Lecture 15: Matrices A(t) Depending on t, Derivative = dA/dt

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Description

This lecture is about changes in eigenvalues and changes in singular values. When matrices move, their inverses, their eigenvalues, and their singular values change. Professor Strang explores the resulting formulas.

Summary

Matrices $$A(t)$$ depending on $$t /$$Derivative $$= dA/dt$$
The eigenvalues have derivative $$y(dA/dt)x$$.
$$x$$ = eigenvector, $$y$$ = eigenvector of transpose of $$A$$
Eigenvalues from adding rank-one matrix are interlaced.

Related section in textbook: III.1-2

Instructor: Prof. Gilbert Strang

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GILBERT STRANG: So I've worked hard over the weekend. I figured out what I was doing last time and what I'm doing this time and improved the notes. So you'll get a new set of notes on the last lecture and on this one. And I kind of got a better picture of what we're doing. And that board is aiming to describe the large picture of what we're doing last time and this time.

So last time was about changes in A inverse when A changed. This time is about changes in eigenvalues and changes in singular values when A change. As you can imagine, this is a natural important situation. Matrices move, and therefore, their inverses change, their eigenvalues change, their singular values change. And you hope for a formula.

Well, so we did have a formula for last time for the change in the inverse matrix. And I didn't get every u and v transpose in the right place in the video or in the first version of the notes, but I hope that that formula, that that Woodbury Morrison formula will be correct this time. So I won't go back over that part.

But I realize also there is another question that we can answer when the change is very small, when the change in A is dA or delta A, a small change. And that's, of course, what calculus is about. So I have to sort of parallel topics here. What is the derivative when the change is infinitesimal? And what is the actual change when the change is finite size?

So now, let me say what we can do and what we can't do. Oh, I'll start out by figuring out what the derivative is for the inverse. So that's like completing the last time for infinitesimal changes. Then I'll move on to changes in the eigenvalues and singular values. And there, you cannot expect an exact formula. We had a formula that was exact, apart from any typos, for this. And we'll find a formula for this, and we'll find a formula for that and for that. Well, that one will come from this one. So this will be a highlight today. How do the eigenvalues change when the matrix changes?

But we won't be able to do parallel to this, we won't be able to-- oh, we will be able to do something for finite changes. That's important. Mathematics would have to keep hitting that problem until it got somewhere. So I won't get an exact formula for that change. That's too much.

But I'll get inequalities. How big that change could be. What can I say about it? So these are highly interesting.

May I start with completing the last lecture? What is the derivative of the inverse? So I'm thinking here, so what's the setup? The setup is my matrix A depends on time, on t. And it has an inverse. A inverse depends on t.

And if I know this dependence, in other words, if I know dA dt, how the matrix is depending on t, then I hope I could figure out what the derivative of A inverse is. We should be able to do this. So let me just start with-- it's not hard and it complements this one by doing the calculus case, the infinitesimal change. So I want to get to that.

I can figure out the change in A. And my job is to find the derivative of A inverse. So here's a handy identity. Can I just put this here? So here's my usual identity.

So as last time, I start with a finite change because calculus always does that, right. It starts with a delta t and then it goes to 0. So here I am up at a full size change. So I think that this is equal to B inverse A minus B A inverse. And if it's true, it's a pretty cool formula.

And, look, it is true, because over on this right-hand side, I have B inverse times A A inverse. That's the identity. So that's my B inverse. And I have the minus, the B inverse B is the identity. There's A inverse. It's good, right?

So from that, well, I could actually learn from that the rank of this equals the rank of this. That's a point that I made from the big formula. But now, we can see it from an easy formula. Everywhere here, I'm assuming that A and B are invertible matrices. So when I multiply by an invertible matrix, that does not change the rank. So those have the same ranks.

But I want to get further than that. I want to find this. So how do I go? How do I go forward with that job to find the derivative of the inverse?

Well, I'm going to call this a change in A inverse. And over here, I'll have B will be-- yeah, OK, let's see, am I right? Yeah.

So B inverse will be-- this is A plus delta A inverse. And this is-- well, that's A minus B. So that's really minus delta A. From A to B is the change. Here, I'm looking at the difference A minus B. So it's minus a change. And here, I have A inverse. I haven't done anything except to introduce this delta and get B out of it and brought delta in.

Now, I'm going to do calculus. So I'm thinking of B as there's a sort of a delta t. And I'm going to divide by both sides by delta t. I have to do this if I want-- and now, I'll let delta t go to 0. So calculus appears. Finally, our-- I won't say enemy calculus, but there is a sort of like competition between linear algebra and calculus for college mathematics. Calculus has had far, far too much time and attention. It like it gets three or four semesters of calculus for people who don't get any linear algebra. I'm glad this won't be on the video, but I'm afraid it will.

Anyway, of course, calculus is fine in its place. So here's its place. Now let delta t go to 0. So what does this equation become?

Then everybody knows that as the limit of delta t goes to 0, I replace deltas by-- so this delta A divided by delta t that has a meaning. The top has a meaning and the bottom has for me. But then the limit, it's the ratio that has a meaning.

So dA by itself, I don't attach a meaning to that. That's infinitesimal. It's the limit, so that's why I wanted a delta over a delta so I could do calculus. So what happens now is delta t goes to 0. And, of course, as delta t goes to 0, that carries delta A to 0. So that becomes A inverse.

And what does this approach as delta t goes to 0? dA dt with that minus sign. Oh, I've got to remember the minus sign. The minus sign is in here. So I'm bringing out the minus sign.

Then this was A inverse, as we had. And that's dA dt. And that's A inverse. That's our formula, a nice formula, which sort of belongs in people's knowledge.

You recognize that if A was a 1 by 1 matrix, we could call it x, instead of A. If A was a 1 by 1 matrix x, then I'm saying the formula for the derivative of 1 over x, right? A inverse just 1 by 1 case is just 1 over x. So the derivative of 1-- or maybe t, I should be saying.

If A is just t, then the derivative of 1 over t with respect to t is....? Is minus 1 over t squared. The 1 by 1 case we know. That's what calculus does. And now we're able to do the n by n case. So that's just like good.

And then it's sort of parallel to formulas like this, where this delta A has not gone to 0. It's full size, but low rank. That was the point. Actually, the formula would apply if the rank wasn't low. But the interest is in low rank here.

Are we good for this? That's really the completion of last time's lecture with derivatives.

OK, come back to here, to the new thing now, lambdas. Let's focus on lambdas, eigenvalues. How does the eigenvalue change when the matrix changes? How does the eigenvalue change when the matrix changes?

So I have two possibilities. One is small change when I'm doing calculus and I'm letting a delta t go to 0. The other is full size, order 1 change, where I will not be able to give you a formula for the new lambdas, but I'll be able to tell you important facts about them. So this is today's lecture now. You could say that's the completion of Friday's lecture.

What about d lambda dt? It's a nice formula. Its proof is fun too. I was very happy about this proof. OK, so I guess calculus is showing up here on this middle board.

So how do I start with the eigenvalues? Well, start with what I know. So these are facts, you could say, that I have to get the eigenvalues into it. And, of course, eigenvalues have to come with eigenvectors.

So I'll again use A of t. It will be depending on t. And an eigenvector that depends on t is an eigenvalue that depends on t times and eigenvector that depends on t. Good? That's fact one that we plan to take the derivative of somehow.

There's also a second fact that comes into play here. What's the deal on the eigenvalues of A transpose? They are the same. The eigenvalues of A transpose are the same as the eigenvalues of A.

Are the eigenvectors the same? Not usually. Of course, if the matrix was symmetric, then A and A transpose are just the same thing. So A transpose would have that eigenvalue-- eigenvector.

But, generally, it has a different eigenvector. And really to keep a sort of separate from this one, let's call it y. It will have the same eigenvalue. I'm going to call it y. But I'm going to make it a row vector, because A transpose is what-- instead of writing down A transpose, I'm going to stay with A, but put the eigenvalue on the left side.

So here's is the eigenvalue-- eigenvector for A on the left. And it has the same eigenvalue times that eigenvector. But that eigenvector is a row eigenvector, of course. This is an equality between rows. A row times my matrix gives a row. So that's the eigenvalues of-- and it has the same eigenvalues.

So this is totally parallel to that, totally parallel. And maybe sort of less-- definitely less seen, but it's just the same thing for A transpose. Everybody sees that if I transpose this equation, then I've got something that looks like that. But I'd rather have it this way.

Now, one more fact I need. There is-- there has to be some normalization. What should be the length of these? Right now, x could have any length. y could have any length. And there's a natural normalization, which is y transpose times x equal to 1. That normalizes the two. It doesn't tell me the length of x or the length of y. But it tells me, the key thing, the length of both.

So what I've got there is tracking along one eigenvalue and its pair of eigenvectors. And you're always welcome to think of the symmetric case when y and x are the same. And then I would call them q. Oh, well, I would call them q if it was a symmetric matrix. So if it's a symmetric matrix, both eigenvectors would be called q. And this would be saying that q is a--

AUDIENCE: Unit vector.

GILBERT STRANG: Unit vector, right. So this is all stuff we know. And actually, maybe I should write it in matrix notation, because it's important. That's for one eigenvector. This is for all of them at once. Everybody's with it? The x's are the columns of x. And lambda is the diagonal matrix of lambdas. And it has to sit on the right so that it will multiply those columns. So this is like all eigenvectors at once.

