Answer:
The Taylor series of f(x) around the point a, can be written as:
Here we have:
f(x) = 4*cos(x)
a = 7*pi
then, let's calculate each part:
f(a) = 4*cos(7*pi) = -4
df/dx = -4*sin(x)
(df/dx)(a) = -4*sin(7*pi) = 0
(d^2f)/(dx^2) = -4*cos(x)
(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4
Here we already can see two things:
the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.
so we only will work with the even powers of the series:
f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....
So we can write it as:
f(x) = ∑fₙ
Such that the n-th term can written as:
Answer: B, the are parallel but not equal. They are parallel line because they move in the same direction and they are not equal because they have different magnitude.
Step-by-step explanation:
I am not sure what you want to calculate as ultimate goal.
but your have an error already in your third line in the picture.
(3-6i)(2-4i) = 3×(2-4i) - 6i(2-4i)
you had there a "+" instead of a "-".
and then you made subsequent mistakes in every line, some of them are funnily bringing you more back to the real result (e.g. -12i + 12i would be 0 and not -24i, but if the third line would have been correct, then yes, -24i is actually needed) - but not completely.
= 3×2 - 3×4i - 6i×2 + 6i×4i =
= 6 - 12i - 12i + 24×-1 = 6 - 24i - 24 = -24i - 18
then in the last 3 lines you kind of lose it completely. I am absolutely not sure, what you are calculating, and where the "+" cube from instead of "×" (multiplications) and such.
(3+6i)(2+4i) = 3×2 + 3×4i + 6i×2 + 6i×4i =
= 6 + 12i + 12i + 24×-1 =
= 6 + 24i - 24 = 24i - 18
is that what you wanted to show ?
