Wikipedia:Reference desk/Archives/Mathematics/2007 February 13

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February 13

exchanging the variable in logarithmic cubic quations

my equation is

(1/T)=a+b(ln R)+c((ln R)*(ln R)*(ln R))

it is of the form T=f(R);

here T and R are variable. a, b , c are known values

i want above equation to be changed to R=f(T) form?? please give the detail procedure..

snehanshu tiwari bangalore india

I don't believe it possible, because of the cube of ln R. If the expression involved the log of the cube of R, that would be different, and straightforward to invert.–81.159.15.39 13:18, 13 February 2007 (UTC)

If we let x=ln(R) then we can re-arrange to get a cubic in x:

$cx^{3}+bx+(a-{\frac {1}{T}})=0$

which can then be solved (in principle - messy in practice) to find x as a function of T, say

$x=g(T)$

and then

$R=e^{x}=e^{g(T)}$. Gandalf61 14:27, 13 February 2007 (UTC)

Solving cubic equations

In school, we learnt how to solve cubic equations of the form ax³ + bx² + cx + d.

To factorise it into (x+r)(x+s)(x+t), we have to find a value for r by trial and error (with the factor theorem), then solve a quadratic equation to get the other two roots, s and t.

In the exams, the value of r will usually be 3 or less, so finding r by trial and error will not take a long time.

However, if r is large, finding r by trial and error will take a long time. Are there any faster, more systematic ways of finding r, besides trial and error? —Preceding unsigned comment added by 218.186.9.3 (talk • contribs) 09:18, 2007 February 13

(Please sign your posts with four tildes, "~~~~".) The roots of a cubic polynomial can be expressed in closed form, so in theory no searching is required. However, the technique is more difficult than using the quadratic formula. When the coefficients, a, b, c, and d, are real numbers (with a nonzero), the cubic is guaranteed to have a real root; but it may have only one. For example,

$x^{3}-3x^{2}+2x-6\,\!$

has only one real root.

But we can say much more. The coefficients are normalized so that the x³ term has coefficient 1. If we list them all, (1,−3,2,−6), the −6 has the largest absolute value. The Cauchy bound says that all real roots must lie between −m and +m, where m is one plus the maximum absolute value. In this example, the roots will be between −7 and 7.

We can do better. The positive coefficients before the −6 are 1 and 2, summing to 3; and before the −3 we have just 1. For an upper bound we can choose the maximum of ⁶⁄₃ and ³⁄₁, which will be 3, and again add 1. That is, the maximum real root is 4. Negating every other term (so we are evaluating the polynomial with −x), we can similarly find a lower bound, which here will be −1.

There is also a method known as Descartes' rule of signs that may help us restrict the number of positive or negative real roots. In this example it is no help with the positive roots (we will have three or one), but it tells us we cannot have a negative root.

This example polynomial has integer coefficients, and may have integer roots. If r is such a root, then it must divide the constant term, so the only possibilities here are 1, 2, 3. We have used the known bounds; and we know that 0 cannot be a root unless the constant term is 0.

We also have ways to bracket roots. If the polynomial evaluates to positive for x = a and negative for x = b, then there must be at least one root between a and b. Using a Sturm sequence, we can learn even more.

This does not exhaust our inventory of tools, but perhaps this is enough for now. Computer programs for solving cubics typically use closed form solutions. Above degree four, however, a famous theorem states that we have no closed form solutions. --KSmrq^T 11:48, 13 February 2007 (UTC)

You might want to read cubic equation. – b_jonas 14:17, 13 February 2007 (UTC)

Anyone bored?

