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741446 | An almost impossible limit
The following limit appeared in a qualification exam: Find the limit of [imath]\lim_{x \to 0} \left( \frac{\tan (\sin (x))-\sin (\tan (x))}{x^7} \right).[/imath] I ended up doing it in Mathematica, is there any other way? Thanks in advance! | 548832 | How find this limit [imath]\lim\limits_{x\to 0^{+}}\frac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}[/imath]
find the limit [imath]\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}[/imath] My try:since [imath]\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+o(x^7)[/imath] [imath]\tan{x}=x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\dfrac{1}{63}x^7+o(x^3)[/imath] so [imath]\sin{(\tan{x})}=\tan{x}-\dfrac{1}{3!}(\tan{x})^3+\dfrac{1}{5!}(\tan{x})^5-\dfrac{1}{7!}(\tan{x})^7+o(x^7)[/imath] But this methods can solve,and I think this problem have nice methods,Thanks. |
742286 | Proof of the Absolute Least Remainder Algorithm
I have managed to grasp and construct the elementary proofs of the division theorem and the modified version were [imath]0 \leq r < \mid b \mid [/imath], but I am not quite there yet with the absolute least remainder algorithm. Theorem: If [imath]a, b \in \mathbb{Z} [/imath] and [imath]a, b \neq 0[/imath], then there are unique [imath]q, r \in \mathbb{Z} [/imath] such that: [imath]a = bq +r, ~~~~ -\frac{|b|}{2} < r \leq \frac{|b|}{2}[/imath] Proof: (here is what I got so far) There are three things to prove: existence, r conditions and uniqueness. Consider these two cases: [imath]0 \leq r < \frac{1}{2} |b| [/imath] [imath]\frac{1}{2} |b| \leq r < |b|[/imath] We apply the original division algorithm to a. In the first case, we have: [imath]a = |b|q' + r[/imath] where [imath]q' = q \frac{|b|}{b} \in \mathbb{Z}[/imath] and [imath]r[/imath] is in the appropriate range. In the second case, we have: [imath]a = |b|q' + r'[/imath] where [imath]q' = (q+1) \frac{|b|}{b} \in \mathbb{Z}[/imath] and [imath]r' = r - |b|[/imath]. This satisfies the r' criteria i.e. [imath]-\frac{1}{2}|b| \leq r' < 0[/imath]. However, I am not sure if this is correct in the first place, or how to connect this to the overarching proof structure of the division algorithm. | 653932 | Prove that if a and b are integers, then there are unique integers q and r such that [imath]a = bq + r[/imath], [imath]-|b|/2 < r \le |b|/2[/imath]
Prove that if a and b are integers, then there are unique integers q and r such that [imath]a = bq + r,[/imath] with the restriction that[imath]-|b|/2 < r \le |b|/2[/imath] |
743010 | Can we have this situation? If so, what examples are there of it?
Let [imath]G[/imath] be a group, let [imath]F[/imath] be a normal subgroup of [imath]G[/imath], and let [imath]E[/imath] be a normal subgroup of [imath]F[/imath]. That is, let [imath]E[/imath] be a normal subgroup of [imath]F[/imath], and let [imath]F[/imath] in turn be a normal subgroup of [imath]G[/imath]. Is [imath]E[/imath] normal in [imath]G[/imath]? If not necessarily, then what example illustrates this situation? | 255274 | Are normal subgroups transitive?
Suppose [imath]G[/imath] is a group and [imath]K\lhd H\lhd G[/imath] are normal subgroups of [imath]G[/imath]. Is [imath]K[/imath] a normal subgroup of [imath]G[/imath], i.e. [imath]K\lhd G[/imath]? If not, what extra conditions on [imath]G[/imath] or [imath]H[/imath] make this possible? Applying the definitions, we know [imath]\{ghg^{-1}\mid h\in H\}=H[/imath] and [imath]\{hkh^{-1}\mid k\in K\}=K[/imath], and want [imath]\{gkg^{-1}\mid k\in K\}=K[/imath]. Clearly, the best avenue for a counterexample is if [imath]gkg^{-1}\not\in K[/imath] for some [imath]k\in K[/imath] and [imath]g\in G-H[/imath]. If no such element exists, [imath]\{gkg^{-1}\mid k\in K\}\subseteq K[/imath] implies [imath]\{gkg^{-1}\mid k\in K\}=K[/imath] because if [imath]k'\in K[/imath], [imath]gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}[/imath]. |
21530 | If a product of relatively prime integers is an [imath]n[/imath]th power, then each is an [imath]n[/imath]th power
Show that if [imath]$n$[/imath], [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] are positive integers with [imath]\gcd(a, b) = 1[/imath] and [imath]ab = c^n[/imath], then there are positive integers [imath]d[/imath], and [imath]e[/imath] such that [imath]a = d^n[/imath] and [imath]b = e^n[/imath]. I know that (by Bezout) [imath]$\gcd\left(a,b\right) = 1$[/imath] implies [imath]$ax + by = 1$[/imath] for some integers [imath]$x$[/imath] and [imath]$y$[/imath], and also that [imath]$\gcd\left(a^n,b^n\right) = 1$[/imath], but this does not help me. | 2348813 | [imath]\gcd=1[/imath] if [imath] m^n=ab \Rightarrow m^n=(a'b')^n[/imath]?
For [imath]a,b,n,m\in \Bbb N[/imath], if [imath]a[/imath] and [imath]b[/imath] are coprime, than for [imath]m^n=a\cdot b[/imath] [imath]\exists a',b'\in \Bbb N: a=a'^n \land b=b'^n[/imath] I thought it would suffices to show that [imath]m=a'\cdot b'[/imath], but I'm not sure that's why I also thought of : As [imath]a,b[/imath] are coprime than we can write [imath]\begin{align}a&=p_1^{u_1}\cdots p_i^{u_i}\\b&=q_1^{v_1}\cdots q_j^{v_j}\end{align}[/imath] where [imath]p_g\neq q_g \forall g[/imath] so[imath]\quad m^n=a\cdot b=p_1^{u_1}\cdots p_i^{u_i}\cdot q_1^{v_1}\cdots q_j^{v_j}[/imath] As the [imath]u_i,v_i[/imath] doesn't necessary need to be distinct, we can rewrite to [imath]m^n=(p_1q_1)^{n_1}\cdots(p_iq_j)^{n_{ij}}[/imath] we define [imath]m:=\prod p_hq_h[/imath] and [imath]n:=\prod n_{ij}[/imath] Not sure how to continue or either it's correct or not :( |
743113 | Finding the summation of the series.
Is there any formula to find out the summation of the series. [imath]\sum_{i=1}^{n} \lfloor \frac{n}{i} \rfloor[/imath] Can someone help me with this. | 740442 | How do I evaluate this sum(involving the floor function)?
[imath] \sum_{i=1}^N\left\lfloor\frac{N}{i}\right\rfloor [/imath] Is there a closed form expression to the above sum? (Mathematica doesn't give me anything) |
743291 | [imath](f_ng_n)[/imath] that does not converge uniformly to [imath]fg[/imath]
Suppose that the sequences [imath](f_n)[/imath] and [imath](g_n)[/imath] converge uniformly to respective functions [imath]f[/imath] and [imath]g[/imath] on some domain [imath]S[/imath]. Can someone provide me with a counterexample (preferably with proof) of sequences [imath](f_n)[/imath] and [imath](g_n)[/imath] converge uniformly to [imath]f[/imath] and [imath]g[/imath] on [imath]S[/imath] but [imath](f_ng_n)[/imath] does NOT converge uniformly to [imath]fg[/imath] on [imath]S[/imath]. | 256323 | Does [imath]\{f_ng_n\}\to fg[/imath] uniformly?
Suppose that [imath]\{f_n\}[/imath] and [imath]\{g_n\}[/imath] are [imath]2[/imath] sequences of unbounded functions on [imath]S\subset\mathbb{R}[/imath] for infinitely many [imath]n\in\mathbb{N}[/imath] such that [imath]\{f_n\}\to f[/imath] and [imath]\{g_n\}\to g[/imath] uniformly. Does [imath]\{f_ng_n\}\to fg[/imath] uniformly? I think it is not possible but I couldn't think of a counter example. |
742473 | Tensor product of quotient rings
[imath]A[/imath] is a commutative ring with unit and [imath]\mathfrak a[/imath], [imath]\mathfrak b[/imath] ideals. I have to show that [imath]A/\mathfrak a \otimes_{A} A/\mathfrak b \cong A/(\mathfrak{a+b}).[/imath] Any hint ? | 720507 | A proof using Yoneda lemma
Martin Brandenburg pointed out elsewhere in the comments that he could give a one line proof, using the Yoneda lemma, of [imath]\frac{\mathbf{C}[x_1,\ldots,x_{n+m}]}{I(X)^e+I(Y)^e} \cong \frac{\mathbf{C}[x_1,\ldots,x_n]}{I(X)} \otimes_\mathbf{C} \frac{\mathbf{C}[x_{n+1},\ldots,x_{n+m}]}{I(Y)}[/imath] (for [imath]X,Y[/imath] affine algebraic varieties), but apparently the proof was too long for his margin. How can this be done? Fundamental question aside: why is there no abstract nonsense tag? |
743508 | question on surds i already asked this question but the answer I got did not match the one in the book
[imath]\sqrt{ 3x }= x + \sqrt {3}[/imath] Give x in the form [imath]A \sqrt {B} + C [/imath] Can you show me how this is done step by step. The answer I have in the book is: [imath]\frac {1}{2} \sqrt{3} + \frac {3}{2} [/imath] this is where I got stuck: [imath] \frac {x^2 +2x \sqrt{3} +3}{3x} [/imath] | 703006 | another question on surds and how to use math symbols in this site
[imath]\sqrt{ 3x }= x + \sqrt {3}[/imath] this is what i tried [imath]\sqrt{ x }= (x + \sqrt {3})^2\\ = x^2 + 3 [/imath] Give x in the form [imath]A \sqrt {B} + C [/imath] Can you show me how this is done step by step. The answer I have in the book is: [imath]\frac {1}{2} \sqrt{3} + \frac {3}{2} [/imath] |
743682 | show that if [imath]\displaystyle\lim_{n \to \infty} f(n+x)=0[/imath] then [imath]\displaystyle\lim_{x \to \infty}f(x)=0[/imath]
Let [imath]f : \left[0,\infty\right]\to \mathbb R[/imath] be uniformly continuous. If [imath]\displaystyle\lim_{n \to \infty} f(n+x)=0[/imath] where [imath]x[/imath] is in [imath][0,1][/imath] then [imath]\displaystyle\lim_{x \to \infty}f(x)=0[/imath] Furthermore if [imath]f[/imath] is just continuous , what happens to the result? | 738292 | [imath]\lim_{n→∞} f(n+x)=0[/imath] results in [imath] \lim_{x→∞} f(x)=0[/imath]
[imath]f\colon [0,\infty) \to \mathbb{R}[/imath] is uniformly continuous. If [imath]\lim_{n\to \infty}f(n+x)=0[/imath] where [imath]x\in [0,1][/imath], then [imath]\lim_{x\to\infty} f(x)=0[/imath]. every my trying to solve this is wrong :( how can i prove this? and if [imath]f[/imath] is just continuous, what happens to the results |
744093 | How many surjective functions?
Let [imath]A[/imath] and [imath]B[/imath] be sets with cardinalities m and n respectively where [imath]m \ge n[/imath] how many surjective functions are there from [imath]A[/imath] to [imath]B[/imath]? Support your answer I have no idea how to go about this one. Any help would be appreciated. Thanks in advance! | 500674 | Number of surjective functions from A to B
Am I on the right track? I am not sure about my reasoning... Number of surjective functions from [imath]A[/imath] to [imath]B[/imath] [imath]A = \{1,2,3,4\} ; B = \{a,b,c\}[/imath] We must count the surjective functions, meaning the functions for which for all [imath]b \in B[/imath], [imath]\exists~a \in A[/imath] such that [imath]f(a) = b[/imath], [imath]f[/imath] being one of those functions. In order for a function [imath]f:A\rightarrow B[/imath] to be a surjective function, all 3 elements of [imath]B[/imath] must be mapped. We need to count how many ways we can map those 3 elements. We will subtract the number of functions from [imath]A[/imath] to [imath]B[/imath] which only maps 1 or 2 elements of [imath]B[/imath] to the number of functions from [imath]A[/imath] to [imath]B[/imath] (computed in 4.c : 81). Only 1 element of [imath]B[/imath] is mapped The first [imath]a \in A[/imath] has three choices of [imath]b \in B[/imath]. The others will then only have one. Total functions from [imath]A[/imath] to [imath]B[/imath] mapping to only one element of [imath]B[/imath] : 3. Exactly 2 elements of [imath]B[/imath] are mapped Similarly, there are [imath]2^4[/imath] functions from [imath]A[/imath] to [imath]B[/imath] mapping to 2 or less [imath]b \in B[/imath]. However, these functions include the ones that map to only 1 element of [imath]B[/imath]. So there are [imath]2^4-3 = 13[/imath] functions respecting the property we are looking for. In the end, there are [imath](3^4) - 13 - 3 = 65[/imath] surjective functions from [imath]A[/imath] to [imath]B[/imath]. |
716983 | Show that [imath]L^1\subsetneq (L^\infty)^*[/imath]
How does one show that [imath]L^1\subsetneq (L^\infty)^*[/imath]? I am having trouble in this. Any help would be appreciated. | 711489 | Show that [imath]L^1[/imath] is strictly contained in [imath](L^\infty)^*[/imath]
How does one show that [imath]L^1[/imath] is strictly contained in [imath](L^\infty)^*[/imath]? Here, [imath](L^\infty)^*[/imath] is the space of linear continuous functionals on [imath]L^\infty[/imath]. |
744571 | How does the empty set work?
