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2893952 | Prove that [imath]f(x)<1+\pi/4[/imath]
Let [imath]f:[1,\infty)\to\mathbb R[/imath] be a twice differentiable function such that [imath] f'(x)=\frac{1}{x^2 + y^2} [/imath] and [imath]f(1)=1[/imath] then prove that [imath] f(x)<1+\frac{\pi}{4} [/imath] for every [imath]x \geqslant 1[/imath]. My process: I am not able to use the given differential equation as I don't know how to do it. If you can help me how to use the differential equation,I think then I can manage to solve this question. | 2268204 | Proving that [imath]f(x)[/imath] is less than or equal to [imath]1+\pi/4[/imath]
Suppose [imath]f[/imath] is a real valued differentiable function defined on [imath][1,\infty)[/imath] with [imath]f(1)=1[/imath]. Suppose also that [imath]f[/imath] satisfies [imath]f'(x)=\frac{1}{x^2+f^2(x)}.[/imath] The question is to prove that [imath]f(x) \leq 1+\pi/4[/imath] for every [imath]x \geq 1[/imath] I tried to solve the differential equation but could not bring it in some known form. I examined the derivative of [imath]\tan^{-1}x[/imath] which looks similar to that in the question. However I could not get any idea with that. Any help shall be highly appreciated. Thanks. |
2893971 | If a function is measurable, then so is its absolute value.
How can I prove this proposition: If a function [imath]f[/imath] is measurable then, its absolute value, [imath]|f|[/imath], is also a measurable function? | 2172368 | Absolute value of a measurable function is measurable
Let [imath](X , \mathscr{M}, \mu)[/imath] be a measure space. Prove that if [imath]f[/imath] is measurable, then [imath]|f|[/imath] is measurable. A function [imath]f[/imath] is measurable if [imath]f^{-1}((a , \infty)) = \{x : f(x) > a\} \in \mathscr{M}[/imath] for every [imath]a \in \mathbb{R}[/imath]. Ok so this is what I have so far: Consider two cases: [imath]a < 0[/imath] and [imath]a \geq 0[/imath]. If [imath]a < 0[/imath], then we have [imath]|f|^{-1}((a , \infty)) = \{x : |f(x)| > a \} = X \in \mathscr{M}[/imath] since [imath]\mathscr{M}[/imath] is a [imath]\sigma[/imath]-algebra. I am confused on how to prove the case when [imath]a \geq 0[/imath]. If [imath]a \geq 0[/imath], we have [imath]|f|^{-1}((a , \infty)) = \{x : |f(x)| > a \} = ?[/imath] I am assuming we will have to use that fact that [imath]f[/imath] is measurable. Any help will be appreciated! |
2885930 | Rational scalar multiplication in Linear transformation using addition property
Linear transformation: Map [imath]T:V\to W[/imath] is said to be Linear transformation when it satisfy property : 1) [imath]L(u+v)=L(u)+L(v)\forall u,v \in V[/imath] 2)[imath]L(cv)=cL(v) \quad \forall c\in \mathbb{F}[/imath] I know when [imath]c \in \mathbb N[/imath] then property 2 can be obtained from property 1, since [imath]L(nv)= \underbrace{L(v)+L(v)......+L(v)}_{\text{$n$ times where $n \in \mathbb{N}$}}[/imath] Now I have read in somewhere that if our scalar [imath]c[/imath] is rational then property 2 also follows from property 1, but I am unable to prove this. How can I show that if [imath]c \in \mathbb{Q}[/imath], when (2) follows from (1)? I know that if our scalar is irrational, then we cannot use property 1 to prove property 2. | 1645660 | If [imath]V_1, V_2[/imath] are [imath]\mathbb{Q}[/imath]-vector spaces and [imath]f \colon V_1 \to V_2[/imath] is additive, then [imath]f[/imath] is [imath]\mathbb{Q}[/imath]-linear
If [imath]V_1[/imath] and [imath]V_2[/imath] are vector spaces over the field of rational numbers and [imath]f \colon V_1 \to V_2[/imath] is a map such that [imath]f(x+y)=f(x)+f(y)[/imath] for all [imath]x, y \in V_1[/imath], show that [imath]f[/imath] is a linear transformation. It is easy to show that [imath]f(0)=0[/imath] and [imath]f(-x)=-x[/imath]. Here scalars are from rationals. If scalar is an integer then clearly [imath]f(\alpha x)=\alpha f(x)[/imath]. Hence [imath]f[/imath] is a linear transformation. Now I have to show that when [imath]\alpha[/imath] is not an integer then [imath]f(\alpha x)=\alpha f(x).[/imath] |
2876770 | Image under [imath]w=e^{1/z}[/imath].
I have been trying to find the image of [imath]S=\{z\in\mathbb{C}; 0<|z|<r\}[/imath], with [imath]r>0[/imath], under [imath]w=e^{1/z}[/imath]. I started this way: if [imath]z=|z|e^{i\theta}\in S[/imath], by Euler's formula [imath] 1/z = |z|^{-1}(\cos\theta-i\sin\theta) = |z|^{-1}\cos\theta-|z|^{-1}i\sin\theta. [/imath] Considering [imath]w=|w|e^{i\alpha}[/imath] as image of [imath]z[/imath], [imath] |w|e^{i\alpha} = e^{|z|^{-1}\cos\theta} e^{-|z|^{-1}i\sin\theta}. [/imath] And then [imath] |w| = e^{|z|^{-1}\cos\theta} \implies \ln|w| = |z|^{-1}\cos\theta \implies \cos\theta = |z| \ln|w|. [/imath] And also [imath] \alpha =-|z|^{-1}i\sin\theta \implies \sin\theta =-\alpha |z|. [/imath] What do I have to do now? | 16909 | Determine the set of values of [imath]\exp(1/z)[/imath] for [imath]0<|z|[/imath]
I can't do this exercise of Conway's Book: For [imath]r>0[/imath] let [imath]A=\{w:w=\exp(1/z), 0<|z|<r\}[/imath], determine the set [imath]A[/imath]. Any hints? |
2894277 | Units within a log after integration
Let [imath]x(t)[/imath] and [imath]a[/imath] be in unit meters, and [imath]t[/imath] unit time. If we look at the indefinite integral [imath] \int\frac{\frac{dx}{dt}}{x(t) + a} dx [/imath] then we note that the units of the above expression are [imath]\frac{meters}{seconds}[/imath]. If we evaluate the integral, we get [imath] \frac{dx}{dt}\log(a + x(t)) [/imath] But the log has units meters in its argument, which, as I understand, transcendental functions' arguments should be dimensionless. Changing the indefinite integral into a definite one does not necessarily fix this, because, as I understand [imath] \int_{x=x_0}^{x=x_1}\frac{\frac{dx}{dt}}{x(t) + a} dx = \frac{dx}{dt}\log(a + x(t))|^{x=x_1}_{x=x_0} = \frac{dx}{dt}(x=x_1)\log(a+x_1) - \frac{dx}{dt}(x=x_0)\log(a+x_0) [/imath] so what gives? | 669440 | units in indefinite integral
When answering this question, I came up with the following: Suppose we have a length [imath]l[/imath] and integrate [imath]\frac 1 l[/imath] over [imath]l[/imath]: [imath]\int \frac {\operatorname d\!l}l=\ln l[/imath] Since the dimension of [imath]l[/imath] is length, and [imath]\ln[/imath] should have a dimensionless argument, this cannot be a 'good' integral in the physical sense, although it would be mathematically (omitting units). A way to solve this would be to reformulate the indefinite integral like this: for [imath]c=-\ln (1 \text{meter})[/imath]. [imath] \int \frac {\operatorname d\!l}l=\ln l+C=\ln l-\ln( 1\text{m})=\ln\left(\frac{l}{1\text{meter}}\right) [/imath] The question is whether the first integral given would be acceptable and how to deal with the two logarithms with bad arguments, which, when combined, aren't a problem anymore. |
2894886 | Prove that when [imath]F[/imath] is a field, [imath]F[x_1 , x_2][/imath] is not a principal ideal ring.
This question is from Herstein’s Topic in Algebra, page 166, question No. 8. Prove that when [imath]F[/imath] is a field, [imath]F[x_1 , x_2][/imath] is not a principal ideal ring. I didn't understand, as no answer has been given in this post: Prove that it is not a Principal Ideal Ring Any hints/solution will be appreciated. Thanks in advance | 54206 | If F is a field, then [imath]F[x,y][/imath] is a Principal Ideal Domain?
Let [imath]F[/imath] be a field, and [imath]F[x,y][/imath] be a ring of polynomials in two variables. Is [imath]F[x,y][/imath] a Principal Ideal Domain? Also show that [imath]F[x,y]/(y^2-x)[/imath] and [imath]F[x,y]/(y^2-x^2)[/imath] are not isomorphic for any field [imath]F[/imath]. |
2881371 | Why are additive limit preserving functors left exact?
I understand that the kernel is a limit and thus preserved by the functor, but why is exactness preserved in the middle? i.e How does the exactness of [imath]0\to A\to B\to C\to 0[/imath] imply the exactness of [imath]0\to F(A)\to F(B)\to F(C)[/imath] at [imath]F(B)[/imath], where [imath]F[/imath] is limit preserving? | 2314626 | Reconciling different definitions of left-exact functor
I've seen three different definitions of what a left-exact functor is: Here as a functor that preserves finite limits Here as a functor [imath]F[/imath] with the property that if [imath]0 \to A \to B \to C \to 0[/imath] is exact then [imath]0 \to F(A) \to F(B) \to F(C)[/imath] is exact Also here as a functor [imath]F[/imath] with the property that if [imath]0 \to A \to B \to C[/imath] is exact then [imath]0 \to F(A) \to F(B) \to F(C)[/imath] is exact I'm having a hard time convincing myself that these three definitions are equivalent. Its clear off the bat that any functor with the third property also has the second (since exactness of [imath]0 \to A \to B \to C \to 0[/imath] implies exactness of [imath]0 \to A \to B \to C[/imath]); but its not immediately obvious why a functor having the second property should also have the third property? And I'm having trouble seeing the link between the second and third definitions and the first definition too. I'm not looking so much for formal proofs as much as just an intuitive idea as to what exactly 'left-exact functors' are; and why (intuitively) these three definitions should be equivalent? |
2895157 | Number of ways to seat [imath]n_i[/imath] people around a round table with [imath]n[/imath] seats [imath]\bigl(n_i \lt n \bigl)[/imath]
So the question says to find the number of ways to seat [imath]n_i[/imath] people around a roundtable with [imath]n[/imath] seats [imath]\bigl(n_i \lt n \bigl)[/imath]. The first thing I did was select [imath]n_i[/imath] seats from [imath]n \Rightarrow \binom {n}{n_i}[/imath] Now the number of ways to arrange these [imath]n_i[/imath] people is obviously [imath]\bigl(n_i -1 \bigl)![/imath] So we get, [imath]\binom {n}{n_i} \cdot \Bigl(n_i -1\Bigl)![/imath] The answer however given is [imath]\left( \frac{n!}{\Bigl(n-n_i\Bigl)!}\right)[/imath] Please tell me where I am going wrong. | 2139932 | Arrange [imath]m[/imath] people in [imath]m+r[/imath] seats around a round table.
The answer is, according to the book: [imath](m-1)!\cdot \binom{m+r-1}{r}[/imath] I get why this is true. You arrange [imath]m[/imath] people in their seats in [imath](m-1)![/imath] ways and then you put [imath]r[/imath] empty spots inbetween them.I was thinking - choose [imath]m[/imath] seats out of [imath]m+r[/imath] to put the people on => [imath]\binom{m+r}{m}[/imath] Arrange them once you've chosen the seats [imath]\to (m-1)![/imath] So in total: [imath]\binom{m+r}{m} \cdot (m-1)![/imath] This is obviously not true. What's wrong with the way I'm thinking? |
2884310 | For a finite commutative ring [imath]R[/imath], [imath]a \in R[/imath] is a root of [imath]p(x)[/imath] iff [imath]p(x)[/imath] can be written as [imath]p(x) = (x-a)g(x)[/imath]
[imath]R[/imath] is a finite commutative ring with identity. Let [imath]p(x) \in R[x][/imath], the ring of polynomials over [imath]R[/imath]. Show that [imath]a \in R[/imath] is a root of [imath]p(x)[/imath] if and only if [imath]p(x)[/imath] can be written as [imath]p(x) = (x-a)g(x)[/imath] with [imath]g(x) \in R[x][/imath] of degree one less than the degree of [imath]g(x)[/imath]. The "if" part is trivial. I'm struggling with the "only if" part since [imath]R[x][/imath] is not an Euclidean domain so we cannot use the Division Algorithm. [imath]R[/imath] may not even be an Integral domain, so I can't view [imath]p(x)[/imath] as a polynomial in its field of fractions. Is there some other way to do it using finiteness of [imath]R[/imath]? | 1599165 | Prove that [imath]p(x)=(x-a)q(x)[/imath] for some [imath]q(x)\in R[x][/imath] of degree= (degree of [imath]p)-1[/imath].
Let [imath]R[/imath] be a finite commutative ring with identity. Let [imath]a\in R[/imath] is a root of [imath]p(x)\in R[x][/imath] . Prove that [imath]p(x)=(x-a)q(x)[/imath] for some [imath]q(x)\in R[x][/imath] of degree= (degree of [imath]p)-1[/imath]. If [imath]R[/imath] were a Euclidean domain then it is obvious. But [imath]R[/imath] is not necessarily a ED. So what I was trying to do is showing [imath]p(x)\in (x-a)[/imath]; the ideal generated by [imath]x-a[/imath]. I tried defining a homomorphism [imath]f:R[x]\rightarrow R[/imath] by [imath]f(x)=a[/imath] and tried to show ker[imath]f=(x-a)[/imath]. But when I was going to show it, I felt like showing the same thing that asks in the question. So no success. Can anybody please give me at least a hint? I feel like this is a standard question. So I searched through MSE for similar questions but could not find. If you find a duplicate please copy a link and then I will delete this question. |
2895741 | How to calculate [imath]\int_{0}^{\pi/2}\dfrac{1}{\sin^3 x + \cos^3 x}dx[/imath]
One way that I tried to solve an integral was to divide integrand by [imath]\cos^3x[/imath] para para obter [imath] \int_{0}^{\pi/2}\dfrac{\sec^3x}{1 + \tan^3x}dx [/imath] Through software, I saw that the result involves [imath]\tanh^{-1}\bigl(\frac{1}{\sqrt{2}}\bigr)[/imath], but I do not know how to get there. | 1694424 | How do I integrate [imath]\int \frac{dx}{\sin^3 x + \cos^3 x}[/imath]?
How do I integrate the following [imath]\int \frac{dx}{\sin^3 x + \cos^3 x}[/imath] ? It appears that I am supposed to break this up into [imath](\sin x + \cos x)(1-\cos x \sin x)[/imath], but the next thing to do is not apparent to me. |
2895873 | Evaluating the function [imath]S(a) = \sum_{n=0}^{\infty} a^n \binom{2n}{n}.[/imath]
Evaluating the function [imath]S(a) = \sum_{n=0}^{\infty} a^n \binom{2n}{n}[/imath]. The first thing that I noticed is that [imath]S(a)[/imath] converges when [imath]a \in [-\frac{1}{4}, \frac{1}{4})[/imath] because [imath] \binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}. [/imath] However, this does not help me actually find [imath]S(a)[/imath]. Furthermore, various strategies such as telescoping and DuSS (differentiation under the summation sign) have been foiled by the pesky binomial coefficient. If we switch to the Gamma function, DuSS is possible but extremely messy. I'm wondering what approach I should utilize in order to find an expression for [imath]S(a)[/imath] in terms of elementary functions. | 1080635 | What is the value of [imath]\sum_{n=0}^{\infty}(-\frac{1}{8})^n\binom{2n}{n}[/imath]
What is the value of [imath]\sum_{n=0}^{\infty}\left(-\frac{1}{8}\right)^n\binom{2n}{n}\;?[/imath] EDIT I bumped into this series when inserting [imath]\overrightarrow{r_1}=\left(\begin{array} {c}0\\0\\1\end{array}\right)[/imath] and [imath]\overrightarrow{r}=\left(\begin{array} {c}1\\1\\0\end{array}\right)[/imath] into [imath]\frac{1}{|\overrightarrow{r}-\overrightarrow{r_1}|}=\sum_{l=0}^\infty\frac{r_{<}^l}{r_{>}^{l+1}}P_{l}(cos\theta)\;.[/imath] See (Series representation of [imath]1/|x-x'|[/imath] using legendre polynomials). So I knew the result, [imath]\sqrt{\frac{2}{3}}[/imath], but wished to find out what other approaches there would be to evaluate the series. It transpires that I overlooked the relatively standard evaluation of [imath]\sum_{n=0}^{\infty}\binom{2n}{n}x^n[/imath] as being equal to [imath]\sqrt{\frac{1}{1-4x}}[/imath], following a calculation similar to that given by achille-hui below, which requires some complex function theory, in particular when proving that [imath]\binom{2n}{n} = \frac{1}{2\pi}\int_0^{2\pi}\left(e^{i\theta}+e^{-i\theta}\right)^{2n}d\theta[/imath], or that of alex.jordan below, requiring no more than Taylor expansion. |
2457359 | show that [imath] \ \sum_{n=0}^{\infty} \binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}} \ [/imath]
If [imath] \ \ \large \sum_{n=0}^{\infty} \frac{1}{n+1} \binom{2n}{n} x^n= \frac{1-\sqrt{1-4x}}{2x} \ [/imath] , then show that [imath] \ \sum_{n=0}^{\infty} \binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}} \ [/imath] . Answer: I have no idea how to start this type of . Any idea ? | 205898 | How to show that [imath]1 \over \sqrt{1 - 4x} [/imath] generate [imath]\sum_{n=0}^\infty \binom{2n}{n}x^n [/imath]
I need to find the values of [imath]a[/imath] and [imath]n[/imath] in the following [imath] (1 + ax)^n = \sum_{i=0}^\infty \binom{2i}{i}x^i[/imath] How can I compare things and show that [imath]a=-4[/imath] and [imath]n = -{1 \over 2}[/imath]. It's easy to expand [imath]1 \over \sqrt{1 -4x}[/imath] see that it's valid but I'm looking other way around. |
2896293 | Show that a connected simple graph for which every edge is in some perfect matching is a block
Show that a connected simple graph for which every edge is in some perfect matching is a block. It seems obvious to prove by contradiction. so Suppose such a graph [imath]G[/imath] has a cut-vertex [imath]v[/imath]. The direction from here seems to be taking two components [imath]H_1, H_2[/imath] of [imath]G-v[/imath], and let [imath]e_1, e_2[/imath] be the edges by which [imath]v[/imath] was connected to those respectively.[imath]e_1[/imath] and [imath]e_2[/imath] are both in some perfect matchings in [imath]G[/imath], but they must be different matchings since they share a vertex. Call them [imath]M_1, M_2[/imath]. How to continue from here? I'm guessing I'm looking for some contradiction to arise when looking at [imath]M_1\cap E(H_2)[/imath] or vice versa, but I'm not sure exactly how. | 184505 | Match covered graph is 2-connected
Seems to be an easy question, but I can't find the right direction. Let [imath]G[/imath] connected graph on at least 4 vertices, such that every edge in it, participates in a perfect matching. Prove that [imath]G[/imath] is 2-connected. Any help will be appreciated. Thanks in advance. |
2896225 | Reduction modulo homomorphism from Z_m to Z_n
Consider positive integers [imath]m>n[/imath] and function [imath]\phi (x)[/imath] from [imath]Z_m[/imath] into [imath]Z_n[/imath] equal to the remainder when [imath]x[/imath] is divided by [imath]n[/imath]. Is [imath]\phi[/imath] a homomorphism? I.e is [imath]\phi(x+y)=\phi(x)+\phi(y)[/imath] for every [imath]x,y \in Z_m[/imath] Fraleigh asks a number of these questions with concrete [imath]m,n[/imath] that can be proved by inspecting cases but this is messy and it seems that if [imath]n|m[/imath] than [imath]\phi[/imath] is a homomorphism but I am not sure how to prove it in the general case. I can prove that if gcd(m,n)=1 that it is not a homomorphism. | 522348 | When are [imath]\mathbb Z_m[/imath] and [imath]\mathbb Z_n[/imath] homomorphic?
