Provide a rough error bound for the following composite simpsons rule.
I am aware that the upper bound is $f$ to the forth derivative evaluated at some $t$ in the open interval $(a,b)\frac{h^4(b-a)}{180}$
However, the $$f(t)= -\frac{500000}{120} \bigg( \exp \bigg(-\frac{t}{120} + \bigg(\frac{t}{120}\bigg)^2 - 2 \bigg( \frac{t}{120} \bigg)^3 \bigg) \bigg)$$ is not easily evaluated at the 4th derivative and because it asks for a rough error bound I am confused as to what to do. The interval is $(0,120)$.
I did think possibly just $h^4 \frac{b-a}{180}$ as the bound but realise this is probably incorrect and I am missing something more obvious.
Please help. Many thanks.
The result of Simpson's rule is between Midpoint and Trapezoidal rules, so the error bound for Trapezoidal rule (which is the larger of the two) is also an error bound for Simpson. You could use this and cut down on the number of derivatives. But I don't think it's necessary: the ordinary error estimate is usable here.
To begin with, I suggest the change of variables $x=t/120$. Then the integral is $$ -500000\int_0^1 e^{f(x)}\,dx,\quad f(x) = -x+x^2-2x^3 $$ The fourth derivative of $e^f$ is $$ \Big(f^{(4)}+4f'''f' + 6 (f')^2 f'' + 3(f'')^2+(f')^4\Big)e^f $$ Note that $0<e^f<1$ because $f<0$ on $(0,1)$. Also, the derivatives of $f$ are quite simple to estimate: $|f'|=|-1+2x-6x^2|\le 9$, $|f''|=|2-12x|\le 12$, $|f'''|=12$, and $f^{(4)}=0$. Throw all this in, and you're done.