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TopNewestA question. When posting solutions or added material to previously posted solutions, is it normal for any such work not "submitted" be lost upon exit? I've had the annoying experience of having spent the better part of an hour developing a solution to a problem (Golden Cubes by W Rose) only to see it vanish while clicking around. I seem to recall that I was able to return to such "work in progress" but maybe my memory isn't right? It has just now happened again. – Michael Mendrin · 2 days, 18 hours ago

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Calvin, another possible bug report, and I seem to recall this has happened before. Here are 2 problems I solved recently

Golden Cube

Co-co-company

In both cases, I found a better solution, entering a smaller number than the "correct" answer. In both cases, my smaller number was declared, "correct!". Well, this shouldn't be happening, don't you think? Anyway, in the case of the Co-co-company, I believe Marla Reece simply overlooked another way of arranging the 5 circles. However, in the case of the Golden Cube, I actually think W Rose was thinking of another problem than the one I solved. I notice that W Rose does have a tendency not to express his problems clearly.

But, mainly, I'm bringing to your attention that in some of these optimization problems, sometimes when I enter an answer that's different from the supposedly "correct" one, my entry is nevertheless declared "correct". Why do you suppose this happens sometimes? – Michael Mendrin · 5 days, 2 hours ago

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Then, you submitted a second answer, that was slightly smaller than our "correcrt answer". Since it is within the 2% error range of the correct answer, hence it was marked correct.

Similarly with Co-co company, the smaller answer that you submitted was within 2% of the error range, which is why you were marked correct.

For these problems, if you're convinced that your value is correct, please submit a report on the problem. Thanks. – Calvin Lin Staff · 5 days, 2 hours ago

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If you believe that the answer is incorrect, please report it with some justification. – Calvin Lin Staff · 4 days, 2 hours ago

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– Michael Mendrin · 5 days, 1 hour ago

Oh, I didn't know that Brilliant accepts answers "within 2%". But I've already posted comments in the reports section in both problems, and right now, W Rose seems to be in agreement with me with both the problem statement and the answer. Can the author of a posted problem that has already been answered change his own "correct" answer? If not, then he needs help from the Brilliant Staff to change his "correct" answer.Log in to reply

– Calvin Lin Staff · 4 days, 2 hours ago

Thanks. I've resolved both reports.Log in to reply

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When I read a solution, I do my best to point out what I feel are valid concerns / issues / errors. I admit that isn't always right, and when I realize that, I try and make it clear that the concern is not valid (typically by editing the first comment). I do not attack the person, but instead focus on the argument that is made. I do my best to pinpoint any misconceptions / fallacies, but those can be hard to convince another person.

In your case, the most recent example that comes to mind in which we had a lengthy discussion, is with regards to riemann integrable functions, which are similar to, but not exactly the same as, the integral that is taught in most high schools calculus concepts. I think that others who saw the problem and solution discussion, gained a lot from seeing these differences.

Just like complex exponentiation, the complex logarithm is a multi-valued function. What you're essentially saying by "without a pre-established domain", is that we haven't chosen a principle branch. In such a scenario, I default to the "typical" branch (which is the same branch that I take for complex exponentiation), namely

As such, this gives us \( \log_{-6} ( -6 ) = 1 \).

In fact, in this particular case, because the terms are the same (and non-1, non-0), no matter what branch we choose, since the value of \( \ln (-6 ) \) is fixed in our given branch, we have \( \log_{-6} { -6} = \frac{ \ln -6 } { \ln -6 } = 1 \).

More generally, when talking about multiple valued functions, the equality sign means that "there is some value on the LHS that is equal to some value of the RHS". Unfortunately, mathematicians do not distinguish between these and the typical "equality sign means that these are exactly equal".

A common misconception is:

\[ i = \frac{ i } { 1} = \frac{ \sqrt{-1} } { \sqrt{1} } = \frac{ \sqrt{ \frac{-1 } { 1 } } } = \sqrt{ \frac{ 1} { -1} } = \frac{ \sqrt{1} } { \sqrt{ -1 } } = \frac{1}{i} = -i \]

In this case, if we abandon the principle square root idea, and instead interpret square roots as a multiple-valued function, then by tracking which equality signs are "actual equal terms" vs "equal in the multiple-valued function sense", then you would be able to see how we arrived at such a contradiction.

Explicitly, the equation is saying

\[ i = i = \{ i, -i \} = \{ i, -i \} = \{ i, -i \} = \{ i, -i \} = -i = -i \]

So, the misconception arises because we're being very sloppy with our notation, and letting the equal sign do double duty. – Calvin Lin Staff · 3 weeks ago

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No, we cannot (yet) say that \( (-1) ^{ \log_{-1} (-1)^x } = ( -1 )^{ x \log_{-1} (-1) } = (-1)^x \).

Reason being that you're now mixing in a lot of "potential misconceptions", without properly accounting for them other than "These rules were true for real numbers and I'm going to assume that they are true for complex numbers", which often they are not especially due to the "multi-valued nature" of complex exponentiation / logarithm.

In particular, the misconception with complex exponentiation that you're sweeping under the rug is that \( (a^x)^y = a^{xy} = (a^y)^x \). Typially, this is expressed as :

\[ (-1) = (-1)^1 = (-1) ^{ 2 \times \frac{1}{2} } = [ (-1)^2 ]^{ \frac{1}{2} } = 1 ^ { \frac{1}{2} } = 1 \]

Once again, using the language of "equal in the multi-valued function sense", do you see how evaluating the function creates the error?

