Question

Find the least integer\(n\)such that\(f(x)\)is\(O({x^n})\)for each of these functions.

a)\(\;f(x) = 2{x^2} + {x^3}\log x\)

b)\(f(x) = 3{x^5} + {(\log x)^4}\)

c)\(f(x) = ({x^4} + {x^2} + 1)/({x^4} + 1)\)

d) \(f(x) = ({x^3} + 5\log x)/({x^4} + 1)\)

Short answer

a)\(n = 4\)

b)\(n = 5\)

c)\(n = 0\)

d) \(n = - 1\)

Step-by-step solution

Subpart (a): \(\;f(x) = 2{x^2} + {x^3}\log x\)Step 1:

When\(x > 0\)we have property\(\log x \le x\)

\( \Rightarrow \;f(x) = 2{x^2} + {x^3}\log x \le 2{x^2} + {x^4}\)

From above equation \(n = 4\) as the largest power of \(x\)in definition of \(f(x)\) with \(\log x\) rounded upto \(x\)is smallest \(n\) for which \(f(x)\)is \(O({x^n})\)

(a) Step 2:

Now, when\(x > 2\)we have property\({x^2} > x > 2\)

Let\(k = 2\)

\(\begin{array}{l} \Rightarrow \left| {f(x)} \right| = \left| {2{x^2} + {x^3}\log x} \right| \le \left| {2{x^2} + {x^4}} \right|\\ \le \left| {2{x^2}} \right| + \left| {{x^4}} \right|\\ = 2{x^2} + {x^4}\end{array}\)

\(\begin{array}{l} \le {x^4} + {x^4}\\ = 2{x^4}\\ = 2\left| {{x^4}} \right|\end{array}\)

\( \Rightarrow C = 2\) and \(k = 2\)

Subpart (b): \(f(x) = 3{x^5} + {(\log x)^4}\)Step 1:

When\(x > 0\)we have property\(\log x \le x\)and when\(x > 1\)we have property\({x^4} < {x^5}\)

From above \(n = 5\) as the largest power of \(x\)in definition of \(f(x)\)is smallest \(n\) for which \(f(x)\)is \(O({x^n})\)

(b) Step 2:

Let\(k = 1\)

\(\begin{array}{l} \Rightarrow \left| {f(x)} \right| = \left| {3{x^5} + {{(\log x)}^4}} \right| \le \left| {3{x^5} + {{(\log x)}^4}} \right|\\ \le \left| {3{x^5}} \right| + \left| {{{(\log x)}^4}} \right|\\ = 3{x^5} + {(\log x)^4}\end{array}\)

\(\begin{array}{l} \le 3{x^5} + {x^4}\\ < 3{x^5} + {x^5}\\ = 4{x^5}\\ = 4\left| {{x^5}} \right|\end{array}\)

\( \Rightarrow C = 4\) and \(k = 1\)

Subpart (c): \(f(x) = ({x^4} + {x^2} + 1)/({x^4} + 1)\)Step 1:

Using long division method in\(f(x) = ({x^4} + {x^2} + 1)/({x^4} + 1)\)we get

Remainder\({x^2}\)

We can write the fraction as

\(f(x) = \frac{{{x^4} + {x^2} + 1}}{{{x^4} + 1}}\)as\(1 + \frac{{{x^2}}}{{{x^4} + 1}}\)

Here, quotient is\(1\)

\( \Rightarrow \)Power of\(x = 0\)

Therefore \(n = 0\) as power of \(x\) of quotient is smallest \(n\) for which \(f(x)\)is \(O({x^n})\)

(c) Step 2:

Let\(k = 0\)

\(\begin{array}{l} \Rightarrow \left| {f(x)} \right| = \left| {1 + \frac{{{x^2}}}{{{x^4} + 1}}} \right| \le \left| 1 \right| + \left| {\frac{{{x^2}}}{{{x^4} + 1}}} \right|\\ = 1 + \frac{{{x^2}}}{{{x^4} + 1}}\\ \le 1 + 1\end{array}\)

\( \Rightarrow C = 2\) and \(k = 0\)

Subpart (d): \(f(x) = ({x^3} + 5\log x)/({x^4} + 1)\)Step 1:

We know, when\(x > 0\)we have property\(\log x \le x\)

\( \Rightarrow \left| {f(x)} \right| = \left| {{x^3} + \frac{{5\log x}}{{{x^4} + 1}}} \right| \le \left| {\frac{{{x^3} + 5x}}{{{x^4} + 1}}} \right|\)

Here,\(n\)will be equal to the power of\(x\)of ratio of largest terms in numerator and denominator i.e.,\(\frac{{{x^3}}}{{{x^4}}}\)

\( = \frac{1}{x}\)

\( = {x^{ - 1}}\)

\( \Rightarrow n = - 1\)

(d) Step 2:

Let\(k = 1\)

\( \Rightarrow \left| {f(x)} \right| = \left| {{x^3} + \frac{{5\log x}}{{{x^4} + 1}}} \right| \le \left| {\frac{{{x^3} + 5x}}{{{x^4} + 1}}} \right|\)

\( \Rightarrow C = 6\) and \(k = 1\)