This may be a strange question; but I've read and re-read the chapter in my textbook  on what exactly a branch of a logarithm is and am having trouble understanding.   What is a branch of a logarithmic function? And what is a branch cut- how does it relate to the analyticity of a function?
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I know this is the most typical example of branches and I think I don't get the concept... Could you help me by giving a detailed development leading to all the required branches? It'd help me understanding more complicated examples...  Thank you very much, this concept really is hard to understand for me...
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When trying to evaluate$$\ln(\ln(\ln(\ln(\cdots\ln(x)\cdots))))$$I noticed that the answer was bound to be complex for any $x$.  Plugging in a very, very large real number in for $x$ will eventually become complex, and the $\log$ of a complex number is always complex or imaginary.  But what happens when we try to evaluate $$\lim_{x\to\infty}\ln(\ln(\ln(\ln(\cdots\ln(x)\cdots))))$$  We see that $$\lim_{x\to\infty}\ln(x)=\infty$$Meaning that if we try to do this repeatedly, we should still get infinite as our final result.  But we also see that for arbitrarily large $x$ where $x$ is real and finite, the result is a complex answer.  Which creates a sort of contradiction.  What does it evaluate to?!?!
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How do I determine this limit&nbsp;?:  $$ \lim_{x \to \infty}\left[x - x^{2}\log\left(1 + {1 \over x}\right)\right] $$  I have tried to decompose the $\log$ function but I don't know how to proceed from there.
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$$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$  Now, $\log(n!) = \Theta (n\log(n))$ so I think we could write,   $$\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) $$  $$\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) $$  $$\log\left(2n\right) - \log\left(n\right) = \log(2)$$  Is this legitimate? I feel this might be wrong since, the $\Theta$ notation conceals a constant factor.
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Let $f$ be an analytic function on $\mathbb{C}$ which takes values in $\mathbb{C}\backslash(-\infty,0]$, i.e. takes values in the complement of the nonpositive part of the real axis. Show that $f$ is constant.  My first observation is that there is a branch of the square root defined on $\mathbb{C}\backslash(-\infty,0]$ which takes positive real values along the positive part of the real axis. But I am not sure what do from thereon out. Any help would be appreciated!
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I need help with a textbook exercise (Stein's Complex Analysis, Chapter 3, Exercises 9). This exercise requires me to show that $$\int_0^1 \log(\sin \pi x)dx=-\log2$$ A hint is given as "Use the contour shown in Figure 9."    Since this is an exercise from Chapter 3, I think I should use the residue formula or something like that. But the function $f(x)=\log(\sin \pi x)$ becomes singular on $x=0$ and $x=1$, which makes the contour illegal for the residue theorem. Can anyone give me a further hint on this problem? Many thanks in advance!  P.S. This is my first time on Math Stack Exchange. If you find my post ambiguous, let me know.
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Show that   $\log_7 n$ is either an integer or an irrational number where n is a positive number.   I assumed that it is rational and tried to get a contradiction for $\log_7 n = a/b$, where b does not divide a, but how can I show that $7^{a/b}$ is not an integer to achieve a contradiction since n is an integer ? If I can exclude rational numbers from the range of log function then it is either integer or irrational.  Or do you suggest other methods ?
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If y is a function of x such that $y(x-y)^2=x$      Statement-I: $$\int\frac{dx}{x-3y}=\frac12\log[(x-y)^2-1]$$   Because      Statement-II: $$\int\frac{dx}{x-3y}=\log(x-3y)+c$$      Question: Is Statement-I true? Is Statement-II true? Is Statement-II a correct explanation for Statement-I?   I can say that II is false because y is not a constant but indeeed a function, I don't know answers to other two questions.I am thinking that probably we have to eleiminate y from the given functional equation or do a clever substitution?
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It is easy to show $\log$ isn't a polynomial (no continuous extension to $\mathbb{R}$). More challenging is showing it isn't rational.   Suppose it were a rational function. Then write, the fraction in lowest terms  $$\log x =\frac{G(x)}{Q(x)} \Longleftrightarrow\frac{G(x)}{\log x} = Q(x)$$  Clearly, as $x \to 0$, $Q(x) \to 0$. However, $Q(x)$ then has $x$ as factor so that   $$\frac{G(x)}{x\log x} = Q_2(x)$$  It is well known that $x \log x \to 0$ as $x \to 0$, so for $Q_2(x)$ to have a finite limit as $x \to 0$, which it must since it is a polynomial, $G(x) \to 0$ as $x \to 0$ so that $x$ is a factor of $G(x)$. This contradicts the assumption that $\frac{G}{Q}$ was in lowest terms.  If there is something wrong with this, please comment, but my main question is      What are some other ways to prove that $\log x$ isn't a rational function?
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I would like to know how logarithms are calculated by computers. The GNU C library, for example, uses a call to the fyl2x() assembler instruction, which means that logarithms are calculated directly from the hardware. So the question is: what algorithm is used by computers to calculate logarithms?
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$\log x =Cx^4$ has only one root. Find C.  I don't know how to solve this problem. Do you take derivative on both sides?  I am thinking C equals 0. Am I correct on that?
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If $\log (a +b +c) =\log a+\log b+\log c$ then show that $$\log \left(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2}\right)= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$$  Trial: If put $a=\frac{2a}{1-a^2}$ and the similar we are done. Can we do this in this way? Otherwise how to do. Please help.
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I came across the following problem that says:  Let $f$ be an analytic function defined on $\mathbb{D}=\{z\in\mathbb{C}:|z|&lt;1\}$ such that the range of $f$ is contained in the set $\mathbb{C}\setminus (-\infty,0]$. Then  $f$ is necessarily a constant function. there exists an analytic function $g$ on $\mathbb{D}$ such that $g(x)$ is a square root of $f(z)$ for each $z\in\mathbb{D}$. there exists an analytic function $g$ on $\mathbb{D}$ such that $\operatorname{Re}g(z)\geq 0$ and $g(z)$ is a square root of $f(z)$ for each $z\in\mathbb{D}$. there exists an analytic function $g$ on $\mathbb{D}$ such that $\operatorname{Re}g(z)\leq 0$ and $g(z)$ is a square root of $f(z)$ for each $z\in\mathbb{D}$.   I have to determine which options are correct. Can someone help in the right direction? Thanks in advance for your time.
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Use the unique factorization for integers theorem and the definition of logarithms to prove that $\log_3 (7)$ is irrational.     I am taking a beginners fundamental mathematics module, no advanced stuff please. Thanks!   My attempt.  Suppose for a contradiction that it is rational, that is $\log_3(7)=\frac{a}{b}$ for some $a,b\in R$ where $b\neq0$. Therefore, by the definition of logarithms, $7=3^{\frac{a}{b}}$. By the theorem of unique factorization, $7=1*7$ is unique.  Ok I'm plain stuck! Any help please?
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How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$ ? i know that its "1", but why? How can i calculate this?  Thank you very very much =)
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I am doing , and am having a few problems with function notation. I understand the form f(x).  What does the form f: x ↦ y mean? Could you also give one or two examples?  And, if possible, state your source. Thank you.
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$f$ is entire without any zeros then there is an entire function $g$ such that $f=e^g$  What I think is since $f$ do not have any zero for some bounded domain, I can define a branch of logarithm $(\log f)$ on that domain which will gives my desired result $f =e^{\log f}$. I don't know if I am doing it right? If this is right I don't know how do I argue $(\log f)$ is entire. Hint please.