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2125289
|
Derivative of series and series of derivatives
Let [imath]f_n(x)[/imath] be real functions. Given that the derivatives exist, and that the sums converge, is the following true? If not, what is a counterexample, and when is it true? [imath]\frac{d}{dx}\sum_{n=1}^{\infty} f_n(x)=\sum_{n=1}^{\infty} \frac{d}{dx}f_n(x)[/imath]
|
147869
|
Interchanging the order of differentiation and summation
Can the order of a differentiation and summation be interchanged, and if so, what is the basis of the justification for this? e.g. is [imath]\frac{\mathrm{d}}{\mathrm{d}x}\sum_{n=1}^{\infty}f_n(x)[/imath] equal to [imath]\sum_{n=1}^{\infty}\frac{\mathrm{d}}{\mathrm{d}x}f_n(x)[/imath] and how can it be proven? My intuition for this is that it should be the case, since in the limit, the summation becomes an integral and this can be interchanged with the differentiation operator, but I don't know how to justify this? Would it suffice to say that [imath]\frac{\mathrm{d}}{\mathrm{d}x}f_1(x)+\frac{\mathrm{d}}{\mathrm{d}x}f_2(x) = \frac{\mathrm{d}}{\mathrm{d}x}(f_1(x)+f_2(x)[/imath] because it is a linear operator and then somehow extend this to an infinite sum? With very many thanks, Froskoy.
|
2123361
|
Prove [imath]\lim_{x\to\infty} f(x)/x = K[/imath]
If f: [imath]\mathbb{R} \rightarrow \mathbb{R}[/imath], [imath]f'[/imath] exists (finite or infinite) for all reals, and [imath]\lim_{x\to\infty} f'(x) = K[/imath]. Prove [imath]\lim_{x\to\infty} \frac{f(x)}{x} = K[/imath] I started trying to prove it for [imath]K = 0[/imath]. By definition of infinity limit: [imath]\lim_{x\to\infty} f'(x) = K \rightarrow[/imath] given [imath]\epsilon > 0[/imath], [imath]\exists x_{0} \in \mathbb{R}[/imath] such that [imath]\forall x > x_{0}[/imath] we have [imath]|f'(x) - 0| < \epsilon[/imath]. By the Mean Value Theorem: for [imath]a,b > x_{0}[/imath], [imath]\exists c \in (a,b)[/imath] such that [imath]f'(c) = \frac{f(b)-f(a)}{b-a}[/imath]. Since [imath]c > x_{0}[/imath], we have that [imath]|f'(c)| = |\frac{f(b)-f(a)}{b-a}| < \epsilon[/imath]. not sure where to go from here ... Conclude: [imath]|f(x)/x| < \epsilon \rightarrow \lim_{x\to\infty} f'(x) = 0 = K[/imath].
|
1895407
|
Prove that [imath]\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}f'(x)[/imath]
I have come up with a theorem. (It probably doesn't qualify as a theorem, but I don't know what else to call it) This is what it states: If [imath]\lim_{x\to\infty}f'(x)[/imath] exists (is finite), then [imath]\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}f'(x)[/imath] I have yet to find a counterexample, and it just seems to make sense to me, but I can't actually prove it. This "theorem" is crucial to an answer to a question that I was asked to give, but my answer won't be a very good one if I can't prove why it works. What I've tried: I have been able to prove it if [imath]f(x)[/imath] has a linear asymptote (I think my terminology is correct). If: [imath]\lim_{x\to\infty}f(x)=\lim_{x\to\infty}mx+b[/imath] Then: [imath]\lim_{x\to\infty}\frac{f(x)}x=\lim_{x\to\infty}\frac{mx+b}x=m[/imath] But that doesn't work if there is no asymptote. Take [imath]f(x)=\ln(x)[/imath] for example. the [imath]b[/imath] in [imath]mx+b[/imath] would be infinite.
|
2125119
|
Find the necessary and sufficient condition for [imath]\mathbb{Q}(\sqrt{D_1})[/imath] to be isomorphic to [imath]\mathbb{Q}(\sqrt{D_2})[/imath]
Find the necessary and sufficient condition on integers [imath]D_1,D_2[/imath] for fields [imath]\mathbb{Q}(\sqrt{D_1})[/imath] to be isomorphic to [imath]\mathbb{Q}(\sqrt{D_2})[/imath] How should I start this problem?
|
2125025
|
What are the conditions for integers [imath]D_1[/imath] and [imath]D_2[/imath] so that [imath]\mathbb{Q}[\sqrt {D_1}] \simeq \mathbb{Q}[\sqrt {D_2}][/imath] as fields.
What are the conditions for integers [imath]D_1[/imath] and [imath]D_2[/imath] so that [imath]\mathbb{Q}[\sqrt {D_1}] \simeq \mathbb{Q}[\sqrt {D_2}][/imath] as fields. Here [imath]\mathbb{Q}[\sqrt {D}] := \{a + b \sqrt D \mid a,b \in \mathbb{Q} \}[/imath] Really not sure where to begin with this sort of problem. I was thinking that I should split into cases where the integer is a square or not.
|
1396872
|
Is there any proper subring of [imath]\mathbb{R}[/imath] with field of fractions equal to [imath]\mathbb{R}[/imath]?
Is there any proper subring of [imath]\mathbb{R}[/imath] with field of fractions equal to [imath]\mathbb{R}[/imath]? Can we construct that proper subring? Is it necessarily an integral domain? Updated: Is there an example to construct the solution?
|
104905
|
Integral domain with fraction field equal to [imath]\mathbb{R}[/imath]
I wonder if there is an integral domain [imath]A\subseteq \mathbb{R}[/imath] which is not a field, and such that the field of fractions of [imath]A[/imath] is equal to [imath]\mathbb{R}[/imath]? Edit: here as a possible direction: it is known that there is no finite field extension [imath]k\subseteq \mathbb{R}[/imath]. Given such an [imath]A[/imath], can we produce such a [imath]k[/imath], contradicting this theorem?
|
2126214
|
Calculate the nature of the following series
Can someone help me determinate the nature of the following series? [imath] \frac {a^n}{ (1 + tg(\frac{3}{2}) ( 1+tg(\frac {3}{4}) + ... (1 + tg(\frac{3}{2n})) }[/imath]
|
2125908
|
Two problems on real number series
Consider the series: [imath]a) \sum_{n=1}^\infty \frac{a^n}{ \prod_{k=1}^n \ (1+\sin\frac{1}{2k})}[/imath] [imath]b) \sum_{n=1}^\infty \frac{a^n}{ \prod_{k=1}^n \ (1+\tan\frac{3}{2k})}[/imath] Showing that these two are convergent (and absolutely convergent) it's no big deal for [imath]a\in R^*[/imath]. But I couldn't figure out how to solve it when [imath]a=1[/imath]. My guess is that the general term has to be brought to another form. And that is my question: how to prove convergence when [imath]a=1[/imath]. Thanks for helping with this!
|
2125193
|
Hartshorne problem 1.8, section 1
In exercise 1.8 of chap I in Hartshorne algebraic geometry, Let [imath]Y[/imath] be an affine variety of dimension [imath]r[/imath] in [imath]\mathbf A^n[/imath]. Let [imath]H[/imath] be a hypersurface in [imath]\mathbf A^n[/imath], and assume that [imath]Y \nsubseteq H[/imath]. Then every irreducible component of [imath]Y \cap H[/imath] has dimension [imath]r-1[/imath]. Let [imath]I(H)={f}[/imath]. I saw a solution in which one of the steps was to prove that [imath]\bar{f}[/imath] is not a unit in [imath]A/I(Y)[/imath]. How to prove this ?
|
1314582
|
Exercise [imath]1.8[/imath] of chapter one in Hartshorne.
In exercise 1.8 of chap I in Hartshorne algebraic geometry, Let [imath]Y[/imath] be an affine variety of dimension [imath]r[/imath] in [imath]\mathbf A^n[/imath]. Let [imath]H[/imath] be a hypersurface in [imath]\mathbf A^n[/imath], and assume that [imath]Y \nsubseteq H[/imath]. Then every irreducible component of [imath]Y \cap H[/imath] has dimension [imath]r-1[/imath]. I refered to a solution. In this solution, why [imath]f[/imath] is not a unit in [imath]B[/imath]?
|
2126146
|
Why can't the Klein bottle embed in [imath]\mathbb{R}^3[/imath]?
I'm working with the definition: A smooth embedding of M into N is an injective immersion [imath]F:M\rightarrow N[/imath] that is also a topological embedding. I'd like to explain why the klein bottle cant embed in [imath]\mathbb{R}^3[/imath]. The immersion is not injective, thus results in self intersections, correct?
|
611834
|
How does one prove that the Klein bottle cannot be embedded in [imath]R^3[/imath]?
How does one prove that the Klein bottle cannot be embedded in [imath]R^3[/imath]? I'm talking about smooth embeddings.
|
2126454
|
Show that [imath](0,1][/imath] is open in [imath][0,1][/imath] but [imath](0,1][/imath] is not open in [imath]\mathbb{R}[/imath]
I am very new to Topology, so excuse my lack knowledge; Show that [imath]A=(0,1][/imath] is open in [imath][0,1][/imath]: I am suppose to choose [imath]x[/imath] in [imath](0,1][/imath]. And construct a ball around with radius [imath]r\leq\min\{ \lvert x-1 \lvert,x\}[/imath]? Should I show that the complement is closed? I could also use a hint on showing that [imath](0,1][/imath] is not open i [imath]\mathbb{R}[/imath].
|
267293
|
show that the interval of the form [imath][0,a)[/imath] or [imath](a, 1][/imath] is open set in metric subspace [imath][0,1][/imath] but not open in [imath]\mathbb R^1[/imath]
On the metric subspace [imath]S = [0,1][/imath] of the Euclidean space [imath]\mathbb R^1 [/imath], every interval of the form [imath]A = [0,a)[/imath] or [imath](a, 1][/imath] where [imath]0<a<1[/imath] is open set in S. These sets are not open in [imath]\mathbb R^1[/imath] Here's what I attempted to show that [imath]A[/imath] is open in [imath]S[/imath]. I have no idea how it is not open in [imath]\mathbb R^1[/imath]. Let [imath]M = \mathbb R^1[/imath], [imath]x \in A = [0,a)[/imath] . If [imath]x = 0[/imath], [imath] r \leq \min \{a, 1-a\}, \\\ B_S(0; r) = B_M(0,r)\cap[0,1] = (-r, r) \cap[0,1] = [0,r) \subseteq A [/imath] If [imath]x \neq 0, r \leq \min \{x,|a-x|, 1-x\}, \\B_S(x; r) = B_M(x,r)\cap[0,1] = (x-r, x+r) \cap[0,1] \subset (0,x) \text{ or } (x, 1) \subset A[/imath]
|
2125952
|
Let [imath]a,b,c[/imath] be positive real numbers such that [imath]abc =1[/imath]
Let [imath]a,b,c[/imath] be positive real numbers such that [imath]abc=1[/imath]. Prove that [imath]a^2+b^2+c^2\geq a+b+c[/imath]. Also, state the condition for equality. My Attempt, [imath]a,b,c[/imath] are real and positive numbers, then [imath](a-1)^2+(b-1)^2+(c-1)^2\ge 0[/imath] [imath]a^2-2a+1+ b^2-2b+1+c^2-2c+1\ge 0[/imath] [imath]a^2+b^2+ c^2-2(a+b+c)+3\ge 0[/imath]. I have made a start in this way, but I am not sure if this works. Please help me, with any simple and beautiful method.
|
581992
|
Let [imath]a, b, c[/imath] be positive real numbers such that [imath]abc = 1[/imath]. Prove that [imath]a^2 + b^2 + c^2 \ge a + b + c[/imath].
Let [imath]a, b, c[/imath] be positive real numbers such that [imath]abc = 1[/imath]. Prove that [imath]a^2 + b^2 + c^2 \ge a + b + c[/imath]. I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?
|
2123497
|
Struggling with a PDE problem. Would like some guidance - not a solution.
I am given a simple river system flowing down a slope where the speed of the flow downstream under the force of gravity builds up until resistive forces [imath]R[/imath] balance the component of gravitational force acting downstream [imath]F[/imath]. [imath]R=av[/imath] [imath]F=bh[/imath] The speed will eventually adjust to when [imath]av=bh \implies v=\frac{b}{a}h[/imath]. We further assume that rain add water and we let seepage into the ground remove water. This will increase the depth of water at any point in the river at the rate [imath]r(h, x, t)[/imath] and the mass conservation law is given by [imath]h_t +(hv)_x = r[/imath] Using Burger's equation and substituting for [imath]v[/imath], I get [imath]u_t +uu_x =f[/imath] where [imath]u=2\frac{b}{a}h[/imath] and [imath]f=2\frac{b}{a}r[/imath]. My first question is how the author got [imath]u[/imath] and [imath]f[/imath] values: how he converted the mass conservation equation to Burger's equation. If [imath]R=av^2[/imath] instead then [imath]v=\sqrt{\frac{b}{a}h}[/imath] and following the same [imath]v[/imath] substitution, I arrive at [imath]h_t +(\sqrt{\frac{b}{a}h}\cdot h)_x = r.[/imath] Is it legal to simply substitute [imath]h[/imath] with [imath]u[/imath] to get [imath]u_t + (\sqrt{\frac{b}{a}}u^{\frac{3}{2}})_x=f\;?[/imath] Again, I'm having trouble identifying the values for [imath]u[/imath] and [imath]f[/imath]. Thank you very much for your time and help!
|
2122772
|
River flowing down a slope
I am given a simple river system flowing down a slope where the speed of the flow downstream under the force of gravity builds up until resistive forces [imath]R[/imath] balance the component of gravitational force acting downstream [imath]F[/imath]. [imath]R=av[/imath] [imath]F=bh[/imath] The speed will eventually adjust to when [imath]av=bh \implies v=\frac{b}{a}h[/imath]. We further assume that rain add water and we let seepage into the ground remove water. This will increase the depth of water at any point in the river at the rate [imath]r(h, x, t)[/imath] and the mass conservation law is given by [imath]h_t +(hv)_x = r[/imath] Using Burger's equation and substituting for [imath]v[/imath], I get [imath]u_t +uu_x =f[/imath] where [imath]u=2\frac{b}{a}h[/imath] and [imath]f=2\frac{b}{a}r[/imath]. My question is how one arrives at the proposed [imath]u[/imath] and [imath]f[/imath] expressions. If [imath]R=av^2[/imath] instead and following the same procedure I arrived at [imath]h_t +(\sqrt{\frac{b}{a}}hh)_x = r[/imath] Can I reduce this to, [imath]u_t +uu_x =f[/imath] where [imath]u=\frac{3}{2}\sqrt{\frac{b}{a}}h\;\text{and}\;f=\frac{3}{2}\sqrt{\frac{b}{a}}r \;?[/imath]
|
2126586
|
For which natural numbers [imath]n[/imath] is [imath]\sqrt n[/imath] irrational? How would you prove your answer?
Can someone help me out here? I'm trying to prove this the same way as whether if [imath]\sqrt 2[/imath] is irrational, but i'm not sure what am i doing. Can someone show me how to prove this?
|
263864
|
Proof that [imath]\sqrt{5}[/imath] is irrational
In my textbook the following proof is given for the fact that [imath]\sqrt{5}[/imath] is irrational: [imath] x = \frac{p}{q}[/imath] and [imath]x^2 = 5[/imath]. We choose [imath]p[/imath] and [imath]q[/imath] so that the have no common factors, so we know that [imath]p[/imath] and [imath]q[/imath] aren't both divisible by [imath]5[/imath]. [imath]\left(\dfrac{p}{q}\right)^2 = 5\\ \text{ so } p^2=5q^2[/imath] This means that [imath]p^2[/imath] is divisble by 5. But this also means that [imath]p[/imath] is divisible by 5. [imath]p=5k[/imath], so [imath]p^2=25k^2[/imath] and so [imath]q^2=5k^2[/imath]. This means that both [imath]q[/imath] and [imath]p[/imath] are divisible by 5, and since that can't be the case, we've proven that [imath]\sqrt{5}[/imath] is irrational. What bothers me with this proof is the beginning, in which we choose a [imath]p[/imath] and [imath]q[/imath] so that they haven't got a common factor. How can we know for sure that there exists a [imath]p[/imath] and [imath]q[/imath] with no common factors such that [imath]x=\dfrac{p}{q} = \sqrt{5}[/imath]? Because it seems that step could be used for every number Edit: I found out what started my confusion: I thought that any fraction with numerator 1 had a common factor, since every integer can be divided by 1. This has given me another question: Are confusions like this the reason why 1 is not considered a prime number?
|
1926799
|
Equivalent criterion for algebraic independence over [imath]\mathbb{Q}[/imath].
How do we show that two distinct complex numbers which are algebraically independent over [imath]\mathbb{Q}[/imath] are also algebraically independent over [imath]\bar{\mathbb{Q}}[/imath]?
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2126321
|
Algebraic independence over [imath]\overline{\mathbb Q}[/imath] and over [imath]\mathbb Q[/imath]
Is it true that if [imath]x_1, \dots, x_n \in \Bbb C[/imath] are algebraically independent over [imath]\mathbb Q[/imath], then they are algebraically independent over [imath]\overline{\mathbb Q}[/imath]? I know how to prove it for [imath]n=1[/imath], by the contrapositive. If [imath]P(x)=0[/imath] and [imath]P \neq 0[/imath] has coefficients [imath]a_0,...,a_n[/imath] in [imath]\overline{\mathbb Q}[/imath], then [imath]x[/imath] is algebraic over [imath]\mathbb Q(a_0,...,a_n)[/imath] which is itself algebraic over [imath]\mathbb Q[/imath], therefore [imath]x[/imath] is algebraic over [imath]\mathbb Q[/imath], so it satisfies some relation [imath]Q(x)=0[/imath] where [imath]Q \neq 0[/imath] is a polynomial with rational coefficients. For [imath]n \geq 2[/imath], I tried induction, but I was not sure how to proceed. Thank you for your help!
