Calculus Taylor Approximation Proof

[Elvin Rivera]

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

 

[HallsofIvy]

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 exponent

Then use parentheses to make that clear: (1+ x)^n= e^(n(x- (1/2)x^2)).

3. 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

You don't- that is not what is asked. You are asked to show that if |nx|< 1 then that equality is true, not the other way around.

 

[Elvin Rivera]

Then use parentheses to make that clear: (1+ x)^n= e^(n(x- (1/2)x^2)).
You don't- that is not what is asked. You are asked to show that if |nx|< 1 then that equality is true, not the other way around.

Any hint on how to start that?