#$&* course MTH 279 3:01 2/25 q_a_09*********************************************
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Given Solution: Self-critique: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Suppose f(t) = e^(2 t) and g(t) = e^(-3 t). Show that the determinant of the 2 x 2 matrix [ f, g ; f ', g ' ] (that is the matrix with first row f g and second row f ' g ' ) is not identically zero. **** #$&* Suppose f(t) = t^2 and g(t) = t^3. Show that the determinant of the 2 x 2 matrix [ f, g; f ', g ' ] is not identically zero. **** #$&* Show that if f is a constant multiple of g, then the determinant of the 2 x 2 matrix [ f, g; f ', g ' ] is identically zero. **** #$&* YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: det [[e^2t, e^-3t][2e^2t, -3e^-3t]]= (e^2t * -3e^-3t) - (e^-3t * 2e^2t) = -2e^-t - 2e^-t = -4e^-t det [[t^2, t^3][2t, 3t^2]] = 3t^2*t^2 - 2t*t^3 = 3t^4- 2t^4= t^4 f= C*g --> det [[f, g][f', g']] ---> det [[Cg, g][Cg', g']] = Cg*g' - Cg'*g = 0 this is clearly holds. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Self-critique: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Recall that the span of two vectors v_1 and v_2 is the set of all their possible linear combinations v = a_1 v_1 + a_2 v_2, where a_1 and a_2 are scalar quantities. Recall also that a set of vectors is linearly independent if there is no nontrivial linear combination of those vectors which is equal to zero, and that this condition is equivalent to saying that none of the vectors can be expressed as a linear combination of the others. Functions can be regarded as vectors, with the scalars being the real numbers. Two functions are said to be linearly independent if neither is a constant multiple of the other. If the functions are f and g, then this condition is equivalent to saying that the determinant of [ f, g; f ', g ' ] is not identically zero. In this case the set of linear combinations A f + B g is said to span a function space of two dimensions. Show that the two functions f(t) = sin(t) and g(t) = cos(t) are linearly independent. **** #$&* Show that the function 2 sin(t) + 4 cos(t) is in the span of the set {f, g}. **** #$&* Show that the function t sin(t) - 2 cos(t) is not in the span of the set {f, g}. **** #$&* Show that any function in the set {e^t, e^(-t) } is linearly independent. **** #$&* Show that any function in the span of the set {e^t, e^(-t)} is a solution of the differential equation y '' - y = 0. **** #$&* YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: det [[sin t, cos t][cos t, -sin t]] = -sin^2 t - cos^2 t ≠ 0 a_1 f + a_2 g--> 2 sin t + 4 cos t where a_1 = 2 and a_2 = 4 Here a_1 = t which is not a constant so this is not a constant multiple. I'm not sure about what this is asking. The set only has two functions that are obviously LinInd and any {Ae^t, Be^-t }
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Given Solution: Self-critique: ------------------------------------------------ Self-critique Rating: `q004. e^(i * theta) = cos(theta) + i sin(theta). This is easily seen by forming the Taylor expansion of e^x, sin(x) and cos(x) and substituting x = i * theta in the first series. The equation e^(i * theta) = cos(theta) + i sin(theta) is called Euler's Identity. Do this. Whether or not you were successful in the preceding, use the laws of exponents to write the product e^(i * alpha) * e^(i * beta) as an expression with a single base and a single exponent. **** #$&* Use Euler's Identity to write this expression in terms of sines and cosines. **** #$&* Use Euler's Identity to write e^(i * alpha) and e^(i * beta) in terms of the sine and cosine function. **** #$&* Write out the product e^(i * alpha) * e^(i * beta) in terms of sines and cosines and simplify. **** #$&* Look at what you've done and see if you haven't derived the formulas for sin( alpha + beta) and cos(alpha + beta). **** #$&* YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I can't seem to quite understand about how to setup the Taylor expansion with the i*theta substitution. e^(i*alpha)*e^(i*beta) = e^(i*alpha+i*beta) = e^(i(alpha+beta)) e^(i(alpha+beta)) = cos (alpha+beta) + i sin (alpha+beta) e^(i*alpha) = cos (alpha) + i sin(alpha) e^(i*beta) = cos (beta) + i sin (beta) e^(i*alpha)*e^(i*beta)= (cos (alpha) + i sin(alpha))*(cos (beta) + i sin (beta)) = (cos(alpha)cos(beta))+ (cos(alpha)i sin (beta)) + i sin(alpha)cos(beta) + i^2 sin(alpha)sin(beta) [cos(alpha)cos(beta) - sin(alpha)sin(beta)] + [i(sin(alpha)cos(beta)+cos(alpha)sin(beta))] This lefthand expression is for cos (alpha+beta) and the imaginary is for sin(alpha+beta)
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Given Solution: Self-critique: ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Show that e^((i + 2) * t) and e^( (2 - i) * t) are linearly independent. **** #$&* If c_1 = a_1 + i * b_1 and c_2 = a_2 + i * b_2, then what is the general form of c_1 e^( (i + 2) * t) + c_2 e^( (2 - i) * t), in terms of sines and cosines? You get something pretty messy, so simplify it as much as you can. **** #$&* YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: det [[e^((i+2)t), e^((2-i)t)][(i+2)e^(i+2)t, (2-i)e^(2-i)t]] = (e^((i+2)t)*(2-i)e^(2-i)t) - (e^((2-i)t)(i+2)*e^((i+2)t) = (2-i)e^4t - (i+2)e^4t = (4-2i)e^4t = 4e^4t - i*2e^4t
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