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2126227
|
Idea to a trig equation
[imath]\tt{tan x =tan(x+10^0)tan(x+20^0)tan(x+30^0)}[/imath] This equation asked from a genius student of my class . I think and solve it like below . My idea based on [imath]cot(x+30) tan(x)=tan(x+10)tan(x+20)\\15=\dfrac{0+30}{2}=\dfrac{10+20}{2}\\so [/imath] [imath]u=x+15^0[/imath] and substitute [imath]u[/imath] in equation . Then my genius student asked for a other Idea to work with this .I show my work , but I can't go on ... please help me to find an Idea . thanks in advanced .
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264927
|
Find the acute angle [imath]x[/imath] for [imath]\tan x = \tan(x+10^\circ)\tan(x+20^\circ)\tan(x+30^\circ)[/imath].
How to solve the following equation? [imath]\tan x= \tan(x+10^\circ)\tan(x+20^\circ)\tan(x+30^\circ)[/imath]
|
2127390
|
equivalence relation proof for all positive integers
The relation [imath]R[/imath] is defined for all positive integers such that [imath](a,b) R (c,d) \longleftrightarrow a+d=b+c[/imath]. Show that [imath]R[/imath] is an equivalence relation.
|
21256
|
Equivalence relation [imath](a,b) R (c,d) \Leftrightarrow a + d = b + c[/imath]
Suppose [imath]A[/imath] is the set composed of all ordered pairs of positive integers. Let [imath]R[/imath] be the relation defined on [imath]A[/imath] where [imath](a,b) R (c,d)[/imath] means that [imath]a + d = b + c[/imath]. (a) Prove that [imath]R[/imath] is an equivalence relation. Here is what I have so far. Is this correct? Reflexive: [imath]a \sim a[/imath] [imath]\implies[/imath] [imath]a+b=a+b[/imath]; [imath](a,b) R (c,d)[/imath] Symmetric: if [imath]a \sim b[/imath] then [imath]b \sim a[/imath] [imath]\implies[/imath] if [imath]a+d=b+c[/imath] then [imath]c+b=d+a[/imath] Transitive: if [imath]a \sim b[/imath] and [imath]b \sim c[/imath] then [imath]a \sim c[/imath] [imath]\implies[/imath] if [imath]a+d=b+c[/imath] and [imath]c+f=d+e[/imath] then [imath]a+d=d+e[/imath] (b) Find [imath][(1,1)][/imath]. I'm not sure how to approach this.
|
2127114
|
Suppose [imath]f[/imath] is differentiable on [imath]\mathbb{R}[/imath] and that [imath]\lim_{x \rightarrow 0} f'(x)=L[/imath]. May we conclude that [imath]f'(0)=L[/imath]
This seems like another good question for consideration. I think the answer is yes just because I cannot think of a way to make it break down because the domain is defined for all of [imath]\mathbb{R}[/imath]. Any thoughts? Thank you in advance. This question is not a duplicate, it deals with the limit of the derivative specifically at 0. The other two cited questions do not. Thank you.
|
1705742
|
If [imath]\lim_{x\to\infty}f(x)[/imath] and [imath]\lim_{x\to\infty}f^{\prime}(x)[/imath] both exist, then [imath]\lim_{x\to\infty}f^{\prime}(x) = 0[/imath]
Suppose [imath]f:\mathbb{R}\to\mathbb{R}[/imath] is everywhere differentiable, and suppose that [imath]\lim_{x\to\infty}f(x)[/imath] and [imath]\lim_{x\to\infty}f^{\prime}(x)[/imath] both exist. I am trying to prove that the latter limit is necessarily [imath]0[/imath]. I have the following argument, but I'm not sure if it's completely sound. Since [imath]f[/imath] is differentiable everywhere, we can apply the Mean Value Theorem to [imath]f[/imath] on [imath][x,x+1][/imath] for all relevant [imath]x[/imath]. This guarantees an [imath]\alpha_{x}\in(x,x+1)[/imath] such that [imath]f^{\prime}(\alpha_{x}) = \frac{f(x+1)-f(x)}{x+1-x} = f(x+1)-f(x).[/imath] Now, the limit as [imath]x\to\infty[/imath] of the right-hand side of this expression must be [imath]0[/imath], since [imath]\lim_{x\to\infty}f(x)[/imath] exists by assumption (and must equal [imath]\lim_{x\to\infty}f(x+1)[/imath]). On the left hand side, we notice that [imath]\alpha_{x}\to\infty[/imath] as [imath]x\to\infty[/imath], since [imath]\alpha_{x}>x[/imath] always, so that: \begin{eqnarray*} 0 & = & \lim_{x\to\infty}[f(x+1)-f(x)]\\ & = & \lim_{x\to\infty}f^{\prime}(\alpha_{x})\\ & = & \lim_{y\to\infty}f^{\prime}(y), \end{eqnarray*} proving the result. I took inspiration for this argument from other sources which use the same trick of "use the Mean Value Theorem to introduce a quantity [imath]\alpha_{x}[/imath] which we have some bounds on, then take limits". However, this style of argument seems dodgy to me: we haven't actually defined a function [imath]\alpha[/imath] to take the limit of as [imath]x\to\infty[/imath], and it's not clear to me that defining such a function is always possible. For example, we can't just say "take the least such value and call it [imath]\alpha_{x}[/imath]", because we haven't shown that there will always be a least such value. Here are my questions: In the above, where have we used the fact that [imath]\lim_{x\to\infty}f^{\prime}(x)[/imath] exists? This is an important assumption: consider for example the function [imath]x\mapsto\sin{(x^{2})}/x[/imath]. My guess is that it's used in the last line, where we must assume this fact to use the chain rule, but I'd like confirmation of this. Does the "[imath]\alpha_{x}[/imath] trick" require something like the Axiom of Choice in general? In particular, the thing which makes me slightly anxious about just saying "choose an [imath]\alpha_{x}[/imath] for every [imath]x[/imath]" is that we have to make (uncountably) infinitely many "choices", and we have no prescribed method of doing this. EDIT: It turns out this has been answered in other questions on this site, see link in the comments below. EDIT: Note that the first question is different to others on related topics because here I am asking very specifically about this argument and why it works.
|
2128032
|
Finding a digit in the sequence of all natural numbers.
I came across a riddle and I wanted to do know if I solved it correctly: If we write every natural number next to each other: Example: [imath]012345678910111213\cdots[/imath] What is the digit at position [imath]10000[/imath]? My attempt: Left side is the amount numbers and right side is the digit size of the number: [imath]10 \qquad 1[/imath] [imath]90 \qquad 2[/imath] [imath]900\qquad 3[/imath] [imath]1777.5\qquad 4[/imath] Now we have [imath]10000=10\cdot 1 + 90\cdot 2+900 \cdot 3+1777.5 \cdot4[/imath] So it should be the second or third digit of [imath]999+1778=2777[/imath] is this correct? And is it possible to write a closed formula to get a specific position?
|
188812
|
[imath]n[/imath]-th digit in the sequence of natural numbers
What's the [imath]n[/imath]-th digit in the sequence [imath]S[/imath] of numbers formed by the natural numbers, i.e., [imath]n[/imath]-th digit in the sequence 1 2 3 4 5 6 7 8 9 10 11 12... ? For example 11th digit in the sequence is 0.
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2128107
|
Proving the existence of a fixed point
Let [imath](X, \rho)[/imath] be a compact metric space. Suppose that [imath]T: X \rightarrow X[/imath] and for all [imath]u \neq v \in X[/imath], [imath] \rho(T(u),T(v)) < \rho(u,v) [/imath] Then show that [imath]T[/imath] has a unique fixed point. I am thinking to first show existence, and then show that it has to be unique. To show existence, there is a hint to use the fact that [imath]x[/imath] is fixed if and only if [imath]\rho(T(x),x) = 0[/imath]. It is similar to the contraction mapping principle, but not enough that you can use it to prove this. Also, if the above strict inequality is replaced by a weak inequality ([imath]\leq[/imath]) then must [imath]T[/imath] have a fixed point? I am thinking no, given that the requirement for a fixed point in the contraction mapping principle is that there be a [imath]c \in (0,1)[/imath] such that [imath] \rho(T(u),T(v)) \leq c\rho(u,v)[/imath]. I am thinking that a counterexample would be enough to show that this is not true.
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1660866
|
Existence of fixed point given a mapping
Let [imath]A[/imath] be a closed and bounded subset of [imath]\mathbb{R}^{n}[/imath]. Given that [imath]f: A \to A[/imath] satisfies [imath]d(f(x),f(y)) < d(x,y)[/imath] for all [imath]x,y \in A[/imath], [imath]x\neq y[/imath] where [imath]d[/imath] is the usual Euclidean metric. Show that [imath]f[/imath] has a unique fixed point. My observations: Since [imath]\mathbb{R^{n}}[/imath] is complete, then [imath](A,d)[/imath] is complete since [imath]A[/imath] is closed; Since [imath]A[/imath] is bounded, then [imath]A \subset B_{R}[/imath] for some [imath]R > 0[/imath]; Banach Fixed Point Theorem guarantees a unique fixed point if [imath]f[/imath] is contraction on [imath]A[/imath]. Please correct me if I am wrong. My quandry here is showing that [imath]f[/imath] is a contraction, i.e. [imath] d(f(x),f(y)) \leqslant k\cdot d(x,y), \hspace{3mm} k \in (0,1) [/imath] I'm looking for some hints or tips. Thank you.
|
2128187
|
Proving the linear map is zero
Let [imath]V[/imath] be a complex innerproduct space and [imath]T[/imath] be a linear operator on [imath]V[/imath]. If [imath]\langle T(v),v \rangle =0[/imath] for every [imath]v \in V[/imath], show that [imath]T[/imath] is a zero operator. I figured out that the only eigenvalue of [imath]T[/imath] is zero, which implies that the monic polynomial of [imath]T[/imath] is [imath]x^n[/imath] for some [imath]n[/imath]. I have to prove that [imath]n=1[/imath], but I don't know how to proceed. I also considered Jordan form and the basis [imath]B=\{v_1, v_2, ...,v_n\}[/imath] such that [imath][T]_B^B[/imath] is the Jordan form of [imath]T[/imath], but it doesn't work. Does anyone have ideas?
|
315942
|
Prove that there is no non-zero linear operator on $\mathbb C^{2}$ such that [imath](\alpha | T\alpha) = 0[/imath]
I'm stuck on this problem for a long time. I can't find a proper solution for this without using complex calculations. I hope some one can help me with this problem: Let [imath] (\;|\;)[/imath] be the standard inner product on [imath]$\mathbb C^{2}$[/imath]. Prove that there is no non-zero linear operator [imath]T[/imath] on [imath]$\mathbb C^{2}$[/imath] such that [imath](\alpha|T\alpha) = 0\;[/imath] for every [imath]\alpha[/imath] in [imath]$\mathbb C^{2}$[/imath]. Generalize.
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2125628
|
Matrices over a PID with equal image only differ by an automorphism.
Let [imath]R[/imath] be a PID and [imath]A,B[/imath] be two [imath]m\times n[/imath] matrices over [imath]R[/imath] with [imath]m\leq n[/imath]. I want to show that if [imath]A[/imath] and [imath]B[/imath] have the same image, they only differ by multiplication by an invertible matrix from the right. Expressing the column vectors of A,B by linear combinations of the column vectors of B,A yields square matrices [imath]X,Y[/imath] with [imath]AX=B,\ BY=A[/imath] and thus [imath]AXY=A,\ BYX=B.[/imath] Using the fact that [imath]R[/imath] is a PID, I can apply the elementary divisor theorem and assume [imath]A[/imath] to be "diagonal", meaning of the form [imath]A=\left(\frac{\textrm{diag}(e_1, \dotsc, e_m)}{0} \right)[/imath] for some [imath]e_i\in R[/imath].
|
1781365
|
Question regarding matrices with same image
Let [imath]A[/imath] and [imath]B[/imath] be [imath]m\times n[/imath] matrices over [imath]\mathbb{Z}[/imath] such that image([imath]A[/imath]) = image([imath]B[/imath]), where [imath]A[/imath] and [imath]B[/imath] are considered as maps [imath]\mathbb{Z}^n \rightarrow \mathbb{Z}^m[/imath]. Does there exist an invertible [imath]n\times n[/imath] matrix [imath]P[/imath] such that [imath]A=BP[/imath]? what happens if we consider the matrices to be over [imath]\mathbb{Q}[/imath] instead of [imath]\mathbb{Z}[/imath]? I would prefer some hints instead of full answer. Thanks in advance.
|
297925
|
How to evaluate this infinite product: [imath]\prod\limits_{n=1}^{\infty }{\left( 1-\frac{1}{{{2}^{n}}} \right)}[/imath]
How to evaluate this one [imath]\prod\limits_{n=1}^{\infty }{\left( 1-\frac{1}{{{2}^{n}}} \right)}[/imath]
|
1200575
|
What is the value of $\prod_{i=1}^\infty \left(1-\frac{1}{2^i}\right)$?
Also, what about in general, for some value p, which has the value 2 in the given formula? MOTIVATION: I was wondering the probability of never getting tails if one forever flipped a coin whose probability of landing tails decreased (in this case, geometrically) each flip. The probability of tails on flip i is [imath]\frac{1}{2^i}[/imath], and the probability of heads on flip i is [imath]1-\frac{1}{2^i}[/imath]. So, first flip the coin is 50-50, next it is 75-25, etc. And the probability of never landing tails is equal to the probability of always landing heads, which is the infinite product of the heads probabilities, yielding [imath]$\prod_{i=1}^\infty \left(1-\frac{1}{2^i}\right)$[/imath]. => ([imath]\frac{1}{2} * \frac{3}{4} * \frac{7}{8} ...[/imath])
|
1481780
|
Is conditional expectation with respect to two sigma algebra exchangeable?
[imath](\Omega, \mathcal{F}, P)[/imath] is a probability space. [imath]X[/imath] is a r.v. defined on it, and [imath]\mathcal{G}_1, \mathcal{G}_2[/imath] are two [imath]\sigma[/imath]-algebra, can we claim the following: [imath] \mathbb{E}\{\mathbb{E}[X|\mathcal{G}_1]|\mathcal{G}_2\}=\mathbb{E}\{\mathbb{E}[X|\mathcal{G}_2]|\mathcal{G}_1\}=\mathbb{E}[X|\mathcal{G}_1\cap\mathcal{G}_2]. [/imath] If this is not true, why the following is true: [imath]X_1,X_2,\ldots,X_n[/imath] are independent r.v. on the same probability space, [imath]Z=f(X_1,X_2,\ldots,X_n)[/imath] is another r.v. And we have: [imath] \mathbb{E}\{\mathbb{E}[Z|X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n]|X_1,\ldots,X_i\}=\mathbb{E}[Z|X_1,\ldots,X_{i-1}]. [/imath]
|
2430515
|
Is [imath]\mathbb E(\mathbb E(X\mid \mathcal F) \mid \mathcal G) = \mathbb E(X \mid \mathcal F\cap \mathcal G)[/imath]?
A proper wanna-be generalization of the "tower property" of conditional expectations would be: [imath]\mathbb E(\mathbb E(X\mid \mathcal F) \mid \mathcal G) = \mathbb E(X \mid \mathcal F\cap \mathcal G)[/imath] regardless of, whether the [imath]\sigma[/imath]-algebras [imath]\mathcal F[/imath] and [imath]\mathcal G[/imath] can be compared with [imath]\subseteq[/imath]. My guess is this is false. What is a counterexample? (or proof?)
|
1958475
|
complex structure compatible with symplectic form and riemannian metric
Given a non-degenerate symplectic form [imath]\omega[/imath] on a finite dimensional vector space [imath]V[/imath] and a complex structure [imath]J[/imath] (that is, [imath]J\in End(V),~J^2=-Id[/imath]). If [imath]J[/imath] is compatible with [imath]\omega[/imath] in the sense that [imath]\omega(Jv,Jw) = \omega(v,w),~ \omega(v,Jv) > 0,[/imath] then we can define a positive definite inner product by setting [imath]g(v,w) = \omega(v,Jw).[/imath] Then [imath]g[/imath] is also Hermitian in the sense that [imath]g(Jv,Jw) = g(v,w).[/imath] Question. Now suppose we start with a symplectic form [imath]\omega[/imath] and a positive definite inner product [imath]g[/imath], under what conditions does there exist an almost complex structure such that [imath]\omega[/imath] and [imath]g[/imath] are related by the above equation? And how does this correspond to the linear algebra statement that [imath]U(n) = GL(n,\mathbb{C})\cap Sp(2n,\mathbb{R})\cap O(2n).[/imath] It is a standard fact that given a metric [imath]g[/imath], there exists a canonical [imath]J[/imath] compatible with [imath]\omega[/imath], but the problem is that the [imath]J[/imath] so found need not satisfy [imath]\omega(v,Jw) = g(v,w).[/imath]
|
424236
|
[imath]2[/imath] out of [imath]3[/imath] property of the unitary group
I am trying to understand the [imath]2[/imath] out of [imath]3[/imath] property of the unitary group. I have almost got it, but I am not completely sure about the interaction between an inner product and a symplectic form to obtain an almost complex structure. Let [imath]V[/imath] be a real vector space. An inner product on [imath]V[/imath] is a positive definite symmetric bilinear form [imath]g[/imath]. An endomorphism [imath]T \in \operatorname{End}(V)[/imath] preserves [imath]g[/imath] if [imath]g(T(u), T(v)) = g(u, v)[/imath] for all [imath]u, v \in V[/imath]; the collection of all such endomorphisms forms a group [imath]O(V, g)[/imath] called the orthogonal group. An almost complex structure on [imath]V[/imath] is an endomorphism [imath]J \in \operatorname{End}(V)[/imath] such that [imath]J^2 = -\operatorname{id}_V[/imath]. An endomorphism [imath]T \in \operatorname{End}(V)[/imath] is complex linear if [imath]T \circ J = J\circ T[/imath]; the collection of all such endomorphisms forms a group [imath]GL(V, J)[/imath] called the complex general linear group. A symplectic form on [imath]V[/imath] is a skew-symmetric non-degenerate bilinear form [imath]\omega[/imath]. An endomorphism [imath]T \in \operatorname{End}(V)[/imath] preserves [imath]\omega[/imath] if [imath]\omega(T(u), T(v)) = \omega(u, v)[/imath] for all [imath]u, v \in V[/imath]; the collection of all such endomorphisms forms a group [imath]Sp(V, \omega)[/imath] called the symplectic group. Almost Complex Structure & Inner Product For an inner product [imath]g[/imath] and a compatible almost complex structure [imath]J[/imath] (i.e. [imath]J \in O(V, g)[/imath]), we obtain a symplectic form by defining [imath]\omega(u, v) := g(u, J(v))[/imath]. It follows that [imath]O(V, g)\cap GL(V, J) \subseteq Sp(V, \omega)[/imath]. Almost Complex Structure & Symplectic Form For a symplectic form [imath]\omega[/imath] and a compatible almost complex structure [imath]J[/imath] (i.e. [imath]J \in Sp(V, \omega)[/imath]) which tames [imath]\omega[/imath] (i.e. [imath]\omega(u, J(u)) > 0[/imath] for all [imath]u \in V\setminus\{0\}[/imath]) we obtain an inner product by defining [imath]g(u, v) := \omega(J(u), v)[/imath]. It follows that [imath]Sp(V, \omega)\cap GL(V, J) \subseteq O(V, g)[/imath]. Inner Product & Symplectic Form This is the part I am unsure about. Denote by [imath]\Phi_g[/imath] the isomorphism [imath]V \to V^*[/imath] induced by [imath]g[/imath]; that is [imath]\Phi_g(v) \in V^*[/imath] is defined by [imath]\Phi_g(v)(u) = g(u, v)[/imath]. Likewise, denote the isomorphism [imath]V \to V^*[/imath] induced by [imath]\omega[/imath] by [imath]\Phi_{\omega}[/imath]; that is [imath]\Phi_{\omega}(v) \in V^*[/imath] is defined by [imath]\Phi_{\omega}(v)(u) = \omega(u, v)[/imath]. Is there any compatibility restriction that we must impose on [imath]\Phi_g[/imath] and [imath]\Phi_{\omega}[/imath]? Is [imath]J = \Phi_g^{-1}\circ\Phi_{\omega}[/imath] an almost complex structure on [imath]V[/imath]? How do we use this to deduce [imath]O(V, g)\cap Sp(V, \omega) \subseteq GL(V, J)[/imath]? For question 3, I can use the previous relationships between the three groups, but I'd like to be able to deduce it from the structures themselves.
|
2129617
|
Is there a better upper bound to the cardinality of the continuum?
