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July 24

Number of solutions

Is it possible to determine the exact number of solutions of the equation $sin{\frac {1}{x}}=0$ in the interval 0≤x≤1? I would be tempted to say infinite, since any solution of the form ${\frac {1}{n\pi }}$ , where n is an integer, would suffice but I'm sure that can't be right. Also, as an aside, what value would $sin{\frac {1}{0}}$ be assigned? Thanks 92.3.252.234 (talk) 01:23, 24 July 2008 (UTC)[reply]

Yes, that equation has infinitely many (more precisely, aleph-null) solutions. There's no standard way of interpreting

\sin {\frac {1}{0}}

, and I can't immediately think of anything sensible except labelling it as nonsense. Algebraist 01:26, 24 July 2008 (UTC)[reply]

(edit conflict) Infinite is right for exactly the reason you stated, what made you think you were wrong?.

sin{\frac {1}{0}}

is undefined, the function oscillates infinitely fast as you reach x=0, so you can't meaningfully assign a value there. Therefore, your interval should be 0<x≤1. You can get a slightly better behaved function by multiplying the whole thing by x (or was it x²?) then the function goes to 0 as x tends to 0, so you can define it to be 0 at 0. --Tango (talk) 01:31, 24 July 2008 (UTC)[reply]

Multiplying by x makes it continuous at zero (with value zero there), by x² makes it differentiable. Algebraist 01:34, 24 July 2008 (UTC)[reply]

In case you're curious, there's an article on this curve: Topologist's sine curve Black Carrot (talk) 05:07, 24 July 2008 (UTC)[reply]

$\sin {\tfrac {1}{x}}=0$ has exactly the same 'number' of solutions for 0<x≤1 as $\sin y=0\;$ for y≥1 (put y=1/x to see the one-to-one correspondence). If you agree the latter equation has infinite set of solutions, the same applies to the former one. --CiaPan (talk) 06:09, 24 July 2008 (UTC)[reply]

Continuously Compounded Interest

So one thing I've come across frequently in the derivation of e is continuously compounded interest calculated as (1+i/n)^n. So a 100% annual interest rate is actually 100% when compounded once, 125% when compounded every six weeks, 144% when compounded quarterly, and 178% as the limit becomes continuous.

Now I understand that for the purposes of getting e, this limit is significant.

However, the idea of continously compounded interest calculated this way strikes me as very contrived. If the bank told me that I would receive 100% annual interest compounded every six months, I would expect 41% interest every six months, not 50%. If the bank told me that I would receive 100% annual interest compounded every quarter, I would expect 19% interest every quarter, not 25%. The fundamental conflict is that i does not equal (1+i/n)^n - 1.

It seems that the calculation of interest as (1+i/n)^n stems from a naive assumption that compound interest accrues as simple interest, where you can just divide the annual yield by the number of compounding intervals.

My question is, is the formula (1+i/n)^n ever used in real life banking?

--MagneticFlux (talk) 04:52, 24 July 2008 (UTC)[reply]

Real life banking is made by real life bankers who do not understand mathematics. If you get r% interest in n days, how much interest should you get in x days? The answer is (100%+r%)^x/n−100%. This is an ugly formula, but it can be beautified stepwise. The first step is to consider the factor of interest F=100%+r% rather than the rate of interest r%. Then the formula reads F_x=F_n^x/n. This is less ugly, as the factor of interest grows exponentially with time. The second step is to consider the natural logarithm of the factor of interest. L=ln(F). Then the formula reads L_x=L_n·x/n. This is quite beautiful as the natural logarithm grows linearly with time. Then the words rate of interest make sense. But apparently bankers use no name for this important natural logarithm of the factor of interest, and the article on interest does not use it. There is approximate equality between ln(100%+r%) and r%, when r% is a few %, so the two concepts of rate of interest are easily confused. There are apparently no standard names for the units of measurement of logarithms. An exchange rate is given as a factor rather than as a natural logarithm, confirming that bankers do not understand mathematics at all. Bo Jacoby (talk) 10:32, 24 July 2008 (UTC).[reply]

:::Seperate answer from a different person below...

YES! It is used extensively in real life banking! The "norm" is semi annual compound interest.

All coupon bonds have a stated yield that fits the following formula.

A $100 bond with a 10% semi-annual yield pays out $5 on June 31st and pays out $5 on Dec 31st

In real life, arbitragers who make a living by skimming fractions of pennies every time that a bond is calculated wrong (or even rounded to 5 digits) will figure out its rate of return assuming continuously compounded interest. They have an excel document setup to convert it from one type of yield to another.

In finance, you must learn how to convert yields using a plain calculator, and when you are done with exam 1, the teacher tells you--aren't you glad you dont need to have to do this one day, since computers do it in a fraction of a second?

