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839525
|
How would I parametrise a straight line?
If I want to parameterise a straight line and I have the equation, eg [imath]y=2x+1[/imath] and I also have two co-ordinates it passes through, would it ok to use the co-ordinates to parameterise in terms of [imath]t[/imath]?
|
1307560
|
How to parameterize a straight line?
Why does the straight line from [imath](x_1,+y_1,+z_1)[/imath] to [imath](x_2,+y_2,+z_2)[/imath] become [imath]r(\vec t)=(1-t)(x_1,+y_1,+z_1)+t(x_2,+y_2,+z_2)[/imath] for [imath]0 \leq t \leq 1[/imath]?
|
2080991
|
Question about normal subgroup that contains Kernel of a homomorphism
Let [imath]G[/imath] be a group, [imath]H[/imath] be an abelian group, [imath]\varphi:G \to H[/imath] be a homomorphism onto [imath]H[/imath]. Let [imath]N[/imath] be a subgroup of [imath]G[/imath] that contains [imath]\operatorname{ker}(\varphi)[/imath]. Show that [imath]N[/imath] is a normal subgroup of [imath]G[/imath]. I don't understand the word "contains", if it says [imath]N=\operatorname{ker}(\varphi)[/imath] then I write: Let [imath]n \in N[/imath], and [imath]g \in G[/imath] be arbitrary, then [imath]\varphi(gng^{-1})=\varphi(g)\varphi(n)\varphi(g)^-1=\varphi(g)\varphi(g)^-1=e[/imath], so [imath]N[/imath] is normal. But in this question, there can be a [imath]y \in N[/imath] and also [imath]y[/imath] is not an element of [imath]\operatorname{ker}(\varphi)[/imath]. Or should I prove first that [imath]N=\operatorname{ker}(\varphi)[/imath]? Thanks for any help in advance.
|
542264
|
Subgroups containing kernel of group morphism to an abelian group are normal.
Let [imath]\varphi:G\rightarrow H[/imath] be a group homomorphism from group [imath]G[/imath] to group [imath]H[/imath]. Show that, if [imath]H[/imath] is abelian, all subgroups of [imath]G[/imath] that contain [imath]\mathrm{ker} (\varphi)[/imath] are normal in [imath]G[/imath].
|
2070941
|
[imath]N[/imath]th root - real meaning when [imath]N[/imath] is not a decimal
Consider the following expression [imath]\sqrt[4]{1296}=6[/imath] To find the 4th root of [imath]1296[/imath], first we write [imath]1296[/imath] as product of prime factors [imath]1296=3^4 \times 2^4[/imath] Now, [imath]\sqrt[4]{1296}=\sqrt[4]{3^4 \times 2^4}=3\times 2=6[/imath] But, I find it confusing when we have a decimal, say find [imath]\sqrt[4.5]{1296}[/imath] What does it really mean? How do we calculate it? I assume there is some real meaning for this because all the calculators calculates these. Note: I know [imath]\sqrt[4.5]{1296}=4.91688[/imath] and [imath]4.91688^{4.5}=1296[/imath]. This is not what I am asking. Actually, [imath]4.5[/imath]th root or [imath]4.5[/imath]th power, all are confusing statements to me and I was trying to understand if this has a real meaning.
|
1790032
|
Exponent of a number is a square root?
Say you have a number [imath]x^{\sqrt 2}[/imath]. Is there any way to represent this number so that there's no root (or irrational) as the exponent (so that it's easier to understand for me)? I just can't wrap my head around this. I was thinking something like [imath]x^{2^{1/2}}[/imath] and extending on that idea, but I don't know how. Also, what about for [imath]x^{\sqrt 3}[/imath]? Or [imath]x^{\sqrt n}[/imath]?
|
2080702
|
factoring [imath]n=pq[/imath] knowing [imath]p+q[/imath] in polynomial time
Lets [imath]n=pq[/imath] with [imath]p[/imath] and [imath]q[/imath] primes If we know the result of [imath]p+q[/imath], can we factorize [imath]n[/imath] in polynomial time ? Best regards.
|
910898
|
Algebraic process to find numbers so that [imath]xy=45[/imath] and [imath]x+y=18[/imath]
Can someone help me with the following question? The sum of two numbers is [imath]18[/imath] and their product is [imath]45[/imath]. Find the numbers. I know that the answer is [imath]15[/imath] and [imath]3[/imath]. But how do I find that answer algebraically? It tried doing it as [imath]x+y=18[/imath] and [imath]xy=45[/imath] and then [imath]x= 18-y[/imath] and I substituted in [imath]xy=45[/imath]. I got it as [imath](18-y)y =45[/imath], and I don't know what to do next.
|
2080950
|
If [imath]\lim_{x \rightarrow a^+}f'(x)[/imath] exists. Then [imath]\lim_{x \rightarrow a^+}f'(x) = f'(a)[/imath]
If a real valued function [imath]f[/imath] is differentiable on a neighborhood of a point [imath]'a'[/imath] and If [imath] \lim_{x \rightarrow a^+}f'(x)[/imath] exists. Then [imath]\lim_{x \rightarrow a^+}f'(x) = f'(a)[/imath] I have written what I thought of. Can that be considered as a proof? I am looking for mistakes (if any) and alternative proofs. (may be mean value theorem can be used but I am not sure about that) My attempt : Let [imath]f[/imath] be differentiable on [imath]I = (a - \delta_0, a + \delta_0)[/imath]. Then for [imath]h \in(0, \delta_0/2)[/imath] [imath] \lim_{h \rightarrow 0^+}\frac{f(x+h) - f(x)}{h} = f'(x) \hspace{4mm} \forall x \in(a - \delta_0/2, a + \delta_0/2)[/imath] hence [imath] \lim_{x \rightarrow a^+} \bigg [\lim_{h \rightarrow 0^+}\frac{f(x+h) - f(x)}{h}\bigg ] = \lim_{x \rightarrow a^+}f'(x) \hspace{4mm} \forall x \in(a - \delta_0/2, a + \delta_0/2)[/imath] Since the [imath]f'(x)[/imath] exist in [imath]I[/imath] we can interchange the limit to get [imath]\implies \lim_{h \rightarrow 0^+}\lim_{x \rightarrow a^+}\frac{f(x+h) - f(x)}{h} = \lim_{x \rightarrow a^+}f'(x) \hspace{4mm} \forall x \in(a - \delta_0/2, a + \delta_0/2)[/imath] [imath] \implies \lim_{h \rightarrow 0^+}\frac{f(a+h) - f(a)}{h} = \lim_{x \rightarrow a^+}f'(x) \hspace{4mm} \forall x \in(a - \delta_0/2, a + \delta_0/2)[/imath] [imath] \implies f'(a) =\lim_{x \rightarrow a^+}f'(x)[/imath] Now I have got an answer to this but why in this attempt why the limits cannot be interchanged as the [imath]f'(x)[/imath] lies inside the I.
|
1833147
|
Is [imath]\lim\limits_{x\to x_0}f'(x)=f'(x_0)[/imath]?
Let [imath]f[/imath] be a function defined in the open interval [imath](a,b)[/imath] and let [imath]x_0\in(a,b)[/imath]. Suppose in addition that [imath]f'(x)[/imath] exists for all [imath]x_0\neq x\in(a,b)[/imath]. Is the following statement true: If [imath]\lim\limits_{x\to x_0}f'(x)[/imath] exists, then [imath]f'(x_0)[/imath] exists and [imath]\lim\limits_{x\to x_0}f'(x)=f'(x_0)[/imath]. Thanks!
|
1484956
|
Integral [imath]\int \sqrt{x+\sqrt{x^2+2}}dx[/imath]
[imath]\int \sqrt{x+\sqrt{x^2+2}}\ dx[/imath] I tried various solving methods but I am not coming forward. I unformed the term also to [imath]{x+\sqrt{x^2+2} \over \sqrt{x+\sqrt{x^2+2}}}[/imath] and even multiplied with [imath]{\sqrt{x-\sqrt{x^2+2}} \over \sqrt{x-\sqrt{x^2+2}}}[/imath]. Trigonometric and hyperbolic substitution didn't help either.
|
1433771
|
Integrate [imath]\int\sqrt{x+\sqrt{x^{2}+2}} dx[/imath] .
Q) [imath]\int\sqrt{x+\sqrt{x^{2}+2}} dx[/imath] . Tried rationalising the numerator twice to get Numerator =-2 but not able to simplify denominator The question reduces to (as per my rationalising) [imath]\int \frac{-2}{\bigl(x - \sqrt{x^2+2}\bigr)\sqrt{x + \sqrt{x^2+2}}}\,dx[/imath]
|
398868
|
Question based on Triangle Inequality [imath]\displaystyle |x+y|\leq |x|+|y|[/imath]
If [imath]x,y,z\in \mathbb{R}-\left\{0\right\}[/imath]. Then prove that [imath]\displaystyle 1\leq \frac{|x+y|}{|x|+|y|}+\frac{| y+z|}{| y |+| z |}+\frac{| z+x|}{| z |+| x |}\leq 3[/imath] My Try:: Using Triangle Inequality [imath]\displaystyle | x+y | \leq | x |+| y |\Rightarrow \frac{| x+y|}{| x |+| y |}\le 1[/imath] Similarly [imath]\displaystyle | y+z | \leq | y |+| z |\Rightarrow \frac{| y+z|}{| y |+| z |}\le 1[/imath] Similarly [imath]\displaystyle | z+x | \leq | z |+| x | \Rightarrow \frac{| z+x|}{| z |+| x |}\le 1[/imath] Now Add all three equations:: [imath]\displaystyle \frac{| x+y|}{| x |+| y |}+\frac{| y+z|}{| y |+| z |}+\frac{| z+x|}{| z |+| x |}\leq 3[/imath] But I did not Understand How can I calculate Lower value of the Given expression.
|
1774162
|
Range of function Involving Modulus Quantity.
If [imath]x,y,z\in \mathbb{R}\;,[/imath] Then Range of [imath]\frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}\,[/imath] [imath]\bf{My\; Try::}[/imath] Here [imath]x,y,z[/imath] Not all Zero Simultaneously. Now Using [imath]\bf{\triangle\; Inequality}\;,[/imath] We get [imath]|x+y|\leq |x|+|y|[/imath] Similarly [imath]|y+z|\leq |y|+|z|[/imath] and [imath]|z+x|\leq |z|+|x|[/imath] So we get [imath]\frac{|x+y|}{|x|+|y|}+\frac{|y+z|}{|y|+|z|}+\frac{|z+x|}{|z|+|x|}\leq 3[/imath] Now I did not understand How can I calculate [imath]\min[/imath] of that expression, Help Required Thanks
|
2081747
|
Cardinality and Rank Nullity Theorem
Let [imath]V[/imath] be a vector space and [imath]f[/imath] a nonzero linear functional on [imath]V[/imath]. Then [imath]\text{dim image}(f)=1[/imath] and [imath]\text{dim null}(f)+\text{dim image}(f)=\text{dim}V[/imath]. And if [imath]\text{dim}V<\infty[/imath], then [imath]\text{dim null}(f)=\text{dim}V-1[/imath]. I'm confused as to why [imath]V[/imath] needs to be finite dimensional in order to conclude [imath]\text{dim null}(f)=\text{dim}V-1[/imath]. Can't we just say that if [imath]\text{dim}V\not<\infty[/imath], we have [imath]\text{dim}V-1=\text{dim}V[/imath]?
|
752056
|
Does the rank-nullity theorem hold for infinite dimensional [imath]V[/imath]?
The rank nullity theorem states that for vector spaces [imath]V[/imath] and [imath]W[/imath] with [imath]V[/imath] finite dimensional, and [imath]T: V \to W[/imath] a linear map, [imath]\dim V = \dim \ker T + \dim \operatorname{im} T.[/imath] Does this hold for infinite dimensional [imath]V[/imath]? According to this, the statement is false. But according to this, page 4, the statement is still true. I'm thoroughly confused.
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1084411
|
If [imath]f: \Bbb{R}\rightarrow\Bbb{R}[/imath] is continuous at [imath]0[/imath] and [imath]f(x)=f(2x)[/imath] for each [imath]x\in\Bbb{R}[/imath] then [imath]f[/imath] is constant.
How do I prove that if [imath]f: \Bbb{R}\rightarrow\Bbb{R}[/imath] is continuous at [imath]0[/imath] and [imath]f(x)=f(2x)[/imath] for each [imath]x\in\Bbb{R}[/imath], then [imath]f[/imath] is constant?
|
1126662
|
Find a function that matches the following conditions.
Find a function that matches the following conditions. (a) [imath]f(x)[/imath] is continuous for all real numbers (b) [imath]f(0)[/imath] = 3 (c) For all real numbers [imath]x[/imath], [imath]f(x) = f(x/2)[/imath] This is from a past paper, and the answer has been given as the following steps: [imath]f(a) = ... = f(a/2n)[/imath] [imath]f(a)[/imath] = lim as n approaches infintiy [imath]f(a)[/imath] = lim as n approaches infinity of [imath]f(a/2n)[/imath] [imath]f[/imath](lim as n approaches infinity of [imath]\frac{a}{2n}[/imath] therefore [imath]f[/imath] is continuous [imath]f(0) = 3[/imath] My question arises due to poor printing of my answer book. In step 1 is [imath]f(a/2n)[/imath] the correct step or should it be [imath](f(\frac{a}{2^n})[/imath]? If so could you please explain why? Because for both possibilities I can't seem to figure out why that's a step in proving the continuity. I'm sorry for the very elementary question. Thank you!
|
2081895
|
Real projective space [imath]\mathbb{R}P^1[/imath] is diffeomorphic to [imath]S^1[/imath]?
I am reading Lee, Manifolds and Differential Geometry, and I am a bit confused. It is seen that the real projective space [imath]\mathbb{R}P^n[/imath] is homeomorphic to [imath]S^n/ \sim[/imath], where [imath]p \sim q[/imath] iff [imath]p = \pm q[/imath]. However, in one of the exercises (1.48) one is asked to prove that [imath]\mathbb{R}P^1[/imath] is diffeomorphic to [imath]S^1[/imath]. Is this really true (is there a simple proof of this) and does this imply something for the relation between [imath]S^1[/imath] and [imath]S^1/\sim[/imath]? Does this also mean that [imath]\mathbb{R}P^1[/imath] is homeomorphic to [imath]S^1[/imath] (since diffeomorphism is a stronger condition)?
|
298879
|
Showing diffeomorphism between [imath]S^1 \subset \mathbb{R}^2[/imath] and [imath]\mathbb{RP}^1[/imath]
I am trying to construct a diffeomorphism between [imath]S^1 = \{x^2 + y^2 = 1; x,y \in \mathbb{R}\}[/imath] with subspace topology and [imath]\mathbb{R P}^1 = \{[x,y]: x,y \in \mathbb{R}; x \vee y \not = 0 \}[/imath] with quotient topology and I am a little stuck. I have shown that both are smooth manifolds, and I used stereographic projection for [imath]S^1[/imath], but now I am runing into trouble when I give the homeomorphism between [imath]S^1[/imath] and [imath]\mathbb{RP}^1[/imath] as the map that takes a line in [imath]\mathbb{RP}^1[/imath] to the point in [imath]S^1[/imath] that you get when letting the parallel line go through the respective pole used in the stereographic projection. If I use the south and north poles I get a potential homeomorphism, but I cannot capture the horizontal line in my image, but when I pick say north and east then my map is not well defined as I get different lines for the same point in [imath]S^1[/imath]. Can somebody give me a hint how to make this construction work, or is it better to move to a different representation of [imath]S^1[/imath] ?
|
2080343
|
Invertibility of an operator [imath]T[/imath] if [imath]\|I-T\|<1[/imath]
Let [imath]X [/imath] be a Banach space [imath](X,\|.\|)[/imath] with [imath]\dim(X)=\infty [/imath] and [imath]T\in B(X) [/imath]. Suppose that [imath]\left\| I-T\right\| < 1[/imath]. Show that [imath]T^{-1}\in B\left( X\right)[/imath] and [imath]\left\| I-T^{-1}\right\| \leq \dfrac {\left\| I-T\right\| } {1-\left\| I-T\right\| }[/imath].
|
2798310
|
is T, near the identity, invertible?
