## ngdiana one year ago Multiple Choice: Find the Taylor polynomial of the indicated order for the function at x=0, and use it to approximate the value of the function at the given value of x. Round to 7 decimal places. Please help, I dont understand. thnx!! 1)lnx, order 4 at x=0.845 a)-0.1676572 b)-0.1440845 c)0.8563919 d)-0.1683981

1. SithsAndGiggles

$$\ln x$$ doesn't have a Taylor polynomial centered at 0. The first term of the Taylor series would be undefined.

2. SithsAndGiggles

Oh, do you mean at $$x=0.845$$? The directions were a bit confusing.

3. ngdiana

I just copied the exact words to the problem and I think my teacher made a mistake. I got confused, because like you said, there is no taylor polynomial at x=0. I guess well go with x=0.845. Where do I go from there?

4. SithsAndGiggles

Find the terms of the polynomial. Manually, if you're not familiar with the power series for $$\ln x$$. $\begin{matrix} f(x)=\ln x&&&f(0.845)\approx-0.1684\\ f'(x)=\frac{1}{x}&&&f'(0.845)\approx1.1834\\ f''(x)=-\frac{1}{x^2}&&&f''(0.845)\approx -1.4005\\ f^{(3)}(x)=\frac{2}{x^3}&&&f^{(3)}(0.845)\approx3.3148\\ f^{(4)}(x)=-\frac{6}{x^4}&&&f^{(4)}(0.845)\approx -11.7686 \end{matrix}$ The Taylor polynomial of order 4 will give you an approximation, i.e. $\ln x\approx \sum_{i=0}^4\frac{f^{(n)}(0.845)}{n!}(x-0.845)^n$ Here's where I start to kick myself for doing more work than necessary. Notice that the $$i=1,2,3,4$$ terms disappear because when you plug in $$x=0.845$$, the terms disappear, leaving you with the $$i=0$$ term. So you have $\ln 0.845\approx -0.1684$

5. SithsAndGiggles

The thing is, we're trying to find an approximation of the number, yet we end up using a calculator to just give us the number. I suspect you're expected to do more work than just plugging into a calculator.

6. ngdiana

Thank you so much for explaining!! :D