), n = 1..25);

>  restart;
>  a := n -> 1-1/exp(exp(n));
>   p :=  n -> product(a(k), k=1..n);
>   product(a(n), n=1..infinity);
>   evalf(%);
>  plot (p(n), n = 4..50);

Exercise:    Examine the infinite products defined by a(n) = 1 + 1/n   and  b(n) = 1 + 1/n².   Graph each sequence of partial products.  Does either infinite product converge?  If so, what is its limit?