I'm having trouble understanding what exactly an empty set is. Does [imath]\varnothing[/imath] mean [imath]\{\}[/imath] ? and what is the difference between [imath]\varnothing[/imath] and [imath]\{\varnothing\}[/imath] ? If someone could shed some light on this, and provide a couple of more examples to help me learn what this weird concept is, I'd greatly appreciate it. Thanks! | 302064 | Empty set does not belong to empty set
Herbert in his book "Elements of set theory" on page no 3 says that we can form the set [imath] \{ \emptyset \} [/imath] whose only member is [imath]\emptyset [/imath]. Note that [imath] \{ \emptyset \} \neq \emptyset [/imath], because [imath] \emptyset \in \{ \emptyset \} [/imath] but [imath]\underline{ \emptyset \notin \emptyset} [/imath]· By the last argument [imath]\emptyset \notin \emptyset[/imath], is he saying that empty set is not a member of or does not belong to empty set OR it is a typo and he wanted it to be [imath] \{\emptyset \} \notin \emptyset [/imath], that set containing empty set is not a member of empty set |
743130 | Property of an irreducible Gaussian integer
[imath]a\in\Bbb{Z}[i][/imath] is an irreducible element such that [imath]a[/imath] divides [imath]p[/imath] ([imath]p[/imath] is a prime number such that [imath]p \equiv 1 \pmod 4[/imath]). Then show that [imath]N (a) = p[/imath]. [imath]N (\alpha) = a^2 + b ^ 2[/imath] with [imath]\alpha = a + ib[/imath] Does somebody know the proof of that ? | 739023 | Question about irreducible Gaussian integers
Let [imath]\pi \in \mathbb{Z}[i][/imath] irreducible and [imath]\pi[/imath] divides [imath]p[/imath] ([imath]p\in\Bbb Z[/imath] prime). Then show that [imath]N(\pi)=p[/imath]. ([imath]N(\alpha)= a^2 +b^2[/imath] where [imath]\alpha = a+ib[/imath].) |
745081 | Characters and conjugacy classes
This comes up in reading David Speyer's answer to this question. Given a finite group [imath]G[/imath] and two non-conjugate elements [imath]x, y,[/imath] how does one construct a unitary representation [imath]\rho[/imath] of [imath]G[/imath] such that [imath]\rho(x)[/imath] and [imath]\rho(y)[/imath] have different traces? (The same question makes sense for infinite groups, but it is far from clear that this is always possible in the infinite setting, even if you drop "unitary"). | 189900 | Some irreducible character separates elements in different conjugacy classes
Let [imath]x[/imath] and [imath]y[/imath] be elements that are not conjugate in [imath]G[/imath]. Then there is some irreducible character [imath]\chi[/imath] such that [imath]\chi(x) \not = \chi(y)[/imath]. Clearly the "irreducible" part isn't important, since any character can be written as the sum of irreducible characters, but I'm having trouble going beyond that. I'd appreciate a good hint over a full answer, and I'd be most interested in a way to construct a group representation [imath]\varphi:G \to GL(V)[/imath] of [imath]G[/imath] such that the character of the representation takes different values on [imath]x[/imath] and [imath]y[/imath]. |
745517 | Geometrical interpretation of tan(x) > x
Is there a geometrical interpretation for [imath]tan(x) > x[/imath] when [imath]x \in (0, \frac{\pi}{2})[/imath] in the unit circle? I can't picture it since [imath]x[/imath] is the angle (although it makes sense when you graph the two functions). | 98998 | Why [imath]x<\tan{x}[/imath] while [imath]0?[/imath]
In proof of [imath]\displaystyle\lim_{x\rightarrow0}\frac{\sin{x}}{x}=1[/imath] is assumed that [imath]\sin{x}\leq{x}\leq\tan{x}[/imath] while [imath]0<x<\frac{\pi}{2}[/imath]. First comparison is clear, arc length must be greater than sine value, but how about [imath]x\leq\tan{x}[/imath], why tangent is longer than arc? |
745674 | A basis for a nilpotent endomorphism
Let [imath]E[/imath] be a complex vector space of dimension 3. Let [imath]f[/imath] be a non zero endomorphism such that [imath]f^2=0[/imath]. I want to show that there is a basis [imath]B=\{b_1,b_2,b_3\}[/imath] of [imath]E[/imath] such that [imath]f(b_1)=0, f(b_2)=b_1,f(b_3)=0[/imath] Edit Here is how i see the answer now: [imath]f[/imath] being non zero there exists [imath]x_0\in E[/imath] such that [imath]f(x_0)\not =0[/imath]. Let [imath]M=span\{f(x_0),x_0\}[/imath]. Since [imath]f^2=0[/imath] we show easily that [imath]f(x_0)[/imath] and [imath]x_0[/imath] are linearly independent hence they form a basis for [imath]M[/imath]. We take [imath]b_1=f(x_0)[/imath], [imath]b_2=x_0[/imath]. Take any [imath]z\not \in M[/imath]. If [imath]z\in \ker f[/imath] then take [imath]b_3=z[/imath]. If [imath]z\not \in \ker f[/imath] then there exists [imath]\beta \not = 0[/imath] such that [imath]f(z)=\beta f(x_0)[/imath] (because [imath]\dim(Im(f))=1[/imath] hence it is spanned by any non zero vector, we take [imath]f(x_0)[/imath] as a spanning vector). Take [imath]z'=\dfrac{1}{\beta}z-f(x_0)[/imath] hence [imath]z'\in \ker f[/imath] and we take [imath]b_3=z'[/imath]. | 742048 | Finding a specific basis for an endomorphism
Let [imath]E[/imath] be [imath]\mathbb C[/imath]-vector space of dimension [imath]3[/imath]. Let [imath]f[/imath] be non zero endomorphism of [imath]E[/imath] such that [imath]f^2=0[/imath]. show that there exists a basis [imath]B=\{b_1,b_2,b_3\} [/imath] of [imath]E[/imath] where the matrix of [imath]f[/imath] is given by : [imath]\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}[/imath] My try : since [imath]f[/imath] is non zero, there exists [imath]x\in E[/imath] such that [imath]f(x)\not =0[/imath]. Take [imath]b_1=f(x)[/imath] and [imath]b_2=x[/imath]. Now [imath]b_1,b_2[/imath] can easily shown to be linearly independent. By the exchange lemma we can find [imath]b_3[/imath] in another generating system of [imath]E[/imath] to make [imath]b_1,b_2,b_3[/imath] into a basis of [imath]E[/imath]. Now clearly [imath]f(b_1)=f^2(x)=0[/imath] and [imath]f(b_2)=f(x)=b_1[/imath]. Now how to find a [imath]b_3[/imath] such that [imath]f(b_3)=0[/imath] and [imath]b_1,b_2,b_3[/imath] is a basis of [imath]E[/imath]? Thank you for your help. Edit: For the linear independence of [imath]b_1,b_2[/imath], if [imath]\alpha x+\beta f(x)=0[/imath] then we apply [imath]f[/imath] to get [imath]\alpha f(x)=0[/imath] and since [imath]f(x)\not =0[/imath] then [imath]\alpha=0[/imath] it remains that [imath]\beta f(x)=0[/imath] and again this gives that [imath]\beta=0[/imath]. |
746067 | Symbol for [imath]\left\{ x \in \mathbb{R} : x > 0 \right\}[/imath]
Is there a symbol for the following set? [imath] \left\{ x \in \mathbb{R} : x > 0 \right\} [/imath] | 27968 | How does one denote the set of all positive real numbers?
What is the "standard" way to denote all positive (or non-negative) real numbers? I'd think [imath] \mathbb R^+ [/imath] but I believe that that is usually used to denote "all real numbers including infinity". So is there a standard way to denote the set [imath] \{x \in \mathbb R : x \geq 0\} \; ?[/imath] |
745898 | What is the difference between these two propositions?
My text says: Let Evens be the set of even integers greater than 2, and let Primes be the set of primes. Then we can write Goldbach’s Conjecture in logic notation as follows: [imath] \forall n \in Evens. \exists p,q \in Primes. n = p + q [/imath] Then later it reports: Swapping quantifiers in Goldbach’s Conjecture creates a patently false state- ment; namely that every even number 2 is the sum of the same two primes: [imath] \exists p,q \in Primes. \forall n \in Evens. n = p+q [/imath] What exactly is the difference between these two notations? They look the same to me, with the same quantifiers -- just written in a different order | 493102 | Nested Quantifiers true or false
I have a concern with nested quantifiers. I have: [imath] \forall x \exists y \forall z(x^2-y+z=0) [/imath] such that [imath] x,y,z \in \Bbb Z^+[/imath] My first question, can it be read like this: [imath] \forall x \forall z \exists y(x^2-y+z=0) [/imath] The way I did it, is I started off with [imath]x=1, z=1 [/imath] [imath] 2-y = 0 [/imath] [imath] y =2 [/imath] Is this a good approach? |
746066 | Volume form for [imath]SE(n)[/imath] and/or [imath]E(n)[/imath].
I wonder what happens when you construct the Tiling spaces considering the natural action of [imath]SE(n)[/imath] or [imath]E(n)[/imath] rather than [imath]\mathbb R^n[/imath]. In order to do that, I need to understand both the riemannian metric, and the volume form of the considered group. Are they unique? Could you suggest an example based reference on Lie groups? | 85514 | metric on the Euclidean Group
I am not an expert in this so I hope this doesn't sound so stupid: what is the common metric used when studying the Euclidean Group [imath]\mathrm{E}(3)[/imath]. One could probably ask the same thing for (homogenous) transformation group [imath]\mathrm{SE}(3)[/imath]. For the latter I have seen use of Riemannian metric. For both I am guessing (as this is not my field) that there are some kind of manifold structure involved, so I can't help to think that there are also good metrics (by good I mean metric that is considered "practical" when one considers rigid body motions). Furthermore, is there a way to visualize the manifolds from [imath]\mathrm{E}(3)[/imath] and [imath]\mathrm{SE}(3)[/imath]? Any good reference in this area? |
746571 | How is [imath]f_{2000}[/imath] equals [imath]0[/imath]?
Let [imath](F_n)_{n\ge 1}[/imath] be the sequence of numbers defined by [imath]F_1=1=F_2[/imath]; [imath]F_{n+1}=F_{n}+F_{n-1}[/imath] for [imath]n\geq2.[/imath] Let [imath]f_n[/imath] be the remainder left when [imath]F_n[/imath] is divided by [imath]5[/imath]. Then [imath]f_{2000}[/imath] equals [imath]0[/imath]. My Try: [imath]F_3=2,F_4=3,F_5=5,F_6=8,F_7=13,\dots[/imath] Then I subtracted [imath]S_n=1+1+2+3+5+8+13+\dots[/imath] from [imath]S_n=1+1+2+3+5+8+13+\dots[/imath] and got [imath]0=1+0+1+1+2+3+5+8+13+\dots[/imath] (Sorry, I can't [imath]\LaTeX[/imath] that properly.) But, how to go further? How can we reach to conclusion that [imath]f_{2000}=0[/imath]? | 344716 | Find remainder of [imath]F_n[/imath] when divided by [imath]5[/imath]
Let [imath]\{ F_n\}[/imath] be the sequence of numbers defined by [imath]F_1=1=F_2;\, F_{n+1}=F_n+F_{n-1}[/imath] for [imath]n \geq 2[/imath]. Let [imath]f_n[/imath] be the remainder left when [imath]F_n[/imath] is divided by [imath]5[/imath]. Then [imath]f_{2000}[/imath] equals (A) [imath]0[/imath] [imath]~~~~~~~~~~~~~~~~[/imath] (B) [imath]1[/imath] [imath]~~~~~~~~~~~~~~~~[/imath] (C) [imath]2[/imath][imath]~~~~~~~~~~~~~~~~[/imath] (D) [imath]3[/imath] I found that [imath]F(1)=1[/imath], [imath]F(2)=1[/imath], [imath]F(3)=2[/imath], [imath]F(4)=3[/imath], [imath]F(5)=5[/imath], [imath]F(6)=8[/imath], [imath]F(7)=13[/imath], [imath]F(8)=21[/imath], [imath]F(9)=34[/imath], and [imath]F(10)=55[/imath]. But I need a systematic pattern to find [imath]F_n[/imath]. |
746750 | Prove that [imath]e^x \gt 0[/imath] for [imath]x \in \mathbb{R}[/imath]
This is a consequence of the exponential rule, but how do I actually prove it to be true? | 626845 | Prove that [imath]\exp(x)>0[/imath] using only formal definition of exp
This problem would be easy if I could use the fact that [imath]\exp(x)=e^x[/imath], but I have to use the following definition: [imath]\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}[/imath] I can also use the fact that [imath]\exp(x+y)=\exp(x)\exp(y)[/imath] So how do I prove, using those two equations, that [imath]\forall x\in \mathbb{R}:\exp(x)>0 [/imath] I mean, I can't just use the definition, because if [imath]x<0[/imath] then it isn't so obvious that [imath]\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}>0[/imath]. Can someone give me a hint or two? Thanks! |
747378 | For an infinite set [imath]A[/imath], does [imath]|A| = |A \times A|[/imath]?
I know that [imath]|\mathbb{N}| = |\mathbb{N}^2|[/imath], and that [imath]|\mathbb{R}| = |\mathbb{R}^2|[/imath]. It seems like this might be true for all sets, but I don't know how to go about proving this. It's easy to prove that [imath]|A| \leq |A \times A|[/imath] (see this question). So to prove the claim it suffices to show that there exists some injective function [imath]f : A \times A \to A[/imath]. For [imath]A = \mathbb{N}[/imath], you can use e.g. [imath]f:p, q \mapsto (p+q)^2 + p[/imath]. For [imath]A = \mathbb{R}[/imath] you can construct various space-filling curves. But I'm not sure how such constructive solutions can be generalized, if at all. Any suggestions on how to approach this? | 180671 | Cardinality of cartesian square
Given an infinite set [imath]A[/imath] - does the cardinality of [imath]A[/imath] equal to the cardinality of [imath]A^2[/imath]? |
747991 | How to prove this indentity [imath]\binom{100}{0}^2-\binom{100}{1}^2+\binom{100}{2}^2-...-\binom{100}{99}^2+\binom{100}{100}^2=\binom{100}{50}[/imath]
I don't know how to prove this identity: [imath]\binom{100}{0}^2-\binom{100}{1}^2+\binom{100}{2}^2-\binom{100}{3}^2+...-\binom{100}{99}^2+\binom{100}{100}^2=\binom{100}{50}[/imath] | 664769 | Show that [imath]\sum\limits_{k=0}^n\binom{2n}{2k}^{\!2}-\sum\limits_{k=0}^{n-1}\binom{2n}{2k+1}^{\!2}=(-1)^n\binom{2n}{n}[/imath]
How can I prove the identity: [imath] \sum_{k=0}^n\binom{2n}{2k}^2-\sum_{k=0}^{n-1}\binom{2n}{2k+1}^2=(-1)^n\binom{2n}{n}? [/imath] Maybe, can we expand [imath] f(x)=(1+x)^{2n}? [/imath] Thank you. |
727852 | In Triangle, [imath]\sin\frac A2\!+\!\sin\frac B2\!+\!\sin\frac C2\!-\!1\!=\!4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4[/imath]
To prove [imath]\sin\frac A 2+\sin\frac B 2+\sin\frac C 2-1=4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4[/imath] My approach : [imath] \begin{align} \text{L.H.S.} & = \sin\frac{A}{2}+\sin \frac{B}{2}+\sin\frac{C}{2} -1 \\[8pt] & = 2\sin\frac{A+B}{4}\cos\frac{A-B}{4}+\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)-1 \\[8pt] & = 2\sin\frac{\pi -C }{4}\cos\frac{A-B}{4} - 2\sin^2\left(\frac{\pi -C}{4} \right) \\[8pt] & =2\sin\frac{\pi -C }{4}\left\{ \cos\frac{A-B}{4} - \sin\left(\frac{\pi -C}{4} \right)\right\} \end{align} [/imath] Unable to move further please help. thanks. | 1897839 | If [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] are the angles of a triangle,....
If [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] are the angles of a triangle, prove that: [imath]sin\frac {A}{2}+sin\frac {B}{2}+sin\frac {C}{2}=1+4sin\frac {B+C}{4}.sin\frac {C+A}{4}.sin\frac {A+B}{4}[/imath] My attempt; Here, [imath]A+B+C=\pi[/imath] Now, [imath]L.H.S=sin\frac {A}{2}+sin\frac {B}{2}+sin\frac {C}{2}[/imath] [imath]= 2sin\frac {\frac {A}{2}+\frac {B}{2}}{2}.cos\frac {\frac {A}{2}-\frac {B}{2}}+sin\frac {C}{2}[/imath] [imath]=2sin\frac {A+B}{4}.cos\frac {A-B}{4}+sin\frac {C}{2}[/imath] [imath]=2sin\frac {\pi-C}{4}.cos\frac {A-B}{4}+sin\frac {C}{2}[/imath] Now, from here I could not.continue, please help me to.complete the proof.. |
748624 | Finding sum of the power series and the sum of the series
(1) Find the sum of the power series [imath]\sum_{n=1}^{\infty} nx^n[/imath] (2) Find the sum of the series [imath]\sum_{n=1}^{\infty} \frac{n}{3^n}[/imath] Any tips on solving the sum of series/power series? | 568701 | Power series representation/calculation
I am struggling a bit with power series at the moment, and I don't quite understand what this question is asking me to do? Am I meant to form a power series from these, or simply evaluate that series? Any explanation/working is appreciated. Using power series representation, calculate [imath]\sum_{n=1}^\infty \frac{n2^n}{3^n}.[/imath] |
748678 | Question about Big O notation for asymptotic behavior in convergent power series
Examples of such use of Big O notation can be found for instance on Wolfram Alpha here. More details on the Wikipedia page. The idea, as I understand it, is that the term between parenthesis in Big O bounds the rest of the series asymptotically (as [imath]x[/imath] goes to zero or infinity) giving and elegant way of writing the series (if you don't like "[imath]+ \dots[/imath]" or "etc") while also providing an error approximation. Taking the example of [imath]e^x[/imath]: [imath] e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+O(x^5) [/imath] When we write [imath]O(x^5)[/imath] we mean that all the rest of the terms (like [imath](x^6)/6!+(x^7)/7!+(x^8)/8!+\dots[/imath]) are bounded by [imath]x^5[/imath] as [imath]x[/imath] goes to [imath]0[/imath]. Giving a more concrete example if we take [imath]e^2.42[/imath] then this is more or less [imath]1 + 2.42 + (1/2)(2.42)^2 + (1/6)(2.42)^3 + (1/24)(2.42)^4[/imath] and the error between the approximation and the actual value is no larger than [imath]2.42^5[/imath]. The question is how do we know this? Certainly [imath]x^5[/imath] bounds, as [imath]x[/imath] goes to [imath]0[/imath], each term individually ([imath]Cx^n=O(x^5)[/imath] as [imath]x\rightarrow 0[/imath] as long as [imath]n>5[/imath], [imath]C[/imath] constant) but how do we prove the sum of all the individual terms is bounded? We have to show that: [imath] O(x^5)=\frac{x^6}{6!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^9}{9!}+\frac{x^{10}}{10!}+\frac{x^{11}}{11!}+...),x\rightarrow 0 [/imath] Falling back on the previous example, maybe the infinite sum will actually be slightly larger than [imath]2.42^5[/imath]. How do we prove this is actually not the case? While I have used the example of [imath]e^x[/imath], feel free to give a more general proof, and not one for this particular function only. Your help with this issue will be greatly appreciated. Thank you for taking the time to read this and have a good day. | 748657 | Question about Big O notation for asymptotic behavior in convergent power series
Examples of such use of Big O notation can be found for instance on Wolfram Alpha here. More details on the Wikipedia page. The idea, as I understand it, is that the term between parenthesis in Big O bounds the rest of the series asymptotically (as [imath]x[/imath] goes to [imath]0[/imath] or [imath]\infty[/imath]) giving and elegant way of writing the series (if you don't like "+ ..." or "etc") while also providing an error approximation. Taking the example of [imath]e^x[/imath]: [imath] e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+O(x^5) [/imath] When we write [imath]O(x^5)[/imath] we mean that all the rest of the terms (like [imath]\frac{x^6}{6!}+\frac{(x^7)}{7!}+\frac{x^8}{8!}+\cdots+\frac{x^y}{y!}[/imath]) are bounded by [imath]x^5[/imath] as [imath]x[/imath] goes to [imath]0[/imath]. Giving a more concrete example if we take [imath]e^{2.42}[/imath] then this is more or less [imath]1 + 2.42 + (\frac12)(2.42)^2 + (\frac16)(2.42)^3 + (\frac{1}{24})(2.42)^4[/imath] and the error between the approximation and the actual value is no larger than [imath]2.42^5[/imath]. The question is how do we know this? Certainly [imath]x^5[/imath] bounds, as [imath]x[/imath] goes to [imath]0[/imath], each term individually ([imath]Cx^n=O(x^5)[/imath] as [imath]\lim_{x\to\infty}[/imath] as long as [imath]n>5, C[/imath] constant) but how do we prove the sum of all the individual terms is bounded? We have to show that: [imath] O(x^5)=\frac{x^6}{6!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^9}{9!}+\frac{x^{10}}{10!}+\frac{x^{11}}{11!}+...),x\rightarrow 0 [/imath] Falling back on the previous example, maybe the infinite sum will actually be slightly larger than [imath]2.42^5[/imath]. How do we prove this is actually not the case? While I have used the example of [imath]e^x[/imath], feel free to give a more general proof, and not one for this particular function only. Your help with this issue will be greatly appreciated. Thank you for taking the time to read this and have a good day. Note: I wrongly asked this on Math Overflow and copied it here. I'm not sure whether the other question gets transferred or if I made the right choice in reasking here. |
744502 | Prove or find a counter example to the claim that for all sets A,B,C if A ∩ B = B ∩ C = A ∩ C = Ø then A∩B∩C ≠ Ø
This is a homework question of mine that Ive answered and I'd really like some feedback on my solution. Prove or find a counter example to the claim that for all sets [imath]A,B,C[/imath], [imath]\text{if }A\cap B=B\cap C=A\cap C=\emptyset \text{ then } A\cap B\cap C\neq \emptyset[/imath]. Since I can only either prove it is true, OR find a counter example I have gone with the latter because I think this statement is not true. My counter example is as follows: Let [imath]A = \{1,2,3 \}[/imath] , [imath]B = \{4,5,6 \}[/imath] and [imath]C = \{7,8,9 \}[/imath]. It is then clear that [imath]A ∩ B = B ∩ C = A ∩ C = \emptyset[/imath] . If we let [imath]x ∈ A\cap B\cap C[/imath] then [imath]x[/imath] must be in [imath]A[/imath] and [imath]B[/imath] and [imath]C[/imath]. Since this cannot be possible because there are no common elements then this implies that [imath]A∩B∩C[/imath] has to be equal to a null set i.e [imath]A∩B∩C = \emptyset[/imath] and thereby showing that (at least for this scenario) if [imath]A \cap B = B \cap C = A \cap C = \emptyset[/imath] then [imath]A\cap B\cap C = \emptyset[/imath]. Is this okay? Any feedback will be greatly appreciated thank you! | 754604 | sets and finding counterexample from empty set
Question Find a counter-example to the claim that for all sets [imath]A,B,C[/imath] if [imath]A\cap B=B\cap C=A\cap C=\emptyset[/imath] then [imath]A\cap B\cap C\neq\emptyset[/imath]. If i have to find a counter-example then It should be an example or fact that is inconsistent with a hypothesis and may be used in argument against it. Can i get help here please. |
749295 | Is there would be any matrices [imath]A[/imath] & [imath]B[/imath] where the relation [imath]AB[/imath]-[imath]BA[/imath]=[imath]I[/imath] holds?
