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2425589 | Limit of [imath]a_1=\sqrt{3}[/imath], [imath]a_2=\sqrt{3\sqrt{3}}[/imath], [imath]a_3=\sqrt{3\sqrt{3\sqrt{3}}}[/imath]...
Find the limit of: [imath]a_1=\sqrt{3}[/imath] [imath]a_2=\sqrt{3\sqrt{3}}[/imath] [imath]a_3=\sqrt{3\sqrt{3\sqrt{3}}}[/imath] ... By using induction, I found that the limit is 3 but it seems pretty strange to me, I thought it would go to a further number. I just want to know if I'm right (proof-verification). | 1839634 | Find a formula for a sequence [imath]\{\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}},...\}[/imath]
I'm trying to find a formula for the following sequence: [imath]\{\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}},...\}[/imath] I thought of solving it recursively and I got this formula: [imath]a_{n}=\sqrt{3*a_{n-1}}[/imath] [imath]a_{0}=1[/imath] Is there a better and non-recursive formula for the given sequence? |
2425706 | [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] continuous and [imath]f(\mathbb{R}[/imath]) is countable, then [imath]f[/imath] is constant.
Let [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] be a continuous function such that [imath]f(\mathbb{R}[/imath]) is countable. Show that [imath]f[/imath] is constant. Some ideas: proving the contrapositive. To suppose that [imath]f[/imath] is not constant and to use the intermediate value theorem to prove that [imath]f(\mathbb{R})[/imath] is not countable. Any help? | 1816632 | Prove that continuous functions mapping irrationals to rationals must be constant
Let [imath]f\colon[0,1] \to \mathbb{R}[/imath] be a continuous function such that any irrational number is mapped to a rational number. Then [imath]f[/imath] must be a constant. Well, the context isn't that much, I was reading some interesting function constructions on Hardy's book. Then raised this question. As the solution suggest it is an elementary exercise. But definitely interesting. |
2426267 | Infinite cardinality [imath]\aleph_2[/imath]
I read in the book 'One two three infinity' by George Gamow that the cardinality of sets of all plane curves is greater than the number of real numbers and so it can be thought of as a strictly larger infinite crdinality [imath]\aleph_2[/imath]. I wanted to know the reasoning or proof of this. | 2078067 | Proof that the set of all possible curves is of cardinality [imath]\aleph_2[/imath]?
I am reading George Gamow's book One Two Three... Infinity and a certain assertion Gamow makes seems a little startling considering what I already know of the subject of infinite cardinality. But the number of all geometrical points, though larger than the number of integer and fractional numbers, is not the largest one known to mathematicians. In fact it was found the variety of all possible curves, including those of the most unusual shapes, has a larger membership than the collection of all geometric points, and thus has to be described by the third number in the infinite sequence. [Gamow, 22] He later does explicitly say the number of all possible curves is [imath]\aleph_2[/imath]. But nowhere else can I find reference to this being true, nor does he provide any sort of reasoning to its validity. My question is how can you prove the set of all possible curves is of cardinality [imath]\aleph_2[/imath] and not [imath]\aleph_1[/imath] or [imath]\aleph_0[/imath]? |
2426368 | If [imath]U[/imath] and [imath]V[/imath] are subspaces of [imath]\mathbb{R^4}[/imath], [imath](U+V)^{\perp}=(U^{\perp}\cap V^{\perp})[/imath]
As the title says, I would like to know if that equality is true if we consider [imath]\mathbb{R^4}[/imath]. I would write down my attempt, but I have no idea how to prove it, or how to choose a counterexample. [imath](U^{\perp}\cap V^{\perp})[/imath] are the vectors which are orthogonal to both [imath]U[/imath] and [imath]V[/imath], while [imath](U+V)^{\perp}[/imath] are the vectors which are orthogonal to the linearly independent vectors of [imath]U[/imath] and [imath]V[/imath] | 246967 | Show [imath](W_1 + W_2 )^\perp = W_1^\perp \cap W_2^\perp[/imath] and [imath]W_1^\perp + W_2^\perp ⊆ (W_1 \cap W_2 )^\perp[/imath]
Given [imath]V[/imath] a inner product space and [imath]W_1[/imath], [imath]W_2[/imath] subspaces of [imath]V[/imath] Show [imath](W_1 + W_2 )^\perp = W_1^\perp \cap W_2^\perp[/imath] and [imath]W_1^\perp + W_2^\perp ⊆ (W_1 \cap W_2 )^\perp[/imath]. |
2423315 | Open axioms of equality
I have a doubt. I need help. Can the basic axioms of equality be presented as "open axioms"? I) (reflexivity) [imath]\qquad x = x[/imath] II) (Substitutivity) [imath]\qquad (x = y) \to \big(F (x, x) \to F (x, y)\big)[/imath] | 2633413 | What is [imath]x = x[/imath]?
Let [imath]x[/imath] be an individual variable. What is [imath]x = x[/imath]? An open sentence? Or a true proposition? Well, I think it's an open sentence, due to the presence of the free variable [imath]x[/imath]. I do not know if I'm wrong. I ask for help. |
2426538 | Is [imath]\mathbb Z[1/2][/imath] a finitely generated [imath]\mathbb Z[/imath]-module?
Consider the subring [imath] \mathbb Z\left[\frac{1}{2}\right] = \left\{ \frac{k}{2^n} : n \ge 0 \right\}. [/imath] As [imath]1/2[/imath] is the root of [imath]2X - 1[/imath], this is an intergral extension. Now according to wikipedia we have the equivalence that if [imath]A \subseteq B[/imath] are rings with [imath]b \in B[/imath] [imath] \mbox{$b$ is integral over $A$} \Leftrightarrow \mbox{the subring $A[b]$ is a finitely generated $A$-module}. [/imath] But why is [imath]$\mathbb Z[1/2]$[/imath] a finitely generated [imath]\mathbb Z[/imath]-module, as I see it the elements [imath]1, 1/2, 1/4, \ldots[/imath] are infinitely many generators? | 1241049 | [imath]\mathbb Z[1/2][/imath] is not finitely generated?
[imath]\mathbb Z[1/2][/imath] is not finitely generated ? Maybe I misunderstood, what finitely generated means. Here it says, we need finitely many elements and I think [imath]1[/imath] and [imath]1/2[/imath] suffices as generators. But here, at the end of Proposition [imath]5.1.4[/imath], from the fact that [imath]\mathbb Z[1/2][/imath] is not finitely generated it is concluded that [imath]1/2[/imath] is not integral. |
2426871 | Integrating [imath]\int_{0}^{+\infty} e^{-st} \cos(at)\,dt[/imath] using the complex exponential
Other similar questions I have found used integration by part of alternative complicated transformations that I do not understand. Using the [imath]\cos(x) = \frac{1}{2} * (e^{ix} + e^{-ix})[/imath] identity, is there a way to find the solution, which should be [imath]a/(a^2 + s^2)[/imath]? | 353833 | Laplace transform of [imath]\cos(at)[/imath]
I need to find the Laplace transform of [imath]\cos(at)[/imath] I know that [imath]L\{\cos(at)\}= \int_{0}^{\infty} e^{-st} \cos (at) dt[/imath] but I am having trouble finding the integral Thank you |
2426531 | calculating the residue of [imath]f(z)=\frac{\cot(z)\coth(z)}{z^3}[/imath]
Find the residue of [imath]f(z)=\frac{\cot(z)\coth(z)}{z^3}[/imath] at [imath]z=0[/imath]. I tried to use the Cauchy's residue formula, but the calculation is very complex. Can it be done in a simpler way? | 552561 | Residue of [imath]f(z)=\frac{\cot(z)\coth(z)}{z^3}[/imath]
How can I find the residue of the following function at the point [imath]z=0[/imath] [imath]f(z)=\frac{\cot(z)\coth(z)}{z^3}[/imath] |
2427248 | Exact value of [imath]\arctan(2)[/imath]
The following problem has been on my mind for a while. Lots of exact values of the arctangent function are known, such as [imath]\arctan 0=0[/imath] [imath]\arctan 1=\frac{\pi}{4}[/imath] [imath]\arctan \frac{1}{\sqrt 3}=\frac{\pi}{6}[/imath] However, I can't seem to find an exact value of [imath]\arctan 2[/imath] How can I find one? Is it possible? NOTE: By exact, I mean that I am looking for an answer in the form [imath]\frac{p\pi}{q}[/imath] with [imath]p,q\in \mathbb Z[/imath]. | 79861 | ArcTan(2) a rational multiple of [imath]\pi[/imath]?
Consider a [imath]2 \times 1[/imath] rectangle split by a diagonal. Then the two angles at a corner are ArcTan(2) and ArcTan(1/2), which are about [imath]63.4^\circ[/imath] and [imath]26.6^\circ[/imath]. Of course the sum of these angles is [imath]90^\circ = \pi/2[/imath]. I would like to know if these angles are rational multiples of [imath]\pi[/imath]. It doesn't appear that they are, e.g., [imath](\tan^{-1} 2 )/\pi[/imath] is computed as 0.35241638234956672582459892377525947404886547611308210540007768713728\ 85232139736632682857010522101960 to 100 decimal places by Mathematica. But is there a theorem that could be applied here to prove that these angles are irrational multiples of [imath]\pi[/imath]? Thanks for ideas and/or pointers! (This question arose thinking about Dehn invariants.) |
2427468 | Proving a Max inequality
Going off of this question: Maximum Proof (Average?) How can we prove that [imath]max(|a|,|b|)\geq \frac{1}{2}(|a+b|)[/imath]? | 2427321 | Maximum Proof (Average?)
The link here: [imath]\max(a,b)=\frac{a+b+|a-b|}{2}[/imath] generalization shows that [imath]\max(a,b)=\frac{a+b+|a-b|}{2}[/imath]. Is it true that [imath]\max(|a|,|b|)=\frac{1}{2}(|a|+|b|)[/imath]? |
2428183 | Find all triples [imath](a,b,c)[/imath] of positive integers such that [imath](1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=3[/imath]
Find all triples [imath](a,b,c)[/imath] of positive integers such that [imath](1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=3[/imath] Supposing [imath]a\ge b\ge c[/imath], [imath](1+\frac{1}{c})^3\ge 3[/imath] [imath]\Rightarrow 1 + \frac{1}{c^3} + 3(1 + \frac{1}{c})\frac{1}{c} \ge3 [/imath] After this how to proceed? | 1887861 | Let [imath]a,b,c[/imath] positive integers such that [imath]\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) = 3[/imath]. Find those triples.
Full question: Let [imath]a[/imath],[imath]b[/imath],[imath]c[/imath] be three positive integers such that [imath]\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) = 3[/imath]. Find those triples. This is actually a national competition question in Vietnam (Violympic), which I have attended (and did poorly, but had a lot of fun). I have understood almost every questions asked that day, but this one really makes my head pop, because I haven't learn much about integer solution equation and how to solve hard cases (like that one) I have solved that a,b,c can't be all the same, because: [imath]a = b = c[/imath] [imath]\implies(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=(1+\frac{1}{a})^3=3[/imath] [imath]⟺1+ \frac{1}{a}=\sqrt[3]{3}[/imath] [imath]⟺\frac{1}{a}=\sqrt[3]{3}-1[/imath] [imath]⟺a=b=c= \frac{1}{(\sqrt[3]{3}-1)}[/imath] And using wolframalpha.com, I found out that [imath](a,b,c) \in \{(1,3,8),(1,4,5),(2,2,3)\}[/imath], but stuck at how to solve it. Thank you in advance for checking out. I really appreciate your effort. |
2427277 | Prove that [imath]x_{n+1}=\frac{1}{2}x_n+\frac{1}{x_n}[/imath] converges [imath]\sqrt{2}[/imath].
I struggle to prove this using the definition only. I can prove it by using some limit laws, but I wanted to use only the definition out of curiosity. I would appreciate some hints. Q: Prove that the sequnece below converges to [imath]\sqrt{2}[/imath]: [imath]x_{n+1}=\frac{1}{2}x_n+\frac{1}{x_n},[/imath] where [imath]x_1 = 2[/imath]. | 2384298 | Find the limit if it exists of [imath]S_{n+1} = \frac{1}{2}(S_n +\frac{A}{S_n})[/imath]
Suppose that [imath]S_0[/imath] and A are positive numbers, let [imath]S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right)[/imath] with [imath]n \geq 0 [/imath]. (a)Show that [imath]S_{n+1} \geq \sqrt{A} [/imath] if [imath]n \geq 0[/imath] (b)Show that [imath]S_{n+1} \leq S_n [/imath] , if [imath]n \geq 1[/imath] (c) Show that [imath]s= \lim\limits_{n \rightarrow \infty} S_n[/imath] exists (d) find s (a) Show that [imath]S_{n+1} \geq \sqrt{A} [/imath] if [imath]n \geq 0[/imath] Given [imath]P_n: S_{n+1} = \frac{1}{2}\left(S_n +\frac{A}{S_n}\right) \geq \sqrt{A}[/imath] [imath]P_0: S_{1} = \frac{1}{2}\left(S_0 +\frac{A}{S_0}\right) \geq \sqrt{A} [/imath] We assume that [imath]P_n[/imath] is true [imath]P_{n+1}: S_{n+2}= \frac{1}{2}\left(S_{n+1} +\frac{A}{S_{n+1}}\right)[/imath] by assumption [imath]S_{n+2}= \frac{1}{2}\left(S_{n+1}\left(1 +\frac{A}{(S_{n+1})^2}\right)\right) \geq \frac{1}{2}\left(\sqrt{A}\left(1 +\frac{A}{(\sqrt{A})^2}\right)\right)[/imath] [imath] S_{n+2}= \frac{1}{2}\left(S_{n+1} +\frac{A}{S_{n+1}}\right) \geq \sqrt{A} [/imath] It follows that [imath]S_{n+1} \geq \sqrt{A} [/imath] (b) Show that [imath]S_{n+1} \leq S_n [/imath] , if [imath]n \geq 1[/imath] [imath]S_{n+1} \leq S_n[/imath] [imath]\frac{1}{2}\left(S_n +\frac{A}{S_n}\right) \leq S_n [/imath] Dividing by [imath]S_n[/imath] [imath]\frac{1}{2}\left(1 +\frac{A}{S_n^2}\right) \leq 1 [/imath] [imath]\frac{A}{2S_n^2} \leq \frac{1}{2}[/imath] [imath]A \leq S_n^2[/imath] [imath]S_n \geq \sqrt{A}[/imath] As [imath]S_{n+1} \leq S_n[/imath] yields a true statement, it follows [imath]S_{n+1} \leq S_n[/imath] is true. (c) Show that [imath]s= \lim\limits_{n \rightarrow \infty} S_n[/imath] exists Since [imath]S_{n+1} \leq S_n[/imath], the sequence is non-increasing, using the non-increasing theorem stating that if [imath]\{S_n\}[/imath] is non-increasing then [imath]\lim\limits_{n \rightarrow > \infty} S_n = \inf\{S_n\} [/imath] (d) find s Is the argumentation in (a) and (b) appropriate? Also, I have to admit I m getting less confident in my argumentation (c) and (d). How to proceed in (c) and (d)? Much appreciated for your input or help. |
2427543 | Proving that E (set of even integers) is equal to F (sum of two odd integers)
Define E to be the set of even integers; E = {[imath]x[/imath] [imath]\in[/imath] [imath]\mathbb{Z}[/imath] : [imath]x[/imath] = 2[imath]k[/imath], where [imath]k[/imath] [imath]\in[/imath] [imath]\mathbb{Z}[/imath]}. Define F to be the set of integers that can be expressed as the sum of two odd integers such that [imath]F = \{y [/imath][imath]\in[/imath] [imath]\mathbb{Z}[/imath] : y = a + b, where [imath]a[/imath] = [imath]2k_1[/imath] + 1 and [imath]b[/imath] = [imath]2k_2 +1\}[/imath]. Prove E = F. My attempt: If [imath]x[/imath] = 2[imath]k[/imath] is even [imath]a[/imath] and [imath]b[/imath] can be written as a = [imath]2k_1[/imath], b = [imath]2k_2[/imath]. so [imath]a + b[/imath] = [imath]2k_1 + 2k_2 + 2[/imath] = 2([imath]k_1 + k_2[/imath] + 1) which is even. | 2427177 | Proving that even numbers equal the sum of two odd numbers.
