This note has been created to ask questions and clarify doubts.

It is solely for the purpose of discussing my doubts.

Thank you.

This note has been created to ask questions and clarify doubts.

It is solely for the purpose of discussing my doubts.

Thank you.

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TopNewest@brian charlesworth , plz help me with following problem whenever you get some free time.

Q. Let

Sbe square with unit are. Consider any quadrilateral which has one vertex on each side ofS. If \(a, b, c, d \) denote the lengths of the quadrilateral, then \(\alpha \le { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }\le \beta \) where(A) \(\alpha =1\)

(B) \(\beta =2\)

(C) \(\alpha =2\)

(D) \(\beta =4\)

I don't know the correct answer. I got this from one my past test papers. I can't even make out which topic does the problem belong to. Whether we will use calculus or some standard inequality or some combination of approaches? Only thing that I could do is to visualize a dynamic diagram of the problem. I don't know how to produce animations so I am trying to describe my approach in words: Let each of the vertex of the variable quadrilateral lie on the midpoint of edges of the given square. Now if each vertex of the variable quadrilateral approaches the vertex of the given square then the value of \({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }\) approaches the value \(4\) ( i.e as an extreme case, each vertex of the variable quadrilateral would be just near each vertex of the given square but won't coincide so that the condition "Consider any quadrilateral which has one vertex on each side of

S" isn't violated. And we would have the variable quadrilateral as approximately congruent to the given square ). So we can have the upper limit as \(\beta =4\). I can't think about \(\alpha \). This is the best I could do. – Soumo Mukherjee · 2 years, 1 month agoLog in to reply

So that was the intuitive approach; now for the formal one. For this, draw a quadrilateral inside the unit square, and then focus on the four right triangles such that \(a,b,c,d\) are the hypotenuses, (starting from the upper right going clockwise), and the legs of each of these triangles lie along the sides of the unit square. Label these legs, starting from the top right leg and going clockwise, \(e,f,g,h,i,j,k,l.\) Then

\(a^{2} + b^{2} + c^{2} + d^{2} = e^{2} + f^{2} + g^{2} + h^{2} + i^{2} + j^{2} + k^{2} + l^{2}\) and \(e + f + g + h + i + j + k + l = 4.\)

Now we use some mathematical inequalities to determine our extremes. For \(\beta\), we can use Cauchy's inequality, which gives us that

\((e^{2} + f^{2} + g^{2} + h^{2} + i^{2} + j^{2} + k^{2} + l^{2})^{2} \le (e + f + g + h + i + j + k + l)^{2} = 4^{2} = 16\)

\(\Longrightarrow (a^{2} + b^{2} + c^{2} + d^{2}) \le 4,\) which as we know can be achieved, and hence \(\beta = 4.\)

To determine \(\alpha\) we can use the Chebyshev Sum Inequality to find that

\(8*(e^{2} + f^{2} + g^{2} + h^{2} + i^{2} + j^{2} + k^{2} + l^{2}) \ge (e + f + g + h + i + j + k + l)^{2} = 4^{2} = 16\)

\(\Longrightarrow (a^{2} + b^{2} + c^{2} + d^{2}) \ge 2,\) which as we know can be achieved, and so \(\alpha = 2.\)

(Note that for Chebyshev to work, we would have to arrange the leg lengths (as well as their squares) in descending order, but since we're adding them all up anyway this condition is easily satisfied.)

So the formal approach confirms the intuitive approach in this instance. There is a lot to digest with this formal approach so don't worry if it doesn't make sense at first glance, but it does use some very useful inequalities. For even more of them, check out this link. – Brian Charlesworth · 2 years, 1 month ago

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– Soumo Mukherjee · 2 years, 1 month ago

The " minimal case of the symmetry " argument was awesome. It will come handy in more problems. I need to learn more about the two inequality used. My stock of standard inequality is very less, that's my fault. But your solution is simple and helpful. And small too. I downloaded the pdf you attached.Thanks.Log in to reply

– Brian Charlesworth · 2 years, 1 month ago

You're welcome. :) As mentioned, the symmetry argument does not always lead to the correct values, but more often that not it will. Some of the inequalities in the pdf are more useful than others, but overall it is an excellent article; I've bookmarked it myself for future reference.Log in to reply

this. – Soumo Mukherjee · 2 years, 1 month ago

Please look intoLog in to reply

@brian charlesworth , if we continue here.

