# A dumb question, but i'm just curious

1. Nov 2, 2004

### bomba923

Well, this problem is kinda dumb; it's in the GIF image file that I attached...
a,b,c,d,...,z are all real constants, and 'n' is natural; I was wondering if
such a function would exist...for the real constants a,b,c,d,...,z
(filename is polyprobgif.gif and sumformul.gif)

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• ###### sumformula.gif
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Last edited: Nov 2, 2004
2. Nov 2, 2004

### JasonRox

How does it get to z?

Is it n-25 = 0 so zx^0 = z?

3. Nov 2, 2004

### AKG

Of course. Let $P_n(\mathbb{R})$ be the space of polynomials of degree less than or equal to n with real coefficients. Since your notation is confusing, instead of a, b, c, ..., z being real constants, let's call them $a_0, a_1, \dots , a_n$. Then f is an operator on $P_n(\mathbb{R})$ which maps a polynomial $b_nx^n + b_{n - 1}x^{n - 1} + \dots + b_0$ to $a_nb_nx^n + \dots + a_0b_0$.

4. Nov 2, 2004

### bomba923

What would be the equation/formula for this function? By the way, can there exist some value 'phi' whose product with the sum will yield this series? <<the attached file phigif.gif has the problem>>--what would be the formula to get the value of phi (possibly some math operations with a,b,c,d,e,f...?)?

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5. Nov 2, 2004

### shmoe

Your notation is still confusing, is n=25? If that's not what you mean, have a look at AKG's post and the use of subscripts to deal with an arbitrary number of coefficients.

You have $$\Phi(x)f(x)=g(x)$$ (can I assume your dot means multiply?) and you want to find $$\Phi(x)$$. Just take $${\mbox\Phi(x)=g(x)/f(x)}$$ for x not a zero of f, whatever you like at the zeros of f, and hope that all the zeros of f are also zeros of g.

edit-You said "value of $$\Phi$$" multiple times, if you were hoping that $$\Phi$$ was a constant (and not dependant on x) then I should say you'd need all the coefficients on the right to be equal (otherwise it's impossible). Then take $$\Phi$$ to be this common value.

Last edited: Nov 2, 2004
6. Nov 3, 2004

### AKG

$$f(b_nx^n + b_{n - 1}x^{n - 1} + \dots + b_0) = a_nb_nx^n + a_{n - 1}b_{n - 1}x^{n - 1} + \dots + a_0b_0,$$

$$b_n = b_{n - 1} = \dots = b_0 = 1$$

and

$$a_n = a,\ a_{n - 1} = b,\ a_{n - 2} = c,\ a_0 = z?$$

$$f(b_nx^n + \dots + b_0) = a_nb_nx^n + \dots + a_0b_0$$

This function takes its argument from the space of n degree (or less) polynomials with real coefficients, and returns another polynomial from the same space but with different coefficients. It is a linear operator. If we choose

$$\{x^n,\ x^{n - 1},\ \dots ,\ x,\ 1\}$$

to be a basis of our space, then we can express polynomials as vectors by their coefficients. So, we'd have:

$$f(b_n,\ b_{n - 1},\ \dots ,\ b_0) = (a_nb_n,\ a_{n - 1}b_{n - 1},\ \dots ,\ a_0b_0)$$

Specifically, if we have the vector (polynomial) like the one in your example, namely, (1, 1, 1, ..., 1), then we get:

$$f(1,\ 1,\ 1,\ \dots ,\ 1) = (a_n,\ a_{n - 1},\ \dots ,\ a_0)$$

Just like you wanted. Just recall that I'm replacing

$$a,\ b,\ c,\ d,\ \dots ,\ z$$

with the less confusing

$$a_n,\ a_{n - 1},\ \dots ,\ a_0,$$

since I assume you didn't mean for there to only be 26 numbers, just n of them.

Last edited: Nov 3, 2004