Overview (MA2730,2812,2815) lecture 18 Lecture slides for MA2730 Analysis I Simon Shaw people.brunel.ac.uk/~icsrsss [email protected]College of Engineering, Design and Physical Sciences bicom & Materials and Manufacturing Research Institute Brunel University November 10, 2015 Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel MA2730, Analysis I, 2015-16
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Reference: The Handbook, Chapter 6, Section 6.1.Homework: Q10, Sheet 5bSeminar: Ad Hoc. Student driven.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Time Management — Tip 4
Begin with the end in mind.
Seek first to understand, then to be understood.
Stephen Covey, The Seven Habits of Highly Effective People
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Time Management — Tip 4
Begin with the end in mind.
Seek first to understand, then to be understood.
Stephen Covey, The Seven Habits of Highly Effective People
Think about this in the context of your assignments.
Don’t rush in. Plan. Gather. Execute.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Reference: Stewart, Chapter 12.10
We now return to Taylor and Maclaurin series.
It’s time to understand when they can be used.
The sums go to infinity. . .
They are infinite series. . .
At the start of term we didn’t know how to deal with them.
Now we do. We’ve come a long way.
Note that The Lectures 18 and 19 material in The Handbook,Chapter 6, Section 6.1 and Section 7.1, explain in some detail howto use the material in that chapter and the next. They are longsections but that is because they contain a lot of worked examples.We’ll cover the main points in the lecture.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Reference: Stewart, Chapter 12.10
We now return to Taylor and Maclaurin series.
It’s time to understand when they can be used.
The sums go to infinity. . .
They are infinite series. . .
At the start of term we didn’t know how to deal with them.
Now we do. We’ve come a long way.
Note that The Lectures 18 and 19 material in The Handbook,Chapter 6, Section 6.1 and Section 7.1, explain in some detail howto use the material in that chapter and the next. They are longsections but that is because they contain a lot of worked examples.We’ll cover the main points in the lecture.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Reference: Stewart, Chapter 12.10
We now return to Taylor and Maclaurin series.
It’s time to understand when they can be used.
The sums go to infinity. . .
They are infinite series. . .
At the start of term we didn’t know how to deal with them.
Now we do. We’ve come a long way.
Note that The Lectures 18 and 19 material in The Handbook,Chapter 6, Section 6.1 and Section 7.1, explain in some detail howto use the material in that chapter and the next. They are longsections but that is because they contain a lot of worked examples.We’ll cover the main points in the lecture.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Reference: Stewart, Chapter 12.10
We now return to Taylor and Maclaurin series.
It’s time to understand when they can be used.
The sums go to infinity. . .
They are infinite series. . .
At the start of term we didn’t know how to deal with them.
Now we do. We’ve come a long way.
Note that The Lectures 18 and 19 material in The Handbook,Chapter 6, Section 6.1 and Section 7.1, explain in some detail howto use the material in that chapter and the next. They are longsections but that is because they contain a lot of worked examples.We’ll cover the main points in the lecture.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Reference: Stewart, Chapter 12.10
We now return to Taylor and Maclaurin series.
It’s time to understand when they can be used.
The sums go to infinity. . .
They are infinite series. . .
At the start of term we didn’t know how to deal with them.
Now we do. We’ve come a long way.
Note that The Lectures 18 and 19 material in The Handbook,Chapter 6, Section 6.1 and Section 7.1, explain in some detail howto use the material in that chapter and the next. They are longsections but that is because they contain a lot of worked examples.We’ll cover the main points in the lecture.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Reference: Stewart, Chapter 12.10
We now return to Taylor and Maclaurin series.
It’s time to understand when they can be used.
The sums go to infinity. . .
They are infinite series. . .
At the start of term we didn’t know how to deal with them.
Now we do. We’ve come a long way.
Note that The Lectures 18 and 19 material in The Handbook,Chapter 6, Section 6.1 and Section 7.1, explain in some detail howto use the material in that chapter and the next. They are longsections but that is because they contain a lot of worked examples.We’ll cover the main points in the lecture.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Reference: Stewart, Chapter 12.10
We now return to Taylor and Maclaurin series.
It’s time to understand when they can be used.
The sums go to infinity. . .
They are infinite series. . .
At the start of term we didn’t know how to deal with them.
Now we do. We’ve come a long way.
