Question

Question #0a436

Answer 1

Maclaurin expansion to the 4th term:

`1 + 1/2x - 1/8x^2 + x^3/16 + O(x^4)`

And using this series to approximate `sqrt(0.5)`: `0.773` which is `95.9%` accurate.

Explanation:

A Maclaurin series is just a Taylor series expansion of a function about 0:

`f(x) = f(0) + f'(0)x + (f''(0)x^2)/(2!) + (f^3(0)x^3)/(3!) + ... + (f^n(0)x^n)/(n!)`

In our case,

`f(x) = sqrt(x+1)` and `f(0) = 1`

To get the other terms, of the series, we must compute derivatives of the function.

To get the first derivative, we should use the power rule

`f'(x) = (x+1)^(1/2)dx = 1/2(x+1)^(-1/2)`

And to get the second term of the expansion, we evaluate the first derivative with `x=0` and multiply it by x.

`f'(0)x = 1/(2(0+1)^(1/2))x = 1/2x`

To get the second derivative, we take the derivative of the first derivative.

`f''(x) = 1/2(x+1)^(-1/2)dx = -1/4(x+1)^(-3/2)`

To get the third term of the expansion, we evaluate the second derivative with `x=0`, multiply it by `x^2` and divide it by `2!`

`f''(0)x^2/(2!) = -1/4(0+1)^(-3/2)x^2/(2!) = -1/4*x^2/(2!) = -1/8x^2`

To get the third derivative, we take the derivative of the second derivative.

`f^3(x) = -1/4(x+1)^(-3/2)dx = 3/8(x+1)^(-5/2)`

To get the fourth term of the expansion, we evaluate the second derivative with `x=0`, multiply it by `x^3` and divide it by `3!`

`f^3(0)x^3/(3!) = 3/8(0+1)^(-5/2)x^3/(3!) = x^3/16`

Thus, the Maclaurin series up to the fourth term is

`1 + 1/2x - 1/8x^2 + x^3/16 + O(x^4)`

Now, to approximate `sqrt(0.5)`, we solve for `x`

`sqrt(0.5) = sqrt(x+1)`

`0.5 = x + 1`

`-0.5 = x`

and we plug `x = -0.5 = -1/2` into our series

`f(-1/2) = 1 + 1/2(-1/2) - 1/8(-1/2)^2 + (-1/2)^3/16 = 1 - (1/4) + (1/32) - (1/128) ~= 0.773`