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Proof of upper and lower bounds theorem for polynomials with real zeros.

On Page 274 of this book:
https://www.stitz-zeager.com/szprecalculus07042013.pdf

It shows the proof for the upper bound theorem. For the lower bound, it says to use f(-x) and then find its upper bound. I am assuming the lower bound of f(x) is equivalent to the upper bound of f(-x) because -x is just a reflection across the y-axis?

As for the proof, I tried this: Let c<0, then f(x)=(x+c)q(x)+r. Now I replace x with -x: f(-x)=(-x+c)q(-x)+r. Here's the problem. The upper bound expressively requires that the final row in the synthetic division tableau be all the same sign. But the quotient polynomial q(-x) alternates signs due to -x. So I really need help here thank you.

Reply 1

mqb2766

Original post

by Jr312s

On Page 274 of this book:
https://www.stitz-zeager.com/szprecalculus07042013.pdf
It shows the proof for the upper bound theorem. For the lower bound, it says to use f(-x) and then find its upper bound. I am assuming the lower bound of f(x) is equivalent to the upper bound of f(-x) because -x is just a reflection across the y-axis?
As for the proof, I tried this: Let c<0, then f(x)=(x+c)q(x)+r. Now I replace x with -x: f(-x)=(-x+c)q(-x)+r. Here's the problem. The upper bound expressively requires that the final row in the synthetic division tableau be all the same sign. But the quotient polynomial q(-x) alternates signs due to -x. So I really need help here thank you.

Theres a reasonable explanation/proof at
https://math.stackexchange.com/questions/1633846/proof-for-theorem-of-upper-and-lower-bounds-on-zeroes-of-polynomials
which doesnt rely on the division tableau. Are you ok with that, or do you want to do it your book way?

Reply 2

Jr312s

I don't understand that at all. Can you show me the book way? Thank you.

Reply 3

mqb2766

Original post

by Jr312s

I don't understand that at all. Can you show me the book way? Thank you.

what year are you in/ (uni)?

Reply 4

Jr312s

1st year

Reply 5

mqb2766

Original post

by Jr312s

On Page 274 of this book:
https://www.stitz-zeager.com/szprecalculus07042013.pdf
It shows the proof for the upper bound theorem. For the lower bound, it says to use f(-x) and then find its upper bound. I am assuming the lower bound of f(x) is equivalent to the upper bound of f(-x) because -x is just a reflection across the y-axis?
As for the proof, I tried this: Let c<0, then f(x)=(x+c)q(x)+r. Now I replace x with -x: f(-x)=(-x+c)q(-x)+r. Here's the problem. The upper bound expressively requires that the final row in the synthetic division tableau be all the same sign. But the quotient polynomial q(-x) alternates signs due to -x. So I really need help here thank you.

If youre first year, then you should be able to work through the other proof, which is arguably a bit simpler (without stating the tableau dependency). Though for c positive or negative, you start with
f(x) = (x-c)q(x) + r
as thats the basic divisor/quotient/remainder, and for c<0 you assume the signs alternate for x<c. So consider f(-x) and -x>-c
f(-x) = (-x-c)q(-x) + r = -(x + c)q(-x) + r
The final line of the division tableau corresponds to the powers x^3, x^2, .... and as you know they alternate for f(x), they must be the same sign for f(-x). So -x is an upper bound for the zeros of f(-x) so x is a lower bound for f(x) by reflection.

(edited 10 months ago)

Reply 6

Jr312s

I understand f(-x)=-(x+c)q(-x)+r. But here's the problem: I'm supposed to find the upper bound from this expression such that
-x > -c. Yet according to the book, I can only apply the upper bound theorem if all the signs in the division tableau are the same. Here with q(-x) they clearly alternate.