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.