how to find the terms through x^5 in the maclaurin series for?
Hint: it may be easiest to use knownMaclaurin series and then perform multiplications, divisions, and so on. For example, tan x = (sin x)/(cos x).
(1)f(x) = tan x,
(2) f(x) = e^2 sin x
(3) 1 + x^2 + X^3, a = 1
To find the terms through x^5 in the Maclaurin series for each of the given functions, we can use known Maclaurin series and perform the necessary multiplications, divisions, and additions.
1. f(x) = tan x:
The Maclaurin series for tan x can be obtained using the known Maclaurin series for sin x and cos x. The Maclaurin series for sin x is x - (x^3)/3! + (x^5)/5! - ... and the Maclaurin series for cos x is 1 - (x^2)/2! + (x^4)/4! - ...
To find the terms through x^5 in the Maclaurin series for tan x, we can divide the Maclaurin series for sin x by the Maclaurin series for cos x. Dividing each term of sin x by cos x, we get:
tan x = (x - (x^3)/3! + (x^5)/5! - ...) / (1 - (x^2)/2! + (x^4)/4! - ...)
To find the terms through x^5, we will perform the necessary multiplications and additions:
tan x = x + ((x^3)/3) + (2/15)x^5 + ...
So the terms through x^5 in the Maclaurin series for f(x) = tan x are: x, (x^3)/3, (2/15)x^5, ...
2. f(x) = e^(2*sin x):
To find the terms through x^5 in the Maclaurin series for e^(2*sin x), we can use the Maclaurin series for e^x and substitute 2*sin x in place of x:
e^(2*sin x) = 1 + (2*sin x) + ((2*sin x)^2)/2! + ((2*sin x)^3)/3! + ((2*sin x)^4)/4! + ...
To simplify this expression, we can apply the binomial theorem to the terms involving (2*sin x)^k. Expanding the terms, we get:
e^(2*sin x) = 1 + 2*sin x + (2^2)*(sin^2(x))/2! + (2^3)*(sin^3(x))/3! + (2^4)*(sin^4(x))/4! + ...
To find the terms through x^5, we will perform the necessary multiplications and additions:
e^(2*sin x) = 1 + 2*sin x + 2*sin^2(x) + (4/3)*sin^3(x) + (2/3)*sin^4(x) + ...
So the terms through x^5 in the Maclaurin series for f(x) = e^(2*sin x) are: 1, 2*sin x, 2*sin^2(x), (4/3)*sin^3(x), (2/3)*sin^4(x), ...
3. f(x) = 1 + x^2 + x^3:
In this case, the given function f(x) is already a polynomial. The Maclaurin series for polynomials is simply the polynomial itself.
So the terms through x^5 in the Maclaurin series for f(x) = 1 + x^2 + x^3 are: 1, x^2, x^3.