Here , the actual question is to find any matrices [imath]A[/imath] and [imath]B[/imath] such that [imath]AB-BA=I[/imath] relation holds. but actually, I dont think that we could not find any such matrices. As, the diagonal elements of the matrices [imath]A[/imath] and [imath]B[/imath] would be same. so, after the subtraction, the diagonal elements of the matrics [imath]AB-BA[/imath] would be all [imath]0[/imath]. so, the trace of the matrics would be actually zero, where in the right hand side, the trace of the matrics [imath]I[/imath] would be [imath]n[/imath]. so, we could not find any such matrices [imath]A[/imath] and [imath]B[/imath].Is my approach is correct? | 284901 | [imath]AB-BA=I[/imath] having no solutions
The following question is from Artin's Algebra. If [imath]A[/imath] and [imath]B[/imath] are two square matrices with real entries, show that [imath]AB-BA=I[/imath] has no solutions. I have no idea on how to tackle this question. I tried block multiplication, but it didn't appear to work. |
742824 | Total Time a ball bounces for from a height of 8 feet and rebounds to a height 5/8
I am trying to solve the following problem. Let's say a ball is dropped from a height of [imath]8[/imath] feet and rebounds to a height [imath]\frac{5}{8}[/imath] of its previous height at each bounce keeps bouncing forever since it takes infinitely many bounces. This is not true! We examine this idea in this problem. A. Show that a ball dropped from a height [imath]h[/imath] feet reaches the floor in [imath]\frac{1}{4}\sqrt{h}[/imath] seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times: This my my answer for part A. I am not 100% if it is correct. time at first bounce =[imath]\frac1{4}\sqrt{8}[/imath] time at second bounce =[imath]\frac1{4}\sqrt{8} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)}[/imath] time at third bounce =[imath]\frac1{4}\sqrt{8} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)^2}[/imath] time at fourth bounce = [imath]\frac1{4}\sqrt{8} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)^2} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)^3}[/imath] B. How long, in seconds, has the ball been bouncing when it hits the floor for the nth time (find a closed form expression)? time at nth bounce = This is my answer for part B. Hopefully it is correct. [imath]\frac1{4}\sqrt{8\left(\frac{5}{8}\right)^{n-1}}[/imath] C. What is the total time that the ball bounces for? I am unsure how to solve this part of the problem. I would appreciate some help on how to solve it. | 1510748 | Need help with finding out when rubber ball is at rest.
in the infinite series section of my book i have the problem: suppose a rubber ball, when dropped from a given height, returns a fraction p of that height. in the absence of air resistance, a ball dropped from a height h requires [imath]\sqrt{\left(\frac{2h}{g}\right)}[/imath] seconds to fall to the ground, where g[imath]\approx{9.8 \frac{m}{s^2}}[/imath] is the acceleration due to gravity. the time it takes to bounce up to a given height equals the time to fall from that height to the ground. how long does it take a ball dropped from 10 m. to come to rest. I am completely stumped with this problem and have no idea on how to do it. can anyone give me a nudge in the right direction? |
749605 | Show that if [imath]p[/imath] is a prime number [imath]> 3[/imath] then [imath]24 \mid p^2-1[/imath]
Hi guys can someone help me with this ?(Without using Modular arithmetic) Show that if [imath]p[/imath] is a prime number [imath]>3[/imath] then [imath]24[/imath] [imath]\mid[/imath] [imath]p^2-1[/imath] | 855 | For any prime [imath]p > 3[/imath], why is [imath]p^2-1[/imath] always divisible by 24?
Let [imath]$p>3$[/imath] be a prime. Prove that [imath]$24 \mid p^2-1$[/imath]. I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know. |
12285 | Isometry in compact metric spaces
Why is the following true? If [imath](X,d)[/imath] is a compact metric space and [imath]f: X \rightarrow X[/imath] is non-expansive (i.e [imath]d(f(x),f(y)) \leq d(x,y)[/imath]) and surjective then [imath]f[/imath] is an isometry. | 749716 | Surjective function on a compact metric space
Assume [imath]f:K\rightarrow K[/imath], is surjective and [imath]K[/imath] is a compact metric space and we have [imath]d(f(x),f(y))\leq d(x,y)\, \forall x,y\in K[/imath]. How can I prove that [imath]d(f(x),f(y))= d(x,y)\, \forall x,y\in K[/imath]? Thank you for your help. |
750119 | Linear Independence of Powers of "roots vector"
Let us be working over the field of complex numbers. Suppose [imath]f(x)= a_n x^n + \cdots +a_1 x + a_0[/imath] is a degree [imath]n[/imath] polynomial with [imath]n[/imath] distinct roots [imath]z_1,\ldots,z_n[/imath]. Is the following matrix always invertible? [imath]\left( \begin{array}{ccc} 1 & 1 & \cdots & 1 \\ z_1 & z_2 & \cdots & z_n \\ \vdots & \vdots & & \vdots \\ z_1^{n-1} & z_2^{n-1} & \cdots & z_n^{n-1} \end{array} \right) [/imath] Motivation: I was learning how to solve linear recurrence relations with constant coefficients, and the standard method involves writing down the characteristic polynomial and then solving for the roots. After which, we use the initial conditions to solve for the constants ->this is the step which I am wondering about: is it always possible to do that? This quesiton amounts to the same question above. Edit: I guess I can extend my question since I am concerned with general linear recurrence relations with constant coefficients. If the roots of the characteristic polynomial are repeated, then we use [imath]z^n, nz^n, n^2z^n,...[/imath] for our "guesses", so the general matrix to invert is something like: [imath]\left( \begin{array}{ccc} 1 & 0 & \cdots & 1 \\ z_1 & z_1 & \cdots & z_k \\ \vdots & \vdots & & \vdots \\ z_1^{n-1} & (n-1)z_1^{n-1} & \cdots & z_k^{n-1} \end{array} \right) [/imath] How do we know this matrix is invertible? | 275202 | Vandermonde Determinant
This is an exercise from Ian Stewart's Galois Theory, [imath]3^{rd}[/imath] edition: If [imath]z_1,z_2,\ldots,z_n[/imath] are distinct complex numbers, show that the determinant [imath]D=\left|\begin{array}[cccc] 11&1&\cdots&1\\ z_1&z_2&\cdots&z_n\\ z_1^2&z_2^2&\cdots&z_n^2\\ \vdots&\vdots&\ddots&\vdots\\ z_1^{n-1}&z_2^{n-1}&\cdots&z_n^{n-1} \end{array}\right|[/imath] is nonzero. Hint: Consider the [imath]z_j[/imath] as independent indeterminates over [imath]\Bbb C[/imath]. Then [imath]D[/imath] is a polynomial in the [imath]z_j[/imath], of total degree [imath]0+1+\cdots+(n-1)=\frac{1}{2}n(n-1)[/imath]. Moreover, [imath]D[/imath] vanishes whenever [imath]z_j=z_k[/imath] for [imath]k\neq j[/imath], as it then has two identical rows. Therefore [imath]D[/imath] is divisible by [imath]z_j-z_k[/imath] for all [imath]j\neq k[/imath], hence it is divisible by [imath]\prod_{j<k}(z_j-z_k)[/imath]. Now compare degrees. My question is how to do this. I follow the hint and I'm thinking of [imath]D[/imath] as a cofactor expansion; is that the idea? Also, the fact that [imath]D[/imath] vanishes whenever [imath]z_j=z_k[/imath] for [imath]k\neq j[/imath] follows since it would have two identical columns, right? I don't see why [imath]z_j=z_k[/imath] would imply it has two identical rows, as the hint suggests. The fact that it is divisible by [imath](z_j-z_k)[/imath] is easy since that would make [imath]z_k[/imath] a root for [imath]k=j[/imath], but I'm not quite sure I follow why the degrees would be different. I think the degree of the product would be [imath]1+2+\cdots+(k-1)[/imath], but then add these up for all [imath]k<n[/imath]? I appreciate any clarification that you can provide! |
750676 | [imath]\sum x_n[/imath] is divergent , [imath] \lim x_n=0[/imath] and the partial sums of [imath]\sum x_n[/imath] are bounded
Does there exist a divergent series [imath]\sum x_n[/imath] such that [imath] \lim x_n=0[/imath] and the partial sums of [imath]\sum x_n[/imath] are bounded ? | 450708 | Proof or Counterexample on the Convergence of a Series
So one of my professors proposed a problem to me and it has stumped me for some time now. Here's how it goes: Suppose you have a sequence [imath]a_n[/imath] of real numbers such that [imath]\lim_{n\to\infty} a_{n} = 0[/imath] and suppose the sequence of partial sums [imath]s_n[/imath] is bounded. Prove that [imath]s_n[/imath] converges or give a counterexample. I'm hoping to figure this out without anyone handing me the complete solution, so if someone could point me in the right direction with a hint, it would be much appreciated. |
751175 | Integrating this complex function, using Residue Theorem
I am having a massive amount of trouble integrating this, I really have no clue how to get the answer in the book: [imath]\int_{-\infty}^{\infty} \frac{x^4}{1+x^8}dx[/imath] I know I need to find the poles on this function, which is basically the value of [imath]x^8 = -1[/imath] or I could split it up like the following: [imath]x^8 + 1 = (x^4 + i)(x^4 - i)[/imath] With this I got the following poles (looking at [imath]x^4 - i[/imath]), all have an order one: \begin{array} & x_1 = e^{\frac{i\pi}{8}} \\ x_2 = e^{\frac{3i\pi}{8}} \\ x_3 = e^{\frac{5i\pi}{8}} \\ x_4 = e^{\frac{7i\pi}{8}} \end{array} Now to calculate the residues I am trying to make use of the result that given a rational function [imath]\frac{F}{G}[/imath] such that both of them are analytic on a disk of radius [imath]r[/imath] with [imath]G(z_0) = 0[/imath] but [imath]G'(z_0) \neq 0[/imath] then we know that: [imath]Res(\frac{F}{G};z_0) = \frac{F(z_0)}{G'(z_0)}[/imath] With that I am having difficulty getting results that are useful, I maybe calculating wrong and I am not sure but I have been spending a long time on this one question. The solution in the book is [imath]\frac{\pi}{4}[sin(\frac{3\pi}{8}]^{-1}[/imath]. I would like a step by step solution to this or at least some guidance because I really need to learn this and I am not sure how to do this properly. Thank you! | 747136 | Compute the Integral
Compute the integral. [imath]\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx[/imath] The answer at the back of the book is [imath]\frac{\pi}{4\sin(\frac{3\pi}{8})}[/imath] |
751283 | Is there a asymptotic formula for product of primes?
[imath]P(x)=\prod_{p\leq x}p[/imath] As you can see P(x) represents the product of primes which are not greater than x. Is there a asymptotic formula for this? | 156509 | Product of all prime numbers upto some prime [imath]p[/imath]
Let [imath]p[/imath] be a prime number. Denote by [imath]P[/imath] the set of all primes which are not greater than [imath]p[/imath]. Is there a well known estimation of the product of all prime numbers in [imath]P[/imath] (i.e. [imath]\prod_{q\in P}q[/imath])? |
751568 | Lipschitz continuity in two variables
Prove that [imath]y \mapsto f(x,y)[/imath] is Lipschitz continuous, where [imath]f(x,y) = \frac{y}{x} \ln{\frac{y}{x}}, \ \ \ |x-1| \leq \frac{1}{2}, |y-1| \leq \frac{1}{2}e[/imath] I tried to solve this, but I find it very difficult to do this in two variables. How do I start solving this? My first attempt was trying to solve this for the four cases ([imath]x > \frac{3}{2}, y > \frac{1}{2}e + 1[/imath]) etc. But this didn't work out very well. | 750540 | Proving a function is Lipschitz continuous
Show that the following function is Lipschitz continuous and find a Lipschitz constant [imath]y\mapsto f(x,y)\\ f(x,y)=\frac{y}{x}\ln(\frac{y}{x})\text{ , } |x-1|\leq\frac{1}{2}\text{ , } |y-e|\leq\frac{e}{2} [/imath] I have no clue on where to begin to prove this. My first question is how I should interpret this function, is [imath]x[/imath] a sort of constant? |
751635 | Prove that [imath]f'_{xy}=f'_{yx}[/imath]
Here is a basic, and probably a bad, question. A fundamental rule of derivatives. Why is [imath]f'_{xy}=f'_{yx}[/imath] true? | 541896 | When is [imath]f_{xy}(x,y)\neq f_{yx}(x,y)?[/imath]
When is [imath]f_{xy}(x,y)\neq f_{yx}(x,y)?[/imath], where [imath]f_{xy}[/imath] and [imath]f_{yx}[/imath] denote the mixed (second) partial derivatives of a multivariable function [imath]z=f(x,y)[/imath]. |
751754 | How to prove that e^x is convex?
For convexity : [imath]e^{ta+(1-t)b} \leq t \cdot e^a+(1-t)e^b \Rightarrow e^{(a-b)^t} \leq te^{a-b} + 1 - t[/imath] Now I'm stuck on how to solve it further. | 702241 | How to prove that [imath]e^x[/imath] is convex?
