So, what's the first order term in x - 1 ?Mr Davis 97 said:Homework Statement
What's the first order term in the expansion ln(x) about x = 1?
Homework Equations
Taylor series formula
The Attempt at a Solution
The question is multiple choice, and the choices are x, 2x, or (1/2)x. However, when I calculate the first order term in the expansion of ln(x) about 1, I get ln(1) + (1/1)(x - 1) = x - 1, which is not one of the options. What am I doing wrong?
Nothing. The problem statement is wrong or, at best, misleading. My guess is the intent was to ask what the linear term of ln(1+x) about x=0 is.Mr Davis 97 said:The question is multiple choice, and the choices are x, 2x, or (1/2)x. However, when I calculate the first order term in the expansion of ln(x) about 1, I get ln(1) + (1/1)(x - 1) = x - 1, which is not one of the options. What am I doing wrong?
The first order term in the Taylor expansion of ln(x) about 1 is simply the first derivative of ln(x) evaluated at x=1. This is represented as ln(1) = 0.
The first order term in the Taylor expansion of ln(x) about 1 is derived using the Taylor series formula, which states that for a function f(x) with continuous derivatives up to the nth order, the nth order Taylor polynomial about x=a is given by: Pn(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + ... + (f^n(a)/n!)(x-a)^n. In this case, the first derivative of ln(x) is used to find the first order term at x=1.
The first order term in the Taylor expansion of ln(x) about 1 represents the linear approximation of the function ln(x) near x=1. It gives us an estimate of the function's value at x=1 and can be used to approximate the function's value at nearby points as well.
Yes, the first order term in the Taylor expansion of ln(x) about 1 can be used to approximate ln(x) at values near 1. However, the accuracy of the approximation decreases as the distance from x=1 increases. This is because the Taylor series is a local approximation and becomes more accurate closer to the point of expansion.
The first order term in the Taylor expansion of ln(x) about 1 is equivalent to the first derivative of ln(x) evaluated at x=1. This shows that the first derivative of a function can be thought of as the linear approximation of the function near a given point, in this case, x=1.