Third-degree Taylor polynomial

**iheartmath** · Mar 30th 2010, 08:26 PM

Problem:

My solution: T3(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7

T3(pi/6) = pi/6 - (pi/6)^3/3 + (pi/6)^5/5 - (pi/6)^7/7 (unsimplified)

Given solution:

I just don't understand the methodology. I thought it was a matter of determining the corresponding Taylor series of the function, writing it out to the nth term specified, and then substituting directly. Yet, that doesn't look like what was done in the given solution. Could someone please help me out?

**Drexel28** · Mar 30th 2010, 10:44 PM

Originally Posted by iheartmath

Problem:

My solution: T3(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7

T3(pi/6) = pi/6 - (pi/6)^3/3 + (pi/6)^5/5 - (pi/6)^7/7 (unsimplified)

Given solution:

I just don't understand the methodology. I thought it was a matter of determining the corresponding Taylor series of the function, writing it out to the nth term specified, and then substituting directly. Yet, that doesn't look like what was done in the given solution. Could someone please help me out?

When it asks for $T_n(x;x_0)$ it's asking for the Taylor polynomial about $x_0$ whose degree is $n$ . You went out too far because you misinterpreted the n to be the number of terms.

**iheartmath** · Mar 30th 2010, 10:56 PM

That makes a lot more sense now. Thanks!

**drumist** · Mar 31st 2010, 12:18 AM

As far as the second part of the question, you are asked to approximate the value of $\frac{\pi}{6}$ . You will have to try to recall some trigonometry to remember that $\tan \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{3}$ . Therefore, $\tan^{-1} \left( \frac{\sqrt{3}}{3} \right) = \frac{\pi}{6}$ .

That's why you should use $x=\frac{\sqrt{3}}{3}$ in order for $f(x) \approx \frac{\pi}{6}$ .

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