In this lecture we investigate a connection between Taylor series - - PowerPoint PPT Presentation

APPROXIMATING TAYLOR SERIES

Lesson 5

• In this lecture we investigate a connection between Taylor series

and Fourier series:

• Taylor series are essentially Fourier series with no negative

coefficients

• We use this connection to calculate functions inside the unit disk

• Recall the Taylor series representation of the function f:

f(z) =

∞

• k=0

ckzk

• Under what conditions does the Taylor series converge?
• Consider z inside the open unit disk

1 –1

D = {z ∈ C : |z| < 1} =

: If |ck| < Ckd for some d, then f converges in the unit disk D. : Since , the Taylor series converges absolutely, hence converges: : If for some , then converges in the unit disk for all . : Follows from induction: which satisfies .

: If |ck| < Ckd for some d, then f converges in the unit disk D. : Since |z| = r < 1, the Taylor series converges absolutely, hence converges:

• k=0
• ckzk

≤ C

• k=0

kdrk < ∞

• : If

for some , then converges in the unit disk for all . : Follows from induction: which satisfies .

: If |ck| < Ckd for some d, then f converges in the unit disk D. : Since |z| = r < 1, the Taylor series converges absolutely, hence converges:

• k=0
• ckzk

≤ C

• k=0

kdrk < ∞

• : If |ck| < Ckd for some d, then f (λ) converges in the unit disk D for all λ.

: Follows from induction: which satisfies .

: If |ck| < Ckd for some d, then f converges in the unit disk D. : Since |z| = r < 1, the Taylor series converges absolutely, hence converges:

• k=0
• ckzk

≤ C

• k=0

kdrk < ∞

• : If |ck| < Ckd for some d, then f (λ) converges in the unit disk D for all λ.

: Follows from induction: f (z) =

• k=0

kckzk1 =

• k=0

(k + 1)ck+1zk which satisfies |(k + 1)ck+1| < Dkd+1.

• Now consider z on the unit circle

1

• Under what conditions does the Taylor series converge on the unit

circle?

U = {z ∈ C : |z| = 1} =

• Let z = t
• Then the Taylor series becomes

f(z) = f(t) =

∞

• k=0

ckkt

• This is a Fourier series in disguise, just now with only positive terms!
• I.e., if

are the Fourier series of

• we have

and

• Thus convergence of the Taylor series on the unit circle is precisely convergence
• f the Fourier series, which we already know

• Let z = t
• Then the Taylor series becomes

f(z) = f(t) =

∞

• k=0

ckkt

• This is a Fourier series in disguise, just now with only positive terms!
• I.e., if ˆ

fk are the Fourier series of f(t) we have . . . , ˆ f−3 = ˆ f−2 = ˆ f−1 = 0 and ˆ f0 = c0, ˆ f1 = c1, . . .

• Thus convergence of the Taylor series on the unit circle is precisely convergence
• f the Fourier series, which we already know

Example: exponential function

• Recall the definition of the exponential function:

exp(z) =

∞

• k=0

zk k!

• Assume we can calculate exp(z) on the unit circle
• We will calculate its Taylor expansion numerically by using the DFT

• 1.0
• 0.5

0.5 1.0

• 1.0
• 0.5

0.5 1.0

• Recall that z denotes the m evenly spaced points on the unit circle

z = θ =

• Then the Fourier sample points of f(θ) are

f = f(θ) = f(z)

• Recall the discrete Fourier transform Fα,β, which maps the m sample points to m

coefficients: Fα,βf =

• αθ, f
• m

. . .

• βθ, f
• m

For f(z) = exp(z), we have:

• 0.166642 + 0. ‰
• 0.0416639 + 0. ‰

0.991667 + 0. ‰ 0.998611 + 0. ‰ 0.499802 + 0. ‰

m = 5

c0 c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15 c16 c17 c18 c19 c20

1. 1. 0.5 0.166667 0.0416667 0.00833333 0.00138889 0.000198413 0.0000248016 2.75573¥10-6 2.75573¥10-7 2.50521¥10-8 2.08768¥10-9 1.6059¥10-10 1.14707¥10-11 7.64716¥10-13 4.77948¥10-14 2.81146¥10-15 1.56192¥10-16 8.22064¥10-18 4.11032¥10-19

= (Exact)

Fα,βf =

14

For f(z) = exp(z), we have: m = 10

0.00833333 + 0. ‰ 0.00138889 + 0. ‰ 0.000198413 + 0. ‰ 0.0000248016 + 0. ‰ 2.75573¥ 10-6 + 0. ‰

