triclino

Prove whether or not the function [math] f(x)=\frac{1}{x+1}[/math] is uniformly continuous in [math](0,\infty)[/math]

Please be so kind as to read my posts again . I am not asking for a geometrical picture where intutively we can say that ,since we cannot fit a disc into a line ,then S is not open in [math]R^2[/math]. WE want an analytical proof . For example, we can easily see that a continuous function in a certain interval ,with f(a)>0 and f(b)<0 for two valeus a,b the function will at least cross the x-axis . But to prove that analytically is quite a job

Given that: 1) A function [math] f: D_{f}\subset R\rightarrow R[/math] is uniformly continuous iff for any pair of sequences {[math]x_{n}[/math]} {[math]y_{n}[/math]} in [math] D_{f}[/math] ( = domain of f), [math]lim_{n\to\infty}|x_{n}-y_{n}| = 0[/math][math]\Longrightarrow lim_{n\to\infty}|f(x_{n})-f(y_{n})| = 0[/math], Determine if the function f(x) = [math]x^2[/math] is uniformly continuous or not.

O.K For the first problem taking ε= min{ x,1-x } it works ,but for the 2nd problem ,wcich you consider as an easy problem ,what would you suggest for the pair [math](x_{1},y_{1})[/math]

. Well what would suggest the "δ" to be chosen ??

Well, did i not write a proof in my post #2 ,apart from the numerical example???. Unless you cannot even recognise what a proof is. I like the expression "proof with variables"

What does your professor ???,actually want you to prove ,if he wants you to prove anything at all Firstly he asks for a proof by using the concept of the GCD. Then he asks for a proof by using the concept of the LCM. To curry on with my example of the air plane: Not only he wants to fly from L.A to New York thru S.Africa ,but now he wants the pilot to get petrol supply in the air. Next ,i suppose, he will ask the pilot to fly the air plane upside down . Do you think he will ever get into an air plane???

Suppose (a,b)=x and (a,c) =y and b/c (=b devides c) . We want to prove : [math]x\leq y[/math] From the definition of the GCD ,if (a,b) =x ,we have that: x>0........................................................................................1 x/a and x/b ..............................................................................2 And if d is any No such that d/a and d/b ,then d/x................................3 Also if (a,c) = y we have that: y>0..............................................................................................4 y/a and y/c......................................................................................5 And if d is any No such that d/a and d/c ,then d/y..............................................................................................6 Put in (6) d = x and we have that: If x/a and x/c ,then x/y.....................................................................................7 But from (2) we have that x/a and x/b, and also is given that b/c ,hence x/a and x/c. And using (7) we have : x/y Now from the definition of x/y that implies that: y=kx ...........................................................................................8 where [math] k\geq 1[/math].......................................................................9 Multiply (9) by x>0 and we have: [math]xk\geq x[/math]..................................................................................10 And substituting (8) into (10) we have: [math]y\geq x[/math] You can easily see the above problem by using a numerical example. Let (a,b) = (4,12) and (4,36) where 12/36 And GCD(=4) of (4,12)[math]\leq[/math] GCD(=4) of (4,36)

Yes you can add [math]m^2[/math] to both sides of [math]A_{3}[/math] by using the following axiom,concerning equality: for all ,x,y,z : [math] x=y\Longrightarrow x+z = y+z[/math] and if we put: x=[math]n^2[/math] , y= 2m+1 ,z = [math]m^2[/math] the above becomes: [math]n^2 = 2m+1\Longrightarrow n^2+m^2 = 2m+1+m^2[/math]. By the way the law that allow us to do that is called Universal Elimination. Now coming back to your exercise ,the right state of affairs is: If n is odd ,then n= 2k+1 and thus [math]n^2 =4k^2 +4k+1 =2(2k^2+2k)+1[/math],hence there exists a natural No m such that [math]n^2=2m+1[/math] ,or [math] m=\frac{n^2-1}{2}[/math] and from here onwards we follow the proof as shown in my post But, by making the substitution n = 2k+1 into [math] m=\frac{n^2-1}{2}[/math] you simply make the proof longer . To give you another simple example. Suppose you want to prove : [math](x+1)^2 =x^2+2x+1[/math]. You can do that the simple way by simple multiplying (x+1) and(x+1) and using the law of distribution obtain the desired result, or you can do it the long way as follows: Let x=y+1 ,then [math](x+1)^2 =[(y+1)+1]^2 = [(y+1)+1][(y+1)+1][/math]= (Y+1)[(y+1)+1] +1[(y+1)+1] =[math]y^2+4y+4[/math] =[math](x-1)^2 +4(x-1) +4 = x^2-2x+1+4x-4+4 =x^2+2x+1[/math] You can make the above simple problem as long as you like. So if there is a direct fly from L.A to New-York,would you fly thru S.Africa??

