Suggested languages for you:

Americas

Europe

Q 3.7-3E

Expert-verifiedFound in: Page 139

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**Determine the recursive formulas for the Taylor method of order 4 for the initial value problem ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{x}}{\mathbf{-}}{\mathbf{y}}{\mathbf{,}}{\mathbf{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}$**.

${\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{h}\mathbf{(}{\mathbf{x}}_{\mathbf{n}}\mathbf{-}{\mathbf{y}}_{\mathbf{n}}\mathbf{)}\mathbf{+}\left(\frac{{h}^{2}}{2}-\frac{{h}^{3}}{6}+\frac{{h}^{4}}{24}\right)\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{x}}_{\mathbf{n}}\mathbf{+}{\mathbf{y}}_{\mathbf{n}}\mathbf{)}$

Here $\mathrm{y}\text{'}=\mathrm{x}-\mathrm{y},\mathrm{y}\left(0\right)=0$

Apply the chain rule.

${\mathrm{f}}_{2}(\mathrm{x},\mathrm{y})=\frac{\partial \mathrm{f}}{\partial \mathrm{x}}(\mathrm{x},\mathrm{y})+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}(\mathrm{x},\mathrm{y})\mathrm{f}(\mathrm{x},\mathrm{y})$

Since $\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{x}-\mathrm{y}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}(\mathrm{x},\mathrm{y})=1\phantom{\rule{0ex}{0ex}}\frac{\partial \mathrm{f}}{\partial \mathrm{y}}(\mathrm{x},\mathrm{y})=-1$

So, the equation is ${\mathrm{f}}_{2}(\mathrm{x},\mathrm{y})=1-\mathrm{x}+\mathrm{y}$

Apply the same procedure as step 1

${\mathrm{f}}_{3}(\mathrm{x},\mathrm{y})=-1+\mathrm{x}-\mathrm{y}\phantom{\rule{0ex}{0ex}}{\mathrm{f}}_{4}(\mathrm{x},\mathrm{y})=1-\mathrm{x}+\mathrm{y}$

The recursive formula is

${\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+\mathrm{h}\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{hf}({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}})+\frac{{\mathrm{h}}^{2}}{2!}{\mathrm{f}}_{2}({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}})+.....\frac{{\mathrm{h}}^{\mathrm{p}}}{\mathrm{p}!}{\mathrm{f}}_{\mathrm{p}}({\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}})$

${\mathrm{x}}_{\mathrm{n}+1}={\mathrm{x}}_{\mathrm{n}}+\mathrm{h}\phantom{\rule{0ex}{0ex}}{\mathrm{y}}_{\mathrm{n}+1}={\mathrm{y}}_{\mathrm{n}}+\mathrm{h}({\mathrm{x}}_{\mathrm{n}}-{\mathrm{y}}_{\mathrm{n}})+\left(\frac{{\mathrm{h}}^{2}}{2}-\frac{{\mathrm{h}}^{3}}{6}+\frac{{\mathrm{h}}^{4}}{24}\right)(1-{\mathrm{x}}_{\mathrm{n}}+{\mathrm{y}}_{\mathrm{n}})$

Where starting points are ${\mathrm{x}}_{\mathrm{o}}=0,{\mathrm{y}}_{0}=0$.

**Hence the solution is role="math" localid="1664316175651" ${{\mathbf{y}}}_{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{=}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{+}\mathbf{h}\mathbf{(}{{\mathbf{x}}}_{{\mathbf{n}}}{\mathbf{-}}{{\mathbf{y}}}_{{\mathbf{n}}}\mathbf{)}\mathbf{+}\mathbf{(}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{3}}}{\mathbf{6}}\mathbf{+}\frac{{\mathbf{h}}^{\mathbf{4}}}{\mathbf{24}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{-}{{\mathbf{x}}}_{{\mathbf{n}}}{\mathbf{+}}{{\mathbf{y}}}_{{\mathbf{n}}}{\mathbf{)}}$**

94% of StudySmarter users get better grades.

Sign up for free