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In the equation y=x+x+x..+x, this is not a continuous function, nor differentiable.

it is differentiable and is continuous - just like \(y=x^2\)

your mistake is that you didn't differentiate the sum of all the x's properly

but I forgot the exact mistake :-P

you can take one and half lots of something

If we put f(x)=y=x+x+..+x (x times), could you tell me what f(0.5) for instance?

f(0.5)=0.5+0.5+... (taken 0.5 times) = 0.25

how can that be?

Yeah, how can that be?

how is f(0.5)= 0.25??

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

or \[\sum_{i=1}^{x}x\]

I didn't what to use that because it implies x must be an integer

oh

What does "taken x times" mean then if not a sum?

but I guess Mr.Math is right - we cannot say x^2=x+x+x+... (x times) for non-integers?

Just quickly 2x=x means x=0, and doesn't mean 2=1.

x cannot be zero?

at ostensibly the same function

*at least

think of the product rule

Mr.Math has hit the nail on the head!

but which rule exactly are we breaking and how?

the product rule?

then df/dx=(1+1+1..+1)+(1+1+..+1) (x times) for each brackets I think.

right

@TuringTest - the being broken is to treat "x" as a constant when it is not

*the rule

Or the chain rule, let x+x+..+x (x times)=u, then we can apply the chain rule I think.

that's what I said, a variable number of terms. I just want to make it concrete in my mind.

I guess I've also learnt a thing or do from your replies on what a continuous function is - thx all

*or two

but at the first place is x+x+x+... xtimes a continuous function to be differentiated?

btw asnaseer who is the guy in your profile pic? I can't stop wondering! :(

that highlights the fact that you cannot treat the x in "x times" as a constant

@agdgdgdgwngo - thats me in the pic - why? do I look odd?

\(h\to 0\)*

no Mr.Math - the first set of "x+h" terms will appear "x+h" times

This is a old problem.

you came late ffm that is why it is old :P

yes FFM - I came across it several years ago and just recalled it :-)

There has been a discussion on this one, let me do some digging.

you mean here on OS?

thats it Mr.Math - well done!

yea get it :D

Here is it http://math.stackexchange.com/questions/1096/

I wasn't aware of that FFM - I ran into this many many years ago at uni

they mention the integer issue as well...

It's okay, I can't imagine why you mean many many and many :D

;-D