I think the right question is : If n is odd then there exists an integer m such that : [math] n^2 + m^2 = (m+1)^2[/math]. Proof: If n is odd ,then [math]n^2[/math] is odd ,hence there exists an integer m and [math]n^2 = 2m+1[/math]. Thus [math] n^2 + m^2 = 2m+1+m^2 = (m+1)^2[/math]. Now the book or your lecturer to make it easier for you they added the extra formula [math] m=\frac{n^2-1}{2}[/math] ,or solving for [math]n^2[/math] is: [math]n^2 = 2m+1[/math],which is the conclusion of the theorem: If n is odd ,then [math]n^2[/math] is odd. Actually one of the facts : "n is odd" or " [math]m=\frac{n^2-1}{2}[/math]" is redundant because either of the two can lead to the conclusion: [math]n^2+m^2 = (m+1)^2[/math] Now in your opening post you ask for a step wise proof : [math]A_{1},A_{2},A_{3}....[/math]. Stepwise proofs can be quite difficult sometimes> Anyway here is a step wise proof: [math]A_{1}[/math] : n is odd......................................................................................................................................................assumption [math]A_{2}[/math] : [math]m=\frac{n^2-1}{2}[/math]..................................................................................................................................assumption ,or derived from [math]A_{1}[/math] [math]A_{3}[/math]: [math]n^2 = 2m+1[/math]....................................................................................................by multiplying [math]A_{2}[/math] by 2 and then adding 1 to both sides [math]A_{4}[/math]: [math]n^2+m^2 = 2m+1+m^2[/math] ...............................................................................by adding [math]m^2[/math] to both sides of [math]A_{3}[/math] [math]A_{5}[/math]: [math] (m+1)^2= 2m+1+m^2[/math].................................................................................................................expanding [math](m+1)^2[/math]. [math]A_{6}[/math] :[math] n^2+m^2 = (m+1)^2[/math]................................................................................by substituting [math]A_{5}[/math] into [math]A_{4}[/math] A more detailed stepwise proof can be done ,where the axioms and the laws of logic involved in the proof can be explicitly mentioned.

O.K S is open in R iff ,for all ,xεS there exists an ε>ο such that for all , y: [math]y\in B(x,\epsilon)\Longrightarrow y\in S[/math]. Here the ball B(x,ε) = {z : |z-x|<ε }. So in this case we are seeking an ε>0 ΑΝD: S is not open in [math]R^2[/math] ,iff there exists an ,xεS such that for all ε>0 ,there exists a pair of [math] (x_{1},y_{1})[/math] belonging to the ball B((x,0),ε) and not belonging to S. Here the ball B((x,0),ε) = {(z,w) : [math]\sqrt{(z-x)^2 + w^2}<\epsilon[/math]}. So in this case we are looking for a pair of [math](x_{1},y_{1})[/math]. Any suggestions??

Prove that the set S = {x: a<x<b } is open in R (=Real Nos ) and not in [math]R^2[/math]

Given that: [math] f: [0,\infty)\rightarrow R[/math],such that [math]f(x)=\sqrt{x+1}[/math] prove that: [math] lim_{x\to 0}\sqrt{x+1} = 1[/math] using the ε-δ definition

Suppose that ; 1)[math] f: R^2\rightarrow R[/math] such that : [math]f(x,y) =xy[/math] 2)The Euclidian norm of a vector [math] v=(u_{1}, u_{2})[/math] is defined as : [math] ||v||_{Eu} =\sqrt{ u_{1}^2 +u_{2}^2}[/math] 3) The maxnorm of a vector [math] v=(u_{1},u_{2})[/math] is defined as : [math] ||v||_{max} = max( |u_{1}|,|u_{2}|)[/math] Where [math] u_{1},u_{2}[/math] belong to the real Nos R Then prove : [math] \lim_{(x,y)\to(1,1)} f(x,y) = 1[/math] ,with respect to both norms

Suppose that ; 1)[math] f: R\rightarrow R^2[/math] such that : [math]f(x) =(2x+1,x^2)[/math] 2)The Euclidian norm of a vector [math] v=(u_{1}, u_{2})[/math] is defined as : [math] ||v||_{Eu} =\sqrt{ u_{1}^2 +u_{2}^2}[/math] 3) The maxnorm of a vector [math] v=(u_{1},u_{2})[/math] is defined as : [math] ||v||_{max} = max( |u_{1}|,|u_{2}|)[/math] Where [math] u_{1},u_{2}[/math] belong to the real Nos R Then prove : [math] \lim_{x\to 0} f(x) = (1,0)[/math] ,with respect to both norms

I suggest you read the opening post again carefully . It does not ask that the sequence ln n does not converge, But, That the sequence ln n diverges to infinity Who said that the negation of convergence to a limit l is : " for a given ε>0,[math]\exists N=N(\epsilon)[/math] such that [math]\forall n>N |x_{n}-l|>\epsilon[/math]" Read all the posts again, nowhere you will find that. A sequence [math] x_{n}[/math] diverges to infinity iff for all ,ε>0 there exists a natural No N ,such that: for all ,n: [math] n\geq N\Longrightarrow x_{n}>\epsilon[/math] This is the definition given by K.G BINMORE in his book :Mathematical A nalysis ,on pages 38-39 To mention one of the books producing that definition

Which definitions??

Prove that the the sequence [math] n^2[/math] diverges to infinity

According to the definition of divergence to infinity: Given an ε>0 we have to find a natural No N ,such that: for all [math]n\geq N\Longrightarrow ln n>\epsilon[/math]. So the central issue here is finding that N

So given ε>0 ,what do you think should be the natural No N chosen ,so that : for all natural Nos [math] n\geq N[/math] , ln n>ε

prove that the sequence ln n diverges to infinity

Suggestion No 1: For any natural No N Let n = N and m = 2N,then |ln n-ln m| = |lnN- ln2N| = |lnN-ln2-lnN|= |ln2| Hence ,choose : [math]0<\epsilon\leq |ln2|[/math] Any other suggestions? Apart from those where : m= 3N,4N,5N...............................kN?

Hence if ln n is not a Cauchy sequence: there exists an ε>0 ,such that : For all natural Nos N ,there exist natural Nos [math]n\geq N,m\geq N[/math] ,such that : [math] |ln n-ln m|\geq\epsilon[/math] Now the question is how do we find ε,n,m to satisfy the above?

prove using the ε-definition,that : ln n is not a Cauchy sequence

In which part of physics ,fuzzy logic applies??