here is the lemma ,
if $p(x)$ has real coefficient , and if $p(a)=0$ where $a>0$ , then $p(x)$ has at least one more sign variation than the quotient $q(x)$.Moreover, when the difference in the number of sign variation is greater than $1$, the difference is always an odd number
here is the part of the proof (from beginning to some part)
Assume the leading coefficient of is positive and consider the synthetic substitution form used to compute $p(a)$ Consider the constant term in $p(x)$ If this constant term is negative , then in the previous column the constant term in $q(x)$ must have been positive in order for the final column numbers to add to make $0$. If the constant term in were positive , then in the previous column the constant term in $q(x)$ must have been negative in order for the final column numbers to add to make $0$. So the constant terms in and must have opposite signs. This argument has depended on the facts that and that $a<0$ and that $p(a)=0$
But $p(x)$ and $q(x)$ both start with the same positive coefficient. Next, reading from left to right, we claim that $q(x)$ cannot change signs until $p(x)$ changes signs. Whenever $q(x)$ change signs from one column to the next, $p(x)$ must also change signs between those same two columns.
but the claim is not necessarily true , because for example take $p(x)=2x^3-x^2-7x+2$ divided by $x-2$.if you do the synthetic substitution the quotient term in 2nd to 3rd column change signs but $p(x)$ in the same column does not , can someone explain about this ( maybe i dont understand the statement of the claim or maybe the claim is wrong?) i am very confuse about it .
Can you give me some recommended Links /books/ paper/whatever to read the Descartes rule proof ?
The question as I understand it concerns the second and third paragraphs on page 15 of "Some Polynomial Theorems" by Some Polynomial Theorems by John Kennedy at Santa Monica College.
The discussion before these paragraphs gives several examples of synthetic division of polynomials, a method for dividing the polynomial $p_n x^n + p_{n-1} x^{n-1} + \cdots + p_2 x^2 + p_1 x + p_0$ by the polynomial divisor $x - a$ in order to obtain the quotient $q_{n-1} x^{n-1} + q_{n-2} x^{n-2} + \cdots + q_2 x^2 + q_1 x + q_0.$ In the linked paper the term $a$ is written above and to the right of the rest of the terms of the synthetic division, but I prefer to write it on the left like this:
\begin{array}{c|ccccc} & p_n & p_{n-1} & p_{n-2} & p_{n-3} & \cdots \\ a & & aq_{n-1} & aq_{n-2} & aq_{n-3} & \cdots \\ \hline & q_{n-1} & q_{n-2} & q_{n-3} & q_{n-4} & \cdots \end{array}
The complete table has $n$ columns to the right of the vertical line; I have shown only the first few columns.
Initially only the $p_i$ values and $a$ would be filled in on this table. To actually perform the division, we copy the leading coefficient $p_n$ to the bottom row (so $q_{n-1} = p_n$). Then, repeatedly, we take the last coefficient of $q$ that we wrote on the bottom row, multiply it by a, put the result in the next column just above the line (so the first time we do this, we write $aq_{n-1}$ in the place shown in the table above), and then add the two numbers above the line to produce a new coefficient of $q$ below the line. For example, $q_{n-2} = p_{n-1} + aq_{n-1}.$
The claim is made that if the coefficients of $q$ in two consecutive columns of the table have opposite signs, the coefficients of $p$ in the same two columns must also have opposite signs. But you have provided a perfectly good counterexample, whose synthetic division comes out like this:
\begin{array}{c|rrrr} & 2 & -1 & -7 & 2 \\ 2 & & 4 & 6 & -2 \\ \hline & 2 & 3 & -1 & \end{array}
We have coefficients $3$ and $-1$ of opposite signs below the line, but the coefficients of the same columns of the first row are $-1$ and $-7$, which have the same sign.
However, the claim that "$q(x)$ cannot change signs until $p(x)$ changes signs" is still essentially correct if you understand "not until" to mean "not before".
Assuming $p_n$ is positive, we start with positive numbers in the first row to the right of the vertical line, and since $a$ also is positive, as long as the $q$ coefficients are positive we will keep carrying up positive values $aq_{n-1},$ $aq_{n-2},$ and so forth in the second row. So as long as we keep finding positive coefficients of $p$ in the first row, we will be adding positive to positive and writing a positive coefficient below the line. In order to produce the first negative coefficient below the line, we must have a negative coefficient on the first row of the same column.
But the implication does not go the other way. The first negative coefficient on the first row might not produce a negative coefficient below the line. We will get the first negative coefficient of $q$ below the line only when the coefficient of $p$ on the first row is negative and its magnitude is larger than the magnitude of the positive number on the second row. That might occur for the first negative coefficient of $p$, or for a later negative coefficient that occurs after another positive coefficient (with any number of other positive and negative coefficients before that), or for a later negative coefficient that occurs after a negative coefficient (as it does in your counterexample). In the first two cases there are sign changes in on both the first and last rows of the same pair of columns, but in the third case there is not.
What we have in all three cases, however, is that there had to be at least one sign change in the coefficients of $p$ at or before the column where the first sign change occurred in the coefficients of $q.$ To put it another way, the coefficient of $q$ can change signs only at or after the first column where the coefficient of $p$ changes signs.
Looking for the next sign change on the bottom row, we are now writing negative numbers (positive $a$ times negative $q_i$) on the second row, so we cannot get a positive coefficient below the line again until we have had a positive coefficient on the first row again -- not necessarily the first time we have a positive coefficient on the first row, and not necessarily simultaneous with a sign change on the first row, but there has to be another sign change in $p$ at or before the next sign change in $q$; that is, the next sign change in $q$ can only occur at or after the next sign change in $p.$
This does not support the next paragraph in the proof, however; in particular, it does not support the statement that "each further time $q(x)$ changes signs, so does $p(x)$." As we saw in your counterexample, the coefficients of $p(x)$ might change signs first, and then the coefficients of $q(x)$ change signs later without any further change in sign in the coefficients of $p(x).$
But we do still have one of two possible situations when we reach the next-to-last column of the table. Either the numbers in the first row have already changed signs more times than the numbers in the last row, in which case the "at least one more sign variation" claim of the lemma is true, or the numbers in the first and last row have changed signs exactly the same number of times, in which case their signs in the next-to-last column are the same. In the second case, the number we write on the second row in the last column will have the same sign as the bottom row of the previous column, and in order for the division to complete without remainder, the number on the first row must have an opposite sign -- which is also opposite the sign of the previous number on the first row, so there's one additional sign change in the coefficients of $p(x).$
The proof is still rather informal and (I think) not particularly rigorous, but rigorously proving Descartes' rule of signs is apparently not a simple business.