I Problem with function approximation

[boniphacy]

We have a function:

$f(x,y)=\sqrt{\frac{1−2x}{1−y^2}} = \frac{\sqrt{1−2x}}{\sqrt{1−y^2}}$

for small x and y, we can use standard approximations:
$1/\sqrt{1−x}=1+x/2+...$
and
$\sqrt{1−x}=1−x/2−...$

Ok. Now we can approximate the whole function f(x,y)

First method:
$\sqrt{\frac{1−2x}{1−y^2}}≈\sqrt{(1−2x)(1+y^2)}=1−x+y^2/2−xy^2$

Second method:
$\frac{\sqrt{1−2x}}{\sqrt{1−y^2}}≈(1−x)(1+y^2/2)=1−x+y^2/2−xy^2/2$

the both approx. are not equal, difference is: $xy^2/2$!

So, which approximation is the correct one?

 

[PeroK]

We have a function:

$f(x,y)=\sqrt{\frac{1−2x}{1−y^2}} = \frac{\sqrt{1−2x}}{\sqrt{1−y^2}}$

for small x and y, we can use standard approximations:
$1/\sqrt{1−x}=1+x/2+...$
and
$\sqrt{1−x}=1−x/2−...$

Ok. Now we can approximate the whole function f(x,y)

First method:
$\sqrt{\frac{1−2x}{1−y^2}}≈\sqrt{(1−2x)(1+y^2)}=1−x+y^2/2−xy^2$

Second method:
$\frac{\sqrt{1−2x}}{\sqrt{1−y^2}}≈(1−x)(1+y^2/2)=1−x+y^2/2−xy^2/2$

the both approx. are not equal, difference is: $xy^2/2$!

So, which approximation is the correct one?

What about putting some small values for $x$ and $y$ into a calculator?

 

[Office_Shredder]

Even worse, if x and y are equally small, you've thrown out at least a $x^2$ term that is more important than the $xy^2$ term.

You took two first order approximations, and then you tried to assume you had a good second order approximation at the end. As you learned, that's not guaranteed to work.

 

[boniphacy]

OK. We assume x ~ y^2, then: what is correct now?

 

[PeroK]

OK. We assume x ~ y^2, then: what is correct now?

I'd say they are both wrong, as you have a cross term in $xy^2$, but no terms in $x^2$ or $y^4$, which should be of the same order.

PS Perhaps the first one is more wrong than the second one!

 

[boniphacy]

for x = y it is:
$\sqrt{\frac{1-2x}{1-x^2}} \approx 1 - x^2 - x^3 - x^4/2 + ...$

so, the first looks is correct.

now try: y^2 = x
$\sqrt{\frac{1-2x}{1-x}} \approx 1 - x/2 - 5/8 x^2 + ...$

?
fantastic.

 

[Office_Shredder]

I think the point we are really trying to get across is, if you want to use the $xy^2$ term then you need to include more terms in each of your original approximations.

Like $\sqrt{1-x}\approx 1-x/2-x^2/8$ instead of just taking the first term.

 

[boniphacy]

This is forbidden. We can use the first terms only.

This function is used in the GR fantastic theory, exactly:
$E = mc^2 \frac{\sqrt{1-2GM/c^2r}}{\sqrt{1 - (r'^2 + h^2/r^2)/c^2}}$

 

[Mark44]

for x = y it is:
$\sqrt{\frac{1-2x}{1-x^2}} \approx 1 - x^2 - x^3 - x^4/2 + ...$

so, the first looks is correct.

now try: y^2 = x
$\sqrt{\frac{1-2x}{1-x}} \approx 1 - x/2 - 5/8 x^2 + ...$

This makes no sense. The original approximation doesn't have any terms involving y, so how can you replace x with $y^2$ or $y^2$ with x, or whatever it is that you're doing?

 

[Office_Shredder]

This makes no sense. The original approximation doesn't have any terms involving y, so how can you replace x with $y^2$ or $y^2$ with x, or whatever it is that you're doing?

The original question had a y in the denominator...

 

[mfb]

A first order approximation cannot give a good second order term. It might produce something that looks like a second order term but it's just meaningless. Use second order approximations if you want a useful second order term.

 

[boniphacy]

$E = mc^2 \frac{\sqrt{1-2GM/c^2r}}{\sqrt{1 - (r'^2 + h^2/r^2)/c^2}}$

OK.
what is correct approximation of this equation, up to second order?

$E = mc^2 + mr'^2/2 + mh^2/2r^2 - mGM/r + the second order term$

we can see 'the first order part' is perfectly Newtonian version.

 

[mfb]

You can calculate the second order from derivatives with the usual formulas. Nothing special about GR here.