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1136340 | Summation of factorial series
I want to calculate the sum of this series: [imath]S = 1\cdot1! + 2\cdot2! + 3\cdot3! + 4\cdot4! +\dots+ n\cdot n![/imath] Is their any formula for finding this sum? | 410290 | Evaluate [imath]\sum_{k=1}^nk\cdot k![/imath]
I discovered that the summation [imath]\displaystyle\sum_{k=1}^n k\cdot k![/imath] equals [imath](n+1)!-1[/imath]. But I want a proof. Could anyone give me one please? Don't worry if it uses very advanced math, I can just check it out on the internet. :) |
1136640 | Studying when [imath]P_n=(p_1\cdot p_2\cdots p_n)+1[/imath] is a square number
Let [imath]p_n[/imath] be the [imath]n[/imath]th prime number. I need to find under which conditions the number [imath]P_n=(p_1\cdot p_2\cdots p_n)+1[/imath] is a square number. So far I have seen that [imath]P_1 = 2+1 =3[/imath] [imath]P_2 = 2\cdot3+1 =7[/imath] [imath]P_3 = 2\cdot3\cdot5+1 =31[/imath] give all prime numbers, this is that [imath]P_n[/imath] is always a prime number, and so it can never be a square number. But I can not find the exact way to prove it. I though of trying something like the prove for the Euclidean Theorem that states that there are infinite prime numbers but I can not figure it out. Thank you. | 368270 | Is my proof correct? [imath]p_1p_2p_3\cdots p_n+1)[/imath] cannot be the square of an integer
Prove that [imath]p_1p_2p_3\cdots p_n+1[/imath], where [imath]p_n[/imath] is the [imath]n^{th}[/imath] prime, cannot be the square of an integer. Let [imath]p_1p_2p_3\cdots p_n+1=Q[/imath] and assume it is the square of an integer, so [imath]\sqrt{p_1p_2p_3\cdots p_n+1}[/imath] is an integer and can therefore be written as [imath]p_ap_bp_cp_d\cdots[/imath] and let [imath]p_ap_bp_cp_d\cdots=R[/imath]. Now [imath]Q/R[/imath] should be equal to [imath]R[/imath], so we have [imath](p_1p_2p_3\cdots p_n+1)/(p_ap_bp_cp_d\cdots)[/imath], or [imath]I+1/(p_ap_bp_cp_d\cdots)[/imath], where [imath]I[/imath] is an integer. However, [imath]I+[/imath] a fraction is clearly not an integer, so this is a contradiction. (Note: I know this can be done with modular arithmetic, but I haven't learned it yet). |
1136413 | Notation for mixed partial derivatives
If [imath]f:\Bbb R^2\to\Bbb R[/imath], what does [imath]\frac{\partial^2 f}{\partial x\, \partial y}[/imath] mean? Does it mean [imath]\dfrac{\partial}{\partial x}\left(\dfrac{\partial f}{\partial y}\right)[/imath] or [imath]\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)[/imath]? (I know that most of the time due to Schwarz theorem it is irrelevant.) | 1130314 | Quick question on multivariate differentiation
Does [imath]\frac{\partial ^2f}{\partial x \partial y}[/imath] mean diff it wrt [imath]x[/imath] THEN wrt [imath]y[/imath]? Or diff wrt [imath]y[/imath] then wrt [imath]x[/imath]? |
765403 | Rationalize a fraction using conjugates
I need help rationalizing the following expression using a conjugate: [imath]\dfrac{1}{\sqrt{3} + \sqrt{2}-\sqrt{5}}[/imath] I have had no luck rationalizing this expression with a conjugate of the denominator. | 765372 | Conjugates of radicals
I am not sure if one exists but is there a conjugate of the following: [imath]\sqrt{3}+\sqrt{2}-\sqrt{5}[/imath] I attempted it many times but can't get anything. |
1058555 | Do the columns of an invertible [imath]n \times n[/imath] matrix form a basis for [imath]\mathbb R^n[/imath]?
The columns of an invertible [imath]n \times n[/imath] matrix form a basis for [imath]\mathbb R^n[/imath]. I follow the definition from the text book, then I guess because the matrix is invertible, each vector in the matrix is linearly independent, thus the basis of column space is span in [imath]\mathbb R^n[/imath]. However, I am still confused. Could someone tell me why or if I've missed something? Thanks. | 889290 | If [imath]A[/imath] is an invertible matrix, then the column span of [imath]A[/imath] equals [imath]F^{n\times1}[/imath]
If [imath]A\in\mathbb{F}^{n\times n}[/imath] is invertible so [imath]\mathbb{F}^{n \times 1}=\text{span}\{C_i(A)\mid1\leq i \leq n\}[/imath] What does being invertible has to do with the column span (unlike any other matrix)? |
16011 | Why [imath]9[/imath] & $11$ are special in divisibility tests using decimal digit sums? (casting out nines & elevens)
I don't know if this is a well-known fact, but I have observed that every number, no matter how large, that is equally divided by [imath]9[/imath], will equal [imath]9[/imath] if you add all the numbers it is made from until there is [imath]1[/imath] digit. A quick example of what I mean: [imath]9*99 = 891[/imath] [imath]8+9+1 = 18[/imath] [imath]1+8 = 9[/imath] This works even with really long numbers like [imath]4376331[/imath] Why is that? This doesn't work with any other number. Similarly for [imath]$11$[/imath] and alternating digits sums. | 1970714 | If [imath]3[/imath] divides the decimal digit sum of [imath]n[/imath] then [imath]3[/imath] divides [imath]n[/imath] (casting out threes)
This is a trick I learnt in primary school, but never gave it much thought. Here's how I formulate it: [imath] n = \sum_{j=0}^{m} x_j 10^{m-j} [/imath] is a decimal expansion of some integer [imath]n[/imath] such that [imath] \sum_{j=0}^{m} x_j = r [/imath] such that [imath]3|r[/imath], then [imath]3|n[/imath]. Or, [imath]r= 3k[/imath] and [imath]n=3i[/imath] with [imath]k \neq i[/imath]. I thought about it for some time, but didn't get any intuition. |
1138116 | Separate form for [imath]f'(x)[/imath]
[imath]\qquad^{\star\star}(b)[/imath] Prove, more generally, that [imath]f'(x) = \lim_{h, k \to 0^+}\frac{f(x + h) - f(x - k)}{h + k}[/imath] ONLY HINTS PLEASE. The denominator is the issue. I thought of [imath]u = h + k[/imath] but that created an issue for the limit bounds. I tried adding and subtraction [imath]f(x)[/imath] in the numerator but the denominator causes issues. ONLY HINTS PLEASE! Attempts: [imath]f'(x) = \lim_{h, k \to 0} \frac{f(x + h) - f(x + h + k) + f(x + h + k) - f(x - k)}{h+k} [/imath] [imath] = \lim_{h, k \to 0} -\frac{f(x + h + k) - f(x + h)}{h+k} + \frac{f(x + h + k) - f(x - k)}{h+k}[/imath] let [imath]u = x + k[/imath]. As [imath]k , h\to 0,[/imath] [imath]u \to x[/imath] and [imath]x - k = u - 2k[/imath] Which makes things weird. | 243488 | Proof for alternative definition of the derivative
I read somewhere recently that you can define the derivative as follows: [imath]f'(x) = \lim_{h, k \to 0^+} \frac{f(x +h) - f(x - k)}{h + k}[/imath] I have been trying to prove this for about 2 hours, and can't seem to get it done. How should I proceed? Edit: Assume [imath]f[/imath] is differentiable for all [imath]x \in (x - \delta, x + \delta)[/imath] for some [imath]\delta > 0[/imath]. |
637130 | Ordered pair definition
Why is the ordered pair [imath](x,y)[/imath] defined to be the set [imath]\{\{x\},\{x,y\}\}[/imath]? My book states that it is simply defined this way and then proceeds to perform a bunch of magic using this definition. | 286380 | Ordered Pair Formal Definition
I understand what an ordered pair is, but I'm having trouble the formal Kuratowski definition which says that [imath]\langle a,b \rangle = \{\{a\},\{a,b\}\}[/imath]. What about this definition imposes order on the pair? |
1138729 | Proving isometries of the plane are bijective
So I'm trying to prove that every isometry [imath]I:\mathbb{R}^2 \to \mathbb{R}^2[/imath] is bijective. I have already proved that I is injective (which is almost immediate) and I also proved [imath]I[/imath] is continuous (because I thought that might be useful) but I'm having trouble with the proof for surjectivity. | 307215 | Prove that every isometry on [imath]\mathbb{R}^2[/imath] is bijective
Let [imath]d(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}[/imath] for [imath]x=(x_1,x_2), y=(y_1,y_2)[/imath]. A isometry of [imath]\mathbb{R^2}[/imath] is an image [imath]f:\mathbb{R^2}\to\mathbb{R^2}:d(x,y)=d(f(x),f(y))[/imath]. Show that every isometry is bijective. I don't know where to start, any hints ? |
647311 | If the dual spaces are isometrically isomorphic are the spaces isomorphic?
Let [imath]X[/imath], [imath]Y[/imath] be Banach spaces such that the duals [imath]X^\ast[/imath] and [imath]Y^\ast[/imath] are isometrically isomorphic. Are [imath]X[/imath] and [imath]Y[/imath] necessarily isomorphic? The answer to the question whether [imath]X[/imath] and [imath]Y[/imath] are automatically also isometrically isomorphic is no as the example [imath]c[/imath], [imath]c_0[/imath] which both have dual [imath]\ell^1[/imath] but are not isometrically isomorphic shows. | 2025500 | If two Banach spaces have isomorphic duals, must they be isomorphic?
Let [imath]X,Y[/imath] be Banach spaces such that there exists an isomorphism between [imath]X^*[/imath] and [imath]Y^*[/imath]. Then, can we claim that there exists an isomorphism between [imath]X[/imath] and [imath]Y[/imath]? |
1138859 | If [imath]E[/imath] is a null set, then [imath]E[/imath] can be covered by a sequence of intervals [imath](I_n)[/imath] such that [imath]\sum\limits_{n=1}^{\infty} m^{*}( I_n)<\infty[/imath].
The following is a problem from Real analysis by N. L. Carothers. 16.19. For [imath]E\subset [a,b][/imath], show that [imath]m^{*}(E) = 0[/imath] if and only if [imath]E[/imath] can be covered by a sequence of intervals [imath](I_n)[/imath] such that [imath]\sum\limits_{n=1}^{\infty} m^{*}(I_n) <\infty[/imath], and such that each [imath]x\in E[/imath] is in infinitely many [imath]I_n[/imath]. I can prove the if-part of the question. This chat discussion suggests how to construct such [imath]I_n[/imath] for countable sets. How can we prove the only if part for a general set? | 226293 | Points of a Measure Zero Sets Covered by Intervals Infinitely Many Times
Given a measure zero set [imath]E[/imath], by definition we have forall [imath]\varepsilon > 0[/imath], a covering of [imath]E[/imath] by intervals whose lengths sum to [imath]< \varepsilon[/imath]. I want to prove that we can cover [imath]E[/imath] in intervals such that the sum of the lengths the intervals is [imath]< 1[/imath] and each point in [imath]E[/imath] is contained in infinitely many of the intervals. Do you know how to prove this? Thank you |
1139203 | General topology
I am interested in a proof of very general theorem, well known: Given a topological space [imath]X[/imath] and a dense subset [imath]A[/imath] (so it has nonempty intersection with any open subset of [imath]X[/imath]), and if we are given two continuous functions [imath]f, g[/imath] which map [imath]X[/imath] to [imath]Y[/imath], where [imath]Y[/imath] is another topological space, Hausdorff, and [imath]f(x) = g(x)[/imath] for all [imath]x[/imath] in [imath]A[/imath], then for all [imath]x[/imath] in [imath]X[/imath] it holds [imath]f(x) = g(x)[/imath]. | 291139 | Suppose [imath]f,g[/imath] are continuous functions from [imath]X[/imath] to [imath]Y[/imath], where [imath]X,Y[/imath] are topologies, and [imath]f=g[/imath]
Specifically, [imath]f = g[/imath] on [imath]S[/imath], where [imath]S[/imath] is a dense subset of [imath]X[/imath]. Is it true that [imath]f = g[/imath] on all of [imath]X[/imath]? I'm quite sure that this holds true in the case where [imath]X = \mathbb{R}[/imath], but I don't think it holds true in general. Is this true? If so, could you provide a counterexample where [imath]f \neq g[/imath] on [imath]X[/imath]? |
1139192 | Optimal strategy with flipping a coin indefinitely
Consider this problem: You are flipping a coin which either heads or tails with equal probabilities. Every time you flip the coin you can end the game or flip once more. This way you flip [imath]N[/imath] times and the coin heads [imath]H[/imath] times. Your goal is to maximize value [imath]H/N[/imath]. Now suppose this simple strategy: Choose a real number [imath]q[/imath] between [imath]0[/imath] and [imath]1[/imath]. If at any moment [imath]H/N \geq q[/imath], then end the game, otherwise proceed (until some very high value of [imath]N[/imath] when [imath]H/N[/imath] will be close to [imath]1/2[/imath]). My question is: What is the best choice of [imath]q[/imath] ? If I am right, when [imath]q=1/2[/imath] or [imath]q=1[/imath], then [imath]H/N[/imath] will be [imath]1[/imath] (if the first flip heads) or [imath]1/2[/imath] otherwise. So [imath]H/N = 1/2*1 + 1/2*1/2=3/4[/imath] on average. But it seems that the optimal [imath]q[/imath] is somewhere between these. Thank you for any help or reference. | 6195 | Why is this coin-flipping probability problem unsolved?
You play a game flipping a fair coin. You may stop after any trial, at which point you are paid in dollars the percentage of heads flipped. So if on the first trial you flip a head, you should stop and earn \[imath]100 because you have 100% heads. If you flip a tail then a head, you could either stop and earn \[/imath]50, or continue on, hoping the ratio will exceed 1/2. This second strategy is superior. A paper by Medina and Zeilberger (arXiv:0907.0032v2 [math.PR]) says that it is an unsolved problem to determine if it is better to continue or stop after you have flipped 5 heads in 8 trials: accept \[imath]62.50 or hope for more. It is easy to simulate this problem and it is clear from even limited experimental data that it is better to continue (perhaps more than 70% chance you'll improve over \[/imath]62.50): My question is basically: Why is this difficult to prove? Presumably it is not that difficult to write out an expression for the expectation of exceeding 5/8 in terms of the cumulative binomial distribution. (5 Dec 2013). A paper on this topic was just published: Olle Häggström, Johan Wästlund. "Rigorous computer analysis of the Chow-Robbins game." (pre-journal arXiv link). The American Mathematical Monthly, Vol. 120, No. 10, December 2013. (Jstor link). From the Abstract: "In particular, we confirm that with 5 heads and 3 tails, stopping is optimal." |
1139430 | How do we know that [imath]\omega_1[/imath] exists in ZF?
In ZF, not all "collections of objects" are sets. For example, there is no set of all sets, and there is no set of all ordinals. So, how do we know that there is a set of all countable ordinals? In other words, how do we know that [imath]\omega_1[/imath] exists? (I'm assuming you don't need Choice.) | 46833 | How do we know an [imath] \aleph_1 [/imath] exists at all?
I have two questions, actually. The first is as the title says: how do we know there exists an infinite cardinal such that there exists no other cardinals between it and [imath] \aleph_0 [/imath]? (We would have to assume or derive the existence of such an object before we label it something like [imath] \aleph_1 [/imath].) My second question is, can we say for certain if there's any limit to the number of cardinals existing between [imath] \aleph_0 [/imath] and continuum (i.e. [imath] 2^{\aleph_0} [/imath])? I mean, how could we know that there's not an infinite number of cardinals between the two - perhaps even more than [imath] \aleph_0 [/imath]? |
1139971 | Prove the inverse of this differentiable function is differentiable?
Suppose we have a differentiable function [imath] g [/imath] that maps from a real interval [imath] I [/imath] to the real numbers and suppose [imath] g'(r)>0[/imath] for all [imath] r[/imath] in [imath] I [/imath]. Then I want to show that [imath] g^{-1}[/imath] is differentiable on [imath]g(I). [/imath] Intuitively this makes sense but I can't come up with a neat proof. I was thinking to use the mean value theorem but I'm not sure if that would get me anywhere. | 251405 | Alternative proof for differentiability of inverse function?
