# third degree taylor polynomials

• May 1st 2010, 04:08 AM
slapmaxwell1
third degree taylor polynomials
ok i did a couple of problems and i was wondering if someone could tell me if i did them right? the first one is on convergence and the second which was a little more involved asked me to find the 3rd degree taylor polynomial...thanks in advanceAttachment 16634
• May 1st 2010, 04:41 AM
Prove It
Quote:

Originally Posted by slapmaxwell1
ok i did a couple of problems and i was wondering if someone could tell me if i did them right? the first one is on convergence and the second which was a little more involved asked me to find the 3rd degree taylor polynomial...thanks in advanceAttachment 16634

The first one is incorrect.

$\displaystyle \sum_{n = 1}^{\infty}\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}{2\cdot 5 \cdot 8 \cdot \dots \cdot (3n - 1)}$ is NOT the same as $\displaystyle \sum_{n = 1}^{\infty}\frac{2n - 1}{3n - 1}$.

If you use the ratio test:

$\displaystyle \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)\cdot (2n + 1)}{2\cdot 5 \cdot 8 \cdot \dots \cdot (3n - 1)\cdot (3n + 2)}}{\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}{2\cdot 5 \cdot 8 \cdot \dots \cdot (3n - 1)}}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{2n + 1}{3n + 2}\right|$

$\displaystyle = \lim_{n \to \infty}\frac{2n + 1}{3n + 2}$

$\displaystyle = \lim_{n \to \infty} \frac{2}{3}$ by L'Hospital's Rule

$\displaystyle = \frac{2}{3}$

$\displaystyle < 1$.

Therefore the series is CONVERGENT.
• May 1st 2010, 05:01 AM
slapmaxwell1
im curious is the ratio test the only test i could have used to prove convergence? i always thought to use the nth term divergence test first if it fails that test then there would be no need to go further...thank you for the correction...
• May 1st 2010, 05:13 AM
Prove It
For the second question, assume that you can write $\displaystyle f(x) = \tan{x}$ as a polynomial centred at $\displaystyle C = \frac{\pi}{4}$ (I assume that's what you meant).

Therefore

$\displaystyle \tan{x} = a_0 + a_1\left(x - \frac{\pi}{4}\right) + a_2\left(x - \frac{\pi}{4}\right)^2 + a_3\left(x - \frac{\pi}{4}\right)^3 + \dots = \sum_{n = 0}^{\infty}a_n\left(x - \frac{\pi}{4}\right)^n$.

At $\displaystyle x = \frac{\pi}{4}$, you find $\displaystyle a_0 = 1$.

Differentiate both sides

$\displaystyle \cos^{-2}{x} = a_1 + 2a_2 \left(x - \frac{\pi}{4}\right) + 3a_3\left(x - \frac{\pi}{4}\right)^2 + 4a_4\left(x - \frac{\pi}{4}\right)^3 + \dots$.

At $\displaystyle x = \frac{\pi}{4}$, you find $\displaystyle \left(\cos{\frac{\pi}{4}}\right)^{-2} = a_1$

$\displaystyle \left(\frac{1}{\sqrt{2}}\right)^{-2} = a_1$

$\displaystyle a_1 = 2$.

Differentiate both sides

$\displaystyle 2\sin{x}\cos^{-3}{x} = 2a_2 + 3\cdot 2a_3\left(x - \frac{\pi}{4}\right) + 4\cdot 3a_4\left(x - \frac{\pi}{4}\right)^2 + 5\cdot 4a_5\left(x - \frac{\pi}{4}\right)^3 + \dots$.

At $\displaystyle x = \frac{\pi}{4}$ you find $\displaystyle \sin{\frac{\pi}{4}}\left(\cos{\frac{\pi}{4}}\right )^{-3} = 2a_2$

$\displaystyle \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\right)^ {-3} = 2a_2$

$\displaystyle \frac{1}{\sqrt{2}}(2\sqrt{2}) = 2a_2$

$\displaystyle 2 = 2a_2$

$\displaystyle a_2 = 1$.

Differentiate both sides

$\displaystyle 2\cos^{-2}{x} + 6\sin^2{x}\cos^{-4}{x} = 3\cdot 2a_3 + 4\cdot 3 \cdot 2a_4\left(x - \frac{\pi}{4}\right) + 5\cdot 4 \cdot 3a_5\left(x - \frac{\pi}{4}\right)^2 + \dots$.

At $\displaystyle x = \frac{\pi}{4}$ you have

$\displaystyle 2\left(\cos{\frac{\pi}{4}}\right)^{-2} + 6\left(\sin{\frac{\pi}{4}}\right)^2\left(\cos{\fra c{\pi}{4}}\right)^{-4} = 3\cdot 2a_3$

$\displaystyle 2\left(\frac{1}{\sqrt{2}}\right)^{-2} + 6\left(\frac{1}{\sqrt{2}}\right)^2 \left(\frac{1}{\sqrt{2}}\right)^{-4} = 6a_3$

$\displaystyle 2\cdot 2 + 6\cdot \frac{1}{2} \cdot 4 = 6a_3$

$\displaystyle 4 + 12 = 6a_3$

$\displaystyle 16 = 6a_3$

$\displaystyle a_3 = \frac{8}{3}$.

Can you finish the last step to find $\displaystyle a_4$?

Also, to answer your question, no the ratio test is not the only way to prove convergence. It IS however an EASY way to prove convergence.

The reason you got the wrong answer is because you made the mistake I pointed out at the beginning of the post.
• May 1st 2010, 05:17 AM
slapmaxwell1
with the tan x problem, c = -pi/4 im gonna have to go over the derivatives. for sec squared x i used the chain rule to find the derivative. so 2 times sec x times the derivative of sec x. is this incorrect?
• May 1st 2010, 05:19 AM
slapmaxwell1
wait wait, i think i see it, cos raised to the -2 is sec squared x. ok i still need to check my derivatives though and thank you, i know now i have alot more work to do before im ready for the final.