Maclaurin expansions

Exercises · Derivatives · with worked solutions · PDF

Exercise 1

Using the Maclaurin expansion of the following functions \(f(x)\), determine the equation of the tangent line to the graph at the point \((0,f(0))\) and the local position of the graph with respect to this tangent

\[ f(x)= 1+\sin(\sqrt{1+x^3}-1) \]

\[ f(x)=2\sqrt{1+\sinh x}-x \]

\[ f(x)=8\sqrt{1+\sin x}+x^2 \]

\[ f(x)=8\sqrt{1+\log(1+ x)}+3x^2 \]

\[ f(x)=8\sqrt{1-2x-x^2}-8+8x \]

Solution

From

\[ \sqrt{1+x^3}-1=\frac{1}{2}x^{3}+o(x^{3}) \]

e

\[ \sin y=y-\frac{1}{6}y^{3}+o(y^{3}) \]

we have

\[ f(x)= 1+\sin(\sqrt{1+x^3}-1)=1+\frac{1}{2}x^{3}+o(x^{3}). \]

The equation of the tangent line is

\[ y=1. \]

The term \(\ds\frac{1}{2}x^{3}+o(x^{3})\) says that there exists a neighborhood \((-\delta,\delta)\), \(\delta>0\), of \(x=0\) where the graph crosses the tangent line from below for \(-\delta<x<0\) to above for \(0<x<\delta\). The point \(x=0\) is an inflection point with horizontal tangent.

Solution

From

\[ \sinh x=x+o(x^{2}) \]

e

\[ 2\sqrt{1+y}=2+y-\frac{1}{4}y^{2}+o(y^{2}) \]

we have

\[ f(x)=2\sqrt{1+\sinh x}-x=2+x-\frac{1}{4}x^{2}+o(x^{2})-x=2-\frac{1}{4}x^{2}+o(x^{2}). \]

The equation of the tangent line is

\[ y=2. \]

The term \(\ds-\frac{1}{4}x^{2}+o(x^{2})\) says that there exists a neighborhood \((-\delta,\delta)\), \(\delta>0\), of \(x=0\) where the graph lies below the tangent line for \(-\delta<x<\delta\), \(x\neq0\). In particular, since the tangent line is horizontal, the point \(x=0\) is a local maximum point.

Solution

From

\[ \sin x=x-\frac{1}{6}x^{3}+o(x^{3}) \]

e

\[ 8\sqrt{1+y}=8+4y-y^{2}+\frac{1}{2}y^{3}+o(y^{3}) \]

we have

\[ \begin{array}{l} \ds f(x)=8\sqrt{1+\sin x}+x^2=\\ \\ \ds8+4x-\frac{2}{3}x^{3}-\left(x-\frac{1}{6}x^{3}\right)^{2} +\frac{1}{2}\left(x-\frac{1}{6}x^{3}\right)^{3}+o(x^{3})+x^{2}=\\ \\ \ds8+4x-\frac{2}{3}x^{3}-x^{2}+\frac{1}{2}x^{3}+x^{2}+o(x^{3})=8+4x-\frac{1}{6}x^{3}+o(x^{3}). \end{array} \]

The equation of the tangent line is

\[ y=8+4x. \]

The term \(\ds-\frac{1}{6}x^{3}+o(x^{3})\) says that there exists a neighborhood \((-\delta,\delta)\), \(\delta>0\), of \(x=0\) where the graph crosses the tangent line from above for \(-\delta<x<0\) to below for \(0<x<\delta\) . The point \(x=0\) is an inflection point.

