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/sci/ - Science & Math

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4496877 No.4496877 [Reply] [Original]

Is there any way to transform this formula so that it shows only y on the left side of equation instead of z(n and y are unknown, z is known)? Wolfram works only if n is specified.

>> No.4496883

Only if n is 4 or less.

>> No.4496884 [DELETED] 

use geometric series formula

>> No.4496886

use geometric series formula

>> No.4496892
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4496892

>>4496883
>>4496886
/sci/, I'm impress.

>> No.4496893

geometric formula only gets to z = [(1+y)^(n+1) -1]/y

>> No.4496896

It worked even with 15 (I typed it as some random number that came to my mind) but only with n (in this case 15) specified. I think that for every n the formula will be different (because of different pow degree). So is my assumption of this being unsolvable right?

>> No.4496910

I don't think you can if n is even. You can easily prove this by considering the derivative of z with respect to 1. It's negative for very low y, and it's positive for large y. From this it follows that z can't be one to one, and hence, an inverse is not well defined.

>> No.4496913

>>4496910
In stead of 1 it should say y. Oops.

>> No.4496922

>>4496910
Y is meant to be between 0 nad 1

>> No.4496930

>>4496896
It worked with 15 because alpha did it numerically.

It has been proven that 5th or higher degree general polynomials can not be solved by radicals, which translates to "no formula for you".
That's not the general polynomial, so i guess it might be possible, but i wouldn't bet on it.
See here http://en.wikipedia.org/wiki/Polynomial#Solving_polynomial_equations for more, if you're interested.

>> No.4496935

>>4496922
as n -> infinity, y = 1/(z - 1)

>> No.4496975
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4496975

>>4496935
as n -> infinity, z -> infinity.

There is a formula for n up to 4. you will have a very hard time finding an analytic formula for n>4, if it exists, which I doubt.
In any case, if <span class="math">y \in [0,1][/spoiler], then there is a numerical solution iff <span class="math">z \in [n,2^(n+1)-2][/spoiler].