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Jocher
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d^1/2(ax+c)^(3/2)/dx^(1/2)=? please help me :)
Half Derivative Of (ax+c)^(3/2)
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In summary, the conversation discusses the use of fractional calculus in computing the derivative of a function. It is noted that taking everything under a common square root does not work for the derivative operator. Instead, the Riemann-Liouville operator is suggested for the definition of the fractional derivative. It is also mentioned that the classical derivative can be computed first, followed by the fractional integration. The conversation concludes by mentioning the use of fractional derivatives in complex impedance analysis and the interest in the fractional derivatives of sinusoidal functions.
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Simon Bridge
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It is tempting to start by taking everything under the common square root, isn't it?
However, you have probably figured out that this does not work for the derivative operator.
Just like when [itex]y=x^{1/2}[/itex], you are actually looking for [itex]y: y^2=x[/itex], when you see [itex]D^{1/2}f[/itex] you are actually looking to use something like [itex]Hf(x)
^2f(x)=Df(x)[/itex].
Look for http://en.wikipedia.org/wiki/Fractional_calculus]fractional[/PLAIN] calculus[/itex].
However, you have probably figured out that this does not work for the derivative operator.
Just like when [itex]y=x^{1/2}[/itex], you are actually looking for [itex]y: y^2=x[/itex], when you see [itex]D^{1/2}f[/itex] you are actually looking to use something like [itex]Hf(x)
^2f(x)=Df(x)[/itex].Look for http://en.wikipedia.org/wiki/Fractional_calculus]fractional[/PLAIN] calculus[/itex].
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JJacquelin
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Jocher said:
In using the Riemann-Liouville operator for the definition of the fractionnal derivative, and in order to avoid somme difficulties of integral convergence, it will be simpler to :
First, compute the fractional integral (order 1/2) of the function.
Then, compute the classical derivative.
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JJacquelin
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You're welcome !
I used fractionnal derivatives in complex impedance analysis.
The particular case of half derivative is common in modeling the electrical properties of many homogeneous compounds (a particular behaviour and impedance, long ago loosely refreed as Warburg's impedance).
Especially, the fractionnal derivatives of the sinusoidal functions are interesting.
For example, in the paper "The Phasance Concept", pp.5-6
http://www.scribd.com/JJacquelin/documents
I used fractionnal derivatives in complex impedance analysis.
The particular case of half derivative is common in modeling the electrical properties of many homogeneous compounds (a particular behaviour and impedance, long ago loosely refreed as Warburg's impedance).
Especially, the fractionnal derivatives of the sinusoidal functions are interesting.
For example, in the paper "The Phasance Concept", pp.5-6
http://www.scribd.com/JJacquelin/documents
1. What is the half derivative of (ax+c)^(3/2)?
The half derivative of (ax+c)^(3/2) is (3/4)a(ax+c)^(1/2).
2. How is the half derivative of (ax+c)^(3/2) calculated?
The half derivative of (ax+c)^(3/2) is calculated using the power rule for half derivatives, which states that the half derivative of x^n is (n/2)x^(n/2-1).
3. Can the half derivative of (ax+c)^(3/2) be simplified further?
Yes, the half derivative of (ax+c)^(3/2) can be simplified to (3/4)√(ax+c).
4. Is the half derivative of (ax+c)^(3/2) always positive?
No, the half derivative of (ax+c)^(3/2) can be positive, negative, or zero depending on the value of a.
5. What does the half derivative of (ax+c)^(3/2) represent?
The half derivative of (ax+c)^(3/2) represents the rate of change of the square root of (ax+c) with respect to x, when the exponent is 3/2.
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