Define E to be the set of even integers; E = {[imath]x[/imath] [imath]\in[/imath] [imath]\mathbb{Z}[/imath] : [imath]x[/imath] = 2[imath]k[/imath], where [imath]k[/imath] [imath]\in[/imath] [imath]\mathbb{Z}[/imath]}. Define F to be the set of integers that can be expressed as the sum of two odd numbers. Prove E = F. My attempt: The only way I can figure out the solution is by providing numbers and examples. It's easy to see that two odd numbers will always equal an even integer. I just don't know how to write the proof for it. |
2428143 | Complex Number Equation: [imath]z^2+(1+i)\overline z+4i=0[/imath]
I am having a hard time solving this equation. A know the solution from WolframAlpha, but the setp-by-step solution is not available. [imath]z^2+(1+i)\,\overline z+4i=0[/imath] http://www.wolframalpha.com/input/?i=z%5E2%2B(1%2Bi)conj(z)%2B4i%3D0 How can I solve this? | 2420558 | Solving complex equation [imath]z^2 + (1+i) \overline{z} + 4i = 0[/imath]
Consider the following equation, where [imath]z \in \mathbb{C}[/imath], [imath]i[/imath] is the imaginary unit and [imath]\overline{z}[/imath] is the conjugate of [imath]z[/imath]: [imath] z^2 + (1+i) \overline{z} + 4i = 0 [/imath] What is the method to deal with equations such as this? I have tried various things: I tried substituting [imath]z[/imath] with [imath]a+bi[/imath], or [imath]re^{i\theta}[/imath], hoping I'd notice something. I thought I could somehow transform this into a quadratic equation, but I couldn't. Now I have no idea what to try. I'd appreciate ideas greatly. |
2018745 | Let [imath]p[/imath] be a prime of the form [imath]3k+2[/imath] that divides [imath]a^2+ab+b^2[/imath]. Prove that [imath]a,b[/imath] are both divisible by [imath]p[/imath]. [Solution verification]
Problem:Let [imath]p[/imath] be a prime of the form [imath]3k+2[/imath] that divides [imath]a^2+ab+b^2[/imath] for some [imath]a,b\in \mathbb{Z}[/imath]. Prove that [imath]a,b[/imath] are both divisible by [imath]p[/imath]. My Attempt: [imath]a^2+ab+b^2\equiv 0 \pmod p\Rightarrow a^3\equiv b^3\pmod p\Rightarrow a^{3k}\equiv b^{3k}\pmod p.[/imath] Next, observe that due to FLT we have [imath]a^{3k+1}\equiv b^{3k+1}\pmod p.[/imath] Now if [imath]p\not|a[/imath] and [imath]p\not|b[/imath], then [imath]\gcd(a,p)=\gcd(b,p)=1.[/imath] Therefore we can use can conclude that [imath]a^{\gcd(3k,3k+1)}\equiv b^{\gcd(3k,3k+1)}\pmod p\Rightarrow a\equiv b\pmod p.[/imath] Therefore [imath]a^2+ab+b^2\equiv 0 \pmod p\Rightarrow 3b^2\equiv 0\pmod p \text{ and } 3a^2\equiv 0\pmod p.[/imath] Which implies that [imath]p|3[/imath] which is a contradiction. Hence Proved. I would like to know whether this proof is correct or not. I am unsure about the use of [imath]\gcd[/imath] in the exponent. Moreover, I acknowledge that this question has been asked before, but I've not seen any answer using this fact explicitly. The fact being: Let [imath]\gcd(a,m)=\gcd(b,m)=1[/imath], then if [imath]a^{x}\equiv b^x\pmod m[/imath] and [imath]a^y\equiv b^y\pmod m\Rightarrow a^{\gcd(x,y)}\equiv b^{\gcd(x,y)}\pmod m.[/imath] | 867413 | Let [imath]p[/imath] be a prime of the form [imath]3k+2[/imath] that divides [imath]a^2+ab+b^2[/imath] for some integers [imath]a,b[/imath]. Prove that [imath]a,b[/imath] are both divisible by [imath]p[/imath].
Let [imath]p[/imath] be a prime of the form [imath]3k+2[/imath] that divides [imath]a^2+ab+b^2[/imath] for some integers [imath]a,b[/imath]. Prove that [imath]a,b[/imath] are both divisible by [imath]p[/imath]. My attempt: [imath]p\mid a^2+ab+b^2 \implies p\mid (a-b)(a^2+ab+b^2)\implies p\mid a^3-b^3[/imath] So, we have, [imath]a^{3k}\equiv b^{3k}\mod p[/imath] and by Fermat's Theorem we have, [imath]a^{3k+1}\equiv b^{3k+1}\mod p[/imath] as [imath]p[/imath] is of the form [imath]p=3k+2[/imath]. I do not know what to do next. Please help. Thank you. |
2428775 | Let [imath]G = {5, 15, 25, 35}[/imath]. Prove that [imath]G[/imath] is a group under multiplication modulo [imath]40[/imath]. (Without using Cayley Table)
Let [imath]G = {5, 15, 25, 35}[/imath]. Prove that G is a group under multiplication modulo [imath]40[/imath]. I know how to solve for associativity and closure but I don't know how to find the identity for the set. All the answers that I've seen are using the Cayley Table but I haven't learned that in school yet. Can someone please help me with another way to solve this? Any help will be appreciated. | 693670 | {5,15,25,35} is a group under multiplication mod 40
Show that the set [imath]\{5,15,25,35\}[/imath] is a group under multiplication modulo [imath]40[/imath]. What is the identity element of this group. Can you see any relationship between this group and [imath]\mathbb U(8)[/imath]? I am very stuck on this question and I think my knowledge of abstract algebra is my liability at the moment. Right now I'm most confused by the fact that I learned: An element of a group of this nature must be relatively prime with the working mod. Obviously here this is not the case, so I don't really know where to begin. Do I need a Cayley Table? Thanks. |
783739 | [imath]M = \max (X,Y)[/imath] with [imath]X[/imath], [imath]Y[/imath] independent uniform variables over [imath][0,1][/imath]. Find CDF of [imath]M[/imath] and [imath]E[M][/imath]
Suppose that [imath]X[/imath] and [imath]Y[/imath] are independent uniform [imath][0,1][/imath] random variables. Find the CDF [imath]F(x)[/imath] for the random variable [imath]M = \max(X,Y)[/imath] and find [imath]E[M][/imath]. So [imath]M[/imath] will be uniform over [imath][0,1][/imath] and is the probability of [imath](0 < M < k)[/imath] the probability that [imath](0 < X < k) + (0 < Y < k)[/imath]? I am not even sure if I should assume these variables are discrete or continuous.. Would the cumulative distribution function be: [imath]0[/imath] when [imath]M < 0[/imath] [imath]2M/1[/imath] when [imath]0 < M < 1[/imath] [imath]1[/imath] when [imath]M \geq 1[/imath] and would [imath]E[M][/imath] simply be [imath]E[X] = E[Y] = 1/2[/imath]? I have a feeling something is going over my head here. Is what I am doing correct? Thank you! | 495958 | maximum of two uniform distributions
I have a question. Let's suppose that the two random variables [imath]X1[/imath] and [imath]X2[/imath] follow two Uniform distributions that are independent but have different parameters: [imath]X1 \sim Uniform(l1, u1)[/imath] [imath]X2 \sim Uniform(l2,u2)[/imath] If we define X3 as the maximum of X1, and X2, i.e., [imath]X3 = max(X1, X2)[/imath], what kind of distribution would it be? It is certainly not a uniform and I can calculate the cumulative probability, i.e., [imath]p(X3 <= a)[/imath], because [imath]p(X3 <= a) = p(X1 <= a) p(X2 <= a)[/imath]. But, I wonder if there exists any density function that can model X3. |
2429894 | Can every complex number be expressed as a zero of a power series with rational coefficients?
It is well-known that not all complex numbers can be written as a zero to a polynomial with rational coefficients (the transcendental numbers). However, I am wondering if we "extend" polynomials to allow for infinitely many terms (i.e. the power series expansion of a holomorphic function) while still restricting the coefficients to be rational, if every complex number could then be expressed as the zero of such a function? [imath]\pi[/imath], for instance, is transcendental, but it is a zero of the [imath]\sin[/imath] function, whose power series expansion centered at [imath]z=0[/imath] has only rational coefficients. | 895449 | Ring of rational-coefficient power series defining entire functions
I'm wondering if anyone has come across the following ring before. Let [imath]R[/imath] be the ring of complex power series [imath]f=\sum_{n \ge 0} a_n t^n[/imath] such that [imath]a_n \in \mathbb{Q} \: \: \forall \: n[/imath] The function [imath]f: \mathbb{C} \rightarrow \mathbb{C}[/imath] is entire Any information about it would be welcome but in order to keep the question specific, I'll list a few things I would like to know in particular. Is [imath]R[/imath] a unique factorisation domain? Are there any elements of [imath]R[/imath] which are known to be prime/irreducible but are not associates of polynomials? If an element [imath]f \in R[/imath] is prime/irreducible, must all its zeroes (in [imath]\mathbb{C}[/imath]) be simple? Is every [imath]\alpha \in \mathbb{C}[/imath] a root of some nonzero [imath]f \in R[/imath] ? |
2430007 | Summation of logarithm [imath]n\log(n)[/imath]
I'm solving a big O problem that resulted a part of the algorithm running in [imath]\sum_{i=0}^{n/2} i \log i[/imath] I tried reducing it, using logarithmic identities into [imath]\log\left(\prod_{i=0}^{n/2} i^i\right)[/imath] I was wondering if a) this was correct and b) if it can be simplified any further. | 2355997 | What is sum of [imath]\sum_{x=1}^n x \log (x)[/imath]
What would be the closed form of [imath]\sum_{x=1}^n x \log (x)[/imath]? I know the summation of [imath]\sum \log x[/imath], but how do I use this series to the [imath]\sum_{x=1}^n x \log (x)[/imath]?$ |
2430123 | Atan(6/10)+Atan(1/4)=Pi/4
Came across the identity, interested in the same observation from others and how have others arrive at it: [imath]\arctan(6/10)+\arctan(1/4) = \frac{\pi}4[/imath] thank you in advance. Probably trivial. I took the long way home. | 326334 | A question about the arctangent addition formula.
In the arctangent formula, we have that: [imath]\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)[/imath] however, only for [imath]uv<1[/imath]. My question is: where does this condition come from? The situation is obvious for [imath]uv=1[/imath], but why the inequality? One of the possibilities I considered was as following: the arctangent addition formula is derived from the formula: [imath]\tan\left(\alpha+\beta\right)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}.[/imath] Hence, if we put [imath]u=\tan{\alpha}[/imath] and [imath]v=\tan{\beta}[/imath] (which we do in order to obtain the arctangent addition formula from the one above), the condition that [imath]uv<1[/imath] would mean [imath]\tan\alpha\tan\beta<1[/imath], which, in turn, would imply (thought I am NOT sure about this), that [imath]-\pi/2<\alpha+\beta<\pi/2[/imath], i.e. that we have to stay in the same period of tangent. However, even if the above were true, I still do not see why we have to stay in the same period of tangent for the formula for [imath]\tan(\alpha+\beta)[/imath] to hold. I would be thankful for a thorough explanation. |
2430177 | I have to prove this inequality. I was just looking for a solution to this problem as I have been stuck on it for a very long time.
Show that if [imath]a[/imath] and [imath]h[/imath] are positive numbers with [imath]h<a^2[/imath] then [imath]\sqrt{a^2+h}-a<\frac{h}{2a}<a-\sqrt{a^2-h}[/imath] | 2422113 | Inequality problem involving square roots
Show that, if [imath]a[/imath] and [imath]h[/imath] are positive numbers, [imath]h < a^2[/imath], then [imath]\sqrt{a^2 + h}-a < \frac{h}{2a} < a - \sqrt{a^2 - h}[/imath] I've been working on this problem for about 2 hours now, but I've made no progress. I'm not looking for an answer, but I just need some help to get me started since we didn't practice inequalities this complex in highschool. Thanks. All I can tell is that we're supposed to take the square root of an expression at some point since one inequality ( h < a^2 ) becomes two. Edit : Thank you guys for the replies, but I'd appreciate only hints in the future (like Robert Israel) so that I can learn. Regardless, I found a different way to do it, so it's cool :) |
2428012 | Showing a function is continuous and analytic
This is an exercise from Conway that I am stuck at. I think the function [imath]φ[/imath] is clearly continuous for points [imath](z, w)[/imath] such that [imath]z[/imath] and [imath]w[/imath] are not equal. However I can't show that [imath]φ[/imath] converges to the derivative of [imath]f[/imath] at z as [imath](z, w)[/imath] goes to points where [imath]z[/imath] and [imath]w[/imath] are equal... Also it is clear that for each fixed [imath]w[/imath], [imath]f[/imath] is analytic for points [imath]z[/imath] not equal to [imath]w[/imath]. But how can I show that [imath]f[/imath] is analytic at [imath]z=w[/imath]? Could anyone please help me with this problem? | 1452848 | Prove the function is continuous, exercise from Conway's "Functions of One Complex Variable I"
For the first proof of Cauchy's integral formula, Conway in his book "Functions of One Complex Variable" (Chapter IV, section 5.4) uses the following claim: Let [imath]G[/imath] be an open subset of [imath]\mathbb C[/imath] and [imath]f : G \to \mathbb C[/imath] an analytic function. Then define [imath]\varphi : G \times G \to \mathbb C[/imath] as [imath] \varphi(z,w) = \begin{cases} \dfrac{f(z)-f(w)}{z-w}, && z\neq w \\ f'(z) && z=w \end{cases}. [/imath] Then [imath]\varphi[/imath] is continuous on [imath]G \times G[/imath]. Unfortunately, he does not give a proof, leaving it as an exercise (Exercise 1). I tried to make a direct proof by taking [imath]|z-z_0| + |w-w_0| < \delta[/imath] as a neighbourhood of [imath](z_0,w_0)[/imath], but it lead to nowhere. I understand, that [imath]z \mapsto \varphi(z,w)[/imath] and [imath]w \mapsto \varphi(z,w)[/imath] are both holomorphic and continuous, but I do not see how one could use it. I also saw this claim in several lecture notes but it goes unproven there either. I would appreciate a hint, we have this problem as a homework. |
2422499 | How to define the exponential function without calculus?
For fun, I would like to define the complex exponential function from these two properties: [imath]\exp(0) = 1[/imath] [imath]\exp(z + w) = \exp(z) \exp(w)[/imath] From here, I would like to find a way to compute values of [imath]\exp(z)[/imath], or at least to compute [imath]\exp(1)[/imath]. So far, I found only two ways: Noting that [imath]\exp'(z) = \exp(z)[/imath] and solving the differential equation, which leads to [imath]\int \frac{\exp'(z)}{\exp(z)} dz = \log(\exp(z)) + C = z[/imath]. Noting that [imath]\exp'(z) = \exp(z)[/imath], computing its Taylor series and checking that what I get is an entire function. The first approach is simply wrong because it involves logarithms, which I have not defined yet. The second approach looks much better. I haven't tried, but I guess I can find a way to manipulate the Taylor series to obtain the limit definition of [imath]e[/imath] and conclude that [imath]\exp(1) = e[/imath], which is my aim. However, I'm struggling to find another way that does not involve differentiation or limits in general. I would be happy to find a way to say [imath]\exp(1) = e[/imath] without calculus. I think that the irrational nature of [imath]e[/imath] forces me to use limits -- am I right? | 833441 | An "elementary" approach to complex exponents?
Is there any way to extend the elementary definition of powers to the case of complex numbers? By "elementary" I am referring to the definition based on [imath]a^n=\underbrace{a\cdot a\cdots a}_{n\;\text{factors}}.[/imath] (Meaning I am not interested in the power series or "compound interest" definitions.) This is extended to negative numbers, fractions, and finally irrationals by letting [imath]a^r=\lim_{n\to\infty} a^{r_n}[/imath] where [imath]r_n[/imath] is rational and approaches [imath]r[/imath]. For a concrete example, how would we interpret [imath]e^i[/imath] in terms of these ideas? |
2430565 | Simple combinatorics formula
i am having some trouble trying to understand this simple formula: [imath]\frac{n!}{\left(n-r!\right)r!}[/imath] Even though i am able to solve exercises easily using this formula i can't understand it. How was this formula derived? Is it possible to understand the formula or is it just the result of some certain mathematical work? The part that i don't understant is how does diving the possible combinations by the redundancies result in the possible number of arragements of something ignoring order. | 1749787 | What is an intuitive explanation of the combinations formula?
I perfectly understand the permutations formula i.e. if you have [imath]n[/imath] things how many ways can you rearrange it if taken [imath]k[/imath] at a time (or if you have [imath]k[/imath] slots)? So you draw the following tree. And the formula comes out naturally. [imath]^nP_k = \frac{n!}{(n-k)!}[/imath] I get that it is including all of the duplicates, too. And to get rid of them we use the combinations formula: [imath]^nC_k = \frac{n!}{k!(n-k)!}[/imath] Can somebody explain why we have to divide by the "number of ways to re-arrange the slots" to get rid of the duplicates? I understand that it works and the division is used to scale a number down but I am not getting that "aha moment" of why we do it. Can somebody explain it? |
894810 | How to understand the combination formula?
I'm reading an algorithm book. In which it mentioned: Imagine (once again) you have n people lined up to see a movie, but there are only k places left in the theater. How many possible subsets of size k could possibly get in? That’s exactly [imath]C(n,k)[/imath], of course, and the metaphor may do some work for us here. We already know that we have [imath]n![/imath] possible orderings of the entire line. What if we just count all these possibilities and let in the first [imath]k[/imath]? The only problem then is that we’ve counted the subsets too many times. A certain group of k friends could stand at the head of the line in a lot of the permutations; in fact, we could allow these friends to stand in any of their [imath]k![/imath] possible permutations, and the remainder of the line could stand in any of their [imath](n–k)![/imath] possible permutations without affecting who’s getting in. Aaaand this gives us the answer! [imath]C(n,k)= \frac{n!}{k!(n −k)!}[/imath] This formula just counts all possible permutations of the line (n!), and divides by the number of times we count each “winning subset,” as explained. I'm not quite understand why the reminder can in any [imath](n-k)![/imath] permutation, if he stands at the end of the queue, won't he let all these people in? And why I should use [imath]n![/imath] to divide by [imath]k!(n-k)![/imath] ? Didn't see that consequence from the metaphor. If you have better ideas to explain this formula, you can use them of course. | 1594622 | The logic behind combinations
As an example, calculate the number of [imath]5[/imath] card hands possible from a standard [imath]52[/imath] card deck. Using the combinations formula, [imath]= \frac{n!}{r!(n-r)!}[/imath] [imath]= \frac{52!}{5!(52-5)!}[/imath] [imath]= \frac{52!}{5!47!}[/imath] [imath]= 2,598,960\text{ combinations}[/imath] I was wondering what the logic is behind combinations? Is it because there are 52 cards to choose from, except we're only selecting [imath]5[/imath] of them, to which the person holding them can rearrange however they please, hence we divide by [imath]5![/imath] to account for the permutations of those 5 cards? Then do we divide by [imath]47![/imath] because the remaining cards are irrelevant? |
2431030 | If [imath]c^2 = 12[/imath], then show that [imath]c[/imath] is an irrational number.
