Suppose $$f(x)=x!$$ for all $x$ in $N$,and $f$ is also continuous in $x&gt;0$,does this mean $f(x)=\Pi (x)$ for all $x&gt;0$?
88840
Usually, when someone says something like $\left(\frac12\right)!$, they are probably referring to the Gamma function, which extends the factorial to any value of $x$.  The usual definition of the factorial is $x!=1\times2\times3\times\dots x$, but for $x\notin\mathbb{N}$, the Gamma function results in $x!=\int_0^\infty t^xe^{-t}dt$.  However, back a while ago, someone mentioned that there may be more than one way to define the factorial for non-integer arguments, and so, I wished to disprove that statement with some assumptions about the factorial function.    the factorial   is a  for $x\in\mathbb{C}$ except at $\mathbb{Z}_{&lt;0}$ because of singularities, which we will see later. is a monotone increasing function that is concave up for $x&gt;1$. satisfies the relation $x!=x(x-1)!$ and lastly $1!=1$     From $3$ and $4$, one can define $x!$ for $x\in\mathbb{N}$, and we can see that for negative integer arguments, the factorial is undefined.  We can also see that $0!=1$.  Since we assumed $2$, we should be able to sketch the factorial for $x&gt;1$, using our points found from $3,4$ as guidelines.  At the same time, when sketching the graph, we remember $1$, so there can be no jumps or gaps from one value of $x$ to the next.  Then we reapply $3$, correcting values for $x\in\mathbb{R}$, since all values of $x$ must satisfy this relationship.  Again, because of $1$, we must re-correct our graph, since having $3$ makes the derivative of $x!$ for $x\in\mathbb N$ undefined.  So, because of $1$ and $3$, I realized that there can only be one way to define the factorial for $x\in\mathbb R$.  Is my reasoning correct?  And can there be only one extension to the factorial?    Oh, and here is a  to how I almost differentiated the factorial only with a few assumptions, like that it is even possible to differentiate.  Putting that in mind, it could be possible to define the factorial with Taylor's theorem?
37886
Assume we have a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that  $f$ is differentiable for all non-zero $x$, $f$ is continuous at $0$ and  $$\lim_{x \uparrow 0} f'(x) = \lim_{x \downarrow 0} f'(x) &lt; \infty. $$  Is it then true that $f$ is also differentiable at $0$?  If we drop the hypothesis that $f$ is continuous, this is not true since the signums function is a simple counterexample. But what if $f$ is continuous? I don't think this is true but I can't find a counterexample...
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Let $f,g\in PC(2\pi)$ continuous. Show that $$\lim_{n\to \infty}\int_{-\pi}^{\pi}f(x)g(nx)\,dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\,dx\int_{-\pi}^{\pi}g(x)\,dx$$ Where if $f\in PC(2\pi)$ then $\int_{-\pi}^{\pi}f(x)\,dx=\int_c^{c+2\pi}f(x)\,dx \quad\forall c\in\mathbb{R}.$ I've been struggling with this proof. Any suggestions would be great!
113748
Possible Duplicates:      I found a nice problem I would like to share. Problem: If $f$ is continuous on $[0,1]$, and if $$\int_0^1 f(x)x^n \ dx =0$$ for every positive integer $n$, prove that $f(x)=0$ on $[0,1]$. Source: W. Rudin, Principles of Mathematical Analysis, Chapter 7, Exercise 20. I have posted a proposed solution in the answers.
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Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0\in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0\neq x\in(a,b)$. Is the following statement true:      If $\lim\limits_{x\to x_0}f'(x)$ exists, then $f'(x_0)$ exists and $\lim\limits_{x\to x_0}f'(x)=f'(x_0)$.   Thanks!
