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How does $\left( 1 - \left(1- \frac{1}{2^{2^k}}\right)\right)$ become $\left(1+ \frac{1}{2^{2^k}}\right)$?

I distributed the former but got negative $-\frac{1}{2^{2^k}}$. So it does not match the latter.

I'm basing the math from this website: http://zimmer.csufresno.edu/~larryc/proofs/proofs.mathinduction.html Please observe the math 2nd part of inductive step for the 2nd example(a recurrence formula)

Nicholas
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    If you want $2^{2^k}$, type 2^{2^k} in dollar signs, and if you want $(2^2)^k$, type (2^2)^k in dollar signs. The problem is not the fraction, but the fact that the expression is ambiguous if you don't group the exponents. – coldnumber Jul 15 '15 at 02:38
  • @ColdNumber +1 for you. Followed your advice. – Nicholas Jul 15 '15 at 02:46
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    There is an issue, mathematically, with your question, regardless of the potential for ambiguity. The two things aren't equal. At all. – Terra Hyde Jul 15 '15 at 02:49
  • Glad it worked :). Now, about your question, I don't understand what you mean by "become". Are you asking whether the expressions are equal? – coldnumber Jul 15 '15 at 02:49
  • Based on your most recent question concerning induction, it seems like you are struggling to master the concept and/or mechanics of proofs by induction. You may find this question and this one to be of use while trying to better understand this proof technique. – Daniel W. Farlow Jul 15 '15 at 02:57

2 Answers2

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EDIT: Looking at the link you provided, you missed a couple of divisions.

$$\left(1-\left(1-\frac 1{2^{2^k}}\right)\Big/2\right) = \left( 1+ \frac 1 {2^{2^k}}\right)\Big/2$$

The line in question, when MathJaxed up a bit, is: $$\begin{align}a_{k+1} & = \left(2 \left(1 - \frac 1{2^{2^k}}\right)\frac 12 \right)\; \left(1 - \left(1 - \frac 1{2^{2^k}}\right)\frac 1 2\right) \\ & = \left(1 - \frac 1 {2^{2^k}}\right)\;\left(1 + \frac 1{2^{2^k}}\right)\frac 12 \\ & = \color{blue}{\left(1 - \left(\frac 1 {2^{2^{k}}}\right)^2\right)\frac 12} \\ & = \color{blue}{\left(1 - \frac 1 {2^{2\times 2^{k}}}\right)\frac 12} \\ & = \left(1 - \frac 1 {2^{2^{k+1}}}\right)\frac 12\end{align}$$

EDIT: Added additional steps.

Graham Kemp
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You have $a_{k+1}=2a_k(1-a_k)$, with $a_k=\frac 12 \left(1-\frac{1}{2^{2^k}}\right).$ You were missing the $\frac 12$, as Graham Kemp notes.

The inductive step should read: \begin{align} a_{k+1}=2a_k(1-a_k)&=2\frac 12 \left(1-\frac{1}{2^{2^k}}\right)\left(1-\frac 12 \left(1-\frac{1}{2^{2^k}}\right)\right)\\ &=\left(1-\frac{1}{2^{2^k}}\right)\left(1-\frac 12 +\frac 12\frac{1}{2^{2^k}}\right)\\ &=\left(1-\frac{1}{2^{2^k}}\right)\left(1 +\frac{1}{2^{2^k}}\right)\frac 12\\ &= \frac 12 \left(1-\left(\frac{1}{2^{2^k}}\right)^2 \right)\\ &= \frac 12 \left(1-\frac{1}{2^{2^{k+1}}} \right) \end{align}

coldnumber
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  • May you explain the last two parts? You multiplied the fraction by itself and got the result. But I cannot comprehend why you got the result you did. – Nicholas Jul 15 '15 at 04:08
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    @Nicholas: The first of the last two parts follows from $(1-a)(1+a)=(1-a^2)$, where $a$ is your fraction. The second one: the denominator of the fraction squared is $2^{2^k}\cdot 2^{2^k}=2^{2^k+2^k}=2^{2\cdot 2^k}=2^{2^{k+1}}$. Were these the parts you were referring to? – coldnumber Jul 15 '15 at 04:26
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    oh okay. now I understand both parts. You broke it down that I understood it 100%! – Nicholas Jul 15 '15 at 04:59
  • Glad I could help :) – coldnumber Jul 15 '15 at 05:01