How to use complex exponential to find higher derivatives of e^x cos(x√3)?

dcl
Messages
54
Reaction score
0
How would one use the complex exponential to find something like this:
<br /> \frac{{d^{10} }}{{dx^{10} }}e^x \cos (x\sqrt 3 )
I'm guessing we'd have to convert the cos into terms of e^{i\theta } but the only thing I can think of doing then is going through each of the derivatives. I am guessing there is another way?

thanks in advance.
 
Mathematics news on Phys.org
e^{i \theta} = \cos{\theta} + i\sin{\theta}

so

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

Then each successive derivative just adds a power to the coefficient of e.

cookiemonster
 
Last edited:
Ahhh, fair enough :)
Thought it was going to be tedious, but isn't nearly that bad..
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top