# Differentiation problem

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

1. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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I reached the 1K post count and this thread alone contributed 3.3% !!!

Thanx xp --- At least there is something good with your vague args

2. ### shabbirAdministratorStaff Member

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Congrats BTW on 1k Posts

3. ### nimeshNew Member

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Congrats Saswat on crossing 1000 posts

4. ### xpi0t0sMentor

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>So, how the hell does you graph become continuous, man ???

That shows your incompetence in maths.
f(x)=x is continuous for all x.
f(x)=x+x is also continuous.
f(x)=Nx is continuous for all N and all x.

Now, let me ask you a question. It's an easy one.

If N=3 and x=3,

Does x=N or not?

5. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Thanx shabbir and nimesh ! :happy: :happy: :happy:

Mr. Genius xpi0t0s ....

f(x) = Nx is ***NOT*** your function by that way. --- there you contradict yourself yet again.
If f(x) = Nx is your function, then when N=x, I can say f(x) = x^2, which you don't agree.

There is a saying :: "Arguing with a fool proves there are two."
So, I end this discussion here.

I know people like you don't give up, although they know they have already lost it.. They can go to any extent ..... Dr.Maths was wrong, mayjune was wrong and may be the whole world is ('coz just google your question and you'll know). Now, I really understand why you called yourself thick-skinned.

6. ### xpi0t0sMentor

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You just can't help with the personal remarks can you? Calling me a fool isn't exactly polite.

I haven't contradicted myself once. I keep contradicting YOU because YOU keep contradicting me based on your misunderstanding of what I'm trying to say.
I haven't lost it because I know I'm right. The only reason for persisting is the hope that you will eventually drop your misunderstanding and listen for once to what I say.

If f(x)=Nx and N=x, then yes, f(x)=x^2. That's the point. *When N=x*. When N does NOT equal x, then f(x) is NOT equal to x^2 and so simply defining f(x) as x^2, which is what you're trying to do, is at odds with my original intent. I offer a single example, and to date despite all your arguing you still have not offered a single counter example to prove I'm wrong: if N=2, then f(x)=x+x. This evaluates to 4 when x=2. And the whole point of the original puzzle was to find out why f(x)=x+x and g(x)=x^2 gave different results (2 and 4 in this case: f'(x)=1+1=2; g'(x)=2x=4) when f(x) and g(x) both evaluated to the same result (x+x=2+2=2*x=4).

7. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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So, you are all set to change the whole world ('coz everyone agrees with me ) .....

and, btw I never called you a fool. May be I am a fool.
I wonder if I mis-interpret things or you do ... ?

8. ### xpi0t0sMentor

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Well how else do I interpret

And you definitely called me crazy more than once.

So I too see little point in continuing. You won't listen to me; you stick religiously to your misunderstanding, quoting Dr Math who is not wrong but who is commenting on a puzzle that is SIMILAR to mine but NOT THE SAME.

9. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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OK. I am sorry for that.

I was a bit hyper....

I hope someday you will realize what I was saying .....

10. ### shabbirAdministratorStaff Member

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Now as I see both of them Stopped so now I would actually Answer the actual question. You would be surprised but I actually read all the 88 posts but now here I am with the solution.

f(x) = x+x+x... (x times)
g(x) = x^2

This both functions are equal no doubt.

But then differentiating both of them are not equal and the main reason is how f(x) is written. You can have multiple ways of writing function and each of them's differentiation would not be same.

Lets take one more example

1. F(x) = x^3
2. G(x) = x^3 * 1

Now in my case as well both the functions are same but their differentiation would not be same and the way I have written G(x) it would always be zero ( by formula )

Coming back to your original example. Now does this mean the way you represent the function changes the differentiation.

Actually yes because the bounds of the function may change.

In your case f(x) is valid only in integral bounds but for g(x) its in real bounds.

I hope this answer your question and I may be wrong as well and correct me if I have missed anything

11. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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I don't accept that, because that is clearly false according to mathematics, 'coz you say the funcs are equal and their derivatives are not !!
If f(x) = g(x), f'(x) = g'(x), as long as the areas they are defined (their domains), are same. But then, if their domains are different, the functions are different themselves.
For example,
f(x) = x
g(x) = sqrt(x)^2
are not the same, because f(x) is defined over the whole real number domain, but g(x) is defined only on the +ve real numbers domain.

As xpi0t0s said, I agree that his definitions of f(x) and g(x) are different and not the same. If they are, then there is not question of their derivatives not being equal.

