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single-valued
single-valued, a. Math. [single a.] Having a unique value for each value of its argument(s); that maps to one and only one point, number, etc. Hence single-valuedness, the property of being single-valued.1879 Maxwell Electr. & Magn. (1881) II. 252 The potential of the magnetic system is single value... Oxford English Dictionary
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Borel Measures: Single-Valued Given the complex plane $\mathbb{C}$. Consider the Dirac measure: $$\mu_\lambda(A):=\chi_A(\lambda)$$ Then it attains only zero and one: $$\mu_\lambda(A)=0,1$$ Are there any other such...
Same proof, less technical. Suppose $\mu$ is a measure taking exactly values $0,1$. We claim $\mu$ is a dirac measure. Let $U$ be the union of all open sets of measure zero. Since $\mathbb C$ is 2nd countable, $U$ is the union of coutably many open sets of measure zero, so $\mu(U) = 0$. Let $F$ be t...
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What is the definition of a single valued function this is potentially a dumb question but I am a touch confused about some terminology. I'm reading Ahlfor's complex analysis, and I am in a section on integrals of har...
Some authors call such functions multi-valued functions. So the text just means it is a "proper" function.
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Why $f(z)=z^2$ is single valued? Why $f(z)=z^2$ is single valued where $z\in\mathbb{C}$? From definition we have $$z^2=e^{2 \log z}=e^{2(\ln|z|+i(2k\pi+Arg(z)))}$$ I dont get it ;/ Maybe it's getting late.
Normally, $z^a$ with $a \in \mathbb Z$ _can_ be defined as a single-valued function without drawbacks, so we do so; it is not defined as $\exp(a \log z is no ambiguity though, because if $a \in \mathbb Z$ we have that $\exp(a \log z) = z^a$ (as defined above); in particular $\exp(a \log z)$ admits a single
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How would yo make a branch cut to define a single-valued branch of the function log(z - 1 + i)? The question says it all. I understand the concept of branch cuts, but I have not quite yet figured out how to find branc...
There are infinite different branch cuts that satisfy this, but for one, you could do $1-i+z$ for $\\{{z\in \mathbb{R}}|z>0\\}$.
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Three-Way Decisions with Single-Valued Neutrosophic Decision ... - Hindawi
The single-valued neutrosophic set (SVNS) can not only depict imperfect information in the real decision system but also handle undetermined and inconformity information flexibly and effectively. Three-way decisions (3WDs) are often used as an effective method to deal with uncertainties, but the conditional probability is given by the decision maker subjectively, which makes the decision ...
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How to find extreme values of an $f(x,y)$ function? I need this for my semester exams, unfortunately I was absent the day this topic was "talked about". My function is the real-valued $$f(x,y)=x-xy+x^2+y^2$$, interpr...
To find the extrems of a function of two (or more) variables, you need: 1. Look for the critical points, that is, points for which $\nabla f(x,y)=(0,0)$. In your case, $\nabla f(x,y)=(1-y+2x,-x+2y)=(0,0) \iff x=-2/3,\ y=-1/3$. 2. Determine if the critical poitn is a maximum, minimum or saddle point,...
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Is there a standard name for this relation property : " aRb --> there is no c different from b such that aRc "? Maybe this property could be called "exclusivity" ? Does it have a standard name? It recalls the defi...
Such relations are (in my experience) called " **functional** ", in analogy with functions. Indeed, such a relation _is_ a partial function (and actually I've heard "$R$ is a partial function" more frequently than I've heard "$R$ is functional"). Similarly, relations such that for every $a$ there is...
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Riemann surface intuition. In my complex variables notes it says that the multivalued $n$-th root function $w=z^{\frac{1}{n}}$ becomes single-valued on an appropriately constructed Riemann surface. It says how to go a...
Algebraic intuition is to note that different values correspond to dividing the argument which is equivalent up to multiples of $2\pi$ by $n$. Define a surface parametrized by $r>0$, the magnitude , and $0<\theta<2\pi n$, the argument. Then map $(r,\theta) \mapsto re^{i\theta/n}$ for which we can vi...
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Can the inverse of set-valued function also be a set-valued function? Can the inverse of set-valued function also be a set-valued function? If so, what we call the set-valued function whose inverse is a single-valued ...
Therefore a set-valued function has a set-valued inverse iff the function is a set-valued set function. You could say, the set-valued function whose inverse is a single-valued function is a single-value-argumented set-valued function, but you should declare
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Domain of definition of $\log \log z$ Find the region where the complex function $\log \log z$ is defined. I don't understand the question. Should I literally just find where the function is defined, or where the map...
$log logz$ can be defined everywhere in the complex plane except at $z=0$ and $z=1$. I am assuming the question asks for the points at which the function can exist.
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Is it okay to ignore all other branches except principal values? I took only a first course of complex analysis. I have learned some multi-valued functions which consistently extend the classical real functions such ...
In such a case, you can pick a branch cut for $\log(z)$ in which the logarithm is single-valued and differentiable.
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Limit of Integral About a Point I'm wondering how one would go about proving rigorously the following statement: Suppose one has an infinitely differentiable (I'm not sure one needs this, but I'm thinking about the T...
For a function $f$ that is bounded on a neighbourhood of $c$, say, $|f(x)|<M$ for all $x\in(c-\delta,c+\delta)$, then $\left|\displaystyle\int_{c-\epsilon}^{c+\epsilon}f(x)dx\right|\leq\displaystyle\int_{c-\epsilon}^{c+\epsilon}|f(x)|dx\leq 2M\epsilon$ for all $\epsilon<\delta$. Now the result follo...
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Request for hint: introductory complex analysis problem _This is not a ‘do-my-homework’ question._ The problem is > For what value of $a$, is $$F(z)=\int^z_{z_0}e^z\left(\frac1z+\frac{a}{z^3}\right)dz$$ single-value...
**Hint:** \begin{align}f\text{ is single-valued}&\iff\text{for every loop }\gamma,\ \displaystyle\int_\gamma e^z\left(\frac 1z+\frac a{z^3}\right)\,\mathrm
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why is each branch of $\log z$ yields a branch of $z^{\alpha}$? $F(z)$ is said to be a $\bf{branch}$ of a multiple-valued function $f(z)$ in a domain D if $F(z)$ is single-valued and continuous in $D$ and has the prop...
If $L(z)$ is a branch of $\log z$, then $e^{\alpha L(z)}$ is a branch of $e^{\alpha \log z}$. That's what the textbook means by "yields".
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