Is It Possible to Uniquely Identify a Shape in $mathbbR^3$ Knowing Its Surface Area and Volume?

Your simple example is in a good direction but too simple. We know the cube is the most efficient parallepiped in terms of volume to surface ratio so you won't get another parallelepiped to match both. But we should be able to find two parallelepipeds, one a flat pancake and one a long needle that have the same surface and volume. Let one be $1 times 1 times 100$ and the other $a times a times b$ with $a gt b$. Then we can write two equations and solve them(thanks Alpha)

$$a^2b1002a^24ab4022a^2frac 400a402afrac 12(3sqrt89-1)approx 13.651$$

More precisely, can an orientable closed compact 2-manifold embedded in $mathbbR^3$ be, up to an isometry, uniquely identified by its surface area and the volume it encloses? Conversely, can a counter example be given to disproof this claim?

In the answer to an earlier question related to the isoperimetric inequality a proof for the statement that "sphere is the only closed surface in $mathbbR^3$ that minimizes the surface area to volume ratio" has been outlined. That and the beginning of this post on Brunn-Minkowski inequality on Terence Tao's What's New made me curious about the existence (or lack thereof) more stringent constraints akin to the isoperimetric inequality.

As a first step towards constructing a counterexample, I examined the simple case of rectangular cube having the same surface area and volume as a cube. I ended up with the following equation:

$$

(underbracea b c_volume)^frac23 frac16timesunderbrace(2ab2bc2ac)_textsurface area,,

$$

where $a$, $b$, and $c$ are the dimensions of the rectangular cube. (For a cube of the sidelength $d$, $(d^3)^2/3frac16,6,d^2$; hence the above equation.) I have no idea how to proceed with the search for any positive definite solutions to the above problem. I suppose it can be simplified by invoking (isotropic) scaling to set one of the dimensions of the rectangular cube to $1$ and the following equation:

$$

(a b)^2 left(frac26right)^3(abba)^3,.

$$

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What Are Some Ways to Keep My Hair Really Healthy?
What Are Some Ways to Keep My Hair Really Healthy?
Use professional shampoo & conditioner. I know it may seem like a rip-off, however all the shampoos like Pantene, Garnier, etc. have a high alcohol content....since the normal pH of your hair is 4.5-5.5, things containing high amounts of alcohol throw your hair all out of whack and cause breakage, split-ends, fly-aways, etc. The reason these shampoos make your hair feel soft is because they contain high amounts of humectants, which are just a cover for how damaging they really are! Another tip is ALWAYS use a thermal protector while applying heat. Heat appliances cause tons of damage to your hair. Usually this is a cream/clear gel you apply before blow drying. Never use a flat iron on hair that is not completely dry!1. Split End Help. ( 10 Points) !!!!!!!!!!!!!!!!!!!!!!?For sure you should get a trim and maybe even cut your hair as short as you are comfortable doing it. You should for no reason straighten or dye your hair. Look for some Garnier Fructis or Pantene shampoos and conditioner for curly hair so that it extra moisturizes it and helps the split ends repair themselves. I say from experience that ppl with curly hair have really bad split ends, which is why we have special stuff in our hair products. The split ends wont magically go away in like a month or a few weeks, but you will see improvement after a while. Oh, and for sure DO NOT PICK THEM OFF!! It will just make it worse. Every two months until your hair is healthy get a trim. just maybe half an inch or an inch will be just fine. Eventually your hair will go back to normal. Good luck with that!.2. Whats the best shampoo conditioner for cheap?clever or V05 are solid inexpensive ($a million) shampoos, yet whilst it is composed of conditioner i could bypass Tresemme ($6) it does an somewhat solid interest. i take advantage of salon shampoo because of the fact i am form of a hair snob, yet I nonetheless use the cheap conditioner. Pantene is undesirable on your hair (too plenty protein and reasons alot of greater build up on the hair) so i does not bypass there. desire this permits! My hair is oily, superb, and colour-taken care of if that helps you any!3. How can I get my curly hair to be nice like Taylor Swift?Haha i have the same hair long and curly..but not pretty curly! im completely obsessed with taylor and her hair so i did so playing with it and what you really NEED to get is the conair hot sticks!! they are these little flexible curlers and they are not very expensive and i would look like a frizz ball without them!! you can like get them at walmart. anyways what i do when iwear my hair curly is like i get up and get in the shower and wash it and condition it and this may sound weird but dont use a whole lot of conditioner just enough to get it done cause it will help your hair hold the curl the just get out and towel dry your hair till it is just damp and then put a curling mouse in it and frizz serum then finish blow drying it straight! and you know have the curlers on like when your gettin in the shower and then just make sure the little light is on meaning that they are hot enough and then seperate your hair into pretty small peices (if you have really thick hair then you might need to buy two sets cause they only have 14 in them and you use small sections) and then start from the botom of your hair and role it through your hair and pin it together youll see the instructions on how to pin them together and make sure you wrap it all the way to the top of your head and leave them for like 15 or 20 minutes while your finishin gettin ready and then use like a light hairspray about 5 minutes before you take them out then your done. i know this thing was really long but its not that hard lol!!!! hope i helped good luck!!!!!!!!
