# i need answer only, question number - 5 please solve this details

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### 1 Answer

The taylor approximations for f(x)=1/x about x=1 are

P1(x)=1

P2(x)=1-(x-1)

P3(x)=1-(x-1)+(x-1)^2

P4(x)=1-(x-1)+(x-1)^3-(x-1)^4

In general we want to know for what values of x these approximations are any good ( in other words what values give us convergence). We want to know how far can we get from x=1 and still have a valid approximation. For that we need to use the convergence test:

lim n->oo |a_n+1/a_n| = r. If r < 1 the series converges, r>1 diverges, and r=1 may or may not converge. Here a_n+1 and a_n are the n+1 and nth terms of the taylor series approximations. In our case,

lim n->oo |(x-1)^(n+1)/(x-1)^n| < 1

this is equal to |x-1| < 1. In other words we need -1 < x-1 < 1. So that 0 < x < 2. So the interval that works, and gives us better approximations as we increase the number of terms would be (0,2) which is an open interval.

As for the interval [1,4] in question (3), x=4 would be outside the interval for which the taylor series would converge. Therefore, it diverges for x=4 as we apply more successive approximations (as we add more terms P1, P2, P3, ... ).