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Topic: Differentiation Disaster (Read 35199 times) |
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-D-
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Differentiation Disaster
It's been way too long since I took Calculus. This problem is killing me.
I think the problem is between converting x*x -> x +x + x (x-times). This method of evaluation would work for constants (eg: 3x -> x + x + x).
But I don't really know, anyone have ideas?
-D-
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| « Last Edit: Aug 28th, 2003, 5:20pm by Icarus » |
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guest
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Re: HARD: Differentiation
« Reply #1 on: Jul 25th, 2002, 4:39pm » |
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One way to look at it would be :
From the first principle of differentiation,
let delxsum=(x+delx)+(x+delx)+... repeated (x+delx) times
let xsum = x+x+x.... repeated x times
then we have
d(x*x)/dx = lim delx -> 0 (delxsum-xsum)/delx
In other words, the "x number of times" operator is a function of x and hence we can't just treat that as a constant while differentiating. We should differentiate this "number of times" operator also.
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David
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Re: HARD: Differentiation
« Reply #2 on: Jul 25th, 2002, 9:24pm » |
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Yeah - the trick is that there is no such thing as just saying x+x+x...(x many times) - this is a function thus the chain rule must be applied. The error would be much more apparent if you write out the summation in big sigma notation.
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Bojinov
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Re: HARD: Differentiation
« Reply #3 on: Jul 26th, 2002, 2:30pm » |
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The point you guys are missing is that the x-es themselves
don't matter when you do the differentiation.
First of all, you can only say "x+x+...+x (x times)" if
"x" is an integer. If it is not, let's say x=y+d, where y
is an integer, and 0<d<1. Then x*x is "(y+d)*(y+d)" and
if you write it out, you will see it is equal to
"y+...+y (y times) +2dy+d^2". Now, the point is that when
you differentiate, the "y+...+y" term disappears. So in this
case, as with calculus in general, the small terms happen to
be the ones that really matter.
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william wu
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Re: HARD: Differentiation
« Reply #4 on: Jul 26th, 2002, 6:15pm » |
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Another observation is that the x2 = x + x + x ... transformation fails if x < 0, even if x is an integer.
But that's kind of minor; I would just say that the key observation is that x2 can only transformed into such a sum if x is a positive integer. While it is true that
42 = 4 + 4 + 4 + 4
what if x = 2.34661? Then you write
2.342 = 2.34 + 2.34 + uh oh (you can't write .34661 of a number)
By applying this transformation, my function is no longer defined on a continuous domain, but on a discontinuous, discrete domain, with isolated points at the positive integer values.
Think of the derivative operator as simply returning a slope: change in y / change in x. Now let's say that K, A, and B are points in the domain of some function f, where A<K<B. If you want to find the slope of f at K, you can approximate it by finding the slope between two points A and B. The closer A and B are together, the more accurate your approximation becomes. Eventually, when A and B are an infinitesimally small distance apart, you have the actual slope at K. Now, if my function is discrete, I can't evaluate such limits because no two points are an infinitesimally small distance apart from each other! All defined points are one integer unit apart.
Bottom line: Differentiation is not defined for discontinuities.
P.S. I stumped a mathematics Master's degree student with this riddle once. It was quite amusing to watch him suffer. His whole universe was temporarily shattered by four lines of pencil scribble. 0wn4g3! 
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NickH
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Re: HARD: Differentiation
« Reply #5 on: Jul 27th, 2002, 4:09am » |
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I don't think William's response quite gets to the heart of the matter. It's true, of course, that differentiation is not defined for a discontinuous function. Consider, though, what happens if we extend the additive notation to cover positive real x.
For example, if x = 2.4, we write x + x + 0.4x. If x = 2.5, we write x + x + 0.5x, and so on.
Now, having restored continuity, we can again pose the question: why is the derivative of this function at x = 2.4 not equal to 1 + 1 + 0.4? The reason, as guest says, is that we are ignoring the fact that the number of x's being added is also changing.
