Math Methods 12. Portfolio Assignment 4 Type II. 1. The general formula for the binomial expansion of ( x + y) n is given by 2! 3!

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1 Math Methods Portfolio Assignment 4 Type II Name: Rajesh Swaminathan Block: D Date: 4-Mar-5 ANALYSIS OF A QUARTIC FUNCTION. The general formula for the binomial epansion of ( + y) n is given by nn ( ) nn ( )( n) ( + y) = + n y+ y + y + + ny + y!! n n n n n n n Substituting = a, y = band n = yields ( a b) a a b ab b + = () The same result may be obtained by multiplying out the factors: a+ b = a+ b a+ b a+ b = ( a + ab+ b )( a+ b) = a + a b+ ab + b ( ) ( )( )( ). Letting a = and b= in () yields ( ) = () + () ( ) + ()( ) + ( ) = Using the result from question, = + ( ) (8 6 ) = Re-arranging the terms in decreasing degrees, we obtain 4 ( ) = which is of the form p q. Comparing coefficients we find p = and q = y = ( ) = y 6 8 dy = + + d Mathematical Methods Portfolio

2 5. According to the chain rule, ( f o g)( ) = f ( g( )) g ( ). Alternately, if y = f( g( )) dy dy du and u = g( ) then y = f( u) and =. d du d Since the function y = ( ) is a composition of two functions, namely f ( ) = ( ) and g ( ) =, we must obey the chain rule while performing the differentiation. If we let u represent the factor in y = ( ), we have y = ( ) y = u dy d = u d d = [ ] ( ) ( ) =( ) The student s result ( ) is incorrect because he has failed to multiply his result by the derivative of the inside function, namely ( ). In doing so, he has inadvertently lost the multiplicand - which is the derivative of( ). 6. We obtain the same result as question 4 if we use a combination of product and chain rules to differentiate y = ( ) : y = ( ) dy d d = + d d d = + ( ) [ ] [( ) ] ( ) [( ) ( )] = y = ( ) dy From questions 4 and 6, we have d = + +. d y The second derivative can be computed by differentiating dy d d. d y So, 6 4 d = + When =, y = ()[ ()] =. Mathematical Methods Portfolio

3 Similar substitutions yield d y and d = Conclusions: = () + 6() 4 =. dy = + + = d = 4() 8() 4() 8 Since y = when =, is a zero of the function f ( ) = ( ). Consequently, the graph of f must touch the -ais at this point. dy Since d = when =, this point is a stationary point of f. In other words, the tangent line at = has gradient and is thus horizontal. d y Since d = when =, it follows that f has an inflection point at =. In other words, f changes the direction of its concavity at the point (, ). dy d y Since = =, it is impossible to say, at this point in time, whether f has a d d relative maimum, a relative minimum, or neither at = by the second derivative test. 8. A rough sketch of y = ( ) can be seen in Figure. The conclusions reached in the previous question are consistent with the graph of f. The last conclusion as to whether f has a relative maimum, a relative minimum, or neither deserves further consideration. Although the second derivative test is inconclusive dy d y since = =, we note that = is a root d d of f and has multiplicity. This is because ( ) divides the polynomial f but 4 ( ) does not. There is a close relationship between the multiplicity of a root of a polynomial and the behavior of the graph in the vicinity of the root. Since the multiplicity m = of the root = is odd (indivisible by ), the graph will be Figure y = ( ) tangent to the -ais at =, will cross the -ais, and will also have an inflection point there. We have already recognized = as being both a stationary as well as an inflection point. However, the fact that the graph crosses the -ais at this point indicates that f has neither a relative maimum nor a relative minimum there. In order for f to have a relative etremum at =, the graph will have to bounce-off the -ais once it touches it. This conclusion is evident in the graph of f in Figure. Mathematical Methods Portfolio

4 9. A sketch graph of the definite integral ( ) d can be seen in Figure. The area enclosed by the graph and the -ais has been shaded and computed to equal.6. Figure y = ( ) d Note: All graphs in this assignment, unless otherwise noted, were generated with GraphCalc, freely available from By the method of u-substitution, it is possible to evaluate the definite integral ( ) d without needing to epand the cubed term. If we make the substitutionu =, it follows that du = d or d = du. This leaves a factor of unresolved in the integrand. However, sinceu =, we have = u. With this substitution so if =, u = = if =, u = = Mathematical Methods Portfolio 4

5 4 = 4 = ( ) 4 5 ( ) d= ( u) u ( du) ( u u ) du u u du u u = 5 5 = 5 8 = 5 =.6 u= which matches the result obtained in question 9. REMARK. The evaluation of the above definite integral is not only tedious but also error-prone since it was performed manually by hand. The same computation was done by the Maima Computer Algebra System (CAS) in less than a millisecond: rajesh@rbo ~ $ maima Maima restarted. (%i) integrate(*(-)^,,, ); 8 (%o) In question, we showed that ( ) = Integration of this quartic polynomial is as follows: Mathematical Methods Portfolio 5