If [imath]X[/imath] is a Banach space and [imath]T:X\to X[/imath] is a linear continuous transformation such that [imath]\|I-T\|<1[/imath] then T is invertible.
|
1998164
|
Complex Least Squares Derivation
Considering the Complex Least Squares Problem, one can write the cost function for estimating H (where Y = XH + Z) as: Reference: Mimo-OFDM Wireless Communications with MATLAB book (page 190) [imath]J(\boldsymbol H) = ||\boldsymbol Y - \boldsymbol X \boldsymbol H ||^2 = (\boldsymbol Y - \boldsymbol X \boldsymbol H)^H(\boldsymbol Y - \boldsymbol X \boldsymbol H) = \boldsymbol Y^H \boldsymbol Y - \boldsymbol Y^H \boldsymbol X \boldsymbol H - \boldsymbol H^H \boldsymbol X^H \boldsymbol Y + \boldsymbol H^H \boldsymbol X^H \boldsymbol X \boldsymbol H[/imath] Calculating the derivative of the above function with respect to H gives: [imath]\frac{\partial J( \boldsymbol H)}{\partial \boldsymbol H} = -2(\boldsymbol X^H \boldsymbol Y)^* + 2(\boldsymbol X^H \boldsymbol X \boldsymbol H)^*[/imath] How do I get to the expression above? Do I need to use some properties? I am trying to do that deriving each term separately.
|
1783397
|
Complex ([imath]\mathbb C[/imath]) least squares derivation
I know how to derive the least squares in the real domain. If a tall matrix [imath]A[/imath] and a column vector [imath]b[/imath] are real, then the solution of the least squares problem [imath]Ax = b[/imath] can be derived as: [imath]\begin{align} \{E(x)\}^2 &= ||Ax - b||^2 \\ &= (Ax-b)^T (Ax-b) \\ &= x^T A^T Ax - x^T A^T b - b^T Ax + b^T b \\ &= x^T A^T Ax - 2 x^T A^T b + b^T b \qquad (\because (Ax)^T b = b^T (Ax)) \end{align}[/imath] Differentiating both sides with respect to [imath]x[/imath], [imath]\begin{align} \frac{d \{E(x)\}^2}{dx} &= 2A^T Ax - 2 A^T b \end{align}[/imath] Setting [imath]\frac{d \{E(x)\}^2}{dx} = 0[/imath] to find when we get the minimum [imath]E(x)[/imath], [imath] 2A^T Ax - 2 A^T b = 0 \\ A^T Ax = A^T b \\ x = (A^T A)^{-1} A^T b [/imath] Now, we turn to the complex-valued situation. Assume [imath]A[/imath] and [imath]b[/imath] are complex, [imath]\begin{align} \{E(x)\}^2 &= ||Ax - b||^2 \\ &= (Ax-b)^H (Ax-b) \\ &= x^H A^H Ax - x^H A^H b - b^H Ax + b^H b \\ \end{align}[/imath] Here, I have some problems. First, [imath]x^H A^H b \neq b^H Ax[/imath] unless [imath](Ax)^H b[/imath] is real. Most of all, I don't know how to differentiate the complex matrices above. How to proceed the derivation? There are plenty of derivations in the real domain in Google, but I couldn't find detailed explanation of the general complex case.
|
2082772
|
Compute average distance between numbers [imath]0[/imath] to [imath]10[/imath]
I'd like to calculate the average difference between two numbers, each between [imath]0[/imath] and [imath]10[/imath]. I calculated this for integers and came up with an average distance of [imath]4[/imath]. My method: there are [imath]10[/imath] ways to obtain a difference of [imath]1[/imath], [imath]9[/imath] ways to obtain a difference of [imath]2[/imath], [imath]3[/imath] of [imath]8[/imath], [imath]4[/imath] of [imath]7[/imath], [imath]5[/imath] of [imath]6[/imath], [imath]6[/imath] of [imath]5[/imath], etc. I took the weighted average of all the possibilities and I ended up with [imath]\frac{220}{55} = 4[/imath]. But I actually have float values between [imath]0[/imath] and [imath]10[/imath]. If I did the integer average distance correctly, is the average distance between [imath]0[/imath] and [imath]10[/imath] continuous still [imath]4[/imath]?
|
195245
|
Average Distance Between Random Points on a Line Segment
Suppose I have a line segment of length [imath]L[/imath]. I now select two points at random along the segment. What is the expected value of the distance between the two points, and why?
|
2083106
|
How to find the convergence of this sequence? [imath]x_n=1+\frac{1}{x_{n-1}}[/imath]
The sequence ([imath]x_n[/imath]) is defined by [imath]x_0=1[/imath] and [imath]x_n=1+\frac{1}{x_{n-1}}[/imath] for [imath]n\in ℕ^*[/imath], then how could I find the limit of it? By trying some, I've found that the answer is [imath]\frac{1+\sqrt5}{2}[/imath], but I'm looking for a "legit" way to find it.
|
1633109
|
limit of the sequence [imath]a_n=1+\frac{1}{a_{n-1}}[/imath] and [imath]a_1=1[/imath]
Problem: Find with proof limit of the sequence [imath]a_n=1+\frac{1}{a_{n-1}}[/imath] with [imath]a_1=1[/imath] or show that the limit does not exist. My attempt: I have failed to determine the existence. However if the limit exists then it is easy to find it. The sequence is not monotonic and I have failed to find any monotonic subsequence subsequence. The sequence is clearly bounded below by [imath]1[/imath]. I have observed that the sequence is a continued fraction so it alternatively increases and decreases. So, please help me.
|
1210723
|
Trigonometric integration [imath] \int \sqrt{1-\sin2x}dx[/imath]
I could not figure out how to solve this problem. Can anyone give me a hint how to solve this: [imath] \int \sqrt{1-\sin2x}dx[/imath]
|
785271
|
How to integrate [imath]\sqrt{1-\sin 2x}[/imath]?
I want to solve the following integral without substitution: [imath]\int{\sqrt{1-\sin2x}} \space dx[/imath] I have: [imath]\int{\sqrt{1-\sin2x}} \space dx = \int{\sqrt{1-2\sin x\cos x}} \space dx = \int{\sqrt{\sin^2x + \cos^2x -2\sin x\cos x}} \space dx[/imath] but this can be written in two ways: [imath]\int{\sqrt{(\sin x - \cos x)^2}} \space dx[/imath] or [imath]\int{\sqrt{(\cos x - \sin x)^2} \space dx}[/imath] and it seems to be pretty far from what the real result is: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ecvsatkj0te&mail=1 Can I get any hints? EDIT: Thank you for your answers! So as you all showed, we have: [imath]\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\sin x - \cos x)^2}} \space dx[/imath] or [imath]\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\cos x - \sin x)^2} \space dx} = \int{\sqrt{(-(\sin x - \cos x))^2} \space dx}[/imath] combined: [imath]\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\pm(\sin x - \cos x))^2} \space dx}[/imath] so we actually have: [imath]\int{|\sin x - \cos x|} \space dx[/imath] I drew myself a trigonometric circle and if I concluded correctly, we have: 1. [imath]\int{\sin x - \cos x} \space dx[/imath] for [imath]x \in [ \frac{\pi}{4} + 2k\pi , \frac{5\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}[/imath] 2. [imath]\int{\cos x - \sin x} \space dx[/imath] for [imath]x \in [ -\frac{3\pi}{4} + 2k\pi , \frac{\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}[/imath] ... which means: 1. [imath]x \in [ \frac{\pi}{4} + 2k\pi , \frac{5\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}[/imath] : [imath]\int{\sqrt{1-\sin2x}} \space dx = \int{\sin x - \cos x} \space dx = -\cos x - \sin x + C[/imath] 2. [imath]x \in [ -\frac{3\pi}{4} + 2k\pi , \frac{\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}[/imath] : [imath]\int{\sqrt{1-\sin2x}} \space dx = \int{\cos x - \sin x} \space dx = \sin x + \cos x + C[/imath]
|
2082731
|
Prove that [imath]\sin (n\beta)[/imath] can get arbitrarily close to any value in [0, 1] if [imath]\frac{\beta}{\pi}[/imath] is irrational
I saw the following statement and I'm not sure how to prove it: Given a constant value [imath]\beta \in \mathbb{R}[/imath], if [imath]\frac{\beta}{\pi}[/imath] is irrational, then for some value [imath]\alpha \in [0, 1][/imath], [imath]\forall\epsilon \in \mathbb{R}:\epsilon > 0[/imath], [imath]\exists n \in \mathbb{N} : |\sin(n\beta) - \alpha| < \epsilon[/imath]
|
341378
|
Can [imath]\sin n[/imath] get arbitrarily close to [imath]1[/imath] for [imath]n\in\mathbb{N}?[/imath]
Or put differently, does [imath]\lim_{n \to \infty}\big(\max \{\sin 1, \sin 2, \ldots ,\sin n\}\big) = 1?[/imath] My intuition says yes, but how can one prove this?
|
2082836
|
Prove that [imath](2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}[/imath] is an integer
When checked in calculator it is 1. But how to prove it? Also it is not a normal addition like [imath]x+ \frac1x[/imath] which needs direct rationalization. So I just need something to proceed.
|
1416720
|
Calculate simple expression: [imath]\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}[/imath]
Tell me please, how calculate this expression: [imath] \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} [/imath] The result should be a number. I try this: [imath] \frac{\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)}{\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)} = [/imath] [imath] = \frac{2 + \sqrt{5} + 2 - \sqrt{5}}{\sqrt[3]{\left(2 + \sqrt{5}\right)^2} + 1 + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}} [/imath] what next?
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2083570
|
multiplicative group modulo n
I've wondered why each multiplicative group I tested has modulo n either 1 or -1. Is there a rule? For example: [imath]\mathbb Z_{10}^*=\{1,3,7,9\}\\ 1\cdot3\cdot7\cdot9=-1 \:(mod\:n)[/imath] or [imath]\mathbb Z_{15}^*=\{ 1, 2, 4, 7, 8, 11, 13, 14 \}\\ 1\cdot2\cdot4\cdot7\cdot8\cdot11\cdot13\cdot14=1 \:(mod\:n)[/imath] or [imath]\mathbb Z_{5}^*=\{1,2,3,4\}\\ 1\cdot2\cdot3\cdot4=-1 \:(mod\:n)[/imath]
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23937
|
Prove that the product of coprimes must be congruent to [imath]-1[/imath] or [imath]1 \pmod{m}[/imath]
I can't figure this problem out. Let [imath]b_1, b_2, \cdots, b_{\phi(m)}[/imath] be the integers between [imath]1[/imath] and [imath]m[/imath] that are relatively prime to [imath]m[/imath]. [imath]B[/imath] is the product of these integers. Prove that either [imath]B[/imath] is congruent to [imath]1[/imath] or [imath]B[/imath] is congruent to [imath]-1 \pmod{m}[/imath]. Could you give me some hint where to start? EDIT: Please avoid use of group theory. I am an undergrad in my first semester of number theory.
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2080961
|
Prove [imath]p\mid\frac{x^{a}-1}{x-1}[/imath] using Fermat's little theorem
Fermat's little theorem states that if [imath]p[/imath] is a prime number, then for any integer [imath]a[/imath], the number [imath]a^{p} − a [/imath] is an integer multiple of [imath]p[/imath]. In the notation of modular arithmetic, this is expressed as [imath]a^{p} \equiv a \pmod p.[/imath] Using this theorem prove: Given a prime number [imath]p[/imath], show that if there are a positive integer [imath]x[/imath] and a prime number [imath]a[/imath] such that [imath]p[/imath] divides [imath]\frac{x^{a}-1}{x-1}[/imath], then either [imath]a = p[/imath] or [imath]p \equiv 1 \pmod a[/imath]. [imath]p\mid\frac{x^{a}-1}{x-1}[/imath] So, I'm thinking: [imath]\frac{x^{a}-1}{x-1} = x^{a-1}+x^{a-2}+...+1[/imath] I tried the telescoping technique but that doesn't work, assuming [imath]a = p[/imath], shows that [imath]x^{p-1}\equiv 1 \pmod p[/imath]. So, what else can I do?
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2079643
|
Can't come even near to the solution.. Can somebody take a look?
Fermat's little theorem states that if [imath]p[/imath] is a prime number, then for any integer [imath]a[/imath], the number [imath]a^p − a[/imath] is an integer multiple of [imath]p[/imath]. In the notation of modular arithmetic, this is expressed as [imath]a^p \equiv a \mod p.[/imath] Use Fermat's little theorem to prove that: Given prime number [imath]p[/imath], show that if there are positive integer [imath]x[/imath] and prime number [imath]a[/imath] such that [imath]p[/imath] divides [imath]\frac{x^a – 1}{x – 1}[/imath], then either [imath]а= p[/imath] or [imath]p \equiv 1 \mod a[/imath]. I tried to connect [imath]\frac{x^a – 1}{x – 1}[/imath] to the theorem, but without any success.. Anything will help.. Thanks in advance Picture in addition same as the text
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2083296
|
Convergence of [imath]\sum_{n\geq 0}\frac{(n!)^d}{(d\cdot n)!}[/imath]
In an exam I have been asked to discuss the convergence of a series regarding a parameter [imath]d[/imath]. Here's the following : [imath]\sum_{n=0}^\infty \frac{(n!)^d}{(d\cdot n)!}[/imath] The answer is that this series converges for [imath]d \geq 2[/imath]. I totally understand that if [imath]d \leq 1[/imath], the series will not converge but I am blocked while trying to use the d'Alembert or Cauchy's rules. Can somebody give me a tip ?
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2082160
|
How to prove the convergence of [imath]\sum_{n\geq 0}\frac{n!^d}{(dn)!}[/imath] for [imath]d\geq 2[/imath]?
How can we find the general formula about the convergence of the serie [imath]\sum_{n=0}^\infty \frac{(n!)^d}{(dn)!}[/imath] for all [imath]d\geq2[/imath] ? I tried using the d'Alembert Criteria but it doesn't simplify itself enough to prove the convergence.
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2082054
|
Solve the matrix equation
So I need to solve [imath]\begin{pmatrix} 1 &2\\ 1 &2 \end{pmatrix}X=\begin{pmatrix} -1 & -5\\ -1 &-5 \end{pmatrix}[/imath]. I took [imath]X=\begin{pmatrix} a &b \\ c& d \end{pmatrix}[/imath] and I got that [imath]a+2c=-1[/imath] and [imath]b+2d=-5[/imath]. What do I do now or do you have any method to solve this and find values for [imath]a,b,c,d[/imath]? P.S. Determinants are [imath]0[/imath] so you can not use the inverse.
|
2081979
|
Solve a matrix equation
I need to find [imath]X[/imath] from [imath]\begin{pmatrix} 1 & 2\\ -3 &-6 \end{pmatrix} X \begin{pmatrix} 1 &2 \\ -1 &-2 \end{pmatrix}=\begin{pmatrix} 2 &4 \\ -6 & -12 \end{pmatrix}[/imath] I wrote [imath]X[/imath] as [imath]X=\begin{pmatrix} a & b\\ c &d \end{pmatrix}[/imath] and I got [imath]a+2c-b-2d=2[/imath] but I do not know what to do next. Please help.
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2083699
|
Peano axioms in FOL - numbers lower than or equal to another
@Moderators: I apologize because this is very similar to my previous post, but I didn't post the exact axioms there from the beginning (I did quite a major change after receiving a reply).. please delete the previous one (Peano/Presburger axioms - "find" numbers lower or equal than another number) if there are any problems. Happy New Year, I am trying to define natural numbers in first-order logic in order to use them in an Automated Theorem Prover. I have tried the Peano axioms on the Wikipedia page with/without multiplication (I don't really need it and thought removing it would make proving faster). I'd like to make the prover "know" (deduce that) the only three numbers lower than or equal to 2 are 0,1,2, but I can't seem to be able to do it. More precisely, I'd like the foolowing statement to be provable (NOTE: 2 is just an example, I'd like it to work for any natural number): [imath]\forall x(nat(x) \to (leq(x, s(s(0))) \to (x=0) \lor (x=s(0)) \lor (x=s(s(0))) ))[/imath] My prover has equality implemented. I define 1,2,3 (where 0,1,2,3 are 0-arity functions.. or I assume I can call them constants) as: [imath]1 = s(0)[/imath] ([imath]equal(1, s(0)[/imath] is what I actually write in the prover) [imath], 2 = s(1), 3 = s(2)[/imath] My axioms are pretty much these: [imath]nat(0)[/imath] [imath]\forall x(nat(x) \to nat(s(x))[/imath] [imath]\forall x,y(nat(x)\land nat(y) \to nat(add(x,y))[/imath] [imath]\forall x( nat(x) \to s(x) \neq 0)[/imath] [imath]\forall x,y (nat(x) \land nat(y) \to (s(x)=s(y) \to s=y))[/imath] [imath]\forall x(nat(x)\land x\neq 0 \to \exists z (nat(z) \land x = s(z)))[/imath] [imath]\forall x(nat(x) \to add(x,0) = 0[/imath] [imath]\forall x,y(nat(x)\land nat(y) \to add(x,s(y)) = s(add(x,y)))[/imath] [imath]\forall x,y(nat(x)\land nat(y) \to (leq(x,y) \leftrightarrow \exists z(nat(z)\land y = x + z )))[/imath] (Maybe for the purpose of this question it would have been better to write the last axiom as "forall x,y leq(x,y) <-> nat(x) /\ nat(y) /\ exists z .....") I know they look quite ugly.. in my prover I'm actually able to type something like this (using what I understand are called "sorts"): [imath]\forall[nat(x)](P(x))[/imath] instead of [imath]\forall x(nat(x) \to P(x)[/imath] so it looks cleaner and easier to follow.
|
2082711
|
Peano/Presburger axioms - "find" numbers lower or equal than another number
[EDIT/CONCLUSION] It turns out it was actually working.. I was just like too stupid to let the prover run for more time and assumed it would take a lot / not be able to prove with what I've provided because it wasn't doing it fast and because it took it ~1 minute with a similar goal asking for numbers lower or equal than 1. However, the goal I gave as an example in this post (leq(x,2)) took ~1 minute as well and for leq(x,5), it was actually able to prove it faster I think (~40 seconds if I recall correctly). Thank you very much to everyone who helped!! [EDIT:] I apologize for not describing my axioms properly.. I have now written them so as they pretty much resemble what I actually wrote. Hello and Happy New Year, I am trying to define natural numbers in order to use them in a First-Order Logic Theorem Prover. The problem is I'm unable to solve the following problem: "Show that the only numbers lower or equal than (for example) 2 are 0,1 and 2". My axioms go like this: [imath]nat(0)[/imath] [imath]\forall x( nat(x) \to nat(s(x)) )[/imath] [imath]\forall x,y( (nat(x)\land nat(y)) \to nat(add(x,y))[/imath] [imath]\forall x( nat(x) \to (s(x) \neq 0) )[/imath] [imath]\forall x,y ( (nat(x) \land nat(y)) \to ( (s(x)=s(y)) \to (s=y)))[/imath] [imath]\forall x( (nat(x)\land x\neq 0) \to (\exists z (nat(z) \land (x = s(z)))))[/imath] [imath]\forall x(nat(x) \to (add(x,0) = x))[/imath] [imath]\forall x,y((nat(x)\land nat(y)) \to (add(x,s(y)) = s(add(x,y))))[/imath] [imath]\forall x,y((nat(x)\land nat(y)) \to ((leq(x,y) \leftrightarrow (\exists z(nat(z)\land (y = x + z) )))))[/imath] I have also added addition axioms for commutativity and associativity of addition. So, what I'd like to be able to prove is something like this (for any number, not just for [imath]2[/imath]): [imath] \forall x(nat(x) \to (x\leq 2 \to (x = 0) \lor(x=1)\lor(x=2))) [/imath]. Are additional axioms needed in order to be able to prove this?
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907353
|
Automorphism of [imath]\mathbb{Q}[/imath]
How can we show that any automorphism of [imath]\mathbb{Q}[/imath] under addition is of the form [imath]x \to qx[/imath] ,for some q in [imath]\mathbb{Q}[/imath]. edited- I found the same question was asked by @tattwamasi-amrutam tagged automorphism from [imath]\mathbb{Q}[/imath] to [imath]\mathbb{Q}[/imath].there is some useful answer.
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511796
|
Automorphism from [imath]\Bbb Q[/imath] to [imath]\Bbb Q[/imath]
Any automorphism of the group [imath]\Bbb Q[/imath] under addition is of the form [imath]x\to qx[/imath] for some [imath]q\in \Bbb Q[/imath]. I don't know how to proceed in this. Even if i say that [imath]1\to q[/imath], I can't claim that [imath]x\to qx[/imath] since [imath]\Bbb Q[/imath] is not cyclic.