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Why is a function f continuous, if it is defined only on integers?   It seems very trivial but I could not find the words to describe the reason. Can smo(someone) give me a hint?  Note: there is no function. This is a verbal question. No equation exists for the function   Thanks
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Possible Duplicate:          Let $f, g:\Bbb{R}\to\Bbb{R}$ to be continuous functions such that $f(x)=g(x)\text{ for all rational numbers}\,x\in\Bbb{Q}$. Does it follow that $f(x)=g(x)$ for all real numbers $x$?  Here is what I think:  f continuous when $\lim\limits_{x\to x_0}f(x)=f(x_0)$ and $\lim\limits_{x\to x_0}g(x)=g(x_0)$  So it does not neccesarily mean that $f(x)=g(x)$ when x is irrational. So I can pick a function f so that   $f(x) = \begin{cases} g(x)  &amp; \text{if $x\in\Bbb{Q}$} \\ x &amp; \text{if $x\in\Bbb{R}\setminus \Bbb{Q}$}  \\ \end{cases}   $
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Let $n$ be a positive integer.  $\;  $  Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be everywhere .  Let $S$ be a subset of $\mathbb{R}^n$ with  zero.  Does it follow that $\{f(s) : s\in S\}$ has Lebesgue measure zero?  (I know it would if $D(f)$ is continuous.)
73794
Suppose that $f:[0,\infty)\rightarrow\mathbb{R}$ is continuous. Is it true that   $$\int_{0}^\infty|f(t)|dt&lt;\infty\Rightarrow \lim_{t\rightarrow\infty}f(t)=0?$$  If so can you provide a proof, otherwise a counter example. Thank you.
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Let $f(x) = \frac{x}{x}$.  Is it correct to say that $f(x) \ne 1$, since $f(x)$ has a discontinuity at $x=0$?
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Please help me to solve this question:  Suppose $f:[a,b] \to \Bbb R$ satisfies:  $f''(x)+f(x)&gt;0$ and $f(x)&gt;0$ for all $x\in(a ,b)$; $f(a)=f(b)=0$.  Prove that $b-a&gt;\pi$.  Thanks in advance.
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Suppose $f(x)$ is continuous on the closed interval $[a,b]$. Define $m(x)=\max_{a\leq s\leq x}\, f(s)$, $a\leq x\leq b$. Is $m(x)$ continuous necessarily? Thank you.
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Let $f:X \rightarrow Y$ be a function. Prove that if $f$ is continuous, then for every convergent sequence $(x_n)$ lim$_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)$  My attempt: Assume $f$ is continuous. Then $\forall \epsilon &gt;0, \exists \delta &gt; 0$ such that $\forall p \in X$ and $q \in X: d(p,q) &lt; \delta \implies d(f(q),f(q)) &lt; \epsilon.$  And also assume that $(x_n)$ converges. Then $\exists L$ such that $\forall \epsilon &gt;0, \exists N$ such that $n&gt;N \implies d(x_n, L) &lt; \epsilon.$  I need to use these definitions to show lim$_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)$ . Any hints?  Edit: lim$_{n \rightarrow \infty}(x_n) \in X$
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If $f$ is continuous from $ [0,1]$ to itself and $f(0)=0$, $f(1)=1$ and $f(f(x))=x$ then show that $f(x)=x$ for every $ x $ in the domain of function.  I'd like to prove this statement, my idea goes like below:  Suppose that the converse of the statement holds. then without loss of generality, assume that there exists some $c$ in $[0,1]$ such that $d=f(c)&gt;c$. By symmetry, $f(d)=f(f(c))&lt;f(c)=d$. And $f$ is bijective because $f$ has an inverse function in this domain. So $f$ should be strictly increasing function. But on $[c,d]$, $f$ is decreasing. This is a contradiction.  So it should be that $f(x)=x$ on $[0,1]$.  Is this argument right? And I'd like to know another argument which prove this statement. Thank you for your answer in advance :)
42752
If $f$ is continuous on $[0,1]$, how can I show that $$\int_0^{n\pi}f(\cos^2(x))\,dx=n\int_0^{\pi}f(\cos^2(x))\,dx?$$  I know how to show that both integrals exist.  I believe that $n$ is a multiple of $\frac12$.  I thought that the identity $\cos^2(x)=\frac{\cos(2x)+1}{2}$ would be useful, but maybe not.  I think I should use some change of variable.
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Let $f$ be a continuous function. If $f$ is uniformly continuous then it maps cauhy sequence into cauchy sequence. Is the converse true? That is if $f$ maps cauchy into cauchy gives the uniform continuity of $f$. I think it's no. Can suggest a counter example
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Let $f:\mathbb R \to \mathbb R$ be a continuous function such that $f(f(f(x)))=x,\forall x \in \mathbb R$ , then is it true that $f(x)=x , \forall x \in \mathbb R$ ? I can see that $f$ is bijective , hence strictly monotone . But I cannot progress further . Please help. Thanks in advance