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Topic: Differentiation Disaster (Read 35085 times) |
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anton
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Re: Differentiation Disaster
« Reply #25 on: Dec 23rd, 2004, 10:10pm » |
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Let f(x) = x + x + x + ... + x (x times) for integer x>=0
Using the definition of the derivative:
f'(x) = limh->0 (f(x+h) - f(x))/h =
= limh->0 (((x+h) + (x+h) + ...x times, since h->0... + (x+h)) - (x + x + ...x times... + x))/h =
= llimh->0 (h + h + ...x times... + h)/h =
= 1 + 1 + ...x times... + 1 = x
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Grimbal
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Re: Differentiation Disaster
« Reply #26 on: Dec 24th, 2004, 7:31am » |
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The problem is that you don't say how you treat non-integers.
If the "x times" includes fractions, you also must compute (x+h) times, and you end up with f(x) = x^2. And the derivate is 2x.
If the "x times" actually means "floor(x) times", the derivate is also floor(x), but you have a problem with a discontinuity if x is an integer.
If you replace floor(x) by round(x), then your formula is true for integers.
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anton
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Re: Differentiation Disaster
« Reply #27 on: Dec 24th, 2004, 2:12pm » |
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Yeah, I see now that there is a flaw in my argument; the mistake was in assuming that (x+h) is added "x times, since h->0". Rounding x in "x times" resolves this, but then the formula is not exactly helpful, since x*round(x) = x2 only for integers; in general the function is completely different (thus the difference of derivatives).
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SomeGuy
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Re: Differentiation Disaster
« Reply #28 on: May 19th, 2005, 8:22pm » |
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I just wanted to point out that the 4th post in this thread gave an excellent, concise reason, and that every other post proceding it has thus been rendered redundant 
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Icarus
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Re: Differentiation Disaster
« Reply #29 on: May 19th, 2005, 9:11pm » |
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You could use a brush-up on the meanings of both "preceding" and "redundant".
Since many later posts concern matters not discussed or even suggested in Bojinov's post (#4), they are hardly redundant. In the 6th post, Nick points out a more fundamental understanding of the problem than Bojinov's explanation.
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prabhakar_misra
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Re: Differentiation Disaster
« Reply #30 on: Jun 23rd, 2006, 9:43am » |
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i think what is the heart of the matter is that we forget that d/dx is the rate of change of f(x) wrt x . its basically the slope of the graph of x^2 . thus you must note that the function x^2 might be equal to x.x but slope of x^2 is not x times the slope of x because the x that you think of multiplying after finding the slope of x has to contribute to the very finding of the slope.
what i write next might not be very mathematical but it is worth a reading.
when you make a dish that involves oil you can't sprinkle the oil after the dish is made as it has a role to play in the very making of the dish .
i hope you all understand 
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Icarus
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Re: Differentiation Disaster
« Reply #31 on: Jun 25th, 2006, 12:41pm » |
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No one forgot the meaning of d/dx. We all know that the calculation is flawed, which is all your argument says. Yet the calculation appears to involve only basic rules concerning the derivative.
The question was not "is this true?". The question was: "what went wrong in the calculation that resulted in this obviously false result?"
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Vespero
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Re: Differentiation Disaster
« Reply #32 on: Aug 22nd, 2006, 7:11am » |
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Uhm...
x^2 = x*x = 'x times'(x)
and
(f(g(x)))' = f'(g(x))*g'(x)
'x times' represents the 'f' function.
Deriving x*x banally as a sum of x-es doesn't take care about composite functions' derivation.
Right?
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| « Last Edit: Aug 22nd, 2006, 7:34am by Vespero » |
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Icarus
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Re: Differentiation Disaster
« Reply #33 on: Aug 22nd, 2006, 4:26pm » |
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"banally" and "doesn't care about" are not mathematical terminologies, and so it is unclear to me what you mean by this.
