$f(x)=e^x(x^2-5x+7)$.Prove $f^{(n)}(x)$ has 2 roots for any $n\in\mathbb N^*$. $f^{(n)}(x)$ means derivative of order n.

Question

$f(x)=e^x(x^2-5x+7)$.Prove $f^{(n)}(x)$ has 2 roots for any $n\in\mathbb N^*$. $f^{(n)}(x)$ means derivative of order n.

I am really confused by this exercise. I have calculated first 10 derivatives and $\Delta$ for each to see if it is positive, and it is indeed. I thought that by doing this eventually I will be able to find a general form for the derivative of order n and then prove it is right by mathematical induction then calculate $\Delta$ and show it is >0. The first and second term are easy to find, first one never changes and second one is $(-5x+2*nx)$, but the third term changes based on the second and last terms of $f^{(n-1)}(x)$ so I am stuck here with this approach.

Thanks in advance to anyone trying to help!

Answer 1

The hint.

Let for $x^2+ax+b$ we have $a^2-4b>0$.

Thus, $$(e^x(x^2+ax+b))'=e^x(x^2+ax+b)+e^x(2x+b)=e^x(x^2+(a+2)x+a+b)$$ and we see that $$(a+2)^2-4(a+b)=a^2-4b+4>0.$$

Now, use induction.

A full solution can be written so.

Easy to see that for all $n$ we can write $f^{(n)}$ in the following form: $$f^{(n)}(x)=e^x(x^2+a_nx+b_n).$$

Now, $$f'(x)=e^x(x-1)(x-2)$$ has two real root. It was a base of the induction.

Let $f^{(n)}(x)=e^x(x^2+a_nx+b_n)$ has two real roots.

Thus, $$a_n^2-4b_n>0.$$

We'll prove that $f^{(n+1)}$ has two real roots.

Indeed, $$f^{(n+1)}(x)=\left(e^x(x^2+a_nx+b_n)\right)'=e^x(x^2+a_nx+b_n)+e^x(2x+a_n)=$$ $$=e^x(x^2+(a_n+2)x+a_n+b_n).$$ Id est, $$a_{n+1}=a_n+2,$$ $$b_{n+1}=a_n+b_n$$ and it's enough to prove that: $$a_{n+1}^2-4b_{n+1}>0$$ or $$(a_n+2)^2-4(a_n+b_n)>0$$ or $$a_n^2-4b_n+4>0,$$ which is true by the assumption of the induction.

Thus, by the induction $f^{(n)}$ has two real roots for all $n\geq1$ and we are done!

Answer 2

If you are looking for a more "explicit" demonstration I believe you will appreciate this answer. Let $g(x)=x^2-5x+7$. We have \begin{align} g^{(1)}(x)=& 2x-5\\ g^{(2)}(x)=& 2\\ g^{(3)}(x)=& 0\\ \end{align} and $$ g^{(n)}(x)= 0 \quad \mbox{ for all } n\geq 3 $$ We will derive the function $f(x)=e^xg(x)$ successively until we find a pattern of recall and then verify the existence of roots for $f(x)=e^xg(x)$.

For $n=1$ we have $$ \begin{array}{rc rl} f(x)=e^xg(x) & \implies &f^{(1)}(x)=&e^xg^{(1)}(x)+e^xg(x) \\ & &= & e^x\Big(g(x)+g^{(1)}(x)\Big) \end{array} $$ For $n=2$ we have $$ \begin{array}{rc rl} f^{(1)}(x)=e^x\Big(g(x)+g^{(1)}(x)\Big) & \implies &f^{(2)}(x)=&e^x\Big(g(x)+g^{(1)}(x)\Big) \\ &&&+e^x\Big(g^{(1)}(x)+g^{(2)}(x)\Big) \\ & &= & e^x\Big(g(x)+2g^{(1)}(x)+g^{(2)}(x)\Big) \end{array} $$ For $n=3$ we have $$ \begin{array}{rc rl} f^{(2)}(x)=e^x\Big(g(x)+2g^{(1)}(x)+g^{(2)}(x)\Big) & \implies &f^{(3)}(x)=&e^x\Big(g(x)+2g^{(1)}(x)+g^{(2)}(x)\Big) \\ &&&+e^x\Big(g^{(1)}(x)+2g^{(2)}(x)\Big) \\ & &= & e^x\Big(g(x)+3g^{(1)}(x)+3g^{(2)}(x)\Big) \end{array} $$ For $n=4$ we have $$ \begin{array}{rc rl} f^{(3)}(x)=e^x\Big(g(x)+3g^{(1)}(x)+3g^{(2)}(x)\Big) & \implies &f^{(4)}(x)=&e^x\Big(g(x)+3g^{(1)}(x)+3g^{(2)}(x)\Big) \\ & &&+e^x\Big(g^{(1)}(x)+3g^{(2)}(x)\Big) \\ & &= & e^x\Big(g(x)+4g^{(1)}(x)+6g^{(2)}(x)\Big) \end{array} $$ For $n=5$ we have $$ \begin{array}{rc rl} f^{(4)}(x)=e^x\Big(g(x)+4g^{(1)}(x)+6g^{(2)}(x)\Big) & \implies &f^{(5)}(x)=&e^x\Big(g(x)+4g^{(1)}(x)+6g^{(2)}(x)\Big) \\ & &&+e^x\Big(g^{(1)}(x)+4g^{(2)}(x)\Big) \\ & &= & e^x\Big(g(x)+5g^{(1)}(x)+10g^{(2)}(x)\Big) \end{array} $$ For $n=6$ we have $$ \begin{array}{rc rl} f^{(5)}(x)=e^x\Big(g(x)+5g^{(1)}(x)+10g^{(2)}(x)\Big) & \implies &f^{(6)}(x)=&e^x\Big(g(x)+5g^{(1)}(x)+10g^{(2)}(x)\Big) \\ & &&+e^x\Big(g^{(1)}(x)+5g^{(2)}(x)\Big) \\ & &= & e^x\Big(g(x)+6g^{(1)}(x)+15g^{(2)}(x)\Big) \end{array} $$ For $A_1=1$ e $B_1=0$ set $$ A_{k+1}=A_k+1 \quad \mbox{ and } \quad B_{k+1}= A_{k}+B_{k} $$ and note that $$ A_{k}=k \quad \mbox{ and } \quad B_{k}= A_{k-1}+\ldots+A_1= \frac{k(k-1)}{2} $$ Furthermore, by pattern of recall above we can prove by induction that \begin{align} f^{(k)}(x)=& e^x\Big( g(x)+A_{k} \cdot g^{(1)}(x)+B_{k}\cdot g^{(2)}(x) \Big)\\ =& e^x\Big( (x^2-5x+7)+A_{k} \cdot(2x-5)+B_{k}\cdot 2 \Big)\\ =& e^x\Big( x^2+(-5+2A_{k})x+(7-5A_k+2B_k)\Big)\\ =& e^x\Big( x^2+(-5+2k)x+(7-5k+k(k-1))\Big)\\ =& e^x\Big( x^2+(2k-5)x+(k^2-6k+7)\Big)\\ \end{align} Now notice that the $f^{(k)}(x)$ function is null if and only if $\Big( x^2+(2k-5)x+(k^2-6k+7)\Big)$ is null. And the expression $$ \Big( x^2+(2k-5)x+(k^2-6k+7)\Big) $$ will have real raises if, and only if, its discriminant $$ (2k-5)^2-4(k^2-6k+7) $$ is greater than zero. In fact we have, $$ (2k-5)^2-4(k^2-6k+7)=14k-3>0 \quad \mbox{ for all } \quad k>0. $$