We know that [imath]\aleph_0[/imath] is smaller than [imath]\mathfrak c[/imath], the cardinality of the continuum. But are there some good upper-bounds? For example, it is trivial that [imath]\mathfrak c<2^{\mathfrak c}[/imath], but I wonder if there are better bounds. Specifically, I have to wonder if there exists [imath]\alpha\in\mathbb N[/imath] such that we know that [imath]\mathfrak c<\aleph_\alpha[/imath]? If not, what about transfinite ordinals [imath]\alpha[/imath]?
|
20385
|
bound on the cardinality of the continuum? I hope not
Suppose we don't believe the continuum hypothesis. Using Von Neumann cardinal assignment (so I guess we believe well-ordering?), is there any "familiar" ordinal number [imath]\alpha[/imath] such that, for non-tautological reasons, [imath]\aleph_\alpha[/imath] is provably larger than the cardinality of the continuum? I would hope not since it would seem pretty silly if something like [imath]\alpha = \omega_0[/imath] worked and we could say "well gee we can't prove that [imath]c = \aleph_1[/imath], but it's definitely one of [imath]\aleph_1, \aleph_2, \ldots , \aleph_{73}, \ldots[/imath]". I (obviously) don't know jack squat about set theory, so this is really just idle curiosity. If a more precise question is desired I guess I would have to make it For any countable ordinal [imath]\alpha[/imath] is the statement: [imath]c < \aleph_\alpha[/imath] independent of ZFC in the same sense as the continuum hypothesis? assuming that even makes sense. Thanks!
|
2129942
|
Prove that [imath]\{v_1, v_2, n\}[/imath] is a basis
Let [imath]P = \text{span } \{ v_1, v_2\}[/imath] be a plane in [imath]\mathbb{R}^3[/imath] with normal vector [imath]n[/imath], show that [imath]\{v_1, v_2, n\}[/imath] is a basis for [imath]\mathbb{R}^3[/imath] It must be that [imath]\{v_1, v_2\}[/imath] is linearly independent (LI) by def of a plane, thus [imath]c_1v_1 + c_2v_2 = \overrightarrow{0}[/imath] for [imath]c_1 = c_2 = 0[/imath]. It must follow that the coefficient of [imath]n[/imath] is 0?
|
2107196
|
Prove set is basis for [imath]R^3[/imath]
Hints only Let [imath]P = \text{Span } \{v_1, v_2\}[/imath] be a plane in [imath]\mathbb{R}^3[/imath] with normal vector [imath]n[/imath]. Show that [imath]\{ v_1, v_2, n\}[/imath] is a basis for [imath]\mathbb{R}^3[/imath]. Hints only Equation for [imath]P[/imath]: [imath]P = c_1v_1 + c_2v_2[/imath]. For real [imath]c_1, c_2[/imath]. We have by definition, [imath]n = v_1 \times v_2[/imath]. To make sure [imath]\{ v_1, v_2, n\}[/imath] is a basis for [imath]\mathbb{R}^3[/imath]. we must have Span [imath]\{ v_1, v_2, n\}[/imath] = [imath]\mathbb{R}^3[/imath] [imath]\{ v_1, v_2, n\}[/imath] Linearly independent. I am having major trouble showing (1). Can I get a SMALL hint? I am not allowed to use dimension. Showing set is LI By definition of a plane, [imath]\{v_1, v_2 \}[/imath] is linearly independent thus [imath]\overrightarrow{0} \not \in \{v_1, v_2 \}[/imath] So [imath]c_1v_1 + c_2v_2 + 0 \overrightarrow{n} = 0[/imath] Thus [imath]\{v_1, v_2, n \}[/imath] is LI. The span part is confusing
|
2130174
|
Probability to pick last four pages
A [imath]125-[/imath]pages document is on the floor randomly spread. If [imath]4[/imath] pages are picked up at random, what is the chance that these are the document's last four pages and in the right order? Page [imath]122, 123, 124[/imath] and [imath]125.[/imath]
|
2128980
|
Chance of selecting the last k pages in correct order from a set of n pages
I have a set of documents numbered 1 to n. If I arrange the documents randomly, what is the chance of selecting the last k pages in the correct order if I select them at random from the documents? Because order is important, I use the following formula: [imath]\displaystyle\frac{n!}{(n-k)!}[/imath] From this I calculate how many ways there are to select k pages from n pages in the correct order. I need to have the last k pages. Which I can do in only one way. So is the chance 1/[imath]\displaystyle\frac{n!}{(n-k)!}[/imath]? It seems unlikely. It seems to me spontaneously that the chance should be smaller.
|
2129978
|
[imath]\phi : A^n \to A^n [/imath] be a surjective [imath]A[/imath]-linear map then [imath]\phi[/imath] is injective as well.
Let [imath]A[/imath] be a commutative ring with [imath]1,[/imath] and [imath]\phi : A^n \to A^n [/imath] be a surjective [imath]A[/imath]-linear map for some natural number [imath]n.[/imath] Then show that [imath]\phi[/imath] is injective as well. Tensoring with [imath]A/m[/imath] for some maximal ideal [imath]m[/imath] in [imath]A[/imath] will give that tensor map is onto and being [imath]A/m[/imath] linear is injective. But from this map I cannot recover [imath]\phi[/imath] and claim that [imath]\phi[/imath] is also injective. Any help will be appreciated. Thanks.
|
153516
|
A surjective homomorphism between finite free modules of the same rank
I know a proof of the following theorem using determinants. For some reason, I'd like to know a proof without using them. Theorem Let [imath]A[/imath] be a commutative ring. Let [imath]E[/imath] and [imath]F[/imath] be finite free modules of the same rank over [imath]A[/imath]. Let [imath]f:E → F[/imath] be a surjective [imath]A[/imath]-homomorphism. Then [imath]f[/imath] is an isomorphism.
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2096779
|
Distribution of [imath](XY)^Z[/imath] for [imath](X,Y,Z)[/imath] i.i.d. uniform on [imath](0,1)[/imath]
Let [imath]X[/imath], [imath]Y[/imath] and [imath]Z[/imath] be i.i.d. uniform (0,1) random variables. What is the distribution of [imath](XY)^Z[/imath]? I've tried to solve it via mgfs, and what I've found is: [imath]E\left(e^{(XY)^Z}\right)=E\left(E\left(e^{(XY)^Z}|Z\right)\right)[/imath] [imath]=E\left(E\left(e^{(XY)}\right)^Z\right)[/imath] [imath]=E\left(2\left(\frac{e^t-1}{t}\right)^Z\right)[/imath] [imath]=2\int_0^1\left(\frac{e^t-1}{t}\right)^zdz[/imath] [imath]=2\int_0^1\exp\left({z\log\left(\frac{e^t-1}{t}\right)}\right)dz[/imath] [imath]=2\left[\frac{\exp\left({z\log\left(\frac{e^t-1}{t}\right)}\right)}{\log\left(\frac{e^t-1}{t}\right)}\right]_0^1[/imath] [imath]=2\frac{{\frac{e^t-1}{t}}-1}{\log\left(\frac{e^t-1}{t}\right)}[/imath] But I don't see where to go from here. It doesn't look like the mgf of any distribution that I know. How can I determine the distribution from this, or is there another way?
|
2130275
|
Prove that random variable is uniformly distributed
The time when I studied probability theory was about three years ago, so I have forgot many things and want to refresh them. I want to prove that if [imath]X,Y,Z[/imath] are independent and uniformly distributed on [imath][0,1][/imath] random variables then [imath](XY)^Z[/imath] is also uniformly distributed on [imath][0,1][/imath]. As I know uniformly distributed on [imath][0,1][/imath] variable has density [imath]\rho (t)= \begin{cases} 1, t \in [0,1]\\ 0, t \notin [0,1] \end{cases}[/imath]. Denote [imath]\xi = XY[/imath]. Since [imath]X[/imath] and [imath]Y[/imath] are independent [imath]\rho_\xi (t) = \rho_X (t) \rho_Y (t) = \begin{cases} 1, t \in [0,1]\\ 0, t \notin [0,1] \end{cases}[/imath], so [imath]\xi[/imath] is uniformly distributed on [imath][0,1][/imath]. Denote [imath]\zeta = (XY)^Z[/imath]. Since [imath]\xi[/imath] and [imath]Z[/imath] are independent, we obtain in the same way that [imath]\rho_\zeta (t) = \rho_\xi (t) \rho_Z (t) = \begin{cases} 1, t \in [0,1]\\ 0, t \notin [0,1] \end{cases}[/imath] and we are done. It seems to me that I'm wrong. Can anyone explain me why it is incorrect (I believe that it is) and show the right solution?
|
2130229
|
Evaluating [imath]\int \frac{1-7\cos^2x}{\sin^7x\cos^2x}dx[/imath]
How do i evaluate [imath]\int \frac{1-7\cos^2x}{\sin^7x\cos^2x}dx[/imath]. I tried using integration by parts and here is my approach [imath]\int \frac{ sinx}{(1-cos^2x)^4\cos^2x} dx[/imath] and then put [imath]cos x=t[/imath] and then tried to use partial fractions.I applied similar logic for the other part.But that made it lengthy to solve as decomposition into partial fractions is very time consuming.This question came in an objective examination in which time was limited.Can anyone help me with a shorter way to solve this problem.Thanks.
|
385636
|
Evaluating this integral : [imath] \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx [/imath]
The question : [imath] \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx [/imath] I tried dividing by [imath]\cos^2 x[/imath] and splitting the fraction. That turned out to be complicated(Atleast for me!) How do I proceed now?
|
2129989
|
How to get ratio of a,b,c from 2 equations in a,b,c
I have 2 equation in terms of [imath]a,b[/imath] and [imath]c[/imath] . [imath]3a + 10b + 5c =0[/imath] and [imath]4a + 6b + 2c =0[/imath] I need to find a:b:c and answer is [imath]\dfrac a5 = \dfrac {b}{-7} = \dfrac {c}{11}[/imath] I want to know how to get that? My attempt: Given equations can be written in form [imath]\left ( \begin{matrix} 3 & 10 & 5 \\ 4 & 6 & 2 \\ \end{matrix} \right ) \left ( \begin{matrix} a\\ b\\ c\\ \end{matrix} \right ) = \left ( \begin{matrix} 0 \\ 0\\ \end{matrix} \right ) [/imath] But here I don't have any idea how to proceed. I can do for 3×3 matrix and I thought it can be done in same way. Thanks.
|
2131625
|
Unable to solve it,will someone help?
Quote: [imath]\begin{aligned}3a+10b+5c&=0\\4a+6b+2c&=0\end{aligned}[/imath] solves to [imath]\frac{a}{30-20}=\frac{b}{6-20}=\frac{c}{-18+40}[/imath] so that [imath]\frac{a}{5}=\frac{b}{-7}=\frac{c}{11}[/imath] Can someone explain in deep how this equations are solved in this way? First I thought it to do with using Cramer's rule but we are all aware that Cramer's rule work only for square matrix. And we can see this are only 2 equations with will not make matrix square. So in my last question I was tempted to do only using Cramer's rule and now it has nothing to do with previous question! I wish someone will find helping more important
|
2131795
|
Evaluating [imath]\lim\limits_{x \to \infty}x\left(\left(1+\frac{1}{x}\right)^x-e \right)[/imath]
I'm trying to find the limit of: [imath]\lim\limits_{x \to \infty}x\left(\left(1+\frac{1}{x}\right)^x-e \right)[/imath] I tried to use L'Hôpital but it lead me nowhere. Any ideas?
|
1962293
|
Compute [imath]\lim_{x\to\infty}x\;\left[\left(1+\frac{1}{x}\right)^x-e\right][/imath]
How can I compute the following limit, and is there a general method to resolve problems of this type? [imath]\lim_{x\to\infty}x\left[\left(1+\frac{1}{x}\right)^x-e\right][/imath]
|
2129747
|
Find all values of [imath]n[/imath] such that [imath]\phi(n)=100[/imath]
How do I go about solving this problem, and similar ones where I have to find the number of values that satisfy this phi function? Thanks!
|
1157163
|
What method would you go through to find an [imath]n[/imath] such that [imath]\phi(n)=100[/imath]?
So, clearly, 101 is a solution. But [imath]n=125[/imath] is also a solution, so is 250. How would one come to that solution without knowing these values beforehand?
|
2132060
|
Definite integration question [imath]\int_0^{\frac{\pi}{2}} \frac{dx}{3+2\sin x+\cos x}[/imath]
Question - [imath]\int_0^{\frac{\pi}{2}} \frac{dx}{3+2\sin x+\cos x}[/imath] I have tried this question and also tried to find online on google. But I only found questions that involves only 2 terms. So i dont have any other way except to ask it here. Please help me.
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1712481
|
Simplifying the integral [imath]\int\frac{dx}{(3 + 2\sin x - \cos x)}[/imath] by an easy approach
[imath]I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}[/imath] If [imath]\tan\left(\frac{x}{2}\right)=u[/imath] or [imath]x=2\cdot\tan^{-1}(u)[/imath] Then, [imath]\sin{x}=\dfrac{2u}{1+u^2}[/imath] [imath]\cos{x}=\dfrac{1-u^2}{1+u^2}[/imath] [imath]dx=\dfrac{2}{1+u^2}[/imath] Substitute [imath]\tan\left(\dfrac{x}{2}\right)=u[/imath] Let us simplify the integrand before integrating [imath]\dfrac{1}{3+2\sin{x}-\cos{x}}[/imath] [imath]=\dfrac{1}{3+2\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}}[/imath] [imath]=\dfrac{1}{3+\frac{4u-1+u^2}{1+u^2}}[/imath] [imath]=\dfrac{1}{\frac{4u-1+u^2+3+3u^2}{1+u^2}}[/imath] [imath]=\dfrac{1+u^2}{4u^2+4u+2}[/imath] [imath]=\dfrac{1+u^2}{(2u+1)^2+1}[/imath] [imath]I=\displaystyle\int\dfrac{1+u^2}{(2u+1)^2+1}\cdot\dfrac{2}{1+u^2}\ du[/imath] [imath]=\displaystyle\int\dfrac{1}{(2u+1)^2+1}\ 2du[/imath] Now, Take : [imath]v=2u+1[/imath] Therefore, [imath]dv=2du[/imath] [imath]I=\displaystyle\int\dfrac{1}{v^2+1}\ dv[/imath] [imath]I=\tan^{-1}(v)[/imath] Substitute everything back [imath]I=\tan^{-1}(2u+1)[/imath] [imath]I=\tan^{-1}\left(2\tan\left(\dfrac{x}{2}\right)+1\right)[/imath] [imath]\boxed{\displaystyle\int\dfrac{dx}{(3 + 2\sin x - \cos x)}=\tan^{-1}\left(2\tan\left(\dfrac{x}{2}\right)+1\right)+C}[/imath] I know that my approach is also not so difficult but still I think there must be an relatively easy approach to this integral. I have tried many different things using trigonometric identities but nothing seems to bring to the solution easily. Kindly help me out.
