5

Show the sequence of functions3.Fn(2) =21-24 dtconverges uniformly to some F(z) for Re(z) > 0. Show F(z) is holomorphic on Re(z) > 0. Find F(z) in polar form:...

Question

Show the sequence of functions3.Fn(2) =21-24 dtconverges uniformly to some F(z) for Re(z) > 0. Show F(z) is holomorphic on Re(z) > 0. Find F(z) in polar form:

Show the sequence of functions 3. Fn(2) = 21-24 dt converges uniformly to some F(z) for Re(z) > 0. Show F(z) is holomorphic on Re(z) > 0. Find F(z) in polar form:



Answers

If f (0) = g(0) = 0 and f", g"are continuous, show that
Z a0f (x)g"(x)dx = f (a)g'(a) ? f'(a)g(a) ?Z a0f"(x)g(x)dx

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.

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.

Hi Ronnie. Okay, so I saw your problem and you need to calculate the MacLaurin series of this function. So the first step that you need to do is start to see what is a pattern that the derivatives follows. So let's take the first derivative of this function. And you will see that these equals to one over three minus eggs. Then if you take another derivative of this function, you will see that the z equals two minus 1/3 minus x square. But if you take another derivative to this function, You will see that this is equal to -2/3 -1. Cute. If you take another derivative because here we start to see a pattern. If we take the fourth derivative of this function, you will see that there's equals 2 -2 times three over three minus x To the four power. So you can see that if we take another derivative, what is going to happen is that we're going to multiply here the four on the numerator and we are going to increase this by by one. So the fifth derivative of this function is equal to -2 times three times 4 over three -X to the five power. So here you can start to observe a pattern that follows the derivative. So you can even for what's going to look the end derivative. So the end derivative of this function first we're going to have the negative value here and then this looks like a factorial. And the maximum value of the factorial in this case is four And the power or derivative that we're taking is the five. So technically we have N -1 factorial. You can see also that here this is three factorial and the power or their derivatives, sorry, Is the 4th derivative here we have two factorial And is the third derivative of the function. So technically here in the numerator we have N -1 factory And this is divided by three -X to the end power. So this is the pattern that the derivative folks. Now, let's remember what's going to be the MacLaurin series of this function. It's going to be the sum from N equals 20 up to infinity of the derivative of the function evaluated at zero because it's the MacLaurin series divided and factorial X to the M power. So let's replace what we have here And we obtain that this is equals two. First we have just the function evaluated at zero and plus the sum from n equals to one up to infinity of minus N -1 Factorial Over. And factorial X to the M power over. And here we are evaluating the derivative at zero. So here, instead of putting X, we put just three. So three to the end power. Right. And what is f evaluated at zero? Will is just logarithms of three. So from this we obtained that the mark. Loring expansion of this function, it's going to be the luck of three minus because we're going to take out this minus sign out of the sun because all the the the elements have this this minus sign here. So it's minus. Yes. The sum from n equals to one up to infinity. And we can simplify this expression. So there's going to be just one over N1 over NX to the end over three to the air. So great. This is the the glittering series of this function. Now. We need to find the radius of convergence. So let's focus on this part on the infinite serious and we need to find the radius of convergence. So we need to apply the ratio test So that means they can limit as an goes to infinity of the absolute value of this expression for n plus one. I mean X two N plus one over And plus one times 3 to the end was born over X to the end over. And 3 to the end of power. Okay. And we need to find the value of X. Such that make this limit less than a one. But let's reduce this expression first. So this will be the limit as goes to infinity off. We're going to simplify here the X N plus one with these X. N. This three end with these three and plus one. So we have wind with X over. So first I'm going we're going to have end over and plus one times X over three. And this Well let's take the limit of this expression. So this limit, it's going to be the square root, the absolute value of X over three, because this limit is going to go, It's going to be one the limit of this expression, and this should be less than one. So from that we can see that the ratio of convergence is that the absolute value of X is less than three, and that's it.

