Wikipedia:Reference desk/Archives/Mathematics/2011 September 1

= September 1 =

No (x) ?
Hi. Some math authors write functions f(x), g(x), etc solely as f or g without (x) in certain cases. For example, Spivak states IBP as: $$\int{fg'}=fg-\int{f'g}$$ When is this allowed, and why? danke. --Meumann — Preceding unsigned comment added by 24.92.85.35 (talk) 00:11, 1 September 2011 (UTC)
 * $$f$$ refers to the function itself whereas $$f(x)$$ refers to the value of $$f$$ evaluated at $$x$$Widener (talk) 00:14, 1 September 2011 (UTC)
 * For example, taking an antiderivative of a function is a transformation applied to the function, so using $$f$$ is preferable to $$f(x)$$ in this case. Widener (talk) 00:18, 1 September 2011 (UTC)
 * Of course, when $$x$$ is an independent variable, $$f(x)$$ is often interpreted as $$ x\mapsto f(x)$$, which by extensionality is the same as $$f$$.--Antendren (talk) 01:41, 1 September 2011 (UTC)
 * Most of the time, this is just a shorthand notation used when it is clear from context where the function is evaluated. -- Meni Rosenfeld (talk) 07:08, 1 September 2011 (UTC)

Uniqueness and existence of the positive nth root
For positive $$r,x$$ and $$n \in \mathbb{N}$$ what is the proof that the polynomial $$r^n=x$$ has exactly one positive root? (as per nth root) Widener (talk) 23:58, 31 August 2011 (UTC)
 * Suppose there are 2. Their quotient is a positive nth root of unity. Invrnc (talk) 00:33, 1 September 2011 (UTC)
 * $$r^n$$ is strictly increasing on $$[0,\infty]$$, goes from 0 to $$\infty$$, and is continuous. Thus, it achieves every positive value (Intermediate value theorem) exactly once (being injective). -- Meni Rosenfeld (talk) 07:04, 1 September 2011 (UTC)

Analytic Functions
Considering the function $$f\left( x \right)=\sum_{k=0}^\infty x^k \sqrt{ \binom{k+t_1}{t_1} \binom{k+t_1}{t_2}}$$ where$$t_1>t_2$$ and $$t_1, t_2$$ are both natural numbers. Is it convergent? How do I tell? and does anyone know if there exists a more concise representation (without the summation)? — Preceding unsigned comment added by 192.76.7.237 (talk) 10:43, 1 September 2011 (UTC)
 * The expression $$\binom{k+a}{b}$$ where a, b are constant is polynomial in k. Thus, the coefficients in your power series have polynomial growth, so it converges for every $$|x|<1$$.
 * Mathematica is unable to find a closed-form expression for this, and I doubt one exists. -- Meni Rosenfeld (talk) 15:45, 1 September 2011 (UTC)

If you want to know how to see if a power series converges then have a look at our radius of convergence article. If you have a power series in a complex variable, say c0 + c1(z – a) + c2(z – a)2 + c3(z – a)3 + &hellip;, then you use the coefficients ck to calculate the radius of convergence. This is a (possibly infinite) positive real number &rho; for which the power series converges for all &thinsp;z – a&thinsp;. If the radius of convergence is infinite then you have an entire function. — Fly by Night  ( talk )  16:06, 2 September 2011 (UTC)