Question

Use Maclaurin theorem to find the series expansion of f(x) =(1-x) up to and including the term containing x4.

Use (i) above to expand In - as a series of descending powers of x as far as the term x4. Use your expansion to find In 1.2 correct to 4 decimal places.

Answer 1

To find the series expansion of f(x) = (1-x) up to and including the term containing x^4, we can use the binomial theorem and substitute a = 1 and b = -x. The resulting series expansion is 1 - nx + n(n-1)x^2 - n(n-1)(n-2)x^3 + n(n-1)(n-2)(n-3)x^4. To find ln 1.2, substitute x = 0.2 into the series expansion and calculate each term to find the value of ln 1.2.

Step-by-step explanation:

To find the series expansion of f(x) = (1-x) using the Maclaurin theorem, we can use the binomial theorem.

The binomial theorem states that (a + b)^n = a^n + n*a^(n-1)*b + n(n-1)*a^(n-2)*b^2 + ... + b^n. In this case, a = 1 and b = -x.

Substituting these values into the binomial theorem, we have (1 - x)^n = 1^n + n*1^(n-1)*(-x) + n(n-1)*1^(n-2)*(-x)^2 + ... + (-x)^n.

Since we want to find the series expansion up to and including the term containing x^4, we can stop at the x^4 term.

Using the binomial theorem and simplifying the terms, we get the series expansion of f(x) = 1 - nx + n(n-1)x^2 - n(n-1)(n-2)x^3 + n(n-1)(n-2)(n-3)x^4.

To find ln 1.2 using this series expansion, we substitute x = 0.2 into the series expansion.

Plugging x = 0.2 into the series expansion, we get ln 1.2 ≈ 1 - nx + n(n-1)x^2 - n(n-1)(n-2)x^3 + n(n-1)(n-2)(n-3)x^4.

Calculating each term of the series expansion and rounding to 4 decimal places, we can find the value of ln 1.2.