In the problem we have This integration 0 to a fx. Just relax. General S X dx. Now this is equal to what. So this is equal to let us consider this as you And this as which is. So this is of the form islet that is inverse logarithmic algebraic, programmatic and exponential. So let us assume you as the first function and we s the second Frandsen according to this. So we have this is effects immigration do double ds X dx minus integration. D upon dx off fx integration. General does X. Dicks hold X. So further this can be read in us Effects judas X minus integration. This is F. S. X. Into judas X. Dicks. Further we have to solve this spot. So it becomes F L. S X. Division G D S X dx minus integration. Day upon the excessive every day. Six integration Due to six DX holy X. So this is equal to every six and two jx minus integration if they were last X into gx dicks. So this is the different integration of the spot. So these every sex into JD six. The ecstatic was this one. So further we have the digit of overall integration. So this is overall integration will become now first of all we neglected so we neglect if true and G row as these are equals to zero. So what all integration become integration and you go to a fx G doubled us X dx at equals two F f. A judas A minus. So this is the stamp F or fe into Judith a minus. This is if does uh, g of a minus integration zero to A. If nevertheless X G of x dx. So this is the overall integration. Hence it is the answer to the problem.


Similar Solved Questions

5 answers
QUESTiON 13Which equation shows a strong acid in water? LiOH(s) 37> Lit (aq) + OH" (aq)HF(aq) + Hzo() <7 H3O+ (aq)F-(aq)HF(aq) 67 H+ (aq)F"(aq)HCI(aq) + HzO(aq) - -> CI-(aq) H3Ot(aq)
QUESTiON 13 Which equation shows a strong acid in water? LiOH(s) 37> Lit (aq) + OH" (aq) HF(aq) + Hzo() <7 H3O+ (aq) F-(aq) HF(aq) 67 H+ (aq) F"(aq) HCI(aq) + HzO(aq) - -> CI-(aq) H3Ot(aq)...
5 answers
Are Provide Are 3 Osteocytes enclosed Are specialized gapjunctions IL chambe muscle for 8 ed acunae E bone muscle tissue: tissue:
Are Provide Are 3 Osteocytes enclosed Are specialized gapjunctions IL chambe muscle for 8 ed acunae E bone muscle tissue: tissue:...
5 answers
Suppose that the bacteria in a colony grow unchecked according to the Law of Exponential Change. The colony starts with bacterium and triples in number every 10 minutes_ How many bacteria will the colony contain at the end of 12 hours?At the end of 12 hours, the colony will contain bacteria. (Use scientific notation. Use the multiplication symbol in the math palette as needed. Round to one decimal place as needed:)
Suppose that the bacteria in a colony grow unchecked according to the Law of Exponential Change. The colony starts with bacterium and triples in number every 10 minutes_ How many bacteria will the colony contain at the end of 12 hours? At the end of 12 hours, the colony will contain bacteria. (Use s...
5 answers
If f(z) find f' (5). 22
If f(z) find f' (5). 22...
5 answers
8040ResourcesHintThc pK_ Of hypochlorous acid is 7.,530 A 57.0) mL solulion of 0114 M sodium hypochlorite (NaOCH) is titrated Will 01.323M HCL Calculate the pH of the solulion after thc addition oC' 6.60 mL o1 0.323 M HCLpH =Calculatc the pH of the solution uller thc Hdidition of 21.0 m.of 0.323 MHCIpH =Calculate thc pH of the solulion at the eqjuivalencc Doint with (1323 M HCI,pH =
