Question

Let ${\mathit{f}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{\text{and}}{\mathit{f}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$be functions from the set of real numbers to the set of positive real numbers. Show that if ${\mathit{f}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{\text{and}}{\mathit{f}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{\text{are both}}\mathbf{\Theta }\mathbf{\left(}\mathit{g}\mathbf{\right(}\mathit{x}\mathbf{\left)}\mathbf{\right)}\mathbf{\text{,}}$, where g(x) is a function from the set of real numbers to the set of positive real numbers, then ${\mathit{f}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}{\mathit{f}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{\text{is}}\mathbf{\Theta }\mathbf{\left(}\mathit{g}\mathbf{\right(}\mathit{x}\mathbf{\left)}\mathbf{\right)}$. Is this still true if ${\mathit{f}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{\text{and}}{\mathit{f}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$can take negative values?

Short answer

Given that $f_1\left(x\right)\text{and}{f}_2\left(x\right)\text{are both}\mathrm{\Theta }\left(g\right(x\left)\right)$then we have to prove that $f_1\left(x\right)+f_2\left(x\right)\text{is}\mathrm{\Theta }\left(g\right(x\left)\right)$ . Also check whether the given condition hold if$f_1\left(x\right)\text{and}{f}_2\left(x\right)$ take negative values.

Step-by-step solution

Step 1:

Assume,$f_1\left(x\right)=a$

$\begin{array}{r}{f}_2\left(x\right)=-a\\ g\left(x\right)=a\end{array}$

Here, both $f_1\left(x\right)=\mathrm{a}\mathrm{\&}{f}_2\left(x\right)=-\mathrm{a}\text{are}\mathrm{\Theta }\left(x\right)$or we can say that they are .

Hence, the given condition $f_1\left(x\right)\text{and}{f}_2\left(x\right)\text{are both}\mathrm{\Theta }\left(g\right(x\left)\right)$has become valid.

Step 2:

For$f_1\left(x\right)+f_2\left(x\right)\text{. We know that}{f}_1\left(x\right)=\mathrm{a},f_2\left(x\right)=-\mathrm{a}$

$\begin{array}{r}so{f}_1\left(x\right)+f_2\left(x\right)\\ =\mathrm{a}-\mathrm{a}\\ =0\end{array}$

As we know that $0\notin \mathrm{\Theta }\left(g\right(x\left)\right)$. Hence, by using the Big-theta notation definition, $f_1\left(x\right)+f_2\left(x\right)\notin \mathrm{\Theta }\left(g\right(x\left)\right)$when any one of $f_1\left(x\right)\text{and}{f}_2\left(x\right)$take negative values.

Hence, $f_1\left(x\right)+f_2\left(x\right)\text{is}\mathrm{\Theta }\left(g\right(x\left)\right)$is is not true when any one of $f_1\left(x\right)\text{and}{f}_2\left(x\right)$and take negative values.