# Finding the 4th order Taylor expansion of $g(t)= t^3 + 2t^2 + 2t + 1$

Given the function $$g(t) = t^3 + 2t^2 + 2t + 1$$

I would like to find the 4th order expansion of $g(t)$ at $t=t_1$.

So far, I have performed the differentiation of $g$, up to $g'''(t)$ w.r.t. $t$, and they are: \begin{align*} g'(t) &= 3t^2 +4t + 2\\ g''(t) &= 6t +4 \\ g'''(t) &= 6 \\ \end{align*} and I know (so far) that the Taylor expansion to 2nd order, $g_2(t)$, is $$g_2(t) = g(t_1) + g'(t_1)(t-t_1)$$ Subsequently, the Taylor expansion to 34d order, $g_3(t)$, is $$g_3(t) = g(t_1) + g'(t_1)(t-t_1)+\frac{g''(t_1)}{2!}(t-t_1)^2$$

What do I do from here?

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I'm confused. Do you want the expansion of the inverse function $t(u)$ at $u=u_1$? Or the value of $u$ about a value of $t$ such that $u(t)=u_1$? –  Ron Gordon Feb 6 '13 at 19:18
I'm very sorry for the confusion, the function has been fixed. –  bryansis2010 Feb 7 '13 at 8:45
I'm not sure why exactly you can't go forward, or the motivation for your work after computing the derivatives. Do you know the general formula for expressing a Taylor series in terms of derivatives? –  Hurkyl Feb 7 '13 at 9:03
I was only given these information by my question, but do not know how to advance from here. My question requires expansion up to 4th order, and in the form in $t_0$ –  bryansis2010 Feb 7 '13 at 9:33
$t_0$ is the same as $t_1$, no? Oh, and the denominator under $g''(t_0)$ should be $2!$, not $3!$. –  Gerry Myerson Feb 7 '13 at 11:32
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