Find power series expansion of f(z) = cos(2z + 3) at the point z = -1.

[warwick]

Find power series expansion of f(z) = cos(2z + 3) at the point z = -1.

z is a complex number. Of course, I want to expand around (z - a) which in this case is (z+1).

f(z) = cos(2z + 3) = cos(2(z+1) +1) = cos(2(z+1))cos1 - sin(2(z+1))sin1 = cos(2(z+1))cos1 - 2sin(1)sin(z+1)cos(z+1) = cos1[1-2sin²(z+1)] - 2sin(1)sin(z+1)cos(z+1)

Of course, I've used the two double angle identities. To finish, I need to use the known sin and cosine power series expansions. This seems a bit cumbersome. Am I missing a simplification?

 

[Steven G]

Find power series expansion of f(z) = cos(2z + 3) at the point z = -1.

z is a complex number. Of course, I want to expand around (z - a) which in this case is (z+1).

f(z) = cos(2z + 3) = cos(2(z+1) +1) = cos(2(z+1))cos1 - sin(2(z+1))sin1 = cos(2(z+1))cos1 - 2sin(1)sin(z+1)cos(z+1) = cos1[1-2sin²(z+1)] - 2sin(1)sin(z+1)cos(z+1)

Of course, I've used the two double angle identities. To finish, I need to use the known sin and cosine power series expansions. This seems a bit cumbersome. Am I missing a simplification?

I thought that a power series is a polynomial that does not end. That is a power series is in the form a0+ a1z + a2z^2 =a3z^3=.... You can also replace each z with (z-a).

 

[warwick]

I thought that a power series is a polynomial that does not end. That is a power series is in the form a0+ a1z + a2z^2 =a3z^3=.... You can also replace each z with (z-a).

Yes, I'm leaving out the infinite series summation substitution for sin and cosine. I'm just wondering if there's a simpler way that I'm missing and/or if my work thus far is correct.

 

[Steven G]

Yes, I'm leaving out the infinite series summation substitution for sin and cosine. I'm just wondering if there's a simpler way that I'm missing and/or if my work thus far is correct.

OK, I looked more closely at your work and you have not made any errors. So continue on...

 

[Ishuda]

Find power series expansion of f(z) = cos(2z + 3) at the point z = -1.

z is a complex number. Of course, I want to expand around (z - a) which in this case is (z+1).

f(z) = cos(2z + 3) = cos(2(z+1) +1) = cos(2(z+1))cos1 - sin(2(z+1))sin1 = cos(2(z+1))cos1 - 2sin(1)sin(z+1)cos(z+1) = cos1[1-2sin²(z+1)] - 2sin(1)sin(z+1)cos(z+1)

Of course, I've used the two double angle identities. To finish, I need to use the known sin and cosine power series expansions. This seems a bit cumbersome. Am I missing a simplification?

Assuming infinite differentiability, the power series expansion for a function f about a point a is
f(z) = \(\displaystyle \Sigma_{n=0}^{n=\infty} \frac{f^{(n)}(a) (z-a)^n}{n!}\)
where f⁽ⁿ⁾(1) is the nth derivative [0 is the function itself] of f at z=1. So start taking derivatives
f⁽⁰⁾(1) = cos(1); f⁽¹⁾(1) = -2 sin(1)
f⁽²⁾(1) = -2² cos(1); f⁽³⁾(1) = 2³ sin(1)
f⁽⁴⁾(1) = 2⁴ cos(1); f⁽⁵⁾(1) = -2⁵ sin(1)
...
which leads to ...

 

[warwick]

Assuming infinite differentiability, the power series expansion for a function f about a point a is
f(z) = \(\displaystyle \Sigma_{n=0}^{n=\infty} \frac{f^{(n)}(a) (z-a)^n}{n!}\)
where f⁽ⁿ⁾(1) is the nth derivative [0 is the function itself] of f at z=1. So start taking derivatives
f⁽⁰⁾(1) = cos(1); f⁽¹⁾(1) = -2 sin(1)
f⁽²⁾(1) = -2² cos(1); f⁽³⁾(1) = 2³ sin(1)
f⁽⁴⁾(1) = 2⁴ cos(1); f⁽⁵⁾(1) = -2⁵ sin(1)
...
which leads to ...

The professor used the substitution method for a simpler example involving sin(1+x) at z = 1. That's why I took this approach. I'll take a look at this one...