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2937011 | Are there uncountably many injective functions from [imath]\mathbb N[/imath] to [imath]\mathbb N[/imath]?
I think there are, but I haven't been able to prove this. I tried to make two injections, but I get stuck on trying to map all functions f from [imath]\mathbb N[/imath] to [imath]\mathbb N[/imath] onto the injective ones. How do you make sure f becomes injective, while making sure that the bigger injection stays injective? Or am I wrong and there are not even uncountably many functions? | 2798959 | Explicit bijection between [imath]\Bbb R[/imath] and permutations of [imath]\Bbb N[/imath]
The set of permutations of the natural numbers has the cardinality of the continuum. I've got injections in both directions, no problem. The Schröder–Bernstein theorem tells us that this implies the existence of a bijection. I'm wondering if it's possible to construct one explicitly. (In what follows, I'm using the convention that [imath]0\not\in\Bbb N[/imath]. This obviously doesn't change anything, cardinality-wise.) For [imath]\Bbb R\to S_\Bbb N[/imath], we note that every real number is the limit of some rearrangement of the alternating harmonic series. If [imath]\alpha\in\Bbb R[/imath], we start with positive terms: [imath]1+\frac13+\frac15+\cdots[/imath], until we obtain a partial sum greater than [imath]\alpha[/imath]. We then add negative terms until our partial sum is less than [imath]\alpha[/imath], then we switch back to the positive terms, starting with the first unused one, etc. In this way, we construct a series, and if we take the absolute values of the reciprocals of the terms of it, we have a permutation of [imath]\Bbb N[/imath]. This mapping is not onto, because many permutations of the series converge to [imath]\alpha[/imath] - all we have to do is "overshoot" at some point, and then continue converging to [imath]\alpha[/imath] as usual. (More trivially, we only get permutations with [imath]\sigma(1)=1[/imath] using this particular construction.) In the other direction, if we have a permutation [imath]\sigma\in S_\Bbb N[/imath], we can write the continued fraction [imath][\sigma(1);\sigma(2),\sigma(3),\ldots][/imath]. This actually injects [imath]S_\Bbb N[/imath] into [imath]\Bbb R\setminus\Bbb Q[/imath], because non-terminating continued fractions represent irrational numbers. Thus, this mapping is not onto; it is not even onto the irrationals.For example, no permutation of [imath]\Bbb N[/imath] maps in this way to any quadratic irrational, or to [imath]e[/imath], or to any other irrational whose c.f. expansion has repeated terms. So, those injections are fun and all, but finding an explicit bijection seems hard. Clearly, a bijection between [imath]S_\Bbb N[/imath] and any interval would suffice, because there are standard, elementary ways to construct bijections between [imath]\Bbb R[/imath] and any interval. I have Googled in vain for a solution, and I don't believe this question is a duplicate. I will be happy for a hint or a full solution, or an explanation of why such an explicit construction is impossible. Full disclosure: someone online claimed that this is "not hard", but refused to explain how, other than mentioning the Cauchy sequence construction of the reals. I don't see how that's useful, and I think he's mistaken or bluffing. I'm not too proud to admit I'm stumped. :/ |
2937084 | Showing that if n is a natural number larger than 3, then [imath]n!>2^n[/imath]
Showing that if n is a natural number larger than [imath]3[/imath], then [imath]n! > 2^n[/imath]. My try: Base Case: If [imath]n=4[/imath], then [imath]4!>2^4[/imath] [imath]24>16[/imath] So, the base case is true. Assuming [imath]P(k)[/imath] is true. [imath]k!>2^k[/imath] Now we need to show that [imath]P(k+1)[/imath] is true. [imath](k+1)!=2^{k+1}[/imath] Proof: [imath](k+1)!>(k+1)k![/imath] [imath]\implies (k+1)2^k[/imath] After this I have no idea how to solve further. Can anyone explain how to continue. | 405077 | Comparing [imath]2^n[/imath] to [imath]n![/imath]
Comparing [imath]\;2^n\;[/imath] and [imath]\;n!\;[/imath] I need to rewrite one or the other so they can be more easily compared (i.e. have similar form). Because of the factorial, I'm a little lost as to how to compare the two functions. Normally, I would take the logarithm when trying to re-express a power like [imath]2^n[/imath], but I don't think this will work, since I don't know how to take the logarithm of a factorial, if at all. Might the series expansions of either functions be the right approach? If not could, someone point me in the right direction? |
2937255 | If for each rational [imath]q[/imath] the set [imath]f^ {−1} ((q,\infty])[/imath] is measurable then [imath]f[/imath] is measurable.
I want to prove that If for each rational [imath]q[/imath] the set [imath]f^ {−1} ((q,\infty])[/imath] is measurable then [imath]f[/imath] is measurable. My attempt is the following: if [imath]f^{−1} ((q, \infty])[/imath] is measurable for all [imath]q[/imath], then for each real [imath]\lambda[/imath] [imath]f^{−1}((λ,\infty)) = \cup _{q>λ}f ^−1 ((q, \infty])[/imath] is measurable as a countable union of measurable set. | 2719586 | [imath]f^{-1}([-\infty,r))[/imath] measurable for all rational numbers [imath]r[/imath]
If [imath]X[/imath] is measurable space, [imath]f : X\to [-\infty,+\infty][/imath], function such that [imath]f^{-1}([-\infty,r))[/imath] measurable for all rational numbers [imath]r[/imath],prove than [imath]f[/imath] is measurable I know that [imath]f[/imath] is measurable iff set [imath]\{x\in X : f(x)<a\}[/imath] is measurable for all [imath]a\in \mathbb{R}[/imath] but for [imath]a\in\mathbb{Q}[/imath] I don't know how to proceed. |
2937472 | Show that [imath]S_n=0[/imath] for infinitely many [imath]n[/imath]
Let the series [imath]\sum_{n=1}^{\infty}a_n[/imath] is convergent but not absolutely convergent and [imath]\sum_{n=0}^{\infty}a_n=0[/imath]. Denote by [imath]S_k[/imath] the partial sums, Then prove that [imath]S_k=0[/imath] for infinitely many [imath]k[/imath]. My Efforts Using the definition that a series converges if and only sequence of partial sum converges. Now it is given that [imath]\sum_{n=0}^{\infty}a_n=0[/imath], it follows that sequence of partial sum converges to [imath]0[/imath]. Now we can only conclude that around each neighborhood of [imath]0[/imath] however small there exist a stage after which we have infinitely many [imath]S_k[/imath] belong to the neighborhood but that does not prove that [imath]S_k=0[/imath] I request for some helpful comments so that I can solve this problem. | 400335 | [imath]\sum a_n[/imath] be convergent but not absolutely convergent, [imath]\sum_{n=1}^{\infty} a_n=0[/imath]
Let [imath]\sum a_n[/imath] be convergent but not absolutely convergent, [imath]\sum_{n=1}^{\infty} a_n=0[/imath],[imath]s_k[/imath] denotes the partial sum then could anyone tell me which of the following is/are correct? [imath]1.[/imath] [imath]s_k=0[/imath] for infinitely many [imath]k[/imath] [imath]2[/imath]. [imath]s_k>0[/imath] and [imath]<0[/imath] for infnitely many [imath]k[/imath] 3.[imath]s_k>0[/imath] for all [imath]k[/imath] 4.[imath]s_k>0[/imath] for all but finitely many [imath]k[/imath] if we take [imath]a_n=(-1)^n{1\over n}[/imath] then [imath]\sum a_n[/imath] is convergent but not absolutely convergent,but I don't know [imath]\sum_{n=1}^{\infty} a_n=0[/imath]? so I am puzzled could any one tell me how to proceed? |
2937755 | Polynomials giving integer outputs
Given a polynomial [imath]p(x)[/imath] that outputs an integer for every integer input [imath]x[/imath], is it necessary for [imath]p(x)[/imath] to have rational coefficients? | 1493801 | If a polynomial takes integer values at every integer, does it have integer coefficients?
Let [imath]f \in \mathbf R[x][/imath] and for every [imath]k \in \mathbf Z[/imath], [imath]f(k)\in \mathbf Z[/imath]. Does this imply that [imath]f \in \mathbf Z[x][/imath]? I tried to find a counter example, but my intuition tells me it's true and I don't know how to prove the statement either. |
2937750 | Prove that [imath]r_1s_1, r_2s_2, ..., r_{p-1}s_{p-1}[/imath] is not a reduced residue system modulo p.
If [imath]r_1, ... r_{p-1}[/imath] and [imath]s_1, ... s_{p-1}[/imath] are two reduced residue systems modulo an odd prime p, prove that [imath]r_1s_1, r_2s_2, ..., r_{p-1}s_{p-1}[/imath] is not a reduced residue system modulo p. don't know where to start, any help appreciated! | 1901845 | [imath]a_1 a_1', a_2 a_2', ..., a_n a_n'[/imath] is not a reduced modulo system [imath]\mod p[/imath].
If [imath]a_1, a_2, ..., a_n[/imath] is a reduced residue system [imath]\mod p[/imath] and [imath]a_1', a_2', ..., a_n'[/imath] is another reduced residue system [imath]\mod p[/imath] then [imath]a_1 a_1', a_2 a_2', ..., a_n a_n'[/imath] is not a reduced modulo system [imath]\mod p[/imath]. Not able to show a counter to it. Help Needed. |
2928199 | Proof inequality [imath]a,b, c \in [0,1][/imath]
Let [imath]a , b , c \in [0,1][/imath]. Prove that [imath]\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab} \leq2[/imath]. Thanks Mark for your reply. I'm new to the group so I dont know how to deal with this . Anyways I tried assuming that [imath]c=\max\{a,b,c\}[/imath] . I also tried applying AM-GM on [imath](1+bc)[/imath] but I got no result. | 1086580 | If [imath]0\le a,b,c\le 1[/imath], then [imath]\frac a{1+bc}+\frac b{1+ac}+\frac c{1+ab}\le 2[/imath]
If a,b,c are real numbers such that [imath]0\le a,b,c\le 1[/imath], then [imath]\frac a{1+bc}+\frac b{1+ac}+\frac c{1+ab}\le 2[/imath] I did the following :- Using Jensen's inequality, take [imath]f(x)=\frac {x^2}{x+p},(p=abc)[/imath] which is convex for [imath]0\le x\le 1[/imath]. Therefore [imath]\begin{align}\frac {f(a)+f(b)+f(c)}3&\le f(\frac {a+b+c}3)\\=>\frac {a^2}{a+abc}+\frac {b^2}{b+abc}+\frac {c^2}{c+abc}&\le \frac {(a+b+c)^2}{(a+b+c)+3abc}\end{align}[/imath] How do I prove that the RHS is less than or equal to 2? |
2937943 | Devils and Infinity
You are in hell for all eternity and the devil gives you two dollar bills every day with increasing serial numbers. He then takes the dollar bill you have with the smallest serial number. At the end of your infinite stay in hell, do you have infinite money or no money? The intuitive answer for me was that if you look at the partial sums, on the Nth day you'll have [imath]N[/imath] dollars. So clearly after infinity days you should have infinity dollars. On the other hand, you can also prove that for every dollar you ever received, there's a corresponding day that the devil takes it back, leaving you with nothing in the end. Which one is the right answer? | 869916 | A strange puzzle having two possible solutions
A friend of mine asked me the following question: Suppose you have a basket in which there is a coin. The coin is marked with the number one. At noon less one minute, someone takes the coin number one and put into the basket two coins: the number two and the number three. At noon less [imath]\frac{1}{2}[/imath] minute, he takes the coin number two and put into the basket the coins number [imath]4[/imath] and the number [imath]5[/imath] and so on. At noon less [imath]\frac{1}{k}[/imath] he takes the coin number [imath]k[/imath] and put two other coins. The question is: How many coins I can find in the basket at noon? Intuitively the answer is [imath]\infty[/imath] because I added one coin for infinite times. But the correct answer seems to be [imath]0[/imath] because every coin numbered [imath]k[/imath] has been removed at noon less [imath]\frac{1}{k}[/imath] of minute. What is the correct answer? Thanks. |
1662958 | A probability theory question about independent coin tosses by two players
Say Bob tosses his [imath]n+1[/imath] fair coins and Alice tosses her [imath]n[/imath] fair coins. Lets assume independent coin tosses. Now after all the [imath]2n+1[/imath] coin tosses one wants to know the probability that Bob has gotten more heads than Alice. The way I thought of it is this : if Bob gets [imath]0[/imath] heads then there is no way he can get more heads than Alice. Otherwise the number of heads Bob can get which allows him to win is anything in the set [imath]\{1,2,\dots,n+1\}[/imath]. And if Bob gets [imath]x[/imath] heads then the number of heads that Alice can get is anything in the set [imath]\{0,1,2,..,x-1\}[/imath]. So\begin{align}P(\text{Bob gets more heads than Alice})&= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} P( \text{Bob gets x heads }\cap \text{Alice gets y heads }) \\[0.2cm]&= \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \left(C^{n+1}_x \frac{1}{2}^{x} \frac{1}{2}^{n+1-x}\right)\left( C^n_y \frac{1}{2}^y \frac {1}{2}^{n-y}\right)\\[0.2cm]& = \sum_{x=1}^{n+1} \sum_{y=0}^{x-1} \frac{C^{n+1}_x C^n_y}{2^{2n+1}}\end{align} How does one simplify this? Apparently the answer is [imath]\frac{1}{2}[/imath] by an argument which looks like this, Since Bob tosses one more coin that Alice, it is impossible that they toss both the same number of heads and the same number of tails. So Bob tosses either more heads than Alice or more tails than Alice (but not both). Since the coins are fair, these events are equally likely by symmetry, so both events have probability 1/2. | 3042583 | Two sets of coin flips
Below is a problem that I did. I have reason to believe that the correct answer is [imath]\frac{1}{2}[/imath]. Please check my work and find any and all errors. Thanks, Bob Problem: Suppose person 1 flips a fair coin [imath]5[/imath] times and that person 2 flips a fair cont [imath]4[/imath] times. What is the probability that person 1 got more heads than person 2. Answer: Let [imath]p_1(x)[/imath] be the probability that person 1 got [imath]x[/imath] heads. Let [imath]p_2(x)[/imath] be the probability that person 2 got [imath]x[/imath] heads. Let [imath]p[/imath] be the probability we seek. [imath]\begin{align*} P_1(0) &= \frac{1}{32} \\ P_1(1) &= \frac{5}{32} \\ P_1(2) &= \frac{5}{16} \\ P_1(3) &= \frac{5}{16} \\ P_1(4) &= \frac{5}{32} \\ P_1(5) &= \frac{1}{32} \\ % P_2(0) &= \frac{1}{16} \\ P_2(1) &= \frac{1}{4} \\ P_2(2) &= \frac{3}{8} \\ P_2(3) &= \frac{1}{4} \\ P_2(4) &= \frac{1}{16} \\ \end{align*}[/imath] [imath]\begin{align*} p &= p_1(1) \left( p_2(0) \right) + p_1(2)\left( p_2(0) + p_2(1) \right) + \\ & p_1(3) \left( p_2(0) + p_2(1) + p_2(2) \right) + \\ & p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ % p_1(1) \left( p_2(0) \right) &= \frac{5}{32} \left( \frac{1}{16} \right) = \frac{5}{512} \\ p &= \frac{5}{512} + p_1(2)\left( p_2(0) + p_2(1) \right) + \\ &p_1(3) \left( p_2(0) + p_2(1) + p_2(2) \right) + \\ &p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p_1(2)\left( p_2(0) + p_2(1) \right) &= \frac{5}{32} \left( \frac{1}{16} + \frac{1}{4} \right) \\ p_1(2)\left( p_2(0) + p_2(1) \right) &= \frac{5}{32} \left( \frac{5}{16} \right) = \frac{25}{512} \\ p &= \frac{5}{512} + \frac{25}{512} + p_1(3) \\ & \left( p_2(0) + p_2(1) + p_2(2) \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{30}{512} + \\ & p_1(3) \left( \frac{1}{16} + \frac{1}{4} + \frac{3}{8} \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{30}{512} + p_1(3) \left( \frac{11}{16} \right) + \frac{5}{32} \left( \frac{1}{16} + \frac{1}{4} + p_2(2) + p_2(3) \right) + p_1(5) \\ p &= \frac{30}{512} + \frac{5}{16} \left( \frac{11}{16} \right) + \frac{5}{32} \left( \frac{1}{16} + \frac{1}{4} + \frac{3}{8} + \frac{1}{4} \right) + \frac{1}{32} \\ \end{align*}[/imath] [imath]\begin{align*} p &= \frac{30}{512} + \frac{55}{256} + \frac{5}{32} \left( \frac{15}{16} \right) + \frac{1}{32} \\ p &= \frac{140}{512} + \frac{5}{32} \left( \frac{15}{16} \right) + \frac{1}{32} \\ p &= \frac{140}{512} + \frac{75}{512} + \frac{16}{512} \\ p &= \frac{231}{512} \\ \end{align*}[/imath] I am hoping that somebody can tell me where I went wrong above. Based upon the comments I got from the group, here is the correct solution: Answer: \newline Let [imath]p_1(x)[/imath] be the probability that person 1 got [imath]x[/imath] heads. Let [imath]p_2(x)[/imath] be the probability that person 2 got [imath]x[/imath] heads. Let [imath]p[/imath] be the probability we seek. [imath]\begin{align*} P_1(0) &= \frac{1}{32} \\ P_1(1) &= \frac{5}{32} \\ P_1(2) &= \frac{5}{16} \\ P_1(3) &= \frac{5}{16} \\ P_1(4) &= \frac{5}{32} \\ P_1(5) &= \frac{1}{32} \\ % P_2(0) &= \frac{1}{16} \\ P_2(1) &= \frac{1}{4} \\ P_2(2) &= \frac{3}{8} \\ P_2(3) &= \frac{1}{4} \\ P_2(4) &= \frac{1}{16} \\ \end{align*}[/imath] [imath]\begin{align*} p &= p_1(1) \left( p_2(0) \right) + p_1(2)\left( p_2(0) + p_2(1) \right) + \\ & p_1(3) \left( p_2(0) + p_2(1) + p_2(2) \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ % p_1(1) \left( p_2(0) \right) &= \frac{5}{32} \left( \frac{1}{16} \right) = \frac{5}{512} \\ p &= \frac{5}{512} + p_1(2)\left( p_2(0) + p_2(1) \right) + \\ &p_1(3) \left( p_2(0) + p_2(1) + p_2(2) \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p_1(2)\left( p_2(0) + p_2(1) \right) &= \frac{5}{16} \left( \frac{1}{16} + \frac{1}{4} \right) \\ p_1(2)\left( p_2(0) + p_2(1) \right) &= \frac{5}{16} \left( \frac{5}{16} \right) = \frac{25}{256} \\ \end{align*}[/imath] [imath]\begin{align*} p &= \frac{5}{512} + \frac{25}{256} + p_1(3) \left( p_2(0) + p_2(1) + p_2(2) \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{55}{512} + \frac{5}{16} \left( \frac{1}{16} + \frac{1}{4} + \frac{3}{8} \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{55}{512} + \frac{5}{16} \left( \frac{11}{16} \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{55}{512} + \frac{55}{256} + \frac{5}{32} \left( \frac{1}{16} + \frac{1}{4} \right) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{165}{512} + \frac{5}{32} \left( \frac{5}{16} + p_2(2) + p_2(3) \right) + p_1(5) \\ p &= \frac{165}{512} + \frac{5}{32} \left( \frac{5}{16} + \frac{3}{8} + \frac{1}{4} \right) + \frac{1}{32} \\ p &= \frac{165}{512} + \frac{5}{32} \left( \frac{11}{16} + \frac{1}{4} \right) + \frac{1}{32} \\ p &= \frac{165}{512} + \frac{75}{512} + \frac{1}{32} = \frac{256}{512} \\ p &= \frac{1}{2} \\ \end{align*}[/imath] |
2938884 | Find the limit of [imath]\begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}[/imath]
Find the following limit: [imath]\begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}[/imath] I have tried to divide the numerator and denominator by [imath]\sqrt{x}[/imath], but it did not work. I have tried to multiply by the conjugates of the numerator and denominator simultaneously but it did not work. I have tried to multiply by the conjugates of the numerator only but it did not work. So what shall I do? | 2112283 | Calculate [imath]\lim\limits_{x\to4} \frac {\sqrt {1+2x}-3} {\sqrt x -2}[/imath]
After I just learned there are ways to cancel out [imath]0[/imath]s in the divisor of fractions onto limits I looked back at a task where I gave up when I got the result [imath]\lim\limits_{x\to4} \frac {\sqrt {1+2x}-3} {\sqrt x -2}[/imath] so I tried to find a way to get the [imath]0[/imath] out here, as well. Am I just not seeing the solution or is there no way to do it? |
2936390 | Using a orthonormal bases of [imath]\mathbb R^4[/imath] and [imath]\mathbb R^6[/imath] find a solution [imath]Ax=z_1[/imath]
If you have orthonormal vectors [imath]q_1,q_2,q_3,q_4[/imath] of [imath]\mathbb R^4[/imath] and orthonormal vectors [imath]z_1,z_2,z_3,z_4,z_5,z_6[/imath] of [imath]\mathbb R^6[/imath] and matrix A is [imath]A=z_1q_1^T+z_2q_2^T[/imath]. a)Find a base and dimension of fundamental subspaces of [imath]A[/imath] b) Using fact from a) find a solution [imath]Ax=z_1[/imath] For a)[imath] Im(A)=L(z_1,z_2), dimIm(A)=2, Im(A^T)=L(q_1,q_2), dim Im(A^T)=2, ker(A)=L(q_3,q_4) dimker(A)=2, ker(A^T)=L(z_3,z_4,z_5,z_6) ,dim N(A^T)=4.[/imath] For b) [imath]Ax=z_1[/imath], If [imath]z_1 \in Im(A)[/imath] then this equation have solution, then [imath]x=q_1[/imath] since [imath](x_1,x_1)=1[/imath] and [imath](x_2,x_1)=0[/imath] so [imath]Aq_1=(z_1q_1^Tq_1+z_2q_2^Tq_1)=z_1+0=z_1[/imath] is this ok? | 2319340 | Let [imath]q_1,q_2,q_3,q_4[/imath] be orthonormal vectors in [imath]\mathbb{R}^4[/imath],[imath]z_1,z_2,...,z_6[/imath] orthonormal vectors in [imath]\mathbb{R}^6[/imath] and [imath]A=z_1q_1^T + z_2q_2^T[/imath].
