M_B
• Member for 7 years, 6 months
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• Reno, NV, United States

Let $f(x)=c$, where $c$ is some constant number, and suppose that the domain of $f$ is the real numbers. Then we can take the limit as $x$ approaches some value, say for example $$\lim_{x\rightarrow0}... View answer 4 votes Books on complex analysis definitely use the topics that you mentioned, but usually assume that the reader is already familiar with some algebra and geometry of complex numbers. The book Visual ... View answer 3 votes You have shown that if there is a period, then it must be infinitesimally small, which implies that the function under consideration must be constant. However, you don't consider the case where there ... View answer 3 votes The definition of a function is not just the rule f. Instead, a function is defined by a domain A and a codomain B together with a rule f that takes every x\in A and returns a unique element ... View answer 3 votes Hint: review properties of exponents, especially that the quotient of two numbers each raised to the the x power is equal to the quotient of the two numbers itself raised to the x power. View answer 2 votes Hint: Assume some x>0 is in the intersection. Can you find a contradiction to this assumption? That is, can you find one of the intervals in the intersection not containing x? Responding to ... View answer 2 votes Assume to the contrary that there is a prime p with f(p) \geq 1. Then f(p^{k})=f(p)^{k}, \forall k \in \mathbb{N}, a contradiction, since then there are infinitely many values of f greater ... View answer 1 votes By the binomial theorem$$ (f_1+f_2)^n -(f_1-f_2)^n = \sum_{k=0}^{n}\binom{n}{k}f_1^{n-k}f_2^k -\sum_{k=0}^{n}\binom{n}{k}f_1^{n-k}(-f_2)^k. $$Notice that since we have (-f_2)^k the second sum is ... View answer Accepted answer 1 votes The dimension theorem for vector spaces states that all bases of a vector space have the same number of elements. So if you can determine the cardinality of one basis, you have determined the ... View answer 1 votes Recall that the argument of a complex number is the angle that z\in\mathbb{C} forms in standard position. Considering the set \{z\in\mathbb{C}:-\pi< \arg (z)<\pi\}, note the radius is not ... View answer 1 votes Let B_{r}(z) be the open ball centered at z\in\mathbb{C} with radius r. Consider the two open balls B_{1}(-1) and B_{1}(1). The first is the set of points with |z+1|<1 and the second is ... View answer 1 votes Note that the minimum is not always well-defined. For example, given the interval (0,1), there is no smallest element in the interval, since we may find numbers in the interval arbitrarily close to ... View answer 0 votes Use the fact that the domain and range have the same number of elements. Since every element in the range is hit by the function when we assume surjectivity and every element in the domain is sent to ... View answer Accepted answer 0 votes I will show you how to get started on a proof by induction. Base case: for n=0, we have a_0=1 by definition, and 2^{0+1}-1=1. Induction step: Suppose a_k=2^{k+1}-1 for some k>0. Then ... View answer 0 votes I will prove that the only topology that satisfies the given property is the discrete topology. Assume that (X,T) is a topological space such that if A and B are open sets with A a proper ... View answer 0 votes Hint: have you tried using the product-to-sum identities? That is,$$ \sin A\sin B =\frac{1}{2}(\cos(A-B)-\cos(A+B))  is one such formula. Note that there are several formulas of this type that can ...