@IG
Still suspecting I haven't fully finished seanieg's question though. Anything else to pick up?
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
@IG
Still suspecting I haven't fully finished seanieg's question though. Anything else to pick up?
^ I think you should try invoking the IVT to prove some things I think you're trying to prove.
(Sorry about delayed response, am wrestling with a problem of my own )
Yep @leehuan, IVT, and EVT are key to rigorous argument here. You are on a nice train of thought and I think you have a good intuitive grasp of why k=2 is impossible, but making this rigorous takes some care.
The "infinitesimal" part of your proof in particular needs to be made more precise, and things like Rolle's theorem are inapplicable here because the function need not be differentiable.
From what you are saying though, it seems to me you have the correct "picture" in your head for your k=2 disproof.
(I am also unconvinced that this "picture" serves to disprove the k>2 possibility so think about explaining that a little clearer).
Yes it is possible. Eg: f(x)=x*sin(x).
This is easy to make rigorous. For large x, x^3+1 < 2x^3 and x^2+1 < 2x^2. So
log(x^3+1)/log(x^2+1) < log(2x^3)/log(x^2) = (log(2)+3log(x))/2log(x).
and
log(x^3+1)/log(x^2+1) > log(x^3)/log(2x^2)=3log(x)/(log(2)+2log(x)).
Squeeze to finish.
Corners do not have to be local extrema. Eg f(x)=max(x,2x) at 0.
Also, lack of differentiability can occur in ways worse than corners. (f(x)-f(a))/(x-a) needs not tend to a limit at all as x tends to a, or it might only tend to a limit from one side.
It is more trouble than it is worth to appeal to theorems about differentiable functions and then treat the exceptional points separately. (In fact a continuous function might be differentiable nowhere!)
Last edited by seanieg89; 24 Apr 2016 at 7:45 PM.
Hmmm
With the IVT I can pretty much find how to apply it. Just get rid of that infinitesimal stuff and properly redefine it (which I might do later). Which gives me some ideas on the EVT but I haven't placed too much thought into it to know what's right and what's flawed.
May continue this when I have more free time. Gah I hate assignment based subjects
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Regarding the continuity part:
Informally, when x≥0, 0≤f(x)≤x^{3} and when x≤0, x^{3}≤f(x)≤0
But in both cases if we take limits to nought, by the squeeze theorem we have lim x->0 f(x) = 0 which incidentally enough equals f(0)
______________________
The answer given to the derivative was wrong because I didn't regard the significance of what h was.
lim x->0 [ f(0+h)-f(0) ] / h
= lim x->0 f(h)/h
but f(h) = h^{3} when h is rational
and f(h) = 0 when h is irrational
Note then that f(h)/h = h^{2} for h in Q
or = 0 for h not in Q
Haven't done questions like this one before yet but just staring at it I'm inclined to say the squeeze theorem can be used again
Yeah it's basically squeeze law. We can use that to show that f'(0) = 0.
Bachelor of Science (Advanced Mathematics) @ USYD
Lol sorry, forgot to say the exponential function was continuous even though I had it written down.
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Yeah you can move the limit inside for a continuous function. If lim as x -> a of g(x) is b and f is continuous at b, then lim as x -> a of f(g(x)) = f(b).
This is not a calculus question, but it's associated so I'll put it here.
Last edited by Paradoxica; 27 Apr 2016 at 8:04 PM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Last edited by seanieg89; 30 Apr 2016 at 12:36 PM.
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