For what values of $\alpha$ would the given conditions hold? I don't know how to proceed.
11127
I've been trying to solve the following problem:  Suppose that $f$ and $f&#39;$ are continuous functions on $\mathbb{R}$, and that $\displaystyle\lim_{x\to\infty}f(x)$ and $\displaystyle\lim_{x\to\infty}f&#39;(x)$ exist. Show that $\displaystyle\lim_{x\to\infty}f&#39;(x) = 0$.  I'm not entirely sure what to do. Since there's not a lot of information given, I guess there isn't very much one can do. I tried using the definition of the derivative and showing that it went to $0$ as $x$ went to $\infty$ but that didn't really work out. Now I'm thinking I should assume $\displaystyle\lim_{x\to\infty}f&#39;(x) = L \neq 0$ and try to get a contradiction, but I'm not sure where the contradiction would come from.   Could somebody point me in the right direction (e.g. a certain theorem or property I have to use?) Thanks
63955
Problem :   If $\tan^2\alpha \tan^2\beta +\tan^2\beta \tan^2\gamma + \tan^2\gamma \tan^2\alpha + 2\tan^2\alpha \tan^2\beta \tan^2\gamma =1$  Then find the value of $\sin^2\alpha + \sin^2\beta +\sin^2\gamma$    Please suggest how to proceed in such problem.
62554
Let $F,G$ be cumulative distributions functions. I need to find a necessary and sufficient condition for $\alpha$ such that $(1-\alpha)F+\alpha G$ is also a cumulative distribution function.   I have found that a sufficient condition is $0 \leq \alpha \leq 1 $, but I'm not sure wether it is a necessity. I'm struggling with proving that given $\alpha &gt;1$ or $\alpha&lt;0$ ,  $(1-\alpha) F+\alpha G$ is not monotonic.  Any ideas?
52385
How do I show that 0.5Y in part (a) down below is an unbiased estimator of alpha? If n, the number of observations, is very large then Y would equal to alpha+alpha=2*alpha an then 0.5Y=alpha so E(0.5Y)=E(alpha)=alpha but the size of n is not stated so how is this question supposed to be solved?
67147
Why is $\alpha \rightarrow \forall x(\alpha)$ not generally correct in first-order logic?  i.e., when there are free occurrences of $x$ in $\alpha$, and, on the same point, why is the formula scheme correct when there are no free occurrences of $x$ in $\alpha$?
75170
Let $\{x_n\}_{n=1}^\infty$ be a squence of nonnegative real numbers such that $\displaystyle \lim_{n \to \infty} x_n =x \in \Bbb R$. If $\alpha \in (0, \infty)$, is it true that $\displaystyle \lim_{n \to \infty} x_n^\alpha =x^\alpha$?  I read the proofs for $\alpha$ integer and for $\alpha=1/2$, but I wonder what happens with the other cases.
32317
I need to "study the limit behavior of the following sequence" and then compute the limit. I can compute the limit and prove the monotonicity, but I am having trouble proving boundedness. I tried to understand from other similar posts how this could be done, but since I didn't understand I'm asking it as a new question. The sequence is $x_{n+1}=\sqrt{\alpha +x_n}, x_0=\sqrt \alpha, \alpha&gt;0$.  It seems to be monotonically increasing.   Proof:   It can be shown by induction that the sequence is monotonically increasing.   Claim that $x_{n+1}\geq x_n$  $$x_1=\sqrt{\alpha+\sqrt\alpha}&gt;\sqrt\alpha=x_0$$ $x_{n+1}\implies x_{n+2}$ $$x_{n+1}=\sqrt{\alpha+x_n}&lt;\sqrt{\alpha+x_{n+1}}=\sqrt{\alpha+\sqrt{\alpha+\sqrt\alpha}}=x_{n+2}$$ Boundedness:  Because the sequence is monotonically increasing and $x_o=\sqrt\alpha$, it is bounded below by $\sqrt\alpha$. Now comes the part where I have to prove that the sequence is bounded above. The problem is that I don't know how to start, so if anyone could give me a bit of a push I'd be grateful.  The limit:  I know that $\lim\ x_n=\lim \ x_{n+1}$, so letting $\lim\ x_n=x$ then $$x=\sqrt{\alpha+x}$$ and solving would give $x=\frac{1+\sqrt{1+4\alpha}}{2}$.  May somebody please confirm my proof thus far, and help me prove the upper bound? Thanks!