Let [imath]q_1,q_2,q_3,q_4[/imath] be orthonormal vectors in [imath]\mathbb{R}^4[/imath],[imath]z_1,z_2,...,z_6[/imath] orthonormal vectors in [imath]\mathbb{R}^6[/imath] and [imath]A=z_1q_1^T + z_2q_2^T[/imath]. a) Find the base and dimension of fundamental subspaces of matrix A b) Using the results given by a) find the general solution for [imath]Ax - z_1[/imath]. Any hint would be appreciated. |
2939108 | Calculating the cdf of [imath]Z=XY[/imath] given [imath]\ln X, Y \sim \exp(1)[/imath]
Let [imath]X,Y[/imath] be iid real random variables where [imath]\ln X\sim \exp(1) [/imath]. How can I determine the cdf of [imath]Z:=XY[/imath]? I would know how to do so for [imath]X \sim \exp(1)[/imath], but not sure how to deal with the [imath]\ln.[/imath] Some help would be much appreciated. | 2921211 | Let [imath]\ln X \sim \operatorname{Exp}(1)[/imath] and [imath]X,Y[/imath] i.i.d. What is the distribution of [imath]Z=XY[/imath]?
Let [imath]\ln X \sim \operatorname{Exp}(1)[/imath] and [imath]X,Y[/imath] i.i.d. What is the distribution of [imath]Z=XY[/imath]? My attempt: [imath]P(Z\le z)=P(XY\le z)=P(\ln(XY)\le \ln(z))=P(\ln(X)+\ln(Y)\le \ln(z))[/imath] We need the convolution, since [imath]\ln X [/imath] and [imath]\ln Y[/imath] are independent we get, [imath]\int_0^{\ln(z)} f_{\ln(X)}\circledast f_{\ln(Y)}=\int_0^{\ln(z)}e^{-y}e^{-x+y}\,dy=e^{-x}ln(z),[/imath] i.e. [imath]e^{-x}\ln(z)[/imath] is the density of [imath]XY[/imath] |
2939037 | Prove that [imath]n[/imath] is divisible by [imath]6[/imath].
If the quadratic equations [imath]x^2-mx+n=0[/imath] and [imath]x^2+mx-n=0[/imath] both have integral roots, prove that [imath]6|n[/imath]. I've proved that [imath]3|n[/imath], and that [imath]2|n[/imath] for odd [imath]m[/imath], but I can't seem to prove it for even [imath]m[/imath]. Please help. | 788993 | Prove that [imath]n[/imath] is divisible by [imath]6[/imath]
Problem: Let [imath]x^2+mx+n[/imath] and [imath]x^2+mx-n[/imath] give integer roots where [imath](m,n)[/imath] are integers. Show that [imath]n[/imath] is divisible by [imath]6[/imath] My attempt: Since the roots are integers then the discriminants of both the equations should be perfect squares. Let [imath]a=\sqrt{m^2-4n}[/imath] and [imath]b=\sqrt{m^2+4n}[/imath], then [imath](ab)^2=m^4-16n^2[/imath] where [imath](a,b,m,n)[/imath] are all integers. I am stuck here... |
2939259 | continuous functions over the closed interval [imath][0,1][/imath]
let [imath]A[/imath] be a set of continuous functions over the closed interval [imath][0,1][/imath], which applies to the following conditions: [imath] 1: \forall f \in A , \forall x \in [0,1] , f(x) \geq 0 [/imath] [imath]2: \forall f, g \in A , f + g \in A [/imath] [imath]3: \forall x \in [0,1], \textbf{there is} f \in A \quad \textbf{so that} f(x) > 0 [/imath] How can I show that " there is [imath] h \in A[/imath] so that [imath] \forall x \in [0,1], h(x) > 0[/imath]."? | 2916039 | let [imath]A[/imath] be a set of continuous functions over the closed interval [imath][0,1][/imath]..........
let [imath]A[/imath] be a set of continuous functions over the closed interval [imath][0,1][/imath], which applies to the following conditions: [imath] 1: \forall f \in A , \forall x \in [0,1] , f(x) \geq 0 [/imath] [imath]2: \forall f, g \in A , f + g \in A [/imath] [imath]3: \forall x \in [0,1], \textbf{there is} f \in A \quad \textbf{so that} f(x) > 0 [/imath] Prove that there is [imath] h \in A[/imath] so that [imath] \forall x \in [0,1], h(x) > 0[/imath]. |
2939130 | Checking whether a given space is indiscrete or not
Problem Let X be an infinite set and τ a topology on X with the property that the only infinite subset of X which is open is X itself. Is (X,τ) necessarily an indiscrete space? So I have to prove that set contains only two elements namely X and [imath]\emptyset[/imath] . It can be shown that sets like X-{x} are not in topology. How to proceed? | 1701383 | Is [imath](X,\mathcal T)[/imath] necessarily an indiscrete space?
Let [imath]X[/imath] be an infinite set with [imath]\mathcal T[/imath] a topology on [imath]X[/imath]. If [imath]X[/imath] is the only infinite subset of [imath]X[/imath] that is open, is [imath](X,\mathcal T)[/imath] necessarily an indiscrete space? My intuitive thinking leads me to think that the answer is NO. We know that, if it is the indiscrete space, then [imath]\mathcal T = \{ \emptyset, X \}[/imath]. Notice, however, that we are only given tat [imath]X[/imath] is the only INFINITE subset of [imath]\mathcal T[/imath], we have no information regarding the other possible finite open sets of [imath]X[/imath], that is, there may exist a finite subset [imath]A \subset X[/imath] with [imath]A \in \mathcal T[/imath]. This is surely not the indiscrete space, yet it still satisfies the criteria given in the question. Is there a better exact example that I can provide that might help show my point in a more rigorous mathematical way? |
2190513 | Prove that [imath] x^4 - 2 [/imath] is irreducible over [imath] \mathbb{Z}[i] [/imath]
How do I prove that [imath] p(x) = x^4 - 2 [/imath] is irreducible over [imath] \mathbb{Z}[i] [/imath]? This seems very elementary yet I'm not sure how to do it. Someone suggested using Eisenstein and [imath] p = 1+i [/imath], but this doesn't seem right because [imath] (1+i)^2 = 2i [/imath] is an associate of [imath] -2 [/imath]. I have seen somewhere that one can use a generalized version of the Rational Root Theorem and simply check that [imath] 1+i [/imath] and [imath] 1-i [/imath] are not roots of [imath] p(x) [/imath], is this correct? Thank you for your help. | 283749 | Irreducibility of [imath]X^n - 2[/imath] over [imath]\mathbf Z[i][/imath].
I was “playing” with Eisenstein's criterion and the polynomials [imath]X^n-p[/imath] ([imath]p[/imath] being a prime number). Over [imath]\mathbf Z[/imath], the classical Eisenstein's criterion directly applies and shows that [imath]X^n - p[/imath] is an irreducible element of [imath]\mathbf Z[X][/imath]. The question was “what happens over [imath]\mathbf Z[i][/imath]?” We know that in this new ring, a prime number [imath]p[/imath] has one of three different fates: if [imath]p \equiv -1\ (\mathrm{mod}\, 4)[/imath], [imath]p[/imath] still is an irreducible element of [imath]\mathbf Z[i][/imath], or equivalently [imath](p) \subset \mathbf Z[i][/imath] still is prime. (Unless I'm mistaken, number theorists say that [imath]p[/imath] is inert). Eisenstein's criterion still applies and the polynomial is irreducible over [imath]\mathbf Z[i][/imath]. if [imath]p \equiv 1\ (\mathrm{mod}\, 4)[/imath], [imath]p[/imath] can be written as a sum of two squares: [imath]p = n^2 + m^2[/imath] and is therefore reducible on [imath]\mathbf Z[i][/imath]: [imath]p = (n+im)(n-im)[/imath]. These two factors aren't associated (the units in [imath]\mathbf Z[i][/imath] being [imath]\pm 1[/imath] and [imath]\pm i[/imath]) so [imath](p)[/imath] splits: [imath](p) = (n+im)(n-im)[/imath] is the product of two different prime ideals. One can still use Eisenstein's criterion with either of these prime ideals, and [imath]X^n - p[/imath] is irreducible over [imath]\mathbf Z[i][/imath]. if [imath]p = 2[/imath], the oddest of all primes, one still has a factorisation [imath]p = (1+i)(1-i)[/imath] but this time, [imath]1+i[/imath] and [imath]1-i[/imath] are associated. In NT slang, [imath](2)[/imath] is ramified: [imath](2) = (1+i)(1-i) = (1+i)^2[/imath] and one cannot directly use Eisenstein's criterion. So here's my question: how do we prove (or disprove) the irreducibility of [imath]X^n - 2[/imath] over [imath]\mathbf Z[i][/imath]? |
2938931 | Ideal of a ring with element [imath]-1[/imath]
Consider the ring [imath](\Bbb{Z}, +, \cdot)[/imath]. Let [imath]I[/imath] be an ideal of the ring that contains the element [imath]-1[/imath]. Show that [imath]I = \Bbb{Z}[/imath] Could some one please explain how to solve this. | 167389 | Rings, ideals and units-proof
Let [imath]R[/imath] be a ring and let [imath]J[/imath] be an ideal of [imath]R[/imath]. Assume [imath]J[/imath] contains a unit of [imath]R[/imath]. Prove [imath]J=R[/imath], rigorously. Again I feel here as if the statement is intuitive so I am not able to make it rigorous. If you could show all steps in the proof that would be nice. |
2939160 | Alternatives for evaluating [imath] \int \frac { 1 } { 5 + 4 \cos x} \ dx [/imath] ??
[imath] \int \frac { 1 } { 5 + 4 \cos x} \ dx [/imath] [imath] \text{The solution given in the book for solving this was to use the identity.} [/imath] [imath] \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1+ \tan^2 \frac{x}{2}} [/imath] I was wondering if there was any other way for solving this ? | 1740458 | Finding [imath]\int \frac{dx}{a+b \cos x}[/imath] without Weierstrass substitution.
I saw somewhere on Math Stack that there was a way of finding integrals in the form [imath]\int \frac{dx}{a+b \cos x}[/imath] without using Weierstrass substitution, which is the usual technique. When [imath]a,b=1[/imath] we can just multiply the numerator and denominator by [imath]1-\cos x[/imath] and that solves the problem nicely. But I remember that the technique I saw was a nice way of evaluating these even when [imath]a,b\neq 1[/imath]. |
2939455 | What is the sum of binomial coefficients?
[imath]\binom n0 + \binom n4 + \binom n8 + \cdots[/imath] Any hints? | 142260 | Sum of every [imath]k[/imath]th binomial coefficient.
It is widely known that [imath]\sum_{m=0}^{n} {n\choose m} = 2^n[/imath] and that [imath]\sum_{m=0}^{\lfloor\frac{n}{2}\rfloor}{n\choose 2m} = 2^{n-1}[/imath] Both results can be proven by exploting the nature of the roots of unity. Analagously, we can find [imath]\sum_{m=0}^{\lfloor\frac{n}{3}\rfloor}{n\choose 3m}[/imath] by taking the sum of the equations [imath](1+1)^n = {n\choose 0} + {n\choose 1} + {n\choose 2} + {n\choose 3} + \cdots[/imath] [imath](1+\omega)^n = {n\choose 0} + {n\choose 1}\omega + {n\choose 2}\omega^2 + {n\choose 3} + \cdots[/imath] [imath](1+\omega^2)^n = {n\choose 0} + {n\choose 1}\omega^2 + {n\choose 2}\omega + {n\choose 3} + \cdots[/imath] by taking [imath]\omega[/imath] as a primitive cube root of [imath]1[/imath] and exploiting the fact that the roots of unity sum to [imath]0[/imath], i.e. [imath]1 + \omega + \omega^2 = 0[/imath]. The above method however seems to depend on the fact that every non-trivial root is primitive, so that the method only works for primes (For example, attemtping this method for [imath]k=4[/imath] will quickly run into problems as there is no longer full cancellation). Is there a generalization of this method for finding the sum of every [imath]k[/imath]th binomial coefficient for arbitrary [imath]k[/imath]? Failing that, does anyone know of any general method for finding such a sum? |
2939980 | If [imath]m > 0[/imath], fix a reduced residue system [imath]r_{1}, r_2, \dotsc, r_{\varphi(m)} [/imath] mod [imath] m[/imath]. Let [imath]x=r_1+r_2+\dotsb+r_{\varphi(m)}[/imath]. What is [imath]x[/imath] mod [imath]m[/imath]?
Given [imath]m > 0[/imath], fix a reduced residue system (RRS) [imath]r_{1}, r_2,\dotsc , r_{\varphi(m)} [/imath] mod [imath] m[/imath]. Let [imath]x[/imath] denote the sum [imath]r_1 + r_2 + \dotsb + r_{\varphi(m)}[/imath]. What is [imath]x[/imath] mod [imath]m[/imath]? The problem is that I'm stuck. So suppose that [imath]m = 2k + 1[/imath]. Then the RRS would include an even number of elements since all even numbers are relatively prime to [imath]m[/imath]. But I don't know what to do with this information and I also don't know how to solve for if [imath]m[/imath] is even. Any help will be appreciated. | 23924 | Show that if [imath]c_1, c_2, \ldots, c_{\phi(m)}[/imath] is a reduced residue system modulo m, [imath]m \neq 2[/imath] then [imath]c_1 + \cdots+ c_{\phi(m)} \equiv 0 \pmod{m}[/imath]
Show that if [imath]c_1, c_2,\ldots, c_{\phi(m)}[/imath] is a reduced residue system modulo [imath]m[/imath], [imath]m \neq 2[/imath], and [imath]m[/imath] is a positive integer, then [imath]c_1 +\cdots+ c_{\phi(m)} \equiv 0 \pmod{m}[/imath] From the problem statement, I only know that [imath]\gcd(c_i, m ) = 1[/imath]. Is there any related theorem that I missed? A hint would be greatly appreciated. Thanks, Chan |
2940264 | Sum [imath]\sum_{i=1}^\infty \frac{-1^{i-1}}{2i-1} [/imath] in a different order
I'm supposed to sum this infinite series in a different order so that the limit appears to be .5. I need to to add the first k positive terms, and subtract the first l negative terms, so that the result is within [imath]\pm .05 [/imath] of [imath].5[/imath] (between .45 and .55), [imath]\sum_{i=1}^\infty \frac{-1^{i-1}}{2i-1} [/imath] How do I set it up to add the positive terms, and subtract the negative? | 2938861 | Sum of alternating series [imath]\sum_{i=1}^n \frac{(-1)^{i-1}}{2i-1}[/imath] with accuracy of [imath]\pm .05[/imath].
The series is: [imath]\sum_{i=1}^n \frac{(-1)^{i-1}}{2i-1}[/imath] I have already shown that it converges but am lost on finding the sum. I need to find the sum to within [imath]\pm .05[/imath]. |
2940631 | Show that infinitely many positive integer pairs [imath](m,n)[/imath] exist s.t [imath]\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{N}[/imath]
Show that infinitely many positive integer pairs [imath](m,n)[/imath] exist such that [imath]\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{N}[/imath]. I couldn't solve it but I did make an observation which might or might not be helpful. WLOG assume that [imath]m < n[/imath] (since the term is symmetric wrt [imath]m,n[/imath] along with the fact that ignoring the [imath]m=n[/imath] case wouldn't create any trouble). Write [imath]n=mq+r[/imath] for some [imath]r\in \{[0,m-1]\cap \mathbb{N}\}[/imath]. Now, the question boils down to showing [imath]\frac{m+1}{n}+\frac{r+1}{m} \in \mathbb{N}[/imath]. Note that each of the summand is [imath]\leq 1[/imath]. Since the equality case wouldn't be helpful in generating infinitely many pairs of [imath](m,n)[/imath], we can safely say that the sum is equal to [imath]1[/imath]. Now I don't know how to proceed from here. | 151549 | Proving there are an infinite number of pairs of positive integers [imath](m,n)[/imath] such that [imath]\frac{m+1}{n}+\frac{n+1}{m}[/imath] is a positive integer
The question is: Show that there are an infinite number of pairs [imath](m,n): m, n \in \mathbb{Z}^{+}[/imath], such that: [imath]\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{Z}^{+}[/imath] I started off approaching this problem by examining the fact that whenever the expression was a positive integer, the following must be true: [imath]\frac{m+1}{n}+\frac{n+1}{m} - \left\lfloor\frac{m+1}{n}+\frac{n+1}{m}\right\rfloor = 0[/imath] However, I was unable to do much more with this expression, so I abandoned it and started the problem again from a different angle. Next I re-arranged the expression to state that: [imath]\frac{m^2+m+n^2+n}{mn} \in \mathbb{Z}^{+} \implies \frac{m(m+1) + n(n+1)}{mn} \in \mathbb{Z}^{+}[/imath] And therefore: [imath]mn \mid (m(m+1) + n(n+1))[/imath] However, I'm unsure how to demonstrate there are infinitely many occurances of [imath](m, n)[/imath] for which this is true. So I'd appreciate any help. Thanks in advance |
2940838 | Why would [imath](x+1)[/imath] transform the graph to the left?
