i tried to solve it

but i didnt get any parameter value

??

http://img403.imageshack.us/img403/5...9192yz6.th.gif

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- Jan 18th 2009, 06:50 AMtransgalacticcontinuety parameter question..
i tried to solve it

but i didnt get any parameter value

??

http://img403.imageshack.us/img403/5...9192yz6.th.gif - Jan 18th 2009, 07:08 AMIsomorphism
http://img403.imageshack.us/img403/5395/25739192yz6.gif

Here's a thought: what happens when m=0 ? - Jan 18th 2009, 07:20 AMtransgalactic
0^0 *bounded is an undefined form

- Jan 18th 2009, 07:46 AMIsomorphism
Look carefully... the definition $\displaystyle x^m \sin \frac1{x}$ holds only for $\displaystyle x \neq 0$. You cant simply substitute 0 in the place of x to check for continuity.

Thus your problem for m = 0 reads:

$\displaystyle f(x) = \begin{cases} \sin \frac1{x} & x\neq 0 \\ 0 & x = 0 \end{cases}$

Is f(x) continuous at x = 0?

(This is a very popular problem. I am sure you would have seen it somewhere in your notes/text) - Jan 18th 2009, 09:29 AMtransgalactic
ok so its 1*bounded wich is not defined either

?? - Jan 18th 2009, 09:34 AMIsomorphism
The function is well defined. The point of my post was to ask if f(x) is continuous at 0. I claim it is not. Can you prove it?

To prove it is discontinuous at 0, observe that sequence $\displaystyle x_n = \frac{2}{(2n+1)\pi}$ converges to 0, but $\displaystyle f(x_n) = (-1)^n$ - Jan 18th 2009, 12:33 PMtransgalactic
how does it finds the values of the m parameter

- Jan 18th 2009, 05:26 PMIsomorphism
First prove the following result:

(*)$\displaystyle \lim_{x \to 0} x^m \sin \frac1{x}$ exists and equal to 0 iff $\displaystyle m > 0$.

Now for part (A) we need to find m for which $\displaystyle \lim_{x \to 0} f(x) = f(0) = 0$, Thus we need m for which $\displaystyle \lim_{x \to 0} x^m \sin \frac1{x} = 0$. From the result (*), this happens iff $\displaystyle m > 0$.

For part (B), we need values of 'm' for which $\displaystyle \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0}$ exists. But $\displaystyle \lim_{x \to 0}\frac{f(x) - f(0)}{x - 0} = x^{m-1} \sin \frac1{x}$. Again by the result, the limit exists iff $\displaystyle m-1 > 0 \implies m > 1$

Do part (C) yourself, its instructive :) - Jan 18th 2009, 11:28 PMtransgalactic
its seems you already answered C

by solving A and B

it is continues on x=0 when m>0

it is differentiable on x=0 when m>1

so its continuously differentiable on m>1

thats the correct solution? - Jan 18th 2009, 11:59 PMIsomorphism
No! A function is continuously differentiable at x = 0 means the derivative at 0 exists and

__the derivative is continuous at 0__! I never proved any result about the derivative!

Ok I will do the last one too... But I will let you prove some statements on your own.

Since we have already proved that m > 1 iff f(x) is differentiable at x = 0, for continuous differentiability we need m > 1.

$\displaystyle f'(x) = x^{m-2} \left(mx \sin \frac1{x} - \cos \frac1{x}\right)$

Now first prove that if m > 2, then the function f'(x) is continuous at 0, using (*) and making a similar rule for $\displaystyle \lim_{x \to 0} x^m \cos \frac1{x}$(prove this too!).

Now the only remaining case is m=2. For this case,$\displaystyle f'(x) = \begin{cases} 2x\sin \frac1{x} - \cos \frac1{x} & x \neq 0 \\ 0 & x = 0 \end{cases}$.

Choose the sequence $\displaystyle x_n = \frac1{n \pi}$, and observe that $\displaystyle x_n \to 0$ but $\displaystyle f(x_n) = -(-1)^n$. Thus f(x) is not continuously differentiable at x = 0.

**So at the end, let me list the exercises you have to do, to complete the proof rigorously:**

(*) http://www.mathhelpforum.com/math-he...9d641f57-1.gif exists and equal to 0 iff $\displaystyle m > 0$

(**) $\displaystyle f'(x) = x^{m-2} \left(mx \sin \frac1{x} - \cos \frac1{x}\right)$

(***) $\displaystyle \lim_{x \to 0} x^m \cos \frac1{x}$ exists and equal to 0 iff $\displaystyle m > 0$

(****) Using (**) and (***), prove that if m > 2, then $\displaystyle f'(x)$ is continuous at $\displaystyle x = 0$

- Jan 19th 2009, 02:30 AMtransgalactic
(*) http://www.mathhelpforum.com/math-he...9d641f57-1.gif exists and equal to 0 iff $\displaystyle m > 0$

the only way to prove it

is by cheching the cases of m.

if m<0 then as x goes to zero and m is negative

we get 0.

so the whole limit would be

0*bouded which is bound

so i think i found a mistake in your theory

its working for m differs 0

- Jan 19th 2009, 03:04 AMtransgalactic
i got a solution to A

and it says

some thing about

if m=0 then

$\displaystyle

x_n = \frac{1}{(2n)\pi}

$

i dont know how they came up with this Xn

and what i need to do with it

?? - Jan 19th 2009, 04:45 AMIsomorphism
- Jan 19th 2009, 05:16 AMtransgalactic
- Jan 19th 2009, 06:42 AMIsomorphism
That solution you got is of no use to part A. To demonstrate discontinuity at $\displaystyle x_0$, we have to find a sequence $\displaystyle x_n \to x_0$, but the sequence $\displaystyle f(x_n) \not \to f(x_0)$.

Your sequence $\displaystyle x_n \to 0$, but $\displaystyle f(x_n) = \sin 2n \pi = 0$. So it is of no use.

As a side note, I am really sorry transgalactic, that I am failing in explaining this concept to you. I have already given you the answer for this in my previous posts. The sequence youuse is $\displaystyle x_n = \frac2{(2n+1)\pi}$**can**

I am telling the following based on our discussion.....If you take it in the right spirit, I suggest you take tutions. Sometimes a live tutor teaching a new concept is an invaluable experience.