Show that, if [imath]x[/imath] satisfies [imath]x^2 = 3[/imath], then [imath]c := 2x[/imath] satisfies [imath]c^2 = 12[/imath]. Using this fact, show that [imath]c[/imath] is an irrational number. | 1607799 | Why is [imath]\sqrt {12} = 2 \sqrt 3[/imath]?
Why [imath]\sqrt {12} = 2 \sqrt 3[/imath]? It is obvious? If we considered the function [imath]f(s) = s^2 [/imath] it is injective on positive numbers so we obtain the conclusion. But in the same time it is an equality between irrational numbers. Suppose that we know just to compute the square roots. |
2431250 | [imath]x^p - t[/imath] has no roots in [imath]\mathbb{F}_p(t)[x][/imath]
I am actually trying to prove that [imath]x^p - t[/imath] is irreducible in [imath]\mathbb{F}_p(t)[x][/imath]. My approach is to consider a splitting field [imath]E[/imath] so that: [imath]x^p - t = (x-\alpha_1)\dots(x-\alpha_p)=(x-\alpha)^p \ \ \ \ \text{for some $\alpha$ in $E$}[/imath] Then assuming by contradiction that [imath]x^p - t = f(x)g(x)[/imath], then [imath]f(x) =(x-\alpha)^r[/imath] with [imath]r<p[/imath]. This implies that [imath]\alpha \in \mathbb{F}_p(t)[/imath]. Should this be a contradiction right? Why can't [imath]\alpha[/imath] be here? | 1304261 | Show [imath]x^p-t[/imath] has no root in the field [imath]\mathbb{F}_p(t)[/imath]
I don't think I fully understand. Let's say there is a root [imath]x_0 \in K=\mathbb{F}_p(t)[/imath], where [imath]p[/imath] is a prime number. Then [imath]x_0 = \frac{P(t)}{Q(t)}[/imath] for some polynomials [imath]P,Q \in \mathbb{F}_p[t][/imath]. We can assume [imath]\gcd(P,Q)=1[/imath] and [imath]x_0^p-t= \frac{(P(t))^p}{(Q(t))^p}-t= \frac{P(t)^p-tQ(t)^p}{Q(t)^p} =0[/imath], so coefficients of [imath]P(t)^p, tQ(t)^p[/imath] must be identical, which contradicts [imath]\gcd(P,Q)=1[/imath], hence such [imath]x_0[/imath] does not exist and the polynomial has no root in [imath]K[/imath]. am I about right? anyway, I would appreciate an explanation about [imath]\mathbb{F}_p(t)[/imath], what is [imath]t[/imath]? what is the meaning of a variable which does not belong to any specific "world"? I cannot use [imath]t[/imath] as if it was a member of [imath]\mathbb{F}_p[/imath] and hence cannot assume [imath]t^{p-1} = 1 \pmod p[/imath]... |
1752068 | A finite difference
I read that the [imath]n[/imath]-th finite difference of the sequence [imath]1^n, 2^n, 3^n,\dots[/imath] is [imath]n![/imath], but I'm not able to prove this. Could someone give an idea of why this is true? | 2093882 | Problem in solving a question related to finite difference.
Show that the [imath]k[/imath]-th finite differences of the sequence [imath]1^k ,2^k,3^k ,...[/imath] are each [imath]k![/imath]. I have tried but I fail when I try it proving using mathematical induction.Please help me. Thank you in advance. |
615968 | If [imath]x,y,z\in\mathbb R\setminus \{1\}[/imath] and [imath]xyz=1[/imath], prove that [imath]\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1[/imath].
If [imath]x,y,z\in\mathbb R\setminus \{1\}[/imath] and [imath]xyz=1[/imath], prove that [imath]\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1[/imath] Without using calculus. There are a few ways I've tried solving this: [imath]1)[/imath] We could try using the Cauchy-Schwarz inequality: [imath]\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge \frac{(x+y+z)^2}{(x-1)^2+(y-1)^2+(z-1)^2}[/imath] But it's apparent that nothing's useful here. [imath]2)[/imath] We could use AM-GM as well: [imath]\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 3\sqrt[3]{\frac{x^2y^2z^2}{(x-1)^2(y-1)^2(z-1)^2}}=\frac{3}{\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}}[/imath] So we only have to prove that: [imath]\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}\le 3[/imath] We could raise both sides to the power of 3: [imath](x-1)^2(y-1)^2(z-1)^2\le 27[/imath] But this inequality doesn't hold. [imath]3)[/imath] We could try cleaning the denominators by multiplying both sides by [imath](x-1)^2(y-1)^2(z-1)^2[/imath]. After a bunch of expanding and simplifying we get that: [imath]x^2y^2+y^2z^2+x^2z^2-6(xy+yz+xz)+2(x+y+z)+9\ge 0[/imath] I can't tell so easily whether the inequality is true or not. You could help me out on this one. Just don't forget that [imath]x,y,z\in\mathbb R\setminus\{1\}[/imath] and so we can't just simply use AM-GM, unless we're using it for squares that have to be non-negative. | 2592422 | Show that [imath]\frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2}\ge 1[/imath] for real numbers [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath] with [imath]xyz=1[/imath].
I've been thinking to prove each one individually then sum them up which seems too complex to because I haven't made any notable success proving [imath]\frac{x^2}{(x-1)^2}> 1[/imath]. I know that [imath](x-1)^2\ge 0[/imath] , but I'm not able to transfer it to what I'm looking to prove. Is there is any techniques to look for when solving these kind of inequalities because I'm totally stuck here. |
2431747 | How to show that [imath]f(A)[/imath] is an open interval.
I was reading Hanhn-Banach separation theorem and it is written that " Since [imath]A[/imath] is convex ,open subset of a normed [imath]\mathbb K[/imath]-linear space and f is [imath]\mathbb R[/imath]-linear and continuous so [imath]f(A)[/imath] is an open interval." Since A is convex f(A) becomes an interval but why open? Please someone help. Thank you.. | 1811328 | If [imath]\varphi \in E'[/imath] and [imath]A[/imath] is convex and open then [imath]\varphi (A)[/imath] is an open interval
Let [imath]E[/imath] be a real normed space and [imath]\varphi \in E'[/imath], [imath]\varphi \neq 0[/imath]. Suppose that [imath]A \subset E[/imath] is an open convex not empty subset. Show that [imath]\varphi(A)[/imath] is an open interval. Since [imath]A[/imath] is connected and [imath]\varphi[/imath] is continuous, [imath]\varphi(A)[/imath] is an interval. How could I show that [imath]\varphi(A)[/imath] is open? I tried to apply the geometric form of Hahn Banach theorem but it didn't work. I appreciate if you could give some hints. Thanks. |
2430075 | Combinatorial proof of [imath]\binom{k}{i}\binom{n}{k}=\binom{n}{i}\binom{n-i}{k-i}[/imath]
[imath]\binom{k}{i}\binom{n}{k}=\binom{n}{i}\binom{n-i}{k-i}[/imath] This identity could be easily shown using algebraic formula of combination. However, I would like to provide a combinatorial proof. I considered applying Pascal's equality, but got stuck. Any advice ? | 2391454 | Combinatorial Proof of [imath]{{n}\choose{k}} {{k}\choose{j}} = {{n}\choose{j}} {{n-j}\choose{k-j}}[/imath]
How do I prove by a combinatorial argument that [imath]{{n}\choose{k}} {{k}\choose{j}} = {{n}\choose{j}} {{n-j}\choose{k-j}}[/imath] |
2431844 | How to show that [imath]e[/imath] is irrational by studying [imath]\sum^{n}_{k=1}\frac{1}{k!}[/imath] and [imath]\sum^{n}_{k=0}\frac{1}{k} + \frac{1}{n \cdot n!}[/imath]?
Suppose you have [imath]a_n = \sum^{n}_{k=0}\frac{1}{k!}[/imath] and [imath]b_n = a_n + \frac{1}{n \cdot n!}. [/imath] By using the fact that [imath]a_n <e <b_n[/imath], which is true [imath]\forall n \in \mathbb{N}[/imath], conclude that [imath]e \not \in \mathbb{Q}[/imath]. And I am unable to do anything. I tried to get something absurd by supposing that [imath]e = \frac{a}{b}[/imath] with [imath]a,b \in \mathbb{Z}[/imath], but didn't get anywhere. A hint would be highly appreciated. | 1483449 | proving "e" irrational using convergent series
So my math professor in college gave us this problem: [imath]U_n=\sum_{k=1}^{k=n}\frac{1}{k!}[/imath] [imath]V_n=U_n+ \frac{1}{n\cdot n!}[/imath] The problem has two questions: 1- Prove that the two series are adjacent (easy) 2- Let [imath]e[/imath] be their limit, prove that [imath]e[/imath] is irrational. I have no idea how to start the second part, since I don't have a professor that does the problems the first professor gives us, so I have a lack in reasoning skills, I tried proof by contradiction, but no result. also I'm new to the site so I don't know how to use the symbols so sorry for that. |
2432004 | [imath]\det(A+I)=1+\text{tr}(A)[/imath], if [imath]\text{rank}(A)=1[/imath] proof?
This was the starred problem in our linear algebra class. So I understand that if a [imath]k[/imath]-by-[imath]k[/imath] matrix has rank [imath]1[/imath], it means that all rows are linearly dependant on lets say the first one. So if I have a matrix, I can write it as: [imath] \begin{pmatrix} a_1 & c_1*a_1 & c_2*a_1 & \cdots& c_k*a_1 \\ a_2 & c_1*a_2 & c_2*a_2 & \cdots & c_k*a_2 \\ a_3 & c_1*a_3 & c_2*a_3 & \cdots & c_k*a_3 \\ \vdots& \vdots& \vdots& \ddots & \vdots\\ a_k & c_1*a_k & c_2*a_k & \cdots& c_k*a_k \\ \end{pmatrix} [/imath] So if I write A+I: [imath] \begin{pmatrix} a_1+1 & c_1*a_1 & c_2*a_1 & \cdots& c_k*a_1 \\ a_2 & c_1*a_2+1 & c_2*a_2 & \cdots & c_k*a_2 \\ a_3 & c_1*a_3 & c_2*a_3+1 & \cdots & c_k*a_3 \\ \vdots& \vdots& \vdots& \ddots & \vdots\\ a_k & c_1*a_k & c_2*a_k & \cdots& c_k*a_k+1 \\ \end{pmatrix} [/imath] And [imath]\det(A+I)=a_1+1[/imath], since I can subtract [imath]c_n[/imath] ([imath]n=1,2,3,\dots k[/imath]) times the first column from all the other columns and I am left with a upper triangular matrix. So I would get [imath]a_1+1=1+\text{tr}(A)[/imath] and [imath]a_1=\text{tr}(A)[/imath], but this can't be right? I can't figure out where I am going wrong. Any help? Of course I read the other one, but fine, no help from here. | 429055 | [imath]\mbox{rank}(A)=1[/imath] implies [imath]\det(A+I)=\mbox{trace}(A)+1[/imath]
This problem comes from exercise from my lecturer Let [imath]A[/imath] be a square matrix such that [imath]\mbox{rank}(A)=1[/imath]. Prove that [imath]\det(A+I)=\mbox{trace}(A)+1[/imath]. ([imath]I[/imath] is the identy matrix) I usually haven't any idea in solving linear algebra problem. |
2431754 | Compute: [imath]\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.[/imath]
Compute: [imath]\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.[/imath] I want to compute this sum by computing one term at a time. It's clear that [imath]\arctan{\frac{1}{7}}=A \Longleftrightarrow\tan{A}=\frac{1}{7}\Longrightarrow A\in\left(0,\frac{\pi}{2}\right).[/imath] Drawing a right triangle with sides [imath]1[/imath], [imath]7[/imath] and [imath]5\sqrt{2},[/imath] I get that [imath]\begin{array}{lcl} \sin{A} & = & \frac{1}{5\sqrt{2}}\Longleftrightarrow A= \arcsin{\frac{1}{5\sqrt{2}}} \\ \cos{A} & = & \frac{7}{5\sqrt{2}}\Longleftrightarrow A= \arccos{\frac{7}{5\sqrt{2}}} \\ \end{array}[/imath] But this will not get me standard angles for [imath]A.[/imath] | 1777652 | Show that, [imath]2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}[/imath]
Show that, [imath]2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}[/imath] There is a mixed of sin and tan, how can I simplify this to [imath]\frac{\pi}{4}[/imath] We know the identity of [imath]\arctan\left(\frac{1}{a}\right)+\arctan\left(\frac{1}{b}\right)=\arctan\left(\frac{a+b}{ab-1}\right)[/imath] |
2432249 | How would one calculate [imath]\lim_{n\to\infty}n((1+\frac{1}{n})^n-e)[/imath]?
What I have thought about this is: we may use L'Hopstal's rule to calculate [imath]\lim_{n\to\infty}\frac{((1+\frac{1}{n})^n-e)}{\frac{1}{n}}[/imath], both the numerator and denominator goes to 0 as n goes to infinity. But calculating the derivative of [imath](1+1/n)^n[/imath] seems to be very complicated. Using Taylor series to calculate the dominant terms of [imath](1+1/n)^n[/imath], but I'm not really sure if it makes sense to let [imath]"n=\infty"[/imath]. Equivalently maybe we can expand [imath](1+x)^{1/x}[/imath] at [imath]x=0[/imath], but it's not defined. Maybe I wasn't on the right track. Even if the solution uses a different approach, I would still love to know how to expand [imath](1+x)^{1/x}[/imath]. Thanks for any suggestions. | 1374577 | Find [imath]\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e][/imath]
[imath]\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e][/imath] I let, [imath]x = \frac{1}{n}[/imath], then as [imath]\lim_{x \to 0} \frac{1}{x}[(1+x)^\frac{1}{x} - e] = \infty[/imath] L'hopital's: [imath]\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{1} = \frac{1}{0}(1+0)^{\frac{1}{0}-1} = \infty[/imath] Again, if we apply L'hopital's: [imath]\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{x+(1+x)}[/imath]. This is also going to [imath]\infty[/imath] as [imath]x \to 0[/imath]. But, I know the answer is [imath]\frac{-e}{2}[/imath], and I am not even close. Can anyone please find me the mistakes here. Oh, I am supposed to use L'Hopital's rule. |
2432079 | Determinant of a matrix that is its own inverse
I need help in showing that when computing the determinant the inverse of an [imath]n \times n[/imath] matrix with the property [imath]M=M^{-1}[/imath] that is [imath]M^2 = I[/imath] the determinant is either [imath]1[/imath] or [imath]-1[/imath]. I've tried showing it in a couple ways and the way I'm trying to show it has me stuck [imath]K^2 = I[/imath] [imath]K^2 - I = 0[/imath] [imath]\det(K^2 - I) = 0[/imath] [imath]\det(I - I) = 0[/imath] I get here and I am hopelessly stuck. Could I go on to prove it this way? Is there any elementary way to prove this? | 380684 | Proving an invertible matrix which is its own inverse has determinant [imath]1[/imath] or [imath]-1[/imath]
Let A be an invertible [imath]n \times n[/imath] matrix whose inverse is itself. Prove that [imath]\det(A)[/imath] is either [imath]1[/imath] or [imath]-1[/imath]. I'm really lost in class. I don't even know where to start. Please help. |
2430338 | Asymptotic probability of a tie in a dice-throwing competition
We throw a dice [imath]n[/imath] times. Each time it falls on 1 I win, each time it falls on 6 you win; otherwise nobody wins. What is the probability that both of us win the same number of times, as a function of [imath]n[/imath], when [imath]n[/imath] is very large? I tried to solve the question using the multidimensional central-limit-theorem. Suppose we throw a single dice. Let [imath]X[/imath] be a random vector of length 6, such that [imath]X_i=1[/imath] if the dice lands on [imath]i[/imath] and 0 otherwise. This is a random vector with (for [imath]p=1/6[/imath]): [imath] \mu = [p,p,p,p,p,p] [/imath] [imath] \Sigma = \begin{pmatrix} p-p^2 & -p^2 & -p^2 & -p^2 & -p^2 & -p^2 \\ -p^2 & p-p^2 & -p^2 & -p^2 & -p^2 & -p^2 \\ -p^2 & -p^2 & p-p^2 & -p^2 & -p^2 & -p^2 \\ -p^2 & -p^2 & -p^2 & p-p^2 & -p^2 & -p^2 \\ -p^2 & -p^2 & -p^2 & -p^2 & p-p^2 & -p^2 \\ -p^2 & -p^2 & -p^2 & -p^2 & -p^2 & p-p^2 \end{pmatrix} [/imath] Let [imath]S[/imath] be a vector that is the sum of [imath]n[/imath] such vectors. Then, for every [imath]i[/imath] between 1 and 6, [imath]S_i[/imath] is the number of times the dice falls on [imath]i[/imath]. When [imath]n[/imath] is sufficiently large, the vector [imath]S[/imath] is distributed like a normal random vector, with mean vector [imath]n\mu[/imath] and covariance matrix [imath]n\Sigma[/imath]. Now, the probability that [imath]S_1=S_6[/imath] is approximately this integral: [imath] \int_{S_1=S_6} f(S) [/imath] where [imath]f[/imath] is the probability-density-function of the normal distribution with mean [imath]n\mu[/imath] and covariance [imath]n\Sigma[/imath]: [imath] f(S) = { \exp\left(-{1\over 2} (S-n\mu)^T(n\Sigma)^{-1}(S-n\mu)\right) \over \sqrt{(2\pi)^6 |n\Sigma|} } [/imath] I calculated [imath]|\Sigma|[/imath] and it is [imath]p^6(1-6 p) = 0[/imath], so [imath]\Sigma[/imath] does not have an inverse. How can I solve the integral then? I do not need an exact solution - just an asymptotic solution for large [imath]n[/imath]. | 734267 | Probability a random walk is back at the origin
I have a symmetric random walk that starts at the origin. With probability [imath]1/6[/imath] it goes right by one and with probability [imath]1/6[/imath] it goes left by one. With probability [imath]4/6[/imath] it stays put. After [imath]n[/imath] time steps, what is the probability that it is at the origin? The answer should be the probability that you go left by the same amount you go right. I am having difficulty working this out however. |
2431367 | Volume obtained by rotating a region
For my calculus class, one of the review questions I'm given is Find the volume V by rotating the region bounded by [imath]y = 5x-x^2[/imath] and [imath]y = x^2 - 5x + 8[/imath] about the [imath]y[/imath]-axis. I've never learned this concept before in high school or previous mathematics courses, can someone explain how to do this? Would appreciate any help! | 2425945 | Find the volume of rotation about the y-axis for the region bounded by [imath]y=5x-x^2[/imath], and [imath]x^2-5x+8[/imath]
Find the volume of rotation about the y-axis for the region bounded by [imath]y=5x-x^2[/imath], and [imath]x^2-5x+8[/imath] Here is an image: Normally I can do this question, but this one is tricky because since we are rotating about the y-axis, and we are quadratic, when I solve for [imath]x[/imath] I get two answers, one positive and one negative. What I mean is that if you look at the graph [imath]5x-x^2[/imath], then there are two halves, and if you look at the green line, that's one half of the curve (positive one), and the red half is the negative one. Since we can see both the negative and positive sections in this graph, and both are being rotated, I wasn't sure how to create the equation that I can later integrate. Usually in the problems I do, the negative doesn't exist, and so it's trivial. In this case I am not sure. |
2432272 | How to complete basis?
