Use binomial theorem to to expand squareroot of 4+x in ascending powers of x to four terms.Give the limits for which the expansion is valid
the general binomial expansion says ...
(1+x)^n = 1 + nx + n(n-1)/2! x^2 + n(n-1)(n-2) + ..
(4+x)^(1/2)
= 4^(1/2) (1 + x/4)^(1/2) = 2(1+x/4)^(1/2)
(1+x/4)^(1/2)
= 1 + (1/2)(x/4) + (1/2)(-1/2)/2! (x/4)^2 + (1/2)(-1/2)(-3/2)/3! (x/4)^3 +
= 1 + (1/8)x - (1/128)x^2 + (1/1024)x^3 - ...
so (4+x)^(1/2)
= 2( 1 + (1/8)x - (1/128)x^2 + (1/1024)x^3 - ...)
= 2 + (1/4)x - (1/64)x^2 + (1/512)x^3
of course the original √(4+x) is defined only for x≥-4
testing:
let x = 5
then √4+5) = 3
my expansion:
= 2 + 5/4 - 25/64 + 125/512 -
= 3.1
appears to converge to 3, as expected
To expand the square root of (4 + x) using the binomial theorem, we start by expressing (4 + x) in the form (a + b)^n. In this case, a = 4, b = x, and n = 1/2 (since we have a square root).
The binomial theorem states that:
(a + b)^n = C(r,0)*a^n*b^0 + C(r,1)*a^(n-1)*b^1 + C(r,2)*a^(n-2)*b^2 + ... + C(r,r)*a^0*b^r
where C(r,k) represents the binomial coefficient, given by C(r,k) = r! / (k! * (r-k)!), and r is the power to which the binomial is raised.
Let's apply this to our problem:
(4 + x)^(1/2) = C(1/2,0)*4^(1/2)*x^0 + C(1/2,1)*4^(1/2-1)*x^1
Now we can simplify:
(4 + x)^(1/2) = 4^(1/2) + (1/2)*4^(-1/2)*x
Further simplifying:
√(4 + x) = 2 + (1/4)*x
So, the expansion of the square root of (4 + x) in ascending powers of x to four terms is given by:
√(4 + x) ≈ 2 + (1/4)*x
Based on the binomial theorem, this expansion is valid when |x| < 4.
Therefore, the limits for which the expansion is valid are:
-4 < x < 4