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February 22

stuck on calculus homework (newton's method)

pic of my work (imgur) the lack writing is my prof's, i just don't get how he got the denominator of x3 to be 157,216. the decimal i got, 2.114816558 does not come anywhere near having a denominator of 157,216. help is appreciated, i gotta go to work but ill check this out later. — Preceding unsigned comment added by 24.139.14.254 (talk) 00:17, 22 February 2013 (UTC)[reply]

The key word here is "as fractions". You need to do this entirely by hand, or with a calculator that has a fraction mode. Sławomir Biały (talk) 00:43, 22 February 2013 (UTC)[reply]

My calculator has a fraction mode, but it only allows 3 digits in the denominator (and 8 digits in all). —Tamfang (talk) 04:47, 31 October 2013 (UTC)[reply]

Using f(x) = x⁴ − 20 and noting that f'(x) = 4x³ and the Newton's method formula

x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}

we get:

x_{1}=2

x_{2}=x_{1}-{\frac {f(x_{1})}{f'(x_{1})}}=2-{\frac {f(2)}{f'(2)}}=2-{\frac {2^{4}-20}{4*2^{3}}}=2-{\frac {16-20}{32}}=2-{\frac {-4}{32}}=2+{\frac {1}{8}}={\frac {2*8+1}{8}}={\frac {17}{8}}

x_{3}=x_{2}-{\frac {f(x_{2})}{f'(x_{2})}}={\frac {17}{8}}-{\frac {f({\frac {17}{8}})}{f'({\frac {17}{8}})}}={\frac {17}{8}}-{\frac {({\frac {17}{8}})^{4}-20}{4*({\frac {17}{8}})^{3}}}={\frac {17}{8}}-{\frac {{\frac {17^{4}}{8^{4}}}-20}{4*{\frac {17^{3}}{8^{3}}}}}={\frac {17}{8}}-{\frac {{\frac {17^{4}}{2^{12}}}-20}{\frac {17^{3}}{2^{7}}}}={\frac {17}{2^{3}}}-{\frac {17^{4}-20*2^{12}}{17^{3}*2^{5}}}={\frac {17*17^{3}*2^{2}-(17^{4}-20*2^{12}}{17^{3}*2^{5}}}={\frac {3*17^{4}+20*2^{12}}{17^{3}*2^{5}}}={\frac {250563+81920}{157216}}={\frac {332483}{157216}}

You can see from the factorisation of the denominator that the fraction will only reduce if the numerator is divisible by 2 or 17, and it isn't. You can also see the improving quality of the approximations as x₁⁴ = 16, x₂⁴ = 20.3908691..., and x₃⁴ = 20.0028007... EdChem (talk) 01:20, 22 February 2013 (UTC)[reply]

Ask yourself why you think "2.114816558 does not come anywhere near having a denominator of 157,216". That's like saying 0.499999999 does not come anywhere near having a denominator of 2, when it is in fact very close to 1/2. It is true that the exact number 2.114816558, as a fraction, has a denominator which is nowhere near 157,216, but that is irrelevant because your decimal value is only an approximation of a number x₃ which cannot be fully represented in a finite number of decimal digits (unlike your x₂ = 17 / 8 = 2.125). You should follow Sławomir Biały's advice above for this problem, but you should also know how to determine the numerator x where x / 157,216 ≈ 2.114816558. What happens when you multiply 2.114816558 * 157,216? Do you get something close to a whole number?

Perhaps someone here can point you to a guide on understanding the limitations of calculators, and how to properly use them in a Calculus class. -- ToE 02:10, 22 February 2013 (UTC)[reply]

As a former calculus teacher, I always disliked this sort of thing. In the real world, a person solving this problem would automatically use a calculator -- or a computer program. Looie496 (talk) 16:40, 22 February 2013 (UTC)[reply]

Yeah, well, some of us really like rationals. I'd use Python's fractions module. —Tamfang (talk) 04:47, 31 October 2013 (UTC)[reply]

What are the house odds of those blackjack variants?

Variant 1: 1-Infinite deck, no surrender, no double down, no split, blackjack pays 1 to 1, dealer show his 2 cards when game start, stand-on-soft-17 variation

Variant 2: 2-Same as 1 but with just one deck. — Preceding unsigned comment added by 177.40.130.246 (talk) 11:29, 22 February 2013 (UTC)[reply]

Model of riffle shuffle done by a person

Is there a model of a riffle shuffle done by a typical person? Not a "perfect shuffle", but what a person normally does? That is, x number of cards from the left hand followed by y number of cards from the right hand, etc, where x (and y) have some probability distribution? Bubba73 ^{You talkin' to me?} 16:23, 22 February 2013 (UTC)[reply]

Dave Bayer and Persi Diaconis used such a model in their paper Trailing the Dovetail Shuffle to Its Lair, which investigates how "close to random" a deck of cards is after n riffle shuffles and thus how many such shuffles are necessary to thoroughly randomize a deck of cards. Their model of shuffling is described in the second paragraph of the paper. —Bkell (talk) 15:00, 23 February 2013 (UTC)[reply]

Thanks, I downloaded that, but Adobe could not open it. However, I looked at it online. Their method is not like I expected - the number of cards that go at one time is proportional to the number in that stack at the time. But maybe they tested that on people. I expected something fixed like 1 card 25% of the time, 2 cards 40%, 3 cards 30%, etc. Bubba73 ^{You talkin' to me?} 15:57, 23 February 2013 (UTC)[reply]

It makes some sense—I think if I had a smaller stack of cards in my left hand and a larger stack in my right, I would try to riffle through the larger stack faster in an attempt to finish both stacks at the same time. That matches the probabilities in the paper. —Bkell (talk) 06:21, 24 February 2013 (UTC)[reply]

Also, that model of the riffle shuffle makes the mathematics in the rest of the paper work out very beautifully, which is another reason to use it. —Bkell (talk) 06:25, 24 February 2013 (UTC)[reply]

But as a person is shuffling, they aren't really aware that there are more cards in one hand, so they let more cards go past that thumb. I'm looking for something that models how people actually shuffle, not beautiful mathematics. Bubba73 ^{You talkin' to me?} 17:32, 24 February 2013 (UTC)[reply]

As one simple experiment, I separated the red and black cards of a deck, and did one riffle shuffle, then counted the runs of each color, starting with the first shuffled. I got: 3 3 2 2 1 1 1 3 4 3 2 4 2 4 4 5 6. Bubba73 ^{You talkin' to me?} 18:15, 24 February 2013 (UTC)[reply]

The paper does say, at the beginning of the third paragraph, that experiments have shown that this model "is a good description of the way real people shuffle real cards," with a reference to Diaconis, Group Representations in Probability and Statistics, 1988. —Bkell (talk) 20:15, 24 February 2013 (UTC)[reply]

OK, I'm going with that.

Resolved

Bubba73 ^{You talkin' to me?} 23:37, 25 February 2013 (UTC)[reply]