mathematicians out there...

[Dacalo]

Please help me on this... I have been looking at this problem for days now, and i cant seem to comprehend it mathematically...

f(x)=e^(-1/x^2) if x does not equal zero and f(0)=0

Now how can x cannot equal zero but f(0)=0 ? I dont understand this logic. It wants me to show that f^n(0)=0 as in all derivatives.

help anyone? 🙁

 

[JayHu]

<< f(x)=e^(-1/x^2) if x does not equal zero and f(0)=0 >>



is this the question?

if so, it seems that they are defining f(0) for you to make the graph continuous (i am assuming ctns, havent looked at the limits yet).

*edit*
just confirmed f(0):=0 does make the graph cnts

 

[khtm]

JayHu,

exactly. that's what I was thinking. If they are defining f(0) = 0, then of course all f^n(0) = 0 because the derivative of a constant (or a point) is = 0.

Dacalo, is there more to this question than we think?...

-khtm-

 

[potatosalad]

<< f(x)=e^(-1/x^2) if x does not equal zero and f(0)=0 >>



It is a piecewise function. Are you familiar with piecewise functions?

Let me give you a simple example.

g(x) is a piecewise function such that:

if x < 10, g(x) = 0
if x = 10, g(x) = 1
if x > 10, g(x) = 0

This would be a piecewise function because for different regions of the domain of x, g(x) takes on specific values.

We can express the original equation as a piecewise function as follows.

f(x) is a piecewise function, such that:

if x < 0, f(x) = f(x)=e^(-1/x^2)
if x = 0, f(x) = 0
if x > 0, f(x) = f(x)=e^(-1/x^2)

Very simple.

 

[Dacalo]

well this pertains to the taylor series and power series...

see the question asks for the first derivative of the equation. then the second derivative...

then it says show why for f^n(0)=0 where n is an integer

i have tried from substituting the 1/x^2 with t to using the substituting in the Maclaurin series for e^x

 

[JayHu]

<< well this pertains to the taylor series and power series...

see the question asks for the first derivative of the equation. then the second derivative...

then it says show why for f^n(0)=0 where n is an integer

i have tried from substituting the 1/x^2 with t to using the substituting in the Maclaurin series for e^x >>



did the actual question tell you that f(0):=0?

then khtm is correct..
but taylor series.. mmm.. its been the big 4 months since i had to do that stuff..

 

[Dacalo]

thank you for all the help!🙂

 

[JayHu]

I hope we answered it..
LMK if there are any other questions.. i like this.. gets me ready for next term.. wait.. no calc next term.. oh well.. stat230..grr

 

[khtm]

<< I hope we answered it..
LMK if there are any other questions.. i like this.. gets me ready for next term.. wait.. no calc next term.. oh well.. stat230..grr >>



Aaaaaaa...
I HATE stats. 🙁 It hates me too. 😉

I also wonder if we answered Dacalo's question. That Taylor series stuff was too long ago for me to remember.

-khtm-

 

[ttn1]

<< exactly. that's what I was thinking. If they are defining f(0) = 0, then of course all f^n(0) = 0 because the derivative of a constant (or a point) is = 0. >>



Actually, I believe the question is asking if all the derivatives of the function, when evaluated at zero, are zero. This is not necessarily so obvious.

I believe you could expand this function, into some series, and then diffentiate each piece and then evaluate all the pieces at zero.

 

[Aelus]

by saying f(0), you mean you set x=0, either i'm wrong or you mean f(x)=0 instead.

also, f(0) != 0; but infinity in the limit

i'm also thinking at least the limit of the first derivate is undefined (inf*inf)

i think i'm just misunderstanding you, translating math stuff inbetween languages is hard.

Aelus

 

[JayHu]

<< also, f(0) != 0; but infinity in the limit >>



hmm..
lets go through with this one...
limx->0 e^(-1/x^2) = lim 1/(e^1/x^2)

so.
as x->0
x^2->0
1/x^2 -> infinity
e^(1/x^2) -> infinity
1/e^(1/x^2) -> 0



<< Actually, I believe the question is asking if all the derivatives of the function, when evaluated at zero, are zero. This is not necessarily so obvious. >>



Actually i find this somewhat obvious..
f'(x) = 2/(x^3) * f(x)
f''(x) = something with even bigger denominator..
f'''(x) = even bigger..

soo..
one could make the generalization that f^n(0) = 0 for all n...