Chapter 2 Problem 8

The problem is to express the series in closed form

f =sin 13

sin215

sin 3⋯

Using the identity

ei=cos i sin

We can rewrite this in terms of exponentials

f =ℑ[e i13e2 i

15e3 i⋯]

Let

g =ei

13e2 i

15e3 i

⋯

Let

z=e i

g z=z13

z215

z3⋯

Let

p= z

And finally we have our series in a form that we can work with

g p=p213

p415

p6⋯

g p=p [ p13

p3

15

p5⋯]

Now lets make one more function definition, let

h p=p13

p315

p5⋯

Taking the derivative we have

d h pdp

=1p2p4

⋯=1

1−p2

Integrating both sides

h p=∫ 1

1−p2 dp=tanh−1 pC1

We can evaluate the constant of integration by using the fact that

h 0=0=C1

So we have

h p=tanh−1 p

Substituting this back in to try to work back up to our original problem we have

g p=p tanh−1 p

g =ei 2 tanh−1e i

2 =cos 2 i sin 2 tanh−1cos 2 isin 2 Just for the sake of my sanity let

x=cos2

y=sin 2

And we have the equation

g xi y = xi y tanh−1 xi y

Thankfully there is an equation that can simplify this if we just look to equation 722.2 in our trusty copy of Dwight's and we realize that the inverse hyperbolic tangent of a complex variable has the simple relation

tanh−1xiy=

14

ln [ 1x 2 y2

1−x 2 y2 ] i2

tan−1[ 2y

1−x2− y2 ]

After a little bit of algebra we end up at the result

g xiy =x4

ln [ 1x 2 y2

1−x 2 y2 ]− y2

tan−1[ 2 y

1−x2− y2 ]

i [ y4 ln [ 1x 2 y2

1−x 2 y2 ] x2

tan−1[ 2 y

1−x2− y2 ]]

Now if we just take the imaginary part of this we are left with

f =sin 2

4ln [ 1cos2

2

sin22

1−cos2 2

sin22 ]

cos 2 2

tan−1[ 2sin 2 1−cos22 −sin22 ]

This can simplify down a bit

f =sin 2

4ln [ 22cos 2

2−2cos 2 ]cos2

2tan−1

∞

And just a bit further

f =14

sin 2 ln [ 22cos 2 2−2cos 2 ]

4cos2

And we finally have a reasonably nice looking closed form solution for the original problem