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Calculus help, anyone? Please :]

T

TrevorRC

Senior member
Having a bit of trouble with Taylor Polynomials.

I understand the concept, but something isn't clicking.

The general equation for a Taylor Polynomial is

f(a)+f'(a)(x-a)^1+(f''(a)/2!) and etc...

I've got a problem in my book as follows...

SqRt.(1+x)
So, first term is 1... [0th term]; 1/1
Second term should be f'(0) [1]/1!*(x-0)^1 [x].
I know this is flawed.... [see below]
Any reason why?

Now, what I'm not getting...
1) Why does it alternate?
2) Bob (Back of the Book) says 1-x/2+x^2/8-x^3/16... and so on.

Anyone mind showing me how to arrive at this conclusion?

I usually pick these things up quickly, there just aren't any decent examples in my book to model after.

Thanks.
--Trevor
 
Wow. I appreciate the bump, thanks Carbo.

Please don't thread crap. I'm ask a serious question; and I'd like a serious answer.
-T
 
this is normal for homework help threads. you'll get ribbed a little and then the help will come in. go with it.
 
it has something to do with (-1)^n, i really forgot, but you can see only odd exponents give you negative.


Oh and I forogt, beware of LoKe, he will rat you out.
 
The kids are saying if you say "bloody Loki" five times, he'll appear and steal your homework (and rat you out).
😛
 
bump since i barely passed pre-calc 🙂
and jeese, i live in the city of the university (Toledo) of the kid that Loke is trying to get expelled. uhohs lol
 
Actually, what you are calculating is the MacLaurin series, a special form of the Taylor series where you fix a=0.

M(x) = f(0) + [ f'(0) / 1! ] x + [ f''(0) / 2! ] x^2 + ...

The reason the series alternates for the function f(x) = (1+x)^.5 is because the succesive values of the derivates of f(x) at x=0 alternate:

f(x) = (1+x)^(1/2), so f(0) = 1
f'(x) = (1/2) * (1+x)^(.5 - 1) = (1/2) * (1+x)^(-1/2), so f'(0) = 1/2.
f''(x) = (-1/2) * (1/2) * (1+x)^(-3/2), so f''(0) = -1/4
f'''(x) = (-3/2) * (-1/2) * (1/2) * (1+x)^(-5/2), so f'''(0) = 3/8
f''''(x) = (-5/2) * (-3/2) * (-1/2) * (1/2) * (1+x)^(-7/2), so f''''(0) = -15/16

etc, etc...
 
Also, it's a lot easier at first if you start the problem like MathMan has above: make a table of values for f(x), f'(x), etc.
Then, stick them into your general equation.
 
Got it. Went to a study session with some friends for an hour, understand it completely.
--closed.--

Side note, thanks for the help from those who had good intentions. I appreciate it.
 
Originally posted by: TrevorRC
Got it. Went to a study session with some friends for an hour, understand it completely.
--closed.--

Side note, thanks for the help from those who had good intentions. I appreciate it.
Click to expand...

It's too late, Loke probably already sent the email to your professor. Your college days are over.
 
Originally posted by: Hough NutZ
Originally posted by: TrevorRC
Got it. Went to a study session with some friends for an hour, understand it completely.
--closed.--

Side note, thanks for the help from those who had good intentions. I appreciate it.
Click to expand...

It's too late, Loke probably already sent the email to your professor. Your college days are over.
Click to expand...

Mm...reading comprehension, FTW. My complaint was that he was paying for the answers, not for the help.

Anyways, the mods agree that it's over now and there's no sense filling others threads with this gargabe. If you've got a problem with what I've done, PM me or make a new thread.
 
Originally posted by: Hough NutZ
Originally posted by: TrevorRC
Got it. Went to a study session with some friends for an hour, understand it completely.
--closed.--

Side note, thanks for the help from those who had good intentions. I appreciate it.
Click to expand...

It's too late, Loke probably already sent the email to your professor. Your college days are over.
Click to expand...

I'd have given serious consideration to doing what Loke did... except I'd have probably opted not to post about it. What that other person did far exceeded simply asking for help (and that's what was provided in this thread - help) The other person was bragging about how much cheating he did. Some of us in this forum are educators... teachers, professors, etc. It was a slap in the face to us to run around bragging about the extent of cheating. Despite what (apparently) a large number of the 14 - 20 year olds on this forum believe, there are some of us who hold academic honesty in high regard.

The situation is over, but it's pretty ridiculous how some of the people in here have 2 sets of standards. So often, they love seeing someone get what's coming to them for trying to cheat the system (i.e. "I pwned someone who was using my picture on ebay") but when the person getting owned for doing something is doing something that a lot of people here have admitted to doing, then it must be striking a little too close to home for comfort.
 
when you take the derivative of a sqrt function, you will end up alternating positive to negative.

f(x) = sqrt(x)
f'(x) = (1/2)[sqrt(x)^-1/2]
f''(x) = (-1/4)[sqrt(x)^-3/2]
f'''(x) = (3/8)[sqrt(x)^-5/2]

etc.

that is why it will alternate
 
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