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April 9th, 2015, 10:06 AM  #1 
Member Joined: Nov 2013 Posts: 62 Thanks: 0  Error in interpreting differentiation.
$ x^{2}1=k *\mathcal{O}(5^{f(x)}) $, k is a constant, $f(x)$ is a welldefined function, $\mathcal{O}$ is Big O notation. Above equation can not have finite solution, because change rate of $\mathcal{O}(5^{f(x)}) $ is much bigger than $ x^{2}1$, $\frac{d}{dx}x^{2} <<\frac{d}{dx} \mathcal{O}(5^{f(x)}) $ So, after certain value of x, the equation will not hold. 0. Is it a correct argument?? If not, then  1. Why the above argument is not correct? 2. What is the proper way to prove above statement? **whatever you think , leave a comment. If you are running short of time, simple "yes","no" will suffice. Last edited by skipjack; April 9th, 2015 at 08:02 PM. 
April 9th, 2015, 10:17 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra 
What if $f(x) = {2 \over \log 5}\log x$? Then $$5^{f(x)} = \mathrm e^{f(x) \log 5} = \mathrm e^{2 \log x} = x^2$$
Last edited by v8archie; April 9th, 2015 at 10:23 AM. 
April 9th, 2015, 11:13 AM  #3 
Member Joined: Nov 2013 Posts: 62 Thanks: 0 
In that case, the argument would not work, but that was not my intention. Say, it is not what you have described, say f(x) is always an integer, can I use above argument?? In general, can I use the argument (at least) for some cases?? Last edited by skipjack; April 9th, 2015 at 08:03 PM. 
April 9th, 2015, 12:03 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra 
If you have functions $f(x)$ and $g(x)$ and $f(a) \gt g(a)$ for some $a \in \mathbb R$ and $f'(x) \gt g'(x)$ for all $x \gt a$, then $f(x) \ne g(x)$ for all $x \ge a$. But this says nothing about what happens for $x \lt a$. For example: $x^2  1 = 5^x$ does have a finite solution at $x = 1.08$, but the reasoning above shows that there are no solutions for $x \gt 1$. (And a modified version would show that there is no solution for $x \lt 1.1$). The notation $\mathcal O(f(x))$ talks about the rate of growth of $f(x)$ for large $x$. So you when you use the notation you are not talking about small or negative $x$. Last edited by v8archie; April 9th, 2015 at 12:06 PM. 
April 9th, 2015, 06:52 PM  #5 
Member Joined: Nov 2013 Posts: 62 Thanks: 0 
I mistyped!! I wanted to write x would not have infinite solution (using above argument), please confirm I can say that and above argument is correct (not rigorous, but ok). Thanks for your time. Last edited by skipjack; April 9th, 2015 at 08:04 PM. 
April 10th, 2015, 06:37 AM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra 
Normally when we write an equation in $x$ it is understood that $x$ is real (or perhaps complex). In standard analysis, there is no $x = \infty$ so the concept of an infinite solution is not defined. The analysis in my post #4 above still holds though and appears to reflect the idea you are considering. The idea of "infinite solutions" would be related to Asymptotic Equivalence which is a concept of Asymptotic Analysis. Your comment that Quote:
 
April 11th, 2015, 12:33 PM  #7  
Member Joined: Nov 2013 Posts: 62 Thanks: 0  Quote:
Earlier you wrote if $f(x) = {2 \over \log 5}\log x$ then f(x) is never an integer after a certain x(eg, x=5), the problem I am trying to solve provides intger f(x) for some x, so , even in your case, the above equation will have finite solution. Last edited by jim198810; April 11th, 2015 at 01:05 PM.  

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differentiation, error, interpreting, number theory., real analysis 
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