Two vectors [imath]a = (1,-2,2,-3)[/imath] and [imath]b=(2,-3,2,4)[/imath] I need to complete the basis for [imath]\Bbb{R}^4[/imath] I know that there's a way of solving 2-equations linear system like [imath]\begin{pmatrix}a \\ b\ \end{pmatrix}\cdot x = 0[/imath] to get vector [imath]c[/imath], and then solving 3-equations linear system to get vector [imath]d[/imath] But are not there other ways? I heard that gram schmid proccess could be helpful, but I am not sure how to apply it here as well ass if this idea is correct at all. | 465870 | how to extend a basis
This is a very elementary question but I can't find the answer in my book at the moment. If I have, for example, two vectors [imath]v_1[/imath] and [imath]v_2[/imath] in [imath]\mathbb R^5[/imath] and knowing that they are linear independent , how can I extend this two vectors to a basis of [imath]\mathbb R^5[/imath]. I know that I just have to add [imath]v_i[/imath] for [imath]3\leq i \leq 5[/imath] such that they are linear independent but how can I do that? Is there an easy algorithm? |
2432634 | Sum of the series up to [imath]n[/imath] terms
Which value does this expression converge to? [imath]\frac{1^2}{2^1}+\frac{2^2}{2^2}+\frac{3^2}{2^3}+\frac{4^2}{2^4}+\frac{5^2}{2^5}+...+\frac{n^2}{2^n}[/imath] The expression above can also be written as [imath]\sum_{n=1}^\infty \frac{n^2}{2^n}[/imath]. | 1685851 | Convergence of [imath]\sum_{n=1}^\infty \frac{n^2}{2^n}[/imath]
I have to decide whether the Series [imath]\sum_{n=1}^\infty \frac{n^2}{2^n}[/imath] Converge or Diverge only by using the Comparison test or by [imath]\frac{1}{n^p}[/imath] test. What I noticed is that it converges by the De'lamber test ([imath]\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|[/imath]). But I'm having trouble finding a way proving it with the provided tests. Would appreciate your advice |
2432857 | Find all positive integer [imath]n[/imath] with the property that there is a partition of the set [imath]({n, n + 1, n + 2, n + 3, n + 4, n + 5})[/imath]
find all positive integers [imath]n[/imath] with the property: there is a partition of the set [imath]({n,n+1,n+2,n+3,n+4,n+5})[/imath] into two disjoint subsets [imath]A[/imath] and [imath]B[/imath] such that the product of element in [imath]A[/imath] is equal to the one in [imath]B[/imath] How do I approach the problem quoted above? I tried using parity arguments but I found nothing. Any help | 1948239 | Find the set of [imath]n\in\Bbb Z^+[/imath] with [imath]M=\{n,n+1,n+2,n+3,n+4,n+5\}[/imath] partitionable into two sets
Find the set of all positive integers [imath]n[/imath] with the property that the set [imath]M=\{n, n + 1,n + 2,n + 3,n + 4,n + 5\}[/imath] can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set. If [imath]n=1[/imath] them [imath]M=\{1,2,3,4,5,6\}[/imath] and there is no such partition, so [imath]n \ge 2[/imath]. If the prime [imath]p|n[/imath] then either [imath]p|2[/imath] or [imath]p|3[/imath] or [imath]p|5[/imath] which means either [imath]p=2[/imath] or [imath]p=3[/imath] or [imath]p=5[/imath]. Suppose [imath]n=2k[/imath]. Then [imath]M=\{2k, 2k + 1,2k + 2,2k + 3,2k + 4,2k + 5\}[/imath]. I have no idea how to proceed. |
2431928 | [imath]a,b,c[/imath] are positive reals and distinct with [imath]a^2+b^2 -ab=c^2[/imath]. Prove [imath](a-c)(b-c)<0[/imath]
[imath]a,b,c[/imath] are positive reals and distinct with [imath]a^2+b^2 -ab=c^2[/imath]. Show that [imath](a-c)(b-c)<0[/imath]. This is a question presented in the "Olimpiadas do Ceará 1987" a math contest held in Brazil. Sorry if this a duplicate. Given the assumptions, it is easy to show that [imath]0<(a-b)^2<a^2+b^2-ab=c^2.[/imath] But could not find a promising route to pursue. Any hint or answer is welcomed. | 2077319 | If [imath]a,b,c[/imath] are positive integers, with [imath]a^2+b^2-ab=c^2[/imath] prove that [imath](a-b)(b-c)\le0[/imath].
I have an inequality problem which is as follow: If [imath]a,b,c[/imath] are positive integers, with [imath]a^2+b^2-ab=c^2[/imath] prove that [imath](a-b)(b-c)\le0[/imath]. I am not so good in inequalities. So, please give me some hints so that I can proceed. Thanks. |
2432788 | Is [imath]Y\cap(X-C) = Y\cap{X} - Y\cap{C}[/imath]?
Is [imath]Y\cap({X-C}) = (Y\cap X) - (Y\cap C)[/imath]? I tries using the definitions of intersections and difference of sets but am not able to prove it. | 172033 | Proving [imath]A\cap(B-C)[/imath] is equal to [imath](A\cap B)-(A\cap C)[/imath]
Prove that: [imath]A\cap (B-C) = (A\cap B)-(A\cap C)[/imath]. Tried to prove this by using Algebra of Classes. Then used idempotent property in [imath]A[/imath] as well as double negation in [imath]A[/imath] but still it didn't work. |
2433341 | Show combinatorially that [imath]S(n+1, k) = \binom n0 S(0,k-1) + \binom n1 S(1,k-1) + \cdots + \binom nn S(n,k-1)[/imath]
I'm trying to show combinatorially that [imath]S(n+1, k) = \binom n0 S(0,k-1) + \binom n1 S(1,k-1) + \cdots + \binom nn S(n,k-1),[/imath] where [imath]S(n,k)[/imath] represents the Stirling number of the second kind. I've received the hint that I should first find a combinatorial argument to show that [imath]\binom {n+1}k = \binom 0{k-1} + \binom 1{k-1} + \cdots + \binom n{k-1},[/imath] but I can only find an algebraic proof using several iterations of [imath]\binom {n+1}k = \binom nk + \binom n{k-1}.[/imath] Could someone please point me in the direction for how I should be thinking about these problems? | 2162787 | Proof of recursion formula for Stirling number of the second kind
Wikipedia has the identity [imath]S(n+1, k+1) = \sum_i {n \choose i} S(i, k)\;.[/imath] I can't seem to find a proof for this and was wondering what it entails. Is it a purely combinatoric argument, or an algebraic manipulation of the explicit formula [imath]S(n, k) = \frac{1}{k!} \sum_{i=0}^k (-1)^i {k \choose i}(k-i)^n [/imath] with the known recurrence formula [imath]S(n+1, k+1) = S(n, k) + (k+1)S(n, k+1)? [/imath] |
2433219 | Non-Separability of Metric Space implies non equivalence of Borel [imath]\sigma[/imath]-algebras on product space.
Let [imath](X_1,d_1)[/imath] and [imath](X_2,d_2)[/imath] be metric spaces. Take [imath]\mathcal{B}(X_1)\times \mathcal{B}(X_2)=\sigma(\{A_1\times A_2 | A_1\in \mathcal{B}(X_1), A_2\in \mathcal{B}(X_2) \})[/imath] Where [imath]\sigma(A)[/imath] denotes the smallest [imath]\sigma[/imath]-algebra generated by [imath]A[/imath], and [imath]\mathcal{B}[/imath] denotes Borel [imath]\sigma[/imath]-algebra. Define the following metric on the product space: [imath]d((x_1,x_2),(y_1,y_2))=\sqrt{d_1(x_1,y_1)^2+d_2(x_2,y_2)^2}[/imath] Let [imath]X=X_1\times X_2[/imath], and [imath]\mathcal{B}(X)=\sigma(\{ \textrm{All open sets in } (X,d)\})[/imath] If [imath](X_1,d_1)[/imath] and [imath](X_2,d_2)[/imath] are separable metric spaces then [imath]\mathcal{B}(X)=\mathcal{B}(X_1)\times \mathcal{B}(X_2)[/imath]. Is it possible to show by counter-example that separability is necessary in that assertion? | 1826264 | Reference request: product Borel [imath]\sigma[/imath]-algebra of non-separable metric spaces
The following is a proposition in Folland's Real Analysis about product sigma algebra: Here [imath]\mathcal{B}_X[/imath] denotes the Borel [imath]\sigma[/imath]-algebra on [imath]X[/imath]. Could anyone come up with an example that [imath] \mathcal{B}_X\setminus\otimes_{1}^n\mathcal{B}_{X_j} [/imath] is nonempty? |
2433409 | Show that a finite group of even order that has a cyclic Sylow [imath]2-[/imath] subgroup is not simple.
Show that a finite group of even order that has a cyclic Sylow [imath]2-[/imath] subgroup is not simple. Attempt: Since [imath]G[/imath] has a Sylow [imath]2[/imath] subgroup so [imath]|G|=2^km[/imath] . I don't know how to use the fact that the Sylow [imath]2-[/imath] subgroup is cyclic. Please give some hints. | 919119 | A Group Having a Cyclic Sylow 2-Subgroup Has a Normal Subgroup.
I want to prove the following: Let [imath]G[/imath] be a group of order [imath]2^nm[/imath], where [imath]m[/imath] is odd, having a cyclic Sylow [imath]2[/imath]-subgroup. Then [imath]G[/imath] has a normal subgroup of order [imath]m[/imath]. ATTEMPT: We will show that [imath]G[/imath] has a subgroup of order [imath]m[/imath]. Let [imath]\theta:G\to \text{Sym}(G)[/imath] be the homomorphism which is defined as [imath] \theta(g)=t_g, \quad \forall g\in G [/imath] where [imath]t_g:G\to G[/imath] is defined as: [imath] t_g(x)=gx, \quad \forall x\in G. [/imath] Let [imath]g[/imath] be an element of order [imath]2^n[/imath] in [imath]G[/imath]. (There exists such an element since [imath]G[/imath] has a cyclic Sylow [imath]2[/imath]-subgroup.) The cyclce representation of [imath]t_g[/imath] is a product of [imath]m[/imath] disjoint cycles each of length [imath]2^n[/imath]. Therefore [imath]t_g[/imath] is an odd permutation. Thus the homomorphism [imath]\epsilon\circ \theta:G\to\{\pm 1\}[/imath], where [imath]\epsilon:\text{Sym}\to\{\pm 1\}[/imath] is the sign homomorphism, is a surjection. By the First Isomprphism Theorem, we conclude that the kernel [imath]K[/imath] of [imath]\epsilon\circ \theta[/imath] is of order [imath]2^{n-1}m[/imath]. If [imath]n[/imath] were equal to [imath]1[/imath] then we are done. If [imath]n>1[/imath] then note that any Sylow [imath]2[/imath]-subgroup of [imath]K[/imath] is also cyclic. This is because each Sylow [imath]2[/imath]-subgroup of [imath]K[/imath] is contained in a Sylow [imath]2[/imath]-subgroup of [imath]G[/imath], where the latter is cyclic. Now we can inductively show that [imath]K[/imath] has a subgroup of order [imath]m[/imath]. What I am struggling with is showing the normality. Can anybody please help me with this. Thanks. |
2433718 | Let [imath]a, b, c[/imath] be positive real numbers such that [imath]a^3 + b^3 = c^3[/imath]. Prove that [imath]a^2 + b^2 - c^2 \gt 6(c - a)(c - b)[/imath] .
Let [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] be positive real numbers such that [imath]a^3 + b^3 = c^3[/imath]. Prove that [imath]a^2 + b^2 - c^2 \gt 6(c - a)(c - b).[/imath] This problem is from the Indian MO 2009, however, I do not know how to approach it, any help? | 1945662 | If [imath]a,b,c>0[/imath] and [imath]a^3+b^3=c^3[/imath] then prove that [imath]a^2+b^2-c^2>6(c-a)(c-b)[/imath].
If [imath]a,b,c>0[/imath] and [imath]a^3+b^3=c^3[/imath] then prove that [imath]a^2+b^2-c^2>6(c-a)(c-b)[/imath]. I tried factoring but things got more complicated. I dont think standard AM-Gm can be applied directly. Can somebody help me to proceed? Thanks a lot. |
2434461 | maximum, minimums and parabolas
Help! Let [imath]f(x,y)[/imath] given by [imath]f(x,y)=ax^2+by^2+cxy+dx+ey+l[/imath] in [imath]R^2[/imath]. If [imath](x_0, y_0)[/imath] is a local maximum point of [imath]f[/imath], show that [imath](x_0, y_0)[/imath] is an global maximum point. The suggestion to consider the function [imath]g[/imath] given by [imath]g (t)=f(x_0+ht,y_0+kt)[/imath] and verify that its graph is a parabola. Even with the suggestion, I could not do it. Excuse, my English is not very good! | 1490301 | if [imath](x_0,y_0)[/imath] is local extrema in [imath]ax^2 + by^2 + cxy + dx + ey + l[/imath] then its global too.
An exercise on a book asks me to prove that if a point [imath](x_0,y_0)[/imath] is a local extrema for the function [imath]f(x,y) = ax^2 + by^2 + cxy + dx + ey + l[/imath] then it's also a global extrema. The exercise asks me to consider that [imath]g(t) = f(x_0+ht,y_0+kt)[/imath] is a parabola, but I couldn't understand how it helps. Maybe since it's a parabola, I know that this derivative has to be [imath]0[/imath] in some point, and also that this function [imath]g(t)[/imath] has a global extrema point, but I couldn't derive anything from this. I also tried to find the derivatives, equal them to [imath]0[/imath] and solve the system. I got: [imath]x_0 = -\frac{2 b d - c e}{4 a b - c^2}, y_0 = -\frac{ c d - 2 a e}{-4 a b + c^2}[/imath] But when I try to find [imath]f\left(-\frac{2 b d - c e}{4 a b - c^2},-\frac{ c d - 2 a e}{-4 a b + c^2}\right)[/imath] so I can try to prove: [imath]f(x,y)-f(x_0,y_0)[/imath] is either greater or less [imath]0[/imath] depending on the case (so I can do [imath]f(x,y) \le f(x_0,y_0)[/imath] and [imath]f(x,y) \ge f(x_0,y_0)[/imath]) things get hairy. |
2434662 | When are the integers extended with the nth root of unity a unique factorization domain?
Let [imath]\zeta_n[/imath] be the [imath]n[/imath]th root of unity. The wikipedia page for unique factorization domain (UFD) states that for [imath]n \in \mathbb Z[/imath], [imath]1 \le n \le 22[/imath], [imath]\mathbb Z[\zeta_n][/imath] is a UFD, but not for [imath]n = 23[/imath]. Is there a general formula for determining when an integer extension with a root of unity is a UFD? Failing that, is it known for higher [imath]n[/imath] than [imath]23[/imath]? | 1496320 | When is [imath]\Bbb{Z[\zeta_n]}[/imath] a PID?
When is [imath]\Bbb{Z[\zeta_n]}[/imath] a PID? I was just wondering if [imath]\Bbb{Z[\zeta_n]}[/imath] is PID or not where [imath]\zeta_n[/imath] is [imath]n[/imath]th primitive root of unity for arbitrary positive [imath]n[/imath] |
2435096 | Find the partial derivatives of [imath]x^TAx[/imath]
[imath]x[/imath] is a vector and [imath]A[/imath] is a matrix and I'm confused as to how to find the partial derivatives of [imath]x^TAx[/imath] with respect to [imath]x_1[/imath], [imath]x_2[/imath], etc. | 189434 | Derivative of Quadratic Form
For the Quadratic Form [imath]X^TAX; X\in\mathbb{R}^n, A\in\mathbb{R}^{n \times n}[/imath] (which simplifies to [imath]\Sigma_{i=0}^n\Sigma_{j=0}^nA_{ij}x_ix_j[/imath]), I tried to take the derivative wrt. X ([imath]\Delta_X X^TAX[/imath]) and ended up with the following: The [imath]k^{th}[/imath] element of the derivative represented as [imath]\Delta_{X_k}X^TAX=[\Sigma_{i=1}^n(A_{ik}x_k+A_{ki})x_i] + A_{kk}x_k(1-x_k)[/imath] Does this result look right? Is there an alternative form? I'm trying to get to the [imath]\mu_0[/imath] of Gaussian Discriminant Analysis by maximizing the log likelihood and I need to take the derivative of a Quadratic form. Either the result I mentioned above is wrong (shouldn't be because I went over my arithmetic several times) or the form I arrived at above is not the terribly useful to my problem (because I'm unable to proceed). I can give more details about the problem or the steps I put down to arrive at the above result, but I didn't want to clutter to start off. Please let me know if more details are necessary. Any link to related material is also much appreciated. |
2434677 | Number of garlands possible
There are [imath]6n[/imath] flower of one kind and [imath]3[/imath] of another kind. Show that number of different garlands made of using all these flowers is [imath]3n^2+3n+1[/imath]. My approach is divide garland in form of [imath]a+b+a=6n+3[/imath]. As it is circular but could not go ahead of it | 1074469 | How many different (circular) garlands can be made using [imath]3[/imath] white flowers and [imath]6m[/imath] red flowers?
