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/sci/ - Science & Math

File: sin.gif (16 KB, 411x411)
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This fucking piecewise function

[eqn]f(x)=x*sin( \frac{1}{x} ), x \neq 0[/eqn]
[eqn]f(x)=0, x=0[/eqn]

has been taunting me for weeks. It's obviously continuous everywhere but at 0, and at 0 the limit is 0 and the value is 0 so it's continuous there, but it seems to have properties that continuous functions shouldn't. When you try to draw it starting, say, 0.01 away from 0, there are an infinite number of wiggles between you and 0. If you finally cross 0, are you going up or down? Either way seems impossible, because you still have an infinite number of wiggles left

>>
Fucked up the latex, haven't used it before. Here's the function
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>>10058033
the value isn't 0, its undefined. therefore its not continuous retard
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>>10058039
I defined a piecewise function where f(x)=0 when x=0. At all other values, f(x)=x*sin(1/x)
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This is one of those times where thinking of continuity as "you can draw it without picking up your hand" is not useful.
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It's not differentiable at 0, but everywhere else. That's whay makes it look odd. You can construct a lot of counterexamples with this as your basis.
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>>10058033
∀ε>0∃δ s.t. |x-0|<δ → |f(x)-0|<ε

since |sin(1/x)|≤1, let δ=ε∎
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>>10058033
>>10058036
That piecewise function is continuous everywhere, including at 0
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>>10058033

think of cos(1/x)
just like your equation, the values don't really approach zero. They don't approach anything since trig functions oscillate forever.

the limits from the left and right don't exist, and neither does the limit @ 0
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>>10058033
The "draw it without picking up your pencil" definition of continuous is only good for teaching dumb freshmen how to pass Calc 1. Continuous functions are actually pretty bizarre and counter-intuitive compared to the analytic almost everywhere functions everyone's familiar with.
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>>10058033
it can be continuos whithout you knowing if its "going up and down" (its irrelevant), many functions behave like that: Weierstrass function,Blancmange function etc
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>>10058033
>If you finally cross 0, are you going up or down
So the derivative? Undefined dummy
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>>10058033
>t. Poincaré
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File: 1539051885926.gif (5 KB, 411x411)
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>>10058033
Optimized.
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>>10058139
Optimized.
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Its has a limit everwhhere, and that limit is equal to the function, therefore its continuous. If sin x is continuous to infinity, then the only issue is at 1/0 which you removed. Simple.
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f(x) ={x>0 : x, x<=0:-x}
definition of derivative from limit from 0- = -1, limit from 0+ = 1
lefthand limit doesn't equal righthand, limit isn't defined, derivative isn't defined at 0

same thing applies here
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>>10058183
The left-hand limit does equal the right-hand limit in OP's equation; they're both 0. You can sandwich it with $y= \mid x \mid$ and $y= - \mid x \mid$ . The former is always larger than the equation, the latter always smaller, and both of them go to 0 at 0, so the limit of the original equation has to as well.
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>>10058144
btfo
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>>10058064
Anything fitting that intuitive understanding is likely smooth (C^inf)
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derivative isn't defined at 0, differentiability only implies continuousness
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the height of the wiggles can be bounded by bounding how close you are to 0.
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>>10058033
You're right in your observations, but wrong in your presuppositions about what properties continuous functions must have. This function is an excellent example of why continuity is a necessary but not sufficient condition for differentiability. That is, although the function is continuous at zero, the derivative is undefined. You can show this graphically or analytically. Look at the first chapter of Advanced Calculus by Fredrick Woods for more info, he discusses almost this exact function.
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>>10058129
underrated

Continuity is not a very strong property OP, there are many pathological functions that are continuous and extremely ugly/impossible to draw with a pen.

The devil's staircase and the weirstrass function are two examples.
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>>10059674
If u substitute the $x$ with $x^2$ you get it to be $C^1$, with 0 derivative at 0 and still have the infinity wiggle around the origin XD
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>>10059921
>>10059674

Hell, you can even get it to be $C^\infty$ by substituting the $x$ with $e^{-\frac{1}{x^2}}$ and still have the infinity wiggle, but at that point it seem so smoot it don't surprise anyone anymore
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>>10058882
This is not true at all. Any function with countably many cusps easily fits that intuitive rule
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>>10058033
Isnt this just a transformation of the sinc function? let z=1/x, xsin(1/x) = sin(z)/z should have the same limits as the original, with 0 and infinity flipped.
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>>10059938
d/dx [f(x^2)] is still undefined, and e^-(x^2) never reaches zero, so the composition doesn't apply. In fact, 1 dimensional limits are path independent.
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>>10058033
Hows calc 1 going you little product rule specialist
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>>10060820
???
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>>10058039
Are you baiting, illiterate or clinically braindead?
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>>10058083
>the limits from the left and right don't exist
How full of shit can you be?
For any ε > 0, there exists a δ > 0 such that for any x: |x| < δ the following is true: | f(x) - 0 | < ε. Just take δ = ε, since sin(x) never exceeds 1 and x approaches 0. And since the function is defined such that f(0)=0, it approaches its limit at 0, so is continuous.
Go take calc 1 or never post here again
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>>10058882
>you can't draw |x| without picking your hand up
It's a rule of thumb for continuity, but definitely not for smoothness
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>>10058033
Just intuitively: the limit of 1/x as x approaches 0 is infinity. Therefore the limit of sin(1/x) as x approaches 0 is equal to the limit of sin(n) as n approaches infinity.

