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asnaseer
How good is your understanding of differentiation? \(y=x^2\) can be written as \(y=x*x\) which can also be written as \(y=x+x+x+...+x\) (i.e. "x" lots of x). using this we get:\[\begin{align} y&=x^2\\ \therefore \frac{dy}{dx}&=2x\\ &\text{but we can also write this as:}\\ y&=x+x+x+...+x\quad\text{i.e. "x" lots of x}\\ \therefore \frac{dy}{dx}&=1+1+1+…+1\quad\text{i.e. "x" lots of 1's}\\ &=x\\ \therefore 2x&=x\\ \therefore 2&=1 \end{align}\]where is the mistake in this?
In the equation y=x+x+x..+x, this is not a continuous function, nor differentiable.
it is differentiable and is continuous - just like \(y=x^2\)
Not really. x here can take only integer values, so it's defined from N to R. While y=x^2 is defined from R to R. So they are tow different functions.
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
Mr.Math - you are probably correct in the strictest sense, but agdgdgdgwngo is closer to where the mistake is.
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
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.
yes Mr.Math - the real question is, how can I get two different results by differentiating the /same/ function?
x cannot be zero?
at ostensibly the same function
think of the product rule
I don't think you can differentiate a variable number of terms like that, but I can't seem to apply the right rule.
f(x)=x+x+..+x (x times). The key is "x" times, since we're differentiating with respect to x. If it's f(x)=x+x+..+x ("n" times), what we did would be correct.
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.
if you re-did this using the definition of differentiation then you would get the right result. i.e.\[f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}\]
@TuringTest - the being broken is to treat "x" as a constant when it is not
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
but at the first place is x+x+x+... xtimes a continuous function to be differentiated?
@Akshay Budhkar - if you use this to evaluate the differential of y = x+x+...+x (x times), then you will get the right answer:\[f'(x)=\lim_{h\rightarrow0} \frac{f(x+h)-f(x)}{h}\]
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?
No i was just wondering if the function was continuous, i know that formula, but if a function is not continuous we cannot differentiate it, and i feel this is not a continuous function
yes - Mr.Math pointed that out earlier - I guess I used a bad example of trying to illustrate a specific point :-(
Wait!! \[f'(x)=\lim_{h\to 0}{x+h+x+h+\dots+x+h-x-x-\dots-x \over h}\] \[=\lim_{x\to 0}{xh \over h}=x\]
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.
Right! \[\lim_{h \to 0}{x+h+x+h+..+x+h-x-x-\dots -x \over h}\] \[=\lim_{h \to 0}{hx+(x+h)h \over h}=2x\]
thats it Mr.Math - well done!
So what is the conclusion of this question ( neglecting the fact that the function is discontinuous) ?
the mistake was in treating the "x" in "x times" as a "constant" when differentiating - this is wrong
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