landdenaw

2022-06-29

Given stepsizes ${h}_{1}$ and ${h}_{2}$ , develop a numerical scheme to approximate ${f}^{\mathrm{\prime}\mathrm{\prime}}({x}_{0})$ with function values $f({x}_{0})$ , $f({x}_{0}+{h}_{1})$ and $f({x}_{0}+{h}_{2})$ . Under what conditions will your method not work?

Lisbonaid

Beginner2022-06-30Added 22 answers

Step 1

If ${h}_{1}=-{h}_{2}$ , you can use second derivative midpoint formula

${f}^{{}^{\u2033}}({x}_{0})=\frac{1}{{h}^{2}}[f({x}_{0}-h)-2f({x}_{0})+f({x}_{0}+h)]+o({h}^{3})$

If ${h}_{1}\ne -{h}_{2}$ , you can do like this:

$f({x}_{0}+{h}_{1})=f({x}_{0})+{h}_{1}{f}^{\prime}({x}_{0})+\frac{{h}_{1}^{2}}{2}{f}^{{}^{\u2033}}({x}_{0})+o({h}_{1}^{3})$

$f({x}_{0}+{h}_{2})=f({x}_{0})+{h}_{2}{f}^{\prime}({x}_{0})+\frac{{h}_{2}^{2}}{2}{f}^{{}^{\u2033}}({x}_{0})+o({h}_{2}^{3})$

Minus the two equations

$f({x}_{0}+{h}_{2})-f({x}_{0}+{h}_{1})=({h}_{2}-{h}_{1}){f}^{\prime}({x}_{0})+\frac{{h}_{2}^{2}-{h}_{1}^{2}}{2}{f}^{{}^{\u2033}}({x}_{0})+o({h}_{2}^{3}-{h}_{1}^{3})$

and

${f}^{\prime}({x}_{0})=\frac{f({x}_{0}+{h}_{2})-f({x}_{0}+{h}_{1})}{{h}_{2}-{h}_{1}}+o({h}_{2}^{2}-{h}_{1}^{2})$

But, considering the accuracy, in general, we pick ${h}_{1}=-{h}_{2}$ , which cancels out the ${f}^{\prime}({x}_{0})$ that has a low accuracy in a certain sense.

An alternative way to find ${f}^{{}^{\u2033}}({x}_{0})$ using only $f({x}_{0})$ , $f({x}_{0}+{h}_{1})$ and $f({x}_{0}+{h}_{2})$ and not involving ${f}^{\prime}({x}_{0})$ is that you should express ${f}^{\prime}({x}_{0})$ with $f({x}_{0})$ , $f({x}_{0}+{h}_{1})$ and $f({x}_{0}+{h}_{2})$ , so a potential formula is

$f({x}_{0}+{h}_{1})=f({x}_{0})+\frac{{h}_{1}}{{h}_{1}+{h}_{2}}[f({x}_{0}+{h}_{1})+f({x}_{0}+{h}_{2})-2f({x}_{0})-(\frac{{h}_{2}^{2}-{h}_{1}^{2}}{2}{f}^{{}^{\u2033}}({x}_{0}))]+\frac{{h}_{1}^{2}}{2}{f}^{{}^{\u2033}}({x}_{0})$

If ${h}_{1}=-{h}_{2}$ , you can use second derivative midpoint formula

${f}^{{}^{\u2033}}({x}_{0})=\frac{1}{{h}^{2}}[f({x}_{0}-h)-2f({x}_{0})+f({x}_{0}+h)]+o({h}^{3})$

If ${h}_{1}\ne -{h}_{2}$ , you can do like this:

$f({x}_{0}+{h}_{1})=f({x}_{0})+{h}_{1}{f}^{\prime}({x}_{0})+\frac{{h}_{1}^{2}}{2}{f}^{{}^{\u2033}}({x}_{0})+o({h}_{1}^{3})$

$f({x}_{0}+{h}_{2})=f({x}_{0})+{h}_{2}{f}^{\prime}({x}_{0})+\frac{{h}_{2}^{2}}{2}{f}^{{}^{\u2033}}({x}_{0})+o({h}_{2}^{3})$

Minus the two equations

$f({x}_{0}+{h}_{2})-f({x}_{0}+{h}_{1})=({h}_{2}-{h}_{1}){f}^{\prime}({x}_{0})+\frac{{h}_{2}^{2}-{h}_{1}^{2}}{2}{f}^{{}^{\u2033}}({x}_{0})+o({h}_{2}^{3}-{h}_{1}^{3})$

and

${f}^{\prime}({x}_{0})=\frac{f({x}_{0}+{h}_{2})-f({x}_{0}+{h}_{1})}{{h}_{2}-{h}_{1}}+o({h}_{2}^{2}-{h}_{1}^{2})$

But, considering the accuracy, in general, we pick ${h}_{1}=-{h}_{2}$ , which cancels out the ${f}^{\prime}({x}_{0})$ that has a low accuracy in a certain sense.

An alternative way to find ${f}^{{}^{\u2033}}({x}_{0})$ using only $f({x}_{0})$ , $f({x}_{0}+{h}_{1})$ and $f({x}_{0}+{h}_{2})$ and not involving ${f}^{\prime}({x}_{0})$ is that you should express ${f}^{\prime}({x}_{0})$ with $f({x}_{0})$ , $f({x}_{0}+{h}_{1})$ and $f({x}_{0}+{h}_{2})$ , so a potential formula is

$f({x}_{0}+{h}_{1})=f({x}_{0})+\frac{{h}_{1}}{{h}_{1}+{h}_{2}}[f({x}_{0}+{h}_{1})+f({x}_{0}+{h}_{2})-2f({x}_{0})-(\frac{{h}_{2}^{2}-{h}_{1}^{2}}{2}{f}^{{}^{\u2033}}({x}_{0}))]+\frac{{h}_{1}^{2}}{2}{f}^{{}^{\u2033}}({x}_{0})$

Kiana Dodson

Beginner2022-07-01Added 5 answers

Step 1

${f}_{1}\approx {f}_{0}+{h}_{1}{f}_{0}^{\prime}+\frac{{h}_{1}^{2}}{2}{f}_{0}^{\u2033},$

${f}_{2}\approx {f}_{0}+{h}_{2}{f}_{0}^{\prime}+\frac{{h}_{2}^{2}}{2}{f}_{0}^{\u2033}.$

Then by elimination of ${f}_{0}^{\prime}$ , you get

${h}_{2}({f}_{1}-{f}_{0})-{h}_{1}({f}_{2}-{f}_{0})=\frac{{h}_{1}{h}_{2}({h}_{1}-{h}_{2})}{2}{f}_{0}^{\u2033}.$

${f}_{1}\approx {f}_{0}+{h}_{1}{f}_{0}^{\prime}+\frac{{h}_{1}^{2}}{2}{f}_{0}^{\u2033},$

${f}_{2}\approx {f}_{0}+{h}_{2}{f}_{0}^{\prime}+\frac{{h}_{2}^{2}}{2}{f}_{0}^{\u2033}.$

Then by elimination of ${f}_{0}^{\prime}$ , you get

${h}_{2}({f}_{1}-{f}_{0})-{h}_{1}({f}_{2}-{f}_{0})=\frac{{h}_{1}{h}_{2}({h}_{1}-{h}_{2})}{2}{f}_{0}^{\u2033}.$

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