Download Taylor Series Solutions and more Summaries Signals and Systems in PDF only on Docsity!
§11.7 (TAYLOR SERIES) NAME: SOLUTIONS
31 July 2018
TAYLOR SERIES
(1) The power series
T (x) =
∞ ∑
n= 0
f (n) (c)
n!
(x − c)
n
is called the Taylor Series for f(x) centered at x = c. If c = 0 , this is called a Maclaurin series.
(2) The N-th partial sum
TN(x) =
N ∑
n= 0
f
(n) (c)
n!
(x − c)
n = f(c) +
f
′ (c)
(x − c) +
f
′′ (c)
(x − c)
2
f
(N) (c)
N!
(x − c)
N
of the Taylor series T (x) is called the N-th Taylor Polynomial for f(x) centered at x = c.
(3) Taylor’s Theorem. The n-th Taylor polynomial Tn(x) centered at x = a approximates the function f(x)
with a remainder
f(x) − Tn(x) =
n!
x
a
(x − u)
n f
(n+ 1 ) (u) du.
Corollary. The n-th Taylor polynomial Tn(x) centered at x = a approximates f(x) with error at most
|f(x) − Tn(x)| ≤ K
|x − a|
n+ 1
(n + 1 )!
where K is a number such that |f (n+ 1 ) (u)| ≤ K for all u ∈ (a, x).
(4) Where functions agree with their Taylor series: Suppose that T (x) is the Taylor series for f(x) centered
at c, with radius of convergence R. If there is a number K such that |f
(n) (x)| ≤ K for all x ∈ (c − R, c + R)
for all n, then f(x) = T (x) for all x ∈ (c − R, c + R).
(5) ( 1 + x)
a = 1 +
∞ ∑
n= 1
a
n
x
n for |x| < 1, where
a
n
a(a − 1 )(a − 2 ) · · · (a − n + 1 )
n!
(6) Some Taylor series:
Function Series Interval of Convergence
e x
∞ ∑
n= 0
x
n
n!
sin(x)
∞ ∑
n= 0
n x 2n+ 1
(2n + 1 )!
cos(x)
∞ ∑
n= 0
n x
2n
(2n)!
1 − x
∞ ∑
n= 0
x
n (− 1 , 1 )
ln( 1 + x)
∞ ∑
n= 0
n x
n+ 1
n + 1
(− 1 , 1 ]
PROBLEMS
(1) Find the Taylor polynomial T 3 (x) for f(x) centered at c = 3 if f( 3 ) = 1 , f ′ ( 3 ) = 2 , f ′′ ( 3 ) = 12 , f ′′′ ( 3 ) = 3.
SOLUTION:
T 3 (x) = f( 3 ) + f
′ ( 3 )(x − 3 ) +
f
′′ ( 3 )
(x − 3 )
2
f
′′′ ( 3 )
(x − 3 )
3
= 1 + 2 (x − 3 ) +
(x − 3 )
2
(x − 3 )
3
= 1 + 2 (x − 3 ) + 6 (x − 3 )
2
(x − 3 )
3
(2) Find the Taylor polynomials T 2 (x) and T 3 (x) for f(x) = 1 1 +x
centered at a = 1.
SOLUTION: We need to take a few derivatives, and then plug in a = 1 to each one.
n n-th derivative f
(n) (x) f
(n) (a)
0 f(x) =
1 + x
f( 1 ) = 1 / 2
1 f
′ (x) =
( 1 + x) 2
f ′ ( 1 ) = − 1 / 4
2 f
′′ (x) =
( 1 + x) 3
f
′′ ( 1 ) = 1 / 4
3 f
′′′ (x) =
( 1 + x)^4
f ′′′ ( 1 ) = − 3 / 8
Then plug these values into the formula for the Taylor polynomial.
T 2 (x) =
(x − 1 )
(x − 1 )
2
T 3 (x) =
(x − 1 )
(x − 1 ) 2
(x − 1 ) 3
(4) (a) Use the fact that arctan(x) is an antiderivative of
1 + x 2
to find a Maclaurin series for arctan(x),
and find the interval of convergence.
SOLUTION: Recall that arctan(x) is an antiderivative of ( 1 + x
2 )
− 1
. We can get a power series
expansion for
1 1 +x^2
by substituting −x
2 into the geometric series formula:
1 + x 2
= 1 − x
2
4 − x
6 +...
This expansion is valid for |x
2 | < 1, or equivalently, |x| < 1. Now integrate term-by-term:
tan
− 1 x =
dx
1 + x 2
1 − x
2
4 − x
6
dx = A + x −
x
3
x
5
x
7
We’re not done yet! We need to find the constant of integration. To do this, plug in x = 0 , so
A = arctan( 0 ) = 0. Therefore,
arctan(x) =
∞ ∑
n= 0
n x 2n+ 1
2n + 1
Integrating term-by-term doesn’t change the radius of convergence, so it still converges for |x| < 1.
But we do need to check the endpoints of this interval: x = ± 1.
For x = 1 , we have the series ∞ ∑
n= 0
n
2n + 1
which converges by the alternating series test.
