Re: [sympy] Introduction to the community

[Gagandeep Singh (B17CS021)]

Please select any issue which you think you can solve. AFAIKA, we don't
assign issues.
Thanks for your interest in the project.

With Regards,
Gagandeep Singh
Github - https://www.github.com/czgdp1807
LinkedIn - https://www.linkedin.com/in/czgdp1807

On Wed, 11 Dec, 2019, 12:47 PM Shubham Sidhartha, 
wrote:

> Hi,
>
> I have gone through the contributing guide and the tutorial. Please guide
> me further and assign me with an issue to work upon to get familiar with
> the codebase.
>
> Regards
> Shubham Sidhartha
> III Year, IMSc Applied Mathematics
> IIT Roorkee, India
>
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> 
> .
>

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Re: [sympy] Introduction to the community

[Shubham Sidhartha]

Hi,

I have gone through the contributing guide and the tutorial. Please guide
me further and assign me with an issue to work upon to get familiar with
the codebase.

Regards
Shubham Sidhartha
III Year, IMSc Applied Mathematics
IIT Roorkee, India

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[sympy] Introduction to the community.

[Nitish Gupta]

Hi! I am Nitish Gupta studying currently pursuing my undergraduate degree 
in Electronics and Communication Engineering from Pranveer Singh Institute 
of Technology, Kanpur, UP, India. I have been programming in Python since 
the past 2.5 years. Maths has always been fun to me. Since my high school 
I've studied mathematics in 3 of my (total 5 completed) semesters. I have 
keen interest in Machine Learning and Deep Learning and have almost a year 
of experience in this domain , working with and studying algorithms. I am 
pretty comfortable with mathematical notations and symbols but never did 
any extra coursework except from my school. I've been an active lurker on 
the sympy community and still getting familiar of the tons of features 
sympy has to offer. Would love to contribute to the community.

Regards
Nitish Gupta

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[sympy] GSoC 2020

[Aaron Meurer]

Google has announced the timeline for the GSoC 2020 program
https://developers.google.com/open-source/gsoc/timeline. The org
applications open in mid January.

If anyone is interested in mentoring, please add your name to the
mentoring section at
https://github.com/sympy/sympy/wiki/GSoC-2020-Ideas. If you have
participated as a student in the past, I encourage you to considering
mentoring.

Also, please update that wiki page with any new GSoC ideas, as well as
updating or removing any ideas which have already been implemented,
such as the ones that were implemented in projects last year.

If you are interested in participating in GSoC as a student, start
here https://github.com/sympy/sympy/wiki/GSoC-2020-Student-Instructions.

Aaron Meurer

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Re: [sympy] Re: How do I use sympy.solve() with greater precision than 15?

[Aaron Meurer]

solve() seems to have a bug here.

Aaron Meurer

On Tue, Dec 10, 2019 at 10:50 AM Damon Turney  wrote:
>
> Update:  An angel on GitHub told me that I can overcome this problem by using 
> solveset() instead of solve().
>
> But why does solve() not work as expected?
>
>
> On Tuesday, December 10, 2019 at 12:41:46 PM UTC-5, Damon Turney wrote:
>>
>> Hi Forum,
>>
>> Help!  sympy.solve() always uses precision of 15 no matter what I try.
>>
>> Recreated problem results:
>>
>> In [164]: import sympy
>>
>> In [165]: w,x,y,z=sympy.symbols('w,x,y,z')
>>
>> In [166]: eqn=sympy.Eq(sympy.Float('1/55500',45)*w,sympy.Float('96500.0')*y)
>>
>> In [167]: eqn.evalf(45)
>>
>> Out[167]: Eq(0.18018018018018018018018018018018018018018018*w, 96500.0*y)
>>
>> In [168]: sympy.solve(eqn,y)[0].evalf(45)
>>
>> Out[168]: 0.0186715212621948369903318027445619613213789734*w
>>
>> In [169]: sympy.solve(eqn.evalf(45),y)[0].evalf(45)
>>
>> Out[169]: 0.0186715212621948369903318027445619613213789734*w
>>
>>
>> The more accurate coefficient (to 45 digit accuracy) is:
>>
>> In [152]: sympy.Float('1/55500',45)/sympy.Float('96500.0',45).evalf(45)
>>
>> Out[152]: 0.0186715212621948373243710031274798114176352518
>>
>> In [174]: 
>> sympy.Float('1',45)/sympy.Float('55500',45)/sympy.Float('96500.0',45).evalf(45)
>>
>> Out[174]: 0.0186715212621948373243710031274798114176352518
>>
>>
>> So you see, the coefficient that results from sympy.solve:
>>
>> 0.0186715212621948369903318027445619613213789734
>>
>> is accurate to only 15 digits of precision.
>>
>>
>> Clearly sympy.solve is using only 15 digit precision, even though I'm 
>> specifying 45 digit precision.
>>
>> How can I force sympy.solve to use 45 digit precision when it solves for y?
>
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[sympy] Re: How do I use sympy.solve() with greater precision than 15?

