Solving Taylor Series

Question

Write the first three terms in the series for:

$y= e^t$

find $y(2)$

I know I need to use:

$1+ t + t^2/2 + t^3/6$

What do I plug in for t though?

Answer 1

You have written y as a function of t.

y(t)=e^t

This you have expanded in your Taylor series, so just plug in 2. i.e. y(2).

Answer 2

You have a function $y(t)=e^t$. The notation $y(2)$ is shorthand for $y(t=2)$. So to find $y(2)$ you could simply substitute in $t=2$ giving you $y(2)=e^2=7.3891...$. You could also use the Taylor series, which is useful if you don't have a calculator.

So, the Taylor series expansion of $y(t)=e^t$ is $$1+t+\frac {t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots$$ So, to find $y(2)=e^2$ plug in $t=2$, knowing that the more terms you calculate in the series the more accurate your answer will be. So we get

$$y(2)=1+2+\frac{4}{2}+\frac{8}{6}+\frac{16}{24}+\cdots=7+\cdots$$

Edit Seeing your comments on the other answer, it seems your confusion lies in the Taylor series. The Taylor series for $y(t)=e^t$ has an infinite number of terms, $$e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}=1+t+\frac{t^2}{2}+\frac{t^3}{3!}+\frac{t^4}{4!}+\frac{t^5}{5!}+\frac{t^6}{6!}+\cdots$$ So if you only use terms up to $t^3/3!$ then you'll get an approximate answer, but you'll need to consider more and more terms in the series to get a more accurate answer.