Can a Linear Operator Define This Specific Polynomial Transformation?

[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)

 

[JasonRox]

How does it get to z?

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

 

[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.

 

[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...?)?

 

[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.

 

[AKG]

Could you please clarify what you're asking? You're asking for a function f such that:

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,

where, in your example,

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

and

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

Is this not what you're asking? If so, then I've already given the answer:

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.