Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 30

MATH100 A December 2023

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Question 30
Find the 2023-rd Maclaurin polynomial for $$f(x)=\left\lbrace {\begin{array}{ll}e^{-{\frac {1}{x^{2}}}}&{\text{if }}x\neq 0\\0&{\text{if }}x=0\end{array}}\right.$$ .

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Maclaurin polynomials use the derivatives of a function evaluated at ${\textstyle 0}$. Is ${\textstyle f}$ differentiable at ${\textstyle 0}$? Can you be sure ${\textstyle f}$ is 2023 times differentiable at ${\textstyle 0}$?

Hint 2
Start by taking a few derivatives of the ${\textstyle x\neq 0}$ part, do you notice a pattern? Are these derivatives continuous at ${\textstyle 0}$?

Hint 3

You want to show that all 2023 derivatives of ${\textstyle f}$ are continuous at ${\textstyle 0}$. You know all 2023 derivatives for the ${\textstyle x=0}$ part, they are all ${\textstyle 0}$. So, you need to show that the derivatives of the ${\textstyle x\neq 0}$ part converge to ${\textstyle 0}$ when you take the limit ${\textstyle x\rightarrow 0}$. Use a pattern in the derivatives of the ${\textstyle x\neq 0}$ part, and that exponential functions change much faster than polynomial functions ever could.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that the ${\textstyle n}$th Maclaurin polynomial of a function ${\textstyle f(x)}$ is $$R_{n}(x)=f(0)+f'(0)x+{\frac {1}{2!}}f''(0)x^{2}+...+{\frac {1}{(n-1)!}}f^{(n-1)}(0)x^{n-1}+{\frac {1}{n!}}f^{(n)}(0)x^{n}.$$ Looking at our function, it is piecewise, with the value at ${\textstyle x=0}$ defined separately. The Maclaurin polynomial requires taking derivatives evaluated at ${\textstyle x=0}$, so that means we would need to determine that this function is indeed differentiable at ${\textstyle 0}$, 2023 times. Showing that ${\textstyle f}$ is differentiable 2023 times is a lot of times, but we have to trust that a pattern will emerge. Let's start with the first derivative. All we have is the definition: $$f'(0)=\lim _{h\rightarrow 0}{\frac {f(0+h)-f(0)}{h}}=\lim _{h\rightarrow 0}{\frac {e^{-h^{-2}}-0}{h}}=\lim _{h\rightarrow 0}{\frac {1/h}{e^{h^{-2}}}}.$$ To evaluate the limit above, recall that the exponential function grows more rapidly than any polynomial. One way to make this precise is to substitute $t={\frac {1}{h^{2}}}$, so the limit becomes $$\lim _{h\rightarrow 0}{\frac {1/h}{e^{h^{-2}}}}=\lim _{t\rightarrow \infty }{\frac {t^{1/2}}{e^{t}}}=0.$$ Computing $f'(0)$ is the technical computation. Away from $x=0$, we can compute the derivatives of $f$ with the chain and product rules: $${\begin{aligned}&f'(x)=2x^{-3}e^{-x^{-2}}=e^{-x^{-2}}\cdot {\frac {2}{x^{3}}},&{\text{ if }}x\neq 0\\&f''(x)=-6x^{-4}e^{-x^{-2}}+2x^{-3}(2x^{-3}e^{-x^{-2}})=e^{-x^{-2}}(-6x^{-4}+4x^{-6})=e^{-x^{-2}}\cdot {\frac {-6x^{2}+4}{x^{6}}},&{\text{ if }}x\neq 0\end{aligned}}$$ Here a pattern is already starting to emerge: the ${\textstyle e^{-x^{-2}}}$ term appears in both derivatives, and it is going to appear in all the derivatives; in fact, it appears as a factor in each summand of every derivative. Once we factor $e^{-x^{-2}}$ out, the remainder is a quotient of polynomials. In other words, the general form of the ${\textstyle n^{\mathrm {th} }}$ derivative of ${\textstyle f}$ for ${\textstyle x\neq 0}$ is: $$f^{(n)}=e^{-x^{-2}}{\frac {P_{n}(x)}{Q_{n}(x)}},$$ where ${\textstyle P_{n}}$ and ${\textstyle Q_{n}}$ are polynomial functions. Now we can compute $f^{(n)}(0)$, again using the limit definition of derivative, (but recalling and hoping to reapply the work we already did). Let's start with the second derivative, to warm up. $$f''(0)=\lim _{h\rightarrow 0}{\frac {f'(h)-f'(0)}{h}}=\lim _{h\rightarrow 0}{\frac {e^{-h^{-2}}{\frac {-2}{h^{3}}}-0}{h}}=\lim _{h\rightarrow 0}{\frac {\frac {P_{2}(h)}{hQ_{2}(h)}}{e^{h^{-2}}}}=0,$$ where $P_{2}(h)=-2$ and $Q_{2}(h)=h^{3}$, which we found when we computed the first derivative above. The last step above once again follows from the fact that the exponential function grows faster than any polynomial. One can make the same substitution $t=1/h^{2}$ that we made to compute $f'(0)$ to rewrite the limit as a limit at infinity that is more familiar: $$\lim _{h\rightarrow 0}{\frac {\frac {P_{2}(h)}{hQ_{2}(h)}}{e^{h^{-2}}}}=\lim _{t\rightarrow \infty }{\frac {\frac {{\sqrt {t}}\cdot P_{2}(1/{\sqrt {t}})}{Q_{2}(1/{\sqrt {t}})}}{e^{t}}}=\lim _{t\rightarrow \infty }{\frac {{\sqrt {t}}\cdot P_{2}(1/{\sqrt {t}})}{Q_{2}(1/{\sqrt {t}})e^{t}}}=0$$ In general, the scheme of the computation is as follows: $$f^{(n+1)}(0)=\lim _{h\rightarrow 0}{\frac {f^{(n)}(h)-f^{(n)}(0)}{h}}=\lim _{h\rightarrow 0}{\frac {e^{-h^{-2}}{\frac {P_{1}(h)}{Q_{n}(h)}}-0}{h}}=\lim _{h\rightarrow 0}{\frac {\frac {P_{n}(h)}{hQ_{n}(h)}}{e^{h^{-2}}}}=0.$$ Thus we have that every derivative of ${\textstyle f}$ evaluated at 0 vanishes, so the 2023rd Maclaurin polynomial is 0.