Is there a theorem or method to prove that we can't express the antiderivative of some functions with elementary functions ? For example Gamma or Zeta or $\dfrac{\sin x}{x!}$? thanks
12405
In the past, I've come across statements  along the lines of &quot;function $f(x)$ has no closed form integral&quot;, which I assume means that there is no combination of the operations:  addition/subtraction multiplication/division raising to powers and roots trigonometric functions exponential functions logarithmic functions  which when differentiated gives the function $f(x)$. I've heard this said about the function $f(x) = x^x$, for example. What sort of techniques are used to prove statements like this? What is this branch of mathematics called?  Merged with &quot;&quot; by : Sometimes we are told that some functions like $\dfrac{\sin(x)}{x}$ don't have an indefinite integral, or that it can't be expressed in term of other simple functions. I wonder how we can prove that kind of assertion?
18915
Show an example where neither $\lim\limits_{x\to c} f(x)$ or $\lim\limits_{x\to c} g(x)$ exists but $\lim\limits_{x\to c} f(x)g(x)$ exists.  Sorry if this seems elementary, I have just started my degree...  Thanks in advance.
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There is a lesser known generalization of Euler's Theorem. Here 'x' and 'm' are NOT COPRIME and $\,n\ge \log_2 m.\,$ I stumbled upon this on this site :   The "derivation" can be found towards the end of the above link; but i can't seem to wrap my head around it. It would be really great if it could be explained a bit better, probably with an example.
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I was wondering if there are functions for which $$f'(x) &gt; f(x)$$ for all $x$. Only examples I could think of were $e^x - c$ and simply $- c$ in which $c &gt; 0$. Also, is there any significance in a function that is always less than its derivative?    Edit: Thank you very much for all the replies. It seems almost all functions that apply are exponential by nature...  Are there more examples like - 1/x?  Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration  is always greater than its velocity]
121842
Friends,  Do you know of a neat proof of the linear independence (over $\mathbb{C}$) of the functions $f(t) = e^{at}$ and $g(t)=e^{bt}$ when $a$ and $b$ are linearly independent over $\mathbb{Q}$?
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This is the question: "Show that the derivative of an even function is odd and that the derivative of an odd function is even. (Write the equation that says f is even, and differentiate both sides, using the chain rule.)"  I already read numerous solutions online.  was the official solution but I didn't quite understand it (particularly, I'm not convinced why exactly $dz/dx=-1$; even though $z=-x$).  Thanks in advance =]
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In general, $\frac{d}{dx}(f(x) \cdot g(x)) \neq  \frac{d}{dx}f(x) \cdot \frac{d}{dx}g(x)$   When does this result hold true? My first try is to use product rule on left side and compare the two sides, but this hasn't helped at all. Any suggestions?
99040
Suppose I was working in radians and took  $$\sin \left(\frac{a\pi}{b}\right)$$  where both $a$ and $b$ are integers.  Is there a proof that the output of this function cannot be transcendental? Or, conversely, that   $$\arcsin \left(t \right)$$  where $t$ is some transcendental number, e.g., $e$, cannot be a rational angle $\left(\frac{a\pi }{b}\right)$?  Sadly, I couldn't get very far with this problem with my maths teacher so any information on the topic would be very useful.
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Find all entire functions $f(z)$ such that $|f(z)|=1$ for $|z|=1$  Hint: First show that $f(z)$ is a polynomial.  Clearly one can not use Cauchy Estimates to prove that $f(z)$ is a polynomial, the other way is to prove that $f(z)$ has a pole at infinity which I am not sure how to prove that, are there any other ways to prove that a given entire function is a polynomial ?  Thank you !
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Show there is no analytic function in $\{|z|&lt;1\}$ such that $f(0)=1$ and $|f(z)|\ge 1+|z|^2$. I guess something went wrong along the way and I would like your assistance.   $f$ does'nt get a maximum in the domain, because if it did, it would be constant in the domain, meaning $|f(z)|=1\ge 1+|z|^2$ which doesn't hold for all z. Since $f$ isn't zero in the domain, ${1\over f}$ is analytic in the domain. but $|{1\over f(z)}|\le {1\over 1+|z|^2}\le 1={1\over f(0)}$, which indicates a local maximum making ${1\over f}$ constant, contradiction the fact the $f$ isn't. What am I really lacking here?
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I'm hoping someone can give me a nudge in the right direction...     Let $F$ be a finite field, and let $f(x)$ be a nonconstant polynomial whose derivative is the zero polynomial.  Prove that $f$ cannot be irreducible over $F$.   I've got that every root of $f$ is a multiple root and that for $F=\mathbb{F}_{p^r}$, the exponent of every term of $f$ is a multiple of $p$.
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I know that $\int \frac{1}{x}~dx = \ln|x| + C$ and I know the antiderivative method works for all powers of $x$ except $-1$. But why is that the case? I am still in high school and teachers aren't really helpful with these questions.   Edit: I do realize dividing by zero is meaningless. That's not the question. Just because dividing by zero is meaningless doesn't mean mathematicians just choose to ignore it and work just make up a new answer. There has to be an explanation, and that's what I am asking for. The explanation.
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Suppose $f$ and $g$ are entire functions with $|f(z)|\leq|g(z)|$ for all $z$.  How can we show that $f=cg$ for some complex constant $c$?  Thanks for any help :)
115888
I need help with the following exercise:  Prove that if $f$ and $g$ are analytic functions on a region $G$ such that $\bar{f}g$ is analytic then $f$ is constant or $g\equiv0$  Thanks!
67634
Let $f:X\to Y$ be a function and $E\subseteq X$ and $F\subseteq X$.   Show that in general $f(E − F)\nsubseteq f(E) − f(F)$.   I have no idea about how to prove this; and could anyone please explain the basic theory of functions(by relating to this question. Thank you.
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I am trying to find a general derivative for the function: $f(x)=x^{x^{x^{...^{x}}}}$however to do that I must find $f^{\prime }$ and $f^{\prime \prime}$...etc. I am now trying to write down a general expression for $f^{\prime \prime}$ and I have stumble opon the series in the picture below, I wonder, do anyone have an idea of what sort of expantion it migth be ? I have tried breaking it down into three different sums, but I wasn't very successful!  EDIT  So since this post has gotten some attention and the title has been change I will tell you what I am on about, although I already have a post just for the question in the title from before. I am as I said earlier trying to find a general derivative for the function $f(x)=x^{x^{x^{...^{x}}}}$ where $n$ is an arbitrary number, natural number and greater or equal to 2. I have already made success when I tried to find the first derivative, and I have also found some interesting pattern emerge when the derivatives as well as n incresses. But my question here was if anyone knew what this expation might be or look similar to be, since I have yet to be lernt series etc. I did not know that the question needed such background since it's straight a forward question.
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Suppose $A=\{z\in \mathbb{C}: 0&lt;|z|&lt;1\}$ and $B=\{z\in \mathbb{C}: 2&lt;|z|&lt;3\}$.   Show that there is no one -to-one analytic function from A to B. Any hints? Thanks!