Section 9.3b
Remainder Estimation Theorem
In the last class, we proved the convergence to a Taylor
series to its generating function (sin(x)), and yet we did
not need to find any actual values for the derivatives of
the function!
Instead, we were able to find an upper bound on the
derivatives, which was enough to ensure that the
remainder converged to zero for all x. This is the
foundation of the new theorem…
Remainder Estimation Theorem
If there are positive constants M and r such that
 t   Mr for all t between a and x, then the
remainder Rn  x  in Taylor’s Theorem satisfies the
f
 n 1
inequality
n 1
Rn  x   M
r
n 1
xa
n 1
 n  1!
If these conditions hold for every n and all the other
conditions of Taylor’s Theorem are satisfied by f, then
the series converges to f (x).
Remainder Estimation Theorem
If there are positive constants M and r such that
 t   Mr for all t between a and x, then the
remainder Rn  x  in Taylor’s Theorem satisfies the
f
 n 1
inequality
n 1
Rn  x   M
r
n 1
xa
n 1
 n  1!
(It doesn’t matter if M and r are huge, the important
thing is that they don’t get any more huge as n
approaches infinity. This allows the factorial growth to
outstrip the power growth and thereby sweep the
remainder to zero.)
Remainder Estimation Theorem
Use the Remainder Estimation Theorem to prove that

k
x for all real x.
e 
k 0 k !
x
By the theorem, we need to find M and r such that
 n 1
t   e
t
is bounded by
and an arbitrary x…
f
Mr
n 1
for t between 0
We know that the exponential function is increasing on
any interval, so it reaches its maximum value at the
right-hand endpoint. We can pick M to be that
maximum value and simply let r = 1.
Remainder Estimation Theorem
Use the Remainder Estimation Theorem to prove that

k
x for all real x.
e 
k 0 k !
x
If the interval is [0, x], we let
M e
If the interval is [x, 0], we let
M  e0  1
x
In either case, we have e  M throughout the interval,
and the Remainder Estimation Theorem guarantees
convergence!
t
Remainder Estimation Theorem
The approximation ln 1  x   x   x 2 2  is used
when x is small. Use the Remainder Estimation Theorem
to get a bound for the maximum error when x  0.1.
f  x   ln 1  x 
We need a bound for
Look at
f
P2  x   x   x 2 
2
R2  x  :
 3
t  
2
1  t 
3
on [–0.1, 0.1]
It is strictly decreasing on the interval, achieving its
maximum value at the left-hand endpoint, –0.1.
Remainder Estimation Theorem
The approximation ln 1  x   x   x 2 2  is used
when x is small. Use the Remainder Estimation Theorem
to get a bound for the maximum error when x  0.1.
P2  x   x   x 2 
f  x   ln 1  x 
We can bound
M
f
 3
t 
2
1   0.1 
3
2
by
2000

729
And we can let r = 1.
Remainder Estimation Theorem
The approximation ln 1  x   x   x 2 2  is used
when x is small. Use the Remainder Estimation Theorem
to get a bound for the maximum error when x  0.1.
f  x   ln 1  x 
P2  x   x   x 2 
2
Conclusion from the Remainder Estimation Theorem:
2000 0.1
2000 x
4
 4.6 10

R2  x  
729
3!
729 3!
3
3
We can even graph this remainder term:
R2  x   ln 1  x    x  x 2 
2
Remainder Estimation Theorem
The approximation ln 1  x   x   x 2 2  is used
when x is small. Use the Remainder Estimation Theorem
to get a bound for the maximum error when x  0.1.
R2  x   ln 1  x    x  x 2 
2
Graph in [–0.12, 0.12] by [–0.0005, 0.0005]
Calculate:
R2  0.1  3.605 10
4
R2  0.1  3.102 10
So the absolute error on the interval:
3.605 10
Which is indeed less than the bound:
4.6 10
4
4
4
Euler’s Formula
We have now seen that sine, cosine, and the exponential
functions equal their respective Maclaurin series for all
real numbers x. Now, let’s see what happens when we
assume that this is also true for all complex numbers…
Recall the powers of
i  1 :
i i
2
i  1
3
i  i
4
i 1
5
i i
1
etc.
Euler’s Formula
Assume that the exponential, cosine, and sine functions
equal their Maclaurin series (as in the table in Section
9.2) for complex numbers as well as for real numbers.
1. Find the Maclaurin series for
e
ix
ix 

 1  ix 
2!
2
2
ix 


3!
3
3

4
e
ix
ix 


n!
5
x
x
x
x
 1  ix   i   i 
2! 3! 4!
5!
n

 i 
n
n
x

n!
Euler’s Formula
2. Use the result of part 1 and the Maclaurin series for
cos(x) and sin(x) to prove that eix  cos x  i sin x
This equation is known as Euler’s formula.
2
3
4
5
n
x
x
x
x
n x
e  1  ix   i   i    i 

2! 3! 4!
5!
n!
2
4
6
2n
x
x
x
n x
 1       1

2! 4! 6!
 2n !
3
5
7
2 n 1

x
x
x
x
n
i  x       1

3! 5! 7!
 2n  1!

 cos x  i sin x
ix



Euler’s Formula
i
3. Use Euler’s formula to prove that e  1  0 . This
beautiful equation, which brings together some of the
most celebrated numbers in mathematics in such a
stunningly unexpected way, is also widely known as
Euler’s formula.
e  cos x  i sin x
ix
So, e
i
 cos   i sin   1  0  1
i
Thus, e  1  0