find-at-least-three-nonzero-terms-including-a_0-and-at-least-two-cosine-terms-and-two-sine-terms--40

Question

Find at least three nonzero terms (including $a_{0}$ and at least two cosine terms and two sine terms if they are not all zero) of the Fourier series for the given functions, and sketch at least three periods of the function.
$$f(x)=\left\{\begin{array}{rr}
0 & -\pi \leq x<0 \
x^{2} & 0 \leq x<\pi
\end{array}ight.$$

Nicholas Barvinok · Accepted Answer

we want to find the 1st 3 turns of the MacLaurin series for Natural Log of Co signed X. Well, if we take the derivative well, let&#x27;s call this f of X. We take the derivative of F of X. We get one over co sign X times, the derivative coast on its, which is minus sign X, and we get minus tangent tax. Now we know that the MacLaurin series for a tangent x is X plus 1/3 X cubed plus 2 15 x to the fifth and we&#x27;re only talking wheeling in the 1st 3 terms from We&#x27;re going to stop there. So the MacLaurin series for minus Tanja necks is just the negative of that minus X minus 1/3 X cubed minus 2 15 Sex to the fifth minus. And now, to find MacLaurin series for F of X, we just integrate turned by term. So yet minus X squared over too minus X to the fourth times for thirds. I&#x27;m sorry. Mine, Uh, times they&#x27;re over 12 right? And then minus. It&#x27;s gonna be a next to the sixth one divided by six. So too, over 90. And so on

Jaida L · Answer

So we need to figure out the MacLaurin series for both for both coastline of X and one divided by one minus X. So we&#x27;ll start with consent of X. So once we find out the derivatives of G of X and the value of those derivatives at X equals zero weaken generate the MacLaurin series, so it&#x27;s gonna look like one plus zero times x minus X squared to barter by two factorial, plus zero times X cubed plus X. The fourth developed for factorial plus dot, dot, dot plus and then a general negative one to the part of end times, actually part to end, divided by two in factorial plus doctor dark. And so this is equal to the Siri&#x27;s form and equals zero or sigma form to infinity. They could have wants to part in intense X, the part you into but by two in factorial and that is just equal to co sign of X or G of X. Other those to the same for one divided by one minus X. So let&#x27;s that h of X equal to one&#x27;s about about one minus X and then figure out the derivatives again at X equals zero and let&#x27;s generate the MacLaurin series. So that is equal to one plus X plus X squared, plus X cubed plus doctor and then in Sigma form. That&#x27;s equal to some from zero to infinity in the next part in and so that is equal to again. Want to ride by one minus X or H backs? And so, since our original function is given as co son of Ex minus two, divided by one minus X, well, that means we can just rewrite this, which is the given in terms of our new calculated Siri&#x27;s So X is equal to remember Coast and an ex. We found that it was Sigma permanent and from n equals zero to infinity, the negative one that wants a part of end times X apart to end bye bye to in factorial Last to remember. It&#x27;s too, because there is a numerator of to appear so last two times the signal that we just found in equal zero to Infinity Ex apartment. And so this sequel, Teoh Negative one must to x slash five hives X squared. You might have started thought once you expand, so this converges on the intersection of negative infinity to positive affinity and negative 1 to 1. And so the interval convergence is negative one one, whereas the 1st 3 terms are up here. Negative one minus two x minus five halves X squared.

