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April 9th, 2015, 10:06 AM   #1
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Question Error in interpreting differentiation.

$ x^{2}-1=k *\mathcal{O}(5^{f(x)}) $, k is a constant, $f(x)$ is a well-defined 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.
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April 9th, 2015, 10:17 AM   #2
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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.
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April 9th, 2015, 11:13 AM   #3
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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.
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April 9th, 2015, 12:03 PM   #4
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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.
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April 9th, 2015, 06:52 PM   #5
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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.
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April 10th, 2015, 06:37 AM   #6
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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:
change rate of $\mathcal{O}(5^{f(x)}) $ is much bigger than $ x^{2}-1$
is conceptually correct (for certain functions $f(x)$), but the formal development of the subject is not usually expressed in that way.
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April 11th, 2015, 12:33 PM   #7
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Quote:
Originally Posted by v8archie View Post
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.
well I mean, there can not be infinite niumber of x as a solution to the given equation.
Earlier you wrote-
Quote:
Originally Posted by v8archie View Post
What if $f(x) = {2 \over \log 5}\log x$? Then $$5^{f(x)} = \mathrm e^{f(x) \log 5} = \mathrm e^{2 \log x} = x^2$$
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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