Note that The Lectures 18 and 19 material in The Handbook,Chapter 6, Section 6.1 and Section 7.1, explain in some detail howto use the material in that chapter and the next. They are longsections but that is because they contain a lot of worked examples.We’ll cover the main points in the lecture.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Reference: Stewart, Chapter 12.10
We now return to Taylor and Maclaurin series.
It’s time to understand when they can be used.
The sums go to infinity. . .
They are infinite series. . .
At the start of term we didn’t know how to deal with them.
Now we do. We’ve come a long way.
Note that The Lectures 18 and 19 material in The Handbook,Chapter 6, Section 6.1 and Section 7.1, explain in some detail howto use the material in that chapter and the next. They are longsections but that is because they contain a lot of worked examples.We’ll cover the main points in the lecture.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Reference: Stewart, Chapter 12.10
We now return to Taylor and Maclaurin series.
It’s time to understand when they can be used.
The sums go to infinity. . .
They are infinite series. . .
At the start of term we didn’t know how to deal with them.
Now we do. We’ve come a long way.
Note that The Lectures 18 and 19 material in The Handbook,Chapter 6, Section 6.1 and Section 7.1, explain in some detail howto use the material in that chapter and the next. They are longsections but that is because they contain a lot of worked examples.We’ll cover the main points in the lecture.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
In Lecture 1 we deduced the general form:
The Taylor and Maclaurin expansion (series)
The infinitely differentiable function f : R → R has the formalexpansion about a ∈ R given by:
f(x) =∞∑
n=0
(x− a)nf (n)(a)
n!
If a = 0 this is called the Maclaurin expansion or series, otherwiseit is called the Taylor expansion or series.
Note the infinite sum — this is now our focus.
Recall the collected IMPORTANT results from Lecture 3. . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
You should know, remember for ever, and be able to derive:
For what values of x does Tnf(x) converge to f(x)?
We answer this question by analysis.
Convergence will occur if and only if
limn→∞
(
f(x)− Tnf(x))
= limn→∞
Rnf(x) = limn→∞
xn+1
1− x= 0.
The set of values for which this happens is the goal of the analysis.
We already know that x = 1 is not allowed.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Theorem
1
1− x=
∞∑
n=0
xn if and only if |x| < 1.
Proof
For x ∈ R \ {1} we have 11−x
= K ∈ R, some real numberindependent of n. Therefore,
Rnf(x) =xn+1
1− x= Kxn+1.
Furthermore we know that xn+1 → 0 as n → ∞ providing that−1 < x < 1, and that xn+1 6→ 0 if x 6 −1 or x > 1.
Hence Rnf(x) = Kxn+1 → 0 as n → ∞ if and only if |x| < 1.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Quite a lot happened there. Let’s look again at what we achieved.
Theorem
1
1− x=
∞∑
n=0
xn if and only if |x| < 1.
Proof — the structure
It is always true, by definition, that f(x)− Tnf(x) = Rnf(x).
Here we found that Rnf(x) = Kxn+1 for K independent of n.
Then, if limn→∞
(
f(x)− Tnf(x)))
= limn→∞
Rnf(x) = 0
we’ll have f(x) = T∞f(x) — the Maclaurin series. We proved that
limn→∞
Rnf(x) = limn→∞
Kxn+1 = 0 if and only if |x| < 1.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
We now have our first rigorously proven Maclaurin series:
1
1− x= 1+ x+ x2 + x3 + x4 + · · · =
∞∑
n=0
xn for − 1 < x < 1
where we can be certain that the ∞ is justified by our analysis.
We were lucky though, we knew Rnf(x) exactly.
Because f(x) = (1− x)−1 we could use the geometric series.
It isn’t usually that easy though. . .
Let’s look at another example and see where the difficulties lie.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
We now have our first rigorously proven Maclaurin series:
1
1− x= 1+ x+ x2 + x3 + x4 + · · · =
∞∑
n=0
xn for − 1 < x < 1
where we can be certain that the ∞ is justified by our analysis.
We were lucky though, we knew Rnf(x) exactly.
Because f(x) = (1− x)−1 we could use the geometric series.
It isn’t usually that easy though. . .
Let’s look at another example and see where the difficulties lie.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
We now have our first rigorously proven Maclaurin series:
1
1− x= 1+ x+ x2 + x3 + x4 + · · · =
∞∑
n=0
xn for − 1 < x < 1
where we can be certain that the ∞ is justified by our analysis.
