Question

Show that \(({x^2} + 1)/(x + 1)\) is \(O(x)\).

Short answer

The given function\(({x^2} + 1)/(x + 1)\)is\(O(x)\)when\(k = 2\)and\(C = 2\).

Step-by-step solution

Step 1:

First we find the quotient using the long division method.

\(x + 1\mathop{\left){\vphantom{1\begin{array}{l}{x^2} + 1\\\underline { - {x^2} - x} \\0 - x + 1\\\underline {0 + x + 1} \\0 + 0 + 2\end{array}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{x^2} + 1\\\underline { - {x^2} - x} \\0 - x + 1\\\underline {0 + x + 1} \\0 + 0 + 2\end{array}}}}

\limits^{\displaystyle\,\,\, {x - 1}}\)

We get,

\(\begin{array}{l}Q = x - 1\\R = 2\end{array}\)

Where,

\(Q = \)Quotient

\(R = \)Remainder

Step 2:

We can now write the fraction as,

\(\begin{array}{l}f(x) = \frac{{{x^2} + 1}}{{x + 1}}\\ = (x - 1) + \frac{2}{{x + 1}}\end{array}\)

We get the property\(x - 1 < x\).

When \(x > 0\), we get the property \(\frac{1}{{x + 1}} < 1\).

Step 3:

We will take\(k = 2\)and get\(x > 2\)

\(\begin{array}{l}\left| {f(x)} \right| = \left| {(x - 1) + \frac{2}{{(x + 1)}}} \right|\\ \le \left| {x - 1} \right| + \left| {\frac{2}{{(x + 1)}}} \right|\\ = 0 \le (x - 1) + \frac{2}{{x + 1}}\\ \le x + 2\end{array}\)

\(\begin{array}{l} \le x + x\\ = 2\left| x \right|\end{array}\)

Therefore taking\(C = 2\)by the definition we get that\(({x^2} + 1)/(x + 1)\)is \(O(x)\)when \(k = 2\) and \(C = 2\).