To make this even clearer, consider that the above extension is equivalent to the following definition: f(x) at 2.4 is defined as 2.4x, at 2.5 it is defined as 2.5x, and so on. The fallacy lies in assuming that, when we calculate from first principles the derivative at x = 2.4, f(x + deltax) = 2.4(x + deltax). The correct formulation is f(x + deltax) = (2.4 + deltax)(x + deltax).
Nick
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| « Last Edit: Jul 27th, 2002, 5:59pm by NickH » |
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william wu
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Re: HARD: Differentiation
« Reply #6 on: Jul 27th, 2002, 4:33am » |
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I now see that I never truly understood why the proof was fallacious. Indeed, my post did not get to the heart of the matter at all ... it's funny, prior to reading this thread, I had thought I really understood the theory behind this riddle. But now I believe do. Many thanks!
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phil m
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Re: HARD: Differentiation
« Reply #7 on: Aug 8th, 2002, 10:08am » |
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x2 = sum(0 to x) [ x] - sum(0 to 0) [ x]
= x*(sum(0 to x) [1] - sum(0 to 0) [1])
= x*(sum(0 to x) [1] - 1).
since sum(0 to 0) [z] = z.
note that
x*( sum(0 to x) [1] -1)
= x*sign(x) * (sum(0 to abs(x)) [1] -1)
= abs(x) * (sum(0 to abs(x)) [1] -1)
This takes care of x<0.
also, if abs(x) is non-integer, and n is the max integer<abs(x), then
sum(0 to abs(x)) [1] -1
= {sum(0 to n) [1] + [abs(x)-n]} -1
= abs(x),
taking care of abs(x) is non-integer.
so, x2 = x* (sum(0 to x) [1] -1)
d/dx(x * (sum(0 to x) [1] -1))
= d/dx(x) *(sum(0 to x) [1] -1) +
x*(sum(0 to d/dx(x)) [1] -1)
= (sum(0 to x) [1] -1) +x*(sum(0 to 1) [1] -1)
= x + x*(2-1)
=2x
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Mongolian_Beef
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Re: HARD: Differentiation
« Reply #8 on: Aug 13th, 2002, 10:43pm » |
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i think the problem is that x+x+x... x number of times cannot be equated to x^2. im not sure how you have a negative number of times but furthermore i think that the reason differentiation gets so screwy is that x number of times in itself is like a minifunction so you have to apply chain rule. and how would you take the derivative of something like that anyway? perhaps i just dont know enough calculus but the fallacy really appears to be (if not equating the functions)equating the derivative of two functions simply because the functions produce the same output.
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Ken Plochinski
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Re: HARD: Differentiation
« Reply #9 on: Aug 16th, 2002, 8:11am » |
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Here's an explanation which works if x is a positive number that is not an integer.
If we use the notation:
[x] = the integer part of x, and {x} = the fractional part of x
(So e.g., [4.2]=4, and {4.2}=0.2)
Then we can write
x2 = x * ([x] + {x}) = (x + x + . . . + x) + x*{x} (where there are [x] terms in the first sum).
Taking derivatives, using the product rule for the last term, and noting that the derivative of {x} is 1 (except where x is integral in which case the derivative does not exist), the derivative of the right hand side is:
(1 + 1 + . . . + 1) + (1*{x} + x*1) (with [x] 1's in the first sum)
which is
[x] + {x} + x = x + x = 2x.
It's interesting that "most" of the derivative is in that last fractional term.
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zarathustra
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Re: HARD: Differentiation
« Reply #10 on: Aug 24th, 2002, 8:58pm » |
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how about presenting the problem in an even simpler way
d/dx[ x ] = 1
but...
d/dx[ x ] = d/dx[ 1 + 1 + 1 + ... (x times) ]
= d/dx[1] + d/dx[1] + d/dx[1] ... (x times)
= 0 + 0 + 0 + ... (x times)
= 0
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Mukul Joshi
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Re: HARD: Differentiation
« Reply #11 on: Aug 28th, 2002, 4:00am » |
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The problem is that "X times" is not independent of x.
You say 5x = x + x + x + x + x
It will work.