Stephen Hobbs · Answer

to find the four year Siris of f of X equals three x will start by evaluating what a subzero is and this is going to be equal to We&#x27;ll call these limits negative pie el toe positive pie el And so it would be, um, won over l which in this case is pie integrated from negative pie or negative L. A positive L and we&#x27;ll evaluate this limit of this integral from three x dx or of the re X dx. And then from here This would actually, since it&#x27;s odd, the area to the left will cancel with the area to the right, so this would actually evaluate to zero. So this first term will be zero from here. Well, you could go ahead and switch over to a sub k. Oh, go and do that underneath and bread and it&#x27;s up, K. And it&#x27;s going to be cool, the integral one over El or one of her pie from negative pi to pie again. And then it&#x27;s just going to multiply. Our function reacts Times Co sign So three x co sign K X and using Wolf Rammer. Dez most we could pull again this integral for K is one and we get zero and K is to and so on and on for each k value that we pull again, they would all evaluate 20 Solve these terms here will go to zero, Which makes sense because co sign, uh would help model things that are even functions. But our function three x is odd. So we really only want to rely on the last terms which are odd. Help us with this so we&#x27;ll go ahead and now evaluate. Be sub K. What about switched? I will go black here, so be sub Okay, It&#x27;s very similar to a so okay except ah replace the co sign with a sign So one over pi from negative pie The pie three x towns sign of X but that k constant in there as well. The axe has left off with the exits for both of those, but they should be there too. And then we&#x27;ll go ahead and evaluate these. So now that we know that this is odd and it should result in a non zero values if we just replace K with one and then Eudes full frame or Desmond&#x27;s to evaluate it will end up evaluating that, um, integral in the area under the curve. And it would end up being six when K is one when K is, too, or, in other words, that be sub to, um, value would be negative. Three. That being said three, you could see that these are actually going to alternate between negatives and positives. The sub three is to Sorry, I was looking ahead. So be Step three is to and then finally be sub four. What the negative 1.5. And we could rewrite the, um, term three. X as a four year Siri&#x27;s with something of signs and CO signs as f of X equals will go and plug in R B sub one value right, that would go in here and it would be six. And then sign K is one. So this would be a woman. So six sign one x or just six sine x. And we ignored the other two terms because they evaluated at zero and our next term be sub to evaluated. Here. It&#x27;s buying this three, and now we&#x27;re just gonna increase the K value right, because it&#x27;s now too. And so would be both blood by sign of two X. So hopefully you&#x27;re starting to see this pattern. Yeah. And each term is just going to increase the, uh, values. Buy one for K each time here. And so next we would have waas. He&#x27;s up three, which is to be plus two sign of forex. Excuse me a sign of three X and eventually we&#x27;re gonna draftees from negative to pi to pie so you could go ahead and do that. Now, if you would like toa kind of verify what this would look like or how close toe the function. Three X That&#x27;s going to be what&#x27;s just four terms and then the last one minus one point by sign up for X. And so this is our four year Siri&#x27;s out to four terms. We&#x27;ll go ahead and expanded up to eight terms in just a moment, but for now, but we&#x27;ll go ahead. Graft. That&#x27;s using, uh, graphing tool of your choice. And as we do so, it would look something along the lines of this from negative two pi to pie. It would come up. Down? Uh, no. Okay, so something like this and it&#x27;s going to be a decent model for this line and blue, which is why equals three X And we know that this Siri&#x27;s approximation is gonna be better with more turns. So you could test these and draftees on your own, and I&#x27;ll show you what it looks like for the next four terms. Out to eat, be some five is gonna about in a girl with changing Kato five is gonna be 1.2. Changing Keita six or be sub sects is going to be thank you for he&#x27;s up. Seven. It&#x27;s 67 and finally beats up eight but the negative three sports And so these air just the coefficient, basically. And so from here, let&#x27;s say plus, then we could use this one. It would be plus 1.2 sign of five x minus. Um, you know, minus one sign of six X plus 6/7. Sign of Southern X, minus 3/4 sign of eight X And then if we graft that we would just give us. It&#x27;s like better approximation for that function. Three X and so it looks very similar to what we just craft, except it&#x27;s a little a little smoother, even though it. Also, it&#x27;s a little more. It kind of hugs the line a little better. So again, that&#x27;s just gonna, um, have reduced a little bit of the error vertically that the previous one did not take care of. And so this is out to eight terms here. And, you know, if we went further and further are approximations would just keep getting better and better. Yeah.

