A Deduction of the Sine and Cosine Series using the Laplace Transform

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A Deduction of the Sine and Cosine Series using the Laplace Transform

J. C. Mohnsam, R. Nunes

To cite this version:

J. C. Mohnsam, R. Nunes. A Deduction of the Sine and Cosine Series using the Laplace Transform.

International Journal of Mathematical Education in Science and Technology, Taylor & Francis, 2018, 8 (1), pp.1 - 4. �hal-02899219�

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c Research India Publications http://www.ripublication.com/

A Deduction of the Sine and Cosine Series using the Laplace Transform

Julio Cesar Mohnsam COMAT-CINAT

IF Sul, Campus Pelotas-RS, Prac¸a 20 de Setembro 455, Pelotas-RS, CEP 96.015-360, Brazil.

E-mail: [email protected]

Ricardo Capiberibe Nunes E.E. Amlio de Caravalho Bas Campo Grande-MS - 79115-020, Brazil.

E-mail: [email protected]

Abstract

In this article, we will present an alternative proof of the Sine and Cosine series using the Laplace Transform. Looking for different tests in Mathematics is very motivating for students in Mathematics because it allows to use other ideas and concepts, and sometimes can bring simpler solutions.

AMS subject classification:44A10; 41A58.

Keywords:Infinite series; Laplace transform; Sine; Cosine.

1. Introduction

Recently the Binomial Theorem was demonstrated using the Laplace Transform method- ology (see [2]). In this work we will prove that the infinite Sine and Cosine series can also be proved using the Laplace transform together with the integration by parts. To arrive at this first we will present the analytical definitions of the Sine and Cosine function.

Proposition 1.1. The Sine function by the Taylor series aroundx=0:

sin(x) =

∞ n=0

∑

(−1)ⁿ x²ⁿ⁺¹

(2n+1)! =x−x³ 3!+x⁵

5!−x⁷

7!+· · ·;∀x∈R (1.1)

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2 Julio Cesar Mohnsam and Ricardo Capiberibe Nunes

Proposition 1.2. The Cosine function is defined by the Taylor formula aroundx=0:

cos(x) =

∞

∑

n=0

(−1)ⁿ x²ⁿ

(2n)! =1−x² 2!+x⁴

4!−x⁶

6!+· · ·;∀x∈R (1.2)

2. Preliminaries

Now let us recall some important concepts for the deductions of said series.

Proposition 2.1. The Laplace Transformation of a function f(t), defined for all real numberst ≥0, is the functionF(s), which is a unilateral transformation defined by

L

^[^f^(t^{)](s) =}

Z _∞

0

f(t)e^−stdt (2.3)

wheresis a complex number parameter [1].

Example: Laplace Transform of Cosine.

L

[cos(at)](s) = Z _∞

0

cos(at)e^−stdt= s s²+a² Example: Laplace Transform of Sine.

L

[sin(at)](s) = Z _∞

0

sin(at)e^−stdt= a s²+a²

Proposition 2.2. The inverse Laplace transform is given by the following complex in- tegral,

f(t) =

L

⁻¹^{{F}(t) =} ¹

2πi lim

T→∞

Z _γ+iT

γ−iT

e^stF(s)ds, (2.4) where is a real number so that the contour path of integration is in the region of conver- gence ofF(s)[1].

Lemma 2.3. (Integration by Parts). Let f andgbe real functions which are contin- uous on the closed interval[a,b]. Let f andghave primitivesF andGrespectively on [a,b].

Then:

Z b a

f(t)G(t)dt= [F(t)G(t)]^b_a− Z b

a

F(t)g(t)dt (2.5)

3. Deduction of series using the Laplace Transform

3.1. Calculation of the Cosine Series

L

^[cost^{](s) =}

Z _∞

0

coste^−stdt

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Using integration by parts we have:

u=cost,du=−sintdx;

dv=e^−stdt,v=−e^−st s .

L

^[cost^{](s) =}

cost

−e^−st s

∞

0

−1 s

Z _∞

0

(sint)e^−stdt= 1 s−1

s Z _∞

0

(sint)e^−stdt

L

[cost](s) =1 s−1

s

sint

−e^−st s

∞

0

+1 s

Z _∞

0

(cost)e^−stdt

=1 s− 1

s² Z _∞

0

(cost)e^−stdt ...

L

[cost](s) = 1 s− 1

s³+ 1 s⁵− 1

s⁷+· · ·=

∞ k=0

∑

(−1)^k 1 s^2k+1

Now let’s apply the inverse Laplace transform cost=

L

⁻¹^{

L

[cost]}(t) =

L

⁻¹

"

∞ k=0

∑

(−1)^k 1 s^2k+1

# (t)

=

∞

∑

k=0

L

⁻¹

(−1)^k 1 s^2k+1

(t) =

∞

∑

k=0

(−1)^k t^2k (2k)!

3.2. Calculation of the Sine Series

L

^[sint^{](s) =}

Z _∞

0

sinte^−stdt

Using integration by parts we have:

u=sint,du=costdx; dv=e^−stdt,v= −e^−st s .

L

^[sin^{t](s) =}

sint

−e^−st s

∞

0

+1 s

Z _∞

0

(cost)e^−stdt=0+1

s

L

^[cos^{t](s) =} ¹_s

L

^[cost](s)

From the previous integration in the cosine part we have

L

^[sint^{](s) =} ¹

s

L

^[cos^{t](s) =}¹

s 1

s− 1 s³+ 1

s⁵− 1 s⁷+· · ·

= 1 s²− 1

s⁴+ 1 s⁶− 1

s⁸+· · ·=

∞ k=0

∑

(−1)^k 1 s^2k+2

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4 Julio Cesar Mohnsam and Ricardo Capiberibe Nunes Now let’s apply the inverse Laplace transform

sint=

L

⁻¹^{

L

[sint]}(t) =

L

⁻¹

"

∞

∑

k=0

(−1)^k 1 s^2k+2

# (t)

=

∞

∑

k=0

(−1)^k

L

⁻¹

1 s^2k+2

(t) =

∞

∑

k=0

(−1)^k t^2k+1 (2k+1)!

Acknowledgement

We would like to thank the teachers of the Mathematics Coordination of the IF-SUL Campus Pelotas for their comments.

References

[1] S. Lipschutz,Mathematical Handbook of Formulas and Tables. Schaum’s Outline Series, McGraw-Hill, 3rd ed, p. 183.

[2] K. Kataria,An Alternate Proof of the Binomial Theorem, The American Mathemat- ical Monthly, Vol 123 - Month 11. 2016, p. 940.