"So..." A hand goes up. It is the hand of a student who does not want to admit weakness, but will do so anyway, under the eyes of Security. "It isn't obvious to me how you got that?"
Keltham starts to answer -
Oh wait never mind it is obvious.
It's too late to say that now, though, so the student will just listen to Keltham speak the awful words that he's just multiplying both sides by infinity.
Factoring out (1/INF)^N from all the terms, that requires summing:
"So..." A hand goes up. It is the hand of a student who does not want to admit weakness, but will do so anyway, under the eyes of Security. "It isn't obvious to me how you got that?"
Keltham starts to answer -
Oh wait never mind it is obvious.
It's too late to say that now, though, so the student will just listen to Keltham speak the awful words that he's just multiplying both sides by infinity.
Factoring out (1/INF)^N from all the terms, that requires summing:
0^N + 1^N + 2^N + 3^N + ... + INF^N
...and getting INF^(N+1) / (N+1).
Retroactive class chatter
A hand goes up, in obedience to stated Orders. "Why are we factoring this out, sir? And why does it require summing this?"
...He's factoring it out because it'll reduce the problem to one that's easier to solve? Then, once they get the solution for an infinite sum of cubes n^N, they'll multiply that whole thing through by (1/INF)^N and by (1/INF), and get back out...
...which represents an infinity of possible hypotheses, for all possible propensities between 0 and 1, for how often a ball goes LEFT, each hypothesis with 1/infinity prior probability.
Look, it's just the same as if they'd used four buckets each with 1/4 prior probability, and asked for the chance of seeing 9 balls going LEFT, that would be:
Somebody then asks whether it ought to be INF+1 in the denominator of the original expression.
Keltham stares at the whiteboard a moment, then sagely announces that, with a sum this large, it won't make any relative difference if they drop any number of finite terms from it, since those contribute 0% of anything. He's just arbitrarily going to declare that one of the ones in the middle is missing.
"So..." A hand goes up. It is the hand of a student who does not want to admit weakness, but will do so anyway, under the eyes of Security. "It isn't obvious to me how you got that?"
Keltham starts to answer -
Oh wait never mind it is obvious.
It's too late to say that now, though, so the student will just listen to Keltham speak the awful words that he's just multiplying both sides by infinity.
Factoring out (1/INF)^N from all the terms, that requires summing:
0^N + 1^N + 2^N + 3^N + ... + INF^N
...and getting INF^(N+1) / (N+1).
Retroactive class chatter
A hand goes up, in obedience to stated Orders. "Why are we factoring this out, sir? And why does it require summing this?"
...Keltham is factoring it out because it'll reduce the problem to one that's easier to solve? Then, once they get the solution for an infinite sum of cubes n^N, they'll multiply that whole thing through by (1/INF)^N and by (1/INF), and get back out...
...which represents an infinity of possible hypotheses, for all possible propensities between 0 and 1, for how often a ball goes LEFT, each hypothesis with 1/infinity prior probability.
Look, it's just the same as if they'd used four buckets each with 1/4 prior probability, and asked for the chance of seeing 9 balls going LEFT, that would be:
...Somebody then asks whether it ought to be INF+1 in the denominator of the original expression.
Keltham stares at the wall for a moment, then sagely announces that, with a sum this large, it won't make any relative difference if they drop any number of finite terms from it, since those contribute 0% of anything. He's just arbitrarily going to declare that one of the ones in the middle is missing.
"So..." A hand goes up. It is the hand of a student who does not want to admit weakness, but will do so anyway, under the eyes of Security. "It isn't obvious to me how you got that?"
Keltham starts to answer -
Oh wait never mind it is obvious.
It's too late to say that now, though, so the student will just listen to Keltham speak the awful words that he's just multiplying both sides by infinity.
Factoring out (1/INF)^N from all the terms, that requires summing:
0^N + 1^N + 2^N + 3^N + ... + INF^N
...and getting INF^(N+1) / (N+1).
Retroactive class chatter
A hand goes up, in obedience to stated Orders. "Why are we factoring this out, sir? And why does it require summing this?"
...Keltham is factoring it out because it'll reduce the problem to one that's easier to solve? Then, once they get the solution for an infinite sum of cubes n^N, they'll multiply that whole thing through by (1/INF)^N and by (1/INF), and get back out...
...which represents an infinity of possible hypotheses, for all possible propensities between 0 and 1, for how often a ball goes LEFT, each hypothesis with 1/infinity prior probability.
Look, it's just the same as if they'd used four buckets each with 1/4 prior probability, and asked for the chance of seeing 9 balls going LEFT, that would be:
...Somebody then asks whether it ought to be INF+1 in the denominator of the original expression.
Keltham stares at the wall for a moment, then sagely announces that, with a sum this large, it won't make any relative difference if they drop any finite number of terms from it, since those contribute 0% of anything. He's just arbitrarily going to declare that one of the ones in the middle is missing.
"So..." A hand goes up. It is the hand of a student who does not want to admit weakness, but will do so anyway, under the eyes of Security. "It isn't obvious to me how you got that?"
Keltham starts to answer -
Oh wait never mind it is obvious.
