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2041254
|
Find all solutions to [imath]x^2\equiv 1\pmod {91}[/imath]
I split this into [imath]x^2\equiv 1\pmod {7}[/imath] and [imath]x^2\equiv 1\pmod {13}[/imath]. For [imath]x^2\equiv 1\pmod {7}[/imath], i did: [imath] (\pm1 )^2\equiv 1\pmod{7}[/imath] [imath](\pm2 )^2\equiv 4\pmod{7}[/imath] [imath](\pm3 )^2\equiv 2\pmod{7}[/imath] Which shows that the solutions to [imath]x^2\equiv 1\pmod {7}[/imath] are [imath]\pm1[/imath]. For [imath]x^2\equiv 1\pmod {13}[/imath], i did: [imath] (\pm1 )^2\equiv 1\pmod{13}[/imath] [imath](\pm2 )^2\equiv 4\pmod{13}[/imath] [imath](\pm3 )^2\equiv 9\pmod{13}[/imath] [imath] (\pm4 )^2\equiv 3\pmod{13}[/imath] [imath](\pm5 )^2\equiv {-1}\pmod{13}[/imath] [imath](\pm6 )^2\equiv 10\pmod{13}[/imath]Which shows that the solutions to [imath]x^2\equiv 1\pmod {13}[/imath] are [imath]\pm1[/imath]. Thus, I concluded that the solutions to [imath]x^2\equiv 1\pmod {91}[/imath] must be [imath]\pm1[/imath]. I thought that [imath]\pm1[/imath] were the only solutions, but apparently I am incorrect! How do I go about finding the other solutions to this congruence?
|
3023847
|
Solving [imath]x^2 \equiv 140 \pmod{221}[/imath]
I'm stuck with the last part of this question: solve [imath]x^2 \equiv 140 \pmod{221}[/imath]. We know that [imath]140 = 7 \times2^2\times5[/imath] and [imath]221 = 13 \times 17[/imath]. We split the original congruence in two, so we have: [imath]x^2 \equiv 140 \pmod{13}[/imath] [imath]x^2 \equiv 140 \pmod{17}[/imath] Applying the properties of moduli we have: [imath]x^2 \equiv 10\pmod{13} \rightarrow x=\pm6[/imath] [imath]x^2 \equiv 4\pmod{17} \rightarrow x=\pm2[/imath] After this point, it's not clear for me how I can arrive to the complete solution. Any advice?
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2094862
|
does a group of order [imath]pqr[/imath] where [imath]p[/imath],[imath]q[/imath] and [imath]r[/imath] are primes and [imath]p always have a normal subgroup of order qr?[/imath]
Theorem A group of order [imath]pqr[/imath] where [imath]p[/imath],[imath]q[/imath] and [imath]r[/imath] are primes and [imath]p<q<r[/imath] always has a normal subgroup of order [imath]qr[/imath]. I found this theorem in a book. But a normal subgroup should have an index of [imath]2[/imath] w.r.t . the group. Now in case of a group of order [imath]3×5×7[/imath], it must have a normal subgroup of order [imath]7×5[/imath] and for this subgroup its index w.r.t. the group is not [imath]2[/imath]. Is it not always necessary for a normal subgroup to have an index of 2 w.r.t. the group ?
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568165
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Groups of order [imath]pqr[/imath] and their normal subgroups
We want to prove that a group, say [imath]G[/imath], of order [imath]pqr[/imath] where [imath]p \gt q \gt r[/imath] has a normal Sylow [imath]p[/imath]-subgroup and deduce that it has a normal subgroup of order [imath]pq[/imath]. I know how to show [imath]G[/imath] has a normal Sylow [imath]p[/imath]-subgroup or normal Sylow [imath]q[/imath]-subgroup, but not the rest. Thanks for your advices.
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2094899
|
How do you prove a triangle with the hypotenuse of length 5 and other sides with lengths 3 and 4 is a right triangle?
I can see that it obviously will be satisfied by the pythagorean theorem, that is: [imath]3^2+4^2=5^2[/imath] But I am sure this isn't the way to prove the statement since you are making the assumption that it is a right triangle. I know I also can't use any of the trig functions like sin, cos, tan. Then given the sides of the triangle, can I prove this to be a right triangle?
|
1911806
|
the converse of Pythagoras Theorem
Is it possible to prove the converse of the Pythagoras theorem with out a geometric proof?.That is from [imath]a^2+b^2=c^2[/imath] to [imath]\Theta=90^\circ[/imath]
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2094484
|
Fermat's Little Theorem's Combinatorial Proof.
The following is a Combinatorial proof of Fermat's Little Theorem from Arthur Engel's book:Problem Solving Strategies. We have pearls with [imath]a[/imath] colors. From these we make necklaces with exactly [imath]p[/imath] pearls. First, we make a string of pearls. There are [imath]a^p[/imath] different strings. If we throw away the a one-colored strings [imath]a^p − a[/imath] strings will remain. We connect the ends of each string to get necklaces. We find that two strings that differ only by a cyclic permutation of its pearls result in indistinguishable necklaces. But there are [imath]p[/imath] cyclic permutations of p pearls on a string. Hence the number of distinct necklaces is [imath](a^p − a)/p[/imath]. Because of its interpretation this is an integer. So [imath]p | (a^p − a)[/imath]. I can't figure out how the primality of [imath]p[/imath] is utilised in this proof. Please Explain.
|
555828
|
Combinatorial Proof Of A Number Theory Theorem--Confusion
I came across a combinatorial proof of the Fermat's Little Theorem which states that If [imath]p[/imath] is a prime number then the number ([imath]a[/imath][imath]p[/imath]-[imath]a[/imath]) is a multiple of [imath]p[/imath] for any natural number [imath]a[/imath]. Let me write down the proof. PROOF We have pearls of [imath]a[/imath] colours . From these we make necklaces of exactly [imath]p[/imath] pearls . First,we make a string of pearls . There are [imath]a[/imath][imath]p[/imath] different strings.If we throw away the [imath]a[/imath] one one -coloured pearls ([imath]a[/imath][imath]p[/imath]-[imath]a[/imath]) strings will remain.We connect the ends of each string to get necklaces.We find that two strings that differ only by a cyclic permutation of its pearls result in indistinguishable necklaces.But there are [imath]p[/imath] cyclic permutations of [imath]p[/imath] pearls on a string . Hence the number of distinct necklaces is [([imath]a[/imath][imath]p[/imath]-[imath]a[/imath])/[imath]p[/imath]]. Because of this interpretation this is an integer.Thus ([imath]a[/imath][imath]p[/imath]-[imath]a[/imath]) is a multiple of [imath]p[/imath] for any natural number [imath]a[/imath]. [HENCE PROVED] MY CONFUSIONS What is the use of the fact that [imath]p[/imath] is a prime in this proof? If this proof is true then why is this not true for every positve integer [imath]p[/imath]?
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2094956
|
Problem in proving that the radical of an ideal is an ideal
Copy from my exercise book: Let [imath]R[/imath] be a commutative ring (not necessarily with identity). For an ideal [imath]A \subseteq R[/imath] we define the radical [imath]\sqrt{A}[/imath] of [imath]A[/imath] by [imath]\sqrt{A}:=\{x \in R : x^n \in A \text{ for some } n \in \mathbb{N}\}[/imath] Now I should show that [imath]\sqrt{A}[/imath] is an ideal in [imath]R[/imath]. Everything is easy except the part where I should show that if [imath]x \in \sqrt{A}[/imath], then also [imath]-x \in \sqrt{A}[/imath]. My idea was to show, that [imath]x^n \in A[/imath] for some [imath]n \in \mathbb{N}[/imath], then also [imath](-x)^n \in A[/imath]. Intuitively we should end up with something like [imath](-x)^n = \pm[/imath] but I am not sure how to prove it. The solutions in the book are rather confusing, since they say something like since [imath]R[/imath] has identity...what was not assumed in the exercise.
|
1793633
|
Prove the Radical of an Ideal is an Ideal
I am given that [imath]R[/imath] is a commutative ring, [imath]A[/imath] is an ideal of [imath]R[/imath], and [imath]N(A)=\{x\in R\,|\,x^n\in A[/imath] for some [imath]n\}[/imath]. I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons. FIRST - We decided to assume [imath]n\in\mathbb{Z}^+[/imath] even though this restriction was not given. We decided [imath]n\ne 0[/imath] because then [imath]x^0=1[/imath] and we are not guaranteed unity. We also decided [imath]n\notin\mathbb{Z}^-[/imath] because [imath]x^{-1}[/imath] has no meaning if there are no multiplicative inverses. Is this a valid argument? SECOND - We want to assume [imath]x,\,y\in N(A)[/imath] which means [imath]x^m,\,y^n\in A[/imath] and use the binomial theorem to expand [imath](x-y)^n[/imath] which we have already proved is valid in a commutative ring and show that each term is in A so [imath]x-y[/imath] is in [imath]N(A)[/imath]. The biggest problem is how to approach the [imath]-y[/imath] if we are not guaranteed unity. Does anyone have any suggestions? Thank you in advance for any insight.
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2094444
|
Discontinuous function on given set
So we know a function has set of points of discontinuity [imath]F_{\sigma}[/imath]. Question: Is any [imath]F_{\sigma}[/imath] the set of discontinuity of some [imath]f[/imath] ? Tried but failed. [imath]\text{Thank you very much}[/imath].
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632652
|
Is every [imath]G_\delta[/imath] set the set of continuity points of some function [imath]f[/imath]?
I can prove that given a function [imath]f:X \rightarrow Y[/imath], where [imath]X,Y[/imath] are metric spaces, the set [imath]A \subseteq X[/imath] of points on which [imath]f[/imath] is continuous, is [imath]G_{\delta}[/imath]. (Take [imath]U_n = \bigcup_{y \in Im(f)} f^{-1}(B_{\frac{1}{n}}(y))[/imath] and [imath]V= \bigcap_{n \in \mathbb{N}}U_n[/imath], and [imath]V[/imath] is [imath]G_\delta[/imath]). My question is: Is the converse direction true? Is it true that, given a [imath]G_\delta[/imath] set [imath]A \subseteq X[/imath], there exists a metric space [imath]Y[/imath], and a function [imath]f:X \rightarrow Y[/imath], such that, the set of points on which [imath]f[/imath] is continuous is [imath]A[/imath]? Thank you! Shir
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2095344
|
[imath]z^2 = \sqrt3+ 3i[/imath] (complex equation)
I know I can take square root on both sides so I get z and [imath]\sqrt{\sqrt3+3i}[/imath] but is there a way of simplifying [imath]\sqrt{\sqrt3+3i}[/imath] ? EDIT: Seems like taking the square root may not be the best thing to do (as answer suggests).
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148871
|
Finding square roots of [imath]\sqrt 3 +3i[/imath]
I was reading an example, where it is calculating the square roots of [imath]\sqrt 3 +3i[/imath]. [imath]w=\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)\\=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})[/imath] Let [imath]z^2=w \Rightarrow r^2(\cos(2\theta)+i\sin(2\theta))=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})[/imath]. But how did they get from [imath]\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})[/imath]? And can one just 'let [imath]z^2=w[/imath]' as above? Edit: [imath]w=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})=z^2\\ \Rightarrow z=\sqrt{2\sqrt 3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})\\ \Rightarrow \sqrt{2\sqrt 3}\frac{\sqrt 3}{2} +i \sqrt{2\sqrt 3} \frac{1}{2}[/imath]
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2095102
|
Prove that [imath]a[/imath] is an eigen value of [imath]p(T)[/imath] where [imath]T[/imath] is a linear operator on [imath]V[/imath] [imath]\iff[/imath] [imath]a=p(\lambda[/imath] )
Let [imath]V[/imath] be a vector space over [imath]\Bbb C[/imath], [imath]p[/imath] is a polynomial over [imath]\Bbb C[/imath] and [imath]a\in \Bbb C[/imath] . Prove that [imath]a[/imath] is an eigen value of [imath]p(T)[/imath] where [imath]T[/imath] is a linear operator on [imath]V[/imath] [imath]\iff[/imath] [imath]a=p(\lambda[/imath] ) for some eigen value [imath]\lambda[/imath] of [imath]T[/imath]. Atempt:[imath]a[/imath] is an eigen value of [imath]p(T)[/imath] where [imath]p(z)=(z-\lambda_1)(z-\lambda_2)\ldots (z-\lambda_n)[/imath] say. [imath]\implies p(T)v=av\implies (z-\lambda_1)(z-\lambda_2)\ldots (z-\lambda_n)(v)=av[/imath] But how to show that [imath]a=p(\lambda)[/imath]. Please help.
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187510
|
Eigenvalue of a polynomial evaluated in a operator
Suppose [imath]T:V\to V[/imath], [imath]p\in \mathcal{P}(\mathbb{C})[/imath] (polynomials with complex coefficients), and [imath]a\in \mathbb{C}[/imath]. Prove that [imath]a[/imath] is an eigenvalue of [imath]p(T)[/imath] if and only if [imath]a=p(\lambda)[/imath] for some eigenvalue [imath]\lambda[/imath] of [imath]T[/imath]. I can prove: if [imath]a=p(\lambda)[/imath] then [imath]a[/imath] is a eigenvalue of [imath]p(T)[/imath] because: [imath]Tv=\lambda v[/imath] [imath]T^kv=\lambda^k v[/imath] [imath]p(T)v=p(\lambda)v=av[/imath] But, how can I justify the other direction? Thanks for your help.
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2095206
|
Show that [imath]\mathbb{R} \backslash \mathbb{Z}[/imath] is open.
Definition. A subset of [imath]\mathbb{R}[/imath] is called open if it is a union of open intervals. Example. [imath]\mathbb{R} \backslash \mathbb{Z}[/imath] is open. Proof (is from my lecture notes).[imath]\mathbb{R} \backslash \mathbb{Z}=\cup\left\{ \left( n,n+1\right) :n\in \mathbb{Z}\right\}[/imath]. My question is: why [imath]n[/imath] in [imath]\mathbb{Z}[/imath]? I think, [imath]n[/imath] cannot be in [imath]\mathbb{Z}[/imath] because we say [imath]\mathbb{R} \backslash \mathbb{Z}[/imath].
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1004001
|
How to prove that a set R\Z is open
So I've been trying to prove that a set [imath]\mathbb{R}[/imath] \ [imath]\mathbb{Z}[/imath] is an open set but I don't quite know where to start since the relative complement is still infinite. So theoretically in [imath]\mathbb{R}[/imath] \ [imath]\mathbb{Z}[/imath] one could pick a real number as close to an integer as needed but you could of course never pick the integer itself. So my idea was that [imath]a-\epsilon<a[/imath] for any [imath]\epsilon>0[/imath] with [imath]a\in \mathbb{Z}[/imath] but I have a feeling I'm doing it wrong. Any ideas or hints on how to solve this properly?
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2095137
|
Show that if [imath]A[/imath] is a nilpotent matrix then there exists [imath]k\leq n[/imath] s.t. [imath]A^k=0[/imath] without using Cayley-Hamilton
I am trying to show that for a nilpotent matrix [imath]A\in M_n(F)[/imath], there exists a natural number [imath]k\leq n[/imath] such that [imath]A^k=0[/imath]. Using the Cayley-Hamilton theorem, it is straightforward since a nilpotent matrix has a characteristic polynomial [imath]p_A(\lambda)=\lambda^n[/imath] so apply Cayley-Hamilton to deduce [imath]A^n=0[/imath]. I'm asking for a solution without using the Theorem. Thank you!
|
108422
|
How to show that the nth power of a [imath]n \times n[/imath] nilpotent matrix equals to zero [imath]A^n=0[/imath]
[imath]A[/imath] is a [imath]n\times n[/imath] matrix such that [imath] A^m = 0 [/imath] for some positive integer [imath]m[/imath]. Show that [imath]A^n = 0[/imath]. My attempt: For [imath]n > m[/imath], it's obvious since matrix multiplication is associative. For [imath]n < m[/imath], [imath]A^n\times A^{m-n} = 0[/imath]; not sure what to do next. Also I know that [imath]\det A = 0[/imath].
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2095861
|
Proofing that [imath]rk(A) \leq r[/imath]
So I have to proof that for an element [imath]A[/imath] of [imath]M_{m\times n}(K)[/imath] r an element of [imath]N[/imath], [imath]rk(A)\leq r[/imath] if there are two matrices [imath]B[/imath] element of [imath]M_{m\times r}(K)[/imath] and [imath]C[/imath] element of [imath]M_{r\times n}(K)[/imath] with [imath]A=BC.[/imath] I would be very happy if someone could help me.
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2094588
|
Connection between rank and matrix product
I have a problem understanding the following: Let [imath]A[/imath] be an [imath]m \times n[/imath] matrix and let t [imath]\in \mathbb{N}[/imath]. Prove that [imath]\operatorname{rank}(A)\leq t[/imath] if and only if there exists an [imath]m \times t[/imath] matrix [imath]B[/imath] and a [imath]t \times n[/imath] matrix [imath]C[/imath] so that [imath]A = BC[/imath]. I know what a rank is but I can't make a connection between the rank and the existence of two matrices such that [imath]A = BC[/imath].
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1961888
|
The Proximal Operator of the [imath] {L}_{1} [/imath] Norm Function
Write down explicitly the optimal solutions to the Moreau-Yosida regularization of the function [imath]f(x)=\lambda\|x\|_1[/imath], where [imath]f:\mathbb{R}^n\to(-\infty,+\infty][/imath]. I have found that the answer is [imath]x_i=sgn(x_i)\max\{|x_i|-1,0\}[/imath] Here is my attempt to get the answer: The proximal operator to [imath]f(x)[/imath] is [imath]\min_{y\in\mathbb{R}^n}\lambda\|y\|_1+\frac{1}{2}\|y-x\|^2_1[/imath]. I need to minimize this over y. I have no idea how to continue. I have read a lot of references but I cannot find an explicit step-by-step solution to this problem. Any help would be appreciated!
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1766811
|
$ {L}_{1} $ Regularized Unconstrained Optimization Problem
I am encountering an unconstrained minimization problem. The problem is of the form [imath]\min_x \frac{\|x-a\|_2^2}{2}+\lambda\|x\|_1[/imath] where [imath]x,a \in R^n[/imath] and [imath]x[/imath] is the optimization variable. [imath]\lambda \in R[/imath]. The problem can be separated in each coordinate as follows, [imath]\min_{x_i} \frac{(x_i-a_i)^2}{2}+\lambda|x_i|[/imath] I found in some resource that the optimal [imath]x_i^*[/imath] has the form [imath] x_i^* = \operatorname{sign}(a_i)\max(0,|a_i|-\lambda)[/imath]. I am having a hard time understanding why is it optimal solution for the problem. Hope anyone helps. Thanks.