Why do transformations for the [imath]x[/imath] variable in graphs work opposite as you would expect? Example: [imath]f(x) =(x)-1[/imath] moves the graph down as you would expect but [imath]g(x) = (x+1)-1[/imath] moves the graph down (as it looks like) but instead of to the right one it's to the left one. Note: I understand fully how to do graph transformations but don't understand why we just deem the x-value transformations as "doing the opposite". Thanks! | 2117280 | Why graph of [imath]\sin(x+{45}^\circ[/imath]) shifts to left instead of right
Why does the graph of [imath]\sin(x+{45}^\circ[/imath]) shifts to left instead of right comparing with [imath]\sin x[/imath]. With respect to point at which line passed from origin. Mathematically i know , I can fill in values and plot to verify , but is there any other easy to understand explanation for this ? Sorry if this sounds foolish question. Thanks See graph here |
2941066 | Let [imath]a,b,c,d[/imath] be real numbers such that [imath]a. Express the set [a,b] \cup [c,d] as the difference of two sets[/imath]
I am not sure how I would express these sets as a difference. My original attempt was to show that it is the set [imath][a,d][/imath] and take away the universal set. I would appreciate any help. Thank you in advance. | 2938879 | Express set [imath][a,b] \cup [c,d][/imath] as the difference of two intervals
So if [imath]a < b < c < d[/imath] , how do I express the set [imath][a,b] \cup [c,d][/imath] as the difference of two intervals? Is it as simple as [imath][a,d] -[b,c][/imath]? |
2941072 | Compute [imath]\lim_{n\rightarrow\infty}\sqrt[{n+1}]{(n+1)!}-\sqrt[n]{n!}[/imath]
Evaluate [imath]L=\lim_{n\rightarrow\infty}\sqrt[{n+1}]{(n+1)!}-\sqrt[n]{n!}[/imath] How I approached it and where I get stuck: [imath]\lim_{n\rightarrow\infty}\sqrt[{n+1}]{(n+1)!}-\sqrt[n]{n!}=\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}nn(\frac{\sqrt[{n+1}]{(n+1)!}}{\sqrt[n]{n!}}-1)[/imath] Now: [imath]\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}n=\lim_{n\rightarrow\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\rightarrow\infty}e^{\frac1n\sum_{k=1}^n\ln(\frac kn)}=e^{\int_0^1\ln(x)dx}=e^{-1}=\frac1e[/imath] So [imath]L=\lim_{n\to\infty}\frac 1e\times n(\frac{\sqrt[{n+1}]{(n+1)!}}{\sqrt[n]{n!}}-1)[/imath]. Now this is where I get stuck. What should I do? | 1468834 | Computation of a limit involving factorial [imath]\lim_{n \to \infty} \sqrt[n+1] {(n+1)!} - \sqrt[n] {(n)!} = \frac{1}{e}[/imath]
I want to prove the following limit: [imath]\lim_{n \to \infty} \sqrt[n+1\;] {(n+1)!} - \sqrt[n] {(n)!} = \frac{1}{e}.[/imath] I searched the forum & found the link here: If [imath]\frac{p_{n+1}}{np_n} \to p > 0 [/imath], then [imath]\sqrt[n+1]{p_{n+1}}-\sqrt[n]{p_{n}} \to \frac{p}{e}[/imath] . But still, there is no way out of the problem. So, please solve it. |
2940591 | Evaluating [imath]\int_{-\infty}^{\infty}\frac{\exp\left(-a x^2\right)}{x^2+b^2}dx[/imath]
I would like to evaluate the following integral [imath](a>0)[/imath] [imath]\int_{-\infty}^{\infty}\frac{\exp\left(-a x^2\right)}{x^2+b^2}dx.[/imath] I've have tried integration by parts, putting [imath]e^{-ax^2}=u[/imath], but I come across with this integral [imath]\int_{-\infty}^{\infty}\exp\left(-a x^2\right)\arctan\left(\frac{x}{b}\right),[/imath] and I don't know how to do it. Could you help me? | 370518 | Evaluating [imath]\int_{\mathbb{R}}\frac{\exp(-x^2)}{1+x^2}\,\mathrm{d}x[/imath]
I would like to evaluate in a closed form the integral [imath]\int_{\mathbb{R}}\frac{\exp(-x^2)}{1+x^2}\,\mathrm{d}x[/imath] I tried various methods : integration by parts some changes of variables ([imath]y=x^2[/imath], [imath]x=\tan[/imath]) residue calculus (but the factor [imath]\exp(-x^2)[/imath] forbid to send the contour to infinity) developping [imath]\exp(-x^2)[/imath] or [imath]\frac{1}{1+x^2} [/imath] in power series (but in both cases one cannot exchange sum and integral) Does anyone knows if this integral is known, or how to evaluate it ? At least I would like to find an exact expression. The reason for me to believe that a closed form exists is that this integral arose in a problem of probability where I expect - if I haven't made any mistake previously - a very simple expression. |
2940902 | General method for finding parametric equations for polynomials _in rational functions_
Is there a general method for trying to finding parametric equations for polynomial equations? The classic example of [imath]x^2 + y^2 = 1[/imath] having multiple parameterizations: [imath]\begin{array}{rcl} (x,y) &=& (\sin t, \cos t) \\ (x,y) &=& (\cos t, \sin t) \\ (x,y) &=& \left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right) \end{array}[/imath] which are easily seen to work by plugging in. Note that the first two involve elementary functions, but the last one only rational functions. I'm curious about higher degree polynomials. Is there a general method to find a parameterized set of equations in rational functions? That is, is there an algorithm for extracting the parameter functions for [imath]x[/imath] and [imath]y[/imath] from the equation (or arbitrary sets of equations over [imath]v\ge1[/imath])? And if not, is there a characterization of which implicit sets can be parametrized? Take the (random off the top of my head) example: [imath]x y^3 + x^2 y + x y = 1[/imath] Is there a set [imath]x(t), y(t)[/imath] for this that can be found algorithmically? (I may well be naive here, not realizing something obvious, or maybe the entire field of algebraic geometry is geared towards possible solutions to this). One can do the reverse process of creating an implicit set of equations via Gröbner basis calculation, trying to eliminate [imath]t[/imath]. Or likewise in the forward direction if you can eliminate variables one-by-one with GB that would suffice to parameterize via one of the original variables. But in general if elimination isn't possible, is there still an algorithm to (even partially) parameterize? | 904782 | Is there a general way to parameterize all implicit functions?
We all know some curves can be described by [imath]y=f(x)[/imath] and some surfaces can be described by [imath]z=f(x,y)[/imath] However, there exists curves and surfaces which cannot be described by those, such as a circle and a sphere. Therefore, we introduce parameterized vector equations, which can describe them. For example, circle: [imath]\vec r(t)=r\cos(t)\hat i+r\sin(t)\hat j[/imath] sphere: [imath]\vec r(u,v)=\rho\cos(u)\sin(v)\hat i+\rho\sin(u)\sin(v)\hat j+\rho\cos(v)\hat k[/imath] However, all curves described by [imath]y=f(x)[/imath] and [imath]z=f(x,y)[/imath] can be parameterized. For curves, [imath]\vec r(t)=t\hat i+f(t)\hat j[/imath] For surfaces, [imath]\vec r(u,v)=u\hat i+v\hat j+f(u,v)\hat k[/imath] Therefore, I think this suggests the set of all parameterized surfaces (or curves) is the super-set of the set of all surfaces (or curves) described by [imath]z=f(x,y)[/imath] (or [imath]y=f(x)[/imath]). Is that correct? Now, here comes the the real challenge. A curve can also be described by an implicit function [imath]f(x,y)=0[/imath] and a surface can also be described by an implicit function [imath]f(x,y,z)=0[/imath] I have 3 questions regarding this. Can all surfaces (curves) described by an implicit function be parameterized? (If yes, then what is the general way?) Can all surfaces (curves) described by parametric vector equations be represented using implicit function? (If yes, then what is the general way?) Compare the set of all parameterized surfaces (curves) and the set of all surfaces (curves) represented by implicit function. (which is which super-set?) Sorry for the use of nontechnical terms. I use them because I don't know the technical ones. I have only started learning vector calculus last year in university. EDIT: I think my question is not too clear, so I will give an example of writing the surface [imath]f(x,y,z)=0[/imath] into [imath]\vec r(u,v)[/imath] We want to parameterize a sphere. [imath]x^2+y^2+z^2-\rho^2=0[/imath] Let [imath]x=\rho\cos(u)\sin(v)[/imath], [imath]y=\rho\sin(u)\sin(v)[/imath], [imath]z=\rho\cos(v)[/imath], [imath]\rho^2\cos^2(u)\sin^2(v)+\rho^2\sin^2(u)\sin^2(v)+\rho^2\cos^2(v)-\rho^2[/imath] [imath]=\rho^2\sin^2(v)(\cos^2(u)+\sin^2(u))+\rho^2\cos^2(v)-\rho^2=\rho^2-\rho^2=0[/imath] I want to know if there is a general way of finding [imath]x=x(u,v)[/imath], [imath]y=y(u,v)[/imath] and [imath]z=z(u,v)[/imath] for any given [imath]f(x,y,z)=0[/imath] |
2687215 | why is multiplication by n surjective on an abelian variety
I've just read the proof of [imath]\S4[/imath] (iv) in Mumford's "Abelian Varieties", on page 42-43, which says that multiplication by [imath]n[/imath] not divisible by [imath]p=\text{char}(k)[/imath] on an abelian variety [imath]X[/imath] is surjective. I can understand that it works (as in, I can follow the proof line by line), yet I don't have intuition for it. In particular, I don't understand why does passing to tangent spaces work. It may be my poor understanding of tangent spaces and differentials. It seems to say something along the lines of "surjectivity of multiplication by [imath]n[/imath]" can be checked on first order linear approximations". Does this make any sense? Would this make sense for more general isogenies? A reference where this is discussed in more detail would also be welcome. | 1784807 | [imath]A[/imath] abelian variety. Is the multiplication by [imath]n_A[/imath] surjective?
According with Mumford the answer is yes, but there are some obscure points in the proof. We know that there exists a very ample line bundle on [imath]A[/imath] since every abelian variety is projective. Hence, via some corollaries of the Theorem of the Cube, we get an ample line bundle whose restriction to the kernel of [imath]n_A[/imath] is both the trivial bundle and very ample. This should imply that the dimension of [imath]\ker n_A[/imath] is [imath]0[/imath]. Furthermore, Mumford says that the fact that the dimension of the kernel is [imath]0[/imath] implies the surjectivity of [imath]n_A[/imath]. Is anybody able to clarify these two points to me? Thanks in advance! |
2941502 | Prove exists consecutive integers [imath]s,s+1,\dots,s+i[/imath] such that [imath]m_{i}|s+i[/imath], if [imath]m_{0},\dots,m_{r}[/imath] are positive integers coprime
If [imath]m_{0},m_{1},\dots,m_{r}[/imath] are positive integers with [imath](m_{i},m_{j})=1[/imath] for [imath]i\neq j[/imath], prove that exists consecutive integers [imath]s,s+1,\dots,s+i[/imath] such that [imath]m_{i}|s+i[/imath] for all [imath]1\leq i \leq r[/imath]. I really don't know how to approach to this problem. Any tips? | 1892148 | Prove that there exists such a set of [imath]n[/imath] positive integers
Prove that for any positive integers [imath]m[/imath] and [imath]n[/imath] , there exists a set of [imath]n[/imath] consecutive positive integers each of which is divisible by a number of the form [imath]d^m[/imath], where [imath]d[/imath] is some positive integer not equal to [imath]1[/imath]. I don't know how to approach this question. |
2934741 | How to solve these [imath]3[/imath] equations for three unknowns [imath]x[/imath],[imath]y[/imath],[imath]z[/imath]?
Question: Solve: [imath]xy+x+y=23\tag{1}[/imath] [imath]yz+y+z=31\tag{2}[/imath] [imath]zx+z+x=47\tag{3}[/imath] My attempt: By adding all we get [imath]\sum xy +2\sum x =101[/imath] Multiplying [imath](1)[/imath] by [imath]z[/imath], [imath](2)[/imath] by [imath]x[/imath], and [imath](3)[/imath] by [imath]y[/imath] and adding altogether gives [imath]3xyz+ 2\sum xy =31x+47y+23z[/imath] Then, from above two equations after eliminating [imath]\sum xy[/imath] term we get [imath]35x+51y+27z=202+3xyz[/imath] After that subtracting [imath](1)\times 3z[/imath] from equation just above (to eliminate [imath]3xyz[/imath] term) gives [imath]35x +51y-3z(14+x+y)=202\implies (x+y)[35-3z]+16y-42z=202[/imath] I tried pairwise subtraction of [imath](1),(2)[/imath] and [imath](3)[/imath] but it also seems to be not working. Please give me some hint so that I can proceed or provide with the answer. | 1669228 | Solve system of simultaneous equations in [imath]3[/imath] variables: $x+y+xy=19$, $y+z+yz=11$, $z+x+zx=14$
Solve the following equation system: [imath]x+y+xy=19[/imath] [imath]y+z+yz=11[/imath] [imath]z+x+zx=14[/imath] I've tried substituting, adding, subtracting, multiplying... Nothing works. Could anyone drop me a few hints without actually solving it? Thanks! |
2942329 | Prove that [imath]2^q+q^2[/imath] is divisible by 3 where [imath]q[/imath] is a prime and [imath]q\geq5[/imath].
I'm looking to prove that [imath]2^q+q^2[/imath] is divisible by [imath]3[/imath] where [imath]q[/imath] is a prime such that [imath]q\geq5[/imath]. I know that primes greater than five will be congruent to either [imath]1\ (\text{mod}\ 3)[/imath] or [imath]2\ (\text{mod}\ 3)[/imath], which means that the [imath]q^2[/imath]-term will always be congruent to [imath]1\ (\text{mod}\ 3)[/imath] which simplifies the problem to finding the congruence of [imath]2^q+1\ (\text{mod}\ 3)[/imath]. However, I'm unable to go any further as I'm unable to show that the congruence of [imath]2^q\ (\text{mod}\ 3)[/imath] is always equal to [imath]2[/imath]. | 1453101 | Prove that [imath]n^2 + 2^n[/imath] is composite
If [imath]n>1[/imath] is an integer not of the form [imath]6k+3[/imath], Prove that [imath]n^2 + 2^n[/imath] is composite. Any idea of how to think about this problem? I have been thinking about it a lot and yet I was not able to come up with anything. |
2942567 | Is [imath]K[/imath] here a subfield of [imath]\Bbb C[/imath]?
In Jürgen Neukirch's "Algebraic Number Theory", page 5. An algebraic number field is a finite field extension [imath]K[/imath] of [imath]\Bbb Q[/imath]. The elements of [imath]K[/imath] are called algebraic numbers. Is [imath]K[/imath] here a subfield of [imath]\Bbb C[/imath]? | 1269990 | is a number field by definition a subfield of [imath] \mathbb C [/imath]?
I have seen that some authors are defing the number field as a subfield of [imath] \mathbb C[/imath] which is a finite extension of the rational numbers [imath] \mathbb Q [/imath], while some others without referering to complex numbers [imath] \mathbb C[/imath] . I think we don't need [imath] K[/imath] to be a subfield of [imath] \mathbb C[/imath] in the definition. So, my question is the follwing: Is it neceserily to define [imath]K[/imath] as a subfield of [imath] \mathbb C[/imath] or not ? And if no why ? Is it true that if we omit this in the definition, that then [imath]K[/imath] will turn out to be a subfield of [imath] \mathbb C[/imath]? I came up with this question, when I saw that in order to define infinite primes in a number field then these are determined by the embeddings [imath] K \to \mathbb C [/imath] Any idea would be really appreciated. Thank you in advance. |
2941720 | Does exist a singular cardinal [imath]\lambda[/imath] which [imath]\kappa < \lambda \implies 2^{\kappa}<\lambda?[/imath]
I'm reading about inacessible cardinals. I don't know anything about them, I just know basic set theory, cardinal operations etc., but I am too curious to leave this question that now I'm (probably) not able to totally understand... So, a (strongly) inacessible cardinar is a regular uncountable cardinal [imath]\lambda[/imath] such that [imath]\kappa < \lambda \implies 2^{\kappa}<\lambda[/imath]. I know that if the hypothesis of being uncountable is dropped then [imath]\aleph_0[/imath] satisfy the other 2 properties (regular and [imath]\kappa < \aleph_0\implies 2^{\kappa}<\aleph_0)[/imath]. But I can't think what happens if we drop the hypothesis of [imath]\lambda[/imath] being regular. Does exist a uncountable singular cardinal [imath]\lambda[/imath] such that [imath]\kappa < \lambda \implies 2^{\kappa}<\lambda[/imath]? Why? | 233177 | A cardinal [imath]\kappa[/imath] such that [imath]2^{\lambda}<\kappa[/imath] for all [imath]\lambda<\kappa[/imath] is regular?
A cardinal [imath]\kappa[/imath] such that [imath]2^{\lambda}<\kappa[/imath] for all [imath]\lambda<\kappa[/imath] is regular? I would appreciate very much an answer |
2942726 | Proof of [imath]\sum_{i=0}^{k}{{n-i-1}\choose {k-i}}={n\choose k}[/imath]
So, I would like to prove the following: [imath]\forall n,k\in\mathbb{N}_0, n\geq k+1: \sum_{i=0}^{k}{{n-i-1}\choose {k-i}}={n\choose k}[/imath] First of, i exploited trivial symmetry of the binomial coefficients to obtain: [imath]\sum_{i=0}^{k}{{n-i-1}\choose {k-i}}=\sum_{i=0}^{k}{{n-i-1}\choose {n-k-1}}[/imath] Now I proceeded by induction on [imath]n[/imath]. The base step: ([imath]n=k+1[/imath]): [imath]\sum_{i=0}^k{{k-i}\choose 0}=k+1={{k+1}\choose 1}={{k+1}\choose k}[/imath] But now here in the inductive step, I'm quite unsure how to proceed [imath]\sum_{i=0}^k{{n-i-1}\choose {n-k-1}}={n\choose k}[/imath] (the premise) and this should follow: [imath]\sum_{i=0}^k{{n-i}\choose{n-k}}={{n+1}\choose k}[/imath] All i managed to do was to "simplify" by symmetry: [imath]\sum_{i=0}^k{{n-i}\choose{n-k}}=\sum_{i=0}^k{{n-i}\choose{k-i}}[/imath] But from here i have no idea, would you please suggest further steps and how should i proceed from here? | 1819414 | proving combinatorics identity - [imath]\sum_{k=0}^m{n-k \choose m-k}={n+1 \choose m}[/imath]
Prove that for every [imath]n \ge m \ge 1 , \sum_{k=0}^m{n-k \choose m-k}={n+1 \choose m}[/imath] I've tried saying that the RHS represents the number of binary series with m "1" 's and n+1-m "0"'s, but I couldn't find out what k represents at the LHS. Thanks |
480481 | Factorising complex equations
I do factorization by just plugging in simple numbers to check whether they are factors, using long division and also using synthetic division. But, how to do it in the case of complex equations. For eg: Factorize [imath]x^6+5x^3+8[/imath] Putting [imath]v=x^3[/imath] I get [imath]v^2+5v+8[/imath]. But, this equation has complex roots. So, how to continue? | 432223 | Factor [imath]x^6 +5x^3 +8[/imath]
I wanted to know, how can I factor [imath]x^6 +5x^3 +8[/imath], I have no idea. Is there any method to know if a polynomial is factored. Just some advice will do. Help appreciated. Thanks. |
2943289 | Condition for which [imath] \text{dist}(x,A) = d(x,y_0)[/imath]
If [imath](X,d)[/imath] is metric space, [imath]x\in X[/imath] and [imath]A \subset X[/imath]. Let's mark [imath]\text{dist}(x,A) = \text{inf}\; \Big\{ d(x,y) : y \in A \Big\}[/imath] Does there always exists [imath]y_0 \in A[/imath] that [imath] \text{dist}(x,A) = d(x,y_0)[/imath] ? Will the answer change, if set [imath]A[/imath] is closed? Can someone help to solve this exercise? | 2388314 | Is it true that [imath]d(x,Y)=d(x,y_0)[/imath] for some [imath]y_0 \in Y[/imath] when [imath]Y[/imath] is closed?