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2132359
|
A curious curiosity
[imath]12^2 = 144[/imath] Now reverse the digits and square it: [imath]21^2 = 441[/imath] The digits in the result have also gotten reversed. The same is true for [imath]13[/imath]: [imath]13^2 = 169[/imath], [imath]31^2 = 961[/imath] As far as I know, this does not hold true for any other number (except, trivially, [imath]0[/imath] through [imath]11[/imath]). Why? Can it be proven? The only thing that occurs to me is that, starting with the next number, [imath]14[/imath], the square of its reverse, [imath]41[/imath], is greater than [imath]1000[/imath]. But I don't know why this would make any difference.
|
1175973
|
reversing digits and squaring
If we reverse the digits of [imath]12[/imath] we will get [imath]21[/imath]. [imath]12^{2}=144[/imath]. If we reverse its digits we will get [imath]441[/imath] which is [imath]21^{2}[/imath]. Here is the puzzle. How many such two digit numbers are there? Digits must be different. We got an algebraic method to solve it if the square is a three digit number.It given here : Let [imath]a>b[/imath] [imath](10a+b)^{2}=100x+10y+z[/imath] and [imath](10b+a)^{2}=100z+10y+x[/imath] Difference between these two equations will lead to [imath]a^{2}-b^{2}=x-z[/imath] [imath]x-z \leq 8[/imath].Then [imath]a^{2}-b^{2} \leq 8[/imath]. Since squares are of three digits [imath]a <4[/imath].But since [imath]a>b ,a \neq 1[/imath]. [imath]b=1 \Longrightarrow a^{2} \leq 9[/imath].Then [imath]a=1,2,3[/imath] but only possibilities are [imath]2 [/imath] and [imath]3[/imath]. [imath]b=2 \Longrightarrow a^{2} \leq 12 \Longrightarrow a= 1,2,3[/imath]. But possibility is [imath]a=3[/imath]. So next such pair will be [imath]13,31[/imath]. Higher values of [imath]b[/imath] will not work since [imath]a[/imath] cannot exceed 3. This is an explanation we obtained from a mathematics group.It works well if the square is a three digit number.What will happen if square is a 4 digit number... Or if we try to extend this problem for numbers having more than two digits ?
|
2131823
|
Prove that [imath]f(7) = 56[/imath] given [imath]f(1)[/imath] and [imath]f(9)[/imath] and [imath]f' \le 6[/imath]
Let [imath]f(x)[/imath] be continues function at [imath][1,9][/imath] and differentiable at [imath](1,9)[/imath] and also [imath]f(1) = 20 , f(9) = 68 [/imath] and [imath] |f'(x)| \le 6[/imath] for every [imath]x \in (1,9)[/imath]. I need to prove that [imath]f(7) = 56[/imath]. I started by using the Lagrange theorem and found that there exist [imath] 1<c<9[/imath] such that [imath]f'(c) = 6[/imath] but I'm not sure how is this relevant and how to proceed.
|
1182448
|
Let [imath]f(x)[/imath] be differentiable at [imath]\mathbb R[/imath], s.t [imath]|f^\prime (x)| \le 6[/imath] . Its given that [imath]f(1)=20[/imath], and [imath]f(9)=68[/imath] , prove that [imath]f(7)=56[/imath].
Let [imath]f(x)[/imath] be differentiable ate [imath]\mathbb R[/imath], s.t [imath]|f^\prime (x)| \le 6[/imath] for every [imath]x[/imath] in [imath]\mathbb R[/imath]. its given also that [imath]f(1)=20[/imath], and [imath]f(9)=68[/imath] , prove that [imath]f(7)=56[/imath]. I'm thinking about applying [imath]\text{Mean Value theorem}[/imath] and [imath]\text{Intermediate Value theorem}[/imath], in here but for some reason I miss something and I can't conclude that [imath]f(7)=56[/imath]. any kind of help would be appreciated.
|
2131706
|
Evaluate a limit involving the inverse of a function
Let [imath]f : [1, \infty) \to [1, \infty)[/imath] be a function such that [imath]f(x) = x(1 + \ln x)[/imath]. Prove that [imath]f[/imath] is bijective and then calculate: [imath]\lim_{x \to \infty} \frac{f^{-1}(x) \ln x}{x}[/imath] I have no difficulties in proving that [imath]f[/imath] is bijective, but I can't calculate the limit. I've tried using l'Hospital's rule but got nothing meaningful. Thank you in advance!
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2120341
|
Finding out the limit [imath]\lim_{a \to \infty} \frac{f(a)\ln a}{a}[/imath]
For any real number [imath]a \geq 1[/imath] let [imath]f(a)[/imath] denote the real solution of the equation [imath]x(1+\ln x)=a[/imath] then the question is to find out [imath] \lim_{a \to \infty} \frac{f(a)\ln a}{a}[/imath]. It is clear that if we denote [imath]h(a)[/imath] by [imath]h(a)=a(1+\ln a)[/imath] then [imath]f(a)[/imath] is the inverse function of [imath]h(a)[/imath]. Also [imath]f(a)[/imath] is increasing function in its domain. Also the limit persuades using lhospital's but I cannot see how to apply it here. Thanks.
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2132966
|
How do I prove this trigonometric inequality?
If [imath]A,B,C \in (0,\frac{\pi}{2})[/imath]. Then prove that [imath]\frac{\sin(A+B+C)}{\sin(A)+\sin(B)+\sin(C)} < 1[/imath]
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1994468
|
If [imath] a,b,c\in \left(0,\frac{\pi}{2}\right)\;,[/imath] Then prove that [imath]\frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1[/imath]
If [imath]\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,[/imath] Then prove that [imath]\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1[/imath] [imath]\bf{My\; Try::}[/imath] Using [imath]\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+c)+\cos a\cdot \sin (b+c)[/imath] [imath] = \sin a\cdot (\cos b\cos c-\sin b\sin c)+\cos a(\sin b\cos c+\cos b\sin c)[/imath] [imath] = \sin a\cos b\cos c-\sin a\sin b\sin c+\cos a \sin b\cos c+\cos a\cos b\sin c[/imath] Now how can i solve it after that , Help required, Thanks
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2133019
|
Disprove or prove [imath]\sum_{n\ge1} \frac{\vert{\sin n}\vert}{n} [/imath] is divergent.
I want to prove or disprove [imath]\displaystyle\sum_{n\ge1} \frac{{\sin n}}{n}[/imath] is absolutely convergent. I can prove [imath]\displaystyle \sum_{n\ge1} \frac{{\sin n}}{n}[/imath] is convergent by dirichlet's test (Can I prove that [imath]\displaystyle \sum_{n\ge1} \frac{{\sin n}}{n}[/imath] converges without dirichlet's test), but I can't prove or disprove that [imath]\displaystyle\sum_{n\ge1} \frac{\vert{\sin n}\vert}{n} [/imath] converges. Any help would be appreciated.
|
1005902
|
Is the Series Sin(k)/k Absolutely Convergent?
I know that the series [imath]\sum_{k=1}^\infty \frac{\sin k}{k}[/imath] converges (to [imath]\frac{\pi - 1}{2}[/imath]), though by crazy stuff with Dirichlet Kernels or by reverse-engineering [imath]\frac{\pi - x}{2} = \sum \frac{\sin (kx)}{k}[/imath] using Fourier Series. My question instead: Is [imath]\sum_{k=1}^\infty \frac{\sin k}{k}[/imath] absolutely convergent? My gut instinct is no: It behaves not-much-better than [imath]\sum \frac1k[/imath], but that's certainly not a rigorous proof.
|
2122166
|
Is there an Intuitive explanation of topological continuity?
I am used to the definition of continuity used in real analysis, based on the idea that if [imath]f[/imath] is continuous at [imath]a[/imath], you can get arbitrarily close to [imath]f(a)[/imath] by going arbitrarily close to [imath]a[/imath]. However, the topological concept of continuity of [imath]f: M \to N[/imath] is the condition that for all [imath]S \subset N[/imath], the preimage of [imath]S[/imath] in [imath]M[/imath] must be open subsets. What does this have to do with the intuitive concept of "continuity"? I cannot even picture how this relates to continuity in Euclidian space, let alone how it would relate to continuity in more abstract topological spaces.
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15963
|
What is the intuition for the point-set topology definition of continuity?
Let [imath]X[/imath] and [imath]Y[/imath] be topological spaces. A function [imath]f: X \rightarrow Y[/imath] is defined as continuous if for each open set [imath]U \subset Y[/imath], [imath]f^{-1}(U)[/imath] is open in [imath]X[/imath]. This definition makes sense to me when [imath]X[/imath] and [imath]Y[/imath] are metric spaces- it is equivalent to the usual [imath]\epsilon-\delta[/imath] definition. But why is this a good definition when [imath]X[/imath] and [imath]Y[/imath] are not metric spaces? How should we think about this definition intuitively?
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2133084
|
Find all real number satisfying [imath]10^x+11^x+12^x = 13^x+14^x[/imath]
Find all real number [imath]x[/imath] satisfying [imath]10^x+11^x+12^x = 13^x+14^x[/imath] My Work Dividing by [imath]13^x[/imath] we get [imath]\left( \frac{10}{13} \right)^x + \left( \frac{11}{13} \right)^x + \left( \frac{12}{13} \right)^x = 1 + \left( \frac{14}{13} \right)^x[/imath] The LHS is a decreasing function of [imath]x[/imath] and the RHS is an increasing function of [imath]x[/imath]. So there is only one intersection in their graph. I am looking for a formal way to find the root. I know that [imath]x=2[/imath] works. But how to formally find this root?
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2074117
|
Solve [imath]10^x+11^x+12^x = 13^x+14^x[/imath]
Find all real numbers [imath]x[/imath] for which [imath]10^x+11^x+12^x = 13^x+14^x.[/imath] I thought about taking the logarithm of both sides, but didn't see how that would help. How can we make the equation simpler?
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2132126
|
Multiplication of ring cosets
Let [imath]R[/imath] be a commutative ring, [imath]A[/imath] an ideal of [imath]R[/imath], and [imath]R/A[/imath] an integral domain. Then, as I can see in my book, [imath](a+A)(b+A)=ab+A=A[/imath]. But how does this multiplication exactly go? If [imath](a+A)(b+A)=ab + aA + Ab + A[/imath] then where does the sum of the terms [imath]aA+Ab[/imath] go? Also, why does [imath](a+A)(b+A)=A[/imath] imply that [imath](a+A)=A[/imath] or [imath](b+A)=A[/imath]? Thank you for your clarifications.
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446492
|
What does "defining multiplication in quotient rings" actually mean?
Say [imath]R[/imath] is a commutative ring and [imath]I\in R[/imath] is an ideal. Let us consider the quotient [imath]R/I[/imath]. It is created by taking every element [imath]a\in R[/imath], and adding all the elements of [imath]I[/imath] to it. The elements of [imath]R/I[/imath] are of the form [imath]a+I[/imath] and [imath]b+I[/imath], [imath]\forall a,b\in R[/imath]. Now [imath](a+I).(b+I)=(a+i_{1}).(b+i_{2}),\forall (i_{1},i_{2})\in I \times I[/imath] [imath](a+i_{1}).(b+i_{2})=ab+a.i_{2}+b.i_{1}+i_{1}i_{2}=ab+I[/imath] We know that [imath]a.i_{2}+b.i_{1}+i_{1}i_{2}\in I[/imath]. If [imath](a+I).(b+I)=ab+I[/imath], then taking suitable [imath](i_{1},i_{2})\in I\times I[/imath], we should be able to prove [imath](a+i_{1}).(b+i_{2})=ab+i_{3},\forall i_{3}\in I[/imath]. However, can every element in [imath]I[/imath] be generated by taking suitable [imath]i_{1},i_{2}\in I[/imath]? And if not, is that the reason why [imath](a+I).(b+I)[/imath] has to be defined as equal to [imath]ab+I[/imath] in violation of the distributive property of ring elements? This has confused me for a long time. EDIT: I figured addition does not have to be defined as such because [imath](a+i_{1})+(b+i_{2})=a+b+i_{1}+i_{2}[/imath], where [imath]i_{1}+i_{2}\in I[/imath]. In fact, every element [imath]i\in I[/imath] can be constructed by taking [imath]i_{1}=0[/imath] and [imath]i_{2}=i[/imath]. Hence, [imath](a+I) + (b+I)=a+b+I[/imath] naturally. This is the logic I followed to determine that [imath](a+I).(b+I)[/imath] doesn't quite work as nicey. I'm not sure if this logic is flawed or not as I haven't had the opportunity to ask anybody. Thanks for your help in advance!
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2133289
|
Simplification of nested module operations.
Can I simplify this formula to use only one module operation? [imath](x + (y \text{ mod } z)) \text{ mod }z[/imath] Intuitively I simplified it so: [imath](x + y) \text{ mod } z[/imath] This seems to do the same as the original. However, I don't know why it works nor if this is right.
|
1996607
|
Prove $(x+y) \text{ mod } n = ((x \text{ mod } n)+(y \text{ mod } n)) \text{ mod } n,\,$ and product analog
For all [imath]n \in \mathbb{N}[/imath], [imath]n > 1[/imath] and [imath]x,y \in \mathbb{Z}[/imath] prove that \begin{equation} (x+y) \text{ mod } n = ((x \text{ mod } n)+(y \text{ mod } n))\text{ mod } n. \end{equation} Would this be a proof if I say: Let [imath]x = x' \text{ mod } n[/imath] and let [imath]y=y' \text{ mod } n[/imath]. Then \begin{equation} x+y = x'+y' \text{ mod } n. \end{equation} Now, if we look at the current thing we want prove, this will follow: \begin{equation} x+y \text { mod } n = (x \text{ mod } n + y \text{ mod } n) \text{ mod } n. \end{equation} I'm not sure if this is allowed or will count as a "proof" at all. Maybe you can show me some other ways of doing this, more precise ones?
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1558013
|
Friend B and C have eaten zero apples. How many more apples has C eaten?
Friend [imath]A[/imath] claims that he has eaten [imath]1[/imath] apple today. Friend [imath]B[/imath] responds. Congrats, I have eaten [imath]0[/imath] apples, so that is [imath]\infty[/imath] more apples than me. Friend [imath]C[/imath] says, but I have also eaten [imath]0[/imath] apples. So how many times more than you have I done? [imath]0[/imath] or [imath]\infty[/imath]?
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86533
|
How many times more than [imath]0[/imath]?
If I have [imath]10[/imath] apples, but you have [imath]5[/imath] apples, then I have [imath]2[/imath] times more apples than you. But what if I have [imath]10[/imath] apples, but you don't have any apples? If you look at the graph [imath]f(x)=\frac{10}{x}[/imath], it shows that when [imath]x[/imath] approaches [imath]0[/imath], [imath]f(x)[/imath] approaches infinity. So it means I have infinity times more apples than you?
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2133901
|
How do I simplify [imath]E[E(ZX\mid Y)\mid Z,Y][/imath]?
[imath]E[E(ZX\mid Y)\mid Z,Y][/imath] can be simplified? Any comment would be much appreciated.
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2133793
|
Conditional probability, [imath]E[E(X\mid Y)\mid Y,Z]=?[/imath]
I know that [imath]E[E(X\mid Y,Z)\mid Y]=E(X\mid Y)[/imath]. What if [imath]E[E(X\mid Y)\mid Y,Z]=?[/imath] It has any meaning?
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1602968
|
Maximal ideal in [imath]\mathfrak{R}[x][/imath]
Let [imath]\mathfrak{R}[/imath] be a commutative ring with identity. Show that if there exists a monic polynomial [imath]p(x)\in \mathfrak{R}[x][/imath] of degree at least one such that the ideal [imath](p(x)) \subseteq\mathfrak{R}[x][/imath] is maximal, then [imath]\mathfrak{R}[/imath] is a field. Thanks.
|
819876
|
Question about fields and quotients of polynomial rings
I don't see how to solve the following problem: Let [imath]R[/imath] be a commutative and unitary ring. If there exists a monic polynomial [imath]f(x) \in R[x][/imath] so that [imath]R[x]/(f(x))[/imath] is a field, show that [imath]R[/imath] is a field. What about if [imath]f(x)[/imath] isn't monic? I don't have a clue about how proving it. I see that if [imath]R[/imath] is a field and [imath]f(x)[/imath] a monic and irreducible polynomial in [imath]R[X][/imath] then [imath]\frac{R[x]}{(f(x))}[/imath] is a field, but that is not the question! Thank you in advance for your help
|
2133958
|
Group actions on topological spaces
There exists a connected topological space such that the permutations group [imath]S_3[/imath] acting without fixed points? I try to consider matrice spaces.
|
2091757
|
Give an example of a topological space [imath]X [/imath] such that [imath]S_3[/imath] acts freely on [imath]X [/imath]
I need an example of a topological space [imath] X[/imath] such that [imath]S_3[/imath] acts freely on [imath]X [/imath]. I tried to find an easy example with google, but I didn't suceed.
|
2134880
|
How to prove isomorphisms of quotient rings
I know that [imath]\mathbb Z[\sqrt{-5}] \cong \mathbb Z[x]/(x^2 + 5)[/imath], but how do I deal with quotients? For example, I've seen that [imath]\mathbb Z[\sqrt{-5}]/(2, 1 + \sqrt{-5}) \cong \mathbb Z[x]/(x^2 + 5, 2, 1 + x)[/imath] but no matter how many other questions I look at, I can't find an answer that explains this rigorously, and in detail. Another example is [imath]\mathbb Z[x]/(2, 1 + x) \cong \mathbb Z/2\mathbb Z[/imath]. Again, I know that [imath]\mathbb Z[x]/(2) \cong \mathbb Z_2[x][/imath] and [imath]\mathbb Z[x] / (1 + x) \cong \mathbb Z[/imath], and I also know by 3rd Isom. Thm that [imath]\mathbb Z[x]/(2, 1 + x) \cong (\mathbb Z[x]/(2))/((2, 1+x)/(2))[/imath], but I what is a rigorous explanation for why the result follows?
|
2134753
|
Explain these ring isomorphisms
I need help understanding the following isomorphisms: [imath]\mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5}) \cong \mathbb Z[X]/(X^2+5,2,1+X) = \mathbb Z[X]/(2,1+X) \cong \mathbb Z/2\mathbb Z[/imath] In general, I am also wondering whether the following are true, and if so briefly why: [imath]\mathbb Z[\sqrt{D}] \cong \mathbb Z[x]/(x^2 - D)[/imath] for all squarefree integers [imath]D[/imath] [imath](R/I)/J \cong R/(I +J)[/imath], ie why do we have things like [imath]\mathbb Z[x]/(2,x^2+5)\simeq \mathbb Z_2[x]/(x^2+5) [/imath] where [imath]\mathbb Z[x] / (2) \cong \mathbb Z_2[x][/imath]? Also for example for something like the following: [imath]\,\mathbb Z[x]/(2,x^2+5)\simeq \mathbb Z_2[x]/(x^2+1)[/imath] Is it true because of the third isomorphism theorem? I'm guessing we set [imath]R = \mathbb Z[x][/imath], [imath]I = (2)[/imath], and [imath]J = (x^2 + 5)[/imath], but please expand more on each step of the reduction.
|
2135070
|
Calculate [imath]\int_{-\infty}^{\infty}{\frac{dx}{x^{2n}+1}}[/imath]
While studying for exam I encountered this problem: [imath]\int_{-\infty}^{\infty}{\frac{dx}{x^{2n}+1}}[/imath]=? As a complex function it has [imath]2n[/imath] singularities : S=[imath] e^{\frac{\pi i(2k+1)}{2n}}|k\in \mathbb{N}, 0\leq k <2n-1 \brace[/imath] [imath]\int_{-\infty}^{\infty}{\frac{dx}{x^{2n}+1}}=\lim_{R\to \infty}\int_{-R}^{R}{\frac{dx}{x^{2n}+1}}[/imath] so using residue theorem : [imath]\int_{-R}^{R}{\frac{dx}{x^{2n}+1}} + \int_{C^{+}_{R}}{\frac{dz}{z^{2n}+1}} = 2\pi i\sum res(f,s_k)\space[/imath] for [imath]s_k \in S[/imath] (for [imath]s_k[/imath] in top half plane). It is easy to see that the second integral vanished as [imath]R \to \infty[/imath]. so: [imath]\int_{-\infty}^{\infty}{\frac{dx}{x^{2n}+1}}= 2\pi i\sum res(f,s_k)\space[/imath] but calculating [imath]res(f,s_k[/imath]) seems complicated... each singularity is of order 1 to [imath]res(f,s_k) = lim_{z\to s_k}\frac{(z-s_k)}{\Pi(z-s_j)}[/imath] haven't got any further... thanks
|
544424
|
Calculation of an integral via residue.