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2095957
|
[imath]R[/imath] a commutative ring, [imath]I[/imath] a f.g. ideal and [imath]I^2=I[/imath] [imath]\implies[/imath] [imath]I=Re[/imath] for some idempotent element
Suppose that [imath]R[/imath] is a commutative ring and [imath]I[/imath] is a finitely generated ideal of [imath]R[/imath] such that [imath]I^2=I[/imath]. Show that [imath]I=Re[/imath] for some idempotent element [imath]e[/imath]. I think I should use induction on the number of generators of [imath]I[/imath]. I have proved that it's true for [imath]n=1[/imath]: Suppose [imath]I=\langle a\rangle[/imath] for some [imath]a \in I[/imath]. Since [imath]I^2=I[/imath] we have [imath]\langle a^2\rangle = \langle a\rangle[/imath] and therefore [imath]\exists r \in R: a=ra^2[/imath]. Multiplying both sides by [imath]r[/imath] and setting [imath]e=ra[/imath] we see that [imath]e^2=e[/imath] and [imath]I=\langle e\rangle=\langle ra\rangle\subseteq \langle a\rangle =I[/imath]. Therefore, [imath]I=Re[/imath]. Now for [imath]n=2[/imath], things look difficult. Suppose [imath]I=\langle a_1,a_2\rangle[/imath]. Again, we have [imath]\langle a_1^2,a_1a_2,a_2^2\rangle=\langle a_1,a_2\rangle[/imath] therefore there exists some coefficients in [imath]R[/imath] such that we have: [imath]a_1 = r_{11}a_1^2 + r_{12}a_1a_2 + r_{22}a_2^2[/imath] [imath]a_2 = s_{11}a_1^2 + s_{12}a_1a_2 + s_{22}a_2^2[/imath] After factorization: [imath]a_1 = a_1(r_{11}a_1 + r_{12}a_2) + r_{22}a_2^2[/imath] [imath]a_2 = s_{11}a_1^2 + (s_{12}a_1 + s_{22}a_2)a_2[/imath] It isn't easy to see an idempotent element in these relations, let alone finding one that generates [imath]I[/imath].
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42727
|
Finitely generated idempotent ideals are principal: proof without using Nakayama's lemma
I am trying to understand Nakayama's lemma. It looks like some "fixed point theorem". Using Nakayama's lemma , I can easily solve the following question. I want another proof. Thanks. Let [imath]A[/imath] be a commutative ring with identity, [imath]I[/imath] be a finitely generated ideal of [imath]A[/imath], such that [imath]I^2=I[/imath]. Show that there exists an element [imath]e\in I[/imath] with [imath]e^2=e[/imath] and [imath]eA=I[/imath]. You can use anything (geometric interpretation are welcomed) except Nakayama's Lemma.
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2095325
|
Trouble in understanding a proof of a theorem related to UFD.
Theorem : In a UFD [imath]R[/imath], for any [imath]a,b \neq 0[/imath] prove that [imath]ab \sim \gcd (a,b) . \operatorname{lcm} (a,b)[/imath]. Proof : Let [imath]d[/imath] be a gcd of [imath]a[/imath] and [imath]b[/imath] respectively. Let [imath]a = dx[/imath] and [imath]b = dy[/imath].We show that [imath]c = dxy[/imath] is the lcm of [imath]a[/imath] and [imath]b[/imath].Clearly [imath]a|c[/imath] and [imath]b|c[/imath].Let [imath]u \in R[/imath] such that [imath]a|u[/imath] and [imath]b|u[/imath].Then [imath]dx|u[/imath] and since [imath]y|b[/imath], we also have [imath]y|u[/imath].Thus [imath]c = dxy | u[/imath]. Therefore [imath]c[/imath] is a lcm of [imath]a[/imath] and [imath]b[/imath]. Now [imath]ab = dx . dy = d^2 xy = dc[/imath]. Also since any two gcds are associates and the same is true for lcms, we have [imath]ab \sim \gcd (a,b) . \operatorname{lcm} (a,b)[/imath] as required. But I fail to understand the logic that why [imath]dx|u[/imath] and [imath]y|u[/imath] [imath]\implies dxy|u[/imath]?I think it should be [imath]dxy | u^{2}[/imath].Would anybody please help me understanding this concept.Then it will help me a lot. Thank you in advance.
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2095021
|
How can I show that [imath]ab \sim \gcd (a,b) {\operatorname{lcm} (a,b)}[/imath] for any [imath]a,b \in R \setminus \{0\}[/imath]?
Let [imath]R[/imath] be a UFD. Then for any two elements [imath]a,b \in R \setminus \{0\}[/imath] [imath]ab \sim \gcd (a,b){\operatorname{lcm} (a,b)}[/imath]. My attempt : Let [imath]d = \gcd (a,b)[/imath].Then [imath]d|a[/imath] and [imath]d|b[/imath].Then [imath]\exists x,y \in R[/imath] such that [imath]a = dx[/imath] and [imath]b = dy[/imath].Now it is to be shown that [imath]c = dxy[/imath] is the lcm of [imath]a,b[/imath].Which I find difficulty to show.It is clear that [imath]a|c[/imath] and [imath]b|c[/imath]. Now if [imath]u[/imath] is any common multiple of [imath]a[/imath] and [imath]b[/imath].Then [imath]dx|u[/imath] and also since [imath]y|b[/imath] we have [imath]y|u[/imath] i.e. [imath]dxy|u^2[/imath] [imath]\implies c|u^2[/imath].But I have to show [imath]c|u[/imath] to complete the proof which I fail to prove.Please help me to complete the proof. Thank you in advance.
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2096285
|
Is the sequence of sum of [imath]\binom{n}{k}^{-1}[/imath] bounded?
Is there a real number [imath]M >0[/imath] such that [imath]S_n \leq M, \forall n \geq 1[/imath] where [imath] S_n = \sum_{k=0}^{n-1}\binom{n}{k}^{-1}, n\geq 1.[/imath] Thank you in advance
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151441
|
Calculate sums of inverses of binomial coefficients
How to calculate the sum of sequence [imath]\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{2}}+\frac{1}{\binom{n}{3}}+\cdots+\frac{1}{\binom{n}{n}}=？[/imath] How about its limit?
|
2096472
|
Proving divisibility of sequences
For [imath]a_n=6^{(2^n)}+1[/imath], how can I show that [imath]a_n\mid(a_{n+1}-2)[/imath] ?
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2096439
|
proving that elements of a sequence are pairwise coprime
For the sequence [imath]a_n = 6^{2^n} + 1[/imath]: How can I prove that the elements of this sequence are pairwise coprime, i.e. prove that if [imath]m[/imath] is not equal to [imath]n[/imath] then [imath]\gcd(a_m, a_n) = 1[/imath]. (begin by proving that [imath]a_n∣ (a_{n+1} - 2)[/imath])
|
2096678
|
An exercise about matrices and arithmetic
Let [imath]n[/imath] an integer, [imath]n>1[/imath]. Suppose there exist two real square matrices [imath]A,B[/imath] of size [imath]n[/imath] such that : [imath]A^2+B^2=AB[/imath] [imath]AB-BA[/imath] is invertible Prove that [imath]n[/imath] is divisible by [imath]3[/imath].
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299651
|
Square matrices satisfying certain relations must have dimension divisible by [imath]3[/imath]
I saw this tucked away in a MathOverflow comment and am asking this question to preserve (and advertise?) it. It's a nice problem! Problem: Suppose [imath]A[/imath] and [imath]B[/imath] are real [imath]n\times n[/imath] matrices with [imath]A^2+B^2=AB[/imath]. If [imath]AB-BA[/imath] is invertible, prove [imath]n[/imath] is a multiple of [imath]3[/imath].
|
2024464
|
Prove that [imath]\prod_{k=0}^{n-1}\sin \left( x + k\frac{\pi}{n} \right) = \frac{\sin nx}{2^{n - 1}}[/imath]
Prove that for any [imath]n \in \mathbb{N}^*[/imath] we have: [imath]\sin x \sin \left( x + \frac{\pi}{n} \right) \sin \left( x + \frac{2\pi}{n} \right)\cdots \sin \left( x + \frac{(n - 1)\pi}{n} \right) = \frac{\sin nx}{2^{n - 1}}[/imath] Here is what I did so far: We know that [imath]\sin x = \Im (e^{ix})[/imath]. So, we can rewrite our product the following way: [imath]\prod_{k = 1}^{n - 1} \sin \left( x + k\frac{\pi}{n} \right) = \prod_{k = 1}^{n - 1} \Im \left( e^{i \left( x + k\frac{\pi}{n} \right)} \right)[/imath] At this point I got stuck, mainly because I'm not sure if I can take the [imath]\Im[/imath] of the whole product or not. Thank you in advance!
|
1814527
|
Proof of [imath]\sin nx=2^{n-1}\prod_{k=0}^{n-1} \sin\left( x + \frac{k\pi}{n} \right)[/imath]
I have seen this identity on Wolfram mathworld and in a comment to another similar trigonometric proof: Prove that [imath]\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}[/imath] I can't seem to find a proof to this identity online anywhere. What method is used to prove this?
|
2096666
|
How can I start with this integral?
I don't know how to start and solve this integral. Can you help me? [imath]\int \dfrac{dx}{1+x^{7}}[/imath]
|
1999869
|
Evaluate [imath]\int\frac1{1+x^n}dx[/imath] for [imath]n\in\mathbb R[/imath]
I was wondering on how to evaluate the following indefinite integral for all [imath]n\in\mathbb R[/imath]. [imath]\int\frac1{1+x^n}dx[/imath] It seems to be peculiar in that we have [imath]\begin{align} \int\frac1{1+x^{-1}}dx&=x-\ln(x+1)+c\\ \int\frac1{1+x^0}dx&=\frac12x+c\\ \int\frac1{1+x^{1/2}}dx&=2\sqrt x-2\ln(1+\sqrt x)+c\\ \int\frac1{1+x^1}dx&=\ln(x+1)+c\\ \int\frac1{1+x^2}dx&=\arctan(x)+c\\ \int\frac1{1+x^3}dx&=\frac13\ln(1+x)-\frac2{3\sqrt3}\arctan\left(\sqrt{\frac43}\left(x-\frac12\right)\right)+c \end{align}[/imath] Naturally, there appears to be some combination of [imath]\ln[/imath] and [imath]\arctan[/imath], though no simple formula arises to solve the general case. It is, however, easy to see that [imath]\int\frac1{1+x^{-n}}dx=\int1-\frac1{1+x^n}dx[/imath] So there is an easy enough connection between positive and negative [imath]n[/imath]. Also, it is easy enough to find the series expansion, taking advantage of the above connection we just made to circumvent problems concerning convergence. [imath]\frac1{1+x^n}=1-x^n+x^{2n}-x^{3n}+\dots\forall\ |x|<1[/imath] [imath]\int\frac1{1+x^n}dx=c+x-\frac1{n+1}x^{n+1}+\frac1{2n+1}x^{2n+1}-\dots[/imath] [imath]=c+\sum_{k=0}^\infty\frac{(-1)^k}{kn+1}x^{kn+1}\ \forall\ |x|<1[/imath] Though this isn't very much along the lines of closed form. For [imath]n=\frac ab[/imath], where [imath]a[/imath] and [imath]b[/imath] are whole numbers, we can use the substitution [imath]x=u^b[/imath] to get [imath]\int\frac1{1+x^n}dx=\int\frac{bu^{b-1}}{1+u^a}du[/imath] though I'm unsure where that could lead. This reduces the integral down to [imath]\int\frac1{1+x^n}dx=b\int P(u)+\frac{u^{b-1-ak}}{1+u^a}du,\quad k\in\mathbb N[/imath] for some polynomial [imath]P(u)[/imath]. Though I'm still clueless as to how this can be advanced. How can I evaluate [imath]\int\frac1{1+x^n}dx\ \forall\ n\in\mathbb R[/imath] in closed form? Can someone prove there at least exists some closed form solution for all [imath]n\in\mathbb Q[/imath] if the above is not possible? If possible, use real numbers.
|
2097028
|
What is wrong with this?
I have a function [imath]f(x)=\frac{x^2-4}{x-2}[/imath] while solving this simply I get [imath]f(1)=3,f(2)=\frac00,f(3)=5,\cdots[/imath] the problem is when I solve this simply I get [imath]f(x)=x+2[/imath] and then I found except [imath]f(2)[/imath] all the values were same. Can someone help me where I did the mistake?
|
2085521
|
[imath]\frac{(x-1)(x+1)}{ (x+1)} \rightarrow (x-1)[/imath] Domain Change
Forgive my ignorance. The below seems 'inconsistent'. If canceling the [imath](x+1)[/imath] is 'legal', how does the domain change? I realize it does, but would someone be so kind as to provide an explanation? [imath] \frac{x^2 - 1}{x + 1} \mbox{ is undefined when } x = -1 [/imath] Its domain (the values that can go into the expression) does not include [imath] -1 [/imath]. Now, we can factor [imath] x^2 - 1 [/imath] into [imath] (x - 1)(x + 1) [/imath] so we get: [imath] \frac{(x - 1)(x + 1)}{(x + 1)} [/imath] It is now tempting to cancel [imath] (x + 1) [/imath] from top and bottom to produce: [imath] x - 1 [/imath] [imath] \mbox{Its domain now } \textbf{does} \mbox{ include } -1 \mbox{.} [/imath] But it is now a different function because it has a different domain. Thanks!
|
2097108
|
Proof of a certain divisibility property
I have been asked to prove that [imath](a+b+c)\big|(a^n + b^n + c^n)[/imath] for infinitely many [imath]n[/imath] given that [imath](a+b+c)\big| (a^2+b^2+c^2)[/imath]. Here | denotes divides (divisibility). I have tried with the strong form of induction and tried to use the fact that [imath]a+b+c[/imath] also divides [imath]2( ab+bc+ca)[/imath], but could not prove it.
|
1408323
|
If [imath] a + b + c \mid a^2 + b^2 + c^2[/imath] then [imath] a + b + c \mid a^n + b^n + c^n[/imath] for infinitely many [imath]n[/imath]
Let [imath] a,b,c[/imath] positive integer such that [imath] a + b + c \mid a^2 + b^2 + c^2[/imath]. Show that [imath] a + b + c \mid a^n + b^n + c^n[/imath] for infinitely many positive integer [imath] n[/imath]. (problem composed by Laurentiu Panaitopol) So far no idea.
|
2097201
|
Complex Limits of [imath]\lim_{z\to0}\frac{1-\cos z}{z^2}[/imath]
How do I calculate the limit [imath]\lim_{z\to0}\frac{1-\cos z}{z^2}[/imath]
|
2096404
|
Compute specific lim without l'hopital rule
how can i compute this limit without L'hopital rule: [imath]\lim \limits _{z\to 0} \frac{1-cos z}{z^2}[/imath] i believe the answer is 1/2 by the rule, thanks
|
2039227
|
Quadratic equation - arithmetic progression
The coefficients of a quadratic equation [imath]ax^2+bx+c=0[/imath] are in arithmetic progression and [imath]a, b, c[/imath] all are positive. If the roots of the equation be [imath]k[/imath] and [imath]l[/imath] are integers.Then [imath]k +l+kl[/imath] =?
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2097100
|
Find the value of [imath]\alpha + \beta + \alpha\beta [/imath] from the given data
Suppose the quadratic polynomial [imath]P(x) = ax^2 + bx +c [/imath] has positive coefficients [imath] a, b, c [/imath] in an arithmetic progression in that order. If [imath]P(x) = 0[/imath] has integer roots [imath]\alpha [/imath] and [imath]\beta [/imath], then [imath]\alpha + \beta + \alpha\beta [/imath] equals ? 1) [imath]3[/imath] 2) [imath]5[/imath] 3) [imath]7[/imath] 4) [imath]14[/imath] It's a single choice correct problem. All genuine answers are welcome :) The question is from a famous Indian Scholarship test 'KVPY' for high school students. The official website doesn't provide any solutions to it that is why I am asking for help. My go on the question we know that in a quadratic equation [imath]ax^2 + bx +c[/imath] with roots [imath]\alpha, \beta[/imath] the sum of roots is [imath]\alpha +\beta = \frac{(-b)}{a} [/imath] and product is [imath]\alpha\beta = \frac{c}{a} [/imath] So [imath]\alpha + \beta + \alpha\beta = \frac{(-b+c)}{a}[/imath] and since [imath]a,b,c[/imath] are in arithmetic progression [imath]\frac{(a+c)}{2} = b[/imath] Something that I tried was also this - [imath] a, b, c [/imath] are in AP (arithmetic progression) [imath]a-b , 0, c-b[/imath] are in AP (subtracting by [imath]b[/imath]) [imath]1-\frac{b}{a}, 0, \frac{(c-b)}{a}[/imath] are in AP (dividing by [imath]a[/imath]) Now I can't proceed further. After a bit of plugging and chugging I still cant get an integer as an answer...
|
2097688
|
GCD of [imath](a+b,a^2-ab+b^2)= 1[/imath] or [imath]3[/imath]
I have been given that [imath](a,b)=1[/imath]. I need to prove that [imath](a+b,a^2-ab+b^2)=1[/imath] or [imath]3[/imath]. I am not able to reduce the expressions [imath]a+b[/imath] and [imath]a^2-ab+b^2[/imath] in a way that I could reach the proof of the given proposition. Any help would be appreciated.
|
759658
|
If [imath]\gcd(a, b) = 1[/imath] then prove that [imath]\gcd(a+b, a^2-ab+b^2) = 1[/imath] or [imath]3[/imath]?
If [imath]\gcd(a, b) = 1[/imath] then prove that [imath]\gcd(a+b, a^2-ab+b^2) = 1[/imath] or [imath]3[/imath]? I am shamelessly asking how to solve the problem? I have no idea how to start and solve. Please help.
|
2097570
|
Can we add, subtract , multiply and divide numbers only using their continued fraction representation?
If we have two real numbers [imath]u,v\ge 0[/imath] with the continued fractions [imath]u=[0,a_1,a_2,a_3,\cdots][/imath] and [imath]v=[0,b_1,b_2,b_3,\cdots][/imath] can we calculate the continued fractions of [imath]u+v,u-v,uv,\frac{u}{v}[/imath] only using the continued fraction representations of [imath]u[/imath] and [imath]v[/imath] ?
|
76036
|
Arithmetic of continued fractions, does it exist?
I'm interested in the arithmmetic of continued fractions and specially in multiplication. Consider [imath] f(x)=\cfrac{f_{0}(x)}{1-\cfrac{f_{1}(x)}{1+f_{1}(x)-\cfrac{f_{2}(x)}{1+f_{2}(x)-\cfrac{f_{3}(x)}{1+f_{3}(x)-\cdots}}}} [/imath] and [imath] g(x)=\cfrac{g_{0}(x)}{1-\cfrac{g_{1}(x)}{1+g_{1}(x)-\cfrac{g_{2}(x)}{1+g_{2}(x)-\cfrac{g_{3}(x)}{1+g_{3}(x)-\cdots}}}} [/imath] Are there arithmetic rules (algorithms) for the multiplication of continued fractions? Specifically, is it possible to obtain a continued fraction [imath]h(x)=f(x)\cdot g(x)[/imath] where [imath] h(x)=\cfrac{h_{0}(x)}{1-\cfrac{h_{1}(x)}{1+h_{1}(x)-\cfrac{h_{2}(x)}{1+h_{2}(x)-\cfrac{h_{3}(x)}{1+h_{3}(x)-\cdots}}}} [/imath] I've found this, but I'd like more. Does anyone here knows of papers, algorithms, etc? Thanks. Update 1: Another link.
