Thanks.

Am Sonntag, 22. Dezember 2019 16:33:29 UTC+1 schrieb Chris Smith:
>
> Another approach for these problems is to use, as a starting point, 
> solutions to a simpler problem which become initial guess to the more 
> difficult problem. e.g. solving sin(x)=0.2 will give you two solutions and 
> as many others as you want by adding or subtraction 2pi. Then these 
> approximate solutions can be used as initial guesses for nsolve as you 
> change the problem to `sin(x)-(a*x+0.2)` with `a` increasing as quickly as 
> possible from 0 to 0.5 (in your case). This is the "continuation" method.
>
> /c
>
> On Saturday, December 21, 2019 at 2:17:52 AM UTC-6, Philipp Gressly 
> Freimann wrote:
>>
>> Hello
>>
>> Well, thanks a lot. Works great. I did not know the "nsolve" command.
>>
>> If I am right, there is no command to find all three solutions?
>>
>> φ
>>
>> Am Freitag, 20. Dezember 2019 11:48:03 UTC+1 schrieb Oscar:
>>>
>>> You can use nsolve to find numerical solutions: 
>>> ``` 
>>> In [10]: nsolve(Eq(sin(x), 0.5*x + 0.2), x, 0) 
>>> Out[10]: 0.425436108484597 
>>> ``` 
>>> This will find one root at a time starting from an initial guess (I've 
>>> used zero). 
>>>
>>> Initial guesses -1 and +1 give two other roots. 
>>> ``` 
>>> In [11]: nsolve(Eq(sin(x), 0.5*x + 0.2), x, -1) 
>>> Out[11]: -2.11307244875263 
>>>
>>> In [12]: nsolve(Eq(sin(x), 0.5*x + 0.2), x, +1) 
>>> Out[12]: 1.59919364642736 
>>> ``` 
>>>
>>> You can get more precise solutions using prec: 
>>> ``` 
>>> In [15]: nsolve(Eq(sin(x), 0.5*x + 0.2), x, 0, prec=50) 
>>> Out[15]: 0.42543610848459725447179186114511470949330179080539 
>>> ``` 
>>>
>>> -- 
>>> Oscar 
>>>
>>> On Fri, 20 Dec 2019 at 10:26, Philipp Gressly Freimann 
>>> <[email protected]> wrote: 
>>> > 
>>> > Hello 
>>> > 
>>> > I want to solve the following equation numerically between -PI and PI: 
>>> > 
>>> > sin(x) = 0.5x + 0.2 
>>> > 
>>> > [which is similar to sin(x) - 0.5x - 0.2 = 0] 
>>> > 
>>> > The graph shows me three solutions. Is there a possibility to solve 
>>> this equation numerically using sympy? 
>>> > 
>>> > Thanks in advance 
>>> > 
>>> > φ 
>>> > 
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>>>  
>>>
>>>
>>

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