Creating Maclaurin series

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

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

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

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

=\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

Question	Answer
What is the general formula for the Maclaurin series of a function f(x)?	f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots
What is the Maclaurin series for \cos x?	\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots
How do you find the Maclaurin series for \cos(2x^3)?	Substitute 2x^3 for x in the series for \cos x: \cos 2x^3 = 1-\frac{(2x^3)^2}{2!}+\frac{(2x^3)^4}{4!}-\frac{(2x^3)^6}{6!}+\cdots
What is the simplified Maclaurin series for \cos(2x^3) up to the x^{18} term?	1-2x^6+\frac{2}{3}x^{12}-\frac{32}{45}x^{18}+\cdots
What are the Maclaurin series for \sin x and e^x used to create the series for e^{\sin x}?	\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!} and e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots
What is the Maclaurin series for \ln(1+x)?	\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots
How do you rewrite \ln(2-3x) to use the known series for \ln(1+x)?	\ln(2-3x) = \ln(2) + \ln\left(1-\frac{3x}{2}\right)
What is the Maclaurin series for \ln(2-3x)?	\ln(2-3x) = \ln(2) - \frac{3x}{2} - \frac{9x^2}{8} - \frac{27x^3}{24} - \frac{81x^4}{64} + \cdots
How do you rewrite \ln(\sqrt{1+2x}) to use a known series?	\ln(\sqrt{1+2x}) = \frac{1}{2}\ln(1+2x)
What is the Maclaurin series for \ln(\sqrt{1+2x})?	\ln(\sqrt{1+2x}) = x - x^2 + \frac{4x^3}{3} - 2x^4 + \cdots