no positive roots for $(y+1)^{7}-2(y+1)^{5}+10(y+1)^{2}-1$

Question

Hint for showing that $(y+1)^{7}-2(y+1)^{5}+10(y+1)^{2}-1$ has no positive roots thanks

Answer 1

Check the value at $y = 0$; look at it's derivative's value for positive $y$.

Answer 2

Expand and you see all coefficients are positive.

$$y^7+7 y^6+19 y^5+25 y^4+15 y^3+11 y^2+17 y+8$$

If you put $y>0$ the whole sum turns out positive.

Expanding (efficiently) the given polynomials in powers of $y$ (by hand):

We want $q(y):=(y+1)^7−2(y+1)^5+10(y+1)^2−1=a_0+a_1y+a_2y^2+...+a_7y^7$, for some $a_0,a_1,...,a_7$.

Notice that $a_0$ is the value of the polynomial on the right evaluated at $y=0$. So we just need to evaluate the polynomial on the left at $y=0$.

This is the same as evaluating the polynomial $p(x):=x^7-2x^5+10x^2-1$ at $x=1$.

Then $(q(y)-a_0)/y=a_1+a_2y+...+a_7y^6$.

We can do the evaluation using Ruffini-Horner algorithm. The advantage is that the algorithm gives us also the coefficient of $(q(y)-a_0)/y$ written as powers of $(y+1)$. So repeating Ruffini-Horner several times we get the coefficients we wanted.

The actual computation to be done by hand: (First read how Ruffini-Horner goes in Wikipedia and how the computation is written in a table).

We get the coefficients of $p(x)=1x^7+0x^6-2x^5+0x^4+0x^3+10x^2+0x-1$ and to Ruffini-Horner to evaluate $p(x)$ at $x_0=1$.

$$\begin{array}{c|rrrrrrrr}x_0&x^7&x^6&x^5&x^4&x^3&x^2&x^1&x^0\\1&1&0&-2&0&0&10&0&-1\\&&1&1&-1&-1&-1&9&9\\\hline\\&1&1&-1&-1&-1&9&9&\boxed{8}\\1&&1&2&1&0&-1&8\\\hline\\&1&2&1&0&-1&8&\boxed{17}\\1&&1&3&4&4&3\\\hline\\&1&3&4&4&3&\boxed{11}\\1&&1&4&8&12\\\hline\\&1&4&8&12&\boxed{15}\\1&&1&5&13\\\hline\\&1&5&13&\boxed{25}\\1&&1&6\\\hline\\&1&6&\boxed{19}\\1&&1\\\hline\\&1&\boxed{7}\\1&&\\\hline\\&\boxed{1}\end{array}$$

In the boxes there are the coefficients of the expansion.

Answer 3

Note that our expression is equal to $$(y+1)^5((y+1)^2 -2)+10(y+1)^2 -1.$$ If $y\ge \sqrt{2}-1$, then the expression is safely positive, indeed pretty big. In particular, if $y\ge 1$ then our expression is positive.

Now suppose that $0\lt y\lt 1$. We have $(y+1)^7 -2(y+1)^5 \gt -(y+1)^5$.

Note that $10(y+1)^2-1-(y+1)^5=2(y+1)^2-1 + (y+1)^2(8-(y+1)^3)$. The part $2(y+1)^2-1$ is $\gt 1$, and the part $8-(y+1)^3$ is positive, since $y\lt 1$.