Let [imath]m[/imath] and [imath]n[/imath] be two given positive integers such that [imath]m<n[/imath]. Then what are the necessary and sufficient conditions for the groups [imath](\mathbb Z_m,+_m)[/imath] and [imath](\mathbb Z_n,+_n)[/imath] to be homomorphic under the map [imath]\phi \colon \mathbb Z_n \to \mathbb Z_m[/imath] such that [imath]\phi(x)[/imath] is the remainder when [imath]x[/imath] is divided by [imath]m[/imath], for each [imath]x[/imath] in [imath]\mathbb Z_n[/imath]. I know that if one of these integers is a divisor of the other, then the groups are homomorphic. |
2896331 | In construction of an inner product that maps from a vector space to a dual, what makes the map considered to be natural?
My question is in reference to the derivation below from the physics site. It shows how the metric tensor raises and lowers indices. I cut it off halfway through because I didn't think the full derivation was relevant to the question, but here it is, just in case. I don't believe my question is a duplicate. My question is, why is the map [imath]\tau[/imath] considered to be "natural"? Does the fact that the vector space has an inner product necessitate that the dual has one? (maybe so that there's an isomorphism?) Edit: I'm asking if the fact that the vector space has an inner product induces the need for its dual to have an inner product to create an isomorphism. | 274959 | Dual Spaces and Natural maps
(I'll explain what I know first and then I'll ask the questions). Given a finite dimensional vector space [imath]V[/imath], it is often remarked that there is no "natural" isomorphism from [imath]V[/imath] to [imath]V^*[/imath] (I guess this means a basis independent isomorphism?). I understand that one typically constructs an isomorphism [imath]V \to V^*[/imath] by fixing a basis for [imath]V[/imath], call this [imath]B = \{v_1, v_2, \ldots, v_n\}[/imath], and then mapping [imath]v_i \mapsto \delta_i[/imath] where [imath]\delta_i[/imath] is the linear functional given by [imath]\delta_i(v_i) = \delta_{ij}[/imath]. Here are my questions: (1) Why is there no natural isomorphism [imath]V \to V^*[/imath]? (2) If [imath]V[/imath] is a finite dimensional inner product space, we can map each [imath]v \mapsto \langle v, \cdot \rangle[/imath]. Is this not "natural"? (I know we can consider sesquilinear forms, but let's keep the discussion to this inner product for simplicity). |
2896397 | Which of the following polynomials has the greatest real root?
Which of the following polynomials has the greatest real root? [imath]\textbf{(A) } x^{19}+2018x^{11}+1 \qquad[/imath] [imath]\textbf{(B) } x^{17}+2018x^{11}+1 \qquad[/imath] [imath]\textbf{(C) } x^{19}+2018x^{13}+1 \qquad[/imath] [imath]\textbf{(D) } x^{17}+2018x^{13}+1 \qquad[/imath] [imath]\textbf{(E) } 2019x+2018[/imath] To 'find' real roots we can set each expression [imath]A[/imath] to [imath]E[/imath] equal to [imath]0[/imath], which makes it pretty clear that the value [imath]x[/imath] must be negative (for else they'd be huge positive numbers), but also pretty small, since the terms with variable [imath]x[/imath] in each expression (despite their coefficients) must equal [imath]-1[/imath], for (A) to (D). [imath]x=-1[/imath] or [imath]x=0[/imath] will not work, however, so we need [imath]x[/imath] to be [imath]-1<x<0[/imath]. For (E) we can just solve for [imath]x[/imath] from [imath]2019x + 2018=0[/imath], but that gives an [imath]x[/imath] pretty close to [imath]-1[/imath] and I need to see if there are any greater roots (i.e. closer to [imath]0[/imath]). However, now I'm stuck and don't know how to move on. Any help would really be appreciated! | 2814099 | 2018 AMC 12A Problem 21: Which of the given polynomials has the greatest real root?
Which of the following polynomials has the greatest real root? [imath]\textbf{(A) } x^{19}+2018x^{11}+1 \qquad \textbf{(B) } x^{17}+2018x^{11}+1 \qquad \textbf{(C) } x^{19}+2018x^{13}+1 \qquad \textbf{(D) } x^{17}+2018x^{13}+1 \qquad \textbf{(E) } 2019x+2018[/imath] The only root that I was able to find was the last one. I have no idea what to do from here. Does anyone have some ideas? |
2896818 | Parabolic Proof
Prove that the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to the vertex. My issue with this problem is that I don't understand it completely. I've tried drawing it a few times, but I can't seem to form a precise image in my head of what it's depicting. Here's my current understanding: The "endpoint" of the perpendicular dropped from the focus to any arbitrary tangent of the parabola also somehow lies on the vertex. The tangent to the vertex of parabola, as I understand, can be described as the line [imath]y = k[/imath] or [imath]x = h[/imath] where [imath]k[/imath] is the ordinate point and [imath]h[/imath] is the abscissa. If my above understanding is correct (which it 99.99% isn't), then I don't think the foot of the perpendicular from the focus to a tangent of a parabola can lie on the tangent to the vertex. If anyone can provide a picture of the situation or give me a hint, that's more than enough for solving this problem. Thank you. | 824027 | Non-brute-force proof of parabola tangent property
I'm working through a classic Calculus book (Morris Kline), and one of the problems is: Prove that the foot of the perpendicular from the focus to any tangent of a parabola lies on the tangent to the vertex. Basically, it's saying that if the parabola [imath]y=x^2[/imath] has a tangent line T at point [imath]P[/imath], and we draw a line [imath]L[/imath] perpendicular to [imath]T[/imath] that goes through the focus [imath]F[/imath], then [imath]T[/imath] meets [imath]L[/imath] at a point where [imath]y=0[/imath]. I managed to prove this by: Calculating the equation of [imath]T[/imath] (based on slope of [imath]y'[/imath] and point [imath]P[/imath]) Calculating the equation of [imath]L[/imath] (based on slope of [imath]\frac{-1}{y'}[/imath] and point [imath]F[/imath]) Putting [imath]y=0[/imath] in both equations and solving for [imath]x[/imath] Observe that they both have the same [imath]x[/imath] at [imath]y=0[/imath] and therefore they meet on the [imath]x[/imath] axis. HOWEVER, this seems like a very brute-force approach. It's almost like I cheated or I simulated it on the computer with [imath]1000[/imath] points and determined that it is so. I don't really understand WHY these [imath]2[/imath] lines must meet in this manner. Is there any sort of geometric or intuitive proof? |
2897786 | A simple question about normal closure of field extensions.
Let [imath]K / F[/imath] be a finite and separable extension of fields and let [imath]L[/imath] a normal closure of [imath]K / F[/imath]. Can I state that [imath]L / F[/imath] is also finite and separable? | 1939472 | Separability and normal closure
I am not sure about this problem. Let [imath]K/F[/imath] be a finite separable extension and let [imath]\widetilde{K}/F[/imath] be a normal closure of [imath]K/F[/imath]. Is [imath]\widetilde{K}/F[/imath] necessarily separable? I tried considering [imath]\alpha \in \widetilde{K} \backslash K[/imath] and assume it is inseparable over [imath]F[/imath]. Then I considered [imath]m(x)[/imath] that is satisfied by [imath]\alpha[/imath], and [imath]\widetilde{K}[/imath] contains all roots of [imath]m(x)[/imath]. Then I am not sure how to proceed. Can anyone help? |
2897912 | Prove that for [imath]z \in\mathbb C\setminus \{1\},\, 1 + z + z^2 +... + z^n = (1-z)^{n+1}/(1 - z)[/imath] for [imath]n = 1,2,3\ldots[/imath]
Prove that for [imath]z \in\mathbb C\setminus \{1\}[/imath], [imath]\ 1 +z+z^2+...+z^n=\frac{1- z^{n+1}}{1-z}\quad (n = 1,2,3\ldots)[/imath] This is a complex analysis problem. I have been out of an official math class for a long time and am struggling to even know where to approach a problem like this, more or less where to start. Any help is greatly appreciated. | 11618 | Algebraic Identity [imath]a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}[/imath]
Prove the following: [imath]\displaystyle a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}[/imath]. So one could use induction on [imath]n[/imath]? Could one also use trichotomy or some type of combinatorial argument? |
2898368 | [imath]n^2 (n^4 - 1)[/imath] is divisible by 60
How does one prove that [imath]n^2 (n^4 - 1)[/imath] is divisible by 60 for any positive integer [imath]n[/imath]? I tried mathematical induction method without success so far. | 771410 | Prove [imath]n^2(n^4-1)[/imath] is divisible by 60 using Mathematical Induction.
Base step: p(2)=4 * 15= 60 Inductive Hypothesis: Assuming p(k) = [imath]k^2(k^4-1)[/imath] = 60q Induction: p(k+1)= [imath](k+1)^2[(k+1)^4-1][/imath] = [imath](k+1)^2[(k+1)^2 + 1][(k+1)^2 - 1][/imath] = [imath](k+1)^2(k^2+2k+2)(k^2+2k+1- 1)[/imath] = [imath](k+1)^2(k^2+2k+2) k (k+2)[/imath] = [imath]k(k+1)(k+2)(k^2+2k+2)(k+1)[/imath] Now [imath]k(k+1)(k+2)[/imath] is the product of 3 consecutive natural numbers. Their product HAS to be a multiple of 2 and 3 and hence also a multiple of 6. So all that I have to do is prove that [imath](k^2+2k+2)(k+1)[/imath] is a multiple of 10. But I can't seem to do so. Also through my induction, I have not at all made use of my hypothesis! How do I go about solving this question through the Principle of Mathematical Induction? |
2898314 | Prove [imath]\ \frac{z-1}{z+1} [/imath] is imaginary no' iff [imath]\ |z| = 1 [/imath]
Let [imath]\ z \not = -1[/imath] be a complex number. Prove [imath]\ \frac{z-1}{z+1} [/imath] is imaginary number iff [imath]\ |z| = 1 [/imath] Assuming [imath]\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 [/imath] and so [imath]\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \cdot \frac{a+1-bi}{a+1-bi} = \frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a} = \frac{bi}{1+a} [/imath] and therefore [imath]\ \frac{z-1}{z+1}[/imath] is imaginary now let me assume [imath]\ \frac{z-1}{z+1} [/imath] is imaginary number, how could I conclude that [imath]\ |z| =1 [/imath] I really can't think of any direction.. Thanks | 2346197 | Prove: if [imath]|z|=1[/imath] than [imath]\frac{z+1}{z-1}[/imath] is an imaginary number
Prove: if [imath]|z|=1[/imath] than [imath]\frac{z+1}{z-1}[/imath] is an imaginary number I have tried to look at [imath]\frac{z+1}{z-1}=\frac{z+1}{z-1}\cdot \overline{\frac{z-1}{\overline{z-1}}}=\frac{z+1}{z-1}\cdot \frac{\overline{z}-1}{\overline{z}-1}[/imath] But did not get far |
1411874 | Proving that a complex number lies on the imaginary axis.
Given that there are two complex numbers - [imath]z, w[/imath] - such that [imath]w\overline{w} = 1[/imath] and [imath]z = \frac{1+w}{1-w}[/imath], how do I deduce that [imath]z[/imath] lies on the imaginary axis? | 118868 | For complex [imath]z[/imath], [imath]|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right) = 0[/imath]
If [imath]|z|=1[/imath], show that: [imath]\mathrm{Re}\left(\frac{1 - z}{1 + z}\right) = 0[/imath] I reasoned that for [imath]z = x + iy[/imath], [imath]\sqrt{x^2 + y^2} = 1\implies x^2 + y ^2 = 1[/imath] and figured the real part would be: [imath]\frac{1 - x}{1 + x}[/imath] I tried a number of manipulations of the equation but couldn't seem to arrive at any point where I could link the two to show that the real part was = 0. |
2898689 | about square elements in finite field
q is power of odd prime.How to show that [imath]-1[/imath] is square in [imath]\mathbb{F_q}[/imath] if [imath] q=1(mod4)[/imath] i.e order of field divide 4 and -1 is not square if [imath] q=3(mod4)[/imath]. I know that since q is power of odd prime; order of [imath]\mathbb{F_q}^*[/imath] is [imath]q-1[/imath] and its cyclic group of even order. I also know only element of order 2 there is -1. Then how to proceed for both this claims? | 881615 | a square in a finite field of odd order
GF(q) is a finite field of order q, where q is odd. Prove that [imath]a\in GF(q), a\neq0[/imath] has a root in [imath]GF(q)[/imath] iff [imath]a^{(q-1)/2}=1[/imath]. I tried to prove it this way: Suppose a has a root in [imath]GF(q)[/imath], so there is some [imath]b\in GF(q)[/imath] so that [imath]b^2=a[/imath]. There is a primitive element [imath]g[/imath] in [imath]GF(q)[/imath], so that [imath]g^x=a, g^y=b[/imath]. Therefore [imath](g^y)^2=g^x\to 2y=x[/imath]. Furthermore [imath]a^{(q-1)/2}=(g^x)^{(q-1)/2}=g^{(x/2)*(q-1)}=g^{y(q-1)}[/imath] and since [imath]y[/imath] is an integer, [imath]g^{y(q-1)}=1^y=1[/imath]. Now suppose [imath]a^{(q-1)/2}=1[/imath], then again [imath](g^x)^{(q-1)/2}=1\to g^{(x/2)*(q-1)}=1[/imath]. That means that [imath]x/2[/imath] is an integer (or else [imath]g^{(x/2)*(q-1)}\neq 1[/imath]), and therefore [imath]0\le x/2\le q-1[/imath] being an integer, [imath]g^{x/2}[/imath] creates a certain element [imath]b\in GF(q)[/imath] so that [imath]b^2=a[/imath]. I am a bit unsure of this proof - first of all since I don't use the fact that [imath]q[/imath] is odd, which means I am wrong somewhere. Second of all, I feel extremely 'weak' with my argument regarding [imath]x/2[/imath] being an integer, yet I have no idea how to strengthen it. Any hints/directions will be extremely appreicated! |
346176 | Ideal commutative rings
Let [imath]R[/imath] a commutative ring and [imath]I[/imath], [imath]J[/imath] ideals of [imath]R[/imath] such that [imath]I + J = R[/imath]. Prove that [imath]IJ = I \cap J [/imath] Is clear that [imath]IJ \subseteq I[/imath] and [imath]IJ \subseteq J[/imath] then [imath]IJ \subseteq I \cap J[/imath] this for any two ideals of a ring. But i not know as proof the other contains, because it is necessary that [imath]I + J = R[/imath]? | 121968 | If [imath]I+J=R[/imath], where [imath]R[/imath] is a commutative rng, prove that [imath]IJ=I\cap J[/imath].
So I basically have to prove what is on the title. Given [imath]R[/imath] a commutative rng (a ring that might not contain a [imath]1[/imath]), with the property that [imath]I+J=R[/imath], (where [imath]I[/imath] and [imath]J[/imath] are ideals) we have to prove that [imath]IJ=I\cap J[/imath]. One inclusion is easy. If [imath]x\in IJ[/imath], then [imath]x=\sum a_ib_i[/imath] where [imath]a_i\in I[/imath] and [imath]b_i\in J[/imath]. Thus for any fixed [imath]i[/imath], we have that since [imath]a_i\in I[/imath], we have that [imath]a_ib_i\in I[/imath], and the same argument shows that [imath]a_ib_i\in J[/imath], thus [imath]\sum a_ib_i\in I[/imath] and [imath]\sum a_ib_i\in J[/imath], this means that [imath]x=\sum a_ib_i\in I\cap J[/imath], and thus [imath]IJ\subset I\cap J[/imath]. I am having troubles proving the other inclusion. Any comments? Thanks |
2899010 | Exercise from Evan's PDE, bounding the norm of a function that is zero on a subset, using Poincare inequality?
This is an extension to Partial differential equations by Lawrence C. Evans, page 291, Q15. Fix [imath]\alpha > 0[/imath], [imath]U[/imath] is a bounded open connected set with smooth boundary. Show that there exists some constant [imath]C[/imath] such that [imath]\forall u \in W^{1, p}(U)[/imath] with trace [imath]Tu = 0[/imath] on a relatively open [imath]\Gamma\subset\partial U[/imath] with [imath]\mathcal{H}^{n-1}(\Gamma)\ge \alpha[/imath], [imath]\|u\|_{L^p(U)} \le C \|\nabla u\|_{L^p(U)}[/imath] [[imath]\mathcal {H}^{n-1}[/imath] is the [imath](n-1)[/imath]-dimensional measure on [imath]\partial U[/imath].] The original question solution uses Poincare inequality, and I think this extension also uses Poincare inequality. Basically, the value of the integral of [imath]u[/imath] is bounded by its average value plus its oscillation, and the oscillation is bounded by its derivative through Poincare inequality. Then it remains to bound the average value of [imath]u[/imath] by its derivative. Since [imath]u[/imath] is anchored at [imath]0[/imath] on [imath]\Gamma[/imath], the average value of [imath]u[/imath] can't deviate too far from zero without causing a big derivative. But how to turn this quantitative is hard. By mimicking the solution to the original question, I have found that, letting [imath]T: W^{1,p}(U) \rightarrow L^p(\partial U)[/imath] be the bounded trace operator. [imath]\|Tu\|_{L^p(\partial U)} \le C\|\nabla(Tu)\|_{L^p(\partial U)}[/imath] This does not help though, since taking the trace gives the wrong direction for the inequality [imath] \|Tu\|_{L^p(\partial U)} \le C\|u\|_{W^{1,p}(U)}[/imath] when I really need to bound [imath]\|u\|_{W^{1,p}(U)}[/imath] from above. | 2050619 | Poincaré inequality for a subspace of [imath]H^1(\Omega)[/imath]
The following is the well known Poincaré inequality for [imath]H_0^1(\Omega)[/imath]: Suppose that [imath]\Omega[/imath] is an open set in [imath]\mathbb{R}^n[/imath] that is bounded. Then there is a constant [imath]C[/imath] such that [imath] \int_\Omega u^2\ dx\leq C\int_\Omega|Du|^2\ dx\quad \textrm{ for all }\ \ \color{red}{u\in H_0^1(\Omega)}. [/imath] Here is my questions: Can the condition [imath]u\in H_0^1(\Omega)[/imath] be relaxed so that the inequality is still true? More precisely, suppose [imath]H[/imath] is a subspace of [imath]H^1(\Omega)[/imath] such that for any [imath]u\in H[/imath], [imath]u|_\Gamma=0[/imath], where [imath]\Gamma[/imath] is a ([Added:]nonempty) closed proper subset of the boundary [imath]\partial\Omega[/imath] and the idenity is in the sense of trace. In other words, [imath]u[/imath] is only zero on a part of the boundary. Do we still have the Poincaré ineqaulity in [imath]H[/imath]? [Added:] This question is motivated by the following exercise |
2898702 | Suppose [imath]U[/imath] is a subspace of [imath]\mathbb{R}^{4}[/imath] defined by [imath]U =\mathrm{span} \,((1,2,3,-4),(-5,4,3,2))[/imath] Find: orthonormal basis of [imath]U[/imath] and [imath]U^{\perp}[/imath]
Suppose [imath]U[/imath] is a supspace of [imath]\mathbb{R}^{4}[/imath] defined by [imath]U = \operatorname{span}((1,2,3,-4),(-5,4,3,2))[/imath] Find: orthonormal basis of [imath]U[/imath] and [imath]U^{\perp}[/imath]. I have no issue finding an orthonormal basis of [imath]U[/imath]. I would just use the Gram Schmidt Process. [imath]U^{\perp}[/imath] is giving me issues. So what I thought of doing was obtain an orthonormal basis for [imath]U[/imath] and then by the relation of inner product spaces I know that: Letting [imath]u_{i} \in U[/imath] and [imath]v\in U^{\perp}[/imath] [imath]<u_1,v> = 0[/imath] and [imath]<u_2,v> = 0[/imath] I would end up with two equations and four unknowns. I debated on if I could pick two of the unknowns and let them be whatever I choose, but that may affect what vectors would be in my subspaces. So I am stuck at this point. Suggestions ? | 1394456 | Find Orthogonal complement
Let [imath]U = Sp\{(3, 3, 1)\}[/imath] How can I find the Orthogonal complement ? I'm not sure how to calculate it. In the book I'm learning from it's saying that I need to write the vectors of [imath]U[/imath] in [imath]Ax = 0[/imath] where the lines of [imath]A[/imath] are the vectors of [imath]U[/imath]. But since [imath]U[/imath] has only one vector I'm not sure how could this help me to find the orthogonal complement |
2899148 | Can infinity be a limiting point?