• 1. + 0. ‰
• 1. + 0. ‰

0.5 + 0. ‰ 0.166667 + 0. ‰ 0.0416667 + 0. ‰

Fα,βf =

c0 c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15 c16 c17 c18 c19 c20

1. 1. 0.5 0.166667 0.0416667 0.00833333 0.00138889 0.000198413 0.0000248016 2.75573¥10-6 2.75573¥10-7 2.50521¥10-8 2.08768¥10-9 1.6059¥10-10 1.14707¥10-11 7.64716¥10-13 4.77948¥10-14 2.81146¥10-15 1.56192¥10-16 8.22064¥10-18 4.11032¥10-19

= (Exact)

15

For f(z) = exp(z), we have: m = 20

• 0.0000248016 + 0. ‰
• 2.75573¥10-6 + 0. ‰
• 2.75573¥10-7 + 0. ‰
• 2.50521¥10-8 + 0. ‰
• 2.08768¥10-9 + 0. ‰
• 1.6059¥10-10 + 0. ‰
• 1.14708¥10-11 + 0. ‰
• 1. + 0. ‰
• 1. + 0. ‰

0.5 + 0. ‰ 0.166667 + 0. ‰ 0.0416667 + 0. ‰ 0.00833333 + 0. ‰ 0.00138889 + 0. ‰ 0.000198413 + 0. ‰

c0 c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15 c16 c17 c18 c19 c20

1. 1. 0.5 0.166667 0.0416667 0.00833333 0.00138889 0.000198413 0.0000248016 2.75573¥10-6 2.75573¥10-7 2.50521¥10-8 2.08768¥10-9 1.6059¥10-10 1.14707¥10-11 7.64716¥10-13 4.77948¥10-14 2.81146¥10-15 1.56192¥10-16 8.22064¥10-18 4.11032¥10-19

= (Exact)

Fα,βf =

For f(z) = exp(z), we have: m = 20

2.75573¥10-7 + 0. ‰ 2.50521¥10-8 + 0. ‰ 2.08768¥10-9 + 0. ‰ 1.60591¥10-10 + 0. ‰ 1.14708¥10-11 + 0. ‰ 7.64677¥10-13 + 0. ‰ 4.77896¥10-14 + 0. ‰ 2.75335¥10-15 + 0. ‰ 1.77636¥10-16 + 0. ‰

• 1.11022¥10-17 + 0. ‰
• 1. + 0. ‰
• 1. + 0. ‰

0.5 + 0. ‰ 0.166667 + 0. ‰ 0.0416667 + 0. ‰ 0.00833333 + 0. ‰ 0.00138889 + 0. ‰ 0.000198413 + 0. ‰ 0.0000248016 + 0. ‰ 2.75573¥10-6 + 0. ‰

c0 c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15 c16 c17 c18 c19 c20

1. 1. 0.5 0.166667 0.0416667 0.00833333 0.00138889 0.000198413 0.0000248016 2.75573¥10-6 2.75573¥10-7 2.50521¥10-8 2.08768¥10-9 1.6059¥10-10 1.14707¥10-11 7.64716¥10-13 4.77948¥10-14 2.81146¥10-15 1.56192¥10-16 8.22064¥10-18 4.11032¥10-19

= (Exact)

Fα,βf =

17

For f(z) = exp(z), we have: m = 40

• 8.13152¥ 10-21 + 0. ‰

4.44089¥ 10-17 + 0. ‰ 4.44089¥ 10-17 + 0. ‰

• 5.55112¥ 10-18 + 0. ‰
• 3.88578¥ 10-17 + 0. ‰
• 5.6205¥ 10-17 + 0. ‰

2.55004¥ 10-17 + 0. ‰ 3.60822¥ 10-17 + 0. ‰

• 1. + 0. ‰
• 1. + 0. ‰

0.5 + 0. ‰ 0.166667 + 0. ‰ 0.0416667 + 0. ‰ 0.00833333 + 0. ‰ 0.00138889 + 0. ‰ 0.000198413 + 0. ‰ 0.0000248016 + 0. ‰ 2.75573¥ 10-6 + 0. ‰ 2.75573¥ 10-7 + 0. ‰ 2.50521¥ 10-8 + 0. ‰ 2.08768¥ 10-9 + 0. ‰ 1.6059¥ 10-10 + 0. ‰ 1.14707¥ 10-11 + 0. ‰ 7.64689¥ 10-13 + 0. ‰ 4.79126¥ 10-14 + 0. ‰ 2.70894¥ 10-15 + 0. ‰ 1.55431¥ 10-16 + 0. ‰

