continuity and differentiability

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

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

Regards
drdolittle

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

Click to expand...

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.

DrDolittle said:

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

Regards
drdolittle

Click to expand...

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.

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

DrDolittle said:

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

Click to expand...

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)

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

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 ...

DrDolittle said:

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

Click to expand...

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

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

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.