Yes the continuously compounded derivation is built into EVERY yield that is quoted. The 10% bond I quoted doesn't actually tell you the yield. It tells you the nCF or the "cash flows" from now until the bond is sold or matures (cash in the $100). So when a yield is quoted, they aren't actually telling you the yield--they are telling you the cash-flow structure of the financial asset.

If a banker sells you a mortage on your house, they do not even care about the interest rate that the cash flows infer. They only look at your credit score and the cash flow structure. If your mortgage is $800 a month for 30 years, isn't that the same thing as a ...

The coupons off of 6 seperate $100 bond at 1600% semi annual yield?

The first bond pays out $800 every Jan 31 and July 31

The second bond pays out $800 every Feb 28 and Aug 31

The third bond pays out $800 every Mar 31 and Sept 30

The fourth bond pays out $800 every Apr 30 and Oct 31

The fifth bond pays out $800 every May 31 and Nov 30

The sixth bond pays out $800 every June 30 and Dec 31

You see the cash-flows dictate the interest rate, not the way you might see. The interest rate is the dependent variable 'y'

y = f(x)

"You tell me the cash flows, and I'll tell you the interest rate" Even if an interest rate you get quoted calls itself one thing, they are not actually telling you the interest rate, they are telling you the cash-flow arrangement under all range of conditions.

When arbitragers can take advantage of a rounding error at 8 decimal places and can go to 12 decimal places, on a multi trillion dollar asset portfolio, they can make millions, just from fractions of fractions of fractions of fractions of a penny, but over a large enough volume to make it worth it. Sentriclecub (talk) 12:39, 24 July 2008 (UTC)[reply]

"You tell me the cash flows, and I'll tell you the interest rate". OK, I'll tell you a cash flow. It is +$100 the first year, −$230 the second year and +$132 the third year, and then we are even. You tell me the interest rate? Bo Jacoby (talk) 05:25, 25 July 2008 (UTC).[reply]

What's your point? If v(t) is a discount factor for t years, then you can set up

100*v(1) + 132*v(3) = 230*v(2), assuming casflows are at the end of the year.

Then decide how many capitalisations per year and compute. Zain Ebrahim (talk) 11:56, 28 July 2008 (UTC)[reply]

Statistics - - What is it called when I multiply 2 elements and then divide by the average?

What is it called when I multiply 2 elements and then divide by the average?

A person weighs 122 kg An object weighs 900 kg

If I Multiply those 2 numbers and then divide by the mean, I get the result...

122*900/511 = 215

What is the operation called?

Also, is the answer a dimensionless unit? Or is it something technical like "(kilogram)-(standard deviations)". —Preceding unsigned comment added by 76.4.128.40 (talk) 12:18, 24 July 2008 (UTC)[reply]

You get the average - the VMR

Hope this helps! Sentriclecub (talk) 13:20, 24 July 2008 (UTC)[reply]

seee also http://en.wikipedia.org/wiki/Variance-to-mean_ratio

It's also twice the reduced mass. -- BenRG (talk) 15:49, 24 July 2008 (UTC)[reply]

It's the harmonic mean. It's not dimensionless, since it's quadratic in the numerator and linear in the denominator. Michael Hardy (talk) 15:59, 24 July 2008 (UTC)[reply]

It's actually twice the harmonic mean. --Random832 (contribs) 17:39, 29 July 2008 (UTC)[reply]

Partial fractions

Is it possible to write ${\frac {1-x^{-n}}{x^{2}-x}}$ as a finite sum of partial fractions? I have to integrate it and partial fractions seems the only way forward but I'm stumped as to how to do it or even if it's possible. I don't want to know how to do it, just if it can be written in the way I want. Thanks 92.3.252.234 (talk) 13:18, 24 July 2008 (UTC)[reply]

{\frac {1-x^{-n}}{x^{2}-x}}={\frac {x^{n}-1}{x^{n}(x^{2}-x)}}={\frac {x^{n}-1}{x^{n+1}(x-1)}}

={A_{1} \over x}+{A_{2} \over x^{2}}+\cdots +{A_{n} \over x^{n}}+{A_{n+1} \over x^{n+1}}+{B \over x-1}.

Michael Hardy (talk) 15:56, 24 July 2008 (UTC)[reply]

Once you have (x^n-1)/(x^(n+1)*(x-1)), you can divide x-1 into x^n-1 evenly to get the result without any more "partial fraction" (systems of equations) work. Simply solving for the A_i is not very hard either, but since n is general, it might be better to go with the shorter solution. JackSchmidt (talk) 16:15, 24 July 2008 (UTC)[reply]

Oh. I did write this in haste, didn't I? OK:

{\frac {x^{n}-1}{x^{n}(x^{2}-x)}}={\frac {(1+x+x^{2}+\cdots +x^{n-1})(x-1)}{x^{n+1}(x-1)}}={1 \over x^{n+1}}+{1 \over x^{n}}+{1 \over x^{n-1}}+\cdots +{1 \over x^{2}}.

