You'd have to write things differently; e.g. 1 + (2*3) = 1 + 6 = 7. I tend to think the present convention is convenient but I could well be biased. I can't see any fundamental changes happening if we switched to a different convention: important changes would only occur if some expressions...
I make a lot of errors myself these days. To get anywhere is a matter of gathering up time and patience and being prepared to check everything a couple of times.
I would stress, though, that when you develop a power series you should investigate when and how fast it will converge.
One more thought. You certainly can't assume the series is converging after two terms unless (x - a)2 is small--meaning much less than unity. It would be instructive to work out a couple of terms in the expansion and consider how they behave when a, x, and (x - a)2 are all greater than unity.
Let's look at the case where x and a are much greater than unity.
Si(∞) is finite, in fact equal to π/2
So Si(a) = π/2 -∫a∞( sinx/ x) dx
Integrating by parts twice gives
Si(x) = π/2 + cosx/x]a∞ + ∫a∞cosx / x2 dx
=π/2 - cosa/a + ∫a∞cosx / x2 dx
=π/2 - cosa/a - sina / a2 + 2∫a∞sinx / x3 dx
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 think I made a sign error, meaning the estimate should be -5.2x10-4..
But it seems clear that there is problem with your formula for this range of values. If A and B are of order 100, Si(A2) and Si(B2) must be of order 10-3 or 10-4 unless A and B are so close that the logarithm in the...
There may be ways of calculating the integrals without doing hundreds of terms or using a computer. But let's try a fairly simple sanity check first.
The denominator in the integral will vary between 10,000 and 12,100, roughly a 20% change. Since (1/x) is a smooth function, we can get a...
If you know how to expand sin(x) in a Taylor series, you can evaluate Si(x) by dividing each term in the series by x and then integrating term by term. After a quick look, my impression is that your result for sin(A.B) works only if A and B are close and not too large.
Try putting B = A + x or...
For the log case, I 'd write
b = a + x
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...
I did make the point about the slopes; probably I should have repunctuated to emphasise it.
But perhaps the main point is that in this case, as for many real-world problems, there is no clean "algebraic" way to get to the solution. You either happen to see a solution (as here) and convince...
Don't know if this will help, but . . .
You have x = log5( 4x+ 1) . This implies x > 0.
In this range, the slope of 5^x is always greater than that of 4^x. (This can be proved by calculus, but it's probably obvious.)
Both are monotonically increasing functions. So, geometrically, it follows...