Interactive Real Analysis
- part of Example 8.4.4 (b): Taylor Series
If the given function had a convergent Taylor series, what would it be:
Back
- f(x) = ex around c = 0 and f(x) = ex around c = 1
- g(x) = cos(x) around c = 0 and g(x) = cos(x) around c = Pi/2
- h(x) = sin(x) around c = 0
- k(x) = 1/1+x around c = 2
BackIn each case we need to find
for the
given function at the specified center. We therefore compute derivatives
until we detect a pattern. Once we see the pattern we have the general
coefficient an and we can write down our
Taylor series.
1. f(x) = ex
The pattern is clear ... it remains to divide by n! to get the an:
c=0 c=1 f(x) = ex f(0) = 1 f(1) = e f '(x) = ex f '(0) = 1 f '(1) = e ... f (n)(x) = ex f (n)(0) = 1 f (n)(1) = e
- Tf(x, 0) =
1/n! xn
- Tf(x, 1) =
2. g(x) = cos(x)
The pattern now repeats so that:
c=0 c=Pi/2 f(x) = cos(x) f(0) = 1 f(Pi/2) = 0 f '(x) = -sin(x) f '(0) = 0 f '(Pi/2) = -1 f ''(x) = -cos(x) f ''(0) = -1 f ''(Pi/2) = 0 f '''(x) = sin(x) f '''(0) = 0 f '''(Pi/2) = 1 ...
- Tf(x, 0) = 1 + 0 x - 1/2! x2 + 0 x3 + 1/4! x4 + ... =
- Tf(x, 1) = 0 - 1/1! x + 0 x2 + 1/3! x3 + 0 x4 + ... = -
3. h(x) = sin(x)
The pattern now repeats so that:
c=0 f(x) = sin(x) f(0) = 0 f '(x) = cos(x) f '(0) = 1 f ''(x) = -sin(x) f ''(0) = 0 f '''(x) = -cos(x) f '''(0) = -1 ...
- Tf(x, 0) = 0 + 1 x + 0 x2 + 1/3! x3 + 0 x4 + ... =
Note that if we substitute x - Pi/2 for x and add a minus sign, we get the second series in the second example. That actually proves a trig identity - which one ?
4. k(x) = 1/1+x
c=2 f(x) = 1/1+x f(0) = 1/3 f '(x) = -1/(1+x)2 f '(0) = -1/32 f ''(x) = 1*2/(1+x)3 f ''(0) = 2!/33 f '''(x) = -1*2*3/(1+x)4 f '''(0) = -3!/34 ... f (n)(x) = (-1)n n!/(1+x)n+1 f (n)(0) = (-1)n n!/3n+1
Thus
Note that we can arrive at the same answer using our - by now familiar - geometric series and some appropriate substitution:
- Tf(x, 2) =
