Countable derivative
f : N → R is defined as
f ( x ) = x + x + x + … ( x times )
Find f ′ ( x ) .
Clarification:
f
′
(
x
)
represents the derivative of
f
(
x
)
with respect to
x
.
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2 solutions
Discussions for this problem are now closed
I suppose x=-1 would produce a good example
My answer is 2x because
f ( x ) = x + x + x + . . . . . . ( x t i m e s )
= > f ( x ) = x ∗ x = x 2
= > f ′ ( x ) = 2 x
can you explain what is wrong in my solution?
The domain of f is the natural numbers, so it is a discrete function. But for a function to be differentiable at some point a then it must also be continuous at a . Since this condition is not met in this instance, we can conclude that f is not differentiable, and hence any discussion on what it's derivative is meaningless. (Even if the domain of a function is all rational numbers it will still not be differentiable.)
If you relabel the function as f ( n ) , then indeed f ( n ) = n 2 , but this only works since n is a natural number. Suppose we were to let the domain be the positive reals; then x + x + x + x + . . . . + x , ( x times), would have no meaning, since we have made an implicit assumption that each term is a "whole" x , and thus have no way of dealing with the cases where x is non-natural, e.g. x = π .
Oh Thank you for clarifying my concept :D
Sir , I couldn't quite understand your solution , can you explain it a bit ? Also what do you mean by a discrete function, sir ?
Edit : My friend Deepanshu has explained it to me sir .
Actually Domain is Natural number , Since our function is : f : N → N f ( x ) = x 2 obviously y = x 2 is an continuous function because it's domain is R But in this particular function we changed the domain (Natural number's) of function keeping rule ( y = x 2 ) same . So this domain make this function as non-continuous or you can say that it is discreet function , means it's range contain's (1 , 4 , 9 , 16 , 25 , 36 ......etc.) .
We defined derivative at a point as "Rate of change of curve at that point " means " Instantaneous Slope " at that point , So for testing differentiation of curve, we test "continuity" of curve first , so due to discreet's value in range , mean's while plotting graph of this rule in it's domain , we got different different point's which are well spaced. So talking of slope at a point is meaning less . Since there is no curve just after and just before that point .
But if domain of function is R , then it is differentiable????....
If the domain were the reals, then the function would not be well-defined. By defining the function as f ( x ) = x + x + x + . . . . . (x times), it is implicit that x is a natural number, since "x times" doesn't make any sense if, say, x = π , or any other non-natural number. So since it is not clear how the function would be defined when x is not natural, it cannot be unambiguously differentiated.
Clearly the graph of this function, when plotted would be discontinuous as input (domain) would only include natural numbers. So as a result graph will have only discrete sharp points, and since at a sharp point on a curve , infinite number of tangents are possible therefore we say that the tangent doesn't exist and hence no derivative exists.
Hope that's useful .
Taking the derivative of a discrete function has no meaning, so by default the answer is "none of these". At first glance, before realizing the issue with taking a derivative of such a function, there does seem to be a paradox; if the "derivative" were taken of the function as written then the result would be x , but if simplified to f ( x ) = x ∗ x = x 2 then the "derivative" would be 2 x . However, this apparent paradox is rendered moot by the fact that the derivative of such a function has no meaning.