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Let’s now look at the second derivative of the function f. f2 = diff(f,2) f2 = 32/(5+4*cos(x))^3*sin(x)^2+4/(5+4*cos(x))^2*cos(x) 1-34 Calculus Equivalently, we can type f2 = diff(f,x,2). The default scaling in ezplot cuts off part of f2’s graph. Set the axes limits manually to see the entire function. ezplot(f2) axis([–2*pi 2*pi –5 2]) 32/(5+4*cos(x))^3*sin(x)^2+4/(5+4*cos(x))^2*cos(x) 2 1 0 −1 −2 −3 −4 −5 −6 −4 −2 0 x 2 4 6 From the graph, it appears that the values of f''(x) lie between -4 and 1.

Adds to 1/(1-x), provided |x| < 1. Three summations are demonstrated below. syms x k s1 = symsum(1/k^2,1,inf) s2 = symsum(x^k,k,0,inf) s1 = 1/6*pi^2 s2 = -1/(x-1) 1-30 Calculus Taylor Series The statement T = taylor(f,8) returns T = 1/9+2/81*x^2+5/1458*x^4+49/131220*x^6 which is all the terms up to, but not including, order eight (O(x8)) in the Taylor series for f(x). ∞ ∑ (x – a) n (n ) f (a) ----------------n! n=0 Technically, T is a MacLaurin series, since its basepoint is a = 0. These commands syms x g = exp(x*sin(x)) t = taylor(g,12,2) generate the first 12 nonzero terms of the Taylor series for g about x = 2.

Mathematical Operation ∫ n+1 x n x dx = ------------n+1 π⁄2 ∫ sin ( 2x ) dx = 1 MATLAB Command int(x^n) or int(x^n,x) int(sin(2*x),0,pi/2) or int(sin(2*x),x,0,pi/2) 0 g = cos ( at + b ) ∫ g ( t ) dt = sin ( at + b ) ⁄ a ∫ J1 ( z ) dz = – J0 ( z ) g = cos(a*t + b) int(g) or int(g,t) int(besselj(1,z)) or int(besselj(1,z),z) In contrast to differentiation, symbolic integration is a more complicated task. A number of difficulties can arise in computing the integral. The antiderivative, F, may not exist in closed form; it may define an unfamiliar function; it may exist, but the software can’t find the antiderivative; the software could find it on a larger computer, but runs out of time or memory on the available machine.

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