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

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1. Feb 20, 2017

### 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.

Last edited: Feb 20, 2017
2. Feb 20, 2017

### 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.

3. Feb 20, 2017

### Kumar8434

How does this work?

4. Feb 22, 2017

### 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.

5. Feb 26, 2017

### 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 $m = \frac{ 2 \int_A^B f(2x) dx}{2B - 2A} = \frac{\int_A^B f(2x) dx}{B-A}$.

6. Feb 26, 2017

### Kumar8434

$A$ 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 $tan^{-1}1.9$.

7. Feb 26, 2017

### Stephen Tashi

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].

8. Feb 26, 2017

### Kumar8434

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

9. Feb 26, 2017

### Stephen Tashi

Yes.

10. Feb 26, 2017

### Kumar8434

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

11. Feb 26, 2017

### Stephen Tashi

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.

Yes, I did.

12. Feb 28, 2017

### 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?

13. Mar 1, 2017

### Kumar8434

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