simplifying a root

[Keiji]

Does anyone know how to simplify an expression of the form:



into an expression of the form:



where k is a integer constant greater than 1 and x₁ to x_n are variables, and y₁ to y_n are either the original variables or products of them?

[pat]

You have some reason to believe this is possible? The only hopes that I have for it involve more than just products of x_js.

[Keiji]

No, I was wondering if it was possible and if so how.

What were you thinking of?

[houserichichi]

What if

y1 = x1+x2+x3+...+xn

and

y2, y3, y4, ... = 0?




And if you're going to be anal, extend the xi list to xj where x(i+1), x(i+2), ... xj = 0. That way my way works just fine.


**EDIT** Nevermind, only products of variables. It would help if I learned to read. My joke just got busted.

[bo198214]

seems quite tedious to work this out, but should be possible. Assume for this time n=2. I.e.
sqrt(x1+x2)=sqrt(y1)+sqrt(y2)
we can this transform (though getting possibly additional solutions) into a polynomial equation:
x1+x2=y1+2sqrt(y1)sqrt(y2)+y2
(x1+x2-y1-y2)^2=4y1y2

Then I would a bit more generally arrange y1 and y2 as polynomials in the variables x1 and x2 of degree m, i.e
y1=sum_{k1+k2<=m} a_{k1,k2} x1^k1 x2^k2
y2=sum_{k1+k2<=m} b_{k1,k2} x1^k1 x2^k2

By putting these polynomials into the equation, expand it and then compare the coefficients for the monomials x1^k1 * x2^k2 for all pairs (k1,k2), we get a bunch of equations for the coefficients a_{i1,i2} and b_{i1,i2}.
It should first turn out that the degree of the polynomials is at most 1 (i.e. a_{i1,i2}=b_{i1,i2}=0 for 1<i1+i2) for and the constants a_{0,0} and b_{0,0} are also zero. I.e.
y1=a_{1,0}x1+a_{0,1}x2
y2=b_{1,0}x1+b_{0,1}x2

and if you have this it should turn out next that either y1=x1+x2 and y2=0 or vice versa.

PS: Maybe Jinydu or PWrong is more interested in actually doing the computations.

[Keiji]

and if you have this it should turn out next that either y1=x1+x2 and y2=0 or vice versa.

But that's pointless, because you end up not doing anything. I need the y variables to be products of the x variables, not sums.

[batmanmg]

im guessing that the root works like a summation and not just random values for x?

well thats not too complicated of a series to work with.

its the k-root of

i was always terreble at these things so i'll have to shake off the rust... (or someone has to beat me to it) before i can get help on the subject

[pat]

where k is a integer constant greater than 1 and x₁ to x_n are variables, and y₁ to y_n are either the original variables or products of them?

Here is evidence (proof with some of the details left to you) that it is impossible to form a general expression of this type where the y_j are products of the x_j.

Consider the case where x_j = 1 for all j. And, n > 1. The y_j can only be 1 since any product of 1's is 1. Further, the k-th root of 1 is 1. So, you're looking for a case where the k-th root of n (the left side) equals n (the right side). The only way that's going to happen with k > 1 is when n = 1.

Now, if you were expecting a different number of y_j than x_j and just failed to indicate that with the subscripts, then it gets a little wonkier. But, probably not impossible. In fact, I suspect it's likely one could use Taylor series to come up with something (infinite).

[Keiji]

Now, if you were expecting a different number of y_j than x_j and just failed to indicate that with the subscripts, then it gets a little wonkier. But, probably not impossible. In fact, I suspect it's likely one could use Taylor series to come up with something (infinite).

Yes, of course I was! Like when expanding (a+b)² you get three terms. There can be as many Ys as necessary.

[PWrong]

For the case n=2, if you don't mind having infinitely many y's, you can use the generalised binomial theorem. The power doesn't have to be an integer, but if it's not, you get a sum over infinitely many products. For some reason, there's no generalised multinomial theorem on wikipedia or mathworld. So I'm not sure if the theorem works for nonintegers.

[Keiji]

if you don't mind having infinitely many y's

I guess it's impossible then.

[bo198214]

and if you have this it should turn out next that either y1=x1+x2 and y2=0 or vice versa.

But that's pointless, because you end up not doing anything. I need the y variables to be products of the x variables, not sums.

A product is a special case of a polynomial in the variables x1, ..., xn.
For the case k=n=2 there are only the products 1,x1, x2, x1x2.
It looks quite trivial to me to show that if you put either into the equation, the equation becomes invalid.
Thatswhy I tried to guess what you mean and thought of sums of products with coefficents and then, still a bit more general, polynomials.

And if the solutions then are only the trivial ones mentioned above, then its shown that you cant do it that way.

[pat]

A product is a special case of a polynomial in the variables x1, ..., xn.
For the case k=n=2 there are only the products 1,x1, x2, x1x2.

No, there is also x₁^s * x₂^t for s and t non-negative integers.

[bo198214]

Now, if you were expecting a different number of y_j than x_j

Yes, of course I was! Like when expanding (a+b)² you get three terms. There can be as many Ys as necessary.

So then next time you mean different numbers of xi and of yi then you should write it and not using n as the number of xi and as the number of yi!

I didnt see any similarity to the binomial formula in the beginning. (x+y)^2 is not a sum of squares of products of x and y! But probably you wanted the positive integer coefficients replaced by that much copies:
(x+y)^2 = x^2 + 2xy + y^2 = x^2 + xy + xy + y^2
And then put somehow sqrt into play. Btw: (x+y)^2 = sqrt(x^4)+sqrt(x^2y^2)+sqrt(x^2y^2)+sqrt(y^4) *lol*

The generalized multinomial formula is an excellent hint, though here we have fractional coefficents before the products of the (x_i)^{1/k} (and of course infinitely many yi).