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ThudnBlunder
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Re: Differentiation Disaster
« Reply #34 on: Aug 22nd, 2006, 8:21pm » |
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on Aug 22nd, 2006, 4:26pm, Icarus wrote:| "banally" and "doesn't care about" are not mathematical terminologies, and so it is unclear to me what you mean by this. |
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I have obtained a provisional decipherment:
I think he is saying that d[x*x]/dx  d[x + x + x + ...(x times)]/dx because the latter does not give you the same answer as the product rule for differentiation when applied to d[x*x]/dx. [But the reason why not seems to have been lost in transmission.]
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| « Last Edit: Feb 18th, 2007, 3:19am by ThudnBlunder » |
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Vespero
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Re: Differentiation Disaster
« Reply #35 on: Aug 23rd, 2006, 3:16am » |
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Actually did not know i had to use a precise mathematical terminology  Just tought i had to express myself with my logical and lingustic 'weapons' 
Uhm...
Got no time right now to explain clarely what i mean, since i'm @ work.
I'll try to take some moment later.
A quick but unclear description of what i mean is that if you treat x no more as a variable, but as an actual parameter ('x times' statement implies you are kinda defining x) derivation rules you aplly will authomatically keep trace of it. Sorry, it almost is an intuition more than a demonstration...
An example of it could be f(x)=nx (n being a parameter, obviously
f'(x) = f'(x+x+x+x+x+..(n times).. +x)=f'('n times'(x))=n
being f(x)=x*x and treating the first x as a parameter
we would have
f'(x)=f'(x+x+x+...x times...+x)=f'('x times'(x))=x
Notice that first x in the one above is treaten as a parameter, and no more as a function, and coherently rules of derivation will result out parameter value.
Is it more 'mathematically' comprehensible?
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| « Last Edit: Aug 23rd, 2006, 3:27am by Vespero » |
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Grimbal
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Re: Differentiation Disaster
« Reply #36 on: Aug 23rd, 2006, 8:43am » |
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on Aug 23rd, 2006, 3:16am, Vespero wrote:| Got no time right now to explain clarely what i mean, since i'm @ work. |
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You might feel better once you have noticed how much the activity on this forum slows down on week-ends.
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Icarus
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Re: Differentiation Disaster
« Reply #37 on: Aug 23rd, 2006, 6:13pm » |
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on Aug 23rd, 2006, 3:16am, Vespero wrote:| Actually did not know i had to use a precise mathematical terminology Just tought i had to express myself with my logical and lingustic 'weapons' |
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Logical and linquistic weapons are blunt and dull when no one can figure out what it is you are trying to say. Others have to be aware of the meanings you are applying with your words for the words to have power. Thus, when speaking about a mathematical conundrum, it is needful to speak in words that have a mathematical meaning known to your audience.
But it appears you have the gist of some of the problem. When you differentiate, you vary the value of x, and this means the value of all x's in the expression. However, the "x times" calculation varies the value of one x while leaving the other constant (i.e., as a "parameter").
There is more, though. The calculation also treats one x as if it were discrete (i.e., an integer) while treating the other as continuous. But this can be overcome, as NickH demonstrates on an earlier page. The key failure is the "one x is variable, the other is a parameter" problem you've described.
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LaCiTy
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Re: Differentiation Disaster
« Reply #38 on: Sep 4th, 2006, 7:18am » |
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This pseudo-riddle is a high offence to mathematics. This should not be is the hard section.
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Sjoerd Job Postmus
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Re: Differentiation Disaster
« Reply #39 on: Sep 4th, 2006, 2:02pm » |
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on Sep 4th, 2006, 7:18am, LaCiTy wrote:| This pseudo-riddle is a high offense to mathematics. This should not be is the hard section. |
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Actually, somehow it should... finding the fault in an erroneous proof isn't that easy. And, pointing the fault out is even harder. Just saying "you can't do that" is easy. Finding out why is the interesting part.
Alright, I'll see if I can think about it.
We have a function f(x) = x^2...
