Past Exam Question (GR8767 #56)

Forum for the GRE subject test in mathematics.
Algebrahs
Posts: 2
Joined: Sun Aug 30, 2015 11:05 am

Past Exam Question (GR8767 #56)

Postby Algebrahs » Sun Aug 30, 2015 11:12 am

Hi,

I was taking a practice exam from several years ago, and got tripped up by this question:

The polynomial p(x)=1+(1/2)(x-1)-(1/8)(x-1)^2 is used to approximate \sqrt(1.01). Which of the following most closely approximates the error \sqrt(1.01)-p(1.01)?

The answer is (1/16)x10^{-6}. But I do not know how they got this answer.

I would think that they want us to approximate p(1.01) by a tangent line approximation at x=1. But I don't know how to approximate the actual difference.

Thanks!

p-adic
Posts: 96
Joined: Fri Mar 27, 2015 6:42 pm

Re: Past Exam Question (GR8767 #56)

Postby p-adic » Sun Aug 30, 2015 10:41 pm

The Taylor series centered at 1 for sqrt(x) truncated after 3 terms is p(x) = 1 + 1/2(x-1) - 1/8(x-1)^2. The alternating series test remainder says that the error in absolute value will be <= the next term. The next coefficient is 3/8*1^(-5/2)/3! = 1/16. And (x-a)^3 = (1.01-1)^3 = .01^3 = 1/100^3 = 1/(10^2)^3 = 1/10^6. So you get 1/16*10^(-6).




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