Suppose that $f\in C^4([-h,h],\Bbb R)$ with $h>0$. Show that $$\left|\int_{-h}^hf(x)\mathrm dx-\frac{h}3(f(-h)+f(h)+4f(0))\right|\le\frac{h^5}{90}\|f^{(4)}\|_\infty$$ HINT: integrate by parts and after use the mean value theorem for integrals.

I tried different approaches with no results, by example I integrate by parts, prior change of variable, using the Bernoulli polynomials and tried to setup something using two different versions of the MVT for integrals.

I tried other more direct approaches without the Bernoulli polynomials with no result. The two versions of the MVT for integrals that I know are these:

MVT for integrals (version I). Let $f,g\in C(I,\Bbb R)$ with $g\ge 0$, then $$\int_\alpha^\beta f(x)g(x)\mathrm dx=f(\xi)\int_\alpha^\beta g(x)\mathrm dx,\quad \xi\in[\alpha,\beta]$$

MVT for integrals (version II). Let $f\in C(I,\Bbb R)$ and $g\in C^1(I,\Bbb R)$ monotone, then $$\int_\alpha^\beta f(x)g(x)\mathrm dx=g(\alpha)\int_\alpha^\xi f(x)\mathrm dx+g(\beta)\int_\xi^\beta f(x)\mathrm dx,\quad \xi\in[\alpha,\beta]$$

Some help will be appreciated, thank you.


As commented, there are several approaches. However, if you want to follow the hint in brute-force fashion, then integrate by parts four times retaining constants of integration. Solve for the constants that give the desired quadrature rule and eliminate terms where $f'$, $f''$ and $f'''$ appear.

Clearly, this is not an elegant approach here, but in other contexts the technique is useful.

For the first, step take

$$\int_{-h}^h f(x) \, dx = \int_{-h}^0 f(x) \, dx + \int_{0}^h f(x) \, dx = \int_{0}^h f(x-h) \, dx + \int_{0}^h f(x) \, dx. $$

Perform integration by parts using $dv = dx \implies v = x + \text{const.}$ to obtain

$$\int_{-h}^h f(x) \, dx \\ = \left. (x + A_1)f(x)\right|_{0}^h + \int_{0}^h (x + A_1) f'(x-h)\,dx + \left. (x + B_1)f(x)\right|_{0}^h - \int_{0}^h (x + B_1) f'(x)\,dx \\ = (h + B_1) f(h) + (h + A_1 - B_1) f(0) - A_1 f(-h) \\ - \int_{0}^h (x + A_1) f'(x-h)\,dx -\int_{0}^h (x + B_1) f'(x)\,dx$$

Put $A_1 = -h/3$ and $B_1 = -2h/3$ to obtain

$$\int_{-h}^h f(x) \, dx = \frac{h}{3} \left(f(-h) + 4 f(0) + f(h) \right) - E,$$


$$E = \int_{0}^h (x + A_1) f'(x-h)\,dx + \int_{0}^h (x + B_1) f'(x)\,dx.$$

Now integrate by parts three more times to obtain

$$\begin{align}E &= \left.\left(\frac{1}{2} x^2 + A_1 x + A_2\right)f'(x-h) \right|_0^h \\ &+ \left.\left(\frac{1}{2} x^2 + B_1 x + B_2\right)f'(x) \right|_0^h \\ &- \left.\left(\frac{1}{6} x^3 + \frac{1}{2}A_1 x^2 + A_2x + A_3\right)f''(x-h) \right|_0^h \\ &- \left.\left(\frac{1}{6} x^3 + \frac{1}{2}B_1 x^2 + B_2x + B_3\right)f''(x) \right|_0^h \\ &+ \left.\left(\frac{1}{24} x^4 + \frac{1}{6}A_1 x^3 + \frac{1}{2}A_2x^2 + A_3x + A_4\right)f'''(x-h) \right|_0^h \\ &+ \left.\left(\frac{1}{24} x^4 + \frac{1}{6}B_1 x^3 + \frac{1}{2}B_2x^2 + B_3x + B_4\right)f'''(x)\right|_0^h \\ &- \int_0^h \left(\frac{1}{24} x^4 + \frac{1}{6}A_1 x^3 + \frac{1}{2}A_2x^2 + A_3x + A_4\right)f^{(4)}(x-h) \, dx \\ &- \int_0^h \left(\frac{1}{24} x^4 + \frac{1}{6}B_1 x^3 + \frac{1}{2}B_2x^2 + B_3x + B_4\right) f^{(4)}(x) \, dx \end{align}$$