By which formula ????????

Look at what Wolfram (Mathematica's creator) Alpha says :
(1) Derivative of x^3 = 3*x^2
(2) Derivative of x^3*1 = 3*x^2
(3) Derivative of (x^3)*1 = 3*x^2
(4) Derivative of (x^3)/1 = 3*x^2

I want to check those links and see that ALL of them have equal derivatives.

Really shabbir, if that what you are telling was to happen, the whole Maths (at least whole calculus) will have to be modified.

Two cases arise ::

(1) Does that mean f'(x) != g'(x) ?
But, you said f(x) = g(x), which means f'(x) == g'(x).

(2) Or, if you say the bounds of the func f(x) changes, then it becomes non-differentiable !!
So, f'(x) does not exist at all.

But, in both these cases, xpi0t0s' proof that f'(x) != g'(x) is unacceptable.

12. ### xpi0t0sMentor

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Would it help to point out that my original intent behind "f(x)=x+x+x...(x times)" was to describe a family of straight lines rather than a single curve? And hence the clarification of f(x)=f1(x)+f2(x)+...fN(x).

And also hence the difference between this and the Dr Math post, because Dr Math reads "f(x)=x+x+x...(x times)" as a single curve exactly equal to g(x)=x^2, NOT as a family of straight lines. So clearly either "f(x)=x+x+x...(x times)" is ambiguous, or I was wrong to use this to represent my original intention. If the latter, then may I just add that at the time I did not know it was wrong and apologise profusely to anyone who was misled by this mistake. In my defence though, my original post stated "Explain why f'(x) != g'(x)." which is of course an impossible question to answer if "f(x)=x+x+x...(x times)" is totally unambiguous and absolutely cannot under any circumstances represent my original meaning. Best response here I suggest would be to point out that the question is unanswerable.

So here is a GraphCalc plot of f(x)=x+x+x...(N times) for N=1..5, and overlaid onto these five lines is the curve y10 which represents g(x)=x^2.
You can see:
at x=N=1, f(x)=x=1 and g(x)=x^2=1 and therefore f(x)=g(x).
at x=N=2, f(x)=2x=4 and g(x)=x^2=4 and therefore f(x)=g(x).
at x=N=3, f(x)=3x=9 and g(x)=x^2=9 and therefore f(x)=g(x).

Let me just clarify "at x=N=3 ... f(x)=g(x)." We say stuff like "At x=0, sin(x)=0". This does not mean sin(x)=x, or "for all x, sin(x)=0", it just means that when x=0 the value of sin(x)=0. In the same way "at x=N=3 f(x)=g(x)" means they have the same value when x and N are both 3; it should not be taken to mean that f(x)=x^2.

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13. ### shabbirAdministratorStaff Member

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Not really. Now here are 2 functions which are defined as

f(x) = x^2; For all positive x. i.e. Its a continuous function for all x > 0
g(x) = x^2; For all negative x i.e. Its a continuous function for all x < 0

f'(x) would be equal to g'(x) but f(x) and g(x) are not equal.

Isnt that true ? Tell me if I am wrong anywhere.

14. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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@ xpi0t0s ::

I have understood that perfectly ...
But, bro ... you still go wrong here :
See, you explicitly defined f(x) = f1(x) + f2(x) + .... + fN(x)
So, there is no question of a function like f2.5(x).
And, as, you are (later) substituting N=x, you can only substitute integral values of x, 'coz N can be 1, 2, 3 or 4 ...
So, as you restrict the values to integers, you are basically making the func discontinuous.

Further ... f(x) = x + x + ... + (x times) is NOT a straight line, actually it's same as x^2.
See this, i am sure you can't say Wolfram is wrong too : http://www.wolframalpha.com/input/?i=sum+x,+j=1+to+x.

Exactly, I was trying to tell that to you...
If f'(Nx) = N, does not mean that f'(x*x) = x.

@ shabbir ::

Ah .... look at what you had posted, and what my question was.
We are not talking abt the case when f'(x) = g'(x).

You said, when f(x) = g(x), f'(x) might not be equal to g'(x).
Show me an example.

Here is your quote, btw :
And, also tell me how G'(x) becomes 0 in the example below ::
BTW, did you read my reply to your post ?

15. ### xpi0t0sMentor

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> So, as you restrict the values to integers, you are basically making the func discontinuous.

No, I restrict N to integers but not x, and this is where you keep going wrong.

