I have a set of points $x_i=\left(i-\frac{1}{2}\right)\Delta x$, for $i=1,...,N$ and the value of a function $f(x)=f(x_i)=f_i$ evaluated at the points $x_i$.

I would like to numerically integrate $f$ from $0$ to $N\Delta x$, i.e.,

$$ \int_0^{N\Delta x}{f(x)dx} $$

Since my points $x_i$ are not endpoints, I can't use the trapezoidal rule, Simpsons rule etc. I tried to use the midpoint rule:

$$ \int_0^{N\Delta x}{f(x)dx}\approx \Delta x\cdot\sum_{i=1}^{N}{f(x_i)} $$

However, the error in the midpoint rule is too big. I would like a better numerical method for the open interval I have, but I am struggling with the open Newton-Cotes formulas given, because the step size of this formulas does not match with what I have. Any suggestions to how should I begin to approach it?


Simpson's Rule would become $$\int_{-\frac32}^{\frac32}f(x)dx\approx\frac98f(-1)+\frac34f(0)+\frac98f(1)$$ I may post better results when I get off of work.

EDIT: Home again, so here goes: $$f(0)=\int_{-\frac12}^{\frac12}f(x)dx-\frac{1}{24}f^{(2)}(\xi)$$ $$f\left(-\frac12\right)+f\left(-\frac12\right)=\int_{-1}^{1}f(x)dx-\frac{1}{12}f^{(2)}(\xi)$$ $$\frac98f(-1)+\frac34f(0)+\frac98f(1)=\int_{-\frac32}^{\frac32}f(x)dx-\frac{21}{640}f^{(4)}(\xi)$$ $$\frac{13}{12}f\left(-\frac32\right)+\frac{11}{12}f\left(-\frac12\right)+\frac{11}{12}f\left(\frac12\right)+\frac{13}{12}f\left(\frac32\right)=\int_{-2}^{2}f(x)dx-\frac{103}{1440}f^{(4)}(\xi)$$ $$\frac{1375}{1152}f(-2)+\frac{125}{288}f(-1)+\frac{335}{192}f(0)+\frac{125}{288}f(1)+\frac{1375}{1152}f(2)=\int_{-\frac52}^{\frac52}f(x)dx-\frac{5575}{193536}f^{(6)}(\xi)$$ $$\frac{741}{640}f\left(-\frac52\right)+\frac{417}{640}f\left(-\frac32\right)+\frac{381}{320}f\left(-\frac12\right)+\frac{381}{320}f\left(\frac12\right)+\frac{417}{640}f\left(\frac32\right)+\frac{741}{640}f\left(\frac52\right)=\int_{-3}^{3}f(x)dx-\frac{1111}{17920}f^{(6)}(\xi)$$ You can get these formulas by just solving $N$ equations in $N$ unknowns for them to be exact for polynomials of degree up to $N-1$. If the number of intervals is not divisible by $3$ one or two copies of the $4$ point formula could be used so as to match the number of intervals much like the usage of Simpson's $3/8$ rule along with his $1/3$ rule when the number of intervals is odd.

I didn't show the $7$ point formula because it has some negative weights, indicating an error-magnifying formula. If you want higher-order formulas you might be better off using formulas that use least-squares fits to the data with less than the highest-order possible polynomial. Due to the lateness of the hour I haven't derived any of these.

  • $\begingroup$ This is not very clear to me. Isn't Simpson's Rule a closed Newton-Cotes formula? i.e., shouldn't I use the points in the extreme of the interval? And changing the limits of integration doesn't change the value of the integral? $\endgroup$ – Thales Dec 19 '17 at 2:46
  • $\begingroup$ I named it after Simpson because it uses the value at 3 equally-spaced points and has error $O(h^4)$. But note that it uses the values of $f(x)$ at the midpoints of the intervals $\left[-\frac32,-\frac12\right]$, $\left[-\frac2,\frac12\right]$, and $\left[\frac12,-\frac32\right]$, just as required. I will post some more formulas with error terms assumed proportional to the first constant derivative of the lowest order polynomial for which they are not exact. $\endgroup$ – user5713492 Dec 19 '17 at 7:04
  • $\begingroup$ Could you hint how did you derived the expressions for the error term? $\endgroup$ – Thales Feb 16 '18 at 22:53
  • 1
    $\begingroup$ Since the formulas are interpolatory by construction the $n$-point formula is exact for polynomials of degree $n-1$ and also of degree $n$ if $n$ is odd due to their symmetry. It can be shown that $\text{error}\le Cf^{(m)}(\xi)$ if $m$ is the first degree for which the formula isn't exact. I have assumed that $\text{error}=Cf^{(m)}(\xi)$ even though I have constructed an example where this isn't true. But on the assumption I just plugged in $f(x)=x^m$ and found the value of $C$. $\endgroup$ – user5713492 Feb 17 '18 at 0:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.