Find the McLauren series expansion (that is, infinite-order approxi-mation) of the function f (x) = 1(1−x) , assuming that x 6 = 1. Underwhat circumstances will the McLauren series expansion of this func-tion be finite? (Hint: You might like to review some of the materialfrom the “sequences, series, and limits” topic.)

The Maclaurin series expansion for a function f(x) is given by:

f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...

The function given is f(x) = 1/(1-x).

The derivative of f(x) is f'(x) = 1/(1-x)^2, the second derivative is f''(x) = 2/(1-x)^3, the third derivative is f'''(x) = 6/(1-x)^4, and so on.

At x = 0, these derivatives become f(0) = 1, f'(0) = 1, f''(0) = 2, f'''(0) = 6, and so on.

So, the Maclaurin series expansion for f(x) = 1/(1-x) is:

f(x) = 1 + x + 2x^2/2! + 6x^3/3! + ...

This simplifies to:

f(x) = 1 + x + x^2 + x^3 + ...

This is a geometric series with a common ratio of x.

The Maclaurin series expansion will be finite if and only if |x| < 1. This is because the geometric series 1 + x + x^2 + x^3 + ... converges to 1/(1-x) if and only if |x| < 1. If |x| >= 1, the series does not converge and the Maclaurin series expansion is not finite.

The Maclaurin series expansion of a function f(x) is a representation of that function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is a special case of the Taylor series expansion where the single point is chosen to be zero.

The function given is f(x) = 1/(1-x).

The Maclaurin series expansion of this function is given by:

f(x) = 1 + x + x^2 + x^3 + x^4 + ... = Σ (from n=0 to ∞) x^n

This is a geometric series with a common ratio of x.

The series converges if the absolute value of the common ratio is less than 1, that is, |x| < 1.

Therefore, the Maclaurin series expansion of the function f(x) = 1/(1-x) is finite if |x| < 1.

If |x| >= 1, the series does not converge and the Maclaurin series expansion is not finite.

In the special case where x = 1, the function f(x) = 1/(1-x) is undefined.

函数f(x) = 1/(1-x)的麦克劳林级数展开是一个无穷级数，其形式为f(x) = 1 + x + x^2 + x^3 + ...，这是一个等比级数，其公比为x。

这个级数在|x| < 1时收敛，也就是说，当x的绝对值小于1时，这个级数有一个有限的和。这是因为等比级数的收敛条件是公比的绝对值小于1。

当x = 1时，函数f(x) = 1/(1-x)变为无穷大，因此在这种情况下，麦克劳林级数展开是无法得到有限的结果的。

总结一下，函数f(x) = 1/(1-x)的麦克劳林级数展开在|x| < 1时是有限的，而在x = 1时是无穷大的。