Expressions of ##log(a+b), tan^{-1}(a+b),sin^{-1}(a+b)##,etc

[Kumar8434]

Hi, I got these:
$$log(a+b)\approx \frac{b*logb-a*loga}{b-a} + log2 -1$$
$$tan^{-1}(a+b)\approx \frac{b*tan^{-1}2b-a*tan^{-1}2a+\frac{1}{4}*ln\frac{1+4a^2}{1+4b^2}}{b-a}$$
$$sin^{-1}(a+b)\approx \frac{b*sin^{-1}2b-a*sin^{-1}2a+\frac{1}{2}*(\sqrt{1-4b^2}-\sqrt{1-4a^2}}{b-a}$$

And, similarly for ##sec^{-1}(a+b)##, ##cosec^{-1}(a+b)##, ##cot^{-1}(a+b)##, etc.
So, you see that the RHS in each of these expressions is the average value of ##f(2x)## between x=a and x=b, i.e. $$\frac{\int_a^bf(2x)dx}{b-a}$$
So, for what values of ##a## and ##b## do these approximations hold good? I checked that these had great accuracy for some pairs I put in my calculator but not so good for others.

 

[Delta2]

I think these formulas are good approximations if b and a are nearby in the real line. Cant exactly define how much nearby they should be.

 

[Kumar8434]

I think these formulas are good approximations if b and a are nearby in the real line. Cant exactly define how much nearby they should be.

How does this work?

 

[John Park]

For the log case, I 'd write
b = a + x
so
b.logb = (a + x).log(a + x) =
(a + x).log( a.(1 + x/a) )
(a + x).( loga + log(1 + x/a) )
≈(a + x).(loga + x/a + . . . )
where I've started an expansion of the second log assuming -1 <x/a <1 .
In this way express your complete formula as a power series in x.

Then do the same for log (a + b) = log(2a + x)
= log( 2a.(1+x/2a) )
≅log(2a) + x/2a + . . .

Compare the results, to see in which power of x = b - a the two expressions differ. A similar approach should work for your other formulae.

 

[Stephen Tashi]

Here's an intuitive way to look at the situation. If ##A## and ##B## are nearly equal then ##A+B \approx 2A \approx 2B##. So a crude approximation is ##f(A+B) \approx f(2A) \approx f(2B)## Instead of using one of those crude approximations, we could use the "average" value of ##f## over the interval ##[2A,2B]##.

That would be ## f(A+B) \approx m = \frac{ \int_{2A}^{2B} f(u) du} {2B- 2A} ##.

Making the change of variable ##2x = u## the integration becomes ##m = \frac{ 2 \int_A^B f(2x) dx}{2B - 2A} = \frac{\int_A^B f(2x) dx}{B-A}##.

 

[Kumar8434]

Here's an intuitive way to look at the situation. If ##A## and ##B## are nearly equal then ##A+B \approx 2A \approx 2B##. So a crude approximation is ##f(A+B) \approx f(2A) \approx f(2B)## Instead of using one of those crude approximations, we could use the "average" value of ##f## over the interval ##[2A,2B]##.

That would be ## f(A+B) \approx m = \frac{ \int_{2A}^{2B} f(u) du} {2B- 2A} ##.

Making the change of variable ##2x = u## the integration becomes ##m = \frac{ 2 \int_A^B f(2x) dx}{2B - 2A} = \frac{\int_A^B f(2x) dx}{B-A}##.

##A## and ##B## need not be nearly equal. For example, when ##A=0.4## and ##B=1.5##. Then ##A+B=1.9##, ##2A=0.8## and ##2B=3##. So, ##A+B## is neither approximately equal to ##2A## nor to ##2B##. The formula still gives a very accurate value of ##tan^{-1}1.9##.

 

[Stephen Tashi]

So, ##A+B## is neither approximately equal to ##2A## nor to ##2B##.

However, A+B is still the average of 2A and 2B which leads to the justification for making an estimate by taking the average value of the function over the interval [2A,2B].

 

[Kumar8434]

However, A+B is still the average of 2A and 2B which leads to the justification for making an estimate by taking the average value of the function over the interval [2A,2B].

2A and 2B will always have A+B as an average no matter what are A and B.

 

[Stephen Tashi]

2A and 2B will always have A+B as an average no matter what are A and B.

Yes.

 

[Kumar8434]

Yes.

Well, that'd mean the formula would always work. You first said that it's supposed to work when ##2A\approx A+B\approx 2B##

 

[Stephen Tashi]

Well, that'd mean the formula would always work.

No, it would suggest that the formula is worth trying. Whether the formula works or not depends on how the function we are approximating behaves.

You first said that it's supposed to work when ##2A\approx A+B\approx 2B##

Yes, I did.

 

[John Park]

I may have more to say later, but here are a few quick thoughts.

I haven't got my notes in front of me but I believe that for large values of a and b (a, b > 1) your arctangent formula is good up to at least terms of order (a + b)^-3. I think terms in (b -a) appear in higher orders.

I expanded your formula about (a + b) = 1; i.e. arctan(a + b ) = π/4. The deviations seemed to be (b - a)²/3 (to be checked).

The limiting case where a = 0 and b is small ( 4b² < 1) is easy to analyse and seems instructive. I believe here your formula first deviates from the true formula in the cubic term.

I do understand your interest in developing these formulae, but where do you think they would be useful?

 

[Kumar8434]

I may have more to say later, but here are a few quick thoughts.

I haven't got my notes in front of me but I believe that for large values of a and b (a, b > 1) your arctangent formula is good up to at least terms of order (a + b)^-3. I think terms in (b -a) appear in higher orders.

I expanded your formula about (a + b) = 1; i.e. arctan(a + b ) = π/4. The deviations seemed to be (b - a)²/3 (to be checked).

The limiting case where a = 0 and b is small ( 4b² < 1) is easy to analyse and seems instructive. I believe here your formula first deviates from the true formula in the cubic term.

I do understand your interest in developing these formulae, but where do you think they would be useful?

It's not about uses, I've just suddenly become interested in thinking about these things.