1. ## Infinite Series

If anyone could explain how the following is done, it would be greatly appreciated!

We have the following info about three functions A, B, and C for all values of T:

A(t))' = B(t)
(B(t))' = A(t)
(A(t))^2 - (B(t))^2 = 1
C(t) = B(t)/A(t)

From this, we can gather that the Maclaurin series for A(t) is 1 + t^2/2! + t^4/4! + t^6/6!... and the Maclaurin series for B(t) is t + t^3/3! + t^5/5! + t^7/7+...

a) What is the Maclaurin series for A(it)?
b) What is the Maclaurin series for B(it)?

2. Originally Posted by clockingly
If anyone could explain how the following is done, it would be greatly appreciated!

We have the following info about three functions A, B, and C for all values of T:

A(t))' = B(t)
(B(t))' = A(t)
(A(t))^2 - (B(t))^2 = 1
C(t) = B(t)/A(t)
A''(t)=A(t),

so A(t) = a e^t + b e^{-t}

and B(t) = a e^t - be^{-t}

A^2+B^2 = a^2 e^t +2 a*b +b^2 e^{-t} -a^2 e^t +2 a*b -b^2 e^{-t}

............. = 4*a*b,

so a*b=1/4.

From this, we can gather that the Maclaurin series for A(t) is 1 + t^2/2! + t^4/4! + t^6/6!... and the Maclaurin series for B(t) is t + t^3/3! + t^5/5! + t^7/7+...
There does not appear to be enough information to conclude the above, as
you need another condition to ensure that a=b=1/2.

If you had such a condition A(t)=cosh(t), and B(t)=sinh(t), and C(t)=tanh(t)

a) What is the Maclaurin series for A(it)?
b) What is the Maclaurin series for B(it)?
A(it) = cos(t), and B(it)=i*sin(t), so:

A(it) = 1 - t^2/2! + t^4/4! - t^6/6!...

and:

B(it) = i[t - t^3/3! + t^5/5! - t^7/7+...]

RonL