# Differentiation problem

Discussion in '\$1 Daily Competition' started by xpi0t0s, Aug 9, 2009.

1. ### xpi0t0sMentor

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There you go avoiding the question yet again. By construction N has to be an integer. You can't say f(x)=x+x+... (3.5 times) because that doesn't make sense. You can have f(x)=x+x+x, which is 3 times, or you have have f(x)=x+x+x+x, which is 4 times.

Would you, in your next post, EITHER answer the question OR concede that I am right.

2. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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x + x + x ... (x times) simply means x*x.

You can't claim f(x) = x + x + x or f(x) = x + x + x + x.
Because you can't fix x and then define f(x).

Even you know you are wrong, 'coz even Dr.Maths says so ... but I know your ego won't stop you from posting another vague argument. Go on ..

3. ### mayjuneNew Member

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4. ### mayjuneNew Member

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according to xp
but according to a mathematician

???

5. ### xpi0t0sMentor

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Why can't I define f(x)=x+x+x? Would you please explain that?

I don't know I'm wrong. In fact I'm sure I'm right because despite asking you repeatedly for an example of N where the equations don't hold you still haven't answered it.

You and Dr Math are both making the same mistake. f(x)=x+x+x...(N times); N=x; does
Code:
```***>>> NOT <<< ***(highlighted for clarity)
```
mean f(x)=x^2.

x+x+x...(N times) = x^2 ONLY when N=x and at no other time. Suppose N=4 and f(x)=x+x+x+x. Then f(x)=16 and g(x)=x^2=16. But if we increase x to 5 while keeping N fixed at 4 then x+x+x+x=20 and x^2=25. So f(x) does not equal g(x) when x!=4.

Here are two examples of N where the equations DO hold.

Reminder: define f(x)=f1(x)+f2(x)+f3(x)+...+fN(x) where fp(x)=x for p=1..N. Also we set x=N.

(Example1) N=3 so f(x)=x+x+x and if x=N then f'(x)=1+1+1=3, g(x)=x^2=9 and g'(x)=2x=6. So f'(x) != g'(x) because 3!=6.

(Example2) N=4 so f(x)=x+x+x+x and if x=N then f'(x)=1+1+1+1=4, g(x)=x^2=16 and g'(x)=2x=8. So again f'(x)!=g'(x) because 4!=8.

So Saswat, mayjune and everyone else, I have given you TWO examples where f(x)=g(x) and f'(x)!=g(x). All I'm asking for is ONE SINGLE VALUE OF N where the above does not work and I will concede that I am wrong.

6. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Yeah .. yeah.

You win. Live happily with your vague arguments.

7. ### mayjuneNew Member

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I think you are contradicting yourself xp....

And now you saying

???

8. ### mayjuneNew Member

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I think you are contradicting yourself xp....

You posted this, you never mentioned any x = N...

And now you saying

???

9. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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@ mayjune :

He has the right to contradict : his own views, mathematics and probably everything else too.
My prediction : he will still ask you for his N ... lol :rofl:

10. ### xpi0t0sMentor

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> You posted this, you never mentioned any x = N...

Yes, that's right. I added the N part in afterwards when it was clear that x+x+x...(x times) was ONLY being interpreted as x*x, so it was for clarity. x+x+x... (N times) and x=N makes the original intent much clearer. I don't think a clarification constitutes a contradiction.

But still no N where it doesn't hold. I wonder why that's not possible, given how obviously wrong and silly and all the rest of it I'm supposed to be.

11. ### mayjuneNew Member

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Ok tell you what, My college is closed 'coz of swine flu,
Whenever it does open, i'll ask the same question to my maths teacher, and whatever she/he says i'll post it here.
I'll ask for the value of N too, and whatever the ans is i'll post it, if he/she says no value i'll post that, if there is i'll post that.

12. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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He did it again ..... :rofl:

He doesn't get the fact that x "need" NOT be an integer. :freak: :goofy:

@ mayjune :

Good work !!
Although we all (even xpi0t0s) know, what your teachers are gonna say ....

13. ### xpi0t0sMentor

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> He doesn't get the fact that x "need" NOT be an integer.

Yes, that's precisely the problem. N (and therefore x, because x=N) does have to be an integer, it's part of the problem definition. If you reject that part of the problem definition then that changes the overall problem and therefore changes the solution.

@mayjune

Try that at college when you get back. When your teacher asks you to integrate sin x, tell him you reject that question and that you intend to answer the question "What colour is the sky", and give the answer "Blue". Then see if he gives you an A or an F. If he gives you an F, argue as vehemently as Saswat is doing here that your answer is completely correct and that the teacher is wrong because he asked the wrong question.

