If f(x+a) = f(x-a) where a is infinitesimally small then....

[Ahmad Kishki]

if f(X+a) = f(X-a) where 'a' is infinitesimally small then does the result f'(X) = 0 follow?

X is a point along the x axis.. Sorry about the fault in the question

 

[PWiz]

Yes it does. You're differentiating a constant with respect to a variable, which will always be 0 no matter how small or large the constant is.

 

[Ahmad Kishki]

Yes it does. You're differentiating a constant with respect to a variable, which will always be 0 no matter how small or large the constant is.

to be sure df(a)/dx means find df(x)/dx then input x=a

 

[Orodruin]

No. Why would you insert the infinitesimal a in the derivative? It is part of the limit in taking the derivative. It is the derivative at x which is zero.

 

[micromass]

if f(x+a) = f(x-a) where 'a' is infinitesimally small then does the result df(a)/dx = 0 follow?

Does $f(x+a) = f(x-a)$ hold for all $x$? For all infinitesimals $a$?

 

[micromass]

It is the derivative at x which is zero.

Not necessarily. Unless perhaps if the equation in the OP holds for all $x$.

 

[HallsofIvy]

Is that really the question you intended to ask? PWiz's point is that f(a) for any specific number a is a constant so its derivative is 0. I suspect that you intended to ask "if f(x_0- a)= f(x_0+ a), for any infinitesimal a, is df/dx, at x= x_0, equal to 0?"

Though I will point out that "df(a)/dx" is a common short form for "df/dx evaluated at x= a" so I think he is stretching the point a little! However, since it is "a" that you are taking to be infinitesimal, you certainly don't what to say "df/dx evaluated at x= a" either.

 

[Ahmad Kishki]

Does $f(x+a) = f(x-a)$ hold for all $x$? For all infinitesimals $a$?

i just realized that my effort to ditch writing the question in latex resulted in a problem... i will edit the question for it to make sense.. Sorry about that.. I meant with x as just some point not the variable.

 

[Ahmad Kishki]

No. Why would you insert the infinitesimal a in the derivative? It is part of the limit in taking the derivative. It is the derivative at x which is zero.

if f(X+a) = f(X-a) where 'a' is infinitesimally small then does the result df(X)/dx = 0 follow?

X is a point along the x axis.. Sorry about the fault in the question

 

[Orodruin]

Not necessarily. Unless perhaps if the equation in the OP holds for all $x$.

Well, to be perfectly frank, the OP is mixing a lot of things.

 

[micromass]

if f(X+a) = f(X-a) where 'a' is infinitesimally small then does the result df(X)/dx = 0 follow?

Assuming that $f(x+a) = f(x-a)$ holds for every infinitesimal, then the answer is no. Take $f(x) = |x|$ at $X=0$.

 

[Ahmad Kishki]

Well, to be perfectly frank, the OP is mixing a lot of things.

Yeah really sorry about that :/

 

[Orodruin]

if f(X+a) = f(X-a) where 'a' is infinitesimally small then does the result df(X)/dx = 0 follow?

X is a point along the x axis.. Sorry about the fault in the question

I strongly suggest that you do not use the notation df(X)/dx to mean the derivative of f wrt x evaluated at X. The more common interpretation would be the derivative of f(X) wrt x, which is zero regardless of x. What was wrong with f'(X)?

 

[PWiz]

if f(X+a) = f(X-a) where 'a' is infinitesimally small then does the result df(X)/dx = 0 follow?

X is a point along the x axis.. Sorry about the fault in the question

Not necessarily. For example, let the function $f(x)$ equal $\frac {sin x}{x}$ . Then for "X" = 0 , you'll approach the same function output as $a$ tends to 0 satisfying the first condition, but the function at "X" will be undefined.

 

[micromass]

If you assume $f$ differentiable at $X$, then the answer is true. Indeed, the Newton quotient for an arbitrary infinitesimal $a$ is

$$\frac{f(X+a) - f(X)}{a} = \frac{f(X+(-a)) - f(X)}{a}$$

Taking standard part of left and right hand sides yields (since the function is differentiable at $X$):

$$f^\prime(X) = - f^\prime(X)$$

which gives us the result.

 

[Orodruin]

Not necessarily. For example, let the function $f(x)$ equal $\frac {sin x}{x}$ . Then for "X" = 0 , you'll approach the same value function output as $a$ tends to 0 satisfying the first condition, but the function at "X" will be undefined.

This function has a well defined limit and can easily be extended to a continuous and differentiable function at x = 0.

$$\frac{f(X+a) - f(X)}{a} = \frac{f(X+a) - f(X - a)}{a} + \frac{f(X-a) - f(X)}{a} = -\frac{f(X+(-a)) - f(X)}{a}$$

Sign error in the last step, but the later conclusion is correct. As it stands, the right hand side would evaluate to f'(X).

 

[PWiz]

This function has a well defined limit and can easily be extended to a continuous and differentiable function at x = 0.

The OP didn't specify the function domain, so I took the liberty of using natural domains.

 

[my2cts]

No. It only means that the function is continuous in x.

 

[micromass]

No. It only means that the function is continuous in x.

No it doesn't.

 

[Orodruin]

No. It only means that the function is continuous in x.

No it does not. Consider the function f(x) = 0 if x ≠ 0 and f(0) = 1.

 

[Ahmad Kishki]

Is that really the question you intended to ask? PWiz's point is that f(a) for any specific number a is a constant so its derivative is 0. I suspect that you intended to ask "if f(x_0- a)= f(x_0+ a), for any infinitesimal a, is df/dx, at x= x_0, equal to 0?"

