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Applied Mathematics
Advanced Calculus and
Methods of Mathematical
Physics
Homework 1

Author: Lirik Maxhuni
Jacobs University

Homework Sheet
Problem 1
Recall the mean value theorem of integral calculus: Suppose g: [a, b] → R is
Riemann integrable and non-negative, and f: [a, b] → R is continuous
...


Problem 2
Recall Taylor’s theorem in the following form
...
Then for all c, x ∈ I,
𝑛

𝑓 (𝑘) (𝑐)
𝑓(𝑥) = ∑
(𝑥 − 𝑐)𝑘 + 𝑅𝑛,𝑐 (𝑥)
𝑘!
𝑘=0

where
𝑥

1
𝑅𝑛,𝑐 (𝑥) = ∫(𝑥 − 𝑡)𝑛 𝑓 (𝑛+1)(𝑡)𝑑𝑡
𝑛!
𝑐

(a) Turn the derivation shown in class into a formal proof by induction
...
)

Problem 3
Compute in a smart way the 4-th order Taylor polynomials around c = 0 of
𝑓(𝑥) = 𝑒 𝑥 𝑠𝑖𝑛(2𝑥) and 𝑔(𝑥) = 𝑒 𝑠𝑖𝑛 𝑥
...

Let f: [a, b] → R be integrable on [a, b] and continuous at ˜x ∈ (a, b)
...
Then F is continuous on [a, b] and
differentiable at ˜x with F ′ (˜x) = f(˜x)
...

Note: Recall the definitions of continuity and differentiability and try to
start from there, using the assumptions of the theorem
...


Problem 1
Part a
If we take f to be continuous on the closed interval [a, b], this means that f
has some min and some max values somewhere in that interval
...
Consequently, g
should be non-negative
...


Problem 2
a)
Proving the integral form formula for the remainder of Taylor series
expansion, using mathematical induction
...
To show that the statement or expression is true for n = 1
2
...
Prove that it is true for n = k + 1

1)
Show that it holds for n = 1
For n = 1, 𝑓 ∈ 𝐶 (2)(𝐼), 𝐼 = (𝑎, 𝑏), 𝑐 ∈ 𝐼
...


3)
Prove that it holds for n = k + 1
𝑥

1
𝑅𝑘+1,𝑐 (𝑥 ) =
∫(𝑥 − 𝑡)𝑘+1 𝑓 (𝑘+2)(𝑡)𝑑𝑡 =
(𝑘 + 1)!
𝑐

"𝑢 = (𝑥 − 𝑡)𝑘+1 ,

∫ 𝑑𝑣 = ∫ 𝑓 (𝑘+2)(𝑡) 𝑑𝑡

𝑑𝑢 = −(𝑘 + 1)(𝑥 − 𝑡)𝑘 𝑑𝑡,

𝑣 = 𝑓 (𝑘+1)(𝑡)"

𝑥

𝑥
1
= [(𝑥 − 𝑡)𝑘+1𝑓 (𝑘+1)(𝑡)| + ∫ 𝑓 (𝑘+1)(𝑡)(𝑘 + 1)(𝑥 − 𝑡)𝑘 𝑑𝑡]
=
(𝑘 + 1)!
𝑐
𝑐

= [−(𝑥 − 𝑐 )𝑘+1 𝑓 (𝑘+1)(𝑐 ) + (𝑘 + 1)
𝑥

+ ∫ 𝑓 (𝑘+1)(𝑡)(𝑘 + 1)(𝑥 − 𝑡)𝑘 𝑑𝑡]
𝑐

1
=
(𝑘 + 1)!

