Derivative

**vernal** · December 26th 2011, 03:01 AM

please help me . i need it .

$\frac{d^{\frac{1}{2}}}{dx^\frac{1}{2}}sinx$

Fractional calculus - Wikipedia, the free encyclopedia

**chisigma** · December 26th 2011, 03:21 AM

Originally Posted by vernal

please help me . i need it .

$\frac{d^{\frac{1}{2}}}{dx^\frac{1}{2}}sinx$

Fractional calculus - Wikipedia, the free encyclopedia

You can apply the general formula...

$D^{\alpha} f(x) = \frac{1}{\Gamma(1-\alpha)} \frac{d}{d x} \int_{0}^{x} \frac{f(t)}{(x-t)^{\alpha}}\ dt$

... setting $\alpha=\frac{1}{2}$ and $f(x)=\sin x$

Kind regards

$\chi$ $\sigma$

**vernal** · December 26th 2011, 03:27 AM

Originally Posted by chisigma

You can apply the general formula...

$D^{\alpha} f(x) = \frac{1}{\Gamma(1-\alpha)} \frac{d}{d x} \int_{0}^{x} \frac{f(t)}{(x-t)^{\alpha}}\ dt$

... setting $\alpha=\frac{1}{2}$ and $f(x)=\sin x$

Kind regards

$\chi$ $\sigma$

thanks, can you solve??

**chisigma** · December 26th 2011, 05:31 AM

A 'brute force approach' probably fails, so that we try a 'step by step' solution. First step is to perform an integration be part obtaining...

$\int_{0}^{x} \frac{\sin t}{\sqrt{x-t}}\ dt = -2\ |\sin t\ \sqrt{x-t}|_{0}^{x} + 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt= 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt$ (1)
From (1) we derive...

$\frac{d}{dx} \int_{0}^{x} \frac{\sin t}{\sqrt{x-t}}\ dt = 2 \frac{d}{dx} \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt= \int_{0}^{x} \frac{\cos t}{\sqrt{x-t}}\ dt$ (2)

Now with a little of patience or using Wolfram if You don't have patience [

...] You find that...

$\int_{0}^{x} \frac{\cos t}{\sqrt{x-t}}\ dt = \sqrt{2 \pi}\ \{ \cos x\ \text{C} (\sqrt{\frac{2 x}{\pi} }) + \sin x\ \text{S} (\sqrt{\frac{2 x}{\pi}})\}$ (3)

… where C and S are the Fresnel Cosine and Fresnel Sine functions, so that, remembering that is $\Gamma(\frac{1}{2})= \sqrt{\pi}$ You obtain finally…

$D^{\frac{1}{2}} \sin x = \sqrt{2}\ \{ \cos x\ \text{C} (\sqrt{\frac{2 x}{\pi} }) + \sin x\ \text{S} (\sqrt{\frac{2 x}{\pi}})\}$ (4)

Kind regards

$\chi$ $\sigma$

**vernal** · December 26th 2011, 05:43 AM

Originally Posted by chisigma

A 'brute force approach' probably fails, so that we try a 'step by step' solution. First step is to perform an integration be part obtaining...

$\int_{0}^{x} \frac{\sin t}{\sqrt{x-t}}\ dt = -2\ |\sin t\ \sqrt{x-t}|_{0}^{x} + 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt= 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt$ (1)
From (1) we derive...

$\frac{d}{dx} \int_{0}^{x} \frac{\sin t}{\sqrt{x-t}}\ dt = 2 \frac{d}{dx} \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt= \int_{0}^{x} \frac{\cos t}{\sqrt{x-t}}\ dt$ (2)

Now with a little of patience or using Wolfram if You don't have patience [

...] You find that...

$\int_{0}^{x} \frac{\cos t}{\sqrt{x-t}}\ dt = \sqrt{2 \pi}\ \{ \cos x\ \text{C} (\sqrt{\frac{2 x}{\pi} }) + \sin x\ \text{S} (\sqrt{\frac{2 x}{\pi}})\}$ (3)

… where C and S are the Fresnel Cosine and Fresnel Sine functions, so that, remembering that is $\Gamma(\frac{1}{2})= \sqrt{\pi}$ You obtain finally…

$D^{\frac{1}{2}} \sin x = \sqrt{2}\ \{ \cos x\ \text{C} (\sqrt{\frac{2 x}{\pi} }) + \sin x\ \text{S} (\sqrt{\frac{2 x}{\pi}})\}$ (4)

Kind regards

$\chi$ $\sigma$

thanks for your help

**vernal** · December 29th 2011, 12:20 PM

Originally Posted by chisigma

A 'brute force approach' probably fails, so that we try a 'step by step' solution. First step is to perform an integration be part obtaining...

$\int_{0}^{x} \frac{\sin t}{\sqrt{x-t}}\ dt = -2\ |\sin t\ \sqrt{x-t}|_{0}^{x} + 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt= 2\ \int_{0}^{x} \cos t\ \sqrt{x-t}\ dt$ (1)
From (1) we derive...

$\chi$ $\sigma$

please say what way did use?

**vernal** · December 29th 2011, 09:43 PM

oh sorry:

$-2\int_{0}^{x}\frac{sin(t)}{-2\sqrt{x-t}}$
then :
$u=sin(t)~~,~dv=\frac{1}{-2\sqrt{x-t}}~~~\Rightarrow ~~~v=\sqrt{x-t}$

and

Is that true?

**chisigma** · December 29th 2011, 10:15 PM

Originally Posted by vernal

oh sorry:

$-2\int_{0}^{x}\frac{sin(t)}{-2\sqrt{x-t}}$
then :
$u=sin(t)~~,~dv=\frac{1}{-2\sqrt{x-t}}~~~\Rightarrow ~~~v=\sqrt{x-t}$

and

Is that true?

Yes, it is!... To understand why that 'little excamotage' is necessary I suggest You to try to apply the 'normal procedure' that consists, given a function...

$F(x)=\int_{\alpha(x)}^{\beta(x)} f(x,t)\ d t$ (1)

... in the computation of...

$\frac{d}{dx} F(x)= \int_{\alpha(x)}^{\beta(x)} f_{x} (x,t)\ dt - \frac{d \alpha}{d x}\ f(x,\alpha) + \frac{d \beta}{d x}\ f(x,\beta)$ (2)

... when $f(x,t)= \frac{\sin t}{\sqrt{x-t}}$ , $\alpha(x)=0$ and $\beta(x)=x$ ...

Kind regards

$\chi$ $\sigma$

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