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Thread: Show that f(x) + f'(x) + f"(x) > 0 for all real x(Complex numbers)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Show that f(x) + f'(x) + f"(x) > 0 for all real x(Complex numbers)?

    If $\displaystyle m^2 + n^2 - mn - m - n - 1\leq 0$, where $\displaystyle m,n\in\mathbb{R}$ and $\displaystyle z_{1},z_{2}$ be the complex numbers such that that $\displaystyle |z_{1}| \leq m, |z_{2}|\leq n$ and $\displaystyle a > (|z_{1}| - |z_{2}|)^2 + (\arg z_{1} - \arg z_{2})^2 - |z_{1} - z_{2}|^2$. If $\displaystyle f(x) = ax^2 + bx + c > 0$ for all $\displaystyle x$, show that $\displaystyle f(x) + f'(x) + f''(x) > 0$ for all real $\displaystyle x$.
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    If $\displaystyle \color{blue}m^2 + n^2 - mn - m - n - 1\leq 0$, where $\displaystyle \color{blue}m,n\in\mathbb{R}$ and $\displaystyle \color{blue}z_{1},z_{2}$ be the complex numbers such that that $\displaystyle \color{blue}|z_{1}| \leq m, |z_{2}|\leq n$ and $\displaystyle \color{blue}a > (|z_{1}| - |z_{2}|)^2 + (\arg z_{1} - \arg z_{2})^2 - |z_{1} - z_{2}|^2$. If $\displaystyle f(x) = ax^2 + bx + c > 0$ for all $\displaystyle x$, show that $\displaystyle f(x) + f'(x) + f''(x) > 0$ for all real $\displaystyle x$.
    You have mixed up two separate problems. The sentence in blue is completely irrelevant to the problem contained in the second sentence.

    The conditions for the quadratic polynomial $\displaystyle f(x) = ax^2 + bx + c$ to be always positive are that $\displaystyle a>0$ and $\displaystyle b^2<4ac$. If f(x) satisfies those conditions then it easy to check that $\displaystyle f(x) + f'(x) + f''(x) = ax^2 + (2a+b)x + (2a+b+c)$ satisfies the same conditions.
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