If and for all and the right hand derivative of at zero exists. Find at

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- June 14th 2009, 07:29 PMChandru1derivative problem
If and for all and the right hand derivative of at zero exists. Find at

- June 15th 2009, 10:17 PMBruno J.

Assuming is differentiable at 0, the limit

converges and is equal to ; which yields . - June 15th 2009, 11:52 PMCaptainBlack
- June 16th 2009, 12:02 AMBruno J.
I thought I had!

- June 16th 2009, 01:26 AMChandru1query
Its not given that g is differentiable

- June 16th 2009, 01:27 AMChandru1query
Its not given that g is differentiable, so u cant assume g' exists

- June 16th 2009, 02:27 PMBruno J.
Here we go then!

You have . Hence either or is identically zero which instantly yields . Hence suppose .

is easy; set x=y.

is also easy.

by the above.

Now consider the complex numbers

.

We have .

In particular , and .

Hence the function is a morphism of groups . Hence we have for some . Clearly since . Hence

Hence ,

so that . - June 16th 2009, 04:02 PMBruno J.
Just a little extra :

In case you are wondering why expressing m as an exponential map is possible, consider

Then for all integers n:

so that

from this it follows easily that for all rational x

.

Hence if is continuous on any interval, it coincides everywhere on that interval with (irrationals included).

But is certainly continuous at 0 (it's differentiable there!). To see that g is continuous at 0, we have

i.e.

and from this it is easy to show (with sequences for example) that is continuous at 0 (using the fact that is continuous at 0 and is equal to 0)