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January 13[edit]

Help with integral[edit]

How do you evaluate this integral? Could someone help?

${\frac {d}{dz}}\left(\int \!{\frac {{\it {f}}\left(z-r\right)}{{r}^{3}}}{dr}\right)$

deeptrivia (talk) 00:12, 13 January 2012 (UTC)[reply]

\int \!{\frac {{\it {f'}}(z-r)dr}{r^{3}}}.

Bo Jacoby (talk) 08:00, 13 January 2012 (UTC).[reply]

Is the integral in the question meant to be a definite integral with r ranging on a fixed interval (which does not depend on z)? – b_jonas 09:07, 13 January 2012 (UTC)[reply]

b_jonas and Bo Jacoby, thanks for your answer. This integral comes out as a part of a general solution to a PDE, where f can be any function. Since the solution you came up with is an integral of a derivative, could this be simplified further?

The PDE I'm trying to solve is

${\frac {\partial ^{2}}{\partial {r}^{2}}}v\left(r,z\right)+{\frac {{\frac {\partial }{\partial r}}v\left(r,z\right)}{r}}-{\frac {v\left(r,z\right)}{{r}^{2}}}+{\frac {\partial ^{2}}{\partial z\partial r}}v\left(r,z\right)-{\frac {{\frac {\partial }{\partial z}}v\left(r,z\right)}{r}}=0$

The solution Maple comes up with is:

$\left(\int \!{\frac {{\it {\_F3}}\left(-r+z\right)}{{r}^{3}}}{dr}+{\it {\_F4}}\left(z\right)\right)r$

Thanks a ton! deeptrivia (talk) 13:31, 13 January 2012 (UTC)[reply]

zero to the power of complex numbers[edit]

How do you raise zero to a complex power? Widener (talk) 15:04, 13 January 2012 (UTC)[reply]

Normally you don't. Zero to a positive power is zero. Zero to the zeroth power is one. Zero to a negative power is undefined (1/0), and so is zero to a nonreal complex power. Bo Jacoby (talk) 16:02, 13 January 2012 (UTC).[reply]

For 0⁰ see Exponentiation#Zero to the zero power. (Quick summary: It is useful in discrete mathematics to define it as 1 (coming from the empty product), but when dealing with limits in analysis, it must be handled as an indeterminate form. Exponentiation#History of differing points of view gives a couple of good quotes.) -- 110.49.224.131 (talk) 08:29, 14 January 2012 (UTC)[reply]

It is not necessary to undefine

0^{0}

, just note that

0=\lim _{y\rightarrow 0^{+}}0^{y}\neq 0^{0}=1

.

Bo Jacoby (talk) 10:20, 14 January 2012 (UTC).[reply]

Did you read the article section? You can't just note that one exception, there are many others. 98.248.42.252 (talk) 15:31, 14 January 2012 (UTC)[reply]

In this context, where the exponential function is clearly assumed to be of the

z^{w}=\exp(w\log z)

variety (by definition!), the expression

0^{0}

is not defined. This isn't because of anyone "undefining" it. It was never defined in the first place. You could come in and add to that definition that it is equal to

0^{0}

, but that would be your own personal convention and not relevant to the subject of the post. (For more on this, see Talk:Exponentiation#more 0^0.) Sławomir Biały (talk) 21:19, 14 January 2012 (UTC)[reply]

One thing that I would like to add to the discussion is that our article on binomial series says that if x = −1 then the binomial series for (1 + x)^α converges if, and only if, either Re(α) > 0 or α = 0. This implies that the binomial series corresponding to (1 − 1)^α = 0^α converges for any complex number α with either Re(α) > 0 or α = 0, and that it converges to 1. — Fly by Night (talk) 21:00, 14 January 2012 (UTC)[reply]

I think you probably mean that the binomial series converges to 0 if Re(α) > 0 and to 1 if α = 0. Sławomir Biały (talk) 21:24, 14 January 2012 (UTC)[reply]

I think you're probably right. — Fly by Night (talk) 01:56, 15 January 2012 (UTC)[reply]

Consider the polynomial $f(x)=\sum _{i=0}^{n}a_{i}x^{i}$ . The value $f(0)=\sum _{i=0}^{n}a_{i}0^{i}=a_{0}$ exactly because you can trust that $0^{0}=1$ . Bo Jacoby (talk) 20:38, 15 January 2012 (UTC).[reply]

Is this true?[edit]

$\lim _{(x,y)\rightarrow (a,b)}x^{y}=(\lim _{x\rightarrow a}x)^{\lim _{y\rightarrow b}y}$ for $(a,b)\neq (0,0)$ ? Widener (talk) 15:09, 13 January 2012 (UTC)[reply]

When

a>0

. Wgunther (talk) 17:09, 13 January 2012 (UTC)[reply]

... and you can see why this restriction is necessary by considering -1^1/3. Gandalf61 (talk) 17:12, 13 January 2012 (UTC)[reply]

Yes, via definition. — Preceding unsigned comment added by 203.11.71.124 (talk) 08:30, 14 January 2012 (UTC)[reply]