That's it. Michael Hardy (talk) 23:00, 24 July 2008 (UTC)[reply]

I'm sure someone has already beaten me to this

So I was lying in bed last night, trying to sleep, and my mind was wandering, just doing its own thing as it tends to do, when I realized something. I'm sure I'm not the first person to recognize this (maybe I even read it on WP), but it seems that there are no numbers (besides 0 and 1, but we'll call those trivial cases) who are the sum of the squares of their digits (or none that I could think of, anyways). I'm sure that a process-of-elimination proof could be written for this (eg, the number can't be larger than 99, since it would seem that a² + b² < 1ab in all cases; it can't end in 0, since no number 1-9 gives a number divisible by 10 when squared; it can't end in 5 for the same reason, etc.) There are several that are close (35, 3² + 5² = 34; 75, 7² + 5² = 74). So has someone already beaten me to this, or should I expect my Nobel Prize in Mathematics in the mail sometime soon? shoy (reactions) 16:17, 24 July 2008 (UTC)[reply]

In other bases there are examples: In base 3, 1*3+2 = 1^2+2^2 and 2*3+2 = 2^2+2^2. In base 5, 2*5+3 = 2^2+3^2 and 3*5+3 = 3^2+3^2. There are examples in base 7, 8, 9, 11, 12, 13, 15, 17, 18, 19, and more for larger bases. JackSchmidt (talk) 16:29, 24 July 2008 (UTC)[reply]

This fact is already known (and in any case would not merit a maths Nobel prize, even if one existed). There's some relevant material at happy number, which implies your statement, though it isn't mentioned explicitly. Algebraist 17:46, 24 July 2008 (UTC)[reply]

Unfortunately, there is no Nobel Prize for Mathematics. All we get is the Fields Medal or the Abel Prize. Paragon12321 (talk) 23:17, 24 July 2008 (UTC)[reply]

I was being amazingly sarcastic wrt the Nobel Prize comment. shoy (reactions) 23:46, 24 July 2008 (UTC)[reply]

We have an article on everything - see narcissistic number. Gandalf61 (talk) 08:51, 25 July 2008 (UTC)[reply]

All base 1 numbers are narcissistic.87.102.86.73 (talk) 09:56, 25 July 2008 (UTC)[reply]

Proving Theorem

How can it be proved that 'The angle subtended by an arc at any point on a circle is half that subtended by the arc on the centre of circle? —Preceding unsigned comment added by Asim Chatterjee (talk • contribs) 19:07, 24 July 2008 (UTC)[reply]

I'm not a geometry expert, but I think arc you are talking about is the curved line "portion" of a circle.

If that's the case, here it goes.

Label the endpoints of the arc X and Y and label the center point C

The chord is the straight line XY that connects the 2 endpoints of the curved line.

The chord is a leg of a triangle formed by the 3 lines XY, YC, and XC.

Now if you make a point M which is the midpoint of the chord. Then the line MC splits the generated angle XCY in half.

A formal geometric proof is a circular argument. If I used trig, I would define sine and cosine using geometry and would then prove the geometric argument using sine and cosine. If I have interpreted your question wrong, I'm sorry and someone will come along soon and correct me. Try these links also... Chord (geometry) and Circle_graph Hope this helps a little bit, until someone can better explain it. Sentriclecub (talk) 19:30, 24 July 2008 (UTC)[reply]

See inscribed angle#proof. Algebraist 19:32, 24 July 2008 (UTC)[reply]

That certainly could do with a diagram - luckily there is one in the links http://www.cut-the-knot.org/pythagoras/Munching/inscribed.shtml

87.102.86.73 (talk) 21:21, 24 July 2008 (UTC)[reply]

The article I linked has three diagrams. Algebraist 21:24, 24 July 2008 (UTC)[reply]

No it doesn't. Eh?87.102.86.73 (talk) 21:58, 24 July 2008 (UTC)[reply]

Yes, it does... have you disabled images in your browser, or something? --Tango (talk) 22:13, 24 July 2008 (UTC)[reply]

It seems I'm missing a plugin for svg images.. Browser didn't even tell me or display anything..87.102.86.73 (talk) 22:55, 24 July 2008 (UTC)[reply]

You are most welcome, friend, for rendering your valuable hands of help. —Preceding unsigned comment added by Asim Chatterjee (talk • contribs) 21:45, 24 July 2008 (UTC)[reply]

Proof

How can we prove that 'the angle of a triangle (with the diameter of its circumcircle as its base) which is opposite to its base is always a right angle' i.e 'the angle on a semicircle is aright angle'?Asim Chatterjee (talk) 21:42, 24 July 2008 (UTC)[reply]

That's a special case of the theorem on inscribed angles I cited in answer to your last question. Algebraist 21:49, 24 July 2008 (UTC)[reply]