I need a help with proving convexity of [imath]e^x[/imath] function. [imath]f(x)[/imath] is a convex function, If it satisfies following condition. [imath]f(c\cdot x_1+(1-c)\cdot x_2) ≤ c\cdot f(x_1)+(1-c)\cdot f(x_2), \quad 0 ≤ c ≤ 1[/imath] |
554836 | Sum of self power
Is there a formula to calculate the sum of a number to the power of this same number, like: [imath]1^1 + 2^2 + 3^3 + 4^4 + 5^5 + ... + n^n[/imath]? or [imath]x^x + (x+1)^{(x+1)} + (x+2)^{(x+2)} + ... + (x+n)^{(x+n)}[/imath] | 1076595 | Calculate [imath]1^1 + 2^2 + 3^3 + ... + n^n[/imath]
Is there a formula to calculate [imath]1^1 + 2^2 + 3^3 + ... + n^n[/imath] I searched but didn't find a formula for increasing powers |
751872 | Solving an Integral With Square Root in the Denominator
Solve for k: [imath]\int_6^{16}\frac{1}{\sqrt{(x^3 + 7x^2 + 8x -16)}}dx = \frac{\pi}{k}[/imath] I have factored the denominator to obtain [imath]x^3 + 7x^2 + 8x - 16 = (x-1)(x+4)^2[/imath] I think that since the answer involves [imath]\pi[/imath], I have to somehow make it look like the derivative of arcsine or something. Or should I do a trigonometric substitution? Please help! | 751286 | Determining k: [imath]\int_{6}^{16} \frac{dx}{\sqrt{x^3 + 7x^2 + 8x - 16}} = \frac{\pi}{k}[/imath]
I have a calculus II final coming up and this question came up in a past final exam: [imath]\int_{6}^{16} \frac{dx}{\sqrt{x^3 + 7x^2 + 8x - 16}} = \frac{\pi}{k},[/imath] where [imath]k[/imath] is a constant. Find [imath]k[/imath]. My progress so far: [imath]\int_{6}^{16} \frac{dx}{\sqrt{(x - 1)}(x + 4)} = \frac{\pi}{k}[/imath] The answer is: [imath]k = 6\sqrt{5}[/imath] I do not know where to from from this step. Any helps or hints will be greatly appreciated! Thank you. EDIT: [imath](x + 4)[/imath] not [imath](x + 2)[/imath] in the denominator. |
751360 | Proof that [imath](1 + x)^n > 1 + nx[/imath] for [imath]x>-1[/imath], [imath]n[/imath] a positive integer
For any positive integer [imath]n[/imath] and real number [imath]x > -1[/imath], show that [imath](1 + x)^n > 1 + nx[/imath]. This is Bernoulli’s inequality but I can't figure out how to start with this. Can someone help? Thanks | 44432 | Proof of Bernoulli's inequality
The question reads [imath]U_n = (1+x)^n - 1 - nx[/imath] Show that [imath]U_2 \geq 0[/imath] Hence or otherwise show that [imath](1+x)^n \geq 1 + nx[/imath] for all [imath]x \gt -1[/imath]. Obviously the [imath]U_2 \geq 0[/imath] is very easy, I can do that without any trouble but I cannot see how it links to the 2nd part of the question. |
751925 | Real part of a holomorphic function is bounded by polynomial then the holomorphic function is a polynomial
Let [imath]u[/imath] be a harmonic function on [imath]\mathbb{R}^2\cong \mathbb{C}[/imath] such that [imath]Ref= u[/imath] where [imath]f[/imath] is an entire function. If [imath]|u(z)|\leq |z|^n[/imath] for any [imath]z\in\mathbb{C}[/imath], then [imath]f[/imath] is a polynomial of degree no greater than [imath]n[/imath]. How to prove this? I am quite confused. Thanks. | 262355 | Polynomial bounded real part of an entire function
Let [imath]f(z)[/imath] be an entire function whose real part is bounded by a polynomial in [imath]|z|[/imath]. Does it follow that [imath]f(z)[/imath] is a polynomial? Or, without loss of generality and more suggestively [imath](f(z)=\sum_{k=0}^\infty a_k z^k \quad \land \quad \operatorname{Re}(f(z))\leq |z|^n)\quad \forall z\in \mathbb{C} \qquad \Rightarrow \qquad f(z)=\sum_{k=0}^n a_k z^k \quad\forall z\in \mathbb{C}[/imath] |
752222 | How to prove [imath] z^n - z^n_0 = (z-z_0) \sum_0^{n-1} z^kz_0^{n-1-k} [/imath]
I want to prove that with [imath]z_0[/imath] a root of [imath]1+z^n[/imath], I have [imath] z^n - z^n_0 = (z-z_0)\sum_0^{n-1} z^kz_0^{n-1-k}[/imath] | 283056 | Induction Proof that [imath]x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})[/imath]
This question is from [Number Theory George E. Andrews 1-1 #3]. Prove that [imath]x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}).[/imath] This problem is driving me crazy. [imath]x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dots +xy^{n-2}+y^{n-1)}[/imath] [imath](x^n-y^n)/(x-y) =[/imath] the sum for the first [imath]n[/imath] numbers and then I added [imath](xy^{(n+1)-2}+y^{(n+1)-1})[/imath] which should equal [imath](x^{n+1}-y^{n+1})/(x-y)[/imath] but I can't figure it out This is a similar problem in the book and I tried this method but it wasn't working out [imath]\quad[/imath]Thereom [imath]\bf1[/imath]-[imath]\bf2[/imath]: [imath]\,\,\,\,[/imath] If [imath]\,x[/imath] is any real number other than [imath]1[/imath], then [imath]\sum_{j=0}^{n-1}x^j=1+x+x^2+\ldots+x^{n-1}=\dfrac{x^n-1}{x-1}.[/imath] [imath]\quad[/imath]Remark: [imath]\displaystyle\sum_{j=0}^{n-1}A_j[/imath] is shorthand for [imath]A_0+A_1+A_2+\ldots+A_{n-1}.[/imath] [imath]\quad[/imath]Proof: Again we proceed by mathematical induction. If [imath]n=1[/imath] then [imath]\displaystyle\sum_{j=0}^{1-1}x^j=x^0=1[/imath] and [imath](x-1)/(x-1)=1[/imath]. Thus the theorem is true for [imath]n=1[/imath]. [imath]\quad[/imath] Assuming that [imath]\displaystyle\sum_{j=0}^{k-1}x^j=(x^k-1)/(x-1)[/imath], we find that [imath] \eqalign{ \sum^{(k+1)-1}_ {j=0}x^j & = \sum^{k-1}_ {j=0}x^j+x^k=\dfrac{x^k-1}{x-1}+x^k \\ &= \dfrac{x^k-1+x^{k+1}-x^k}{x-1}\\ &= \dfrac{x^{k+1}-1}{x-1}. }[/imath] Hence condition [imath](\rm ii)[/imath] is fulfilled, and we have established the theorem. [imath]\quad[/imath]Corollary [imath]\bf1[/imath]-[imath]\bf1[/imath]: [imath]\,\,[/imath] If [imath]\,m[/imath] and [imath]n[/imath] are positive integers and if [imath]m>1[/imath], then [imath]n<m^n.[/imath] |
753113 | How to find an equation of the plane, given its normal vector and a point on the plane?
I have a question regarding vectors: Find the equation of the plane perpendicular to the vector [imath]\vec{n}\space=(2,3,6)[/imath] and which goes through the point [imath] A(1,5,3)[/imath]. (A cartesian and parametric equation). Also find the distance between the beginning of axis and this plane. I'm not really sure where to start. Any help would be appreciated. | 752252 | Find plane by normal and instance point + distance between origin and plane
I have a question regarding Vectors; Find the equation of the plane perpendicular to the vector [imath]\vec{n}\space=(2,3,6)[/imath] and which goes through the point [imath] A(1,5,3)[/imath]. (A cartesian and parametric equation). Also find the distance between the beginning of axis and this plane. I'm not really sure where to start. Any help would be appreciated. |
753571 | Does [imath]\int_a^b f(z)\ \overline{dz} = \int_a^b f(z)\ dz[/imath]?
Question: Attempted Answer: Yes, for if [imath]f = u + iv[/imath] where [imath]u[/imath] and [imath]v[/imath] are real-valued functions, then we have that [imath] \int_a^b f(z)\ \overline{dz} = \overline{\int_a^b \overline{f(z)}\ dz} =\overline{ \int_a^b u(t)\ dt - i v(t)\ dt} [/imath] so that [imath] \int_a^b f(z)\ \overline{dz} =\overline{ \int_a^b u(t)\ dt + i\int_a^b - v(t)\ dt} = \overline{ \int_a^b u(t)\ dt - i\int_a^b v(t)\ dt} [/imath] and hence [imath] \int_a^b f(z)\ \overline{dz} = \int_a^b u(t)\ dt + i\int_a^b v(t)\ dt = \int_a^b f(z)\ dz [/imath] Is my reasoning correct? If so, wouldn't my answer extend to any line integral along a horizontal arc in the complex plane? That is, if we lifted [imath][a,b][/imath] to the straight line from say [imath]a+z[/imath] to [imath]b+z[/imath] (for [imath]z \in \mathbb{C}[/imath]), then wouldn't [imath]\int_\gamma f(z)\ dz = \int_\gamma f(z)\ \overline{dz}[/imath]? | 734402 | Understanding the Definition of [imath]\int_\gamma f\ \overline{dz}[/imath]
Definitionally, we have that [imath] \int_\gamma f\ \overline{dz} = \overline{\int_\gamma \overline{f}\ dz} [/imath] Now let [imath]\int_\gamma f\ dz = w = x +yi[/imath]. Question 1: Is it not the case that [imath]\int_\gamma \overline{f}\ dz = \bar{w} = x - iy[/imath]? Question 2: Assuming that [imath]\int_\gamma \overline{f}\ dz = \bar{w} = x - iy[/imath], then is it not the case that [imath] \overline{\int_\gamma \overline{f}\ dz} = \overline{x - iy} = x+iy = \int_\gamma f\ dz? [/imath] Basically, I'm failing to see how [imath]\overline{\int_\gamma \overline{f}\ dz}[/imath] isn't just [imath]\int_\gamma f\ dz[/imath]. |
753804 | How to show [imath]\int_{0}^{\infty}e^{-x}\ln^{2}x\:\mathrm{d}x=\gamma ^{2}+\frac{\pi ^{2}}{6}[/imath]?
How to show this equation below is true? [imath]\int_{0}^{\infty}e^{-x}\ln^{2}x\:\mathrm{d}x=\gamma ^{2}+\frac{\pi ^{2}}{6}[/imath] Where [imath]\gamma[/imath] is the Euler-Mascheroni constant.... | 188418 | Elementary derivation of certian identites related to the Riemannian Zeta function and the Euler-Mascheroni Constant
Is the proof of these identities possible, only using elementary differential and integral calculus? If it is, can anyone direct me to the proofs? ( or give a hint for the solution ) 1)[imath]\int_0^\infty { e^{-x^2} \ln x }\,dx = -\tfrac14(\gamma+2 \ln 2) \sqrt{\pi} [/imath] 2)[imath]\int_0^\infty { e^{-x} \ln^2 x }\,dx = \gamma^2 + \frac{\pi^2}{6} [/imath] 3) [imath]\gamma = \int_0^1 \frac{1}{1+x} \sum_{n=1}^\infty x^{2^n-1} \, dx[/imath] and lastly, 4) [imath]\zeta(s) = \frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)} \prod_\rho \left(1 - \frac{s}{\rho} \right) e^{s/\rho}\![/imath] I personally think the last is obtained from a simple use of the Weierstrass factorization theorem. I'm unsure as to what substitution is used. [imath]\gamma[/imath] is the Euler-Mascheroni constant and [imath]\zeta(s)[/imath] is the Riemannian Zeta function. Thanks in advance. |
754237 | a continuous function, satisfying [imath]f(α) = f(β) +f(α −β)[/imath] for any [imath]α, β ∈ \mathbb{R}[/imath]
Hi need some help with this problem: Assume [imath]f : \mathbb{R} → \mathbb{R}[/imath] is a continuous function, satisfying [imath]f(α) = f(β) +f(α −β)[/imath] for any [imath]α, β ∈ \mathbb{R}[/imath], and [imath]f(0) = 0[/imath]. Then [imath]f(α) = α f(1)[/imath]. any hints, thank you. | 356645 | I need to find all functions [imath]f:\mathbb R \rightarrow \mathbb R[/imath] which are continuous and satisfy [imath]f(x+y)=f(x)+f(y)[/imath]
I need to find all functions [imath]f:\mathbb R \rightarrow \mathbb R[/imath] such that [imath]f(x+y)=f(x)+f(y)[/imath]. I know that there are other questions that are asking the same thing, but I'm trying to figure this out by myself as best as possible. Here is how I started out: Try out some cases: [imath]x=0:[/imath] [imath]f(0+y)=f(0)+f(y) \iff f(y)=f(0)+f(y) \iff 0=f(0) [/imath] The same result is for when [imath]y=0[/imath] [imath]x=-y:[/imath] [imath]f(-y+y)=f(-y)+f(y) \iff f(0)=f(-y)+f(y) \iff 0=f(-y)+f(y)\iff \quad f(-y)=-f(y)[/imath] I want to extend the result of setting [imath]x=-y[/imath] to numbers other that [imath]-1[/imath], perhaps all real numbers or all rational numbers. I got a little help from reading other solutions on the next part: Let [imath]q=1+1+1+...+1[/imath]. Then [imath]f(qx)=f((1+1+...+1)x)=f(x+x+...+x)=f(x)+f(x)+...+f(x)=qf(x)[/imath] I understood this part, but I don't understand why this helps me find all the functions that satisfy the requirement that [imath]f(x+y)=f(x)+f(y)[/imath], but here is how I went on: Thus [imath]f(qx)=qf(x)[/imath] and it should follow that [imath]f \bigg (\frac {1}{q} x\bigg)= \frac{1}{q}f(x)[/imath] where [imath]q\not =0[/imath], then it further follows that [imath]f \bigg (\frac {p}{q} x\bigg)= \frac{p}{q}f(x)[/imath] where [imath]\frac{p}{q}[/imath] is rational, and lastly it further follows that [imath]f (ax)= af(x)[/imath] where [imath]a[/imath] is real. Thus functions of the form [imath]f(ax)[/imath] where [imath]a[/imath] is real satisfies the requirement of [imath]f(x+y)=f(x)+f(y)[/imath]. I don't know how much of what I did is correct\incorrect, and any help would be greatly appreciated. Also is there any way that I can say that functions of the form [imath]f(ax)[/imath] where [imath]a[/imath] is real are the only functions that satisfy the requirement of [imath]f(x+y)=f(x)+f(y)[/imath]? Or do other solutions exist? Again, thanks a lot for any help! (Hints would be appreciated, I'll really try to understand the hints!) |
702414 | What Is Exponentiation?