Solution

From

\[ \log(1+x)=x-\frac{1}{2}x^{2}+\frac{1}{3}x^{3}+o(x^{3}) \]

e

\[ 8\sqrt{1+y}=8+4y-y^{2}+\frac{1}{2}y^{3}+o(y^{3}) \]

we have

\[ \begin{array}{l} \ds f(x)=8\sqrt{1+\log(1+ x)}+3x^2=\\ \\ \ds8+4\left(x-\frac{1}{2}x^{2}+\frac{1}{3}x^{3}\right)-\left(x-\frac{1}{2}x^{2}+ \frac{1}{3}x^{3}\right)^{2}+\\ \\ \ds\frac{1}{2}\left(x-\frac{1}{2}x^{2}+\frac{1}{3}x^{3}\right)^{3}+o(x^{3})+3x^{2}=\\ \\ \ds8+4x-2x^{2}+\frac{4}{3}x^{3}-x^{2}+x^{3}+\frac{1}{2}x^{3}+3x^{2}+o(x^{3})=\\ \\ \ds8+4x+\frac{17}{6}x^{3}+o(x^{3}). \end{array} \]

The equation of the tangent line is

\[ y=8+4x. \]

The term \(\ds\frac{17}{6}x^{3}+o(x^{3})\) says that there exists a neighborhood \((-\delta,\delta)\), \(\delta>0\), of \(x=0\) where the graph crosses the tangent line from below for \(-\delta<x<0\) to above for \(0<x<\delta\). The point \(x=0\) is an inflection point.

Exercise 2

Write the Maclaurin expansion of order 3 of the function

\[ f(x)=e^x-e^{-x^2}-\sin x \]

and determine its order of infinitesimal. Based only on the expansion obtained, say whether the function has at \(x=0\) a relative minimum point, a relative maximum point, an inflection point, or none of these. Justify your answer.

Solution

Since we have to expand \(f(x)\) to the third order, we can stop at the second order in the expansion of \(e^{-x^2}\). That is:

\[ e^x=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+o(x^3) \]

\[ e^{-x^2}=1-x^2+o(x^3) \]

\[ \sin x=x-\frac{1}{6}x^3+o(x^3) \]

We then have

\[\begin{align*} f(x)&=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+o(x^3)-(1-x^2+o(x^3))-\left(x-\frac{1}{6}x^3+o(x^3)\right)\\ &=\frac{3}{2}x^2+\frac{1}{3}x^3+o(x^3) \end{align*}\]

In particular, we have

\[ f(x) \sim \frac{3}{2}x^2 \quad \text{as } x\to 0 \]

from which we can immediately state that the order of infinitesimal of \(f(x)\) is \(\alpha=2\) and the principal part is \(\frac{3}{2}x^2\). The point \(x=0\) is a local minimum point; indeed, the term \(\frac{3}{2}x^2+o(x^{2})\) says that there exists a neighborhood \((-\delta,\delta)\), \(\delta>0\), of \(x=0\) where the graph lies entirely above the horizontal tangent line \(y=0\).

Exercise 3

Determine the Taylor polynomial of order 3, centered at the point \(x_0=\frac{\pi}{3}\), of the function \(f(x)=\cos x\).

Solution

In general, the Taylor polynomial of order \(n\), centered at \(x_0\), of a function \(f(x)\) is

\[ T_{n,f,x_0}(x)=\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k \]

with \(f^{(0)}(x_0)=f(x_0)\). In this case, we need the first, second and third derivatives, evaluated at \(x_0=\frac{\pi}{3}\). We have

\[ f'\left(\frac{\pi}{3}\right)=-\sin\left(\frac{\pi}{3}\right)=-\frac{\sqrt 3}{2} \]

\[ f''\left(\frac{\pi}{3}\right)=-\cos\left(\frac{\pi}{3}\right)=-\frac{1}{2} \]

\[ f'''\left(\frac{\pi}{3}\right)=\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt 3}{2} \]

and the Taylor polynomial of order 3, centered at the point \(x_0=\frac{\pi}{3}\), of the function \(f(x)=\cos x\) is

\[ T_{3,f,\frac{\pi}{3}}(x)=\frac{1}{2}-\frac{\sqrt 3}{2}\left(x-\frac{\pi}{3}\right)-\frac{1}{4}\left(x-\frac{\pi}{3}\right)^2+\frac{\sqrt 3}{12}\left(x-\frac{\pi}{3}\right)^3 \]