It's too late to say that now, though, so the student will just listen to Keltham speak the awful words that he's just multiplying both sides by infinity.
Factoring out (1/INF)^N from all the terms, that requires summing:
0^N + 1^N + 2^N + 3^N + ... + INF^N
...and getting INF^(N+1) / (N+1).
Retroactive class chatter
A hand goes up, in obedience to stated Orders. "Why are we factoring this out, sir? And why does it require summing this?"
...Keltham is factoring it out because it'll reduce the problem to one that's easier to solve? Then, once they get the solution for an infinite sum of cubes n^N, they'll multiply that whole thing through by (1/INF)^N and by (1/INF), and get back out...
...which represents an infinity of possible hypotheses, for all possible propensities between 0 and 1, for how often a ball goes LEFT, each hypothesis with 1/infinity prior probability.
Look, it's just the same as if they'd used four buckets each with 1/4 prior probability, and asked for the chance of seeing 9 balls going LEFT, that would be:
...Somebody then asks whether it ought to be INF+1 in the denominator of the original expression.
Keltham stares at the wall for a moment, then sagely announces that, with a sum this large, it won't make any relative difference if they drop any finite number of terms from it, since those contribute 0% of anything. He's just going to arbitrarily declare that one of the ones in the middle is missing.
"So..." A hand goes up. It is the hand of a student who does not want to admit weakness, but will do so anyway, under the eyes of Security. "It isn't obvious to me how you got that?"
Keltham starts to answer -
Oh wait never mind it is obvious.
It's too late to say that now, though, so the student will just listen to Keltham speak the awful words that he's just multiplying both sides by infinity.
Factoring out (1/INF)^N from all the terms (n/INF)^N reduces the problem to one of summing:
0^N + 1^N + 2^N + 3^N + ... + INF^N
...and getting INF^(N+1) / (N+1).
Retroactive class chatter
A hand goes up, in obedience to stated Orders. "Why are we factoring this out, sir? And why does it require summing this?"
...they're taking (0/INF)^N + ... and factoring out (1/INF)^N from all terms (n/INF)^N to yield n^N * (1/INF)^N, for each term, and then dividing both sides to yield the sum over n^N equalling [ INF/(N+1) / (1/INF)^N ] = INF^(N+1) / (N+1).
Then, once they get the solution for an infinite sum of cubes n^N, they'll multiply that whole thing through by (1/INF)^N and by (1/INF), and get back out...
...which represents an infinity of possible hypotheses, for all possible propensities between 0 and 1, for how often a ball goes LEFT, each hypothesis with 1/infinity prior probability.
Look, it's just the same as if they'd used four buckets each with 1/4 prior probability, and asked for the chance of seeing 9 balls going LEFT, that would be:
...Somebody then asks whether it ought to be INF+1 in the denominator of the original expression.
Keltham stares at the wall for a moment, then sagely announces that, with a sum this large, it won't make any relative difference if they drop any finite number of terms from it, since those contribute 0% of anything. He's just going to arbitrarily declare that one of the ones in the middle is missing.
"So..." A hand goes up. It is the hand of a student who does not want to admit weakness, but will do so anyway, under the eyes of Security. "It isn't obvious to me how you got that?"
Keltham starts to answer -
Oh wait never mind it is obvious.
It's too late to say that now, though, so the student will just listen to Keltham speak the awful words that he's just multiplying both sides by infinity.
Factoring out (1/INF)^N from all the terms (n/INF)^N reduces the problem to one of summing:
0^N + 1^N + 2^N + 3^N + ... + INF^N
...and getting INF^(N+1) / (N+1).
Retroactive parenthetical class chatter
A hand goes up, in obedience to stated Orders. "Why are we factoring this out, sir? And why does it require summing this?"
...they're taking (0/INF)^N + ... and factoring out (1/INF)^N from all terms (n/INF)^N to yield n^N * (1/INF)^N, for each term, and then dividing both sides to yield the sum over n^N equalling [ INF/(N+1) / (1/INF)^N ] = INF^(N+1) / (N+1).
Then, once they get the solution for an infinite sum of cubes n^N, they'll multiply that whole thing through by (1/INF)^N and by (1/INF), and get back out...
...which represents an infinity of possible hypotheses, for all possible propensities between 0 and 1, for how often a ball goes LEFT, each hypothesis with 1/infinity prior probability.
Look, it's just the same as if they'd used four buckets each with 1/4 prior probability, and asked for the chance of seeing 9 balls going LEFT, that would be:
...Somebody then asks whether it ought to be INF+1 in the denominator of the original expression.
Keltham stares at the wall for a moment, then sagely announces that, with a sum this large, it won't make any relative difference if they drop any finite number of terms from it, since those contribute 0% of anything. He's just going to arbitrarily declare that one of the ones in the middle is missing.
Retroactive parenthetical class chatter
"So..." A hand goes up. It is the hand of a student who does not want to admit weakness, but will do so anyway, under the eyes of Security. "It isn't obvious to me how you got that?"
Keltham starts to answer -
Oh wait never mind it is obvious.
It's too late to say that now, though, so the student will just listen to Keltham speak the awful words that he's just multiplying both sides by infinity.
