Is it possible to solve \cos(x) + 2e^{x} = 0

wuntsongo0cy

wuntsongo0cy

Answered question

2022-04-23

Is it possible to solve cos(x)+2ex=0 analytically?

Answer & Explanation

Colonninisxi

Colonninisxi

Beginner2022-04-24Added 16 answers

f(xn)=cos(xn)+2exn
f(xn)=2exnsin(xn)
now use the iterative formula:
xn+1=xnf(xn)f(xn)
xn+1=xn2exn+cos(xn)2exnsin(xn)
this is probably the easiest numerical way to solve it
Edit:
In terms of using:
eix+eix+4ex=0
we can express this as:
(ex)i+(ex)i+4ex=0
so if we let u=ex we get:
ui+ui+4u=0
which I do not think can be solved

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