[imath]\int_{-\infty}^{\infty}{{\rm d}x \over 1 + x^{2n}}[/imath] How to calculate this integral? I guess I need to use residue. But I looked at its solution. But it seems too complicated to me. Thus, I asked here. Thank you for help.
|
1691770
|
|[imath]deg^+ (v) - deg^- (v)[/imath]| [imath]\le 1[/imath].
Prove or disproved that for every undirected graph [imath]G[/imath], there is an orientation [imath]D[/imath] on [imath]G[/imath] such that |[imath]deg^+ (v) - deg^- (v)[/imath]| [imath]\le 1[/imath] ??
|
396009
|
Prove the edges of a multigraph may be oriented such that the net-degree of any vertex is [imath]\leq 1[/imath].
The net-degree of a vertex [imath]v[/imath], denoted [imath]\text{netdeg}(v)[/imath], in a digraph [imath]G[/imath] is defined by [imath] \text{netdeg}(v)=| ~ \text{outdeg}(v) - \text{indeg}(v) ~| [/imath] where [imath]\text{outdeg}(v)[/imath] and [imath]\text{indeg}(v)[/imath] are the out-degree and in-degree of [imath]v[/imath]. Show that the edges of any multigraph may be oriented so that [imath]\text{netdeg}(v) \leq 1[/imath] for all its vertices [imath]v[/imath]. My first thought was to order the vertices and alternate edges in and out beginning at the first vertex then move on to the next vertex and orient the remaining edges depending on the net-degree of the remaining vertices, but this approach does not seem to work.
|
2131632
|
Is [imath]\mathbb Z[i/2]:=\{f(i/2) : f(x) \in \mathbb Z[x] \}[/imath] dense in [imath]\mathbb C[/imath]?
Is [imath]\mathbb Z[i/2]:=\{f(i/2) : f(x) \in \mathbb Z[x] \}[/imath] (smallest subring containing [imath]\mathbb Z[/imath] and [imath]i/2[/imath] ) dense in [imath]\mathbb C[/imath] ? NOTE : [imath]\mathbb Z [i/2] \ne \mathbb Z + \dfrac i2 \mathbb Z [/imath]
|
808355
|
When is [imath]\mathbb{Z}[\alpha][/imath] dense in [imath]\mathbb{C}[/imath]?
Let [imath]\alpha[/imath] be a nonreal algebraic number. I'm interested in conditions that imply that [imath]\mathbb{Z}[\alpha][/imath] is dense in [imath]\mathbb{C}[/imath]. I'm particularly interested algebraic integer [imath]\alpha[/imath]. This is what I know so far: if there is a [imath]n \in \mathbb{N}[/imath] such that [imath]\alpha^n \in \mathbb{R} \setminus \mathbb{Z}[/imath], then [imath]\mathbb{Z}[\alpha][/imath] is dense in [imath]\mathbb{C}[/imath]; algebraic integers of degree two don't satisfy the condition, although algebraic nonintegers of degree two may. Many thanks in advance.
|
2135136
|
Could be this : [imath]7131372917538397234773191167617941438959[/imath] written as [imath]x^{2}+y^{2}[/imath] with [imath]x, y[/imath] integers?
I have constructed the number [imath]7131372917538397234773191167617941438959[/imath], which is prime as shown here, using all primes under [imath]100[/imath] which contains two digits by randomly ordering them as [imath]31,37,29,\ldots, 59[/imath] except [imath]13[/imath]. I have got that number [imath]7131372917538397234773191167617941438959[/imath] satisfies the following properties: 1.- The sum of its digits is also prime: it is equal to [imath]193[/imath]. 3.- The number is of the form [imath]6n+1[/imath] 4.- This number can't be written as a sum of [imath]3[/imath] squares. Now my question here is: Could be this : [imath]7131372917538397234773191167617941438959[/imath] written as [imath]x^{2}+y^{2}[/imath] with [imath]x, y[/imath] integers?
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966517
|
How to select the right modulus to prove that there do not exist integers [imath]a[/imath] and [imath]b[/imath] such that [imath]a^2+b^2=1234567[/imath]?
I understand the solution but I don't know how the author decided to start with modulo 4 instead of something else? What is it about the expression [imath]a^2+b2=1234567[/imath] that would trigger us to select modulo 4 instead of something else. I tried to solve the question in a similar manner using modulo 2 but eventually got stuck. This leads me to believe that this question can only be solved using modulo 4. Is this true?
|
2135717
|
Cyclic group [imath]G[/imath] with generator [imath]ab[/imath].
Let [imath]G[/imath] be an Abelian group of order [imath]mn[/imath] where [imath]\gcd(m,n)=1[/imath]. Assume that [imath]G[/imath] contains an element of [imath]a[/imath] of order [imath]m[/imath] and an element [imath]b[/imath] of order [imath]n[/imath]. Prove [imath]G[/imath] is cyclic with generator [imath]ab[/imath]. The idea is that [imath](ab)^k[/imath] for [imath]k \in [0, \dots , mn-1][/imath] will make distinct elements but do not know how to argue it. Could I say something like [imath]<a>=A[/imath], [imath]<b>=B[/imath], somehow [imath]AB=\{ ab : a \in A , b \in B \}[/imath] and that has order [imath]|A||B|=mn[/imath]? Don't know if it's the same exact or similar to Finite group of order [imath]mn[/imath] with [imath]m,n[/imath] coprime.
|
652377
|
gcd and order of elements of group
Suppose a and b are elements of group G and both a and b have finite order. How do I show if [imath]ab = ba[/imath], [imath]gcd(|a|, |b|)= 1 \implies |ab| = |a||b|[/imath]
|
986231
|
Is a trigonometric function applied to a rational multiple of [imath]\pi[/imath] always algebraic?
Specifically, just to talk about cosine, is it true that [imath]\cos(\frac{a\pi}{b})[/imath] is algebraic for integers [imath]a[/imath] and [imath]b[/imath]? Looking at this post and the link to trigonometric constants in the comments, it seems likely that this is true. But most of the values calculated there are the result of sum/difference of angle formulas for existing algebraic values of sine and cosine. This came up when looking at [imath]\cos(\frac{\pi}{7})[/imath]. If this is algebraic, how can we calculate it? If it is not, for which arguments will sine and cosine be algebraic and for which arguments will they be transcendental?
|
1933177
|
Proof that a trigonometric function of a rational angle must be non-transcendental
Suppose I was working in radians and took [imath]\sin \left(\frac{a\pi}{b}\right)[/imath] where both [imath]a[/imath] and [imath]b[/imath] are integers. Is there a proof that the output of this function cannot be transcendental? Or, conversely, that [imath]\arcsin \left(t \right)[/imath] where [imath]t[/imath] is some transcendental number, e.g., [imath]e[/imath], cannot be a rational angle [imath]\left(\frac{a\pi }{b}\right)[/imath]? Sadly, I couldn't get very far with this problem with my maths teacher so any information on the topic would be very useful.
|
2135257
|
Rigorous Proof of a Limit, where f(x) ≡ A.
Suppose that [imath]f[/imath](x) is a function on (0, +∞), satisfying that [imath]f[/imath](2x) = [imath]f[/imath](x). If lim x→∞ [imath]f[/imath](x) = A, prove that [imath]f[/imath](x) ≡ A for all x ∈ (0, +∞). Hi, I am trying to understand this question but I have no idea how to approach this. I also don't fully understand why the 'identical to' sign is there rather than the equal sign. Any help would be appreciated!
|
2135143
|
What does this notation mean? [imath]f(x) \equiv A[/imath]?
If [imath]\lim_{x\rightarrow\infty}f(x) = A[/imath], prove that [imath]f(x)\equiv A[/imath]. Any help would be appreciated! I just don't even understand the question.
|
2135972
|
Let [imath]x > 0[/imath]. Prove that the value of [imath]\int^{x}_{0} \frac{1}{1+t^2}dt + \int^{\frac{1}{x}}_{0} \frac{1}{1+t^2}dt[/imath] does not depend on [imath]x[/imath].
Let [imath]x > 0[/imath]. Prove that the value of [imath]\int^{x}_{0} \frac{1}{1+t^2}dt + \int^{\frac{1}{x}}_{0} \frac{1}{1+t^2}dt[/imath] does not depend on [imath]x[/imath]. I really have no idea how to start this problem. How do I show something doesnt "depend" on something?
|
2134430
|
How to do this question that talks about dependency of x
Let [imath]x > 0[/imath]. Prove that the value of the following expression doesn't depend on x [imath]\int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt[/imath] Attempt: Left: f'(x) = [imath]\frac{1}{1+x^2}[/imath] Right: f'(x) = [imath]\frac{1}{1+(\frac{1}{x})^2} -\frac{1}{x^2}[/imath] [imath]= \frac{1}{(1+\frac{1}{x^2})} - \frac{1}{x^2}[/imath] [imath]=\frac{x^2}{1+x^2} - \frac{1}{x^2}[/imath] [imath]=\frac{x^4 - x^2 - 1}{1+x^2}[/imath] Yeah I don't know what I am doing, I tried to remove the integral but failed miserably
|
2136129
|
Proving uniform boundedness of sequence of functions
Let [imath](f_{n})[/imath] be a sequence in [imath]C[0,1][/imath] that is equicontinuous on [imath][0,1][/imath], and let [imath]p\in [0,1][/imath] be given. Show that if [imath](f_{n}(p))^{\alpha}_{n=1}[/imath] is bounded, then [imath](f_{n})[/imath] is uniformly bounded. Can I use Arzela Ascoli theorem to prove the above problem?
|
1993217
|
Why is this sequence of equicontinuous functions uniformly bounded?
Let [imath]\left\{f_{n}\right\}[/imath] be a sequence of equicontinuous functions where [imath]f_n: [0,1] \rightarrow \mathbf{R}[/imath]. If [imath]\{f_n(0)\}[/imath] is bounded, why is [imath]\left\{f_{n}\right\}[/imath] uniformly bounded?
|
2135304
|
Which quaternions are solutions of [imath]x^2+1=0[/imath]?
What would the final Quaternion Solutions look like for [imath]x^2+1=0[/imath]? I substituted in [imath]x = a+bi+cj+dk[/imath] and came up with a very long +/- square root.
|
1934479
|
Stuff which squares to [imath]-1[/imath] in the quaternions, thinking geometrically.
How can we think of the set[imath]\{x \in \mathbb{H} : x^2 = -1\}[/imath]geometrically? Is this set finite or infinite? Are there some more geometric ways of thinking about than meets the eye? Here, [imath]\mathbb{H}[/imath] denotes the quaternions.
|
2136221
|
To find product [imath]\cos{\theta}\cos{2\theta}\cos{3\theta}...\cos{999\theta}[/imath]
How to find out value of product [imath]\cos{\theta}\cos{2\theta}\cos{3\theta}...\cos{999\theta}[/imath] where [imath]\theta=\frac{2\pi}{1999}[/imath] i tried to multiply and divide by sin\theta to convert to sin2A, but couldnot do it. i would like to know many methods, if possible, to do this question thanks
|
1284744
|
Find the value of : [imath]\cos x \cos 2x...\cos 999x[/imath] given that [imath]x=\frac {2\pi}{1999}[/imath]
Given [imath]x=\frac {2\pi}{1999}[/imath] Find the value of [imath]\cos x \cos 2x \cos 3x ...\cos 999x[/imath] So I tried expanding [imath]\sin {2000x}=2\sin 1000x \cos 1000x[/imath] Then rewriting [imath]\cos 1000x= \cos {(999x+x)}[/imath] No luck so far.
|
2136634
|
The set [imath]\{\frac{\varphi(n)}n:n\in \Bbb N\}[/imath]
Let [imath]f(n)=\varphi(n)/n[/imath], where [imath]\varphi[/imath] is the totient function. Since [imath]0<\varphi(n)\le n[/imath] and [imath]\lim_{n\to \infty}f(p_n)=1[/imath] (where [imath]\{p_n\}[/imath] is the increasing sequence of primes) and [imath]\lim_{n\to\infty}f(n\#)=0[/imath] we know that [imath]\limsup f(n)=1[/imath] and [imath]\liminf f(n)=0[/imath]. But is the set [imath]\{f(n):n\in\Bbb N\}[/imath] dense in [imath][0,1][/imath]? Warning: this is not a problem from a book, so it might be very hard (honestly, I have no idea).
|
1076723
|
Prove that [imath]\{\frac{\phi (n)}{n}\}_{n \in \Bbb N}[/imath] is dense in [imath][0,1][/imath]
suppose that [imath]\phi(n)[/imath] is Euler function. prove that, [imath]\{\frac{\phi (n)}{n}\}_{n \in \Bbb N}[/imath] is dense in [imath][0,1][/imath] (if [imath]$A_n=\{1 \leq m \leq n \mid m \in \Bbb N ; \gcd(n,m)=1\}$[/imath] then [imath]\phi(n)=|A_n|[/imath]) I think : If [imath]p[/imath] is prime then [imath]\phi(p)=p-1[/imath]. for any [imath] \varepsilon > 0 [/imath] there is a prime number [imath]p[/imath] such that [imath]1-\varepsilon \leq \frac{p-1}{p} \leq 1[/imath]. for other elements i don't know what can i do.
|
2136734
|
the largest integer [imath]n[/imath] for which [imath]n+5[/imath] divivides [imath]n^5+5[/imath]?
When we divide [imath]n^5+5[/imath] by [imath]n+5[/imath], we get a remainder [imath]-620[/imath], i.e., [imath]n^5+5=K(n+5)-620[/imath] , now how to proceed further?
|
2132151
|
[imath]\max(\{n: n+5 | n^5 + 5\})[/imath]
What is the maximum value of [imath]n[/imath] if [imath]n + 5[/imath] divides [imath]n^5 + 5[/imath], [imath]n[/imath] being a natural number? I tried to solve by binomial theorem but it failed.
|
2134032
|
Prove that [imath]\lim_{x→0} f(x) = A[/imath].
Suppose that [imath]\lim_{x→0} f(x^3) = A[/imath], where [imath]A[/imath] is a constant number. Prove that [imath]\lim_{x→0} f(x) = A[/imath]. The domain of f is R. But that is the only information provided. I am completely lost on how to approach this. Any help would be greatly appreciated!
|
332859
|
Prove that if one of [imath]\lim_{x \rightarrow 0 }{f(x)}[/imath] and [imath]\lim_{x \rightarrow 0}{f(x^3)}[/imath] exists, then the other one also exists.
Prove that if one of [imath]\displaystyle\lim_{x \rightarrow 0 }{f(x)}[/imath] and [imath]\displaystyle\lim_{x \rightarrow 0}{f(x^3)}[/imath] exists, then the other one also exists. Can anyone guide me on this ? I have no idea on how to start .
|
2137373
|
Linear independence of the set of vectors
Let [imath]\{(a_{ij},...,a_{in}) \in T^n \mid i = 1,...,s \le n\}[/imath] be a set of vectors. Prove that if [imath]|a_{jj}| \gt \sum_{i=1, i\neq j}^s|a_{ij}|, \quad 1 \leq j \leq s[/imath] then the set is linearly independent. My attempt: I was trying an example with [imath]n=s=3[/imath], $|a_{11}| \gt [imath]\sum_{i=1, i\neq j}^{n=3}|a_{i1}|= |a_{21}|+|a_{31}|$ [/imath] $|a_{22}| \gt \sum_{i=1, i\neq j}^{n=3}|a_{i1}|= |a_{12}|+|a_{32}|$ $|a_{33}| \gt [imath]\sum_{i=1, i\neq j}^{n=3}|a_{i1}|= |a_{13}|+|a_{23}|$[/imath] For the linear independence we would have to satisfy, that $x=y=z=0$ \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} $[imath][/imath] But I just don't know what's next. Like the |a_{jj}|$ are on the places where are pivots and if these elements are after Gauss elimination nonzero, then our given set in linearly independent. But I don't know how to prove it. Thanks for help.
|
1733489
|
If [imath]\mid a_{jj}\mid \gt \sum_{i \neq j} \mid a_{ij} \mid[/imath] then vectors [imath]a_1,\dots ,a_n \in \mathbb{R}^n[/imath] are linearly indendent.
Let [imath]a_1,\dots ,a_n \in \mathbb{R}^n[/imath] with [imath]a_i = (a_{i1}, \dots , a_{in})[/imath]. Show that when [imath]\mid a_{jj}\mid \gt \sum_{i \neq j} \mid a_{ij} \mid \; \forall \; j\in \{1,\dots ,n\}[/imath], then [imath]a_1,\dots ,a_n[/imath] are linearly independent. I want to show that the matrix \begin{matrix} a_{11} & a_{21} & \cdots & a_{n1} \\ a_{12} & a_{22} & \cdots & a_{n2} \\ \vdots & & \ddots & \\ a_{1n} & a_{2n} & \cdots & a_{nn} \end{matrix} can be transformed into an upper triangular matrix with all entries in the main diagonal [imath]\neq 0[/imath]. For the cases [imath]n=1[/imath] and [imath]n=2[/imath] this is easy. However, I fail to show this for the general case; the problem being that the tranformations mess up the coefficients such that I can't use the assumption on the [imath]a_{jj}[/imath]'s anymore. I'm pretty sure that this has already been solved somewhere on here, but I don't know how to search for it.
|
2137280
|
Show that [imath]\int_a^b f(x)\,dx>0[/imath].