Problem Let [imath]f[/imath] be a continuous one-one function defined on an interval and suppose that [imath]f[/imath] is differentiable at [imath]f^{-1}(b)[/imath] with the derivative [imath]f'(f^{-1}(x)) \neq 0[/imath]. Prove that [imath]f^{-1}(x)[/imath] is also differentiable at [imath]b[/imath]. I read the proof in Spivak Calculus book, but it's quite confusing to me. Then I looked up online, and I saw exact the same proof everywhere. So I wonder is there an alternative proof to this problem that I'm not aware of? Here is the proof from Spivak book Theorem 5 (Calculus by Spivak 4th edition, page 237-238) Proof. Let [imath]b = f(a)[/imath]. Then [imath]\displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - f^{-1}(b)}{h} = \displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - a}{h}[/imath] Now every number [imath]b + h[/imath] in the domain of [imath]f^{-1}[/imath] can be written in the form [imath]b + h = f(a + k)[/imath] for a unique [imath]k[/imath]. Then [imath]\displaystyle\lim_{h\to 0} \dfrac{f^{-1}(b + h) - a}{h} = \displaystyle\lim_{h\to 0}\dfrac{f^{-1}(f(a + k)) - a}{f(a + k) - b} = \displaystyle\lim_{h\to 0} \dfrac{k}{f(a + k) - f(a)}[/imath] We have [imath]f^{-1}(b + h) = a + k \Leftrightarrow k = f^{-1}(b + h) - f^{-1}(b)[/imath] Since [imath]f[/imath] is continuous and one-one, [imath]f^{-1}[/imath] is also continuous at [imath]b[/imath]. This means that [imath]k[/imath] approaches [imath]0[/imath] as [imath]h[/imath] approaches [imath]0[/imath]. Since [imath]\displaystyle\lim_{k\to 0} \dfrac{f(a + k) - f(a)}{k} = f'(a) = f'(f^{-1}(b)) \neq 0[/imath] this implies that [imath](f^{-1})'(b) = \dfrac{1}{f'(f^{-1}(b))}[/imath] |
158054 | Complex matrices with null trace
I'm trying to prove the following: Let [imath]A\in \mathbb{C}^{n\times n}[/imath] be a matrix with null trace; then [imath]A[/imath] is similar to a matrix [imath]B[/imath] such that [imath]B_{jj}=0[/imath] (i.e. it has zeroes on its diagonal). Any ideas? Induction on [imath]n[/imath] sounded feasible but I wasn't able to put together anything. | 1140438 | If [imath]Tr(A)=0[/imath] then [imath]T=R^{-1}AR[/imath] has all entries on its main diagonal equal to [imath]0[/imath]
Prove that if [imath]A[/imath] is a square matrix and [imath]Tr(A)=0[/imath], then there exixts an invertible matrix [imath]R[/imath] such that the matrix [imath]T=R^{-1}AR[/imath] has all entries on its main diagonal equal to [imath]0[/imath]. It seems like the formula [imath]A=S\Lambda S^{-1}[/imath], but maybe it does not help. Thanks so much. |
260211 | Prove by induction that [imath]n! > n^2[/imath]
How does one prove by induction that [imath]n! > n^2[/imath] for [imath]n \geq 4[/imath] | 1140396 | Prove by induction that [imath]n^2[/imath]
How can I show that [imath]n^2<n![/imath] for all [imath]n\geq 4[/imath] Step 1 For [imath]n=1[/imath], the LHS=[imath]4^2=16[/imath] and RHS=[imath]4!=24[/imath]. So LHS[imath]<[/imath] RHS. Step 2 Suppose the result be true for [imath]n=k[/imath] i.e., [imath]k^2<k![/imath] Step 3 For [imath]n=k+1[/imath] [imath](k+1)^2=k^2+2k+1[/imath] What will be the next step? |
519960 | the compact Hausdorff [imath]\beta X[/imath]
The bellow example show that the compact Hausdorff [imath]\beta X[/imath] does not have non-trivial subsequence: Assume [imath]R=(\mathcal{U}_n)_n[/imath] is a sequence of distinct ultrafilters on some set [imath]X[/imath]. Since every Hausdorff space has an infinite discrete subspace, there is a subsequence [imath]R=(\mathcal{V}_n)_n[/imath] of [imath]R=(\mathcal{U}_n)_n[/imath] such that [imath]\{\mathcal{V}_n|n \in \mathbb{N} \}[/imath] is a discrete subspace of [imath]\beta X[/imath]. In particular, there is a sequence [imath](A_n)_n[/imath] of sets with [imath]A_n \in \mathcal{V}_m[/imath] if and only if [imath]m =n[/imath]. If we let [imath]B_n=A_n - (A_0 \cup...\cup A_{n−1})[/imath], then the sequence [imath](B_n)_n[/imath] is pairwise disjoint and [imath]B_n \in \mathcal{V}_m[/imath] iff [imath]m=n[/imath]. If we let [imath]B =\cup_n B_{2n}[/imath], then [imath]B \in \mathcal{V}_m[/imath] if and only if [imath]m[/imath] is even. In other words, if [imath]B =\{ \mathcal{V} \in βX|B \in \mathcal{V}\}[/imath], then [imath]B[/imath] is a clopen set with [imath]\mathcal{V}_m \in B[/imath] iff [imath]m[/imath] is even. Therefore, we conclude that the sequence [imath](\mathcal{V}_m)_m[/imath] cannot converge to any point, so the sequence [imath](\mathcal{U}_n)_n[/imath] cannot converge to any point either. I would like to know that: (1) :What does it mean "sequence of distinct ultrafilters on some set [imath]X[/imath]"?I mean, is it possible to define different ultrafilters on some set [imath]X[/imath]? (2) Why "[imath]B_n \in \mathcal{V}_m[/imath] iff [imath]m=n[/imath]"? And " [imath]B =\cup_n B_{2n}[/imath], then [imath]B \in \mathcal{V}_m[/imath] if and only if [imath]m[/imath] is even "? (3)[imath]B[/imath] is a clopen set with [imath]\mathcal{V}_m \in B[/imath] iff [imath]m[/imath] is even | 522514 | Question about Stone-Čech compactification
Assume [imath]R=(\mathcal{U}_n)_n[/imath] is a sequence of distinct ultrafilters on some set [imath]X[/imath]. Since every Hausdorff space has an infinite discrete subspace, there is a subsequence [imath]R=(\mathcal{V}_n)_n[/imath] of [imath]R=(\mathcal{U}_n)_n[/imath] such that [imath]\{\mathcal{V}_n|n \in \mathbb{N} \}[/imath] is a discrete subspace of [imath]\beta X[/imath]. In particular, there is a sequence [imath](A_n)_n[/imath] of sets with [imath]A_n \in \mathcal{V}_m[/imath] if and only if [imath]m =n[/imath]. If we let [imath]B_n=A_n - (A_0 \cup...\cup A_{n−1})[/imath], then the sequence [imath](B_n)_n[/imath] is pairwise disjoint and [imath]B_n \in \mathcal{V}_m[/imath] iff [imath]m=n[/imath]. If we let [imath]B =\cup_n B_{2n}[/imath], then [imath]B \in \mathcal{V}_m[/imath] if and only if [imath]m[/imath] is even. In other words, if [imath]B =\{ \mathcal{V} \in βX|B \in \mathcal{V}\}[/imath], then [imath]B[/imath] is a clopen set with [imath]\mathcal{V}_m \in B[/imath] iff [imath]m[/imath] is even. Therefore, we conclude that the sequence [imath](\mathcal{V}_m)_m[/imath] cannot converge to any point, so the sequence [imath](\mathcal{U}_n)_n[/imath] cannot converge to any point either. I would like to know that: (1) Why "[imath]B_n \in \mathcal{V}_m[/imath] iff [imath]m=n[/imath]"? And " [imath]B =\cup_n B_{2n}[/imath], then [imath]B \in \mathcal{V}_m[/imath] if and only if [imath]m[/imath] is even "? (2)[imath]B[/imath] is a clopen set with [imath]\mathcal{V}_m \in B[/imath] iff [imath]m[/imath] is even |
1140903 | Prove that [imath]\sum_{n=1}^{\infty}\frac{|\cos n|}{n}[/imath] diverges
Prove that the series [imath]\displaystyle \sum_{n=1}^{\infty}\frac{|\cos n|}{n}[/imath] diverges. I knew that the series [imath]\displaystyle \sum_{n=1}^{\infty}\frac{|\sin n|}{n}[/imath] diverges, but it seems the proof for the case of sine cannot be applied to the case of cosine. Thanks so much. | 396000 | Methods for determining the convergence of [imath]\sum\frac{\cos n}{n}[/imath] or [imath]\sum\frac{\sin n}{n}[/imath]
As far as I know, the textbook approach to determining the convergence of series like [imath]\sum_{n=1}^\infty\frac{\cos n}{n}[/imath] and [imath]\sum_{n=1}^\infty\frac{\sin n}{n}[/imath] uses Dirichlet's test, which involves bounding the partial sums of the cosine or sine terms. I have two questions: Are there any other approaches to seeing that these series are convergent? I'm mostly just interested to see what other kinds of arguments might be made. What's the best way to show that these two series are only conditionally convergent? I don't even know the textbook approach to that question. |
1141247 | Is [imath]\Bbb Q[/imath] homeomorphic to [imath]\Bbb Q^2[/imath]?
It's an easy excercise in set theory to exhibit a bijection [imath]\Bbb Q \cong \Bbb Q\times \Bbb Q[/imath]. However, none of the bijections I'm aware of respect the topologies on [imath]\Bbb Q[/imath] and [imath]\Bbb Q^2[/imath], inherited from their respective embeddings into [imath]\Bbb R[/imath] and [imath]\Bbb R^2[/imath]. Therefore, I'm asking whether there exists a homeomorphism [imath]\phi: \Bbb Q^2 \to \Bbb Q[/imath]. I don't believe that there is any such map, but since the standard techniques of algebraic topology don't enable one to discern between [imath]\Bbb Q[/imath] and a discrete space, I wasn't able to prove it. Maybe Cech cohomology provides a means to attack this problem, but I haven't even got the slightest Iiea how to calculate [imath]H^1(\Bbb Q,\Bbb Z)[/imath]. | 559193 | Why this two spaces do not homeomorphic?
Consider [imath]\Bbb Q[/imath] with subspace topology and [imath]\Bbb Q\times \Bbb Q[/imath] with product topology. Why this two spaces are not homeomorphic?([imath]\Bbb Q[/imath] is the rational numbers) |
1141179 | Finding the volume of a torus
A torus (a doughnut-shaped object) is formed by revolving the circle [imath]x^2 + y^2 = a^2[/imath] about the vertical line [imath]x = b[/imath], where [imath]0 < a < b[/imath]. Find its volume. I have tried to create an integral that makes sense but I keep getting it wrong. We are learning shell method and disk method if that helps. I'm just stuck at where I should even start with this problem. Any help would be appreciated! | 104876 | Volume of the torus with an integral
I am stuck on the following question: Find the volume of the torus obtained by rotating the circle [imath](x-a)^2+y^2=b^2[/imath], where [imath]a>b[/imath], around the y-axis. |
1141164 | [imath]x^4-x \in Z(R)[/imath] implies commutativity
Let [imath]R[/imath] be a ring and [imath]Z(R)[/imath] the centre of [imath]R[/imath]. There exist elementary proofs (that is, proofs not using the structure theory of rings) of the fact that if [imath]x^n-x \in Z(R)[/imath], then [imath]R[/imath] is commutative for [imath]n=2,3[/imath]. However I am not aware of such a proof in the case of [imath]n=4[/imath]. Does anybody know of an elementary proof of this result? Incidentally, I imagine that this is the sort of question that an automated proof system might be well suited to tackle, but I know nothing about these things. Would this be an easy thing for a system like Coq to deal with? Edit: This question being has been marked as a duplicate of this one. However, isn't [imath]x^4-x \in Z(R)[/imath] for all [imath]x[/imath] a weaker condition than [imath]x^4=x[/imath] for all [imath]x[/imath]? | 942189 | Rings with [imath]a^5=a[/imath] are commutative
Let [imath]R[/imath] be a ring such that [imath]a^5=a[/imath] for all [imath]a \in R[/imath]. Then it follows that [imath]R[/imath] is commutative. This is part of a more general well-known theorem by Jacobson for arbitrary exponents ([imath]a^n=a[/imath]), which appeared on math.stackexchange a couple of times. But what I would like to see is a proof which is (1) direct / elementary and (2) in equational language (in particular, first-order). (As far as I know, no published proof of Jacobson's Theorem is equational, but such a proof has to exist). See here and there for such proofs for the exponents [imath]3[/imath] and [imath]4[/imath]. I hope that these proofs make the "rules" clear. What I've done so far: Reduced to the case that [imath]R[/imath] has characteristic [imath]5[/imath]. Any element of the form [imath]a^4[/imath] is central, because it is idempotent and [imath]R[/imath] is reduced. Also, it suffices to prove [imath](ab)^2= b^2 a^2[/imath] for [imath]a,b \in R[/imath]. In fact, this implies [imath]ab=(ab)^5=a(ba)^4 b = (ba)^4 ab = b a b (ab)^2 a^2 b[/imath] [imath] = b a b (b^2 a^2) a^2 b = b a b^3 a^4 b = b a a^4 b^3 b = b a b^4 = b b^4 a = ba.[/imath] Perhaps someone can feed automatic theorem provers with this problem. But my experience is that their proofs are not so easy to follow, quite long and not intuitive. So I am actually looking for hand-made proofs which eventually might also work for rings satisfying [imath]a^p=a[/imath] where [imath]p[/imath] is any prime. Edit. Here is some progress, inspired by the proof by Nagahara and Tominaga. Consider [imath]a \in R[/imath] and decompose the finite commutative reduced ring [imath]\langle a \rangle[/imath] into a product of fields - this is just the Chinese Remainder Theorem. Hence, [imath]a=a_1+\dotsc+a_n[/imath] where each [imath]\langle a_i \rangle[/imath] is a field with unit [imath]e_i=a_i^4[/imath]. The only fields satisfying the equation are [imath]\mathbb{F}_2[/imath], [imath]\mathbb{F}_3[/imath] and [imath]\mathbb{F}_5[/imath], which are prime fields. Hence, [imath]a_i = z_i e_i[/imath] for some [imath]z \in \mathbb{Z}[/imath] is central and therefore [imath]a[/imath] is central. (This incredibly quick proof for the [imath](a^n=a)[/imath]-problem works for all exponents [imath]n[/imath] where every prime power [imath]q[/imath] such that [imath]q-1|n-1[/imath] is actually a prime number.) 2nd Edit. This leads to an equational proof. For simplicity, let us assume that [imath]R[/imath] has characteristic [imath]5[/imath]. (This reduction is not in equational language, but the cases of characteristic [imath]2[/imath] and [imath]3[/imath] can be dealt separately and in the end everything may be put into a single equational proof.) Then [imath]\langle a \rangle[/imath] is a quotient of [imath]\mathbb{F}_5[T]/(T^4-1)[/imath], which is isomorphic to [imath]\mathbb{F}_5 \times \mathbb{F}_5 \times \mathbb{F}_5 \times \mathbb{F}_5[/imath] via [imath]f \mapsto (f(1),f(2),f(3),f(4))[/imath]. We can compute the corresponding idempotents: [imath]e_1 = (t-2)^2 (t-3)(t-4), e_2 = (t-1) (t-3) (t-4)^2, \\e_3 = (t-1)^2 (t-2) (t-4), e_4 = (t-1) (t-2) (t-3)^2.[/imath] Back in [imath]R[/imath] we obtain the equation [imath]a = e_1(a) + 2 e_2(a) + 3 e_3(a) + 4 e_4(a).[/imath] Each [imath]e_i(a)[/imath] is idempotent, hence central, and therefore [imath]a[/imath] is central. This proof may be carried out in equational language - but then we would have to verify the equation and that the [imath]e_i(a)[/imath] are idempotent, which is obviously very tedious. Is there also a proof which is (3) not tedious? |
1141239 | How many numbers from 1 to 2^n have '11' as a substring in binary representation?
For example: say [imath]n = 2[/imath]. The numbers from [imath]1[/imath] to [imath]2^2[/imath] are [imath]1, 2, 3, 4[/imath]. i.e. [imath]1, 10, 11, 100[/imath] in binary. So the result is [imath]1[/imath], because only one number i.e. [imath]3[/imath] is there such that it has 11 in it. For [imath]n = 3[/imath], [imath]3, 6, 7[/imath] have '11', so the result is [imath]3[/imath]. | 1140898 | How many numbers from [imath]1[/imath] to [imath]2^n[/imath] will have [imath]``11"[/imath] as substring in binary representation?
For example say, [imath]n = 2[/imath]. So our set is [imath]\{1, 2, 3, 4\}[/imath] in base [imath]10[/imath] and [imath]\{1, 10, 11, 100\}[/imath] in base [imath]2[/imath]. So Output [imath]1[/imath], because only one number i.e. [imath]3[/imath] is there such that it has [imath]``11"[/imath] in it. For [imath]n = 3[/imath], we get [imath]\{3, 6, 7\}[/imath] as having [imath]``11"[/imath], so output [imath]3[/imath]. |
1142074 | Analysis Question Involving Real Numbers and Sets
I've been sick and I've missed a couple of lectures Analysis. But I didn't want to be too behind in lecture, so I was trying to catch up by reading my textbook and solving some problems. Here's a question I was having some trouble with: Prove [imath]\sqrt{2} \in \mathbb{R}[/imath] because [imath]x^2=2[/imath] where [imath]x=A\mid B[/imath] is a cut in [imath]\mathbb{Q}[/imath] with [imath]A=\{r\in \mathbb{Q}: r\le 0 \text{ or } r^2<2\}.[/imath] Any help would be appreciated! Thank you! :) | 1141559 | Analysis Question Involving Real Numbers
So I've been sick and I've missed a couple of lectures of my Analysis class. But I didn't want to be too behind in lecture tomorrow, so I was trying to catch up by reading my textbook and solving some problems. Here's one I was having some trouble with: Prove [imath]\sqrt2\in\mathbb R[/imath] by showing [imath]x^2 = 2[/imath] where [imath]x = A|B[/imath] is a cut w/ [imath]A = \{r \in \mathbb Q : r \leqslant 0 \text{ or } r^2 < 2\}[/imath]. Any help would be appreciated! Thank you! :) |
1141874 | Prove that [imath]\sin x, if x>0[/imath]
Prove that [imath]\sin x<x[/imath], if [imath]x>0[/imath]. May it seems a silly question, but I don't know how to prove it analytically (i.e., without comparing graph of [imath]\sin x[/imath] and [imath]x[/imath]) and without more advanced topics, e.g. Taylor series, or even derivation. Please help me how can I prove it for a middle-school student, using definition of [imath]\sin x[/imath] or maybe some trigonometric identities. Thank you. | 125298 | how to strictly prove [imath]\sin x for 0[/imath]
[imath]\sin x<x\,(0<x<\frac{\pi}{2})[/imath] In most textbooks, to prove this inequality is based on geometry illustration (draw a circle, compare arc length and chord ), but I think that strict proof should be based on analysis reasoning without geometry illustration. Who can prove it? Thank you very much. ps: By differentiation, monotonicity and Taylor formula, all are wrong, because [imath](\sin x)'=\cos x[/imath] must use [imath]\lim_{x \to 0}\frac{\sin x}{x}=1[/imath], and this formula must use [imath]\sin x< x[/imath]. This is vicious circle. If we use Taylor series of [imath]\sin x[/imath] to define [imath]\sin x[/imath], strictly prove [imath]\sin x<x[/imath] is very easy, but how can we obtain geometry meaning of [imath]\sin x[/imath]? |
1142378 | Number of natural solutions to the equation [imath]x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}=1000000[/imath]
I'm trying to tackle some question and I think I solved it, but I'm not sure and would like someone to check my work. The question goes like this: Find the number of non-negative solutions of the equation [imath]x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}=1000000[/imath]. My try: First I have written the equation as [imath]x_1\cdot{x_2}\cdot{x_3}\cdot{x_4}=10^6=2^6\cdot{5^6}[/imath], hence [imath]x_i[/imath] is of the form [imath]\displaystyle 2^{a_i}\cdot{5^{b_i}}[/imath] where [imath]0\le a_i,b_i \le 6[/imath]. Now we can write [imath]2^{a_1}5^{b_1}\cdot{2^{a_2}5^{b_2}}\cdot{2^{a_3}5^{b_3}}\cdot{2^{a_4}5^{b_4}}=2^{a_1+a_2+a_3+a_4}\cdot{5^{b_1+b_2+b_3+b_4}}=2^6\cdot{5^6}[/imath] Thus, we are looking to find the number of non-negative solutions to the equations [imath]\begin{cases}a_1+a_2+a_3+a_4=6\\b_1+b_2+b_3+b_4=6\end{cases}[/imath] We know that [imath]0 \le a_i,b_i \le 6[/imath], hence the generating function would be [imath]g(x)=(1+x+...+x^6)^4=\left(\frac{1-x^7}{1-x}\right)^4=(1-x^7)^4\cdot{\frac{1}{(1-x)^4}}\\g(x)=(1-x^7)^4\cdot{\sum_{n=0}^{\infty}{{n+3}\choose 3}x^n}[/imath] Now, [imath]\displaystyle (1-x^7)^4=x^0-4x^7+6x^{14}-4x^{21}+x^{28}[/imath], hence the only relavant term is [imath]x^0[/imath] (we want to find the coefficient of [imath]x^6[/imath]). We find that the required coefficient is [imath]\displaystyle {9\choose 3}=84[/imath], so we have 84 optional solutions to any of the equations. Edit: There is no dependence between the equations, hence the number of solutions is [imath]84^2[/imath]. Is my reasoning correct? Please help me fix my errors if there any. Thanks! | 298755 | the number of positive-integer solutions that satisfy [imath]x_1\cdot x_2\cdot x_3\cdot x_4=1,000,000[/imath]?
