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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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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#Definition and notation) Widener (talk) 23:58, 31 August 2011 (UTC)[reply]

Suppose there are 2. Their quotient is a positive nth root of unity. Invrnc (talk) 00:33, 1 September 2011 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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 + …, then you use the coefficients c_k to calculate the radius of convergence. This is a (possibly infinite) positive real number ρ for which the power series converges for all | z – a | < ρ. If the radius of convergence is infinite then you have an entire function. — Fly by Night (talk) 16:06, 2 September 2011 (UTC)[reply]