Difference Between Epimers and Diastereomers
This scheme I just drew up specifically for you should answer your question.Diastereomers are stereoisomers that are not enantiomers of each other. That includes conformers (geometric isomers that derive from single bond rotation; usually interconverting rapidly) and atropisomers (under which I would subsume E/Z isomers; they derive from hindered rotation around a typically single bond and are separable) and anything with differeces in asymmetric carbons.If you have exactly one asymmetric carbon with inverted stereochemistry, then the diastereomer is a special one: an epimer. Anomers (e.g. - and -D-glucose) should be considered special cases of epimers.Let me use the menthol isomers (image taken from Wikipedia, where a full list of authors is available) to clarify the concept:Two vertically aligned structures are always enantiomers. Two structures that are not vertically aligned are diastereomers.()-Isomenthol is ()-menthols C4-epimer, while ($-$)-neoisomenthol is ()-menthols C1-epimer. And ($-$)-neomenthol is ()-menthols C2-epimer. All of these are also diastereomers.()-Neomenthol is not an epimer of ()-menthol. It is still a diastereomer, howeverMy lecture notes describe epimers as compounds which differ byconfiguration at only one carbon and gives the example of D-erythrose and D-threose as being epimers.However isn't that the same thing as diastereomers? My understanding is that epimers are supposed to be some kind of subset of diastereomers, however I am not sure how or why. I tried searching the web but I got more confused with one site saying that carbohydrates in open form can have two different chirality centres and still be epimers.
Direct Proof That There Is No Modular Form of Weight $2$ for $SL_2(mathbbZ)$.
I think we can try a more algebraic answer in the case $Gamma SL_2(mathbbZ)$, $f in M_2(Gamma)$.$f in M_2(Gamma)$ means $f$ is analytic on $mathcalH z in mathbbC, Im(z) > 0$ and $forall gamma in Gamma$,$gamma(z) fracazbczd$, $f(gamma (z)) (czd)^2 f(z)$. Since $ad-bc1$ then $d gamma (z) fracdz(czd)^2$ so that $f(z)dz f(gamma( z))dgamma (z)$ and hence $$F(z) int_i^z f(s)ds$$is an holomorphic map $mathcalH to mathbbC$ satisfying $$F(gamma (i))int_i^gamma(i)f(s)dsint_i^gamma(i)f(gamma_2(s))dgamma_2(s)int_gamma_2(i)^gamma_2(gamma(i))f(s)ds F(gamma_2(gamma (i)))-F(gamma_2 (i))$$ Thus $phi : gamma to F(gamma (i))$ is an homomorphism $Gamma to mathbbC$, since $mathbbC$ is an abelian group, $phi$ factors through the abelianization $Gamma^ab$. Here they show $Gamma,Gamma$ is a subgroup of index $12$ so that the abelianization $Gamma^ab Gamma/Gamma,Gamma$ is a finite group. Therefore $forall gamma in Gamma, F(gamma (i)) in phi(Gamma^ab)$ which is a finite set of complex values and $F$ is now an holomorphic map $mathcalH^*/Gamma,Gamma to mathbbC$. Since $mathcalH^*/Gamma,Gamma$ is compact we can take $z in mathcalH^*/Gamma,Gamma$ where $|F(z)|$ is maximum, and hence $F$ is constant by the maximum modulus principle, so that $f 0$.If the above argument is correct, it would be nice to explain how the commutator is different for $Gamma$ a congruence subgroup (allowing in those case $f in S_2(Gamma)$ to be an holomorphic map $mathcalH^* / Gamma to mathbbC/Lambda cong E$ an elliptic curve)I want to show that there are no Modular forms of weight $2$ for $SL_2(mathbbZ)$ without using the dimension formular or the $frack12$- formular. I was given some hints, too. However, there are two things missing.The idea is to look at a antiderivative of $f in M_2(SL_2(mathbbZ))$ which is given by $$F(tau)a_0taufrac12pi i sum_n1^infty fraca_nnq^n$$ where $fsum_n0^infty a_nq^n$. First I want to show that $F(gammatau)F(tau)C(gamma)$ where $gammatau$ denotes a mobius transformation and $C(gamma)$ does not depend on $tau$.My approach: $$F'(gamma tau)f(gamma tau)*(ctaud)^-2 (chain rule)$$ Since $f(gammatau)f(tau)*(ctaud)^2$ it follows that $$F'(gamma tau)f(tau)$$ And after integrating both sides I got $$F(gamma tau)F(tau) C(gamma).