Matematické Fórum

stuart clark

If $f(x)$ is twice differentiable function on $[a,e]$ and for $a<b<c<d<e,f(a)=0,f(b)=2,f(c)=1,f(d)=2,f(e)=0$

Then minimum number of roots of the equation $f(x)f''(x)+(f'(x))^2=0$ is

laszky · 28. 04. 2018 21:09 — Editoval laszky (02. 05. 2018 01:05)

Hi

Autor skryl část textu. Zobrazit/skrýt!

stuart clark · 01. 05. 2018 10:56

↑ laszky: Thanks .

stuart clark · 01. 05. 2018 10:58

Please explain

If $f(x)$ is twice differentiable function on $[a,e]$ and for $a<b<c<d<e,f(a)=0,f(b)=2,f(c)=-1,f(d)=2,f(e)=0$

Then minimum number of roots of the equation $f(x)f''(x)+(f'(x))^2=0$ is

laszky · 02. 05. 2018 01:30 — Editoval laszky (02. 05. 2018 13:21)

↑ stuart clark:

Since $\frac{1}{2}f^2(x)$ is non-negative and has zeroes at $a$ , $e$ and somewhere between $b$ , $c$ and $c$ , $d$ , there exist at least 6 roots of $g(x)=\frac{1}{2}f^2(x)-\frac{1}{2}$ in $[a,e]$ . Consequently, as in the previous case, there exist at least 4 roots of $g''(x)=\left(\frac{1}{2}f^2(x)-\frac{1}{2}\right)''= f(x)f''(x)+(f'(x))^2=0$ in $[a,e]$ .

stuart clark

Thanks ↑ laszky: But answer Given as $6$ .

i have one doubt How can we assume function $g(x)=\frac{(f(x))^2}{2}-1$ in first question. (We can assume other function also like $g(x)=\frac{(f(x))^2}{2}-2$ type

Similarly How can we assume function $g(x)=\frac{(f(x))^2}{2}-\frac{1}{2}.$ for second question. We can also assume other function.

please clearfy me , thanks in advanced.

laszky · 03. 05. 2018 10:43

↑ stuart clark:

The crucial words are "at least". At first you have to notice that $\left(\frac{1}{2}f^2(x)\right)''=f(x)f''(x)+(f'(x))^2$ , then you observe that $\left(\frac{1}{2}f^2(x)\right)''=\left(\frac{1}{2}f^2(x)-ax-b\right)''$ for arbitrary $a,b\in\mathbb{R}$ and you choose $a,b$ in such a way that $g(x)=\frac{1}{2}f^2(x)-ax-b$ has maximal possible zeros. If you choose them wrong, you prove only partial result (not the optimal one) - like me in my previous answer. (I forgot that you can substract any linear function and substracted only constant). My approach is: draw a picture of the presumable graph of the function $\frac{1}{2}f^2(x)$ and then try to draw a line which intersects the function $\frac{1}{2}f^2(x)$ in maximal possible points.

The following picture shows my first attempt with 6 zeros of the function g and my second attempt with 7 zeros of the function g (due to the zero derivative of $\frac{1}{2}f^2(x)$ there are two intersections near the point $a$ ). Hence, using this second attempt, we can prove that there exists at least 5 roots of $f(x)f''(x)+(f'(x))^2$ . But 6 is still problem for me.