Question 1153897
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Here are the first few terms of the expansion of (a+b)^n using the binomial theorem.<br>
{{{C(n,0)*((a^n)(b^0)) + C(n,1)*((a^(n-1))(b^1)) + C(n,2)*((a^(n-2))(b^2)) + C(n,3)*((a^(n-3))(b^3))}}}<br>
When n is not a positive integer, we want to write the {{{C(n,r)}}} numbers in expanded form:<br>
{{{C(n,0) = 1}}}
{{{C(n,1) = n/1!}}}
{{{C(n,2) = (n(n-1))/2!)}}}
etc....<br>
Then the first few terms of the expansion are<br>
{{{(a+b)^n = a^n + ((n)/1!)((a^(n-1))(b^1)) + (((n)(n-1))/2!)((a^(n-2))(b^2)) + (((n)(n-1)(n-2))/3!)((a^(n-3))(b^3))}}}<br>
Now we can plug in n=0.5, a=16, and b=0.32 to find the square root of 16.32 to several decimal places.<br>
{{{a^n = 16^0.5 = 4}}}<br>
{{{((n)/1!)((a^(n-1))(b^1)) = (.5/1)(((16)^(-0.5))(.32^1)) = .04}}}<br>
{{{(((n)(n-1))/2!)((a^(n-2))(b^2)) = (((.5)(-.5))/2)(((16)^(-1.5))(.32^2)) = -0.0002}}}<br>
{{{(((n)(n-1)(n-2))/3!)((a^(n-3))(b^3)) = (((.5)(-.5)(-1.5))/6)(((16)^(-2.5))(.32^3)) = 0.000002}}}<br>
So we have the square root of 16.32, using the first four terms of the binomial expansion of {{{(16+.32)^0.5}}}, as being<br>
{{{4+.04-.0002+.000002 = 4.039802}}}<br>
The actual value to 9 decimal places is 4.039801975, so our approximation is good to 6 decimal places.<br>