Okay, since I lost my original link, it seems hard to trace it back. But you can goggle search about women's multitasking ability and read one or two article which appear to you convincing. This might be a good read BBC News. I have read your link. Thanks

And that about education. I agree with you on that too. Not as strict as Moore method but yes students must be given enough time and space for independent work. On the other hand, life is too short for us to find through our own means the maths one must know as basics and fundamentals to move to further realms. And yes there is this exam pressure. Students need to score such and such marks on such and such topic. We need a system which would help student to do as much as independent work as possible and then reward them for their works, also help them overcome their difficulties in work but only when required. Can we have one? What about an individual, can he built a better system for himself.

Well it is very much possible that a student is inclined more to one topic. Moreover age plays a crucial role too. Example it's very hard to convey the how-to to a small child. It's better that he learns them for some reward to which he can attach a value as a small child. Another example take me, I like geometry more than other topics, but it may happen that because of my age I am attaching more value to ''visual thinking", and geometry has it more.I don't know but it may be possible that later, as my experience in mathematics grows, I may get involved much with more abstract areas, visual or not . But I must be given freedom to enjoy what I do. You were also right about this.

Role of Psychology in learning must not be overlooked. I can't say about neuroscience. What is your opinion?

There's also the magic effect of a teacher. Many successful mathematicians who wrote research papers) were greatly influenced by their teachers. We must not overlook the education and resources being spent on them. How are they being trained?

I don't know what I wanted to tell here. But I just wanted to tell it. This isn't so about mathematics but more of a conversation I want to have with someone whose views I find simple (if not easy) \(\stackrel{\wedge\,\wedge}{\smile}\) and interesting...

We also need you here – Soumo Mukherjee · 2 years, 2 months ago

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What I had in mind was the educational philosophy of student-centered learning, as opposed to the standard practice of teacher-centered learning. Those jurisdictions, (at least within Canada), which have adopted this approach have seen impressive results, not only in the emotional well-being of students but also in their scores on proficiency tests. Now not all students learn the same way, and for some the traditional approach may still be best, (at least for some period of their development), but if parents were to have a choice as to which school programs they can enter their children into I think we would all be better served.

Now your point about teachers is indeed a critical one. The training of teachers to facilitate this philosophy is more involved than normal, but I also think that it would be more fulfilling for them in the long run to work in this system. So I believe that once this option becomes more available, more teachers will see its benefits and will want to become involved. I suspect that it would attract the more capable educators, thus ensuring the quality of the instruction and guidance the children would receive.

I don't see this philosophy taking the educational world by storm any time soon, however; the status quo of prescriptive, standardized learning is strongly entrenched in most societies. But revolutions can be quiet, so it can occur over the next century, one school at a time.

Having said all this, regardless of the educational system involved, there is also the factor of luck. The right person, whether it be a teacher, professor, family member, counselor or fellow student, at the right time in a child's development can make all the difference in the world. As the saying goes, it takes a village to raise a child, so the more people there are that have a child's best interests at heart the more likely that child's potential will be fulfilled.

As for the fascinating plane photo thread, I'm still not sure if it's photoshopped or not. Guiseppi's photo shows that weird inversions can occur, but to have the plane going in different directions in every drop is a bit of a stretch. There's nothing I can add to the discussion that hasn't already been said, so I'm just waiting to find out if anyone knows for certain if this was a fake or not. :) – Brian Charlesworth · 2 years, 2 months ago

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I had a dream last night. As I was in Brilliant today, it struck me that I had a dream about Brilliant. May be I am spending too much time here. I saw that wiki section has introduced like/dislike feature and Calvin has posted a note about it and many of us are getting engaged in it. Lol... I can't construct what exactly I saw. Or may be that was a memory intermixing with ideas and all. 'cuz we do have a rating feature here...anyways.. that's away from our topic

That controversial image seems quite real to me. I can't find enough reasons to believe it's not real. Someone could try to take a photo of rain drops inverting image of an object. But that would require expensive gadgets and camera. So, I am going to take a look at photos from sites like National Geographic and Discovery, where I think I can find some extremely hard to take shot, but real shots. If I find one similar to the rain drop inversion effect, I will notify. I know I might be pushing this too much... – Soumo Mukherjee · 2 years, 2 months ago

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No, you do not need any expensive gadgets for that effect. Just a rounded glass,

andthe right focal length.With a little physics knowledge, viola! – Calvin Lin Staff · 2 years, 2 months agoLog in to reply

haha... yeah, this site has got many things to get attached to especially peoples.