$$ The next step is to show that $C(gamma_1gamma_2)C(gamma_1)C(gamma_2)$ (for $gamma_1,gamma_2 in SL_2(mathbbZ$)) but I have no idea how to show that.Additionally, I am not sure how to show that $F$ is holomorphic. I see that $F(tau1)F(tau)$, so $F$ is bounded along the real axis. However, what happens if $lim y rightarrow infty$, especially since $a_0 tau$ might be unbounded? Thanks for your help·OTHER ANSWER:I want to show that there are no Modular forms of weight $2$ for $SL_2(mathbbZ)$ without using the dimension formular or the $frack12$- formular. I was given some hints, too. However, there are two things missing.The idea is to look at a antiderivative of $f in M_2(SL_2(mathbbZ))$ which is given by $$F(tau)a_0taufrac12pi i sum_n1^infty fraca_nnq^n$$ where $fsum_n0^infty a_nq^n$. First I want to show that $F(gammatau)F(tau)C(gamma)$ where $gammatau$ denotes a mobius transformation and $C(gamma)$ does not depend on $tau$.My approach: $$F'(gamma tau)f(gamma tau)*(ctaud)^-2 (chain rule)$$ Since $f(gammatau)f(tau)*(ctaud)^2$ it follows that $$F'(gamma tau)f(tau)$$ And after integrating both sides I got $$F(gamma tau)F(tau) C(gamma).$$ The next step is to show that $C(gamma_1gamma_2)C(gamma_1)C(gamma_2)$ (for $gamma_1,gamma_2 in SL_2(mathbbZ$)) but I have no idea how to show that.Additionally, I am not sure how to show that $F$ is holomorphic. I see that $F(tau1)F(tau)$, so $F$ is bounded along the real axis. However, what happens if $lim y rightarrow infty$, especially since $a_0 tau$ might be unbounded? Thanks for your help
Find Wrapping Angle of Helix on a Torus
Ted Shifrin answer to my original question is incomplete and I can't use it the way it is. The almost-complete version is like this:$$alpha(t)beginpmatrixxBig(R_1R_2cos(t)Big)cosbiggr(carctanBig(dfrac(R_1-R_2)tan(t/2)sqrtR_1^2-R_2^2Big)biggr)yBig(R_1R_2cos(t)Big)cosbiggr(carctanBig(dfrac(R_1-R_2)tan(t/2)sqrtR_1^2-R_2^2Big)biggr)zR_2sin(t)endpmatrix$$I have injuries for scratching my head trying to more or less do semi-educated guess what extra terms goes where by trail and error (https://en.wikipedia.org/wiki/Trial_and_error)That's because I'm an engineer not a mathematician which should make sense to as why I did it like this.Using "consecrated" math terminology from here: https://en.wikipedia.org/wiki/Toroidal_and_poloidal(I hope it's ok to use the word "consecrated" in the meaning to "to devote to a purpose with deep dedication" (according to Merriam-Webster dictionary) english is not my native language so I'm asking for leniency)Let there be the term : "toroidal direction step angle" or the angle done in toroidal direction by only 1 poloidal turn of my curve (a STEP in the right way - literally).Therefore I can't show mathematically, but have proven undoubtedly in numerous tests and experiments that:$$cfractextstep angle180 $$$$text1 poloidal turn wire lengthfrac2pi R_2cos(textwrapping angle) $$$$textstep anglefrac360R_2tan(textwrapping angle)sqrtR_1^2-R_2^2$$$$textwrapping anglearctanbiggr(fractextstep anglesqrtR_1^2-R_2^2360R_2biggr)$$Again, emphasis on how these equations were determined: semi-educated guesses, trial and error a dose of madness and insanity and lots