Read the other replies but this is what clicked it for me:
Between step 2 and 3, you applied the derivative to all of the x’s in the sum (x+x+x…) but ignored the x in the “x times”.
This nonstandard notation helps to hide that. If you wrote this in sigma notation, you’d have:
If you differentiate this with respect to x, you can’t ignore the x in the sigma limit. When differentiating a summation where the limits are a function of the target variable I believe you need to use Leibniz rule(?), but I’ll leave it there
You cannot differentiate a sum when the variable being differentiated is used to define the number of terms in the sum — unless you rewrite the sum as a closed-form, continuous expression. Even in Sigma notation as you used.
The act of summing “x terms” as you expressed in your sum is not itself a differentiable process.
Once you turn it into a continuous function (in this case x^2), then you can differentiate it.
The Leibnitz rule doesn’t do anything here because you still have an unextractable “x” that’s defining your summation.
That makes sense, thanks. I knew someone would know better than me how to interpret that.
i see what you’re doing here.
i typed it out in latex for you:
what you’re forgetting is to add the derivative of the integral limits into your calculation, basically.
message me if you have any questions.
Because differentiation needs the expression to be continuous. And x + x + x … x times is not continuous.
Derivates are built on limits, which assume the function is continuous. ie. it gives an output for every single Real number, even decimals.
This is the case for x^2 but not x + x + x … x times. Here you’re only adding one x every full number. So the function is discrete, a value only exists every full number. You cannot express something you wish to differentiate this way.
ELI5: x^2 and x + x + x … x times are only equal if you use full positive numbers, ie. 1, 17,
if you use decimal numbers or negative numbers x + x + x … x times is undefined.
This means it cannot be differentiated.
Is d/dx x^2 really equal to d/dx (x+x+x+…+x)? Inserting some value for x seems to break that part. Say x=5:
d/dx 5^2=10 and
d/dx (x+x+x+x+x)=d/dx 5x=5
10 ≠ 5
I understand that x^2, (x+x+x+…+x), and x*x are the same. But its equivalent to doing d/dx x*x =x as d/dx n*x=n. Again this results in 2x=x or 2=1.
The mistake is to treat a variable as a constant and deriving in that way. Doing a sum x times means you have an unaccounted variable when you do d/dx (x+x+x+…+x), this is not 1+1+1+…+1 x times. i.e. The rate of change of one of the x is not incorporated into the derivative.
Someone ping me when someone drops the answer.
ping
Ping
Thanks, kind man!
Ping
The issue here is that the operation of derivation is not an equivalence tranformation. Equivalence transformations must be reversible, but reversing the derivation would be an integration. But the integration introduces an integration constant of which the actual value is unknown. So, by deriving and integrating one expression, we basically add a constant to it and have changed it. This is why the derivation is no equivalence tranformation and messes up our equation.
this true, but in physics (and in maths too, especially in real analysis), we use differentiation as a equivalent transformation, in a generalised sense. we often come up with ways to define it with something like - finitely many discontinous points or something, which we can iron it. but major problem here (i think) is order of operation - we can not interchange sum and diff when they both depend on x
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What the hell is this “x times” notation? I guess you could rewrite this using sum notation (Sigma)
At the bottom that derivative isn’t distributed to the “X times portion”, it should be summed up d/dx(x) times and apply the chain rule.
if i am not wrong, there is only one problem - sum to x (x times)
when we write it in a bit more concise but equivalent notation - d/dx sum_1^x x it basically becomes x^2 again. It is kinda a a=a proof, so not very interesting but that is the only problem. since sum is happening a variable number of times, we can not really let it loose.
in a more reasonable wording - what the image showed was interchanging differentiation and sum (like instead of doing sum after diff, instead of before), and such operations are allowed, but if the sum and diff are independent (not based on same variables).
this interchange works if you used some other variable which is independent of x