[Damon Turney]

Update:  An angel on GitHub told me that I can overcome this problem by 
using solveset() instead of solve().

But why does solve() not work as expected?


On Tuesday, December 10, 2019 at 12:41:46 PM UTC-5, Damon Turney wrote:
>
> Hi Forum,
>
> Help!  sympy.solve() always uses precision of 15 no matter what I try.
>
> Recreated problem results:
>
> In [*164*]: *import* *sympy*  
> 
>  
>
> In [*165*]: w,x,y,z=sympy.symbols('w,x,y,z') 
> 
>  
>
> In [*166*]: 
> eqn=sympy.Eq(sympy.Float('1/55500',45)*w,sympy.Float('96500.0')*y) 
> 
>
>
> In [*167*]: eqn.evalf(45) 
> 
> 
>
> Out[*167*]: Eq(0.18018018018018018018018018018018018018018018*w, 
> 96500.0*y)
>
> In [*168*]: sympy.solve(eqn,y)[0].evalf(45)   
> 
> 
>
> Out[*168*]: 0.0186715212621948369903318027445619613213789734*w
>
> In [*169*]: sympy.solve(eqn.evalf(45),y)[0].evalf(45) 
> 
> 
>
> Out[*169*]: 0.0186715212621948369903318027445619613213789734*w
>
> The more accurate coefficient (to 45 digit accuracy) is:
>
> In [*152*]: sympy.Float('1/55500',45)/sympy.Float('96500.0',45).evalf(45) 
> 
>  
>
> Out[*152*]: 0.0186715212621948373243710031274798114176352518
>
> In [*174*]: sympy.Float('1',45)/sympy.Float('55500',45)/sympy.Float(
> '96500.0',45).evalf(45)   
>  
>
> Out[*174*]: 0.0186715212621948373243710031274798114176352518
>
>
> So you see, the coefficient that results from sympy.solve: 
>
> 0.0186715212621948369903318027445619613213789734
>
> is accurate to only 15 digits of precision.
>
>
> Clearly sympy.solve is using only 15 digit precision, even though I'm 
> specifying 45 digit precision. 
>
> How can I force sympy.solve to use 45 digit precision when it solves for 
> y?
>

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[sympy] How do I use sympy.solve() with greater precision than 15?

[Damon Turney]

Hi Forum,

Help!  sympy.solve() always uses precision of 15 no matter what I try.

Recreated problem results:

In [*164*]: *import* *sympy*

   

In [*165*]: w,x,y,z=sympy.symbols('w,x,y,z')   

   

In [*166*]: eqn=sympy.Eq(sympy.Float('1/55500',45)*w,sympy.Float('96500.0')*y) 

   

In [*167*]: eqn.evalf(45)   

  

Out[*167*]: Eq(0.18018018018018018018018018018018018018018018*w, 
96500.0*y)

In [*168*]: sympy.solve(eqn,y)[0].evalf(45) 

  

Out[*168*]: 0.0186715212621948369903318027445619613213789734*w

In [*169*]: sympy.solve(eqn.evalf(45),y)[0].evalf(45)   

  

Out[*169*]: 0.0186715212621948369903318027445619613213789734*w

The more accurate coefficient (to 45 digit accuracy) is:

In [*152*]: sympy.Float('1/55500',45)/sympy.Float('96500.0',45).evalf(45)   

   

Out[*152*]: 0.0186715212621948373243710031274798114176352518

In [*174*]: sympy.Float('1',45)/sympy.Float('55500',45)/sympy.Float(
'96500.0',45).evalf(45) 
   

Out[*174*]: 0.0186715212621948373243710031274798114176352518


So you see, the coefficient that results from sympy.solve: 

0.0186715212621948369903318027445619613213789734

is accurate to only 15 digits of precision.


Clearly sympy.solve is using only 15 digit precision, even though I'm 
specifying 45 digit precision. 

How can I force sympy.solve to use 45 digit precision when it solves for y?

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