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April 22

Natural logarithms, limits and sequences of functions

Hello, I was inspired by a previous question about reconciling the integral definition of the natural logarithm

\ln(x):=\int _{1}^{x}{\frac {\mathrm {d} t}{t}},

with the limit expression for Euler's number

e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}.

I thought I might try and do this by considering the sequence of functions

f_{n}(x):=\int _{1}^{x}t^{{\frac {1}{n}}-1}\mathrm {d} t.

If we further define $a_{n}$ as the solution to $f_{n}(a_{n})=1$ , we can show that

a_{n}=\left(1+{\frac {1}{n}}\right)^{n}.

Now I'd like to show that in the limit $n\to \infty$ that $f_{n}(x)\to \ln(x)$ and therefore, by defining Euler's number by $\ln(e)=1$ , that

e=\lim _{n\to \infty }a_{n}

.

My problem is in making these statements rigorous. I tried to show that the sequence of functions converges uniformly to $\ln(x)$ , but I don't think it does. Is it possible to say that this sequence converges in some other way? If it does converge, is it possible to extend the convergence of a function sequence to some convergence of its roots? Can anybody help me complete this argument? Thank you in advance. 129.234.53.19 (talk) 12:16, 22 April 2012 (UTC)[reply]

f_{n}

doesn't converge uniformly to ln on

(0,\infty )

, but it does on

[1,3]

, which is all you need once you've established that

a_{n}

and

e

are all in [1,3], which is technical (and can be made easier by taking a looser bound). ln is also uniformly continuous in this domain. From the fact that ln is increasing (and thus one-to-one) it follows that to show that

e=\lim _{n\to \infty }a_{n}

it is sufficient to show that

1=\ln(e)=\ln(\lim _{n\to \infty }a_{n})

. I think it is true (and easy to prove) that if

f_{n}\to f

uniformly,

f

is uniformly continuous, and

a_{n}\to a

then

f(a)=\lim _{n\to \infty }f_{n}(a_{n})

. Apply this and you're done. -- Meni Rosenfeld (talk) 20:13, 23 April 2012 (UTC)[reply]

Thank you, Meni. I'll have to wait until I have time to work through it, but your comments look like they'll carry me most of the way. 129.234.53.19 (talk) 12:58, 25 April 2012 (UTC)[reply]

real integrals by residue theorem

I am stuck wid this prblm as my answer isnt matching wid d book. integration of (sinsqarex divided by 5+4cosx) from 0 to 2pi. sorry 4 poor scripting. im using the substitution 4 sinx nd then sqaring it, i also tried sinx= 1-cos2x/2 but in both cases im not getting the correct answer. im getting poles to be 0 nd -1/2 and the numerator to be a biquadratic in z. in one solution one used, 1-cos2x/2= 1-zsqare/2 and i am still not getting past it.please highlight where im missin.59.165.108.89 (talk) 12:24, 22 April 2012 (UTC)[reply]

I just want to note that writing in this twitter style means making it easier for you to write while making it harder for other people to read. Why should we work to answer the question when you're not willing to do the minimal work of asking the question properly? Regards, Looie496 (talk) 18:30, 22 April 2012 (UTC)[reply]

lolz n00b Der iz pol @

x=\pi +i\log 2

!!1! Clclate by using rect contour arnd pol, then snd top 2 &infinity;. Sławomir Biały (talk) 20:20, 22 April 2012 (UTC)[reply]

thnk u stawomicr.but pls tell how to integrate it using unit circle as the contour.

LMAO — Fly by Night (talk) 22:19, 23 April 2012 (UTC)[reply]

thanks for reply looie and if i wrote in this style then it doesnt mean that i was twittering .but,again commonsense is very uncommon.

Polynomials in Zp

Hi,

Are there criteria for determining a given polynomial with no constant term is a zero polynomial in Zp?

Are there criteria for determining a given polynomial with a constant term is reducible in Zp (not irreducible)?

I'm not necessarily concerned with all cases. I'm more interested in special cases. Nkot (talk) 17:51, 22 April 2012 (UTC)[reply]

Are you interested in polynomials in one variable or several variables? By "a zero polynomial" do you mean the actual zero element of the polynomial ring, or maybe a polynomial which evaluates to zero at every point? Rckrone (talk) 19:15, 22 April 2012 (UTC)[reply]

Initially, I was envisioning one variable, but a polynomial with more variables might also be interesting if it had special properties relating to above. Also, by "zero polynomial," I did mean a polynomial that evaluates to zero at every point. Nkot (talk) 19:25, 22 April 2012 (UTC)[reply]

For question 1, Euler's theorem tells us a^p = a for all a in Z_p, so if you have a polynomial f with terms of degree p or higher, you can reduce those terms to get an equivalent (as a function) polynomial with degree less than p. The only polynomial with degree less than p which always evaluates to zero is the zero polynomial. (You can show that polynomials with degree less than p are in bijection with the functions Z_p → Z_p.) I think this works the same way for multiple variables.

I don't know a good answer for the second question. Rckrone (talk) 20:56, 22 April 2012 (UTC)[reply]

An algorithm similar to the Sieve of Eratosthenes can be used to find the irreducible polynomials in

\mathbb {Z} _{p}[x]

. Since there are finitely many polynomials below a given degree, if you want to determine whether

f\in \mathbb {Z} _{p}[x]

is irreducible, you can just try all possible factors below, and you can optimize such an algorithm to only try the candidates which must show up in a factorization of