Would a situation where a=6300 pi and x=pi meet these constraints? This satisfies a >> x, but in this case, cos( 6300pi + pi ) is nowhere near cos( 6300 pi )...
Or did you mean to say pi >> x?
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Just use the standard method to work out the Taylor expansion about a. Yes, yes, this means taking derivatives, but this is sci.math not sci.sums.
Phil
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Since cosine is 2pi periodic, you need to explain what you mean by a>>x here. For example, if a is any multiple of 2pi, then cos (a+x) = cos (x), so by choosing an appropriate multiple of 2pi, I can always have a >> x, and just look at cos (x) instead.
for that matter, I can even have a dependant on x: a(x)=2000pi *(1+ | [[x]] | ), where [[x]] is the greatest-integer function. If you need a to be even larger, increase the coefficient of pi until it is big enough for you.
That is the direction I was looking at. For very small (x), cos (x) is pretty much unity and sin(x) in radians is pretty much x.
Which degenerates to the two term Taylor expansion cos(a+x) = cos(a) - x*sin(a)
I'm trying to convert a rapidly converging Magic Sinewave solution into an instant one. More at
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cos(a) - x*sin(a) won't quite hack it.
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I'm not clear why you think the suggestion an "awful idea".
We get the same power series expansion your way, do we not, as by Narasimham's suggestion? The advantage of the latter being that the MacLaurin series for sin,cos are probably well known enough that a calculus student can do it by inspection:
Everything happens in the first quadrant. Sorry I was not more specific on this.
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Many thanks,
Don Lancaster voice phone: (928)428-4073
Synergetics 3860 West First Street Box 809 Thatcher, AZ 85552
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... I think you degenerated a term or two too far. cos(a+x) = cos(a)cos(x) - sin(a)sin(x) sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ... cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ... so cos(a+x) ~ cos(a)(1 - x^2/2 + x^4/24) - sin(a)(x - x^3/6). Whether you can drop the x^3/6 and higher terms depends on a and x. Substitute typical values of a and an extreme value of x and see if your accuracy criterion is met.
and, of course, the knowledge of cos(a+x)=cos(a)*cos(x)-sin(a)*sin (x), giving the first terms as cos(a)-x*sin(a)-(x^2/2)*cos(a)+(x^3/6)*sin(a) could help too..0
As I misinterpreted the freedom of the variable in "expand sin(x) and cos(x)", infering "expand sin() and cos()". Thus cos(a) and sin(a), being cos(x) and sin(x) at x=a, would also be expanded, yielding noise.
Phil
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"Home taping is killing big business profits. We left this side blank
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