𝑥

𝑓 (𝑘+1)(𝑐 )
1
(𝑥 − 𝑐 )𝑘+1 + ∫ 𝑓 (𝑘+1)(𝑡)(𝑘 + 1)(𝑥 − 𝑡)𝑘 𝑑𝑡 =
=−
(𝑘 + 1)!
𝑘!
𝑐

𝑓 (𝑘+1)(𝑐 )
(𝑥 − 𝑐 )𝑘+1 + 𝑅𝑘,𝑐 (𝑥) =
=−
(𝑘 + 1)!
𝑓 (𝑘+1)(𝑐 )
(𝑥 − 𝑐 )𝑘+1 =
= 𝑓(𝑥) − 𝑃𝑘,𝑐 (𝑥) −
(𝑘 + 1)!
= 𝑓(𝑥) − 𝑃𝑘+1,𝑐 (𝑥) = 𝑅𝑘+1,𝑐 (𝑥)

The equality of the two sides is reached and therefore the theorem holds
for any 𝑘 ∈ 𝑁
...

𝑥

1
∫(𝑥 − 𝑡)𝑛 𝑓 (𝑛+1) (𝑡)𝑑𝑡
𝑛!
𝑐

Let,
𝑡 = 𝑐 + 𝑠(𝑥 − 𝑐 )
Then we have
⇒ 𝑡 − 𝑐 = 𝑠(𝑥 − 𝑐) ⇒ 𝑠 =

𝑡−𝑐
𝑥−𝑐

𝑑𝑠
𝑑
1
1
1
=
(
(𝑡 − 𝑐)) =
⇒ 𝑑𝑠 =
𝑑𝑡 ⇒ 𝑑𝑡 = (𝑥 − 𝑐) 𝑑𝑠
𝑑𝑡 𝑑𝑥 𝑥 − 𝑐
𝑥−𝑐
𝑥−𝑐

Above we did a substitution, 𝑡 → 𝑠, however is yet incomplete, since we
also must substitute the limits of the integral that are dependant on t, to be
dependent on s
...


1

1
∫(𝑥 − 𝑐)(𝑥 − (𝑐 + 𝑠(𝑥 − 𝑐)))𝑛 𝑓 (𝑛+1)(𝑐 + 𝑠(𝑥 − 𝑐)) 𝑑𝑠 =
𝑛!
0

1

1
=
∫(𝑥 − 𝑐)(𝑥 − 𝑐 − 𝑠𝑥 + 𝑠𝑐)𝑛 𝑓 (𝑛+1)(𝑐 + 𝑠(𝑥 − 𝑐)) 𝑑𝑠 =
𝑛!
0

1

1
=
∫(𝑥 − 𝑐)((𝑥 − 𝑠𝑥) + (𝑠𝑐 − 𝑐))𝑛 𝑓 (𝑛+1)(𝑐 + 𝑠(𝑥 − 𝑐)) 𝑑𝑠 =
𝑛!
0

1

1
=
∫(𝑥 − 𝑐)(𝑥(1 − 𝑠) − 𝑐(1 − 𝑠))𝑛 𝑓 (𝑛+1)(𝑐 + 𝑠(𝑥 − 𝑐)) 𝑑𝑠 =
𝑛!
0

1

1
=
∫(𝑥 − 𝑐)(𝑥 − 𝑐)𝑛 (1 − 𝑠)𝑛 𝑓 (𝑛+1)(𝑐 + 𝑠(𝑥 − 𝑐)) 𝑑𝑠 =
𝑛!
0

1

(𝑥 − 𝑐)𝑛+1
=
∫(1 − 𝑠)𝑛 𝑓 (𝑛+1) (𝑐 + 𝑠(𝑥 − 𝑐)) 𝑑𝑠
𝑛!
0

Thus, we conclude that the integral form of the Taylor series expansion
remainder can be expressed like below:
1

(𝑥 − 𝑐 )𝑛+1
𝑅𝑛,𝑐 (𝑥) =
∫(1 − 𝑠)𝑛 𝑓 (𝑛+1)(𝑐 + 𝑠(𝑥 − 𝑐)) 𝑑𝑠
𝑛!
0

c)
The remainder can also be conveyed in the Lagrange form of the
remainder
...

1 (𝑛+1)
(𝑥 − 𝑐)𝑛+1
𝑅𝑛,𝑐 (𝑥) = ∙ 𝑓
(𝜉)
𝑛!
𝑛+1
𝑅𝑛,𝑐 (𝑥) = 𝑓

(𝑛+1)

(𝑥 − 𝑐)𝑛+1
(𝜉)
(𝑛 + 1)!