Jaida L · Answer

So the model in Syria&#x27;s generated by sign of eggs is equal to and the Sigma from an equal zero to infinity Negative 12 part end about a bite to impossible on factorial times x the power of to in plus one This serious converges on negative infinity to infinity he series generated by, um X time Senate X squared is going to be so let me just read it. Also f X equals x times, um, sort of X and in squared. So you&#x27;re going to replace son of ex with what we found right up here. And so it is gonna look like X times. So we&#x27;re just refining the, uh, serious. So an equal zero to infinity Negative wants the party end by the bite to and plus one factorial times exit, part two in plus one. Remember, there&#x27;s a squared right here, so we&#x27;re going to add a squared. And so once we simplify, we get X cubed minus 1/3 x tooth there plus two divided by 45 times x, the seventh minus one divided by 3 15 Excellent. Nine plus dot dot Doc, This convergence on negative infinity to infinity and so here are the first few terms. The first few terms we&#x27;re going to be X cubed minus one. That expects, plus to to buy the 45 access seventh. So right up here, These are the 1st 3 terms, and the interval of convergence is negative. Infinity to infinity.

Stephen Hobbs · Answer

to find the four year Siris of three X on this interval will start with finding a sub zero, which is going to be one over and then our limits here from negative L toe L, which is negative. 222 That&#x27;s gonna be one over L, which is to the integral from negative l toe l negative 222 of three x dx. And since this is an odd function and we&#x27;re integrating it equally from the left and the right areas, they&#x27;re going to cancel. And, uh, it&#x27;s going to result in and then ruled zero, which knocks out this first term. So next we&#x27;re gonna focus on what a sub K term will be. And similarly, it&#x27;ll start with one over l, which is to, and a girl from negative to to Oh, the reax Times Co sign. And, of course, the Cape I X several of the general form or as we were just using, uh K X. But that was one or limits had pie in them. And so this is just a more general version of that. Okay, so this is going to be OK. Hi X over two dx and then Let&#x27;s just go ahead and change these two to I already subconsciously changed this one over here to already, even though it should have started as an l. I&#x27;ve been changed to it too. Um and then if we point in K is 1234 and so on. Each of those and the girls are going to evaluate 20 and we do not end up with anything in the middle term, which makes sense because everything in the middle represents and even type of function that&#x27;s symmetric with respect to the Y axis and what we&#x27;re looking at, it is three x, which is odd and so really the odd part is only going to show up in the last terms of this expression or for your Siri&#x27;s. So lastly, we could look at, uhm evaluating what BK is, and it&#x27;s identical to what a K is. It&#x27;s just we&#x27;re going to replace co sign with sign, so 1/2 and so on. And I&#x27;ve written out the general form of it as the answer here. But we can also look at it turn by turn, which is what I&#x27;m going to do here to get a good idea of how closely this graph mimics the function. Three x So again, Once you set this up, you could go ahead and change K Tau one and then evaluate and then to evaluate. And of course, this would have a DX. I&#x27;m just leaving that out. And what will end up getting here? Let&#x27;s scroll back up a little. Is that b of one changing cato? One integrating They&#x27;re all gonna be some pretty lengthy constants. 3.8 one 972 it&#x27;s up to or changing cater to would result in a negative. It&#x27;s gonna alternate between positives and negatives. Negative. 909 86 Um, and so on be sub three. We&#x27;ll go out to four terms because this will give a pretty decent approximation 1.27 3 to 4 and finally be so before. Uh, negative. 0.9 493 So let&#x27;s discuss what before your Siri&#x27;s will look like Well, first off, it&#x27;s going to be this term multiplied by sine of Cape I X Over to There would be this times sign of hi X over two minus. This term is the constant outfront kind sign of pi X plus this turn times sine of three pi X over two minus this term time sign of two pi to pie X. And so instead of kind of writing out that whole thing, I just wanted toe explain what parts I was using two, well, again and find a rough graph of this thing. So it&#x27;s actually going have heights that go to six and down to negative six. And we&#x27;re evaluating that someone under girl on an interval from negative to to and graphing that function that a just mentioned would give something that looks somewhat like this. And so it would in fact, have values very, very close to what they should be. Right, because the original graph reacts would look something like that green line. So this is behaving very similar toe what that for your Siri&#x27;s would be and making it a more general. I also have included the answer of a more generalized version of this with leaving K alone and not replacing them with their integers one and going forward. But just with the general form of the beasts up Kay and then the, um respective for your Siris For that

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Find at least three nonzero terms (including $a_{…

Problem 12 Easy Difficulty

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