Now, we have another function. Let's split it up.
g(x) = x * x
h(x) = x
[g(h(x))]' = g'(h(x)) * h(x) = 2(h(x)) * 1 = 2x
So just plainly using the chain-rule doesn't bother the outcome.
Let's recite it again (just to get my mind thinking)
f(x) = x^2
= x + x + x + x + x ... (x times)
f'(x) = x' + x' + x' ... = 1 + 1 + 1 + 1 (x times)
f'(X) = x
Which is clearly erroneous, as we know the hoped-for outcome. 2x...
Problem 1: in one, x is discrete.
Solution: Add frac(x) to the equation
f(x) = x + x + x ...(* floor(x))... + frac(x)
Problem 2: One x is parameter, one x is variable
Solution: None.
We should actually write
fx(x) = x + x + x ...(* floor(x))... + frac(x)
Which raises questions... Let's call the parameter x p(arameter) instead
fp(x) = x + x + x ...(* floor(p))... + frac(p)
Which is equal to:
fp(x) = p*x
The new formulation is totally different from the x^2 one. This causes us to get another result.
fp'(x) = p
We exchanged the parameter x with p, so let's restore it.
fp=x'(x) = x
(don't know if the previous equation is legal math notation, the p=x part, or if I should just use x)
Let's re-coup the problem:
When going from x*x -> SUM(1,x) (x) + frac(x), we have to do with two types of x's. One is a parameter (re: constant, like a and b in y=ax+b) and the other is variable. Instead of looking at one function, we are looking at a set of functions, fp(x)... and are causing a lot of confusion by saying p=x...
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Icarus
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Re: Differentiation Disaster
« Reply #40 on: Sep 8th, 2006, 6:48pm » |
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on Sep 4th, 2006, 7:18am, LaCiTy wrote:| This pseudo-riddle is a high offence to mathematics. This should not be is the hard section. |
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Mathematics is not a "person" to be offended or to take offence. And as a mathematician, I do not find this offensive at all. Instead, it should be taken in the same spirit as the various 2=1 "proofs" (in which form it could also be cast). By making use of a poor conception to produce an obviously bogus result, it challenges us to figure out what is poor about the conception, and therefore to correct our understanding. Indeed, when NickH first suggested the interpretation I will reproduce below, it actually surprised William Wu and several others who never thought about the solution in this fashion.
As for being a "pseudo-puzzle", it has puzzled many, so there is nothing "pseudo" about it.
on Sep 4th, 2006, 2:02pm, Sjoerd Job Postmus wrote:Problem 2: One x is parameter, one x is variable
Solution: None. |
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No. This too has a solution, most of which you have arrived at (see NickH's post where he does the same thing).
You can treat the other x as a variable too. What you get can be expressed this way, using your notations: f(x) = fp(x)|p=x. More particularly, define the two-variable function
F(x,p) = floor(p)*x + frac(p)*x to explain what we mean by "adding p x's together" when p is not integer (note that this differs from your expression by multiplying the frac(p) by x). In addition to the floor(p) x's everyone knows adds up to floor(p)*x, we add a fraction of another x according to the size of the fractional part of p.
Then we find that f(x) = F(x,p)|p=x. To differentiate, we turn to the two-variable change rule:
df/dx = (dF/dx)(dx/dx) + (dF/dp)(dp/dx)
where the derivatives of F should be partials.
dF/dx = floor(p)*1 + frac(p)*1 = p
dF/dp = 0*x + 1*x = x
Since p=x, dx/dx = dp/dx = 1. Hence df/dx = p*1 + x*1 = p + x = x + x = 2x.
Hence the reason the problem fails is that it fails to account for the contribution of frac(p) in differentiating x + x + ... + x (p times).
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LaCiTy
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Re: Differentiation Disaster
« Reply #41 on: Sep 22nd, 2006, 7:01am » |
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I mean one can easily find more subtle appearant paradox with more precise formalisation.
ex.:
Let (Bt, t>0) be a standard brownian motion.
