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May 10

Transcendental Functions

Why are trigonometric functions transcendental? Besides a formal proof, is there some intutition behind the reason? Thanks-Shahab (talk) 01:54, 10 May 2011 (UTC)[reply]

I think maybe the intuition is, why should they be algebraic? Being an algebraic function is a very special thing. If there's no reason you can elucidate that a function should be algebraic, then it's probably not. --Trovatore (talk) 02:01, 10 May 2011 (UTC)[reply]

It all boils down to why is

π

transcendental. Trigonometric functions of a rational number times

π

are algebraic. Dmcq (talk) 08:45, 10 May 2011 (UTC)[reply]

Dmcq - not sure if I understand you correctly. Are you saying that

\sin(x)

is a transcendental function but

\sin(\pi x)

is an algebraic function ? So we can transform a transcendental function into an algebraic function or vice versa by rescaling the x axis ?? Gandalf61 (talk) 09:18, 10 May 2011 (UTC)[reply]

I took him to mean that the value of a trig function at a rational multiple of

\pi

was algebraic as a number. Icthyos (talk) 09:51, 10 May 2011 (UTC)[reply]

Yes exactly, the function itself can't be transformed that way into an algebraic one, for one thing sin has an infinity of zeroes. Dmcq (talk) 11:57, 10 May 2011 (UTC)[reply]

I think a trigonometric function grows too fast (exponentially) for an algebraic function. – b_jonas 14:22, 10 May 2011 (UTC)[reply]

Presumably you mean in the imaginary axis? --COVIZAPIBETEFOKY (talk) 17:27, 10 May 2011 (UTC)[reply]

Well, it doesn't grow exponentially on the real axis, does it? But to stay in real numbers, I think it's also true that a nonconstant algebraic function cannot be periodic. -- Meni Rosenfeld (talk) 10:31, 11 May 2011 (UTC)[reply]

An algebraic function having a taylor expansion which is convergent for all values of x, is a polynomial. The functions e^x and sin(x) has taylor expansions which are convergent for all values of x, but they are not polynomials. So e^x and sin(x) are transcendental functions. This is my intuitive reason. Bo Jacoby (talk) 21:13, 10 May 2011 (UTC).[reply]

Thank you all. My purpose was to find out an intuitive reason to explain the meaning of a transcendental function to someone who barely knows calculus. Maybe, I can explain a transcendental function as follows: All the nice functions are made up of sums of powers of x. (This is true for the complex case certainly as analytic functions are precisely the true functions of z, but I cannot tell that). So all the nice functions that we have are polynomials or sort of "infinite polynomials". In the former case they are called algebraic (for obvious reasons) and in the latter since they go beyond finite sums and products they are called transcendental (the non-nice ones are also put in this category since they too transcend algebra). Trigonometric functions fall in the latter category. However is there some intuitive way to relate the Taylor expansions of the sine with the y-coordinates of the unit circle? Preferably without using any calculus?-Shahab (talk) 09:18, 11 May 2011 (UTC)[reply]

I'd say this gives the wrong impression of what an algebraic function is. If you want to say that sin x is not a polynomial just say so. But

{\sqrt {x^{2}+1}}

is algebraic and it doesn't have a finite power series. -- Meni Rosenfeld (talk) 10:31, 11 May 2011 (UTC)[reply]

Forgive my ignorance but I based my understanding on the remark "An algebraic function having a taylor expansion which is convergent for all values of x, is a polynomial." above. Maybe I need to understand things properly yet-Shahab (talk) 17:37, 11 May 2011 (UTC)[reply]

Well, that's not a contradiction. The power series for

{\sqrt {x^{2}+1}}

, near (say) x=0, has a radius of convergence of 1; you can tell because it can't work at x=i, where the function has a branch point. So even on the real line, the series diverges for x greater than 1 or less than −1. You can get series with larger radii of convergence by taking a different center, but the radius will always be finite. --Trovatore (talk) 19:14, 11 May 2011 (UTC)[reply]

Kelly criterion vs logarithmic utility function

Can someone please explain to me why, assuming we have a bankroll of 1 and are betting x, the Kelly fraction f is also the maximum of the expected utility calculated using logarithmic utility. i.e. if we're betting x with a win probability of p, our expected value (EV) is

EV = (1+x)*p + (1-x)*(1-p)

while our expected utility (u) is

log(u) = log [(1+x)^p] + log [(1-x)^(1-p)]
log(u) = p log (1+x) + (1-p) log (1-x)
u = (1+x)^p * (1-x)^(1-p)

Why is this particular utility function the best one in maximizing long-term profits over a series of fractional bets? Thank you. 161.53.179.232 (talk) 11:34, 10 May 2011 (UTC)[reply]

Well it's only best if you don't have a spread of bets, this is one of the reasons people have a spread of shares - they do better than this. Try thinking about the log of the amount, it will go up or down by a constant amount and your intuition would be correct about the utility of the logs. Dmcq (talk) 12:04, 10 May 2011 (UTC)[reply]

You have some problems with your notation. If we denote our bankroll by

r

, then our utility is

u=\log r

and we have

E[r] = r_0*[(1+x)*p + (1-x)*(1-p)]
E[u] = u_0 + p log (1+x) + (1-p) log (1-x)

Now, a utility function cannot be "best for" anything. A utility function is primary and embodies the desires of an individual - optimization comes after the function was decided. If an individual's utility function is linear in his bankroll then he should go for broke on every bet, but that's not realistic because of the diminishing marginal returns of money. A logarithmic function embodies the assumption that a person who has $10,000 is as happy about getting $100 as a person who has $100,000 is about getting $1,000. If that's the utility function used, you can observe that the final utility is the sum of independent random variables, one for each bet (assuming the fraction wagered is chosen independently of previous wins), so the central limit theorem kicks in. -- Meni Rosenfeld (talk) 10:08, 11 May 2011 (UTC)[reply]

Very sorry I misread the question. In fact I'd worked the Kelly criterion myself years ago so I should have known better, but I didn't known it was called that and just assumed the question was about volatility in shares. I've been meaning to have a good read of Black–Scholes even if it did lead to a financial collapse and see what all this area is like nowadays. Dmcq (talk) 14:56, 11 May 2011 (UTC)[reply]