Matematické Fórum

Nevíte-li si rady s jakýmkoliv matematickým problémem, toto místo je pro vás jako dìlané.

Nástìnka
🔒 23. 3. 2019 Pøešli jsme na HTTPS. Prosíme o kontrolu funkènosti fóra.
!! 17.06.2018 (Jel.) Khanova škola zve nadšence ke spolupráci na pøekladech návodù pro uèitele a rodièe.
! 04.11.2016 (Jel.) Ètete, prosím, pøed vložení dotazu, dìkuji!
17.01.2016 (Jel.) Rok 2016 s novými a novìjšími krystaly od kolegy Pavla!
17.01.2016 (Jel.) Nabídka knih z oborù matematiky, fyziky, chemie
23.10.2013 (Jel.) Zkuste pøed zadáním dotazu použít nìkterý z online-nástrojù, konzultovat použití mùžete v sekci CAS.

Nejste pøihlášen(a). Pøihlásit

#1 28. 04. 2018 20:41 — Editoval stuart clark (01. 05. 2018 10:57)

stuart clark
Pøíspìvky: 935
Reputace:

roots of function

If $f(x)$ is twice differentiable function on $[a,e]$ and for $a

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

Offline

#2 28. 04. 2018 21:09 — Editoval laszky (02. 05. 2018 01:05)

laszky
Pøíspìvky: 1372
Škola: MFF UK, FJFI CVUT
Reputace:   105

Hi

Offline

#3 01. 05. 2018 10:56

stuart clark
Pøíspìvky: 935
Reputace:

Re: roots of function

↑ laszky: Thanks .

Offline

#4 01. 05. 2018 10:58

stuart clark
Pøíspìvky: 935
Reputace:

Re: roots of function

If $f(x)$ is twice differentiable function on $[a,e]$ and for $a

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

Offline

#5 02. 05. 2018 01:30 — Editoval laszky (02. 05. 2018 13:21)

laszky
Pøíspìvky: 1372
Škola: MFF UK, FJFI CVUT
Reputace:   105

Re: roots of function

↑ 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 $[a,e]$.

Offline

#6 03. 05. 2018 07:10 — Editoval stuart clark (03. 05. 2018 09:12)

stuart clark
Pøíspìvky: 935
Reputace:

Re: roots of function

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.

Offline

#7 03. 05. 2018 10:43

laszky
Pøíspìvky: 1372
Škola: MFF UK, FJFI CVUT
Reputace:   105

Re: roots of function

↑ 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.

Offline

#8 05. 05. 2018 04:11

stuart clark
Pøíspìvky: 935
Reputace:

Re: roots of function

Thanks ↑ laszky:.

Offline