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~ Б0ЯИ Τ0 С0δЭ ~
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
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Go4Expert Founder
Quote:
Originally Posted by SaswatPadhi
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
Congrats BTW on 1k Posts
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Invasive contributor
Congrats Saswat on crossing 1000 posts
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Mentor
>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?
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~ Б0ЯИ Τ0 С0δЭ ~
Thanx shabbir and nimesh !

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.
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Mentor
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).
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~ Б0ЯИ Τ0 С0δЭ ~
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 ... ?
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Mentor
Well how else do I interpret

Quote:
Originally Posted by SaswatPadhi
There is a saying :: "Arguing with a fool proves there are two."
So, I end this discussion here.
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.
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~ Б0ЯИ Τ0 С0δЭ ~
Quote:
Originally Posted by xpi0t0s
And you definitely called me crazy more than once.
OK. I am sorry for that.

I was a bit hyper....

Quote:
Originally Posted by xpi0t0s
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.
I hope someday you will realize what I was saying .....
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Go4Expert Founder
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