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# math DVQ

Using Euler’s formula, exp(ix)=cos(x)+isin(x), and the usual rules of exponents, establish De Moivre’s formula,(cos(nθ)+isin(nθ))=(cos(θ)+isin(θ))n.Use DeMoivre’s formula to write the following in terms of sin(θ) and cos(θ).cos(6θ)sin(6θ)One of the properties of thesin(x)andcos(x)that I hope you recall from trigonometry is thatcos(x)is an even function, i.e.cos(−x)=cos(x),whilesin(x)is an odd function, i.e.sin(−x)=−sin(x).We will see that any function can be split into pieces with these symmetries.Given a general function f(x), define fe(x)=f(x)+f(−x)2 and fo(x)=f(x)−f(−x)2.. Showf(x)=fe(x)+fo(x), andfe(x) is an even function and fo(x) is an odd function.So every function can be split into even and odd pieces.Given that f(x)=fe(x)+fo(x) where fe(x) is an even function and fo(x) is an odd function, show thatfe(x)=f(x)+f(−x)2, andfo(x)=f(x)−f(−x)2, andThis shows the decomposition in the previous problem is the unique way to cut a function into even and odd pieces.If we apply the decomposition developed in the previous two problems to the exponential function, we get thehyperbolic functions,cosh(x)sinh(x)=exp(x)+exp(−x)2=exp(x)−exp(−x)2These functions are covered in your calculus text, but sometimes that section is skipped. They are closely related to the usual trigonometric functions and you can define hyperbolic tangent, secant, cosecant, and cotangent in the obvious way (e.g.tanh(x)=sinh(x)cosh(x)). The next few problems give a quick overview of some of their properties.Showcosh(ix)=cos(x), andsinh(ix)=isin(x).So the hyperbolic functions are just rotations of the usual trigonometric functions in the complex plane.Verify the following hyperbolic trig identities.cosh2(x)−sinh2(x)=1cosh(s+t)=cosh(s)cosh(t)+sinh(s)sinh(t)Note that these are almost the same as the corresponding identities for the regular trig functions, except for changes in the signs. You can derive hyperbolic identities corresponding to all the different identities you learned in trigonometry.Invert the formula sinh(x)=exp(x)−exp(−x)2 to write sinh−1(x) in terms of log(x). Note that you will need to use the quadratic formula to get the inverse.Verify the following differentiation rules for the hyperbolic functionsdsinh(x)dx=cosh(x), anddcosh(x)dx=sinh(x).So for the hyperbolic functions you don’t have to try to remember which derivative gets a minus sign.Use the substitution x=sinh(u) to evaluate the integral∫dx1+x2‾‾‾‾‾‾√.This integral can also be evaluated with a trig substitution, but using a hyperbolic substitution simplifies the process.Suppose L is the operator defined by Ly=Dy−3y. Compute L(exp(x)).Suppose L is the operator defined by Ly=D2y+3Dy+2y. Compute L(exp(2x)).Suppose L is the operator defined by Ly=D2y−3Dy+y. Compute L(sin(x)).Suppose L is the operator defined by Ly=yDy. Compute L(2×3).Show Ly=Dy−xy is a linear operator. Remember you must show both parts of the definition hold for all functions y.Show Ly=D2y+3Dy−4y is a linear operator. Remember you must show both parts of the definition hold for all functions y.Show Ly=Dy−xy2 is not a linear operator. Hint: It is sufficient to show L(y+z)≠Ly+Lz. for some choice of y and z.Show Ly=yDy is not a linear operator. Hint: It is sufficient to show L(y+z)≠Ly+Lz. for some choice of y and z.Suppose L is a linear operator and we have three partial particular solutions,yp1 with Lyp1=f1yp2 with Lyp2=f2yp3 with Lyp3=f3Show y=ayp1+byp2+cyp3 is a particular solution to Ly=af1+bf2+cf3.Show exp(x) and exp(2x) are linearly independent.Show exp(rx) and exp(sx) are linearly independent if r≠s.Match the equations on the right with the factorings on the left.(D+x)(D+2)y=0(D+2)(D+x)y=0(D+x)(D+x)y=0y″+(x+2)y′+(2x+1)y=0y″+(x+2)y′+2xy=0y″+2xy′+(x2+1)y=0Match the equations on the right with the factorings on the left.(D+x)(D−x)y=0(D−x)(D+x)y=0(D+x2)(D−1)y=0y″+(x2−1)y′−x2y=0y″−(x2+1)y=0y″−(x2−1)y=0