Saswat, is f(x)=x discontinuous? What about f(x)=2x? What about f(x)=3x? You see, this is a FAMILY of STRAIGHT LINES, generalised as f(x)=Nx.

Let's take one member of this family as an example: f(x)=3x. So N=3. We draw f(x) out as a straight line. You insist you know what that looks like (despite claiming it to be discontinuous) so I won't bother with a screenshot. Now, look carefully. The line passes through a number of points where x has a value and y has a value, and the line relates x-values to y-values. Where x=0, y=0, for example, and where x=1, y=3; where x=10, y=30; where x=123, y=369. And so on, and as it's continuous, where x=pi, y=3pi; where x=sqrt(2), y=3sqrt(2) and there's a whole host of real numbers we could add to the list.

Now, look even more carefully at this next bit. Where x=3, y=9. Now, don't go off on this "haha, that proves f(x)=x^2" stuff again. We're still on the straight line f(x)=3x. x=3, y=9, because y=3*x, which is 3*3, which is 9 (which, barring major accidents, you should be able to figure out with your fingers). Now, because N=3, (take this slowly because we are STILL on the straight line y=3x), when x=3, (startling conclusion following) x equals N! (We're still on y=3x here, remember, which is a continuous straight line function.) Now, just because x=N at that point does not mean the whole line is now defined as y=x^2.

Now, the same is true for all other members of the same family. For the line y=2x, N=2, and where x=2, x=N, although the value of N has changed. Note that when x is some value other than 2, including all positive integers, x=N is not a true statement. N is a constant for the straight line, so x=N only where x=2. Where x=3, x no longer =N because 3!=2. And where x=4 (and N still hasn't changed, because if you can remember stuff longer than a goldfish then you'll know we're still talking about y=2x here) then x still doesn't =N, because x=4 and N=2 and 4!=2.

So let's recap what I did.
For y=x, N=1, and we're interested in the point along the straight line where x=1 (and because N=1, x=N, or N=x, which is actually the same thing).
For y=2x, N=2, and we're interested in the point along the straight line (which is a different straight line than in the previous statement) where x=2, and again because N=2 in this line, x=N and N=x.
For y=3x, N=3 and ... x=3 so x=N.
For y=4x, N=4 and ... x=4 so x=N
For y=5x, N=5 and ... x=5 so x=N
For y=6x, N=6 and ... x=6 so x=N
Got it yet? I'll add a couple more just in case you still haven't spotted the pattern yet.
For y=7x, N=7 and ... x=7 so x=N
For y=8x, N=8 and ... x=8 so x=N
These are all straight line graphs. Not one of them is a curve. If you try every positive integer I bet you won't find a single member of this family that is a curve, or that is discontinuous.

> Further ... f(x) = x + x + ... + (x times) is NOT a straight line, actually it's same as x^2.

Well, if you look back you'll see that because of your insistence it can only mean that, I've clarified my original intention as f(x)=x+x+x...(N times). You still don't appear to have figured that out. I'm happy to accept that you're completely unable to understand f(x)=x+x+x...(x times) means anything other than f(x)=x^2 despite 10 pages of clarification and insistence that that is not what I meant.

So this is why I've stopped saying f(x)=x+x+x...(x times) and have started saying, actually since quite early on, that f(x)=x+x+x...(N times). So if possible could you please stop talking about f(x)=x+x+x...(x times) because that is no longer the topic of discussion; we are now talking about f(x)=x+x+x...(N times).

In fact you did seem to have got the point on page 3:
N _is_ independent of x, that's what I've been saying all along. What gets you confused is the point along each straight line where the value of x corresponds to the value of N which is just a long way of saying x=N. So in the straight line graph y=4x there is a point where x=4 and y=16, and despite this, the graph is still y=4x and has not suddenly mutated into y=x^2. And if we generalise this there is a point along the straight line graph y=Nx where x=N and y=N*N.

But as soon as I generalise this that is where you completely lose the plot and go off on this "f(x)=x^2" stuff, insisting that N must be dependent on x, just because at some point in the graph x=N. That is your mistake.

16. ### xpi0t0sMentor

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Okidoki, if you want WolframAlpha quotes, try this one:
http://www.wolframalpha.com/input/?i=plot+x%2C+2x%2C+3x%2C+4x%2C+5x%2C+x^2+from+x%3D-1+to+6
See where x^2 crosses the other lines?
If the other lines are of the form y=Nx, what do you notice about the value of x where the x^2 line crosses them? (Hint: x=N)

17. ### shabbirAdministratorStaff Member

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Saswat what I am trying to say is

If f(x) = g(x) then f'(x) = g'(x) is always true provided the bounds of x is same for f(x) and g(x). Do you agree on this or not.

18. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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@ xpi0t0s ::

I am tired of this.
Man, I am truly sick of your generalizations now....

Why don't you get a fact so simple as this, that x is a variable and N is a constant.
You define the func in terms of a constant,
calculate the damn derivative and,
then substitute a variable i.e. x, in place of a constant ... incredible.

Do whatever you want; you are just incorrigible.

@ shabbir ::
All I am trying to say is ... if bounds are not same for f(x) and g(x), they are NOT equal.
So, you don't have to explicitly mention :
And, when the bounds are not the same, the functions are not the same (you can verify that : if domains of funcs are unequal, the funcs are also unequal).

And, I wanted to know how derivative of G(x) = x^3 * 1 becomes 0 ??

See, the way I write funcs won't change the derivative, unless I not only change the way of writing but also the function it self.
Let's see for example, say I take f(x) = 1 => f'(x) = 0.
If, I would rewrite f(x) = (sin x)^2 + (cos x)^2,
f'(x) = 2 * sin x * cos x + 2 * cos x * (- sin x)
=> f'(x) = 0 is still true !

If just changing the way would change the funcs, then every theorem in Maths will contradict the other ...

19. ### xpi0t0sMentor

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Yes, I know x is a variable. What's the problem with observing that at some point in the graph, x=N?
After all, if a function is valid for -Inf to +Inf, and N=3, then why isn't the function valid at x=3 and N=3?
Or is the problem with you that if x=3 and N=3 that you really can't stand to see "x=N", and if so, on what basis? I have already quoted the mathematical axiom that if a=b and b=c, then a=c. So if f'(x)=3 and x=3 then f'(x)=x by EXACTLY THE SAME AXIOM, correct?

I'm not substituting x for anything. You really need to start understanding that.
I'm simply pointing out that WHEN x=2 for f(x)=2x, WHEN x=3 for f(x)=3x, WHEN x=4 for f(x)=4x and so on, x=N.

Tell me honestly, and I'll use different letters to save confusion:

if m=3 and n=3

does m=n or not?

Does observing that m=n make m dependent on n, or the other way round, or are we just observing that the two variables have the same value at that point?

> then substitute a variable i.e. x, in place of a constant ... incredible.

No, what's incredible here is that after I've told you 10,000,000 times that x is not dependent on N, or vice versa, you still insist that's what I'm saying.

> Why don't you get a fact so simple as this, that x is a variable and N is a constant.

Yeah, I've got that fact. I knew that in my first reference to N.

> You define the func in terms of a constant,

Yep. f(x)=Nx.

> calculate the damn derivative and,

Yep. f'(x)=N

> then substitute a variable i.e. x, in place of a constant ... incredible.

NO! No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No. No.

Got it yet? N. O. NO. YOU ARE WRONG.

I am not substituting x for N at all.

Is there a substitution here? if f'(x)=N and N=3, where x=0, x!=N?
What about here: if f'(x)=N and N=3, where x=1, x!=N?
If no to both the above, then why is there a problem when I make the OBSERVATION (not substitution):
if f'(x)=N and N=3, where x=3, x=N?

Why is that a substitution?

If f'(x)=N and N=3 and x=3, why is it a substitution and not an OBSERVATION to OBSERVE that f(x)=x?

> I am tired of this.

Then why don't you just open your thick skull for 5 minutes and entertain the remotest possibility that I could have been right all along and you could have been wrong all this time. Or is there too much loss of face in admitting stuff like that in your part of the world? Would accepting that mean you had to commit suicide or something daft? Is the problem here in fact that I have to give you an honourable way out of this? There is one: simply admit you are wrong. Just as you are trying to force shabbir to do with the d/dx (x^3*1)=0 stuff (which I think is based on a mistaken form of the product rule, i.e. the mistake is taking the product rule as "d/dx[f(x)g(x)]=f(x)g'(x)" - where g(x)=1 and so g'(x)=0. In fact the product rule is "d/dx[f(x)g(x)]=f(x)g'(x)+f'(x)g(x)", so d/dx((3x^2)(1))=0*3x^2+6x=6x).

Getting through to you is like trying to stone a donkey to death with ripe figs.

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