Great444
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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.
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mqb2766
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(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.
Last edited by mqb2766; 10 months ago
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Great444
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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
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mqb2766
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(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?
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Great444
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when the second term is less than the first term and third less than second and so on
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mqb2766
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(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?
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Great444
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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
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mqb2766
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(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.
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Great444
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so the first two terms become 1 + x/2e - x^2/8e^2
which is now valid for x< e ( e is epsilon here )
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mqb2766
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(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.
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Great444
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(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
Last edited by Great444; 10 months ago
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mqb2766
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(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.
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Great444
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Thank you for the help, I will go through it again to understand it more.
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