I can't solve the following: Find the number of positive-integer solutions that satisfy [imath]x_1\cdot x_2\cdot x_3\cdot x_4=1,000,000[/imath]. Thanks. |
1142939 | Primitive roots modulo primes congruent to n!
for [imath]N \ge 4[/imath]. Show for prime numbers, [imath]p \equiv 1[/imath] mod [imath](N!)[/imath] that none of the numbers [imath]1,2,...,N[/imath] are primitive roots modulo [imath]p[/imath] I can't figure out where to start with this question, all I can think to use is the Legendre symbol and Euler's Criterion but I haven't been able to do it. Any help would be much appreciated. | 1143131 | If [imath]p \equiv 1 \pmod{N!}[/imath] prove there are no primitive roots [imath]\pmod p[/imath], [imath]g[/imath], that are less than [imath]N[/imath]
I've recently been looking at different questions and proofs in my book and one eludes me for the 2nd day in the row. Let [imath]N \geq 4[/imath]. Show that if [imath]p[/imath] is a prime such that [imath]p \equiv 1 \pmod{N!}[/imath] then none of the numbers [imath]1, 2, \ldots, N[/imath] are primitive roots modulo [imath]p[/imath]. I've tried using CRT, the fact that [imath]\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)[/imath] and Gaussian sum [imath]G(p)[/imath] properties, yet to no avail. This proof has been really bugging me, could you please point me in the right direction? |
1143519 | [imath]F[/imath] is isomorphic to [imath]\Bbb Z_p[/imath] for some prime number [imath]p[/imath].
Suppose [imath]F[/imath] is a field and there is a ring homomorphism from [imath]\Bbb Z[/imath] onto [imath]F[/imath]. Then show that [imath]F[/imath] is isomorphic to [imath]\Bbb Z_p[/imath] for some prime number [imath]p[/imath]. I am facing difficulty to do the proof. I think that we have to make a map so that the kernel of the homomorphism goes to class [imath]0[/imath] of [imath]\Bbb Z_p[/imath]. | 1139913 | Let [imath]F[/imath] be a field and [imath]f: \mathbb{Z} \to F[/imath] be a ring epimorphism.
Here we have [imath]F[/imath] a field and [imath]f: \mathbb{Z} \to F[/imath] a ring epimorphism. We are to prove that [imath]F[/imath] is a finite field with non-zero characteristic. I know that since [imath]f[/imath] is an epimorphism, we have that [imath]\forall x\in F[/imath], [imath]\exists y \in \mathbb{Z}[/imath] such that [imath]f(x) = y[/imath]. (As an epimorphism is a surjective ring homomorphism). I am unsure about two things: Am I to prove that F is finite, or just prove that it has a non-zero characteristic. Either way, I could use some guidance. Thank you :) PS This is for HW, so I only require a good hint about how to approach the problem. |
1143689 | Show [imath]\operatorname{Hom}_{\mathbb{Z}}\left ( \mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z} \right )\cong \mathbb{Z}/\left ( n,m \right )\mathbb{Z}[/imath]
Show that [imath]\operatorname{Hom}_{\mathbb{Z}}\left ( \mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z} \right )\cong \mathbb{Z}/\left ( n,m \right )\mathbb{Z}[/imath] I think that the hom-set (of [imath]\mathbb{Z}[/imath] module homomorphisms ) is isomorphic to [imath]\left \{ a\in \mathbb{Z}/m\mathbb{Z},na=m\mathbb{Z} \right \}=\left \{ k+m\mathbb{Z} \right \}[/imath] where [imath]m\mid (nk)[/imath] but I can't show that it's isomorphic to [imath]\mathbb{Z}/\left ( n,m \right )\mathbb{Z}[/imath] | 955678 | Describe explicitly [imath]\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m)[/imath].
Describe explicitly [imath]\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m) := \{\varphi:\mathbb{Z}_n \rightarrow \mathbb{Z}_m \mid \mathbb{Z}\text{-linear homomorphism}\}[/imath] There is an answer to this question that says [imath]\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m) \cong \mathbb{Z}_{(n,m)}[/imath] where [imath](n,m)=\gcd(n,m)[/imath]. But I am having some trouble seeing this. For the sake of an argument, suppose the above answer is true, that [imath]\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m) \cong \mathbb{Z}_{(n,m)}[/imath]. Suppose [imath]n=4[/imath] and [imath]m=8[/imath]. Then [imath]\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_4, \mathbb{Z}_8) \cong \mathbb{Z}_4[/imath]. Now consider each of the following maps in [imath]\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_4, \mathbb{Z}_8)[/imath]: [imath]\varphi_1(1)=1[/imath] [imath]\varphi_2(1)=2[/imath] [imath]\varphi_3(1)=3[/imath] [imath]\varphi_4(1)=4[/imath] [imath]\varphi_5(1)=5[/imath] [imath]\varphi_6(1)=6[/imath] [imath]\varphi_7(1)=7[/imath] Are each of these maps not unique? If there are two that are identical, could you please explain which two and why? Thank you! My alternate answer is that [imath]\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m) \cong \mathbb{Z}_m[/imath], since I think each [imath]\varphi \in \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m)[/imath] is uniquely determined by where it maps [imath]\varphi(1) \in \mathbb{Z}_m[/imath]. |
1143578 | Equilateral triangle in a plane
If I have the plane [imath]\pi:x+y-z-1=0[/imath] and the two points [imath]A:(1,1,1), B:(2,1,2)[/imath] as two vertices of an equilateral triangle in the plane [imath]\pi[/imath]. How can I find all (I'm assuming only 2?) sets of coordinates for the vertex [imath]C[/imath]. Note that [imath]A,B,\text{ and }C[/imath] lies in [imath]\pi[/imath]. I've noted that [imath]\vec{AB} = (1,0,1)\iff ||\vec{AB}||=\sqrt2[/imath] Let [imath]Q[/imath] be midpoint for [imath]\vec{AB}[/imath], i.e. its coordinates are [imath]\frac{1}{2}(A+B) = (\frac{3}{2},1,\frac{3}{2})[/imath]. The plane has the normal vector [imath]\vec{n} = (1,1,-1)[/imath] [imath]\therefore[/imath] The line which goes through [imath]Q[/imath] and [imath]C[/imath] is orthogonal to [imath]\vec{n}[/imath] and [imath]\vec{AB}[/imath]. We call this line [imath]L[/imath]. [imath]\vec{n} \times \vec{AB} = (1,-2,-1)[/imath] Using this and that the point [imath]Q[/imath] lies on [imath]L[/imath] we get the equation for [imath]L:[/imath] [imath]L = (\frac{3}{2}+t,\,1-2t,\,\frac{3}{2}-t)[/imath] So combining this with the fact that [imath]||\vec{QC}||=\frac{\sqrt{3}}{2}||\vec{AB}|| = \sqrt{\frac{3}{2}}[/imath] as [imath]||\vec{AB}|| = \sqrt{2}[/imath]. We can now put [imath]||L(t)|| = ||\vec{QC}||[/imath]. [imath]||L(t)|| = \sqrt{\left(\frac{3}{2}+t\right)^2 + \left(1-2t\right)^2 + \left(\frac{3}{2}-t\right)^2} = \sqrt{6t^2-4t+\frac{11}{2}} = ||\vec{QC}|| = \sqrt{\frac{3}{2}}[/imath] [imath]\therefore 6t^2-4t+\frac{11}{2} = \frac{3}{2}[/imath] This equation has no real solutions for [imath]t[/imath]. What am I to do with this result? Did I do something wrong, or must the solutions contain imaginary units? | 1142012 | Linear algebra - find all possible positions of the third corner?
An equilateral triangle lies in the plane [imath]x + y - z = 1[/imath] and corners in points [imath](1, 1, 1)[/imath] and [imath](2, 1, 2)[/imath]. Determine all possible positions of the third corner? |
1028329 | Continuity of Inner product
Let [imath]f[/imath] and [imath]g[/imath] be continuous mapping of [imath]S \subseteq \mathbb{R}^n[/imath] into [imath]\mathbb{R}^m[/imath]. Show that the inner product [imath]h(x) = \langle f(x),g(x)\rangle[/imath] is continuous. | 284494 | Inner product of two continuous maps is continuous
Let f,g be continuous maps of S in [imath]\mathbb{R}^n[/imath], to [imath]\mathbb{R}^m[/imath]. Show the inner product h(x) = [imath]\langle f(x), g(x)\rangle[/imath] is continuous. My attempt: Since f,g are continuous, their product is a continuous mapping, so for [imath]\epsilon \gt 0\: t.e\: \delta \gt 0[/imath] such that [imath]\parallel x - a\parallel \lt \delta [/imath] implies [imath]\parallel \langle f(x),g(x)\rangle - \langle f(a)g(a)\rangle \parallel \le\parallel f(x)g(x) - f(a)g(a)\parallel \lt \epsilon[/imath] by schwartz inequality. |
1120966 | How do I specify a function without a defined argument?
A function [imath]f[/imath] with the argument [imath]x[/imath] is commonly written [imath]f_x : A\to B, x\mapsto f(x)[/imath], or [imath]f_x : \mathbb{R} \to \mathbb{R}, x\mapsto x^2[/imath], but say I don't want to specify the argument, how would I write this ? [imath]f : \cdot\to\cdot, \mapsto ^2[/imath] seems unnatural. | 82473 | Notation for functions vs. numbers
I understand that [imath]f[/imath] represents a function while [imath]f(x)[/imath] represents the value of a function, but while I can easily see how to apply this convention to work with functions instead of numbers in some circumstances, e.g., [imath]\sin(x)\text{ becomes } \sin\text{,}[/imath] [imath]\cos(x)\text{ becomes } \cos\text{, and}[/imath] [imath]g(x)\text{ becomes } g\text{,}[/imath] it isn't clear to me how to to express the functions corresponding to the numbers [imath]x^{2}\text{, or}[/imath] [imath]2\cdot x\text{.}[/imath] Short of defining new symbols, like [imath]S[/imath] or [imath]I[/imath] (which I've seen for the function corresponding to [imath]x \mapsto x[/imath]), can one say things like [imath]\left(\cdot\right)^{2}\text{ for }x^{2}\text{, or}[/imath] [imath]2\cdot \left(\cdot\right)\text{ or }2\left(\cdot\right)\text{ or }2\cdot\text{ for }2\cdot x\text{?}[/imath] What is the general idiom for expressing such functions concisely? |
1143683 | Finding the inverse of an element of [imath]S_n[/imath] and it's order
I have two questions, 1) What are the ways to find the inverse of an element of [imath]S_n[/imath]? 2) What are the ways to find the order of an element of [imath]S_n[/imath]? | 973250 | Computing the inverse of a permutation
I didn't understand the permutation and of course, I got this question wrong. Compute the inverse of the following permutation: [imath] \begin{pmatrix} 1&2&3&4&5&6\\ 2&5&4&1&6&3 \end{pmatrix} [/imath] Could you please help me with this question and provide an explanation? Furthermore, is it possible to compute the direct permutation, not the inverse one? If yes, how? Thank you in advance! |
1144385 | finding the center of a circle given 3 points
Suppose we have three complex numbers [imath]a_1,a_2,a_3[/imath] which are non-collinear. What is the best way to find the center of the circle that contains these three points ? | 213658 | Get the equation of a circle when given 3 points
Get the equation of a circle through the points [imath](1,1), (2,4), (5,3) [/imath]. I can solve this by simply drawing it, but is there a way of solving it (easily) without having to draw? |
1144436 | Midpoint with a connected subset and continuous function
Given a connected subset [imath]A[/imath] of [imath]\mathbb{R}^p[/imath] and a continuous function [imath]f:A\to \mathbb{R}[/imath]. If I have points [imath]b,c \in A[/imath] and a point [imath]x \in \mathbb{R}[/imath] that satisfies [imath]f(a) < x < f(b)[/imath], how do I show that there is a point [imath]c \in A[/imath] such that [imath]f(c) = x[/imath]? | 1129766 | Continuous function from a connected set?
If [imath] f [/imath] is a continuous mapping from a connected set to the real numbers and there exists a real number s that nothing maps to, then the image is either greater than or less than s. this is clear to me intuitively but I can't think of how to prove it. |
1144643 | Can you help me solve this series: [imath]\sum_{k=0}^\infty p^kk[/imath]?
I have the infinite series: [imath]\sum_{k=0}^\infty p^kk[/imath] where [imath]0<p<1[/imath]. I can see from a computer calculation that the series converges to [imath]\frac{p}{(1-p)^2}[/imath], but I can't see why. Thanks in advance for any help. | 352150 | Differentiating an infinite sum
Some other summation formulas can be obtained by differentiating the above equations on both sides. Assume [imath]|r|<1[/imath]. Show that [imath] a+2ar+3ar^2+\cdots = \frac{a}{(1-r)^2} [/imath] by starting with the geometric series formula. This seems to be trivial to prove by differentiation of both sides of the infinited geometric series formula. Is this a legal operation? |
1143012 | Modular Exponentiation 8^5^4
I am trying to find the last digit of [imath]8^{5^4}[/imath] (or [imath]8^{(5^4)}[/imath], if you will) using modular exponentiation. What I know is that the value I want is: [imath]8^{5^4} \mod 10[/imath]. Normally I would find a pattern that [imath]8^{n} \mod 10[/imath] depends upon and use that information to know how to approach the [imath]5^{4}[/imath] term, but [imath]8^{n} \mod 10[/imath] doesn't seem to have a useful pattern. I know that the answer is [imath]8[/imath] (via WolframAlpha) but I don't know how to connect the dots here. Any help would be appreciated. Thanks! | 1142941 | Modular Exponentiation 3^5^7
I am trying to find the last digit of [imath]3^{5^7}[/imath] (or [imath]3^{(5^7)}[/imath], if you will) using modular exponentiation. Here's what I've figured out: The value I want is [imath]3^{5^7} \mod 10[/imath]. [imath]5^n \mod 10 = 5[/imath] (if [imath]n[/imath] is a positive natural number). The residues of [imath]3^n \mod 10[/imath] repeat themselves every time [imath]n[/imath] increases by [imath]4[/imath]. The pattern is [imath]1, 3, 9, 7, 1, 3, 9, 7, \ldots[/imath] for [imath]n = 0, 1, 2, 3, 4, 5, 6, 7, \ldots[/imath] Now, I don't know how to use these facts to get what I need. I know that the answer [imath]3[/imath] (via WolframAlpha) but I don't know how to connect the dots here. Any help would be appreciated. Thanks! |
880589 | Minimal polynomial of a matrix whose elements have a certain form
Find the minimal polynomial of the [imath]n[/imath]-dimensional matrix [imath](a_{ij})[/imath] when the matrix elements [imath]a_{ij}[/imath] have the form [imath]a_{ij} = u_i v_j.[/imath] Let [imath]A=uv^T[/imath] where [imath]u,v[/imath] are column vectors. Then rank[imath](A)\leq[/imath]rank[imath](u)\leq1.[/imath] So kernal[imath](A)\geq n-1.[/imath] That is, the geometric multiplicity [imath]\geq n-1.[/imath] According to Jordan decomposition theorem, the number of Jordan blocks w.r.t. [imath]0\ \geq n-1.[/imath] Therefore, the algebraic multiplicity of [imath]0[/imath] [imath]\geq n-1.[/imath] Suppose rank[imath](A)=1,[/imath] how do I find the other eigenvalue? | 52395 | Determining possible minimal polynomials for a rank one linear operator
I have come across a question about determining possible minimal polynomials for a rank one linear operator and I am wondering if I am using the correct proof method. I think that the facts needed to solve this problem come from the section on Nilpotent operators from Hoffman and Kunze's "Linear Algebra". Question: Let [imath]V[/imath] be a vector space of dimension [imath]n[/imath] over the field [imath]F[/imath] and consider a linear operator [imath]T[/imath] on [imath]V[/imath] such that [imath]\mathrm{rank}(T) = 1[/imath]. List all possible minimal polynomials for [imath]T[/imath]. Sketch of Proof: If [imath]T[/imath] is nilpotent then the minimal polynomial of [imath]T[/imath] is [imath]x^k[/imath] for some [imath]k\leq n[/imath]. So suppose [imath]T[/imath] is not nilpotent, then we can argue that [imath]T[/imath] is diagonalizable based on the fact that [imath]T[/imath] must have one nonzero eigenvalue otherwise it would be nilpotent (I am leaving details of the proof of diagonalization but it is the observation that the characteristic space of the nonzero eigenvalue is the range of T and has dimension [imath]1[/imath]). Thus the minimal polynomial of [imath]T[/imath] is just a linear term [imath]x-c[/imath]. Did I make a mistake assuming that [imath]T[/imath] can have only one nonzero eigenvalue? Thanks for your help |
1144918 | Dimension of the vector space given by the quotient of an Artin ring by the product of all its maximal ideals
Let [imath]\mathcal{M}_1,\dots,\mathcal{M}_r[/imath] be all the maximal ideals of an Artin ring [imath]A[/imath] which is a finite [imath]\mathbb{K}[/imath]-algebra; so let [imath]A/\mathcal{M}_1\cdots\mathcal{M}_r[/imath] be a [imath]\mathbb{K}[/imath]-vector space. Which is its dimension? Is it a general case that the number of maximal ideals in [imath]A[/imath] is equal to [imath]\operatorname{dim}_{\mathbb{K}} A+1[/imath]? | 919463 | A finite dimensional algebra over a field has only finitely many prime ideals and all of them are maximal
Let [imath]K[/imath] be a field and let [imath]R[/imath] be a [imath]K[/imath]-algebra with unity which is finite dimensional as a [imath]K[/imath]-vector space. Prove that [imath]R[/imath] has only finitely many prime ideals all of which are maximal. (Hint: Chinese remainder theorem) My attempt: We have [imath]R \simeq K^n[/imath], for some [imath]n[/imath], as vector spaces. I claim that there are finitely many ideals of [imath]K^n[/imath]. Let [imath]I[/imath] be an ideal of [imath]K^n[/imath] and [imath](a_1, a_2,...,a_n) \in I[/imath]. For every entry [imath]a_i[/imath], we have two choices, either [imath]a_i=0[/imath] or not. So we get [imath]2^n[/imath] many cases, and any case give a particular ideal of [imath]K^n[/imath] which are distinct from others. Is my proof correct or am I missing some point? If it is true, how can we give a proof by using the Chinese remainder theorem? Thanks! |
414120 | Two problems about rings.
