Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence.

Question

As the question states:

"Determine the Taylor series for $f(x) = x^3 \cdot \ln{\sqrt{x}}$ around the point $a = 1$ and determine its radius of convergence."

I have consulted this related question, and understand the steps to be:

1. find the first few terms of the Taylor polynomial.
2. Generalize the terms by making use of an infinite sum to represent the function as the Taylor series.
3. use the infinite sum in the ratio test to find the radius of convergence.

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Progress so far:

1. The first 6 terms (n = 0 to n = 5) of the Taylor polynomial I have calculated to be:

$x^3 \cdot \ln{(\sqrt{x})} + \frac{1}{2}(x-a) + \frac{5}{4}(x-a)^2 + \frac{11}{12}(x-a)^3 + \frac{1}{8}(x-a)^4 - \frac{1}{40}(x-a)^5$

It is at this point however that I fall over. It is not intuitive to me how I can write the c-terms as a function without utilizing some sort of online maths engine for fitting the data to a curve.

Is there some sort of first-year-student-friendly technique for modelling these data points systematically? Alternatively, does someone have an intuition they would be willing to share for solving this problem?

Answer 1

In fact a lot of the work needed is already done. Let's recapitulate. We have $f(x)=x^3\ln\left(\sqrt{ x}\right)$ and the Taylor expansion at $a=1$ is given as \begin{align*} f(x)=\sum_{n=0}^\infty \frac{1}{n!}f^{(n)}(a)(x-a)^n \end{align*}

We obtain \begin{align*} f(x)&=x^3\ln\left(\sqrt{ x}\right)=\frac{1}{2}x^3\ln(x)\\ f^{\prime}(x)&=\frac{3}{2}x^2\ln(x)+\frac{1}{2}x^2\\ f^{\prime\prime}(x)&=3x\ln(x)+\frac{5}{2}x\\ f^{\prime\prime\prime}(x)&=3\ln(x)+\frac{11}{2}\\ f^{(4)}(x)&=\frac{3}{x}\\ f^{(5)}(x)&=-\frac{3}{x^2},\qquad f^{(6)}(x)=3\cdot\frac{2!}{x^3},\qquad f^{(7)}=-3\cdot \frac{3!}{x^4}\\ &\vdots\\ f^{(n)}(x)&=3(-1)^n\frac{(n-4)!}{x^{n-3}}\qquad\qquad n\geq 4\tag{1} \end{align*}

From the fourth derivative $\frac{3}{x}$ we can relatively easy obtain higher derivatives and assume the general formula (1) which can be shown by induction.

Evaluated at $a=1$ we have \begin{align*} f(1)=0, f^{\prime}(1)=\frac{1}{2}, f^{\prime\prime}(1)=\frac{5}{2}, f^{\prime\prime\prime}(1)=\frac{11}{2}, f^{(n)}(1)=3(-1)^n(n-4)!\qquad n\geq 4 \end{align*}

We obtain from the derivatives above the Taylor series \begin{align*} \color{blue}{f(x)}&\color{blue}{=x^3\ln\sqrt{ x}}\\ &\color{blue}{=\frac{1}{2}(x-1)+\frac{5}{4}(x-1)^2+\frac{11}{12}(x-1)^3+3\sum_{n=4}^\infty\frac{(-1)^n}{n(n-1)(n-2)(n-3)}(x-1)^n} \end{align*}

The radius $R$ of convergence is \begin{align*} R&=\lim_{n\to \infty}\left|\frac{a_{n}}{a_{n+1}}\right|=\lim_{n\to \infty}\left|\frac{3(-1)^n(n+1)n(n-1)(n-2)}{3(-1)^{n+1}n(n-1)(n-2)(n-3)}\right|\\ &=\lim_{n\to\infty}\left|\frac{n+1}{n-3}\right|\\ &=1 \end{align*}

Answer 2

$$f(x+1)=(x+1)^3\ln(\sqrt{x+1})=\frac{(x+1)^3}{2}\ln(x+1)$$ Now write $$\ln(x+1) = \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n} $$ And multiply the infinite sum by the polynomial. Then you can switch back to $f(x)$. Your approach is good in some cases, but often it's just guessing.

Easiest way to derive the Taylor series for $\ln(1+x)$ is, i think, as follows. We know that $$\int\limits_0^x\frac{1}{1-t}dt = -\ln(1-x) $$ But we also know the Taylor series for $\frac{1}{1-t}$. $$ \frac{1}{1-t}=\sum\limits_{n=0}^\infty t^n $$ We can integrate the sum, and we can do it term by term, and we get: $$ -\ln(1-x)=\sum\limits_{n=1}^\infty \frac{x^n}{n} $$ So that $$ \ln(1+x)=\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n} $$ You can also try your guessing method to derive the Taylor series for $\ln(x+1)$, that will be more straightforward.

Answer 3

Write $x=1+t$, so you need to find the Taylor series of $$ \frac{1}{2}(1+t)^3\ln(1+t) $$ at $t=0$. Since $$ \ln(1+t)=\sum_{n>0}\frac{(-1)^{n+1}t^n}{n} $$ your Taylor series is $$ \frac{1}{2}\sum_{n>0}\frac{(-1)^{n+1}t^n}{n}+ \frac{3}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+1}}{n}+ \frac{3}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+2}}{n}+ \frac{1}{2}\sum_{n>0}\frac{(-1)^{n+1}t^{n+3}}{n} $$ that can be rewritten as $$ \frac{1}{2}\sum_{n>0}\frac{(-1)^{n+1}t^n}{n}+ \frac{3}{2}\sum_{n>1}\frac{(-1)^{n}t^{n}}{n-1}+ \frac{3}{2}\sum_{n>2}\frac{(-1)^{n-1}t^{n}}{n-2}+ \frac{1}{2}\sum_{n>3}\frac{(-1)^{n-2}t^{n}}{n-3} $$ Now isolate the only term for $n=1$, the two terms for $n=2$ and the three terms for $n=3$: $$ \frac{t}{2}+\frac{1}{2}\left(-\frac{t^2}{2}+3t^2\right)+ \frac{1}{2}\left(\frac{t^3}{3}-\frac{3t^3}{2}+3t^3\right)+\\ \frac{1}{2}\sum_{n>3}\left(\frac{(-1)^{n+1}}{n}+\frac{(-1)^n3}{n-1}+\frac{(-1)^{n-1}3}{n-2}+\frac{(-1)^{n-2}}{n-3}\right)t^n $$ The last summation can be rewritten as $$ \frac{1}{2}\sum_{n>3}\left(-\frac{1}{n}+\frac{3}{n-1}-\frac{3}{n-2}+\frac{1}{n-3}\right)(-1)^nt^n= \sum_{n>3}\frac{3(-1)^nt^n}{n(n-1)(n-2)(n-3)} $$ so the end result is $$ \frac{t}{2}+\frac{5}{4}t^2+\frac{11}{12}t^3+\sum_{n>3}\frac{3(-1)^nt^n}{n(n-1)(n-2)(n-3)} $$ and you can check that the coefficient for $t^4$ is indeed $1/8$; the coefficient for $t^5$ is $-1/40$ as you computed.

The Taylor series for $f(x)=x^3\ln\sqrt{x}$ around $1$ is obtained by substituting $t$ with $x-1$. The radius of convergence is the same as for $\ln(1+t)$, that is, $1$. You can also compute it with the ratio test, as further check.