Is there an intuitive definition of exponentiation? In elementary school, we learned that [imath] a^b = a \cdot a \cdot a \cdot a \cdots (b\ \textrm{ times}) [/imath] where [imath]b[/imath] is an integer. Then later on this was expanded to include rational exponents, so that [imath] a^{\frac{b}{c}} = \sqrt[c]{a^b} [/imath] From there we could evaluate decimal exponents like [imath]4^{3.24}[/imath] by first converting to a fraction. However, even after learning Euler's Identity, I feel as though there is no discussion on what exponentiation really means. The definitions I found are either overly simplistic or unhelpfully complex. Once we stray from the land of rational powers into real powers in general, is there an intuitive definition or explanation of exponentiation? I am thinking along the lines of, for example, [imath]2^\pi[/imath] or [imath]3^{\sqrt2}[/imath] (or any other irrational power, really). What does this mean? Or, is there no real-world relationship? To draw a parallel to multiplication: If we consider the expression [imath]e\cdot \sqrt5[/imath], I could tell you that this represents the area of a rectangle with side lengths [imath]e[/imath] cm and [imath]\sqrt5[/imath] cm. Or maybe [imath]e \cdot \pi[/imath] is the cost of [imath]\pi[/imath] kg of material that costs [imath]e[/imath] dollars per kg. Of course these quantities would not be exact, but the underlying intuition does not break down. The idea of repeated addition still holds, just that fractional parts of terms, rather than the entire number, are being added. So does such an intuition for exponentiation exist? Or is this one of the many things we must accept with proof but not understanding? This question stems from trying to understand complex exponents including Euler's identity and [imath]2^i[/imath], but I realized that we must first understand reals before moving on the complex numbers. | 2859866 | The concept of an real irrational power
One can understand the concept of natural power, as [imath]x^n[/imath], being a product of a number by itself [imath]n[/imath] times: [imath]x^n=\underbrace{x\cdot x\cdot\dotsb\cdot x}_n[/imath] We can also get the idea of a rational power, [imath]x^{\frac{p}{q}}=\sqrt[q]{x^p}[/imath] , looking for the [imath]q[/imath]-th root of a number, where [imath]q[/imath] is a integer, so it is the same idea as a power of a natural number. But how can we understand a power of a irrational number [imath]x^r[/imath]? Of course we can define it as the limit of the power of the rational number who is tending to the number [imath]r[/imath], but is there a better more intuitive way of explaining this, staying with the number [imath]r[/imath], instead of its approximation? |
754263 | [imath] \sum_{j=1}^{\infty} \frac{(2^j)+ j}{(3^j) - j} [/imath] converges
Show the series [imath] \sum_{j=1}^{\infty} \frac{(2^j)+ j}{(3^j) - j} [/imath] converges. I have looked at an answer here, but I do not understand what these results give us. For example, in the first answer: [imath]\frac{2^j + j}{3^j - j} \le \frac{2^j + 2^j}{3^j - j} \le \frac{2^j + 2^j}{3^j - \frac{1}{2} 3^j}[/imath] What do we do with the last expression? I understand we want to compare it with something, but don't see what. | 571119 | Show the series [imath] \sum_{j=1}^{\infty} \frac{(2^j)+ j}{(3^j) - j} [/imath] converges
Show the following series converges [imath] \sum_{j=1}^{\infty} \frac{(2^j)+ j}{(3^j) - j} . [/imath] I tried to use the comparison test and tried to compare it with the series of [imath]\dfrac{2^j}{3^j}[/imath] because this is the geometric series. However, [imath]\dfrac {2^j}{3^j}[/imath] is smaller than [imath]\dfrac{2^j+j}{3^j-j}[/imath], so by comparison test, the series [imath]\sum_{j=1}^{\infty}\dfrac{2^j}{3^j}[/imath] converges does NOT indicate that the series [imath]\sum_{j=1}^{\infty}\dfrac{2^j+j}{3^j-j}[/imath] converges... Thus I'm confused. Thanks! Thanks! |
754851 | A basic question on limit calculation
How to prove that the following limit exist without calculating its value [imath] \lim_{t \to\infty} \int_{0}^{t}\frac{\sin x}{x} dx [/imath] | 547090 | Proof about boundedness of [imath]\rm Si[/imath]
[imath]\def\Si{{\rm Si}}[/imath] I want to prove the boundedness of [imath]\Si(x) := \int_0^x \frac {\sin \xi} \xi d\xi[/imath] as part of a homework (about the non-surjectivity of [imath]\mathcal F : L^1(\mathbb R) \to C_0^0(\mathbb R)[/imath]). As a first step, I found (by means of plotting) [imath]\frac{\sin x}{x} \leq \cos \frac x 2 \qquad \forall\ x \in (0, \pi)[/imath] but couldn't easily prove that. If that is proved, we had, including some other arguments [imath]\Si(x) \leq \int_0^\pi \frac {\sin \xi} \xi d\xi \leq \int_0^\pi \cos\frac\xi 2 d\xi = 2[/imath] And I'd be done. How can I prove [imath]\frac{\sin x}{x} \leq \cos \frac x 2 \qquad \forall\ x \in (0, \pi)[/imath] in a simple way, or do you have any alternative proof for my objective? Thanks in advance Okay with the hints provided by @HaraldHanche-Olsen in his answer, it became very clear: [imath]\frac{\sin x} x \stackrel{\text{addition thm.}}{=} \frac{2 \sin\frac x 2 \cos\frac x 2}{x} = \underbrace{\frac{\sin\frac x 2}{\frac x 2}}_{\leq 1} \cos \frac x 2 \leq \cos\frac x 2[/imath] |
718457 | Functional weakly lower-semicontinuous
If [imath]X[/imath] is a topological space, then a functional [imath]\varphi:X\to\mathbb{R}[/imath] is lower-semicontinuous (l.s.c) if [imath]\varphi^{-1}(a,\infty)[/imath] is open in [imath]X[/imath] for any [imath]a\in\mathbb{R}[/imath]. If [imath]X[/imath] is a Hilbert space, then [imath]\varphi[/imath] is weakly l.s.c if it is l.s.c on [imath]X[/imath] with its weak topology. My question: If [imath]X[/imath] is a Hilbert space and [imath]\varphi:X\to\mathbb{R}[/imath], then [imath]\varphi(x)\leq \liminf{x_n}[/imath] whenever [imath]x_n[/imath] converges weakly to [imath]x[/imath] [imath]\Rightarrow[/imath] [imath]\varphi[/imath] weakly l.s.c? I read this in "A invitation to Variational Methods in Differential Equations". It isn't a exercise, maybe a definition. | 482684 | Lower Semicontinuity Concepts
Let [imath]X[/imath] be a real Banach space, let [imath]f:X\rightarrow \overline{\mathbb{R}}[/imath] be a functional. We have known that: If [imath]f[/imath] is weakly lower semicontinuous then [imath]f[/imath] is weakly sequentially lower semicontinuous; If [imath]f[/imath] is weakly sequentially lower semicontinuous then [imath]f[/imath] is lower semicontinuous. a) I would like to construct two counterexamples to show that the reverse is not true. (It is interesting to find counterexamples in Hilbert space.) b) When do these concepts coincide? Thank you for all help and comments. |
755150 | Let [imath]p: E\to B[/imath] be a covering map. If [imath]B[/imath] is compact and [imath]p^{-1}(b)[/imath] is finite, then [imath]E[/imath] is compact.
So I start off and assume that some [imath]\{U_\alpha\}[/imath] is a cover of [imath]E[/imath]. I want to reduce this cover to a finite subcover of [imath]E[/imath]. Since [imath]p[/imath] is a covering map it is also an open map, therefore [imath]p(\{U_\alpha\})[/imath] is an open cover of [imath]B[/imath] denote it [imath]\{W_\alpha\}[/imath]. But since [imath]B[/imath] is compact there is a finite subcover of [imath]\{W_\alpha\}[/imath], [imath]\cup_{i=1}^nW_i[/imath]. But then since [imath]p[/imath] is continuous [imath]p^{-1}(\cup_{i=1}^nW_i)[/imath] is an open cover of [imath]E[/imath] which must stay finite since [imath]p^{-1}(b)[/imath] is finite for all [imath]b[/imath]. | 310353 | covering map with finite fibres and preimage of a compact set
Let [imath]f:X\to Y[/imath] be a covering map (covering maps are surjective) , Y be compact set. And suppose that [imath]f^{-1}(y) [/imath] is finite for each [imath]y\in Y[/imath]. Prove that [imath]X[/imath] is also compact. I think that this problem does not need the hyphotesis of covering map , I think that only is necessary that [imath]f[/imath] is an open map (covering maps are open maps) Well I dunno how to start this problem :/ If I consider an open covering [imath] X = \bigcup U_a [/imath] , then [imath]f(X)=Y = f(\bigcup U_a) = \bigcup f(U_a) [/imath] I don't know if it's convenient to consider a finite subcovering of [imath]Y[/imath] at this moment, or I need some tricky argument ... Please help me )= |
757371 | Interpreting this Riemann Intergral problem
I saw this on the message board and am very interested in this problem. The question reads as the following. Let [imath]f[/imath] be continuous on interval [imath][0,1][/imath]. Prove that [imath]\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^nf(\frac{k}{n})= \int_0^1f(x)dx[/imath] The person then stated this Again any help would be greatly appreciate. I have attempted throwing the definition of continuity at it cause it states that it is continuous. I then used a uniform partition and stated my [imath]M_k[/imath], [imath]m_k[/imath] based on the equation. I had my partition looking like this [imath]P=({0}\frac{1}{n},\frac{2}{n},...,\frac{n}{n})[/imath]. Now I'm having a hard time interpreting the sum on the left. Is this a collection of rational numbers on the interval [imath][0,1][/imath]? I know that the rationals are dense so there are an infinite amount of points in this interval? Am I reading this correctly? | 757274 | Riemann Intergral proof question
I am currently studying for a grad test and I have came across one problem that I cannot even begin to solve. I have thrown definitions and theorems at it and just keep running into road blocks. A little help or maybe a shove in the right direction would be appreciated. The question reads as the following. Let f be continuous on interval [0,1]. Prove that [imath]\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^nf(\frac{k}{n})= \int_0^1f(x)dx[/imath] Again any help would be greatly appreciate. I have attempted throwing the definition of continuity at it cause it states that it is continuous. I then used a uniform partition and stated my [imath]M_k[/imath], [imath]m_k[/imath] based on the equation. I had my partion looking like this [imath]P=(\frac{1}{n}+\frac{2}{n}+...+\frac{n}{n})[/imath]. From here it was not apparent where to go. |
757549 | If [imath]\sum_na_n=\infty[/imath] and [imath]a_n\downarrow 0[/imath] then [imath]\sum\limits_n\min(a_{n},\frac{1}{n})=\infty[/imath]
If [imath]\sum_na_n=\infty[/imath] and [imath]a_n\downarrow 0[/imath] then [imath]\sum\limits_n\min(a_{n},\frac{1}{n})=\infty[/imath] My idea: [imath]b_{n}=\min(a_{n},\tfrac{1}{n})\Longrightarrow b_{n}\le a_{n},b_{n}\le\tfrac{1}{n}[/imath] then I don't know how to go on. | 739975 | Show that if [imath]\sum_na_n=\infty[/imath] and [imath]a_n\downarrow 0[/imath] then [imath]\sum\limits_n\min(a_{n},\frac{1}{n})=\infty[/imath]
Let [imath]a_n[/imath] be a sequence decreasing to [imath]0[/imath], and [imath]\sum {{a_n} = \infty } [/imath]. Show that:[imath]\sum {\min \left( {{a_n},{1 \over n}} \right)} = \infty [/imath] If there's [imath]N_0[/imath] such that [imath]\forall n>N_0: \min(a_n, {1\over n}) = {1\over n}[/imath] or [imath]\forall n>N_0: \min(a_n, {1\over n}) = a_n[/imath] then, the problem is trivial. Otherwise, let us define [imath]b_n = \min(a_n, {1\over n})[/imath] and two subsequences: [imath]b_{n_k} = {1 \over n_k},\quad b_{n_l} = a_{n_l}[/imath] [imath]\sum {{b_n} = } \sum {{b_{{n_k}}}} + \sum {{b_{{n_l}}}} = \infty + \infty [/imath] Is that right? |
459714 | How to prove Gauss’s Multiplication Formula?
How to prove Gauss’s Multiplication Formula? [imath]\Gamma(nz)=(2\pi)^{(1-n)/2}n^{nz-(1/2)}\prod_{k=0}^{n-1}\Gamma\left(z+\frac{k}{n}\right)[/imath] (original image) Any help like an answer or link would be appreciated. Thanks for all help. | 752895 | Ahlfors "Prove the formula of Gauss"
He says: Prove the formula of Gauss: [imath] (2\pi)^\frac{n-1}{2} \Gamma(z) = n^{z - \frac{1}{2}}\Gamma(z/n)\Gamma(\frac{z+1}{n})\cdots\Gamma(\frac{z+n-1}{n}) [/imath] This is an exercise out of Ahlfors. By taking the logarithmic derivative, it's easy to show the left & right hand sides are the the same up to a multiplicative constant. After that I'm lost. It's easy using another identity when [imath]n[/imath] is even to use induction. But when [imath]n[/imath] is odd I am lost. It's obvious when [imath]n[/imath] is a power of 2. |
757764 | Simple demonstration for [imath]\lim_{n\to\infty}(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}) = \frac{1}{e}[/imath]
What is the simple demonstration with elementary means for Lalescu Sequence: [imath]\lim_{n\to\infty}(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}) = \frac{1}{e}?[/imath] (Traian Lalescu-romanian mathematician (1882-1929)) | 161682 | Find the limit of: [imath]\lim_{n\to\infty} \frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}}[/imath]
Could be the following limit computed without using Stirling's approximation formula? [imath]\lim_{n\to\infty} \frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}}[/imath] I know that the limit is [imath]e[/imath], but I'm looking for some alternative ways that doesn't require to resort to the use of Stirling's approximation. I really appreciate any support at this limit. Thanks. |
756385 | [imath]C^\infty[/imath] approximations of [imath]f(r) = |r|^{m-1}r[/imath]
Consider [imath]f(r) = |r|^{m-1}r[/imath] where [imath]m \geq 1[/imath]. Is it possible to find [imath]C^\infty[/imath] functions [imath]f_n[/imath], such that [imath]f_n \to f[/imath] uniformly on compact subsets of [imath]\mathbb{R}[/imath], [imath]f_n' \to f'[/imath] uniformly on compact subsets of [imath]\mathbb{R}[/imath], [imath]f_n' > 0[/imath], [imath]f_n(0) = 0[/imath] and do we get any upper/lower bounds on [imath]f_n[/imath] and [imath]f_n'[/imath], and if so, how do they depend on [imath]n[/imath]? Ideally these should not depend on the argument. The natural choice (thanks to daw in the linked thread below) [imath]f_n(r) = \left(\sqrt{r^2 + \frac{1}{n}}\right)^{m-1}r[/imath], for [imath]m > 1[/imath], satisfies these properties but the [imath]f_n' \to f'[/imath] uniformly requires [imath]m \geq 2[/imath]. How to remove these restrictions? Is anything possible for [imath]0<m\leq 1[/imath]? I have seearched around for literature (I am aware this comes up in the porous medium equation) but nothing good appears. (From [imath]C^\infty[/imath] approximations of [imath]f(r) = |r|^{m-1}r[/imath]) | 748125 | [imath]C^\infty[/imath] approximations of [imath]f(r) = |r|^{m-1}r[/imath]
Consider [imath]f(r) = |r|^{m-1}r[/imath] where [imath]m \geq 1[/imath]. Is it possible to find [imath]C^\infty[/imath] functions [imath]f_n[/imath], such that [imath]f_n \to f[/imath] uniformly on compact subsets of [imath]\mathbb{R}[/imath], [imath]f_n' \to f'[/imath] uniformly on compact subsets of [imath]\mathbb{R}[/imath], [imath]f_n'(0) > 0[/imath], [imath]f_n(0) = 0[/imath] and do we get any upper/lower bounds on [imath]f_n[/imath] and [imath]f_n'[/imath], and if so, how do they depend on [imath]n[/imath]? It is possible to find [imath]f_n \in C^\infty[/imath] such that 1,3,4 hold. It is possible also to find [imath]\tilde f_n \in C^\infty[/imath] such that 1 and 2 hold. But how about all four together? |
689575 | Proof that every polynomial of odd degree has one real root
I want to prove that every real polynomial of odd degree has at least one real root, using the intermediate value theorem. Let [imath]P(x) = x^{2n+1} + a_n x^{2n} + . . . + a_0[/imath] for each [imath]a_i \in \mathbb{R}[/imath] and [imath]n \in \mathbb{N}[/imath]. By the fundamental theorem of algebra I know that [imath]P(x)[/imath] has exactly [imath]2n+1[/imath] complex roots, so [imath]P(x) = (x+r_1)(x+r_2) . . . (x+r_{2n+1})[/imath] for each [imath]r_i \in \mathbb{C}[/imath] I do not know how to complete this but I do know that, at some point, I probably have to show that each root with imaginary part non zero has to come in conjugate pairs, and since [imath]2n+1[/imath] is odd there is at least [imath]1[/imath] root that is imaginary part [imath]0[/imath] and thus real. | 1138327 | Odd-degree polynomials have roots (Intermediate Value Theorem)
Let [imath]f(x)[/imath] be a monic polynomial of odd degree. Prove that [imath]\exists A\in \mathbb{R}[/imath] s.t. [imath]f(A)<0[/imath] and there exists [imath]B \in \mathbb{R}[/imath] such that [imath]f(B)>0[/imath]. Deduce that every polynomial of odd degree has a real root. There are questions that answer the final part, but they do not do so by proving the first part. I am fairly sure that this involves the intermediate value theorem, but not sure how to implement it in this case. |
354351 | Union of a finite set and a countably infinite set is countably infinite
Ok, here is the problem statement: Prove that if [imath]S[/imath] is any finite set of real numbers, then the union of [imath]S[/imath] and the integers is countably infinite. This seems pretty obvious to me, knowing that 2 countable sets are countable. But is there some step by step way to prove this? Like do I need to prove bijectivity or something? Thanks! | 773017 | Prove [imath]|\mathbb{N}|=|\mathbb{N}| + 1[/imath]
The problem is to show [imath]|\mathbb{N}|=|\mathbb{N}|+1[/imath], where [imath]\mathbb{N}[/imath] is the set of natural numbers. The way I had it (which is the wrong way of proving it) was: because [imath]|\mathbb{N}|=\aleph_0[/imath], so the problem becomes [imath]\aleph_0=\aleph_0+1[/imath]. My teacher says I am "showing it" and not "proving it"... How do I prove it? |
757394 | Proving that [imath]f: N\times N \rightarrow N[/imath] is surjective
I am having trouble proving that the function [imath]f: N\times N \rightarrow N, \ \ f(i,j)=2^{i-1}(2j-1)[/imath] is surjective. Work: I know that using the theorem in which [imath]n[/imath] is the product of prime numbers play an important role. For instance, [imath]n=2^{i-1}p[/imath], [imath]p[/imath] is the product of all prime factors of [imath]n[/imath]. However, from here I don't know how to proceed. What makes it difficult for me is that how am I suppose to find an arbitrary [imath]n\in N[/imath] such that [imath]f(i,j)=n[/imath]? | 757313 | Proving that [imath]f[/imath] is a bijection from [imath]N[/imath]x[imath]N[/imath] to [imath]N[/imath].
