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Taylor Series Solutions, Summaries of Signals and Systems

Solutions to problems related to Taylor Series, including finding Taylor polynomials, determining intervals of convergence, and finding the Taylor series of certain functions. It also covers Maclaurin series, the geometric series formula, and the binomial series formula.

What you will learn

  • What is the interval of convergence of a power series?
  • How do you find the Taylor series of a function f(x) centered at a point c?
  • What is the radius of convergence of a Taylor series?
  • How do you find the Taylor polynomial Tn(x) for a given function f(x) centered at a point c?

Typology: Summaries

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§11.7 (TAYLOR SERIES) NAME:SOLUTIONS
31 July 2018
TAYLOR SERIES
(1) The power series
T(x) =
X
n=0
f(n)(c)
n!(xc)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
X
n=0
f(n)(c)
n!(xc)n=f(c) + f0(c)
1!(xc) + f00(c)
2!(xc)2+·· · +f(N)(c)
N!(xc)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) = 1
n!Zx
a
(xu)nf(n+1)(u)du.
Corollary. The n-th Taylor polynomial Tn(x)centered at x=aapproximates f(x)with error at most
|f(x) Tn(x)|K|xa|n+1
(n+1)! ,
where Kis a number such that |f(n+1)(u)|Kfor 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(cR
,
c+R)
for all n, then f(x) = T(x)for all x(cR,c+R).
(5) (1+x)a=1+
X
n=1a
nxnfor |x|< 1, where a
n=a(a1)(a2)·· · (an+1)
n!
(6) Some Taylor series:
Function Series Interval of Convergence
ex
X
n=0
xn
n!(−,)
sin(x)
X
n=0
(−1)nx2n+1
(2n +1)! (−,)
cos(x)
X
n=0
(−1)nx2n
(2n)! (−,)
1
1x
X
n=0
xn(−1,1)
ln(1+x)
X
n=0
(−1)nxn+1
n+1(−1,1]
1
pf3
pf4
pf5

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§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

  • x

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

  • x

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

  • 2

n

4

  • 2

x n

n

4

  • 2

(n + 1 ) 4

  • 2

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