I have a function
$$g(t)=t^{8}\rm e^{t-1}$$
I need to (by hand) find the 2nd order Taylor polynomial for $g(t)$ in $(1,g(1))$
I get: $$g(1)=1$$ $$g'(1)=8\cdot1^7\rm e^{1-1}+t^8\cdot\rm e^{1-1}=9$$ $$g''(1)=56\cdot1^6\rm e^{1-1}+16\cdot1^7\rm e^{1-1}+t^8\rm e^{1-1}=73$$
I use the Taylor formula and get:
$$\tilde{g}_2(t)=1+9(t-1)+\frac{73}{2}=8+9\,t+{\frac {73\, \left( t-1 \right) ^{2}}{2}}$$
When I use Maple's mtayloer function, however, the first term 8 is negative.
Can anyone see what I've done wrong?
The Taylor polynomial in question is
$$1+9\,(t-1)+{\frac {73\, \left( t-1 \right) ^{2}}{2}}.$$