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Expert-verified Found in: Page 140 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # If the Taylor method of order p is used in Problem 17, show that $${{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)^{\bf{n}}}$$, n = 1, 2, …

proved

See the step by step solution

## Step 1: Find the value of $${{\bf{f}}_{\bf{2}}}{\bf{(x,y)}}$$

Here $${\bf{f(x,y) = y}}$$, $${{\bf{x}}_{\bf{o}}}{\bf{ = 0,}}{{\bf{y}}_{\bf{o}}}{\bf{ = 1}}$$, $${\rm{h = }}\frac{{\rm{1}}}{{\rm{n}}}$$

Apply the chain rule.

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

Now

$$\begin{array}{l}\frac{{\partial {\bf{f}}}}{{\partial {\bf{x}}}}{\bf{(x,y) = }}0\\\frac{{\partial {\bf{f}}}}{{\partial {\bf{y}}}}{\bf{(x,y) = }}1\end{array}$$

So, the equation is $${{\bf{f}}_{\bf{2}}}{\bf{(x,y) = y}}$$

## Step 2: Find the other values of $${{\bf{f}}_{\bf{n}}}{\bf{(x,y)}}{\bf{.}}$$

$$\begin{array}{c}{{\bf{f}}_{\bf{3}}}{\bf{(x,y) = y}}\\{{\bf{f}}_{\bf{4}}}{\bf{(x,y) = y}}\\{\bf{.}}\\{\bf{.}}\\{{\bf{f}}_{\bf{p}}}{\bf{(x,y) = y}}\end{array}$$

## Step 2: Apply the recursive formulas for order n

The recursive formula is

$$\begin{array}{l}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ = }}{{\bf{x}}_{\bf{n}}}{\bf{ + h}}\\{{\bf{y}}_{{\bf{n + 1}}}}{\bf{ = }}{{\bf{y}}_{\bf{n}}}{\bf{ + hf(}}{{\bf{x}}_{\bf{n}}}{\bf{ + }}{{\bf{y}}_{\bf{n}}}{\bf{) + }}\frac{{{{\bf{h}}^{^{\bf{2}}}}}}{{{\bf{2!}}}}{{\bf{f}}_{\bf{2}}}{\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{ + }}{{\bf{y}}_{\bf{n}}}{\bf{) + }}.....\frac{{{{\bf{h}}^{\bf{p}}}}}{{{\bf{p!}}}}{{\bf{f}}_{\bf{p}}}{\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{ + }}{{\bf{y}}_{\bf{n}}}{\bf{)}}\end{array}$$$$\begin{array}{c}{{\bf{y}}_{\bf{1}}}{\bf{ = }}\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)\\{{\bf{y}}_{\bf{2}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)^{\bf{2}}}\\{\bf{.}}\\{\bf{.}}\\{{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)^{\bf{n}}}\\{{\bf{y}}_{{\bf{n + 1}}}}{\bf{ = }}\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right){{\bf{y}}_{\bf{n}}}\;{\bf{n}} \in {\bf{N}}\end{array}$$

Hence it is proved that $${{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)^{\bf{n}}}$$ ### Want to see more solutions like these? 