But x times is not being accounted during differentiation.
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Mukul
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Re: HARD: Differentiation
« Reply #12 on: Aug 29th, 2002, 10:23pm » |
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Is x + x + x + ... ( x times ) continuous? 
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local
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Re: HARD: Differentiation
« Reply #13 on: Nov 17th, 2002, 3:59am » |
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hi
my english is not pro,
and i'm not the real expert at this things but i think i know whats the problem.. x is not accounted during deriving.. thats for sure..
but lets look problem more algebraicly (algebra?)
derivation is linear transformation.. we all know derivating rules.. lambda - a scalar(number, skalar?) i will use the letter l for lambda, f - function, ' - derivation - lin. transf. is defined by:
(l*f)' = l*(f)' and
(f+g)' = f' + g'
and when it comes to derivating, the lambda (skalar) is understood as a constant, and we can move these constants out of the derivation..
d/dx[x] + d/dx[x] + d/dx[x] ... (x times)
if we sum up these.. we get (x)* d/dx[x] ... so x is a constant..and u cannot move x inside .. to get
d/dx[x^2] !! u change function..
u cannot move objects from different algebraic structures..
function is function, scalars are scalars.. (i dont know how u call it)
the funtion x^2 is not the same as x times x (first x is a constant, and we can rename it to k)..
so in 1 case u actualy derive x^2, in other case u derive kx,
em. so when doing this..
d/dx [x^2] = x d/dx[x].. this is wrong.. (1st rule)
u can only move out the constants of derivation, but x is not a constant..
because + doesnt have same em, features? as *
(f + g)' = f' +g' but,
(f*g)' =! f' * g' (=! - not equal), we all know
(f*g)' = f(g)' + (f)'g
d/dx[x^2] = d/dx[x*x] = x(x)' + (x)'x = x+x = 2x
cu
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dogfriend_ltk
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Re: HARD: Differentiation
« Reply #14 on: Jan 30th, 2003, 2:06am » |
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I've the initial riddle and the answers above but still some questions have come to mind:
Ok, so basically there are two problems : the function that takes x as a parameter and returns x + x + ... + x (x times) can only be defined on N (unsigned integers). But one can define the limit of a function in a point if and only if that point is a point of accumulation for the function's domain [x point of accumulation for D <=> (for any r>0, (x-r , x+r)\{x} intersected with D != void)]. But the only accumulation point for N is infinity, so the problem ends.
Furthermore, if we were to "bend" the function so that it would become continous, william's formula remains bogus, as "x times" isn't taken into account in differentiation.
There is still a question worth asking. Let's define as "countable" a domain D with the property that one can define a "one-to-one" function between D and N (or Z, it's the same). Let x be a member of D. Can one define the limit of a function f : D-> R in x? What about differentiability?
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Icarus
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Re: HARD: Differentiation
« Reply #15 on: Jan 30th, 2003, 6:33pm » |
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on Jan 30th, 2003, 2:06am, dogfriend_ltk wrote:| Let's define as "countable" a domain D with the property that one can define a "one-to-one" function between D and N (or Z, it's the same). Let x be a member of D. Can one define the limit of a function f : D-> R in x? What about differentiability? |
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Easily. After all the set of rational numbers is also countable.
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dogfriend_ltk
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Re: HARD: Differentiation
« Reply #16 on: Feb 3rd, 2003, 12:55pm » |
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Yeah, you're kindda right...I'm afraid the way I put the problem was wrong. Try this one...
Let f : [0,1] -> [0,1] be a continous, one to one function.
Let A = { f(x) - f(y) | x, y members of [0,1]\Q }.
Determine A.
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Nerd
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Here's the solution:
2x = x + x
3x = x + x + x
x*x = x + x +x ... x times (where x in x times is constant)
therefore, d/dx (x*x) = 1*x = x
derivative of constant is constant, so that x does not change.
