Re: Mathematical models for loop calculations

In article <f72dnU8Dh4Rt5mfbnZ2dnUVZ8sCsnZ2d@xxxxxx>, rjh@xxxxxxxxxxxxxxx says...
Tim Frink said:

On Wed, 26 Sep 2007 13:46:20 +0000, Richard Heathfield wrote:

<snip>

Overall:

a += I;
b = pow(2, I);
c = pow(2, pow(2, I));
d = I * (I + 1) /2 + 1;
e += I;

I just see that these cases work since most of the variables
were assigned the value '2'. If you change them to '3', some
formulas like the one for 'b' will not work any more.

Correct.

Are there also any general formulas that depend on the current
value of the identifiers before entering the loop?

Sometimes, yes. For example, b = b * pow(2, I - 1); (and although I haven't
checked it out, a similar technique probably exists for c). a and e are
the same. d can be done, too, with thought and care. Hint: x + x+1 + x+2 +
x+3 + x+4 + ... + x+n = n * (n + 1) / 2 - x

Shouldn't that be: x + x+1 + x+2 + x+3 + x+4 + ... + x+n = n * ( n + 1 ) / 2 + n * x
or = n * ( n + 1 + 2 * x ) / 2.

And the formula for b will be something like b = B0 * power( 2, n) were B0 is the starting value for b.
Similarly c = power( C0, power( 2, n)) with C0 as the starting value for c.

As an obs-puzzle, is there a simple function f(x,i), (and starting values for i and x) for which looping n times
through
i := i+1;
x = f(x,i);
will end up with x = power(3,n)?


Mike
.

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