I am having trouble with the following problem: [imath]f: N\times N\rightarrow N[/imath] and [imath]f(i,j)=2^{i-1}(2j-1)[/imath]. Prove that [imath]f[/imath] is a bijection thus [imath]N\times N[/imath] and [imath]N[/imath] are numerically equivalent. Work: I am told that Euclid's Lemma will be useful to proving injection and also that for any positive integer [imath]n[/imath], [imath]n[/imath] is a product of prime numbers. For injection, I set [imath]f(a,b)=f(x,y)\rightarrow 2^{a-1}(2b-1)=2^{x-1}(2y-1)[/imath]. Then I distributed across to get [imath]2^ab-2^{a-1}=2^xy-2^{x-1} \rightarrow 2^ab-2^xy=2^{a-1}-2^{x-1} \rightarrow 2(2^{a-1}b-2^{x-1}y)=2^{a-1}-2^{x-1}[/imath]. From here, I am unsure how to apply Euclid's Lemma to arrive at the injection result. I am even more unsure how to prove surjection. |
758532 | Proving [imath]P(A\cap \overline{B})=P(A)-P(A\cap B)[/imath]
how does one formula prove that [imath]P(A\cap \overline{B})=P(A)-P(A\cap B)[/imath]. I can show with Venn diagrams that [imath](A\cap \overline{B})=A\backslash(A\cap B)[/imath] but how does one formally prove it? And is it sufficient to prove this identity to get the identity for the probabilities? | 400575 | Does [imath]P(A\cap B) + P(A\cap B^c) = P(A)[/imath]?
Based purely on intuition, it would seem that the following statement is true, when thinking of the events as sets: [imath]P(A\cap B) + P(A\cap B^c) = P(A)[/imath] However, I am not sure if this is true, and cannot find out how to prove it, or describe a straight forward intuition of it. Is this equation true? And if not, is there another way to link [imath]P(A\cap B)[/imath] and [imath]P(A\cap B^c)[/imath] with [imath]P(A)[/imath]? |
755008 | Question on the Prime Number Theorem (the Tchebychev Function)
This has been giving me nothing but a headache: Let the Tchebychev Function, [imath]\psi (x)[/imath] be defined: [imath]\psi (x) = \sum_{p^m \le x}\log p \space \space \space , \space \space \space p \in \mathbb P[/imath] Prove that the integrated Tchebychev Function, [imath]\psi_1 (x)[/imath], is as follows: [imath]\psi_1 (x) = \frac{x^2}2 - \sum_{\rho}\frac{x^{\rho + 1}}{\rho(\rho +1)} - E(x)[/imath] Where [imath]\rho[/imath] is a zero of the Riemann Zeta Function in the critical strip and: [imath]E(x) = \left(\frac{\zeta ' (0)}{\zeta(0)}\right)x + \left(\frac{\zeta ' (-1)}{\zeta(-1)}\right) + \sum_{k=1}^{\infty}\frac{x^{1-2k}}{2k(2k-1)}[/imath] Can anyone point me to the right direction? Not really sure where to begin. Thanks! | 281334 | Estimating the integrated Tchebychev function and calculating its error
I would like to understand how to derive (2) from (1) below. Problem: If [imath]\psi_1[/imath] is the integrated Tchebychev function below [imath]\psi_1(x)=\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty}\frac{x^{s+1}}{s(s+1)} \left(\frac{-\zeta '(s)}{\zeta(s)}\right) \, ds \tag 1[/imath] where [imath]c[/imath] > 1, then [imath]\psi_1(x)=\frac{x^2}{2}-\sum_\rho\frac{x^{\rho+1}}{\rho(\rho+1)}-E(x)\quad(2)[/imath] where the sum is taken over all zeros of zeta function in the critical strip. Which error term will be [imath]E(x)=c_1x+c_0+\sum_{k=1}^{\infty}\frac{x^{1-2k}}{2k(2k-1)}[/imath] where [imath]c_1=\frac{\zeta '(0)}{\zeta(0)}[/imath] and [imath]c_0=\frac{\zeta '(-1)}{\zeta(-1)}[/imath]. |
759040 | How can one determine whether the following series converges or diverges
[imath] \sum_{n = 2}^{\infty}\frac{(-1)^{n}}{\sqrt{n} + (-1)^{n}} [/imath] Wolfram Alpha returns nothing useful, except that the ratio test was inconclusive. | 71253 | Series Divergence - Apostol Calculus Vol I, Section 10.20 #7
Prove that [imath]\displaystyle\sum_{n=2}^\infty\frac {(-1)^n}{\sqrt{n}+(-1)^n} [/imath] diverges. It is easy to see that this absolutely diverges, however how can it be proven to diverge in general? The idea as posted in another forum was to group the terms in the sequence as follows: [imath]\sum_{n=2}^\infty \frac{(-1)^n}{\sqrt{n}+(-1)^n}=-\sum_{n=1}^\infty \left(\frac{1}{\sqrt{2n+1}-1}-\frac 1 {\sqrt{2n}+1}\right)[/imath] and then show that this sequence diverges. I did end up getting an answer from that forum (thank you!) however it depended on results from WolframAlpha which seem extremely difficult to solve by hand. This question is from Apostol's Calculus, written in 1969, so I'd like to have a solution which doesn't depend on WolframAlpha. |
759182 | Problem with a sequence with multiple integrals
How to compute the following limit, [imath]\displaystyle \lim\limits_{n \to \infty} \int_0^1 \int_0^1 \ldots \int_0^1 \sin \bigg(\frac{x_1+x_2+\ldots+x_n}{n}\bigg)\,dx_1 \,dx_2 \ldots \,dx_n[/imath] ? I will greatly appreciate it if a solution only uses analysis. Thank you. | 757330 | How prove this integral limit [imath]=f(\frac{1}{2})[/imath]
Let [imath]f[/imath] be a continuous function on the unit interval [imath][0,1][/imath]. Show that [imath]\lim_{n\to\infty}\int_{0}^{1}\cdots\int_0^1\int_{0}^{1}f\left(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=f\left(\dfrac{1}{2}\right)[/imath] This problem is from Selected Problems in Real Analysis. But the author doesn't include a solution. Maybe there is a method for this sort of problem? Maybe we can use this: [imath]\int_0^1\cdots\int_{0}^{1}\int_{0}^{1}\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{2}?[/imath] I also found a similar problem in this question. |
759185 | Order of groups and group elements?
Let G be a group and let p be a prime. Let g and h be elements of G with order p. I am wondering how I can use group theory to find the possible orders of the intersection between [imath]\def\subgroup#1{\langle#1\rangle}\subgroup g[/imath] and [imath]\subgroup h[/imath] and also to prove that the number of elements of order [imath]p[/imath] in [imath]G[/imath] is a multiple of [imath]p-1[/imath]. I've been looking for the path for ages and got nothing really. These are presented as typical applications to group theory and I'm not at ease with the subject so I'd like to see how you think on this example (in order to get a better idea). Can you hint me? Thank you. | 92613 | Number of elements of order [imath]p[/imath] is a multiple of [imath]p-1[/imath] (finite group).
This is a question in Pinter's A Book of Abstract Algebra. Let [imath]S=\{g\in G\mid \operatorname{ord}(g)=p\}[/imath]. Prove the order of [imath]S[/imath] is a multiple of [imath]p-1[/imath]. In his solution Pinter says [imath]a \in S[/imath] implies that [imath]a[/imath] generates a subgroup with [imath]p-1[/imath] elements. Shouldn't there be [imath]p[/imath] elements [imath]\{1,a^1,\dots,a^{p-1}\}[/imath]? Or is it typical to only count the non-trivial elements in a subgroup? |
749097 | When does [imath](uv)'=u'v'?[/imath]
In any calculus course, one of the first thing we learn is that [imath](uv)'=u'v+v'u[/imath] rather than the what I've written in the title. This got me wondering: when is this dream product rule true? There are of course trivial examples, and also many instances where the equality is true at a handful of points. Less obvious though, is the following: Are there non-constant [imath]u,v[/imath] such that there exists an interval [imath]I[/imath] where [imath](uv)'=u'v'[/imath] over [imath]I?[/imath] I have a feeling there should be, but I am having trouble constructing such a pair. | 684001 | When does product of derivatives equals derivative of products?
In general, [imath]\frac{d}{dx}(f(x) \cdot g(x)) \neq \frac{d}{dx}f(x) \cdot \frac{d}{dx}g(x)[/imath] When does this result hold true? My first try is to use product rule on left side and compare the two sides, but this hasn't helped at all. Any suggestions? |
759585 | How to prove a function from [imath]\mathbb N\times \mathbb N[/imath] to [imath]\mathbb N[/imath] is bijective.
I am having trouble with this problem: [imath]f\colon \mathbb N\times \mathbb N \rightarrow \mathbb N[/imath] is defined by [imath]f(i,j)=\dfrac{(i+j-1)(i+j-2)}{2}+i[/imath]. How do you prove that [imath]f[/imath] is a bijection from [imath]\mathbb N\times \mathbb N[/imath] to [imath]\mathbb N[/imath]? Work: I tried to set [imath]f(i,j)=f(a,b)[/imath]. Assuming that [imath]f(i,j)\neq f(a,b)[/imath], then I assume [imath]f(i,j)< f(a,b)[/imath]and that [imath]i+j=m[/imath] and [imath]a+b=m+r[/imath] for some remainder [imath]r[/imath]. I replaced these into my equation and try to obtain a contradiction thus proving that [imath]i=a, j=b[/imath]. However, I get stuck after this part and I do not know of a better way of doing this. Please help. | 758606 | Proving that a function from [imath]N\times N[/imath] to [imath]N[/imath] is bijective.
I am stuck on this problem: Define [imath]f: N\times N \rightarrow N[/imath] by [imath]f(i,j)=\frac{(i+j-1)(i+j-2)}{2}+i[/imath]. How do you prove that [imath]f[/imath] is a bijection thus [imath]N\times N[/imath] and [imath]N[/imath] are numerically equivalent? Work: First I tried to prove that [imath]f[/imath] is injective by setting [imath]f(i,j)=f(x,y)[/imath]. However, from here I am stuck and I cannot reduce it properly. My work really does not get anywhere significant and its really messy. |
760252 | Prove that [imath]\cos^2\theta+\sin^2\theta=1[/imath]
I try to find the question but I didn't How do you do it? I'm really stuck on this proof. Can someone please explain? | 607103 | Prove [imath]\sin^2\theta + \cos^2\theta = 1[/imath]
How do you prove the following trigonometric identity: [imath] \sin^2\theta+\cos^2\theta=1[/imath] I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine. |
135351 | Sum of GCD(k,n)
I want to find this [imath] \sum_{k=1}^n \gcd(k,n)[/imath] but I don't know how to solve. Does anybody can help me to finding this problem. Thanks. | 1990320 | How do I simplify [imath]\sum_{k=1}^n \gcd(k,n)[/imath]?
For a given positive integer [imath]n[/imath], how do I simplify [imath]\sum_{k=1}^n \gcd(k,n)?[/imath] By simplification I mean is there any formula which makes it easier to compute it (rather than just compute each term and sum up)? |
760338 | Assume [imath]f : [0, 1] \rightarrow [0, 1][/imath] is continuous. Show that there must be a point [imath]x \in [0, 1][/imath] such that [imath]f(x) = x[/imath]
Assume [imath]f : [0, 1] \rightarrow [0, 1][/imath] is continuous. Show that there must be a point [imath]x \in [0, 1][/imath] such that [imath]f(x) = x[/imath] I am not even sure how to begin with this problem, I know that [imath]f[/imath] is continuous so for all [imath]\epsilon > 0[/imath] there is a [imath]\delta > 0[/imath] such that whenever [imath]|x-c|< \delta[/imath] it follows that [imath]|f(x) - f(c)| < \epsilon[/imath] Any advice? | 255403 | proof of that a continuous function has a fixed point
Can you please help me to understand this proof: Consider [imath]g(x)=f(x)-x[/imath]. [imath]f(a)\ge a[/imath] so [imath]g(a)=f(a)-a\ge 0[/imath]. [imath]f(b)\le b[/imath] so [imath]g(b)=f(b)-b\le 0[/imath]. By the Intermediate Value Theorem, since [imath]g[/imath] is continuous and [imath]0\in[g(b),g(a)][/imath] there exists [imath]c\in[a,b][/imath] such that [imath]g(c)=f(c)-c=0[/imath], so [imath]f(c)=c[/imath] for some [imath]c\in[a,b][/imath]." My question is why [imath]f(a)\ge a[/imath] and [imath]f(b)\le b[/imath]?? Thanks a lot! |
760454 | Prove by induction that [imath](1+x)^n \geq 1+nx[/imath]
Prove by induction that [imath]\forall x \in \mathbb{R}, x \geq -1, \forall n \in \mathbb{N},n \geq 0[/imath] that [imath](1+x)^n \geq 1+nx[/imath] First of all I have a problem with x being a real number, how can I use induction over a real number? Can I simply assume that it holds for all [imath]y \in [-1,x][/imath] and then show that it also holds for [imath]x + \epsilon, \epsilon \in \mathbb{R},\epsilon \gt 0[/imath]? If that were possible I would try to prove over n. Prove the base case for n, use induction over x and then prove the induction step over n. But I have some problems with the induction step over n. Assuming that it works for all [imath]l \in \mathbb{N}, l \leq n-1[/imath]. I need to show [imath](1+x)^n=(1+x)^{n-1}(1+x) \geq 1+(n-1)x+x[/imath] which basically results in showing that [imath]z(1+x) \geq z + x[/imath] for some quantity z, correct? But then I'd get [imath]zx \geq x[/imath], which would obviously depend on the value of z. I'm kind of confused at this point, any help would be appreciated. | 475309 | Proof by induction of Bernoulli's inequality: [imath](1 + x)^n \geq 1 + nx[/imath]
I'm asked to used induction to prove Bernoulli's Inequality: If [imath]1+x>0[/imath], then [imath](1+x)^n\geq 1+nx[/imath] for all [imath]n\in\mathbb{N}[/imath]. This what I have so far: Let [imath]n=1[/imath]. Then [imath]1+x\geq 1+x[/imath]. This is true. Now assume that the proposed inequality holds for some arbitrary [imath]k[/imath], namely that [imath]1+x>0\implies (1+x)^k\geq 1+kx,~\forall~k\in\mathbb{N}\setminus\{1\}[/imath] is true. We want to show that the proposed inequality holds for [imath]k+1[/imath]. Thus multiplication of [imath](1+x)[/imath] on each side of the above inequality produces the following result: [imath](1+x)(1+x)^k\geq (1+kx)(1+x)\implies (1+x)^{k+1}\geq 1+x+kx+kx^2\cdots\cdots\cdots[/imath] I'm not sure where to go from here. |
761132 | [imath]\sum\limits_{n=1}^\infty n(\frac{1}{2})^{n}[/imath]
I am trying to find the expected value of the number of even numbers rolled before the first odd number when rolling a fair die until an odd number comes up. I arrived at [imath]\sum\limits_{n=1}^\infty n(\frac{1}{2})^{n+1}[/imath], then I factored out a [imath]\frac{1}{2}[/imath] to give me [imath]\sum\limits_{n=1}^\infty n(\frac{1}{2})^{n}[/imath]. What I now lack is knowledge of infinite series. I know it converges to 2; what I want to know is why? | 757263 | How to find answer to the sum of series [imath]\sum_{n=1}^{\infty}\frac{n}{2^n} [/imath]
I have put his on wolfram and obtained answer as follows: [imath]\sum_{n=1}^{\infty}\frac{n}{2^n} = 2[/imath] And the series is convergent too because [imath]\lim_{n\to\infty} \frac {n}{2^n} = 0[/imath] However I am wondering if there is a convenient way to solve this; I don't think you can represent it by a geometric progression either. So how do we have to do it on paper? |
761204 | Is [imath]hh^T[/imath] positive semi-definite ([imath]h[/imath] is a column non-negative vector)?
Is [imath]hh^T[/imath] positive semi-definite? It seems to be positive semi-definite, but I cannot prove it. Please help:) | 678693 | Positive semidefinite cone is generated by all rank-[imath]1[/imath] matrices.
