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# Help - Taylor watch

1. Hey guys, thanks for clicking on a thread that was asking for help, would really appriciate it if you could help me out. Basically I decided to help my friend out with some of his degree resit work that he couldnt do, but I am also struggling.

The question is
4.

write down the first 4 terms of the Taylor series for f(x) = sin^2(x), expanding about x = pi/4

use your answer to estimate the value of sin^2(pi/3)

Now I used my general formula for Taylor, using my a as a=pi/4, this was my working:

So we need to write down the first 4 terms of the Taylor series for f(x)=sin^2(x)
We know our general formula for taylors series is

Pn(x) = f(a) + (x-a)f'(a) + ((x-a)^2/2!)f''(a) + ..... + ((x-a)^n/n!)f(n)(a)

So here we need
f,f',f'',f''' :
f=sin^2(x)
f'=2sin(x)cos(x)=sin(2x)
f''=2cos(2x)
f'''=-4sin(2x)

f(pi/4) = sin^2(pi/4) = 1/2
(x-a)f'(pi/4) = (x-pi/4)sin(pi/2)=(x-pi/4)*1=x-pi/4
((x-a)^2/2!)f''(a) = ([(x-pi/4)^2]/2).2cos(pi/2)=0
((x-a)^3/3!)f'''(a)= ([(x-pi/4)^3]/6).-4sin(pi/2)=-4((x-pi/4)^3)) / 6

write down the first 4 terms of the Taylor series for f(x) = sin^2(x), expanding about x = pi/4

So we get
P(x) = 1/2 + (x-pi/4) - 4((x-pi/4)^3)) / 6

So I thought that was the expansion, but then the second part asks for

use your answer to estimate the value of sin^2(pi/3)

So obviously I cant just put in pi/3 in place of x as its then still using the stuff centred about pi/4, so have I made an error. Was it wrong to use the a=pi/4, should a have been 0 and then replaced x with pi/4? However that would have given a number and not a series.

So I am very confused, if someone could explain to me how to do these in general, I (and my friend) would be very grateful.

Thanks alot, Joe x
2. There is no issue here... Replace x with pi/3...
3. (Original post by K..W.Studd)
So obviously I cant just put in pi/3 in place of x as its then still using the stuff centred about pi/4, so have I made an error.
Wait a minute. Don't you just x=pi/3.

P.S. Damn you v-zero. Yeah, what v-zero said. You should really correct your logic that you was using to deduce that.
4. It's fine to substitute pi/3 in the expression you've worked out. Whatever number you expand your function about, the sum is always the same. The only difference is how many terms you need to take to make it a good approximation (if you're close to the point you've expanded about, a polynomial expansion, by taking the first few terms of the sum, will be good).
5. Ok sorry guys but slow down a little. So the fact that I expanded around pi/4 makes no difference to finding an estimate for sin^2(x), I could use that expansion I worked out for say pi/9?
All I do is sub in x for the value I want to estimate.
That is so helpful to know, so I made no errors?
6. Also, dont worry about looking at this one now guys as I know its late, but if you get a chance tomorrow or something:
5. evaluate

lim x -> 0 (sinx - tanx)/x^3

Ive completely forgotten how to go about these limits questions, any hints and tips would be really helpful. Thanks!
7. (Original post by K..W.Studd)
Also, dont worry about looking at this one now guys as I know its late, but if you get a chance tomorrow or something:
5. evaluate

lim x -> 0 (sinx - tanx)/x^3

Ive completely forgotten how to go about these limits questions, any hints and tips would be really helpful. Thanks!
http://www.thestudentroom.co.uk/show...l#post19986898 (2nd problem)
8. You could start by expanding sin x and tan x as a taylor series. Probably best to expand about 0. Then see what the RHS looks like.

Or look up L'hopitals rule on wikipedia.
9. Yeah, I was wondering why you can't just use the L'hospital rule three times and then plug in zero.
10. (Original post by Simplicity)
Yeah, I was wondering why you can't just use the L'hospital rule three times and then plug in zero.
You can, but it's easier to use the taylor series for sin and cos (IMHO).
11. I havent learned L'hopitals rule in my first year, and Id prefer to use things Ive done. I have done limits and Taylors but our limits questions so far involved only polynomials. So does that link suggest using the expansion of Tan and Sin, and then factorising and simplifying? How do we know how many terms for sin and tan to use, in general, not just refering to this question.

Also, was I right in what I said regarding my first problem.

Thanks guys, can I do anything to repay all this help?
12. (Original post by K..W.Studd)
I havent learned L'hopitals rule in my first year, and Id prefer to use things Ive done. I have done limits and Taylors but our limits questions so far involved only polynomials. So does that link suggest using the expansion of Tan and Sin, and then factorising and simplifying?
Once you find the Taylor expansion of Tan, the question is trivial. You'll find

sin x = x - x^3/6 + ... and tan x = x - Kx^3 (for a constant K that you need to find).

Then sin x - tan x = (K-1/6)x^3 + ...

and so the limit is (K-1/6).

How do we know how many terms for sin and tan to use, in general, not just refering to this question.
You to need as many terms as it takes. It's largely experience (i.e. in this question you know that if the limit exists and is non-zero, the first "non-cancelling" term in the expansion of the top must be a term in x^3, so you need to use terms up to x^3 for sin and tan).

Also, was I right in what I said regarding my first problem.
The method looks right. I confess that unless you LaTeX it, I'm not going to be bothered to check you've made no algebraic slips.

Incidentally, if you look at the "guide to posting" thread at the top of the forum, there's some advice on how to use LaTeX to typeset mathematics on here.
13. It looks all right, except for one thing: usually, when a question asks for the first 4 terms of an expansion, it means the first 4 non-zero terms. However, I realise that this is an aside to your problem.
14. So the fact that I expanded around pi/4 makes no difference to finding an estimate for sin^2(x), I could use that expansion I worked out for say pi/9, by replacing x by pi/9?
All I do is sub in x for the value I want to estimate, into the taylor expansion centred around pi/4/

So for 5, can I explain what I have understood by what you lot have said, and you correct me if I go wrong.

So we want
lim x -> 0 (sinx - tanx)/x^3

Now youve said use Taylors expansion, is this always the process when we have a limits question involving trig terms, always try to use taylors expansions.
Now weve said to use Taylors centred around 0, so this is Maclaurins right, why do we choose 0, is it just because its easiest and it works with whatever centre value we choose?
So using Taylors we find that
sinx= x - x^3/6
tanx= x + x^3/3

Now DFranklin, Im still confused when you say you take as many terms as it takes, what does this mean, as it takes to what?

So sinx-tanx = [x - x^3/6] - [x + x^3/3] = -x^3 / 2

Now once we put this back into the actual limit question we have

lim x -> 0 (sinx - tanx)/x^3 = lim x -> 0 [ {-x^3 / 2} / x^3 ] = lim x -> 0 ( -1/2 ) = -1/2

So the answer, the limit as x tends to 0 is -1/2
15. Now I think the answer there is right, but Im more concerned as how I got there, in terms of applying it to other questions, was it the right process. My main problem is not understanding how many polynomial terms to take in the Taylor expansions.

Thanks for all this help, I really appriciate it.
16. Well, if you'd only taken one term, your "error" term would be O(x^2), which is bigger than x^3, so the error is too big.
By taking terms up to x^3, your error term is O(x^4) which is negligible compared with x^3, so you're OK.
17. So the correct process then? You think its all along the right lines?
18. Yes.

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