of dedication and perseverance.These formulas never failed me so far for any value of $R_1$ , $R_2$, c, step angle or wrapping angle.Of course for the answer to be complete, I would have to replace $R_1R_2cos(t)$ with $R_1R_2cos(at)$ for some constant "a" representing how many turns in the poloidal direction my curve does in the same step angle, the final goal being to generate a special family of closed curves tori knots (with curve tangent(aka velocity vector) always constant against all meridians and parallels of the torus)I was asked to provide more context to my question:https://www.researchgate.net/publication/311742883_Visualization_in_3D_of_Dynamics_of_Toroidal_Helical_Coils_in_quest_of_optimum_designs_for_a_Concordian_MandalaI would like to add some sentences (which most likely will be deleted by someone):Nothing less than a miracle or a fluke for a professor like Ted Shifrin answering my question.My question was written with sloppy terminology.It would have been amazing to have seen the math he did behind the scenes for the parametrization of such curves.Can I hope for yet another miracle?Why is speaking like this forbidden on this forum? Reading between the lines 1st sentence shows gratitude towards altruism, respect and awe. 2nd- introspection, constructive self-criticism and honesty. 3rd mild regret, call for compassion and desire. and 4th expectation and hopeWhy speaking (in and with) these terms illegal on this forum? Aren't these aspects extremely important in any form of communication? As a newbie on this forum I'm very confused... Yes... stick to the math and disregard and delete everything else...monstrously strict and rigid.I need some help in calculating the wrapping angle of a spiral helix wrapped on a torus with constant angle against all the meridians of the torus.The wrapping angle (or the angle measured around and/or against the torus circular cross-section See here a (13,6)tori knot) always remains constant as the helix curve spirals around the torus. In layman's terms means it spirals with the same angle always. if I were to know the arc length of one such turn, my solution will be trivial: $$Arclen 2pi R_2over cos(wrapping_angle) .(1)$$ I know this formula is counterintuitive to be true and valid regardless of R1 (torus major radius), but it is. I have an algorithm to test it out and it is true beyond any shadow of a doubt. It took me a while to figure it out, but indeed it does seem to suggest that the helix is on a cylinder, but it's on a curved cylinder (a torus). It doesn't make sense, but the formula is correct. I have a way to test it out. I'm trying to see that the tangent of any point on the curve is constant (derivative0?? i don't know) so it projects(unwinds) as a straight line in a 2D plane. All i know is that equation 1 is correct. From here i can find my wrapping angle, where R2 is torus minor radius. This formula is valid only for 1 turn spiral helix and ONLY if I have constant wrapping angle. The thing is I don't know the arc length of one turn, but I do know R1, R2 and the azimuth angle of 1 turn (or 1 turn step-angle or the angle measured around torus central axis see it here). I want to find the arc length of 1 turn of the helix as a function of step-angle and using such formula in combination with the one I already have, I can find my wrapping angle. I don't know how to integrate but I see something in my mind along these lines: $$Arclen int_0^stp pi int_0^2pi(R_1-R_2cos(theta) )dtheta over 180 beta dbeta something .