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2097611
|
Problem with floor function and sequence [imath]a_{n+1}= \lfloor a_n \rfloor \{ a_n \}[/imath]
I found this problem in a Romanian Magazine. Let [imath](a_n)_{n \ge 1}[/imath] be a sequence such that [imath]a_1 \lt 0[/imath] and [imath]a_{n+1}= \lfloor a_n \rfloor \{ a_n \}[/imath]. Show that there exists [imath]n_0 \gt 1[/imath] such that [imath]a_{n+2}=a_n[/imath], [imath] \forall n \ge n_0[/imath].
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1872851
|
Proving that a sequence has [imath]a_n = a_{n+2}[/imath] for [imath]n[/imath] sufficiently large.
The problem states: a sequence of real numbers [imath]a_0, a_1, \dots [/imath] is defined as follows. [imath]a_0[/imath] is an arbitrary real number and for [imath]n \ge 0, a_{n +1} = \lfloor a_n \rfloor \{ a_n \}[/imath] (where [imath]\{x\} = x - \lfloor x \rfloor[/imath]). Prove that for [imath]a_n = a_{n+2}[/imath] for [imath]n[/imath] large enough. my attempt: I notice that this is a decreasing sequence because [imath]\frac{a_{n+2}}{a_{n+1}} = \frac{\lfloor a_{n+1} \rfloor \{ a_{n+1} \}}{ a_{n+1}} < 1[/imath] since [imath]\{ a_{n+1} \} \le 1[/imath] (we can exclude the case where it's equal to one since then it becomes the null sequence that satisfies the thesis). Moreover this sequence is bounded between [imath]0[/imath] and [imath]a_0[/imath] so by the monotone convergence theorem it must have a limit. Taking the limit I obtain [imath]\ell = \lim \lfloor a_n \rfloor (\ell - \lim \lfloor a_n \rfloor) \iff \ell = - \lim \lfloor a_n \rfloor^2 / (1 - \lim \lfloor a_n \rfloor) [/imath] And so if I prove that there exists [imath]n[/imath] s.t. [imath] \lfloor a_n \rfloor < 1[/imath] I am done. Because then the limit could only be [imath]0[/imath] and this implies that there exist infinite [imath]a_n = a_{n+2} = 0[/imath]. But I can't seem to prove it. Is my reasoning up to now ok? How could I proceed? Edit: Add original text of the problem.
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1615708
|
Is [imath]x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}[/imath] bounded?
Consider the sequence: [imath]x_1=3, x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}[/imath]. This sequence is bounded or is unbounded? Attempt Checking a few terms we get [imath]x_1 = 3, x_2 = \dfrac{5}{6}, x_3 = -\dfrac{119}{60},x_4 = \dfrac{239}{14280},\cdots[/imath]. I will prove by contradiction that it is bounded. Suppose that the sequence is bounded. Then there exists some [imath]x_k[/imath] such that [imath]x_k \geq \dfrac{x_n}{2}-\dfrac{2}{x_n}[/imath] or [imath]x_k \leq \dfrac{x_n}{2}-\dfrac{2}{x_n}[/imath] for all [imath]n[/imath]. I seem to get stuck here.
|
970480
|
The sequence [imath]x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}, x_0>0[/imath] is bounded?
Consider the sequence [imath]x_{n+1}=\frac{x_n}{2}-\frac{2}{x_n}, x_0>0.[/imath] How can you prove that this sequence is bounded or unbounded for those values of [imath]x_0[/imath] for which it is defined. With a number generator I noticed that all terms of the positive out of range [imath][-2,2][/imath] form finite sets in strictly decreasing order. We found that it can be periodically. For [imath]x_0=\frac{2}{\sqrt{3}}[/imath] its terms repeat of 2 by 2. I tried to prove boundedness to the method of reduction to absurdity but we did. Thanks so much for any suggestion
|
2098043
|
Why does [imath]\vert z\vert = z^5 [/imath] have 6 solutions in [imath]\mathbb{C}[/imath]
Why does [imath]\vert z\vert = z^5 [/imath] have 6 solutions in [imath]\mathbb{C}[/imath] I tried to solve it with the polar method and I only found 5 solutions. [imath] z_{(n+1)} = e^{\left(\frac{2\cdot\pi}{5} +\frac{2\cdot\pi\cdot n}{5} \right)} \quad with\quad n=0,1,2,3,4 [/imath] There must be something special with the [imath]\vert z\vert[/imath] , but I can't find what.
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2097868
|
Solutions of [imath]|z| = z^5[/imath] in [imath]\Bbb{C}[/imath]?
Why does this equation have 6 solutions in [imath]\Bbb{C}[/imath], [imath]|z| = z^5[/imath]
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2097798
|
If two groups [imath]G,G^{'} [/imath] of order [imath]n[/imath] are isomorphic?
If we have two groups [imath]G,G^{'} [/imath] of order [imath]n[/imath] and both have the same number of elements of a given order then are the two groups isomorphic? I meant that if [imath]o(G)=o(G^{'})=n[/imath] and if [imath]G[/imath] has [imath]p[/imath] elements of order [imath]m[/imath] then [imath]G^{'}[/imath] has also [imath]p[/imath] elements of order [imath]m[/imath] and this holds for each [imath]m\in \Bbb N[/imath] , then are [imath]G,G^{'} [/imath] isomorphic?
|
729611
|
Is a finite group uniquely determined by the orders of its elements?
Suppose we are given a finite group [imath]G[/imath], and suppose we know the order of [imath]G[/imath] as well as the orders of each of its elements. Does this information alone uniquely determine the group up to isomorphism? What if we add extra hypotheses about [imath]G[/imath] (solvable, simple, etc)? If not, can we at least extract some useful information regarding its structure? For example, if [imath]|G| = 8[/imath] and [imath]G[/imath] contains three elements of order [imath]2[/imath] and four elements of order [imath]4[/imath], then I know [imath]G \cong \mathbb{Z}_4\times\mathbb{Z}_2[/imath].
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558028
|
Limit of product with unbounded sequence [imath]\lim_{n\to\infty} \sqrt{n}(\sqrt[n]{n}-1)=0[/imath]
I have to show that [imath]\lim_{n\to\infty} \sqrt{n}(\sqrt[n]{n}-1)=0[/imath] We proofed that [imath]\lim_{n\to\infty} \sqrt[n]{n}=1[/imath] My problem is, that I do not know how to solve that. That [imath]\lim_{n\to\infty}(\sqrt[n]{n}-1)=0[/imath] is clear. But [imath]\lim_{n\to\infty} \sqrt{n}[/imath] is not bounded. So I can not simply calculate: [imath]\lim_{n\to\infty}\sqrt{n}\cdot \lim_{n\to\infty}(\sqrt[n]{n}-1)[/imath] right? I would be thankfull for every hint. :-)
|
73403
|
How to prove [imath]\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n} - 1) = 0[/imath]?
I want to show that [imath]\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n}-1) = 0[/imath] and my assistant teacher gave me the hint to find a proper estimate for [imath]\sqrt[n]{n}-1[/imath] in order to do this. I know how one shows that [imath]\lim_{n \to \infty} \sqrt[n]{n} = 1[/imath], to do this we can write [imath]\sqrt[n]{n} = 1+x_n[/imath], raise both sides to the n-th power and then use the binomial theorem (or to be more specific: the term to the second power). However, I don't see how this or any other trivial term (i.e. the first or the n-th) could be used here. What estimate am I supposed to find or is there even a simpler way to show this limit? Thanks for any answers in advance.
|
2097656
|
Pigeonhole principle: From any ten two-digit numbers we can take two subsets with the same sum
Given a set [imath]B \subseteq \{10,11,12,\dots,99\}[/imath] such that [imath]\lvert B\rvert = 10[/imath] show there are at least two disjoint non empty subsets of [imath]B[/imath] such that their sum is equal. I want to use pigeonhole principle, but a little confused, How can I know how many disjoint subsets of [imath]B[/imath] exists? Am I in the right direction? Thanks.
|
2087594
|
Any [imath]B \subset \{10,11,...,99\}[/imath], with [imath]|B| =10[/imath] has two subsets such that [imath]\sum_{a \in B_1} a = \sum_{b \in B_2} b[/imath]
Let there be [imath]B \subset \{10,11,...,99\}[/imath], with [imath]|B| =10[/imath]. I need to prove, using the pigeonhole principle, that there are at least two disjoint subsets of [imath]B[/imath] (and none empty) where the sums of their elements are identical. I.e. [imath]\exists B_1 \ne \varnothing, B_2 \ne \varnothing[/imath] such that [imath]B_1 \bigcup B_2 = B[/imath], [imath]B_1 \bigcap B_2 =\varnothing[/imath] and [imath]\sum_{a \in B_1} a = \sum_{b \in B_2} b[/imath]
|
443874
|
Prove that MCD is 1 or 3 knowing (a,b) = 1
I'm don't know from where to start in this problem. I have to prove that [imath](a+b,a^2-ab+b^2)=1 \text{ or } 3[/imath] knowing that [imath](a,b) = 1[/imath]. I've tried using the method they taught us on class, so I've written that: [imath]d\ |\ a+b \Rightarrow d\ | a^2 + 2ab + b^2[/imath] and [imath]d\ |\ a^2-ab+b^2 [/imath] [imath]\Rightarrow d\ | \ 3ab[/imath] But from here, I dont know where to go. Usually, I should be able to find that d divides something in terms of a, and something in terms of b, and therefore, using that [imath](a:b)=1[/imath] I would get the solution. But this is not the case. Any operation I've tried between the two terms, gives me an other term, but always in terms of a and b, and no a or b, so I'm not able to prove what I'm asked. Any ideas?
|
1190930
|
If [imath]\gcd(m,n)=1[/imath], find [imath]\gcd(m+n,m^2-mn+n^2)[/imath]?
If [imath]\gcd(m,n)=1[/imath], find [imath]\gcd(m+n,m^2-mn+n^2)[/imath], I don't know how to find it, I tried with linear combination [imath]-(m+n)*(m+n)+m^2-mn+n^2=-3mn [/imath], but it doesn't help me...
|
1485620
|
How to calculate the following limit: [imath]\lim_{x\to+\infty} \sqrt{n}(\sqrt[n]{x}-1)[/imath]?
I have to solve a series of limits and I can't find out this one. [imath]\lim_{n\to+\infty} \sqrt{n}(\sqrt[n]{x}-1)= \,?[/imath] I have the feeling that this is equal to [imath]0[/imath], but I don't know how to prove it. Note that it may be a good idea to use [imath]\lim_{x\to+\infty} \sqrt[n]{x}=1[/imath].
|
1981222
|
Limit [imath]\lim_{n\to\infty}\sqrt n(\sqrt[n]x-1)[/imath]
[imath]\lim_{n\to\infty}\sqrt n(\sqrt[n]x-1)[/imath] I have never previously encountered functional sequences, so I am a newbie in this matter. I know that intuitively, [imath]x^{1/n}\to1[/imath] and that [imath]\infty\cdot0=1[/imath]. As a result, the limit should be 1. However, can this reasoning be considered correct and rigorous enough? I'm pretty sure that if I write this in my first year bachelor exam, my answer will be considered incorrect. What is the more formal way of doing this? (I think that the squeeze theorem should work, but can't choose suitable bounds). Thanks in advance.
|
2097885
|
What points in the complex plane satisfy [imath]|z|=arg(z)[/imath].
Question: What points in the complex plane satisfies [imath]|z|=arg(z)[/imath]. Work thus far: Let [imath]z=a+bi[/imath] and [imath]arg(z)[/imath] is the angle between the vector [imath]z[/imath] and the axis. So [imath]|z|=\sqrt{a^2+b^2}[/imath] and [imath]a=|z|\cos(\theta),b=|z|\sin(\theta)[/imath]. I would expect from intuition that the shape traced out by the points is a spiral. From here I don't how to proceed. Any hints would be appreciated.
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1952356
|
Solve [imath]|z|=\arg z[/imath]
Im trying to find all set of points on the complex plane for which [imath]|z|=\arg z[/imath]. I rewrote [imath]|z|= \sqrt{x^2 + y^2}[/imath] and [imath]\arg z[/imath] as [imath]\tan^{-1}(y/x)[/imath].I set them equal But im not sure what to do next.
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2098211
|
How to proceed this definite integral?
[imath]\int_{\frac{1}{2014}}^{2014}\frac{\arctan x}{x}\text{d}x[/imath]
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1774081
|
Definite integral [imath]\int_{\frac{1}{a}}^a \frac{\arctan(x)}{x}[/imath]
I have to fund the value of the above integral [imath]\int_{\frac{1}{a}}^a \frac{\arctan(x)}{x}[/imath] for [imath]a=2014[/imath]. I just saw that integral of [imath]\frac{\arctan(x)}{x}[/imath] does not have a closed form. So I used the substitution [imath]x=tan(\theta)[/imath] and then used integration by parts but did not get a result.
|
2098267
|
Let [imath]z[/imath] be a complex number, and let [imath]n[/imath] be a positive integer such that [imath]z^n = (z + 1)^n = 1[/imath]. Prove that [imath]n[/imath] is divisible by 6.
Let [imath]z[/imath] be a complex number, and let [imath]n[/imath] be a positive integer such that [imath]z^n = (z + 1)^n = 1[/imath]. Prove that [imath]n[/imath] is divisible by 6. I feel like [imath]|z|=1,[/imath] but I don't know how to prove it. Even if I found that, how would it help? Solutions are greatly appreciated. Thanks in advance! *I believe there's multiple ways to solve this question. And I would like to solve it using the one without much calculus. I only know up to intermediate algebra.
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1222722
|
[imath]z^n=(z+1)^n=1[/imath], show that [imath]n[/imath] is divisible by [imath]6[/imath].
we are given [imath]z^n=(z+1)^n=1[/imath], [imath]z[/imath] complex number. we want to prove that [imath]n[/imath] is divisible by [imath]6[/imath]. I showed that [imath]|z|=|z+1|=1[/imath]. Hence [imath]z[/imath] is on the intersection of two unit circles, one centered at [imath](0,0)[/imath] and the other one centered at [imath](-1,0)[/imath]. Then [imath]z[/imath] can take two values [imath]z=\operatorname{cis}(2 \pi /3)[/imath] or [imath]z=\operatorname{cis}(4 \pi/3)[/imath]. Then from [imath]z^n=1=\operatorname{cis}(2 \pi k)[/imath], [imath]k[/imath] integer I showed that [imath]n[/imath] should be divisible by [imath]3[/imath]. I am left to show that [imath]n[/imath] should be divisible by [imath]2[/imath]. I think if I figure out a way to show that [imath]\operatorname{arg}(z+1)=\pi/3[/imath] or [imath]-\pi/3[/imath] then I would be done, by using [imath](z+1)^n=\operatorname{cis}(2\pi k)[/imath]. but how can I do this? Any ideas are welcome! Thanks in advance!
|
643024
|
Complex numbers - Exponential numbers - Proof
Let [imath]z[/imath] be a complex number, and let [imath]n[/imath] be a positive integer such that [imath]z^n = (z + 1)^n = 1[/imath]. Prove that [imath]n[/imath] is divisible by 6. For this problem I am stumped...how should I begin? Also there's a hint for it: From [imath]z^n = 1[/imath], prove that [imath]|z| = 1[/imath]. What does the equation [imath](z + 1)^n = 1[/imath] tell you? What do the resulting equations tell you about [imath]z[/imath]? Could someone give me a hint on where to begin? thanks in advance
|
2692031
|
Is this question correct?
Suppose [imath]A[/imath] is a complex number and [imath]n\in \mathbb{N}[/imath] such that [imath]A^n = (A+1)^n= 1[/imath], then the least value of [imath]n[/imath] is? I really doubt this question is correct because if [imath]A= A+1[/imath] then [imath]1=0[/imath], which is obviously false. If it correct, I'd like to know the reason(s) and how do I begin solving the problem (full solution not needed).
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2098201
|
Show that sum [imath]1+z+z^2+...+z^{n-1}=0[/imath].
Question: Let [imath]z=\cos(2\pi /n)+i\sin(2\pi /n)[/imath] for an integer [imath]n\geq2[/imath]. Show that [imath]1+z+z^2+...+z^{n-1}=0[/imath]. Work thus far: My initial idea is was to use euler's formula, but this leads to a sum of exponentials that can't be simplified easily. If I can separate the sum into two sums (one real, one imaginary) which both add to zero then I can show that the equality holds. Using DeMoivre's formula [imath]z^a=\cos(a2\pi /n)+i\sin(a2\pi /n)[/imath]. So we get [imath]1+\cos(2\pi /n)+\cos(4\pi /n)+...+\cos(2n\pi/n)[/imath] [imath]i\sin(2\pi /n)+i\sin(4\pi /n)+...+i\sin(2n\pi/n)[/imath] From there however I do not know how to find the sum of both series.
|
517966
|
Show that if [imath]r[/imath] is an nth root of [imath]1[/imath] and [imath]r\ne1[/imath], then [imath]1 + r + r^2 + ... + r^{n-1} = 0[/imath].
Show that if [imath]r[/imath] is an nth root of [imath]1[/imath] and [imath]r\ne1[/imath], then [imath]1 + r + r^2 + ... + r^{n-1} = 0[/imath]. I think I can represent all the roots of 1 as follows: [imath]r = 1^{\frac{1}{n}} ( \frac{\cos{2\pi k}}{n} + i\frac{\sin{2\pi k}}{n} )[/imath] From there I am not sure how to get to [imath]1 + r + r^2 + ... + r^{n-1} = 0[/imath].
|
2098376
|
How many solutions for the equation [imath]Y_1 + Y_2 + Y_3 + Y_4 + Y_5 = 6[/imath]?
How many solutions for the equation [imath]Y_1 + Y_2 + Y_3 + Y_4 + Y_5 = 6[/imath] ? where [imath]0\leq Y_1 \leq 3[/imath] and [imath]1\leq Y_2 \leq3[/imath] and [imath]Y_3,Y_4,Y_5\geq 0[/imath] and [imath]Y1,Y2,Y3,Y4,Y5 [/imath] are integers. I tried to apply brute force and got following equations : [imath]Y3 + Y4 + Y5 = 5\quad[/imath] has [imath]21[/imath] Solutions [imath]Y3 + Y4 + Y5 = 4\quad[/imath] has [imath]15[/imath] Solutions [imath]Y3 + Y4 + Y5 = 3\quad[/imath] has [imath]10[/imath] Solutions [imath]Y3 + Y4 + Y5 = 2\quad[/imath] has [imath]6[/imath] Solutions [imath]Y3 + Y4 + Y5 = 1\quad[/imath] has [imath]3[/imath] Solutions [imath]Y3 + Y4 + Y5 = 0\quad[/imath] has [imath]1[/imath] Solution Am I going right here ? EDIT: Using generating functions gives [imath]106[/imath], but what is the error in above method.
|
1879619
|
How many solutions are there to [imath]x_1 + x_2 + ... + x_5 = 21[/imath]?
