Asked by ceasar
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
Answers
Answered by
Reiny
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
(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
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