(2)$$ I don't know what this is. I believe it's some sort of a double integration in polar coordinates(?) of infinitesimal circle-arcs which i know nothing about. stp is my step_angle, $theta$ is the infinitesimal angle made by R2 in respect to equatorial plane of torus, and $beta$ is supposed to be the step_angle chopped down into infinitesimal bits. I know equation (2) is wrong, but let's say it's not, then i write: $$equation(1)equation(2) > wrapping_anglearccos(2pi R_2 over something)$$ Since equation(2) is wrong, I also see in my mind something like this: $$Arclenint_0^stp (sqrt(pi [int_0^2pi(R_1-R_2cos(theta)) dtheta]*beta over 180)^2(int_0^2pipi R_2alpha over180dalpha)^2)dbeta . (3)$$ I don't know how to integrate. Please don't ask me where i got these equations. Maybe they can provide a place to start for you to help me. I just need a little bit of help to correct this mess. Can you help?·OTHER ANSWER:I need some help in calculating the wrapping angle of a spiral helix wrapped on a torus with constant angle against all the meridians of the torus.The wrapping angle (or the angle measured around and/or against the torus circular cross-section See here a (13,6)tori knot) always remains constant as the helix curve spirals around the torus. In layman's terms means it spirals with the same angle always. if I were to know the arc length of one such turn, my solution will be trivial: $$Arclen 2pi R_2over cos(wrapping_angle) .(1)$$ I know this formula is counterintuitive to be true and valid regardless of R1 (torus major radius), but it is. I have an algorithm to test it out and it is true beyond any shadow of a doubt. It took me a while to figure it out, but indeed it does seem to suggest that the helix is on a cylinder, but it's on a curved cylinder (a torus). It doesn't make sense, but the formula is correct. I have a way to test it out. I'm trying to see that the tangent of any point on the curve is constant (derivative0?? i don't know) so it projects(unwinds) as a straight line in a 2D plane. All i know is that equation 1 is correct. From here i can find my wrapping angle, where R2 is torus minor radius. This formula is valid only for 1 turn spiral helix and ONLY if I have constant wrapping angle. The thing is I don't know the arc length of one turn, but I do know R1, R2 and the azimuth angle of 1 turn (or 1 turn step-angle or the angle measured around torus central axis see it here). I want to find the arc length of 1 turn of the helix as a function of step-angle and using such formula in combination with the one I already have, I can find my wrapping angle. I don't know how to integrate but I see something in my mind along these lines: $$Arclen int_0^stp pi int_0^2pi(R_1-R_2cos(theta) )dtheta over 180 beta dbeta something .(2)$$ I don't know what this is. I believe it's some sort of a double integration in polar coordinates(?) of infinitesimal circle-arcs which i know nothing about. stp is my step_angle, $theta$ is the infinitesimal angle made by R2 in respect to equatorial plane of torus, and $beta$ is supposed to be the step_angle chopped down into infinitesimal bits. I know equation (2) is wrong, but let's say it's not, then i write: $$equation(1)equation(2) > wrapping_anglearccos(2pi R_2 over something)$$ Since equation(2) is wrong, I also see in my mind something like this: $$Arclenint_0^stp (sqrt(pi [int_0^2pi(R_1-R_2cos(theta)) dtheta]*beta over 180)^2(int_0^2pipi R_2alpha over180dalpha)^2)dbeta . (3)$$ I don't know how to integrate. Please don't ask me where i got these equations. Maybe they can provide a place to start for you to help me. I just need a little bit of help to correct this mess. Can you help?
