Formula

7. Taylor Series method `h=x-x_0` `y_1 = y_0 + hy_0' + h^2/(2!) y_0'' + h^3/(3!) y_0''' + h^4/(4!) y_0'''' + ...`

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Examples
1. Find y(0.2) for `y'=(x-y)/2`, `x_0=0, y_0=1`, with step length 0.1 using Taylor Series method (1st order derivative)

Solution:
Given `y'=(x-y)/(2), y(0)=1, h=0.1, y(0.2)=?`

Here, `x_0=0,y_0=1,h=0.1,x_n=0.2`

Differentiating successively, we get

Derivative steps

`d/(dx)(x/2-y/2)`

`=d/(dx)(x/2)-d/(dx)(y/2)`

`=(1/2)-((y')/2)`

`=1/2-(y')/2`

Now, `d^2/(dx^2)(x/2-y/2)=d/(dx)(1/2-(y')/2)`

`=d/(dx)(1/2)-d/(dx)((y')/2)`

`=0-((y'')/2)`

`=0-(y'')/2`

`=-(y'')/2`

Now, `d^3/(dx^3)(x/2-y/2)=d/(dx)(-(y'')/2)`

`=-(y''')/2`

`y'=(x-y)/(2)`

`y''=1/2-(y')/2`

`y'''=-(y'')/2`

`y''''=-(y''')/2`

Now substituting, we get
`y_0'=(x_0-y_0)/(2)=-0.5`

`y_0''=1/2-(y_0')/2=0.75`

`y_0'''=-(y_0'')/2=-0.375`

`y_0''''=-(y_0''')/2=0.1875`

Putting these values in Taylor Series, we have
`y_1 = y_0 + hy_0' + h^2/(2!) y_0'' + h^3/(3!) y_0''' + h^4/(4!) y_0'''' + ...`

`=1+0.1*(-0.5)+(0.1)^2/(2!)*(0.75)+(0.1)^3/(3!)*(-0.375)+(0.1)^4/(4!)*(0.1875)+...`

`=1-0.05+0.0038+0+0+...`

`=0.9537`

`:.y(0.1)=0.9537`

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Again taking `(x_1,y_1)` in place of `(x_0,y_0)` and repeat the process

Now substituting, we get
`y_1'=(x_1-y_1)/(2)=-0.4268`

`y_1''=1/2-(y_1')/2=0.7134`

`y_1'''=-(y_1'')/2=-0.3567`

`y_1''''=-(y_1''')/2=0.1784`

Putting these values in Taylor Series, we have
`y_2 = y_1 + hy_1' + h^2/(2!) y_1'' + h^3/(3!) y_1''' + h^4/(4!) y_1'''' + ...`

`=0.9537+0.1*(-0.4268)+(0.1)^2/(2!)*(0.7134)+(0.1)^3/(3!)*(-0.3567)+(0.1)^4/(4!)*(0.1784)+...`

`=0.9537-0.0427+0.0036+0+0+...`

`=0.9145`

`:.y(0.2)=0.9145`

`n`	`x_n`	`y_n`	`y_n'`	`y_n''`	`y_n'''`	`y_n''''`	`x_(n+1)`	`y_(n+1)`
0	0	1	-0.5	0.75	-0.375	0.1875	0.1	0.9537
1	0.1	0.9537	-0.4268	0.7134	-0.3567	0.1784	0.2	0.9145

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