calculus
For part a i get that its asking to use the midpoint trap and simpsons rule
I dont understand how to solve part b. Do i have to plug in the error somewhere and solve for n? or do i want the answer to be less than 10^-4? Why is it mentioning the 4th derivative of the function as a hint?
some clarification on what the question is asking and/or a push in the right direction would really help me.
Simpson's Rule is exact for polynomials of degree 3 or less, so the error term can't depend on any derivative less than the 4th. Put another way: the 3rd derivative of a 3rd degree polynomial is nonzero, so it can't be involved in the error term for Simpson, since the error for 3rd degree polynomials is zero.
An upper bound estimate would be:
$$E \le h | f(b) - f(a) | = \dfrac{1}{6} |f(1) - f(0)| = \dfrac{1}{6}$$
If we draw the picture using six-rectangles and superimpose the parabola, we get:
Using the Right-Hand Riemann sum, we have:
$$I = h \sum_{n=1}^6 f(n/6) = \dfrac{1}{6} \left(f(1/6) + f(2/6) + f(3/6) + f(4/6) + f(5/6) + f(6/6) \right) = 0.421296$$
See if you can derive the two items above.
The actual integral result is:
$$\displaystyle \int_0^1 x^2 dx = \dfrac{1}{3}$$
Now compare the Riemann result to actual and see if that matches what the error estimate $E$ provided.
Best Answer
I will do it for midpoint rule and leave other for you. Denote $$ h= \frac{2-0}{N}=\frac{2}{4} = \frac12 $$ and $$ x_i = 0 + \frac12 (2i -1)h = \frac12(2i-1)\frac12, \quad i =1,2,3,4 $$ The sum for midpoint rule for given function is then $$ M_N = h\sum_{i=1}^N \cos(x_i^3) = \frac12 (\cos(\frac1{64}) + \cos(\frac{27}{64}) + \cos(\frac{125}{64}) + \cos(\frac{343}{64})) $$ Without derivation (via Taylor expansion) the formula to bound the midpoint rule error $E_M$ is $$ E_M \leq \frac{h^2(2-0)}{24}\Bigg| \frac{d^2}{dx^2}\cos(x^3) \Bigg| $$ It follows ($h=\frac{2}{N}$) $$ E_M \leq \frac{1}{6N^2} \Bigg| \frac{d}{dx}(-\sin(x^3)3x^2) \Bigg| = \frac{1}{6N^2} \Bigg| -\cos(x^3)9x^4 - \sin(x^3)6x \Bigg| $$ The absolute value of the second derivative reaches its maximum value $144.27$ at $x=2$ (checked by Mathematica). Your error is therefore bounded by $$ E_M \leq \frac{24.4}{N^2} \leq 10^{-4} $$ This translates to $$ N^2 \geq 244000 $$ or $$ N \geq \sqrt{244000} = 493.97 $$ Consider a bit of rounding error...your $N$ should be at least $494$.
You will need the approximation of fourth derivative when calculating error of the Simpson's rule.