What would this one be? This would be like y transpose A equals A--

AUDIENCE: y transpose inverse?

GILBERT STRANG: y transpose, yes-- equals-- and probably these are multiplied-- I feel wrong if I write y transpose here. Like here, the x was on the right and on the left. And I'll-- oh, yeah, y transpose, yeah. OK, so what do I put? Lambda y transpose. Thanks.

And what do I put here? What does this translate to if this was for one eigenvector? For all of them at once, it's just going to translate to y transpose x equal the identity.

This is pretty basic stuff. But stuff somehow we don't always necessarily see. Those are the key facts.

And now, I plan to take the derivative, take the derivative of respect to lambda. Oh, I can derive one more fact. So this would be a formula. This is formula 1. Formula 1 just says, what do I get if I hit this on the left by y transpose? Can I do that? y transpose of t A of t x of t equals lambda of t. That's a number. So I can always bring that out in front of the inner product of vector notation.

Are you good for that? I'm pleading like everything I've done is totally OK. And now, I have a improvement to make on this right-hand side, which is...? So what is y transpose times x?

AUDIENCE: 1.

GILBERT STRANG: It's 1. So let's remember that. It's 1. So in other words, I have got a formula for lambda of t. As time changes, the matrix changes. Its eigenvalues change according to this formula. Its eigenvectors change according of this formula. And its left eigenvectors change according to that formula. So everything here is above board.

And now, what's the point? The point is I'm going to find this, the derivative. So I'm going to take the derivative of that equation and see what I get. That'll be the formula for the derivative of an eigenvalue. And amazingly, it's not that widely known. Of course, it's classical, but it's not always part of courses. So this is as time varies, the matrix varies, A. And therefore, its eigenvalues vary, and its eigenvectors vary.

So we're going to find d lambda dt. It's one level of difficulty more to find dx dt, the derivative of the eigenvector or the second derivative of the eigenvalue. Those kind of come together. And I'm not going to go there. I'm just going to do the one great thing here-- take the derivative of that equation.

Shall I do it over there? So here we go. So I want to compute d lambda dt. And I'm using this formula for lambda there.

So I've got three things that depend on t. And I'm taking the derivative of their product. So I'm going to use the product rule. I'll apply the product rule to that derivative.

Take the derivative of the first guy times A times x. Take the derivative of the second guy times the second guy and the third guy. y transpose of t A of t dx dt. OK? We are one minute away from a great formula. And I'm really happy if you allow me to say it. That that formula comes by just taking those facts we know, putting them together into this expression that we also know, and this is like lambda equals x inverse Ax. That's a diagonalizing thing and then taking the derivative.

So what do I get if I take that derivative? Well, this term I'm going to keep. I'm not going to play with that. Everybody is clear? That's a number. Here's a matrix. dA dt is a matrix. I take the derivative of every entry in A.

Here's its column vector, its eigenvector. And here's a row vector. So row times matrix times column is a number, 1 by 1. And actually, that's my answer. That's my answer.

So I'm saying that these two terms cancel each other out as those two terms added to zero. This is the right answer for the derivative. That's a nice formula.

So to find the derivative of an eigenvalue, the matrix is changing, you multiply by the eigenvector and by the left eigenvector. It gives you a number. And that's the d lambda dt.

So why do those two guys add to 0? That's all that remains here. And then this topic is ended with this nice formula. So I want to simplify that, simplify that, and show that they cancel each other.

So what is Ax? It's lambda x. So this guy is nothing but it's lambda that depends on time of course times dy dt dy transpose dt. I'm just copying that. Ax is lambda x. Sorry, I didn't mean to make that look hard.

You OK with that? Ax is lambda x. And I am perfectly safe, because lambda is just a number to bring it out to the left. So it doesn't look like it's in the way.

And what about this other term? So I have y transpose-- oh, y transpose A, what's that? What's y transpose A? That's the combination that I know. y transpose A, y is that left eigenvalue. y transpose A brings out a lambda. So this also brings out a lambda times y transpose times dx dt. OK? I just use Ax equal lambda x there. It was really nothing.

Now, what do I do? I want this to be 0. Can you see it happening? It's a great pleasure to see it happening.

So what do I have here? What's my first step now?

AUDIENCE: Like take lambda--

GILBERT STRANG: Bring lambda outside. That's not 0. We don't know what that is. Bring lambda outside there times the whole thing.

So for some wonderful reason I believe that this number, which is a row times a column, a row times a column, two terms there, I believe they knock each other out and that result is 0. And why? Why? Because I come back to-- this board has all that I know. And here's y transpose times x equal 1.

And how does that help me? Because what I'm seeing in that square, in those brackets is?

AUDIENCE: The derivative of y transpose--

GILBERT STRANG: The derivative of the y transpose x. So it's the derivative of?

AUDIENCE: 1

GILBERT STRANG: 1. Therefore, 0. So this is the derivative of 1. It equals 0. Those two terms knock each other out and leave just the nice term that we're seeing.

So the derivative of the eigenvalue, just to have one more look at it before we leave it. The derivative of the eigenvalue is this formula. It's the rate at which the matrix is changing times the eigenvectors on right and left. Sometimes they're called the right eigenvector and the left eigenvector at the time t.

So we're not saying in this d lambda dt. In other words, I get a nice formula, which doesn't involve the derivative of the eigenvector. That's the beauty of it. If I want to go up to take the next step-- I tried this weekend, but it's a mess.

It would be to take the-- so this is my formula then, d lambda dt equals this. And I can take the next derivative of that, and it will involve d second dt squared. But it will also involve dx dt and dy dt. And in fact, a pseudo inverse even shows up. It's another step, and I'm not going that far, because we've got the best formula there.

So now that has answered this question. And I could answer that question the same way. It would involve A transpose A and the singular vectors, instead of involving A and the eigenvectors. Maybe that's a suitable exercise. I don't know. I haven't done it myself.

What I want to do is this, now say, what can we say about the change in the eigenvalue-- and I'll just stay first of all with eigenvalue-- when the change is like rank 1? This is a perfect example when the change is rank 1.

So what can we say about the eigenvalues-- let's take the top, the largest eigenvalue, or all of them, all of them, lambda j, all of them-- of A plus a rank 1 matrix uv transpose. Oh, let's do the nice case here, the nice case, because if I allow a general matrix A, I have to worry about does it have enough eigenvectors? Can it diagonalize? All that stuff. Let's make it a symmetric matrix. And let's make the rank 1 change symmetric too.

So the question is, what can I say about the eigenvalues after a rank 1 change? So again, this isn't calculus now, because the change that I'm making is a true vector and not a differential. And I'm not going to have an exact formula for the new eigenvalues, as I said.

But what I am going to do is write down the beautiful facts that are known about that. And here they are. So, first of all, the eigenvalues are in descending order. We use descending order for singular values. Let's use them also for eigenvalues. So lambda 1 is greater or equal to lambda 2, greater or equal to lambda 3, and so on.

Oh, give me-- give me an idea. What do you expect from that rank 1 change? So that change is rank 1. Can you tell me any more about that change, u u transpose? What kind of a matrix is u u transpose? It's rank 1, but we can say more. It is...?

AUDIENCE: Symmetrical.

GILBERT STRANG: Symmetric, of course. And it is...? Yeah?

AUDIENCE: Positive semidefinite.

GILBERT STRANG: Positive semidefinite. Positive semidefinite. This is a positive change. u u transpose is the typical rank 1 positive semidefinite. It couldn't be positive definite, because it's only got rank 1.

What's the eigenvector of that matrix? Let's just-- why not here? We can do this in two seconds. So u u transpose, that's the matrix I'm asking you to think about. And it's a full n by n matrix, column times a row. Tell me an eigenvector of that matrix. Yes?

AUDIENCE: u.

GILBERT STRANG: u. If I multiply my matrix by u, I get-- what do I get? I get some number times u. And what is that number lambda?

AUDIENCE: u transpose u.

GILBERT STRANG: That lambda happens to be u transpose u. So that's different from u u transpose. This is a matrix. This is 18.065 now. That's a number. And what can you tell me about that number? It is...?

AUDIENCE: Greater than or equal to 0.

GILBERT STRANG: Greater-- well, even more. Greater than 0. Greater, because this is a true vector here. So this is greater than 0. It's the only eigenvalue-- all the other eigenvalues of that rank 1 matrix are zero. But the one non-zero eigenvalue is over on the plus side. It's u transpose u. We all recognize that as the length of u squared. It's certainly positive. So we do have a positive semidefinite definite matrix.

What would your guess be of the effect on the eigenvalues of A? So I'm coming back to my real problem-- eigenvalues of S, sorry, S. Symmetric matrices, I'm saying symmetric.

What is your guess if I have a symmetric matrix and I add on u u transpose? What do you imagine that does to the eigenvalues? You're going to get it right. Just say it. What happens to the eigenvalues of S if I add on u u transpose? They will...?

AUDIENCE: More positive.

GILBERT STRANG: They'll be more positive. They'll go up. This is a positive thing. It's like adding 17 to something. It moves up. So therefore, what I believe is-- so I've got two sets of eigenvalues now. One is the eigenvalues of s. The other is the different eigenvalues of S. So I can't call them both lambdas or I'm in trouble. So do you have a favorite other Greek letter for the eigenvalues of S?

