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 10 months 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
 10 months 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

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SithsAndGiggles
 10 months ago
Best ResponseYou've already chosen the best response.1\(\ln x\) doesn't have a Taylor polynomial centered at 0. The first term of the Taylor series would be undefined.

SithsAndGiggles
 10 months ago
Best ResponseYou've already chosen the best response.1Oh, do you mean at \(x=0.845\)? The directions were a bit confusing.

ngdiana
 10 months ago
Best ResponseYou've already chosen the best response.0I 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?

SithsAndGiggles
 10 months ago
Best ResponseYou've already chosen the best response.1Find the terms of the polynomial. Manually, if you're not familiar with the power series for \(\ln x\). \[\begin{matrix} f(x)=\ln x&&&f(0.845)\approx0.1684\\ f'(x)=\frac{1}{x}&&&f'(0.845)\approx1.1834\\ f''(x)=\frac{1}{x^2}&&&f''(0.845)\approx 1.4005\\ f^{(3)}(x)=\frac{2}{x^3}&&&f^{(3)}(0.845)\approx3.3148\\ f^{(4)}(x)=\frac{6}{x^4}&&&f^{(4)}(0.845)\approx 11.7686 \end{matrix}\] The Taylor polynomial of order 4 will give you an approximation, i.e. \[\ln x\approx \sum_{i=0}^4\frac{f^{(n)}(0.845)}{n!}(x0.845)^n\] 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 \[\ln 0.845\approx 0.1684\]

SithsAndGiggles
 10 months ago
Best ResponseYou've already chosen the best response.1The 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.

ngdiana
 10 months ago
Best ResponseYou've already chosen the best response.0Thank you so much for explaining!! :D
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