This is my first question here. I'm given [imath]3[/imath] white flowers and [imath]6m[/imath] red flowers, for some [imath]m \in \mathbb{N}[/imath]. I want to make a circular garland using all of the flowers. Two garlands are considered the same if they can be rotated to be the same. My question is: how many different garlands can I make? I tried an approach in which I divided the question into three parts: No. of ways in which the white flowers are all together No. of ways in which two white flowers are together No. of ways in which no two white flowers are together Can you please help? The answer is [imath]3m^2+3m+1[/imath] in the book that I saw Thanks |
2435977 | [imath]|f| + |g| \equiv c[/imath] in set Ω where [imath]f,g[/imath] holomorphic in Ω
Prove that if [imath]|f| + |g| \equiv c[/imath] ([imath]c\in \Re[/imath]) in set [imath]Ω\subseteq C[/imath] where [imath]f,g[/imath] holomorphic in Ω then [imath]f\equiv a[/imath] , [imath]g\equiv b[/imath] where [imath]a,b\in C[/imath]. Well the real problem was to show it for [imath]\sum_{i=1}^{n}|f_{i}| \equiv c[/imath] but I think i should starting by proving it for 2 of them.(This was once given to us on an exam as the only problem in it and it was valuated with 100% of the grade) | 2082670 | If [imath]|f|+|g|[/imath] is constant on [imath]D,[/imath] prove that holomorphic functions [imath]f,~g[/imath] are constant on [imath]D[/imath].
Let [imath]D\subseteq \mathbb{C}[/imath] be open and connected and [imath]f,~g:D \rightarrow \mathbb{C}[/imath] holomorphic functions such that [imath]|f|+|g|[/imath] is constant on [imath]D.[/imath] Prove that [imath]f,~g[/imath] are constant on [imath]D.[/imath] Attempt. I noticed first that this problem is already set before, and is considered to be duplicate (If [imath]|f|+|g|[/imath] is constant then each of [imath]f, g[/imath] is constant) and we are sent directly to problem sum of holomorphic functions. However, the last reference deals with the sum [imath]|f|^2+|g|^2[/imath], which I do not see how is connected to the sum [imath]|f|+|g|[/imath], as stated in our title (for example, the equality [imath]|f|^2+|g|^2=(|f|+|g|)^2-2|f||g|[/imath] is not helpful here). Thanks in advance! |
2435433 | If [imath]R[/imath] is a ring and [imath]M[/imath] be any maximal ideal with [imath]a\notin M[/imath], then there exists [imath]q\in M[/imath] and [imath]s\in R[/imath] such that forall [imath]r\in R[/imath] [imath]r=sar+qr[/imath]
If [imath]R[/imath] is a ring and [imath]M[/imath] be any maximal ideal with [imath]a\notin M[/imath], then there exists [imath]q\in M[/imath] and [imath]s\in R[/imath] such that forall [imath]r\in R[/imath] we have [imath]r=sar+qr[/imath]. Also show that this is true only when [imath]M[/imath] is maximal ideal. I am assuming [imath]R[/imath] is commutative and [imath]1\in R[/imath]. Please help me. I am unable to proceed. | 2432879 | Prove that for every [imath]a\in A[/imath] [imath]\exists[/imath] [imath]q\in M,s\in A[/imath] s.t. [imath]a=sba+qa[/imath] where [imath]M[/imath] is maximal ideal in [imath]A[/imath].
Let A be a commutative ring with unity ring, M be a maximal ideal and [imath]b\notin M[/imath] . Show that there exists [imath]q \in M, s\in A[/imath] such that for every [imath]a \in A[/imath] we have [imath]a=sba+qa[/imath]. Show that this is true only when [imath]M[/imath] is a maximal ideal. Suppose if we take [imath]A=2\mathbb Z[/imath] and [imath]M=4\mathbb Z[/imath], [imath]a=6.[/imath] Then by the given statement [imath]\exists[/imath] [imath]s\in A[/imath], [imath]q\in M[/imath] s.t. [imath]6=6(2s+q)[/imath] [imath]\implies[/imath][imath]s=\frac{1-q}{2}\implies q \text { is odd}[/imath]. But [imath]q \in 4\mathbb Z.[/imath] Please someone give some hints. how can solve this. Thank you.. |
2435711 | Can this [imath]x^2+y^2=\frac{dy}{dx}[/imath] be solved by integrating factor method?
[imath]x^2+y^2=\frac{dy}{dx}[/imath] Originally, this is a question I asked on Quora. https://www.quora.com/How-do-I-solve-dfrac-dy-dx-x-2+y-2?filter=all&nsrc=1&snid3=1500842823 Maybe this question has been solved but I was wondering if it can be done by integrating factor? [imath]x^2=\frac{dy}{dx}-y^2[/imath] Using method of integrating factor [imath]\text{I.F}= e^{\int-ydx}[/imath] [imath]\text{I.F}= e^{-yx}[/imath] Can it be done in this way? | 446926 | Riccati differential equation [imath]y'=x^2+y^2[/imath]
[imath]y'=x^2+y^2[/imath] I know, that this is a kind of Riccati equation, but is it possible to solve it with only simple methods? Thank you |
2436586 | Proving [imath]\mathbb{N} \times \mathbb{N}[/imath] is countable without theorem of arithmetic.
I am trying to show that there exists a bijective function [imath]f:\mathbb{N}\rightarrow \mathbb{N} \times \mathbb{N}[/imath] without using prime factorization. I have pieces here and there, but unable to put togther a complete proof. It would be great someone could help me out. This is what I roughly got. I choose a sequence of postive integer [imath]a_n=\frac{n(n+1)}{2}[/imath] and I know they will be "captured once" by the interval [imath]I_n=(a_{n-1},a_n][/imath] for all [imath]n\geq 2[/imath]. I will let [imath]I_1=\{1\}[/imath]. Then once show that there is indeed a bijective function between the set of intervals and the set of number defined by the sequence. I will invoke the property that any infintie subset of a countable set is countable to complete the proof. | 278679 | Does anyone know a closed-form expression for a bijection between [imath]\mathbb{N}^k[/imath] and [imath]\mathbb{N}[/imath]?
I want to publish an article and one of its results is a simple closed-form expression for a natural bijection between [imath]\mathbb{N}^k[/imath] and [imath]\mathbb{N}[/imath]. I wish to know whether it is already known. |
1475123 | If [imath]\sum_{n=1}^\infty a_n[/imath] converges, prove that [imath]\lim_{n\to \infty} (1/n) \sum_{k=1}^n ka_k = 0[/imath].
the series [imath]a_n[/imath] is any arbitrary converging series. My thought process was that the [imath]1/n[/imath] will definitely go to zero as n approaches infinity; however, the series [imath]k*a_k[/imath] seems to approach infinity at the same rate. This confuses me because if both the numerator and denominator approach infinity, I think that the overall equation will equal 1, not zero. I also considered that the sum is geometric, but I do not believe that is the case, because both [imath]a_k[/imath] and [imath]k[/imath] are variables that are not constant. I was thinking that I could argue if the series [imath]a_k[/imath] converges to a point, eventually the terms of the sequence will be so close to each other that it resembles a constant multiplier, but that argument is not very strong. Any tips, hints, or leads at the solution are appreciated! | 2439931 | Show that if [imath]\sum \limits_{n=1}^\infty a_n[/imath] converges then [imath]\lim\limits_{n\to\infty}\frac{1}{n}\sum _{k=1}^n ka_k=0[/imath]
Show that if [imath]\sum \limits_{n=1}^\infty a_n[/imath] converges then [imath]\lim\limits_{n\to\infty}\frac{1}{n}\sum \limits_{k=1}^n ka_k=0[/imath] since given that [imath]\sum\limits _{n=1}^\infty a_n[/imath] converges then we can say that [imath]\lim _{n\to \infty} a_n=0[/imath] but i don't get any idea to prove this result how to solve this problem |
2435472 | Show the determinant of a matrix is divisible by 17 without evaluating the determinant
Use the fact that 11322, 13209, 29189, 56661, and 15096 are all divisible by 17 to show that [imath]\begin{vmatrix} 1&1&2&5&1\\ 1&3&9&6&5\\ 3&2&1&6&0\\ 2&0&8&6&9\\ 2&9&9&1&6 \end{vmatrix}[/imath] is divisible by 17 without directly evaluating the determinant. | 1567302 | Determine a determinant is divisible by 23 or not
Consider the fact that [imath]25875, 46552, 41354, 48691, 95818[/imath] are all divisible by [imath]23[/imath]. Use this fact to determine if \begin{vmatrix} 2 &5 &8 &7 &5 \\ 4 &6 &5 &5 &2 \\ 4 &1 &3 &5 &4 \\ 4 &8 &6 &9 &1 \\ 9 &5 &8 &1 &8 \end{vmatrix} is divisible by 23 or not, without directly evaluating the determinant. |
2436956 | Logical argument for [imath](1+2+\dots+n)^2=1^3+2^3+\dots+n^3[/imath]
I know that [imath]1+2+\dots+n=\dfrac{n(n+1)}{2}[/imath] and [imath]1^3+2^3+\dots+n^3=\dfrac{n^2(n+1)^2}{4}[/imath] and can show by algebraic calculation. But I am wondering if there is any logical argument or any intuitive argument to show that: [imath](1+2+\dots+n)^2=1^3+2^3+\dots+n^3[/imath] Thanks. | 1305097 | Geometrical interpretation of [imath](\sum_{k=1}^n k)^2=\sum_{k=1}^n k^3[/imath]
Using induction it is straight forward to show [imath]\left(\sum_{k=1}^n k\right)^2=\sum_{k=1}^n k^3.[/imath] But is there also a geometrical interpretation that "proves" this fact? By just looking at those formulas I don't see why they should be equal. |
2436971 | Proving the AM-GM inequality for 3 variables
I was able to prove the AGM inequality for four variables, however when asked to prove it for 3 I don't understand how to correctly do it. [imath]\sqrt[3]{xyz} ≤ \frac{x+y+z}{3}[/imath] But I'm supposed to solve it using the AGM inequality for four variables, with [imath]w = \sqrt[3]{xyz}[/imath] Step by step would be really helpful. | 2435537 | Proving the AGM inequality with 3 variables
Prove that for any [imath]x,y,z \ge 0[/imath] [imath]\sqrt[3]{xyz} \le \frac{x+y+z}{3}[/imath] I've done it with four variables, but my textbook is asking me to do it with three and I'm not sure where to start. It also says to use the four variable method, but have [imath]w=(xyz)^\frac{1}{3}[/imath] which confuses me further. |
2436220 | Proving [imath]\int_{0}^{\infty} u^{x-1}e^{-u} du = 2 \int_{0}^{\infty} u^{2x-1}e^{-u^2} du[/imath]
I am trying to solve exercise 1.1 in Neural Networks for Pattern Recognition by Bishop. They give the gamma function as: [imath]\int_{0}^{\infty} u^{x-1}e^{-u} du[/imath] I can only solve it using the form of the gamma function: [imath]2 \int_{0}^{\infty} u^{2x-1}e^{-u^2} du[/imath] Which is equation 3 here It is very unclear to me how these two forms of the gamma function are equal. Is there a well-known proof? | 2436182 | Need help with the integral [imath]\int_{0}^\infty e^{-x^{2}}x^{2n+1}dx [/imath]
Problem: Prove that [imath]I = \int_{0}^\infty e^{-x^{2}}x^{2n+1}dx = \frac{n!}{2} [/imath] Source: A problem I found on an integral test. The problem bugged me for long and I did end up leaving it on the exam. Back home I decided to tackle it again. Here's my go on it. My try: I have never encountered such a problem before, not even in my assignments and workbooks. Here's how I tried it We can first take an indefinite integral and then we can work up towards a reduction formula.We can plug in the limits later. I don't know if it's any good but I did write this up on the exam: [imath]I_{2n+1} = \int{e^{x^{-2}}}{x^{2n+1}}dx[/imath] On applying Integration by parts: [imath]I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{2}{2n+2} \int{e^{-x^{2}}}{x^{2n+3}}dx[/imath] or [imath]I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{I_{2n+3}}{n+1} [/imath] or [imath]I_{2n+1} = {e^{-x^{2}}\over 2}\Bigl(\frac{x^{2n+2}}{n+1}\Bigl) +\frac{I_{2n+3}}{n+1} [/imath] What can I do next? Am I doing it incorrectly? Is this the wrong way? Please help me as I can't stop thinking about this problem! Thanks! Edit 1: As suggested (already tried by me) lets put [imath]u = -x^2[/imath] [imath] du = -2xdx[/imath] plug the above expression in the problem [imath]I_{n} = -{1 \over 2}\int{e^{u}}{u^{n}}du[/imath] [imath]I_{n} = -{1 \over 2}e^{-u}\frac{u^{n+1}}{n+1} + \frac{I_(n+1)}{n+1}[/imath] Certainly helps. Now what? Edit 2: Use the gamma function to get the result. Now unfortunately I didn't read it for the test that's why I didn't get it. Thank you all for pointing out the right direction! |
2437345 | Maximum value of [imath]\sin (x)\;\sin (2x)\;\sin (3x)[/imath]
To prove : [imath]\sin (x)\;\sin (2x)\;\sin (3x) \lt 9/16[/imath] With some transformations i was able to prove the above result using calculus but I am not getting the way to solve it without the use of calculus I've tried grouping [imath]\sin(x)\;\sin(3x)[/imath] and then using C-D formula Also I've tried to find the maximum value of sinxsin3x and then using the fact that sin2x is a fraction. Although I was able ti prove that sinxsin3x is less than 9/16 but the minimum value was less than -9/16 and since sin2x can be negative so i could not draw any conclusion. | 1804426 | Prove that [imath]\sin x \cdot \sin (2x) \cdot \sin(3x) < \tfrac{9}{16}[/imath] for all [imath]x[/imath]
Prove that [imath] \sin (x) \cdot \sin (2x) \cdot \sin(3x) < \dfrac{9}{16} \quad \forall \ x \in \mathbb{R}[/imath] I thought about using derivatives, but it would be too lengthy. Any help will be appreciated. Thanks. |
2436992 | If [imath]x>y>0[/imath] why is [imath]\sqrt{x} > \sqrt {y}[/imath]?
If [imath]x>y>0[/imath] why is [imath]\sqrt{x} > \sqrt {y}[/imath]? I know this to be true, but I don't understand why because can't the square root of [imath]x[/imath] be negative and the square root of [imath]y[/imath] be positive thus disproving it? | 376730 | Proving if [imath]x then \sqrt{x} < \sqrt{y}[/imath]
I am stuck on this homework problem! Prove that if [imath]x[/imath] and [imath]y[/imath] are real numbers such that [imath]0<x<y[/imath] , then [imath]\sqrt{x}<\sqrt{y}[/imath]. This is in a chapter involving the least upper bound axiom and the Archimedean Principle, but I cannot figure how to use either of those to prove this! Any help would be most appreciated! Thanks |
2434499 | Formula for expanding powers
I can't seem to understand the following: [imath]A^n-B^n=\left(A-B\right)\cdot \left(A^{n-1}+A^{n-2}\cdot B+\cdots+A\cdot B^{n-2}+B^{n-1}\right)[/imath] How can i derive this formula ? Also, when does the [imath]A^\text{something}[/imath] end and when does [imath]B^\text{something}[/imath] start. Thank you. | 2031059 | Prove that [imath]a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})[/imath].