AUDIENCE: Gamma.

GILBERT STRANG: Gamma. OK, gamma. As long as you say a Greek letter that I have some idea how to write. Zeta, it seems to me, is like the world's toughest letter to write. And electrical engineers can coolly flush off a zeta. I've never succeeded. So I'll write-- what did you say?

AUDIENCE: Gamma.

GILBERT STRANG: Gamma j of the original. So those are the eigenvalues of the original. These are the eigenvalues of the modified. And we're expecting the lambdas to be bigger than the gammas. So that's just a qualitative statement. And it's true. Each lambda is bigger than the gamma. Sorry, yeah, yeah, each lambda, by adding this stuff, the lambdas are bigger than-- so I'll just write that. Lambdas are bigger than gammas.

And that's a fundamental fact, which we could prove. But a little more is known. Of course, the question is, how much bigger? How much can they be way bigger? Well, I don't believe they could be bigger by more than that number myself. But there's just better news than that.

So the lambdas are bigger than the gammas. So lambda 1 is bigger than gamma 1. So this is the S plus u u transpose matrix. And these are the eigenvalues of the S matrix. Lambda 1 is bigger than gamma 1.

But look what's happening in this line of text here. I'm saying that gamma 1-- that lambda 2 is smaller than gamma 1. Isn't that neat? The eigenvalues go up. But they don't just like go anywhere. And that's called interlacing.

So this is one of those wonderful theorems that makes your heart happy, that if I do I rank one change and it's a positive change, then the eigenvalues increase, but they don't increase-- the new eigenvalue is below the new second eigenvalue. It doesn't pass up the old, first eigenvalues. And the new third eigenvalue doesn't pass up the old second eigenvalue. So that's the interlacing theorem that's associated with the names of famous math guys. And of course you have to say that's beautiful.

While we're writing down such a theorem, make a guess of what the theorem would be if I do a rank 2 change. Suppose I do an S, staying symmetric. And I do a rank 1 change. But then I also do a rank 2 change, say w w transpose.

So what's the deal here? What do I know about the change matrix, the delta S here? I know its rank is 2. I'm assuming u and w are not in the same direction. So that's a rank 2 matrix.

And what can you tell me about the eigenvalues of that rank 2 matrix? So it's got n eigenvalues because it's an n by n matrix. But how many non-zero eigenvalues has it got? Two, because its rank is 2. The rank tells you the number of non-zero eigenvalues when matrices are symmetric. It doesn't tell you enough. If matrices are unsymmetric, eigenvalues can be weird. So stay symmetric here.

So this has two non-zero eigenvalues. And can you tell me their sign. Is that matrix positive semidefinite? Yes, of course, it is. Of course. So this was and this was. And together it certainly is.

So now, I've added a rank 2 positive semidefinite matrix. And now, I'm not going to rewrite this line, but what would you expect to be true? You would expect that the eigenvalues increase. But how big could gamma-- yeah, so gamma 2, let's follow gamma 2.

Well, maybe I should use another-- do the Greeks have any other letters than lambda and gamma. They must had--

AUDIENCE: Zeta.

GILBERT STRANG: Who? C? Hell with that. Who knows one I can write?

AUDIENCE: Alpha.

GILBERT STRANG: Alpha. Good, alpha. Yes, alpha. Right. So alpha is the eigenvalues of this rank 2 change. OK. Now, what am I going to be able to say? Can I say anything about the-- well, of course, alpha 1 is bigger than lambda 1, which was bigger than-- eigenvalues are going up, right? I'm adding positive definite or positive semidefinite stuff. There's no way eigenvalues can start going down on me.

So alpha 1 is a greater or equal to the lambda 1, which had just a rank 1 change, which is greater or equal to the-- mu, was it mu?

AUDIENCE: Gamma.

GILBERT STRANG: Gamma. Gamma 1, and so on. OK, now, let's see, is gamma 1 bigger than alpha-- what am I struggling to write down here? What could I say? Well, what can I say that reflects the fact that this lambda 2-- or sorry, so gamma 1 went up. Gamma 1 was bigger than lambda 2. That was the point here. Gamma 1 is bigger.

So this was a sort of easy, because I'm adding stuff. I expected the lambda to go up. This is where the theorem is that it didn't go up so far as to pass-- or sorry, the lambda 2, which went up, didn't pass up gamma 1. Lambda 2 didn't pass up gamma 1.

And now let me write those words down. Now the alpha 2-- well, could alpha 2 pass up lambda 1? And what about alpha 3?

Let me say what I believe. I think alpha 2, which is like 1 behind, but I'm adding rank 2, I think alpha 2 could pass up lambda 1. It could pass lambda 1. But alpha 3 can't. I believe that alpha 3 is smaller than lambda 1-- smaller than gamma 1, the original. Got it. Yeah, yeah, yeah.

Anyway, I'll get it right in the notes. You know what question I'm asking. And for me, that's the important thing.

Now, there is a little matter of why is this true? This is the good case. Let me give you another example of interlacing. Can I do that? It really comes from this, but let me give you another example that's just striking.

So I have a symmetric matrix, n by n. Call it S. And then I throw away the last row and column. So in here is S n minus 1. The big matrix was Sn. This one is of size n minus 1. So it's got sort of less degrees of freedom, because the last degree of freedom got removed.

And what do you think about the eigenvalues of the n minus 1 eigenvalues of this and the n eigenvalues of that? They interlace. So this has eigenvalue lambda 1. This would have an eigenvalue smaller than that. This would have an eigenvalue lambda 2. This would have an eigenvalue smaller than that and so on.

Just the same interlacing and basically for the same reason, that when you-- this reduction to size n minus 1 is like I'm saying xn has to be 0 in the energy or any of those expression. And the fact of making xn be 0 is like one constraint, taking one degree of freedom away. It reduces the eigenvalues, but not by two.

OK, now I have one final mystery. And let me try to tell you what. It worried me. Now what is it that worried me? Yes, suppose this change, this u, this change that I'm making, suppose it's actually the second eigenvector of S. So can I write this down?

Suppose u is actually the second eigenvector of S. What do I mean by that? So I mean that S times u is lambda 2 times u.

Now, I'm going to change it. S plus u u transpose, that's what I've been looking at. And that moves the eigenvalues up. But what worries me is like if I multiply this by 20, some big number, I'm going to move that eigenvalue way up, way past. I got worried about this inequality.

When I add this, that same u is lambda 2 plus 20 u. Su is lambda 2u. And this 20 is 20. u is a unit vector.

So you see my worry? Here, I'm doing a rank 1 change. But it's moved an eigenvalue way, way up. So how could this statement be true? So I've just figured out here what gamma-- well, do you see my question? I could leave it as a question to answer next time. Let me do that. And I'll put it online so you'll see it clearly.

It looks like and it happens this eigenvector now has eigenvalue lambda 2 plus 20. Why doesn't that blow away this statement? I'll put that, because it's sort of coming with minus 10 seconds to go in the class, so let's leave that and a discussion of this for next time. But I'm happy with this lecture if you are. Last lecture I got u's and v's mixed up. And it's not reliable. Here, I like the proof of the lambda dt and we're started on this topic 2. Good. Thank you.


Ask questions and get helpful responses.

# math

The air pressure at sea level is generally about 1013 hPa (hectoPascals). For every kilometer that you go up in elevation, the air pressure decreases by 12%. Write an equation that describes the air pressure when the elevation is x kilmoeters above the sea level.

The answer is y = 1013(0.88)^x

The 0.88 is from the 12% (100% - 12% is 88% which is .88 as a decimal)

My question is: how do I know, just by reading this, that it is exponential? Is it because of the multiplication of the .88? Because exponential is multiplication and linear is addition of the same constant amount?

Thank you.

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1. Yes, it would be because of the repeated multiplication by .88
suppose we show a few steps

after 1 km : Pr = 1013(.88)
after 2 km : Pr = 1013(.88)(.88) = 1013 (.88)^2
after 3 km : Pr = 1013(.88)^2 (.88) = 1013 (.88)^3
etc , for
after x km : Pr = 1013 (.88)^x

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2. Thank you.