The positive semidefinite cone is generated by all rank-[imath]1[/imath] matrices [imath]xx^T[/imath], which form the extreme rays of the cone. Positive definite matrices lie in the interior of the cone. Positive semidefinite matrices with at least one zero eigenvalue are on the boundary. I am unable to justify why the statements above are true. Please give some hints. |
761526 | Prove that [imath]|x+y| \leq |x|+|y|[/imath]
How to Prove the triangle inequality which says for all x (no matter how big or small) and for all y (no matter its size) in the set of irrational+rational numbers, this holds: [imath]|x+y| \leq |x|+|y|[/imath] | 307348 | Proof of triangle inequality
I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that [imath]|a+b| \leq |a|+|b|[/imath]. Any help would be appreciated :) |
761447 | To construct a right triangle given the hypotenuse and sum of two legs
NOTE: I want a hint only. A compass and a straightedge construction:Given a hypotenuse and the sum of lengths of the legs,we need to construct a right triangle. MY TRY: From any ray [imath]BE[/imath], ,let us cut off [imath]BC[/imath] equal to our hypotenuse.Now we draw a circle with the hypotenuse as diameter.Now,if we take any point on the circle and join it with the end points of the hypotenuse,we will have a right triangle by Thales' theorem. BACKTRACKING: As is often the case,first assuming that we have constructed our figure and then trying to reverse-engineer the steps helps in construction of the required figure.Suppose we have [imath]\Delta ABC[/imath] inscribed in a circle with [imath]\angle A[/imath] the right angle.We extend one of the legs [imath]BA[/imath] to [imath]K[/imath] such that [imath]BK[/imath] is equal to the sum of the legs of our triangle.Then [imath]AK=AC[/imath] and therefore [imath]\angle AKC=\angle ACK[/imath]. Similarly,we extend [imath]AC[/imath] to [imath]L[/imath] such that it is equal to the sum of the legs.Then we can argue about angle equality the same way.But this is not helping me in constructing the right triangle. A hint will be appreciated. | 209342 | Construction of a right triangle
It's a high school level question which we can't seem to solve. Here it is: Given [imath]2[/imath] lines, one of the length of the hypotenuse and the other with the length of the sum of the [imath]2[/imath] legs, construct with straightedge and compass the corresponding right triangle I didn't make much progress. It seems that there's a theorem or a few basic facts about right triangles that I'm missing. What path do you suggest I take? Thanks for your help. |
754979 | Isometry <=> Adjoint left inverse
Is it true that: [imath]T\text{ isometric}\iff T^*\text{ left inverse}[/imath] Obviously: [imath]\text{"}\Rightarrow\text{": }\langle x,\tilde{x}\rangle=\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle[/imath] [imath]\text{"}\Leftarrow\text{": }\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle=\langle x,\tilde{x}\rangle[/imath] My problem is that I'm not sure about if the adjoint is left inverse then the operator necessarily is bounded so that domain issues might come into play... | 754909 | Isometric <=> Left Inverse Adjoint
Is it true that: [imath]T\text{ isometric}\iff T^*\text{ left inverse}[/imath] Obviously: [imath]\text{"}\Rightarrow\text{": }\langle x,\tilde{x}\rangle=\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle[/imath] [imath]\text{"}\Leftarrow\text{": }\langle Tx,T\tilde{x}\rangle=\langle x,T^*T\tilde{x}\rangle=\langle x,\tilde{x}\rangle[/imath] My problem is that I'm not sure about if the adjoint is left inverse then the operator necessarily is bounded so that domain issues might come into play... |
761838 | Extremely hard sum of infinite series problem
I found this in a previous years iit paper. It seems really interesting, but incredibly difficult. Can someone please help [imath] \sum_{i = 1}^\infty \arctan\frac{1}{2i^2} [/imath] This whole expression is equal to [imath]t[/imath] Find [imath]\tan(t)[/imath] | 144944 | Finding [imath]\tan t[/imath] if [imath]t=\sum\tan^{-1}(1/2t^2)[/imath]
I am solving this problem. If [imath]\sum\limits_{i=1}^{\infty} \tan^{-1}\biggl(\frac{1}{2i^{2}}\biggr)= t[/imath] then find the value of [imath]\tan{t}[/imath]. My solution is like the following: I can rewrite: \begin{align*} \tan^{-1}\biggl(\frac{1}{2i^{2}}\biggr) & = \tan^{-1}\biggl[\frac{(2i+1) - (2i-1)}{1+(2i+1)\cdot (2i-1)}\biggr] \\\ &= \tan^{-1}(2i+1) - \tan^{-1}(2i-1) \end{align*} and when I take the summation the only term which remains is [imath]-\tan^{-1}(1)[/imath], from which I get the value of [imath]\tan{t}[/imath] as [imath]-1[/imath]. But the answer appears to be [imath]1[/imath]. Can anyone help me on this. |
762243 | Range of f(x) = [imath]\frac{\sqrt3\,\sin x}{2 + \cos x}[/imath]
Can you give any idea about the range of the following function? [imath]f(x) = \frac{\sqrt{3}\,\sin x}{2 + \cos x}[/imath] | 382136 | A problem on range of a trigonometric function: what is the range of [imath]\frac{\sqrt{3}\sin x}{2+\cos x}[/imath]?
What is the range of the function [imath]\frac{\sqrt{3}\sin x}{2+\cos x}[/imath] |
762625 | [imath]\lim_{n\to \infty}\left(1 - \frac {1}{n^2}\right)^n =?[/imath]
Can you give any idea regarding the evaluation of the following limit? [imath]\lim_{n\to \infty}\left(1 - \frac {1}{n^2}\right)^n[/imath] We know that [imath]\lim_{n\to \infty}\left(1 - \frac {1}{n}\right)^n = e^{-1}[/imath], but how do I use that here? | 576619 | Limit of [imath]\left(1-\frac{1}{n^2}\right)^n[/imath]
I'm trying to find the limit of [imath]a_n = \left(1-\frac{1}{n^2}\right)^n[/imath] for [imath]n \rightarrow \infty[/imath]. It seems that the limit is [imath]1[/imath], since [imath]a_n = 0.999...[/imath] for large [imath]n[/imath]. The presentation [imath]a_n = \frac{(n^2-1)^n}{n^{2n}}[/imath] and expanding was my first idea, but I couldn't get the result from there. Any ideas? |
762692 | Classifying functions that satisfy [imath]|f(x)-f(y)| \leq M|x-y|^{\alpha}[/imath]
If [imath]f: [0,1] \to \mathbb{R}[/imath] is a (not necessarily continuous) function satisfying [imath] |f(x)-f(y)| \leq M|x-y|^{\alpha} [/imath] where [imath]M[/imath] and [imath]\alpha[/imath] are fixed real numbers and [imath]\alpha > 1[/imath]. Classify all such functions [imath]f[/imath]. | 603291 | How to show that every [imath]\alpha[/imath]-Hölder function, with [imath]\alpha>1[/imath], is constant?
Suppose [imath]f:(a,b) \to \mathbb{R} [/imath] satisfy [imath]|f(x) - f(y) | \le M |x-y|^\alpha[/imath] for some [imath]\alpha >1[/imath] and all [imath]x,y \in (a,b) [/imath]. Prove that [imath]f[/imath] is constant on [imath](a,b)[/imath]. I'm not sure which theorem should I look to prove this question. Can you guys give me a bit of hint? First of all how to prove some function [imath]f(x)[/imath] is constant on [imath](a,b)[/imath]? Just show [imath]f'(x) = 0[/imath]? |
762467 | Axioms for Set Theory
I'm reading Zorich's "Mathematical Analysis I", Ed 4, 2004. Ch "1 .4 . 1 The Cardinality of a Set (Cardinal Numbers)" says: 3° [imath]\forall X \forall Y (\text{card} X \le \text{card} Y) \vee (\text{card} Y \le \text{card} X)[/imath] (Cantor's theorem). My question is, how to prove this Cantor's theorem? -- the proof is not provided by Zorich. Note: I suppose the "Cantor's theorem" referred by Zorich is not this one: [imath]\text{card} X < \text{card} \mathcal{P}(X)[/imath]. Though this is mentioned as Cantor's theorem in wiki, it is immediately mentioned by Zorich in the text, simply put as "Theorem". I can't see how [imath]\forall X \forall Y (\text{card} X \le \text{card} Y) \vee (\text{card} Y \le \text{card} X)[/imath] can be derived by [imath]\text{card} X < \text{card} \mathcal{P}(X)[/imath]. | 268942 | For any two sets [imath]A,B[/imath] , [imath]|A|\leq|B|[/imath] or [imath]|B|\leq|A|[/imath]
Let [imath]A,B[/imath] be any two sets. I really think that the statement [imath]|A|\leq|B|[/imath] or [imath]|B|\leq|A|[/imath] is true. Formally: [imath]\forall A\forall B[\,|A|\leq|B| \lor\ |B|\leq|A|\,][/imath] If this statement is true, what is the proof ? |
763040 | Prove that the only sets in [imath]R[/imath] which are both open and closed are the empty set and [imath]R[/imath] itself.
I am trying to prove the above proposition. I tried to prove it by way of contradiction letting [imath]S[/imath] be such nonempty proper subset of [imath]R[/imath]. Then [imath]T=R-S[/imath] would also be a nonempty proper subset of [imath]R[/imath] which is both open and closed. The method I thought of to lead contradiction is to create a sequence like this. Take [imath]s_0[/imath] from [imath]S[/imath] and [imath]t_0[/imath] from [imath]T[/imath]. Then consider the point [imath]\frac {s_0+t_0}{2}[/imath] and if this point belongs to [imath]S[/imath] call it [imath]s_1[/imath] otherwise [imath]t_1[/imath]. Repeat this process then eventually there is a sequence [imath]s_m[/imath] and [imath]t_m[/imath] which both converge to the same limit. But then since both sets are closed the limit point belongs to both sets, which is a contradiction. It's clear that such a sequence exists intuitively but I'm having trouble rigorously constructing such a sequence. Since I have no idea which set each of the newly constructed point belongs to, I don't know how to prove by [imath]\epsilon-N[/imath] definition of a sequence that both sequences converge to the same number. Can anybody help me? Moreover, how may I be able to generalize this proposition to the [imath]R^n[/imath] setting? Thanks. | 751886 | If a nonempty set of real numbers is open and closed, is it [imath]\mathbb{R}[/imath]? Why/Why not?
In other words, are [imath]\emptyset[/imath] and [imath]\mathbb{R}[/imath] the only open and closed sets in [imath]\mathbb{R}[/imath]? Why/Why not? I tried by assuming a set is equal to its interior points and contains its limit points. A bounded set will not do since stuff like [imath][1,4][/imath] and [imath]\{5\}[/imath] will not work, though that is not really proof. Help please? Anyway, it must then be unbounded. If [imath]a[/imath] is a real number then [imath](a,\infty)[/imath], [imath](-\infty,a)[/imath], [imath][a,\infty)[/imath] and [imath](-\infty,a][/imath] don't seem to cut it so it must be [imath]\mathbb{R}[/imath]. |
763217 | The only sets in [imath]R^n[/imath] which are both open and closed are the empty set and [imath]R^n[/imath] itself.
Proposition : The only sets in [imath]R^n[/imath] which are both open and closed are the empty set and [imath]R^n[/imath] itself. I came up with the proof of this claim and I'd like to know if my proof is correct. Proof: Let [imath]S[/imath] be the nonempty proper subset of [imath]R^n[/imath] which is both open and closed. Then [imath]T = R^n-S[/imath] is also a nonempty proper subset of [imath]R^n[/imath] which is both open and closed. Now take a point [imath]s_0[/imath] in [imath]S[/imath] and [imath]t_0[/imath] in [imath]T[/imath]. Then we can consider a closed box [imath]J_1[/imath] which contains both [imath]s_0[/imath] and [imath]t_0[/imath]. The n-dimensional interval [imath]I_1^k[/imath] is defined by the inequalities [imath]s_k\le x_k \le t_k[/imath] (k=1, 2, [imath]...[/imath], n ), where each [imath]s_k[/imath] and [imath]t_k[/imath] are the k-th coordinates of [imath]s_0[/imath] and [imath]t_0[/imath]. Without loss of generality, assume that for all k, [imath]s_k \le t_k[/imath]. Here [imath]J_1[/imath] denotes the catesian product [imath]J_1[/imath] = [imath]I_1^1[/imath] x [imath]\cdots[/imath] x [imath]I_n^1[/imath] Now, bisect all of the n 1-dimensional intervals and the point [imath]\frac {s_0+t_0}{2}[/imath] will lie in the center of the bisected box. This point belongs to either [imath]S[/imath] or [imath]T[/imath]. Say it belongs to [imath]S[/imath] then, rename it [imath]s_1[/imath] and the original [imath]t_0[/imath] to be [imath]t_1[/imath]. Now again consider a box that has both points as vertices like the previous case and repeat the process. Writing [imath]I_k^m[/imath] = [[imath]s_k^m[/imath],[imath]t_k^m[/imath]], we have [imath]t_k^m[/imath] - [imath]s_k^m[/imath] = [imath]\frac {t_k-s_k}{2^{m-2}}[/imath] Then for each fixed k, the sup of all left endpoints must therefore be equal to the inf of all right endpoints, and their common value we donote by [imath]b_k[/imath] (actually we know this by Bolzano-Weierstrass). Now we assert that b=([imath]b_1[/imath],[imath]...[/imath],[imath]b_n[/imath]) is a limit point of [imath]S[/imath] and [imath]T[/imath]. To see this, take any n-ball B(b; [imath]r[/imath]). The point b of course belongs to each of the boxes [imath]J_1[/imath], [imath]J_2[/imath], [imath]...[/imath] constructed above, and when [imath]m[/imath] is such that [imath]max_k\frac {t_k-s_k}{2^{m-2}}[/imath] < r/2, this neighborhood will include [imath]J_m[/imath] and since [imath]J_m[/imath] contains both points of [imath]S[/imath] and [imath]T[/imath], b is a limit point of both sets. But since both sets are closed, they both contain b, which is a contradiction. Hence, the proposition is true. | 2216227 | prove that if a nonempty subset [imath]S[/imath] of [imath]\mathbb{R}^n[/imath] is both open and closed, then [imath]S=\mathbb{R}^n[/imath]
I have to prove that if a nonempty subset [imath]S[/imath] of [imath]\mathbb{R}^n[/imath] is both open and closed, then [imath]S=\mathbb{R}^n[/imath] Does anyone have an idea I am new to this stuff. |
763597 | Let [imath]\{f_n(x)\}[/imath] be a sequence of polynomials defined as [imath]f_1(x) = (x - 2)^2[/imath], [imath]f_{n+1}(x) = (f_n(x) - 2)^2[/imath]; [imath]n \ge 1[/imath].
Let [imath]\{f_n(x)\}[/imath] be a sequence of polynomials defined as [imath]f_1(x) = (x - 2)^2[/imath] [imath]f_{n+1}(x) = (f_n(x) - 2)^2[/imath]; [imath]n \ge 1[/imath]. How to find the constant term and the coefficient of [imath]x[/imath] in [imath]f_n(x)[/imath]. | 459533 | What are the coefficients of the polynomial inductively defined as [imath]f_1=(x-2)^2\,\,\,;\,\,\,f_{n+1}=(f_n-2)^2[/imath]?
Let [imath]\{f_n(x)\}_{n\in \Bbb N}[/imath] be a sequence of polynomials defined inductively as [imath]\begin{matrix} f_1(x) & = & (x-2)^2 & \\ f_{n+1}(x)& = & (f_n(x)-2)^2, &\text{ for all }n\ge 1.\end{matrix}[/imath] Let [imath]a_n[/imath] and [imath]b_n[/imath] respectively denote the constant term and the coefficient of [imath]x[/imath] in [imath]f_n(x)[/imath]. Then [imath](\text A)\,\, a_n=4, b_n=-4^n[/imath] [imath](\text B)\,\,a_n=4, b_n=-4n^2[/imath] [imath](\text C)\,\, a_n=4^{(n-1)!}, b_n=-4^n[/imath] [imath](\text C)\,\, a_n=4^{(n-1)!}, b_n=-4n^2[/imath] Please help me how to solve this problem. I am not able to solve it. The problem can be found on this link. |
590653 | finding a recurrence relation for tile covering problem
for [imath]n \ge 1[/imath] let [imath]t_n[/imath] be the number of ways to.cover the squares of a 2xn xheckerboard using 1x2 tiles which can be rotated (ie 2x1 tile) and 2x2 tiles. 1x2 tile comes in 5 different colors and 2x2 tile in 4 different colors. find and solve a recurrence relation for [imath]t_n[/imath] | 591863 | Generating and solving recurrence relations
I am trying to do this question but don't know where to go from here: The question: For [imath]n\ge1[/imath] let [imath]t_n[/imath] be the number of ways to tile the squares of a 2xn checkerboard using 1x2(which can be rotated to be 2x1 tiles) and 2x2 tiles, if the 1x2 tiles come in 5 different colors and the 2x2 tiles in 4 different colors. Find and solve a recurrence relation for [imath]t_n[/imath]. I think I first part of 1x2 tiles to be : [imath]t_n=3t_{n-1} +9t_{n-2}[/imath] Is that part right? and how do i go from here |
764009 | Prove:[imath]a|bc[/imath] and [imath]\gcd(a,b)=1[/imath] implies [imath]a|c[/imath]
I need help with the following. Prove:[imath]a|bc[/imath] and [imath]\gcd(a,b)=1[/imath] implies [imath]a|c[/imath] following these writing guidelines http://i.imgur.com/qpIYqPp.png What I know so far: By the Euclidean algorithm there are integers [imath]x[/imath] and [imath]y[/imath] with [imath]ax + by = gcd(a,b) \implies ax + by = 1[/imath], in our case. Multiplying both sides by [imath]c[/imath] we have [imath]acx + bcy = c[/imath]. Then I get stuck trying to finish it. Update: By the Euclidean algorithm there are integers [imath]x[/imath] and [imath]y[/imath] with [imath]ax+by=\gcd(a,b)[/imath] thus [imath]ax+by=1[/imath]. Multiplying both sides by [imath]c[/imath] we get [imath]cax+cby=c[/imath] since [imath]a|bc[/imath] [imath]∃k[/imath] such that [imath]ak=bc[/imath] thus [imath]acx+aky=c[/imath] therefore [imath]a(cx+ky)=c[/imath], so [imath]a|c[/imath] for some [imath](cx+ky) \in \mathbb{Z}[/imath] Can someone verify that this is written properly? | 434164 | Prove the following: If [imath]a \mid bc[/imath], then [imath]a \mid \gcd(a, b)c[/imath].
