find first three terms in the expansion of ( 1 –x )^-1/4
Use (a+b)^n
= a^n + [n(n-1)/1!] a^(n-1) (b) + [n(n-1)(n-2)/2!] a^(n-2) b^2 + ..
(1 - x)^(-1/4)
= 1 +(-1/4) (-x) + (-1/4)(-5/4)/2 (-x)^2 + ...
= 1 + (1/4)x - (5/32)x^2 + ...
check my algebra
To find the first three terms in the expansion of (1-x)^(-1/4), we can use the binomial expansion formula:
(1 + x)^n = 1 + nx + (n(n-1)/2)x^2 + ...
In this case, we have (1-x) instead of (1+x), but we can still use the formula by replacing x with -x:
(1-x)^n = 1 + (-x)n + (n(n-1)/2)(-x)^2 + ...
Let's apply this formula to our problem:
(1-x)^(-1/4) = 1 + (-1/4)x + (-1/4)(-1/4-1)/2 * x^2 + ...
Simplifying this expression, we have:
(1-x)^(-1/4) = 1 - x/4 + (3/32)x^2 + ...
Therefore, the first three terms in the expansion of (1-x)^(-1/4) are: 1, -x/4, and (3/32)x^2.
To find the first three terms in the expansion of (1 - x)^(-1/4), we can use the binomial series expansion formula.
The binomial series expansion formula states that for any real number r and a real number x such that |x| < 1, the expression (1 + x)^r can be expanded as a power series:
(1 + x)^r = 1 + rx + [(r)(r - 1)(x^2)]/2! + [(r)(r - 1)(r - 2)(x^3)]/3! + ...
In our case, we need to find the expansion of (1 - x)^(-1/4). So we substitute -x for x and -1/4 for r in the formula:
(1 - x)^(-1/4) = 1 + (-1/4)(-x) + [(-1/4)(-1/4 - 1)(-x^2)]/2! + [(-1/4)(-1/4 - 1)(-1/4 - 2)(-x^3)]/3! + ...
Simplifying, we have:
(1 - x)^(-1/4) = 1 + (1/4)x + (3/32)x^2 + (5/128)x^3 + ...
Now we have the expansion of (1 - x)^(-1/4) in terms of a power series.
To find the first three terms, we will keep only the terms with the powers of x up to x^3:
(1 - x)^(-1/4) = 1 + (1/4)x + (3/32)x^2
Therefore, the first three terms in the expansion of (1 - x)^(-1/4) are 1, (1/4)x, and (3/32)x^2.