Countable Orthonormal Basis for Hilbert Space
Actually, we have a non-constructive proof.Lemma : Let $mathcalH$ be a Hilbert space. Then it is impossiblethat $mathcalH$ has a countable dense subset $D$ and an uncountableorthonormal set $e_imid iin I$.Proof: Prove by contradiction. Suppose the contrary that $mathcalH$has a countable dense subset $D$ and an uncountable orthonormal set$e_imid iin I.$ Let $rfrac14$ and for each $iin I$,let $B_iB(e_i,r)$, then open ball centered at $e_i$ withradius $r$. Note that $B_icap B_jemptyset$ whenever $ineq j$.(For, if there exists $xin B_icap B_j$, then $||e_i-e_j||leq||e_i-x||||x-e_j||e_i$, where the sum is independentof ordering. (I leave the proof to you)./////////////////////////////////////////////////////////////I expand the last line, state it as a theorem and give a self-contained proof. This is known as the Fourier expansion in Hilbert space.////////////////////////////////////////////////////////////Theorem: Let $mathcalH$ be a Hilbert space (over $mathbbR$or $mathbbC$) and let $e_imid iin I$ be a maximal (withrespect to $subseteq$) orthonormal set (where the index $I$ maybe uncountable). Then for each $xinmathcalH$, we have $xsum_iin Ilangle x,e_irangle e_i$,where the series converges in unorder sense (explained in Claim (4)below).Proof: Claim (1) (Bessel's inequality): For any finite subset $I_1subseteq I$and any $xinmathcalH$, $sum_iin I_1|langle x,e_irangle|^2leq||x||^2$. Proof of Claim (1): Denote $Pxsum_iin I_1langle x,e_irangle e_i$.Observe that $Pxbot x-Px$, so $||x||^2||Px||^2||x-Px||^2geq||Px||^2sum_iin I_1|langle x,e_irangle|^2$.Claim (2): For any $xinmathcalH$, $iin Imidlangle x,e_irangleneq0$is a countable set.Proof of Claim (2): Let $I'iin Imidlangle x,e_irangleneq0$.Prove by contradiction. Suppose the contrary that $I'$ is uncountable.For each $ninmathbbN$, let $I_niin Imid|langle x,e_irangle|geqfrac1n$.Observe that $I'cup_nI_n$, so there exists $n$ such that $I_n$is uncountable. Choose $k$ such that $frackn^2>||x||^2$.Choose a finite subset $I'_nsubseteq I_n$ that contains $k$elements, then $sum_iin I'_n|langle x,e_irangle|^2geqfrackn^2>||x||^2$,contradicting to claim (1).Claim (3): Given $xinmathcalH$ and let $I_xiin Imidlangle x,e_irangleneq0$.Fix an enumeration for $I_x$, say $I_xi_1,i_2,ldots$(finite or infinite), then the series $sum_klangle x,e_i_krangle e_i_k$is convergent.Proof of Claim (3): If $I_x$ is a finite set, we are done. Supposethat $I_x$ is a countably infinite set. Let $s_nsum_k1^nlangle x,e_i_krangle e_i_k$.By Claim (1), for each $n$, $sum_k1^n|langle x,e_i_krangle|^2leq||x||^2$,so the series $sum_k1^infty|langle x,e_i_krangle|^2$is convergent. We show that $s_n$ is a Cauchy sequence in $mathcalH$.Let $varepsilon>0$. Choose $N$ such that $sum_kN^infty|langle x,e_i_krangle|^20$be arbitrary. Let $U_yB(y,varepsilon)$ be the open ball centeredat $y$ with radius $varepsilon$. Choose $N$ such that $||sum_k1^Nlangle x,e_i_krangle e_i_k-y||leqfracvarepsilon4$.