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2083789
|
Decomposition of finite cyclic group into direct product
Let [imath]m, n \in \mathbb{N}[/imath]. Then [imath]C_{mn} \cong C_m \times C_n \iff \gcd(m, n) = 1[/imath]. I'm struggling to understand proof of this fact. Can someone explain it step by step?
|
1271223
|
Isomorphism of a product [imath]C_n \times C_m[/imath] of cyclic groups with the cyclic group [imath]C_{mn}[/imath]
Given that [imath]C_n[/imath] is a cyclic group of order [imath]n[/imath], what conditions must integers [imath]n[/imath] and [imath]m[/imath] satisfy such that the group [imath]C_n \times C_m[/imath] is isomorphic to C[imath]_{mn}[/imath]? So I attempted to investigate a few lower-order cyclic groups to find a pattern. In particular, I looked at [imath]C_2 \times C_3[/imath] and compared it to [imath]C_6[/imath]. Here are a couple of diagrams: [imath]C_2 \times C_3[/imath] - http://escarbille.free.fr/group/?g=6_2b [imath]C_6[/imath] - http://escarbille.free.fr/group/?g=6_2a They Cayley tables look pretty different (identity elements are all over the place for [imath]C_2 \times C_3[/imath]) so I concluded that they are not isomorphic. I did the same for [imath]C_2 \times C_4[/imath] and [imath]C_8[/imath]: [imath]C_2 \times C_4[/imath] - http://escarbille.free.fr/group/?g=8_2 [imath]C_8[/imath] - http://escarbille.free.fr/group/?g=8_1 Again, I arrived at the conclusion that the 2 groups are not isomorphic due to their Cayley table structure. It seems I'm stuck now. Are my conclusions wrong for the examples I examined? If so, how do I find the conditions for the general case of comparing [imath]C_m \times C_n[/imath] with [imath]C_{mn}[/imath]?
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487011
|
Showing that a homomorphism between groups of units is surjective.
Let [imath]n[/imath] be a positive integer and let [imath]d[/imath] be some divisor of [imath]n[/imath]. Consider the group of units modulo [imath]n[/imath], which we shall denote by [imath]U(n)[/imath]. Likewise, denote the group of units modulo [imath]d[/imath] by [imath]U(d)[/imath]. Consider the homomorphism [imath]f:\ U(n)\rightarrow U(d)[/imath] given by reducing modulo [imath]d[/imath], [imath]f(u) = u \pmod{d}[/imath] I wish to show that this map is surjective. Can anyone supply a simple proof of this fact? Alternatively, it is sufficient to prove that the set [imath]S = \left\{u\in U(n) \mid u\equiv 1 \pmod{d} \right\}[/imath] has cardinality [imath]|S| = \phi(n)/\phi(d)[/imath] where [imath]\phi[/imath] is the totient function.
|
70717
|
Why does the natural ring homomorphism induce a surjective group homomorphism of units?
I'm trying an old problem here: http://www.math.dartmouth.edu/archive/m111s09/public_html/homework-posted/hw1.pdf Suppose [imath]n\mid m[/imath], and I have a natural ring homomorphism [imath]\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}[/imath] defined by [imath]\varphi(j)=j\mod{n}[/imath]. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups [imath](\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times[/imath]? My question in particular is why is it surjective? I take [imath]k\in(\mathbb{Z}/n\mathbb{Z})^\times[/imath]. Then there exist integers [imath]s,t[/imath] such that [imath]sn+tk=1[/imath]. Since [imath]n\mid m[/imath], I can also write [imath]m=na[/imath]. So multiplying through I get [imath]sm+tka=a[/imath]. I'm lost here. How can I find a unit in [imath](\mathbb{Z}/m\mathbb{Z})^\times[/imath] which maps to [imath]k[/imath] to show surjectivity?
|
2084355
|
Inequality satisfied by coefficients
Let [imath]g(x)=a_1\sin x+a_2\sin 2x+\ldots+a_n\sin nx[/imath], where [imath]a_1,a_2,\ldots,a_n\in \mathbb{R}[/imath] such that [imath]|g(x)|<|\sin x|\;\forall x\in\mathbb{R}[/imath]. Then, can we prove that [imath]|a_1+2a_2+\ldots+na_n|<1[/imath]. I think yes, because we may apply the triangle inequality repeatedly and the boundedness of [imath]\sin[/imath] function. Any ideas. Thanks beforehand.
|
584509
|
triangular series and inequality
Let [imath]a_1,a_2,\ldots,a_n\in \mathbb R[/imath] and [imath]f(x)=a_1\sin x+a_2\sin 2x+\ldots+a_n\sin nx[/imath] such that [imath]|f(x)|\leq|\sin x|[/imath] for every [imath]x\in \mathbb R[/imath]. Prove that [imath]|a_1+2a_2+3a_3+\ldots+na_n|\leq1[/imath]. Solve: I solved it very simple. [imath]|a_1+2a_2+\ldots+na_n|=\lim_{x\to 0}\frac{|a_1\sin x+a_2\sin 2x+\ldots+a_n\sin nx|}{|\sin x|}=\lim_{x\to 0}\frac{|f(x)|}{|\sin x|}\leq1.[/imath]
|
2083608
|
From the given figure
From the given figure, prove that [imath]\cot\ \theta=\cot\ A+\cot\ B+\cot\ C[/imath] My work. I have got this solution from one of my friends. But I didn't understand all the process. I understood the application of ceva theorem in the first step but the steps after that I can’t understand. Also how has be [imath]\prod [/imath] used here, why? How it works here? Please make me understand
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1984128
|
Geometry with Trigonometry
From the given figure, prove that [imath]\cot\ \theta=\cot\ A+\cot\ B+\cot\ C[/imath] I could not even start to the first point. Please help me to solve this. Thanks in advance
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2084591
|
How to compute [imath]\frac{d}{dt} \int_{a(t)}^{b(t)} f(x,t) dx[/imath]
I want to differentiate [imath] \int_{a(t)}^{b(t)} f(x,t) dx[/imath] this integral with respect to [imath]t[/imath] [imath]i.e[/imath], \begin{align} \frac{d}{dt} \int_{a(t)}^{b(t)} f(x,t) dx \end{align} I only know for the case of constant [imath]a,b[/imath]. For this case how can i differentiatie?
|
58594
|
How do I differentiate this integral?
That is: [imath]\left(\int_{a(x)}^{b(x)}\!f(x,t)\,dt\right)'[/imath] I don't know how to differentiate a integral if functions of [imath]x[/imath] are at its limits. Can you guys show me how to do this?
|
2084698
|
How to find out the algebraic integer in some fields.
As shown in the title: How to find all the algebraic integer in [imath]\mathbb{Q}[\sqrt{-1}][/imath] and [imath]\mathbb{Q}[\sqrt{-3}][/imath]
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654202
|
Determining ring of integers for [imath]\mathbb{Q}[\sqrt{17}][/imath]
I'm trying to find the ring of integers of [imath]\mathbb{Q}[\sqrt{17}][/imath], and it comes down to determining the set [imath]\{(a,b)\in\mathbb{Q}^2\mid 2a\in \mathbb{Z}, a^2-17b^2\in\mathbb{Z}\}[/imath]. How can I determine this set?
|
61478
|
Seemingly simple system of equations [imath]x^{2} + y = 31[/imath] and [imath]x + y^{2} = 41[/imath]
I have the following system: [imath]x^{2} + y = 31[/imath] [imath]x + y^{2} = 41[/imath] As I try to solve it via simple substitution, I get into 4-th power equations, which I can simplify to [imath](x-5)(x^{3}+5x^{2}-37x-184)[/imath] (and I am not sure how to get the cubic here). Is there a simpler way to solve this? There are 4 pairs of answers, I have got one (5 and 6).
|
94957
|
What are the common solutions of [imath]x^2+y=31[/imath] and [imath]y^2+x=41[/imath]?
A friend asked me if I have a certain algorithm to solve [imath]x^2+y = 31[/imath] and [imath]y^2+x=41[/imath] simultanously. We found the solutions but we didn't find a way to solve both equations. Any ideas?
|
2084852
|
Is [imath]\sum_{p} \frac{1}{2^p}[/imath] irrational?
[imath] \sum_{p \, \in \text{ primes}} \frac{1}{2^p} [/imath] My friend came across a book which claimed this to be irrational and claimed it could be solved using the pigeonhole principle.
|
678985
|
Irrationality of [imath]\sum_{p\in\mathbb{P}} \frac{1}{2^{p}}[/imath]
Let [imath]\mathbb{P}[/imath] be the set of prime numbers, and consider [imath]m=\displaystyle\sum_{p\in\mathbb{P}} \frac{1}{2^{p}}[/imath]. Is [imath]m[/imath] irrational? In the following paper, the author recalls several sufficient criteria for irrationality. When applying some of this criteria to [imath]m[/imath] I arrive that the condition of some of the criteria are not satisfied by means of Bertrand´s Postulate. I found a similar result by Sándor: Let [imath]\lbrace a_m\rbrace[/imath], [imath]m\geq 1[/imath] be a sequence of positive integers such that [imath]\text{lim sup}\frac{a_{m+1}}{a_1a_2\cdots a_m}=\infty \;\;\;\;\text{and}\;\;\;\;\;\;\text{lim inf}\frac{a_{m+1}}{a_m}>1[/imath] Then the series [imath]\displaystyle\sum_{m} \frac{1}{a_m}[/imath] is an irrational number. I have yet to prove that if [imath]f[/imath] is a continuous function then [imath]\text{lim sup} \;f(a_m)=f(\text{lim sup}\, a_m)[/imath]. Assuming this; [imath]\text{lim inf}\; P_{m+1}-P_{m}>1[/imath] where [imath]P_m[/imath] is the [imath]m[/imath]-th prime. But I have trouble with [imath]\text{lim sup}\; P_{m+1}-\sum_{j=1}^{m}P_{j}[/imath] my guess is that [imath]\text{lim sup}\; P_{m+1}-\sum_{j=1}^{m}P_{j}\neq\infty[/imath] so [imath]\text{lim sup}\frac{2^{P_{m+1}}}{2^{P_1}\cdots 2^{P_m}}\neq\infty[/imath] and this theorem will result useless to tell if [imath]m[/imath] is irrational. How can I prove (or disprove) that [imath]m[/imath] is irrational (are there any other simpler criteria)? How can I use Dirichlet criterion or Hurwitz criterion? What is its irrationality measure? Any help is highly appreciated. Edit: That [imath]m[/imath] is irrational is clear by the criteria provided in the comments, namely that if [imath]x[/imath] is rational then the binary representation of [imath]x[/imath] is periodic. The number [imath]m[/imath] in binary expansion has [imath]1[/imath] in the [imath]P[/imath]-th position and zero elsewhere. As there are arbitrarily large gaps between primes; then the binary representation fails to be periodic. So the question that remains unsolved is: What is its irrationality measure?
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2084977
|
Convergence of series depending on convergent series
I have a convergent series of positive terms [imath]\sum_{n=1}^{\infty} a_n[/imath] and have to check if series [imath] \sum_{n=1}^{\infty}\frac{\sqrt{a_n}}{\log(n)}\left(n^{a_n}-1\right)[/imath] converges. I think it does, but don't know how to prove that, I tried using Cauchy criterion, but couldn't really make it work.
|
2050131
|
If [imath]\sum_{n=2}^{\infty} a_n[/imath] is convergent, is [imath]\sum_{n=2}^{\infty}{\sqrt{a_n} \over \ln{n}}\left( n^{a_n}-1 \right)[/imath] convergent as well?
Suppose [imath]\sum_{n=2}^{\infty} a_n[/imath] is a convergent series. Is the following series convergent as well? [imath]\sum_{n=2}^{\infty}{\sqrt{a_n} \over \ln{n}}\left( n^{a_n}-1 \right)[/imath] The part with [imath]n^{a_n}-1 [/imath] looks quite similar to [imath]\sqrt[n]{n}-1[/imath] which behaves like [imath]\ln{n}[/imath], however, I'm not able to tie this fact to the behavior of the whole expression. Any suggestions?
|
1069623
|
Show that [imath]\lim_{n\rightarrow\infty}{\sum_{i=1}^{n}{\frac{F_n}{2^n}}}=2[/imath]
I need help to show that [imath]\lim_{n\rightarrow\infty}{\displaystyle\sum_{i=1}^{n}{\frac{F_n}{2^n}}}=2[/imath], where [imath]F_n[/imath] is the n-th number in the Fibonacci sequence. I know how to prove this by putting that [imath]A_n={\displaystyle\sum_{i=1}^{n}{\frac{F_n}{2^n}}}[/imath] and than finding a closed form for [imath]A_n[/imath] (I can't remember how the closed form looks like bcs I did this problem 2 years ago). Now that I started to learn limits at school I wanted to know if it is possible to solve this problem in another way using some tricks with limits or something similar.
|
88529
|
How to prove the Fibonacci sum [imath]\sum \limits_{n=0}^{\infty}\frac{F_n}{p^n} = \frac{p}{p^2-p-1}[/imath]
We are familiar with the nifty fact that given the Fibonacci series [imath]F_n = 0, 1, 1, 2, 3, 5, 8,\dots[/imath] then [imath]0.0112358\dots\approx 1/89[/imath]. In fact, [imath]\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}[/imath] How do we prove that, more generally, for [imath]p > 1[/imath] then, [imath]\sum_{n=0}^{\infty}\frac{F_n}{p^n} = \frac{p}{p^2-p-1}[/imath] (The above simply was the case [imath]p = 10[/imath].)
|
2084831
|
How to prove that if [imath]I+A[/imath] and [imath]I-A[/imath] are nonsingular and [imath](I−A)(I+A)^{−1}[/imath] is orthogonal then [imath]A[/imath] is skew symmetric?
Let [imath]A[/imath] be an [imath]n\times n[/imath] matrix with real elements so that [imath]I+A[/imath] and [imath]I-A[/imath] are nonsingular and [imath]B=(I−A)(I+A)^{−1}[/imath] is orthogonal. Is [imath]A[/imath] skew symmetric?
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1315959
|
If [imath](I-A)(I+A)^{-1}[/imath] is orthogonal then prove that A is skew symmetric.
Question from Determinants.Can't solve !
|
2084713
|
The Number of different [imath]n \times n[/imath] Skew-symmetric matrices with each element being equal to either [imath]0[/imath] or [imath]1[/imath] or [imath]-1[/imath]
The Number of different [imath]n \times n[/imath] skew-symmetric matrices with each element being equal to either [imath]0[/imath] or [imath]1[/imath] or [imath]-1[/imath] where [imath]n=5[/imath] is ?
|
2079125
|
Finding the no. of possible matrices given the order and limited no. of entries
The no. of all possible matrices of order [imath]3\times 3[/imath] with each entry [imath]0[/imath] and [imath]1[/imath] and [imath] -1[/imath].
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2084551
|
Which of these definations is correct for proximal operator?
For given function [imath]g(W)[/imath], where [imath]W \in R^{M \times T}[/imath]. I have seen two different definition of proximal operator of it, but I don't know which one is correct ? One with [imath]L_2[/imath]-norm, and second one with Frobenius norm [imath]prox_g(W)={arg\,min}_u(g(u) + \frac{1}{2} |||u-W||_2^2 )[/imath] [imath]prox_g(W)={arg\,min}_u(g(u) + \frac{1}{2} |||u-W||_F^2 )[/imath]
|
878770
|
Proximal operator fixed point property for matrices
[imath]\newcommand{\prox}{\operatorname{prox}}[/imath] [imath]\newcommand{\argmin}{\operatorname{argmin}}[/imath] [imath]\newcommand{\dom}{\operatorname{dom}}[/imath] Recall again that the proximal operator for vectors [imath]\prox_{f}: R^n \rightarrow R^n[/imath] of [imath]f[/imath] is defined as: [imath]\prox_f(v) := \argmin_{x} \left(f(x) +(1/2)\|x-v\|_2^2 \right) [/imath], where [imath]f: R^n \rightarrow R \cup \infty[/imath] is a closed proper convex function and [imath]\|\cdot \|_2[/imath] is the Euclidean norm. [imath]\prox[/imath] is strongly convex and not everywhere infinite, so there it will have a unique minimizer for every [imath]v \in R^n[/imath] The crucial property of proximal operator is that [imath]x^*[/imath] minimizes [imath]f(x)[/imath] iff [imath]x^* = \prox_f(x^*)[/imath], i.e. [imath]x^*[/imath] is a fixed point of [imath]\prox_f[/imath]. Let us consider the proof of this property. The first part, namely, that if [imath]x^*[/imath] minimizes [imath]f[/imath] then [imath]\prox_f(x^*)=x^*[/imath], is trivial: [imath]f(x) +(1/2)\|x-x^*\|_2^2 \le f(x^*) = f(x^*) +(1/2)\|x^*-x^*\|_2^2[/imath] The second part uses the notion of subdifferential. In the proof authors say that [imath]\tilde{x}[/imath] minimizes [imath]f(x) +(1/2)\|x-v\|_2^2[/imath] i.e. [imath]\tilde{x} = \prox_f(v)[/imath] iff [imath]0 \in \partial f(\tilde{x}) + (\tilde{x}-v)[/imath], where the sum is of a set and a point. Recall the definition of the subdifferential: [imath]\partial f(x) = \{ y[/imath] | [imath] f(z)\le f(x) + y^T(z-x)[/imath] [imath]\forall z \in \dom{f}\}[/imath] Take [imath]\tilde{x}=v=x^*[/imath], it follows that [imath]0 \in \partial f(x^*)[/imath] so [imath]x^*[/imath] minimizes [imath]f[/imath]. Question 1: Consider proximal operator defined for matrices now: [imath]\prox_f(Y) := \argmin_{X} \left(f(X) +(1/2)\|X-Y\|^2 \right)[/imath], where [imath]X[/imath] is some real [imath]m[/imath] by [imath]n[/imath] matrix. What norm function should be taken in the definition of the [imath]\prox[/imath] in this case? Frobenius? Or spectral norm (induced 2nd norm) which is the largest eigenvalue (is spectral norm even differentiable?)? Or something else? Question 2: Can the norm for the definition of [imath]\prox[/imath] in case of matrices be chosen in different ways? What conditions does the proof above impose on the norm function?
|
2085259
|
Why [imath]O(n)[/imath] has exactly two connected components?
It is well know that [imath]SO(n)[/imath] is connected and [imath]O(n)[/imath] has two connected components: [imath]O^+(n)=\{A\in O(n):\det A=+1\}[/imath] and [imath]O^-(n)=\{A\in O(n):\det A=-1\}[/imath]. In what book can I find this property?
|
2278359
|
Path-components of [imath]O(n)[/imath]
I have look here in order to find an answer why the orthogonal group [imath]O(n)[/imath] has two path-components, given by [imath]\det =1[/imath] and [imath]\det =-1[/imath]. But what is a possible proof for that fact? I can easily show that there are at least two components (not path-components) by using the determinant-function.
|
2085602
|
Is my proof for all these statements are equivalent correct?