• 4.44089¥ 10-17 + 0. ‰

c0 c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15 c16 c17 c18 c19 c20

1. 1. 0.5 0.166667 0.0416667 0.00833333 0.00138889 0.000198413 0.0000248016 2.75573¥10-6 2.75573¥10-7 2.50521¥10-8 2.08768¥10-9 1.6059¥10-10 1.14707¥10-11 7.64716¥10-13 4.77948¥10-14 2.81146¥10-15 1.56192¥10-16 8.22064¥10-18 4.11032¥10-19

= (Exact)

Fα,βf =

DISCRETE TAYLOR TRANSFORM

19

For f(z) = exp(z), we have: m = 20

2.75573¥ 10-7 + 0. ‰ 2.50521¥ 10-8 + 0. ‰ 2.08768¥ 10-9 + 0. ‰ 1.60591¥ 10-10 + 0. ‰ 1.14708¥ 10-11 + 0. ‰ 7.64677¥ 10-13 + 0. ‰ 4.77896¥ 10-14 + 0. ‰ 2.75335¥ 10-15 + 0. ‰ 1.77636¥ 10-16 + 0. ‰

• 1.11022¥ 10-17 + 0. ‰
• 1. + 0. ‰
• 1. + 0. ‰

0.5 + 0. ‰ 0.166667 + 0. ‰ 0.0416667 + 0. ‰ 0.00833333 + 0. ‰ 0.00138889 + 0. ‰ 0.000198413 + 0. ‰ 0.0000248016 + 0. ‰ 2.75573¥ 10-6 + 0. ‰

c0 c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15 c16 c17 c18 c19 c20

1. 1. 0.5 0.166667 0.0416667 0.00833333 0.00138889 0.000198413 0.0000248016 2.75573¥10-6 2.75573¥10-7 2.50521¥10-8 2.08768¥10-9 1.6059¥10-10 1.14707¥10-11 7.64716¥10-13 4.77948¥10-14 2.81146¥10-15 1.56192¥10-16 8.22064¥10-18 4.11032¥10-19

= (Exact)

Fα,βf =

20

For f(z) = exp(z), we have:

c0 c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15 c16 c17 c18 c19 c20

1. 1. 0.5 0.166667 0.0416667 0.00833333 0.00138889 0.000198413 0.0000248016 2.75573¥10-6 2.75573¥10-7 2.50521¥10-8 2.08768¥10-9 1.6059¥10-10 1.14707¥10-11 7.64716¥10-13 4.77948¥10-14 2.81146¥10-15 1.56192¥10-16 8.22064¥10-18 4.11032¥10-19

= (Exact)

Fα,βf =

• 2.50521¥10-8 + 0. ‰
• 2.08768¥10-9 + 0. ‰
• 1.6059¥10-10 + 0. ‰
• 1.14708¥10-11 + 0. ‰
• 7.64632¥10-13 + 0. ‰
• 4.77079¥10-14 + 0. ‰
• 2.83371¥10-15 + 0. ‰
• 2.74912¥10-16 + 0. ‰
• 8.45884¥10-17 + 0. ‰
• 0. + 0. ‰
• 1. + 0. ‰
• 1. + 0. ‰

0.5 + 0. ‰ 0.166667 + 0. ‰ 0.0416667 + 0. ‰ 0.00833333 + 0. ‰ 0.00138889 + 0. ‰ 0.000198413 + 0. ‰ 0.0000248016 + 0. ‰ 2.75573¥10-6 + 0. ‰ 2.75573¥10-7 + 0. ‰

m = 21

21

• For

Fα,βf =

• αθ, f
• m

. . .

• βθ, f
• m
• we noticed the following surprising behaviour:
• αθ, f
• m ≈ (−)m ˆ

fβ+1

• (α+1)θ, f
• m ≈ (−)m ˆ

fβ+2 . . .

• −θ, f
• m ≈ (−)m ˆ

fm−1

• Why is this? Recall from last lecture
• On the other hand,
• When

is nonzero, this is a bad (non converging) approximation to

• However, when

, we have

22

• For

Fα,βf =

• αθ, f
• m

. . .

• βθ, f
• m
• we noticed the following surprising behaviour:
• αθ, f
• m ≈ (−)m ˆ

fβ+1

• (α+1)θ, f
• m ≈ (−)m ˆ

fβ+2 . . .