38308
So basically I want to show:  $\aleph_{\alpha}^{\aleph_{1}} = \aleph_{\alpha}^{\aleph_{0}}\cdot2^{\aleph_1}$ , Where $\omega\leq\alpha&lt;\omega_1$  There is a proof that I found;    But I don't understand his proof for the limit case of $\alpha$.  To be more specific, I dont understand why $ \left( \sup_{\omega \le \beta &lt; \alpha} \aleph_{\beta}\right)^{\aleph_1} \le \left( \prod_{\omega \le \beta &lt; \alpha} \aleph_{\beta}\right)^{\aleph_1}$  Any insights or help is deeply appreciated.  Cheers
44541
I have looked extensively for a proof on the internet but all of them were too obscure. I would appreciate if someone could lay out a simple proof for this important result. Thank you.
27462
I am trying to prove the following statement:  Suppose $ \alpha $ is a limit ordinal. Then $ 1 ^\alpha + 2^\alpha = 3^\alpha$.  I'm not sure how to grasp this. Obviously induction won't work, since the statement doesn't hold for non-limit $ \alpha $.  I tried to solve this for $ \alpha = \omega $, but then the statement gets trivial since both sides equal $ \omega $.  I would appreciate a hint
79272
Please refer the attached image. I want to write \omega \in \mathbb{R} below sup as shown in the equation in the image. How should I do it?
11391
If $$\alpha=\frac {5}{2!3}+\frac {5\cdot 7}{3!3^2}+\frac {5\cdot 7\cdot 9}{4!3^3}+\dots $$  Find the value of $\alpha^2+4\alpha $.   The possible options are:   21 23 25 27   I think $$\alpha=\sum \frac {(2n+4)!}{2^{n+2}(n+1)!(n+2)!3^{n+1}}$$ but cannot think of what else to do.
59316
This was  on Quora. I thought about it a little bit but didn't make much progress beyond some obvious upper and lower bounds. The answer probably depends on AC and perhaps also GCH or other axioms. A quick search also failed to provide answers.
101201
The question is :     Let $f : [0,1] \longrightarrow \mathbb {R}$ be defined by    $f(x) = x^{\alpha} \sin {\frac {1} {x^{\beta}}}$ , whenever $x \in (0,1]$ and $=0$ , whenever $x = 0$.  Then find the values of $\alpha$ and $\beta$ for which $f$ becomes a function of bounded variation on $[0,1]$.  Please give me a right way to proceed.Thank you in advance.  I have just obtained the result which is  $\alpha &gt; 2$ and $\beta &lt; \alpha - 2$.Is it correct at all?Please verify it.
111304
So if I have the following generator and an initial condition:  $$A(f)(x) = \alpha x f'(x) + f''(x) \\ X_0 = x \in \mathbb{R}^+$$  I've been asked to find $X_t$ and assume that $\alpha$ is a constant.  From the definition of the infinitesimal generator, I get that:  $$dX_t = \alpha X_t dt + \sqrt{2} dB_t$$  How do I go about setting up an application of Ito's Lemma to this diffusion in order to solve for $X_t$?  Separation of variables doesn't seem to apply in this situation.
55527
If $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, and $(\beta_1,\beta_2,\beta_3) = (\alpha_1,\alpha_2,\alpha_3)C$. Prove that if $\mid C\mid \neq 0$, then $\beta_1,\beta_2,\beta_3$ are linearly independent.   This statement seems to be trivial, if $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, then $\det{(\alpha_1,\alpha_2,\alpha_3)} \neq0$, and $\det{(\alpha_1,\alpha_2,\alpha_3) C} = \det{(\alpha_1,\alpha_2,\alpha_3)}*\det{}C\neq 0$, which implies $\det(\beta_1,\beta_2,\beta_3) \neq0$, and hence the columns are linearly independent.  I'm not sure how to prove it rigorously, or if I'm missing any steps here. Would really appreciate any help! Thank you!
21748
I am having issues solving these problems: I tried using $(|a|+|b|)^2 \geq 0$ and $(|a|-|b|)^2 \geq 0$ but I am having problems constructing the proof. I need to gain some intuition on how to proceed.     Show that for $a,b\in (-\infty,+\infty)$,   the following inequalities apply:   a) $2|ab| \leq a^2 + b^2$  b) $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$