Mathematics desk
< October 13	<< Sep \| October \| Nov >>	October 15 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

October 14

Equation from graph data

hello! i have a set of data that i have graphed, what i intend is to hopefully find an equation for it. it is here unfortunately the data is 'loosish' (approximated) and so its not a case of joining the dots and going from there, but the data does have a logarithmic (i think) trend. Can anyone point me in the direction of how to derive an equation from this? sorry if my jargon is a bit lo-brow, its been years since this sorta stuff has occupied any mental real estate of mine. thank you!Boomshanka 02:01, 14 October 2007 (UTC)[reply]

The Curve fitting article should get you on your way. Looking at the picture, however, I'm not sure you have enough data points to find a good fit. If you can get more, I'd certainly try that before continuing. risk 02:52, 14 October 2007 (UTC)[reply]

Try plotting log x versus log(−y), which might suggest a (linear?) relationship between the two. If that relationship is indeed linear, there is a power law relationship between x and y. --Lambiam 10:09, 16 October 2007 (UTC)[reply]

spot my error

ʃ1/x dx = ln (x) + C

but integrating by parts gives

ʃ1/x dx = x/x - ʃx d/dx(1/x) dx  = x/x - ʃx (-1/x²) dx 

ʃ1/x dx = x/x - ʃx (-1/x²) dx = 1 + ʃx/x² dx = 1 + ʃ1/x dx

Where does it go wrong?

I also got ʃ1/(a+x) dx = Sum (n=1 to n=infinity) (x/a+x)ⁿ1/n

this is fine for a=1 with x between 0 and 1 - so no problems calculating logs - but when a=0 a similar problem to above occurs giving ln(x)-ln(1)=Sum (1/n) - Sum (1/n)

It's late and I know there's a mistake here - but if anyone can point out the error (especially in the first example) - I'd appreciate it as I can't see were I've gone wrong. 87.102.19.106 04:13, 14 October 2007 (UTC)[reply]

I doesn't go wrong. If I understand correctly you assume that

ʃ1/x dx = 1 + ʃ1/x dx (A)

implies

0 = 1 (B)

and that you thus have a contradiction. This is not the case. In fact, (A) is correct and (B) doesn't follow from (A). The reason for this is that ʃf dx is a notation that doesn't represent a particular primitive of f but rather the family of all the primitives of f, and is therefore not subject to the ordinary laws of arithmetic.

The expression g = ʃf dx is really a shortcut for dg/dx = f. (A) is thus equivalent with 1/x = d1/dx + d(ʃ1/x dx)/dx <=> 1/x = 0 + 1/x <=> 1/x = 1/x <=> 0 = 0. Morana 06:14, 14 October 2007 (UTC)[reply]

See Invalid proof#Proof that 0 = 1 --Spoon! 08:45, 14 October 2007 (UTC)[reply]

Yes thanks - I realised where I made the error later on but was too tired to come and delete.. also understand any oddities in the second part too.87.102.82.26 17:59, 14 October 2007 (UTC)[reply]

Solution of x*cos(x) = c

Is there a closed form solution to x*cos(x) = c ? Or maybe a series solution. How to write all solutions of this equation as a general expression? deeptrivia (talk) 19:01, 14 October 2007 (UTC)[reply]

I doubt there is a closed form solution, but their is a series. It uses the Taylor series for cosine. Look up that series, and then multiply it by x, this results in a new series, which is equal to $x\cdot cos(x)$ . A math-wiki 00:00, 15 October 2007 (UTC)[reply]

I should note that in order to find your solutions, you will to apply the general solution to polynomials, since your polynomial is technically of infinite degree. A math-wiki 00:02, 15 October 2007 (UTC)[reply]

I'm not too sure what A math-wiki meant by the general solution to polynomials, however, once you have a series describing c in terms of x, the Lagrange inversion theorem should allow you to find a series of x in terms of c. I am fairly certain that, indeed, no closed form solution in elementary functions exists (though you may be able to do something with the Lambert W function). -- Meni Rosenfeld (talk) 00:28, 15 October 2007 (UTC)[reply]

The first few terms, valid for

x\approx 0

, are

x=c+{\frac {c^{3}}{2}}+{\frac {17c^{5}}{24}}+{\frac {961c^{7}}{720}}+\cdots

. For different regions you will have to use different series. -- Meni Rosenfeld (talk) 00:53, 15 October 2007 (UTC)[reply]

Question 2: Thanks for your replies. How can I numerically find out the first n solutions? deeptrivia (talk) 02:48, 15 October 2007 (UTC)[reply]

That depends... In general I would suggest using Newton's method with different initial points. Choosing the correct points that will guarantee not skipping any root can be tricky, but if you can plot a graph of