Search results

  1. J

    I Convention of order of operations

    Cute. (But several decades after it was introduced on calculators, RPN doesn't seem to have really caught on for some reason.)
  2. J

    I Convention of order of operations

    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...
  3. J

    I Expression of ##sin(A*B)##

    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.
  4. J

    I Expression of ##sin(A*B)##

    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.
  5. J

    I Expression of ##sin(A*B)##

    I can't find my earlier calculation and my algebra skills aren't what they were, but I get 0.59 from your formula: Si(4) ≈ 1.767; Si(1) ≈ 0.953; 2ln(2) = 1.386.
  6. J

    I Expression of ##sin(A*B)##

    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 The...
  7. J

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

    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...
  8. J

    I Expression of ##sin(A*B)##

    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...
  9. J

    I Expression of ##sin(A*B)##

    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...
  10. J

    I Expression of ##sin(A*B)##

    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...
  11. J

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

    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...
  12. J

    B How to solve this specific logarithm problem

    Look at the original equation: x=log5(4x+1) (Sorry for the typo above.) The quantity 4x must be greater than zero; therefore the argument of the log is greater than 1, and the log is positive.
  13. J

    B How to solve this specific logarithm problem

    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...
  14. J

    B How to solve this specific logarithm problem

    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...
Top