Factoring out (1/INF)^N from all the terms (n/INF)^N reduces the problem to one of summing:
0^N + 1^N + 2^N + 3^N + ... + INF^N
...and getting INF^(N+1) / (N+1).
Retroactive parenthetical class chatter
A hand goes up, in obedience to stated Orders. "Why are we factoring this out, sir? And why does it require summing this?"
...they're taking (0/INF)^N + ... and factoring out (1/INF)^N from all terms (n/INF)^N to yield n^N * (1/INF)^N, for each term, and then dividing both sides to yield the sum over n^N equalling [ INF/(N+1) / (1/INF)^N ] = INF^(N+1) / (N+1).
Then, once they get the solution for an infinite sum of cubes n^N, they'll multiply that whole thing through by (1/INF)^N and by (1/INF), and get back out...
...which represents an infinity of possible hypotheses, for all possible propensities between 0 and 1, for how often a ball goes LEFT, each hypothesis with 1/infinity prior probability.
Look, it's just the same as if they'd used four buckets each with 1/4 prior probability, and asked for the chance of seeing 9 balls going LEFT, that would be:
...Somebody then asks whether it ought to be INF+1 in the denominator of the original expression.
Keltham stares at the wall for a moment, then sagely announces that, with a sum this large, it won't make any relative difference if they drop any finite number of terms from it, since those contribute 0% of anything. He's just going to arbitrarily declare that one of the ones in the middle is missing.
(Retroactive parenthetical class chatter)
"So..." A hand goes up. It is the hand of a student who does not want to admit weakness, but will do so anyway, under the eyes of Security. "It isn't obvious to me how you got that?"
Keltham starts to answer -
Oh wait never mind it is obvious.
It's too late to say that now, though, so the student will just listen to Keltham speak the awful words that he's just multiplying both sides by infinity.
Factoring out (1/INF)^N from all the terms (n/INF)^N reduces the problem to one of summing:
0^N + 1^N + 2^N + 3^N + ... + INF^N
...and getting INF^(N+1) / (N+1).
(Retroactive parenthetical class chatter)
A hand goes up, in obedience to stated Orders. "Why are we factoring this out, sir? And why does it require summing this?"
...they're taking (0/INF)^N + ... and factoring out (1/INF)^N from all terms (n/INF)^N to yield n^N * (1/INF)^N, for each term, and then dividing both sides to yield the sum over n^N equalling [ INF/(N+1) / (1/INF)^N ] = INF^(N+1) / (N+1).
Then, once they get the solution for an infinite sum of cubes n^N, they'll multiply that whole thing through by (1/INF)^N and by (1/INF), and get back out...
...which represents an infinity of possible hypotheses, for all possible propensities between 0 and 1, for how often a ball goes LEFT, each hypothesis with 1/infinity prior probability.
Look, it's just the same as if they'd used four buckets each with 1/4 prior probability, and asked for the chance of seeing 9 balls going LEFT, that would be:
...Somebody then asks whether it ought to be INF+1 in the denominator of the original expression.
Keltham stares at the wall for a moment, then sagely announces that, with a sum this large, it won't make any relative difference if they drop any finite number of terms from it, since those contribute 0% of anything. He's just going to arbitrarily declare that one of the ones in the middle is missing.
(Retroactive parenthetical class chatter)
"So..." A hand goes up. It is the hand of a student who does not want to admit weakness, but will do so anyway, under the eyes of Security. "It isn't obvious to me how that reduces?"
Keltham starts to answer -
Oh wait never mind it is obvious.
It's too late to say that now, though, so the student will just listen to Keltham speak the awful words that he's just multiplying both the original sum and the original result 1/(N+1) by infinity.
Factoring out (1/INF)^N from all the terms (n/INF)^N reduces the problem to one of summing:
0^N + 1^N + 2^N + 3^N + ... + INF^N
...and getting INF^(N+1) / (N+1).
(Retroactive parenthetical class chatter)
A hand goes up, in obedience to stated Orders. "Why are we factoring this out, sir? And why does it require summing this?"
...they're taking (0/INF)^N + ... and factoring out (1/INF)^N from all terms (n/INF)^N to yield n^N * (1/INF)^N, for each term, and then dividing both sides to yield the sum over n^N equalling [ INF/(N+1) / (1/INF)^N ] = INF^(N+1) / (N+1).
Then, once they get the solution for an infinite sum of cubes n^N, they'll multiply that whole thing through by (1/INF)^N and by (1/INF), and get back out...
...which represents an infinity of possible hypotheses, for all possible propensities between 0 and 1, for how often a ball goes LEFT, each hypothesis with 1/infinity prior probability.
Look, it's just the same as if they'd used four buckets each with 1/4 prior probability, and asked for the chance of seeing 9 balls going LEFT, that would be:
...Somebody then asks whether it ought to be INF+1 in the denominator of the original expression.
Keltham stares at the wall for a moment, then sagely announces that, with a sum this large, it won't make any relative difference if they drop any finite number of terms from it, since those contribute 0% of anything. He's just going to arbitrarily declare that one of the ones in the middle is missing.