My initial guess would be no, as infinity is not a point. Using the same definition of a limiting point from What is and how to find a limiting point? , and say for the sequence of negative integers [imath]s_n=[/imath]{-1,-2,-3,-4...}. [imath]\lim_{n \to \infty }s_n = -\infty[/imath] and since [imath]-\infty[/imath] is not a point in the negative integers I am guessing that means (negative)infinity is a limiting point of [imath]s_n[/imath]. So what would be the correct interpretation here? Is infinity a (divergent)limit of the sequence or a limiting point? | 777031 | How can infinity be an accumulation point?
I can't wrap my mind around this: An accumulation point is a point of a set, which in every yet so small neighborhood (of itself) contains infinitely many points of the set, right? So if (in my case) [imath]-\infty[/imath] is an accumulation point, then we could find a point of the set, that is finitely distant from infintity?! Here's the task: Let [imath](a_n)_{n∈\mathbb N}[/imath] be a set in [imath]\mathbb R[/imath]. For every [imath]r ∈ \mathbb R[/imath] exits an accumulation point [imath]b ∈ R[/imath] ∪ {−∞} of [imath](a_n)_{n∈\mathbb N}[/imath] (+∞ und −∞ are allowed as accumulation points) with b < r. Show, that −∞ is an accumulation point of [imath](a_n)_{n∈\mathbb N}[/imath]. |
2898751 | How to get the equation where a circle goes through three points
If I have the equation [imath]ax^2+ay^2+bx+cy+d=0[/imath] how do I get the equation where the circumference goes through the points P = (1,1), Q = (−1,−1) and R = (−1,1) I have it in mind to solve it with a matrix, but the instructions seem confusing to me, can I get some help? | 919057 | find the equation of the circle [imath]x^2 +y^2 +ax +by = c[/imath] passing through points [imath](6,8), (8,4), (3,9)[/imath]
Find the equation of the circle [imath]x^2 +y^2 +ax +by = c[/imath] passing through points [imath](6,8), (8,4), (3,9)[/imath]. How do I go about solving this? I don't have a textbook assigned so I'm not even sure what this relates to, but its the beginning of linear algebra. so I guess I'm supposed to construct some kind of system of equations... |
2899305 | How to prove that [imath] a^2+b^2+c^2=1[/imath] implies [imath]ab+bc+ca \in [-\frac12,1][/imath]?
If [imath]a,b,c[/imath] belong to set of real numbers and [imath] a^2+b^2+c^2=1[/imath] then prove that [imath]ab+bc+ca[/imath] belongs to [imath][-\frac12,1][/imath] I have tried AM>GM>HM(progressions mean inequality) but I am unable to do anything. I have even attempted to assume a,b and c to be sides of an triangle and using [imath]|a-b|<|c|[/imath] and squaring but still I don't get the right-hand limit right. Please tell me how to prove it. | 1119879 | If [imath]a^2+b^2+c^2=1[/imath] then prove the following.
If [imath]a^2+b^2+c^2=1[/imath], prove that [imath]\frac{-1}{2}\le\ ab+bc+ca\le 1[/imath]. I was able to prove that [imath] ab+bc+ca\le 1[/imath]. But I am unable to gain an equation to prove that [imath] \frac{-1}{2}\le\ ab+bc+ca[/imath] . Thanks in advance ! |
2899665 | [imath]R[/imath] is an associative ring with no zero divisors that has an element [imath]a\neq 0[/imath], such that [imath]a*a=a[/imath]. Prove [imath]a[/imath] is the identity element of [imath]R[/imath]
Let [imath]R(+,*)[/imath] be an associative ring with no zero divisors that has an element [imath]a\neq 0[/imath], such that [imath]a*a=a[/imath]. Prove that [imath]a[/imath] is the multiplicative identity element of [imath]R[/imath]. I found similar questions posted here, but none of them had [imath]a*a=a[/imath]. Any hints? | 663518 | Idempotents in a ring without unity (rng) and no zero divisors.
Question: Given a ring without unity and with no zero-divisors, is it possible that there are idempotents other than zero? Def: [imath]a[/imath] is idempotent if [imath]a^2 = a[/imath]. Originally the problem was to show that [imath]1[/imath] and [imath]0[/imath] are the only idempotents in a ring with unity and no zero-divisors, but I wonder what happens if we remove the unity condition. I am trying to find a ring with idempotents not equal to [imath]0[/imath] or [imath]1[/imath]. So far my biggest struggle has been coming up with examples of rings with the given properties. Does anyone have any hints? How should I attack this problem? |
2899318 | Borel sets and Lebesgue-Stieltjes measures
Let [imath]F : \mathbb R\to\mathbb R[/imath] be a distribution function (increasing right continuous), and [imath]m_F : M_F\to[0,\infty][/imath] the Lebesgue-Stieltjes measure generated by [imath]F[/imath]. In most cases I've come across there are sets in [imath]M_F[/imath] that are not Borel sets. Is the general statement "For every choice of [imath]F[/imath], [imath]M_F[/imath] contains a set which isn't a Borel set" a theorem? | 815949 | Why is any Lebesgue-Stieltjes measure on [imath](\mathbb{R},\mathcal{B}_\mathbb{R})[/imath] not complete?
Let [imath]F: \mathbb{R} \to \mathbb{R}[/imath] be any increasing, right-continuous function. Then we define the Lebesgue-Stieltjes measure associated to [imath]F[/imath] to be the unique measure [imath]\mu_F[/imath] on [imath](\mathbb{R},\mathcal{B}_\mathbb{R})[/imath] such that [imath]\mu_F((a.b])=F(b)-F(a)[/imath] for all [imath]a, b[/imath]. We wish to show that the completion of [imath]\mathcal{B}_\mathbb{R}[/imath] with respect to [imath]\mu_F[/imath], denoted by [imath]M_F[/imath], is strictly larger than [imath]\mathcal{B}_\mathbb{R}[/imath]. Since [imath]\text{card}(\mathcal{B}_\mathbb{R})=\mathfrak{c}[/imath], it will suffice to exhibit a set [imath]K[/imath] in [imath]M_F[/imath] such that [imath]\mu_F(K)=0[/imath] and [imath]\text{card}(K)=\mathfrak{c}[/imath], for then [imath]\text{card}(M_F)>\mathfrak{c}[/imath]. |
2899802 | 4 normals can be drawn from any point to an ellipse proof doubt: How do we know the key quartic has 4 distinct real roots?
Theorem: 4 normals can be drawn from any point to the ellipse [imath]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/imath]. [imath]( a > b > 0)[/imath] I had a doubt in the proof of the above statement given in my book. Proof given in Book. The normal at the point with eccentric angle [imath]\phi[/imath] is [imath]ax\sec\phi - by \csc\phi = a^2 - b^2 [/imath]. Let P[imath](h,k)[/imath] be any point in [imath]\mathbb{R^2}[/imath]. If the normal passes through P: [imath]ah\sec\phi - bk \csc\phi = a^2 - b^2 = a^2e^2 [/imath] Let [imath] t = \tan \frac{\phi}{2}[/imath], then [imath]\cos \phi = \frac{1 - t^2}{1 + t^2}[/imath] and [imath]\sin \phi = \frac{2t}{1 + t^2}[/imath]. Substituting and rearranging, gives: [imath]bkt^4 + 2t^3(ah + a^2e^2) + 2t(ah - a^2e^2) - bk = 0 [/imath] This is a quartic in [imath]t[/imath], hence has 4 roots, corresponding to 4 points on the ellipse which are conormal. Doubt: My question is how do we know that the roots of that quartic equation are real and distinct? Because if not, then the theorem is not yet proved. I looked up the discriminant of a quartic on Wikipedia (https://en.wikipedia.org/wiki/Quartic_function), but the inequalities are too difficult to handle. How can I show that the roots are all real and distinct? | 609351 | Number of normals from a point to an ellipse
From a typical point [imath]P[/imath] inside an ellipse, how many points [imath]Q_i[/imath] on the ellipse have [imath]PQ_i[/imath] normal to the ellipse? Someone asked me at school many years ago but I don't think I worked it out. |
2900208 | What is the relation to [imath]\sinh{x},\cosh{x}[/imath] and [imath]\sin{x},\cos{x}[/imath]
I've learned what [imath]\sinh{x},\cosh{x}[/imath] (the hyperbolic trig functions) are defined as formula, but how is it related to [imath]\sin{x},\cos{x}?[/imath] The only thing I've noticed is that [imath]\cosh^2(x)-\sinh^2(x)=1.[/imath] | 1736068 | What's the intuition behind the identities [imath]\cos(z)= \cosh(iz)[/imath] and [imath]\sin(z)=-i\sinh(iz)[/imath]?
I'm trying to understand in an intuitive manner the relationship between the circular and hyperbolic functions in the complex plane, i.e.: [imath]\cos(z)= \cosh(iz)[/imath] [imath]\sin(z)=-i\sinh(iz)[/imath] where [imath]z[/imath] is a complex number. From a geometric point of view, what I understand is that cos is the composition of a rotation through [imath]\frac{\pi}{2}[/imath], followed by cosh, and sin is the composition of a rotation through [imath]\frac{\pi}{2}[/imath], followed by sinh, followed by a rotation through [imath]-\frac{\pi}{2}[/imath] (where sin, cos, sinh, cosh are defined as complex functions). Where does this connection come from? Is there some way it can be visualized in terms of complex mappings? (I'm not asking for a proof of the identities, I already know one). |
2900082 | Maximum Modulus Principle Intuition
I understand that the Maximum Modulus Principle works, but I'm a little baffled as to why. To be more precise, the picture I have in my head is something like this: for a compact set [imath]K \subset \mathbb{C}[/imath], since [imath]|f|[/imath] (for [imath]f[/imath] holomorphic) can only attain its maximum on the boundary [imath]\partial K[/imath], if you consider a disk centered on the origin, no matter what value [imath]|f|[/imath] obtains on the boundary of this disc, you can just increase the radius of the disc a little bit and find a higher value of [imath]|f|[/imath]; in other words, the absolute value of function just keeps growing without bound. What exactly is driving/forcing this growth? I'm looking for some kind of explanation, geometric or otherwise, that could aid my intuition. In particular, why does [imath]\mathbb{C}[/imath] behave so differently from [imath]\mathbb{R}[/imath] here? EDIT. In response to the "duplicate" tag: I am looking for something a little deeper than the answers there. As mentioned above, an explanation of the difference in behaviours of [imath]\mathbb{C}[/imath] and [imath]\mathbb{R}[/imath] in this regard, perhaps with a tie-in to the Cauchy-Riemann equations...it seems like there's something that causes functions to behave fundamentally differently over [imath]\mathbb{C}[/imath], and I would like to understand why, with the Maxiumum Modulus Principle as a concrete example. | 762280 | Intuition Behind Maximum Principle (Complex Analysis)
Let [imath]D[/imath] be an open set in the complex plane and [imath]f(z)[/imath] be a non-constant holomorphic function on D. Then [imath]|f(z)|[/imath] has no local maximum on D. I can follow the proof fine - usually if I don't understand a theorem intuitively beforehand, the proof will offer the insight necessary. Here, however, I can't see the reason for the Maximum Principle to hold - or perhaps I was just too shallow in my grasp of the proof. Does anybody have any shillings of wisdom that they would be willing to offer? Cheers. |
2900575 | quadratic grid in which orthogonal triangle formed by grid points
Determine all natural numbers [imath]n[/imath] with [imath]n>1[/imath] that applies: If each grid point of a quadratic grid in the plane is colored with one of n given Colors, then there are always three grid points of the same color, that form a orthogonal triangle form whose catheters are parallel to the grid lines. It is a Cartesian coordinate system. So I tried to find a solution to this question with Ramsey theory, but so far with no success. | 2898846 | Triangles defined on an infinite Go board by same-colored stones
You start with an infinite Go board. On every point of the board you place one colored stone. There are [imath]n>1[/imath] different colors. Find all natural numbers [imath]n[/imath] that no matter how the stones are colored, three stones of the same color form the vertices of a right-angled triangle. The catheti (legs) of the right triangle must be on the lines of the board. Any ideas how to solve this kind of problem and to which area of mathematics this question belongs? |
2901128 | If [imath]A^3 = A[/imath], prove that [imath]\operatorname{Ker}(A - I) \oplus\operatorname{ Im}(A - I) = V[/imath]
I just can't wrap my head around this proof. How to do it? Here is what I tried: [imath]A^3 = A[/imath] [imath]A^3 - A = 0[/imath] [imath]A(A-I)(A+I) = 0[/imath] Matrix [imath]A[/imath] satisfies the polynomial [imath]p(x)=x(x-1)(x+1)[/imath]. Zeroes of [imath]p[/imath] are [imath]-1, 0 ,1[/imath]. Minimal polynomial of the matrix [imath]A[/imath] has at least one root, which can only take value of roots of [imath]p[/imath]. Maximum multiplicity of that root in the minimal polynomial is one. One can then take a look at the eigenspaces that correspond to the specific eigenvalues. What then...? Is that the path that can lead to a solution? Is there a better solution? I don't really get it... Please, be kind and explain this if you can spare the time. I found this question, but there seems to be a leap of logic in the question, and the accepted answer. [imath]A^3 = A[/imath] doesn't imply [imath]A^2 = I[/imath] For instance, take matrix [imath] \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} [/imath] Also, if we take a linear operator [imath]L:V \to V[/imath], it is obvious that [imath]Ker(L) \cap Im(L) = \{0\}[/imath] ? Do I also need to prove that sum of kernel and image is direct sum? | 2117662 | If [imath] A^3=A[/imath] prove that [imath]Ker\left(A-I\right)+Im\left(A-I\right)=V[/imath]
If [imath] A^3=A[/imath] prove that [imath]Ker\left(A-I\right)+Im\left(A-I\right)=V[/imath] I am not sure how to approach this problem, but first things first if we have [imath]A^3=A[/imath] that is [imath]A^2=I[/imath], what does that tell me (what does it imply) about [imath]A[/imath]? The only thing I can tell at this point is that [imath]A[/imath] is it's own inverse and if a matrice is invertible it has a full rank which implies that the dimension of the image space is [imath]n[/imath] and the dimension of the null-space aka kernel is zero. Thus we would have that [imath]Im(A)+Ker(A)=V[/imath], but subtracting [imath]I[/imath] confuses me here. P.S. [imath]A[/imath] is the matrix representation of a linear operator and [imath]V[/imath] is the vector space for which the operator is defined. |
2901087 | derivative of inseparable equation
Is there a way to find the derivative of an equation where I cannot separate the independent variable from the dependent variable. For example, what is [imath]dy/dx[/imath] given \begin{equation*} \cos(y) = \dfrac{x}{x+y} \end{equation*} | 39414 | Implicit Differentiation Help
I have to use implicit differentiation to find [imath]\frac{dy}{dx}[/imath] given: [imath]x^2 \cos(y) + \sin(2y) = xy[/imath] I don't even know where to begin, I missed the class where we went over implicit differentiation, and because of that, I am completely stuck. Thank you everyone. Edit: I don't know how to make the equation look all nice and whatnot, so sorry about that |
2900828 | Probability of real roots for [imath]x^2 + Bx + C = 0[/imath]
Question: The numbers [imath]B[/imath] and [imath]C[/imath] are chosen at random between [imath]-1[/imath] and [imath]1[/imath], independently of each other. What is the probability that the quadratic equation [imath]x^2 + Bx + C = 0[/imath] has real roots? Also, derive a general expression for this probability when B and C are chosen at random from the interval [imath](-q, q)[/imath] for and [imath]q>0[/imath]. My approach: since we're trying to find the probability of real roots. We should first realize when it has imaginary roots. So, [imath]B^2 - 4aC < 0[/imath] [imath]a = 1[/imath] [imath]B^2 < 4C [/imath] I'm not sure where to go from here. How do I now find the probability of [imath]B[/imath] being greater than [imath]4C[/imath]? | 2431208 | Finding probability that an equation has real roots when two numbers are chosen from a uniform distribution
The Question reads as follows: Choose independently two numbers B and C at random from the interval [-1,1] with uniform distribution and consider the quadratic equation: [imath]x^2+Bx+C=0[/imath] find the probability that the roots of this equation a) are both real b) are both positive I know a) requires that [imath]0\le B^2-4C[/imath] and b) requires [imath]0\le B^2-4C[/imath] , [imath]B \le 0[/imath] , [imath]0 \le C[/imath]. I know what this looks like geometrically and I believe for part a I should be able to integrate something from -1 to 1 but I am not sure what. |
2901876 | Textbook has a strange section on absolute value that I don't quite understand
My textbook says [imath]|a| = -a[/imath] for [imath]a \le 0[/imath]. What does it mean by this? I'm confused and think that the absolute value of a negative number like [imath]|-5|[/imath] would be positive [imath]5[/imath]? | 2369950 | My teacher describes absolute value confusingly: [imath]|x|=\pm x,\quad \text{if}\enspace x>0 [/imath].
He says (direct quote): "In higher mathematics the absolute value of a number, [imath]|x|[/imath], is equal to positive and negative [imath]x[/imath], if [imath]x[/imath] is a positive number." Then he wrote: [imath]|x|=\pm x,\quad \text{if}\enspace x>0 [/imath]. I think he misunderstood the definition of absolute value, or did I? From what I understand, absolute value of a number is the distance of a number from zero, so it is always positive. Am I wrong? |
2901688 | If the tangent at [imath](x_0,y_0)[/imath] to the curve [imath]x^3+y^3=a^3[/imath] meets the curve again at [imath](x_1,x_1)[/imath] then prove that [imath]\frac{x_0}{x_1}+\frac{y_0}{y_1}=1[/imath]
Problem : If the tangent at [imath](x_0,y_0)[/imath] to the curve [imath]x^3+y^3=a^3[/imath] meets the curve again at [imath](x_1,x_1)[/imath] then prove that [imath]\frac{x_0}{x_1}+\frac{y_0}{y_1}=1[/imath] We have tangent to the given curve at [imath](x_0,y_0)[/imath] is [imath]xx_0^2+yy_0^2=a^3[/imath]. Since it passes through [imath](x_1,y_1)[/imath], so we get [imath]x_1x_0^2+y_1y_0^2=a^3\tag{1}[/imath] Again, [imath](x_1,y_1)[/imath] lies on the curve, so. [imath]x_1^3+y_1^3=a^3\tag{2}[/imath]. How can I get the required result from above? | 897196 | Tangent to curve [imath]x^3+y^3=a^3[/imath] meets it again.
Tangent to curve [imath]x^3+y^3=a^3[/imath] at [imath](x_1,y_1)[/imath] meets it again at [imath](x_2,y_2)[/imath].How to prove that [imath]\frac{x_2}{x_1}+\frac{y_2}{y_1}+1=0[/imath] Since [imath]y'=-\frac{x_1^2}{y_1^2}[/imath] [imath]\frac{y_2-y_1}{x_2-x_1}=-\frac{x_1^2}{y_1^2}=\frac{x_1^2+x_1x_2+x_2^2}{y_1^2+y_1y_2+y_2^2}[/imath] Solving we get: [imath]\frac{x_2}{x_1}+\frac{y_2}{y_1}+\frac{x_2^2}{x_1^2}+\frac{y_2^2}{y_1^2}+2=0[/imath] Or [imath]\left(\frac{x_2}{x_1}+\frac{y_2}{y_1}\right)+\left(\frac{x_2^2}{x_1^2}+\frac{y_2^2}{y_1^2}\right)^2=2\frac{x_2y_2-x_1y_1}{x_1y_1}[/imath] Which isn't what is nedded to prove. can someone guide me? |
2902154 | Sums of consecutive primes
Seems to me that for 3 consecutive primes [imath]p_1, p_2, p_3, p_1>2[/imath], it is always the case that [imath]p_1+p_2 > p_3[/imath]. Do you know a proof? (Initially I thought there must be large primes for which this is the case, but the limit as [imath]n \to \infty[/imath] of the ratio of [imath]n^{th}[/imath] prime gap to the nth prime is 0...) | 413163 | Do 3 consecutive primes always form a triangle?
