Find a good numerical approximation to F(1) for the function with the properties that F'(x) = e^-((x)^2)/5 and F(0) = 4.

Question

Find a good numerical approximation to F(1) for the function with the properties that

F'(x) = e^{-((x)^2)/5} and F(0) = 4.

(The answer is not 5.)

Answer 1

Mike gives a reasonable approach. We could use the nth order approximation from sums.

Another would be to do a Taylor series approximation of the function and integrate this approximation from 0 to 1.

As the Taylor series for e^x is ∑x^n/n! , we substitute -x^2/5 in this expression.

Then, we have ∑(-1)^n x^(2n)/(5^n n!)

As we integrate this from 0 to 1, we get ∑(-1)^n /((2n+1)5^n n!) where the sum is for n = 0 to infinity.

Note that the absolute value of the ratio between successive terms is less than 1/(5n)

The term for n = 11 is -2.2 * 10^-17, which means that, as this is an alternating series, our error is less than this for n = 10.

Of course, we add 4 to the result.

It turns out that the cumulative distribution for the normal is on most calculators. The density function shown is that of the normal with standard deviation sqrt(5/2) with a necessary multiplication correction (as the normal density has a constant 1/(sqrt(2π)σ) of sqrt(5π)

Then, sqrt(5π) (norm.dist(1,0,sqrt(5/2),1)-.5) = 0.937150028797979, and 4 + 0.937150028797979 = 4.937150028797979

Our result at n = 10 matches this.

Answer 2

Zena

Crude answer. F(1) ≈ F(0) + 1. F'(0), which gives 5.

Better. F(0.5) ≈ F(0) + 0.5 . F'(0)

F(1) ≈ F(0.5) + 0.5 F' (0.5)

And so on .. you can keep breaking the interval down into 4, 8 , .. parts until you get a 'good' approximation whatever you define that as.

Mike