We have liking, sharing, and upvoting system. Which are real motivating . Because it gives us 'feeling of importance'. So Brilliant.org did consider the psychological side of humans. Which deserves a lot of appreciation.

Then we have this discussion feature. where we can share our view points...and lol crazy dreams :D. To have a different perspective may be sometimes quite difficult to get only through our own effeort (e.g. recently there was note on 'why boys are more inclined to maths than girls'). And if we can reach out to people who are willing to listen as well as respond to our doubts, that 's a precious thing. And who doesn't like precious things ;)

This site has taken into consideration many things, which are easy to overlook, and sometimes may be hard to spot. You and your team are really working hard into it and we all know that. There's no site, I think, which has got such wonderful staff-user relationship. I feel like brilliant is more democratic than other sites :D

The corresponding color that you put on level of problem isn't arbitrary at all. They are color of the flames :D I like admiring small details. They tell us how much dedicated people are to their works that they use good ideas from beautiful things .

Well, I think, if people just browse through it and just have look at any discussion. They may have a feeling: "I should join this site." We must conduct an experiment where people will be asked to just have a look at it and then scan their brains lol... And about ideas, give me some more days to 'sleep on it' and we might get some crazy stuff lol and then someone can refine it :) :D XD

Yes, today I was browsing internet for similar photos...and eureka! .. I found them...and I will post them soon. So that people can revise their views.

Thanks for replying. – Soumo Mukherjee · 2 years, 2 months ago

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As for the photo, after staring at it for a few more minutes, I agree that it's probably real. To photoshop something with this complexity and sophistication would be too difficult. – Brian Charlesworth · 2 years, 2 months ago

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Have a look at this...

Inversions – Soumo Mukherjee · 2 years, 2 months ago

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– Brian Charlesworth · 2 years, 2 months ago

You've found some great pictures. Given this "evidence", I don't think there's any reason now to doubt the authenticity of the inverted plane photo. :)Log in to reply

The experiment – Megh Choksi · 2 years, 2 months ago

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@megh choksi :) that was helpful as well as entertaining. – Soumo Mukherjee · 2 years, 2 months ago

ThanksLog in to reply

I believe that the photo was taken by Guiseppi himself, (and is real). Even if you do not believe it is real, this is something that you can easily reproduce with (almost) any wine glass that is filled with water. Because it is rounded left to right and up to down, you get it flipped horizontally and vertically.

You can reproduce a similar effect with a glass cylinder (like a glass or water, or even a vase) Start with the cylinder right at your eye with the image pretty clear. Then move the cylinder away from you and it will start to get blur. At the focal point, you will see that the image has switched around (just left to right). – Calvin Lin Staff · 2 years, 2 months ago

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With all the photos flying around I now feel inspired to look through random glass containers filled with liquid. Optics wasn't my favourite topic in school, but I think that had more to do with how it was taught than the subject itself. – Brian Charlesworth · 2 years, 2 months ago

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If \(y=-x\) & if \(x\) can attain only values of negative integers. Then can we say that \(y\) is inversely proportional to \(x\) ?

I have had my name changed from 'Math Philic' to 'Soumo Mukherjee'

Brian Charlesworth

:) – Soumo Mukherjee · 1 year, 12 months ago

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Normally we would say that \(y\) is "inversely proportional" to \(x\) if \(y = \dfrac{k}{x}\) for some real constant \(k\), which would not be the case here. With \(y = kx\) for some real constant \(k\), we would say that \(y\) is proportional to \(x\), regardless of whether \(k\) is positive or negative. – Brian Charlesworth · 1 year, 12 months ago

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So, there is nothing to do with the values of \(y\)? I meant to say that in our function (\(y=-x\)) the domain is negative integers, whereas the range is positive integers, i.e.,

for \(x=-1\) we have \(y=1\)

for \(x=-2\) we have \(y=2\)

for \(x=-3\) we have \(y=3\)

with decreasing values of \(x\) we have increasing values of \(y\). So, I though \(y\) is inversely proportional to \(x\).

But according to the form \(y=kx\), We cannot say \(y\) is inversely proportional to \(x\).

I came across these two informal definition :

Inversely Proportional: when one value decreases at the same rate that the other increases.

Directly proportional: as one amount increases, another amount increases at the same rate.

This is why I am confused. – Soumo Mukherjee · 1 year, 12 months ago

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– Brian Charlesworth · 1 year, 12 months ago

Ah, o.k.. From those definitions you've made a valid interpretation, but those "informal" definitions only really apply if both values are always positive. What I have stated are the formal definitions, so it would probably be best to stick with them to avoid any misinterpretation. :)Log in to reply

for \(y=-x\) is a straight line, even with the restriction on domain and range. :) – Soumo Mukherjee · 1 year, 12 months ago

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@brian charlesworth , I am having a doubt concerning this problem. If you have some time plz reply.

