The Solutions to last week's problems:
1) Let f(x) = y = x 3 - 3 x 2 + 5x - 4
And: g(u) = x = u 2 + u
Use the chain rule to show: d y/ du = dy/dx (dx/du)
Soln.: x = u 2 + u
But: dy/ dx = 3 x 2 - 6x + 5
=> 3 (u 2 + u) 2 - 6 (u 2 + u) + 5
dx/ du = 2u + 1
Then: dy/dx (dx/du) =
[3 (u 2 + u) 2 - 6 (u 2 + u) + 5] (2u + 1)
2) Find the fractional derivatives: D 1/2 (x 0) and D 1/2 (x 2)
Solns.:
D 1/2 (x 0) = G(n + 1)/ G(n + 1 - a)
D 1/2 (x 0) = G(0 + 1)/ G(0 + 1 - ½) = G(1)/ G( ½)
D 1/2 (x 2) Where: n = 2, a = ½
D 1/2 (x 2 ) = G(2 + 1)/ G(2 + 1 - ½) = G(3)/ G(2 ½)
= 1.505
3) Find the fractional derivative: D 1/2 (sin (x))
Soln.
D a (sin (x)) = sin (x + a p / 2) and: a = ½
So: D 1/2 (sin (x)) = sin (x + ( ½) p / 2) = sin (x + p / 4)
4) Is D 1/2 (1) the same as: d q (1)/ [d(x - a) ]q
For q = 1 in the 2nd? Explain.
Ans.
The two forms are the same, provided:
i) The interval of interest a < x < b is integrable. and
ii) The latter form is not confused with the Heaviside function for which:
f = 0, - ¥ < x < a, f = 1
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