Problem 3
In order to find the fourth order polynomial of the functions given:

a) 𝑓(𝑥) = 𝑒 𝑥 𝑠𝑖𝑛(2𝑥)
b) 𝑔(𝑥) = 𝑒 𝑠𝑖𝑛 (𝑥)
we split the functions in factors and then used the separate Taylor series
expansion formulas for each of the factors, as they were functions
themselves
...

∞

𝑓(𝑥 ) = 𝑒 𝑠𝑖𝑛(𝑥) = 𝑒𝑥𝑝 (∑
𝑘=0

(−1)𝑘
(2𝑘 + 1)!
𝑥

=𝑒 ∙𝑒

𝑥3
−6

(2𝑥 )2𝑘+1) = 𝑒

∙ 𝑒 𝑂(𝑥

5)

(𝑥−

𝑥3
+𝑂(𝑥 5 ))
6

=

Problem 4
Let us define a function such that for a given f(t) dt , we have
𝑥

𝐹(𝑥) = ∫ 𝑓(𝑡) 𝑑𝑡
𝑎

So, for any number in [a, b] we can obtain the same form as above
...
Therefore
𝜏

𝐹(𝜏) = ∫ 𝑓(𝑡) 𝑑𝑡
𝑎

and
𝜏+ℎ

𝐹(𝜏 + ℎ) = ∫ 𝑓(𝑡) 𝑑𝑡
𝑎

If the Fundamental Theorem of Calculus holds, then the following equation
is true:
𝜏+ℎ

𝜏

𝐹 (𝜏 + ℎ) − 𝐹 (𝜏) = ∫ 𝑓 (𝑡)𝑑𝑡 − ∫ 𝑓 (𝑡)𝑑𝑡 =
𝑎
𝜏

𝜏+ℎ

𝑎
𝜏

= ∫ 𝑓 (𝑡)𝑑𝑡 + ∫ 𝑓 (𝑡)𝑑𝑡 − ∫ 𝑓 (𝑡)𝑑𝑡 =
𝑎

𝜏

𝑎

By means of Mean Value Theorem for Integrals, there exists a 𝑐 ∈ [𝜏, 𝜏 + ℎ]
such that
𝜏+ℎ

∫ 𝑓 (𝑡)𝑑𝑡 = 𝑓 (𝑐 ) ∙ ℎ
𝜏

𝐹 (𝜏 + ℎ) − 𝐹 (𝜏) = 𝑓 (𝑐 ) ∙ ℎ
𝑓 (𝑐 ) =

𝐹 (𝜏 + ℎ) − 𝐹 (𝜏)
ℎ

If the limit of both sides of the equation as ℎ → 0 is taken, we have
𝐹 (𝜏 + ℎ) − 𝐹 (𝜏)
ℎ→0
ℎ

𝑙𝑖𝑚 𝑓 (𝑐 ) = 𝑙𝑖𝑚
ℎ→0

It is clear that the left-hand side is the derivative of the function 𝐹 (𝜏),
which means that
𝑙𝑖𝑚 𝑓(𝑐 ) = 𝐹′(𝜏)
ℎ→0

For the theorem to hold true, the right-hand side should be equal to the
function under the integral
...
By Squeeze Law: 𝑐 ∈ [𝜏, 𝜏 + ℎ], 𝑠𝑜 𝜏 ≤ 𝑐 ≤ 𝜏 + ℎ

𝑙𝑖𝑚 𝜏 = 𝜏
ℎ→0

and
𝑙𝑖𝑚 𝜏 + ℎ = 𝜏 ,
ℎ→0

Then this means that
𝑙𝑖𝑚 𝑐 = 𝜏
ℎ→0

Since 𝑓 is a continuous function at 𝜏
𝑙𝑖𝑚 𝑓 (𝑐 ) = 𝑓 (𝜏)
ℎ→0

and this leads us to
𝐹′(𝜏) = 𝑓(𝜏)
Consequently, the proof is concluded

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