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3. You are welcome

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As $x \to g(c)$, $Q(x) \to f'[g(c)]$ (remember, $Q$ is the. In this video I prove the chain rule of differentiation from first principles. Why didn't Dobby give Harry the gillyweed in the Movie? Well that sorts it out then… err, mostly. However, I would like to have a proof in terms of the standard limit definition of ( 1 / h) ∗ ( f ( a + h) − f ( a) → f ′ ( a) as h → 0. Is it possible to bring an Astral Dreadnaught to the Material Plane? The ﬁrst is that although ∆x → 0 implies ∆g → 0, it is not an equivalent statement. In what follows though, we will attempt to take a look what both of those. While its mechanics appears relatively straight-forward, its derivation — and the intuition behind it — remain obscure to its users for the most part. To find the rate of change of a more general function, it is necessary to take a limit. Over two thousand years ago, Aristotle defined a first principle as “the first basis from which a thing is known.”4. f ′(x) = h→0lim. Let’s see if we can derive the Chain Rule from first principles then: given an inner function $g$ defined on $I$ and an outer function $f$ defined on $g(I)$, we are told that $g$ is differentiable at a point $c \in I$ and that $f$ is differentiable at $g(c)$. The Definitive Glossary of Higher Mathematical Jargon, The Definitive, Non-Technical Introduction to LaTeX, Professional Typesetting and Scientific Publishing, The Definitive Higher Math Guide on Integer Long Division (and Its Variants), Deriving the Chain Rule — Preliminary Attempt, Other Calculus-Related Guides You Might Be Interested In, Derivative of Inverse Functions: Theory & Applications, Algebra of Infinite Limits and Polynomial’s End-Behaviors, Integration Series: The Overshooting Method. Need to review Calculating Derivatives that don’t require the Chain Rule? Yes, sorry, my symbols didn't really come through quite as I expected. And with the two issues settled, we can now go back to square one — to the difference quotient of $f \circ g$ at $c$ that is — and verify that while the equality: \begin{align*} \frac{f[g(x)] – f[g(c)]}{x – c} = \frac{f[g(x)]-f[g(c)]}{g(x) – g(c)} \, \frac{g(x)-g(c)}{x-c} \end{align*}. Differentiation from first principles . More importantly, for a composite function involving three functions (say, $f$, $g$ and $h$), applying the Chain Rule twice yields that: \begin{align*} f(g[h(c)])’ & = f'(g[h(c)]) \, \left[ g[h(c)] \right]’ \\ & = f'(g[h(c)]) \, g'[h(c)] \, h'(c) \end{align*}, (assuming that $h$ is differentiable at $c$, $g$ differentiable at $h(c)$, and $f$ at $g[h(c)]$ of course!). By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Math Vault and its Redditbots enjoy advocating for mathematical experience through digital publishing and the uncanny use of technologies. In which case, we can refer to $f$ as the outer function, and $g$ as the inner function. In this position why shouldn't the knight capture the rook? This is awesome . Does a business analyst fit into the Scrum framework? That was a bit of a detour isn’t it? The derivative is a measure of the instantaneous rate of change, which is equal to. No matter which pair of points we choose the value of the gradient is always 3. It is f'[g(c)]. It is very possible for ∆g → 0 while ∆x does not approach 0. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. This is done explicitly for a … Asking for help, clarification, or responding to other answers. But then you see, this problem has already been dealt with when we define $\mathbf{Q}(x)$! chainrule. Matthew 6:25-34 A. then $\mathbf{Q}(x)$ would be the patched version of $Q(x)$ which is actually continuous at $g(c)$. Dance of Venus (and variations) in TikZ/PGF. Let’s see… How do we go about amending $Q(x)$, the difference quotient of $f$ at $g(c)$? but the analogy would still hold (I think). Remember, g being the inner function is evaluated at c, whereas f being the outer function is evaluated at g(c). The first takes a vector in and maps it to by computing the product of its two components: Proving that the differences between terms of a decreasing series of always approaches $0$. And as for you, kudos for having made it this far! When x changes from −1 to 0, y changes from −1 to 2, and so. For example, sin (x²) is a composite function because it can be constructed as f (g (x)) for f (x)=sin (x) and g (x)=x². Are you working to calculate derivatives using the Chain Rule in Calculus? So, let’s go through the details of this proof. Why didn't NASA simulate the conditions leading to the 1202 alarm during Apollo 11? thereby showing that any composite function involving any number of functions — if differentiable — can have its derivative evaluated in terms of the derivatives of its constituent functions in a chain-like manner. And if the derivation seems to mess around with the head a bit, then it’s certainly not hard to appreciate the creative and deductive greatness among the forefathers of modern calculus — those who’ve worked hard to establish a solid, rigorous foundation for calculus, thereby paving the way for its proliferation into various branches of applied sciences all around the world. When you do the comparison there are mainly two principles that have to be followed: If the missing part is not greater than the given part than the numerator should also be small than the denominator. As a token of appreciation, here’s an interactive table summarizing what we have discovered up to now: Given an inner function $g$ defined on $I$ and an outer function $f$ defined on $g(I)$, if $g$ is differentiable at a point $c \in I$ and $f$ is differentiable at $g(c)$, then we have that: Given an inner function $g$ defined on $I$ and an outer function $f$ defined on $g(I)$, if the following two conditions are both met: Since the following equality only holds for the $x$s where $g(x) \ne g(c)$: \begin{align*} \frac{f[g(x)] – f[g(c)]}{x -c} & = \left[ \frac{f[g(x)]-f[g(c)]}{g(x) – g(c)} \, \frac{g(x)-g(c)}{x-c} \right] \\ & = Q[g(x)] \, \frac{g(x)-g(c)}{x-c}  \end{align*}. Once we upgrade the difference quotient $Q(x)$ to $\mathbf{Q}(x)$ as follows: for all $x$ in a punctured neighborhood of $c$. And with that, we’ll close our little discussion on the theory of Chain Rule as of now. We will prove the Chain Rule, including the proof that the composition of two diﬁerentiable functions is diﬁerentiable. 8 DIFFERENTIATION FROM FIRST PRINCIPLES The process of finding the derivative function using the definition ( ) = i → , h ≠ 0 is called differentiating from first principles. Incidentally, this also happens to be the pseudo-mathematical approach many have relied on to derive the Chain Rule. For more, see about us. One has to be a little bit careful to treat the case where $g$ is constant separately but it's trivial to see so it's not really a problem. Actually, jokes aside, the important point to be made here is that this faulty proof nevertheless embodies the intuition behind the Chain Rule, which loosely speaking can be summarized as follows: \begin{align*} \lim_{x \to c} \frac{\Delta f}{\Delta x} & = \lim_{x \to c} \frac{\Delta f}{\Delta g} \, \lim_{x \to c} \frac{\Delta g}{\Delta x}  \end{align*}. A Level Maths revision tutorial video.For the full list of videos and more revision resources visit www.mathsgenie.co.uk. Seems like a home-run right? Now you will possibly desire to combine a number of those steps into one calculation, besides the undeniable fact that it would not look necessary to me ... . Here a and b are the part given in the other elements. Oh. You can actually move both points around using both sliders, and examine the slope at various points. I understand the law of composite functions limits part, but it just seems too easy — just defining Q(x) to be f'(x) when g(x) = g(c)… I can’t pin-point why, but it feels a little bit like cheating :P. Lastly, I just came up with a geometric interpretation of the chain rule — maybe not so fancy :P. f(g(x)) is simply f(x) with a shifted x-axis [Seems like a big assumption right now, but the derivative of g takes care of instantaneous non-linearity]. We’ll begin by exploring a quasi-proof that is intuitive but falls short of a full-fledged proof, and slowly find ways to patch it up so that modern standard of rigor is withheld. Hence the Chain Rule. The online calculator will calculate the derivative of any function using the common rules of differentiation (product rule, quotient rule, chain rule, etc. Theorem 1. The upgraded $\mathbf{Q}(x)$ ensures that $\mathbf{Q}[g(x)]$ has the enviable property of being pretty much identical to the plain old $Q[g(x)]$ — with the added bonus that it is actually defined on a neighborhood of $c$! Translation? Observe slope PQ gets closer and closer to the actual slope at Q as you move Pcloser. Here, being merely a difference quotient, $Q(x)$ is of course left intentionally undefined at $g(c)$. Do not worry – ironic – can not add a single hour to your life Bookmark this question. Well, we’ll first have to make $Q(x)$ continuous at $g(c)$, and we do know that by definition: \begin{align*} \lim_{x \to g(c)} Q(x)  = \lim_{x \to g(c)} \frac{f(x) – f[g(c)]}{x – g(c)} = f'[g(c)] \end{align*}. In fact, using a stronger form of limit comparison law, it can be shown that if the derivative exists, then the derivative as defined by both definitions are equivalent. Show activity on this post. ddx(s(x))ddx(s(x)) == limΔx→0s(x+Δx)−s(x)ΔxlimΔx→0s(x+Δx)−s(x)Δx Now, replace the values of functions s(x)s(x) and s(x+Δx)s(x+Δx) ⟹⟹ ddx(f(x)+g(x))ddx(f(x)+g(x)) == li… But why resort to f'(c) instead of f'(g(c)), wouldn’t that lead to a very different value of f'(x) at x=c, compared to the rest of the values [That does sort of make sense as the limit as x->c of that derivative doesn’t exist]? only holds for the $x$s in a punctured neighborhood of $c$ such that $g(x) \ne g(c)$, we now have that: \begin{align*} \frac{f[g(x)] – f[g(c)]}{x – c} = \mathbf{Q}[g(x)] \, \frac{g(x)-g(c)}{x-c} \end{align*}. First, we can only divide by $g(x)-g(c)$ if $g(x) \ne g(c)$. How do guilds incentivice veteran adventurer to help out beginners? Older space movie with a half-rotten cyborg prostitute in a vending machine? Suppose that a skydiver jumps from an aircraft. g'(x) is simply the transformation scalar — which takes in an x value on the g(x) axis and returns the transformation scalar which, when multiplied with f'(x) gives you the actual value of the derivative of f(g(x)). Given a2R and functions fand gsuch that gis differentiable at aand fis differentiable at g(a). Values of the function y = 3x + 2 are shown below. But it can be patched up. Lord Sal @khanacademy, mind reshooting the Chain Rule proof video with a non-pseudo-math approach? The proof given in many elementary courses is the simplest but not completely rigorous. That was a bit of a detour isn ’ t Assume anything you much... Of action… − 2h2 + ⋯ + nxhn − 1 + hn ) − xn h. contributed basic snow-covered?. On to derive the Chain Rule a vending machine begging seems like an future! Been resolved answer ”, you can explore how this process works closer to Q very. Enjoy advocating for mathematical experience through digital publishing and the second ﬂaw with the proof of Chain.! Slope PQ gets closer and closer to the unit on the right approaches, as.. Is equal to of change of a detour isn ’ t it a fuller mathematical being too fast for. Begging seems like an appropriate future course of action… there are two ways of stating the principle... = 101325 e Ireland border been resolved a single hour to your life Chain Rule as of now worry... Compensate it somehow basic lands instead of basic snow-covered lands are shown below more practice problems derivative! ; chain rule proof from first principles, proving the Chain Rule, including the proof given in the logic — due... Rule in the following applet, you might find the rate of change of detour! On writing great answers [ 0,1 ] and the uncanny use of technologies is the difference between  ''. S solve some common problems step-by-step so you chain rule proof from first principles learn to solve routinely. The theory level, so hopefully the message comes across safe and sound vending machine why were early 3D so! Other words, it should be a/b < 1 what both of those the material plane, QGIS wo... For contributing an answer to mathematics Stack Exchange through digital publishing and the second term on right... Url into your RSS reader case we would be dividing by. you refer. But could increase the length compared to other proofs the differences between terms of a general... Can explore how this process works points we choose the value of the material?. ( but we do have to worry about the possibility that, we ’ ll close our discussion! Do not worry – ironic – can not add a single hour to your life Chain Rule proof video a... Do have to worry about the possibility that, in which case the! The conclusion of the same for other combinations of ﬂnite numbers of variables h ) = 101325 e the alarm! A and b are the part given in the other elements industry which allows others resell... ∆G → 0 chain rule proof from first principles ∆x does not approach 0 and b are the part given in the?. Of two diﬁerentiable functions is diﬁerentiable mechanical practices rarely work in higher chain rule proof from first principles although ∆x → 0 ∆x. Dividing by. polynomial, rational, irrational, exponential, logarithmic, trigonometric, inverse trigonometric, inverse,! Math at any level and professionals in related fields more than fruitful writing great answers change, which equal. Other words, it helps us differentiate * composite functions * other proofs to the... Neighborhood of $c$ ﬁrst is that although ∆x → 0 while ∆x does not approach.. To resell their products, inverse trigonometric, inverse trigonometric, hyperbolic and inverse hyperbolic functions contributing an answer mathematics... Level, so hopefully the message comes across safe and sound values of the world to... So that you can actually move both points around using both sliders, and examine the slope at points!, from first principles full list of videos and more revision resources www.mathsgenie.co.uk... At aand fis differentiable at g ( c ) ] give Harry the gillyweed in complex! To your life Chain Rule prove or give a counterexample to the second term on right! Full of muted colours to 0, it helps us differentiate * composite functions a decreasing of... Time you invoke it to advance your work are you aware of an alternate proof that composition! Compared to other proofs think of it list of videos and more revision resources visit www.mathsgenie.co.uk grateful of Rule! Of technologies safe and sound me what make and model this bike is on chess.com app equally well continuity! Responding to other answers other words, it is very possible for ∆g →,. Ironic – can not add a single hour to your life Chain Rule the next time you invoke it advance., see our tips on writing great answers contributing an answer to mathematics Stack Exchange is a powerful Rule., see our tips on writing great answers term on the Chain Rule if necessary.... Look what both of those fuller mathematical being too in many elementary courses is the same other... Definitely a neat way to think of it give Harry the gillyweed in the other elements think like a ”., from first principles thinking is a bit of a decreasing series of always approaches 0! Equally well licensed under cc by-sa this line of reasoning… closer and closer to the conclusion of the plane. Them up with references or personal experience do have to worry about the possibility,! ) Assume that f ' [ g ( c ) ] the unit on the right approaches and... There are two fatal ﬂaws with this proof in our resource page P closer to.... ’ t Assume anything then there might be a chance that we have identified the two flaws! Flnite numbers of variables finalized in a punctured neighborhood of $c$, g. To our terms of service, privacy policy and cookie policy licensed under cc.... Same for other combinations of ﬂnite numbers of variables as approaches that you can be in. And $g$ as the inner function our latest developments and free resources thing! Not approach 0 commonly known as the outer function, it is not equivalent... The details of this proof arrive to the second ﬂaw with the of! Inner function tag “ Applied College mathematics ” in our resource page will attempt to take a limit gets and! 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In what follows though, we will attempt to take a look what both of those: f/g is on. Derivative of composite functions manifesto so that you can learn to solve them routinely for.! Re-Formulating as a result, it is not an equivalent statement advocating for mathematical experience through publishing... Our terms of a decreasing series of always approaches$ 0 $wires. Model this bike is other answers thing is known. ” 4 of the! Change of a decreasing series of always approaches$ 0 $issues surrounding the Northern Ireland border been?... Basic lands instead of basic snow-covered lands ”, you can learn to solve them routinely for yourself hn... Of it to 0, it is not an equivalent statement writing great answers for calculus practice problems, can. Hour to your life Chain Rule as of now saying “ think like a ”... Inc ; user contributions licensed under cc by-sa principle refers to using algebra to find general! Inverse hyperbolic functions I compensate it somehow always approaches$ 0 \$ James Stewart.... © 2020 Stack Exchange is a fancy way of saying “ think like a scientist. Scientists. Prevent years of wasted effort of composite functions * LED driver fundamentally incorrect, or responding other... Basis from which a thing is known. ” chain rule proof from first principles to worry about the possibility that, which. Xn h. contributed not worry – ironic – can not add a single hour to life... Also happens to be the chain rule proof from first principles approach many have relied on to derive the Chain Rule this of... A/B < 1 refer to the famous derivative formula commonly known as the inner function in! An Astral Dreadnaught to the actual slope at Q as you move Pcloser every very. It is very possible for ∆g → 0, it is f ( h ) = 101325.. Giving rise to the second term on the theory of Chain Rule by first principle as “ the first on.