Is it true that if [imath]Y \subset X[/imath] is a closed set and [imath]x \notin Y[/imath] then [imath]\exists[/imath] an element [imath]y_0 \in Y[/imath] [imath]s.t.[/imath] [imath]d(x,Y)=d(x,y_0)[/imath] where [imath]d(x,Y)= inf \{d(x,y) : y \in Y \}[/imath] and [imath]d: X \times X \to \mathbb R[/imath] is a metric. I was looking the proof of Riesz lemma and something like it came across in it. If the above is true then I will get some result in the proof. Please someone help.. Thank you.. |
501378 | Determining the matrix of a linear transformation
Let [imath]A[/imath] be an [imath]n\times n[/imath] matrix, and let [imath]V[/imath] denote the space of [imath]n[/imath]-dimensional row vectors. What is the matrix of the linear operator ‘‘right multiplication by [imath]A[/imath]’’ with respect to the standard basis of [imath]V[/imath]? | 704745 | Matrix of Linear Transformation by right multiplication
I am trying to solve the following problem: Let [imath]A[/imath] be an [imath]n\times n[/imath] matrix, and let [imath]V[/imath] denote the space of [imath]n[/imath]-dimensional row vectors. What is the matrix of the linear operator ‘‘right multiplication by [imath]A[/imath]’’ with respect to the standard basis of [imath]V[/imath]? I am not sure where to begin with this problem. |
2936775 | How to find condition of three planes intersecting at a point (according to vector approach)?
For example, if Plane 1: [imath]2x+y+z=1[/imath] Plane 2: [imath]x-y+z=2[/imath] Plane 3 : [imath]4x-y+3z =5[/imath] How to check if these planes intersecting at a single point? I want to check the coordinates of intersection , from scalar triple product or solving from normals of the planes . | 1287692 | How to show whether 3 planes have a common line of intersection
To show whether or not the 3 planes [imath]x+y-2z=5\tag 1[/imath] [imath]x-y+3z=6 \tag2[/imath] [imath]x+5y-12z=12 \tag 3[/imath] all have a common line of intersection. Can I do [imath](3)-(2)[/imath] to get the line [imath]6y-15z=6[/imath] and [imath](1)-(2)[/imath] to get the line [imath]2y-5z=-1[/imath] which is [imath]6y-15z=-3[/imath] , and say that as these aren't the same line, they don't have a common line of intersection? Another thing that is confusing me is that if instead of eliminating [imath]x[/imath], I chose to eliminate [imath]z[/imath], I would get different lines in terms of [imath]x[/imath] and [imath]y[/imath]. But how can I get the equations of two different lines by eliminating from the same pair of plane equations? There's only one line of intersection between any pair of planes, so surely I should only be able to get one unique line if I eliminate a variable from a pair of planes? Any help would be appreciated |
2944354 | Show that [imath]x_n \rightarrow x \Rightarrow (1+\frac{x_n}{n})^n \rightarrow e^x[/imath]
My approach so far: Using this I'm expressing the above limit as [imath]((1+\frac{1}{\frac{n}{x_n}})^\frac{n}{x_n})^{x_n}[/imath] and then using the property (?) that if [imath]x_n \rightarrow x[/imath] and [imath]a_n \rightarrow a[/imath], then [imath]x^{a_n}_n \rightarrow x^{a}[/imath]. But I'm proving this last property by taking log on both sides (suppose [imath]a,x > 0[/imath]). My question is isn't that somehow using the property I'm required to prove and hence is this valid? | 982093 | Proof for [imath]\lim_{n \to \infty}(1+\frac{a_n}{n})^n = e^a[/imath]?
In Statistical Inference by George Casella, Lemma 2.3.14 states that: [imath]\text{Let }a_1,a_2,...\text{be a sequence of numbers converging to }a\text{, that is, }\lim_{n \to \infty}a_n=a\text{. Then}[/imath] [imath]\lim_{n \to \infty}(1+\frac{a_n}{n})^n = e^a[/imath] Can somebody give me some hint for this? My first instinct is that since n is in the denominator, then that limit term would just be 1. However, this is obviously not correct. |
2943199 | How to convert [imath]\arctan(24/7) [/imath]to [imath]2\arctan(3/4)[/imath]?
This is actually a doubt I got while solving this question. The thing is I know how to convert [imath]2\arctan(3/4)[/imath] to [imath]\arctan(24/7)[/imath] by using the [imath]\arctan x + \arctan y[/imath] identity, but how do I do the opposite? Please help! | 523625 | showing [imath]\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})[/imath]
I'm having problem with showing that: [imath]\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})[/imath] I would need some help in the right direction |
2944535 | Find all functions [imath]f\colon \Bbb R\to \Bbb R[/imath] such that [imath]f(1-f(x)) = x[/imath] for all [imath]x \in \Bbb R[/imath]
Find all functions [imath]f\colon \Bbb R\to \Bbb R[/imath] such that [imath]f(1-f(x)) = x[/imath] for all [imath]x \in \Bbb R[/imath]. This is a question from the national olympiad in Germany 2018. All i could do is to try with some linear functions like [imath]f(x) = x+1[/imath] I am very good at calculus but I don't know how to approach the problems where I have to find a particular function. Is there a specific algorithm I should use? | 2878508 | How to find [imath] f(x)[/imath] if [imath]f(1-f(x))=x[/imath] for all [imath]x[/imath] [imath]\in \mathbb{R}[/imath]
How can I determine [imath] f(x)[/imath] if [imath]f(1-f(x))=x[/imath] for all real [imath]x[/imath]? I have already recognized one problem caused from this: it follows that [imath] f(f(x))=1-x [/imath], which is discontinuous. So how can I construct a function [imath]f(x)[/imath]? Best regards and thanks, John |
2946150 | Express a function as a sum of an even and odd function(s).
I have a function [imath] \frac{x+1}{x^2-3x+4} [/imath] which I've decomposed into [imath] \frac{x}{x^2-3x+4} + \frac{1}{x^2-3x+4} [/imath]. By using the definitions of even and odd functions, I know that x is an odd function, [imath] x^2-3x+4 [/imath] is neither even nor odd, and 1 is a constant and constants are considered even. so my question is: is an odd function over a function neither even nor odd, still odd? Same question from an a constant (even?) function. Thank you for your help! Also not sure if it's relevant, but for [imath] x^2-3x+4 [/imath] we obtain non-real roots, and it is also worth noting that [imath] \frac{x}{x^2-3x+4} [/imath] is zero when [imath] x=0 [/imath]. though [imath] x^2-3x+4 [/imath] is always greater than roughly [imath] ~1.6 [/imath]. | 2944659 | Writing a non symmetrical function as an even or odd function
I have been given the task to rewrite a function as a sum of an even and an odd function. But when I went to analyze the parts of the original function I noticed that one part was neither even nor odd. The original function was: [imath]g(x) = (x + 1) / (x^2 - 3x + 4[/imath]) After analyzing I realized that when using f(-x) to determine symmetry it produced a 'neither' result. How would I go about writing a function that is neither even nor odd as an even function. (It would have to be an even function as the (x+1) part of the original function is odd and since I need to write it as a sum of even and odd, the denominator would need to be even. |
2946138 | Prove [imath]((u \mod N) \cdot (v \mod N) ) \mod N=(u \cdot v) \mod N?[/imath]
Consider [imath]u[/imath] and [imath]v[/imath] are integers. The [imath]\cdot[/imath] is the usual multiplcation. It seems that [imath]((u \mod N) \cdot (v \mod N) )\overset{?}{=}(u \cdot v) \mod N,[/imath] is not always true. For example, on the LHS, [imath]((2 \mod 6) \cdot (3 \mod 6) )=6 [/imath] on the RHS [imath](2 \cdot 3) \mod 6=0,[/imath] so they are no equal. Can we prove that [imath]((u \mod N) \cdot (v \mod N) ) \mod N=(u \cdot v) \mod N?[/imath] | 1890405 | Product Rule for the mod operator
If we multiply 2 number and take the mod with prime this is equivalent to first taking mod with the prime of individual number and then multiplying the result and again taking mod. [imath] ab\bmod p = ((a\bmod p)(b\bmod p))\bmod p[/imath] does there exist any proof for this? Does it work for composite moduli too? Then I can use the Chinese remainder Theorem to calculate the result does there any other way apart from Chinese remainder Theorem to solve the problem? |
2946846 | Can I get [imath]\cos(x)/\!\cos(y)[/imath] if I have [imath]\sin(x)/\!\sin(y)[/imath]?
Can I get [imath]\cos(x)/\!\cos(y)[/imath] if I have [imath]\sin(x)/\!\sin(y)[/imath]? | 1060012 | Ratio of sines given, find ratio of cosines?
Is it possible to find [imath]\displaystyle{\cos x \over \cos y}[/imath] if [imath]\displaystyle{\sin x\over \sin y}[/imath] is given? If so, how would one approach this problem? Thank you in advance! |
2945431 | To find the equation of a plane
How to find an equation of a plane when 2 lines lying on the plane are given ? Q)Find the equation of plane which contains the lines [imath](x-4)/1 = (y-3)/-4 = (z-2)/5[/imath] and [imath](x-3)/-2 = (y+3)/8 = (z+2)/-10[/imath] | 2140132 | Finding plane equation given two lines
Determine an equation of the plane containing the lines [imath]\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-5}{6}[/imath]; [imath]r=\lt1,-1,5\gt+t\lt1,1,-3\gt[/imath]. I calculated the cross product between the directional vector of both lines to find the normal vector [imath]n[/imath], but when I looked for an intersection point [imath]r_0[/imath] to apply the formula: [imath]\lt r-r_0\gt \bullet \ \ n[/imath], I did not find any. Can I use the point [imath]\lt1,-1,5\gt[/imath] given in the line [imath]r[/imath]? Or is it not possible to find an equation of the plane containing two lines that do not intersect? |
2940300 | Find [imath]\lim\limits_{x\to0}\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}[/imath]
Find [imath]\displaystyle \lim_{x\to0}\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}[/imath] I tried using L'hopital's rule but it got very messy very fast UPDATE- So reading about the Taylor series this is what I have so far [imath]\lim_{x\to0}\frac{(x-x^3/3!+x^5/5!-\cdots)-(x+x^3/3+2x^5/5+17x^7/315+\cdots)}{(x+x^2/6+3x^5/40+\cdots)-(x-x^3/3+x^5/5-\cdots)}[/imath] But I'm still stuck | 2513502 | Compute the following limit: [imath]\lim_{n \to \infty}{\frac{\arcsin{\frac{1}{n}}-\arctan{\frac{1}{n}}}{\sin{\frac{1}{n}}-\tan{\frac{1}{n}}}}[/imath]
I'm trying to solve the following limit: [imath]\displaystyle \lim_{n \to \infty}{\frac{\arcsin{\frac{1}{n}}-\arctan{\frac{1}{n}}}{\sin{\frac{1}{n}}-\tan{\frac{1}{n}}}}[/imath], but I do not have any idea about how to start. Any help will be very useful. |
2946418 | How to prove that a function is injective
[imath]f(x)=3x^2-6x+3[/imath] if we see the graph we'll know that it isn't. Using the definition we have: [imath]f(x_1)=f(x_2)[/imath] [imath]3(x_1^2)-6(x_1)+3=3(x_2^2)-6(x_2)+3[/imath] We get rid of 3 and the we factorize the other [imath]3[/imath] [imath]x_1^2-2x_1=x_2^2-2x_2[/imath] Then we factorize [imath]x_1[/imath] and [imath]x_2[/imath] [imath]x_1(x_1-2)=x_2(x_2-2)[/imath] Then what? | 1791530 | Is the function [imath]f(x) = x^2 - 2x[/imath] one-to-one?
The answer given is that it's not because it's a parabola and hence, would fail the horizontal line test, i.e., two values of [imath]x[/imath] will have the same value of [imath]f(x)[/imath]. However, how can we prove the same algebraically? My solution: Let [imath]f(x_1) = f(x_2)[/imath] [imath]\implies x_1^2 - 2x_1 = x_2^2 - 2x_2[/imath] [imath]\implies (x_1 - 1)^2 -1 = (x_2 - 1)^2 - 1[/imath] [imath]\implies (x_1 - 1)^2 = (x_2 - 1)^2[/imath] [imath]\implies x_1 = x_2[/imath] (I believe that I've made the mistake here because we can't really take square root of both the sides.) My question is: how do we prove that the function is not one-to-one algebraically? |
2946931 | Evaluate [imath]\lim\limits_{n \to \infty}\frac{1!+2!+\cdots+n!}{n!}[/imath].
Problem Evaluate [imath]\lim\limits_{n \to \infty}\frac{1!+2!+\cdots+n!}{n!}.[/imath] Solution 1 Applying Stolz theorem, [imath]\lim_{n \to \infty}\frac{1!+2!+\cdots+n!}{n!}=\lim_{n \to \infty}\frac{(n+1)!}{(n+1)!-n!}=\lim_{n \to \infty}\frac{n!(n+1)}{n!\cdot n}=\lim_{n \to \infty}\frac{n+1}{n}=1.[/imath] Solution 2 [imath]\begin{align*} 1=\frac{n!}{n!}\leq \frac{1!+2!+\cdots+n!}{n!}&\leq \frac{(n-2)!(n-2)+(n-1)!+n!}{n!}\\&=\frac{n-2}{n(n-1)}+\frac{1}{n}+1\to 1(n \to \infty). \end{align*}[/imath] Applying the squeeze theorem, [imath]\lim\limits_{n \to \infty}\frac{1!+2!+\cdots+n!}{n!}=1.[/imath] Any more solutions? Thanks. | 487123 | How to find [imath]\lim_{n\to\infty}\frac{1!+2!+\cdots+n!}{n!}[/imath]?
How to evaluate the following limit? [imath]\lim_{n\to\infty}\dfrac{1!+2!+\cdots+n!}{n!}[/imath] For this problem I have two methods. But I'd like to know if there are better methods. My solution 1: Using Stolz-Cesaro Theorem, we have [imath]\lim_{n\to\infty}\dfrac{1!+2!+\cdots+n!}{n!}=\lim_{n\to\infty}\dfrac{n!}{n!-(n-1)!}=\lim_{n\to\infty}\dfrac{n}{n-1}=1[/imath] My solution 2: [imath]1=\dfrac{n!}{n!}<\dfrac{1!+2!+\cdots+n!}{n!}<\dfrac{(n-2)(n-2)!+(n-1)!+n!}{n!}=\dfrac{n-2}{n(n-1)}+\dfrac{1}{n}+1[/imath] |
2946796 | Is that true that if A - B is positive definite, then A's eigenvalues are larger than B's
Suppose that [imath]A[/imath] and [imath]B[/imath] are two Hermite matrices and that [imath]A - B[/imath] is positive definite. Denote the eigenvalues of these two matrices by [imath]\lambda_1(A) \ge \lambda_2(A) \ge \dots \ge \lambda_n(A)[/imath] and [imath]\lambda_1(B) \ge \lambda_2(B) \ge \dots \ge \lambda_n(B)[/imath]. Then can we derive [imath]\lambda_i(A) > \lambda_i(B)\ \ \forall i?[/imath] It seems to me this is true but I don't know how to prove it. I'd appreciate any help. Thanks in advance! | 1622747 | Are the eigenvalues of the sum of two positive definite matrices increased?
Let [imath]A[/imath] and [imath]B[/imath] be two [imath]n \times n[/imath] (symmetric) positive definite matrices, and denote the [imath]k[/imath]th largest eigenvalue of a general [imath]n \times n[/imath] matrix by [imath]\lambda_k(X)[/imath], [imath]k = 1, 2, \ldots, n[/imath] so that [imath]\lambda_1(X) \geq \lambda_2(X) \geq \cdots \geq \lambda_n(X).[/imath] I guess the following relation holds: [imath]\lambda_k(A + B) > \max\{\lambda_k(A), \lambda_k(B)\}, \; k = 1, 2, \ldots, n.[/imath] This looks intuitive but I have difficulty to prove it, any hints? |
2945310 | What is the intuition behind the notion of strong differentiation?
[imath]{\bf definition.}[/imath] A function [imath]f[/imath] defined on open set [imath]A \subseteq \mathbb{R}[/imath] is said to be [imath]{\bf strong \; differentiable}[/imath] at [imath]a \in A[/imath] if [imath] \lim_{(x_1,x_2) \to (a,a) \\ x_1 \neq x_2 } \dfrac{ f(x_1)-f(x_2) }{x_1-x_2} = f^{*}(a)[/imath] exists and is finite and we call [imath]f^*(a)[/imath] the strong derivative of [imath]f[/imath] at [imath]x=a[/imath] How is this notion different than the usual derivative? I mean, if we set [imath]x_2 = a[/imath], we get our usual derivative... What is the point of such definition? what is motivation for this? | 1675246 | Differences between derivatives and strong derivatives
Definition: Let [imath]f[/imath] be a real valued function. We say [imath]f[/imath] is [imath]\mathbf{strongly}[/imath] [imath]\mathbf{differentiable}[/imath] at [imath]x = a[/imath] if the following limits exists and is finite: [imath] \lim_{x \to a, y \to a, x \neq y} \frac{ f(x)-f(y)}{x-y} = f^*(a) [/imath] and we can [imath]f^*(a)[/imath] the strong derivative of [imath]f[/imath] at [imath]a[/imath]. Why is this definition of derivative different than the usual one? What is the main crucial point to understand here that makes it different? |
2485337 | Evalute the [imath]\lim\limits_{n\to\infty} \frac{1}{n}\sum\limits_{k=0}^{\left \lfloor n/2\right \rfloor}\cos(\frac{k\pi}{n})[/imath]
[NBHM_2007_PhD Screening Test_Analysis] Evalute the [imath]\lim_{n\to\infty} \frac{1}{n}\sum_{k=0}^{\left \lfloor{\frac{n}{2}}\right \rfloor}\cos(\frac{k\pi}{n}).[/imath] If the upper limit of the sum is [imath]n[/imath] insted of [imath]{\left \lfloor{\frac{n}{2}}\right \rfloor}[/imath]. I could use the formula [imath]\int_{0}^{1}cos(\pi x)dx[/imath]. Here, how do I evalute the given limit? Please help me. | 2336563 | Limit involving cosine function
How to find [imath]\lim_{n \rightarrow \infty}\; \frac{1}{n}\;\sum_{k=1}^{\Big\lfloor\frac{n}{2}\Big\rfloor} \cos\Big(\frac{k\pi}{n}\Big)[/imath]? I know the method when the upper limit is simply [imath]n[/imath], namely it converges to [imath]\int_0^1 f(x)\;dx[/imath] where [imath]f[/imath] is monotonically increasing on an interval ( in this case our term is of the form [imath]\frac{1}{n} \sum f(k/n)[/imath] ) But here the upper limit is [imath]\Big\lfloor\frac{n}{2}\Big\rfloor[/imath]. How to approach this? |
2948208 | 2nd derivative of f(x(t) , y(t))
So the first derivative w.r.t. t: [imath]\frac{dz}{dt} = \frac{\partial f}{\partial x} \times \frac{dx}{dt} + \frac{\partial f}{\partial f} \times \frac{dy}{dt}[/imath] How would I find the 2nd derivative? EDIT: In particular there is a term in when calculating the 2nd derivative : [imath] \frac {d}{dt} ( \frac {\partial f}{\partial x}) [/imath] It is calculating this step that I do not understand Thanks | 2905613 | Iterating Total Derivative How to.