Prove the following: If [imath]a \mid bc[/imath], then [imath]a \mid \gcd(a, b)c[/imath]. I tried to set [imath]\gcd(a, b)[/imath] to [imath]b[/imath] and used the fundamental theorem of arithmetic to prove that it is divisible by [imath]a[/imath], but I can't prove that [imath]a \mid bc[/imath], if and only if [imath]a\mid b[/imath] and [imath]a\mid c[/imath]. Please help. Thanks. |
763969 | How to evaluate the following series
Determine the sum of [imath]\sum_n^\infty \frac{k}{3^k}[/imath] Can someone teach me how to solve this please thanks. | 337937 | Why [imath]\sum_{k=1}^{\infty} \frac{k}{2^k} = 2[/imath]?
Can you please explain why [imath] \sum_{k=1}^{\infty} \dfrac{k}{2^k} = \dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots = 2 [/imath] I know [imath]1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}[/imath] |
764248 | tensor product of infinite dimensional vector spaces
Let [imath]V,W[/imath] be vector space over field [imath]k[/imath]. Then \begin{eqnarray*} V^*\otimes W &=&V^*(\oplus_{i\in I} k_i)\\ &=&\oplus_{i\in I}(V^*\otimes k_i)\\ &=& \oplus_{i\in I}V^*_i\\ &=& \oplus_{i\in I}Hom(V,k)_i, \end{eqnarray*} \begin{eqnarray*} Hom(V,W)=Hom(V,\oplus_{i\in I}k_i). \end{eqnarray*} When [imath]W[/imath] is finite dimensional, we have [imath]Hom(V,\oplus_{i\in I}k_i)=\oplus_{i\in I}Hom(V,k)_i[/imath]. Thus [imath]V^*\otimes W=Hom(V,W)[/imath]. Does the equality holds when [imath]W[/imath] is infinite dimensional and [imath]V[/imath] finite dimensional? Is the equality holds when [imath]W[/imath] is infinite dimensional and [imath]V[/imath] also infinite dimensional? I am confused. By universal property I only obtain [imath] Hom(M,\Pi_{i\in I}N_i)=\Pi_{i\in I}Hom(M,N_i)[/imath]. Can we obtain [imath]Hom(M,\oplus_{i\in I}N_i)=\oplus_{i\in I}Hom(M,N_i)[/imath] in general? Thanks. | 573378 | [imath]U^*\otimes V[/imath] versus [imath]L(U,V)[/imath] for infinite dimensional spaces
It's well known that [imath]U^*\otimes V\cong L(U,V)[/imath] for finite dimensional spaces. However, why people say that [imath]U^*\otimes V[/imath] is not isomorphic to [imath]L(U,V)[/imath] for infinite dimensional spaces and many books don't say anything related to this? |
545032 | Sum of squares of the quadratic nonresidues modulo [imath]p[/imath] is divisible by [imath]p[/imath]
Let [imath]p[/imath] be a prime number with [imath]p > 5[/imath]. Prove that the sum of the squares of the quadratic nonresidues modulo [imath]p[/imath] is divisible by [imath]p[/imath]. My idea is to use the fact that any quadratic residue is congruent modulo [imath]p[/imath] to one integer in the set [imath]\{1^2,2^2,\ldots,\left( \frac{p-1}{2} \right)^2 \}[/imath]. And none of the quadratic residues are congruent to each other. So the quadratic nonresidues must all be in the set [imath]\{ \left( \frac{p+1}{2} \right)^2 , \left( \frac{p+3}{2}\right)^2 \ldots, (p-1)^2\}[/imath]. So the sum of the quadratic nonresidues must be given by [imath]\sum_{k=1}^{p-1} k^2 - \sum_{k=1}^{\frac{p-1}{2}}k^2 = \frac{(p-1)p(2p-1)}{6} - \frac{\left( \frac{p-1}{2} \right) \left( \frac{p+1}{2} \right) p}{6} = \frac{p}{6} \left( (p-1)(2p-1) - \frac{1}{4}(p-1)(p+1) \right).[/imath] Is this the correct approach? Prove that the term in parenthesis is an integer divisible by [imath]6[/imath]? | 451467 | Sum of squared quadratic non-residues
Can you prove that if [imath]p[/imath] is a prime greater than [imath]5[/imath], then the sum of the squares of the quadratic nonresidues modulo [imath]p[/imath] is divisible by [imath]p[/imath]? Note that I have just proved that the sum of the quadratic residues modulo [imath]p[/imath] is divisible by [imath]p[/imath] for [imath]p[/imath] greater than [imath]3[/imath]. |
764499 | [imath]\frac {a_{n+1}}{a_n} \le \frac {b_{n+1}}{b_n}[/imath] If [imath]\sum_{n=1}^\infty b_n[/imath] converges then [imath]\sum_{n=1}^\infty a_n[/imath] converges as well
We have two positive series: [imath]\displaystyle\sum_{n=1}^\infty a_n[/imath], [imath]\displaystyle\sum_{n=1}^\infty b_n[/imath] and we know that: [imath]\frac {a_{n+1}}{a_n} \le \frac {b_{n+1}}{b_n}[/imath] (from a certain index). Show that if [imath]\displaystyle\sum_{n=1}^\infty b_n[/imath] converges then [imath]\displaystyle\sum_{n=1}^\infty a_n[/imath] converges as well. I think I know how to do it with contrapositive (suppose [imath]a_n[/imath] diverges, show it's bounded by [imath]b_n[/imath] thus it can't diverge) but I don't know how to show that [imath]\frac {a_{n+1}}{a_n} \le \frac {b_{n+1}}{b_n} \Rightarrow a_n \le b_n \ \forall n[/imath]. | 329003 | Prove if [imath]|a_{n+1}|/|a_n| \leq |b_{n+1}|/|{b_n}|[/imath] and [imath]\sum |b_n|[/imath] is convergent, then [imath]\sum |a_n|[/imath] is convergent
How do I prove that if [imath]|a_{n+1}|/|a_n| \leq |b_{n+1}|/|{b_n}|[/imath] and [imath]\sum |b_n|[/imath] is convergent, then [imath]\sum |a_n|[/imath] is convergent? I know that if I can get [imath]|b_{n+1}/{b_n}| < 1[/imath] then I can use the ratio test to say [imath]\sum a_n[/imath] is convergent, but I don't know if this the proper way to approach the problem. |
764570 | Prove that [imath]r^n/n![/imath] converges?
I need to show that [imath]r^n/n![/imath] converges where [imath]n\ge r[/imath]. Which is basically showing [imath]\lim_{n\to inf}\frac{r^n}{n!}=0.[/imath]. Yotas Trejos told I need to do this Let [imath]N[/imath] be an integer number such that [imath]N> r[/imath]. Then for [imath]n>N[/imath] the following holds: [imath]\displaystyle\frac{r^n}{n!}=\displaystyle\frac{r}{1}\cdots\displaystyle\frac{r}{N-1}\displaystyle\frac{r}{N}\cdots\displaystyle\frac{r}{n}<\displaystyle\frac{r}{1}\cdots\displaystyle\frac{r}{N-1}(\displaystyle\frac{r}{N})^{n-N} [/imath] you have that [imath]\displaystyle\frac{r}{1}\cdots\displaystyle\frac{r}{N-1}[/imath] is a constant and [imath]\displaystyle\frac{r}{N}<1[/imath]. But I'm having trouble concluding it. Is it just because it is a geometric sequence then after that? | 763919 | Prove that [imath]r^n/n![/imath] converges where [imath]n\ge r[/imath]
The answer is in the title of the question. I need to show it converges to 0 and [imath]r>0[/imath]. I am sorry if this is a bad question, I'm having trouble explaining it. So essentially this Do the [imath]\lim_{n\to inf}\frac{r^n}{n!}=0.[/imath] |
764633 | prove that the number [imath]38^n+31[/imath] is composite
Prove that for every positive integer [imath]n[/imath], [imath]38^n+31[/imath] is a composite number. for example [imath]38+31=69[/imath] is composite. [imath]38^2+31=1475[/imath] is also composite. I have tried modulo but it didn't work. | 597234 | Least prime of the form [imath]38^n+31[/imath]
I search the least n such that [imath]38^n+31[/imath] is prime. I checked the [imath]n[/imath] upto [imath]3000[/imath] and found none, so the least prime of that form must have more than [imath]4000[/imath] digits. I am content with a probable prime, it need not be a proven prime. |
765013 | how do i prove that the the set of irrationals cannot be a countable union of closed subsets?
Let [imath]\mathbb{R}[/imath] be equipped with the standard topology. Let [imath]E[/imath] be the set of irrational numbers. How do i prove that [imath]E[/imath] is not a countable union of closed subsets, using Baire Category Theorem? | 26311 | Proof That [imath]\mathbb{R} \setminus \mathbb{Q}[/imath] Is Not an [imath]F_{\sigma}[/imath] Set
I am trying to prove that the set of irrational numbers [imath]\mathbb{R} \setminus \mathbb{Q}[/imath] is not an [imath]F_{\sigma}[/imath] set. Here's my attempt: Assume that indeed [imath]\mathbb{R} \setminus \mathbb{Q}[/imath] is an [imath]F_{\sigma}[/imath] set. Then we may write it as a countable union of closed subsets [imath]C_i[/imath]: [imath] \mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty} \ C_i [/imath] But [imath]\text{int} ( \mathbb{R} \setminus \mathbb{Q}) = \emptyset[/imath], so in fact each [imath]C_i[/imath] has empty interior as well. But then each [imath]C_i[/imath] is nowhere dense, hence [imath] \mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty} \ C_i[/imath] is thin. But we know [imath]\mathbb{R} \setminus \mathbb{Q}[/imath] is thick, a contradiction. This seems a bit too simple. I looked this up online, and although I haven't found the solution anywhere, many times there is a hint: Use Baire's Theorem. Have I skipped an important step I should explain further or is Baire's Theorem used implicitly in my proof? Or is my proof wrong? Thanks. EDIT: Thin and thick might not be the most standard terms so: Thin = meager = 1st Baire category |
765050 | Please explain uniform distribution to me
I am confused about Uniform Distribution why does [imath]P(v < 2b1)[/imath] equal 2b1 ? | 764959 | Game Theory and Uniform Distribution question?
In an Auction , two players are bidding. Their bids will be a unknown fraction of their valuations. The valuations come from a uniform distribution [imath][0,1] [/imath] If Player 2 bids [imath] v/2 [/imath] and Player 1 bids [imath]b1<1/2[/imath] What is the probability player 1 wins ? Clearly for player 1 to win,players 2 bid has to be less than player 1 bids. [imath]P(v/2 < b1)[/imath] [imath]P(v < 2b1)[/imath] I follow the question up to this stage. Now it says since its uniformly distributed the probability player 1 wins is [imath]2b1[/imath] Im confused how can you just get 2b1 from the inequality ? Thanks in advance |
765142 | How do i show that if every continuous function on [imath]X[/imath] is bounded, then [imath]X[/imath] is compact?
Let [imath](X,d)[/imath] be a metric space. Assume every continuous function on [imath]X[/imath] is bounded. Prove that [imath]X[/imath] is compact. Well, i don't know which continuous function should i fix to start an argument. I tried to fix a function [imath]f(x)=d(x,x_0)[/imath], but i think it doesn't work. How do i prove this? | 668905 | If every real-valued continuous function is bounded on [imath]X[/imath] (metric space), then [imath]X[/imath] is compact.
Let [imath]X[/imath] be a metric space. Prove that if every continuous function [imath]f: X \rightarrow \mathbb{R}[/imath] is bounded, then [imath]X[/imath] is compact. This has been asked before, but all the answers I have seen prove the contrapositive. Realistically, this may be the way to go, but is there way to exhibit an unbounded continuous function (under the assumption [imath]X[/imath] is not compact) without appealing to results beyond introductory real analysis (e.g., the solution I've seen involves the Tietze Extension theorem)? Because we're working with metric spaces, it's clear that the assumption that [imath]X[/imath] is noncompact (towards proving contrapositive) will lead us to extract a sequence of points in the space with no convergent subsequences. But how can we infer the existence of an unbounded continuous on the space knowing only about some sequence of points in this space (without using anything too advanced)? |
765053 | Meaning symbol [imath]\overline{\lim_{n\rightarrow \infty}}[/imath]
I found this symbol on a math book that I'm studying. It had never happened before. What does this mean? [imath]\overline{\lim_{n\rightarrow \infty}}[/imath] | 21839 | What does limit notation with an underline or an overline mean?
I've never seen this notation before, and I'm having trouble finding a reference through search. Could someone explain what these notations mean for me? In context, the statement they're in is the following: a bounded [imath]f[/imath] is Riemann integrable iff [imath]\varliminf_{||C||\to 0}\mathcal{L}(f; C)\ge\varlimsup_{||C||\to 0}\mathcal{U}(f;C)[/imath] where [imath]C[/imath] is a non-overlapping, finite, exact cover of a rectangular region [imath]J[/imath] in [imath]\mathbb{R}^N[/imath], [imath]||C||[/imath] denotes mesh size, and [imath]\mathcal{L}, \mathcal{U}[/imath] represent the lower and upper Riemann sums, respectively. |
765086 | Show [imath]\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{y}{1+y^2}e^{-ay} dy =0 [/imath]
How to use LDC and MCT to prove? | 764230 | Show [imath]\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{1}{1+y^2}e^{-ay} dy =0 [/imath]
Need to prove [imath]\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{1}{1+y^2}e^{-ay} dy =0 [/imath] and [imath]\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{y}{1+y^2}e^{-ay} dy =0 [/imath] Can someone solve using dominated convergence theorem? I want to know how LDC is applied. |
765207 | How prove [imath]\binom{n}{m}\le\left(\frac{en}{m}\right)^m[/imath]
Show that [imath]\binom{n}{m}\le\left(\dfrac{en}{m}\right)^m[/imath] where [imath]0<m\le n,m,n\in N^{+}[/imath] My idea: since [imath](\dfrac{n}{m}-\dfrac{m-1}{m})(\dfrac{n}{m}-\dfrac{m-2}{m})\cdots\dfrac{n}{m}\le\left(\dfrac{en}{m}\right)^m[/imath] [imath]\sum_{i=1}^{m}\ln{\left(\dfrac{n}{m}-\dfrac{m-i}{m}\right)}\le m\ln{\dfrac{en}{m}}[/imath] then I can't,Thank you | 528852 | Simple upper bound for [imath]\binom{n}{k}[/imath]
I remember seeing an upper bound for the binomial [imath]\binom{n}{k}[/imath] with an exponential function, something like [imath]\binom{n}{k}\leq \left(ne/k\right)^k[/imath]. What exactly is it, and are there other similar good upper bounds for [imath]\binom{n}{k}[/imath]? Edit: As the link in Macavity's comment shows, the bound is indeed [imath]\binom{n}{k}\leq \left(ne/k\right)^k[/imath]. How can we prove this? |
765464 | Proving that [imath]S_k = \{A \subset \mathbb{N} : |A| = k\}[/imath] for [imath]k\in\mathbb{N}[/imath] is denumerable.
I am having trouble with this problem for quite some time. I posted this question before but I still can not figure out this problem. So far,from the suggestion of user134824, I have tried to define an injective function [imath]f[/imath] from [imath]S_k[/imath] to [imath]\mathbb{N}[/imath] as [imath]f_k({n_1,n_2,...n_k})=p_1^{n_1}p_2^{n_2}...p_k^{n_k}[/imath]. I am able to prove that this function is injective. However, after this point, I am stuck. I can either prove that [imath]f[/imath] is bijective by proving it is surjective or I can find an injective function [imath]g:\mathbb{N}\rightarrow S_k[/imath] and use the Schroeder-Bernstein theorem. | 765170 | Proving that these two sets are denumerable.
(a) [imath]S_k=\{A\subset\mathbb{N}: |A|=k\}[/imath] for [imath]k\in\mathbb{N}[/imath] (b) [imath]S = \bigcup_{k=1}^\infty S_k[/imath] Work: For (a), I am not too sure about what approach I should use. I think finding a bijective function between the [imath]S_k[/imath] and [imath]\mathbb{N}[/imath] is hard to do . I also thought about using the Schroeder-Bernstein theorem and find injective functions [imath]f:\mathbb{N}\rightarrow S_k[/imath]and [imath]g:S_k\rightarrow\mathbb{N}[/imath] in order to prove that [imath]S_k[/imath] and [imath]\mathbb{N}[/imath] are numerically equivalent. However, I have a hard time finding and proving these injective functions. I haven't really looked at (b) as I have yet to complete (a). |
766243 | Integrate the function [imath]x^x[/imath]?
Solve it in standard mathematical functions or any other? [imath]\int x^x dx[/imath] I don't know how to do it I tried many ways but couldn't. | 107870 | Proof that [imath]\int x^x dx[/imath] can't be done in terms of elementary functions?
Is there any easy proof that [imath]\int x^x dx[/imath] can't be done in terms of elementary functions? I know this is true, because the Risch algorithm gives a decision process for integration in terms of elementary functions, Axiom provides a complete software implementation of the Risch algorithm, and Axiom can't do the integral. However, it would be nicer to have a human-readable proof. If it could be reduced to a standard special function such as a hypergeometric function, then we could reduce the proof to a proof that that function can't be expressed in terms of elementary functions. But neither Axiom nor Wolfram Alpha can reduce it to any other form. |
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