i tried to solve it

but i didnt get any parameter value

??

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

Printable View

- January 18th 2009, 07:50 AMtransgalacticcontinuety parameter question..
i tried to solve it

but i didnt get any parameter value

??

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

Here's a thought: what happens when m=0 ? - January 18th 2009, 08:20 AMtransgalactic
0^0 *bounded is an undefined form

- January 18th 2009, 08:46 AMIsomorphism
Look carefully... the definition holds only for . You cant simply substitute 0 in the place of x to check for continuity.

Thus your problem for m = 0 reads:

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) - January 18th 2009, 10:29 AMtransgalactic
ok so its 1*bounded wich is not defined either

?? - January 18th 2009, 10:34 AMIsomorphism
- January 18th 2009, 01:33 PMtransgalactic
how does it finds the values of the m parameter

- January 18th 2009, 06:26 PMIsomorphism
First prove the following result:

(*) exists and equal to 0 iff .

Now for part (A) we need to find m for which , Thus we need m for which . From the result (*), this happens iff .

For part (B), we need values of 'm' for which exists. But . Again by the result, the limit exists iff

Do part (C) yourself, its instructive :) - January 19th 2009, 12:28 AMtransgalactic
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? - January 19th 2009, 12:59 AMIsomorphism
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.

Now first prove that if m > 2, then the function f'(x) is continuous at 0, using (*) and making a similar rule for (prove this too!).

Now the only remaining case is m=2. For this case, .

Choose the sequence , and observe that but . 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

(**)

(***) exists and equal to 0 iff

(****) Using (**) and (***), prove that if m > 2, then is continuous at

- January 19th 2009, 03:30 AMtransgalactic
(*) http://www.mathhelpforum.com/math-he...9d641f57-1.gif exists and equal to 0 iff

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

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

and it says

some thing about

if m=0 then

i dont know how they came up with this Xn

and what i need to do with it

?? - January 19th 2009, 05:45 AMIsomorphism
- January 19th 2009, 06:16 AMtransgalactic
- January 19th 2009, 07:42 AMIsomorphism
That solution you got is of no use to part A. To demonstrate discontinuity at , we have to find a sequence , but the sequence .

Your sequence , but . 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**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.