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Elvin Rivera
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1. The question is. Show that if |nx| <1, the following is exact up to (and including) the x^2 order. The hint giving says to use the Taylor Expansion for both sides of the equation2. (1+x)^n = e^n(x-(1/2)x^2) ; the n(x-(1/2)x^2) is all an exponent3. My first attempt was to take the taylor series of both sides. The left hand side is pretty easy.
1 + nx + (n/2)(n-1)x^2 = LHS
the right hand side gives the same thing though
f(0) = RHS = 1
f'(0) = n(1-x)e^n(x-(1/2)x^2) = n
f''(0) = -n*e^n(x-(1/2)x^2 + n^2 (1+x)^2 * e^n(x-(1/2)x^2) = -n + n^2
which in the taylor series gives the same as the LHS
1 + nx + (n/2)(n-1)x^2 = RHS
but now I'm stuck... how do I show that |nx| < 1
1 + nx + (n/2)(n-1)x^2 = LHS
the right hand side gives the same thing though
f(0) = RHS = 1
f'(0) = n(1-x)e^n(x-(1/2)x^2) = n
f''(0) = -n*e^n(x-(1/2)x^2 + n^2 (1+x)^2 * e^n(x-(1/2)x^2) = -n + n^2
which in the taylor series gives the same as the LHS
1 + nx + (n/2)(n-1)x^2 = RHS
but now I'm stuck... how do I show that |nx| < 1
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