I wanted to know , is that question complete, without any restriction on \(a\) and \(b\) or would it be better to have some, at-least indirect, restriction on them? I have a solution ready, but I think this might get a bit controversial due to arbitrary nature \(a\) and \(b\).

:) – Soumo Mukherjee · 2 years, 2 months ago

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– Brian Charlesworth · 2 years, 2 months ago

This is a great question, (and fun, too). I think it might be a good idea to specify that the coefficients \(a,b\) are real. Also, option (P) should probably read "It has at most three real roots", just to avoid any disputes involving complex solutions. I'm still looking at what happens when we look at the equation over the field of complex numbers, but for your problem I think it's best to just keep it real. :)Log in to reply

For removing the modulus sign or absolute value sign for \(\left| x \right|\), we have to consider real domain (\(x<0,x=0,x>0\)). And based on our previous discussion complex numbers cannot be ordered. So, that could serve as an indirect indication that we are dealing with real numbers. Can we reason that way?

Still for a number like \(x=a+ib\), \(\left| x \right|\) can re-written without modulus or absolute value sign.

One more thing what is Pell's equation? I just guessed the answer of your question with a very silly reasoning. :) – Soumo Mukherjee · 2 years, 2 months ago

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\((c + id)*\sqrt{c^{2} + d^{2}} + a + b*(c + id) = 0\).

Since the right side is real, we would require that the left side is as well, thus requiring that

\(d*\sqrt{c^{2} + d^{2}} + bd = 0\).

When I got to this point, it wasn't immediately clear to me how many solutions there might be if \(d \ne 0\), so I decided to advise you to specify "real roots" to avoid any possible issues. – Brian Charlesworth · 2 years, 2 months ago

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Still I think we are in trouble because doing so we would restrict the values \(a\) and \(b\) can attain and hence this might affect the discriminant. I wanted to say that I think this also might affect option (S). If it won't then we would go for 'three real roots' – Soumo Mukherjee · 2 years, 2 months ago

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I notice that Calvin LIn has re-shared your problem, implying that he seems to approve of the wording, so for now it's probably best if you leave it as is. Anyway, you have an idea now as to what questions people might raise and what answers you can give them. I'm guessing that it will be a popular problem :)

P.S.. Yes, Pell's equation could be used to find integer solutions; good observation. However, I don't think that we'd need it. For \(d \ne 0\) we would have \(\sqrt{c^{2} + d^{2}} = -b\), so for starters we would only have solutions for \(b \lt 0\). Then we would end up with \(c^{2} + d^{2} = b^2\), which would give us a circle in the complex plane of radius \(|b|\), i.e., an infinite solution set. – Brian Charlesworth · 2 years, 2 months ago

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Calvin has gone through our discussion and thinks you are right about keeping the problem real :D see the comments of this problem – Soumo Mukherjee · 2 years, 2 months ago

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And if your problem were to get controversial, then people would learn things in the discussion, which is always a good thing. – Brian Charlesworth · 2 years, 2 months ago

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I had just posted my problem (this controversial problem), suddenly I just, somehow, saw the problem posted by Calvin. And then I went to solve it, the options were bit tricky. And, the format of the problem, the statement as well as the options were very similar to the kind of trickiness I always want to put in my problems. So instead writing a solution I posted, in excitement, a meme and wrote " this problem in an ideal one for me and it's very tricky. Very carefully formulated, etc. etc.". Calvin said to post a proper solution concerning the problem as people would expect to see a solution of the problem. Afterwards I wrote a solution. And I deleted the things I posted earlier. And then Calvin said it's ok to post such stuff but we also need to have a proper solution to the problems. And that was it all about . I think this was the part which might have appeared a bit awkward and would appear to anyone.

