BUSI3119-E1 ADVANCED CALCULUS FOR BUSINESS, ECONOMICS AND FINANCE 2020-2021
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BUSI3119-E1
A LEVEL 3 MODULE, AUTUMN SEMESTER 2020-2021
ADVANCED CALCULUS FOR BUSINESS, ECONOMICS AND FINANCE
1. (a) Prove that the intersection of any two subspaces of V is a subspace of V. (13 Marks)
Sol: Suppose that and are any two subspaces of . Because and are subspaces of , so we have the additive identity 0 is in both and , hence it is in the intersection of and .
Suppose that . Then, it is clear that and That implies and as both U and W are subspaces of V with both being closed under addition. Hence, we have . This shows that is closed under addition.
Suppose that . Then, we have and . Because U and W are subspaces of V, so we have and where . That means we have . This shows that is closed under scalar multiplication.
Therefore, by showing that 0 is in , and is closed under addition as well as under scalar multiplication, we proved that the intersection of any two subspaces of V is a subspace of V.
(b) Suppose and are subspaces of such that , and =. Prove that . (7 Marks)
Sol: By the fundamental theorem of linear mapping, we know that
Because , this implies that
Hence, we have .
2. (a) Find all eigenvalues and eigenvectors of defined by
(12 Marks)
Sol: Suppose is an eigenvalue of . For this particular, the eigenvalue-eigenvector equation becomes the system of equations
Based on the above equations, we see that can be chosen freely and the other variables can be defined in terms of . So we have
Thus, for each being an eigenvalue of , we can find the corresponding eigenvector as
(b) Suppose is invertible and is a nonzero constant. Prove that is an eigenvalue of if and only if is an eigenvalue of . (8 Marks)
Sol: Suppose is an eigenvalue of . There exists a nonzero vector such that
. Multiplied both sides of the equation by , we obtain Thus, we have . That is to say that is an eigenvalue of .
To prove the implication in the other direction, replace by , and by , and then apply the result from the paragraph above.
3. Solve the following differential equation
(20 Marks)
Sol: The general solution of the homogeneous equation is found by first solving its associated characteristic equation
.
The above equation has a repeated root which equals to 5. Hence, the general solution of the differential equation is of the following form
where are constant.
While the particular solution of the given differential equation, when it is substituted into the differential equation it should produce the terms on the right hand side of the given differential equation. Hence, we have
where , and are constant.
Substitute the PS into the differential equation, by equalizing the coefficients for each of the corresponding terms on both sides of the resulting equation, one can find the values for , and hence the particular solution
for .
Therefore, for , the solution of the differential equation is given by
.
When, the solution of the differential equation is given by
.
4. Solve the initial value problems
(a) , (8 Marks)
Sol: This question can be solved by using separation of variables method, which is to rewrite the above equation so that each of two variables occurs on a different side of the equation. Next, one could integrate the terms on both side of the equation with the lower bound of the integrals set to be zero while its upper bound equals to t. Then, it is not difficult to find the solution equals to
(b) , (12 Marks)
Sol: This question can be solved by using separation of variables method, which is to rewrite the above equation so that each of two variables occurs on a different side of the equation. Next, one could integrate the terms on both side of the equation with the lower bound of the integrals set to be 1 while its upper bound equals to t. Then, it requires bit more work than part a) to find the solution which equals
5. Find the solution of the variational problem
(20 Marks)
Sol:
2022-08-02