Nerd
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Icarus
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Re: HARD: Differentiation
« Reply #18 on: Mar 30th, 2003, 11:10am » |
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Sorry Nerd, but you haven't got it. "Constant" is a relative term. It is defined in situations where you have multiple variables which are not independent of each other. To say that a particular variable is "constant" in such a situation means that it's value does not change when then values of the other variables are changed.
In this case the only variable around is x. And since the differentiation is with respect to x, x changes. The idea that "x is constant with respect to x" is nonsensical.
Besides which, d(x*x)/dx = 2x, not x.
The problem here, as you will see if you read through the thread, is that the puzzle tries to mix treating x as a discrete variable, defined only for integers (in the formula x*x = x + x + x +...+x x times) with treating x as a continuous variable, defined on the whole real numbers (in the differentiation).
If you extend the concept of "adding x times" in the only reasonable fashion to include non-integer values of x, then you get the correct formula for d(x*x)/dx.
If you instead define a discrete version D of d/dx, you discover that for it the formula D(x*x) = x is correct.
It is only when you try to mix discrete and continuous in the same formula that you get garbage.
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nerde
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Re: HARD: Differentiation
« Reply #19 on: Mar 30th, 2003, 2:58pm » |
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Hey,
Thats what I am saying... d (x(variable)*x(constant))/dx
is x(constant).d/dx (x).
i.e x
1 + 1 + 1..n times = 1*n (here n is a variable and 1 is constant).
similarily 1 + 1 + 1..x times = 1*x.
and x + x + x + .. x times = x * x.
1st x is constant and second x is not.
I hope I cud have been a bit more clear.
nerd.
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Icarus
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Re: HARD: Differentiation
« Reply #20 on: Mar 30th, 2003, 8:17pm » |
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Please read my whole post and not just the last 2 lines. You cannot have 1 x constant and the other not! They are the same thing! When you vary the value of x, EVERY instance of x changes - not just the ones that are convenient! Also
d(x*x)/dx = 2x.
IT DOES NOT = x, EVER!
I wrote an equation saying you could define a discrete equivalent operation D so that D(x*x) = x, but there are two things you should note: (1) D is NOT the same thing as d/dx. The latter is not even definable for discrete variables. (2) My equation was wrong. I was misremembering something from the calculus of finite differences, and it was only after I posted that I really thought about it and realized my mistake. There are actually three such operators D, the formulas for them are:
Forward Difference: D(x*x) = 2x + 1
Backward Difference: D(x*x) = 2x - 1
Middle Difference: D(x*x) = 2x
(The middle difference is the average of the forward and backward differences.)
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Ahmed
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Re: HARD: Differentiation
« Reply #21 on: May 5th, 2003, 1:36pm » |
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Hum. isn't Nerd right though, the fact that we are treating one of the x*x as a constent is where the error comes from?
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ThudnBlunder
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Re: HARD: Differentiation
« Reply #22 on: May 5th, 2003, 2:04pm » |
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on Mar 30th, 2003, 8:17pm, Icarus wrote: d(x*x)/dx = 2x.
IT DOES NOT = x, EVER!
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Yeah, I think Icarus was a bit hard on Nerd there.
He didn't even allow that d(x*x)/dx = x when x = 0 
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| « Last Edit: Jun 24th, 2006, 1:43pm by ThudnBlunder » |
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davut
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Re: HARD: Differentiation
« Reply #23 on: Aug 16th, 2003, 7:18pm » |
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what's wrong is the following:
d/dx[ x + x + x + ... (x times) ]
IS NOT EQUAL TO
d/dx[x] + d/dx[x] + d/dx[x] ... (x times)
because x is a variable. let's write the first equation as;
to=x
d/dx[ x + x + x + ... (x times) ] = d/dx[SUM x ]
from=1
to=x
which is NOT equal to SUM 1
from=1
because the upper limit of the sum is not a constant but the x itself.
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davut
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Re: HARD: Differentiation
« Reply #24 on: Aug 16th, 2003, 7:22pm » |
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somehow the formatting in my previous post was messed up. what i meant by the sums and limits are the following.
to=x
SUM x
from=1
and
to=x
SUM 1
from=1
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