@Saswat

This I think is why you won't give me a straight answer to "What is N, where these equations don't hold". Because there isn't one, and therefore I'm right. You can end this right here and right now by showing me a SINGLE value of N, which by "reduction ad absurdum" I will happily accept as proof that I am wrong. Because I think it holds for all N.

14. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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OK. Lets handle things your way now.
( But I first recommend you to revise your calculus lessons. )

So, you say that you have f(x) = x + x + ..(x times),
which does not mean x*x = x^2....
rather it means x multiplied x times, which is an integral quantity, right ??

Now, b4 going to derivatives and all that ( which you are not strong in ), I assume you know what limits, continuity and differentiability of a function means. You know what is differentiability right ??
( If not, refer : http://en.wikipedia.org/wiki/Differentiability )

So, as per your definition of f(x), it is only defined for integral points like 3, 4, 5 and not anything like 3.5, which you had said earlier. That means, f(x) is not defined for any value of x such that N < x < N+1, where N is a natural number. Don't you say so ... ?

Now, if f(x) is not defined for 1<x<2, 2<x<3, 3<x<4 ... etc.. f(x) is discontinuous in the real numbers domain. As f(x) is discontinuous, it is non-differentiable too !! DO YOU GET THAT, Mr Genius ??? :shout:

Hence, you are trying to differentiate a non-differentiable function f(x), which is simply absurd !

So, either define x+x+x... (x times) as x^2 which will be valid for all x in real domain or just stop posting rubbish.
PLEASE HAVE A STRONG HOLD ON WHAT YOU ARE ARGUING, BEFORE ARGUING ON IT.

15. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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That applies to you, xpi0t0s.
You are simply crazy, now. Think b4 you speak ... just don't go begging for your "N".

I must have misunderstood the meaning of the word mentor. It seems like :

• a wise and trusted guide and advisor
Wise and trusted !! LOL. :rofl:

16. ### xpi0t0sMentor

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> So, you say that you have f(x) = x + x + ..(x times),
> which does not mean x*x = x^2....
> rather it means x multiplied x times, which is an integral quantity, right ??

Wrong. This is why you have a problem: you do not understand the problem definition and keep changing it to a different one. I have repeatedly defined and clarified the problem but you keep replying based on your misunderstanding and not based on the problem itself.

> I must have misunderstood the meaning of the word mentor

Yep, you seem to be doing a lot of misunderstanding in this thread. This seems to be something you need to work on.

What strikes me as odd is that after deducing that f'(x)=g'(x), you didn't take the question "why does f'(x) != g'(x)" as indicating that your understanding of the problem was wrong, you took this as meaning I was wrong and your understanding of the problem was not.

Let me use the expanded definition of the problem, and expand it yet further just to be completely clear.

Let us DEFINE: f(x)=f1(x)+f2(x)+f3(x)+...+fN(x).

Alternatively we can use capital sigma notation, but this doesn't work well with ASCII and is tedious to draw out. See http://en.wikipedia.org/wiki/Summation and please consider carefully the part "Successive values of i are found by adding 1 to the previous value of i, stopping when i = n" and the implications then on whether or not i and n, or N in this case, are integers:
Code:
```N
---
\
>   fp(x)
/
---
p=1
```
Let us also DEFINE fp(x)=x, and g(x)=x^2, and consider the corner case where N=x.

So, from this we can see that since d/dx (x)=1, fp'(x)=1 and according to Wolfram "differential of a sum is equal to the sum of differentials", so f'(x)=
Code:
```N
---
\
>   1
/
---
p=1
```
which is equal to N, and in the corner case we are looking at (where N=x) that this is equal to x.

But d/dx(x^2)=2x, so g'(x)=2x.

This is the PROBLEM DEFINITION. Of course, you can change any part of this, and the problem will change, and so the answer may change. The answer to the modified problem will not necessarily be the same as the answer to the original problem. For some reason you seem to be changing the definition of fp(x) from x to x^2 and insisting that N must be 1 but cannot be an integer (while simultaneously accusing me of contradicting myself), and ending up with f(x)=g(x) and differentiating what is basically g(x), instead of differentiating f(x) as per the problem definition, and (obviously) ending up with f'(x)=g'(x) and therefore completely unable to answer the question "why does f'(x) != g'(x)".

17. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Look man, instead of yelling that no body understands your question, try to show where exactly we go wrong in our understanding, and as you didn't do that, in any of your posts where you say I have misunderstood, I guess, you are simply avoiding things and diverting the attention (towards the same vague things).