[imath](i)[/imath] [imath]x[/imath] is irrational [imath](ii)[/imath] [imath]3x + 2[/imath] is irrational [imath](iii)[/imath] [imath]\frac{x}{2}[/imath] is irrational My working : [imath](i) \to (ii) \land (ii \to iii) \land (iii \to i)[/imath] [imath](i) \to (ii)[/imath]: Proof By contraposition: x = p/q y = c/d 3x + 2 = y 3(p/q) + 2 = c/d 3p+2q/q = c / d c = 3p + 2q d = q c/d = c/d [imath](ii) \to (iii)[/imath]: Proof by Contraposition: x/2 = c/d -> 3x+2 = a/b Now from the previous example we know that 3x+2 = a/b is true. [imath](iii) \to (i) [/imath] Proof by contraposition: x = a/b -> x / 2 = c/d (a/b) / 2 = c/d a/2b = c/d c = a d = 2b c / d = c / d
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2085544
|
Why do I have to solve for [imath]x[/imath] in this proof, if I choose not to find a counterexample?
If [imath]x[/imath] is irrational then [imath]3x+2[/imath] is irrational. My Working: Proof by contraposition: [imath]3x + 2[/imath] is rational and [imath]x[/imath] is rational [imath]x = a/b[/imath] [imath]y = c/d[/imath] [imath]3x + 2 = c/d[/imath] [imath]3x = (c/d) - 2[/imath] [imath]x = c-2d/3d[/imath] [imath]a/b = c-2d / 3d[/imath] I am having a hard time understanding as to why do we end up solving for [imath]x[/imath] in this equation to determine weather [imath]3x+2[/imath] is rational or not if [imath]x[/imath] is rational.
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2084309
|
Circumfrence to diameter ratio of a Circle scribed on a sphere?
On the face of it this might just seem silly, but I wonder if I scribe a circle on a sphere will the ratio of the circumference of such a circle to its diameter would still be [imath]\pi[/imath]. If so, is there a way I could prove it or disprove otherwise?
|
280789
|
Circle on sphere
Foreword This question was inspired by initial mistakes in this question. I wanted to explore the strange circle with [imath]A>\pi r^2[/imath] and got lost into geometrical jungle. A spherical cap is usually described by it's height [imath]h[/imath], radius of the base [imath]a[/imath] and radius of the sphere [imath]r[/imath]. Besides we have a relation [imath]a^2+(r-h)^2=r^2[/imath] so the cap actually only has two degrees of freedom. When considering a case of circle on a sphere (like the Earth), it's the surface of a spherical cap, but we can't see the height or other classic parameters of this spherical cap. Instead we encounter parameters like area [imath]A[/imath], perimeter [imath]C[/imath] and diameter [imath]d[/imath]. I will try to write formulae for these: [imath]A=2\pi rh[/imath] [imath]C=2\pi a[/imath] [imath]d=2r\theta[/imath] Where [imath]\theta[/imath] is introduced with [imath]\sin\theta=\frac{a}{r}[/imath] or [imath]\cos\theta=(1-\frac{h}{r})[/imath]. As the cap only has two degrees of freedom, it should be a good problem with unique solution if only two values are known. A-C Problem The simple example: [imath]A[/imath] and [imath]C[/imath] are given. I show this only as example before the other tasks. We can easily find [imath]a=\frac{C}{2\pi}[/imath] and then solve: [imath]A=2\pi rh = 2\pi r (r-\sqrt{r^2-a^2})[/imath] to obtain [imath]r=\frac{A}{2\sqrt{\pi(A-\pi a^2)}}[/imath] Parameters [imath]a[/imath] and [imath]r[/imath] are now known and it is straightforward to find [imath]h[/imath], [imath]d[/imath], [imath]\theta[/imath] and anything else. C-d Problem [imath]C[/imath] and [imath]d[/imath] are given. First step once again is easy: [imath]a=\frac{C}{2\pi}[/imath]. The knowing of [imath]a[/imath] suggests that we might prefer the [imath]\theta[/imath] definition with the sine. We obtain: [imath]d=2r\theta=2r\arcsin\frac{a}{r}[/imath] That's where I got stuck. Is there an easy way to solve this? Is there a hard way to solve this? I tried solving numerically with Wolfram Mathematica and it went into complex numbers. When solving the geometrically equivalent [imath]\sin\frac{d}{2r}=\frac{a}{r}[/imath] I got some roots, but the search seemed very unstable and not finding the root closest to initial value. Besides it was finding a lot of different roots while I am pretty sure that there should only be one [imath]r[/imath] for each pair of [imath]d[/imath] and [imath]a[/imath]. What would be the right formulation to solve this problem numerically? A-d problem [imath]A[/imath] and [imath]d[/imath] are given. This seems the hardest one. We obtain a system of equations: [imath]A=2\pi r h[/imath] [imath]d=2r\arccos\left(1-\frac{h}{r}\right)[/imath] (The first equation suggested solving the problem in [imath]h[/imath] and [imath]r[/imath]). Is this system solvable analytically? Or numerically? My experiments with Wolfram Mathematica never converged. Of course, I could express [imath]h[/imath] or [imath]r[/imath] from the first equation and leave only the second in one of these forms: [imath]d = 2r\arccos\left(1-\frac{A}{2\pi r^2}\right)[/imath] [imath]d = \frac{A}{\pi h}\arccos\left(1-\frac{2\pi h^2}{A}\right)[/imath] But I have no idea how to solve these either.
|
2084832
|
negative gaussian curvature surface
I am looking for examples of surfaces [imath]M[/imath] with gaussian curvature [imath]K\lt0[/imath] Additionally, a closed geodesic lies on the surface [imath]M[/imath] Thanks
|
1652720
|
Determining if there can be smooth closed geodesic given its curvature
In my class on differential geometry I have been given the following question on which I am stuck: Let S be a regular orientable surface in [imath] R^3 [/imath] with Gaussian curvature [imath]K[/imath] (not necessarily constant) and we are to check if there can be a smooth closed geodesic on S in the following case: [imath] K>0 [/imath] [imath] K=0 [/imath] [imath] K<0 [/imath] and we are to give an example when it is possible that the closed geodesic bounds a simply connected region I know for starters about the first one with positive curvature I may take the sphere and a great circle which is a geodesic and obviously bounds a simply connected subset of the sphere. But with the zero and negative curvatures I have no idea how to tell if a closed geodesic exists I think the Gauss Bonnet theorem has something to do with this but I cannot really proceed from there and I am posting here in the hope of getting help ******* Progress: I can work out what happens with Gauss Bonnet if I have the constraint that the geodesic bounds a simply connected region as this is simple but the fact of the matter is where I am stuck is the general existence or non existence if no constraint is given on the region bounded
|
2085742
|
Is it possible to write [imath]\tan^{-1}(x)[/imath] as a power series of [imath]\tanh(x)[/imath]?
[imath]\tan^{-1}(x)[/imath] looks very similar to [imath]\tanh(x)[/imath] if [imath]x[/imath] is small enough. Look. But they diverge from each other as [imath]x[/imath] grows. And for very big [imath]x[/imath]'s, They almost represent the constant functions [imath]1[/imath] and [imath]\frac \pi 2[/imath] (for [imath]\tanh(x)[/imath] and [imath]\tan^{-1}(x)[/imath], respectively). Is it possible to write [imath]\tan^{-1}(x)[/imath] as a power expansion of [imath]\tanh(x)[/imath]? I mean can we say this? [imath]\tan^{-1}(x)=\sum^{\infty}_{i=0} \alpha_i \tanh^i(x)[/imath] The power series is the thing I want. Not the resemblance between them.
|
92533
|
Relationship between [imath]\tanh x[/imath] and [imath]\arctan x[/imath]
The functions [imath]\tanh x[/imath] and [imath]\arctan x[/imath] have a similar graph. Is there a formula to transform [imath]\tanh x[/imath] to [imath]\arctan x[/imath]?
|
2086066
|
Calculate [imath]\lim_{n\to\infty}{\sqrt{5+\sqrt{5+...+\sqrt 5}}}[/imath]
Calculate [imath]\lim_{n\to\infty}{\sqrt{5+\sqrt{5+...+\sqrt 5}}}[/imath] I have a similar example solved in my textbook so I tried to use the same technique though I'm not entirely sure what I'm doing. [imath]a_n:=\lim_{n\to\infty}{\sqrt{5+\sqrt{5+...+\sqrt 5}}}[/imath] [imath]a_1=\sqrt5[/imath] [imath]a_{n+1}=\sqrt{5+a_n}[/imath] [imath]L=\sqrt{5+L} \implies L=\frac{1+\sqrt{21}}{2}[/imath] This is what I'm confused about, where did [imath]L=\sqrt{5+L}[/imath] come from? And after that I have to prove that [imath]a_n[/imath] is bounded above with [imath]\frac{1+\sqrt{21}}{2}[/imath], right? So I use induction to prove that: [imath]a_1=\sqrt5\leq\frac{1+\sqrt{21}}{2}[/imath] Let's assume that for some [imath]n\in \Bbb N[/imath] it's true that [imath]a_n\leq\frac{1+\sqrt{21}}{2}[/imath] Then [imath]a_{n+1}=\sqrt{5+a_n}\leq\sqrt{5+\frac{1+\sqrt{21}}{2}}=\sqrt\frac{11+\sqrt{21}}{2}[/imath] Well, this doesn't seem right. What did I prove here? After that I proved that [imath]a_n[/imath] is strictly increasing, so I should now be able to find the limit only I don't know how to proceed.
|
1376667
|
Proposition: [imath]\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = \frac{1 + \sqrt{1 + 4x}}{2}[/imath].
Proposition: [imath]\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = \frac{1 + \sqrt{1 + 4x}}{2}[/imath] I believe that this is true, and, using Desmos Graphing Calculator, it seems to be true. I will add how I derived the formula in a moment, if you would like. Working I will be honest; all that I did was use the Desmos Graphing Calculator, and let [imath]y = \sqrt{x + \sqrt{x + \sqrt{x + ...}}},[/imath] let [imath]x = 1, 2, 3, ...[/imath], looked at the point at which the two graphs meet, and searched for the number in Google. It turned up an interesting website, which you may access here, which seemed to show a pattern. I used this pattern to derive the formula that I stated earlier.
|
2086091
|
Show that set of Real numbers has the same cardinality as a set of Irrational Numbers.
How can I show that sets [imath]| \mathbb R-\mathbb Q |[/imath] and [imath] | \mathbb R |[/imath] have same cardinality. My Solution: As long as for element in one set I can match it with unique element in another set they have same cardinality: After removing [imath]\mathbb Q [/imath] from [imath]\mathbb R[/imath] I can shift elements of Irrational numbers to correspond to elements of two sets match like this: [imath] \begin{array}{c|c|c|c|c|c} \mathbb R&0&1&\cdots&n&\cdots\\\hline \mathbb R\setminus\mathbb Q & \sqrt{2} &\sqrt{5} &\cdots & \sqrt{n+1} &\cdots \end{array} [/imath] Note that I use n as Real number for simplicity. Even if this solution is correct, I feel like Professor won't accept it on exam. How would formal solution look like?
|
695470
|
Cardinality of Irrational Numbers
I know and I have proved more than once that the set of irrational numbers ([imath]\mathbb{I}[/imath]) is uncountable, but now I'm given to solve this problem: Show that [imath]|\mathbb{I}|=|\mathbb{R}|[/imath], How can I do that? Do I need to assume the Continuum Hypothesis in order to that statement to be true? Thanks
|
2085918
|
Compute [imath]\lim_{n \rightarrow \infty} \sin (n {\pi} {\sqrt[3] {n^{3} + 3n^{2} + 4n - 5}})[/imath]
How can I evaluate the limit? [imath]\lim_{n \rightarrow \infty} \sin (n {\pi} {\sqrt[3] {n^{3} + 3n^{2} + 4n - 5}})[/imath] I think the limit is [imath]0[/imath] but I fail to work it out properly. Please help me. Thank you in advance.
|
1069742
|
Does [imath]\lim_{n\rightarrow\infty}\sin\left(\pi\sqrt[3]{n^{3}+1}\right)[/imath] exist?
I have this limit: [imath]\lim_{n\rightarrow\infty}\sin\left(\pi\sqrt[3]{n^{3}+1}\right)[/imath] I don't even know if it exists. If so, what its value ? Really don't have any idea..
|
2086207
|
An inequality with a load of variables: [imath] (1-a_1)(1-a_2)...(1-a_n) \ge 1/2[/imath]
It is known about numbers [imath]a_1, a_2, ... a_n[/imath] that [imath]a_1 + a_2 +...+a_n \le 1/2.[/imath] Prove that [imath] (1-a_1)(1-a_2)...(1-a_n) \ge 1/2[/imath] I have tried using [imath]a^2 \geq 0[/imath], it led to nothing. How can I make my inequality look like in the possible duplicate?
|
216721
|
How to show [imath]x_1,x_2, \dots ,x_n \geq 0 [/imath] and [imath] x_1 + x_2 + \dots + x_n \leq \frac{1}{2} \implies (1-x_1)(1-x_2) \cdots (1-x_n) \geq \frac{1}{2}[/imath]
How to show [imath]x_1,x_2, \dots ,x_n \geq 0 [/imath] and [imath] x_1 + x_2 + \dots + x_n \leq \frac{1}{2} \implies (1-x_1)(1-x_2) \cdots (1-x_n) \geq \frac{1}{2}[/imath]
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2086828
|
Prove that for all real numbers x, the inequality [imath]2^x[/imath] + [imath]3^x[/imath] - [imath]4^x[/imath] + [imath]6^x[/imath] -[imath]9^x[/imath] < 1 holds
Prove that for all [imath]x \in \mathbb{R}[/imath] the inequality [imath]2^x + 3^x - 4^x + 6^x - 9^x < 1[/imath] holds. Firstly, I rewrote the inequality as: [imath]2^x + 3^x - (2^x)^2 + (2^x \cdot 3^x) - (3^x)^2 < 1[/imath] and then tried to analyze a function for [imath]x > 1[/imath] and [imath]x < 0[/imath]. Thank you for your responses.
|
838857
|
[imath]2^x + 3^x - 4^x + 6^x - 9^x ≤ 1[/imath] [imath]\forall x \in R[/imath]
How can i prove that [imath]2^x + 3^x - 4^x + 6^x - 9^x ≤ 1[/imath] [imath]\forall x \in R[/imath]. I tried [imath]log(2^x + 3^x - 4^x + 6^x - 9^x) = log (1296^x) = x log(1296)[/imath] i don't know if im correct here i stuck some help please
|
2086739
|
In an integral domain [imath]R[/imath] associates of an irreducible element are irreducible.
My attempt : Let [imath]p[/imath] be an irreducible element in [imath]R[/imath].Let [imath]q[/imath] be an associate of [imath]p[/imath].Then there exists a unit [imath]u \in R[/imath] such that [imath]q = pu[/imath].Now if [imath]q[/imath] were reducible then [imath]q[/imath] can be written as [imath]q = ab[/imath] where none of [imath]a[/imath] and [imath]b[/imath] are units in [imath]R[/imath].Then we have [imath]pu = ab[/imath] and so [imath]p = (u^{-1} a) b[/imath].Since [imath]p[/imath] is irreducible one of them should be a unit.Since [imath]b[/imath] is not a unit.So [imath]u^{-1} a[/imath] is a unit.Now I have stuck and can't proceed further. Please give me a hint to complete it. Thank you in advance.
|
835436
|
If [imath]p[/imath] is an irreducible element of an integral domain [imath]D[/imath], and if [imath]e[/imath] belongs to [imath]D^\ast[/imath] prove that [imath]ep[/imath] is also ireducible
If [imath]p[/imath] is an irreducible element of an integral domain [imath]D[/imath], and if [imath]e[/imath] belongs to [imath]D^\ast[/imath], prove that [imath]ep[/imath] is also ireducible . To me it seems so profound, but I cannot get the proof . I'm trying to say that it is not irreducible, so there exists [imath]a,b[/imath] so that [imath]ep=ab[/imath] where [imath]a,b[/imath] don't belong to [imath]D^\ast[/imath]....
|
2085955
|
Sum of a power series [imath]\sum_{n=0}^{\infty}n3^nx^n[/imath]
How to evaluate the sum of the series: [imath]\sum_{n=0}^{\infty}(n \cdot 3^n\cdot x^n)[/imath] I have tried with integration and derivation, but that doesn't lead to any intuitive (known) series.
|
377277
|
Calculating the sum of [imath]f(x) = \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^n}[/imath]
In Calculus, how do I calculate this sum? [imath]f(x) = \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^n}[/imath] This is what I did so far: [imath] f(x) = 2x \cdot \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^{n-1}} [/imath] Therefore: [imath] \frac{\int{f(x)}}{2x} = \sum_{n=0}^{\infty} {2^n \cdot x^n}[/imath] But I have no idea where to continue from here!
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2086954
|
Hints for [imath]\lim_{z \to \infty} \frac{\operatorname{Re} f(z)}{z}=0 [/imath] implies [imath]f[/imath] is constant.
[imath]\lim_{z \to \infty} \frac{\operatorname{Re} f(z)}{z}=0 [/imath] implies [imath]f[/imath] is constant. [imath]f[/imath] is entire function on complex plane I tried to bound some derivative of [imath]e^{f(z)}[/imath] by cauchy integral inequality hence [imath]e^{f(z)}[/imath] is polynomial. But since it has exponential term depending on radius, and denominator has [imath]R^n[/imath] for radius [imath]R[/imath] so i can not bound it. Should I try some other way?
|
528517
|
Hints for a complex limit: Prove if [imath]\lim_{z \to \infty} f(z)/z = 0[/imath] then [imath]f(z)[/imath] is constant.
