Use binomial series to find the Taylor series about 0 for the function f(x)=(1+x)^-3/5 giving all terms up to the one in x^4. Then use this series and Taylor series for sin x to find the quartic Taylor polynomial about 0 for the function
f(x)= sinx/(1+x)^3/5
To find the Taylor series about 0 for the function f(x) = (1+x)^(-3/5), we can use the binomial series. The binomial series is given by the formula:
(1+x)^n = 1 + nx + (n(n-1)x^2)/2! + (n(n-1)(n-2)x^3)/3! + ...
In this case, we have n = -3/5. Let's find the terms up to the one in x^4.
Term 1: (1+x)^(-3/5) = 1
Term 2: (-3/5)x(1+x)^(-8/5) = -3/5 x -3/5 x x = 9/25 x^2
Term 3: (-3/5)(-8/5)x^2(1+x)^(-13/5) = 24/25 x^3
Term 4: (-3/5)(-8/5)(-13/5)x^3(1+x)^(-18/5) = -312/125 x^4
Putting it all together, the Taylor series about 0 for f(x) = (1+x)^(-3/5) up to the term in x^4 is:
f(x) = 1 + (9/25) x^2 + (24/25) x^3 - (312/125) x^4
Now, to find the quartic Taylor polynomial about 0 for the function f(x) = sin(x)/(1+x)^(-3/5), we need to multiply the Taylor series for sin(x) by the Taylor series for (1+x)^(-3/5) up to the term in x^4.
The Taylor series for sin(x) about 0 is:
sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
Multiplying each term by the terms we found in the Taylor series for (1+x)^(-3/5) up to the term in x^4:
Term 1: x * 1 = x
Term 2: -3/5 * x * (9/25) x^2 = -27/125 x^3
Term 3: x^3 * (24/25) x^3 = 24/25 x^6
Term 4: -3/5 * x^3 * (-312/125) x^4 = -7488/625 x^7
Putting it all together, the quartic Taylor polynomial about 0 for f(x) = sin(x)/(1+x)^(-3/5) is:
f(x) = x - (27/125) x^3 + (24/25) x^6 - (7488/625) x^7
To find the Taylor series for the function f(x) = (1+x)^(-3/5) about 0, we can use the binomial series. The binomial series states that for any real number n and a real number x such that |x| < 1, we have:
(1 + x)^n = 1 + nx + (n(n-1)x^2)/2! + (n(n-1)(n-2)x^3)/3! + ... + (n(n-1)(n-2)...(n-k+1)x^k)/k! + ...
In this case, we have f(x) = (1+x)^(-3/5), and we want to find the terms up to x^4. Using the binomial series, we can expand f(x) as follows:
f(x) = (1+x)^(-3/5) = 1 + (-3/5)x + (-3/5)(-3/5-1)x^2/2! + (-3/5)(-3/5-1)(-3/5-2)x^3/3! + ...
To simplify the expression, let's expand the coefficients:
f(x) = 1 - (3/5)x + (3/5)(8/5)x^2/2! - (3/5)(8/5)(13/5)x^3/3! + ...
Now, let's find the terms up to x^4:
f(x) = 1 - (3/5)x + (3/5)(8/5)x^2/2! - (3/5)(8/5)(13/5)x^3/3! + (3/5)(8/5)(13/5)(18/5)x^4/4! + ...
To find the quartic Taylor polynomial for the function f(x) = sin(x)/(1+x)^(-3/5), we need to multiply the terms of the Taylor series we found for f(x) by the corresponding terms in the Taylor series for sin(x).
The Taylor series for sin(x) about 0 is:
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
To find the quartic Taylor polynomial, we will multiply the terms up to x^4 in the series for f(x) by the corresponding terms in the series for sin(x):
quartic Taylor polynomial for f(x) = (1)x - (3/5)(x^2)/2! + (3/5)(8/5)(x^3)/3! - (3/5)(8/5)(13/5)(x^4)/4! + ...
Simplifying the expression, we get:
quartic Taylor polynomial for f(x) = x - (3/10)x^2 + (24/50)x^3 - (624/1000)x^4 + ...
And this is the desired quartic Taylor polynomial for the function f(x) = sin(x)/(1+x)^(-3/5) about 0.