# Tag Info

5

Some ideas: $$\frac{x^n}{x^n+1}=1-\frac1{x^n+1}\xrightarrow[n\to\infty]{}1\;,\;\;\forall\;x\in (1,2]$$ Yet, for $\;x=1\;$ , we have $$\frac{1^n}{1^n+1}=\frac12\xrightarrow[n\to\infty]{}\frac12$$ So the limit function isn't continuous...yet for all $\;n\; ,\;\ f_n\;$ is, and then...

5

$$\bigg|\frac{f(x)-f(y) }{x-y} -0 \bigg|\leq |x-y|$$ So if $x$ goes to $y$ then $f'(y)=0$ for all $y$

3

Use the elementary identity $\dfrac{1}{n^2} < \dfrac{1}{n(n-1)}$ for $n \ge 2$ to make the estimate $$s - s_n = \sum_{k=n+1}^\infty \frac 1{k^2} < \sum_{k=n+1}^\infty \frac{1}{k(k-1)}.$$ Since $$\frac{1}{k(k-1)} = \frac{1}{k-1} - \frac{1}{k}$$ you have $$\sum_{k=n+1}^\infty \frac{1}{k(k-1)} = \sum_{k=n+1}^\infty \left[ \frac 1{k-1} - \frac 1{k} ... 3 As other answers show, there is a large class of solutions to your functional equation if one does not restrict f more. As per your request, we shall try the restriction that f should be a polynomal. Two polynomial solutions jump into our eyes, namely the zero polynomial f(x)=0 and f(x)=2. Neither of these has f(4)=65, though. If f is a ... 3 Suppose we work on a compact set: The uniform convergence theorem uses a much stronger assumption on the nature of the convergence of your sequence of functions. It roughly says that if the sequence f_n converges uniformly to f you can exchange limit and integration. The dominated convergence theorem requires f_n only to converge pointwise almost ... 2 Hints. In a. If x or y vanishes for some t_0, then they stay zero, and due to uniqueness, they are zero everywhere - contraction. In b. Note that if x(t_0)= 0, for some t_0, then x\equiv 0, again due to uniqueness. In c. (x^2+y^2)'=-2x(x^2+y^2-1)=0. 2 HINT: For all n\in\Bbb Z^+,$$\begin{align*} |f(x)-f(x+a)|&\le\sum_{k=0}^{n-1}\left|f\left(x+\frac{ka}n\right)-f\left(x+\frac{(k+1)a}n\right)\right|\\ &\le\sum_{k=0}^{n-1}\left(\frac{a}n\right)^2\\ &=\frac{a^2}n\;. \end{align*}$$1 Think about a few things: The limit definition of the derivative, The derivative of a constant function And let x tend to 0 1 If \phi is a step function, i.e., \phi= \Sigma_{i=1}^n c_i\chi_i , then \phi  is Riemann integrable, so it is Lebesgue integrable, so that \phi \in L^1, so Cauchy-Schwarz applies:$$ \int (\phi) (\phi)= \int \phi^2 \leq \int \phi \int \phi= (\int \phi )^2 $$If you have not seen Lebesgue integration, then you can use the fact that the space ... 1 Suppose that x\notin A. Since A is closed we can find some open, maximal interval I\ni x disjoint from A. (Why?). Since A is not empty, there will be some y\in A. Suppose, for example, that y>x. Then I has an upper bound, and let z=\sup I. Then z must be in A. But every open neighbourhood of z intersects I, and this contradicts ... 1 The series \sum_{n\geq 1}\frac{1}{n^2} is converging since:$$0\leq \frac{1}{n^2}\leq \frac{1}{n-\frac{1}{2}}-\frac{1}{n+\frac{1}{2}}=\frac{1}{n^2-\frac{1}{4}}\tag{1} $$gives:$$ 0\leq \sum_{n\geq N}\frac{1}{n^2} \leq \sum_{n\geq N}\left(\frac{1}{n-\frac{1}{2}}-\frac{1}{n+\frac{1}{2}}\right) = \frac{1}{N-\frac{1}{2}}\tag{2}$$so:$$\sum_{n\geq ...