How many solutions are there to the equation [imath]x_1 + x_2 + x_3 + x_4 + x_5 = 21[/imath] where [imath]x_1, i = 1,2,3,4,5[/imath] is a nonnegative integer such that [imath]0 ≤ x_1 ≤ 3, 1 ≤ x_2 < 4[/imath], and [imath]x_3 ≥ 15[/imath]? I have correctly completed the previous parts of the question but am having trouble with this particular restriction. The total number of solutions to the equation is [imath]C(25,21) = 12650[/imath]. I have tried to do each restriction individually: For [imath]0 ≤ x_1 ≤ 3[/imath], I solved this by subtracting the number of solutions when [imath]x_1 ≥ 4[/imath] from the total number of solutions. This is [imath]C(25,21) - C(21,17) = 5985[/imath]. For [imath]1 ≤ x_2 < 4[/imath], I have [imath]C(24,20) - C(21,17) = 5985[/imath]. For [imath]x_3 ≥ 15[/imath], I have [imath]C(10,6) = 210[/imath]. However, the restriction is when these are all together. I am not sure how to proceed. The answer is [imath]106[/imath].
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2098224
|
If [imath]a^a[/imath] divides [imath]b^b[/imath], then [imath]a[/imath] divides [imath]b[/imath]?
Consider [imath]a[/imath] and [imath]b[/imath], both positive integers. Is it true that [imath]a^a[/imath] divides [imath]b^b[/imath] implies [imath]a[/imath] divides [imath]b[/imath]? I can't seem to figure out this proof. My intuition is to use the fundamental theorem of arithmetic to break each number into its prime components, however I haven't been able to come up with a solution. Any help would be appreciated.
|
52583
|
How do we prove [imath]n^n \mid m^m \Rightarrow n \mid m[/imath]?
I'm not sure I've got this right. When proving [imath]a^n \mid b^n \Rightarrow a \mid b[/imath], can we do this indirectly? In short, "Suppose [imath]a[/imath] does not divide [imath]b[/imath], this implies that [imath]a^n[/imath] does not divide [imath]b^n[/imath]. But [imath]a^n \mid b^n[/imath], hence [imath]a[/imath] divides [imath]b[/imath]." How about [imath]n^n \mid m^m \Rightarrow n \mid m[/imath]? Can we do this the same way?
|
2098475
|
Prove the inequality....
Let [imath]x,y[/imath] be positive reals such that [imath]x+y=2[/imath]. Prove that : [imath]x^3y^3(x^3+y^3) \leq 2[/imath] Source : INMO 2002 My attempt : I started with the left side of the inequality to be proved. [imath]x^3y^3(x^3+y^3) = x^3y^3(x+y)(x^2+y^2-xy) = 2 x^3y^3(x^2+y^2+2xy-3xy)[/imath] [imath]=2x^3y^3(4-3xy)[/imath] How to proceed ? Do I have to some AM-GM or Cauchy-Schwarz on particular set of values ?
|
1649315
|
Show that [imath]x^3y^3(x^3+y^3) \leq 2[/imath]
If [imath]x,y[/imath] are positive reals such that [imath]x+y = 2[/imath] show that [imath]x^3y^3(x^3+y^3) \leq 2[/imath]. I thought about working backwards on this one. We can try to prove that [imath]x^3y^3(x^3+y^3) \leq 2[/imath] is true by first realizing that [imath]x^3+y^3 = (x+y)(x^2-xy+y^2)[/imath] and substituting in for [imath]x+y = 2[/imath]. We then get [imath]x^3y^3(x^2-xy+y^2) \leq 1[/imath]. How do I prove this is true?
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2097351
|
Find all integer value of [imath]a[/imath] ...
Find all integer values of [imath]a[/imath] such that the quadratic expression [imath](x+a)(x+1991)+1[/imath] can be factored as a product [imath](x+b)(x+c)[/imath] where [imath]b[/imath] and [imath]c[/imath] are integers. I tried expanding the brackets and equating it with [imath](x+b)(x+c)[/imath] : [imath]x^2 + 1991x + ax+1991a + 1 = x^2 + cx + bx + bc[/imath] [imath]=>(1991+a)x + (1991a+1) = (c+b)x + bc[/imath] If I equate corresponding coefficients, I get two equations with three unknowns. What is the "proper" approach to solve such problems ? EDIT : This problem has been taken from a math contest "RMO" held in the year 1991. So I think this problem can be generalized for the expression [imath](x+a)(x+\lambda)+1[/imath]. How to solve it in that case ?
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1949062
|
Find [imath]a[/imath] such that [imath](x+a)(x+1991)+1=(x+b)(x+c)[/imath] with [imath]a,b,c\in\Bbb Z[/imath]
Find all integer values of [imath]a[/imath] such that the quadratic expression [imath](x+a)(x+1991)+1[/imath] can be factored as a product [imath](x+b)(x+c)[/imath] where b and c are integers. I tried to do it by comparing the two expressions but I can't proceed.
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2097639
|
Universal property of free abelian group implies [imath]\phi(S)[/imath] generates F
I saw some definition of free abelian group of an arbitrary set [imath]S[/imath]: An free abelian group on [imath]S[/imath] is a abelian group [imath]F[/imath] and a map [imath]\phi:S\to F[/imath], s.t. for any ableian group [imath]G[/imath] and any map [imath]\rho:S\to G[/imath], there exists an unique homomorphism [imath]f:F\to G[/imath] s.t. [imath]f\circ \phi=\rho[/imath]. A problem ask me to prove that [imath]\phi(S)[/imath] generates [imath]F[/imath]. The hint is consider [imath]\phi(S)[/imath] as a subgroup of [imath]F[/imath]. This implies that there exists an unique homomorphism [imath]f:F\to \phi(S)[/imath] s.t. [imath]f|_{\phi(S)}[/imath] is identity function. But I don't know what to do next. Could you give me some hints?
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1499460
|
Showing a free abelian group is generated by its basis
I'm following Massey's "Basic Course in Algebraic Topology" and I'm stuck on his section explaining free abelian groups. He seems to be using a definition of free abelian groups that differs from most others' definitions, and then he leaves it to the reader to show that they're the same, but I can't figure out the proof. More specifically: He defines a free abelian group on an arbitrary set [imath]S[/imath] as an abelian group [imath]F[/imath] together with a function [imath]\phi:S\to F[/imath] such that for any abelian group [imath]A[/imath] and any function [imath]\psi:S\to A[/imath], there exists a unique homomorphism [imath]f:F\to A[/imath] such that [imath]f\circ\phi=\psi[/imath]. He shows that free abelian groups over a given set are unique up to isomorphism, and then leaves the following as an exercise: "Prove directly from the definition that [imath]\phi(S)[/imath] generates [imath]F[/imath]. [Hint: Assume not; consider the subgroup [imath]F'[/imath] generated by [imath]\phi(S)[/imath].]" I can't figure out this exercise, but it seems really important to understanding this particular definition of free abelian groups. My guess is to let [imath]\langle\phi(S)\rangle[/imath] be [imath]A[/imath] and let [imath]\psi[/imath] be [imath]\phi[/imath] in the above definition, but I can't produce a contradiction. Thank you for your help. Edit: I just realized this question is very related to this other question, but I'm not convinced by the responses. The responses show that [imath]\langle\phi(S)\rangle[/imath] is isomorphic to [imath]F[/imath], but that doesn't seem (to me...) to show that they are equal, since [imath]2\mathbb{Z}[/imath] is isomorphic to [imath]\mathbb{Z}[/imath] without being equal.
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2097561
|
Polynomial ring [imath]\Bbb Z[X]/(2, X^2+1)[/imath]
How does [imath]\Bbb Z[X]/(2, X^2+1)[/imath] look like? I've got a problem that asks a proof for the fact that this ring has [imath]4[/imath] elements, yet it's not isomorphic with [imath]\Bbb Z_2 \times \Bbb Z_2[/imath], but I can't even understand how this set looks like.
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1586458
|
Determining the structure of the quotient ring [imath]\mathbb{Z}[x]/(x^2+3,p)[/imath]
I'm interested in the following problem from Artin's Algebra text: Determine the structure of the ring [imath]\mathbb Z[x]/(x^2 + 3,p)[/imath], where (a) p = 3, (b) p = 5. I know that by the isomorphism theorems for rings we can take the quotients successively, and so [imath]\mathbb{Z}[x]/(p) \cong (\mathbb{Z}/p \mathbb{Z})[x] [/imath] as the map [imath]\mathbb{Z}[x] \to (\mathbb{Z}/p \mathbb{Z})[x][/imath] defined by [imath]\sum_{n} a_n x^n \mapsto \sum_{n} \overline{a_n} x^n[/imath] is a surjective ring homomorphism with kernel [imath](p)[/imath]. Thus it remains to study the quotients [imath](\mathbb{Z}/p \mathbb{Z})[x]/(x^2+3) [/imath] for [imath]p \in \{3,5\}[/imath]. If [imath]p=3[/imath], [imath](x^2+3)=(x^2)[/imath] in [imath](\mathbb{Z}/3 \mathbb{Z})[x][/imath], and by using polynomial division all distinct coset representatives can be reduced to the following list of 9 elements [imath]\{0,1,2,x,1+x,2+x,2x,1+2x,2+2x\}. [/imath] Moreover, it can shown that the list above gives 9 distinct cosets, as no difference of two distinct elements of the list is a multiple of [imath]x^2[/imath]. Since [imath]1[/imath] and [imath]x[/imath] generate two distinct additive groups of order [imath]3[/imath], the additive group of our quotient ring is not cyclic. Elementary group theory then shows [imath](\mathbb{Z}/3 \mathbb{Z})[x]/(x^2)^+ \cong (\mathbb{Z}/ 3 \mathbb{Z})^2 [/imath] as additive groups. I was then about to conclude that the multiplication on the quotient is then compatible with the usual one in [imath](\mathbb{Z}/3\mathbb{Z})^2[/imath], but this is wrong! It can be seen that the quotient is not isomorphic to [imath](\mathbb{Z}/3\mathbb{Z})^2[/imath] as a ring, because the former contains a nonzero element (represented by [imath]x[/imath]) whose square is zero, while the latter contains no such elements. If [imath]p=5[/imath], a full list of coset representatives is of length 25 [imath]\{0,1,2,3,4,x,1+x,2+x,3+x,4+x,2x,1+2x,2+2x,3+2x,4+2x,3x,1+3x,2+3x,3+3x,4+3x,4x,1+4x,2+4x,3+4x,4+4x \} .[/imath] And once again, one can see that these represent 25 distinct cosets. Similarly to the [imath]p=5[/imath] case, I've managed to prove that the additive group of this ring is isomorphic to [imath](\mathbb{Z}/5 \mathbb{Z})^2[/imath]. My questions: Have I made any mistakes in my argument? What exactly am I supposed to do in this question? determine the number of elements? Write down the tables for addition and multiplication? Any further information about these quotients will be appreciated, thanks!
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2098531
|
How does [imath](a,b)[/imath] where [imath]a,b\in \mathbb{Q}[/imath] form a basis for [imath]\mathbb{R}[/imath]?
Well I can't think how the open set [imath](\sqrt{2},\sqrt{3})[/imath] can be formed from this basis.
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974130
|
Set of open intervals in R with rational endpoints is a basis for standard topology on R
Show that the set [imath]\mathcal{B} = \{(a,b) \subset \mathbb{R}: a,b \in \mathbb{Q}\}[/imath] is a basis for the standard topology on [imath]\mathbb{R}[/imath] First I'll show that [imath]\mathcal{B}[/imath] is a basis on [imath]\mathbb{R}[/imath] and then i'll do a fineness comparison via bases: 1) Since [imath]\cup_{p,q \in \mathbb{Q}\;for \;p < q}(p,q) = \mathbb{R}[/imath], [imath]\mathcal{B}[/imath] covers [imath]\mathbb{R}[/imath]. Since the intersection of open intervals is an open interval, every point in the intersection of two open intervals is contained in an open subinterval. Thus, [imath]\mathcal{B}[/imath] is a basis for some toplology [imath]\tau[/imath] on [imath]\mathbb{R}[/imath].2) Let[imath]\;\mathcal{B_{stand}}\;[/imath]be the standard basis for the standard topology [imath]\tau_{stand}[/imath] on [imath]\mathbb{R}[/imath]. Let [imath](p, q)\in \mathcal{B}[/imath]. Since [imath]\cup_{x \in \mathbb{R},\;x > p} (x,q) = (p,q)[/imath], every basis element in [imath]\mathcal{B}[/imath] can be written as the union of basis elements in [imath]\mathcal{B_{stand}}[/imath]. Thus, [imath]\tau_{stand}[/imath] is finer than [imath]\tau[/imath].Let [imath](x, y)\in \mathcal{B}[/imath]. Since [imath]\cup_{p,q \in \mathbb{Q}, \;x< p< q < y\;} (p,q) = (x,y)[/imath], every basis element in [imath]\mathcal{B_{stand}}[/imath] can be written as the union of basis elements in [imath]\mathcal{B}[/imath]. Thus, [imath]\tau[/imath] is finer than [imath]\tau_{stand}[/imath].[imath]\therefore\;[/imath] [imath]\mathcal{B} = \{(a,b) \subset \mathbb{R}: a,b \in \mathbb{Q}\}[/imath] is a basis for the standard topology on [imath]\mathbb{R}[/imath] Feedback on my proof and alternative ways to solve the problem would be appreciated. (I also found a post where someone asked the same question, but I'm looking for proof verification/alternative proofs so I'm not sure what the protocol is.) Collection of open intervals [imath](a,b) \ a,b\in \mathbb{Q}[/imath] is a basis for euclidean topology on [imath]\mathbb{R}[/imath]
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2096102
|
Uniform continuity of [imath]f(x)=x + \frac{x}{x+1}[/imath] using definition
I need to prove that the following function is uniformly continuous on the interval [imath][0,\infty)[/imath] : [imath]f(x)=x + \frac{x}{x+1}[/imath] I want to prove it by defenition, any help?
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1784535
|
Show that [imath]f(x)=\frac{x}{1+|x|}[/imath] is uniformly continuous.
I tried to use the definition and arrived this far: [imath]|f(x)-f(y)|=\left|\frac{x}{1+|x|}-\frac{y}{1+|y|}\right|=\frac{|x-y+x|y|-y|x||}{(1+|x|)(1+|y|)}\leq|x-y+x|y|-y|x||[/imath]. Any suggestion for ending the proof? I also tried to prove that [imath]\frac{x}{1+x}[/imath] is uniformly continuous on [imath][0,\infty[[/imath] and that [imath]\frac{x}{1-x}[/imath] is uniformly continuous on [imath][-\infty,0[[/imath], but I wonder if we can use the definition with the function [imath]f(x)=\frac{x}{1+|x|}[/imath] itself.
|
769985
|
Prove that the product spaces [imath][0, 1] \times (0, 1][/imath] and [imath](0, 1] \times (0, 1][/imath] are homeomorphic.
Let [imath][0, 1][/imath] and [imath](0, 1][/imath] be given in the standard topology. Prove that the product spaces [imath][0, 1] \times (0, 1][/imath] and [imath](0, 1] \times (0, 1][/imath] are homeomorphic.
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521033
|
Show that [imath][0, 1)\times[0, 1)[/imath] is homeomorphic to [imath][0, 1]\times[0, 1)[/imath] but not to [imath][0, 1]\times[0, 1][/imath].
Show that [imath][0, 1)\times[0, 1)[/imath] is homeomorphic to [imath][0, 1]\times[0, 1)[/imath] but not to [imath][0, 1]\times[0, 1][/imath]. When I sketch these spaces it this statement makes sense intuitively because [imath][0, 1]\times[0, 1][/imath] is closed and [imath][0, 1)\times[0, 1)[/imath] and [imath][0, 1]\times[0, 1)[/imath] are each "missing" part of their boundary which would make them closed. I also know, [imath][0, 1]\times[0, 1][/imath] is compact and compact spaces are not homeomorphic to non-compact spaces. I am having trouble saying this explicitly, as is pretty obvious in the clumsiness of the wording above.
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2097303
|
Which of the following value(s) of [imath]t^3[/imath] are not possible?
Let [imath]t[/imath] be a real number such that [imath]t^2 = at + b[/imath] for some positive integers [imath]a[/imath] and [imath]b[/imath]. Then for any choice of positive integers [imath]a[/imath] and [imath]b[/imath], [imath]t^3[/imath] is not equal to - 1) [imath]4t+3[/imath] 2) [imath]8t+5[/imath] 3) [imath]10t+3[/imath] 4) [imath]6t+5[/imath] This question is from a prestigious Indian Scholarship exam, and I'm solving it for fun. But I can't get at the bottom of the problem. I'm also a bit shaky on which concept to use in this problem. My try on this question [imath]t^2 =at +b[/imath] [imath]t = \sqrt(at+b)[/imath] For [imath]t \in R[/imath] [imath]at+b \ge 0[/imath] Since [imath]a[/imath] and [imath]b[/imath] are positive integers [imath]at \ge -b[/imath] and [imath]t \ge \frac{-b}{a}[/imath] I also tried differentiating the function [imath]t^2 = at + b[/imath] to get [imath]\frac{d}{dt}(t^2) = a\frac{d}{dt}(t) + 0[/imath] [imath]2t = a[/imath] Alas it leads nowhere. But [imath]\frac{a}{2} \ge \frac{-b}{a}[/imath] Since [imath]a[/imath] and [imath]2[/imath] are positive integers [imath]a^2 \ge -2b[/imath] What has to done next? Please help
|
2071932
|
[imath]t^3[/imath] is never equal to
Let [imath]t[/imath] be a real number such that [imath]t^2=at+b[/imath] for some positive integers [imath]a[/imath] and [imath]b[/imath]. Then for any choice of positive integers [imath]a[/imath] and [imath]b[/imath], [imath]t^3[/imath] is never equal to: (A). [imath]4t+3[/imath] (B). [imath]8t+5[/imath] (C). [imath]10t+3[/imath] (D). [imath]6t+5[/imath] We have [imath]t^3 = at^2+bt[/imath] and so for (A) we have [imath]t^3 = 4t+3 = 4(at^2+bt)+3[/imath]. I don't see how to continue since this is a cubic equation. Is there an easier way of finding a contradiction?