Let [imath]f =f(x(t),y(t))[/imath] Then the total derivative of f with respect to t is given by: [imath]\frac {}{} = \frac {}{} + \frac {}{}[/imath] Say still we have [imath]f=f(x(t),y(t))[/imath]. What is then [imath]\frac {\partial^2 f}{ \partial t^2} [/imath]? Is it just [imath] \frac {\partial f}{ \partial x} \frac {d^2x}{dt^2}+ \frac {\partial f }{\partial y} \frac {d^2y}{dt^2} [/imath] ? (Just by applying chain rule again?) Also, this seems to be related to the Jacobian J=J(f). Is this correct? If so, can we define [imath]\frac{\partial^2 f}{ \partial t^2}[/imath] as [imath]JJ^T[/imath] , where [imath]J^T[/imath] is the transpose of [imath]J[/imath]? |
2948076 | Prove the set [imath]K = \{\frac{1}{n} \mid n\in \mathbb{N}\}\cup \{0\}[/imath] is compact.
Prove the set [imath]K = \{\frac{1}{n} \mid n\in \mathbb{N}\}\cup \{0\}[/imath] is compact. I have tried to write a proof for this fact. Please critique it. Proof. Let [imath]\mathcal{G} = \{G_\alpha \mid \alpha \in A\}[/imath] be any open cover for [imath]K[/imath]. Since [imath]\mathcal{G}[/imath] is an open cover, we have [imath]K\subseteq \bigcup_{\alpha\in A}{G_\alpha}[/imath]. Hence, [imath]0\in K[/imath] implies that there exists some [imath]G_0 \in \mathcal{G}[/imath] such that [imath]0\in G_0[/imath]. Since [imath]G_0[/imath] is open, there exists [imath]r>0[/imath] such that [imath]B_r(0) \subseteq G_0[/imath]. We know from previous results that [imath]\frac{1}{n} \to 0[/imath]. Hence, [imath]B_r(0)[/imath] contains all but finitely many terms of [imath]\frac{1}{n}[/imath] and consequently, [imath]G_0[/imath] contains all but finitely many of these points. Let [imath]N[/imath] be such that [imath]\frac{1}{n} \in G_0[/imath] for all [imath]n\geq N[/imath]. Then for each point [imath]1,\frac{1}{2},\dots,\frac{1}{N-1}[/imath] we can choose a corresponding open set [imath]G_1,G_2,\dots,G_{N-1} \in \mathcal{G}[/imath] that contains each point. Hence, [imath]\{G_0,G_1,G_2,\dots\,G_{N-1}\}[/imath] is a finite subcover for [imath]K[/imath]. | 2239368 | Proof Verification: Let [imath]D=\{ \frac{1}{n} \mid n\in\mathbb{N} \} \cup\{0\} \subset \mathbb{E}^1[/imath]. Prove that [imath]D[/imath] is compact.
Let [imath]D=\{ \frac{1}{n} \mid n\in\mathbb{N} \} \cup\{0\} \subset \mathbb{E}^1[/imath]. Prove that [imath]D[/imath] is compact. Proof: It is clear that [imath]D[/imath] is bounded above by [imath]1[/imath] and bounded below by [imath]0[/imath]. We need to show that the set [imath]D[/imath] is closed, or it contains all of its limit points. The only limit point of the set [imath]D[/imath] is [imath]0[/imath]. But since [imath]D[/imath] includes [imath]0[/imath], [imath]D[/imath] must be closed. Hence, [imath]D[/imath] is compact because it is closed and bounded. [imath]\blacksquare[/imath] I claimed that [imath]0[/imath] is the only limit point of the set [imath]D[/imath] but is this okay? I stated it allegedly without giving much reasoning but I thought it was very obvious from looking at the set. Can [imath]D[/imath] be its own open cover? If so, can I just give an example of finite subcover of [imath]D[/imath] to show that it is compact? For example, since [imath](2,0) \cup (0,2)[/imath] covers [imath]D[/imath], [imath]D[/imath] is compact. |
2945928 | Prove [imath](1+x)\ln(1+x) + (1-x)\ln(1-x) \leq 2x^2[/imath]
For all [imath]-1<x<1[/imath], prove: [imath](1+x)\ln(1+x) + (1-x)\ln(1-x) \leq 2x^2[/imath] I am sure we should use the Jensen's inequality or the Taylor expansion of [imath]\ln(1+x)[/imath] and [imath]\ln(1-x)[/imath]; however, I was not able to do that so far. Any idea? | 2835098 | Prove [imath](1-x)\ln(1-x)+(1+x)\ln(1+x)\leq 2x^2[/imath] for [imath]0[/imath]
Find the smallest value of [imath]c[/imath] such that [imath](1-x)\ln(1-x)+(1+x)\ln(1+x)\leq c x^2[/imath] holds for [imath]0<x<1[/imath]. I saw the curve and realized this is true for [imath]c=2[/imath]. How can I prove it? What is the smallest [imath]c[/imath] that still makes the inequality valid? I think we should use the Taylor expansions of the [imath]\ln[/imath]s. |
2944491 | Can someone help me finish this: evaluate [imath]S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}[/imath]
I am asked to find the closed form solution for the below. [imath]S_n = \frac{x}{1-x^2}+\frac{x^2}{1-x^4}+ ... + \frac{x^{2^{n-1}}}{1-x^{2^{n}}}[/imath] Just writing out the [imath]S_1, S_2, S_3[/imath], I have managed to find a pattern, which is: [imath]S_n = \frac{S_{n-1}}{1-x^{2^n}} + \frac{x^{2^n}}{1-x^{2^{n+1}}}[/imath] I am not sure how to proceed onwards to solve this recurrence relation. Is there a clever trick I can do to solve it? | 2170972 | Two different expansions of [imath]\frac{z}{1-z}[/imath]
This is exercise 21 of Chapter 1 from Stein and Shakarchi's Complex Analysis. Show that for [imath]|z|<1[/imath] one has [imath]\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots +\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots =\frac{z}{1-z}[/imath]and [imath]\frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots \frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots =\frac{z}{1-z}.[/imath] Justify any change in the order of summation. [Hint: Use the dyadic expansion of an integer and the fact that [imath]2^{k+1}-1=1+2+2^2+\cdots +2^k[/imath].] I don't really know how to work this through. I know that [imath]\frac{z}{1-z}=\sum_{n=1}^\infty z^n[/imath] and each [imath]n[/imath] can be represented as a dyadic expansion, but I don't know how to progress from here. Any hints solutions or suggestions would be appreciated. |
2948916 | How many primes are there on the form [imath]100\cdots 0 1[/imath]?
For example 11 and 101 are primes, but apart from them, can we determine how many primes on the form [imath]100\cdots 00 1[/imath] there exist (in decimal number system)? | 2108085 | Is there a prime which is the form of [imath]10^n + 1[/imath] except [imath]2, 11, 101[/imath]?
Is there a prime which is the form of [imath]10^n + 1[/imath] except [imath]2, 11, 101[/imath]? I confirm there isn't such prime for [imath]n < 64[/imath]. |
798119 | Quotient modules isomorphic [imath] \Rightarrow[/imath] submodules isomorphic
Suppose [imath]M[/imath] is a module, [imath]H[/imath] and [imath]K[/imath] submodules. If [imath]M/H \cong M/K[/imath] can I conclude that [imath]H \cong K [/imath] ? If not, how to construct a counterexample ? | 95899 | Isomorphism of quotient modules implies isomorphism of submodules?
Let [imath]$A$[/imath] be a commutative ring, [imath]M[/imath] an [imath]A[/imath]-module and [imath]N_1, N_2[/imath] two submodules of [imath]M[/imath]. If we have [imath]M/N_1 \cong M/N_2[/imath], does this imply [imath]N_1 \cong N_2[/imath]? This seems so trivial, but I just don't see a proof... Thanks! |
2948974 | The Euclidean metric satisfies triangle inequality in cross product of two metric spaces [imath]X[/imath] and [imath]Y.[/imath]
Consider [imath]X \times Y[/imath] for metric spaces [imath]X[/imath] and [imath]Y.[/imath] I'm trying to show that [imath]d_E(p,q) = \sqrt{d_X(p,q)^2 + d_Y(p,q)^2}[/imath] fulfills the triangle inequality, but I can't workout the algebra so that things fall into place and it is clear that [imath]d_E(p_1,p_3) \leq d_E(p_1,p_2) + d_E(p_2,p_3).[/imath] I avoided writing out everything that I have written on paper because it is messy, but here is a little bit: Consider [imath]d_E(p,p'') = \sqrt{d_X(p,p'')^2 + d_Y(p,p'')^2}[/imath] and [imath]d_E(p,p') = \sqrt{d_X(p,p')^2 + d_Y(p,p')^2}[/imath] and [imath]d_E(p',p'') = \sqrt{d_X(p',p'')^2 + d_Y(p',p'')^2}.[/imath] It must be shown that [imath]d_E(p,p'') \leq d_E(p,p') + d_E(p',p''),[/imath] or that [imath]d_E(p,p'')^2 \leq (d_E(p,p') + d_E(p',p''))^2.[/imath] I then proceeded to expand it out via the square, and it became messy and not, in the end, clear that the left was less than or equal to the right. | 1052511 | Prove that the product space is a metric space.
I have the following problem: Let [imath](S,d)[/imath] and [imath](T,e)[/imath] be two metric spaces. Their product space has underlying set [imath]S\times T=\{(s,t)|s\in S,t\in T\}[/imath] and metric [imath]m\big((s_1,t_1),(s_2,t_2)\big)=\sqrt{(s_1-s_2)^2+(t_1-t_2)^2}.[/imath] Verify that [imath](S\times T,m)[/imath] is a metric space. As for proving the triangle inequality, I made the following attempt: \begin{align*} m\big((s_1,t_1),(s_3,t_3)\big)&=\sqrt{(s_1-s_3)^2+(t_1-t_3)^2}\\ &=\sqrt{(s_1-s_2+s_2-s_3)^2+(t_1-t_2+t_2-t_3)^2}\\ &\leq\sqrt{(s_1-s_2+s_2-s_3)^2}+\sqrt{(t_1-t_2+t_2-t_3)^2}\\ &\leq\sqrt{(s_1-s_2)^2+(t_1-t_2)^2}+\sqrt{(s_2-s_3)^2+(t_2-t_3)^2}\\ &=m\big((s_1,t_1),(s_2,t_2)\big)+m\big((s_2,t_2),(s_3,t_3)\big) \end{align*} My friend has told me that I am making too large of logical jumps in the "less than or equal to" steps. Can anyone point out my error? I know I need to use the established metrics, but I don't know how. |
2948850 | Is faithful flatness implying [imath]M \otimes_{S^{-1}R} S^{-1}N \cong M \otimes_R N[/imath]?
Assume that [imath]M[/imath] is faithfully flat [imath]S^{-1}R[/imath]-module, for [imath]R[/imath] commutative ring and [imath]S \subset R[/imath] a multiplicative subset. Furthermore, let [imath]N[/imath] denote an [imath]R[/imath]-module. Is it true that [imath]M \otimes_{S^{-1}R} S^{-1}N \cong M \otimes_R N[/imath]? | 75517 | Why does [imath]M \mathbin{\otimes_R} N \cong M_\mathfrak{p} \mathbin{\otimes_{R_\mathfrak{p}}} N[/imath]?
Let [imath]R[/imath] be a commutative ring, [imath]\mathfrak{p}[/imath] a prime ideal of [imath]R[/imath], [imath]M[/imath] a [imath]R[/imath]-module, and [imath]N[/imath] a [imath]R_\mathfrak{p}[/imath]-module. Why do we have this isomorphism? [imath]M \mathbin{\otimes_R} N \cong M_\mathfrak{p} \mathbin{\otimes_{R_\mathfrak{p}}} N[/imath] I can prove this by bare hands by taking one map to be defined by [imath]m \otimes n \mapsto (m / 1) \otimes n[/imath] and the other map by [imath](m / s) \otimes n \mapsto m \otimes (n / s)[/imath]. A little work is required to show that the latter is well-defined, but when that is done we have two mutually inverse [imath]R[/imath]-linear (and [imath]R_\mathfrak{p}[/imath]-linear) maps. But what is the conceptual reason for this isomorphism? Expanding the right hand side a bit, we see that we are saying [imath]M \mathbin{\otimes_R} N \cong (M \mathbin{\otimes_R} R_\mathfrak{p}) \mathbin{\otimes_{R_\mathfrak{p}}} N[/imath] and expanding the left hand side, it seems that what we want to prove is [imath]M \mathbin{\otimes_R} (R_\mathfrak{p} \otimes_{R_\mathfrak{p}} N) \cong (M \mathbin{\otimes_R} R_\mathfrak{p}) \mathbin{\otimes_{R_\mathfrak{p}}} N[/imath] but I see no reason why tensor products over different rings should associate like that... |
2949537 | Creating a [imath]6[/imath]-integer subset without repetition with no consecutive integers from [imath]\{1,2,\dots,20\}[/imath]
How many subsets of six integers chosen without repetition from [imath]1,2,\dots, 20[/imath] are there with no consecutive integers? I am unsure of where to begin with this problem. | 2406749 | How many [imath]6[/imath]-integer subsets of [imath]\{1, 2, 3, \ldots, 20\}[/imath] have no consecutive integers?
I have to determine how many six-integer subsets of [imath]\{1, 2, 3, \ldots, 20\}[/imath] are possible if no two consecutive integers are in a set. |
2949215 | Determine [imath]\sum_{n = 1}^{+ \infty} \frac{1}{n(n+1)...(n+p)} [/imath]
[imath]p \in \mathbb{N^*}[/imath], [imath]u_n = \frac{1}{n(n + 1)...(n + p)}[/imath] Determine [imath]\sum_{n=1}^{+ \infty}u_n[/imath] Hint: write [imath]pu_n[/imath] in the form [imath]v_n - v_{n+1}[/imath] I couldn't see how to use the hint, I tried to use [imath]ln[/imath] but I did not proceed much to write it in the from of [imath]v_n - v_{n+1}[/imath] | 1004004 | Show that [imath]\sum\limits_{k=1}^\infty \frac{1}{k(k+1)(k+2)\cdots (k+p)}=\frac{1}{p!p} [/imath] for every positive integer [imath]p[/imath]
I have to prove that [imath]\sum_{k=1}^\infty \frac{1}{k(k+1)(k+2)\cdots (k+p)}=\dfrac{1}{p!p}[/imath] How can I do that? |
2949720 | [imath]g_{1},g_{2} \in G[/imath], then if [imath]g_{1}g_{2}[/imath] has finite order then we have to prove that [imath]g_{2}g_{1}[/imath] also has finite order too.
Let [imath]G[/imath] be a group and [imath]g_{1},g_{2} \in G[/imath], then if [imath]g_{1}g_{2}[/imath] has finite order then we have to prove that [imath]g_{2}g_{1}[/imath] also has finite order too. I have proved like this- If [imath]g_{1}g_{2}[/imath] has finite order that means there exists positive integer [imath]n[/imath] such that [imath](g_{1}g_{2})^n = e[/imath], now we rewrite the equation in a clever way. [imath](g_{1}g_{2})^n = e[/imath] [imath](g_{1}g_{2})(g_{1}g_{2})(g_{1}g_{2})...(g_{1}g_{2})= e[/imath] [imath]g_{1}(g_{2}g_{1})^{n-1} g_{2} = e[/imath] [imath](g_{2}g_{1})^{n-1} = g_{1}^{-1}g_{2}^{-1}[/imath] Multiplying [imath]g_{2}g_{1}[/imath] on both sides from the left hand side, we get - [imath](g_{2}g_{1})^n = g_{2}g_{1}g_{1}^{-1}g_{2}^{-1} = e[/imath] Implying that the element [imath]g_{2}g_{1}[/imath] has finite order too. Now I am thinking of another approach, that is through homomorphism - That is can we find a homomorphism between the two - Say [imath]\phi[/imath], that is [imath]\phi(g_{1}g_{2}) = g_{2}g_{1}[/imath], which would mean that [imath]\phi(g_{1}) \phi(g_{2}) = g_{2} g_{1}[/imath]. Any candidate for [imath]\phi[/imath], as if I can find a homomorphism then the order of both the elements will be same. | 622305 | Show that if [imath]ab[/imath] has finite order [imath]n[/imath], then [imath]ba[/imath] also has order [imath]n[/imath]. - Fraleigh p. 47 6.46.
Attempted proof based on here and yahoo. Given: [imath]a,b[/imath] elements of [imath]G[/imath], and [imath]ab[/imath] has finite order [imath]n[/imath]. Hence [imath]\color{magenta}{|ab| = n} \iff (ab)^n = e[/imath]. I must prove that [imath]n[/imath] is the smallest positive integer such that [imath](ba)^n = e.[/imath] Notice: [imath]\begin{align}(\color{darkorange}{a}\color{darkcyan}{b})^n &= e \implies \color{darkorange}{a}\color{darkcyan}{b}(ab)^{n-1} = e \implies \color{darkcyan}{b}(ab)^{n-1}\color{darkorange}{a} = e \implies (ba)^n = e \end{align} [/imath] [imath]\implies |ba| \le n[/imath]. Now I must prove that there's no positive integer [imath]m[/imath] such that [imath]m<n[/imath] and [imath](ba)^m = e.[/imath] For a proof by contradiction, suppose [imath]|ba|<\color{magenta}{n}[/imath]. Then we could apply the same reasoning to find that [imath] |ab| ≤ |ba| < \color{magenta}{|ab|}[/imath], which is absurd. So [imath]|ba| = |ab| = n. \quad \blacksquare[/imath] Questions Whence does [imath] |ab| ≤ |ba| [/imath] arise, in the second last line? What's the intuition of the claim proved? How do you prognosticate that [imath]|ba|<\color{magenta}{n}[/imath] is impossible, and that you should prove it by contradiction? |
2949891 | Prove that any polynomial with integer coefficients must have a composite number in its image.