The limit of Sin(x) as x approaches infinity is undefined (it's bounded but not convergent). Therefore, the limit of x*Sin(1/x) does not exist.

The issue your are having is mistaking boundedness for convergence.
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>>10061834
>Therefore, the limit of x*Sin(1/x) does not exist.
But it does. It will always be less than the absolute value of x, and always be more than the negative absolute value of x. Both of those limits at 0 are 0, so x*sin(1/x) gets squeezed and has to be 0. You know the squeeze theorem, right?
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>>10061840
*less than or equal to, and greater than or equal to
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>>10061834
You sir are absolutely retarded. Just stop posting.
For an explanation, see
>>10061827
>>10061840
>>10058078
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>>10061840
>squeeze theorem
Big homosexual, just use the definition for this one.
>>10058033
>I need to imagine things to do maths
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>>10058033
It's non-differentiable at 0 but continuous. That's why the derivative (see: "are you going up or down?") acts wacky.
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>>10061833
Retard, the point he was trying to make was that our intuition is of functions that are at least smooth almost everywhere, which is the reason why the jordan curve theorem for smooth curves was proved like 50 years before that for continuous curves, retard
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>>10058139
based
>>10058144
double based
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>>10058033
>If you finally cross 0, are you going up or down? Either way seems impossible, because you still have an infinite number of wiggles left
>If you finally cross 0
You don't.
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>>10058033
Reminds me to this
http://www.ams.org/journals/proc/1973-038-02/S0002-9939-1973-0310824-0/S0002-9939-1973-0310824-0.pdf
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>>10062029
Thanks anon. That makes sense.
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Check out the book Elementary Analysis: The Theory of Calculus, by Kenneth A. Ross. He talks about this function
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This is high school analysis. The rule for continuity in R to R functions is that the function is continuous in a point x=a if it is defined (this one is), its limit exists (it does) and both the function and its limit in that point are equal.

Trying to calculate the limit seems tricky because 1/x goes to infinity when x is close to zero, and the sine of something very big will oscillate between -1 and 1 very wildly. But it will only oscillate in that range, so whatever the value the sine takes, its absolute value isn't gonna be bigger than 1. On the other side, you have an x multiplying that sin(1/x), so the limit of x*sin(1/x) when x goes to zero must be zero, for the sine is bounded and you're multiplying it times a really small value. The limit of the function when x goes to zero EXISTS and is equal to zero.

Therefore, since the limit exists and is equal to the value of the function in x=0, the function is continuous at x=0. Note that all of this is just an "argument" I constructed using words: if you want to do it right you should write out a more compelling mathematical line of reasoning:

lim (x*sin(1/x)) ≤ lim (x) = 0, so lim x*sin(1/x)=0
If we're talking about limits when x goes to 0, of course.
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>>10059979
sin(1/x) does have countably many cusp though...
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>>10064048
Please provide a mapping onto the integers, then. I'll wait..

Alternatively, how many are there in (0,1] ?
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>>10064090
the amount of cusps is equal to the amount of minima and maxima who occur when cos(1/x)=0, and those are so when 1/x=pi/2+k*pi or x=1/(pi/2+k*pi) where k is any integer.
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>>10064090
it has at most M cusps, where M is the largest integer
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>>10060032
this guy gets it
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>>10064115
That's a pretty slick solutiom
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>>10064842
lol
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>>10058033
>trying to solve this
solve what? it's just a function. there's nothing to solve. it just is.
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>>10065309
The fact that it's continuous but has properties that don't fit the common sense definitions of continuity
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>>10058033
It's continuous, just not differentiable.
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>>10061874 He proved continuity at 0 by definition which is what OP asked for, whats your problem?
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>>10066065
...are you quoting the right guy? The post I was replying to was a guy saying it's not continuous
t. >>10061874
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>>10065317
How do they not fit the common sense definitions of continuity? The only problem with it is that the left and right limits at 0 dont match up
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>>10066196
>left and right limits at 0 dont match up
The fuck are you retard on about
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One weird function is the function f(x) = x for x rational, f(x) = 0 for x irrational.
This function is actually continuous at 0 (in a similar way to your function) even though it's discontinuous literally everywhere else.
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>The things pure mathematicians think about
Kneel before the glorious applied math master race, autists
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>>10065297
That's not slick, just obvious.
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>>10058033
Just because the limit is 0, doesn't mean it converges to 0. You're looking at something that is less and less divergent, but still diverges. However the function is Bounded, so not wildly divergent.
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>>10067845
You, sir, are another retard who didn't even study calc 1 and decided to ignore every single continuity proof ITT.

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