:D – Soumo Mukherjee · 2 years, 2 months ago

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As for the conversation with Calvin, that's pretty much what I guessed had happened. Calvin Lin is like a shepherd making sure his flock doesn't go too far astray. :D – Brian Charlesworth · 2 years, 2 months ago

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Lol.................... >:O..................Lol XD

– Soumo Mukherjee · 2 years, 2 months agoLog in to reply

@Calvin Lin will appreciate the sentiment. :) – Brian Charlesworth · 2 years, 2 months ago

This is hilarious. I'm sure thatLog in to reply

– Calvin Lin Staff · 2 years, 2 months ago

That's the boy who cried wolf! So not me. Boo hoo hoo.Log in to reply

– Brian Charlesworth · 2 years, 2 months ago

OH! Erps.... I thought that he was just calling for his dog, but on second glance that does look like a wolf in the background. So much for my Aesop awareness. :( For the sake of the poor sheep I hope this isn't THE boy who cried wolf, and that the villagers come to his aid before the carnage ensues. Maybe Math Philic can find a less foreboding and more pastoral image of a shepherd minding his flock when he gets a chance.Log in to reply

Well, Mr/Mrs Google provides a very different kind of 'shepherd' in the search result ;D . This one is more appropriate than that.

:D – Soumo Mukherjee · 2 years, 2 months ago

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– Brian Charlesworth · 2 years, 2 months ago

I see what you mean; most of the images are biblical in nature. The one you posted is far more amusing. :)Log in to reply

But I was talking about G. Shepherd, entirely different species :p

And congrats my Guru ;) , you are one of the most admired persons on Brilliant. They even changed your name to "

BrainCharlesworth" \(\stackrel{{\Large\wedge\,\wedge}}{{\smile}}\) – Soumo Mukherjee · 2 years, 2 months agoLog in to reply

Thank you for asking questions that make me think and for appreciating what I have to say. I think my year-long streak started the day I joined Brilliant. I guess that makes me a bit of an addict. I know that there others who have even longer streaks, but there probably aren't too many who have made it to 365 days. Feels good. :) – Brian Charlesworth · 2 years, 2 months ago

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From Firefly/Serenity, one of my favorite sci-fi shows – Calvin Lin Staff · 2 years, 2 months ago

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– Brian Charlesworth · 2 years, 2 months ago

Hahaha. O.k., I'm really going to have to binge on that show some time. :)Log in to reply

@Math Philic @brian charlesworth I remember seeing you guys discuss about the power tower, but I can't find it anymore so I'll post it on the official page of doubts.

You've talked about infinite power towers, but have you considered what happens when there's only a finite number of powers?

Here's what I'm heading for: \(x^x\) passes through \((1,1)\) and \((0,1)\). However, \(x^{x^x}\) passes through \((0,0)\) instead of \((0,1)\). \(x^{x^{x^x}}\) passes through \((1,1)\) and \((0,1)\) again, while \(x^{x^{x^{x^x}}}\) passes through \((0,0)\) instead of \((0,1)\).

Can you prove this general fact?

Another question I am curious about:

Let's say that a power tower with \(x\) powers is denoted as \(T_n\). Then, what's \[\lim\limits_{n\to \infty}\text{min}(T_{2n})\]

I noticed that the minimum as we approach infinity power tower seems to reach a single point, around \((0.09,0.397)\) – Daniel Liu · 2 years, 2 months ago

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tetrations, with the extension to infinite heights discussed here. I suppose the exponential tower could not be directly evaluated at \(x = 0\) regardless of the height, (due to the indeterminate nature of \(0^{0}\)), but would rather converge to the two points you have mentioned. In this last sub-link you'll note the range of convergence for an infinite tetration is \(\frac{1}{e^{e}} \le x \le e^{\frac{1}{e}}\), a rather beautiful result.

I guess the more "official" term would be(The specific issue that came up before was what happens when \(x = -1\), but as we see from Euler's result there won't be convergence for this value of \(x\), even though a look at successive finite tetrations would suggest that it would converge to \(-1\). I still find this a bit weird, frankly.)

Anyway, I had made a mental note to investigate the same specific questions you have asked last time Math Philic and I were discussing this topic, and I never got around to it. I'll give them a try later, but for now I just wanted to post this short reply so that you have a little more information to work with. – Brian Charlesworth · 2 years, 2 months ago

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@brian charlesworth I have created a note where we can discuss.

I have a doubt: \( \quad Is\quad i>0,\quad where\quad { i }^{ 2 }=-1.\) . If you get some time, plz reply to my doubt. – Soumo Mukherjee · 2 years, 2 months ago

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Now say we're back in the real plane. It wouldn't make sense to say that the point \((-1, -3)\) "is greater than" the point \((1,1)\), but it would make sense to say that the magnitude of the former is greater than that of the latter.

The same goes for complex numbers. It wouldn't mean anything to say that \(z_{1} = a + ib\) is greater than \(z_{2} = c + id\), but we could say that \(|z_{1}| \gt |z_{2}|\) if and only if \(\sqrt{a^{2} + b^{2}} \gt \sqrt{c^{2} + d^{2}}\).