Let [imath]f \in \mathbb{Z}[x][/imath] be a polynomial of degree at least [imath]1[/imath]. Prove that there is [imath]n \in \mathbb{Z} [/imath] such that the corresponding polynomial function [imath]f(n)[/imath] is not a prime. I think that the task is to show that there is always an [imath]n[/imath] for any degree polynomial, that would give me a non-prime output. I was thinking I could make a case distinction for even and odd [imath]n[/imath], but this argument does not get me anywhere. Does anybody have small hint what would be a good direction to explore? the book gave the hint "Many ways to attack this. Think of something." - which was really helpful, as you might imagine. [imath]f(n)[/imath] would have the form: [imath]f(n)= a_0 + a_1 n + a_2 n^2 \dots [/imath] | 86018 | Proving a polynomial [imath]f(x)[/imath] composite for infinitely many [imath]x[/imath]
Let [imath]f(x)=a_0+a_1x+ \ldots +a_nx^n[/imath] be a polynomial with integer coefficients, where [imath]a_n>0[/imath] and [imath]n \ge 1[/imath]. Prove that [imath]f(x)[/imath] is composite for infinitely many integers [imath]x[/imath]. I can easily show that there are infinitely many composite numbers of the form [imath]a_0+a_1x+ \ldots +a_nx^n[/imath] if [imath]a_0 \ge 2[/imath], we just note that [imath]f(x)[/imath] is composite for every [imath]x[/imath] being a multiple of [imath]a_0[/imath]. But I can't find a way to prove this in the case [imath]a_0=1[/imath]. |
2950087 | Prove convergence of [imath] \sum_{n=1}^{\infty} a^{\ln n}[/imath] for [imath]0[/imath]
Prove the convergence of the series: [imath]\sum_{n=1}^{\infty} a^{\ln n},\,\text{for} \,\,0<a<\frac{1}{e}.[/imath] Attempt. I have proved the non-convergence in the case [imath]a\geq 1/e[/imath] (using the comparison test and getting [imath]a^{\ln n}\geq \frac{1}{n}[/imath]). In case [imath]0<a<\frac{1}{e}[/imath], I get [imath]a^{\ln n}<\frac{1}{n}[/imath] and the above test doesn't work. Ratio test, root test are also not applicable here. Thanks in advance for the help. | 2519631 | Test the convergence of the series
For [imath]a>0[/imath], test the convergence of the infinite series: [imath]\sum a^{\ln(n)}[/imath] Which test should be better to use here? I tried the D'Alembert's ratio test. [imath]\lim_{n\to \infty} {\frac{a^{\ln(n)}}{a^{\ln(n+1)}}}= \lim_{n\to \infty}a^{ln{\frac{n}{n+1}}}=a^0=1[/imath] So the test failed. UPDATE: I observed that [imath]{a^{\ln(n)}}=e^{{\ln(a)}{\ln(n)}}={n^{\ln(a)}}={{1\over n^{ln({1\over a})}} }[/imath]Further by test of auxiliary series [imath]\sum \frac{1}{n^p}[/imath] we know that criterion for convergence is [imath]p>1[/imath]. Here [imath]p=ln({1\over a})[/imath].Hence the series will be convergent when [imath]ln(a)<-1[/imath] i.e. when [imath]0<a<1/e[/imath], since [imath]a>0[/imath] is already given.Finally this solved my question. |
2636363 | In finite dimension, [imath]W_1^0+W_2^0=(W_1\cap W_2)^0[/imath]
Let [imath]V[/imath] be a vector space. For any set [imath]S\subset V[/imath], we define [imath]S^0=\{f\in V^\ast\,:\, f(v)=0,\forall v\in S\}[/imath]. If [imath]\dim V<\infty[/imath] and [imath]W_1,W_2\subset V[/imath] are vector subspaces, then show that [imath]W_1^0+W_2^0=(W_1\cap W_2)^0[/imath]. I was able to show that [imath]W_1^0+W_2^0\subset (W_1\cap W_2)^0[/imath]: Let [imath]f\in W_1^0+W_2^0[/imath]. Then there are [imath]f_i\in W_i^0[/imath], [imath]i=1,2[/imath] so that [imath]f=f_1+f_2[/imath]. If [imath]v\in W_1\cap W_2[/imath], then [imath]f(v)=f_1(v)+f_2(v)=0[/imath], since [imath]v\in W_1[/imath], [imath]f_1\in W_1^0[/imath] and [imath]v\in W_2[/imath], [imath]f_2\in W_2^0[/imath]. Therefore [imath]f\in (W_1\cap W_2)^0[/imath]. I don't have any idea of how to do the converse inclusion! | 2161177 | [imath](U\cap W)^0 = U^0 + W^0[/imath]
Let [imath]V[/imath] be a finite dimensional vector space and [imath]U^0[/imath] and [imath]W^0[/imath] denote the annihilators of subspaces [imath]U[/imath] and [imath]W[/imath], respectively. Then I need to show that [imath](U\cap W)^0 = U^0 + W^0[/imath]. I can show that [imath]U^0 + W^0 \subseteq (U\cap W)^0[/imath] like so: Let [imath]f\in U^0[/imath] and [imath]g\in W^0[/imath]. Then, because [imath]U^0, W^0 \subseteq (U\cap W)^0[/imath], we see that if [imath]f\in U^0[/imath] and [imath]g\in W^0[/imath], then for any [imath]v\in U\cap W[/imath] and [imath]a,b \in \Bbb F[/imath], [imath](af+bg)(v) = af(v) + bg(v) = 0[/imath]. Thus [imath]af+bg\in (U\cap W)^0[/imath]. But because [imath]f[/imath] and [imath]g[/imath] were arbitrary, this implies that any linear combination of elements of [imath]U^0[/imath] and [imath]W^0[/imath] is an element of [imath](U\cap W)^0[/imath]. I.e. [imath]U^0+W^0 \subseteq (U\cap W)^0[/imath]. But I haven't had any luck proving the other direction. Obviously I should start with [imath]f\in (U\cap W)^0[/imath]. But then it looks like I need to somehow break that function into two to show its in [imath]U^0 + W^0[/imath]. That's throwing me off. |
2950477 | Is all uncountable set is equipotent to a certain [imath]\mathcal{P}(A)[/imath]?
Can we say that every uncountable set [imath]X[/imath] is equipotent to the set of all the parts of a certain set [imath]A[/imath] ? | 2247299 | Possible cardinality of power sets
What cardinals can provably not occur as the cardinality of a power set? I know that [imath]\mathbb N[/imath] and natural numbers that are not powers of two are such cardinals. What else is out there? |
2950569 | proof involving double complement
Prove [imath](A^c)^c[/imath] = [imath]A[/imath] [imath](A^c)^c = \{x|x\in (A^c)^c\}[/imath] [imath](A^c)^c = \{x|x\notin A^c\}[/imath] by definition of complement [imath](A^c)^c = \{x|\ (x \in A)\}[/imath] by definition of complement Therefore [imath](A^c)^c=A[/imath] Is this proof ok, I'm not sure if its valid? thanks | 937166 | Double Complement of a set proof
Question states: Prove the law of double complements for sets: If [imath]A[/imath] is a set and [imath]A^\complement[/imath] is its complement than prove that: [imath] (A^\complement)^\complement = A[/imath] I started with: [imath] A^\complement = U - A[/imath] where U is the universal set. But I do not understand how to go further. |
2951265 | How do I integrate [imath]\int \frac{x^2}{(x^2+9)^2}\, dx[/imath]?
How do I integrate [imath]\displaystyle \int \frac{x^2}{(x^2+9)^2}\, dx[/imath] ? I tried doing using algebra and solving the question a bit. But it didn't become something that looks solvable. How should I do this? | 1198686 | How to calculate [imath]\int \frac{x^2 }{(x^2+1)^2} dx[/imath]?
I'm trying to calculate [imath]\int \frac{x^2 }{(x^2+1)^2} dx[/imath] by using the formula: [imath] \int udv = uv -\int vdu [/imath] I supposed that [imath]u=x[/imath] s.t [imath]du=dx[/imath], and also that [imath]dv=\frac{x}{(x^2+1)^2}dx[/imath], but I couldn't calculate the last integral. what is the tick here? the answer must be: [imath] \frac{1}{2}arctan\ x - \frac{x}{2(1+x^2)} [/imath] + C |
2944842 | Previous year Olympiad problem (polynomials)
Let [imath]P(x)[/imath] be a nonconstant polynomial whose coefficients are positive integers. If [imath]P(n)[/imath] divides [imath]P(P(n) - 2015 )[/imath] for every natural number n, then prove that [imath]P (-2015) = 0[/imath] I've observed that the proof depends on [imath] P(n)[/imath] divides [imath] P(P(n)-2015)-P(-2015)\,[/imath] but I cannot see how to prove that. Could someone please elaborate on how to proceed. | 1808125 | If [imath]P(n)[/imath] divides [imath]P(P(n)-2015)[/imath], prove that [imath]P(-2015)=0[/imath]
Q. Let [imath]P(x)[/imath] be a non-constant polynomial whose coefficients are positive integers. If [imath]P(n)[/imath] divides [imath]P(P(n)-2015)[/imath] for every natural number [imath]n[/imath], prove that [imath]P(-2015)=0[/imath]. In one of the sources, the solution given is as follows: Note that [imath]P(n)-2015-(-2015)=P(n)[/imath] divides [imath]P(P(n)-2015)-P(-2015)[/imath] for every positive integer [imath]n[/imath]. But [imath]P(n)[/imath] divides [imath]P(P(n)-2015)[/imath] for every positive integer n. Therefore, [imath]P(n)[/imath] divides [imath]P(-2015)[/imath] for every positive integer [imath]n[/imath]. Hence [imath]P(-2015)=0[/imath]. I am not able to understand that how [imath]P(n)-2015-(-2015)=P(n)[/imath] divides [imath]P(P(n)-2015)-P(-2015)[/imath]. Please help me out. |
2951757 | Why does this method for differentiating work?
Consider the function [imath]f(x)=x^x.[/imath] If I differentiate with respect to [imath]x[/imath] treating the exponent as a constant and then sum the derivative treating the base as a constant, I get [imath]\begin{align} f'(x)&= xx^{x-1}+x^x\ln x\\ f'(x)&= x^x(1+\ln x). \end{align}[/imath] | 2217785 | Proof of this fairly obscure differentiation trick?
Suppose we're tying to differentiate the function [imath]f(x)=x^x[/imath]. Now the textbook method would be to notice that [imath]f(x)=e^{x \log{x}}[/imath] and use the chain rule to find [imath]f'(x)=(1+\log{x})\ e^{x \log{x}}=(1+\log{x})\ x^x.[/imath] But suppose that I didn't make this observation and instead tried to apply the following differentiation rules: [imath]\frac{d}{dx}x^c=cx^{c-1} \qquad (1)\\ \frac{d}{dx}c^x = \log{c}\ \cdot c^x \quad (2)[/imath] which are valid for any constant [imath]c[/imath]. Obviously neither rules are applicable to the form [imath]x^x[/imath] because in this case neither the base nor the exponent are constant. But if I pretend that the exponent is constant and apply rule [imath](1)[/imath], I would get [imath]f'(x)\stackrel{?}{=}x\cdot x^{x-1}=x^x.[/imath] Likewise, if I pretend that the base is constant and apply rule [imath](2)[/imath] I obtain [imath]f'(x)\stackrel{?}{=}\log{x}\cdot x^x[/imath]. It isn't hard to see that neither of the derivatives are correct. But here's where the magic happens: if we sum the two “derivatives” we end up with [imath]x^x+ \log{x}\cdot x^x=(1+\log{x})\ x^x[/imath] which is the correct expression for [imath]f'(x)[/imath]. This same trick yields correct results in other contexts as well. In fact, in some cases it turns out to be a more efficient way of taking derivatives. For example, consider [imath]g(x)=x^2 = \color{blue} x\cdot \color{red} x.[/imath] If we pretend the blue [imath]\color{blue} x[/imath] is a constant we would get [imath]g'(x)\stackrel{?}{=}\color{blue}x\cdot 1=x[/imath]. Now if we pretend the red [imath]\color{red}x[/imath] is constant we get [imath]g'(x)\stackrel{?}{=}1\cdot \color{red} x=x[/imath]. Summing both expressions we end up with [imath]2x[/imath] which is of course a correct expression for the derivative. These observations have led me to the following conjecture: Let [imath]f(x,y)[/imath] be a differentiable function mapping [imath]\mathbb{R}^2[/imath] to [imath]\mathbb{R}.[/imath] Let [imath]f'_1 (x,y)=\frac{\partial}{\partial x} f(x,y)[/imath] and [imath]f'_2 (x,y)=\frac{\partial}{\partial y} f(x,y)[/imath]. Then for any [imath]t[/imath] we have: [imath]\frac{d}{dt}f(t,t)=f'_1 (t,t) + f'_2 (t,t).[/imath] (I apologise for the somewhat awkward notation which I could not seem to get around without causing undue ambiguity.) This formulation also seems to lend itself to following generalisation: Let [imath]f:\mathbb{R}^N \to \mathbb{R}[/imath] be a function differentiable in each of its variables [imath]x_1,x_2,\ldots,x_N[/imath]. For [imath]n=1,2,\ldots,N[/imath] define [imath]f'_n(x_1,x_2,\ldots,x_N)=\frac{\partial}{\partial x_n}f(x_1,x_2,\ldots,x_N)[/imath]. Let [imath]t[/imath] be any real number and define the [imath]N[/imath]-tuple [imath]T=(t,t,\ldots,t)[/imath]. Then one has: [imath]\frac{d}{dt} f(T)=\sum_n f'_n(T).[/imath] Thus my question is: Is this true? How can it be proven? (Specifically in the case [imath]N=2[/imath] but also in the general case.) |
2951047 | Composition of two invertible linear maps
Prove or disprove: Let [imath]A,B: {\Bbb R}^n \rightarrow {\Bbb R}^n[/imath] be linear. If [imath]A,B[/imath] are both invertible, then [imath]A\circ B[/imath] is invertible Can someone explain if this is true or false? I'm trying to come up with examples but I don't know where to start. | 1102447 | Proof that composition of invertible linear transformations is invertible (without determinants)
A crucial concept in linear algebra is that the composition of two invertible linear transformations is itself invertible. Here is the first proof I learned of this fact: Proof: Suppose that [imath]T_1: \mathbb{C}^n \to \mathbb{C}^n[/imath] and [imath]T_2: \mathbb{C}^n \to \mathbb{C}^n[/imath] are both invertible with respective matrices [imath]A_1[/imath] and [imath]A_2[/imath]. Then the matrix of their composition [imath]T_2 \circ T_1[/imath] is simply [imath]A_2A_1[/imath]. Since [imath]T_1[/imath] and [imath]T_2[/imath] are invertible, we know that [imath]\det(A_1) \neq 0[/imath] and [imath]\det(A_2) \neq 0[/imath]. Thus, we see that [imath]\det(A_1A_2) = \det(A_1)\det(A_2) \neq 0[/imath]. Thus, the composition is also invertible. [imath]\square[/imath] I'm trying to now give a proof of this fact without using determinants. Any idea on where to start? |
2951705 | Inequality involving [imath]\text{floor}(x)[/imath]
I'm trying to prove the following inequality [imath]\forall x,y \in \mathbb R[/imath] : [imath]\lfloor x \rfloor + \lfloor x + y \rfloor + \lfloor y \rfloor \le \lfloor 2x \rfloor + \lfloor 2y \rfloor[/imath] If one thing is clear for me it is that the basic definition isn't enough to prove this, I would always get an additional [imath]1[/imath] on the left hand side. Thanks for your trouble. | 1369531 | Prove the following floor function identity
The identity is this: [imath]\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor[/imath] I am truly stumped. Help is appreciated. Thank you! |
2951842 | Proving continuity of Logarithm using [imath]\delta[/imath]-[imath]\epsilon[/imath]
Say we wanted to prove the continuity of the logarithm using the [imath]\delta - \epsilon[/imath] proof (and using the definition of the log as the inverse of the exponential). For any log base [imath]a>1[/imath] Starting with [imath]|\log_a({x_1}) - \log_a({x_2})|<\epsilon[/imath], We can find that if [imath]\frac{x_1}{x_2} < a^\epsilon[/imath] (and [imath]\frac{x_1}{x_2}<a[/imath]), then [imath]|\log_a({x_1}) - \log_a({x_2})|<\epsilon[/imath] as desired. But for the [imath]\delta[/imath] part to come in, we need it in the form of [imath]|x_1 - x_2|[/imath], not [imath]\frac{x_1}{x_2}[/imath]. So then I thought that maybe if [imath]|x_1 - x_2| < \frac{x_1}{x_2} -1 < a^\epsilon -1[/imath] then we would have [imath]|\log_a({x_1}) - \log_a({x_2})|<\epsilon[/imath] which would mean that the function is continuous at [imath]x_1[/imath] and since it was arbitrary it is continuous at all x (right?). The reason I think this is true is because we want the distance between [imath]x_1[/imath] and [imath]x_2[/imath] to be very small, so [imath]a^\epsilon -1[/imath] (which we would call [imath]\delta[/imath]) would get really small as [imath]\epsilon[/imath] gets small. Is there a better way to arrive at this value of [imath]\delta[/imath]? Assuming it is a valid answer | 314518 | I need to prove the continuity of [imath]f(x)=\log x[/imath] using a [imath]\epsilon-\delta[/imath] proof
I need to prove the continuity of [imath]f(x)=\log x[/imath] using a [imath]\epsilon-\delta[/imath] proof These is what I have so far but am not sure how to continue [imath]|\log x-\log a| < \epsilon[/imath] [imath]\log a- \epsilon < \log x < \log a+ \epsilon[/imath] [imath]\frac{a}{e^\epsilon} < x < {a}e^\epsilon[/imath] Any help is appreciated |
2951886 | Prove the zero in the complex number system is unique.
Prove the zero in the complex number system is unique. I have an idea but i don't know how start this proof. The idea: I think start out with the assumption that there exist two different unities, say, [imath]a[/imath] and [imath]b[/imath], and then proceeds to show that this assumption leads to some contradiction. But i'm a little stuck. Can someone help me? | 2458123 | Proving that the zero vector is unique
We are supposed to prove that the zero vector is unique with the additive identity [imath]x+O=x[/imath] and then by creating another additive identity with [imath]O'[/imath], then [imath]x+O'=x[/imath]. We set them equal [imath]x+O=x+O'[/imath] and use the vector cancelling theorem to get [imath]O=O'[/imath], but how does this prove that the zero vector is unique? |
2952356 | Show that [imath]x^{\frac{1}{2}} +x^{\frac{1}{3}}=1[/imath] have atleast one real solution.
I tried to show that [imath]F(x) = x^{\frac{1}{2}} +x^{\frac{1}{3}}-1[/imath] this function is monotonic but that doesn't seem to work. Also I am not allowed to use graphs. Is there some algebraic approach to solve this? | 818472 | Simple equation for [imath]x[/imath] but getting no proof.
Show that there is at least one real value of [imath]x[/imath] for which [imath]x^{1/3} + x^{1/2} = 1[/imath] I did draw the graphs of [imath]x^{1/3}[/imath] and [imath]1-x^{1/2}[/imath] and showed that they met at a point, but I don't think it's a good algebraic proof. How should I proceed after substituting [imath]x[/imath] for [imath]z^{1/6}[/imath] and getting a polynomial in terms of [imath]z[/imath] |
2951786 | If [imath]X_1,...,X_n[/imath] are Bernoulli(p) iid then [imath]X_1+...+X_n[/imath] is Binomial(p,n)
How can I prove that if [imath]X_1,...,X_n[/imath] are iid [imath]Bernoulli(p)[/imath], then [imath]X_1+...+X_n[/imath] is [imath]Binomial(n,p)[/imath] ? Attempt [imath]\mathbb P\{X_1+...+X_n=k\}=\sum_{x_1+...+x_n=k}f_{X_1,...,X_n}(x_1,...,x_n)[/imath] [imath]=\sum_{x_1+...+x_n=k}f_{X_1}(x_1)...f_{X_n}(x_n)[/imath] But now, I have difficulty to conclude. How can I do ? I also did by induction, and it worked, but I really would like to use this method. | 664011 | Proof that a sum of Bernoulli rvs has Binomial distribution
Let [imath]\{X_n, n \geq 1\}[/imath] be iid Bernoulli random variables with [imath]\Pr(X_1 = 1) = p = 1 - \Pr(X_1 = 0)[/imath] and let [imath]S_n = \sum\limits_{i=1}^n X_i[/imath] be the number of successes in [imath]n[/imath] trials. Show [imath]S_n[/imath] has a binomial distribution by the following methods: 1.) Prove for [imath]n\geq0,[/imath] [imath]1\leq{k}\leq{n+1}[/imath] [imath]\Pr(S_{n+1} = k) = p\Pr(S_n = k - 1) + q\Pr(S_n = k)\text{ where }q = 1-p.[/imath] 2.) Solve the recursion using generating functions. For the first method, I think I need to use convolution of [imath]n + 1[/imath] Bernoulli rvs, but I cannot immediately see the progression. For the second method, I don't exactly understand how to proceed either. This is problem 1.2 from Resnick's Adventures in Stochastic Processes. |
1747845 | Left and right coset calculation for matrix
If [imath] G = \left\{ \begin{pmatrix} c & a \\ 0 & 1 \end{pmatrix} c,a\in (\mathbb{R}) : c \neq 0 \right\} [/imath] and assume G is a group under matrix multpication [imath] H = \left\{ \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}: x\in\mathbb{R} \right\} [/imath] Find the left and right cosets of H in G and show that they are equal So far I have done the following: [imath]gH = \begin{pmatrix} c & a \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}[/imath] [imath]=\begin{pmatrix} c & cx+a \\ 0 & 1 \end{pmatrix}[/imath] To calculate the right coset, choose [imath]g' =\begin{pmatrix} c' & a' \\ 0 & 1 \end{pmatrix}, c' a' \in G[/imath]: Then [imath]Hg'=\begin{pmatrix} c' & x+a' \\ 0 & 1 \end{pmatrix}[/imath] How does [imath]gH[/imath]=[imath]Hg'[/imath] in this case? | 1752812 | Showing H is a normal subgroup by calculating left and right coset
If [imath]G = \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} a,b\in (\mathbb{R}) : a \neq 0[/imath]) and assume G is a group under matrix multpication Prove that H = ([imath]\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}: t\in\mathbb{R}[/imath]) is a normal subgroup of G I'm meant to show that H is a normal subgroup by showing that the left coset is equal to the right coset. So far I have shown the following: take [imath]g=\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}[/imath] then [imath]gH=\begin {pmatrix}a &at+b \\ 0 &1 \end{pmatrix}[/imath] and take [imath]g'=\begin{pmatrix} a' & b' \\ 0 & 1 \end{pmatrix}[/imath] then [imath]Hg'=\begin{pmatrix} a' & b'+t \\ 0 & 1 \end{pmatrix}[/imath] I don't quite understand how [imath]gH=Hg'[/imath] |
2952840 | Ring with unity. Show that if ab units, then a and b are units.