Suppose that [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] are any three consecutive primes other than the triple [imath]2[/imath], [imath]3[/imath], and [imath]5[/imath]. Do they satisfy the triangle inequalities: [imath]a + b > c[/imath]; [imath]b + c > a[/imath]; [imath]c + a > b[/imath]? In other words, can we always form a triangle with sides being the [imath]3[/imath] successive prime numbers? Is this a well-known result? Where can I read about its proof or refutation? Thanks in advance. |
2902076 | [imath]\{1/2\}\times (1/2,1][/imath] is not open in the ordered square
Consider the square [imath][0,1]^2[/imath] with the dictionary order (the restriction of the dictionary order on [imath]\mathbb R^2[/imath] to the subset [imath][0,1]^2[/imath]. I'm trying to understand why the set [imath]A=\{1/2\}\times (1/2,1][/imath] is not open in the order topology on [imath][0,1]^2[/imath]. Munkres give this picture that should clarify things: The order topology on the square is given by the basis that consists of all elements of the form [imath][a_0,b)[/imath] where [imath]a_0=(-1,1)[/imath], [imath](a,b_0][/imath] where [imath]b_0=[1,1][/imath], and [imath](a,b)[/imath] (w.r.t. the dictionary order). The definition of [imath]A[/imath] being open is this: for every every [imath]t\in A[/imath] there is a basis element [imath]U[/imath] s.t. [imath]t\in U\subset A[/imath]. Its negation is: there is [imath]t\in A[/imath] such that for all basis elements [imath]U[/imath] it is not true that [imath]t\in U\subset A[/imath]. But I couldn't find such [imath]t[/imath] and unravel "it is not true that [imath]t\in U\subset A[/imath]". | 374998 | Subspace topology and order topology (2)
Yesterday I asked a very similar question but now I am confused again... Let [imath]I=[0,1][/imath]. The dictionary order on [imath]I\times I[/imath] is just the restriction to [imath]I\times I[/imath] of the dictionary order on the plane [imath]\mathbb{R}\times\mathbb{R}[/imath]. now, how can I show that the set [imath]\{1/2\}\times (1/2,1][/imath] is not open in the order topology? The maximum element of [imath]I[/imath] is 1. So the order topology on [imath]I\times I[/imath] has intervals of the form [imath](a,1]\times (b,1][/imath] as basis elements. Isn't [imath]\{1/2\}\times (1/2,1][/imath] itself a basis element of the order topology? |
2903083 | Proof of 1-1 under an interval (0,1)
How do I prove that the function [imath]f(x) = \frac{-2x+1}{(2x-1)^2-1}[/imath] is one-to-one from the interval (0,1)? I have simplified the function [imath]f(x)=f(y)[/imath] and have a multivariable quadratic equation equal to zero. How do I conclude that the function is one to one? | 2903034 | How to prove that [imath]f(x) = \frac{-2x+1}{(2x-1)^2-1}[/imath] is one-to-one on [imath](0,1)[/imath]?
How do I prove that the function [imath]f(x) = \frac{-2x+1}{(2x-1)^2-1}[/imath] is one-to-one on the interval [imath](0,1)[/imath]? I have simplified the function [imath]f(x)=f(y)[/imath] and have a multivariable quadratic equation equal to zero. How do I conclude that the function is one to one? |
2903555 | Sum the series [imath]\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}[/imath]
[imath]\frac{n}{1⋅2⋅3}+\frac{n-1}{2⋅3⋅4}+ \frac{n-2}{3⋅4⋅5}+...\text{(upto n terms)}[/imath] Clearly, we can see that [imath]T_r=\frac{n-r+1}{r(r+1)(r+2)}[/imath] Now, somehow, we have to make this telescoping. But we do not have [imath]n[/imath] as a factor of the denominator, so we have to multiply the denominator by [imath]n[/imath] and after our manipulation, multiply the whole sum by [imath]n[/imath]. But that too did not work. PLease help | 2079081 | sum of series [imath]\frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots \cdots n[/imath] terms
sum of series [imath]\displaystyle \frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots \cdots n[/imath] terms assuming [imath]\displaystyle S_{n} =\frac{n}{1\cdot 2 \cdot 3}+\frac{n-1}{2\cdot 3\cdot 4}+\frac{n-2}{3\cdot 4 \cdot 5}+\cdots +\frac{n-(n-1)}{n\cdot n+1 \cdot n+2}[/imath] [imath]\displaystyle S_{n} = \sum^{n-1}_{r=0}\frac{n-r}{(r+1)\cdot (r+2) \cdot (r+3)}[/imath] wan, t be able to solve after that , could some help me with this |
2903760 | If matrices [imath]A[/imath] and [imath]B[/imath] are orthogonal then [imath]A-B[/imath] or [imath]A+B[/imath] are singular
If [imath]A[/imath] and [imath]B[/imath] are orthogonal matrices and they have same odd row, prove that one of matrices [imath]A+B[/imath] and [imath]A-B[/imath] are singular. I have no idea, can you help me? I tried with [imath]\det(A+B)(A-B)=\det(A^2-AB+BA-B^2)[/imath], but I do not get anything. | 1805790 | Prove that, at least one of the matrices [imath]A+B[/imath] and [imath]A-B[/imath] has to be singular
Problem: Let [imath]A[/imath] and [imath]B[/imath] be real orthogonal matrices, [imath]n[/imath]x[imath]n[/imath], where [imath]n[/imath] is an odd number. Prove that, at least one of the matrices [imath]A+B[/imath] and [imath]A-B[/imath] has to be singular. What have I done so far: -Since matrices [imath]A[/imath] and [imath]B[/imath] are real orthogonal, it means that their determinants are [imath]-1[/imath] or [imath]+1[/imath] First I observed matrix [imath]A+B[/imath]: [imath]A+B=AI+BI=ABB^T+BAA^T=AB(B^T+A^T)=AB(A+B)^T\Rightarrow[/imath] [imath]det(A+B)=det(AB)det(A+B)^T=det(A)det(B)det(A+B)[/imath] So, if [imath]detA=detB[/imath] we have [imath]det(A+B)=det(A+B)[/imath] which tells me nothing. But, if [imath]detA=-detB[/imath] we have [imath]det(A+B)=-det(A+B)[/imath] which can only hapen if [imath]det(A+B)=0[/imath] making [imath]A+B[/imath] singular. Then, I tried the same for [imath]A-B[/imath]: [imath]A-B=AI-BI=ABB^T-BAA^T=AB(B^T-A^T)=AB(A-B)^T\Rightarrow[/imath] [imath]det(A-B)=det(AB)det(A-B)^T=det(A)det(B)det(A-B)[/imath] So, if [imath]detA=-detB[/imath] we have [imath]det(A-B)=-det(A-B)[/imath] which can happen if [imath]det(A-B)=0[/imath] making [imath]A-B[/imath] singular. Is this correct or should I do matrix [imath]A-B[/imath] differently? I am also a little confused why does it have to be said that [imath]n[/imath] is an odd number? What should that tell me? Any help is greatly appreciated. |
2903447 | How to evaluate the integral [imath]\int_{0}^{1} \frac{\log x}{\sqrt {1+x^2}}dx[/imath]
[imath]I=\int_{0}^{1} \frac{\log x}{\sqrt {1+x^2}}dx[/imath] My attempt:[imath]I=\int_{0}^{1}\log x d(\log(x+\sqrt{1+x^2}))[/imath] [imath]=\log x\log(x+\sqrt{1+x^2})|_0^1-\int_{0}^{1}\frac{\log(x+\sqrt{1+x^2})}{x}dx[/imath] I don't know how to proceed below, please help me. That is different to me. | 2746439 | Integral [imath]\int_{\frac{1}{2}}^1 \frac{\ln x}{\sqrt{x^2+1}}dx[/imath]
I am trying to evaluate [imath]I=\int_{1/2}^1 \frac{\log(x)}{\sqrt{x^2+1}}\,dx[/imath] My attempt was to integrate by parts with [imath]u=\log(x)[/imath] and [imath]v=\log{(x+\sqrt{1+x^2}})[/imath] to get[imath]I=-\log{\frac{1}{2}}\log{\left(\frac{1}{2}+\sqrt{1+\frac{1}{4}}\right)}-\int_{1/2}^1 \frac{\log{(x+\sqrt{1+x^2}})}{x}\,dx[/imath] so simplifying the integral is [imath]I=\log{(2\phi)}-I_1[/imath] where [imath]I_1=\int_{\frac{1}{2}}^1 \frac{\sinh^{-1}(x)}{x}\,dx[/imath] If possible I would love to get some help in evaluating those. I also thought about considering [imath]I(k)=\int_{\frac{1}{2}}^1 \frac{x^k} {\sqrt{x^2+1}}\,dx[/imath] writting the denominator as a binomial series, integrate it then derivate and plug [imath]k=0[/imath] to get the answer but the series is not nice. |
2903917 | Evaluating a general solution to [imath]5\cos \theta -12\sin \theta = 13[/imath] using vectors
Given that [imath]5\cos \theta -12\sin \theta = 13[/imath] I'm trying to evaluate a general solution for this equation. It appears I'll be using vector product. My equation is equivalent to [imath]\langle (5,12), (\cos\theta, \sin\theta)\rangle = 13[/imath] which yields (by Cauch Schwarz Inequality) [imath]|\langle (5,12), (\cos\theta, \sin\theta)\rangle| \le \|(5,12)\|\|(\cos\theta, \sin\theta)\| = 13[/imath] This is where I'm stuck. Regards | 2903190 | Determining the general solution for the trigonometric equation [imath] 5\cos(x)-12\sin (x) = 13 [/imath]
Given that [imath]5\cos(x)-12\sin (x) = 13 [/imath] I'm trying to evaluate the general solution for that expression. It reminds me of [imath]5-12-13[/imath] triangle. Since we don't know the degree of [imath]x[/imath], I couldn't proceed further. Specifically, let's take its derivate, which yields [imath]\dfrac{d}{dx} 5\cos(x)-12\sin (x) = 13 = 0[/imath] |
1936277 | Prove that dim[(span[imath](v_1 + w,...,v_m + w)] \geq m - 1. [/imath]
Suppose [imath]v_1,...,v_m[/imath] is linearly independent in [imath]V[/imath] and [imath]w \in V. [/imath] Prove that dim[(span[imath](v_1 + w,...,v_m + w)] \geq m - 1. [/imath] attempt: Let [imath]v_1,...,v_m[/imath] be linear independent. So there is [imath]a_1,..,a_m \in F,[/imath] such that [imath]a_1v_1 + .... + a_mv_m = 0[/imath]. where [imath]a_1=...=a_m = 0.[/imath] Then [imath](v_1 - v_2,...,v_{m-1} - v_m)[/imath] span and are linear independent in [imath]V[/imath]. And dim[imath]span (v_1,...,v_m) = m.[/imath] Can someone please help me? I am stuck. Any help would be really appreciated. Thanks! | 1269123 | If [imath]v_1,...,v_m[/imath] are linearly independent, then the span [imath]v_1+w,...,v_m+w[/imath] has dimension [imath]\ge m-1[/imath]
Suppose [imath]v_1,...,v_m[/imath] is linearly independent in [imath]V[/imath] and [imath]w\in V[/imath]. Prove that [imath] \dim (\operatorname{span}(v_1+w,...,v_m+w)) \ge m-1[/imath] It's an exercise in the book Linear Algebra Done Right. I'm wondering if I can write [imath]U_1 =\operatorname{span}(v_1,...,v_m)[/imath] and [imath]U_2=\operatorname{span}(w)[/imath] then write [imath] \dim(\operatorname{span}(v_1+w,...,v_m+w)) = \dim(U_1+U_2)[/imath] Would you please help me with this problem, I really want a rigorous proof, thanks. |
2604247 | Prove that [imath]R[x][/imath] is an integral domain if and only if [imath]R[/imath] is an integral domain.
I want to prove that for a ring [imath]R[/imath], [imath]R[x][/imath] is an integral domain if and only if [imath]R[/imath] is an integral domain. I have one direction of the proof ([imath]R[/imath] an integral domain implies [imath]R[x][/imath]) an integral domain, but I am having trouble proving the other direction. Here is the first direction: Let [imath]R[/imath] be an integral domain, and suppose on the contrary that [imath]R[x][/imath] is not an integral domain, meaning that it has zero divisors. In particular, let [imath]f(x),g(x)\neq 0\in R[x][/imath] such that [imath]\deg(f)=n[/imath] and [imath]\deg(g)=m[/imath] and [imath]f(x)g(x)=0[/imath]. Specifically, [imath]f(x)=a_nx^n+...a_1x+a_0[/imath] and [imath]g(x)=b_mx^m+...b_1x+b_0[/imath] where [imath]a_i,b_j\in R[/imath] for all [imath]0\leq i,j\leq n,m[/imath]. Now, consider the polynomial term [imath]a_nb_mx^{n+m}\in f(x)g(x)[/imath]. Because [imath]f(x)g(x)=0[/imath], this means that each polynomial coefficient must be zero, so in particular [imath]a_nb_m=0[/imath]. But, [imath]R[/imath] is an integral domain so either [imath]a_n=0[/imath] or [imath]b_m=0[/imath], contradicting the degree of [imath]f(x)[/imath] or [imath]g(x)[/imath]. Thus, [imath]R[x][/imath] is an integral domain. And here is what I have so far for the other direction, but I'm not sure if I have the right set-up. Any hints or comments are appreciated. Now, suppose conversely that [imath]R[x][/imath] is an integral domain, and suppose on the contrary that [imath]R[/imath] is not, meaning that it has zero divisors. In particular, let [imath]ab=0[/imath] where [imath]a\neq 0[/imath] and [imath]b\neq 0[/imath] for [imath]a_0,b_0\in R[/imath]. Furthermore, consider the polynomial product [imath]f(x)g(x)=0[/imath] for [imath]f(x),g(x)\in R[x][/imath], where [imath]f(x)[/imath] and [imath]g(x)[/imath] are defined in the same way as above. Because [imath]R[x][/imath] is an integral domain, either [imath]f(x)=0[/imath] or [imath]g(x)=0[/imath]. | 1415641 | Prove that [imath]R[/imath] is an integral domain [imath]\Leftrightarrow[/imath] [imath]R[x][/imath] is an integral domain
Here is an exercise(p.129, ex.1.15) from Algebra: Chapter 0 by P.Aluffi. Prove that [imath]R[x][/imath] is an integral domain if and only if [imath]R[/imath] is an integral domain. The implication part makes no problems, because [imath]R[/imath] is a subring of [imath]R[x][/imath]. For the ''if'' part, however.. Let [imath]R[/imath] be an integral domain. Now let's review pairs [imath]f(x)=\sum \nolimits a_ix^i,g(x)= \sum \nolimits b_ix_i \in R[x][/imath] such that [imath]f(x)g(x) = 0[/imath] So, we have [imath]\sum \limits_{k=0}^{\infty} \sum \nolimits_{i+j=k} a_ib_jx^{i+j} = 0[/imath]. Now, I'm not sure how to deduce that [imath]f(x) = 0 \vee g(x) = 0[/imath] If I look at [imath]f(x),g(x)[/imath] such that, for example, [imath]deg((f(x))=3, deg((g(x))=2)[/imath], it makes sense. I begin by something like that "if [imath]f(x)g(x)=0[/imath] then [imath]a_0b_0=0[/imath] then [imath]a_0 = 0 \vee b_0 = 0[/imath]. If [imath]a_0 =0 [/imath], then [imath]a_0b_1 = a_0b_2 = 0[/imath]" and so on. As I understand, it comes down to proving the following implication: [imath](\forall k \in \mathbb{N} \sum \nolimits_{i+j=k} a_ib_j = 0)(1) \Rightarrow ((\forall n \in \mathbb{N} \ \ a_n = 0) \vee (\forall m \in \mathbb{N} \ \ b_m = 0))(2)[/imath] We can say that [imath](1)[/imath] is a system of equations in [imath]R[/imath]. And [imath](2)[/imath] is it solution. |
2902696 | which of the following function are reimann integrable on the interval [imath][0,1].[/imath]?
which of the following function are reimann integrable on the interval [imath][0,1].[/imath]? [imath]1)[/imath] [imath]f(x) =\begin{cases} 1, &\text{if x is rational }\\ 0, &\text{if x is irrational } \end{cases}[/imath] [imath]2)[/imath] [imath]f(x) =\begin{cases} 1, &\text{if x } \in \{\alpha_1,\alpha_2,.......,\alpha_n\}\\ 0, &\text{otherwise } \end{cases}[/imath] i know that option [imath]1) [/imath] will not reimann integrable because it is not bounded. im confused about option [imath]2)[/imath] Any hints/solution | 2274177 | Riemann integrabilty of an indicator function
Consider the following function on [imath][0,1][/imath]: [imath]f(x)=\begin{cases}1&x\in\{a_1,a_2,\dots,a_n\}\\0 &\text{otherwise}\end{cases}[/imath] The [imath]a_i[/imath] are fixed and all in [imath][0,1][/imath]. Determine whether [imath]f[/imath] is Riemintegrable or not. By the Riemann–Lebesgue criterion [imath]f[/imath] is Riemann integrable. How to prove directly? |
2903717 | (Parameter Choice) Using Differentiation under integral for [imath]e^{-x^2}[/imath]
I have the integral: [imath]\int_{-\infty}^\infty e^{-x^2} \, dx[/imath] And I’d like to solve it using differentiation under the integral sign. I understand that I must convert [imath]e^{-x^2}[/imath] to [imath]e^{-x^2}g(x,t)[/imath], where [imath]g(x,t)[/imath] is just some term involving a new parameter [imath]t[/imath]. I tried multiplying by [imath]e^{-2xt}[/imath], which doesn’t really seem very logical, however, I just do not know what to put. How do I find the appropriate term to make the whole process work? | 390850 | Integrating [imath]\int^{\infty}_0 e^{-x^2}\,dx[/imath] using Feynman's parametrization trick
I stumbled upon this short article on last weekend, it introduces an integral trick that exploits differentiation under the integral sign. On its last page, the author, Mr. Anonymous, left several exercises without any hints, one of them is to evaluate the Gaussian integral [imath] \int^\infty_0 e^{-x^2} \,dx= \frac{\sqrt{\pi}}{2} [/imath] using this parametrization trick. I had been evaluating it through trial and error using different paramatrizations, but no luck so far. Here are what I have tried so far: A first instinct would be do something like:[imath] I(b) = \int^\infty_0 e^{-f(b)x^2}\,dx [/imath] for some permissible function [imath]f(\cdot)[/imath], differentiating it will lead to a simple solvable ode: [imath] \frac{I'(b)}{I(b)} = -\frac{f'(b)}{2f(b)} [/imath] which gives: [imath] I(b) = \frac{C}{\sqrt{f(b)}}. [/imath] However, finding this constant [imath]C[/imath] basically is equivalent to evaluating the original integral, we are stuck here without leaving this parametrization trick framework. A second try involves an exercise on the same page: [imath] I(b) = \int^\infty_0 e^{-\frac{b^2}{x^2}-x^2}dx. [/imath] Taking derivative and rescaling the integral using change of variable we have: [imath] I'(b) = -2I(b). [/imath] This gives us another impossible to solve constant [imath]C[/imath] in: [imath] I(b) = C e^{-2b} [/imath] without leaving this framework yet again. The third try is trying modify Américo Tavares's answer in this MSE question: [imath] I(b) = \int^\infty_0 be^{-b^2x^2}\,dx. [/imath] It is easy to show that: [imath] I'(b) = \int^\infty_0 e^{-b^2x^2}\,dx - \int^\infty_0 2b^2 x^2 e^{-b^2x^2}\,dx = 0 [/imath] by an integration by parts identity: [imath] \int^\infty_0 x^2 e^{- c x^2}\,dx = \frac{1}{2c}\int^\infty_0 e^{- c x^2}\,dx . [/imath] Then [imath]I(b) = C[/imath], ouch, stuck again at this constant. Notice in that Proving [imath]\displaystyle\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}[/imath] question, Bryan Yocks's answer is somewhat similar to the idea of parametrization, however he has to introduce another parametric integration to produce a definite integral leading to [imath]\arctan[/imath]. Is there such a one shot parametrization trick solution like the author Anonymous claimed to be "creative parameterizations and a dose of differentiation under the integral"? |
1970825 | Find the determinant of [imath]I + A[/imath]
What is the determinant of an [imath]n \times n[/imath] matrix [imath]B=A+I[/imath], where [imath]I[/imath] is the [imath]n\times n[/imath] identity matrix and [imath]A[/imath] is the [imath]n\times n[/imath] matrix [imath] A=\begin{pmatrix} a_1 & a_2 & \dots & a_n \\ a_1 & a_2 & \dots & a_n \\ \vdots & \vdots & \ddots & \vdots \\ a_1 & a_2 & \dots & a_n \\ \end{pmatrix} [/imath] | 2911469 | Evaluate the determinant of a rank-1 matrix plus [imath]I[/imath]
let [imath]a_1, .......a_n \in \mathbb{R}[/imath]. Evaluate the determinant [imath]\begin{bmatrix} 1+a_1 &a_2 &....&a_n\\a_1 & 1+a_2 &....&a_n \\...&...&....&...\\a_1 &a_2& ....&1 +a_n \end{bmatrix}[/imath] First i take [imath]2\times2[/imath] matrix A =[imath]\begin{bmatrix} 1+a_1 &a_2 \\ a_1&1 +a_2 \end{bmatrix}[/imath] As Determinant of [imath]A = (1+a_1)(1+a_2) -a_1a_2= 1+a_2+a_1a_2 -a_1a_2= 1+ a_1 + a_2[/imath] I don't know how find the determinant of [imath]n\times n[/imath] matrix. Any hints/solution thanks u |
2904725 | Differential functional equation
I'm struggling to solve this problem: Let [imath]f[/imath] be a continuous function defined on [imath][0,+\infty[[/imath], derivable in every point such that [imath] f'(x)=2f(2x)-f(x), \ \ \ \ \ \ \forall x>0, [/imath] and [imath] M_n=\int_{0}^{\infty}x^nf(x)dx<+\infty, \ \ \ \ \ \ \forall n\in\Bbb{N}. [/imath] Prova that exists a non zero function that satisfies this conditions. For which sequences of real numbers [imath]\{a_n\}[/imath] it's true that [imath]a_n=M_n, \ \ \forall n\in\Bbb{N}[/imath]. All I managed to get is that [imath] M_n=n!\prod_{i=1}^{n}\frac{2^i}{2^i-1}M_0=n!\prod_{i=1}^{n}\frac{2^i}{2^i-1}\int_{0}^{\infty}f(x)dx. [/imath] | 2553131 | Solving functional differential equation [imath]f'(x)=2f(2x)-f(x)[/imath]
Show that there is at least a nonzero function [imath]f[/imath], differentiable on [imath][0,+\infty)[/imath], satisfying [imath]f'(x)=2f(2x)-f(x) \qquad \forall x>0 [/imath] [imath]M_n:=\int_{0}^{\infty}x^nf(x)dx<\infty \qquad \forall n\in \mathbb{N} [/imath] My best idea so far is to assume that the solution is a power series, i.e. [imath] f(x)=\sum_{n=0}^{\infty}a_nx^n\qquad \forall x>0[/imath] Then the equation becomes [imath] \sum_{n=0}^{\infty}na_nx^{n-1}=2\sum_{n=0}^{\infty}a_n2^nx^n-\sum_{n=0}^{\infty}a_nx^n[/imath] equating all the coefficients of the same degree I get [imath]na_n=(2^{n}-1)a_{n-1}\qquad \forall n\geq 1[/imath] So setting [imath]a_0=1[/imath], I get [imath]a_{n}=\frac{1}{n!}\prod_{k=1}^{n}(2^k-1) \qquad \forall n[/imath] But does the power series actually converge? Using Hadamard's formula, and that [imath]2^{k}-1\geq 2^{k-1}[/imath], [imath] |a_n|^{1/n}\geq\frac{1}{(n!)^{1/n}}\left[2^{n(n-1)/2}\right]^{1/n}\sim\frac{e}{n(2\pi n)^{1/2n}}2^{(n-1)/2}\to \infty[/imath] so the radius of converge of the series is [imath]0[/imath], so it doesn't actually define a solution on [imath][0,+\infty)[/imath]. |
2905018 | [imath]\sum_{k=0}^n \binom{n}{k}x^k(1-x)^{n-k}k[/imath]?
