- #1

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[tex]f(x_0)=0[/tex]

does it follow that

[tex] f'(x_0)=0[/tex] [tex] f''(x_0)=0[/tex] [tex] f'''(x_0)=0[/tex] [tex]...[/tex] [tex] f^n(x_0)=0[/tex]?

- Thread starter Zacarias Nason
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- #1

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[tex]f(x_0)=0[/tex]

does it follow that

[tex] f'(x_0)=0[/tex] [tex] f''(x_0)=0[/tex] [tex] f'''(x_0)=0[/tex] [tex]...[/tex] [tex] f^n(x_0)=0[/tex]?

- #2

Samy_A

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Look at the sine function. ##\sin(0)=0##. What about the derivatives of the sine function in 0?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?

Another example ##f(x)=x##.

- #3

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The sine function example makes plenty sense to me, but the f(x) one doesn't. If you have the point [tex]f(x_0)=0, \ \ \ x_0=0,[/tex][tex]f'(x_0)=0[/tex]

and so will its other derivatives at f(x)=0.

and so will its other derivatives at f(x)=0.

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- #4

Samy_A

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If ##f(x)=x## (for all ##x##), then ##f'(x)=1##.The sine function example makes plenty sense to me, but the f(x) one doesn't. If you have the point [tex]f(x_0)=0, \ \ \ x=0,[/tex][tex]f'(x_0)=0[/tex]

and so will its other derivatives at f(x)=0.

- #5

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

- #6

Samy_A

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The derivative of a function ##f## in a point ##x_0## doesn't depend only on the value of the function in that point.Yes, but I explicitly didn't say x, I said x_0.

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.)The sine function example makes plenty sense to me, but the f(x) one doesn't. If you have the point [tex]f(x_0)=0, \ \ \ x=0,[/tex][tex]f'(x_0)=0[/tex]

and so will its other derivatives at f(x)=0.

- #7

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

- #8

Samy_A

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Even then, the two examples I gave show that ##f(x_0)=0## doesn't imply ##f'(x_0)=0##.Oh, this is a notation problem, that lies with me. That was supposed to be x_0=0, not x=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

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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,Even then, the two examples I gave show that f(x0)=0f(x0)=0f(x_0)=0 doesn't imply f′(x0)=0f′(x0)=0f'(x_0)=0.

- #10

Samy_A

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Frankly I don't understand what you ask.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 [tex]f(x)=x, [/tex] because in the instance that [tex]f(x_0)=0 [/tex] and [tex]x_0=0, f'(x_0)=0 [/tex] 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.

The answer to the OP:

is no.

[tex]f(x_0)=0[/tex]

does it follow that

[tex] f'(x_0)=0[/tex] [tex] f''(x_0)=0[/tex] [tex] f'''(x_0)=0[/tex] [tex]...[/tex] [tex] f^n(x_0)=0[/tex]?

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

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- #12

HallsofIvy

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- #13

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My original,

[tex] \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 [/tex]

The

[tex] \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 [/tex]

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

- #14

Svein

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- #15

SammyS

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Thecorrectly stated version of an already incorrect ideashould have been:

[tex] \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 [/tex]

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.

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

- #16

pbuk

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Taking the derivative of f(x)=x, evaluated at x_0=0, willnotshowcase 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

- #17

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What a massive, massive waste of time.

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