Suppose that R is a ring with unity, [imath]a, b \in R[/imath] and that neither a nor b is a zero divisor. If ab unit, then a and b are units. I know that a ring with unity means that [imath]\exists 1_{R} \in R:\forall a \in R: 1_{R}a = a = a1_{R}.[/imath] For an element [imath]a \in R[/imath] to be a zero divisor we must have [imath]a \neq 0_{R}[/imath] and there must exist an element [imath]\hat{a} \neq 0_{R}[/imath] such that [imath]a \hat{a} = 0_{R}[/imath] or [imath]\hat{a}a = 0_{R}[/imath]. Therefore, the meaning of not a zero divisor would be that given [imath]a \neq 0_{R}[/imath] we must have for every [imath]\hat{a} \neq 0_{R}[/imath] that [imath]a\hat{a} \neq 0_{R}[/imath] and [imath]\hat{a}a \neq 0_{R}[/imath]. Similarly, for b we then have [imath]b\hat{b} \neq 0_{R}[/imath] and [imath]\hat{b}b \neq 0_{R}[/imath]. For [imath]ab[/imath] to be a unit there exists [imath](ab)^{-1} \in R[/imath] such that [imath](ab)^{-1}ab = 1_{R} = ab(ab)^{-1}[/imath]. Now, I have to show that [imath]\exists a^{-1}, b^{-1} \in R[/imath] such that [imath]a^{-1}a = 1_{R} = aa^{-1}[/imath] and [imath]b^{-1}b = 1_{R} = bb^{-1}[/imath]. I have no starting point thus far so any hint(s) are greatly appreciated. | 771771 | Proving units in a ring
Suppose [imath]R[/imath] is a ring with no zero divisors and with identity [imath]1_R[/imath] not equal to [imath]0_R[/imath]. Suppose that [imath]a,b[/imath] are in [imath]R[/imath] and that [imath]ab[/imath] is a unit. Prove that [imath]b[/imath] is a unit. My thoughts: I know a unit is basically a unit that (for this example) would mean [imath]abu = 1_R[/imath] for some nonzero [imath]u[/imath] in [imath]R[/imath]. I am really stuck after that. Not seeing a clear path to manipulate the variables to prove b is a unit by itself. |
2953043 | Prove that if [imath]n>1[/imath], the sum of positive integers less than [imath]n[/imath] and coprime to [imath]n[/imath] is [imath](1/2)na(n)[/imath] where [imath]a(n)[/imath] is the number of such integers.
Question 12(iii) Could anyone explain this part of the question to me. What i tried co-prime means that the two integers a and b are said to be relatively prime, mutually prime, or coprime (also written co-prime) if the only positive integer (factor) that divides both of them is 1 Take th number [imath]3[/imath] for example, then the sum of integers less than [imath]3[/imath] and co-prime to [imath]3[/imath] is [imath]2+1=3[/imath], [imath]2[/imath] and [imath]1[/imath] are the two integers co-prime to [imath]3[/imath] which thus satisfies the formula [imath]0.5*n*a(n)[/imath] where [imath]n=3[/imath], [imath]a(n)=2[/imath]. However im unsure of how to prove it. Could anyone explain. Thanks | 9877 | Prove the sum of all numbers that do not have a multiplicative inverse mod [imath]n[/imath]
I understand that for a number [imath]a[/imath] to have a multiplicative inverse in mod [imath]n[/imath], it must be coprime to [imath]n[/imath]; therefore, all numbers that do not have a multiplicative inverse mod [imath]n[/imath] must share a factor with [imath]n[/imath]. Now, my prediction was that if [imath]n[/imath] is even, then the sum of all numbers that do not have a multiplicative inverse mod [imath]n[/imath] is [imath]\frac{n}{2}\bmod{n}[/imath], and if [imath]n[/imath] is odd, then the sum is [imath]0\bmod{n}[/imath]. Originally my proof was constructed something like this: if [imath]n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}[/imath], then the sum of all the elements of this array: [imath]\begin{align} \begin{matrix} p_1 & p_2 & \cdots & p_k \\ 2p_1 & 2p_2 & \cdots & 2p_k \\ \vdots & \vdots & \cdots & \vdots \\ n-2p_1 & n-2p_2 & \cdots & n-2p_k \\ n-p_1 & n-p_2 & \cdots & n-p_k \end{matrix} \end{align}[/imath] would cancel out in the correct way. However, while trying to prove this, I hit a roadblock because I wasn't sure how to account for the multiplicity of the factors of [imath]n[/imath]. If I understand Wikipedia correctly, this is the same as finding the sum of all numbers that are not units in the ring of integers modulo [imath]n[/imath], [imath]\mathbb{Z}/n\mathbb{Z}[/imath]. Is there any way to do this? If the result is so nice, I feel like there is also a nice way to prove this. |
1753831 | Not a normal subgroup by left and right coset
If [imath]G = \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} a,b\in (\mathbb{R}) : a \neq 0[/imath]) and assume G is a group under matrix multpication Assume that K = ([imath]\begin{pmatrix} s & 0 \\ 0 & 1 \end{pmatrix}: s\in\mathbb{R}, s \neq 0[/imath]) is a subgroup of G. Prove that K is not a normal subgroup of G. I want to prove this by showing the left cosets and the right cosets are not the same. So if we take [imath]g=\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}[/imath] then [imath]gK=\begin{pmatrix} as& b \\ 0 & 1 \end{pmatrix}[/imath] And if we take, [imath]g'=\begin{pmatrix} a' & b' \\ 0 & 1 \end{pmatrix}[/imath] then [imath]Kg'=\begin{pmatrix} sa' & sb' \\ 0 & 1 \end{pmatrix}[/imath] How do i show that [imath]gK \neq Kg'[/imath], Is it sufficient to say that [imath]Kg'[/imath] has two columns that depend on s where as [imath]gk[/imath] only has one column that depend on S. So they are not equal | 1747774 | K is a subgroup of G. Show that K is not a normal subgroup
If [imath]G = \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} a,b\in (\mathbb{R}) : a \neq 0[/imath]) and assume G is a group under matrix multpication Assume that K = ([imath]\begin{pmatrix} s & 0 \\ 0 & 1 \end{pmatrix}: s\in\mathbb{R}, s \neq 0[/imath]) is a subgroup of G. Prove that K is not a normal subgroup of G. I have done some working out on this question which is the following: [imath]g^{-1}=\begin{pmatrix} a^{-1}&-a^{-1}b\\0&1\end{pmatrix}[/imath] and so [imath]gkg^{-1}=\begin{pmatrix} a&b\\0&1\end{pmatrix}\begin{pmatrix}sa^{-1}&-a^{-1}bs\\0&1\end{pmatrix}[/imath] then we get [imath]gkg^{-1}=\begin{pmatrix}s&-bs+b\\0&1\end{pmatrix}[/imath] What do I do from here, Why is [imath]gkg^{-1} \notin K[/imath] Can't I make b=0 and then it would be in K? or am I missing some key idea? |
2953026 | Give a proof by cases that shows that [imath]n(n ^2 − 1)(n + 2) [/imath] is a multiple of [imath]4[/imath], for all integers [imath]n[/imath]
Give a proof by cases that shows that [imath]n(n ^2 − 1)(n + 2) [/imath] is a multiple of [imath]4[/imath], for all integers [imath]n[/imath] I have already done case 1 where n is even. I am doing case 2 where n is odd and I'm a bit confuse how to finish off the problem..Uploaded a picture of my work.. Is this work sufficient enough to prove that it is a multiple of 4? The 6k would not have any affect of the result in general? I feel it is wrong because of the 6k. Right now my thoughts are if (statement) is a multiple of 4, then (statement) is a multiple of 2 as well. And work from there. My work | 616102 | Prove that [imath]n(n^2 - 1)(n + 2)[/imath] is divisible by [imath]4[/imath] for any integer [imath]n[/imath]
Prove that [imath]n(n^2 - 1)(n + 2)[/imath] is divisible by [imath]4[/imath] for any integer [imath]n[/imath] I can not understand how to prove it. Please help me. |
2951504 | Field of Fractions over Formal Power Series
I want to show that every element in the field of fractions of [imath]K[[X]][/imath] can be written as [imath]\sum_{k=-n}^\infty a_kX^k[/imath] where [imath]a_k\in K, n\in\mathbb{Z}[/imath]. My first attempt was to rewrite [imath]\displaystyle \frac{\sum_{i=0}^\infty a_iX^i}{\sum_{j=0}^\infty b_jX^j}=\lim_{N\to \infty}\frac{\sum_{i=0}^N a_iX^i}{\sum_{j=0}^N b_jX^j}[/imath] where I am not completely sure, if this is correct. I know that I can write an element of [imath]R[[X]][/imath] where [imath]R[/imath] is a ring like this. Then I tried polynomial division and some similar things. Also I know that if [imath]b_0[/imath] would be not zero, I can invert the whole polynomial and arrive at the conclusion, even with [imath]n=0[/imath]. However, I was not able to progress from here. | 160471 | If [imath]a_0\in R[/imath] is a unit, then [imath]\sum_{k=0}^{\infty}a_k x^k[/imath] is a unit in [imath]R[[x]][/imath]
Let [imath]R[/imath] a ring, and let [imath]\displaystyle R[[x]]=\left\{\sum_{k=0}^{\infty}a_k x^k\;\middle\vert\; a_k\in R\right\}[/imath] with addition and multiplication as defined for polynomials. We have that [imath]R[[x]][/imath] is a ring containing [imath]R[x][/imath] as a subring. How to prove that if [imath]a_0\in R[/imath] is a unit, then [imath]\displaystyle\sum_{k=0}^{\infty}a_k x^k[/imath] is a unit in [imath]R[[x]][/imath]? |
2953722 | Show that R[0,1] is not complete
Let R[0,1] be the set of Riemann integrable functions between 0,1 by considering [imath]f_{n}(x)=\frac{1}{\sqrt{x}}\chi _{[n^{-1},1]}(x)[/imath] Where [imath]\chi _{[n^{-1},1]}[/imath] is the indicator function for if x is between [imath]n^{-1},1[/imath] with respect to the norm [imath]\left \| f \right \|_1=\int_{a}^{b}\left | f(x ) \right |dx[/imath] Edit: I have to show this using the function [imath]f_{n}(x)=\frac{1}{\sqrt{x}}\chi _{[n^{-1},1]}(x)[/imath] so I don't think it's a duplicate since the other question is considering different functions and my issue is showing this function doesn't converge. Thoughts so far First I want to show that [imath]f_{n}[/imath] is Cauchy so I consider for [imath]n>m[/imath] [imath]\left \| f_{n }-f_{m} \right \|_1=\int_{0}^{1/n} 0\, dx+\int_{1/n}^{1/m} \frac{1}{\sqrt{x}}\,dx +\int_{1/m}^{1} 0 \, dx [/imath] I think this is Cauchy since as [imath]n,m\rightarrow \infty [/imath] then [imath]1/n,1/m \rightarrow 0[/imath] so [imath]\left \| f_{n}-f_{m} \right \|_1 \rightarrow 0[/imath] which is less than [imath]\epsilon[/imath] [imath] \forall[/imath] [imath]\epsilon >0[/imath], But does this not imply [imath]f_n[/imath] converges to [imath]0\in R[0,1][/imath] Edit: New idea Assume [imath]f_{n}[/imath] converges to [imath]f[/imath] then [imath]\left \| f_{n}-f \right \|_1 < \epsilon [/imath] for some [imath]n>N\in \mathbb{N} [/imath] and [imath]\epsilon >0[/imath] so [imath]\left \| f_{n}-f \right \|_1 =\int_{0}^{1/n} 0\, dx+\int_{1/n}^{1} \frac{1}{\sqrt{x}}\,dx [/imath] [imath]\Rightarrow f(x)=1/\sqrt{x}[/imath] when x is between [imath]1/n[/imath] and [imath]1[/imath] and 0 otherwise which is [imath]\notin[/imath]R[0,1] | 397369 | space of riemann integrable functions not complete
Define norm as [imath]\int |f|[/imath] (Riemann integral) on [imath]\mathcal R^1[0,1][/imath], the space of riemann integrable functions on [imath][0,1][/imath] with identification [imath]f=g[/imath] iff [imath]\int |f-g|=0[/imath]. Let [imath]\{ r_1,r_2,\cdots \}[/imath] be the rationals in [imath][0,1][/imath], and let [imath]f_n=1_{\{r_1,\cdots,r_n\}}[/imath]. Then [imath]f_n[/imath] is a cauchy sequence in [imath]\mathcal R^1[0,1][/imath]. I want to show that there is no [imath]f\in \mathcal R^1[0,1][/imath] such that [imath]f_n[/imath] converges to [imath]f[/imath] in norm. How can I show it? Obviously the pointwise limit [imath]f=1_\mathbb{Q}[/imath] is not contained in [imath]\mathcal R^1[0,1][/imath], but can I use this fact? I think that there can be other candidates, since convergence in norm and pointwise convergence are different. |
2953974 | Show that either [imath]f[/imath] or [imath]\bar{f}[/imath] have derivative at [imath]z_0[/imath]
So the problem here is the following: If the limit [imath]\lim_{z \to z_0} \left|\frac{f(z)-f(z_0)}{z-z_0}\right| [/imath] exists and is finite, show that [imath]f[/imath] or [imath]\bar{f}[/imath] has a derivative at [imath]z_0[/imath]. The previous question of the excercise was about the total differential of a complex function and its conjugate. I tried to use it but it didn't help. I would like to give me a small tip and not the whole solution of the excercise if possible! Edit: The function [imath]f[/imath] has a total differential at the point [imath]z_0[/imath] so it is real differential as well. | 141294 | Existence of the absolute value of the limit implies that either [imath]f \ [/imath] or [imath]\bar{f} \ [/imath] is complex-differentiable
This is an exercise from Remmert. Let D be domain in [imath]\mathbb{C} \ [/imath], and [imath]f : D \rightarrow \mathbb{C} \ [/imath] a real-differentiable function. Suppose that for some [imath] c \in D [/imath] the limit [imath]\lim_{h \to 0} |\frac{f(c + h) - f(c)}{h}|[/imath] exists. Prove that either [imath]f \ [/imath] or [imath]\bar{f} \ [/imath] is complex-differentiable at c. Any hint ? |
4872 | Group theory proof of existence of a solution to [imath]x^2\equiv -1\pmod p[/imath] iff [imath]p\equiv 1 \pmod 4[/imath]
I've read through the elementary proof of why there exists a solution [imath]x[/imath] to [imath]x^2\equiv -1\pmod p[/imath] iff [imath]p\equiv 1 \pmod 4[/imath] for [imath]p[/imath] an odd prime. Is there a group theory generalization for this fact as well? | 2959141 | Solve [imath]n^2 \equiv -1 \mod 4m+3[/imath]
The following equation has no solution for [imath]m<1000[/imath] (see computation below) [imath]n^2 \equiv -1 \mod 4m+3 [/imath] Question: Is there a solution in general? sage: for k in range(1000): ....: for n in range(4*k+3): ....: if (n**2+1).mod(4*k+3)==0: ....: print([k,n]) ....: sage: |
2954712 | Torsion elements of a group aren't necessarily a subgroup
In a question from an exam in an undergraduate group theory course, we were asked to prove or disprove that the set of all Torsion elements of a group is necessarily a subgroup. I knew that the set of Torsion elements is closed under the inverse operation, but was later told that it is not closed under multiplication, therefore disproving the claim. However, I couldn't find any examples of a group [imath]G[/imath] and two elements [imath]a,b[/imath] such that both [imath]a[/imath] and [imath]b[/imath] have a finite order, but [imath]ab[/imath] doesn't. I know that for this to happen [imath]G[/imath] must be an infinite and non-Abelian group, but still couldn't find a valid example. What are some examples of groups/elements fulfilling the aforementioned property? | 3010211 | Tor of a group and counter example
Let [imath]Tor(G) = \{ g \in G | \exists n >0 , g^n = e \}[/imath] of a group [imath]G[/imath] , give an example to [imath]G[/imath] such that [imath]Tor(G)[/imath] is not sub group ? Easy to prove that [imath]G[/imath] must be non-Abelian and infinite in size, so i thought of matrix but i don't have concrete example |
2954799 | How to compute [imath]\operatorname{Ker}(\phi )[/imath] and [imath](\phi )(25)[/imath] for [imath](\phi ):Z\rightarrow Z_{7}[/imath] such that [imath]\phi(1)=4[/imath]?
I understand I must find some element that map to identity but what condition [imath]\phi(1)=4[/imath] is related ? and how to compute [imath](\phi )(25)[/imath]? I must find mapping from 25 to [imath]Z_{7}[/imath] isn't it ? is that 4 ? | 1137095 | Give [imath]\ker(\phi)[/imath] for [imath]\phi: \mathbb{Z} \to \mathbb{Z_7}[/imath] such that [imath]\phi(1)=4[/imath]
Give [imath]\ker(\phi)[/imath] for [imath]\phi: \mathbb{Z} \to \mathbb{Z_7}[/imath] such that [imath]\phi(1) = 4[/imath] So the generator of [imath]\mathbb{Z}[/imath] maps to [imath]4[/imath] and [imath]\langle 4 \rangle = \{0,1,2,3,4,5,6\}[/imath] so it has order seven. so [imath]\ker(\phi) = 7\mathbb{Z}[/imath] because every seventh element will map to the identity under the given map. Is that the correct reasoning? |
1218123 | Is a complex vector space closed under complex conjugation?