(To clarify, I would just like a hint. Please do not give me the answer to this problem. ) The solution to the following problem has really evaded me here: Problem: Assume that [imath]f[/imath] is entire and that [imath]\lim_{z \to \infty} f(z)/z = 0.[/imath] Prove that [imath]f(z)[/imath] is constant. My Thoughts and Work So Far: We know that proving that [imath]f'(z) = 0[/imath] or that for some fixed [imath]c \in \mathbb{C}[/imath], [imath]f(z) = c[/imath] for all [imath]z\in \mathbb{C}[/imath]. My first approach was to use Liouville's Theorem; If I could could show that [imath]f[/imath] is bounded then I am done. Since [imath]\lim_{z \to \infty} f(z)/z = 0[/imath], for all [imath]\varepsilon > 0[/imath] there exists a [imath]N \in \mathbb{C}[/imath] so large that if [imath]z \geq N[/imath] then [imath]|f(z)/z| \leq \varepsilon[/imath]. Thus, if [imath]C_R[/imath] is the circle of radius [imath]R[/imath] centered at the point [imath]z[/imath], then, as long as z is large enough by Cauchy's Inequality [imath] |f'(z)| \leq \frac{1}{2\pi i} \oint_{C_R} \frac{f(\zeta)}{(\zeta - z)^2}\ d\zeta \leq \bigg | \frac{1}{2\pi i} \bigg | \oint_{C_R} \bigg | \frac{f(\zeta)}{(\zeta - z)^2} \bigg | d \zeta \leq \frac{1}{2\pi} \frac{|\zeta|\varepsilon 2\pi R}{R^2} = \frac{|\zeta|\varepsilon}{R}. [/imath] Now taking the limit as [imath]R \to \infty[/imath] (which to me says, "let our circle around our point z dilate to an infinite radius so that it covers all of [imath]\mathbb{C}[/imath]) [imath]|f'(z)| \leq \lim_{R \to \infty} \frac{|\zeta|\varepsilon}{R} = 0.[/imath] Thus [imath]f'(z) = 0[/imath] and [imath]z[/imath] was arbitrary, so [imath]f[/imath] must be constant. Why I Think Im Wrong: I say [imath]z[/imath] was arbitrary, but really it is "any [imath]z \geq N[/imath]" which really isn't all that arbitrary. This is where I am stuck. Am I right, wrong, close, or totally lost? Any hints would be great. Edit: I am very sorry but I accidentally posted this before I was done typing the problem.
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2087276
|
Sets of equal cardinality
how to show that sets [imath]\{0,1\}^{\mathbb{N}}[/imath] and [imath]\mathbb{R}^{\mathbb{N}}[/imath] have an equal cardinality? I tried to use Cantor-Bernstein theorem but it seems to be hard.
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413836
|
cardinality of all real sequences
I was wondering what the cardinality of the set of all real sequences is. A random search through this site says that it is equal to the cardinality of the real numbers. This is very surprising to me, since the cardinality of all rational sequences is the same as the cardinality of reals, and it seemed fairly intuitive to me that if cardinality of a set [imath]A[/imath] is strictly greater than the cardinality of the set [imath]B[/imath], then cardinality of [imath]A^{\mathbb{N}}[/imath] should be strictly greater than cardinality of [imath]B^{\mathbb{N}}[/imath]. It turns out to be false. Some technical answers have appeared in this forum elsewhere but I do not understand them. As I am not an expert in this topic, could some one explain me in simple terms why this is happening? Also is the cardinality of all functions from reals to reals also the same as the cardinality of reals?
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2085123
|
Can every Banach space be given an inner product homeomorphically?
Let [imath](X,||.||)[/imath] be a real Banach space , then is it true that there exists a norm [imath]||.||_1[/imath] on [imath]X[/imath] coming from an inner product such that [imath]||.|| , ||.||_1[/imath] generates same topology on [imath]X[/imath] or atleast [imath](X,||.||)[/imath] and [imath](X,||.||_1)[/imath] are homeomorphic ?
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1312845
|
Showing that two Banach spaces are homeomorphic when their dimensions are equal.
Let [imath]X[/imath] and [imath]Y[/imath] be Banach spaces. It is quite easy to show that they are homeomorphic when their dimensions are finite and equal. However, I find it difficult to show that they are homeomorphic when their dimensions are infinite and equal. Here, the statement that their dimensions are equal must mean that a basis of [imath]X[/imath] and a basis of [imath]Y[/imath] have the same cardinality. (I know that any vector space has a basis by Zorn's Lemma.) Could anyone help me how to show that they are homeomorphic when their dimensions are infinite?
|
1381994
|
[imath]A,B[/imath] are two real positive matrices then [imath]\det (A+B) > \max(\det A , \det B)[/imath]
Let [imath]A,B[/imath] two square-real-positive matrices. Prove that [imath]\det (A+B) > \max(\det A , \det B)[/imath] So I found this solution: https://math.stackexchange.com/a/41478/160028 Basically, if [imath]A=I_n[/imath] and [imath]B[/imath] is diagonal then the proof is immediate. Now, I know that if [imath]M[/imath] is positive-symmetric then: [imath]M[/imath] is conjugate to [imath]I_n[/imath]. [imath]M[/imath] is conjugate to [imath]\text{Diag}(c_1,\ldots,c_n)[/imath] where [imath]c_i > 0[/imath]. but as far as I understand it doesn't have to be the same [imath]P[/imath]. Anyhow, how do I utilize it in order to prove the inequality? Thanks.
|
41463
|
[imath]A,B\in M_{n}(\mathbb{R})[/imath] so that [imath]A>0, B>0[/imath], prove that [imath]\det (A+B)>\max (\det(A), \det(B))[/imath]
Let [imath]A,B\in M_{n}(\mathbb{R})[/imath] be symmetric, with [imath]A>0[/imath] and [imath]B>0[/imath]. I need to prove that [imath]\det (A+B)>\max (\det(A), \det(B))[/imath]. I want to use Sylvester theorem of having a matrix [imath]D[/imath] so that [imath]D=\operatorname{diag}(1,1,\ldots, 1,-1,-1, \ldots-1,0,0,\ldots,0)[/imath]. Do I need to use it? How do I use it here? Thank you.
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2087529
|
Given [imath]a_1=1, a_2=2, a_{n+1}=n(a_n+a_{n-1})[/imath] find the general term
Given recursively defined sequence [imath](a_n)[/imath]: [imath]a_1=1[/imath] [imath]a_2=2[/imath] [imath]a_{n+1}=n(a_n+a_{n-1}), n\geq 2[/imath] Find the formula for the general term [imath]a_n[/imath]. This is what I did: So the first few terms are: [imath]a_1=1, a_2=2,a_3=6,a_4=24,a_5=120,...[/imath] I guess the general term can be written as [imath]n![/imath]. Let's prove it by mathematical induction: Base case: [imath]a_1=1!=1[/imath] so it's true for [imath]n=1[/imath]. Let's assume it's true for some [imath]n[/imath],i.e. [imath]a_n=n![/imath]. Then [imath]a_{n+1}=n(a_n+a_{n-1})=n(n!+(n-1)!)=n(n(n-1)!+(n-1)!)=n(n-1)!(n+1)=(n+1)n(n-1)!=(n+1)![/imath] Using the principle of mathematical induction we've proved that the general term of this sequence is [imath]n![/imath]. Is this proof valid? I'm not sure whether it's allowed to put [imath]a_{n-1}=(n-1)![/imath] here.
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964136
|
Finding [imath]a_n[/imath] if [imath]a_0=a_1=1,a_{n+1}=n(a_n+a_{n-1})\ \ (n\ge 1).[/imath]
The problem states: Suppose [imath]a_0,a_1,a_2,...[/imath] is a sequence such that [imath]a_0=a_1=1,\ \ \ a_{n+1}=n(a_n+a_{n-1})\ \ \ (n\ge 1).[/imath] Guess a formula for [imath]a_n[/imath], valid for [imath]n\ge 1[/imath], and use mathematical induction to prove that your guess is correct. I used this equation (and check with WolframAlpha) to find the results of a few different cases. [imath]a_2=2,\ a_3=6,\ a_4=24,\ a_5=120.[/imath] Unfortunately, I'm not really sure how I can prove this. I would appreciate any help in solving this!
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2087704
|
Factoring [imath]n[/imath] when [imath]\phi(n)[/imath] is given?
I am stuck on a problem where I am given the value of [imath]\phi(n)=31615577098574867424[/imath] for [imath]n=pq=31615577110997599711[/imath] where [imath]p[/imath] and [imath]q[/imath] are prime numbers. Now, I need to find the values of [imath]p[/imath] and [imath]q[/imath] but I am not able to get an idea as to how I can solve it. I know that [imath]\phi(n)[/imath] gives the number of relative prime to a given number [imath]n[/imath] but I can use Euler-totient function to factorize [imath]n[/imath]?
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222125
|
Prime factorization knowing n and Euler's function
My task is to factorize [imath]n[/imath] knowing it has two factors and [imath]\varphi(n)[/imath]. How can I do it? Thanks for any advice.
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2087936
|
Bounded sequence. What's the best constant [imath]c[/imath]?
Let [imath](x_n)_{n \geq 1}[/imath] be a sequence with [imath]x_1=1[/imath], defined by \begin{align*} x_{n+1}=x_n+\frac{1}{3x_n^2} \end{align*} Prove that there exists a fixed [imath]c>0[/imath] such that [imath]x_n<\sqrt[3]{n+c}[/imath] I cubed the relation in order to prove that [imath]x_n \geq \sqrt[3]{n}[/imath], and I guess that [imath]c[/imath] may be [imath]1[/imath], but I couldn't prove it.
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2078405
|
Sequence problem involving inequalities
I am triyng to solve this interesting problem, but I have no idea. I tried induction, but it didn't work. Let [imath](x_n)_{n \ge 1}[/imath] be a sequence, where [imath]x_1=1[/imath] and defined as follows: [imath]x_{n+1}=x_n+ \frac {1} {3x_n^2},[/imath] for every [imath]n \ge 1[/imath]. Show that there exists a real number [imath]c \gt 0[/imath] so as [imath]x_n \lt \sqrt[3] {n+c}.[/imath]
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1921161
|
A problem on rank of [imath]A[/imath] and [imath]\operatorname{adj}(A)[/imath]
Let [imath]A \in M_n(\mathbf{C})[/imath] and let [imath]B = \operatorname{adj}(A)[/imath] be the adjugate matrix. a) Prove that if [imath]A[/imath] is invertible, then so is [imath]B[/imath]. b) Prove that if [imath]A[/imath] has rank [imath]n-1[/imath], then [imath]B[/imath] has rank [imath]1[/imath]. c) Prove that if [imath]A[/imath] has rank at most [imath]n-2[/imath], then [imath]B = O_n[/imath]. a) We have that [imath]A[/imath] is invertible, so [imath]\det(A) \neq 0[/imath]. The rank of [imath]A[/imath] is [imath]n[/imath]. Also [imath] A \operatorname{adj}(A) = \operatorname{adj}(A) A = \det(A) I. [/imath] But what next? Please help me to solve the problem using elementary theory of matrix algebra (also please do not use the rank–nullity theorem). EDIT: b) Clearly [imath]\det(A) = 0[/imath]. Then [imath] A \operatorname{adj}(A) = \operatorname{adj}(A) A = \det(A) I = 0, [/imath] i.e. [imath]A \operatorname{adj}(A) = 0[/imath]. But how does this imply that [imath]\operatorname{adj}(A)[/imath] has rank [imath]1[/imath]? May I request for any alternative solution of the problem? Can we conclude the result from [imath]A \operatorname{adj}(A) = 0[/imath]?
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2964917
|
Why is the adjugate matrix the null matrix?
I struggle to understand why if [imath]A\in M_n(\mathbb{C}) [/imath], [imath]\det A=0[/imath] and [imath]rank A \le n-2[/imath], then [imath]A^*=O_n[/imath]. Could you please tell me why this claim holds? My textbook offers no proof for this.
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2085977
|
The limit [imath]\lim_\limits{n\to{\infty}}\int_0^{1}\sqrt[n]{x^n+(1-x)^n} dx[/imath]
This is Problem 11941 in the American Mathematical Monthly. What can we say about the limit [imath]\lim_\limits{n\to{\infty}}\int_0^{1}\sqrt[n]{x^n+(1-x)^n} dx\ ?[/imath] I think we need to use gamma and beta functions here. Again, the result may involve the Euler's constant.
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2079437
|
Limit question related to integration
Find the limit, [imath]L=\lim_{n\to \infty}\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx[/imath] My try: [imath] \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}xdx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}(1-x)dx< \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx< \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}(1-x)dx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}xdx[/imath] Now taking the limit I get that, [imath]\frac{1}{4}<L<\frac{3}{4}[/imath] But, how can I get the exact answer!! This is Problem 11941 from the American Mathematical Monthly.
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2083217
|
The sum of series with natural logarithm: [imath]\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)[/imath]
Calculate the sum of series: [imath]\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)[/imath] I tried to spread this logarithm, but I'm not seeing any method for this exercise.
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1676164
|
Prove that [imath]\sum_{n=1}^\infty \ln\left(\frac{n(n+2)}{(n+1)^2}\right)[/imath] converges and find its sum
Since this is not a geometric series, I know that I should use the definition of a convergent series, so [imath]S_n = \sum_{i=1}^n \ln\left(\frac{i(i+2)}{(i+1)^2}\right)[/imath] After this, I tried two different ways: 1) I simplified the fraction to read [imath]\frac{\ln(i^2+2i)}{i^2+2i+1}[/imath] and then I used long division to get [imath]\ln((1)-\frac{1}{(i+1)^2})[/imath] However, once I start plugging in i starting at 1, I don't know where to go from there. 2) I simplified the equation to read [imath]\ln(i^2+2i)-\ln(i^2+2i+1)[/imath], but again, once I start plugging in i starting at 1, I don't know where to go from there. What can I do? Note: Originally the series was presented as [imath]\sum_{n=1}^\infty \frac{\ln(n(n+2))}{(n+1)^2}[/imath] which is what some of the answers below are adressing.
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2088671
|
the question about the closed nest
For each [imath]\xi>0[/imath], [imath]S+\xi B[/imath] is a closed bounded set, where [imath]S[/imath] is a closed bounded subset of [imath]int ~dom f[/imath], [imath]B[/imath] is the unit ball, [imath]domf[/imath] denotes the effective domain of the proper convex function [imath]f[/imath]. My question is that why the nest of sets [imath] ( S+\xi B ) \cap (int~ dom f)^{C}[/imath] is empty.
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1864620
|
Prove that a nest of sets has an empty intersection
Let [imath]f[/imath] be a real convex function and [imath]S[/imath] an arbitrary closed bounded subset of the relative interior of the effective domain of [imath]f[/imath]. Let [imath]B[/imath] be a closed Euclidean unit ball. The nest of sets [imath](S + \varepsilon B) \cap (\mathbb{R}^n \backslash \text{int}(\text{dom} f)), \quad \varepsilon > 0[/imath] has an empty intersection. Why? Is it because [imath]\text{cl}S=\bigcap \{S + \varepsilon B \big | \varepsilon > 0\}[/imath] and [imath]S \subset \text{int}(\text{dom} f))[/imath]?
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2088251
|
What is [imath]\lim_{x \rightarrow 0} x\log(x)[/imath]?
Is it true that: [imath]\lim_{x \rightarrow 0} x\log_{n}(x) = 0 \quad ; \quad n \in \mathbb{R}[/imath]
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303976
|
How do I calculate [imath]\lim_{x\rightarrow 0} x\ln x[/imath]
I was thinking about the reasons behind [imath]0^0=1[/imath] and I remember one of my friends studying math arguing about the continuity of the function [imath]x^x[/imath] in [imath]0[/imath]. But when I write as [imath]x^x=e^{x\ln x}[/imath] I am now looking at [imath]\lim_{x\rightarrow 0} x\ln x[/imath] Graphically I can see in Mathematica that it goes to [imath]0.[/imath] But I can't calculate by using a Taylor expansion, because I can't expand log around [imath]0[/imath]. How do you prove that?
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2088556
|
Power set P(S) = 2^S
Let [imath]\mathcal{P}(S)[/imath] denote the power set of [imath]S[/imath]. Why is [imath]\mathcal{P}(S) = 2^S[/imath]? I know that the power set of [imath]S = \{x,y,z\}[/imath] is [imath]\mathcal{P}(S) = \{\{\}, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\},[/imath] so how does this relate to [imath]\mathcal{P}(S) = 2^S[/imath]? I cannot understand why [imath]\mathcal{P}(S) = \mathcal{P}(\{x,y,z\}) = 2^{\{x,y,z\}} = \{\{\}, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\},[/imath] I assume it is somehow related tot he fact that the set [imath]\{x,y,z\}[/imath] has 3 elements, and [imath]\mathcal{P}(\{x,y,z\})[/imath] is a set of [imath]2^3 = 8[/imath] sets.
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1115767
|
Power Set Explanation
The following is from the Wikipedia page on the Power set: By identifying a function in [imath]2^S[/imath] with the corresponding preimage of [imath]1[/imath], we see that there is a bijection between [imath]2^S[/imath] and [imath]\mathcal P(S)[/imath], where each function is the characteristic function of the subset in [imath]\mathcal P(S)[/imath] with which it is identified. I don't understand the start of the sentence,"By identifying a function in [imath]2^S[/imath] with the corresponding preimage of [imath]1[/imath]". I can make out some sense out of the rest. Please help.
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2087611
|
Let [imath]\mathbb{F}[x][/imath] be the ring polynomials in one variable [imath]x[/imath] over a field [imath]\mathbb{F}[/imath] with the relation [imath]x^n=0[/imath]
Let [imath]\mathbb{F}[x][/imath] be the ring polynomials in one variable [imath]x[/imath] over a field [imath]\mathbb{F}[/imath] with the relation [imath]x^n=0[/imath],for a fixed [imath]n[/imath],is a natural number.Then What is dimension of [imath]\mathbb{F}[x][/imath] over [imath]\mathbb{F}[/imath]? a) [imath]1[/imath] b) [imath]n-1[/imath] c) [imath]n[/imath] d) [imath]\infty[/imath]
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199694
|
The Uniqueness of a Coset of [imath]R[x]/\langle f\rangle[/imath] where [imath]f[/imath] is a Polynomial of Degree [imath]d[/imath] in [imath]R[x][/imath]
Suppose [imath]R[/imath] is a field and [imath]f[/imath] is a polynomial of degree [imath]d[/imath] in [imath]R[x][/imath]. How do you show that each coset in [imath]R[x]/\langle f\rangle[/imath] may be represented by a unique polynomial of degree less than [imath]d[/imath]? Secondly, if [imath]R[/imath] is finite with [imath]n[/imath] elements, how do you show that [imath]R[x]/I[/imath] has exactly [imath]n^d[/imath] cosets?
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2088940
|
Show that [imath]y^2=x^3+23[/imath] has no integer solutions (hint: use [imath]y^2+4[/imath])
I saw a previous proof on this site that addresses this question, but the proof leaves off with a number of form [imath]4k-1[/imath] can't divide [imath]t^2+1[/imath]. Why is this? Thanks.
|
579928
|
Find all solutions of the equality [imath]y^2=x^3+23[/imath] for integers [imath]x,y[/imath]
Find all solutions of the equality [imath]y^2=x^3+23[/imath] for integers [imath]x,y[/imath] I guess, that x cannot be even. because we can apply mod 4 test say [imath]x=2k[/imath] then [imath]y^2=8k^3+23\equiv3\mod(4)[/imath] but, this is not possible, since the square of an integer is congruent to [imath]0[/imath] or [imath]1[/imath] [imath]\mod (4)[/imath] so [imath]x[/imath] is of the form [imath]4k+3[/imath] or [imath]4k+1[/imath] [imath]x=4k+3[/imath] is also not possible for the same reason but for the last case the test fails. any hints ?
