# Differentiation problem

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

1. ### xpi0t0sMentor

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OK, you have in part. But you have missed some significant questions.

(1) Is f(x)=x+x+x differentiable, or not?

(2) If it is, what is its value?
(3) Does the value differ when x=3 from when x!=3?

Also please answer questions 2 and 3 for f(x)=x and x=1, and for f(x)=x+x and x=2.

2. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Even I do know that.
But, may be I focus more on brain instead of mechanical things

In maths you can NOT write N = x, in the present context; 'coz that would restrict the domain of x from Real numbers to Natural numbers.
Secondly, how on earth can you assign a variable to a constant ?????

Even in maths, when limit of x tends to infinity, x-1 = x = x+1

3. ### xpi0t0sMentor

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Sorry, just realised there is more ambiguity in that and as you're in 120% pedantic mode this could be a problem. Let me restate the questions.

Is f(x)=x differentiable, or not? If it is, what is the value of f'(x)? Is that value dependent on x? Does the value of f'(x) differ when x=1 from when x!=1?

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

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

4. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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LOL, you have asked a sequence of such questions earlier too ...

Why the hell, do you keep diverting the topic ??

Anyway : f(x) = Kx => f'(x) = K, which is independent of x.

5. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Ask them to yourself man, that might clear some fundamentals.

One line is sufficient : f(x) = Kx => f'(x) = K for all x.

6. ### xpi0t0sMentor

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and of course I should add for f(x)=x+x and f(x)=x+x+x: Is that value (i.e. the value of f'(x)) dependent on x?

7. ### xpi0t0sMentor

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> Why the hell, do you keep diverting the topic ??

Why the hell do you keep avoiding the questions? Are you unable to answer exactly the questions asked?
Are they difficult?

8. ### xpi0t0sMentor

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> Ask them to yourself man, that might clear some fundamentals.

OK, will do, let's see if we agree.

Is f(x)=x differentiable, or not? --> YES
If it is, what is the value of f'(x)? --> 1
Is that value dependent on x? --> NO
Does the value of f'(x) differ when x=1 from when x!=1? --> NO

Now you try the questions for f(x)=x+x and f(x)=x+x+x. This should be really easy, and is important in establishing where I'm going with this. If you just want to feed me the answers without any structure (e.g. YES 2 NO NO YES 3 NO NO) then that's fine; I will match them up myself to save you the difficulty.

9. ### xpi0t0sMentor

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So the next step is to generalise this. f(x)=x+x+x...N times. Or f(x)=Nx if preferred. N is a constant.
f'(x) is then N, regardless of x. f'(x)=N when x!=N, and f'(x)=N when x=N. f'(x) is not dependent on the value of x as it is also constant.

Therefore when x=N, x+x+x...(N times) = x+x+x...(x times) (because N and x are equal), and f'(x)=x (because f'(x)=N and N=x, and because of the mathematical axiom "if a=b and b=c then a=c"). Simple.

This doesn't hold when x!=N, of course. If x=2 and N=3 for instance, f(x)=x+x+x=6; g(x)=x^2=4 and x+x+x != x^2. x+x+x=x^2 only when x=3.

10. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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As mayjune suggested me, I will give you time to realize things, for 2 reasons,

(1) I am tired of this, now.
(2) I don't really care if you don't get the correct thing. Truth remains truth.

But, I would just give you a hint :
We know for every number A, A/A = 1.
But we can't generalize 0/0 = 1.

And, yeah ... thanx for all this vague arguments, replying to which made my post count close to 1K.

11. ### xpi0t0sMentor

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Seems you're determined (a) not to understand what I'm saying and (b) to keep insisting that you are not wrong but that I am.

No idea how 0/0 is related to this puzzle. Also I don't know why you won't answer the simple questions I've asked you; these are designed to clarify the reasoning so that you will understand what I've said.

Simply "giving me time to realise things" isn't going to change my mind - I don't think I'm wrong, but if you would come along with me in the direction I'm going in, then you could perhaps point out exactly where my error comes in, and that would need to be based on what I'm saying, not your misinterpretation of what I'm saying.

If you are prepared to do this then please answer this: do you agree with the answers YES 2 NO NO YES 3 NO NO? I've tried to keep those questions precise and non-vague as possible and to be honest I really can't see how a question like "Is f(x)=x+x differentiable?" is vague.

12. ### xpi0t0sMentor

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heh, just an afterthought Saswat, you *CAN* extrapolate from A/A=1 to 0/0=1; that's what calculus is all about. Look:

d/dx x = lim (h->0) ((x+h)-x)/h = h/h = 1. But the limit where h=0 is where differentiation works, and where 0/0=1 is assumed to hold. Otherwise lim (h->0) h/h is just as undefined as 0/0 is.

Still, I suppose you will also dismiss that as vague nonsense.

13. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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That's a limit man.

Think what you are saying, b4 posting.

lim x tends to y f(x) != f(y).

14. ### xpi0t0sMentor

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Anyway what about the answers YES 2 NO NO YES 3 NO NO? Do you agree with them or not?

15. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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(1) YES. f'(x) = 1. NO NO

(2) YES. f'(x) = 2. NO

(3) YES. f'(x) = 3. NO

Go on... then generalize again. That's all you know, and let me tell you that's exactly where you go wrong.
You are so blind with your vague arguments .... blive me take some time and think abt it.
When you know what the real thing is and what you were arguing on, you'll feel bad Mr.Genius.

BTW, You didn't answer one of my questions : Were the graphs of f(x) and g(x) alright ?

16. ### xpi0t0sMentor

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> BTW, You didn't answer one of my questions : Were the graphs of f(x) and g(x) alright ?

Yes, they were absolutely spot on, according to *your* definition of f(x), that is, which differs from mine. Your graph of f(x) does not apply to my definition of f(x), because our definitions differ. So in that sense, i.e. is your graph of f(x) correct for my definition of f(x): no, it is not, due to the different definition of the function.

OK let's generalise again and I will take this slowly so that you can spot the error precisely where it happens.

f(x)=x is identical to f(x)=Nx where N=1. This is differentiable and equal to 1, which is equal to N.
f(x)=x+x is identical to f(x)=Nx where N=2. This is differentiable and equal to 2, which is equal to N.
f(x)=x+x+x is identical to f(x)=Nx where N=3. This is differentiable and equal to 3, which is equal to N.

So, (deep breath), we have a family of lines here which can be summarised:

Where N is any positive integer, f(x)=x+x+x...(N times) is identical to f(x)=Nx; both are differentiable and equal to N.

Correct so far? Or am I wrong at this point? Is this vague or precise?

17. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Precise.

But, I didn't get where my graph of f(x) differed from your definition ??
Clarify that first.

18. ### xpi0t0sMentor

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Basically you say that f(x)=x+x+x (x times) can ONLY be interpreted as "f(x)=x^2", meaning you end up with f(x)=g(x) and obviously f'(x)=g'(x), which leaves the question "why is f'(x)!=g'(x)" unanswerable.

Can I take "Precise" to mean you agree with the generalisation? If so then I'll proceed to the next step.

19. ### xpi0t0sMentor

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OK, here are some graphs (if the attachment works). GraphCalc is freeware; anyone can download it, and you can try the equations yourself if you think I've doctored the screenshot at all.

For no obvious reason I've picked N=3 and the graphs are as follows:

y1: f(x)=x+x.. (N times) (N=3) = x+x+x
y2: f'(x)=3
y3: g(x)=x^2
y4: g'(x)=2x

So now have a close look at the graph where x=3 and in particular the behaviour of y1 and y3 at that point. To my vision they appear to have the same value, but feel free to correct me if you think I'm wrong.

At x=3 f(x)=9 and g(x)=9
But have a look at the other two lines: at x=3 f'(x)=3 and g'(x)=6

So at x=3, f(x)=x+x...(x times), which is x+x...(3 times), which is x+x+x.
At x=3, f(x)=g(x)
At x=3, f'(x) != g'(x).

If in your mind "x+x...(x times)" can *only* be interpreted as x^2 (which is not my original intent and you should stop trying to force me to say it was), then let's introduce a constant N that takes this problem away:

So at x=3, f(x)=x+x...(N times, where N=3), which is x+x...(3 times), which is x+x+x.
At x=3, f(x)=g(x)
At x=3, f'(x) != g'(x).

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20. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Oh Wow......
I am a kid and you are teaching me graphs of :

y1: f(x)=x+x.. (N times) (N=3) = x+x+x
y2: f'(x)=3
y3: g(x)=x^2
y4: g'(x)=2x.

Come on man, you know you are losing this. Face the reality .... for how many days more, will you keep contradicting yourself ?

Look you say *****your***** f(x) = x + x + x .. (N times) but f(x) != x^2.
OK. As I said earlier, I agree to that.

Now listen carefully, 'coz you seem to either ignore this; or pretend that you don't understand it :

If you define f(x) = x + x + ... (N Times) and then generalize it for N=x;
you restrict the domain of x to Natural numbers only, because N can take only natural number values and so can x, when N=x (LOL, 6=x).
So, how the hell does you graph become continuous, man ???

If you had defined f(x) = x + x + x + ... (x times) = x^2, the graph would have been continuous, 'coz then you don't restrict the domain to natural numbers. The domain is real numbers.
I would also like to remind you ... I never again said f(x) = g(x). I am trying to disprove you by your own (vague) conditions and arguments, but you seem to keep circling around N and never getting to the point.

And btw, the 0/0 was an example of how vague your generalizations can be.