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Let $f(x)=\ln(3x+1)$. I computed the third order Taylor polynomial of $f$ around $0$:

$$T_3(x)=3x-\frac{9}{2}x^2+9x^3. $$

So, I need to evaluate an approximation of $\ln(1.03)$ and $\ln(0.97)$ using this, and I need to prove that the error in both cases is less that $10^{-6}.$

$\ln(1.03)=f(0.1).$So, the approximation is $T_3(0.1)=0.264$

$\ln(0.97)=f(-0.1)$. So, the approximation is $ T_3(-0.1)=-0.264$

My problem is the proof about the error.

I want to use the Lagrange's formula for the error:

$$R_3(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$ for some $c\in(0,x)$ in the first case and $c\in(x,0)$ in the second case.

In the first case, I tried to use the maximum of $f^{(n+1)}$ in the interval $(0,0.1).$

$$f^{(4)}(x)=-\frac{486}{(3x+1)^4} $$

The function $f^{(4)}$ is increasing on the interval $(0,0.1),$ so an upper bound for it is $$f^{(4)}(0.1)=-170.162109\dots\geq-171.$$

So, if $M=-171$,

$$R_3(x)\leq \frac{M}{4!}(0.1)^4=-0.000709008\dots>10^{-4}>10^{-6}.$$

So, I couldn't prove what I want.

What can I do?

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  • $\begingroup$ The series continues $\ln(1+3\cdot10^{-2})=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{3^k10^{-2k}}{k}$. This series is alternating with decreasing absolute values of the terms. Then the partial sum zigzag around the sum getting closer at each step. So, if you sum the first three terms, the sum is going to be at a distance from the sum of the first three terms that is smaller than the absolute value of the fourth term. The fourth term is $\frac{3^410^{-8}}{5}<10^{-6}$. $\endgroup$
    – user758994
    Commented Mar 14, 2020 at 1:18
  • $\begingroup$ "$f(x) = \ln (3 x + 1)$" and "$\ln (1.03) = f(0.1)$." Huh?? $\endgroup$ Commented Mar 14, 2020 at 1:23

3 Answers 3

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To estimate $\ln(1.03)$ and $\ln(0.97)$, we plug in $x=\pm 0.01$ and not $x=\pm 0.1$, as you've done. That gives us the estimates $T_3(0.01)=0.029559$ and $T_3(-0.01)=-0.030459$. Next, $f^{(4)}$ is increasing on the interval $[-0.01,0.01]$. Hence, $$\left|f^{(4)}(x)\right|\leq \left|f^{(4)}(-0.01)\right|=M\approx 548.971 $$ and $$\left|R_3(x)\right|\leq\frac{M}{4!}x^4\leq\frac{M}{24}10^{-8} $$ since $|x|\leq 10^{-2}$. So, all you need to show is that $M/24<10^2$.

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The main issue is that you have plugged in the wrong values:

$\ln(1.03)=f(0.01)\\\ln(0.97)=f(-0.01)$

Secondly, you need to consider the maximum value of the magnitude of $f^{(n+1)}$ to get a proper upper bound to the error:

$$|f^{(4)}(x)|=\frac{486}{(3x+1)^4}$$

This is decreasing in $x$, so we should use the left bounds, opposed to the right bounds as you had:

$$|R_3(0.01)|\le\frac{|f^{(4)}(0)|}{4!}|0.01|^4=2.025\times10^{-7}$$

$$|R_3(-0.01)|\le\frac{|f^{(4)}(-0.01)|}{4!}|-0.01|^4\simeq1.799\times10^{-7}$$

which are both below $10^{-6}$.

The main use of the form without taking magnitudes is that we can determine the sign of the error, which in this case can be seen to show they are both negative and hence the provided results of the Taylor expansion are overestimates.


As a sidenote, the Maclaurin expansion of $f$ converges to $f$, and when $x>0$, the terms are alternating and decreasing in magnitude, allowing us to use the simpler bound of just taking the next term:

$$f(x)=3x-\frac92x^2+9x^3-\frac{81}4x^4+\dots$$

We thus have

$$|R_3(0.01)|\le\frac{81}4|0.01|^4=2.025\times10^{-7}$$

as was already derived.

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for finding $f(x)$ for 1.03 or .97, expand $f(x)$ in Taylor series about $x=1$ as $$f(x)=\ln(3x+1)=\ln(3x-3+4)=\ln 4+\ln[1+\frac{3}{4}(x-1)]=\ln 4+\ln(1+t)$$ $$\implies =\ln 4+t-t^2/2+t^3/3-...= \ln 4+(3/4)(x-1)-(9/32)(x-1)^2+(9/64)(x-1)^2+...$$

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  • $\begingroup$ I don't see how this is helpful in any way. Perhaps reread the question? $\endgroup$ Commented Mar 14, 2020 at 19:53

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