# Approximations. Newton's method - composite Simpson's rule

Can you help me please to solve this problems and if you can give me some helpful information.

Thanks!

We are asked to find the value of $x$ where:

$$\int_0^x \dfrac{1}{\sqrt{2 \pi}} e^{-t^2/2}~dt = 0.45$$

We need two numerical approaches here. One to find the zeros of the function $f(x)$, Newton's Method, and one to estimate the integral, Composite Simpson and Composite Trapezoidal, above where:

$$f(x) = \int_0^x \dfrac{1}{\sqrt{2 \pi}} e^{-t^2/2}~dt - 0.45 = 0$$

The derivative wrt $x$ of this function is (there is a slight typo in problem specification):

$$f'(x) = \dfrac{1}{\sqrt{2 \pi}} e^{-x^2/2}$$

The Newton-Raphson method is given by:

$$\displaystyle x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)} = x_n - \dfrac{\displaystyle \int_0^{x_n} \dfrac{1}{\sqrt{2 \pi}} e^{-t^2/2}~dt - 0.45}{\dfrac{1}{\sqrt{2 \pi}} e^{-x_n^2/2}}$$

At each iteration, we have to use the Composite Simpson's Rule to find the value of that integral for the next $x_n$.

$$s = \int_a^b f(x) \approx \dfrac{h}{3} \left( f(a) + f(b) + 4 \sum_{i=1}^{n/2}~f(a + (2i - 1)h)+2 \sum_{i=1}^{(n-2)/2} f(a+2 ih) \right)$$

The initial starting point is $x_0 = 0.5$ with a desired accuracy of $10^{-5}$.

The iterations are:

• $x_0 = 0.5$
• Using Composite Simpson, with $n=4$: $s$ evaluated between $(0, 0.5)$ gives $s = 0.191463$
• Using Newton's iteration: $x_1 = x_0 - \dfrac{f(x_0)}{f'(x_0)} = 0.5 - \dfrac{0.191463 - 0.45}{0.352065} = 1.23435$
• $s$ evaluated between $(0, 1.23435)$ gives $s = 0.391464$
• $x_2 = 1.54866$
• $s = 0.439269$
• $x_3 = 1.63789$
• $s = 0.449278$
• $x_4 = 1.64481$
• $s = 0.449278$
• $x_5 = 1.64485$
• $s = 0.45$
• $x_6 = 1.64485$
• It took five iterations to converge to the desired accuracy.

Lets compare this to the exact result and validate we found the correct value of $x$. We have:

$$\int \dfrac{1}{\sqrt{2 \pi}} e^{-x^2/2}~dx = \frac{1}{2} \text{erf}\left(\frac{x}{\sqrt{2}}\right)$$

Evaluating this at $x = 1.64485$ yields:

$$\frac{1}{2} \text{erf}\left(\frac{1.64485}{\sqrt{2}}\right) = 0.45$$

Repeat this exact same procedure using the Composite Trapezoidal Rule for calculating the values of $s$.

Curious question, is it possible to calculate the value of $n$ for the desired accuracy apriori to doing the iterative steps when using these two numerical approaches? Probably, but I will leave that for you to ponder.

• That's right! It's POETS Day! TGIF!! Are you feeling better today? – Namaste Apr 4 '14 at 12:36