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2088694
|
How to calculate dice roll probability with more dice?
I'm no math graduate, so be patient. For two dice it is easy, because of the small number of possibilities, and it’s still easy for three, but how can I work out the case for ten (with some formula)? How to calculate probability that: a) Rolling three dice, the sum of them is greater than [imath]$8$[/imath]. b) The same, but with more dice, let's say [imath]$10$[/imath], where the sum has to be greater than [imath]$27$[/imath]. Can you explain it and can you also send some useful resources?
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2068036
|
Probability that [imath]X \ge Y[/imath] given N rolls on a d20
Probability that [imath]X \ge Y[/imath] given N rolls on a [imath]d20[/imath] Example: Originally, we wanted to roll a [imath]13[/imath] or higher [imath]7[/imath] times in a row. In and attempt to improve the odds and make it easier to understand we changed it to be [imath]7(N)[/imath] [imath]d20s[/imath] where one must be greater than or equal to [imath]91(Y)[/imath]. So, what is the probability that [imath]X[/imath](sum of rolls) is greater than or equal to [imath]Y(91)[/imath] given [imath]N(7 rolls)?[/imath] We think the answer is [imath]0.16%[/imath] or [imath]1/625[/imath] because in order to achieve the minimum of [imath]91[/imath], one must roll at least a [imath]13[/imath] on every roll. What we're wondering is does the fact that it's now cumulative change the probability? Thanks for your time.
|
2089028
|
What is the sum of [imath]\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(2n+1)}{3^{n}}[/imath]
What I got figured out is that I should pair up [imath]\frac{(-1)^{n}}{3^{n}}[/imath] giving me [imath]\sum_{n=1}^{\infty} (\frac{1}{3})^{n}(-2n-1)[/imath]. I know that the first part converges to [imath]\frac{3}{4}[/imath], but the other part is divergent. How should I proceed?
|
56677
|
Find the sum for [imath]\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{2n+1}{3^n} [/imath]
I have been trying to find the sum [imath]\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{2n+1}{3^n} [/imath]. After some calculation, I got here: [imath]\frac{-6}{8}+\frac{1}{4}+8\sum\frac{k}{9^k}[/imath]. I know the result is [imath]\frac{5}{8}[/imath] , and I verified it with Wolfram Alpha.I saw that [imath]\sum_{k=1}^{\infty}\frac{k}{9^k}=\frac{9}{64}[/imath]. But I don't know how to prove the last equation: [imath]\sum_{k=1}^{\infty}\frac{k}{9^k}=\frac{9}{64}[/imath]. I hope someone could help me or show me another method to find the sum for my initial series.
|
2088712
|
Number theoretic olympiad question involving a diophantine equation
I don't know the exact source of this question, but here it goes: Find all ordered pairs [imath](x,y)[/imath] such that [imath]x,y \in \Bbb{N}[/imath] and satisfy the relation: [imath]y^3=x^2+2[/imath] Of course [imath](x,y)=(5,3)[/imath] is a solution but proving that infinitely many exist or none other exist is the part I'm stuck at. I tried to work an infinite descent with congruent modulo but due to difference in degree one coefficient always vanishes. I am not that good at olympiad number theory, so do help me out here.
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1863227
|
Finding integer solutions to [imath]y^2=x^3-2[/imath]
I have the equation: [imath]y^2=x^3-2[/imath] It seems to be deceivingly simple, yet I simply cannot crack it. It is obviously equivalent to finding a perfect cube that is two more than a perfect square, and a brute force check shows no solutions other than [imath]y=5[/imath] and [imath]x=3[/imath] under 10,000. However, I can't prove it. Are there other integer solutions to this equation? If so, how many? If not, can you prove that there aren't? Bonus: What about the more general equation" [imath]y^2=x^3-c[/imath] Where [imath]c[/imath] is a positive integer?
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2089248
|
Prove that if [imath]n\ge4[/imath] then [imath]n,n+2,n+4[/imath] cannot all be primes.
My question is well written in title. I thought that something like divisibility by [imath]3[/imath] will work but it is not. Please help.
|
1653536
|
Show that we cannot have a prime triplet of the form [imath]p[/imath], [imath]p + 2[/imath], [imath]p + 4[/imath] for [imath]p >3[/imath]
Show that we cannot have a prime triplet of the form [imath]p[/imath], [imath]p + 2[/imath], [imath]p + 4[/imath] for [imath]p >3[/imath] I was a bit lost with this proof until I found a similar looking proof-based question from a previous homework assignment in this class which said "If [imath]a[/imath] is an integer, prove that one of the numbers [imath]a[/imath], [imath]a + 2[/imath], [imath]a + 4[/imath] is divisible by 3. These problems seem very similar to me and would lead me to assume that they would be proven similarly, although I am unsure as to how I would approach this for prime triplets as I am brand new to them. Any help is appreciated
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546721
|
Proving that [imath]f(x,y) = \frac{xy^2}{x^2 + y^2}[/imath] with [imath]f(0,0)=0[/imath] is a continuous function using epsilon-delta.
THE QUESTION: Use the metric [imath](x,y)[/imath] = [imath]\rho(x,y)=|x-y|[/imath] for the reals and use the metric [imath]\rho((x_1,y_1),(x_2,y_2)) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}[/imath] for the plane. Define [imath]f:R\times R \to R[/imath] as [imath]f(x,y)=\frac{xy^2}{(x^2 + y^2)}[/imath] for [imath]f(x,y)[/imath] [imath]\neq[/imath] [imath](0,0)[/imath] and set [imath]f(0,0) = 0[/imath]. Determine whether [imath]f[/imath] is continuous using [imath]\epsilon[/imath]-[imath]\delta[/imath] proof. WHAT I'VE DONE: I've tried using a direct approach by using the definition of a continuous function: [imath]f:R\times R \to R[/imath] is continuous at [imath](x_1,y_1) \in R \times R[/imath], [imath]\forall \epsilon>0, \exists \delta >0[/imath] such that [imath]\forall (x_2,y_2) \in R \times R[/imath] if [imath]\sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2 } <\delta[/imath], then [imath]|f(x_1,y_1)-f(x_2,y_2)|<\epsilon[/imath] However, with the [imath]f(x,y)[/imath] as defined above, it seems impossible with the messy algebra since I want to find a [imath]\delta[/imath] that satisfies [imath]|\frac{x_1y_1^2}{(x_1^2 + y_1^2)}-\frac{x_2y_2^2}{(x_2^2 + y_2^2)}| < \epsilon[/imath] using the metrics defined above. Is there a trick I am supposed to be seeing? Would really appreciate if anyone could show me the proof for this question. Thank you!
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2088999
|
For which [imath]a\in \mathbb{R}[/imath] is [imath]f(x,y) = \frac{xy^a}{x^2+y^2}[/imath] with [imath]f(0,0)=0[/imath] continuous?
My function is defined as: [imath]f(x,y) = \begin{cases} \dfrac{xy^a}{x^2+y^2}&(x,y) \neq (0,0)\\ \ 0&(x,y) = (0,0) \end{cases} [/imath] I want to find out for which [imath]a\in \mathbb{R}[/imath] the function is continuous. I am not quite sure how to solve something like this. I know that for [imath]a=1[/imath] the function is definitely not continuous, which can be checked, since [imath](\frac1n,\frac1n)[/imath] converges to [imath](0,0)[/imath] but [imath]f(\frac1n,\frac1n)[/imath] does not. Now if I want [imath]f[/imath] to be steady in [imath](0,0)[/imath], I figured that this is the case if [imath]a>1[/imath]. But this is where I get stuck. How to I prove or check, if for [imath]a>1[/imath] the rest of the function is continuous as well? Any help is greatly appreciated!
|
1197246
|
Subspace of metrizable and separable space is separable
I need to show (using the fact that for metrizable space: space is separable [imath]\iff[/imath] it has got a countable base) that if [imath]X[/imath] is metrizable and separable, then every subspace [imath]Y \subset X [/imath] is separable. My idea: If [imath]X[/imath] is separable (and is metrizable), then [imath]X[/imath] has got a countable base. Let's call it [imath]\mathcal{B}[/imath]. Let [imath]U=\bigcup_{i \in I} Bi[/imath] such that [imath]B_i \in \mathcal{B}[/imath]. So [imath]U[/imath] is open in [imath]X[/imath]. Then [imath]U\cap Y[/imath] is open in [imath]Y[/imath]. So [imath]U \cap Y=\bigcup_{i \in I} Bi \cap Y[/imath] such that [imath]B_i \in \mathcal{B}[/imath]. So I want to say that [imath]\mathcal{C}=\{B \cap Y: B\in\mathcal{B}\}[/imath] is the base of [imath]Y[/imath]. And becasue [imath]\mathcal{B}[/imath] is countable, then [imath]\mathcal{C}[/imath] is also countable. Then [imath]Y[/imath] is also separable. Is it a correct proof (if not how can I improve it)? How should I show that [imath]\mathcal{C}[/imath] is really a base?
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1628253
|
Subspace of a separable space is separable
Let [imath](X,d)[/imath] be a separable space and [imath]Y \subset X[/imath]. Show that [imath](Y,d)[/imath] is also separable. My approach is as follows: Let [imath](X,d)[/imath] be a separable space and [imath]Y \subset X[/imath]. Since [imath]X[/imath] is separable, by definition there exists a countable dense subset in [imath]X[/imath], call it [imath]K[/imath]. We want to show that [imath]K \subset Y[/imath]. Is this a valid approach to take to the proof? P.S. This is an analysis course, not strictly a topology course.
|
2088643
|
Show matrix A is invertible if and only if [imath]a_0 \neq 0[/imath] with [imath]A^n+a_{n-1}A^{n-1}+....+a_1A+a_0I=O[/imath].
Let [imath]A[/imath] be a real square matrix en let [imath]I[/imath] be the identity matrix with the same size. Let [imath]n>0[/imath] be the smallest natural number for which there exist [imath]a_0,a_1,...a_{n-1} \in {R}[/imath] such that [imath]A^n+a_{n-1}A^{n-1}+....+a_1A+a_0I=O[/imath]. Show that A is invertible if and only if [imath]a_0\neq0 [/imath]. So i need to proof [imath]A[/imath] invertible [imath]\rightarrow[/imath][imath]a_0 \neq 0[/imath] [imath]a_0 \neq0 \rightarrow[/imath] [imath]A[/imath] invertible I don't really know where to start with 1, and with 2 I tried to prove it with a contradiction but I don't think it's right either.
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252733
|
Minimal polynomial, determinants and invertibility
I need to prove the following: Let [imath]A[/imath] be a square matrix over a field. If the matrix [imath]$A$[/imath] is invertible, then the minimal polynomial [imath]m_A[/imath] satisfies [imath]m_A(0) \neq 0[/imath]. There is one definition I am unsure of or need help making more clear. I will proceed with proof by contraposition: We must show that if [imath]m_A(0) = 0[/imath] then [imath]A[/imath] is not invertible. By definition of minimum polynomial of [imath]A[/imath] we have: [imath]m_A(x) = x^r - \lambda_{r-1} x^{r-1} - \ldots - \lambda_1 x + \det(A)[/imath]. Not sure about the determinant term here So, [imath]m_A(0) = \det(A) = 0[/imath]. We know [imath]\det(A) = 0 \implies A[/imath] is not invertible.
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205630
|
Evaluating the limit of multivariable equation [imath]\frac{x^2y}{x^2+y^2}[/imath]
[imath] \lim_{(x,y)\to(0,0)}{\frac{x^2y}{x^2+y^2}} [/imath] If you substitute in [imath](0,0)[/imath] for [imath]x[/imath] and [imath]y[/imath]. It becomes [imath] \frac{0}{0} [/imath] usually you would apply the L'Hopital's rule. However, I think it is not possible with the multivariable equation. If I cannot apply the L'Hopital's rule here, what should be my first step in solving this limit?
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899256
|
Finding multivariable limits for the function [imath]\frac{3x^2y}{x^2+y^2}[/imath]
Could anyone help me with this Find [imath]\lim_{(x,y) \to (0,0)} \frac{3x^2y}{x^2+y^2}[/imath] if this limit exists I tried using the squeeze theorem, but I could not find a suitable expression for the squeeze theorem as I could not find any number that allow me to squeeze the limit in between and cancel the denominator
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2088887
|
convergence of [imath]\sum_{n=1}^{\infty }\frac{(-1)^{[\sqrt{n}]}}{n}[/imath] where [imath][a][/imath] means floor function of [imath]a[/imath]
I have a problem with solving the convergence of [imath]\sum_{n=1}^{\infty }\frac{(-1)^{[\sqrt{n}]}}{n}[/imath] where [imath][a][/imath] means floor function of [imath]a[/imath]. I've already made the first thep and rewrote it with the [imath]\sum_{n=1}^{\infty }(-1)^{n}\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{k}[/imath], which converges if and only if the previous one converges. Now I want to use Leibniz criterion, but I cannot prove that the second sum in the previous expression is monotonic. Does anyone know how to do it? (I cannot use Riemann integral, we haven't defined it yet)
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535784
|
How find series [imath]\sum_{n=1}^{\infty}\frac{(-1)^{[\sqrt[m]{n}]}}{n^a}[/imath]?
let [imath]m[/imath] is give a positive integers, Determine for which values of [imath]a[/imath],the series [imath]\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}[/imath] converges where [imath][x][/imath] is the largest integer not greater than [imath]x[/imath]. and I have see this not hard problem :show that [imath]\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}[/imath] converges, solution: note [imath]\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}=\sum_{n=1}^{\infty}(-1)^n\left(\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+2n}\right)[/imath] and we have let [imath]a_{n}=\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+2n}[/imath] [imath]0<a_{n}<\dfrac{2n+1}{n^2}<\dfrac{3}{n}\longrightarrow 0[/imath] and [imath]a_{n}-a_{n+1}=\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}+\dfrac{1}{n^2+1}-\dfrac{1}{(n+1)^2+1}+\cdots+\dfrac{1}{n^2+2n}-\dfrac{1}{(n+1)^2+2n}-\left(\dfrac{1}{(n+1)^2+2n+1}+\dfrac{1}{(n+1)^2+2(n+1)}\right)[/imath] so [imath]a_{n}-a_{n+1}=(2n+1)\left(\dfrac{1}{n^2(n+1)^2}+\cdots+\dfrac{1}{(n^2+2n)(n^2+4n+1)}\right)-\dfrac{1}{n^2+4n+2}-\dfrac{1}{n^2+4n+3}[/imath] so [imath]a_{n}-a_{n+1}>\dfrac{(2n+1)^2}{(n^2+2n)(n^2+4n+1)}-\dfrac{2}{n^2+4n+1}=\dfrac{2n^2+1}{(n^2+2n)(n^2+4n+1)}>0[/imath] so [imath]\sum_{n=1}^{\infty}(-1)^na_{n}=\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}[/imath] converges But for my problem,I can't My try: [imath]\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}=\sum_{n=1}^{\infty}(-1)^n\sum_{k=n^m}^{(n+1)^m-1}\dfrac{1}{k^a}[/imath] let [imath]b_{n}=\sum_{k=n^m}^{(n+1)^m-1}\dfrac{1}{k^a}[/imath] then how prove [imath]b_{n}\longrightarrow 0[/imath]and [imath]b_{n}>b_{n+1}[/imath]
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2089464
|
Irreducibility of reciprocal polynomials
Given that [imath]f(x)=\sum\limits_{k=0}^n a_kx^k[/imath] is an irreducible polynomial, show that [imath]g(x)=\sum\limits_{k=0}^n a_kx^{n-k}[/imath] is irreducible too. I don't see how this specific coefficient exchange would preserve the irreducibility. Is there something else I should be looking at?
|
1758745
|
Prove that [imath]f(x)[/imath] is irreducible iff its reciprocal polynomial [imath]f^*(x)[/imath] is irreducible.
This is what I'm trying to prove: Let [imath]f(x)\in\mathbb{Q}[x][/imath] and [imath]\deg(f(x))>1[/imath]. Prove that [imath]f(x)[/imath] is irreducible in [imath]\mathbb{Q}[x][/imath] iff its reciprocal polynomial [imath]f^*(x)[/imath] is irreducible in [imath]\mathbb{Q}[x][/imath]. Note: The reciprocal polynomial [imath]f^*(x)=x^nf(1/x) \in \mathbb{Q}[x][/imath]. So my thoughts are to prove the contrapositive, i.e. [imath]f^*(x)[/imath] is reducible iff [imath]f(x)[/imath] is reducible. So to prove the ([imath]\Rightarrow[/imath]) direction I assume that [imath]f^*(x)=g(x)h(x)[/imath], so I get that [imath]g(x)h(x)=x^nf(1/x)[/imath] which implies [imath]f(1/x)=(1/x^n)g(x)h(x)[/imath]. I want to say this somehow makes [imath]f(x)[/imath] reducible but I am unable to proceed as [imath]f(1/x)\notin \mathbb{Q}[x] [/imath]. I thought about substituting [imath]y=1/x[/imath], which gives [imath]f(y)=y^ng(1/y)h(1/y)[/imath], but I am unable to show that this is irreducible in [imath]\mathbb{Q}[y][/imath]. If I could, then it would be irreducible in [imath]\mathbb{Q}[x][/imath] since [imath]\mathbb{Q}[y]\cong \mathbb{Q}[x][/imath] Any suggestions or hints will be appreciated.
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1046196
|
If [imath] f(f(f(x)))=x[/imath], does[imath] f(x)=x[/imath] necessarily follow?
Does [imath]f(f(f(x)))=x \implies f(x)=x[/imath]? Is it necessary for it to follow? How do we prove this? Do we have to substitute some special [imath]x[/imath]? Or is it some other consideration of its properties? Edit: does it follow in the special case when f is surjective? Or would we wind up with the same thing?
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2258564
|
[imath]f(f(f(x))) = x[/imath]. Prove or disprove that f is the identity function
Let [imath]f[/imath] be a continuous function on [imath]\mathbb R[/imath] satisfying the relation [imath]f(f(f(x))) = x\ \text{for all}\ x \in \mathbb R[/imath] Prove or disprove that [imath]f[/imath] is the identity function. I tried taking the derivative. From the derivative, I'm not sure about it, but I concluded it had to be of degree 1 if it is a polynomial since if it'd have been of degree 2 or higher.. there needed to be terms of [imath]x[/imath] in the derivative.. which are not there.