8040 Resources Hint Thc pK_ Of hypochlorous acid is 7.,530 A 57.0) mL solulion of 0114 M sodium hypochlorite (NaOCH) is titrated Will 01.323M HCL Calculate the pH of the solulion after thc addition oC' 6.60 mL o1 0.323 M HCL pH = Calculatc the pH of the solution uller thc Hdidition of 21.0 m.of...
5 answers
Lt U = {G.U_0}9 R" whero Show that dependent _ (For Fun} Lct U {u5,", 1,1 < R" lecto {0.63U,"} linenly depentent . (For Moro Fuun) 14t U {uuaW44 i5} wbute M,h 0,1w linetuly depx-mlent_any @ € R" . UU{"} {4,04.7 "n. u} is linearkyShowM U € R" s[Mn(U) U U {"}Shan tnt Ict anY M € R" #Pa(u) Uu{"}
Lt U = {G.U_0}9 R" whero Show that dependent _ (For Fun} Lct U {u5,", 1,1 < R" lecto {0.63U,"} linenly depentent . (For Moro Fuun) 14t U {uuaW44 i5} wbute M,h 0,1w linetuly depx-mlent_ any @ € R" . UU{"} {4,04.7 "n. u} is linearky Show M U € R" ...
1 answers
A coil of wire, which has a total length of $7.50 \mathrm{m},$ is moved perpendicularly to Earth's magnetic field at $5.50 \mathrm{m} / \mathrm{s} .$ What is the size of the current in the wire if the total resistance of the wire is $5.0 \times 10^{-2} \mathrm{m} \Omega ?$ Assume Earth's magnetic field is $5 \times 10^{-5} \mathrm{T}.$
A coil of wire, which has a total length of $7.50 \mathrm{m},$ is moved perpendicularly to Earth's magnetic field at $5.50 \mathrm{m} / \mathrm{s} .$ What is the size of the current in the wire if the total resistance of the wire is $5.0 \times 10^{-2} \mathrm{m} \Omega ?$ Assume Earth's m...
1 answers
Use the Pythagorean Identities to write the expression as an integer. (a) $7 \sec ^{2} y-7 \tan ^{2} y$ (b) $7 \sec ^{2}(\gamma / 3)-7 \tan ^{2}(\gamma / 3)$
Use the Pythagorean Identities to write the expression as an integer. (a) $7 \sec ^{2} y-7 \tan ^{2} y$ (b) $7 \sec ^{2}(\gamma / 3)-7 \tan ^{2}(\gamma / 3)$...
5 answers
Translating Write each phrase as an algebraic expression. Use the variable $x$ to represent each unknown number. See Example 8 .Two and three-fourths less than a number.
Translating Write each phrase as an algebraic expression. Use the variable $x$ to represent each unknown number. See Example 8 . Two and three-fourths less than a number....
5 answers
D0 c) A particle moves along the curve r = [ - 2cose SO thatFind the value of2t at 0 = and interpret your answer in terms of the motion of the particle
d0 c) A particle moves along the curve r = [ - 2cose SO that Find the value of 2t at 0 = and interpret your answer in terms of the motion of the particle...
5 answers
Question 51ptsThe number of row exchanges is:None
Question 5 1pts The number of row exchanges is: None...
5 answers
(5 points each) Given below are subsets of different vector spaces Determine whether or not each subset is a subspace You must prove your claim either way_The subset of Mzz consisting of all invertible 2 x 2 matrices_The set of all polynomials a + bx + cx2 in Pz such that a + b + c = 0.
(5 points each) Given below are subsets of different vector spaces Determine whether or not each subset is a subspace You must prove your claim either way_ The subset of Mzz consisting of all invertible 2 x 2 matrices_ The set of all polynomials a + bx + cx2 in Pz such that a + b + c = 0....
5 answers
3 1 3 Find ( Point 2JH the following IU 2((-3, "2) "(Sm and ((2, 11). (1,2)} 18)2)
3 1 3 Find ( Point 2 JH the following IU 2 ((-3, "2) "(Sm and ((2, 11). (1,2)} 1 8)2)...
5 answers
U In the directlon of the vector V-4 (2, Find the Unit vector72 A06 B 039 V0 015 E0
U In the directlon of the vector V-4 (2, Find the Unit vector 72 A0 6 B 0 3 9 V0 0 15 E0...

-- 0.022909--