How do you use the binomial theorem to approximate #1.08^(1/2)# and hence find #sqrt(3)# to #4# significant figures?

Question

Kennedy Adams · Accepted Answer

Cutting the binomial series short,

#(1+x)^(1/2) ~= 1+x/2-x^2/8#

Then #sqrt(3) = 5/3sqrt(1.08) ~= 1+0.08/2-(0.08^2)/8 = 1.732# #(1+x)^(1/2) = 1+(1/2)x+(1/(2!))(1/2)(1/2-1)x^2+(1/(3!))(1/2)(1/2-1)(1/2-2)x^3+...# #=1+x/2-x^2/8+x^3/16-...# If #x# is small this will converge rapidly So with #x = 0.08# we get: #sqrt(1.08) = (1+0.08)^(1/2) ~= 1 + 0.08/2 - (0.08^2)/8# #=1+0.04-0.0008 = 1.0392# Now #1.08 = 3^3/(5^2)# So #sqrt(1.08) = sqrt(3^3/(5^2)) = sqrt(3*(3/5)^2) = 3/5sqrt(3)# So #sqrt(3) = 5/3sqrt(1.08) ~= 5/3*1.0392 = 1.732#

Kennedy Adams · Answer

undefined To approximate $1.08^{1/2}$ using the binomial theorem, we can express $1.08$ as $(1+x)$, where $x = 0.08$. Then, applying the binomial theorem to $(1+x)^{1/2}$, we have:

$(1+x)^{1/2} \approx 1 + \frac{1}{2}x$

Substituting $x = 0.08$, we get:

$1.08^{1/2} \approx 1 + \frac{1}{2}(0.08) = 1 + 0.04 = 1.04$

To find $\sqrt{3}$ to 4 significant figures using this approximation, we can calculate:

$\sqrt{3} \approx 1.08^{1/2} \approx 1.04$

Therefore, $\sqrt{3} \approx 1.04$ to 4 significant figures.