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2090133
|
How to prove that [imath]3^i5^j[/imath] can never be a perfect number?
I need to prove that [imath]3^i5^j[/imath] can never be a perfect number. How do I tackle this problem. I know that a perfect number is a positive number which is equal to the sum of its positive divisors excluding the number itself. But how do I prove that the number [imath]3^i5^j[/imath] is not perfect?
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1148588
|
Prove that [imath]p^j q^i[/imath] cannot be a perfect number for [imath]p, q[/imath] odd, distinct primes.
Define [imath]\sigma(m) = \sum[/imath] d : d|n. Prove that $p^j[imath]q^i$ cannot be a perfect number for $p, q$ odd, distinct primes.[/imath] Attempt at Solution: I have shown that $p^k$ can never be a perfect number, and im trying to use the multiplicative property of $\sigma$ to generalize to $p^jq^i$
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668155
|
Let [imath]n[/imath] be a positive integer. If [imath]2+2\sqrt{28n^2+1}[/imath] is an integer, then it is a perfect square.
Let [imath]n[/imath] be a positive integer. If [imath]2+2\sqrt{28n^2+1}[/imath] is an integer, then it is a perfect square. My work: [imath]2+2\sqrt{28n^2+1}=m \implies 4(28n^2+1)=m^2-4m+4 \implies m=2k[/imath] [imath]28n^2+1=k^2-2k+1 \implies 28n^2=k^2-2k \implies k=2q \implies 28n^2=4q^2-4q[/imath] [imath]7n^2=q^2-q \implies 7n^2=q(q-1)[/imath] Now, we have [imath]q[/imath] and [imath]q-1[/imath] to be co-prime. Here, two cases arises. (i) [imath]q=7x^2,q-1=y^2 \implies 7x^2-y^2=1[/imath]. This cannot occur because [imath]y^2 \equiv 6 \mod7[/imath] is not possible. (ii) [imath]q=x^2,q-1=7y^2[/imath]. Now, what am I supposed to do? Please help.
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2090665
|
Show that if [imath]2+2\sqrt{28n^2+1}[/imath] is an integer then it must be perfect square.
As written in title, I want to prove that If [imath]n[/imath] is an integer, show that if [imath]2+2\sqrt{28n^2+1}[/imath] is an integer than it must be perfect square. I m struggling in making a start . Please help.
|
2090819
|
Possible values for the determinant of [imath]A[/imath] given that [imath]A^3-A^2-3A+2I=0[/imath]
We have to determine as a function of [imath]n[/imath], the number of possible values for [imath]\det(A)[/imath] given that [imath]A[/imath] is an [imath]n\times n[/imath] real matrix with [imath]A^3-A^2-3A+2I=0[/imath]. I think the solution is just to notice that this polynomial has one real root [imath]a[/imath] and two complex roots [imath]b,\overline b[/imath]. Clearly all of the complex roots of [imath]A[/imath] must be [imath]0,a,b[/imath] or [imath]\overline b[/imath]. Since the determinant is the product of the complex roots with multiplicity we just have to find all the suitable combinations for the eigenvalue's multiplicities. The first option is when [imath]0[/imath] is an eigenvalue. Otherwise there are [imath]\lfloor\frac{n}{2}\rfloor[/imath] options for the multiplicity of [imath]a[/imath], and this determines the multiplicities of [imath]b[/imath] and [imath]\overline b[/imath] which must be equal. Hence the answer is [imath]\lfloor \frac{n}{2} \rfloor+1[/imath]. Is this correct? How would you make it fancier?
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1291006
|
Determine the number of possible values for [imath]\det(A)[/imath], given that [imath]A[/imath] is an [imath]n \times n[/imath] matrix with real entries such that [imath]A^3 - A^2 -3A +2I=0[/imath].
Determine the number of possible values for [imath]\det(A)[/imath], given that [imath]A[/imath] is an [imath]n \times n[/imath] matrix with real entries such that [imath]A^3 - A^2 -3A +2I=0[/imath]. here is the source of the problem. In the last comment, I don't understand why we can say for sure that [imath]A[/imath] has only three distinct eigenvalues. Can anyone explain to me? Remark: I understand that the roots of the equation [imath]x^3-x^2-3x+2=0[/imath] are eigenvalues of [imath]A[/imath]. My problem is why the matrix [imath]A[/imath] cannot have another eigenvalue?
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1310230
|
Does [imath]IJ=IK\implies J=K[/imath] always hold for integral domain and finitely generated nonzero ideal [imath]I[/imath]?
Let [imath]R[/imath] be a commutative integral domain, [imath]I,J,K[/imath] three ideals of [imath]R[/imath] with [imath]I\neq (0)[/imath] being finitely generated. Then does [imath]IJ=IK[/imath] imply [imath]J=K[/imath]? With Nakayama lemma, I can prove it if one of [imath]J[/imath] and [imath]K[/imath] equals to [imath]R[/imath]. And I also know it holds when [imath]R[/imath] is a Prüfer domain or [imath]I[/imath] is singly generated.
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2936798
|
Product of two different ideals by a third may be equal
It is well-known that, for a commutative ring [imath]A[/imath] which is not a PID, there may exist a triple of ideals, [imath]0\neq I, J_1, J_2[/imath] such that [imath]J_1\neq J_2[/imath] but [imath]IJ_1=IJ_2[/imath]. I would like to present an explicit example of this fenomenon to my students at a basic undergraduate course in commutative algebra. Do you have any suggestion? Essentially the unique examples they will understand will be about polinomial rings and matrix rings.
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2091029
|
Suppose [imath]f : [0,1]\mapsto R[/imath] is continuous, [imath]f[/imath] is differentiable on [imath](0,1) [/imath]and [imath]f(0)=0[/imath].
Assume [imath]|f'(x)|\le |f(x)|[/imath] for all [imath]x\in(0,1).[/imath] Prove that [imath]f(x)=0[/imath] for all [imath]x\in[0,1][/imath]. I thought maybe I could use the limit but since there is no equation it won't work.
|
509876
|
Prove [imath]f(x) = 0 [/imath]for all [imath]x \in [0, \infty)[/imath] when [imath]|f'(x)| \leq |f(x)|[/imath]
mathematicians! I want to ask to all wise people about a problem I met at the quiz to obtain some ideas. The problem is following. It may not be accurate since the problem is dependent on my memory Let [imath]f : [0, \infty) \rightarrow \mathbb{R}[/imath] and [imath]f[/imath] is differentiable. [imath]|f'(x) |\; \leq\ |f(x)|[/imath] for all [imath]x \in [0, \infty)[/imath]. [imath]f(0) = 0[/imath]. Then prove that [imath]f(x) = 0[/imath] for all [imath]x \in [0,\infty)[/imath]. At the quiz, I applied the Mean Value Theorem and the Cauchy-Schwarz inequality. For any [imath]x[/imath], [imath]|f(x)| = |\{f(x) - f(0)\}(x-0)| = |f'(c_{1})(x)|[/imath] for some [imath]c_{1} \in (0,x)[/imath], and then [imath]|f'(c_{1})(x)| \leq |f'(c_{1})||x| \leq |f(c_{1})||x|[/imath] by Cauchy-Schwarz and given assumption. By doing so consecutively, we can obtain the inequality that [imath] |f(x)| \leq |f(c_{n})||x||c_{1}|\ldots |c_{n-1}| [/imath] for any [imath]x \in [0,\infty)[/imath]. As [imath]\{c_{i}\}, \; i\in \mathbb{N}[/imath] is decreasing sequence and [imath]f[/imath] is continuous, [imath]|f(c_{n})|[/imath] can be arbitrary close to [imath]0[/imath] when [imath]c_{n} \rightarrow 0[/imath]. Thus the inequality above can be bounded by [imath]\epsilon[/imath]. I think my answer is right, but a bit uncertain. Could you give me a certainty? Thank you very much.
|
2091197
|
Is there a closed form formula for [imath]\sum\limits_{n=0}^\infty e^{-n^2}[/imath]?
I am curious about the value of [imath]\sum_{n=0}^\infty e^{-n^2}.[/imath] It comes to my mind by observing Gauss integral that it is equal to [imath]\int_{0}^\infty e^{-t^2}dt=\dfrac{\sqrt{\pi}}{2}.[/imath]
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174877
|
Evaluation of [imath]\sum_{x=0}^\infty e^{-x^2}[/imath]
Most of us are aware of the classic Gaussian Integral [imath]\int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}[/imath] I would be interested in evaluating the similar sum [imath]\sum_{x=0}^\infty e^{-x^2}[/imath] Now, because [imath]\exp(-\lfloor x \rfloor^2) \ge \exp(-x)[/imath], we find [imath]\sum_{x=0}^\infty e^{-x^2}= \int_0^\infty e^{-\lfloor x \rfloor^2}\, dx \ge \int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}[/imath] Does a closed form for this sum exist? If so, what would it be? I would be very interested in how a closed form would be found for this function.
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2091711
|
Characterization of the number [imath]e[/imath]
The number [imath]e[/imath] is characterized in calculus as: [imath]e:=\lim_{n\to\infty}\left(1+\frac1n\right)^n[/imath] My question is this: As [imath]n[/imath] heads to infinity, [imath]1/n[/imath] gets closer and closer to zero. So what we're really doing is to add [imath]1[/imath] to a number that's getting closer and closer to zero and then raising that expression to an increasingly large number. So, the limit as [imath]n[/imath] moves towards infinity is [imath]1[/imath]. Why is it [imath]e[/imath]?
|
550485
|
Why don't I get [imath]e[/imath] when I solve [imath]\lim_{n\to \infty}(1 + \frac{1}{n})^n[/imath]?
If I were given [imath]\lim_{n\to \infty}(1 + \frac{1}{n})^n[/imath], and asked to solve, I would do so as follows: [imath]\lim_{n\to \infty}(1 + \frac{1}{n})^n[/imath] [imath]=(1 + \frac{1}{\infty})^\infty[/imath] [imath]=(1 + 0)^\infty[/imath] [imath]=1^\infty[/imath] [imath]=1[/imath] I'm aware that this limit is meant to equal to [imath]e[/imath], and so I ask: why don't I get [imath]e[/imath] when I solve [imath]\lim_{n\to \infty}(1 + \frac{1}{n})^n[/imath]?
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2091804
|
Infinite probability of natural numbers.
Two natural numbers [imath]x[/imath] and [imath]y[/imath] are selected at random. Find the probability that [imath]x^2 + y^2[/imath] is divisible by [imath]10[/imath]. I have no clue how to do this.
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271437
|
Probability of two integers' square sum divisible by [imath]10[/imath]
Given two random integers [imath]a[/imath], [imath]b[/imath] calculate probability of [imath]a^2[/imath] + [imath]b^2[/imath] is divisible by [imath]10[/imath]. I've tried to simulate this process and got a result about [imath]0.18[/imath] (maybe incorrect), but have no idea why.
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2092046
|
Prove by contradiction that any prime number is either in the from of [imath]ab+1[/imath] or [imath]ab+5[/imath].
It says: Prove by contradiction that any prime number is either of the form of [imath]ab+1[/imath] or [imath]ab+5[/imath]. And this was all. But it seems both belong to [imath]\mathbb{N} \cup \{0\}[/imath]. Because otherwise it wouldn't hold for [imath]2[/imath] which is prime. What the problem says is to say: [imath](p\neq ab+1)\wedge (p\neq ab+5)\implies p\text{ is not prime}[/imath] I tried to show it would then be written as the product of two numbers and thus not prime. But up to what number should I do that? Stuck.
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756547
|
Prove that every prime larger than [imath]3[/imath] gives a remainder of [imath]1[/imath] or [imath]5[/imath] if divided by [imath]6[/imath]
Can we prove that every prime larger than [imath]3[/imath] gives a remainder of [imath]1[/imath] or [imath]5[/imath] if divided by [imath]6[/imath] and if so, which formulas can be used while proving?
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2091833
|
Color edges of [imath]K_6[/imath], there is a cycle of length 4
If we color the edges of [imath]K_6[/imath] (complete graph on [imath]6[/imath] vertices) red or blue, how to prove that there is a cycle of length [imath]4[/imath] with monochromatic edges ?
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1751954
|
Color the edges of [imath]K_6[/imath] red or blue. Prove that there is a cycle of length 4 with monochromatic edges.
Color the edges of [imath]K_6[/imath] red or blue. Prove that there is a cycle of length 4 with monochromatic edges. Attempt: I know that i have to... prove that there must be TWO vertices with “red-degree” at least 3, or two with "blue-degree" at least 3. Call these U and V. (We'll assume we're in the red case; the blue case is similar.) Now, ask if they’re connected by a red edge, and how many red-adjacent neighbors they share. If you analyze this, there should be only a few cases to consider – and each should be relatively straightforward. For each case, you’ll show either that that case can’t actually happen in [imath]K_6[/imath], or that it gives rise to a monochromatic 4-cycle.
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2091483
|
Show that a group of order [imath]108[/imath] has a normal subgroup of order [imath]9[/imath] or [imath]27[/imath].
Exercise from Herstein:Abstract Algebra. Please do it without group action as I don't know it. In Herstein there is no mention of Group Action as in the answer given Show that a group of order [imath]108[/imath] has a normal subgroup of order [imath]9[/imath] or [imath]27[/imath]. Attempt: [imath]108=2^2\times 3^3[/imath]. If [imath]n_2[/imath] denotes the number of Sylow [imath]2[/imath] subgroups then [imath]n_2=1+2k| 27\implies n_2=1,3,9,27.[/imath] If [imath]n_3[/imath] denotes the number of Sylow [imath]3[/imath] subgroups then [imath]n_3=1+3k| 8\implies n_3=1,2,4,8.[/imath] If [imath]n_3=1\implies [/imath] the group has a normal subgroup of order [imath]27[/imath] but how to neglect the other choices. I am confused totally .Please help.
|
178214
|
A group of order [imath]108[/imath] has a proper normal subgroup of order [imath]\geq 6[/imath].
Problem: Let [imath]G[/imath] be a group of order [imath]108 = 2^23^3[/imath]. Prove that [imath]G[/imath] has a proper normal subgroup of order [imath]n \geq 6[/imath]. My attempt: From the Sylow theorems, if [imath]n_3[/imath] and [imath]n_2[/imath] denote the number of subgroups of order [imath]27[/imath] and [imath]4[/imath], respectively, in [imath]G[/imath], then [imath]n_3 = 1[/imath] or [imath]4[/imath], since [imath]n_3 \equiv 1[/imath] (mod [imath]3[/imath]) and [imath]n_3~|~2^2[/imath], and [imath]n_2 = 1, 3, 9[/imath] or [imath]27[/imath], because [imath]n_2~|~3^3[/imath]. Now, I don't know what else to do. I tried assuming [imath]n_3 = 4[/imath] and seeing if this leads to a contradiction, but I'm not even sure that this can't happen. I'm allowed to use only the basic results of group theory (the Sylow theorems being the most sophisticated tools). Any ideas are welcome; thanks!
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2091132
|
Computing an infinite product [imath]\prod_{n=1}^{\infty} e(\frac{n}{n + 1})^n * \sqrt\frac{n}{n + 1}[/imath]
The answer to the question is [imath]\frac{\sqrt{2\pi}}e[/imath], although I do not know how to come to this solution. This is a problem from a math contest from the year 2016.
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1038054
|
Evaluation of [imath]\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}[/imath]
During my calculations I ended up with the following product: [imath]P=\prod_{n=1}^\infty e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}[/imath] I tried to express it in terms of series by taking the logarithm [imath]S=\ln P=\sum_{n=1}^\infty \ln\left(e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}}\right)[/imath] but I also got stuck. Numerical calculation suggests that it is equal to [imath]P\stackrel{?}=\frac{\sqrt{2\pi}}{e}[/imath] but I am not able to prove the conjecture. Any idea about how to evaluate the product? Any help would be appreciated. Thanks in advance.
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2091389
|
Is the topologist's sine curve a manifold?
I'm studying about manifolds. I'm an engineer and want to grasp the concept of manifold without too deep dive into rigorous mathematics. My question is whethere the "topologist's sine curve" defined as [imath]S = \{(0, y) : y \in (-1,1)\} \cup \{(x, \sin (1/x)) : x \in (0, 2/\pi)\}[/imath] can be a manifold. I tried to find some explnations in books and web. However, what I found are all about (path/local) connectedness. I'm not familiar with topology and not want to study it extensively. So I tried to understand the question with very minimal knowledge. As far as I understand, to say that a set is a manifold, I need to set a topology (open subsets) on the set. As a subset of [imath]\mathbb{R}^2[/imath], every open set in [imath]S[/imath] should have a form [imath]S \cap A[/imath] for an open set [imath]A \in \mathbb{R}^2[/imath]. Any interval in the vertical axis part of [imath]S[/imath], e.g. [imath]I = \{(0, y) : 0 < y < 1/2\}[/imath], cannot be an open set in [imath]S[/imath]. It is because every open set in [imath]\mathbb{R}^2[/imath] containing [imath]I[/imath] also contains some other part of [imath]S[/imath] : there is no way to express [imath]I[/imath] in the form of [imath]S \cap A[/imath] where [imath]A[/imath] is open in [imath]\mathbb{R}^2[/imath]. Then there is no homeomorphism from an open subset of [imath]S[/imath] containing [imath]I[/imath] to an open set in [imath]\mathbb{R}[/imath]. It is because, if there is one, the preimage of an open interval in [imath]\mathbb{R}[/imath], to which [imath]I[/imath] is mapped, must be open but [imath]I[/imath] is not open. So there cannot be a coordinate chart for [imath]I[/imath] and [imath]S[/imath] is not a manifold. This is what I understand. My first question is whether or not this understanding is correct. The second question is whether [imath]S[/imath] can be a manifold if I give a different topology. I cannot express precisly how to define the topology but intuitively the sine function part is homeomorphic to an open interval in the real line and the vertical part is also homeomorphic to an open interval in the real line, if I consider [imath]S[/imath] as a space of its own, not a subspace of [imath]\mathbb{R}^2[/imath]. Is this true? If true, how can I express the topology precisley? This question is not a duplicate of Topologist's Sine Curve not a regular submanifold of [imath]\mathbb{R^2}[/imath]?. I read it. However, the concept of connectedness confuses me. As I said in the first paragraph, I want to understand my question with minimal knowledge on topology such as just the definition of open sets, continuous functions and manifolds. Moreover the linked question does not concern my second question.
