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uniformly valid expansion

For (x+epsilon)^1/2, using the binomial expansion we can see that the expansion is valid if x>>epsilon, rescale using (epsilon X) with X=O(1) as epsilon goes to 0 to give an asymptotic expansion that is valid for x< epsilon

What did I do, I substitute x = epsilon X above so ( epsilon X +epsilon) ^1/2, then I rearrange it I applied binomial expansion and the answer is

epsilon ^1/2(( 1 + x/2 - x^2/8 + x^3/16 - (5 x^4)/128 + (7 x^5)/256 + O(x^6))

Is this correct? and to prove that this asymptotic expansion is valid when x<< epsilon.
Reply 1
Avatar for mqb2766
mqb2766
21
Original post by meme12
For (x+epsilon)^1/2, using the binomial expansion we can see that the expansion is valid if x>>epsilon, rescale using (epsilon X) with X=O(1) as epsilon goes to 0 to give an asymptotic expansion that is valid for x< epsilon

What did I do, I substitute x = epsilon X above so ( epsilon X +epsilon) ^1/2, then I rearrange it I applied binomial expansion and the answer is

epsilon ^1/2(( 1 + x/2 - x^2/8 + x^3/16 - (5 x^4)/128 + (7 x^5)/256 + O(x^6))

Is this correct? and to prove that this asymptotic expansion is valid when x<< epsilon.


Are the two bold bits what you intended? Agree with the second.
(edited 4 years ago)
Reply 2
Avatar for Great444
Great444
OP
14
Not sure what do you mean
I want to rescale (x+epsilon)^1/2 by ................ (1) suing x = epsilon X to make expansion (1) valid when x<< epsilon

So I substitute x= Epsilon X and applied the binomial expansion so the answer is
epsilon ^1/2(( 1 + x/2 - x^2/8 + x^3/16 - (5 x^4)/128 + (7 x^5)/256 + O(x^6))

The first question is this valid for x<< epsilon and
Second, how did you know? Do I substitute just general values and compare
Thanks
Reply 3
Avatar for mqb2766
mqb2766
21
Original post by meme12
Not sure what do you mean
I want to rescale (x+epsilon)^1/2 by ................ (1) suing x = epsilon X to make expansion (1) valid when x<< epsilon

So I substitute x= Epsilon X and applied the binomial expansion so the answer is
epsilon ^1/2(( 1 + x/2 - x^2/8 + x^3/16 - (5 x^4)/128 + (7 x^5)/256 + O(x^6))

The first question is this valid for x<< epsilon and
Second, how did you know? Do I substitute just general values and compare
Thanks

When is the standard binomial (1+X)^1/2 = ... valid?
Reply 4
Avatar for Great444
Great444
OP
14
when the second term is less than the first term and third less than second and so on
Reply 5
Avatar for mqb2766
mqb2766
21
Original post by meme12
when the second term is less than the first term and third less than second and so on

But what is the test on X?
Reply 6
Avatar for Great444
Great444
OP
14
test that this expansion (1) is valid when x<<epsilon , means when each term is less than the previous term .
This how I understood it
Reply 7
Avatar for mqb2766
mqb2766
21
Original post by meme12
test that this expansion (1) is valid when x<<epsilon , means when each term is less than the previous term .
This how I understood it

For the standard binomial (1+X)^(1/2), it's when
|X| < 1
It's easy enough to look up. Now sub X = x/epsilon to give the result you want. The << doesn't really have a meaning as a test.
Reply 8
Avatar for Great444
Great444
OP
14
so the first two terms become 1 + x/2e - x^2/8e^2
which is now valid for x< e ( e is epsilon here )
Reply 9
Avatar for mqb2766
mqb2766
21
Original post by meme12
so the first two terms become 1 + x/2e - x^2/8e^2
which is now valid for x< e ( e is epsilon here )

Yes. But remember you've factored an epsilon^(1/2) out as well.
Reply 10
Avatar for Great444
Great444
OP
14
Original post by mqb2766
Yes. But remember you've factored an epsilon^(1/2) out as well.

yes but we did the same for ( x+e)^1/2 ... it becomes x^1/2(1+e/x)^1/2 and we said it's valid only when x>e
(edited 4 years ago)
Original post by meme12
yes but we did the same for ( x+e)^1/2 ... it becomes x^1/2(1+e/x)^1/2 and we said it's valid only when x>e

Not really sure what you're saying here. But as long as you understand it now.
Reply 12
Avatar for Great444
Great444
OP
14
Thank you for the help, I will go through it again to understand it more.

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