Solve for $A_2, A_3, A_4, B_2, B_3, B_4$ to eliminate the boundary terms and apply the bound $\|f^{(4)} \|_\infty$ inside the integrals to obtain the desired result

$$|E| \leqslant \frac{h^5}{90} \|f^{(4)} \|_\infty.$$

  • $\begingroup$ Thank you very much, I realized all these steps but I thought this system of equations is not solvable, I mean, I cant see how to eliminate all these terms at once. Are you sure that we can eliminate all but the error term? The only difference, when I tried, was that I dont divided the integrals in two parts, I was trying to solve the polynomials for $\int_{-h}^h$. Well, I will try again. $\endgroup$
    – Masacroso
    Mar 5 '17 at 2:56
  • 1
    $\begingroup$ @Masacroso: Yes, they are all linear functions of the constants with similar powers of $x$ as factors. I did this exercise once long ago. I'll work another step and show you what it looks like when I get a chance. $\endgroup$
    – RRL
    Mar 5 '17 at 3:00
  • $\begingroup$ its ok, thank you again. $\endgroup$
    – Masacroso
    Mar 5 '17 at 3:01
  • $\begingroup$ I made a correction to $A_1$ and replaced $F_2, G_2,$, etc, with the actual expressions. $\endgroup$
    – RRL
    Mar 5 '17 at 6:14

Just for the record I will add the "ugly" brute-force solution. Integrating by parts recursively we get

$$\int_{-h}^hf(x)\mathrm dx=\int_0^h f(x)\mathrm dx+\int_{-h}^0f(x)\mathrm dx=\\=p_1(x)f(x)|_{-h}^0+q_1(x)f(x)|_0^h-\int_{-h}^0p_1(x)f'(x)\mathrm dx-\int_0^h q_1(x)f'(x)\mathrm dx=\\=\sum_{k=1}^4(-1)^{k+1}\Bigg(p_k(x)f^{(k-1)}(x)\Big|_{-h}^0+q_k(x)f^{(k-1)}(x)\Big|_{0}^h\Bigg)+\int_{-h}^0 p_4(x)f^{(4)}(x)\mathrm dx+\int_0^h q_4(x)f^{(4)}(x)\mathrm dx$$

where the $p_k$ and $q_k$ are polynomials of degree $k$ such that $[p'_k]=p_{k-1}$ and $[q_k]'=q_{k-1}$. We must set the coefficients of these polynomials to fit the desired result, that is, we need that


and that the other terms, but the last integral, will be zero. Hence setting $p_1(x)=x+\frac{2h}3$ and $q_1(x)=x-\frac{2h}3$ above we get the RHS. Now we set

$$p_2(x)=\frac{x^2}2+h\frac{2x}3+h^2\frac16,\quad q_2(x)=\frac{x^2}2-h\frac{2x}3+h^2\frac16$$

$$p_3(x)=\frac{x^3}6+h\frac{x^2}3+h^2\frac{x}6,\quad q_3(x)=\frac{x^3}6-h\frac{x^2}3+h^2\frac{x}6$$

$$p_4(x)=\frac{x^4}{24}+h\frac{x^3}9+h^2\frac{x^2}{12}-\frac{h^4}{72},\quad q_4(x)=\frac{x^4}{24}-h\frac{x^3}9+h^2\frac{x^2}{12}-\frac{h^4}{72}$$

Under these conditions we have that

$$p_k(x)f^{(k-1)}(x)|_{-h}^0+ q_k(x)f^{(k-1)}(x)|_0^h=0,\quad k\in\{2,3,4\}$$

Then we have the identity

$$\int_{-h}^hf(x)\mathrm dx=\frac{h}3\big(f(-h)+4f(0)+f(h)\big)+\int_{-h}^0 f^{(4)}(x)p_4(x)\mathrm dx+\int_0^h f^{(4)}(x)q_4(x)\mathrm dx$$


$$\begin{align}\left|\int_{-h}^0 f^{(4)}(x)p_4(x)\mathrm dx+\int_0^h f^{(4)}(x)q_4(x)\mathrm dx\right|&\le\int_{-h}^0|f^{(4)}(x)p_4(x)|\mathrm dx+\int_0^h|f^{(4)}(x)q_4(x)|\mathrm dx\\&\le\|f^{(4)}\|_\infty\left(\int_{-h}^0|p_4(x)|\mathrm dx+\int_0^h|q_4(x)|\mathrm dx\right)\\&\le2\|f^{(4)}\|_\infty\int_0^h\left|\frac{x^4}{24}-h\frac{x^3}9+h^2\frac{x^2}{12}-\frac{h^4}{72}\right|\mathrm dx\\&\le\|f^{(4)}\|_\infty\frac{h^5}{90}\end{align}$$

where we used the fact that $p_4(x)=q_4(-x)$ and that $q_4(x)\le0$ in $[0,h]$.$\Box$

  • 1
    $\begingroup$ - admirable effort in completing this. $\endgroup$
    – RRL
    Mar 6 '17 at 2:18

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