So \(i = (0 + i*1) \gt 0\) doesn't mean anything, but \(|i| = \sqrt{0^{2} + 1^{2}} = 1 \gt 0\) is a valid statement. – Brian Charlesworth · 2 years, 2 months ago

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First, show that \( - 1 < 0 \).

This is actually the crux of the issue, as you will see below. Find a way to convince yourself that this is true, other than "because someone told me so".

Case 1. \( i = 0 \)

If so, then \( - 1 = i^2 = 0 ^ 2 = 0 \) which is a contradiction.

Case 2. \( i > 0 \).

If so, then \( -1 = i^2 > 0 \) which is a contradiction.

Case 3. \( i < 0 \).

If so, then \( -1 = I^2 > 0 \) which is a contradiction. – Calvin Lin Staff · 2 years, 2 months ago

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– Michael Mendrin · 2 years, 2 months ago

The fact that complex numbers can't be ordered has been used as an argument why complex numbers play such a significant role in quantum mechanics. It's already been pointed out that the difference between the classical heat diffusion equation and the Schrodinger wave equation is the introduction of complex time--and while in classical mechanics, things can always be ordered in spacetime, in quantum mechanics not so much.Log in to reply

Also, are events, such as The Big Bang, which are singularities in real time, necessarily singularities in complex time? – Brian Charlesworth · 2 years, 2 months ago

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Edit: Here it is:

Higher Dimensions – Michael Mendrin · 2 years, 2 months ago

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– Brian Charlesworth · 2 years, 2 months ago

Ah, o.k., I hadn't realized that its status as even a true mathematical singularity was still in question. And thanks for the link as well; the theory that our universe may be ejecta from a collapsed 4-D star is beyond cool.Log in to reply

– Megh Choksi · 2 years, 2 months ago

Just a small doubt can we represent real axis as x and y along with complex axis as zLog in to reply

– Michael Mendrin · 2 years, 2 months ago

In fact, all kinds of mathematics has been done using complex numbers as components of multidimensional space, probably the most famous being the Calabi-Yau spaces, where all 3 axes, x, y, z, are represented by complex numbers. There's nothing improper or impermissible about mixing them up, complex numbers behave just like real numbers in many respects. We're all familiar with the "complex plane", but we're hardly limited to that.Log in to reply

As for showing that \(-1 \lt 0\), a couple of thoughts. We could just take it as an axiom that \(1 \lt 2\), and then subtract \(2\) from both sides to get the desired result. If we want to get fancy, I guess we could use a proof by contradiction. Start by assuming that \(1 \lt 0\) and that we are working under the confines of an ordered field. Add \(-1\) to both sides to obtain \(0 \lt -1\). Then since in an ordered field the product of two positives is a positive, we would end up with \(1 = (-1)(-1) > 0\), contradicting our original assumption. Thus \(1 \gt 0\), and so \(-1 \lt 0\)..... Or something like that. This kind of proof always makes me feel like I'm a dog chasing my own tail. – Brian Charlesworth · 2 years, 2 months ago

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So if I replace \(x\) by \(\left| x \right| \) in my problem then the answer would definitely be \(-1\)

Yes, you may delete it, out of 50 people attempting only 10 got the answer. (^_^) – Soumo Mukherjee · 2 years, 2 months ago

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Another way you could have phrased your problem would have been ".... given that \(x\) is real and \(x + \frac{1}{x} = 1\)". Essentially the same trick but without raising the issue we've been discussing regarding complex numbers. – Brian Charlesworth · 2 years, 2 months ago

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Anyways, I learned a new thing today about complex number. And I can almost explain it to anybody else.

Thanks.

And now with newly unlocked knowledge I would frame another similar question, this time stating that \(\left| x \right| >0\).

Ok next time I would mention my doubt here. Since it is a note I think I won't face any trouble mentioning names unlike in solution. – Soumo Mukherjee · 2 years, 2 months ago

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– Brian Charlesworth · 2 years, 2 months ago

You're welcome. This is the official "Page of Doubts", then. :)Log in to reply

Official page of doubts" haha yes and "you" are the "official doubt resolver"I am about to post a problem based on our discussion. Lets see what happens. – Soumo Mukherjee · 2 years, 2 months ago

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@Brian Charlesworth Plz luk into this problem

The correct option according to my book is

option (B) smaller steps ensure smaller friction. Some are saying option (A) is the correct option. I have requested Calvin to undo it. But he might be flooded with @'s.I also request @John Muradeli and @Michael Mendrin to help me resolve this doubt.

