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Compute the Maclaurin series (i.e., the Taylor series about 0) of
f(x)=x^2 + arcsin(x)
up to and including terms of order two. Then: evaluate (a-zero)^2 + (a-one)^2 + (a-two)^2.
Hint: d/dx arcsin(x) = 1/sqrt(1 - x^2)
I'm getting 9/4 by just computing the Series to the second derivative and then using the rest of the formula. It's incorrect.
Also, can someone explain what the bolded part.
Thanks!
This is meaningless because "a-zero", "a-one", and "a-two" have not been defined. What does the problem really say?
If g(x)= arcsin(x), then g(0)= 0. As given, g'= 1/sqrt(1- x^2), so g'(0)= 1. That is the same as g'= (1- x^2)^(1/2) so g''= (1/2)(1- x^2)^(-1/2)(-2x)= -x(1- x^2)^(-1/2) and g''(0)= 0.Hint: d/dx arcsin(x) = 1/sqrt(1 - x^2)
Since you are asked only for the MacLaurin series up to "order 2", that is sufficient. The MacLaurin series for arcsin, up to degree 2, is 0+ x+ 0 so the MacLaurin series, up to degree 2, for x^2+ arcsin(x) isx+ x^2. I still have no idea how to answer the given questio because I do not know what "a-zero", "a_one", and "a_two" mean. I might guess that they are the coefficents of x^0, x^1, and x^2, respectively, in which case "[itex]a_0^2+ a_1^2+ a_2^2= 0+ 1+ 1= 2[/itex]", but I don't know that.
I'm getting 9/4 by just computing the Series to the second derivative and then using the rest of the formula. It's incorrect.
Also, can someone explain what the bolded part.
Thanks!