continuity and differentiability

[DrDolittle]

**how the function (modulus(x)) discontinuous.

**Somebody please explain this statement with an example "not all continuos functions are differentiable".

Regards
drdolittle

 

[Piccol0z]

**Somebody please explain this statement with an example "not all continuos functions are differentiable".

Well, lets take the closed-form function f(x) = |x|. It is continuous at every real number (including x = 0), but not differentiable at x = 0.

 

[Learner]

**Somebody please explain this statement with an example "not all continuos functions are differentiable".

Regards
drdolittle

Another way of looking at Piccol0z's explaination, when a function has a sharp peak it is not possible to derive the gradient of the peak. therefore eventhough it is a continuous function it is not differentiable.

If someone do not agree to this, please feel free to correct me. I would be appreciated.

 

[DrDolittle]

I slept on it and came out with this explantion.

As x -> o, f'(x) = f(0+h) - f(0)/ h ==> modulus(h)/h. this has the value +1 if h>0 and -1 if h<0 and hence does not tend to a limit as h->0.

My next question differentiation of sin(x). using limits we can say it is cos(x). can sombody give the geometrical interpretation of it.

Regards
drdolittle

 

[claudiocamera]

My next question differentiation of sin(x). using limits we can say it is cos(x). can sombody give the geometrical interpretation of it.

Well, I will not give you a geometric interpretation, I will give you a frequency domain interpretation.

The differentiation property of the Fourier transform tell us that diffentiation in time consist of multiplying the equivalent in frequency by Jw .

Multiplying the signal in frequency by Jw consist of a displacement in phase of 90°.

For our case sin(wt+90°) = cos(wt)

 

[DrDolittle]

I appreciate your effort. Explaining it using fourier transform is a piece of cake. Even the proof of fourier transform is constructed with the help of limits. I am looking for explanations at this level. Thanks again. no offence intended.

Regards
drdolittle

 

[LouisSheffield]

Might the orthogonality properties of sin(nx) and cos(nx) be a step toward a geometrical interpretation for you?

Added after 13 minutes:

... or a right triangle on a unit circle (hypotenuse = 1), where sin(x) is the opposite side and cos(x) is the adjacent - both by definition locked at right angles to each other ...

 

[claudiocamera]

I appreciate your effort. Explaining it using fourier transform is a piece of cake. Even the proof of fourier transform is constructed with the help of limits. I am looking for explanations at this level. Thanks again. no offence intended.

Regards
drdolittle

Don't worry it was not a offence, the explanation given using Fourier was an attempt for a practical approach, since we can easily construct diferenciators circuits, applying a sinusoidal in its input and observe that the output is 90 degree phased. There is no problem in providing the classical mathematicall proof.

The mathematical proof using limits is following:

d(sinx)/dx = lim Δx->infinit of [ sinx -sin(x+Δx)] / Δx

developing this we have:

d(sinx)/dx = lim Δx->infinit of [ sinx -( sinx*cosΔx+ sinΔx*cosx)] / Δx

We can infer that :

when Δx --> 0

cosΔx --> 1 and sinΔx-->Δx

replacing these values in the limit we get:

d(sinx)/dx = lim Δx->infinit of [ sinx - sinx*1 - Δx*cosx)] / Δx

what results in

d(sinx)/dx = lim Δx->infinit of [ - Δx*cosx)] / Δx

cancelling Δx in numerator with Δx in denominator we finally come across with:

d(sinx)/dx = - cosx

I think this is the explanation at the level of limits, is it what you wanted?

Similar approach is done in order to find the derivative of cosx.

Cheers

 

[DrDolittle]

ok guys! Gladly i found out atlast. I am using flash to create a movie like file. I will give the link once i am done with it...

Regards
drdolittle

 

[LouisSheffield]

Greetngs all ... (and nicely done)
Another way would be to look at the infinite series expansions for sin(x) and cos(x). The derivative/integral relationship of these comes out directly, just using polynomials.