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February 8

Difficulty in finding Indefinite integral

I am a graduate student of computer science and I pursue integration as a hobby /passion.I find it impossible to find Integral f(x)dx where f(x)=(x^4-1)^(-1/4) by a way other than the following f(x)=(x^-1)(1-x^(-4)) or f(x)=x^4/(x^5)(1-x^(-4)) or Integral f(x)dx=4*Integral f(x)/4 dx =4* Integral [ x^4/(4*x^5)(1-x^(-4)) dx .I feel that this is a very clumsy idiosyncratic way .I want the solution or hints to the solution of another integral:- Integral f(x)dx where f(x)=[sin(x)]^-1/2Is there any elegant and logical way of solving the problem by method of substitution or integration by parts.More specifically is there any good heuristic way of determining what substitution function to be used in the method of substitution.Ichgab (talk) 15:04, 8 February 2011 (UTC)[reply]

Let's try and put this together. I think the lack of replies is because it's quite hard to understand mathematics when it's written using a keyboard. Am I right in assuming that you want to calculate the following indefinite integral:

\int \!{\frac {\operatorname {d} x}{\sqrt[{4}]{x^{4}-1}}}\,.

As far as I remember, this doesn't have an explicit anti-derivative in terms of elementary functions. That means that there is no expression using elementary functions whose derivative gives you (x⁴ – 1)^–1/4. In fact, sadly, the same is true for

\int {\frac {\operatorname {d} x}{\sqrt {\sin x}}}\,.

The anti-derivatives of these two integrands are expressible in terms of hypergeometric functions and elliptic integrals, but not as elementary functions. There may be techniques to evaluate definite integrals with these integrands, but there are no elementary anti-derivatives. — Fly by Night (talk) 21:41, 8 February 2011 (UTC)[reply]

The Risch algorithm is a decision procedure (incorporated into some symbolic algebra software) that tells you exactly whether an integral can done as elementary functions. The wider topic area for differential equations is differential Galois theory, which I just came across a few days ago and don't know anything about. 71.141.88.54 (talk) 22:26, 8 February 2011 (UTC)[reply]

Thank you for the answer,my doubt has been cleared.Ichgab (talk) 05:26, 9 February 2011 (UTC)[reply]

The integral is elementary, albeit complicated. Try http://www.wolframalpha.com integrate[(x^4-1)^(-1/4),x] . Bo Jacoby (talk) 09:43, 9 February 2011 (UTC).[reply]

That's a hand link that you supply; I'll have to use that in the future. It says that

\int \!{\frac {\operatorname {d} x}{\sqrt[{4}]{x^{4}-1}}}={\frac {x{\sqrt[{4}]{1-x^{4}}}\,{}_{2}F_{1}\!\left({\frac {1}{4}},{\frac {1}{4}};{\frac {5}{4}};x^{4}\right)}{\sqrt[{4}]{1-x^{4}}}}\,.

The ₂F₁ is a hypergeometric function. I didn't think that they were elementary functions. It seems that some are and some aren't; it depends on the arguments. So the question is: Is ₂F₁(1/4,1/4;5/4;x⁴) one of the special hypergeometric functions that reduces to an elementary function? I would guess not; for if it did then that link would hopefully present the elementary function and not ₂F₁(1/4,1/4;5/4;x⁴). — Fly by Night (talk) 14:28, 9 February 2011 (UTC)[reply]

Having said that, we have ₂F₁(1/4,1/4;5/4;x⁴). — Fly by Night (talk) 14:48, 9 February 2011 (UTC)[reply]

The integral is expressed in terms of the elementary functions logarithm and arctangent like this

integral 1/(x^4-1)^(1/4) dx = ((1-x)^(1/4) x (x+1)^(1/4) (x^2+1)^(1/4) (-(tan^(-1)(1-x/(sqrt(2) (1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4)), -x/(sqrt(2) (1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4))))/(2 sqrt(2) x)-(tan^(-1)(x/(sqrt(2) (1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4))+1, -x/(sqrt(2) (1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4))))/(2 sqrt(2) x)-(log(x^2/(sqrt(1-x) sqrt(x+1) sqrt(x^2+1))-(sqrt(2) x)/((1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4))+1))/(4 sqrt(2) x)+(log(x^2/(sqrt(1-x) sqrt(x+1) sqrt(x^2+1))+(sqrt(2) x)/((1-x)^(1/4) (x+1)^(1/4) (x^2+1)^(1/4))+1))/(4 sqrt(2) x)))/(x^4-1)^(1/4)+constant

Bo Jacoby (talk) 01:27, 10 February 2011 (UTC).[reply]

Yes, we know. I already linked to that 11 hours ago. Here it is again, in case you missed it: ₂F₁(1/4,1/4;5/4;x⁴). This is somewhat unusual, however, since not all hypergeometric functions reduce to elementary functions. I mentioned that too. — Fly by Night (talk) 01:56, 10 February 2011 (UTC)[reply]

People have a right to read a discussion without following the links, and it is wrong to make them a prerequisite to understanding what you're saying. If you said you didn't think the function was elementary and then wanted to retract it, a better form would be "Turns out that it is elementary - see ₂F₁(1/4,1/4;5/4;x⁴)". Since you did not explicitly say it was elementary, Bo was in the right to do so. -- Meni Rosenfeld (talk) 09:48, 10 February 2011 (UTC)[reply]