Thnks

I am not allowed to post any solutions for that problem. I am posting an explanation here.

Let the angle between our legs be \(2\theta\) while we move forward (by stretching it). The angle between a leg and the vertical is \(\theta\) as shown in the attached pic. Now for equilibrium \(F\cos { \theta } =N\) and \(F\sin { \theta } =f\). According to the latter equation smaller \(\theta\) means smaller friction (\(f\)). Also friction on icy surface is less. And smaller \(\theta\) also means smaller steps. . – Soumo Mukherjee · 2 years ago

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The next question would be: what if the surface you are on offers absolutely no resistance? – Brian Charlesworth · 2 years ago

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Lets see if Calvin and Mr. Mendrin are convinced or not. I also tried a to add a bit to the discussion

World without friction. Interesting

;) – Soumo Mukherjee · 2 years ago

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You've made some good observations regarding the frictional force being "self-adjusting", so there is an argument to be made to keep the answer options, and the answer, as they presently are. (The coefficients of friction, both static and kinetic, are the only constants.) But the suggested changes, i.e., focussing on the horizontal component being lesser or greater, would probably be more a more accurate phrasing. Regardless, the discussion has being very informative for all participants and thus the question has served its purpose. It will be interesting, though, to see what Calvin decides to do before removing the 'report flag' from the question. – Brian Charlesworth · 2 years ago

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I couldn't remove frictional to have a no resistance world, but I switched off earth's gravity ;). And now what would happen if we don't have both ? – Soumo Mukherjee · 2 years ago

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Something similar

This time I honestly don't know the solution :(

@Michael Mendrin and @John Muradeli – Soumo Mukherjee · 2 years ago

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In reality, though, if the Earth wasn't rotating about its axis, but was rotating about the Sun, and its gravity suddenly shut off, then we would all become independent satellites of the Sun, with each of our trajectories being slightly different depending on where we were on the Earth at the time of the shut-off. We would collectively form a sort-of human asteroid belt. :P The Earth's atmosphere would disappear too, so if anyone managed to hold on to a tree or rock in time they would end up suffocating anyway. A really sad state of affairs. :( haha

Anyway, for these reasons I don't really know how to answer "Unattractive Earth" as presently worded. If the Earth were rotating about its axis and its surface were entirely frictionless then we might end up sliding for a while depending on where we were at the time of shut-off. I think the key is that when gravity shuts off we become satellites of the Sun rather than "prisoners" of Earth's gravity. I prefer being a prisoner, though; I like to breathe. :) – Brian Charlesworth · 2 years ago

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;) – Soumo Mukherjee · 2 years ago

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– Brian Charlesworth · 2 years ago

O.k., thanks, I've just answered "correctly", so I can now enjoy reading the conversation. :)Log in to reply

– Michael Mendrin · 2 years ago

See my reply to John Muradeli 's answer to "Unattractive Earth".Log in to reply

– Michael Mendrin · 2 years ago

See my reply to Caleb Townsend. Following your recommendation, I entered "B" and got a "Nice Try!" But at least I was able to reply to Caleb. Yes, very BADLY explained physics. An excellent example of the blind leading the blind.Log in to reply

A and B have literally nothing to do with the answer. The answer is D.

The person who posted the solution argued that taking smaller steps ensures one retaining static friction, which is greater than kinetic friction. However, as you may know from cars, it does not matter if you increase the parameter of speed (i.e. rpm of rolling tires or steps per second); the coefficient used only applies to the type of contact one makes with the ground.

This being said, we take smaller steps because we can better control the amount of force we exert on the ground. And the amount of force is the key to overcoming any type of friction; if you take larger steps, you are more prone to exerting more normal force than the surface of contact can oppose with friction. Also, the angle might have something to do with it; if you take larger steps, you are more prone to landing your foot on an angle and thus distributing your force a bit more forward, thus lowering your dependence on friction and slipping (imagine the extreme case in which you lay your foot almost horizontally; then, pretty much all resistance is done by air, not the ground).

Thus, by taking smaller steps, you land your foot more perpendicularly on the ground, and control your body's motion so that you don't make any rough movements that may cause you to unevenly distribute your mass (and force), and stay within the bounds of the frictional force.

