# Solving a certain polynomial

1. Oct 28, 2009

### robbycon

So I have an equation:
h(x) = $$\sum a_{i}x^{i}$$ from i=0 to d.

I know $$a_{i}$$ and x.
I am trying to prove that there is a y where g(x) = $$\sum a_{i}y^{i}$$ from i=0 to d, g(x) = h(x), and y does not = x.

How do I do this? Sorry for the bad use of Latex.

2. Oct 28, 2009

### Office_Shredder

Staff Emeritus
In general? You can't, for example, the polynomial x3 is 1-1. It's unclear what you mean here, g(x) is not actually a function of x, and in fact seems to be h(y). So what you really want is to find x and y so that h(x)=h(y) right?

3. Oct 28, 2009

### robbycon

Yes, sorry. That was a typo. That is g(y), not g(x).

4. Oct 28, 2009

### HallsofIvy

Staff Emeritus
It's still not clear what you want to prove. If you have $h(x)= \sum_{i=0}^d a_i x^i$ and you replace the variable x by any y, you get $h(y)= \sum_{i=0}^d a_iy^i$. It is the same function, just written differently.

If you mean x and y to be specific numbers and want to prove that there exist $y\ne x$ such that h(y)= h(x), you can't- it is not, in general, true. As Office Shredder says, polynomials can be one-to-one. His example of h(x)= x3/sup] shows that.

5. Oct 28, 2009

### robbycon

I am looking for a y in terms of $$a_{i}$$ and $$x^{i}$$ in a solution that contains no polynomial equation. The most important thing I think is to figure out the properties of the summation of $$a_{i}$$ (which I don't know at all). I know I can't just divide that out, but there has to be a way to separate it from the summation of y.

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