Somebody can to help me in such exercices: (1) A ring R such that [imath]a^2 = a[/imath] for all [imath]a\in R[/imath] is called a Boolean ring. Prove that every Boolean ring R is commutative and [imath]a + a = 0[/imath] for all [imath]a \in R[/imath]. (2) Let R be a ring with more than one element such that for each nonzero [imath]a\in R[/imath] there is a unique [imath]b \in R[/imath] such that [imath]aba = a[/imath]. Prove: (a) R has no zero divisors. (b) [imath]bab = b[/imath]. (c) R has an identity. (d) R is a division ring. My greatest difficulty in the question 2 is to prove that R has no zero divisors. Thanks! | 785614 | [imath]R[/imath] is a ring. [imath]\forall 0\ne a\in R \exists ! b\in R; aba=a[/imath]. Can I conclude that [imath]R[/imath] is a ring with identity? How?
[imath]R[/imath] is a ring, containing more than one element, such that [imath]\forall 0\ne a\in R \exists ! b\in R; aba=a[/imath]. Here's what I did in the very first place: [imath](aba)a^{-1}=aa^-1 \text{ and } a^{-1}(aba)=a^{-1}a\Longrightarrow \left\{ \begin{array}{rl} ab=1 & \forall0\ne a\in R \exists !b\in R \\ ba=1 & \forall0\ne a\in R \exists !b\in R \end{array} \right.[/imath] Therefore, since [imath]ab\in R\Longrightarrow 1\in R[/imath]. So [imath]R[/imath] is a ring with identity. But then I realized, how one can say [imath]aa^{-1}=1[/imath] when one doesn't know whether [imath]R[/imath] is a ring with identity. |
1145350 | Is [imath]j_n(x)=\frac{x}{1+n^2x^2}[/imath] uniformly convergent on [imath][0,1][/imath]?
Is [imath]j_n(x)=\frac{x}{1+n^2x^2}[/imath] uniformly continuous on [imath][0,1][/imath]? I have basically been working with the following theorem: Let E be a non-empty subset of [imath]\mathbb{R}[/imath] Let [imath]f_n, f: E \rightarrow \mathbb{R}[/imath]. Then the following are equivalent i. [imath]f_n\rightarrow f[/imath] uniformly on E ii. [imath]\exists N[/imath] s.t. [imath]\forall n >N[/imath], [imath]m_n :=sup_{x\in E}|f_n(x)-f(x)|[/imath] exists and [imath]m_n \rightarrow 0[/imath] as [imath]n \rightarrow\infty[/imath] I am not sure whether the given function is increasing or decreasing on the interval since [imath]j_n(x)=\frac{x}{1+n^2x^2}=\frac{1}{\frac{1}{x}+n^2x}[/imath] The reason that I am concerned with the increasing/decreasing behavior of this function is because I am looking for a way to meaningful describe [imath]m_n[/imath]. Another tool I was considering, but am less comfortable with is Cauchy's Criterion. | 614457 | Showing [imath]f_n(x):=\frac{x}{1+n^2x^2}[/imath] uniformly convergent in [imath]\mathbb R[/imath] using [imath]\epsilon-n_0[/imath]
I want to show [imath]f_n(x):=\frac{x}{1+n^2x^2}[/imath] is uniformly convergent in [imath]\mathbb R[/imath] using [imath]\epsilon-n_0[/imath] argument. So [imath]|f_n-f|\leq\frac{|x|}{1+n^2|x|^2}[/imath] I know [imath]|f_n-f|\leq\frac1n[/imath] since [imath]\frac1n[/imath] is the maximum of [imath]f_n[/imath]. I can show this using the derivatives of [imath]f_n[/imath]. My question is: Can you show this or any other equation (independent of [imath]x[/imath]) without using derivatives in any calculation? I've tried [imath]|f_n-f|\leq \frac1n \iff -\frac1n\leq\frac{x}{1+n^2x^2}\leq\frac1n \iff\ldots[/imath] but I get no inequality which is obviously right. |
1145474 | isomorphism between [imath]k[[x]][/imath] into [imath]\varprojlim_n k[x]/(x^n)[/imath]
i want to find isomorphism between [imath]k[[x]][/imath] and [imath]\varprojlim_n k[x]/(x^n)[/imath] but I cant.please help me to find this. | 1145211 | inverse limit of [imath]k[x]/(x^n)[/imath]
I know the inverse limit of [imath]k[x]/(x^n)[/imath] is [imath]k[[x]][/imath] but i can't show the homomorphism between these is onto. For example [imath]\alpha:k[[X]]\to\varprojlim k[x]/(x^n)[/imath] by F(x)=family of[imath] F(x)+(x^n)[/imath]. for example [imath](1+(x),0,5+7x^2+(x^3),3+11x+132x^2+(x^4),0,...)[/imath] is image of things in k[[x]]? |
1145525 | Proving [imath]adj(A)A=det(A)I[/imath]
I am reading a proof of the Cayley-Hamilton theorem which uses the fact that [imath]adj(A)A=det(A)I[/imath], where I is the [imath]n\times n[/imath] identity matrix and [imath]A\in M_n(\mathbb{F})[/imath] for a field [imath]\mathbb{F}[/imath]. Could you indicate me a quick way of proving this or to a reference proving it ? And is is true for a matrix with coefficients over any commutative ring ? I would guess so because C-H is true for any commutative ring. In this case, is the proof the same ? Thanks | 345517 | Why is it true that [imath]\mathrm{adj}(A)A = \det(A) \cdot I[/imath]?
This is a statement in linear algebra that I can't seem to understand the proof behind. For a square matrix [imath]A[/imath], why is: [imath]\mathrm{adj}(A)A = \det(A) \cdot I[/imath] Any explanation would be greatly appreciated. |
152236 | Localisation is isomorphic to a quotient of polynomial ring
I am having trouble with the following problem. Let [imath]R[/imath] be an integral domain, and let [imath]a \in R[/imath] be a non-zero element. Let [imath]D = \{1, a, a^2, ...\}[/imath]. I need to show that [imath]R_D \cong R[x]/(ax-1)[/imath]. I just want a hint. Basically, I've been looking for a surjective homomorphism from [imath]R[x][/imath] to [imath]R_D[/imath], but everything I've tried has failed. I think the fact that [imath]f(a)[/imath] is a unit, where [imath]f[/imath] is our mapping, is relevant, but I'm not sure. Thanks | 1145685 | Showing that [imath]R[X]/(Xf-1) \cong R[1/f][/imath]
Let [imath]R[/imath] be an integral domain with quotient field [imath]K[/imath]. Let [imath]0 \neq f \in R[/imath]. I want to prove Statement: [imath]R[X]/(Xf-1) \cong R[1/f][/imath]. Argument: Consider the epimorphism [imath]\phi: R[X] \rightarrow R[1/f][/imath] which sends [imath]X[/imath] to [imath]1/f[/imath]. The goal is to show that the kernel of [imath]\phi[/imath] is the ideal generated by [imath]Xf-1[/imath]. If [imath]\phi(p[X])=0[/imath] then [imath]p(1/f)=0[/imath]. This shows that [imath]1/f[/imath] is a root of [imath]p(X)[/imath]. Viewing [imath]p[X][/imath] as an element of [imath]K[X][/imath], this implies that there exists some [imath]g(X) \in K[X][/imath] such that [imath]p(X) = (X-1/f) g(X) = (Xf-1) \left[g(X) /f\right][/imath]. I will be done if i can show that [imath]g(X)/f \in R[X][/imath]. Question 1: How to complete the argument? Question 2 (optional): Is there any other proof to the statement? |
1145775 | Similar matrices over [imath]\mathbb{C}[/imath] and [imath]\mathbb{R}[/imath].
Let's assume that real [imath]A,B[/imath] are similar over [imath]\mathbb{C}[/imath]. Is it true that [imath]A, B[/imath] are similar over [imath]\mathbb{R}[/imath]? The first step should look like this: [imath]P=U+iV[/imath], we have [imath]PA=BP[/imath], then [imath](U+iV)(A)=B(U+iV)[/imath], [imath]UA+iVA=BU+iBV[/imath], according to the fact that [imath]A, B, U, V[/imath] are real, we state that [imath]UA=BU, VA=BV[/imath], [imath]A=U^{-1}BU, A=V^{-1}BV[/imath]. Actually, the solution above doesn't work if [imath]U, V[/imath] are not invertible. According to the fact that [imath]P[/imath] is invertible and [imath]P=U+iV[/imath] --- how to prove that [imath]U, V[/imath] are also invertible? | 1129628 | Similarity of real matrices over [imath]\mathbb{C}[/imath]
[imath]A \underset{\mathbb{C}}{\sim} B \overset{\text{def}}{\iff} A=C^{-1}BC, \space C\in M_{n}(\mathbb{C})[/imath] and similarly for [imath]\underset{\mathbb{R}}{\sim}[/imath]. I want to prove that [imath] A \underset{\mathbb{C}}{\sim} B[/imath] for [imath]A,B \in M_{n}(\mathbb{R})[/imath] therefore [imath]A \underset{\mathbb{R}}{\sim} B[/imath]. My idea is that elementary divisors of [imath]A,B[/imath] over [imath]\mathbb{C}[/imath] are the same, and if [imath](x-z)^k[/imath] is elementary divisor than [imath](x-\overline{z})^k[/imath] is also elementary divisor [imath]\implies[/imath] [imath]A,B[/imath] have same elementary divisors over [imath]\mathbb{R}[/imath]. But i think it's not clear. |
1146102 | Sum of [imath]1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots[/imath]
My problem is to find the sum of the series [imath] S = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots [/imath] where the terms are the reciprocals of the positive integers whose only prime factors are [imath]2[/imath]s and [imath]3[/imath]s. I can see that every term in [imath]S[/imath] is of the form [imath]\dfrac{1}{2^m3^n}[/imath] where [imath]m,n\geq 0[/imath]. I can also see that each term only occurs once, but I'm not really sure how to use this information effectively. Any ideas (apparently geometric series was a hint)? | 827361 | What is the sum of this? [imath] 1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\cdots[/imath]
I'm in trouble with this homework. Find the sum of the series [imath] 1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\cdots[/imath], where the terms are the inverse of the positive integers whose only prime factors are 2 and 3. Hint: write the series as a product of two geometric series. Ok, I found some patterns in these numbers, but I can't find the two series. |
525415 | Contraction and Fixed Point
How do I show that for [imath]T: X \rightarrow X[/imath] where X is complete and [imath]T^m[/imath] is a contraction that T has a unique fixed point [imath]x_0 \in X[/imath]. I know there exists [imath]\lambda_1 \in (0,1)[/imath] for [imath]x, y \in X[/imath] such that [imath]d(T^mx, T^my) \leq \lambda_1 d(x, y)[/imath] and I need to show that T is a contraction and then apply the fixed point theorem but how do I do that? | 629098 | If [imath]T^n[/imath] is [imath]q[/imath]-contractive, [imath]T[/imath] exactly has one fixed point
Consider a complete metric space [imath](X,d)[/imath] and [imath]T\colon X\to X[/imath]. Suppose there exists [imath]n\in\mathbb{N}[/imath] such that the n-th power of [imath]T[/imath] is [imath]q[/imath]-contractive. Show that then [imath]T[/imath] has exactly one fixed point [imath]\overline{x}\in X[/imath]. The n-th power is defined inductively: [imath] T^{n+1}(x):=T(T^{n}(x)), n\in\mathbb{N}. [/imath] And for a function [imath]T\colon X\to X[/imath], [imath]q[/imath]-contractive means [imath] \exists 0\leq q<1~\forall~x,y\in X: d(T(x),T(y))\leq q d(x,y). [/imath] Now to the proof. I think I have to show, that [imath]T[/imath] is q-contractive, because then it follows with Banach, that [imath]T[/imath] does have exactly one fixed point. So I have to show, that there exists a [imath]0\leq q<1[/imath] so that for all [imath]x,y\in X[/imath] it is [imath] d(T(x),T(y))\leq q\cdot d(x,y). [/imath] And most likely I have to use the q-contractivity of [imath]T^n[/imath], i.e. that there exists a [imath]0\leq q <1[/imath] so that for all [imath]x,y\in X[/imath] it is [imath] d(T^n(x),T^n(y))\leq q\cdot d(x,y). [/imath] Can you help me? |
1146443 | Is there any connection between the symbol [imath]\supset[/imath] when it means implication and its meaning as superset?
A rather old-fashioned symbol for logical implication is [imath]\supset[/imath] (see list of logic symbols). For example [imath]p \supset q[/imath] means [imath]p \implies q[/imath] or [imath]p \rightarrow q[/imath] in more recent notations. Is there any semantic connection between using the symbol to show implication and using it to represent subset/superset relationships? I can see a connection myself but the problem is I see it in the opposite direction ([imath]\subset[/imath] instead of [imath]\supset[/imath]). For example, I see [imath]x \in A \implies x \in B[/imath] is (somewhat) equivalent to [imath]A \subset B[/imath]. Therefore, if I had to reuse the subset/super set symbol for logical implication, I would have used [imath]p \subset q[/imath] to denote [imath]p \implies q[/imath]. Is there any historical or semantic reason for using [imath]\supset[/imath]? | 1106001 | Why is there this strange contradiction between the language of logic and that of set theory?
In standard probability theory events are represented by sets consisting of elementary events. Consider two events for which (as sets) [imath]A \subset B[/imath]. If an elementary event [imath]x \in A[/imath] takes places then we say that [imath]$B$[/imath] takes place as well. Since [imath]A \subset B[/imath], [imath]x \in A[/imath] implies that [imath]B[/imath] takes place too. It seems that when for sets (representing events) [imath]A \subset B[/imath] [imath]then \ A \ implies \ B .[/imath] On the other hand, if we use the language of logic for the events then [imath]A \supset B[/imath] [imath] \ means\ that \ A\ implies\ B .[/imath] Why is this strange virtual contradiction between the language of sets and the language of logic? (In order to avoid down votes and unplesant comments I reveal that I happen to know that the true translation of the sentence [imath]A \supset B[/imath] of logic to the language of sets (representing events) is [imath]\overline{A \cap \overline B}[/imath].) |
1146887 | Is [imath](\Bbb{Q},+)[/imath] isomorphic to [imath](\Bbb{Q}_{>0},\cdot)[/imath]?
I want to know if [imath](\Bbb{Q},+)\cong (\Bbb{Q}_{>0},\cdot)[/imath], but have no idea where to begin my proof. Can anyone offer a hint? Thanks! | 334686 | Prove that the additive group [imath]ℚ[/imath] is not isomorphic with the multiplicative group [imath]ℚ^*[/imath].
