logarithm problem: 2^N -2^(-N) = C (constant) Solve for N.

[tstone]

2^N -2^(-N) = C (constant) Solve for N.

I can solve this iteratively for a given C, but does it have a closed form solution? TIA.

 

[tkhunny]

Usually, a little transformation will do the trick on this sort.

\(\displaystyle 2^{N} - \dfrac{1}{2^{N}} = C\)

\(\displaystyle 2^{2N} - 1 = C\cdot 2^{N}\)

Do you see it, yet?

 

[stapel]

2^N -2^(-N) = C (constant) Solve for N.

I can solve this iteratively for a given C, but does it have a closed form solution? TIA.

The "trick" for these is to multiply through by 2^N to convert this to a quadratic form. (No, I didn't figure this out on my own, either; somebody had to show me.) It looks like this:

. . . . .\(\displaystyle 2^N\, -\, 2^{-N}\, =\, C\)

. . . . .\(\displaystyle 2^N\, \big(2^N\, -\, 2^{-N}\big)\, =\, 2^N\, (C)\)

. . . . .\(\displaystyle 2^{N+N}\, -\, 2^{N-N}\, =\, 2^N\, C\)

. . . . .\(\displaystyle 2^{2N}\, -\, 1\, =\, 2^N\, C\)

. . . . .\(\displaystyle 2^{2N}\, -\, 2^N\, C\, -\, 1\, =\, 0\)

. . . . .\(\displaystyle (2^N)^2\, -\, (2^N)\,C\, -\, 1\, =\, 0\)

This is a quadratic in 2^N, with a = 1, b = -C, and c = -1. You can solve (using the Quadratic Formula) for 2^N. Then solve the two exponential equations for N in terms of C.

If you get stuck, please reply showing your steps, starting with the last step I displayed above. Thank you!

 

[tstone]

Thank you!

That was a great help. The answer I get is

N = (log(C +/- sqrt(C^2 + 4)))/log2 - 1

which gives the answers I was expecting.