1) Bt is also a martingale, in particular, for all bounded stopping time T we have : E[BT] = E[Bt] = E[B0] = 0, by optional stopping theorem. (E denotes expectation of a random variable).
2) Now consider the stopping time :
T := inf { t : Bt > a }, for fixed a > 0
3) One have, almost surely BT = a, by continuity of the paths (and the fact that T is finite almost surely) and thus E[BT] = a.
4) But applying optional sampling theorem à time T yields :
E[BT] = E[B0] = 0, and thus a = 0 which is absurd.
Rem.: 1), 2), 3) are true statements.
Rem.: All the information is provided with accepted formalisation. You do not need any particular background to suspect where the problem should come in.
Modif. Ok, forget about it. Statement 4 is not true because the stopping time is not bounded, even if it is almost surely finite that's different. So optional sampling theorem does not apply. I think this would have been more simpler with a random walk. Nevermind.
What I wanted to say is that I don't like to try to explain why something does not work when the true question is why it should work. Most of time a counter-example suffices without any farfetched explanation trying to create a deep reason when it is clear that theorems or definitions have been misused.
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| « Last Edit: Sep 23rd, 2006, 4:06pm by LaCiTy » |
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Greg
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Re: Differentiation Disaster
« Reply #42 on: Feb 18th, 2007, 2:16am » |
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Let's review the statement of the riddle and point out where it went wrong. Then I'll offer an alternative to the faulty notation we'll discover.
dx2/dx = d(x1 + x2 + ... + xx)/dx
= (dx1/dx + dx2/dx + ... + dxx/dx)
= (11 + 12 + ... + 1x)
= x
- Note that I distinguish the various xes and number of xes in the proof by applying subscripts, and that the subscripts don't change the value of the xes. Also, I changed the notation of the derivative of something with respect to x. For example, d/dx[x] from the question now would now read dx/dx, which is more accurate to normal American mathematical notation.
The problem is in assuming the second line. The derivative of x2 is not
(dx1/dx + dx2/dx + ... + dxx/dx).
Many people probably induce the second line to be true based on the following accepted notation:
x*n = (x1 + x2 + ... + xn), where n is a constant.
When we differentiate the function x*n with respect to x, we are in fact right in assuming that it is equal to the following:
(dx1/dx+ dx2/dx + ... + dxn/dx) = (11 + 12 + ... + 1n) = n
However, we cannot replace x for n in such logic, because this notation of deducing a derivative is not true if n is a variable that changes as x does. The basic notation for describing n*x as a series of sums may sometimes be convenient for simple proofs involving arithmetic (or any other mathematical technique that it does support), but that does not mean it can be treated like an ordinary operation or term.
In fact, perhaps the notation we see above in d(x*n)/dx is simplified. By that I mean that the notation works only for constants because it is only part of the notation necessary for any term. Let's try to expand our notation to work for both variables and constants in place of n. We'll start by using the chain rule on the derivative of the function x*z
d(x*z)/dx = dx/dx*z + x*dz/dx = z + x*d(z)/dx
Then:
dx/dx*z = (dx1/dx) + dx2/dx + ... + dxz/dx) = (11 + 12 + ... + 1z) = z
And similarly:
dz/dx*x = (dz1/dx + dz2/dx + ... + dzx/dx) = x*dz/dx
So that:
d(x*z)/dx = d(x1 + x2 + ... + xz)/dx = (dx1/dx + dx2/dx + ... + dxz/dx) + (dz1/dx + dz2/dx + ... + dzx/dx) = (11 + 12 + ... + 1z) + x*dz/dx = z + x*dz/dx
if we allow z to equal x, we get the final set of equations:
x + x*dx/dx = x + x = 2x
Or from the error in the original problem:
d(x1 + x2 + ... + xx)/dx = (dx1/dx + dx2/dx + ... + dxx/dx) + (dx1/dx + dx2/dx + ... + dxx/dx) = (11 + 12 + 1x) + (11 + 12 + 1x) = (x) + (x) = 2x.