I any of you [lovely] people have too much spare time on your hands could you give me a hand. I have the formula to work out reps until exhaustion(y) from % of max load (x), and either would like the formula rearanging, or the location of the formula the other way around, its a really nasty formula!:

$y=173.525-6.31x+(9.576\times 10^{-2})x^{2}-(6.742\times 10^{-4})x^{3}+(1.75\times 10^{-6})x^{4}$

MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 21:46, 13 February 2007 (UTC)

I can't make heads or tails out of your question. What are reps . Until what exhaustion? 202.168.50.40 22:00, 13 February 2007 (UTC)

It looks like the question is about weight training, and the equation is supposed to describe the number of repetitions the trainer can perform, as a function of the weight he is lifting (expressed as a percentage of the maximal weight he can lift). However, I cannot imagine how this formula was arrived at, or what the OP wants to do with it. -- Meni Rosenfeld (talk) 22:51, 13 February 2007 (UTC)

I did an interpolation for the pairs (50,25), (70,12), (90,5), (100,1) and arrived at something like

$102-2y-0.1y^{2}+4\times 10^{-3}y^{3}$

which gives 76% for 10 reps. Ok?--80.136.135.63 23:03, 13 February 2007 (UTC)

That's quite close, thanks! I didn't come up with it myself, it would have been easyer to rearange if I had, I got it from a sprt science website, and wanted to be able to work it backwards, if only excel had a function to work out inverse functions (or does it, maybe a qu for IT desk...), anyway many thanks for your efforts!MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 23:13, 13 February 2007 (UTC)

That $x(y)$ is of course not quite the inverse function of what you started with; if you actually want the inverse, you can get it from applying the quartic formula, but it won't be pretty: I get ${\begin{array}{l}{\frac {1}{2}}{\sqrt {{\frac {12.3344-5.0397y}{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}}+151181.0526{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}+625.7665}}\\-{\frac {1}{2}}{\sqrt {{\frac {5.0397y-12.3344}{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}}-151181.0526{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}+{\frac {425419.3114}{\sqrt {{\frac {12.3344-5.0397y}{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}}+151181.0526{\sqrt[{3}]{-2.0697\times 10^{-7}y+1.9247\times 10^{-7}{\sqrt {(y+10.7684)((y-16.9544)y+122.1164)}}+7.0182\times 10^{-6}}}+625.7665}}}+1251.5331}}+96.3143\end{array}}$ which gives 74.6442% for 10 reps. --Tardis 17:13, 14 February 2007 (UTC)

Please tell me that you used Mathematica or Maxima or something for that :) Doing to TeX alone would be murder! Oskar 17:42, 17 February 2007 (UTC)

Mathematica solved the quartic (with some help from me to pick a solution branch), and generated approximate TeX with TeXForm; I used Emacs to reformat the result in a WP-friendly manner: round the numbers, pretty-print the scientific notation, etc. --Tardis 16:32, 20 February 2007 (UTC)

Can someone give me a question please?

I am currently studying maths at GCSE level (first year) and am finding it relatively easy and relatively boring ("Yes miss, we know how to get a term-to-term rule. Yes, you don't need to give us 15 exercises on it, we get it already. You're not listening are you miss? You're going to give us those 15 exercises anyway aren't you miss, even the the answers are obvious. Oh look miss, you've put the answers on the board and got three wrong. Oh look, you've tried to correct it and got stuck and now you're asking us what the answer is. Well done. Oh look, now you're giving us some exercises that we don't know how to do and which neither the book nor you explain to us. It just gives us the answers without explaining why they are correct. Thanks for that."). As such, I would appreciate it if anyone can give me an equation to solve the includes a bit of everything up to the level of calculus. I've been trying to get my algebra really good, but the textbook I have doesn't really help and only gives basic questions. I am sort of looking for a question with stuff life binomial expansion, negative powers, fractional powers, rearranging divisions and multiplications, the whole lot. I most probably won't be able to solve it, but it will give me a starting point to ask questions about how to do certain sections of it. Thanks for any help you can give. P.S. I apologise for the off topic rant in between the brackets there. --80.229.152.246 22:55, 13 February 2007 (UTC)

...which is why I took my GCSE a year early and progressed on to As! Without knowing exactly what you are capable of, I darent just give you something from As (which isn't much harder than GCSE, by the way!), try looking here or here for some good qus, which come with explanations MHDIV ɪŋglɪʃnɜː(r)d(Suggestion?|wanna chat?) 23:18, 13 February 2007 (UTC)

I mean just an equation to solve that has a bit of everything in it. I don't care if I can't solve it with my knowledge, I just want to use it as set-off point to learn new things.

