Ah, yes, that is helpful.

In general I'm looking for an automated solution for problems like this.
 I'd like to solve this problem without being smart :)


On Thu, Feb 21, 2013 at 10:35 AM, Tom Bachmann <[email protected]> wrote:

> Multiplying through by lambda and taking terms independent of i from the
> sum, I seem to get 4 \sum x_i = n(lambda^2 - 6\lambda). This is just a
> quadratic equation for lambda.
>
> Am I missing something?
>
> Best,
> Tom
>
>
> On 21.02.2013 18:31, Matthew Rocklin wrote:
>
>> I need to solve the following problem for lambda.
>>
>>    n
>>   ____
>>   ╲
>>    ╲      2
>>     ╲  - λ  + 6⋅λ + 4⋅xᵢ
>>     ╱  ───────────────── = 0
>>    ╱          4⋅λ
>>   ╱
>>   ‾‾‾‾
>> i = 1
>>
>> This is the end of a simplification around a data-oriented problem.  In
>> the future I will be given a set of many `x_i`s as data and will be
>> asked to solve for lambda.  I'm trying to make this future computation
>> as simple as possible.
>>
>> Can SymPy help me to reduce this problem any further?  (it has already,
>> by the way, done wonders.)
>>
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