It is important to check the nitty-gritty details like verifying assumptions, conditions of the theorem, etc. Common examples where people forget about doing so are

Sometimes, people say that this is being "pedantic" or even "stick-up-the-ass". However, you can see why this is important in scenarios like the above. We do have to pedantically checking whether this instance of complex exponentiation is actually valid. – Calvin Lin Staff · 3 weeks ago

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The comment of "have to pedantically check whether this instance of complex exponentiation is actually valid" is still relevant. This is something that you would have to do, to justify the complex exponentiation step. – Calvin Lin Staff · 2 weeks, 6 days ago

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In particular, the misconception with complex exponentiation is that it is not always true that \( (a^x)^y = a^{xy} = (a^y)^x \), especially when \(x, y \) are non-real values. The cases when we have an identity include (but are not limited to)

We have to "pedantically check whether this instance of complex exponentiation is actually valid", and unfortunately neither of these cases apply in the example you wrote, so more work has to be done. – Calvin Lin Staff · 2 weeks, 6 days ago

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\[ -1 = (-1)^1 = ( ( e ^ { i \pi } )^ 2 ) ^ { \frac{1}{2} } = ( e^{2 i \pi } )^{ \frac{1}{2} } = 1 ^ { \frac{1}{2}} = 1 \]

So, it is not alway true that \( ( e^x ) ^ y = e^{xy} = (e^y)^x \) for all complex numbers.

If you write up your proof, that might help you identify what assumptions you are making that would indicate why you're having this misconception. – Calvin Lin Staff · 2 weeks, 6 days ago

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Can you add your proof of \( ( e^x)^y = e^{xy} = e^{yx} = (e^y)^x \)? – Calvin Lin Staff · 2 weeks, 6 days ago

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Points to take note of regarding the statement \( 1 = 1 ^ { \frac{1}{2} } \).

1) If we are "using the main branch of the multivalued logarithmic/exponential function" implies that we are evaluating the function. We evaluate \( \sqrt{1} \times e ^ {i0} = 1 \). So, it is true that

\[ 1 = 1 ^ \frac{1}{2}. \]

However, note that

\[ -1 \neq 1^ { \frac{1}{2} }. \]

2) If we"Consider the multivalued function" means that we're taking equality of multivalued functions, meaning that "one of the values in the set is equal to one of the values in the other set" (which is also why we do not have the transitive value of equality, and hence it doesn't really actually behave like the usual equality sign we're used to). Since \( 1 ^ { \frac{1}{2} } \) is the set \( \{ 1, -1 \} \), hence it is true that

\[ 1 = 1 ^ { \frac{1}{2} }. \]

Furthermore, in this case, we have

\[ - 1 = 1 ^ { \frac{1}{2} }. \]

So, while it ultimately matters whether we're talking about "principle branch evaluation" or "multi-valued functions", it is always true that \( 1 = 1 ^ \frac{1}{2} \). – Calvin Lin Staff · 2 weeks, 4 days ago

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Discussion The subject of complex exponentiation is horribly messy, and Calvin has been right in pointing out that many identities that hold elsewhere don't necessarily hold when there's complex exponentiation is involved. Quite literally, the subject branches out everywhere. – Michael Mendrin · 2 weeks, 6 days ago

Guillermo, for something like this, there should be aLog in to reply

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attempt to explain (one of) the hurdles to \( ( e^x)^y = e^{xy} = e^{yx} = (e^y)^x \) . – Calvin Lin Staff · 2 weeks, 4 days ago

Let's continue the discussion here, where ILog in to reply

Is the dress White and Gold, or is it Blue and Black? Please don't start here! – Michael Mendrin · 2 weeks, 6 days ago

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@Calvin Lin,

Sir, which GEOMETRY book you used during your preparation for IMO?

How you prepared for it? What helped you the most at IMO? – Priyanshu Mishra · 2 days, 9 hours ago

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Calvin, sorry, we have had many problems.... My apologies to you.

Now, my apologies and sorry to everybody here( again ). I have to be calm, I have to be quite and I have to be silent.... Really, my apologies to everyone, I have finished in this message-board.

Sincerely,

Guillermo Templado. – Guillermo Templado · 4 days, 9 hours ago

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What about the "Activity Tab"? When are you planning to (re)introduce it? – Samuel Jones · 1 month ago

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– Nihar Mahajan · 1 month ago

You can observe the activity of the users under their "Contributions tab". It was introduced recently, so may be you are not familiar with it.Log in to reply

– Samuel Jones · 1 month ago

Ah I see! It still doesn't show what solution the users upvoted (like it did before). I hope that too gets added.Log in to reply

– Calvin Lin Staff · 1 month ago

We are currently not adding "like problem" or "upvoted solution" as a contributor activity.Log in to reply

– Samuel Jones · 1 month ago

That's a shame. Is it because it takes too much space?Log in to reply

Sir, how does a problem gets rated? – Rohit Udaiwal · 1 year, 3 months ago

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1. The problem creator or moderator gives it an initial seed level. If a problem gets 100++ views, we will also give it a seed level.

2. The community works on the problem.

3. After enough people have worked on it and we have become more confident that the rating is accurate, we will release the level of the problem.

You can help your problems get rated by:

1. Adding a relevant seed level when you post the problem

2. Making it interesting and engaging for others to work on the problem – Calvin Lin Staff · 1 year, 3 months ago

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– Rohit Udaiwal · 1 year, 3 months ago

OK thanks!Log in to reply

Sir, what is the way to post images along posting questions?

Is there any markdown or something else which would help me to draw pictures?

You can see that in all my posts, none of any one posts consists of pictures, geometrical diagrams.

Please help – Priyanshu Mishra · 1 year ago

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sir does wrong answers affect my leveling up? – InkWhite 2000 · 9 months, 1 week ago

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stats page, you can see how your ratings change as you answer problems. – Calvin Lin Staff · 8 months, 3 weeks ago

Yes. On theLog in to reply