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QUESTION 28The Maclaurin series for cos(x) converges for all x TrueFalseQUESTION 29The sum of the series 2 1 n70 3"nlTrueFalseQUESTION 30(-1)"We could use...

Question

QUESTION 28The Maclaurin series for cos(x) converges for all x TrueFalseQUESTION 29The sum of the series 2 1 n70 3"nlTrueFalseQUESTION 30(-1)"We could use the Ratio Test to deterrine if the series n =0 Vn?+2n -3convergent:TrueFalse

QUESTION 28 The Maclaurin series for cos(x) converges for all x True False QUESTION 29 The sum of the series 2 1 n70 3"nl True False QUESTION 30 (-1)" We could use the Ratio Test to deterrine if the series n =0 Vn?+2n -3 convergent: True False



Answers

Decide if the statements are true or false. Give an explanation for your answer. If the power series $\sum C_{n} x^{n}$ converges for $x=2$, then it converges for $x$

In this problem, we need to determine whether or not a given statement is true or false. Now the giving statement is that the power series C n x to the power end converges for X equals to two. Then we can conclude that the power series was also converts for X is equal to one. Now in order to determine whether or not this is true or false. First of all, let us consider I to be the interval of convergence for this. Power sees. Now the series has extra deep our head. So the interval of convergence will be centered at the point X equals to zero. Also, since the power series converges for X equals to two, We can say that the interval open -2 and closed at two. This interval will be a subset of the interval of convergence. This interval is centered at 0 -2. May or may not be included in it. So we can consider that to be open and two will be included in this interval. So this interval is a subset of the interval of convergence. Now we need to show that we need to determine whether it converges for X equals to one and know that the number one is uh element of this interval minus two, coma, too. And since this interval is a subset of I, this will imply That the .1 belongs to the interval of convergence, I since one belongs to the undeveloped convergence, hence the power series converges for X equals to one, and thus the given statement is true.

In this problem, we need to determine whether a given statement is true or false. Now in this problem, the given statement is that if a power series summation C n X to the power end converges for X is equal to one, then the power issues will also convert for X equals to two. Now, in order to determine whether this is true or false, let us consider that the radius of convergence of this series is are in that case this series will converts for all values of excess. That models of X is less than our. Now it is said that the series converges for X equals to one, so one will satisfy this inequality and thus will have models of one is less than R or R is greater than models of one, which means that our is greater than one. Now we need to determine whether the power series converges for X equals to two. Now it will convert if this inequality is satisfied for X equals to two. Now, if we substitute X equals to two, then we'll get that The model is of two is less than up and that will imply that R is greater than the models of two, which means that our is greater than two. Now, this may or may not be true. We know for a fact that the series converges and X equals to one And hence we know that our is better than one, but this does not imply that are will be greater than to for example, if R is equal to 1.5, then it will be greater than one, but it will not be greater than two. So this means that we cannot guarantee that the point x equals to two lies within the interval of convergence. And hence we cannot conclude that the power series will for certain convert for X equals two. Hence the given statement is false.

In this problem, we need to determine whether a given statement or is true or false. Now in this question, the given statement is that if the power series submission, see and actually Power End converges for X equals to one, then the power series converges for X equals to two. Now, for determining whether or not this is true or false, let us consider I to be the interval of convergence of this power series. Now, since this is X to the power and so the interval of convergence I will be centered at the point X equals to zero. Also, it has been said that the series converges for X is equal to one. So that means that the interval open minus one and closed at one. This will be a subset of the interval of convergence. Now we do not know about the point minus one, but we do know about the 10.1. That it convert this. So this interval which is centered at zero must be a subset of the interval of convergence. I now we need to figure out whether the part series converges for X equals to two, and we can see that too does not belong to this interval, it lies out of this interval. And since it lies beyond this interval we cannot guarantee that you will belong to the interval of convergence. I it is possible that the interval of convergence maybe supposed minus 1.5 comma 1.5 in this case too, will not belong to that interval of convergence, but minus one comma one is a sub set of this interval of convergence. So that means that we cannot guarantee that you will belong to the interval of convergence and hence we cannot conclude that the parties will converse for extra close to two. Hence the given statement is false.

For this problem, we want to determine if the statement is true or false, assuming that the taylor series for a function converges to that function. We want to give an explanation for the answer where the statement is to find the taylor series for sine X plus kovacs about any point. You add the taylor series for syntax and cossacks about that point. So to begin, we can think about what the definition of a taylor series is. Now, it's going to be some from any equal zero up to infinity of the 10th derivative of our function evaluated at a point. Let's call that a times X minus a divided by and factorial. Now we know that differentiation is a linear operation. So if we take the derivative of F of X plus G fx, that's the same thing as the derivative of F plus the derivative of G. So if we imagine saying that F is going to be Sin X plus Cossacks, then our taylor series of Sin X plus Cossacks, it's going to be an equal, or the sum from n equals zero up to infinity of the 10th derivative of sine of X, which I'll write as Synnex time or to the power of bracket. And so that's the end derivative plus the anti derivative of casa backs. Yeah, right times X minus A All divided by N factorial. Which we can then right as the sum from n E equals zero up to infinity of the anti derivative of sine of X. Uh with the way that I have things written, I'm not able to specify that we're evaluating at A but we are um so we have and derivative of sine of X times X minus a. Mhm over N factorial plus the sum from an equal zero up to infinity of the end derivative of cosine X evaluated at a times x minus a over N factorial, which is just equal to the taylor series for sine of X. Or sign of facts about a just equal sign of a plus Kosovo. Or actually, it would be accurate to say sign of exposed coast of acts expanded out about A. So we can say that that statement is in fact true.


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