For x = − 1 , notice that (− 1 ) 2n+ 1 = − 1 , so we have the series
∞ ∑
n= 0
n+ 1
2n + 1
which again converges by the alternating series test.
Therefore, the interval of convergence is [− 1 , 1 ].
(b) Use the fact that tan
π
and your answer to the previous part to find a series that
converges to π.
SOLUTION: We have arctan( 1 /
3 ) = π/ 6. Since x = 1 /
3 is inside the radius of convergence,
so we can plug in 1 /
3 into the series from the previous part:
π
= arctan
∞ ∑
n= 0
n
2n+ 1
2n + 1
∞ ∑
n= 0
n
n+ 1 / 2 (2n + 1 )
Thus we have
π =
∞ ∑
n= 0
n
n+ 1 / 2 (2n + 1 )
(5) Find the interval of convergence of the following power series.
(a)
∞ ∑
n= 0
x
n
n^4 + 2
SOLUTION: Start with the ratio test:
∣ ∣ ∣ ∣
x
n+ 1
(n + 1 ) 4
n
4
x n
n
4
(n + 1 ) 4
x
→ |x|
So the series converges when |x| < 1 and diverges when |x| > 1. We must now check the cases
when |x| = 1 manually: when x = 1 and x = − 1 , the resulting series converges by limit comparison
to
( 1 /n
4 ). Hence the interval of convergence is [− 1 , 1 ].
(b)
∞ ∑
n= 0
n
3n
(x + 3 )
n
SOLUTION: Ratio test:
∣ ∣ ∣ ∣
n+ 1 (x + 3 )
n+ 1
3 (n + 1 )
3n
n (x + 3 ) n
3n
3 (n + 1 )
· 2 (x + 3 )
→ | 2 (x + 3 )|.
Thus the series converges for |x + 3 | < 1/ 2. Check the endpoints: when x + 3 = 1 / 2 then the
series is ∞ ∑
n= 0
n
3n
n
∞ ∑
n= 0
3n
which diverges, and when x + 3 = − 1 / 2 the series is the alternating version of the above, which
converges. Hence the interval of convergence is [− 3 − 1 / 2 , − 3 + 1 / 2 ) = [− 7 / 2 , − 5 / 2 ).
(c)
∞ ∑
n= 0
(x + 4 ) n
(n ln n) 2
SOLUTION: Ratio test:
(x + 4 ) n+ 1
((n + 1 ) ln(n + 1 )) 2
(n ln n) 2
(x + 4 ) n
n
n + 1
ln n
ln(n + 1 )
2
(x + 4 )
→ |x + 4 |.
(Use L’H opital’s rule if you are not confident with the limit.)ˆ So the series converges when
|x + 4 | < 1, that is for x ∈ (− 5 , − 3 ). Checking the endpoints, we find that when x = − 5 , we have
∞ ∑
n= 0
n
(n ln n)^2
which converges by the Alternating Series Test, and when x = − 3 we have
∞ ∑
n= 0
(n ln n) 2
which converges by limit comparison to
1 /n
2
. Therefore the interval of convergence is [− 5 , − 3 ].
(d) f(x) =
3x − 2
, centered at c = − 1.
SOLUTION: Rewrite the function as follows:
3x − 2
− 5 + 3 (x + 1 )
3 (x+ 1 ) 5
Now use the geometric series formula, valid for |x| < 1.
3x − 2
∞ ∑
n= 0
3 (x + 1 )
n
= −
∞ ∑
n= 0
n 5
n (x + 1 )
n = −
∞ ∑
n= 0
n
n+ 1
(x + 1 )
n
This formula is now valid for
3 (x+ 1 ) 5
∣ < 1, or |x + 1 | <
5 3
. So the radius of convergence is
5 3
(e) f(x) = ( 1 + x)
1 / 3 , centered at c = 0.
SOLUTION: Use the binomial series formula with a =
1 3
( 1 + x)
1 (^3) = 1 +
∞ ∑
n= 1
1 3
n
x
n
The radius of convergence is 1 , since the formula is valid for |x| < 1.
(f) f(x) =
x, centered at c = 4.
SOLUTION: First rewrite the function
x =
4 + (x − 4 ) =
x − 4
x − 4
Now find the MacLaurin series of
1 + u by setting a =
1 2
in the binomial series formula.
( 1 + u)
1 (^2) =
1 + u = 1 +
∞ ∑
n= 1
1 2
n
u
n .
This is valid for |u| < 1. Now replace u by
x− 4 4 to get
x − 4
∞ ∑
n= 1
1 2
n
x − 4
)n
∞ ∑
n= 1
1 2
n
n
(x − 4 )
n
This is valid for
x− 4 4
∣ < 1 or |x − 4 | < 4. So the radius of convergence is 4.
The final answer is:
√ x = 2 +
∞ ∑
n= 1
1 2
n
n
(x − 4 )
n
If you’re willing to do a lot of simplifying, you can eventually get to:
x = 2 +
∞ ∑
n= 1
n− 1 n(2n − 2 )!
4n− 2 (n!) 2
(x − 4 )
n
