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I've decided to dive in Concrete Mathematics despite only doing a couple of years of undergraduate maths many years ago. I'm looking to work through all the material whilst plugging gaps in my knowledge no matter how large they are.

However, solving recurrence relation 1.1 - The Towers of Hanoi is causing me issues.

The recurrence relation is as follows:

$T_0 = 0$
$T_n=2T_{n-1}+1$ for $n>0$

To find the closed form the book adds 1 to both sides of the equations to get:

$T_0+1=1$
$T_n+1=2T_{n-1}+2$

Then we let $U_n=T_n+1$. Now I get lost. How do I get from the line above to the second line below after the substitution of $U_n$. It seems like I'm missing something simple.

$U_0=1$
$U_n=2U_{n-1}$

Then the book goes on to say:

It doesn't take genius to discover that the solution to this recurrence is just $U_n=2^n$; hence $T_n=2^n-1$.

Again, I'm lost, how do I get to $U_n=2^n$?

A bit disheartening considering this is meant to be so easy.

Any help is appreciated. Thanks.

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  • $\begingroup$ Wow... the book says that? Did the author wrote it against his will? $\endgroup$ – Git Gud Nov 25 '13 at 23:38
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    $\begingroup$ @GitGud: Three authors, Ronald L. Graham, Donald E. Knuth, and Oren Patashnik. It’s one of the best mathematics textbooks I’ve ever encountered. $\endgroup$ – Brian M. Scott Nov 25 '13 at 23:47
  • $\begingroup$ A very detailed answer can be seen here perkss.github.io/#/MathFundamentals/… $\endgroup$ – perkss Jan 21 '18 at 11:23
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You’re just missing a little algebra. You have $U_n=T_n+1$ for all $n\ge 0$, so $U_{n-1}=T_{n-1}+1$, and therefore $2T_{n-1}+2=2(T_{n-1}+1)=2U_{n-1}$. Combine this with $T_n+1=2T_{n-1}+2$, and $U_n=T_n+1$, and you get $U_n=2U_{n-1}$, with $U_0=1$.

Now notice that $U_n$ is just doubling each time $n$ is increased by $1$:

$$\begin{align*} U_1&=2U_0\\ U_2&=2U_1=2^2U_0\\ U_3&=2U_2=2^3U_0\\ U_4&=2U_3=2^4U_0 \end{align*}$$

The pattern is clearly going to persist, so we conjecture that $U_n=2^nU_0$ for each $n\ge 0$. This is certainly true for $n=0$. Suppose that it’s true for some $n\ge 0$; then $$U_{n+1}=2U_n=2\cdot2^nU_0=2^{n+1}U_0\;,$$ and the conjecture follows by mathematical induction.

Now we go back and use the fact that $U_0=1$ to say that $U_n=2^n$ for each $n\ge 0$, and hence $T_n=U_n-1=2^n-1$.

In my answer to this question I solved another problem using this technique; you might find the explanation there helpful.

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  • $\begingroup$ Thank you for the detail here. Very helpful. $\endgroup$ – Ramp Nov 26 '13 at 21:35
  • $\begingroup$ @Ramp: You’re very welcome. $\endgroup$ – Brian M. Scott Nov 26 '13 at 21:38
  • $\begingroup$ I know it has been a while since you answered this question, but how do you develop the mathematical intuition to notice such a pattern? Additionally, how would one trying to solve this even think to add one to both sides or to define $U_n = T_n + 1$? Thanks. $\endgroup$ – Jamie Corkhill Dec 27 '18 at 17:18
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    $\begingroup$ @JamieCorkhill you can do it by finding a pattern like \begin{align*} H_n&=2H_{n-1}+1\\ &=2(2H_{n-2}+1)+1 \\ &=2^{(2-0)}H_{n-2}+2^{(2-1)}+2^{(2-2)}\\ &=2^2(2H_{n-3}+1)+2+1\\ &=2^{(3-0)}H_{n-3}+2^{(3-1)}+2^{(3-2)}+2^{(3-3)}\\ &\vdots\\ &=2^{((n-1)-0)}H_{(n-(n-1))}+2^{(n-2)}+\cdots+2^{(n-(n-1))}+2^{(n-n)}\\ &=2^{n-1}.1+2^{n-2}+\cdots+2+1\\ &=2^n-1 \end{align*} $\endgroup$ – emonHR Jul 11 '19 at 19:56

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