Suppose [imath]f(x)[/imath] is continuous on [imath][a, b][/imath] with [imath]f(x)\geq 0[/imath] and such that [imath]f(x) > 0[/imath] for some [imath]x\in\mathbb [a,b][/imath]. Show that [imath]\int_a^b f(x)\,dx>0[/imath].
|
1102845
|
Suppose [imath]\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b][/imath]
Let [imath]a<b[/imath] be real numbers. Let [imath]f:[a,b]\to\mathbb{R}[/imath] be a continuous non-negative function. Suppose [imath]\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b][/imath] Proof: Suppose for the sake of contradiction [imath]\exists x_0\in[a,b] \text{ such that } f(x_0)= D> 0[/imath] Since the function is continuous at [imath]x_0[/imath], we have [imath]\forall \epsilon>0, \exists \delta>0 \text{ such that } |f(x)-f(x_0)|\leq\epsilon\text{ whenever } x\in[a,b]\cap(x_0-\delta,x_0+\delta)[/imath] choose [imath]\delta=\delta_1[/imath] such that [imath]\epsilon=D/2[/imath], then we can say: [imath]-D/2 \leq f(x)-f(x_0)\leq D/2[/imath] [imath]-D/2 \leq f(x) - D[/imath] [imath]f(x)\geq D/2>0[/imath] This implies that [imath]\int_{[a,b]\cap(x_0-\delta_1,x_0+\delta_1)}f >0[/imath] Now since [imath]\int_{[a,b]} f = 0[/imath]. we have [imath]\int_{[a,b]}f=\int_{[a,x_0-\delta_1]}f+ \int_{(x_0-\delta_1, x_0+\delta_1)} f + \int_{[x_0+\delta_1,b]}f=0[/imath] [imath]\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f<-D/2,[/imath] but this is a contradiction since [imath]f(x)\geq 0[/imath] and hence [imath]\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f\geq 0[/imath] Is my proof correct?
|
2131589
|
How to prove [imath]a^ma^n = a^{m+n}[/imath] using field axioms?
I came across this while doing questions on field axioms. I need this proved for where [imath]m,n \in \mathbb Z[/imath] and [imath]a \neq 0[/imath]. Please do it in complete steps. I looked this up on the site but I couldn't understand what's been done.
|
443840
|
Suggestions on how to prove the following equality. [imath]a^{m+n}=a^m a^n[/imath]
Let [imath]a[/imath] be a nonzero number and [imath]m[/imath] and [imath]n[/imath] be integers. Prove the following equality: [imath]a^{m+n}=a^{m}a^{n}[/imath] I'm not really sure what direction to go in. I'm not sure if I need to show for [imath]n[/imath] positive and negative separately or is there an easier way. Can you use induction on integers? My attempt: 1) Base case [imath]m=0[/imath]. Prove that [imath]a^{m+n}=a^m a^n[/imath]. Is true. 2) Assume the result holds for [imath]m[/imath]. So I want to prove it holds for [imath]m+1[/imath]. So I know that [imath]a^{m+n}=a^ma^n[/imath]. So does this imply that: [imath]a^{m+n+1}= a^{m+1+n}= a^{m+1}a^n[/imath]? Am I going in the right direction? I'm not sure what to do next...
|
2137490
|
Are These Two Formulas Interchangeable?
When creating this question, StackExchange told me that this question appears subjective and is likely to be closed, but I'm going to ask it anyway to see if anyone can help me answer this. Sorry if this is not a valid question for Math StackExchange In previous courses (and looking online) the method I've always been taught to find a confidence interval with [imath]c\%[/imath] for [imath]\mu_1 - \mu_2[/imath] when both [imath]\sigma_1[/imath] and [imath]\sigma_2[/imath] are unknown is [imath](\bar{x}_1 - \bar{x}_2) \pm t_c \sqrt{\frac{s^2_p}{n_1}+\frac{s^2_p}{n_2}} [/imath] With degrees of freedom [imath]n_1+n_2-2[/imath] and [imath]s^2_p = \frac{(n_1-1)(s^2_1)+(n_2-1)(s^2_2)}{n_1+n_2-2}[/imath] However in my class this year my professor told me that the formula is: [imath](\bar{x}_1 - \bar{x}_2) \pm t_c \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} [/imath] With degrees of freedom being the smaller of [imath]n_1-1[/imath] and [imath]n_2-1[/imath] My question is this: Are these two formulas interchangeable? Or is one better than the other? Thank you.
|
1257855
|
Two different formulas for standard error of difference between two means
I mostly see this formula when searching for a formula for the estimate of the standard error in difference between two means, and it is also used in this video. [imath]\Delta=\sqrt{\left(\dfrac{s_1}{N_1}\right)^2+\left(\dfrac{s_2}{N_2}\right)^2}[/imath] But I've also seen this one (and this is the one my book uses): [imath]\Delta'=\sqrt{\dfrac{\left(N_1-1\right)s_1^2+\left(N_2-1\right)s_2^2}{N_1+N_2-2}\left(\dfrac{1}{N_1}+\dfrac{1}{N_2}\right)}[/imath] As these are two very different formulas, how come they are used seemingly interchangeably?
|
2135915
|
Möbius maps preserve cross-ratios
My question is different from this question since my question asks to clarify the proof, while the other question asks to check if the proof is correct. Let [imath]z_1, z_2, z_3\in\mathbb{C}_\infty[/imath] be distinct points. Then [imath]\alpha(z):=[z; w_1, w_2, w_3]=[\mu(z); \mu(w_1), \mu(w_2), \mu(w_3)]=\frac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}[/imath] for all [imath]z\in\mathbb{C}_\infty[/imath] (where [imath]\mu[/imath] is a Möbius map). This means that Möbius maps preserve cross-ratios. Here's the proof I have in the notes: [imath]\mu^{-1}[/imath] is a Möbius map, and: [imath]\alpha\circ \mu^{-1}(\mu(w_1))=\alpha(w_1)=0 [/imath] [imath]\alpha\circ \mu^{-1}(\mu(w_2))=\alpha(w_2)=1 [/imath] [imath]\alpha\circ \mu^{-1}(\mu(w_3))=\alpha(w_1)=\infty [/imath] Thus [imath]\alpha \circ \mu^{-1}(z)=[z;\mu(w_1), \mu(w_2), \mu(w_3)], \forall z\in \mathbb{C}_\infty[/imath]. What I don't understand is how this proves that Möbius maps preserve cross-ratios. I would assume that we would have to take [imath]\alpha(\mu(w))[/imath] to show the preservation. Moreover, if we show that some map is equal to [imath]\alpha[/imath] at just three points, how does this prove that two cross-ratios are equal? Perhaps I don't understand something about the meaning of cross-ratios.
|
625578
|
Invariance of cross ratio under Möbius transformation and another problem related to cross ratio.
Problem statement: Given [imath]z_1,z_2,z_3,z_4[/imath] different points of [imath]\overline {\mathbb C}[/imath], we define the cross ratio [imath](z_1,z_2,z_3,z_4)[/imath] as [imath](z_1,z_2,z_3,z_4)=\dfrac{z_1-z_2}{z_1-z_4}\dfrac{z_3-z_4}{z_3-z_2}[/imath]. Note that [imath](z_1,z_2,z_3,z_4)[/imath] is the image of [imath]z_1[/imath] under the Möbius transformation [imath]T[/imath] such that [imath]T(z_2)=0[/imath], [imath]T(z_3)=1[/imath], [imath]T(z_4)=\infty[/imath]. a) Prove that if [imath]T \in \mathcal H[/imath] then [imath](T(z_1),T(z_2),T(z_3),T(z_4))=(z_1,z_2,z_3,z_4)[/imath]. b) Show that [imath]z_1,z_2,z_3,z_4[/imath] lie in a line or circle if and only if [imath](z_1,z_2,z_3,z_4) \in \mathbb R[/imath] My attempt at a solution: For a), using the "hint" they give, if [imath]T \in \mathcal H[/imath], I did the following: If I call [imath]H=(z,,z_2,z_3,z_4)[/imath], I can consider [imath]H \circ T^{-1} (z)[/imath]. Note that [imath]H \circ T^{-1}(T(z_2))=0[/imath], [imath]H \circ T^{-1} (T(z_3))=1[/imath] and [imath]H \circ T^{-1} (T(z_4))=\infty[/imath]. This means that [imath](z_1,z_2,z_3,z_4)=H \circ T^{-1}(T(z_1))=(T(z_1),T(z_2),T(z_3),T(z_4))[/imath]. I don't know if my answer is correct, I would like to check it, and if anyone has a better or different answer, he/she is very welcome to post it. For b) I am lost, for the forward implication, I've tried to show that [imath](z_1,z_2,z_3,z_4)=\overline {(z_1,z_2,z_3,z_4)}[/imath] or that [imath]arg((z_1,z_2,z_3,z_4))[/imath] is a multiple of [imath]\pi[/imath] but I couldn't conclude anything. I would appreciate some help with this point.
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2137358
|
Roots Of Unity Proof On Concurrent Lines
Let [imath]P_1 P_2 \dotsb P_{18}[/imath] be a regular 18-gon. Show that [imath]P_1 P_{10}[/imath], [imath]P_2 P_{13}[/imath], and [imath]P_3 P_{15}[/imath] are concurrent. When positioned on the unit circle, I know that [imath]P_1 P_{10}[/imath] is the diameter. I also know that if [imath]p[/imath] and [imath]q[/imath] are points on the unit circle such that the line through [imath]p[/imath] and [imath]q[/imath] intersects the real axis and if [imath]z[/imath] is the point where this line intersects the real axis, then [imath]z = \dfrac{p+q}{pq+1}[/imath]. So I should let $P_1 P_{10} be on the real axis. Where should I go from now? Thanks.
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2126918
|
Concurrent lines proof for a regular 18-gon
Let [imath]X_1 X_2 \dotsb X_{18}[/imath] be a regular 18-gon. Show that [imath]X_1 X_{10}[/imath], [imath]X_2 X_{13}[/imath], and [imath]X_3 X_{15}[/imath] are concurrent. What would be the best way to prove this? I am actually struggling understanding 'concurrent' as I am not sure how they are concurrent in a regular 18-gon
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2138348
|
Is the Lebesgue measure of any open ball in Euclidean space zero?
Let [imath]B(x,r)[/imath] be an open ball in [imath]\mathbb R^n[/imath] , then is the Lebesgue measure of the boundary of [imath]B(x,r)[/imath] zero ?
|
751734
|
Boundary of Ball of radius R has zero measure
If [imath]\mu[/imath] is a Radon measure on [imath]\mathbb{R}^n[/imath] and [imath]B_r[/imath] is a closed ball of radius [imath]r[/imath]. Why is [imath]\mu(\partial B_r) = 0[/imath]? Or how can I prove that there is at least one [imath]r_0 > 0[/imath] such that [imath]\mu(\partial B_{r_0}) = 0[/imath]?
|
2138778
|
Given the matrix [imath]A[/imath] such that [imath]A^3=0[/imath]. Determine if matrix [imath]A^2-A+I[/imath] is invertible, and if it is, find it's inverse.
Given the matrix [imath]A[/imath] such that [imath]A^3=0[/imath]. Determine if matrix [imath]A^2-A+I[/imath] is invertible, and if it is, find it's inverse. Is it possible to directly evaluate matrix [imath]A[/imath] from the condition [imath]A^3=0[/imath]? If not, can we use Cayley-Hamilton theorem?
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2114927
|
If [imath]A^3=O[/imath] is the matrice [imath]A^2-A+I[/imath] invertible?
Is the matrice [imath]A^2-A+I[/imath] invertible? [imath]A^2=A-I /\cdot A[/imath] [imath] A^3 = A^2 - A[/imath] [imath]A^2=A[/imath] Thus I conclude that [imath]\lambda=0[/imath] is one of the eigenvalues of this matrice and it isn't invertible, my conclusion comes from the fact that the eigenvalues of the matrices [imath]A,A^2,...[/imath] are all connected, and if [imath]A^3=O[/imath] that means that those two have the same eigenvalues.
|
2139034
|
Example of function that cannot be expressed as a power series
I want to find an example of a [imath]\textbf{continuous everywhere}[/imath] function but can't be expressed as a convergent power series [imath]\sum_{n=0}^{\infty}c_n(x-a)^n[/imath] near a point [imath]a[/imath] (i.e. on [imath](a−\epsilon,a+\epsilon)[/imath] for all [imath]\epsilon[/imath], a is constant). Do I need to consider some Fourier series?
|
207217
|
Power series of a function
I am wondering if there are any functions [imath]f(x)[/imath] such that it cannot be expressed as a power series of [imath]x[/imath]? This might turn out to be a silly question, but I can't think of one at the moment. Thanks!
|
2134045
|
Prove that [imath]a^n-b^n \geq (a-b)nb^{n-1}[/imath] where [imath]a \geq b \geq 0[/imath]
I am having trouble with this proof. I believe the way to do it is through induction. This is what I have so far. Proof: We begin by induction on n. For the case that n = 1, we have [imath]a^1-b^1= (a-b) \geq (a-b)(1)(b^0) = (a-b) [/imath]. [imath](a-b) \geq (a-b)[/imath] Now we assume that this is true for some natural number k. [imath]a^k-b^k \geq (a-b)kb^{k-1}[/imath] Now we must show it is true for k + 1. So [imath]a^{k+1} - b^{k+1} = a^ka-b^kb.[/imath] I am not really sure how to proceed from this point. Where can I use the inductive hypothesis?
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368725
|
Proving Inequality using Induction [imath]a^n-b^n \leq na^{n-1}(a-b)[/imath]
I was trying to prove this inequality using induction, but couldn't do. Question: Suppose [imath]a[/imath] and [imath]b[/imath] are real numbers with [imath]0 < b < a[/imath]. Prove that if [imath]n[/imath] is a positive integer, then: [imath]a^n-b^n \leq na^{n-1}(a-b)[/imath]
|
2138997
|
Prove that [imath]\sin\dfrac{\pi}n·\sin\dfrac{2\pi}n···\sin\dfrac{(n-1)\pi}n=\dfrac{n}{2^{n-1}}[/imath]
How to prove that [imath] \sin\dfrac{\pi}n·\sin\dfrac{2\pi}n···\sin\dfrac{(n-1)\pi}n=\dfrac{n}{2^{n-1}} [/imath] using the roots of [imath](z+1)^n-1=0[/imath]? My rough idea is to solve [imath](z+1)^n-1=0[/imath] and use De Moivre's Theorem to find the product of roots to prove the equality.
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1253739
|
Proving [imath]\prod_k \sin \pi k / n = n / 2^{n-1}[/imath]
I am stuck trying to prove [imath]\prod_{k=1}^{n-1} \sin {\pi k \over n} = {n \over 2^{n-1}}[/imath] and I'd appreciate help. What I have done so far: [imath]z^n - 1 = \prod_{k=1}^n (z - \xi^k)[/imath] where [imath]\xi = e^{2 \pi i k\over n}[/imath]. Dividing both sides by [imath]z-1[/imath] we get [imath] {z^n - 1 \over z - 1} = \prod_{k=1}^{n-1} (z - \xi^k)[/imath] Taking the limit [imath]z \to 1[/imath] on both sides, [imath] n = \prod_{k=1}^{n-1} (1 - \xi^k)[/imath] So I'm getting close to what I want to prove but unfortunately, this is where I am stuck! How to proceed form here?
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2139016
|
How to find the max value of [imath]a^2+b^2[/imath] if [imath]x^4+ax^3+3x^2+bx+1\ge0[/imath]
Let [imath]f(x)=x^4+ax^3+3x^2+bx+1[/imath], where [imath]a,b \in \mathbb{R}[/imath]. If [imath]f(x) \ge 0[/imath] for all [imath]x \in \mathbb{R}[/imath], what is the maximum possible value of [imath]a^2+b^2[/imath]? I don't know how to proceed. Hints or help will be appreciated.
|
1169302
|
Polynomial maximization: If [imath]x^4+ax^3+3x^2+bx+1 \ge 0[/imath], find the maximum value of [imath]a^2+b^2[/imath]
If [imath]x^4+ax^3+3x^2+bx+1 \ge 0[/imath] for all real [imath]x[/imath] where [imath]a,b \in R[/imath]. Find the maximum value of [imath](a^2+b^2)[/imath]. I tried setting up inequalities in [imath]a[/imath] and [imath]b[/imath] but in the end had a hideous two variable expression whose maxima I had to calculate using Partial Derivatives. There must be a better way of doing it. Thanks!
|
2139288
|
Expressing tensor product as direct sum
Tensor products can sometimes be decomposed. For example [imath]2\otimes 2 = 3 \oplus 1[/imath] [imath]4\otimes 3 = 6\oplus 4\oplus 2[/imath] Where the numbers represent the dimension of a representation of [imath]SU(2)[/imath] (I think) and so are equal to [imath]2j+1[/imath]. I don't understand how this was done, and most internet resources use mathematical notation that is too complicated for me. I can see that multiplying the two on the LHS gives the same number as adding on the RHS, but I don't know how to find the numbers for the RHS. Thanks for any help!
|
95797
|
Product of [imath]SU(2)[/imath] representations
I am familiar with the interpretation of the irreducible representations (finite dimensional) of [imath]SU(2)[/imath] in terms of homogeneous polynomials of degree [imath]n[/imath]. If I take two of these irreducible representations, say [imath]\rho_{i}[/imath] and [imath]\rho_{j}[/imath] of dimensions [imath]i+1[/imath] and [imath]j+1[/imath] respectively, and take their product [imath]\rho_{i}\rho_{j}[/imath], the resulting representation will be the direct sum of other [imath]\rho_{k}[/imath]'s. For a fixed [imath]k[/imath], how do I calculate the multiplicity of [imath]\rho_{k}[/imath] as a summand in the product representation in terms of [imath]i[/imath] and [imath]j[/imath]?
|
2139322
|
[imath]f[/imath] is analytic in the unit circle, except a simple pole. Prove that [imath]\displaystyle \lim_{n\to\infty} |a_{n}|>0[/imath]
I'm learning complex analysis and encountered this question: Denote [imath]\mathbb{D}[/imath] to be the unit circle. [imath]f[/imath] is analytic in [imath]{\mathbb{D}}\subseteq U[/imath] except a simple pole [imath]z_{0}\in \partial\mathbb{D}[/imath]. Let [imath]\displaystyle f=\sum_{n=0}^\infty a_{n}z^n[/imath] be the Taylor representation of [imath]f[/imath] around [imath]z=0[/imath]. Prove that [imath]\displaystyle \lim_{n\to\infty} |a_{n}|>0[/imath] and compute [imath]\displaystyle \lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}[/imath]. My attempt: Write [imath]\displaystyle f(z)=\frac{g(z)}{z_{0}-z}[/imath] and using Taylor representations of both understand [imath]a_{n}[/imath] better. I got stuck though, so I would love some help. And follow-up questions: Can this be generalized to poles (not just simple ones)?
|
195767
|
An analytic function with a simple pole
Let [imath]f(z)[/imath] be analytic in the disk [imath]|z|<R \ \ \ (R>1)[/imath] except for a simple pole at a point [imath]z_0[/imath], [imath]|z_0|=1[/imath]. Consider the expansion [imath]f(z)=a_0+a_1 z+ \cdots[/imath], and show that [imath]\lim_{n \to \infty} \frac {a_n} {a_{n+1}}=z_0[/imath] All my attempts failed. I wanted to use the Laurent series at [imath]z_0[/imath] but the problem needs expansion at [imath]0[/imath].
|
2139578
|
How do I go about finding some [imath]f(x)[/imath] that satisfies the following first-order DE?