(If $I_x$ in Claim (3) is a finite set, let $N$ be the numberof elements in $I_x$.) We adopt the notation in Claim (3) and continueto work with the enumeration $I_xi_1,i_2,ldots$. Let$I_1i_1,i_2,ldots,i_N$. Clearly $I_1inmathcalC$.Consider the case that $I_x$ is an infinite set (the finite caseis trivial). By continuity of norm (i.e., the continuity of the map$xmapsto||x||$), we have$$lim_n||sum_k1^Nlangle x,e_i_krangle e_i_k-sum_k1^nlangle x,e_i_krangle e_i_k||||sum_k1^Nlangle x,e_i_krangle e_i_k-y||leqfracvarepsilon4.$$On the other hand, for any $n>N$, we have $||sum_k1^Nlangle x,e_i_krangle e_i_k-sum_k1^nlangle x,e_i_krangle e_i_k||^2sum_kN1^n|langle x,e_i_krangle|^2$.Hence $sum_kN1^infty|langle x,e_i_krangle|^2leqleft(fracvarepsilon4right)^2$.Let $I_2inmathcalC$ be arbitrary such that $I_1preceq I_2$.Then$$||theta(I_2)-y||leq||sum_k1^Nlangle x,e_i_krangle e_i_k-y||||sum_iin I_2setminus I_1langle x,e_irangle e_i||leqfracvarepsilon4||sum_iin I_2setminus I_1langle x,e_irangle e_i||.$$Observe that$$||sum_iin I_2setminus I_1langle x,e_irangle e_i||^2sum_iin I_2setminus I_1|langle x,e_irangle|^2leqsum_kN1^infty|langle x,e_i_krangle|^2leqleft(fracvarepsilon4right)^2$$ because for any $iin I_2setminus I_1$, if $inotini_N1,i_N2,ldots$,then $langle x,e_irangle0$.Now we have: $||theta(I_2)-y||leqfracvarepsilon2$. Thisshows that $sum_iin Ilangle x,e_irangle e_iy$ in unorderedsense.Claim (5): The $y$ defined in Claim (3) and Claim (4) is $x$. Thatis $xsum_iin Ilangle x,e_irangle e_i$.Proof of Claim (5): Let $zx-y$. Prove by contradiction. Supposethe contrary that $zneq0$. We adopt the notation in Claim (3) andClaim (4). Recall that for each $ainmathcalH$, the map $xmapstolangle x,arangle$is continuous. Let $iin I$ be arbitrary. Conside the case that $I_x$is infinite (The finite case is trivial.). We have$$langle z,e_iranglelim_nrightarrowinftylangle x-sum_k1^nlangle x,e_i_krangle e_i_k,e_irangle.$$ If $inotin I_x$, we have $langle x,e_irangle0$ and $langle e_i_k,e_irangle0$for all $k$, and hence $langle z,e_irangle0$. Suppose that$iin I_0$, say $ii_k'$ for some $k'inmathbbN$. Then forany $ngeq k'$, we have$$langle x-sum_k1^nlangle x,e_i_krangle e_i_k,e_iranglelangle x,e_irangle-langlesum_k1^nlangle x,e_i_krangle e_i_k,e_iranglelangle x,e_i_k'rangle-langle x,e_i_k'rangle0.$$ Therefore $langle z,e_irangle0$ in all cases. Define $tildezz/||z||$,then $e_imid iin Icuptildez$ is an orthonormal set,containing $e_imid iin I$ as a proper subset. This contradictsto the maximality of $e_imid iin I$.If $Dx_n : nin mathbbN$ is a countable set dense in a Hilbert space $mathcalH$, how can I show that Gram-Schmidt algorithm applied to $D$ (or a subset of $D$) produces an orthonormal numerable basis for $mathcalH$?So far I have been able to prove that every ortonormal basis of $mathcalH$ has to be numerable.
How Was the First Sale of Vivo X60 Series?
How Was the First Sale of Vivo X60 Series?