--80.229.152.246 17:40, 14 February 2007 (UTC)

Have you tried any of the unsolved problems in mathematics? − Twas Now ( talk • contribs • e-mail ) 05:38, 15 February 2007 (UTC)

Bearing in mind I haven't even got a GCSE in maths yet, I think attempting one of those would be pretty futile. For example, I have now taught myself the binomial theorem and would like to practice rearranging equations with negative powers, fractional powers, brackets, square roots, fractions multiplied by variable, etc. Just give me anything. I don't really mind if I can solve it or not, I would just like to know what to look up for the bits I get stuck on. --80.229.152.246 17:20, 15 February 2007 (UTC)

OK here's some things to try:

1. Find the intersection of line with a

a. sphere

b. ellipsoid

c. cone

d. plane in 3 dimensions.

2. Find the normal to this point and express it in terms of the equation of the line and the equation of the sphere/ellipsoid etc.

3. Find an expression for the closest approach of two lines in three dimensions

4. Read about integration by parts - then learn to do it.

5. Calculate the distance from the centre of a tetrahedron to a point (the tetrahedron has a known side length l)

6. Show that the differential of sin(x) is cos(x)

7. Show that the differential with respect to x of xⁿ is nx^n-1 (n is integer)

8. Find an expression/equation that differentiates to give the same equation.

I don't know if these are too hard/easy for you. But good luck.213.249.237.49 18:40, 15 February 2007 (UTC)

Thanks very much for those ideas. I'll see if I can do some of them. --80.229.152.246 18:07, 16 February 2007 (UTC)

By the way, I don't understand the first 3 questions. I don't really get what you want me to do. --80.229.152.246 18:15, 16 February 2007 (UTC)

For 1.a. Take a line in 3 dimensions eg y=4x+2, z=4y+3 (or use vectors) then find the point(s) at which it crosses through the surface of a sphere eg (x-2)²+(y-3)²+(z+1)²=r² is the equation of a sphere centred on (2,3,-1) of radius r. The others are similar - you might need to look up some of the terms to do these then.87.102.7.220 14:15, 17 February 2007 (UTC)

How to pay back infinite dollars for a finite debt

$\sum _{n=1}^{\infty }{\frac {1}{n}}=\infty$

Now this does not make sense. For example: If I borrow $1000 off you and promised to pay you back infinite dollars, it would sounds like an absolute great deal.

So year 0 : borrow $1000 from you

and year 1 : pay you back $1.00

and year 2 : pay you back $0.50

and year 3 : pay you back $0.33

and so forth. But this does not make sense at all.

I think we can solve America's national debt. 202.168.50.40 23:39, 13 February 2007 (UTC)

I would be paying back an infinite total amount of money, but it would have a finite present value. In fact, at current interest rates, you should be able to buy a promise to pay you one dollar a year, forever, for about fifty dollars. --Trovatore 00:30, 14 February 2007 (UTC)

You assume that the interest rate is greater than zero. What if the interest rate is exactly zero. Say I borrow 1000 tonnes of uranium from you and pays back in uranium. Does your argument still holds? 202.168.50.40 00:43, 14 February 2007 (UTC)

I can't imagine why you think I'd be willing to lend you uranium at an interest rate of zero. --Trovatore 01:01, 14 February 2007 (UTC)

Because usury is such a dirty word. − Twas Now ( talk • contribs • e-mail ) 05:35, 15 February 2007 (UTC)

The partial sums of the infinite series are the harmonic numbers. They increase very slowly. You need to add the first 12367 terms to even reach the value 10. To get to 100, you need about 15092688622113829863255886615135332918918094 terms, give or take a few. The universe will have evaporated long before you reach 1000.  --Lambiam^Talk 02:26, 14 February 2007 (UTC)

It's your analogy that doesn't make sense. Depending on where you live, there will be a lower bound on the amount of money that you can make up, but you will need to pay back amounts less than this lower bound, which you obviously can't do. —The preceding unsigned comment was added by 203.49.220.190 (talk) 09:29, 14 February 2007 (UTC).

If you're interested in similar situations to yours, you should check out the article on the St. Petersburg paradox Oskar 16:09, 16 February 2007 (UTC)