I cant wrap my head around simplifying the following sum: [imath]\sum_{k=0}^n \binom{n}{k}x^k(1-x)^{n-k}k,[/imath] where [imath]0<x<1[/imath]. I tried to apply standard formulas here. | 703653 | variation of the binomial theorem
Why does: [imath] \sum_{k=0}^{n} k \binom nk p^k (1-p)^{n-k} = np [/imath] ? Taking the derivative of: [imath] \sum_{k=0}^{n} \binom nk p^k (1-p)^{n-k} = (1 + [1-P])^n = 1 [/imath] does not seem useful, since you would get zero. And induction hasn't yet worked for me, since – during the inductive step – I am unable to prove that: [imath] \sum_{k=0}^{n+1} k \binom {n+1}{k} p^k (1-p)^{n+1-k} = (n+1)p [/imath] assuming that: [imath] \sum_{k=0}^{n} k \binom {n}{k} p^k (1-p)^{n-k} = np [/imath] |
1406741 | Prove that if [imath]|a_1\sin x+a_2\sin2x+a_3\sin3x+\cdots+a_n\sin nx|\leq|\sin x|[/imath] for [imath]x\in R,[/imath]then [imath]|a_1+2a_2+3a_3+....+na_n|\leq1[/imath]
Prove that if [imath]|a_1\sin x+a_2\sin2x+a_3\sin3x+\cdots+a_n\sin nx|\leq|\sin x|[/imath] for [imath]x\in R,[/imath]then [imath]|a_1+2a_2+3a_3+\cdots+na_n|\leq1[/imath] When we try to differentiate it on both sides wrt [imath]x[/imath],then modulus sign comes,how will one prove this question?Please help me. | 92999 | how can I show this question : function
Let [imath]f\colon\mathbb R\to\mathbb R[/imath] defined by [imath]f(x)=a_1\sin x+a_2\sin2x+a_3\sin3x+\cdots+a_n\sin nx,[/imath] for some values [imath]a_1,a_2,a_3,\cdots,a_n\in\mathbb R.[/imath] Prove that [imath] |f(x)|\le|\sin x|\quad \forall x\in \mathbb{R}[/imath] implies: [imath]|a_1+2a_2+3a_3+\cdots+na_n|\le1[/imath] |
2839841 | eigenvalues of [imath]AA^T[/imath] and [imath]A^TA[/imath]
Is it true (and under which conditions) that the products of an non-square matrix [imath]A[/imath] and its transpose and vice versa (so the product of the transpose and [imath]A[/imath]) share the same eigenvalues (multiplicities omitted)? | 2672419 | Let [imath]A[/imath] be an [imath]m \times n[/imath] matrix. Show [imath]A^TA[/imath] and [imath]AA^T[/imath] have the same eigenvalues
Let [imath]A[/imath] be an [imath]m \times n[/imath] matrix. Show [imath]A^TA[/imath] and [imath]AA^T[/imath] have the same eigenvalues. I'm unsure how to approach this. I'm trying to assume that [imath]\lambda[/imath] is an eigenvalue of [imath]A^TA[/imath] with its eigenvector [imath]\neq 0[/imath] and use that to prove that [imath]\lambda[/imath] is also an eigenvalue of [imath]AA^T[/imath] with eigenvector [imath]Ax \neq 0[/imath], but I'm lost on how to actually state it with notation and build off it. Should I be trying something else? |
2905172 | Distance making [imath](\mathbb{R} \setminus \mathbb{Z},d) [/imath] complete and homeomorphic to [imath]\mathbb{R}\times \mathbb{Z}[/imath]
As in the title I have to construct a metric [imath]d[/imath] on [imath]\mathbb{R} \setminus \mathbb{Z}[/imath] s.t. [imath]( \mathbb{R} \setminus \mathbb{Z}, d)[/imath] is complete and homeomorphic to [imath] \mathbb{R} \times \mathbb{Z}[/imath]. I don't know how to proceed! | 685550 | Construct such [imath]d[/imath] that [imath](\mathbb{R} \setminus \mathbb{Z}, d)[/imath] is complete metric space
Good evening! I had topology exam yesterday and I had a question that gave me problems. Lets consider [imath]\mathrm{G} = \mathbb{R} \setminus \mathbb{Z} [/imath] , i.e. real line without integrals, where [imath]\rho[/imath] is natural metric on [imath]\mathrm{G}[/imath]. Find metric [imath]d[/imath] equivalent to [imath]\rho[/imath], where [imath](\mathrm{G},d)[/imath] is complete metric space. It is obvious that [imath](\mathrm{G},\rho)[/imath] is not complete, but I don't know how to construct [imath]d[/imath]. Can I do something like get rid of Cauchy's property only in integral points of real line and maintain in the other ones? |
2904898 | Why not convex?
I am reading Nonnegative Matrix Factorization(NMF) and encounter following sentences: Although the functions [imath]||V - W H||^2[/imath] are convex in W only or H only, they are not convex in both variables together Can someone explain why it's not convex in both variables together? Thanks. | 393447 | Why does the non-negative matrix factorization problem non-convex?
Supposing [imath]\mathbf{X}\in\mathbb{R}_+^{m\times n}[/imath], [imath]\mathbf{Y}\in\mathbb{R}_+^{m\times r}[/imath], [imath]\mathbf{W}\in\mathbb{R}_+^{r\times n}[/imath], the non-negative matrix factorization problem is defined as: [imath]\min_{\mathbf{Y},\mathbf{W}}\left\|\mathbf{X}-\mathbf{Y}\mathbf{W}\right\|_F^2[/imath] Why is this problem non-convex? |
270305 | Pick out the case(s) which ensure that the polynomial [imath]p(\cdot)[/imath] has a root in the interval [imath][0, 1][/imath]
Please help me to solve the problem below. Consider the polynomial [imath]p(x) = a_0 + a_1x + a_2x^2+\dots+ a_nx^n[/imath], with real coefficients. Pick out the case(s) which ensure that the polynomial [imath]p(\cdot)[/imath] has a root in the interval [imath][0, 1][/imath]. (a) [imath]a_0 < 0[/imath] and [imath]a_0 + a_1 +\dots+ a_n > 0[/imath]. (b) [imath]a_0 +a_1/2+ \dots +a_n/(n + 1)= 0[/imath]. (c)[imath]\frac{a_0}{1.2}+\frac{a_1}{2.3}+ \dots+\frac{a_n}{(n + 1)(n + 2)}= 0[/imath]. | 554262 | sufficient condition for a polynomial to have roots in [imath][0,1][/imath]
Question is to check : which of the following is sufficient condition for a polynomial [imath]f(x)=a_0 +a_1x+a_2x^2+\dots +a_nx^n\in \mathbb{R}[x] [/imath] to have a root in [imath][0,1][/imath]. [imath]a_0 <0[/imath] and [imath]a_0+a_1+a_2+\dots +a_n >0[/imath] [imath]a_0+\frac{a_1}{2}+\frac{a_2}{3}+\dots +\frac{a_n}{n+1}=0[/imath] [imath]\frac{a_0}{1.2}+\frac{a_1}{2.3}+\dots+\frac{a_n}{(n+1).(n+2)} =0[/imath] First of all i tried by considering degree [imath]1[/imath] polynomial and then degree [imath]2[/imath] polynomial and then degree [imath]3[/imath] polnomial hoping to see some patern but could not make it out. And then, I saw that [imath]a_0= f(0)[/imath] and [imath]f(1)=a_0+a_1+a_2+\dots +a_n[/imath]. So, if [imath]f(0)<0[/imath] and [imath]f(1)>0[/imath] it would be sufficient for [imath]f[/imath] to have root in [imath][0,1][/imath] In first case we have [imath]a_0 <0[/imath] i.e., [imath]f(0)<0[/imath] and [imath]f(1)>1>0[/imath]. So, first condition should be implying existence of a root in [imath][0,1][/imath] for second case, let [imath]f(x)[/imath] be a linear polynomial i.e., [imath]f(x)=a_0+a_1x[/imath] Now, [imath]a_0+\frac{a_1}{2}=0[/imath] implies [imath]0\leq x=\frac{-a_0}{a_1}=\frac{1}{2}< 1[/imath] So, this might be possibly give existence in case of linear polynomials. Now, [imath]\frac{a_0}{1.2}+\frac{a_1}{2.3}=0[/imath]implies [imath]0\leq x=\frac{-a_0}{a_1}=\frac{1}{3}< 1[/imath] So, this might be possibly give existence in case of linear polynomials. So, for linear polynomials all the three conditions imply existence of a root in [imath][0,1][/imath]. But, i guess this can not be generalized for higher degree polynomial. I think there should be some "neat idea" than checking for roots and all. I am sure about first case but I have no idea how to consider the other two cases. please provide some hints to proceed further. |
2906152 | Prove uniqueness of limit of convergent sequence in [imath]\mathbb{R^n}[/imath]
Using the definition, that for every [imath]\epsilon>0[/imath], there exists a [imath]N\in\mathbb{N}[/imath], so that [imath]d(v_i,L)[/imath] for all [imath]i\geq N[/imath], where [imath]v_i[/imath] is the [imath]i[/imath]th vector in the sequence. And [imath]L[/imath] is the limit. I assumed that let there be two limits, [imath]L[/imath] and [imath]M[/imath]. Now I choose my epsilon to be [imath]\frac13d(L,M)[/imath], but after that, how do I proceed? Plain algebra doesn't help. Note that [imath]d(a,b)[/imath] is the distance between vectors [imath]a[/imath] and [imath]b[/imath] in [imath]\mathbb{R^n}[/imath]. | 1519941 | Every convergent sequence in a metric space has a unique limit.
I'm reading through a proof in my lecture notes: Every convergent sequence of a metric space has a unique limit. Suppose [imath]\lim_{n \to \infty} x_n = l[/imath] and [imath]\lim_{n \to \infty} x_n = m.[/imath] Then for every [imath]\epsilon > 0,[/imath] there exists [imath]N[/imath] such that for every [imath]n \geqslant N,[/imath] [imath]d(l,m) \leqslant d(l,x_n) + d(x_n,m) < \epsilon + \epsilon = 2 \epsilon. [/imath] So [imath](\star) \hspace{1mm} 0 \leqslant \frac{d(l,m)}{2} < \epsilon \Longrightarrow \hspace{1mm} (\star^\prime) \hspace{1mm} \frac{d(l,m)}{2} =0 \Longrightarrow l = m.[/imath] I can follow the proof up until the jump from [imath](\star)[/imath] to [imath](\star^\prime)[/imath]. Why does [imath]\frac{d(l,m)}{2}=0[/imath]? |
2905900 | Find [imath]\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)[/imath]
Find [imath]\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)[/imath] , without using squeeze theorem. I have done the solution as below using squeeze theorem ... [imath]Let \left[\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)\right]=f(x)\implies \\ \left({x \over x^2+x}+{x \over x^2+x}+\cdots +{x\over x^2+x}\right)\lt f(x)\lt \left({x \over x^2+1}+{x \over x^2+1}+\cdots +{x\over x^2+1}\right) \\ {x^2 \over x+x^2}\lt f(x) \lt {x^2\over 1+x^2}\\ \text{applying limit on both sides }\\ \implies\lim_{x\to \infty}{x^2 \over x+x^2}= \lim_{x\to \infty}{x^2\over 1+x^2}=1\\ \implies \lim_{x\to \infty}f(x)=1[/imath] Can we do this without squeeze theorem? | 1995860 | Sum of [imath]\lim_{n\rightarrow \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots \cdots \cdots +\frac{n}{n^2+n}\right)[/imath]
Sum of [imath]\lim_{n\rightarrow \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots \cdots \cdots +\frac{n}{n^2+n}\right)[/imath] [imath]\bf{My\; Try::}[/imath] I have solved it using Squeeze theorem [imath]\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+n}\leq \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+r}\leq \lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+1}[/imath] So we get [imath]\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+r} = 1[/imath] My question is can we solve above limit without using Squeeze Theorem, If yes then plz explian me, Thanks |
1030249 | Inequality: showing that [imath]p(1-p)\leq \frac{1}{4}[/imath] if [imath]0.[/imath]
I was wondering if there was another way to derive the inequality shown in the title of this question apart from calculus. Doing it with calculus is pretty straightforward: Let [imath]f:(0,1)\to\mathbb R[/imath] be a function defined by [imath]f(p)=p(1-p)[/imath]. Taking the first derivative [imath]f'(p) = 1-2p=0\Rightarrow p=1/2[/imath] is a critical point. Then the maximum is [imath]1/4[/imath] and it is reached when [imath]p=1/2[/imath]. Hence [imath]p(1-p)\leq 1/4,[/imath] [imath]\forall p\in(0,1)[/imath]. | 682093 | Upper Bound for [imath]p(1-p) [/imath] where [imath]0 \le p \le 1[/imath]
In a book on statistics, I've seen the upper bound of [imath]p(1-p) [/imath] to be [imath] p(1-p) \le \frac{1}{4} [/imath] for [imath]0 \le p \le 1[/imath] which seemed correct. I tried to duplicate this with a simple derivation, [imath]p(1-p) = C[/imath] [imath]p-p^2 = C[/imath] Taking derivatives of both sides, gives [imath]1-2p=0[/imath] [imath]p=\frac{1}{2}[/imath] which is the value of [imath]p[/imath] at the maximum, and [imath]1/2*1/2 = 1/4[/imath]. I obtained this without using the constraint [imath]0 \le p \le 1[/imath] though. How would I include this constraint? Through Lagrange? Is that overkill? |
2906174 | Express a matrix [imath]A[/imath] as the sum of a symmetric and a skew symmetric matrix
I'm trying to express [imath]A = \left[ \begin{array} { r r r } { 2 } & { - 2 } & { - 4 } \\ { - 1 } & { 3 } & { 4 } \\ { 1 } & { - 2 } & { - 3 } \end{array} \right][/imath] as the sum of a symmetric and a skew symmetric matrix. So far I have tried this: converting into transpose [imath]A ^ { \prime } = \left[ \begin{array} { c c c } { 2 } & { - 1 } & { 1 } \\ { - 2 } & { 3 } & { 4 } \\ { - 4 } & { 4 } & { - 3 } \end{array} \right][/imath] well this dont seem to be symmteric matrix according to me. are there steps to be taken. and for skew symmteric matrix i tried to do like this transpose of equation [imath]A ^ { \prime } = \left[ \begin{array} { c c c } { 2 } & { - 1 } & { 1 } \\ { - 2 } & { 3 } & { 4 } \\ { - 4 } & { 4 } & { - 3 } \end{array} \right][/imath] took minus as common [imath]A ^ { \prime } = \left[ \begin{array} { c c c } { - 2 } & { 1 } & { - 1 } \\ { 2 } & { - 3 } & { 2 } \\ { 4 } & { - 4 } & { 3 } \end{array} \right][/imath] can you guide me how to proceed with them | 1817421 | Any [imath]n \times n[/imath] matrix [imath]A[/imath] can be written as [imath]A = B + C[/imath] with [imath]B[/imath] is symmetric and [imath]C[/imath] skew-symmetric.
How can I proof the following statement? Any [imath]n \times n[/imath] matrix [imath]A[/imath] can be written as a sum [imath] A = B + C [/imath] where [imath]B[/imath] is symmetric and [imath]C[/imath] is skew-symmetric. I tried to work out the properties of a matrix to be symmetric or skew-symmetric, but I could not prove this. Does someone know a way to prove it? Thank you. PS: The question Prove: Square Matrix Can Be Written As A Sum Of A Symmetric And Skew-Symmetric Matrices may be similiar, in fact gives a hint to a solution, but if someone does not mind in expose another way, our a track to reach to what is mentioned in the question of the aforementioned link. |
2905989 | Characterisation of compactness in [imath]\mathbb{R}^n[/imath]
Let [imath]E[/imath] be a bounded measurable set of [imath]\mathbb{R}^n[/imath]. If every countinuous function [imath]f:E\to\mathbb{R}[/imath] is also uniformly continuous, then [imath]E[/imath] is compact. Since I do not know topological property of [imath]f(E)[/imath], it seems that I must use some measure property to show [imath]E[/imath] is closed. However, measure and topology seems not quite relevant. Thank you! | 316143 | Inverse of Heine–Cantor theorem
We have by Heine–Cantor theorem that: If [imath]M[/imath] and [imath]N[/imath] are metric spaces and [imath]M[/imath] is compact then every continuous function [imath]f : M \to N[/imath], is uniformly continuous. Is the inverse of this theorem is satisfied? |
2906153 | Topology generated by a basis or subbasis is the intersection of all topologies containing it
I am self studying Munkres' Topology and was struck on this exercise question of section 13. I don't have any help from teacher or friends. So, I am asking for any explanation here. The question is Show that if [imath]A[/imath] is a basis for a topology on [imath]X[/imath], then the topology generated by [imath]A[/imath] equals the intersection of all topologies on [imath]X[/imath] that contain [imath]A[/imath]. Prove the same if [imath]A[/imath] is a subbasis. The intuition is clear to me but I don't know how to write a proper proof. Can you please tell me for both parts. | 1169981 | Problem 5, sec. 13 in Munkres' Topology, 2nd ed.: How to prove the assertion if [imath]\mathscr{A}[/imath] is a subbasis?
