# `4(e^(-x^2))sin(x) = x^2 - x + 1` Use Newton's method to find all roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations.

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`4(e^(-x^2))sin(x)=x^2-x+1`

`f(x)=4(e^(-x^2))sin(x)-x^2+x-1=0`

To solve equation using Newton's method apply the formula,

`x_(n+1)=x_n-f(x_n)/(f'(x_n))`

`f'(x)=4((e^(-x^2))cos(x)+sin(x)(-2xe^(-x^2)))-2x+1`

`f'(x)=(4cos(x))/e^(x^2)-(8xsin(x))/(e^(x^2))-2x+1`

Plug in f(x) and f'(x) in the formula,

`x^(n+1)=x_n-(4e^((-x_n)^2)sin(x_n)-(x_n)^2+x_n-1)/((4cos(x_n))/(e^((x_n)^2))-(8x_nsin(x_n))/(e^((x_n)^2))-2x_n+1)`

See the attached graph for getting the initial values of x . The curve intersects the x axis at `~~` 0.20 and 1.1.

Let's solve for the first zero x_1=0.2, carry out iteration until we have same approximations at decimal places.

`x_2=0.2-(e^((-0.2)^2)sin(0.2)-(0.2)^2+0.2-1)/((4cos(0.2))/e^((0.2^2))-(8(0.2)sin(0.2))/e^((0.2)^2)-2(0.2)+1)`

`x_2~~0.21883273`

`x_3~~0.21916357`

`x_4~~0.21916367`

`x_5~~0.21916367`

Now we have two approximations that have same decimal places,

Now let's solve for second zero x_1=1.1,

`x_2=1.1-(4e^(-1.1^2)sin(1.1)-1.1^2+1.1-1)/((4cos(1.1))/(e^(1.1^2))-(8(1.1)sin(1.1))/(e^(1.1^2))-2(1.1)+1)`

`x_2~~1.08432830`

`x_3~~1.08422462`

`x_4~~1.08422461`

`x_5~~1.08422461`

So the roots of the equation to the eight decimal places are 0.21916367 , 1.08422461