How do you find the maclaurin series expansion of #e^(7x)ln((1-x)/3)#?

Question

Emilia Aguilar · Accepted Answer

#f(x) = ln(1/3) + (7ln(1/3) - 1)x + (49ln(1/3) - 15)/(2)x^2 +...#

Note I've left the logs as #ln(1/3)# but could easily rewrite as #-ln(3)# The maclaurin series is a special case of the Taylor expansion with #x_0 = 0#. Taylor Expansion: #f(x) = f(x_0) + (f'(x_0))/(1!)(x-x_0) + (f''(x_0))/(2!)(x-x_0)^2 + (f'''(x_0))/(3!)(x-x_0)^3+...# So Maclaurin series will be: #f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ...# We have #f(x) = e^(7x)ln((1-x)/3)# I'm only going to compute up to the second derivative, ie quadratic terms of x because this function will just get so messy it'll be easy to make a simple mistake. If you need more either work them out or use one of the various online tools. Using product rule and chain rule: #f'(x) = d/(dx)(e^(7x))ln((1-x)/3) + e^(7x)d/(dx)(ln((1-x)/3)))# #f'(x) = 7e^(7x)ln((1-x)/3) + e^(7x) * (3)/(1-x) * (-1/3)# #f'(x) = e^(7x)(7ln((1-x)/3) + 1/(x-1))# #f''(x) = d/(dx)(e^(7x))(7ln((1-x)/3) + 1/(x-1)) + e^(7x)d/(dx)(7ln((1-x)/3) + 1/(x-1)))# #f''(x) = 7e^(7x)(7ln((1-x)/3) + 1/(x-1)) + e^(7x)(7/(x-1) - 1/(x-1)^2)# #f''(x) = e^(7x)[49ln((1-x)/3) + 14/(x-1) - 1/(x-1)^2]# So now that we have these, calculate #f(0), f'(0) and f''(0)# #f(0) = e^0ln(1/3) = ln(1/3)# #f'(0) = e^0[7ln(1/3) - 1] = 7ln(1/3) -1# #f''(0) = e^0[49ln(1/3) - 14 - 1] = 49ln(1/3) - 15# Hence: #f(x) = ln(1/3) + (7ln(1/3) - 1)x + (49ln(1/3) - 15)/(2)x^2 +...#

Charles Anctil · Answer

undefined To find the Maclaurin series expansion of $ e^{7x} \ln\left(\frac{1-x}{3}\right) $, we use the formula for the Maclaurin series expansion of a composite function. The formula states that if $ f(x) $ and $ g(x) $ have Maclaurin series expansions, then the Maclaurin series expansion of $ f(g(x)) $ is obtained by substituting the series expansion of $ g(x) $ into $ f(x) $, term by term.

Here, $ f(x) = e^x $ and $ g(x) = 7x \ln\left(\frac{1-x}{3}\right) $.

The Maclaurin series expansion of $ e^x $ is $ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $.

For $ g(x) = 7x \ln\left(\frac{1-x}{3}\right) $, we need to find its Maclaurin series expansion. We'll use the properties of logarithms and the geometric series to expand $ \ln\left(\frac{1-x}{3}\right) $.

$ \ln\left(\frac{1-x}{3}\right) = \ln(1-x) - \ln(3) $

$ = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots\right) - \ln(3) $

Now, multiply by $ 7x $:

$ 7x \ln\left(\frac{1-x}{3}\right) = -7x^2 - \frac{7x^3}{2} - \frac{7x^4}{3} - \ldots - 7x\ln(3) $

Now, substitute this series expansion into the Maclaurin series expansion of $ e^{7x} $:

$ e^{7x} \ln\left(\frac{1-x}{3}\right) = e^{7x}(-7x^2 - \frac{7x^3}{2} - \frac{7x^4}{3} - \ldots - 7x\ln(3)) $

This gives us the Maclaurin series expansion of $ e^{7x} \ln\left(\frac{1-x}{3}\right) $.