1

If you restrict the inner product on $\mathbb{R}^3$ to the sphere $\mathbb{S}^2$, the result is a Riemannian metric. Every Riemannian metric comes with a canonical top-dimensional form, the volume form, which tells you which bases for a tangent space have unit volume. This 2-form is the volume form for the metric on the sphere. It's called an "area form" ...

1

Hint: $$\bigcap_{k=1}^{\infty} B\Big(a ; \frac{1}{k}\Big) = \{a\}$$ Where $B\Big(a ; \frac{1}{k}\Big) \subset M$ is the open ball centered at $a$ with radius $\frac{1}{k}$. Any finite set $F = \{a_1, \ldots , a_n\} \subset M$ is closed in $M$. Take any $a \notin F$ and conclude that $d (a, F) > 0$. (Assuming $M$ has a metric).

1

You are correct. We generally need the axiom of choice for this, at least countable choice. Lucky for you, most analysis is done at least in the presence of countable choice. But you're in luck, because in the case of the complex plane we can avoid this. Indeed in general, without the axiom of choice, it is possible that there is a subspace of $\Bbb C$ ...

1

"Choosing" dense subsets of compact sets in plane (or $\mathbb{R}^n$) can be done uniformly as follows: Given $K$ compact, let $\{r_n : n < \omega \}$ list all rational points in plane. For each $n$, let $d_n$ be the distance of $r_n$ from $K$. If $d_n = 0$, let $x_n = r_n$, otherwise choose $x_n \in K$ to be the unique point in $K$ whose distance from ...

1

From context it sounds like you are talking about Jordan measure. In this case the answer to your question is no; this is one of the major reasons analysts tend to prefer the Lebesgue measure over the Jordan measure. You are already on the right track for building a counterexample. Consider $\{ q_i \}_{i=1}^\infty$ an enumeration of $A=\mathbb{Q} \cap ... 1 No, there are no other limit points: this set is closed. Suppose$(x, y)$is a limit point. Then there is a sequence$x_n$of real numbers such that$x_n \to x$and$\frac{1}{x_n} \to y$. Then$xy = \lim x_n \frac{1}{x_n} = 1$, because the multiplication map$\mathbb{R}^2 \to \mathbb{R}$is continuous. So$x \neq 0$and$y = \frac{1}{x}$. (It's clear that ... 1 Note that plugging in$x=1$, we obtain $$f(1)^2 = 2f(1) \implies f(1) = 0 \text{ or }2$$ Similarly, plugging in$x=-1$, we obtain $$f(-1) = 0 \text{ or }2$$ We have $$f(1/x)(f(x) - 1) = f(x) \implies f(1/x) = \dfrac{f(x)}{f(x)-1}$$ Hence, for$\vert x \vert > 1$define$f(x) = g(x)$such that$g(x) \neq 1$for all$\vert x \vert >1$. Then define ... 1 Assuming that$f$is nonzero, we can rewrite this equation as: $$\frac{1}{f(x)} + \frac{1}{f(1/x)} = 1$$ Making the substitution$g(y) = -\frac{1}{2} + \frac{1}{f(e^y)}$, we get the functional equation$g(y) + g(-y) = 0$, which simply says that$g$is odd. This equation can be solved, with the initial condition, by setting$g(y) = cy$for a suitable ... 1 Following up on @user17762's answer, suppose$f(x)=\sum_{i=0}^n a_i x^i$where$a_n\ne 0$and$n\ge 1$, then$f=g$for$|x|>1$, we have for$|x|<1$, $$\sum_{i=0}^n a_i x^i=\frac{\sum_{i=0}^n a_i x^{-i}}{\sum_{i=0}^n a_i x^{-i}-1}=\frac{\sum_{i=0}^n a_i x^{n-i}}{\sum_{i=0}^n a_i x^{n-i}-x^n}$$ For RHS to be a polynomial you have$a_0=1$or$a_0\ne 1$... 1$\bf{My\; Solution}::$Given$\displaystyle f(x)+f\left(\frac{1}{x}\right) = f(x)\cdot f\left(\frac{1}{x}\right)......................(\star.)$Now We can write$(\star)$as $$\displaystyle f(x) = \frac{f\left(\frac{1}{x}\right)}{f\left(\frac{1}{x}\right)-1}..........................................(\star\star)$$ Now again we can write$(\star)\$ as ...

Only top voted, non community-wiki answers of a minimum length are eligible