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Consider the periodic functionsin t for 0 < t<3 f(t) (o for $ < <t f(t+#) =f():(a) Sketch this function over the interval -27 < t < 21 [4] (b) Sta...

Question

Consider the periodic functionsin t for 0 < t<3 f(t) (o for $ < <t f(t+#) =f():(a) Sketch this function over the interval -27 < t < 21 [4] (b) State whether this function is even. odd, neither . [1] (c) Find the Fourier series for this function To obtain full marks. your final answer should be simplified so that it does not contain any terms of the form cos 5 sin $ where kis an integer, [20] (d) By considering pointwise convergence of this Fourier series at t = evaluate the sum

Consider the periodic function sin t for 0 < t<3 f(t) (o for $ < <t f(t+#) =f(): (a) Sketch this function over the interval -27 < t < 21 [4] (b) State whether this function is even. odd, neither . [1] (c) Find the Fourier series for this function To obtain full marks. your final answer should be simplified so that it does not contain any terms of the form cos 5 sin $ where kis an integer, [20] (d) By considering pointwise convergence of this Fourier series at t = evaluate the sum of the series ~" (5]



Answers

Find the Taylor series and radius of convergence.

Okay. So this question is asking us to evaluate the integral shown below. So before we get started with, just right off the central for you. Yeah. Okay so we're dealing with a definite integral here and we can make use of an integral property to make this question a bit easier. This central property is the phone police force so we can split up uh foreign scroll. If we have two functions that are subtracting we can rewrite it like this. So knowing this and using this property, we can easily make use of it in our question. So let's rewrite are in school. Mhm. Mhm. Okay. So now let's evaluate this definite integral. 1st let's pull it out over here, solve the central role. We're going to have to make use of a U. Substitution. So let's let new equal five x. Yes. It's going to be equal five X. And dx. Mhm. Okay. Now let's substitute you into our different integral. Yeah. Mhm. You Uh huh And this is just a constant. So we have we can rewrite this like this. We have 1/5 times the co sign of you. Do you? It's on a bit. Take it from the top here. Uh We can take out this constant to the front of our of your control this and we know that the integral of coastline. Effexor coastline you here is simply a sign of you. So we have five farms evaluated for two, you know your father. So we got to remember what you as you is actually 5X. So let's bring that back in. Okay so now let's evaluate this. We're gonna have 1/5 times like it. So five times pi over two minus. So five times negative pi over to we do the mouth quickly and we see that we get our answer to be 0.4 4 to 5. Alright so that's half of our problem here. And now we have to solve this other integral. We split up the sign for max. So let's bring that down over here or other have two of the question. Is this? Yeah. No so we do the same thing we did last time we see that we have to make use of a U. Substitution. So it's like equal five X. Here. Yeah. Cool. Looks can we can't to the department to sign you Do you over five. We can take out the constant again to 1/5 over 22 sign you d you. And of course the integral of sine uh is negative coasts. So let's break that out over here. 1/5 times negative post sign. You know, you know coat. Uh huh. Now we bring back resolved this year. So one of the five times five X. Sorry. Yeah. Mm. Oops. We have 1/5. Bye bye. Get of course 552 minus negative coasts. five times -5 two. This is equal to zero. Our final answer when we put everything together. Yeah that this was two or 5 and The other integral was signed five X 0. So putting it all together, we have the integral harbor too. We'll sign by. That's minus sign. Yes is equal to To over 5 0, which is equal to 2/5. We're 0.4. Okay, Hope to stop.

