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@ @ + @]ul 100%1.52 PMMATH2O4Home My courses Lecture Notes Math204 Numerical AnalysisQuestion 7Answer savedMarked out ofFlag questionConsider f(a) = 3 cos(2x) over ...

Question

@ @ + @]ul 100%1.52 PMMATH2O4Home My courses Lecture Notes Math204 Numerical AnalysisQuestion 7Answer savedMarked out ofFlag questionConsider f(a) = 3 cos(2x) over (0.5,1). Find the error bound for Ej (w). (Error for the Linear Lagrange polynomialsSelect one; a,0.125586*0 *q0.012d. 1.253None of themClear my_choicePrevious pageNext page

@ @ + @] ul 100% 1.52 PM MATH2O4 Home My courses Lecture Notes Math204 Numerical Analysis Question 7 Answer saved Marked out of Flag question Consider f(a) = 3 cos(2x) over (0.5,1). Find the error bound for Ej (w). (Error for the Linear Lagrange polynomials Select one; a,0.125 586*0 *q 0.012 d. 1.253 None of them Clear my_choice Previous page Next page



Answers

Use the remainder to find a bound on the error in approximating the following quantities with the nth-order Taylor polynomial centered at 0. Estimates are not unique. $$\cos 0.45, n=3$$

For this problem, we are we are told to suppose that we use P two of ax equals one minus x squared over two. To approximate coast of 0.5. Were then asked to give a bound on the error of the approximation based on in part a the alternating series error bound and in part B, the lagrange error formula. So for part A. Using the alternating series bound, we know that the error is going to be bounded above By the subsequent term in the series. So we know then that it's going to be bounded above by 0.5 to the power of four over for factorial. And in part B, we know that using the lagrange error formula, The error that should be our two FX is going to be bounded above by the maximum value of coast on that interval, which would just be one. Uh it would be bounded above by Ux X or by 0.5 to the power of three divided by three factorial.

Yeah. In per day we want to find a bound and the arrow incurring approximating the function F. X equals log acts. By the fourth taylor party nominee first went to find a capital M. Which is greater than Echo two. The absolute value Of the 5th order derivative of the function. And the 5th started. The relative echoes 24 over Eric's to the 5th. This derivative riches it's maximo at one Enhance Capital. Leon Michels 24. So the residual At 1.5 equals capital am over five. Factorial Times 0.5 to the power so far. Medicals .006. Too far. Then for part B we know that the fourth order taylor ponte phenomenal for the function he calls X minus one miners half X -1 Squared plus one third Ex miners one cubed miners. 1/4 x -1 to the 4th enhance. It's integral from 1 to 1.5. First refines the anti derivative of this function. Dhs a half x minus one squared miners. 16 That's -1 Cube Plus 1/12 That's -1 to the 4th Miners. 1/20 X -1 to the Fix. And this value is 0.10 78125. And then for percy the arrow. The bounds of the arrow for the approximation of the integral is just The Indian Girl from 1 to 1.5. And then this arrow Which is just .5 times this arrow which equals point 00 straight 125 And in part D. Yeah. The real integral of the function from 1 to 1.5 1st. We use the integration by parts and we can have this results. Richie calls three half Slog three halves miners, one calf, which is approximately 0.1 zero 81 977 So the arrow is a difference of this number and this number which echoes zero 385

In part A. We want to find abound in the arrow incurred in approximating Fx by the third. Taylor point phenomena. Yes we need to find a capital M. Which is greater than Echo two. The force ordered derivative of FX. Which is E. To the power of minors efforts. And When Eric Seiko zero this force order derogative reaches its maximum Which is what hence capital er Michels one. So the residual at one Because Cafeteria over four factorial times while to the force magical 0.0 4167. And then for barbie From Exercise one We know that the third order taylor part phenomenal for this function echoes minus X plus a half X squared My nurse 1 6 x cubed. So it's into a girl. From There we go to one first we find its anti derivative function which is ex miners a half X squared. Close one six x cubed -1 24th. Now it's 24th from went to zero Which equals .6- five. In Percy the bounce in the arrow in the approximation of this integral is just Integral from 0 to 1. And the bounds of the arrow, which has a point zero 41 6 7 Articles one times this arrow, which is just the same as this. Every in party. The integral of this function from 0 to 1 first the anti derivative function is miners. E. To the power of minor thanks and this equals miners E. To the power minus one plus one, which is approximately point 63 21 enhance the arrow echoes points 6321 miners point six 25 Kg calls point 00 Silver want to

For this problem we are told to suppose that we use P three of x equals x minus x cubed over six to approximate sign of 0.5. Were then asked to give a bound on the error of the approximation based on in part a the alternating series error bound and in be the Lagrange error formula. So in part a using the alternating series error bound. Uh we know that sign of ex uh satisfies the criteria for using the alternating series bound. So we can say then that the error is going to be less than or equal to what would be the third term in the series. So we can say then that the error, it's going to be less than or equal to the absolute value of x power of five over five factorial. Which in this case that would be a sign of zero or not sign of but that would be 0.5 to the power of 5/5 factorial. Which I'm going to leave in that form. You can calculate that in decimal form if you so choose in part B. Were asked to use the La granja era formula. So first we need to figure out what the fourth derivative of sine is going to be. We know that it would go sign coast, negative coast or excuse me, negative sign negative coast positive signs. So that's 1234. So actually we would want actually That would be 01234. So we know that our fourth derivative of our function Evaluated at 0.5 is going to be and equal then, or is going to be bounded above by sign. So that is itself going to be all. We can say that 20 Bounded above in magnitude. Just buy one. So we can say then that the error at the third term, there are three of X is going to be less than or equal to One over for factorial. That's not for I it's for factorial times. We want to be evaluating this at 0.5 times 0.5 zero, or just times 0.5 To the power of four.


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