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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



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