We were lucky though, we knew Rnf(x) exactly.
Because f(x) = (1− x)−1 we could use the geometric series.
It isn’t usually that easy though. . .
Let’s look at another example and see where the difficulties lie.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
We now have our first rigorously proven Maclaurin series:
1
1− x= 1+ x+ x2 + x3 + x4 + · · · =
∞∑
n=0
xn for − 1 < x < 1
where we can be certain that the ∞ is justified by our analysis.
We were lucky though, we knew Rnf(x) exactly.
Because f(x) = (1− x)−1 we could use the geometric series.
It isn’t usually that easy though. . .
Let’s look at another example and see where the difficulties lie.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
We now have our first rigorously proven Maclaurin series:
1
1− x= 1+ x+ x2 + x3 + x4 + · · · =
∞∑
n=0
xn for − 1 < x < 1
where we can be certain that the ∞ is justified by our analysis.
We were lucky though, we knew Rnf(x) exactly.
Because f(x) = (1− x)−1 we could use the geometric series.
It isn’t usually that easy though. . .
Let’s look at another example and see where the difficulties lie.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Here’s another familiar result: with f(x) = cos(x) we have,
Tnf(x) =
n∑
m=0
(−1)mx2m
(2m)!= 1−
x2
2!+
x4
4!−
x6
6!+ · · · +
(−1)nx2n
(2n)!
We would like to proceed as before and say that:
limn→∞
Rnf(x) = 0 so Tnf(x) → cos(x) as n → ∞.
But we don’t know what Rnf(x) is for f(x) = cos(x).
All we know so far about the remainder is in Definition 1.8 andProposition 1.9 from Lecture 5. . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
From Lecture 5. . .
Definition 1.8 in The Handbook
Let I ⊆ R be an open interval, a ∈ I and f : I → R be an n-timesdifferentiable function, then Ra
nf(x) = f(x)− T anf(x) is called the
n-th order remainder (or error) term of the Taylor polynomial T anf
of f . We write R0nf simply as Rnf for Maclaurin polynomials.
Proposition 1.9 in The Handbook
Let I ⊆ R be an open interval, a ∈ I and f : I → R be an n-timesdifferentiable function, then there exists a neighbourhood ofa ∈ J ⊆ I and a function hn : J → R such that the error termRa
Let I ⊆ R be an open interval, a ∈ I and f : I → R be an n-timesdifferentiable function, then there exists a neighbourhood ofa ∈ J ⊆ I and a function hn : J → R such that the error termRa
So, in general, for a Maclaurin polynomial Tnf(x) (where a = 0),
Rnf(x) = f(x)− Tnf(x) and Rnf(x) → 0 as x → 0.
But. . . This tells us nothing about convergence: limn→∞
Rnf(x) =?
We need to know much more about the remainder, Rnf(x), todeal with the convergence of Maclaurin series.Here’s a very general form. . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Proposition 1.9 in The Handbook
Let I ⊆ R be an open interval, a ∈ I and f : I → R be an n-timesdifferentiable function, then there exists a neighbourhood ofa ∈ J ⊆ I and a function hn : J → R such that the error termRa
So, in general, for a Maclaurin polynomial Tnf(x) (where a = 0),
Rnf(x) = f(x)− Tnf(x) and Rnf(x) → 0 as x → 0.
But. . . This tells us nothing about convergence: limn→∞
Rnf(x) =?
We need to know much more about the remainder, Rnf(x), todeal with the convergence of Maclaurin series.Here’s a very general form. . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Proposition 1.9 in The Handbook
Let I ⊆ R be an open interval, a ∈ I and f : I → R be an n-timesdifferentiable function, then there exists a neighbourhood ofa ∈ J ⊆ I and a function hn : J → R such that the error termRa
So, in general, for a Maclaurin polynomial Tnf(x) (where a = 0),
Rnf(x) = f(x)− Tnf(x) and Rnf(x) → 0 as x → 0.
But. . . This tells us nothing about convergence: limn→∞
Rnf(x) =?
We need to know much more about the remainder, Rnf(x), todeal with the convergence of Maclaurin series.Here’s a very general form. . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Proposition 1.9 in The Handbook
Let I ⊆ R be an open interval, a ∈ I and f : I → R be an n-timesdifferentiable function, then there exists a neighbourhood ofa ∈ J ⊆ I and a function hn : J → R such that the error termRa
So, in general, for a Maclaurin polynomial Tnf(x) (where a = 0),
Rnf(x) = f(x)− Tnf(x) and Rnf(x) → 0 as x → 0.