Exercise Prove that [imath]a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})[/imath]. I've posted my solution below. In case someone has a more clever solution, feel free to post it! (TBH, I was surprised that there was no question on Math.SE regarding this equation!) |
2437967 | Prove series [imath]\frac{1}{x\ln x}[/imath] diverges
Prove series [imath]\sum\limits_{n=2}^{\infty}\frac{1}{n\ln n}[/imath] doesn't converge without integral test. I really don't know any hint for [imath]\ln n[/imath]. So I understand I have to show [imath]\frac{1}{n\ln n} > \frac{1}{x^b}[/imath] where [imath]b<1[/imath]. But I really don't know how | 2323143 | Convergence/Divergence of some series
If I have [imath]\sum_{n=2}^{\infty} \frac {1}{n\log n}[/imath] and want to prove that it diverges, can I use following? [imath]\frac {1}{n\log n} \lt \frac {1}{n}[/imath] [imath]\sum_{n=1}^\infty \frac 1 n[/imath] diverges, but the limit of [imath]\frac 1 n[/imath] equals to zero so the comparasion I think isn't useful. Or can I say it diverges because [imath]\sum_{n=1}^\infty \frac 1{\log n}[/imath] also diverges? And If I have [imath]\sum_{n=2}^\infty \frac {(-1)^n}{(n)}[/imath] I can use the Leibniz rule. [imath]\sum_{n=1}^\infty \frac 1 n[/imath] diverges, but the limit of it equals to zero, so I am confused if this series diverges/converges again. I understand that [imath]\sum_{n=1}^\infty \frac 1 n[/imath] has an infinite sum, but what to do in this case? |
1844643 | Simple proof of Newton identities
The functions [imath]s_1=x_1+x_2+\cdots +x_n[/imath], [imath]s_2=\sum_{i<j} x_ix_j[/imath], [imath]\cdots[/imath], [imath]s_n=x_1x_2\cdots x_n[/imath] are elementary symmetric functions in [imath]x_1,x_2,\cdots,x_n[/imath] (or more precisely, elementary symmetric polynomials). There are other natural symmetric functions in [imath]x_1,x_2,\cdots,x_n[/imath]: [imath]t_1=x_1+\cdots + x_, \hskip5mm t_2=x_1^2+x_2^2+\cdots + x_n^2, \cdots ,t_n=x_1^n+x_2^n+\cdots+x_n^n.[/imath] The two sets of symmetric functions are related by some identities, known as Newton's identities. What are the simple proofs of these Newton identities? How do we prove them with some intuition or motivation? | 844571 | Proving Newton's identities
Assume [imath]F[/imath] is a field of zero characteristic. Denote the elementary symmetric polynomials of [imath]n[/imath] variables by [imath]e_k[/imath], [imath]\quad k=\overline{1,n}[/imath]. Let the symbol [imath]\sum ax_1^{i_1}\dots x_n^{i_n}[/imath] denote the sum of all different monomials which can be obtained from [imath]ax_1^{i_1}\dots x_n^{i_n}[/imath] by permuting its variables. For instance, if [imath]n=3[/imath], [imath]\sum x_1^2x_2=x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2[/imath] The polynomials [imath]S_k=\sum_{i=1}^n x_i^k[/imath] are called power sum symmetric polynomials. Obviously [imath]S_1=e_1[/imath] and [imath]S_2=e_1^2-2e_2[/imath]. In order to prove Newton's identities one can see that for [imath]2<k<n+1[/imath] [imath]e_iS_{k-i}=\sum x_1^{k-i+1}x_2\dots x_i+\sum x_1^{k-i}x_2\dots x_ix_{i+1}[/imath] How can we see the above identity? |
2438046 | show invertibility of linear transformations
My understanding of this part of linear algebra is somewhat clouded. I fail to see how these two problems are related and/or how to prove them: 1) Show that if [imath]A[/imath] is a linear transformation with [imath]A^2 - A + 1 = 0[/imath], then [imath]A[/imath] is invertible. 2) If [imath]A[/imath] and [imath]B[/imath] are linear transformations on the same vector space and [imath]AB = 1[/imath]. Prove that if A has exactly one [imath]right\space inverse\space B[/imath], then A is invertible. (Consider [imath]BA + B - 1[/imath]) My Attempt: 1) Show that if [imath]Ax = 0[/imath], then [imath]x[/imath] must be [imath]0[/imath]. i.e. [imath]A^2 - A + 1 = 0[/imath] implies [imath]A^2 + 1 = A[/imath], implies [imath](A^2 + 1)x = Ax[/imath] imples [imath]A(A(x)) + x = A(x)[/imath] imples [imath]A(0) + x = 0[/imath]. Since A is linear ([imath]A(0) = 0[/imath]), [imath]x = 0[/imath]. 2) No idea. | 2436359 | Showing a Linear Transformation is invertible
I have the following question: Suppose A is a linear transformation. Show that if [imath]A^2 - A + I = O[/imath], then A is invertible. To show [imath]A[/imath] is invertible, I can show if [imath]Ax = 0[/imath] then [imath]x = 0[/imath]. But, if [imath]Ax = 0[/imath], then [imath](Ax)^2 + (Ax) + 1 = 0 + 0 + 1 \neq 0[/imath] so the equation in the question is not satisfied. What gives? |
2438292 | If [imath]A[/imath] is a semilocal ring and [imath]f : A \twoheadrightarrow B[/imath], then [imath]f(\text{rad}(A)) = \text{rad}(B)[/imath]?
I know that if [imath]f[/imath] is surjective then [imath]f(\mathfrak{a})[/imath] is an ideal for every ideal [imath]\mathfrak{a}[/imath] in [imath]A[/imath]. I want to first prove that if [imath]\mathfrak{m}[/imath] is maximal in [imath]A[/imath] then [imath]f(\mathfrak{m}) = \mathfrak{m'}[/imath] is maximal in [imath]B[/imath]. [imath]A/\mathfrak{m}[/imath] is a field by maximality of [imath]\mathfrak{m}[/imath] and [imath]f[/imath] induces [imath]f^* : A/\mathfrak{m} \twoheadrightarrow B / \mathfrak{m'}[/imath]. If [imath]I' \leqslant B / \mathfrak{m}'[/imath] is an ideal and [imath]f^{*-1}(I')[/imath] must either be [imath](0 + \mathfrak{m})[/imath] or [imath](1 + \mathfrak{m}) = I[/imath]. If the latter then [imath]f^*(I) = B/\mathfrak{m'}[/imath] because [imath]f^*[/imath] is surjective, but [imath]f^*(f^{*-1}(I')) \subset I'[/imath] so [imath]I' = B/\mathfrak{m'}[/imath] the whole ring. And if [imath]I = (0 + \mathfrak{m})[/imath] then [imath]f^*(f^{*-1}(I')) = 0[/imath] so that [imath]I'[/imath] can only be [imath]0[/imath]. So, [imath]B / \mathfrak{m'}[/imath] is a field and [imath]\mathfrak{m'}[/imath] is maximal. Is the above proof valid? We want to show that [imath]f(\bigcap\limits_{\mathfrak{m} \text{ maximal}} \mathfrak{m}) \subset \bigcap\limits_{\mathfrak{m}' \text{ maximal}} \mathfrak{m'}[/imath]. Next, the inverse map [imath]f^{-1}[/imath] takes maximal ideals in [imath]B[/imath] over to maximal ideals in [imath]A[/imath] so [imath]f(\bigcap \mathfrak{m}) \subset \bigcap f(\mathfrak{m}) = \text{rad}(B)[/imath]. Now how do I prove the question in the title? A semilocal ring is one with a finite number of maximal ideals [imath]\mathfrak{m}_1, \dots, \mathfrak{m}_n[/imath]. Each one of these ideals is pairwise coprime so that [imath]\phi : A / \mathfrak{m}_1 \times \dots \times A / \mathfrak{m}_n \xrightarrow{\sim} A / (\mathfrak{m}_1 \cap \dots \cap \mathfrak{m}_n)[/imath] is an isomorphism. But in the first proof, those field homomorphism, since they weren't identically [imath]0[/imath] must be injective so that [imath]A/\mathfrak{m}_i \simeq B/\mathfrak{m'}_i[/imath] so we have an isomorphism: [imath]A / (\mathfrak{m}_1 \cap \dots \cap \mathfrak{m}_n) \simeq B/(\mathfrak{m}'_1 \cap \dots \cap \mathfrak{m}'_n)[/imath] where [imath]f(\mathfrak{m}_1 \cap \dots \cap \mathfrak{m}_n) \subset f(\mathfrak{m}_1) \cap \dots \cap f(\mathfrak{m}_n) = \mathfrak{m}_1' \cap \dots \cap \mathfrak{m}_n'[/imath]. Now how do I prove that that is not an isomorphism if [imath]\subset[/imath] can't be replaced by [imath]=[/imath]? | 1010740 | If [imath]A[/imath] is a semilocal ring and [imath]f:A\rightarrow B[/imath] is a surjective homomorphism, then [imath]f(\operatorname{rad}A)=\operatorname{rad}B[/imath]
If [imath]A[/imath] is a semilocal ring and [imath]f:A\rightarrow B[/imath] is a surjective homomorphism, then [imath]f(\operatorname{rad}A)=\operatorname{rad}B[/imath]. I know that if [imath]A[/imath] is a semilocal ring and if [imath]I_{1},\dots, I_{n}[/imath] are all of its maximal ideals, then [imath]\operatorname{rad}A=I_{1}\cdots I_{n}=I_{1}\cap\cdots\cap I_{n}[/imath]. Let [imath]y\in\operatorname{rad}B[/imath], then there is [imath]x\in A[/imath] such that [imath]f(x)=y[/imath]. I would like to show that [imath]x\in\operatorname{rad}A[/imath]. Any hint? |
2438269 | Evaluate [imath]\lim_{x\to 0} \sin (\frac {1}{x})[/imath]
Evaluate: [imath]\lim_{x\to 0} \sin (\dfrac {1}{x})[/imath] My Attempt: let [imath]\dfrac {1}{x}=k[/imath] As [imath]\dfrac {1}{x}\to 0[/imath], [imath]k\to \infty[/imath] Now, [imath]=\lim_{k\to \infty} \sin k[/imath] | 509678 | How prove this [imath]\lim_{n\to \infty}\sin{n^m}[/imath] divergent.
show that [imath]\lim_{n\to \infty}\sin{(n^m)}[/imath] divergent,where [imath]m\in N^{+}[/imath] I have kown [imath]\lim_{n\to\infty}\sin{n^2}[/imath] to be divergent and dense in [imath][-1,1][/imath]. This is very famous problem,the problem is first is post [imath]AMM[/imath]( The American Mathematical Monthly,1970-1975) problem,and after some years [imath]AMM[/imath] post [imath]\sin{(n^{14})}[/imath] in [imath][-1,1][/imath] was dense? and this is open problem.so I Guess [imath]\sin{(n^m)}[/imath] is dense in [imath][-1,1][/imath]. |
2438390 | Find the probability density function for the function [imath]E = d^2[/imath]
I'm struggling with this text book question "Suppose [imath]d[/imath] is a Gaussian random variable with zero mean and unit variance. What is the probability density function of: [imath]E[/imath] = [imath]d^2[/imath]? Hint: Since the sign of d gets lost when it is squared, you can assume that p(d) is one-sided, that is, defined for only [imath]d \geq 0[/imath] and with twice the amplitude of the usual Gaussian." Intuitively this is the way I would attempt it: if [imath]E=d^2[/imath] then [imath]d=\sqrt{E}[/imath] (considering only when d is positive), then substituting this into the equation for the normal distribution [imath]2*N(1,0)[/imath] returns: [imath]\large{ \frac{2}{(\sqrt(2*\pi))}\exp{(-x/2)}}[/imath] Taking the derivative of this would give me p(d). This sort of feels right since the derivative will be negative everywhere and tends to zero. If some one could shed some light on the concepts here it would be greatly appreciated I don't have much if any experience with probability (or latex for that matter). Thanks if advance! | 71537 | Derivation of chi-squared pdf with one degree of freedom from normal distribution pdf
How can we derive the chi-squared probability density function (pdf) using the pdf of normal distribution? I mean, I need to show that [imath]f(x)=\frac{1}{2^{r/2}\Gamma(r/2)}x^{r/2-1}e^{-x/2} \>, \qquad x > 0\>.[/imath] |
698528 | How can I find [imath]\int_0^{\pi/2}x\cot x\,dx[/imath]?
Can [imath]\displaystyle\int_0^{\pi/2}x\cot x\,dx[/imath] be found using elementary functions? If so how could I possibly do it? Is there any other way to calculate above definite integral? | 381976 | Tricky elementary integral [imath]\int_{0}^{\frac{\pi}{2}}x\cot(x)dx[/imath].
[imath]\int_{0}^{\frac{\pi }{2}}x\cot(x)dx[/imath] I tried integration by parts and got [imath]\dfrac{1}{2}\int_{0}^{\frac{\pi }{2}}x^{2} \csc^{2}x dx[/imath] which doesn't help at all. I don't really know what to do. Any help will be greatly appreciated. |
2438914 | Connected set in [imath]\mathbb{S}^1[/imath]
Let [imath]A[/imath] be a connected set in [imath]\mathbb{S}^1[/imath]. Show that [imath]A^c[/imath] is also connected. My idea: suppose [imath]A^c[/imath] is not connected. Then, there exist [imath]U,V \subset \mathbb{S}^1[/imath] open and disjoint such that [imath]U \cup V =A^c [/imath]. Therefore [imath]\mathbb{S}^1 = A \cup U \cup V[/imath]. If I had that [imath]A\cup U[/imath] was open I could arrive at contradiction, since [imath]\mathbb{S}^1[/imath] is connected, but I can't say that... | 2460287 | Show that the complement of every connected subset of [imath]S^1[/imath] is also connected.
Show that the complement of every connected subset of [imath]S^1[/imath] is also connected. Here [imath]S^1[/imath] means [imath]S^1 = \{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}[/imath]. Can someone give some hints on this? It seems that arcs are the only connected subset of [imath]S^1[/imath]. If it is true, the the complement of this subset is also a arc, which of course is also connected. Is this intuition true? Can someone give some hints on how to proceed? Thanks in advance. |
2439201 | Identity function (Topology)
Let [imath]X[/imath] be Hausdorff and [imath]f:X \to X[/imath] be continuous. Prove [imath]\{x \mid f(x)=x\}[/imath] is closed in [imath]X[/imath]. I know that the points in a Hausdorff space are closed, I was thinking about the diagonal but this is with cartesian product. Could you give me any ideas about it, please? Thanks for your help | 1651081 | Let [imath]F : X → X[/imath] be continuous. Prove that the set [imath]\{x ∈ X : F(x) = x\}[/imath] of fixed points of F is closed in X
Here X is a Hausdorff Space. I know that singleton sets, {x}, are closed in a Hausdorff space. Although Im not sure if thats how to use the Hausdorff property. Should I investigate [imath]h=F(x)-x[/imath]? Can anyone give a hint |
2439900 | How can I approximate the solutions to systems of differential
I'm new to differential equations and I was wondering if I could use the Euler method and the midpoint method to approximate the solution to these three differential equations, as I cannot solve this set of differential equations at my skill level. In addition, are any other ways to approximate these solutions? These equations are the SIR equations used to model an infectious disease. [imath]{\frac {dS}{dt}}=-{\frac {\beta IS}{N}}[/imath] [imath]{\frac {dI}{dt}}={\frac {\beta IS}{N}}-\gamma I[/imath] [imath]{\displaystyle {\frac {dR}{dt}}=\gamma I}[/imath] If you are interested in the specifics of the model | 2230547 | Euler method for SIR model
I know how to solve SIR model with ode45, but I just wanted to try it out using Euler Method. The SIR model is: [imath]\dot S=-\beta IS \\ \dot I = \beta IS - \gamma I \\ \dot R = \gamma I[/imath] The code that I wrote is function Euler deltaT=1; s=1; i=1e-6; r=1-(s+i); beta=1.4247; gamma=0.14286-1e-6; maxTime=70; t=[0]; for n=1:maxTime t(n+1)=t(n)+deltaT; s(n+1)=s(n)-beta*s(n)*i(n)*deltaT; i(n+1)=i(n)+(beta*s(n)*i(n)-gamma*i(n))*deltaT; r(n+1)=r(n)+gamma*i(n)*deltaT; end figure plot(0:maxTime,s) hold on plot(0:maxTime,i) hold on plot(0:maxTime,r) when I make the step size to be 1, I get the following figure which looks like the expected behaviour. But when I make the step size small= 0.001, I get a very different figure. Why does the behaviour change when I change the step sizes. What have I done wrong in the code? |
2440410 | How to solve [imath]f'(x)=f(x+1)[/imath]?
Are there non-trivial solutions [imath]f:\Bbb R\to\Bbb R[/imath] of the following "differential equation"? [imath]f'(x)=f(x+1).[/imath] I consider [imath]f(x)=0[/imath] as trivial solution. I tried to express [imath]f[/imath] as a power series, i.e. [imath]f(x)=\sum_n a_n/n!\cdot x^n[/imath]. This brought me to the following identity for the coefficients: [imath]a_{n+1}=\sum_{k=0}^\infty \frac {a_{k+n}}{k!}.[/imath] This does not really help, as the right side contains infinitely many coefficients itself. And infinitely many must be non-zero. I expect that a possible solution looks something like this: | 61818 | How to solve differential equations of the form [imath]f'(x) = f(x + a)[/imath]
What could one do to find analytic solutions for [imath]f'(x) = f(x + a)[/imath] for various values of [imath]a[/imath]? I know that [imath]c_1\sin(x + c_2)[/imath] is solution when [imath]a = \frac{1}{2}\pi[/imath], and of course [imath]c_1e^x[/imath] when [imath]a = 0[/imath]. For instance, is there a function satisfying [imath]f'(x) = f(x + 1)[/imath]? What about negative or imaginary [imath]a[/imath]? Is there possibly a generalization? |
2440821 | How is the set [imath]\emptyset^\emptyset[/imath] defined?
By [imath]\emptyset^{\large\emptyset}[/imath] i mean the set of functions from [imath]\emptyset[/imath] to [imath]\emptyset[/imath]. Is it the empty set or the set containing the empty set? I just dont know how to prove it. | 2440458 | It is posible to have a function [imath]f[/imath] with the following property?
I am considering the function [imath]f[/imath] defined as [imath]f:\emptyset\to\emptyset[/imath] My thought is that a function maps elements from a set to another one, but the empty set has no elements to map, so I think there cannot exist a function [imath]f[/imath] that has this property, but I don't know how to prove it and I'd like some light on this. There is a function [imath]I[/imath] though which can map the empty set to the empty set if its domain and codomain are the set containing the empty set. But it is not the same question. |
2441122 | [imath]\sum\limits_{n=1}^∞ \frac{|\sin(n\theta)|}{n}[/imath] diverges?
Let [imath]\theta[/imath] be any fixed constant in [imath](0,\pi/2)[/imath]. Then the seires [imath]\sum_{n=1}^∞ \frac{|\sin(n\theta)|}{n}[/imath] diverges. I don't have any idea how to prove it. Maybe equidistribution theorem is related? | 938894 | How to prove that [imath]\sum_{n=1}^{\infty}\frac{|\sin{(n\theta_0)}|}{n}[/imath] diverges for given [imath]\theta_0\in (0,\pi)[/imath]?
How to prove [imath]\sum_{n=1}^{\infty}\frac{|\sin{(n\theta_0)}|}{n}[/imath] diverges for given [imath]\theta_0\in (0,\pi)[/imath]? I would appreciate any help. |
2441638 | algebra How to prove?
