MATH 2FM3 Lecture Notes - Lecture 13: Accrued Interest, Discounting

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ivoryprairie-dog119

7 Nov 2015

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3. 1 the amortization method of loan repayment: suppose a loan l is taken at time 0 and is to be repaid with payments of amounts. K1, k2, kn at times 1, 2, , n. if the interest rate per period is i, the equation value at time 0 for this loan is. L = k1 + k2 2 + + kn n, with = 1. Ob1 = l(1 + i) k1 = l (k1 l i), with l i being the accrued interest on the original principal for rst period; k1 is the. Rst payment so the remainder k1 l i ( rst payment minus accrued interest also called principal repaid) is applied to reduce principal; let us denote ob0 = l and. I1 = ob0 i, p r1 = k1 i1, then. Ob1 = ob0 (k1 i1) = ob0 p r1. This algorithm can be extended to the next periods:

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Below is my question and a solution to this question with k = 8c. Ineed to find the solution with k = 2c. I have tried using thisexample solution to answer the problem with k = 2c, but nothing Iput is accepted. Please help me figure this out with k = 2c, not k= 8c.

A smokestack deposits soot on the ground with a concentrationinversely proportional to the square of the distance from thestack. With two smokestacksdmilesapart, the concentration of the combined deposits on the linejoining them, at a distancexfromone stack, is given by

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taupellama68

Consider the equation (z+1)^(2n)-1=0.

1) Fink the solution set for Z(k) (k=0,1,2,...) andfor what values of k are the solutions for z purely imaginary.

Question, I think I have a reasonable solution to thefirst part of the question, but I'm not totally convinced for mysolutions for purely imaginary numbers my answer seems to beindependant of n which doesn't sit well with me. I'v included somepics of my demonstrations, for some reason I couldn't upload mysecond pic.Any help where I went wrong would be awesome.

First part, see photo

for second part;

Z(k)=(+/-)1Exp(i(theta))theat=Arg(Z(k))=(Pi/2n)(4k-1)

Z(k)=(+/-)( (cos(Arg(Z(k))-1)+isin(Arg(Z(k))) so thereal portions needs to be nul for Z(k) to be purely imaginarythen

cos(Arg(Z(k))=1, so that Arg(Z(k)) has to be in mPi,m Natural

then isolating for k,

k=( (2n)m-1)/4

Subsituting this in Arg(Z(k)) yields m(Pi), but Ireally don't think that's right...

MATH 2FM3 Lecture Notes - Lecture 13: Accrued Interest, Discounting

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