But. . . This tells us nothing about convergence: limn→∞
Rnf(x) =?
We need to know much more about the remainder, Rnf(x), todeal with the convergence of Maclaurin series.Here’s a very general form. . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Proposition 1.9 in The Handbook
Let I ⊆ R be an open interval, a ∈ I and f : I → R be an n-timesdifferentiable function, then there exists a neighbourhood ofa ∈ J ⊆ I and a function hn : J → R such that the error termRa
So, in general, for a Maclaurin polynomial Tnf(x) (where a = 0),
Rnf(x) = f(x)− Tnf(x) and Rnf(x) → 0 as x → 0.
But. . . This tells us nothing about convergence: limn→∞
Rnf(x) =?
We need to know much more about the remainder, Rnf(x), todeal with the convergence of Maclaurin series.Here’s a very general form. . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Taylor’s theorem, Theorem 6.6 in The Handbook
Denote the mth derivative of a function f : R → R by f (m). Letf (n+1) be integrable on R, for some n ∈ N, then
f(x) =n∑
m=0
(x− a)m
m!f (m)(a) +Ra
nf(x),
for all x, a ∈ R, and where
Ra
nf(x) =1
n!
∫
x
a
(x− ξ)nf (n+1)(ξ) dξ
is the remainder.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Proof
We use induction. Assume the theorem is true forn = k ∈ {0, 1, 2, . . .} and use the definitions:
u = f (k+1)(ξ) =⇒ du = f (k+2)(ξ) dξ,
dv = (x− ξ)k dξ =⇒ v = −(x− ξ)k+1
k + 1,
to integrate Ra
kf by parts. This gives:
Ra
kf =(x− a)k+1
(k + 1)!f (k+1)(y) +Ra
k+1f,
and demonstrates that if the theorem holds for n = k, then it alsoholds for n = k + 1. For n = 0 Taylor’s theorem simply asserts thefundamental theorem of calculus and this completes the proof.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
So, we have from this very general result that,
f(x) = T a
nf(x) = Ra
nf =1
n!
∫
x
a
(x− ξ)nf (n+1)(ξ) dξ,
but we can do better than this. . .
Recall that if G takes non-negative values then
∫
b
a
G(ξ) dξ =
the area between thegraph of G and the x axisbetween x = a and x = b.
So:
∫
b
a
G(ξ) dξ 6 M(b−a) where M = max{G(x) : a 6 x 6 b}
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
If M = max{G(x) : a 6 x 6 b} then, because M is a constant, bycomparing areas we have,
∫
b
a
G(ξ) dξ 6
∫
b
a
M dξ = M(b− a)
0 1 2 3 4 5 6 7 8 9 10 11 12
0
20
40
60
0 1 2 3 4 5 6 7 8 9 10 11 12
0
20
40
60
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
If M = max{G(x) : a 6 x 6 b} then, because M is a constant, bycomparing areas we have,
∫
b
a
G(ξ) dξ 6
∫
b
a
M dξ = M(b− a)
0 1 2 3 4 5 6 7 8 9 10 11 12
0
20
40
60
0 1 2 3 4 5 6 7 8 9 10 11 12
0
20
40
60
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Furthermore, because graphs below the x axis create cancellationthrough negative areas, we get for any function that,. . .∫
b
a
G(ξ) dξ 6
∣
∣
∣
∣
∫
b
a
G(ξ) dξ
∣
∣
∣
∣
6
∫
b
a
|G(ξ)| dξ 6
∫
b
a
M dξ = M(b−a)
where, this time, M = max{|G(x)| : a 6 x 6 b}
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
Applying this to our remainder gives (if a 6 x),
|Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ . . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Furthermore, because graphs below the x axis create cancellationthrough negative areas, we get for any function that,. . .∫
b
a
G(ξ) dξ 6
∣
∣
∣
∣
∫
b
a
G(ξ) dξ
∣
∣
∣
∣
6
∫
b
a
|G(ξ)| dξ 6
∫
b
a
M dξ = M(b−a)
where, this time, M = max{|G(x)| : a 6 x 6 b}
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
Applying this to our remainder gives (if a 6 x),
|Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ . . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Furthermore, because graphs below the x axis create cancellationthrough negative areas, we get for any function that,. . .∫
b
a
G(ξ) dξ 6
∣
∣
∣
∣
∫
b
a
G(ξ) dξ
∣
∣
∣
∣
6
∫
b
a
|G(ξ)| dξ 6
∫
b
a
M dξ = M(b−a)
where, this time, M = max{|G(x)| : a 6 x 6 b}
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
