MHB Sequence of b_{k} with Explicit Formula: Proving by Math Induction

hkhgg
Messages
1
Reaction score
0
$$b_{k} = b_{k - 1}/2 +b_{k-1} $$
$$b_{0} = 1 $$

What would be the sequence for this expression, I calculated it to be 1, 2/3, 2/5, 2/7 ...

Is it right?

My explicit formula is $$b_{n} = 2/n+2 $$

What would be the explicit formula in your view and how can that formula be proved by mathematical induction? thanks
 
Physics news on Phys.org
First, what is the formula defining the sequence? You wrote $b_k= b_{k-1}/2+ b_{k-1}$ but since that would be more simply written as $3b_{k-1}/2$ I wonder if you did not intend $b_k= \frac{b_{k-1}}{2+ b_{k-1}}$.

Assuming what you wrote was correct, since $b_0= 1$, $b_1= 1/2+ 1= 3/2$, $b_2= (3/2)/2+ (3/2)= 3/4+ 3/2=3/4+ 6/4=10/4=5/2$... That is not what you write so apparently that is not what you intend.

With $b_k= \frac{b_{k-1}}{2+b_{k-1}}$ and $b_0=1$, $b_1= \frac{1}{2+ 1}= \frac{2}{3}$, $b_2= \frac{\frac{2}{3}}{2+ \frac{2}{3}}= \frac{\frac{2}{3}}{\frac{6}{3}+ \frac{2}{3}}= \frac{\frac{2}{3}}{\frac{8}{3}}= \frac{2}{3}\frac{3}{8}= \frac{2}{8}= \frac{1}{4}$.

That is also not what you have so, no, you are not correct.

I can't respond to the rest of you questions because I do not know whether you intend $b_k= \frac{b_{k-1}}{2}+ b_k= \left(\frac{3}{2}\right)b_{k-1}$ or $b_k= \frac{b_{k-1}}{2+ b_{k-1}}$.
 
$b_k = \dfrac{b_{k-1}}{2+b_{k-1}}$

$\displaystyle \bigg\{ 1, \dfrac{1}{3}, \dfrac{1}{7}, \dfrac{1}{15}, … , \dfrac{1}{2^{n+1}-1} , … \bigg \}$
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top