6 4 ( ) d= ( ) d 5 4 = = () () = + 4() + 4() 5 = = 5 =.6 which once again agrees with the result obtained in questions 9 and.. The trapezoidal approimation of a definite integral is an average of the left-hand and right-hand approimations. This approimation is given by b a b a f ( d ) y + y+ + yn + y n [ ] n With f ( ) ( ) =, a =, b = and n = 4 the definite integral may be approimated as follows (the values for the trapezoids being incremented by =.5 each time): 4 ( ) d f() + ( f (.5 ) + f() + f (.5 )) + f() 4 [ + ( ) + ] The absolute error (or margin of error) in this approimation is given by b E = f( ) dt T a n Using this, we find the error in approimation to be =.65 or.65 %.%.6 Mathematical Methods Portfolio 6

7 . The easiest way to find out the number of solutions to the equation ( ) = would be to graph the function y = ( ) and count the number of intersections with the -ais. Since this is a quartic function, we epect to have at most 4 solutions. The graph of y = ( ) is shown in Figure. Figure y = ( ) It is evident from the above graph that the equation solutions. ( ) = has eactly two real Another method to determine the number of solutions would be to graph the function y = over the graph of y = ( ) that was graphed in Figure and count the number of intersections (see Figure 4). It can be seen that the two curves intersect twice and so the equation ( ) = has eactly two real solutions. Figure 4 y = y = ( ) Mathematical Methods Portfolio 7

8 REMARK. If a graphing utility is unavailable, the Maima CAS can give us the number of solutions quickly, thanks to the built-in NROOTS() or REALROOTS() functions. For instance: (%i) REALROOTS(*(-)^-); 5964 (%o) [ = , = ] 5544 We find two real roots for this equation. To confirm, we could use the NROOTS() function. (%i) NROOTS(*(-)^-); (%o) Since we want all roots between = and =, we specify the lower and upper bounds. However, since both solutions are between these bounds, we still obtain. (%i) NROOTS(*(-)^-, -, ); (%o) The equation ( ) = can be re-arranged in two ways: Dividing both sides by If equaled ( ) p ( ) = ( ) = ( ) = = ( ) [we can do this since, then p =. ], we get = ( ). where the second step is justified by recognizing the fact that. If equaled q ( ), then q =. Using the methods of simple iteration and taking =.5 : Mathematical Methods Portfolio 8

9 = =.96 M [ (.5)] = =. [ (.96)] = =.7 [ (.)] = =.67 [ (.68)] 8 = (.5) =.74 M = (.74) =.8945 = (.8945) =.96 = (.9999) =. These two results (accurate to within 4 decimal places) are approimations of the solutions to the equation ( ) = as can be seen in Figure. Each form of the / original equation ( ) =, namely = /( ) and =, yields one of the two solutions. Using other values for, we find that the results converge to the same answer every time. This convergence can be seen graphically in Figure 5. It is helpful to remind ourselves of what actually happens as we perform the iteration. Each term of the iterative sequence n is computed from the previous term using the iterative function. Mathematically, this means that = n+ f( n), where f ( ) is any one of the above two iterative functions. If this process is repeated continuously and the limit taken, we will eventually obtain a solution to the original equation. This is to say that lim n+ = lim f( n), where f ( ) is an iterative function (the re-arranged equation n n solved for ). In our case, lim = lim.67, and n n+ n ( n ) n+ n n lim = lim ( ) =. n The Newton-Raphson method of approimation is given by the following iterative function: f( n ) = n n, n,,,... + f ( ) = n () In order to determine if the Newton-Raphson method can be used to yield the same results, we check to see whether the absolute value of the derivative evaluated at some test value is in between and. In other words, if f ( ) is a differentiable function Mathematical Methods Portfolio 9

10 un ( ) = ( un ( )) un ( ) = un ( ) Figure 5 Mathematical Methods Portfolio

11 Note: The graphs in Figure 5 were generated with Gnuplot, freely available from and we started with some test value, then f ( ) must be less than for the iterative function to converge and yield an approimation. If this condition is not satisfied, we will have a diverging iterative sequence and will never reach an approimation. Furthermore, if f ( ) is positive, we will have a monotonic convergence. Conversely, if f ( ) is negative, we will have an oscillating convergence. The meaning of these two terms is left uneplained for the interested to pursue. Since we are attempting to solve the equation ( ) =, we let so f ( ) = (verify) and () becomes f( ) = ( ), 4 n+ n =, or n+ = n () If we started with a test value =.5, then f (.5) = 4(.5) + 8(.5) 4(.5) + 8 = and can therefore not be used to give a result since the denominator in () evaluates to and is thus undefined. If we used different test values such as =.,.,., we still find that we are unable to use the Newton-Raphson method since in each case f ( ) >. For what value of will f ( ) be less than but not equal to? The answer to this problem can be obtained by graphing f ( ) [see Figure 6] and looking for values of whose y values are not zero and are in the interval (-, ). For this, we solve the equations f ( ) = and f ( ) =. This is a Figure 6 y = straight-forward task that can be achieved with the Maima CAS: (%i) f():= *(-)^-; (%i) define(df(), epand(diff(f(), ))); Mathematical Methods Portfolio

12 (%i) realroots(df()-); (%o) [ = ] 5544 (%i4) realroots(df()+); (%o4) [ = , = -, = ] ,.64, We can thus conclude that in order to be able to solve the equation ( ) = using the Newton-Raphson method, we must use an initial value that is in the interval (.4,.5) U(.5,.64) U(.5,.) U (.,.66). Using any initial value from this interval and (), we obtain the same solutions to the equation ( ) =, namely =.67 and = (verify). Mathematical Methods Portfolio