What does k<<x and y<<k make?

[maryhann2007]

Hello,

ich have some function, where:

f(x,y)=(2xy)/(k^2+2xk+2yk)+(2xy)/(k^2+2xk+2yk+4xy)+ln((k^2+2xk+2yk)/(k^2+2xk+2yk+4xy)), where k is a constant.

If k<<x and k<<y, then

f(x,y)=(2xy)/(2xk+2yk), WHY??????

I understand, that k^2=0, but what about the second and third fraction? Why have they been dissapeared?

Thank you for any clue,

Mary

[VBAhack]

Doesn't make any sense to me.

The 2nd fraction can be rewritten as 2xy/(ε + 4xy). This reduces to 2xy/4xy = 1/2 as ε gets smaller.

The 3rd fraction can be rewritten as ln[ε/(ε + 4xy)] = ln(ε) - ln(ε + 4xy). Seems to me this goes to -infinity as ε gets smaller.

[maryhann2007]

Thank you for the answer. It is not the first time, thay you help me. Thank you again.

But I am sure, that this solution f(x,y)=(2xy)/(2xk+2yk) is correct, because I have seen it in many books. Maybe we can find together an explanation. I have to understand this symbol k<<x, because I need it for other equations.

Every number, which has something in common with k, you took as ε. Hmm, it is not good, that k disapears for ever. Have you got any other clue?

Mary

[VBAhack]

k<<x just means k is not only less than x, but MUCH less than x (the implication is orders of magnitude less).

If you've seen the equation in books, it must be a popular topic. What is the topic/subject? Maybe a web search on the topic can help.

A practical approach is to use Excel to calculate the values of the equations for various values of x, y, and k. You can reduce k progressively and see what the result is.

If you are sure the answer is correct, then I don't see it. Maybe others in this forum can help. Sorry.

[zaza]

Doesn't make any sense to me.

The 2nd fraction can be rewritten as 2xy/(ε + 4xy). This reduces to 2xy/4xy = 1/2 as ε gets smaller.

The 3rd fraction can be rewritten as ln[ε/(ε + 4xy)] = ln(ε) - ln(ε + 4xy). Seems to me this goes to -infinity as ε gets smaller.

You're almost there. The second fraction does indeed reduce to 1/2. Considering the third, you could write it as:


ln((k²+2xk+2yk / (k+2x)(k+2y)) => ln(1 - (4xy / (k+2x)(k+2y)))

=> ln(1 - D) where D is close to but less than 1 for k > 0

You can write out the Taylor series for this if it makes things easier; for ln(1+ε) the Taylor series is:

x - x²/2 + x³/3 -...


Clearly as k -> 0, D -> 1 and hence the ln tends to -inf.

But the question is, how quickly does it do so? Because by looking at the first part of the function, you will find that

(2xy)/(k²+2xk+2yk) -> +inf as k -> 0


So what you need to decide is, which gets there first? There are standard ways to do this and you will find that the ln gets there much less rapidly than the first part. If you want to convince yourself that this is the right course of action, then try filling in x=y=1, k= 1x10^-15 to both bits and see what you get.

So the upshot will be that the first bit by far dominates the equation, and the conclusion follows shortly afterwards.


zaza