# University of Florida/Egm4313/s12.team8.dupre/R2.7

R2.7

## Problem Statement

Develop the Maclaurin series (Taylor Series at t=0) for ${\displaystyle \displaystyle e^{t},cos(t),sin(t)}$

a) ${\displaystyle \displaystyle e^{t}}$ (7-1)

b)${\displaystyle \displaystyle cos(t)}$ (7-2)

c)${\displaystyle \displaystyle sin(t)}$ (7-3)

### Solution

For a Taylor series at any point t, the general form is as follows (to make it more clear, I have changed the usual t in (t-a) to a capital T):

${\displaystyle \sum _{n=0}^{\infty }{\frac {f^{n}(a)}{n!}}(T-a)^{n}=f(a)+{\frac {f'(a)}{1!}}(T-a)+{\frac {f^{2}(a)}{2!}}(T-a)^{2}+{\frac {f^{3}(a)}{3!}}(T-a)^{3}+{\frac {f^{4}(a)}{4!}}(T-a)^{4}...(7-4)}$

#### Part a solution

Using (7-4) as applied to (7-1) at time t=0 gives us the Maclaurin series for (7-1):

${\displaystyle \displaystyle e^{t}=\sum _{n=0}^{\infty }{\frac {T^{n}}{n!}}=e^{0}+{\frac {e^{0}}{1!}}(T)+{\frac {e^{0}}{2!}}(T)^{2}+{\frac {e^{0}}{3!}}(T)^{3}+{\frac {e^{0}}{4!}}(T)^{4}...(7-6)}$

Simplifying this result, we obtain our final solution:

${\displaystyle \displaystyle e^{t}=1+T+{\frac {T^{2}}{2}}+{\frac {T^{3}}{6}}+{\frac {T^{4}}{24}}...(7-7)}$


#### Part b solution

Using (7-4) as applied to (7-2) at time t=0 gives us the Maclaurin series for (7-2):

${\displaystyle \displaystyle cos(t)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}T^{2n}}{(2n)!}}=cos(0)+{\frac {-sin(0)}{1!}}(T)+{\frac {-cos(0)}{2!}}(T)^{2}+{\frac {sin(0)}{3!}}(T)^{3}+{\frac {cos(0)}{4!}}(T)^{4}+...(7-8)}$

Simplifying this result, we obtain our final solution:

${\displaystyle \displaystyle cos(t)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}T^{2n}}{(2n)!}}=1-{\frac {T^{2}}{4}}+{\frac {T^{4}}{24}}...(7-9)}$


#### Part c solution

Using (7-5) as applied to (7-3) at time t=0 gives us the Maclaurin series for (7-3):

${\displaystyle \displaystyle sin(t)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}T^{2n+1}}{(2n+1)!}}=sin(0)+{\frac {cos(0)}{1!}}(T)+{\frac {-sin(0)}{2!}}(T)^{2}+{\frac {-cos(0)}{3!}}(T)^{3}+{\frac {sin(0)}{4!}}(T)^{4}+...(7-10)}$

Simplifying this result, we obtain our final solution:

${\displaystyle \displaystyle sin(t)=T-{\frac {T^{3}}{6}}+...(7-11)}$