Show that if [imath]\mathscr{A}[/imath] is a basis for a topology on [imath]X[/imath], then the topology generated by [imath]\mathscr{A}[/imath] equals the intersection of all topologies on [imath]X[/imath] that contain [imath]\mathscr{A}[/imath]. This is what I've managed. However, the following stumps me: Prove the same if [imath]\mathscr{A}[/imath] is a subbasis. That is, how to show that the topology generated by [imath]\mathscr{A}[/imath] as a subbasis is the same as the intersection of all topologies on [imath]X[/imath] that contain [imath]\mathscr{A}[/imath]? Let [imath]\mathscr{T}[/imath] be the topology that consists of all unions of finite intersections of sets in [imath]\mathscr{A}[/imath], and let [imath]\mathscr{T}^\prime[/imath] be the topology that is the intersection of all topologies that contain [imath]\mathscr{A}[/imath]. Now [imath]\mathscr{A}[/imath] is a collection of subsets of [imath]X[/imath] whose union is all of [imath]X[/imath], and each set [imath]A \in \mathscr{A}[/imath] is in both [imath]\mathscr{T}[/imath] and [imath]\mathscr{T}^\prime[/imath]. Am I right so far? And if so, then what next? |
2905746 | Can a relation be transitive when it is symmetric but not reflexive?
Pretty much what the title asks. But here's some context: Suppose [imath]X[/imath] is finite and [imath]R[/imath] is a relation on [imath]X[/imath] that is not reflexive but it is symmetric. Also, suppose we can rule out [imath]xRy[/imath] and [imath]yRz[/imath] both occurring [imath]\forall x,y,z\in X[/imath] s.t. [imath]x\neq y[/imath] and [imath]y\neq z[/imath] and [imath]x\neq z[/imath]. So we can rule out that [imath]xRy[/imath] and [imath]yRz[/imath] when [imath]x,y,z[/imath] are distinct. At this point, it seems that [imath]R[/imath] might be vacuously transitive. However, since [imath]R[/imath] is symmetric, [imath]xRy[/imath] and [imath]yRx[/imath] are both fine. But since [imath]R[/imath] is not reflexive, we cannot have [imath]xRx[/imath], which seems to be a violation of transitivity if [imath]xRy[/imath] and [imath]yRx[/imath]. Is this a "legitimate" counter-example? | 1592652 | Example of a relation that is symmetric and transitive, but not reflexive
Can you give an example of a relation that is symmetric and transitive, but not reflexive? By definition, [imath]R[/imath], a relation in a set [imath]$X$[/imath], is reflexive if and only if [imath]\forall x\in X[/imath], [imath]x\,R\,x[/imath]. [imath]R[/imath] is symmetric if and only if [imath]\forall x, y\in X[/imath], [imath]x\,R\,y\implies y\,R\,x[/imath]. [imath]R[/imath] is transitive if and only if [imath]\forall x, y, z\in X[/imath], [imath]x\,R\,y\land y\,R\,z\implies x\,R\,z[/imath]. I can give a relation [imath]\leqslant[/imath], in a set of real numbers, as an example of reflexive and transitive, but not symmetric. But I can't think of a relation that is symmetric and transitive, but not reflexive. |
2906866 | Sum calculation of combination with floor
How could I calculate this sum: [imath]\sum\limits_{k=0}^{\lfloor \frac{n}{2}\rfloor} {n \choose 2k}[/imath] | 1896994 | Evaluate [imath]\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+...+\binom{n}{2k}[/imath]
What is the closed formula for finding [imath] \sum_{i=0}^{k} \binom{n}{2i}=?\quad ,k\leq \lfloor{n/2}\rfloor [/imath] I need summation only up to some intermediate [imath]k[/imath] where [imath]k\leq \lfloor{n/2}\rfloor[/imath] . SAY I need a closed formula to calculate [imath]\binom{8}{0}+ \binom{8}{2}+\binom{8}{4} +\binom{8}{6}[/imath] HERE [imath]n=8[/imath] and [imath]k=6[/imath] |
2906434 | Volume of revolution by the region bounded by the parabolas [imath]y^2=x, y^2=8x, x^2=y, x^2=8y[/imath] about the [imath]x[/imath] axis is [imath]\frac{279\pi}{2}[/imath].
Show that the volume generated by revolving the region in the first quadrant bounded by the parabolas [imath]y^2=x, y^2=8x, x^2=y, x^2=8y[/imath] about the [imath]x[/imath] axis is [imath]\frac{279\pi}{2}[/imath]. I've used the substitution [imath]y^2=ux, x^2=vy[/imath]. The domain I'm supposed to integrate over is something like this leaf shaped region with another leaf cut out from it. So I've computed the inverse Jacobian to be [imath]\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}[/imath] = [imath]-\frac{y^2}{x^2}(-\frac{x^2}{y^2})-\frac{2x}{y}\frac{2y}{x}[/imath]=[imath]1-4[/imath]= [imath]-3[/imath] So we get the Jacobian of this transformation to be [imath]\frac{-1}{3}[/imath]. The problem is that I don't know what to plug into my limits of integration and the function I have to integrate over. I have the general sense of where I'm supposed to go from here. For instance, I can visualise values of [imath]u[/imath] taking on a family of curves 'fitted' in between [imath]x=y^2[/imath] to [imath]x=8y^2[/imath] and likewise for values of [imath]v[/imath]. Increasing values of [imath]u[/imath] and [imath]v[/imath], intutively seems like going along these family of curves until you get to u=8 and v=8 respectively. But I have no idea how to translate what I know about calculating the volume of a region rotated into this kind of [imath]u[/imath]-[imath]v[/imath] 'coordinate system', if that's even a possible way of putting it. Intutively, it seems like [imath]\int^8_1 \int^8_1 u^2-v^2 (-\frac{1}{3}) du dv[/imath] would work, but it doesn't. Can someone help me from here? Also, the negative value of the Jacobian seems a little fishy to me. I'd be grateful if someone could help me clarify any mistakes/misconceptions. | 2830650 | Volume generated by revolving the region in the first quadrant bounded by the parabolas [imath]y^2=x[/imath], [imath]y^2=8x[/imath], [imath]x^2=y[/imath], [imath]x^2=8y[/imath] about the [imath]x[/imath]-axis.
I came across this question from Schaum's Outlines of Advanced Calculus: Show that the volume generated by revolving the region in the first quadrant bounded by the parabolas [imath]y^2=x[/imath], [imath]y^2=8x[/imath], [imath]x^2=y[/imath], [imath]x^2=8y[/imath] about the [imath]x[/imath]-axis is [imath]\dfrac{279\pi}{2}[/imath]. My solution was to divide the region into three parts: [imath]\pi\int_{1}^{2}(x^2)^2 - (\sqrt{x})^2\;dx = 4.7\pi[/imath] [imath]\pi\int_{4}^{8}(\sqrt{8x})^2 - (\sqrt{x})^2\;dx = 42\pi[/imath] [imath]\pi\int_{2}^{4} (x^2)^2 - \left(\frac{1}{8}x^2 \right)^2\;dx = 44.9\pi.[/imath] Hence the total volume will be [imath]\pi(4.7 + 42 + 44.9) = 91.6\pi[/imath] which is wrong. Can someone point out my mistake? |
2907113 | Compute [imath]\lim\limits_{n \to \infty} {\frac{1 \cdot 3 \cdot 5 \cdots(2n - 1)}{2 \cdot 4 \cdot 6 \cdots (2n)}}[/imath]
EDIT: @Holo has kindly pointed out that my concept of ln rules used in this question is wrong. However, the intuition behind using the tangent of a curve to find the sum to infinity of a series still stands. Therefore, I won't be editing the post. I tried expanding out the equation from the question and got [imath]{a_{n}} = \frac{1}{2} .\frac{3}{4} . \frac{5}{6} ...\frac{2n -1}{2n}[/imath] I then tried taking the ln of the equation which works out to [imath]\ln(1 - \frac{1}{2}) + \ln(1 - \frac{1}{4}) + ...[/imath] Here is my question. I used the rules of ln functions and took [imath]ln(\frac{1}{1/2}) + ln(\frac{1}{\frac{1}{4}})[/imath] + ... Since ln(1) is 0, shouldn't the limit work out to 0 as n tends to infinity? Then [imath]ln(L) = 0, where L = limit[/imath] [imath]L = e^0[/imath] [imath]L = 1[/imath] However, the limit is actually 0, and my tutor used a method which I couldn't understand as he tried approximating the function [imath]y = ln(x)[/imath] to [imath]y = x - 1[/imath], as he says that the linear equation is actually a tangent to the ln curve. Could someone please explain this intuition behind it? He differentiated the equation, and since the curve cuts the x-axis at x = 1, he got the linear curve [imath]y = x - 1[/imath]. He did mention that the linear equation is just an approximation, but said such an approximation would be more than sufficient. | 2402508 | Convergence point of [imath]\frac{\prod_{k=1}^n (2k-1)} {\prod_{k=1}^n 2k}[/imath]
Let [imath]a_n = \frac{\prod_{k=1}^n (2k-1)} {\prod_{k=1}^n 2k}[/imath] is a sequence of numbers . Is [imath]a_n[/imath] monotone and bounded ? If it's a convergent sequence find convergence point . My try : It's obvious that [imath]a_n[/imath] is decreasing and bounded. So we can deduce it's a convergent sequence but I'm not able to find [imath]\lim_{n \to \infty} a_n[/imath] . |
2906838 | Fundamental Group of complement of Integers
Let [imath]X=\mathbb{R}\times (-1,1)-\mathbb{Z}\times \{0\}[/imath]. I want to compute the fundamental group of that space. I would be tempted to do some division like the following to apply Van Kampen, but to do it all the sets must intersect in the basepoint of [imath]\pi(X,x)[/imath], so I can't really do it. If I incorrectly use Van Kampen, I will get to the correct result that it's free in a countable number of generators, though. Is there any other way to do that, I mean, with the basic theory of Hatcher's of Massey's book (It's an exercise from Massey's book)? | 2126722 | Fundamental group of [imath]\mathbb{R}^2 \setminus (\mathbb{Z} \times \{0\})[/imath]
Good evening, I need to compute [imath]\pi_1(\mathbb{R}^2 \setminus (\mathbb{Z} \times \{0\}))[/imath]. I see that the set [imath]\mathbb{R}^2 \setminus \mathbb{Z} \times \{0\}[/imath] has a countable set of points less than whole [imath]\mathbb{R}^2[/imath], so at each point we can retract the space into circles, and finally the space can retract to a wedge of infinite countable circles. Aplying Seifert-van Kampen theorem we can conclude that the fundamental group is free product of [imath]\mathbb{Z}[/imath] with itself countable times. Is this idea correct? Thanks a lot. |
2477610 | Proving that [imath]a=2[/imath] and [imath]n[/imath] is a prime given [imath]p[/imath] is a prime in [imath]p= a^n - 1[/imath]
Let [imath]a[/imath] and [imath]n[/imath] be integers greater than [imath]1[/imath]. Suppose that [imath]a^n - 1[/imath] is prime. Show that [imath]a = 2[/imath] and that [imath]n[/imath] is prime. First of all let [imath]p = a^n - 1[/imath] Since both [imath]a[/imath] and [imath]n[/imath] are integers greater than one then [imath]p \ne 2[/imath] Thus [imath]p[/imath] is an odd prime Rearanging: [imath]a^n = p + 1[/imath] Thus [imath]a^n[/imath] is an even integer, and then [imath]a[/imath] is and even integer. Taking [imath]\log_2[/imath] of both sides: [imath] n\log_2(a) = \log_2(p+1) [/imath] and [imath]n = \frac{\log_2(p+1)}{\log_2(a)}[/imath] Since n is an integer then [imath]\log_2(a) = 1[/imath] and [imath]a=2[/imath] I managed to get thus far, but when it comes to proving n is a prime I don't seem to know how to approach it. | 328671 | For [imath]a,n \in \Bbb N[/imath], how to prove if [imath]a^n-1[/imath] is prime then [imath]a=2[/imath] and [imath]n[/imath] is prime?
Suppose [imath]a,n \in \Bbb N [/imath]. How do you prove that if [imath]a^n-1[/imath] is prime, then [imath]a=2[/imath] and [imath]n[/imath] is prime . Is the converse true I.E : (if [imath]p[/imath] is prime then [imath]2^p-1[/imath] is prime)? |
189724 | What is the necessary and sufficient condition for the matrix [imath]A = I − 2xx^T [/imath]to be orthogonal?
Let [imath]x[/imath] be a non-zero (column) vector in [imath]\mathbb{R}^n[/imath]. What is the necessary and sufficient condition for the matrix [imath]A = I − 2xx^T[/imath] to be orthogonal? | 143695 | Necessary and sufficient condition for the matrix $A = I - 2 x x^t$ to be orthogonal
Let [imath]x[/imath] be a non zeo (column) vector in [imath]\mathbb{R}^n[/imath]. What is the necessary and sufficient condition for the matrix [imath]A = I-2xx^t[/imath] to be orthogonal? |
2907420 | Does the position of the [imath]1[/imath] in a Jordan block matters?
I am studying the Jordan Canonical Form of a matrix and I noticed that most of the books put the [imath]1's[/imath] of the Jordan blocks on the superdiagonal like this for example : \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix} But my professor puts it on the subdiagonal : \begin{pmatrix} \lambda & 0 \\ 1& \lambda \end{pmatrix} Does it make any difference ? I know this might change the position of the vectors of the Jordan basis on the matrix [imath]P[/imath] such that [imath]J = P^{-1} A P[/imath] , but can can I say that one of this constructions is better than the other or that one of them is more correct than the other ? | 1847498 | Two different definitions of Jordan canonical form
I am currently reading two linear algebra books. One is Hoffman/Kunze's and the other one is Friedberg/Insel/Spence's. They define Jordan canonical form of linear operator in different ways. In Hoffman's book, Jordan forms are obtained by the primary decomposition and cyclic decomposition. In this case, 1's can appear some sub-diagonal entries. For example, [imath]M = \begin{array}{cc} 2 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 0 & 3 \\ \end{array}[/imath] But in Friedberg's (And almost any other linear algebra books for engineers..), 1's can appear some super-diagonal entries. For example, [imath]M' = \begin{array}{cc} 2 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 3 \\ \end{array}[/imath] Can someone explain why they define differently and relation of each other? |
2015920 | Multivariable limit - perhaps a trickier problem I am stuck on.
I am trying to solve the following limit: [imath]\lim_{(x,y) \to (0,0)} \frac{x^4y^4}{(x^2 + y^4)^3}[/imath] (This is a more challenging problem from Folland Calculus, it seems). I am pretty sure this limit does not exist (however, this is just a guess, and I am not 100% sure.) I am trying to approach via the paths y = mx and x = my but... they don't seem to work. I have tried simpler cases such as x = 0 and y = 0. Any help appreciate. | 855395 | Check if limit exists and its value
How to check if this limit exists: [imath]\lim_{(x,y)\rightarrow (0,0)}{ \frac{x^4y^4}{(x^2 + y^4)^3}}[/imath] Can I convert it to polar form ? [imath]\frac{r^8 (\cos^4 (\theta)\sin^4(\theta)}{r^6 (\cos^6 (\theta) + r^6 \sin^{12} (\theta))} = \frac{r^2}{\cos^6(\theta) + 0} = 0[/imath] |
2901216 | How would the graph of [imath]y=(-1)^x[/imath] look like?
Question When typing in [imath]y=(-1)^x[/imath] into a graphing calculator, it does not display anything at all. And once you think about it, you begin to realize that the f(x) values flip from positive to negative. How would this graph look like? | 2329298 | Graph of [imath](-1)^x[/imath]
What would the graph of [imath](-1)^{x}[/imath] look like? I know that the value of the function alternates between [imath]1[/imath] and [imath]-1[/imath] when it is defined so I think it would just be points spread over the lines [imath]x=1[/imath] and [imath]x=-1[/imath]. Is this correct? Also, will there be any definite pattern of the points. |
2907612 | Execute the term [imath]P=A(A^TA)^{-1}A^T[/imath]
Let [imath]P:\mathbb R^m \to \mathbb R^m[/imath] is orthogonal linear transformation on some subspace [imath]U[/imath] of [imath]\mathbb R^m[/imath]. Execute the term for matrix of linear transformation P using standard base. Explain the procedure. I know matrix for orthogonal linear transformation is P=[imath]A(A^TA)^{-1}A^T[/imath], I saw proof that this matrix is matrix of projection and it is orthogonal but,how I get that matrix, I could not find anything how to get that matrix, can you help me? | 2716442 | Writing projection in terms of projection matrix
I've been reading this document. The goal is to find projection of [imath]b[/imath] on the line [imath]L[/imath] which is determined by vector [imath]a[/imath]. ([imath]Proj_L(b)[/imath]) In the document, It is mentioned that If [imath]p[/imath] is thought as approximation to [imath]b[/imath], then [imath]e=b-p[/imath] is the error in that approximation (the word approximation is little confusing though, Isn't [imath]e[/imath] the exact value? Although this is not the problem). Then we know that If [imath]p[/imath] lies on the line through [imath]a[/imath], then [imath]p = xa[/imath] for some [imath]x[/imath]. We also know that, [imath]p[/imath] is orthogonal to [imath]e[/imath], therefore their dot product equates to zero: [imath]a^T(b-xa)=0[/imath] [imath]a^Tb - a^Txa = 0[/imath] [imath]xa^Ta = a^Tb[/imath] [imath]x = \frac{a^Tb}{a^Ta}[/imath] Solving for [imath]p[/imath]: [imath]p = ax = a\frac{a^Tb}{xa^Ta}[/imath] First part was almost completely understandable, but in the second part this projection is written in the terms of projection matrix ("[imath]P: p = Pb[/imath]"): [imath]p = xa = \frac{aa^Ta}{a^Ta}[/imath] where did [imath]b[/imath] go? [imath]x[/imath] and [imath]a[/imath] have changed places, but isn't dot product commutative? Then, [imath]P[/imath] is solved: [imath]P = \frac{aa^T}{a^Ta}[/imath] Somehow, [imath]aa^T[/imath] is 3x3 matrix. How was this concluded? From my knowledge, The general definition of projection matrix is [imath]A(A^TA)^-1A^T\vec{x}[/imath] (where [imath]A[/imath] is matrix) Does the definition above has any relations with the projection matrix that was represented in document? If not, how was it derived? Thank you! |
589001 | Abstract algebra: homomorphism and kernel from [imath]\Bbb Z[/imath] to [imath]S_{3}[/imath]
1) Assume that there exist homomorphisms [imath]f:\mathbb{Z} \rightarrow S_{3}[/imath] where [imath]\ker(f)\neq \mathbb{Z}[/imath]? What are the possible kernels? Explain. 2) Lest at least two non-trivial homomorphisms [imath]f:\mathbb{Z} \rightarrow S_{3}[/imath] whose kernel are not [imath]\mathbb{Z}[/imath]. | 592121 | Homomorphisms from [imath]\mathbb{Z} \rightarrow S_3[/imath].
