???Code:1=0
Let x = 1
Then, x^2=x
x^2-1 = x-1
Dividing both sides by (x-1), we get
x+1 = 1
x+1-1 = 0
Since x=1,
1=0
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???Code:1=0
Let x = 1
Then, x^2=x
x^2-1 = x-1
Dividing both sides by (x-1), we get
x+1 = 1
x+1-1 = 0
Since x=1,
1=0
so:Quote:
x^2-1 = x-1
X^2=X
So what you're doing is dividing 0/0 and this can be anything. For example: 0/0 = 27.43 just because 0 = 0*27.43Quote:
Originally Posted by thegreatone
becaus x-1=0
Again the 1=0 proofs :) I saw a cool Calculus one a while back, lemme see if I can remember it
12=1
22=2+2=4
32=3+3+3=9
x2=x+x+x...[x times]
d (x2) = 2x
dx
d (x+x+x...[x times]) = 1+1+1... [x times] = x.
dx
2x = x, defined at the least for all nonnegative integers x. Where's the error? (I actually don't know concretely myself)
For the calculus one:
d(x + x + x + ... + x) [x times]
dx
this is the same as:
d (cx)
dx
where c is a constant. As when x changes, the number of x's in the addition is still the same. It's not the same as x^2, except for the value at x = c.
Is this inspired by the recently-dredged-up "2=1" post from the pre-Jurassic thread, c/o bver?
Shall we just put a "Links to 2=1, 0.9r =1 etc" sticky at the top of this forum? That way everybody can help themselves to another nauseating spoonful every couple of years when they have forgotten the last set of proofs or when yet another smart 11 year old noob joins VBF.
I have been up for about 36hrs now. Top tips: Don't smoke, don't drink and drive, and don't fly BWIA and post. Take no offence, anybody.
VB Code:
Forgive me if I'm wrong. But isn't differentiation only valid when the function is continuous?Quote:
Originally Posted by jemidiah
f(x) = x2 above is clearly discrete.
I'm not sure what you mean by discrete. By x2 is a continuous function.
Consider the domain of f(x) = x<sup>2</sup> in question:
1<sup>2</sup> = 1
2<sup>2</sup> = 2 + 2
...
x<sup>2</sup> = x + x + ... + x
it only contains positive integers, which makes it a discrete function.
When you graph it, you will see dots and dots, not connected curve.
Differentiation determines the gradient of the tangent at any particular point on the graph. There is no tangent to a discrete graph (tangent to a point?), and it's meaningless to ask for one.
So by differentiating f(x) = x<sup>2</sup> in question, you're not getting anywhere.
Hope that is right. :p
I'm kinda new to a lot of the stuff you are mentioning so please bare with me...
How is 22 = 2 + 2... isn't it 2 * 2?
Also you said there is no tangent to a discrete graph. Is that a well-known fact that I just don't know about.
I guess if you define a discrete function as only graphing points (is that what it is), then you are correct about differentiating a point. It wouldn't make any sense or work. I just don't see why x2 is discrete (just points?), unless you choose to define it that way for the currenct function. In which case you are bringing the lack of continiuity upon yourself by defining it that way.
Maybe someone can clear up where I am going wrong here?
eyeR, don't get into these threads. Get out now while you still have a chance.
2^2 = 2*2 = 2+2
3^2 = 3*3 = 3+3+3
4^2 = 4*4 = 4+4+4+4 etc.
What lambchopz is saying, I believe, is that this function is defined only for positive integers in the case of the "cool calculus "puzzle"" given above. Clearly the function y=x^2 is continuous, generally. However, if in the "puzzle" given above you were to try putting in 1.2^2, you would not be able to define this as x+x+... because you can't add 1.2 to itself 1.2 times. Hence the whole argument falls down because the approximation has implicitly been made that the function only applies to integers, whilst also claiming that it is differentiable. I think you seem to have the right idea.
sql_lall is along the right track - the function x^2 is not the same as the function nx, except at the point n=x. So you can't equate the function x^2 with x+x+x+... because this is not the same function for all values of x, so all your points on the graph belong to different functions. Clearly the way to define the function in terms of the sum is to set n=x, at which point we have
x*x = x^2, exactly what we wanted.
I hope that clears things up a bit...
zaza
I see now. Thanks zaza. I still feel like the problem isn't that logical just because you are putting so many limitations on the function (only positive integers) that it obviously won't be differentiable. Thats just my 2 cents and you guys are way above me to begin with so don't mind me. ;)
After reading though these posts, I think if we simply go back to the limit definition of the derivative, we'll see why d(x2)/dx can't be defined as I've defined it.
f(x) = Sum[i=1..x, x]
lim h-->0 (f(x + h)-f(x))/h will get to a closer and closer approximation for smaller and smaller values of h which get ever closer to 0. However, the only values for which h is defined are when h is an integer; the basic value that any limit describes is not fulfilled because the function is not continuous for a finite distance around the point, so the limit itself can't exist in the current form.
However :)
I got to thinking about this, and thought I'd define f(x) for rational numbers as well.
1.22 = ?
(1+2/10)2=
(A+B/10)2=A2+AB/5+B2/100 = A+A+A...A times + B+B+B...B times/100 + A+A+A...B times / 5
Yay, it works!
1+2/5+4/100 = 1.44 = 1.22
So, let's redefine the function to be
f(x) = f(A, B) = Sum[i=1..A, A]+Sum[i=1..B, B]/100+Sum[i=1..A, B]/5
Granted, this doesn't change the invalidity of differentiating f(x), but it makes it more difficult to spot the mistake :)
d/dx (f(x)) = d/dA (f(A, B)) + d/dB (f(A, B)) 'Note: partial derivatives there
= 2A + B/5 + 2B + A/5 with the A2 version, and
= (1+1+1... A times) + (1+1+1...B times)/5 + (1+1+1...B times)/100 = A+B/5+B/100 {or A+A/5+B/100 if you feel like it}. These two equations should be equal, and the function is defined at all rational numbers. They of course are not equal to each other, and math is wrong yet again (more or less)!
Note: I may have taken d/dx in terms of A and B wrong, but in any case, they should be equal and they won't be.
Now all that's left is to define f for irrationals, yay!