Zeiss lens has a history of 175 years. Its biggest advantage is its ultra-clear imaging ability. Generally, only using high-end cameras has become the dream of camera lovers around the world with its unique selling points such as color restoration, high resolution and almost no corner distortion. Why do you talk about dreams here? Because of its ultra-high price, ordinary people can only think but can not become a reality.On December 29, 2020, vivo X60 series, the first product jointly developed by vivo and Zeiss, was officially launched. All the series are equipped with Zeiss optical lenses as standard, which can be owned for a minimum of 3498 yuan. On January 8, vivo X60 series was officially launched. We can see that it won the double champions of JD platform sales sales, tmall platform sales sales, Suning platform sales sales, and also the double champions of online sales and sales. Besides Zeiss blessing, why is vivo X60 series so popular with the market and consumers?Ultra lightweight appearanceThe thickness of vivo X60 does not exceed 8mm, and the weight of vivo X60 does not exceed 180g (excluding the unpublished Super Cup). Among them, the vivo X60 is only 7.36mm thick, making it the lightest 5g mobile phone in the world, 0.04mm thinner than the iPhone 12 series. The slim and fashionable appearance and vivo's strict pursuit of technology and design create the unparalleled hand feel of vivo X60 series. If you don't believe it, you can try it in the offline experience store in person. You can't forget the feeling of holding it in your hand.Top imaging systemThe vivo X60 is equipped with Zeiss optical lens as standard. Whether it is the main camera, ultra wide angle or professional image lens, it is an optical lens jointly developed by both sides. The whole system is equipped with the second generation micro PTZ as standard. With these two hardware blessings and the image system jointly developed by vivo Zeiss, you can take a good picture at any light source and scene. You can learn from the comments of professional photographers on the imaging effect of vivo X60 series.Performance of flagship aircraftVivo X60 is the world's first Samsung Orion 1080 processor. The overall performance of this processor is lower than that of Xiaolong 865. It is much stronger than what MediaTek Tianji 1000 and Xiaolong 765G are used in the same level models. With the blessing of Samsung Orion 1080, vvio X60 is far more competitive than its competitors.
Huami Technology Held a Press Conference on August 27, and Amazfit Smart Sports Watch 3 Officially A
Huami Technology Held a Press Conference on August 27, and Amazfit Smart Sports Watch 3 Officially A
This morning, huami technology officially announced that it would hold the 2019 new product Conference on August 27. Then, Huang Wang, CEO of huami technology, posted the invitation letter for the new product Conference on his microblog. Huami technology distributed amazfit one-way wet guide short sleeved T-shirts and amazfit antelope running shoes to the media as physical invitations. The gift box vacated a watch position, which triggered extensive discussion in the media. Huang Wang also explained that this vacancy was actually reserved for the new amazfit intelligent movement watch 3.(invitation letter from Huang Wang, CEO of huami Technology)The theme of huami technology's 2019 new product launch is "run, amazfit! Run!", which is adapted from the classic line "run, Forrest! Run!" of Forrest Gump. With his pure love for sports, Forrest Gump, the protagonist in the film, runs across the United States. No matter how many difficulties and ridicules he encounters, he also doesn't give up. "Run, Forrest! Run!" is the symbolic line that Forrest Gump's mother encourages him, and Forrest Gump can persevere and succeed step by step. This spirit also inspires many people. According to Huang Wang's microblog, the amazfit smart sports watch 3 not only symbolizes huami technology's understanding and adherence to the sports spirit, but also has a taste of paying tribute to Forrest Gump.(theme poster of huami technology's 2019 new product launch)At present, huami technology has become a leading enterprise in the wearable field. With the power of technology, it promotes everyone in the world to enjoy better sports, health and medical services. Sports is undoubtedly an important part of the mission of huami technology. In the past amazfit smart watch products, we can also see a lot of improvements in sports functions.Taking the amazfit smart watch GTR released last month as an example, it provides 12 sports modes such as outdoor running, walking, outdoor cycling, indoor running, indoor cycling, open water swimming, swimming pool swimming, elliptical machine, mountaineering, cross-country running, skiing and exercise. Turn on the corresponding professional sports mode before sports, you can always understand your sports status and help improve sports efficiency.(huami technology amazfit smart watch GTR)Facing the professional sports market, huami technology has also launched several products of amazfit smart sports watches 1, 2 and 2S. It is worth mentioning that it has been a long time since huami technology last updated the smart sports watch series. With more and more attention to sports and healthy life, huami technology still needs to bring more stimulation to the sports watch market and further implement the mission of huami technology. This may also be an important reason why huami technology updates amazfit smart sports watch 3.In terms of function, at present, amazfit smart sports watch 2 has been equipped with 16 sports modes. According to Xiaobian's guess, amazfit smart sports watch 3 will be richer in sports modes to meet the sports monitoring needs of different users. With more powerful functions, the price of amazfit smart sports watch 3 may reach 2500 yuan, which is also the price level of mainstream sports watches at present. As for whether the new product has more professional sports function improvement, we still need to wait for the official answer on August 27.
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