• −θ, f
• m ≈ (−)m ˆ

fm−1

• Why is this? Recall from last lecture
• −θ, f
• m = · · · + (−)m ˆ

f−m−1 + ˆ f−1 + (−)m ˆ fm−1 + ˆ f2m−1 + · · ·

• On the other hand,
• When

is nonzero, this is a bad (non converging) approximation to

• However, when

, we have

23

• For

Fα,βf =

• αθ, f
• m

. . .

• βθ, f
• m
• we noticed the following surprising behaviour:
• αθ, f
• m ≈ (−)m ˆ

fβ+1

• (α+1)θ, f
• m ≈ (−)m ˆ

fβ+2 . . .

• −θ, f
• m ≈ (−)m ˆ

fm−1

• Why is this? Recall from last lecture
• −θ, f
• m = · · · + (−)m ˆ

f−m−1 + ˆ f−1 + (−)m ˆ fm−1 + ˆ f2m−1 + · · ·

• On the other hand,
• (m−1)θ, f
• m = · · · + ˆ

f−m−1 + (−)m ˆ f−1 + ˆ fm−1 + (−)m ˆ f2m−1 + · · · = (−)m −θ, f

• m
• When ˆ

f−1 is nonzero, this is a bad (non converging) approximation to ˆ fm−1

• However, when . . . , ˆ

f−2, ˆ f−1 = 0, we have

• −θ, f
• m = (−)m ˆ

fm−1 + ˆ f2m−1 + · · · ≈ (−)m ˆ fm−1

• We thus define the discrete Taylor transform as the standard discrete Fourier trans-

form: T f := F0,m−1f =

• 1, fm

. . .

• (m−1)θ, f
• m

CONVERGENCE IN THE DISK

• Because of the connection with Fourier series, we know that the approximate

Taylor series fm(z) = 1 | · · · | zm−1 T f =

m−1

• k=0
• kθ, f
• m zk

converges for |z| = 1 (for nice f)

• What about |z| < 1? It must converge from Taylor series theory!

10 20 30 40 10-14 10-11 10-8 10-5 0.01

z = rei/10

r = 1,1/2,1/10

n

• What about |z| > 1? Not so good:

10 20 30 40 10-11 10-6 0.1 104 109

z = rei/10

r = 1,2,5

• This is due to round-off error

RATE OF CONVERGENCE

• Recall that the rate of convergence of approximate Fourier series is tied directly to

the rate of decay of the Fourier coefficients

• In the last lecture, we showed that f ∈ λ+2[T] implied

Ff ∈ ∞

λ+2 (i.e. ˆ

fk = O(k−λ−2)) which implied Ff ∈ 1

λ

which implied that Fourier series converges uniformly at the rate O(n−λ)

• In particular,
• implied the convergence was faster than
• for all
• This carries over to Taylor series: i.e., if

arises from a Taylor expansion and

• implies

for all

• Recall that the rate of convergence of approximate Fourier series is tied directly to

the rate of decay of the Fourier coefficients

• In the last lecture, we showed that f ∈ λ+2[T] implied

Ff ∈ ∞

λ+2 (i.e. ˆ

fk = O(k−λ−2)) which implied Ff ∈ 1

λ

which implied that Fourier series converges uniformly at the rate O(n−λ)

• In particular, f ∈ ∞[T] implied the convergence was faster than O
• k−λ

for all

• This carries over to Taylor series: i.e., if f arises from a Taylor expansion and

f ∈ ∞[U] implies ck = ˆ fk = O

• k−λ

for all

Taylor coefficients

5 10 15 20 10-14 10-11 10-8 10-5 0.01 100 200 300 400 10-14 10-11 10-8 10-5 0.01 10 20 30 40 50 60 10-13 10-10 10-7 10-4 0.1

ez ez z − 2 ez z − 1.1

The convergence rate is dictated by the domain of analyticity!

coefficient coefficient coefficient

: Suppose f is analytic in the disk RD and |f| is bounded by M in RU. Then

• ˆ

fk

• =
• f (λ)(0)
• λ!

≤ M Rλ :

: Suppose f is analytic in the disk RD and |f| is bounded by M in RU. Then

• ˆ

fk

• =
• f (λ)(0)
• λ!

≤ M Rλ : ˆ fλ = 1 2π π

−π

f(θ)−kθ θ

: Suppose f is analytic in the disk RD and |f| is bounded by M in RU. Then

• ˆ

fk

• =
• f (λ)(0)
• λ!

≤ M Rλ : ˆ fλ = 1 2π π

−π

f(θ)−kθ θ = 1 2π

• U

f(z) zλ+1 z

• = 1

2π

• RU

f(z) zλ+1 z

• π

: Suppose f is analytic in the disk RD and |f| is bounded by M in RU. Then

• ˆ

fk

• =
• f (λ)(0)
• λ!