In this problem we have to find interval of convergence for the given in finite series. So first thing that we can note here, this whole term represents any term of the series, let us say anyth term is represented by B N. So bien is six X. Raised to the power and divided by fifth root of and therefore and plus one term will be we can we have to replace and by end plus one. So this is going to be six times X raised to the power and plus one divided by fifth root of and plus one. No, we will perform raise your taste here. So from ratio taste what raise your taste says? We have to find limit off absolute value of be in divided by B N plus one at an approaches to infinity. If this value is less than one then the city's converse's. And at the end we would have to check the value at the end points. So first we will calculate limit of the function here and then we will calculate the interval. So what is going to be limited? So limit as an approaches to infinity models of being we will put its value six x rays to the power and divided by fifth root of and and what is B N plus one since it is in denominator. So we will reverse the a numerator and denominator term of be off and plus one. So it will be fifth the root of and plus one divided by six times X raised to the power and plus one. Now a few terms will get canceled here and we'll get canceled. Now we can write X raised to the power and plus one X raised to the power and plus one as X in X raised to the power end into X raised to the power one. So simply X. Now this X raised to the power and will get cancelled with the numerator term here. So the left terms will be limit of and approaches to infinity. The numerator terms will be fifth root of n plus one divided by fifth root of and into once upon X. So limit of and approaches to infinity. We can combine these two terms, fifth, the root of endless one upon and into one by X. We can take an comin from the numerator. So limit of this function will be as an approaches to infinity. Fifth wrote off and in 21 plus one by N divided by N into one upon X. Now as here the animal gets canceled now as the end approaches to infinity so this term will approach to zero. Therefore limit value of this entire function will be one therefore, if you perform limit here. So this function reduces to models of one upon X. Now for the conversions we know that the value of the limits should be less than one. Therefore model Fund by X should be less than one. Now from here we get to solutions first solution is minus one by X minus one way X is greater than one and second solution is one by X is less than one, so if minus one where X is greater than one, then X will be less than minus one. And from here we will get X is greater than one. Therefore the interval of convergence is going to be X belongs to minus infinity to minus one union, one to infinity. But we have to check the endpoints here at minus one and one. So accordingly we will modify this result. No, if we dig if exit the calls to one so what will be our cities? So series was energy equals to eight to infinity six and two x rays to the power and divided by fifth root of And so if you put the value of x rays to the power in here. So if you put value of X equals to one here. So this will reduce to sigma. Energy costs 218 to infinity six and 21 raised to the power N divided by fifth root of. And now we know that power of power any number to the power of one will give one. Therefore this will be shake him off and equals to eight to infinity six divided by fifth root of and no we can use we can use divergence taste here. So what divergence taste? It says if you perform divergence test and we take the limit. If we take the limit of the any term of the function here. Six divided by fifth root of n. As N approaches to infinity. So we see that this is a close to zero. It means which is less than one. So if it is, if the limit of the function at any and approaches to infinity is less than one, this implies that the function is conversion. Therefore X equals to one is going to be convergent. So we will use equality at X equals to one. So our so our interval will modify too. X belongs to one to infinity. Now we will check the scene for X equals two minus one. So if exes equals two minus one, so at X equals two minus one. Our cities will modify us. What was our series? It was sigma from any close to eight to infinity. Six x rays to the power and upon fifth root of and that is equals two, six and two minus one raised to the power and divided by fifth root of and and limit as N approaches to infinity. Now if we now if we go for divergence taste then limit as an approaches to infinity of the term that is minus one raised to the power N into six, divided by fifth root of. And we observe here that here the term present is minus one raised to the power end, which is an oscillating limit, which will give an oscillating limit. This implies that limit does not exist. Limit does not exist if limited does not exist. This implies that a taxi cost to minus one. The cities will not be convergent. Therefore that there is the engine of the range of convergence will be X belongs to minus infinity to minus one. There will be an open bracket at taxes equals two minus one union, one to infinity, and one will be included in our solution. Therefore, there will be closed interpreter, so this is going to be our final answer.

