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November 29th, 2010, 02:17 PM  #1 
Member Joined: Nov 2009 Posts: 67 Thanks: 0  Upper and Lower Bound Theorems I ran across an exercise that has me a bit puzzled. Consider: Find an integral upper bound using the Upper Bound Theorem and find an integral lower bound using the Lower Bound Theorem. An upper bound is an integer greater than or equal to the greatest real zero. Upper Bound Theorem: If you divide a polynomial function f(x) by (x  c), where c > 0, using synthetic division and this yields all nonnegative numbers, then c is an upper bound to the real roots of the equation f(x) = 0. Using synthetic division, I found the integral upper bound to be 2. All the coefficients and the remainder are nonnegative. Now on to the lower bound. A lower bound is an integer less than or equal to the least real zero. Lower Bound Theorem: If you divide a polynomial function f(x) by (x  c), where c < 0, using synthetic division and this yields alternating signs, then c is a lower bound to the real roots of the equation f(x) = 0. Special note that zeros can be either positive or negative. Using synthetic division again, I tested 1, 2, 3, 4, and 5. Only 5 yielded alternating signs in the coefficients and remainder of the quotient. This says that 5 is a lower bound and all the real zeros of can be found in the interval Now, 4 failed the lower bound theorem test because the quotient did not produce alternating signs. This would suggest that 4 is not a lower bound, right? However, upon further inspection, the actual zeros are {3.62, 1.38, 2) This would seem to indicate that a lower bound (in fact the greatest lower bound) should have been 4. Why did 4 fail the Lower Bound Theorem test? 
November 29th, 2010, 07:54 PM  #2 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: Upper and Lower Bound Theorems
If P, then Q. Not all bounds pass the test. The problem is just in the logic of the theorem. 
November 29th, 2010, 10:37 PM  #3 
Senior Member Joined: Nov 2010 Posts: 502 Thanks: 0  Re: Upper and Lower Bound Theorems
The point here is that this is not exclusive. The theorem does not say that, to use Chaz's notation, if not P, then not Q. Instead, this is just a onesided logic test. This is similar to the idea that if a function is differentiable at a point, then the function is continuous at that point. You can of course have functions that are continuous somewhere but not differentiable there (like a cusp, an angle, etc.). But we do not say, if it is not differentiable then it is not continuous. 
November 30th, 2010, 10:36 AM  #4  
Member Joined: Nov 2009 Posts: 67 Thanks: 0  Re: Upper and Lower Bound Theorems Quote:
Given the function , is 4 a lower bound? Even though the answer is "yes", this cannot be shown using the Lower Bound Theorem.  
December 23rd, 2015, 06:49 AM  #5 
Newbie Joined: Dec 2015 From: USA Posts: 4 Thanks: 0 
Is it true that the theorem only provides conditions in cases where the polynomial has at least one positive root and at least one negative root? I.e., if a polynomial has at least one positive root, then the upper bound for the roots of the polynomial will be the number c, such that when the polynomial is divided by c the coefficients of the quotient and the remainder will be nonnegative, and also the analogous conditions will apply to the lower bound. And if the polynomial has two positive roots then the theorem can only be used to show an upper bound, not a lower bound? (Similarly in the case of two negative roots the theorem can be used to find a lower bound.) Is this accurate?


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