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I know how to compute finite exponentiation limit ordinal expressed in normal form. However, how do i compute for nonlimit ordinal?

I know this statement is true. If $r$ is a limit ordinal and $b$ is a finite ordinal >0, then $r^b = \omega^{a_1b-1}r$

How do i compute ordinal such as $\omega^\omega + 12 = b$ then $b^5$?

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Is $w$ in your notation really $\omega$ (\omega)? –  Asaf Karagila Jun 2 '12 at 19:18
    
Yes i meant $\omega$ –  Katlus Jun 2 '12 at 19:19
    
@Asaf I have proved how to express normal form for nonlimit ordinal and the result shows that 'if b>0 is a finite ordinal and $r$ is a nonlimit ordinal and $a_n$ is the last term for normal form representation for $r$ then the last term for normal form representation of $r^b$ is $a_n$. Is it true? –  Katlus Jun 2 '12 at 21:24
    
Above holds for n>1 –  Katlus Jun 2 '12 at 22:19
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1 Answer 1

up vote 3 down vote accepted

For finite $a$ you have

$$\begin{align*} \left(\omega^\omega+a\right)^2&=\left(\omega^\omega+a\right)\cdot\left(\omega^\omega+a\right)\\ &=\left(\omega^\omega+a\right)\cdot\omega^\omega+\left(\omega^\omega+a\right)\cdot a\\ &=\omega^{\omega+\omega}+\omega^\omega\cdot a+a\;, \end{align*}$$

$$\begin{align*} \left(\omega^\omega+a\right)^3&=\left(\omega^{\omega+\omega}+\omega^\omega\cdot a+a\right)\cdot\left(\omega^\omega+a\right)\\ &=\left(\omega^{\omega+\omega}+\omega^\omega\cdot a+a\right)\cdot\omega^\omega+\left(\omega^{\omega+\omega}+\omega^\omega\cdot a+a\right)\cdot a\\ &=\omega^{\omega\cdot3}+\left(\omega^{\omega\cdot2}\cdot a+\omega^\omega\cdot a+a\right)\\ &=\omega^{\omega\cdot3}+\omega^{\omega\cdot2}\cdot a+\omega^\omega\cdot a+a\;, \end{align*}$$

and

$$\begin{align*} \left(\omega^\omega+a\right)^4&=\left(\omega^{\omega\cdot3}+\omega^{\omega\cdot2}\cdot a+\omega^\omega\cdot a+a\right)\cdot\left(\omega^\omega+a\right)\\ &=\left(\omega^{\omega\cdot3}+\omega^{\omega\cdot2}\cdot a+\omega^\omega\cdot a+a\right)\cdot\omega^\omega+\left(\omega^{\omega\cdot3}+\omega^{\omega\cdot2}\cdot a+\omega^\omega\cdot a+a\right)\cdot a\\ &=\omega^{\omega\cdot4}+\left(\omega^{\omega\cdot3}\cdot a+\omega^{\omega\cdot2}\cdot a+\omega^\omega\cdot a+a\right)\\ &=\omega^{\omega\cdot4}+\omega^{\omega\cdot3}\cdot a+\omega^{\omega\cdot2}\cdot a+\omega^\omega\cdot a+a\;, \end{align*}$$

and the induction is pretty clear:

$$\left(\omega^\omega+a\right)^n=\omega^{\omega\cdot n}+\omega^{\omega\cdot(n-1)}\cdot a+\omega^{\omega\cdot(n-2)}\cdot a+\ldots+\omega^{\omega\cdot2}\cdot a+\omega^\omega\cdot a+a\;.$$

In particular,

$$\left(\omega^\omega+12\right)^5=\omega^{\omega\cdot5}+\omega^{\omega\cdot4}\cdot12+\omega^{\omega\cdot3}\cdot12+\omega^{\omega\cdot2}\cdot12+\omega^\omega\cdot12+12\;.$$

In general, if $\eta$ is a limit ordinal, you’ll have

$$(\eta+a)^n=\eta^n+\eta^{n-1}\cdot a+\eta^{n-2}\cdot a+\ldots+\eta^2\cdot a+\eta\cdot a+a\;.$$

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That's exactly how i proved it. Thanks :) –  Katlus Jun 2 '12 at 23:05
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