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

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

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

Congrats Saswat on crossing 1000 posts

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?

Mr. Genius

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.

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).

and, btw I never called you a fool. May be I am a fool.

I wonder if I mis-interpret things or you do ... ?

Quote:

Originally Posted bySaswatPadhiThere is a saying :: "Arguing with a fool proves there are two."

So, I end this discussion here.

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 was a bit hyper....

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

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