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2099161
|
Equation [imath]x+e^x=2[/imath]
Show that the equation [imath]x+e^x=2[/imath] has only one solution in [imath]\mathbb{R}[/imath]. Which solution is this? I don't know how to show this formally but if we take [imath]e^x=2-x[/imath], than the left part ist increasing and the right part decreasing. This implies that we have only one solution in [imath]\mathbb{R}[/imath]. Right? How to show this formally and how to find this solution?
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380983
|
Solutions to exp(x) + x = 2
I am extremely new to mathematics, and I don't have much training except for the basics so please excuse my rather basic question. The question itself: If I have the relationship [imath]e^x + x - 2 = 0[/imath]; and [imath]k[/imath] is the number of solutions in [imath][0,1][/imath] and [imath]n[/imath] is the number of solutions not in [imath][0,1][/imath] what is [imath]k[/imath]? what is [imath]n[/imath]? My Questions (sorry if they are many) (1) What is the proper name for this type of equation. It doesn't seem to be a function, just a relationship of one variable to some constants. (2) How do you tell how many solutions an equation like this has just from looking at it? It seems like if you have an equation like [imath]x^2 + 2x + 5 = 0[/imath]; you have at least two solutions. But what about the [imath]e^x[/imath]? How do you predict how many solutions there will be then? (3) Once I re-arrange the equation to [imath]e^x + x = 2[/imath]; what algebraic steps could I follow if I wanted to find the exact solution?
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974762
|
Extinction probability of binomial branching process tends to poisson one.
The folowing is stated and proved in the random graphs book by Luczak, Janson, Rucinski and this is on page 108 in the Giant component section. I can't understand why the conclusion follows from the last statement in the proof. Question: We consider a series of branching processes with immediate offspring distribution [imath]X_n \sim \mathrm{Bin}(n,p)[/imath] where [imath]np \to c[/imath] as [imath]n\to \infty[/imath] for some constant [imath]c>1[/imath]. We also consider a random variable [imath]X \sim \mathrm{Poiss}(c)[/imath]. If we denote by [imath]\rho_{X_i}[/imath] the extinction probability. Show [imath]\rho_{X_i} \to \rho_X[/imath]. Solution: This is the solution given in my textbook, it seems wrong to me and I don't seem to be able to fix it, it is quite important to me as the remainder of the chapter is based on it. We know that generating functions are [imath]f_{X_n}(x)=(1-p+xp)^n[/imath] and [imath]f_X(x)=e^{c(x-1)}[/imath] It is now a standard result that for a fixed [imath]x[/imath] [imath]f_{X_n}(x) \to f_X(x)[/imath] so [imath]f_n[/imath] converge pointwise to [imath]f[/imath]. As [imath]c<1[/imath] we have [imath]f_{X_n}(\rho_{X_n})=\rho_{X_n}[/imath] and [imath]f_{X}(\rho_{X})=\rho_{X}[/imath] Now the book just states that this implies [imath]\rho_{X_n} \to \rho_X[/imath]. And I just don't see why.
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2098667
|
Extinction probabilities of binomial tends to Poisson distribution
I am stuck on exercise 11.2 From Grimmett's probability on graphs. Here is a link to the pdf on his website. Consider a branching process whose family-sizes have the binomial distribution bin[imath](n, \frac{\lambda}{n})[/imath]. Show that the extinction probability converges to [imath]\eta(\lambda)[/imath] as [imath]n \rightarrow \infty[/imath], where [imath]\eta(\lambda)[/imath] is the extinction probability of a branching process with family-sizes distributed as [imath]\text{Po}(\lambda)[/imath]. To solve this exercise, we use a theorem from his other book, which states that if you have a branching process whose family sizes are [imath]X[/imath] distributed then the extinction probability of this branching process is the smallest non-negative fixed point of the equation [imath]s = G(s)[/imath] where [imath]G[/imath] is the probability generating function [imath]G(s) = \sum_{k = 0}^\infty \mathbb{P}(X = k)s^k[/imath]. My next step was computing these probability generating functions for [imath]\text{bin}(n,\lambda)[/imath], these are [imath]G_n(s) = (1+\frac{\lambda( s -1)}{n})^n[/imath]. When the family sizes are Poisson distributed with parameter [imath]\lambda[/imath] we get [imath]G(s) = e^{\lambda(s-1)}[/imath]. and so clearly [imath]G_n[/imath] converges pointwise to [imath]G[/imath]. This however is not enough, as we want to show that the extinction probabilities also converge, so we have [imath]s_n[/imath] extinction probabilities of the [imath]n[/imath]-th process, i.e. [imath]s_n[/imath] is the smallest non-negative solution to [imath]s = G_n(s)[/imath], and we want to show that this converges the smallest non-negative solution of [imath]s = G(s)[/imath]. I hoped I could use some theorem which states that if you have pointwise convergence then the fixed points also converge, but this is false, see this counterexample. During writing this I thought of an attempt to solve this using Hurwitz theorem, namely, show that the smallest non-negative fixed point of [imath]G(s)[/imath] is smaller than 1, then show that the function [imath]G'[/imath] is complex differentiable with non-zero derrivative, use inverse function theorem, get an open neighbourhood of our fixed point, find sequence of fixed points which converge to the smallest non-negative fixed point [imath]\eta(\lambda)[/imath] of [imath]G[/imath]. Here I am stuck trying to show that [imath]s_n[/imath] are the smallest non-negative fixed points of [imath]G_n[/imath]. How to proceed?
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2037508
|
maximum value of the expression : [imath]2x+3y+z[/imath] as [imath](x,y,z)[/imath] varies over the sphere [imath]x^2+y^2+z^2=1[/imath]
I have to find out the maximum value of the expression : [imath]2x+3y+z[/imath] as [imath](x,y,z)[/imath] varies over the sphere [imath]x^2+y^2+z^2=1.[/imath]
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276861
|
Optimization of [imath]2x+3y+z[/imath] under the constraint [imath]x^2+ y^2+ z^2= 1[/imath]
Let [imath]x, y[/imath] and [imath]z[/imath] be real numbers such that [imath]x^2+ y^2+ z^2= 1[/imath]. Find the maximum and minimum values of [imath]2x + 3y + z[/imath] . How can I able to solve the problem? Thanks for your help.
|
2099980
|
Prove that [imath]{\sqrt {n} }^{\sqrt {n+1}} > {\sqrt {n+1}}^{\sqrt {n}};n=7,8,9...[/imath]
Prove that [imath]{\sqrt {n} }^{\sqrt {n+1}} > {\sqrt {n+1}}^{\sqrt {n}};n=7,8,9...[/imath] SOURCE : "Inequalities proposed in CRUX Mathematicorum" I tried induction, but could not prove it. Can anybody provide a hint ?
|
317383
|
Prove that [imath]{\sqrt {n} }^{\sqrt {n+1}} > {\sqrt {n+1}}^{\sqrt {n}}[/imath].
Which is greater? [imath]\sqrt{n}^{\sqrt{n+1}}[/imath] or [imath]\sqrt{n+1}^\sqrt{n}[/imath] I know that [imath]\sqrt{n}^{\sqrt{n+1}}[/imath] is greater but I tried using induction and I couldn't figure it out. Thanks for the help.
|
2087808
|
Slick proofs that if [imath]\sum\limits_{k=1}^\infty \frac{a_k}{k}[/imath] converges then [imath]\lim\limits_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^n a_k=0[/imath]
I'm looking for slick proofs that if [imath]a_n[/imath] is a sequence of complex numbers such that [imath]\sum\limits_{k=1}^\infty \frac{a_k}{k}[/imath] converges then [imath]\lim\limits_{n\to\infty} \frac{1}{n}\sum\limits_{k=1}^n a_k=0[/imath]. My not so slick proof: Let [imath]A_x=\sum\limits_{k=1}^{x} a_k[/imath] and apply an Abel sum with the function [imath]f(x)=\frac{1}{x}[/imath]. We get [imath]\sum\limits_{k=1}^n \frac{a_k}{k}-\int\limits_1^n\frac{A_x}{x^2}dx=\frac{A_n}{n}.[/imath] Of course we can split the integral into chunks to get [imath]\sum\limits_{k=1}^n \frac{a_k}{k}-\sum\limits_{i=1}^n \left( \frac{1}{k+1}-\frac{1}{k} \right) A_k=\frac{A_n}{n}.[/imath] If we expand [imath]A_k[/imath] on the left side, it telescopes and yields [imath]\sum\limits_{k=1}^n \frac{a_k}{k} + \sum\limits_{k=1}^n \left( 1-\frac{1}{k+1}\right) a_k=\frac{A_n}{n}\iff \frac{(n-1)A_n}{n}=\sum\limits_{k=1}^n\frac{a_k}{k(k+1)}.[/imath] The norm of the right side of the equation is clearly [imath]\mathcal O(\log(n))[/imath] (using the triangle inequality and the fact that [imath]\frac{a_k}{k}[/imath] is bounded). The desired result follows after dividing both sides by [imath]n-1[/imath]. I would also appreciate some proof verification. Initially, I thought that the problem was going to be really easy, but it took me a bit of effort. Am I missing something major? The original problem is for real sequences, but I don't think this helps. Obviously if the sequence converged absolutely then it would also be absolutely trivial.
|
2312163
|
If [imath]\sum_{k = 1}^\infty \frac{a_k}{k} < + \infty[/imath], then [imath]\frac{1}{n} \sum_{k = 1}^n a_k \to 0[/imath]
I'm currently reading a paper, where they assert that for a nonnegative sequence [imath]a_k[/imath] of real numbers with [imath]\sum_{k = 1}^\infty \frac{a_k}{k} < + \infty[/imath], we have [imath]\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^n a_k = 0[/imath]. My attempt to prove this was using the following idea: If [imath]a_k > \varepsilon > 0[/imath] for all [imath]k \geq k_0[/imath], then [imath] \sum_{k = 1}^\infty \frac{a_k}{k} > \varepsilon \sum_{k = k_0}^\infty \frac{1}{k} = + \infty.[/imath] So we should have some decay condition on the [imath]a_k[/imath]. However, we could have [imath]a_k[/imath] to be the characteristic function of [imath]\{ n^2 : n \geq 1 \}[/imath] for instance; then we can't compare the series [imath]\sum_{k = 1}^\infty \frac{a_k}{k}[/imath] to the harmonic numbers; but then we can equally deduce the fact since square numbers have density zero. I know that this is very vague. The reason for this is, that I couldn't get the idea to work. How can I prove the fact? Thanks!
|
2100321
|
Show that [imath]f(x + 1/n) = f(x)[/imath] for some [imath]x[/imath], given [imath]f(0) = f(1)[/imath]
Let [imath]f: [0,1] \to \mathbb{R}[/imath] be continuous and such that [imath]f(0) = f(1)[/imath]. For each [imath]n \in \mathbb{N}[/imath], prove that there exists [imath]x \in [0,1][/imath] such that [imath]x + 1/n \in [0,1][/imath] and [imath]f(x + 1/n) = f(x)[/imath]. Don't even know where to start.
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570017
|
Show that there is [imath]\xi[/imath] s.t. [imath]f(\xi)=f\left(\xi+\frac{1}{n}\right)[/imath]
Let [imath]f:[0,1]\to\mathbb{R}[/imath] be continuous and [imath]f(0)=f(1)[/imath]. Show that for any integer [imath]n\geqslant 2[/imath], there is [imath]\xi\in(0,1)[/imath] s.t.[imath]f(\xi)=f\left(\xi+\frac{1}{n}\right).[/imath] I think this requires the intermediate value theorem somewhere.
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2099300
|
Irreducible Polynomials: How many elements are in E?
Consider the irreducible polynomial [imath]g = X^4 + X + 1[/imath] over [imath]F_2[/imath] and let [imath]E[/imath] be the extension of [imath]F_2 =[/imath] {0, 1} with a root [imath]α[/imath] of [imath]g[/imath]. How many elements does [imath]E[/imath] have? Im really not sure how to go by this question so any help will be appreciated
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2101390
|
Could [imath]E[/imath] have a subfield of order 8
Consider the irreducible polynomial [imath]g = X^4 + X + 1[/imath] over [imath]F_2[/imath] and let [imath]E[/imath] be the extension of [imath]F_2 =[/imath] {0, 1} with a root [imath]α[/imath] of [imath]g.[/imath] Could [imath]E[/imath] have a subfield of order 8 When working through my work Ive come across this question which has stumped me and I cant get my head aound if this is true of not so any help will be appreciated.
|
1820543
|
When are quantities outside of the real numbers considered equal, and when do they exist?
From the context of the real numbers we know that the square root of [imath]-1[/imath] is not a real number. Because of this the historical way we approach this is that the concept of complex numbers were created to give meaning to the nonexistence of [imath]\sqrt{-1} = i[/imath] and study the potential implications of such a value. We even knew beforehand how the square root of [imath]-1[/imath] as it existed as a concept. It was merely that it did not exist as an intermediate vale. When that set was created, certain properties were different from the set of real numbers and other properties (such as ordering) ceased to exist. If some superset of the complex plane were created to give meaning to another currently 'nonexistent' operation, what algebraic properties would we expect to retain? For instance, if one wanted to extend nonexsistent limits into being for which the one sided limits exist, would you only assume them equal to themselves or would you make them equal to limits with similar shapes around the point? Of course that is a hypothetical example, not to be taken as an actual request. The details of that would depend on whether geometric equivalence was the equalizing factor or the sequence used to compute the limit. I know I have written two questions here. So to summarize: First: what extensions upon real numbers currently exist other than the complex plane? Second: How would one go about creating a new extension, and how do people usually work out their properties aside from creative thinking? Is there a formal method that can be used to determine the properties of an extension coming from attempting to give meaning to an intermediate value that is "undefined"? Note: I couldn't come up with the proper tag so some guidance would be appreciated. I was thinking the tag "equality" or "extensions upon real numbers" or "monexistence" would be good, but those are not available.
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441956
|
Definition of complex number
In many situations (problems as well as solutions) I encounter the complex number [imath]i[/imath] which many times is defined as [imath]i^2=-1[/imath] instead of [imath]i=\sqrt{-1}[/imath], since it is "more preferred". Does anyone know why? Is it because there are two solutions to the equation [imath]x^2+a=0, \ \ a>0[/imath]?
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1829108
|
Derivative/integral relationship appears to disprove the fundamental theorem of calculus!!!
Consider the floor function: [imath]f(x) = \lfloor x \rfloor[/imath] The indefinite integral of f is: [imath]\int_0^x f(x) dx = x\lfloor x \rfloor - \frac {\lfloor x \rfloor^2 + \lfloor x \rfloor} 2[/imath] This should be an antiderivative of floor, right? Nope! If you take the derivative of the integral you find that sharp corners cause the derivative to not exist. So then this would mean that the integral of floor is not an antiderivative right? Therefore, I have found a case where the antiderivative does not equal the indefinite integral. In that case I must be flawed as that violates the first fundamental theorem of calculus. Where is the mistake in my logic and why does it appear to disprove the first fundamental theorem's relationship between integral and derivative? This isn't part of the above question per se, but I noticed interesting enough, that the derivative of the integral above is: [imath]\lfloor x \rfloor \frac {x - \lfloor x \rfloor}{x - \lfloor x \rfloor}[/imath] I wonder what sort of properties would be altered within integration/differentiation if one were to IDK... redefine the derivative by canceling the terms in that fraction? Or for that matter, canceling all fractions of that nature?
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1444851
|
Fundamental Theorem of Calculus.
Fundamental Theorem of Calculus says Let [imath]f[/imath] be an integrable function on [imath][a,b].[/imath] For [imath]x[/imath] in [imath][a,b],[/imath] let [imath]F(x)=\int_{a}^{x}f(t)dt.[/imath] Then [imath]F[/imath] is continuous on [imath][a,b].[/imath] If [imath]f[/imath] is continuous at [imath]x_{0}[/imath] in [imath](a,b),[/imath] then [imath]F[/imath] is differentiable at [imath]x_{0}[/imath] and [imath]F^{'}(x_{0})=f(x_{0}).[/imath] Now my question is why we are saying that this is "Fundamental Theorem of whole Calculus". It is clear that it connects intergal and differential calculus. But i am not getting the exact concept, why this is so important as its name is fundamental theorem of calculus. Why we says that it is fundamental theorem of calculus? Please suggest me. Thanks in advance.
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1942637
|
What is the indefinite integral of [imath]\lfloor x^{-1} \rfloor[/imath]?
I am trying to find the indefinite integral of [imath]\lfloor \frac 1x \rfloor[/imath] My attempt Please be aware that what comes after this sentence is nothing more than my work. It makes to me, and a lot of it contains made up notations, a lot of which I have made up and are very difficult for me to explain and/or define clearly. If it makes no sense to you, feel free to ask, but since those things are tangential to the question, feel free to also ignore them and refer purely to my above queston. So, since this is a piecewise constant function, I used the method of jump series to integrate it. I found the implied integral to be [imath]x\lfloor x \rfloor[/imath]. This means that the integral now equals [imath]x\lfloor \frac 1x \rfloor - JS(x \lfloor \frac 1x \rfloor)[/imath] Now, the piecewsie constant component troubles me. Because the function is infinite at 0, I cannot shift all of the disconnected pieces to line up with x = 0 held fixed (physically I mean). That would make the function infinite everywhere. Plus, I am not sure if the integral is divergent. I think the floor chops away enough area to make it finite. However, it throws me off. Extra question Assuming that the answerer understands my writing (or at least can understand my progression of computation), could the answerer find and compute JS(x * floor(1/x))? It shouldnt be too hard. (After all, it is just a matter of solving for it in the expression earlier mentioned involving it). If you cannot find it (doesnt reveal itself explicitly) or the weird notation confused you, feel free to ignore this "extra question". Its not relevant to the original question at the very top of this post, and as I said, these later sections pertain to my attempts, not neccessarily the solution that one finds. As they say, more than one way to skin a cat. Dont have to like my method to skin the cat. :p Definitions (to help clarify things) Implied Integral: A form of symbolic integration that holds all symbols of the form [imath]\lfloor f(x) \rfloor[/imath] constant. Implied Integration varies by piece-wise constants rather than constants. Eric Stucky more formally defined this concept here (note, I did not write that paper. 'Assistance' is merely credit to me making up this integral.) Jump Series: The portion of a piece-wise continuous function (bounded or unbounded) that consists purely of jump discontinuities. The Jump Series will be a piece-wise constant function varying by a real number constant. If there is a jump from an infinite value to a finite value that cannot in some way be reduced down, we still consider the Jump Series to be existing, but either 'diverging' or 'unbounded'. Integration Method of Jump Series: A method of integration where regular integration is split into an implied integral of the function being integrated minus the jump series of the aforementioned implied integral. This method is guaranteed to work so long as the final integral has a closed form. All of the operators involved are guaranteed to exist so long as the original function being integrated exists; however, the degree to which they can be expressed is... unknown.
|
360323
|
How to solve an definite integral of floor valute function?