Applying this to our remainder gives (if a 6 x),
|Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ . . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Furthermore, because graphs below the x axis create cancellationthrough negative areas, we get for any function that,. . .∫
b
a
G(ξ) dξ 6
∣
∣
∣
∣
∫
b
a
G(ξ) dξ
∣
∣
∣
∣
6
∫
b
a
|G(ξ)| dξ 6
∫
b
a
M dξ = M(b−a)
where, this time, M = max{|G(x)| : a 6 x 6 b}
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
Applying this to our remainder gives (if a 6 x),
|Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ . . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Furthermore, because graphs below the x axis create cancellationthrough negative areas, we get for any function that,. . .∫
b
a
G(ξ) dξ 6
∣
∣
∣
∣
∫
b
a
G(ξ) dξ
∣
∣
∣
∣
6
∫
b
a
|G(ξ)| dξ 6
∫
b
a
M dξ = M(b−a)
where, this time, M = max{|G(x)| : a 6 x 6 b}
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
Applying this to our remainder gives (if a 6 x),
|Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ . . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Furthermore, because graphs below the x axis create cancellationthrough negative areas, we get for any function that,. . .∫
b
a
G(ξ) dξ 6
∣
∣
∣
∣
∫
b
a
G(ξ) dξ
∣
∣
∣
∣
6
∫
b
a
|G(ξ)| dξ 6
∫
b
a
M dξ = M(b−a)
where, this time, M = max{|G(x)| : a 6 x 6 b}
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8 9 10 11 12
−0.5
0
0.5
1
1.5
Applying this to our remainder gives (if a 6 x),
|Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ . . .
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Assuming that a 6 x we have,
|Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ,
and defining Mn = max{|f (n)(ξ)| : a 6 ξ 6 x},
gives: |Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ,
6Mn+1
n!
∫
x
a
(x− ξ)n dξ,
=Mn+1
n!
[
(x− ξ)n+1
n+ 1
]a
x
,
=Mn+1|x− a|n+1
(n+ 1)!.
(Challenge: what happens if a > x?)This is the Lagrange bound of Ra
nf(x). We now have a theorem. . .Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Assuming that a 6 x we have,
|Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ,
and defining Mn = max{|f (n)(ξ)| : a 6 ξ 6 x},
gives: |Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ,
6Mn+1
n!
∫
x
a
(x− ξ)n dξ,
=Mn+1
n!
[
(x− ξ)n+1
n+ 1
]a
x
,
=Mn+1|x− a|n+1
(n+ 1)!.
(Challenge: what happens if a > x?)This is the Lagrange bound of Ra
nf(x). We now have a theorem. . .Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Assuming that a 6 x we have,
|Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ,
and defining Mn = max{|f (n)(ξ)| : a 6 ξ 6 x},
gives: |Ra
nf | 61
n!
∫
x
a
(x− ξ)n|f (n+1)(ξ)| dξ,
6Mn+1
n!
∫
x
a
(x− ξ)n dξ,
=Mn+1
n!
[
(x− ξ)n+1
n+ 1
]a
x
,
=Mn+1|x− a|n+1
(n+ 1)!.
(Challenge: what happens if a > x?)This is the Lagrange bound of Ra
nf(x). We now have a theorem. . .Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Taylor’s theorem (Lagrange bound of the remainder)Theorem 6.6 in The Handbook
Denote the mth derivative of a function f : R → R by f (m). Letf (n+1) be bounded on R for some n ∈ N then
f(x) =n∑
m=0
(x− a)m
m!f (m)(a) +Ra
nf(x),
for all x, a ∈ R, and where,
|Ra
nf(x)| 6Mn+1|x− a|n+1
(n+ 1)!
with Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between a and x}.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
That was all pretty tough.
But it is real progress.
The next step is to put this result to work for us.
The central question remains:
Given a function, f , and its Maclaurin polynomial, Tnf(x), when isit true that
f(x) = limn→∞
Tnf(x)?
Taking a = 0, this is what we now know for x ∈ R:
Rnf(x) = f(x)− Tnf(x) and |Rnf(x)| 6Mn+1|x|
n+1
(n+ 1)!.
where Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between 0 and x}.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
That was all pretty tough.
