expand 3sin(4x)+4cos(10x) in to functional series of e(exp)ix
Recall that
sin(x) = (e^ix - e^-ix)/2i
cos(x) = (e^ix + e^-ix)/2
3 sin(4x) = 3(e^4ix - e^-4ix)/2i
4 cos(10x) = 4(e^10x + e^-10x)/2
Simplify as you will. Not sure what you mean by "functional series" involving e^ix. Using those definitions, there is no series involved.
If you just want series expansions, recall that
sin(x) = x - x^3/3! + x^5/5! - ...
cos(x) = 1 - x^2/2! + x^4/4! - ...
make substitutions for your values, and rewrite as needed.
that should be
4 cos(10x) = 4(e^10ix + e^-10ix)/2
To expand the given expression into a functional series of e^(ix), we can use the Euler's formula, which states that e^(ix) = cos(x) + i*sin(x), where i is the imaginary unit.
First, let's expand 3sin(4x):
Using Euler's formula, we can write sin(4x) as:
sin(4x) = (1/2i) * (e^(4ix) - e^(-4ix))
Next, let's expand 4cos(10x):
Using Euler's formula, we can write cos(10x) as:
cos(10x) = (1/2) * (e^(10ix) + e^(-10ix))
Now, let's substitute these expansions back into the given expression:
3sin(4x) + 4cos(10x) = 3 * (1/2i) * (e^(4ix) - e^(-4ix)) + 4 * (1/2) * (e^(10ix) + e^(-10ix))
Simplifying further, we can rewrite this as:
(3/2i) * (e^(4ix) - e^(-4ix)) + 2 * (e^(10ix) + e^(-10ix))
Next, we can distribute the fractions and simplify:
(3/2i) * e^(4ix) - (3/2i) * e^(-4ix) + 2 * e^(10ix) + 2 * e^(-10ix)
Now, let's combine the terms with the same powers of e^(ix):
= (3/2i) * e^(4ix) + 2 * e^(10ix) - (3/2i) * e^(-4ix) + 2 * e^(-10ix)
Finally, we can use Euler's formula to rewrite these terms in terms of e^(ix):
= (3/2i) * (cos(4x) + i*sin(4x)) + 2 * (cos(10x) + i*sin(10x)) - (3/2i) * (cos(-4x) + i*sin(-4x)) + 2 * (cos(-10x) + i*sin(-10x))
Simplifying further:
= (3/2i) * (cos(4x) + i*sin(4x)) + 2 * (cos(10x) + i*sin(10x)) - (3/2i) * (cos(4x) - i*sin(4x)) + 2 * (cos(10x) - i*sin(10x))
And finally, combining like terms, we have:
= (3/2i) * 2i * sin(4x) + 2 * (cos(10x) + cos(-10x)) + 2i * (cos(10x) - cos(-10x))
= 3 * sin(4x) + 2 * (cos(10x) + 1)
Therefore, the expanded functional series of 3sin(4x) + 4cos(10x) in terms of e^(ix) is 3*sin(4x) + 2*(cos(10x) + 1).