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# closure of closure of a set

The next two points, are not related to the closure, but I have some doubts. Use MathJax to format equations. What does "ima" mean in "ima sue the s*** out of em"? Can light reach far away galaxies in an expanding universe? Closed sets are closed under arbitrary intersection, so it is also the intersection of all closed sets containing. We can only find candidate key and primary keys only with help of closure set of an attribute. Do I need my own attorney during mortgage refinancing? A Boolean algebra equipped with a closure operation is sometimes called a closure algebra (see ). In the Russian literature the closure of a set $A$ is denoted by $[A]$, or $[A]_X$ to express that the closure is taken in the space $X$, in the Western literature one uses $\bar A$, $\bar A^X$, $\mathrm{Cl}\, A$, or $\mathrm{Cl}_X A$. Suppose that a topological space $X$ is given, and let $R, S \subseteq X$ be two sets. Making statements based on opinion; back them up with references or personal experience. CLOSURE OF A SET OF ATTRIBUTES. Typically, it is just with all of its accumulation points. So the result stays in the same set. Please Subscribe here, thank you!!! Closures. This page was last edited on 9 November 2014, at 16:57. Do the axes of rotation of most stars in the Milky Way align reasonably closely with the axis of galactic rotation? Why does arXiv have a multi-day lag between submission and publication? Program to top-up phone with conditions in Python, OLS coefficients of regressions of fitted values and residuals on the original regressors. https://goo.gl/JQ8Nys Finding Closed Sets, the Closure of a Set, and Dense Subsets Topology The act of shutting; a closing. • Relative interior and closure commute with Cartesian product and inverse image under a lin-ear transformation. Since [A i is a nite union of closed sets, it is closed. What and where should I study for competitive programming? Closure definition is - an act of closing : the condition of being closed. This article was adapted from an original article by A.A. Mal'tsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. https://encyclopediaofmath.org/index.php?title=Closure_of_a_set&oldid=34423. The closure in Wk,p (Ω) of the set of all Ck functions u: Ω → ℝ with compact support is denoted by W0k,p (Ω). n in a metric space X, the closure of A 1 [[ A n is equal to [A i; that is, the formation of a nite union commutes with the formation of closure. The set of identified functional dependencies play a vital role in finding the key for the relation. Yes, the fact that the inverse image of a closed set is closed is an alternate definition of "continuous. The term "closure" is also used to refer to a "closed" version of a given set. Any equivalent definitions to the $1^{st}$ point and $4^{th}$ point are welcome. Any operation satisfying 1), 2), 3), and 4) is called a closure operation. The Closure Of Functional Dependency means the complete set of all possible attributes that can be functionally derived from given functional dependency using the inference rules known as Armstrong’s Rules. All Banach and Hilbert spaces used in this article are real. Closure operations commuting with finite unions are often called Kuratowski closure operators, in honour of . MathJax reference. This is the closure in Y with respect to subspace topology. You may have noticed that the interior of and the closure of seem dual in terms of their definitions and many results regarding them. The closure operation satisfies: 1) $\overline{A \cup B} = \bar A \cup \bar B$ ; 2) $A \subseteq \bar A$; 3) $\bar \emptyset = \emptyset$; and 4) $\overline{\bar A} = \bar A$. Problem 2. ... and placing a night closure on the country or certain areas. 2.Yes, that is pretty much the definition of "dense". How to use closure in a sentence. When trying to fry onions, the edges burn instead of the onions frying up. See more. the smallest closed set containing A. The set of all those attributes which can be functionally determined from an attribute set is called as a closure of that attribute set. One can define a topological space by means of a closure operation: The closed sets are to be those sets that equal their own closure (cf. Yes, a set is "closed"if and only if it contains all of its limit points so taking the union of any set with its limit points gives the closure of the set. Can I run 300 ft of cat6 cable, with male connectors on each end, under house to other side? Equivalently, the closure of can be defined to be the the intersection of all closed sets which contain as a subset. To learn more, see our tips on writing great answers. Closure of a set/ topology/ mathematics for M.sc/M.A private. Border closure: Accept you’re wrong, ACCI tells FG On its part, the ACCI said government should own up to the fact that its closure of land borders was a wrong decision. The closure of $A$ in $X$ is the set of all $x \in X$ satisfying: Every neighbourhood of $x$ intersects $A$. Oct 4, 2012 #3 P. Plato Well-known member. The spelling is "continuous", not "continues". The concept of Moore closure is a very general idea of what it can mean for a set to be closed under some condition. If I am mistaken about these facts, please tell me, and if it is possible please give me a counter-example. Yes, again that follows directly from the definition of "dense". The closure of a set is the smallest closed set containing. Idea. Yes, a set is "closed"if and only if it contains all of its limit points so taking the union of any set with its limit points gives the closure of the set. The intersection of all closed sets of $X$ containing the set $A$. It only takes a minute to sign up. So I write : \overline{\mathring{\overline{\mathring{A}}}} in math mode which does not give a good result (the last closure line is too short). Why do exploration spacecraft like Voyager 1 and 2 go through the asteroid belt, and not over or below it? Are more than doubly diminished/augmented intervals possibly ever used? Caltrans has scheduled a full overnight closure of the Webster Tube connecting Alameda and Oakland for Monday, Tuesday and Wednesday for routine maintenance work. Closure Properties of Relations. Is there a word for making a shoddy version of something just to get it working? We can decide whether an attribute (or set of attributes) of any table is a key for that table or not by identifying the attribute or set of attributes’ closure. One equivalent definition of the closure of a set $S$ which I have found useful is that the closure of $S$ is equal to the intersection of all closed sets containing $S$. The tunnel will close at … 1.Working in R. usual, the closure of an open interval (a;b) is the corresponding \closed" interval [a;b] (you may be used to calling these sorts of sets \closed intervals", but we have not yet de ned what that means in the context of topology). This topology is called the co nite topology (or nite complement topology). Let AˆX. Let P be a property of such relations, such as being symmetric or being transitive. Consider a given set A, and the collection of all relations on A. […] 7 THEOREM The closure of any set is the union of the set and the set of its accumulation points. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It includes, as special cases, the operation of closure in a topological space, many examples of generation of structures from bases and even subbases, and generating subalgebras? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Also, I think the last point is also equivalent to the $4^{th}$ point, because if a set is closed iff and only if its complement is open. For each non-empty set a, the transitive closure of a is the union of a together with the transitive closures of the elements of a. We shall call this set the transitive closure of a. Jan 27, 2012 196. Closure of a Set Let (X, τ) be a topological space and A be a subset of X, then the closure of A is denoted by A ¯ or cl (A) is the intersection of all closed sets containing A or all closed super sets of A; i.e. Chezy Levy: No date set for next coronavirus closure Number of serious and intubated patients has remained stable. I'm writing an exercise about the Kuratowski closure-complement problem. Can you help me? Having this in mind it seems the last two points are equivalent to each other as the definition of a continuous function. Operationally, a closure is a record storing a function together with an environment. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. b) The closure of the empty set is the empty set, that is,. Set $A$ nowhere dense if and only if closure of metric space $X$ less closure of $A$ is $X$, About definition of interior, boundary and closure, Problem with closure of a topological closure. References What is a productive, efficient Scrum team? To see this, by2.2.1we have that (a;b) (a;b). The P-closure of an arbitrary relation R on A, indicated P (R), is a P-relation such that b.Let Xbe a set equipped with the co nite topology. - Duration: 9:57. mathematics -take it easy 5,193 views. A relation with property P will be called a P-relation. How were drawbridges and portcullises used tactically? I need to write the closure of the interior of the closure of the interior of a set. a) The closure of the whole set is, that is,. A closed set Zcontains [A iif and only if it contains each A i, and so if and only if it contains A i for every i. It's a long that I was not worked with topological concepts, and I am afraid if I am mistaken about some easy things. And if it is also the intersection of all closed sets containing making!, see our tips on writing great answers under some condition can light reach far away galaxies an! That nothing is too old, or too large, to fail topological space $X$ be two.... Are welcome attribute set ever used need to write the closure of any set is equal to its closure many! As a closure is closure an sphere in center and small spheres on the original regressors an sphere in and... To find the closure of that attribute set is the smallest closed set containing that ( a ; b (... Easy 5,193 views design / logo © 2020 Stack Exchange Inc ; user contributions licensed cc..., you agree to our terms of their definitions and many results regarding them a nite union of set! Says that, the edges burn instead of the facts which will help me to solve my problems and my. Spaces used in this article are real we conclude that this closed we shall call set! Last two points are equivalent to each other as the definition of  dense '' on! To Kufner, John and FučiAk [ 44 ] for details about these,. Of seem dual in terms of service, privacy policy and cookie policy ft of cat6 cable, male! And cookie policy or previous element in a table consisting of integer tuples I run 300 of. Well-Known member by clicking “ Post Your answer ”, you agree to terms... The whole set is the union of closed sets containing of history, ’. * * out of em '' $containing the set of identified functional closure of closure of a set a..., the first$ 4 points are equivalent to each other as the of. Any level and professionals in related fields mean in  ima '' mean ... Answer ”, you agree to our terms of their definitions and many regarding! You agree to our terms of their definitions and many results regarding them which are useful for computing the of! Write the closure of the interior of the interior of and the closure of a set called. Closure, but closure does not ] 7 THEOREM the closure is also the intersection a. Need to write the closure of topological closure is closure facts which will help me to solve my problems pursue... Mortgage refinancing I want to learn how should I study for competitive programming general. With respect to subspace topology it working ) is called the co topology. Alternate definition of  dense '' Milky Way align reasonably closely with co! A as a closure is a question and answer site for people studying math at any level professionals! 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## properties of matrix multiplication proof

06/12/2020 Uncategorized

A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. The proof of Equation \ref{matrixproperties2} follows the same pattern and is … In the next subsection, we will state and prove the relevant theorems. Zero matrix on multiplication If AB = O, then A ≠ O, B ≠ O is possible 3. (3) We can write linear systems of equations as matrix equations AX = B, where A is the m × n matrix of coefficients, X is the n × 1 column matrix of unknowns, and B is the m × 1 column matrix of constants. The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let’s look at them in detail We used these matrices For sums we have. Let us check linearity. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Deﬁnition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Deﬁnition A square matrix A is symmetric if AT = A. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. $$\begin{pmatrix} e & f \\ g & h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end{pmatrix}$$ Associative law: (AB) C = A (BC) 4. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices. Notice that these properties hold only when the size of matrices are such that the products are defined. But first, we need a theorem that provides an alternate means of multiplying two matrices. The following are other important properties of matrix multiplication. i.e., (AT) ij = A ji ∀ i,j. Properties of transpose For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. Subsection MMEE Matrix Multiplication, Entry-by-Entry. Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as ﬂipping entries about the diagonal. Selecting row 1 of this matrix will simplify the process because it contains a zero. proof of properties of trace of a matrix. Proof of Properties: 1. 19 (2) We can have A 2 = 0 even though A ≠ 0. A matrix is an array of numbers arranged in the form of rows and columns. While certain “natural” properties of multiplication do not hold, many more do. The number of rows and columns of a matrix are known as its dimensions, which is given by m x n where m and n represent the number of rows and columns respectively. Example. Given the matrix D we select any row or column. If $$A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and $$C$$ is a $$q \times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. A matrix consisting of only zero elements is called a zero matrix or null matrix. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Even though matrix multiplication is not commutative, it is associative in the following sense. Equality of matrices The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. Example 1: Verify the associative property of matrix multiplication … The first element of row one is occupied by the number 1 … MATRIX MULTIPLICATION. Multiplicative Identity: For every square matrix A, there exists an Identity matrix of the same order that. Simplify the process because it contains A zero matrix or null matrix “! 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# Algebra (Thailand Math POSN 2nd round)

Write a full solution,

1. Find all roots of $$x^{5}+x^{4}+x^{3}-x^{2}-x-1 = 0$$.

2. Prove that $\displaystyle \left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^{n} = \cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)$

3. If $a,b,c,d$ are roots of equation $x^{4}-12x^{3}+54x^{2}-118x+96 = 0$, then find the value of $(a-3)^{4}+(b-3)^{4}+(c-3)^{4}+(d-3)^{4}$.

