1. Using Newton's forward/backward differentiation method to find solution

x	f(x)
0.0	1.0000
0.1	0.9975
0.2	0.9900
0.3	0.9776
0.4	0.8604

x = 0

Solution:
Numerical differentiation method to find solution.
The value of table for `X` and `Y`

X	0 `x_0=0`	0.1 `x_1=0.1`	0.2 `x_2=0.2`	0.3 `x_3=0.3`	0.4 `x_4=0.4`
Y	1 `y_0=1`	0.9975 `y_1=0.9975`	0.99 `y_2=0.99`	0.9776 `y_3=0.9776`	0.8604 `y_4=0.8604`

Newton's forward differentiation table is as follows.

X	Y(X)	`DeltaY`	`Delta^2Y`	`Delta^3Y`	`Delta^4Y`
0 `X_0=0`	1 `Y_0=1`
		-0.0025 `-0.0025=0.9975-1` `DeltaY_0=Y_1-Y_0`
0.1 `X_1=0.1`	0.9975 `Y_1=0.9975`		-0.005 `-0.005=-0.0075--0.0025` `Delta^2Y_0=DeltaY_1-DeltaY_0`
		-0.0075 `-0.0075=0.99-0.9975` `DeltaY_1=Y_2-Y_1`		0.0001 `0.0001=-0.0049--0.005` `Delta^3Y_0=Delta^2Y_1-Delta^2Y_0`
0.2 `X_2=0.2`	0.99 `Y_2=0.99`		-0.0049 `-0.0049=-0.0124--0.0075` `Delta^2Y_1=DeltaY_2-DeltaY_1`		-0.1 `-0.1=-0.0999-0.0001` `Delta^4Y_0=Delta^3Y_1-Delta^3Y_0`
		-0.0124 `-0.0124=0.9776-0.99` `DeltaY_2=Y_3-Y_2`		-0.0999 `-0.0999=-0.1048--0.0049` `Delta^3Y_1=Delta^2Y_2-Delta^2Y_1`
0.3 `X_3=0.3`	0.9776 `Y_3=0.9776`		-0.1048 `-0.1048=-0.1172--0.0124` `Delta^2Y_2=DeltaY_3-DeltaY_2`
		-0.1172 `-0.1172=0.8604-0.9776` `DeltaY_3=Y_4-Y_3`
0.4 `X_4=0.4`	0.8604 `Y_4=0.8604`

The value of `x` at you want to find `f(x) : x_0 = 0`

`h = x_1 - x_0 = 0.1 - 0 = 0.1`


`[(dy)/(dx)]_(x=x_0) = 1/h * ( Delta Y_0 - 1/2 * Delta^2 Y_0 + 1/3 * Delta^3 Y_0 - 1/4 * Delta^4 Y_0)`

`:.[(dy)/(dx)]_(x=0) = 1/0.1 * (-0.0025 - 1/2 xx -0.005 + 1/3 xx 0.0001 - 1/4 xx -0.1)`

`:.[(dy)/(dx)]_(x=0) = 0.25033`


`[(d^2y)/(dx^2)]_(x=x_0) = 1/h^2 * ( Delta^2 Y_0 - Delta^3 Y_0 + 11/12 * Delta^4 Y_0 )`

`:.[(d^2y)/(dx^2)]_(x=0) = 1/0.01 * (-0.005 - 0.0001 + 11/12 xx -0.1)`

`:.[(d^2y)/(dx^2)]_(x=0) = -9.67667`


Solution for `Pn'(0) = 0.25033`

Solution for `Pn''(0) = -9.67667`