I have no idea where to begin or what to do because I have no notes, etc... on this stuff. I really need this as an example to work with. Thanks.

(a) Determine the second Taylor Polynomial of f(x) = x at x = 16.

(b) Use the polynomial found in part (a) to estimate 16.1 to the fifth decimal place.

2. Google. Best place to start if you have no clue about it at all.
Taylor series - Wikipedia, the free encyclopedia

As taylor polynomial is based on set of terms basic on the nth derivative of a function you are trying to find a approximation for in a series.

The general taylor polynomial is as such, a general taylor polynomial starts at the 0th degree and you continue on till the degree which you were given

X = A = 16

$P_n=\frac{f^{n}(a)}{n!}(x-a)^n$

You are taking the n'th derivative of the function you were given to a degree over the nth factorial center at x - a to the nth power.

Here is your example, it helps to build a table of each nth derivative you take
Your problem states to the 2nd degree, will do as such and it is centered at x = 16

Making a table of each nth derivative as such evulated at a which is equal to x which is 16

$f^{0}(a) = (a)^{\frac{1}{2}}$

$f^{1}(a) = \frac{1}{2(a)^\frac{1}{2}}$

$f^{2}(a) = - \frac{1}{4(a)^\frac{3}{2}}$

Now evulate at a which equals 16 and set up your polynomial keeping in mind the nth factiorial and $(x-16)^n$

3. im struggling with these as well.. is there anyone here who can tell me if im anywhere close to the right track

f(x)+f'(x)(x-16)+f''(x)/2! * (x-16)^2

and get: 4+ 1/8 (x-16) -1/128 (x-16)^2

i don't know how to do the 2nd part at all, but is that remotle right?

4. Check out my edited post above

5. is that what mine said, or no?

i think theyre the same...

6. Yes correct, but your last term in incorrect, you divided the denominator on the top by 2, when you should be multiplying the whole fraction by 1/2

$\frac{\frac{1}{256}}{2} \ne \frac{1}{128}$

If you visualize the expression as such

$\frac{\frac{1}{256}}{\frac{2}{1}}$

to divide a fraction by a fraction you simply multiply the top by the reciprocal of the bottom

As for the second part, what you have here is an alternating series, from your polynomial form a general expression of the polynomial as a power series, and using the theorem of how to estimate an a value of the series with in a given error tolerance, in your case .00001

7. does that have anything to do with "little o" notation or is that something completely different? now that i got the basics down , those are next on the chopping block

ps. sorry to the op for highjacking his thread

8. What part are you asking pertains to the "little" notation, your next step is to approx the function to 5 decimals place, aka, your tolerance

The standard of how to approximate an sum of an alternating series with in a given tolerance is as such

$|S-S_n| \le a_{n+1} \le .00001$

Leaving you only to work with

$a_{n+1} \le .00001$

Where you solve for n, to find how many terms you must evaluate to get the approximation within 5 decimal places