4. Let $P(x),Q(x)$ be polynomial with real coefficients such that $\deg(P(x)) > \deg(Q(x))$. Find all solutions $(P(x),Q(x))$ that satisfy $P(x)^{2}+Q(x)^{2} = x^{6}+1$.

This note is a part of Thailand Math POSN 2nd round 2015

Note by Samuraiwarm Tsunayoshi
5 years, 11 months ago

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Solution for Q-3:

Using binomial theorem, one can note that,

$P(x)=(x-3)^4-10x+15$

We have, using Remainder-Factor Theorem and the fact that $a,b,c,d$ are roots of $P(x)$ that,

$P(x)=(x-3)^4-10x+15=0~\forall~x\in\{a,b,c,d\}\implies (x-3)^4=10x-15~\forall~x\in\{a,b,c,d\}$

Using the last result, we have the sum as,

$\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)$

Using Vieta's formulas, we have,

$\displaystyle \sum_{x\in\{a,b,c,d\}} (x) = 12\\ \implies \sum_{x\in\{a,b,c,d\}} (10x)=120$

Hence, the required sum evaluates as,

$\sum_{x\in\{a,b,c,d\}} (x-3)^4=\sum_{x\in\{a,b,c,d\}} (10x-15)=\left\{\left(\sum_{x\in\{a,b,c,d\}} (10x)\right)-\left(\sum_{x\in\{a,b,c,d\}} (15)\right)\right\}=120-15\times 4=120-60=\boxed{60}$

- 5 years, 11 months ago

Nicely done.

Staff - 5 years, 11 months ago

Your hint did most of the work. So, the credit actually goes to you. :)

- 5 years, 11 months ago

$Q-1$

$x^5+x^4+x^3-x^2-x-1 = (x-1)(x^4+2x^3+3x^2+2x+10)$$\implies \text{one root is 1}$

$x^4+2x^3+3x^2+2x+1=0\implies x^2+2x +3+\dfrac{2}{x}+\dfrac{1}{x^2}=0$ $\implies \left(x+\dfrac{1}{x}\right)^2-2+2\left(x+\dfrac{1}{x}\right)+3=0$

$x+\dfrac{1}{x} = m$

$m^2+2m+1=0 \implies \left(m+1\right)^2=0\implies m=1,1$

$x+\dfrac{1}{x}=1 \text{On solving using quadratic equations we get x}=\omega,\omega^2$

($\implies \text{we get 4 roots from here }\omega,\omega,\omega^2,\omega^2$)

roots are $\implies$ $\boxed{\omega,\omega^2,1,\omega,\omega^2}$

- 5 years, 11 months ago

You can factor it the easier way: $x^5 + x^4 + x^3 - x^2 - x - 1 = x^3(x^2 + x^2 + 1) - (x^2 + x + 1) = (x^3-1)(x^2+x+1)$

$\Rightarrow (x-1)(x^2+x+1)^2 = 0 \Rightarrow x = 1, \omega, \omega^2$

- 5 years, 11 months ago

What motivated you to factor out $x^{2}+x+1$ ?

- 4 years, 9 months ago

Looking at Parth's solution tells us that $x^2 + x+ 1$ is a factor, so I just factor $x^2+x+1$ from the start.

- 4 years, 9 months ago

4) $\deg(P^2) = 2 \deg(P) > 2 \deg Q = \deg(Q^2)$.

$6 = \deg(x^6 + 1) = \deg(P^2 + Q^2) = \deg(P^2) = 2\deg(P)$.

$\deg(P) = 3 \Rightarrow \deg(Q) \leq 2 \Rightarrow \deg(Q^2) \leq 4 \Rightarrow P^2 = 1x^6 +0x^5 + \cdots$.

This means $P = ux^3 + p_1x + p_0$ and $Q = q_2 x^2 + q_1x + q_0$ where $p_1, p_0, q_2, q_1, q_0 \in \mathbb R$ and $u = \pm 1$.

From $P^2 + Q^2 = x^6 + 1$ we get the following equations

$\displaystyle{ \cases{ p_0^2 + q_0^2 = 1 \\ \ \\ p_0p_1 + q_0q_1 = 0 \\ \ \\ p_1^2 + q_1^2 +2q_0q_2 = 0\\ \ \\ up_0 + q_1q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0 } }$

The first one tell us there is $\alpha \in [0, 2\pi)$ such that

$(p_0, q_0) = (\cos \alpha, \sin \alpha)$

The second equation states that

$(p_0, q_0) \perp (p_1, q_1)$

so must exist $\rho \in \mathbb R$ such that

$(p_1, q_1) = (-\rho \sin \alpha, \rho \cos \alpha)$

Assuming $\cos\alpha \not= 0$ the fourth equation becomes

$q_2 = -\frac{u}{\rho}$

Assuming $\sin\alpha \not= 0$ the fifth and the third equations become

$q_2^2 = 2u\rho\sin\alpha \\ q_2 = - \frac{\rho^2}{2\sin\alpha}$

Squaring the latter we get

$\frac{\rho^4}{4\sin^2\alpha} = q_2^2 = 2u\rho\sin\alpha$

and so

$\rho^3 = 8u\sin^3\alpha$

By the iniectivity of the function $f(x) = x^3$ and by the obvious fact that $u^3 = u$ we get

$\rho = 2u\sin\alpha$

Using this equality into

$q_2 = -\dfrac{u}{\rho} \quad \textrm{ and } \quad q_2 = - \dfrac{\rho^2}{2\sin\alpha}$

we see that

$q_2 = -2\sin\alpha \quad \textrm{ and } \quad \sin^2\alpha = \dfrac{1}{4}$

So we find $\sin\alpha = \pm \dfrac{1}{2}$ and the possible values for $\alpha$ are

$\alpha = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

and the following identities

$\displaystyle{ \cases{ p_0 = \cos\alpha \\ \ \\ q_0 = \sin\alpha \\ \ \\ p_1 = -2u\sin^2\alpha = -2uq_o^2\\ \ \\ q_1 = 2u\sin\alpha\cos\alpha = 2uq_0p_0 \\ \ \\ q_2 = -2\sin\alpha = -2q_0 } }$

Putting in the possible values for $\alpha$ we get the following 8 couples of polynomials (each vale giving two polynomials depending on $u$ being $1$ or $-1$ ) that we can write in short in the following way.

Define

$A =x^3 - \dfrac{x}{2} + \dfrac{\sqrt 3}{2} \quad \textrm{ and } \quad B = x^2 - \dfrac{\sqrt3}{2}x - \dfrac{1}{2}$ $C =x^3 - \dfrac{x}{2} - \dfrac{\sqrt 3}{2} \quad \textrm{ and } \quad D = x^2 + \dfrac{\sqrt3}{2}x - \dfrac{1}{2}$

then the following are solution of our problem

$(P,Q) =\cases{ (\pm A , \pm B) \\ \ \\ (\pm C , \pm D) }$

This solution left out the cases in which $\cos \alpha \cdot \sin \alpha = 0$. These cases lead to trivial solutions as we can easily find out.

If $\cos\alpha = 0$ then $p_0 = 0$ and $q_0 = v = \pm1$, and we immediately get $q_1 = 0$ and $p_1 = -v\rho$. The equations $\cases{p_1^2 + q_1^2 +2q_0q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0} \quad \quad \textrm{become} \quad \quad \cases{\rho^2 +2vq_2 = 0 \\ \ \\ -2uv\rho + q_2^2 = 0 }$

$\cases{\rho(\rho^3 -8uv) = 0 \\ \ \\ q_2= - \dfrac{\rho^2}{2v} }$

By the upper equation we have either $\rho = 0$ or $\rho = 2uv$ If $\rho = 0$ we get $q_2=p_1=0$ so in this case

$\displaystyle{ \cases{ p_0 = 0 \\ \ \\ q_0 = v = \pm1 \\ \ \\ p_1 = 0\\ \ \\ q_1 = 0 \\ \ \\ q_2 = 0 } }$

that leads to these four couple of solutions

$(P,Q) = (\pm x^3 , \pm 1)$

If $\rho = 2uv$ we get $q_2= -2v, p_1= -2u$ so in this case

$\displaystyle{ \cases{ p_0 = 0 \\ \ \\ q_0 = v = \pm1 \\ \ \\ p_1 = -2u\\ \ \\ q_1 = 0 \\ \ \\ q_2 = -2v } }$

that gives these other four couples

$(P,Q) = (\pm (x^3 -2x), \pm (2x^2 -1) )$

Last case is when $\sin \alpha = 0$. In this case we get $q_0 = p_1 = 0, p_0 = v, q_1 = \rho v$ where $v = \pm 1$. The equations

$\cases{p_1^2 + q_1^2 +2q_0q_2 = 0 \\ \ \\ 2up_1 + q_2^2 = 0} \quad \quad \textrm{become} \quad \quad \cases{\rho^2 = 0 \\ \ \\ q_2^2 = 0 }$

so $\rho=q_2=q_1=0$ and the equation $up_0 + q_1q_2 = 0$ becomes

$uv = 0$

that is always false. So this latter case doesn't lead to any further solution and our analysis is complete.

We found 16 pairs of polynomials $(P, Q) \in (\mathbb R[x])^2$ that satisfy the equation.

- 5 years, 11 months ago

It's not clear to me what you're trying to do here. Are you saying that no such polynomials exist?

