James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

ExamplE 3 If f 0 sxd − xysx 1 1d and f n11 − f 0 8 f n for n − 0, 1, 2, . . . , find a formula
for f n sxd.
PS Analogy: Try a similar, simpler
problem
SolutION We start by finding formulas for f n sxd for the special cases n − 1, 2, and 3.
f 1 sxd − s f 0 8 f 0dsxd − f 0( f 0 sxd) − f 0S x
x 1 1D
−
x
x 1 1
x
x 1 1 1 1 −
x
x 1 1
2x 1 1
x 1 1
−
x
2x 1 1
f 2 sxd − s f 0 8 f 1dsxd − f 0( f 1 sxd) − f 0S
x
−
x
2x 1 1
x
2x 1 1 1 1 −
x
2x 1 1
3x 1 1
2x 1 1
−
f 3 sxd − s f 0 8 f 2dsxd − f 0( f 2 sxd) − f 0S x
2x 1 1D
x
3x 1 1
3x 1 1D
PS Look for a pattern
−
x
3x 1 1
x
3x 1 1 1 1 −
x
3x 1 1
4x 1 1
3x 1 1
−
x
4x 1 1
We notice a pattern: The coefficient of x in the denominator of f n sxd is n 1 1 in the
three cases we have computed. So we make the guess that, in general,
4 f n sxd −
x
sn 1 1dx 1 1
To prove this, we use the Principle of Mathematical Induction. We have already verified
that (4) is true for n − 1. Assume that it is true for n − k, that is,
x
f k sxd −
sk 1 1dx 1 1
Then f k11 sxd − s f 0 8 f kdsxd − f 0( f k sxd) − f 0S
x
sk 1 1dx 1 1D
−
x
sk 1 1dx 1 1
−
x
sk 1 1dx 1 1 1 1
x
sk 1 1dx 1 1
sk 1 2dx 1 1
sk 1 1dx 1 1
−
x
sk 1 2dx 1 1
This expression shows that (4) is true for n − k 1 1. Therefore, by mathematical
induction, it is true for all positive integers n.
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