Formula

Taylor Series method `h=x-x_n` `y_(n+1) = y_n + hy_n' + h^2/(2!) y_n'' + h^3/(3!) y_n''' + h^4/(4!) y_n^(iv) + h^5/(5!) y_n^(v) + ...`

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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)(0.5x-0.5y)`

`=d/(dx)(0.5x)-d/(dx)(0.5y)`

`=0.5-0.5y'`

Now, `d^2/(dx^2)(0.5x-0.5y)=d/(dx)(0.5-0.5y')`

`=d/(dx)(0.5)-d/(dx)(0.5y')`

`=0-0.5y''`

`=-0.5y''`

Now, `d^3/(dx^3)(0.5x-0.5y)=d/(dx)(-0.5y'')`

`=-0.5y'''`

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

`y''=0.5-0.5y'`

`y'''=-0.5y''`

`y^(iv)=-0.5y'''`

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

`y_0''=0.5-0.5y_0'=0.75`

`y_0'''=-0.5y_0''=-0.375`

`y_0^(iv)=-0.5y_0'''=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^(iv) + ...`

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for `n=0,x_0=0,y_0=1`

`=1+0.1*(-0.5)+(0.1)^2/(2)*(0.75)+(0.1)^3/(6)*(-0.375)+(0.1)^4/(24)*(0.1875)+...`

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

`=0.9537`

`x_1=x_0+h=0+0.1=0.1`

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

`y_1''=0.5-0.5y_1'=0.7134`

`y_1'''=-0.5y_1''=-0.3567`

`y_1^(iv)=-0.5y_1'''=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^(iv) + ...`

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for `n=1,x_1=0.1,y_1=0.9537`

`=0.9537+0.1*(-0.4268)+(0.1)^2/(2)*(0.7134)+(0.1)^3/(6)*(-0.3567)+(0.1)^4/(24)*(0.1784)+...`

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

`=0.9145`

`x_2=x_1+h=0.1+0.1=0.2`

`:.y(0.2)=0.9145`

`n`	`x_n`	`y_n`	`y_n'`	`y_n''`	`y_n'''`	`y_n^(iv)`	`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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