How do you prove this identity: [imath]\int_0^{n^2}\lfloor\sqrt{t}\rfloor dt = \frac{1}{6}n(n-1)(4n+1)[/imath] I'd very much appreciate your help on this one!
|
1998322
|
A composition of a periodic function that is somehow not periodic.
I have this conjecture: [imath]\forall f: R \to R(\exists g:R \to R (x \cdot f(x) = \int f(x) dx + g(f(x))))[/imath] Basically, my problem is that I am very very sure this is true. I have no way to prove it, but I strongly believe it to be true. However, [imath]f(x) = (x \mod 1)[/imath] seems like a counter-example. The difference between the two expressions is not a periodic function. Is there some g for which g(x mod 1) is not periodic? If so, please tell me what it is.
|
1992503
|
Specific examples of sets of functions that fulfill these qualities, and can therefore redefine "anti-differentiation" "nicely".
I want to know all sets of single-variate functions that fulfill the following qualities: [imath]\forall_{f:R^2 \to R^2} \forall_{g \in CAF} \forall_{h \in CAF} \exists_{k\in CAF} k(x) = f(g(x),h(x))[/imath], where CAF (closed-algebraic functions) is the set I am wishing to find. I know that constants and piece wise constants both fulfill these qualities. Interestingly enough, any CAF set can be used to define an alternate "anti-derivative" with similar enough properties to serve useful. This is done by changing the identity [imath]A_x f(x)= \int f(x) + C(x)[/imath], where C(x) is now a function in the set of functions CAF. This means that this "anti-derivative" operator will vary by an arbitrary CAF function. These anti-derivatives only change in that now it is legal to under that "anti-derivative" operator to treat CAF functions as constants. As such, knowing what functions fulfill this quality might lead to ways of simplifying various integrals. I know for sure that many differential equations might not be easily solvable (possible but would require more work) without the anti-derivative with piece wise constants used as C(x).
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2008912
|
Someone needs to set the record straight on what exactly antiderivatives vary by: constants functions or piecewise constant functions.
My problem Alright, so this is my dilemma and it all boils down to the very troublesome function floor ([imath]\lfloor x \rfloor[/imath]) and piecewise constant functions in general ([imath]\lfloor f(x) \rfloor[/imath]). Everyone knows what the indefinite integral is. Just in case it is not well known this is what it is: [imath]F(x) = \int_{c}^{x} f(t)dt[/imath] The indefinite integral is an accumulation function with some base value that it starts out with. It varies by an arbitrary constant. Now let me define a second operator that we will refer to as an implied integral ([imath]\int^*[/imath]). It is best defined by example: [imath]\int^* \lfloor x \rfloor dx = x \lfloor x \rfloor + C(x)[/imath] [imath]\int^* x \lfloor x \rfloor dx = \frac 12 x^2 \lfloor x \rfloor + C(x)[/imath] [imath]\int^* \lfloor f(x) \rfloor dx = x \lfloor f(x) \rfloor + C(x)[/imath] Basically, it is an indefinite integral that varies be piecewise constant functions (C(x)) and treats peicewise constant functions as constants. Now,there is ambiguousness with the term "antiderivative" that I recently stumbled upon. The antiderivative seems to refer to both of these operations. For instance, see my question about the bug in matlab. The answer there claims that the implied integral is a valid antiderivative. I remember from elementary calculus that antiderivatives vary by constants. However, I am willing to give the benefit of the doubt since that user is a professor and probably knows way more than me. Why is it important? Well, while the area function is a set-in-stone idea differential equations rely upon use of the antiderivative and not the indefinite integral. Therefore, depending on which one the antiderivative follows... the result will change the solutions to differential equations. Specifically: It will decide whether or not their solutions must be continuous or if they are allowed to be discontinuous. A more concrete example is that we would no longer be able to just solve homogenous linear equations with constant coefficients. We would be able solve one's with piecewise constant coefficients trivially (it is possible to solve them for continuous solutions by doing things). To sum it all up: Do antiderivatives vary by constant functions or piecewise constant functions? It truly is an important question.
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2006176
|
Matlab's "buggy" symbolic integration function seems to be equivalent to the Implied Integral. Coincedence or intentional?
In matlab there is a function called "int(f,x)" which takes in a symbolic function and a variable and integrates it. I decided to try using it today to play around and do a few integrals and I got a pretty interesting surprise. I integrated the following and got the following results from matlab: [imath]\int floor(x) dx = x*floor(x)[/imath] [imath]\int floor(\sqrt x) dx = x*floor(\sqrt x)[/imath] [imath]\int x*floor(x) dx = \frac {x^2}2*floor(x)[/imath] I think you see the pattern. Matlab's integral seems to treat symbols of the form floor(f(x)) as constants. Oddly enough, I actually have a name for and use this operator in self-developed integration methods. I call it an implied integral. Is there some actual reason why matlab's integration function does implied integration or is it just a "bug" in the lack of algorithms for certain integrals? This is the first instance I've seen of anything mimicking the sort of things I've been doing over the past year or so, so if matlab has followed down the same path I am and is doing something I'm not yet aware of that would interesting to see. On the flipside I now know of a working implied integral calculator, which is awesome.
|
2100651
|
prove that for a continious function if f(q) = 0 for all q where q is rational then all f(x) = 0 when x is real
it is given that f is Continious. and [imath]\forall q\in\mathbb{Q}[/imath] [imath]f(q)=0[/imath] how do I prove that: for all [imath]x\in\mathbb{R}[/imath] : [imath]f(x) = 0[/imath]
|
1781199
|
Prove that [imath]f(x)=0[/imath] for all [imath]x\in\mathbb{R}[/imath].
The question is: Prove that if [imath]f:\mathbb{R}\to\mathbb{R}[/imath] is continuous and [imath]f(x) = 0[/imath] whenever [imath]x[/imath] is rational, then [imath]f(x) = 0[/imath] for all [imath]x\in\mathbb{R}[/imath]. My proof: Let [imath] x\in\mathbb{R} [/imath]. If [imath] x\in\mathbb{Q}[/imath], then [imath] f(x)=0 [/imath]. Otherwise [imath] x\in\mathbb{R\backslash Q} [/imath]. Then there exists a sequence of rational numbers [imath] (q_n)_{n\in\mathbb{N}}[/imath] (for example [imath]q_n= \frac{\lfloor{nx\rfloor}}{n} ,\forall n\in\mathbb{N}[/imath]) such that [imath]q_n\to x[/imath] as [imath]n\to \infty[/imath]. By the properties of [imath]f[/imath], we have [imath]f(q_n)=0[/imath] for all [imath]n\in\mathbb{N}[/imath]. So [imath]f(q_n)\to 0[/imath] as [imath]n\to \infty[/imath]. Since [imath]f[/imath] is continuous, [imath]q_n\to x[/imath] implies that [imath]f(q_n)\to f(x)=0[/imath]. So [imath]f(x) = 0[/imath] for all [imath]x\in\mathbb{R}[/imath]. Thank you for any help.
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2101001
|
How do you solve [imath](i^{i})^{i}[/imath]?
I am trying to solve this problem, and show that it does not necessarily equal [imath]i^{-1}[/imath]. My method so far is [imath](i)^{i} \\ =e^{(i)\ln{(i)} } i = \cos(\theta)+i\sin(\theta) \Rightarrow \theta = \frac{\pi}{2}\pm2n\pi, n \in \mathbb{Z} \\ \Rightarrow i = e^{i\frac{\pi}{2}\pm2in\pi} \\ \Rightarrow \ln{(i)} = i\frac{\pi}{2}\pm2in\pi \\ \Rightarrow e^{(i)\ln{(i)} } = e^{(i)(i\frac{\pi}{2}\pm2in\pi)}\\ =e^{(-\frac{\pi}{2}\pm2n\pi)}[/imath] However when I raise this to the power of [imath]i[/imath] once more, the answer ends up becoming [imath]i^{-1}[/imath] once more. What am I missing here? If someone can please show me where I'm going wrong that would be fantastic!
|
1732211
|
What is [imath]i^i[/imath] and [imath](i^i)^i[/imath]
What is [imath]i^i[/imath] and [imath](i^i)^i[/imath]. I fond some answers online but is there a clearer explanation of this? Maybe something help me to understand.
|
2099488
|
For every [imath]R>0[/imath], prove that there is a positive integer [imath]n[/imath] such that [imath]1 + z + \frac {z^2}{2!} +...+\frac {z^n}{n!}[/imath] has no zeros in [imath]|z|.[/imath]
The problem is exactly that of the title. I attempted to apply Rouché's theorem, by setting another function [imath]f(z)=e^z[/imath] and comparing the modulus of [imath]f(z)[/imath] and [imath]f(z)-P_n (z)[/imath] at the circle [imath]|z|=R[/imath], but I failed to proceed anymore. Is my approach correct? Or is there another solution?
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131479
|
Complex zeros of the polynomials [imath]\sum_{k=0}^{n} z^k/k![/imath], inside balls
this is a question from a Temple prelim exam, and i'm trapped in it! We have [imath]p_n(z)=\sum_{k=0}^n\frac{z^k}{k!}[/imath] and we have to prove that [imath]\forall r>0 \quad \exists N\in\mathbb{N}[/imath] s.t. [imath]p_n(z)[/imath] has no zeros in [imath]B_r(0)[/imath] for [imath]n>N[/imath]. I tried to use Rouche's theorem observing that [imath]1+\frac{z^n}{n!}[/imath] has the nth roots of [imath]n![/imath] as complex zeros, which are outside a fixed ball for n sufficiently large, but i'm having trouble proving that [imath]|1+\frac{z^n}{n!}|>|z+\cdots+\frac{z^{n-1}}{(n-1)!}|[/imath] in [imath]|z|=r[/imath] (if this is true). If anyone could help..
|
2099081
|
Summation of [imath]\arcsin [/imath] series.
What is [imath]a [/imath] if [imath]\sum _{n=1} ^{\infty} \arcsin \left(\frac {\sqrt {n}-\sqrt {n-1}}{\sqrt {n (n+1)}}\right) =\frac {\pi }{a} \,?[/imath] Attempt: What I tried is to convert the series to [imath]\arctan[/imath] and then convert it telescoping series. So in terms of [imath]\arctan [/imath] it becomes [imath]\arctan \left(\frac {\sqrt {n}-\sqrt {n-1}}{\sqrt {n}+\sqrt {n-1}}\right) [/imath] but now if I divide by [imath]n[/imath] it simplifies as [imath]n\frac {\pi}4-\sum _1^{\infty} \arctan \left(\frac {\sqrt {n-1}}{\sqrt {n}}\right) [/imath] but as [imath]n[/imath] is tending towards infinity it will lead to infinity which seems wrong. Also note that [imath]a[/imath] is an integer . Thanks!
|
2066984
|
Limit with inverse trignometry
We have to find the value of [imath]\lim_{n\to \infty}\sum_{k=2}^n\arccos\left(\frac{1+\sqrt{(k-1)k(k+1)(k+2)}}{k(k+1)}\right)[/imath] I tried it a lot but could not find any start. Please help me in this .
|
482551
|
An exponential "rearrangement" inequality: [imath]x^x+y^y>x^y+y^x[/imath]
Let [imath]x,y[/imath] be distinct real numbers greater than [imath]0[/imath]. Prove [imath]x^x+y^y>x^y+y^x .[/imath] Source: I think it comes from a Russian test given in 1991, but I haven't been able to verify this.
|
410895
|
Inequality with exponents [imath]x^x+y^y \ge x^y +y^x[/imath]
Let [imath]x,y[/imath] be positive numbers. Prove that [imath]x^x+y^y \ge x^y +y^x[/imath]. This question appeared in Summer 1991 Russian Olympiad team test. Apparently, I tried to come up with different approach such as Jensen, Karamata's inequality and nothing works so far. I just need a discussion here. Hints are not necessary.
|
710679
|
Interpolation inequality on Holder space
Let [imath]0< \beta < \gamma <1[/imath]. Show that the interpolation inequality holds. [imath]||U||_{C^{0,\gamma}(U)} \le ||U||^{\frac{1-\gamma}{1-\beta}}_{C^{0,\beta}(U)} ||U||^{\frac{\gamma-\beta}{1-\beta}}_{C^{0,1}(U)}.[/imath]
|
1098530
|
Prove an interpolation inequality
Assume [imath]0 < \beta < \gamma \le 1[/imath]. Prove the interpolation inequality [imath]\|u\|_{C^{0,\gamma}(U)} \le \|u\|_{C^{0,\beta}(U)}^{\frac{1-\gamma}{1-\beta}} \|u\|_{C^{0,1}(U)}^\frac{\gamma-\beta}{1-\beta}.[/imath] From PDE Evans, 2nd edition: Chapter 5, Exercise 2 I would love to employ Hölder's inequality which can easily justify this inequality. But Hölder's inequality requires [imath]u \in L^p(U), v \in L^q (U)[/imath]. Instead, this problem has [imath]u \in C^{0,\beta}(U) \cap C^{0,1}(U)[/imath]. The textbook gives the definition of \begin{align} \|u\|_{C^{0,\gamma}(\bar{U})} &:= \|u\|_{C(\bar{U})}+[u]_{C^{0,\gamma}(\bar{U})} \\ &= \sup_{x \in U} |u(x)|+\sup_{\substack{x,y \in U \\ x \not= y}} \left\{\frac{|u(x)-u(y)|}{|x-y|^\gamma} \right\} \end{align} All I know so far is that, given [imath]0 < \beta < \gamma \le 1[/imath] ... If [imath]|x-y|<1[/imath], then [imath]\frac 1{|x-y|^\gamma}<\frac 1{|x-y|^\beta}[/imath], which means [imath]\|u\|_{C^{0,\gamma}(U)} < \|u\|_{C^{0,\beta}(U)}[/imath]. If [imath]|x-y|\ge 1[/imath], then [imath]\frac 1{|x-y|^\gamma}\le \frac 1{|x-y|}[/imath], which means [imath]\|u\|_{C^{0,\gamma}(U)} \le \|u\|_{C^{0,1}(U)}[/imath].
|
2102153
|
Prove: if [imath]x,y\in\mathbb{R}[/imath] then [imath]2xy\le x^2+y^2[/imath]
prove: if [imath]x,y\in\mathbb{R}[/imath] then [imath]2xy\le x^2+y^2[/imath] if one or both of [imath]x,y[/imath] is [imath]0[/imath] or one is negative, the result follows. if both are negative the negatives cancel and it is the same as if they were both positive. both positive or both negative: [imath]2xy\le x^2+y^2\Leftrightarrow 2y\le{x+{y^2\over x}}\Leftrightarrow 2\le {x\over y}+{y\over x}[/imath].[imath]\space\space[/imath]if [imath]x=y[/imath] the result follows with [imath]{x\over y}+{y\over x}=2[/imath]. if [imath]x\gt y\space[/imath] then [imath]x=y+i,i\in\mathbb{R_+} \space[/imath] and [imath]\space{x\over y}+{y\over x}={{y+i}\over y}+{y\over {y+i}}={y\over y}+{i\over y}+{y\over {y+i}}\gt {y\over y}+{i\over {y+i}}+{y\over {y+i}}[/imath] [imath]=1+{{i+y}\over{i+y}}=2[/imath].[imath]\space\space[/imath]similar argument can be used for [imath]y\gt x[/imath]. Is this proof valid for real numbers? especially irrational numbers. thanks.
|
357272
|
How can I prove that [imath]xy\leq x^2+y^2[/imath]?
How can I prove that [imath]xy\leq x^2+y^2[/imath] for all [imath]x,y\in\mathbb{R}[/imath] ?
|
2102159
|
On the assumption that [imath]f[/imath] is [imath]C^1[/imath], prove that if [imath]D=\{x:f'(x)=0\}[/imath], then measure of [imath]f(D)[/imath] is 0.
We have [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] which is continunous and its derivative is also continuous. We define set [imath]D=\{x:f'(x)=0\}[/imath]. Prove that set [imath]f(D)[/imath] is of 0 (Lebesgue) measure. If we assume that [imath]f[/imath] is monotonic then every maximal interval on which [imath]f[/imath] is constant is landing on single point. Between such interval there is always a gap, so there are no intervals in [imath]f(D)[/imath]. However, I believe that [imath]f(D)[/imath] can have "more" that countably infinite elements. If that's true, then I do not know how to proceed.
|
648237
|
Measure of image of critical points set is equal 0
Let [imath]f:\mathbb{R}\to \mathbb{R}[/imath] be [imath]C^1[/imath] function and [imath]K = \{x : f'(x) = 0 \} [/imath]. Show that [imath]\mu \left(f\left(K\right)\right) = 0[/imath], where [imath]\mu[/imath] is Lebesgue measure. My attempt was following: [imath]\mu \left(f\left(K\right)\right)= \int_{f(k)} 1 dy \stackrel{(*)}{=} \int_{K}f'(x) dx = \mu\left(K\right) \cdot 0 = 0[/imath] but we cannot substitute [imath]y = f(x)[/imath] at [imath](*)[/imath] like that. I was told that there exists quite elementary proof (not using Sard's theorem) so I'm looking for it.
|
2101455
|
Last two digits of [imath]2^{100}[/imath]
I can calculate the number with computer and read the digits, or I can say PowerMod[2, 100, 100] in Mathematica, or I can even determine the digits myself, pencil-and-paper, by considering congruences modulo 4 and modulo 25 separately. But if I determine the last digit first -- it's 6 -- and then set this congruent equation for the penultimate digit [imath]y[/imath]: [imath]\frac{2^{100}-6}{10} \equiv y\ (mod\ 10)[/imath] Which simplifies to [imath]2^{99} - 3 \equiv 5\ y\ (mod\ 10)[/imath], or [imath]8 \equiv 5\ y + 3\ (mod\ 10)[/imath], or [imath]y \equiv 1\ (mod\ 2)[/imath]. This says this digit is odd. But how can I determine its exact value from here? I wonder why sequential modulo 10 congruences don't yield a specific answer.
|
1844558
|
How to find last two digits of [imath]2^{2016}[/imath]
What should the 'efficient' way of finding the last two digits of [imath]2^{2016}[/imath] be? The way I found them was by multiplying the powers of [imath]2[/imath] because [imath]2016=1024+512+256+128+64+32[/imath]. I heard that one way would be with the Chinese Remainder Lemma but I don't really know how I should start?
|
2102479
|
Can we recover [imath]X[/imath] from [imath]K = X^T X[/imath]?
