1. 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 advanceIMG.pdf

2. 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 advanceIMG.pdf
The first one is incorrect.

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

If you use the ratio test:

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

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

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

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

$= \frac{2}{3}$

$< 1$.

Therefore the series is CONVERGENT.

3. 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...

4. For the second question, assume that you can write $f(x) = \tan{x}$ as a polynomial centred at $C = \frac{\pi}{4}$ (I assume that's what you meant).

Therefore

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

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

Differentiate both sides

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

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

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

$a_1 = 2$.

Differentiate both sides

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

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

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

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

$2 = 2a_2$

$a_2 = 1$.

Differentiate both sides

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

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

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

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

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

$4 + 12 = 6a_3$

$16 = 6a_3$

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

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

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.

5. 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?

6. 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.