Let [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] be positive real numbers such that [imath](1+a+b+c)\big(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)=16.[/imath] Prove that [imath]a+b+c=3.[/imath] I am not able to get how to prove that? Thanks for help in advance . | 2053861 | Condition on a,b and c satisfying an equation(TIFR GS 2017)
Let [imath]a,b,c[/imath] be positive real numbers satisfying [imath](1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16,[/imath] then [imath]a+b+c=3[/imath]. I thought about the application of the AM-GM-HM inequality, but in vain. I also thought about splitting 16 into factors and comparing but went nowhere. Any ideas. Thanks beforehand. |
2442001 | Prove that for each set [imath]S[/imath] that [imath]S\notin S[/imath] with regularity axiom.
I want to prove for each set [imath]S[/imath] that [imath]S\notin S[/imath]. I know that I can prove this with the Regularity Axiom which says: [imath]\exists x(x\in S) \implies (\exists y\in S)(\forall z \in S)(z \notin y)[/imath] However if I take a simple example, let's say S = {a,{a},S} Then I can pick [imath]y = a[/imath] so that [imath]a \notin a[/imath] and {a}[imath] \notin a[/imath] and [imath]S \notin a[/imath]. Where is my logic wrong? | 271555 | How does the axiom of regularity forbid self containing sets?
The axiom of regularity basically says that a set must be disjoint from at least one element. I have heard this disproves self containing sets. I see how it could prevent [imath]A=\{A\}[/imath], but it would seem to do nothing about [imath]B=\{B,\emptyset\}[/imath]. It is disjoint from [imath]\emptyset[/imath]. What is a disproof of the existence of [imath]B[/imath], and how is it related to the axiom of regularity? |
2441404 | Knowing [imath]\lim_{n\to\infty}x_ny_n = 0[/imath],try to prove [imath]\lim_{n\to\infty}x_n = 0[/imath] or [imath]\lim_{n\to\infty}y_n = 0[/imath]
Knowing [imath]\lim_{n\to\infty}x_ny_n = 0[/imath]try to prove [imath]\lim_{n\to\infty}x_n = 0[/imath]or[imath]\lim_{n\to\infty}y_n = 0[/imath] | 1236007 | If [imath]x_n\cdot y_n\to 0[/imath] then [imath]x_n \to 0[/imath] or [imath]y_n \to 0 [/imath]
Assume [imath]\lim_{n\to \infty}{x_ny_n}=0[/imath] then [imath]\lim_{n\to \infty}{x_n}=0[/imath] or [imath]\lim_{n\to \infty}{y_n}=0[/imath] I couldnt find two series who disprove that.. just hint please thanks |
2439405 | Infinite truth assignments
I've been trying to solve the following problem, but I am running out of ideas. The problem is: Is there a set [imath]\Gamma \subset \mathcal{L}_0[/imath] and an enumeration of an infinite set of truth assignments [imath]v_1, v_2, \ldots[/imath] such that for every truth assignment [imath]v[/imath], [imath]v[/imath] satisfies all the formulas in [imath]\Gamma[/imath] if and only if there is an [imath]i[/imath] such that [imath]v = v_i[/imath]? I tried to generate a contradiction using an argument similar to that one of Cantor to prove that rational numbers are countable, but it failed. Please I would appreciate any help. Thanks! | 2438285 | Countable set of truth assignments satisfying set of well formed formulas
I am trying to answer the following question posed in my logic class: (throughout we work in propositional logic) is there a set [imath]S[/imath] of well-formed-formulas (wffs) satisfied by countable set [imath]V[/imath] of truth assignments such that any truth assignment satisfying [imath]S[/imath] lies in [imath]V[/imath]? My guess is no. My first reaction to this was to do a diagonalization. Present [imath]V[/imath] as an array with [imath]v_{i, j}[/imath] the truth value that assignment [imath]v_i[/imath] gives to the [imath]j[/imath]-th propositional letter. Obviously the hard case is when [imath]S[/imath] is infinite; if [imath]S[/imath] is finite, then only finitely many propositional letters occur in the [imath]S[/imath]-formulae. Then there is some maximal [imath]k[/imath] for which all [imath]A_n[/imath] with [imath]n \geq k[/imath] do not occur in V. Then the diagonalization is trivial: just diagonalize from the [imath]k[/imath]-th letter onward, since changing [imath]A_k[/imath] in this range doesn't affect anything in [imath]S[/imath]. But I have little clue about how to approach the main case when [imath]S[/imath] is infinite. I thought maybe I should try using the fact that [imath]S[/imath] is consistent (since it's satisfiable). If it's maximally consistent, then I think this is simple too, since for every propositional letter [imath]A_n[/imath], either [imath]A_n[/imath] is in [imath]S[/imath] or [imath](\neg A_n)[/imath] is in [imath]S[/imath], implying that there is only one truth assignment satisfying [imath]S[/imath]--since every truth assignment must agree on each propositional letter. But if [imath]S[/imath] is not maximally consistent, this isn't available to me. Somehow I would like to show that infinitely many propositional letters do not occur in [imath]S[/imath]. Then I can apply the trick above and just diagonalize through the superfluous letters. So, yes, just looking for some hints or corrections if you have any please, though maybe if I don't get anywhere on my own I'll give up and ask for more. Thanks. (By the way, by referring to letters that "occur in [imath]S[/imath]" I mean letters that occur in some formula in [imath]S[/imath]). |
2442084 | Evaluate [imath]\sum_{k=2}^{\infty} k^{-1} x^{3k+1}[/imath]
I need to evaluate the series [imath]\sum_{k=2}^{\infty} k^{-1} x^{3k+1}[/imath] for [imath]|x| < 1[/imath], and then say something about the series for when [imath]x=1[/imath] and [imath]x \neq 1[/imath]. So far I have the following work: we note that [imath]\sum_{k=2}^{\infty} k^{-1} x^{3k+1} = x \sum_{k=2}^{\infty} k^{-1} x^{3k}[/imath]. Now consider the series [imath]\sum_{k=2}^{\infty} k^{-1} x^\alpha[/imath]. Then \begin{align} \sum_{k=2}^{\infty} k^{-1} x^\alpha = \sum_{k=2}^{\infty} (k^{-1} +k - k) x^{\alpha} = \sum_{k=2}^{\infty} (k^{-1} - k) x^{\alpha} + \sum_{k=2}^{\infty} k x^{\alpha}. \end{align} This is where I'm stuck. The term [imath]\sum_{k=2}^{\infty} k x^{\alpha}[/imath] is easy to handle, but I can't see what to do with the first term. | 347200 | Evaluate: [imath]\sum_{n=1}^{\infty}\frac{1}{n k^n}[/imath]
How to evaluate this series for [imath]k > 1[/imath]? [imath]\sum_{n=1}^{\infty}\frac{1}{n k^n}[/imath] For [imath]k = 2[/imath], I tried to evaluate [imath]\displaystyle \sum_{n = 0}^\infty \int_{1}^{2} x^{-(n+1)}dx = \int_{1}^{2} \sum_{n = 0}^\infty x^{-(n+1)}dx = \int_1^{2}\frac{1}{x(x-1)}dx[/imath] [imath]\displaystyle = \int_{1}^{2}\frac{1}{x-1}dx - \int_{1}^{2}\frac 1 x dx = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n2^n}\right)[/imath] But both [imath]\displaystyle \int_{1}^{2}\frac{1}{x-1}dx[/imath] and [imath]\displaystyle \sum_{n=1}^\infty\frac1n[/imath] diverges. The answer is [imath]\ln 2[/imath], are these divergent terms equal? |
2440877 | Can't there be an infinite number of prime triplets separated by 2?
I came across a question on brilliant.org where it had been mentioned that [imath]X, X + 2, X + 4[/imath] are all prime numbers for [imath]X = 3[/imath]. So is there any other [imath]X[/imath] for which [imath]X, X + 2, X + 4[/imath] are all primes and the answer was no. So, can we say that there cannot be infinite number of prime triplets separated by [imath]2[/imath]. Please explain why? | 2137627 | There are infinitely many prime numbers [imath]p[/imath] for which [imath]p + 2[/imath] and [imath]p + 4[/imath] are also prime numbers
I need help with this question: There are infinitely many prime numbers [imath]p[/imath] for which [imath]p + 2[/imath] and [imath]p + 4[/imath] are also primes. Where should I start? what proof techniques will be useful? Any input will be very useful to me. |
2243646 | For a given condition is it true that [imath]a+b+c=3[/imath].
Suppose a, b, c are positive real numbers such that [imath](1+a+b+c)\left(1+\frac 1a+\frac 1b+\frac 1c\right)=16[/imath] Then is it true that we must have [imath]a+b+c=3[/imath] ? Please help me to solve this. Thanks in advance. | 2163597 | Let [imath]a,b,c\in \Bbb R^+[/imath] such that [imath](1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16[/imath]. Find [imath](a+b+c)[/imath]
Let [imath]a,b,c\in \Bbb R^+[/imath] such that [imath](1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16[/imath]. Find [imath](a+b+c)[/imath]. I computed the whole product ;If [imath](a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16[/imath]. Unable to view how to proceed further. Please help. |
1110538 | How to determine the acute angle bisector of given two lines
If we have two lines given by the equations: [imath]ax+by+c=0[/imath] [imath]px+qy+r=0[/imath] We know that the two angle bisectors are represented by the equations [imath]\frac{ax+by+c}{\sqrt{a^2+b^2}}=\pm\frac{px+qy+r}{\sqrt{p^2+q^2}}[/imath] Is there any simple way to distinguish between the two angle bisectors? That is, can we tell which sign will give the acute angle bisector and which the obtuse? I tried using the fact that the angle between the acute angle bisector and one of the lines will be less than 45 degrees,but the formula for that is too hairy. I remember reading somewhere that if [imath]c, r[/imath] are of the same sign and [imath]ap +bq[/imath] is negative then the positive sign gives the acute cangle bisector. I tried it out for some values in a graphing programme and it does seem to work. But I was not able to come up with a proof. | 2403530 | How do I prove this method of determining the sign for acute or obtuse angle bisector in the angle bisector formula works?
The formula for finding the angular bisectors of two lines [imath]ax+by+c=0[/imath] and [imath]px+qy+r=0[/imath] is [imath]\frac{ax+by+c}{\sqrt{a^2+b^2}} = \pm\frac{px+qy+r}{\sqrt{p^2+q^2}}[/imath] I understand the proof of this formula but I do not understand how to determine which sign is for acute bisector and which one for obtuse. I can find the angle between a bisector and a line, and if it comes less than [imath]45^\circ[/imath] then it is acute bisector. But that is a lengthy method and involves calculation. My book says, if [imath]ap+bq[/imath] is positive then the negative sign in the formula is for acute bisector. I want a proof of this method. Edit: Using the method for finding the position of two points with respect to a line is okay for the proof. |
2442921 | If a prime (of the form [imath]4k+3[/imath]) divides the sum of two squares, then it divides each of the two numbers
Statement : A prime [imath]p[/imath] is of the form : [imath]3 \pmod 4[/imath]. Show that if [imath]p \mid a^2+b^2[/imath], then [imath]p \mid a[/imath] and [imath]p\mid b[/imath]. I don't know exactly how to proceed. I got the following proof from somewhere: Assuming that [imath]p\mid a^2+b^2[/imath] and [imath]p\nmid a[/imath], we get that [imath]\exists k \ni ak\equiv 1 \pmod p \implies (ak)^2+(bk)^2\equiv k^2(a^2+b^2)\equiv 0 \pmod p \implies (bk)^2 \equiv -1 \pmod p[/imath]. And this is a contradiction. Please explain me why [imath](bk)^2\equiv -1 \pmod p[/imath] leads to a contradiction. | 2391516 | Primes that can't be written as sum of two squares. Mod
I came to this conclusion: the primes that can't be written as the sum of two squares (lets use[imath]p\prime[/imath] to denote) can be written as [imath]{p}\prime \equiv 3 \;(\bmod\; 4)[/imath]. I need a simple proof to show this and am stuck. I would appreciate guidance! |
2443484 | Integer Solutions of [imath]x^4+y^4=z^2[/imath]
Show that there do not exist any non-trivial integer solutions of [imath]x^4+y^4=z^2[/imath] This was asked in a number theory test, so I tried showing that there can't exist any Pythagorean Triple of the form [imath](x^2, y^2, z)[/imath] and using possible congruence modulos with [imath]3, 4, 5[/imath] etc. but every time I get some case that remains to be disproved. Any help or hint is appreciated. | 2087679 | Proving [imath]x^4+y^4=z^2[/imath] has no integer solutions
I need solution check to see if I overlooked something: If [imath]x^4+y^4=z^2[/imath] has an integer solution then [imath](\frac{x}{y})^4+1=(\frac{z}{y^2})^2[/imath] has a solution in rationals. Second equation is equivalent to [imath]X^4+1=Y^2[/imath] which can be written as cubic Weierstrass form [imath]v^2=u^3-4u[/imath] where [imath]u=\frac{2(Y+1)}{X^2}[/imath] and [imath]v=\frac{4(Y+1)}{X^3}[/imath]. By using SAGE we can see that Mordell-Weil group of this elliptic curve is empty (SAGE returns empty set, but I'm not sure if that is enough to conclude this), so we can conclude that there are no rational solutions to the second equation and no integer solution to the first one. edit: No integer solution in positive integers. |
2433851 | Let [imath]\frak{g}[/imath] a nilpotent Lie álgebra. Prove that there is an ideal [imath]\frak{h}\subseteq\frak{g}[/imath] such that [imath]\dim(\frak{g}) = \dim(\frak{h}) + 1[/imath]
Could someone give me a suggestion to solve this problem? Let [imath]\frak{g}[/imath] a nilpotent Lie álgebra. Prove that there is an ideal [imath]\frak{h}\subseteq\frak{g}[/imath] such that [imath]\dim(\frak{g}) = \dim(\frak{h}) + 1[/imath] | 374032 | Solvable Lie algebra with codimension 1 ideal
There is an exercise in Humphreys's An Introduction to Lie Algebras and Representation Theory: "Any nilpotent Lie algebra contains a codimension 1 ideal". The proof I am thinking of is the following. Suppose the Lie algebra [imath]L[/imath] satisfies [imath]L\neq[L,L][/imath]. Then [imath]\pi\colon L\to L/[L,L][/imath] is a projection onto an abelian algebra. Choose a codimension 1 subspace [imath]I\subset L/[L,L][/imath]. Then [imath]I[/imath] is an ideal and its preimage [imath]\pi^{-1}(I)[/imath] is necessarily an ideal in [imath]L[/imath], because any subspace containing the derived algebra is an ideal. Moreover, [imath]\pi^{-1}(I)[/imath] has codimension 1. [imath]\square[/imath] The only condition on [imath]L[/imath] is that [imath]L\neq[L,L][/imath], which holds for solvable Lie algebras as well. Is this proof correct? How come Humphreys didn't set the exercise with "nilpotent" replaced by "solvable"? |
2443540 | Let [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] be integers. Prove that if [imath]4|(a+bc)[/imath] and [imath]6|(b+ac)[/imath], then [imath]2|(a^2-b^2)[/imath].
This is my work so far. How can I isolate the [imath](a^2-b^2)[/imath]? I am not asking for a full solution, just a hint and next step. All done! Thanks Michael. | 2442542 | If [imath]4\mid a+bc[/imath] and [imath]6\mid b+ac[/imath] prove that [imath]2\mid a^2-b^2[/imath]
If [imath]4\mid a+bc[/imath] and [imath]6\mid b+ac[/imath] prove that [imath]2\mid a^2-b^2[/imath] This is as far as I get: [imath]a + bc = 4k\qquad\text{for some $k\in\Bbb Z$} \\b+ac =6l\qquad\text{for some $l\in\Bbb Z$}\\\implies (a^2-b^2)(1-c^2) = 16k^2 + 36l^2[/imath] |
1349249 | How do I evaluate [imath]\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}[/imath]?
How do i evaluate this limit : [imath]\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+............+\frac{1}{\sqrt{n}}\right)[/imath] ? Thank you for any help . | 1815371 | Evaluate the limit [imath]\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)[/imath]
Evaluate the limit [imath]\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)[/imath], without using a Riemann sum [imath]\bf{My\; Try:}[/imath] Using the graph of [imath]\displaystyle f(x) = \frac{1}{\sqrt{x}}\;,[/imath] we get [imath]\int_{1}^{n+1}\frac{1}{\sqrt{x}}dx <\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<1+\int_{1}^{n}\frac{1}{\sqrt{x}}dx[/imath] So we get [imath]2\sqrt{n+1}-2<\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<2\sqrt{n}-1[/imath] So [imath]\lim_{n\rightarrow \infty}\frac{2\sqrt{n+1}-2}{\sqrt{n}}=2<\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)<\lim_{n\rightarrow \infty}\frac{2\sqrt{n}-1}{\sqrt{n}} = 2[/imath] So using the Sandwich Theorem, we get [imath]\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)=2[/imath] My question is can we solve it by using any other method? If yes then please explain it here. |
2443931 | If [imath]R[/imath] is commutative, the left [imath]R-[/imath]module can be given the structure of right [imath]R-[/imath]module.