Though I will point out that "df(a)/dx" is a common short form for "df/dx evaluated at x= a" so I think he is stretching the point a little! However, since it is "a" that you are taking to be infinitesimal, you certainly don't what to say "df/dx evaluated at x= a" either.

Check corrected question... Its a question that i wrote down with a + epsilon, and since i wanted to avoid latex, it got me confused... In short, check the corrected question... Sorry about that

 

[Ahmad Kishki]

Thank you all for the answes provided. The answer to the question is that the claim is true if f'(X) exists.

 

[Ahmad Kishki]

Follow up:

I was talkig about this with a friend and he told me that he suspects if this is true and gave me an example of f(x) = x^2 and said f(x-a) = f(x+a) for say x=2 when a is infinitesimally small, but the derivative of x^2 at x=2 is obviously not zero. I answered him by saying that f(x+a) is in fact not equal to f(x-a) they must differ by a small distance delta such that delta/2a is equivalent to the derivative at the point in question. I am not very solid with the epsilon-delta definition so is this the right way to answer this?

 

[micromass]

$f(2 -a ) = (2 - a)^2 = 4 - 2a + a^2$ and $f(2+a) = (2+a)^2 = 4 + 2a + a^2$. Hence $f(2-a) - f(2+a) = -4a$. This is not zero, so there is indeed an infinitesimal distance between them.

 

[Ahmad Kishki]

$f(2 -a ) = (2 - a)^2 = 4 - 2a + a^2$ and $f(2+a) = (2+a)^2 = 4 + 2a + a^2$. Hence $f(2-a) - f(2+a) = -4a$. This is not zero, so there is indeed an infinitesimal distance between them.

Interesting point: i would expect 4a/2a=2 to be the derivative of x^2 at x=2 but this isn't the case the derivative is 4 not 2, so what does this 2 mean and why don't we get the derivative?

(Since the deivative is f(2+a) - f(2-a) = 2a * f'(2) leading to 4a/2a = 2)

What went wrong?

 

[my2cts]

No it does not. Consider the function f(x) = 0 if x ≠ 0 and f(0) = 1.

This is mathematical pettifoggery. But yes, the case a=0 was excluded by speaking of "a" as "infinitesimally small".

Not necessarily. For example, let the function $f(x)$ equal $\frac {sin x}{x}$ . Then for "X" = 0 , you'll approach the same function output as $a$ tends to 0 satisfying the first condition, but the function at "X" will be undefined.

Everybody except a mathematician or a pocket calculator knows that sin(x)/x=1. ;-)

 

[micromass]

This is mathematical pettifoggery. But yes, the case a=0 was excluded by speaking of "a" as "infinitesimally small".

Yes, but that is besides the point.

 

[PWiz]

Everybody except a mathematician or a pocket calculator knows that sin(x)/x=1. ;-)

If the domain is not carefully specified, then the value remains undefined "at" x=0 as the natural domain of the function is then considered.

 

[Orodruin]

But yes, the case a=0 was excluded by speaking of "a" as "infinitesimally small".

Even if $a = 0$, the quoted function satisfies the requirement $f(x+a) = f(x-a)$ at $x = 0$ (in fact, it is satisfied for all $a$, be it finite, infinitesimal, or zero).

For $a = 0$, any function will satisfy the relation, regardless of whether the function is continuous or not.

This is mathematical pettifoggery.

This is a mathematics subforum after all. There was one statement in your post and it was not a logical conclusion from the given fact. I think it is worth pointing this out in order not to confuse the OP.

 

[sgphysics]

No it does not. Consider the function f(x) = 0 if x ≠ 0 and f(0) = 1.

It is always true for a function that is continuous around x. You can not say that if it is discontinuous. But sure, you may find examples of discontinuous functions where it also holds.

 

[micromass]

It is always true for a function that is continuous around x. You can not say that if it is discontinuous. But sure, you may find examples of discontinuous functions where it also holds.

What is always true for a continuous function? That $f(x+a) = f(x-a)$? That would be false.

 

[Orodruin]

What is always true for a continuous function? That $f(x+a) = f(x-a)$? That would be false.

Well, seeing that he is quoting my reply to #18, the logical conclusion would be that it is referring to #18 and thus saying that a continuous function is always continuous, but that seems like a tautology. I would say he is referring to $f'(x) = 0$, but then that has already been established in this thread.

 

[sgphysics]

What is always true for a continuous function? That $f(x+a) = f(x-a)$? That would be false.

The OPs question, as I understand iit, is if limit f(x+a)=f(x-a) as a-->0. It is obviously true for a function cont. around x. Always. It is mererly the definition of a continuous function.

 

[Orodruin]

The OPs question, as I understand iit, is if limit f(x+a)=f(x-a) as a-->0. It is obviously true for a function cont. around x. Always. It is mererly the definition of a continuous function.

This is not what the OP is asking. The OP is asking if f(X+a) = f(X-a) implies f'(X) = 0. It does if the function is differentiable at X, continuity is not sufficient as the derivative might not exist. An example which is continuous and satisfies the relation at X = 0 would be Weierstrass' function, which is not differentiable at X = 0 (or anywhere else for that matter). A more mundane example would be $f(x) = |x|$ (also at $x = 0$), which is continuous but not differentiable at $x = 0$.

The OP's question has been answered, I am closing therefore closing this thread.