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Julianne L.
Mathematics major, Education minor at Binghamton University
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Linear Algebra
TutorMe
Question:

Let A and v be the matrix and vector given by: A=|0 0 2| |-3 1 6| |0 0 1 | v= |1| |3| |0| Find Av. What does this say about eigenvectors and eigenvalues of A? Find the characteristic polynomial of A. Find all eigenvalues of A.

Julianne L.
Answer:

Av is the product of A and v. It is a 3x1 matrix where the entry in each row is the product of the corresponding row of A dot multiplied with the column on v. The first row is (0)(1)+(0)(3)+(2)(0)=0. The second row is (-3)(1)+(1)(3)+(6)(0)=0. The third row is (0)(1)+(0)(3)+(1)(0)=0. Therefore Av= | 0 0 0 | Because Av=0v, v is an eigenvector of A with eigenvalue 0 by the definitions of eigenvector and eigenvalue The characteristic polynomial of A is the determinant of the matrix (A-MI) I will be using M in place of the commonly used lambda. This matrix is | -M 0 2 | | -3 1-M 6 | | 0 0 1-M | The determinant is found by taking (-1)^(1+row#)*(the first entry in the row)*det(matrix obtained by deleting that row and the first column) for each row and adding these values. So the determinant is -M*det |1-M 6 | + 0 + 2*det |-3 1-M | | 0 1-M | | 0 0 | =-M *((1-M)*(1-M)) = -M (1-2M+M^2) = -M+2M^2-M^3 = -M*(1-M)^2 Any of these are proper expressions of the characteristic polynomial. The eigenvalues are the roots of the characteristic polynomial so all you need to do to find them is to set -M*(1-M)^2=0 and solve for M. We find that in this case M=0 and M=1. These are the eigenvalues of A.

Calculus
TutorMe
Question:

Find two linearly independent power series solutions of the differential equation y''-xy'-2y=0 Write the answer to at least x^5.

Julianne L.
Answer:

We will be substituting in y=(summation n=0->inf) ((Cn)*x^n) Cn is a coefficient C subscript n By differentiating we find that y'=(summation n=1->inf)( (Cn)*n*x^(n-1) ) and that y''=(summation n=2->inf) ( (Cn)*n*(n-1)*x^(n-2) ) By substituting these into the original equation, our new equation is: (sum n=2->inf)((Cn)*n*(n-1)*x^(n-2)) -x*(sum n=1->inf)((Cn)*n*x^(n-1)) -2*(sum n=0->inf) ((Cn)*x^n) =0 the (-x) and (-2) can be moved inside the summations: (sum n=2->inf)((Cn)*n*(n-1)*x^(n-2)) +(sum n=1->inf)(-(Cn)*n*x^n) +(sum n=0->inf) (-2(Cn)*x^n)=0 We want to add the summations together so we will reindex the first using k=n-2 and the second two by k=n so that all of the summations contain x^k: (sum k=0->inf)((C(k+2))(k+2)(k+1)x^k) +(sum k=1->inf)(-(Ck)*k*x^k) +(sum k=0->inf) (-2(Ck)*x^k) =0 Now if we take out the k=0 term for the first and third summations, we can add all of the summations into one: 2*C2-2*C0+ (sum k=1->inf)( [(C(K+2))(k+2)(k+1)-(Ck)*k-2(Ck)]x^k ) =0 Using 2C2-2C0=0 We find that C0=C2 must be true Now we plug in values for k starting at 1 to find equations for the other coefficients, I will skip showing the computations for these as it is just plugging in integers for the Ks within the summation. Using k=1 we find that C3=(1/2)C1 with k=2 C4=(1/3)C2 and k=3 C5=(1/4)C3 Now all we need to do is pick a C0 and C1 for our two power series y1 and y2 For y1 we will choose C0=0 and C1=1. Then C2=0 C3=1/2 C4=0 C5=1/8 For y2 choose C0=1 and C1=0. Then C2=1 C3=0 C4=1/3 C5=0 Finally, we can put together our solutions. The Cn are the coefficients for each x^n term so y1= x + (1/2)x^3 + (1/8)x^5 and y2=1 + x^2 + (1/3)x^4

Calculus
TutorMe
Question:

What is the triple integral of e^z over E, the space inclosed by the paraboloid z=1+x^2+y^2, the cylinder x^2+y^2=25, and the (x,y) plane. Use cylindrical coordinates.

Julianne L.
Answer:

By subtracting (25=x^2+y^2) from (z=1+x^2+y^2) and simplifying, we find that z=24 is where the cylinder and paraboloid overlap. Drawing a picture of the space E shows that the paraboloid lies above the cylinder under z=24, therefore the top bound for z is the paraboloid and the bottom bound for z is the (x,y) plane or z=0. To find the triple integral using cylindrical coordinates, the limits must be in terms of z, r, and theta. The lower limit for z is z=0 and the upper limit is z=1+x^2+y^2 which is equivalent to z=1+r^2 in cylindrical coordinates since r^2=x^2+y^2. The largest radius is the radius of the cylinder x^2+y^2=25 which is 5, therefore the radius spans from 0 to 5. Theta spans from 0 to 2(pi) since the cylinder and paraboloid span a full 2(pi) radians. You additionally need to multiply an "r" within the integral since it is the Jacobian of cylindrical coordinates. The integral is then (0->2(pi)) (0->5) (0->(1+r^2)) r*e^z dz dr d(theta) Beginning with the innermost integral, r*e^z integrated with respect to z is still r*e^z and after substituting in the limits, this comes out to be (r*e^(1+r^2)-r) (r*e^(1+r^2)-r) integrated with respect to r is (1/2)e^(1+r^2)-(1/2)r^2 and this becomes (1/2)e^26-(1/2)e-(25/2) after substituting in the limits (1/2)e^26-(1/2)e-(25/2) integrated with respect to theta simply adds a theta to each element and substituting in the limits give us the final answer: (e^26)(pi)-25(pi)-e(pi)

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