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November 25

Polynomial coefficients generated by a sum of powers

Is there a general algorithm to generate the coefficients of the expansion of $f_{k}(n)=\sum _{i=1}^{n}i^{k}$ ? For example, $f_{7}(n)={\frac {n^{8}}{8}}+{\frac {n^{7}}{2}}+{\frac {7n^{6}}{12}}-{\frac {7n^{4}}{24}}+{\frac {n^{2}}{12}}$ . I could use polynomial interpolation on the first $k+2$ terms of the series, but that gets impractical quickly if $k$ is large. 24.255.17.182 (talk) 21:47, 25 November 2016 (UTC)[reply]

One interesting thing I noticed is that

{\frac {n^{k-i}(i+1)!}{\operatorname {C} (k,i-1)}}

is invariant with respect to

k

, and starts out

1,{\frac {1}{2}},0,-1,0,20,0,-1512,0,302400,0,-131345280,0,108972864000,0,157661939635200,0,371498581149696000...

, but I don't see an obvious pattern here and this series isn't in OEIS. 24.255.17.182 (talk) 22:13, 25 November 2016 (UTC)[reply]

(ec)Use Binomial_coefficient#Binomial_coefficients_as_a_basis_for_the_space_of_polynomials and the Hockey-stick identity.

\sum _{i=1}^{n}i^{k}=\sum _{i=1}^{n}\sum _{m=0}^{k}a_{m}{\binom {i}{m}}=\sum _{m=0}^{k}a_{m}\sum _{i=1}^{n}{\binom {i}{m}}=\sum _{m=0}^{k}a_{m}{\binom {n+1}{m+1}}

Bo Jacoby (talk) 22:24, 25 November 2016 (UTC).[reply]

Sorry if it's obvious but could you explain how

a_{m}

is to be computed? 24.255.17.182 (talk) 23:03, 25 November 2016 (UTC)[reply]

I think what you're looking for is Faulhaber's formula. A generalization is the Euler–Maclaurin formula. --RDBury (talk) 01:43, 26 November 2016 (UTC)[reply]

The formula is found in the link I gave you.

a_{m}=\sum _{i=0}^{m}(-1)^{m-i}{\binom {m}{i}}i^{k}.

Bo Jacoby (talk) 07:18, 26 November 2016 (UTC).[reply]

November 26

Martingale betting system

If I apply a Martingale betting strategy to a random process (dice, roulette, coins) I'll end up broke. Variations of it (not doubling after each loss, but halving after each loss, for example) won't make a difference.

However, what would happen if I apply it (or some variation of it) to a process where the results are not independent? That is, a process where there are tendencies, bad and good stretches? E.g. sports, politics, horses.

Would Martingale work here? --Hofhof (talk) 14:29, 26 November 2016 (UTC)[reply]

With a martingale betting system for coin tosses, you won't necessarily end up broke if you stop as soon as you are ahead. For example, if you have enough initial wealth to finance 6 tosses, then the chance of going broke is the chance of throwing 6 straight tails, or

1/2^{6}=1/64.

Now if the tosses are positively serially correlated, then one tail is more likely to be followed by another tail, so you still could go broke (I would think with a higher probability than before). On the other hand, if the tosses are negatively serially correlated, except in the extreme case where a tail is necessarily followed by a head, there is still a chance of a sufficiently long string of tails to make you go broke. Loraof (talk) 20:56, 26 November 2016 (UTC)[reply]

The problem is that "as soon as you are ahead" will almost surely happen with the "double when lose" martingale, but you will win only little. If you repeat until you either (say) double your initial account or go bankrupt, your success chances are not improved (they actually deteriorate, compared to betting your whole money at once, if the house takes a percentage on each toss). I would assume Hofhof is asking for a way to (say) play until they make double or bankrupt with a probability of winning > 0.5. Tigraan^{Click here to contact me} 21:13, 27 November 2016 (UTC)[reply]

It would depend on whether the odds you're given accurately reflect the correlation and whether you know the effects of the correlation. In an extreme case where a head is always followed by a tail and vice versa, if the person you're betting against always sets the odds at 1:1 based on the average, then you can bet according to the rule and win each time. On the other hand suppose that there is a deterministic rule for the next result, but it depends on the previous 100 results. In that case you'd need at least 2¹⁰⁰ results to figure out what the rule is and you'd be back where you started. In real life, if there was a way to figure the odds better than the person you're betting against then that person would soon go broke and would no longer accept bets. An example is card counting where you could, with a sufficiently good memory, win against the house in Blackjack. But the axiom "The house always wins," Dustin Hoffman aside, holds because if the house didn't win it would no longer be in business. --RDBury (talk) 00:32, 27 November 2016 (UTC)[reply]