Suppose we have a real matrix [imath]K[/imath] which is a gram matrix of [imath]X[/imath], [imath]K = X^TX[/imath], for some unknown [imath]X[/imath]. In this case, would [imath]X[/imath] be unique so that we may have a function [imath] f(K) = X? [/imath]
|
832384
|
Solve for $ A $ from $ A {A}^{T} $
I'm sure that I knew how to do this once many moons ago and that it's really simple. I have a matrix [imath]$ X $[/imath] which is defined as: [imath]$$ X = A {A}^{T} $$[/imath] How do I find [imath]$ A $[/imath] given [imath]$ X $[/imath]?
|
2102081
|
[imath]\sum_{n=1}^{\infty} \frac{n}{(n+1)!}[/imath] converges
I am looking for some help with the following question. How do I prove that the series [imath]\displaystyle \sum_{n=1}^{\infty} \frac{n}{(n+1)!}[/imath] converges, and then how would I find its sum? I see that this series [imath]a_n[/imath] converges if the sequence of partial sums [imath]s_n[/imath] converges. So [imath]s_n = \frac{1}{2} + \frac{1}{3} + \frac{1}{8} + \frac{1}{30}+... + \frac{n}{(n+1)!}[/imath] I see that [imath]\lim\limits_{n\to\infty}\frac{n}{(n+1)!}= 0 [/imath] as [imath](n+1)![/imath] approaches infinity faster than [imath]n[/imath]
|
1895587
|
Convergence of [imath]\sum_{n=1}^\infty\frac{n}{(n+1)!}[/imath]
Can someone give an explanation using the definition of convergence in partial sum to show how the above infinite sum converges to 1? Thanks
|
2102385
|
convergence of [imath]\sum_{n=1}^{\infty} \frac{1}{\sqrt n + \sqrt{n+1}}[/imath]
I need to determine if the series [imath]\sum_{n=1}^{\infty} \frac{1}{\sqrt n + \sqrt{n+1}}[/imath] converges or diverges. My work so far: Using the comparison test, [imath]\frac{1}{\sqrt n + \sqrt{n+1}} < \frac{1}{\sqrt n } [/imath] and [imath]\frac {1}{\sqrt n }[/imath] is a divergent series
|
162434
|
Does [imath]\sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}}[/imath] converge?
Does the following series converge or diverge? [imath] \sum\limits_{n=1}^\infty\frac{1}{\sqrt{n}+\sqrt{n+1}} [/imath] The methods I have at my disposal are geometric and harmonic series, comparison test, limit comparison test, and the ratio test.
|
2103144
|
Inequality with one equality constraint ....
Let [imath]x,y,z[/imath] be positive real numbers satisfying [imath]x^2+y^2+z^2=1[/imath]. Prove that : [imath]\frac {1}{x} + \frac {1}{y}+\frac {1}{z} \geq 3{\sqrt{3}}.[/imath] I derived the equality case easily. I was able to prove the inequality with the help of Lagrange Multipliers, which made it look very easy. Is there any other way to prove the same inequality without calculus ? I tried AM-GM and Cauchy-Schwarz but could not find a proper set of values to apply these on so as to obtain the inequality. Any help would be appreciated . :)
|
1208269
|
Inequality with condition [imath]x^2+y^2+z^2=1[/imath].
let [imath]x,y,z>0[/imath] such that [imath]x^2+y^2+z^2=1[/imath]. Find the minimum of [imath]\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.[/imath] Is the answer [imath]3\sqrt{3}[/imath] by any chance?
|
2100383
|
Fundamental group of [imath]S^2\cup\{xyz = 0\}[/imath]
We want to compute the fundamental group of [imath]S^2 \cup \{xyz=0\}[/imath]. It is easy to see that it retracts to a sphere joined with three disks. How can I show that its fundamental group is trivial?
|
1401377
|
Help finding the fundamental group of [imath]S^2 \cup \{xyz=0\}[/imath]
let [imath]X=S^2 \cup \{xyz=0\}\subset\mathbb{R}^3[/imath] be the union of the unit sphere with the 3 coordinate planes. I'd like to find the fundamental group of [imath]X[/imath]. These are my ideas: I think the first thing I should do is to retract all the points outside the sphere to the sphere (is that possible? how?) then using spherical coordinates I could make the following deformation: [imath](1,\phi,\theta)\to ((\sin\phi \sin\theta \cos \phi\cos \theta)^t,\phi,\theta)[/imath]. This collapses all the point in [imath]S^2 \cap \{xyz=0\}[/imath] to [imath]0[/imath] obtaining [imath]8[/imath] deformed spheres touching each other in [imath]0[/imath] (how can I prove rigorously that they are simply connected), using Van Kampen theorem we can say that [imath]X[/imath] is simply connected.
|
1949754
|
Surjectivity of [imath]{(\mathbb{Z}/m\mathbb{Z})}^{\times}\rightarrow {(\mathbb{Z}/n\mathbb{Z})}^{\times}[/imath]
Let [imath]R[/imath] be a commutative ring with identity. Let [imath]I,J[/imath] be ideals of [imath]R[/imath], such that [imath]I\subseteq J[/imath]. Then the canonical map [imath]\Phi: R/I\rightarrow R/J[/imath] defined as [imath]\Phi(I+x)=J+x[/imath] is a surjective ring homomorphism. Now take [imath]R=\mathbb{Z}[/imath] and [imath]I=m\mathbb{Z}[/imath] and [imath]J=n\mathbb{Z}[/imath] where [imath]n\mid m[/imath]. It is clear that the image of [imath]{(\mathbb{Z}/m\mathbb{Z})}^{\times}[/imath] under [imath]\Phi[/imath] is contained in [imath]{(\mathbb{Z}/n\mathbb{Z})}^{\times}[/imath]. If we denote [imath]\phi[/imath] as the restriction of [imath]\Phi[/imath] to [imath]{(\mathbb{Z}/m\mathbb{Z})}^{\times}[/imath] then we have [imath]\phi:{(\mathbb{Z}/m\mathbb{Z})}^{\times}\rightarrow {(\mathbb{Z}/n\mathbb{Z})}^{\times}[/imath] I checked a few cases and it seems [imath]\phi[/imath] is surjective. Is it true in general ?
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885216
|
Why [imath]f\colon \mathbb{Z}_n^\times \to \mathbb{Z}_m^\times[/imath] is surjective?
If [imath]m|n[/imath]. Why the map [imath]f\colon \mathbb{Z}_n^\times \to \mathbb{Z}_m^\times[/imath] given by [imath]a \mod{n}\mapsto a \mod m[/imath] is a surjective homomorphism of groups? Attempt: I proved it is well a well defined homomorphism because the canonical projection [imath]\overline{f}\colon \mathbb{Z}_n \to \mathbb{Z}_m[/imath] is a surjective ring homomorphism, it extends [imath]f[/imath] and maps units to units. But I can't prove surjectiveness. Given [imath]a\in\mathbb{Z}[/imath] such that [imath](a,m)=1[/imath] I'm looking for a [imath]k\in\mathbb{Z}[/imath] such that [imath](a+k\cdot m,n)=1[/imath]. I don't know why one of the [imath]n/m[/imath] candidates for such [imath]k[/imath] makes the number [imath]a+k\cdot m[/imath] an unit modulo [imath]n[/imath].
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2103013
|
Can any 2x2 matrix be written like sum of two squared matrices?
How to prove that for any [imath]A[/imath] is a [imath]2\times2[/imath] matrix with real elements exist [imath]B[/imath] and [imath]C[/imath] so that [imath]A=B^2+C^2[/imath]? So far, I used Cayley-Hamilton theorem and I have: [imath]A =[/imath] [imath]\frac{1}{Tr(A)}A^2 + \frac{det(A)}{Tr(A)}I_n[/imath]. I know that I need a positive trace, so I choose [imath]A_1 = A + tI_n[/imath] and [imath]\lim_{t\to∞}(A + tI_n) = \infty[/imath]
|
738270
|
Is it true that all matrices in [imath]M_2(\mathbb R)[/imath] is the sum of two squares?
I recently show that every polynomial with real coefficient and [imath]P[/imath] is always positive. is a sum of two squares of polynomials. These questions also appear often in arithmetic. What if we change polynomials by matrices? I ask my professor and unfortunately He had no idea if this result is already shown or not. I tried, unsuccessfully, several tracks : First denote that [imath]\forall M\in M_n(\mathbb R) , M=1/2[(A+A{}^t)+(A-A{}^t)].[/imath] and perhaps used the following result : Any symmetric positive matrix admits a square root Proof. By spectral theorem ,there exist [imath]\Omega \in \mathbb O_n(\mathbb R)[/imath] and [imath](\lambda_1,...\lambda_n)\in (\mathbb R^{+})^n[/imath] such that [imath]S=\Omega D \Omega^{-1}[/imath] with [imath]D[/imath] the diagonal matrix formed by the positive eigenvalues. Let [imath]D'[/imath] the diagonal matrix formed of the roots of the eigenvalues. Then , [imath]D'²=S[/imath] and [imath]D'{}^t=D'[/imath] thus she is a symmetric positive matrix. Question : Is it possible to believe that all matrices (say in [imath]M_2(\mathbb R)[/imath]) is the sum of two squares?
|
2103112
|
Determine limit: [imath]\lim_{x \to -\infty} (\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}) [/imath]
Determine the limit of: [imath]\lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) [/imath] I've tried a few times, most notably the following two versions. I'm looking for a comment on both, since both amount to a wrong answer. First attempt [imath]L = \lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) [/imath] I thought it'd be nice to get rid of as much ugliness as possible by moving [imath]x[/imath] out of the roots: [imath] \begin{split} L &= \lim_{x \to -\infty} \left(x\left(\sqrt{1 + \frac{2}{x}}\right) - x\left(\sqrt{1 - \frac{2}{x}}\right)\right)\\ &= \lim_{x \to -\infty} \left(x\left(\sqrt{1 + \frac{2}{x}} - \sqrt{1 - \frac{2}{x}}\right)\right) \end{split} [/imath] Seems simple enough, since [imath]\frac{1}{x}[/imath] should be an (infinitely) small number I thought it could be discarded with regard to [imath]\sqrt1[/imath], yielding: [imath]L = \lim_{x \to -\infty} \left(x\left(\sqrt{1} - \sqrt{1}\right)\right) = 0[/imath] Wrong, Second attempt Multply both sides with the conjugate: [imath] \begin{split} L &= \lim_{x \to -\infty} \sqrt{x^2 +2x} - \sqrt{x^2 - 2x}) \\ &= \lim_{x \to -\infty} \frac{4x}{(\sqrt{x^2 +2x} + \sqrt{x^2 - 2x})} \\ &= \lim_{x \to -\infty} \frac{4x}{x\sqrt{1 + \frac{2}{x}} + x\sqrt{1 - \frac{2}{x}}} \\ &= \lim_{x \to -\infty} \frac{4}{\sqrt{1 + \frac{2}{x}} + \sqrt{1 - \frac{2}{x}}}\\ &= \frac{4}{\sqrt{1} + \sqrt{1}} = \frac{4}{2} = 2 \end{split} [/imath] Wrong as well... The answer should be [imath]-2[/imath], but I don't get how to 'get the negative' in...
|
675516
|
Find the value of [imath]\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)[/imath]
I am stuck on this. I would like the algebraic explanation or trick(s) that shows that the equation below has limit of [imath]-2[/imath] (per the book). The wmaxima code of the equation below. [imath] \lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right) [/imath] I've tried factoring out an [imath]x[/imath] using the [imath]\sqrt{x^2} = |x|[/imath] trick. That doesn't seem to work. I get [imath]1 - 1 = 0[/imath] for the other factor meaning the limit is zero...but that's obviously not correct way to go about it :( Thanks.
|
1208933
|
Understanding the ideal [imath]IJ[/imath] in [imath]R[/imath]
I'm a little confused why our book defines an ideal [imath]IJ[/imath] in [imath]R[/imath] (where [imath]I[/imath] and [imath]J[/imath] are ideals) in such a complicated way: [imath]IJ=\left\{ \displaystyle\sum\limits_{i=1}^n a_i b_i\mid n\geq 1 ,a_i\in I, b_i\in J \right\}[/imath] How is this different that defining more simply as: [imath]\{ab\mid a\in I,b\in J\}[/imath] Since [imath]\{ab\}[/imath] is closed on addition then elements of the form [imath]a_1b_1+\cdots +a_mb_m\in\{ab\}[/imath], which sure does seem like it is equal to [imath]IJ[/imath]. Am I missing something here? Thanks.
|
2341926
|
Understanding the definition of t IJ when I AND J are ideals
if I is ideal and J is ideal then [imath]IJ=\left\{ \displaystyle\sum\limits_{i=1}^n a_i b_i\mid a_i\in I, b_i\in J \right\}[/imath] i am not undetsnaging the defination of it, what is n some n that belong to N? n = |I| or |J|? according to this definitiation it look like that |I| = |J| so each ideals inside some ring have the same size?
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2103950
|
Quadratic Recurrence : [imath]f(n) = f(n-1) + f(n-2) + f(n-1) f(n-2)[/imath] Solution? How?
I have encountered this question in a coding competition and want the formula for [imath]f(n)[/imath]. Is there any way to do this?
|
2104519
|
Solution to recurrence
Can anyone tell me how to solve this recurrence efficiently [imath]F(n) = F(n-1) + F(n-2) + F(n-1)*F(n-2)[/imath]. I will be provided value of [imath]F(0)[/imath] and [imath]F(1)[/imath] and have to calculate [imath]F(n)[/imath]. [imath]0 \le N \le 10^9[/imath]
|
2105196
|
Is [imath]\sqrt{4 + 2\sqrt{3}} - \sqrt{3}[/imath] rational?
Is there a method to proving whether nested roots like [imath]\sqrt{4 + 2\sqrt{3}} - \sqrt{3}[/imath] are rational or not? For simpler radicals, I can try using a contradiction, assume that they do equal some rational number and use algebra to simplify and show how one side of the equation is clearly rational but the other reduced to some well known root, like 2 or 3, and conclude irrationality.
|
2076737
|
Show that [imath]\sqrt{4 + 2\sqrt{3}} - \sqrt{3}[/imath] is rational.
Show that [imath]\sqrt{4 + 2\sqrt{3}} - \sqrt{3}[/imath] is rational. I've tried to attempt algebra on this problem. I noticed that there is some kind of nesting effect when trying to solve this. Please help me to understand how to attempt to denest this number. Any help would be greatly appreciated.
|
2105253
|
Find a limit related to a convergent series.
If [imath]\sum a_n[/imath] is a convergent series and [imath]S=\lim{s_n}[/imath] where [imath]s_n[/imath] is the [imath]n[/imath]th partial sum, how would I prove that [imath]\lim_{n\to \infty} \frac{s_1+...+s_n}{n} = S[/imath]? I'm not sure where to start, I've been having trouble understanding partial sums. Any tips on where to begin?
|
2103817
|
If [imath]\sum a_n[/imath] is a convergent series with [imath]S = \lim s_n[/imath], where [imath]s_n[/imath] is the nth partial sum, then [imath]\lim_{n \to \infty} \frac{s_1+...+s_n}{n} = S[/imath]
Let [imath]\sum a_n[/imath] be a convergent series, and let [imath]S = \lim s_n[/imath], where [imath]s_n[/imath] is the nth partial sum. I need to prove the following: [imath]\lim_{n \to \infty} \frac{s_1+...+s_n}{n} = S[/imath] How do I go about proving that proof? Definition of a limit [imath]\lim_{n \to \infty} f(x) = L[/imath] if for every number [imath]\epsilon>0[/imath] there is some number [imath]\sigma >0[/imath] such that [imath]|f(x)-L| <\epsilon[/imath] whenever [imath]0<|x-a|<\sigma[/imath]
|
1314142
|
Trace of AB = Trace of BA
We can define trace if [imath]A =\sum_{i} \langle e_i, Ae_i\rangle[/imath] where [imath]e_i[/imath]'s are standard column vectors, and [imath]\langle x, y\rangle =x^t y[/imath] for suitable column vectors [imath]x, y[/imath]. With this set up, I want to prove trace of AB and BA are same, so it's enough to prove that [imath]\sum_{i} \langle e_i, ABe_i\rangle =\sum_{i} \langle e_i, BAe_i\rangle[/imath] but how to conclude that?
|
1099745
|
How to prove [imath]\operatorname{Tr}(AB) = \operatorname{Tr}(BA)[/imath]?
there is a similar thread here Coordinate-free proof of [imath]\operatorname{Tr}(AB)=\operatorname{Tr}(BA)[/imath]?, but I'm only looking for a simple linear algebra proof.
|
2104264
|
Irreducible factors of number in [imath]\Bbb Z[\sqrt{3}][/imath]
So I have to give a irreducible factor decomposition of [imath]11[/imath] in [imath]\Bbb Z[\sqrt3][/imath]. I know that I have to use factors of the form [imath](a+b\sqrt{3})[/imath] but I don't know how to do it :(
|
105015
|
Method for determining irreducibles and factorising in [imath]\mathbb Z[\sqrt{d}][/imath]
I know that [imath]\mathbb Z[\sqrt{7}][/imath] is a UFD, and I can write the equation [imath](2 + \sqrt{7})(3 - 2\sqrt{7}) = (5 - 2\sqrt{7})(18 + 7\sqrt{7}) [/imath]. So clearly these are not all irreducibles. How do I determine which of these aren't irreducible? How do I factor those into irreducibles? It seems difficult when dealing with [imath]\mathbb Z [\sqrt{d}] [/imath] with [imath]d[/imath] positive, since the norm map isn't as nice (it's not always positive). Thanks
|
1463656
|
Order in matrix products, Complex conjugation, Transpose, dagger, ...
First, state what i know about products of matrices \begin{align} & (AB)^{-1} = B^{-1} A^{-1} \\ &(AB)^T = B^T A^T \\ & (AB)^{\dagger} = B^{\dagger} A^{\dagger} \end{align} Where [imath]-1[/imath] represents the inverse, [imath]T[/imath] transpose [imath]\dagger[/imath] dagger (which means the complex conjugate and transpose) Now, I got wonder the complex conjugation of matrices It seems to me that complex conjugation itself does not change the order of products, \begin{align} (AB)^{*} = A^* B^* \end{align} Is this statement correct?
|
2105172
|
[imath](XY)^*=X^* Y^*[/imath]?
Is it true for complex matrices [imath]X,Y[/imath] that [imath] (XY)^*=X^* Y^*? [/imath] where [imath]^*[/imath] refers to complex conjugation. How can we prove this if so? Thanks! Note: I am referring to complex conjugation, not the hermitan transpose. The answer below refers to hermitan tranpose.
|
2104915
|
First decimal digit of a very large number.
