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Let Qn; n 0,1,2, be the polynomials of best approximation to f € C*[a,6]. Show that the series Qo Xa_o[Qn+1 Qn] converges in the topology of the space, that i...

Question

Let Qn; n 0,1,2, be the polynomials of best approximation to f € C*[a,6]. Show that the series Qo Xa_o[Qn+1 Qn] converges in the topology of the space, that is the series Xao[Qn+1 QaJla) uniformly converges on [a,b] for each q € N: Hint: use the Markov inequality and the Jackson theorem_

Let Qn; n 0,1,2, be the polynomials of best approximation to f € C*[a,6]. Show that the series Qo Xa_o[Qn+1 Qn] converges in the topology of the space, that is the series Xao[Qn+1 QaJla) uniformly converges on [a,b] for each q € N: Hint: use the Markov inequality and the Jackson theorem_



Answers

Let $(s_{n})$ be a sequence such that \[








|s_{n+1}-s_{n}| < 2^{-n} \qquad \text{for all $n \in \mathbb{N}$.}






\]






Prove $(s_{n})$ is a Cauchy sequence and hence a convergent sequence.

In this problem, but told to you steering 9.11 to determine in this series converges or diverges using the P series. So we have a series of, We have a series of one over and Set of power 1.4, 1.4. So we know the piece series. We know P is equal to 1.4, and the P series tells us that if P, Which is 1.4 Is greater than one, we know that P 1.4 Is greater than one, then therefore this series converges To the series one of the end To the 1.4 and emerges.

Hi here it is. Given that a random variable Y in follows a binomial distributions with number of trials and and property of success P. Now we have to first we have to prove. First problem is Y. And delivery by N converges in probably to do small P. That is property of success. And in other words one capital Y. And delivered via in conversation priority to one minus small people. Okay. And third problem is why in Hollywood it brian into one minus Y. In Hollywood brian converges in probability to Being do one -P. Okay, now going forward to the first problem see as Y and follows financial institutions parameters Mp. So expectation of Y. N equals two. Np great expectation of why and who liberated by N equals to peak. Okay, now variants of Ryan equals two N. B into one minus P. Right? No very unsolved Y and Hollywood by N. Because two np one minus B. Hollywood by N square. This is paying to 1 -4. Who delivered by an right now variants of this means expectation of buy and hold a word by end minus expectation of this country. This is speak whole square equals to this. Okay, now here we use championships equally to prove Y. And holy weighted by N converges improved too small people. Okay, so by chair reception equality we have jimmy ships inequality by championships inequality we have expect sorry, probability mud? Why and who liberated by end minus small P more close together than equals two. Excellent is less than equals two expectation al mud, Y and Holy war brian minus p. Not closed the square. Holy word by excellent square where Excellent change to zero and this is a positive quantity. Okay, Now see as it is a square quantity, so more of these square means only 1 -2 whole square. So this is equal to expectation of why? In minus and minus b whole square holding very, very excellent square. No, this is equals two variants of Y and holy very brian. Well, there will be a solemn square Great. Now this is equal to P into one minus P delivered by N into excellent scratch. Now when intends 2050 this tends to zero as intends to infinity. Now we can ride probability maude why and who delivered by end minus b Not close get it then equals to have sarin is less than equals to zero when it is a limit limit Intends to infinity. Now as it is a probability as this is a probability super ability must get up then equals to zero Because it is the Portobello priorities between 0-1. So from here to satisfy this inequality this limit, this quantity must be equal to zero. This in place limit intends to infinity probability marred why unholy worried by end minus b mod clothes, Get it then equals two. Absalon equals to zero training. So this is from one of the definitions of convergence improbability. Okay, so we can write The random variable y and deliver it by N converges in probably two small p as N tends to infinity. This is our first problem. Right now, going, going forward to the second problem let us take a function G X equals to one minus X. For all eggs belongs to our Now clearly X the G F. The function X is continuous at X belongs to her. So we can write G X. The function G X is continuous and X equals two b. Because limit extends to b minus G X Equals two. GP equals to limit extends to Exchange two p plus GX. Deep is the parametric function. Right? So as Y and whole Everett by N converges in probability to pee. So this implies G. Of Y. And holy worried by N converges in probability to G P. Right? So it can recognize one minus Y. And hold over it by n. Converges improbability to one minus small. Peak has entrance to infinity. Right? These are second problem. Now going forward to the large problem, it does take a function F X Equals two x into 1 -6 where eggs belongs to our now clearly fx is continuous and X equals to peak because limit extends to being my nurse If X equals two F P equals two limit exchange to be plus effects. Okay, so and Why and who delivered by N converges in probably two small b. We can write one up by one of the definitions of convergence. One of the cure aims of convergence improbability, F or Y. N by N converges to F. P improbabilities. So this means why and by N into one minus Y. And by N converges in probability to be into one minus speak. Yes intends to infinity. This is our time. Thank you.

Very serious here, So this is equivalent to one over end to the 1/5. How are here? That means P is equal to 1/5, which is less than Able to one hear between zero and what. So therefore, this is the serious diverges for the pieces.

In this problem were given that a sequence essen is such that The magnitude of difference between the end plus one term and the entire term is always less than do you raise to the power of negative and so we can consider the maximum difference rather the magnitude of the maximum difference. Mhm. Being equal to s to the Brother, 2 to the power negative end. Now a cautious sequence is defined as a sequence in which the subject subsequent terms become arbitrarily close to each other. So If a seven is a sequence than the magnitude of a plus and And a seven plus 1 minus a seven, mhm is equal to zero as N approaches infinity. So they become arbitrarily close to each other. Now to prove that this sequence is a cautious sequence. We can dig the limit as N tends to infinity on both sides. Yeah. Mhm Okay, okay, Okay. Mhm. What people saw limit uh and tends to infinity of due to the negative end and we can solve for the limit. What the fuck? Mhm. Okay. And it's actually a pretty easy limit to solve. So that's the limit as intends to infinity of two to the end. Now, as N approaches infinity, due to the end also approaches infinity, so and is in the in the exponent and as and increases due to the end blows up to infinity and that means that one over to the end. Yeah, Ghost zero. So our limit evaluates 20. Yeah. So the limit as n tends to infinity As 7-plus 1 minus S seven Rather their magnitude equal zero and therefore as intends to infinity the subsequent terms right, subsequent terms Of the sequence, S seven become arbitrarily close, become are bit rarely close, and therefore the given sequence is a cause she sequence. And since cautious sequences always can converge so cause she sequences in words follows that the sequence S N. Don't forget that also. Yeah. Okay. And bridges and there is our solution to this broad.


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