Well, let me explain, I saw you definition of f(x) as sigma fp(x) in one of your previous posts, and I already AM TALKING ABOUT THAT.

Look :
Code:
```[COLOR=Wheat]####[/COLOR]p=N
[COLOR=Wheat]###[/COLOR]+-------
[COLOR=Wheat]####[/COLOR]\
f(x) =  >      fp(x)
[COLOR=Wheat]####[/COLOR]/
[COLOR=Wheat]###[/COLOR]+-------
[COLOR=Wheat]####[/COLOR]p=1
```
You only mentioned that in the previous post.

Now, what I am saying is, when you define f(x) as above, you consider N to be a constant quantity, 'coz if you are defining f(x) as f1(x)+f2(x), it is different from f1(x)+f2(x)+f3(x). As, the func definition changes, when you change N, N HAS TO BE A CONSTANT.

Now, if you look at the case N = x. It is against maths,
Why ? : 'coz you are assigning a constant to a variable.
It's like writing int x; 6 = x; in C++.
You know abt semantics right ?
[[We follow some rules for everything, not as arbitrarily as you.]]

So, even if you ignore (rather don't respect) the rules of maths, yet I can prove you false. How ? Look at this :
Suppose you define the func this way :
Code:
```[COLOR=Wheat]####[/COLOR]p=x
[COLOR=Wheat]###[/COLOR]+-------
[COLOR=Wheat]####[/COLOR]\
f(x) =  >      fp(x)
[COLOR=Wheat]####[/COLOR]/
[COLOR=Wheat]###[/COLOR]+-------
[COLOR=Wheat]####[/COLOR]p=1
```
As you know sigma increases the value p by 1 in each step. So, f(x) simply does not exist for x = 0.5, sqrt(2), log 2, pi, e etc ...

As, f(x) is undefined at these numbers, look at my previous post where I proved f(x) is discontinuous and hence NON-DIFFERENTIABLE AT INTEGRAL POINTS.

The graph for your f(x) looks something like this :

Clearly you can see the discontinuity in the graph (unless you have some defect in vision).
( If you think I have posted the wrong graph, please feed your func into Mathematica, and re-post the graph)

And, the graph of g(x) as defined by you looks like :

You can clearly see the smooth continuous curve.

So, g(x) is continuous and differentiable for ALL REAL X.
But, f(x) is discontinuous and NON-DIFFERENTIABLE.

That's where you make the FUNDAMENTAL MISTAKE OF differentiating f(x).

Next time don't try to divert attention of the members by saying no one understands what you say. We all know English too

Truth always triumphs

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18. ### xpi0t0sMentor

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> Suppose you define the func this way [i.e. some other way than I define it]
> [other stuff, graphs and all]

Sure, there is nothing wrong with your reasoning after you've redefined the function from the way I defined it to the way you want to define it. I disagree with absolutely nothing you say except for the redefinition of the problem.

> f(x) is discontinuous and NON-DIFFERENTIABLE.

Your f(x). Yes, you are absolutely correct. YOUR (and that is significant) version of f(x) is non-differentiable and I agree completely with all of this, except for the small point of you trying to force your definition of f(x) into my definition.

Tell me Saswat, if we define:

f(x)=x

is that differentiable, or not? If it is, what is its value? Is that value dependent on x? Does the value differ when x=1 from when x!=1?

Is f(x)=x+x differentiable, or not? If it is, what is its value? Does the value differ when x=2 from when x!=2?

Is f(x)=x+x+x differentiable, or not? If it is, what is its value? Does the value differ when x=3 from when x!=3?

19. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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f(x) = x + x
and, f(x) = x + x + x are both differentiable, because of the point I mentioned : N is a constant.

In the first f(x), you took N=2 and in the second, N=3.

When N depends on x and then you restrict N (or x) to integers, your function is a quantized(discrete) one.

20. ### xpi0t0sMentor

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> You know abt semantics right ?

Yes I do, and it is clear you are confusing maths with programming. (Also, I know how to spell "about", and where to find O and U on the keyboard.)

> It's like writing 6 = x; in C++.

No, it's not. In maths, if a=b then b=a. This is a basic axiom of the entire system, but in programming these mean substantially different things (assign the value of variable b to variable a, and assign the value of variable a to variable b). In maths, x=x+1 is impossible, but very few programs would work without that ability.

Anyway, answer my previous post please. I have tried once to explain your error to you; I will now try once more, but you seem determined not to get it.