Let p be a real number and c=0 an integer such that c−0.1<xp(1+(1+x)101−(1+x)10)<c+0.1 for all (positive) real numbers x with 0<x<10−100. Find the ordered pair (p,c).
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
We are essentially studying the rational function f(x):=1+(1+x)101−(1+x)10=2+O(x)−10x+O(x2). Intuitively, f(x)≈2−10x=−5x for "small nonzero x ". So g(x):=xpf(x)≈−5xp+1 for "small nonzero x ". If p+1=0,g≈−5 becomes approximately constant as x→0. Since c is an integer, we must have c=−5 (as -5 is the only integer within 0.1 of -5 ).
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