≤ M Rλ : ˆ fλ = 1 2π π

−π

f(θ)−kθ θ = 1 2π

• U

f(z) zλ+1 z

• = 1

2π

• RU

f(z) zλ+1 z

• =

1 2πRλ

• π

−π

f(Rθ)−kθ θ

• π

: Suppose f is analytic in the disk RD and |f| is bounded by M in RU. Then

• ˆ

fk

• =
• f (λ)(0)
• λ!

≤ M Rλ : ˆ fλ = 1 2π π

−π

f(θ)−kθ θ = 1 2π

• U

f(z) zλ+1 z

• = 1

2π

• RU

f(z) zλ+1 z

• =

1 2πRλ

• π

−π

f(Rθ)−kθ θ

• ≤

M 2πRλ π

−π

θ = M Rλ

LAURENT SERIES

• Laurent series

f(z) =

∞

• k=−∞

ˆ fkzk are precisely Fourier series under the transformation z = θ

• Therefore the coefficients of Laurent series can also be calculated using the FFT
• If f is analytic in an annulus surrounding the unit circle, then the Fourier series will

converge in the annulus

• The size of the annulus dictates the decay rate of ˆ

fk, and hence the convergence rate of Fourier series

1

r R

Computed Laurent coefficients for

f(z) = ez+1/z (z + 1/2)(z − 3)

• 40
• 20

20 40 10-15 10-12 10-9 10-6 0.001 1

Warning: theoretical equality doesn’t imply numerical equality

• 3
• 2
• 1

1 2 3 0.005 0.010 0.015 0.020

n = 10

f(z) = ez+1/z

Error in approximating by approximate Fourier series

z ∈ U

f(z) = ez+1/z

Error in approximating by approximate Fourier series

z ∈ U

• 3
• 2
• 1

1 2 3

• 1. ¥10-7
• 2. ¥10-7
• 3. ¥10-7
• 4. ¥10-7
• 5. ¥10-7
• 6. ¥10-7

n = 20

f(z) = ez+1/z

Error in approximating by approximate Fourier series

z ∈ U

• 3
• 2
• 1

1 2 3

• 5. ¥10-16
• 1. ¥10-15

1.5 ¥10-15

• 2. ¥10-15

2.5 ¥10-15

• 3. ¥10-15

3.5 ¥10-15

n = 60

f(z) = ez+1/z

Error in approximating by approximate Fourier series

z ∈ U

• 3
• 2
• 1

1 2 3

• 1. ¥10-15
• 2. ¥10-15
• 3. ¥10-15
• 4. ¥10-15

n = 80

z ∈ 2U f(z) = ez+1/z

Error in approximating by approximate Fourier series

• 3
• 2
• 1

1 2 3 0.1 0.2 0.3 0.4

n = 10

z ∈ 2U f(z) = ez+1/z

Error in approximating by approximate Fourier series

• 3
• 2
• 1

1 2 3 1.¥10-8 2.¥10-8 3.¥10-8 4.¥10-8 5.¥10-8 6.¥10-8

n = 60

z ∈ 2U f(z) = ez+1/z

Error in approximating by approximate Fourier series

• 3
• 2
• 1

1 2 3

• 5. ¥10-6

0.00001 0.000015 0.00002 0.000025 0.00003

n = 80

z ∈ 2U f(z) = ez+1/z

Error in approximating by approximate Fourier series

• 3
• 2
• 1

1 2 3 0.01 0.02 0.03 0.04 0.05 0.06

n = 100

f(z) = ez+1/z

Error in approximating by approximate Fourier series

z ∈ U 2

• 3
• 2
• 1

1 2 3 0.05 0.10 0.15 0.20 0.25 0.30 0.35

n = 10

f(z) = ez+1/z

Error in approximating by approximate Fourier series

z ∈ U 2

• 3
• 2
• 1

1 2 3 0.00005 0.00010 0.00015 0.00020 0.00025 0.00030 0.00035

n = 20

f(z) = ez+1/z

Error in approximating by approximate Fourier series

z ∈ U 2

• 3
• 2
• 1

1 2 3

• 2. ¥10-11
• 4. ¥10-11
• 6. ¥10-11
• 8. ¥10-11
• 1. ¥10-10

1.2 ¥10-10

n = 40

f(z) = ez+1/z

Error in approximating by approximate Fourier series

z ∈ U 2

• 3
• 2
• 1

1 2 3 0.00005 0.00010 0.00015 0.00020 0.00025

n = 80