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

1. May 23, 2015

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

Last edited: May 23, 2015
2. May 23, 2015

### 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.

3. May 23, 2015

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

4. May 23, 2015

### Orodruin

Staff Emeritus
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.

5. May 23, 2015

### micromass

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

6. May 23, 2015

### micromass

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

7. May 23, 2015

### HallsofIvy

Staff Emeritus
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.

8. May 23, 2015

i just realised 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.

9. May 23, 2015

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

10. May 23, 2015

### Orodruin

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

11. May 23, 2015

### micromass

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

12. May 23, 2015

Yeah really sorry about that :/

13. May 23, 2015

### Orodruin

Staff Emeritus
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)?

14. May 23, 2015

### PWiz

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.

15. May 23, 2015

### micromass

Staff Emeritus
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.

Last edited: May 23, 2015
16. May 23, 2015

### Orodruin

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

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

17. May 23, 2015

### PWiz

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

18. May 23, 2015

### my2cts

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

19. May 23, 2015

### micromass

Staff Emeritus
No it doesn't.

20. May 23, 2015

### Orodruin

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