Thank you Meni! Fly by Night, how many hours earlier did I tell you that "The integral is elementary, albeit complicated"? If you follow my link, and wait a while, then the solution in terms of the hypergeometric function is overwritten by the elementary solution. Bo Jacoby (talk) 10:29, 10 February 2011 (UTC).[reply]

But note that the computation can time out, in which case you'll have to click "try again with more time". -- Meni Rosenfeld (talk) 11:11, 10 February 2011 (UTC)[reply]

I said that "I would guess that it is not; for if it did [reduce to an elementary function] then that link would hopefully present the elementary function and not ₂F₁(1/4,1/4;5/4;x⁴)." And by starting the next sentence with "Having said that…" it would be clear to all that I was about to retract that last conjecture. The truth is that Bo didn't finish reading my post and just flew into a rage: "I'll show that Fly by Night...Grrr!" I don't get your point about the link; Bo used a link "to justify" that the integral was elementary, even though that link didn't prove it to be elementary and it was just luck that the hypergeometric function turned out to be elementary. And Meni; if you prefer that spam like post of Bo's over a nice tidy blue link then, well,… — Fly by Night (talk) 15:08, 10 February 2011 (UTC)[reply]

It was not clear to me if Fly was about to retract his last conjecture or not. The easier thing for Fly to do would be to remove his incorrect conjecture from his contribution. My link did prove the integral to be elementary, but Fly apparently did not believe it, so I included the elementary solution. Bo Jacoby (talk) 21:39, 10 February 2011 (UTC).[reply]

No, "having said that" does not at all mean that you were going to retract something.

If you insist to ascribe malicious intents to everyone else's actions don't be surprised to always be getting into arguments.

There's a world of difference between "The function is elementary; here's a link for its elementary form" (what Bo did), and "here's a link" where one has to follow the link to have a clue what you're saying (what you did).

There was some miscommunication between you two because you got a timeout in Wolfram Alpha and Bo didn't - Bo's original comment is perfectly understandable in this light, and you shouldn't keep dwelling on it even after I've clarified it.

I agree that the expression Bo put here is a bit cumbersome, but if we can't rely on a link to show the same result consistently then it's a necessary clarification. -- Meni Rosenfeld (talk) 05:46, 11 February 2011 (UTC)[reply]

There is a similar problem that I find intriguing:: $\int \!{\frac {\operatorname {d} x}{\sqrt[{4}]{x^{4}+1}}}\,.$ Can someone show how to solve this problem or explain if solution does not exists.Ichgab (talk) 11:43, 10 February 2011 (UTC)[reply]

integral 1/(1+x^4)^(1/4) dx = x _2F_1(1/4, 1/4;5/4;-x^4)+constant Bo Jacoby (talk) 12:16, 10 February 2011 (UTC).[reply]

Can game theory help with reverse auctions?

Hello. Does game theory provide any "best strategy" in a reverse auction? Let's imagine the following case:

A public administration puts to tender the construction of a stadium for an approximate value of $25 million. Its main criterion for assessing bids is cost (it will choose the lowest bid);
All bidders provide the same service with the same level of quality, so that variable is irrelevant here;
Each bidder wants to win the auction, but they also want to maximise the price they will be paid for the works if they win;
Bidders do not know what price the other bidders have submitted.

If I were a bidder, what would be the sweet spot which would give me the most chances of winning the auction, while at the same time getting a high enough price to make a decent profit? Is there such a sweet spot at all? Thanks in advance. Leptictidium (mt) 16:26, 8 February 2011 (UTC)[reply]

I think there is something optimal about the different auction types. Indeed if you look at Auction theory there is the Revenue equivalence theorem which states all auction types will give the same result for the seller. I think the upshot is to bid what you think it is worth, bid less and another bidder will win.--Salix (talk): 17:47, 8 February 2011 (UTC)[reply]

I think that as far as the theory is concerned, reverse auctions are completely equivalent to regular auctions.

The key issue here is what each contractor knows about the project cost for himself and for others. If all costs are known the result is simple - the contractor with the lowest cost wins, with a bid slightly less than the second best's cost. Otherwise each contractor needs to estimate (with probability distributions, preferably) his own cost and what the others will bid, and solve the resulting optimization problem. -- Meni Rosenfeld (talk) 20:17, 8 February 2011 (UTC)[reply]

You also need to know the cost of performing the service to a bidder (you said they're equivalent); no one will bid below that because they would do better not to bid at all. You could also incorporate the cost of not getting the bid: salaries may need to be paid anyway, and another bid will have to be prepared. In that case you can actually get bids slightly below cost because the alternative is (say) paying your staff to do nothing. --Tardis (talk) 21:39, 8 February 2011 (UTC)[reply]

Conversely, there may be an opportunity cost of taking on the job at a low price, if it ties up resources that could otherwise be used on more profitable work. AndrewWTaylor (talk) 09:46, 9 February 2011 (UTC)[reply]

I would be surprised if people behave the same way in reverse auctions as in regular auctions. Many auctions that are theoretically equivalant have known behavioral tendencies in participants, while prospect theory demonstrates that people's economic decisions under uncertainty are affected by context (here, whether the bids increase or decrease over time). Vernon Smith has written much about this; if this is more than an academic question, I would check out his writings. —Preceding unsigned comment added by 12.186.80.1 (talk) 15:34, 9 February 2011 (UTC)[reply]