Original post by Glavien
How would I do this question? I have tried most things including Taylor expansion and can’t seem to get the answer of 0.5. Any help is appreciated.
Edit: See below
Did you extract an x^2, x^3 from under the roots prior to the expansion?
If so, post your working, if not, try that.
(edited 6 years ago)
Original post by ghostwalker
Taylor expanions would do it.
Did you extract an x^2, x^3 from under the roots prior to the expansion?
If so, post your working, if not, try that.
Did you extract an x^2, x^3 from under the roots prior to the expansion?
If so, post your working, if not, try that.
So shall I do this?
Original post by Glavien
So shall I do this?
Yep.
When you use an expansion you want your terms to reflect the original function, with the remaining unspecified terms, the error, to be as small as possible.
Since x is going off to infinity, any terms in x^n for n positive will go off to infinity - not nice. So, we extract the x^2, x^3, and now our terms are in x^-n for n positive and their value gets smaller as n increases.
Edit: It's actually a binomial expansion.
(edited 6 years ago)
Original post by ghostwalker
Yep.
When you use an expansion you want your terms to reflect the original function, with the remaining unspecified terms, the error, to be as small as possible.
Since x is going off to infinity, any terms in x^n for n positive will go off to infinity - not nice. So, we extract the x^2, x^3, and now our terms are in x^-n for n positive and their value gets smaller as n increases.
Edit: It's actually a binomial expansion.
When you use an expansion you want your terms to reflect the original function, with the remaining unspecified terms, the error, to be as small as possible.
Since x is going off to infinity, any terms in x^n for n positive will go off to infinity - not nice. So, we extract the x^2, x^3, and now our terms are in x^-n for n positive and their value gets smaller as n increases.
Edit: It's actually a binomial expansion.
When I try do the Taylor expansion I get undefined values.
Original post by Glavien
...
Sorry, should have said standard binomial expansion is what you want.
Original post by ghostwalker
Sorry, should have said standard binomial expansion is what you want.
Thanks makes sense now.
Also, will the following method work?
Original post by Glavien
Thanks makes sense now.
Also, will the following method work?
Also, will the following method work?
Can't see anything wrong in principle with doing that for one expansion.
However, you may encounter a problem with your first expansion having , and your second expansion needing .
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