How do you find the taylor series series at x=0 of #f(x) = 1/(1-2x)#?

Question

Christopher Agresta · Accepted Answer

Rather than differentiating, just write out a power series that when multiplied by #1-2x# gives #1#.

#f(x) = sum_(n=0)^oo 2^nx^n = 1 + 2x + 4x^2 + 8x^3 + ...# Then we find: #(sum_(n=0)^oo 2^nx^n)(1-2x)# #= (sum_(n=0)^oo 2^nx^n) - 2x(sum_(n=0)^oo 2^nx^n)# #= (sum_(n=0)^oo 2^nx^n) - (sum_(n=1)^oo 2^nx^n)# #=2^0x^0 = 1# Of course this only works when the sums converge - which works for #abs(x) < 1/2# at least.

Scarlett Allison · Answer

undefined To find the Taylor series of $ f(x) = \frac{1}{1 - 2x} $ centered at $ x = 0 $, we can use the formula for the Taylor series expansion, which is given by:

$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \cdot x^n $$

First, let's find the derivatives of $ f(x) $:

$$ f'(x) = \frac{d}{dx}\left(\frac{1}{1 - 2x}\right) = \frac{2}{(1 - 2x)^2} $$
$$ f''(x) = \frac{d^2}{dx^2}\left(\frac{1}{1 - 2x}\right) = \frac{d}{dx}\left(\frac{2}{(1 - 2x)^2}\right) = \frac{8}{(1 - 2x)^3} $$
$$ f'''(x) = \frac{d^3}{dx^3}\left(\frac{1}{1 - 2x}\right) = \frac{d}{dx}\left(\frac{8}{(1 - 2x)^3}\right) = \frac{48}{(1 - 2x)^4} $$

From the pattern, we can see that the nth derivative of $ f(x) $ evaluated at $ x = 0 $ is $ 2^n $.

Now, let's plug these derivatives into the formula for the Taylor series:

$$ f(0) = 1 $$
$$ f'(0) = 2 $$
$$ f''(0) = 8 $$
$$ f'''(0) = 48 $$

Thus, the Taylor series expansion of $ f(x) = \frac{1}{1 - 2x} $ centered at $ x = 0 $ is:

$$ f(x) = 1 + 2x + 8x^2 + 48x^3 + \ldots $$

Therefore, the Taylor series at $ x = 0 $ of $ f(x) = \frac{1}{1 - 2x} $ is $ \sum_{n=0}^{\infty} 2^n \cdot x^n $.