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August 20

Loan repayment problem

Bold text I would like to know how long it would take to pay off a loan of $2.6M AT $55,000 PA at an interest rate of 5%.

Thanks Andrew — Preceding unsigned comment added by 202.177.218.43 (talk) 06:55, 20 August 2012 (UTC)[reply]

5% of $2.6m is $130,000, so you need to be repaying at least $130,000 per annum just to cover interest. So simple answer is that you can't pay off a loan of $2.6m at 5% interest by repaying only $55,000 per annum. Gandalf61 (talk) 08:07, 20 August 2012 (UTC)[reply]

Perhaps you meant that they should pay off the interest each year and an additional $55K towards the principal, for $185K per year, initially ? StuRat (talk) 08:24, 20 August 2012 (UTC)[reply]

In which case, 25 years.←86.139.64.77 (talk) 10:38, 20 August 2012 (UTC)[reply]

Hmmm, I get over 47 years: $2600000 / $55000 = 47.272727... StuRat (talk) 22:47, 20 August 2012 (UTC)[reply]

You are assuming annual payment is annual interest + $55,000, so annual payments start at $185,000, but get smaller each year, keeping a constant repayment of $55,000 of principal each year. 86.139.64.77 is assuming annual payments stay constant at $185,000, so each year a smaller amount of the annual payment goes towards paying interest and a greater amount goes towards repayment of principal. Whether the OP meant one or the other of these scenarios or something else is anyone's guess. Gandalf61 (talk) 10:19, 21 August 2012 (UTC)[reply]

Yes, that's the diff. Thanks for the clarification. StuRat (talk) 10:30, 21 August 2012 (UTC)[reply]

coprime

If a and b are coprime show that there exists c such that $ca\equiv 1\mod b$ --150.203.114.14 (talk) 20:29, 20 August 2012 (UTC)[reply]

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know.

That said, try considering Euclid's algorithm. Straightontillmorning (talk) 20:34, 20 August 2012 (UTC)[reply]

Have you studied our article coprime ? Bo Jacoby (talk) 22:23, 21 August 2012 (UTC).[reply]

sin of integers (interesting problem)

Given $1>c>0$ , Show that there are infinitely many integers $n$ such that $\sin(n)>c$ . Widener (talk) 20:47, 20 August 2012 (UTC)[reply]

Widener, a lot of times you post things that are indistinguishable from homework questions, except perhaps in that people may remember you personally and that you have said in the past they are not. I think this is a problem, even if they are not homework, because people may get into the habit of answering homework problems, and we don't want to become that sort of resource.

I think it would be better if you followed the guidelines for asking homework questions, even if in your personal situation that is not what they are. --Trovatore (talk) 20:59, 20 August 2012 (UTC)[reply]

You know I don't agree with the homework policy anyway Trovatore (actually, did you know that?). Widener (talk) 21:15, 20 August 2012 (UTC)[reply]

I didn't know it. Nevertheless it has general support here. In my opinion, if you refuse to comply with it, people should stop answering your questions and remove any answers that are given. --Trovatore (talk) 21:17, 20 August 2012 (UTC)[reply]

Well fine. One might naïvely try to show

\sin(n)=c

but that's not true. Then one might try to show that any interval in the set

\sin ^{-1}[c,1]

contains an integer, but that's not try either. To me, the statement is not obvious at all. Widener (talk) 21:42, 20 August 2012 (UTC)[reply]

The hard case is when c is large (close to 1), right? So you need to show that there are infinitely many n such that sin(n) is arbitrarily close to 1.

When is sin(x) close to 1? When x is close to π/2. Or 5π/2. Or....

So start looking at whether lots of values of n are close to these values. Put another way, whether lots of these values are close to integers. --Trovatore (talk) 21:49, 20 August 2012 (UTC)[reply]

Ok. You can approximate

{\frac {\pi }{2}}

with a rational number

{\frac {a}{b}}

to arbitrary precision and you can make

a

divisible by 2. Then you want integers such that

n=(2m+1){\frac {a}{b}}

so you take the infinite set

S

of integers which are divisible by 2b and for each

j

of those you apply the transformation

{\frac {j-a}{2b}}

to get n phew. Widener (talk) 22:06, 20 August 2012 (UTC)[reply]

It's a good start, but as you said in a comment you later rmoved, it doesn't quite work because the error might be getting bigger with m. Suppose for example that π/2 were equal to

{\frac {25}{16}}

. Then no odd multiple of π/2 could ever be closer than

{\frac {1}{16}}

to an integer.

So you need to figure out what's the important property that π/2 has but

{\frac {25}{16}}

lacks, and figure out why it changes the answer. --Trovatore (talk) 22:46, 20 August 2012 (UTC)[reply]

Instead of approximating

{\frac {\pi }{2}}

with a rational number, you could instead approximate

(2m+1){\frac {\pi }{2}}

with a rational number (as a function of m). Somehow. I'm not really sure what you trying to say. Widener (talk) 23:34, 20 August 2012 (UTC)[reply]

If π/2 were

{\frac {25}{16}}

, would the argument you have in mind change? If not, then that argument has to be wrong, because it gives a false conclusion. We're looking at values of the form

(4m+1){\frac {\pi }{2}}

, and if π/2 were equal to

{\frac {25}{16}}

, then

(4m+1){\frac {\pi }{2}}

would always be an odd number of sixteenths, and therefore at least

{\frac {1}{16}}

away from any integer. --Trovatore (talk) 04:39, 21 August 2012 (UTC)[reply]

Basically because

\pi /2

is irrational, right? So you can start with some rational approximation depending on

c

and you could take some sequence of rationals

\{q\}

indexed by

m

with the first term being equal to your first approximation and with

\lim\{q\}={\frac {\pi }{2}}

so the approximation gets better with m compensating for multiplying the error by m. The denominator will get bigger (actually approach infinity) as you go along the sequence so eventually you will become within

{\frac {1}{k}}

th of an integer.Widener (talk) 11:45, 25 August 2012 (UTC)[reply]

You would want the error to be sort of constant. Maybe you can truncate

(2m+1){\frac {\pi }{2}}

to a finite number of digits. 23:41, 20 August 2012 (UTC)

Actually that way of doing things probably is a good idea after all. Widener (talk) 22:06, 20 August 2012 (UTC)[reply]

This is from Stein and Shakarchi's book on Fourier Analysis by the way (well, sort of. Inspired by it). I can highly recommend it. Widener (talk) 22:18, 20 August 2012 (UTC)[reply]

Equidistribution theorem may be useful. Ssscienccce (talk) 22:58, 20 August 2012 (UTC)[reply]

It seems like this follows pretty directly out of Liouville's criterion for transcendental numbers.Phoenixia1177 (talk) 04:23, 21 August 2012 (UTC)[reply]

Transcendence is overkill. You can do it with just irrationality. More interesting than applying big theorems is to try to figure it out directly (that's how to get insight into why the big theorems are true, which is much more useful than just following a canned proof and agreeing that each step is valid). --Trovatore (talk) 04:33, 21 August 2012 (UTC)[reply]

I suppose I see your point, but I didn't think I was citing a "big theorem", that's the first thing that came to mind when I looked at it, it seemed the obvious thing. And it seems immediate once you start thinking about rationals getting close to pi/2. Phoenixia1177 (talk) 11:06, 21 August 2012 (UTC)[reply]

A more general theorem holds: for any $c\in [-1,1]$ there is a subsequence of $\left(\sin(n)\right)_{n\in N}$ which converges to $c$ . --CiaPan (talk) 05:49, 21 August 2012 (UTC)[reply]