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26

Elementary functions are finite sums, differences, products, quotients, compositions, and $n$th roots of constants, polynomials, exponentials, logarithms, trig functions, and all of their inverse functions. The reason they are defined this way is because someone, somewhere thought they were useful. And other people believed him. Why, for example, don't we ...


17

I would approach the question this way. We can think of our "library" of functions being built up recursively: Start with a few basic functions (polynomials, exponentials, logarithms, trig functions, etc.), and start composing, concatenating, integrating, etc. At each stage you have a collection of functions that have been defined "so far". What ...


11

Generalized Laguerre polynomials have the generating function $$\sum_{n=0}^{\infty}\xi^n L_n^{(t)}(x)=(1-\xi)^{-t-1}e^{-\frac{\xi x}{1-\xi}}.$$ Multiplying this identity by $\xi^{t-1}$ and integrating w.r.t. $\xi$ from $0$ to $1$, one finds \begin{align}\sum_{n=0}^{\infty}\frac{L_n^{(t)}(x)}{t+n}&=\int_0^1\left(\frac{\xi}{1-\xi}\right)^{t-1} ...


8

The motivation here is similar to that in elementary Galois theory where you might study whether or not you can write the roots of a polynomial using the standard arithmetic functions along with the $n$th root function. You might wonder why anyone cares about solving polynomial equations using these restricted functions when you could just use any of a ...


8

Some functions that have counter-intuitive properties on $\mathbb R$ that you should know : The Weierstrass function, continuous everywhere, differentiable nowhere function. The Cantor function, a uniformly continuous function that is not absolutely continuous. The Minkowski's question mark function, an increasing continuous function, non differentiable on ...


6

I dont think there is a list of functions that "every mathematician should know". There are simply to many functions, and besides the elementary functions (polynomials, rational functions, log, exp, power functions, trig and inverse trig functions) one must learn what is needed. In statistics, for instance, we use a lot special functions such as the gamma ...


5

Hint: Let $x=\dfrac1{t^2+1}$ and then recognize the expression of the beta function in the new integral. But first, using the parity of the integrand, write $\displaystyle\int_{-\infty}^\infty f(t)~dt~=~2\int_0^\infty f(t)~dt$.


3

Try to substitute $$x=(b-a)t+a \;\text{ where }t\in [0,1]$$ Then use the beta function $$B(x,y) =\int_{0}^{1} t^{x-1}(1-t)^{y-1} dt =\cfrac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$$


3

I should add that proving an indefinite integral is not one of a given "library" of elementary functions is very, very difficult. Computer algebra systems use the Risch algorithm, hope i have the spelling correct. And you can define any new function you like, but only the ones found widely useful keep such names.


3

The other answer has not answered your question, “Why is $e$ equal to $2.718281828\cdots$?”. Let’s form the number $E=\lim_n(1+\frac1n)^n$, and evaluate it knowing the continuity of the log function and what its derivative is. Of course this number $E$ is computable, even if slowly, directly by hand. And if you take $n$ large enough, you will indeed find ...


3

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3

The equation is equivalent to $$\binom{n+m}{n}^{-1}=(n+m+1)\int_{0}^{1}x^{n}(1-x)^{m}dx$$ with $$ m\ge0 , n\ge 0$$ Or $${1\over(n+m+1)!}=\int_{0}^{1}{x^{n}(1-x)^{m}\over n!m!}dx$$ And hence one can multiply the two series and integrate from 0 to 1: $$e^{tx} = \sum_{n=0}^\infty{t^nx^n\over n!}$$ $$e^{s(1-x)} = \sum_{m=0}^\infty{s^m(1-x)^m\over m!}$$ ...


3

You should be able to apply the Frobenius method here. Writing $f(x)=\sum_{n=0}^{\infty}a_n x^n$ and substituting into your differential equation gives the following: $$ (x^2-1)\sum_{n=0}^{\infty}n(n-1)a_n x^{n-2} + x(8x^2-7)\sum_{n=0}^{\infty}na_n x^{n-1} - 4(C+1)\sum_{n=0}^{\infty}a_n x^n=0. $$ The coefficient of $x^n$ is $$ \left(-n(n+6)-4(C+1)\right)a_n ...


2

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2

Anticlimax ahead: $$\sum_{k=0}^\infty \frac{1}{4^k(k!)^2} < \sum_{k=0}^\infty \frac{1}{4^k\,k!}$$ follows since $4^k(k!)^2 \geqslant 4^k\, k!$ for all $k$ and the inequality is strict for $k\geqslant 2$. For the inequality $$\sum_{k=0}^\infty \frac{1}{5^k\,k!} < \sum_{k=0}^\infty \frac{1}{4^k(k!)^2},$$ we note that the terms for $k = 0$ are equal ...


