Skal vise at
[tex]\frac{d}{dz} \sin ^{-1}z = \frac{1}{(1-z^2)^{1/2}[/tex]
når
[tex]\sin ^{-1}z = -i \log[iz + (1-z^2)^{1/2}][/tex].
Jeg får et uttrykk som jeg ikke ser ut til å klare å forenkle til det ønskelige.
Deriverte av kompleks arcsin.
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- Ramanujan
- Posts: 260
- Joined: 16/04-2009 21:41
[tex]\small{\text{atm: fys1120, ast1100, mat1120, mat2410 \ . Prev: mat1110, fys-mek1110, mek1100, mat1100, mat-inf1100, inf1100}}[/tex]
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- Ramanujan
- Posts: 260
- Joined: 16/04-2009 21:41
Allright 
[tex]\frac{d}{dz}\sin ^{-1}z =\frac{d}{dz} -i \log[iz + (1-z^2)^{1/2}][/tex]
[tex]= \frac{-i}{iz + (1+z^2)^{1/2}} \left( i + 1/2 (1-z^2)^{-1/2}(-2z)\right) = \frac{1 + iz(1-z^2)^{-1/2}}{iz + (1-z^2)^{1/2}}[/tex]

[tex]\frac{d}{dz}\sin ^{-1}z =\frac{d}{dz} -i \log[iz + (1-z^2)^{1/2}][/tex]
[tex]= \frac{-i}{iz + (1+z^2)^{1/2}} \left( i + 1/2 (1-z^2)^{-1/2}(-2z)\right) = \frac{1 + iz(1-z^2)^{-1/2}}{iz + (1-z^2)^{1/2}}[/tex]
[tex]\small{\text{atm: fys1120, ast1100, mat1120, mat2410 \ . Prev: mat1110, fys-mek1110, mek1100, mat1100, mat-inf1100, inf1100}}[/tex]
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- Ramanujan
- Posts: 260
- Joined: 16/04-2009 21:41
Da kommer jeg i mål. Takk for det 

[tex]\small{\text{atm: fys1120, ast1100, mat1120, mat2410 \ . Prev: mat1110, fys-mek1110, mek1100, mat1100, mat-inf1100, inf1100}}[/tex]