Find the first digit (the left one) of the number [imath]2016^{2016}[/imath], not by actually compute it. I know the solution is 7, thanks to Wolfram Alpha's power, but I did not succeeded in finding it. Question number two: how may i calculate log values used in solving this?
|
1136486
|
Finding the first digit of [imath]2015^{2015}[/imath]
Can anyone help me find the first digit of [imath]2015^{2015}[/imath]? It is easy to find the last digit but I have no idea for the first digit.
|
2104877
|
[imath]f(x)[/imath] is continuous on [imath][0,\pi][/imath].Prove that the following limit is [imath]\frac{2}{\pi}\int_{0}^{\pi}f(x)[/imath]
[imath]f(x)[/imath] is continuous on [imath][0,\pi][/imath].Prove that [imath]\lim_{n\to\infty}\int_0^\pi |\sin(nx)|f(x)\ dx = \frac 2\pi \int_0^\pi f(x) dx[/imath] I made the substitution [imath]x=\frac{u}{n} \implies \text{dx}=\frac{1}{n} \text{du}[/imath]. So, we have [imath]\lim_{n \rightarrow \infty} \int_{0}^{\frac{\pi}{n}} \sin u f(\frac{u}{n}) du[/imath] Now i have no idea how to proceed.
|
328314
|
Computing [imath]\lim_{n \rightarrow\infty} \int_{a}^{b}\left ( f(x)\left | \sin(nx) \right | \right )[/imath] with [imath]f[/imath] continuous on [imath][a,b][/imath]
Let [imath]a,b \in \mathbb{R}[/imath] and [imath]\textit{f} :[a,b] \rightarrow \mathbb{R}[/imath] continuous on [imath][a,b][/imath]. Does the sequence [imath]\left (\int_{a}^{b} f(x)\left |\sin(nx) \right |dx \right )[/imath] converge? If it does, what is its limit ? I know how to solve this for [imath]\left(\int_{a}^{b} f(x)\sin(nx)dx \right )[/imath] with integration by parts when [imath]f[/imath] is a class [imath]C^1[/imath] function. Here, I don't know how to deal with the absolute value and the non-differentiability of [imath]\textit{f}[/imath]. Any help is appreciated, thanks in advance.
|
2105741
|
How to evaluate [imath]\int_0^1\arctan (1-x+x^2)dx[/imath]?
I tried using [imath]\int_a^bf(x)dx=\int_a^bf(a+b-x)dx[/imath] but it's giving back the same result. I tried using integration by parts but it's giving a very long answer. Is there any simple way to do this? Btw the answer to this question is [imath]\log 2[/imath].
|
617032
|
calculation of [imath]\int_{0}^{1}\tan^{-1}(1-x+x^2)dx[/imath]
Compute the definite integral [imath] \int_{0}^{1}\tan^{-1}(1-x+x^2)\,dx [/imath] Failed Attempt: Let [imath]1-x+x^2=t[/imath]. Then [imath] \begin{align} (2x-1)\,dx &= dt\\ dx &= \frac{1}{(2x-1)}dt \end{align} [/imath] Changing the limits of integration, we get [imath]\int_{1}^{1}\tan^{-1}(t)\cdot \frac{1}{(2x-1)}dt = \int_{1}^{1}\tan^{-1}(t)\cdot f(t)dt = 0 [/imath] where [imath]f(t)=\frac{1}{(2x-1)}[/imath]. Is it true that [imath]\int_{a}^{a}f(x)dx = 0[/imath]? If not, then where have I made a mistake in my attempted solution?
|
2105505
|
Tourists attending same place in groups
There are 5 tourists going to London. There are 7 places to visit in London. What is the probability that two attend one (same) location and 3 attend another (same) location. So this is asking what is the probability that the 2 groups of people attend different places, but everyone from the groups is there. ex. [imath]AB | C D E[/imath] for the groups. There are [imath]\binom{5}{2}[/imath] ways to choose 2 groups. There are [imath]7^5[/imath] total choices considering each person has seven choices. There are [imath]\binom{7}{2}[/imath] ways to pick 2 locations from the 7. Thus, [imath]P = \frac{\binom{5}{2} \binom{7}{2}}{7^5}[/imath] But the answer the book has is [imath]P = \frac{7*6 \binom{5}{2}}{7^5}[/imath] Why are the locations suddenly distinguishable?
|
2094321
|
Probability of people attending same and different places
[imath]5[/imath] tourists plan to attend Octoberfest. Each attends a location at random from the choices: Alpine, Bingeman, Concordia, Kitchener, Queens, Schwaben, Transylvania. Q What is the probability that two attend one location and three attend another same location? I get that there are [imath]\binom{5}{2}[/imath] possibilities to create 2 groups. There are [imath]\binom{7}{2}[/imath] ways to choose 2 places so shouldn't it be [imath]\frac{\binom{7}{2}\binom{5}{2}}{7^5}[/imath]?
|
2105240
|
Proving an alternate quadratic formula
It is well known that the quadratic formula for [imath]ax^2+bx+c=0[/imath] is given by[imath]x=\dfrac {-b\pm\sqrt{b^2-4ac}}{2a}\tag1[/imath] Where [imath]a\ne0[/imath]. However, I read somewhere else that given [imath]ax^2+bx+c=0[/imath], we have another solution for [imath]x[/imath] as[imath]x=\dfrac {-2c}{b\pm\sqrt{b^2-4ac}}\tag2[/imath] Where [imath]c\ne0[/imath]. In fact, [imath](2)[/imath] gives solutions for [imath]0x^2+bx+c=0[/imath]! Question: How would you prove [imath](2)[/imath]? Why is [imath](2)[/imath] somewhat similar to [imath]\dfrac 1{(1)}[/imath] but with [imath]2a[/imath] replaced with [imath]-2c[/imath]?
|
2072174
|
"Citardauq" formula derivation?
I'm trying to understand how we got to the "citardauq" formula (note: "quadratic", reversed) I found this question here, first answer by Andre says Multiply "top" and "bottom" by [imath]-b\mp\sqrt{b^2-4ac}[/imath]. After the smoke clears, we obtain [imath]\frac{2c}{-b \mp \sqrt{b^2-4ac}}.[/imath] Question is, how does the smoke actually clear? I was able to get to the final result by just distributing and canceling out terms but I wasn't sure I went about it the right way, the [imath]\mp[/imath] confused me a bit. How do you multiply [imath]\pm \sqrt{b^2-4ac}[/imath] with [imath]\mp \sqrt{b^2-4ac}[/imath]?
|
2106184
|
Algebraic numbers in the interval [imath][0,1][/imath]
A number [imath]x\in\mathbb{R}[/imath] is said to be algebraic if it is the root of a polynomial over [imath]\mathbb{Q}[/imath]. Define [imath]A:[0,1]\rightarrow \{0,1\}[/imath] as [imath]A(x)=1 \text { if } x \text{ is algebraic, } A(x)=0\text{ if } x \text{ is not algebraic, }[/imath] What is the value of [imath]\int_{0}^{1} A(x)dx[/imath] ?
|
631359
|
evaluation of an integral involving algebraic numbers
Define [imath]A\colon[0,1]\to \mathbb R[/imath] by [imath]A(x) =\begin{cases} 1, &\text{if }x \text{ is algebraic}\\ 0, &\text{otherwise} \end{cases}[/imath] Evaluate : [imath]\int_{0}^{1}A(x)dx[/imath] Is it riemann Integrable in the first place?? Intuitively it looks like [imath]0[/imath] which is also the answer. But I can't justify.
|
2103252
|
Is [imath]\mathbb{Z}[t,t^{-1}][/imath] a PID? What about [imath]\mathbb{Z}[\sqrt{2}i][/imath]?
Is [imath]\mathbb{Z}[t,t^{-1}][/imath] a PID? What about [imath]\mathbb{Z}[\sqrt{2}i][/imath]? I don't know how to prove that a set IS a PID. I only know how to prove when it is NOT (by proving it is not UFD, for example). How can I show an ideal can only be generated by a single element? In [imath]\mathbb{Z}[/imath] I understand, there is a minimality argument. But in those sets up there I have no idea how to start.
|
2096860
|
Do the Laurent polynomials over [imath]\mathbb{Z}[/imath] form a principal ideal domain?
I'm trying to prove whether or not the Laurent polynomials [imath]\mathbb{Z}[x, x^{-1}][/imath] with coefficients in [imath]\mathbb{Z}[/imath] form a principal ideal domain. I know that [imath]\mathbb{F}[x, x^{-1}][/imath] is a PID when [imath]\mathbb{F}[/imath] is a field, but clearly [imath]\mathbb{Z}[/imath] is not a field so I cannot appeal to this result. And my intuitions are not serving me very well at the moment. Can anyone provide a hint or direction to take?
|
2104455
|
Convergence of a sequence in R
If [imath]a_n[/imath] is a sequence of real numbers, then I know the following result: If [imath]\underset{n\to\infty}{\lim} a_{n}[/imath] exist then [imath]\lim_{n\rightarrow\infty} a_{2n} \quad \text{and} \quad \lim_{n\to\infty} a_{2n-1} \quad \text{and} \quad \lim_{n\rightarrow\infty} a_{3n}[/imath] exist and equal. Now what can I say about the converse? i.e. if I know that [imath]\lim_{n\rightarrow\infty} a_{2n} \quad \text{and} \quad \lim_{n\to\infty} a_{2n-1} \quad \text{and} \quad \lim_{n\rightarrow\infty} a_{3n}[/imath] exist, what can i say about [imath]\underset{n\to\infty}{\lim} a_{n}[/imath]?
|
1035250
|
Does [imath]a_n[/imath] converges if and only if [imath]a_{2n},a_{3n},a_{2n-1}[/imath] converge?
Does [imath]a_n[/imath] converges if and only if [imath]a_{2n},a_{3n},a_{2n-1}[/imath] converge? [imath]a_{2n}[/imath] is a subsequence of [imath]a_n[/imath] and [imath]a_{3n},a_{2n}[/imath] are subsequences of [imath]a_{6n}[/imath] so they all have the same limit But what about [imath]a_{2n-1}[/imath] can I say it is a subsequence of [imath]a_{2n}[/imath] and therefore they all have the same limit?
|
802675
|
Quotient group of the non-zero complex numbers
Let [imath]\mathbb{C}[/imath] denote the multiplicative group of non-zero complex numbers and let [imath]P[/imath] denote the subgroup of positive (real) numbers. I am trying to find the quotient group [imath]\mathbb{C}/P[/imath]. Please help.
|
650941
|
Identifying a quotient group (NBHM-[imath]2014[/imath])
Let [imath]\mathbb C^*[/imath] denote the multiplicative group of non-zero complex numbers and let [imath]P[/imath] denote the subgroup of positive real numbers. Identify the quotient group. My thought [imath]\frac{\mathbb C^*}{P}=\{P,-P,iP,-iP\}.[/imath] Is it right?
|
2089490
|
How to prove that [imath]|\cos x-\cos y|\leq |x-y| \ \forall x,y \in \mathbb{R}[/imath]?
As above, how would you go about proving this inequality?
|
2101931
|
[imath]\forall x,y \in \mathbb{R}:\vert\cos x- \cos y\vert\leq\vert x-y\vert[/imath]
Prove that [imath]\forall x,y \in \mathbb{R}:\vert\cos x- \cos y\vert\leq\vert x-y\vert.[/imath] My try : [imath] f(x) =\cos x[/imath] and use the mean value theorem.
|
2104037
|
For which [imath]n[/imath] the polynomial [imath]x^n + y^n[/imath] will be irreducible over [imath]\mathbb Z[/imath]?
For which [imath]n[/imath] the polynomial [imath]x^n + y^n[/imath] will be irreducible over [imath]\mathbb Z[/imath]? I think when n is multiple of 2. Can you please correct me if I am wrong and explain why.
|
695266
|
Factoring [imath]x^n + y^n[/imath] over the integers
Here's what i know (or think i know) about the factoring. For integer [imath]n> 1 [/imath] 1) If [imath]n[/imath] is a positive power of [imath]2[/imath] then it is irreducible. 2) If [imath]n[/imath] is an odd prime then [imath]x^n + y^n = (x + y)(x^{n-1} - x^{n-2}y + \cdots - xy^{n-2} + y^{n-1} ) [/imath] 3) If [imath]n[/imath] has an odd prime factor then it is factorable but the factorization is more complicated , for example [imath]x^{14} + y^{14}[/imath] has 2 distinct irreducible factors and [imath]x^{15} + y^{15}[/imath] has [imath]4[/imath] distinct irreducible factors. Is there a connection between the prime factorization of [imath]n[/imath] and the number of distinct irreducible factors of [imath]x^n + y^n[/imath] ? Is there a connection between [imath]n[/imath], AND the number of distinct irreducible factors , AND the highest power occurring in each factor? Example: [imath]x^{15} + y^{15} = (x + y)(x^2 - \cdots)(x^4 - \cdots)(x^8 + \cdots)[/imath] In other words , i'm also asking if there is a connection between [imath]n = 15[/imath] , the number of factors [imath]4[/imath] , and the powers [imath]\{1 , 2 , 4 , 8\}[/imath] For this particular example the numbers work out nicely but i'm not sure the pattern is so obvious in general. Thank you for your consideration in this matter.
|
1771931
|
Roots of the Taylor approximation of the exponential
While answering another question, I looked at the roots of the [imath]n^{th}[/imath] degree Taylor approximation of the exponential. [imath]e^x\approx E_n(x):=\sum_{k=0}^n\frac{x^k}{k!}.[/imath] Apparently, these root are aligned on a parabola-like smooth curve depending on [imath]n[/imath]. How would you address the problem of finding the equation of this curve, ignoring the exact positions of the roots along it ?
|
109360
|
Roots of the incomplete gamma function
Is there any way that one can describe all the roots of the incomplete gamma function [imath]\Gamma(n,z)[/imath], for [imath]n\in \mathbb{N}[/imath], analytically?
|
2107076
|
Product of two i)non-differentiable functions and ii)differentiable functions in Complex plane.
Is the product of two non-differentiable functions in [imath]\mathbb{C}[/imath] always non-differentiable ? similarly for differentiable functions ? and if [imath]f[/imath] is not differentiable in [imath]\mathbb{C}[/imath] does that imply that [imath]f[/imath] will always be differentiable in [imath]\mathbb{R}[/imath] ? If Yes , how should i approach this ?
|
948421
|
Is the product of two non-holomorphic function always non-holomorphic?
Suppose [imath]h(z)[/imath] is a complex function. I have noticed that [imath]h(z) = f(z)\cdot g(z)[/imath], where [imath]f(z)[/imath] and [imath]g(z)[/imath] are non-holomorphic. Can [imath]h(z)[/imath] be holomorphic? Can a similar statement be made, if [imath]f(z)[/imath] is holomorphic, but [imath]g(z)[/imath] is not?
|
2107367
|
Find solution in integers of [imath]x^3+x-y^2=1[/imath]
Find integers [imath]x[/imath] and [imath]y[/imath] such that [imath]x^3+x-y^2=1.[/imath] My try: [imath]x^3+x-y^2=1 \implies x^3+x-1=y^2.[/imath] Now, when [imath]x^3+x-1[/imath] is a perfect square?
|
703137
|
Cubic diophantine equation
How can be solved the equation [imath]x^3+x-1=y^2[/imath] in positive integers? I know this equation defines an elliptic curve, but this seems to be a non-elementary way to solve this question. Is there a more elementary solution? By the way I found three solutions: [imath](1,1), (2,3)[/imath] and [imath](13,47)[/imath]. Is this related to the law group property of elliptic curves? Thanks for further answers.
|
2105850
|
Jordan decomposition, finding [imath]S[/imath]
let [imath]A = \left( \begin{array}{} 1 & 0 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array} \right)[/imath] [imath]A[/imath] has one eigenvalue [imath]\lambda=1[/imath] with algebraic multiplicity [imath]3[/imath] but it has only two linearly independent eigenvectors [imath]\left( \begin{array}{} 1 \\ 0 \\ 0 \end{array} \right),\left( \begin{array}{} 0 \\ 1 \\ 0 \end{array} \right)[/imath](this can be easily checked since [imath]A-\lambda I=A-I=\left( \begin{array}{} 0 & 0 & 3 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{array} \right)[/imath] and both vectors are in [imath]ker(A-I)[/imath]), so the geometric multiplicity is [imath]2[/imath] which is less than 3, the algebraic multiplicity. and from here, I was told that I need to compute something like [imath](A-I)w=v[/imath] to find [imath]w[/imath] where [imath]v[/imath] is the eigenvector. but I have two [imath]v[/imath]'s, so I don't know which [imath]v[/imath] I should take. and as I put [imath]A[/imath] in the Wolfram Alpha, I found its Jordan decomposition and there was no [imath]\left( \begin{array}{} 1 \\ 0 \\ 0 \end{array} \right)[/imath] in the column of [imath]S[/imath], such that [imath]A=SJS^{-1}[/imath], where [imath]J[/imath] is the Jordan form. exactly how to find [imath]S[/imath]? I'm so confused.
|
2058079
|
Finding [imath]P[/imath] in [imath]A = P^{-1}JP[/imath] (Jordan Form)
I'm having a lot of trouble understanding the process of finding a basis for the Jordan canonical form (the "algorithm"). My textbook (Friedberg 4E) isn't very clear, and I can't seem to find anything online. If we consider the example [imath]A[/imath] = [imath]\begin{bmatrix}6 & -2 & -1\\3 & 1 & -1\\2 & -1 & 2\end{bmatrix}[/imath], we get the characteristic polynomial [imath]p(t)[/imath] = ([imath]3 - t[/imath])[imath]^3[/imath]. I understand the process of finding the matrix [imath]J[/imath] in [imath]A = P^{-1}JP[/imath], but I just can't seem to grasp finding [imath]P[/imath]. Any clarification/explanation on the steps to do so would be incredibly helpful. Thank you.
|
2100120
|
Show that [imath]\alpha(x+y) = \alpha(x)α(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)[/imath] for every [imath]x, y \in\mathbb R[/imath].
[imath]\alpha(x) =\sum_{j=0}^\infty \frac{x^{3j}}{(3j)!}[/imath] [imath]\beta(x) = \sum_{j=0}^\infty \frac{x^{3j+2}}{(3j+2)!}[/imath] [imath]\gamma(x) = \sum_{j=0}^\infty \frac{x^{3j+1}}{(3j+1)!}[/imath] Show that [imath]\alpha(x+y) = \alpha(x)α(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)[/imath] for every [imath]x, y \in\mathbb R[/imath]. Any help will be appreciated.
|
2098942
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Power Series proofs
[imath]\alpha(x) =\sum_{j=0}^\infty \frac{x^{3j}}{(3j)!}[/imath] [imath]\beta(x) = \sum_{j=0}^\infty \frac{x^{3j+2}}{(3j+2)!}[/imath] [imath]\gamma(x) = \sum_{j=0}^\infty \frac{x^{3j+1}}{(3j+1)!}[/imath] Show that [imath]\alpha(x+y) = \alpha(x)α(y) + \beta(x)\gamma(y) + \beta(y)\gamma(x)[/imath] for every [imath]x, y \in\mathbb R[/imath]. Show that [imath]\alpha(x)^3 + \beta(x)^3 + \gamma(x)^3 − 3\alpha(x)\beta(x)\gamma(x) = 1[/imath] for every [imath]x\in\mathbb R[/imath]. My work: Ive noticed that [imath]\sinh(x)=\sum_{j=0}^\infty \frac{x^{2j+1}}{(2j+1)!}[/imath] and [imath]\cosh(x)=\sum_{j=0}^\infty \frac{x^{2j}}{(2j)!}[/imath] Can hyperbolic trig identies be used to solve these and if so how?
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