# De nition: Continuously Differentiable Function in Java Make QR Code 2d barcode in Java De nition: Continuously Differentiable Function

De nition: Continuously Differentiable Function
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Consider an open region D in the complex plane and a function f : D If this function is continuous and if the partial derivatives f / x and f / y exist and are continuous, we say that f is continuously differentiable in D If f is k times continuously differentiable where k = 0,1, 2, (that is, k derivatives of f exist and are continuous) we say that f is C k If f is C 0 , this is a continuous function which is not differentiable Now, how do we determine if a function is analytic If f is a continuously differentiable function on some region D then it is analytic if it has no dependence on z That function is analytic in a domain D provided that df (314) =0 dz This condition must hold for all points in D Note that a function which is analytic is also called holomorphic Later we will see that we can form a local power series expansion of a holomorphic or analytic function Now let s go back to writing a function of a complex variable in terms of real and imaginary parts as in Eq (313) We want to think about how to compute the derivative d/dz in terms of derivatives with respect to the real variables x and y Let s go back to basics Remember that z = x + iy This tells us that z =1 x and z =i y (315)
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Since z = x iy , it is the case that z =1 x and z = i y (316)
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These formulas can be inverted Recalling from Chap 1 that x = ( z + z )/2 and y = ( z z ) /2i we nd that x 1 x = = z 2 z and i y = 2 z i y =+ 2 z (318) (317)
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(remember that 1 /i = i) Using these results we can write the derivatives / z and / z in terms of the derivatives / x and / y In the rst case we have x y 1 i = + = z z x z y 2 x 2 y Similarly x y 1 i = + + = z z x z y 2 x 2 y (320) (319)
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Now we can use these results to write the derivatives df /dz and df /d z in terms of derivatives with respect to the real variables x and y: 1 f 1 = i f = i ( u + iv ) 2 x y y z 2 x = 1 u v i v u + + 2 x y 2 x y (321)
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1 f 1 = + i f = + i ( u + iv ) 2 x y y z 2 x = 1 u v i v u + + 2 x y 2 x y (322)
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Now we are in a position to determine whether or not a function is analytic that is, if it has no dependence on z by examining how it depends on the real variables
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x and y No dependence on z implies that the real and imaginary parts of Eq (322) must independently vanish This gives us the Cauchy-Riemann equations
De nition: The Cauchy-Riemann Equations
Using Eq (322) the requirement that f / z = 0 leads to ( u /dx ) ( v / y) = 0 and ( v / x ) + ( u / y) = 0 This gives the Cauchy-Riemann equations: u v = x y u v = y x
(323)
The Cauchy-Riemann equations can also be derived by looking at limits We do this by writing everything in terms of real and imaginary parts to that once again we take f ( z ) = u( x , y) + iv ( x , y), z 0 = x 0 + iy0 , and z = x + i y Now following what we did earlier and taking w = f ( z ) for notational convenience, we have w = f ( z0 + z ) f ( z0 ) = u( x 0 + x , y0 + y) u( x 0 , y0 ) + i[ v ( x 0 + x , y0 + y) v ( x 0 , y0 )] If the function f ( z ) is differentiable, we can approach the origin in any way we like So we try this in two ways, rst along the x axis (and hence setting y = 0) and then along the y axis (and setting x = 0) Going with the rst case, we set y = 0 and get w u( x 0 + x , y0 ) u( x 0 , y0 ) [ v ( x 0 + x , y0 ) v ( x 0 , y0 )] = +i z x x Now, taking the limit x 0 we see that these expressions are nothing other than partial derivatives That is:
x 0
[ v ( x 0 + x , y0 ) v ( x 0 , y0 )] u( x 0 + x , y0 ) u( x 0 , y0 ) w + i lim = lim x 0 x 0 x z x v u +i = x x
If the derivative f ( z ) exists, we must obtain the same answer even if we choose another way for ( x , y) to go to zero Now we use the other option, approaching the origin along the y axis Hence we set x = 0 This time we have w u( x 0 , y0 + y) u( x 0 , y0 ) [ v ( x 0 , y0 + y) v ( x 0 , y0 )] = + z i y y