So the first thing you're being asked to do to do in this problem is to find the first four terms for this infinite geometric sequence. So the first thing to do because on the bottom and is equal to one. To find the first term of the sequence. We're going to substitute one in place event. So I'm negative four times 1 30 being raised to the one minus one power. Well one minus one is zero. So we have negative four raised negative four times one third to the zero power. Or remember anything to the zero power is equal to one. So we're often negative four times one which is just equal to negative four. So negative four is the first term in the sequence. So now to find the second term of the sequence we substitute and two friends. So we're gonna have negative four times one third raised to the tu minus one power. Well two minus one is one. So we have negative four to the one times one third to the first power. Well one third to the first power is just one third. So we have negative four times one third which is equal to negative four thirds. So that's the second term of the sequence to find the third term. We substituted three for N. So I'm negative four times one. Third race to the three minus one power, well three minus one is two. So we have negative four times one third square Well now we have to square one third. Well that's equal to 19 So we have negative four times 19 and negative four times one. Ninth is negative 4/9. And lastly to find the fourth term we substituted for for N. So we'll have negative four times one third raised to the four minus one power. Well four minus one is three. So we have negative four times one third to the third power. And then when we're going to add one third, that would be 1/27. So we have negative four times 1/27 and negative four times 1/27 is negative for over 27. Alright, perfect. So now we've found the first four terms for this infinite geometric series. Now in part be the first thing we're being asked to do is determine if the series is going to converge or diverge and that's all based off our base value. So remember when our base value is less than the absolute value of it. So they usually refer to the bases are when the absolute value of art is less than one, this is when are infinite geometric series will converge and when the absolute value of that base is greater than one, that's when it's going to diverge. So in our particular case R. Is equal to one third. Well the absolute value of one third is less than one. Therefore this geometric series is going to converge and therefore we need to go ahead and find the sum. Well remember the formula to find the sum of an infinite geometric seek of the infinite geometric series is S equals A. Someone divided by one minus R. Well remember ace of one refers to the first term in the sequence in the series which we found to be negative for. So I can have negative four divided by one minus R which is one third. Well one minus one third is two thirds. So we have negative four divided by two thirds. Well remember this just means negative four divided by two thirds. And remember when you multiply or divide fractions you need to multiply by the reciprocal. so you're gonna have negative four times 3/2. Well negative four times 3/2 is equal to negative six. So the some of this infinite geometric series is negative six.

Okay to find a taylor series first you need some derivatives. So we get minus Synnex minus cosine X. Mhm. Synnex. And then they start over again. All right a is Pi over six. I had to get it out of the way so we need to know the derivatives at five or six. Well five or six is 30 degrees, 30°. Looking like 1 to Squirrel 2 3. So the coastline is square to 3/2 and the sign is one half somebody's would have minus the square to three or to one half. Then it starts over again. Okay so then the taylor series will go um Yeah the function was um the mhm. The derivative at five or 6 times X -A Over one factorial to the one plus the function or the derivative X -5 or six squared over two factorial plus the number X -5 or six Cubed over three factorial etcetera. Oops. Okay so really the only thing you could do to make it look any nicer you can factor out of one half because they all have a two in the denominator and then you get square 23 minus x minus piper six -23 X minus pi over six squared over two factorial plus x minus pi over six to the third over three factorial minus plus. And then if you wanted to write it in series form you can have trouble because some are square to three and some are um ones. Okay so let's now I know what to do. Let's break it up into and take a square 23 out. So we got one minus x minus pi over six squared over two factorial plus x minus pi over six. So all the odd ones are all the even ones will be in this little bit with the square root plus then all the odd ones will be in here so minus X minus five or six to the one plus x minus five or six to the third over three factorial uh minus x minus five or six to the fifth over five factorial. Okay so then now I can't write it like this one half. Let's just call it square to 3/2 for the first one summation X minus pi over six. We want the even one so two in Over two in factorial From an equal 0 to infinity. Um Well I forgot to make an alternate. So right in the front here you would want to -1 to the end And then -1 half because the first one starts with minus here so we want to mine assist. Okay now we want the odd powered one. So X -5 or six To the two n plus one over two. N plus one factorial. I forgot the alternating thing again -1 to the end goes right there. Okay so those two together will write the co sign. Um And there's no way the other way to put them together because of the square to three and the one Okay, sorry, that was so disorganized. Hope you understand it.


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