Prove that the additive group [imath]ℚ[/imath] is not isomorphic with the multiplicative group [imath]ℚ^*[/imath]. Prove that [imath]ℚ^*_{>0}[/imath] is not isomorphic with [imath]ℚ[/imath]. |
1147507 | Evaluating [imath]\int\sqrt{1 + \frac1{x^2}}\mathrm dx[/imath]
How do I solve this integral? [imath]\int \sqrt{1 + \frac1{x^2}}\mathrm dx[/imath] I tried substituting [imath]x=\cot t[/imath] but ended up getting another tricky integral, so any help would be appreciated. | 286118 | Integral of [imath]\frac{\sqrt{x^2+1}}{x}[/imath]
So I have to do the integration of [imath]\frac{\sqrt{x^2+1}}{x}[/imath]. Give me a hint. What should I replace? Should I do it with integration by parts? |
1147425 | Fallacy of affirming the conclusion question
I'm a little unsure about the exact reason why/when you get "fallacy of affirming the conclusion" In this simple example: [imath]p\rightarrow q [/imath] [imath]q[/imath] [imath]----[/imath] [imath]\therefore p[/imath] Not exactly sure how I would format that but anyways, for the argument to be valid if [imath]p \rightarrow q[/imath] and [imath]q[/imath] are both true then [imath]p[/imath] must also be true. Is it an invalid statement because if [imath]p[/imath] is false and [imath]q[/imath] is true then [imath]p \rightarrow q[/imath] is true therefore, the fact that [imath]p[/imath] can be either false or true instead of it needing to be true makes it invalid? Also, could someone explain to me why [imath]p \rightarrow (q \rightarrow r)[/imath] [imath]q \rightarrow (p \rightarrow r)[/imath] [imath]-------[/imath] [imath]\therefore (p \vee q) \rightarrow r[/imath] is an invalid statement? | 49993 | Distinguishing between valid and fallacious arguments (propositional calculus)
I am having some difficulties understanding logical arguments. I was taught that the notion of a valid argument is formalized as follows: "An argument [imath]P_1, P_2,\cdots , P_n ⊢ Q [/imath] is said to be valid if [imath]Q[/imath] is true whenever all the premises [imath]P_1, P_2,\cdots , P_n[/imath] are true. An argument which does not satisfy this condition is a fallacy." But (according to my book) the argument [imath]p \rightarrow q,\quad q\;⊢ \;p\; [/imath] is a fallacy, but I don't see why. If we construct a truth table [imath]p[/imath] | [imath]q\;[/imath] | [imath]p \rightarrow q[/imath] | T | T | T | T | F | F | F | T | T | F | F | T | we see that when [imath]p[/imath] and [imath]q[/imath] are true, [imath]p \rightarrow q[/imath] is also true (line 1). So then, why isn't this valid? I do know that for an argument to be valid, with premises [imath]P_1, P_2,\cdots , P_n[/imath], the proposition [imath](P_1 \land P_2 \land \cdots \land P_n) \rightarrow Q[/imath] should be a tautology, but that doesn't dispel my confusion. What am I missing? |
1147747 | Why is [imath]| √2 - 1| = √2-1[/imath] and [imath]|3-π| = π - 3[/imath]?
The absolute value of a number is the distance that number is from [imath]0[/imath] on the real number line. Distances are always positive or [imath]0[/imath]. [imath]|\sqrt 2 - 1| = \sqrt 2-1[/imath]. Why? This is an expression. Should I be setting it to [imath]0[/imath]? However, [imath]|3-π| = π - 3[/imath]. Why? (I have asked my calculus teacher to explain and he has done a terrible job at doing so.) | 1126560 | Why is the Absolute Value of [imath]3-π[/imath] equal to [imath]π - 3[/imath]
Why if the absolute value is the distance from [imath]0[/imath] on the number line, is [imath]\lvert 3-\pi \rvert = \pi - 3[/imath] and [imath]\left\lvert\sqrt{2}-1\right\rvert= \sqrt{2}-1[/imath] |
1147937 | Prove that if [imath] n > 2 [/imath] then between [imath]n[/imath] and [imath]n![/imath] is at least one prime.
Prove that if [imath] n > 2 [/imath] then between [imath]n[/imath] and [imath]n![/imath] is at least one prime. Ok I can see that it's obviously true, but what to use to prove it? | 946517 | Prime number between [imath]n[/imath] and [imath]n!+1[/imath]
I am trying to prove that ([imath]\forall \ n\in\mathbb{N}[/imath]) there exists a prime number [imath]q[/imath] such that [imath]n < q \le 1 + n![/imath] I have made a graph with [imath]n=0[/imath] through [imath]n=10[/imath] and found solutions to all of them looking for a pattern and I see that [imath]n![/imath] gets enormous fast and it becomes quite obvious that there is a prime number in between them. I have considered trying to prove by contradiction that [imath]q[/imath] does not exist on that interval, but I don't know where to go from that statement. Could anybody help me figure it out? I have been staring at it for hours and I can't figure out where to go. Thank you. http://mathforum.org/library/drmath/view/62825.html |
1147987 | How to prove [imath]\sum_{k=0}^{\infty}k^2x^{k} = \frac{x(1+x)}{(1-x)^3}\text{, }|x| < 1[/imath]?
How do I prove that the summation [imath]\sum_{k=0}^{\infty}k^2x^{k} = \dfrac{x(1+x)}{(1-x)^3}\text{, }|x| < 1\text{?}[/imath] | 338852 | Find a closed form of the series [imath]\sum_{n=0}^{\infty} n^2x^n[/imath]
The question I've been given is this: Using both sides of this equation: [imath]\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n[/imath] Find an expression for [imath]\sum_{n=0}^{\infty} n^2x^n[/imath] Then use that to find an expression for [imath]\sum_{n=0}^{\infty}\frac{n^2}{2^n}[/imath] This is as close as I've gotten: \begin{align*} \frac{1}{1-x} & = \sum_{n=0}^{\infty} x^n \\ \frac{-2}{(x-1)^3} & = \frac{d^2}{dx^2} \sum_{n=0}^{\infty} x^n \\ \frac{-2}{(x-1)^3} & = \sum_{n=2}^{\infty} n(n-1)x^{n-2} \\ \frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n(n-1)\frac{x^n}{x}(x+1) \\ \frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} (n^2x + n^2 - nx - n)\frac{x^n}{x} \\ \frac{-2x(x+1)}{(x-1)^3} & = \sum_{n=0}^{\infty} n^2x^n + n^2\frac{x^n}{x} - nx^n - n\frac{x^n}{x} \\ \end{align*} Any help is appreciated, thanks :) |
1147668 | Vector bundle is homotopy equivalence
If [imath]\pi:E\rightarrow B[/imath] is a vector bundle (I allow it to be of nonconstant rank), then I want to prove that it is a homotopy equivalence. As a homotopy inverse I propose the zero section [imath]s:B\rightarrow E[/imath]. So [imath]\pi s=id_B[/imath] and [imath]s\pi[/imath] is homotopic to [imath]id_E[/imath] by [imath]f:E\times[0,1]\rightarrow E[/imath], [imath]f(e,t)=te+(1-t)s\pi(e)[/imath]. Is this correct? | 1134741 | De Rham Cohomology of the tangent bundle of a manifold
I would like to compute the de Rham cohomology of the tangent bundle [imath]TM[/imath] of a manifold [imath]M[/imath]. It seems to me that we can just homotopy each fibre to a point, and that this would give a homotopy equivalence between [imath]TM[/imath] and [imath]M[/imath]. Since de Rham cohomology is a homotopy invariant, this would imply that [imath]H^*_{\text{dR}}(TM) \cong H^*_{\text{dR}}(M)[/imath]. Is it always true that the tangent bundle of a manifold is homotopic to the manifold itself by the homotopy of each fibre with a point? |
1148384 | Prove that [imath]f[/imath] has derivatives of all orders at [imath]x=0[/imath]
Let [imath]\displaystyle f(x) = \begin{cases}e^{- \frac{1}{x^2}} &\text{for } x \neq 0 \\ 0 & \text{when } x=0 \end{cases}.[/imath] Prove that [imath]f[/imath] has derivatives of all orders at [imath]x=0[/imath], and that [imath]f^n(0)=0[/imath] for [imath]n=1,2,3,...[/imath] I found this problem from Rudin's principles of Mathematical Analysis book (p.196). My problem is, can I use the formula [imath]e^x=1+x+\frac{1}{2!}x^2+...[/imath] and use directly Theorem 8.1 and Corollary on page 173 in the same book? If not can anyone please explain me the reason? | 459911 | Prove that [imath]f[/imath] has derivatives of all orders at [imath]x=0[/imath], and that [imath]f^{(n)}(0)=0[/imath] for [imath]n=1,2,3\ldots[/imath]
Define that [imath]f(x) = e^{-1/x^2}[/imath] when [imath]x[/imath] is not [imath]0[/imath], and [imath]f(x) = 0[/imath] when [imath]x=0[/imath]. Prove that [imath]f[/imath] has derivatives of all orders at [imath]x=0[/imath], and that [imath]f^{(n)}(0)=0[/imath] for [imath]n=1,2,3,\ldots[/imath] |
1148484 | Ordinals that are not cardinals
I am reading Jech's set theory and he defines a cardinal number as an ordinal [imath]\alpha[/imath] (a cardinal) if [imath]|\alpha| \neq |\beta|[/imath] for all [imath]\beta < \alpha[/imath], and he says that all infinite cardinals are limit ordinals. My question is: are there any ordinals that are not cardinals? Although I know that an ordinal describe ordering and a cardinal the size of a set, I am a bit confused here. Thanks | 592823 | Why [imath]\omega+1[/imath] and [imath]\omega^2[/imath] are not cardinal numbers?
I see the following definition of cardinal number in notes: An ordinal [imath]\alpha[/imath] is a cardinal number if [imath]|\beta|<|\alpha|[/imath] for all [imath]\beta\in\alpha[/imath]. Why [imath]\omega+1[/imath] and [imath]\omega^2[/imath] are not cardinal numbers? For [imath]\omega+1[/imath], is it because [imath]\omega\in\omega+1[/imath] but [imath]|\omega|=|\omega+1|[/imath]? |
1148581 | Determinant evaluation for matrix with [imath]-1, 2, -1[/imath] below/on/above diagonal
What is the trick for evaluating the determinant of this matrix? [imath]\begin{bmatrix} 2 & -1 \\ -1 & 2 & -1 \\ & -1 & 2 & -1 \\ && -1 & 2 & -1 \\ &&& -1 & 2 & -1 \\ &&&& -1 & 2 \end{bmatrix}[/imath] [imath]%Commenting out original picture ![enter image description here][1][/imath] | 995779 | Proving the determinant of a tridiagonal matrix with [imath]-1, 2, -1[/imath] on diagonal.
Let [imath]A_n[/imath] denote an [imath]n \times n[/imath] tridiagonal matrix. [imath]A_n=\begin{pmatrix}2 & -1 & & & 0 \\ -1 & 2 & -1 & & \\ & \ddots & \ddots & \ddots & \\ & & -1 & 2 & -1 \\ 0 & & & -1 & 2 \end{pmatrix} \quad\text{for }n \ge 2[/imath] Set [imath]D_n = \det(A_n)[/imath] Prove that [imath]D_n = 2D_{n-1} - D_{n-2}[/imath] for [imath]n \ge 4[/imath]. |
1149036 | Verification on [imath]\phi[/imath] proof
I need to find a positive integer [imath]n[/imath] such that [imath]ϕ(n) = ϕ(n + 1) = ϕ(n + 2)[/imath] where [imath]ϕ(n)[/imath] denotes Euler's totient function. What I am given: (1) You may take [imath]ϕ(n) = 2592[/imath]. (2) [imath]ϕ(2n) = ϕ(n)[/imath] provided that [imath]n[/imath] is odd. (3) [imath]ϕ(p) = p − 1[/imath] for p a prime. What I did: Taking the hint (1), we observe that [imath]p=2592+1[/imath] is prime and also [imath]\frac{p+1}2=1297[/imath] is prime. Hence together with your computation [imath]\begin{align}\phi(2p+0)&=\phi(p)=2592\\ \phi(2p+1)&=\phi(5187)=2592\\ \phi(2p+2)&=\phi\left(4\cdot \tfrac{p+1}2\right)=2\phi\left(\tfrac{p+1}2\right)=p-1=2592\end{align}[/imath] So does this mean that my n is 5187 or 5186? | 1143468 | Find a positive integer [imath]n[/imath] such that [imath]ϕ(n) = ϕ(n + 1) = ϕ(n + 2)[/imath]
I need to find a positive integer [imath]n[/imath] such that [imath]ϕ(n) = ϕ(n + 1) = ϕ(n + 2)[/imath] where [imath]ϕ(n)[/imath] denotes Euler's totient function. What I am given: (1) You may take [imath]ϕ(n) = 2592[/imath]. (2) [imath]ϕ(2n) = ϕ(n)[/imath] provided that [imath]n[/imath] is odd. (3) [imath]ϕ(p) = p − 1[/imath] for p a prime. What I did: I thought that I can take in a value for [imath]ϕ(n)[/imath], so I tried to take [imath]ϕ(n) = 2592[/imath]. I found that [imath]ϕ (5187) = 2·6·12·18 = 2592[/imath] So can someone verify that I meet all the requirements that I am give, and that my answer is correct. |
1148719 | Identify the graph that shows the region [imath]x\ge 5[/imath] and [imath]y\ge 5[/imath]
I draw the equation in wolframalpha.com but I don't know how can determine correct answer please I really need help to understand the difference: | 1148696 | Please guide me to draw this : graph of [imath]x \geq 5[/imath] and [imath]y \geq 5[/imath]
how can I draw the graph for this the graph of [imath]x \geq 5[/imath] and [imath]y \geq5[/imath] is? please guide me to draw it thanks in advance |
1149130 | Let [imath]A[/imath] be a [imath]3[/imath]X[imath]3[/imath] matrix whose eigenvalues are [imath]1[/imath], [imath]2[/imath], [imath]3[/imath]. Find [imath]\det(B)[/imath] where [imath]B = A^2 + A^T[/imath].
Let [imath]A[/imath] be a [imath]3[/imath]X[imath]3[/imath] matrix whose eigenvalues are [imath]1[/imath], [imath]2[/imath], [imath]3[/imath]. Find [imath]\det(B)[/imath] where [imath]B = A^2 + A^T[/imath]. I know that [imath]\det(A) = 6[/imath], but I cannot proceed after [imath]|A^2 + A^T|[/imath]. Any hints as to how to approach the problem? | 1147566 | Find [imath]\det(A^{2}+A^{T})[/imath] when eigenvalues are [imath]1,2,3[/imath]
We have to find [imath]\det(A^{2}+A^{T})[/imath]. It is given that eigenvalues of [imath]A[/imath] are [imath]1,2,3[/imath]. My attempt: Since the question implicitly states that the answer would be same for all [imath]A[/imath] with eigenvalues [imath]1,2,3[/imath] we find that the answer is [imath](1+1)*(2+4)*(3+9)[/imath] by taking [imath]A[/imath] as a diagonal matrix :P I am unable to prove it for general [imath]A[/imath]. Any help will be appreciated. |
765273 | prove that [imath]3[/imath] does not divide [imath]n^2+1[/imath]
How do I prove that [imath]3[/imath] does not divide [imath]n^2+1[/imath], for all [imath]n\in\mathbb{Z}[/imath], thought of in separate cases, but did not get, induction also was unable to .... | 620153 | [imath]3[/imath] never divides [imath]n^2+1[/imath]
Problem: Is it true that [imath]3[/imath] never divides [imath]n^2+1[/imath] for every positive integer [imath]n[/imath]? Explain. Explanation: If [imath]n[/imath] is odd, then [imath]n^2+1[/imath] is even. Hence [imath]3[/imath] never divides [imath]n^2+1[/imath], when [imath]n[/imath] is odd. If [imath]n[/imath] is even, then [imath]n^2+1[/imath] is odd. So [imath]3[/imath] could divide [imath]n^2+1[/imath]. And that is where I am stuck. I try to plug in numbers for [imath]n[/imath] but I want a more general form of showing that [imath]3[/imath] can't divide [imath]n^2+1[/imath] when [imath]n[/imath] is even. |
1148894 | Seating arrangement round table
Given a group of [imath]p_1[/imath] ladies and [imath]p_2[/imath] gents and a round table with exactly [imath]p_1+p_2[/imath] places. For reasons which I do not dare to mention, I want to find a seating arrangement such that nobody is sitting in-between two ladies. 1) Say that [imath]p_1[/imath]=[imath]p_2[/imath]=[imath]p \geq 10[/imath]. For which values of [imath]p[/imath] it is possible to make a seating arrangement? 2) Give an expression for the number of possible seating arrangements for those values of [imath]p[/imath] for which a seating arrangement is possible. In this case, a host and hostess have fixed places abreast each other and all the ladies and gents are distinguishable. Results so far: I know that there a no possibilities if [imath]p_1[/imath]>[imath]p_2[/imath]. It seems to me that a seating arrangement is possible if [imath]p[/imath] is odd. But I don't know if this is true. | 1147616 | Pigeonhole principle application
Say there are [imath]p_{1}[/imath] red balls and [imath]p_{2}[/imath] green balls. We put all the balls in a circle with [imath]p_{1}+p_{2}[/imath] places in total. It is forbidden that a ball (red or green) is placed between two red balls. Show that this is impossible if [imath]p_{1}>p_{2}[/imath]. Further on, say [imath]p_{1}=p_{2}=p\geq10[/imath]. For which values of [imath]p[/imath] the ordination is possible? Can somebody help me? Thanx in advance! |
1009637 | Prove that if [imath]3\mid n^2 [/imath] then [imath]3\mid n [/imath].
[imath]n \in \mathbb{N}[/imath] Prove that if [imath]3\mid n^2 [/imath] then [imath]3\mid n [/imath] I want to prove this in a accepted formal way, I thought about the fact that every integer can be written as multiplication of prime numbers. and if we raise this integer to the power two we will get multiples of each prime. [imath]3\mid n^2 [/imath], it means that this [imath]"3"[/imath] must have been in [imath]"n"[/imath] before raising it to the power. I am going to the right direction? and if yes, can someone hint how can I write it formally? | 992181 | Prove that if [imath]n^2[/imath] is divided by 3, then also [imath]n[/imath] can also be divided by 3.
[imath]n\in \Bbb N[/imath] Prove that if [imath]n^2[/imath] is divided by 3, then also n can also be divided by 3. I started solving this by induction, but I'm not sure that I'm going in the right direction, any suggestions? |
1146460 | Is there [imath]a,b,c,d\in \mathbb N[/imath] so that [imath]a^2+b^2=c^2[/imath], [imath]b^2+c^2=d^2[/imath]?
Question: Are there [imath]a,b,c,d \in \mathbb N[/imath] such that [imath]a^2 + b^2 = c^2 \ \ \text{and} \ \ b^2 + c^2 = d^2[/imath] I'm a bit lost here. | 960816 | Are there any positive integers [imath]a, b, c, d[/imath] such that both [imath](a, b, c)[/imath] and [imath](b, c, d)[/imath] are Pythagorean triples?
