# Does (f(x_0))=0 guarantee that d^n/dx^n(f(x_0)=0?

1. Feb 20, 2016

### Zacarias Nason

I just realized that I've never gotten anything like a firm or analytical answer to this question: Does the value of a function being zero at a given point necessitate that the value of the function at the same point is zero for all of its derivatives? If
$$f(x_0)=0$$
$$f'(x_0)=0$$ $$f''(x_0)=0$$ $$f'''(x_0)=0$$ $$...$$ $$f^n(x_0)=0$$?

2. Feb 20, 2016

### Samy_A

Look at the sine function. $\sin(0)=0$. What about the derivatives of the sine function in 0?
Another example $f(x)=x$.

3. Feb 20, 2016

### Zacarias Nason

The sine function example makes plenty sense to me, but the f(x) one doesn't. If you have the point $$f(x_0)=0, \ \ \ x_0=0,$$$$f'(x_0)=0$$
and so will its other derivatives at f(x)=0.

Last edited: Feb 20, 2016
4. Feb 20, 2016

### Samy_A

If $f(x)=x$ (for all $x$), then $f'(x)=1$.

5. Feb 20, 2016

### Zacarias Nason

Yes, but I explicitly didn't say x, I said x_0.

6. Feb 20, 2016

### Samy_A

The derivative of a function $f$ in a point $x_0$ doesn't depend only on the value of the function in that point.
All you give here is the value of $f$ in $x_0$. You can't compute the derivative without specifying $f$ in an open neighborhood of $x_0$. (I don't understand what you mean by $x=0$ in this context.)

7. Feb 20, 2016

### Zacarias Nason

Oh, this is a notation problem, that lies with me. That was supposed to be x_0=0, not x=0.

8. Feb 20, 2016

### Samy_A

Even then, the two examples I gave show that $f(x_0)=0$ doesn't imply $f'(x_0)=0$.

There is no generally valid relation between the value of a function in one point, and the value of the derivative at that same point.

Let's take two simple examples (where the function definition applies to all real $x$, and $A$ is an arbitrary real number):
1) $f(x)=x+A$.
Then $f(0)=A, f'(0)=1$.

2 $f(x)=Ax$ .
Then $f(0)=0, f'(0)=A$

9. Feb 20, 2016

### Zacarias Nason

I know and I agree. We're talking about two different things, heh. I agree that the two examples you gave show that it isn't necessary all the derivatives aren't zero, if my question had been written correctly in the first place. If the notation was correct in the first place the counterexample could not have included $$f(x)=x,$$ because in the instance that $$f(x_0)=0$$ and $$x_0=0, f'(x_0)=0$$ as well. The only possible value of f(x) where f(x_0) is 0 is x_0 = 0, with which case the derivative would always be zero. It's my mistake that this happened, but I get it.

10. Feb 20, 2016

### Samy_A

Frankly I don't understand what you ask.
is no.

If the OP doesn't present your question correctly, please state your question in a new post in a clear and unambiguous way.

Last edited: Feb 20, 2016
11. Feb 20, 2016

### Zacarias Nason

The answer to the OP, stated incorrectly, is no. The answer to the OP, stated correctly, is no. The counterexamples to justify that "no" in each case overlap but are not identical.

12. Feb 20, 2016

### HallsofIvy

Staff Emeritus
You seem to be confusing the fact that $f(x_0)$ is a number with f being a constant function. Any function is, of course, a constant a fixed value of $x_0$ but the derivative at $x= x_0$ depends upon value of f(x) close to $x_0$ as well as at $x_0$.

13. Feb 20, 2016

### Zacarias Nason

This has just spiraled out of control at this point into some growing thread based on misunderstanding of what other parties know at this point rather than actual misunderstanding. I get it, and got it after the second statement from Samy_A, my statements after the cos(x) and x posts related to me incorrectly expressing an already incorrect idea and Samy's counterexamples only applying wholly to the "correctly" expressed incorrect idea, and partly to the incorrectly expressed incorrect idea. There is no confusion at this point.

$$\text{If you have a function} \ f(x) \ \text{evaluated at} \ f(x_0)=0, \ \text{where} \ x=0, \ \text{this must mean that}\ f'(x_0)=0, \ f''(x_0)=0, \ ... \ , \ f^n(x_0)=0$$
The correctly stated version of an already incorrect idea should have been:
$$\text{If you have a function} \ f(x) \ \text{evaluated at} \ f(x_0)=0, \ \text{where} \ x_0=0, \ \text{this must mean that}\ f'(x_0)=0, \ f''(x_0)=0, \ ... \ , \ f^n(x_0)=0$$
Samy's first counterexample f(x)=sin(x) demonstrates failure of the first and second version, but his second counterexample f(x)=x does not demonstrate failure of the second version because the second version was not the original post, and the original post was "doubly wrong" while the second was only "partly wrong". One was wrong in both expression and ideas, the other was wrong only in idea.

Taking the derivative of f(x)=sin(x), evaluated at x_0=0, will showcase that just because the zeroth derivative evaluated at x_0=0 is zero does not mean any of the other derivatives evaluated at the same point will by necessity be zero.

Taking the derivative of f(x)=x, evaluated at x_0=0, will not showcase that just because the zeroth derivative evaluated at x_0=0 is zero does not mean any of the other derivatives evaluated at the same point. will by necessity be zero, because for f(x)=x evaluated at x_0=0, f'(x_0) DOES equal zero, f''(x_0) DOES equal zero, and so on and so forth. This is a

14. Feb 20, 2016

### Svein

Another counterexample: Let x0 = 0 and let $f(x)=e^{x}-1$. Then f(0)=0, but f'(0)=1 etc.

15. Feb 20, 2016

### SammyS

Staff Emeritus
SamyA's counter example of ƒ(x) = x, is a valid counter example for this correctly stated version , at least for the first derivative, $\ f'(x_0)\$.

It gives $\ f(x_0)= x_0=0\$ and $\ f'(x_0)=1\$.

16. Feb 20, 2016

### MrAnchovy

NO IT DOESN'T. For f(x) = x, f'(x) = 1 everywhere.

17. Feb 20, 2016

### Zacarias Nason

I'm nearly ready to slam my head through drywall, not because I think you're all wrong BUT I JUST REALIZED YOU'RE ALL RIGHT AND I DONE GOOFED x3

What a massive, massive waste of time.
I'm not sleeping much and am cramming a huge amount of information in a short time period, I owe you all a sincere apology. My bad, folks. If I was able to give out IOU's that were in any way meaningfully redeemable, you'd all have one now.