[imath]f'(\sin^2x) = \cos^2x + \tan^2x,\quad[/imath] subject to [imath]\quad0<x<1[/imath] Many thanks.
|
2124335
|
First-order trigonometric differential equation
Find [imath]f(x)[/imath] that satisfies the first-order differential equation: [imath]f'(\sin^2x)=\cos^2x+\tan^2x\tag{for $0<x<1$}[/imath]
|
1003011
|
CDF, PDF, and integral with respect to a function
In a certain textbook, I see the Cumulative Distribution Function (CDF) of a continuous random variable X defined as [imath]\int_{-\infty}^{x'} dp(x)[/imath] where p(x) is the Probability Density Function of X. Usually, I see the CDF defined thus: [imath]\int_{-\infty}^{x'} p(x)dx[/imath] I have only ever seen and solved integrals with respect to a variable, not with respect to a function of a variable, so I don't know if those two are equivalent. Are they? Or is the first expression wrong?
|
380785
|
What does it mean to integrate with respect to the distribution function?
If [imath]f(x)[/imath] is a density function and [imath]F(x)[/imath] is a distribution function of a random variable [imath]X[/imath] then I understand that the expectation of x is often written as: [imath]E(X) = \int x f(x) dx[/imath] where the bounds of integration are implicitly [imath]-\infty[/imath] and [imath]\infty[/imath]. The idea of multiplying x by the probability of x and summing makes sense in the discrete case, and it's easy to see how it generalises to the continuous case. However, in Larry Wasserman's book All of Statistics he writes the expectation as follows: [imath]E(X) = \int x dF(x)[/imath] I guess my calculus is a bit rusty, in that I'm not that familiar with the idea of integrating over functions of [imath]x[/imath] rather than just [imath]x[/imath]. What does it mean to integrate over the distribution function? Is there an analogous process to repeated summing in the discrete case? Is there a visual analogy? UPDATE: I just found the following extract from Wasserman's book (p.47): The notation [imath]\int x d F(x)[/imath] deserves some comment. We use it merely as a convenient unifying notation so that we don't have to write [imath]\sum_x x f(x)[/imath] for discrete random variables and [imath]\int x f(x) dx[/imath] for continuous random variables, but you should be aware that [imath]\int x d F(x)[/imath] has a precise meaning that is discussed in a real analysis course. Thus, I would be interested in any insights that could be shared about what is the precise meaning that would be discussed in a real analysis course?
|
2139559
|
counter example for zorn lemma.
Let [imath](E,\leq)[/imath] a partially ordered set. Zorn lemma says that if all chain of [imath]E[/imath] has a supremum, then [imath]E[/imath] has a maximal element. So if I consider, [imath]\Big((0,1),\leq \Big)[/imath], it has no maximal element but all chain is upper bounded by [imath]1[/imath], so it doesn't work here, no ?
|
1397991
|
Why doesn't Zorn's lemma apply to [imath][0,1)[/imath]?
Consider the real interval [imath][0,1)[/imath], this is partially ordered set (totally ordered actually). This set has a upper bound like [imath]1[/imath], and according to Zorn's Lemma each partially ordered set with a upper bound should have at least one maximal element. However, in this set there is no maximal element, i.e., element that is greater than every element of the set because you can be as close as to [imath]1[/imath]. I am should I don't understand Zorn's Lemma. Please Help!!
|
2141127
|
draw [imath]\triangle ABC[/imath] in which [imath]AB=5.5[/imath]cm, [imath]\angle C =40^{\circ}[/imath] and [imath]BC-AC=2.5[/imath]cm
I tried this question whole day but could not find its solution I have also not got a clue please help me in doing it
|
2142282
|
draw triangle [imath]\Delta ABC[/imath] in which [imath]\overline{AB}=5.5[/imath] cm, [imath]\widehat{C} =40^{\circ}[/imath] and [imath]\overline{BC}-\overline{AC}=2.5[/imath] cm
Help me doing this question I tried this question whole day but could not find the answer
|
2140570
|
Initial value problem, show y(x) = 1-x is a solution
I'm not really sure on how to go about solving this problem... Consider the initial value problem. [imath]y' (x) = \frac{1}{2}(-x + \sqrt{x^2 + 4y})[/imath] [imath]y(2) = -1[/imath] [imath]a)[/imath] Show that [imath]y(x) = 1-x[/imath] and [imath]y(x) = -\frac{x^2}{4}[/imath] are two solutions to the above IVP.
|
1952299
|
[imath]y'=[/imath] [imath]{-x+\sqrt{(x^2+4y)}}\over 2[/imath] , [imath]y(2)=-1[/imath]
Prove that the solutions [imath]y_1[/imath] and [imath]y_2[/imath] for the initial value problem [imath]y'=[/imath][imath]{-x+\sqrt{(x^2+4y)}}\over 2[/imath] , [imath]y(2)=-1[/imath] are: [imath]y_1=1-x[/imath] [imath]y_2={{-x^2}\over 4}[/imath] And explain why the two solutions not contradiction with the theory of existence and uniqueness. ( sorry I don't speak English well ) My answer: For the initial condition : [imath]y_1=1-x[/imath] Satisfy the initial condition [imath]y(2)=-1[/imath] [imath]y_1=1-2=-1[/imath] And Satisfy the differential equation [imath]{y_1}^{\prime}=-1[/imath] [imath]{{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4(1-x)}}\over 2}[/imath] [imath]={{-x+\sqrt{(x-2)^2}}\over 2}[/imath] [imath]={{-x+x-2}\over 2}=-1[/imath] L.h.s.=R.h.s. [imath]y^{\prime}={{-x+\sqrt{x^2+4y}}\over 2}[/imath] Also [imath]y_2[/imath] Satisfy the initial condition [imath]y_2={{-x^2}\over4}={-4\over4}=-1[/imath] And satisfy the differential equation. [imath]y_2^{\prime}={-x\over 2}[/imath] [imath]{{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4({-x^2\over 4})}}\over 2}={-x\over 2}[/imath] And for second part of exercise: [imath]{\partial f\over \partial y}={1\over \sqrt{x^2+4y}}[/imath] and this not defined when [imath]x=2, \ y=-1[/imath] So [imath]f(x,y)[/imath] no satisfy lipshtize condition in any rectangular has (2,-1), Lipshtize condition including uniqueness of solution, so no contradict with the existence and uniqueness. True ? If this also wrong I well be delet it , I answered by the same way of my teacher :(
|
2142111
|
Sum of floor with irrational number multiplication [imath]\sum_{k=1}^{n} \left\lfloor {k*\sqrt{2}} \right\rfloor[/imath]
As part of a Google foobar challenge I have been asked to calculate: [imath]\sum_{k=1}^{n} \left\lfloor {k*\sqrt{2}} \right\rfloor[/imath] for numbers up to [imath]10^{100}[/imath] My attempt so far resulted in imprecise results: Using the formula [imath]\sum_{k=1}^{n-1} \left\lfloor \frac{km}{n} \right\rfloor = \frac{1}{2}(m - 1)(n - 1)[/imath] I made this calculation: [imath] \frac{1}{2} ((n+1)*\sqrt{2}-1)*n [/imath] The results needed to be rounded (I suppose because of the irrationality of [imath]\sqrt{2}[/imath]) and were incorrect in about 10% of test cases. Thanks in advance! Note: I've looked at many of the threads regarding sum of floors and wasn't able to find a closed form that yields better accuracy.
|
2052179
|
How to find [imath]\sum_{i=1}^n\left\lfloor i\sqrt{2}\right\rfloor[/imath] A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)).
[A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)).][1] If [imath]n = 5[/imath] then [imath]\left\lfloor1\sqrt{2}\right\rfloor+ \left\lfloor2\sqrt{2}\right\rfloor + \left\lfloor3\sqrt{2}\right\rfloor +\left\lfloor4 \sqrt{2}\right\rfloor+ \left\lfloor5\sqrt{2}\right\rfloor = 1+2+4+5+7 = 19[/imath] Sequence from [imath]1[/imath] to [imath]20[/imath] is: [imath]S=\{1,2,4,5,7,8,9,11,12,14,15,16,18,19,21,22,24,25,26,28\}[/imath] I want to find answer for [imath]10^{100} n[/imath]?
|
2142318
|
Limit [imath]\lim_{ x \to \infty }\frac{\sqrt{x+1}+\sqrt{x+2}-\sqrt{4x+12}}{\sqrt{x}+\sqrt{x+2}-\sqrt{4x+12}}[/imath]
find the limits :[imath]\lim_{ x \to \infty}\frac{\sqrt{x+1}+\sqrt{x+2}-\sqrt{4x+12}}{\sqrt{x}+\sqrt{x+2}-\sqrt{4x+12}}[/imath] WolframAlpha gives me: [imath]\lim_{ x \to \infty }\frac{\sqrt{x+1}+\sqrt{x+2}-\sqrt{4x+12}}{\sqrt{x}+\sqrt{x+2}-\sqrt{4x+12}}=3/4[/imath] What is a simple way for the limits?
|
1854348
|
Limit [imath]\lim_{x\to \infty} \frac{\sqrt{x+1}+\sqrt{x+2}-\sqrt{4x+12}}{\sqrt{x+2}+\sqrt{x}-\sqrt{4x+12}}[/imath]
Using the equivalency, I got the answer equal to [imath]1[/imath], however the answer is given as [imath]\frac34[/imath]. WolframAlpha also gives [imath]\frac34[/imath]. Could someone tell me that the given answer is wrong or I made a mistake!
|
2142905
|
Finding all triangles which have an inradius of x
How would you be able to find all triangles with a given inradius and all three sides of the triangle have integer lengths? More specifically how would you do it without randomly guessing at numbers? For example, if you were given an inradius of 2, you could find a triangle with sides of 5, 12, 13 or 7, 15, 20. If you wanted to go the work out the inradius from a triangle with known sides it would be fairly simple using the following formula. [imath]R = \frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}[/imath] I tried a range of methods, such as rearranging Heron's formula to find specific sides. However, that left me with equations such as the following. [imath]a = \frac{A^2}{s(s-b)(s-c)} + s[/imath] This equation still has an equation full of unknown values and showed me I was attempting to solve the problem incorrectly. The problem now is that I don't know what else I can do to solve the initial question.
|
2028918
|
Find possible number of triangles with integer sides for a given inradius
inradius = [imath]5[/imath] possible triangle sides [imath](25, 20, 15)[/imath] [imath](37, 35, 12) (39, 28, 17)[/imath] ... Find formula to find other possible sides of triangles.
|
2142205
|
Solution of the differential equation of first order and second degree implicit equation.
Find the solution of the differential equation [imath](\frac{dy}{dx})^2-x(\frac{dy}{dx})+y=0[/imath].
|
1643067
|
solution of differential equation [imath]\left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0[/imath]
The solution of differential equation [imath]\displaystyle \left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0[/imath] [imath]\bf{My\; Try::}[/imath] Let [imath]\displaystyle \frac{dy}{dx} = t\;,[/imath] Then Diferential equation convert into [imath]t^2-xt+y=0[/imath] So Its solution is given by [imath]\displaystyle t=\frac{x\pm \sqrt{x^2-4y}}{2}[/imath] So we get [imath]\frac{dy}{dx} = \frac{x\pm \sqrt{x^2-4y}}{2}[/imath] Now How can I solve after that, Help me Thanks
|
2141918
|
Linear independence over Q of logarithmic powers of prime numbers
I am trying to solve the following question but without any result so far: Let [imath]p_1, p_2, \ldots, p_t[/imath] be different primes. Prove that [imath]\log (p_1), \log (p_2), \ldots, \log (p_t)[/imath] are linearly independent over [imath]\mathbb{Q}[/imath] that is, if [imath]x_1,x_2,\ldots, x_t[/imath] are rational numbers with [imath] x_1 \log (p_1) + x_2 \log (p_2) + \cdots + x_t \log (p_t) = 0 [/imath] then [imath]x_1 = x_2 = \cdots = x_t = 0[/imath]. We are supposed to use the following result: Let [imath]x \in \mathbb{Q}[/imath] with [imath]x > 0[/imath]. Then there is a unique sequence of integers [imath](n_2,n_3,n_5, \ldots)[/imath], almost all equal to [imath]0[/imath], such that [imath] x = \prod_{p \text{ prime}} p^{n_p}. [/imath] Any tips are welcome for solving it, thanks.
|
2134770
|
Prove linear combinations of logarithms of primes over [imath]\mathbb{Q}[/imath] is independent
Suppose we have a set of primes [imath]p_1,\dots,p_t[/imath]. Prove that [imath]\log p_1,\dots,\log p_t[/imath] is linear independent over [imath]\mathbb{Q}[/imath]. Now, this implies [imath] \sum_{j=1}^{t}x_j\log(p_j)=0 \iff x_1=\dots=x_t=0[/imath]. I think I have to use that fact that every [imath]q\in\mathbb{Q}[/imath] can be written as [imath]\prod_{\mathcal{P}}[/imath], where [imath]n_p[/imath] is a unique sequence ([imath]n_2[/imath],[imath]n_3[/imath],[imath]\dots[/imath]) with domain [imath]\mathbb{Z}[/imath]. Here, [imath]\mathcal{P}[/imath] denotes the set of all integers. Now how can I use this to prove the linear independency?
|
2137037
|
algebraic closure of [imath]\mathbb{Q}_p[/imath] is not complete
A paper I'm reading says that the algebraic closure of [imath]\mathbb{Q}_p[/imath] is not complete, by using for example the Baire theorem. Wikipedia says the Baire theorem says that a complete metric space is a Baire space (meaning a countable intersection of dense open sets remains dense). I don't see how this can be used to show the claim that [imath]\overline{\mathbb{Q}_p}[/imath] is not complete. Any other argument is also welcome.
|
123925
|
Is the algebraic closure of a [imath]p[/imath]-adic field complete
Let [imath]K[/imath] be a finite extension of [imath]\mathbf{Q}_p[/imath], i.e., a [imath]p[/imath]-adic field. (Is this standard terminology?) Why is (or why isn't) an algebraic closure [imath]\overline{K}[/imath] complete? Maybe this holds more generally: Let [imath]K[/imath] be a complete Hausdorff discrete valuation field. Then, why is [imath]\overline{K}[/imath] complete? I think I can show that finite extensions of complete discrete valuation fields are complete.
|
2143432
|
Find the value [imath]C[/imath] such that [imath]\int^{\infty}_{0}(\frac{x}{x^2+1} - \frac{C}{3x+1})dx[/imath] converges
Find the value [imath]C[/imath] such that [imath]\int^{\infty}_{0}(\frac{x}{x^2+1} - \frac{C}{3x+1})dx[/imath] converges [imath]\int^{\infty}_{0}(\frac{x}{x^2+1} - \frac{C}{3x+1})dx[/imath] How do I do this? The only thing I can think of is creating a common denominator? [imath]\displaystyle\int\limits^{\infty}_{0} \dfrac{x^2\left(3-c\right)+x-c}{\left(x^2+1\right)\left(3x+1\right)}\,\mathrm{d}x[/imath] Something tells me [imath]C = 3[/imath] is right, but I don't know why I think that or if it's even right.
|
2143154
|
Finding constant to make integrals converge
Find the value of the constant C for which the following integral [imath]\int_{0}^{\infty} \bigg(\frac{x}{x^2+1} - \frac{C}{3x+1}\bigg)dx[/imath] converges. Evaluate the integral for that value of C. Make sure to fully justify your answer. \ My solution: [imath]\lim_{A\to\infty}\int_{0}^{A} \frac{x}{x^2+1} dx - \lim_{A\to\infty} \int_{0}^{A} \frac{C}{3x+1} dx[/imath] [imath]= \lim_{A\to\infty} \frac{1}{2} ln(x^2+1) \bigg|_{0}^{A} - \lim_{A\to\infty} \frac{C}{3} ln|3x+1| \bigg|_{0}^{A}[/imath] [imath]= \frac{1}{2}(\infty + 0) - \frac{C}{3} (\infty + 0)[/imath] Idk how to make it converge.
|
2143543
|
infinite sum of (-1)^n/(x+πn) = csc x
I was playing with some graphs and I noticed that [imath]\frac{1}{x}-\frac{1}{x+\pi}+\frac{1}{x+2\pi}[/imath] looked similar to [imath]\csc x[/imath]. I then plugged in (to desmos.com) [imath]\sum_{n=-500}^{500}\frac{(-1)^n}{x+nπ}[/imath] and found that it was pretty much [imath]\csc{x}[/imath] on the interval near [imath](-500π,500π)[/imath]. So, my question is, why is it that [imath]\sum_{-\infty}^\infty \frac{(-1)^n}{x+nπ} = \csc x[/imath]
|
1963263
|
Prove [imath]\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}[/imath]
Prove [imath]\csc(x)=\sum_{k=-\infty}^{\infty}\frac{(-1)^k}{x+k\pi}[/imath] Hardy uses this fact without proof in a monograph on different ways to evaluate [imath]\int_0^{\infty}\frac{\sin(x)}{x} dx[/imath].
|
2143632
|
Dimension of Subspaces
Consider two subspaces V and W of [imath]\mathbb{R}^n[/imath]. Show that dim(V) + dim(W) = dim(V ∩ W) + dim(V + W). Can anyone provide me with the proof for this?
|
482872
|
dimension of a subspace spanned by two subspaces
If [imath]M[/imath] and [imath] N[/imath] be the subspaces of a vector space [imath]V[/imath] then prove that [imath](\dim M) + (\dim N) = (\dim M+N) + (\dim M \cap N)[/imath].
|
2143606
|
Cartesian product of connected spaces with a common factor
Suppose [imath]U[/imath], [imath]V[/imath] and [imath]W[/imath] are connected topological spaces. Is it possible for [imath]U \times V[/imath] and [imath]U \times W[/imath] to be homeomorphic but for [imath]V[/imath] and [imath]W[/imath] not to be? It seems like it should not always be true but I am unable to find a counterexample.
|
1057793
|
Cancellation in topological product
I was wondering whether [imath]M\times \mathbb{R}[/imath] is homeomorphic to [imath]N\times \mathbb{R}[/imath] implies [imath]M[/imath] is homeomorphic to [imath]N[/imath], where let us say [imath]M,N[/imath] are smooth manifolds. (They are certainly homotopy equivalent.) More generally what if [imath]\mathbb{R}[/imath] is replaced by some other manifold say [imath]\Sigma[/imath]? Any possible approach/idea?
|
2144088
|
Showing that these two rings [imath]\frac{R[T]}{(aT-1)}, R[a^{-1}][/imath] are isomorphism.