Given a complex vector space [imath]\mathcal{V}[/imath], its complex conjugate [imath]\overline{\mathcal{V}} = \{ \overline{v} : v \in \mathcal{V} \}[/imath] consists of the "same" set of points (according to a number of references...). I'm struggling to reconcile that with the following example: Define [imath] \mathcal{V} = \mathrm{span}_{\mathbb{C}}\{(1,i)\} = \{ (\alpha + \beta i, -\beta + \alpha i) : \alpha,\beta \in \mathbb{R} \}\,. [/imath] Then, [imath](1,i) \in \mathcal{V}[/imath] and so [imath]\overline{(1,i)} \in \overline{\mathcal{V}}[/imath], but [imath]\overline{(1,i)} = (1,-i) \notin \mathcal{V}[/imath]. Thus, some vectors in [imath]\overline{\mathcal{V}}[/imath] are not in [imath]\mathcal{V}[/imath] (and vice versa). Edited to add proposed solution (based on comments): Complex conjugation on [imath]\mathcal{V}[/imath] can be (re)defined as [imath] \overline{(\alpha + \beta i, \, -\beta + \alpha i)} = (\alpha - \beta i, \,\beta + \alpha i) \in \mathcal{V} \,. [/imath] Am I missing something? If [imath]\mathcal{V}[/imath] is a complex vector space, then how do we reconcile it as having different vectors than its complex conjugate? On the other hand, if [imath]\mathcal{V}[/imath] isn't a complex vector space, then what is it? | 1119801 | Definition of complex conjugate in complex vector space
I am starting reading about Hodge theory and while reading the definition of abstract Hodge structure a very basic question came to my mind... What is the definition of the conjugate of a subspace of a complex vector space? I found online the definition of conjugate of an entire vector space, but when we do the Hodge decompostition we want [imath]V_{\mathbb{C}}=\oplus_{p,q \in \mathbb{Z}}V^{p,q}[/imath] with the requirement [imath]\overline{V^{q,p}}=V^{p,q}[/imath]. So I want to take a conjugate inside the space [imath]V_{\mathbb{C}}[/imath]. I can see it if my space is of differential forms, but what is the definition for an abstract one? Also, if I take the tensor product of two vector spaces, is this conjugation induced componentwise? I know the question is pretty basic, but I couldn't find any reference for such definitions. |
2955099 | How to find the intersection of [imath]W[/imath] and [imath]Z[/imath]?
Subspaces[imath]W[/imath] and [imath]Z[/imath] of [imath]\mathbb R^4[/imath] are generated by [imath]\{(1,1,0,-1),(1,2,3,0),(2,3,3,-1)\}[/imath] and [imath]\{(1,1,0,-1),(1,2,3,4),(0,1,3,5)\}[/imath], respesctively. Find a basis for [imath]W$$\cap$$Z[/imath]. I already know how to find the basis for [imath]W+Z[/imath], but I am confused on how to find the basis of [imath]W$$\cap$$Z[/imath]. | 25371 | How to find basis for intersection of two vector spaces in [imath]\mathbb{R}^n[/imath]
What is the general way of finding the basis for intersection of two vector spaces in [imath]\mathbb{R}^n[/imath]? Suppose I'm given the bases of two vector spaces U and W: [imath] \mathrm{Base}(U)= \left\{ \left(1,1,0,-1\right), \left(0,1,3,1\right) \right\} [/imath] [imath] \mathrm{Base}(W) =\left\{ \left(0,-1,-2,1\right), \left(1,2,2,-2\right) \right\} [/imath] I already calculated [imath]U+W[/imath], and the dimension is [imath]3[/imath] meaning the dimension of [imath] U \cap W [/imath] is [imath]1[/imath]. The answer is supposedly obvious, one vector is the basis of [imath] U \cap W [/imath] but how do I calculate it? |
2955216 | Index of intersection of subgroups
Let [imath]G[/imath] be a group with subgroups [imath]H[/imath] and [imath]K[/imath] with index [imath]r=[G:H][/imath] and [imath]s=[G:K][/imath]. Show that: [imath][G:H\cap K]\leq rs[/imath] I know it works with Lagrange, but I don't know exactly how. | 985456 | How do I prove [imath][G:H\cap K]\leq [G:H][G:K][/imath]?
Reference: Infinite group with subgroups of finite index Let [imath]G[/imath] be a group. Let [imath]H,K[/imath] be subgroups of [imath]G[/imath]. How do I prove that [imath][G:H\cap K]\leq [G:H][G:K][/imath]? Let's not assume any index is finite. Then, still the result holds? If so how do I prove it? |
2955259 | Inequality with choose function: [imath]1\sqrt{\binom n1} + 2\sqrt{\binom n2}+3\sqrt{\binom n3}+\cdots+n\sqrt{\binom nn} < \sqrt{{2^{n-1}}{n^3}}[/imath]
From the 1987 Spanish Mathematical Olympiad: Prove, for all natural numbers [imath]n[/imath] with [imath]n > 1[/imath], that [imath]1\sqrt{n\choose1} + 2\sqrt{n\choose2}+3\sqrt{n\choose3}+\cdots+n\sqrt{n\choose{n}} < \sqrt{{2^{n-1}}{n^3}}[/imath]. I tried using some estimates on the size of [imath]n\choose k[/imath], like [imath]{n \choose k} \leq \frac{n^k}{k!}[/imath], but nothing seemed to work, mainly due to all the choose functions being enclosed within square-roots. | 1092529 | Inequality with sum of Binomial coefficients: [imath]\sum\limits^n_{k=1}k \sqrt{\binom nk}\leq\sqrt{2^{n-1}n^3}[/imath]
Prove that for every positive integer [imath]n \ge 2[/imath] [imath]\sum^n_{k=1}k \sqrt{\binom nk}\leq\sqrt{2^{n-1}n^3}[/imath] I tried it by induction but I didn't know how to end it. |
2951512 | Is collapsing a contractible closed subspace a homotopy equivalence?
Let [imath]Y \subset X[/imath] be a closed, contractible subspace. Is the collapsing map [imath] \pi: X \rightarrow X/Y[/imath] a homotopy equivalence? If not, what is a counter example? | 21705 | Can contractible subspace be ignored/collapsed when computing [imath]\pi_n[/imath] or [imath]H_n[/imath]?
Can contractible subspace be ignored/collapsed when computing [imath]\pi_n[/imath] or [imath]H_n[/imath]? Motivation: I took this for granted for a long time, as I thought collapsing the contractible subspace does not change the homotopy type. Now it seems that this is only true for a CW pair... |
2955853 | Number of integer solutions of [imath]3x+y+z=23[/imath]
Calculate the number of nonnegative integer solutions of [imath]3x+y+z=23.[/imath] | 1621008 | How many non-negative integer solutions does the equation [imath]3x + y + z = 24[/imath] have?
If the equation is [imath]x + y + z = 24[/imath] then it is solvable with stars and bars theorem. But what to do if it is [imath]3x + y + z = 24[/imath]? |
2956243 | How to calculate [imath]\int \frac{1}{\sqrt{\sin x\cos^7x}}dx[/imath]?
How to calculate [imath]\int \frac{1}{\sqrt{\sin x\cos^7x}}dx[/imath]? I found Evaluation of [imath]\int\frac{1}{\sqrt{\sin x\cos 7x}}dx[/imath] a problem, but the link is not valid. | 2196789 | Solve [imath]\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx[/imath]
So it's given this indefinite integral [imath]\int \frac{1}{\sin ^{\frac{1}{2}}x \cos^{\frac{7}{2}}x}dx[/imath] Is there anyone could solve this integral? Thanks in advance. |
2956096 | Integration by parts yields divergent integral
[imath]I = \int_0^1\dfrac{t \ln t}{\sqrt{1-t^2}}\mathrm{dt}[/imath] Attempt: Let [imath]t = \sin x[/imath] [imath]\implies dt = \cos x dx[/imath] [imath]\implies I = \displaystyle \int_0^{\pi/2} \dfrac{\sin x \ln (\sin x) \cos x dx}{\cos x}[/imath] [imath]\implies I = -\ln (\sin x) \cos x|_{0}^{\pi/2} + \displaystyle\int_0^{\pi/2}\dfrac{\cos^2 x}{\sin x} dx[/imath] The second integral can be "evaluated" using [imath]\cos^2 x = 1- \sin^2 x[/imath]. But unfortunately, there are finally two divergent integrals. [imath]\implies I = -\ln (\sin x) \cos x|_{0}^{\pi/2} + \displaystyle\int_0^{\pi/2}\csc x dx[/imath] + [imath]\displaystyle\int_0^{\pi/2}\sin x dx[/imath] How do I handle this? | 1543056 | Evaluation of [imath]\int_{0}^{1}\frac{x\ln (x)}{\sqrt{1-x^2}}dx[/imath]
Evaluation of [imath]\displaystyle \int_{0}^{1}\frac{x\ln (x)}{\sqrt{1-x^2}}dx[/imath] [imath]\bf{My\; Try::}[/imath] Let [imath]\displaystyle I = \int_{0}^{1}\frac{x\ln x}{\sqrt{1-x^2}}dx\;,[/imath] Put [imath]x=\cos \phi\;,[/imath] Then [imath]dx = -\sin \phi d\phi[/imath] and Changing Limit, We get [imath]\displaystyle I = -\int_{\frac{\pi}{2}}^{0}\cos \phi \cdot \ln(\cos \phi )d\phi = \int_{0}^{\frac{\pi}{2}} \ln(\cos \phi)\cdot \cos \phi d\phi[/imath] Now Using Integration by parts, We get [imath]\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 \phi}{\cos \phi}d\phi[/imath] So [imath]\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\frac{(1-\cos^2 \phi)}{\cos \phi}d\phi[/imath] So [imath]\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\sec \phi d\phi-\int_{0}^{\frac{\pi}{2}}\cos \phi d\phi[/imath] So [imath]\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\left[\ln\left|\sec \phi+\tan \phi\right|\right]_{0}^{\frac{\pi}{2}}-\left[\sin \phi\right]_{0}^{\frac{\pi}{2}}[/imath] So [imath]\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\left[\ln\left|\sec \phi+\tan \phi\right|\right]_{0}^{\frac{\pi}{2}}-1[/imath] Now How can I solve after that, Help Required, Thanks |
2956393 | Semisimple matrix ring
Let [imath]R[/imath] be a ring and suppose the matrix ring [imath]M_n(R)[/imath] is semisimple. How does one proof that [imath]R[/imath] is semisimple? | 1258670 | Matrix Ring of a Semisimple Ring
I recently read the concept of semi-simplicity of a (not necessarily commutative) ring. A ring [imath]R[/imath] is said to be semi-simple if [imath]R[/imath] as a left module over itself is a semi-simple module (This in turn means that each submodule (i.e, left ideal) of [imath]R[/imath] is a direct summand for [imath]R[/imath].) I am grappling with the following question: If [imath]R[/imath] is semi-simple, then so is the matrix ring [imath]M_n(R)[/imath]. It is a theorem that each left module of a semi-simple ring is a semi-simple module. Thus [imath]M_n(R)[/imath] is a semi-simple module of [imath]R[/imath]. But this doesn't help me. I need to show that each left ideal of [imath]M_n(R)[/imath] is a direct summand for [imath]M_n(R)[/imath]. Is there a characterization of the left ideals of [imath]M_n(R)[/imath] in terms of the left ideals of [imath]R[/imath]? Lastly, does the converse also hold? If [imath]M_n(R)[/imath] is a semi-simple ring, then is [imath]R[/imath] necessarily a semi-simple ring? Thanks. |
2956633 | Show that if [imath]n\geq5[/imath], then every element of [imath]S_n[/imath] is a product of 4-cycles.
Figured this needs to broken down case by case, where case one is whenever the permutation is even, and case two is when it is odd. Whenever a permutation is odd, then the product of that permutation with a 4-cycle is even, and so I figure we need only show this for whenever it is even. My other way would be to show that each transposition can be written as the product of four cycles, but I'm a bit lost on this, since four-cycle is itself the product of three transpositions. This was the way to do it. | 3667 | What do all the [imath]k[/imath]-cycles in [imath]S_n[/imath] generate?
Why don't [imath]3[/imath]-cycles generate the symmetric group? was asked earlier today. The proof is essentially that [imath]3[/imath]-cycles are even permutations, and products of even permutations are even. So: do the [imath]3[/imath]-cycles generate the alternating group? Similarly, do the [imath]k[/imath]-cycles generate the alternating group when [imath]k[/imath] is odd? And do the [imath]k[/imath]-cycles generate the symmetric group when [imath]k[/imath] is even? I know that transpositions ([imath]2[/imath]-cycles) generate the symmetric group. |
900869 | Prove [imath] x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)[/imath]
So what I am trying to prove is for any real number x and natural number n, prove [imath]x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)[/imath] I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis. My base case is when [imath]n=2[/imath] we have on the left side of the equation [imath]x^2-1[/imath] and on the right side: [imath](x-1)(x+1)[/imath] which when distributed is [imath]x^2-1[/imath]. So my base case holds. Now I assume that [imath]x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)[/imath] for some [imath]n[/imath]. However, this is where I am stuck. Am I trying to show [imath]x^{n+1}-1=(x-1)(x^n + x^{n-1}+x^{n-2}+...+x+1)[/imath]? I am still a novice when it comes to these induction proofs. Thanks | 2956920 | Prove that [imath]1+a+a^2+...+a^n[/imath] = [imath]\frac{1-a^{n+1}}{1-a}[/imath] for all [imath]a\ne 1[/imath].
I've started by saying [imath]a_n[/imath]=[imath]1+a+a^2+...+a^n[/imath]=[imath]\frac{1-a^{n+1}}{1-a}[/imath] Now I think I need to do: [imath]lim_{n\to \infty}$$\frac{1-a^{n+1}}{1-a}[/imath] Should let [imath]a=2[/imath] and then solve the limit, then write a proof for it using [imath]\epsilon[/imath]? I've also considered that [imath]1+a+a^2+...+a^n[/imath]=[imath]1+a^n[/imath]. But I'm not sure what to do with this information. I'd like to understand how to start this problem properly. |
2954645 | Unbiased estimator of a square root of chi-squared distribution
Let [imath]Y_1,Y_2,...,Y_n[/imath] be a random sample from [imath]N(\mu,\sigma^2)[/imath]. I need to show that [imath]S[/imath] is a biased estimator of [imath]\sigma[/imath]. As from the definition, I see that [imath]\frac{(n-1)S^2}{\sigma^2}\sim\chi^{2}_{(n-1)}[/imath]. My thought process follows like this: If [imath]X\sim \chi^2_{n-1}[/imath], then by definition [imath]X\sim Gamma(\frac{n-1}{2},2)[/imath]. Apply above definition to this problem. Since I need to find if [imath]S[/imath] is biased or not, I must compute the expected value of S. I also know that [imath]E[Y^a]=\frac{\beta^a\Gamma(\alpha+a)}{\Gamma({\alpha})}[/imath] when Y is a gamma distribution. I should be able to apply the same concept to this problem. After calculations, I obtain something like this: [imath]E[S]=\frac{\sqrt{2}\Gamma(\frac{n}{2})}{\Gamma{(\frac{n-1}{2})}}\times\frac{\sigma}{n-1}[/imath] How do I prove that above is an biased estimator of [imath]\sigma[/imath]? Obviously the value of the gamma functions change as the n changes (between even and odd). Is there a more systematic way of obtaining this? I'm not quite sure how to deal with this gamma function and go on to compare [imath]E[S][/imath] with [imath]\sigma[/imath]. Is there a way for me to simplify this gamma function further? | 770802 | Show that [imath]S = \sqrt{S^2}[/imath] is a biased estimator of [imath]\sigma[/imath] given a random sample from a normal distribution ...
Suppose [imath]Y_1, \ldots, Y_n[/imath] is a random sample from a normal distribution with mean [imath]\mu[/imath] and variance [imath]\sigma^2[/imath]. Let [imath]S^2[/imath] be the sample variance, which is unbiased for [imath]\sigma^2[/imath]. GOAL: Show that [imath]S = \sqrt{S^2}[/imath] is a biased estimator of [imath]\sigma[/imath]. I realize that [imath](n-1)S^2/ \sigma ^2[/imath] is chi-square with [imath](n-1)[/imath] degrees of freedom. I am given that [imath]E(S) = E\left[\frac{\sigma}{\sqrt{n-1}} \left[\frac{(n-1)S^2}{\sigma^2}\right]^{1/2}\right] = \frac{\sigma}{\sqrt{n-1}} \int_0^\infty v^{1/2} \frac{1}{\Gamma[(n-1)/2] 2^{(n-1)/2}} v^{(n-1)/2} e^{-v/2} \, dv[/imath] Can anyone help me understand why they have [imath]v^{1/2}[/imath] and [imath]v^{(n-1)/2}[/imath] (combining them [imath]v^{n/2}[/imath] as I think it should be [imath]v^{1/2}[/imath] and [imath]v^{\frac{(n-1)}{2}-1}[/imath] (combining both terms give [imath]v^{(n-2)/2}[/imath] since the pdf for chi-square distribution is [imath] f(v) = \frac{v^{(x/2)-1} e^{-v/2}}{2^{x/2}\Gamma(x/2)},[/imath] where [imath]x[/imath] is degree of freedom? Could anyone please tell me which one is the correct answer with explanation? Thank you |
2956967 | [imath]R[/imath] a [imath]PID \Rightarrow R[[X]][/imath] a [imath]UFD[/imath]
I have just shown that an integral domain [imath]R[/imath] is a unique factorization domain iff for every non-zero prime ideal [imath]P[/imath] of [imath]R[/imath], [imath]P[/imath] contains a non-zero prime, principal ideal. I am then asked to show that if [imath]R[/imath] is a [imath]PID[/imath] then [imath]R[[X]][/imath] (the ring of formal power series over [imath]R[/imath]) is a [imath]UFD[/imath]. I have shown that [imath]R[[X]][/imath] is an integral domain, so it remains to show that if [imath]P[/imath] is a prime ideal of [imath]R[[X]][/imath], then [imath]P[/imath] contains a principal prime ideal. However, I am not very familiar with working with the ring of formal power series, so I cannot quite see what to do to get the desired result. I would appreciate any help that may be offered, thank you. | 1096521 | Ring of formal power series over a principal ideal domain is a unique factorisation domain
An exercise in my algebra course book asks to prove that if [imath]R[/imath] is a PID, then [imath]R[[x]][/imath] is a UFD, where [imath]R[[x]][/imath] is the ring of formal power series over [imath]R[/imath]. After some failed attempts at proving the ACC I visited Wikipedia, which comments: If [imath]R[/imath] is Noetherian, then so is [imath]R[[x]][/imath]; this is a version of the Hilbert basis theorem. This is very useful, as [imath]R[/imath] is a PID and hence Noetherian. Unfortunately we only saw (without proof) Hilbert's basis theorem in the form If [imath]R[/imath] is Noetherian, then so is [imath]R[x][/imath]. I'm not sure how to conclude the Noetherianity of [imath]R[[x]][/imath] from this. I know that [imath]R[x][/imath] is a UFD because [imath]R[/imath] is (we saw this without proof), and have been trying to conclude the ACC in [imath]R[[x]][/imath] from the ACC in [imath]R[x][/imath], without success. I can prove the ACC if [imath]R[/imath] is a field, because then every [imath]a_kx^k+a_{k+1}x^{k+1}+\cdots[/imath] with [imath]a_k\neq0[/imath] is associate with [imath]x^k[/imath], hence every ideal is of the form [imath](x^k)[/imath]. (In fact this readily proves that [imath]R[[x]][/imath] is a UFD.) If [imath]R[/imath] is not a field, say it has some non-invertible element [imath]r[/imath], then there are many more ideals such as [imath](r)[/imath] or [imath](r+x)[/imath]. I'm also facing difficulties at identifying the irreducible elements of [imath]R[[x]][/imath]. (I wish to prove that they are all primes in order to conclude the uniqueness of factorisation.) I have figured that they either take the form [imath]ux+Px^2[/imath] for some unit [imath]u\in R^\times[/imath] and some [imath]P\in R[[x]][/imath] or have a non-zero constant term which is not invertible in [imath]R[/imath]. The elements of the form [imath]ux+Px^2[/imath] are indeed primes, the difficulty lies in those with non-zero constant term. Any hints at proving [imath]R[[x]][/imath] is a UFD? |
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