Thx for teh cool prob, Philic ;) – John Muradeli · 2 years ago

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– Michael Mendrin · 2 years ago

John, the normal force is the straight downward force by the foot onto the ground. Hence, at some point in time as you go through your steps, this normal force is going to equal your weight--no matter whether big or small steps are taken. Caleb Townsend has given the best analysis of this problem so far.Log in to reply

BAM! I REFUTED MICHAEL MENDRIN!

WOOOOOOOOOOOOO!!!!– John Muradeli · 2 years agoLog in to reply

– Michael Mendrin · 2 years ago

And what is the color of that blue-and-black dress? The actual physics involved in walking, on dry ground or ice, is pretty messy, but as Caleb Townsend pointed out, the factor to keep an eye on is static friction, and how that relates to the horizontal force that is exerted by the leg at an angle--while the body's center of gravity is in motion.Log in to reply

This Problem says Gaussian surfaces are two-dimensional. I disagree. Gaussian surface (used in calculation of electric field) are three dimensional.

Either I am missing some vital information about this or cannot correctly interpret the problem.

@Brian Charlesworth @Michael Mendrin @John Muradeli. Help. – Soumo Mukherjee · 1 year, 11 months ago

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I see Mr. Mathopedia has got your back. Thanks for mentioning me I will come back to this to learn more.

Later! – John Muradeli · 1 year, 11 months ago

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– Michael Mendrin · 1 year, 11 months ago

Toroidal equipotential surfaces are pretty common in EM theory, isn't it?Log in to reply

– Soumo Mukherjee · 1 year, 11 months ago

Yes, I had my doubt clarified. I will keep mentioning all three of you in my doubts. We all can learn something that way. After studying it, you can reply anytime.Log in to reply

– Brian Charlesworth · 1 year, 11 months ago

Thanks for keeping me in the loop, Soumo. I always learn something from Michael's comments. :)Log in to reply

– Soumo Mukherjee · 1 year, 11 months ago

All three of you help me in learning lots of things. I keep bringing your arguments in discussion with my friends. I say, "I got this fact/concept from Brian." My friends ask "Who is he." Then the discussion turns into a story, "Once upon a time in Brilliant.org there lived three wise men: Brian, Michael & John, etc., etc.,...."Log in to reply

– Brian Charlesworth · 1 year, 11 months ago

Haha. I've never thought of myself as wise, but I appreciate the compliment, nonetheless. I think Michael has the corner on that attribute, (and not just because he's a bit older than us). :)Log in to reply

– Soumo Mukherjee · 1 year, 11 months ago

BTW, Thanks for liking and sharing my set AYWC? . I will try keeping your interest alive by posting nice problems (especially in geometry) :)Log in to reply

This is just a confusion about mathematical nomenclature. Sometimes a physicist is looking at the whole solid of the planet Earth, say, which is a \(3D\) solid, but sometimes he's just looking at the surface of it only, which is a \(2D\) thing,. You only need TWO coordinates to know where you are on the surface of the planet. – Michael Mendrin · 1 year, 11 months ago

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new note where I will post only my doubts in Physics. We can continue this there. This page is taking too much time, on my network, to reload. Plz subscribe to it because I will turn to three of you for help. Thanks – Soumo Mukherjee · 1 year, 11 months ago

I have created aLog in to reply

If the confusion is caused by mathematical nomenclature (or wordings). I think we must avoid it. And try something unambiguous. Should I request for editing?

Moreover the answer to that problem is 2-dimensional (plane figure). How come? – Soumo Mukherjee · 1 year, 11 months ago

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– Michael Mendrin · 1 year, 11 months ago

Since a single parameter defines a circle, it's actually a one dimensional object. As long as we stick to the definition that the dimensionality of an geometrical object is the number of independent parameters to define it, then we have consistent mathematical nomenclature. If you study \(4D\) general relativity, for example, it sticks to that definition as well. The dimensionality of a circle doesn't depend on what kind of space you find it in. For example, both a circle and a spherical surface can exist in \(3D\) space---does that mean both are \(3D\) objects? No, the circle is still a \(1D\) object and the sphere is still a \(2D\) object.Log in to reply

plz reply here – Soumo Mukherjee · 1 year, 11 months ago

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– Michael Mendrin · 1 year, 11 months ago

Yes, a torus is definitely a two-parameter surface! Therefore it's a \(2D\) thing.Log in to reply

Thanks. – Soumo Mukherjee · 1 year, 11 months ago

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I don't think my book is wrong. But the book doesn't have any solution, only the answer has been given. We need to rethink on it. – Soumo Mukherjee · 2 years ago

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