MATH100 A December 2023
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Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Recall that the Taylor series about $x=c$ for a function $f$ is the series
$\sum _{n=0}^{\infty }{\frac {f^{(n)}(c)}{n!}}(xc)^{n}={\frac {f(c)}{0!}}+{\frac {f'(c)}{1!}}(xc)+{\frac {f''(c)}{2!}}(xc)^{2}+{\frac {f^{(3)}(c)}{3!}}(xc)^{3}+\cdots$

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
 If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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Solution

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Using the first hint, if $f$ has a Taylor series centred at 0, then
$f(x)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}x^{n},$
so the $x^{3}$ term is precisely ${\frac {f^{(3)}(0)}{3!}}$. We can compute this. The first derivative of $f$ can be computed using the product rule:
$f'(x)=e^{x}\cdot {\frac {1}{1x}}+e^{x}\cdot {\frac {1}{(1x)^{2}}}=f(x)+f(x)\cdot {\frac {1}{1x}}=f(x)\cdot \left(1+{\frac {1}{1x}}\right).$
Although not necessary, we notice in the last steps above that the function $f(x)$ shows up again in the derivative. Again, the product rule allows us to compute the second derivative:
${\begin{aligned}f''(x)&=f'(x)\cdot \left(1+{\frac {1}{1x}}\right)+f(x)\cdot \left({\frac {1}{(1x)^{2}}}\right)\\&=f(x)\cdot \left(1+{\frac {1}{1x}}\right)\cdot \left(1+{\frac {1}{1x}}\right)+f(x)\cdot \left({\frac {1}{(1x)^{2}}}\right)\\&=f(x)\cdot \left(\left(1+{\frac {1}{1x}}\right)^{2}+{\frac {1}{(1x)^{2}}}\right)\end{aligned}}$
And the third:
${\begin{aligned}f'''(x)&=f'(x)\cdot \left(\left(1+{\frac {1}{1x}}\right)^{2}+\left({\frac {1}{(1x)^{2}}}\right)\right)+f(x)\cdot \left(2\left(1+{\frac {1}{1x}}\right)\cdot {\frac {1}{(1x)^{2}}}+{\frac {2}{(1x)^{3}}}\right)\\\end{aligned}}$
Finally, using the formulae we found for $f'''$ and $f'$, we can compute the coefficient:
${\begin{aligned}{\frac {f'''(0)}{3!}}&={\frac {1}{6}}\cdot \left(f'(0)\cdot \left(\left(1+{\frac {1}{10}}\right)^{2}+{\frac {1}{10^{2}}}\right)+f(0)\cdot \left(2\left(1+{\frac {1}{10}}\right)\cdot {\frac {1}{(10)^{2}}}+{\frac {2}{(10)^{3}}}\right)\right)\\&={\frac {1}{6}}\cdot \left(f(0)\cdot (1+1)\cdot \left(2^{2}+1\right)+{\frac {e^{0}}{1}}\cdot (2(1+1)\cdot 1+2)\right)\\&={\frac {1}{6}}\cdot (1\cdot 2\cdot 5+1\cdot 6)\\&={\frac {16}{6}}={\frac {8}{3}}\end{aligned}}$


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