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Solve 1 = (x^2)(e^x) using a numerical method
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Solve by a numerical method, eg the Newton-Raphson iterative method.
This improves an estimate `x_n` to `x_(n+1)` by the formula
`x_(n+1) = x_n - f(x_n)/(f'(x_n))`
where `f(x)` is the function to be solved (equated to zero) and `f'(x)` is its derivative with respect to `x`.
Here `f(x) = x^2e^x - 1`
so that `f'(x) = x^2e^x + 2xe^x - 1 = xe^x(x+2) - 1`
and `f(x)/(f'(x)) = (x^2e^x-1)/(xe^x(x+2)-1)`
First choose a starting point/estimate `x_0` of the solution. Let's try `x_0=1` . Using the iterative formula given above
`x_1 = 0.760`, `x_2 = 0.693`, `x_3 = 0.708`, `x_4 = 0.702` ,
`x_5 = 0.704`, `x_6 = 0.703`, `x_7 = 0.704`, `x_8 = 0.703`,
`x_9 = 0.703`
` `When the answers repeat we stop.
Therefore ` ``x = 0.703` to 3sf
Check there isn't another solution by looking at the graph
The graph tends to infinity for larger x and eventually tends to -1 for smaller x after a maximum of -0.5 at x approx -2.
Using the iterative method of Newton-Raphson x = 0.703 to 3sf
Posted by mathsworkmusic on June 9, 2013 at 10:47 AM (Answer #3)
Let us assume `x>0`
Let consider a function f(x)=2 log(x)+x .
Solving (i) , and find the zero of f(x) stands same.
x=.8 ,red line above x axis
x=.69 green line below x axis
Here approximately we can say x=.705 ( one can estimate solution by Newton Raphson method).
Posted by pramodpandey on May 25, 2013 at 6:09 AM (Answer #2)
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