Question

What is the first approximation of the function f(x)=(1−x)100 near x=0?

Answer 1

Let us use the Taylor expansion of function into series.

We know that for function $f$ at point $x$ we may write the formula

$f(x) = f(x_0) + \dfrac{f'(x_0) } {1!}(x-x_0) + \dfrac{f''(x_0) } {2!}(x-x_0)^2 +\ldots\\ + \dfrac{f^{(n)} (x_0) } {n!}(x-x_0)^n + \ldots$

We should use only the first approximation, so we may write

$f(x) \approx f(x_0) + \dfrac{f'(x_0)} {1!}(x-x_0).$ (1)

We know that $x_0=0$ . Next, we obtain the formula for the derivative of $f$

$f'(x) = \big((1-x)^{100}\big)' = -100(1-x)^{99}=100(x-1)^{99}$ .

Therefore,

$f'(x_0)=f'(0)=100(0-1)^{99}=-100.$

Next we use (1) to write

$f(x) \approx f(x_0) + \dfrac{f'(x_0)} {1!}(x-x_0) = 1 + \dfrac{-100}{1!}(x-0) = 1-100x.$