Pythagorean triple is a triple of integers [imath](a, b, c)[/imath] such that [imath]a^2+b^2=c^2[/imath]. Is there any Pythagorean triple such that, not only [imath]a^2+b^2[/imath], but also [imath]b^2+c^2[/imath] is a square number? If not, how to prove it? I tried to prove non-existence the following way: If true, it would mean that there is a pair of integers such that both sum and difference of their squares is a square number. Let's call these integers [imath]a[/imath] and [imath]b[/imath] and [imath]a<b[/imath]. Then, there are integers [imath]c[/imath] and [imath]d[/imath] such that: \begin{align} &a^2+b^2=c^2 \\ &a^2-b^2=d^2 \end{align} Multiplying those equations gives: \begin{equation} a^4=(cd)^2+b^4 \end{equation} This is similar to Fermat's Last Theorem for [imath]n=4[/imath], but using it only shows that [imath]cd[/imath] can't be square number, not that there are no integer solutions. |
1149313 | four different integers exist, what is the least product value?
Four different positive integers [imath]a, b, c, d[/imath] are such that [imath]a^2 + b^2 = c^2 + d^2[/imath]. What is the smallest possible value of [imath]abcd[/imath]? [imath]a^2 - c^2 = d^2 - b^2[/imath] [imath](a-c)(a+c) = (d-b)(d+b)[/imath] [imath](a-c)(a+c) - (d-b)(d+b) = 0[/imath] So there is one pair I see: [imath]a -c = d - b, a + c = d + b[/imath] Suppose WLOG, [imath]a > b > c > d > 0[/imath] But that doesnt strike anything. | 1117884 | Four different positive integers a, b, c, and d are such that [imath]a^2 + b^2 = c^2 + d^2[/imath]
Four different positive integers [imath]a, b, c[/imath], and [imath]d[/imath] are such that [imath]a^2 + b^2 = c^2 + d^2[/imath] What is the smallest possible value of [imath]abcd[/imath]? I just need a few hints, nothing else. How should I begin? Number theory? |
1149516 | Show that the closed ball around some point in a metric space is a closed set.
The closed ball is the set of [imath]y[/imath] around some [imath]x[/imath] such that [imath]d(x,y) \leq r[/imath] for some [imath]r>0[/imath] where [imath]d(x,y)[/imath] is a metric on the space containing [imath]x[/imath] and [imath]y[/imath]. | 661759 | A closed ball in a metric space is a closed set
Prove that a closed ball in a metric space is a closed set My attempt: Suppose [imath]D(x_0, r)[/imath] is a closed ball. We show that [imath]X \setminus D [/imath] is open. In other words, we need to find an open ball contained in [imath]X \setminus D[/imath]. Pick [imath]t \in X-D \implies d(t,x_0) > r \implies d(t,x_0) - r > 0 [/imath] Let [imath]B(y, r_1)[/imath] be an open ball, and pick [imath]z \in B(y,r_1)[/imath]. Then, we must have [imath]d(y,z) < r_1 [/imath]. We need to choose [imath]r_1[/imath] so that [imath]d(z,x_0) > r[/imath]. Notice by the triangle inequality [imath] d(x_0,t) \leq d(x_0,z) + d(z,t) \implies d(z,x_0) \geq d(x_0,t) - d(z,t) > d(x_0,t) - r_1.[/imath] Notice, if we pick [imath]r_1 = d(t,x_0)-r[/imath] then we are done. Is this correct? |
1148924 | Show that [imath]a^2 + b^2 = c^3 [/imath] has infinitely many solutions
Show that [imath]a^2 + b^2 = c^3 [/imath] has infinitely many solutions in [imath] \{ (a,b,c) \in \Bbb Z ^3 | (a,b)=1, (a,c)=1, (b,c)=1 \}[/imath] . Describe all these solutions. I don't know how to approach this question. I'm thinking about something like [imath]a^2+b^2=(a-bi)(a+bi)[/imath], so we get [imath]xy=c^3[/imath] where [imath]x,y \in \Bbb Z [i][/imath]. I'm not sure it's the right way to do this. | 334839 | Variation of Pythagorean triplets: [imath]x^2+y^2 = z^3[/imath]
I need to prove that the equation [imath]x^2 + y^2 = z^3[/imath] has infinitely many solutions for positive [imath]x, y[/imath] and [imath]z[/imath]. I got to as far as [imath]4^3 = 8^2[/imath] but that seems to be of no help. Can some one help me with it? |
1149290 | Factoring Polynomials (x^4) - using completing squares
Exercise 6. By viewing the polynomials as a difference of two squares, factorise the following polymomials. [imath]x^4+x^2+1[/imath], [imath]x^4+3x^2+4[/imath], [imath]x^4+4[/imath]. To solve difference of two squares isnt it necessary to have negatives? im confused about how to start solving these questions. I cant seem to factorise by grouping either /: Information on how to factorise these types of questions would be appreciated. thank You! | 1148984 | By viewing the polynomials as a difference of two squares, factorise the following polynomials.
By viewing the polynomials as a difference of two squares, factorise the following polynomial: [imath]x^4+x^2+1.[/imath] I searched but couldn't find a way to solve this Edit: By using Hans Lundmark hint, I get: [imath](x^2+1)^2-x^2[/imath] Is it fully factorized? |
1149913 | Let [imath](x_n)[/imath] be a sequence that converges to a number x. Show that [imath](s_n)[/imath] converges to x as well.
Let [imath](x_n)[/imath] be a sequence that converges to a number x. Define a new sequence [imath](s_n)[/imath] by [imath]s_n= \frac{1}{n}\sum\limits_{k=1}^n x_k[/imath] Show that [imath]s_n[/imath] converges to [imath]x[/imath] as well. | 533626 | Convergence of the arithmetic mean
Let [imath](a_n)_{n \in \mathbb{N}}[/imath] be a convergent sequence with limit [imath]a \in \mathbb{R}[/imath]. Show that the arithmetic mean given by: [imath]s_n:= \frac{1}{n}\sum_{i=1}^n a_i \tag{A.M.} [/imath] also converges to [imath]a[/imath]. I have read: arithmetic mean of a sequence converges but unfortunately the answers there don't help me much because I don't understand their substitutions and most of it all, why their substitutions seem to work. What I know from the problem is that since [imath]a_n[/imath] is convergent and [imath]\epsilon >0[/imath] is given, I can say that: [imath]\exists N_1 \in \mathbb{N}, \forall n \geq N_1: |a_n-a|<\epsilon_1 [/imath] I also know that since [imath]a_n[/imath] is convergent, it is bound, so [imath](a_n) < M, \ \forall n \in \mathbb{N}[/imath] I need to show that: [imath]\exists N_2 \in \mathbb{N}, \forall n \geq N_2: |s_n-a|< \epsilon_2[/imath] I started as follows: [imath] \left|s_n-a \right| = \left|\frac{1}{n}\sum_{i=1}^na_i -a\right|= \left|\frac{1}{n}\sum_{i=1}^m(a_i-a)+\frac{1}{n}\sum_{i=m+1}^n(a_i-a) \right| \\ \leq \frac{1}{n} \sum_{i=1}^m|(a_i-a)|+\frac{1}{n}\sum_{i=m+1}^n|a_i-a|[/imath] I believe to understand that the left sum after the [imath]\leq[/imath] is finite, bound and doesn't depend on the values that [imath]n[/imath] takes on. However, I don't understand where all the substitutions come from and make this proof so seemingly easy to complete. Is there a general idea I can follow to complete such proofs? Because I know that the last step is to show that the given sum is smaller than [imath]\epsilon[/imath]. I also know that I should bring the condition [imath]a_n < M[/imath] into place somewhere, but I don't know where. If I choose an [imath]n \geq N: |a_n-a|< \epsilon'[/imath] what does that tell me about [imath]|a_i-a|[/imath]? I know that they are seemingly the same just with a different index. |
1150187 | Evaluating the definite integral [imath]\int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1} dx[/imath]
I have two integral questions listed below: [imath]\int_0^{\infty} \frac{\ln x}{x^2 + 1} dx \qquad (1)[/imath] [imath]\int_0^{\infty} \frac{(\ln x)^2}{x^2 + 1} dx \qquad (2)[/imath] The first one, I've solved it, by separating the integral into integrals over [imath]0[/imath] to [imath]1[/imath] and over [imath]1[/imath] to [imath]\infty[/imath]; with substitution, it becomes zero. But the second one wasn't that easy. Can someone give me help? Wolfram alpha says the answer is [imath]\frac{\pi^3}{8}.[/imath] | 1030246 | Evaluate [imath]\int_0^\infty \frac{(\log x)^2}{1+x^2} dx[/imath] using complex analysis
How do I compute [imath]\int_0^\infty \frac{(\log x)^2}{1+x^2} dx[/imath] What I am doing is take [imath]f(z)=\frac{(\log z)^2}{1+z^2}[/imath] and calculating [imath]\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}[/imath] which came out to be [imath]\dfrac{\pi}{2}-\dfrac{i\pi^2}{8}+\dfrac{i\pi}{2}[/imath] Im not too sure how to move on from here. the given answer is [imath]\dfrac{\pi^3}{8}[/imath] Any help will be appreciated. thank you in advanced. |
1150337 | What is the importance of last axiom of vector space
What is the importance of last axiom of vector space. Axiom is [imath]1u=u[/imath]. | 777622 | Examples of 'almost' vector spaces where unitary law fails
I was looking at the definition on wikipedia of a vector space (similar/equivalent definitions are everywhere, but I thought I'd list it here for completion): A vector space over a field [imath]F[/imath] is a set [imath]V[/imath] together with two binary operations that satisfy the eight axioms listed below. Elements of [imath]V[/imath] are called vectors. Elements of [imath]F[/imath] are called scalars. In the two examples above, our set consists of the planar arrows with fixed starting point and of pairs of real numbers, respectively, while our field is the real numbers. The first operation, vector addition, takes any two vectors [imath]v[/imath] and [imath]w[/imath] and assigns to them a third vector which is commonly written as [imath]v + w[/imath], and called the sum of these two vectors. The second operation takes any scalar a and any vector [imath]v[/imath] and gives another vector [imath]av[/imath]. In view of the first example, where the multiplication is done by rescaling the vector [imath]v[/imath] by a scalar [imath]a[/imath], the multiplication is called scalar multiplication of [imath]v[/imath] by [imath]a[/imath]. To qualify as a vector space, the set [imath]V[/imath] and the operations of addition and multiplication must adhere to a number of requirements called axioms. In the list below, let [imath]u[/imath], [imath]v[/imath] and [imath]w[/imath] be arbitrary vectors in [imath]V[/imath], and [imath]a[/imath] and [imath]b[/imath] scalars in [imath]F[/imath]. Associativity of addition: [imath]u + (v + w) = (u + v) + w[/imath] Commutativity of addition: [imath]u + v = v + u[/imath] Identity element of addition: There exists an element [imath]0 ∈ V[/imath], called the zero vector, such that [imath]v + 0 = v[/imath] for all [imath]v ∈ V[/imath]. Inverse elements of addition: For every [imath]v ∈ V[/imath], there exists an element [imath]−v ∈ V[/imath], called the additive inverse of [imath]v[/imath], such that [imath]v + (−v) = 0[/imath] Compatibility of scalar multiplication with field multiplication: [imath]a(bv) = (ab)v[/imath] Identity element of scalar multiplication: [imath]1v = v[/imath], where [imath]1[/imath] denotes the multiplicative identity in [imath]F[/imath]. Distributivity of scalar multiplication with respect to vector addition: [imath]a(u + v) = au + av[/imath] Distributivity of scalar multiplication with respect to field addition: [imath](a + b)v = av + bv[/imath] I was wondering if there were any examples of vector spaces where only the law for the identity element of scalar multiplication fails? The only one I could think of would be to redefine a true vector space's scalar multiplication to multiply the result by a constant factor. Are there others? |
1150361 | Show if [imath]R[/imath] is Noetherian, then [imath]R_S[/imath] is Noetherian
Show if [imath]R[/imath] is Noetherian, then [imath]R_S[/imath] is Noetherian. here is what I have read from somewhere else. Suppose [imath]R[/imath] is Noetherian and [imath]J[/imath] is an ideal [imath]R_S[/imath]. Then [imath]J=IR_S[/imath] for some ideal [imath]I[/imath] of [imath]R[/imath]. Since [imath]R[/imath] is Noetherian, [imath]I[/imath] is finitely generated, say [imath]I=(r_1,r_2,…r_n)[/imath] Thus [imath]J=(r_1/1,r_2/1,…r_n/1)[/imath]. Hence every ideal in [imath]R_S[/imath] is finitely generated and so [imath]R_S[/imath] is Noetherian. My confusion is how to get [imath]J=(r_1/1,r_2/1,…r_n/1)[/imath] from [imath]I=(r_1,r_2,…r_n)[/imath]. And I also confuse about the notation of [imath]r_1/1[/imath], what are they, are they suppose just be like [imath]r_1/1=r_1[/imath] ? Thank you. | 110735 | Why is the localization of a commutative Noetherian ring still Noetherian?
This is an unproven proposition I've come across in multiple places. Suppose [imath]A[/imath] is a commutative Noetherian ring, and [imath]S[/imath] a multiplicative subset of [imath]A[/imath]. Then [imath]S^{-1}A[/imath] is Noetherian. Why is this? I thought about taking some chain of submodules [imath] S^{-1}M_1\subset S^{-1}M_2\subset\cdots [/imath] and pulling back to a chain [imath] M_1\subset M_2\subset\cdots [/imath] of submodules of [imath]A[/imath] which must eventually stablize. Is there more to it than this? I kind of wary of assuming all submodules of [imath]S^{-1}A[/imath] have form [imath]S^{-1}M[/imath] for [imath]M\leq A[/imath]. |
1148885 | Prove that every triangle-free graph is [imath]\lceil 2\sqrt{n}\rceil[/imath] colorable
Prove that every triangle-free graph is [imath]\lceil 2\sqrt{n}\rceil[/imath] colorable | 571308 | Upper bound on [imath]\chi(G)[/imath] for a triangle-free graph
I'm struggling with the following question; For every graph [imath]G[/imath] such that [imath]K_3 \not\subseteq G[/imath] (i.e. [imath]G[/imath] does not contain a triangle), prove that [imath]\chi(G) \leq 2\sqrt{n} +1[/imath] (where [imath]\chi(G)[/imath] denotes the chromatic number). Any hints? I really cannot see the connection between the triangle-free nature of [imath]G[/imath] and a chromatic number being bounded above by [imath]O(\sqrt{n})[/imath]. |
1150819 | Euclidean Domain and Divison Algorithm
In ED definition we need for all [imath]a, b[/imath] not zero then there exist [imath]q[/imath] and [imath]r[/imath] such that [imath]a = bq+r[/imath] with some condition with respect to norm. Actually I am interested whether [imath]q[/imath] and [imath]r[/imath] are unique in ED or not Further Questions: 1. Can we define some norm in ED with respect to which q and r are unique And in how many ways we can define such norms If it is not possible what algebraic structure like this i.e. ED with uniqueness of q and r, has some special properties. | 323235 | Euclidean domain in which the quotient and remainder are always unique
Let [imath]R[/imath] be a Euclidean domain in which the quotient and remainder are always unique. Does it follow that the ring [imath]R[/imath] is either a field or a polynomial ring [imath]F[X][/imath] for some field [imath]F[/imath]? |
1151342 | Counting all possible ways to choose [imath]M[/imath] numbers from [imath]1[/imath] to [imath]N[/imath] given some conditions .
Let [imath]A=\{ 1,2,\ldots,N\}[/imath] and let [imath]B=\prod^M A[/imath] be the cartesian product of [imath]A[/imath] with itself [imath]M[/imath] times. I want to find the cardinality of the subset [imath]\{(a_1 ,a_2,\ldots,a_M), a_1 \leq a_2\leq\cdots\leq a_M\}[/imath] . I tried to solve this problem by constructing a tree. If you choose [imath]1\leq k\leq N[/imath] and put it in the first slot , You can imagine that it's the root of a tree. Then , there are [imath]N-k+1[/imath] branches and so on .You can then count all the branches and sum over [imath]k[/imath] but I can't go anywhere using this method .I'm sure there are standard methods to solve it but I don't know much beyond permutations and combinations . | 658250 | number of strictly increasing sequences of length [imath]K[/imath] with elements from [imath]\{1, 2,\cdots,N\}[/imath]?
What is the number of strictly incremental sequences of length [imath]K[/imath] with elements from [imath]\{1, 2,\cdots,N\}[/imath] ? Is there any exact value? How about approximations? |
1151176 | Show that [imath]P_n=p(1-P_{n-1})+(1-p)P_{n-1}[/imath] [imath]n \ge 1[/imath]
Independent trials that result in a success with probability p and a failure with probability 1-p are called Bernoulli trials. Let [imath]P_n[/imath] denote the probability that n Bernoulli trials result in an even number of success (0 being considered an even number). Show that [imath]P_n=p(1-P_{n-1})+(1-p)P_{n-1}[/imath] [imath]n \ge 1[/imath] and use this to prove (by induction) that [imath]P_n=\frac{1+(1-2p)^n}{2}[/imath] So [imath]P_n[/imath] denotes the probability that n Bernoulli trials result in an even number of successes p = probability of success p-1 = probability of failure [imath]P_n=p(1-P_{n-1})+(1-p)P_{n-1}[/imath] [imath]n \ge 1[/imath] I can see that the first part of the equation is the probability of success multiplied by the probability that n-1 trials result in failure, plus the probability of failure multiplied by probability that n-1 trials result in success. But I don;t know how to show this mathematically.. So I think for the second part, for induction I need to start with the base case, n=1: [imath]P_n=p(1-P_{n-1})+(1-p)P_{n-1}[/imath] [imath]P_1=p(1-P_{1-1})+(1-p)P_{1-1}[/imath] [imath]P_0=p(1-P_{0})+(1-p)P_{0}[/imath] Then I get kind of stuck, well..[imath]P_0[/imath] is probability of n=0 trials, so would that be 0? [imath]P_0=p(1)=0[/imath] Then for induction, I need to show that n+1 is true right? | 1149270 | probability that a random variable is even
Suppose [imath]X[/imath] is binomial [imath]B(n,p)[/imath]. How can I find the probability that [imath]X[/imath] is even ? I know [imath]P(X = k ) = \frac{ n!}{(n-k)!k!} p^k(1-p)^{n-k} [/imath] where [imath]X=1,....,n[/imath]. Are they just asking to find [imath]P(X = 2m )[/imath] for some [imath]m > 0[/imath] ? |
1148862 | Finding the value of the infinite sum [imath]1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... [/imath]
Can anyone help me to find what is the value of [imath]1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... [/imath] when it tends to infinity The first i wanna find the pattern but it seems do not have any unique pattern can anyone help me? | 1066465 | Prove that [imath]\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}[/imath]
Prove that [imath]\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}[/imath] I tried to look at [imath] f_n(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^n [/imath] And maybe taking it's derivative but it didn't work out well. Any ideas? |
334028 | What is true for a ring with exactly two right ideals
This is question #66 from http://www.ets.org/s/gre/pdf/practice_book_math.pdf Let [imath]R[/imath] be a ring with a multiplicative identity. If [imath]U[/imath] is an additive subgroup of [imath]R[/imath] such that [imath]ur \in U[/imath] for all [imath]u \in U[/imath] and for all [imath]r \in R[/imath] , then [imath]U[/imath] is said to be a right ideal of [imath]R[/imath]. If [imath]R[/imath] has exactly two right ideals, which of the following must be true? I. [imath]R[/imath] is commutative. II. [imath]R[/imath] is a division ring (that is, all elements except the additive identity have multiplicative inverses). III. [imath]R[/imath] is infinite. Here is my reasoning: Because [imath]R[/imath] is a ring, [imath]R[/imath] is also a additive group with some identity element [imath]0[/imath]. We have a theorem that says [imath]0r = 0 [/imath] in any ring, so [imath]\{0\}[/imath] is a right ideal of [imath]R[/imath]. Also, [imath]R[/imath] is a right ideal of [imath]R[/imath]. Now I have found two different right ideals and there mustn't be any more. Edit: As mentioned in the comments, the example below is not a ring, so it is not applicable to the problem. I could not fix it by taking additive closure because that introduced more than two ideals. A possible candidate for [imath]R[/imath] could be the set of [imath]2\times2[/imath] matrices [imath]\{0,I,-I,a,-a\}[/imath] where [imath]a = [[^1_0] ,[^0_0]][/imath]. The only right ideals are [imath]R[/imath] and [imath]\{0\}[/imath]. This ring satisfies only property I, but the answer key says that II is the correct answer. | 2051178 | Show that R is a skew field.