But it is real progress.
The next step is to put this result to work for us.
The central question remains:
Given a function, f , and its Maclaurin polynomial, Tnf(x), when isit true that
f(x) = limn→∞
Tnf(x)?
Taking a = 0, this is what we now know for x ∈ R:
Rnf(x) = f(x)− Tnf(x) and |Rnf(x)| 6Mn+1|x|
n+1
(n+ 1)!.
where Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between 0 and x}.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
That was all pretty tough.
But it is real progress.
The next step is to put this result to work for us.
The central question remains:
Given a function, f , and its Maclaurin polynomial, Tnf(x), when isit true that
f(x) = limn→∞
Tnf(x)?
Taking a = 0, this is what we now know for x ∈ R:
Rnf(x) = f(x)− Tnf(x) and |Rnf(x)| 6Mn+1|x|
n+1
(n+ 1)!.
where Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between 0 and x}.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
That was all pretty tough.
But it is real progress.
The next step is to put this result to work for us.
The central question remains:
Given a function, f , and its Maclaurin polynomial, Tnf(x), when isit true that
f(x) = limn→∞
Tnf(x)?
Taking a = 0, this is what we now know for x ∈ R:
Rnf(x) = f(x)− Tnf(x) and |Rnf(x)| 6Mn+1|x|
n+1
(n+ 1)!.
where Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between 0 and x}.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
That was all pretty tough.
But it is real progress.
The next step is to put this result to work for us.
The central question remains:
Given a function, f , and its Maclaurin polynomial, Tnf(x), when isit true that
f(x) = limn→∞
Tnf(x)?
Taking a = 0, this is what we now know for x ∈ R:
Rnf(x) = f(x)− Tnf(x) and |Rnf(x)| 6Mn+1|x|
n+1
(n+ 1)!.
where Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between 0 and x}.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
That was all pretty tough.
But it is real progress.
The next step is to put this result to work for us.
The central question remains:
Given a function, f , and its Maclaurin polynomial, Tnf(x), when isit true that
f(x) = limn→∞
Tnf(x)?
Taking a = 0, this is what we now know for x ∈ R:
Rnf(x) = f(x)− Tnf(x) and |Rnf(x)| 6Mn+1|x|
n+1
(n+ 1)!.
where Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between 0 and x}.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
That was all pretty tough.
But it is real progress.
The next step is to put this result to work for us.
The central question remains:
Given a function, f , and its Maclaurin polynomial, Tnf(x), when isit true that
f(x) = limn→∞
Tnf(x)?
Taking a = 0, this is what we now know for x ∈ R:
Rnf(x) = f(x)− Tnf(x) and |Rnf(x)| 6Mn+1|x|
n+1
(n+ 1)!.
where Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between 0 and x}.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Taking a = 0, this is what we now know for x ∈ R:
Rnf(x) = f(x)− Tnf(x) and |Rnf(x)| 6Mn+1|x|
n+1
(n+ 1)!.
where Mn+1 = max{|f (n+1)(ξ)| : for ξ varying between 0 and x}.
So, we’ll have,f(x) = lim
n→∞
Tnf(x)
and we can rigorously write f(x) = T∞f(x), if and only if
limn→∞
|Rnf(x)| 6 limn→∞
Mn+1|x|n+1
(n+ 1)!= 0.
In the next lecture we will use this for some familiar functions.Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
Lecture 18
Summary
We can:
Give a complete analysis of the Maclaurin series for (1− x)−1.
Prove Taylor’s theorem with integral remainder.
Derive the Lagrange form of the remainder.
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
End of Lecture
Computational andαpplie∂ Mathematics
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
End of Lecture
Computational andαpplie∂ Mathematics
Begin with the end in mind.
Seek first to understand, then to be understood.
Stephen Covey, The Seven Habits of Highly Effective People
Shaw bicom, mathematics, CEDPS, IMM, CI, Brunel
MA2730, Analysis I, 2015-16
Overview (MA2730,2812,2815) lecture 18
End of Lecture
Computational andαpplie∂ Mathematics
Begin with the end in mind.
Seek first to understand, then to be understood.
Stephen Covey, The Seven Habits of Highly Effective People
Reference: The Handbook, Chapter 6, Section 6.1.Homework: Q10, Sheet 5bSeminar: Ad Hoc. Student driven.