Approximating ln(1.75) with Taylor/Maclaurin Polynomials Using 6 Terms

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SUMMARY

The discussion focuses on approximating ln(1.75) using a Taylor polynomial with 6 terms. Participants clarify that a Maclaurin polynomial cannot be used since ln(x) is not defined at x=0. Instead, they recommend expanding the function around x=1, where it is differentiable. The correct approach involves calculating the derivatives of ln(x) at x=1 and constructing the Taylor series accordingly.

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  • Familiarity with the natural logarithm function ln(x)
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Homework Statement



Find a Taylor or Maclaurin polynomial to apporximate ln(1.75) using 6 terms.

Homework Equations





The Attempt at a Solution



I now that a MacLaurin polynomial is as follows.. c=0

)+f%27(c)x+\frac{f%27%27(c)}{2!}x^{2}+\frac{f%27%27%27(c)}{3!}x^{3}+...+\frac{f^{n}(c)}{n!}x^{n}.gif


and a Talyor polynomial is as follows..
frac{f%27%27(c)}{2!}(x-c)^{2}+\frac{f%27%27%27(c)}{3!}(x-c)^{3}+...+\frac{f^{n}(c)}{n!}(x-c)^{n}.gif



so do I assume I'm working with ln(x)? I'm not really sure where to go with a value.. I have worked with questions such as.. estimate the sin(x) with a 5th degree taylor polynomial.. but this is a little bit different, maybe someone can push me in the right direction!
 
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You can expand around a point where the function is sufficently often differentiable. So, you cannot expand around the point x = 0 (i.e. take c = 0, as Log(x) is undefined there and certainly not differentiable at that point), but you can expand around x = 1.
 

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