I recently had my second abstract algebra exam returned to me and I missed this particular question (unfortunately I left the problem blank, just a bad test day). I've been trying to figure out the problem though and could use some help. a) Does there exist a surjective homomphism [imath]\phi: \mathbb{Z} \rightarrow S_3[/imath]? Explain. What I'm thinking: Let [imath]\phi: \mathbb{Z} \rightarrow S_3[/imath] be a homomorphism.[imath]\mathbb{Z}[/imath] is a cyclic group and thus [imath]\phi(\mathbb{Z})[/imath] is cyclic. However, [imath]S_3[/imath] is not cyclic. Thus [imath]\phi(\mathbb{Z}) \neq S_3[/imath] and [imath]\phi[/imath] cannot be surjective. b) Assume that there exist homomorphisms [imath]\phi: \mathbb{Z} \rightarrow S_3[/imath] where Ker[imath](\phi) \neq \mathbb{Z}[/imath]. What are the possible kernels? Explain. What I'm thinking: Since [imath]1[/imath] generates [imath]\mathbb{Z}[/imath], if [imath]1\in[/imath] Ker[imath](\phi)[/imath], then [imath]\mathbb{Z}=[/imath]Ker[imath](\phi)[/imath]. So could Ker[imath](\phi)[/imath] = [imath]n\mathbb{Z}[/imath] for [imath]n \in \mathbb{Z}, \; n\neq1[/imath]? c)List at least two non-trivial homomorphisms [imath]\phi: \mathbb{Z} \rightarrow S_3[/imath] whose kernals are not [imath]\mathbb{Z}[/imath]. [imath]\phi_1(n) = (1\; 2)^n[/imath] [imath]\phi_2(n) = (1\; 3)^n[/imath] Any input is much appreciated! |
395202 | How to show the given equality
Show that [imath]\dfrac{3}{1\cdot2\cdot4}+\dfrac{4}{2\cdot3\cdot5}+\dfrac{5}{3\cdot4\cdot6}+\cdots+\dfrac{n+2}{n\cdot(n+1)\cdot(n+3)}\\=\dfrac{1}{6}\left[\dfrac{29}{6}-\dfrac{4}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3} \right][/imath] Trial: I proceed in this way [imath]\dfrac{3}{1\cdot2\cdot4}+\dfrac{4}{2\cdot3\cdot5}+\dfrac{5}{3\cdot4\cdot6}+\cdots+\dfrac{n+2}{n\cdot(n+1)\cdot(n+3)}\\=\dfrac{1}{6} \left[\dfrac{18}{1\cdot2\cdot4}+\dfrac{24}{2\cdot3\cdot5}+\dfrac{30}{3\cdot4\cdot6}+\cdots+\dfrac{6n+12}{n\cdot(n+1)\cdot(n+3)} \right] [/imath]Then stuck how to break [imath]6n+12[/imath] so that we get the following result. | 2904182 | Sum the series [imath]\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}[/imath]
[imath]\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}[/imath] The general term seems to be [imath]T_r= \frac{r+2}{r(r+1)(r+3)}.[/imath] I see no way to telescope this because the factors of the denominator of the general term are not in arithmetic progression. Do I have to use something else? Or am I missing some easy manipulation? |
2907949 | theorem of existence of an orthogonal complement
are there any theorems that give us any conditions to know if a linear subspace [imath]E[/imath] plus its orthogonal complement span the whole vector space? For exemple, I know that in [imath]\mathbb{R}[X][/imath], the complement orthogonal of the hyperspace [imath]Span(1+X, 1+X^2, ...)[/imath] is [imath]\{0\}[/imath] and thus the sum does not span the whole space Thanks ! | 24763 | Proving that [imath]\mathbf{W}[/imath]+[imath]\mathbf{W^{\perp}}[/imath]=[imath]\mathbb{R^{n}}[/imath]
I am trying to prove that given a subspace [imath]\mathbf{W}[/imath] in [imath]\mathbb{R^{n}}[/imath], the subspace and its orthogonal complement 'cover' whole of [imath]\mathbb{R^{n}}[/imath] through '+' where we define [imath]\mathbf{W}[/imath]+[imath]\mathbf{W^{\perp}}[/imath] as linear combinations of vectors both in the subspace and in its orthogonal complement. It seems intuitively right, and I can prove that the sum of their dimensions adds up to n, but I am not sure how to prove the question I am looking at. Thanks! |
2905062 | Syntax vs semantics in logic and metalogic
Trying to better understand syntax vs semantics (structure vs meaning) in logic and metalogic. I am using this as my reference point (and the example(s) below on this link): https://en.wikipedia.org/wiki/Propositional_calculus#Generic_description_of_a_propositional_calculus e.g. let's say we have the connectives/operators: [imath]\Omega = \Omega_0 \cup \Omega_1 \cup \Omega_2 = \{T, F\} \cup \{\lnot\} \cup \{\to\}[/imath] And the following rule of inference (modus ponens): [imath]P, P \to Q \vdash Q[/imath] And the following axioms (Jan Łukasiewicz system, picking these arbitrarily): I. [imath](p \to (q \to p))[/imath] II. [imath]((p \to (q \to r)) \to ((p \to q) \to (p \to r)))[/imath] III. [imath]((\lnot p \to \lnot q) \to (q \to p))[/imath] Does this fully describe the entire logical system, or only in terms of syntax? Do the semantics follow or do we have to impose those on top of all this? Is imposing semantics a metalogical level concept? To my current understanding, we'd use truth tables to define various operators and such to give semantic meaning but are these optional / more of a convenience, or are they necessary? Are truth tables still metalogical? What about other definitions, like introducing operators such as [imath]\land[/imath], [imath]\lor[/imath], etc, are these still metalogical or would they need to be embedded in [imath]\Omega[/imath] up front? For example, how do we use all this to describe how [imath]\lnot[/imath] and [imath]\to[/imath] work? i.e. showing that: [imath]\lnot T = F[/imath] [imath]\lnot F = T[/imath] [imath]F \to F = T[/imath] [imath]F \to T = T[/imath] [imath]T \to F = F[/imath] [imath]T \to T = T[/imath] Or law of excluded middle: [imath]p \lor \lnot p = T[/imath] (after defining [imath]\land[/imath] and [imath]\lor[/imath], etc) Do we explicitly define these with truth tables, or derive them from the axioms and inferences somehow? How do the axioms and rules of inference play with each other here and allow us to ultimately go from something involving [imath]\vdash[/imath] to something with [imath]=[/imath] or [imath]\to[/imath]? Trying to understand how we start off with a system and start unrolling a logical system as we might normally use it. | 1218354 | Why are [imath]\vdash[/imath] and [imath]\vDash[/imath] symbols from metalanguage?
I've read in some textbooks that [imath]\vdash[/imath] and [imath]\vDash[/imath] are symbols present only in metalanguage. From this, I infer that their use in object language is unacceptable. I would like to know why. Can't we define them as relation symbols in a structure? Or introduce them in statements for the sake of formal proofs? |
2907817 | [imath]\lim_{x \to 2} \frac{\cos{\left(\frac{\pi}{x}\right)}}{x-2}[/imath] without using De L'Hospital
[imath]\lim_{x \to 2} \frac{\cos{\left(\frac{\pi}{x}\right)}}{x-2}[/imath] This limit is supposed be found without L'Hospital's Rule, but I have not been able to get close to the answer using conjugates, squares, Pythagorean Identity, half angle formulas or Squeeze Theorem. For each attempt, I expected at least one of the well-known limits [imath]\lim_{x \to 0} \frac{\sin{x}}{x}[/imath] or [imath]\lim_{x \to 0} \frac{\cos{x} - 1}{x}[/imath] to come into use. They frequently did in my attempts but did not go anywhere. I have also been experimented substitutions but in vain. How can this be solved without using L'Hospital's Rule? | 1449177 | Limit as [imath]x\to 2[/imath] of [imath]\frac{\cos(\frac \pi x)}{x-2} [/imath]
I am a kid trying to teach myself Calculus in order to prepare for next year. I have the expression [imath]\lim_{x\to 2}\frac{\cos(\frac \pi x)}{x-2} [/imath] There is a hint that says to substitute t for [imath](\frac \pi2 - \frac \pi x)[/imath] and WolframAlpha evaluates this expression as [imath]\frac \pi4[/imath]. However, I got the answer of [imath]1[/imath]. Can someone clarify the steps to solving this problem. |
2908151 | Link infinite series to finite series
In my econometric book, it says [imath]\sum\limits_{i=1}^n a^{i-1}=(1-a^n)/(1-a)[/imath] because [imath]\sum\limits_{i=1}^n a^{i-1}=(1-a^n)\sum\limits_{i=0}^\infty a^i[/imath]. I understand the proof of [imath]\sum\limits_{i=0}^\infty a^i=1/(1-a)[/imath] and [imath]\sum\limits_{i=1}^n a^{i-1}=(1-a^n)/(1-a)[/imath] separately. But how do we know [imath]\sum\limits_{i=1}^n a^{i-1}=(1-a^n)\sum\limits_{i=0}^\infty a^i[/imath], assuming we don't know the representation of [imath]\sum\limits_{i=0}^\infty a^i[/imath] and [imath]\sum\limits_{i=1}^n a^{i-1}[/imath]? | 2060166 | Deriving Sum of a Geometric Progression
I was trying to derive the sum of a geometric progression: [imath]\sum_{i=0}^{n-1}{ar^i}[/imath] This is equal to: [imath]ar^0+ar^1+ar^2+\cdots+ar^{n-1}[/imath] Factorising with [imath]a[/imath], I get: [imath]a \sum_{i=0}^{n-1}{r^i}[/imath] Assuming [imath]r[/imath] is [imath]6[/imath] for example, I get: [imath]a\left[r^0(1+r(1+r(1+r(1+r(1+r)))))\right][/imath] The number of [imath](1+r)[/imath]s inside is equal to [imath]n-1[/imath]. I have no idea how to expand the above equation, and simplify it. Can someone please give me a step-by-step expansion and simplification of the equation. Something a [imath]9^{th}[/imath] grader would understand. EDIT In general, I don't understand how to expand nested braces(recursion isn't the most intuitive thing for me) so a very simple explanation of how this is done will be much appreciated. |
2908295 | If [imath]f(x+y)=f(x)+f(y) ,\forall\;x,y\in\Bbb{R}[/imath], then if [imath]f[/imath] is continuous at [imath]0[/imath], then it is continuous on [imath]\Bbb{R}.[/imath]
I know that this question has been asked here before but I want to use a different approach. Here is the question. A function [imath]f:\Bbb{R}\to\Bbb{R}[/imath] is such that \begin{align} f(x+y)=f(x)+f(y) ,\;\;\forall\;x,y\in\Bbb{R}\qquad\qquad\qquad(1)\end{align} I want to show that if [imath]f[/imath] is continuous at [imath]0[/imath], it is continuous on [imath]\Bbb{R}.[/imath] MY WORK Since [imath](1)[/imath] holds for all [imath]x\in \Bbb{R},[/imath] we let \begin{align} x=x-y+y\end{align} Then, \begin{align} f(x-y+y)=f(x-y)+f(y)\end{align} \begin{align} f(x-y)=f(x)-f(y)\end{align} Let [imath]x_0\in \Bbb{R}, \;\epsilon>[/imath] and [imath]y=x-x_0,\;\;\forall\,x\in\Bbb{R}.[/imath] Then, \begin{align} f(x-(x-x_0))=f(x)-f(x-x_0)\end{align} \begin{align} f(x_0)=f(x)-f(x-x_0)\end{align} \begin{align} f(y)=f(x_0)-f(x)\end{align} HINTS BY MY PDF: Let [imath]x_0\in \Bbb{R}, \;\epsilon>[/imath] and [imath]y=x-x_0,\;\;\forall\,x\in\Bbb{R}.[/imath] Then, show that \begin{align} \left|f(x_0)-f(x)\right|=\left|f(y)-f(0)\right|\end{align} Using this equation and the continuity of [imath]f[/imath] at [imath]0[/imath], establish properly that \begin{align}\left|f(y)-f(0)\right|<\epsilon,\end{align} in some neighbourhood of [imath]0[/imath]. My problem is how to put this hint together to complete the proof. Please, I need assistance, thanks! | 2924837 | continuity at zero implies continuity everywhere
Suppose [imath]f(x)[/imath] is defined on the entire line and continuous at the origin with the property that [imath]f(\alpha+\beta) = f(\alpha) + f(\beta)[/imath] where [imath]\alpha,\beta \in \mathbb{R}[/imath]. Prove that [imath]f(x)[/imath] must be continuous at every point [imath]x=a[/imath]. Try: First notice that [imath]f(0) = f(0+0)=f(0)+f(0)=2f(0) \implies f(0)=0[/imath]. Since [imath]f[/imath] continuous at zero, then [imath]\lim_{x \to 0} f(x) = f(0) = 0[/imath]. Next, let [imath]a [/imath] be arbitary real number. Then, [imath] \lim_{ x \to a} f(x) = \lim_{x \to a} f(x-a)+f(a)=_{h = x-a} \lim_{h \to 0} f(h) + f(a) = 0 + f(a)=f(a)[/imath] So f is continuous everywhere. Is this a correct solution? |
2908314 | Solve for [imath]\theta[/imath]: [imath]x = \theta - \sin\theta[/imath]
Solve for [imath]\theta[/imath]: [imath]x = \theta - \sin\theta[/imath] Is this type of isolation a matter of identities? If so, which one(s)? | 1053472 | How to solve Kepler's equation [imath]M=E-\varepsilon \sin E[/imath] for [imath]E[/imath]?
I'm trying to create a program to solve a set of Kepler's Equation and I cannot isolate the single variable to use the expression in my program. The Kepler Equation is [imath]M = E - \varepsilon \sin(E)[/imath] I will enter the values of [imath]M[/imath] and [imath]\varepsilon[/imath] and wish to find the value of [imath]E[/imath]. The website Wolfram Alpha could find a solution for this input [imath]30(\frac{\pi}{180})=x-0.3 \sin(x)[/imath] How can I find [imath]E[/imath]? EDIT: I would like to propose this algorithm (Javascript) to solve the equation. It might require some adjustments depending on the programming language used. I did some basic tests with it, and would like feedback on it. The following algorithm doesn't have an user input method and is treating the expression: [imath]0 = -x + \sin(-x)[/imath] which is related to: [imath]x = \sin(-x)[/imath] I used this algorithm to solve [imath]E[/imath] in Kepler's Equation rewriting it as (And swapping M and e for numerical constants, that in my case will be user inputs): [imath]0 = -M + E - e*sin(E)[/imath] <script> var start = Number.MIN_SAFE_INTEGER; end = Number.MAX_VALUE; var result = null; while(result === null) { var resultStart = Math.abs(-start + Math.sin(-start)); var resultEnd = Math.abs(-end + Math.sin(-end)); if(resultStart === 0) { result = start; } else if (resultEnd === 0) { result = end; } else { if (resultStart > resultEnd) { var startOld = start; start = (start+end)/2; if(startOld === start) { // underflow result = start; } } else { var endOld = end; end = (start+end)/2; if(endOld === end) { // underflow result = end; } } } } console.log(result); </script> |
2908811 | Let [imath]f[/imath] be a nonnegative continuous function on [imath][a,b][/imath], then [imath] \lim\limits_{n\to \infty} \left[\int^{b}_{a}f^{n}(x)dx\right]^{\frac{1}{n}}=M[/imath]
Let [imath]f[/imath] be a nonnegative continuous function on [imath][a,b].[/imath] Let [imath]M=\max \{f(x):x\in[a,b]\}[/imath]. Show that \begin{align} \lim\limits_{n\to \infty} \left[\int^{b}_{a}f^{n}(x)dx\right]^{\frac{1}{n}}=M\end{align} MY TRIAL My idea is to estimate and arrive at [imath]\epsilon.[/imath] \begin{align} \left|\left(\int^{b}_{a}f^{n}(x)dx\right)^{\frac{1}{n}}-M\right|&\leq \left|\int^{b}_{a}f(x)dx-M\right| \\&= \left|\int^{b}_{a}f(x)dx-\int^{b}_{a}\frac{M}{(a-b)}dx\right| \\&= \int^{b}_{a}\left|f(x)-\frac{1}{a-b}M\right|dx\end{align} I'm stuck here but I sense that the above idea of mine, sounds quite unscholarly of me but I could do nothing more? Please, are there any better explanations out there? Thanks! | 2073959 | Prove that the sequence [imath]\{v_n\}[/imath] converges to the maximum value of [imath]f[/imath] on [imath][a,b][/imath]
Question : Assume that [imath]f[/imath] is a continuous not negative function on any point defined on the interval [imath][a,b][/imath]. For each natural number [imath]n[/imath] , Assume that [imath]v_n[/imath] is the [imath]n[/imath]'th root of [imath]\int_a^b f^n[/imath]. Prove that the sequence [imath]\{v_n\}[/imath] converges to the maximum value of [imath]f[/imath] on [imath][a,b][/imath]. Note : The problem is how to connect these things in a formal way. I know that many important things happen on the roots. ( Specially maxima and minima ) But I don't know how to relate them to convergence of a sequence like [imath]\{v_n\}[/imath]. Please, If you have the time, explain your answer a bit more. I'm new to integration. Thanks in advance. |
2909067 | Solving Complex Equation for both sides
I'm having a little trouble with this problem - Find all z such that [imath]z^7=(z+1)^7[/imath]. I've moved [imath](z+1)[/imath] to the other side to obtain [imath](\frac {z}{z+1})^7=1,[/imath] but I'm not sure how to carry on. I've tried converting to complex exponential form, but how would I simplify and solve? Thanks! | 2905854 | How to solve the equation [imath](z+1)^7 = z^7[/imath] for all [imath]z[/imath]?
Find all [imath]z[/imath] for the equation [imath](z+1)^7 = z^7[/imath] The different solutions can be unsimplified and both rectangular or exponential. I have the lead that I subsitute the [imath]z+1[/imath] term with a root of unity. Then, I got lost when I looked at that [imath]z^7[/imath]. A thorough explanation would be great! |
2900263 | For a complex function [imath]f(z)[/imath] which continuous in [imath]\overline{D}[/imath] and holomorphic in [imath]D[/imath], can the zeros of [imath]f[/imath] on [imath]\partial D[/imath] have limit point?
As title goes, one knows that for a holomorphism [imath]f(z)[/imath], zeros of [imath]f(z)[/imath] cannot have limit point unless [imath]f(z)[/imath] vanishes everywhere. But I am confused at the behaviors of holomorphism at boundary. My question is Let [imath]\mathbb{D}=\{z\in \mathbb{C}: |z|<1\}[/imath] to be the unit disc. If [imath]f: \mathbb{D}\to \mathbb{C}[/imath] is continuous in [imath]\overline{\mathbb{D}}[/imath] and holomorphic inside [imath]\mathbb{D}[/imath]. Can the zeros of [imath]f[/imath] in [imath]\partial \mathbb{Z}[/imath], [imath]Z_0=\{z\in \partial\mathbb{D}: f(z)=0\}[/imath], have limit point for nonzero [imath]f[/imath]? Here are some remark on it. Firstly, if [imath]Z_0[/imath] is the complete unit circle, then by maximal module principle, [imath]f[/imath] equals to zero inside [imath]\mathbb{D}[/imath]. And, if [imath]Z_0[/imath] consists a piece of arc of unit circle, then [imath]f[/imath] vanishes everywhere. Since one can take [imath]F=\prod_n f(\mathrm{e}^{i\theta_n}z)[/imath] for some suitable [imath]\theta_n[/imath] such that [imath]F[/imath] vanishes all over the unit circle. Note that given a boundary value of [imath]f(z)[/imath], then by Cauchy's formula, [imath]f(z)[/imath] is given by [imath]f(z)=\frac{1}{2\pi i}\int_{C} \frac{f(\zeta)\textrm{d}\zeta}{\zeta-z}[/imath] Naive thought is to give a continuous function at boundary whose zeros have limit point. But there are two problems, firstly, the integral may fails to be continuous at boundary; secondly, the function should at least satisfy [imath]\int_C f(\zeta)\textrm{d}\zeta=0[/imath], so not all continuous functions can derive a holomorphism from this method. Poisson's integral determines [imath]f(z)[/imath] by the real part of boundary value of [imath]f(z)[/imath], which may not help in this problem. Even more, I am confused that What is the condition of a function on unit circle being the boundary value of some holomorphism? Of course, it is essentially a PDE problem. | 560784 | Holomorphic function having finitely many zeros in the open unit disc
Suppose [imath]f[/imath] is continuous on the closed unit disc [imath]\overline{\mathbb{D}}[/imath] and is holomorphic on the open unit disc [imath]\mathbb{D}[/imath]. Must [imath]f[/imath] have finitely many zeros in [imath]\mathbb{D}[/imath]? I know that this is true if [imath]f[/imath] is holomorphic in [imath]\overline{\mathbb{D}}[/imath] (by compactness of the closed unit disc), but I'm not sure of what happens when I just consider [imath]\mathbb{D}[/imath]. |
2760757 | Show that the series converges to 1
I want to show the following: [imath] \sum_{r=1}^\infty \frac{1}{r(r+1)} = 1. [/imath] I've found this series as part of a calculation to prove a formula for the Gamma function. I know it converges to 1 because of the result of this calculation, but otherwise I wouldn't even know how to find this value, so any help is appreciated with how to find out the series converges to 1 and/or how to prove it converges to 1 (in case these steps are done separately). | 1027110 | Infinite Series [imath]\sum 1/(n(n+1))[/imath]
I am confused on the following series: [imath]\sum\limits_{n=1}^{\infty}\frac{1}{n(n+1)} = 1[/imath] My calculator reveals that the answer found when evaluating this series is 1. However, I am not sure how it arrives at this conclusion. I understand that partial fractions will be used creating the following equation. I just don't understand how to proceed with the problem. [imath]\sum\limits_{n=1}^{\infty}(\frac{1}{n}-\frac{1}{n+1}) = 1[/imath] |
1372092 | Sum of an infinite series [imath](1 - \frac 12) + (\frac 12 - \frac 13) + \cdots[/imath] - not geometric series?
