Re: New to Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg126399] Re: New to Mathematica*From*: "Oleksandr Rasputinov" <oleksandr_rasputinov at ymail.com>*Date*: Tue, 8 May 2012 04:07:14 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <jntga4$2c$1@smc.vnet.net> <jo2nld$mds$1@smc.vnet.net>

On Mon, 07 May 2012 01:30:43 +0100, J.Jack.J. <jack.j.jepper at googlemail.com> wrote: > On May 6, 8:25 am, David Bailey <d... at removedbailey.co.uk> wrote: >> On 05/05/2012 09:16, J.Jack.J. wrote: >> >> > On May 3, 9:40 am, "J.Jack.J."<jack.j.jep... at googlemail.com> wrote: >> >> Hi, >> >> >> I have just downloaded my free trial version of Mathematica. >> >> I only need it for one thing (as yet, anyway), and I wonder whether >> >> someone can help me in step-by-step fashion to get what I want. I >> need >> >> the solution to the following (I might make amendments to the >> values): >> >> >> "For any integer k, let r(k) be x such that >> >> >> (x/ln(x))*(1 + 1/ln(x)) = 108.2 + k) >> >> >> product (k = 0 to 3000) (1-1/r(k))" >> >> >> With many thanks in advance. >> >> > Can nobody help me with this? Just for a Newbie? >> > As most will realise, the first lines define my function and the >> > product (k = 0 to 3000) (1-1/r(k)) >> > is my desired calculation. I need to know what inputs to use. >> >> > With thanks in advance. >> >> If you press F1, you will enter the help system. Once there, you could >> try looking up log and product. The examples will give you the basics of >> Mathematica syntax, and you can cut and paste them into your notebook, >> and alter them as desired. >> >> David Baileyhttp://www.dbaileyconsultancy.co.uk > > > Have tried and failed, tried and failed, with this method! Can > somebody give me the requisite inputs? Note that I need to add the > condition that x be the highest integer for which <definition of > r(k)>. > > With thanks. > Just give up. There is no such integer x for any integer k between 0 and 3000 inclusive, and Mathematica can prove it: Exists[ x, x \[Element] Integers, (x/Log[x]) (1 + 1/Log[x]) == 541/5 + k && k \[Element] Integers && 0 <= k <= 3000 ] // Resolve False However, your problem was not well stated from the beginning, so I doubt if you really meant what you wrote above. Here is one of the 3^3001 possible solutions (type this into Mathematica; it is an absurdly large number, too big to post here) to your problem consistent with the definition of r(k) originally given, and with the additional assumption that r(k) is the largest *real* x such that (x/Log[x]) (1 + 1/Log[x]) == 541/5 + k: r[k_?NumericQ] := Block[{x}, x /. FindRoot[ (x/Log[x]) (1 + 1/Log[x]) == 541/5 + k, {x, 5}, WorkingPrecision -> $MachinePrecision ] ]; Product[1 - 1/r[k], {k, 0, 3000}] 0.6308341356354897 I don't know if you will be able to find a closed form for this, but if an approximation good to 100 places will do, you could use the second root of: 0 == -70 - 13047 x + 9183 x^2 - 5019 x^3 + 35491 x^4 - 3680 x^5 - 14978 x^6 + 6707 x^7 + 28462 x^8 + 9675 x^9 + 30267 x^10 + 19638 x^11 + 3179 x^12 + 15750 x^13 - 21197 x^14 - 5200 x^15 + 37078 x^16 - 4243 x^17 + 10128 x^18 - 15288 x^19 + 27082 x^20

**Re: New to Mathematica**

**Re: New to Mathematica**

**Re: New to Mathematica**

**Re: New to Mathematica**