Question

how can this be simplified
\[\cos (\ln x)\] ?

Answer 1

You can make it into a hyperbolic I believe or make close approximations to it using sums.

Answer 2

can you suggest a substitution ?

Answer 3

I apologize I put the answer, not a substitution. Remember how to to translate from regular into hyperbolic?

Answer 4

not really, i something

Answer 5

Remember Euler's formula. That should help.

Answer 6

\[re^{i\theta}=r(\cos\theta+i\sin \theta)\]

Answer 7

If you still stuck I'll give you another hint.

Answer 8

\[\cos\theta=\frac{e^{i\theta}-e^{i\theta}}2\]

Answer 9

You have got the wrong formula. You have sin(lnx). Also one of your i's is wrong.

Answer 10

ops i should have a negative somewhere

Answer 11

\[\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}2\]\[\cos (\ln x)=\frac{e^{i\ln x}+e^{-i\ln x}}2\]

Answer 12

Exactly. Now you should be able to write that in form of cosh correct?

Answer 13

Now use the Log Property:

\[\large e^{ilnx} \implies e^{lnx^i}\]

\[\large \implies e^{lnx} = x\]

Answer 14

\[\cos (\ln x)=\frac{i+-i}2=0?\]

Answer 15

I mean use the second formula there..

Answer 16

No..

Answer 17

\[\large \cos(lnx) = \frac{x^i + x^{-i}}{2}\]

Answer 18

oh yeah

Answer 19

Getting??

Want to solve further??

Answer 20

\[\large \cos(lnx) = \frac{x^{2i} + 1}{2x^i}\]

Answer 21

hmm..

Answer 22

is it getting simpler?

Answer 23

say i have this equation
\[y(x)=A\cos (\ln x)+B\sin (\ln x)+\frac12+\frac {\ln x}2\]
is it a good idea to 'simplify' \[y(x)=A\left(\frac{e^{-i \ln x}+e^{i \ln x}}2\right)+B\left(\frac{ie^{i\ln x}-ie^{-i\ln x}}2\right)+\frac12+\frac {\ln x}2\]
\[y(x)=\frac A2\left(x^{i}+x^{-i}\right)+\frac B2\left({ix^{-i^2}-ix^{i^2}}\right)+\frac12+\frac {\ln x}2\]to;
\[y(x)=\frac A2\left(x^{i}+x^{-i}\right)+\frac B2\left({ix-ix^{-1}}\right)+\frac12+\frac {\ln x}2\]

Answer 24

or was it simpler before?

Answer 25

I believe it was simpler before, unless you like using hyperbolics instead.

Answer 26

\[y(x)=A'\left(x^{i}+x^{-i}\right)+B'\left({ix-ix^{-1}}\right)+\frac12+\frac {\ln x}2\]

Answer 27

how does it look in hyperbolic form/

Answer 28

\[y(x)=A\cosh(i\ln x)+B\sinh (i\ln x)+\frac12+\frac {\ln x}2\]?

Answer 29

i think it was simpler before

Answer 30

Previous one is better than this actually..

You can also write it in Hyperbolic form..

\[\large \cos (\ln x)=\frac{e^{i\ln x}+e^{-i\ln x}}2 \implies \cos(lnx) = \cos(ilnx)\]

Answer 31

\[\large \sin(lnx) = \frac{-i(e^{ilnx} - e^{-ilnx})}{2} \implies \frac{-i(x^i - x^{-i})}{2}\]