I'm a bit confused as to this problem: Consider the infinite series: [imath]\left(1 - \frac 12\right) + \left(\frac 12 - \frac 13\right) + \left(\frac 13 - \frac 14\right) \cdots[/imath] a) Find the sum [imath]S_n[/imath] of the first [imath]n[/imath] terms. b) Find the sum of this infinite series. I can't get past part a) - or rather I should say I'm not sure how to do it anymore. Because the problem asks for a sum of the infinite series, I'm assuming the series must be Geometric, so I tried to find the common ratio based on the formula [imath]r = S_n / S_{n-1} = \frac{\frac 12 - \frac 13}{1 - \frac 12} = \frac 13[/imath] Which is fine, but when I check the ratio of the 3rd and the 2nd terms: [imath]r = \frac{\frac 13 - \frac 14}{\frac 12 - \frac 13} =\frac 12[/imath] So the ratio isn't constant... I tried finding a common difference instead, but the difference between 2 consecutive terms wasn't constant either. I feel like I must be doing something wrong or otherwise missing something, because looking at the problem, I notice that the terms given have a pattern: [imath]\left(1 - \frac 12\right) + \left(\frac 12 - \frac 13\right) + \left(\frac 13 - \frac 14\right) + \cdots + \left(\frac 1n - \frac{1}{n+1}\right)[/imath] Which is reminiscent of the textbook's proof of the equation that yields the sum of the first [imath]n[/imath] terms of a geometric sequence, where every term besides [imath]a_1[/imath] and [imath]a_n[/imath] cancel out and yield [imath]a_1 \times \frac{1 - r^n}{1 - r}.[/imath] But I don't know really know how to proceed at this point, since I can't find a common ratio or difference. | 2909507 | Show that the series converges and find its sum
Show that [imath] \sum_{n=1}^\infty \left( \frac{1}{n(n+1)} \right) = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+ \;... [/imath] converges and find its sum. My solution so far: I am thinking about finding the partial sum first and show that the series converges since its finite partial sum converges. Now [imath] S_N=\sum_{n=1}^N \left( \frac{1}{n(n+1)} \right)=\sum_{n=1}^N \left( \frac{1}{n}-\frac{1}{n+1} \right)=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\;...= \left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right) + \;...+ \left( \frac{1}{N}-\frac{1}{N+1} \right)[/imath] but I don't know how to go on with this. Now [imath] \lim_{N \to\infty} \left( \frac{1}{N}-\frac{1}{N+1} \right)=0[/imath] but the right answer should be [imath]1[/imath]. |
2909692 | A rabbit on an integer
Let [imath]a[/imath] and [imath]b[/imath] be two positve integers. An invisible rabbit is standing on the number line at the number [imath]a[/imath]. Every step we tell an integer to a magician and he tells us if the rabbit is there. After every step, the rabbit jumps to the number [imath]c+b[/imath], where [imath]c[/imath] denotes the number where the rabbit was. Find a strategy to find the rabbit with finite many steps (i.e. whatever numbers [imath]a,b[/imath] are, we need finite many steps)! Please help! Thanks! This is from a Hungarian book: “Strategy games with numbers” | 1085980 | Puzzle: Give an algorithm for finding a frog that jumps along the number line
You are playing a game, your goal in this game is to catch a frog that's leaping between natural numbers. At first, the frog is found at the number [imath]a \in \mathbb N[/imath] which is not known to you. Each turn, you take a guess at where the frog is found. If you are right - you win. If you are wrong - the frog leaps [imath]b \in \mathbb N[/imath] numbers to the right. Meaning, if you got the first guess wrong, the frog is now at [imath]a+b[/imath]. If you get the guess wrong again, it's now at [imath]a+2b[/imath]. Neither [imath]a[/imath] or [imath]b[/imath] are known to you. All you know is that they are natural numbers. Propose an algorithm that will find the frog in a finite number of steps, regardless of what [imath]a[/imath] and [imath]b[/imath] are. Additional challenge: Same question, but now [imath]a,b \in \mathbb Z[/imath]. |
2870314 | Finding a set of continuous functions with a certain property
I need help finding the set of continuous functions [imath]f : \Bbb R \to \Bbb R[/imath] such that for all [imath]x \in \Bbb R[/imath], the following integral converges: [imath]\int_0^1 \frac {f(x+t) - f(x)} {t^2} \ \mathrm dt[/imath] I am thinking it could be the set of constant functions but i havent been able to prove it :( I have also noticed that you can kind of take any two functions and stick them together (continuously extend one into the other) the resulting function verifies the property in question. I hope you can provide some insight and thank you . | 2898983 | Finding a set of continuous functions with a certain property 2
I need help finding the set of continuous functions [imath]f : \Bbb R \to \Bbb R[/imath] such that for all [imath]x \in \Bbb R[/imath], the following integral converges: [imath]\int_0^1 \frac {f(x+t) - f(x)} {t^2} \ \mathrm dt[/imath] I think it might be the set of constant functions but i havent been able to prove it :( I was thinking that you can use the stone weiestrass theorem considering the set of continuous functions on a closed interval(non trivial) ,and a subset which contains the set of continuous functions whose integral above diverges in some point in that interval along with with the set of constant functions. So in order to solve the problem i need only to prove that if two functions do not meet the condition of the problem then their product does not as well . I hope you can provide some insight and thank you . |
2909383 | A field with [imath]2=0[/imath]
Is there any field other than [imath]\mathbb{F}_2[/imath] which has [imath]1+1=0[/imath] ? My initial feeling was no but I don't know how to prove it. Can someone help ? | 438107 | Is there any field of characteristic two that is not [imath]\mathbb{Z}/2\mathbb{Z}[/imath]
Is there any field of characteristic two that is not [imath]\mathbb{Z}/2\mathbb{Z}[/imath]? That is, if a field is of characteristic 2, then does this field have to be [imath]\{0,1\}[/imath]? |
2909951 | How to write [imath]4-3i[/imath] in the form [imath]e^z[/imath] with [imath]z=x+yi[/imath]
I need to write [imath]4-3i[/imath] in the form of [imath]e^z[/imath] with [imath]z=x+yi[/imath]. Important is that it shouldn't be like [imath]z=\ln(\cdot)[/imath]. The exercise demands this. The problem is that I do not know how to evade the [imath]\ln()[/imath] form. I've only managed to do this: [imath]|4-3i|= 5[/imath]. [imath]\text{Arg}(4-3i) = \tan^{-1}(-3/4)[/imath]. So [imath]e^x=5 \hspace{10mm} e^y = \tan^{-1}(-3/4)[/imath] [imath]x=\ln(5) \hspace{10mm} y=\ln(-3/4)[/imath] Anyone have any idea how to evade the [imath]\ln(\cdot)[/imath]? | 627920 | Express [imath]-1+i[/imath] in exponential form.
Express [imath]-1+i[/imath] in exponential form. My attempt so far Let [imath]z=-1+i[/imath] [imath]r=|z|=\sqrt2[/imath] [imath]\theta=\tan^{-1}(-1)=-\frac{\pi}{4}[/imath] Now, this is where I go wrong (I don't know why it's wrong!): So in exponential form: [imath]-1+i=\sqrt2 e^{-i\pi/4}[/imath] According to the solutions, [imath]\theta=3\pi/4[/imath]. Thanks in advance |
2908585 | How to evaluate the following definite integral [imath]\int_0^1\frac{\arctan(ax)}{x\sqrt{1-x^2}}dx[/imath]?
[imath]\int_0^1\frac{\arctan(ax)}{x\sqrt{1-x^2}}dx[/imath] My working is as follows; Let [imath]I(a)=\int_0^1\frac{\arctan(ax)}{x\sqrt{1-x^2}}dx[/imath] and thus [imath]I'(a)=0-0+\int_0^1\frac{x}{(1+a^2x^2)(x\sqrt{1-x^2})}dx[/imath] (I did this using Leibniz-Newton's formula. I took the partial derivative of the integrand with respect to [imath]a[/imath], and since the two limits of the integral are constants, the first two terms are zero.) Now, I'm stuck... I've been trying this one for the last hour. Can someone hint me towards the right answer? (which I looked up to be [imath]\frac{\pi}{2}\sinh^{-1}a[/imath]) P.S: I am aware of this site's norm of keeping MathJax out of the titles, but I'm getting a message that a question with this title already exists... Can someone please help? | 2805010 | Integrating [imath]\frac{\arctan x}{x\sqrt{\smash[b]{1-x^2}}}[/imath]
How to evaluate [imath]\int_0^1\frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx\text{?}[/imath] The steps I can think of is integration by parts as [imath]\int_0^1\frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx=\int_0^1\frac{\arctan x}{x}\,\mathrm d(\arcsin x)[/imath] or integration by substitution using [imath]x=\sin t[/imath] as [imath]\int_0^1 \frac{\arctan x}{x\sqrt{1-x^2}}\,\mathrm dx = \int_0^{\frac{\pi}{2}}\frac{\arctan(\sin t)}{\sin t}\,\mathrm dt,[/imath] but both seems to make the problem more complicated. Thanks a lot. |
2910243 | How to prove that [imath]\sin(\sqrt{x})[/imath] is not periodic?
How to prove that [imath]\sin(\sqrt{x})[/imath] is not periodic? THe definition of a periodic function is [imath]f(x+P)=f(x)[/imath]. So I assume that [imath]\sin(\sqrt{x+P})=\sin(\sqrt{x})[/imath]. This is equivalent to [imath]\sin(\sqrt{x+P})-\sin(\sqrt{x})=0[/imath]. This implies [imath]2cos(\frac{\sqrt{x+P}+\sqrt{x}}{2})\sin(\frac{\sqrt{x+P}-\sqrt{x}}{2})[/imath]. What should I do next? | 891233 | Prove that [imath]\sin(\sqrt x)[/imath] not periodic
[imath]\sin\sqrt x[/imath] is not a periodic function. How can one prove this? |
2903406 | find the number of ordered triples [imath](x,y,z)[/imath]
find the number of ordered triples [imath](x,y,z)[/imath] satisfying [imath]x\mid yz-1[/imath] [imath]y\mid xz-1[/imath] [imath]z\mid xy-1[/imath] and [imath]0<x,y,z<2014[/imath] note: [imath]x,y,z[/imath] are integers I found [imath](2,3,5)[/imath] as one set of solution (actually [imath]6[/imath]), and I tried change [imath]x\mid yz-1\Longrightarrow xk_1=yz-1[/imath]but it didn't work. Any idea/hint of how can I make progress? Thank you in advance. Note:I DO NOT want a computer testing all combinations. | 646205 | If [imath]p,q,r[/imath] are all primes,and [imath]p|qr-1[/imath],[imath]q|pr-1[/imath] and [imath]r|pq-1[/imath],find all possible values of [imath]pqr[/imath].
If [imath]p,q,r[/imath] are all primes,and $p|qr-1[imath]~~~~~~~~~~~~~[/imath]q|pr-1$[imath]~~~~~~~~~~[/imath] and [imath]~r|pq-1[/imath]. Find all possible values of [imath]pqr[/imath]. My work: [imath]qr-1=pk_1[/imath] [imath]pr-1=qk_2[/imath] [imath]pq-1=rk_3[/imath] From the above equations, we can conclude that either [imath]k_1,k_2,k_3[/imath] are all even or one of the primes is [imath]2[/imath]. I can also get that, [imath]p^2q^2r^2=(pk_1+1)(qk_2+1)(rk_3+1)[/imath] Now, I cannot proceed. Please help! |
2910729 | Show that for positive numbers a,b,c,d, [imath]\sum_{cyc} ab \leq \frac{1}{4}\left(\sum_{cyc} a \right)^2[/imath] and ...
Let a,b,c,d be four positive real numbers. Show that [imath]\sum_{cyc} ab \leq \frac{1}{4}\left(\sum_{cyc} a \right)^2[/imath] and [imath]\sum_{cyc} abc \leq \frac{1}{16}\left(\sum_{cyc} a \right)^3[/imath] My textbook said these inequalities can be referred from the AM-GM inequality but didn't say how to derive it. I can find only looser bounds. That is, for each term in the left hand side of the first inequality, by AM-GM inequality, [imath]ab \leq \left(\frac{a+b}{2}\right)^2 \leq \frac{1}{4} \left(a+b+c+d\right)^2[/imath] [imath]\Longrightarrow \sum_{cyc} ab \leq \frac{4}{4} \left(\sum_{cyc} a \right)^2[/imath] Using the same techniques, we get another loose bound for [imath]\sum_{cyc} abc[/imath]. How could I get the bounds in the original inequalities? Please suggest. PS. It seems like if these inequalities are true, we may get a general relationship for n positive variables as [imath]\sum_{cyc} a_1 ... a_k \leq \frac{1}{n^{k-1}}\left(\sum_{cyc} a_i \right)^k[/imath]. PS2. Sorry for double posting. There was an old thread here: Showing [imath]\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2[/imath] One answer points to Maclaurin's inequality, but I still cannot figure out how to prove it using AM-GM. | 2009517 | Showing [imath]\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2[/imath]
[imath]a,b,c,d>0[/imath] with [imath]a+b+c+d=4[/imath] then [imath]\sum\limits_{cyc}ab\le\frac14\left(\sum\limits_{cyc}a\right)^2[/imath] How can I apply AM-GM here, is there a generalization, say we have [imath]n[/imath] variables summing up to [imath]n[/imath] then, [imath]\sum\limits_{cyc}a_1\cdot a_2\dots a_k\le\frac1{n^{k-1}}\left(\sum\limits_{cyc}a\right)^k[/imath] because the following inequality should be also true [imath]\sum\limits_{cyc}abc\le\frac1{16}\left(\sum\limits_{cyc}a\right)^3[/imath] |
2911166 | How to solve the following problem - A ∩ B ⊂ A ⊂ A ∪ B?
Prove that [imath](A \cap B) \subset A \subset (A \cup B).[/imath] I understand [imath]\cap[/imath], [imath]\subset [/imath], [imath]\cup[/imath] concepts, but i wonder how to solve this problem. | 1023898 | Show that [imath]A\cap B\subseteq A[/imath] and [imath]A\subseteq A\cup B[/imath]
[imath]A \cap B \subseteq A[/imath] My first step would be to write it as [imath](x \in A \land x \in B) \subseteq A[/imath]. Then I know by the following implication that is always true [imath]P \land Q \implies P[/imath]. But I am not sure how to write it down mathematically correct. [imath]A \subseteq A \cup B [/imath] I would write it as [imath]A \subseteq (x \in A \lor x \in B)[/imath] . Then I know by following implication that is always true [imath]P \implies P \lor Q[/imath]. But again I do not know how to write in down mathematically correct. May you could help a beginner? |
2911355 | Why has [imath]\pi[/imath] to be surjective in the definition of the quotient topology?
The quotient topology is defined as follows: Let [imath]X[/imath] be a topological space and [imath]\pi: X\to Y[/imath] a surjective function. We call the by [imath]\pi[/imath] induced topology [imath]\tau_{Y,\pi}[/imath] quotient topology where [imath]U\subseteq Y[/imath] open iff [imath]\pi^{-1}(U)\subseteq X[/imath] open. It is easy to see, that this defines indeed a topology: We have [imath]\emptyset, Y\in\tau_{Y,\pi}[/imath]: [imath]\emptyset, Y[/imath] are open, since [imath]\pi^{-1}(\emptyset)=\emptyset[/imath] open and [imath]\pi^{-1}(Y)=X[/imath] open. For [imath]U, V\subseteq Y[/imath] open, we have [imath]U\cap V[/imath] open: [imath]\pi^{-1}(U\cap V)=\underbrace{\pi^{-1}(U)}_{\text{open}}\cap\underbrace{\pi(V)}_{\text{open}}[/imath] open. Let [imath]I[/imath] be an index set and [imath]i\in I[/imath]. Let [imath]U_i\subseteq Y[/imath] open for every [imath]i\in I[/imath]. Then [imath]\bigcup_{i\in I} U_i[/imath] open: [imath]\pi^{-1}(\bigcup_{i\in I} U_i)=\bigcup_{i\in I}\underbrace{\pi^{-1}(U_i)}_{\text{open}}[/imath] open. Hence [imath]\tau_{Y,\pi}[/imath] is indeed a topology. But why has [imath]\pi[/imath] to be a surjective function? One does not need this property to show, that the condition above defines a topology. Is the proof correct? What do we need the condition, that [imath]\pi[/imath] is surjective for? Is it just so we have the equivalent relation [imath]x\sim y\Leftrightarrow f(x)\sim f(y)[/imath]. And then we take [imath]Y=X/\sim[/imath]. But there is no such relation defined. I know that quotient maps have to be surjective in general. But what would happen if we had just a function [imath]f: X\to Y[/imath] and define a topology in the same way? Thanks in advance. | 2711112 | Why do we require quotient to be surjective?
Definition. Let [imath]X[/imath] and [imath]Y[/imath] be topological spaces; let [imath]p : X \to Y[/imath] a surjective map. The map [imath]p[/imath] is said to be a quotient map provided a subset [imath]U[/imath] of [imath]Y[/imath] is open in [imath]Y[/imath] if and only if [imath]p^{-1}(U)[/imath] is open in [imath]X[/imath]. Definition. If [imath]X[/imath] is a space and [imath]A[/imath] is a set and if [imath]p : X\to A[/imath] is a surjective map, then there exists exactly one topology [imath]\tau[/imath] on [imath]A[/imath] relative to which [imath]p[/imath] is a quotient map; it is called the quotient topology induced by [imath]p[/imath]. Definition. Let X be a topological space, and let [imath]X^*[/imath] be a partition of [imath]X[/imath] into disjoint subsets whose union is [imath]X[/imath]. Let [imath]p : X \to X^*[/imath] be the surjective map that carries each point of [imath]X[/imath] to the element of [imath]X^*[/imath] containing it. In the quotient topology induced by [imath]p[/imath], the space [imath]X^ *[/imath] is called a quotient space of [imath]X[/imath] It seems that to induce a quotient topology by using [imath]p[/imath], we don't need the property of [imath]p[/imath] being surjective. Why do we require a quotient map to be surjective? Are there some important applications of the surjection property? To prove a function is a quotient map, do we need to explicitly prove the function is surjective ? Does the following picture give some intuition? It's from Wikipedia. The next stage of this gif is the [imath]S^2[/imath] obtained by gluing the boundary (in blue) of the disk [imath]D^2[/imath] together to a single point. |
2911183 | Residue of f(z) with pole of 2nd order
Generally finding a Residue of a function with [imath]n^{th}[/imath] order pole is done with \begin{equation}\label{eq:1} Res(f(z),z_0) = \dfrac{1}{(n-1)!} \lim_{z \to z_0} \dfrac{d^{n-1}}{dz^{n-1}} (z-z_0)^n f(z) \end{equation} For poles of [imath]1^{st}[/imath] order it's simply [imath] Res(f(z),z_0) =\lim_{z \to z_0} (z-z_0) f(z) [/imath]. This can be written as [imath] Res(f(z),z_0) = \dfrac{g(z_0)}{h'(z_0)}[/imath] if the original function was in the shape [imath]\dfrac{g(z)}{h(z)}[/imath] Now. I am looking for such an expression of [imath]2^{nd}[/imath] order poles. For the [imath]1^{st}[/imath] order case because [imath]h(z)[/imath] has zeros of 1st order so [imath]h(z)[/imath] can be written as [imath]h(z)=h(z)-h(z_0)[/imath] and [imath]h(z_0)=0[/imath].So it all falls on the same place and gives you the simplified equation for residue. But for [imath]2^{nd}[/imath] order pole there is much complication, i.e there is additional derivative term and I can't simplify. | 552492 | Calculating residue of pole of order [imath]2[/imath]
Is there a good way to compute the residue of [imath]f(z)=\dfrac{1+z}{1-\sin z}[/imath] at [imath]z=\pi/2[/imath], which is a pole of order [imath]2[/imath]? Using the residue calculation formula yields [imath]\text{Res}_{z=\pi/2}f(z)=\lim_{z\rightarrow\pi/2}\dfrac{d}{dz}\left(\left(z-\dfrac\pi2\right)^2f(z)\right)[/imath] The derivative is quite ugly, and calculating the limit requires L'Hospital probably twice (or more). The calculation is just too much. Is there a better way? |
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