I propose that our initial assumption in going from line 1 to 2 of the proof of the riddle (going from the original proof's 2nd and 3rd equation) is wrong, and that it should instead be replaced by my above equation. Notice how it also yields the correct derivative of x*z with respect to x if z is a constant (simply z).
I don't see how this problem touches upon the heart of any problem central to calculus. The notation used to represent and differentiate x2 was simply faulty. When corrected, the main feature of calculus that comes to my mind is the simplicity of the power rule.
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| « Last Edit: Feb 27th, 2007, 9:41pm by Greg » |
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deolig
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Re: Differentiation Disaster
« Reply #43 on: Jul 16th, 2007, 11:31pm » |
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An alternative way of thinking of this problem is the following:
The function f(x)=x2 is nonlinear, (ie when you plot x2 versus x, you obtain a parabola passing through the origin.)
However in going from f(x)=x2 to f(x)=x+x+...x (x-times), we have transformed the original function to a sum of linear functions, since by definition g(x)=x is linear. A sum of linear functions is itself linear. And thus there is a problem with this transformation.
As has been pointed out in previous threads, this transformation may be correct for positive integers, but it cannot be applied anywhere else.
Hence the error occurs in assuming that
x2 = x+x+...+x (x-times)
is valid for all real numbers.
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Sameer
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Re: Differentiation Disaster
« Reply #44 on: Jul 17th, 2007, 10:50am » |
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on Jul 16th, 2007, 11:31pm, deolig wrote: we have transformed the original function to a sum of linear functions, since by definition g(x)=x is linear. A sum of linear functions is itself linear.
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What if you let those linear functions approach to zero and sum over it, will it be still linear?
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abcbcdcdef
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Re: Differentiation Disaster
« Reply #45 on: Aug 24th, 2007, 6:59pm » |
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it's quite simple actually, if I haven't got it wrong.
d(c)=0, where c is a constant
d/dx(x^2)=/=d/dx(x+x+x... (x times))
This (x+x+x... (x times)) can only work if x is a constant, but if it is, it immediately becomes 0.
x^2 can only be a function and cannot be written as (x+x+x... (x times))
I know manny people have made this statement already, I just want to help make it clearer. 
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srn437
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Re: Differentiation Disaster
« Reply #46 on: Aug 28th, 2007, 12:15pm » |
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d/dx(x^2)=d/dx(x)...(x times). Applying it to the terms individually doesn't work.
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pex
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Re: Differentiation Disaster
« Reply #47 on: Aug 28th, 2007, 12:31pm » |
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on Aug 28th, 2007, 12:15pm, srn347 wrote:| d/dx(x^2)=d/dx(x)...(x times). Applying it to the terms individually doesn't work. |
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Clearly; but the point of the riddle is figuring out why it doesn't work.
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| « Last Edit: Aug 28th, 2007, 12:32pm by pex » |
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Gary
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Re: Differentiation Disaster
« Reply #48 on: Dec 8th, 2008, 3:59pm » |
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Anton was nearly correct. He was the only one who went to the definition of the derivative.
f'(x) = lim h ->0 [f(x+h) - f(x) ] / h
This definition doesn't fail you--- not even here (integer or not -- not under consideration).
f(x)=x^2 = x + x + x + ...+x ; (x times)
f(x+h)=(x+h)^2 = (x+h) + (x+h) + ...+ (x+h) ; (x+h times)
The "x times" that Anton had in his first post was incorrect and he caught it but didn't review it.
f(x+h) - f(x) = (x+h) - x + (x+h) -x ... ; x times
+ x+h + x+h ... ; h times
= h + h + h + ...+h ; x times
+ x+h + x+h + ...+ x+h; h times
We're smart enough to fold that back together...
f(x+h) - f(x) = h*x + (x+h)*h
= 2h*x + h^2 (shoot, we could have guessed that).
divide by h and let the limit of h go to zero and we get.... 2x.
Go back to the definitions and they'll work for you.
Gary
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