Let [imath]R[/imath] be a ring with unity such that the only left ideals of [imath]R[/imath] are [imath]\{0\}[/imath] and [imath]R[/imath]. Show that [imath]R[/imath] is a skew field. Please help me to solve this problem. |
1138180 | If there is no cycle through [imath]e[/imath] and [imath]g[/imath], then all paths go through the same vertex
Suppose there is no cycle that goes through edges [imath]e[/imath] and [imath]g[/imath]. The question is to prove that for every path that begins at one of the vertices incident to [imath]e[/imath] and and ends at a vertex incident to [imath]g[/imath], that there exists some vertex [imath]u[/imath] that is always on that path. My work: Pick the edge closest to beginning vertex such that [imath]e[/imath] and this edge, call it edge [imath]f_{j-1}[/imath] (connection [imath]v_{j-1}[/imath] and [imath]v_j[/imath]) do not form a cycle on the graph [imath]G[/imath]. Then using this cycle and a path joining a vertex on the cycle to one of the vertices [imath]v_{j+1}, \ldots, v_n[/imath], construct some contradiction by creating a cycle with edge [imath]f_j[/imath]. Any thoughts? | 1141534 | No cycle containing edges [imath]e[/imath] and [imath]g[/imath] implies there is a vertex [imath]u[/imath] so that every path sharing one end with [imath]e[/imath] and another with [imath]g[/imath] contains [imath]u[/imath]
There is a proof in my textbook for the following claim, which doesn't make a whole lot of sense to me. My annotations are in bold. Could someone perhaps elaborate on what's going on? Claim. If there does not exist a cycle containing edges [imath]e[/imath] and [imath]g[/imath] then there exists a vertex [imath]u \in V (G)[/imath] such that every path in [imath]G[/imath] sharing one end with e and another with [imath]g[/imath] contains [imath]u[/imath]. Proof: The claim trivially holds if [imath]e[/imath] or [imath]g[/imath] is a loop OK, so we assume that neither is. Let [imath]P[/imath] with vertex set [imath]v_1, v_2, . . . , v_k,[/imath] in order, be a path with [imath]e[/imath] joining [imath]v_1[/imath] to [imath]v_2[/imath] and [imath]g[/imath] joining [imath]v_{k−1}[/imath] and [imath]v_k[/imath]. Let [imath]f_i \in E(P_i)[/imath] be the edge with ends [imath]v_i[/imath] and [imath]v_{i+1}[/imath]. Let [imath]j[/imath] be chosen minimum so that no cycle in [imath]G[/imath] contains [imath]e[/imath] and [imath]f_j[/imath] We can do this because we know that at least the edge [imath]g[/imath] will not create a cycle by assumption, right?. We will show that [imath]u = v_j[/imath] satisfies the claim. Suppose not. Let [imath]C[/imath] be a cycle containing [imath]e[/imath] and [imath]f_{j−1}[/imath] What if [imath]f_j = e[/imath]? then what cycle? a single vertex? and let [imath]P′[/imath] be a path from an end of [imath]e[/imath] to an end of [imath]f[/imath] Not sure what the book meant by [imath]f[/imath] here, any guesses? avoiding [imath]u[/imath]. Choose a subpath [imath]Q[/imath] of [imath]P′[/imath] with one end in [imath]V (C)[/imath] and another in [imath]{v_{j+1}, v_{j+2}, . . . , v_k}[/imath] as short as possible. Then [imath]C \cup Q \cup P[/imath] contains a cycle containing both [imath]e[/imath] and [imath]f_j[/imath], a contradiction. (The last statement requires some case checking.) This ending seems a bit abrupt and non-obvious to me Thanks for the help |
1153491 | Show that these rings are not isomorphic
Prove that the rings [imath]2 \mathbb{Z}[/imath] and [imath]3\mathbb{Z}[/imath] are not isomorphic. I find that if i make a map [imath]\phi[/imath]:[imath]2 \mathbb{Z}[/imath] [imath]\longmapsto[/imath] [imath]3\mathbb{Z}[/imath] the map homomorphism works. so the only thing that i need is to show that [imath]\phi[/imath] is not surjective. How do i show that. | 1146568 | How are [imath]2\mathbb{Z}\ncong3\mathbb{Z}[/imath] different as rings?
How are [imath]2\mathbb{Z}\ncong3\mathbb{Z}[/imath] different as rings? What interesting properties does one have that the other doesn't? |
428185 | Isomorphism between Hom and tensor product
I am looking for an explicit isomorphism [imath]Hom(V,V^*)\rightarrow V^*\otimes V^*[/imath] where [imath]V[/imath] is a vector space. I thought of: [imath]\phi\rightarrow ((u,v)\rightarrow \phi(u)(v))[/imath] But I'm not sure this works. Does anyone have a suggestion? | 679584 | Why is [imath]\text{Hom}(V,W)[/imath] the same thing as [imath]V^* \otimes W[/imath]?
I have a couple of questions about tensor products: Why is [imath]\text{Hom}(V,W)[/imath] the same thing as [imath]V^* \otimes W[/imath]? Why is an element of [imath]V^{*\otimes m}\otimes V^{\otimes n}[/imath] the same thing as a multilinear map [imath]V^m \to V^{\otimes n}[/imath]? What is the general formulation of this principle? |
1153092 | If [imath]A, B[/imath] are open dense subsets of a metric space [imath]X[/imath], is their intersection dense??
If [imath]A, B[/imath] are open dense subsets of a metric space [imath]X[/imath], is their intersection dense?? We know normally intersection of two dense sets are not dense. | 1143211 | Intersection of two open dense sets is dense
Let [imath]X[/imath] be a topological space and suppose that [imath]H[/imath] and [imath]G[/imath] are open dense subsets of [imath]X[/imath].Then show that [imath]G \bigcap H[/imath] is also an open dense subset of [imath]X[/imath]. My attempt : Well since the finite intersection of open sets is open therefore, [imath]G \bigcap H[/imath] is also open. I was trying to prove the other part by the method of contradiction : Suppose on the contrary that [imath]G \bigcap H[/imath] is not dense. Then that implies that [imath]Int(Cl(G \bigcap H)) = \phi[/imath] . Now, [imath]Cl(G \bigcap H) \subset Cl(G) \bigcap Cl(H)[/imath] so, [imath]Int(Cl(G \bigcap H)) \subset Int(Cl(G) \bigcap Cl(H)) = Int(Cl(G)) \bigcap Int(Cl(H)) = X [/imath]. How do i proceed further to get a contradiction ? |
1153998 | Working with sets, set difference, union and intersection
I am asked to prove that [imath] A \setminus (B \cap C) = (A\setminus B) \cup (A\setminus C). [/imath] However I am only finding this to be not true. How would I prove this? | 597499 | de morgan law [imath]A\setminus (B \cap C) = (A\setminus B) \cup (A\setminus C) [/imath]
First part : I want to prove the following De Morgan's law : ref.(dfeuer) [imath]A\setminus (B \cap C) = (A\setminus B) \cup (A\setminus C) [/imath] Second part: Prove that [imath](A\setminus B) \cup (A\setminus C) = A\setminus (B \cap C) [/imath] Proof: Let [imath]y\in (A\setminus B) \cup (A\setminus C)[/imath] [imath](A\setminus B) \cup (A\setminus C) = (y \in A\; \land y \not\in B\;) \vee (y \in A\; \land y \not\in C\;)[/imath] [imath]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= y \in A\ \land ( y \not\in B\; \vee \; y \not\in C ) [/imath] [imath]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= y \in A \land (\lnot ( y\in B) \lor \lnot( y\in C) )[/imath] According to De-Morgan's theorem : [imath]\lnot( B \land C) \Longleftrightarrow (\lnot B \lor \lnot C)[/imath] thus [imath]y \in A \land (\lnot ( y\in B) \lor \lnot( y\in C) ) = y \in A \land y \not\in (B \land C)[/imath] We can conclude that [imath](A\setminus B) \cup (A\setminus C) = A\setminus (B \cap C)[/imath] |
1153978 | Proof that if [imath]f[/imath] and [imath]g[/imath] are uniformly continous on an interval [imath](a,b)[/imath] then so is [imath]f+g[/imath].
I'm trying to prove that if [imath]f[/imath] and [imath]g[/imath] are uniformly continuous on an interval [imath](a,b)[/imath] then so is [imath]f+g[/imath]. I have a picture in my head but I can't seem to make it into a formal proof. | 786527 | If [imath]f,g[/imath] are uniformly continuous prove [imath]f+g[/imath] is uniformly continuous but $fg$ and $\dfrac{f}{g}$ are not
Suppose [imath]f:\mathbb{R} \supset E \rightarrow \mathbb{R}[/imath] and [imath]g: \mathbb{R} \supset E \rightarrow \mathbb{R}[/imath] are uniformly continuous. Show that [imath]f+g[/imath] is uniformly continuous. What about [imath]fg[/imath] and [imath]\dfrac{f}{g}[/imath]? My Attempt Firstly let's state the definition; a function is uniformly continuous if [imath]\forall \varepsilon >0\ \ \exists \ \ \delta >0 \ \ \text{such that} \ \ |f(x)-f(y)|< \varepsilon \ \ \forall \ \ x,y \in \mathbb{R} \ \ \text{such that} \ \ |x-y|<\delta[/imath] Sum [imath]f+g[/imath] Now to to prove [imath]f+g[/imath] is uniformly continuous; [imath]\bullet[/imath] Choose [imath]\delta_1 >0[/imath] such that [imath]\forall[/imath] [imath]x,y \in \mathbb{R}[/imath] [imath]|x-y|<\delta_1[/imath] [imath]\implies[/imath] [imath]|f(x)-f(y)|< \dfrac{\epsilon}{2}[/imath] [imath]\bullet[/imath] Choose [imath]\delta_2 >0[/imath] such that [imath]\forall[/imath] [imath]x,y \in \mathbb{R}[/imath] [imath]|x-y|<\delta_2[/imath] [imath]\implies[/imath] [imath]|g(x)-g(y)|< \dfrac{\varepsilon}{2}[/imath] [imath]\bullet[/imath] Now take [imath]\delta := min\{ \delta_1, \delta_2\}[/imath] Then we obtain for all [imath]x,y \in \mathbb{R}[/imath] [imath] |x-y|<\delta \implies |f(x)+g(x)-f(y)+g(y)| < |f(x)-f(y)| + |g(x)-g(y)| < \dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}= \varepsilon[/imath] Product [imath]fg[/imath] Now for [imath]fg[/imath] for this to hold both [imath]f:E \rightarrow \mathbb{R}[/imath] and [imath]g:E \rightarrow \mathbb{R}[/imath] must be bounded , if not it doesn't hold. [imath]\bullet[/imath] [imath]\exists \ \ M>0 \ \ such \ that \ \ |f(x)| [/imath] [imath]\bullet[/imath] Choose [imath]\delta_1 >0[/imath] such that [imath]\forall[/imath] [imath]x,y \in \mathbb{R}[/imath] [imath]|x-y|<\delta_1[/imath] [imath]\implies[/imath] [imath]|f(x)-f(y)|< \dfrac{\epsilon}{2M}[/imath] [imath]\bullet[/imath] Choose [imath]\delta_2 >0[/imath] such that [imath]\forall[/imath] [imath]x,y \in \mathbb{R}[/imath] [imath]$|x-y|<\delta_2$[/imath] [imath]\implies[/imath] [imath]|g(x)-g(y)|< \dfrac{\epsilon}{2M}[/imath] [imath]\bullet[/imath] Now take [imath]\delta := min\{ \delta_1, \delta_2\}[/imath]. Then, [imath]|x-y|<\delta[/imath] implies for all [imath]x,y \in \mathbb{R}[/imath], that [imath]|f(x)g(x)-f(y)g(y)| \leq |g(x)||f(x)+f(y)|+|f(y)||g(x)+g(y)| \leq [/imath] [imath] M|f(x)+f(y)| + M|g(x)+g(y)| < M \dfrac{\epsilon}{2M} + M \dfrac{\epsilon}{2M} = \epsilon[/imath] Are these proofs correct? I am not sure how to approach the [imath]\dfrac{f}{g}[/imath] case. |
309692 | How to factor [imath]a^n - b^n[/imath]?
Wikipedia provides a proof, but I don't understand how: [imath]a^n - b^n = (a-b)(a^{n-1} + ba^{n-2} +\cdots + b^{n-1})[/imath] follows from [imath]x^{n-1} + x^{n-2} +\cdots + x + 1 = \frac{x^n - 1}{x-1}[/imath] Could someone explain to me how the summation of the the geometric series explains the factorization? | 1316174 | Factoring [imath]a^{k} - b^{k}[/imath]
I am a bit lost how to factor [imath]a^{k} - b^{k}[/imath]. I know it links to the binomial theorem but I can't remember how to do it. Could anyone explain? |
1154302 | Differentiability of a function.
Suppose a function [imath]f(x)[/imath] is continuous on [imath]x=a[/imath]. If [imath]\lim_{x \rightarrow a} f'(x)[/imath] exists, is it also differentiable at [imath]x=a[/imath]? I mean is [imath]f'(a)[/imath] exists? I don't know how to deal with this problem. If [imath]a[/imath] is contained in some open interval [imath](c,d)[/imath] with differentiable except [imath]x=a[/imath], then it is true. But with "only" [imath]x=a[/imath], I don't have any clue.. | 1070858 | Show that if [imath]\lim_{x\to a}f'(x) = A[/imath] then [imath]f'(a)[/imath] exists and equals A
I ran across this problem in my Analysis class and can't seem to come up with a good solution. Here's the question: Show that if [imath]\lim_{x\to a}f'(x) = A[/imath] then [imath]f'(a)[/imath] exists and equals [imath]A[/imath]. [imath]f[/imath] is continuous on [imath][a,b][/imath] and differentiable on [imath](a,b)[/imath]. I tried applying the Mean Value theorem like this: There exists a [imath]c \in (a, x)[/imath] st [imath]f(x)-f(a)=f'(c)(x-a) \implies \frac{f(x)-f(a)}{x-a}=f'(c) \implies f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{c\to a}f'(c)=A[/imath] However, I think this is wrong. I am pretty sure the Mean Value theorem is needed, but my application of it seems to be incorrect. Would somebody be able to point me in the right direction? |
1150257 | Finding Radical of an Ideal
Given the ideal [imath]J^\prime=\langle xy,xz-yz\rangle[/imath], find it's radical. I know that the ideal [imath]\langle xy,yz,zx\rangle[/imath] is radical ideal but that's not the case. How can I compute the radical here? | 1140439 | Let K be a field, and [imath]I=(XY,(X-Y)Z)⊆K[X,Y,Z][/imath]. Prove that [imath]√I=(XY,XZ,YZ)[/imath].
