SOLUTION: Please Help Me With This: Show That Sqrt{{(9+x^2)}}= 3 + x^2/6 - x^4/216. For What Value Of x is the expansion valid?

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Question 1024021: Please Help Me With This: Show That Sqrt{{(9+x^2)}}= 3 + x^2/6 - x^4/216. For What Value Of x is the expansion valid?
Found 2 solutions by ankor@dixie-net.com, robertb:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Show That Sqrt{{(9+x^2)}}= 3 + x^2/6 - x^4/216. For What Value Of x is the expansion valid?
:
sqrt%289%2Bx%5E2%29=+3+%2B+x%5E2%2F6+-+x%5E4%2F216
multiply by 216, cancel the denominators
216%2Asqrt%289%2Bx%5E2%29=+3%28216%29+%2B+36x%5E2+-+x%5E4
216%2Asqrt%289%2Bx%5E2%29=+648+%2B+36x%5E2+-+x%5E4
this is a mess, let's assume that x=0 and see what we have
216%2Asqrt%289%29=+648+%2B+0+-+0
216(3) = 648, That works! No other value for x does

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
If the right-hand side of the equation is supposed to be an infinite series, then the equation is true by using the binomial theorem:
+....
=3+%2B+%281%2F6%29x%5E2+-%281%2F216%29x%5E4+...
For this particular series the radius of convergence is x%5E2%3C=9, or -3%3C=x%3C=3, the only valid values for x.