The definition of the module is given in Hungerford's Algebra by: And I can not understand the following paragraph, could anyone explain this for me please especially why commutativity is needed for (iii)? Thanks!! | 773402 | Modules over commutative rings
I'm trying to get my head around modules, and there's a problem that's bothering me regarding scalar multiplication from the left vs from the right. In many books/articles I've read, the author may refer to a left [imath]R[/imath]-module [imath]M[/imath], and then continue to talk of terms such as [imath]xr[/imath], where [imath]x\in M[/imath], [imath]r\in R[/imath]. What is right scalar multiplication in a left module? Does it even make sense to talk about it? I've been trying to show that over a commutative ring, right scalar multiplication and left scalar multiplication are the same thing, but I've not managed it so far. I've googled it to no avail, and I can't see why the distinction between left and right modules disappears when the base ring is commutative. Thanks for any replies! |
2444000 | Sum of squares of integers question
Use congruence's to prove the following result. If [imath]n \cong 3 [/imath] mod 4, then n cannot be written as the sum of the squares of two integers. Im not even really sure where to start this a hint would be great. | 280345 | Proof that if [imath]p\equiv3\,\left(\mbox{mod 4}\right)[/imath] then [imath]p[/imath] can't be written as a sum of two squares
I'd appreciate your help showing that if [imath]p\equiv3\,\left(\mbox{mod 4}\right)[/imath] then p can't be written as a sum of two squares. Thanks! |
2444109 | Use mean value theorem to find [imath]f(\xi)+f''(\xi)=0[/imath]
Question: [imath]f\in C^2(-\infty,+\infty)[/imath], [imath]|f(x)|\leq 1[/imath], [imath][f(0)]^2+[f'(0)]^2=4[/imath]. Prove that there exists [imath]\xi[/imath] s.t. [imath] f(\xi)+f''(\xi)=0. [/imath] I find this question in the chapter about mean value theorem, and I set up function [imath]G=f^2+f'^2[/imath] as usual. WLOG set [imath]f(0)\geq 0[/imath] and [imath]f'(0)\geq 0[/imath]. Use Taylor mean value theorem I get [imath] f(1)=f(0)+ f'(0)+\frac{1}{2}f''(\eta). [/imath] Set [imath]x=f(0)\geq 0[/imath] I get [imath] f''(\eta)\leq 1-x-\sqrt{4-x^2}\leq -1. [/imath] So I get [imath] f(\eta)+f''(\eta)\leq 0 [/imath] I want to prove by contradition， then I assume [imath]\forall x\in(-\infty,+\infty)[/imath], [imath]f(x)+f''(x)<0[/imath]. Then I have no idea. My question is: Is this idea correct? How to use the function [imath]f^2+f'^2[/imath](I haven't use it yet)? | 717060 | If [imath]\{f(0)\}^{2} + \{f'(0)\}^{2} = 4[/imath] then there is a [imath]c[/imath] with [imath]f(c) + f''(c) = 0[/imath]
This is from Putnam: If a function [imath]f: \mathbb{R} \to [-1, 1][/imath] is such that [imath]f''(x)[/imath] exists for all [imath]x \in \mathbb{R}[/imath] and [imath]\{f(0)\}^{2} + \{f'(0)\}^{2} = 4[/imath] then prove that there is a point [imath]c[/imath] such that [imath]f(c) + f''(c) = 0[/imath]. Now the condition concerning values of [imath]f(0), f'(0)[/imath] (and the fact that range of [imath]f[/imath] is subset of [imath][-1, 1][/imath]) implies that [imath]f'(0) \neq 0[/imath] so that function is not a constant. If we take [imath]g(x) = \{f(x)\}^{2} + \{f'(x)\}^{2}[/imath] then we can see that [imath]g' = 2f'(f + f'')[/imath]. One hope of solving the problem is to show that [imath]g[/imath] attains local extrema at some point different from an extrema of [imath]f[/imath]. This would ensure that [imath]g'[/imath] vanishes without making [imath]f'[/imath] vanish and that will lead to [imath]f + f'' = 0[/imath]. However the constraint on [imath]f(0), f'(0)[/imath] does not help in analyzing the extrema of [imath]f[/imath] or [imath]g[/imath]. Another line of thought which could be helpful here is to consider [imath]F(x) = f'(x)\sin x - f(x)\cos x[/imath] so that [imath]F'(x) = \sin x\{f(x) + f''(x)\}[/imath] but in this case I am not able to see how to use the condition on values [imath]f(0), f'(0)[/imath] to impose some constraint on [imath]F(x)[/imath]. Perhaps both the approaches which came to my mind are not in right direction (or may be they are, but I am unable to see). Please provide any hints. Update: Based on the answer by Sandeep Thilakan, I have dropped the requirement for continuity of [imath]f''(x)[/imath]. |
2443482 | If [imath]x,y,z[/imath] are positive numbers, then prove that [imath]\frac {x}{x+y} + \frac{y}{y+z} +\frac {z}{z+x} \le 2[/imath]
If [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] are positive numbers, then prove that [imath]\frac {x}{x+y} + \frac{y}{y+z} +\frac {z}{z+x} \le 2.[/imath] Though I have solved a lot of problems on AM-GM inequality, I am unable to solve this one. I am also not showing my working because I do not think that they will be of any help. | 980751 | Proof of Nesbitt's Inequality: [imath]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}[/imath]?
I just thought of this proof but I can't seem to get it to work. Let [imath]a,b,c>0[/imath], prove that [imath]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}[/imath] Proof: Since the inequality is homogeneous, WLOG [imath]a+b+c=1[/imath]. So [imath]b+c=1-a[/imath], and similarly for [imath]b,c[/imath]. Hence it suffices to prove that [imath]\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}[/imath]. From [imath]a+b+c=1[/imath] and [imath]a,b,c>0[/imath] we have [imath]0<a,b,c<1[/imath], so we have [imath]\frac{a}{1-a}=a+a^2+a^3+...[/imath] and similarly for [imath]b,c[/imath], so it suffices to prove that [imath]\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}[/imath], or equivalently (by [imath]a+b+c=1[/imath]) [imath]\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}[/imath], where [imath]\sum_{cyc} a=a+b+c[/imath] similarly for [imath]\sum_{cyc}a^n=a^n+b^n+c^n[/imath]. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the [imath]a^2+b^2+c^2\ge (a+b+c)^2/3[/imath] inequality. "stuff below": Now, from [imath]0<a<1[/imath] we have [imath]a^3>a^4[/imath], [imath]a^5>a^8[/imath], [imath]a^7>a^8[/imath], [imath]a^9>a^{16}[/imath], and so on, so it suffices to prove that [imath]\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}[/imath], or, multiplying by 2, [imath]2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1[/imath], which by a simple inequality (i.e. recursively using [imath]a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}[/imath]) is equivalent to [imath]2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1[/imath]. But then http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B5859879%7D+2%5Ei+*%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed. |
2444553 | Finding the limit of [imath]\left(\sum\limits_{k=2}^n\frac1{k\log k}\right)-\left(\log \log n\right)[/imath]
There is this well known limit: [imath]\lim_n \sum_{k=1}^n \frac 1k -\log n=\gamma[/imath] Where [imath]\log[/imath] is the natural logarithm and [imath]\gamma[/imath] is Euler constant. I was wondering if my similar situation yelds to a similar result: [imath]\lim_n \sum_{k=2}^n \frac 1{k\log k}-\log \log n=?[/imath] I know there is a formula (always due to Euler) which can be used in those situations but I can't see if there is a way to put the result in "closed form" rather than having only an approximation (supposing the limit exists and is finite at first) | 1584319 | What is known about the 'Double log Eulers constant', [imath]\lim_{n \to \infty}{\sum_{k=2}^n\frac{1}{k\ln{k}}-\ln\ln{n}}[/imath]?
The Euler constant is defined as [imath]\gamma = \lim_{n \to \infty}{\sum_{k=1}^n\frac{1}{k}-\ln{n}}[/imath] Let [imath]q = \lim_{n \to \infty}{\sum_{k=2}^n\frac{1}{k\ln{k}}-\ln\ln{n}}[/imath] I managed to prove that [imath]\frac{1}{3\ln{3}}+\frac{1}{2\ln{2}}-\ln\ln{3} \geq q \geq \frac{1}{2\ln{2}}-\ln\ln{3}[/imath] Is there something known about the constant [imath]q[/imath]? For instance, is [imath]q[/imath] expressible in terms of [imath]\gamma[/imath]? |
2172441 | Hint needed to show that [imath]a\cos t+b\sin t\leq \sqrt{a^2+b^2}[/imath] for [imath]t\in [0,2\pi)[/imath]
And that the upper bound is achieved for some choice of [imath]\theta[/imath]. This exercise shows up in the Cauchy-Schwarz section of a textbook I am looking through but I don't see how to apply CS to prove. I would prefer a hint towards how to use this ineq. specifically. Through standard techniques, you can see that the maximum of [imath] f(t)=a\cos t+b\sin t-\sqrt{a^2+b^2} [/imath] occurs for [imath]\arctan\frac{a}{b}[/imath] provided [imath]t\ne \frac{\pi}{2},\frac{3\pi}{2}[/imath]. Not sure if I can do much from there though. | 1834356 | Prove: [imath]|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}[/imath]
[imath]|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}[/imath] I have tried: [imath]|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}[/imath] enough to prove: [imath]|a+b|\leq \sqrt{a^2+b^2}[/imath] But I can find how to continue from here |
2445022 | If [imath]f[/imath] is convex then [imath]f\left(\sum_{i=1}^n\lambda_i x_i\right) \leq \sum_{i=1}^n\lambda_i f (x_i)[/imath]
Show by induction that if [imath]f:\mathbb{R}\to\mathbb{R}[/imath] is convex, then for any [imath]x_1,\dots ,x_n[/imath] and [imath]\lambda_1,\dots ,\lambda_n[/imath] with [imath]\sum_{i=1}^n\lambda_i = 1[/imath], [imath] f\left(\sum_{i=1}^n\lambda_i x_i\right) \leq \sum_{i=1}^n\lambda_i f (x_i) . [/imath] I think this is similar to the Jensen's inequality, but am not sure how to formally proof this. Thank you so much! | 1278761 | Prove [imath]f(\sum^k_{i=1} \alpha_i x_i) \leq \sum^k_{i=1} \alpha_i f(x_i) [/imath] for a convex function f
I'm learning for an exam in calculus and have come across this question which I can't seem to prove: Let [imath]C \subseteq V[/imath] be a convex set. Let there be a function >[imath]f:C\rightarrow \mathbb{R}[/imath] a convex function. Now let there be scalars [imath]\alpha_1,\alpha_2 \dots \alpha_k >0[/imath] such that >[imath]\sum^k_{i=1} \alpha_i = 1[/imath], and also let there be a group of points >[imath]x_1,x_2\dots x_k \in C[/imath]. Show that the following inequality exists: [imath]f(\sum^k_{i=1} \alpha_i x_i) \leq \sum^k_{i=1} \alpha_i f(x_i) [/imath] I thought about using induction to show the inequality, whereby the basis k=2 is the actual definition of a convex function i.e: [imath]f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y)[/imath] I've ran into a problem using induction, since the scalars won't sum up to 1 in the induction hypothesis. I also don't really use the fact that C is convex, which I'm probably supposed to... Either there is some algebraic trick that can be done so I can use induction, or I'm doing something wrong (and induction is not the way to go with this inequality) Any help will be appreciated, thank you! |
2444540 | Showing that the winding number is continuous
This is an exercise from Conway I am stuck at. Here [imath]n([/imath][imath]\gamma[/imath] [imath],[/imath] [imath]w[/imath]) is the winding number and [imath](6.2)[/imath] is like below. I need to prove that [imath]h[/imath] is continuous using only the definition of winding number, not Cauchy Theorem. In fact I have to use the result of the exercise 9 to prove the Cauchy Theorem. So, I have to use some [imath]\delta[/imath] [imath]-[/imath] [imath]\epsilon[/imath] argument. But, I can't find a way to do so. Could anyone help me how to solve the exercise 9? The definition of winding number is like below. | 1025633 | Winding numbers are continuous: The proof was too easy
There's a question in my complex analysis book: Let [imath]G[/imath] be a region and let [imath]\gamma_0[/imath] and [imath]\gamma_1[/imath] be two closed smooth curves in [imath]G[/imath]. Suppose [imath]\gamma_0\sim\gamma_1[/imath] and [imath]\Gamma[/imath] is a homotopy between them. Also suppose that [imath]\gamma_t(s)=\Gamma(s,t)[/imath] is smooth for each [imath]t[/imath]. If [imath]w\in \mathbb{C}-G[/imath], define [imath]h(t)=n(\gamma_t;w)[/imath] and show that [imath]h:[0,1]\rightarrow\mathbb{Z}[/imath] is continuous. This is an exercise right after the section that proves the following version of Cauchy's theorem: If [imath]\gamma_0[/imath] and [imath]\gamma_1[/imath] are two closed rectifiable curves in [imath]G[/imath] and [imath]\gamma_0 \sim \gamma_1[/imath] then [imath]\int_{\gamma_{0}}f=\int_{\gamma_1} f[/imath] for every function [imath]f[/imath] analytic on [imath]G.[/imath] I must be missing something. It seems like you can just say the following: 'Proof:' Since [imath]\frac{1}{z-w}[/imath] is analytic in [imath]G[/imath] ([imath]w[/imath] isn't in [imath]G[/imath]), and [imath]\gamma_0\sim\gamma_t\ \forall t\in[0,1][/imath] (consider a rescaled version of [imath]\Gamma[/imath]), then by Cauchy's theorem, [imath]n(\gamma_t;w)=n(\gamma_0;w)=n(\gamma_1;w)[/imath]. Thus, the winding number is constant along [imath]t[/imath] and thus certainly continuous. My problem with that is that this doesn't require the assumption that [imath]\Gamma[/imath] is a smooth homotopy. I think the author had meant for the reader to work harder for this result. What is the proof 'supposed' to look like? |
2444918 | Obtaining the equation of circle in a 3-dimentional space.
My thinking: To obtain the equation of a circle in a three dimentional space we need to get the intersection of a sphere and a plane. According to me this can be done by : [imath](equation\ of\ sphere)^2+(equation \ of \ plane)^2=0[/imath] but to test this I don't have the adequate software. So my question is: Am I thinking in the right way? If not, then where I am going wrong? If yes, then can you help me with a sample of the graph of the equation. Thank you :) | 1184038 | What is the equation of a general circle in 3-D space?
I know that [imath](x-x_0)^2+(y-y_0)^2-r^2=0[/imath] is a general planar circle and [imath](x-x_0)^2+(y-y_0)^2+(z-z_0)^2-r^2=0[/imath] is a general sphere. I want to know the general expression of a circle in space. Can anyone please give me some advice here? |
2444498 | How can I prove this by induction? [imath]\sum _{k=1}^n\left(\frac{1}{\sqrt{k}}\right)\ge 2\left(\sqrt{n+1}-1\right)[/imath]
I am given this expression and I have to prove it using mathematical induction [imath]{\displaystyle \forall }n\ge 1 [/imath] , [imath] \sum _{k=1}^n\left(\frac{1}{\sqrt{k}}\right)\ge 2\left(\sqrt{n+1}-1\right)[/imath] I have proved it is true for [imath]n=1[/imath], then suposed it was true until [imath]n[/imath] and wanted to prove it was also true until [imath]n+1[/imath]. I have operated a bit and arrived to this expression I want to prove: [imath]{\frac{1}{2\sqrt{n+1}}}\ge (\sqrt{n+2}-\sqrt{n+1})[/imath] , and I don't know how to continue to clear that inequality to prove it. Can you help me please?? (I have just started studying this method so I'm a bit lost :/ ) | 1515400 | Proving a summation inequality with induction
The exact question: Prove: [imath]\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}}\gt2(\sqrt{n+1}-1)[/imath] I have looked at similar problems but still don't understand how to prove this inequality by induction. So far I have this: Induction basis: Let n=1 [imath]\displaystyle\sum_{k=1}^n \frac{1}{\sqrt{k}} = \frac{1}{\sqrt{1}} = 1 > 2(\sqrt{1+1}-1) = ~.828[/imath] [imath]1>.828[/imath] So it proves the inequality true when n=1. Now i really don't know how to continue even with all the examples i have browsed through. One of them i came across showed that the induction hypothesis should let P(n) equal the equation above and do something with P(n+1). I am not looking for the answer I just need help on how to continue with the problem. What other steps are necessary for me to complete this proof by induction. |
2445507 | Does there exist a function satisfying the given conditions?[TIFR Mathematics GS-2017]
Does there exist a function [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] satisfying [imath]f(1) = 1[/imath], [imath]f(-1) = -1[/imath] and [imath]|f(x)-f(y)|\leq|x-y|^{3/2}[/imath] [imath]\forall x,y\in\mathbb{R}[/imath]? I tried polynomial functions, [imath]\sin(\frac{\pi}{2}x)[/imath] etc. None is fitting the third condition. In competitive exams, searching for functions is a bad idea. Can anyone give advice? What should I do when I get this type of question? How to prove or disprove? Please explain. | 361400 | Function on [imath][a,b][/imath] that satisifies a Hölder condition of order [imath]\alpha > 1 [/imath] is constant
I want to show that if a function [imath]f:[a,b]\rightarrow \mathbb R[/imath] satisfies a Hölder condition of order [imath]\alpha > 1 [/imath] then it is constant. The way I think of it is as follows: [imath]|f(x) - f(y)| < K|x-y|^\alpha[/imath] [imath]\frac{|f(x) - f(y)|} {|x-y]} < K|x-y|^{\alpha -1}[/imath] [imath]\lim_{y\rightarrow x} \frac{|f(x) - f(y)|} {|x-y]} \le \lim_{y\rightarrow x} K|x-y|^{\alpha -1} =0 [/imath] As the limit is [imath]0[/imath], we can remove the modulus, so we get: [imath]\lim_{y\rightarrow x} \frac{f(x) - f(y)} {x-y} = 0[/imath] So [imath]f[/imath] is derivable and [imath]f'(x) = 0[/imath] for all [imath]x[/imath] in [imath][a,b][/imath]. Note that the reason we can add the limit [imath]y\rightarrow x[/imath] is because [imath][a,b][/imath] is closed in [imath]\mathbb R[/imath]. However, the question gives as a hint using the mean value theorem. I am not sure why one should do that. You would first have to prove that [imath]f[/imath] is derivable in a similar manner to what I did, and then prove that [imath]f[/imath] is constant. Or is there a simpler way to do it and I am missing it? Also please inform me of any mistakes I did in the proof (if any)/ Thank you! |
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