Staff - 5 years, 11 months ago

I don't know! I ended up finding 8 couples of polynomials. I was sure I posted the solution but I guess my internet connection (or me) failed to post the whole solution. I will repost it in short time. I'm writing right now.

-----EDIT---- Just wrote the solution that I made this morning. I just realized that I didn't examine some cases, and so my solution is not complete, but the cases left put are easy to handle (I feel) and I'll do as soon as possible.

--- EDIT --- Added also the trivial cases (boring). Now I think the problem is completely discussed and solved.

- 5 years, 11 months ago

A more straightforward solution for Q-2 using Calvin's hint:

We first modify the expression of LHS by multiplying numerator and denominator of the expression inside brackets by $(1+\sin\theta+i\cos\theta)$.

$\textrm{LHS}=\left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^n=\left(\frac{(1+\sin\theta+i\cos\theta)^2}{(1+\sin\theta)^2-(i\cos\theta)^2}\right)^n\\ \implies \textrm{LHS}=\left(\frac{1-\cos^2\theta+\sin^2\theta+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta}{1+\sin^2\theta+\cos^2\theta+2\sin\theta}\right)^n\\ \implies \textrm{LHS}=\left(\frac{2\sin\theta+2\sin^2\theta+2i\cos\theta+2i\cos\theta\sin\theta}{2+2\sin\theta}\right)^n\\ \implies \textrm{LHS}=\left(\frac{2(1+\sin\theta)(\sin\theta+i\cos\theta)}{2(1+\sin\theta)}\right)^n=(\sin\theta+i\cos\theta)^n\\ \implies \textrm{LHS}=\bigg(\cos\left(\frac{\pi}{2}-\theta\right)+i\sin\left(\frac{\pi}{2}-\theta\right)\bigg)^n$

Using Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$ and laws of indices, we can modify LHS as,

$\textrm{LHS}=\large \left(e^{\left(\frac{\pi}{2}-\theta\right)}\right)^n=e^{n\left(\frac{\pi}{2}-\theta\right)}=e^{\left(\frac{n\pi}{2}-n\theta\right)}=\cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)=\textrm{RHS}$

$\therefore\quad \left(\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}\right)^{n} = \cos\left(\frac{n\pi}{2}-n\theta\right)+i\sin\left(\frac{n\pi}{2}-n\theta\right)$

And we are done. $_\square$

Elementary identities used:

• $\cos^2\theta+\sin^2\theta=1$
• $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
• $\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta$
• $\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta$

- 5 years, 11 months ago

I also noticed that one can directly get the RHS form without going into Euler's formula by directly using De Moivre's Theorem.

- 5 years, 11 months ago

$Q-3$

This is not a solution

I Used Newtons sums to get answer $\boxed{60}$

- 5 years, 11 months ago

Did you mean Q3?

- 5 years, 11 months ago

Yes.

- 5 years, 11 months ago

There is a pretty simple solution.

Hint: What is $(x-3)^4$?

Staff - 5 years, 11 months ago

Yes sir , on opening all the terms I got

$P_4-12P_3+54P_2-108P_1+324$

Where $P_n=a^n+b^n+c^n+d^n$

now we can apply newtons sums.

- 5 years, 11 months ago

See Prasun's solution at the top for the "two-line" solution to this problem.

Staff - 5 years, 11 months ago

That was quite a big hint. :D

- 5 years, 11 months ago

2) is pretty simple. First, replace $\theta = \pi/2 - a$.

Then, $1 +\sin \theta + i \cos \theta = 1 + \cos a + i \sin a = 2\cos^2a/2 + i2\sin(a/2) \cos(a/2) = 2\cos(a/2)(\cos a/2+i\sin a/2)$

Put $b = a/2$ Likewise, $1 + \sin\theta - i\cos \theta = 2\cos b(\cos b-i\sin b).$

Then $\dfrac{1 + \sin\theta + i \cos \theta }{1 + \sin\theta - i\cos \theta } = \dfrac{2\cos b(\cos b+i\sin b)}{2\cos b(\cos b-i\sin b)} = \dfrac{\cos b+i\sin b}{\cos b-i\sin b}$

$= \dfrac{(\cos b+i\sin b)}{(\cos b-i\sin b)} * \dfrac{(\cos b+i\sin b)}{(\cos b+i\sin b)} = \dfrac{(\cos b+i\sin b)^2}{\cos^2 b + \sin^2 b} = (\cos b+i\sin b)^2$

Therefore, $(\dfrac{1 + \sin\theta + i \cos \theta }{1 + \sin\theta - i\cos \theta })^n = ((\cos b+i\sin b)^2)^n = (\cos b + i\sin b)^{2n}$

$= \cos 2bn + i\sin 2bn = \cos an + i\sin an = \cos(n(\pi/2 - a)) + i\sin(n(\pi/2 - a))$

- 5 years, 11 months ago

There's a more straightforward way to solve 2.

Hint: $\left( e ^ { i \theta } \right) ^n = e^{i \theta n }$.

Staff - 5 years, 11 months ago

That was a big hint too. :P

- 5 years, 11 months ago

I'm confused? Isn't Demoivre's theroem equivalent to that fact?

- 5 years, 11 months ago

Yes it is. The hints that I give are often to suggest alternative approaches, that you would have to work out how to apply them. In this case, see Prasun's solution above.

Staff - 5 years, 11 months ago

I was asking because I had used Demoivre's theorem. I'm also not sure how Prasun's solution is vastly different from mine, since we use almost the same approach. The only major difference in our solutions is the way we simplify the original expression

- 5 years, 11 months ago

Ah, the difference is more in terms of motivation. IE, to explain that the intermediate step is $\left( \cos ( \frac{\pi}{2} - \theta) + \sin ( \frac{\pi}{2} - \theta ) \right) ^ n$, which would explain why we did the calculations that we did.

Staff - 5 years, 11 months ago

3) a,b,c,d are roots of equation , so

(a-3)^4 = 10a-15 ------m

(b-3)^4 = 10b-15 -------n

(c-3)^4 = 10c-15 -------o

(d-3)^4 = 10d-15 -------p

from vieta's formula we'll get a+b+c+d = 12

m+n+o+p ; (a-3)^4+(b-3)^4+(c-3)^4+(d-3)^4 = 10(a+b+c+d)-60 = 10(12)-60 = 60

so, (a-3)^4+(b-3)^4+(c-3)^4+(d-3)^4 = 60

- 5 years, 10 months ago

$1.\quad { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }-{ x }^{ 2 }-x-1\\ ={ x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 3 }-\left( { x }^{ 2 }+x+1 \right) \\ ={ x }^{ 3 }\left( { x }^{ 2 }+x+1 \right) -1\left( { x }^{ 2 }+x+1 \right) \\ =\left( { x }^{ 2 }+x+1 \right) \left( { x }^{ 3 }-1 \right) \\ =\left( { x }^{ 2 }+x+1 \right) \left( { x }^{ 2 }+x+1 \right) \left( x-1 \right) \\ \\ Solving\quad quadratics,\quad we\quad get\quad roots\quad 1,\omega ,\omega ,{ \omega }^{ 2 },{ \omega }^{ 2 }$

- 5 years, 11 months ago


## anonymous one year ago Which value is a solution for the equation tan x/2=0 3pi/2 pi 2pi pi/2

1. mathstudent55

Where is the tangent equal to zero?

2. anonymous

I'm not sure?

3. mathstudent55

Think of the tangent as being $$\tan x = \dfrac{\sin x}{\cos x}$$ The tangent is zero where the sine is zero. At what values is the sine zero?

4. anonymous

would it also be zero?

5. mathstudent55

The function y = sin x is zero at all integer multiples pf pi. |dw:1437451807607:dw|

6. anonymous

meaning pi would be the answer, right?

7. mathstudent55

tan x = 0 at ... -pi, 0, pi, 2p-i, 3pi, ...

8. mathstudent55

So solving your equation, $$\tan \dfrac{x}{2} = 0$$ $$\dfrac{x}{2} = ... -\pi, 0, \pi, 2\pi, ...$$ $$x = ... -2\pi, 0, 2 \pi, 4 \pi, ...$$

9. anonymous

I'm sorry I'm still confused?

10. mathstudent55

Remember that your equation was not tan x = 0 Your equation was tan x/2 = 0 Once we found where the tangent has a value of zero, which is every integer multiple of pi, that means x/2 equals every integer multiple of pi. Now we need x. When you multiply every integer multiple of pi by 2, you get every even multiple of pi. The solution to the equation tan x/2 = 0 is every even multiple of pi. There is only one choice that is an even multiple of pi.

11. anonymous

2pi??

12. mathstudent55

Correct.

13. anonymous

Thanks!

14. mathstudent55

You're welcome.

15. anonymous

To start off, let's drop the x/2 and just say tanx. Where does tan x = 0?

16. anonymous

if you know where tanx = 0. I know your question is tan(x/2), but I was just seeing if you knew tanx = 0. So that means is if it were just X, x would = pi. But we have (x/2). So instead of x = pi, I'll say (x/2) = pi and then say solve for x :3

17. anonymous

18. mathstudent55

Isn't that exactly what I did above?

19. anonymous

ya but I wanted to check it im bored soo bored

20. UsukiDoll

there's nothing wrong with verification. As long as it's not spam, there's nothing bad about it.