Let [imath]K[/imath] be a field, and let [imath]I=(XY,(X-Y)Z) \subset K[X,Y,Z][/imath]. Prove that [imath]\sqrt{I}=(XY,XZ,YZ)[/imath]. I have no idea how to start with this question, can anybody give me some hint? Thanks a lot. |
100999 | [imath]f(x)= 0[/imath] for [imath]x \notin\mathbb{Q}[/imath], and [imath]f(x)=1/q[/imath] for [imath]x=p/q[/imath]. Prove: [imath]f[/imath] is integrable
[imath]f(x)= 0[/imath] , if [imath]x \notin \mathbb{Q}[/imath], otherwise [imath]f(x)=1/q[/imath] for [imath]x=p/q[/imath] such that [imath]p[/imath] and [imath]q[/imath] don't share common divisor. I 'd love your help proving that [imath]f[/imath] is integrable and that [imath]\int_{0}^{1}f=0[/imath]. I showed that the lower Darboux sum is [imath]0[/imath] and I basically need to show that for every epsilon we can find division such that the upper Darboux sum is smaller than the given epsilon. The upper darboux sum is [imath]\bar{S}=\sum_{1}^{n}f(x_i) \Delta x_i[/imath] for all [imath]x_i[/imath] of the partition, I tried to replace the [imath]f(x_i)[/imath] in [imath]1/q_i[/imath], and check if this series converges to [imath]0[/imath]. Thanks a lot. | 479965 | [imath]f(x)=1/q[/imath] for [imath]x=p/q[/imath] is integrable
Let [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] be defined by setting [imath]f(x)=1/q[/imath] if [imath]x=p/q[/imath], where [imath]p[/imath] and [imath]q[/imath] are positive integers with no common factor, and [imath]f(x)=0[/imath] otherwise. Show that [imath]f[/imath] is integrable over [imath][0,1][/imath]. I'm using the Darboux definition of integration, so I want to prove that for any [imath]\epsilon>0[/imath] there exists a partition [imath]P[/imath] of [imath][0,1][/imath] such that [imath]U(f,P)-L(f,P)<\epsilon[/imath]. Equivalently, there exists a partition [imath]P[/imath] of [imath][0,1][/imath] such that [imath]\sum_Rv(R)(M_R(f)-m_R(f)) < \epsilon[/imath] where [imath]M_R(f)[/imath] is the supremum of [imath]f[/imath] inside interval [imath]R[/imath], [imath]m_R(f)[/imath] is the infimum of [imath]f[/imath] inside interval [imath]R[/imath], and [imath]R[/imath] ranges over all intervals in the partition. So I tried taking [imath]P=[0,\dfrac1n,\dfrac2n,\ldots,1][/imath]. The sum in question becomes [imath]\dfrac1n\sum_{i=0}^{n-1}(M_{[\frac{i}{n},\frac{i+1}{n}]}(f)-m_{[\frac{i}{n},\frac{i+1}{n}]}(f))[/imath] I know that [imath]m_{[\frac{i}{n},\frac{i+1}{n}]}(f)=0[/imath], because in the interval [imath][\dfrac{i}{n},\dfrac{i+1}{n}][/imath] there is an irrational number, so the sum reduces to [imath]\dfrac1n\sum_{i=0}^{n-1}M_{[\frac{i}{n},\frac{i+1}{n}]}(f)[/imath] I don't really know anything about the fraction with lowest denominator inside [imath][\dfrac{i}{n},\dfrac{i+1}{n}][/imath]. How can I prove that this sum goes to [imath]0[/imath] as [imath]n\rightarrow\infty[/imath]? |
1151254 | First order partial derivatives and ball centered at [imath]a[/imath]
Suppose [imath]r>0[/imath], [imath]a\in\mathbb{R}^n[/imath], and [imath]f\colon B_r(a)\to\mathbb{R}^m[/imath]. If all first order partial derivatives of [imath]f[/imath] exist on [imath]B_r (a)[/imath], and [imath]f_{x_j}(x) = 0[/imath] for all [imath]x\in B_r (a)[/imath] and all [imath]j=1,2,\ldots ,n[/imath]. Prove [imath]f[/imath] has only one value on [imath]B_r(a)[/imath]. I guess I don't quite understand the problem so I can't solve it. If there is a ball of radius [imath]r>0[/imath] centered at [imath]a\in\mathbb{R}^n[/imath], its not clear to me how [imath]f[/imath] can only have one value on [imath]B_r(a)[/imath]. Also, what does every partial derivative being equal to [imath]0[/imath] for all [imath]x\in B_r(a)[/imath] tell me? Can anyone give me some hints and/or solve this problem? | 69294 | How does one prove if a multivariate function is constant?
Suppose we are given a function [imath]f(x_{1}, x_{2})[/imath]. Does showing that [imath]\frac{\partial f}{\partial x_{i}} = 0[/imath] for [imath]i = 1, 2[/imath] imply that [imath]f[/imath] is a constant? Does this hold if we have [imath]n[/imath] variables instead? |
1154801 | Integer values of [imath]a[/imath] for which [imath]\lim_{x\rightarrow 1}\left(\frac{-ax+\sin (x-1)+a}{x+\sin (x-1)-1}\right)^{\frac{1-x}{1-\sqrt{x}}} = \frac{1}{4}[/imath]
Find Largest Integer values of [imath]a[/imath] for which [imath]\displaystyle \lim_{x\rightarrow 1}\Bigg(\frac{-ax+\sin (x-1)+a}{x+\sin (x-1)-1}\Bigg)^{\frac{1-x}{1-\sqrt{x}}} = \frac{1}{4}\;,[/imath] is [imath]\bf{My\; Solution::}[/imath] We can write the above Limit as [imath]\displaystyle \lim_{x\rightarrow 1}\Bigg(\frac{-a(x-1)+\sin(x-1)}{(x-1)+\sin(x-1)}\Bigg)^{1+\sqrt{x}} = \frac{1}{4}[/imath] So [imath]\displaystyle \lim_{x\rightarrow 1}\Bigg(\frac{-a+\frac{\sin(x-1)}{x-1}}{1+\frac{\sin (x-1)}{x-1}}\Bigg)^2 = \frac{1}{4}\Rightarrow \bigg(\frac{-a+1}{2}\bigg)^2 = \bigg(\frac{1}{2}\bigg)^2[/imath] So We get [imath]\displaystyle (-a+1) = \pm 1\Rightarrow a = 0[/imath] or [imath]a=2[/imath] But answer given as [imath]a = 0[/imath] and I did not understand why it can not be [imath]a = 2[/imath] Please explain the reasoning behind it, Thanks | 818138 | A limit question (JEE [imath]2014[/imath])
The following is a JEE (A national level entrance test) question: Find the largest value of the non-negative integer ( a ) for which: [imath] \displaystyle \lim_{x \to 1} \left( \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } \right)^{\dfrac{1-x}{1-\sqrt x} } = \dfrac 1 4 [/imath] On solving this, we get [imath] a = 0 [/imath] or [imath] a = 2 [/imath]. So we take the answer as [imath] 2 [/imath]. But the official answer key says that the answer is [imath] 0 [/imath]. The reason given (not in the official key - they simple contain the answer. I saw the reason in an "unofficial" solution to the question paper) is that if [imath] a = 2 [/imath], then the term [imath]\displaystyle \dfrac{-ax + \sin(x-1) + a} { x + \sin(x-1) -1 } [/imath] tends to a negative value ( [imath]-0.5[/imath] ). So [imath] a = 0[/imath]. BUT: By putting [imath]a = 2[/imath] in the limit, and inputting that to wolframalpha (http://www.wolframalpha.com/input/?i=lim(+(+(-2x+%2B+2+%2B+sin(x-1)+)+%2F+(+x+-+1+%2B+sin(+x+-+1)+)+)+%5E+(+(1-x)%2F(1-sqrt(x))+)+)+as+x+tends+to+1), the answer turns out to be [imath]0.25[/imath]. So, by putting [imath] a = 2[/imath] also, we get the same value of the limit - [imath]0.25[/imath]. How to solve this dispute? Also, could you please check with Mathematica? Unfortunately, I don't have access to it! (This is very important as the change in the answer key will affect the rankings in a dramatic way) |
1154544 | Why not add something to both sides of a purported identity to prove it?
A section in my precalculus book is devoted to establishing (=proving) trigonometric identities, and a typical problem in the book presents a purported identity and asks students to establish it. The book recommends this method for doing so: Consider the more complicated-looking side of the purported identity. Use rules of algebra and known trigonometric identities to manipulate that side until it matches the other side. (Sometimes you'll need to manipulate both sides until they match one another.) The book then has this warning: Be careful not to handle identities to be established as if they were equations. You cannot establish an identity by such methods as adding the same expression to each side and obtaining a true statement. This practice is not allowed, because the original statement is precisely the one that you are trying to establish. You do not know until it has been established that it is, in fact, true. Huh? I mean, I understand that you need to be careful. I understand that you can't manipulate your purported identity thus:[imath]\{\textrm{purported identity}\}\Rightarrow\{\textrm{something else}\}[/imath]I understand that every implication must be instead like this:[imath]\{\textrm{purported identity}\}\Leftarrow\{\textrm{something else}\}[/imath]And therefore, for example, one cannot raise both sides of the purported identity to an even power, or multiply both sides by [imath]0[/imath]. Fine. But what's wrong with "adding the same expression to each side and obtaining a true statement"?? | 101053 | Why is it that when proving trig identities, one must work both sides independently?
Suppose that you have to prove the trig identity: [imath]\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}=\sin\theta[/imath] I have always been told that I should manipulate the left and right sides of the equation separately, until I have transformed them each into something identical. So I would do: [imath]\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}[/imath] [imath]=\frac{\sin\theta(1 - \sin^2\theta)}{\cos^2\theta}[/imath] [imath]=\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}[/imath] [imath]=\sin\theta[/imath] And then, since the left side equals the right side, I have proved the identity. My problem is: why can't I manipulate the entire equation? In this situation it probably won't make things any easier, but for certain identities, I can see ways to "prove" the identity by manipulating the entire equation, but cannot prove it by keeping both sides isolated. I understand, of course, that I can't simply assume the identity is true. If I assume a false statement, and then derive from it a true statement, I still haven't proved the original statement. However, why can't I do this: [imath]\frac{\sin\theta - \sin^3\theta}{\cos^2\theta}\not=\sin\theta[/imath] [imath]\sin\theta - \sin^3\theta\not=(\sin\theta)(\cos^2\theta)[/imath] [imath]\sin\theta(1 - \sin^2\theta)\not=(\sin\theta)(\cos^2\theta)[/imath] [imath](\sin\theta)(\cos^2\theta)\not=(\sin\theta)(\cos^2\theta)[/imath] Since the last statement is obviously false, is this not a proof by contradiction that the first statement is false, and thus the identity is true? Or, why can't I take the identity equation, manipulate it, arrive at [imath](\sin\theta)(\cos^2\theta)=(\sin\theta)(\cos^2\theta)[/imath], and then work backwards to arrive at the trig identity. Now, I start with a statement which is obviously true, and derive another statement (the identity) which must also be true - isn't that correct? Another argument that I have heard for keeping the two sides isolated is that manipulating an equation allows you to do things that are not always valid in every case. But the same is true when manipulating just one side of the equation. In my first proof, the step [imath]\frac{\sin\theta(\cos^2\theta)}{\cos^2\theta}[/imath] [imath]=\sin\theta[/imath] is not valid when theta is [imath]\pi/2[/imath], for example, because then it constitutes division by zero. |
1155422 | Evaluate [imath]\sum\limits_{k=1}^{n} \frac{k}{2^k}[/imath]
Evaluate [imath]\sum\limits_{k=1}^{n} \frac{k}{2^k}[/imath] | 1657690 | Compute the sum [imath]\sum_{k=1}^{10}{\dfrac{k}{2^k}}[/imath]
Compute the sum [imath]\sum_{k=1}^{10}{\dfrac{k}{2^k}}[/imath] This question is taken from SMO junior (I can't remember which year it is). I have no idea how to start. Can anyone give some hint? By writing out the sum, one has [imath]\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + ... + \frac{10}{1024}[/imath] but I don't know how to proceed from here. |
1155579 | Proof involving fibonacci number: [imath] \binom{n}{1}F_1+\binom{n}{2}F_2+\binom{n}{3}F_3+\cdots+\binom{n}{n-1}F_{n-1}+F_n=F_{2n}, [/imath]
Problem: Prove that [imath] \binom{n}{1}F_1+\binom{n}{2}F_2+\binom{n}{3}F_3+\cdots+\binom{n}{n-1}F_{n-1}+F_n=F_{2n}, [/imath] where [imath]F_n[/imath] denotes the [imath]n[/imath]th Fibonacci number. I tried induction, but I didn't know what to do for [imath]n=k+1[/imath]. | 569873 | Fibonacci combinatorial identity: [imath]F_{2n} = {n \choose 0} F_0 + {n\choose 1} F_1 + ... {n\choose n} F_n[/imath]
Can someone explain how to prove the following identity involving Fibonacci sequence [imath]F_n[/imath] [imath]F_{2n} = {n \choose 0} F_0 + {n\choose 1} F_1 + ... {n\choose n} F_n[/imath] ? |
1155889 | Assuming [imath]a_k + b_k = 1[/imath] (Putnam 2003)
I do not understand as I wrote in a previous question: solution: I see that we can scale: [imath]u_k[/imath] but I do not understand why it is legal to say [imath]a_k + b_k = 1[/imath] what is [imath]a_1 = 20[/imath] and [imath]b_1 = 1[/imath] then [imath]a_1 + b_1 \ne 1[/imath]? | 1152994 | Putnam and Beyond AM-GM help
From Putnam and Beyond: The Solution is: The only part I do NOT understand is how: [imath]a_k + b_k = 1[/imath] for every [imath]k[/imath]? The problem just specifies nonnegative numbers? |
1156199 | How to prove this function is actually constant?
Let [imath]f[/imath] be function [imath]\:f:\mathbb{R}\rightarrow \mathbb{R}[/imath] which for any [imath]x[/imath], [imath]f\left(x\right)=f\left(\frac{x}{2}\right)[/imath]. Also, [imath]f[/imath] continuous at [imath]x=0[/imath]. Show that [imath]f[/imath] is constant function. So far i saw something here Does [imath]f(x) = f(2x)[/imath] for all real [imath]x[/imath], imply that [imath]f(x)[/imath] is a constant function?, but i don't understand the solution that he was given. I thought about assuming that there exist [imath]a,b[/imath] such that [imath]f\left(a\right)\ne f\left(b\right)[/imath] and get a contradict. Any ideas about what give me that [imath]f[/imath] is continuous at [imath]0[/imath]? and any ideas about a way to prove that? tnx in advance! | 374507 | Does [imath]f(x) = f(2x)[/imath] for all real [imath]x[/imath], imply that [imath]f(x)[/imath] is a constant function?
If a Continuous function [imath]f(x)[/imath] satisfies [imath]f(x) = f(2x)[/imath], for all real [imath]x[/imath], then does [imath]f(x)[/imath] necessarily have to be constant function? If so, how do you prove it? If not any counter examples? |
1141696 | How to compute [imath]\int_0^{\infty} \ln (1 + e^{-x})\, dx[/imath] and [imath]\int_0^{\infty} \ln (1 - e^{-x})\, dx[/imath]?
Bierenes de Haan's book (page 377) shows that [imath]\int_0^{\infty} \ln (1 + e^{-x})\, dx = \frac{\pi^2}{12}[/imath], and [imath]\int_0^{\infty} \ln (1 - e^{-x})\, dx = -\frac{\pi^2}{6}[/imath]. Anybody know how to compute them? Thanks. | 60478 | Evaluating the integral, [imath]\int_{0}^{\infty} \ln\left(1 - e^{-x}\right) \,\mathrm dx [/imath]
I recently got stuck on evaluating the following integral. I do not know an effective substitution to use. Could you please help me evaluate: [imath]\int_{0}^{\infty} \ln\left(1 - e^{-x}\right) \,\mathrm dx [/imath] |
709569 | wave question with piecewise initial conditions
My problem is regarding the wave equation of the form: [imath]u_{tt}=u_{xx}[/imath] in the domain [imath]D = \{ (x,t) \mid - \infty<x< \infty\ ,t>0\} [/imath] subject to the initial conditions: [imath]u(x,0)=\left\{ \begin{array}{c l} x^3-x, & |x|\leq1\\ 0, & |x|\geq1 \end{array}\right.[/imath] [imath]u_t(x,0)=\left\{ \begin{array}{c l} 1-x^2, & |x|\leq1\\ 0, & |x|\geq1 \end{array}\right.[/imath] I'm pretty sure this can be solved using D'Alambert's solution for a general wave equation, however I'm slightly unsure as to how to do this question when there are piecewise initial conditions. I know the initial conditions are continuous. How do I tackle this? Any help would be greatly appreciated! | 708385 | Solving Wave Equation with Initial Values
I am trying to solve the wave equation: [imath]u_{tt}[/imath] = [imath]u_{xx}[/imath] With initial values: [imath]u(x,0) =\begin{cases} x^3 - x, &\text{for }|x|\le 1,\ \\0, &\text{for }|x|\ge1\end{cases}[/imath] [imath]u_t(x,0) =\begin{cases} 1 - x^2, \text{for } |x|\le 1, \\ 0, \text{for } |x|\ge1 \end{cases}[/imath] In the domain [imath]D= \{(x,t) | - \infty < x<\infty, \space\space t>0\} [/imath] I am using D'Alembert's Formula and after inputting the values in I have: [imath]u(x,t) = \frac{1}{2}[\frac{2}{3}(x+t)^3 + \frac{4}{3}(x-t)^3 - 2(x-t)] [/imath] I am not sure where to go from here. I think because the initial conditions are not differentiable at some points then there will be points in the solution that are not differentiable So I'd need to consider the different cases of [imath]|x+t|\le 1[/imath]? I could be going down the complete wrong path so any help is appreciated. Thanks |
467755 | Counter examples in group theory
Let [imath]G[/imath] be a finite group with normal subgroups [imath]N_{1}[/imath] and [imath]N_{2}[/imath]. Find counter examples to the following statements 1) If [imath]N_{1}\cong N_{2}[/imath] then [imath]G/N_{1}\cong G/N_{2}[/imath] 2) If [imath]G/N_{1}\cong G/N_{2}[/imath] then [imath]N_{1}\cong N_{2}[/imath]. Thanks for the help. | 543839 | Counterexample for [imath]A,B\triangleleft G,G/A\cong B \Rightarrow G/B\cong A[/imath]
Let [imath]A,B\triangleleft G[/imath]. Give counterexample for the claims: a. [imath]G/A\cong B \Rightarrow G/B\cong A[/imath] b. [imath]G/A\cong G/B\Leftrightarrow A\cong B[/imath] I don't know from where to start. Can you give some counterexamples? addiotionally, what is the intuition of solving question in algebra which requires giving counter-examples |
347261 | Finding the limit [imath]\lim\limits_{n \to \infty} \int_1^a \frac{n}{1+x^n} dx[/imath] with [imath]a > 1[/imath]
[imath]I_n[/imath] is given by [imath]I_n=\int_1^a \dfrac{n \ dx}{1+x^n}, \qquad a>1.[/imath] Find [imath]\lim\limits_{n\to\infty}I_n[/imath]. | 339685 | [imath]\lim_{n \to \infty} \displaystyle \int_1^a \dfrac{n}{1+x^n}dx[/imath]
[imath]\lim_{n \to \infty} \displaystyle \int_1^a \dfrac{n}{1+x^n}dx[/imath], where parameter [imath]a>1[/imath]. Some trick is probably needed here, but i don't see it. |
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