For the meaning , see Prove that for all .(Hint induction using parts (a) and (b

Question

For the meaning , see Prove that for all .(Hint induction using parts (a) and (b) of exercise 18, or you can ( part (c) of exercise 18 with exercise 10 of Section 6.2, or plug something in for x and y in part (d) of exercise 18.) Prove that for all . A sequence a0, a1, a2,... is defined recursively as follows: a0 = 0; for every Prove that for all . Explain the paradox in the proof of Theorem 6.3.4, in which we i proof easier by changing the goal to a statement that looked like be harder to prove.

Explanation / Answer

Let's first prove that an>0 for n>0. Rearrange the given equation to read:

an+1-an=(1/4)*(2an-1)2 which is clearly positive. So an+1>an, and since a1>0, so are the rest. This proves the first inequality.

Now consider a=x (for brevity), which is the infinity-th term of the series. Let c=1/4 also for brevity.

Clearly (((c2+c)2+c)2+c+...)2+c=x. But the term within brackets itself equals a (or x)!

So x2+1/4 =x. Solve to get x=1/2 which is clearly < 1. Now we have already proved that an+1>an, so any term in the series with finite subscript is less than x=a, i.e. am<1/2 for any m. So we have proved a stricter upper bound than requested in the second inequality.

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