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459164
|
Topologist's Sine Curve not a regular submanifold of [imath]\mathbb{R^2}[/imath]?
I am trying to work out the details of following example from page 101 of Tu's An Introduction to Manifolds: Example 9.3. Let [imath]\Gamma[/imath] be the graph of the function [imath]f(x) = \sin(1/x)[/imath] on the interval [imath]]0, 1[[/imath], and let [imath]S[/imath] be the union of [imath]\Gamma[/imath] and the open interval [imath]I=\{(0,y)\in \mathbb{R^2} |−1<y<1\}[/imath]. The subset [imath]S[/imath] of [imath]\mathbb{R^2}[/imath] is not a regular submanifold for the following reason: if [imath]p[/imath] is in the interval [imath]I[/imath], then there is no adapted chart containing [imath]p[/imath], since any sufficiently small neighborhood [imath]U[/imath] of [imath]p[/imath] in [imath]\mathbb{R^2}[/imath] intersects [imath]S[/imath] in infinitely many components. Using Tu's definitions, I need to show that given a neighborhood [imath]U[/imath] of [imath]p[/imath], there exists no homeomorphism [imath]\phi[/imath] from [imath]U[/imath] onto an open set of [imath]V \subset\mathbb{R^2}[/imath] with the property that [imath]U \cap S[/imath] is the pre-image with respect to [imath]\phi[/imath] of the [imath]x[/imath] or [imath]y[/imath] axes intersected with [imath]V[/imath]. I am not sure where the fact that [imath]U \cap S[/imath] has infinitely many components comes into play. I tried to derive a contradiction using this fact; but even if restrict my attention to connected neighborhoods [imath]U[/imath] of [imath]p[/imath], the intersection of the connected set [imath]\phi(U)[/imath] with the [imath]x[/imath] or [imath]y[/imath] axes might have infinitely many components (I think), so there's no contradiction with the fact that homeomorphism preserves components. I would appreciate any help!
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2090907
|
Give an example of a topological space [imath]X[/imath] such that [imath]\pi_1(X)=S_3[/imath].
Give an example of a topological space [imath]X[/imath] such that [imath]\pi_1(X)=S_3[/imath], where [imath]\pi_1(X)[/imath] is the fundamental group.
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939856
|
Every Group is a Fundamental Group
I am studying elementary Algebraic Topology recently. I have seen that a topological space is identified with a group. We are telling the group as Fundamental Group. So every topological space [imath]X[/imath] and for any [imath]x \epsilon X[/imath] there is a fundamental group [imath]\pi_1 (X,x)[/imath]. I have a question about the converse. If we consider any arbitrary group [imath]G[/imath], then does there exist a topological space [imath]X[/imath] and an element [imath]x\epsilon X[/imath] such that [imath]\pi_1 (X,x)=G[/imath]?
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2091763
|
fieldtheory questions
Show that: a) The field Extension [imath]\mathbb{F_p}(X,Y)|\mathbb{F_p}(X^p,Y^p)[/imath] is not simple. b)Find a primitive element of the field extension [imath]\mathbb{Q}(\sqrt2+i,\sqrt3-i)|\mathbb{Q}[/imath] c) Let [imath]L|K[/imath] be a separable field extension and there exists a [imath]n\in \mathbb{N}[/imath] so that [imath][K(x):K]\le n[/imath] for all [imath]x \in L[/imath]. Show that: [imath]|L:K| \le n[/imath]. Now in a) I don't have a clue. For be I could find the minimal polynoms, but I don't know how to continue... . It would be nice if you gave me some advice how to start. Thank you for taking your time.
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276627
|
A finite field extension that is not simple
Let [imath]F=\mathbb{F}_p(X,Y)[/imath] be the field of rational functions in variables [imath]X,Y[/imath] over the finite field of [imath]p[/imath] elements. Let [imath]K=\mathbb{F}_p(X^p,Y^p)[/imath] be a subfield. Note that for any [imath]f\in F[/imath], [imath]f^p\in K[/imath]. Deduce from this that [imath]F/K[/imath] is not a simple extension.
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1605100
|
Value of [imath]\lim_{n\to\infty}\sum_{k=1}^n \frac{n}{n^2+k^2}[/imath]
[imath]\lim_{n\to\infty}\sum_{k=1}^n \frac{n}{n^2+k^2}[/imath] I tried this using powerseries just putting as [imath]x=1[/imath] there ,even tried thinking subtracting [imath]s_{n+1} - s_{n}[/imath] would be of some help, also I thought of writing the denominator as product of two complex numbers and then doing the partial fractions but it did not help. Any method guys? Thanks in advance for guiding to think about these kind of problems.
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1502956
|
To show that the limit of the sequence [imath]\sum\limits_{k=1}^n \frac{n}{n^2+k^2}[/imath] is [imath]\frac{\pi}{4}[/imath]
Show that [imath]\lim_{n \to \infty} \sum\limits_{k=1}^n \frac{n}{n^2+k^2} = \frac{\pi}{4}.[/imath] I am familiar with Taylor series and Fourier series of the standard functions. I tried to compare with those and see if there is a relation, I don't seen to find any. How do I tackle this?
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2092083
|
Let [imath]A[/imath] be a Hermitian square complex matrix of size [imath]n[/imath] then [imath]\langle Ax,x\rangle=0[/imath] for all [imath]x \in \Bbb C^n[/imath]. Then [imath]A=0[/imath]
Let [imath]A[/imath] be a Hermitian square complex matrix of size [imath]n[/imath] then [imath]\langle Ax,x\rangle=0[/imath] for all [imath]x \in \Bbb C^n[/imath]. Then [imath]A=0[/imath] I need a hint to prove it, if possible Thank you
|
2091819
|
Let [imath]A[/imath] be a square complex matrix of size [imath]n[/imath] then [imath]\langle Ax,x\rangle=0[/imath] for [imath]\forall x \in \Bbb C^n[/imath] [imath]\iff A=0[/imath]
Let [imath]A[/imath] be a Hermitian square complex matrix of size [imath]n[/imath] then [imath]\langle Ax,x \rangle=0[/imath], [imath]\forall x \in \Bbb C^n[/imath] if and only if [imath]A=0[/imath] I want to know if the condition that [imath]A[/imath] is Hermitian is required for the statement to still hold, please? Thank you
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2092687
|
Inductive proof that there are infinitely many pairs of consecutive powerful integers?
One way to prove that there is an infinity of consecutive powerful integers is to just note that there are infinite solutions to the equation [imath]x^2-8y^2=1[/imath], because clearly [imath]x^2[/imath] and [imath]8y^2[/imath] are clearly powerful. But this uses a bit of theory. I found this problem under an induction worksheet, and all of the surrounding problems are easy. Is there really an elementary solution to the problem? All of my attempts have been fruitless. Note: a number is powerful if whenever a prime [imath]p[/imath] divides it we have that [imath]p^2[/imath] also divides it.
|
80801
|
Proving there are infinitely many pairs of square-full consecutive integers
So the question goes like this : A positive integer [imath]n[/imath] is called square-full if for every prime [imath]p[/imath], [imath]p \, | \, n[/imath] implies [imath]p^2 \, | \, n[/imath], i.e. every prime power factor of [imath]n[/imath] occurs at least at the second power. This is equivalent to saying that there is no prime [imath]p[/imath] such that [imath]p \, \| \, n[/imath]. Show that there are infinitely many consecutive pairs of square-full numbers, i.e. infinitely many [imath]n[/imath] such that [imath]n[/imath] and [imath]n+1[/imath] are square-full. The context of the question highly suggests that this can be proved by induction and it is the solution that I am looking for. It is not a homework, just something that caught my attention because I couldn't find any natural answer to it... I hope I'm not missing something trivial! I've tried assuming [imath]n_1, n_2, \dots, n_k[/imath] were the first [imath]n_k[/imath] integers such that [imath]n_i[/imath] and [imath]n_i+1[/imath] are square-full, and then generate some polynomial in those [imath]2k[/imath] integers to get a new couple [imath]n_{k+1}[/imath], [imath]n_{k+1} + 1[/imath] but after a few natural tries I got nothing good out of it. Any ideas?
|
2092517
|
Finding a context-sensitive grammar for L
I need help for finding a context-sensitive grammar for the following language: [imath] L = \left\{ ww \left| w \in \left\{a, b\right\}^* \right. \right\} [/imath] Would that be a correct context-sensitive grammar? S -> ε | AUAV | BUBV (1) U -> AUC | BUD (2) CA -> AC (3) CB -> BC (4) CV -> AV (5) DA -> AD (6) DB -> BD (7) DV -> BV (8) U -> ε (9) V -> ε (10) A -> a (11) B -> b (12)
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163830
|
Context-sensitive grammar for the copy language
I need to find a context-sensitive grammar for the copy language [imath]L = \{ww \; | w \in \{0,1\}^* \}[/imath] This is what I got so far: [imath]\begin{eqnarray} S & \Rightarrow & \lambda \; | \; X \\ X & \Rightarrow & 0XA \; | \; 1XB \\ XB & \Rightarrow & CX \\ XA & \Rightarrow & DX \\ CX & \Rightarrow & X0 \\ DX & \Rightarrow & X1 \\ 0X & \Rightarrow & 0 \\ 1X & \Rightarrow & 1 \end{eqnarray}[/imath] This works fine for 0101 but fails even for 00 or 11. Could you please help me to find a solution? Thanks in advance!
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2093023
|
If [imath]f(z)^N[/imath] is analytic then [imath]f(z)[/imath] is analytic
I am a beginner in Complex Analysis: Question from Gamelin Complex Analysis: If [imath]f[/imath] is continuous on a domain [imath]D[/imath] and [imath]f(z)^N[/imath] is analytic for some [imath]N\in \Bbb N,[/imath] then show that [imath]f(z)[/imath] is analytic too. HINT:Use zeros of analytic function are isolated. Attempt: Let [imath]z_0\in D[/imath] then either [imath]f(z_0)=0[/imath] or [imath]f(z_0)\neq 0[/imath]. CASE I : [imath]f(z_0)=0[/imath]. If [imath]f(z_0)= 0\implies f(z_0)^N=0\implies z_0[/imath] is a zero of [imath]f^N[/imath] and hence [imath]\exists r>0[/imath] such that [imath]f^N(z_0)\neq 0\forall 0<|z-z_0|<r[/imath]. What should I do now? CASE II :[imath]f(z_0)\neq 0[/imath]. I am totally confused how to proceed.
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831121
|
Show [imath]f[/imath] is analytic if [imath]f^8[/imath] is analytic
This is from Gamelin's book on Complex Analysis. Problem: Show that if [imath]f(z)[/imath] is continuous on a domain [imath]D[/imath], and if [imath]f(z)^8[/imath] is analytic on [imath]D[/imath], then [imath]f(z)[/imath] is analytic on [imath]D[/imath]. (I assume the intention is that [imath]D[/imath] is nice, i.e. open and connected) I'm not exactly sure how to approach this. I'm guessing it has something to do with zeroes of [imath]f[/imath]. For example, around non-zeroes, [imath]f[/imath] is analytic since [imath]z^{1/8}[/imath] is nonzero in a neighborhood. Or something along those lines. The details are eluding me. Any help would be appreciated!
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2093253
|
Prove that [imath]\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2[/imath]
Prove that: [imath]\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2[/imath] My Work: I guess that I have to use the formula : [imath]\tan A = \frac {2 \tan(\frac {A}{2})}{1-\tan^2 (\frac {A}{2})}[/imath] But, I am not being able to use it. Please help me.
|
472594
|
Prove that: [imath] \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6[/imath]
How to prove the following trignometric identity? [imath] \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6[/imath] Using half angle formulas, I am getting a number for [imath]\cot7\frac12 ^\circ [/imath], but I don't know how to show it to equal the number [imath]\sqrt2 + \sqrt3 + \sqrt4 + \sqrt6[/imath]. I would however like to learn the technique of dealing with surds such as these, especially in trignometric problems as I have a lot of similar problems and I don't have a clue as to how to deal with those. Hints please! EDIT: What I have done using half angles is this: (and please note, for convenience, I am dropping the degree symbols. The angles here are in degrees however). I know that [imath] \cos 15 = \dfrac{\sqrt3+1}{2\sqrt2}[/imath] So, [imath]\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}[/imath] [imath]\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} [/imath] [imath]\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 - \sqrt3 + 1}} [/imath]
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2093033
|
Proof convergence of [imath]\sum\limits_{n = 1}^\infty\frac {n!} {n^n}[/imath]
To prove the convergence of [imath]\sum\limits_{n = 1}^\infty\frac {n!} {n^n}[/imath] I used that [imath]\lim\limits_{n\to\infty}|\frac {a_{n+1}} {a_n}|[/imath] has to be [imath]<1[/imath]: [imath]\lim\limits_{n\to\infty}|\frac {a_{n+1}} {a_n}|[/imath] [imath]=\lim\limits_{n\to\infty}|\frac {\frac {(n+1)!} {(n+1)^{n+1}}} {\frac {n!} {n^n}}|[/imath] [imath]=\lim\limits_{n\to\infty}|\frac {(n+1)!\cdot n^n} {n!\cdot (n+1)^{n+1}}|[/imath] [imath]=\lim\limits_{n\to\infty}|\frac {(n+1)\cdot n^n} {(n+1)^{n+1}}|[/imath] [imath]=\lim\limits_{n\to\infty}|\frac {n^n} {(n+1)^n}|[/imath] [imath]=\lim\limits_{n\to\infty}|{(\frac {n} {n+1})}^n|=\frac 1 e[/imath] Okay... problem solved after thankful advice
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354487
|
Does the series [imath]\sum n!/n^n[/imath] converge or diverge?
so I used the root test, but i'm not quite sure if i'm allowed to. I think im performing the operations correctly,a and i keep ending up with [imath](1)^{\infty}[/imath]. So really my question is am i performing the operations wrong or do i have to use a different test?
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2093799
|
Do the following positive series converge?
[imath](d_k)[/imath] is a positive sequence in [imath]\mathbb{R}[/imath] with: [imath]\lim_{n\to \infty} \sum_{k=1}^n d_k = \infty.[/imath] Do the following series converge or not? a) [imath]\sum_{n\geq 1} \frac{d_n}{1+d_n}[/imath] b) [imath]\sum_{n\geq 1} \frac{ d_n}{1 + n d_n}[/imath], c) [imath]\sum_{n\geq1} \frac{d_n}{1 + n^2 d_n}[/imath] d) [imath]\sum_{n\geq 1} \frac{d_n}{1+d_n^2}[/imath]
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1276016
|
Do the following series converge if [imath]a_n>0[/imath] and [imath] \sum_{n=1}^{\infty}a_n[/imath] diverges?
Do the following series converge if [imath]a_n>0[/imath] and [imath]\sum_{n=1}^{\infty}a_n[/imath] diverges ? a.) [imath]\sum_{n=1}^{\infty}{a_n \over 1+ a_n}[/imath] b.)[imath]\sum_{n=1}^{\infty}{a_n\over 1+ a_n ^2}[/imath] c.)[imath]\sum_{n=1}^{\infty}{a_n\over 1+ na_n }[/imath] d.)[imath]\sum_{n=1}^{\infty}{a_n\over 1+ n^2a_n }[/imath] What I thought was using Alembert criteria , but it just not seem to work. I also now if [imath]a_n[/imath] converges so does [imath]a_n^2[/imath]. Ive tried everything to no avail, can anyone please help ?
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2093959
|
A topological space [imath]X[/imath] which can be [imath]drawn[/imath], whose fundamental group is [imath]S_3[/imath]
Does there exist an example of a topological space [imath]X[/imath] which can be drawn, whose fundamental group is [imath]S_3[/imath]? I know that every group is the fundamental group of a topological space, but I need a concrete example.
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2093715
|
A topological space [imath]X[/imath] who can be [imath]drawn[/imath], whose fundamental group is [imath]S_3[/imath]
There exists an example of a topological space [imath]X[/imath] who can be [imath]drawn[/imath], whose fundamental group is [imath]S_3[/imath]? I know that every group is the fundamental group of a topological space, but I need a concrete example.
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2093466
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Why aren't the functions [imath]f(x) = \frac{x-1}{x-1}[/imath] and [imath]f(x) = 1[/imath] the same?
I understand that division by zero isn't allowed, but we merely just multiplied [imath]f(x) = 1[/imath] by [imath]\frac{x-1}{x-1}[/imath] to get [imath]f(x) = \frac{x-1}{x-1}[/imath] and [imath]a\cdot 1 = 1\cdot a = a[/imath] so they're the same function but with different domain how is this possible? Or in other words why don't we simplify [imath]f(x) = \frac{x-1}{x-1}[/imath] to [imath]f(x) = 1[/imath] before plotting the points. Is it just defined this way or is there a particular reason ? Note: my book says the domain of [imath]f(x) = 1[/imath] is [imath]\mathbb{R}[/imath] and the domain of [imath]f(x) = \frac{x-1}{x-1}[/imath] is [imath]\mathbb{R}[/imath] except [imath]1[/imath].
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3018559
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How is [imath]\frac {x^2 + x - 6}{x - 2}[/imath] different from [imath](x+3)[/imath]?
Let's consider the function: [imath] f(x) = \frac {x^2 + x - 6}{x - 2} [/imath] At x = 2, the value fo the function is undefined, because [imath] x - 2 [/imath] = 0. But if factor the expression on the numerator, we get [imath] \frac {(x + 3) (x-2)}{x-2} [/imath], which on simplifying yields [imath] x + 3 [/imath]. At [imath] x = 2 [/imath], the function gives 5, which is not what we get before simplifying. If all we do too the initial function is simple algebraic manipulation, shouldn't both functions yield the same value?
|
2094742
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Find the 4th root of the following complex number
find all [imath]z[/imath] in polar form satisfying the following [imath]z^4 = -1 +i\sqrt{3}[/imath] What I did was the following. Let [imath]z = r(\cos\phi\ +\ i\sin\phi)[/imath] Therefore [imath]z^4 = 2(\cos\frac{\pi}{3} + i\sin(\frac{\pi}{3}))= r^4(\cos4\phi\ +\ i\sin4\phi)[/imath] Therefore by comparism, [imath]r = 2^{1/4}, \ \phi = \frac{\pi}{12} + \frac{n\pi}{2}[/imath] where [imath]n \in \{0,1,-1,2\}[/imath] But my answer seemed to be wrong and I was hoping if you guys can point out my mistake. Any help or insight is deeply appreciated.
|
2091619
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Roots of a equation on a complex plane
How do you get the roots of the equation in algebraic form? I have no idea after changing to a trigonometric form. The angle was not simple. [imath]z^4- \sqrt3 i+1=0[/imath]
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