Differential calculus: If f (x) is given by (cosx + i sinx)(cos3x + i sin3x)...
- Thread starter Ishan
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I will guess that you are referring to Exercise 12.
12. If f (x) is given by f (x) = (cos x + i sin x)(cos 3x + i sin 3x) [cut off in image]
. . . . . . . . . ....(cos (2n - 1)x + i sin (2n - 1)x),
then f "(x) is equal to:
(a) n3 f (x) . . .(b) -n4 f (x) . . .(c) -n2 f (x) . . .(d) n4 f (x)
Please be complete. Thank you!
12. If f (x) is given by f (x) = (cos x + i sin x)(cos 3x + i sin 3x) [cut off in image]
. . . . . . . . . ....(cos (2n - 1)x + i sin (2n - 1)x),
then f "(x) is equal to:
(a) n3 f (x) . . .(b) -n4 f (x) . . .(c) -n2 f (x) . . .(d) n4 f (x)
What was contained in the cut-off portion of the image? What did you get when you applied de Moivre's Theorem (here)? Where are you stuck?
Please be complete. Thank you!
YesI will guess that you are referring to Exercise 12.
12. If f (x) is given by f (x) = (cos x + i sin x)(cos 3x + i sin 3x) [cut off in image]
. . . . . . . . . ....(cos (2n - 1)x + i sin (2n - 1)x),
then f "(x) is equal to:
(a) n3 f (x) . . .(b) -n4 f (x) . . .(c) -n2 f (x) . . .(d) n4 f (x)
What was contained in the cut-off portion of the image? What did you get when you applied de Moivre's Theorem (here)? Where are you stuck?
Please be complete. Thank you!
Sent from my m2 note using Tapatalk
I didnt knew this bu applying this theoram i gor the answer. -n^4f(x)I will guess that you are referring to Exercise 12.
12. If f (x) is given by f (x) = (cos x + i sin x)(cos 3x + i sin 3x) [cut off in image]
. . . . . . . . . ....(cos (2n - 1)x + i sin (2n - 1)x),
then f "(x) is equal to:
(a) n3 f (x) . . .(b) -n4 f (x) . . .(c) -n2 f (x) . . .(d) n4 f (x)
What was contained in the cut-off portion of the image? What did you get when you applied de Moivre's Theorem (here)? Where are you stuck?
Please be complete. Thank you!
Sent from my m2 note using Tapatalk
Thnx for ur helpI will guess that you are referring to Exercise 12.
12. If f (x) is given by f (x) = (cos x + i sin x)(cos 3x + i sin 3x) [cut off in image]
. . . . . . . . . ....(cos (2n - 1)x + i sin (2n - 1)x),
then f "(x) is equal to:
(a) n3 f (x) . . .(b) -n4 f (x) . . .(c) -n2 f (x) . . .(d) n4 f (x)
What was contained in the cut-off portion of the image? What did you get when you applied de Moivre's Theorem (here)? Where are you stuck?
Please be complete. Thank you!
Sent from my m2 note using Tapatalk
pka
Elite Member
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- 11,978
Thank you, thank you MS stapel. I would never have guessed that. I wish monitors would ban unreadable images.
\(\displaystyle \cos(kx)+{\bf{i}}\sin(kx)=\exp({\bf{i}}kx) \) thus \(\displaystyle f(x)=\prod\limits_{k = 1}^n {\exp ({\bf{i}}(2k-1)x)} = \exp ({\bf{i}}Ax)\) where \(\displaystyle A \) is the sum of the first \(\displaystyle n \) odd integers.