2

The $\log$ function in the order 2 Clausen function is the natural logarithm. $$\mathrm{Cl}_2(\phi) = -\int\limits_{0}^{\phi} \log_e |2 \sin \frac{x}{2}| \mathrm{d}x$$ Logarithms in calculus expressions are always base $e$, rather than base $10$, unless explicitly stated otherwise.  The subscript is often omitted by lazy authors because this ...


2

One function I would love to mention: The Greatest integer function or the Floor function. $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by $f(x)=[x]$ where $[x]$ denotes the greatest integer less than or equal to $x$


2

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2

A reference: Chaudhry, M.A. et al. Asymptotics and closed form of a generalized incomplete gamma function.


2

If we ignore all "coincidences", including associativity and commutativity where it applies, we can enumerate all valid exprsssions by arranging the $n$ numbers in one of $n!$ orders insertingof parentheses among them in $C(n-1)$$=\frac{(2n-2)!}{(n-1)!n!}$ ways inserting operators among the subexpressions in $4^{n-1}$ ways That gives us a total of ...


2

A closed form for the integral is the Incomplete Beta function : $$\int{ \frac{1}{\sqrt{x+1}}{x^n}} dx = (-1)^{n+1}B_{-x}(n+1 , 1/2)+constant$$ The inverse function needs numerical calculus to be evaluated : http://www.dtic.mil/dtic/tr/fulltext/u2/a467901.pdf


1

Any function with "canonical" in the name. And the identity function. (I've been getting perversely much mileage out of the identity function lately.) canonical projection takes an element to its equivalence class (really set theory, but usually first seen in group theory) canonical projection morphism takes an element of a product to a specific one of ...


1

I always though the exponential function (and its generalizations) was widely regarded as the most mathematically significant function. From a complex analysis point of view, the trig. functions all come under the umbrella of the exponential function. Also, the exponential map is central in many areas closely connected to differential geometry, there's the ...


1

1) Polynomial functions 2) rational functions 3) trigonometric functions 4) hyperbolic functions 5) $e^x$ (also the complex e-function) and $log_a(b)$ And also many special functions like the dirichlet function, because its not riemann-integrable, but lebesgue integrable. (Nowhere continous), the squared function, which is not lipschitz-continous, ...


1

As I see it, the important functions that every mathematician should know are: $\Gamma(s)$ $\zeta(s)$ $sinh(x)$ $ch(x)$ $ln(x)$ (Where $s$ is a complex variable, and $x$ is a real number.) Because they are important in pure mathematics and important in number theory, and we know that almost all unsolved mathematics are focused in number theory.


1

$\qquad\qquad\qquad\qquad \qquad\qquad$ $\zeta(s)$, $\delta(x)$, $\vartheta(z,q)$ and $_2F_1(a,b;c;t)$.


1

No you can't do this: if you substitute your $A$ and $B$ into the RHS, then bring it all to a common denominator, you will find that you do not get the LHS. Actually if you think about it this is fairly obvious: the numerator on the RHS after combining terms will be a constant times $x$ plus a constant, and this is not what the numerator looks like on the ...


1

Define $f(x):= \ln \left(K_0(\sqrt{x})\right)$ and compute (e.g. with the help of a CAS) $$f'(x)=-\frac{1}{2}\frac{K_1(\sqrt{x})}{\sqrt{x}K_0(\sqrt{x})}$$ $$f''(x) = \frac{1}{4} \frac{\sqrt{x}K_0(\sqrt{x})^2+2K_0(\sqrt{x})K_1(\sqrt{x})-\sqrt{x}K_1(\sqrt{x})^2} {x^{3/2}K_0(\sqrt{x})^2}$$ From http://dlmf.nist.gov/10.37 we have $K_1(x)>K_0(x)>0\;$ for ...


1

You also need to impose $\log(1) = 0$; otherwise, it's only determined up to a constant. The inverse $f^{-1}$ has derivative $$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} = f^{-1}(x)$$ and $f^{-1}(0) = 1$ ; this is exactly the definition (well, one definition) of $e^x$.


1

If I understood your question correctly, the easiest way to do this is $$ f(\pi_k)=\frac{\pi_k}{\sum_{k=1}^{n} \pi_k} $$ another one: $$ f(\pi_k) = \frac{e^{-\pi_k}}{\sum_{k=1}^{n} e^{-\pi_k}} $$



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