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I am working through some problems and I have come across one I do not understand. Could someone clarify why

$$2x^3 + 3x^2\log(x) + 7x + 1$$

is $O(x^3\log(x))$ for $x>0$?

I guess I am missing some sort of knowledge to be able to answer this question :(

lulu
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decipher
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    https://rob-bell.net/2009/06/a-beginners-guide-to-big-o-notation/ may help. –  Sep 24 '17 at 12:38
  • Welcome to Math SX! Well, for me it's $O(x^3)$, not $O(x^3\log x)$. – Bernard Sep 24 '17 at 12:46
  • I have the problem set in both answered and non-answered version using the answered as feedback when I have gone through a whole exercise. I was thinking the same as you, and also answered so when I was working my way through it, but when I looked to compare, I apparently got it wrong. No idea why :(

    Edit: Thank you! :D

     – decipher Sep 24 '17 at 12:49
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    Perhaps the second term ought to have read $3x^3\log (x)$ ? – lulu Sep 24 '17 at 13:07
  • @lulu I guess I'll have to assume that is the case. I can't find any logarithm rules that would make it true. – decipher Sep 24 '17 at 13:55

1 Answers1

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if f(x) = $2x^3 + 3x^2\log(x) + 7x + 1$, f(x) = $x^3$$(2 + \frac{3\log(x)}{x} + \frac{7}{x^2} + \frac{1}{x^3})$. Then if x approaches the infinity, ($\frac{3\log(x)}{x} + \frac{7}{x^2} + \frac{1}{x^3}$) is negligible comparing to $x^3$.

And, if $f(x)$ = $2x^3 + 3x^3\log(x) + 7x + 1$, f(x) = $x^3log(x)$$(\frac{2}{log(x)} + 3 + \frac{7}{x^2} + \frac{1}{x^3})$. Then if x approaches the infinity, ($3+ \frac{7}{x^2} + \frac{1}{x^3}$) is negligible comparing to $x^3log(x)$.

kimi Tanaka
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