Probability/statistics question: estimating indoor climbing safety

Does anyone have micromort figures for indoor climbing (as opposed to mountaineering) in terms of say, micromorts per climbing visit? The nearest I can find is this citing this paper, which suggests a really low figure. But is that figure statistically sound, given how rare an event a one-in-a-million chance is? Clearly, even though the study given recorded no deaths, the actual risk cannot be zero: see this report. Does anyone here have the relevant knowledge to work out what we can tell from these results using, say, a Bayesian framework? The Anome (talk) 18:28, 26 November 2016 (UTC)[reply]

November 28

Qualities or properties of Quadratic equations (parabola) and linear equations

For quadratic functions, I already know that we need to know the vertex, axis of symmetry, x-intercepts, y-intercepts, whether it goes down or up, and whether it has minimum or maximum value. What else do we need to know? Also, for the linear equations, I already know that we need to know the x-intercepts, the y-intercepts, whether it goes from bottom right to top left or it goes from bottom right to top left, and the slope. What else do we need to know about the linear equation? Donmust90 (talk) 02:03, 28 November 2016 (UTC)Donmust90Donmust90 (talk) 02:03, 28 November 2016 (UTC)[reply]

What one "needs to know" is context-dependent, conditional on what one needs it for. In your case, the most likely answer is "go ask your teacher." If that's not the right context then you should try to ask a better question. --JBL (talk) 03:52, 28 November 2016 (UTC)[reply]

With a linear function we can determine, just by observation, the slope, the x-axis intercept and the y-axis intercept. (If the slope is zero, the image of the function is parallel to the x-axis. If the slope is positive, the image of the function slopes from low-left to high-right. If the slope is negative, the image of the function slopes from high-left to low-right. If the slope is undefined, it is parallel to the y-axis.)

In some questions we are not told the function but instead are told the coordinates of two points that lie on the image of the function; or we are told the slope, and the coordinates of one point. In such questions we can determine the function.

With a quadratic polynomial function we can proceed in a similar way to the above, but there are added challenges. See your text book. Dolphin (t) 06:19, 28 November 2016 (UTC)[reply]

November 29

Can ${\sqrt[{4}]{2}}$ or ${\sqrt[{2n}]{2}}$ be constructed? יהודה שמחה ולדמן (talk) 17:28, 29 November 2016 (UTC)[reply]

You can always construct the geometric mean of two given lengths. Which means that if you are given a unit interval, you can construct the square root of any given number. This means you can construct

{\sqrt {2}},{\sqrt[{4}]{2}},{\sqrt[{8}]{2}}

and in general

{\sqrt[{2^{n}}]{2}}

. But I believe you can't construct any other roots of 2. For example,

{\sqrt[{3}]{2}}

is famously non-constructible, so

{\sqrt[{6}]{2}}

can't be constructed either. -- Meni Rosenfeld (talk) 18:11, 29 November 2016 (UTC)[reply]

So how do we construct

{\sqrt[{4}]{2}}

as height in the right triangle with hypotenuse

1+{\sqrt {2}}

? יהודה שמחה ולדמן (talk) 18:54, 29 November 2016 (UTC)[reply]

You can see the details of the construction at http://math.stackexchange.com/a/708/153429. -- Meni Rosenfeld (talk) 19:35, 29 November 2016 (UTC)[reply]

"Equivalent to" vs "Approximately" symbols

What is the difference between these symbols: ≍ ≈ ? I've seen "≈" used many times before (i.e. Pi ≈ 3.14). But "≍" is new to me. It seems they have a similar meaning? --209.203.125.162 (talk) 21:06, 29 November 2016 (UTC)[reply]

The ≍ symbol is sometimes used to denote equivalent "order of magnitude" in asymptotic analysis, for example, in Asymptotic notations. --Mark viking (talk) 21:49, 29 November 2016 (UTC)[reply]

November 30

Wolfram Alpha indefinite integral

According to Wolfram Alpha, $\int x^{\sqrt {x}}dx={\frac {x^{{\sqrt {x}}+1}}{{\sqrt {x}}+1}}+C$ . Numerically and algebraically this doesn't seem right (I don't think the differential of the right hand side simplifies to the left). Does Mathematica also give this result? If this is wrong, is there a nice closed form, or what is WA doing here? 24.255.17.182 (talk) 04:58, 30 November 2016 (UTC)[reply]