Let [imath]R[/imath] be an integral domain and [imath]0 \neq a \in R[/imath]. I need to show that: [imath]\frac{R[T]}{(aT-1)} \cong R[a^{-1}].[/imath] My Works: Let us define the ring homomorphism [imath]\phi: R[T] \to R[a^{-1}][/imath] which is defined by [imath]T \mapsto a^{-1}[/imath]. It is easy to show that [imath]\phi[/imath] is an epimorphism and [imath](aT-1) \subset \mathrm{ker} (\phi)[/imath]. I do not know how to get the inverse inclusion. In the case that [imath]R[/imath] is a field, we may use the Euclidean algorithm to show that every [imath]g \in \mathrm{ker} (\phi)[/imath] is a multiple of [imath]aT-1[/imath]. But since [imath]R[/imath] is not a field, I do not know how to show that [imath]g[/imath] must be a multiple of [imath]aT-1[/imath]. Thanks for your help.
|
88810
|
Factorizing a polynomial [imath]f[/imath] in [imath]A[x][/imath] (with [imath]A[/imath] commutative), where [imath]f[/imath] has a zero in its field of fractions
Let [imath]A[/imath] be a commutative ring and [imath]S[/imath] a multiplicative subset of [imath]A[/imath] generated by [imath]s\in A[/imath] which is not a zero-divisor. Consider the polynomial ring [imath]A[x][/imath]. Given a polynomial [imath]f\in A[x][/imath], suppose that [imath]f(1/s)=0[/imath] for [imath]s\in S[/imath]. Does it follow by some application of the first isomorphism theorem that [imath]f(x)=(sx-1)g(x)[/imath], with [imath]g(x)\in A[x][/imath]? On the one hand I think this can be proven by constructing an isomorphism from [imath]A[x][/imath] modulo the relation [imath]x=1/s[/imath] (really the replacement of [imath]x[/imath] with [imath]1/s[/imath]). This is evidently an isomorphism from [imath]A[x]/(x=1/s)[/imath] into [imath]S^{-1}A[/imath]. It is tempting to conclude (though I'm not certain that this should work), that there must be some surjective homomorphism [imath]\phi: A[X]\to S^{-1}A[/imath] such that [imath]\ker(\phi)=(sx-1)[/imath]. [Since generally [imath](A/B\cong A/C) \nRightarrow (B\cong C)[/imath], it is not sufficient to conclude that, say, the kernel of [imath]\psi: f\to f(1/s)[/imath] is [imath](sx-1)[/imath].] I am especially interested in answers as to whether, say, [imath]\psi[/imath] has kernel [imath](sx-1)[/imath]. But for anyone who can follow my own attempt at a proof, I'm also interested in whether my argument breaks down or can be made to work rigorously.
|
2141418
|
The dual version of trichotomy of cardinality without AC
Let [imath]A[/imath], [imath]B[/imath] be sets and denote by [imath]A\leq^* B[/imath] an assertion that if [imath]A[/imath] is non-empty, then there's a surjection from [imath]B[/imath] onto [imath]A[/imath]. This might be understood as a dual definition of cardinality. (Typically we use injections to capture the notion of A not being bigger than B. The [imath]\leq^*[/imath] notation is to be read as B not being smaller than A.) Is this relation trichotomous without using the axiom of choice? Meaning for every two sets [imath]A[/imath] and [imath]B[/imath], it holds [imath]A\leq^* B[/imath] or [imath]B\leq^* A[/imath]. (I would be surprised). Would trichotomy imply some interesting extra-ZF axiom? More broadly, is there anything to be said about [imath]\leq^*[/imath]? For example how does it relate to the usual comparisons of cardinality by injections? I suspect there's no clear answer, since it is somehow reminiscent of Partition Principle ([imath]A\leq^* B \implies A\leq B[/imath]) and Weak Partition Principle ([imath]A\leq^* B \implies B\not< A [/imath]) which both are independent of ZF.
|
499066
|
In ZF, how would the structure of the cardinal numbers change by adopting this definition of cardinality?
In ZFC, a good way of ordering sets by cardinality is by leveraging the notion of an injection. We define: [imath]X \lesssim Y \leftrightarrow \mbox{ there exists an injection } X \rightarrow Y.[/imath] Alternatively, we can leverage the notion of a partial surjection, by which I mean a partial function that hits every point of its codomain at least once. [imath]Y \succsim X \leftrightarrow \mbox{ there exists a partial surjection } Y \rightarrow X.[/imath] Now in ZFC, these approaches are equivalent, in the sense that [imath]X \lesssim Y \iff Y \succsim X.[/imath] But if we just consider ZF (that is, we drop choice) then the equivalence breaks down. However, we've still got the forward implication, because the converse of an injection is always a partial surjection. A specific question. Define that two sets are weakly equipotent iff [imath]X \succsim Y[/imath] and [imath]Y \succsim X[/imath]. Question: does weak equipotence imply genuine equipotence? I'm guessing "no." A general question: In broad terms, how would the structure of the cardinal numbers in ZF be altered by adopting [imath]\succsim[/imath] rather than [imath]\lesssim[/imath] as the fundamental order relation, and taking equality of cardinal numbers to correspond to weak equipotence of sets? I'm guessing this would hide a lot of the (usually very complicated) structure of the cardinal numbers in ZF. Remark. The order symbols can be written \lesssim and \succsim.
|
1291012
|
how to proof formula for general addition rule of three events
can somebody please help to prove the formula for general additional rule of three events? [imath]P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)[/imath]
|
669249
|
Probability of the union of [imath]3[/imath] events?
I need some clarification for why the probability of the union of three events is equal to the right side in the following: [imath]P(E\cup F\cup G)=P(E)+P(F)+P(G)-P(E\cap F)-P(E\cap G)-P(F\cap G)+P(E\cap F\cap G)[/imath] What I don't understand is, why is the last term(intersection of all) added back just once, when it was subtracted three times as it appears from a Venn Diagram? Here on page 3, this is explained but not in enough details that I can understand it: http://www.math.dartmouth.edu/archive/m19w03/public_html/Section6-2.pdf
|
2144098
|
linear matrix problem
How to prove the equation below: [imath](A^TA + \mu I)^{-1}A^T[/imath] = [imath]A^T(AA^T + \mu I)^{-1}[/imath] no ideas at all, i major in signal processing and it needs some math techniques so i am not familiar with this part of knowledge, really need help please.
|
2136333
|
Matrix Identity [imath](I-G_1G_2)^{-1}G_1=G_1(I-G_2G_1)^{-1}[/imath]?
I am reading a book on control theory for multiple input multiple output (MIMO) systems. The author claims that for the identity matrix [imath]I[/imath] and the complex transfer function matrices [imath]G_1[/imath] and [imath]G_2[/imath] (let us assume they are square matrices) the following identity holds: [imath](I-G_1G_2)^{-1}G_1=G_1(I-G_2G_1)^{-1}.[/imath] Is this statement true? And how can I derive it?
|
2144690
|
Analysis - differentiation
Suppose that [imath]f : [a, b] \rightarrow \mathbb{R}[/imath] is a strictly increasing continuous function which is twice differentiable at [imath]c \in (a, b)[/imath], with [imath]f(c)[/imath] not equal to [imath]0[/imath]. Show that the second derivative of the inverse function [imath]g[/imath] at [imath]f(c)[/imath] exists and find a formula for it. Any help would be greatly appreciated. Thanks!!
|
249253
|
second derivative of the inverse function
I know that the derivative of the inverse function of [imath]f(x)[/imath] is [imath]g'(y) = \frac{1}{f'(x)}[/imath] But how to derive the formula for the second derivative of g(y) knowing that [imath]\left[\frac{1}{f(x)}\right]' = -\frac{f'(x)}{(f(x))^2}[/imath] ? I just started studying this chapter, so please try to be as simple as possible ;-) Thank you.
|
2145046
|
How do I prove that if [imath]A\setminus C = B\setminus C[/imath] and [imath]A\cap C = B\cap C[/imath] then [imath]A = B[/imath]?
I am stuck. I know that I need to prove [imath]A[/imath] is a subset of [imath]B[/imath] and that [imath]B[/imath] is a subset of [imath]A[/imath], and that in each case all elements in A are elements in [imath]B[/imath], but I don't know where to go from there. Any help is appreciated!
|
2144738
|
Prove that if [imath]A – C = B – C[/imath] and [imath]A\cap C = B\cap C[/imath] then [imath]A = B[/imath]
I am trying to prove that if [imath]A – C = B – C[/imath] and [imath]A\cap C = B\cap C[/imath] then [imath]A = B[/imath]. I have tried using Venn Diagrams as a proof technique, but we are not able to use proof by Venn Diagrams.
|
2145330
|
Every number is the sum of three squares with signs
The question. Can every [imath]n\in \mathbb N[/imath] can be written: [imath]n=a^2\pm b^2\pm c^2[/imath] where [imath]\pm[/imath] are signs of your choice? We know with Lagrange's four-square theorem that every integer can be written as the sum of four squares. Plus, with have Legendre's three-square theorem stated that an integer can not be written as the sum of three squares if, and only if, it is of the form: [imath]4^k(8n+7).[/imath] So we just have to prove (or disprove) it for every number of this form. I have checked it until [imath]55[/imath], and it seems to work so far. So the number we have to check are these ones. For instance: [imath]31=6^2-2^2-1^2[/imath] and [imath]39=6^2+2^2-1^2.[/imath] The issue here is that [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] can be arbitrarily large. For instance: [imath]183=14542^2-14541^2-170^2.[/imath] So I don't really know how to prove or disprove this result, and I think it could go either way.
|
894777
|
Representation of integers by ternary quadratic form [imath]x^2+y^2-z^2[/imath]
Let [imath]Q[/imath] be the ternary quadratic form [imath]Q(x,y,z)=x^2+y^2-z^2[/imath]. Since [imath]Q(0,p+1,p)=2p+1[/imath] and [imath]Q(1,p+1,p)=2p+3[/imath], we see that for every integer [imath]k[/imath], the equation [imath]E_k:Q(x,y,z)=k[/imath] always has a solution. Is it known for which integers [imath]k[/imath] there are infinitely many solutions to [imath]E_k[/imath] ?
|
2145498
|
Proof that [imath]\sigma (p(A)) = p(\sigma (A))[/imath] for any polynomial and matrix
My problem: Prove that for any matrix [imath]A[/imath] and for any polynomial [imath]p(x)[/imath] we have that [imath]\sigma (p(A)) = p(\sigma (A))[/imath], where [imath]\sigma[/imath] is the spectrum of a matrix. Prove that [imath]A[/imath] is an invertible matrix if no root of [imath]p(x)[/imath] is an eigenvalue. My solution: For example, if I have the polynomial [imath]\;[/imath] [imath]p(x) = x^2 + x[/imath] [imath]\;[/imath] and the matrix [imath] A= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} [/imath] then [imath] p(A)= \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}^{2}+ {\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}}= {\begin{pmatrix} 2 & 0 \\ 0 & 2 \\ \end{pmatrix}} [/imath] [imath] [/imath] We can see that the spectrum of [imath]p(A)[/imath] contains the eigenvalues [imath]\lambda_{1,2} = 2[/imath] The spectrum of the matrix [imath]A[/imath] contains eigenvalues [imath]\lambda_{1,2} = 1[/imath] and if we substitude it into [imath]p(x)[/imath], then [imath]p(\lambda) = p(1) = 1^2 + 1 = 2[/imath] The spectrum of [imath]A[/imath] is the set of its eigenvalues and the eigenvalues are the roots of the characteristic polynomial, but I am not sure how to prove [imath]\sigma (p(A)) = p(\sigma (A))[/imath] generally. The second proof I think I understand. Can anyone help me?
|
555075
|
What is the relationship between the spectrum of a matrix and its image under a polynomial function?
Clearly if [imath]\lambda[/imath] is an eigenvalue of [imath]A[/imath] then [imath]p(\lambda)[/imath] is an eigenvalue of [imath]p(A)[/imath] where [imath]p[/imath] is a polynomial. And there are cases where [imath]A[/imath] may have eigenvalues other than these. Is there a general rule for finding all eigenvalues of [imath]p(A)[/imath]? Does it depend on the field from which the elements of [imath]A[/imath] are taken? Are their any partial solutions to this problem?
|
2141079
|
No bounded linear surjectivity from [imath]\ell^2(\mathbb{N})[/imath] to [imath]\ell^1(\mathbb{N})[/imath]
I would like to show that a bounded linear operator [imath]T: \ell^2(\mathbb{N}) \rightarrow \ell^1(\mathbb{N})[/imath] cannot be surjective. If we assume surjectivity, then by the Open Mapping Theorem, [imath]T[/imath] is an open map, i.e. for every open [imath]O \subset \ell^2(\mathbb{N})[/imath], its image [imath]T(O)[/imath] is open in [imath]\ell^1(\mathbb{N})[/imath]. Is it then possible to contradict Baire's Theorem by constructing a sequence of open dense subsets of [imath]\ell^1(\mathbb{N})[/imath], whose intersection is not dense? Any help pointing me into the right direction would be appreciated.
|
1312825
|
No surjective bounded linear map from [imath]\ell^2(\mathbf{N})[/imath] to [imath]\ell^1(\mathbf{N})[/imath]
The second part of an exercise asks that I show: There does not exist a surjective bounded linear map from [imath]\ell^2(\mathbf{N})[/imath] to [imath]\ell^1(\mathbf{N})[/imath]. The first part was to show that these two spaces are separable in their respective [imath]L^p[/imath] norms. I've really hit a wall on this one. I know, in fact, that there does exist a surjective bounded linear map from [imath]\ell^1(\mathbf{N})[/imath] to [imath]\ell^2(\mathbf{N})[/imath]. So my intuition says that if there were also one going the other way, then the two spaces would have to be isomorphic, which is false . But I don't know of any such theorem that I could apply to "the category of Banach spaces". My only other idea is to argue by contradiction using the open mapping theorem, but nothing is coming to me... My guess is that separability is also going to be important, given the first part of the exercise. -Thanks.
|
2145215
|
Intuition of a non-differentiable function graph as a differentiable manifold?
A [imath]d[/imath]-dimensional topological manifold [imath]M[/imath] is a topological space for which there exists continuous maps (charts) to [imath]R^d[/imath], such that every point [imath]p \in M[/imath] is mapped by at least one of those charts. A differentiable manifold [imath]M[/imath] is a topological manifold with the additional requirement that its Atlas contains only charts whose transition maps are all differentiable. I find this second definition (of differentiable manifolds) very strange, because it only makes reference to the charts in the Atlas of manifold [imath]M[/imath], but not to the manifold itself. If I'm correct, it means we can construct a differentiable manifold out of a non-differentiable continuous function: `Take the graph [imath]M[/imath] of a non-differentiable function (e.g. [imath]f(x)=|x|[/imath]). This clearly is a [imath]1[/imath]-dimensional topological manifold. Now we can simply take the Atlas that consists of the single chart map that maps every point on the graph to the [imath]x \in \mathbb R[/imath] that produced it. This clearly is a chart map, and it clearly has a chart transition map to itself that is differentiable. So this means that manifolds that have "kinks" in them, like the graphs of non-differentiable functions, can still be differentiable manifolds. Could even a function like the Weierstrass function be a differentiable manifold? What does "differentiability" of a manifold intuitively mean, since it is not intuitively the same thing as the differentiability of a real function?
|
470370
|
example of a non differentiable manifold
Take the manifold: The graph of [imath]|x|[/imath] on [imath](-1,1)[/imath], with the induced topology from [imath]\mathbb{R}^2[/imath]. This is a topological manifold, which is homeomorphic to [imath](-1,1)[/imath] by projection. Is it a differentiable manifold? I believe it is, because we can take an atlas with only the projection, and then we will have only one transition map which is the identity, and therefore differentiable. But I'm not sure I get the definition of a differentiable manifold right... Thanks!
|
2146049
|
Monotony problem with sequences
Let us consider the sequence [imath](a_n)_{n \ge 1}[/imath] such that [imath]a_n=\frac {1}{\sqrt {n^2+1}}+ \frac {1}{\sqrt {n^2+2}} + \dots +\frac {1}{\sqrt {n^2+n}}.[/imath] Show that the sequence is not monotone. I found the problem, but the solution is wrong.
|
2088985
|
How can I solve the following sequence
Let we have the following sequence [imath]y_n= \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\ \sqrt{n^2+n}}[/imath] find the limit of the sequence [imath]y_n[/imath] decide whether it is increasing or decreasing
|
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