Finding the Maclaurin series for a function

Finding the Maclaurin series for a function

f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots

Finding a series from standard results

Create a series for \cos2x^3

We know the maclaurin series for \cos x:

\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots

Now we can just substitute 2x^3 in place of x:

\cos 2x^3 = 1-\frac{(2x^3)^2}{2!}+\frac{(2x^3)^4}{4!}-\frac{(2x^3)^6}{6!}+\cdots

Next, we can expand + simplify the powers:

\cos 2x^3 = 1-\frac{4x^6}{2!}+\frac{16x^{12}}{4!}-\frac{64x^{18}}{6!}+\cdots

And then simplify the factorials and coefficient fractions:

\cos 2x^3 = 1-2x^6+\frac{2}{3}x^{12}-\frac{32}{45}x^{18}+\cdots

Create a series for e^{\sin x}

• We know the maclaurin series for e^x and \sin x:
  • \sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}
  • e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}
• What if we substitute the series for \sin x into the series for e^x?
  • e^{x-\frac{x^3}6+\frac{x^5}{120}+\cdots} = e^x \cdot e^{-\frac{x^3}6} \cdot e^{\frac{x^5}{120}} \cdots
• Again, we can substitute the series for e^x into each of these:
  • e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots
  • e^{-\frac{x^3}6} = 1-\frac{x^3}6+\frac{\left(-\frac{x^3}6\right)^2}{2!}+\cdots = 1-\frac{x^3}6+\frac{x^6}{72}+\cdots
  • e^x+e^{-\frac{x^3}6}= \left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right)\left(1-\frac{x^3}6+\frac{x^6}{72}+\cdots\right)

TODO: finish this stupid question, I can’t bear even looking at it anymore

Create a series for \ln(2-3x)

• We know the maclaurin series for \ln(1+x):
  • \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots
• We can rewrite \ln(2-3x) as:
  • \ln(2-3x) = \ln\left(2\left(1-\frac{3x}2\right)\right) = \ln(2) + \ln\left(1-\frac{3x}2\right)
• Now we substitute -\frac{3x}2 in place of x in the series for \ln(1+x):
  • \ln\left(1-\frac{3x}2\right) = -\frac{3x}2 - \frac{\left(-\frac{3x}2\right)^2}{2} + \frac{\left(-\frac{3x}2\right)^3}{3} - \frac{\left(-\frac{3x}2\right)^4}{4} + \cdots
• Expand and simplify the powers:
  • \ln\left(1-\frac{3x}2\right) = -\frac{3x}2 - \frac{9x^2}{8} - \frac{27x^3}{24} - \frac{81x^4}{64} + \cdots
• At the end, we combine this with \ln(2) to get the final series:
  • \ln(2-3x) = \ln(2) - \frac{3x}2 - \frac{9x^2}{8} - \frac{27x^3}{24} - \frac{81x^4}{64} + \cdots

Create a series for the expression \ln(\sqrt{1+2x})

• =\ln((1+2x)^{\frac12})
• =\frac12\ln(1+2x)
• We know the maclaurin series for \ln(1+x):
  • \ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots
• We can substitute 2x in place of x in the series for \ln(1+x):
  • \ln(1+2x) = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \frac{(2x)^4}{4} + \cdots
• Expand and simplify the powers:
  • \ln(1+2x) = 2x - 2x^2 + \frac{8x^3}{3} - 4x^4 + \cdots
• Finally, we multiply the series by \frac12 to get the final result:
  • \ln(\sqrt{1+2x}) = x - x^2 + \frac{4x^3}{3} - 2x^4 + \cdots